research article a collocation method for numerical...
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Research ArticleA Collocation Method for Numerical Solution of HyperbolicTelegraph Equation with Neumann Boundary Conditions
R C Mittal and Rachna Bhatia
Department of Mathematics IIT Roorkee Roorkee Uttarakhand 247667 India
Correspondence should be addressed to Rachna Bhatia rachnabhsinghyahoocom
Received 24 June 2014 Revised 8 September 2014 Accepted 10 September 2014 Published 28 September 2014
Academic Editor Zhijie Xu
Copyright copy 2014 R C Mittal and R Bhatia This is an open access article distributed under the Creative Commons AttributionLicense which permits unrestricted use distribution and reproduction in any medium provided the original work is properlycited
We present a technique based on collocation of cubic B-spline basis functions to solve second order one-dimensional hyperbolictelegraph equation with Neumann boundary conditions The use of cubic B-spline basis functions for spatial variable and itsderivatives reduces the problem into system of first order ordinary differential equations The resulting system subsequently hasbeen solved by SSP-RK54 schemeThe accuracy of the proposed approach has been confirmed with numerical experiments whichshows that the results obtained are acceptable and in good agreement with the exact solution
1 Introduction
Thehyperbolic partial differential equationsmodel the vibra-tions of structures (eg building beams and machines) andare the basis for fundamental equations of atomic physicsIn this paper we consider nonlinear second order one-dimensional hyperbolic telegraph equation
119906119905119905(119909 119905) + 2120572119906
119905(119909 119905) + 120573
2119906 (119909 119905)
= 119906119909119909
(119909 119905) + 119891 (119909 119905) + 119892 (119906)
119886 le 119909 le 119887 119905 ge 0
(1)
with initial conditions
119906 (119909 0) = 1198911(119909) 119906
119905(119909 0) = 119891
2(119909) (2)
and Neumann boundary conditions
119906119909(119886 119905) = 119908
1(119905) 119906
119909(119887 119905) = 119908
2(119905) 119905 ge 0 (3)
where 120572 and 120573 are known real constants If 119892(119906) = 0 (1)represents a linear hyperbolic telegraph equation in whichboth the electric voltage and current in a double conductorsatisfy the equation where 119909 is distance and 119905 is time For120572 gt 0 120573 = 0 it represents a damped wave equation
and for 120572 gt 120573 gt 0 it is called a telegraph equationLinear second order hyperbolic telegraph equation arise inthe study of propagation of electric signal in a cable oftransmission line and wave phenomena Interaction betweenconvection and diffusion or reciprocal action of reaction anddiffusion describes a number of nonlinear phenomena inphysical and biological process In fact telegraph equation ismore suitable than ordinary diffusion equation in modelingreaction diffusion for such branches of sciences
Numerical solutions of one-dimensional linear hyper-bolic equation with Dirichlet boundary conditions have beenstudied by many authors El-Azab and El-Ghamel [1] haveused Routhe-wavelet method for the numerical solutionof telegraph equation Dehghan and Shokri [2] presenteda meshless method based on collocation with radial basisfunctions Spline solutions of hyperbolic telegraph equationhave been studied in [3ndash6] L B Liu and H W Liu [3] usedthe quartic spline method In [4] H-W Liu and L-B Liuapplied an unconditionally stable spline difference methodDosti andNazemi [5] presented a quartic119861-spline collocationmethod modified cubic 119861-spline collocation method hasbeen proposed by Mittal and Bhatia [6] to find the numer-ical solution of one-dimensional linear hyperbolic telegraphequation Numerical solution of linear and nonlinear one-dimensional hyperbolic telegraph equation with variable
Hindawi Publishing CorporationInternational Journal of Computational MathematicsVolume 2014 Article ID 526814 9 pageshttpdxdoiorg1011552014526814
2 International Journal of Computational Mathematics
coefficient has been presented in [7 8] Unconditionallystable finite difference schemes have been proposed in [9 10]Dehghan and Lakestani [11] used the Chebyshev cardinalfunctions whereas Saadatmandi and Dehghan [12] usedthe Chebyshev tau method for expanding the approximatesolution of one-dimensional telegraph equation Mohebbiand Dehghan [13] reported a higher order compact finitedifference approximation of fourth order in space and usedcollocation method for time direction Other techniquesused for numerical solutions of one-dimensional hyperbolictelegraph equation with Dirichlet boundary conditions areinterpolating scaling function technique [14] and radial basisfunction technique [15] Thus much work has been done tosolve (1) with Dirichlet boundary conditions Little has beendone for numerical solution of one-dimensional hyperbolictelegraph equation with Neumann boundary conditions In[16] Dehghan and Ghesmati reported a dual reciprocityboundary integral equation (DRBIE) method in whichthree different types of radial basis functions have beenused to approximate the solution of one-dimensional linearhyperbolic telegraph equation Recently L B Liu and HW Liu [17] developed an unconditionally stable compactdifference scheme for one-dimensional telegraphic equationswith Neumann boundary conditions
In the literature the119861-spline collocationmethod has beensuccessfully applied to find the numerical solutions of variouslinear and nonlinear partial differential equations [6 18 19]119861-splines have some special properties like local supportsmoothness and capability of handling local phenomenawhich make them suitable to solve linear and nonlinear par-tial differential equations easily and elegantlyTheuse of cubic119861-spline basis functions leads to a global solution for whichnot only the functions but also their first and second deriva-tives are continuous The present collocation method hastwo great advantages the set-up procedure does not involveintegrations and the resulting matrix system is banded withsmall band width Hence 119861-spline when combined with thecollocation provides a simple solution procedure of linearand nonlinear partial differential equations In this paper wehave developed a collocationmethod based on cubic119861-splinebasis function to solve hyperbolic telegraph equation withNeumann boundary conditions Equation (1) is convertedinto a system of equations using the transformation 119906
119905= V
Then for approximating the solution we use the collocationof cubic 119861-spline basis functions Finally we get a systemof first order systems of ordinary differential equationswhich have been solved by five-stage fourth-order strongstability preserving time stepping Runge Kutta (SSP-RK54)[20] schemeThemethod needs less storage space and causesless accumulation of numerical errors
The outline of the paper is as follows In Section 2 cubic119861-spline basis functions are introduced In Section 3 numer-ical scheme is presented for telegraph equation using cubic119861-spline functions based collocation method In Section 4initial vector is computed for given initial conditions InSection 5 stability of the scheme is discussed and found tobe unconditionally stable In Section 6 computational resultsfor telegraph equations (1)ndash(3) for some test problems areillustrated and finally conclusions are included in Section 6
Table 1 Values of cubic 119861-spline and its derivatives at differentknots
119909 119909119895minus2
119909119895minus1
119909119895
119909119895+1
119909119895+2
119861119895(119909) 0 1 4 1 0
1198611015840
119895(119909) 0 3ℎ 0 minus3ℎ 0
11986110158401015840
119895(119909) 0 6ℎ
2minus12ℎ
26ℎ2 0
2 Cubic B-Spline Collocation Method
The solution domain 119886 le 119909 le 119887 is partitioned into a meshof uniform length ℎ = 119909
119894+1minus 119909119894by the knot 119909
119894 where 119894 =
0 1 2 119873 minus 1119873 such that 119886 = 1199090
lt 1199091
lt sdot sdot sdot lt 119909119873minus1
lt
119909119873
= 119887In the cubic 119861-spline collocation method the approxi-
mate solution can be written as the linear combination ofcubic 119861-spline basis functions for the approximation spaceunder consideration
Our numerical treatment for solving (1)ndash(3) using the col-locationmethodwith cubic119861-spline is to find an approximatesolution 119880(119909 119905) to the exact solution 119906(119909 119905) in the form
119880 (119909 119905) =
119873+1
sum
119895=minus1
119888119895(119905) 119861119895(119909) (4)
where 119888119895(119905) are the time dependent quantities to be deter-
mined from boundary conditions and collocation from thedifferential equation
The cubic 119861-spline 119861119895(119909) at the knots is given by
119861119895(119909)
=
1
ℎ3
(119909 minus 119909119895minus2
)
3
119909 isin [119909119895minus2
119909119895minus1
)
(119909 minus 119909119895minus2
)
3
minus 4(119909 minus 119909119895minus1
)
3
119909 isin [119909119895minus1
119909119895)
(119909119895+2
minus 119909)
3
minus 4(119909119895+1
minus 119909)
3
119909 isin [119909119895 119909119895+1
)
(119909119895+2
minus 119909)
3
119909 isin [119909119895+1
119909119895+2
)
0 otherwise(5)
where the set of functions 119861minus1
1198610 1198611 119861
119873minus1 119861119873 119861119873+1
forms a basis for the function defined over the region 119886 le 119909 le
119887with the obvious adjustment of the boundary base functionsto avoid undefined knots Each cubic 119861-spline covers fourelements so that an element is covered by four cubic 119861-splines
The values of 119861119895(119909) and its derivatives are tabulated in
Table 1 Since 119861119895(119909) = 0 outside the interval (119909
119895minus2 119909119895+2
) sothere is no need to tabulate 119861
119895(119909) for other values of 119909
Then using approximate function (4) and Table 1 theapproximate values of 119880(119909 119905) and its two derivatives at the
International Journal of Computational Mathematics 3
knots are determined in terms of the time parameters 119888119895as
follows
119880119895= 119888119895minus1
+ 4119888119895+ 119888119895+1
1198801015840
119895=
3
ℎ
(119888119895+1
minus 119888119895minus1
)
11988010158401015840
119895=
6
ℎ2(119888119895minus1
minus 2119888119895+ 119888119895+1
)
(6)
3 Numerical Scheme
The telegraph equation (1) is decomposed into a system ofpartial differential equations using the following transforma-tion
Let 119906119905(119909 119905) = V(119909 119905)
Then the transformed form of (1) is
119906119905= V
V119905= minus2120572V minus 120573
2119906 + 119906119909119909
+ 119891 (119909 119905) + 119892 (119906)
(7)
For solving the coupled equations (7) we assume our approx-imate solution as the linear combination of cubic 119861-splinebasis function
119880 (119909 119905) =
119873+1
sum
119895=minus1
119888119895(119905) 119861119895(119909) (8)
Using approximate solution (8) approximate values of119880119905(119909 119905) can be written as follows
119880119905(119909 119905) =
119873+1
sum
119895=minus1
119888119895(119905) 119861119895(119909) (9)
where 119888119895(119905) is the derivative of 119888
119895(119905) with respect to time 119905
Using approximate solution (8) and Neumann boundaryconditions (3) the approximate solution at the boundarypoints can be written as
119880119909(1199090 119905) =
1
sum
119895=minus1
1198881198951198611015840
119895(1199090) = 1199081(119905)
119880119909(119909119873 119905) =
119873+1
sum
119895=119873minus1
1198881198951198611015840
119895(119909119873) = 1199082(119905)
(10)
Using Table 1
3
ℎ
(1198881minus 119888minus1
) = 1199081(119905)
3
ℎ
(119888119873+1
minus 119888119873minus1
) = 1199082(119905)
(11)
From (11) we have
1198881minus 119888minus1
=
ℎ
3
1(119905)
119888119873+1
minus 119888119873minus1
=
ℎ
3
2(119905)
(12)
Now using (8) and (9) in the coupled system (7) we have
119873+1
sum
119895=minus1
119888119895119861119895(119909) = V
119895 0 le 119895 le 119873
V119895= minus2120572V
119895minus 1205732
119873+1
sum
119895=minus1
119888119895119861119895(119909) +
119873+1
sum
119895=minus1
11988811989511986110158401015840
119895(119909)
+ 119891 (119909119895 119905) + 119892(
119873+1
sum
119895=minus1
119888119895119861119895(119909)) 0 le 119895 le 119873
(13)
Using (6) and Table 1 in (13) we have
119888119895minus1
+ 4 119888119895+ 119888119895+1
= V119895 0 le 119895 le 119873
V119895= minus2120572V
119895minus 1205732(119888119895minus1
+ 4119888119895+ 119888119895+1
)
+
6
ℎ2(119888119895minus1
minus 2119888119895+ 119888119895+1
) + 119891 (119909119895 119905)
+ 119892 (119888119895minus1
+ 4119888119895+ 119888119895+1
) 0 le 119895 le 119873
(14)
Eliminating 119888minus1 119888119873+1
119888minus1 119888119873+1
in (14) by using (11)and (12) we get following system of first order differentialequations
119872 119888 = 119883
= 119884
(15)
[
[
[
[
[
[
[
[
4 2 sdot sdot sdot sdot sdot sdot
1 4 1 sdot sdot sdot
sdot sdot sdot sdot sdot sdot sdot sdot sdot
sdot sdot sdot sdot sdot sdot sdot sdot sdot
1 4 1
2 4
]
]
]
]
]
]
]
]
[
[
[
[
[
[
[
[
1198880
1198881
sdot sdot sdot
sdot sdot sdot
119888119873minus1
119888119873
]
]
]
]
]
]
]
]
=
[
[
[
[
[
[
[
[
1205830
1205831
sdot sdot sdot
sdot sdot sdot
120583119873minus1
120583119873
]
]
]
]
]
]
]
]
[
[
[
[
[
[
[
[
V0
V1
sdot sdot sdot
sdot sdot sdot
V119873minus1
V119873
]
]
]
]
]
]
]
]
=
[
[
[
[
[
[
[
[
1205940
1205941
sdot sdot sdot
sdot sdot sdot
120594119873minus1
120594119873
]
]
]
]
]
]
]
]
(16)
where
1205830= V0+
ℎ
3
1(119905)
120583119895= V119895
120583119873
= V119873
minus
ℎ
3
2(119905)
1205940= minus2120572V
0minus 1205732(41198880+ 21198881minus
ℎ
3
1199081(119905))
+
6
ℎ2(minus21198880+ 21198881minus
ℎ
3
1199081(119905)) + 119891 (119909
0 119905)
4 International Journal of Computational Mathematics
+ 119892(41198880+ 21198881minus
ℎ
3
1199081(119905))
120594119895= minus2120572V
119895minus 1205732(119888119895minus1
+ 4119888119895+ 119888119895+1
)
+
6
ℎ2(119888119895minus1
minus 2119888119895+ 119888119895+1
) + 119891 (119909119895 119905)
+ 119892 (119888119895minus1
+ 4119888119895+ 119888119895+1
)
120594119873
= minus2120572V119873
minus 1205732(2119888119873minus1
+ 4119888119873
+
ℎ
3
1199082(119905))
+
6
ℎ2(minus2119888119873
+ 2119888119873minus1
+
ℎ
3
1199082(119905)) + 119891 (119909
119873 119905)
+ 119892(2119888119873minus1
+ 4119888119873
+
ℎ
3
1199082(119905))
(17)
119872 is (119873 + 1) times (119873 + 1) tridiagonal matrix is columnvector of order (119873+1) and119883 and 119884 are (119873+1) order vectorrepresents the rhs of system (14)
Once the vector 119888 = [1198880 1198881 119888
119873]119879 is computed at a
specific time level the approximate solution 119880(119909 119905) can becomputed at the required knots
System (16) represents a system of (2119873 + 2) first orderordinary differential equations which is solved using SSP-RK54 [20] scheme with a variant of Thomas algorithm andconsequently the approximate solution119880
119873(119909 119905) is completely
known
4 Initial Vectors
To solve the resulting systems of first order ordinary differ-ential equations we need initial vectors 119888
0 and V0 whichare computed from the initial and boundary conditions asfollows
41 Initial Vector 1198880 Initial vector 119888
0 is computed from theinitial condition and boundary values of the derivatives as thefollowing expressions
119880119909(1199090 0) = 119908
1(0) 119895 = 0
119880 (119909119895 0) = 119891
1(119909119895) 119895 = 2 3 119873 minus 1
119880119909(119909119873 0) = 119908
2(0) 119895 = 119873
(18)
Using (10)ndash(12) in (18) we get a (119873 + 1) times (119873 + 1) system ofequations of the form
1198721198880= 119867 (19)
where
119872 =
[
[
[
[
[
[
[
[
4 2 sdot sdot sdot sdot sdot sdot
1 4 1 sdot sdot sdot
sdot sdot sdot sdot sdot sdot sdot sdot sdot
sdot sdot sdot sdot sdot sdot sdot sdot sdot
1 4 1
2 4
]
]
]
]
]
]
]
]
1198880=
[
[
[
[
[
[
[
[
[
[
[
[
1198880
0
1198880
1
sdot sdot sdot
sdot sdot sdot
1198880
119873minus1
1198880
119873
]
]
]
]
]
]
]
]
]
]
]
]
119867 =
[
[
[
[
[
[
[
[
[
[
[
[
[
[
[
[
[
[
[
1198911(1199090) +
ℎ
3
1199081(0)
1198911(1199091)
sdot sdot sdot
sdot sdot sdot
1198911(119909119873minus1
)
1198911(119909119873) minus
ℎ
3
1199082(0)
]
]
]
]
]
]
]
]
]
]
]
]
]
]
]
]
]
]
]
(20)
The matrix 119872 is tridiagonal matrix System (19) issolved using Thomas algorithm and hence initial vector 119888
0 iscomputed
42 Initial Vector V0 Initial vector V0 can be computed usingthe initial condition (2) as
119880119905(119909119895 0) = 119891
2(119909119895) (21)
We have
V (119909119895 0) = 119891
2(119909) 119895 = 0 1 119873 minus 1119873 (22)
5 Stability of Scheme
The stability of system (16) is very important since it is relatedto the stability of numerical scheme for solving it If thesystem of ordinary differential equations (16) is unstablethen stable numerical scheme for temporal discretizationmaynot generate converged solution The stability of this systemdepends on the eigenvalues of coefficient matrix since itsexact solution can be directly found using its eigenvalues
For linear case that is if 119892(119906) = 0 we consider
119860
119889119862
119889119905
= 119861119862 + 119865 (119905) (23)
where 119862 = [1198881 1198882 119888
119873minus1 V1 V2 V
119873minus1] is the unknown
vector and119865(119905) is a column vector of order 2(119873minus1) Consider
[
1198791
119874
119874 119868] [
119888
V] = [
119874 119868
1198792
minus2120572119868] [
119888
V] + 119865 (119905) (24)
where 1198791and 119879
2are symmetric tridiagonal matrix of order
119873 minus 1 given by
International Journal of Computational Mathematics 5
1198791=
[
[
[
[
[
[
[
[
4 1 sdot sdot sdot sdot sdot sdot
1 4 1 sdot sdot sdot
sdot sdot sdot sdot sdot sdot sdot sdot sdot
sdot sdot sdot sdot sdot sdot sdot sdot sdot
1 4 1
1 4
]
]
]
]
]
]
]
]
1198792=
[
[
[
[
[
[
[
[
[
[
[
[
[
[
[
[
[
[
minus41205732minus
12
ℎ2
minus1205732+
6
ℎ2
sdot sdot sdot sdot sdot sdot
minus1205732+
6
ℎ2
minus41205732minus
12
ℎ2
minus1205732+
6
ℎ2
sdot sdot sdot
sdot sdot sdot sdot sdot sdot sdot sdot sdot
sdot sdot sdot sdot sdot sdot sdot sdot sdot
minus1205732+
6
ℎ2
minus41205732minus
12
ℎ2
minus1205732+
6
ℎ2
0 minus1205732+
6
ℎ2
minus41205732minus
12
ℎ2
]
]
]
]
]
]
]
]
]
]
]
]
]
]
]
]
]
]
(25)
where 119868 and 119874 are identity and null matrix of order 119873 minus 1respectively
Stability of system (23) depends on the eigenvalues ofthe coefficient matrix 119860
minus1119861 If all the eigenvalues are having
negative real part then the system (23) will be stableLet 120582 be any eigenvalue of 119860
minus1119861 and 119909
1and 119909
2are
two components each of order (119873 minus 1) of eigenvectorcorresponding to eigenvalue 120582 We have
[
[
119874 119879minus1
1
1198792
minus2120572119868
]
]
[
1199091
1199092
] = 120582 [
1199091
1199092
] (26)
From (26) we have
119879minus1
11199092= 1205821199091
11987921199091minus 2120572119909
2= 1205821199092
(27)
From the above system of equations we get
1198792119879minus1
11199092= (1205822+ 2120572120582) 119909
2 (28)
rArr 120582(120582 + 2120572) is an eigenvalue of 1198792119879minus1
1
Since 1198791and 119879
2are tridiagonal matrices so their eigen-
values are
120582119904(1198791) = 4 + 2 cos 120587119904
119873
119904 = 1 2 119873 minus 1
120582119904(1198792) = minus2120573
2minus 41205732cos2 120587119904
2119873
minus
24
ℎ2sin2 120587119904
2119873
119904 = 1 2 119873 minus 1
(29)
1198791and 119879
2are symmetric matrices so eigenvalues of
matrix 1198792119879minus1
1are
120582119904(1198792119879minus1
1) = (minus2120573
2minus 41205732cos2 120587119904
2119873
minus
24
ℎ2sin2 120587119904
2119873
)
times (4 + 2 cos 120587119904
119873
)
minus1
119904 = 1 2 119873 minus 1
(30)
which is negative and realLet 120582 = 119909 + 119894119910 where 119909 and 119910 are real numbersWe have that (119909 + 119894119910)(119909 + 119894119910 + 2120572) is real and negative
which implies
119909 (119909 + 2120572) minus 1199102lt 0
119910 (119909 + 120572) = 0
(31)
From the above set of equations we get the solutions asfollows
(i) 119909 + 120572 = 0 and 119910 is arbitrary real number
rArr 119909 is negative real number since 120572 is real and posi-tive
(ii) 119910 = 0
rArr 119909(119909 + 2120572) lt 0
rArr (119909 + 120572)2lt 1205722
rArr 119909 is negative and real since 120572 is real and positive
Hence the real part of eigenvalues of 119860minus1119861 should always benegative for stability
6 International Journal of Computational Mathematics
Table 2 The errors of numerical solutions with Δ119905 = 001
119905
ℎ = 05 ℎ = 02
1198712
119871infin
RMS error 1198712
Łinfin
RMS error1 51145119864 minus 4 49491119864 minus 4 20328119864 minus 4 17865119864 minus 4 16756119864 minus 4 71176119864 minus 5
2 32011119864 minus 4 24541119864 minus 4 12722119864 minus 4 15356119864 minus 4 10723119864 minus 4 61178119864 minus 5
3 19952119864 minus 4 13436119864 minus 4 78599119864 minus 4 10689119864 minus 4 66156119864 minus 5 42587119864 minus 5
Table 3 Comparison of RMS errors at 119905 = 3
ℎ
The proposed method Dehghan and Ghesmati [16]RMS error CPU time (s) RMS error CPU time (s)
05 78599119864 minus 4 70 3010119864 minus 4 mdash
02 42587119864 minus 5 17 7128119864 minus 5 mdash
