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Triple Positive Solutions for a Type of Second-Order Singular Boundary Problems
Boundary Value Problems volume 2010, Article number: 376471 (2010)
Abstract
This paper deals with the existence of triple positive solutions for a type of second-order singular boundary problems with general differential operators. By using the Leggett-Williams fixed point theorem, we establish an existence criterion for at least three positive solutions with suitable growth conditions imposed on the nonlinear term.
1. Introduction
In this paper, we study the existence of triple positive solutions for the following second-order singular boundary value problems with general differential operators:
where , , and with
It is easy to see that and may be singular at and/or
When or and , the two kinds of singular boundary value problems have been discussed extensively in the literature; see [1–10] and the references therein. Hence, the problem that we consider is more general and is different from those in previous work.
Furthermore, we will see in the later that the presence of brings us three main difficulties:
(1) the Green's function cannot be explicitly expressed;
(2) the equivalence between BVP (1.1) and its associated integral equation has to be proved;
(3) the compactness of associated integral operator has to be verified.
We will overcome the above mentioned difficulties in Section 2. Also, although the Leggett-William fixed point theorem is used extensively in the study of triple positive differential equations, the method has not been used to study this type of second-order singular boundary value problem with general differential operators. We are concerned with solving these problems in this paper.
To state our main tool used in this paper, we give some definitions and notations.
Let be a real Banach space with a cone . A map is said to be a nonnegative continuous concave functional on if is a continuous and
for all and . Let be two numbers such that and a nonnegative continuous concave functional on . We define the following convex sets:
Theorem 1.1 (Leggett-Williams fixed point theorem).
Let be completely continuous, and let be a nonnegative continuous concave functional on such that for all . Suppose that there exist such that
(i) and , for ;
(ii) , for ;
(iii) , for with .
Then has at least three fixed points in satisfying and .
Remark 1.2.
We note the existence of triple positive solutions of other kind of boundary value problems; see He and Ge [11], Zhao et al. [12], Zhang and Liu [13], Graef et al. [14], and the references therein.
The rest of the paper is organized as follows. In Section 2, we overcome the above-mentioned difficulties in this work. The main results are formulated and proved in Section 3. Finally, an example is presented to demonstrate the application of the main theorems in Section 4.
2. Preliminaries and Lemmas
Throughout this paper, we assume the following:
(H1) ;
(H2) , and ;
(H3) is continuous and does not vanish identically on any subinterval of , and ;
(H4) is continuous.
Lemma 2.1.
Suppose that (H1) and (H2) hold. Then
(i) the initial value problem
has a unique solution and ;
(ii) the initial value problem
has a unique solution and .
Proof.
We only prove (i). (ii) can be treated in the same way.
Suppose that and is a solution of (2.1), that is,
Let
Multiplying both sides of (2.3) by , then
Since and , integrating (2.5) on , we have
Moreover, integrating (2.6) on , , we have
Let
Clearly, , and (2.7) reduces to
By using Fubini's theorem, we have
Therefore,
which implies that is a solution of integral equation (2.11).
Conversely, if is a solution of (2.11) with , by reversing the above argument we could deduce that the function is a solution of (2.1) and satisfy and . Therefore, to prove that (2.1) has a unique solution, , and is equivalent to prove that (2.11) has a unique solution .
To do this, we endow the following norm in :
Let be operator defined by
Since
then, is well defined. Set
Then, for any ,
and subsequently,
Thus,
Since , has a unique fixed point by Banach contraction principle. That is, (2.11) has a unique solution .
Remark 2.2.
Lemma 2.1 generalizes Theorem of [1], where .
Lemma 2.3.
Suppose that (H1) and (H2) hold. Then
(i) is nondecreasing in ;
(ii) is nonincreasing in .
Proof.
We only prove (i). (ii) can be treated in the same way.
Suppose on the contrary that is not nondecreasing in . Then there exists such that
This together with the equation implies that
which is a contradiction!
Remark 2.4.
From Lemmas 2.1 and 2.3, there exist positive constants , , , and such that
In fact, since
we have that and . Then, there exist constants and , such that
that is
In the following, we will show that . Suppose on the contrary, if there exist , such that
then, , which is a contradiction!
