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Volume 2011, Article ID 416416, 15 pagesdoi:10.1155/2011/416416 Research Article Existence of Solutions to a Nonlocal Boundary Value Problem with Nonlinear Growth Xiaojie Lin School of M

Volume 2011, Article ID 416416, 15 pages

doi:10.1155/2011/416416

Research Article

Existence of Solutions to a Nonlocal Boundary

Value Problem with Nonlinear Growth

Xiaojie Lin

School of Mathematical Sciences, Xuzhou Normal University, Xuzhou, Jiangsu 221116, China

Correspondence should be addressed to Xiaojie Lin,linxiaojie1973@163.com

Received 17 July 2010; Accepted 17 October 2010

Academic Editor: Feliz Manuel Minh ´os

Copyrightq 2011 Xiaojie Lin This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited

This paper deals with the existence of solutions for the following differential equation: xt 

f t, xt, xt, t ∈ 0, 1, subject to the boundary conditions: x0  αxξ, x1 1

0xsdgs, where α ≥ 0, 0 < ξ < 1, f : 0, 1 × R2 → R is a continuous function, g : 0, 1 → 0, ∞ is a nondecreasing function with g0  0 Under the resonance condition g1  1, some existence

results are given for the boundary value problems Our method is based upon the coincidence degree theory of Mawhin We also give an example to illustrate our results

1 Introduction

In this paper, we consider the following second-order differential equation:

xt  ft, x t, xt, t ∈ 0, 1, 1.1 subject to the boundary conditions:

x 0  αxξ, x1 

1

0

xsdgs, 1.2

where α ≥ 0, 0 < ξ < 1, f : 0, 1 × R2 → R is a continuous function, g : 0, 1 → 0, ∞ is a nondecreasing function with g0  0 In boundary conditions 1.2, the integral is meant in the Riemann-Stieltjes sense

2 Boundary Value Problems

We say that BVP1.1, 1.2 is a problem at resonance, if the linear equation

xt  0, t ∈ 0, 1, 1.3

with the boundary condition 1.2 has nontrivial solutions Otherwise, we call them a problem at nonresonance

Nonlocal boundary value problems were first considered by Bicadze and Samarski˘ı

1 and later by Il’pin and Moiseev 2,3 In a recent paper 4, Karakostas and Tsamatos studied the following nonlocal boundary value problem:

xt  qtfx t, xt 0, t ∈ 0, 1,

x 0  0, x1 

1

0

xsdgs. 1.4

Under the condition 0 g0 ≤ g1 < 1 i.e., nonresonance case, they used Krasnosel’skii’s fixed point theorem to show that the operator equation x  Ax has at least one fixed point, where operator A is defined by

Axt  t

1− g1

1

0

1

s

q rfx r, xrdr dg s 

t

0

1

s

q rfx r, xrdr ds. 1.5

However, if g1  1 i.e., resonance case, then the method in 4 is not valid

As special case of nonlocal boundary value problems, multipoint boundary value problems at resonance case have been studied by some authors5 11

The purpose of this paper is to study the existence of solutions for nonlocal BVP1.1,

1.2 at resonance case i.e., g1  1 and establish some existence results under nonlinear growth restriction of f Our method is based upon the coincidence degree theory of Mawhin

12

2 Main Results

We first recall some notation, and an abstract existence result

Let Y , Z be real Banach spaces, let L : dom L ⊂ Y → Z be a linear operator which

is Fredholm map of index zeroi.e., Im L, the image of L, Ker L, the kernel of L are finite dimensional with the same dimension as the Z/ Im L, and let P : Y → Y, Q : Z → Z

be continuous projectors such that Im P  Ker L, Ker Q  Im L and Y  Ker L ⊕ Ker P, Z 

Im L ⊕ Im Q It follows that L| dom L∩Ker P : dom L ∩ Ker P → Im L is invertible; we denote the inverse by KP LetΩ be an open bounded, subset of Y such that dom L ∩ Ω / ∅, the map

N : Y → Z is said to be L-compact on Ω if QNΩ is bounded, and KP I − QN : Ω → Y is compact Let J : Im Q → Ker L be a linear isomorphism.