01 37601119864 minus 5 31 4320119864 minus 5 mdash
When 119892(119906) = 0 that is when (1) is nonlinear we have thefollowing system
119860
119889119862
119889119905
= 119865 (119905 119888) (32)
The stability can be discussed by finding the eigenvalues ofthe matrix 119860
minus1119869 where 119869 = 120597119865120597119888 is the Jacobian matrix For
stability eigenvalues of matrix 119860minus1
119869 should be negative
6 Numerical Experiments
In this section three experiments including linear and non-linear form of problem (1)ndash(3) are considered to demonstratethe accuracy and applicability of the proposed method Thenumerical efficiency is shown by calculating the 119871
infin 1198712 and
root mean square error norms
Example 1 We consider the following linear telegraph equa-tion
119906119905119905(119909 119905) + 8119906
119905(119909 119905) + 4119906 (119909 119905)
= 119906119909119909
(119909 119905) minus 2119890minus119905 sin119909 0 le 119909 le 2120587 119905 ge 0
(33)
with initial conditions
119906 (119909 0) = sin119909 119906119905(119909 0) = minus sin119909 (34)
and boundary conditions
119906119909(0 119905) = 119906
119909(2120587 119905) = 119890
minus119905 (35)
The exact solution is given [16] by
119906 (119909 119905) = 119890minus119905 sin119909 (36)
1198712 119871infin and RMS errors are reported in Table 2 with
ℎ = 02 ℎ = 05 and Δ119905 = 001 RMS errors with differentspace step sizes are reported in Table 3 and compared withthe results given by Dehghan and Ghesmati [16] It is noticedthat our results are in good agreement with the results of [16]From Figure 1 it is clear that approximate solution coincides
0 1 2 3 4 5 6
x
minus04
minus03
minus02
minus01
0
01
02
03
04
u(xt)
Numerical solution at t = 1
Exact solution at t = 1
Numerical solution at t = 2
Exact solution at t = 2
Numerical solution at t = 3
Exact solution at t = 3
Figure 1 Numerical and exact solutions with ℎ = 05 Δ119905 = 001
005
115
225
3
t
minus1
minus05
0
05
1
u(xt)
0 12
34 5
6
x
Figure 2 Space-time graph of numerical solutions with ℎ = 05Δ119905 = 001
with the exact solution at 119905 = 1 2 3 with ℎ = 05 and Δ119905 =
001 Figure 2 depicts the space-time graph of approximatesolution up to time 119905 = 3 with ℎ = 05 and Δ119905 = 001Similar figures have been depicted in [16 17] It is clear thatthe obtained numerical results are accurate and comparablefavorably with the exact solutions and earlier studies
Example 2 We consider the following linear telegraph equa-tion
119906119905119905(119909 119905) + 12119906
119905(119909 119905) + 4119906 (119909 119905)
= 119906119909119909
(119909 119905) minus 12 sin 119905 sin119909 + 4 cos 119905 cos119909
0 le 119909 le 4 119905 ge 0
(37)
with initial conditions
119906 (119909 0) = sin119909 119906119905(119909 0) = 0 (38)
International Journal of Computational Mathematics 7
Table 4 The errors of numerical solutions with Δ119905 = 001
119905
ℎ = 05 ℎ = 02
1198712
119871infin
RMS error 1198712
Łinfin
RMS error1 67752119864 minus 4 66070119864 minus 4 33667119864 minus 4 52299119864 minus 4 38981119864 minus 4 26084119864 minus 4
2 60042119864 minus 4 51361119864 minus 4 29835119864 minus 4 57309119864 minus 4 43320119864 minus 4 28588119864 minus 4
3 17361119864 minus 4 17448119864 minus 4 86268119864 minus 5 78085119864 minus 5 66723119864 minus 5 38945119864 minus 5
Table 5 Comparison of RMS errors at 119905 = 2 for Example 2
ℎ
The proposed method Dehghan and Ghesmati [16]RMS error CPU time (s) RMS error CPU time (s)
05 29835119864 minus 4 56 2153119864 minus 4 mdash
01 28758119864 minus 4 24 7012119864 minus 5 mdash
005 28819119864 minus 4 42 4003119864 minus 5 mdash
0 05 1 15 2 25 3 35 4
x
minus1
minus08
minus06
minus04
minus02
0
02
04
06
08
u(xt)
Numerical solution at t = 1
Exact solution at t = 1
Numerical solution at t = 2
Exact solution at t = 2
Numerical solution at t = 3
Exact solution at t = 3
Figure 3 Numerical and exact solutions with ℎ = 05 Δ119905 = 001
and boundary conditions
119906119909(0 119905) = cos 119905 119906
119909(4 119905) = cos 119905 cos 4 (39)
The exact solution is given [16] by
119906 (119909 119905) = cos 119905 sin119909 (40)
1198712 119871infin and RMS errors are reported in Table 4 with
ℎ = 05 and Δ119905 = 001 In Table 5 RMS errors are reportedwith different space step sizes and compared with the resultsgiven by Dehghan and Ghesmati [16] Figure 3 shows thatthe approximate solution and exact solution coincide for 119905 =
1 2 3 with ℎ = 05 and Δ119905 = 001 Space-time graph ofapproximate solution is depicted in Figure 4 with ℎ = 05 andΔ119905 = 001 up to 119905 = 3 Similar figures have been depicted in[16 17]
Example 3 In this example we consider the nonlinear tele-graph equation
119906119905119905(119909 119905) + 119906
119905(119909 119905)
= 119906119909119909
(119909 119905) minus 2 sin 119906 minus 1205872119890minus119905 cos (120587119909)
+ 2 sin (119890minus119905
(1 minus cos (120587119909))) 0 le 119909 le 2 119905 ge 0
(41)
0051
152
253
t
minus1
minus05
0
05
1
u(xt)
005 1 15
2 25 335
4
x
Figure 4 Space-time graph of numerical solutions with ℎ = 05Δ119905 = 001
0 02 04 06 08 1 12 14 16 18 2
x
minus01
0
01
02
03
04
05
06
07
08
u(xt)
Numerical solution at t = 1
Exact solution at t = 1
Numerical solution at t = 2
Exact solution at t = 2
Numerical solution at t = 3
Exact solution at t = 3
Numerical solution at t = 5
Exact solution at t = 5
Figure 5 Numerical and exact solutions with ℎ = 05 Δ119905 = 001
with initial conditions
119906 (119909 0) = 1 minus cos (120587119909) 119906119905(119909 0) = minus1 + cos (120587119909)
(42)
and boundary conditions
119906119909(0 119905) = 119906
119909(2 119905) = 0 (43)
The exact solution is given [17] by
119906 (119909 119905) = 119890minus119905
(1 minus cos (120587119909)) (44)
1198712 119871infin and RMS errors are computed for different values
of 119905 and reported in Table 6 with ℎ = 05 and Δ119905 = 001 InTable 7 error norms are reported for ℎ = 1 and Δ119905 = 01
and compared with the results given by L B Liu and H WLiu [17] FromFigure 5 we observe that approximate solutionand exact solution coincide for 119905 = 1 2 3 with ℎ = 05
8 International Journal of Computational Mathematics
Table 6 The errors of numerical solutions with Δ119905 = 001
119905
ℎ = 05 ℎ = 02
1198712
119871infin
RMS error 1198712
Łinfin
RMS error1 13341119864 minus 3 14954119864 minus 3 93182119864 minus 4 28200119864 minus 4 30385119864 minus 4 19841119864 minus 4
3 31344119864 minus 4 38812119864 minus 4 21892119864 minus 4 16958119864 minus 4 15103119864 minus 4 11939119864 minus 4
5 42496119864 minus 5 49131119864 minus 5 29681119864 minus 5 56791119864 minus 5 50866119864 minus 5 39958119864 minus 5
7 21773119864 minus 5 26530119864 minus 5 15207119864 minus 5 24272119864 minus 5 19499119864 minus 5 17500119864 minus 5
10 10059119864 minus 5 11691119864 minus 5 70261119864 minus 6 42303119864 minus 6 37155119864 minus 6 38945119864 minus 6
15 51178119864 minus 7 66345119864 minus 7 35574119864 minus 7 23315119864 minus 7 24379119864 minus 7 38945119864 minus 7
Table 7 Comparison of RMS errors at different time levels for Example 3
119905
The proposed method L B Liu and H W Liu [17]ℎ = 1 Δ119905 = 01 ℎ = 05 Δ119905 = 01
1198712
119871infin
RMS error CPU time RMS1 28083119864 minus 3 33349119864 minus 3 19379119864 minus 3 03 7591119864 minus 6
2 24692119864 minus 3 22557119864 minus 3 17038119864 minus 3 04 1833119864 minus 6
5 38689119864 minus 4 27429119864 minus 4 26698119864 minus 4 06 6976119864 minus 7
7 15951119864 minus 4 15796119864 minus 4 11007119864 minus 4 08 mdash10 38193119864 minus 5 43924119864 minus 5 26355119864 minus 5 11 5177119864 minus 8
15 22855119864 minus 6 21111119864 minus 6 11584119864 minus 6 14 3943119864 minus 9
005
115
2
x01
23
45
0
05
1
15
2
t
u(xt)
Figure 6 Space-time graph of numerical solutions with ℎ = 05Δ119905 = 001
and Δ119905 = 001 Space-time graph of approximate solution isdepicted in Figure 6 with ℎ = 05 andΔ119905 = 001 which showsthe approximate solution profile up to 119905 = 3 Similar figurehas been depicted in [17]
It should be noticed that in this case we obtain a system ofnonlinear first order ordinary differential equations Unlikethe numerical methods [17] which use finite differencemethods to solve nonlinear hyperbolic wave equation to geta system of nonlinear algebraic equations and finally solvedusing the Newton-Raphson method we do not need to doany extra effort to handle the nonlinear term and solutionsare easily computed by solving the obtained system of ODEsusing SSP-RK54 scheme Hence the present scheme reducesthe computational cost
7 Conclusions
The main idea of the present paper is to first convert thegiven problem into a coupled system of partial differential
equations and then using cubic 119861-spline basis functionsfor spatial variable and derivatives the coupled system ofequations is converted into system of first order ordinarydifferential equations The resulting system of first orderordinary differential equations has been solved by SSP-RK54schemeThe stability of the scheme is discussed using matrixstability analysis and found to be unconditionally stable Themethod is also capable of solving nonlinear telegraph equa-tion with Neumann boundary conditions It has been noticedthat obtained numerical solutions are in good agreementwiththe exact solution and earlier studies These facts illustratethat the proposed method is a reliable valid and powerfultool for solving telegraph equation with Neumann boundaryconditions
Conflict of Interests
The authors declare that there is no conflict of interestsregarding to the publication of this paper
References
[1] M S El-Azab and M El-Gamel ldquoA numerical algorithm forthe solution of telegraph equationsrdquo Applied Mathematics andComputation vol 190 no 1 pp 757ndash764 2007
[2] M Dehghan and A Shokri ldquoA numerical method for solvingthe hyperbolic telegraph equationrdquo Numerical Methods forPartial Differential Equations vol 24 no 4 pp 1080ndash1093 2008
[3] L B Liu and H W Liu ldquoQuartic spline methods for solvingone-dimensional telegraphic equationsrdquo Applied Mathematicsand Computation vol 216 no 3 pp 951ndash958 2010
[4] H-W Liu and L-B Liu ldquoAn unconditionally stable splinedifference scheme of (k2 + h4) for solving the second-order1D linear hyperbolic equationrdquo Mathematical and ComputerModelling vol 49 no 9-10 pp 1985ndash1993 2009
International Journal of Computational Mathematics 9
[5] M Dosti and A Nazemi ldquoQuartic B-spline collocation methodfor solving one-dimensional hyperbolic telegraph equationrdquoThe Journal of Information and Computing Science vol 7 no2 pp 83ndash90 2012
[6] R C Mittal and R Bhatia ldquoNumerical solution of second orderone dimensional hyperbolic telegraph equation by cubic B-spline collocation methodrdquo Applied Mathematics and Compu-tation vol 220 pp 496ndash506 2013
[7] R K Mohanty ldquoAn unconditionally stable finite differenceformula for a linear second order one space dimensional hyper-bolic equation with variable coefficientsrdquo Applied Mathematicsand Computation vol 165 no 1 pp 229ndash236 2005
[8] R K Mohanty M K Jain and K George ldquoOn the use ofhigh order difference methods for the system of one spacesecond order nonlinear hyperbolic equations with variablecoefficientsrdquo Journal of Computational and Applied Mathemat-ics vol 72 no 2 pp 421ndash431 1996
[9] R KMohanty ldquoAnunconditionally stable difference scheme forthe one-space-dimensional linear hyperbolic equationrdquoAppliedMathematics Letters vol 17 no 1 pp 101ndash105 2004
[10] F Gao and C Chi ldquoUnconditionally stable difference schemesfor a one-space-dimensional linear hyperbolic equationrdquoApplied Mathematics and Computation vol 187 no 2 pp 1272ndash1276 2007
[11] M Dehghan andM Lakestani ldquoThe use of Chebyshev cardinalfunctions for solution of the second-order one-dimensionaltelegraph equationrdquo Numerical Methods for Partial DifferentialEquations vol 25 no 4 pp 931ndash938 2009
[12] A Saadatmandi and M Dehghan ldquoNumerical solution ofhyperbolic telegraph equation using the Chebyshev taumethodrdquo Numerical Methods for Partial Differential Equationsvol 26 no 1 pp 239ndash252 2010
[13] A Mohebbi and M Dehghan ldquoHigh order compact solution ofthe one-space-dimensional linear hyperbolic equationrdquoNumer-ical Methods for Partial Differential Equations vol 24 no 5 pp1222ndash1235 2008
[14] M Lakestani and B N Saray ldquoNumerical solution of telegraphequation using interpolating scaling functionsrdquo Computers ampMathematics with Applications vol 60 no 7 pp 1964ndash19722010
[15] M Esmaeilbeigi M M Hosseini and S T Mohyud-Din ldquoAnew approach of the radial basis functionsmethod for telegraphequationsrdquo International Journal of Physical Sciences vol 6 no6 pp 1517ndash1527 2011
[16] M Dehghan and A Ghesmati ldquoSolution of the second-orderone-dimensional hyperbolic telegraph equation by using thedual reciprocity boundary integral equation (DRBIE) methodrdquoEngineering Analysis with Boundary Elements vol 34 no 1 pp51ndash59 2010
[17] L B Liu and H W Liu ldquoCompact difference schemes forsolving telegraphic equations with Neumann boundary condi-tionsrdquo Applied Mathematics and Computation vol 219 no 19pp 10112ndash10121 2013
[18] R CMittal and R K Jain ldquoCubic B-splines collocationmethodfor solving nonlinear parabolic partial differential equationswith Neumann boundary conditionsrdquoCommunications in Non-linear Science andNumerical Simulation vol 17 no 12 pp 4616ndash4625 2012
[19] R C Mittal and R K Jain ldquoNumerical solutions of nonlinearBurgersrsquo equation with modified cubic B-Splines collocationmethodrdquo Applied Mathematics and Computation vol 218 no15 pp 7839ndash7855 2012
[20] R J Spiteri and S J Ruuth ldquoA new class of optimal high-orderstrong-stability-preserving time discretization methodsrdquo SIAMJournal on Numerical Analysis vol 40 no 2 pp 469ndash491 2002
Submit your manuscripts athttpwwwhindawicom
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
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Mathematical Problems in Engineering
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Differential EquationsInternational Journal of
Volume 2014
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International Journal of
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Operations ResearchAdvances in
Journal of
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Function Spaces
Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014
International Journal of Mathematics and Mathematical Sciences
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Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Algebra
Discrete Dynamics in Nature and Society
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
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Decision SciencesAdvances in
Discrete MathematicsJournal of
Hindawi Publishing Corporationhttpwwwhindawicom
Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Stochastic AnalysisInternational Journal of
2 International Journal of Computational Mathematics
coefficient has been presented in [7 8] Unconditionallystable finite difference schemes have been proposed in [9 10]Dehghan and Lakestani [11] used the Chebyshev cardinalfunctions whereas Saadatmandi and Dehghan [12] usedthe Chebyshev tau method for expanding the approximatesolution of one-dimensional telegraph equation Mohebbiand Dehghan [13] reported a higher order compact finitedifference approximation of fourth order in space and usedcollocation method for time direction Other techniquesused for numerical solutions of one-dimensional hyperbolictelegraph equation with Dirichlet boundary conditions areinterpolating scaling function technique [14] and radial basisfunction technique [15] Thus much work has been done tosolve (1) with Dirichlet boundary conditions Little has beendone for numerical solution of one-dimensional hyperbolictelegraph equation with Neumann boundary conditions In[16] Dehghan and Ghesmati reported a dual reciprocityboundary integral equation (DRBIE) method in whichthree different types of radial basis functions have beenused to approximate the solution of one-dimensional linearhyperbolic telegraph equation Recently L B Liu and HW Liu [17] developed an unconditionally stable compactdifference scheme for one-dimensional telegraphic equationswith Neumann boundary conditions
In the literature the119861-spline collocationmethod has beensuccessfully applied to find the numerical solutions of variouslinear and nonlinear partial differential equations [6 18 19]119861-splines have some special properties like local supportsmoothness and capability of handling local phenomenawhich make them suitable to solve linear and nonlinear par-tial differential equations easily and elegantlyTheuse of cubic119861-spline basis functions leads to a global solution for whichnot only the functions but also their first and second deriva-tives are continuous The present collocation method hastwo great advantages the set-up procedure does not involveintegrations and the resulting matrix system is banded withsmall band width Hence 119861-spline when combined with thecollocation provides a simple solution procedure of linearand nonlinear partial differential equations In this paper wehave developed a collocationmethod based on cubic119861-splinebasis function to solve hyperbolic telegraph equation withNeumann boundary conditions Equation (1) is convertedinto a system of equations using the transformation 119906
119905= V
Then for approximating the solution we use the collocationof cubic 119861-spline basis functions Finally we get a systemof first order systems of ordinary differential equationswhich have been solved by five-stage fourth-order strongstability preserving time stepping Runge Kutta (SSP-RK54)[20] schemeThemethod needs less storage space and causesless accumulation of numerical errors
The outline of the paper is as follows In Section 2 cubic119861-spline basis functions are introduced In Section 3 numer-ical scheme is presented for telegraph equation using cubic119861-spline functions based collocation method In Section 4initial vector is computed for given initial conditions InSection 5 stability of the scheme is discussed and found tobe unconditionally stable In Section 6 computational resultsfor telegraph equations (1)ndash(3) for some test problems areillustrated and finally conclusions are included in Section 6
Table 1 Values of cubic 119861-spline and its derivatives at differentknots
119909 119909119895minus2
119909119895minus1
119909119895
119909119895+1
119909119895+2
119861119895(119909) 0 1 4 1 0
1198611015840
119895(119909) 0 3ℎ 0 minus3ℎ 0
11986110158401015840
119895(119909) 0 6ℎ
2minus12ℎ
26ℎ2 0
2 Cubic B-Spline Collocation Method
The solution domain 119886 le 119909 le 119887 is partitioned into a meshof uniform length ℎ = 119909
119894+1minus 119909119894by the knot 119909
119894 where 119894 =
0 1 2 119873 minus 1119873 such that 119886 = 1199090
lt 1199091
lt sdot sdot sdot lt 119909119873minus1
lt
119909119873
= 119887In the cubic 119861-spline collocation method the approxi-
mate solution can be written as the linear combination ofcubic 119861-spline basis functions for the approximation spaceunder consideration
Our numerical treatment for solving (1)ndash(3) using the col-locationmethodwith cubic119861-spline is to find an approximatesolution 119880(119909 119905) to the exact solution 119906(119909 119905) in the form
119880 (119909 119905) =
119873+1
sum
119895=minus1
119888119895(119905) 119861119895(119909) (4)
where 119888119895(119905) are the time dependent quantities to be deter-
mined from boundary conditions and collocation from thedifferential equation
The cubic 119861-spline 119861119895(119909) at the knots is given by
119861119895(119909)
=
1
ℎ3
(119909 minus 119909119895minus2
)
3
119909 isin [119909119895minus2
119909119895minus1
)
(119909 minus 119909119895minus2
)
3
minus 4(119909 minus 119909119895minus1
)
3
119909 isin [119909119895minus1
119909119895)
(119909119895+2
minus 119909)
3
minus 4(119909119895+1
minus 119909)
3
119909 isin [119909119895 119909119895+1
)
(119909119895+2
minus 119909)
3
119909 isin [119909119895+1
119909119895+2
)
0 otherwise(5)
where the set of functions 119861minus1
1198610 1198611 119861
119873minus1 119861119873 119861119873+1
forms a basis for the function defined over the region 119886 le 119909 le
119887with the obvious adjustment of the boundary base functionsto avoid undefined knots Each cubic 119861-spline covers fourelements so that an element is covered by four cubic 119861-splines
The values of 119861119895(119909) and its derivatives are tabulated in
Table 1 Since 119861119895(119909) = 0 outside the interval (119909
119895minus2 119909119895+2
) sothere is no need to tabulate 119861
119895(119909) for other values of 119909
Then using approximate function (4) and Table 1 theapproximate values of 119880(119909 119905) and its two derivatives at the
International Journal of Computational Mathematics 3
knots are determined in terms of the time parameters 119888119895as
follows
119880119895= 119888119895minus1
+ 4119888119895+ 119888119895+1
1198801015840
119895=
3
ℎ
(119888119895+1
minus 119888119895minus1
)
11988010158401015840
119895=
6
