By using the cone theory and the Banach contraction mapping principle, the existence and uniqueness results are established for singular thirdorder boundary value problems. The theorems obtained are very general and complement previous known results.
1. Introduction
Thirdorder differential equations arise in a variety of different areas of applied mathematics and physics, such as the deflection of a curved beam having a constant or varying cross section, threelayer beam, electromagnetic waves, or gravitydriven flows [1]. Recently, thirdorder boundary value problems have been studied extensively in the literature (see, e.g., [2–13], and their references). In this paper, we consider the following thirdorder boundary value problem:
where , .
Threepoint boundary value problems (BVPs for short) have been also widely studied because of both practical and theoretical aspects. There have been many papers investigating the solutions of threepoint BVPs, see [2–5, 10, 12] and references therein. Recently, the existence of solutions of thirdorder threepoint BVP (1.1) has been studied in [2, 3]. Guo et al. [2] show the existence of positive solutions for BVP (1.1) when and is separable by using cone expansioncompression fixed point theorem. In [3], the singular thirdorder threepoint BVP (1.1) is considered under some conditions concerning the first eigenvalues corresponding to the relevant linear operators, where , is separable and is not necessary to be nonnegative, and the existence results of nontrivial solutions and positive solutions are given by means of the topological degree theory. Motivated by the above works, we consider the singular thirdorder threepoint BVP (1.1). Here, we give the unique solution of BVP (1.1) under the conditions that and is mixed nonmonotone in and does not need to be separable by using the cone theory and the Banach contraction mapping principle.
2. Preliminaries
Let , . By [2, Lemma 2.1], we have that is a solution of (1.1) if and only if
where
It is shown in [2] that is the Green's function to , , and .
Let
It is easy to see that .
is generating if and only if there exists a constant such that every element can be represented in the form , where and , .
3. Singular ThirdOrder Boundary Value Problem
This section discusses singular thirdorder boundary value problem (1.1).
Let . Obviously, is a normal solid cone of Banach space ; by [16, Lemma 2.1.2], we have that is a generating cone in .
Theorem 3.1.
Suppose that , and there exist two positive linear bounded operators and with such that for any , , , , we have
and there exists , such that
Then (1.1) has a unique solution in . And moreover, for any , the iterative sequence
converges to .
Remark 3.2.
Recently, in the study of BVP (1.1), almost all the papers have supposed that the Green's function is nonnegative. However, the scope of is not limited to in Theorem 3.1, so, we do not need to suppose that is nonnegative.
Remark 3.3.
The function in Theorem 3.1 is not monotone or convex; the conclusions and the proof used in this paper are different from the known papers in essence.
Proof.
It is easy to see that, for any , can be divided into finite partitioned monotone and bounded function on , and then by (3.2), we have
For any , let , , then , by (3.1), we have
Hence
Following the former inequality, we can easily have
thus
Similarly, by and being converged, we have that
Define the operator by
Then is the solution of BVP (1.1) if and only if . Let
By (3.1) and (3.10), for any , , , we have
so we can choose an , which satisfies , and so there exists a positive integer such that
Since is a generating cone in , from Lemma 2.1, there exists such that every element can be represented in
This implies
Let
By (3.16), we know that is well defined for any . It is easy to verify that is a norm in . By (3.15)–(3.17), we get
On the other hand, for any which satisfies , we have . Thus , where denotes the normal constant of . Since is arbitrary, we have
It follows from (3.18) and (3.19) that the norms and are equivalent.
Now, for any and which satisfies , let
then , , , , and .
It follows from (3.12) that
Subtracting (3.22) from (3.21) + (3.23), we obtain
Let ; then we have
As and are both positive linear bounded operators, so, is a positive linear bounded operator, and therefore . Hence, by mathematical induction, it is easy to know that for natural number in (3.14), we have
Since , we see that
which implies by virtue of the arbitrariness of that
By , we have . Thus the Banach contraction mapping principle implies that has a unique fixed point in , and so has a unique fixed point in ; by the definition of has a unique fixed point in , that is, is the unique solution of (1.1). And, for any , let ; we have . By the equivalence of and again, we get . This completes the proof.
Example 3.4.
In this paper, the results apply to a very wide range of functions, we are following only one example to illustrate.
Consider the following singular thirdorder boundary value problem:
where and there exists , such that for any , , , we have
Applying Theorem 3.1, we can find that (3.29) has a unique solution provided . And moreover, for any , the iterative sequence
converges to .
To see that, we put
Then (3.1) is satisfied for any , , , and .
In fact, if , then
If , then
Similarly,
Next, for any , by (3.30) and (3.32), we get
Then, from (3.32) and (3.36), we have
so it is easy to know by induction, for any , we get
thus
so
then we get
Let ; then
Thus all conditions in Theorem 3.1 are satisfied.
Acknowledgment
The author is grateful to the referees for valuable suggestions and comments.
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