Research ArticleComputation of Spectral Parameter of Discontinuous DiracSystems with a Gaussian Multiplier

Mohammed M Tharwat12 and Mohammed A Alghamdi1

1 Department of Mathematics Faculty of Science King Abdulaziz University Jeddah 21589 Saudi Arabia2Department of Mathematics Faculty of Science Beni-Suef University Beni-Suef 62511 Egypt

Correspondence should be addressed to Mohammed MTharwat zahraa26yahoocom

Received 9 March 2014 Revised 21 April 2014 Accepted 21 April 2014 Published 13 May 2014

Academic Editor Ali H Bhrawy

Copyright copy 2014 M MTharwat and M A Alghamdi This is an open access article distributed under the Creative CommonsAttribution License which permits unrestricted use distribution and reproduction in any medium provided the original work isproperly cited

We aim in this paper to apply a sinc-Gaussian technique to compute the eigenvalues of a Dirac system which has a discontinuity atone point and contains a spectral parameter in all boundary conditions We establish the needed properties of eigenvalues of ourproblemThe error of this method decays exponentially in terms of the number of involved samples Therefore the accuracy of thenew technique is higher than the classical sinc-method Numerical worked examples with tables and illustrative figures are givenat the end of the paper

1 Introduction

Problems with a spectral parameter in equations and bound-ary conditions form an important part of spectral theoryof linear differential operators A bibliography of papers inwhich such problems were considered in connection withspecific physical processes can be found in [1 2] Let 120590 gt 0The Paley-Wiener spaceB2

120590is the space of all entire functions

of exponential type 120590which lie in 1198712(R)when restricted toR

that is

B2

120590= 119891 119891 entire 100381610038161003816

1003816119891 (120583)

1003816100381610038161003816le 119862119890120590|I120583|

int

R

1003816100381610038161003816119891 (120583)

1003816100381610038161003816

2

119889120583 lt infin

(1)

Assume that 119891 isin B2120590 and Whittaker-Kotelrsquonikov-Shannon

sampling theorem (WKS) states that any function 119891 can bereconstructed via the classical sampling expansion [3ndash5]

119891 (120582) = sum

119899isinZ

119891(

119899120587

120590

) sinc (120590120582 minus 119899120587) 120582 isin C (2)

where

sinc (120582) =

sin (120582)

120582

120582 = 0

1 120582 = 0

(3)

In (2) the series is convergent uniformly on R and isconvergent absolutely and uniformly on compact subsetsof C cf [6] The WKS sampling series is widely used inapproximation theory It is used to approximate functionsand their derivatives solutions of differential and integralequations and integral transforms (Fourier Laplace Hankeland Mellin) and to approximate the eigenvalues of boundaryvalue problems see for example [7ndash10] The use of WKSsampling theory is called sinc methods cf [11ndash14] The sinc-method has a slow rate of decay at infinity which is as slowas 119874(|120582

minus1|) There are several attempts to improve the rate

of decay One of the interesting ways is to multiply the sinc-function in (2) by a kernel function see for example [15ndash17]Let ℎ isin (0 120587120590] and 120574 isin (0 120587 minus ℎ120590) Assume that Φ isin B2

120574

Hindawi Publishing CorporationAbstract and Applied AnalysisVolume 2014 Article ID 462698 12 pageshttpdxdoiorg1011552014462698

2 Abstract and Applied Analysis

such that Φ(0) = 1 then for 119891 isin B2120590we have the expansion

[18]

119891 (120582) =

infin

sum

119899=minusinfin

119891 (119899ℎ) sinc (ℎminus1120587120582 minus 119899120587)Φ (ℎminus1120582 minus 119899) (4)

The speed of convergence of the series in (4) is determinedby the decay of |Φ(120582)| But the decay of an entire functionof exponential type cannot be as fast as 119890minus119888|119909| as |119909| rarr infinfor some positive 119888 [18] For 120590 gt 0 ℎ isin (0 120587120590] and 120596 =

(120587minusℎ120590)2 Schmeisser and Stenger [18] defined the operator

(Gℎ119873

119891) (120582) = sum

119899isinZ119873(120582)

119891 (119899ℎ) sinc (ℎminus1120587120582 minus 119899120587)

times 119866(

radic120596 (120582 minus 119899ℎ)

radic119873ℎ

)

(5)

where 119866(119905) = exp(minus1199052) which is called the Gaussianfunction Z

119873(120582) = 119899 isin Z |[ℎ

minus1R120582 + 12] minus 119899| le 119873119873 isin N and [120582] denotes the integer part of 120582 isin R see also[19ndash21] Note that the summation limits in (5) depend on thereal part of 120582 The authors [18] proved that if 119891 is an entirefunction such that

1003816100381610038161003816119891 (120585 + 119894120578)

1003816100381610038161003816le 120601 (

10038161003816100381610038161205851003816100381610038161003816) 119890120590|120578|

120585 120578 isin R (6)

where 120601 is a nondecreasing nonnegative function on [0infin)

and 120590 ge 0 then for ℎ isin (0 120587120590) 120596 = (120587 minus ℎ120590)2119873 isin N and|I120582| lt 119873 we have

1003816100381610038161003816119891 (120582) minus (G

ℎ119873119891) (120582)

1003816100381610038161003816

le 2

10038161003816100381610038161003816sin (ℎ

minus1120587120582)

10038161003816100381610038161003816120601 (|R120582| + ℎ (119873 + 1))

times

119890minus120596119873

radic120587120596119873

120573119873(ℎminus1I120582) 120582 isin 119862

(7)

where

120573119873(119905) = cosh (2120596119905) +

21198901205961199052119873

radic120587120596119873[1 minus (119905119873)2]

+

1

2

[

1198902120596119905

1198902120587(119873minus119905)

minus 1

+

119890minus2120596119905

1198902120587(119873+119905)

minus 1

]

(8)

In [22] the authors derived an estimate for amplitudeerror They proved that if

sup119899isinZ119899(120582)

10038161003816100381610038161003816119891 (119899ℎ) minus

119891 (119899ℎ)

10038161003816100381610038161003816lt 120576 (9)

holds then for |I120582| lt 119873 we have10038161003816100381610038161003816(Gℎ119873

119891) (120582) minus (Gℎ119873

119891) (120582)

10038161003816100381610038161003816le 119860120576119873

(I120582) (10)

where 119891(119899ℎ) is the approximations of the exact values 119891(119899ℎ)

of (5) 120576 gt 0 is sufficiently small

119860120576119873

(I120582) = 2120576119890minus1205964119873

(1 + radic119873

120596120587

) exp ((120596 + 120587) ℎminus1

|I120582|)

(11)

Consider the Dirac system which consists of the systemof differential equations

(u10158402(119909) minus 119901

1(119909)u1(119909)

u10158401(119909) + 119901

2(119909)u2(119909)

) = 120582(u1(119909)

minusu2(119909)

)

119909 isin [minus1 0) cup (0 1]

(12)

with boundary conditions

L1(u) = (120572

1+ 120582 sin 120579

1)u1(minus1)

minus (1205722+ 120582 cos 120579

1)u2(minus1) = 0

(13)

L2(u) = (120573

1+ 120582 sin 120579

2)u1(1)

minus (1205732+ 120582 cos 120579

2)u2(1) = 0

(14)

and transmission conditions

L3(u) = u

1(0minus) minus 120575u

1(0+) = 0

L4(u) = u

2(0minus) minus 120575u

2(0+) = 0

(15)

Here 120582 is the spectral parameter u = (u1u2 ) the real valued

function 1199011(sdot) and 119901

2(sdot) are continuous in [minus1 0) and (0 1]

and have finite limits 1199011(0plusmn) = lim

119909rarr0plusmn1199011(119909) 119901

2(0plusmn) =

lim119909rarr0

plusmn1199012(119909) 120572

1 1205722 1205731 1205732 120575 isin R 120579

1 1205792isin [0 120587) and 120575 = 0

Throughout the following we assume that

1205881= 1205722sin 1205791minus 1205721cos 1205791gt 0

1205882= 1205731cos 1205792minus 1205732sin 1205792gt 0

(16)

The aim of the present work is to compute the eigenvaluesof (12)ndash(15) numerically by the sinc-Gaussian techniquewith errors analysis truncation error and amplitude errorIn [23] Annaby and Tharwat computed the eigenvalues ofa second-order operator pencil by sinc-Gaussian methodAlso the authors introduced some examples illustrating thesinc-Gaussian method accompanied by comparison withthe sinc-method which explained that the sinc-Gaussianmethod gives remarkably better results see also [24 25]Tharwat and Al-Harbi [26] computed the eigenvalues ofdiscontinuous Dirac system but with eigenparameter in oneboundary condition In [14 27] Tharwat et al computedthe eigenvalues of discontinuousDirac system approximatelywith eigenparameter appears in any of boundary conditionsby Hermite interpolations and regularized sinc-methodsrespectively In regularized sinc-method also the same inHermite interpolations method the basic idea is that theeigenvalues are characterised as the zeros of an analyticfunction 119891(120582) which can be written in the form 119891(120582) =

119896(120582) + 119906(120582) where 119896(120582) is known part The ingenuity of theapproach is in trying to choose the function 119891(120582) so that119906(120582) isin B2

120590(unknown part) and can be approximated by the

sampling theorem if its values at some equally spaced pointsare known see [11ndash14] Recall that in regularized sinc andHermite interpolations methods it is necessary that 119906(120582) is1198712-function In this paper wewill use sinc-Gaussian sampling

formula (5) to compute eigenvalues of (12)ndash(15) numerically

Abstract and Applied Analysis 3

The proposed method reduces the error bounds remarkably(see examples of Section 4) The basic idea is to write thefunction of eigenvalues as the sum of two terms one knownand the other unknown but an entire function of exponentialtype which satisfies (6) In other words the unknown termis not necessarily an 119871

2-function Then we approximate theunknown part using (5) and obtain better results We wouldlike to mention that the papers in computing eigenvaluesby sinc-Gaussian method are few see [22ndash26] In Sections2 and 3 we discuss some properties of the eigenvalues ofthe boundary value problems (12)ndash(15) and also we derivethe sinc-Gaussian technique to compute the eigenvalues of(12)ndash(15) with error estimates The last section involves someillustrative examples

2 Some Important Results

In the following we study some properties of the eigenvaluesof the problems (12)ndash(15) which need in our method see [2628] For functions u(119909) which defined on [minus1 0) cup (0 1] andhas finite limitu(plusmn0) = lim

119909rarrplusmn0u(119909) byu

(1)(119909) andu

(2)(119909)

we denote the functions

u(1)

(119909) =

u (119909) 119909 isin [minus1 0)

u (0minus) 119909 = 0

u(2)

(119909) =

u (119909) 119909 isin (0 1]

u (0+) 119909 = 0

(17)

which are defined on Γ1

= [minus1 0] and Γ2

= [0 1]respectively

In the following lemma wewill prove that the eigenvaluesof the problems (12)ndash(15) are real see [29 30]

Lemma 1 The eigenvalues of the problems (12)ndash(15) are real

Proof Assume the contrary that 1205820is a nonreal eigenvalue of

problems (12)ndash(15) Let ( u1(119909)

u2(119909)) be a corresponding (nontriv-

ial) eigenfunction By (12) we have

119889

119889119909

u1(119909)u2(119909) minus u

1(119909)u2(119909)

= (1205820minus 1205820)

1003816100381610038161003816u1(119909)

1003816100381610038161003816

2

+1003816100381610038161003816u2(119909)

1003816100381610038161003816

2

119909 isin [minus1 0) cup (0 1]

(18)

Integrating the above equation through [minus1 0] and [0 1] weobtain

(1205820minus 1205820) [int

0

minus1

(1003816100381610038161003816u1(119909)

1003816100381610038161003816

2

+1003816100381610038161003816u2(119909)

1003816100381610038161003816

2

) 119889119909]

= u1(0minus) 1199062(0minus) minus u1(0minus)u2(0minus)

minus [u1(minus1)u

2(minus1) minus u

1(minus1)u

2(minus1)]

(1205820minus 1205820) [int

1

0

(1003816100381610038161003816u1(119909)

1003816100381610038161003816

2

+1003816100381610038161003816u2(119909)

1003816100381610038161003816

2

) 119889119909]

= u1(1)u2(1) minus u

1(1)u2(1)

minus [u1(0+)u2(0+) minus u1(0+)u2(0+)]

(19)

Then from (13) (14) and transmission conditions we haverespectively

u1(minus1)u

2(minus1) minus u

1(minus1)u

2(minus1)

=

1205881(1205820minus 1205820)1003816100381610038161003816u2(minus1)

1003816100381610038161003816

2

10038161003816100381610038161205721+ 1205820sin 1205791

1003816100381610038161003816

2

u1(1)u2(1) minus u

1(1)u2(1)

= minus

1205882(1205820minus 1205820)1003816100381610038161003816u2(1)

1003816100381610038161003816

2

10038161003816100381610038161205731+ 1205820sin 1205792

1003816100381610038161003816

2

u1(minus0) 119906

2(minus0) minus u

1(minus0)u

2(minus0)

= 1205752[u1(+0) 119906

2(+0) minus u

1(+0)u

2(+0)]

(20)

Since 1205820

= 1205820 it follows from the last three equations and (19)

that

int

0

minus1

(1003816100381610038161003816u1(119909)

1003816100381610038161003816

2

+1003816100381610038161003816u2(119909)

1003816100381610038161003816

2

) 119889119909

+ 1205752int

1

0

(1003816100381610038161003816u1(119909)

1003816100381610038161003816

2

+1003816100381610038161003816u2(119909)

1003816100381610038161003816

2

) 119889119909

= minus

1205881

1003816100381610038161003816u2(minus1)

1003816100381610038161003816

2

10038161003816100381610038161205721+ 1205820sin 1205791

1003816100381610038161003816

2minus

120588212057521003816100381610038161003816u2(1)

1003816100381610038161003816

2

10038161003816100381610038161205731+ 1205820sin 1205792

1003816100381610038161003816

2

(21)

This contradicts the conditions int0minus1(|u1(119909)|2+ |u2(119909)|2)119889119909 +

1205752int

1

0(|u1(119909)|2+ |u2(119909)|2)119889119909 gt 0 and 120588

119894gt 0 119894 = 1 2

Consequently 1205820must be real

Now we shall construct a special fundamental system ofsolutions of (12) for 120582 not being an eigenvalue Let us considerthe next initial value problem

u1015840

2(119909) minus 119901

1(119909)u1(119909) = 120582u

1(119909)

u1015840

1(119909) + 119901

2(119909)u2(119909) = minus120582u

2(119909)

119909 isin (minus1 0)

(22)

u1(minus1) = 120572

2+ 120582 cos 120579

1 u

2(minus1) = 120572

1+ 120582 sin 120579

1 (23)

By virtue of Theorem 11 in [31] this problem has a uniquesolution u = (

12059311(119909120582)

12059321(119909120582)) which is an entire function of 120582 isin C

for each fixed 119909 isin [minus1 0] Similarly employing the same

4 Abstract and Applied Analysis

method as in proof of Theorem 11 in [31] we see that theproblem

u1015840

2(119909) minus 119901

1(119909)u1(119909) = 120582u

1(119909)

u1015840

1(119909) + 119901

2(119909)u2(119909) = minus120582u

2(119909)

119909 isin (0 1)

u1(1) = 120573

2+ 120582 cos 120579

2 u

2(1) = 120573

1+ 120582 sin 120579

2

(24)

has a unique solution u = (12059512(119909120582)

12059522(119909120582)) which is an entire

function of parameter 120582 for each fixed 119909 isin [0 1]Now the functions 120593

1198942(119909 120582) and 120595

1198941(119909 120582) are defined in

terms of 1205931198941(119909 120582) and 120595

1198942(119909 120582) 119894 = 1 2 respectively as

follows the initial-value problem

u1015840

2(119909) minus 119901

1(119909)u1(119909) = 120582u

1(119909)

u1015840

1(119909) + 119901

2(119909)u2(119909) = minus120582u

2(119909)

119909 isin (0 1)

(25)

u1(0) =

1

120575

12059311

(0 120582) u2(0) =

1

120575

12059321

(0 120582) (26)

has unique solution u = (12059312(119909120582)

12059322(119909120582)) for each 120582 isin C

Similarly the following problem also has a unique solu-tion u = (

12059511(119909120582)

12059521(119909120582))

u1015840

2(119909) minus 119901

1(119909)u1(119909) = 120582u

1(119909)

u1015840

1(119909) + 119901

2(119909)u2(119909) = minus120582u

2(119909)

119909 isin (minus1 0)

(27)

u1(0) = 120575120595

12(0 120582) u

2(0) = 120575120595

22(0 120582) (28)

Let us construct two basic solutions of (12) as

120593 (sdot 120582) = (1205931(sdot 120582)

1205932(sdot 120582)

) 120595 (sdot 120582) = (1205951(sdot 120582)

1205952(sdot 120582)

) (29)

where

1205931(119909 120582) =

12059311

(119909 120582) 119909 isin [minus1 0)

12059312

(119909 120582) 119909 isin (0 1]

1205932(119909 120582) =

12059321

(119909 120582) 119909 isin [minus1 0)

12059322

(119909 120582) 119909 isin (0 1]

(30)

1205951(119909 120582) =

12059511

(119909 120582) 119909 isin [minus1 0)

12059512

(119909 120582) 119909 isin (0 1]

1205952(119909 120582) =

12059521

(119909 120582) 119909 isin [minus1 0)

12059522

(119909 120582) 119909 isin (0 1]

(31)

By virtue of (26) and (28) these solutions satisfy bothtransmission conditions (15) These functions are entire in 120582

for all 119909 isin [minus1 0) cup (0 1]

Let W(120593 120595)(sdot 120582) denote the Wronskian of 120593(sdot 120582) and120595(sdot 120582) defined in [32 page 194] that is

W (120593 120595) (sdot 120582) =

10038161003816100381610038161003816100381610038161003816

1205931(sdot 120582) 120593

2(sdot 120582)

1205951(sdot 120582) 120595

2(sdot 120582)

10038161003816100381610038161003816100381610038161003816

(32)

It is obvious that the Wronskians

Ω119894(120582) = W (120593 120595) (119909 120582)

= 1205931119894(119909 120582) 120595

2119894(119909 120582)

minus 1205932119894(119909 120582) 120595

1119894(119909 120582) 119909 isin Γ

119894 119894 = 1 2

(33)

are independent on variable 119909 isin Γ119894(119894 = 1 2) and 120593

119894(119909 120582)

and 120595119894(119909 120582) are entire functions of the parameter 120582 for each

119909 isin Γ119894(119894 = 1 2) Taking into account (26) and (28) a short

calculation gives

Ω1(120582) = 120575

2Ω2(120582) (34)

for each 120582 isin C

Corollary 2 The zeros of the functions Ω1(120582) and Ω

2(120582)

coincide

Then we may introduce to the consideration the charac-teristic functionΩ(120582) as

Ω (120582) = Ω1(120582) = 120575

2Ω2(120582) (35)

Note that all eigenvalues of problems (12)ndash(15) are just zerosof the function Ω(120582) Indeed since the functions 120593

1(119909 120582)

and 1205932(119909 120582) satisfy the boundary condition (13) and both

transmission conditions (15) to find the eigenvalues ofthe (12)ndash(15) we have to insert the functions 120593

1(119909 120582) and

1205932(119909 120582) in the boundary condition (14) and find the roots

of this equation In the following lemma we show that alleigenvalues of the problems (12)ndash(15) are simple

Lemma 3 The eigenvalues of the boundary value problems(12)ndash(15) form an at most countable set without finite limitpoints All eigenvalues of the boundary value problems (12)ndash(15) (of Ω(120582)) are simple

Proof The eigenvalues are the zeros of the entire functionoccurring on the left-hand side in see (35)

(1205731+ 120582 sin 120579

2) 12059312

(1 120582) minus (1205732+ 120582 cos 120579

2) 12059322

(1 120582) = 0

(36)

We have shown (see Lemma 1) that this function doesnot vanish for nonreal 120582 In particular it does not vanishidentically Therefore its zeros form an at most countable setwithout finite limit points

By (12) we obtain for 120582 120583 isin C 120582 = 120583

119889

119889119909

1205931(119909 120582) 120593

2(119909 120583) minus 120593

1(119909 120583) 120593

2(119909 120582)

= (120583 minus 120582) 1205931(119909 120582) 120593

1(119909 120583) + 120593

2(119909 120582) 120593

2(119909 120583)

(37)

Abstract and Applied Analysis 5

Integrating the above equation through [minus1 0] and [0 1] andtaking into account the initial conditions (23) (26) and (28)we obtain

1205752(12059312

(1 120582) 12059322

(1 120583) minus 12059312

(1 120583) 12059322

(1 120582))

minus (120583 minus 120582) 1205881

= (120583 minus 120582) (int

0

minus1

(12059311

(119909 120582) 12059311

(119909 120583)

+ 12059321

(119909 120582) 12059321

(119909 120583)) 119889119909

+ 1205752int

1

0

(12059312

(119909 120582) 12059312

(119909 120583)

+ 12059322

(119909 120582) 12059322

(119909 120583)) 119889119909)

(38)

Dividing both sides of (38) by (120582 minus 120583) and by letting 120583 rarr 120582we arrive to the relation

1205752(12059322

(1 120582)

12059712059312

(1 120582)

120597120582

minus 12059312

(1 120582)

12059712059322

(1 120582)

120597120582

) + 1205881

= minus(int

0

minus1

(100381610038161003816100381612059311

(119909 120582)1003816100381610038161003816

2

+100381610038161003816100381612059321

(119909 120582)1003816100381610038161003816

2

) 119889119909

+ 1205752int

1

0

(100381610038161003816100381612059312

(119909 120582)1003816100381610038161003816

2

+100381610038161003816100381612059322

(119909 120582)1003816100381610038161003816

2

) 119889119909)

(39)

We show that equation

Ω (120582) = W (120593 120595) (1 120582)

= 1205752((1205731+ 120582 sin 120579

2) 12059312

(1 120582)

minus (1205732+ 120582 cos 120579

2) 12059322

(1 120582)) = 0

(40)

has only simple roots Assume the converse that is (40) hasa double root 120582lowast Then the following two equations hold

(1205731+ 120582lowast sin 120579

2) 12059312

(1 120582lowast) minus (120573

2+ 120582lowast cos 120579

2) 12059322

(1 120582lowast) = 0

sin 120579212059312

(1 120582lowast) + (120573

1+ 120582lowast sin 120579

2)

12059712059312

(1 120582lowast)

120597120582

minus cos 120579212059322

(1 120582lowast) minus (120573

2+ 120582lowast cos 120579

2)

12059712059322

(1 120582lowast)

120597120582

= 0

(41)

Since 1205882

= 0 and 120582lowast is real then (120573

1+ 120582lowast sin 120579

2)2

+

(1205732+ 120582lowast cos 120579

2)2

= 0 Let 1205731+ 120582lowast sin 120579

2= 0 From (41)

12059312

(1 120582lowast) =

(1205732+ 120582lowast cos 120579

2)

(1205731+ 120582lowast sin 120579

2)

12059322

(1 120582lowast)

12059712059312

(1 120582lowast)

120597120582

=

120588212059322

(1 120582lowast)

(1205731+ 120582lowast sin 120579

2)2

+

(1205732+ 120582lowast cos 120579

2)

(1205731+ 120582lowast sin 120579

2)

12059712059322

(1 120582lowast)

120597120582

(42)

Combining (42) and (39) with 120582 = 120582lowast we obtain

12058821205752(12059322

(1 120582lowast))2

(1205731+ 120582lowast sin 120579

2)2

+ 1205881

= minus(int

0

minus1

(100381610038161003816100381612059311

(119909 120582)1003816100381610038161003816

2

+100381610038161003816100381612059321

(119909 120582)1003816100381610038161003816

2

) 119889119909

+ 1205752int

1

0

(

1003816100381610038161003816100381612059312

(119909 120582) |2+

1003816100381610038161003816100381612059322

(119909 120582) |2) 119889119909)

(43)

contradicting the assumption 120588119894gt 0 119894 = 1 2 The other case

when 1205732+120582lowast cos 120579

2= 0 can be treated similarly and the proof

is complete

Recall that boundary value problems (12)ndash(15) have denu-merable set of real and simple eigenvalues see the abovesection From (23) and (26) the solution of (12)

120593 (sdot 120582) = (1205931(sdot 120582)

1205932(sdot 120582)

)

120593119894(119909 120582) =

1205931198941(119909 120582) 119909 isin [minus1 0)

1205931198942(119909 120582) 119909 isin (0 1]

119894 = 1 2

(44)

satisfies the following initial conditions

(12059311

(minus1 120582) 12059312

(0+ 120582)

12059321

(minus1 120582) 12059322

(0+ 120582)

)

= (1205722+ 120582 cos 120579

1120575minus112059311

(0minus 120582)

1205721+ 120582 sin 120579

1120575minus112059321

(0minus 120582)

)

(45)

Since 120593(sdot 120582) satisfies (13) then the eigenvalues of the prob-lems (12)ndash(15) are the zeros of the function

Ω (120582) = 1205752((1205731+ 120582 sin 120579

2) 12059312

(1 120582)

minus (1205732+ 120582 cos 120579

2) 12059322

(1 120582)) = 0

(46)

We shall transform (12) (23) (26) and (30) into the integralequations see [32]

12059311

(119909 120582) = 1205722cos 120582 (119909 + 1) minus 120572

1sin 120582 (119909 + 1)

+ 120582 cos (120582 (119909 + 1) + 1205791) minusTminus11

12059311

(119909 120582)

minusTminus12

12059321

(119909 120582)

(47)

12059321

(119909 120582) = 1205721cos 120582 (119909 + 1) + 120572

2sin 120582 (119909 + 1)

+ 120582 sin (120582 (119909 + 1) + 1205791) +

Tminus11

12059311

(119909 120582)

minusTminus12

12059321

(119909 120582)

(48)

12059312

(119909 120582) =

1

120575

12059311

(0minus 120582) cos (120582119909) minus 1

120575

12059321

(0minus 120582) sin (120582119909)

minusT01

12059312

(119909 120582) minusT02

12059322

(119909 120582)

(49)

6 Abstract and Applied Analysis

12059322

(119909 120582) =

1

120575

12059311

(0minus 120582) sin (120582119909) +

1

120575

12059321

(0minus 120582) cos (120582119909)

+T01

12059312

(119909 120582) minusT02

12059322

(119909 120582)

(50)

where Tminus1119894

Tminus1119894

T0119894

and T0119894 119894 = 1 2 are the Volterra

integral operators defined by

Tminus1119894

120593 (119909 120582) = int

119909

minus1

sin 120582 (119909 minus 119905) 119901119894(119905) 120593 (119905 120582) 119889119905

Tminus1119894

120593 (119909 120582) = int

119909

minus1

cos 120582 (119909 minus 119905) 119901119894(119905) 120593 (119905 120582) 119889119905

T0119894120593 (119909 120582) = int

119909

0

sin 120582 (119909 minus 119905) 119901119894(119905) 120593 (119905 120582) 119889119905

T0119894120593 (119909 120582) = int

119909

0

cos 120582 (119909 minus 119905) 119901119894(119905) 120593 (119905 120582) 119889119905

(51)

For convenience we define the constants

1198881= max 100381610038161003816

10038161205721

1003816100381610038161003816+10038161003816100381610038161205722

10038161003816100381610038161003816100381610038161003816sin 1205791

1003816100381610038161003816+1003816100381610038161003816cos 1205791

1003816100381610038161003816

1198882= int

0

minus1

[10038161003816100381610038161199011(119905)

1003816100381610038161003816+10038161003816100381610038161199012(119905)

1003816100381610038161003816] 119889119905 119888

3= 11988811198882exp (119888

2)

1198884= int

1

0

[10038161003816100381610038161199011(119905)

1003816100381610038161003816+10038161003816100381610038161199012(119905)

1003816100381610038161003816] 119889119905 119888

5= 1198883+

2

|120575|

(1198881+ 1198883)

1198886= max 100381610038161003816

10038161205721

1003816100381610038161003816+10038161003816100381610038161205722

10038161003816100381610038161003816100381610038161003816sin 1205792

1003816100381610038161003816+1003816100381610038161003816cos 1205792

1003816100381610038161003816

(52)

Now we define yminus1119894

(sdot 120582) and y0119894(sdot 120582) 119894 = 1 2 to be

yminus11

(119909 120582) = Tminus11

12059311

(119909 120582) +Tminus12

12059321

(119909 120582)

yminus12

(119909 120582) =Tminus11

12059311

(119909 120582) minusTminus12

12059321

(119909 120582)

y01

(119909 120582) = T01

12059312

(119909 120582) +T02

12059322

(119909 120582)

y02

(119909 120582) =T01

12059312

(119909 120582) minusT02

12059322

(119909 120582)

(53)

Lemma 4 The functions yminus11

(119909 120582) and yminus12

(119909 120582) are entirein 120582 for any fixed 119909 isin [minus1 0) and satisfy the growth condition

1003816100381610038161003816yminus11

(119909 120582)1003816100381610038161003816

1003816100381610038161003816yminus12

(119909 120582)1003816100381610038161003816le 21198883(1 + |120582|) 119890

|I120582|(119909+1)

120582 isin C

(54)

Proof Since yminus11

(119909 120582) = Tminus11

12059311(119909 120582) +

Tminus12

12059321(119909 120582)

then from (47) and (48) we obtain

yminus11

(119909 120582)

= 1205722Tminus11

cos 120582 (119909 + 1) minus 1205721Tminus11

sin 120582 (119909 + 1)

+ 120582Tminus11

cos (120582 (119909 + 1) + 1205791) minusTminus11

yminus11

(119909 120582)

+ 1205721Tminus12

cos 120582 (119909 + 1) + 1205722Tminus12

sin 120582 (119909 + 1)

+ 120582Tminus12

sin (120582 (119909 + 1) + 1205791) +

Tminus12

yminus12

(119909 120582)

(55)

Using the inequalities | sin 119911| le 119890|I119911| and | cos 119911| le 119890

|I119911| for119911 isin C leads for 120582 isin C to1003816100381610038161003816yminus11

(119909 120582)1003816100381610038161003816

le10038161003816100381610038161205722

1003816100381610038161003816

1003816100381610038161003816Tminus11

cos 120582 (119909 + 1)1003816100381610038161003816+10038161003816100381610038161205721Tminus11

sin 120582 (119909 + 1)1003816100381610038161003816

+ |120582|1003816100381610038161003816Tminus11

cos (120582 (119909 + 1) + 1205791)1003816100381610038161003816+1003816100381610038161003816Tminus11

yminus11

(119909 120582)1003816100381610038161003816

+10038161003816100381610038161205721

1003816100381610038161003816

10038161003816100381610038161003816

Tminus12

cos 120582 (119909 + 1)

10038161003816100381610038161003816+10038161003816100381610038161205722

1003816100381610038161003816

10038161003816100381610038161003816

Tminus12

sin 120582 (119909 + 1)

10038161003816100381610038161003816

+ |120582|

10038161003816100381610038161003816

Tminus12

sin (120582 (119909 + 1) + 1205791)

10038161003816100381610038161003816+

10038161003816100381610038161003816

Tminus12

yminus12

(119909 120582)

10038161003816100381610038161003816

le 21198881(1 + |120582|) 119890

|I120582|(119909+1)int

119909

minus1

[10038161003816100381610038161199011(119905)

1003816100381610038161003816+10038161003816100381610038161199012(119905)

1003816100381610038161003816] 119889119905

+ 119890|I120582|(119909+1)

int

119909

minus1

[10038161003816100381610038161199011(119905)

1003816100381610038161003816

1003816100381610038161003816yminus11

(119905 120582)1003816100381610038161003816

+10038161003816100381610038161199012(119905)

1003816100381610038161003816

1003816100381610038161003816yminus12

(119905 120582)1003816100381610038161003816] 119890minus|I120582|(119905+1)

119889119905

le 211988811198882(1 + |120582|) 119890

|I120582|(119909+1)

+ 119890|I120582|(119909+1)

int

119909

minus1

[10038161003816100381610038161199011(119905)

1003816100381610038161003816

1003816100381610038161003816yminus11

(119905 120582)1003816100381610038161003816

+10038161003816100381610038161199012(119905)

1003816100381610038161003816

1003816100381610038161003816yminus12

(119905 120582)1003816100381610038161003816] 119890minus|I120582|(119905+1)

119889119905

(56)

The above inequality can be reduced to

119890minus|I120582|(119909+1) 1003816

100381610038161003816yminus11

(119909 120582)1003816100381610038161003816

le 211988811198882(1 + |120582|)

+ int

119909

minus1

[10038161003816100381610038161199011(119905)

1003816100381610038161003816

1003816100381610038161003816yminus11

(119905 120582)1003816100381610038161003816

+10038161003816100381610038161199012(119905)

1003816100381610038161003816

1003816100381610038161003816yminus12

(119905 120582)1003816100381610038161003816] 119890minus|I120582|(119905+1)

119889119905

(57)

Similarly we can prove that

119890minus|I120582|(119909+1) 1003816

100381610038161003816yminus12

(119909 120582)1003816100381610038161003816

le 211988811198882(1 + |120582|)

+ int

119909

minus1

[10038161003816100381610038161199011(119905)

1003816100381610038161003816

1003816100381610038161003816yminus11

(119905 120582)1003816100381610038161003816

+10038161003816100381610038161199012(119905)

1003816100381610038161003816

1003816100381610038161003816yminus12

(119905 120582)1003816100381610038161003816] 119890minus|I120582|(119905+1)

119889119905

(58)

Then from (57) (58) and Lemma 31 of [32 pp 204] weobtain (54)

In a similar manner we will prove the following lemmafor y01

(sdot 120582) and y02

(sdot 120582)

Lemma 5 The functions y01

(119909 120582) and y02

(119909 120582) are entire in120582 for any fixed 119909 isin (0 1] and satisfy the growth condition

1003816100381610038161003816y01

(119909 120582)1003816100381610038161003816

1003816100381610038161003816y02

(119909 120582)1003816100381610038161003816le 211988841198885(1 + |120582|) 119890

|I120582|(119909+1)

120582 isin C

(59)

Abstract and Applied Analysis 7

Proof Since y01

(119909 120582) = T01

12059312(119909 120582) +

T02

12059322(119909 120582) then

from (49) and (50) we obtain

y01

(119909 120582) =

1

120575

12059311

(0minus 120582)T

01cos (120582119909)

minus

1

120575

12059321

(0minus 120582)T

01sin (120582119909)

minusT01yminus12

(119909 120582) +

1

120575

12059311

(0minus 120582)

T02

sin (120582119909)

+

1

120575

12059321

(0minus 120582)

T02

cos (120582119909) + T02yminus12

(119909 120582)

(60)

Then from (47) and (48) and Lemma 4 we get

1003816100381610038161003816y01

(119909 120582)1003816100381610038161003816le

1

|120575|

100381610038161003816100381612059311

(0minus 120582)

1003816100381610038161003816

1003816100381610038161003816T01

cos (120582119909)1003816100381610038161003816

+

1

|120575|

100381610038161003816100381612059321

(0minus 120582)

1003816100381610038161003816

1003816100381610038161003816T01

sin (120582119909)1003816100381610038161003816

+1003816100381610038161003816T01yminus12

(119909 120582)1003816100381610038161003816

+

1

|120575|

100381610038161003816100381612059311

(0minus 120582)

1003816100381610038161003816

10038161003816100381610038161003816

T02

sin (120582119909)

10038161003816100381610038161003816

+

1

|120575|

100381610038161003816100381612059321

(0minus 120582)

1003816100381610038161003816

10038161003816100381610038161003816

T02

cos (120582119909)10038161003816100381610038161003816

+

10038161003816100381610038161003816

T02yminus12

(119909 120582)

10038161003816100381610038161003816

le 21198884(1 + |120582|)

times (1198883+

2

|120575|

(1198881+ 1198883)) 119890|I120582|(119909+1)

= 211988841198885(1 + |120582|) 119890

|I120582|(119909+1)

(61)

Similarly we can prove that1003816100381610038161003816y02

(119909 120582)1003816100381610038161003816le 211988841198885(1 + |120582|) 119890

|I120582|(119909+1) (62)

3 The Numerical Scheme

In this section we derive the method of computing eigen-values of problems (12)ndash(15) numerically The basic idea ofthe scheme is to split Ω(120582) into two parts a known partK(120582) and an unknown oneU(120582)Thenwe approximateU(120582)

using (5) to get the approximate Ω(120582) and then compute theapproximate zeros We first splitΩ(120582) into two parts

Ω (120582) = K (120582) +U (120582) (63)

whereU(120582) is the unknown part involving integral operators

U (120582) = 120575 [(1205732+ 120582 cos 120579

2) sin 120582

minus (1205731+ 120582 sin 120579

2) cos 120582] y

minus11(0minus 120582)

minus 120575 [(1205731+ 120582 sin 120579

2) sin 120582

+ (1205732+ 120582 cos 120579

2) cos 120582] y

minus12(0minus 120582)

minus 1205752[(1205731+ 120582 sin 120579

2) y01

(1 120582)

+ (1205732+ 120582 cos 120579

2) y02

(1 120582)]

(64)

andK(120582) is the known part

K (120582) = 120575 (1205731+ 120582 sin 120579

2)

times [1205722cos 2120582 minus 120572

1sin 2120582 + 120582 cos (2120582 + 120579

1)]

minus 120575 (1205732+ 120582 cos 120579

2)

times [1205722sin 2120582 + 120572

1cos 2120582 + 120582 sin (2120582 + 120579

1)]

(65)

Then from Lemmas 4 and 5 we have the following result

Lemma 6 The function U(120582) is entire in 120582 and the followingestimate holds

|U (120582)| le 120601 (120582) 1198902|I120582|

(66)

where

120601 (120582) = 119872(1 + |120582|)2

119872 = 4 |120575| 1198886 (21198883 + |120575| 11988841198885)

(67)

Proof From (64) we have

|U (120582)| le |120575| [(10038161003816100381610038161205732

1003816100381610038161003816+ |120582|

1003816100381610038161003816cos 1205792

1003816100381610038161003816) |sin 120582|

+ (10038161003816100381610038161205731

1003816100381610038161003816+ |120582|

1003816100381610038161003816sin 1205792

1003816100381610038161003816) |cos 120582|] 100381610038161003816

1003816yminus11

(0minus 120582)

1003816100381610038161003816

+ |120575| [(10038161003816100381610038161205731

1003816100381610038161003816+ |120582|

1003816100381610038161003816sin 1205792

1003816100381610038161003816) |sin 120582|

+ (10038161003816100381610038161205732

1003816100381610038161003816+1003816100381610038161003816120582 cos 120579

2

1003816100381610038161003816) |cos 120582|] 100381610038161003816

1003816yminus12

(0minus 120582)

1003816100381610038161003816

+ 1205752[(10038161003816100381610038161205731

1003816100381610038161003816+ |120582|

1003816100381610038161003816sin 1205792

1003816100381610038161003816)1003816100381610038161003816y01

(1 120582)1003816100381610038161003816

+ (10038161003816100381610038161205732

1003816100381610038161003816+ |120582|

1003816100381610038161003816cos 1205792

1003816100381610038161003816)1003816100381610038161003816y02

(1 120582)1003816100381610038161003816]

(68)

Using the inequalities | sin 120582| le 119890|I120582| and | cos 120582| le 119890

|I120582| for120582 isin C and Lemmas 4 and 5 imply (66)

ThusU(120582) is an entire function of exponential type 120590 = 2In the following we let 120582 isin R since all eigenvalues are realNow we approximate the function U(120582) using the operator(5) where ℎ isin (0 1205872) and 120596 = (120587 minus 2ℎ)2 and then from(7) we obtain

1003816100381610038161003816U (120582) minus (G

ℎ119873U) (120582)

1003816100381610038161003816le 119879ℎ119873

(120582) (69)

where

119879ℎ119873

(120582) = 2

10038161003816100381610038161003816sin (ℎ

minus1120587120582)

10038161003816100381610038161003816

times 120601 (|R120582| + ℎ (119873 + 1))

119890minus120596119873

radic120587120596119873

120573119873(0) 120582 isin R

(70)

The samplesU(119899ℎ) = Ω(119899ℎ) minusK(119899ℎ) 119899 isin Z119873(120582) cannot be

computed explicitly in the general caseWe approximate these

8 Abstract and Applied Analysis

samples numerically by solving the initial-value problemsdefined by (12) and (45) to obtain the approximate valuesU(119899ℎ) 119899 isin Z

119873(120582) that is U(119899ℎ) = Ω(119899ℎ) minus K(119899ℎ)

Here we use a computer algebra system mathematica toobtain the approximate solutions with the required accuracyHowever a separate study for the effect of different numericalschemes and the computational costs would be interestingAccordingly we have the explicit expansion

(Gℎ119873

U) (120582) = sum

119899isinZ119873(120582)

U (119899ℎ) sinc (ℎminus1120587119909 minus 119899120587)

times 119866(

radic120596 (120582 minus 119899ℎ)

radic119873ℎ

)

(71)

Therefore we get cf (10)10038161003816100381610038161003816(Gℎ119873

U) (120582) minus (Gℎ119873

U) (120582)

10038161003816100381610038161003816le 119860120576119873

(0) 120582 isin R (72)

Now let Ω119873(120582) = K(120582) + (G

ℎ119873U)(120582) From (69) and (72)

we obtain10038161003816100381610038161003816Ω (120582) minus Ω

119873(120582)

10038161003816100381610038161003816le 119879ℎ119873

(120582) + 119860120576119873

(0) 120582 isin R (73)

Let 120582lowast be an eigenvalue and 120582119873its desired approximation

that is Ω(120582lowast) = 0 and Ω

119873(120582119873) = 0 From (73) we have

|Ω119873(120582lowast)| le 119879

ℎ119873(120582lowast) + 119860120576119873

(0) Define the curves

119886plusmn(120582) = Ω

119873(120582) plusmn 119879

ℎ119873(120582) + 119860

120576119873(0) (74)

The curves 119886+(120582) 119886minus(120582) enclose the curve ofΩ(120582) for suitably

large 119873 Hence the closure interval is determined by solving119886plusmn(120582) = 0 giving an interval

119868120576119873

= [119886minus 119886+] (75)

It is worthwhile to mention that the simplicity of the eigen-values guarantees the existence of approximate eigenvaluesthat is the 120582

119873rsquos for which Ω

119873(120582119873) = 0 Next we estimate the

error |120582lowast minus 120582119873| for the eigenvalue 120582lowast

Theorem 7 Let 120582lowast be an eigenvalue of (12)ndash(15) and let 120582119873

be its approximation Then for 120582 isin R we have the followingestimate

1003816100381610038161003816120582lowastminus 120582119873

1003816100381610038161003816lt

119879ℎ119873

(120582119873) + 119860120576119873

(0)

inf120577isin119868120576119873

1003816100381610038161003816Ω1015840(120577)

1003816100381610038161003816

(76)

where the interval 119868120576119873

is defined above

Proof Replacing 120582 by 120582119873in (73) we obtain

1003816100381610038161003816Ω (120582119873) minus Ω (120582

lowast)1003816100381610038161003816lt 119879ℎ119873

(120582119873) + 119860120576119873

(0) (77)

where we have used Ω119873(120582119873) = Ω(120582

lowast) = 0 Using the

mean value theorem yields that for some 120577 isin 119869120576119873

=

[min(120582lowast 120582119873)max(120582lowast 120582

119873)]

10038161003816100381610038161003816(120582lowastminus 120582119873)Ω1015840(120577)

10038161003816100381610038161003816le 119879ℎ119873

(120582119873) + 119860120576119873

(0) 120577 isin 119869120576119873

sub 119868120576119873

(78)

Since 120582lowast is simple and 119873 is sufficiently large then

inf120577isin119868120576119873

|Ω1015840(120577)| gt 0 and we get (76)

4 Numerical Examples

In this section we introduce two examples illustrating the theabove method Also in the following examples we observethat the exact solutions 120582

119896are all inside the interval [119886

minus 119886+]

In these two examples we indicate the effect of the amplitudeerror in the method by determining enclosure intervals fordifferent values of 120576 We also indicate the effect of 119873 and ℎ

by several choices The eigenvalues of the following examplescannot computed concretely then we use mathematicato obtain the exact values mathematica is also used inrounding off the exact eigenvalues which are square rootsBoth numerical results and the associated figures prove thecredibility of the method

Example 1 Consider the system

1199101015840

2(119909) minus 119901 (119909) 119910

1(119909) = 120582119910

1(119909)

1199101015840

1(119909) + 119901 (119909) 119910

2(119909) = minus120582119910

2(119909)

119909 isin [minus1 0) cup (0 1]

1205821199101(minus1) minus (radic2 + 120582) 119910

2(minus1) = 0

(radic2 + 120582) 1199101(1) minus 120582119910

2(1) = 0

1199101(0minus) minus 2119910

1(0+) = 0 119910

2(0minus) minus 2119910

2(0+) = 0

(79)

Here

1199011(119909) = 119901

2(119909) = 119903 (119909) =

119909 119909 isin [minus1 0)

1 (0 1]

(80)

1205791= 1205792= 1205874 120572

1= 1205732= 0 120572

2= 1205731= 1 and 120575 = 2 Direct

calculations give

K (120582) = 2 ((radic2120582 + 1) cos (2120582) minus 120582 (120582 + radic2) sin (2120582))

Ω (120582) = 2 ((radic2120582 + 1) cos(2120582 +

1

2

)

minus120582 (120582 + radic2) sin(2120582 +

1

2

))

(81)

As is clearly seen the eigenvalues cannot be computedexplicitly Tables 1 2 and 3 indicate the application of ourtechnique to this problem and the effect of 120576 By exact wemean the zeros of Δ(120582) computed by Mathematica

Figures 1 and 2 illustrate the enclosure intervals domi-nating 120582

0for 119873 = 25 ℎ = 01 and 120576 = 10

minus2 120576 = 10minus5

respectively The middle curve represents Δ(120582) while theupper and lower curves represent the curves of 119886

+(120582) 119886minus(120582)

respectively We notice that when 120576 = 10minus5 the two curves

are almost identical Similarly Figures 3 and 4 illustrate theenclosure intervals dominating 120582

1for ℎ = 01 119873 = 25 and

120576 = 10minus2 120576 = 10

minus5 respectively

Abstract and Applied Analysis 9

Table 1 The approximation 120582119896119873

and the exact solutions 120582119896for different choices of ℎ and119873

120582119896

120582minus2

120582minus1

1205820

1205821

Exact 120582119896

minus12061856254879546 minus05151195632138896 034274095892547224 1616237903407542

120582119896119873

ℎ = 06 119873 = 15 minus12061855965725146 minus05151190347168934 034274053593937165 1616237692038969120596 = 09708 119873 = 25 minus12061856254843495 minus05151195631493376 0342740958869788 16162379033820231ℎ = 01 119873 = 15 minus12061856254844203 minus05151195632158179 03427409589350932 16162379034134882

120596 = 14708 119873 = 25 minus12061856254879548 minus05151195632138894 034274095892547236 16162379034075425

Table 2 Absolute error |120582119896minus 120582119896119873

|

120582119896

120582minus2

120582minus1

1205820

1205821

ℎ = 06119873 = 15 289154 times 10

minus8528497 times 10

minus7422986 times 10

minus7211369 times 10

minus7

119873 = 25 360512 times 10minus12

64552 times 10minus11

556842 times 10minus11

255189 times 10minus11

ℎ = 01119873 = 15 353428 times 10

minus12192824 times 10

minus12962097 times 10

minus12594613 times 10

minus12

119873 = 25 222045 times 10minus16

222045 times 10minus16

555112 times 10minus17

444089 times 10minus16

Table 3 For119873 = 25 and ℎ = 01 the exact solutions 120582119896are all inside the interval [119886

minus 119886+] for different values of 120576

120582119896

120582minus2

120582minus1

1205820

1205821

Exact 120582119896

minus12061856254879546 minus05151195632138896 034274095892547224 1616237903407542119868120576119873

120576 = 10minus2

[minus12335696 minus11771922] [minus05486790 minus04826816] [03246478 03601346] [16100119 16224066]

119868120576119873

120576 = 10minus5

[minus12062137 minus12061575] [minus05151525 minus05150867] [03427232 03427586] [16162317 16162441]

030 032 034 036 038

00

02

04

aminus

a+

Ω(120582)

minus02

minus04

minus06

Figure 1The enclosure interval dominating 1205820for ℎ = 01119873 = 25

and 120576 = 10minus2

Example 2 In this example we consider the system

1199101015840

2(119909) minus 119901 (119909) 119910

1(119909) = 120582119910

1(119909)

1199101015840

1(119909) + 119901 (119909) 119910

2(119909) = minus120582119910

2(119909)

