International Mathematical Forum, 5, 2010, no. 43, 2131 - 2136

Positive Solutions for Boundary Value Problems

of Higher Order Differential Equations

S. N. Odda

Department of Mathematics, Faculty of Computer Science Qassim University, Burieda, Saudi Arabia

nabhan100@yahoo.com

Abstract In this paper, we investigate the problem of existence and nonexistence of positive solutions for the nonlinear boundary value problem: 1,t0 ,0))(()()()( <<=+ tuftatu n λ 0(1)u ,0)0()0()0()0()0( )2( =′′′===′′′=′′=′= −nuuuuu LL . Our analysis relies on Krasnoselskii’s fixed theorem of cone. An example is also given to illustrate the main results. Mathematics Subject Classification: 34B15, 34B18 Keywords: Higher-order differential equations, Boundary value problems, Krasnoselskii’s fixed point theorem

1. Introduction The purpose of this paper is to establish the existence of positive solutions to nonlinear higher-order boundary value problems: 1,t0 ,0))(()()()( <<=+ tuftatu n λ (1) 0(1)u ,0)0()0()0()0()0( )2( =′′′===′′′=′′=′= −nuuuuu LL , (2) Where λ is a positive parameter. Throughout the paper, we assume that

),0[),0[::1 ∞→∞fC is continuous

2132 S. N. Odda

)[0,)1(0, :a :2 ∞→C is continuous function such that ∫ >1

0

0)( dtta .

We present here some notations and lemmas that will be used in the proof our main results. These definitions and lemmas can be found in the recent literature [2,3,5]. Definition 1. Let E be a real Banach space. A nonempty closed set EK ⊂ is called a cone of E if it satisfies the following conditions: (1) 0, ≥∈ σKx implies Kx∈σ ; (2) KxKx ∈−∈ , implies 0=x . Definition 2. An operator is called completely continuous if it is continuous and maps bounded sets into precompact sets. Lemma1. Let E be a Banach space and EK ⊂ is a cone in E. Assume that 1Ω and 2Ω are open subsets of E with 10 Ω∈ and 21 Ω⊂Ω . Let KKT →ΩΩ∩ )\(: 12 be a completely continuous operator. In addition suppose either: (H1) 1, Ω∂∩∈∀≤ KuuuT and 2, Ω∂∩∈∀≥ KuuuT or

(H2) 2, Ω∂∩∈∀≤ KuuuT and 1, Ω∂∩∈∀≥ KuuuT

holds. Then T has a fixed point in )\( 12 ΩΩ∩K . 2. Green functions and their properties Lemma 2. Let ]1,0[Cy∈ then the boundary value problem 1,t0 ,0)()()( <<=+ tytu n (3) 0(1)u ,0)0()0()0()0()0( )2( =′′′===′′′=′′=′= −nuuuuu LL , (4) has a unique solution

, )(),()(1

0∫= dssystGtu

where

1;st0 if

)!1()1(

1;ts0 if )!1(

)( - )!1()1(

),()4()1(

)1()4()1(

⎪⎪⎩

⎪⎪⎨

⎧

≤≤≤−−

≤≤≤−

−−−

=−−

−−−

nst

nst

nst

stGnn

nnn

Proof: Applying the Laplace transform to Eq. (2.1) in the light of Eq. (2.2) we get

Positive solutions for boundary value problems 2133

)()0()0()0()0()0()( 14321 syuusususussus nnnnnn −===′′′=′′−′−− −−−−− LL , (5) Where )(su and )(sy is the Laplace transform of )(tu and )(ty respectively. The Laplace inversion of Eq. (5) gives final solution as:

dssyn

stdssyn

sttut nnn

)()!1(

)()()!1()1()(

0

11

0

41

∫∫ −−

−−−

=−−−

. (6)

The proof is complete. It is obvious that 0),( ≥stG and .1,0),,(),1( ≤≤≥ ststGsG (7) Lemma 3. ),1()(),( sGtqstG ≥ for 1,0 ≤≤ st , where 1)( −= nttq . Proof: If st ≤ , then

1

),1(),( −= nt

sGstG .

If st ≥ , then

114

141

)1()1()()1(

),1(),( −

−−

−−−

≥−−−−−−

= nnn

nnn

tss

ststsGstG

The proof is complete. 3. Main results In this section, we will apply Krasnoselskii’s fixed-point theorem to the eigenvalue problem (1). We note that )(tu is a solution of (1), (2) if and only if

10,))(()(),()(1

0

≤≤= ∫ tdssufsastGtu λ .

For our constructions, we shall consider the Banach space ]1,0[CX = equipped with standard norm Xutuu

t∈=

≤≤,)(max

10. Define a cone P by

]}1,0[,)()(,0)({ ∈≥≥∈= tutqtutuXuP .

