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Existence Results of ThreePoint Boundary Value Problems for Secondorder Ordinary Differential Equations
Boundary Value Problems volume 2011, Article number: 901796 (2010)
Abstract
We establish existence results of the following threepoint boundary value problems: , , and , where and . The approach applied in this paper is upper and lower solution method associated with basic degree theory or Schauder's fixed point theorem. We deal with this problem with the function which is Carathéodory or singular on its domain.
1. Introduction
In this paper, we consider threepoint boundary value problem
where and .
In the mathematical literature, a number of works have appeared on nonlocal boundary value problems, and one of the first of these was [1]. Il'in and Moiseev initiated the research of multipoint boundary value problems for secondorder linear ordinary differential equations, see [2, 3], motivated by the study [4–6] of Bitsadze and Samarskii.
Recently, nonlinear multipoint boundary value problems have been receiving considerable attention, and have been studied extensively by using iteration scheme (e.g., [7]), fixed point theorems in cones (e.g., [8]), and the LeraySchauder continuation theorem (e.g., [9]). We refer more detailed treatment to more interesting research [10, 11] and the references therein.
The theory of upper and lower solutions is also a powerful tool in studying boundary value problems. For the existence results of twopoint boundary value problem, there already are lots of interesting works by applying this essential technique (see [12, 13]). Recently, it is shown that this method plays an important role in proving the existence of solutions for threepoint boundary value problems (see [14–16]).
Last but not least, as the singular source term appearing in twopoint problems, singular threepoint boundary value problems also attract more attention (e.g., [17]).
In this paper, we will discuss the existence of solutions of some general types on threepoint boundary value problems by using upper and lower solution method associated with basic degree theory or Schauder's fixed point theorem.
This paper is organized as follows. In Section 2, we give two lemmas which will be extensively used later. In Section 3, when the source term is a Carathéodory function, we consider the Sobolev space defined by
and obtain the existence of solution in Theorems 3.5 and 3.11. In Section 4, we discuss the singular case, that is, maybe singular at the end points or , or at . We will introduce the class of functions and another space (see [18, 19]) as follows:
and prove the existence of solution in Theorems 4.1 and 4.4. Some sufficient conditions for constructing upper and lower solutions are given in each section for applications.
2. Preliminaries
Define by
where and are given as (1.2) and
By direct computations, we get the following results.
Lemma 2.1.

(i)
The function defined by (2.1), is the Green function corresponding for the problem
(2.3)
(ii) The function defined by (2.1), is continuous.

(iii)
In the case , we have
(2.4)
Lemma 2.2.
If , then the problem
with boundary condition (1.2) has a unique solution such that
where is defined by (2.1).
3. Carathéodory Case
In this section we first introduce the Carathéodory function as follows.
Definition 3.1.
A function defined on is called a Carathéodory function on if
(i)for almost every is continuous on ;
(ii)for any the function is measurable on ;
(iii)for any , there exists such that for any and for almost every with , we have .
We in this section assume that is a Carathéodory function and discuss the existence of solution by assuming the existence of upper and lower solutions.
3.1. Existence of Solutions
We first introduce the definitions of upper and lower solutions as below.
Definition 3.2.
A function is called a lower solution of problem (1.1) and (1.2) if it satisfies
(i), , and
(ii)for any , either , or there exists an open interval containing such that and, for almost every , we have
Definition 3.3.
A function is called a upper solution of problem (1.1) and (1.2) if it satisfies
(i), , and
(ii)for any , either , or there exists an open interval containing such that and, for almost every , we have
Before proving our main results, we first consider such a modified problem given as follows:
with boundary condition (1.2), where is defined by
Proposition 3.4.
Let and be respective lower and upper solutions of problem (1.1) and (1.2) with on . If is a solution of problem (3.3) and (1.2), then , for any .
Proof.
Suppose there exists such that
Case 1.
If , we have , which implies Hence, by Definition 3.2 and the continuity of at , there exist an open interval with , and a neighborhood of contained in such that for almost every ,
Furthermore, it follows from that for , , we have
This implies that the minimum of cannot occur at , a contradiction.