119909 isin [minus1 0) cup (0 1]

radic31199101(minus1) minus 119910

2(minus1) = 0

(1 +

1

2

120582)1199101(1) minus (1 +

radic3

2

120582)1199102(1) = 0

1199101(0minus) minus 3119910

1(0+) = 0 119910

2(0minus) minus 3119910

2(0+) = 0

(82)

030 032 034 036 038

00

02

aminus

a+

Ω(120582)

minus02

minus04

Figure 2The enclosure interval dominating 1205820for ℎ = 01119873 = 25

and 120576 = 10minus5

where

1199011(119909) = 119901

2(119909) = 119901 (119909) =

119909 119909 isin [minus1 0)

1199092 (0 1]

(83)

1205791= 1205792= 0 120572

1= 1205732= 1 120572

2= 1205731= 2 and 120575 = 12 Direct

calculations give

K (120582) =

1

2

((120582 + 3) cos (2120582) minus (1205822+ 3120582 + 4) sin (2120582))

Ω (120582) =

1

2

((1205822+ 3120582 + 4) sin(

1

6

minus 2120582)

+ (120582 + 3) cos(1

6

minus 2120582))

(84)

10 Abstract and Applied Analysis

Table 4 The approximation 120582119896119873

and the exact solutions 120582119896for different choices of ℎ and119873

120582119896

120582minus2

120582minus1

1205820

1205821

Exact 120582119896

minus30661256186938894 minus10959703083466883 03689030343814017 18338437675275228

120582119896119873

ℎ = 08 119873 = 15 minus3066123822200742 minus10959640944020903 036890463046491767 18338437607680411120596 = 007708 119873 = 25 minus30661256172349627 minus10959703026494807 036890303588091056 18338437676561417

ℎ = 02 119873 = 15 minus30661256186844446 minus10959703083278305 03689030343916001 1833843767517251120596 = 13708 119873 = 25 minus306612561869389 minus1095970308346688 03689030343814024 18338437675275225

Table 5 Absolute error |120582119896minus 120582119896119873

|

120582119896

120582minus2

120582minus1

1205820

1205821

ℎ = 08119873 = 15 179649 times 10

minus6621394 times 10

minus6159608 times 10

minus6675948 times 10

minus9

119873 = 25 145893 times 10minus9

569721 times 10minus9

149951 times 10minus9

128619 times 10minus10

ℎ = 02119873 = 15 944489 times 10

minus12188578 times 10

minus12101984 times 10

minus11102718 times 10

minus11

119873 = 25 444089 times 10minus16

222045 times 10minus16

666134 times 10minus16

222045 times 10minus16

Table 6 For119873 = 25 and ℎ = 02 the exact solutions 120582119896are all inside the interval [119886

minus 119886+] for different values of 120576

120582119896

120582minus2

120582minus1

1205820

1205821

Exact 120582119896

minus30661256186938894 minus10959703083466883 03689030343814017 18338437675275228119868120576119873

120576 = 10minus2

[minus31098742 minus30188665] [minus11631502 minus10339754] [03437434 03933486] [18220764 18454828]

119868120576119873

120576 = 10minus5

[minus30661709 minus30660803] [minus10960346 minus10959060] [03688783 03689278] [18338321 18338555]

158 159 160 161 162 163 164 165

00

05

10

minus05

minus10

aminus

a+

Ω(120582)

Figure 3The enclosure interval dominating 1205821for ℎ = 01119873 = 25

and 120576 = 10minus2

Tables 4 and 6 give the exact eigenvalues 1205821198961

119896=minus2and their

approximate ones for different values of ℎ 119873 and 120576 InTable 5 we give the absolute error for different values of ℎ and119873

Here Figures 5 6 7 and 8 illustrate the enclosureintervals dominating 120582

minus2and 120582

minus1for ℎ = 02 119873 = 25 and

120576 = 10minus2 120576 = 10

minus5 respectively

158 159 160 161 162 163 164 165

00

05

minus05

aminus

a+

Ω(120582)

Figure 4The enclosure interval dominating 1205821for ℎ = 01119873 = 25

and 120576 = 10minus5

5 Conclusion

With a simple analysis and with values of solutions ofinitial value problems computed at a few values of theeigenparameter we have computed the eigenvalues of aDirac system which has a discontinuity at one point andcontains a spectral parameter in all boundary conditionswith a certain estimated error The method proposed is

Abstract and Applied Analysis 11

00

05

10

minus05

minus33 minus32 minus31 minus30 minus29 minus28

aminus

a+

Ω(120582)

Figure 5 The enclosure interval dominating 120582minus2

for ℎ = 02 119873 =

25 and 120576 = 10minus2

00

05

10

minus05

minus33 minus32 minus31 minus30 minus29 minus28

aminus

a+

Ω(120582)

Figure 6 The enclosure interval dominating 120582minus2

for ℎ = 02 119873 =

25 and 120576 = 10minus5

a shooting procedure that is the problem is reformulatedas two initial value ones due to the interior discontinuity ofsize two and a miss-distance is defined at the right end of theinterval of integration whose roots are the eigenvalues to becomputed The unknown partU(120582) of the miss-distance canbewritten in terms of a functionwhich is an entire function ofexponential typeTherefore we propose to approximate suchterm by means of a truncated cardinal series with samplingvalues approximated by solving numerically correspondingsuitable initial value problems Finally in Section 4 weintroduced two instructive examples where both numericalresults and the associated figures have proved the credibilityof the method

00

05

10

minus05

minus14 minus13 minus12 minus11 minus10 minus09 minus08 minus07

aminus

a+

Ω(120582)

Figure 7 The enclosure interval dominating 120582minus1

for ℎ = 02 119873 =

25 and 120576 = 10minus2

00

05

10

minus05

minus14 minus13 minus12 minus11 minus10 minus09 minus08 minus07

aminus

a+

Ω(120582)

Figure 8 The enclosure interval dominating 120582minus1

for ℎ = 02 119873 =

25 and 120576 = 10minus5

Conflict of Interests

The authors declare that there is no conflict of interestsregarding the publication of this paper

Acknowledgment

This paper was funded by the Deanship of Scientific Research(DSR) King Abdulaziz University Jeddah The authorstherefore acknowledge with thanks DSR technical and finan-cial support

12 Abstract and Applied Analysis

References

[1] C T Fulton ldquoTwo-point boundary value problems witheigenvalue parameter contained in the boundary conditionsrdquoProceedings of the Royal Society of Edinburgh A Mathematicsvol 77 no 3-4 pp 293ndash308 1977

[2] JWalter ldquoRegular eigenvalue problems with eigenvalue param-eter in the boundary conditionrdquoMathematische Zeitschrift vol133 pp 301ndash312 1973

[3] V Kotelnikov ldquoOn the carrying capacity of the ether and wirein telecommunications material for the first All-Unionrdquo inConference on Questions of Communications vol 55 pp 55ndash64Izd Red Upr Svyazi RKKA 1933

[4] C E Shannon ldquoCommunication in the presence of noiserdquoProceedings of the IRE vol 37 pp 10ndash21 1949

[5] E TWhittaker ldquoOn the functions which are represented by theexpansion of the interpolation theoryrdquo Proceedings of the RoyalSociety of Edinburgh A vol 35 pp 181ndash194 1915

[6] P L Butzer G Schmeisser and R L Stens ldquoAn introductionto sampling analysisrdquo in Non Uniform Sampling Theory andPractices F Marvasti Ed pp 17ndash121 Kluwer New York NYUSA 2001

[7] M A Kowalski K A Sikorski and F Stenger Selected Topics inApproximation andComputation OxfordUniversity Press NewYork NY USA 1995

[8] J Lund and K L Bowers Sinc Methods for Quadrature andDifferential Equations SIAM Philadelphia Pa USA 1992

[9] F Stenger ldquoNumericalmethods based onWhittaker cardinal orsinc functionsrdquo SIAM Review vol 23 no 2 pp 165ndash224 1981

[10] F Stenger Numerical Methods Based on Sinc and AnalyticFunctions Springer New York NY USA 1993

[11] A Boumenir ldquoHigher approximation of eigenvalues by thesampling methodrdquo BIT Numerical Mathematics vol 40 no 2pp 215ndash225 2000

[12] A Boumenir ldquoSampling and eigenvalues of non-self-adjointSturm-Liouville problemsrdquo SIAM Journal on Scientific Comput-ing vol 23 no 1 pp 219ndash229 2001

[13] M M Tharwat A H Bhrawy and A Yildirim ldquoNumericalcomputation of eigenvalues of discontinuous Sturm-Liouvilleproblems with parameter dependent boundary conditionsusing sinc methodrdquoNumerical Algorithms vol 63 no 1 pp 27ndash48 2013

[14] M M Tharwat A H Bhrawy and A Yildirim ldquoNumericalcomputation of the eigenvalues of a discontinuousDirac systemusing the sincmethodwith error analysisrdquo International Journalof Computer Mathematics vol 89 no 15 pp 2061ndash2080 2012

[15] P L Butzer and R L Stens ldquoA modification of the Whittaker-Kotelrsquonikov-Shannon sampling seriesrdquo Aequationes Mathemat-icae vol 28 no 3 pp 305ndash311 1985

[16] R Gervais Q I Rahman and G Schmeisser ldquoA bandlimitedfunction simulating a duration-limited onerdquo in ApproximationTheory andFunctionalAnalysis P L Butzer andR L Stens Edspp 355ndash362 Birkhaauser Boston Mass USA 1984

[17] R L Stens ldquoSampling by generalized kernelsrdquo in SamplingThe-ory in Fourier and Signal Analysis Advanced Topics J RHigginsand R L Stens Eds pp 130ndash157 Oxford University PressOxford UK 1999

[18] G Schmeisser and F Stenger ldquoSinc approximation with a Gaus-sianmultiplierrdquo SamplingTheory in Signal and Image Processingvol 6 no 2 pp 199ndash221 2007

[19] L Qian ldquoOn the regularized Whittaker-Kotelrsquonikov-Shannonsampling formulardquo Proceedings of the American MathematicalSociety vol 131 no 4 pp 1169ndash1176 2003

[20] L Qian and D B Creamer ldquoA modification of the samplingseries with aGaussianmultiplierrdquo SamplingTheory in Signal andImage Processing vol 5 no 1 pp 1ndash20 2006

[21] L Qian and D B Creamer ldquoLocalized sampling in the presenceof noiserdquoAppliedMathematics Letters vol 19 no 4 pp 351ndash3552006

[22] M H Annaby and R M Asharabi ldquoComputing eigenvaluesof boundary-value problems using sinc-Gaussian methodrdquoSamplingTheory in Signal and Image Processing vol 7 no 3 pp293ndash312 2008

[23] M H Annaby and M MTharwat ldquoA sinc-Gaussian techniquefor computing eigenvalues of second-order linear pencilsrdquoApplied Numerical Mathematics vol 63 pp 129ndash137 2013

[24] A H Bhrawy M M Tharwat and A Al-Fhaid ldquoNumericalalgorithms for computing eigenvalues of discontinuous Diracsystem using sinc-Gaussian methodrdquo Abstract and AppliedAnalysis vol 2012 Article ID 925134 13 pages 2012

[25] M M Tharwat A H Bhrawy and A S Alofi ldquoApproximationof eigenvalues of discontinuous sturm-liouville problems witheigenparameter in all boundary conditionsrdquo Boundary ValueProblems vol 2013 article 132 2013

[26] M M Tharwat and S M Al-Harbi ldquoApproximation of eigen-values of boundary value problemsrdquo Boundary Value Problemsvol 2014 article 51 2014

[27] MMTharwat and A H Bhrawy ldquoComputation of eigenvaluesof discontinuous Dirac system using Hermite interpolationtechniquerdquo Advances in Difference Equations vol 2012 article59 2012

[28] M M Tharwat A Yildirim and A H Bhrawy ldquoSampling ofdiscontinuous Dirac systemsrdquo Numerical Functional Analysisand Optimization vol 34 no 3 pp 323ndash348 2013

[29] M M Tharwat ldquoDiscontinuous Sturm-Liouville problems andassociated sampling theoriesrdquo Abstract and Applied Analysisvol 2011 Article ID 610232 30 pages 2011

[30] MMTharwat ldquoOn sampling theories and discontinuousDiracsystems with eigenparameter in the boundary conditionsrdquoBoundary Value Problems vol 2013 article 65 2013

[31] B M Levitan and I S Sargsjan ldquoIntroduction to spectral the-ory self adjoint ordinary differential operatorsrdquo in Translationof Mthematical Monographs vol 39 American MathematicalSociety Providence RI USA 1975

[32] B M Levitan and I S Sargsjan Sturm-Liouville and Dirac oper-ators vol 59 Kluwer Academic Publishers Group DordrechtThe Netherlands 1991

Submit your manuscripts athttpwwwhindawicom

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical Problems in Engineering

Hindawi Publishing Corporationhttpwwwhindawicom

Differential EquationsInternational Journal of

Volume 2014

Applied MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Probability and StatisticsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical PhysicsAdvances in

Complex AnalysisJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

OptimizationJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

CombinatoricsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Operations ResearchAdvances in

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Function Spaces

Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of Mathematics and Mathematical Sciences

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

The Scientific World JournalHindawi Publishing Corporation httpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Algebra

Discrete Dynamics in Nature and Society

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Decision SciencesAdvances in

Discrete MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom

Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Stochastic AnalysisInternational Journal of

2 Abstract and Applied Analysis

such that Φ(0) = 1 then for 119891 isin B2120590we have the expansion

[18]

119891 (120582) =

infin

sum

119899=minusinfin

119891 (119899ℎ) sinc (ℎminus1120587120582 minus 119899120587)Φ (ℎminus1120582 minus 119899) (4)

The speed of convergence of the series in (4) is determinedby the decay of |Φ(120582)| But the decay of an entire functionof exponential type cannot be as fast as 119890minus119888|119909| as |119909| rarr infinfor some positive 119888 [18] For 120590 gt 0 ℎ isin (0 120587120590] and 120596 =

(120587minusℎ120590)2 Schmeisser and Stenger [18] defined the operator

(Gℎ119873

119891) (120582) = sum

119899isinZ119873(120582)

119891 (119899ℎ) sinc (ℎminus1120587120582 minus 119899120587)

times 119866(

radic120596 (120582 minus 119899ℎ)

radic119873ℎ

)

(5)

where 119866(119905) = exp(minus1199052) which is called the Gaussianfunction Z

119873(120582) = 119899 isin Z |[ℎ

minus1R120582 + 12] minus 119899| le 119873119873 isin N and [120582] denotes the integer part of 120582 isin R see also[19ndash21] Note that the summation limits in (5) depend on thereal part of 120582 The authors [18] proved that if 119891 is an entirefunction such that

1003816100381610038161003816119891 (120585 + 119894120578)

1003816100381610038161003816le 120601 (

10038161003816100381610038161205851003816100381610038161003816) 119890120590|120578|

120585 120578 isin R (6)

where 120601 is a nondecreasing nonnegative function on [0infin)

and 120590 ge 0 then for ℎ isin (0 120587120590) 120596 = (120587 minus ℎ120590)2119873 isin N and|I120582| lt 119873 we have

1003816100381610038161003816119891 (120582) minus (G

ℎ119873119891) (120582)

1003816100381610038161003816

le 2

10038161003816100381610038161003816sin (ℎ

minus1120587120582)

10038161003816100381610038161003816120601 (|R120582| + ℎ (119873 + 1))

times

119890minus120596119873

radic120587120596119873

120573119873(ℎminus1I120582) 120582 isin 119862

(7)

where

120573119873(119905) = cosh (2120596119905) +

21198901205961199052119873

radic120587120596119873[1 minus (119905119873)2]

+

1

2

[

1198902120596119905

1198902120587(119873minus119905)

minus 1

+

119890minus2120596119905

1198902120587(119873+119905)

minus 1

]

(8)

In [22] the authors derived an estimate for amplitudeerror They proved that if

sup119899isinZ119899(120582)

10038161003816100381610038161003816119891 (119899ℎ) minus

119891 (119899ℎ)

10038161003816100381610038161003816lt 120576 (9)

holds then for |I120582| lt 119873 we have10038161003816100381610038161003816(Gℎ119873

119891) (120582) minus (Gℎ119873

119891) (120582)

10038161003816100381610038161003816le 119860120576119873

(I120582) (10)

where 119891(119899ℎ) is the approximations of the exact values 119891(119899ℎ)

of (5) 120576 gt 0 is sufficiently small

119860120576119873

(I120582) = 2120576119890minus1205964119873

(1 + radic119873

120596120587

) exp ((120596 + 120587) ℎminus1

|I120582|)

(11)

Consider the Dirac system which consists of the systemof differential equations

(u10158402(119909) minus 119901

1(119909)u1(119909)

u10158401(119909) + 119901

2(119909)u2(119909)

) = 120582(u1(119909)

minusu2(119909)

)

119909 isin [minus1 0) cup (0 1]

(12)

with boundary conditions

L1(u) = (120572

1+ 120582 sin 120579

1)u1(minus1)

minus (1205722+ 120582 cos 120579

1)u2(minus1) = 0

(13)

L2(u) = (120573

1+ 120582 sin 120579

2)u1(1)

minus (1205732+ 120582 cos 120579

2)u2(1) = 0

(14)

and transmission conditions

L3(u) = u

1(0minus) minus 120575u

1(0+) = 0

L4(u) = u

2(0minus) minus 120575u

2(0+) = 0

(15)

Here 120582 is the spectral parameter u = (u1u2 ) the real valued

function 1199011(sdot) and 119901

2(sdot) are continuous in [minus1 0) and (0 1]

and have finite limits 1199011(0plusmn) = lim

119909rarr0plusmn1199011(119909) 119901

2(0plusmn) =

lim119909rarr0

plusmn1199012(119909) 120572

1 1205722 1205731 1205732 120575 isin R 120579

1 1205792isin [0 120587) and 120575 = 0

Throughout the following we assume that

1205881= 1205722sin 1205791minus 1205721cos 1205791gt 0

1205882= 1205731cos 1205792minus 1205732sin 1205792gt 0

(16)

The aim of the present work is to compute the eigenvaluesof (12)ndash(15) numerically by the sinc-Gaussian techniquewith errors analysis truncation error and amplitude errorIn [23] Annaby and Tharwat computed the eigenvalues ofa second-order operator pencil by sinc-Gaussian methodAlso the authors introduced some examples illustrating thesinc-Gaussian method accompanied by comparison withthe sinc-method which explained that the sinc-Gaussianmethod gives remarkably better results see also [24 25]Tharwat and Al-Harbi [26] computed the eigenvalues ofdiscontinuous Dirac system but with eigenparameter in oneboundary condition In [14 27] Tharwat et al computedthe eigenvalues of discontinuousDirac system approximatelywith eigenparameter appears in any of boundary conditionsby Hermite interpolations and regularized sinc-methodsrespectively In regularized sinc-method also the same inHermite interpolations method the basic idea is that theeigenvalues are characterised as the zeros of an analyticfunction 119891(120582) which can be written in the form 119891(120582) =

119896(120582) + 119906(120582) where 119896(120582) is known part The ingenuity of theapproach is in trying to choose the function 119891(120582) so that119906(120582) isin B2

120590(unknown part) and can be approximated by the

sampling theorem if its values at some equally spaced pointsare known see [11ndash14] Recall that in regularized sinc andHermite interpolations methods it is necessary that 119906(120582) is1198712-function In this paper wewill use sinc-Gaussian sampling

formula (5) to compute eigenvalues of (12)ndash(15) numerically

Abstract and Applied Analysis 3

The proposed method reduces the error bounds remarkably(see examples of Section 4) The basic idea is to write thefunction of eigenvalues as the sum of two terms one knownand the other unknown but an entire function of exponentialtype which satisfies (6) In other words the unknown termis not necessarily an 119871

2-function Then we approximate theunknown part using (5) and obtain better results We wouldlike to mention that the papers in computing eigenvaluesby sinc-Gaussian method are few see [22ndash26] In Sections2 and 3 we discuss some properties of the eigenvalues ofthe boundary value problems (12)ndash(15) and also we derivethe sinc-Gaussian technique to compute the eigenvalues of(12)ndash(15) with error estimates The last section involves someillustrative examples

2 Some Important Results

In the following we study some properties of the eigenvaluesof the problems (12)ndash(15) which need in our method see [2628] For functions u(119909) which defined on [minus1 0) cup (0 1] andhas finite limitu(plusmn0) = lim

119909rarrplusmn0u(119909) byu

(1)(119909) andu

(2)(119909)

we denote the functions

u(1)

(119909) =

u (119909) 119909 isin [minus1 0)

u (0minus) 119909 = 0

u(2)

(119909) =

u (119909) 119909 isin (0 1]

u (0+) 119909 = 0

(17)

which are defined on Γ1

= [minus1 0] and Γ2

= [0 1]respectively

In the following lemma wewill prove that the eigenvaluesof the problems (12)ndash(15) are real see [29 30]

Lemma 1 The eigenvalues of the problems (12)ndash(15) are real

Proof Assume the contrary that 1205820is a nonreal eigenvalue of

problems (12)ndash(15) Let ( u1(119909)

u2(119909)) be a corresponding (nontriv-

ial) eigenfunction By (12) we have

119889

119889119909

u1(119909)u2(119909) minus u

1(119909)u2(119909)

= (1205820minus 1205820)

1003816100381610038161003816u1(119909)

1003816100381610038161003816

2

+1003816100381610038161003816u2(119909)

1003816100381610038161003816

2

119909 isin [minus1 0) cup (0 1]

(18)

Integrating the above equation through [minus1 0] and [0 1] weobtain

(1205820minus 1205820) [int

0

minus1

(1003816100381610038161003816u1(119909)

1003816100381610038161003816

2

+1003816100381610038161003816u2(119909)

1003816100381610038161003816

2

) 119889119909]

= u1(0minus) 1199062(0minus) minus u1(0minus)u2(0minus)

minus [u1(minus1)u

2(minus1) minus u

1(minus1)u

2(minus1)]

(1205820minus 1205820) [int

1

0

(1003816100381610038161003816u1(119909)

1003816100381610038161003816

2

+1003816100381610038161003816u2(119909)

1003816100381610038161003816

2

) 119889119909]

= u1(1)u2(1) minus u

1(1)u2(1)

minus [u1(0+)u2(0+) minus u1(0+)u2(0+)]

(19)

Then from (13) (14) and transmission conditions we haverespectively

u1(minus1)u

2(minus1) minus u

1(minus1)u

2(minus1)

=

1205881(1205820minus 1205820)1003816100381610038161003816u2(minus1)

1003816100381610038161003816

2

10038161003816100381610038161205721+ 1205820sin 1205791

1003816100381610038161003816

2

u1(1)u2(1) minus u

1(1)u2(1)

= minus

1205882(1205820minus 1205820)1003816100381610038161003816u2(1)

1003816100381610038161003816

2

10038161003816100381610038161205731+ 1205820sin 1205792

1003816100381610038161003816

2

u1(minus0) 119906

2(minus0) minus u

1(minus0)u

2(minus0)

= 1205752[u1(+0) 119906

2(+0) minus u

1(+0)u

2(+0)]

(20)

Since 1205820

= 1205820 it follows from the last three equations and (19)

that

int

0

minus1

(1003816100381610038161003816u1(119909)

1003816100381610038161003816

2

+1003816100381610038161003816u2(119909)

1003816100381610038161003816

2

) 119889119909

+ 1205752int

1

0

(1003816100381610038161003816u1(119909)

1003816100381610038161003816

2

+1003816100381610038161003816u2(119909)

1003816100381610038161003816

2

) 119889119909

= minus

1205881

1003816100381610038161003816u2(minus1)

1003816100381610038161003816

2

10038161003816100381610038161205721+ 1205820sin 1205791

1003816100381610038161003816

2minus

120588212057521003816100381610038161003816u2(1)

1003816100381610038161003816

2

10038161003816100381610038161205731+ 1205820sin 1205792

1003816100381610038161003816

2

(21)

This contradicts the conditions int0minus1(|u1(119909)|2+ |u2(119909)|2)119889119909 +

1205752int

1

0(|u1(119909)|2+ |u2(119909)|2)119889119909 gt 0 and 120588

119894gt 0 119894 = 1 2

Consequently 1205820must be real

Now we shall construct a special fundamental system ofsolutions of (12) for 120582 not being an eigenvalue Let us considerthe next initial value problem

u1015840

2(119909) minus 119901

1(119909)u1(119909) = 120582u

1(119909)

u1015840

1(119909) + 119901

2(119909)u2(119909) = minus120582u

2(119909)

119909 isin (minus1 0)

(22)

u1(minus1) = 120572

2+ 120582 cos 120579

1 u

2(minus1) = 120572

1+ 120582 sin 120579

1 (23)

By virtue of Theorem 11 in [31] this problem has a uniquesolution u = (

12059311(119909120582)

12059321(119909120582)) which is an entire function of 120582 isin C

for each fixed 119909 isin [minus1 0] Similarly employing the same

4 Abstract and Applied Analysis

method as in proof of Theorem 11 in [31] we see that theproblem

u1015840

2(119909) minus 119901

1(119909)u1(119909) = 120582u

1(119909)

u1015840

1(119909) + 119901

2(119909)u2(119909) = minus120582u

2(119909)

119909 isin (0 1)

u1(1) = 120573

2+ 120582 cos 120579

2 u

2(1) = 120573

1+ 120582 sin 120579

2

(24)

has a unique solution u = (12059512(119909120582)

12059522(119909120582)) which is an entire

function of parameter 120582 for each fixed 119909 isin [0 1]Now the functions 120593

1198942(119909 120582) and 120595

1198941(119909 120582) are defined in

terms of 1205931198941(119909 120582) and 120595

1198942(119909 120582) 119894 = 1 2 respectively as

follows the initial-value problem

u1015840

2(119909) minus 119901

1(119909)u1(119909) = 120582u

1(119909)

u1015840

1(119909) + 119901

2(119909)u2(119909) = minus120582u

2(119909)

119909 isin (0 1)

(25)

u1(0) =

1

120575

12059311

(0 120582) u2(0) =

1

120575

12059321

(0 120582) (26)

has unique solution u = (12059312(119909120582)

12059322(119909120582)) for each 120582 isin C

Similarly the following problem also has a unique solu-tion u = (

12059511(119909120582)

12059521(119909120582))

u1015840

2(119909) minus 119901

1(119909)u1(119909) = 120582u

1(119909)

u1015840

1(119909) + 119901

2(119909)u2(119909) = minus120582u

2(119909)

119909 isin (minus1 0)

(27)

u1(0) = 120575120595

12(0 120582) u

2(0) = 120575120595

22(0 120582) (28)

Let us construct two basic solutions of (12) as

120593 (sdot 120582) = (1205931(sdot 120582)

1205932(sdot 120582)

) 120595 (sdot 120582) = (1205951(sdot 120582)

1205952(sdot 120582)

) (29)

where

1205931(119909 120582) =

12059311

(119909 120582) 119909 isin [minus1 0)

12059312

(119909 120582) 119909 isin (0 1]

1205932(119909 120582) =

12059321

(119909 120582) 119909 isin [minus1 0)

12059322

(119909 120582) 119909 isin (0 1]

(30)

1205951(119909 120582) =

12059511

(119909 120582) 119909 isin [minus1 0)

12059512

(119909 120582) 119909 isin (0 1]

1205952(119909 120582) =

12059521

(119909 120582) 119909 isin [minus1 0)

12059522

(119909 120582) 119909 isin (0 1]

(31)

By virtue of (26) and (28) these solutions satisfy bothtransmission conditions (15) These functions are entire in 120582

for all 119909 isin [minus1 0) cup (0 1]

Let W(120593 120595)(sdot 120582) denote the Wronskian of 120593(sdot 120582) and120595(sdot 120582) defined in [32 page 194] that is

W (120593 120595) (sdot 120582) =

10038161003816100381610038161003816100381610038161003816

1205931(sdot 120582) 120593

2(sdot 120582)

1205951(sdot 120582) 120595

2(sdot 120582)

10038161003816100381610038161003816100381610038161003816

(32)

It is obvious that the Wronskians

Ω119894(120582) = W (120593 120595) (119909 120582)

= 1205931119894(119909 120582) 120595

2119894(119909 120582)

minus 1205932119894(119909 120582) 120595

1119894(119909 120582) 119909 isin Γ

119894 119894 = 1 2

(33)

are independent on variable 119909 isin Γ119894(119894 = 1 2) and 120593

119894(119909 120582)

and 120595119894(119909 120582) are entire functions of the parameter 120582 for each

119909 isin Γ119894(119894 = 1 2) Taking into account (26) and (28) a short

calculation gives

Ω1(120582) = 120575

2Ω2(120582) (34)

for each 120582 isin C

Corollary 2 The zeros of the functions Ω1(120582) and Ω

2(120582)

coincide

Then we may introduce to the consideration the charac-teristic functionΩ(120582) as

Ω (120582) = Ω1(120582) = 120575

2Ω2(120582) (35)

Note that all eigenvalues of problems (12)ndash(15) are just zerosof the function Ω(120582) Indeed since the functions 120593

1(119909 120582)

and 1205932(119909 120582) satisfy the boundary condition (13) and both

transmission conditions (15) to find the eigenvalues ofthe (12)ndash(15) we have to insert the functions 120593

1(119909 120582) and

1205932(119909 120582) in the boundary condition (14) and find the roots

of this equation In the following lemma we show that alleigenvalues of the problems (12)ndash(15) are simple

Lemma 3 The eigenvalues of the boundary value problems(12)ndash(15) form an at most countable set without finite limitpoints All eigenvalues of the boundary value problems (12)ndash(15) (of Ω(120582)) are simple

Proof The eigenvalues are the zeros of the entire functionoccurring on the left-hand side in see (35)

(1205731+ 120582 sin 120579

2) 12059312

(1 120582) minus (1205732+ 120582 cos 120579

2) 12059322

(1 120582) = 0

(36)

We have shown (see Lemma 1) that this function doesnot vanish for nonreal 120582 In particular it does not vanishidentically Therefore its zeros form an at most countable setwithout finite limit points

By (12) we obtain for 120582 120583 isin C 120582 = 120583

119889

119889119909

1205931(119909 120582) 120593

2(119909 120583) minus 120593

1(119909 120583) 120593

2(119909 120582)

= (120583 minus 120582) 1205931(119909 120582) 120593

1(119909 120583) + 120593

2(119909 120582) 120593

2(119909 120583)

(37)

Abstract and Applied Analysis 5

Integrating the above equation through [minus1 0] and [0 1] andtaking into account the initial conditions (23) (26) and (28)we obtain

1205752(12059312

(1 120582) 12059322

(1 120583) minus 12059312

(1 120583) 12059322

(1 120582))

minus (120583 minus 120582) 1205881

= (120583 minus 120582) (int

0

minus1

(12059311

(119909 120582) 12059311

(119909 120583)

+ 12059321

(119909 120582) 12059321

(119909 120583)) 119889119909

+ 1205752int

1

0

(12059312

(119909 120582) 12059312

(119909 120583)

+ 12059322

(119909 120582) 12059322

(119909 120583)) 119889119909)

(38)

Dividing both sides of (38) by (120582 minus 120583) and by letting 120583 rarr 120582we arrive to the relation

1205752(12059322

(1 120582)

12059712059312

(1 120582)

120597120582

minus 12059312

(1 120582)

12059712059322

(1 120582)

120597120582

) + 1205881

= minus(int

0

minus1

(100381610038161003816100381612059311

(119909 120582)1003816100381610038161003816

2

+100381610038161003816100381612059321

(119909 120582)1003816100381610038161003816

2

) 119889119909

+ 1205752int

1

0

(100381610038161003816100381612059312

(119909 120582)1003816100381610038161003816

2

+100381610038161003816100381612059322

(119909 120582)1003816100381610038161003816

2

) 119889119909)

(39)

We show that equation

Ω (120582) = W (120593 120595) (1 120582)

= 1205752((1205731+ 120582 sin 120579

2) 12059312

(1 120582)

minus (1205732+ 120582 cos 120579

2) 12059322

(1 120582)) = 0

(40)

has only simple roots Assume the converse that is (40) hasa double root 120582lowast Then the following two equations hold

(1205731+ 120582lowast sin 120579

2) 12059312

(1 120582lowast) minus (120573

2+ 120582lowast cos 120579

2) 12059322

(1 120582lowast) = 0

sin 120579212059312

(1 120582lowast) + (120573

1+ 120582lowast sin 120579

2)

12059712059312

(1 120582lowast)

120597120582

minus cos 120579212059322

(1 120582lowast) minus (120573

2+ 120582lowast cos 120579

2)

12059712059322

(1 120582lowast)

120597120582

= 0

(41)

Since 1205882

= 0 and 120582lowast is real then (120573

1+ 120582lowast sin 120579

2)2

+

(1205732+ 120582lowast cos 120579

2)2

= 0 Let 1205731+ 120582lowast sin 120579

2= 0 From (41)

12059312

(1 120582lowast) =

(1205732+ 120582lowast cos 120579

2)

(1205731+ 120582lowast sin 120579

2)

12059322

(1 120582lowast)

12059712059312

(1 120582lowast)

120597120582

=

120588212059322

(1 120582lowast)

(1205731+ 120582lowast sin 120579

2)2

+

(1205732+ 120582lowast cos 120579

2)

(1205731+ 120582lowast sin 120579

2)

12059712059322

(1 120582lowast)

120597120582

(42)

Combining (42) and (39) with 120582 = 120582lowast we obtain

12058821205752(12059322

(1 120582lowast))2

(1205731+ 120582lowast sin 120579

2)2

+ 1205881

= minus(int

0

minus1

(100381610038161003816100381612059311

(119909 120582)1003816100381610038161003816

2

+100381610038161003816100381612059321

(119909 120582)1003816100381610038161003816

2

) 119889119909

+ 1205752int

1

0

(

1003816100381610038161003816100381612059312

(119909 120582) |2+

1003816100381610038161003816100381612059322

(119909 120582) |2) 119889119909)

(43)

contradicting the assumption 120588119894gt 0 119894 = 1 2 The other case

when 1205732+120582lowast cos 120579

2= 0 can be treated similarly and the proof

is complete

Recall that boundary value problems (12)ndash(15) have denu-merable set of real and simple eigenvalues see the abovesection From (23) and (26) the solution of (12)

120593 (sdot 120582) = (1205931(sdot 120582)

1205932(sdot 120582)

)

120593119894(119909 120582) =

1205931198941(119909 120582) 119909 isin [minus1 0)

1205931198942(119909 120582) 119909 isin (0 1]

119894 = 1 2

(44)

satisfies the following initial conditions

(12059311

(minus1 120582) 12059312

(0+ 120582)

12059321

(minus1 120582) 12059322

(0+ 120582)

)

= (1205722+ 120582 cos 120579

1120575minus112059311

(0minus 120582)

1205721+ 120582 sin 120579

1120575minus112059321

(0minus 120582)

)

(45)

Since 120593(sdot 120582) satisfies (13) then the eigenvalues of the prob-lems (12)ndash(15) are the zeros of the function

Ω (120582) = 1205752((1205731+ 120582 sin 120579

2) 12059312

(1 120582)

minus (1205732+ 120582 cos 120579

2) 12059322

(1 120582)) = 0

(46)

We shall transform (12) (23) (26) and (30) into the integralequations see [32]

12059311

(119909 120582) = 1205722cos 120582 (119909 + 1) minus 120572

1sin 120582 (119909 + 1)

+ 120582 cos (120582 (119909 + 1) + 1205791) minusTminus11

12059311

(119909 120582)

minusTminus12

12059321

(119909 120582)

(47)

12059321

(119909 120582) = 1205721cos 120582 (119909 + 1) + 120572

2sin 120582 (119909 + 1)

+ 120582 sin (120582 (119909 + 1) + 1205791) +

Tminus11

12059311

(119909 120582)

minusTminus12

12059321

(119909 120582)

(48)

12059312

(119909 120582) =

1

120575

12059311

(0minus 120582) cos (120582119909) minus 1

120575

12059321

(0minus 120582) sin (120582119909)

minusT01

12059312

(119909 120582) minusT02

12059322

(119909 120582)

(49)

6 Abstract and Applied Analysis

12059322

(119909 120582) =

1

120575

12059311

(0minus 120582) sin (120582119909) +

1

120575

12059321

(0minus 120582) cos (120582119909)

+T01

12059312

(119909 120582) minusT02

12059322

(119909 120582)

(50)

where Tminus1119894

Tminus1119894

T0119894

and T0119894 119894 = 1 2 are the Volterra

integral operators defined by

Tminus1119894

120593 (119909 120582) = int

119909

minus1

sin 120582 (119909 minus 119905) 119901119894(119905) 120593 (119905 120582) 119889119905

Tminus1119894

120593 (119909 120582) = int

119909

minus1

cos 120582 (119909 minus 119905) 119901119894(119905) 120593 (119905 120582) 119889119905

T0119894120593 (119909 120582) = int

119909

0

sin 120582 (119909 minus 119905) 119901119894(119905) 120593 (119905 120582) 119889119905

T0119894120593 (119909 120582) = int

119909

0

cos 120582 (119909 minus 119905) 119901119894(119905) 120593 (119905 120582) 119889119905

(51)

For convenience we define the constants

1198881= max 100381610038161003816

10038161205721

1003816100381610038161003816+10038161003816100381610038161205722

10038161003816100381610038161003816100381610038161003816sin 1205791

1003816100381610038161003816+1003816100381610038161003816cos 1205791

1003816100381610038161003816

1198882= int

0

minus1

[10038161003816100381610038161199011(119905)

1003816100381610038161003816+10038161003816100381610038161199012(119905)

1003816100381610038161003816] 119889119905 119888

3= 11988811198882exp (119888

2)

1198884= int

1

0

[10038161003816100381610038161199011(119905)

1003816100381610038161003816+10038161003816100381610038161199012(119905)

1003816100381610038161003816] 119889119905 119888

5= 1198883+

2

|120575|

(1198881+ 1198883)

1198886= max 100381610038161003816

10038161205721

1003816100381610038161003816+10038161003816100381610038161205722

10038161003816100381610038161003816100381610038161003816sin 1205792

1003816100381610038161003816+1003816100381610038161003816cos 1205792

1003816100381610038161003816

(52)

Now we define yminus1119894

(sdot 120582) and y0119894(sdot 120582) 119894 = 1 2 to be

yminus11

(119909 120582) = Tminus11

12059311

(119909 120582) +Tminus12

12059321

(119909 120582)

yminus12

(119909 120582) =Tminus11

12059311

(119909 120582) minusTminus12

12059321

(119909 120582)

y01

(119909 120582) = T01

12059312

(119909 120582) +T02

12059322

(119909 120582)

y02

(119909 120582) =T01

12059312

(119909 120582) minusT02

12059322

(119909 120582)

(53)

Lemma 4 The functions yminus11

(119909 120582) and yminus12

(119909 120582) are entirein 120582 for any fixed 119909 isin [minus1 0) and satisfy the growth condition

1003816100381610038161003816yminus11

(119909 120582)1003816100381610038161003816

1003816100381610038161003816yminus12

(119909 120582)1003816100381610038161003816le 21198883(1 + |120582|) 119890

|I120582|(119909+1)

120582 isin C

(54)

Proof Since yminus11

(119909 120582) = Tminus11

12059311(119909 120582) +

Tminus12

12059321(119909 120582)

then from (47) and (48) we obtain

yminus11

(119909 120582)

= 1205722Tminus11

cos 120582 (119909 + 1) minus 1205721Tminus11

sin 120582 (119909 + 1)

+ 120582Tminus11

cos (120582 (119909 + 1) + 1205791) minusTminus11

yminus11

(119909 120582)

+ 1205721Tminus12

cos 120582 (119909 + 1) + 1205722Tminus12

sin 120582 (119909 + 1)

+ 120582Tminus12

sin (120582 (119909 + 1) + 1205791) +

Tminus12

yminus12

(119909 120582)

(55)

Using the inequalities | sin 119911| le 119890|I119911| and | cos 119911| le 119890

|I119911| for119911 isin C leads for 120582 isin C to1003816100381610038161003816yminus11

(119909 120582)1003816100381610038161003816

le10038161003816100381610038161205722

1003816100381610038161003816

1003816100381610038161003816Tminus11

cos 120582 (119909 + 1)1003816100381610038161003816+10038161003816100381610038161205721Tminus11

sin 120582 (119909 + 1)1003816100381610038161003816

+ |120582|1003816100381610038161003816Tminus11

cos (120582 (119909 + 1) + 1205791)1003816100381610038161003816+1003816100381610038161003816Tminus11

yminus11

(119909 120582)1003816100381610038161003816

+10038161003816100381610038161205721

1003816100381610038161003816

10038161003816100381610038161003816

Tminus12

cos 120582 (119909 + 1)

10038161003816100381610038161003816+10038161003816100381610038161205722

1003816100381610038161003816

10038161003816100381610038161003816

Tminus12

sin 120582 (119909 + 1)

10038161003816100381610038161003816

+ |120582|

10038161003816100381610038161003816

Tminus12

sin (120582 (119909 + 1) + 1205791)

10038161003816100381610038161003816+

10038161003816100381610038161003816

Tminus12

yminus12

(119909 120582)

10038161003816100381610038161003816

le 21198881(1 + |120582|) 119890

|I120582|(119909+1)int

119909

minus1

[10038161003816100381610038161199011(119905)

1003816100381610038161003816+10038161003816100381610038161199012(119905)

1003816100381610038161003816] 119889119905

+ 119890|I120582|(119909+1)

int

119909

minus1

[10038161003816100381610038161199011(119905)

1003816100381610038161003816

1003816100381610038161003816yminus11

(119905 120582)1003816100381610038161003816

+10038161003816100381610038161199012(119905)

1003816100381610038161003816

1003816100381610038161003816yminus12

(119905 120582)1003816100381610038161003816] 119890minus|I120582|(119905+1)

119889119905

le 211988811198882(1 + |120582|) 119890

|I120582|(119909+1)

+ 119890|I120582|(119909+1)

int

119909

minus1

[10038161003816100381610038161199011(119905)

1003816100381610038161003816

1003816100381610038161003816yminus11

(119905 120582)1003816100381610038161003816

+10038161003816100381610038161199012(119905)

1003816100381610038161003816

1003816100381610038161003816yminus12

(119905 120582)1003816100381610038161003816] 119890minus|I120582|(119905+1)

119889119905

(56)

The above inequality can be reduced to

119890minus|I120582|(119909+1) 1003816

100381610038161003816yminus11

(119909 120582)1003816100381610038161003816

le 211988811198882(1 + |120582|)

+ int

119909

minus1

[10038161003816100381610038161199011(119905)

1003816100381610038161003816

1003816100381610038161003816yminus11

(119905 120582)1003816100381610038161003816

+10038161003816100381610038161199012(119905)

1003816100381610038161003816

1003816100381610038161003816yminus12

(119905 120582)1003816100381610038161003816] 119890minus|I120582|(119905+1)

119889119905

(57)

Similarly we can prove that

119890minus|I120582|(119909+1) 1003816

100381610038161003816yminus12

(119909 120582)1003816100381610038161003816

le 211988811198882(1 + |120582|)

+ int

119909

minus1

[10038161003816100381610038161199011(119905)

1003816100381610038161003816

1003816100381610038161003816yminus11

(119905 120582)1003816100381610038161003816

+10038161003816100381610038161199012(119905)

1003816100381610038161003816

1003816100381610038161003816yminus12

(119905 120582)1003816100381610038161003816] 119890minus|I120582|(119905+1)

119889119905

(58)

Then from (57) (58) and Lemma 31 of [32 pp 204] weobtain (54)

In a similar manner we will prove the following lemmafor y01

(sdot 120582) and y02

(sdot 120582)

Lemma 5 The functions y01

(119909 120582) and y02

(119909 120582) are entire in120582 for any fixed 119909 isin (0 1] and satisfy the growth condition

1003816100381610038161003816y01

(119909 120582)1003816100381610038161003816

1003816100381610038161003816y02

(119909 120582)1003816100381610038161003816le 211988841198885(1 + |120582|) 119890

|I120582|(119909+1)

120582 isin C

(59)

Abstract and Applied Analysis 7

Proof Since y01

(119909 120582) = T01

12059312(119909 120582) +

T02

12059322(119909 120582) then

from (49) and (50) we obtain

y01

(119909 120582) =

1

120575

12059311

(0minus 120582)T

01cos (120582119909)

minus

1

120575

12059321

(0minus 120582)T

01sin (120582119909)

minusT01yminus12

(119909 120582) +

1

120575

12059311

(0minus 120582)

T02

sin (120582119909)

+

1

120575

12059321

(0minus 120582)

T02

cos (120582119909) + T02yminus12

(119909 120582)

(60)

Then from (47) and (48) and Lemma 4 we get

1003816100381610038161003816y01

(119909 120582)1003816100381610038161003816le

1

|120575|

100381610038161003816100381612059311

(0minus 120582)

1003816100381610038161003816

1003816100381610038161003816T01

cos (120582119909)1003816100381610038161003816

+

1

|120575|

100381610038161003816100381612059321

(0minus 120582)

1003816100381610038161003816

1003816100381610038161003816T01

sin (120582119909)1003816100381610038161003816

+1003816100381610038161003816T01yminus12

(119909 120582)1003816100381610038161003816

+

1

|120575|

100381610038161003816100381612059311

(0minus 120582)

1003816100381610038161003816

10038161003816100381610038161003816

T02

sin (120582119909)

10038161003816100381610038161003816

+

1

|120575|

100381610038161003816100381612059321

(0minus 120582)

1003816100381610038161003816

10038161003816100381610038161003816

T02

cos (120582119909)10038161003816100381610038161003816

+

10038161003816100381610038161003816

T02yminus12

(119909 120582)

10038161003816100381610038161003816

le 21198884(1 + |120582|)

times (1198883+

2

|120575|

(1198881+ 1198883)) 119890|I120582|(119909+1)

= 211988841198885(1 + |120582|) 119890

|I120582|(119909+1)

(61)

Similarly we can prove that1003816100381610038161003816y02

(119909 120582)1003816100381610038161003816le 211988841198885(1 + |120582|) 119890

|I120582|(119909+1) (62)

3 The Numerical Scheme

In this section we derive the method of computing eigen-values of problems (12)ndash(15) numerically The basic idea ofthe scheme is to split Ω(120582) into two parts a known partK(120582) and an unknown oneU(120582)Thenwe approximateU(120582)

using (5) to get the approximate Ω(120582) and then compute theapproximate zeros We first splitΩ(120582) into two parts

Ω (120582) = K (120582) +U (120582) (63)

whereU(120582) is the unknown part involving integral operators

U (120582) = 120575 [(1205732+ 120582 cos 120579

2) sin 120582

minus (1205731+ 120582 sin 120579

2) cos 120582] y

minus11(0minus 120582)

minus 120575 [(1205731+ 120582 sin 120579

2) sin 120582

+ (1205732+ 120582 cos 120579

2) cos 120582] y

minus12(0minus 120582)

minus 1205752[(1205731+ 120582 sin 120579

2) y01

(1 120582)

+ (1205732+ 120582 cos 120579

2) y02

(1 120582)]

(64)

andK(120582) is the known part

K (120582) = 120575 (1205731+ 120582 sin 120579

2)

times [1205722cos 2120582 minus 120572

1sin 2120582 + 120582 cos (2120582 + 120579

1)]

minus 120575 (1205732+ 120582 cos 120579

2)

times [1205722sin 2120582 + 120572

1cos 2120582 + 120582 sin (2120582 + 120579

1)]

(65)

Then from Lemmas 4 and 5 we have the following result

Lemma 6 The function U(120582) is entire in 120582 and the followingestimate holds

|U (120582)| le 120601 (120582) 1198902|I120582|

(66)

where

120601 (120582) = 119872(1 + |120582|)2

119872 = 4 |120575| 1198886 (21198883 + |120575| 11988841198885)

(67)

Proof From (64) we have

|U (120582)| le |120575| [(10038161003816100381610038161205732

1003816100381610038161003816+ |120582|

1003816100381610038161003816cos 1205792

1003816100381610038161003816) |sin 120582|

+ (10038161003816100381610038161205731

1003816100381610038161003816+ |120582|

1003816100381610038161003816sin 1205792

1003816100381610038161003816) |cos 120582|] 100381610038161003816

1003816yminus11

(0minus 120582)

1003816100381610038161003816

+ |120575| [(10038161003816100381610038161205731

1003816100381610038161003816+ |120582|

1003816100381610038161003816sin 1205792

1003816100381610038161003816) |sin 120582|

+ (10038161003816100381610038161205732

1003816100381610038161003816+1003816100381610038161003816120582 cos 120579

2

1003816100381610038161003816) |cos 120582|] 100381610038161003816

1003816yminus12

(0minus 120582)

