Research ArticleDistributions of Zeros of Solutions for Third-OrderDifferential Equations with Variable Coefficients

Samir H Saker and Mohammed A Arahet

Department of Mathematics Faculty of Science Mansoura 35516 Egypt

Correspondence should be addressed to Mohammed A Arahet malroheetyahoocom

Received 9 November 2014 Accepted 11 December 2014

Academic Editor Zhen-Lai Han

Copyright copy 2015 S H Saker and M A ArahetThis is an open access article distributed under the Creative Commons AttributionLicense which permits unrestricted use distribution and reproduction in anymedium provided the originalwork is properly cited

For the third-order linear differential equations of the form (119903(119905)11990910158401015840(119905))1015840 + 119901(119905)1199091015840(119905) + 119902(119905)119909(119905) = 0 we will establish lowerbounds for the distance between zeros of a solution andor its derivatives The main results will be proved by making use ofHardyrsquos inequality and some generalizations of Opial and Wirtinger type inequalities

1 Introduction

The mathematical description of some physical systemsdemands that we often solve linear differential equationssubject to some boundary conditions Some of these prob-lems are the mathematical models of the deflection ofbeams These beams which appear in many structuresdeflect under their own weight or under the influence ofsome external forces For example if a load is applied to thebeam in a vertical plane containing the axis of symmetrythe beam undergoes a distortion and the curve connectingthe centroids of all cross sections is called the deflectioncurve or elastic curve In elasticity it is shown that thedeflection of the curve say 119910(119909) measured from the 119909-axis approximates the shape of the beam and satisfies alinear fourth-order differential equation on an interval say[120572 120573] with some boundary conditions The distributionof boundary conditions which is the distribution of zerosof solutions of differential equations has been started byPicard [1 2] who derived some uniqueness results forsolutions of the second-order nonlinear differential equationwith two-point boundary conditions when the nonlinearfunction satisfies the Lipschitz condition The distributionof zeros of 119899th-order differential equations with more thantwo points has been considered by Niccoletti [3] Moti-vated by the work of Picard and Niccoletti de la Vallee

Poussin [4] considered the general 119899th-order linear differ-ential equation

119910(119899)

(119905) + 1199011(119905)119910(119899minus1)

(119905) + sdot sdot sdot + 119901119899(119905)119910(119905) = 0

119905 isin I = [119886 119887](1)

with real coefficients that are locally integrable inside I andstudied the disconjugacy of solutions

Equation (1) is said to be disconjugate on an interval Iif every nontrivial solution has less than 119899 zeros on I withmultiple zeros being counted according to their multiplic-ity More precisely disconjugate on a connected set meansthat the number of zeros of a nontrivial solution cannotequal the order of the equation Equation (1) is said to be(119896 119899minus119896)-disconjugate on an interval I if no nontrivial solutionhas a zero of order 119896 followed by a zero of order 119899 minus 119896This means that for every pair of points 120572 120573 isin I 120572 lt

120573 there does not exist a nontrivial solution of (1) whichsatisfies

119910(119894)

(120572) = 0 119894 = 0 119896 minus 1

119910(119895)

(120573) = 0 119895 = 0 119899 minus 119896 minus 1(2)

The last value of 120573 such that there exists a nontrivial solutionwhich satisfies (2) is called the (119896 119899 minus 119896)-conjugate pointof 120572 The first work that has been published by de la Vallee

Hindawi Publishing CorporationMathematical Problems in EngineeringVolume 2015 Article ID 158460 9 pageshttpdxdoiorg1011552015158460

2 Mathematical Problems in Engineering

Poussin in 1929 was on the evaluation of the length ofthe interval [0 ℎ] in which the boundary problem 119910(119905

1) =

119910(1199052) = sdot sdot sdot = 119910(119905

119899) = 0 (0 le 119905

1lt 1199052lt sdot sdot sdot lt 119905

119899le ℎ)

for the linear differential equation (1) only admits the nullsolution

Following the way indicated by de la Vallee Poussinone has tried to evaluate the length ℎ in a function of theupper bounds of the coefficients The precise evaluation ofthe maximal length of the considered interval has only beenobtained for second-order differential equations [5 6] withthe best possible constant 14 Coming then to (1) de la ValleePoussin proved that this equation is disconjugate in [0 ℎ]

if the coefficients and the length of the interval satisfy theinequality

119899

sum119894=1

119875119894

ℎ119894

119894lt 1 119875

119894= max119886le119905le119887

1003816100381610038161003816119901119894 (119905)1003816100381610038161003816 for 119894 = 1 2 119899 (3)

This is the part of his paper which has motivated the manyrefinements and generalizations described in the analysisThe importance of Poussinrsquos work has been emphasized andtestified by many authors in the literature we refer the readerto the papers [7ndash16] and the references cited therein Forcompleteness we recall some of the related results whichmotivate the contents of this paper Lasota [8] considered thethird-order differential equation

119910101584010158401015840

(119905) + 1199011(119905)11991010158401015840

(119905) + 1199012(119905)1199101015840

(119905) + 1199013(119905)119910(119905) = 0

119905 isin [119886 119887](4)

and proved that if

1198751

ℎ

4+ 1198752

ℎ2

1205872+ 1198753

ℎ3

21205872le 1 (5)

then (4) is disconjugate where 119875119894= max |119901

119894(119905)| and ℎ = 119887minus119886

Mathsen [9] proved that (4) is disconjugate if 1199012(119909) le 0 on

[119886 119887] and

(2ℎ1+ 1)1198753

[exp(21198751ℎ1) minus exp(119875

1ℎ1) minus ℎ11198751]

11987521

le 1 (6)

where 1198751= max |119901

1(119905)| 119875

3= max |119901

3(119905)| on [119886 119887] and ℎ

1=

(119887 minus 119886)2 Casadei [7] also proved that if

1198751

ℎ

2+ 1198752

ℎ2

8+ 1198753

ℎ3

24le 1 (7)

then (4) is disconjugate Agarwal and Krishnamurthy [17]also proved that if

1198751

2ℎ

3+ 1198752

ℎ2

6+ 1198753

2ℎ3

81le 1 (8)

then (4) is disconjugate

Motivated by these papers we will study the distribu-tion of zeros of the solutions of the third-order differen-tial equation

(119903(119905)11990910158401015840(119905))1015840

+ 119901 (119905) 1199091015840

(119905) + 119902 (119905) 119909 (119905) = 0 119905 isin I (9)

where I is an interval of reals and 119903(119905) 119901(119905) and 119902(119905) are real-valued functions defined on I such that 119903(119905) gt 0

By a solution of (9) on the interval J sube I we mean anontrivial real-valued function 119909 isin 119862

2(J) which has theproperty that 119903(119905)11990910158401015840(119905) isin 1198621(J) and satisfies (9) on J Thenontrivial solution 119909(119905) of (9) is said to oscillate or to beoscillatory if it has arbitrarily large zeros Equation (9) isoscillatory if one of its nontrivial solutions is oscillatory Anequation of the form (9) is said to be disconjugate on aninterval I if no nontrivial solution has more than two zeroson I counting multiplicities We say that (9) is right disfocal(left disfocal) on the interval [119886 119887] (119886 lt 119887) if the solutions of(9) such that 1199091015840(119886) = 0 119909(119886) = 0 (1199091015840(119887) = 0 and 119909(119887) = 0) donot have two zeros counting multiplicities in (119886 119887] ([119886 119887))

The paper is organized as follows In Section 2 we presentHardyrsquos inequality the extensions of Opialrsquos inequality andWirtingerrsquos inequality that will be used to prove our mainresults In Section 3 we are concerned with the lower boundsof the distance between zeros of a nontrivial solution andorits derivatives for the third-order differential equation (9)subject to two sets of boundary conditions In particular wewill prove the following

(i) Obtain lower bounds for the spacing 120573minus120572 where 119909 isa nontrivial solution of (9) which satisfies

119909 (120572) = 1199091015840

(120572) = 1199091015840(120573) = 0 or

119909 (120573) = 1199091015840(120573) = 119909

1015840

(120572) = 0

(10)

(ii) Obtain lower bounds for the spacing 120573minus120572 where 119909 isa solution of (9) which satisfies

119909 (120572) = 1199091015840

(120572) = 11990910158401015840(120573) = 0 or

119909 (120573) = 1199091015840(120573) = 119909

10158401015840

(120572) = 0

(11)

2 Hardy Opial and Wirtinger Inequalities

In this section we present Hardyrsquos inequality and somegeneralizations of Opial and Wirtinger type inequalitiesthat will be needed in the proof of the main results Thegeneralizations of Opial and Wirtinger type inequalities areadapted from Agarwal and Pang [18 19] and the papers dueto Beesack and Das [20] Clark and Hinton [21] and Fink[22] The Hardy inequality is adapted from the book due toKufner and Persson [23] We begin with Hardyrsquos inequalitywhich states the following If 119910 is absolutely continuous on(120572 120573) such that 119910(120572) = 0 or (119910(120573) = 0) then

(int120573

120572

119901(119905)1003816100381610038161003816119910(119905)

1003816100381610038161003816119899

119889119905)

1119899

le 119862(int120573

120572

119903(119905)100381610038161003816100381610038161199101015840(119905)

10038161003816100381610038161003816

119898

119889119905)

1119898

(12)

Mathematical Problems in Engineering 3

where the weighted functions 119901 119903 are positive functionsdefined on (120572 120573) and 119899 119898 are real parameters that satisfy0 lt 119899 le infin and 1 le 119898 le infin The constant 119862 is given by

119862 le 1198981119898

(1198981)11198981119860(120572 120573) for 1 lt 119898 le 119899 119898

1=

119898

119898 minus 1

(13)

where

119860(120572 120573) = sup120572le119905le120573

(int120573

119905

119901(119905)119889119905)

1119899

(int119905

120572

1199031minus1198981(119904)119889119904)

11198981

if 119910(120572) = 0

119860(120572 120573) = sup120572le119905le120573

(int119905

120572

119901(119905)119889119905)

1119899

(int120573

119905

1199031minus1198981(119904)119889119904)

11198981

if 119910(120573) = 0

(14)

Note that inequality (12) has an immediate application to thecase when 119910(120572) = 119910(120573) = 0 In this case inequality (12) issatisfied if and only if

119860(120572 120573)

= sup(119888119889)sub(120572120573)

(int119889

119888

119901(119905)119889119905)

1119899

timesmin(int119888

120572

1199031minus1198981(119904)119889119904)

11198991

(int120573

119889

1199031minus1198981(119904)119889119904)

11198981

(15)

exists and is finite The Opial type inequality due to Beesackwhich is a generalization of Opialrsquos inequality states that if 119910is absolutely continuous on [120572 120573] with 119910(120572) = 0 then thefollowing inequality holds

int120573

120572

119902 (119905)1003816100381610038161003816119910(119905)

1003816100381610038161003816119898100381610038161003816100381610038161199101015840(119905)

10038161003816100381610038161003816

119899

119889119905 le 1198701(119898 119899) int

120573

120572

119901 (119905)100381610038161003816100381610038161199101015840(119905)

10038161003816100381610038161003816

119898+119899

119889119905

(16)

where 119898 119899 are real numbers such that 119898119899 gt 0 119898 + 119899 gt 1119902(119905) and 119901(119905) are nonnegative measurable functions definedon (120572 120573) such that int119905

120572(119901(119904))minus1(119898+119899minus1)119889119904 lt infin and

1198701(119898 119899)

= (119899

119898 + 119899)119899(119898+119899)

times [int120573

120572

119902(119898+119899)119899

(119905)119901minus119899119898

(119905)

times (int119905

120572

(119901(119904))minus1(119898+119899minus1)

119889119904)

119898+119899minus1

119889119905]

119898(119898+119899)

(17)

If instead 119910(120573) = 0 then (16) holds where 1198701(119898 119899) is

replaced by

1198702(119898 119899)

= (119899

119898 + 119899)119899(119898+119899)

times [int120573

120572

119902(119898+119899)119899

(119905) 119901minus(119899119898)

(119905)

times(int120573

119905

(119901(119904))minus1(119898+119899minus1)

119889119904)

119898+119899minus1

119889119905]

119898(119898+119899)

(18)

TheOpial type inequality due to Agarwal and Pang states thatif 119910(119905) isin 119862(119899minus1)[120572 120573] and satisfies119910(119894)(120572) = 0 0 le 119896 le 119894 le 119899minus1

(119899 ge 1) and 119910(119899minus1)(119905) absolutely continuous on (120572 120573) then

int120573

120572

120601 (119905)10038161003816100381610038161003816119910(119896)(119905)

10038161003816100381610038161003816

11989710038161003816100381610038161003816119910(119899)(119905)

10038161003816100381610038161003816

119898

119889119905

le 1198671[int120573

120572

120593(119905)10038161003816100381610038161003816119910(119899)(119905)

10038161003816100381610038161003816

119888

119889119905]

(119897+119898)119888

(19)

where 120601 and 120593 are nonnegative and measurable functiondefined on (120572 120573) 119898 119899 are real numbers such that 119888119898 gt 1and

1198671=

(119898(119898 + 119897))119898119888

(119899 minus 119896 minus 1)

times [int120573

120572

(120601119888

(119905)120593minus119898

(119905))1(119888minus119898)

times (1198661119896(119905))119897(119888minus1)(119888minus119898)

119889119905]

(119888minus119898)119888

1198661119896

(119905) = int119905

120572

(119905 minus 119904)(119899minus119896minus1)119888(119888minus1)

(120593(119904))minus1(119888minus1)

119889119904

(20)

If instead 119910(119894)(120573) = 0 0 le 119896 le 119894 le 119899 minus 1 (119899 ge 1) then (19)holds where119867

1is replaced by

1198672=

(119898(119898 + 119897))119898119888

(119899 minus 119896 minus 1)

times [int120573

120572

(120601119888(119905)120593minus119898

(119905))1(119888minus119898)

times (1198662119896(119905))119897(119888minus1)(119888minus119898)

119889119905]

(119888minus119898)119888

1198662119896

(119905) = int120573

119905

(119904 minus 119905)(119899minus119896minus1)119888(119888minus1)

(120593(119904))minus1(119888minus1)

119889119904

(21)

4 Mathematical Problems in Engineering

We also need the following inequality which is the specialcase of an inequality proved by Agarwal and Pang [18] withtwo functions

int120573

120572

119888(119905)10038161003816100381610038161003816119910(119896)

(119905)10038161003816100381610038161003816

10038161003816100381610038161003816119910(119896+1)

(119905)10038161003816100381610038161003816119889119905 le 119862

120572int120573

120572

119903 (119905)10038161003816100381610038161003816119910(119899)(119905)

10038161003816100381610038161003816

2

119889119905

(22)

where 119910(119905) isin 119862119899minus1[120572 120573] and satisfies 119910(119894)(120572) = 0 119896 le 119894 le 119899minus10 le 119896 le 119899 minus 1 (119899 ge 1) and 119910(119899minus1) is absolutely continuous on(120572 120573) 119888(119905) and 119903(119905) being nonnegative measurable functionsdefined on (120572 120573) and

119862120572=

1

2((119899 minus 119896 minus 1))2max119905isin[120572120573]

119888 (119905) int120573

120572

(119904 minus 120572)2(119899minus119896minus1)

119903 (119904)119889119904

(23)

If 119910(119894)(120573) = 0 119896 le 119894 le 119899 minus 1 then (22) holds where 119862120572is

replaced

119862120573=

1

2((119899 minus 119896 minus 1))2max119905isin[120572120573]

119888 (119905) int120573

120572

(120573 minus 119904)2(119899minus119896minus1)

119903 (119904)119889119904

(24)

TheWirtinger type inequality due to Agarwal and Pang statesthat if 119910(120572) = 119910(120573) = 0 then

int120573

120572

119910120574+1

(119905)119889119905 le(120573 minus 120572)

120574+1

Γ2 ((120574 + 2)2)

2Γ(120574 + 2)

times int120573

120572

(1199101015840(119905))120574+1

(119905) 119889119905

(25)

TheOpial type inequality due to Clark and Hinton inequalitystates that if 119910 isin 119862

2[120572 120573] where 119910(120572) = 1199101015840(120572) = 0 then

(int120573

120572

1003816100381610038161003816119910(119905)100381610038161003816100381621003816100381610038161003816100381611991010158401015840(119905)

10038161003816100381610038161003816

2

119889119905)

1119903

le(120573 minus 120572)

32

radic3int120573

120572

1003816100381610038161003816100381611991010158401015840

(119905)10038161003816100381610038161003816

2

119889119905

(26)

The Opial type inequality due to Fink inequality states that if119910(119894)(120572) = 0 0 le 119894 le 119899 minus 1 120583 ge 1 (1120583) + (1]) = 1 then

int120573

120572

10038161003816100381610038161003816119910(119896)

(119905) 119910(119903)

(119905)10038161003816100381610038161003816119889119905

le 119862 (119899 119896 119903 120583) (120573 minus 120572)2119899minus119896minus119903+1minus2120583

(int120573

120572

10038161003816100381610038161003816119910(119899)(119905)

10038161003816100381610038161003816

120583

119889119905)

2120583

(27)

where 0 le 119896 lt 119903 lt 119899 (119899 ge 2) and

119862 (119899 119896 119903 120583) =1

2((119899 minus 119896 minus 1))2[(119899 minus 119896 minus 1)] + 1]

2]

(28)

3 Main Results

In this section we state and prove the main results Forsimplicity we introduce the following notations

1198671(120572 120573 119866

10) =

1

radic2[int120573

120572

1198762

1(119905)

1198661(119905)

119903(119905)119889119905]

12

1198661(119905) = int

119905

120572

(119905 minus 119904)2

119903 (119904)119889119904 119876

1(119905) = int

120573

119905

119902 (119904) 119889119904

(29)

1198672(120572 120573 119866

2) =

1

radic2[int120573

120572

1198762

2(119905)

1198662(119905)

119903(119905)119889119905]

12

1198662(119905) = int

120573

119905

(119905 minus 119904)2

119903 (119904)119889119904 119876

2(119905) = int

119905

120572

119902 (119904) 119889119904

(30)

119860119894(120572 120573) = 119869

119894(119905)min(int

119888

120572

1

119903(119904)119889119904)

12

(int120573

119888

1

119903(119904)119889119904)

12

(31)

where 119869119894(119905) = sup

(119888119889)sub(120572120573)(int120573

120572119863119894(119905)119889119905)12 and |119863

119894(119905)| =

|119876119894(119905)| + |119901(119905)| 119894 = 1 2

1198611(120572 120573) = sup

120572le119905le120573

(int120573

119905

119901(119905)119889119905)

12

(int119905

120572

1

119903(119904)119889119904)

12

119862120572=

1

2max120572le119905le120573

119902 (119905) int120573

120572

(119904 minus 120572)2

119903 (119904)119889119904

(32)

1198612(120572 120573) = sup

120572le119905le120573

(int119905

120572

119901(119905)119889119905)

12

(int120573

119905

1

119903(119904)119889119904)

12

119862120573=

1

2max120572le119905le120573

119902 (119905) int120573

120572

(120573 minus 119904)2

119903 (119904)119889119904

(33)

120595(120572 120573) = sup120572le119905le120573

(int120573

119905

119903(119905)119889119905)

12

(int119905

120572

1

119903(119904)119889119904)

12

(34)

Now we are ready to state and prove the main results

Theorem 1 Assume that 119909(119905) is a nontrivial solution of (9) If119909(120572) = 1199091015840(120572) = 1199091015840(120573) = 0 then

1198671(120572 120573 119866

1) + 4119860

2

1(120572 120573) ge 1 (35)

If 1199091015840(120572) = 119909(120573) = 1199091015840(120573) = 0 then

1198672(120572 120573 119866

2) + 4119860

2

2(120572 120573) ge 1 (36)

Proof Weprove (35)Multiplying (9) by 1199091015840(119905) and integratingthe new equation from 120572 to 120573 we obtain

int120573

120572

1199091015840(119903(11990910158401015840))1015840

119889119905 = minusint120573

120572

119901(1199091015840)2

119889119905 minus int120573

120572

1199021199091199091015840119889119905 (37)

Mathematical Problems in Engineering 5

Integrating by parts the left-hand side we get that

11990311990910158401199091015840101584010038161003816100381610038161003816

120573

120572minus int120573

120572

119903(11990910158401015840)2

119889119905 = minusint120573

120572

119901(1199091015840)2

119889119905 minus int120573

120572

1199021199091199091015840119889119905

(38)

Using the assumptions that 1199091015840(120572) = 1199091015840(120573) = 0 and 1198761(119905) =

int120573

119905119902(119904)119889119904 we have

int120573

120572

119903(119905)(11990910158401015840

(119905))2

119889119905 = int120573

120572

119901 (119905) (1199091015840(119905))2

119889119905

+ int120573

120572

1198761015840

1(119905) 119909 (119905) 119909

1015840

(119905) 119889119905

(39)

Integrating the term int120573

12057211987610158401(119905)119909(119905)1199091015840(119905)119889119904 by parts and using

the assumption that 1199091015840(120572) = 1199091015840(120573) = 0 we obtain

int120573

120572

1198761015840

1(119905) 119909 (119905) 119909

1015840

(119905) 119889119904

= minusint120573

120572

1198761(119905) (1199091015840

(119905))2

119889119905 minus int120573

120572

1198761(119905) 119909 (119905) 119909

10158401015840

(119905) 119889119905

(40)

Substituting (40) into (39) we get

int120573

120572

119903(119905)1003816100381610038161003816100381611990910158401015840

(119905)10038161003816100381610038161003816

2

119889119905 le int120573

120572

10038161003816100381610038161198631 (119905)1003816100381610038161003816100381610038161003816100381610038161199091015840(119905)

10038161003816100381610038161003816

2

119889119905

+ int120573

120572

10038161003816100381610038161198761 (119905)1003816100381610038161003816|119909 (119905)|

1003816100381610038161003816100381611990910158401015840

(119905)10038161003816100381610038161003816119889119905

(41)

where |1198631(119905)| = |119876

1(119905)| + |119901(119905)| Applying inequality (19) on

the integral

int120573

120572

10038161003816100381610038161198761 (119905)1003816100381610038161003816|119909 (119905)|

1003816100381610038161003816100381611990910158401015840

(119905)10038161003816100381610038161003816119889119905 (42)

with 120601(119905) = |1198761(119905)| 120593(119905) = 119903(119905) 119896 = 0119898 = 119897 = 1 119899 = 2 119888 = 2

and 119909(120572) = 1199091015840(120572) = 0 we get

int120573

120572

10038161003816100381610038161198761 (119905)1003816100381610038161003816|119909 (119905)|

1003816100381610038161003816100381611990910158401015840

(119905)10038161003816100381610038161003816119889119905

le 1198671(120572 120573 119866

1) int120573

120572

119903 (119905)1003816100381610038161003816100381611990910158401015840(119905)

10038161003816100381610038161003816

2

119889119905

(43)

where 1198671(120573 120572 119866

1) is defined as in (29) Applying inequality

(12) on the integral int120573120572|1198631(119905)||1199091015840(119905)|2119889119905 with 119910(119905) = 1199091015840(119905) and

119910(120572) = 119910(120573) = 0 we have that

int120573

120572

10038161003816100381610038161198631 (119905)1003816100381610038161003816100381610038161003816100381610038161199091015840(119905)

10038161003816100381610038161003816

2

119889119905 le 41198602

1(120572 120573) int

120573

120572

119903 (119905)1003816100381610038161003816100381611990910158401015840(119905)

10038161003816100381610038161003816

2

119889119905

(44)

where1198601(120572 120573) is defined as in (31) Substituting (44) and (43)

into (41) we get

int120573

120572

|119903 (119905)|1003816100381610038161003816100381611990910158401015840(119905)

10038161003816100381610038161003816

2

119889119905 le 41198602

1(120572 120573) int

120573

120572

119903 (119905)1003816100381610038161003816100381611990910158401015840(119905)

10038161003816100381610038161003816

2

119889119905

+ 1198671(120572 120573 119866

1) int120573

120572

119903 (119905)1003816100381610038161003816100381611990910158401015840(119905)

10038161003816100381610038161003816

2

119889119905

(45)

Cancelling the term int120573

120572119903(119905)|11990910158401015840(119905)|2119889119905 we get

41198602

1(120572 120573) + 119867

1(120572 120573 119866

1) ge 1 (46)

which is the desired inequality (35) The proof of (36) issimilar to the proof of (35) by replacing 119867

1(120572 120573 119866

1) by

1198672(120572 120573 119866

2) and 119860

1(120572 120573) by 119860

2(120572 120573) The proof is com-

plete

In the following we apply the Clark and Hinton inequal-ity (26) to get a new result

Theorem 2 Assume that 119903(119905) is a nonincreasing function andsuppose that 119909(119905) is a solution of (9) If 119909(120572) = 1199091015840(120572) = 1199091015840(120573) =

0 then

41198602

1(120572 120573) +

(120573 minus 120572)32

radic3119903 (120573)(int120573

120572

10038161003816100381610038161198761(119905)10038161003816100381610038162

119889119905)

12

ge 1 (47)

where 1198601(120572 120573) is defined as in (31) for 119894 = 1 and 119876

1(119905) is

defined as in (29)

Proof Multiply (9) by 1199091015840(119905) and proceed as in the proof ofTheorem 1 to get

int120573

120572

119903(119905)1003816100381610038161003816100381611990910158401015840

(119905)10038161003816100381610038161003816

2

119889119905 le int120573

120572

10038161003816100381610038161198631 (119905)1003816100381610038161003816100381610038161003816100381610038161199091015840(119905)

10038161003816100381610038161003816

2

119889119905

+ int120573

120572

10038161003816100381610038161198761 (119905)1003816100381610038161003816|119909 (119905)|

1003816100381610038161003816100381611990910158401015840

(119905)10038161003816100381610038161003816119889119905

(48)

where |1198631(119905)| = |119876

1(119905)|+ |119901(119905)| Applying the Schwarz inequal-

ity

int120573

120572

1003816100381610038161003816119891 (119905) 119892 (119905)1003816100381610038161003816119889119905 le (int

120573

120572

1003816100381610038161003816119891(119905)10038161003816100381610038162

119889119905)

12

(int120573

120572

1003816100381610038161003816119892(119905)10038161003816100381610038162

119889119905)

12

(49)

on the integral int120573120572|1198761(119905)||119909(119905)||11990910158401015840(119905)|119889119905 we have that

int120573

120572

10038161003816100381610038161198761 (119905)1003816100381610038161003816|119909 (119905)|

1003816100381610038161003816100381611990910158401015840

(119905)10038161003816100381610038161003816119889119905 le (int

120573

120572

10038161003816100381610038161198761(119905)10038161003816100381610038162

119889119905)

12

times (int120573

120572

|119909(119905)|21003816100381610038161003816100381611990910158401015840(119905)

10038161003816100381610038161003816

2

119889119905)

12

(50)Applying inequality (26) and using the assumption 119909(120572) =

1199091015840(120572) = 0 we get that

(int120573

120572

|119909(119905)|21003816100381610038161003816100381611990910158401015840(119905)

10038161003816100381610038161003816

2

119889119905)

12

le(120573 minus 120572)

32

radic3int120573

120572

1003816100381610038161003816100381611990910158401015840(119905)

10038161003816100381610038161003816

2

119889119905

(51)Substituting (51) into (50) and using the assumption that 119903(119905)is a nonincreasing function we have

int120573

120572

10038161003816100381610038161198761(119905)1003816100381610038161003816|119909(119905)|

1003816100381610038161003816100381611990910158401015840

(119905)10038161003816100381610038161003816

le(120573 minus 120572)

32

radic3119903 (120573)(int120573

120572

10038161003816100381610038161198761(119905)10038161003816100381610038162

119889119905)

12

int120573

120572

119903 (119905)1003816100381610038161003816100381611990910158401015840(119905)

10038161003816100381610038161003816

2

119889119905

(52)

6 Mathematical Problems in Engineering

Applying Hardyrsquos inequality (12) on the integralint120573

120572|1198631(119905)||1199091015840(119905)

2|119889119905 we obtain

int120573

120572

10038161003816100381610038161198631 (119905)1003816100381610038161003816100381610038161003816100381610038161199091015840(119905)

10038161003816100381610038161003816

2

119889119905 le 41198602

1(120572 120573) int

120573

120572

119903 (119905)1003816100381610038161003816100381611990910158401015840(119905)

10038161003816100381610038161003816

2

119889119905 (53)

where 1199091015840(120572) = 1199091015840(120573) = 0 Substituting (52) and (53) into (48)we obtain

int120573

120572

|119903(119905)|1003816100381610038161003816100381611990910158401015840

(119905)10038161003816100381610038161003816

2

119889119905

le 41198602

1(120572 120573) int

120573

120572

119903 (119905)1003816100381610038161003816100381611990910158401015840(119905)

10038161003816100381610038161003816

2

119889119905

+(120573 minus 120572)

32

radic3119903 (120573)(int120573

120572

10038161003816100381610038161198761(119905)10038161003816100381610038162

119889119905)

12

int120573

120572

119903 (119905)1003816100381610038161003816100381611990910158401015840(119905)

10038161003816100381610038161003816

2

119889119905

(54)

The desired inequality (47) followed by cancelling the termint120573

120572|119903(119905)||11990910158401015840(119905)|2119889119905 The proof is complete

The proof of the following theorem is similar to the proofof Theorem 2

Theorem 3 Assume that 119903(119905) is a nonincreasing function andsuppose that 119909(119905) is a solution of (9) If 1199091015840(120572) = 119909(120573) = 1199091015840(120573) =

0 then

41198602

2(120572 120573) +

(120573 minus 120572)32

radic3119903 (120573)(int120573

120572

10038161003816100381610038161198761(119905)10038161003816100381610038162

119889119905)

12

ge 1 (55)

where 1198602(120572 120573) is defined as in (31) for 119894 = 2 and 119876

2(119905) is

defined as in (30)

One can use Theorems 2 and 3 to obtain some differentspecial cases For example as a special case of Theorem 2when 119903(119905) = 1 and 119901(119905) = 0 we have the following result

Corollary 4 If 119909(119905) is a nontrivial solution of

119909101584010158401015840

(119905) + 119902 (119905) 119909 (119905) = 0 119905 isin [120572 120573] (56)

which satisfies 119909(120572) = 1199091015840(120572) = 1199091015840(120573) = 0 then

int120573

120572

10038161003816100381610038161198761(119905)10038161003816100381610038162

119889119905 ge3

(120573 minus 120572)3 (57)

where 1198761(119905) is defined as in (29)

Now we will prove a new result when 119903(119905) = 1

Theorem 5 Suppose that 119903(119905) = 1 and assume that 119909(119905) is asolution of (9) If 119909(120572) = 1199091015840(120572) = 1199091015840(120573) = 0 or 119909(120573) = 1199091015840(120573) =

1199091015840(120572) = 0 then

119875120587(120573 minus 120572)

2

16+ 119876

(120573 minus 120572)3

6ge 1 (58)

where 119875 = max120572le119905le120573

|119901(119905)| and 119876 = max120572le119905le120573

|119902(119905)|

Proof Multiplying (9) by 1199091015840 and integrating by parts the left-hand side we have

1199091015840

(119905)11990910158401015840

(119905)10038161003816100381610038161003816

120573

120572minus int120573

120572

(11990910158401015840

(119905))2

119889119905

= minusint120573

120572

119901(119905)(1199091015840

(119905))2

119889119905 minus int120573

120572

119902(119905)119909(119905)1199091015840

(119905)119889119905

(59)

Using the assumptions that 1199091015840(120572) = 1199091015840(120573) = 0 we obtain

int120573

120572

(11990910158401015840)2

119889119905 = int120573

120572

119901(1199091015840)2

119889119905 + int120573

120572

1199021199091199091015840119889119905 (60)

By using themaximumvalues of |119901(119905)| and |119902(119905)| we canwrite(60) as

int120573

120572

1003816100381610038161003816100381611990910158401015840

(119905)10038161003816100381610038161003816

2

119889119905 le 119875int120573

120572

100381610038161003816100381610038161199091015840

(119905)10038161003816100381610038161003816

2

119889119905 + 119876int120573

120572

|119909(119905)|100381610038161003816100381610038161199091015840

(119905)10038161003816100381610038161003816119889119905

(61)

For the first term on the right-hand side int120573

120572|1199091015840(119905)|2119889119905 we

apply inequality (25) with 119910(119905) = 1199091015840(119905) (note that 1199091015840(120572) =

1199091015840(120573) = 0) and 120574 = 1 to obtain

int120573

120572

10038161003816100381610038161003816119909101584010038161003816100381610038161003816

2

119889119905 le120587(120573 minus 120572)

2

16int120573

120572

100381610038161003816100381610038161199091015840101584010038161003816100381610038161003816

2

119889119905 (62)

Applying Fink inequality (27) on int120573

120572|119909||1199091015840|119889119905 with 119896 = 0 119903 =

1 120583 = ] = 2 and 119899 = 2 we have

int120573

120572

|119909|10038161003816100381610038161003816119909101584010038161003816100381610038161003816119889119905 le