The theorem we use in the following is Theorem IV.13 of12

Theorem 2.1 Let L be a Fredholm operator of index zero, and let N be L-compact on Ω Assume that

the following conditions are satisfied:

i Lx / λNx for every x, λ ∈ dom L \ Ker L ∩ ∂Ω × 0, 1,

ii Nx /∈ Im L for every x ∈ Ker L ∩ ∂Ω,

iii degJQN| Ker L , Ω ∩ Ker L, 0 / 0,

where Q : Z → Z is a projection with Im L  Ker Q Then the equation Lx  Nx has at least one

solution in dom L ∩ Ω.

For x ∈ C10, 1, we use the norms x∞  maxt∈0,1 |xt| and x  max{x∞, x∞}

and denote the norm in L10, 1 by · 1 We will use the Sobolev space W 2,1 0, 1 which may be

defined by

W 2,1 0, 1 x : 0, 1 −→ R | x, xare absolutely continuous on 0, 1 with x∈ L10, 1.

2.1

Let Y  C10, 1, Z  L10, 1 L : dom L ⊂ Y → Z is a linear operator defined by

Lx  x, x ∈ dom L, 2.2

where

dom L



x ∈ W 2,1 0, 1 : x0  αxξ, x1 

1

0

xsdgs

. 2.3

Let N : Y → Z be defined as

Nx  ft, x t, xt, t ∈ 0, 1. 2.4

Then BVP1.1, 1.2 is Lx  Nx.

We will establish existence theorems for BVP1.1, 1.2 in the following two cases: casei: α  0, g1  1,1

0s dg s / 1;

caseii: α  1, g1  1,1

0s dg s / 1.

H1 there exist functions a, b, c, r ∈ L10, 1 and constant θ ∈ 0, 1 such that for all x, y ∈

R2, t ∈ 0, 1, it holds that

f

|x| θ y θ

 rt, 2.5

4 Boundary Value Problems

H2 there exists a constant M > 0, such that for x ∈ dom L, if |xt| > M, for all t ∈ 0, 1,

then

1

0

f

s, x s, xsds−

1

0

s

0

f

v, x v, xvdv dg s / 0, 2.6

H3 there exists a constant M∗> 0, such that either

0

f s, ds, dds −

1

0

s

0

f v, dv, ddv dgs



< 0, for any |d| > M∗, 2.7

or else

0

f s, ds, dds −

1

0

s

0

f v, dv, ddv dgs



> 0, for any |d| > M∗. 2.8

Then BVP1.1, 1.2 with α  0, g1  1, and1

0s dg s / 1 has at least one solution in C10, 1

provided that

a1 b1< 1

Theorem 2.2 is satisfied, and

H4 there exists a constant M > 0, such that for x ∈ dom L, if |xt| > M, for all t ∈ 0, 1,

then

1

0

f

s, x s, xsds−

1

0

s

0

f

v, x v, xvdv dg s / 0, 2.10

H5 there exists a constant M∗> 0, such that either

0

f s, e, 0ds −

1

0

s

0

f v, e, 0dv dgs



< 0, for any |e| > M∗, 2.11

or else

0

f s, e, 0ds −

1

0

s

0

f v, e, 0dv dgs



> 0, for any |e| > M∗. 2.12

Then BVP1.1, 1.2 with α  1, g1  1, and1

0s dg s / 1 has at least one solution in C10, 1

provided that

a1 b1< 1

3 Proof of Theorems 2.2 and 2.3

We first proveTheorem 2.2via the following Lemmas

0s dg s / 1, then L : dom L ⊂ Y → Z is a Fredholm operator

of index zero Furthermore, the linear continuous projector operator Q : Z → Z can be defined by

1−1

0s dg s

1 0

y sds −

1

0

s

0

y vdv dgs



, 3.1

and the linear operator K P : Im L → dom L ∩ Ker P can be written by

K P y

t

0

s

0

y vdv ds. 3.2

Furthermore,

K P y  ≤ y1, for every y ∈ Im L. 3.3

Proof It is clear that

Ker L  {x ∈ dom L : x  dt, d ∈ R, t ∈ 0, 1}. 3.4 Obviously, the problem

has a solution xt satisfying x0  0, x1 1

0xsdgs, if and only if

1

0

y sds −

1

0

s

0

y vdv dgs  0, 3.6 which implies that

Im L



y ∈ Z :