ℎ2(119888119895minus1
minus 2119888119895+ 119888119895+1
)
(6)
3 Numerical Scheme
The telegraph equation (1) is decomposed into a system ofpartial differential equations using the following transforma-tion
Let 119906119905(119909 119905) = V(119909 119905)
Then the transformed form of (1) is
119906119905= V
V119905= minus2120572V minus 120573
2119906 + 119906119909119909
+ 119891 (119909 119905) + 119892 (119906)
(7)
For solving the coupled equations (7) we assume our approx-imate solution as the linear combination of cubic 119861-splinebasis function
119880 (119909 119905) =
119873+1
sum
119895=minus1
119888119895(119905) 119861119895(119909) (8)
Using approximate solution (8) approximate values of119880119905(119909 119905) can be written as follows
119880119905(119909 119905) =
119873+1
sum
119895=minus1
119888119895(119905) 119861119895(119909) (9)
where 119888119895(119905) is the derivative of 119888
119895(119905) with respect to time 119905
Using approximate solution (8) and Neumann boundaryconditions (3) the approximate solution at the boundarypoints can be written as
119880119909(1199090 119905) =
1
sum
119895=minus1
1198881198951198611015840
119895(1199090) = 1199081(119905)
119880119909(119909119873 119905) =
119873+1
sum
119895=119873minus1
1198881198951198611015840
119895(119909119873) = 1199082(119905)
(10)
Using Table 1
3
ℎ
(1198881minus 119888minus1
) = 1199081(119905)
3
ℎ
(119888119873+1
minus 119888119873minus1
) = 1199082(119905)
(11)
From (11) we have
1198881minus 119888minus1
=
ℎ
3
1(119905)
119888119873+1
minus 119888119873minus1
=
ℎ
3
2(119905)
(12)
Now using (8) and (9) in the coupled system (7) we have
119873+1
sum
119895=minus1
119888119895119861119895(119909) = V
119895 0 le 119895 le 119873
V119895= minus2120572V
119895minus 1205732
119873+1
sum
119895=minus1
119888119895119861119895(119909) +
119873+1
sum
119895=minus1
11988811989511986110158401015840
119895(119909)
+ 119891 (119909119895 119905) + 119892(
119873+1
sum
119895=minus1
119888119895119861119895(119909)) 0 le 119895 le 119873
(13)
Using (6) and Table 1 in (13) we have
119888119895minus1
+ 4 119888119895+ 119888119895+1
= V119895 0 le 119895 le 119873
V119895= minus2120572V
119895minus 1205732(119888119895minus1
+ 4119888119895+ 119888119895+1
)
+
6
ℎ2(119888119895minus1
minus 2119888119895+ 119888119895+1
) + 119891 (119909119895 119905)
+ 119892 (119888119895minus1
+ 4119888119895+ 119888119895+1
) 0 le 119895 le 119873
(14)
Eliminating 119888minus1 119888119873+1
119888minus1 119888119873+1
in (14) by using (11)and (12) we get following system of first order differentialequations
119872 119888 = 119883
= 119884
(15)
[
[
[
[
[
[
[
[
4 2 sdot sdot sdot sdot sdot sdot
1 4 1 sdot sdot sdot
sdot sdot sdot sdot sdot sdot sdot sdot sdot
sdot sdot sdot sdot sdot sdot sdot sdot sdot
1 4 1
2 4
]
]
]
]
]
]
]
]
[
[
[
[
[
[
[
[
1198880
1198881
sdot sdot sdot
sdot sdot sdot
119888119873minus1
119888119873
]
]
]
]
]
]
]
]
=
[
[
[
[
[
[
[
[
1205830
1205831
sdot sdot sdot
sdot sdot sdot
120583119873minus1
120583119873
]
]
]
]
]
]
]
]
[
[
[
[
[
[
[
[
V0
V1
sdot sdot sdot
sdot sdot sdot
V119873minus1
V119873
]
]
]
]
]
]
]
]
=
[
[
[
[
[
[
[
[
1205940
1205941
sdot sdot sdot
sdot sdot sdot
120594119873minus1
120594119873
]
]
]
]
]
]
]
]
(16)
where
1205830= V0+
ℎ
3
1(119905)
120583119895= V119895
120583119873
= V119873
minus
ℎ
3
2(119905)
1205940= minus2120572V
0minus 1205732(41198880+ 21198881minus
ℎ
3
1199081(119905))
+
6
ℎ2(minus21198880+ 21198881minus
ℎ
3
1199081(119905)) + 119891 (119909
0 119905)
4 International Journal of Computational Mathematics
+ 119892(41198880+ 21198881minus
ℎ
3
1199081(119905))
120594119895= minus2120572V
119895minus 1205732(119888119895minus1
+ 4119888119895+ 119888119895+1
)
+
6
ℎ2(119888119895minus1
minus 2119888119895+ 119888119895+1
) + 119891 (119909119895 119905)
+ 119892 (119888119895minus1
+ 4119888119895+ 119888119895+1
)
120594119873
= minus2120572V119873
minus 1205732(2119888119873minus1
+ 4119888119873
+
ℎ
3
1199082(119905))
+
6
ℎ2(minus2119888119873
+ 2119888119873minus1
+
ℎ
3
1199082(119905)) + 119891 (119909
119873 119905)
+ 119892(2119888119873minus1
+ 4119888119873
+
ℎ
3
1199082(119905))
(17)
119872 is (119873 + 1) times (119873 + 1) tridiagonal matrix is columnvector of order (119873+1) and119883 and 119884 are (119873+1) order vectorrepresents the rhs of system (14)
Once the vector 119888 = [1198880 1198881 119888
119873]119879 is computed at a
specific time level the approximate solution 119880(119909 119905) can becomputed at the required knots
System (16) represents a system of (2119873 + 2) first orderordinary differential equations which is solved using SSP-RK54 [20] scheme with a variant of Thomas algorithm andconsequently the approximate solution119880
119873(119909 119905) is completely
known
4 Initial Vectors
To solve the resulting systems of first order ordinary differ-ential equations we need initial vectors 119888
0 and V0 whichare computed from the initial and boundary conditions asfollows
41 Initial Vector 1198880 Initial vector 119888
0 is computed from theinitial condition and boundary values of the derivatives as thefollowing expressions
119880119909(1199090 0) = 119908
1(0) 119895 = 0
119880 (119909119895 0) = 119891
1(119909119895) 119895 = 2 3 119873 minus 1
119880119909(119909119873 0) = 119908
2(0) 119895 = 119873
(18)
Using (10)ndash(12) in (18) we get a (119873 + 1) times (119873 + 1) system ofequations of the form
1198721198880= 119867 (19)
where
119872 =
[
[
[
[
[
[
[
[
4 2 sdot sdot sdot sdot sdot sdot
1 4 1 sdot sdot sdot
sdot sdot sdot sdot sdot sdot sdot sdot sdot
sdot sdot sdot sdot sdot sdot sdot sdot sdot
1 4 1
2 4
]
]
]
]
]
]
]
]
1198880=
[
[
[
[
[
[
[
[
[
[
[
[
1198880
0
1198880
1
sdot sdot sdot
sdot sdot sdot
1198880
119873minus1
1198880
119873
]
]
]
]
]
]
]
]
]
]
]
]
119867 =
[
[
[
[
[
[
[
[
[
[
[
[
[
[
[
[
[
[
[
1198911(1199090) +
ℎ
3
1199081(0)
1198911(1199091)
sdot sdot sdot
sdot sdot sdot
1198911(119909119873minus1
)
1198911(119909119873) minus
ℎ
3
1199082(0)
]
]
]
]
]
]
]
]
]
]
]
]
]
]
]
]
]
]
]
(20)
The matrix 119872 is tridiagonal matrix System (19) issolved using Thomas algorithm and hence initial vector 119888
0 iscomputed
42 Initial Vector V0 Initial vector V0 can be computed usingthe initial condition (2) as
119880119905(119909119895 0) = 119891
2(119909119895) (21)
We have
V (119909119895 0) = 119891
2(119909) 119895 = 0 1 119873 minus 1119873 (22)
5 Stability of Scheme
The stability of system (16) is very important since it is relatedto the stability of numerical scheme for solving it If thesystem of ordinary differential equations (16) is unstablethen stable numerical scheme for temporal discretizationmaynot generate converged solution The stability of this systemdepends on the eigenvalues of coefficient matrix since itsexact solution can be directly found using its eigenvalues
For linear case that is if 119892(119906) = 0 we consider
119860
119889119862
119889119905
= 119861119862 + 119865 (119905) (23)
where 119862 = [1198881 1198882 119888
119873minus1 V1 V2 V
119873minus1] is the unknown
vector and119865(119905) is a column vector of order 2(119873minus1) Consider
[
1198791
119874
119874 119868] [
119888
V] = [
119874 119868
1198792
minus2120572119868] [
119888
V] + 119865 (119905) (24)
where 1198791and 119879
2are symmetric tridiagonal matrix of order
119873 minus 1 given by
International Journal of Computational Mathematics 5
1198791=
[
[
[
[
[
[
[
[
4 1 sdot sdot sdot sdot sdot sdot
1 4 1 sdot sdot sdot
sdot sdot sdot sdot sdot sdot sdot sdot sdot
sdot sdot sdot sdot sdot sdot sdot sdot sdot
1 4 1
1 4
]
]
]
]
]
]
]
]
1198792=
[
[
[
[
[
[
[
[
[
[
[
[
[
[
[
[
[
[
minus41205732minus
12
ℎ2
minus1205732+
6
ℎ2
sdot sdot sdot sdot sdot sdot
minus1205732+
6
ℎ2
minus41205732minus
12
ℎ2
minus1205732+
6
ℎ2
sdot sdot sdot
sdot sdot sdot sdot sdot sdot sdot sdot sdot
sdot sdot sdot sdot sdot sdot sdot sdot sdot
minus1205732+
6
ℎ2
minus41205732minus
12
ℎ2
minus1205732+
6
ℎ2
0 minus1205732+
6
ℎ2
minus41205732minus
12
ℎ2
]
]
]
]
]
]
]
]
]
]
]
]
]
]
]
]
]
]
(25)
where 119868 and 119874 are identity and null matrix of order 119873 minus 1respectively
Stability of system (23) depends on the eigenvalues ofthe coefficient matrix 119860
minus1119861 If all the eigenvalues are having
negative real part then the system (23) will be stableLet 120582 be any eigenvalue of 119860
minus1119861 and 119909
1and 119909
2are
two components each of order (119873 minus 1) of eigenvectorcorresponding to eigenvalue 120582 We have
[
[
119874 119879minus1
1
1198792
minus2120572119868
]
]
[
1199091
1199092
] = 120582 [
1199091
1199092
] (26)
From (26) we have
119879minus1
11199092= 1205821199091
11987921199091minus 2120572119909
2= 1205821199092
(27)
From the above system of equations we get
1198792119879minus1
11199092= (1205822+ 2120572120582) 119909
2 (28)
rArr 120582(120582 + 2120572) is an eigenvalue of 1198792119879minus1
1
Since 1198791and 119879
2are tridiagonal matrices so their eigen-
values are
120582119904(1198791) = 4 + 2 cos 120587119904
119873
119904 = 1 2 119873 minus 1
120582119904(1198792) = minus2120573
2minus 41205732cos2 120587119904
2119873
minus
24
ℎ2sin2 120587119904
2119873
119904 = 1 2 119873 minus 1
(29)
1198791and 119879
2are symmetric matrices so eigenvalues of
matrix 1198792119879minus1
1are
120582119904(1198792119879minus1
1) = (minus2120573
2minus 41205732cos2 120587119904
2119873
minus
24
ℎ2sin2 120587119904
2119873
)
times (4 + 2 cos 120587119904
119873
)
minus1
119904 = 1 2 119873 minus 1
(30)
which is negative and realLet 120582 = 119909 + 119894119910 where 119909 and 119910 are real numbersWe have that (119909 + 119894119910)(119909 + 119894119910 + 2120572) is real and negative
which implies
119909 (119909 + 2120572) minus 1199102lt 0
119910 (119909 + 120572) = 0
(31)
From the above set of equations we get the solutions asfollows
(i) 119909 + 120572 = 0 and 119910 is arbitrary real number
rArr 119909 is negative real number since 120572 is real and posi-tive
(ii) 119910 = 0
rArr 119909(119909 + 2120572) lt 0
rArr (119909 + 120572)2lt 1205722
rArr 119909 is negative and real since 120572 is real and positive
Hence the real part of eigenvalues of 119860minus1119861 should always benegative for stability
6 International Journal of Computational Mathematics
Table 2 The errors of numerical solutions with Δ119905 = 001
119905
ℎ = 05 ℎ = 02
1198712
119871infin
RMS error 1198712
Łinfin
RMS error1 51145119864 minus 4 49491119864 minus 4 20328119864 minus 4 17865119864 minus 4 16756119864 minus 4 71176119864 minus 5
2 32011119864 minus 4 24541119864 minus 4 12722119864 minus 4 15356119864 minus 4 10723119864 minus 4 61178119864 minus 5
3 19952119864 minus 4 13436119864 minus 4 78599119864 minus 4 10689119864 minus 4 66156119864 minus 5 42587119864 minus 5
Table 3 Comparison of RMS errors at 119905 = 3
ℎ
The proposed method Dehghan and Ghesmati [16]RMS error CPU time (s) RMS error CPU time (s)
05 78599119864 minus 4 70 3010119864 minus 4 mdash
02 42587119864 minus 5 17 7128119864 minus 5 mdash
01 37601119864 minus 5 31 4320119864 minus 5 mdash
When 119892(119906) = 0 that is when (1) is nonlinear we have thefollowing system
119860
119889119862
119889119905
= 119865 (119905 119888) (32)
The stability can be discussed by finding the eigenvalues ofthe matrix 119860
minus1119869 where 119869 = 120597119865120597119888 is the Jacobian matrix For
stability eigenvalues of matrix 119860minus1
119869 should be negative
6 Numerical Experiments
In this section three experiments including linear and non-linear form of problem (1)ndash(3) are considered to demonstratethe accuracy and applicability of the proposed method Thenumerical efficiency is shown by calculating the 119871
infin 1198712 and
root mean square error norms
Example 1 We consider the following linear telegraph equa-tion
119906119905119905(119909 119905) + 8119906
119905(119909 119905) + 4119906 (119909 119905)
= 119906119909119909
(119909 119905) minus 2119890minus119905 sin119909 0 le 119909 le 2120587 119905 ge 0
(33)
with initial conditions
119906 (119909 0) = sin119909 119906119905(119909 0) = minus sin119909 (34)
and boundary conditions
119906119909(0 119905) = 119906
119909(2120587 119905) = 119890
minus119905 (35)
The exact solution is given [16] by
119906 (119909 119905) = 119890minus119905 sin119909 (36)
1198712 119871infin and RMS errors are reported in Table 2 with
ℎ = 02 ℎ = 05 and Δ119905 = 001 RMS errors with differentspace step sizes are reported in Table 3 and compared withthe results given by Dehghan and Ghesmati [16] It is noticedthat our results are in good agreement with the results of [16]From Figure 1 it is clear that approximate solution coincides
0 1 2 3 4 5 6
x
minus04
minus03
minus02
minus01
0
01
02
03
04
u(xt)
Numerical solution at t = 1
Exact solution at t = 1
Numerical solution at t = 2
Exact solution at t = 2
Numerical solution at t = 3
Exact solution at t = 3
Figure 1 Numerical and exact solutions with ℎ = 05 Δ119905 = 001
005
115
225
3
t
minus1
minus05
0
05
1
u(xt)
0 12
34 5
6
x
Figure 2 Space-time graph of numerical solutions with ℎ = 05Δ119905 = 001
with the exact solution at 119905 = 1 2 3 with ℎ = 05 and Δ119905 =
001 Figure 2 depicts the space-time graph of approximatesolution up to time 119905 = 3 with ℎ = 05 and Δ119905 = 001Similar figures have been depicted in [16 17] It is clear thatthe obtained numerical results are accurate and comparablefavorably with the exact solutions and earlier studies
Example 2 We consider the following linear telegraph equa-tion
119906119905119905(119909 119905) + 12119906
119905(119909 119905) + 4119906 (119909 119905)
= 119906119909119909
(119909 119905) minus 12 sin 119905 sin119909 + 4 cos 119905 cos119909
0 le 119909 le 4 119905 ge 0
(37)
with initial conditions
119906 (119909 0) = sin119909 119906119905(119909 0) = 0 (38)
International Journal of Computational Mathematics 7
Table 4 The errors of numerical solutions with Δ119905 = 001
119905
ℎ = 05 ℎ = 02
1198712
119871infin
RMS error 1198712
Łinfin
RMS error1 67752119864 minus 4 66070119864 minus 4 33667119864 minus 4 52299119864 minus 4 38981119864 minus 4 26084119864 minus 4
2 60042119864 minus 4 51361119864 minus 4 29835119864 minus 4 57309119864 minus 4 43320119864 minus 4 28588119864 minus 4
3 17361119864 minus 4 17448119864 minus 4 86268119864 minus 5 78085119864 minus 5 66723119864 minus 5 38945119864 minus 5
Table 5 Comparison of RMS errors at 119905 = 2 for Example 2
ℎ
The proposed method Dehghan and Ghesmati [16]RMS error CPU time (s) RMS error CPU time (s)
05 29835119864 minus 4 56 2153119864 minus 4 mdash
01 28758119864 minus 4 24 7012119864 minus 5 mdash
005 28819119864 minus 4 42 4003119864 minus 5 mdash
0 05 1 15 2 25 3 35 4
x
minus1
minus08
minus06
minus04
minus02
0
02
04
06
08
u(xt)
Numerical solution at t = 1
Exact solution at t = 1
Numerical solution at t = 2
Exact solution at t = 2
Numerical solution at t = 3
Exact solution at t = 3
Figure 3 Numerical and exact solutions with ℎ = 05 Δ119905 = 001
and boundary conditions
119906119909(0 119905) = cos 119905 119906
119909(4 119905) = cos 119905 cos 4 (39)
The exact solution is given [16] by
119906 (119909 119905) = cos 119905 sin119909 (40)
1198712 119871infin and RMS errors are reported in Table 4 with
ℎ = 05 and Δ119905 = 001 In Table 5 RMS errors are reportedwith different space step sizes and compared with the resultsgiven by Dehghan and Ghesmati [16] Figure 3 shows thatthe approximate solution and exact solution coincide for 119905 =
1 2 3 with ℎ = 05 and Δ119905 = 001 Space-time graph ofapproximate solution is depicted in Figure 4 with ℎ = 05 andΔ119905 = 001 up to 119905 = 3 Similar figures have been depicted in[16 17]
Example 3 In this example we consider the nonlinear tele-graph equation
119906119905119905(119909 119905) + 119906
119905(119909 119905)
= 119906119909119909
(119909 119905) minus 2 sin 119906 minus 1205872119890minus119905 cos (120587119909)
+ 2 sin (119890minus119905
(1 minus cos (120587119909))) 0 le 119909 le 2 119905 ge 0
(41)
0051
152
253
t
minus1
minus05
0
05
1
u(xt)
005 1 15
2 25 335
4
x
Figure 4 Space-time graph of numerical solutions with ℎ = 05Δ119905 = 001
0 02 04 06 08 1 12 14 16 18 2
x
minus01
0
01
02
03
04
05
06
07
08
u(xt)
Numerical solution at t = 1
Exact solution at t = 1
Numerical solution at t = 2
Exact solution at t = 2
Numerical solution at t = 3
Exact solution at t = 3
Numerical solution at t = 5
Exact solution at t = 5
Figure 5 Numerical and exact solutions with ℎ = 05 Δ119905 = 001
with initial conditions
119906 (119909 0) = 1 minus cos (120587119909) 119906119905(119909 0) = minus1 + cos (120587119909)
(42)
and boundary conditions
119906119909(0 119905) = 119906
119909(2 119905) = 0 (43)
The exact solution is given [17] by
119906 (119909 119905) = 119890minus119905
(1 minus cos (120587119909)) (44)
1198712 119871infin and RMS errors are computed for different values
of 119905 and reported in Table 6 with ℎ = 05 and Δ119905 = 001 InTable 7 error norms are reported for ℎ = 1 and Δ119905 = 01
and compared with the results given by L B Liu and H WLiu [17] FromFigure 5 we observe that approximate solutionand exact solution coincide for 119905 = 1 2 3 with ℎ = 05
8 International Journal of Computational Mathematics
Table 6 The errors of numerical solutions with Δ119905 = 001
119905
ℎ = 05 ℎ = 02
1198712
119871infin
RMS error 1198712
Łinfin
RMS error1 13341119864 minus 3 14954119864 minus 3 93182119864 minus 4 28200119864 minus 4 30385119864 minus 4 19841119864 minus 4
3 31344119864 minus 4 38812119864 minus 4 21892119864 minus 4 16958119864 minus 4 15103119864 minus 4 11939119864 minus 4
5 42496119864 minus 5 49131119864 minus 5 29681119864 minus 5 56791119864 minus 5 50866119864 minus 5 39958119864 minus 5
7 21773119864 minus 5 26530119864 minus 5 15207119864 minus 5 24272119864 minus 5 19499119864 minus 5 17500119864 minus 5
10 10059119864 minus 5 11691119864 minus 5 70261119864 minus 6 42303119864 minus 6 37155119864 minus 6 38945119864 minus 6
15 51178119864 minus 7 66345119864 minus 7 35574119864 minus 7 23315119864 minus 7 24379119864 minus 7 38945119864 minus 7
Table 7 Comparison of RMS errors at different time levels for Example 3
119905
The proposed method L B Liu and H W Liu [17]ℎ = 1 Δ119905 = 01 ℎ = 05 Δ119905 = 01
1198712
119871infin
RMS error CPU time RMS1 28083119864 minus 3 33349119864 minus 3 19379119864 minus 3 03 7591119864 minus 6
2 24692119864 minus 3 22557119864 minus 3 17038119864 minus 3 04 1833119864 minus 6
5 38689119864 minus 4 27429119864 minus 4 26698119864 minus 4 06 6976119864 minus 7
7 15951119864 minus 4 15796119864 minus 4 11007119864 minus 4 08 mdash10 38193119864 minus 5 43924119864 minus 5 26355119864 minus 5 11 5177119864 minus 8
15 22855119864 minus 6 21111119864 minus 6 11584119864 minus 6 14 3943119864 minus 9
005
115
2
x01
23
45
0
05
1
15
2
t
u(xt)
Figure 6 Space-time graph of numerical solutions with ℎ = 05Δ119905 = 001
and Δ119905 = 001 Space-time graph of approximate solution isdepicted in Figure 6 with ℎ = 05 andΔ119905 = 001 which showsthe approximate solution profile up to 119905 = 3 Similar figurehas been depicted in [17]
It should be noticed that in this case we obtain a system ofnonlinear first order ordinary differential equations Unlikethe numerical methods [17] which use finite differencemethods to solve nonlinear hyperbolic wave equation to geta system of nonlinear algebraic equations and finally solvedusing the Newton-Raphson method we do not need to doany extra effort to handle the nonlinear term and solutionsare easily computed by solving the obtained system of ODEsusing SSP-RK54 scheme Hence the present scheme reducesthe computational cost
7 Conclusions
The main idea of the present paper is to first convert thegiven problem into a coupled system of partial differential
equations and then using cubic 119861-spline basis functionsfor spatial variable and derivatives the coupled system ofequations is converted into system of first order ordinarydifferential equations The resulting system of first orderordinary differential equations has been solved by SSP-RK54schemeThe stability of the scheme is discussed using matrixstability analysis and found to be unconditionally stable Themethod is also capable of solving nonlinear telegraph equa-tion with Neumann boundary conditions It has been noticedthat obtained numerical solutions are in good agreementwiththe exact solution and earlier studies These facts illustratethat the proposed method is a reliable valid and powerfultool for solving telegraph equation with Neumann boundaryconditions
Conflict of Interests
The authors declare that there is no conflict of interestsregarding to the publication of this paper
References
[1] M S El-Azab and M El-Gamel ldquoA numerical algorithm forthe solution of telegraph equationsrdquo Applied Mathematics andComputation vol 190 no 1 pp 757ndash764 2007
[2] M Dehghan and A Shokri ldquoA numerical method for solvingthe hyperbolic telegraph equationrdquo Numerical Methods forPartial Differential Equations vol 24 no 4 pp 1080ndash1093 2008
[3] L B Liu and H W Liu ldquoQuartic spline methods for solvingone-dimensional telegraphic equationsrdquo Applied Mathematicsand Computation vol 216 no 3 pp 951ndash958 2010
[4] H-W Liu and L-B Liu ldquoAn unconditionally stable splinedifference scheme of (k2 + h4) for solving the second-order1D linear hyperbolic equationrdquo Mathematical and ComputerModelling vol 49 no 9-10 pp 1985ndash1993 2009
