A class of second-order three-point integral boundary value problems at resonance is investigated in this paper. Using intermediate value theorems, we obtain a sufficient condition for the existence of the solution for the equation. An example is given to demonstrate our main results.
MSC: 34B10, 34B16, 34B18.
Keywords:integral boundary value problem; resonance; fixed point theorem; intermediate value theorem
We are interested in the existence of the solutions for the following second-order three-point integral boundary value problems at resonance:
where , and .
In the last few decades, many authors have studied the multi-point boundary value problems for linear and nonlinear ordinary differential equations by using various methods, such as Leray-Schauder fixed point theorem, coincidence degree theory, Krasnosel’skii fixed point theorem, the shooting method and Leggett-Williams fixed point theorem. We refer the readers to [1-10] and references therein. Also, there are a lot of papers dealing with the resonant case for multi-point boundary value problems, see [11-17].
In , Infante and Zima studied the existence of solutions for the following n-point boundary value problem with resonance:
where and . Using the Leggett-Williams norm-type theorem, they obtained the existence of a positive solution for problem (1.3)-(1.4).
Problem (1.1)-(1.2) with and was studied by Tariboon and Sitthiwirattham in . They obtained the existence of at least one positive solution. In this paper, we are interested in the existence of the solution for problem (1.1)-(1.2) under the condition , which is a resonant case.
In this paper, using some properties of the Green function and intermediate value theorems, we establish a sufficient condition for the existence of positive solutions of problem (1.1)-(1.2).
The rest of the paper is organized as follows. The main results for problem (1.1)-(1.2) under the condition are given in Section 2. In Section 3, we give some lemmas for our results. We prove our main result in Section 4, and finally an example is given to illustrate our result.
2 Some lemmas and main results
In this section, we first introduce some lemmas which will be useful in the proof of our main results.
Let , equipped with the norm
then Ω is a Banach space.
LetXbe a Banach space with closed and convex. Assume thatUis a relatively open subset ofCwith and is completely continuous. Then either
(i) Thas a fixed point in , or
(ii) there exist and with .
Lemma 2.2Problem (1.1)-(1.2) is equivalent to the following integral equation:
Proof Assume that is a solution of problem (1.1)-(1.2), then it satisfies the following integral equation:
where , are constants. By the boundary value condition (1.2), we obtain
Combining (2.3) with (2.4), we have
According to (2.5) it is easy to see that (2.1) holds.
On the other hand, if is a solution of equation (2.1), deriving both sides of (2.1) two order, it is easy to show that is also a solution of problem (1.1)-(1.2).
Therefore, problem (1.1)-(1.2) is equivalent to the integral equation (2.1) with the function defined in (2.2). The proof is completed. □
Lemma 2.3For any , is continuous, and for any .
Proof The continuity of for any is obvious. Let
Here we only need to prove that for , the rest of the proof is similar. So, from the definition of , and the resonant condition , we have
for . The proof is completed. □
Thus, problem (1.1)-(1.2) is equivalent to the following integral equation:
By a simple computation, the new Green function has the following properties.
Lemma 2.4For any , is continuous, and for any . Furthermore,
Lemma 2.5For any , is nonincreasing with respect to , and for any , , and for . That is, , where
Then , and equation (2.8) gives
Now we let
Then , and equation (2.13) gives
We replace by any real number μ, then (2.15) can be rewritten as
To present our result, we assume that satisfies the following:
(H) and there exist two positive continuous functions such that
where . Furthermore,
for any .
Our results are the following theorems.
Theorem 2.1Assume that (H) holds. If
then problem (1.1)-(1.2) has at least one solution, where
We define an operator T on the set Ω as follows:
Lemma 2.6Assume that and (2.19) hold. Then the operatorTis completely continuous in Ω.
Proof It is not difficult to check that T maps Ω into itself. Next, we divide the proof into three steps.
Step 1. is continuous with respect to .