(120573 minus 120572)3

6int120573

120572

1003816100381610038161003816100381611990910158401015840(119905)

10038161003816100381610038161003816

2

119889119905 (63)

Substituting (62) and (63) into (61) we get after cancellingthe term int

120573

120572|11990910158401015840(119905)|2119889119905 that

119875120587(120573 minus 120572)

2

16+ 119876

(120573 minus 120572)3

6ge 1 (64)

which is the desired inequality (58) The proof is complete

As a special case of Theorem 5 when 119901(119905) = 0 we havethe following result

Corollary 6 Let 119909(119905) be a nontrivial solution of (56) If 119909(120572) =1199091015840(120572) = 1199091015840(120573) = 0 then

max120572le119905le120573

1003816100381610038161003816119902 (119905)1003816100381610038161003816 ge

6

(120573 minus 120572)3 (65)

In the following we prove some results related to theboundary conditions presented in (ii)

Theorem 7 Assume that 119909(119905) is a nontrivial solution of (9) If119909(120572) = 1199091015840(120572) = 11990910158401015840(120573) = 0 then

41198612

1(120572 120573) + 119862

120572ge 1 (66)

If 119909(120573) = 1199091015840(120573) = 11990910158401015840(120572) = 0 then

41198612

2(120572 120573) + 119862

120573ge 1 (67)

Mathematical Problems in Engineering 7

Proof Multiplying (9) by 1199091015840(119905) and integrating by parts theleft-hand side from 120572 to 120573 we get

11990311990910158401199091015840101584010038161003816100381610038161003816

120573

120572minus int120573

120572

119903(11990910158401015840)2

119889119905 = minusint120573

120572

119901(1199091015840)2

119889119905 minus int120573

120572

1199021199091199091015840119889119905

(68)

Using the assumption that 1199091015840(120572) = 11990910158401015840(120573) = 0 we have

int120573

120572

119903(119905)1003816100381610038161003816100381611990910158401015840

(119905)10038161003816100381610038161003816

2

119889119905 le int120573

120572

1003816100381610038161003816119901 (119905)1003816100381610038161003816100381610038161003816100381610038161199091015840(119905)

10038161003816100381610038161003816

2

119889119905

+ int120573

120572

1003816100381610038161003816119902 (119905)1003816100381610038161003816|119909 (119905)|

100381610038161003816100381610038161199091015840

(119905)10038161003816100381610038161003816119889119905

(69)

Applying inequality (12) on the integral int120573120572|119901(119905)||1199091015840(119905)|2119889119905

with 119899 = 119898 = 2 we get

int120573

120572

1003816100381610038161003816119901 (119905)1003816100381610038161003816100381610038161003816100381610038161199091015840(119905)

10038161003816100381610038161003816

2

119889119905 le 41198612

1(120572 120573) int

120573

120572

|119903 (119905)|1003816100381610038161003816100381611990910158401015840

(119905)210038161003816100381610038161003816119889119905 (70)

where1198611(120572 120573) is defined as in (32) Again applying inequality

(22) on the integral int120573120572|119902(119905)||119909(119905)||1199091015840(119905)|119889119905 we get

int120573

120572

1003816100381610038161003816119902 (119905)1003816100381610038161003816|119909 (119905)|

100381610038161003816100381610038161199091015840

(119905)10038161003816100381610038161003816119889119905 le 119862

120572int120573

120572

|119903 (119905)|1003816100381610038161003816100381611990910158401015840

(119905)210038161003816100381610038161003816119889119905 (71)

where 119862120572is defined as in (32) Substituting (70) and (71) into

(69) we obtain

int120573

120572

|119903 (119905)|1003816100381610038161003816100381611990910158401015840(119905)

10038161003816100381610038161003816

2

119889119905

le 41198612

1int120573

120572

|119903 (119905)|1003816100381610038161003816100381611990910158401015840

(119905)210038161003816100381610038161003816119889119905 + 119862

120572int120573

120572

|119903 (119905)|1003816100381610038161003816100381611990910158401015840

(119905)210038161003816100381610038161003816119889119905

(72)

By cancelling the term int120573

120572|119903(119905)||119909

10158401015840(119905)2|119889119905 we get that

41198612

1(120572 120573) + 119862

120572ge 1 (73)

which is the desired result (66) The proof (67) is similar tothe proof of (66) by using 119861

2(120572 120573) and 119862

120573instead of 119861

1(120572 120573)

and 119862120572and hence is omitted The proof is complete

In the following we apply an inequality due to Boyd [24]to obtain new results The Boyd inequality states that if 119910 isin

1198621[120572 120573] with 119910(120572) = 0 (or 119910(120573) = 0) then

int120573

120572

1003816100381610038161003816119910(119905)1003816100381610038161003816]100381610038161003816100381610038161199101015840(119905)

10038161003816100381610038161003816

120578

119889119905

le 119873 (] 120578 119904) (120573 minus 120572)][int119887

119886

100381610038161003816100381610038161199101015840(119905)

10038161003816100381610038161003816

119904

119889119905]

(]+120578)119904

(74)

where ] gt 0 119904 gt 1 0 le 120578 lt 119904 and

119873(] 120578 119904) =(119904 minus 120578)]]119903]+120578minus119904

(119904 minus 1) (] + 120578) (119868(] 120578 119904))]

119903 = ](119904 minus 1) + (119904 minus 120578)

(119904 minus 1)(] + 120578)

1119904

(75)

119868(] 120578 119904) = int1

0

1 +119904(120578 minus 1)

119904 minus 120578119905

minus(]+120578+119904])119904]

times [1 + (120578 minus 1)]1199051(]minus1)

119889119905

(76)

Note that an inequality of type (34) also holds when 119910(120572) =

119910(120573) = 0 Choose 119888 = (120572 + 120573)2 and apply (34) to [120572 119888] and[119888 120573] and by addition we obtain

int120573

120572

1003816100381610038161003816119910(119905)1003816100381610038161003816]100381610038161003816100381610038161199101015840(119905)

10038161003816100381610038161003816

120578

119889119905

le 119873 (] 120578 119904) (120573 minus 120572

2)

]

[int120573

120572

100381610038161003816100381610038161199101015840(119905)

10038161003816100381610038161003816

119904

119889119905]

(]+120578)119904

(77)

where 119873(] 120578 119904) is defined as in (75) An inequality of type(74) holds when 120578 = 119904 and 119910(120572) = 0 (or 119910(120573) = 0) In thiscase (74) becomes

int120573

120572

1003816100381610038161003816119910(119905)1003816100381610038161003816]100381610038161003816100381610038161199101015840(119905)

10038161003816100381610038161003816

120578

119889119905

le 119871 (] 120578) (120573 minus 120572)][int120573

120572

100381610038161003816100381610038161199101015840(119905)

10038161003816100381610038161003816

120578

119889119905]

(]+120578)120578

(78)

where

119871 (] 120578) =120578]120578

] + 120578(

]] + 120578

)

]120578

(Γ(((120578 + 1)120578) + (1]))Γ((120578 + 1)120578)Γ(1])

)

]

(79)

and Γ is the gamma function

Theorem 8 Assume that 119903(119905) is nonincreasing function and119909(119905) is a nontrivial solution of (9) If119909(120572) = 1199091015840(120572) = 11990910158401015840(120573) = 0then

41198612

1(120572 120573) +

8 (120573 minus 120572)

120587119903 (120573)1205952(120572 120573)(int

120573

120572

1003816100381610038161003816119902(119905)10038161003816100381610038162

119889119905)

12

ge 1 (80)

If 119909(120573) = 1199091015840(120573) = 11990910158401015840(120572) = 0 then

41198612

2(120572 120573) +

8 (120573 minus 120572)

120587119903 (120573)1205952(120572 120573)(int

120573

120572

1003816100381610038161003816119902(119905)10038161003816100381610038162

119889119905)

12

ge 1 (81)

Proof Multiplying (9) by 1199091015840(119905) and integrating by parts theleft-hand side we have that

11990311990910158401199091015840101584010038161003816100381610038161003816

120573

120572minus int120573

120572

119903(11990910158401015840)2

119889119905 = minusint120573

120572

119901(1199091015840)2

119889119905 minus int120573

120572

1199021199091199091015840119889119905

(82)

8 Mathematical Problems in Engineering

Using the assumption 1199091015840(120572) = 11990910158401015840(120573) = 0 we get that

int120573

120572

|119903 (119905)|1003816100381610038161003816100381611990910158401015840(119905)

10038161003816100381610038161003816

2

119889119905 le int120573

120572

1003816100381610038161003816119901 (119905)1003816100381610038161003816100381610038161003816100381610038161199091015840(119905)

10038161003816100381610038161003816

2

119889119905

+ int120573

120572

1003816100381610038161003816119902 (119905)1003816100381610038161003816|119909 (119905)|

100381610038161003816100381610038161199091015840

(119905)10038161003816100381610038161003816119889119905

(83)

Applying inequality (12) on the integral int120573120572|119901(119905)||1199091015840(119905)|2119889119905

with119898 = 119899 = 2 we get

int120573

120572

1003816100381610038161003816119901 (119905)1003816100381610038161003816100381610038161003816100381610038161199091015840(119905)

10038161003816100381610038161003816

2

119889119905

le 41205952(120572 120573) int

120573

120572

|119903 (119905)|1003816100381610038161003816100381611990910158401015840

(119905)210038161003816100381610038161003816119889119905

(84)

where 120595(120572 120573) is defined as in (34) Applying Schwarzrsquosinequality on the term int

120573

120572|119902(119905)||119909(119905)||1199091015840(119905)|119889119905 we get that

int120573

120572

1003816100381610038161003816119902 (119905)1003816100381610038161003816|119909 (119905)|

100381610038161003816100381610038161199091015840

(119905)10038161003816100381610038161003816119889119905

le (int120573

120572

1003816100381610038161003816119902(119905)10038161003816100381610038162

119889119905)

12

(int120573

120572

|119909|210038161003816100381610038161003816119909101584010038161003816100381610038161003816

2

119889119905)

12

(85)

Applying again inequality (78) on the integral int120573120572|119909|2|1199091015840|2119889119905

with ] = 120578 = 2 (note that 119909(120572) = 0) we obtain

int120573

120572

|119909|210038161003816100381610038161003816119909101584010038161003816100381610038161003816

2

119889119905 le4(120573 minus 120572)

2

12058721199032 (120573)[int120573

120572

119903(119905)10038161003816100381610038161003816119909101584010038161003816100381610038161003816

2

119889119905]

2

(86)

where 119903(119905) is a nonincreasing function Substituting (86) into(85) we get

int120573

120572

1003816100381610038161003816119902 (119905)1003816100381610038161003816|119909 (119905)|

100381610038161003816100381610038161199091015840

(119905)10038161003816100381610038161003816119889119905

le (int120573

120572

1003816100381610038161003816119902(119905)10038161003816100381610038162

119889119905)

12

2 (120573 minus 120572)

120587119903 (120573)int120573

120572

119903 (119905)10038161003816100381610038161003816119909101584010038161003816100381610038161003816

2

119889119905

(87)

Applying inequality (12) on the integral int120573120572119903(119905)|1199091015840|2119889119905 with

119901(119905) = 119902(119905) = 119903(119905) and119898 = 119899 = 2 we get

int120573

120572

|119903 (119905)|100381610038161003816100381610038161199091015840(119905)

10038161003816100381610038161003816

2

119889119905 le 41198612

1(120572 120573) int

120573

120572

|119903 (119905)|1003816100381610038161003816100381611990910158401015840(119905)

10038161003816100381610038161003816

2

119889119905 (88)

Substituting (88) into (87) we have

int120573

120572

1003816100381610038161003816119902 (119905)1003816100381610038161003816|119909 (119905)|

100381610038161003816100381610038161199091015840

(119905)10038161003816100381610038161003816119889119905

le (int120573

120572

1003816100381610038161003816119902(119905)10038161003816100381610038162

119889119905)

12

8(120573 minus 120572)

120587119903(120573)1205952(120572 120573)

times int120573

120572

|119903 (119905)|1003816100381610038161003816100381611990910158401015840(119905)

10038161003816100381610038161003816

2

119889119905

(89)

Substituting (84) and (89) into (83) we obtain

int120573

120572

|119903 (119905)|1003816100381610038161003816100381611990910158401015840(119905)

10038161003816100381610038161003816

2

119889119905

le 41198612

1(120572 120573) int

120573

120572

|119903(119905)|1003816100381610038161003816100381611990910158401015840

(119905)10038161003816100381610038161003816

2

119889119905

+(int120573

120572

1003816100381610038161003816119902(119905)10038161003816100381610038162

119889119905)

12

8 (120573 minus 120572)

120587119903 (120573)1205952(120572 120573)

times int120573

120572

|119903 (119905)|1003816100381610038161003816100381611990910158401015840(119905)

10038161003816100381610038161003816

2

119889119905

(90)

Cancelling the term int120573

120572|119903(119905)||11990910158401015840(119905)|2119889119905 we obtain

41198612

1(120572 120573) +

8 (120573 minus 120572)

120587119903 (120573)1205952(120572 120573)(int

120573

120572

1003816100381610038161003816119902(119905)10038161003816100381610038162

119889119905)

12

ge 1

(91)

which is the desired inequality (80) The proof of (81) issimilar to the proof of (80) by using 119861

2(120572 120573) instead of

1198611(120572 120573) The proof is complete

Conflict of Interests

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

References

[1] E Picard ldquoMemoire sur la theorie des equations aux derivespartielleset la methode desapproximations successivesrdquo Journalde Mathematiques Pures et Appliquees vol 6 pp 145ndash210 1890

[2] E Picard ldquoSur lrsquoapplication des methodes drsquoapproximationssuccessives a lrsquoetude de certaines equations differentielles ordi-nairesrdquo Journal deMathematiques Pures et Appliquees vol 9 pp217ndash271 1893

[3] O Niccoletti ldquoSulle consizioni iniziali che determinano gliintegrale delle equazioni diff eren ziali ordinarierdquo Atti dellaAccademia delle Scienze di Torino vol 33 pp 746ndash759 1897

[4] C de la Vallee Poussin ldquoSur lrsquoequation differentielle lineaire dusecond ordrerdquo Journal de Mathematiques Pures et Appliqueesvol 8 pp 125ndash144 1929

[5] A M Lyapunov ldquoProbleme general de la stabilite du mouve-mentrdquo Annales de la Faculte des sciences de Toulouse vol 9 pp203ndash474 1907

[6] S H Saker ldquoSome new disconjugacy criteria for second orderdifferential equations with a middle termrdquo Bulletin Mathema-tique de la Societe des Sciences Mathematiques de Roumanie vol57 no 1 pp 109ndash120 2014

[7] G Casadei ldquoSul teorema di unicita di De La Vallee Poussinper equazioni differenziali del terzo ordinerdquo Rendiconti delSeminario Matematico della Universita di Padova vol 41 pp300ndash315 1968

[8] A Lasota ldquoSur la distance entre les zeros de lrsquoequation dif-ferentielle lineaire du troisieme ordrerdquo Annales Polonici Mathe-matici vol 13 pp 129ndash132 1963

Mathematical Problems in Engineering 9

[9] RMMathsen ldquoA disconjugacy condition for119910101584010158401015840+119886211991010158401015840+119886

11199101015840+

1198860119910 = 0rdquo Proceedings of the AmericanMathematical Society vol

17 pp 627ndash632 1966[10] B G Pachpatte ldquoOn the zeros of solutions of certain differential

equationsrdquo Demonstratio Mathematica vol 25 no 4 pp 825ndash833 1992

[11] B G Pachpatte ldquoLyapunov type integral inequalities for certaindifferential equationsrdquo Georgian Mathematical Journal vol 4no 2 pp 139ndash148 1997

[12] S Panigrahi ldquoLyapunov-type integral inequalities for certainhigher-order differential equationsrdquo Electronic Journal of Differ-ential Equations vol 2009 no 28 pp 1ndash14 2009

[13] N Parhi and S Panigrahi ldquoOn Liapunov-type inequality forthird-order differential equationsrdquo Journal of MathematicalAnalysis and Applications vol 233 no 2 pp 445ndash460 1999

[14] N Parhi and S Panigrahi ldquoDisfocality and Liapunov-typeinequalities for third-order equationsrdquo Applied MathematicsLetters vol 16 no 2 pp 227ndash233 2003

[15] U Richard ldquoMetodi diversi per ottenere disequaglianze allade la Vallee Poussin nelle equazioni differenziali ordinarie delsecondo e terzo ordinerdquo Rendiconti del Seminario MatematicoUniversita e Politecnico di Torino vol 27 pp 35ndash68 1968

[16] X Yang ldquoOn Liapunov-type inequality for certain higher-orderdifferential equationsrdquo Applied Mathematics and Computationvol 134 no 2-3 pp 307ndash317 2003

[17] R P Agarwal and P R Krishnamurthy ldquoOn the uniquenessof solution of nonlinear boundary value problemsrdquo Journal ofMathematical Sciences vol 10 no 1 pp 17ndash31 1976

[18] R P Agarwal and P Y PangOpial Inequalities with Applicationsin Differential and Difference Equation Kluwer AcademicDordrecht The Netherlands 1995

[19] R P Agarwal and P Y Pang ldquoRemarks on the generalizationsof Opialrsquos inequalityrdquo Journal of Mathematical Analysis andApplications vol 190 no 2 pp 559ndash577 1995

[20] P R Beesack and K M Das ldquoExtensions of Opialrsquos inequalityrdquoPacific Journal of Mathematics vol 26 pp 215ndash232 1968

[21] S Clark and D Hinton ldquoSome disconjugacy criteria for dif-ferential equations with oscillatory coefficientsrdquoMathematischeNachrichten vol 278 no 12-13 pp 1476ndash1489 2005

[22] A M Fink ldquoOn Opials inequality for 119891(119899)rdquo Proceedings of theAmerican Mathematical Society vol 155 pp 177ndash181 1992

[23] A Kufner and L-E Persson Weighted Inequalities of HardyType World Scientific River Edge NJ USA 2003

[24] D W Boyd ldquoBest constants in a class of integral inequalitiesrdquoPacific Journal of Mathematics vol 30 pp 367ndash383 1969

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

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The Scientific World JournalHindawi Publishing Corporation httpwwwhindawicom Volume 2014

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Algebra

Discrete Dynamics in Nature and Society

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Decision SciencesAdvances in

Discrete MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom

Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Stochastic AnalysisInternational Journal of

2 Mathematical Problems in Engineering

Poussin in 1929 was on the evaluation of the length ofthe interval [0 ℎ] in which the boundary problem 119910(119905

1) =

119910(1199052) = sdot sdot sdot = 119910(119905

119899) = 0 (0 le 119905

1lt 1199052lt sdot sdot sdot lt 119905

119899le ℎ)

for the linear differential equation (1) only admits the nullsolution

Following the way indicated by de la Vallee Poussinone has tried to evaluate the length ℎ in a function of theupper bounds of the coefficients The precise evaluation ofthe maximal length of the considered interval has only beenobtained for second-order differential equations [5 6] withthe best possible constant 14 Coming then to (1) de la ValleePoussin proved that this equation is disconjugate in [0 ℎ]

if the coefficients and the length of the interval satisfy theinequality

119899

sum119894=1

119875119894

ℎ119894

119894lt 1 119875

119894= max119886le119905le119887

1003816100381610038161003816119901119894 (119905)1003816100381610038161003816 for 119894 = 1 2 119899 (3)

This is the part of his paper which has motivated the manyrefinements and generalizations described in the analysisThe importance of Poussinrsquos work has been emphasized andtestified by many authors in the literature we refer the readerto the papers [7ndash16] and the references cited therein Forcompleteness we recall some of the related results whichmotivate the contents of this paper Lasota [8] considered thethird-order differential equation

119910101584010158401015840

(119905) + 1199011(119905)11991010158401015840

(119905) + 1199012(119905)1199101015840

(119905) + 1199013(119905)119910(119905) = 0

119905 isin [119886 119887](4)

and proved that if

1198751

ℎ

4+ 1198752

ℎ2

1205872+ 1198753

ℎ3

21205872le 1 (5)

then (4) is disconjugate where 119875119894= max |119901

119894(119905)| and ℎ = 119887minus119886

Mathsen [9] proved that (4) is disconjugate if 1199012(119909) le 0 on

[119886 119887] and

(2ℎ1+ 1)1198753

[exp(21198751ℎ1) minus exp(119875

1ℎ1) minus ℎ11198751]

11987521

le 1 (6)

where 1198751= max |119901

1(119905)| 119875

3= max |119901

3(119905)| on [119886 119887] and ℎ

1=

(119887 minus 119886)2 Casadei [7] also proved that if

1198751

ℎ

2+ 1198752

ℎ2

8+ 1198753

ℎ3

24le 1 (7)

then (4) is disconjugate Agarwal and Krishnamurthy [17]also proved that if

1198751

2ℎ

3+ 1198752

ℎ2

6+ 1198753

2ℎ3

81le 1 (8)

then (4) is disconjugate

Motivated by these papers we will study the distribu-tion of zeros of the solutions of the third-order differen-tial equation

(119903(119905)11990910158401015840(119905))1015840

+ 119901 (119905) 1199091015840

(119905) + 119902 (119905) 119909 (119905) = 0 119905 isin I (9)

where I is an interval of reals and 119903(119905) 119901(119905) and 119902(119905) are real-valued functions defined on I such that 119903(119905) gt 0

By a solution of (9) on the interval J sube I we mean anontrivial real-valued function 119909 isin 119862

2(J) which has theproperty that 119903(119905)11990910158401015840(119905) isin 1198621(J) and satisfies (9) on J Thenontrivial solution 119909(119905) of (9) is said to oscillate or to beoscillatory if it has arbitrarily large zeros Equation (9) isoscillatory if one of its nontrivial solutions is oscillatory Anequation of the form (9) is said to be disconjugate on aninterval I if no nontrivial solution has more than two zeroson I counting multiplicities We say that (9) is right disfocal(left disfocal) on the interval [119886 119887] (119886 lt 119887) if the solutions of(9) such that 1199091015840(119886) = 0 119909(119886) = 0 (1199091015840(119887) = 0 and 119909(119887) = 0) donot have two zeros counting multiplicities in (119886 119887] ([119886 119887))

The paper is organized as follows In Section 2 we presentHardyrsquos inequality the extensions of Opialrsquos inequality andWirtingerrsquos inequality that will be used to prove our mainresults In Section 3 we are concerned with the lower boundsof the distance between zeros of a nontrivial solution andorits derivatives for the third-order differential equation (9)subject to two sets of boundary conditions In particular wewill prove the following

(i) Obtain lower bounds for the spacing 120573minus120572 where 119909 isa nontrivial solution of (9) which satisfies

119909 (120572) = 1199091015840

(120572) = 1199091015840(120573) = 0 or

119909 (120573) = 1199091015840(120573) = 119909

1015840

(120572) = 0

(10)

(ii) Obtain lower bounds for the spacing 120573minus120572 where 119909 isa solution of (9) which satisfies

119909 (120572) = 1199091015840

(120572) = 11990910158401015840(120573) = 0 or

119909 (120573) = 1199091015840(120573) = 119909

10158401015840

(120572) = 0

(11)

2 Hardy Opial and Wirtinger Inequalities

In this section we present Hardyrsquos inequality and somegeneralizations of Opial and Wirtinger type inequalitiesthat will be needed in the proof of the main results Thegeneralizations of Opial and Wirtinger type inequalities areadapted from Agarwal and Pang [18 19] and the papers dueto Beesack and Das [20] Clark and Hinton [21] and Fink[22] The Hardy inequality is adapted from the book due toKufner and Persson [23] We begin with Hardyrsquos inequalitywhich states the following If 119910 is absolutely continuous on(120572 120573) such that 119910(120572) = 0 or (119910(120573) = 0) then

(int120573

120572

119901(119905)1003816100381610038161003816119910(119905)

1003816100381610038161003816119899

119889119905)

1119899

le 119862(int120573

120572

119903(119905)100381610038161003816100381610038161199101015840(119905)

10038161003816100381610038161003816

119898

119889119905)

1119898

(12)

Mathematical Problems in Engineering 3

where the weighted functions 119901 119903 are positive functionsdefined on (120572 120573) and 119899 119898 are real parameters that satisfy0 lt 119899 le infin and 1 le 119898 le infin The constant 119862 is given by

119862 le 1198981119898

(1198981)11198981119860(120572 120573) for 1 lt 119898 le 119899 119898

1=

119898

119898 minus 1

(13)

where

119860(120572 120573) = sup120572le119905le120573

(int120573

119905

119901(119905)119889119905)

1119899

(int119905

120572

1199031minus1198981(119904)119889119904)

11198981

if 119910(120572) = 0

119860(120572 120573) = sup120572le119905le120573

(int119905

120572

119901(119905)119889119905)

1119899

(int120573

119905

1199031minus1198981(119904)119889119904)

11198981

if 119910(120573) = 0

(14)

Note that inequality (12) has an immediate application to thecase when 119910(120572) = 119910(120573) = 0 In this case inequality (12) issatisfied if and only if

119860(120572 120573)

= sup(119888119889)sub(120572120573)

(int119889

119888

119901(119905)119889119905)

1119899

timesmin(int119888

120572

1199031minus1198981(119904)119889119904)

11198991

(int120573

119889

1199031minus1198981(119904)119889119904)

11198981

(15)

exists and is finite The Opial type inequality due to Beesackwhich is a generalization of Opialrsquos inequality states that if 119910is absolutely continuous on [120572 120573] with 119910(120572) = 0 then thefollowing inequality holds

int120573

120572

119902 (119905)1003816100381610038161003816119910(119905)

1003816100381610038161003816119898100381610038161003816100381610038161199101015840(119905)

10038161003816100381610038161003816

119899

119889119905 le 1198701(119898 119899) int

120573

120572

119901 (119905)100381610038161003816100381610038161199101015840(119905)

10038161003816100381610038161003816

119898+119899

119889119905

(16)

where 119898 119899 are real numbers such that 119898119899 gt 0 119898 + 119899 gt 1119902(119905) and 119901(119905) are nonnegative measurable functions definedon (120572 120573) such that int119905

120572(119901(119904))minus1(119898+119899minus1)119889119904 lt infin and

1198701(119898 119899)

= (119899

119898 + 119899)119899(119898+119899)

times [int120573

120572

119902(119898+119899)119899

(119905)119901minus119899119898

(119905)

times (int119905

120572

(119901(119904))minus1(119898+119899minus1)

119889119904)

119898+119899minus1

119889119905]

119898(119898+119899)

(17)

If instead 119910(120573) = 0 then (16) holds where 1198701(119898 119899) is

replaced by

1198702(119898 119899)

= (119899

119898 + 119899)119899(119898+119899)

times [int120573

120572

119902(119898+119899)119899

(119905) 119901minus(119899119898)

(119905)

times(int120573

119905

(119901(119904))minus1(119898+119899minus1)

119889119904)

119898+119899minus1

119889119905]

119898(119898+119899)

(18)

TheOpial type inequality due to Agarwal and Pang states thatif 119910(119905) isin 119862(119899minus1)[120572 120573] and satisfies119910(119894)(120572) = 0 0 le 119896 le 119894 le 119899minus1

(119899 ge 1) and 119910(119899minus1)(119905) absolutely continuous on (120572 120573) then

int120573

120572

120601 (119905)10038161003816100381610038161003816119910(119896)(119905)

10038161003816100381610038161003816

11989710038161003816100381610038161003816119910(119899)(119905)

10038161003816100381610038161003816

119898

119889119905

le 1198671[int120573

120572

120593(119905)10038161003816100381610038161003816119910(119899)(119905)

10038161003816100381610038161003816

119888

119889119905]

(119897+119898)119888

(19)

where 120601 and 120593 are nonnegative and measurable functiondefined on (120572 120573) 119898 119899 are real numbers such that 119888119898 gt 1and

1198671=

(119898(119898 + 119897))119898119888

(119899 minus 119896 minus 1)

times [int120573

120572

(120601119888

(119905)120593minus119898

(119905))1(119888minus119898)

times (1198661119896(119905))119897(119888minus1)(119888minus119898)

119889119905]

(119888minus119898)119888

1198661119896

(119905) = int119905

120572

(119905 minus 119904)(119899minus119896minus1)119888(119888minus1)

(120593(119904))minus1(119888minus1)

119889119904

(20)

If instead 119910(119894)(120573) = 0 0 le 119896 le 119894 le 119899 minus 1 (119899 ge 1) then (19)holds where119867

1is replaced by

1198672=

(119898(119898 + 119897))119898119888

(119899 minus 119896 minus 1)

times [int120573

120572

(120601119888(119905)120593minus119898

(119905))1(119888minus119898)

times (1198662119896(119905))119897(119888minus1)(119888minus119898)

119889119905]

(119888minus119898)119888

1198662119896

(119905) = int120573

119905

(119904 minus 119905)(119899minus119896minus1)119888(119888minus1)

(120593(119904))minus1(119888minus1)

119889119904

(21)

4 Mathematical Problems in Engineering

We also need the following inequality which is the specialcase of an inequality proved by Agarwal and Pang [18] withtwo functions

int120573

120572

119888(119905)10038161003816100381610038161003816119910(119896)

(119905)10038161003816100381610038161003816

10038161003816100381610038161003816119910(119896+1)

(119905)10038161003816100381610038161003816119889119905 le 119862

120572int120573

120572

119903 (119905)10038161003816100381610038161003816119910(119899)(119905)

10038161003816100381610038161003816

2

119889119905

(22)

where 119910(119905) isin 119862119899minus1[120572 120573] and satisfies 119910(119894)(120572) = 0 119896 le 119894 le 119899minus10 le 119896 le 119899 minus 1 (119899 ge 1) and 119910(119899minus1) is absolutely continuous on(120572 120573) 119888(119905) and 119903(119905) being nonnegative measurable functionsdefined on (120572 120573) and

119862120572=

1

2((119899 minus 119896 minus 1))2max119905isin[120572120573]

119888 (119905) int120573

120572

(119904 minus 120572)2(119899minus119896minus1)

119903 (119904)119889119904

(23)

If 119910(119894)(120573) = 0 119896 le 119894 le 119899 minus 1 then (22) holds where 119862120572is

replaced

119862120573=

1

2((119899 minus 119896 minus 1))2max119905isin[120572120573]

119888 (119905) int120573

120572

(120573 minus 119904)2(119899minus119896minus1)

119903 (119904)119889119904

(24)

TheWirtinger type inequality due to Agarwal and Pang statesthat if 119910(120572) = 119910(120573) = 0 then

int120573

120572

119910120574+1

(119905)119889119905 le(120573 minus 120572)

120574+1

Γ2 ((120574 + 2)2)

2Γ(120574 + 2)

times int120573

120572

(1199101015840(119905))120574+1

(119905) 119889119905

(25)

TheOpial type inequality due to Clark and Hinton inequalitystates that if 119910 isin 119862

2[120572 120573] where 119910(120572) = 1199101015840(120572) = 0 then

(int120573

120572

1003816100381610038161003816119910(119905)100381610038161003816100381621003816100381610038161003816100381611991010158401015840(119905)

10038161003816100381610038161003816

2

119889119905)

1119903

le(120573 minus 120572)

32

radic3int120573

120572

1003816100381610038161003816100381611991010158401015840

(119905)10038161003816100381610038161003816

2

119889119905

(26)

The Opial type inequality due to Fink inequality states that if119910(119894)(120572) = 0 0 le 119894 le 119899 minus 1 120583 ge 1 (1120583) + (1]) = 1 then

int120573

120572

10038161003816100381610038161003816119910(119896)

(119905) 119910(119903)

(119905)10038161003816100381610038161003816119889119905

le 119862 (119899 119896 119903 120583) (120573 minus 120572)2119899minus119896minus119903+1minus2120583

(int120573

120572

10038161003816100381610038161003816119910(119899)(119905)

10038161003816100381610038161003816

120583

119889119905)

2120583

(27)

where 0 le 119896 lt 119903 lt 119899 (119899 ge 2) and

119862 (119899 119896 119903 120583) =1

2((119899 minus 119896 minus 1))2[(119899 minus 119896 minus 1)] + 1]

2]

(28)

3 Main Results

In this section we state and prove the main results Forsimplicity we introduce the following notations

1198671(120572 120573 119866

10) =

1

radic2[int120573

120572

1198762

1(119905)

1198661(119905)

119903(119905)119889119905]

12

1198661(119905) = int

119905

120572

(119905 minus 119904)2

119903 (119904)119889119904 119876

1(119905) = int

120573

119905

119902 (119904) 119889119904

(29)

1198672(120572 120573 119866

2) =

1

radic2[int120573

120572

1198762

2(119905)

1198662(119905)

119903(119905)119889119905]