International Journal of Computational Mathematics 9
[5] M Dosti and A Nazemi ldquoQuartic B-spline collocation methodfor solving one-dimensional hyperbolic telegraph equationrdquoThe Journal of Information and Computing Science vol 7 no2 pp 83ndash90 2012
[6] R C Mittal and R Bhatia ldquoNumerical solution of second orderone dimensional hyperbolic telegraph equation by cubic B-spline collocation methodrdquo Applied Mathematics and Compu-tation vol 220 pp 496ndash506 2013
[7] R K Mohanty ldquoAn unconditionally stable finite differenceformula for a linear second order one space dimensional hyper-bolic equation with variable coefficientsrdquo Applied Mathematicsand Computation vol 165 no 1 pp 229ndash236 2005
[8] R K Mohanty M K Jain and K George ldquoOn the use ofhigh order difference methods for the system of one spacesecond order nonlinear hyperbolic equations with variablecoefficientsrdquo Journal of Computational and Applied Mathemat-ics vol 72 no 2 pp 421ndash431 1996
[9] R KMohanty ldquoAnunconditionally stable difference scheme forthe one-space-dimensional linear hyperbolic equationrdquoAppliedMathematics Letters vol 17 no 1 pp 101ndash105 2004
[10] F Gao and C Chi ldquoUnconditionally stable difference schemesfor a one-space-dimensional linear hyperbolic equationrdquoApplied Mathematics and Computation vol 187 no 2 pp 1272ndash1276 2007
[11] M Dehghan andM Lakestani ldquoThe use of Chebyshev cardinalfunctions for solution of the second-order one-dimensionaltelegraph equationrdquo Numerical Methods for Partial DifferentialEquations vol 25 no 4 pp 931ndash938 2009
[12] A Saadatmandi and M Dehghan ldquoNumerical solution ofhyperbolic telegraph equation using the Chebyshev taumethodrdquo Numerical Methods for Partial Differential Equationsvol 26 no 1 pp 239ndash252 2010
[13] A Mohebbi and M Dehghan ldquoHigh order compact solution ofthe one-space-dimensional linear hyperbolic equationrdquoNumer-ical Methods for Partial Differential Equations vol 24 no 5 pp1222ndash1235 2008
[14] M Lakestani and B N Saray ldquoNumerical solution of telegraphequation using interpolating scaling functionsrdquo Computers ampMathematics with Applications vol 60 no 7 pp 1964ndash19722010
[15] M Esmaeilbeigi M M Hosseini and S T Mohyud-Din ldquoAnew approach of the radial basis functionsmethod for telegraphequationsrdquo International Journal of Physical Sciences vol 6 no6 pp 1517ndash1527 2011
[16] M Dehghan and A Ghesmati ldquoSolution of the second-orderone-dimensional hyperbolic telegraph equation by using thedual reciprocity boundary integral equation (DRBIE) methodrdquoEngineering Analysis with Boundary Elements vol 34 no 1 pp51ndash59 2010
[17] L B Liu and H W Liu ldquoCompact difference schemes forsolving telegraphic equations with Neumann boundary condi-tionsrdquo Applied Mathematics and Computation vol 219 no 19pp 10112ndash10121 2013
[18] R CMittal and R K Jain ldquoCubic B-splines collocationmethodfor solving nonlinear parabolic partial differential equationswith Neumann boundary conditionsrdquoCommunications in Non-linear Science andNumerical Simulation vol 17 no 12 pp 4616ndash4625 2012
[19] R C Mittal and R K Jain ldquoNumerical solutions of nonlinearBurgersrsquo equation with modified cubic B-Splines collocationmethodrdquo Applied Mathematics and Computation vol 218 no15 pp 7839ndash7855 2012
[20] R J Spiteri and S J Ruuth ldquoA new class of optimal high-orderstrong-stability-preserving time discretization methodsrdquo SIAMJournal on Numerical Analysis vol 40 no 2 pp 469ndash491 2002
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Differential EquationsInternational Journal of
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Function Spaces
Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014
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Algebra
Discrete Dynamics in Nature and Society
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Decision SciencesAdvances in
Discrete MathematicsJournal of
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Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Stochastic AnalysisInternational Journal of
International Journal of Computational Mathematics 3
knots are determined in terms of the time parameters 119888119895as
follows
119880119895= 119888119895minus1
+ 4119888119895+ 119888119895+1
1198801015840
119895=
3
ℎ
(119888119895+1
minus 119888119895minus1
)
11988010158401015840
119895=
6
ℎ2(119888119895minus1
minus 2119888119895+ 119888119895+1
)
(6)
3 Numerical Scheme
The telegraph equation (1) is decomposed into a system ofpartial differential equations using the following transforma-tion
Let 119906119905(119909 119905) = V(119909 119905)
Then the transformed form of (1) is
119906119905= V
V119905= minus2120572V minus 120573
2119906 + 119906119909119909
+ 119891 (119909 119905) + 119892 (119906)
(7)
For solving the coupled equations (7) we assume our approx-imate solution as the linear combination of cubic 119861-splinebasis function
119880 (119909 119905) =
119873+1
sum
119895=minus1
119888119895(119905) 119861119895(119909) (8)
Using approximate solution (8) approximate values of119880119905(119909 119905) can be written as follows
119880119905(119909 119905) =
119873+1
sum
119895=minus1
119888119895(119905) 119861119895(119909) (9)
where 119888119895(119905) is the derivative of 119888
119895(119905) with respect to time 119905
Using approximate solution (8) and Neumann boundaryconditions (3) the approximate solution at the boundarypoints can be written as
119880119909(1199090 119905) =
1
sum
119895=minus1
1198881198951198611015840
119895(1199090) = 1199081(119905)
119880119909(119909119873 119905) =
119873+1
sum
119895=119873minus1
1198881198951198611015840
119895(119909119873) = 1199082(119905)
(10)
Using Table 1
3
ℎ
(1198881minus 119888minus1
) = 1199081(119905)
3
ℎ
(119888119873+1
minus 119888119873minus1
) = 1199082(119905)
(11)
From (11) we have
1198881minus 119888minus1
=
ℎ
3
1(119905)
119888119873+1
minus 119888119873minus1
=
ℎ
3
2(119905)
(12)
Now using (8) and (9) in the coupled system (7) we have
119873+1
sum
119895=minus1
119888119895119861119895(119909) = V
119895 0 le 119895 le 119873
V119895= minus2120572V
119895minus 1205732
119873+1
sum
119895=minus1
119888119895119861119895(119909) +
119873+1
sum
119895=minus1
11988811989511986110158401015840
119895(119909)
+ 119891 (119909119895 119905) + 119892(
119873+1
sum
119895=minus1
119888119895119861119895(119909)) 0 le 119895 le 119873
(13)
Using (6) and Table 1 in (13) we have
119888119895minus1
+ 4 119888119895+ 119888119895+1
= V119895 0 le 119895 le 119873
V119895= minus2120572V
119895minus 1205732(119888119895minus1
+ 4119888119895+ 119888119895+1
)
+
6
ℎ2(119888119895minus1
minus 2119888119895+ 119888119895+1
) + 119891 (119909119895 119905)
+ 119892 (119888119895minus1
+ 4119888119895+ 119888119895+1
) 0 le 119895 le 119873
(14)
Eliminating 119888minus1 119888119873+1
119888minus1 119888119873+1
in (14) by using (11)and (12) we get following system of first order differentialequations
119872 119888 = 119883
= 119884
(15)
[
[
[
[
[
[
[
[
4 2 sdot sdot sdot sdot sdot sdot
1 4 1 sdot sdot sdot
sdot sdot sdot sdot sdot sdot sdot sdot sdot
sdot sdot sdot sdot sdot sdot sdot sdot sdot
1 4 1
2 4
]
]
]
]
]
]
]
]
[
[
[
[
[
[
[
[
1198880
1198881
sdot sdot sdot
sdot sdot sdot
119888119873minus1
119888119873
]
]
]
]
]
]
]
]
=
[
[
[
[
[
[
[
[
1205830
1205831
sdot sdot sdot
sdot sdot sdot
120583119873minus1
120583119873
]
]
]
]
]
]
]
]
[
[
[
[
[
[
[
[
V0
V1
sdot sdot sdot
sdot sdot sdot
V119873minus1
V119873
]
]
]
]
]
]
]
]
=
[
[
[
[
[
[
[
[
1205940
1205941
sdot sdot sdot
sdot sdot sdot
120594119873minus1
120594119873
]
]
]
]
]
]
]
]
(16)
where
1205830= V0+
ℎ
3
1(119905)
120583119895= V119895
120583119873
= V119873
minus
ℎ
3
2(119905)
1205940= minus2120572V
0minus 1205732(41198880+ 21198881minus
ℎ
3
1199081(119905))
+
6
ℎ2(minus21198880+ 21198881minus
ℎ
3
1199081(119905)) + 119891 (119909
0 119905)
4 International Journal of Computational Mathematics
+ 119892(41198880+ 21198881minus
ℎ
3
1199081(119905))
120594119895= minus2120572V
119895minus 1205732(119888119895minus1
+ 4119888119895+ 119888119895+1
)
+
6
ℎ2(119888119895minus1
minus 2119888119895+ 119888119895+1
) + 119891 (119909119895 119905)
+ 119892 (119888119895minus1
+ 4119888119895+ 119888119895+1
)
120594119873
= minus2120572V119873
minus 1205732(2119888119873minus1
+ 4119888119873
+
ℎ
3
1199082(119905))
+
6
ℎ2(minus2119888119873
+ 2119888119873minus1
+
ℎ
3
1199082(119905)) + 119891 (119909
119873 119905)
+ 119892(2119888119873minus1
+ 4119888119873
+
ℎ
3
1199082(119905))
(17)
119872 is (119873 + 1) times (119873 + 1) tridiagonal matrix is columnvector of order (119873+1) and119883 and 119884 are (119873+1) order vectorrepresents the rhs of system (14)
Once the vector 119888 = [1198880 1198881 119888
119873]119879 is computed at a
specific time level the approximate solution 119880(119909 119905) can becomputed at the required knots
System (16) represents a system of (2119873 + 2) first orderordinary differential equations which is solved using SSP-RK54 [20] scheme with a variant of Thomas algorithm andconsequently the approximate solution119880
119873(119909 119905) is completely
known
4 Initial Vectors
To solve the resulting systems of first order ordinary differ-ential equations we need initial vectors 119888
0 and V0 whichare computed from the initial and boundary conditions asfollows
41 Initial Vector 1198880 Initial vector 119888
0 is computed from theinitial condition and boundary values of the derivatives as thefollowing expressions
119880119909(1199090 0) = 119908
1(0) 119895 = 0
119880 (119909119895 0) = 119891
1(119909119895) 119895 = 2 3 119873 minus 1
119880119909(119909119873 0) = 119908
2(0) 119895 = 119873
(18)
Using (10)ndash(12) in (18) we get a (119873 + 1) times (119873 + 1) system ofequations of the form
1198721198880= 119867 (19)
where
119872 =
[
[
[
[
[
[
[
[
4 2 sdot sdot sdot sdot sdot sdot
1 4 1 sdot sdot sdot
sdot sdot sdot sdot sdot sdot sdot sdot sdot
sdot sdot sdot sdot sdot sdot sdot sdot sdot
1 4 1
2 4
]
]
]
]
]
]
]
]
1198880=
[
[
[
[
[
[
[
[
[
[
[
[
1198880
0
1198880
1
sdot sdot sdot
sdot sdot sdot
1198880
119873minus1
1198880
119873
]
]
]
]
]
]
]
]
]
]
]
]
119867 =
[
[
[
[
[
[
[
[
[
[
[
[
[
[
[
[
[
[
[
1198911(1199090) +
ℎ
3
1199081(0)
1198911(1199091)
sdot sdot sdot
sdot sdot sdot
1198911(119909119873minus1
)
1198911(119909119873) minus
ℎ
3
1199082(0)
]
]
]
]
]
]
]
]
]
]
]
]
]
]
]
]
]
]
]
(20)
The matrix 119872 is tridiagonal matrix System (19) issolved using Thomas algorithm and hence initial vector 119888
0 iscomputed
42 Initial Vector V0 Initial vector V0 can be computed usingthe initial condition (2) as
119880119905(119909119895 0) = 119891
2(119909119895) (21)
We have
V (119909119895 0) = 119891
2(119909) 119895 = 0 1 119873 minus 1119873 (22)
5 Stability of Scheme
The stability of system (16) is very important since it is relatedto the stability of numerical scheme for solving it If thesystem of ordinary differential equations (16) is unstablethen stable numerical scheme for temporal discretizationmaynot generate converged solution The stability of this systemdepends on the eigenvalues of coefficient matrix since itsexact solution can be directly found using its eigenvalues
For linear case that is if 119892(119906) = 0 we consider
119860
119889119862
119889119905
= 119861119862 + 119865 (119905) (23)
where 119862 = [1198881 1198882 119888
119873minus1 V1 V2 V
119873minus1] is the unknown
vector and119865(119905) is a column vector of order 2(119873minus1) Consider
[
1198791
119874
119874 119868] [
119888
V] = [
119874 119868
1198792
minus2120572119868] [
119888
V] + 119865 (119905) (24)
where 1198791and 119879
2are symmetric tridiagonal matrix of order
119873 minus 1 given by
International Journal of Computational Mathematics 5
1198791=
[
[
[
[
[
[
[
[
4 1 sdot sdot sdot sdot sdot sdot
1 4 1 sdot sdot sdot
sdot sdot sdot sdot sdot sdot sdot sdot sdot
sdot sdot sdot sdot sdot sdot sdot sdot sdot
1 4 1
1 4
]
]
]
]
]
]
]
]
1198792=
[
[
[
[
[
[
[
[
[
[
[
[
[
[
[
[
[
[
minus41205732minus
12
ℎ2
minus1205732+
6
ℎ2
sdot sdot sdot sdot sdot sdot
minus1205732+
6
ℎ2
minus41205732minus
12
ℎ2
minus1205732+
6
ℎ2
sdot sdot sdot
sdot sdot sdot sdot sdot sdot sdot sdot sdot
sdot sdot sdot sdot sdot sdot sdot sdot sdot
minus1205732+
6
ℎ2
minus41205732minus
12
ℎ2
minus1205732+
6
ℎ2
0 minus1205732+
6
ℎ2
minus41205732minus
12
ℎ2
]
]
]
]
]
]
]
]
]
]
]
]
]
]
]
]
]
]
(25)
where 119868 and 119874 are identity and null matrix of order 119873 minus 1respectively
Stability of system (23) depends on the eigenvalues ofthe coefficient matrix 119860
minus1119861 If all the eigenvalues are having
negative real part then the system (23) will be stableLet 120582 be any eigenvalue of 119860
minus1119861 and 119909
1and 119909
2are
two components each of order (119873 minus 1) of eigenvectorcorresponding to eigenvalue 120582 We have
[
[
119874 119879minus1
1
1198792
minus2120572119868
]
]
[
1199091
1199092
] = 120582 [
1199091
1199092
] (26)
From (26) we have
119879minus1
11199092= 1205821199091
11987921199091minus 2120572119909
2= 1205821199092
(27)
From the above system of equations we get
1198792119879minus1
11199092= (1205822+ 2120572120582) 119909
2 (28)
rArr 120582(120582 + 2120572) is an eigenvalue of 1198792119879minus1
1
Since 1198791and 119879
2are tridiagonal matrices so their eigen-
values are
120582119904(1198791) = 4 + 2 cos 120587119904
119873
119904 = 1 2 119873 minus 1
120582119904(1198792) = minus2120573
2minus 41205732cos2 120587119904
2119873
minus
24
ℎ2sin2 120587119904
2119873
119904 = 1 2 119873 minus 1
(29)
1198791and 119879
2are symmetric matrices so eigenvalues of
matrix 1198792119879minus1
1are
120582119904(1198792119879minus1
1) = (minus2120573
2minus 41205732cos2 120587119904
2119873
minus
24
ℎ2sin2 120587119904
2119873
)
times (4 + 2 cos 120587119904
119873
)
minus1
119904 = 1 2 119873 minus 1
(30)
which is negative and realLet 120582 = 119909 + 119894119910 where 119909 and 119910 are real numbersWe have that (119909 + 119894119910)(119909 + 119894119910 + 2120572) is real and negative
which implies
119909 (119909 + 2120572) minus 1199102lt 0
119910 (119909 + 120572) = 0
(31)
From the above set of equations we get the solutions asfollows
(i) 119909 + 120572 = 0 and 119910 is arbitrary real number
rArr 119909 is negative real number since 120572 is real and posi-tive
(ii) 119910 = 0
rArr 119909(119909 + 2120572) lt 0
rArr (119909 + 120572)2lt 1205722
rArr 119909 is negative and real since 120572 is real and positive
Hence the real part of eigenvalues of 119860minus1119861 should always benegative for stability
6 International Journal of Computational Mathematics
Table 2 The errors of numerical solutions with Δ119905 = 001
119905
ℎ = 05 ℎ = 02
1198712
119871infin
RMS error 1198712
Łinfin
RMS error1 51145119864 minus 4 49491119864 minus 4 20328119864 minus 4 17865119864 minus 4 16756119864 minus 4 71176119864 minus 5
2 32011119864 minus 4 24541119864 minus 4 12722119864 minus 4 15356119864 minus 4 10723119864 minus 4 61178119864 minus 5
3 19952119864 minus 4 13436119864 minus 4 78599119864 minus 4 10689119864 minus 4 66156119864 minus 5 42587119864 minus 5
Table 3 Comparison of RMS errors at 119905 = 3
ℎ
The proposed method Dehghan and Ghesmati [16]RMS error CPU time (s) RMS error CPU time (s)
05 78599119864 minus 4 70 3010119864 minus 4 mdash
02 42587119864 minus 5 17 7128119864 minus 5 mdash
01 37601119864 minus 5 31 4320119864 minus 5 mdash
When 119892(119906) = 0 that is when (1) is nonlinear we have thefollowing system
119860
119889119862
119889119905
= 119865 (119905 119888) (32)
The stability can be discussed by finding the eigenvalues ofthe matrix 119860
minus1119869 where 119869 = 120597119865120597119888 is the Jacobian matrix For
stability eigenvalues of matrix 119860minus1
119869 should be negative
6 Numerical Experiments
In this section three experiments including linear and non-linear form of problem (1)ndash(3) are considered to demonstratethe accuracy and applicability of the proposed method Thenumerical efficiency is shown by calculating the 119871
infin 1198712 and
root mean square error norms
Example 1 We consider the following linear telegraph equa-tion
119906119905119905(119909 119905) + 8119906
119905(119909 119905) + 4119906 (119909 119905)
= 119906119909119909
(119909 119905) minus 2119890minus119905 sin119909 0 le 119909 le 2120587 119905 ge 0
(33)
with initial conditions
119906 (119909 0) = sin119909 119906119905(119909 0) = minus sin119909 (34)
and boundary conditions
119906119909(0 119905) = 119906
119909(2120587 119905) = 119890
minus119905 (35)
The exact solution is given [16] by
119906 (119909 119905) = 119890minus119905 sin119909 (36)
1198712 119871infin and RMS errors are reported in Table 2 with
ℎ = 02 ℎ = 05 and Δ119905 = 001 RMS errors with differentspace step sizes are reported in Table 3 and compared withthe results given by Dehghan and Ghesmati [16] It is noticedthat our results are in good agreement with the results of [16]From Figure 1 it is clear that approximate solution coincides
0 1 2 3 4 5 6
x
minus04
minus03
minus02
minus01
0
01
02
03
04
u(xt)
Numerical solution at t = 1
Exact solution at t = 1
Numerical solution at t = 2
Exact solution at t = 2
Numerical solution at t = 3
Exact solution at t = 3
Figure 1 Numerical and exact solutions with ℎ = 05 Δ119905 = 001
005
115
225
3
t
minus1
minus05
0
05
1
u(xt)
0 12
34 5
6
x
Figure 2 Space-time graph of numerical solutions with ℎ = 05Δ119905 = 001
with the exact solution at 119905 = 1 2 3 with ℎ = 05 and Δ119905 =
001 Figure 2 depicts the space-time graph of approximatesolution up to time 119905 = 3 with ℎ = 05 and Δ119905 = 001Similar figures have been depicted in [16 17] It is clear thatthe obtained numerical results are accurate and comparablefavorably with the exact solutions and earlier studies
Example 2 We consider the following linear telegraph equa-tion
119906119905119905(119909 119905) + 12119906
119905(119909 119905) + 4119906 (119909 119905)
= 119906119909119909
(119909 119905) minus 12 sin 119905 sin119909 + 4 cos 119905 cos119909
0 le 119909 le 4 119905 ge 0
(37)
with initial conditions
119906 (119909 0) = sin119909 119906119905(119909 0) = 0 (38)
International Journal of Computational Mathematics 7
Table 4 The errors of numerical solutions with Δ119905 = 001
119905
ℎ = 05 ℎ = 02
1198712
119871infin
RMS error 1198712
Łinfin
RMS error1 67752119864 minus 4 66070119864 minus 4 33667119864 minus 4 52299119864 minus 4 38981119864 minus 4 26084119864 minus 4
2 60042119864 minus 4 51361119864 minus 4 29835119864 minus 4 57309119864 minus 4 43320119864 minus 4 28588119864 minus 4
3 17361119864 minus 4 17448119864 minus 4 86268119864 minus 5 78085119864 minus 5 66723119864 minus 5 38945119864 minus 5
Table 5 Comparison of RMS errors at 119905 = 2 for Example 2
ℎ
The proposed method Dehghan and Ghesmati [16]RMS error CPU time (s) RMS error CPU time (s)
05 29835119864 minus 4 56 2153119864 minus 4 mdash
01 28758119864 minus 4 24 7012119864 minus 5 mdash
005 28819119864 minus 4 42 4003119864 minus 5 mdash
0 05 1 15 2 25 3 35 4
x
minus1
minus08
minus06
minus04
minus02
0
02
04
06
08
u(xt)
Numerical solution at t = 1
Exact solution at t = 1
Numerical solution at t = 2
Exact solution at t = 2
Numerical solution at t = 3
Exact solution at t = 3
Figure 3 Numerical and exact solutions with ℎ = 05 Δ119905 = 001
and boundary conditions
119906119909(0 119905) = cos 119905 119906
119909(4 119905) = cos 119905 cos 4 (39)
The exact solution is given [16] by
119906 (119909 119905) = cos 119905 sin119909 (40)
1198712 119871infin and RMS errors are reported in Table 4 with
ℎ = 05 and Δ119905 = 001 In Table 5 RMS errors are reportedwith different space step sizes and compared with the resultsgiven by Dehghan and Ghesmati [16] Figure 3 shows thatthe approximate solution and exact solution coincide for 119905 =
1 2 3 with ℎ = 05 and Δ119905 = 001 Space-time graph ofapproximate solution is depicted in Figure 4 with ℎ = 05 andΔ119905 = 001 up to 119905 = 3 Similar figures have been depicted in[16 17]
Example 3 In this example we consider the nonlinear tele-graph equation
119906119905119905(119909 119905) + 119906
119905(119909 119905)
= 119906119909119909
(119909 119905) minus 2 sin 119906 minus 1205872119890minus119905 cos (120587119909)
+ 2 sin (119890minus119905
(1 minus cos (120587119909))) 0 le 119909 le 2 119905 ge 0
(41)
0051
152
253
t
minus1
minus05
0
05
1
u(xt)
005 1 15
2 25 335
4
x
Figure 4 Space-time graph of numerical solutions with ℎ = 05Δ119905 = 001
0 02 04 06 08 1 12 14 16 18 2
x
minus01
0
01
02
03
04
05
06
07
08
u(xt)
Numerical solution at t = 1
Exact solution at t = 1
Numerical solution at t = 2
Exact solution at t = 2
Numerical solution at t = 3
Exact solution at t = 3
Numerical solution at t = 5
Exact solution at t = 5
Figure 5 Numerical and exact solutions with ℎ = 05 Δ119905 = 001
with initial conditions
119906 (119909 0) = 1 minus cos (120587119909) 119906119905(119909 0) = minus1 + cos (120587119909)
(42)
and boundary conditions
119906119909(0 119905) = 119906
119909(2 119905) = 0 (43)
The exact solution is given [17] by
119906 (119909 119905) = 119890minus119905
(1 minus cos (120587119909)) (44)
