Abstract
In this paper, we consider the existence of multiple solutions for second-order equation with Stieltjes integral boundary condition using the three-critical-point theorem and variational method. Firstly, a novel space is established and proved to be Hilbert one. Secondly, based on the above work, we obtain the existence of multiple solutions for our problem. Finally, in order to illustrate the effectiveness of our problem better, the example is listed.
1. Introduction
We are concerned with the following general second-order differential equation where is a positive parameter, is a continuous function, and is a nondecreasing function, , .
In the past few decades, the boundary value problems have appealed to many scholars in the mathematical field. Generally speaking, the boundary value problems mostly involve in two-point [1–5], three-point [6–8], and multipoint [9–11]. Many physical phenomena were formulated as nonlocal mathematical models with integral boundary conditions [12–21], such as fractional differential equation [22–30], nonlinear singular parabolic equation [31], and general second-order equation [19, 32–40].
For the integral boundary value problems such as (1), many researchers studied mainly by the method of topological degree. For examples, using the fixed point theorem, Ma [39] studied ordinary second-order equation as below where and are either superlinear or sublinear and obtained the existence of positive solutions; Karakostas et al. [19] studied the existence of three positive solutions of the following problem by the theory of fixed point index on a cone; Benchohra et al. [33] investigated the following second-order equation via contraction principle and Leray-Schauder alternative theorem; Galvis [41] considered the nonlinear second-order problem by Schauder’s fixed point theorem. Some other works on fractional equation with integral boundary conditions can be found in [26, 28, 29].
To the best of my knowledge, no one use the theory of critical point and variational method to deal with the existence of solution of problem (1).
In order to discuss problem (1), we introduce a new space as follows: endowed with the norm .
The paper is organized as follows. In Section 2, we introduce some concepts related to solve problem (1) more conveniently. In Section 3, we prove that constructed space is a Hilbert. Section 4 demonstrates the existence of at least three solutions for problem (1) mostly via a three-critical-point theorem. Section 5 gives an example to explain efficacy of our method. Our method is different from those in [19, 33, 39, 41], and the nonlinear term is neither superlinear nor sublinear in our paper. Some ideas of our proof come from [42, 43].
2. Preliminaries
Definition 1. Let and be metric spaces, and the operator meets the conditions as follows: (1)The mapping is surjective(2), for any Then, is isometric isomorphic to ; moreover, is named as an isometric isomorphic mapping.
Lemma 2. Let be linear normed space. Suppose that for any , there exist ()0,, satisfying then, space is uniformly convex.
Lemma 3 (Milman theorem [44]). A uniformly convex Banach space is reflexive.
Lemma 4 (Clarkson inequality [44]). For ((0,1)), , one has
Lemma 5 ([45]). Let be a nonempty set. and are two real functions on . Assume that there are and , such that then, for all , satisfying one has
Theorem 6 (three-critical-point theorem [46]). Let be a separable and reflexive Banach space, is a continuously Gteaux differentiable and sequentially weakly lower semicontinuous functional whose Gteaux derivative generates a continuous inverse on , and is a continuously Gteaux differentiable function whose Gteaux derivative is compact. Assume (1)for all , (2)there exists a continuous concave function such that Then, there exist an open interval and a positive real number such that, for each , the equation has at least three solutions in whose norms are less than .
3. Separability and Reflexivity for Space
In this section, we illustrate that is a separable and reflexive real Banach space to guarantee our main results. The following theorem is given.
Theorem 7. is a separable and reflexive real Banach space.
Proof. The proof is divided into three parts.
Part 1. Space is a Banach space.
It is well known that is a normed linear space. Let {} be an arbitrary cauchy sequence of space . According to Morrey’s inequality [47], one has
where is a constant, . So, for any , is uniform convergence in space , which means
By Lebesgue control convergence theorem, we have
From the second of (16) and (17), we obtain . Therefore, Part 1. holds.
Part 2. Space is separable.
Let be an enumerable subset. Due to the separability of space , there exists with , for any .
Define mapping satisfying
where . Constants , , , and in (3) are denote by
Let , and it is easy to obtain . Consider set ; so, is an enumerable subset.
Next, let us demonstrate that there exists with , for any . Owing to , there exists sequence such that . Let and write
Hence, Part 2. holds.
Part 3. is reflexive.
Define operator :
That is to say, for all Let set
endowed with the norm
we obtain
Consequently,
Next, we will prove operator is an isometric isomorphic mapping. Evidently, mapping is surjective. By Definition 1, we only illustrate as isometric. According to mapping , we deduce
Hence, operator is an isometric isomorphic mapping.
Then, we demonstrate that is uniformly convex. For , such that choosing On account of Lemma 4, particularly , we deduce
By Lemma 2, is uniformly convex. Owing to is an isometric isomorphic mapping, and is also uniformly convex. Part 3. holds by Lemma 3.
We complete the proof of the theorem.
4. Main Results
In this section, we will show there are at least three solutions for problem (1) mainly by Theorem 6. Define function for any , and is a continuous function. For our main results, the following lemma is first given.
Lemma 8. Suppose that there exist six positive constants , , , , and with and , such that (1), ;(2),then, there exist , such that and
Proof. Define Notice that space is normed . It is readily found that . In a simple calculation, we have where By the hypotheses , choosing , , any number and , we have . According to assumption (a), one has Thus, inequality (29) holds.
Theorem 9. Assume that there exist eight positive constants , , , , , , , and with , and , such that (1), ;(2);(3), for and Then, there exists an open interval and a positive real number such that for each , problem (1) has at least three solutions belonging to whose norms are less than .
Proof. Consider the following functions.
for any .
Notice that the critical points of are the generalized solutions of problem (1). Therefore, we just validate that and accord with the conditions of Theorem 6. It is obvious that is a continuously Gteaux differentiable and sequentially weakly lower semicontinuous functional whose Gteaux derivative yields a continuous inverse on , and is a cntinuously Gteaux differentiable functional whose Gteaux derivative is compact.
In addition, by condition and Poincar inequality, one readily has
for any
Next, we prove that there exist satisfying
From Lemma 8, , one has ; on the other hand, . Thus, for all . Thus,
for any . Moreover,
From Lemma 8, one has
So, inequality (36) holds.
By Lemma 5, selecting , we have
Therefore, the proof is complete by Theorem 6.
For the purpose of explaining the validity of our results, an example is given as follows.
Example. Consider the special second-order problem as below. where , . Choosing , defined by
Proof. According to the integrability of and for , where , we have
which means for So, the assumption (a) of Theorem 9 is satisfied.
Define
Thus,
Owing to , one has
Choosing , , , , , and , we guarantee that conditions (b) and (c) of Theorem 9 are satisfied.
Data Availability
No data were used to support the findings of study.
Conflicts of Interest
The authors declare that they have no conflicts of interest.
Acknowledgments
This research is supported by the NSFC of China (62073203) and the Fund of Natural Science of Shandong Province (ZR2018MA022).