12

1198662(119905) = int

120573

119905

(119905 minus 119904)2

119903 (119904)119889119904 119876

2(119905) = int

119905

120572

119902 (119904) 119889119904

(30)

119860119894(120572 120573) = 119869

119894(119905)min(int

119888

120572

1

119903(119904)119889119904)

12

(int120573

119888

1

119903(119904)119889119904)

12

(31)

where 119869119894(119905) = sup

(119888119889)sub(120572120573)(int120573

120572119863119894(119905)119889119905)12 and |119863

119894(119905)| =

|119876119894(119905)| + |119901(119905)| 119894 = 1 2

1198611(120572 120573) = sup

120572le119905le120573

(int120573

119905

119901(119905)119889119905)

12

(int119905

120572

1

119903(119904)119889119904)

12

119862120572=

1

2max120572le119905le120573

119902 (119905) int120573

120572

(119904 minus 120572)2

119903 (119904)119889119904

(32)

1198612(120572 120573) = sup

120572le119905le120573

(int119905

120572

119901(119905)119889119905)

12

(int120573

119905

1

119903(119904)119889119904)

12

119862120573=

1

2max120572le119905le120573

119902 (119905) int120573

120572

(120573 minus 119904)2

119903 (119904)119889119904

(33)

120595(120572 120573) = sup120572le119905le120573

(int120573

119905

119903(119905)119889119905)

12

(int119905

120572

1

119903(119904)119889119904)

12

(34)

Now we are ready to state and prove the main results

Theorem 1 Assume that 119909(119905) is a nontrivial solution of (9) If119909(120572) = 1199091015840(120572) = 1199091015840(120573) = 0 then

1198671(120572 120573 119866

1) + 4119860

2

1(120572 120573) ge 1 (35)

If 1199091015840(120572) = 119909(120573) = 1199091015840(120573) = 0 then

1198672(120572 120573 119866

2) + 4119860

2

2(120572 120573) ge 1 (36)

Proof Weprove (35)Multiplying (9) by 1199091015840(119905) and integratingthe new equation from 120572 to 120573 we obtain

int120573

120572

1199091015840(119903(11990910158401015840))1015840

119889119905 = minusint120573

120572

119901(1199091015840)2

119889119905 minus int120573

120572

1199021199091199091015840119889119905 (37)

Mathematical Problems in Engineering 5

Integrating by parts the left-hand side we get that

11990311990910158401199091015840101584010038161003816100381610038161003816

120573

120572minus int120573

120572

119903(11990910158401015840)2

119889119905 = minusint120573

120572

119901(1199091015840)2

119889119905 minus int120573

120572

1199021199091199091015840119889119905

(38)

Using the assumptions that 1199091015840(120572) = 1199091015840(120573) = 0 and 1198761(119905) =

int120573

119905119902(119904)119889119904 we have

int120573

120572

119903(119905)(11990910158401015840

(119905))2

119889119905 = int120573

120572

119901 (119905) (1199091015840(119905))2

119889119905

+ int120573

120572

1198761015840

1(119905) 119909 (119905) 119909

1015840

(119905) 119889119905

(39)

Integrating the term int120573

12057211987610158401(119905)119909(119905)1199091015840(119905)119889119904 by parts and using

the assumption that 1199091015840(120572) = 1199091015840(120573) = 0 we obtain

int120573

120572

1198761015840

1(119905) 119909 (119905) 119909

1015840

(119905) 119889119904

= minusint120573

120572

1198761(119905) (1199091015840

(119905))2

119889119905 minus int120573

120572

1198761(119905) 119909 (119905) 119909

10158401015840

(119905) 119889119905

(40)

Substituting (40) into (39) we get

int120573

120572

119903(119905)1003816100381610038161003816100381611990910158401015840

(119905)10038161003816100381610038161003816

2

119889119905 le int120573

120572

10038161003816100381610038161198631 (119905)1003816100381610038161003816100381610038161003816100381610038161199091015840(119905)

10038161003816100381610038161003816

2

119889119905

+ int120573

120572

10038161003816100381610038161198761 (119905)1003816100381610038161003816|119909 (119905)|

1003816100381610038161003816100381611990910158401015840

(119905)10038161003816100381610038161003816119889119905

(41)

where |1198631(119905)| = |119876

1(119905)| + |119901(119905)| Applying inequality (19) on

the integral

int120573

120572

10038161003816100381610038161198761 (119905)1003816100381610038161003816|119909 (119905)|

1003816100381610038161003816100381611990910158401015840

(119905)10038161003816100381610038161003816119889119905 (42)

with 120601(119905) = |1198761(119905)| 120593(119905) = 119903(119905) 119896 = 0119898 = 119897 = 1 119899 = 2 119888 = 2

and 119909(120572) = 1199091015840(120572) = 0 we get

int120573

120572

10038161003816100381610038161198761 (119905)1003816100381610038161003816|119909 (119905)|

1003816100381610038161003816100381611990910158401015840

(119905)10038161003816100381610038161003816119889119905

le 1198671(120572 120573 119866

1) int120573

120572

119903 (119905)1003816100381610038161003816100381611990910158401015840(119905)

10038161003816100381610038161003816

2

119889119905

(43)

where 1198671(120573 120572 119866

1) is defined as in (29) Applying inequality

(12) on the integral int120573120572|1198631(119905)||1199091015840(119905)|2119889119905 with 119910(119905) = 1199091015840(119905) and

119910(120572) = 119910(120573) = 0 we have that

int120573

120572

10038161003816100381610038161198631 (119905)1003816100381610038161003816100381610038161003816100381610038161199091015840(119905)

10038161003816100381610038161003816

2

119889119905 le 41198602

1(120572 120573) int

120573

120572

119903 (119905)1003816100381610038161003816100381611990910158401015840(119905)

10038161003816100381610038161003816

2

119889119905

(44)

where1198601(120572 120573) is defined as in (31) Substituting (44) and (43)

into (41) we get

int120573

120572

|119903 (119905)|1003816100381610038161003816100381611990910158401015840(119905)

10038161003816100381610038161003816

2

119889119905 le 41198602

1(120572 120573) int

120573

120572

119903 (119905)1003816100381610038161003816100381611990910158401015840(119905)

10038161003816100381610038161003816

2

119889119905

+ 1198671(120572 120573 119866

1) int120573

120572

119903 (119905)1003816100381610038161003816100381611990910158401015840(119905)

10038161003816100381610038161003816

2

119889119905

(45)

Cancelling the term int120573

120572119903(119905)|11990910158401015840(119905)|2119889119905 we get

41198602

1(120572 120573) + 119867

1(120572 120573 119866

1) ge 1 (46)

which is the desired inequality (35) The proof of (36) issimilar to the proof of (35) by replacing 119867

1(120572 120573 119866

1) by

1198672(120572 120573 119866

2) and 119860

1(120572 120573) by 119860

2(120572 120573) The proof is com-

plete

In the following we apply the Clark and Hinton inequal-ity (26) to get a new result

Theorem 2 Assume that 119903(119905) is a nonincreasing function andsuppose that 119909(119905) is a solution of (9) If 119909(120572) = 1199091015840(120572) = 1199091015840(120573) =

0 then

41198602

1(120572 120573) +

(120573 minus 120572)32

radic3119903 (120573)(int120573

120572

10038161003816100381610038161198761(119905)10038161003816100381610038162

119889119905)

12

ge 1 (47)

where 1198601(120572 120573) is defined as in (31) for 119894 = 1 and 119876

1(119905) is

defined as in (29)

Proof Multiply (9) by 1199091015840(119905) and proceed as in the proof ofTheorem 1 to get

int120573

120572

119903(119905)1003816100381610038161003816100381611990910158401015840

(119905)10038161003816100381610038161003816

2

119889119905 le int120573

120572

10038161003816100381610038161198631 (119905)1003816100381610038161003816100381610038161003816100381610038161199091015840(119905)

10038161003816100381610038161003816

2

119889119905

+ int120573

120572

10038161003816100381610038161198761 (119905)1003816100381610038161003816|119909 (119905)|

1003816100381610038161003816100381611990910158401015840

(119905)10038161003816100381610038161003816119889119905

(48)

where |1198631(119905)| = |119876

1(119905)|+ |119901(119905)| Applying the Schwarz inequal-

ity

int120573

120572

1003816100381610038161003816119891 (119905) 119892 (119905)1003816100381610038161003816119889119905 le (int

120573

120572

1003816100381610038161003816119891(119905)10038161003816100381610038162

119889119905)

12

(int120573

120572

1003816100381610038161003816119892(119905)10038161003816100381610038162

119889119905)

12

(49)

on the integral int120573120572|1198761(119905)||119909(119905)||11990910158401015840(119905)|119889119905 we have that

int120573

120572

10038161003816100381610038161198761 (119905)1003816100381610038161003816|119909 (119905)|

1003816100381610038161003816100381611990910158401015840

(119905)10038161003816100381610038161003816119889119905 le (int

120573

120572

10038161003816100381610038161198761(119905)10038161003816100381610038162

119889119905)

12

times (int120573

120572

|119909(119905)|21003816100381610038161003816100381611990910158401015840(119905)

10038161003816100381610038161003816

2

119889119905)

12

(50)Applying inequality (26) and using the assumption 119909(120572) =

1199091015840(120572) = 0 we get that

(int120573

120572

|119909(119905)|21003816100381610038161003816100381611990910158401015840(119905)

10038161003816100381610038161003816

2

119889119905)

12

le(120573 minus 120572)

32

radic3int120573

120572

1003816100381610038161003816100381611990910158401015840(119905)

10038161003816100381610038161003816

2

119889119905

(51)Substituting (51) into (50) and using the assumption that 119903(119905)is a nonincreasing function we have

int120573

120572

10038161003816100381610038161198761(119905)1003816100381610038161003816|119909(119905)|

1003816100381610038161003816100381611990910158401015840

(119905)10038161003816100381610038161003816

le(120573 minus 120572)

32

radic3119903 (120573)(int120573

120572

10038161003816100381610038161198761(119905)10038161003816100381610038162

119889119905)

12

int120573

120572

119903 (119905)1003816100381610038161003816100381611990910158401015840(119905)

10038161003816100381610038161003816

2

119889119905

(52)

6 Mathematical Problems in Engineering

Applying Hardyrsquos inequality (12) on the integralint120573

120572|1198631(119905)||1199091015840(119905)

2|119889119905 we obtain

int120573

120572

10038161003816100381610038161198631 (119905)1003816100381610038161003816100381610038161003816100381610038161199091015840(119905)

10038161003816100381610038161003816

2

119889119905 le 41198602

1(120572 120573) int

120573

120572

119903 (119905)1003816100381610038161003816100381611990910158401015840(119905)

10038161003816100381610038161003816

2

119889119905 (53)

where 1199091015840(120572) = 1199091015840(120573) = 0 Substituting (52) and (53) into (48)we obtain

int120573

120572

|119903(119905)|1003816100381610038161003816100381611990910158401015840

(119905)10038161003816100381610038161003816

2

119889119905

le 41198602

1(120572 120573) int

120573

120572

119903 (119905)1003816100381610038161003816100381611990910158401015840(119905)

10038161003816100381610038161003816

2

119889119905

+(120573 minus 120572)

32

radic3119903 (120573)(int120573

120572

10038161003816100381610038161198761(119905)10038161003816100381610038162

119889119905)

12

int120573

120572

119903 (119905)1003816100381610038161003816100381611990910158401015840(119905)

10038161003816100381610038161003816

2

119889119905

(54)

The desired inequality (47) followed by cancelling the termint120573

120572|119903(119905)||11990910158401015840(119905)|2119889119905 The proof is complete

The proof of the following theorem is similar to the proofof Theorem 2

Theorem 3 Assume that 119903(119905) is a nonincreasing function andsuppose that 119909(119905) is a solution of (9) If 1199091015840(120572) = 119909(120573) = 1199091015840(120573) =

0 then

41198602

2(120572 120573) +

(120573 minus 120572)32

radic3119903 (120573)(int120573

120572

10038161003816100381610038161198761(119905)10038161003816100381610038162

119889119905)

12

ge 1 (55)

where 1198602(120572 120573) is defined as in (31) for 119894 = 2 and 119876

2(119905) is

defined as in (30)

One can use Theorems 2 and 3 to obtain some differentspecial cases For example as a special case of Theorem 2when 119903(119905) = 1 and 119901(119905) = 0 we have the following result

Corollary 4 If 119909(119905) is a nontrivial solution of

119909101584010158401015840

(119905) + 119902 (119905) 119909 (119905) = 0 119905 isin [120572 120573] (56)

which satisfies 119909(120572) = 1199091015840(120572) = 1199091015840(120573) = 0 then

int120573

120572

10038161003816100381610038161198761(119905)10038161003816100381610038162

119889119905 ge3

(120573 minus 120572)3 (57)

where 1198761(119905) is defined as in (29)

Now we will prove a new result when 119903(119905) = 1

Theorem 5 Suppose that 119903(119905) = 1 and assume that 119909(119905) is asolution of (9) If 119909(120572) = 1199091015840(120572) = 1199091015840(120573) = 0 or 119909(120573) = 1199091015840(120573) =

1199091015840(120572) = 0 then

119875120587(120573 minus 120572)

2

16+ 119876

(120573 minus 120572)3

6ge 1 (58)

where 119875 = max120572le119905le120573

|119901(119905)| and 119876 = max120572le119905le120573

|119902(119905)|

Proof Multiplying (9) by 1199091015840 and integrating by parts the left-hand side we have

1199091015840

(119905)11990910158401015840

(119905)10038161003816100381610038161003816

120573

120572minus int120573

120572

(11990910158401015840

(119905))2

119889119905

= minusint120573

120572

119901(119905)(1199091015840

(119905))2

119889119905 minus int120573

120572

119902(119905)119909(119905)1199091015840

(119905)119889119905

(59)

Using the assumptions that 1199091015840(120572) = 1199091015840(120573) = 0 we obtain

int120573

120572

(11990910158401015840)2

119889119905 = int120573

120572

119901(1199091015840)2

119889119905 + int120573

120572

1199021199091199091015840119889119905 (60)

By using themaximumvalues of |119901(119905)| and |119902(119905)| we canwrite(60) as

int120573

120572

1003816100381610038161003816100381611990910158401015840

(119905)10038161003816100381610038161003816

2

119889119905 le 119875int120573

120572

100381610038161003816100381610038161199091015840

(119905)10038161003816100381610038161003816

2

119889119905 + 119876int120573

120572

|119909(119905)|100381610038161003816100381610038161199091015840

(119905)10038161003816100381610038161003816119889119905

(61)

For the first term on the right-hand side int120573

120572|1199091015840(119905)|2119889119905 we

apply inequality (25) with 119910(119905) = 1199091015840(119905) (note that 1199091015840(120572) =

1199091015840(120573) = 0) and 120574 = 1 to obtain

int120573

120572

10038161003816100381610038161003816119909101584010038161003816100381610038161003816

2

119889119905 le120587(120573 minus 120572)

2

16int120573

120572

100381610038161003816100381610038161199091015840101584010038161003816100381610038161003816

2

119889119905 (62)

Applying Fink inequality (27) on int120573

120572|119909||1199091015840|119889119905 with 119896 = 0 119903 =

1 120583 = ] = 2 and 119899 = 2 we have

int120573

120572

|119909|10038161003816100381610038161003816119909101584010038161003816100381610038161003816119889119905 le

(120573 minus 120572)3

6int120573

120572

1003816100381610038161003816100381611990910158401015840(119905)

10038161003816100381610038161003816

2

119889119905 (63)

Substituting (62) and (63) into (61) we get after cancellingthe term int

120573

120572|11990910158401015840(119905)|2119889119905 that

119875120587(120573 minus 120572)

2

16+ 119876

(120573 minus 120572)3

6ge 1 (64)

which is the desired inequality (58) The proof is complete

As a special case of Theorem 5 when 119901(119905) = 0 we havethe following result

Corollary 6 Let 119909(119905) be a nontrivial solution of (56) If 119909(120572) =1199091015840(120572) = 1199091015840(120573) = 0 then

max120572le119905le120573

1003816100381610038161003816119902 (119905)1003816100381610038161003816 ge

6

(120573 minus 120572)3 (65)

In the following we prove some results related to theboundary conditions presented in (ii)

Theorem 7 Assume that 119909(119905) is a nontrivial solution of (9) If119909(120572) = 1199091015840(120572) = 11990910158401015840(120573) = 0 then

41198612

1(120572 120573) + 119862

120572ge 1 (66)

If 119909(120573) = 1199091015840(120573) = 11990910158401015840(120572) = 0 then

41198612

2(120572 120573) + 119862

120573ge 1 (67)

Mathematical Problems in Engineering 7

Proof Multiplying (9) by 1199091015840(119905) and integrating by parts theleft-hand side from 120572 to 120573 we get

11990311990910158401199091015840101584010038161003816100381610038161003816

120573

120572minus int120573

120572

119903(11990910158401015840)2

119889119905 = minusint120573

120572

119901(1199091015840)2

119889119905 minus int120573

120572

1199021199091199091015840119889119905

(68)

Using the assumption that 1199091015840(120572) = 11990910158401015840(120573) = 0 we have

int120573

120572

119903(119905)1003816100381610038161003816100381611990910158401015840

(119905)10038161003816100381610038161003816

2

119889119905 le int120573

120572

1003816100381610038161003816119901 (119905)1003816100381610038161003816100381610038161003816100381610038161199091015840(119905)

10038161003816100381610038161003816

2

119889119905

+ int120573

120572

1003816100381610038161003816119902 (119905)1003816100381610038161003816|119909 (119905)|

100381610038161003816100381610038161199091015840

(119905)10038161003816100381610038161003816119889119905

(69)

Applying inequality (12) on the integral int120573120572|119901(119905)||1199091015840(119905)|2119889119905

with 119899 = 119898 = 2 we get

int120573

120572

1003816100381610038161003816119901 (119905)1003816100381610038161003816100381610038161003816100381610038161199091015840(119905)

10038161003816100381610038161003816

2

119889119905 le 41198612

1(120572 120573) int

120573

120572

|119903 (119905)|1003816100381610038161003816100381611990910158401015840

(119905)210038161003816100381610038161003816119889119905 (70)

where1198611(120572 120573) is defined as in (32) Again applying inequality

(22) on the integral int120573120572|119902(119905)||119909(119905)||1199091015840(119905)|119889119905 we get

int120573

120572

1003816100381610038161003816119902 (119905)1003816100381610038161003816|119909 (119905)|

100381610038161003816100381610038161199091015840

(119905)10038161003816100381610038161003816119889119905 le 119862

120572int120573

120572

|119903 (119905)|1003816100381610038161003816100381611990910158401015840

(119905)210038161003816100381610038161003816119889119905 (71)

where 119862120572is defined as in (32) Substituting (70) and (71) into

(69) we obtain

int120573

120572

|119903 (119905)|1003816100381610038161003816100381611990910158401015840(119905)

10038161003816100381610038161003816

2

119889119905

le 41198612

1int120573

120572

|119903 (119905)|1003816100381610038161003816100381611990910158401015840

(119905)210038161003816100381610038161003816119889119905 + 119862

120572int120573

120572

|119903 (119905)|1003816100381610038161003816100381611990910158401015840

(119905)210038161003816100381610038161003816119889119905

(72)

By cancelling the term int120573

120572|119903(119905)||119909

10158401015840(119905)2|119889119905 we get that

41198612

1(120572 120573) + 119862

120572ge 1 (73)

which is the desired result (66) The proof (67) is similar tothe proof of (66) by using 119861

2(120572 120573) and 119862

120573instead of 119861

1(120572 120573)

and 119862120572and hence is omitted The proof is complete

In the following we apply an inequality due to Boyd [24]to obtain new results The Boyd inequality states that if 119910 isin

1198621[120572 120573] with 119910(120572) = 0 (or 119910(120573) = 0) then

int120573

120572

1003816100381610038161003816119910(119905)1003816100381610038161003816]100381610038161003816100381610038161199101015840(119905)

10038161003816100381610038161003816

120578

119889119905

le 119873 (] 120578 119904) (120573 minus 120572)][int119887

119886

100381610038161003816100381610038161199101015840(119905)

10038161003816100381610038161003816

119904

119889119905]

(]+120578)119904

(74)

where ] gt 0 119904 gt 1 0 le 120578 lt 119904 and

119873(] 120578 119904) =(119904 minus 120578)]]119903]+120578minus119904

(119904 minus 1) (] + 120578) (119868(] 120578 119904))]

119903 = ](119904 minus 1) + (119904 minus 120578)

(119904 minus 1)(] + 120578)

1119904

(75)

119868(] 120578 119904) = int1

0

1 +119904(120578 minus 1)

119904 minus 120578119905

minus(]+120578+119904])119904]

times [1 + (120578 minus 1)]1199051(]minus1)

119889119905

(76)

Note that an inequality of type (34) also holds when 119910(120572) =

119910(120573) = 0 Choose 119888 = (120572 + 120573)2 and apply (34) to [120572 119888] and[119888 120573] and by addition we obtain

int120573

120572

1003816100381610038161003816119910(119905)1003816100381610038161003816]100381610038161003816100381610038161199101015840(119905)

10038161003816100381610038161003816

120578

119889119905

le 119873 (] 120578 119904) (120573 minus 120572

2)

]

[int120573

120572

100381610038161003816100381610038161199101015840(119905)

10038161003816100381610038161003816

119904

119889119905]

(]+120578)119904

(77)

where 119873(] 120578 119904) is defined as in (75) An inequality of type(74) holds when 120578 = 119904 and 119910(120572) = 0 (or 119910(120573) = 0) In thiscase (74) becomes

int120573

120572

1003816100381610038161003816119910(119905)1003816100381610038161003816]100381610038161003816100381610038161199101015840(119905)

10038161003816100381610038161003816

120578

119889119905

le 119871 (] 120578) (120573 minus 120572)][int120573

120572

100381610038161003816100381610038161199101015840(119905)

10038161003816100381610038161003816

120578

119889119905]

(]+120578)120578

(78)

where

119871 (] 120578) =120578]120578

] + 120578(

]] + 120578

)

]120578

(Γ(((120578 + 1)120578) + (1]))Γ((120578 + 1)120578)Γ(1])

)

]

(79)

and Γ is the gamma function

Theorem 8 Assume that 119903(119905) is nonincreasing function and119909(119905) is a nontrivial solution of (9) If119909(120572) = 1199091015840(120572) = 11990910158401015840(120573) = 0then

41198612

1(120572 120573) +

8 (120573 minus 120572)

120587119903 (120573)1205952(120572 120573)(int

120573

120572

1003816100381610038161003816119902(119905)10038161003816100381610038162

119889119905)

12

ge 1 (80)

If 119909(120573) = 1199091015840(120573) = 11990910158401015840(120572) = 0 then

41198612

2(120572 120573) +

8 (120573 minus 120572)

120587119903 (120573)1205952(120572 120573)(int

120573

120572

1003816100381610038161003816119902(119905)10038161003816100381610038162

119889119905)

12

ge 1 (81)

Proof Multiplying (9) by 1199091015840(119905) and integrating by parts theleft-hand side we have that

11990311990910158401199091015840101584010038161003816100381610038161003816

120573

120572minus int120573

120572

119903(11990910158401015840)2

119889119905 = minusint120573

120572

119901(1199091015840)2

119889119905 minus int120573

120572

1199021199091199091015840119889119905

(82)

8 Mathematical Problems in Engineering

Using the assumption 1199091015840(120572) = 11990910158401015840(120573) = 0 we get that

int120573

120572

|119903 (119905)|1003816100381610038161003816100381611990910158401015840(119905)

10038161003816100381610038161003816

2

119889119905 le int120573

120572

1003816100381610038161003816119901 (119905)1003816100381610038161003816100381610038161003816100381610038161199091015840(119905)

10038161003816100381610038161003816

2

119889119905

+ int120573

120572

1003816100381610038161003816119902 (119905)1003816100381610038161003816|119909 (119905)|

100381610038161003816100381610038161199091015840

(119905)10038161003816100381610038161003816119889119905

(83)

Applying inequality (12) on the integral int120573120572|119901(119905)||1199091015840(119905)|2119889119905

with119898 = 119899 = 2 we get

int120573

120572

1003816100381610038161003816119901 (119905)1003816100381610038161003816100381610038161003816100381610038161199091015840(119905)

10038161003816100381610038161003816

2

119889119905

le 41205952(120572 120573) int

120573

120572

|119903 (119905)|1003816100381610038161003816100381611990910158401015840

(119905)210038161003816100381610038161003816119889119905

(84)

where 120595(120572 120573) is defined as in (34) Applying Schwarzrsquosinequality on the term int

120573

120572|119902(119905)||119909(119905)||1199091015840(119905)|119889119905 we get that

int120573

120572

1003816100381610038161003816119902 (119905)1003816100381610038161003816|119909 (119905)|

100381610038161003816100381610038161199091015840

(119905)10038161003816100381610038161003816119889119905

le (int120573

120572

1003816100381610038161003816119902(119905)10038161003816100381610038162

119889119905)

12

(int120573

120572

|119909|210038161003816100381610038161003816119909101584010038161003816100381610038161003816

2

119889119905)

12

(85)

Applying again inequality (78) on the integral int120573120572|119909|2|1199091015840|2119889119905

with ] = 120578 = 2 (note that 119909(120572) = 0) we obtain

int120573

120572

|119909|210038161003816100381610038161003816119909101584010038161003816100381610038161003816

2

119889119905 le4(120573 minus 120572)

2

12058721199032 (120573)[int120573

120572

119903(119905)10038161003816100381610038161003816119909101584010038161003816100381610038161003816

2

119889119905]

2

(86)

where 119903(119905) is a nonincreasing function Substituting (86) into(85) we get

int120573

120572

1003816100381610038161003816119902 (119905)1003816100381610038161003816|119909 (119905)|

100381610038161003816100381610038161199091015840

(119905)10038161003816100381610038161003816119889119905

le (int120573

120572

1003816100381610038161003816119902(119905)10038161003816100381610038162

119889119905)

12

2 (120573 minus 120572)

120587119903 (120573)int120573

120572

119903 (119905)10038161003816100381610038161003816119909101584010038161003816100381610038161003816

2

119889119905

(87)

Applying inequality (12) on the integral int120573120572119903(119905)|1199091015840|2119889119905 with

119901(119905) = 119902(119905) = 119903(119905) and119898 = 119899 = 2 we get

int120573

120572

|119903 (119905)|100381610038161003816100381610038161199091015840(119905)

10038161003816100381610038161003816

2

119889119905 le 41198612

1(120572 120573) int

120573

120572

|119903 (119905)|1003816100381610038161003816100381611990910158401015840(119905)

10038161003816100381610038161003816

2

119889119905 (88)

Substituting (88) into (87) we have

int120573

120572

1003816100381610038161003816119902 (119905)1003816100381610038161003816|119909 (119905)|

100381610038161003816100381610038161199091015840

(119905)10038161003816100381610038161003816119889119905

le (int120573

120572

1003816100381610038161003816119902(119905)10038161003816100381610038162

119889119905)

12

8(120573 minus 120572)

120587119903(120573)1205952(120572 120573)

times int120573

120572

|119903 (119905)|1003816100381610038161003816100381611990910158401015840(119905)

10038161003816100381610038161003816

2

119889119905

(89)

Substituting (84) and (89) into (83) we obtain

int120573

120572

|119903 (119905)|1003816100381610038161003816100381611990910158401015840(119905)

10038161003816100381610038161003816

2

119889119905

le 41198612

1(120572 120573) int

120573

120572

|119903(119905)|1003816100381610038161003816100381611990910158401015840

(119905)10038161003816100381610038161003816

2

119889119905

+(int120573

120572

1003816100381610038161003816119902(119905)10038161003816100381610038162

119889119905)

12

8 (120573 minus 120572)

120587119903 (120573)1205952(120572 120573)

times int120573

120572

|119903 (119905)|1003816100381610038161003816100381611990910158401015840(119905)

10038161003816100381610038161003816

2

119889119905

(90)

Cancelling the term int120573

120572|119903(119905)||11990910158401015840(119905)|2119889119905 we obtain

41198612

1(120572 120573) +

8 (120573 minus 120572)

120587119903 (120573)1205952(120572 120573)(int

120573

120572

1003816100381610038161003816119902(119905)10038161003816100381610038162

119889119905)

12

ge 1

(91)

which is the desired inequality (80) The proof of (81) issimilar to the proof of (80) by using 119861

2(120572 120573) instead of

1198611(120572 120573) The proof is complete

Conflict of Interests

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

References

[1] E Picard ldquoMemoire sur la theorie des equations aux derivespartielleset la methode desapproximations successivesrdquo Journalde Mathematiques Pures et Appliquees vol 6 pp 145ndash210 1890

[2] E Picard ldquoSur lrsquoapplication des methodes drsquoapproximationssuccessives a lrsquoetude de certaines equations differentielles ordi-nairesrdquo Journal deMathematiques Pures et Appliquees vol 9 pp217ndash271 1893

[3] O Niccoletti ldquoSulle consizioni iniziali che determinano gliintegrale delle equazioni diff eren ziali ordinarierdquo Atti dellaAccademia delle Scienze di Torino vol 33 pp 746ndash759 1897

[4] C de la Vallee Poussin ldquoSur lrsquoequation differentielle lineaire dusecond ordrerdquo Journal de Mathematiques Pures et Appliqueesvol 8 pp 125ndash144 1929

[5] A M Lyapunov ldquoProbleme general de la stabilite du mouve-mentrdquo Annales de la Faculte des sciences de Toulouse vol 9 pp203ndash474 1907

[6] S H Saker ldquoSome new disconjugacy criteria for second orderdifferential equations with a middle termrdquo Bulletin Mathema-tique de la Societe des Sciences Mathematiques de Roumanie vol57 no 1 pp 109ndash120 2014

[7] G Casadei ldquoSul teorema di unicita di De La Vallee Poussinper equazioni differenziali del terzo ordinerdquo Rendiconti delSeminario Matematico della Universita di Padova vol 41 pp300ndash315 1968

[8] A Lasota ldquoSur la distance entre les zeros de lrsquoequation dif-ferentielle lineaire du troisieme ordrerdquo Annales Polonici Mathe-matici vol 13 pp 129ndash132 1963

Mathematical Problems in Engineering 9

[9] RMMathsen ldquoA disconjugacy condition for119910101584010158401015840+119886211991010158401015840+119886

11199101015840+

1198860119910 = 0rdquo Proceedings of the AmericanMathematical Society vol

17 pp 627ndash632 1966[10] B G Pachpatte ldquoOn the zeros of solutions of certain differential

equationsrdquo Demonstratio Mathematica vol 25 no 4 pp 825ndash833 1992

[11] B G Pachpatte ldquoLyapunov type integral inequalities for certaindifferential equationsrdquo Georgian Mathematical Journal vol 4no 2 pp 139ndash148 1997

[12] S Panigrahi ldquoLyapunov-type integral inequalities for certainhigher-order differential equationsrdquo Electronic Journal of Differ-ential Equations vol 2009 no 28 pp 1ndash14 2009

[13] N Parhi and S Panigrahi ldquoOn Liapunov-type inequality forthird-order differential equationsrdquo Journal of MathematicalAnalysis and Applications vol 233 no 2 pp 445ndash460 1999

[14] N Parhi and S Panigrahi ldquoDisfocality and Liapunov-typeinequalities for third-order equationsrdquo Applied MathematicsLetters vol 16 no 2 pp 227ndash233 2003

[15] U Richard ldquoMetodi diversi per ottenere disequaglianze allade la Vallee Poussin nelle equazioni differenziali ordinarie delsecondo e terzo ordinerdquo Rendiconti del Seminario MatematicoUniversita e Politecnico di Torino vol 27 pp 35ndash68 1968

[16] X Yang ldquoOn Liapunov-type inequality for certain higher-orderdifferential equationsrdquo Applied Mathematics and Computationvol 134 no 2-3 pp 307ndash317 2003

[17] R P Agarwal and P R Krishnamurthy ldquoOn the uniquenessof solution of nonlinear boundary value problemsrdquo Journal ofMathematical Sciences vol 10 no 1 pp 17ndash31 1976

[18] R P Agarwal and P Y PangOpial Inequalities with Applicationsin Differential and Difference Equation Kluwer AcademicDordrecht The Netherlands 1995

[19] R P Agarwal and P Y Pang ldquoRemarks on the generalizationsof Opialrsquos inequalityrdquo Journal of Mathematical Analysis andApplications vol 190 no 2 pp 559ndash577 1995

[20] P R Beesack and K M Das ldquoExtensions of Opialrsquos inequalityrdquoPacific Journal of Mathematics vol 26 pp 215ndash232 1968

[21] S Clark and D Hinton ldquoSome disconjugacy criteria for dif-ferential equations with oscillatory coefficientsrdquoMathematischeNachrichten vol 278 no 12-13 pp 1476ndash1489 2005

[22] A M Fink ldquoOn Opials inequality for 119891(119899)rdquo Proceedings of theAmerican Mathematical Society vol 155 pp 177ndash181 1992