1198712 119871infin and RMS errors are computed for different values
of 119905 and reported in Table 6 with ℎ = 05 and Δ119905 = 001 InTable 7 error norms are reported for ℎ = 1 and Δ119905 = 01
and compared with the results given by L B Liu and H WLiu [17] FromFigure 5 we observe that approximate solutionand exact solution coincide for 119905 = 1 2 3 with ℎ = 05
8 International Journal of Computational Mathematics
Table 6 The errors of numerical solutions with Δ119905 = 001
119905
ℎ = 05 ℎ = 02
1198712
119871infin
RMS error 1198712
Łinfin
RMS error1 13341119864 minus 3 14954119864 minus 3 93182119864 minus 4 28200119864 minus 4 30385119864 minus 4 19841119864 minus 4
3 31344119864 minus 4 38812119864 minus 4 21892119864 minus 4 16958119864 minus 4 15103119864 minus 4 11939119864 minus 4
5 42496119864 minus 5 49131119864 minus 5 29681119864 minus 5 56791119864 minus 5 50866119864 minus 5 39958119864 minus 5
7 21773119864 minus 5 26530119864 minus 5 15207119864 minus 5 24272119864 minus 5 19499119864 minus 5 17500119864 minus 5
10 10059119864 minus 5 11691119864 minus 5 70261119864 minus 6 42303119864 minus 6 37155119864 minus 6 38945119864 minus 6
15 51178119864 minus 7 66345119864 minus 7 35574119864 minus 7 23315119864 minus 7 24379119864 minus 7 38945119864 minus 7
Table 7 Comparison of RMS errors at different time levels for Example 3
119905
The proposed method L B Liu and H W Liu [17]ℎ = 1 Δ119905 = 01 ℎ = 05 Δ119905 = 01
1198712
119871infin
RMS error CPU time RMS1 28083119864 minus 3 33349119864 minus 3 19379119864 minus 3 03 7591119864 minus 6
2 24692119864 minus 3 22557119864 minus 3 17038119864 minus 3 04 1833119864 minus 6
5 38689119864 minus 4 27429119864 minus 4 26698119864 minus 4 06 6976119864 minus 7
7 15951119864 minus 4 15796119864 minus 4 11007119864 minus 4 08 mdash10 38193119864 minus 5 43924119864 minus 5 26355119864 minus 5 11 5177119864 minus 8
15 22855119864 minus 6 21111119864 minus 6 11584119864 minus 6 14 3943119864 minus 9
005
115
2
x01
23
45
0
05
1
15
2
t
u(xt)
Figure 6 Space-time graph of numerical solutions with ℎ = 05Δ119905 = 001
and Δ119905 = 001 Space-time graph of approximate solution isdepicted in Figure 6 with ℎ = 05 andΔ119905 = 001 which showsthe approximate solution profile up to 119905 = 3 Similar figurehas been depicted in [17]
It should be noticed that in this case we obtain a system ofnonlinear first order ordinary differential equations Unlikethe numerical methods [17] which use finite differencemethods to solve nonlinear hyperbolic wave equation to geta system of nonlinear algebraic equations and finally solvedusing the Newton-Raphson method we do not need to doany extra effort to handle the nonlinear term and solutionsare easily computed by solving the obtained system of ODEsusing SSP-RK54 scheme Hence the present scheme reducesthe computational cost
7 Conclusions
The main idea of the present paper is to first convert thegiven problem into a coupled system of partial differential
equations and then using cubic 119861-spline basis functionsfor spatial variable and derivatives the coupled system ofequations is converted into system of first order ordinarydifferential equations The resulting system of first orderordinary differential equations has been solved by SSP-RK54schemeThe stability of the scheme is discussed using matrixstability analysis and found to be unconditionally stable Themethod is also capable of solving nonlinear telegraph equa-tion with Neumann boundary conditions It has been noticedthat obtained numerical solutions are in good agreementwiththe exact solution and earlier studies These facts illustratethat the proposed method is a reliable valid and powerfultool for solving telegraph equation with Neumann boundaryconditions
Conflict of Interests
The authors declare that there is no conflict of interestsregarding to the publication of this paper
References
[1] M S El-Azab and M El-Gamel ldquoA numerical algorithm forthe solution of telegraph equationsrdquo Applied Mathematics andComputation vol 190 no 1 pp 757ndash764 2007
[2] M Dehghan and A Shokri ldquoA numerical method for solvingthe hyperbolic telegraph equationrdquo Numerical Methods forPartial Differential Equations vol 24 no 4 pp 1080ndash1093 2008
[3] L B Liu and H W Liu ldquoQuartic spline methods for solvingone-dimensional telegraphic equationsrdquo Applied Mathematicsand Computation vol 216 no 3 pp 951ndash958 2010
[4] H-W Liu and L-B Liu ldquoAn unconditionally stable splinedifference scheme of (k2 + h4) for solving the second-order1D linear hyperbolic equationrdquo Mathematical and ComputerModelling vol 49 no 9-10 pp 1985ndash1993 2009
International Journal of Computational Mathematics 9
[5] M Dosti and A Nazemi ldquoQuartic B-spline collocation methodfor solving one-dimensional hyperbolic telegraph equationrdquoThe Journal of Information and Computing Science vol 7 no2 pp 83ndash90 2012
[6] R C Mittal and R Bhatia ldquoNumerical solution of second orderone dimensional hyperbolic telegraph equation by cubic B-spline collocation methodrdquo Applied Mathematics and Compu-tation vol 220 pp 496ndash506 2013
[7] R K Mohanty ldquoAn unconditionally stable finite differenceformula for a linear second order one space dimensional hyper-bolic equation with variable coefficientsrdquo Applied Mathematicsand Computation vol 165 no 1 pp 229ndash236 2005
[8] R K Mohanty M K Jain and K George ldquoOn the use ofhigh order difference methods for the system of one spacesecond order nonlinear hyperbolic equations with variablecoefficientsrdquo Journal of Computational and Applied Mathemat-ics vol 72 no 2 pp 421ndash431 1996
[9] R KMohanty ldquoAnunconditionally stable difference scheme forthe one-space-dimensional linear hyperbolic equationrdquoAppliedMathematics Letters vol 17 no 1 pp 101ndash105 2004
[10] F Gao and C Chi ldquoUnconditionally stable difference schemesfor a one-space-dimensional linear hyperbolic equationrdquoApplied Mathematics and Computation vol 187 no 2 pp 1272ndash1276 2007
[11] M Dehghan andM Lakestani ldquoThe use of Chebyshev cardinalfunctions for solution of the second-order one-dimensionaltelegraph equationrdquo Numerical Methods for Partial DifferentialEquations vol 25 no 4 pp 931ndash938 2009
[12] A Saadatmandi and M Dehghan ldquoNumerical solution ofhyperbolic telegraph equation using the Chebyshev taumethodrdquo Numerical Methods for Partial Differential Equationsvol 26 no 1 pp 239ndash252 2010
[13] A Mohebbi and M Dehghan ldquoHigh order compact solution ofthe one-space-dimensional linear hyperbolic equationrdquoNumer-ical Methods for Partial Differential Equations vol 24 no 5 pp1222ndash1235 2008
[14] M Lakestani and B N Saray ldquoNumerical solution of telegraphequation using interpolating scaling functionsrdquo Computers ampMathematics with Applications vol 60 no 7 pp 1964ndash19722010
[15] M Esmaeilbeigi M M Hosseini and S T Mohyud-Din ldquoAnew approach of the radial basis functionsmethod for telegraphequationsrdquo International Journal of Physical Sciences vol 6 no6 pp 1517ndash1527 2011
[16] M Dehghan and A Ghesmati ldquoSolution of the second-orderone-dimensional hyperbolic telegraph equation by using thedual reciprocity boundary integral equation (DRBIE) methodrdquoEngineering Analysis with Boundary Elements vol 34 no 1 pp51ndash59 2010
[17] L B Liu and H W Liu ldquoCompact difference schemes forsolving telegraphic equations with Neumann boundary condi-tionsrdquo Applied Mathematics and Computation vol 219 no 19pp 10112ndash10121 2013
[18] R CMittal and R K Jain ldquoCubic B-splines collocationmethodfor solving nonlinear parabolic partial differential equationswith Neumann boundary conditionsrdquoCommunications in Non-linear Science andNumerical Simulation vol 17 no 12 pp 4616ndash4625 2012
[19] R C Mittal and R K Jain ldquoNumerical solutions of nonlinearBurgersrsquo equation with modified cubic B-Splines collocationmethodrdquo Applied Mathematics and Computation vol 218 no15 pp 7839ndash7855 2012
[20] R J Spiteri and S J Ruuth ldquoA new class of optimal high-orderstrong-stability-preserving time discretization methodsrdquo SIAMJournal on Numerical Analysis vol 40 no 2 pp 469ndash491 2002
Submit your manuscripts athttpwwwhindawicom
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Function Spaces
Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014
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Discrete Dynamics in Nature and Society
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Decision SciencesAdvances in
Discrete MathematicsJournal of
Hindawi Publishing Corporationhttpwwwhindawicom
Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Stochastic AnalysisInternational Journal of
4 International Journal of Computational Mathematics
+ 119892(41198880+ 21198881minus
ℎ
3
1199081(119905))
120594119895= minus2120572V
119895minus 1205732(119888119895minus1
+ 4119888119895+ 119888119895+1
)
+
6
ℎ2(119888119895minus1
minus 2119888119895+ 119888119895+1
) + 119891 (119909119895 119905)
+ 119892 (119888119895minus1
+ 4119888119895+ 119888119895+1
)
120594119873
= minus2120572V119873
minus 1205732(2119888119873minus1
+ 4119888119873
+
ℎ
3
1199082(119905))
+
6
ℎ2(minus2119888119873
+ 2119888119873minus1
+
ℎ
3
1199082(119905)) + 119891 (119909
119873 119905)
+ 119892(2119888119873minus1
+ 4119888119873
+
ℎ
3
1199082(119905))
(17)
119872 is (119873 + 1) times (119873 + 1) tridiagonal matrix is columnvector of order (119873+1) and119883 and 119884 are (119873+1) order vectorrepresents the rhs of system (14)
Once the vector 119888 = [1198880 1198881 119888
119873]119879 is computed at a
specific time level the approximate solution 119880(119909 119905) can becomputed at the required knots
System (16) represents a system of (2119873 + 2) first orderordinary differential equations which is solved using SSP-RK54 [20] scheme with a variant of Thomas algorithm andconsequently the approximate solution119880
119873(119909 119905) is completely
known
4 Initial Vectors
To solve the resulting systems of first order ordinary differ-ential equations we need initial vectors 119888
0 and V0 whichare computed from the initial and boundary conditions asfollows
41 Initial Vector 1198880 Initial vector 119888
0 is computed from theinitial condition and boundary values of the derivatives as thefollowing expressions
119880119909(1199090 0) = 119908
1(0) 119895 = 0
119880 (119909119895 0) = 119891
1(119909119895) 119895 = 2 3 119873 minus 1
119880119909(119909119873 0) = 119908
2(0) 119895 = 119873
(18)
Using (10)ndash(12) in (18) we get a (119873 + 1) times (119873 + 1) system ofequations of the form
1198721198880= 119867 (19)
where
119872 =
[
[
[
[
[
[
[
[
4 2 sdot sdot sdot sdot sdot sdot
1 4 1 sdot sdot sdot
sdot sdot sdot sdot sdot sdot sdot sdot sdot
sdot sdot sdot sdot sdot sdot sdot sdot sdot
1 4 1
2 4
]
]
]
]
]
]
]
]
1198880=
[
[
[
[
[
[
[
[
[
[
[
[
1198880
0
1198880
1
sdot sdot sdot
sdot sdot sdot
1198880
119873minus1
1198880
119873
]
]
]
]
]
]
]
]
]
]
]
]
119867 =
[
[
[
[
[
[
[
[
[
[
[
[
[
[
[
[
[
[
[
1198911(1199090) +
ℎ
3
1199081(0)
1198911(1199091)
sdot sdot sdot
sdot sdot sdot
1198911(119909119873minus1
)
1198911(119909119873) minus
ℎ
3
1199082(0)
]
]
]
]
]
]
]
]
]
]
]
]
]
]
]
]
]
]
]
(20)
The matrix 119872 is tridiagonal matrix System (19) issolved using Thomas algorithm and hence initial vector 119888
0 iscomputed
42 Initial Vector V0 Initial vector V0 can be computed usingthe initial condition (2) as
119880119905(119909119895 0) = 119891
2(119909119895) (21)
We have
V (119909119895 0) = 119891
2(119909) 119895 = 0 1 119873 minus 1119873 (22)
5 Stability of Scheme
The stability of system (16) is very important since it is relatedto the stability of numerical scheme for solving it If thesystem of ordinary differential equations (16) is unstablethen stable numerical scheme for temporal discretizationmaynot generate converged solution The stability of this systemdepends on the eigenvalues of coefficient matrix since itsexact solution can be directly found using its eigenvalues
For linear case that is if 119892(119906) = 0 we consider
119860
119889119862
119889119905
= 119861119862 + 119865 (119905) (23)
where 119862 = [1198881 1198882 119888
119873minus1 V1 V2 V
119873minus1] is the unknown
vector and119865(119905) is a column vector of order 2(119873minus1) Consider
[
1198791
119874
119874 119868] [
119888
V] = [
119874 119868
1198792
minus2120572119868] [
119888
V] + 119865 (119905) (24)
where 1198791and 119879
2are symmetric tridiagonal matrix of order
119873 minus 1 given by
International Journal of Computational Mathematics 5
1198791=
[
[
[
[
[
[
[
[
4 1 sdot sdot sdot sdot sdot sdot
1 4 1 sdot sdot sdot
sdot sdot sdot sdot sdot sdot sdot sdot sdot
sdot sdot sdot sdot sdot sdot sdot sdot sdot
1 4 1
1 4
]
]
]
]
]
]
]
]
1198792=
[
[
[
[
[
[
[
[
[
[
[
[
[
[
[
[
[
[
minus41205732minus
12
ℎ2
minus1205732+
6
ℎ2
sdot sdot sdot sdot sdot sdot
minus1205732+
6
ℎ2
minus41205732minus
12
ℎ2
minus1205732+
6
ℎ2
sdot sdot sdot
sdot sdot sdot sdot sdot sdot sdot sdot sdot
sdot sdot sdot sdot sdot sdot sdot sdot sdot
minus1205732+
6
ℎ2
minus41205732minus
12
ℎ2
minus1205732+
6
ℎ2
0 minus1205732+
6
ℎ2
minus41205732minus
12
ℎ2
]
]
]
]
]
]
]
]
]
]
]
]
]
]
]
]
]
]
(25)
where 119868 and 119874 are identity and null matrix of order 119873 minus 1respectively
Stability of system (23) depends on the eigenvalues ofthe coefficient matrix 119860
minus1119861 If all the eigenvalues are having
negative real part then the system (23) will be stableLet 120582 be any eigenvalue of 119860
minus1119861 and 119909
1and 119909
2are
two components each of order (119873 minus 1) of eigenvectorcorresponding to eigenvalue 120582 We have
[
[
119874 119879minus1
1
1198792
minus2120572119868
]
]
[
1199091
1199092
] = 120582 [
1199091
1199092
] (26)
From (26) we have
119879minus1
11199092= 1205821199091
11987921199091minus 2120572119909
2= 1205821199092
(27)
From the above system of equations we get
1198792119879minus1
11199092= (1205822+ 2120572120582) 119909
2 (28)
rArr 120582(120582 + 2120572) is an eigenvalue of 1198792119879minus1
1
Since 1198791and 119879
2are tridiagonal matrices so their eigen-
values are
120582119904(1198791) = 4 + 2 cos 120587119904
119873
119904 = 1 2 119873 minus 1
120582119904(1198792) = minus2120573
2minus 41205732cos2 120587119904
2119873
minus
24
ℎ2sin2 120587119904
2119873
119904 = 1 2 119873 minus 1
(29)
1198791and 119879
2are symmetric matrices so eigenvalues of
matrix 1198792119879minus1
1are
120582119904(1198792119879minus1
1) = (minus2120573
2minus 41205732cos2 120587119904
2119873
minus
24
ℎ2sin2 120587119904
2119873
)
times (4 + 2 cos 120587119904
119873
)
minus1
119904 = 1 2 119873 minus 1
(30)
which is negative and realLet 120582 = 119909 + 119894119910 where 119909 and 119910 are real numbersWe have that (119909 + 119894119910)(119909 + 119894119910 + 2120572) is real and negative
which implies
119909 (119909 + 2120572) minus 1199102lt 0
119910 (119909 + 120572) = 0
(31)
From the above set of equations we get the solutions asfollows
(i) 119909 + 120572 = 0 and 119910 is arbitrary real number
rArr 119909 is negative real number since 120572 is real and posi-tive
(ii) 119910 = 0
rArr 119909(119909 + 2120572) lt 0
rArr (119909 + 120572)2lt 1205722
rArr 119909 is negative and real since 120572 is real and positive
Hence the real part of eigenvalues of 119860minus1119861 should always benegative for stability
6 International Journal of Computational Mathematics
Table 2 The errors of numerical solutions with Δ119905 = 001
119905
ℎ = 05 ℎ = 02
1198712
119871infin
RMS error 1198712
Łinfin
RMS error1 51145119864 minus 4 49491119864 minus 4 20328119864 minus 4 17865119864 minus 4 16756119864 minus 4 71176119864 minus 5
2 32011119864 minus 4 24541119864 minus 4 12722119864 minus 4 15356119864 minus 4 10723119864 minus 4 61178119864 minus 5
3 19952119864 minus 4 13436119864 minus 4 78599119864 minus 4 10689119864 minus 4 66156119864 minus 5 42587119864 minus 5
Table 3 Comparison of RMS errors at 119905 = 3
ℎ
The proposed method Dehghan and Ghesmati [16]RMS error CPU time (s) RMS error CPU time (s)
05 78599119864 minus 4 70 3010119864 minus 4 mdash
02 42587119864 minus 5 17 7128119864 minus 5 mdash
01 37601119864 minus 5 31 4320119864 minus 5 mdash
When 119892(119906) = 0 that is when (1) is nonlinear we have thefollowing system
119860
119889119862
119889119905
= 119865 (119905 119888) (32)
The stability can be discussed by finding the eigenvalues ofthe matrix 119860
minus1119869 where 119869 = 120597119865120597119888 is the Jacobian matrix For
stability eigenvalues of matrix 119860minus1
119869 should be negative
6 Numerical Experiments
In this section three experiments including linear and non-linear form of problem (1)ndash(3) are considered to demonstratethe accuracy and applicability of the proposed method Thenumerical efficiency is shown by calculating the 119871
infin 1198712 and
root mean square error norms
Example 1 We consider the following linear telegraph equa-tion
119906119905119905(119909 119905) + 8119906
119905(119909 119905) + 4119906 (119909 119905)
= 119906119909119909
(119909 119905) minus 2119890minus119905 sin119909 0 le 119909 le 2120587 119905 ge 0
(33)
with initial conditions
119906 (119909 0) = sin119909 119906119905(119909 0) = minus sin119909 (34)
and boundary conditions
119906119909(0 119905) = 119906
119909(2120587 119905) = 119890
minus119905 (35)
The exact solution is given [16] by
119906 (119909 119905) = 119890minus119905 sin119909 (36)
1198712 119871infin and RMS errors are reported in Table 2 with
ℎ = 02 ℎ = 05 and Δ119905 = 001 RMS errors with differentspace step sizes are reported in Table 3 and compared withthe results given by Dehghan and Ghesmati [16] It is noticedthat our results are in good agreement with the results of [16]From Figure 1 it is clear that approximate solution coincides
0 1 2 3 4 5 6
x
minus04
minus03
minus02
minus01
0
01
02
03
04
u(xt)
Numerical solution at t = 1
Exact solution at t = 1
Numerical solution at t = 2
Exact solution at t = 2
Numerical solution at t = 3
Exact solution at t = 3
Figure 1 Numerical and exact solutions with ℎ = 05 Δ119905 = 001
005
115
225
3
t
minus1
minus05
0
05
1
u(xt)
0 12
34 5
6
x
Figure 2 Space-time graph of numerical solutions with ℎ = 05Δ119905 = 001
with the exact solution at 119905 = 1 2 3 with ℎ = 05 and Δ119905 =
001 Figure 2 depicts the space-time graph of approximatesolution up to time 119905 = 3 with ℎ = 05 and Δ119905 = 001Similar figures have been depicted in [16 17] It is clear thatthe obtained numerical results are accurate and comparablefavorably with the exact solutions and earlier studies
Example 2 We consider the following linear telegraph equa-tion
119906119905119905(119909 119905) + 12119906
119905(119909 119905) + 4119906 (119909 119905)
= 119906119909119909
(119909 119905) minus 12 sin 119905 sin119909 + 4 cos 119905 cos119909
0 le 119909 le 4 119905 ge 0
(37)
with initial conditions
119906 (119909 0) = sin119909 119906119905(119909 0) = 0 (38)
International Journal of Computational Mathematics 7
Table 4 The errors of numerical solutions with Δ119905 = 001
119905
ℎ = 05 ℎ = 02
1198712
119871infin
RMS error 1198712
Łinfin
RMS error1 67752119864 minus 4 66070119864 minus 4 33667119864 minus 4 52299119864 minus 4 38981119864 minus 4 26084119864 minus 4
2 60042119864 minus 4 51361119864 minus 4 29835119864 minus 4 57309119864 minus 4 43320119864 minus 4 28588119864 minus 4
3 17361119864 minus 4 17448119864 minus 4 86268119864 minus 5 78085119864 minus 5 66723119864 minus 5 38945119864 minus 5
Table 5 Comparison of RMS errors at 119905 = 2 for Example 2
ℎ
The proposed method Dehghan and Ghesmati [16]RMS error CPU time (s) RMS error CPU time (s)
05 29835119864 minus 4 56 2153119864 minus 4 mdash
01 28758119864 minus 4 24 7012119864 minus 5 mdash
005 28819119864 minus 4 42 4003119864 minus 5 mdash
0 05 1 15 2 25 3 35 4
x
minus1
minus08
minus06
minus04
minus02
0
02
04
06
08
u(xt)
Numerical solution at t = 1
Exact solution at t = 1
Numerical solution at t = 2
Exact solution at t = 2
Numerical solution at t = 3
Exact solution at t = 3
Figure 3 Numerical and exact solutions with ℎ = 05 Δ119905 = 001
and boundary conditions
119906119909(0 119905) = cos 119905 119906
119909(4 119905) = cos 119905 cos 4 (39)
The exact solution is given [16] by
119906 (119909 119905) = cos 119905 sin119909 (40)
1198712 119871infin and RMS errors are reported in Table 4 with
ℎ = 05 and Δ119905 = 001 In Table 5 RMS errors are reportedwith different space step sizes and compared with the resultsgiven by Dehghan and Ghesmati [16] Figure 3 shows thatthe approximate solution and exact solution coincide for 119905 =
1 2 3 with ℎ = 05 and Δ119905 = 001 Space-time graph ofapproximate solution is depicted in Figure 4 with ℎ = 05 andΔ119905 = 001 up to 119905 = 3 Similar figures have been depicted in[16 17]
Example 3 In this example we consider the nonlinear tele-graph equation
119906119905119905(119909 119905) + 119906
119905(119909 119905)
= 119906119909119909
(119909 119905) minus 2 sin 119906 minus 1205872119890minus119905 cos (120587119909)
+ 2 sin (119890minus119905
(1 minus cos (120587119909))) 0 le 119909 le 2 119905 ge 0
(41)
0051
152
253
t
minus1
minus05
0
05
1
u(xt)
005 1 15
2 25 335
4
x
Figure 4 Space-time graph of numerical solutions with ℎ = 05Δ119905 = 001