1003816100381610038161003816

+ 1205752[(10038161003816100381610038161205731

1003816100381610038161003816+ |120582|

1003816100381610038161003816sin 1205792

1003816100381610038161003816)1003816100381610038161003816y01

(1 120582)1003816100381610038161003816

+ (10038161003816100381610038161205732

1003816100381610038161003816+ |120582|

1003816100381610038161003816cos 1205792

1003816100381610038161003816)1003816100381610038161003816y02

(1 120582)1003816100381610038161003816]

(68)

Using the inequalities | sin 120582| le 119890|I120582| and | cos 120582| le 119890

|I120582| for120582 isin C and Lemmas 4 and 5 imply (66)

ThusU(120582) is an entire function of exponential type 120590 = 2In the following we let 120582 isin R since all eigenvalues are realNow we approximate the function U(120582) using the operator(5) where ℎ isin (0 1205872) and 120596 = (120587 minus 2ℎ)2 and then from(7) we obtain

1003816100381610038161003816U (120582) minus (G

ℎ119873U) (120582)

1003816100381610038161003816le 119879ℎ119873

(120582) (69)

where

119879ℎ119873

(120582) = 2

10038161003816100381610038161003816sin (ℎ

minus1120587120582)

10038161003816100381610038161003816

times 120601 (|R120582| + ℎ (119873 + 1))

119890minus120596119873

radic120587120596119873

120573119873(0) 120582 isin R

(70)

The samplesU(119899ℎ) = Ω(119899ℎ) minusK(119899ℎ) 119899 isin Z119873(120582) cannot be

computed explicitly in the general caseWe approximate these

8 Abstract and Applied Analysis

samples numerically by solving the initial-value problemsdefined by (12) and (45) to obtain the approximate valuesU(119899ℎ) 119899 isin Z

119873(120582) that is U(119899ℎ) = Ω(119899ℎ) minus K(119899ℎ)

Here we use a computer algebra system mathematica toobtain the approximate solutions with the required accuracyHowever a separate study for the effect of different numericalschemes and the computational costs would be interestingAccordingly we have the explicit expansion

(Gℎ119873

U) (120582) = sum

119899isinZ119873(120582)

U (119899ℎ) sinc (ℎminus1120587119909 minus 119899120587)

times 119866(

radic120596 (120582 minus 119899ℎ)

radic119873ℎ

)

(71)

Therefore we get cf (10)10038161003816100381610038161003816(Gℎ119873

U) (120582) minus (Gℎ119873

U) (120582)

10038161003816100381610038161003816le 119860120576119873

(0) 120582 isin R (72)

Now let Ω119873(120582) = K(120582) + (G

ℎ119873U)(120582) From (69) and (72)

we obtain10038161003816100381610038161003816Ω (120582) minus Ω

119873(120582)

10038161003816100381610038161003816le 119879ℎ119873

(120582) + 119860120576119873

(0) 120582 isin R (73)

Let 120582lowast be an eigenvalue and 120582119873its desired approximation

that is Ω(120582lowast) = 0 and Ω

119873(120582119873) = 0 From (73) we have

|Ω119873(120582lowast)| le 119879

ℎ119873(120582lowast) + 119860120576119873

(0) Define the curves

119886plusmn(120582) = Ω

119873(120582) plusmn 119879

ℎ119873(120582) + 119860

120576119873(0) (74)

The curves 119886+(120582) 119886minus(120582) enclose the curve ofΩ(120582) for suitably

large 119873 Hence the closure interval is determined by solving119886plusmn(120582) = 0 giving an interval

119868120576119873

= [119886minus 119886+] (75)

It is worthwhile to mention that the simplicity of the eigen-values guarantees the existence of approximate eigenvaluesthat is the 120582

119873rsquos for which Ω

119873(120582119873) = 0 Next we estimate the

error |120582lowast minus 120582119873| for the eigenvalue 120582lowast

Theorem 7 Let 120582lowast be an eigenvalue of (12)ndash(15) and let 120582119873

be its approximation Then for 120582 isin R we have the followingestimate

1003816100381610038161003816120582lowastminus 120582119873

1003816100381610038161003816lt

119879ℎ119873

(120582119873) + 119860120576119873

(0)

inf120577isin119868120576119873

1003816100381610038161003816Ω1015840(120577)

1003816100381610038161003816

(76)

where the interval 119868120576119873

is defined above

Proof Replacing 120582 by 120582119873in (73) we obtain

1003816100381610038161003816Ω (120582119873) minus Ω (120582

lowast)1003816100381610038161003816lt 119879ℎ119873

(120582119873) + 119860120576119873

(0) (77)

where we have used Ω119873(120582119873) = Ω(120582

lowast) = 0 Using the

mean value theorem yields that for some 120577 isin 119869120576119873

=

[min(120582lowast 120582119873)max(120582lowast 120582

119873)]

10038161003816100381610038161003816(120582lowastminus 120582119873)Ω1015840(120577)

10038161003816100381610038161003816le 119879ℎ119873

(120582119873) + 119860120576119873

(0) 120577 isin 119869120576119873

sub 119868120576119873

(78)

Since 120582lowast is simple and 119873 is sufficiently large then

inf120577isin119868120576119873

|Ω1015840(120577)| gt 0 and we get (76)

4 Numerical Examples

In this section we introduce two examples illustrating the theabove method Also in the following examples we observethat the exact solutions 120582

119896are all inside the interval [119886

minus 119886+]

In these two examples we indicate the effect of the amplitudeerror in the method by determining enclosure intervals fordifferent values of 120576 We also indicate the effect of 119873 and ℎ

by several choices The eigenvalues of the following examplescannot computed concretely then we use mathematicato obtain the exact values mathematica is also used inrounding off the exact eigenvalues which are square rootsBoth numerical results and the associated figures prove thecredibility of the method

Example 1 Consider the system

1199101015840

2(119909) minus 119901 (119909) 119910

1(119909) = 120582119910

1(119909)

1199101015840

1(119909) + 119901 (119909) 119910

2(119909) = minus120582119910

2(119909)

119909 isin [minus1 0) cup (0 1]

1205821199101(minus1) minus (radic2 + 120582) 119910

2(minus1) = 0

(radic2 + 120582) 1199101(1) minus 120582119910

2(1) = 0

1199101(0minus) minus 2119910

1(0+) = 0 119910

2(0minus) minus 2119910

2(0+) = 0

(79)

Here

1199011(119909) = 119901

2(119909) = 119903 (119909) =

119909 119909 isin [minus1 0)

1 (0 1]

(80)

1205791= 1205792= 1205874 120572

1= 1205732= 0 120572

2= 1205731= 1 and 120575 = 2 Direct

calculations give

K (120582) = 2 ((radic2120582 + 1) cos (2120582) minus 120582 (120582 + radic2) sin (2120582))

Ω (120582) = 2 ((radic2120582 + 1) cos(2120582 +

1

2

)

minus120582 (120582 + radic2) sin(2120582 +

1

2

))

(81)

As is clearly seen the eigenvalues cannot be computedexplicitly Tables 1 2 and 3 indicate the application of ourtechnique to this problem and the effect of 120576 By exact wemean the zeros of Δ(120582) computed by Mathematica

Figures 1 and 2 illustrate the enclosure intervals domi-nating 120582

0for 119873 = 25 ℎ = 01 and 120576 = 10

minus2 120576 = 10minus5

respectively The middle curve represents Δ(120582) while theupper and lower curves represent the curves of 119886

+(120582) 119886minus(120582)

respectively We notice that when 120576 = 10minus5 the two curves

are almost identical Similarly Figures 3 and 4 illustrate theenclosure intervals dominating 120582

1for ℎ = 01 119873 = 25 and

120576 = 10minus2 120576 = 10

minus5 respectively

Abstract and Applied Analysis 9

Table 1 The approximation 120582119896119873

and the exact solutions 120582119896for different choices of ℎ and119873

120582119896

120582minus2

120582minus1

1205820

1205821

Exact 120582119896

minus12061856254879546 minus05151195632138896 034274095892547224 1616237903407542

120582119896119873

ℎ = 06 119873 = 15 minus12061855965725146 minus05151190347168934 034274053593937165 1616237692038969120596 = 09708 119873 = 25 minus12061856254843495 minus05151195631493376 0342740958869788 16162379033820231ℎ = 01 119873 = 15 minus12061856254844203 minus05151195632158179 03427409589350932 16162379034134882

120596 = 14708 119873 = 25 minus12061856254879548 minus05151195632138894 034274095892547236 16162379034075425

Table 2 Absolute error |120582119896minus 120582119896119873

|

120582119896

120582minus2

120582minus1

1205820

1205821

ℎ = 06119873 = 15 289154 times 10

minus8528497 times 10

minus7422986 times 10

minus7211369 times 10

minus7

119873 = 25 360512 times 10minus12

64552 times 10minus11

556842 times 10minus11

255189 times 10minus11

ℎ = 01119873 = 15 353428 times 10

minus12192824 times 10

minus12962097 times 10

minus12594613 times 10

minus12

119873 = 25 222045 times 10minus16

222045 times 10minus16

555112 times 10minus17

444089 times 10minus16

Table 3 For119873 = 25 and ℎ = 01 the exact solutions 120582119896are all inside the interval [119886

minus 119886+] for different values of 120576

120582119896

120582minus2

120582minus1

1205820

1205821

Exact 120582119896

minus12061856254879546 minus05151195632138896 034274095892547224 1616237903407542119868120576119873

120576 = 10minus2

[minus12335696 minus11771922] [minus05486790 minus04826816] [03246478 03601346] [16100119 16224066]

119868120576119873

120576 = 10minus5

[minus12062137 minus12061575] [minus05151525 minus05150867] [03427232 03427586] [16162317 16162441]

030 032 034 036 038

00

02

04

aminus

a+

Ω(120582)

minus02

minus04

minus06

Figure 1The enclosure interval dominating 1205820for ℎ = 01119873 = 25

and 120576 = 10minus2

Example 2 In this example we consider the system

1199101015840

2(119909) minus 119901 (119909) 119910

1(119909) = 120582119910

1(119909)

1199101015840

1(119909) + 119901 (119909) 119910

2(119909) = minus120582119910

2(119909)

119909 isin [minus1 0) cup (0 1]

radic31199101(minus1) minus 119910

2(minus1) = 0

(1 +

1

2

120582)1199101(1) minus (1 +

radic3

2

120582)1199102(1) = 0

1199101(0minus) minus 3119910

1(0+) = 0 119910

2(0minus) minus 3119910

2(0+) = 0

(82)

030 032 034 036 038

00

02

aminus

a+

Ω(120582)

minus02

minus04

Figure 2The enclosure interval dominating 1205820for ℎ = 01119873 = 25

and 120576 = 10minus5

where

1199011(119909) = 119901

2(119909) = 119901 (119909) =

119909 119909 isin [minus1 0)

1199092 (0 1]

(83)

1205791= 1205792= 0 120572

1= 1205732= 1 120572

2= 1205731= 2 and 120575 = 12 Direct

calculations give

K (120582) =

1

2

((120582 + 3) cos (2120582) minus (1205822+ 3120582 + 4) sin (2120582))

Ω (120582) =

1

2

((1205822+ 3120582 + 4) sin(

1

6

minus 2120582)

+ (120582 + 3) cos(1

6

minus 2120582))

(84)

10 Abstract and Applied Analysis

Table 4 The approximation 120582119896119873

and the exact solutions 120582119896for different choices of ℎ and119873

120582119896

120582minus2

120582minus1

1205820

1205821

Exact 120582119896

minus30661256186938894 minus10959703083466883 03689030343814017 18338437675275228

120582119896119873

ℎ = 08 119873 = 15 minus3066123822200742 minus10959640944020903 036890463046491767 18338437607680411120596 = 007708 119873 = 25 minus30661256172349627 minus10959703026494807 036890303588091056 18338437676561417

ℎ = 02 119873 = 15 minus30661256186844446 minus10959703083278305 03689030343916001 1833843767517251120596 = 13708 119873 = 25 minus306612561869389 minus1095970308346688 03689030343814024 18338437675275225

Table 5 Absolute error |120582119896minus 120582119896119873

|

120582119896

120582minus2

120582minus1

1205820

1205821

ℎ = 08119873 = 15 179649 times 10

minus6621394 times 10

minus6159608 times 10

minus6675948 times 10

minus9

119873 = 25 145893 times 10minus9

569721 times 10minus9

149951 times 10minus9

128619 times 10minus10

ℎ = 02119873 = 15 944489 times 10

minus12188578 times 10

minus12101984 times 10

minus11102718 times 10

minus11

119873 = 25 444089 times 10minus16

222045 times 10minus16

666134 times 10minus16

222045 times 10minus16

Table 6 For119873 = 25 and ℎ = 02 the exact solutions 120582119896are all inside the interval [119886

minus 119886+] for different values of 120576

120582119896

120582minus2

120582minus1

1205820

1205821

Exact 120582119896

minus30661256186938894 minus10959703083466883 03689030343814017 18338437675275228119868120576119873

120576 = 10minus2

[minus31098742 minus30188665] [minus11631502 minus10339754] [03437434 03933486] [18220764 18454828]

119868120576119873

120576 = 10minus5

[minus30661709 minus30660803] [minus10960346 minus10959060] [03688783 03689278] [18338321 18338555]

158 159 160 161 162 163 164 165

00

05

10

minus05

minus10

aminus

a+

Ω(120582)

Figure 3The enclosure interval dominating 1205821for ℎ = 01119873 = 25

and 120576 = 10minus2

Tables 4 and 6 give the exact eigenvalues 1205821198961

119896=minus2and their

approximate ones for different values of ℎ 119873 and 120576 InTable 5 we give the absolute error for different values of ℎ and119873

Here Figures 5 6 7 and 8 illustrate the enclosureintervals dominating 120582

minus2and 120582

minus1for ℎ = 02 119873 = 25 and

120576 = 10minus2 120576 = 10

minus5 respectively

158 159 160 161 162 163 164 165

00

05

minus05

aminus

a+

Ω(120582)

Figure 4The enclosure interval dominating 1205821for ℎ = 01119873 = 25

and 120576 = 10minus5

5 Conclusion

With a simple analysis and with values of solutions ofinitial value problems computed at a few values of theeigenparameter we have computed the eigenvalues of aDirac system which has a discontinuity at one point andcontains a spectral parameter in all boundary conditionswith a certain estimated error The method proposed is

Abstract and Applied Analysis 11

00

05

10

minus05

minus33 minus32 minus31 minus30 minus29 minus28

aminus

a+

Ω(120582)

Figure 5 The enclosure interval dominating 120582minus2

for ℎ = 02 119873 =

25 and 120576 = 10minus2

00

05

10

minus05

minus33 minus32 minus31 minus30 minus29 minus28

aminus

a+

Ω(120582)

Figure 6 The enclosure interval dominating 120582minus2

for ℎ = 02 119873 =

25 and 120576 = 10minus5

a shooting procedure that is the problem is reformulatedas two initial value ones due to the interior discontinuity ofsize two and a miss-distance is defined at the right end of theinterval of integration whose roots are the eigenvalues to becomputed The unknown partU(120582) of the miss-distance canbewritten in terms of a functionwhich is an entire function ofexponential typeTherefore we propose to approximate suchterm by means of a truncated cardinal series with samplingvalues approximated by solving numerically correspondingsuitable initial value problems Finally in Section 4 weintroduced two instructive examples where both numericalresults and the associated figures have proved the credibilityof the method

00

05

10

minus05

minus14 minus13 minus12 minus11 minus10 minus09 minus08 minus07

aminus

a+

Ω(120582)

Figure 7 The enclosure interval dominating 120582minus1

for ℎ = 02 119873 =

25 and 120576 = 10minus2

00

05

10

minus05

minus14 minus13 minus12 minus11 minus10 minus09 minus08 minus07

aminus

a+

Ω(120582)

Figure 8 The enclosure interval dominating 120582minus1

for ℎ = 02 119873 =

25 and 120576 = 10minus5

Conflict of Interests

The authors declare that there is no conflict of interestsregarding the publication of this paper

Acknowledgment

This paper was funded by the Deanship of Scientific Research(DSR) King Abdulaziz University Jeddah The authorstherefore acknowledge with thanks DSR technical and finan-cial support

12 Abstract and Applied Analysis

References

[1] C T Fulton ldquoTwo-point boundary value problems witheigenvalue parameter contained in the boundary conditionsrdquoProceedings of the Royal Society of Edinburgh A Mathematicsvol 77 no 3-4 pp 293ndash308 1977

[2] JWalter ldquoRegular eigenvalue problems with eigenvalue param-eter in the boundary conditionrdquoMathematische Zeitschrift vol133 pp 301ndash312 1973

[3] V Kotelnikov ldquoOn the carrying capacity of the ether and wirein telecommunications material for the first All-Unionrdquo inConference on Questions of Communications vol 55 pp 55ndash64Izd Red Upr Svyazi RKKA 1933

[4] C E Shannon ldquoCommunication in the presence of noiserdquoProceedings of the IRE vol 37 pp 10ndash21 1949

[5] E TWhittaker ldquoOn the functions which are represented by theexpansion of the interpolation theoryrdquo Proceedings of the RoyalSociety of Edinburgh A vol 35 pp 181ndash194 1915

[6] P L Butzer G Schmeisser and R L Stens ldquoAn introductionto sampling analysisrdquo in Non Uniform Sampling Theory andPractices F Marvasti Ed pp 17ndash121 Kluwer New York NYUSA 2001

[7] M A Kowalski K A Sikorski and F Stenger Selected Topics inApproximation andComputation OxfordUniversity Press NewYork NY USA 1995

[8] J Lund and K L Bowers Sinc Methods for Quadrature andDifferential Equations SIAM Philadelphia Pa USA 1992

[9] F Stenger ldquoNumericalmethods based onWhittaker cardinal orsinc functionsrdquo SIAM Review vol 23 no 2 pp 165ndash224 1981

[10] F Stenger Numerical Methods Based on Sinc and AnalyticFunctions Springer New York NY USA 1993

[11] A Boumenir ldquoHigher approximation of eigenvalues by thesampling methodrdquo BIT Numerical Mathematics vol 40 no 2pp 215ndash225 2000

[12] A Boumenir ldquoSampling and eigenvalues of non-self-adjointSturm-Liouville problemsrdquo SIAM Journal on Scientific Comput-ing vol 23 no 1 pp 219ndash229 2001

[13] M M Tharwat A H Bhrawy and A Yildirim ldquoNumericalcomputation of eigenvalues of discontinuous Sturm-Liouvilleproblems with parameter dependent boundary conditionsusing sinc methodrdquoNumerical Algorithms vol 63 no 1 pp 27ndash48 2013

[14] M M Tharwat A H Bhrawy and A Yildirim ldquoNumericalcomputation of the eigenvalues of a discontinuousDirac systemusing the sincmethodwith error analysisrdquo International Journalof Computer Mathematics vol 89 no 15 pp 2061ndash2080 2012

[15] P L Butzer and R L Stens ldquoA modification of the Whittaker-Kotelrsquonikov-Shannon sampling seriesrdquo Aequationes Mathemat-icae vol 28 no 3 pp 305ndash311 1985

[16] R Gervais Q I Rahman and G Schmeisser ldquoA bandlimitedfunction simulating a duration-limited onerdquo in ApproximationTheory andFunctionalAnalysis P L Butzer andR L Stens Edspp 355ndash362 Birkhaauser Boston Mass USA 1984

[17] R L Stens ldquoSampling by generalized kernelsrdquo in SamplingThe-ory in Fourier and Signal Analysis Advanced Topics J RHigginsand R L Stens Eds pp 130ndash157 Oxford University PressOxford UK 1999

[18] G Schmeisser and F Stenger ldquoSinc approximation with a Gaus-sianmultiplierrdquo SamplingTheory in Signal and Image Processingvol 6 no 2 pp 199ndash221 2007

[19] L Qian ldquoOn the regularized Whittaker-Kotelrsquonikov-Shannonsampling formulardquo Proceedings of the American MathematicalSociety vol 131 no 4 pp 1169ndash1176 2003

[20] L Qian and D B Creamer ldquoA modification of the samplingseries with aGaussianmultiplierrdquo SamplingTheory in Signal andImage Processing vol 5 no 1 pp 1ndash20 2006

[21] L Qian and D B Creamer ldquoLocalized sampling in the presenceof noiserdquoAppliedMathematics Letters vol 19 no 4 pp 351ndash3552006

[22] M H Annaby and R M Asharabi ldquoComputing eigenvaluesof boundary-value problems using sinc-Gaussian methodrdquoSamplingTheory in Signal and Image Processing vol 7 no 3 pp293ndash312 2008

[23] M H Annaby and M MTharwat ldquoA sinc-Gaussian techniquefor computing eigenvalues of second-order linear pencilsrdquoApplied Numerical Mathematics vol 63 pp 129ndash137 2013

[24] A H Bhrawy M M Tharwat and A Al-Fhaid ldquoNumericalalgorithms for computing eigenvalues of discontinuous Diracsystem using sinc-Gaussian methodrdquo Abstract and AppliedAnalysis vol 2012 Article ID 925134 13 pages 2012

[25] M M Tharwat A H Bhrawy and A S Alofi ldquoApproximationof eigenvalues of discontinuous sturm-liouville problems witheigenparameter in all boundary conditionsrdquo Boundary ValueProblems vol 2013 article 132 2013

[26] M M Tharwat and S M Al-Harbi ldquoApproximation of eigen-values of boundary value problemsrdquo Boundary Value Problemsvol 2014 article 51 2014

[27] MMTharwat and A H Bhrawy ldquoComputation of eigenvaluesof discontinuous Dirac system using Hermite interpolationtechniquerdquo Advances in Difference Equations vol 2012 article59 2012

[28] M M Tharwat A Yildirim and A H Bhrawy ldquoSampling ofdiscontinuous Dirac systemsrdquo Numerical Functional Analysisand Optimization vol 34 no 3 pp 323ndash348 2013

[29] M M Tharwat ldquoDiscontinuous Sturm-Liouville problems andassociated sampling theoriesrdquo Abstract and Applied Analysisvol 2011 Article ID 610232 30 pages 2011

[30] MMTharwat ldquoOn sampling theories and discontinuousDiracsystems with eigenparameter in the boundary conditionsrdquoBoundary Value Problems vol 2013 article 65 2013

[31] B M Levitan and I S Sargsjan ldquoIntroduction to spectral the-ory self adjoint ordinary differential operatorsrdquo in Translationof Mthematical Monographs vol 39 American MathematicalSociety Providence RI USA 1975

[32] B M Levitan and I S Sargsjan Sturm-Liouville and Dirac oper-ators vol 59 Kluwer Academic Publishers Group DordrechtThe Netherlands 1991

Submit your manuscripts athttpwwwhindawicom

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

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Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical Problems in Engineering

Hindawi Publishing Corporationhttpwwwhindawicom

Differential EquationsInternational Journal of

Volume 2014

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Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Probability and StatisticsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical PhysicsAdvances in

Complex AnalysisJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

OptimizationJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

CombinatoricsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Operations ResearchAdvances in

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Function Spaces

Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of Mathematics and Mathematical Sciences

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

The Scientific World JournalHindawi Publishing Corporation httpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Algebra

Discrete Dynamics in Nature and Society

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Decision SciencesAdvances in

Discrete MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom

Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Stochastic AnalysisInternational Journal of

Abstract and Applied Analysis 3

The proposed method reduces the error bounds remarkably(see examples of Section 4) The basic idea is to write thefunction of eigenvalues as the sum of two terms one knownand the other unknown but an entire function of exponentialtype which satisfies (6) In other words the unknown termis not necessarily an 119871

2-function Then we approximate theunknown part using (5) and obtain better results We wouldlike to mention that the papers in computing eigenvaluesby sinc-Gaussian method are few see [22ndash26] In Sections2 and 3 we discuss some properties of the eigenvalues ofthe boundary value problems (12)ndash(15) and also we derivethe sinc-Gaussian technique to compute the eigenvalues of(12)ndash(15) with error estimates The last section involves someillustrative examples

2 Some Important Results

In the following we study some properties of the eigenvaluesof the problems (12)ndash(15) which need in our method see [2628] For functions u(119909) which defined on [minus1 0) cup (0 1] andhas finite limitu(plusmn0) = lim

119909rarrplusmn0u(119909) byu

(1)(119909) andu

(2)(119909)

we denote the functions

u(1)

(119909) =

u (119909) 119909 isin [minus1 0)

u (0minus) 119909 = 0

u(2)

(119909) =

u (119909) 119909 isin (0 1]

u (0+) 119909 = 0

(17)

which are defined on Γ1

= [minus1 0] and Γ2

= [0 1]respectively

In the following lemma wewill prove that the eigenvaluesof the problems (12)ndash(15) are real see [29 30]

Lemma 1 The eigenvalues of the problems (12)ndash(15) are real

Proof Assume the contrary that 1205820is a nonreal eigenvalue of

problems (12)ndash(15) Let ( u1(119909)

u2(119909)) be a corresponding (nontriv-

ial) eigenfunction By (12) we have

119889

119889119909

u1(119909)u2(119909) minus u

1(119909)u2(119909)

= (1205820minus 1205820)

1003816100381610038161003816u1(119909)

1003816100381610038161003816

2

+1003816100381610038161003816u2(119909)

1003816100381610038161003816

2

119909 isin [minus1 0) cup (0 1]

(18)

Integrating the above equation through [minus1 0] and [0 1] weobtain

(1205820minus 1205820) [int

0

minus1

(1003816100381610038161003816u1(119909)

1003816100381610038161003816

2

+1003816100381610038161003816u2(119909)

1003816100381610038161003816

2

) 119889119909]

= u1(0minus) 1199062(0minus) minus u1(0minus)u2(0minus)

minus [u1(minus1)u

2(minus1) minus u

1(minus1)u

2(minus1)]

(1205820minus 1205820) [int

1

0

(1003816100381610038161003816u1(119909)

1003816100381610038161003816

2

+1003816100381610038161003816u2(119909)

1003816100381610038161003816

2

) 119889119909]

= u1(1)u2(1) minus u

1(1)u2(1)

minus [u1(0+)u2(0+) minus u1(0+)u2(0+)]

(19)

Then from (13) (14) and transmission conditions we haverespectively

u1(minus1)u

2(minus1) minus u

1(minus1)u

2(minus1)

=

1205881(1205820minus 1205820)1003816100381610038161003816u2(minus1)

1003816100381610038161003816

2

10038161003816100381610038161205721+ 1205820sin 1205791

1003816100381610038161003816

2

u1(1)u2(1) minus u

1(1)u2(1)

= minus

1205882(1205820minus 1205820)1003816100381610038161003816u2(1)

1003816100381610038161003816

2

10038161003816100381610038161205731+ 1205820sin 1205792

1003816100381610038161003816

2

u1(minus0) 119906

2(minus0) minus u

1(minus0)u

2(minus0)

= 1205752[u1(+0) 119906

2(+0) minus u

1(+0)u

2(+0)]

(20)

Since 1205820

= 1205820 it follows from the last three equations and (19)

that

int

0

minus1

(1003816100381610038161003816u1(119909)

1003816100381610038161003816

2

+1003816100381610038161003816u2(119909)

1003816100381610038161003816

2

) 119889119909

+ 1205752int

1

0

(1003816100381610038161003816u1(119909)

1003816100381610038161003816

2

+1003816100381610038161003816u2(119909)

1003816100381610038161003816

2

) 119889119909

= minus

1205881

1003816100381610038161003816u2(minus1)

1003816100381610038161003816

2

10038161003816100381610038161205721+ 1205820sin 1205791

1003816100381610038161003816

2minus

120588212057521003816100381610038161003816u2(1)

1003816100381610038161003816

2

10038161003816100381610038161205731+ 1205820sin 1205792

1003816100381610038161003816

2

(21)

This contradicts the conditions int0minus1(|u1(119909)|2+ |u2(119909)|2)119889119909 +

1205752int

1

0(|u1(119909)|2+ |u2(119909)|2)119889119909 gt 0 and 120588

119894gt 0 119894 = 1 2

Consequently 1205820must be real

Now we shall construct a special fundamental system ofsolutions of (12) for 120582 not being an eigenvalue Let us considerthe next initial value problem

u1015840

2(119909) minus 119901

1(119909)u1(119909) = 120582u

1(119909)

u1015840

1(119909) + 119901

2(119909)u2(119909) = minus120582u

2(119909)

119909 isin (minus1 0)

(22)

u1(minus1) = 120572

2+ 120582 cos 120579

1 u

2(minus1) = 120572

1+ 120582 sin 120579

1 (23)

By virtue of Theorem 11 in [31] this problem has a uniquesolution u = (

12059311(119909120582)

12059321(119909120582)) which is an entire function of 120582 isin C

for each fixed 119909 isin [minus1 0] Similarly employing the same

4 Abstract and Applied Analysis

method as in proof of Theorem 11 in [31] we see that theproblem

u1015840

2(119909) minus 119901

1(119909)u1(119909) = 120582u

1(119909)

u1015840

1(119909) + 119901

2(119909)u2(119909) = minus120582u

2(119909)

119909 isin (0 1)

u1(1) = 120573

2+ 120582 cos 120579

2 u

2(1) = 120573

1+ 120582 sin 120579

2

(24)

has a unique solution u = (12059512(119909120582)

12059522(119909120582)) which is an entire

function of parameter 120582 for each fixed 119909 isin [0 1]Now the functions 120593

1198942(119909 120582) and 120595

1198941(119909 120582) are defined in

terms of 1205931198941(119909 120582) and 120595

1198942(119909 120582) 119894 = 1 2 respectively as

follows the initial-value problem

u1015840

2(119909) minus 119901

1(119909)u1(119909) = 120582u

1(119909)

u1015840

1(119909) + 119901

2(119909)u2(119909) = minus120582u

2(119909)

119909 isin (0 1)

(25)

u1(0) =

1

120575

12059311

(0 120582) u2(0) =

1

120575

12059321

(0 120582) (26)

has unique solution u = (12059312(119909120582)

12059322(119909120582)) for each 120582 isin C

Similarly the following problem also has a unique solu-tion u = (

12059511(119909120582)

12059521(119909120582))

u1015840

2(119909) minus 119901

1(119909)u1(119909) = 120582u

1(119909)

u1015840

1(119909) + 119901

2(119909)u2(119909) = minus120582u

2(119909)

119909 isin (minus1 0)

(27)

u1(0) = 120575120595

12(0 120582) u

2(0) = 120575120595

22(0 120582) (28)

Let us construct two basic solutions of (12) as

120593 (sdot 120582) = (1205931(sdot 120582)

1205932(sdot 120582)

) 120595 (sdot 120582) = (1205951(sdot 120582)

1205952(sdot 120582)

) (29)

where

1205931(119909 120582) =

12059311

(119909 120582) 119909 isin [minus1 0)

12059312

(119909 120582) 119909 isin (0 1]

1205932(119909 120582) =

12059321

(119909 120582) 119909 isin [minus1 0)

12059322

(119909 120582) 119909 isin (0 1]

(30)

1205951(119909 120582) =

12059511

(119909 120582) 119909 isin [minus1 0)

12059512

(119909 120582) 119909 isin (0 1]

1205952(119909 120582) =

12059521

(119909 120582) 119909 isin [minus1 0)

12059522

(119909 120582) 119909 isin (0 1]

(31)

By virtue of (26) and (28) these solutions satisfy bothtransmission conditions (15) These functions are entire in 120582

for all 119909 isin [minus1 0) cup (0 1]

Let W(120593 120595)(sdot 120582) denote the Wronskian of 120593(sdot 120582) and120595(sdot 120582) defined in [32 page 194] that is

W (120593 120595) (sdot 120582) =

10038161003816100381610038161003816100381610038161003816

1205931(sdot 120582) 120593

2(sdot 120582)

1205951(sdot 120582) 120595

2(sdot 120582)

10038161003816100381610038161003816100381610038161003816

(32)

It is obvious that the Wronskians

Ω119894(120582) = W (120593 120595) (119909 120582)

= 1205931119894(119909 120582) 120595

2119894(119909 120582)

minus 1205932119894(119909 120582) 120595

1119894(119909 120582) 119909 isin Γ

119894 119894 = 1 2

(33)

are independent on variable 119909 isin Γ119894(119894 = 1 2) and 120593

119894(119909 120582)

and 120595119894(119909 120582) are entire functions of the parameter 120582 for each

119909 isin Γ119894(119894 = 1 2) Taking into account (26) and (28) a short

calculation gives

Ω1(120582) = 120575

2Ω2(120582) (34)

for each 120582 isin C

Corollary 2 The zeros of the functions Ω1(120582) and Ω

2(120582)

coincide

Then we may introduce to the consideration the charac-teristic functionΩ(120582) as

Ω (120582) = Ω1(120582) = 120575

2Ω2(120582) (35)

Note that all eigenvalues of problems (12)ndash(15) are just zerosof the function Ω(120582) Indeed since the functions 120593

1(119909 120582)

and 1205932(119909 120582) satisfy the boundary condition (13) and both

transmission conditions (15) to find the eigenvalues ofthe (12)ndash(15) we have to insert the functions 120593

1(119909 120582) and

1205932(119909 120582) in the boundary condition (14) and find the roots

of this equation In the following lemma we show that alleigenvalues of the problems (12)ndash(15) are simple

Lemma 3 The eigenvalues of the boundary value problems(12)ndash(15) form an at most countable set without finite limitpoints All eigenvalues of the boundary value problems (12)ndash(15) (of Ω(120582)) are simple

Proof The eigenvalues are the zeros of the entire functionoccurring on the left-hand side in see (35)

(1205731+ 120582 sin 120579

2) 12059312

(1 120582) minus (1205732+ 120582 cos 120579

2) 12059322

(1 120582) = 0

(36)

We have shown (see Lemma 1) that this function doesnot vanish for nonreal 120582 In particular it does not vanishidentically Therefore its zeros form an at most countable setwithout finite limit points

By (12) we obtain for 120582 120583 isin C 120582 = 120583

119889

119889119909

1205931(119909 120582) 120593

2(119909 120583) minus 120593

1(119909 120583) 120593

2(119909 120582)

= (120583 minus 120582) 1205931(119909 120582) 120593

1(119909 120583) + 120593

2(119909 120582) 120593

2(119909 120583)

(37)

Abstract and Applied Analysis 5

Integrating the above equation through [minus1 0] and [0 1] andtaking into account the initial conditions (23) (26) and (28)we obtain

1205752(12059312

(1 120582) 12059322

(1 120583) minus 12059312

(1 120583) 12059322

(1 120582))

minus (120583 minus 120582) 1205881

= (120583 minus 120582) (int

0

minus1

(12059311

(119909 120582) 12059311

(119909 120583)

+ 12059321

(119909 120582) 12059321

(119909 120583)) 119889119909

+ 1205752int

1

0

(12059312

(119909 120582) 12059312

(119909 120583)

+ 12059322

(119909 120582) 12059322

(119909 120583)) 119889119909)

(38)

Dividing both sides of (38) by (120582 minus 120583) and by letting 120583 rarr 120582we arrive to the relation

1205752(12059322

(1 120582)

12059712059312

(1 120582)

120597120582

minus 12059312

(1 120582)

12059712059322

(1 120582)

120597120582

) + 1205881

= minus(int

0

minus1

(100381610038161003816100381612059311

(119909 120582)1003816100381610038161003816

2

+100381610038161003816100381612059321

(119909 120582)1003816100381610038161003816

2

) 119889119909

+ 1205752int

1

0

(100381610038161003816100381612059312

(119909 120582)1003816100381610038161003816

2

+100381610038161003816100381612059322

(119909 120582)1003816100381610038161003816

2

) 119889119909)

(39)

We show that equation

Ω (120582) = W (120593 120595) (1 120582)

= 1205752((1205731+ 120582 sin 120579

2) 12059312

(1 120582)

minus (1205732+ 120582 cos 120579

2) 12059322

(1 120582)) = 0

(40)

has only simple roots Assume the converse that is (40) hasa double root 120582lowast Then the following two equations hold

(1205731+ 120582lowast sin 120579

2) 12059312

(1 120582lowast) minus (120573

2+ 120582lowast cos 120579

2) 12059322

(1 120582lowast) = 0

sin 120579212059312

(1 120582lowast) + (120573

1+ 120582lowast sin 120579

2)

12059712059312

(1 120582lowast)

120597120582

minus cos 120579212059322

(1 120582lowast) minus (120573

2+ 120582lowast cos 120579

2)

12059712059322

(1 120582lowast)

120597120582

= 0

(41)

Since 1205882

= 0 and 120582lowast is real then (120573

1+ 120582lowast sin 120579

2)2

+

(1205732+ 120582lowast cos 120579

2)2

= 0 Let 1205731+ 120582lowast sin 120579

2= 0 From (41)

12059312

(1 120582lowast) =

(1205732+ 120582lowast cos 120579

2)

(1205731+ 120582lowast sin 120579

2)

12059322

(1 120582lowast)

12059712059312

(1 120582lowast)

120597120582

=

120588212059322

(1 120582lowast)

(1205731+ 120582lowast sin 120579

2)2

+

(1205732+ 120582lowast cos 120579

2)

(1205731+ 120582lowast sin 120579

2)

12059712059322

(1 120582lowast)

120597120582

(42)

Combining (42) and (39) with 120582 = 120582lowast we obtain

12058821205752(12059322

(1 120582lowast))2

(1205731+ 120582lowast sin 120579

2)2

+ 1205881

= minus(int

0

minus1

(100381610038161003816100381612059311

(119909 120582)1003816100381610038161003816

2

+100381610038161003816100381612059321

(119909 120582)1003816100381610038161003816

2

) 119889119909

+ 1205752int

1

0

(

1003816100381610038161003816100381612059312

(119909 120582) |2+

1003816100381610038161003816100381612059322

(119909 120582) |2) 119889119909)

(43)

contradicting the assumption 120588119894gt 0 119894 = 1 2 The other case

when 1205732+120582lowast cos 120579

2= 0 can be treated similarly and the proof

is complete

Recall that boundary value problems (12)ndash(15) have denu-merable set of real and simple eigenvalues see the abovesection From (23) and (26) the solution of (12)

120593 (sdot 120582) = (1205931(sdot 120582)

1205932(sdot 120582)

)

120593119894(119909 120582) =

1205931198941(119909 120582) 119909 isin [minus1 0)

1205931198942(119909 120582) 119909 isin (0 1]

119894 = 1 2

(44)

satisfies the following initial conditions

(12059311

(minus1 120582) 12059312

(0+ 120582)

12059321

(minus1 120582) 12059322

(0+ 120582)

)

= (1205722+ 120582 cos 120579

1120575minus112059311

(0minus 120582)

1205721+ 120582 sin 120579

1120575minus112059321

(0minus 120582)

)

(45)

Since 120593(sdot 120582) satisfies (13) then the eigenvalues of the prob-lems (12)ndash(15) are the zeros of the function

Ω (120582) = 1205752((1205731+ 120582 sin 120579

2) 12059312

(1 120582)

minus (1205732+ 120582 cos 120579

2) 12059322

(1 120582)) = 0

(46)

We shall transform (12) (23) (26) and (30) into the integralequations see [32]

12059311

(119909 120582) = 1205722cos 120582 (119909 + 1) minus 120572

1sin 120582 (119909 + 1)

+ 120582 cos (120582 (119909 + 1) + 1205791) minusTminus11

12059311

(119909 120582)

minusTminus12

12059321

(119909 120582)

(47)

12059321

(119909 120582) = 1205721cos 120582 (119909 + 1) + 120572

2sin 120582 (119909 + 1)

+ 120582 sin (120582 (119909 + 1) + 1205791) +

Tminus11

12059311

(119909 120582)

minusTminus12

12059321

(119909 120582)

(48)

12059312

(119909 120582) =

1

120575

12059311

(0minus 120582) cos (120582119909) minus 1

120575

12059321

(0minus 120582) sin (120582119909)

minusT01

12059312

(119909 120582) minusT02

12059322

(119909 120582)

(49)

6 Abstract and Applied Analysis

12059322

(119909 120582) =

1

120575

12059311

(0minus 120582) sin (120582119909) +

1

120575

12059321

(0minus 120582) cos (120582119909)

+T01

12059312

(119909 120582) minusT02

12059322

(119909 120582)

(50)

where Tminus1119894

Tminus1119894

T0119894

and T0119894 119894 = 1 2 are the Volterra

integral operators defined by

Tminus1119894

120593 (119909 120582) = int

119909

minus1

sin 120582 (119909 minus 119905) 119901119894(119905) 120593 (119905 120582) 119889119905

Tminus1119894

120593 (119909 120582) = int

119909

minus1

cos 120582 (119909 minus 119905) 119901119894(119905) 120593 (119905 120582) 119889119905

T0119894120593 (119909 120582) = int

119909

0

sin 120582 (119909 minus 119905) 119901119894(119905) 120593 (119905 120582) 119889119905

T0119894120593 (119909 120582) = int

119909

0

cos 120582 (119909 minus 119905) 119901119894(119905) 120593 (119905 120582) 119889119905

(51)

For convenience we define the constants

1198881= max 100381610038161003816

10038161205721

1003816100381610038161003816+10038161003816100381610038161205722

10038161003816100381610038161003816100381610038161003816sin 1205791

1003816100381610038161003816+1003816100381610038161003816cos 1205791

1003816100381610038161003816

1198882= int

0

minus1

[10038161003816100381610038161199011(119905)

1003816100381610038161003816+10038161003816100381610038161199012(119905)

1003816100381610038161003816] 119889119905 119888

3= 11988811198882exp (119888

2)

1198884= int

1

0

[10038161003816100381610038161199011(119905)

1003816100381610038161003816+10038161003816100381610038161199012(119905)

1003816100381610038161003816] 119889119905 119888

5= 1198883+

2

|120575|

(1198881+ 1198883)

1198886= max 100381610038161003816

10038161205721

1003816100381610038161003816+10038161003816100381610038161205722

10038161003816100381610038161003816100381610038161003816sin 1205792

1003816100381610038161003816+1003816100381610038161003816cos 1205792

1003816100381610038161003816

(52)

Now we define yminus1119894

(sdot 120582) and y0119894(sdot 120582) 119894 = 1 2 to be

yminus11

(119909 120582) = Tminus11

12059311

(119909 120582) +Tminus12

12059321

(119909 120582)

yminus12

(119909 120582) =Tminus11

12059311

(119909 120582) minusTminus12

12059321

(119909 120582)

y01

(119909 120582) = T01

12059312

(119909 120582) +T02

12059322

(119909 120582)

y02

(119909 120582) =T01

12059312

(119909 120582) minusT02

12059322

(119909 120582)

(53)

Lemma 4 The functions yminus11

(119909 120582) and yminus12

(119909 120582) are entirein 120582 for any fixed 119909 isin [minus1 0) and satisfy the growth condition

1003816100381610038161003816yminus11

(119909 120582)1003816100381610038161003816

1003816100381610038161003816yminus12

(119909 120582)1003816100381610038161003816le 21198883(1 + |120582|) 119890

|I120582|(119909+1)

120582 isin C

(54)

Proof Since yminus11

(119909 120582) = Tminus11

12059311(119909 120582) +

Tminus12

12059321(119909 120582)

then from (47) and (48) we obtain

yminus11

(119909 120582)

= 1205722Tminus11

cos 120582 (119909 + 1) minus 1205721Tminus11

sin 120582 (119909 + 1)

+ 120582Tminus11

cos (120582 (119909 + 1) + 1205791) minusTminus11

yminus11

(119909 120582)

+ 1205721Tminus12

cos 120582 (119909 + 1) + 1205722Tminus12

sin 120582 (119909 + 1)

+ 120582Tminus12

sin (120582 (119909 + 1) + 1205791) +

Tminus12

yminus12

(119909 120582)

(55)

Using the inequalities | sin 119911| le 119890|I119911| and | cos 119911| le 119890

|I119911| for119911 isin C leads for 120582 isin C to1003816100381610038161003816yminus11

(119909 120582)1003816100381610038161003816

le10038161003816100381610038161205722

1003816100381610038161003816

1003816100381610038161003816Tminus11

cos 120582 (119909 + 1)1003816100381610038161003816+10038161003816100381610038161205721Tminus11

sin 120582 (119909 + 1)1003816100381610038161003816

+ |120582|1003816100381610038161003816Tminus11

cos (120582 (119909 + 1) + 1205791)1003816100381610038161003816+1003816100381610038161003816Tminus11

yminus11

(119909 120582)1003816100381610038161003816

+10038161003816100381610038161205721

1003816100381610038161003816

10038161003816100381610038161003816

Tminus12

cos 120582 (119909 + 1)

10038161003816100381610038161003816+10038161003816100381610038161205722

1003816100381610038161003816

10038161003816100381610038161003816

Tminus12

sin 120582 (119909 + 1)

10038161003816100381610038161003816

+ |120582|

10038161003816100381610038161003816

Tminus12

sin (120582 (119909 + 1) + 1205791)

10038161003816100381610038161003816+

10038161003816100381610038161003816

Tminus12

yminus12

(119909 120582)

10038161003816100381610038161003816

le 21198881(1 + |120582|) 119890

|I120582|(119909+1)int

119909

minus1

[10038161003816100381610038161199011(119905)

1003816100381610038161003816+10038161003816100381610038161199012(119905)

1003816100381610038161003816] 119889119905

+ 119890|I120582|(119909+1)

int

119909

minus1

[10038161003816100381610038161199011(119905)

1003816100381610038161003816

1003816100381610038161003816yminus11

(119905 120582)1003816100381610038161003816

+10038161003816100381610038161199012(119905)

1003816100381610038161003816

1003816100381610038161003816yminus12

(119905 120582)1003816100381610038161003816] 119890minus|I120582|(119905+1)

119889119905

le 211988811198882(1 + |120582|) 119890

|I120582|(119909+1)

+ 119890|I120582|(119909+1)

int

119909

minus1

[10038161003816100381610038161199011(119905)

1003816100381610038161003816

1003816100381610038161003816yminus11

(119905 120582)1003816100381610038161003816

+10038161003816100381610038161199012(119905)

1003816100381610038161003816

1003816100381610038161003816yminus12

(119905 120582)1003816100381610038161003816] 119890minus|I120582|(119905+1)

119889119905

(56)

The above inequality can be reduced to

119890minus|I120582|(119909+1) 1003816

100381610038161003816yminus11

(119909 120582)1003816100381610038161003816

le 211988811198882(1 + |120582|)

+ int

119909

minus1

[10038161003816100381610038161199011(119905)

1003816100381610038161003816

1003816100381610038161003816yminus11

(119905 120582)1003816100381610038161003816

+10038161003816100381610038161199012(119905)

1003816100381610038161003816

1003816100381610038161003816yminus12

(119905 120582)1003816100381610038161003816] 119890minus|I120582|(119905+1)

119889119905

(57)

Similarly we can prove that

119890minus|I120582|(119909+1) 1003816

100381610038161003816yminus12

(119909 120582)1003816100381610038161003816

le 211988811198882(1 + |120582|)

+ int

119909

minus1

[10038161003816100381610038161199011(119905)

1003816100381610038161003816

1003816100381610038161003816yminus11

(119905 120582)1003816100381610038161003816

+10038161003816100381610038161199012(119905)

1003816100381610038161003816

1003816100381610038161003816yminus12

(119905 120582)1003816100381610038161003816] 119890minus|I120582|(119905+1)

119889119905

(58)

Then from (57) (58) and Lemma 31 of [32 pp 204] weobtain (54)

In a similar manner we will prove the following lemmafor y01

(sdot 120582) and y02

(sdot 120582)

Lemma 5 The functions y01

(119909 120582) and y02

(119909 120582) are entire in120582 for any fixed 119909 isin (0 1] and satisfy the growth condition

1003816100381610038161003816y01

(119909 120582)1003816100381610038161003816

1003816100381610038161003816y02

(119909 120582)1003816100381610038161003816le 211988841198885(1 + |120582|) 119890

|I120582|(119909+1)

120582 isin C

(59)

Abstract and Applied Analysis 7

Proof Since y01

(119909 120582) = T01

12059312(119909 120582) +

T02

12059322(119909 120582) then

from (49) and (50) we obtain

y01

(119909 120582) =

1

120575

12059311

(0minus 120582)T

01cos (120582119909)

minus

1

120575

12059321

(0minus 120582)T

01sin (120582119909)

minusT01yminus12

(119909 120582) +

1

120575

12059311

(0minus 120582)

T02

sin (120582119909)

+

1

120575

12059321

(0minus 120582)

T02

cos (120582119909) + T02yminus12

(119909 120582)

(60)

Then from (47) and (48) and Lemma 4 we get

1003816100381610038161003816y01

(119909 120582)1003816100381610038161003816le

1

|120575|

100381610038161003816100381612059311

(0minus 120582)

1003816100381610038161003816

1003816100381610038161003816T01

cos (120582119909)1003816100381610038161003816

+

1

|120575|

100381610038161003816100381612059321

(0minus 120582)

1003816100381610038161003816

1003816100381610038161003816T01

sin (120582119909)1003816100381610038161003816

+1003816100381610038161003816T01yminus12

(119909 120582)1003816100381610038161003816

+

1

|120575|

100381610038161003816100381612059311

(0minus 120582)