[23] A Kufner and L-E Persson Weighted Inequalities of HardyType World Scientific River Edge NJ USA 2003

[24] D W Boyd ldquoBest constants in a class of integral inequalitiesrdquoPacific Journal of Mathematics vol 30 pp 367ndash383 1969

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Mathematical Problems in Engineering

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Stochastic AnalysisInternational Journal of

Mathematical Problems in Engineering 3

where the weighted functions 119901 119903 are positive functionsdefined on (120572 120573) and 119899 119898 are real parameters that satisfy0 lt 119899 le infin and 1 le 119898 le infin The constant 119862 is given by

119862 le 1198981119898

(1198981)11198981119860(120572 120573) for 1 lt 119898 le 119899 119898

1=

119898

119898 minus 1

(13)

where

119860(120572 120573) = sup120572le119905le120573

(int120573

119905

119901(119905)119889119905)

1119899

(int119905

120572

1199031minus1198981(119904)119889119904)

11198981

if 119910(120572) = 0

119860(120572 120573) = sup120572le119905le120573

(int119905

120572

119901(119905)119889119905)

1119899

(int120573

119905

1199031minus1198981(119904)119889119904)

11198981

if 119910(120573) = 0

(14)

Note that inequality (12) has an immediate application to thecase when 119910(120572) = 119910(120573) = 0 In this case inequality (12) issatisfied if and only if

119860(120572 120573)

= sup(119888119889)sub(120572120573)

(int119889

119888

119901(119905)119889119905)

1119899

timesmin(int119888

120572

1199031minus1198981(119904)119889119904)

11198991

(int120573

119889

1199031minus1198981(119904)119889119904)

11198981

(15)

exists and is finite The Opial type inequality due to Beesackwhich is a generalization of Opialrsquos inequality states that if 119910is absolutely continuous on [120572 120573] with 119910(120572) = 0 then thefollowing inequality holds

int120573

120572

119902 (119905)1003816100381610038161003816119910(119905)

1003816100381610038161003816119898100381610038161003816100381610038161199101015840(119905)

10038161003816100381610038161003816

119899

119889119905 le 1198701(119898 119899) int

120573

120572

119901 (119905)100381610038161003816100381610038161199101015840(119905)

10038161003816100381610038161003816

119898+119899

119889119905

(16)

where 119898 119899 are real numbers such that 119898119899 gt 0 119898 + 119899 gt 1119902(119905) and 119901(119905) are nonnegative measurable functions definedon (120572 120573) such that int119905

120572(119901(119904))minus1(119898+119899minus1)119889119904 lt infin and

1198701(119898 119899)

= (119899

119898 + 119899)119899(119898+119899)

times [int120573

120572

119902(119898+119899)119899

(119905)119901minus119899119898

(119905)

times (int119905

120572

(119901(119904))minus1(119898+119899minus1)

119889119904)

119898+119899minus1

119889119905]

119898(119898+119899)

(17)

If instead 119910(120573) = 0 then (16) holds where 1198701(119898 119899) is

replaced by

1198702(119898 119899)

= (119899

119898 + 119899)119899(119898+119899)

times [int120573

120572

119902(119898+119899)119899

(119905) 119901minus(119899119898)

(119905)

times(int120573

119905

(119901(119904))minus1(119898+119899minus1)

119889119904)

119898+119899minus1

119889119905]

119898(119898+119899)

(18)

TheOpial type inequality due to Agarwal and Pang states thatif 119910(119905) isin 119862(119899minus1)[120572 120573] and satisfies119910(119894)(120572) = 0 0 le 119896 le 119894 le 119899minus1

(119899 ge 1) and 119910(119899minus1)(119905) absolutely continuous on (120572 120573) then

int120573

120572

120601 (119905)10038161003816100381610038161003816119910(119896)(119905)

10038161003816100381610038161003816

11989710038161003816100381610038161003816119910(119899)(119905)

10038161003816100381610038161003816

119898

119889119905

le 1198671[int120573

120572

120593(119905)10038161003816100381610038161003816119910(119899)(119905)

10038161003816100381610038161003816

119888

119889119905]

(119897+119898)119888

(19)

where 120601 and 120593 are nonnegative and measurable functiondefined on (120572 120573) 119898 119899 are real numbers such that 119888119898 gt 1and

1198671=

(119898(119898 + 119897))119898119888

(119899 minus 119896 minus 1)

times [int120573

120572

(120601119888

(119905)120593minus119898

(119905))1(119888minus119898)

times (1198661119896(119905))119897(119888minus1)(119888minus119898)

119889119905]

(119888minus119898)119888

1198661119896

(119905) = int119905

120572

(119905 minus 119904)(119899minus119896minus1)119888(119888minus1)

(120593(119904))minus1(119888minus1)

119889119904

(20)

If instead 119910(119894)(120573) = 0 0 le 119896 le 119894 le 119899 minus 1 (119899 ge 1) then (19)holds where119867

1is replaced by

1198672=

(119898(119898 + 119897))119898119888

(119899 minus 119896 minus 1)

times [int120573

120572

(120601119888(119905)120593minus119898

(119905))1(119888minus119898)

times (1198662119896(119905))119897(119888minus1)(119888minus119898)

119889119905]

(119888minus119898)119888

1198662119896

(119905) = int120573

119905

(119904 minus 119905)(119899minus119896minus1)119888(119888minus1)

(120593(119904))minus1(119888minus1)

119889119904

(21)

4 Mathematical Problems in Engineering

We also need the following inequality which is the specialcase of an inequality proved by Agarwal and Pang [18] withtwo functions

int120573

120572

119888(119905)10038161003816100381610038161003816119910(119896)

(119905)10038161003816100381610038161003816

10038161003816100381610038161003816119910(119896+1)

(119905)10038161003816100381610038161003816119889119905 le 119862

120572int120573

120572

119903 (119905)10038161003816100381610038161003816119910(119899)(119905)

10038161003816100381610038161003816

2

119889119905

(22)

where 119910(119905) isin 119862119899minus1[120572 120573] and satisfies 119910(119894)(120572) = 0 119896 le 119894 le 119899minus10 le 119896 le 119899 minus 1 (119899 ge 1) and 119910(119899minus1) is absolutely continuous on(120572 120573) 119888(119905) and 119903(119905) being nonnegative measurable functionsdefined on (120572 120573) and

119862120572=

1

2((119899 minus 119896 minus 1))2max119905isin[120572120573]

119888 (119905) int120573

120572

(119904 minus 120572)2(119899minus119896minus1)

119903 (119904)119889119904

(23)

If 119910(119894)(120573) = 0 119896 le 119894 le 119899 minus 1 then (22) holds where 119862120572is

replaced

119862120573=

1

2((119899 minus 119896 minus 1))2max119905isin[120572120573]

119888 (119905) int120573

120572

(120573 minus 119904)2(119899minus119896minus1)

119903 (119904)119889119904

(24)

TheWirtinger type inequality due to Agarwal and Pang statesthat if 119910(120572) = 119910(120573) = 0 then

int120573

120572

119910120574+1

(119905)119889119905 le(120573 minus 120572)

120574+1

Γ2 ((120574 + 2)2)

2Γ(120574 + 2)

times int120573

120572

(1199101015840(119905))120574+1

(119905) 119889119905

(25)

TheOpial type inequality due to Clark and Hinton inequalitystates that if 119910 isin 119862

2[120572 120573] where 119910(120572) = 1199101015840(120572) = 0 then

(int120573

120572

1003816100381610038161003816119910(119905)100381610038161003816100381621003816100381610038161003816100381611991010158401015840(119905)

10038161003816100381610038161003816

2

119889119905)

1119903

le(120573 minus 120572)

32

radic3int120573

120572

1003816100381610038161003816100381611991010158401015840

(119905)10038161003816100381610038161003816

2

119889119905

(26)

The Opial type inequality due to Fink inequality states that if119910(119894)(120572) = 0 0 le 119894 le 119899 minus 1 120583 ge 1 (1120583) + (1]) = 1 then

int120573

120572

10038161003816100381610038161003816119910(119896)

(119905) 119910(119903)

(119905)10038161003816100381610038161003816119889119905

le 119862 (119899 119896 119903 120583) (120573 minus 120572)2119899minus119896minus119903+1minus2120583

(int120573

120572

10038161003816100381610038161003816119910(119899)(119905)

10038161003816100381610038161003816

120583

119889119905)

2120583

(27)

where 0 le 119896 lt 119903 lt 119899 (119899 ge 2) and

119862 (119899 119896 119903 120583) =1

2((119899 minus 119896 minus 1))2[(119899 minus 119896 minus 1)] + 1]

2]

(28)

3 Main Results

In this section we state and prove the main results Forsimplicity we introduce the following notations

1198671(120572 120573 119866

10) =

1

radic2[int120573

120572

1198762

1(119905)

1198661(119905)

119903(119905)119889119905]

12

1198661(119905) = int

119905

120572

(119905 minus 119904)2

119903 (119904)119889119904 119876

1(119905) = int

120573

119905

119902 (119904) 119889119904

(29)

1198672(120572 120573 119866

2) =

1

radic2[int120573

120572

1198762

2(119905)

1198662(119905)

119903(119905)119889119905]

12

1198662(119905) = int

120573

119905

(119905 minus 119904)2

119903 (119904)119889119904 119876

2(119905) = int

119905

120572

119902 (119904) 119889119904

(30)

119860119894(120572 120573) = 119869

119894(119905)min(int

119888

120572

1

119903(119904)119889119904)

12

(int120573

119888

1

119903(119904)119889119904)

12

(31)

where 119869119894(119905) = sup

(119888119889)sub(120572120573)(int120573

120572119863119894(119905)119889119905)12 and |119863

119894(119905)| =

|119876119894(119905)| + |119901(119905)| 119894 = 1 2

1198611(120572 120573) = sup

120572le119905le120573

(int120573

119905

119901(119905)119889119905)

12

(int119905

120572

1

119903(119904)119889119904)

12

119862120572=

1

2max120572le119905le120573

119902 (119905) int120573

120572

(119904 minus 120572)2

119903 (119904)119889119904

(32)

1198612(120572 120573) = sup

120572le119905le120573

(int119905

120572

119901(119905)119889119905)

12

(int120573

119905

1

119903(119904)119889119904)

12

119862120573=

1

2max120572le119905le120573

119902 (119905) int120573

120572

(120573 minus 119904)2

119903 (119904)119889119904

(33)

120595(120572 120573) = sup120572le119905le120573

(int120573

119905

119903(119905)119889119905)

12

(int119905

120572

1

119903(119904)119889119904)

12

(34)

Now we are ready to state and prove the main results

Theorem 1 Assume that 119909(119905) is a nontrivial solution of (9) If119909(120572) = 1199091015840(120572) = 1199091015840(120573) = 0 then

1198671(120572 120573 119866

1) + 4119860

2

1(120572 120573) ge 1 (35)

If 1199091015840(120572) = 119909(120573) = 1199091015840(120573) = 0 then

1198672(120572 120573 119866

2) + 4119860

2

2(120572 120573) ge 1 (36)

Proof Weprove (35)Multiplying (9) by 1199091015840(119905) and integratingthe new equation from 120572 to 120573 we obtain

int120573

120572

1199091015840(119903(11990910158401015840))1015840

119889119905 = minusint120573

120572

119901(1199091015840)2

119889119905 minus int120573

120572

1199021199091199091015840119889119905 (37)

Mathematical Problems in Engineering 5

Integrating by parts the left-hand side we get that

11990311990910158401199091015840101584010038161003816100381610038161003816

120573

120572minus int120573

120572

119903(11990910158401015840)2

119889119905 = minusint120573

120572

119901(1199091015840)2

119889119905 minus int120573

120572

1199021199091199091015840119889119905

(38)

Using the assumptions that 1199091015840(120572) = 1199091015840(120573) = 0 and 1198761(119905) =

int120573

119905119902(119904)119889119904 we have

int120573

120572

119903(119905)(11990910158401015840

(119905))2

119889119905 = int120573

120572

119901 (119905) (1199091015840(119905))2

119889119905

+ int120573

120572

1198761015840

1(119905) 119909 (119905) 119909

1015840

(119905) 119889119905

(39)

Integrating the term int120573

12057211987610158401(119905)119909(119905)1199091015840(119905)119889119904 by parts and using

the assumption that 1199091015840(120572) = 1199091015840(120573) = 0 we obtain

int120573

120572

1198761015840

1(119905) 119909 (119905) 119909

1015840

(119905) 119889119904

= minusint120573

120572

1198761(119905) (1199091015840

(119905))2

119889119905 minus int120573

120572

1198761(119905) 119909 (119905) 119909

10158401015840

(119905) 119889119905

(40)

Substituting (40) into (39) we get

int120573

120572

119903(119905)1003816100381610038161003816100381611990910158401015840

(119905)10038161003816100381610038161003816

2

119889119905 le int120573

120572

10038161003816100381610038161198631 (119905)1003816100381610038161003816100381610038161003816100381610038161199091015840(119905)

10038161003816100381610038161003816

2

119889119905

+ int120573

120572

10038161003816100381610038161198761 (119905)1003816100381610038161003816|119909 (119905)|

1003816100381610038161003816100381611990910158401015840

(119905)10038161003816100381610038161003816119889119905

(41)

where |1198631(119905)| = |119876

1(119905)| + |119901(119905)| Applying inequality (19) on

the integral

int120573

120572

10038161003816100381610038161198761 (119905)1003816100381610038161003816|119909 (119905)|

1003816100381610038161003816100381611990910158401015840

(119905)10038161003816100381610038161003816119889119905 (42)

with 120601(119905) = |1198761(119905)| 120593(119905) = 119903(119905) 119896 = 0119898 = 119897 = 1 119899 = 2 119888 = 2

and 119909(120572) = 1199091015840(120572) = 0 we get

int120573

120572

10038161003816100381610038161198761 (119905)1003816100381610038161003816|119909 (119905)|

1003816100381610038161003816100381611990910158401015840

(119905)10038161003816100381610038161003816119889119905

le 1198671(120572 120573 119866

1) int120573

120572

119903 (119905)1003816100381610038161003816100381611990910158401015840(119905)

10038161003816100381610038161003816

2

119889119905

(43)

where 1198671(120573 120572 119866

1) is defined as in (29) Applying inequality

(12) on the integral int120573120572|1198631(119905)||1199091015840(119905)|2119889119905 with 119910(119905) = 1199091015840(119905) and

119910(120572) = 119910(120573) = 0 we have that

int120573

120572

10038161003816100381610038161198631 (119905)1003816100381610038161003816100381610038161003816100381610038161199091015840(119905)

10038161003816100381610038161003816

2

119889119905 le 41198602

1(120572 120573) int

120573

120572

119903 (119905)1003816100381610038161003816100381611990910158401015840(119905)

10038161003816100381610038161003816

2

119889119905

(44)

where1198601(120572 120573) is defined as in (31) Substituting (44) and (43)

into (41) we get

int120573

120572

|119903 (119905)|1003816100381610038161003816100381611990910158401015840(119905)

10038161003816100381610038161003816

2

119889119905 le 41198602

1(120572 120573) int

120573

120572

119903 (119905)1003816100381610038161003816100381611990910158401015840(119905)

10038161003816100381610038161003816

2

119889119905

+ 1198671(120572 120573 119866

1) int120573

120572

119903 (119905)1003816100381610038161003816100381611990910158401015840(119905)

10038161003816100381610038161003816

2

119889119905

(45)

Cancelling the term int120573

120572119903(119905)|11990910158401015840(119905)|2119889119905 we get

41198602

1(120572 120573) + 119867

1(120572 120573 119866

1) ge 1 (46)

which is the desired inequality (35) The proof of (36) issimilar to the proof of (35) by replacing 119867

1(120572 120573 119866

1) by

1198672(120572 120573 119866

2) and 119860

1(120572 120573) by 119860

2(120572 120573) The proof is com-

plete

In the following we apply the Clark and Hinton inequal-ity (26) to get a new result

Theorem 2 Assume that 119903(119905) is a nonincreasing function andsuppose that 119909(119905) is a solution of (9) If 119909(120572) = 1199091015840(120572) = 1199091015840(120573) =

0 then

41198602

1(120572 120573) +

(120573 minus 120572)32

radic3119903 (120573)(int120573

120572

10038161003816100381610038161198761(119905)10038161003816100381610038162

119889119905)

12

ge 1 (47)

where 1198601(120572 120573) is defined as in (31) for 119894 = 1 and 119876

1(119905) is

defined as in (29)

Proof Multiply (9) by 1199091015840(119905) and proceed as in the proof ofTheorem 1 to get

int120573

120572

119903(119905)1003816100381610038161003816100381611990910158401015840

(119905)10038161003816100381610038161003816

2

119889119905 le int120573

120572

10038161003816100381610038161198631 (119905)1003816100381610038161003816100381610038161003816100381610038161199091015840(119905)

10038161003816100381610038161003816

2

119889119905

+ int120573

120572

10038161003816100381610038161198761 (119905)1003816100381610038161003816|119909 (119905)|

1003816100381610038161003816100381611990910158401015840

(119905)10038161003816100381610038161003816119889119905

(48)

where |1198631(119905)| = |119876

1(119905)|+ |119901(119905)| Applying the Schwarz inequal-

ity

int120573

120572

1003816100381610038161003816119891 (119905) 119892 (119905)1003816100381610038161003816119889119905 le (int

120573

120572

1003816100381610038161003816119891(119905)10038161003816100381610038162

119889119905)

12

(int120573

120572

1003816100381610038161003816119892(119905)10038161003816100381610038162

119889119905)

12

(49)

on the integral int120573120572|1198761(119905)||119909(119905)||11990910158401015840(119905)|119889119905 we have that

int120573

120572

10038161003816100381610038161198761 (119905)1003816100381610038161003816|119909 (119905)|

1003816100381610038161003816100381611990910158401015840

(119905)10038161003816100381610038161003816119889119905 le (int

120573

120572

10038161003816100381610038161198761(119905)10038161003816100381610038162

119889119905)

12

times (int120573

120572

|119909(119905)|21003816100381610038161003816100381611990910158401015840(119905)

10038161003816100381610038161003816

2

119889119905)

12

(50)Applying inequality (26) and using the assumption 119909(120572) =

1199091015840(120572) = 0 we get that

(int120573

120572

|119909(119905)|21003816100381610038161003816100381611990910158401015840(119905)

10038161003816100381610038161003816

2

119889119905)

12

le(120573 minus 120572)

32

radic3int120573

120572

1003816100381610038161003816100381611990910158401015840(119905)

10038161003816100381610038161003816

2

119889119905

(51)Substituting (51) into (50) and using the assumption that 119903(119905)is a nonincreasing function we have

int120573

120572

10038161003816100381610038161198761(119905)1003816100381610038161003816|119909(119905)|

1003816100381610038161003816100381611990910158401015840

(119905)10038161003816100381610038161003816

le(120573 minus 120572)

32

radic3119903 (120573)(int120573

120572

10038161003816100381610038161198761(119905)10038161003816100381610038162

119889119905)

12

int120573

120572

119903 (119905)1003816100381610038161003816100381611990910158401015840(119905)

10038161003816100381610038161003816

2

119889119905

(52)

6 Mathematical Problems in Engineering

Applying Hardyrsquos inequality (12) on the integralint120573

120572|1198631(119905)||1199091015840(119905)

2|119889119905 we obtain

int120573

120572

10038161003816100381610038161198631 (119905)1003816100381610038161003816100381610038161003816100381610038161199091015840(119905)

10038161003816100381610038161003816

2

119889119905 le 41198602

1(120572 120573) int

120573

120572

119903 (119905)1003816100381610038161003816100381611990910158401015840(119905)

10038161003816100381610038161003816

2

119889119905 (53)

where 1199091015840(120572) = 1199091015840(120573) = 0 Substituting (52) and (53) into (48)we obtain

int120573

120572

|119903(119905)|1003816100381610038161003816100381611990910158401015840

(119905)10038161003816100381610038161003816

2

119889119905

le 41198602

1(120572 120573) int

120573

120572

119903 (119905)1003816100381610038161003816100381611990910158401015840(119905)

10038161003816100381610038161003816

2

119889119905

+(120573 minus 120572)

32

radic3119903 (120573)(int120573

120572

10038161003816100381610038161198761(119905)10038161003816100381610038162

119889119905)

12

int120573

120572

119903 (119905)1003816100381610038161003816100381611990910158401015840(119905)

10038161003816100381610038161003816

2

119889119905

(54)

The desired inequality (47) followed by cancelling the termint120573

120572|119903(119905)||11990910158401015840(119905)|2119889119905 The proof is complete

The proof of the following theorem is similar to the proofof Theorem 2

Theorem 3 Assume that 119903(119905) is a nonincreasing function andsuppose that 119909(119905) is a solution of (9) If 1199091015840(120572) = 119909(120573) = 1199091015840(120573) =

0 then

41198602

2(120572 120573) +

(120573 minus 120572)32

radic3119903 (120573)(int120573

120572

10038161003816100381610038161198761(119905)10038161003816100381610038162

119889119905)

12

ge 1 (55)

where 1198602(120572 120573) is defined as in (31) for 119894 = 2 and 119876

2(119905) is

defined as in (30)

One can use Theorems 2 and 3 to obtain some differentspecial cases For example as a special case of Theorem 2when 119903(119905) = 1 and 119901(119905) = 0 we have the following result

Corollary 4 If 119909(119905) is a nontrivial solution of

119909101584010158401015840

(119905) + 119902 (119905) 119909 (119905) = 0 119905 isin [120572 120573] (56)

which satisfies 119909(120572) = 1199091015840(120572) = 1199091015840(120573) = 0 then

int120573

120572

10038161003816100381610038161198761(119905)10038161003816100381610038162

119889119905 ge3

(120573 minus 120572)3 (57)

where 1198761(119905) is defined as in (29)

Now we will prove a new result when 119903(119905) = 1

Theorem 5 Suppose that 119903(119905) = 1 and assume that 119909(119905) is asolution of (9) If 119909(120572) = 1199091015840(120572) = 1199091015840(120573) = 0 or 119909(120573) = 1199091015840(120573) =

1199091015840(120572) = 0 then

119875120587(120573 minus 120572)

2

16+ 119876

(120573 minus 120572)3

6ge 1 (58)

where 119875 = max120572le119905le120573

|119901(119905)| and 119876 = max120572le119905le120573

|119902(119905)|

Proof Multiplying (9) by 1199091015840 and integrating by parts the left-hand side we have

1199091015840

(119905)11990910158401015840

(119905)10038161003816100381610038161003816

120573

120572minus int120573

120572

(11990910158401015840

(119905))2

119889119905

= minusint120573

120572

119901(119905)(1199091015840

(119905))2

119889119905 minus int120573

120572

119902(119905)119909(119905)1199091015840

(119905)119889119905

(59)

Using the assumptions that 1199091015840(120572) = 1199091015840(120573) = 0 we obtain

int120573

120572

(11990910158401015840)2

119889119905 = int120573

120572

119901(1199091015840)2

119889119905 + int120573

120572

1199021199091199091015840119889119905 (60)

By using themaximumvalues of |119901(119905)| and |119902(119905)| we canwrite(60) as

int120573

120572

1003816100381610038161003816100381611990910158401015840

(119905)10038161003816100381610038161003816

2

119889119905 le 119875int120573

120572

100381610038161003816100381610038161199091015840

(119905)10038161003816100381610038161003816

2

119889119905 + 119876int120573

120572

|119909(119905)|100381610038161003816100381610038161199091015840

(119905)10038161003816100381610038161003816119889119905

(61)

For the first term on the right-hand side int120573

120572|1199091015840(119905)|2119889119905 we

apply inequality (25) with 119910(119905) = 1199091015840(119905) (note that 1199091015840(120572) =

1199091015840(120573) = 0) and 120574 = 1 to obtain

int120573

120572

10038161003816100381610038161003816119909101584010038161003816100381610038161003816

2

119889119905 le120587(120573 minus 120572)

2

16int120573

120572

100381610038161003816100381610038161199091015840101584010038161003816100381610038161003816

2

119889119905 (62)

Applying Fink inequality (27) on int120573

120572|119909||1199091015840|119889119905 with 119896 = 0 119903 =

1 120583 = ] = 2 and 119899 = 2 we have

int120573

120572

|119909|10038161003816100381610038161003816119909101584010038161003816100381610038161003816119889119905 le

(120573 minus 120572)3

6int120573

120572

1003816100381610038161003816100381611990910158401015840(119905)

10038161003816100381610038161003816

2

119889119905 (63)

Substituting (62) and (63) into (61) we get after cancellingthe term int

120573

120572|11990910158401015840(119905)|2119889119905 that

119875120587(120573 minus 120572)

2

16+ 119876

(120573 minus 120572)3

6ge 1 (64)

which is the desired inequality (58) The proof is complete

As a special case of Theorem 5 when 119901(119905) = 0 we havethe following result

Corollary 6 Let 119909(119905) be a nontrivial solution of (56) If 119909(120572) =1199091015840(120572) = 1199091015840(120573) = 0 then

max120572le119905le120573

1003816100381610038161003816119902 (119905)1003816100381610038161003816 ge

6

(120573 minus 120572)3 (65)

In the following we prove some results related to theboundary conditions presented in (ii)

Theorem 7 Assume that 119909(119905) is a nontrivial solution of (9) If119909(120572) = 1199091015840(120572) = 11990910158401015840(120573) = 0 then

41198612

1(120572 120573) + 119862

120572ge 1 (66)

If 119909(120573) = 1199091015840(120573) = 11990910158401015840(120572) = 0 then

41198612

2(120572 120573) + 119862

120573ge 1 (67)

Mathematical Problems in Engineering 7

Proof Multiplying (9) by 1199091015840(119905) and integrating by parts theleft-hand side from 120572 to 120573 we get

11990311990910158401199091015840101584010038161003816100381610038161003816

120573

120572minus int120573

120572

119903(11990910158401015840)2

119889119905 = minusint120573

120572

119901(1199091015840)2

119889119905 minus int120573

120572

1199021199091199091015840119889119905

(68)

Using the assumption that 1199091015840(120572) = 11990910158401015840(120573) = 0 we have

int120573

120572

119903(119905)1003816100381610038161003816100381611990910158401015840

(119905)10038161003816100381610038161003816

2

119889119905 le int120573

120572

1003816100381610038161003816119901 (119905)1003816100381610038161003816100381610038161003816100381610038161199091015840(119905)

10038161003816100381610038161003816

2

119889119905

+ int120573

120572

1003816100381610038161003816119902 (119905)1003816100381610038161003816|119909 (119905)|

100381610038161003816100381610038161199091015840

(119905)10038161003816100381610038161003816119889119905

(69)

Applying inequality (12) on the integral int120573120572|119901(119905)||1199091015840(119905)|2119889119905

with 119899 = 119898 = 2 we get

int120573

120572

1003816100381610038161003816119901 (119905)1003816100381610038161003816100381610038161003816100381610038161199091015840(119905)

10038161003816100381610038161003816

2

119889119905 le 41198612

1(120572 120573) int

120573

120572

|119903 (119905)|1003816100381610038161003816100381611990910158401015840

(119905)210038161003816100381610038161003816119889119905 (70)

where1198611(120572 120573) is defined as in (32) Again applying inequality

(22) on the integral int120573120572|119902(119905)||119909(119905)||1199091015840(119905)|119889119905 we get

int120573

120572

1003816100381610038161003816119902 (119905)1003816100381610038161003816|119909 (119905)|

100381610038161003816100381610038161199091015840

(119905)10038161003816100381610038161003816119889119905 le 119862

120572int120573

120572

|119903 (119905)|1003816100381610038161003816100381611990910158401015840

(119905)210038161003816100381610038161003816119889119905 (71)

where 119862120572is defined as in (32) Substituting (70) and (71) into

(69) we obtain

int120573

120572

|119903 (119905)|1003816100381610038161003816100381611990910158401015840(119905)

10038161003816100381610038161003816

2

119889119905

le 41198612

1int120573

120572

|119903 (119905)|1003816100381610038161003816100381611990910158401015840

(119905)210038161003816100381610038161003816119889119905 + 119862

120572int120573

120572

|119903 (119905)|1003816100381610038161003816100381611990910158401015840

(119905)210038161003816100381610038161003816119889119905

(72)

By cancelling the term int120573

120572|119903(119905)||119909

10158401015840(119905)2|119889119905 we get that

41198612

1(120572 120573) + 119862

120572ge 1 (73)

which is the desired result (66) The proof (67) is similar tothe proof of (66) by using 119861

2(120572 120573) and 119862

120573instead of 119861

1(120572 120573)

and 119862120572and hence is omitted The proof is complete

In the following we apply an inequality due to Boyd [24]to obtain new results The Boyd inequality states that if 119910 isin

1198621[120572 120573] with 119910(120572) = 0 (or 119910(120573) = 0) then

int120573

120572

1003816100381610038161003816119910(119905)1003816100381610038161003816]100381610038161003816100381610038161199101015840(119905)

10038161003816100381610038161003816

120578

119889119905

le 119873 (] 120578 119904) (120573 minus 120572)][int119887

119886

100381610038161003816100381610038161199101015840(119905)

10038161003816100381610038161003816

119904

119889119905]

(]+120578)119904

(74)

where ] gt 0 119904 gt 1 0 le 120578 lt 119904 and

119873(] 120578 119904) =(119904 minus 120578)]]119903]+120578minus119904

(119904 minus 1) (] + 120578) (119868(] 120578 119904))]

119903 = ](119904 minus 1) + (119904 minus 120578)

(119904 minus 1)(] + 120578)

1119904

(75)

119868(] 120578 119904) = int1

0

1 +119904(120578 minus 1)

119904 minus 120578119905

minus(]+120578+119904])119904]

times [1 + (120578 minus 1)]1199051(]minus1)

119889119905

(76)

Note that an inequality of type (34) also holds when 119910(120572) =

119910(120573) = 0 Choose 119888 = (120572 + 120573)2 and apply (34) to [120572 119888] and[119888 120573] and by addition we obtain

int120573

120572

1003816100381610038161003816119910(119905)1003816100381610038161003816]100381610038161003816100381610038161199101015840(119905)

10038161003816100381610038161003816

120578

119889119905

le 119873 (] 120578 119904) (120573 minus 120572

2)

]

[int120573

120572

100381610038161003816100381610038161199101015840(119905)

10038161003816100381610038161003816

119904

119889119905]

(]+120578)119904

(77)

where 119873(] 120578 119904) is defined as in (75) An inequality of type(74) holds when 120578 = 119904 and 119910(120572) = 0 (or 119910(120573) = 0) In thiscase (74) becomes

int120573

120572

1003816100381610038161003816119910(119905)1003816100381610038161003816]100381610038161003816100381610038161199101015840(119905)

10038161003816100381610038161003816

120578

119889119905

le 119871 (] 120578) (120573 minus 120572)][int120573

120572

100381610038161003816100381610038161199101015840(119905)

10038161003816100381610038161003816

120578

119889119905]

(]+120578)120578

(78)

where

119871 (] 120578) =120578]120578

] + 120578(

]] + 120578

)

]120578

(Γ(((120578 + 1)120578) + (1]))Γ((120578 + 1)120578)Γ(1])

)

]

(79)

and Γ is the gamma function

Theorem 8 Assume that 119903(119905) is nonincreasing function and119909(119905) is a nontrivial solution of (9) If119909(120572) = 1199091015840(120572) = 11990910158401015840(120573) = 0then

41198612

1(120572 120573) +

8 (120573 minus 120572)

120587119903 (120573)1205952(120572 120573)(int

120573

120572

1003816100381610038161003816119902(119905)10038161003816100381610038162

119889119905)

12

ge 1 (80)

If 119909(120573) = 1199091015840(120573) = 11990910158401015840(120572) = 0 then

41198612

2(120572 120573) +

8 (120573 minus 120572)

120587119903 (120573)1205952(120572 120573)(int

120573

120572

1003816100381610038161003816119902(119905)10038161003816100381610038162

119889119905)

12

ge 1 (81)

Proof Multiplying (9) by 1199091015840(119905) and integrating by parts theleft-hand side we have that

11990311990910158401199091015840101584010038161003816100381610038161003816

120573

120572minus int120573

120572

119903(11990910158401015840)2

119889119905 = minusint120573

120572

119901(1199091015840)2

119889119905 minus int120573

120572

1199021199091199091015840119889119905

(82)

8 Mathematical Problems in Engineering

Using the assumption 1199091015840(120572) = 11990910158401015840(120573) = 0 we get that

int120573

120572

|119903 (119905)|1003816100381610038161003816100381611990910158401015840(119905)

10038161003816100381610038161003816

2

119889119905 le int120573

120572

1003816100381610038161003816119901 (119905)1003816100381610038161003816100381610038161003816100381610038161199091015840(119905)

10038161003816100381610038161003816

2

119889119905

+ int120573

120572

1003816100381610038161003816119902 (119905)1003816100381610038161003816|119909 (119905)|