Case 2.
If , by the definition of lower solution , we then have
And we get a contradiction.
Case 3.
If , it follows from the conclusion of Case 1 that
which is impossible.
Consequently, we obtain on . By the similar arguments as above, we also have
Theorem 3.5.
Let and be lower and upper solutions of problem (1.1) and (1.2) such that on and let be a Carathéodory function on , where
Then problem (1.1) and (1.2) has at least one solution such that, for all ,
Proof.
We consider the modified problem (3.3) and (1.2) with respect to the given and . Consider the Banach space with supremum and the operator by
for , where is defined as (2.1). Since is a Carathéodory function on , for almost all and for all , there exists a function , we have
Define
where
It is clear that is a closed, bounded and convex set in and one can show that is a completely continuous mapping by ArzelàAscoli theorem and Lebesgue dominated convergence theorem. By applying Schauder's fixed point theorem, we obtain that has a fixed point in which is a solution of problem (3.3) and (1.2). From Proposition 3.4, this fixed point is also a solution of problem (1.1) and (1.2). Hence, we complete the proof.
We further illustrate the use of Theorem 3.11 in the following secondorder differential equation:
with the boundary condition (1.2).
Corollary 3.6.
Assume that is a Carathéodory function satisfying is essentially bounded for , where is a constant large enough. Assume further that and there exists a constant such that
Then, problem (3.18) and (1.2) has at least one solution.
Proof.
By hypothesis, for any given small enough such that and for almost all , for any large enough, we have
We now choose an upper solution of the form
To this end, we compute
Clearly, one can choose such that
that is,
and choose , where , which is a positive solution of
Hence, if is large enough, we can show that and , where , which implies that is a positive upper solution. In the same way we construct a lower solution on .
3.2. Nontangency Solution
In this subsection, we afford another stronger lower and upper solutions to get a strict inequality of the solution between them.
Definition 3.7.
A function is a strict lower solution of problem (1.1) and (1.2), if it is not a solution of problem (1.1) and (1.2), , and for any , one of the following is satisfied:
(i);
(ii)there exist an interval and such that int, and for almost every , for all we have
Definition 3.8.
A function is a strict upper solution of problem (1.1) and (1.2), if it is not a solution of problem (1.1) and (1.2), , and for any , one of the following is satisfied:
(i),
(ii)there exist an interval and such that int, and for almost every , for all we have
Remark 3.9.
Every strict lower(upper) solution of problem (1.1) and (1.2) is a lower(upper) solution.
Now we are going to show that the solution curve of problem (1.1) and (1.2) cannot be tangent to upper or lower solutions from below or above.
Proposition 3.10.
Let and be respective strict lower and upper solutions of problem (1.1) and (1.2) with on . If is a solution of problem (1.1) and (1.2) with on , then , for any .
Proof.
As is not a solution, is not identical to . Assume, the conclusion does not hold, then
exists. Hence, has minimum at , that is, .
Case 1.
Set . Since has minimum at , we have . According to the Definition 3.7, there exist , and with such that, for every , , and for a.e.
Hence, we have the contradiction since
Case 2.
If , by the definition of strict lower solution that , we then have
And we get a contradiction.
Case 3.
If , repeat the same arguments in Case 3 of the proof of Proposition 3.4. Therefore, we obtain on . The inequality on can be proved by the similar arguments as above.
Theorem 3.11.
Let and be strict lower and upper solutions of problem (1.1) and (1.2) such that on and let be a Carathéodory function, where
Then, problem (1.1) and (1.2) has at least one solution such that, for any ,
Proof.
This is a consequence of Theorem 3.5 and Proposition 3.10 and hence, we omits this proof.
4. Singular Case
In this section we give a more general existence result than Theorem 3.11 by assuming the existence of lower and upper solutions. This makes us to deal with problem (1.1) and (1.2), where the function is singular at the end point and .
Theorem 4.1.
Let and be lower and upper solutions of problem (1.1) and (1.2) such that on and let satisfy the following conditions:
(i)for almost every is continuous on ;
(ii)for any the function is measurable on ;
(iii)there exists a function such that, for all ,
where
Then problem (1.1) and (1.2) has at least one solution such that, for all ,
Proof.