Suppose that is a sequence in Ω, and converges to . Because of being continuous with respect to and from Lemma 2.4, it is obvious that is uniformly continuous with respect to . Then, for any positive number ε, there exists an integer N. When , we have
It follows from (2.21) and (2.22) that
Thus the operator T is continuous in Ω.
Step 2. T maps a bounded set in Ω into a bounded set.
Assume that is a bounded set with for any . Then we have from (2.17) and (2.21) that
This implies that the operator T maps a bounded set into a bounded set in Ω.
Step 3. T is equicontinuous in Ω.
It suffices to show that for any and any , as . There are the following three possible cases:
Case (i) ;
Case (ii) ;
Case (iii) .
We only need to consider case (i) because the proofs of the other two are similar. Since D is bounded, then there exists such that . From (2.21), for any , we have
Because of Step 1 to Step 3, it follows that the operator T is completely continuous in Ω. The proof is completed. □
Lemma 2.7Assume that and (2.17) and (2.19) hold. Then the integral equation (2.16) has at least one solution for any real numberμ.
Proof We only need to present that the operator T is a priori bounded. Set
and define a set as follows:
To use Lemma 2.1 to prove the existence of a fixed point of the operator T, we need to show that the second possibility of Lemma 2.1 should not happen.
In fact, assume that there exists with and such that . It follows that
Here we use the inequality
Obviously, (2.25) contradicts our assumption that . Therefore, by Lemma 2.1, it follows that T has a fixed point . Hence, the integral equation (2.21) has at least a solution . The proof is completed. □
3 The proof of Theorem 2.1
In this section, we prove Theorem 2.1 by using Lemmas 2.5-2.7 and the intermediate value theorem.
Proof of Theorem 2.1 From the right-hand side of (2.21), we know that (2.21) is continuously dependent on the parameter μ. So, we just need to find μ such that , which implies that .
We rewrite (2.16) for any given real number μ as follows:
From (3.1), it suffices to show that there exists μ such that
Obviously, is continuously dependent on the parameter μ. Our aim here is to prove that there exists such that , we only need to prove that and .
Firstly, we prove that . On the contrary, we suppose that . Then there exists a sequence with such that , which implies that the sequence is bounded. Notice that the function is continuous with respect to and . So, it is impossible to have
as is large enough. Indeed, assume that (3.3) is true. Then by (3.1) we have
Thus we get that
Since we have from (H) that
by (3.2), (3.5) and (3.6), we have
which contradicts our assumption.
Now, for large , we define
Then is not empty.
Secondly, we divide the set into set and set as follows:
Obviously, we get that , . So, we have from (H) that is not empty.
From (H) again, the function is bounded below by a constant for and . Thus, there exists a constant M (<0), independent of t and , such that
From the definitions of and , we have
and it follows that as (since if is bounded below by a constant as , then (3.7) holds). Therefore, we can choose large enough such that
for . From (H), (3.1), (3.8) and (3.9) and the definitions of and , for any , we have
from which it follows that
which implies that
This contradicts (3.9). Thus, we have proved that . By a similar method, we can also prove that .
Notice that is continuous with respect to . It follows from the intermediate value theorem  that there exists such that , that is, , which satisfies the second boundary value condition of (1.2). The proof is completed. □
In this section, we give an example to illustrate our main result.
Example Consider the boundary value problem
So, we have
Now we take
It is easy to check that
Thus the conditions of Theorem 2.1 are satisfied. Therefore problem (4.1)-(4.2) has at least a nontrivial solution.
The authors declare that they have no competing interest.
Each of the authors HL and ZO contributes to each part of this study equally and read and approved the final vision of the manuscript.
The work was partially supported by the Natural Science Foundation of Hunan Province (No. 13JJ3074), the Foundation of Science and Technology of Hengyang city (No. J1) and the Scientific Research Foundation for Returned Scholars of University of South China (No. 2012XQD43).
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