100381610038161003816100381610038161199091015840

(119905)10038161003816100381610038161003816119889119905

(83)

Applying inequality (12) on the integral int120573120572|119901(119905)||1199091015840(119905)|2119889119905

with119898 = 119899 = 2 we get

int120573

120572

1003816100381610038161003816119901 (119905)1003816100381610038161003816100381610038161003816100381610038161199091015840(119905)

10038161003816100381610038161003816

2

119889119905

le 41205952(120572 120573) int

120573

120572

|119903 (119905)|1003816100381610038161003816100381611990910158401015840

(119905)210038161003816100381610038161003816119889119905

(84)

where 120595(120572 120573) is defined as in (34) Applying Schwarzrsquosinequality on the term int

120573

120572|119902(119905)||119909(119905)||1199091015840(119905)|119889119905 we get that

int120573

120572

1003816100381610038161003816119902 (119905)1003816100381610038161003816|119909 (119905)|

100381610038161003816100381610038161199091015840

(119905)10038161003816100381610038161003816119889119905

le (int120573

120572

1003816100381610038161003816119902(119905)10038161003816100381610038162

119889119905)

12

(int120573

120572

|119909|210038161003816100381610038161003816119909101584010038161003816100381610038161003816

2

119889119905)

12

(85)

Applying again inequality (78) on the integral int120573120572|119909|2|1199091015840|2119889119905

with ] = 120578 = 2 (note that 119909(120572) = 0) we obtain

int120573

120572

|119909|210038161003816100381610038161003816119909101584010038161003816100381610038161003816

2

119889119905 le4(120573 minus 120572)

2

12058721199032 (120573)[int120573

120572

119903(119905)10038161003816100381610038161003816119909101584010038161003816100381610038161003816

2

119889119905]

2

(86)

where 119903(119905) is a nonincreasing function Substituting (86) into(85) we get

int120573

120572

1003816100381610038161003816119902 (119905)1003816100381610038161003816|119909 (119905)|

100381610038161003816100381610038161199091015840

(119905)10038161003816100381610038161003816119889119905

le (int120573

120572

1003816100381610038161003816119902(119905)10038161003816100381610038162

119889119905)

12

2 (120573 minus 120572)

120587119903 (120573)int120573

120572

119903 (119905)10038161003816100381610038161003816119909101584010038161003816100381610038161003816

2

119889119905

(87)

Applying inequality (12) on the integral int120573120572119903(119905)|1199091015840|2119889119905 with

119901(119905) = 119902(119905) = 119903(119905) and119898 = 119899 = 2 we get

int120573

120572

|119903 (119905)|100381610038161003816100381610038161199091015840(119905)

10038161003816100381610038161003816

2

119889119905 le 41198612

1(120572 120573) int

120573

120572

|119903 (119905)|1003816100381610038161003816100381611990910158401015840(119905)

10038161003816100381610038161003816

2

119889119905 (88)

Substituting (88) into (87) we have

int120573

120572

1003816100381610038161003816119902 (119905)1003816100381610038161003816|119909 (119905)|

100381610038161003816100381610038161199091015840

(119905)10038161003816100381610038161003816119889119905

le (int120573

120572

1003816100381610038161003816119902(119905)10038161003816100381610038162

119889119905)

12

8(120573 minus 120572)

120587119903(120573)1205952(120572 120573)

times int120573

120572

|119903 (119905)|1003816100381610038161003816100381611990910158401015840(119905)

10038161003816100381610038161003816

2

119889119905

(89)

Substituting (84) and (89) into (83) we obtain

int120573

120572

|119903 (119905)|1003816100381610038161003816100381611990910158401015840(119905)

10038161003816100381610038161003816

2

119889119905

le 41198612

1(120572 120573) int

120573

120572

|119903(119905)|1003816100381610038161003816100381611990910158401015840

(119905)10038161003816100381610038161003816

2

119889119905

+(int120573

120572

1003816100381610038161003816119902(119905)10038161003816100381610038162

119889119905)

12

8 (120573 minus 120572)

120587119903 (120573)1205952(120572 120573)

times int120573

120572

|119903 (119905)|1003816100381610038161003816100381611990910158401015840(119905)

10038161003816100381610038161003816

2

119889119905

(90)

Cancelling the term int120573

120572|119903(119905)||11990910158401015840(119905)|2119889119905 we obtain

41198612

1(120572 120573) +

8 (120573 minus 120572)

120587119903 (120573)1205952(120572 120573)(int

120573

120572

1003816100381610038161003816119902(119905)10038161003816100381610038162

119889119905)

12

ge 1

(91)

which is the desired inequality (80) The proof of (81) issimilar to the proof of (80) by using 119861

2(120572 120573) instead of

1198611(120572 120573) The proof is complete

Conflict of Interests

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

References

[1] E Picard ldquoMemoire sur la theorie des equations aux derivespartielleset la methode desapproximations successivesrdquo Journalde Mathematiques Pures et Appliquees vol 6 pp 145ndash210 1890

[2] E Picard ldquoSur lrsquoapplication des methodes drsquoapproximationssuccessives a lrsquoetude de certaines equations differentielles ordi-nairesrdquo Journal deMathematiques Pures et Appliquees vol 9 pp217ndash271 1893

[3] O Niccoletti ldquoSulle consizioni iniziali che determinano gliintegrale delle equazioni diff eren ziali ordinarierdquo Atti dellaAccademia delle Scienze di Torino vol 33 pp 746ndash759 1897

[4] C de la Vallee Poussin ldquoSur lrsquoequation differentielle lineaire dusecond ordrerdquo Journal de Mathematiques Pures et Appliqueesvol 8 pp 125ndash144 1929

[5] A M Lyapunov ldquoProbleme general de la stabilite du mouve-mentrdquo Annales de la Faculte des sciences de Toulouse vol 9 pp203ndash474 1907

[6] S H Saker ldquoSome new disconjugacy criteria for second orderdifferential equations with a middle termrdquo Bulletin Mathema-tique de la Societe des Sciences Mathematiques de Roumanie vol57 no 1 pp 109ndash120 2014

[7] G Casadei ldquoSul teorema di unicita di De La Vallee Poussinper equazioni differenziali del terzo ordinerdquo Rendiconti delSeminario Matematico della Universita di Padova vol 41 pp300ndash315 1968

[8] A Lasota ldquoSur la distance entre les zeros de lrsquoequation dif-ferentielle lineaire du troisieme ordrerdquo Annales Polonici Mathe-matici vol 13 pp 129ndash132 1963

Mathematical Problems in Engineering 9

[9] RMMathsen ldquoA disconjugacy condition for119910101584010158401015840+119886211991010158401015840+119886

11199101015840+

1198860119910 = 0rdquo Proceedings of the AmericanMathematical Society vol

17 pp 627ndash632 1966[10] B G Pachpatte ldquoOn the zeros of solutions of certain differential

equationsrdquo Demonstratio Mathematica vol 25 no 4 pp 825ndash833 1992

[11] B G Pachpatte ldquoLyapunov type integral inequalities for certaindifferential equationsrdquo Georgian Mathematical Journal vol 4no 2 pp 139ndash148 1997

[12] S Panigrahi ldquoLyapunov-type integral inequalities for certainhigher-order differential equationsrdquo Electronic Journal of Differ-ential Equations vol 2009 no 28 pp 1ndash14 2009

[13] N Parhi and S Panigrahi ldquoOn Liapunov-type inequality forthird-order differential equationsrdquo Journal of MathematicalAnalysis and Applications vol 233 no 2 pp 445ndash460 1999

[14] N Parhi and S Panigrahi ldquoDisfocality and Liapunov-typeinequalities for third-order equationsrdquo Applied MathematicsLetters vol 16 no 2 pp 227ndash233 2003

[15] U Richard ldquoMetodi diversi per ottenere disequaglianze allade la Vallee Poussin nelle equazioni differenziali ordinarie delsecondo e terzo ordinerdquo Rendiconti del Seminario MatematicoUniversita e Politecnico di Torino vol 27 pp 35ndash68 1968

[16] X Yang ldquoOn Liapunov-type inequality for certain higher-orderdifferential equationsrdquo Applied Mathematics and Computationvol 134 no 2-3 pp 307ndash317 2003

[17] R P Agarwal and P R Krishnamurthy ldquoOn the uniquenessof solution of nonlinear boundary value problemsrdquo Journal ofMathematical Sciences vol 10 no 1 pp 17ndash31 1976

[18] R P Agarwal and P Y PangOpial Inequalities with Applicationsin Differential and Difference Equation Kluwer AcademicDordrecht The Netherlands 1995

[19] R P Agarwal and P Y Pang ldquoRemarks on the generalizationsof Opialrsquos inequalityrdquo Journal of Mathematical Analysis andApplications vol 190 no 2 pp 559ndash577 1995

[20] P R Beesack and K M Das ldquoExtensions of Opialrsquos inequalityrdquoPacific Journal of Mathematics vol 26 pp 215ndash232 1968

[21] S Clark and D Hinton ldquoSome disconjugacy criteria for dif-ferential equations with oscillatory coefficientsrdquoMathematischeNachrichten vol 278 no 12-13 pp 1476ndash1489 2005

[22] A M Fink ldquoOn Opials inequality for 119891(119899)rdquo Proceedings of theAmerican Mathematical Society vol 155 pp 177ndash181 1992

[23] A Kufner and L-E Persson Weighted Inequalities of HardyType World Scientific River Edge NJ USA 2003

[24] D W Boyd ldquoBest constants in a class of integral inequalitiesrdquoPacific Journal of Mathematics vol 30 pp 367ndash383 1969

Submit your manuscripts athttpwwwhindawicom

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical Problems in Engineering

Hindawi Publishing Corporationhttpwwwhindawicom

Differential EquationsInternational Journal of

Volume 2014

Applied MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Probability and StatisticsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Mathematical PhysicsAdvances in

Complex AnalysisJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

OptimizationJournal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

CombinatoricsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Operations ResearchAdvances in

Journal of

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Function Spaces

Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of Mathematics and Mathematical Sciences

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

The Scientific World JournalHindawi Publishing Corporation httpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Algebra

Discrete Dynamics in Nature and Society

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Decision SciencesAdvances in

Discrete MathematicsJournal of

Hindawi Publishing Corporationhttpwwwhindawicom

Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Stochastic AnalysisInternational Journal of

4 Mathematical Problems in Engineering

We also need the following inequality which is the specialcase of an inequality proved by Agarwal and Pang [18] withtwo functions

int120573

120572

119888(119905)10038161003816100381610038161003816119910(119896)

(119905)10038161003816100381610038161003816

10038161003816100381610038161003816119910(119896+1)

(119905)10038161003816100381610038161003816119889119905 le 119862

120572int120573

120572

119903 (119905)10038161003816100381610038161003816119910(119899)(119905)

10038161003816100381610038161003816

2

119889119905

(22)

where 119910(119905) isin 119862119899minus1[120572 120573] and satisfies 119910(119894)(120572) = 0 119896 le 119894 le 119899minus10 le 119896 le 119899 minus 1 (119899 ge 1) and 119910(119899minus1) is absolutely continuous on(120572 120573) 119888(119905) and 119903(119905) being nonnegative measurable functionsdefined on (120572 120573) and

119862120572=

1

2((119899 minus 119896 minus 1))2max119905isin[120572120573]

119888 (119905) int120573

120572

(119904 minus 120572)2(119899minus119896minus1)

119903 (119904)119889119904

(23)

If 119910(119894)(120573) = 0 119896 le 119894 le 119899 minus 1 then (22) holds where 119862120572is

replaced

119862120573=

1

2((119899 minus 119896 minus 1))2max119905isin[120572120573]

119888 (119905) int120573

120572

(120573 minus 119904)2(119899minus119896minus1)

119903 (119904)119889119904

(24)

TheWirtinger type inequality due to Agarwal and Pang statesthat if 119910(120572) = 119910(120573) = 0 then

int120573

120572

119910120574+1

(119905)119889119905 le(120573 minus 120572)

120574+1

Γ2 ((120574 + 2)2)

2Γ(120574 + 2)

times int120573

120572

(1199101015840(119905))120574+1

(119905) 119889119905

(25)

TheOpial type inequality due to Clark and Hinton inequalitystates that if 119910 isin 119862

2[120572 120573] where 119910(120572) = 1199101015840(120572) = 0 then

(int120573

120572

1003816100381610038161003816119910(119905)100381610038161003816100381621003816100381610038161003816100381611991010158401015840(119905)

10038161003816100381610038161003816

2

119889119905)

1119903

le(120573 minus 120572)

32

radic3int120573

120572

1003816100381610038161003816100381611991010158401015840

(119905)10038161003816100381610038161003816

2

119889119905

(26)

The Opial type inequality due to Fink inequality states that if119910(119894)(120572) = 0 0 le 119894 le 119899 minus 1 120583 ge 1 (1120583) + (1]) = 1 then

int120573

120572

10038161003816100381610038161003816119910(119896)

(119905) 119910(119903)

(119905)10038161003816100381610038161003816119889119905

le 119862 (119899 119896 119903 120583) (120573 minus 120572)2119899minus119896minus119903+1minus2120583

(int120573

120572

10038161003816100381610038161003816119910(119899)(119905)

10038161003816100381610038161003816

120583

119889119905)

2120583

(27)

where 0 le 119896 lt 119903 lt 119899 (119899 ge 2) and

119862 (119899 119896 119903 120583) =1

2((119899 minus 119896 minus 1))2[(119899 minus 119896 minus 1)] + 1]

2]

(28)

3 Main Results

In this section we state and prove the main results Forsimplicity we introduce the following notations

1198671(120572 120573 119866

10) =

1

radic2[int120573

120572

1198762

1(119905)

1198661(119905)

119903(119905)119889119905]

12

1198661(119905) = int

119905

120572

(119905 minus 119904)2

119903 (119904)119889119904 119876

1(119905) = int

120573

119905

119902 (119904) 119889119904

(29)

1198672(120572 120573 119866

2) =

1

radic2[int120573

120572

1198762

2(119905)

1198662(119905)

119903(119905)119889119905]

12

1198662(119905) = int

120573

119905

(119905 minus 119904)2

119903 (119904)119889119904 119876

2(119905) = int

119905

120572

119902 (119904) 119889119904

(30)

119860119894(120572 120573) = 119869

119894(119905)min(int

119888

120572

1

119903(119904)119889119904)

12

(int120573

119888

1

119903(119904)119889119904)

12

(31)

where 119869119894(119905) = sup

(119888119889)sub(120572120573)(int120573

120572119863119894(119905)119889119905)12 and |119863

119894(119905)| =

|119876119894(119905)| + |119901(119905)| 119894 = 1 2

1198611(120572 120573) = sup

120572le119905le120573

(int120573

119905

119901(119905)119889119905)

12

(int119905

120572

1

119903(119904)119889119904)

12

119862120572=

1

2max120572le119905le120573

119902 (119905) int120573

120572

(119904 minus 120572)2

119903 (119904)119889119904

(32)

1198612(120572 120573) = sup

120572le119905le120573

(int119905

120572

119901(119905)119889119905)

12

(int120573

119905

1

119903(119904)119889119904)

12

119862120573=

1

2max120572le119905le120573

119902 (119905) int120573

120572

(120573 minus 119904)2

119903 (119904)119889119904

(33)

120595(120572 120573) = sup120572le119905le120573

(int120573

119905

119903(119905)119889119905)

12

(int119905

120572

1

119903(119904)119889119904)

12

(34)

Now we are ready to state and prove the main results

Theorem 1 Assume that 119909(119905) is a nontrivial solution of (9) If119909(120572) = 1199091015840(120572) = 1199091015840(120573) = 0 then

1198671(120572 120573 119866

1) + 4119860

2

1(120572 120573) ge 1 (35)

If 1199091015840(120572) = 119909(120573) = 1199091015840(120573) = 0 then

1198672(120572 120573 119866

2) + 4119860

2

2(120572 120573) ge 1 (36)

Proof Weprove (35)Multiplying (9) by 1199091015840(119905) and integratingthe new equation from 120572 to 120573 we obtain

int120573

120572

1199091015840(119903(11990910158401015840))1015840

119889119905 = minusint120573

120572

119901(1199091015840)2

119889119905 minus int120573

120572

1199021199091199091015840119889119905 (37)

Mathematical Problems in Engineering 5

Integrating by parts the left-hand side we get that

11990311990910158401199091015840101584010038161003816100381610038161003816

120573

120572minus int120573

120572

119903(11990910158401015840)2

119889119905 = minusint120573

120572

119901(1199091015840)2

119889119905 minus int120573

120572

1199021199091199091015840119889119905

(38)

Using the assumptions that 1199091015840(120572) = 1199091015840(120573) = 0 and 1198761(119905) =

int120573

119905119902(119904)119889119904 we have

int120573

120572

119903(119905)(11990910158401015840

(119905))2

119889119905 = int120573

120572

119901 (119905) (1199091015840(119905))2

119889119905

+ int120573

120572

1198761015840

1(119905) 119909 (119905) 119909

1015840

(119905) 119889119905

(39)

Integrating the term int120573

12057211987610158401(119905)119909(119905)1199091015840(119905)119889119904 by parts and using

the assumption that 1199091015840(120572) = 1199091015840(120573) = 0 we obtain

int120573

120572

1198761015840

1(119905) 119909 (119905) 119909

1015840

(119905) 119889119904

= minusint120573

120572

1198761(119905) (1199091015840

(119905))2

119889119905 minus int120573

120572

1198761(119905) 119909 (119905) 119909

10158401015840

(119905) 119889119905

(40)

Substituting (40) into (39) we get

int120573

120572

119903(119905)1003816100381610038161003816100381611990910158401015840

(119905)10038161003816100381610038161003816

2

119889119905 le int120573

120572

10038161003816100381610038161198631 (119905)1003816100381610038161003816100381610038161003816100381610038161199091015840(119905)

10038161003816100381610038161003816

2

119889119905

+ int120573

120572

10038161003816100381610038161198761 (119905)1003816100381610038161003816|119909 (119905)|

1003816100381610038161003816100381611990910158401015840

(119905)10038161003816100381610038161003816119889119905

(41)

where |1198631(119905)| = |119876

1(119905)| + |119901(119905)| Applying inequality (19) on

the integral

int120573

120572

10038161003816100381610038161198761 (119905)1003816100381610038161003816|119909 (119905)|

1003816100381610038161003816100381611990910158401015840

(119905)10038161003816100381610038161003816119889119905 (42)

with 120601(119905) = |1198761(119905)| 120593(119905) = 119903(119905) 119896 = 0119898 = 119897 = 1 119899 = 2 119888 = 2

and 119909(120572) = 1199091015840(120572) = 0 we get

int120573

120572

10038161003816100381610038161198761 (119905)1003816100381610038161003816|119909 (119905)|

1003816100381610038161003816100381611990910158401015840

(119905)10038161003816100381610038161003816119889119905

le 1198671(120572 120573 119866

1) int120573

120572

119903 (119905)1003816100381610038161003816100381611990910158401015840(119905)

10038161003816100381610038161003816

2

119889119905

(43)

where 1198671(120573 120572 119866

1) is defined as in (29) Applying inequality

(12) on the integral int120573120572|1198631(119905)||1199091015840(119905)|2119889119905 with 119910(119905) = 1199091015840(119905) and

119910(120572) = 119910(120573) = 0 we have that

int120573

120572

10038161003816100381610038161198631 (119905)1003816100381610038161003816100381610038161003816100381610038161199091015840(119905)

10038161003816100381610038161003816

2

119889119905 le 41198602

1(120572 120573) int

120573

120572

119903 (119905)1003816100381610038161003816100381611990910158401015840(119905)

10038161003816100381610038161003816

2

119889119905

(44)

where1198601(120572 120573) is defined as in (31) Substituting (44) and (43)

into (41) we get

int120573

120572

|119903 (119905)|1003816100381610038161003816100381611990910158401015840(119905)

10038161003816100381610038161003816

2

119889119905 le 41198602

1(120572 120573) int

120573

120572

119903 (119905)1003816100381610038161003816100381611990910158401015840(119905)

10038161003816100381610038161003816

2

119889119905

+ 1198671(120572 120573 119866

1) int120573

120572

119903 (119905)1003816100381610038161003816100381611990910158401015840(119905)

10038161003816100381610038161003816

2

119889119905

(45)

Cancelling the term int120573

120572119903(119905)|11990910158401015840(119905)|2119889119905 we get

41198602

1(120572 120573) + 119867

1(120572 120573 119866

1) ge 1 (46)

which is the desired inequality (35) The proof of (36) issimilar to the proof of (35) by replacing 119867

1(120572 120573 119866

1) by

1198672(120572 120573 119866

2) and 119860

1(120572 120573) by 119860

2(120572 120573) The proof is com-

plete

In the following we apply the Clark and Hinton inequal-ity (26) to get a new result

Theorem 2 Assume that 119903(119905) is a nonincreasing function andsuppose that 119909(119905) is a solution of (9) If 119909(120572) = 1199091015840(120572) = 1199091015840(120573) =

0 then

41198602

1(120572 120573) +

(120573 minus 120572)32

radic3119903 (120573)(int120573

120572

10038161003816100381610038161198761(119905)10038161003816100381610038162

119889119905)

12

ge 1 (47)

where 1198601(120572 120573) is defined as in (31) for 119894 = 1 and 119876

1(119905) is

defined as in (29)

Proof Multiply (9) by 1199091015840(119905) and proceed as in the proof ofTheorem 1 to get

int120573

120572

119903(119905)1003816100381610038161003816100381611990910158401015840

(119905)10038161003816100381610038161003816

2

119889119905 le int120573

120572

10038161003816100381610038161198631 (119905)1003816100381610038161003816100381610038161003816100381610038161199091015840(119905)

10038161003816100381610038161003816

2

119889119905

+ int120573

120572

10038161003816100381610038161198761 (119905)1003816100381610038161003816|119909 (119905)|

1003816100381610038161003816100381611990910158401015840

(119905)10038161003816100381610038161003816119889119905

(48)

where |1198631(119905)| = |119876

1(119905)|+ |119901(119905)| Applying the Schwarz inequal-

ity

int120573

120572

1003816100381610038161003816119891 (119905) 119892 (119905)1003816100381610038161003816119889119905 le (int

120573

120572

1003816100381610038161003816119891(119905)10038161003816100381610038162

119889119905)

12

(int120573

120572

1003816100381610038161003816119892(119905)10038161003816100381610038162

119889119905)

12

(49)

on the integral int120573120572|1198761(119905)||119909(119905)||11990910158401015840(119905)|119889119905 we have that

int120573

120572

10038161003816100381610038161198761 (119905)1003816100381610038161003816|119909 (119905)|

1003816100381610038161003816100381611990910158401015840

(119905)10038161003816100381610038161003816119889119905 le (int

120573

120572

10038161003816100381610038161198761(119905)10038161003816100381610038162

119889119905)

12

times (int120573

120572

|119909(119905)|21003816100381610038161003816100381611990910158401015840(119905)

10038161003816100381610038161003816

2

119889119905)

12

(50)Applying inequality (26) and using the assumption 119909(120572) =

1199091015840(120572) = 0 we get that

(int120573

120572

|119909(119905)|21003816100381610038161003816100381611990910158401015840(119905)

10038161003816100381610038161003816

2

119889119905)

12

le(120573 minus 120572)

32

radic3int120573

120572

1003816100381610038161003816100381611990910158401015840(119905)

10038161003816100381610038161003816

2

119889119905

(51)Substituting (51) into (50) and using the assumption that 119903(119905)is a nonincreasing function we have

int120573

120572

10038161003816100381610038161198761(119905)1003816100381610038161003816|119909(119905)|

1003816100381610038161003816100381611990910158401015840

(119905)10038161003816100381610038161003816

le(120573 minus 120572)

32

radic3119903 (120573)(int120573

120572

10038161003816100381610038161198761(119905)10038161003816100381610038162

119889119905)

12

int120573

120572

119903 (119905)1003816100381610038161003816100381611990910158401015840(119905)

10038161003816100381610038161003816

2

119889119905

(52)

6 Mathematical Problems in Engineering

Applying Hardyrsquos inequality (12) on the integralint120573

120572|1198631(119905)||1199091015840(119905)

2|119889119905 we obtain

int120573

120572

10038161003816100381610038161198631 (119905)1003816100381610038161003816100381610038161003816100381610038161199091015840(119905)

10038161003816100381610038161003816

2

119889119905 le 41198602

1(120572 120573) int

120573

120572

119903 (119905)1003816100381610038161003816100381611990910158401015840(119905)

10038161003816100381610038161003816

2

119889119905 (53)

where 1199091015840(120572) = 1199091015840(120573) = 0 Substituting (52) and (53) into (48)we obtain

int120573

120572

|119903(119905)|1003816100381610038161003816100381611990910158401015840

(119905)10038161003816100381610038161003816

2

119889119905

le 41198602

1(120572 120573) int

120573

120572

119903 (119905)1003816100381610038161003816100381611990910158401015840(119905)

10038161003816100381610038161003816

2

119889119905

+(120573 minus 120572)

32

radic3119903 (120573)(int120573

120572

10038161003816100381610038161198761(119905)10038161003816100381610038162

119889119905)

12

int120573

120572

119903 (119905)1003816100381610038161003816100381611990910158401015840(119905)

10038161003816100381610038161003816

2

119889119905

(54)

The desired inequality (47) followed by cancelling the termint120573

120572|119903(119905)||11990910158401015840(119905)|2119889119905 The proof is complete

The proof of the following theorem is similar to the proofof Theorem 2

Theorem 3 Assume that 119903(119905) is a nonincreasing function andsuppose that 119909(119905) is a solution of (9) If 1199091015840(120572) = 119909(120573) = 1199091015840(120573) =

0 then

41198602

2(120572 120573) +

(120573 minus 120572)32

radic3119903 (120573)(int120573

120572

10038161003816100381610038161198761(119905)10038161003816100381610038162

119889119905)

12

ge 1 (55)

where 1198602(120572 120573) is defined as in (31) for 119894 = 2 and 119876

2(119905) is

defined as in (30)

One can use Theorems 2 and 3 to obtain some differentspecial cases For example as a special case of Theorem 2when 119903(119905) = 1 and 119901(119905) = 0 we have the following result

Corollary 4 If 119909(119905) is a nontrivial solution of

119909101584010158401015840

(119905) + 119902 (119905) 119909 (119905) = 0 119905 isin [120572 120573] (56)

which satisfies 119909(120572) = 1199091015840(120572) = 1199091015840(120573) = 0 then

int120573

120572

10038161003816100381610038161198761(119905)10038161003816100381610038162

119889119905 ge3

(120573 minus 120572)3 (57)

where 1198761(119905) is defined as in (29)

Now we will prove a new result when 119903(119905) = 1

Theorem 5 Suppose that 119903(119905) = 1 and assume that 119909(119905) is asolution of (9) If 119909(120572) = 1199091015840(120572) = 1199091015840(120573) = 0 or 119909(120573) = 1199091015840(120573) =

1199091015840(120572) = 0 then

119875120587(120573 minus 120572)

2

16+ 119876

(120573 minus 120572)3

6ge 1 (58)

where 119875 = max120572le119905le120573

|119901(119905)| and 119876 = max120572le119905le120573

|119902(119905)|

Proof Multiplying (9) by 1199091015840 and integrating by parts the left-hand side we have

1199091015840

(119905)11990910158401015840

(119905)10038161003816100381610038161003816

120573

120572minus int120573

120572

(11990910158401015840

(119905))2

119889119905

= minusint120573

120572

119901(119905)(1199091015840

(119905))2

119889119905 minus int120573

120572

119902(119905)119909(119905)1199091015840

(119905)119889119905

(59)

Using the assumptions that 1199091015840(120572) = 1199091015840(120573) = 0 we obtain

int120573

120572

(11990910158401015840)2

119889119905 = int120573

120572

119901(1199091015840)2

119889119905 + int120573

120572

1199021199091199091015840119889119905 (60)

By using themaximumvalues of |119901(119905)| and |119902(119905)| we canwrite(60) as

int120573

120572

1003816100381610038161003816100381611990910158401015840

(119905)10038161003816100381610038161003816

2

119889119905 le 119875int120573

120572

100381610038161003816100381610038161199091015840

(119905)10038161003816100381610038161003816

2

119889119905 + 119876int120573

120572

|119909(119905)|100381610038161003816100381610038161199091015840

(119905)10038161003816100381610038161003816119889119905

(61)

For the first term on the right-hand side int120573

120572|1199091015840(119905)|2119889119905 we

apply inequality (25) with 119910(119905) = 1199091015840(119905) (note that 1199091015840(120572) =

1199091015840(120573) = 0) and 120574 = 1 to obtain

int120573

120572

10038161003816100381610038161003816119909101584010038161003816100381610038161003816

2

119889119905 le120587(120573 minus 120572)

2

16int120573

120572

100381610038161003816100381610038161199091015840101584010038161003816100381610038161003816

2

119889119905 (62)

Applying Fink inequality (27) on int120573

120572|119909||1199091015840|119889119905 with 119896 = 0 119903 =

1 120583 = ] = 2 and 119899 = 2 we have

int120573

120572

|119909|10038161003816100381610038161003816119909101584010038161003816100381610038161003816119889119905 le

(120573 minus 120572)3

6int120573

120572

1003816100381610038161003816100381611990910158401015840(119905)

10038161003816100381610038161003816

2

119889119905 (63)

Substituting (62) and (63) into (61) we get after cancellingthe term int

120573

120572|11990910158401015840(119905)|2119889119905 that

119875120587(120573 minus 120572)

2

16+ 119876

(120573 minus 120572)3

6ge 1 (64)

which is the desired inequality (58) The proof is complete

As a special case of Theorem 5 when 119901(119905) = 0 we havethe following result

Corollary 6 Let 119909(119905) be a nontrivial solution of (56) If 119909(120572) =1199091015840(120572) = 1199091015840(120573) = 0 then

max120572le119905le120573

1003816100381610038161003816119902 (119905)1003816100381610038161003816 ge

6

(120573 minus 120572)3 (65)

In the following we prove some results related to theboundary conditions presented in (ii)

Theorem 7 Assume that 119909(119905) is a nontrivial solution of (9) If119909(120572) = 1199091015840(120572) = 11990910158401015840(120573) = 0 then

41198612

1(120572 120573) + 119862

120572ge 1 (66)

If 119909(120573) = 1199091015840(120573) = 11990910158401015840(120572) = 0 then

41198612

2(120572 120573) + 119862

120573ge 1 (67)

Mathematical Problems in Engineering 7

Proof Multiplying (9) by 1199091015840(119905) and integrating by parts theleft-hand side from 120572 to 120573 we get

11990311990910158401199091015840101584010038161003816100381610038161003816

120573

120572minus int120573

120572

119903(11990910158401015840)2

119889119905 = minusint120573

120572

119901(1199091015840)2

119889119905 minus int120573

120572

1199021199091199091015840119889119905

(68)

Using the assumption that 1199091015840(120572) = 11990910158401015840(120573) = 0 we have

int120573

120572

119903(119905)1003816100381610038161003816100381611990910158401015840

(119905)10038161003816100381610038161003816

2

119889119905 le int120573

120572

1003816100381610038161003816119901 (119905)1003816100381610038161003816100381610038161003816100381610038161199091015840(119905)

10038161003816100381610038161003816

2

119889119905

+ int120573

120572

1003816100381610038161003816119902 (119905)1003816100381610038161003816|119909 (119905)|

100381610038161003816100381610038161199091015840

(119905)10038161003816100381610038161003816119889119905

(69)

Applying inequality (12) on the integral int120573120572|119901(119905)||1199091015840(119905)|2119889119905

with 119899 = 119898 = 2 we get

int120573

120572

1003816100381610038161003816119901 (119905)1003816100381610038161003816100381610038161003816100381610038161199091015840(119905)

10038161003816100381610038161003816

2

119889119905 le 41198612

1(120572 120573) int

120573

120572

|119903 (119905)|1003816100381610038161003816100381611990910158401015840

(119905)210038161003816100381610038161003816119889119905 (70)

where1198611(120572 120573) is defined as in (32) Again applying inequality

(22) on the integral int120573120572|119902(119905)||119909(119905)||1199091015840(119905)|119889119905 we get

int120573

120572

1003816100381610038161003816119902 (119905)1003816100381610038161003816|119909 (119905)|

100381610038161003816100381610038161199091015840

(119905)10038161003816100381610038161003816119889119905 le 119862

120572int120573

120572

|119903 (119905)|1003816100381610038161003816100381611990910158401015840

(119905)210038161003816100381610038161003816119889119905 (71)

where 119862120572is defined as in (32) Substituting (70) and (71) into

(69) we obtain

int120573

120572

|119903 (119905)|1003816100381610038161003816100381611990910158401015840(119905)