1003816100381610038161003816

10038161003816100381610038161003816

T02

sin (120582119909)

10038161003816100381610038161003816

+

1

|120575|

100381610038161003816100381612059321

(0minus 120582)

1003816100381610038161003816

10038161003816100381610038161003816

T02

cos (120582119909)10038161003816100381610038161003816

+

10038161003816100381610038161003816

T02yminus12

(119909 120582)

10038161003816100381610038161003816

le 21198884(1 + |120582|)

times (1198883+

2

|120575|

(1198881+ 1198883)) 119890|I120582|(119909+1)

= 211988841198885(1 + |120582|) 119890

|I120582|(119909+1)

(61)

Similarly we can prove that1003816100381610038161003816y02

(119909 120582)1003816100381610038161003816le 211988841198885(1 + |120582|) 119890

|I120582|(119909+1) (62)

3 The Numerical Scheme

In this section we derive the method of computing eigen-values of problems (12)ndash(15) numerically The basic idea ofthe scheme is to split Ω(120582) into two parts a known partK(120582) and an unknown oneU(120582)Thenwe approximateU(120582)

using (5) to get the approximate Ω(120582) and then compute theapproximate zeros We first splitΩ(120582) into two parts

Ω (120582) = K (120582) +U (120582) (63)

whereU(120582) is the unknown part involving integral operators

U (120582) = 120575 [(1205732+ 120582 cos 120579

2) sin 120582

minus (1205731+ 120582 sin 120579

2) cos 120582] y

minus11(0minus 120582)

minus 120575 [(1205731+ 120582 sin 120579

2) sin 120582

+ (1205732+ 120582 cos 120579

2) cos 120582] y

minus12(0minus 120582)

minus 1205752[(1205731+ 120582 sin 120579

2) y01

(1 120582)

+ (1205732+ 120582 cos 120579

2) y02

(1 120582)]

(64)

andK(120582) is the known part

K (120582) = 120575 (1205731+ 120582 sin 120579

2)

times [1205722cos 2120582 minus 120572

1sin 2120582 + 120582 cos (2120582 + 120579

1)]

minus 120575 (1205732+ 120582 cos 120579

2)

times [1205722sin 2120582 + 120572

1cos 2120582 + 120582 sin (2120582 + 120579

1)]

(65)

Then from Lemmas 4 and 5 we have the following result

Lemma 6 The function U(120582) is entire in 120582 and the followingestimate holds

|U (120582)| le 120601 (120582) 1198902|I120582|

(66)

where

120601 (120582) = 119872(1 + |120582|)2

119872 = 4 |120575| 1198886 (21198883 + |120575| 11988841198885)

(67)

Proof From (64) we have

|U (120582)| le |120575| [(10038161003816100381610038161205732

1003816100381610038161003816+ |120582|

1003816100381610038161003816cos 1205792

1003816100381610038161003816) |sin 120582|

+ (10038161003816100381610038161205731

1003816100381610038161003816+ |120582|

1003816100381610038161003816sin 1205792

1003816100381610038161003816) |cos 120582|] 100381610038161003816

1003816yminus11

(0minus 120582)

1003816100381610038161003816

+ |120575| [(10038161003816100381610038161205731

1003816100381610038161003816+ |120582|

1003816100381610038161003816sin 1205792

1003816100381610038161003816) |sin 120582|

+ (10038161003816100381610038161205732

1003816100381610038161003816+1003816100381610038161003816120582 cos 120579

2

1003816100381610038161003816) |cos 120582|] 100381610038161003816

1003816yminus12

(0minus 120582)

1003816100381610038161003816

+ 1205752[(10038161003816100381610038161205731

1003816100381610038161003816+ |120582|

1003816100381610038161003816sin 1205792

1003816100381610038161003816)1003816100381610038161003816y01

(1 120582)1003816100381610038161003816

+ (10038161003816100381610038161205732

1003816100381610038161003816+ |120582|

1003816100381610038161003816cos 1205792

1003816100381610038161003816)1003816100381610038161003816y02

(1 120582)1003816100381610038161003816]

(68)

Using the inequalities | sin 120582| le 119890|I120582| and | cos 120582| le 119890

|I120582| for120582 isin C and Lemmas 4 and 5 imply (66)

ThusU(120582) is an entire function of exponential type 120590 = 2In the following we let 120582 isin R since all eigenvalues are realNow we approximate the function U(120582) using the operator(5) where ℎ isin (0 1205872) and 120596 = (120587 minus 2ℎ)2 and then from(7) we obtain

1003816100381610038161003816U (120582) minus (G

ℎ119873U) (120582)

1003816100381610038161003816le 119879ℎ119873

(120582) (69)

where

119879ℎ119873

(120582) = 2

10038161003816100381610038161003816sin (ℎ

minus1120587120582)

10038161003816100381610038161003816

times 120601 (|R120582| + ℎ (119873 + 1))

119890minus120596119873

radic120587120596119873

120573119873(0) 120582 isin R

(70)

The samplesU(119899ℎ) = Ω(119899ℎ) minusK(119899ℎ) 119899 isin Z119873(120582) cannot be

computed explicitly in the general caseWe approximate these

8 Abstract and Applied Analysis

samples numerically by solving the initial-value problemsdefined by (12) and (45) to obtain the approximate valuesU(119899ℎ) 119899 isin Z

119873(120582) that is U(119899ℎ) = Ω(119899ℎ) minus K(119899ℎ)

Here we use a computer algebra system mathematica toobtain the approximate solutions with the required accuracyHowever a separate study for the effect of different numericalschemes and the computational costs would be interestingAccordingly we have the explicit expansion

(Gℎ119873

U) (120582) = sum

119899isinZ119873(120582)

U (119899ℎ) sinc (ℎminus1120587119909 minus 119899120587)

times 119866(

radic120596 (120582 minus 119899ℎ)

radic119873ℎ

)

(71)

Therefore we get cf (10)10038161003816100381610038161003816(Gℎ119873

U) (120582) minus (Gℎ119873

U) (120582)

10038161003816100381610038161003816le 119860120576119873

(0) 120582 isin R (72)

Now let Ω119873(120582) = K(120582) + (G

ℎ119873U)(120582) From (69) and (72)

we obtain10038161003816100381610038161003816Ω (120582) minus Ω

119873(120582)

10038161003816100381610038161003816le 119879ℎ119873

(120582) + 119860120576119873

(0) 120582 isin R (73)

Let 120582lowast be an eigenvalue and 120582119873its desired approximation

that is Ω(120582lowast) = 0 and Ω

119873(120582119873) = 0 From (73) we have

|Ω119873(120582lowast)| le 119879

ℎ119873(120582lowast) + 119860120576119873

(0) Define the curves

119886plusmn(120582) = Ω

119873(120582) plusmn 119879

ℎ119873(120582) + 119860

120576119873(0) (74)

The curves 119886+(120582) 119886minus(120582) enclose the curve ofΩ(120582) for suitably

large 119873 Hence the closure interval is determined by solving119886plusmn(120582) = 0 giving an interval

119868120576119873

= [119886minus 119886+] (75)

It is worthwhile to mention that the simplicity of the eigen-values guarantees the existence of approximate eigenvaluesthat is the 120582

119873rsquos for which Ω

119873(120582119873) = 0 Next we estimate the

error |120582lowast minus 120582119873| for the eigenvalue 120582lowast

Theorem 7 Let 120582lowast be an eigenvalue of (12)ndash(15) and let 120582119873

be its approximation Then for 120582 isin R we have the followingestimate

1003816100381610038161003816120582lowastminus 120582119873

1003816100381610038161003816lt

119879ℎ119873

(120582119873) + 119860120576119873

(0)

inf120577isin119868120576119873

1003816100381610038161003816Ω1015840(120577)

1003816100381610038161003816

(76)

where the interval 119868120576119873

is defined above

Proof Replacing 120582 by 120582119873in (73) we obtain

1003816100381610038161003816Ω (120582119873) minus Ω (120582

lowast)1003816100381610038161003816lt 119879ℎ119873

(120582119873) + 119860120576119873

(0) (77)

where we have used Ω119873(120582119873) = Ω(120582

lowast) = 0 Using the

mean value theorem yields that for some 120577 isin 119869120576119873

=

[min(120582lowast 120582119873)max(120582lowast 120582

119873)]

10038161003816100381610038161003816(120582lowastminus 120582119873)Ω1015840(120577)

10038161003816100381610038161003816le 119879ℎ119873

(120582119873) + 119860120576119873

(0) 120577 isin 119869120576119873

sub 119868120576119873

(78)

Since 120582lowast is simple and 119873 is sufficiently large then

inf120577isin119868120576119873

|Ω1015840(120577)| gt 0 and we get (76)

4 Numerical Examples

In this section we introduce two examples illustrating the theabove method Also in the following examples we observethat the exact solutions 120582

119896are all inside the interval [119886

minus 119886+]

In these two examples we indicate the effect of the amplitudeerror in the method by determining enclosure intervals fordifferent values of 120576 We also indicate the effect of 119873 and ℎ

by several choices The eigenvalues of the following examplescannot computed concretely then we use mathematicato obtain the exact values mathematica is also used inrounding off the exact eigenvalues which are square rootsBoth numerical results and the associated figures prove thecredibility of the method

Example 1 Consider the system

1199101015840

2(119909) minus 119901 (119909) 119910

1(119909) = 120582119910

1(119909)

1199101015840

1(119909) + 119901 (119909) 119910

2(119909) = minus120582119910

2(119909)

119909 isin [minus1 0) cup (0 1]

1205821199101(minus1) minus (radic2 + 120582) 119910

2(minus1) = 0

(radic2 + 120582) 1199101(1) minus 120582119910

2(1) = 0

1199101(0minus) minus 2119910

1(0+) = 0 119910

2(0minus) minus 2119910

2(0+) = 0

(79)

Here

1199011(119909) = 119901

2(119909) = 119903 (119909) =

119909 119909 isin [minus1 0)

1 (0 1]

(80)

1205791= 1205792= 1205874 120572

1= 1205732= 0 120572

2= 1205731= 1 and 120575 = 2 Direct

calculations give

K (120582) = 2 ((radic2120582 + 1) cos (2120582) minus 120582 (120582 + radic2) sin (2120582))

Ω (120582) = 2 ((radic2120582 + 1) cos(2120582 +

1

2

)

minus120582 (120582 + radic2) sin(2120582 +

1

2

))

(81)

As is clearly seen the eigenvalues cannot be computedexplicitly Tables 1 2 and 3 indicate the application of ourtechnique to this problem and the effect of 120576 By exact wemean the zeros of Δ(120582) computed by Mathematica

Figures 1 and 2 illustrate the enclosure intervals domi-nating 120582

0for 119873 = 25 ℎ = 01 and 120576 = 10

minus2 120576 = 10minus5

respectively The middle curve represents Δ(120582) while theupper and lower curves represent the curves of 119886

+(120582) 119886minus(120582)

respectively We notice that when 120576 = 10minus5 the two curves

are almost identical Similarly Figures 3 and 4 illustrate theenclosure intervals dominating 120582

1for ℎ = 01 119873 = 25 and

120576 = 10minus2 120576 = 10

minus5 respectively

Abstract and Applied Analysis 9

Table 1 The approximation 120582119896119873

and the exact solutions 120582119896for different choices of ℎ and119873

120582119896

120582minus2

120582minus1

1205820

1205821

Exact 120582119896

minus12061856254879546 minus05151195632138896 034274095892547224 1616237903407542

120582119896119873

ℎ = 06 119873 = 15 minus12061855965725146 minus05151190347168934 034274053593937165 1616237692038969120596 = 09708 119873 = 25 minus12061856254843495 minus05151195631493376 0342740958869788 16162379033820231ℎ = 01 119873 = 15 minus12061856254844203 minus05151195632158179 03427409589350932 16162379034134882

120596 = 14708 119873 = 25 minus12061856254879548 minus05151195632138894 034274095892547236 16162379034075425

Table 2 Absolute error |120582119896minus 120582119896119873

|

120582119896

120582minus2

120582minus1

1205820

1205821

ℎ = 06119873 = 15 289154 times 10

minus8528497 times 10

minus7422986 times 10

minus7211369 times 10

minus7

119873 = 25 360512 times 10minus12

64552 times 10minus11

556842 times 10minus11

255189 times 10minus11

ℎ = 01119873 = 15 353428 times 10

minus12192824 times 10

minus12962097 times 10

minus12594613 times 10

minus12

119873 = 25 222045 times 10minus16

222045 times 10minus16

555112 times 10minus17

444089 times 10minus16

Table 3 For119873 = 25 and ℎ = 01 the exact solutions 120582119896are all inside the interval [119886

minus 119886+] for different values of 120576

120582119896

120582minus2

120582minus1

1205820

1205821

Exact 120582119896

minus12061856254879546 minus05151195632138896 034274095892547224 1616237903407542119868120576119873

120576 = 10minus2

[minus12335696 minus11771922] [minus05486790 minus04826816] [03246478 03601346] [16100119 16224066]

119868120576119873

120576 = 10minus5

[minus12062137 minus12061575] [minus05151525 minus05150867] [03427232 03427586] [16162317 16162441]

030 032 034 036 038

00

02

04

aminus

a+

Ω(120582)

minus02

minus04

minus06

Figure 1The enclosure interval dominating 1205820for ℎ = 01119873 = 25

and 120576 = 10minus2

Example 2 In this example we consider the system

1199101015840

2(119909) minus 119901 (119909) 119910

1(119909) = 120582119910

1(119909)

1199101015840

1(119909) + 119901 (119909) 119910

2(119909) = minus120582119910

2(119909)

119909 isin [minus1 0) cup (0 1]

radic31199101(minus1) minus 119910

2(minus1) = 0

(1 +

1

2

120582)1199101(1) minus (1 +

radic3

2

120582)1199102(1) = 0

1199101(0minus) minus 3119910

1(0+) = 0 119910

2(0minus) minus 3119910

2(0+) = 0

(82)

030 032 034 036 038

00

02

aminus

a+

Ω(120582)

minus02

minus04

Figure 2The enclosure interval dominating 1205820for ℎ = 01119873 = 25

and 120576 = 10minus5

where

1199011(119909) = 119901

2(119909) = 119901 (119909) =

119909 119909 isin [minus1 0)

1199092 (0 1]

(83)

1205791= 1205792= 0 120572

1= 1205732= 1 120572

2= 1205731= 2 and 120575 = 12 Direct

calculations give

K (120582) =

1

2

((120582 + 3) cos (2120582) minus (1205822+ 3120582 + 4) sin (2120582))

Ω (120582) =

1

2

((1205822+ 3120582 + 4) sin(

1

6

minus 2120582)

+ (120582 + 3) cos(1

6

minus 2120582))

(84)

10 Abstract and Applied Analysis

Table 4 The approximation 120582119896119873

and the exact solutions 120582119896for different choices of ℎ and119873

120582119896

120582minus2

120582minus1

1205820

1205821

Exact 120582119896

minus30661256186938894 minus10959703083466883 03689030343814017 18338437675275228

120582119896119873

ℎ = 08 119873 = 15 minus3066123822200742 minus10959640944020903 036890463046491767 18338437607680411120596 = 007708 119873 = 25 minus30661256172349627 minus10959703026494807 036890303588091056 18338437676561417

ℎ = 02 119873 = 15 minus30661256186844446 minus10959703083278305 03689030343916001 1833843767517251120596 = 13708 119873 = 25 minus306612561869389 minus1095970308346688 03689030343814024 18338437675275225

Table 5 Absolute error |120582119896minus 120582119896119873

|

120582119896

120582minus2

120582minus1

1205820

1205821

ℎ = 08119873 = 15 179649 times 10

minus6621394 times 10

minus6159608 times 10

minus6675948 times 10

minus9

119873 = 25 145893 times 10minus9

569721 times 10minus9

149951 times 10minus9

128619 times 10minus10

ℎ = 02119873 = 15 944489 times 10

minus12188578 times 10

minus12101984 times 10

minus11102718 times 10

minus11

119873 = 25 444089 times 10minus16

222045 times 10minus16

666134 times 10minus16

222045 times 10minus16

Table 6 For119873 = 25 and ℎ = 02 the exact solutions 120582119896are all inside the interval [119886

minus 119886+] for different values of 120576

120582119896

120582minus2

120582minus1

1205820

1205821

Exact 120582119896

minus30661256186938894 minus10959703083466883 03689030343814017 18338437675275228119868120576119873

120576 = 10minus2

[minus31098742 minus30188665] [minus11631502 minus10339754] [03437434 03933486] [18220764 18454828]

119868120576119873

120576 = 10minus5

[minus30661709 minus30660803] [minus10960346 minus10959060] [03688783 03689278] [18338321 18338555]

158 159 160 161 162 163 164 165

00

05

10

minus05

minus10

aminus

a+

Ω(120582)

Figure 3The enclosure interval dominating 1205821for ℎ = 01119873 = 25

and 120576 = 10minus2

Tables 4 and 6 give the exact eigenvalues 1205821198961

119896=minus2and their

approximate ones for different values of ℎ 119873 and 120576 InTable 5 we give the absolute error for different values of ℎ and119873

Here Figures 5 6 7 and 8 illustrate the enclosureintervals dominating 120582

minus2and 120582

minus1for ℎ = 02 119873 = 25 and

120576 = 10minus2 120576 = 10

minus5 respectively

158 159 160 161 162 163 164 165

00

05

minus05

aminus

a+

Ω(120582)

Figure 4The enclosure interval dominating 1205821for ℎ = 01119873 = 25

and 120576 = 10minus5

5 Conclusion

With a simple analysis and with values of solutions ofinitial value problems computed at a few values of theeigenparameter we have computed the eigenvalues of aDirac system which has a discontinuity at one point andcontains a spectral parameter in all boundary conditionswith a certain estimated error The method proposed is

Abstract and Applied Analysis 11

00

05

10

minus05

minus33 minus32 minus31 minus30 minus29 minus28

aminus

a+

Ω(120582)

Figure 5 The enclosure interval dominating 120582minus2

for ℎ = 02 119873 =

25 and 120576 = 10minus2

00

05

10

minus05

minus33 minus32 minus31 minus30 minus29 minus28

aminus

a+

Ω(120582)

Figure 6 The enclosure interval dominating 120582minus2

for ℎ = 02 119873 =

25 and 120576 = 10minus5

a shooting procedure that is the problem is reformulatedas two initial value ones due to the interior discontinuity ofsize two and a miss-distance is defined at the right end of theinterval of integration whose roots are the eigenvalues to becomputed The unknown partU(120582) of the miss-distance canbewritten in terms of a functionwhich is an entire function ofexponential typeTherefore we propose to approximate suchterm by means of a truncated cardinal series with samplingvalues approximated by solving numerically correspondingsuitable initial value problems Finally in Section 4 weintroduced two instructive examples where both numericalresults and the associated figures have proved the credibilityof the method

00

05

10

minus05

minus14 minus13 minus12 minus11 minus10 minus09 minus08 minus07

aminus

a+

Ω(120582)

Figure 7 The enclosure interval dominating 120582minus1

for ℎ = 02 119873 =

25 and 120576 = 10minus2

00

05

10

minus05

minus14 minus13 minus12 minus11 minus10 minus09 minus08 minus07

aminus

a+

Ω(120582)

Figure 8 The enclosure interval dominating 120582minus1

for ℎ = 02 119873 =

25 and 120576 = 10minus5

Conflict of Interests

The authors declare that there is no conflict of interestsregarding the publication of this paper

Acknowledgment

This paper was funded by the Deanship of Scientific Research(DSR) King Abdulaziz University Jeddah The authorstherefore acknowledge with thanks DSR technical and finan-cial support

12 Abstract and Applied Analysis

References

[1] C T Fulton ldquoTwo-point boundary value problems witheigenvalue parameter contained in the boundary conditionsrdquoProceedings of the Royal Society of Edinburgh A Mathematicsvol 77 no 3-4 pp 293ndash308 1977

[2] JWalter ldquoRegular eigenvalue problems with eigenvalue param-eter in the boundary conditionrdquoMathematische Zeitschrift vol133 pp 301ndash312 1973

[3] V Kotelnikov ldquoOn the carrying capacity of the ether and wirein telecommunications material for the first All-Unionrdquo inConference on Questions of Communications vol 55 pp 55ndash64Izd Red Upr Svyazi RKKA 1933

[4] C E Shannon ldquoCommunication in the presence of noiserdquoProceedings of the IRE vol 37 pp 10ndash21 1949

[5] E TWhittaker ldquoOn the functions which are represented by theexpansion of the interpolation theoryrdquo Proceedings of the RoyalSociety of Edinburgh A vol 35 pp 181ndash194 1915

[6] P L Butzer G Schmeisser and R L Stens ldquoAn introductionto sampling analysisrdquo in Non Uniform Sampling Theory andPractices F Marvasti Ed pp 17ndash121 Kluwer New York NYUSA 2001

[7] M A Kowalski K A Sikorski and F Stenger Selected Topics inApproximation andComputation OxfordUniversity Press NewYork NY USA 1995

[8] J Lund and K L Bowers Sinc Methods for Quadrature andDifferential Equations SIAM Philadelphia Pa USA 1992

[9] F Stenger ldquoNumericalmethods based onWhittaker cardinal orsinc functionsrdquo SIAM Review vol 23 no 2 pp 165ndash224 1981

[10] F Stenger Numerical Methods Based on Sinc and AnalyticFunctions Springer New York NY USA 1993

[11] A Boumenir ldquoHigher approximation of eigenvalues by thesampling methodrdquo BIT Numerical Mathematics vol 40 no 2pp 215ndash225 2000

[12] A Boumenir ldquoSampling and eigenvalues of non-self-adjointSturm-Liouville problemsrdquo SIAM Journal on Scientific Comput-ing vol 23 no 1 pp 219ndash229 2001

[13] M M Tharwat A H Bhrawy and A Yildirim ldquoNumericalcomputation of eigenvalues of discontinuous Sturm-Liouvilleproblems with parameter dependent boundary conditionsusing sinc methodrdquoNumerical Algorithms vol 63 no 1 pp 27ndash48 2013

[14] M M Tharwat A H Bhrawy and A Yildirim ldquoNumericalcomputation of the eigenvalues of a discontinuousDirac systemusing the sincmethodwith error analysisrdquo International Journalof Computer Mathematics vol 89 no 15 pp 2061ndash2080 2012

[15] P L Butzer and R L Stens ldquoA modification of the Whittaker-Kotelrsquonikov-Shannon sampling seriesrdquo Aequationes Mathemat-icae vol 28 no 3 pp 305ndash311 1985

[16] R Gervais Q I Rahman and G Schmeisser ldquoA bandlimitedfunction simulating a duration-limited onerdquo in ApproximationTheory andFunctionalAnalysis P L Butzer andR L Stens Edspp 355ndash362 Birkhaauser Boston Mass USA 1984

[17] R L Stens ldquoSampling by generalized kernelsrdquo in SamplingThe-ory in Fourier and Signal Analysis Advanced Topics J RHigginsand R L Stens Eds pp 130ndash157 Oxford University PressOxford UK 1999

[18] G Schmeisser and F Stenger ldquoSinc approximation with a Gaus-sianmultiplierrdquo SamplingTheory in Signal and Image Processingvol 6 no 2 pp 199ndash221 2007

[19] L Qian ldquoOn the regularized Whittaker-Kotelrsquonikov-Shannonsampling formulardquo Proceedings of the American MathematicalSociety vol 131 no 4 pp 1169ndash1176 2003

[20] L Qian and D B Creamer ldquoA modification of the samplingseries with aGaussianmultiplierrdquo SamplingTheory in Signal andImage Processing vol 5 no 1 pp 1ndash20 2006

[21] L Qian and D B Creamer ldquoLocalized sampling in the presenceof noiserdquoAppliedMathematics Letters vol 19 no 4 pp 351ndash3552006

[22] M H Annaby and R M Asharabi ldquoComputing eigenvaluesof boundary-value problems using sinc-Gaussian methodrdquoSamplingTheory in Signal and Image Processing vol 7 no 3 pp293ndash312 2008

[23] M H Annaby and M MTharwat ldquoA sinc-Gaussian techniquefor computing eigenvalues of second-order linear pencilsrdquoApplied Numerical Mathematics vol 63 pp 129ndash137 2013

[24] A H Bhrawy M M Tharwat and A Al-Fhaid ldquoNumericalalgorithms for computing eigenvalues of discontinuous Diracsystem using sinc-Gaussian methodrdquo Abstract and AppliedAnalysis vol 2012 Article ID 925134 13 pages 2012

[25] M M Tharwat A H Bhrawy and A S Alofi ldquoApproximationof eigenvalues of discontinuous sturm-liouville problems witheigenparameter in all boundary conditionsrdquo Boundary ValueProblems vol 2013 article 132 2013

[26] M M Tharwat and S M Al-Harbi ldquoApproximation of eigen-values of boundary value problemsrdquo Boundary Value Problemsvol 2014 article 51 2014

[27] MMTharwat and A H Bhrawy ldquoComputation of eigenvaluesof discontinuous Dirac system using Hermite interpolationtechniquerdquo Advances in Difference Equations vol 2012 article59 2012

[28] M M Tharwat A Yildirim and A H Bhrawy ldquoSampling ofdiscontinuous Dirac systemsrdquo Numerical Functional Analysisand Optimization vol 34 no 3 pp 323ndash348 2013

[29] M M Tharwat ldquoDiscontinuous Sturm-Liouville problems andassociated sampling theoriesrdquo Abstract and Applied Analysisvol 2011 Article ID 610232 30 pages 2011

[30] MMTharwat ldquoOn sampling theories and discontinuousDiracsystems with eigenparameter in the boundary conditionsrdquoBoundary Value Problems vol 2013 article 65 2013

[31] B M Levitan and I S Sargsjan ldquoIntroduction to spectral the-ory self adjoint ordinary differential operatorsrdquo in Translationof Mthematical Monographs vol 39 American MathematicalSociety Providence RI USA 1975

[32] B M Levitan and I S Sargsjan Sturm-Liouville and Dirac oper-ators vol 59 Kluwer Academic Publishers Group DordrechtThe Netherlands 1991

Submit your manuscripts athttpwwwhindawicom

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical Problems in Engineering

Hindawi Publishing Corporationhttpwwwhindawicom

Differential EquationsInternational Journal of

Volume 2014

Applied MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Probability and StatisticsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical PhysicsAdvances in

Complex AnalysisJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

OptimizationJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

CombinatoricsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Operations ResearchAdvances in

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Function Spaces

Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of Mathematics and Mathematical Sciences

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

The Scientific World JournalHindawi Publishing Corporation httpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Algebra

Discrete Dynamics in Nature and Society

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Decision SciencesAdvances in

Discrete MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom

Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Stochastic AnalysisInternational Journal of

4 Abstract and Applied Analysis

method as in proof of Theorem 11 in [31] we see that theproblem

u1015840

2(119909) minus 119901

1(119909)u1(119909) = 120582u

1(119909)

u1015840

1(119909) + 119901

2(119909)u2(119909) = minus120582u

2(119909)

119909 isin (0 1)

u1(1) = 120573

2+ 120582 cos 120579

2 u

2(1) = 120573

1+ 120582 sin 120579

2

(24)

has a unique solution u = (12059512(119909120582)

12059522(119909120582)) which is an entire

function of parameter 120582 for each fixed 119909 isin [0 1]Now the functions 120593

1198942(119909 120582) and 120595

1198941(119909 120582) are defined in

terms of 1205931198941(119909 120582) and 120595

1198942(119909 120582) 119894 = 1 2 respectively as

follows the initial-value problem

u1015840

2(119909) minus 119901

1(119909)u1(119909) = 120582u

1(119909)

u1015840

1(119909) + 119901

2(119909)u2(119909) = minus120582u

2(119909)

119909 isin (0 1)

(25)

u1(0) =

1

120575

12059311

(0 120582) u2(0) =

1

120575

12059321

(0 120582) (26)

has unique solution u = (12059312(119909120582)

12059322(119909120582)) for each 120582 isin C

Similarly the following problem also has a unique solu-tion u = (

12059511(119909120582)

12059521(119909120582))

u1015840

2(119909) minus 119901

1(119909)u1(119909) = 120582u

1(119909)

u1015840

1(119909) + 119901

2(119909)u2(119909) = minus120582u

2(119909)

119909 isin (minus1 0)

(27)

u1(0) = 120575120595

12(0 120582) u

2(0) = 120575120595

22(0 120582) (28)

Let us construct two basic solutions of (12) as

120593 (sdot 120582) = (1205931(sdot 120582)

1205932(sdot 120582)

) 120595 (sdot 120582) = (1205951(sdot 120582)

1205952(sdot 120582)

) (29)

where

1205931(119909 120582) =

12059311

(119909 120582) 119909 isin [minus1 0)

12059312

(119909 120582) 119909 isin (0 1]

1205932(119909 120582) =

12059321

(119909 120582) 119909 isin [minus1 0)

12059322

(119909 120582) 119909 isin (0 1]

(30)

1205951(119909 120582) =

12059511

(119909 120582) 119909 isin [minus1 0)

12059512

(119909 120582) 119909 isin (0 1]

1205952(119909 120582) =

12059521

(119909 120582) 119909 isin [minus1 0)

12059522

(119909 120582) 119909 isin (0 1]

(31)

By virtue of (26) and (28) these solutions satisfy bothtransmission conditions (15) These functions are entire in 120582

for all 119909 isin [minus1 0) cup (0 1]

Let W(120593 120595)(sdot 120582) denote the Wronskian of 120593(sdot 120582) and120595(sdot 120582) defined in [32 page 194] that is

W (120593 120595) (sdot 120582) =

10038161003816100381610038161003816100381610038161003816

1205931(sdot 120582) 120593

2(sdot 120582)

1205951(sdot 120582) 120595

2(sdot 120582)

10038161003816100381610038161003816100381610038161003816

(32)

It is obvious that the Wronskians

Ω119894(120582) = W (120593 120595) (119909 120582)

= 1205931119894(119909 120582) 120595

2119894(119909 120582)

minus 1205932119894(119909 120582) 120595

1119894(119909 120582) 119909 isin Γ

119894 119894 = 1 2

(33)

are independent on variable 119909 isin Γ119894(119894 = 1 2) and 120593

119894(119909 120582)

and 120595119894(119909 120582) are entire functions of the parameter 120582 for each

119909 isin Γ119894(119894 = 1 2) Taking into account (26) and (28) a short

calculation gives

Ω1(120582) = 120575

2Ω2(120582) (34)

for each 120582 isin C

Corollary 2 The zeros of the functions Ω1(120582) and Ω

2(120582)

coincide

Then we may introduce to the consideration the charac-teristic functionΩ(120582) as

Ω (120582) = Ω1(120582) = 120575

2Ω2(120582) (35)

Note that all eigenvalues of problems (12)ndash(15) are just zerosof the function Ω(120582) Indeed since the functions 120593

1(119909 120582)

and 1205932(119909 120582) satisfy the boundary condition (13) and both

transmission conditions (15) to find the eigenvalues ofthe (12)ndash(15) we have to insert the functions 120593

1(119909 120582) and

1205932(119909 120582) in the boundary condition (14) and find the roots

of this equation In the following lemma we show that alleigenvalues of the problems (12)ndash(15) are simple

Lemma 3 The eigenvalues of the boundary value problems(12)ndash(15) form an at most countable set without finite limitpoints All eigenvalues of the boundary value problems (12)ndash(15) (of Ω(120582)) are simple

Proof The eigenvalues are the zeros of the entire functionoccurring on the left-hand side in see (35)

(1205731+ 120582 sin 120579

2) 12059312

(1 120582) minus (1205732+ 120582 cos 120579

2) 12059322

(1 120582) = 0

(36)

We have shown (see Lemma 1) that this function doesnot vanish for nonreal 120582 In particular it does not vanishidentically Therefore its zeros form an at most countable setwithout finite limit points

By (12) we obtain for 120582 120583 isin C 120582 = 120583

119889

119889119909

1205931(119909 120582) 120593

2(119909 120583) minus 120593

1(119909 120583) 120593

2(119909 120582)

= (120583 minus 120582) 1205931(119909 120582) 120593

1(119909 120583) + 120593

2(119909 120582) 120593

2(119909 120583)

(37)

Abstract and Applied Analysis 5

Integrating the above equation through [minus1 0] and [0 1] andtaking into account the initial conditions (23) (26) and (28)we obtain

1205752(12059312

(1 120582) 12059322

(1 120583) minus 12059312

(1 120583) 12059322

(1 120582))

minus (120583 minus 120582) 1205881

= (120583 minus 120582) (int

0

minus1

(12059311

(119909 120582) 12059311

(119909 120583)

+ 12059321

(119909 120582) 12059321

(119909 120583)) 119889119909

+ 1205752int

1

0

(12059312

(119909 120582) 12059312

(119909 120583)

+ 12059322

(119909 120582) 12059322

(119909 120583)) 119889119909)

(38)

Dividing both sides of (38) by (120582 minus 120583) and by letting 120583 rarr 120582we arrive to the relation

1205752(12059322

(1 120582)

12059712059312

(1 120582)

120597120582

minus 12059312

(1 120582)

12059712059322

(1 120582)

120597120582

) + 1205881

= minus(int

0

minus1

(100381610038161003816100381612059311

(119909 120582)1003816100381610038161003816

2

+100381610038161003816100381612059321

(119909 120582)1003816100381610038161003816

2

) 119889119909

+ 1205752int

1

0

(100381610038161003816100381612059312

(119909 120582)1003816100381610038161003816

2

+100381610038161003816100381612059322

(119909 120582)1003816100381610038161003816

2

) 119889119909)

(39)

We show that equation

Ω (120582) = W (120593 120595) (1 120582)

= 1205752((1205731+ 120582 sin 120579

2) 12059312

(1 120582)

minus (1205732+ 120582 cos 120579

2) 12059322

(1 120582)) = 0

(40)

has only simple roots Assume the converse that is (40) hasa double root 120582lowast Then the following two equations hold

(1205731+ 120582lowast sin 120579

2) 12059312

(1 120582lowast) minus (120573

2+ 120582lowast cos 120579

2) 12059322

(1 120582lowast) = 0

sin 120579212059312

(1 120582lowast) + (120573

1+ 120582lowast sin 120579

2)

12059712059312

(1 120582lowast)

120597120582

minus cos 120579212059322

(1 120582lowast) minus (120573

2+ 120582lowast cos 120579

2)

12059712059322

(1 120582lowast)

120597120582

= 0

(41)

Since 1205882

= 0 and 120582lowast is real then (120573

1+ 120582lowast sin 120579

2)2

+

(1205732+ 120582lowast cos 120579

2)2

= 0 Let 1205731+ 120582lowast sin 120579

2= 0 From (41)

12059312

(1 120582lowast) =

(1205732+ 120582lowast cos 120579

2)

(1205731+ 120582lowast sin 120579

2)

12059322

(1 120582lowast)

12059712059312

(1 120582lowast)

120597120582

=

120588212059322

(1 120582lowast)

(1205731+ 120582lowast sin 120579

2)2

+

(1205732+ 120582lowast cos 120579

2)

(1205731+ 120582lowast sin 120579

2)

12059712059322

(1 120582lowast)

120597120582

(42)

Combining (42) and (39) with 120582 = 120582lowast we obtain

12058821205752(12059322

(1 120582lowast))2

(1205731+ 120582lowast sin 120579

2)2

+ 1205881

= minus(int

0

minus1

(100381610038161003816100381612059311

(119909 120582)1003816100381610038161003816

2

+100381610038161003816100381612059321

(119909 120582)1003816100381610038161003816

2

) 119889119909

+ 1205752int

1

0

(

1003816100381610038161003816100381612059312

(119909 120582) |2+

1003816100381610038161003816100381612059322

(119909 120582) |2) 119889119909)

(43)

contradicting the assumption 120588119894gt 0 119894 = 1 2 The other case

when 1205732+120582lowast cos 120579

2= 0 can be treated similarly and the proof

is complete

Recall that boundary value problems (12)ndash(15) have denu-merable set of real and simple eigenvalues see the abovesection From (23) and (26) the solution of (12)

120593 (sdot 120582) = (1205931(sdot 120582)

1205932(sdot 120582)

)

120593119894(119909 120582) =

1205931198941(119909 120582) 119909 isin [minus1 0)

1205931198942(119909 120582) 119909 isin (0 1]

119894 = 1 2

(44)

satisfies the following initial conditions

(12059311

(minus1 120582) 12059312

(0+ 120582)

12059321

(minus1 120582) 12059322

(0+ 120582)

)

= (1205722+ 120582 cos 120579

1120575minus112059311

(0minus 120582)

1205721+ 120582 sin 120579

1120575minus112059321

(0minus 120582)

)

(45)

Since 120593(sdot 120582) satisfies (13) then the eigenvalues of the prob-lems (12)ndash(15) are the zeros of the function

Ω (120582) = 1205752((1205731+ 120582 sin 120579

2) 12059312

(1 120582)

minus (1205732+ 120582 cos 120579

2) 12059322

(1 120582)) = 0

(46)

We shall transform (12) (23) (26) and (30) into the integralequations see [32]

12059311

(119909 120582) = 1205722cos 120582 (119909 + 1) minus 120572

1sin 120582 (119909 + 1)

+ 120582 cos (120582 (119909 + 1) + 1205791) minusTminus11

12059311

(119909 120582)

minusTminus12

12059321

(119909 120582)

(47)

12059321

(119909 120582) = 1205721cos 120582 (119909 + 1) + 120572

2sin 120582 (119909 + 1)

+ 120582 sin (120582 (119909 + 1) + 1205791) +

Tminus11

12059311

(119909 120582)

minusTminus12

12059321

(119909 120582)

(48)

12059312

(119909 120582) =

1

120575

12059311

(0minus 120582) cos (120582119909) minus 1

120575

12059321

(0minus 120582) sin (120582119909)

minusT01

12059312

(119909 120582) minusT02

12059322

(119909 120582)

(49)

6 Abstract and Applied Analysis

12059322

(119909 120582) =

1

120575

12059311

(0minus 120582) sin (120582119909) +

1

120575

12059321

(0minus 120582) cos (120582119909)

+T01

12059312

(119909 120582) minusT02

12059322

(119909 120582)

(50)

where Tminus1119894

Tminus1119894

T0119894

and T0119894 119894 = 1 2 are the Volterra

integral operators defined by

Tminus1119894

120593 (119909 120582) = int

119909

minus1

sin 120582 (119909 minus 119905) 119901119894(119905) 120593 (119905 120582) 119889119905

Tminus1119894

120593 (119909 120582) = int

119909

minus1

cos 120582 (119909 minus 119905) 119901119894(119905) 120593 (119905 120582) 119889119905

T0119894120593 (119909 120582) = int

119909

0

sin 120582 (119909 minus 119905) 119901119894(119905) 120593 (119905 120582) 119889119905

T0119894120593 (119909 120582) = int

119909

0

cos 120582 (119909 minus 119905) 119901119894(119905) 120593 (119905 120582) 119889119905

(51)

For convenience we define the constants

1198881= max 100381610038161003816

10038161205721

1003816100381610038161003816+10038161003816100381610038161205722

10038161003816100381610038161003816100381610038161003816sin 1205791

1003816100381610038161003816+1003816100381610038161003816cos 1205791

1003816100381610038161003816

1198882= int

0

minus1

[10038161003816100381610038161199011(119905)

1003816100381610038161003816+10038161003816100381610038161199012(119905)

1003816100381610038161003816] 119889119905 119888

3= 11988811198882exp (119888

2)

1198884= int

1

0

[10038161003816100381610038161199011(119905)

1003816100381610038161003816+10038161003816100381610038161199012(119905)

1003816100381610038161003816] 119889119905 119888

5= 1198883+

2

|120575|

(1198881+ 1198883)

1198886= max 100381610038161003816

10038161205721

1003816100381610038161003816+10038161003816100381610038161205722

10038161003816100381610038161003816100381610038161003816sin 1205792

1003816100381610038161003816+1003816100381610038161003816cos 1205792

1003816100381610038161003816

(52)

Now we define yminus1119894

(sdot 120582) and y0119894(sdot 120582) 119894 = 1 2 to be

yminus11

(119909 120582) = Tminus11

12059311

(119909 120582) +Tminus12

12059321

(119909 120582)

yminus12

(119909 120582) =Tminus11

12059311

(119909 120582) minusTminus12

12059321

(119909 120582)

y01

(119909 120582) = T01

12059312

(119909 120582) +T02

12059322

(119909 120582)

y02

(119909 120582) =T01

12059312

(119909 120582) minusT02

12059322

(119909 120582)

(53)

Lemma 4 The functions yminus11

(119909 120582) and yminus12

(119909 120582) are entirein 120582 for any fixed 119909 isin [minus1 0) and satisfy the growth condition

1003816100381610038161003816yminus11

(119909 120582)1003816100381610038161003816

1003816100381610038161003816yminus12

(119909 120582)1003816100381610038161003816le 21198883(1 + |120582|) 119890

|I120582|(119909+1)

120582 isin C

(54)

Proof Since yminus11

(119909 120582) = Tminus11

12059311(119909 120582) +

Tminus12

12059321(119909 120582)

then from (47) and (48) we obtain

yminus11

(119909 120582)

= 1205722Tminus11

cos 120582 (119909 + 1) minus 1205721Tminus11

sin 120582 (119909 + 1)

+ 120582Tminus11

cos (120582 (119909 + 1) + 1205791) minusTminus11

yminus11

(119909 120582)

+ 1205721Tminus12

cos 120582 (119909 + 1) + 1205722Tminus12

sin 120582 (119909 + 1)

+ 120582Tminus12

sin (120582 (119909 + 1) + 1205791) +

Tminus12

yminus12

(119909 120582)

(55)

Using the inequalities | sin 119911| le 119890|I119911| and | cos 119911| le 119890

|I119911| for119911 isin C leads for 120582 isin C to1003816100381610038161003816yminus11

(119909 120582)1003816100381610038161003816

le10038161003816100381610038161205722

1003816100381610038161003816

1003816100381610038161003816Tminus11

cos 120582 (119909 + 1)1003816100381610038161003816+10038161003816100381610038161205721Tminus11

sin 120582 (119909 + 1)1003816100381610038161003816

+ |120582|1003816100381610038161003816Tminus11

cos (120582 (119909 + 1) + 1205791)1003816100381610038161003816+1003816100381610038161003816Tminus11

yminus11

(119909 120582)1003816100381610038161003816

+10038161003816100381610038161205721

1003816100381610038161003816

10038161003816100381610038161003816

Tminus12

cos 120582 (119909 + 1)

10038161003816100381610038161003816+10038161003816100381610038161205722

1003816100381610038161003816

10038161003816100381610038161003816

Tminus12

sin 120582 (119909 + 1)

10038161003816100381610038161003816

+ |120582|

10038161003816100381610038161003816

Tminus12

sin (120582 (119909 + 1) + 1205791)

10038161003816100381610038161003816+

10038161003816100381610038161003816

Tminus12

yminus12

(119909 120582)

10038161003816100381610038161003816

le 21198881(1 + |120582|) 119890

|I120582|(119909+1)int

119909

minus1

[10038161003816100381610038161199011(119905)

1003816100381610038161003816+10038161003816100381610038161199012(119905)

1003816100381610038161003816] 119889119905

+ 119890|I120582|(119909+1)

int

119909

minus1

[10038161003816100381610038161199011(119905)

1003816100381610038161003816

1003816100381610038161003816yminus11

(119905 120582)1003816100381610038161003816

+10038161003816100381610038161199012(119905)

1003816100381610038161003816

1003816100381610038161003816yminus12

(119905 120582)1003816100381610038161003816] 119890minus|I120582|(119905+1)

119889119905

le 211988811198882(1 + |120582|) 119890

|I120582|(119909+1)

+ 119890|I120582|(119909+1)

int

119909

minus1

[10038161003816100381610038161199011(119905)

1003816100381610038161003816

1003816100381610038161003816yminus11

(119905 120582)1003816100381610038161003816

+10038161003816100381610038161199012(119905)

1003816100381610038161003816

1003816100381610038161003816yminus12

(119905 120582)1003816100381610038161003816] 119890minus|I120582|(119905+1)

119889119905

(56)

The above inequality can be reduced to

119890minus|I120582|(119909+1) 1003816

100381610038161003816yminus11

(119909 120582)1003816100381610038161003816

le 211988811198882(1 + |120582|)

+ int

119909

minus1

[10038161003816100381610038161199011(119905)

1003816100381610038161003816

1003816100381610038161003816yminus11

(119905 120582)1003816100381610038161003816

+10038161003816100381610038161199012(119905)

1003816100381610038161003816

1003816100381610038161003816yminus12

(119905 120582)1003816100381610038161003816] 119890minus|I120582|(119905+1)

119889119905

(57)

Similarly we can prove that

119890minus|I120582|(119909+1) 1003816

100381610038161003816yminus12

(119909 120582)1003816100381610038161003816

le 211988811198882(1 + |120582|)

+ int

119909

minus1

[10038161003816100381610038161199011(119905)

1003816100381610038161003816

1003816100381610038161003816yminus11

(119905 120582)1003816100381610038161003816

+10038161003816100381610038161199012(119905)

1003816100381610038161003816

1003816100381610038161003816yminus12

(119905 120582)1003816100381610038161003816] 119890minus|I120582|(119905+1)

119889119905

(58)

Then from (57) (58) and Lemma 31 of [32 pp 204] weobtain (54)

In a similar manner we will prove the following lemmafor y01

(sdot 120582) and y02

(sdot 120582)

Lemma 5 The functions y01

(119909 120582) and y02

(119909 120582) are entire in120582 for any fixed 119909 isin (0 1] and satisfy the growth condition

1003816100381610038161003816y01

(119909 120582)1003816100381610038161003816

1003816100381610038161003816y02

(119909 120582)1003816100381610038161003816le 211988841198885(1 + |120582|) 119890

|I120582|(119909+1)

120582 isin C

(59)

Abstract and Applied Analysis 7

Proof Since y01

(119909 120582) = T01

12059312(119909 120582) +

T02

12059322(119909 120582) then

from (49) and (50) we obtain

y01

(119909 120582) =

1

120575

12059311

(0minus 120582)T

01cos (120582119909)

minus

1

120575

12059321

(0minus 120582)T

01sin (120582119909)

minusT01yminus12

(119909 120582) +

1

120575

12059311

(0minus 120582)

T02

sin (120582119909)

+

1

120575

12059321

(0minus 120582)

T02

cos (120582119909) + T02yminus12

(119909 120582)

(60)

Then from (47) and (48) and Lemma 4 we get

1003816100381610038161003816y01

(119909 120582)1003816100381610038161003816le

1

|120575|

100381610038161003816100381612059311

(0minus 120582)

1003816100381610038161003816

1003816100381610038161003816T01

cos (120582119909)1003816100381610038161003816

+

1

|120575|

100381610038161003816100381612059321

(0minus 120582)

1003816100381610038161003816

1003816100381610038161003816T01

sin (120582119909)1003816100381610038161003816

+1003816100381610038161003816T01yminus12

(119909 120582)1003816100381610038161003816

+

1

|120575|

100381610038161003816100381612059311

(0minus 120582)

1003816100381610038161003816

10038161003816100381610038161003816

T02

sin (120582119909)

10038161003816100381610038161003816

+

1

|120575|

100381610038161003816100381612059321

(0minus 120582)

1003816100381610038161003816

10038161003816100381610038161003816

T02

cos (120582119909)10038161003816100381610038161003816

+

10038161003816100381610038161003816

T02yminus12

(119909 120582)

10038161003816100381610038161003816

le 21198884(1 + |120582|)

times (1198883+

2

|120575|

(1198881+ 1198883)) 119890|I120582|(119909+1)

= 211988841198885(1 + |120582|) 119890

|I120582|(119909+1)

(61)

Similarly we can prove that1003816100381610038161003816y02

(119909 120582)1003816100381610038161003816le 211988841198885(1 + |120582|) 119890

|I120582|(119909+1) (62)

3 The Numerical Scheme

In this section we derive the method of computing eigen-values of problems (12)ndash(15) numerically The basic idea ofthe scheme is to split Ω(120582) into two parts a known partK(120582) and an unknown oneU(120582)Thenwe approximateU(120582)

using (5) to get the approximate Ω(120582) and then compute theapproximate zeros We first splitΩ(120582) into two parts

Ω (120582) = K (120582) +U (120582) (63)

whereU(120582) is the unknown part involving integral operators

U (120582) = 120575 [(1205732+ 120582 cos 120579

2) sin 120582

minus (1205731+ 120582 sin 120579

2) cos 120582] y

minus11(0minus 120582)

minus 120575 [(1205731+ 120582 sin 120579

2) sin 120582

+ (1205732+ 120582 cos 120579

2) cos 120582] y

minus12(0minus 120582)

minus 1205752[(1205731+ 120582 sin 120579

2) y01

(1 120582)