0 02 04 06 08 1 12 14 16 18 2
x
minus01
0
01
02
03
04
05
06
07
08
u(xt)
Numerical solution at t = 1
Exact solution at t = 1
Numerical solution at t = 2
Exact solution at t = 2
Numerical solution at t = 3
Exact solution at t = 3
Numerical solution at t = 5
Exact solution at t = 5
Figure 5 Numerical and exact solutions with ℎ = 05 Δ119905 = 001
with initial conditions
119906 (119909 0) = 1 minus cos (120587119909) 119906119905(119909 0) = minus1 + cos (120587119909)
(42)
and boundary conditions
119906119909(0 119905) = 119906
119909(2 119905) = 0 (43)
The exact solution is given [17] by
119906 (119909 119905) = 119890minus119905
(1 minus cos (120587119909)) (44)
1198712 119871infin and RMS errors are computed for different values
of 119905 and reported in Table 6 with ℎ = 05 and Δ119905 = 001 InTable 7 error norms are reported for ℎ = 1 and Δ119905 = 01
and compared with the results given by L B Liu and H WLiu [17] FromFigure 5 we observe that approximate solutionand exact solution coincide for 119905 = 1 2 3 with ℎ = 05
8 International Journal of Computational Mathematics
Table 6 The errors of numerical solutions with Δ119905 = 001
119905
ℎ = 05 ℎ = 02
1198712
119871infin
RMS error 1198712
Łinfin
RMS error1 13341119864 minus 3 14954119864 minus 3 93182119864 minus 4 28200119864 minus 4 30385119864 minus 4 19841119864 minus 4
3 31344119864 minus 4 38812119864 minus 4 21892119864 minus 4 16958119864 minus 4 15103119864 minus 4 11939119864 minus 4
5 42496119864 minus 5 49131119864 minus 5 29681119864 minus 5 56791119864 minus 5 50866119864 minus 5 39958119864 minus 5
7 21773119864 minus 5 26530119864 minus 5 15207119864 minus 5 24272119864 minus 5 19499119864 minus 5 17500119864 minus 5
10 10059119864 minus 5 11691119864 minus 5 70261119864 minus 6 42303119864 minus 6 37155119864 minus 6 38945119864 minus 6
15 51178119864 minus 7 66345119864 minus 7 35574119864 minus 7 23315119864 minus 7 24379119864 minus 7 38945119864 minus 7
Table 7 Comparison of RMS errors at different time levels for Example 3
119905
The proposed method L B Liu and H W Liu [17]ℎ = 1 Δ119905 = 01 ℎ = 05 Δ119905 = 01
1198712
119871infin
RMS error CPU time RMS1 28083119864 minus 3 33349119864 minus 3 19379119864 minus 3 03 7591119864 minus 6
2 24692119864 minus 3 22557119864 minus 3 17038119864 minus 3 04 1833119864 minus 6
5 38689119864 minus 4 27429119864 minus 4 26698119864 minus 4 06 6976119864 minus 7
7 15951119864 minus 4 15796119864 minus 4 11007119864 minus 4 08 mdash10 38193119864 minus 5 43924119864 minus 5 26355119864 minus 5 11 5177119864 minus 8
15 22855119864 minus 6 21111119864 minus 6 11584119864 minus 6 14 3943119864 minus 9
005
115
2
x01
23
45
0
05
1
15
2
t
u(xt)
Figure 6 Space-time graph of numerical solutions with ℎ = 05Δ119905 = 001
and Δ119905 = 001 Space-time graph of approximate solution isdepicted in Figure 6 with ℎ = 05 andΔ119905 = 001 which showsthe approximate solution profile up to 119905 = 3 Similar figurehas been depicted in [17]
It should be noticed that in this case we obtain a system ofnonlinear first order ordinary differential equations Unlikethe numerical methods [17] which use finite differencemethods to solve nonlinear hyperbolic wave equation to geta system of nonlinear algebraic equations and finally solvedusing the Newton-Raphson method we do not need to doany extra effort to handle the nonlinear term and solutionsare easily computed by solving the obtained system of ODEsusing SSP-RK54 scheme Hence the present scheme reducesthe computational cost
7 Conclusions
The main idea of the present paper is to first convert thegiven problem into a coupled system of partial differential
equations and then using cubic 119861-spline basis functionsfor spatial variable and derivatives the coupled system ofequations is converted into system of first order ordinarydifferential equations The resulting system of first orderordinary differential equations has been solved by SSP-RK54schemeThe stability of the scheme is discussed using matrixstability analysis and found to be unconditionally stable Themethod is also capable of solving nonlinear telegraph equa-tion with Neumann boundary conditions It has been noticedthat obtained numerical solutions are in good agreementwiththe exact solution and earlier studies These facts illustratethat the proposed method is a reliable valid and powerfultool for solving telegraph equation with Neumann boundaryconditions
Conflict of Interests
The authors declare that there is no conflict of interestsregarding to the publication of this paper
References
[1] M S El-Azab and M El-Gamel ldquoA numerical algorithm forthe solution of telegraph equationsrdquo Applied Mathematics andComputation vol 190 no 1 pp 757ndash764 2007
[2] M Dehghan and A Shokri ldquoA numerical method for solvingthe hyperbolic telegraph equationrdquo Numerical Methods forPartial Differential Equations vol 24 no 4 pp 1080ndash1093 2008
[3] L B Liu and H W Liu ldquoQuartic spline methods for solvingone-dimensional telegraphic equationsrdquo Applied Mathematicsand Computation vol 216 no 3 pp 951ndash958 2010
[4] H-W Liu and L-B Liu ldquoAn unconditionally stable splinedifference scheme of (k2 + h4) for solving the second-order1D linear hyperbolic equationrdquo Mathematical and ComputerModelling vol 49 no 9-10 pp 1985ndash1993 2009
International Journal of Computational Mathematics 9
[5] M Dosti and A Nazemi ldquoQuartic B-spline collocation methodfor solving one-dimensional hyperbolic telegraph equationrdquoThe Journal of Information and Computing Science vol 7 no2 pp 83ndash90 2012
[6] R C Mittal and R Bhatia ldquoNumerical solution of second orderone dimensional hyperbolic telegraph equation by cubic B-spline collocation methodrdquo Applied Mathematics and Compu-tation vol 220 pp 496ndash506 2013
[7] R K Mohanty ldquoAn unconditionally stable finite differenceformula for a linear second order one space dimensional hyper-bolic equation with variable coefficientsrdquo Applied Mathematicsand Computation vol 165 no 1 pp 229ndash236 2005
[8] R K Mohanty M K Jain and K George ldquoOn the use ofhigh order difference methods for the system of one spacesecond order nonlinear hyperbolic equations with variablecoefficientsrdquo Journal of Computational and Applied Mathemat-ics vol 72 no 2 pp 421ndash431 1996
[9] R KMohanty ldquoAnunconditionally stable difference scheme forthe one-space-dimensional linear hyperbolic equationrdquoAppliedMathematics Letters vol 17 no 1 pp 101ndash105 2004
[10] F Gao and C Chi ldquoUnconditionally stable difference schemesfor a one-space-dimensional linear hyperbolic equationrdquoApplied Mathematics and Computation vol 187 no 2 pp 1272ndash1276 2007
[11] M Dehghan andM Lakestani ldquoThe use of Chebyshev cardinalfunctions for solution of the second-order one-dimensionaltelegraph equationrdquo Numerical Methods for Partial DifferentialEquations vol 25 no 4 pp 931ndash938 2009
[12] A Saadatmandi and M Dehghan ldquoNumerical solution ofhyperbolic telegraph equation using the Chebyshev taumethodrdquo Numerical Methods for Partial Differential Equationsvol 26 no 1 pp 239ndash252 2010
[13] A Mohebbi and M Dehghan ldquoHigh order compact solution ofthe one-space-dimensional linear hyperbolic equationrdquoNumer-ical Methods for Partial Differential Equations vol 24 no 5 pp1222ndash1235 2008
[14] M Lakestani and B N Saray ldquoNumerical solution of telegraphequation using interpolating scaling functionsrdquo Computers ampMathematics with Applications vol 60 no 7 pp 1964ndash19722010
[15] M Esmaeilbeigi M M Hosseini and S T Mohyud-Din ldquoAnew approach of the radial basis functionsmethod for telegraphequationsrdquo International Journal of Physical Sciences vol 6 no6 pp 1517ndash1527 2011
[16] M Dehghan and A Ghesmati ldquoSolution of the second-orderone-dimensional hyperbolic telegraph equation by using thedual reciprocity boundary integral equation (DRBIE) methodrdquoEngineering Analysis with Boundary Elements vol 34 no 1 pp51ndash59 2010
[17] L B Liu and H W Liu ldquoCompact difference schemes forsolving telegraphic equations with Neumann boundary condi-tionsrdquo Applied Mathematics and Computation vol 219 no 19pp 10112ndash10121 2013
[18] R CMittal and R K Jain ldquoCubic B-splines collocationmethodfor solving nonlinear parabolic partial differential equationswith Neumann boundary conditionsrdquoCommunications in Non-linear Science andNumerical Simulation vol 17 no 12 pp 4616ndash4625 2012
[19] R C Mittal and R K Jain ldquoNumerical solutions of nonlinearBurgersrsquo equation with modified cubic B-Splines collocationmethodrdquo Applied Mathematics and Computation vol 218 no15 pp 7839ndash7855 2012
[20] R J Spiteri and S J Ruuth ldquoA new class of optimal high-orderstrong-stability-preserving time discretization methodsrdquo SIAMJournal on Numerical Analysis vol 40 no 2 pp 469ndash491 2002
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
International Journal of Computational Mathematics 5
1198791=
[
[
[
[
[
[
[
[
4 1 sdot sdot sdot sdot sdot sdot
1 4 1 sdot sdot sdot
sdot sdot sdot sdot sdot sdot sdot sdot sdot
sdot sdot sdot sdot sdot sdot sdot sdot sdot
1 4 1
1 4
]
]
]
]
]
]
]
]
1198792=
[
[
[
[
[
[
[
[
[
[
[
[
[
[
[
[
[
[
minus41205732minus
12
ℎ2
minus1205732+
6
ℎ2
sdot sdot sdot sdot sdot sdot
minus1205732+
6
ℎ2
minus41205732minus
12
ℎ2
minus1205732+
6
ℎ2
sdot sdot sdot
sdot sdot sdot sdot sdot sdot sdot sdot sdot
sdot sdot sdot sdot sdot sdot sdot sdot sdot
minus1205732+
6
ℎ2
minus41205732minus
12
ℎ2
minus1205732+
6
ℎ2
0 minus1205732+
6
ℎ2
minus41205732minus
12
ℎ2
]
]
]
]
]
]
]
]
]
]
]
]
]
]
]
]
]
]
(25)
where 119868 and 119874 are identity and null matrix of order 119873 minus 1respectively
Stability of system (23) depends on the eigenvalues ofthe coefficient matrix 119860
minus1119861 If all the eigenvalues are having
negative real part then the system (23) will be stableLet 120582 be any eigenvalue of 119860
minus1119861 and 119909
1and 119909
2are
two components each of order (119873 minus 1) of eigenvectorcorresponding to eigenvalue 120582 We have
[
[
119874 119879minus1
1
1198792
minus2120572119868
]
]
[
1199091
1199092
] = 120582 [
1199091
1199092
] (26)
From (26) we have
119879minus1
11199092= 1205821199091
11987921199091minus 2120572119909
2= 1205821199092
(27)
From the above system of equations we get
1198792119879minus1
11199092= (1205822+ 2120572120582) 119909
2 (28)
rArr 120582(120582 + 2120572) is an eigenvalue of 1198792119879minus1
1
Since 1198791and 119879
2are tridiagonal matrices so their eigen-
values are
120582119904(1198791) = 4 + 2 cos 120587119904
119873
119904 = 1 2 119873 minus 1
120582119904(1198792) = minus2120573
2minus 41205732cos2 120587119904
2119873
minus
24
ℎ2sin2 120587119904
2119873
119904 = 1 2 119873 minus 1
(29)
1198791and 119879
2are symmetric matrices so eigenvalues of
matrix 1198792119879minus1
1are
120582119904(1198792119879minus1
1) = (minus2120573
2minus 41205732cos2 120587119904
2119873
minus
24
ℎ2sin2 120587119904
2119873
)
times (4 + 2 cos 120587119904
119873
)
minus1
119904 = 1 2 119873 minus 1
(30)
which is negative and realLet 120582 = 119909 + 119894119910 where 119909 and 119910 are real numbersWe have that (119909 + 119894119910)(119909 + 119894119910 + 2120572) is real and negative
which implies
119909 (119909 + 2120572) minus 1199102lt 0
119910 (119909 + 120572) = 0
(31)
From the above set of equations we get the solutions asfollows
(i) 119909 + 120572 = 0 and 119910 is arbitrary real number
rArr 119909 is negative real number since 120572 is real and posi-tive
(ii) 119910 = 0
rArr 119909(119909 + 2120572) lt 0
rArr (119909 + 120572)2lt 1205722
rArr 119909 is negative and real since 120572 is real and positive
Hence the real part of eigenvalues of 119860minus1119861 should always benegative for stability
6 International Journal of Computational Mathematics
Table 2 The errors of numerical solutions with Δ119905 = 001
119905
ℎ = 05 ℎ = 02
1198712
119871infin
RMS error 1198712
Łinfin
RMS error1 51145119864 minus 4 49491119864 minus 4 20328119864 minus 4 17865119864 minus 4 16756119864 minus 4 71176119864 minus 5
2 32011119864 minus 4 24541119864 minus 4 12722119864 minus 4 15356119864 minus 4 10723119864 minus 4 61178119864 minus 5
3 19952119864 minus 4 13436119864 minus 4 78599119864 minus 4 10689119864 minus 4 66156119864 minus 5 42587119864 minus 5
Table 3 Comparison of RMS errors at 119905 = 3
ℎ
The proposed method Dehghan and Ghesmati [16]RMS error CPU time (s) RMS error CPU time (s)
05 78599119864 minus 4 70 3010119864 minus 4 mdash
02 42587119864 minus 5 17 7128119864 minus 5 mdash
01 37601119864 minus 5 31 4320119864 minus 5 mdash
When 119892(119906) = 0 that is when (1) is nonlinear we have thefollowing system
119860
119889119862
119889119905
= 119865 (119905 119888) (32)
The stability can be discussed by finding the eigenvalues ofthe matrix 119860
minus1119869 where 119869 = 120597119865120597119888 is the Jacobian matrix For
stability eigenvalues of matrix 119860minus1
119869 should be negative
6 Numerical Experiments
In this section three experiments including linear and non-linear form of problem (1)ndash(3) are considered to demonstratethe accuracy and applicability of the proposed method Thenumerical efficiency is shown by calculating the 119871
infin 1198712 and
root mean square error norms
Example 1 We consider the following linear telegraph equa-tion
119906119905119905(119909 119905) + 8119906
119905(119909 119905) + 4119906 (119909 119905)
= 119906119909119909
(119909 119905) minus 2119890minus119905 sin119909 0 le 119909 le 2120587 119905 ge 0
(33)
with initial conditions
119906 (119909 0) = sin119909 119906119905(119909 0) = minus sin119909 (34)
and boundary conditions
119906119909(0 119905) = 119906
119909(2120587 119905) = 119890
minus119905 (35)
The exact solution is given [16] by
119906 (119909 119905) = 119890minus119905 sin119909 (36)
1198712 119871infin and RMS errors are reported in Table 2 with
ℎ = 02 ℎ = 05 and Δ119905 = 001 RMS errors with differentspace step sizes are reported in Table 3 and compared withthe results given by Dehghan and Ghesmati [16] It is noticedthat our results are in good agreement with the results of [16]From Figure 1 it is clear that approximate solution coincides
0 1 2 3 4 5 6
x
minus04
minus03
minus02
minus01
0
01
02
03
04
u(xt)
Numerical solution at t = 1
Exact solution at t = 1
Numerical solution at t = 2
Exact solution at t = 2
Numerical solution at t = 3
Exact solution at t = 3
Figure 1 Numerical and exact solutions with ℎ = 05 Δ119905 = 001
005
115
225
3
t
minus1
minus05
0
05
1
u(xt)
0 12
34 5
6
x
Figure 2 Space-time graph of numerical solutions with ℎ = 05Δ119905 = 001
with the exact solution at 119905 = 1 2 3 with ℎ = 05 and Δ119905 =
001 Figure 2 depicts the space-time graph of approximatesolution up to time 119905 = 3 with ℎ = 05 and Δ119905 = 001Similar figures have been depicted in [16 17] It is clear thatthe obtained numerical results are accurate and comparablefavorably with the exact solutions and earlier studies
Example 2 We consider the following linear telegraph equa-tion
119906119905119905(119909 119905) + 12119906
119905(119909 119905) + 4119906 (119909 119905)
= 119906119909119909
(119909 119905) minus 12 sin 119905 sin119909 + 4 cos 119905 cos119909
0 le 119909 le 4 119905 ge 0
(37)
with initial conditions
119906 (119909 0) = sin119909 119906119905(119909 0) = 0 (38)
International Journal of Computational Mathematics 7
Table 4 The errors of numerical solutions with Δ119905 = 001
119905
ℎ = 05 ℎ = 02
1198712
119871infin
RMS error 1198712
Łinfin
RMS error1 67752119864 minus 4 66070119864 minus 4 33667119864 minus 4 52299119864 minus 4 38981119864 minus 4 26084119864 minus 4
2 60042119864 minus 4 51361119864 minus 4 29835119864 minus 4 57309119864 minus 4 43320119864 minus 4 28588119864 minus 4
3 17361119864 minus 4 17448119864 minus 4 86268119864 minus 5 78085119864 minus 5 66723119864 minus 5 38945119864 minus 5
Table 5 Comparison of RMS errors at 119905 = 2 for Example 2
ℎ
The proposed method Dehghan and Ghesmati [16]RMS error CPU time (s) RMS error CPU time (s)
05 29835119864 minus 4 56 2153119864 minus 4 mdash
01 28758119864 minus 4 24 7012119864 minus 5 mdash
005 28819119864 minus 4 42 4003119864 minus 5 mdash
0 05 1 15 2 25 3 35 4
x
minus1
minus08
minus06
minus04
minus02
0
02
04
06
08
u(xt)
Numerical solution at t = 1
Exact solution at t = 1
Numerical solution at t = 2
Exact solution at t = 2
Numerical solution at t = 3
Exact solution at t = 3
Figure 3 Numerical and exact solutions with ℎ = 05 Δ119905 = 001
and boundary conditions
119906119909(0 119905) = cos 119905 119906
119909(4 119905) = cos 119905 cos 4 (39)
The exact solution is given [16] by
119906 (119909 119905) = cos 119905 sin119909 (40)
1198712 119871infin and RMS errors are reported in Table 4 with
ℎ = 05 and Δ119905 = 001 In Table 5 RMS errors are reportedwith different space step sizes and compared with the resultsgiven by Dehghan and Ghesmati [16] Figure 3 shows thatthe approximate solution and exact solution coincide for 119905 =
1 2 3 with ℎ = 05 and Δ119905 = 001 Space-time graph ofapproximate solution is depicted in Figure 4 with ℎ = 05 andΔ119905 = 001 up to 119905 = 3 Similar figures have been depicted in[16 17]
Example 3 In this example we consider the nonlinear tele-graph equation
119906119905119905(119909 119905) + 119906
119905(119909 119905)
= 119906119909119909
(119909 119905) minus 2 sin 119906 minus 1205872119890minus119905 cos (120587119909)
+ 2 sin (119890minus119905
(1 minus cos (120587119909))) 0 le 119909 le 2 119905 ge 0
(41)
0051
152
253
t
minus1
minus05
0
05
1
u(xt)
005 1 15
2 25 335
4
x
Figure 4 Space-time graph of numerical solutions with ℎ = 05Δ119905 = 001
0 02 04 06 08 1 12 14 16 18 2
x
minus01
0
01
02
03
04
05
06
07
08
u(xt)
Numerical solution at t = 1
Exact solution at t = 1
Numerical solution at t = 2
Exact solution at t = 2
Numerical solution at t = 3
Exact solution at t = 3
Numerical solution at t = 5
Exact solution at t = 5
Figure 5 Numerical and exact solutions with ℎ = 05 Δ119905 = 001
with initial conditions
119906 (119909 0) = 1 minus cos (120587119909) 119906119905(119909 0) = minus1 + cos (120587119909)
(42)
and boundary conditions
119906119909(0 119905) = 119906
119909(2 119905) = 0 (43)
The exact solution is given [17] by
119906 (119909 119905) = 119890minus119905
(1 minus cos (120587119909)) (44)
1198712 119871infin and RMS errors are computed for different values
of 119905 and reported in Table 6 with ℎ = 05 and Δ119905 = 001 InTable 7 error norms are reported for ℎ = 1 and Δ119905 = 01
and compared with the results given by L B Liu and H WLiu [17] FromFigure 5 we observe that approximate solutionand exact solution coincide for 119905 = 1 2 3 with ℎ = 05
8 International Journal of Computational Mathematics
Table 6 The errors of numerical solutions with Δ119905 = 001
119905
ℎ = 05 ℎ = 02
1198712
119871infin
RMS error 1198712
Łinfin
RMS error1 13341119864 minus 3 14954119864 minus 3 93182119864 minus 4 28200119864 minus 4 30385119864 minus 4 19841119864 minus 4
3 31344119864 minus 4 38812119864 minus 4 21892119864 minus 4 16958119864 minus 4 15103119864 minus 4 11939119864 minus 4
5 42496119864 minus 5 49131119864 minus 5 29681119864 minus 5 56791119864 minus 5 50866119864 minus 5 39958119864 minus 5
7 21773119864 minus 5 26530119864 minus 5 15207119864 minus 5 24272119864 minus 5 19499119864 minus 5 17500119864 minus 5
10 10059119864 minus 5 11691119864 minus 5 70261119864 minus 6 42303119864 minus 6 37155119864 minus 6 38945119864 minus 6
15 51178119864 minus 7 66345119864 minus 7 35574119864 minus 7 23315119864 minus 7 24379119864 minus 7 38945119864 minus 7
Table 7 Comparison of RMS errors at different time levels for Example 3
119905
The proposed method L B Liu and H W Liu [17]ℎ = 1 Δ119905 = 01 ℎ = 05 Δ119905 = 01
1198712
119871infin
RMS error CPU time RMS1 28083119864 minus 3 33349119864 minus 3 19379119864 minus 3 03 7591119864 minus 6
2 24692119864 minus 3 22557119864 minus 3 17038119864 minus 3 04 1833119864 minus 6
5 38689119864 minus 4 27429119864 minus 4 26698119864 minus 4 06 6976119864 minus 7
7 15951119864 minus 4 15796119864 minus 4 11007119864 minus 4 08 mdash10 38193119864 minus 5 43924119864 minus 5 26355119864 minus 5 11 5177119864 minus 8
15 22855119864 minus 6 21111119864 minus 6 11584119864 minus 6 14 3943119864 minus 9
005
115
2
x01
23
45
0
05
1
15
2
t
u(xt)
Figure 6 Space-time graph of numerical solutions with ℎ = 05Δ119905 = 001
and Δ119905 = 001 Space-time graph of approximate solution isdepicted in Figure 6 with ℎ = 05 andΔ119905 = 001 which showsthe approximate solution profile up to 119905 = 3 Similar figurehas been depicted in [17]
It should be noticed that in this case we obtain a system ofnonlinear first order ordinary differential equations Unlikethe numerical methods [17] which use finite differencemethods to solve nonlinear hyperbolic wave equation to geta system of nonlinear algebraic equations and finally solvedusing the Newton-Raphson method we do not need to doany extra effort to handle the nonlinear term and solutionsare easily computed by solving the obtained system of ODEsusing SSP-RK54 scheme Hence the present scheme reducesthe computational cost
7 Conclusions
The main idea of the present paper is to first convert thegiven problem into a coupled system of partial differential