10038161003816100381610038161003816

2

119889119905

le 41198612

1int120573

120572

|119903 (119905)|1003816100381610038161003816100381611990910158401015840

(119905)210038161003816100381610038161003816119889119905 + 119862

120572int120573

120572

|119903 (119905)|1003816100381610038161003816100381611990910158401015840

(119905)210038161003816100381610038161003816119889119905

(72)

By cancelling the term int120573

120572|119903(119905)||119909

10158401015840(119905)2|119889119905 we get that

41198612

1(120572 120573) + 119862

120572ge 1 (73)

which is the desired result (66) The proof (67) is similar tothe proof of (66) by using 119861

2(120572 120573) and 119862

120573instead of 119861

1(120572 120573)

and 119862120572and hence is omitted The proof is complete

In the following we apply an inequality due to Boyd [24]to obtain new results The Boyd inequality states that if 119910 isin

1198621[120572 120573] with 119910(120572) = 0 (or 119910(120573) = 0) then

int120573

120572

1003816100381610038161003816119910(119905)1003816100381610038161003816]100381610038161003816100381610038161199101015840(119905)

10038161003816100381610038161003816

120578

119889119905

le 119873 (] 120578 119904) (120573 minus 120572)][int119887

119886

100381610038161003816100381610038161199101015840(119905)

10038161003816100381610038161003816

119904

119889119905]

(]+120578)119904

(74)

where ] gt 0 119904 gt 1 0 le 120578 lt 119904 and

119873(] 120578 119904) =(119904 minus 120578)]]119903]+120578minus119904

(119904 minus 1) (] + 120578) (119868(] 120578 119904))]

119903 = ](119904 minus 1) + (119904 minus 120578)

(119904 minus 1)(] + 120578)

1119904

(75)

119868(] 120578 119904) = int1

0

1 +119904(120578 minus 1)

119904 minus 120578119905

minus(]+120578+119904])119904]

times [1 + (120578 minus 1)]1199051(]minus1)

119889119905

(76)

Note that an inequality of type (34) also holds when 119910(120572) =

119910(120573) = 0 Choose 119888 = (120572 + 120573)2 and apply (34) to [120572 119888] and[119888 120573] and by addition we obtain

int120573

120572

1003816100381610038161003816119910(119905)1003816100381610038161003816]100381610038161003816100381610038161199101015840(119905)

10038161003816100381610038161003816

120578

119889119905

le 119873 (] 120578 119904) (120573 minus 120572

2)

]

[int120573

120572

100381610038161003816100381610038161199101015840(119905)

10038161003816100381610038161003816

119904

119889119905]

(]+120578)119904

(77)

where 119873(] 120578 119904) is defined as in (75) An inequality of type(74) holds when 120578 = 119904 and 119910(120572) = 0 (or 119910(120573) = 0) In thiscase (74) becomes

int120573

120572

1003816100381610038161003816119910(119905)1003816100381610038161003816]100381610038161003816100381610038161199101015840(119905)

10038161003816100381610038161003816

120578

119889119905

le 119871 (] 120578) (120573 minus 120572)][int120573

120572

100381610038161003816100381610038161199101015840(119905)

10038161003816100381610038161003816

120578

119889119905]

(]+120578)120578

(78)

where

119871 (] 120578) =120578]120578

] + 120578(

]] + 120578

)

]120578

(Γ(((120578 + 1)120578) + (1]))Γ((120578 + 1)120578)Γ(1])

)

]

(79)

and Γ is the gamma function

Theorem 8 Assume that 119903(119905) is nonincreasing function and119909(119905) is a nontrivial solution of (9) If119909(120572) = 1199091015840(120572) = 11990910158401015840(120573) = 0then

41198612

1(120572 120573) +

8 (120573 minus 120572)

120587119903 (120573)1205952(120572 120573)(int

120573

120572

1003816100381610038161003816119902(119905)10038161003816100381610038162

119889119905)

12

ge 1 (80)

If 119909(120573) = 1199091015840(120573) = 11990910158401015840(120572) = 0 then

41198612

2(120572 120573) +

8 (120573 minus 120572)

120587119903 (120573)1205952(120572 120573)(int

120573

120572

1003816100381610038161003816119902(119905)10038161003816100381610038162

119889119905)

12

ge 1 (81)

Proof Multiplying (9) by 1199091015840(119905) and integrating by parts theleft-hand side we have that

11990311990910158401199091015840101584010038161003816100381610038161003816

120573

120572minus int120573

120572

119903(11990910158401015840)2

119889119905 = minusint120573

120572

119901(1199091015840)2

119889119905 minus int120573

120572

1199021199091199091015840119889119905

(82)

8 Mathematical Problems in Engineering

Using the assumption 1199091015840(120572) = 11990910158401015840(120573) = 0 we get that

int120573

120572

|119903 (119905)|1003816100381610038161003816100381611990910158401015840(119905)

10038161003816100381610038161003816

2

119889119905 le int120573

120572

1003816100381610038161003816119901 (119905)1003816100381610038161003816100381610038161003816100381610038161199091015840(119905)

10038161003816100381610038161003816

2

119889119905

+ int120573

120572

1003816100381610038161003816119902 (119905)1003816100381610038161003816|119909 (119905)|

100381610038161003816100381610038161199091015840

(119905)10038161003816100381610038161003816119889119905

(83)

Applying inequality (12) on the integral int120573120572|119901(119905)||1199091015840(119905)|2119889119905

with119898 = 119899 = 2 we get

int120573

120572

1003816100381610038161003816119901 (119905)1003816100381610038161003816100381610038161003816100381610038161199091015840(119905)

10038161003816100381610038161003816

2

119889119905

le 41205952(120572 120573) int

120573

120572

|119903 (119905)|1003816100381610038161003816100381611990910158401015840

(119905)210038161003816100381610038161003816119889119905

(84)

where 120595(120572 120573) is defined as in (34) Applying Schwarzrsquosinequality on the term int

120573

120572|119902(119905)||119909(119905)||1199091015840(119905)|119889119905 we get that

int120573

120572

1003816100381610038161003816119902 (119905)1003816100381610038161003816|119909 (119905)|

100381610038161003816100381610038161199091015840

(119905)10038161003816100381610038161003816119889119905

le (int120573

120572

1003816100381610038161003816119902(119905)10038161003816100381610038162

119889119905)

12

(int120573

120572

|119909|210038161003816100381610038161003816119909101584010038161003816100381610038161003816

2

119889119905)

12

(85)

Applying again inequality (78) on the integral int120573120572|119909|2|1199091015840|2119889119905

with ] = 120578 = 2 (note that 119909(120572) = 0) we obtain

int120573

120572

|119909|210038161003816100381610038161003816119909101584010038161003816100381610038161003816

2

119889119905 le4(120573 minus 120572)

2

12058721199032 (120573)[int120573

120572

119903(119905)10038161003816100381610038161003816119909101584010038161003816100381610038161003816

2

119889119905]

2

(86)

where 119903(119905) is a nonincreasing function Substituting (86) into(85) we get

int120573

120572

1003816100381610038161003816119902 (119905)1003816100381610038161003816|119909 (119905)|

100381610038161003816100381610038161199091015840

(119905)10038161003816100381610038161003816119889119905

le (int120573

120572

1003816100381610038161003816119902(119905)10038161003816100381610038162

119889119905)

12

2 (120573 minus 120572)

120587119903 (120573)int120573

120572

119903 (119905)10038161003816100381610038161003816119909101584010038161003816100381610038161003816

2

119889119905

(87)

Applying inequality (12) on the integral int120573120572119903(119905)|1199091015840|2119889119905 with

119901(119905) = 119902(119905) = 119903(119905) and119898 = 119899 = 2 we get

int120573

120572

|119903 (119905)|100381610038161003816100381610038161199091015840(119905)

10038161003816100381610038161003816

2

119889119905 le 41198612

1(120572 120573) int

120573

120572

|119903 (119905)|1003816100381610038161003816100381611990910158401015840(119905)

10038161003816100381610038161003816

2

119889119905 (88)

Substituting (88) into (87) we have

int120573

120572

1003816100381610038161003816119902 (119905)1003816100381610038161003816|119909 (119905)|

100381610038161003816100381610038161199091015840

(119905)10038161003816100381610038161003816119889119905

le (int120573

120572

1003816100381610038161003816119902(119905)10038161003816100381610038162

119889119905)

12

8(120573 minus 120572)

120587119903(120573)1205952(120572 120573)

times int120573

120572

|119903 (119905)|1003816100381610038161003816100381611990910158401015840(119905)

10038161003816100381610038161003816

2

119889119905

(89)

Substituting (84) and (89) into (83) we obtain

int120573

120572

|119903 (119905)|1003816100381610038161003816100381611990910158401015840(119905)

10038161003816100381610038161003816

2

119889119905

le 41198612

1(120572 120573) int

120573

120572

|119903(119905)|1003816100381610038161003816100381611990910158401015840

(119905)10038161003816100381610038161003816

2

119889119905

+(int120573

120572

1003816100381610038161003816119902(119905)10038161003816100381610038162

119889119905)

12

8 (120573 minus 120572)

120587119903 (120573)1205952(120572 120573)

times int120573

120572

|119903 (119905)|1003816100381610038161003816100381611990910158401015840(119905)

10038161003816100381610038161003816

2

119889119905

(90)

Cancelling the term int120573

120572|119903(119905)||11990910158401015840(119905)|2119889119905 we obtain

41198612

1(120572 120573) +

8 (120573 minus 120572)

120587119903 (120573)1205952(120572 120573)(int

120573

120572

1003816100381610038161003816119902(119905)10038161003816100381610038162

119889119905)

12

ge 1

(91)

which is the desired inequality (80) The proof of (81) issimilar to the proof of (80) by using 119861

2(120572 120573) instead of

1198611(120572 120573) The proof is complete

Conflict of Interests

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

References

[1] E Picard ldquoMemoire sur la theorie des equations aux derivespartielleset la methode desapproximations successivesrdquo Journalde Mathematiques Pures et Appliquees vol 6 pp 145ndash210 1890

[2] E Picard ldquoSur lrsquoapplication des methodes drsquoapproximationssuccessives a lrsquoetude de certaines equations differentielles ordi-nairesrdquo Journal deMathematiques Pures et Appliquees vol 9 pp217ndash271 1893

[3] O Niccoletti ldquoSulle consizioni iniziali che determinano gliintegrale delle equazioni diff eren ziali ordinarierdquo Atti dellaAccademia delle Scienze di Torino vol 33 pp 746ndash759 1897

[4] C de la Vallee Poussin ldquoSur lrsquoequation differentielle lineaire dusecond ordrerdquo Journal de Mathematiques Pures et Appliqueesvol 8 pp 125ndash144 1929

[5] A M Lyapunov ldquoProbleme general de la stabilite du mouve-mentrdquo Annales de la Faculte des sciences de Toulouse vol 9 pp203ndash474 1907

[6] S H Saker ldquoSome new disconjugacy criteria for second orderdifferential equations with a middle termrdquo Bulletin Mathema-tique de la Societe des Sciences Mathematiques de Roumanie vol57 no 1 pp 109ndash120 2014

[7] G Casadei ldquoSul teorema di unicita di De La Vallee Poussinper equazioni differenziali del terzo ordinerdquo Rendiconti delSeminario Matematico della Universita di Padova vol 41 pp300ndash315 1968

[8] A Lasota ldquoSur la distance entre les zeros de lrsquoequation dif-ferentielle lineaire du troisieme ordrerdquo Annales Polonici Mathe-matici vol 13 pp 129ndash132 1963

Mathematical Problems in Engineering 9

[9] RMMathsen ldquoA disconjugacy condition for119910101584010158401015840+119886211991010158401015840+119886

11199101015840+

1198860119910 = 0rdquo Proceedings of the AmericanMathematical Society vol

17 pp 627ndash632 1966[10] B G Pachpatte ldquoOn the zeros of solutions of certain differential

equationsrdquo Demonstratio Mathematica vol 25 no 4 pp 825ndash833 1992

[11] B G Pachpatte ldquoLyapunov type integral inequalities for certaindifferential equationsrdquo Georgian Mathematical Journal vol 4no 2 pp 139ndash148 1997

[12] S Panigrahi ldquoLyapunov-type integral inequalities for certainhigher-order differential equationsrdquo Electronic Journal of Differ-ential Equations vol 2009 no 28 pp 1ndash14 2009

[13] N Parhi and S Panigrahi ldquoOn Liapunov-type inequality forthird-order differential equationsrdquo Journal of MathematicalAnalysis and Applications vol 233 no 2 pp 445ndash460 1999

[14] N Parhi and S Panigrahi ldquoDisfocality and Liapunov-typeinequalities for third-order equationsrdquo Applied MathematicsLetters vol 16 no 2 pp 227ndash233 2003

[15] U Richard ldquoMetodi diversi per ottenere disequaglianze allade la Vallee Poussin nelle equazioni differenziali ordinarie delsecondo e terzo ordinerdquo Rendiconti del Seminario MatematicoUniversita e Politecnico di Torino vol 27 pp 35ndash68 1968

[16] X Yang ldquoOn Liapunov-type inequality for certain higher-orderdifferential equationsrdquo Applied Mathematics and Computationvol 134 no 2-3 pp 307ndash317 2003

[17] R P Agarwal and P R Krishnamurthy ldquoOn the uniquenessof solution of nonlinear boundary value problemsrdquo Journal ofMathematical Sciences vol 10 no 1 pp 17ndash31 1976

[18] R P Agarwal and P Y PangOpial Inequalities with Applicationsin Differential and Difference Equation Kluwer AcademicDordrecht The Netherlands 1995

[19] R P Agarwal and P Y Pang ldquoRemarks on the generalizationsof Opialrsquos inequalityrdquo Journal of Mathematical Analysis andApplications vol 190 no 2 pp 559ndash577 1995

[20] P R Beesack and K M Das ldquoExtensions of Opialrsquos inequalityrdquoPacific Journal of Mathematics vol 26 pp 215ndash232 1968

[21] S Clark and D Hinton ldquoSome disconjugacy criteria for dif-ferential equations with oscillatory coefficientsrdquoMathematischeNachrichten vol 278 no 12-13 pp 1476ndash1489 2005

[22] A M Fink ldquoOn Opials inequality for 119891(119899)rdquo Proceedings of theAmerican Mathematical Society vol 155 pp 177ndash181 1992

[23] A Kufner and L-E Persson Weighted Inequalities of HardyType World Scientific River Edge NJ USA 2003

[24] D W Boyd ldquoBest constants in a class of integral inequalitiesrdquoPacific Journal of Mathematics vol 30 pp 367ndash383 1969

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

Mathematical Problems in Engineering 5

Integrating by parts the left-hand side we get that

11990311990910158401199091015840101584010038161003816100381610038161003816

120573

120572minus int120573

120572

119903(11990910158401015840)2

119889119905 = minusint120573

120572

119901(1199091015840)2

119889119905 minus int120573

120572

1199021199091199091015840119889119905

(38)

Using the assumptions that 1199091015840(120572) = 1199091015840(120573) = 0 and 1198761(119905) =

int120573

119905119902(119904)119889119904 we have

int120573

120572

119903(119905)(11990910158401015840

(119905))2

119889119905 = int120573

120572

119901 (119905) (1199091015840(119905))2

119889119905

+ int120573

120572

1198761015840

1(119905) 119909 (119905) 119909

1015840

(119905) 119889119905

(39)

Integrating the term int120573

12057211987610158401(119905)119909(119905)1199091015840(119905)119889119904 by parts and using

the assumption that 1199091015840(120572) = 1199091015840(120573) = 0 we obtain

int120573

120572

1198761015840

1(119905) 119909 (119905) 119909

1015840

(119905) 119889119904

= minusint120573

120572

1198761(119905) (1199091015840

(119905))2

119889119905 minus int120573

120572

1198761(119905) 119909 (119905) 119909

10158401015840

(119905) 119889119905

(40)

Substituting (40) into (39) we get

int120573

120572

119903(119905)1003816100381610038161003816100381611990910158401015840

(119905)10038161003816100381610038161003816

2

119889119905 le int120573

120572

10038161003816100381610038161198631 (119905)1003816100381610038161003816100381610038161003816100381610038161199091015840(119905)

10038161003816100381610038161003816

2

119889119905

+ int120573

120572

10038161003816100381610038161198761 (119905)1003816100381610038161003816|119909 (119905)|

1003816100381610038161003816100381611990910158401015840

(119905)10038161003816100381610038161003816119889119905

(41)

where |1198631(119905)| = |119876

1(119905)| + |119901(119905)| Applying inequality (19) on

the integral

int120573

120572

10038161003816100381610038161198761 (119905)1003816100381610038161003816|119909 (119905)|

1003816100381610038161003816100381611990910158401015840

(119905)10038161003816100381610038161003816119889119905 (42)

with 120601(119905) = |1198761(119905)| 120593(119905) = 119903(119905) 119896 = 0119898 = 119897 = 1 119899 = 2 119888 = 2

and 119909(120572) = 1199091015840(120572) = 0 we get

int120573

120572

10038161003816100381610038161198761 (119905)1003816100381610038161003816|119909 (119905)|

1003816100381610038161003816100381611990910158401015840

(119905)10038161003816100381610038161003816119889119905

le 1198671(120572 120573 119866

1) int120573

120572

119903 (119905)1003816100381610038161003816100381611990910158401015840(119905)

10038161003816100381610038161003816

2

119889119905

(43)

where 1198671(120573 120572 119866

1) is defined as in (29) Applying inequality

(12) on the integral int120573120572|1198631(119905)||1199091015840(119905)|2119889119905 with 119910(119905) = 1199091015840(119905) and

119910(120572) = 119910(120573) = 0 we have that

int120573

120572

10038161003816100381610038161198631 (119905)1003816100381610038161003816100381610038161003816100381610038161199091015840(119905)

10038161003816100381610038161003816

2

119889119905 le 41198602

1(120572 120573) int

120573

120572

119903 (119905)1003816100381610038161003816100381611990910158401015840(119905)

10038161003816100381610038161003816

2

119889119905

(44)

where1198601(120572 120573) is defined as in (31) Substituting (44) and (43)

into (41) we get

int120573

120572

|119903 (119905)|1003816100381610038161003816100381611990910158401015840(119905)

10038161003816100381610038161003816

2

119889119905 le 41198602

1(120572 120573) int

120573

120572

119903 (119905)1003816100381610038161003816100381611990910158401015840(119905)

10038161003816100381610038161003816

2

119889119905

+ 1198671(120572 120573 119866

1) int120573

120572

119903 (119905)1003816100381610038161003816100381611990910158401015840(119905)

10038161003816100381610038161003816

2

119889119905

(45)

Cancelling the term int120573

120572119903(119905)|11990910158401015840(119905)|2119889119905 we get

41198602

1(120572 120573) + 119867

1(120572 120573 119866

1) ge 1 (46)

which is the desired inequality (35) The proof of (36) issimilar to the proof of (35) by replacing 119867

1(120572 120573 119866

1) by

1198672(120572 120573 119866

2) and 119860

1(120572 120573) by 119860

2(120572 120573) The proof is com-

plete

In the following we apply the Clark and Hinton inequal-ity (26) to get a new result

Theorem 2 Assume that 119903(119905) is a nonincreasing function andsuppose that 119909(119905) is a solution of (9) If 119909(120572) = 1199091015840(120572) = 1199091015840(120573) =

0 then

41198602

1(120572 120573) +

(120573 minus 120572)32

radic3119903 (120573)(int120573

120572

10038161003816100381610038161198761(119905)10038161003816100381610038162

119889119905)

12

ge 1 (47)

where 1198601(120572 120573) is defined as in (31) for 119894 = 1 and 119876

1(119905) is

defined as in (29)

Proof Multiply (9) by 1199091015840(119905) and proceed as in the proof ofTheorem 1 to get

int120573

120572

119903(119905)1003816100381610038161003816100381611990910158401015840

(119905)10038161003816100381610038161003816

2

119889119905 le int120573

120572

10038161003816100381610038161198631 (119905)1003816100381610038161003816100381610038161003816100381610038161199091015840(119905)

10038161003816100381610038161003816

2

119889119905

+ int120573

120572

10038161003816100381610038161198761 (119905)1003816100381610038161003816|119909 (119905)|

1003816100381610038161003816100381611990910158401015840

(119905)10038161003816100381610038161003816119889119905

(48)

where |1198631(119905)| = |119876

1(119905)|+ |119901(119905)| Applying the Schwarz inequal-

ity

int120573

120572

1003816100381610038161003816119891 (119905) 119892 (119905)1003816100381610038161003816119889119905 le (int

120573

120572

1003816100381610038161003816119891(119905)10038161003816100381610038162

119889119905)

12

(int120573

120572

1003816100381610038161003816119892(119905)10038161003816100381610038162

119889119905)

12

(49)

on the integral int120573120572|1198761(119905)||119909(119905)||11990910158401015840(119905)|119889119905 we have that

int120573

120572

10038161003816100381610038161198761 (119905)1003816100381610038161003816|119909 (119905)|

1003816100381610038161003816100381611990910158401015840

(119905)10038161003816100381610038161003816119889119905 le (int

120573

120572

10038161003816100381610038161198761(119905)10038161003816100381610038162

119889119905)

12

times (int120573

120572

|119909(119905)|21003816100381610038161003816100381611990910158401015840(119905)

10038161003816100381610038161003816

2

119889119905)

12

(50)Applying inequality (26) and using the assumption 119909(120572) =

1199091015840(120572) = 0 we get that

(int120573

120572

|119909(119905)|21003816100381610038161003816100381611990910158401015840(119905)

10038161003816100381610038161003816

2

119889119905)

12

le(120573 minus 120572)

32

radic3int120573

120572

1003816100381610038161003816100381611990910158401015840(119905)

10038161003816100381610038161003816

2

119889119905

(51)Substituting (51) into (50) and using the assumption that 119903(119905)is a nonincreasing function we have

int120573

120572

10038161003816100381610038161198761(119905)1003816100381610038161003816|119909(119905)|

1003816100381610038161003816100381611990910158401015840

(119905)10038161003816100381610038161003816

le(120573 minus 120572)

32

radic3119903 (120573)(int120573

120572

10038161003816100381610038161198761(119905)10038161003816100381610038162

119889119905)

12

int120573

120572

119903 (119905)1003816100381610038161003816100381611990910158401015840(119905)

10038161003816100381610038161003816

2

119889119905

(52)

6 Mathematical Problems in Engineering

Applying Hardyrsquos inequality (12) on the integralint120573

120572|1198631(119905)||1199091015840(119905)

2|119889119905 we obtain

int120573

120572

10038161003816100381610038161198631 (119905)1003816100381610038161003816100381610038161003816100381610038161199091015840(119905)

10038161003816100381610038161003816

2

119889119905 le 41198602

1(120572 120573) int

120573

120572

119903 (119905)1003816100381610038161003816100381611990910158401015840(119905)

10038161003816100381610038161003816

2

119889119905 (53)

where 1199091015840(120572) = 1199091015840(120573) = 0 Substituting (52) and (53) into (48)we obtain

int120573

120572

|119903(119905)|1003816100381610038161003816100381611990910158401015840

(119905)10038161003816100381610038161003816

2

119889119905

le 41198602

1(120572 120573) int

120573

120572

119903 (119905)1003816100381610038161003816100381611990910158401015840(119905)

10038161003816100381610038161003816

2

119889119905

+(120573 minus 120572)

32

radic3119903 (120573)(int120573

120572

10038161003816100381610038161198761(119905)10038161003816100381610038162

119889119905)

12

int120573

120572

119903 (119905)1003816100381610038161003816100381611990910158401015840(119905)

10038161003816100381610038161003816

2

119889119905

(54)

The desired inequality (47) followed by cancelling the termint120573

120572|119903(119905)||11990910158401015840(119905)|2119889119905 The proof is complete

The proof of the following theorem is similar to the proofof Theorem 2

Theorem 3 Assume that 119903(119905) is a nonincreasing function andsuppose that 119909(119905) is a solution of (9) If 1199091015840(120572) = 119909(120573) = 1199091015840(120573) =

0 then

41198602

2(120572 120573) +

(120573 minus 120572)32

radic3119903 (120573)(int120573

120572

10038161003816100381610038161198761(119905)10038161003816100381610038162

119889119905)

12

ge 1 (55)

where 1198602(120572 120573) is defined as in (31) for 119894 = 2 and 119876

2(119905) is

defined as in (30)

One can use Theorems 2 and 3 to obtain some differentspecial cases For example as a special case of Theorem 2when 119903(119905) = 1 and 119901(119905) = 0 we have the following result

Corollary 4 If 119909(119905) is a nontrivial solution of

119909101584010158401015840

(119905) + 119902 (119905) 119909 (119905) = 0 119905 isin [120572 120573] (56)

which satisfies 119909(120572) = 1199091015840(120572) = 1199091015840(120573) = 0 then

int120573

120572

10038161003816100381610038161198761(119905)10038161003816100381610038162

119889119905 ge3

(120573 minus 120572)3 (57)

where 1198761(119905) is defined as in (29)

Now we will prove a new result when 119903(119905) = 1

Theorem 5 Suppose that 119903(119905) = 1 and assume that 119909(119905) is asolution of (9) If 119909(120572) = 1199091015840(120572) = 1199091015840(120573) = 0 or 119909(120573) = 1199091015840(120573) =

1199091015840(120572) = 0 then

119875120587(120573 minus 120572)

2

16+ 119876

(120573 minus 120572)3

6ge 1 (58)

where 119875 = max120572le119905le120573

|119901(119905)| and 119876 = max120572le119905le120573

|119902(119905)|

Proof Multiplying (9) by 1199091015840 and integrating by parts the left-hand side we have

1199091015840

(119905)11990910158401015840

(119905)10038161003816100381610038161003816

120573

120572minus int120573

120572

(11990910158401015840

(119905))2

119889119905

= minusint120573

120572

119901(119905)(1199091015840

(119905))2

119889119905 minus int120573

120572

119902(119905)119909(119905)1199091015840

(119905)119889119905

(59)

Using the assumptions that 1199091015840(120572) = 1199091015840(120573) = 0 we obtain

int120573

120572

(11990910158401015840)2

119889119905 = int120573

120572

119901(1199091015840)2

119889119905 + int120573

120572

1199021199091199091015840119889119905 (60)

By using themaximumvalues of |119901(119905)| and |119902(119905)| we canwrite(60) as

int120573

120572

1003816100381610038161003816100381611990910158401015840

(119905)10038161003816100381610038161003816

2

119889119905 le 119875int120573

120572

100381610038161003816100381610038161199091015840

(119905)10038161003816100381610038161003816

2

119889119905 + 119876int120573

120572

|119909(119905)|100381610038161003816100381610038161199091015840

(119905)10038161003816100381610038161003816119889119905

(61)

For the first term on the right-hand side int120573

120572|1199091015840(119905)|2119889119905 we

apply inequality (25) with 119910(119905) = 1199091015840(119905) (note that 1199091015840(120572) =

1199091015840(120573) = 0) and 120574 = 1 to obtain

int120573

120572

10038161003816100381610038161003816119909101584010038161003816100381610038161003816

2

119889119905 le120587(120573 minus 120572)

2

16int120573

120572

100381610038161003816100381610038161199091015840101584010038161003816100381610038161003816

2

119889119905 (62)

Applying Fink inequality (27) on int120573

120572|119909||1199091015840|119889119905 with 119896 = 0 119903 =

1 120583 = ] = 2 and 119899 = 2 we have

int120573

120572

|119909|10038161003816100381610038161003816119909101584010038161003816100381610038161003816119889119905 le

(120573 minus 120572)3

6int120573

120572

1003816100381610038161003816100381611990910158401015840(119905)

10038161003816100381610038161003816

2

119889119905 (63)

Substituting (62) and (63) into (61) we get after cancellingthe term int

120573

120572|11990910158401015840(119905)|2119889119905 that

119875120587(120573 minus 120572)

2

16+ 119876

(120573 minus 120572)3

6ge 1 (64)

which is the desired inequality (58) The proof is complete

As a special case of Theorem 5 when 119901(119905) = 0 we havethe following result

Corollary 6 Let 119909(119905) be a nontrivial solution of (56) If 119909(120572) =1199091015840(120572) = 1199091015840(120573) = 0 then

max120572le119905le120573

1003816100381610038161003816119902 (119905)1003816100381610038161003816 ge

6

(120573 minus 120572)3 (65)

In the following we prove some results related to theboundary conditions presented in (ii)

Theorem 7 Assume that 119909(119905) is a nontrivial solution of (9) If119909(120572) = 1199091015840(120572) = 11990910158401015840(120573) = 0 then

41198612

1(120572 120573) + 119862

120572ge 1 (66)

If 119909(120573) = 1199091015840(120573) = 11990910158401015840(120572) = 0 then

41198612

2(120572 120573) + 119862

120573ge 1 (67)

Mathematical Problems in Engineering 7

Proof Multiplying (9) by 1199091015840(119905) and integrating by parts theleft-hand side from 120572 to 120573 we get

11990311990910158401199091015840101584010038161003816100381610038161003816

120573

120572minus int120573

120572

119903(11990910158401015840)2

119889119905 = minusint120573

120572

119901(1199091015840)2

119889119905 minus int120573

120572

1199021199091199091015840119889119905

(68)

Using the assumption that 1199091015840(120572) = 11990910158401015840(120573) = 0 we have

int120573

120572

119903(119905)1003816100381610038161003816100381611990910158401015840

(119905)10038161003816100381610038161003816

2

119889119905 le int120573

120572

1003816100381610038161003816119901 (119905)1003816100381610038161003816100381610038161003816100381610038161199091015840(119905)

10038161003816100381610038161003816

2

119889119905

+ int120573

120572

1003816100381610038161003816119902 (119905)1003816100381610038161003816|119909 (119905)|

100381610038161003816100381610038161199091015840

(119905)10038161003816100381610038161003816119889119905

(69)

Applying inequality (12) on the integral int120573120572|119901(119905)||1199091015840(119905)|2119889119905

with 119899 = 119898 = 2 we get

int120573

120572

1003816100381610038161003816119901 (119905)1003816100381610038161003816100381610038161003816100381610038161199091015840(119905)

10038161003816100381610038161003816

2

119889119905 le 41198612

1(120572 120573) int

120573

120572

|119903 (119905)|1003816100381610038161003816100381611990910158401015840

(119905)210038161003816100381610038161003816119889119905 (70)

where1198611(120572 120573) is defined as in (32) Again applying inequality

(22) on the integral int120573120572|119902(119905)||119909(119905)||1199091015840(119905)|119889119905 we get

int120573

120572

1003816100381610038161003816119902 (119905)1003816100381610038161003816|119909 (119905)|

100381610038161003816100381610038161199091015840

(119905)10038161003816100381610038161003816119889119905 le 119862

120572int120573

120572

|119903 (119905)|1003816100381610038161003816100381611990910158401015840

(119905)210038161003816100381610038161003816119889119905 (71)

where 119862120572is defined as in (32) Substituting (70) and (71) into

(69) we obtain

int120573

120572

|119903 (119905)|1003816100381610038161003816100381611990910158401015840(119905)

10038161003816100381610038161003816

2

119889119905

le 41198612

1int120573

120572

|119903 (119905)|1003816100381610038161003816100381611990910158401015840

(119905)210038161003816100381610038161003816119889119905 + 119862

120572int120573

120572

|119903 (119905)|1003816100381610038161003816100381611990910158401015840

(119905)210038161003816100381610038161003816119889119905

(72)

By cancelling the term int120573

120572|119903(119905)||119909

10158401015840(119905)2|119889119905 we get that

41198612

1(120572 120573) + 119862

120572ge 1 (73)

which is the desired result (66) The proof (67) is similar tothe proof of (66) by using 119861

2(120572 120573) and 119862

120573instead of 119861

1(120572 120573)

and 119862120572and hence is omitted The proof is complete

In the following we apply an inequality due to Boyd [24]to obtain new results The Boyd inequality states that if 119910 isin

1198621[120572 120573] with 119910(120572) = 0 (or 119910(120573) = 0) then

int120573

120572

1003816100381610038161003816119910(119905)1003816100381610038161003816]100381610038161003816100381610038161199101015840(119905)

10038161003816100381610038161003816

120578

119889119905

le 119873 (] 120578 119904) (120573 minus 120572)][int119887

119886

100381610038161003816100381610038161199101015840(119905)

10038161003816100381610038161003816

119904

119889119905]

(]+120578)119904

(74)

where ] gt 0 119904 gt 1 0 le 120578 lt 119904 and

119873(] 120578 119904) =(119904 minus 120578)]]119903]+120578minus119904

(119904 minus 1) (] + 120578) (119868(] 120578 119904))]

119903 = ](119904 minus 1) + (119904 minus 120578)

(119904 minus 1)(] + 120578)

1119904

(75)

119868(] 120578 119904) = int1

0

1 +119904(120578 minus 1)

119904 minus 120578119905

minus(]+120578+119904])119904]

times [1 + (120578 minus 1)]1199051(]minus1)