1

0

y sds −

1

0

s

0

y vdv dgs  0

. 3.7

6 Boundary Value Problems

In fact, if3.5 has solution xt satisfying x0  0, x1 1

0xsdgs, then from 3.5 we have

x t  x0t 

t

0

s

0

y vdv ds. 3.8

According to x1 1

0xsdgs, g1  1, we obtain

x1  x0 

1

0

y sds 

1

0

xsdgs



1

0



x0 

s

0

y vdv



dg s

 x0g1 

1

0

s

0

y vdv dgs,

3.9

then

1

0

y sds −

1

0

s

0

y vdv dgs  0. 3.10

On the other hand, if3.6 holds, setting

x t  dt 

t

0

s

0

y vdv ds, 3.11

where d is an arbitrary constant, then xt is a solution of 3.5, and x0  0, and from

g1  1 and 3.6, we have

x1 −

1

0

xsdgs  d 

1

0

y sds −

1

0



d

s

0

y vdv



dg s

 d1− g1

1

0

y sds −

1

0

s

0

y vdv dgs

 0.

3.12

Then x1 1

0xsdgs Hence 3.7 is valid

For y ∈ Z, define

1−1

0s dg s

1 0

y sds −

1

0

s

0

y vdv dgs



, 0≤ t ≤ 1. 3.13

Let y1 y − Qy, and we have

1−

1

0

s dg s



Qy1

1

0



y − Qysds −

1

0

s

0



y − Qyvdv dgs



1

0

y sds − Qy −

1

0

s

0

y vdv dgs  Qy

1

0

s dg s



1

0

y sds −

1

0

s

0

y vdv dgs − Qy 1−

1

0

s dg s



 0,

3.14

then Qy1  0, thus y1 ∈ Im L Hence, Z  Im L  Z1, where Z1 {xt ≡ d : t ∈ 0, 1, d ∈ R}, also Im L ∩ Z1 {0} So we have Z  Im L ⊕ Z1, and

dim Ker L  dim Z1 co dim Im L  1. 3.15

Thus, L is a Fredholm operator of index zero.

We define a projector P : Y → Ker L by Pxt  x0t Then we show that KP

defined in3.2 is a generalized inverse of L : dom L ∩ Y → Z.

In fact, for y ∈ Im L, we have

LKP yt K P y

t yt, 3.16

and, for x ∈ dom L ∩ Ker P, we know

KP L xt 

t

0

s

0

xvdv ds  xt − x0 − x0t. 3.17

In view of x ∈ dom L ∩ Ker P, x0  0, and Px  0, thus

KP L xt  xt. 3.18

This shows that K P  L| dom L∩Ker P−1 Also we have

K P y

∞≤

1

0

y v dv dsy

1, KP y

∞≤y

1, 3.19 thenKP y ≤ y1 The proof ofLemma 3.1is finished

satisfying

f

8 Boundary Value Problems

Proof Without loss of generality, we suppose that c1  1

0|ct|dt  β > 0 Take γ ∈

0, 1/2β1/2 − a1 b1, then there exists M > 0 such that

|x| θ ≤ γ|x|  M, y θ ≤ γ y 3.21 Let

a t  at  γct, b t  bt  γct, r t  rt  2Mct. 3.22

Obviously, a, b, r ∈ L10, 1, and

a1≤ a1 γc1,



b

1≤ b1 γc1. 3.23 Then

a1b

1≤ a1 b1 2βγ < 1

2, 3.24 and from2.5 and 3.21, we have

f

t, x, y

 at|x|  bt y

3.25

Hence we can take a, b, 0, and r to replace a, b, c, and r, respectively, in2.5, and for the

convenience omit the bar above a, b, and r, that is,

f

0s dg s / 1 hold, then the set

Ω1 {x ∈ dom L \ Ker L : Lx  λNx for some λ ∈ 0, 1} is a bounded subset of Y.