Consider the modified problem (3.3) and (1.2) with respect to the given and and define by (3.13). Note that by Lemma 2.2, is well defined. Define
where
and is defined by (3.17). The rest arguments are similar to the proof of Theorem 3.5.
Remark 4.2.
We have similar results of Theorems 3.5–4.1, respectively, for (1.1) equipped with
where is a constant and , are given as (1.2).
Example 4.3.
Consider the problem (4.7), for , , ,
Clearly, is a lower solution of (4.7) and
where
From Lemma 2.1, we have and define . Since, for ,
that is, , we have, from Lemma 2.2, and exists. Let
and, by Lemma 2.2 again, choose such that
Note that according to the direct computation, we see that is welldefined and is bounded by . Next, let . By Young's inequality, it follows that
Hence, such is a upper solution of (4.7) and on . Clearly, satisfies (i), (ii) of Theorem 4.1. By using Young's inequality again, for , we have
and . Therefore, satisfies the assumption (iii) of Theorem 4.1. Consequently, we conclude that this problem has at least one solution such that, for all ,
Notice that in Theorem 4.1, one can only deal with the case that is singular at end points , . However, when is singular at , there is no hope to obtain the solutions directly from Theorem 4.1. We will establish the following theorem to deal with this case by constructing upper and lower solutions to solve this problem.
Theorem 4.4.
Assume
the function is continuous;
there exists and for any compact set , there is such that
for some and , there is such that
where is defined as in Lemma 2.1.
for any compact set , there is such that
Then problem (1.1) and (1.2) with has at least one solution
Remark 4.5 (see [12, Remark ]).
Assumption is equivalent to the assumption that there exists and a function such that:
(i) for all ,
(ii), for all , ,
(iii), for all ,
where
Proof.
Step 1.
Construction of lower solutions. Consider such that and the function
where is chosen small enough so that
Next, we choose from the Remark 4.5, and let
where is small enough so that for some points , , we have:
Notice that by (4.24) and (4.25), for any such that
we have:
Step 2.
Approximation problems. We define for each , ,
and set
We have that, for each index , is continuous and
where
Hence, the sequence of functions converges to uniformly on any set , where is an arbitrary compact subset of . Next we define
Each of the functions is a continuous function defined on , moreover
and the sequence converges to uniformly on the compact subsets of since
Define now a decreasing sequence such that
and consider a sequence of the following approximation problems:
where .
Step 3.
A lower solution of ( ). It is clear that for any ,
As the sequence is decreasing, we also have
Clearly, satisfies
It follows from (4.25) and (4.27) that is a lower solution of ( ).
Step 4.
Existence of a solution of (4.7) such that
From assumption , we can find and such that, for all , ,
Also, one has
where is a suitable constant. Hence, we obtain, for such and ,
Let be a constant such that
Choose such that
that is,
where is defined by (2.1). Note that is welldefined and since . It is easy to see that
So by Remark 4.2, there is a solution of (4.7) such that
Step 5.
The problem ( ) has at least one solution such that
Notice that is an upper solution of ( ), since
Step 6.
Existence of a solution. Consider the pointwise limit
It is clear that, for any ,
and therefore on . Let be a compact interval. There is an index such that for all and therefore for these ,
Moreover, we have
By ArzeláAscoli theorem it is standard to conclude that is a solution of problem (1.1) and (1.2) on the interval . Since is arbitrary, we find that and for all ,
Since
it remains only to check the continuity of at . This can be deduced from the continuity of and the fact that as .
Example 4.6.
Consider the following problem , for ,, ,
Let , where . Obviously, satisfies and . Moreover, for any given and for any compact set , for small enough, we have
Hence, holds. Furthermore, for large enough, , we have, from Young's inequality by choosing and ,
where . Hence, holds. By Theorem 4.4, has at least one solution
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Keywords
 Fixed Point Theorem
 Open Interval
 Lower Solution
 Nonlocal Boundary
 Linear Ordinary Differential Equation