119889119905

(76)

Note that an inequality of type (34) also holds when 119910(120572) =

119910(120573) = 0 Choose 119888 = (120572 + 120573)2 and apply (34) to [120572 119888] and[119888 120573] and by addition we obtain

int120573

120572

1003816100381610038161003816119910(119905)1003816100381610038161003816]100381610038161003816100381610038161199101015840(119905)

10038161003816100381610038161003816

120578

119889119905

le 119873 (] 120578 119904) (120573 minus 120572

2)

]

[int120573

120572

100381610038161003816100381610038161199101015840(119905)

10038161003816100381610038161003816

119904

119889119905]

(]+120578)119904

(77)

where 119873(] 120578 119904) is defined as in (75) An inequality of type(74) holds when 120578 = 119904 and 119910(120572) = 0 (or 119910(120573) = 0) In thiscase (74) becomes

int120573

120572

1003816100381610038161003816119910(119905)1003816100381610038161003816]100381610038161003816100381610038161199101015840(119905)

10038161003816100381610038161003816

120578

119889119905

le 119871 (] 120578) (120573 minus 120572)][int120573

120572

100381610038161003816100381610038161199101015840(119905)

10038161003816100381610038161003816

120578

119889119905]

(]+120578)120578

(78)

where

119871 (] 120578) =120578]120578

] + 120578(

]] + 120578

)

]120578

(Γ(((120578 + 1)120578) + (1]))Γ((120578 + 1)120578)Γ(1])

)

]

(79)

and Γ is the gamma function

Theorem 8 Assume that 119903(119905) is nonincreasing function and119909(119905) is a nontrivial solution of (9) If119909(120572) = 1199091015840(120572) = 11990910158401015840(120573) = 0then

41198612

1(120572 120573) +

8 (120573 minus 120572)

120587119903 (120573)1205952(120572 120573)(int

120573

120572

1003816100381610038161003816119902(119905)10038161003816100381610038162

119889119905)

12

ge 1 (80)

If 119909(120573) = 1199091015840(120573) = 11990910158401015840(120572) = 0 then

41198612

2(120572 120573) +

8 (120573 minus 120572)

120587119903 (120573)1205952(120572 120573)(int

120573

120572

1003816100381610038161003816119902(119905)10038161003816100381610038162

119889119905)

12

ge 1 (81)

Proof Multiplying (9) by 1199091015840(119905) and integrating by parts theleft-hand side we have that

11990311990910158401199091015840101584010038161003816100381610038161003816

120573

120572minus int120573

120572

119903(11990910158401015840)2

119889119905 = minusint120573

120572

119901(1199091015840)2

119889119905 minus int120573

120572

1199021199091199091015840119889119905

(82)

8 Mathematical Problems in Engineering

Using the assumption 1199091015840(120572) = 11990910158401015840(120573) = 0 we get that

int120573

120572

|119903 (119905)|1003816100381610038161003816100381611990910158401015840(119905)

10038161003816100381610038161003816

2

119889119905 le int120573

120572

1003816100381610038161003816119901 (119905)1003816100381610038161003816100381610038161003816100381610038161199091015840(119905)

10038161003816100381610038161003816

2

119889119905

+ int120573

120572

1003816100381610038161003816119902 (119905)1003816100381610038161003816|119909 (119905)|

100381610038161003816100381610038161199091015840

(119905)10038161003816100381610038161003816119889119905

(83)

Applying inequality (12) on the integral int120573120572|119901(119905)||1199091015840(119905)|2119889119905

with119898 = 119899 = 2 we get

int120573

120572

1003816100381610038161003816119901 (119905)1003816100381610038161003816100381610038161003816100381610038161199091015840(119905)

10038161003816100381610038161003816

2

119889119905

le 41205952(120572 120573) int

120573

120572

|119903 (119905)|1003816100381610038161003816100381611990910158401015840

(119905)210038161003816100381610038161003816119889119905

(84)

where 120595(120572 120573) is defined as in (34) Applying Schwarzrsquosinequality on the term int

120573

120572|119902(119905)||119909(119905)||1199091015840(119905)|119889119905 we get that

int120573

120572

1003816100381610038161003816119902 (119905)1003816100381610038161003816|119909 (119905)|

100381610038161003816100381610038161199091015840

(119905)10038161003816100381610038161003816119889119905

le (int120573

120572

1003816100381610038161003816119902(119905)10038161003816100381610038162

119889119905)

12

(int120573

120572

|119909|210038161003816100381610038161003816119909101584010038161003816100381610038161003816

2

119889119905)

12

(85)

Applying again inequality (78) on the integral int120573120572|119909|2|1199091015840|2119889119905

with ] = 120578 = 2 (note that 119909(120572) = 0) we obtain

int120573

120572

|119909|210038161003816100381610038161003816119909101584010038161003816100381610038161003816

2

119889119905 le4(120573 minus 120572)

2

12058721199032 (120573)[int120573

120572

119903(119905)10038161003816100381610038161003816119909101584010038161003816100381610038161003816

2

119889119905]

2

(86)

where 119903(119905) is a nonincreasing function Substituting (86) into(85) we get

int120573

120572

1003816100381610038161003816119902 (119905)1003816100381610038161003816|119909 (119905)|

100381610038161003816100381610038161199091015840

(119905)10038161003816100381610038161003816119889119905

le (int120573

120572

1003816100381610038161003816119902(119905)10038161003816100381610038162

119889119905)

12

2 (120573 minus 120572)

120587119903 (120573)int120573

120572

119903 (119905)10038161003816100381610038161003816119909101584010038161003816100381610038161003816

2

119889119905

(87)

Applying inequality (12) on the integral int120573120572119903(119905)|1199091015840|2119889119905 with

119901(119905) = 119902(119905) = 119903(119905) and119898 = 119899 = 2 we get

int120573

120572

|119903 (119905)|100381610038161003816100381610038161199091015840(119905)

10038161003816100381610038161003816

2

119889119905 le 41198612

1(120572 120573) int

120573

120572

|119903 (119905)|1003816100381610038161003816100381611990910158401015840(119905)

10038161003816100381610038161003816

2

119889119905 (88)

Substituting (88) into (87) we have

int120573

120572

1003816100381610038161003816119902 (119905)1003816100381610038161003816|119909 (119905)|

100381610038161003816100381610038161199091015840

(119905)10038161003816100381610038161003816119889119905

le (int120573

120572

1003816100381610038161003816119902(119905)10038161003816100381610038162

119889119905)

12

8(120573 minus 120572)

120587119903(120573)1205952(120572 120573)

times int120573

120572

|119903 (119905)|1003816100381610038161003816100381611990910158401015840(119905)

10038161003816100381610038161003816

2

119889119905

(89)

Substituting (84) and (89) into (83) we obtain

int120573

120572

|119903 (119905)|1003816100381610038161003816100381611990910158401015840(119905)

10038161003816100381610038161003816

2

119889119905

le 41198612

1(120572 120573) int

120573

120572

|119903(119905)|1003816100381610038161003816100381611990910158401015840

(119905)10038161003816100381610038161003816

2

119889119905

+(int120573

120572

1003816100381610038161003816119902(119905)10038161003816100381610038162

119889119905)

12

8 (120573 minus 120572)

120587119903 (120573)1205952(120572 120573)

times int120573

120572

|119903 (119905)|1003816100381610038161003816100381611990910158401015840(119905)

10038161003816100381610038161003816

2

119889119905

(90)

Cancelling the term int120573

120572|119903(119905)||11990910158401015840(119905)|2119889119905 we obtain

41198612

1(120572 120573) +

8 (120573 minus 120572)

120587119903 (120573)1205952(120572 120573)(int

120573

120572

1003816100381610038161003816119902(119905)10038161003816100381610038162

119889119905)

12

ge 1

(91)

which is the desired inequality (80) The proof of (81) issimilar to the proof of (80) by using 119861

2(120572 120573) instead of

1198611(120572 120573) The proof is complete

Conflict of Interests

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

References

[1] E Picard ldquoMemoire sur la theorie des equations aux derivespartielleset la methode desapproximations successivesrdquo Journalde Mathematiques Pures et Appliquees vol 6 pp 145ndash210 1890

[2] E Picard ldquoSur lrsquoapplication des methodes drsquoapproximationssuccessives a lrsquoetude de certaines equations differentielles ordi-nairesrdquo Journal deMathematiques Pures et Appliquees vol 9 pp217ndash271 1893

[3] O Niccoletti ldquoSulle consizioni iniziali che determinano gliintegrale delle equazioni diff eren ziali ordinarierdquo Atti dellaAccademia delle Scienze di Torino vol 33 pp 746ndash759 1897

[4] C de la Vallee Poussin ldquoSur lrsquoequation differentielle lineaire dusecond ordrerdquo Journal de Mathematiques Pures et Appliqueesvol 8 pp 125ndash144 1929

[5] A M Lyapunov ldquoProbleme general de la stabilite du mouve-mentrdquo Annales de la Faculte des sciences de Toulouse vol 9 pp203ndash474 1907

[6] S H Saker ldquoSome new disconjugacy criteria for second orderdifferential equations with a middle termrdquo Bulletin Mathema-tique de la Societe des Sciences Mathematiques de Roumanie vol57 no 1 pp 109ndash120 2014

[7] G Casadei ldquoSul teorema di unicita di De La Vallee Poussinper equazioni differenziali del terzo ordinerdquo Rendiconti delSeminario Matematico della Universita di Padova vol 41 pp300ndash315 1968

[8] A Lasota ldquoSur la distance entre les zeros de lrsquoequation dif-ferentielle lineaire du troisieme ordrerdquo Annales Polonici Mathe-matici vol 13 pp 129ndash132 1963

Mathematical Problems in Engineering 9

[9] RMMathsen ldquoA disconjugacy condition for119910101584010158401015840+119886211991010158401015840+119886

11199101015840+

1198860119910 = 0rdquo Proceedings of the AmericanMathematical Society vol

17 pp 627ndash632 1966[10] B G Pachpatte ldquoOn the zeros of solutions of certain differential

equationsrdquo Demonstratio Mathematica vol 25 no 4 pp 825ndash833 1992

[11] B G Pachpatte ldquoLyapunov type integral inequalities for certaindifferential equationsrdquo Georgian Mathematical Journal vol 4no 2 pp 139ndash148 1997

[12] S Panigrahi ldquoLyapunov-type integral inequalities for certainhigher-order differential equationsrdquo Electronic Journal of Differ-ential Equations vol 2009 no 28 pp 1ndash14 2009

[13] N Parhi and S Panigrahi ldquoOn Liapunov-type inequality forthird-order differential equationsrdquo Journal of MathematicalAnalysis and Applications vol 233 no 2 pp 445ndash460 1999

[14] N Parhi and S Panigrahi ldquoDisfocality and Liapunov-typeinequalities for third-order equationsrdquo Applied MathematicsLetters vol 16 no 2 pp 227ndash233 2003

[15] U Richard ldquoMetodi diversi per ottenere disequaglianze allade la Vallee Poussin nelle equazioni differenziali ordinarie delsecondo e terzo ordinerdquo Rendiconti del Seminario MatematicoUniversita e Politecnico di Torino vol 27 pp 35ndash68 1968

[16] X Yang ldquoOn Liapunov-type inequality for certain higher-orderdifferential equationsrdquo Applied Mathematics and Computationvol 134 no 2-3 pp 307ndash317 2003

[17] R P Agarwal and P R Krishnamurthy ldquoOn the uniquenessof solution of nonlinear boundary value problemsrdquo Journal ofMathematical Sciences vol 10 no 1 pp 17ndash31 1976

[18] R P Agarwal and P Y PangOpial Inequalities with Applicationsin Differential and Difference Equation Kluwer AcademicDordrecht The Netherlands 1995

[19] R P Agarwal and P Y Pang ldquoRemarks on the generalizationsof Opialrsquos inequalityrdquo Journal of Mathematical Analysis andApplications vol 190 no 2 pp 559ndash577 1995

[20] P R Beesack and K M Das ldquoExtensions of Opialrsquos inequalityrdquoPacific Journal of Mathematics vol 26 pp 215ndash232 1968

[21] S Clark and D Hinton ldquoSome disconjugacy criteria for dif-ferential equations with oscillatory coefficientsrdquoMathematischeNachrichten vol 278 no 12-13 pp 1476ndash1489 2005

[22] A M Fink ldquoOn Opials inequality for 119891(119899)rdquo Proceedings of theAmerican Mathematical Society vol 155 pp 177ndash181 1992

[23] A Kufner and L-E Persson Weighted Inequalities of HardyType World Scientific River Edge NJ USA 2003

[24] D W Boyd ldquoBest constants in a class of integral inequalitiesrdquoPacific Journal of Mathematics vol 30 pp 367ndash383 1969

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Mathematical Problems in Engineering

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Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Stochastic AnalysisInternational Journal of

6 Mathematical Problems in Engineering

Applying Hardyrsquos inequality (12) on the integralint120573

120572|1198631(119905)||1199091015840(119905)

2|119889119905 we obtain

int120573

120572

10038161003816100381610038161198631 (119905)1003816100381610038161003816100381610038161003816100381610038161199091015840(119905)

10038161003816100381610038161003816

2

119889119905 le 41198602

1(120572 120573) int

120573

120572

119903 (119905)1003816100381610038161003816100381611990910158401015840(119905)

10038161003816100381610038161003816

2

119889119905 (53)

where 1199091015840(120572) = 1199091015840(120573) = 0 Substituting (52) and (53) into (48)we obtain

int120573

120572

|119903(119905)|1003816100381610038161003816100381611990910158401015840

(119905)10038161003816100381610038161003816

2

119889119905

le 41198602

1(120572 120573) int

120573

120572

119903 (119905)1003816100381610038161003816100381611990910158401015840(119905)

10038161003816100381610038161003816

2

119889119905

+(120573 minus 120572)

32

radic3119903 (120573)(int120573

120572

10038161003816100381610038161198761(119905)10038161003816100381610038162

119889119905)

12

int120573

120572

119903 (119905)1003816100381610038161003816100381611990910158401015840(119905)

10038161003816100381610038161003816

2

119889119905

(54)

The desired inequality (47) followed by cancelling the termint120573

120572|119903(119905)||11990910158401015840(119905)|2119889119905 The proof is complete

The proof of the following theorem is similar to the proofof Theorem 2

Theorem 3 Assume that 119903(119905) is a nonincreasing function andsuppose that 119909(119905) is a solution of (9) If 1199091015840(120572) = 119909(120573) = 1199091015840(120573) =

0 then

41198602

2(120572 120573) +

(120573 minus 120572)32

radic3119903 (120573)(int120573

120572

10038161003816100381610038161198761(119905)10038161003816100381610038162

119889119905)

12

ge 1 (55)

where 1198602(120572 120573) is defined as in (31) for 119894 = 2 and 119876

2(119905) is

defined as in (30)

One can use Theorems 2 and 3 to obtain some differentspecial cases For example as a special case of Theorem 2when 119903(119905) = 1 and 119901(119905) = 0 we have the following result

Corollary 4 If 119909(119905) is a nontrivial solution of

119909101584010158401015840

(119905) + 119902 (119905) 119909 (119905) = 0 119905 isin [120572 120573] (56)

which satisfies 119909(120572) = 1199091015840(120572) = 1199091015840(120573) = 0 then

int120573

120572

10038161003816100381610038161198761(119905)10038161003816100381610038162

119889119905 ge3

(120573 minus 120572)3 (57)

where 1198761(119905) is defined as in (29)

Now we will prove a new result when 119903(119905) = 1

Theorem 5 Suppose that 119903(119905) = 1 and assume that 119909(119905) is asolution of (9) If 119909(120572) = 1199091015840(120572) = 1199091015840(120573) = 0 or 119909(120573) = 1199091015840(120573) =

1199091015840(120572) = 0 then

119875120587(120573 minus 120572)

2

16+ 119876

(120573 minus 120572)3

6ge 1 (58)

where 119875 = max120572le119905le120573

|119901(119905)| and 119876 = max120572le119905le120573

|119902(119905)|

Proof Multiplying (9) by 1199091015840 and integrating by parts the left-hand side we have

1199091015840

(119905)11990910158401015840

(119905)10038161003816100381610038161003816

120573

120572minus int120573

120572

(11990910158401015840

(119905))2

119889119905

= minusint120573

120572

119901(119905)(1199091015840

(119905))2

119889119905 minus int120573

120572

119902(119905)119909(119905)1199091015840

(119905)119889119905

(59)

Using the assumptions that 1199091015840(120572) = 1199091015840(120573) = 0 we obtain

int120573

120572

(11990910158401015840)2

119889119905 = int120573

120572

119901(1199091015840)2

119889119905 + int120573

120572

1199021199091199091015840119889119905 (60)

By using themaximumvalues of |119901(119905)| and |119902(119905)| we canwrite(60) as

int120573

120572

1003816100381610038161003816100381611990910158401015840

(119905)10038161003816100381610038161003816

2

119889119905 le 119875int120573

120572

100381610038161003816100381610038161199091015840

(119905)10038161003816100381610038161003816

2

119889119905 + 119876int120573

120572

|119909(119905)|100381610038161003816100381610038161199091015840

(119905)10038161003816100381610038161003816119889119905

(61)

For the first term on the right-hand side int120573

120572|1199091015840(119905)|2119889119905 we

apply inequality (25) with 119910(119905) = 1199091015840(119905) (note that 1199091015840(120572) =

1199091015840(120573) = 0) and 120574 = 1 to obtain

int120573

120572

10038161003816100381610038161003816119909101584010038161003816100381610038161003816

2

119889119905 le120587(120573 minus 120572)

2

16int120573

120572

100381610038161003816100381610038161199091015840101584010038161003816100381610038161003816

2

119889119905 (62)

Applying Fink inequality (27) on int120573

120572|119909||1199091015840|119889119905 with 119896 = 0 119903 =

1 120583 = ] = 2 and 119899 = 2 we have

int120573

120572

|119909|10038161003816100381610038161003816119909101584010038161003816100381610038161003816119889119905 le

(120573 minus 120572)3

6int120573

120572

1003816100381610038161003816100381611990910158401015840(119905)

10038161003816100381610038161003816

2

119889119905 (63)

Substituting (62) and (63) into (61) we get after cancellingthe term int

120573

120572|11990910158401015840(119905)|2119889119905 that

119875120587(120573 minus 120572)

2

16+ 119876

(120573 minus 120572)3

6ge 1 (64)

which is the desired inequality (58) The proof is complete

As a special case of Theorem 5 when 119901(119905) = 0 we havethe following result

Corollary 6 Let 119909(119905) be a nontrivial solution of (56) If 119909(120572) =1199091015840(120572) = 1199091015840(120573) = 0 then

max120572le119905le120573

1003816100381610038161003816119902 (119905)1003816100381610038161003816 ge

6

(120573 minus 120572)3 (65)

In the following we prove some results related to theboundary conditions presented in (ii)

Theorem 7 Assume that 119909(119905) is a nontrivial solution of (9) If119909(120572) = 1199091015840(120572) = 11990910158401015840(120573) = 0 then

41198612

1(120572 120573) + 119862

120572ge 1 (66)

If 119909(120573) = 1199091015840(120573) = 11990910158401015840(120572) = 0 then

41198612

2(120572 120573) + 119862

120573ge 1 (67)

Mathematical Problems in Engineering 7

Proof Multiplying (9) by 1199091015840(119905) and integrating by parts theleft-hand side from 120572 to 120573 we get

11990311990910158401199091015840101584010038161003816100381610038161003816

120573

120572minus int120573

120572

119903(11990910158401015840)2

119889119905 = minusint120573

120572

119901(1199091015840)2

119889119905 minus int120573

120572

1199021199091199091015840119889119905

(68)

Using the assumption that 1199091015840(120572) = 11990910158401015840(120573) = 0 we have

int120573

120572

119903(119905)1003816100381610038161003816100381611990910158401015840

(119905)10038161003816100381610038161003816

2

119889119905 le int120573

120572

1003816100381610038161003816119901 (119905)1003816100381610038161003816100381610038161003816100381610038161199091015840(119905)

10038161003816100381610038161003816

2

119889119905

+ int120573

120572

1003816100381610038161003816119902 (119905)1003816100381610038161003816|119909 (119905)|

100381610038161003816100381610038161199091015840

(119905)10038161003816100381610038161003816119889119905

(69)

Applying inequality (12) on the integral int120573120572|119901(119905)||1199091015840(119905)|2119889119905

with 119899 = 119898 = 2 we get

int120573

120572

1003816100381610038161003816119901 (119905)1003816100381610038161003816100381610038161003816100381610038161199091015840(119905)

10038161003816100381610038161003816

2

119889119905 le 41198612

1(120572 120573) int

120573

120572

|119903 (119905)|1003816100381610038161003816100381611990910158401015840

(119905)210038161003816100381610038161003816119889119905 (70)

where1198611(120572 120573) is defined as in (32) Again applying inequality

(22) on the integral int120573120572|119902(119905)||119909(119905)||1199091015840(119905)|119889119905 we get

int120573

120572

1003816100381610038161003816119902 (119905)1003816100381610038161003816|119909 (119905)|

100381610038161003816100381610038161199091015840

(119905)10038161003816100381610038161003816119889119905 le 119862

120572int120573

120572

|119903 (119905)|1003816100381610038161003816100381611990910158401015840

(119905)210038161003816100381610038161003816119889119905 (71)

where 119862120572is defined as in (32) Substituting (70) and (71) into

(69) we obtain

int120573

120572

|119903 (119905)|1003816100381610038161003816100381611990910158401015840(119905)

10038161003816100381610038161003816

2

119889119905

le 41198612

1int120573

120572

|119903 (119905)|1003816100381610038161003816100381611990910158401015840

(119905)210038161003816100381610038161003816119889119905 + 119862

120572int120573

120572

|119903 (119905)|1003816100381610038161003816100381611990910158401015840

(119905)210038161003816100381610038161003816119889119905

(72)

By cancelling the term int120573

120572|119903(119905)||119909

10158401015840(119905)2|119889119905 we get that

41198612

1(120572 120573) + 119862

120572ge 1 (73)

which is the desired result (66) The proof (67) is similar tothe proof of (66) by using 119861

2(120572 120573) and 119862

120573instead of 119861

1(120572 120573)

and 119862120572and hence is omitted The proof is complete

In the following we apply an inequality due to Boyd [24]to obtain new results The Boyd inequality states that if 119910 isin

1198621[120572 120573] with 119910(120572) = 0 (or 119910(120573) = 0) then

int120573

120572

1003816100381610038161003816119910(119905)1003816100381610038161003816]100381610038161003816100381610038161199101015840(119905)

10038161003816100381610038161003816

120578

119889119905

le 119873 (] 120578 119904) (120573 minus 120572)][int119887

119886

100381610038161003816100381610038161199101015840(119905)

10038161003816100381610038161003816

119904

119889119905]

(]+120578)119904

(74)

where ] gt 0 119904 gt 1 0 le 120578 lt 119904 and

119873(] 120578 119904) =(119904 minus 120578)]]119903]+120578minus119904

(119904 minus 1) (] + 120578) (119868(] 120578 119904))]

119903 = ](119904 minus 1) + (119904 minus 120578)

(119904 minus 1)(] + 120578)

1119904

(75)

119868(] 120578 119904) = int1

0

1 +119904(120578 minus 1)

119904 minus 120578119905

minus(]+120578+119904])119904]

times [1 + (120578 minus 1)]1199051(]minus1)

119889119905

(76)

Note that an inequality of type (34) also holds when 119910(120572) =

119910(120573) = 0 Choose 119888 = (120572 + 120573)2 and apply (34) to [120572 119888] and[119888 120573] and by addition we obtain

int120573

120572

1003816100381610038161003816119910(119905)1003816100381610038161003816]100381610038161003816100381610038161199101015840(119905)

10038161003816100381610038161003816

120578

119889119905

le 119873 (] 120578 119904) (120573 minus 120572

2)

]

[int120573

120572

100381610038161003816100381610038161199101015840(119905)

10038161003816100381610038161003816

119904

119889119905]

(]+120578)119904

(77)

where 119873(] 120578 119904) is defined as in (75) An inequality of type(74) holds when 120578 = 119904 and 119910(120572) = 0 (or 119910(120573) = 0) In thiscase (74) becomes

int120573

120572

1003816100381610038161003816119910(119905)1003816100381610038161003816]100381610038161003816100381610038161199101015840(119905)

10038161003816100381610038161003816

120578

119889119905

le 119871 (] 120578) (120573 minus 120572)][int120573

120572

100381610038161003816100381610038161199101015840(119905)

10038161003816100381610038161003816

120578

119889119905]

(]+120578)120578

(78)

where

119871 (] 120578) =120578]120578

] + 120578(

]] + 120578

)

]120578

(Γ(((120578 + 1)120578) + (1]))Γ((120578 + 1)120578)Γ(1])

)

]

(79)

and Γ is the gamma function

Theorem 8 Assume that 119903(119905) is nonincreasing function and119909(119905) is a nontrivial solution of (9) If119909(120572) = 1199091015840(120572) = 11990910158401015840(120573) = 0then

41198612

1(120572 120573) +

8 (120573 minus 120572)

120587119903 (120573)1205952(120572 120573)(int

120573

120572

1003816100381610038161003816119902(119905)10038161003816100381610038162

119889119905)

12

ge 1 (80)

If 119909(120573) = 1199091015840(120573) = 11990910158401015840(120572) = 0 then

41198612

2(120572 120573) +

8 (120573 minus 120572)

120587119903 (120573)1205952(120572 120573)(int

120573

120572

1003816100381610038161003816119902(119905)10038161003816100381610038162

119889119905)

12

ge 1 (81)

Proof Multiplying (9) by 1199091015840(119905) and integrating by parts theleft-hand side we have that

11990311990910158401199091015840101584010038161003816100381610038161003816

120573

120572minus int120573

120572

119903(11990910158401015840)2

119889119905 = minusint120573

120572

119901(1199091015840)2

119889119905 minus int120573

120572

1199021199091199091015840119889119905

(82)

8 Mathematical Problems in Engineering

Using the assumption 1199091015840(120572) = 11990910158401015840(120573) = 0 we get that

int120573

120572

|119903 (119905)|1003816100381610038161003816100381611990910158401015840(119905)

10038161003816100381610038161003816

2

119889119905 le int120573

120572

1003816100381610038161003816119901 (119905)1003816100381610038161003816100381610038161003816100381610038161199091015840(119905)

10038161003816100381610038161003816

2

119889119905

+ int120573

120572

1003816100381610038161003816119902 (119905)1003816100381610038161003816|119909 (119905)|

100381610038161003816100381610038161199091015840

(119905)10038161003816100381610038161003816119889119905

(83)

Applying inequality (12) on the integral int120573120572|119901(119905)||1199091015840(119905)|2119889119905

with119898 = 119899 = 2 we get

int120573

120572

1003816100381610038161003816119901 (119905)1003816100381610038161003816100381610038161003816100381610038161199091015840(119905)

10038161003816100381610038161003816

2

119889119905

le 41205952(120572 120573) int

120573

120572

|119903 (119905)|1003816100381610038161003816100381611990910158401015840

(119905)210038161003816100381610038161003816119889119905

(84)

where 120595(120572 120573) is defined as in (34) Applying Schwarzrsquosinequality on the term int

120573

120572|119902(119905)||119909(119905)||1199091015840(119905)|119889119905 we get that

int120573

120572

1003816100381610038161003816119902 (119905)1003816100381610038161003816|119909 (119905)|

100381610038161003816100381610038161199091015840

(119905)10038161003816100381610038161003816119889119905

le (int120573

120572

1003816100381610038161003816119902(119905)10038161003816100381610038162

119889119905)

12

(int120573

120572

|119909|210038161003816100381610038161003816119909101584010038161003816100381610038161003816

2

119889119905)

12

(85)

Applying again inequality (78) on the integral int120573120572|119909|2|1199091015840|2119889119905

with ] = 120578 = 2 (note that 119909(120572) = 0) we obtain

int120573

120572

|119909|210038161003816100381610038161003816119909101584010038161003816100381610038161003816

2

119889119905 le4(120573 minus 120572)

2

12058721199032 (120573)[int120573

120572

119903(119905)10038161003816100381610038161003816119909101584010038161003816100381610038161003816

2

119889119905]

2

(86)

where 119903(119905) is a nonincreasing function Substituting (86) into(85) we get

int120573

120572

1003816100381610038161003816119902 (119905)1003816100381610038161003816|119909 (119905)|

100381610038161003816100381610038161199091015840

(119905)10038161003816100381610038161003816119889119905

le (int120573

120572

1003816100381610038161003816119902(119905)10038161003816100381610038162

119889119905)

12

2 (120573 minus 120572)

120587119903 (120573)int120573

120572

119903 (119905)10038161003816100381610038161003816119909101584010038161003816100381610038161003816

2

119889119905

(87)

Applying inequality (12) on the integral int120573120572119903(119905)|1199091015840|2119889119905 with

119901(119905) = 119902(119905) = 119903(119905) and119898 = 119899 = 2 we get

int120573

120572

|119903 (119905)|100381610038161003816100381610038161199091015840(119905)

10038161003816100381610038161003816

2

119889119905 le 41198612

1(120572 120573) int

120573

120572

|119903 (119905)|1003816100381610038161003816100381611990910158401015840(119905)

10038161003816100381610038161003816

2

119889119905 (88)

Substituting (88) into (87) we have

int120573

120572

1003816100381610038161003816119902 (119905)1003816100381610038161003816|119909 (119905)|

100381610038161003816100381610038161199091015840

(119905)10038161003816100381610038161003816119889119905

le (int120573

120572

1003816100381610038161003816119902(119905)10038161003816100381610038162

119889119905)

12

8(120573 minus 120572)

120587119903(120573)1205952(120572 120573)

times int120573

120572

|119903 (119905)|1003816100381610038161003816100381611990910158401015840(119905)

10038161003816100381610038161003816

2

119889119905

(89)

Substituting (84) and (89) into (83) we obtain

int120573

120572

|119903 (119905)|1003816100381610038161003816100381611990910158401015840(119905)

10038161003816100381610038161003816

2

119889119905

le 41198612

1(120572 120573) int

120573

120572

|119903(119905)|1003816100381610038161003816100381611990910158401015840

(119905)10038161003816100381610038161003816

2

119889119905

+(int120573

120572

1003816100381610038161003816119902(119905)10038161003816100381610038162

119889119905)

12

8 (120573 minus 120572)

120587119903 (120573)1205952(120572 120573)

times int120573

120572

|119903 (119905)|1003816100381610038161003816100381611990910158401015840(119905)

10038161003816100381610038161003816

2

119889119905

(90)

Cancelling the term int120573

120572|119903(119905)||11990910158401015840(119905)|2119889119905 we obtain

41198612

1(120572 120573) +

8 (120573 minus 120572)

120587119903 (120573)1205952(120572 120573)(int

120573

120572

1003816100381610038161003816119902(119905)10038161003816100381610038162

119889119905)

12

ge 1

(91)

which is the desired inequality (80) The proof of (81) issimilar to the proof of (80) by using 119861

2(120572 120573) instead of

1198611(120572 120573) The proof is complete

Conflict of Interests

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

References

[1] E Picard ldquoMemoire sur la theorie des equations aux derivespartielleset la methode desapproximations successivesrdquo Journalde Mathematiques Pures et Appliquees vol 6 pp 145ndash210 1890

[2] E Picard ldquoSur lrsquoapplication des methodes drsquoapproximationssuccessives a lrsquoetude de certaines equations differentielles ordi-nairesrdquo Journal deMathematiques Pures et Appliquees vol 9 pp217ndash271 1893

[3] O Niccoletti ldquoSulle consizioni iniziali che determinano gliintegrale delle equazioni diff eren ziali ordinarierdquo Atti dellaAccademia delle Scienze di Torino vol 33 pp 746ndash759 1897

[4] C de la Vallee Poussin ldquoSur lrsquoequation differentielle lineaire dusecond ordrerdquo Journal de Mathematiques Pures et Appliqueesvol 8 pp 125ndash144 1929

[5] A M Lyapunov ldquoProbleme general de la stabilite du mouve-mentrdquo Annales de la Faculte des sciences de Toulouse vol 9 pp203ndash474 1907

[6] S H Saker ldquoSome new disconjugacy criteria for second orderdifferential equations with a middle termrdquo Bulletin Mathema-tique de la Societe des Sciences Mathematiques de Roumanie vol57 no 1 pp 109ndash120 2014

[7] G Casadei ldquoSul teorema di unicita di De La Vallee Poussinper equazioni differenziali del terzo ordinerdquo Rendiconti delSeminario Matematico della Universita di Padova vol 41 pp300ndash315 1968

[8] A Lasota ldquoSur la distance entre les zeros de lrsquoequation dif-ferentielle lineaire du troisieme ordrerdquo Annales Polonici Mathe-matici vol 13 pp 129ndash132 1963