+ (1205732+ 120582 cos 120579

2) y02

(1 120582)]

(64)

andK(120582) is the known part

K (120582) = 120575 (1205731+ 120582 sin 120579

2)

times [1205722cos 2120582 minus 120572

1sin 2120582 + 120582 cos (2120582 + 120579

1)]

minus 120575 (1205732+ 120582 cos 120579

2)

times [1205722sin 2120582 + 120572

1cos 2120582 + 120582 sin (2120582 + 120579

1)]

(65)

Then from Lemmas 4 and 5 we have the following result

Lemma 6 The function U(120582) is entire in 120582 and the followingestimate holds

|U (120582)| le 120601 (120582) 1198902|I120582|

(66)

where

120601 (120582) = 119872(1 + |120582|)2

119872 = 4 |120575| 1198886 (21198883 + |120575| 11988841198885)

(67)

Proof From (64) we have

|U (120582)| le |120575| [(10038161003816100381610038161205732

1003816100381610038161003816+ |120582|

1003816100381610038161003816cos 1205792

1003816100381610038161003816) |sin 120582|

+ (10038161003816100381610038161205731

1003816100381610038161003816+ |120582|

1003816100381610038161003816sin 1205792

1003816100381610038161003816) |cos 120582|] 100381610038161003816

1003816yminus11

(0minus 120582)

1003816100381610038161003816

+ |120575| [(10038161003816100381610038161205731

1003816100381610038161003816+ |120582|

1003816100381610038161003816sin 1205792

1003816100381610038161003816) |sin 120582|

+ (10038161003816100381610038161205732

1003816100381610038161003816+1003816100381610038161003816120582 cos 120579

2

1003816100381610038161003816) |cos 120582|] 100381610038161003816

1003816yminus12

(0minus 120582)

1003816100381610038161003816

+ 1205752[(10038161003816100381610038161205731

1003816100381610038161003816+ |120582|

1003816100381610038161003816sin 1205792

1003816100381610038161003816)1003816100381610038161003816y01

(1 120582)1003816100381610038161003816

+ (10038161003816100381610038161205732

1003816100381610038161003816+ |120582|

1003816100381610038161003816cos 1205792

1003816100381610038161003816)1003816100381610038161003816y02

(1 120582)1003816100381610038161003816]

(68)

Using the inequalities | sin 120582| le 119890|I120582| and | cos 120582| le 119890

|I120582| for120582 isin C and Lemmas 4 and 5 imply (66)

ThusU(120582) is an entire function of exponential type 120590 = 2In the following we let 120582 isin R since all eigenvalues are realNow we approximate the function U(120582) using the operator(5) where ℎ isin (0 1205872) and 120596 = (120587 minus 2ℎ)2 and then from(7) we obtain

1003816100381610038161003816U (120582) minus (G

ℎ119873U) (120582)

1003816100381610038161003816le 119879ℎ119873

(120582) (69)

where

119879ℎ119873

(120582) = 2

10038161003816100381610038161003816sin (ℎ

minus1120587120582)

10038161003816100381610038161003816

times 120601 (|R120582| + ℎ (119873 + 1))

119890minus120596119873

radic120587120596119873

120573119873(0) 120582 isin R

(70)

The samplesU(119899ℎ) = Ω(119899ℎ) minusK(119899ℎ) 119899 isin Z119873(120582) cannot be

computed explicitly in the general caseWe approximate these

8 Abstract and Applied Analysis

samples numerically by solving the initial-value problemsdefined by (12) and (45) to obtain the approximate valuesU(119899ℎ) 119899 isin Z

119873(120582) that is U(119899ℎ) = Ω(119899ℎ) minus K(119899ℎ)

Here we use a computer algebra system mathematica toobtain the approximate solutions with the required accuracyHowever a separate study for the effect of different numericalschemes and the computational costs would be interestingAccordingly we have the explicit expansion

(Gℎ119873

U) (120582) = sum

119899isinZ119873(120582)

U (119899ℎ) sinc (ℎminus1120587119909 minus 119899120587)

times 119866(

radic120596 (120582 minus 119899ℎ)

radic119873ℎ

)

(71)

Therefore we get cf (10)10038161003816100381610038161003816(Gℎ119873

U) (120582) minus (Gℎ119873

U) (120582)

10038161003816100381610038161003816le 119860120576119873

(0) 120582 isin R (72)

Now let Ω119873(120582) = K(120582) + (G

ℎ119873U)(120582) From (69) and (72)

we obtain10038161003816100381610038161003816Ω (120582) minus Ω

119873(120582)

10038161003816100381610038161003816le 119879ℎ119873

(120582) + 119860120576119873

(0) 120582 isin R (73)

Let 120582lowast be an eigenvalue and 120582119873its desired approximation

that is Ω(120582lowast) = 0 and Ω

119873(120582119873) = 0 From (73) we have

|Ω119873(120582lowast)| le 119879

ℎ119873(120582lowast) + 119860120576119873

(0) Define the curves

119886plusmn(120582) = Ω

119873(120582) plusmn 119879

ℎ119873(120582) + 119860

120576119873(0) (74)

The curves 119886+(120582) 119886minus(120582) enclose the curve ofΩ(120582) for suitably

large 119873 Hence the closure interval is determined by solving119886plusmn(120582) = 0 giving an interval

119868120576119873

= [119886minus 119886+] (75)

It is worthwhile to mention that the simplicity of the eigen-values guarantees the existence of approximate eigenvaluesthat is the 120582

119873rsquos for which Ω

119873(120582119873) = 0 Next we estimate the

error |120582lowast minus 120582119873| for the eigenvalue 120582lowast

Theorem 7 Let 120582lowast be an eigenvalue of (12)ndash(15) and let 120582119873

be its approximation Then for 120582 isin R we have the followingestimate

1003816100381610038161003816120582lowastminus 120582119873

1003816100381610038161003816lt

119879ℎ119873

(120582119873) + 119860120576119873

(0)

inf120577isin119868120576119873

1003816100381610038161003816Ω1015840(120577)

1003816100381610038161003816

(76)

where the interval 119868120576119873

is defined above

Proof Replacing 120582 by 120582119873in (73) we obtain

1003816100381610038161003816Ω (120582119873) minus Ω (120582

lowast)1003816100381610038161003816lt 119879ℎ119873

(120582119873) + 119860120576119873

(0) (77)

where we have used Ω119873(120582119873) = Ω(120582

lowast) = 0 Using the

mean value theorem yields that for some 120577 isin 119869120576119873

=

[min(120582lowast 120582119873)max(120582lowast 120582

119873)]

10038161003816100381610038161003816(120582lowastminus 120582119873)Ω1015840(120577)

10038161003816100381610038161003816le 119879ℎ119873

(120582119873) + 119860120576119873

(0) 120577 isin 119869120576119873

sub 119868120576119873

(78)

Since 120582lowast is simple and 119873 is sufficiently large then

inf120577isin119868120576119873

|Ω1015840(120577)| gt 0 and we get (76)

4 Numerical Examples

In this section we introduce two examples illustrating the theabove method Also in the following examples we observethat the exact solutions 120582

119896are all inside the interval [119886

minus 119886+]

In these two examples we indicate the effect of the amplitudeerror in the method by determining enclosure intervals fordifferent values of 120576 We also indicate the effect of 119873 and ℎ

by several choices The eigenvalues of the following examplescannot computed concretely then we use mathematicato obtain the exact values mathematica is also used inrounding off the exact eigenvalues which are square rootsBoth numerical results and the associated figures prove thecredibility of the method

Example 1 Consider the system

1199101015840

2(119909) minus 119901 (119909) 119910

1(119909) = 120582119910

1(119909)

1199101015840

1(119909) + 119901 (119909) 119910

2(119909) = minus120582119910

2(119909)

119909 isin [minus1 0) cup (0 1]

1205821199101(minus1) minus (radic2 + 120582) 119910

2(minus1) = 0

(radic2 + 120582) 1199101(1) minus 120582119910

2(1) = 0

1199101(0minus) minus 2119910

1(0+) = 0 119910

2(0minus) minus 2119910

2(0+) = 0

(79)

Here

1199011(119909) = 119901

2(119909) = 119903 (119909) =

119909 119909 isin [minus1 0)

1 (0 1]

(80)

1205791= 1205792= 1205874 120572

1= 1205732= 0 120572

2= 1205731= 1 and 120575 = 2 Direct

calculations give

K (120582) = 2 ((radic2120582 + 1) cos (2120582) minus 120582 (120582 + radic2) sin (2120582))

Ω (120582) = 2 ((radic2120582 + 1) cos(2120582 +

1

2

)

minus120582 (120582 + radic2) sin(2120582 +

1

2

))

(81)

As is clearly seen the eigenvalues cannot be computedexplicitly Tables 1 2 and 3 indicate the application of ourtechnique to this problem and the effect of 120576 By exact wemean the zeros of Δ(120582) computed by Mathematica

Figures 1 and 2 illustrate the enclosure intervals domi-nating 120582

0for 119873 = 25 ℎ = 01 and 120576 = 10

minus2 120576 = 10minus5

respectively The middle curve represents Δ(120582) while theupper and lower curves represent the curves of 119886

+(120582) 119886minus(120582)

respectively We notice that when 120576 = 10minus5 the two curves

are almost identical Similarly Figures 3 and 4 illustrate theenclosure intervals dominating 120582

1for ℎ = 01 119873 = 25 and

120576 = 10minus2 120576 = 10

minus5 respectively

Abstract and Applied Analysis 9

Table 1 The approximation 120582119896119873

and the exact solutions 120582119896for different choices of ℎ and119873

120582119896

120582minus2

120582minus1

1205820

1205821

Exact 120582119896

minus12061856254879546 minus05151195632138896 034274095892547224 1616237903407542

120582119896119873

ℎ = 06 119873 = 15 minus12061855965725146 minus05151190347168934 034274053593937165 1616237692038969120596 = 09708 119873 = 25 minus12061856254843495 minus05151195631493376 0342740958869788 16162379033820231ℎ = 01 119873 = 15 minus12061856254844203 minus05151195632158179 03427409589350932 16162379034134882

120596 = 14708 119873 = 25 minus12061856254879548 minus05151195632138894 034274095892547236 16162379034075425

Table 2 Absolute error |120582119896minus 120582119896119873

|

120582119896

120582minus2

120582minus1

1205820

1205821

ℎ = 06119873 = 15 289154 times 10

minus8528497 times 10

minus7422986 times 10

minus7211369 times 10

minus7

119873 = 25 360512 times 10minus12

64552 times 10minus11

556842 times 10minus11

255189 times 10minus11

ℎ = 01119873 = 15 353428 times 10

minus12192824 times 10

minus12962097 times 10

minus12594613 times 10

minus12

119873 = 25 222045 times 10minus16

222045 times 10minus16

555112 times 10minus17

444089 times 10minus16

Table 3 For119873 = 25 and ℎ = 01 the exact solutions 120582119896are all inside the interval [119886

minus 119886+] for different values of 120576

120582119896

120582minus2

120582minus1

1205820

1205821

Exact 120582119896

minus12061856254879546 minus05151195632138896 034274095892547224 1616237903407542119868120576119873

120576 = 10minus2

[minus12335696 minus11771922] [minus05486790 minus04826816] [03246478 03601346] [16100119 16224066]

119868120576119873

120576 = 10minus5

[minus12062137 minus12061575] [minus05151525 minus05150867] [03427232 03427586] [16162317 16162441]

030 032 034 036 038

00

02

04

aminus

a+

Ω(120582)

minus02

minus04

minus06

Figure 1The enclosure interval dominating 1205820for ℎ = 01119873 = 25

and 120576 = 10minus2

Example 2 In this example we consider the system

1199101015840

2(119909) minus 119901 (119909) 119910

1(119909) = 120582119910

1(119909)

1199101015840

1(119909) + 119901 (119909) 119910

2(119909) = minus120582119910

2(119909)

119909 isin [minus1 0) cup (0 1]

radic31199101(minus1) minus 119910

2(minus1) = 0

(1 +

1

2

120582)1199101(1) minus (1 +

radic3

2

120582)1199102(1) = 0

1199101(0minus) minus 3119910

1(0+) = 0 119910

2(0minus) minus 3119910

2(0+) = 0

(82)

030 032 034 036 038

00

02

aminus

a+

Ω(120582)

minus02

minus04

Figure 2The enclosure interval dominating 1205820for ℎ = 01119873 = 25

and 120576 = 10minus5

where

1199011(119909) = 119901

2(119909) = 119901 (119909) =

119909 119909 isin [minus1 0)

1199092 (0 1]

(83)

1205791= 1205792= 0 120572

1= 1205732= 1 120572

2= 1205731= 2 and 120575 = 12 Direct

calculations give

K (120582) =

1

2

((120582 + 3) cos (2120582) minus (1205822+ 3120582 + 4) sin (2120582))

Ω (120582) =

1

2

((1205822+ 3120582 + 4) sin(

1

6

minus 2120582)

+ (120582 + 3) cos(1

6

minus 2120582))

(84)

10 Abstract and Applied Analysis

Table 4 The approximation 120582119896119873

and the exact solutions 120582119896for different choices of ℎ and119873

120582119896

120582minus2

120582minus1

1205820

1205821

Exact 120582119896

minus30661256186938894 minus10959703083466883 03689030343814017 18338437675275228

120582119896119873

ℎ = 08 119873 = 15 minus3066123822200742 minus10959640944020903 036890463046491767 18338437607680411120596 = 007708 119873 = 25 minus30661256172349627 minus10959703026494807 036890303588091056 18338437676561417

ℎ = 02 119873 = 15 minus30661256186844446 minus10959703083278305 03689030343916001 1833843767517251120596 = 13708 119873 = 25 minus306612561869389 minus1095970308346688 03689030343814024 18338437675275225

Table 5 Absolute error |120582119896minus 120582119896119873

|

120582119896

120582minus2

120582minus1

1205820

1205821

ℎ = 08119873 = 15 179649 times 10

minus6621394 times 10

minus6159608 times 10

minus6675948 times 10

minus9

119873 = 25 145893 times 10minus9

569721 times 10minus9

149951 times 10minus9

128619 times 10minus10

ℎ = 02119873 = 15 944489 times 10

minus12188578 times 10

minus12101984 times 10

minus11102718 times 10

minus11

119873 = 25 444089 times 10minus16

222045 times 10minus16

666134 times 10minus16

222045 times 10minus16

Table 6 For119873 = 25 and ℎ = 02 the exact solutions 120582119896are all inside the interval [119886

minus 119886+] for different values of 120576

120582119896

120582minus2

120582minus1

1205820

1205821

Exact 120582119896

minus30661256186938894 minus10959703083466883 03689030343814017 18338437675275228119868120576119873

120576 = 10minus2

[minus31098742 minus30188665] [minus11631502 minus10339754] [03437434 03933486] [18220764 18454828]

119868120576119873

120576 = 10minus5

[minus30661709 minus30660803] [minus10960346 minus10959060] [03688783 03689278] [18338321 18338555]

158 159 160 161 162 163 164 165

00

05

10

minus05

minus10

aminus

a+

Ω(120582)

Figure 3The enclosure interval dominating 1205821for ℎ = 01119873 = 25

and 120576 = 10minus2

Tables 4 and 6 give the exact eigenvalues 1205821198961

119896=minus2and their

approximate ones for different values of ℎ 119873 and 120576 InTable 5 we give the absolute error for different values of ℎ and119873

Here Figures 5 6 7 and 8 illustrate the enclosureintervals dominating 120582

minus2and 120582

minus1for ℎ = 02 119873 = 25 and

120576 = 10minus2 120576 = 10

minus5 respectively

158 159 160 161 162 163 164 165

00

05

minus05

aminus

a+

Ω(120582)

Figure 4The enclosure interval dominating 1205821for ℎ = 01119873 = 25

and 120576 = 10minus5

5 Conclusion

With a simple analysis and with values of solutions ofinitial value problems computed at a few values of theeigenparameter we have computed the eigenvalues of aDirac system which has a discontinuity at one point andcontains a spectral parameter in all boundary conditionswith a certain estimated error The method proposed is

Abstract and Applied Analysis 11

00

05

10

minus05

minus33 minus32 minus31 minus30 minus29 minus28

aminus

a+

Ω(120582)

Figure 5 The enclosure interval dominating 120582minus2

for ℎ = 02 119873 =

25 and 120576 = 10minus2

00

05

10

minus05

minus33 minus32 minus31 minus30 minus29 minus28

aminus

a+

Ω(120582)

Figure 6 The enclosure interval dominating 120582minus2

for ℎ = 02 119873 =

25 and 120576 = 10minus5

a shooting procedure that is the problem is reformulatedas two initial value ones due to the interior discontinuity ofsize two and a miss-distance is defined at the right end of theinterval of integration whose roots are the eigenvalues to becomputed The unknown partU(120582) of the miss-distance canbewritten in terms of a functionwhich is an entire function ofexponential typeTherefore we propose to approximate suchterm by means of a truncated cardinal series with samplingvalues approximated by solving numerically correspondingsuitable initial value problems Finally in Section 4 weintroduced two instructive examples where both numericalresults and the associated figures have proved the credibilityof the method

00

05

10

minus05

minus14 minus13 minus12 minus11 minus10 minus09 minus08 minus07

aminus

a+

Ω(120582)

Figure 7 The enclosure interval dominating 120582minus1

for ℎ = 02 119873 =

25 and 120576 = 10minus2

00

05

10

minus05

minus14 minus13 minus12 minus11 minus10 minus09 minus08 minus07

aminus

a+

Ω(120582)

Figure 8 The enclosure interval dominating 120582minus1

for ℎ = 02 119873 =

25 and 120576 = 10minus5

Conflict of Interests

The authors declare that there is no conflict of interestsregarding the publication of this paper

Acknowledgment

This paper was funded by the Deanship of Scientific Research(DSR) King Abdulaziz University Jeddah The authorstherefore acknowledge with thanks DSR technical and finan-cial support

12 Abstract and Applied Analysis

References

[1] C T Fulton ldquoTwo-point boundary value problems witheigenvalue parameter contained in the boundary conditionsrdquoProceedings of the Royal Society of Edinburgh A Mathematicsvol 77 no 3-4 pp 293ndash308 1977

[2] JWalter ldquoRegular eigenvalue problems with eigenvalue param-eter in the boundary conditionrdquoMathematische Zeitschrift vol133 pp 301ndash312 1973

[3] V Kotelnikov ldquoOn the carrying capacity of the ether and wirein telecommunications material for the first All-Unionrdquo inConference on Questions of Communications vol 55 pp 55ndash64Izd Red Upr Svyazi RKKA 1933

[4] C E Shannon ldquoCommunication in the presence of noiserdquoProceedings of the IRE vol 37 pp 10ndash21 1949

[5] E TWhittaker ldquoOn the functions which are represented by theexpansion of the interpolation theoryrdquo Proceedings of the RoyalSociety of Edinburgh A vol 35 pp 181ndash194 1915

[6] P L Butzer G Schmeisser and R L Stens ldquoAn introductionto sampling analysisrdquo in Non Uniform Sampling Theory andPractices F Marvasti Ed pp 17ndash121 Kluwer New York NYUSA 2001

[7] M A Kowalski K A Sikorski and F Stenger Selected Topics inApproximation andComputation OxfordUniversity Press NewYork NY USA 1995

[8] J Lund and K L Bowers Sinc Methods for Quadrature andDifferential Equations SIAM Philadelphia Pa USA 1992

[9] F Stenger ldquoNumericalmethods based onWhittaker cardinal orsinc functionsrdquo SIAM Review vol 23 no 2 pp 165ndash224 1981

[10] F Stenger Numerical Methods Based on Sinc and AnalyticFunctions Springer New York NY USA 1993

[11] A Boumenir ldquoHigher approximation of eigenvalues by thesampling methodrdquo BIT Numerical Mathematics vol 40 no 2pp 215ndash225 2000

[12] A Boumenir ldquoSampling and eigenvalues of non-self-adjointSturm-Liouville problemsrdquo SIAM Journal on Scientific Comput-ing vol 23 no 1 pp 219ndash229 2001

[13] M M Tharwat A H Bhrawy and A Yildirim ldquoNumericalcomputation of eigenvalues of discontinuous Sturm-Liouvilleproblems with parameter dependent boundary conditionsusing sinc methodrdquoNumerical Algorithms vol 63 no 1 pp 27ndash48 2013

[14] M M Tharwat A H Bhrawy and A Yildirim ldquoNumericalcomputation of the eigenvalues of a discontinuousDirac systemusing the sincmethodwith error analysisrdquo International Journalof Computer Mathematics vol 89 no 15 pp 2061ndash2080 2012

[15] P L Butzer and R L Stens ldquoA modification of the Whittaker-Kotelrsquonikov-Shannon sampling seriesrdquo Aequationes Mathemat-icae vol 28 no 3 pp 305ndash311 1985

[16] R Gervais Q I Rahman and G Schmeisser ldquoA bandlimitedfunction simulating a duration-limited onerdquo in ApproximationTheory andFunctionalAnalysis P L Butzer andR L Stens Edspp 355ndash362 Birkhaauser Boston Mass USA 1984

[17] R L Stens ldquoSampling by generalized kernelsrdquo in SamplingThe-ory in Fourier and Signal Analysis Advanced Topics J RHigginsand R L Stens Eds pp 130ndash157 Oxford University PressOxford UK 1999

[18] G Schmeisser and F Stenger ldquoSinc approximation with a Gaus-sianmultiplierrdquo SamplingTheory in Signal and Image Processingvol 6 no 2 pp 199ndash221 2007

[19] L Qian ldquoOn the regularized Whittaker-Kotelrsquonikov-Shannonsampling formulardquo Proceedings of the American MathematicalSociety vol 131 no 4 pp 1169ndash1176 2003

[20] L Qian and D B Creamer ldquoA modification of the samplingseries with aGaussianmultiplierrdquo SamplingTheory in Signal andImage Processing vol 5 no 1 pp 1ndash20 2006

[21] L Qian and D B Creamer ldquoLocalized sampling in the presenceof noiserdquoAppliedMathematics Letters vol 19 no 4 pp 351ndash3552006

[22] M H Annaby and R M Asharabi ldquoComputing eigenvaluesof boundary-value problems using sinc-Gaussian methodrdquoSamplingTheory in Signal and Image Processing vol 7 no 3 pp293ndash312 2008

[23] M H Annaby and M MTharwat ldquoA sinc-Gaussian techniquefor computing eigenvalues of second-order linear pencilsrdquoApplied Numerical Mathematics vol 63 pp 129ndash137 2013

[24] A H Bhrawy M M Tharwat and A Al-Fhaid ldquoNumericalalgorithms for computing eigenvalues of discontinuous Diracsystem using sinc-Gaussian methodrdquo Abstract and AppliedAnalysis vol 2012 Article ID 925134 13 pages 2012

[25] M M Tharwat A H Bhrawy and A S Alofi ldquoApproximationof eigenvalues of discontinuous sturm-liouville problems witheigenparameter in all boundary conditionsrdquo Boundary ValueProblems vol 2013 article 132 2013

[26] M M Tharwat and S M Al-Harbi ldquoApproximation of eigen-values of boundary value problemsrdquo Boundary Value Problemsvol 2014 article 51 2014

[27] MMTharwat and A H Bhrawy ldquoComputation of eigenvaluesof discontinuous Dirac system using Hermite interpolationtechniquerdquo Advances in Difference Equations vol 2012 article59 2012

[28] M M Tharwat A Yildirim and A H Bhrawy ldquoSampling ofdiscontinuous Dirac systemsrdquo Numerical Functional Analysisand Optimization vol 34 no 3 pp 323ndash348 2013

[29] M M Tharwat ldquoDiscontinuous Sturm-Liouville problems andassociated sampling theoriesrdquo Abstract and Applied Analysisvol 2011 Article ID 610232 30 pages 2011

[30] MMTharwat ldquoOn sampling theories and discontinuousDiracsystems with eigenparameter in the boundary conditionsrdquoBoundary Value Problems vol 2013 article 65 2013

[31] B M Levitan and I S Sargsjan ldquoIntroduction to spectral the-ory self adjoint ordinary differential operatorsrdquo in Translationof Mthematical Monographs vol 39 American MathematicalSociety Providence RI USA 1975

[32] B M Levitan and I S Sargsjan Sturm-Liouville and Dirac oper-ators vol 59 Kluwer Academic Publishers Group DordrechtThe Netherlands 1991

Submit your manuscripts athttpwwwhindawicom

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Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Stochastic AnalysisInternational Journal of

Abstract and Applied Analysis 5

Integrating the above equation through [minus1 0] and [0 1] andtaking into account the initial conditions (23) (26) and (28)we obtain

1205752(12059312

(1 120582) 12059322

(1 120583) minus 12059312

(1 120583) 12059322

(1 120582))

minus (120583 minus 120582) 1205881

= (120583 minus 120582) (int

0

minus1

(12059311

(119909 120582) 12059311

(119909 120583)

+ 12059321

(119909 120582) 12059321

(119909 120583)) 119889119909

+ 1205752int

1

0

(12059312

(119909 120582) 12059312

(119909 120583)

+ 12059322

(119909 120582) 12059322

(119909 120583)) 119889119909)

(38)

Dividing both sides of (38) by (120582 minus 120583) and by letting 120583 rarr 120582we arrive to the relation

1205752(12059322

(1 120582)

12059712059312

(1 120582)

120597120582

minus 12059312

(1 120582)

12059712059322

(1 120582)

120597120582

) + 1205881

= minus(int

0

minus1

(100381610038161003816100381612059311

(119909 120582)1003816100381610038161003816

2

+100381610038161003816100381612059321

(119909 120582)1003816100381610038161003816

2

) 119889119909

+ 1205752int

1

0

(100381610038161003816100381612059312

(119909 120582)1003816100381610038161003816

2

+100381610038161003816100381612059322

(119909 120582)1003816100381610038161003816

2

) 119889119909)

(39)

We show that equation

Ω (120582) = W (120593 120595) (1 120582)

= 1205752((1205731+ 120582 sin 120579

2) 12059312

(1 120582)

minus (1205732+ 120582 cos 120579

2) 12059322

(1 120582)) = 0

(40)

has only simple roots Assume the converse that is (40) hasa double root 120582lowast Then the following two equations hold

(1205731+ 120582lowast sin 120579

2) 12059312

(1 120582lowast) minus (120573

2+ 120582lowast cos 120579

2) 12059322

(1 120582lowast) = 0

sin 120579212059312

(1 120582lowast) + (120573

1+ 120582lowast sin 120579

2)

12059712059312

(1 120582lowast)

120597120582

minus cos 120579212059322

(1 120582lowast) minus (120573

2+ 120582lowast cos 120579

2)

12059712059322

(1 120582lowast)

120597120582

= 0

(41)

Since 1205882

= 0 and 120582lowast is real then (120573

1+ 120582lowast sin 120579

2)2

+

(1205732+ 120582lowast cos 120579

2)2

= 0 Let 1205731+ 120582lowast sin 120579

2= 0 From (41)

12059312

(1 120582lowast) =

(1205732+ 120582lowast cos 120579

2)

(1205731+ 120582lowast sin 120579

2)

12059322

(1 120582lowast)

12059712059312

(1 120582lowast)

120597120582

=

120588212059322

(1 120582lowast)

(1205731+ 120582lowast sin 120579

2)2

+

(1205732+ 120582lowast cos 120579

2)

(1205731+ 120582lowast sin 120579

2)

12059712059322

(1 120582lowast)

120597120582

(42)

Combining (42) and (39) with 120582 = 120582lowast we obtain

12058821205752(12059322

(1 120582lowast))2

(1205731+ 120582lowast sin 120579

2)2

+ 1205881

= minus(int

0

minus1

(100381610038161003816100381612059311

(119909 120582)1003816100381610038161003816

2

+100381610038161003816100381612059321

(119909 120582)1003816100381610038161003816

2

) 119889119909

+ 1205752int

1

0

(

1003816100381610038161003816100381612059312

(119909 120582) |2+

1003816100381610038161003816100381612059322

(119909 120582) |2) 119889119909)

(43)

contradicting the assumption 120588119894gt 0 119894 = 1 2 The other case

when 1205732+120582lowast cos 120579

2= 0 can be treated similarly and the proof

is complete

Recall that boundary value problems (12)ndash(15) have denu-merable set of real and simple eigenvalues see the abovesection From (23) and (26) the solution of (12)

120593 (sdot 120582) = (1205931(sdot 120582)

1205932(sdot 120582)

)

120593119894(119909 120582) =

1205931198941(119909 120582) 119909 isin [minus1 0)

1205931198942(119909 120582) 119909 isin (0 1]

119894 = 1 2

(44)

satisfies the following initial conditions

(12059311

(minus1 120582) 12059312

(0+ 120582)

12059321

(minus1 120582) 12059322

(0+ 120582)

)

= (1205722+ 120582 cos 120579

1120575minus112059311

(0minus 120582)

1205721+ 120582 sin 120579

1120575minus112059321

(0minus 120582)

)

(45)

Since 120593(sdot 120582) satisfies (13) then the eigenvalues of the prob-lems (12)ndash(15) are the zeros of the function

Ω (120582) = 1205752((1205731+ 120582 sin 120579

2) 12059312

(1 120582)

minus (1205732+ 120582 cos 120579

2) 12059322

(1 120582)) = 0

(46)

We shall transform (12) (23) (26) and (30) into the integralequations see [32]

12059311

(119909 120582) = 1205722cos 120582 (119909 + 1) minus 120572

1sin 120582 (119909 + 1)

+ 120582 cos (120582 (119909 + 1) + 1205791) minusTminus11

12059311

(119909 120582)

minusTminus12

12059321

(119909 120582)

(47)

12059321

(119909 120582) = 1205721cos 120582 (119909 + 1) + 120572

2sin 120582 (119909 + 1)

+ 120582 sin (120582 (119909 + 1) + 1205791) +

Tminus11

12059311

(119909 120582)

minusTminus12

12059321

(119909 120582)

(48)

12059312

(119909 120582) =

1

120575

12059311

(0minus 120582) cos (120582119909) minus 1

120575

12059321

(0minus 120582) sin (120582119909)

minusT01

12059312

(119909 120582) minusT02

12059322

(119909 120582)

(49)

6 Abstract and Applied Analysis

12059322

(119909 120582) =

1

120575

12059311

(0minus 120582) sin (120582119909) +

1

120575

12059321

(0minus 120582) cos (120582119909)

+T01

12059312

(119909 120582) minusT02

12059322

(119909 120582)

(50)

where Tminus1119894

Tminus1119894

T0119894

and T0119894 119894 = 1 2 are the Volterra

integral operators defined by

Tminus1119894

120593 (119909 120582) = int

119909

minus1

sin 120582 (119909 minus 119905) 119901119894(119905) 120593 (119905 120582) 119889119905

Tminus1119894

120593 (119909 120582) = int

119909

minus1

cos 120582 (119909 minus 119905) 119901119894(119905) 120593 (119905 120582) 119889119905

T0119894120593 (119909 120582) = int

119909

0

sin 120582 (119909 minus 119905) 119901119894(119905) 120593 (119905 120582) 119889119905

T0119894120593 (119909 120582) = int

119909

0

cos 120582 (119909 minus 119905) 119901119894(119905) 120593 (119905 120582) 119889119905

(51)

For convenience we define the constants

1198881= max 100381610038161003816

10038161205721

1003816100381610038161003816+10038161003816100381610038161205722

10038161003816100381610038161003816100381610038161003816sin 1205791

1003816100381610038161003816+1003816100381610038161003816cos 1205791

1003816100381610038161003816

1198882= int

0

minus1

[10038161003816100381610038161199011(119905)

1003816100381610038161003816+10038161003816100381610038161199012(119905)

1003816100381610038161003816] 119889119905 119888

3= 11988811198882exp (119888

2)

1198884= int

1

0

[10038161003816100381610038161199011(119905)

1003816100381610038161003816+10038161003816100381610038161199012(119905)

1003816100381610038161003816] 119889119905 119888

5= 1198883+

2

|120575|

(1198881+ 1198883)

1198886= max 100381610038161003816

10038161205721

1003816100381610038161003816+10038161003816100381610038161205722

10038161003816100381610038161003816100381610038161003816sin 1205792

1003816100381610038161003816+1003816100381610038161003816cos 1205792

1003816100381610038161003816

(52)

Now we define yminus1119894

(sdot 120582) and y0119894(sdot 120582) 119894 = 1 2 to be

yminus11

(119909 120582) = Tminus11

12059311

(119909 120582) +Tminus12

12059321

(119909 120582)

yminus12

(119909 120582) =Tminus11

12059311

(119909 120582) minusTminus12

12059321

(119909 120582)

y01

(119909 120582) = T01

12059312

(119909 120582) +T02

12059322

(119909 120582)

y02

(119909 120582) =T01

12059312

(119909 120582) minusT02

12059322

(119909 120582)

(53)

Lemma 4 The functions yminus11

(119909 120582) and yminus12

(119909 120582) are entirein 120582 for any fixed 119909 isin [minus1 0) and satisfy the growth condition

1003816100381610038161003816yminus11

(119909 120582)1003816100381610038161003816

1003816100381610038161003816yminus12

(119909 120582)1003816100381610038161003816le 21198883(1 + |120582|) 119890

|I120582|(119909+1)

120582 isin C

(54)

Proof Since yminus11

(119909 120582) = Tminus11

12059311(119909 120582) +

Tminus12

12059321(119909 120582)

then from (47) and (48) we obtain

yminus11

(119909 120582)

= 1205722Tminus11

cos 120582 (119909 + 1) minus 1205721Tminus11

sin 120582 (119909 + 1)

+ 120582Tminus11

cos (120582 (119909 + 1) + 1205791) minusTminus11

yminus11

(119909 120582)

+ 1205721Tminus12

cos 120582 (119909 + 1) + 1205722Tminus12

sin 120582 (119909 + 1)

+ 120582Tminus12

sin (120582 (119909 + 1) + 1205791) +

Tminus12

yminus12

(119909 120582)

(55)

Using the inequalities | sin 119911| le 119890|I119911| and | cos 119911| le 119890

|I119911| for119911 isin C leads for 120582 isin C to1003816100381610038161003816yminus11

(119909 120582)1003816100381610038161003816

le10038161003816100381610038161205722

1003816100381610038161003816

1003816100381610038161003816Tminus11

cos 120582 (119909 + 1)1003816100381610038161003816+10038161003816100381610038161205721Tminus11

sin 120582 (119909 + 1)1003816100381610038161003816

+ |120582|1003816100381610038161003816Tminus11

cos (120582 (119909 + 1) + 1205791)1003816100381610038161003816+1003816100381610038161003816Tminus11

yminus11

(119909 120582)1003816100381610038161003816

+10038161003816100381610038161205721

1003816100381610038161003816

10038161003816100381610038161003816

Tminus12

cos 120582 (119909 + 1)

10038161003816100381610038161003816+10038161003816100381610038161205722

1003816100381610038161003816

10038161003816100381610038161003816

Tminus12

sin 120582 (119909 + 1)

10038161003816100381610038161003816

+ |120582|

10038161003816100381610038161003816

Tminus12

sin (120582 (119909 + 1) + 1205791)

10038161003816100381610038161003816+

10038161003816100381610038161003816

Tminus12

yminus12

(119909 120582)

10038161003816100381610038161003816

le 21198881(1 + |120582|) 119890

|I120582|(119909+1)int

119909

minus1

[10038161003816100381610038161199011(119905)

1003816100381610038161003816+10038161003816100381610038161199012(119905)

1003816100381610038161003816] 119889119905

+ 119890|I120582|(119909+1)

int

119909

minus1

[10038161003816100381610038161199011(119905)

1003816100381610038161003816

1003816100381610038161003816yminus11

(119905 120582)1003816100381610038161003816

+10038161003816100381610038161199012(119905)

1003816100381610038161003816

1003816100381610038161003816yminus12

(119905 120582)1003816100381610038161003816] 119890minus|I120582|(119905+1)

119889119905

le 211988811198882(1 + |120582|) 119890

|I120582|(119909+1)

+ 119890|I120582|(119909+1)

int

119909

minus1

[10038161003816100381610038161199011(119905)

1003816100381610038161003816

1003816100381610038161003816yminus11

(119905 120582)1003816100381610038161003816

+10038161003816100381610038161199012(119905)

1003816100381610038161003816

1003816100381610038161003816yminus12

(119905 120582)1003816100381610038161003816] 119890minus|I120582|(119905+1)

119889119905

(56)

The above inequality can be reduced to

119890minus|I120582|(119909+1) 1003816

100381610038161003816yminus11

(119909 120582)1003816100381610038161003816

le 211988811198882(1 + |120582|)

+ int

119909

minus1

[10038161003816100381610038161199011(119905)

1003816100381610038161003816

1003816100381610038161003816yminus11

(119905 120582)1003816100381610038161003816

+10038161003816100381610038161199012(119905)

1003816100381610038161003816

1003816100381610038161003816yminus12

(119905 120582)1003816100381610038161003816] 119890minus|I120582|(119905+1)

119889119905

(57)

Similarly we can prove that

119890minus|I120582|(119909+1) 1003816

100381610038161003816yminus12

(119909 120582)1003816100381610038161003816

le 211988811198882(1 + |120582|)

+ int

119909

minus1

[10038161003816100381610038161199011(119905)

1003816100381610038161003816

1003816100381610038161003816yminus11

(119905 120582)1003816100381610038161003816

+10038161003816100381610038161199012(119905)

1003816100381610038161003816

1003816100381610038161003816yminus12

(119905 120582)1003816100381610038161003816] 119890minus|I120582|(119905+1)

119889119905

(58)

Then from (57) (58) and Lemma 31 of [32 pp 204] weobtain (54)

In a similar manner we will prove the following lemmafor y01

(sdot 120582) and y02

(sdot 120582)

Lemma 5 The functions y01

(119909 120582) and y02

(119909 120582) are entire in120582 for any fixed 119909 isin (0 1] and satisfy the growth condition

1003816100381610038161003816y01

(119909 120582)1003816100381610038161003816

1003816100381610038161003816y02

(119909 120582)1003816100381610038161003816le 211988841198885(1 + |120582|) 119890

|I120582|(119909+1)

120582 isin C

(59)

Abstract and Applied Analysis 7

Proof Since y01

(119909 120582) = T01

12059312(119909 120582) +

T02

12059322(119909 120582) then

from (49) and (50) we obtain

y01

(119909 120582) =

1

120575

12059311

(0minus 120582)T

01cos (120582119909)

minus

1

120575

12059321

(0minus 120582)T

01sin (120582119909)

minusT01yminus12

(119909 120582) +

1

120575

12059311

(0minus 120582)

T02

sin (120582119909)

+

1

120575

12059321

(0minus 120582)

T02

cos (120582119909) + T02yminus12

(119909 120582)

(60)

Then from (47) and (48) and Lemma 4 we get

1003816100381610038161003816y01

(119909 120582)1003816100381610038161003816le

1

|120575|

100381610038161003816100381612059311

(0minus 120582)

1003816100381610038161003816

1003816100381610038161003816T01

cos (120582119909)1003816100381610038161003816

+

1

|120575|

100381610038161003816100381612059321

(0minus 120582)

1003816100381610038161003816

1003816100381610038161003816T01

sin (120582119909)1003816100381610038161003816

+1003816100381610038161003816T01yminus12

(119909 120582)1003816100381610038161003816

+

1

|120575|

100381610038161003816100381612059311

(0minus 120582)

1003816100381610038161003816

10038161003816100381610038161003816

T02

sin (120582119909)

10038161003816100381610038161003816

+

1

|120575|

100381610038161003816100381612059321

(0minus 120582)

1003816100381610038161003816

10038161003816100381610038161003816

T02

cos (120582119909)10038161003816100381610038161003816

+

10038161003816100381610038161003816

T02yminus12

(119909 120582)

10038161003816100381610038161003816

le 21198884(1 + |120582|)

times (1198883+

2

|120575|

(1198881+ 1198883)) 119890|I120582|(119909+1)

= 211988841198885(1 + |120582|) 119890

|I120582|(119909+1)

(61)

Similarly we can prove that1003816100381610038161003816y02

(119909 120582)1003816100381610038161003816le 211988841198885(1 + |120582|) 119890

|I120582|(119909+1) (62)

3 The Numerical Scheme

In this section we derive the method of computing eigen-values of problems (12)ndash(15) numerically The basic idea ofthe scheme is to split Ω(120582) into two parts a known partK(120582) and an unknown oneU(120582)Thenwe approximateU(120582)

using (5) to get the approximate Ω(120582) and then compute theapproximate zeros We first splitΩ(120582) into two parts

Ω (120582) = K (120582) +U (120582) (63)

whereU(120582) is the unknown part involving integral operators

U (120582) = 120575 [(1205732+ 120582 cos 120579

2) sin 120582

minus (1205731+ 120582 sin 120579

2) cos 120582] y

minus11(0minus 120582)

minus 120575 [(1205731+ 120582 sin 120579

2) sin 120582

+ (1205732+ 120582 cos 120579

2) cos 120582] y

minus12(0minus 120582)

minus 1205752[(1205731+ 120582 sin 120579

2) y01

(1 120582)

+ (1205732+ 120582 cos 120579

2) y02

(1 120582)]

(64)

andK(120582) is the known part

K (120582) = 120575 (1205731+ 120582 sin 120579

2)

times [1205722cos 2120582 minus 120572

1sin 2120582 + 120582 cos (2120582 + 120579

1)]

minus 120575 (1205732+ 120582 cos 120579

2)

times [1205722sin 2120582 + 120572

1cos 2120582 + 120582 sin (2120582 + 120579

1)]

(65)

Then from Lemmas 4 and 5 we have the following result

Lemma 6 The function U(120582) is entire in 120582 and the followingestimate holds

|U (120582)| le 120601 (120582) 1198902|I120582|

(66)

where

120601 (120582) = 119872(1 + |120582|)2

119872 = 4 |120575| 1198886 (21198883 + |120575| 11988841198885)

(67)

Proof From (64) we have

|U (120582)| le |120575| [(10038161003816100381610038161205732

1003816100381610038161003816+ |120582|

1003816100381610038161003816cos 1205792

1003816100381610038161003816) |sin 120582|

+ (10038161003816100381610038161205731

1003816100381610038161003816+ |120582|

1003816100381610038161003816sin 1205792

1003816100381610038161003816) |cos 120582|] 100381610038161003816

1003816yminus11

(0minus 120582)

1003816100381610038161003816

+ |120575| [(10038161003816100381610038161205731

1003816100381610038161003816+ |120582|

1003816100381610038161003816sin 1205792

1003816100381610038161003816) |sin 120582|

+ (10038161003816100381610038161205732

1003816100381610038161003816+1003816100381610038161003816120582 cos 120579

2

1003816100381610038161003816) |cos 120582|] 100381610038161003816

1003816yminus12

(0minus 120582)

1003816100381610038161003816

+ 1205752[(10038161003816100381610038161205731

1003816100381610038161003816+ |120582|

1003816100381610038161003816sin 1205792

1003816100381610038161003816)1003816100381610038161003816y01

(1 120582)1003816100381610038161003816

+ (10038161003816100381610038161205732

1003816100381610038161003816+ |120582|

1003816100381610038161003816cos 1205792

1003816100381610038161003816)1003816100381610038161003816y02

(1 120582)1003816100381610038161003816]

(68)

Using the inequalities | sin 120582| le 119890|I120582| and | cos 120582| le 119890

|I120582| for120582 isin C and Lemmas 4 and 5 imply (66)

ThusU(120582) is an entire function of exponential type 120590 = 2In the following we let 120582 isin R since all eigenvalues are realNow we approximate the function U(120582) using the operator(5) where ℎ isin (0 1205872) and 120596 = (120587 minus 2ℎ)2 and then from(7) we obtain

1003816100381610038161003816U (120582) minus (G

ℎ119873U) (120582)

1003816100381610038161003816le 119879ℎ119873

(120582) (69)

where

119879ℎ119873

(120582) = 2

10038161003816100381610038161003816sin (ℎ

minus1120587120582)

10038161003816100381610038161003816

times 120601 (|R120582| + ℎ (119873 + 1))

119890minus120596119873

radic120587120596119873

120573119873(0) 120582 isin R

(70)

The samplesU(119899ℎ) = Ω(119899ℎ) minusK(119899ℎ) 119899 isin Z119873(120582) cannot be

computed explicitly in the general caseWe approximate these

8 Abstract and Applied Analysis

samples numerically by solving the initial-value problemsdefined by (12) and (45) to obtain the approximate valuesU(119899ℎ) 119899 isin Z

119873(120582) that is U(119899ℎ) = Ω(119899ℎ) minus K(119899ℎ)

Here we use a computer algebra system mathematica toobtain the approximate solutions with the required accuracyHowever a separate study for the effect of different numericalschemes and the computational costs would be interestingAccordingly we have the explicit expansion

(Gℎ119873

U) (120582) = sum

119899isinZ119873(120582)

U (119899ℎ) sinc (ℎminus1120587119909 minus 119899120587)

times 119866(

radic120596 (120582 minus 119899ℎ)

radic119873ℎ

)

(71)

Therefore we get cf (10)10038161003816100381610038161003816(Gℎ119873

U) (120582) minus (Gℎ119873

U) (120582)

10038161003816100381610038161003816le 119860120576119873

(0) 120582 isin R (72)

Now let Ω119873(120582) = K(120582) + (G

ℎ119873U)(120582) From (69) and (72)

we obtain10038161003816100381610038161003816Ω (120582) minus Ω

119873(120582)

10038161003816100381610038161003816le 119879ℎ119873

(120582) + 119860120576119873

(0) 120582 isin R (73)

Let 120582lowast be an eigenvalue and 120582119873its desired approximation

that is Ω(120582lowast) = 0 and Ω

119873(120582119873) = 0 From (73) we have

|Ω119873(120582lowast)| le 119879

ℎ119873(120582lowast) + 119860120576119873

(0) Define the curves

119886plusmn(120582) = Ω

119873(120582) plusmn 119879

ℎ119873(120582) + 119860

120576119873(0) (74)

The curves 119886+(120582) 119886minus(120582) enclose the curve ofΩ(120582) for suitably

large 119873 Hence the closure interval is determined by solving119886plusmn(120582) = 0 giving an interval

119868120576119873

= [119886minus 119886+] (75)

It is worthwhile to mention that the simplicity of the eigen-values guarantees the existence of approximate eigenvaluesthat is the 120582

119873rsquos for which Ω

119873(120582119873) = 0 Next we estimate the

error |120582lowast minus 120582119873| for the eigenvalue 120582lowast

Theorem 7 Let 120582lowast be an eigenvalue of (12)ndash(15) and let 120582119873

be its approximation Then for 120582 isin R we have the followingestimate

1003816100381610038161003816120582lowastminus 120582119873

1003816100381610038161003816lt

119879ℎ119873

(120582119873) + 119860120576119873

(0)

inf120577isin119868120576119873

1003816100381610038161003816Ω1015840(120577)

1003816100381610038161003816

(76)

where the interval 119868120576119873

is defined above

Proof Replacing 120582 by 120582119873in (73) we obtain

1003816100381610038161003816Ω (120582119873) minus Ω (120582

lowast)1003816100381610038161003816lt 119879ℎ119873

(120582119873) + 119860120576119873

(0) (77)

where we have used Ω119873(120582119873) = Ω(120582

lowast) = 0 Using the

mean value theorem yields that for some 120577 isin 119869120576119873

=

[min(120582lowast 120582119873)max(120582lowast 120582

119873)]

10038161003816100381610038161003816(120582lowastminus 120582119873)Ω1015840(120577)

10038161003816100381610038161003816le 119879ℎ119873

(120582119873) + 119860120576119873

(0) 120577 isin 119869120576119873

sub 119868120576119873

(78)

Since 120582lowast is simple and 119873 is sufficiently large then

inf120577isin119868120576119873

|Ω1015840(120577)| gt 0 and we get (76)

4 Numerical Examples

In this section we introduce two examples illustrating the theabove method Also in the following examples we observethat the exact solutions 120582

119896are all inside the interval [119886

minus 119886+]

In these two examples we indicate the effect of the amplitudeerror in the method by determining enclosure intervals fordifferent values of 120576 We also indicate the effect of 119873 and ℎ

by several choices The eigenvalues of the following examplescannot computed concretely then we use mathematicato obtain the exact values mathematica is also used inrounding off the exact eigenvalues which are square rootsBoth numerical results and the associated figures prove thecredibility of the method

Example 1 Consider the system

1199101015840

2(119909) minus 119901 (119909) 119910

1(119909) = 120582119910

1(119909)

1199101015840

1(119909) + 119901 (119909) 119910

2(119909) = minus120582119910

2(119909)

119909 isin [minus1 0) cup (0 1]