equations and then using cubic 119861-spline basis functionsfor spatial variable and derivatives the coupled system ofequations is converted into system of first order ordinarydifferential equations The resulting system of first orderordinary differential equations has been solved by SSP-RK54schemeThe stability of the scheme is discussed using matrixstability analysis and found to be unconditionally stable Themethod is also capable of solving nonlinear telegraph equa-tion with Neumann boundary conditions It has been noticedthat obtained numerical solutions are in good agreementwiththe exact solution and earlier studies These facts illustratethat the proposed method is a reliable valid and powerfultool for solving telegraph equation with Neumann boundaryconditions
Conflict of Interests
The authors declare that there is no conflict of interestsregarding to the publication of this paper
References
[1] M S El-Azab and M El-Gamel ldquoA numerical algorithm forthe solution of telegraph equationsrdquo Applied Mathematics andComputation vol 190 no 1 pp 757ndash764 2007
[2] M Dehghan and A Shokri ldquoA numerical method for solvingthe hyperbolic telegraph equationrdquo Numerical Methods forPartial Differential Equations vol 24 no 4 pp 1080ndash1093 2008
[3] L B Liu and H W Liu ldquoQuartic spline methods for solvingone-dimensional telegraphic equationsrdquo Applied Mathematicsand Computation vol 216 no 3 pp 951ndash958 2010
[4] H-W Liu and L-B Liu ldquoAn unconditionally stable splinedifference scheme of (k2 + h4) for solving the second-order1D linear hyperbolic equationrdquo Mathematical and ComputerModelling vol 49 no 9-10 pp 1985ndash1993 2009
International Journal of Computational Mathematics 9
[5] M Dosti and A Nazemi ldquoQuartic B-spline collocation methodfor solving one-dimensional hyperbolic telegraph equationrdquoThe Journal of Information and Computing Science vol 7 no2 pp 83ndash90 2012
[6] R C Mittal and R Bhatia ldquoNumerical solution of second orderone dimensional hyperbolic telegraph equation by cubic B-spline collocation methodrdquo Applied Mathematics and Compu-tation vol 220 pp 496ndash506 2013
[7] R K Mohanty ldquoAn unconditionally stable finite differenceformula for a linear second order one space dimensional hyper-bolic equation with variable coefficientsrdquo Applied Mathematicsand Computation vol 165 no 1 pp 229ndash236 2005
[8] R K Mohanty M K Jain and K George ldquoOn the use ofhigh order difference methods for the system of one spacesecond order nonlinear hyperbolic equations with variablecoefficientsrdquo Journal of Computational and Applied Mathemat-ics vol 72 no 2 pp 421ndash431 1996
[9] R KMohanty ldquoAnunconditionally stable difference scheme forthe one-space-dimensional linear hyperbolic equationrdquoAppliedMathematics Letters vol 17 no 1 pp 101ndash105 2004
[10] F Gao and C Chi ldquoUnconditionally stable difference schemesfor a one-space-dimensional linear hyperbolic equationrdquoApplied Mathematics and Computation vol 187 no 2 pp 1272ndash1276 2007
[11] M Dehghan andM Lakestani ldquoThe use of Chebyshev cardinalfunctions for solution of the second-order one-dimensionaltelegraph equationrdquo Numerical Methods for Partial DifferentialEquations vol 25 no 4 pp 931ndash938 2009
[12] A Saadatmandi and M Dehghan ldquoNumerical solution ofhyperbolic telegraph equation using the Chebyshev taumethodrdquo Numerical Methods for Partial Differential Equationsvol 26 no 1 pp 239ndash252 2010
[13] A Mohebbi and M Dehghan ldquoHigh order compact solution ofthe one-space-dimensional linear hyperbolic equationrdquoNumer-ical Methods for Partial Differential Equations vol 24 no 5 pp1222ndash1235 2008
[14] M Lakestani and B N Saray ldquoNumerical solution of telegraphequation using interpolating scaling functionsrdquo Computers ampMathematics with Applications vol 60 no 7 pp 1964ndash19722010
[15] M Esmaeilbeigi M M Hosseini and S T Mohyud-Din ldquoAnew approach of the radial basis functionsmethod for telegraphequationsrdquo International Journal of Physical Sciences vol 6 no6 pp 1517ndash1527 2011
[16] M Dehghan and A Ghesmati ldquoSolution of the second-orderone-dimensional hyperbolic telegraph equation by using thedual reciprocity boundary integral equation (DRBIE) methodrdquoEngineering Analysis with Boundary Elements vol 34 no 1 pp51ndash59 2010
[17] L B Liu and H W Liu ldquoCompact difference schemes forsolving telegraphic equations with Neumann boundary condi-tionsrdquo Applied Mathematics and Computation vol 219 no 19pp 10112ndash10121 2013
[18] R CMittal and R K Jain ldquoCubic B-splines collocationmethodfor solving nonlinear parabolic partial differential equationswith Neumann boundary conditionsrdquoCommunications in Non-linear Science andNumerical Simulation vol 17 no 12 pp 4616ndash4625 2012
[19] R C Mittal and R K Jain ldquoNumerical solutions of nonlinearBurgersrsquo equation with modified cubic B-Splines collocationmethodrdquo Applied Mathematics and Computation vol 218 no15 pp 7839ndash7855 2012
[20] R J Spiteri and S J Ruuth ldquoA new class of optimal high-orderstrong-stability-preserving time discretization methodsrdquo SIAMJournal on Numerical Analysis vol 40 no 2 pp 469ndash491 2002
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
6 International Journal of Computational Mathematics
Table 2 The errors of numerical solutions with Δ119905 = 001
119905
ℎ = 05 ℎ = 02
1198712
119871infin
RMS error 1198712
Łinfin
RMS error1 51145119864 minus 4 49491119864 minus 4 20328119864 minus 4 17865119864 minus 4 16756119864 minus 4 71176119864 minus 5
2 32011119864 minus 4 24541119864 minus 4 12722119864 minus 4 15356119864 minus 4 10723119864 minus 4 61178119864 minus 5
3 19952119864 minus 4 13436119864 minus 4 78599119864 minus 4 10689119864 minus 4 66156119864 minus 5 42587119864 minus 5
Table 3 Comparison of RMS errors at 119905 = 3
ℎ
The proposed method Dehghan and Ghesmati [16]RMS error CPU time (s) RMS error CPU time (s)
05 78599119864 minus 4 70 3010119864 minus 4 mdash
02 42587119864 minus 5 17 7128119864 minus 5 mdash
01 37601119864 minus 5 31 4320119864 minus 5 mdash
When 119892(119906) = 0 that is when (1) is nonlinear we have thefollowing system
119860
119889119862
119889119905
= 119865 (119905 119888) (32)
The stability can be discussed by finding the eigenvalues ofthe matrix 119860
minus1119869 where 119869 = 120597119865120597119888 is the Jacobian matrix For
stability eigenvalues of matrix 119860minus1
119869 should be negative
6 Numerical Experiments
In this section three experiments including linear and non-linear form of problem (1)ndash(3) are considered to demonstratethe accuracy and applicability of the proposed method Thenumerical efficiency is shown by calculating the 119871
infin 1198712 and
root mean square error norms
Example 1 We consider the following linear telegraph equa-tion
119906119905119905(119909 119905) + 8119906
119905(119909 119905) + 4119906 (119909 119905)
= 119906119909119909
(119909 119905) minus 2119890minus119905 sin119909 0 le 119909 le 2120587 119905 ge 0
(33)
with initial conditions
119906 (119909 0) = sin119909 119906119905(119909 0) = minus sin119909 (34)
and boundary conditions
119906119909(0 119905) = 119906
119909(2120587 119905) = 119890
minus119905 (35)
The exact solution is given [16] by
119906 (119909 119905) = 119890minus119905 sin119909 (36)
1198712 119871infin and RMS errors are reported in Table 2 with
ℎ = 02 ℎ = 05 and Δ119905 = 001 RMS errors with differentspace step sizes are reported in Table 3 and compared withthe results given by Dehghan and Ghesmati [16] It is noticedthat our results are in good agreement with the results of [16]From Figure 1 it is clear that approximate solution coincides
0 1 2 3 4 5 6
x
minus04
minus03
minus02
minus01
0
01
02
03
04
u(xt)
Numerical solution at t = 1
Exact solution at t = 1
Numerical solution at t = 2
Exact solution at t = 2
Numerical solution at t = 3
Exact solution at t = 3
Figure 1 Numerical and exact solutions with ℎ = 05 Δ119905 = 001
005
115
225
3
t
minus1
minus05
0
05
1
u(xt)
0 12
34 5
6
x
Figure 2 Space-time graph of numerical solutions with ℎ = 05Δ119905 = 001
with the exact solution at 119905 = 1 2 3 with ℎ = 05 and Δ119905 =
001 Figure 2 depicts the space-time graph of approximatesolution up to time 119905 = 3 with ℎ = 05 and Δ119905 = 001Similar figures have been depicted in [16 17] It is clear thatthe obtained numerical results are accurate and comparablefavorably with the exact solutions and earlier studies
Example 2 We consider the following linear telegraph equa-tion
119906119905119905(119909 119905) + 12119906
119905(119909 119905) + 4119906 (119909 119905)
= 119906119909119909
(119909 119905) minus 12 sin 119905 sin119909 + 4 cos 119905 cos119909
0 le 119909 le 4 119905 ge 0
(37)
with initial conditions
119906 (119909 0) = sin119909 119906119905(119909 0) = 0 (38)
International Journal of Computational Mathematics 7
Table 4 The errors of numerical solutions with Δ119905 = 001
119905
ℎ = 05 ℎ = 02
1198712
119871infin
RMS error 1198712
Łinfin
RMS error1 67752119864 minus 4 66070119864 minus 4 33667119864 minus 4 52299119864 minus 4 38981119864 minus 4 26084119864 minus 4
2 60042119864 minus 4 51361119864 minus 4 29835119864 minus 4 57309119864 minus 4 43320119864 minus 4 28588119864 minus 4
3 17361119864 minus 4 17448119864 minus 4 86268119864 minus 5 78085119864 minus 5 66723119864 minus 5 38945119864 minus 5
Table 5 Comparison of RMS errors at 119905 = 2 for Example 2
ℎ
The proposed method Dehghan and Ghesmati [16]RMS error CPU time (s) RMS error CPU time (s)
05 29835119864 minus 4 56 2153119864 minus 4 mdash
01 28758119864 minus 4 24 7012119864 minus 5 mdash
005 28819119864 minus 4 42 4003119864 minus 5 mdash
0 05 1 15 2 25 3 35 4
x
minus1
minus08
minus06
minus04
minus02
0
02
04
06
08
u(xt)
Numerical solution at t = 1
Exact solution at t = 1
Numerical solution at t = 2
Exact solution at t = 2
Numerical solution at t = 3
Exact solution at t = 3
Figure 3 Numerical and exact solutions with ℎ = 05 Δ119905 = 001
and boundary conditions
119906119909(0 119905) = cos 119905 119906
119909(4 119905) = cos 119905 cos 4 (39)
The exact solution is given [16] by
119906 (119909 119905) = cos 119905 sin119909 (40)
1198712 119871infin and RMS errors are reported in Table 4 with
ℎ = 05 and Δ119905 = 001 In Table 5 RMS errors are reportedwith different space step sizes and compared with the resultsgiven by Dehghan and Ghesmati [16] Figure 3 shows thatthe approximate solution and exact solution coincide for 119905 =
1 2 3 with ℎ = 05 and Δ119905 = 001 Space-time graph ofapproximate solution is depicted in Figure 4 with ℎ = 05 andΔ119905 = 001 up to 119905 = 3 Similar figures have been depicted in[16 17]
Example 3 In this example we consider the nonlinear tele-graph equation
119906119905119905(119909 119905) + 119906
119905(119909 119905)
= 119906119909119909
(119909 119905) minus 2 sin 119906 minus 1205872119890minus119905 cos (120587119909)
+ 2 sin (119890minus119905
(1 minus cos (120587119909))) 0 le 119909 le 2 119905 ge 0
(41)
0051
152
253
t
minus1
minus05
0
05
1
u(xt)
005 1 15
2 25 335
4
x
Figure 4 Space-time graph of numerical solutions with ℎ = 05Δ119905 = 001
0 02 04 06 08 1 12 14 16 18 2
x
minus01
0
01
02
03
04
05
06
07
08
u(xt)
Numerical solution at t = 1
Exact solution at t = 1
Numerical solution at t = 2
Exact solution at t = 2
Numerical solution at t = 3
Exact solution at t = 3
Numerical solution at t = 5
Exact solution at t = 5
Figure 5 Numerical and exact solutions with ℎ = 05 Δ119905 = 001
with initial conditions
119906 (119909 0) = 1 minus cos (120587119909) 119906119905(119909 0) = minus1 + cos (120587119909)
(42)
and boundary conditions
119906119909(0 119905) = 119906
119909(2 119905) = 0 (43)
The exact solution is given [17] by
119906 (119909 119905) = 119890minus119905
(1 minus cos (120587119909)) (44)
1198712 119871infin and RMS errors are computed for different values
of 119905 and reported in Table 6 with ℎ = 05 and Δ119905 = 001 InTable 7 error norms are reported for ℎ = 1 and Δ119905 = 01
and compared with the results given by L B Liu and H WLiu [17] FromFigure 5 we observe that approximate solutionand exact solution coincide for 119905 = 1 2 3 with ℎ = 05
8 International Journal of Computational Mathematics
Table 6 The errors of numerical solutions with Δ119905 = 001
119905
ℎ = 05 ℎ = 02
1198712
119871infin
RMS error 1198712
Łinfin
RMS error1 13341119864 minus 3 14954119864 minus 3 93182119864 minus 4 28200119864 minus 4 30385119864 minus 4 19841119864 minus 4
3 31344119864 minus 4 38812119864 minus 4 21892119864 minus 4 16958119864 minus 4 15103119864 minus 4 11939119864 minus 4
5 42496119864 minus 5 49131119864 minus 5 29681119864 minus 5 56791119864 minus 5 50866119864 minus 5 39958119864 minus 5
7 21773119864 minus 5 26530119864 minus 5 15207119864 minus 5 24272119864 minus 5 19499119864 minus 5 17500119864 minus 5
10 10059119864 minus 5 11691119864 minus 5 70261119864 minus 6 42303119864 minus 6 37155119864 minus 6 38945119864 minus 6
15 51178119864 minus 7 66345119864 minus 7 35574119864 minus 7 23315119864 minus 7 24379119864 minus 7 38945119864 minus 7
Table 7 Comparison of RMS errors at different time levels for Example 3
119905
The proposed method L B Liu and H W Liu [17]ℎ = 1 Δ119905 = 01 ℎ = 05 Δ119905 = 01
1198712
119871infin
RMS error CPU time RMS1 28083119864 minus 3 33349119864 minus 3 19379119864 minus 3 03 7591119864 minus 6
2 24692119864 minus 3 22557119864 minus 3 17038119864 minus 3 04 1833119864 minus 6
5 38689119864 minus 4 27429119864 minus 4 26698119864 minus 4 06 6976119864 minus 7
7 15951119864 minus 4 15796119864 minus 4 11007119864 minus 4 08 mdash10 38193119864 minus 5 43924119864 minus 5 26355119864 minus 5 11 5177119864 minus 8
15 22855119864 minus 6 21111119864 minus 6 11584119864 minus 6 14 3943119864 minus 9
005
115
2
x01
23
45
0
05
1
15
2
t
u(xt)
Figure 6 Space-time graph of numerical solutions with ℎ = 05Δ119905 = 001
and Δ119905 = 001 Space-time graph of approximate solution isdepicted in Figure 6 with ℎ = 05 andΔ119905 = 001 which showsthe approximate solution profile up to 119905 = 3 Similar figurehas been depicted in [17]
It should be noticed that in this case we obtain a system ofnonlinear first order ordinary differential equations Unlikethe numerical methods [17] which use finite differencemethods to solve nonlinear hyperbolic wave equation to geta system of nonlinear algebraic equations and finally solvedusing the Newton-Raphson method we do not need to doany extra effort to handle the nonlinear term and solutionsare easily computed by solving the obtained system of ODEsusing SSP-RK54 scheme Hence the present scheme reducesthe computational cost
7 Conclusions
The main idea of the present paper is to first convert thegiven problem into a coupled system of partial differential
equations and then using cubic 119861-spline basis functionsfor spatial variable and derivatives the coupled system ofequations is converted into system of first order ordinarydifferential equations The resulting system of first orderordinary differential equations has been solved by SSP-RK54schemeThe stability of the scheme is discussed using matrixstability analysis and found to be unconditionally stable Themethod is also capable of solving nonlinear telegraph equa-tion with Neumann boundary conditions It has been noticedthat obtained numerical solutions are in good agreementwiththe exact solution and earlier studies These facts illustratethat the proposed method is a reliable valid and powerfultool for solving telegraph equation with Neumann boundaryconditions
Conflict of Interests
The authors declare that there is no conflict of interestsregarding to the publication of this paper
References
[1] M S El-Azab and M El-Gamel ldquoA numerical algorithm forthe solution of telegraph equationsrdquo Applied Mathematics andComputation vol 190 no 1 pp 757ndash764 2007
[2] M Dehghan and A Shokri ldquoA numerical method for solvingthe hyperbolic telegraph equationrdquo Numerical Methods forPartial Differential Equations vol 24 no 4 pp 1080ndash1093 2008
[3] L B Liu and H W Liu ldquoQuartic spline methods for solvingone-dimensional telegraphic equationsrdquo Applied Mathematicsand Computation vol 216 no 3 pp 951ndash958 2010
[4] H-W Liu and L-B Liu ldquoAn unconditionally stable splinedifference scheme of (k2 + h4) for solving the second-order1D linear hyperbolic equationrdquo Mathematical and ComputerModelling vol 49 no 9-10 pp 1985ndash1993 2009
International Journal of Computational Mathematics 9
[5] M Dosti and A Nazemi ldquoQuartic B-spline collocation methodfor solving one-dimensional hyperbolic telegraph equationrdquoThe Journal of Information and Computing Science vol 7 no2 pp 83ndash90 2012
[6] R C Mittal and R Bhatia ldquoNumerical solution of second orderone dimensional hyperbolic telegraph equation by cubic B-spline collocation methodrdquo Applied Mathematics and Compu-tation vol 220 pp 496ndash506 2013
[7] R K Mohanty ldquoAn unconditionally stable finite differenceformula for a linear second order one space dimensional hyper-bolic equation with variable coefficientsrdquo Applied Mathematicsand Computation vol 165 no 1 pp 229ndash236 2005
[8] R K Mohanty M K Jain and K George ldquoOn the use ofhigh order difference methods for the system of one spacesecond order nonlinear hyperbolic equations with variablecoefficientsrdquo Journal of Computational and Applied Mathemat-ics vol 72 no 2 pp 421ndash431 1996
[9] R KMohanty ldquoAnunconditionally stable difference scheme forthe one-space-dimensional linear hyperbolic equationrdquoAppliedMathematics Letters vol 17 no 1 pp 101ndash105 2004
[10] F Gao and C Chi ldquoUnconditionally stable difference schemesfor a one-space-dimensional linear hyperbolic equationrdquoApplied Mathematics and Computation vol 187 no 2 pp 1272ndash1276 2007
[11] M Dehghan andM Lakestani ldquoThe use of Chebyshev cardinalfunctions for solution of the second-order one-dimensionaltelegraph equationrdquo Numerical Methods for Partial DifferentialEquations vol 25 no 4 pp 931ndash938 2009
[12] A Saadatmandi and M Dehghan ldquoNumerical solution ofhyperbolic telegraph equation using the Chebyshev taumethodrdquo Numerical Methods for Partial Differential Equationsvol 26 no 1 pp 239ndash252 2010
[13] A Mohebbi and M Dehghan ldquoHigh order compact solution ofthe one-space-dimensional linear hyperbolic equationrdquoNumer-ical Methods for Partial Differential Equations vol 24 no 5 pp1222ndash1235 2008
[14] M Lakestani and B N Saray ldquoNumerical solution of telegraphequation using interpolating scaling functionsrdquo Computers ampMathematics with Applications vol 60 no 7 pp 1964ndash19722010
[15] M Esmaeilbeigi M M Hosseini and S T Mohyud-Din ldquoAnew approach of the radial basis functionsmethod for telegraphequationsrdquo International Journal of Physical Sciences vol 6 no6 pp 1517ndash1527 2011
[16] M Dehghan and A Ghesmati ldquoSolution of the second-orderone-dimensional hyperbolic telegraph equation by using thedual reciprocity boundary integral equation (DRBIE) methodrdquoEngineering Analysis with Boundary Elements vol 34 no 1 pp51ndash59 2010
[17] L B Liu and H W Liu ldquoCompact difference schemes forsolving telegraphic equations with Neumann boundary condi-tionsrdquo Applied Mathematics and Computation vol 219 no 19pp 10112ndash10121 2013
[18] R CMittal and R K Jain ldquoCubic B-splines collocationmethodfor solving nonlinear parabolic partial differential equationswith Neumann boundary conditionsrdquoCommunications in Non-linear Science andNumerical Simulation vol 17 no 12 pp 4616ndash4625 2012
[19] R C Mittal and R K Jain ldquoNumerical solutions of nonlinearBurgersrsquo equation with modified cubic B-Splines collocationmethodrdquo Applied Mathematics and Computation vol 218 no15 pp 7839ndash7855 2012
[20] R J Spiteri and S J Ruuth ldquoA new class of optimal high-orderstrong-stability-preserving time discretization methodsrdquo SIAMJournal on Numerical Analysis vol 40 no 2 pp 469ndash491 2002
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
International Journal of Computational Mathematics 7
Table 4 The errors of numerical solutions with Δ119905 = 001
119905
ℎ = 05 ℎ = 02
1198712
119871infin
RMS error 1198712
Łinfin
RMS error1 67752119864 minus 4 66070119864 minus 4 33667119864 minus 4 52299119864 minus 4 38981119864 minus 4 26084119864 minus 4
2 60042119864 minus 4 51361119864 minus 4 29835119864 minus 4 57309119864 minus 4 43320119864 minus 4 28588119864 minus 4
3 17361119864 minus 4 17448119864 minus 4 86268119864 minus 5 78085119864 minus 5 66723119864 minus 5 38945119864 minus 5
Table 5 Comparison of RMS errors at 119905 = 2 for Example 2
ℎ
The proposed method Dehghan and Ghesmati [16]RMS error CPU time (s) RMS error CPU time (s)
05 29835119864 minus 4 56 2153119864 minus 4 mdash
01 28758119864 minus 4 24 7012119864 minus 5 mdash
005 28819119864 minus 4 42 4003119864 minus 5 mdash
0 05 1 15 2 25 3 35 4
x
minus1
minus08
minus06
minus04
minus02
0
02
04
06
08
u(xt)
Numerical solution at t = 1
Exact solution at t = 1
Numerical solution at t = 2
Exact solution at t = 2
Numerical solution at t = 3
Exact solution at t = 3
Figure 3 Numerical and exact solutions with ℎ = 05 Δ119905 = 001
and boundary conditions
119906119909(0 119905) = cos 119905 119906
119909(4 119905) = cos 119905 cos 4 (39)
The exact solution is given [16] by
