Existence of Solutions to a Nonlocal Boundary Value Problem with Nonlinear Growth

Abstract

This paper deals with the existence of solutions for the following differential equation: , , subject to the boundary conditions: , , where , , is a continuous function, is a nondecreasing function with . Under the resonance condition , 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:

(1.1)

subject to the boundary conditions:

(1.2)

where , , is a continuous function, is a nondecreasing function with . In boundary conditions (1.2), the integral is meant in the Riemann-Stieltjes sense.

We say that BVP (1.1), (1.2) is a problem at resonance, if the linear equation

(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 Samarskiĭ [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:

(1.4)

Under the condition (i.e., nonresonance case), they used Krasnosel'skii's fixed point theorem to show that the operator equation has at least one fixed point, where operator is defined by

(1.5)

However, if (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 authors [5–11].

The purpose of this paper is to study the existence of solutions for nonlocal BVP (1.1), (1.2) at resonance case (i.e., ) and establish some existence results under nonlinear growth restriction of . 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 , be real Banach spaces, let be a linear operator which is Fredholm map of index zero (i.e., , the image of , , the kernel of are finite dimensional with the same dimension as the ), and let , be continuous projectors such that = , = and , . It follows that is invertible; we denote the inverse by . Let be an open bounded, subset of such that , the map is said to be -compact on if is bounded, and is compact. Let be a linear isomorphism.

The theorem we use in the following is Theorem of [12].

Theorem 2.1.

Let be a Fredholm operator of index zero, and let be -compact on . Assume that the following conditions are satisfied:

(i) for every ,

(ii) for every ,

(iii),

where is a projection with . Then the equation has at least one solution in .

For , we use the norms and and denote the norm in by . We will use the Sobolev space which may be defined by

(2.1)

Let , . is a linear operator defined by

(2.2)

where

(2.3)

Let be defined as

(2.4)

Then BVP (1.1), (1.2) is .

We will establish existence theorems for BVP (1.1), (1.2) in the following two cases:

case (i): , ;

case (ii): , .

Theorem 2.2.

Let be a continuous function and assume that

(H1) there exist functions and constant such that for all , , it holds that

(2.5)

(H2) there exists a constant , such that for , if , for all , then

(2.6)

(H3) there exists a constant , such that either

(2.7)

or else

(2.8)

Then BVP (1.1), (1.2) with , , and has at least one solution in provided that

(2.9)

Theorem 2.3.

Let be a continuous function. Assume that assumption (H1) of Theorem 2.2 is satisfied, and

(H4) there exists a constant , such that for , if , for all , then

(2.10)

(H5) there exists a constant , such that either

(2.11)

or else

(2.12)

Then BVP (1.1), (1.2) with , and has at least one solution in provided that

(2.13)

3. Proof of Theorems 2.2 and 2.3

We first prove Theorem 2.2 via the following Lemmas.

Lemma 3.1.

If , , and , then is a Fredholm operator of index zero. Furthermore, the linear continuous projector operator can be defined by

(3.1)

and the linear operator can be written by

(3.2)

Furthermore,

(3.3)

Proof.

It is clear that

(3.4)

Obviously, the problem

(3.5)

has a solution satisfying , , if and only if

(3.6)

which implies that

(3.7)

In fact, if (3.5) has solution satisfying , , then from (3.5) we have

(3.8)

According to , , we obtain

(3.9)

then

(3.10)

On the other hand, if (3.6) holds, setting

(3.11)

where is an arbitrary constant, then is a solution of (3.5), and , and from and (3.6), we have

(3.12)

Then . Hence (3.7) is valid.

For , define

(3.13)

Let , and we have

(3.14)

then , thus . Hence, , where , also . So we have , and

(3.15)

Thus, is a Fredholm operator of index zero.

We define a projector by . Then we show that defined in (3.2) is a generalized inverse of .

In fact, for , we have

(3.16)

and, for , we know

(3.17)

In view of , , and , thus

(3.18)

This shows that . Also we have

(3.19)

then . The proof of Lemma 3.1 is finished.

Lemma 3.2.

Under conditions (2.5) and (2.9), there are nonnegative functions satisfying

(3.20)

Proof.

