This paper is concerned with the following thirdorder boundary value problem with integral boundary conditions , where and . By using the GuoKrasnoselskii fixedpoint theorem, some sufficient conditions are obtained for the existence and nonexistence of monotone positive solution to the above problem.
1. Introduction
Thirdorder differential equations arise in a variety of different areas of applied mathematics and physics, for example, in the deflection of a curved beam having a constant or varying cross section, a threelayer beam, electromagnetic waves or gravity driven flows and so on [1].
Recently, thirdorder twopoint or multipoint boundary value problems (BVPs for short) have attracted a lot of attention [2–17]. It is known that BVPs with integral boundary conditions cover multipoint BVPs as special cases. Although there are many excellent works on thirdorder twopoint or multipoint BVPs, a little work has been done for thirdorder BVPs with integral boundary conditions. It is worth mentioning that, in 2007, Anderson and Tisdell [18] developed an interval of values whereby a positive solution exists for the following thirdorder BVP with integral boundary conditions
by using the GuoKrasnoselskii fixedpoint theorem. In 2008, Graef and Yang [19] studied the thirdorder BVP with integral boundary conditions
For secondorder or fourthorder BVPs with integral boundary conditions, one can refer to [20–24].
In this paper, we are concerned with the following thirdorder BVP with integral boundary conditions
Throughout this paper, we always assume that and . Some sufficient conditions are established for the existence and nonexistence of monotone positive solution to the BVP (1.3). Here, a solution of the BVP (1.3) is said to be monotone and positive if , and for . Our main tool is the following GuoKrasnoselskii fixedpoint theorem [25].
Theorem 1.1.
Let be a Banach space and let be a cone in . Assume that and are bounded open subsets of such that , and let be a completely continuous operator such that either
(1) for and for , or
(2) for and for .
Then has a fixed point in .
2. Preliminaries
For convenience, we denote .
Lemma 2.1.
Let . Then for any , the BVP
has a unique solution
where
Proof.
Let be a solution of the BVP (2.1). Then, we may suppose that
By the boundary conditions in (2.1), we have
Therefore, the BVP (2.1) has a unique solution
Lemma 2.2 (see [12]).
For any ,
Lemma 2.3 (see [26]).
For any ,
In the remainder of this paper, we always assume that , and .
Lemma 2.4.
If and for , then the unique solution of the BVP (2.1) satisfies
(1), ,
(2), and , where .
Proof.
Since (1) is obvious, we only need to prove (2). By (2.2), we get
which indicates that for .
On the one hand, by (2.9) and Lemma 2.3, we have
On the other hand, in view of (2.2) and Lemma 2.2, we have
It follows from (2.10) and (2.11) that
which together with Lemma 2.2 implies that
Let be equipped with the norm . Then is a Banach space. If we denote
then it is easy to see that is a cone in . Now, we define an operator on by
Obviously, if is a fixed point of , then is a monotone nonnegative solution of the BVP (1.3).
Lemma 2.5.
is completely continuous.
Proof.
First, by Lemma 2.4, we know that .
Next, we assume that is a bounded set. Then there exists a constant such that for any . Now, we will prove that is relatively compact in . Suppose that . Then there exist such that . Let
Then for any , by Lemma 2.2, we have
which implies that is uniformly bounded. At the same time, for any , in view of Lemma 2.3, we have
which shows that is also uniformly bounded. This indicates that is equicontinuous. It follows from ArzelaAscoli theorem that has a convergent subsequence in . Without loss of generality, we may assume that converges in . On the other hand, by the uniform continuity of , we know that for any , there exists such that for any with , we have
Let . Then for any , with , we have
which implies that is equicontinuous. Again, by ArzelaAscoli theorem, we know that has a convergent subsequence in . Therefore, has a convergent subsequence in . Thus, we have shown that is a compact operator.
Finally, we prove that is continuous. Suppose that and . Then there exists such that for any , . Let
Then for any and , in view of Lemmas 2.2 and 2.3, we have
By applying Lebesgue Dominated Convergence theorem, we get
which indicates that is continuous. Therefore, is completely continuous.
3. Main Results
For convenience, we define
Theorem 3.1.
If , then the BVP (1.3) has at least one monotone positive solution.
Proof.
In view of , there exists such that
By the definition of , we may choose so that
Let . Then for any , in view of (3.2) and (3.3), we have
By integrating the above inequality on , we get
which together with (3.4) implies that
On the other hand, since , there exists such that
By the definition of , we may choose , so that
Let . Then for any , in view of (3.7) and (3.8), we have
which implies that
Therefore, it follows from (3.6), (3.10), and Theorem 1.1 that the operator has one fixed point , which is a monotone positive solution of the BVP (1.3).
Theorem 3.2.
If , then the BVP (1.3) has at least one monotone positive solution.
Proof.
The proof is similar to that of Theorem 3.1 and is therefore omitted.
Theorem 3.3.
If for and , then the BVP (1.3) has no monotone positive solution.
Proof.
Suppose on the contrary that is a monotone positive solution of the BVP (1.3). Then and for , and
By integrating the above inequality on , we get
which together with (3.11) implies that
This is a contradiction. Therefore, the BVP (1.3) has no monotone positive solution.
Similarly, we can prove the following theorem.
Theorem 3.4.
If for and , then the BVP (1.3) has no monotone positive solution.
Example 3.5.
Consider the following BVP:
Since and , if we choose , then it is easy to compute that
which shows that
So, it follows from Theorem 3.1 that the BVP (3.14) has at least one monotone positive solution.
Acknowledgment
This work was supported by the National Natural Science Foundation of China (10801068).
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