The other inequality can be treated in the same manner.
Lemma 2.5.
Suppose that (H1), (H2), and (H3) hold. Then
Proof.
We only prove the first equality; the other can be treated in the same way. From Remark 2.4 and (H3), we have
Lemma of [2] together with the facts that and (H3) implies that
Combining (2.27) and (2.28), we have
Lemma 2.6.
Suppose that (H1), (H2), and (H3) hold. Then the problem
has a unique solution
where
Moreover, on .
Proof.
By Lemma 2.3 and (2.32), we have
This together with Remark 2.4 implies that the right side of (2.31) is well defined.
Now we check that the function
satisfies (2.30). In fact,
Therefore,
Equation (2.34) and Lemma 2.5 imply that
Since for , then
Let with the norm
and let be a cone in defined by
Lemma 2.7.
Suppose that (H1)–(H3) hold and is a positive solution of (2.30). Then
where
Furthermore, for any , there exists corresponding such that
Proof.
In fact, if , then
and if , then
Combining this and , we have
Take
Then Lemma 2.3 guarantees that , and Lemma 2.7 guarantees that (2.43) holds.
Remark 2.8.
From Lemma 2.7 and Remark 2.4, we have
Now, for any , we can define the operator by
Lemma 2.9.
Let (H1)–(H4) hold. Then is a completely continuous operator.
Proof.
From (H3) and (H4) and Lemma 2.6, it is easy to see that , and is continuous by the Lebesgues dominated convergence theorem.
Let be any bounded set. Then (H3) and (H4) imply that is a bounded set in .
Since
then this together with the similar proof of Lemma of [2] yields
From this fact, it is easy to verify that is equicontinuous. Therefore, by the Arzela-Ascoli theorem, is a completely continuous operator.
3. Main Result
Let be nonnegative continuous concave functional defined by
We notice that, for each , , and also that by Lemma 2.6, is a solution of (1.1) if and only if is a fixed point of the operator .
For convenience we introduce the following notations. Let
Theorem 3.1.
Assume that (H1)–(H4) hold. Let , and suppose that satisfies the following conditions:
(S1) , for ;
(S2) , for ;
(S3) , for .
Then the boundary value problem (1.1) has at least three positive solutions in satisfying , and .
Proof.
From Lemma 2.9, is a completely continuous operator. If , then , and assumption (S3) implies that . Therefore
Hence, . In the same way, if , then . Therefore, condition (ii) of Leggett-williams fixed-point theorem holds.
To check condition (i) of Leggett-Williams fixed-point theorem, choose . It is easy to see that and . so,
Hence, if , then . From assumption (S2) and Remark 2.8, we have
Finally, we assert that if and , then . To see this, suppose that and , then
To sum up, all the conditions of Leggett-williams fixed-point theorem are satisfied. Therefore, has at least three fixed points, that is, problem (1.1) has at least three positive solutions in satisfying and .
Theorem 3.2.
Assume that (H1)–(H4) hold. Let , and suppose that satisfies the following conditions:
(A1) , for ;
(A2) , for .
Then the boundary value problem (1.1) has at least positive solutions.
Proof.
When , it follows from condition (A1) that , which means that has at least one fixed point by the Schauder fixed-point Theorem. When , it is clear that Theorem 3.1 holds (with ). Then we can obtain at least three positive solutions , and satisfying and with . Following this way, we finish the proof by the induction method.
4. Example
Consider the following boundary value problem:
where
Then, by computation, we have
Furthermore, for ,
In fact, let , then , and . It is easy to compute that
Then, , that is
The other inequalities in (4.4) can be proved by the same method.
Thus, we can choose that , and . By computation, we have
Let , and . Then, we can compute
Consequently,
Therefore, all the conditions of Theorem 3.1 are satisfied, then problem (4.1) has at least three positive solutions , and satisfying
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The first author was partially supported by NNSF of China (10901075).
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Li, J., Wang, J. Triple Positive Solutions for a Type of Second-Order Singular Boundary Problems. Bound Value Probl 2010, 376471 (2010). https://doi.org/10.1155/2010/376471
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DOI: https://doi.org/10.1155/2010/376471