1205821199101(minus1) minus (radic2 + 120582) 119910

2(minus1) = 0

(radic2 + 120582) 1199101(1) minus 120582119910

2(1) = 0

1199101(0minus) minus 2119910

1(0+) = 0 119910

2(0minus) minus 2119910

2(0+) = 0

(79)

Here

1199011(119909) = 119901

2(119909) = 119903 (119909) =

119909 119909 isin [minus1 0)

1 (0 1]

(80)

1205791= 1205792= 1205874 120572

1= 1205732= 0 120572

2= 1205731= 1 and 120575 = 2 Direct

calculations give

K (120582) = 2 ((radic2120582 + 1) cos (2120582) minus 120582 (120582 + radic2) sin (2120582))

Ω (120582) = 2 ((radic2120582 + 1) cos(2120582 +

1

2

)

minus120582 (120582 + radic2) sin(2120582 +

1

2

))

(81)

As is clearly seen the eigenvalues cannot be computedexplicitly Tables 1 2 and 3 indicate the application of ourtechnique to this problem and the effect of 120576 By exact wemean the zeros of Δ(120582) computed by Mathematica

Figures 1 and 2 illustrate the enclosure intervals domi-nating 120582

0for 119873 = 25 ℎ = 01 and 120576 = 10

minus2 120576 = 10minus5

respectively The middle curve represents Δ(120582) while theupper and lower curves represent the curves of 119886

+(120582) 119886minus(120582)

respectively We notice that when 120576 = 10minus5 the two curves

are almost identical Similarly Figures 3 and 4 illustrate theenclosure intervals dominating 120582

1for ℎ = 01 119873 = 25 and

120576 = 10minus2 120576 = 10

minus5 respectively

Abstract and Applied Analysis 9

Table 1 The approximation 120582119896119873

and the exact solutions 120582119896for different choices of ℎ and119873

120582119896

120582minus2

120582minus1

1205820

1205821

Exact 120582119896

minus12061856254879546 minus05151195632138896 034274095892547224 1616237903407542

120582119896119873

ℎ = 06 119873 = 15 minus12061855965725146 minus05151190347168934 034274053593937165 1616237692038969120596 = 09708 119873 = 25 minus12061856254843495 minus05151195631493376 0342740958869788 16162379033820231ℎ = 01 119873 = 15 minus12061856254844203 minus05151195632158179 03427409589350932 16162379034134882

120596 = 14708 119873 = 25 minus12061856254879548 minus05151195632138894 034274095892547236 16162379034075425

Table 2 Absolute error |120582119896minus 120582119896119873

|

120582119896

120582minus2

120582minus1

1205820

1205821

ℎ = 06119873 = 15 289154 times 10

minus8528497 times 10

minus7422986 times 10

minus7211369 times 10

minus7

119873 = 25 360512 times 10minus12

64552 times 10minus11

556842 times 10minus11

255189 times 10minus11

ℎ = 01119873 = 15 353428 times 10

minus12192824 times 10

minus12962097 times 10

minus12594613 times 10

minus12

119873 = 25 222045 times 10minus16

222045 times 10minus16

555112 times 10minus17

444089 times 10minus16

Table 3 For119873 = 25 and ℎ = 01 the exact solutions 120582119896are all inside the interval [119886

minus 119886+] for different values of 120576

120582119896

120582minus2

120582minus1

1205820

1205821

Exact 120582119896

minus12061856254879546 minus05151195632138896 034274095892547224 1616237903407542119868120576119873

120576 = 10minus2

[minus12335696 minus11771922] [minus05486790 minus04826816] [03246478 03601346] [16100119 16224066]

119868120576119873

120576 = 10minus5

[minus12062137 minus12061575] [minus05151525 minus05150867] [03427232 03427586] [16162317 16162441]

030 032 034 036 038

00

02

04

aminus

a+

Ω(120582)

minus02

minus04

minus06

Figure 1The enclosure interval dominating 1205820for ℎ = 01119873 = 25

and 120576 = 10minus2

Example 2 In this example we consider the system

1199101015840

2(119909) minus 119901 (119909) 119910

1(119909) = 120582119910

1(119909)

1199101015840

1(119909) + 119901 (119909) 119910

2(119909) = minus120582119910

2(119909)

119909 isin [minus1 0) cup (0 1]

radic31199101(minus1) minus 119910

2(minus1) = 0

(1 +

1

2

120582)1199101(1) minus (1 +

radic3

2

120582)1199102(1) = 0

1199101(0minus) minus 3119910

1(0+) = 0 119910

2(0minus) minus 3119910

2(0+) = 0

(82)

030 032 034 036 038

00

02

aminus

a+

Ω(120582)

minus02

minus04

Figure 2The enclosure interval dominating 1205820for ℎ = 01119873 = 25

and 120576 = 10minus5

where

1199011(119909) = 119901

2(119909) = 119901 (119909) =

119909 119909 isin [minus1 0)

1199092 (0 1]

(83)

1205791= 1205792= 0 120572

1= 1205732= 1 120572

2= 1205731= 2 and 120575 = 12 Direct

calculations give

K (120582) =

1

2

((120582 + 3) cos (2120582) minus (1205822+ 3120582 + 4) sin (2120582))

Ω (120582) =

1

2

((1205822+ 3120582 + 4) sin(

1

6

minus 2120582)

+ (120582 + 3) cos(1

6

minus 2120582))

(84)

10 Abstract and Applied Analysis

Table 4 The approximation 120582119896119873

and the exact solutions 120582119896for different choices of ℎ and119873

120582119896

120582minus2

120582minus1

1205820

1205821

Exact 120582119896

minus30661256186938894 minus10959703083466883 03689030343814017 18338437675275228

120582119896119873

ℎ = 08 119873 = 15 minus3066123822200742 minus10959640944020903 036890463046491767 18338437607680411120596 = 007708 119873 = 25 minus30661256172349627 minus10959703026494807 036890303588091056 18338437676561417

ℎ = 02 119873 = 15 minus30661256186844446 minus10959703083278305 03689030343916001 1833843767517251120596 = 13708 119873 = 25 minus306612561869389 minus1095970308346688 03689030343814024 18338437675275225

Table 5 Absolute error |120582119896minus 120582119896119873

|

120582119896

120582minus2

120582minus1

1205820

1205821

ℎ = 08119873 = 15 179649 times 10

minus6621394 times 10

minus6159608 times 10

minus6675948 times 10

minus9

119873 = 25 145893 times 10minus9

569721 times 10minus9

149951 times 10minus9

128619 times 10minus10

ℎ = 02119873 = 15 944489 times 10

minus12188578 times 10

minus12101984 times 10

minus11102718 times 10

minus11

119873 = 25 444089 times 10minus16

222045 times 10minus16

666134 times 10minus16

222045 times 10minus16

Table 6 For119873 = 25 and ℎ = 02 the exact solutions 120582119896are all inside the interval [119886

minus 119886+] for different values of 120576

120582119896

120582minus2

120582minus1

1205820

1205821

Exact 120582119896

minus30661256186938894 minus10959703083466883 03689030343814017 18338437675275228119868120576119873

120576 = 10minus2

[minus31098742 minus30188665] [minus11631502 minus10339754] [03437434 03933486] [18220764 18454828]

119868120576119873

120576 = 10minus5

[minus30661709 minus30660803] [minus10960346 minus10959060] [03688783 03689278] [18338321 18338555]

158 159 160 161 162 163 164 165

00

05

10

minus05

minus10

aminus

a+

Ω(120582)

Figure 3The enclosure interval dominating 1205821for ℎ = 01119873 = 25

and 120576 = 10minus2

Tables 4 and 6 give the exact eigenvalues 1205821198961

119896=minus2and their

approximate ones for different values of ℎ 119873 and 120576 InTable 5 we give the absolute error for different values of ℎ and119873

Here Figures 5 6 7 and 8 illustrate the enclosureintervals dominating 120582

minus2and 120582

minus1for ℎ = 02 119873 = 25 and

120576 = 10minus2 120576 = 10

minus5 respectively

158 159 160 161 162 163 164 165

00

05

minus05

aminus

a+

Ω(120582)

Figure 4The enclosure interval dominating 1205821for ℎ = 01119873 = 25

and 120576 = 10minus5

5 Conclusion

With a simple analysis and with values of solutions ofinitial value problems computed at a few values of theeigenparameter we have computed the eigenvalues of aDirac system which has a discontinuity at one point andcontains a spectral parameter in all boundary conditionswith a certain estimated error The method proposed is

Abstract and Applied Analysis 11

00

05

10

minus05

minus33 minus32 minus31 minus30 minus29 minus28

aminus

a+

Ω(120582)

Figure 5 The enclosure interval dominating 120582minus2

for ℎ = 02 119873 =

25 and 120576 = 10minus2

00

05

10

minus05

minus33 minus32 minus31 minus30 minus29 minus28

aminus

a+

Ω(120582)

Figure 6 The enclosure interval dominating 120582minus2

for ℎ = 02 119873 =

25 and 120576 = 10minus5

a shooting procedure that is the problem is reformulatedas two initial value ones due to the interior discontinuity ofsize two and a miss-distance is defined at the right end of theinterval of integration whose roots are the eigenvalues to becomputed The unknown partU(120582) of the miss-distance canbewritten in terms of a functionwhich is an entire function ofexponential typeTherefore we propose to approximate suchterm by means of a truncated cardinal series with samplingvalues approximated by solving numerically correspondingsuitable initial value problems Finally in Section 4 weintroduced two instructive examples where both numericalresults and the associated figures have proved the credibilityof the method

00

05

10

minus05

minus14 minus13 minus12 minus11 minus10 minus09 minus08 minus07

aminus

a+

Ω(120582)

Figure 7 The enclosure interval dominating 120582minus1

for ℎ = 02 119873 =

25 and 120576 = 10minus2

00

05

10

minus05

minus14 minus13 minus12 minus11 minus10 minus09 minus08 minus07

aminus

a+

Ω(120582)

Figure 8 The enclosure interval dominating 120582minus1

for ℎ = 02 119873 =

25 and 120576 = 10minus5

Conflict of Interests

The authors declare that there is no conflict of interestsregarding the publication of this paper

Acknowledgment

This paper was funded by the Deanship of Scientific Research(DSR) King Abdulaziz University Jeddah The authorstherefore acknowledge with thanks DSR technical and finan-cial support

12 Abstract and Applied Analysis

References

[1] C T Fulton ldquoTwo-point boundary value problems witheigenvalue parameter contained in the boundary conditionsrdquoProceedings of the Royal Society of Edinburgh A Mathematicsvol 77 no 3-4 pp 293ndash308 1977

[2] JWalter ldquoRegular eigenvalue problems with eigenvalue param-eter in the boundary conditionrdquoMathematische Zeitschrift vol133 pp 301ndash312 1973

[3] V Kotelnikov ldquoOn the carrying capacity of the ether and wirein telecommunications material for the first All-Unionrdquo inConference on Questions of Communications vol 55 pp 55ndash64Izd Red Upr Svyazi RKKA 1933

[4] C E Shannon ldquoCommunication in the presence of noiserdquoProceedings of the IRE vol 37 pp 10ndash21 1949

[5] E TWhittaker ldquoOn the functions which are represented by theexpansion of the interpolation theoryrdquo Proceedings of the RoyalSociety of Edinburgh A vol 35 pp 181ndash194 1915

[6] P L Butzer G Schmeisser and R L Stens ldquoAn introductionto sampling analysisrdquo in Non Uniform Sampling Theory andPractices F Marvasti Ed pp 17ndash121 Kluwer New York NYUSA 2001

[7] M A Kowalski K A Sikorski and F Stenger Selected Topics inApproximation andComputation OxfordUniversity Press NewYork NY USA 1995

[8] J Lund and K L Bowers Sinc Methods for Quadrature andDifferential Equations SIAM Philadelphia Pa USA 1992

[9] F Stenger ldquoNumericalmethods based onWhittaker cardinal orsinc functionsrdquo SIAM Review vol 23 no 2 pp 165ndash224 1981

[10] F Stenger Numerical Methods Based on Sinc and AnalyticFunctions Springer New York NY USA 1993

[11] A Boumenir ldquoHigher approximation of eigenvalues by thesampling methodrdquo BIT Numerical Mathematics vol 40 no 2pp 215ndash225 2000

[12] A Boumenir ldquoSampling and eigenvalues of non-self-adjointSturm-Liouville problemsrdquo SIAM Journal on Scientific Comput-ing vol 23 no 1 pp 219ndash229 2001

[13] M M Tharwat A H Bhrawy and A Yildirim ldquoNumericalcomputation of eigenvalues of discontinuous Sturm-Liouvilleproblems with parameter dependent boundary conditionsusing sinc methodrdquoNumerical Algorithms vol 63 no 1 pp 27ndash48 2013

[14] M M Tharwat A H Bhrawy and A Yildirim ldquoNumericalcomputation of the eigenvalues of a discontinuousDirac systemusing the sincmethodwith error analysisrdquo International Journalof Computer Mathematics vol 89 no 15 pp 2061ndash2080 2012

[15] P L Butzer and R L Stens ldquoA modification of the Whittaker-Kotelrsquonikov-Shannon sampling seriesrdquo Aequationes Mathemat-icae vol 28 no 3 pp 305ndash311 1985

[16] R Gervais Q I Rahman and G Schmeisser ldquoA bandlimitedfunction simulating a duration-limited onerdquo in ApproximationTheory andFunctionalAnalysis P L Butzer andR L Stens Edspp 355ndash362 Birkhaauser Boston Mass USA 1984

[17] R L Stens ldquoSampling by generalized kernelsrdquo in SamplingThe-ory in Fourier and Signal Analysis Advanced Topics J RHigginsand R L Stens Eds pp 130ndash157 Oxford University PressOxford UK 1999

[18] G Schmeisser and F Stenger ldquoSinc approximation with a Gaus-sianmultiplierrdquo SamplingTheory in Signal and Image Processingvol 6 no 2 pp 199ndash221 2007

[19] L Qian ldquoOn the regularized Whittaker-Kotelrsquonikov-Shannonsampling formulardquo Proceedings of the American MathematicalSociety vol 131 no 4 pp 1169ndash1176 2003

[20] L Qian and D B Creamer ldquoA modification of the samplingseries with aGaussianmultiplierrdquo SamplingTheory in Signal andImage Processing vol 5 no 1 pp 1ndash20 2006

[21] L Qian and D B Creamer ldquoLocalized sampling in the presenceof noiserdquoAppliedMathematics Letters vol 19 no 4 pp 351ndash3552006

[22] M H Annaby and R M Asharabi ldquoComputing eigenvaluesof boundary-value problems using sinc-Gaussian methodrdquoSamplingTheory in Signal and Image Processing vol 7 no 3 pp293ndash312 2008

[23] M H Annaby and M MTharwat ldquoA sinc-Gaussian techniquefor computing eigenvalues of second-order linear pencilsrdquoApplied Numerical Mathematics vol 63 pp 129ndash137 2013

[24] A H Bhrawy M M Tharwat and A Al-Fhaid ldquoNumericalalgorithms for computing eigenvalues of discontinuous Diracsystem using sinc-Gaussian methodrdquo Abstract and AppliedAnalysis vol 2012 Article ID 925134 13 pages 2012

[25] M M Tharwat A H Bhrawy and A S Alofi ldquoApproximationof eigenvalues of discontinuous sturm-liouville problems witheigenparameter in all boundary conditionsrdquo Boundary ValueProblems vol 2013 article 132 2013

[26] M M Tharwat and S M Al-Harbi ldquoApproximation of eigen-values of boundary value problemsrdquo Boundary Value Problemsvol 2014 article 51 2014

[27] MMTharwat and A H Bhrawy ldquoComputation of eigenvaluesof discontinuous Dirac system using Hermite interpolationtechniquerdquo Advances in Difference Equations vol 2012 article59 2012

[28] M M Tharwat A Yildirim and A H Bhrawy ldquoSampling ofdiscontinuous Dirac systemsrdquo Numerical Functional Analysisand Optimization vol 34 no 3 pp 323ndash348 2013

[29] M M Tharwat ldquoDiscontinuous Sturm-Liouville problems andassociated sampling theoriesrdquo Abstract and Applied Analysisvol 2011 Article ID 610232 30 pages 2011

[30] MMTharwat ldquoOn sampling theories and discontinuousDiracsystems with eigenparameter in the boundary conditionsrdquoBoundary Value Problems vol 2013 article 65 2013

[31] B M Levitan and I S Sargsjan ldquoIntroduction to spectral the-ory self adjoint ordinary differential operatorsrdquo in Translationof Mthematical Monographs vol 39 American MathematicalSociety Providence RI USA 1975

[32] B M Levitan and I S Sargsjan Sturm-Liouville and Dirac oper-ators vol 59 Kluwer Academic Publishers Group DordrechtThe Netherlands 1991

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Stochastic AnalysisInternational Journal of

6 Abstract and Applied Analysis

12059322

(119909 120582) =

1

120575

12059311

(0minus 120582) sin (120582119909) +

1

120575

12059321

(0minus 120582) cos (120582119909)

+T01

12059312

(119909 120582) minusT02

12059322

(119909 120582)

(50)

where Tminus1119894

Tminus1119894

T0119894

and T0119894 119894 = 1 2 are the Volterra

integral operators defined by

Tminus1119894

120593 (119909 120582) = int

119909

minus1

sin 120582 (119909 minus 119905) 119901119894(119905) 120593 (119905 120582) 119889119905

Tminus1119894

120593 (119909 120582) = int

119909

minus1

cos 120582 (119909 minus 119905) 119901119894(119905) 120593 (119905 120582) 119889119905

T0119894120593 (119909 120582) = int

119909

0

sin 120582 (119909 minus 119905) 119901119894(119905) 120593 (119905 120582) 119889119905

T0119894120593 (119909 120582) = int

119909

0

cos 120582 (119909 minus 119905) 119901119894(119905) 120593 (119905 120582) 119889119905

(51)

For convenience we define the constants

1198881= max 100381610038161003816

10038161205721

1003816100381610038161003816+10038161003816100381610038161205722

10038161003816100381610038161003816100381610038161003816sin 1205791

1003816100381610038161003816+1003816100381610038161003816cos 1205791

1003816100381610038161003816

1198882= int

0

minus1

[10038161003816100381610038161199011(119905)

1003816100381610038161003816+10038161003816100381610038161199012(119905)

1003816100381610038161003816] 119889119905 119888

3= 11988811198882exp (119888

2)

1198884= int

1

0

[10038161003816100381610038161199011(119905)

1003816100381610038161003816+10038161003816100381610038161199012(119905)

1003816100381610038161003816] 119889119905 119888

5= 1198883+

2

|120575|

(1198881+ 1198883)

1198886= max 100381610038161003816

10038161205721

1003816100381610038161003816+10038161003816100381610038161205722

10038161003816100381610038161003816100381610038161003816sin 1205792

1003816100381610038161003816+1003816100381610038161003816cos 1205792

1003816100381610038161003816

(52)

Now we define yminus1119894

(sdot 120582) and y0119894(sdot 120582) 119894 = 1 2 to be

yminus11

(119909 120582) = Tminus11

12059311

(119909 120582) +Tminus12

12059321

(119909 120582)

yminus12

(119909 120582) =Tminus11

12059311

(119909 120582) minusTminus12

12059321

(119909 120582)

y01

(119909 120582) = T01

12059312

(119909 120582) +T02

12059322

(119909 120582)

y02

(119909 120582) =T01

12059312

(119909 120582) minusT02

12059322

(119909 120582)

(53)

Lemma 4 The functions yminus11

(119909 120582) and yminus12

(119909 120582) are entirein 120582 for any fixed 119909 isin [minus1 0) and satisfy the growth condition

1003816100381610038161003816yminus11

(119909 120582)1003816100381610038161003816

1003816100381610038161003816yminus12

(119909 120582)1003816100381610038161003816le 21198883(1 + |120582|) 119890

|I120582|(119909+1)

120582 isin C

(54)

Proof Since yminus11

(119909 120582) = Tminus11

12059311(119909 120582) +

Tminus12

12059321(119909 120582)

then from (47) and (48) we obtain

yminus11

(119909 120582)

= 1205722Tminus11

cos 120582 (119909 + 1) minus 1205721Tminus11

sin 120582 (119909 + 1)

+ 120582Tminus11

cos (120582 (119909 + 1) + 1205791) minusTminus11

yminus11

(119909 120582)

+ 1205721Tminus12

cos 120582 (119909 + 1) + 1205722Tminus12

sin 120582 (119909 + 1)

+ 120582Tminus12

sin (120582 (119909 + 1) + 1205791) +

Tminus12

yminus12

(119909 120582)

(55)

Using the inequalities | sin 119911| le 119890|I119911| and | cos 119911| le 119890

|I119911| for119911 isin C leads for 120582 isin C to1003816100381610038161003816yminus11

(119909 120582)1003816100381610038161003816

le10038161003816100381610038161205722

1003816100381610038161003816

1003816100381610038161003816Tminus11

cos 120582 (119909 + 1)1003816100381610038161003816+10038161003816100381610038161205721Tminus11

sin 120582 (119909 + 1)1003816100381610038161003816

+ |120582|1003816100381610038161003816Tminus11

cos (120582 (119909 + 1) + 1205791)1003816100381610038161003816+1003816100381610038161003816Tminus11

yminus11

(119909 120582)1003816100381610038161003816

+10038161003816100381610038161205721

1003816100381610038161003816

10038161003816100381610038161003816

Tminus12

cos 120582 (119909 + 1)

10038161003816100381610038161003816+10038161003816100381610038161205722

1003816100381610038161003816

10038161003816100381610038161003816

Tminus12

sin 120582 (119909 + 1)

10038161003816100381610038161003816

+ |120582|

10038161003816100381610038161003816

Tminus12

sin (120582 (119909 + 1) + 1205791)

10038161003816100381610038161003816+

10038161003816100381610038161003816

Tminus12

yminus12

(119909 120582)

10038161003816100381610038161003816

le 21198881(1 + |120582|) 119890

|I120582|(119909+1)int

119909

minus1

[10038161003816100381610038161199011(119905)

1003816100381610038161003816+10038161003816100381610038161199012(119905)

1003816100381610038161003816] 119889119905

+ 119890|I120582|(119909+1)

int

119909

minus1

[10038161003816100381610038161199011(119905)

1003816100381610038161003816

1003816100381610038161003816yminus11

(119905 120582)1003816100381610038161003816

+10038161003816100381610038161199012(119905)

1003816100381610038161003816

1003816100381610038161003816yminus12

(119905 120582)1003816100381610038161003816] 119890minus|I120582|(119905+1)

119889119905

le 211988811198882(1 + |120582|) 119890

|I120582|(119909+1)

+ 119890|I120582|(119909+1)

int

119909

minus1

[10038161003816100381610038161199011(119905)

1003816100381610038161003816

1003816100381610038161003816yminus11

(119905 120582)1003816100381610038161003816

+10038161003816100381610038161199012(119905)

1003816100381610038161003816

1003816100381610038161003816yminus12

(119905 120582)1003816100381610038161003816] 119890minus|I120582|(119905+1)

119889119905

(56)

The above inequality can be reduced to

119890minus|I120582|(119909+1) 1003816

100381610038161003816yminus11

(119909 120582)1003816100381610038161003816

le 211988811198882(1 + |120582|)

+ int

119909

minus1

[10038161003816100381610038161199011(119905)

1003816100381610038161003816

1003816100381610038161003816yminus11

(119905 120582)1003816100381610038161003816

+10038161003816100381610038161199012(119905)

1003816100381610038161003816

1003816100381610038161003816yminus12

(119905 120582)1003816100381610038161003816] 119890minus|I120582|(119905+1)

119889119905

(57)

Similarly we can prove that

119890minus|I120582|(119909+1) 1003816

100381610038161003816yminus12

(119909 120582)1003816100381610038161003816

le 211988811198882(1 + |120582|)

+ int

119909

minus1

[10038161003816100381610038161199011(119905)

1003816100381610038161003816

1003816100381610038161003816yminus11

(119905 120582)1003816100381610038161003816

+10038161003816100381610038161199012(119905)

1003816100381610038161003816

1003816100381610038161003816yminus12

(119905 120582)1003816100381610038161003816] 119890minus|I120582|(119905+1)

119889119905

(58)

Then from (57) (58) and Lemma 31 of [32 pp 204] weobtain (54)

In a similar manner we will prove the following lemmafor y01

(sdot 120582) and y02

(sdot 120582)

Lemma 5 The functions y01

(119909 120582) and y02

(119909 120582) are entire in120582 for any fixed 119909 isin (0 1] and satisfy the growth condition

1003816100381610038161003816y01

(119909 120582)1003816100381610038161003816

1003816100381610038161003816y02

(119909 120582)1003816100381610038161003816le 211988841198885(1 + |120582|) 119890

|I120582|(119909+1)

120582 isin C

(59)

Abstract and Applied Analysis 7

Proof Since y01

(119909 120582) = T01

12059312(119909 120582) +

T02

12059322(119909 120582) then

from (49) and (50) we obtain

y01

(119909 120582) =

1

120575

12059311

(0minus 120582)T

01cos (120582119909)

minus

1

120575

12059321

(0minus 120582)T

01sin (120582119909)

minusT01yminus12

(119909 120582) +

1

120575

12059311

(0minus 120582)

T02

sin (120582119909)

+

1

120575

12059321

(0minus 120582)

T02

cos (120582119909) + T02yminus12

(119909 120582)

(60)

Then from (47) and (48) and Lemma 4 we get

1003816100381610038161003816y01

(119909 120582)1003816100381610038161003816le

1

|120575|

100381610038161003816100381612059311

(0minus 120582)

1003816100381610038161003816

1003816100381610038161003816T01

cos (120582119909)1003816100381610038161003816

+

1

|120575|

100381610038161003816100381612059321

(0minus 120582)

1003816100381610038161003816

1003816100381610038161003816T01

sin (120582119909)1003816100381610038161003816

+1003816100381610038161003816T01yminus12

(119909 120582)1003816100381610038161003816

+

1

|120575|

100381610038161003816100381612059311

(0minus 120582)

1003816100381610038161003816

10038161003816100381610038161003816

T02

sin (120582119909)

10038161003816100381610038161003816

+

1

|120575|

100381610038161003816100381612059321

(0minus 120582)

1003816100381610038161003816

10038161003816100381610038161003816

T02

cos (120582119909)10038161003816100381610038161003816

+

10038161003816100381610038161003816

T02yminus12

(119909 120582)

10038161003816100381610038161003816

le 21198884(1 + |120582|)

times (1198883+

2

|120575|

(1198881+ 1198883)) 119890|I120582|(119909+1)

= 211988841198885(1 + |120582|) 119890

|I120582|(119909+1)

(61)

Similarly we can prove that1003816100381610038161003816y02

(119909 120582)1003816100381610038161003816le 211988841198885(1 + |120582|) 119890

|I120582|(119909+1) (62)

3 The Numerical Scheme

In this section we derive the method of computing eigen-values of problems (12)ndash(15) numerically The basic idea ofthe scheme is to split Ω(120582) into two parts a known partK(120582) and an unknown oneU(120582)Thenwe approximateU(120582)

using (5) to get the approximate Ω(120582) and then compute theapproximate zeros We first splitΩ(120582) into two parts

Ω (120582) = K (120582) +U (120582) (63)

whereU(120582) is the unknown part involving integral operators

U (120582) = 120575 [(1205732+ 120582 cos 120579

2) sin 120582

minus (1205731+ 120582 sin 120579

2) cos 120582] y

minus11(0minus 120582)

minus 120575 [(1205731+ 120582 sin 120579

2) sin 120582

+ (1205732+ 120582 cos 120579

2) cos 120582] y

minus12(0minus 120582)

minus 1205752[(1205731+ 120582 sin 120579

2) y01

(1 120582)

+ (1205732+ 120582 cos 120579

2) y02

(1 120582)]

(64)

andK(120582) is the known part

K (120582) = 120575 (1205731+ 120582 sin 120579

2)

times [1205722cos 2120582 minus 120572

1sin 2120582 + 120582 cos (2120582 + 120579

1)]

minus 120575 (1205732+ 120582 cos 120579

2)

times [1205722sin 2120582 + 120572

1cos 2120582 + 120582 sin (2120582 + 120579

1)]

(65)

Then from Lemmas 4 and 5 we have the following result

Lemma 6 The function U(120582) is entire in 120582 and the followingestimate holds

|U (120582)| le 120601 (120582) 1198902|I120582|

(66)

where

120601 (120582) = 119872(1 + |120582|)2

119872 = 4 |120575| 1198886 (21198883 + |120575| 11988841198885)

(67)

Proof From (64) we have

|U (120582)| le |120575| [(10038161003816100381610038161205732

1003816100381610038161003816+ |120582|

1003816100381610038161003816cos 1205792

1003816100381610038161003816) |sin 120582|

+ (10038161003816100381610038161205731

1003816100381610038161003816+ |120582|

1003816100381610038161003816sin 1205792

1003816100381610038161003816) |cos 120582|] 100381610038161003816

1003816yminus11

(0minus 120582)

1003816100381610038161003816

+ |120575| [(10038161003816100381610038161205731

1003816100381610038161003816+ |120582|

1003816100381610038161003816sin 1205792

1003816100381610038161003816) |sin 120582|

+ (10038161003816100381610038161205732

1003816100381610038161003816+1003816100381610038161003816120582 cos 120579

2

1003816100381610038161003816) |cos 120582|] 100381610038161003816

1003816yminus12

(0minus 120582)

1003816100381610038161003816

+ 1205752[(10038161003816100381610038161205731

1003816100381610038161003816+ |120582|

1003816100381610038161003816sin 1205792

1003816100381610038161003816)1003816100381610038161003816y01

(1 120582)1003816100381610038161003816

+ (10038161003816100381610038161205732

1003816100381610038161003816+ |120582|

1003816100381610038161003816cos 1205792

1003816100381610038161003816)1003816100381610038161003816y02

(1 120582)1003816100381610038161003816]

(68)

Using the inequalities | sin 120582| le 119890|I120582| and | cos 120582| le 119890

|I120582| for120582 isin C and Lemmas 4 and 5 imply (66)

ThusU(120582) is an entire function of exponential type 120590 = 2In the following we let 120582 isin R since all eigenvalues are realNow we approximate the function U(120582) using the operator(5) where ℎ isin (0 1205872) and 120596 = (120587 minus 2ℎ)2 and then from(7) we obtain

1003816100381610038161003816U (120582) minus (G

ℎ119873U) (120582)

1003816100381610038161003816le 119879ℎ119873

(120582) (69)

where

119879ℎ119873

(120582) = 2

10038161003816100381610038161003816sin (ℎ

minus1120587120582)

10038161003816100381610038161003816

times 120601 (|R120582| + ℎ (119873 + 1))

119890minus120596119873

radic120587120596119873

120573119873(0) 120582 isin R

(70)

The samplesU(119899ℎ) = Ω(119899ℎ) minusK(119899ℎ) 119899 isin Z119873(120582) cannot be

computed explicitly in the general caseWe approximate these

8 Abstract and Applied Analysis

samples numerically by solving the initial-value problemsdefined by (12) and (45) to obtain the approximate valuesU(119899ℎ) 119899 isin Z

119873(120582) that is U(119899ℎ) = Ω(119899ℎ) minus K(119899ℎ)

Here we use a computer algebra system mathematica toobtain the approximate solutions with the required accuracyHowever a separate study for the effect of different numericalschemes and the computational costs would be interestingAccordingly we have the explicit expansion

(Gℎ119873

U) (120582) = sum

119899isinZ119873(120582)

U (119899ℎ) sinc (ℎminus1120587119909 minus 119899120587)

times 119866(

radic120596 (120582 minus 119899ℎ)

radic119873ℎ

)

(71)

Therefore we get cf (10)10038161003816100381610038161003816(Gℎ119873

U) (120582) minus (Gℎ119873

U) (120582)

10038161003816100381610038161003816le 119860120576119873

(0) 120582 isin R (72)

Now let Ω119873(120582) = K(120582) + (G

ℎ119873U)(120582) From (69) and (72)

we obtain10038161003816100381610038161003816Ω (120582) minus Ω

119873(120582)

10038161003816100381610038161003816le 119879ℎ119873

(120582) + 119860120576119873

(0) 120582 isin R (73)

Let 120582lowast be an eigenvalue and 120582119873its desired approximation

that is Ω(120582lowast) = 0 and Ω

119873(120582119873) = 0 From (73) we have

|Ω119873(120582lowast)| le 119879

ℎ119873(120582lowast) + 119860120576119873

(0) Define the curves

119886plusmn(120582) = Ω

119873(120582) plusmn 119879

ℎ119873(120582) + 119860

120576119873(0) (74)

The curves 119886+(120582) 119886minus(120582) enclose the curve ofΩ(120582) for suitably

large 119873 Hence the closure interval is determined by solving119886plusmn(120582) = 0 giving an interval

119868120576119873

= [119886minus 119886+] (75)

It is worthwhile to mention that the simplicity of the eigen-values guarantees the existence of approximate eigenvaluesthat is the 120582

119873rsquos for which Ω

119873(120582119873) = 0 Next we estimate the

error |120582lowast minus 120582119873| for the eigenvalue 120582lowast

Theorem 7 Let 120582lowast be an eigenvalue of (12)ndash(15) and let 120582119873

be its approximation Then for 120582 isin R we have the followingestimate

1003816100381610038161003816120582lowastminus 120582119873

1003816100381610038161003816lt

119879ℎ119873

(120582119873) + 119860120576119873

(0)

inf120577isin119868120576119873

1003816100381610038161003816Ω1015840(120577)

1003816100381610038161003816

(76)

where the interval 119868120576119873

is defined above

Proof Replacing 120582 by 120582119873in (73) we obtain

1003816100381610038161003816Ω (120582119873) minus Ω (120582

lowast)1003816100381610038161003816lt 119879ℎ119873

(120582119873) + 119860120576119873

(0) (77)

where we have used Ω119873(120582119873) = Ω(120582

lowast) = 0 Using the

mean value theorem yields that for some 120577 isin 119869120576119873

=

[min(120582lowast 120582119873)max(120582lowast 120582

119873)]

10038161003816100381610038161003816(120582lowastminus 120582119873)Ω1015840(120577)

10038161003816100381610038161003816le 119879ℎ119873

(120582119873) + 119860120576119873

(0) 120577 isin 119869120576119873

sub 119868120576119873

(78)

Since 120582lowast is simple and 119873 is sufficiently large then

inf120577isin119868120576119873

|Ω1015840(120577)| gt 0 and we get (76)

4 Numerical Examples

In this section we introduce two examples illustrating the theabove method Also in the following examples we observethat the exact solutions 120582

119896are all inside the interval [119886

minus 119886+]

In these two examples we indicate the effect of the amplitudeerror in the method by determining enclosure intervals fordifferent values of 120576 We also indicate the effect of 119873 and ℎ

by several choices The eigenvalues of the following examplescannot computed concretely then we use mathematicato obtain the exact values mathematica is also used inrounding off the exact eigenvalues which are square rootsBoth numerical results and the associated figures prove thecredibility of the method

Example 1 Consider the system

1199101015840

2(119909) minus 119901 (119909) 119910

1(119909) = 120582119910

1(119909)

1199101015840

1(119909) + 119901 (119909) 119910

2(119909) = minus120582119910

2(119909)

119909 isin [minus1 0) cup (0 1]

1205821199101(minus1) minus (radic2 + 120582) 119910

2(minus1) = 0

(radic2 + 120582) 1199101(1) minus 120582119910

2(1) = 0

1199101(0minus) minus 2119910

1(0+) = 0 119910

2(0minus) minus 2119910

2(0+) = 0

(79)

Here

1199011(119909) = 119901

2(119909) = 119903 (119909) =

119909 119909 isin [minus1 0)

1 (0 1]

(80)

1205791= 1205792= 1205874 120572

1= 1205732= 0 120572

2= 1205731= 1 and 120575 = 2 Direct

calculations give

K (120582) = 2 ((radic2120582 + 1) cos (2120582) minus 120582 (120582 + radic2) sin (2120582))

Ω (120582) = 2 ((radic2120582 + 1) cos(2120582 +

1

2

)

minus120582 (120582 + radic2) sin(2120582 +

1

2

))

(81)

As is clearly seen the eigenvalues cannot be computedexplicitly Tables 1 2 and 3 indicate the application of ourtechnique to this problem and the effect of 120576 By exact wemean the zeros of Δ(120582) computed by Mathematica

Figures 1 and 2 illustrate the enclosure intervals domi-nating 120582

0for 119873 = 25 ℎ = 01 and 120576 = 10

minus2 120576 = 10minus5

respectively The middle curve represents Δ(120582) while theupper and lower curves represent the curves of 119886

+(120582) 119886minus(120582)

respectively We notice that when 120576 = 10minus5 the two curves

are almost identical Similarly Figures 3 and 4 illustrate theenclosure intervals dominating 120582

1for ℎ = 01 119873 = 25 and

120576 = 10minus2 120576 = 10

minus5 respectively

Abstract and Applied Analysis 9

Table 1 The approximation 120582119896119873

and the exact solutions 120582119896for different choices of ℎ and119873

120582119896

120582minus2

120582minus1

1205820

1205821

Exact 120582119896

minus12061856254879546 minus05151195632138896 034274095892547224 1616237903407542

120582119896119873

ℎ = 06 119873 = 15 minus12061855965725146 minus05151190347168934 034274053593937165 1616237692038969120596 = 09708 119873 = 25 minus12061856254843495 minus05151195631493376 0342740958869788 16162379033820231ℎ = 01 119873 = 15 minus12061856254844203 minus05151195632158179 03427409589350932 16162379034134882

120596 = 14708 119873 = 25 minus12061856254879548 minus05151195632138894 034274095892547236 16162379034075425

Table 2 Absolute error |120582119896minus 120582119896119873

|

120582119896

120582minus2

120582minus1

1205820

1205821

ℎ = 06119873 = 15 289154 times 10

minus8528497 times 10

minus7422986 times 10

minus7211369 times 10

minus7

119873 = 25 360512 times 10minus12

64552 times 10minus11

556842 times 10minus11

255189 times 10minus11

ℎ = 01119873 = 15 353428 times 10

minus12192824 times 10

minus12962097 times 10

minus12594613 times 10

minus12

119873 = 25 222045 times 10minus16

222045 times 10minus16

555112 times 10minus17

444089 times 10minus16

Table 3 For119873 = 25 and ℎ = 01 the exact solutions 120582119896are all inside the interval [119886

minus 119886+] for different values of 120576

120582119896

120582minus2

120582minus1

1205820

1205821

Exact 120582119896

minus12061856254879546 minus05151195632138896 034274095892547224 1616237903407542119868120576119873

120576 = 10minus2

[minus12335696 minus11771922] [minus05486790 minus04826816] [03246478 03601346] [16100119 16224066]

119868120576119873

120576 = 10minus5

[minus12062137 minus12061575] [minus05151525 minus05150867] [03427232 03427586] [16162317 16162441]

030 032 034 036 038

00

02

04

aminus

a+

Ω(120582)

minus02

minus04

minus06

Figure 1The enclosure interval dominating 1205820for ℎ = 01119873 = 25

and 120576 = 10minus2

Example 2 In this example we consider the system

1199101015840

2(119909) minus 119901 (119909) 119910

1(119909) = 120582119910

1(119909)

1199101015840

1(119909) + 119901 (119909) 119910

2(119909) = minus120582119910

2(119909)

119909 isin [minus1 0) cup (0 1]

radic31199101(minus1) minus 119910

2(minus1) = 0

(1 +

1

2

120582)1199101(1) minus (1 +

radic3

2

120582)1199102(1) = 0

1199101(0minus) minus 3119910

1(0+) = 0 119910

2(0minus) minus 3119910

2(0+) = 0

(82)

030 032 034 036 038

00

02

aminus

a+

Ω(120582)

minus02

minus04

Figure 2The enclosure interval dominating 1205820for ℎ = 01119873 = 25

and 120576 = 10minus5

where

1199011(119909) = 119901

2(119909) = 119901 (119909) =

119909 119909 isin [minus1 0)

1199092 (0 1]

(83)

1205791= 1205792= 0 120572

1= 1205732= 1 120572

2= 1205731= 2 and 120575 = 12 Direct

calculations give

K (120582) =

1

2

((120582 + 3) cos (2120582) minus (1205822+ 3120582 + 4) sin (2120582))

Ω (120582) =

1

2

((1205822+ 3120582 + 4) sin(

1

6

minus 2120582)

+ (120582 + 3) cos(1

6

minus 2120582))

(84)

10 Abstract and Applied Analysis

Table 4 The approximation 120582119896119873

and the exact solutions 120582119896for different choices of ℎ and119873

120582119896

120582minus2

120582minus1

1205820

1205821

Exact 120582119896

minus30661256186938894 minus10959703083466883 03689030343814017 18338437675275228

120582119896119873

ℎ = 08 119873 = 15 minus3066123822200742 minus10959640944020903 036890463046491767 18338437607680411120596 = 007708 119873 = 25 minus30661256172349627 minus10959703026494807 036890303588091056 18338437676561417

ℎ = 02 119873 = 15 minus30661256186844446 minus10959703083278305 03689030343916001 1833843767517251120596 = 13708 119873 = 25 minus306612561869389 minus1095970308346688 03689030343814024 18338437675275225

Table 5 Absolute error |120582119896minus 120582119896119873

|

120582119896

120582minus2

120582minus1

1205820

1205821

ℎ = 08119873 = 15 179649 times 10

minus6621394 times 10

minus6159608 times 10

minus6675948 times 10

minus9

119873 = 25 145893 times 10minus9

569721 times 10minus9

149951 times 10minus9

128619 times 10minus10

ℎ = 02119873 = 15 944489 times 10

minus12188578 times 10

minus12101984 times 10

minus11102718 times 10

minus11

119873 = 25 444089 times 10minus16

222045 times 10minus16

666134 times 10minus16

222045 times 10minus16

Table 6 For119873 = 25 and ℎ = 02 the exact solutions 120582119896are all inside the interval [119886

minus 119886+] for different values of 120576

120582119896

120582minus2

120582minus1

1205820

1205821

Exact 120582119896

minus30661256186938894 minus10959703083466883 03689030343814017 18338437675275228119868120576119873

120576 = 10minus2

[minus31098742 minus30188665] [minus11631502 minus10339754] [03437434 03933486] [18220764 18454828]

119868120576119873

120576 = 10minus5

[minus30661709 minus30660803] [minus10960346 minus10959060] [03688783 03689278] [18338321 18338555]

158 159 160 161 162 163 164 165

00

05

10

minus05

minus10

aminus

a+

Ω(120582)

Figure 3The enclosure interval dominating 1205821for ℎ = 01119873 = 25

and 120576 = 10minus2

Tables 4 and 6 give the exact eigenvalues 1205821198961

119896=minus2and their

approximate ones for different values of ℎ 119873 and 120576 InTable 5 we give the absolute error for different values of ℎ and119873

Here Figures 5 6 7 and 8 illustrate the enclosureintervals dominating 120582

minus2and 120582

minus1for ℎ = 02 119873 = 25 and

120576 = 10minus2 120576 = 10

minus5 respectively

158 159 160 161 162 163 164 165

00

05

minus05

aminus

a+

Ω(120582)

Figure 4The enclosure interval dominating 1205821for ℎ = 01119873 = 25

and 120576 = 10minus5

5 Conclusion

With a simple analysis and with values of solutions ofinitial value problems computed at a few values of theeigenparameter we have computed the eigenvalues of aDirac system which has a discontinuity at one point andcontains a spectral parameter in all boundary conditionswith a certain estimated error The method proposed is

Abstract and Applied Analysis 11

00

05

10

minus05

minus33 minus32 minus31 minus30 minus29 minus28

aminus

a+

Ω(120582)

Figure 5 The enclosure interval dominating 120582minus2

for ℎ = 02 119873 =

25 and 120576 = 10minus2

00

05

10

minus05

minus33 minus32 minus31 minus30 minus29 minus28

aminus

a+

Ω(120582)

Figure 6 The enclosure interval dominating 120582minus2

for ℎ = 02 119873 =

25 and 120576 = 10minus5

a shooting procedure that is the problem is reformulatedas two initial value ones due to the interior discontinuity ofsize two and a miss-distance is defined at the right end of theinterval of integration whose roots are the eigenvalues to becomputed The unknown partU(120582) of the miss-distance canbewritten in terms of a functionwhich is an entire function ofexponential typeTherefore we propose to approximate suchterm by means of a truncated cardinal series with samplingvalues approximated by solving numerically correspondingsuitable initial value problems Finally in Section 4 weintroduced two instructive examples where both numericalresults and the associated figures have proved the credibilityof the method

00

05

10

minus05

minus14 minus13 minus12 minus11 minus10 minus09 minus08 minus07

aminus

a+

Ω(120582)

Figure 7 The enclosure interval dominating 120582minus1

for ℎ = 02 119873 =

25 and 120576 = 10minus2

00

05

10

minus05

minus14 minus13 minus12 minus11 minus10 minus09 minus08 minus07

aminus

a+

Ω(120582)

Figure 8 The enclosure interval dominating 120582minus1

for ℎ = 02 119873 =

25 and 120576 = 10minus5

Conflict of Interests

The authors declare that there is no conflict of interestsregarding the publication of this paper

Acknowledgment

This paper was funded by the Deanship of Scientific Research(DSR) King Abdulaziz University Jeddah The authorstherefore acknowledge with thanks DSR technical and finan-cial support

12 Abstract and Applied Analysis

References

[1] C T Fulton ldquoTwo-point boundary value problems witheigenvalue parameter contained in the boundary conditionsrdquoProceedings of the Royal Society of Edinburgh A Mathematicsvol 77 no 3-4 pp 293ndash308 1977

[2] JWalter ldquoRegular eigenvalue problems with eigenvalue param-eter in the boundary conditionrdquoMathematische Zeitschrift vol133 pp 301ndash312 1973

[3] V Kotelnikov ldquoOn the carrying capacity of the ether and wirein telecommunications material for the first All-Unionrdquo inConference on Questions of Communications vol 55 pp 55ndash64Izd Red Upr Svyazi RKKA 1933

[4] C E Shannon ldquoCommunication in the presence of noiserdquoProceedings of the IRE vol 37 pp 10ndash21 1949

[5] E TWhittaker ldquoOn the functions which are represented by theexpansion of the interpolation theoryrdquo Proceedings of the RoyalSociety of Edinburgh A vol 35 pp 181ndash194 1915

[6] P L Butzer G Schmeisser and R L Stens ldquoAn introductionto sampling analysisrdquo in Non Uniform Sampling Theory andPractices F Marvasti Ed pp 17ndash121 Kluwer New York NYUSA 2001

[7] M A Kowalski K A Sikorski and F Stenger Selected Topics inApproximation andComputation OxfordUniversity Press NewYork NY USA 1995

[8] J Lund and K L Bowers Sinc Methods for Quadrature andDifferential Equations SIAM Philadelphia Pa USA 1992

[9] F Stenger ldquoNumericalmethods based onWhittaker cardinal orsinc functionsrdquo SIAM Review vol 23 no 2 pp 165ndash224 1981

[10] F Stenger Numerical Methods Based on Sinc and AnalyticFunctions Springer New York NY USA 1993

[11] A Boumenir ldquoHigher approximation of eigenvalues by thesampling methodrdquo BIT Numerical Mathematics vol 40 no 2pp 215ndash225 2000

[12] A Boumenir ldquoSampling and eigenvalues of non-self-adjointSturm-Liouville problemsrdquo SIAM Journal on Scientific Comput-ing vol 23 no 1 pp 219ndash229 2001

[13] M M Tharwat A H Bhrawy and A Yildirim ldquoNumericalcomputation of eigenvalues of discontinuous Sturm-Liouvilleproblems with parameter dependent boundary conditionsusing sinc methodrdquoNumerical Algorithms vol 63 no 1 pp 27ndash48 2013

[14] M M Tharwat A H Bhrawy and A Yildirim ldquoNumericalcomputation of the eigenvalues of a discontinuousDirac systemusing the sincmethodwith error analysisrdquo International Journalof Computer Mathematics vol 89 no 15 pp 2061ndash2080 2012

[15] P L Butzer and R L Stens ldquoA modification of the Whittaker-Kotelrsquonikov-Shannon sampling seriesrdquo Aequationes Mathemat-icae vol 28 no 3 pp 305ndash311 1985

[16] R Gervais Q I Rahman and G Schmeisser ldquoA bandlimitedfunction simulating a duration-limited onerdquo in ApproximationTheory andFunctionalAnalysis P L Butzer andR L Stens Edspp 355ndash362 Birkhaauser Boston Mass USA 1984

[17] R L Stens ldquoSampling by generalized kernelsrdquo in SamplingThe-ory in Fourier and Signal Analysis Advanced Topics J RHigginsand R L Stens Eds pp 130ndash157 Oxford University PressOxford UK 1999