It is easy to see that if Pu∈ , then )1(uu = . Define an integral operator by:

.,10,))(()(),()(1

0

PutdssufsastGtuT ∈≤≤= ∫λ (8)

Lemma 4. PPT ⊂)( . Proof: Notice from lemma (3) that, for Pu∈ , 0)( ≥tuT on ]1,0[ and

2134 S. N. Odda

))(()(),(max)(

))(()(),1()(

))(()(),()(

1

010

1

0

1

0

dssufsastGtq

dssufsasGtq

dssufsastGtuT

t ∫

∫

∫

≤≤≥

≥

=

λ

λ

λ

,)()( tuTtq= for all ]1,0[, ∈st . Thus .)( PPT ⊂ By standard argument, it is easy to see that PPT →: is a completely continuous. Following Sun and Wen [4], we define some important constants:

∫∫ ==1

0

1

0

,)(),1(,)()(),1( dssasGBdssqsasGA

uuff

uufF

uuff

uufF

uu

uu

)(inflim,)(suplim

.)(inflim,)(suplim0

00

0

+∞→∞+∞→∞

→→

==

==++

Here we assume that 01=

∞fAif ∞→∞f and ∞=

0

1FB

if 00 →F and 01

0

=fA

if

∞→0f and ∞=∞FB

1 if 0→∞F .

Theorem 1. Suppose that 0FBfA >∞ . Then for each )1,1(0FBfA ∞

∈λ the problem (1)

and (2) has at least one positive solution. Proof: Following Sun and Wen [4], we choose 0>ε sufficiently small such that 1)( 0 ≤+ BF λε . Then we have uuT ≤ . By definition of 0F , we see that there exists

an 01 >l , such that uFuf )()( 0 ε+≤ for 10 lu ≤< . If Pu∈ with 1lu = , we have

.)(

)(),1()(

)())((),1(

))(()(),1()1)((

0

1

00

1

00

1

0

uuFB

dssasGuF

dssuFsasG

dssufsasGuTuT

≤+≤

+≤

+=

==

∫

∫

∫

ελ

ελ

ελ

λ

Positive solutions for boundary value problems 2135 Then we have uuT ≤ . Thus if we let }{ 11 luXu <∈=Ω , then uuT ≤ for

1Ω∂∩∈Pu .

Following Yang [5], we choose 0>ε and 1(0, )4

c ∈ , such that

1)()(),1()(1

≥⎟⎟⎠

⎞⎜⎜⎝

⎛− ∫∞

c

dssqsasGf ελ .

There exists 03 >l , such that ufuf )()( ε−≥ ∞ for 3lu > . Let ⎭⎬⎫

⎩⎨⎧

= 13

2 2,)(

max lcq

ll . If

Pu∈ with 2lu = , then we have 322 )()()( llcqltqtu ≥≥≥ . Therefore, for each If Pu∈ with 2lu = , we have

.)()(),1()(

)()()(),1(

))(()(),1()1)((

1

0

1

0

1

0

udssqsasGuf

dssufsasG

dssufsasGuTuT

∫

∫

∫

≥−≥

−≥

==

∞

∞

ελ

ελ

λ

Thus if we let }{ 22 luXu <∈=Ω , then 21 Ω⊂Ω and uuT ≥ for 2Ω∂∩∈Pu . Condition (H1) of Krasnoselskii’s fixed-point theorem is satisfied. So there exists a fixed point of T in P . This completes the proof.

Theorem 2. Suppose that ∞> FBfA 0 . Then for each )1,1(0 ∞

∈FBfA

λ the problem (1)

and (2) has at least one positive solution. The Proof of theorem 2 is very similar to that of theorem 1 and therefore omitted. Theorem 3. Suppose that uufB <)(λ for ),0( ∞∈u . Then the problem (1), (2) has no positive solution. Proof: Following El- Shahed[1], assume to the contrary that u is a positive solution of (1), (2). Then

)1()(),1()1()()(),1(1))(()(),1()1(1

0

1

0

1

0

udssasGB

udssusasGB

dssufsasGu =≤<= ∫∫∫λ .

This is a contradiction and completes the proof.

2136 S. N. Odda Theorem 4. Suppose that uufA >)(λ for ),0( ∞∈u . Then the problem (1), (2) has no positive solution. The Proof of theorem 4 is very similar to that of theorem 3 and therefore omitted. Example 1. Consider the equation

1t0 ,0)sin6(4

62

)(2

)4( ≤≤=+++

+ uu

uuttu λ (9)

0(1)u ),0()0()0( =′′′′′=′= uuu . (10) Then ,600 == fF ,42=∞F 30=∞f and uufu 42)(6 << . By direct calculations, we obtain that 016369.0=A and 0375.0=B . From theorem 1 we see that if

)44444.4,0363636.2(∈λ , then the problem (9), (10) has a positive solution. From theorem 3 we have that if ,6349206.0<λ then the problem (9), (10) has no positive solution. By theorem 4 we have that if ,181818.10>λ then the problem (9), (10) has no positive solution. REFERENCES [1] El-Shahed, M. (2009), Positive solutions for nonlinear singular third order boundary value problem, Communications in Nonlinear Science and Numerical Simulations. 14 424-429. [2] Guo, D and Lakshmikantham, V. 1988, Nonlinear problems in abstract cones,

Academic Press, San Diego. [3] Li, S. 2006, Positive solutions of nonlinear singular third-order two-point boundary value problem, J. Math. Anal. Appl., 323,413–425. [4] Sun, H. and Wen, W. 2006, On the Number of Positive solutions for a nonlinear third order boundary value problem, International Journal of Difference Equations, 1,165-176. [5] Yang, B. 2005, Positive solutions for a fourth order boundary value problem, Electronic Journal of Qualitative Theory of Differential Equations,3, 1-17. Received: January, 2010