Proof Suppose that x∈ Ω1and Lx  λNx Thus λ / 0 and QNx  0, so that

1

0

y sds −

1

0

s

0

y vdv dgs  0, 3.27 thus by assumptionH2, there exists t0∈ 0, 1, such that |xt0| ≤ M In view of

x0  xt0 −

t0

0

xtdt, 3.28

then, we have

1 M  Lx1 ≤ M  Nx1. 3.29

Again for x ∈ Ω1, x ∈ dom L \ Ker L, then I − Px ∈ dom L ∩ Ker P, LPx  0 thus from

Lemma 3.1, we know

I − Px  KP L I − Px ≤ LI − Px1 Lx1≤ Nx1. 3.30 From3.29 and 3.30, we have

x ≤ Px  I − Px  x0 1 M. 3.31

If2.5 holds, from 3.31, and 3.26, we obtain

x ≤ 2



a1x∞ b1x

∞ r1 M

2



. 3.32 Thus, fromx∞≤ x and 3.32, we have

x∞≤ 2

1− 2a1



b1x

∞ r1 M

2



Fromx∞≤ x, 3.32, and 3.33, one has

x

∞≤ x ≤ 2



1 2a1

1− 2a1



b1x

∞ r1M

2



 2

1− 2a1



b1x

∞ r1M

2



,

3.34

that is,

x∞≤ 2

1− 2a1 b1



r1M 2

 : M1. 3.35 From3.35 and 3.33, there exists M2> 0, such that

x∞≤ M2. 3.36 Thus

x  maxx∞,x

∞



≤ max{M1, M2}. 3.37

10 Boundary Value Problems

Again from2.5, 3.35, and 3.36, we have

x

1 Lx1≤ Nx1≤ a1M2 b1M1 r1. 3.38 Then we show thatΩ1is bounded

Proof Let x ∈ Ω2, then x ∈ Ker L  {x ∈ dom L : x  dt, d ∈ R, t ∈ 0, 1} and QNx  0;

therefore,

1

0

f s, ds, dds −

1

0

s

0

f v, dv, ddv dgs  0, 3.39

From assumptionH2, x∞ |d| ≤ M, so x  |d| ≤ M, clearly Ω2is bounded

1−1

0s dg s

1 0

f s, ds, dds −

1

0

s

0

f v, dv, ddv dgs



< 0, 3.40

for all |d| > M∗ Let

Ω3 {x ∈ Ker L : −λx  1 − λJQNx  0, λ ∈ 0, 1}, 3.41

where J : Im Q → Ker L is the linear isomorphism given by Jd  dt, for all d ∈ R, t ∈ 0, 1 Then

Ω3is bounded.

Proof Suppose that x  d0t∈ Ω3, then we obtain

λd0t 1 − λt

1−1

0s dg s

1 0

f s, d0s, d0ds −

1

0

s

0

f v, d0v, d0dv dgs



, 0≤ t ≤ 1, 3.42

or equivalently

1−1

0s dg s

1 0

f s, d0s, d0ds −

1

0

s

0

f v, d0v, d0dv dgs



. 3.43

If λ  1, then d0 0 Otherwise, if |d0| > M∗, in view of3.40, one has

λd20  d01 − λ

1−1

0s dg s

1 0

f s, d0s, d0ds −

1

0

s

0

f v, d0v, d0dv dgs



< 0, 3.44

which contradicts λd2

0 ≥ 0 Then |x|  |d0t | ≤ |d0| ≤ M∗and we obtainx ≤ M∗; therefore,

Ω3⊂ {x ∈ Ker L : x ≤ M∗} is bounded

The proof of Theorem 2.2 is now an easy consequence of the above lemmas and

Theorem 2.1

Proof of Theorem 2.2 Let Ω  {x ∈ Y : x < δ} such that3

i1Ωi ⊂ Ω By the Ascoli-Arzela

theorem, it can be shown that KP I − QN : Ω → Y is compact; thus N is L-compact on Ω.

Then by the above Lemmas, we have the following

i Lx / λNx for every x, λ ∈ dom L \ Ker L ∩ ∂Ω × 0, 1.

ii Nx /∈ Im L for every x ∈ Ker L ∩ ∂Ω.

iii Let Hx, λ  −λx  1 − λJQNx, with J as inLemma 3.5 We know H x, λ / 0, for

x ∈ Ker L ∩ ∂Ω Thus, by the homotopy property of degree, we get

degJQN|Ker L , Ω ∩ Ker L, 0  degH·, 0, Ω ∩ Ker L, 0

 degH·, 1, Ω ∩ Ker L, 0

 deg−I, Ω ∩ Ker L, 0.