[18] G Schmeisser and F Stenger ldquoSinc approximation with a Gaus-sianmultiplierrdquo SamplingTheory in Signal and Image Processingvol 6 no 2 pp 199ndash221 2007

[19] L Qian ldquoOn the regularized Whittaker-Kotelrsquonikov-Shannonsampling formulardquo Proceedings of the American MathematicalSociety vol 131 no 4 pp 1169ndash1176 2003

[20] L Qian and D B Creamer ldquoA modification of the samplingseries with aGaussianmultiplierrdquo SamplingTheory in Signal andImage Processing vol 5 no 1 pp 1ndash20 2006

[21] L Qian and D B Creamer ldquoLocalized sampling in the presenceof noiserdquoAppliedMathematics Letters vol 19 no 4 pp 351ndash3552006

[22] M H Annaby and R M Asharabi ldquoComputing eigenvaluesof boundary-value problems using sinc-Gaussian methodrdquoSamplingTheory in Signal and Image Processing vol 7 no 3 pp293ndash312 2008

[23] M H Annaby and M MTharwat ldquoA sinc-Gaussian techniquefor computing eigenvalues of second-order linear pencilsrdquoApplied Numerical Mathematics vol 63 pp 129ndash137 2013

[24] A H Bhrawy M M Tharwat and A Al-Fhaid ldquoNumericalalgorithms for computing eigenvalues of discontinuous Diracsystem using sinc-Gaussian methodrdquo Abstract and AppliedAnalysis vol 2012 Article ID 925134 13 pages 2012

[25] M M Tharwat A H Bhrawy and A S Alofi ldquoApproximationof eigenvalues of discontinuous sturm-liouville problems witheigenparameter in all boundary conditionsrdquo Boundary ValueProblems vol 2013 article 132 2013

[26] M M Tharwat and S M Al-Harbi ldquoApproximation of eigen-values of boundary value problemsrdquo Boundary Value Problemsvol 2014 article 51 2014

[27] MMTharwat and A H Bhrawy ldquoComputation of eigenvaluesof discontinuous Dirac system using Hermite interpolationtechniquerdquo Advances in Difference Equations vol 2012 article59 2012

[28] M M Tharwat A Yildirim and A H Bhrawy ldquoSampling ofdiscontinuous Dirac systemsrdquo Numerical Functional Analysisand Optimization vol 34 no 3 pp 323ndash348 2013

[29] M M Tharwat ldquoDiscontinuous Sturm-Liouville problems andassociated sampling theoriesrdquo Abstract and Applied Analysisvol 2011 Article ID 610232 30 pages 2011

[30] MMTharwat ldquoOn sampling theories and discontinuousDiracsystems with eigenparameter in the boundary conditionsrdquoBoundary Value Problems vol 2013 article 65 2013

[31] B M Levitan and I S Sargsjan ldquoIntroduction to spectral the-ory self adjoint ordinary differential operatorsrdquo in Translationof Mthematical Monographs vol 39 American MathematicalSociety Providence RI USA 1975

[32] B M Levitan and I S Sargsjan Sturm-Liouville and Dirac oper-ators vol 59 Kluwer Academic Publishers Group DordrechtThe Netherlands 1991

Submit your manuscripts athttpwwwhindawicom

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical Problems in Engineering

Hindawi Publishing Corporationhttpwwwhindawicom

Differential EquationsInternational Journal of

Volume 2014

Applied MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Probability and StatisticsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical PhysicsAdvances in

Complex AnalysisJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

OptimizationJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

CombinatoricsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Operations ResearchAdvances in

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Function Spaces

Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of Mathematics and Mathematical Sciences

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

The Scientific World JournalHindawi Publishing Corporation httpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Algebra

Discrete Dynamics in Nature and Society

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Decision SciencesAdvances in

Discrete MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom

Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Stochastic AnalysisInternational Journal of

Abstract and Applied Analysis 7

Proof Since y01

(119909 120582) = T01

12059312(119909 120582) +

T02

12059322(119909 120582) then

from (49) and (50) we obtain

y01

(119909 120582) =

1

120575

12059311

(0minus 120582)T

01cos (120582119909)

minus

1

120575

12059321

(0minus 120582)T

01sin (120582119909)

minusT01yminus12

(119909 120582) +

1

120575

12059311

(0minus 120582)

T02

sin (120582119909)

+

1

120575

12059321

(0minus 120582)

T02

cos (120582119909) + T02yminus12

(119909 120582)

(60)

Then from (47) and (48) and Lemma 4 we get

1003816100381610038161003816y01

(119909 120582)1003816100381610038161003816le

1

|120575|

100381610038161003816100381612059311

(0minus 120582)

1003816100381610038161003816

1003816100381610038161003816T01

cos (120582119909)1003816100381610038161003816

+

1

|120575|

100381610038161003816100381612059321

(0minus 120582)

1003816100381610038161003816

1003816100381610038161003816T01

sin (120582119909)1003816100381610038161003816

+1003816100381610038161003816T01yminus12

(119909 120582)1003816100381610038161003816

+

1

|120575|

100381610038161003816100381612059311

(0minus 120582)

1003816100381610038161003816

10038161003816100381610038161003816

T02

sin (120582119909)

10038161003816100381610038161003816

+

1

|120575|

100381610038161003816100381612059321

(0minus 120582)

1003816100381610038161003816

10038161003816100381610038161003816

T02

cos (120582119909)10038161003816100381610038161003816

+

10038161003816100381610038161003816

T02yminus12

(119909 120582)

10038161003816100381610038161003816

le 21198884(1 + |120582|)

times (1198883+

2

|120575|

(1198881+ 1198883)) 119890|I120582|(119909+1)

= 211988841198885(1 + |120582|) 119890

|I120582|(119909+1)

(61)

Similarly we can prove that1003816100381610038161003816y02

(119909 120582)1003816100381610038161003816le 211988841198885(1 + |120582|) 119890

|I120582|(119909+1) (62)

3 The Numerical Scheme

In this section we derive the method of computing eigen-values of problems (12)ndash(15) numerically The basic idea ofthe scheme is to split Ω(120582) into two parts a known partK(120582) and an unknown oneU(120582)Thenwe approximateU(120582)

using (5) to get the approximate Ω(120582) and then compute theapproximate zeros We first splitΩ(120582) into two parts

Ω (120582) = K (120582) +U (120582) (63)

whereU(120582) is the unknown part involving integral operators

U (120582) = 120575 [(1205732+ 120582 cos 120579

2) sin 120582

minus (1205731+ 120582 sin 120579

2) cos 120582] y

minus11(0minus 120582)

minus 120575 [(1205731+ 120582 sin 120579

2) sin 120582

+ (1205732+ 120582 cos 120579

2) cos 120582] y

minus12(0minus 120582)

minus 1205752[(1205731+ 120582 sin 120579

2) y01

(1 120582)

+ (1205732+ 120582 cos 120579

2) y02

(1 120582)]

(64)

andK(120582) is the known part

K (120582) = 120575 (1205731+ 120582 sin 120579

2)

times [1205722cos 2120582 minus 120572

1sin 2120582 + 120582 cos (2120582 + 120579

1)]

minus 120575 (1205732+ 120582 cos 120579

2)

times [1205722sin 2120582 + 120572

1cos 2120582 + 120582 sin (2120582 + 120579

1)]

(65)

Then from Lemmas 4 and 5 we have the following result

Lemma 6 The function U(120582) is entire in 120582 and the followingestimate holds

|U (120582)| le 120601 (120582) 1198902|I120582|

(66)

where

120601 (120582) = 119872(1 + |120582|)2

119872 = 4 |120575| 1198886 (21198883 + |120575| 11988841198885)

(67)

Proof From (64) we have

|U (120582)| le |120575| [(10038161003816100381610038161205732

1003816100381610038161003816+ |120582|

1003816100381610038161003816cos 1205792

1003816100381610038161003816) |sin 120582|

+ (10038161003816100381610038161205731

1003816100381610038161003816+ |120582|

1003816100381610038161003816sin 1205792

1003816100381610038161003816) |cos 120582|] 100381610038161003816

1003816yminus11

(0minus 120582)

1003816100381610038161003816

+ |120575| [(10038161003816100381610038161205731

1003816100381610038161003816+ |120582|

1003816100381610038161003816sin 1205792

1003816100381610038161003816) |sin 120582|

+ (10038161003816100381610038161205732

1003816100381610038161003816+1003816100381610038161003816120582 cos 120579

2

1003816100381610038161003816) |cos 120582|] 100381610038161003816

1003816yminus12

(0minus 120582)

1003816100381610038161003816

+ 1205752[(10038161003816100381610038161205731

1003816100381610038161003816+ |120582|

1003816100381610038161003816sin 1205792

1003816100381610038161003816)1003816100381610038161003816y01

(1 120582)1003816100381610038161003816

+ (10038161003816100381610038161205732

1003816100381610038161003816+ |120582|

1003816100381610038161003816cos 1205792

1003816100381610038161003816)1003816100381610038161003816y02

(1 120582)1003816100381610038161003816]

(68)

Using the inequalities | sin 120582| le 119890|I120582| and | cos 120582| le 119890

|I120582| for120582 isin C and Lemmas 4 and 5 imply (66)

ThusU(120582) is an entire function of exponential type 120590 = 2In the following we let 120582 isin R since all eigenvalues are realNow we approximate the function U(120582) using the operator(5) where ℎ isin (0 1205872) and 120596 = (120587 minus 2ℎ)2 and then from(7) we obtain

1003816100381610038161003816U (120582) minus (G

ℎ119873U) (120582)

1003816100381610038161003816le 119879ℎ119873

(120582) (69)

where

119879ℎ119873

(120582) = 2

10038161003816100381610038161003816sin (ℎ

minus1120587120582)

10038161003816100381610038161003816

times 120601 (|R120582| + ℎ (119873 + 1))

119890minus120596119873

radic120587120596119873

120573119873(0) 120582 isin R

(70)

The samplesU(119899ℎ) = Ω(119899ℎ) minusK(119899ℎ) 119899 isin Z119873(120582) cannot be

computed explicitly in the general caseWe approximate these

8 Abstract and Applied Analysis

samples numerically by solving the initial-value problemsdefined by (12) and (45) to obtain the approximate valuesU(119899ℎ) 119899 isin Z

119873(120582) that is U(119899ℎ) = Ω(119899ℎ) minus K(119899ℎ)

Here we use a computer algebra system mathematica toobtain the approximate solutions with the required accuracyHowever a separate study for the effect of different numericalschemes and the computational costs would be interestingAccordingly we have the explicit expansion

(Gℎ119873

U) (120582) = sum

119899isinZ119873(120582)

U (119899ℎ) sinc (ℎminus1120587119909 minus 119899120587)

times 119866(

radic120596 (120582 minus 119899ℎ)

radic119873ℎ

)

(71)

Therefore we get cf (10)10038161003816100381610038161003816(Gℎ119873

U) (120582) minus (Gℎ119873

U) (120582)

10038161003816100381610038161003816le 119860120576119873

(0) 120582 isin R (72)

Now let Ω119873(120582) = K(120582) + (G

ℎ119873U)(120582) From (69) and (72)

we obtain10038161003816100381610038161003816Ω (120582) minus Ω

119873(120582)

10038161003816100381610038161003816le 119879ℎ119873

(120582) + 119860120576119873

(0) 120582 isin R (73)

Let 120582lowast be an eigenvalue and 120582119873its desired approximation

that is Ω(120582lowast) = 0 and Ω

119873(120582119873) = 0 From (73) we have

|Ω119873(120582lowast)| le 119879

ℎ119873(120582lowast) + 119860120576119873

(0) Define the curves

119886plusmn(120582) = Ω

119873(120582) plusmn 119879

ℎ119873(120582) + 119860

120576119873(0) (74)

The curves 119886+(120582) 119886minus(120582) enclose the curve ofΩ(120582) for suitably

large 119873 Hence the closure interval is determined by solving119886plusmn(120582) = 0 giving an interval

119868120576119873

= [119886minus 119886+] (75)

It is worthwhile to mention that the simplicity of the eigen-values guarantees the existence of approximate eigenvaluesthat is the 120582

119873rsquos for which Ω

119873(120582119873) = 0 Next we estimate the

error |120582lowast minus 120582119873| for the eigenvalue 120582lowast

Theorem 7 Let 120582lowast be an eigenvalue of (12)ndash(15) and let 120582119873

be its approximation Then for 120582 isin R we have the followingestimate

1003816100381610038161003816120582lowastminus 120582119873

1003816100381610038161003816lt

119879ℎ119873

(120582119873) + 119860120576119873

(0)

inf120577isin119868120576119873

1003816100381610038161003816Ω1015840(120577)

1003816100381610038161003816

(76)

where the interval 119868120576119873

is defined above

Proof Replacing 120582 by 120582119873in (73) we obtain

1003816100381610038161003816Ω (120582119873) minus Ω (120582

lowast)1003816100381610038161003816lt 119879ℎ119873

(120582119873) + 119860120576119873

(0) (77)

where we have used Ω119873(120582119873) = Ω(120582

lowast) = 0 Using the

mean value theorem yields that for some 120577 isin 119869120576119873

=

[min(120582lowast 120582119873)max(120582lowast 120582

119873)]

10038161003816100381610038161003816(120582lowastminus 120582119873)Ω1015840(120577)

10038161003816100381610038161003816le 119879ℎ119873

(120582119873) + 119860120576119873

(0) 120577 isin 119869120576119873

sub 119868120576119873

(78)

Since 120582lowast is simple and 119873 is sufficiently large then

inf120577isin119868120576119873

|Ω1015840(120577)| gt 0 and we get (76)

4 Numerical Examples

In this section we introduce two examples illustrating the theabove method Also in the following examples we observethat the exact solutions 120582

119896are all inside the interval [119886

minus 119886+]

In these two examples we indicate the effect of the amplitudeerror in the method by determining enclosure intervals fordifferent values of 120576 We also indicate the effect of 119873 and ℎ

by several choices The eigenvalues of the following examplescannot computed concretely then we use mathematicato obtain the exact values mathematica is also used inrounding off the exact eigenvalues which are square rootsBoth numerical results and the associated figures prove thecredibility of the method

Example 1 Consider the system

1199101015840

2(119909) minus 119901 (119909) 119910

1(119909) = 120582119910

1(119909)

1199101015840

1(119909) + 119901 (119909) 119910

2(119909) = minus120582119910

2(119909)

119909 isin [minus1 0) cup (0 1]

1205821199101(minus1) minus (radic2 + 120582) 119910

2(minus1) = 0

(radic2 + 120582) 1199101(1) minus 120582119910

2(1) = 0

1199101(0minus) minus 2119910

1(0+) = 0 119910

2(0minus) minus 2119910

2(0+) = 0

(79)

Here

1199011(119909) = 119901

2(119909) = 119903 (119909) =

119909 119909 isin [minus1 0)

1 (0 1]

(80)

1205791= 1205792= 1205874 120572

1= 1205732= 0 120572

2= 1205731= 1 and 120575 = 2 Direct

calculations give

K (120582) = 2 ((radic2120582 + 1) cos (2120582) minus 120582 (120582 + radic2) sin (2120582))

Ω (120582) = 2 ((radic2120582 + 1) cos(2120582 +

1

2

)

minus120582 (120582 + radic2) sin(2120582 +

1

2

))

(81)

As is clearly seen the eigenvalues cannot be computedexplicitly Tables 1 2 and 3 indicate the application of ourtechnique to this problem and the effect of 120576 By exact wemean the zeros of Δ(120582) computed by Mathematica

Figures 1 and 2 illustrate the enclosure intervals domi-nating 120582

0for 119873 = 25 ℎ = 01 and 120576 = 10

minus2 120576 = 10minus5

respectively The middle curve represents Δ(120582) while theupper and lower curves represent the curves of 119886

+(120582) 119886minus(120582)

respectively We notice that when 120576 = 10minus5 the two curves

are almost identical Similarly Figures 3 and 4 illustrate theenclosure intervals dominating 120582

1for ℎ = 01 119873 = 25 and

120576 = 10minus2 120576 = 10

minus5 respectively

Abstract and Applied Analysis 9

Table 1 The approximation 120582119896119873

and the exact solutions 120582119896for different choices of ℎ and119873

120582119896

120582minus2

120582minus1

1205820

1205821

Exact 120582119896

minus12061856254879546 minus05151195632138896 034274095892547224 1616237903407542

120582119896119873

ℎ = 06 119873 = 15 minus12061855965725146 minus05151190347168934 034274053593937165 1616237692038969120596 = 09708 119873 = 25 minus12061856254843495 minus05151195631493376 0342740958869788 16162379033820231ℎ = 01 119873 = 15 minus12061856254844203 minus05151195632158179 03427409589350932 16162379034134882

120596 = 14708 119873 = 25 minus12061856254879548 minus05151195632138894 034274095892547236 16162379034075425

Table 2 Absolute error |120582119896minus 120582119896119873

|

120582119896

120582minus2

120582minus1

1205820

1205821

ℎ = 06119873 = 15 289154 times 10

minus8528497 times 10

minus7422986 times 10

minus7211369 times 10

minus7

119873 = 25 360512 times 10minus12

64552 times 10minus11

556842 times 10minus11

255189 times 10minus11

ℎ = 01119873 = 15 353428 times 10

minus12192824 times 10

minus12962097 times 10

minus12594613 times 10

minus12

119873 = 25 222045 times 10minus16

222045 times 10minus16

555112 times 10minus17

444089 times 10minus16

Table 3 For119873 = 25 and ℎ = 01 the exact solutions 120582119896are all inside the interval [119886

minus 119886+] for different values of 120576

120582119896

120582minus2

120582minus1

1205820

1205821

Exact 120582119896

minus12061856254879546 minus05151195632138896 034274095892547224 1616237903407542119868120576119873

120576 = 10minus2

[minus12335696 minus11771922] [minus05486790 minus04826816] [03246478 03601346] [16100119 16224066]

119868120576119873

120576 = 10minus5

[minus12062137 minus12061575] [minus05151525 minus05150867] [03427232 03427586] [16162317 16162441]

030 032 034 036 038

00

02

04

aminus

a+

Ω(120582)

minus02

minus04

minus06

Figure 1The enclosure interval dominating 1205820for ℎ = 01119873 = 25

and 120576 = 10minus2

Example 2 In this example we consider the system

1199101015840

2(119909) minus 119901 (119909) 119910

1(119909) = 120582119910

1(119909)

1199101015840

1(119909) + 119901 (119909) 119910

2(119909) = minus120582119910

2(119909)

119909 isin [minus1 0) cup (0 1]

radic31199101(minus1) minus 119910

2(minus1) = 0

(1 +

1

2

120582)1199101(1) minus (1 +

radic3

2

120582)1199102(1) = 0

1199101(0minus) minus 3119910

1(0+) = 0 119910

2(0minus) minus 3119910

2(0+) = 0

(82)

030 032 034 036 038

00

02

aminus

a+

Ω(120582)

minus02

minus04

Figure 2The enclosure interval dominating 1205820for ℎ = 01119873 = 25

and 120576 = 10minus5

where

1199011(119909) = 119901

2(119909) = 119901 (119909) =

119909 119909 isin [minus1 0)

1199092 (0 1]

(83)

1205791= 1205792= 0 120572

1= 1205732= 1 120572

2= 1205731= 2 and 120575 = 12 Direct

calculations give

K (120582) =

1

2

((120582 + 3) cos (2120582) minus (1205822+ 3120582 + 4) sin (2120582))

Ω (120582) =

1

2

((1205822+ 3120582 + 4) sin(

1

6

minus 2120582)

+ (120582 + 3) cos(1

6

minus 2120582))

(84)

10 Abstract and Applied Analysis

Table 4 The approximation 120582119896119873

and the exact solutions 120582119896for different choices of ℎ and119873

120582119896

120582minus2

120582minus1

1205820

1205821

Exact 120582119896

minus30661256186938894 minus10959703083466883 03689030343814017 18338437675275228

120582119896119873

ℎ = 08 119873 = 15 minus3066123822200742 minus10959640944020903 036890463046491767 18338437607680411120596 = 007708 119873 = 25 minus30661256172349627 minus10959703026494807 036890303588091056 18338437676561417

ℎ = 02 119873 = 15 minus30661256186844446 minus10959703083278305 03689030343916001 1833843767517251120596 = 13708 119873 = 25 minus306612561869389 minus1095970308346688 03689030343814024 18338437675275225

Table 5 Absolute error |120582119896minus 120582119896119873

|

120582119896

120582minus2

120582minus1

1205820

1205821

ℎ = 08119873 = 15 179649 times 10

minus6621394 times 10

minus6159608 times 10

minus6675948 times 10

minus9

119873 = 25 145893 times 10minus9

569721 times 10minus9

149951 times 10minus9

128619 times 10minus10

ℎ = 02119873 = 15 944489 times 10

minus12188578 times 10

minus12101984 times 10

minus11102718 times 10

minus11

119873 = 25 444089 times 10minus16

222045 times 10minus16

666134 times 10minus16

222045 times 10minus16

Table 6 For119873 = 25 and ℎ = 02 the exact solutions 120582119896are all inside the interval [119886

minus 119886+] for different values of 120576

120582119896

120582minus2

120582minus1

1205820

1205821

Exact 120582119896

minus30661256186938894 minus10959703083466883 03689030343814017 18338437675275228119868120576119873

120576 = 10minus2

[minus31098742 minus30188665] [minus11631502 minus10339754] [03437434 03933486] [18220764 18454828]

119868120576119873

120576 = 10minus5

[minus30661709 minus30660803] [minus10960346 minus10959060] [03688783 03689278] [18338321 18338555]

158 159 160 161 162 163 164 165

00

05

10

minus05

minus10

aminus

a+

Ω(120582)

Figure 3The enclosure interval dominating 1205821for ℎ = 01119873 = 25

and 120576 = 10minus2

Tables 4 and 6 give the exact eigenvalues 1205821198961

119896=minus2and their

approximate ones for different values of ℎ 119873 and 120576 InTable 5 we give the absolute error for different values of ℎ and119873

Here Figures 5 6 7 and 8 illustrate the enclosureintervals dominating 120582

minus2and 120582

minus1for ℎ = 02 119873 = 25 and

120576 = 10minus2 120576 = 10

minus5 respectively

158 159 160 161 162 163 164 165

00

05

minus05

aminus

a+

Ω(120582)

Figure 4The enclosure interval dominating 1205821for ℎ = 01119873 = 25

and 120576 = 10minus5

5 Conclusion

With a simple analysis and with values of solutions ofinitial value problems computed at a few values of theeigenparameter we have computed the eigenvalues of aDirac system which has a discontinuity at one point andcontains a spectral parameter in all boundary conditionswith a certain estimated error The method proposed is

Abstract and Applied Analysis 11

00

05

10

minus05

minus33 minus32 minus31 minus30 minus29 minus28

aminus

a+

Ω(120582)

Figure 5 The enclosure interval dominating 120582minus2

for ℎ = 02 119873 =

25 and 120576 = 10minus2

00

05

10

minus05

minus33 minus32 minus31 minus30 minus29 minus28

aminus

a+

Ω(120582)

Figure 6 The enclosure interval dominating 120582minus2

for ℎ = 02 119873 =

25 and 120576 = 10minus5

a shooting procedure that is the problem is reformulatedas two initial value ones due to the interior discontinuity ofsize two and a miss-distance is defined at the right end of theinterval of integration whose roots are the eigenvalues to becomputed The unknown partU(120582) of the miss-distance canbewritten in terms of a functionwhich is an entire function ofexponential typeTherefore we propose to approximate suchterm by means of a truncated cardinal series with samplingvalues approximated by solving numerically correspondingsuitable initial value problems Finally in Section 4 weintroduced two instructive examples where both numericalresults and the associated figures have proved the credibilityof the method

00

05

10

minus05

minus14 minus13 minus12 minus11 minus10 minus09 minus08 minus07

aminus

a+

Ω(120582)

Figure 7 The enclosure interval dominating 120582minus1

for ℎ = 02 119873 =

25 and 120576 = 10minus2

00

05

10

minus05

minus14 minus13 minus12 minus11 minus10 minus09 minus08 minus07

aminus

a+

Ω(120582)

Figure 8 The enclosure interval dominating 120582minus1

for ℎ = 02 119873 =

25 and 120576 = 10minus5

Conflict of Interests

The authors declare that there is no conflict of interestsregarding the publication of this paper

Acknowledgment

This paper was funded by the Deanship of Scientific Research(DSR) King Abdulaziz University Jeddah The authorstherefore acknowledge with thanks DSR technical and finan-cial support

12 Abstract and Applied Analysis

References

[1] C T Fulton ldquoTwo-point boundary value problems witheigenvalue parameter contained in the boundary conditionsrdquoProceedings of the Royal Society of Edinburgh A Mathematicsvol 77 no 3-4 pp 293ndash308 1977

[2] JWalter ldquoRegular eigenvalue problems with eigenvalue param-eter in the boundary conditionrdquoMathematische Zeitschrift vol133 pp 301ndash312 1973

[3] V Kotelnikov ldquoOn the carrying capacity of the ether and wirein telecommunications material for the first All-Unionrdquo inConference on Questions of Communications vol 55 pp 55ndash64Izd Red Upr Svyazi RKKA 1933

[4] C E Shannon ldquoCommunication in the presence of noiserdquoProceedings of the IRE vol 37 pp 10ndash21 1949

[5] E TWhittaker ldquoOn the functions which are represented by theexpansion of the interpolation theoryrdquo Proceedings of the RoyalSociety of Edinburgh A vol 35 pp 181ndash194 1915

[6] P L Butzer G Schmeisser and R L Stens ldquoAn introductionto sampling analysisrdquo in Non Uniform Sampling Theory andPractices F Marvasti Ed pp 17ndash121 Kluwer New York NYUSA 2001

[7] M A Kowalski K A Sikorski and F Stenger Selected Topics inApproximation andComputation OxfordUniversity Press NewYork NY USA 1995

[8] J Lund and K L Bowers Sinc Methods for Quadrature andDifferential Equations SIAM Philadelphia Pa USA 1992

[9] F Stenger ldquoNumericalmethods based onWhittaker cardinal orsinc functionsrdquo SIAM Review vol 23 no 2 pp 165ndash224 1981

[10] F Stenger Numerical Methods Based on Sinc and AnalyticFunctions Springer New York NY USA 1993

[11] A Boumenir ldquoHigher approximation of eigenvalues by thesampling methodrdquo BIT Numerical Mathematics vol 40 no 2pp 215ndash225 2000

[12] A Boumenir ldquoSampling and eigenvalues of non-self-adjointSturm-Liouville problemsrdquo SIAM Journal on Scientific Comput-ing vol 23 no 1 pp 219ndash229 2001

[13] M M Tharwat A H Bhrawy and A Yildirim ldquoNumericalcomputation of eigenvalues of discontinuous Sturm-Liouvilleproblems with parameter dependent boundary conditionsusing sinc methodrdquoNumerical Algorithms vol 63 no 1 pp 27ndash48 2013

[14] M M Tharwat A H Bhrawy and A Yildirim ldquoNumericalcomputation of the eigenvalues of a discontinuousDirac systemusing the sincmethodwith error analysisrdquo International Journalof Computer Mathematics vol 89 no 15 pp 2061ndash2080 2012

[15] P L Butzer and R L Stens ldquoA modification of the Whittaker-Kotelrsquonikov-Shannon sampling seriesrdquo Aequationes Mathemat-icae vol 28 no 3 pp 305ndash311 1985

[16] R Gervais Q I Rahman and G Schmeisser ldquoA bandlimitedfunction simulating a duration-limited onerdquo in ApproximationTheory andFunctionalAnalysis P L Butzer andR L Stens Edspp 355ndash362 Birkhaauser Boston Mass USA 1984

[17] R L Stens ldquoSampling by generalized kernelsrdquo in SamplingThe-ory in Fourier and Signal Analysis Advanced Topics J RHigginsand R L Stens Eds pp 130ndash157 Oxford University PressOxford UK 1999

[18] G Schmeisser and F Stenger ldquoSinc approximation with a Gaus-sianmultiplierrdquo SamplingTheory in Signal and Image Processingvol 6 no 2 pp 199ndash221 2007

[19] L Qian ldquoOn the regularized Whittaker-Kotelrsquonikov-Shannonsampling formulardquo Proceedings of the American MathematicalSociety vol 131 no 4 pp 1169ndash1176 2003

[20] L Qian and D B Creamer ldquoA modification of the samplingseries with aGaussianmultiplierrdquo SamplingTheory in Signal andImage Processing vol 5 no 1 pp 1ndash20 2006

[21] L Qian and D B Creamer ldquoLocalized sampling in the presenceof noiserdquoAppliedMathematics Letters vol 19 no 4 pp 351ndash3552006

[22] M H Annaby and R M Asharabi ldquoComputing eigenvaluesof boundary-value problems using sinc-Gaussian methodrdquoSamplingTheory in Signal and Image Processing vol 7 no 3 pp293ndash312 2008

[23] M H Annaby and M MTharwat ldquoA sinc-Gaussian techniquefor computing eigenvalues of second-order linear pencilsrdquoApplied Numerical Mathematics vol 63 pp 129ndash137 2013

[24] A H Bhrawy M M Tharwat and A Al-Fhaid ldquoNumericalalgorithms for computing eigenvalues of discontinuous Diracsystem using sinc-Gaussian methodrdquo Abstract and AppliedAnalysis vol 2012 Article ID 925134 13 pages 2012

[25] M M Tharwat A H Bhrawy and A S Alofi ldquoApproximationof eigenvalues of discontinuous sturm-liouville problems witheigenparameter in all boundary conditionsrdquo Boundary ValueProblems vol 2013 article 132 2013

[26] M M Tharwat and S M Al-Harbi ldquoApproximation of eigen-values of boundary value problemsrdquo Boundary Value Problemsvol 2014 article 51 2014

[27] MMTharwat and A H Bhrawy ldquoComputation of eigenvaluesof discontinuous Dirac system using Hermite interpolationtechniquerdquo Advances in Difference Equations vol 2012 article59 2012

[28] M M Tharwat A Yildirim and A H Bhrawy ldquoSampling ofdiscontinuous Dirac systemsrdquo Numerical Functional Analysisand Optimization vol 34 no 3 pp 323ndash348 2013

[29] M M Tharwat ldquoDiscontinuous Sturm-Liouville problems andassociated sampling theoriesrdquo Abstract and Applied Analysisvol 2011 Article ID 610232 30 pages 2011

[30] MMTharwat ldquoOn sampling theories and discontinuousDiracsystems with eigenparameter in the boundary conditionsrdquoBoundary Value Problems vol 2013 article 65 2013

[31] B M Levitan and I S Sargsjan ldquoIntroduction to spectral the-ory self adjoint ordinary differential operatorsrdquo in Translationof Mthematical Monographs vol 39 American MathematicalSociety Providence RI USA 1975

[32] B M Levitan and I S Sargsjan Sturm-Liouville and Dirac oper-ators vol 59 Kluwer Academic Publishers Group DordrechtThe Netherlands 1991

Submit your manuscripts athttpwwwhindawicom

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

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Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical Problems in Engineering

Hindawi Publishing Corporationhttpwwwhindawicom

Differential EquationsInternational Journal of

Volume 2014

Applied MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Probability and StatisticsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical PhysicsAdvances in

Complex AnalysisJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

OptimizationJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

CombinatoricsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Operations ResearchAdvances in

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Function Spaces

Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of Mathematics and Mathematical Sciences

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

The Scientific World JournalHindawi Publishing Corporation httpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Algebra

Discrete Dynamics in Nature and Society

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Decision SciencesAdvances in

Discrete MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom

Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Stochastic AnalysisInternational Journal of

8 Abstract and Applied Analysis

samples numerically by solving the initial-value problemsdefined by (12) and (45) to obtain the approximate valuesU(119899ℎ) 119899 isin Z

119873(120582) that is U(119899ℎ) = Ω(119899ℎ) minus K(119899ℎ)

Here we use a computer algebra system mathematica toobtain the approximate solutions with the required accuracyHowever a separate study for the effect of different numericalschemes and the computational costs would be interestingAccordingly we have the explicit expansion

(Gℎ119873

U) (120582) = sum

119899isinZ119873(120582)

U (119899ℎ) sinc (ℎminus1120587119909 minus 119899120587)

times 119866(

radic120596 (120582 minus 119899ℎ)

radic119873ℎ

)

(71)

Therefore we get cf (10)10038161003816100381610038161003816(Gℎ119873

U) (120582) minus (Gℎ119873

U) (120582)

10038161003816100381610038161003816le 119860120576119873

(0) 120582 isin R (72)

Now let Ω119873(120582) = K(120582) + (G

ℎ119873U)(120582) From (69) and (72)

we obtain10038161003816100381610038161003816Ω (120582) minus Ω

119873(120582)

10038161003816100381610038161003816le 119879ℎ119873

(120582) + 119860120576119873

(0) 120582 isin R (73)

Let 120582lowast be an eigenvalue and 120582119873its desired approximation

that is Ω(120582lowast) = 0 and Ω

119873(120582119873) = 0 From (73) we have

|Ω119873(120582lowast)| le 119879

ℎ119873(120582lowast) + 119860120576119873

(0) Define the curves

119886plusmn(120582) = Ω

119873(120582) plusmn 119879

ℎ119873(120582) + 119860

120576119873(0) (74)

The curves 119886+(120582) 119886minus(120582) enclose the curve ofΩ(120582) for suitably

large 119873 Hence the closure interval is determined by solving119886plusmn(120582) = 0 giving an interval

119868120576119873

= [119886minus 119886+] (75)

It is worthwhile to mention that the simplicity of the eigen-values guarantees the existence of approximate eigenvaluesthat is the 120582

119873rsquos for which Ω

119873(120582119873) = 0 Next we estimate the

error |120582lowast minus 120582119873| for the eigenvalue 120582lowast

Theorem 7 Let 120582lowast be an eigenvalue of (12)ndash(15) and let 120582119873

be its approximation Then for 120582 isin R we have the followingestimate

1003816100381610038161003816120582lowastminus 120582119873

1003816100381610038161003816lt

119879ℎ119873

(120582119873) + 119860120576119873

(0)

inf120577isin119868120576119873

1003816100381610038161003816Ω1015840(120577)

1003816100381610038161003816

(76)

where the interval 119868120576119873

is defined above

Proof Replacing 120582 by 120582119873in (73) we obtain

1003816100381610038161003816Ω (120582119873) minus Ω (120582

lowast)1003816100381610038161003816lt 119879ℎ119873

(120582119873) + 119860120576119873

(0) (77)

where we have used Ω119873(120582119873) = Ω(120582

lowast) = 0 Using the

mean value theorem yields that for some 120577 isin 119869120576119873

=

[min(120582lowast 120582119873)max(120582lowast 120582

119873)]

10038161003816100381610038161003816(120582lowastminus 120582119873)Ω1015840(120577)

10038161003816100381610038161003816le 119879ℎ119873

(120582119873) + 119860120576119873

(0) 120577 isin 119869120576119873

sub 119868120576119873

(78)

Since 120582lowast is simple and 119873 is sufficiently large then

inf120577isin119868120576119873

|Ω1015840(120577)| gt 0 and we get (76)

4 Numerical Examples

In this section we introduce two examples illustrating the theabove method Also in the following examples we observethat the exact solutions 120582

119896are all inside the interval [119886

minus 119886+]

In these two examples we indicate the effect of the amplitudeerror in the method by determining enclosure intervals fordifferent values of 120576 We also indicate the effect of 119873 and ℎ

by several choices The eigenvalues of the following examplescannot computed concretely then we use mathematicato obtain the exact values mathematica is also used inrounding off the exact eigenvalues which are square rootsBoth numerical results and the associated figures prove thecredibility of the method

Example 1 Consider the system

1199101015840

2(119909) minus 119901 (119909) 119910

1(119909) = 120582119910

1(119909)

1199101015840

1(119909) + 119901 (119909) 119910

2(119909) = minus120582119910

2(119909)

119909 isin [minus1 0) cup (0 1]

1205821199101(minus1) minus (radic2 + 120582) 119910

2(minus1) = 0

(radic2 + 120582) 1199101(1) minus 120582119910

2(1) = 0

1199101(0minus) minus 2119910

1(0+) = 0 119910

2(0minus) minus 2119910

2(0+) = 0

(79)

Here

1199011(119909) = 119901

2(119909) = 119903 (119909) =

119909 119909 isin [minus1 0)

1 (0 1]

(80)

1205791= 1205792= 1205874 120572

1= 1205732= 0 120572

2= 1205731= 1 and 120575 = 2 Direct

calculations give

K (120582) = 2 ((radic2120582 + 1) cos (2120582) minus 120582 (120582 + radic2) sin (2120582))

Ω (120582) = 2 ((radic2120582 + 1) cos(2120582 +

1

2

)

minus120582 (120582 + radic2) sin(2120582 +

1

2

))

(81)

As is clearly seen the eigenvalues cannot be computedexplicitly Tables 1 2 and 3 indicate the application of ourtechnique to this problem and the effect of 120576 By exact wemean the zeros of Δ(120582) computed by Mathematica

Figures 1 and 2 illustrate the enclosure intervals domi-nating 120582

0for 119873 = 25 ℎ = 01 and 120576 = 10

minus2 120576 = 10minus5

respectively The middle curve represents Δ(120582) while theupper and lower curves represent the curves of 119886

+(120582) 119886minus(120582)

respectively We notice that when 120576 = 10minus5 the two curves

are almost identical Similarly Figures 3 and 4 illustrate theenclosure intervals dominating 120582

1for ℎ = 01 119873 = 25 and

120576 = 10minus2 120576 = 10

minus5 respectively

Abstract and Applied Analysis 9

Table 1 The approximation 120582119896119873

and the exact solutions 120582119896for different choices of ℎ and119873

120582119896

120582minus2

120582minus1

1205820

1205821

Exact 120582119896

minus12061856254879546 minus05151195632138896 034274095892547224 1616237903407542

120582119896119873

ℎ = 06 119873 = 15 minus12061855965725146 minus05151190347168934 034274053593937165 1616237692038969120596 = 09708 119873 = 25 minus12061856254843495 minus05151195631493376 0342740958869788 16162379033820231ℎ = 01 119873 = 15 minus12061856254844203 minus05151195632158179 03427409589350932 16162379034134882

120596 = 14708 119873 = 25 minus12061856254879548 minus05151195632138894 034274095892547236 16162379034075425

Table 2 Absolute error |120582119896minus 120582119896119873

|

120582119896

120582minus2

120582minus1

1205820

1205821

ℎ = 06119873 = 15 289154 times 10

minus8528497 times 10

minus7422986 times 10

minus7211369 times 10

minus7

119873 = 25 360512 times 10minus12

64552 times 10minus11

556842 times 10minus11

255189 times 10minus11

ℎ = 01119873 = 15 353428 times 10

minus12192824 times 10

minus12962097 times 10

minus12594613 times 10

minus12

119873 = 25 222045 times 10minus16

222045 times 10minus16

555112 times 10minus17

444089 times 10minus16

Table 3 For119873 = 25 and ℎ = 01 the exact solutions 120582119896are all inside the interval [119886

minus 119886+] for different values of 120576

120582119896

120582minus2

120582minus1

1205820

1205821

Exact 120582119896

minus12061856254879546 minus05151195632138896 034274095892547224 1616237903407542119868120576119873

120576 = 10minus2

[minus12335696 minus11771922] [minus05486790 minus04826816] [03246478 03601346] [16100119 16224066]

119868120576119873

120576 = 10minus5

[minus12062137 minus12061575] [minus05151525 minus05150867] [03427232 03427586] [16162317 16162441]

030 032 034 036 038

00

02

04

aminus

a+

Ω(120582)

minus02

minus04

minus06

Figure 1The enclosure interval dominating 1205820for ℎ = 01119873 = 25

and 120576 = 10minus2

Example 2 In this example we consider the system

1199101015840

2(119909) minus 119901 (119909) 119910

1(119909) = 120582119910

1(119909)

1199101015840

1(119909) + 119901 (119909) 119910

2(119909) = minus120582119910

2(119909)

119909 isin [minus1 0) cup (0 1]

radic31199101(minus1) minus 119910

2(minus1) = 0

(1 +

1

2

120582)1199101(1) minus (1 +

radic3

2

120582)1199102(1) = 0

1199101(0minus) minus 3119910

1(0+) = 0 119910

2(0minus) minus 3119910

2(0+) = 0

(82)

030 032 034 036 038

00

02

aminus

a+

Ω(120582)

minus02

minus04

Figure 2The enclosure interval dominating 1205820for ℎ = 01119873 = 25

and 120576 = 10minus5

where

1199011(119909) = 119901

2(119909) = 119901 (119909) =

119909 119909 isin [minus1 0)

1199092 (0 1]

(83)

1205791= 1205792= 0 120572

1= 1205732= 1 120572

2= 1205731= 2 and 120575 = 12 Direct

calculations give

K (120582) =

1

2

((120582 + 3) cos (2120582) minus (1205822+ 3120582 + 4) sin (2120582))

Ω (120582) =

1

2

((1205822+ 3120582 + 4) sin(

1

6

minus 2120582)

+ (120582 + 3) cos(1

6

minus 2120582))

(84)

10 Abstract and Applied Analysis

Table 4 The approximation 120582119896119873

and the exact solutions 120582119896for different choices of ℎ and119873

120582119896

120582minus2

120582minus1

1205820

1205821

Exact 120582119896

minus30661256186938894 minus10959703083466883 03689030343814017 18338437675275228

120582119896119873

ℎ = 08 119873 = 15 minus3066123822200742 minus10959640944020903 036890463046491767 18338437607680411120596 = 007708 119873 = 25 minus30661256172349627 minus10959703026494807 036890303588091056 18338437676561417

ℎ = 02 119873 = 15 minus30661256186844446 minus10959703083278305 03689030343916001 1833843767517251120596 = 13708 119873 = 25 minus306612561869389 minus1095970308346688 03689030343814024 18338437675275225

Table 5 Absolute error |120582119896minus 120582119896119873

|

120582119896

120582minus2

120582minus1

1205820

1205821

ℎ = 08119873 = 15 179649 times 10

minus6621394 times 10

minus6159608 times 10

minus6675948 times 10

minus9

119873 = 25 145893 times 10minus9

569721 times 10minus9

149951 times 10minus9

128619 times 10minus10

ℎ = 02119873 = 15 944489 times 10

minus12188578 times 10

minus12101984 times 10

minus11102718 times 10

minus11

119873 = 25 444089 times 10minus16

222045 times 10minus16

666134 times 10minus16

222045 times 10minus16

Table 6 For119873 = 25 and ℎ = 02 the exact solutions 120582119896are all inside the interval [119886

minus 119886+] for different values of 120576

120582119896

120582minus2

120582minus1

1205820

1205821

Exact 120582119896

minus30661256186938894 minus10959703083466883 03689030343814017 18338437675275228119868120576119873

120576 = 10minus2

[minus31098742 minus30188665] [minus11631502 minus10339754] [03437434 03933486] [18220764 18454828]

119868120576119873

120576 = 10minus5

[minus30661709 minus30660803] [minus10960346 minus10959060] [03688783 03689278] [18338321 18338555]

158 159 160 161 162 163 164 165

00

05

10

minus05

minus10

aminus

a+

Ω(120582)

Figure 3The enclosure interval dominating 1205821for ℎ = 01119873 = 25

and 120576 = 10minus2

Tables 4 and 6 give the exact eigenvalues 1205821198961

119896=minus2and their

approximate ones for different values of ℎ 119873 and 120576 InTable 5 we give the absolute error for different values of ℎ and119873

Here Figures 5 6 7 and 8 illustrate the enclosureintervals dominating 120582

minus2and 120582

minus1for ℎ = 02 119873 = 25 and

120576 = 10minus2 120576 = 10

minus5 respectively

158 159 160 161 162 163 164 165

00

05

minus05

aminus

a+

Ω(120582)

Figure 4The enclosure interval dominating 1205821for ℎ = 01119873 = 25

and 120576 = 10minus5

5 Conclusion

With a simple analysis and with values of solutions ofinitial value problems computed at a few values of theeigenparameter we have computed the eigenvalues of aDirac system which has a discontinuity at one point andcontains a spectral parameter in all boundary conditionswith a certain estimated error The method proposed is

Abstract and Applied Analysis 11

00

05

10

minus05

minus33 minus32 minus31 minus30 minus29 minus28

aminus

a+

Ω(120582)

Figure 5 The enclosure interval dominating 120582minus2

for ℎ = 02 119873 =

25 and 120576 = 10minus2

00

05

10

minus05

minus33 minus32 minus31 minus30 minus29 minus28

aminus

a+

Ω(120582)

Figure 6 The enclosure interval dominating 120582minus2

for ℎ = 02 119873 =

25 and 120576 = 10minus5

a shooting procedure that is the problem is reformulatedas two initial value ones due to the interior discontinuity ofsize two and a miss-distance is defined at the right end of theinterval of integration whose roots are the eigenvalues to becomputed The unknown partU(120582) of the miss-distance canbewritten in terms of a functionwhich is an entire function ofexponential typeTherefore we propose to approximate suchterm by means of a truncated cardinal series with samplingvalues approximated by solving numerically correspondingsuitable initial value problems Finally in Section 4 weintroduced two instructive examples where both numericalresults and the associated figures have proved the credibilityof the method

00

05

10

minus05

minus14 minus13 minus12 minus11 minus10 minus09 minus08 minus07

aminus

a+

Ω(120582)

Figure 7 The enclosure interval dominating 120582minus1

for ℎ = 02 119873 =

25 and 120576 = 10minus2

00

05

10

minus05

minus14 minus13 minus12 minus11 minus10 minus09 minus08 minus07

aminus

a+

Ω(120582)

Figure 8 The enclosure interval dominating 120582minus1

for ℎ = 02 119873 =

25 and 120576 = 10minus5

Conflict of Interests

The authors declare that there is no conflict of interestsregarding the publication of this paper

Acknowledgment

This paper was funded by the Deanship of Scientific Research(DSR) King Abdulaziz University Jeddah The authorstherefore acknowledge with thanks DSR technical and finan-cial support

12 Abstract and Applied Analysis

References

[1] C T Fulton ldquoTwo-point boundary value problems witheigenvalue parameter contained in the boundary conditionsrdquoProceedings of the Royal Society of Edinburgh A Mathematicsvol 77 no 3-4 pp 293ndash308 1977

[2] JWalter ldquoRegular eigenvalue problems with eigenvalue param-eter in the boundary conditionrdquoMathematische Zeitschrift vol133 pp 301ndash312 1973

[3] V Kotelnikov ldquoOn the carrying capacity of the ether and wirein telecommunications material for the first All-Unionrdquo inConference on Questions of Communications vol 55 pp 55ndash64Izd Red Upr Svyazi RKKA 1933

[4] C E Shannon ldquoCommunication in the presence of noiserdquoProceedings of the IRE vol 37 pp 10ndash21 1949

[5] E TWhittaker ldquoOn the functions which are represented by theexpansion of the interpolation theoryrdquo Proceedings of the RoyalSociety of Edinburgh A vol 35 pp 181ndash194 1915

[6] P L Butzer G Schmeisser and R L Stens ldquoAn introductionto sampling analysisrdquo in Non Uniform Sampling Theory andPractices F Marvasti Ed pp 17ndash121 Kluwer New York NYUSA 2001

[7] M A Kowalski K A Sikorski and F Stenger Selected Topics inApproximation andComputation OxfordUniversity Press NewYork NY USA 1995

[8] J Lund and K L Bowers Sinc Methods for Quadrature andDifferential Equations SIAM Philadelphia Pa USA 1992

[9] F Stenger ldquoNumericalmethods based onWhittaker cardinal orsinc functionsrdquo SIAM Review vol 23 no 2 pp 165ndash224 1981

[10] F Stenger Numerical Methods Based on Sinc and AnalyticFunctions Springer New York NY USA 1993

[11] A Boumenir ldquoHigher approximation of eigenvalues by thesampling methodrdquo BIT Numerical Mathematics vol 40 no 2pp 215ndash225 2000

[12] A Boumenir ldquoSampling and eigenvalues of non-self-adjointSturm-Liouville problemsrdquo SIAM Journal on Scientific Comput-ing vol 23 no 1 pp 219ndash229 2001

[13] M M Tharwat A H Bhrawy and A Yildirim ldquoNumericalcomputation of eigenvalues of discontinuous Sturm-Liouvilleproblems with parameter dependent boundary conditionsusing sinc methodrdquoNumerical Algorithms vol 63 no 1 pp 27ndash48 2013

[14] M M Tharwat A H Bhrawy and A Yildirim ldquoNumericalcomputation of the eigenvalues of a discontinuousDirac systemusing the sincmethodwith error analysisrdquo International Journalof Computer Mathematics vol 89 no 15 pp 2061ndash2080 2012