Without loss of generality, we suppose that . Take , then there exists such that

(3.21)

Let

(3.22)

Obviously, , and

(3.23)

Then

(3.24)

and from (2.5) and (3.21), we have

(3.25)

Hence we can take , , 0, and to replace , , , and , respectively, in (2.5), and for the convenience omit the bar above , , and , that is,

(3.26)

Lemma 3.3.

If assumptions (H1), (H2) and , , and hold, then the set for some is a bounded subset of .

Proof.

Suppose that and . Thus and , so that

(3.27)

thus by assumption (H2), there exists , such that . In view of

(3.28)

then, we have

(3.29)

Again for , , then , thus from Lemma 3.1, we know

(3.30)

From (3.29) and (3.30), we have

(3.31)

If (2.5) holds, from (3.31), and (3.26), we obtain

(3.32)

Thus, from and (3.32), we have

(3.33)

From , (3.32), and (3.33), one has

(3.34)

that is,

(3.35)

From (3.35) and (3.33), there exists , such that

(3.36)

Thus

(3.37)

Again from (2.5), (3.35), and (3.36), we have

(3.38)

Then we show that is bounded.

Lemma 3.4.

If assumption (H2) holds, then the set is bounded.

Proof.

Let , then and ; therefore,

(3.39)

From assumption (H2), , so , clearly is bounded.

Lemma 3.5.

If the first part of condition (H3) of Theorem 2.2 holds, then

(3.40)

for all . Let

(3.41)

where is the linear isomorphism given by , for all , . Then is bounded.

Proof.

Suppose that , then we obtain

(3.42)

or equivalently

(3.43)

If , then . Otherwise, if , in view of (3.40), one has

(3.44)

which contradicts . Then = and we obtain ; therefore, 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 such that . By the Ascoli-Arzela theorem, it can be shown that is compact; thus is -compact on . Then by the above Lemmas, we have the following.

(i) for every .

(ii) for every .

(iii)Let , with as in Lemma 3.5. We know , for . Thus, by the homotopy property of degree, we get

(3.45)

According to definition of degree on a space which is isomorphic to , , and

(3.46)

We have

(3.47)

and then

(3.48)

Then by Theorem 2.1, has at least one solution in , so that the BVP (1.1), (1.2) has at least one solution in . The proof is completed.

Remark 3.6.

If the second part of condition (H3) of Theorem 2.2 holds, that is,

(3.49)

for all , then in Lemma 3.5, we take

(3.50)

and exactly as there, we can prove that is bounded. Then in the proof of Theorem 2.2, we have

(3.51)

since . The remainder of the proof is the same.

By using the same method as in the proof of Theorem 2.2 and Lemmas 3.1–3.5, we can show Lemma 3.7 and Theorem 2.3.

Lemma 3.7.

If , , and , then is a Fredholm operator of index zero. Furthermore, the linear continuous projector operator can be defined by

(3.52)

and the linear operator can be written by

(3.53)

Furthermore,

(3.54)

Notice that

(3.55)

Proof of Theorem 2.3.

Let

(3.56)

Then, for , ; thus , ; hence

(3.57)

thus, from assumption (H4), there exists , such that and in view of , we obtain

(3.58)

From , there exists , such that . Thus, from , one has

(3.59)

We let ; hence from (3.58) and (3.59), we have

(3.60)

thus, by using the same method as in the proof of Lemmas 3.2 and 3.3, we can prove that is bounded too. Similar to the other proof of Lemmas 3.4–3.7 and Theorem 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:

(3.61)

where ,

(3.62)

and satisfying , , and , then we can choose , , and , for ; thus

(3.63)

Since

(3.64)

and has the same sign as when , we may choose , and then the conditions (H1)–(H3) of Theorem 2.2 are satisfied. Theorem 2.2 implies that BVP (3.61) has at least one solution, .

Example 3.9.

Consider the following boundary value problem:

(3.65)

where ,

(3.66)

and satisfying , and , then we can choose , , and , for ; thus

(3.67)

Similar to Example 3.8, we have

(3.68)

and has the same sign as when , we may choose , and then all conditions of Theorem 2.3 are satisfied. Theorem 2.3 implies that BVP (3.65) has at least one solution .