Mathematical Problems in Engineering 9

[9] RMMathsen ldquoA disconjugacy condition for119910101584010158401015840+119886211991010158401015840+119886

11199101015840+

1198860119910 = 0rdquo Proceedings of the AmericanMathematical Society vol

17 pp 627ndash632 1966[10] B G Pachpatte ldquoOn the zeros of solutions of certain differential

equationsrdquo Demonstratio Mathematica vol 25 no 4 pp 825ndash833 1992

[11] B G Pachpatte ldquoLyapunov type integral inequalities for certaindifferential equationsrdquo Georgian Mathematical Journal vol 4no 2 pp 139ndash148 1997

[12] S Panigrahi ldquoLyapunov-type integral inequalities for certainhigher-order differential equationsrdquo Electronic Journal of Differ-ential Equations vol 2009 no 28 pp 1ndash14 2009

[13] N Parhi and S Panigrahi ldquoOn Liapunov-type inequality forthird-order differential equationsrdquo Journal of MathematicalAnalysis and Applications vol 233 no 2 pp 445ndash460 1999

[14] N Parhi and S Panigrahi ldquoDisfocality and Liapunov-typeinequalities for third-order equationsrdquo Applied MathematicsLetters vol 16 no 2 pp 227ndash233 2003

[15] U Richard ldquoMetodi diversi per ottenere disequaglianze allade la Vallee Poussin nelle equazioni differenziali ordinarie delsecondo e terzo ordinerdquo Rendiconti del Seminario MatematicoUniversita e Politecnico di Torino vol 27 pp 35ndash68 1968

[16] X Yang ldquoOn Liapunov-type inequality for certain higher-orderdifferential equationsrdquo Applied Mathematics and Computationvol 134 no 2-3 pp 307ndash317 2003

[17] R P Agarwal and P R Krishnamurthy ldquoOn the uniquenessof solution of nonlinear boundary value problemsrdquo Journal ofMathematical Sciences vol 10 no 1 pp 17ndash31 1976

[18] R P Agarwal and P Y PangOpial Inequalities with Applicationsin Differential and Difference Equation Kluwer AcademicDordrecht The Netherlands 1995

[19] R P Agarwal and P Y Pang ldquoRemarks on the generalizationsof Opialrsquos inequalityrdquo Journal of Mathematical Analysis andApplications vol 190 no 2 pp 559ndash577 1995

[20] P R Beesack and K M Das ldquoExtensions of Opialrsquos inequalityrdquoPacific Journal of Mathematics vol 26 pp 215ndash232 1968

[21] S Clark and D Hinton ldquoSome disconjugacy criteria for dif-ferential equations with oscillatory coefficientsrdquoMathematischeNachrichten vol 278 no 12-13 pp 1476ndash1489 2005

[22] A M Fink ldquoOn Opials inequality for 119891(119899)rdquo Proceedings of theAmerican Mathematical Society vol 155 pp 177ndash181 1992

[23] A Kufner and L-E Persson Weighted Inequalities of HardyType World Scientific River Edge NJ USA 2003

[24] D W Boyd ldquoBest constants in a class of integral inequalitiesrdquoPacific Journal of Mathematics vol 30 pp 367ndash383 1969

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

Mathematical Problems in Engineering 7

Proof Multiplying (9) by 1199091015840(119905) and integrating by parts theleft-hand side from 120572 to 120573 we get

11990311990910158401199091015840101584010038161003816100381610038161003816

120573

120572minus int120573

120572

119903(11990910158401015840)2

119889119905 = minusint120573

120572

119901(1199091015840)2

119889119905 minus int120573

120572

1199021199091199091015840119889119905

(68)

Using the assumption that 1199091015840(120572) = 11990910158401015840(120573) = 0 we have

int120573

120572

119903(119905)1003816100381610038161003816100381611990910158401015840

(119905)10038161003816100381610038161003816

2

119889119905 le int120573

120572

1003816100381610038161003816119901 (119905)1003816100381610038161003816100381610038161003816100381610038161199091015840(119905)

10038161003816100381610038161003816

2

119889119905

+ int120573

120572

1003816100381610038161003816119902 (119905)1003816100381610038161003816|119909 (119905)|

100381610038161003816100381610038161199091015840

(119905)10038161003816100381610038161003816119889119905

(69)

Applying inequality (12) on the integral int120573120572|119901(119905)||1199091015840(119905)|2119889119905

with 119899 = 119898 = 2 we get

int120573

120572

1003816100381610038161003816119901 (119905)1003816100381610038161003816100381610038161003816100381610038161199091015840(119905)

10038161003816100381610038161003816

2

119889119905 le 41198612

1(120572 120573) int

120573

120572

|119903 (119905)|1003816100381610038161003816100381611990910158401015840

(119905)210038161003816100381610038161003816119889119905 (70)

where1198611(120572 120573) is defined as in (32) Again applying inequality

(22) on the integral int120573120572|119902(119905)||119909(119905)||1199091015840(119905)|119889119905 we get

int120573

120572

1003816100381610038161003816119902 (119905)1003816100381610038161003816|119909 (119905)|

100381610038161003816100381610038161199091015840

(119905)10038161003816100381610038161003816119889119905 le 119862

120572int120573

120572

|119903 (119905)|1003816100381610038161003816100381611990910158401015840

(119905)210038161003816100381610038161003816119889119905 (71)

where 119862120572is defined as in (32) Substituting (70) and (71) into

(69) we obtain

int120573

120572

|119903 (119905)|1003816100381610038161003816100381611990910158401015840(119905)

10038161003816100381610038161003816

2

119889119905

le 41198612

1int120573

120572

|119903 (119905)|1003816100381610038161003816100381611990910158401015840

(119905)210038161003816100381610038161003816119889119905 + 119862

120572int120573

120572

|119903 (119905)|1003816100381610038161003816100381611990910158401015840

(119905)210038161003816100381610038161003816119889119905

(72)

By cancelling the term int120573

120572|119903(119905)||119909

10158401015840(119905)2|119889119905 we get that

41198612

1(120572 120573) + 119862

120572ge 1 (73)

which is the desired result (66) The proof (67) is similar tothe proof of (66) by using 119861

2(120572 120573) and 119862

120573instead of 119861

1(120572 120573)

and 119862120572and hence is omitted The proof is complete

In the following we apply an inequality due to Boyd [24]to obtain new results The Boyd inequality states that if 119910 isin

1198621[120572 120573] with 119910(120572) = 0 (or 119910(120573) = 0) then

int120573

120572

1003816100381610038161003816119910(119905)1003816100381610038161003816]100381610038161003816100381610038161199101015840(119905)

10038161003816100381610038161003816

120578

119889119905

le 119873 (] 120578 119904) (120573 minus 120572)][int119887

119886

100381610038161003816100381610038161199101015840(119905)

10038161003816100381610038161003816

119904

119889119905]

(]+120578)119904

(74)

where ] gt 0 119904 gt 1 0 le 120578 lt 119904 and

119873(] 120578 119904) =(119904 minus 120578)]]119903]+120578minus119904

(119904 minus 1) (] + 120578) (119868(] 120578 119904))]

119903 = ](119904 minus 1) + (119904 minus 120578)

(119904 minus 1)(] + 120578)

1119904

(75)

119868(] 120578 119904) = int1

0

1 +119904(120578 minus 1)

119904 minus 120578119905

minus(]+120578+119904])119904]

times [1 + (120578 minus 1)]1199051(]minus1)

119889119905

(76)

Note that an inequality of type (34) also holds when 119910(120572) =

119910(120573) = 0 Choose 119888 = (120572 + 120573)2 and apply (34) to [120572 119888] and[119888 120573] and by addition we obtain

int120573

120572

1003816100381610038161003816119910(119905)1003816100381610038161003816]100381610038161003816100381610038161199101015840(119905)

10038161003816100381610038161003816

120578

119889119905

le 119873 (] 120578 119904) (120573 minus 120572

2)

]

[int120573

120572

100381610038161003816100381610038161199101015840(119905)

10038161003816100381610038161003816

119904

119889119905]

(]+120578)119904

(77)

where 119873(] 120578 119904) is defined as in (75) An inequality of type(74) holds when 120578 = 119904 and 119910(120572) = 0 (or 119910(120573) = 0) In thiscase (74) becomes

int120573

120572

1003816100381610038161003816119910(119905)1003816100381610038161003816]100381610038161003816100381610038161199101015840(119905)

10038161003816100381610038161003816

120578

119889119905

le 119871 (] 120578) (120573 minus 120572)][int120573

120572

100381610038161003816100381610038161199101015840(119905)

10038161003816100381610038161003816

120578

119889119905]

(]+120578)120578

(78)

where

119871 (] 120578) =120578]120578

] + 120578(

]] + 120578

)

]120578

(Γ(((120578 + 1)120578) + (1]))Γ((120578 + 1)120578)Γ(1])

)

]

(79)

and Γ is the gamma function

Theorem 8 Assume that 119903(119905) is nonincreasing function and119909(119905) is a nontrivial solution of (9) If119909(120572) = 1199091015840(120572) = 11990910158401015840(120573) = 0then

41198612

1(120572 120573) +

8 (120573 minus 120572)

120587119903 (120573)1205952(120572 120573)(int

120573

120572

1003816100381610038161003816119902(119905)10038161003816100381610038162

119889119905)

12

ge 1 (80)

If 119909(120573) = 1199091015840(120573) = 11990910158401015840(120572) = 0 then

41198612

2(120572 120573) +

8 (120573 minus 120572)

120587119903 (120573)1205952(120572 120573)(int

120573

120572

1003816100381610038161003816119902(119905)10038161003816100381610038162

119889119905)

12

ge 1 (81)

Proof Multiplying (9) by 1199091015840(119905) and integrating by parts theleft-hand side we have that

11990311990910158401199091015840101584010038161003816100381610038161003816

120573

120572minus int120573

120572

119903(11990910158401015840)2

119889119905 = minusint120573

120572

119901(1199091015840)2

119889119905 minus int120573

120572

1199021199091199091015840119889119905

(82)

8 Mathematical Problems in Engineering

Using the assumption 1199091015840(120572) = 11990910158401015840(120573) = 0 we get that

int120573

120572

|119903 (119905)|1003816100381610038161003816100381611990910158401015840(119905)

10038161003816100381610038161003816

2

119889119905 le int120573

120572

1003816100381610038161003816119901 (119905)1003816100381610038161003816100381610038161003816100381610038161199091015840(119905)

10038161003816100381610038161003816

2

119889119905

+ int120573

120572

1003816100381610038161003816119902 (119905)1003816100381610038161003816|119909 (119905)|

100381610038161003816100381610038161199091015840

(119905)10038161003816100381610038161003816119889119905

(83)

Applying inequality (12) on the integral int120573120572|119901(119905)||1199091015840(119905)|2119889119905

with119898 = 119899 = 2 we get

int120573

120572

1003816100381610038161003816119901 (119905)1003816100381610038161003816100381610038161003816100381610038161199091015840(119905)

10038161003816100381610038161003816

2

119889119905

le 41205952(120572 120573) int

120573

120572

|119903 (119905)|1003816100381610038161003816100381611990910158401015840

(119905)210038161003816100381610038161003816119889119905

(84)

where 120595(120572 120573) is defined as in (34) Applying Schwarzrsquosinequality on the term int

120573

120572|119902(119905)||119909(119905)||1199091015840(119905)|119889119905 we get that

int120573

120572

1003816100381610038161003816119902 (119905)1003816100381610038161003816|119909 (119905)|

100381610038161003816100381610038161199091015840

(119905)10038161003816100381610038161003816119889119905

le (int120573

120572

1003816100381610038161003816119902(119905)10038161003816100381610038162

119889119905)

12

(int120573

120572

|119909|210038161003816100381610038161003816119909101584010038161003816100381610038161003816

2

119889119905)

12

(85)

Applying again inequality (78) on the integral int120573120572|119909|2|1199091015840|2119889119905

with ] = 120578 = 2 (note that 119909(120572) = 0) we obtain

int120573

120572

|119909|210038161003816100381610038161003816119909101584010038161003816100381610038161003816

2

119889119905 le4(120573 minus 120572)

2

12058721199032 (120573)[int120573

120572

119903(119905)10038161003816100381610038161003816119909101584010038161003816100381610038161003816

2

119889119905]

2

(86)

where 119903(119905) is a nonincreasing function Substituting (86) into(85) we get

int120573

120572

1003816100381610038161003816119902 (119905)1003816100381610038161003816|119909 (119905)|

100381610038161003816100381610038161199091015840

(119905)10038161003816100381610038161003816119889119905

le (int120573

120572

1003816100381610038161003816119902(119905)10038161003816100381610038162

119889119905)

12

2 (120573 minus 120572)

120587119903 (120573)int120573

120572

119903 (119905)10038161003816100381610038161003816119909101584010038161003816100381610038161003816

2

119889119905

(87)

Applying inequality (12) on the integral int120573120572119903(119905)|1199091015840|2119889119905 with

119901(119905) = 119902(119905) = 119903(119905) and119898 = 119899 = 2 we get

int120573

120572

|119903 (119905)|100381610038161003816100381610038161199091015840(119905)

10038161003816100381610038161003816

2

119889119905 le 41198612

1(120572 120573) int

120573

120572

|119903 (119905)|1003816100381610038161003816100381611990910158401015840(119905)

10038161003816100381610038161003816

2

119889119905 (88)

Substituting (88) into (87) we have

int120573

120572

1003816100381610038161003816119902 (119905)1003816100381610038161003816|119909 (119905)|

100381610038161003816100381610038161199091015840

(119905)10038161003816100381610038161003816119889119905

le (int120573

120572

1003816100381610038161003816119902(119905)10038161003816100381610038162

119889119905)

12

8(120573 minus 120572)

120587119903(120573)1205952(120572 120573)

times int120573

120572

|119903 (119905)|1003816100381610038161003816100381611990910158401015840(119905)

10038161003816100381610038161003816

2

119889119905

(89)

Substituting (84) and (89) into (83) we obtain

int120573

120572

|119903 (119905)|1003816100381610038161003816100381611990910158401015840(119905)

10038161003816100381610038161003816

2

119889119905

le 41198612

1(120572 120573) int

120573

120572

|119903(119905)|1003816100381610038161003816100381611990910158401015840

(119905)10038161003816100381610038161003816

2

119889119905

+(int120573

120572

1003816100381610038161003816119902(119905)10038161003816100381610038162

119889119905)

12

8 (120573 minus 120572)

120587119903 (120573)1205952(120572 120573)

times int120573

120572

|119903 (119905)|1003816100381610038161003816100381611990910158401015840(119905)

10038161003816100381610038161003816

2

119889119905

(90)

Cancelling the term int120573

120572|119903(119905)||11990910158401015840(119905)|2119889119905 we obtain

41198612

1(120572 120573) +

8 (120573 minus 120572)

120587119903 (120573)1205952(120572 120573)(int

120573

120572

1003816100381610038161003816119902(119905)10038161003816100381610038162

119889119905)

12

ge 1

(91)

which is the desired inequality (80) The proof of (81) issimilar to the proof of (80) by using 119861

2(120572 120573) instead of

1198611(120572 120573) The proof is complete

Conflict of Interests

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

References

[1] E Picard ldquoMemoire sur la theorie des equations aux derivespartielleset la methode desapproximations successivesrdquo Journalde Mathematiques Pures et Appliquees vol 6 pp 145ndash210 1890

[2] E Picard ldquoSur lrsquoapplication des methodes drsquoapproximationssuccessives a lrsquoetude de certaines equations differentielles ordi-nairesrdquo Journal deMathematiques Pures et Appliquees vol 9 pp217ndash271 1893

[3] O Niccoletti ldquoSulle consizioni iniziali che determinano gliintegrale delle equazioni diff eren ziali ordinarierdquo Atti dellaAccademia delle Scienze di Torino vol 33 pp 746ndash759 1897

[4] C de la Vallee Poussin ldquoSur lrsquoequation differentielle lineaire dusecond ordrerdquo Journal de Mathematiques Pures et Appliqueesvol 8 pp 125ndash144 1929

[5] A M Lyapunov ldquoProbleme general de la stabilite du mouve-mentrdquo Annales de la Faculte des sciences de Toulouse vol 9 pp203ndash474 1907

[6] S H Saker ldquoSome new disconjugacy criteria for second orderdifferential equations with a middle termrdquo Bulletin Mathema-tique de la Societe des Sciences Mathematiques de Roumanie vol57 no 1 pp 109ndash120 2014

[7] G Casadei ldquoSul teorema di unicita di De La Vallee Poussinper equazioni differenziali del terzo ordinerdquo Rendiconti delSeminario Matematico della Universita di Padova vol 41 pp300ndash315 1968

[8] A Lasota ldquoSur la distance entre les zeros de lrsquoequation dif-ferentielle lineaire du troisieme ordrerdquo Annales Polonici Mathe-matici vol 13 pp 129ndash132 1963

Mathematical Problems in Engineering 9

[9] RMMathsen ldquoA disconjugacy condition for119910101584010158401015840+119886211991010158401015840+119886

11199101015840+

1198860119910 = 0rdquo Proceedings of the AmericanMathematical Society vol

17 pp 627ndash632 1966[10] B G Pachpatte ldquoOn the zeros of solutions of certain differential

equationsrdquo Demonstratio Mathematica vol 25 no 4 pp 825ndash833 1992

[11] B G Pachpatte ldquoLyapunov type integral inequalities for certaindifferential equationsrdquo Georgian Mathematical Journal vol 4no 2 pp 139ndash148 1997

[12] S Panigrahi ldquoLyapunov-type integral inequalities for certainhigher-order differential equationsrdquo Electronic Journal of Differ-ential Equations vol 2009 no 28 pp 1ndash14 2009

[13] N Parhi and S Panigrahi ldquoOn Liapunov-type inequality forthird-order differential equationsrdquo Journal of MathematicalAnalysis and Applications vol 233 no 2 pp 445ndash460 1999

[14] N Parhi and S Panigrahi ldquoDisfocality and Liapunov-typeinequalities for third-order equationsrdquo Applied MathematicsLetters vol 16 no 2 pp 227ndash233 2003

[15] U Richard ldquoMetodi diversi per ottenere disequaglianze allade la Vallee Poussin nelle equazioni differenziali ordinarie delsecondo e terzo ordinerdquo Rendiconti del Seminario MatematicoUniversita e Politecnico di Torino vol 27 pp 35ndash68 1968

[16] X Yang ldquoOn Liapunov-type inequality for certain higher-orderdifferential equationsrdquo Applied Mathematics and Computationvol 134 no 2-3 pp 307ndash317 2003

[17] R P Agarwal and P R Krishnamurthy ldquoOn the uniquenessof solution of nonlinear boundary value problemsrdquo Journal ofMathematical Sciences vol 10 no 1 pp 17ndash31 1976

[18] R P Agarwal and P Y PangOpial Inequalities with Applicationsin Differential and Difference Equation Kluwer AcademicDordrecht The Netherlands 1995

[19] R P Agarwal and P Y Pang ldquoRemarks on the generalizationsof Opialrsquos inequalityrdquo Journal of Mathematical Analysis andApplications vol 190 no 2 pp 559ndash577 1995

[20] P R Beesack and K M Das ldquoExtensions of Opialrsquos inequalityrdquoPacific Journal of Mathematics vol 26 pp 215ndash232 1968

[21] S Clark and D Hinton ldquoSome disconjugacy criteria for dif-ferential equations with oscillatory coefficientsrdquoMathematischeNachrichten vol 278 no 12-13 pp 1476ndash1489 2005

[22] A M Fink ldquoOn Opials inequality for 119891(119899)rdquo Proceedings of theAmerican Mathematical Society vol 155 pp 177ndash181 1992

[23] A Kufner and L-E Persson Weighted Inequalities of HardyType World Scientific River Edge NJ USA 2003

[24] D W Boyd ldquoBest constants in a class of integral inequalitiesrdquoPacific Journal of Mathematics vol 30 pp 367ndash383 1969

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 Mathematical Problems in Engineering

Using the assumption 1199091015840(120572) = 11990910158401015840(120573) = 0 we get that

int120573

120572

|119903 (119905)|1003816100381610038161003816100381611990910158401015840(119905)

10038161003816100381610038161003816

2

119889119905 le int120573

120572

1003816100381610038161003816119901 (119905)1003816100381610038161003816100381610038161003816100381610038161199091015840(119905)

10038161003816100381610038161003816

2

119889119905

+ int120573

120572

1003816100381610038161003816119902 (119905)1003816100381610038161003816|119909 (119905)|

100381610038161003816100381610038161199091015840

(119905)10038161003816100381610038161003816119889119905

(83)

Applying inequality (12) on the integral int120573120572|119901(119905)||1199091015840(119905)|2119889119905

with119898 = 119899 = 2 we get

int120573

120572

1003816100381610038161003816119901 (119905)1003816100381610038161003816100381610038161003816100381610038161199091015840(119905)

10038161003816100381610038161003816

2

119889119905

le 41205952(120572 120573) int

120573

120572

|119903 (119905)|1003816100381610038161003816100381611990910158401015840

(119905)210038161003816100381610038161003816119889119905

(84)

where 120595(120572 120573) is defined as in (34) Applying Schwarzrsquosinequality on the term int

120573

120572|119902(119905)||119909(119905)||1199091015840(119905)|119889119905 we get that

int120573

120572

1003816100381610038161003816119902 (119905)1003816100381610038161003816|119909 (119905)|

100381610038161003816100381610038161199091015840

(119905)10038161003816100381610038161003816119889119905

le (int120573

120572

1003816100381610038161003816119902(119905)10038161003816100381610038162

119889119905)

12

(int120573

120572

|119909|210038161003816100381610038161003816119909101584010038161003816100381610038161003816

2

119889119905)

12

(85)

Applying again inequality (78) on the integral int120573120572|119909|2|1199091015840|2119889119905

with ] = 120578 = 2 (note that 119909(120572) = 0) we obtain

int120573

120572

|119909|210038161003816100381610038161003816119909101584010038161003816100381610038161003816

2

119889119905 le4(120573 minus 120572)

2

12058721199032 (120573)[int120573

120572

119903(119905)10038161003816100381610038161003816119909101584010038161003816100381610038161003816

2

119889119905]

2

(86)

where 119903(119905) is a nonincreasing function Substituting (86) into(85) we get

int120573

120572

1003816100381610038161003816119902 (119905)1003816100381610038161003816|119909 (119905)|

100381610038161003816100381610038161199091015840

(119905)10038161003816100381610038161003816119889119905

le (int120573

120572

1003816100381610038161003816119902(119905)10038161003816100381610038162

119889119905)

12

2 (120573 minus 120572)

120587119903 (120573)int120573

120572

119903 (119905)10038161003816100381610038161003816119909101584010038161003816100381610038161003816

2

119889119905

(87)

Applying inequality (12) on the integral int120573120572119903(119905)|1199091015840|2119889119905 with

119901(119905) = 119902(119905) = 119903(119905) and119898 = 119899 = 2 we get

int120573

120572

|119903 (119905)|100381610038161003816100381610038161199091015840(119905)

10038161003816100381610038161003816

2

119889119905 le 41198612

1(120572 120573) int

120573

120572

|119903 (119905)|1003816100381610038161003816100381611990910158401015840(119905)

10038161003816100381610038161003816

2

119889119905 (88)

Substituting (88) into (87) we have

int120573

120572

1003816100381610038161003816119902 (119905)1003816100381610038161003816|119909 (119905)|

100381610038161003816100381610038161199091015840

(119905)10038161003816100381610038161003816119889119905

le (int120573

120572

1003816100381610038161003816119902(119905)10038161003816100381610038162

119889119905)

12

8(120573 minus 120572)

120587119903(120573)1205952(120572 120573)

times int120573

120572

|119903 (119905)|1003816100381610038161003816100381611990910158401015840(119905)

10038161003816100381610038161003816

2

119889119905

(89)

Substituting (84) and (89) into (83) we obtain

int120573

120572

|119903 (119905)|1003816100381610038161003816100381611990910158401015840(119905)

10038161003816100381610038161003816

2

119889119905

le 41198612

1(120572 120573) int

120573

120572

|119903(119905)|1003816100381610038161003816100381611990910158401015840

(119905)10038161003816100381610038161003816

2

119889119905

+(int120573

120572

1003816100381610038161003816119902(119905)10038161003816100381610038162

119889119905)

12

8 (120573 minus 120572)

120587119903 (120573)1205952(120572 120573)

times int120573

120572

|119903 (119905)|1003816100381610038161003816100381611990910158401015840(119905)

10038161003816100381610038161003816

2

119889119905

(90)

Cancelling the term int120573

120572|119903(119905)||11990910158401015840(119905)|2119889119905 we obtain

41198612

1(120572 120573) +

8 (120573 minus 120572)

120587119903 (120573)1205952(120572 120573)(int

120573

120572

1003816100381610038161003816119902(119905)10038161003816100381610038162

119889119905)

12

ge 1

(91)

which is the desired inequality (80) The proof of (81) issimilar to the proof of (80) by using 119861

2(120572 120573) instead of

1198611(120572 120573) The proof is complete

Conflict of Interests

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

References

[1] E Picard ldquoMemoire sur la theorie des equations aux derivespartielleset la methode desapproximations successivesrdquo Journalde Mathematiques Pures et Appliquees vol 6 pp 145ndash210 1890

[2] E Picard ldquoSur lrsquoapplication des methodes drsquoapproximationssuccessives a lrsquoetude de certaines equations differentielles ordi-nairesrdquo Journal deMathematiques Pures et Appliquees vol 9 pp217ndash271 1893

[3] O Niccoletti ldquoSulle consizioni iniziali che determinano gliintegrale delle equazioni diff eren ziali ordinarierdquo Atti dellaAccademia delle Scienze di Torino vol 33 pp 746ndash759 1897

[4] C de la Vallee Poussin ldquoSur lrsquoequation differentielle lineaire dusecond ordrerdquo Journal de Mathematiques Pures et Appliqueesvol 8 pp 125ndash144 1929

[5] A M Lyapunov ldquoProbleme general de la stabilite du mouve-mentrdquo Annales de la Faculte des sciences de Toulouse vol 9 pp203ndash474 1907

[6] S H Saker ldquoSome new disconjugacy criteria for second orderdifferential equations with a middle termrdquo Bulletin Mathema-tique de la Societe des Sciences Mathematiques de Roumanie vol57 no 1 pp 109ndash120 2014

[7] G Casadei ldquoSul teorema di unicita di De La Vallee Poussinper equazioni differenziali del terzo ordinerdquo Rendiconti delSeminario Matematico della Universita di Padova vol 41 pp300ndash315 1968

[8] A Lasota ldquoSur la distance entre les zeros de lrsquoequation dif-ferentielle lineaire du troisieme ordrerdquo Annales Polonici Mathe-matici vol 13 pp 129ndash132 1963

Mathematical Problems in Engineering 9

[9] RMMathsen ldquoA disconjugacy condition for119910101584010158401015840+119886211991010158401015840+119886

11199101015840+

1198860119910 = 0rdquo Proceedings of the AmericanMathematical Society vol

17 pp 627ndash632 1966[10] B G Pachpatte ldquoOn the zeros of solutions of certain differential

equationsrdquo Demonstratio Mathematica vol 25 no 4 pp 825ndash833 1992

[11] B G Pachpatte ldquoLyapunov type integral inequalities for certaindifferential equationsrdquo Georgian Mathematical Journal vol 4no 2 pp 139ndash148 1997

[12] S Panigrahi ldquoLyapunov-type integral inequalities for certainhigher-order differential equationsrdquo Electronic Journal of Differ-ential Equations vol 2009 no 28 pp 1ndash14 2009

[13] N Parhi and S Panigrahi ldquoOn Liapunov-type inequality forthird-order differential equationsrdquo Journal of MathematicalAnalysis and Applications vol 233 no 2 pp 445ndash460 1999

[14] N Parhi and S Panigrahi ldquoDisfocality and Liapunov-typeinequalities for third-order equationsrdquo Applied MathematicsLetters vol 16 no 2 pp 227ndash233 2003

[15] U Richard ldquoMetodi diversi per ottenere disequaglianze allade la Vallee Poussin nelle equazioni differenziali ordinarie delsecondo e terzo ordinerdquo Rendiconti del Seminario MatematicoUniversita e Politecnico di Torino vol 27 pp 35ndash68 1968

[16] X Yang ldquoOn Liapunov-type inequality for certain higher-orderdifferential equationsrdquo Applied Mathematics and Computationvol 134 no 2-3 pp 307ndash317 2003

[17] R P Agarwal and P R Krishnamurthy ldquoOn the uniquenessof solution of nonlinear boundary value problemsrdquo Journal ofMathematical Sciences vol 10 no 1 pp 17ndash31 1976

[18] R P Agarwal and P Y PangOpial Inequalities with Applicationsin Differential and Difference Equation Kluwer AcademicDordrecht The Netherlands 1995

[19] R P Agarwal and P Y Pang ldquoRemarks on the generalizationsof Opialrsquos inequalityrdquo Journal of Mathematical Analysis andApplications vol 190 no 2 pp 559ndash577 1995

[20] P R Beesack and K M Das ldquoExtensions of Opialrsquos inequalityrdquoPacific Journal of Mathematics vol 26 pp 215ndash232 1968

[21] S Clark and D Hinton ldquoSome disconjugacy criteria for dif-ferential equations with oscillatory coefficientsrdquoMathematischeNachrichten vol 278 no 12-13 pp 1476ndash1489 2005

[22] A M Fink ldquoOn Opials inequality for 119891(119899)rdquo Proceedings of theAmerican Mathematical Society vol 155 pp 177ndash181 1992

[23] A Kufner and L-E Persson Weighted Inequalities of HardyType World Scientific River Edge NJ USA 2003

[24] D W Boyd ldquoBest constants in a class of integral inequalitiesrdquoPacific Journal of Mathematics vol 30 pp 367ndash383 1969

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

Mathematical Problems in Engineering 9

[9] RMMathsen ldquoA disconjugacy condition for119910101584010158401015840+119886211991010158401015840+119886

11199101015840+

1198860119910 = 0rdquo Proceedings of the AmericanMathematical Society vol

17 pp 627ndash632 1966[10] B G Pachpatte ldquoOn the zeros of solutions of certain differential

equationsrdquo Demonstratio Mathematica vol 25 no 4 pp 825ndash833 1992

[11] B G Pachpatte ldquoLyapunov type integral inequalities for certaindifferential equationsrdquo Georgian Mathematical Journal vol 4no 2 pp 139ndash148 1997

[12] S Panigrahi ldquoLyapunov-type integral inequalities for certainhigher-order differential equationsrdquo Electronic Journal of Differ-ential Equations vol 2009 no 28 pp 1ndash14 2009

[13] N Parhi and S Panigrahi ldquoOn Liapunov-type inequality forthird-order differential equationsrdquo Journal of MathematicalAnalysis and Applications vol 233 no 2 pp 445ndash460 1999

[14] N Parhi and S Panigrahi ldquoDisfocality and Liapunov-typeinequalities for third-order equationsrdquo Applied MathematicsLetters vol 16 no 2 pp 227ndash233 2003

[15] U Richard ldquoMetodi diversi per ottenere disequaglianze allade la Vallee Poussin nelle equazioni differenziali ordinarie delsecondo e terzo ordinerdquo Rendiconti del Seminario MatematicoUniversita e Politecnico di Torino vol 27 pp 35ndash68 1968

[16] X Yang ldquoOn Liapunov-type inequality for certain higher-orderdifferential equationsrdquo Applied Mathematics and Computationvol 134 no 2-3 pp 307ndash317 2003

[17] R P Agarwal and P R Krishnamurthy ldquoOn the uniquenessof solution of nonlinear boundary value problemsrdquo Journal ofMathematical Sciences vol 10 no 1 pp 17ndash31 1976

[18] R P Agarwal and P Y PangOpial Inequalities with Applicationsin Differential and Difference Equation Kluwer AcademicDordrecht The Netherlands 1995

[19] R P Agarwal and P Y Pang ldquoRemarks on the generalizationsof Opialrsquos inequalityrdquo Journal of Mathematical Analysis andApplications vol 190 no 2 pp 559ndash577 1995

[20] P R Beesack and K M Das ldquoExtensions of Opialrsquos inequalityrdquoPacific Journal of Mathematics vol 26 pp 215ndash232 1968

[21] S Clark and D Hinton ldquoSome disconjugacy criteria for dif-ferential equations with oscillatory coefficientsrdquoMathematischeNachrichten vol 278 no 12-13 pp 1476ndash1489 2005

[22] A M Fink ldquoOn Opials inequality for 119891(119899)rdquo Proceedings of theAmerican Mathematical Society vol 155 pp 177ndash181 1992

[23] A Kufner and L-E Persson Weighted Inequalities of HardyType World Scientific River Edge NJ USA 2003

[24] D W Boyd ldquoBest constants in a class of integral inequalitiesrdquoPacific Journal of Mathematics vol 30 pp 367ndash383 1969

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