3.45

According to definition of degree on a space which is isomorphic to R n , n <∞, and

Ω ∩ Ker L  {dt : |d| < δ}. 3.46

We have

deg−I, Ω ∩ Ker L, 0  deg−J−1IJ, J−1Ω ∩ Ker L, J−1{0}

 deg−I, −δ, δ, 0  −1 / 0,

3.47

and then

degJQN|Ker L , Ω ∩ Ker L, 0 / 0. 3.48

Then byTheorem 2.1, Lx  Nx has at least one solution in dom L ∩ Ω, so that the BVP 1.1,

1.2 has at least one solution in C10, 1 The proof is completed.

Remark 3.6 If the second part of conditionH3 ofTheorem 2.2holds, that is,

1−1

0s dg s

1 0

f s, ds, dds −

1

0

s

0

f v, dv, ddv dgs



> 0, 3.49

for all|d| > M∗, then inLemma 3.5, we take

Ω3 {x ∈ Ker L : λx  1 − λJQNx  0, λ ∈ 0, 1}, 3.50

12 Boundary Value Problems

and exactly as there, we can prove thatΩ3is bounded Then in the proof ofTheorem 2.2, we have

degJQN|Ker L , Ω ∩ Ker L, 0  degI, Ω ∩ Ker L, 0  1, 3.51 since 0∈ Ω ∩ Ker L The remainder of the proof is the same.

By using the same method as in the proof ofTheorem 2.2and Lemmas3.1–3.5, we can showLemma 3.7andTheorem 2.3

0s dg s / 1, then L : dom L ⊂ Y → Z is a Fredholm operator

of index zero Furthermore, the linear continuous projector operator Q : Z → Z can be defined by

1−1

0s dg s

1 0

y sds −

1

0

s

0

y vdv dgs



, 3.52

and the linear operator K P : Im L → dom L ∩ Ker P can be written by

K P y −t

ξ

ξ

0

s

0

y vdv ds 

t

0

s

0

y vdv ds. 3.53

Furthermore,

K P y  ≤ 2y1, ∀y ∈ Im L. 3.54

Notice that

Ker L  {x ∈ dom L : x  e, e ∈ R},

Im L



y ∈ Z :

1

0

y sds −

1

0

s

0

y vdv dgs  0

. 3.55

Proof of Theorem 2.3 Let

Ω1 {x ∈ dom L \ Ker L : Lx  λNx for some λ ∈ 0, 1}. 3.56

Then, for x∈ Ω1, Lx  λNx; thus λ / 0, Nx ∈ Im L  Ker Q; hence

1

0

y sds −

1

0

s

0

y vdv dgs  0, 3.57

thus, from assumptionH4, there exists t0 ∈ 0, 1, such that |xt0| < M and in view of

x 0  xt0 −t0

0 xtdt, we obtain

|x0| ≤ M x

From x0  xξ, there exists t1 ∈ 0, ξ, such that xt1  0 Thus, from xt  xt1 

t

t1xtdt, one has

x

∞≤x

We let P x  x0; hence from 3.58 and 3.59, we have

Px  |x0| ≤ M x

∞≤ M x

1

 M  Lx1≤ M  Nx1, 3.60

thus, by using the same method as in the proof of Lemmas3.2and3.3, we can prove thatΩ1

is bounded too Similar to the other proof of Lemmas3.4–3.7andTheorem 2.2, we can verify

Theorem 2.3

Finally, we give two examples to demonstrate our results

Example 3.8 Consider the following boundary value problem:

x t3 8  sin x31

9t  1x, t ∈ 0, 1,

x 0  0, x1 

1

0

xsdgs,

3.61

where α 0,

f

t, x, y

 t3 8  sin x31

9t  1y, t ∈ 0, 1, 3.62

and gs  s2 satisfying g0  0, g1  1, and1

0s dg s  2/3 / 1, then we can choose

a t  0, bt  2/9, and rt  10, for t ∈ 0, 1; thus

f

t, x, y

9 y

a1 b1 2

9 <

1

2.

3.63

Since

1

0

f

s, x s, xsds−

1

0

s

0

f

v, x v, xvdv dg s



1

0

f

v, x v, xvdv dg s −

1

0

s

0

f

v, x v, xvdv dg s



1

0

1

s

f

v, x v, xvdv dg s,

3.64

6 Boundary Value Problems

In fact, if3.5 has solution xt satisfying x0  0, x1...

satisfying

f

8 Boundary Value Problems

Proof Without loss of. ..