[15] P L Butzer and R L Stens ldquoA modification of the Whittaker-Kotelrsquonikov-Shannon sampling seriesrdquo Aequationes Mathemat-icae vol 28 no 3 pp 305ndash311 1985

[16] R Gervais Q I Rahman and G Schmeisser ldquoA bandlimitedfunction simulating a duration-limited onerdquo in ApproximationTheory andFunctionalAnalysis P L Butzer andR L Stens Edspp 355ndash362 Birkhaauser Boston Mass USA 1984

[17] R L Stens ldquoSampling by generalized kernelsrdquo in SamplingThe-ory in Fourier and Signal Analysis Advanced Topics J RHigginsand R L Stens Eds pp 130ndash157 Oxford University PressOxford UK 1999

[18] G Schmeisser and F Stenger ldquoSinc approximation with a Gaus-sianmultiplierrdquo SamplingTheory in Signal and Image Processingvol 6 no 2 pp 199ndash221 2007

[19] L Qian ldquoOn the regularized Whittaker-Kotelrsquonikov-Shannonsampling formulardquo Proceedings of the American MathematicalSociety vol 131 no 4 pp 1169ndash1176 2003

[20] L Qian and D B Creamer ldquoA modification of the samplingseries with aGaussianmultiplierrdquo SamplingTheory in Signal andImage Processing vol 5 no 1 pp 1ndash20 2006

[21] L Qian and D B Creamer ldquoLocalized sampling in the presenceof noiserdquoAppliedMathematics Letters vol 19 no 4 pp 351ndash3552006

[22] M H Annaby and R M Asharabi ldquoComputing eigenvaluesof boundary-value problems using sinc-Gaussian methodrdquoSamplingTheory in Signal and Image Processing vol 7 no 3 pp293ndash312 2008

[23] M H Annaby and M MTharwat ldquoA sinc-Gaussian techniquefor computing eigenvalues of second-order linear pencilsrdquoApplied Numerical Mathematics vol 63 pp 129ndash137 2013

[24] A H Bhrawy M M Tharwat and A Al-Fhaid ldquoNumericalalgorithms for computing eigenvalues of discontinuous Diracsystem using sinc-Gaussian methodrdquo Abstract and AppliedAnalysis vol 2012 Article ID 925134 13 pages 2012

[25] M M Tharwat A H Bhrawy and A S Alofi ldquoApproximationof eigenvalues of discontinuous sturm-liouville problems witheigenparameter in all boundary conditionsrdquo Boundary ValueProblems vol 2013 article 132 2013

[26] M M Tharwat and S M Al-Harbi ldquoApproximation of eigen-values of boundary value problemsrdquo Boundary Value Problemsvol 2014 article 51 2014

[27] MMTharwat and A H Bhrawy ldquoComputation of eigenvaluesof discontinuous Dirac system using Hermite interpolationtechniquerdquo Advances in Difference Equations vol 2012 article59 2012

[28] M M Tharwat A Yildirim and A H Bhrawy ldquoSampling ofdiscontinuous Dirac systemsrdquo Numerical Functional Analysisand Optimization vol 34 no 3 pp 323ndash348 2013

[29] M M Tharwat ldquoDiscontinuous Sturm-Liouville problems andassociated sampling theoriesrdquo Abstract and Applied Analysisvol 2011 Article ID 610232 30 pages 2011

[30] MMTharwat ldquoOn sampling theories and discontinuousDiracsystems with eigenparameter in the boundary conditionsrdquoBoundary Value Problems vol 2013 article 65 2013

[31] B M Levitan and I S Sargsjan ldquoIntroduction to spectral the-ory self adjoint ordinary differential operatorsrdquo in Translationof Mthematical Monographs vol 39 American MathematicalSociety Providence RI USA 1975

[32] B M Levitan and I S Sargsjan Sturm-Liouville and Dirac oper-ators vol 59 Kluwer Academic Publishers Group DordrechtThe Netherlands 1991

Submit your manuscripts athttpwwwhindawicom

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical Problems in Engineering

Hindawi Publishing Corporationhttpwwwhindawicom

Differential EquationsInternational Journal of

Volume 2014

Applied MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Probability and StatisticsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical PhysicsAdvances in

Complex AnalysisJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

OptimizationJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

CombinatoricsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Operations ResearchAdvances in

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Function Spaces

Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of Mathematics and Mathematical Sciences

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

The Scientific World JournalHindawi Publishing Corporation httpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Algebra

Discrete Dynamics in Nature and Society

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Decision SciencesAdvances in

Discrete MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom

Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Stochastic AnalysisInternational Journal of

Abstract and Applied Analysis 9

Table 1 The approximation 120582119896119873

and the exact solutions 120582119896for different choices of ℎ and119873

120582119896

120582minus2

120582minus1

1205820

1205821

Exact 120582119896

minus12061856254879546 minus05151195632138896 034274095892547224 1616237903407542

120582119896119873

ℎ = 06 119873 = 15 minus12061855965725146 minus05151190347168934 034274053593937165 1616237692038969120596 = 09708 119873 = 25 minus12061856254843495 minus05151195631493376 0342740958869788 16162379033820231ℎ = 01 119873 = 15 minus12061856254844203 minus05151195632158179 03427409589350932 16162379034134882

120596 = 14708 119873 = 25 minus12061856254879548 minus05151195632138894 034274095892547236 16162379034075425

Table 2 Absolute error |120582119896minus 120582119896119873

|

120582119896

120582minus2

120582minus1

1205820

1205821

ℎ = 06119873 = 15 289154 times 10

minus8528497 times 10

minus7422986 times 10

minus7211369 times 10

minus7

119873 = 25 360512 times 10minus12

64552 times 10minus11

556842 times 10minus11

255189 times 10minus11

ℎ = 01119873 = 15 353428 times 10

minus12192824 times 10

minus12962097 times 10

minus12594613 times 10

minus12

119873 = 25 222045 times 10minus16

222045 times 10minus16

555112 times 10minus17

444089 times 10minus16

Table 3 For119873 = 25 and ℎ = 01 the exact solutions 120582119896are all inside the interval [119886

minus 119886+] for different values of 120576

120582119896

120582minus2

120582minus1

1205820

1205821

Exact 120582119896

minus12061856254879546 minus05151195632138896 034274095892547224 1616237903407542119868120576119873

120576 = 10minus2

[minus12335696 minus11771922] [minus05486790 minus04826816] [03246478 03601346] [16100119 16224066]

119868120576119873

120576 = 10minus5

[minus12062137 minus12061575] [minus05151525 minus05150867] [03427232 03427586] [16162317 16162441]

030 032 034 036 038

00

02

04

aminus

a+

Ω(120582)

minus02

minus04

minus06

Figure 1The enclosure interval dominating 1205820for ℎ = 01119873 = 25

and 120576 = 10minus2

Example 2 In this example we consider the system

1199101015840

2(119909) minus 119901 (119909) 119910

1(119909) = 120582119910

1(119909)

1199101015840

1(119909) + 119901 (119909) 119910

2(119909) = minus120582119910

2(119909)

119909 isin [minus1 0) cup (0 1]

radic31199101(minus1) minus 119910

2(minus1) = 0

(1 +

1

2

120582)1199101(1) minus (1 +

radic3

2

120582)1199102(1) = 0

1199101(0minus) minus 3119910

1(0+) = 0 119910

2(0minus) minus 3119910

2(0+) = 0

(82)

030 032 034 036 038

00

02

aminus

a+

Ω(120582)

minus02

minus04

Figure 2The enclosure interval dominating 1205820for ℎ = 01119873 = 25

and 120576 = 10minus5

where

1199011(119909) = 119901

2(119909) = 119901 (119909) =

119909 119909 isin [minus1 0)

1199092 (0 1]

(83)

1205791= 1205792= 0 120572

1= 1205732= 1 120572

2= 1205731= 2 and 120575 = 12 Direct

calculations give

K (120582) =

1

2

((120582 + 3) cos (2120582) minus (1205822+ 3120582 + 4) sin (2120582))

Ω (120582) =

1

2

((1205822+ 3120582 + 4) sin(

1

6

minus 2120582)

+ (120582 + 3) cos(1

6

minus 2120582))

(84)

10 Abstract and Applied Analysis

Table 4 The approximation 120582119896119873

and the exact solutions 120582119896for different choices of ℎ and119873

120582119896

120582minus2

120582minus1

1205820

1205821

Exact 120582119896

minus30661256186938894 minus10959703083466883 03689030343814017 18338437675275228

120582119896119873

ℎ = 08 119873 = 15 minus3066123822200742 minus10959640944020903 036890463046491767 18338437607680411120596 = 007708 119873 = 25 minus30661256172349627 minus10959703026494807 036890303588091056 18338437676561417

ℎ = 02 119873 = 15 minus30661256186844446 minus10959703083278305 03689030343916001 1833843767517251120596 = 13708 119873 = 25 minus306612561869389 minus1095970308346688 03689030343814024 18338437675275225

Table 5 Absolute error |120582119896minus 120582119896119873

|

120582119896

120582minus2

120582minus1

1205820

1205821

ℎ = 08119873 = 15 179649 times 10

minus6621394 times 10

minus6159608 times 10

minus6675948 times 10

minus9

119873 = 25 145893 times 10minus9

569721 times 10minus9

149951 times 10minus9

128619 times 10minus10

ℎ = 02119873 = 15 944489 times 10

minus12188578 times 10

minus12101984 times 10

minus11102718 times 10

minus11

119873 = 25 444089 times 10minus16

222045 times 10minus16

666134 times 10minus16

222045 times 10minus16

Table 6 For119873 = 25 and ℎ = 02 the exact solutions 120582119896are all inside the interval [119886

minus 119886+] for different values of 120576

120582119896

120582minus2

120582minus1

1205820

1205821

Exact 120582119896

minus30661256186938894 minus10959703083466883 03689030343814017 18338437675275228119868120576119873

120576 = 10minus2

[minus31098742 minus30188665] [minus11631502 minus10339754] [03437434 03933486] [18220764 18454828]

119868120576119873

120576 = 10minus5

[minus30661709 minus30660803] [minus10960346 minus10959060] [03688783 03689278] [18338321 18338555]

158 159 160 161 162 163 164 165

00

05

10

minus05

minus10

aminus

a+

Ω(120582)

Figure 3The enclosure interval dominating 1205821for ℎ = 01119873 = 25

and 120576 = 10minus2

Tables 4 and 6 give the exact eigenvalues 1205821198961

119896=minus2and their

approximate ones for different values of ℎ 119873 and 120576 InTable 5 we give the absolute error for different values of ℎ and119873

Here Figures 5 6 7 and 8 illustrate the enclosureintervals dominating 120582

minus2and 120582

minus1for ℎ = 02 119873 = 25 and

120576 = 10minus2 120576 = 10

minus5 respectively

158 159 160 161 162 163 164 165

00

05

minus05

aminus

a+

Ω(120582)

Figure 4The enclosure interval dominating 1205821for ℎ = 01119873 = 25

and 120576 = 10minus5

5 Conclusion

With a simple analysis and with values of solutions ofinitial value problems computed at a few values of theeigenparameter we have computed the eigenvalues of aDirac system which has a discontinuity at one point andcontains a spectral parameter in all boundary conditionswith a certain estimated error The method proposed is

Abstract and Applied Analysis 11

00

05

10

minus05

minus33 minus32 minus31 minus30 minus29 minus28

aminus

a+

Ω(120582)

Figure 5 The enclosure interval dominating 120582minus2

for ℎ = 02 119873 =

25 and 120576 = 10minus2

00

05

10

minus05

minus33 minus32 minus31 minus30 minus29 minus28

aminus

a+

Ω(120582)

Figure 6 The enclosure interval dominating 120582minus2

for ℎ = 02 119873 =

25 and 120576 = 10minus5

a shooting procedure that is the problem is reformulatedas two initial value ones due to the interior discontinuity ofsize two and a miss-distance is defined at the right end of theinterval of integration whose roots are the eigenvalues to becomputed The unknown partU(120582) of the miss-distance canbewritten in terms of a functionwhich is an entire function ofexponential typeTherefore we propose to approximate suchterm by means of a truncated cardinal series with samplingvalues approximated by solving numerically correspondingsuitable initial value problems Finally in Section 4 weintroduced two instructive examples where both numericalresults and the associated figures have proved the credibilityof the method

00

05

10

minus05

minus14 minus13 minus12 minus11 minus10 minus09 minus08 minus07

aminus

a+

Ω(120582)

Figure 7 The enclosure interval dominating 120582minus1

for ℎ = 02 119873 =

25 and 120576 = 10minus2

00

05

10

minus05

minus14 minus13 minus12 minus11 minus10 minus09 minus08 minus07

aminus

a+

Ω(120582)

Figure 8 The enclosure interval dominating 120582minus1

for ℎ = 02 119873 =

25 and 120576 = 10minus5

Conflict of Interests

The authors declare that there is no conflict of interestsregarding the publication of this paper

Acknowledgment

This paper was funded by the Deanship of Scientific Research(DSR) King Abdulaziz University Jeddah The authorstherefore acknowledge with thanks DSR technical and finan-cial support

12 Abstract and Applied Analysis

References

[1] C T Fulton ldquoTwo-point boundary value problems witheigenvalue parameter contained in the boundary conditionsrdquoProceedings of the Royal Society of Edinburgh A Mathematicsvol 77 no 3-4 pp 293ndash308 1977

[2] JWalter ldquoRegular eigenvalue problems with eigenvalue param-eter in the boundary conditionrdquoMathematische Zeitschrift vol133 pp 301ndash312 1973

[3] V Kotelnikov ldquoOn the carrying capacity of the ether and wirein telecommunications material for the first All-Unionrdquo inConference on Questions of Communications vol 55 pp 55ndash64Izd Red Upr Svyazi RKKA 1933

[4] C E Shannon ldquoCommunication in the presence of noiserdquoProceedings of the IRE vol 37 pp 10ndash21 1949

[5] E TWhittaker ldquoOn the functions which are represented by theexpansion of the interpolation theoryrdquo Proceedings of the RoyalSociety of Edinburgh A vol 35 pp 181ndash194 1915

[6] P L Butzer G Schmeisser and R L Stens ldquoAn introductionto sampling analysisrdquo in Non Uniform Sampling Theory andPractices F Marvasti Ed pp 17ndash121 Kluwer New York NYUSA 2001

[7] M A Kowalski K A Sikorski and F Stenger Selected Topics inApproximation andComputation OxfordUniversity Press NewYork NY USA 1995

[8] J Lund and K L Bowers Sinc Methods for Quadrature andDifferential Equations SIAM Philadelphia Pa USA 1992

[9] F Stenger ldquoNumericalmethods based onWhittaker cardinal orsinc functionsrdquo SIAM Review vol 23 no 2 pp 165ndash224 1981

[10] F Stenger Numerical Methods Based on Sinc and AnalyticFunctions Springer New York NY USA 1993

[11] A Boumenir ldquoHigher approximation of eigenvalues by thesampling methodrdquo BIT Numerical Mathematics vol 40 no 2pp 215ndash225 2000

[12] A Boumenir ldquoSampling and eigenvalues of non-self-adjointSturm-Liouville problemsrdquo SIAM Journal on Scientific Comput-ing vol 23 no 1 pp 219ndash229 2001

[13] M M Tharwat A H Bhrawy and A Yildirim ldquoNumericalcomputation of eigenvalues of discontinuous Sturm-Liouvilleproblems with parameter dependent boundary conditionsusing sinc methodrdquoNumerical Algorithms vol 63 no 1 pp 27ndash48 2013

[14] M M Tharwat A H Bhrawy and A Yildirim ldquoNumericalcomputation of the eigenvalues of a discontinuousDirac systemusing the sincmethodwith error analysisrdquo International Journalof Computer Mathematics vol 89 no 15 pp 2061ndash2080 2012

[15] P L Butzer and R L Stens ldquoA modification of the Whittaker-Kotelrsquonikov-Shannon sampling seriesrdquo Aequationes Mathemat-icae vol 28 no 3 pp 305ndash311 1985

[16] R Gervais Q I Rahman and G Schmeisser ldquoA bandlimitedfunction simulating a duration-limited onerdquo in ApproximationTheory andFunctionalAnalysis P L Butzer andR L Stens Edspp 355ndash362 Birkhaauser Boston Mass USA 1984

[17] R L Stens ldquoSampling by generalized kernelsrdquo in SamplingThe-ory in Fourier and Signal Analysis Advanced Topics J RHigginsand R L Stens Eds pp 130ndash157 Oxford University PressOxford UK 1999

[18] G Schmeisser and F Stenger ldquoSinc approximation with a Gaus-sianmultiplierrdquo SamplingTheory in Signal and Image Processingvol 6 no 2 pp 199ndash221 2007

[19] L Qian ldquoOn the regularized Whittaker-Kotelrsquonikov-Shannonsampling formulardquo Proceedings of the American MathematicalSociety vol 131 no 4 pp 1169ndash1176 2003

[20] L Qian and D B Creamer ldquoA modification of the samplingseries with aGaussianmultiplierrdquo SamplingTheory in Signal andImage Processing vol 5 no 1 pp 1ndash20 2006

[21] L Qian and D B Creamer ldquoLocalized sampling in the presenceof noiserdquoAppliedMathematics Letters vol 19 no 4 pp 351ndash3552006

[22] M H Annaby and R M Asharabi ldquoComputing eigenvaluesof boundary-value problems using sinc-Gaussian methodrdquoSamplingTheory in Signal and Image Processing vol 7 no 3 pp293ndash312 2008

[23] M H Annaby and M MTharwat ldquoA sinc-Gaussian techniquefor computing eigenvalues of second-order linear pencilsrdquoApplied Numerical Mathematics vol 63 pp 129ndash137 2013

[24] A H Bhrawy M M Tharwat and A Al-Fhaid ldquoNumericalalgorithms for computing eigenvalues of discontinuous Diracsystem using sinc-Gaussian methodrdquo Abstract and AppliedAnalysis vol 2012 Article ID 925134 13 pages 2012

[25] M M Tharwat A H Bhrawy and A S Alofi ldquoApproximationof eigenvalues of discontinuous sturm-liouville problems witheigenparameter in all boundary conditionsrdquo Boundary ValueProblems vol 2013 article 132 2013

[26] M M Tharwat and S M Al-Harbi ldquoApproximation of eigen-values of boundary value problemsrdquo Boundary Value Problemsvol 2014 article 51 2014

[27] MMTharwat and A H Bhrawy ldquoComputation of eigenvaluesof discontinuous Dirac system using Hermite interpolationtechniquerdquo Advances in Difference Equations vol 2012 article59 2012

[28] M M Tharwat A Yildirim and A H Bhrawy ldquoSampling ofdiscontinuous Dirac systemsrdquo Numerical Functional Analysisand Optimization vol 34 no 3 pp 323ndash348 2013

[29] M M Tharwat ldquoDiscontinuous Sturm-Liouville problems andassociated sampling theoriesrdquo Abstract and Applied Analysisvol 2011 Article ID 610232 30 pages 2011

[30] MMTharwat ldquoOn sampling theories and discontinuousDiracsystems with eigenparameter in the boundary conditionsrdquoBoundary Value Problems vol 2013 article 65 2013

[31] B M Levitan and I S Sargsjan ldquoIntroduction to spectral the-ory self adjoint ordinary differential operatorsrdquo in Translationof Mthematical Monographs vol 39 American MathematicalSociety Providence RI USA 1975

[32] B M Levitan and I S Sargsjan Sturm-Liouville and Dirac oper-ators vol 59 Kluwer Academic Publishers Group DordrechtThe Netherlands 1991

Submit your manuscripts athttpwwwhindawicom

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical Problems in Engineering

Hindawi Publishing Corporationhttpwwwhindawicom

Differential EquationsInternational Journal of

Volume 2014

Applied MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Probability and StatisticsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical PhysicsAdvances in

Complex AnalysisJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

OptimizationJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

CombinatoricsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Operations ResearchAdvances in

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Function Spaces

Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of Mathematics and Mathematical Sciences

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

The Scientific World JournalHindawi Publishing Corporation httpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Algebra

Discrete Dynamics in Nature and Society

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Decision SciencesAdvances in

Discrete MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom

Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Stochastic AnalysisInternational Journal of

10 Abstract and Applied Analysis

Table 4 The approximation 120582119896119873

and the exact solutions 120582119896for different choices of ℎ and119873

120582119896

120582minus2

120582minus1

1205820

1205821

Exact 120582119896

minus30661256186938894 minus10959703083466883 03689030343814017 18338437675275228

120582119896119873

ℎ = 08 119873 = 15 minus3066123822200742 minus10959640944020903 036890463046491767 18338437607680411120596 = 007708 119873 = 25 minus30661256172349627 minus10959703026494807 036890303588091056 18338437676561417

ℎ = 02 119873 = 15 minus30661256186844446 minus10959703083278305 03689030343916001 1833843767517251120596 = 13708 119873 = 25 minus306612561869389 minus1095970308346688 03689030343814024 18338437675275225

Table 5 Absolute error |120582119896minus 120582119896119873

|

120582119896

120582minus2

120582minus1

1205820

1205821

ℎ = 08119873 = 15 179649 times 10

minus6621394 times 10

minus6159608 times 10

minus6675948 times 10

minus9

119873 = 25 145893 times 10minus9

569721 times 10minus9

149951 times 10minus9

128619 times 10minus10

ℎ = 02119873 = 15 944489 times 10

minus12188578 times 10

minus12101984 times 10

minus11102718 times 10

minus11

119873 = 25 444089 times 10minus16

222045 times 10minus16

666134 times 10minus16

222045 times 10minus16

Table 6 For119873 = 25 and ℎ = 02 the exact solutions 120582119896are all inside the interval [119886

minus 119886+] for different values of 120576

120582119896

120582minus2

120582minus1

1205820

1205821

Exact 120582119896

minus30661256186938894 minus10959703083466883 03689030343814017 18338437675275228119868120576119873

120576 = 10minus2

[minus31098742 minus30188665] [minus11631502 minus10339754] [03437434 03933486] [18220764 18454828]

119868120576119873

120576 = 10minus5

[minus30661709 minus30660803] [minus10960346 minus10959060] [03688783 03689278] [18338321 18338555]

158 159 160 161 162 163 164 165

00

05

10

minus05

minus10

aminus

a+

Ω(120582)

Figure 3The enclosure interval dominating 1205821for ℎ = 01119873 = 25

and 120576 = 10minus2

Tables 4 and 6 give the exact eigenvalues 1205821198961

119896=minus2and their

approximate ones for different values of ℎ 119873 and 120576 InTable 5 we give the absolute error for different values of ℎ and119873

Here Figures 5 6 7 and 8 illustrate the enclosureintervals dominating 120582

minus2and 120582

minus1for ℎ = 02 119873 = 25 and

120576 = 10minus2 120576 = 10

minus5 respectively

158 159 160 161 162 163 164 165

00

05

minus05

aminus

a+

Ω(120582)

Figure 4The enclosure interval dominating 1205821for ℎ = 01119873 = 25

and 120576 = 10minus5

5 Conclusion

With a simple analysis and with values of solutions ofinitial value problems computed at a few values of theeigenparameter we have computed the eigenvalues of aDirac system which has a discontinuity at one point andcontains a spectral parameter in all boundary conditionswith a certain estimated error The method proposed is

Abstract and Applied Analysis 11

00

05

10

minus05

minus33 minus32 minus31 minus30 minus29 minus28

aminus

a+

Ω(120582)

Figure 5 The enclosure interval dominating 120582minus2

for ℎ = 02 119873 =

25 and 120576 = 10minus2

00

05

10

minus05

minus33 minus32 minus31 minus30 minus29 minus28

aminus

a+

Ω(120582)

Figure 6 The enclosure interval dominating 120582minus2

for ℎ = 02 119873 =

25 and 120576 = 10minus5

a shooting procedure that is the problem is reformulatedas two initial value ones due to the interior discontinuity ofsize two and a miss-distance is defined at the right end of theinterval of integration whose roots are the eigenvalues to becomputed The unknown partU(120582) of the miss-distance canbewritten in terms of a functionwhich is an entire function ofexponential typeTherefore we propose to approximate suchterm by means of a truncated cardinal series with samplingvalues approximated by solving numerically correspondingsuitable initial value problems Finally in Section 4 weintroduced two instructive examples where both numericalresults and the associated figures have proved the credibilityof the method

00

05

10

minus05

minus14 minus13 minus12 minus11 minus10 minus09 minus08 minus07

aminus

a+

Ω(120582)

Figure 7 The enclosure interval dominating 120582minus1

for ℎ = 02 119873 =

25 and 120576 = 10minus2

00

05

10

minus05

minus14 minus13 minus12 minus11 minus10 minus09 minus08 minus07

aminus

a+

Ω(120582)

Figure 8 The enclosure interval dominating 120582minus1

for ℎ = 02 119873 =

25 and 120576 = 10minus5

Conflict of Interests

The authors declare that there is no conflict of interestsregarding the publication of this paper

Acknowledgment

This paper was funded by the Deanship of Scientific Research(DSR) King Abdulaziz University Jeddah The authorstherefore acknowledge with thanks DSR technical and finan-cial support

12 Abstract and Applied Analysis

References

[1] C T Fulton ldquoTwo-point boundary value problems witheigenvalue parameter contained in the boundary conditionsrdquoProceedings of the Royal Society of Edinburgh A Mathematicsvol 77 no 3-4 pp 293ndash308 1977

[2] JWalter ldquoRegular eigenvalue problems with eigenvalue param-eter in the boundary conditionrdquoMathematische Zeitschrift vol133 pp 301ndash312 1973

[3] V Kotelnikov ldquoOn the carrying capacity of the ether and wirein telecommunications material for the first All-Unionrdquo inConference on Questions of Communications vol 55 pp 55ndash64Izd Red Upr Svyazi RKKA 1933

[4] C E Shannon ldquoCommunication in the presence of noiserdquoProceedings of the IRE vol 37 pp 10ndash21 1949

[5] E TWhittaker ldquoOn the functions which are represented by theexpansion of the interpolation theoryrdquo Proceedings of the RoyalSociety of Edinburgh A vol 35 pp 181ndash194 1915

[6] P L Butzer G Schmeisser and R L Stens ldquoAn introductionto sampling analysisrdquo in Non Uniform Sampling Theory andPractices F Marvasti Ed pp 17ndash121 Kluwer New York NYUSA 2001

[7] M A Kowalski K A Sikorski and F Stenger Selected Topics inApproximation andComputation OxfordUniversity Press NewYork NY USA 1995

[8] J Lund and K L Bowers Sinc Methods for Quadrature andDifferential Equations SIAM Philadelphia Pa USA 1992

[9] F Stenger ldquoNumericalmethods based onWhittaker cardinal orsinc functionsrdquo SIAM Review vol 23 no 2 pp 165ndash224 1981

[10] F Stenger Numerical Methods Based on Sinc and AnalyticFunctions Springer New York NY USA 1993

[11] A Boumenir ldquoHigher approximation of eigenvalues by thesampling methodrdquo BIT Numerical Mathematics vol 40 no 2pp 215ndash225 2000

[12] A Boumenir ldquoSampling and eigenvalues of non-self-adjointSturm-Liouville problemsrdquo SIAM Journal on Scientific Comput-ing vol 23 no 1 pp 219ndash229 2001

[13] M M Tharwat A H Bhrawy and A Yildirim ldquoNumericalcomputation of eigenvalues of discontinuous Sturm-Liouvilleproblems with parameter dependent boundary conditionsusing sinc methodrdquoNumerical Algorithms vol 63 no 1 pp 27ndash48 2013

[14] M M Tharwat A H Bhrawy and A Yildirim ldquoNumericalcomputation of the eigenvalues of a discontinuousDirac systemusing the sincmethodwith error analysisrdquo International Journalof Computer Mathematics vol 89 no 15 pp 2061ndash2080 2012

[15] P L Butzer and R L Stens ldquoA modification of the Whittaker-Kotelrsquonikov-Shannon sampling seriesrdquo Aequationes Mathemat-icae vol 28 no 3 pp 305ndash311 1985

[16] R Gervais Q I Rahman and G Schmeisser ldquoA bandlimitedfunction simulating a duration-limited onerdquo in ApproximationTheory andFunctionalAnalysis P L Butzer andR L Stens Edspp 355ndash362 Birkhaauser Boston Mass USA 1984

[17] R L Stens ldquoSampling by generalized kernelsrdquo in SamplingThe-ory in Fourier and Signal Analysis Advanced Topics J RHigginsand R L Stens Eds pp 130ndash157 Oxford University PressOxford UK 1999

[18] G Schmeisser and F Stenger ldquoSinc approximation with a Gaus-sianmultiplierrdquo SamplingTheory in Signal and Image Processingvol 6 no 2 pp 199ndash221 2007

[19] L Qian ldquoOn the regularized Whittaker-Kotelrsquonikov-Shannonsampling formulardquo Proceedings of the American MathematicalSociety vol 131 no 4 pp 1169ndash1176 2003

[20] L Qian and D B Creamer ldquoA modification of the samplingseries with aGaussianmultiplierrdquo SamplingTheory in Signal andImage Processing vol 5 no 1 pp 1ndash20 2006

[21] L Qian and D B Creamer ldquoLocalized sampling in the presenceof noiserdquoAppliedMathematics Letters vol 19 no 4 pp 351ndash3552006

[22] M H Annaby and R M Asharabi ldquoComputing eigenvaluesof boundary-value problems using sinc-Gaussian methodrdquoSamplingTheory in Signal and Image Processing vol 7 no 3 pp293ndash312 2008

[23] M H Annaby and M MTharwat ldquoA sinc-Gaussian techniquefor computing eigenvalues of second-order linear pencilsrdquoApplied Numerical Mathematics vol 63 pp 129ndash137 2013

[24] A H Bhrawy M M Tharwat and A Al-Fhaid ldquoNumericalalgorithms for computing eigenvalues of discontinuous Diracsystem using sinc-Gaussian methodrdquo Abstract and AppliedAnalysis vol 2012 Article ID 925134 13 pages 2012

[25] M M Tharwat A H Bhrawy and A S Alofi ldquoApproximationof eigenvalues of discontinuous sturm-liouville problems witheigenparameter in all boundary conditionsrdquo Boundary ValueProblems vol 2013 article 132 2013

[26] M M Tharwat and S M Al-Harbi ldquoApproximation of eigen-values of boundary value problemsrdquo Boundary Value Problemsvol 2014 article 51 2014

[27] MMTharwat and A H Bhrawy ldquoComputation of eigenvaluesof discontinuous Dirac system using Hermite interpolationtechniquerdquo Advances in Difference Equations vol 2012 article59 2012

[28] M M Tharwat A Yildirim and A H Bhrawy ldquoSampling ofdiscontinuous Dirac systemsrdquo Numerical Functional Analysisand Optimization vol 34 no 3 pp 323ndash348 2013

[29] M M Tharwat ldquoDiscontinuous Sturm-Liouville problems andassociated sampling theoriesrdquo Abstract and Applied Analysisvol 2011 Article ID 610232 30 pages 2011

[30] MMTharwat ldquoOn sampling theories and discontinuousDiracsystems with eigenparameter in the boundary conditionsrdquoBoundary Value Problems vol 2013 article 65 2013

[31] B M Levitan and I S Sargsjan ldquoIntroduction to spectral the-ory self adjoint ordinary differential operatorsrdquo in Translationof Mthematical Monographs vol 39 American MathematicalSociety Providence RI USA 1975

[32] B M Levitan and I S Sargsjan Sturm-Liouville and Dirac oper-ators vol 59 Kluwer Academic Publishers Group DordrechtThe Netherlands 1991

Submit your manuscripts athttpwwwhindawicom

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical Problems in Engineering

Hindawi Publishing Corporationhttpwwwhindawicom

Differential EquationsInternational Journal of

Volume 2014

Applied MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Probability and StatisticsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical PhysicsAdvances in

Complex AnalysisJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

OptimizationJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

CombinatoricsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Operations ResearchAdvances in

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Function Spaces

Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of Mathematics and Mathematical Sciences

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

The Scientific World JournalHindawi Publishing Corporation httpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Algebra

Discrete Dynamics in Nature and Society

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Decision SciencesAdvances in

Discrete MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom

Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Stochastic AnalysisInternational Journal of

Abstract and Applied Analysis 11

00

05

10

minus05

minus33 minus32 minus31 minus30 minus29 minus28

aminus

a+

Ω(120582)

Figure 5 The enclosure interval dominating 120582minus2

for ℎ = 02 119873 =

25 and 120576 = 10minus2

00

05

10

minus05

minus33 minus32 minus31 minus30 minus29 minus28

aminus

a+

Ω(120582)

Figure 6 The enclosure interval dominating 120582minus2

for ℎ = 02 119873 =

25 and 120576 = 10minus5

a shooting procedure that is the problem is reformulatedas two initial value ones due to the interior discontinuity ofsize two and a miss-distance is defined at the right end of theinterval of integration whose roots are the eigenvalues to becomputed The unknown partU(120582) of the miss-distance canbewritten in terms of a functionwhich is an entire function ofexponential typeTherefore we propose to approximate suchterm by means of a truncated cardinal series with samplingvalues approximated by solving numerically correspondingsuitable initial value problems Finally in Section 4 weintroduced two instructive examples where both numericalresults and the associated figures have proved the credibilityof the method

00

05

10

minus05

minus14 minus13 minus12 minus11 minus10 minus09 minus08 minus07

aminus

a+

Ω(120582)

Figure 7 The enclosure interval dominating 120582minus1

for ℎ = 02 119873 =

25 and 120576 = 10minus2

00

05

10

minus05

minus14 minus13 minus12 minus11 minus10 minus09 minus08 minus07

aminus

a+

Ω(120582)

Figure 8 The enclosure interval dominating 120582minus1

for ℎ = 02 119873 =

25 and 120576 = 10minus5

Conflict of Interests

The authors declare that there is no conflict of interestsregarding the publication of this paper

Acknowledgment

This paper was funded by the Deanship of Scientific Research(DSR) King Abdulaziz University Jeddah The authorstherefore acknowledge with thanks DSR technical and finan-cial support

12 Abstract and Applied Analysis

References

[1] C T Fulton ldquoTwo-point boundary value problems witheigenvalue parameter contained in the boundary conditionsrdquoProceedings of the Royal Society of Edinburgh A Mathematicsvol 77 no 3-4 pp 293ndash308 1977

[2] JWalter ldquoRegular eigenvalue problems with eigenvalue param-eter in the boundary conditionrdquoMathematische Zeitschrift vol133 pp 301ndash312 1973

[3] V Kotelnikov ldquoOn the carrying capacity of the ether and wirein telecommunications material for the first All-Unionrdquo inConference on Questions of Communications vol 55 pp 55ndash64Izd Red Upr Svyazi RKKA 1933

[4] C E Shannon ldquoCommunication in the presence of noiserdquoProceedings of the IRE vol 37 pp 10ndash21 1949

[5] E TWhittaker ldquoOn the functions which are represented by theexpansion of the interpolation theoryrdquo Proceedings of the RoyalSociety of Edinburgh A vol 35 pp 181ndash194 1915

[6] P L Butzer G Schmeisser and R L Stens ldquoAn introductionto sampling analysisrdquo in Non Uniform Sampling Theory andPractices F Marvasti Ed pp 17ndash121 Kluwer New York NYUSA 2001

[7] M A Kowalski K A Sikorski and F Stenger Selected Topics inApproximation andComputation OxfordUniversity Press NewYork NY USA 1995

[8] J Lund and K L Bowers Sinc Methods for Quadrature andDifferential Equations SIAM Philadelphia Pa USA 1992

[9] F Stenger ldquoNumericalmethods based onWhittaker cardinal orsinc functionsrdquo SIAM Review vol 23 no 2 pp 165ndash224 1981

[10] F Stenger Numerical Methods Based on Sinc and AnalyticFunctions Springer New York NY USA 1993

[11] A Boumenir ldquoHigher approximation of eigenvalues by thesampling methodrdquo BIT Numerical Mathematics vol 40 no 2pp 215ndash225 2000

[12] A Boumenir ldquoSampling and eigenvalues of non-self-adjointSturm-Liouville problemsrdquo SIAM Journal on Scientific Comput-ing vol 23 no 1 pp 219ndash229 2001

[13] M M Tharwat A H Bhrawy and A Yildirim ldquoNumericalcomputation of eigenvalues of discontinuous Sturm-Liouvilleproblems with parameter dependent boundary conditionsusing sinc methodrdquoNumerical Algorithms vol 63 no 1 pp 27ndash48 2013

[14] M M Tharwat A H Bhrawy and A Yildirim ldquoNumericalcomputation of the eigenvalues of a discontinuousDirac systemusing the sincmethodwith error analysisrdquo International Journalof Computer Mathematics vol 89 no 15 pp 2061ndash2080 2012

[15] P L Butzer and R L Stens ldquoA modification of the Whittaker-Kotelrsquonikov-Shannon sampling seriesrdquo Aequationes Mathemat-icae vol 28 no 3 pp 305ndash311 1985

[16] R Gervais Q I Rahman and G Schmeisser ldquoA bandlimitedfunction simulating a duration-limited onerdquo in ApproximationTheory andFunctionalAnalysis P L Butzer andR L Stens Edspp 355ndash362 Birkhaauser Boston Mass USA 1984

[17] R L Stens ldquoSampling by generalized kernelsrdquo in SamplingThe-ory in Fourier and Signal Analysis Advanced Topics J RHigginsand R L Stens Eds pp 130ndash157 Oxford University PressOxford UK 1999

[18] G Schmeisser and F Stenger ldquoSinc approximation with a Gaus-sianmultiplierrdquo SamplingTheory in Signal and Image Processingvol 6 no 2 pp 199ndash221 2007

[19] L Qian ldquoOn the regularized Whittaker-Kotelrsquonikov-Shannonsampling formulardquo Proceedings of the American MathematicalSociety vol 131 no 4 pp 1169ndash1176 2003

[20] L Qian and D B Creamer ldquoA modification of the samplingseries with aGaussianmultiplierrdquo SamplingTheory in Signal andImage Processing vol 5 no 1 pp 1ndash20 2006

[21] L Qian and D B Creamer ldquoLocalized sampling in the presenceof noiserdquoAppliedMathematics Letters vol 19 no 4 pp 351ndash3552006

[22] M H Annaby and R M Asharabi ldquoComputing eigenvaluesof boundary-value problems using sinc-Gaussian methodrdquoSamplingTheory in Signal and Image Processing vol 7 no 3 pp293ndash312 2008

[23] M H Annaby and M MTharwat ldquoA sinc-Gaussian techniquefor computing eigenvalues of second-order linear pencilsrdquoApplied Numerical Mathematics vol 63 pp 129ndash137 2013

[24] A H Bhrawy M M Tharwat and A Al-Fhaid ldquoNumericalalgorithms for computing eigenvalues of discontinuous Diracsystem using sinc-Gaussian methodrdquo Abstract and AppliedAnalysis vol 2012 Article ID 925134 13 pages 2012

[25] M M Tharwat A H Bhrawy and A S Alofi ldquoApproximationof eigenvalues of discontinuous sturm-liouville problems witheigenparameter in all boundary conditionsrdquo Boundary ValueProblems vol 2013 article 132 2013

[26] M M Tharwat and S M Al-Harbi ldquoApproximation of eigen-values of boundary value problemsrdquo Boundary Value Problemsvol 2014 article 51 2014

[27] MMTharwat and A H Bhrawy ldquoComputation of eigenvaluesof discontinuous Dirac system using Hermite interpolationtechniquerdquo Advances in Difference Equations vol 2012 article59 2012

[28] M M Tharwat A Yildirim and A H Bhrawy ldquoSampling ofdiscontinuous Dirac systemsrdquo Numerical Functional Analysisand Optimization vol 34 no 3 pp 323ndash348 2013

[29] M M Tharwat ldquoDiscontinuous Sturm-Liouville problems andassociated sampling theoriesrdquo Abstract and Applied Analysisvol 2011 Article ID 610232 30 pages 2011

[30] MMTharwat ldquoOn sampling theories and discontinuousDiracsystems with eigenparameter in the boundary conditionsrdquoBoundary Value Problems vol 2013 article 65 2013

[31] B M Levitan and I S Sargsjan ldquoIntroduction to spectral the-ory self adjoint ordinary differential operatorsrdquo in Translationof Mthematical Monographs vol 39 American MathematicalSociety Providence RI USA 1975

[32] B M Levitan and I S Sargsjan Sturm-Liouville and Dirac oper-ators vol 59 Kluwer Academic Publishers Group DordrechtThe Netherlands 1991

Submit your manuscripts athttpwwwhindawicom

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical Problems in Engineering

Hindawi Publishing Corporationhttpwwwhindawicom

Differential EquationsInternational Journal of

Volume 2014

Applied MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Probability and StatisticsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical PhysicsAdvances in

Complex AnalysisJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

OptimizationJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

CombinatoricsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Operations ResearchAdvances in

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Function Spaces

Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of Mathematics and Mathematical Sciences

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

The Scientific World JournalHindawi Publishing Corporation httpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Algebra

Discrete Dynamics in Nature and Society

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Decision SciencesAdvances in

Discrete MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom

Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Stochastic AnalysisInternational Journal of

12 Abstract and Applied Analysis

References

[1] C T Fulton ldquoTwo-point boundary value problems witheigenvalue parameter contained in the boundary conditionsrdquoProceedings of the Royal Society of Edinburgh A Mathematicsvol 77 no 3-4 pp 293ndash308 1977

[2] JWalter ldquoRegular eigenvalue problems with eigenvalue param-eter in the boundary conditionrdquoMathematische Zeitschrift vol133 pp 301ndash312 1973

[3] V Kotelnikov ldquoOn the carrying capacity of the ether and wirein telecommunications material for the first All-Unionrdquo inConference on Questions of Communications vol 55 pp 55ndash64Izd Red Upr Svyazi RKKA 1933

[4] C E Shannon ldquoCommunication in the presence of noiserdquoProceedings of the IRE vol 37 pp 10ndash21 1949

[5] E TWhittaker ldquoOn the functions which are represented by theexpansion of the interpolation theoryrdquo Proceedings of the RoyalSociety of Edinburgh A vol 35 pp 181ndash194 1915

[6] P L Butzer G Schmeisser and R L Stens ldquoAn introductionto sampling analysisrdquo in Non Uniform Sampling Theory andPractices F Marvasti Ed pp 17ndash121 Kluwer New York NYUSA 2001

[7] M A Kowalski K A Sikorski and F Stenger Selected Topics inApproximation andComputation OxfordUniversity Press NewYork NY USA 1995

[8] J Lund and K L Bowers Sinc Methods for Quadrature andDifferential Equations SIAM Philadelphia Pa USA 1992

[9] F Stenger ldquoNumericalmethods based onWhittaker cardinal orsinc functionsrdquo SIAM Review vol 23 no 2 pp 165ndash224 1981

[10] F Stenger Numerical Methods Based on Sinc and AnalyticFunctions Springer New York NY USA 1993

[11] A Boumenir ldquoHigher approximation of eigenvalues by thesampling methodrdquo BIT Numerical Mathematics vol 40 no 2pp 215ndash225 2000

[12] A Boumenir ldquoSampling and eigenvalues of non-self-adjointSturm-Liouville problemsrdquo SIAM Journal on Scientific Comput-ing vol 23 no 1 pp 219ndash229 2001

[13] M M Tharwat A H Bhrawy and A Yildirim ldquoNumericalcomputation of eigenvalues of discontinuous Sturm-Liouvilleproblems with parameter dependent boundary conditionsusing sinc methodrdquoNumerical Algorithms vol 63 no 1 pp 27ndash48 2013

[14] M M Tharwat A H Bhrawy and A Yildirim ldquoNumericalcomputation of the eigenvalues of a discontinuousDirac systemusing the sincmethodwith error analysisrdquo International Journalof Computer Mathematics vol 89 no 15 pp 2061ndash2080 2012

[15] P L Butzer and R L Stens ldquoA modification of the Whittaker-Kotelrsquonikov-Shannon sampling seriesrdquo Aequationes Mathemat-icae vol 28 no 3 pp 305ndash311 1985

[16] R Gervais Q I Rahman and G Schmeisser ldquoA bandlimitedfunction simulating a duration-limited onerdquo in ApproximationTheory andFunctionalAnalysis P L Butzer andR L Stens Edspp 355ndash362 Birkhaauser Boston Mass USA 1984

[17] R L Stens ldquoSampling by generalized kernelsrdquo in SamplingThe-ory in Fourier and Signal Analysis Advanced Topics J RHigginsand R L Stens Eds pp 130ndash157 Oxford University PressOxford UK 1999

[18] G Schmeisser and F Stenger ldquoSinc approximation with a Gaus-sianmultiplierrdquo SamplingTheory in Signal and Image Processingvol 6 no 2 pp 199ndash221 2007

[19] L Qian ldquoOn the regularized Whittaker-Kotelrsquonikov-Shannonsampling formulardquo Proceedings of the American MathematicalSociety vol 131 no 4 pp 1169ndash1176 2003

[20] L Qian and D B Creamer ldquoA modification of the samplingseries with aGaussianmultiplierrdquo SamplingTheory in Signal andImage Processing vol 5 no 1 pp 1ndash20 2006

[21] L Qian and D B Creamer ldquoLocalized sampling in the presenceof noiserdquoAppliedMathematics Letters vol 19 no 4 pp 351ndash3552006

[22] M H Annaby and R M Asharabi ldquoComputing eigenvaluesof boundary-value problems using sinc-Gaussian methodrdquoSamplingTheory in Signal and Image Processing vol 7 no 3 pp293ndash312 2008

[23] M H Annaby and M MTharwat ldquoA sinc-Gaussian techniquefor computing eigenvalues of second-order linear pencilsrdquoApplied Numerical Mathematics vol 63 pp 129ndash137 2013

[24] A H Bhrawy M M Tharwat and A Al-Fhaid ldquoNumericalalgorithms for computing eigenvalues of discontinuous Diracsystem using sinc-Gaussian methodrdquo Abstract and AppliedAnalysis vol 2012 Article ID 925134 13 pages 2012

[25] M M Tharwat A H Bhrawy and A S Alofi ldquoApproximationof eigenvalues of discontinuous sturm-liouville problems witheigenparameter in all boundary conditionsrdquo Boundary ValueProblems vol 2013 article 132 2013

[26] M M Tharwat and S M Al-Harbi ldquoApproximation of eigen-values of boundary value problemsrdquo Boundary Value Problemsvol 2014 article 51 2014

[27] MMTharwat and A H Bhrawy ldquoComputation of eigenvaluesof discontinuous Dirac system using Hermite interpolationtechniquerdquo Advances in Difference Equations vol 2012 article59 2012

[28] M M Tharwat A Yildirim and A H Bhrawy ldquoSampling ofdiscontinuous Dirac systemsrdquo Numerical Functional Analysisand Optimization vol 34 no 3 pp 323ndash348 2013

[29] M M Tharwat ldquoDiscontinuous Sturm-Liouville problems andassociated sampling theoriesrdquo Abstract and Applied Analysisvol 2011 Article ID 610232 30 pages 2011

[30] MMTharwat ldquoOn sampling theories and discontinuousDiracsystems with eigenparameter in the boundary conditionsrdquoBoundary Value Problems vol 2013 article 65 2013

[31] B M Levitan and I S Sargsjan ldquoIntroduction to spectral the-ory self adjoint ordinary differential operatorsrdquo in Translationof Mthematical Monographs vol 39 American MathematicalSociety Providence RI USA 1975

[32] B M Levitan and I S Sargsjan Sturm-Liouville and Dirac oper-ators vol 59 Kluwer Academic Publishers Group DordrechtThe Netherlands 1991

Submit your manuscripts athttpwwwhindawicom

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical Problems in Engineering

Hindawi Publishing Corporationhttpwwwhindawicom

Differential EquationsInternational Journal of

Volume 2014

Applied MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Probability and StatisticsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical PhysicsAdvances in

Complex AnalysisJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

OptimizationJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

CombinatoricsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Operations ResearchAdvances in

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Function Spaces

Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of Mathematics and Mathematical Sciences

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

The Scientific World JournalHindawi Publishing Corporation httpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Algebra

Discrete Dynamics in Nature and Society

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Decision SciencesAdvances in

Discrete MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom

Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Stochastic AnalysisInternational Journal of

Submit your manuscripts athttpwwwhindawicom

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical Problems in Engineering

Hindawi Publishing Corporationhttpwwwhindawicom

Differential EquationsInternational Journal of

Volume 2014

Applied MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Probability and StatisticsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical PhysicsAdvances in

Complex AnalysisJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

OptimizationJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

CombinatoricsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Operations ResearchAdvances in

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Function Spaces

Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of Mathematics and Mathematical Sciences

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

The Scientific World JournalHindawi Publishing Corporation httpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Algebra

Discrete Dynamics in Nature and Society

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Decision SciencesAdvances in

Discrete MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom

Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Stochastic AnalysisInternational Journal of