119906 (119909 119905) = cos 119905 sin119909 (40)
1198712 119871infin and RMS errors are reported in Table 4 with
ℎ = 05 and Δ119905 = 001 In Table 5 RMS errors are reportedwith different space step sizes and compared with the resultsgiven by Dehghan and Ghesmati [16] Figure 3 shows thatthe approximate solution and exact solution coincide for 119905 =
1 2 3 with ℎ = 05 and Δ119905 = 001 Space-time graph ofapproximate solution is depicted in Figure 4 with ℎ = 05 andΔ119905 = 001 up to 119905 = 3 Similar figures have been depicted in[16 17]
Example 3 In this example we consider the nonlinear tele-graph equation
119906119905119905(119909 119905) + 119906
119905(119909 119905)
= 119906119909119909
(119909 119905) minus 2 sin 119906 minus 1205872119890minus119905 cos (120587119909)
+ 2 sin (119890minus119905
(1 minus cos (120587119909))) 0 le 119909 le 2 119905 ge 0
(41)
0051
152
253
t
minus1
minus05
0
05
1
u(xt)
005 1 15
2 25 335
4
x
Figure 4 Space-time graph of numerical solutions with ℎ = 05Δ119905 = 001
0 02 04 06 08 1 12 14 16 18 2
x
minus01
0
01
02
03
04
05
06
07
08
u(xt)
Numerical solution at t = 1
Exact solution at t = 1
Numerical solution at t = 2
Exact solution at t = 2
Numerical solution at t = 3
Exact solution at t = 3
Numerical solution at t = 5
Exact solution at t = 5
Figure 5 Numerical and exact solutions with ℎ = 05 Δ119905 = 001
with initial conditions
119906 (119909 0) = 1 minus cos (120587119909) 119906119905(119909 0) = minus1 + cos (120587119909)
(42)
and boundary conditions
119906119909(0 119905) = 119906
119909(2 119905) = 0 (43)
The exact solution is given [17] by
119906 (119909 119905) = 119890minus119905
(1 minus cos (120587119909)) (44)
1198712 119871infin and RMS errors are computed for different values
of 119905 and reported in Table 6 with ℎ = 05 and Δ119905 = 001 InTable 7 error norms are reported for ℎ = 1 and Δ119905 = 01
and compared with the results given by L B Liu and H WLiu [17] FromFigure 5 we observe that approximate solutionand exact solution coincide for 119905 = 1 2 3 with ℎ = 05
8 International Journal of Computational Mathematics
Table 6 The errors of numerical solutions with Δ119905 = 001
119905
ℎ = 05 ℎ = 02
1198712
119871infin
RMS error 1198712
Łinfin
RMS error1 13341119864 minus 3 14954119864 minus 3 93182119864 minus 4 28200119864 minus 4 30385119864 minus 4 19841119864 minus 4
3 31344119864 minus 4 38812119864 minus 4 21892119864 minus 4 16958119864 minus 4 15103119864 minus 4 11939119864 minus 4
5 42496119864 minus 5 49131119864 minus 5 29681119864 minus 5 56791119864 minus 5 50866119864 minus 5 39958119864 minus 5
7 21773119864 minus 5 26530119864 minus 5 15207119864 minus 5 24272119864 minus 5 19499119864 minus 5 17500119864 minus 5
10 10059119864 minus 5 11691119864 minus 5 70261119864 minus 6 42303119864 minus 6 37155119864 minus 6 38945119864 minus 6
15 51178119864 minus 7 66345119864 minus 7 35574119864 minus 7 23315119864 minus 7 24379119864 minus 7 38945119864 minus 7
Table 7 Comparison of RMS errors at different time levels for Example 3
119905
The proposed method L B Liu and H W Liu [17]ℎ = 1 Δ119905 = 01 ℎ = 05 Δ119905 = 01
1198712
119871infin
RMS error CPU time RMS1 28083119864 minus 3 33349119864 minus 3 19379119864 minus 3 03 7591119864 minus 6
2 24692119864 minus 3 22557119864 minus 3 17038119864 minus 3 04 1833119864 minus 6
5 38689119864 minus 4 27429119864 minus 4 26698119864 minus 4 06 6976119864 minus 7
7 15951119864 minus 4 15796119864 minus 4 11007119864 minus 4 08 mdash10 38193119864 minus 5 43924119864 minus 5 26355119864 minus 5 11 5177119864 minus 8
15 22855119864 minus 6 21111119864 minus 6 11584119864 minus 6 14 3943119864 minus 9
005
115
2
x01
23
45
0
05
1
15
2
t
u(xt)
Figure 6 Space-time graph of numerical solutions with ℎ = 05Δ119905 = 001
and Δ119905 = 001 Space-time graph of approximate solution isdepicted in Figure 6 with ℎ = 05 andΔ119905 = 001 which showsthe approximate solution profile up to 119905 = 3 Similar figurehas been depicted in [17]
It should be noticed that in this case we obtain a system ofnonlinear first order ordinary differential equations Unlikethe numerical methods [17] which use finite differencemethods to solve nonlinear hyperbolic wave equation to geta system of nonlinear algebraic equations and finally solvedusing the Newton-Raphson method we do not need to doany extra effort to handle the nonlinear term and solutionsare easily computed by solving the obtained system of ODEsusing SSP-RK54 scheme Hence the present scheme reducesthe computational cost
7 Conclusions
The main idea of the present paper is to first convert thegiven problem into a coupled system of partial differential
equations and then using cubic 119861-spline basis functionsfor spatial variable and derivatives the coupled system ofequations is converted into system of first order ordinarydifferential equations The resulting system of first orderordinary differential equations has been solved by SSP-RK54schemeThe stability of the scheme is discussed using matrixstability analysis and found to be unconditionally stable Themethod is also capable of solving nonlinear telegraph equa-tion with Neumann boundary conditions It has been noticedthat obtained numerical solutions are in good agreementwiththe exact solution and earlier studies These facts illustratethat the proposed method is a reliable valid and powerfultool for solving telegraph equation with Neumann boundaryconditions
Conflict of Interests
The authors declare that there is no conflict of interestsregarding to the publication of this paper
References
[1] M S El-Azab and M El-Gamel ldquoA numerical algorithm forthe solution of telegraph equationsrdquo Applied Mathematics andComputation vol 190 no 1 pp 757ndash764 2007
[2] M Dehghan and A Shokri ldquoA numerical method for solvingthe hyperbolic telegraph equationrdquo Numerical Methods forPartial Differential Equations vol 24 no 4 pp 1080ndash1093 2008
[3] L B Liu and H W Liu ldquoQuartic spline methods for solvingone-dimensional telegraphic equationsrdquo Applied Mathematicsand Computation vol 216 no 3 pp 951ndash958 2010
[4] H-W Liu and L-B Liu ldquoAn unconditionally stable splinedifference scheme of (k2 + h4) for solving the second-order1D linear hyperbolic equationrdquo Mathematical and ComputerModelling vol 49 no 9-10 pp 1985ndash1993 2009
International Journal of Computational Mathematics 9
[5] M Dosti and A Nazemi ldquoQuartic B-spline collocation methodfor solving one-dimensional hyperbolic telegraph equationrdquoThe Journal of Information and Computing Science vol 7 no2 pp 83ndash90 2012
[6] R C Mittal and R Bhatia ldquoNumerical solution of second orderone dimensional hyperbolic telegraph equation by cubic B-spline collocation methodrdquo Applied Mathematics and Compu-tation vol 220 pp 496ndash506 2013
[7] R K Mohanty ldquoAn unconditionally stable finite differenceformula for a linear second order one space dimensional hyper-bolic equation with variable coefficientsrdquo Applied Mathematicsand Computation vol 165 no 1 pp 229ndash236 2005
[8] R K Mohanty M K Jain and K George ldquoOn the use ofhigh order difference methods for the system of one spacesecond order nonlinear hyperbolic equations with variablecoefficientsrdquo Journal of Computational and Applied Mathemat-ics vol 72 no 2 pp 421ndash431 1996
[9] R KMohanty ldquoAnunconditionally stable difference scheme forthe one-space-dimensional linear hyperbolic equationrdquoAppliedMathematics Letters vol 17 no 1 pp 101ndash105 2004
[10] F Gao and C Chi ldquoUnconditionally stable difference schemesfor a one-space-dimensional linear hyperbolic equationrdquoApplied Mathematics and Computation vol 187 no 2 pp 1272ndash1276 2007
[11] M Dehghan andM Lakestani ldquoThe use of Chebyshev cardinalfunctions for solution of the second-order one-dimensionaltelegraph equationrdquo Numerical Methods for Partial DifferentialEquations vol 25 no 4 pp 931ndash938 2009
[12] A Saadatmandi and M Dehghan ldquoNumerical solution ofhyperbolic telegraph equation using the Chebyshev taumethodrdquo Numerical Methods for Partial Differential Equationsvol 26 no 1 pp 239ndash252 2010
[13] A Mohebbi and M Dehghan ldquoHigh order compact solution ofthe one-space-dimensional linear hyperbolic equationrdquoNumer-ical Methods for Partial Differential Equations vol 24 no 5 pp1222ndash1235 2008
[14] M Lakestani and B N Saray ldquoNumerical solution of telegraphequation using interpolating scaling functionsrdquo Computers ampMathematics with Applications vol 60 no 7 pp 1964ndash19722010
[15] M Esmaeilbeigi M M Hosseini and S T Mohyud-Din ldquoAnew approach of the radial basis functionsmethod for telegraphequationsrdquo International Journal of Physical Sciences vol 6 no6 pp 1517ndash1527 2011
[16] M Dehghan and A Ghesmati ldquoSolution of the second-orderone-dimensional hyperbolic telegraph equation by using thedual reciprocity boundary integral equation (DRBIE) methodrdquoEngineering Analysis with Boundary Elements vol 34 no 1 pp51ndash59 2010
[17] L B Liu and H W Liu ldquoCompact difference schemes forsolving telegraphic equations with Neumann boundary condi-tionsrdquo Applied Mathematics and Computation vol 219 no 19pp 10112ndash10121 2013
[18] R CMittal and R K Jain ldquoCubic B-splines collocationmethodfor solving nonlinear parabolic partial differential equationswith Neumann boundary conditionsrdquoCommunications in Non-linear Science andNumerical Simulation vol 17 no 12 pp 4616ndash4625 2012
[19] R C Mittal and R K Jain ldquoNumerical solutions of nonlinearBurgersrsquo equation with modified cubic B-Splines collocationmethodrdquo Applied Mathematics and Computation vol 218 no15 pp 7839ndash7855 2012
[20] R J Spiteri and S J Ruuth ldquoA new class of optimal high-orderstrong-stability-preserving time discretization methodsrdquo SIAMJournal on Numerical Analysis vol 40 no 2 pp 469ndash491 2002
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
8 International Journal of Computational Mathematics
Table 6 The errors of numerical solutions with Δ119905 = 001
119905
ℎ = 05 ℎ = 02
1198712
119871infin
RMS error 1198712
Łinfin
RMS error1 13341119864 minus 3 14954119864 minus 3 93182119864 minus 4 28200119864 minus 4 30385119864 minus 4 19841119864 minus 4
3 31344119864 minus 4 38812119864 minus 4 21892119864 minus 4 16958119864 minus 4 15103119864 minus 4 11939119864 minus 4
5 42496119864 minus 5 49131119864 minus 5 29681119864 minus 5 56791119864 minus 5 50866119864 minus 5 39958119864 minus 5
7 21773119864 minus 5 26530119864 minus 5 15207119864 minus 5 24272119864 minus 5 19499119864 minus 5 17500119864 minus 5
10 10059119864 minus 5 11691119864 minus 5 70261119864 minus 6 42303119864 minus 6 37155119864 minus 6 38945119864 minus 6
15 51178119864 minus 7 66345119864 minus 7 35574119864 minus 7 23315119864 minus 7 24379119864 minus 7 38945119864 minus 7
Table 7 Comparison of RMS errors at different time levels for Example 3
119905
The proposed method L B Liu and H W Liu [17]ℎ = 1 Δ119905 = 01 ℎ = 05 Δ119905 = 01
1198712
119871infin
RMS error CPU time RMS1 28083119864 minus 3 33349119864 minus 3 19379119864 minus 3 03 7591119864 minus 6
2 24692119864 minus 3 22557119864 minus 3 17038119864 minus 3 04 1833119864 minus 6
5 38689119864 minus 4 27429119864 minus 4 26698119864 minus 4 06 6976119864 minus 7
7 15951119864 minus 4 15796119864 minus 4 11007119864 minus 4 08 mdash10 38193119864 minus 5 43924119864 minus 5 26355119864 minus 5 11 5177119864 minus 8
15 22855119864 minus 6 21111119864 minus 6 11584119864 minus 6 14 3943119864 minus 9
005
115
2
x01
23
45
0
05
1
15
2
t
u(xt)
Figure 6 Space-time graph of numerical solutions with ℎ = 05Δ119905 = 001
and Δ119905 = 001 Space-time graph of approximate solution isdepicted in Figure 6 with ℎ = 05 andΔ119905 = 001 which showsthe approximate solution profile up to 119905 = 3 Similar figurehas been depicted in [17]
It should be noticed that in this case we obtain a system ofnonlinear first order ordinary differential equations Unlikethe numerical methods [17] which use finite differencemethods to solve nonlinear hyperbolic wave equation to geta system of nonlinear algebraic equations and finally solvedusing the Newton-Raphson method we do not need to doany extra effort to handle the nonlinear term and solutionsare easily computed by solving the obtained system of ODEsusing SSP-RK54 scheme Hence the present scheme reducesthe computational cost
7 Conclusions
The main idea of the present paper is to first convert thegiven problem into a coupled system of partial differential
equations and then using cubic 119861-spline basis functionsfor spatial variable and derivatives the coupled system ofequations is converted into system of first order ordinarydifferential equations The resulting system of first orderordinary differential equations has been solved by SSP-RK54schemeThe stability of the scheme is discussed using matrixstability analysis and found to be unconditionally stable Themethod is also capable of solving nonlinear telegraph equa-tion with Neumann boundary conditions It has been noticedthat obtained numerical solutions are in good agreementwiththe exact solution and earlier studies These facts illustratethat the proposed method is a reliable valid and powerfultool for solving telegraph equation with Neumann boundaryconditions
Conflict of Interests
The authors declare that there is no conflict of interestsregarding to the publication of this paper
References
[1] M S El-Azab and M El-Gamel ldquoA numerical algorithm forthe solution of telegraph equationsrdquo Applied Mathematics andComputation vol 190 no 1 pp 757ndash764 2007
[2] M Dehghan and A Shokri ldquoA numerical method for solvingthe hyperbolic telegraph equationrdquo Numerical Methods forPartial Differential Equations vol 24 no 4 pp 1080ndash1093 2008
[3] L B Liu and H W Liu ldquoQuartic spline methods for solvingone-dimensional telegraphic equationsrdquo Applied Mathematicsand Computation vol 216 no 3 pp 951ndash958 2010
[4] H-W Liu and L-B Liu ldquoAn unconditionally stable splinedifference scheme of (k2 + h4) for solving the second-order1D linear hyperbolic equationrdquo Mathematical and ComputerModelling vol 49 no 9-10 pp 1985ndash1993 2009
International Journal of Computational Mathematics 9
[5] M Dosti and A Nazemi ldquoQuartic B-spline collocation methodfor solving one-dimensional hyperbolic telegraph equationrdquoThe Journal of Information and Computing Science vol 7 no2 pp 83ndash90 2012
[6] R C Mittal and R Bhatia ldquoNumerical solution of second orderone dimensional hyperbolic telegraph equation by cubic B-spline collocation methodrdquo Applied Mathematics and Compu-tation vol 220 pp 496ndash506 2013
[7] R K Mohanty ldquoAn unconditionally stable finite differenceformula for a linear second order one space dimensional hyper-bolic equation with variable coefficientsrdquo Applied Mathematicsand Computation vol 165 no 1 pp 229ndash236 2005
[8] R K Mohanty M K Jain and K George ldquoOn the use ofhigh order difference methods for the system of one spacesecond order nonlinear hyperbolic equations with variablecoefficientsrdquo Journal of Computational and Applied Mathemat-ics vol 72 no 2 pp 421ndash431 1996
[9] R KMohanty ldquoAnunconditionally stable difference scheme forthe one-space-dimensional linear hyperbolic equationrdquoAppliedMathematics Letters vol 17 no 1 pp 101ndash105 2004
[10] F Gao and C Chi ldquoUnconditionally stable difference schemesfor a one-space-dimensional linear hyperbolic equationrdquoApplied Mathematics and Computation vol 187 no 2 pp 1272ndash1276 2007
[11] M Dehghan andM Lakestani ldquoThe use of Chebyshev cardinalfunctions for solution of the second-order one-dimensionaltelegraph equationrdquo Numerical Methods for Partial DifferentialEquations vol 25 no 4 pp 931ndash938 2009
[12] A Saadatmandi and M Dehghan ldquoNumerical solution ofhyperbolic telegraph equation using the Chebyshev taumethodrdquo Numerical Methods for Partial Differential Equationsvol 26 no 1 pp 239ndash252 2010
[13] A Mohebbi and M Dehghan ldquoHigh order compact solution ofthe one-space-dimensional linear hyperbolic equationrdquoNumer-ical Methods for Partial Differential Equations vol 24 no 5 pp1222ndash1235 2008
[14] M Lakestani and B N Saray ldquoNumerical solution of telegraphequation using interpolating scaling functionsrdquo Computers ampMathematics with Applications vol 60 no 7 pp 1964ndash19722010
[15] M Esmaeilbeigi M M Hosseini and S T Mohyud-Din ldquoAnew approach of the radial basis functionsmethod for telegraphequationsrdquo International Journal of Physical Sciences vol 6 no6 pp 1517ndash1527 2011
[16] M Dehghan and A Ghesmati ldquoSolution of the second-orderone-dimensional hyperbolic telegraph equation by using thedual reciprocity boundary integral equation (DRBIE) methodrdquoEngineering Analysis with Boundary Elements vol 34 no 1 pp51ndash59 2010
[17] L B Liu and H W Liu ldquoCompact difference schemes forsolving telegraphic equations with Neumann boundary condi-tionsrdquo Applied Mathematics and Computation vol 219 no 19pp 10112ndash10121 2013
[18] R CMittal and R K Jain ldquoCubic B-splines collocationmethodfor solving nonlinear parabolic partial differential equationswith Neumann boundary conditionsrdquoCommunications in Non-linear Science andNumerical Simulation vol 17 no 12 pp 4616ndash4625 2012
[19] R C Mittal and R K Jain ldquoNumerical solutions of nonlinearBurgersrsquo equation with modified cubic B-Splines collocationmethodrdquo Applied Mathematics and Computation vol 218 no15 pp 7839ndash7855 2012
[20] R J Spiteri and S J Ruuth ldquoA new class of optimal high-orderstrong-stability-preserving time discretization methodsrdquo SIAMJournal on Numerical Analysis vol 40 no 2 pp 469ndash491 2002
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
International Journal of Computational Mathematics 9
[5] M Dosti and A Nazemi ldquoQuartic B-spline collocation methodfor solving one-dimensional hyperbolic telegraph equationrdquoThe Journal of Information and Computing Science vol 7 no2 pp 83ndash90 2012
[6] R C Mittal and R Bhatia ldquoNumerical solution of second orderone dimensional hyperbolic telegraph equation by cubic B-spline collocation methodrdquo Applied Mathematics and Compu-tation vol 220 pp 496ndash506 2013
[7] R K Mohanty ldquoAn unconditionally stable finite differenceformula for a linear second order one space dimensional hyper-bolic equation with variable coefficientsrdquo Applied Mathematicsand Computation vol 165 no 1 pp 229ndash236 2005
[8] R K Mohanty M K Jain and K George ldquoOn the use ofhigh order difference methods for the system of one spacesecond order nonlinear hyperbolic equations with variablecoefficientsrdquo Journal of Computational and Applied Mathemat-ics vol 72 no 2 pp 421ndash431 1996
[9] R KMohanty ldquoAnunconditionally stable difference scheme forthe one-space-dimensional linear hyperbolic equationrdquoAppliedMathematics Letters vol 17 no 1 pp 101ndash105 2004
[10] F Gao and C Chi ldquoUnconditionally stable difference schemesfor a one-space-dimensional linear hyperbolic equationrdquoApplied Mathematics and Computation vol 187 no 2 pp 1272ndash1276 2007
[11] M Dehghan andM Lakestani ldquoThe use of Chebyshev cardinalfunctions for solution of the second-order one-dimensionaltelegraph equationrdquo Numerical Methods for Partial DifferentialEquations vol 25 no 4 pp 931ndash938 2009
[12] A Saadatmandi and M Dehghan ldquoNumerical solution ofhyperbolic telegraph equation using the Chebyshev taumethodrdquo Numerical Methods for Partial Differential Equationsvol 26 no 1 pp 239ndash252 2010
[13] A Mohebbi and M Dehghan ldquoHigh order compact solution ofthe one-space-dimensional linear hyperbolic equationrdquoNumer-ical Methods for Partial Differential Equations vol 24 no 5 pp1222ndash1235 2008
[14] M Lakestani and B N Saray ldquoNumerical solution of telegraphequation using interpolating scaling functionsrdquo Computers ampMathematics with Applications vol 60 no 7 pp 1964ndash19722010
[15] M Esmaeilbeigi M M Hosseini and S T Mohyud-Din ldquoAnew approach of the radial basis functionsmethod for telegraphequationsrdquo International Journal of Physical Sciences vol 6 no6 pp 1517ndash1527 2011
[16] M Dehghan and A Ghesmati ldquoSolution of the second-orderone-dimensional hyperbolic telegraph equation by using thedual reciprocity boundary integral equation (DRBIE) methodrdquoEngineering Analysis with Boundary Elements vol 34 no 1 pp51ndash59 2010
[17] L B Liu and H W Liu ldquoCompact difference schemes forsolving telegraphic equations with Neumann boundary condi-tionsrdquo Applied Mathematics and Computation vol 219 no 19pp 10112ndash10121 2013
[18] R CMittal and R K Jain ldquoCubic B-splines collocationmethodfor solving nonlinear parabolic partial differential equationswith Neumann boundary conditionsrdquoCommunications in Non-linear Science andNumerical Simulation vol 17 no 12 pp 4616ndash4625 2012
[19] R C Mittal and R K Jain ldquoNumerical solutions of nonlinearBurgersrsquo equation with modified cubic B-Splines collocationmethodrdquo Applied Mathematics and Computation vol 218 no15 pp 7839ndash7855 2012
[20] R J Spiteri and S J Ruuth ldquoA new class of optimal high-orderstrong-stability-preserving time discretization methodsrdquo SIAMJournal on Numerical Analysis vol 40 no 2 pp 469ndash491 2002
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