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 secondorder differential equation:
subject to the boundary conditions:
where , , is a continuous function, is a nondecreasing function with . In boundary conditions (1.2), the integral is meant in the RiemannStieltjes sense.
We say that BVP (1.1), (1.2) is a problem at resonance, if the linear equation
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:
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
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
Let , . is a linear operator defined by
where
Let be defined as
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
(H2) there exists a constant , such that for , if , for all , then
(H3) there exists a constant , such that either
or else
Then BVP (1.1), (1.2) with , , and has at least one solution in provided that
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
(H5) there exists a constant , such that either
or else
Then BVP (1.1), (1.2) with , and has at least one solution in provided that
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
and the linear operator can be written by
Furthermore,
Proof.
It is clear that
Obviously, the problem
has a solution satisfying , , if and only if
which implies that
In fact, if (3.5) has solution satisfying , , then from (3.5) we have
According to , , we obtain
then
On the other hand, if (3.6) holds, setting
where is an arbitrary constant, then is a solution of (3.5), and , and from and (3.6), we have
Then . Hence (3.7) is valid.
For , define
Let , and we have
then , thus . Hence, , where , also . So we have , and
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
and, for , we know
In view of , , and , thus
This shows that . Also we have
then . The proof of Lemma 3.1 is finished.
Lemma 3.2.
Under conditions (2.5) and (2.9), there are nonnegative functions satisfying
Proof.
Without loss of generality, we suppose that . Take , then there exists such that
Let
Obviously, , and
Then
and from (2.5) and (3.21), we have
Hence we can take , , 0, and to replace , , , and , respectively, in (2.5), and for the convenience omit the bar above , , and , that is,
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
thus by assumption (H2), there exists , such that . In view of
then, we have
Again for , , then , thus from Lemma 3.1, we know
From (3.29) and (3.30), we have
If (2.5) holds, from (3.31), and (3.26), we obtain
Thus, from and (3.32), we have
From , (3.32), and (3.33), one has
that is,
From (3.35) and (3.33), there exists , such that
Thus
Again from (2.5), (3.35), and (3.36), we have
Then we show that is bounded.
Lemma 3.4.
If assumption (H2) holds, then the set is bounded.
Proof.
Let , then and ; therefore,
From assumption (H2), , so , clearly is bounded.
Lemma 3.5.
If the first part of condition (H3) of Theorem 2.2 holds, then
for all . Let
where is the linear isomorphism given by , for all , . Then is bounded.
Proof.
Suppose that , then we obtain
or equivalently
If , then . Otherwise, if , in view of (3.40), one has
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 AscoliArzela 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
According to definition of degree on a space which is isomorphic to , , and
We have
and then
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,
for all , then in Lemma 3.5, we take
and exactly as there, we can prove that is bounded. Then in the proof of Theorem 2.2, we have
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
and the linear operator can be written by
Furthermore,
Notice that
Proof of Theorem 2.3.
Let
Then, for , ; thus , ; hence
thus, from assumption (H4), there exists , such that and in view of , we obtain
From , there exists , such that . Thus, from , one has
We let ; hence from (3.58) and (3.59), we have
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:
where ,
and satisfying , , and , then we can choose , , and , for ; thus
Since
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:
where ,
and satisfying , and , then we can choose , , and , for ; thus
Similar to Example 3.8, we have
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 .
Acknowledgment
This work was sponsored by the National Natural Science Foundation of China (11071205), the NSF of Jiangsu Province Education Department, NFS of Xuzhou Normal University.
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