Existence of Multiple Solutions for Second-Order Problem with Stieltjes Integral Boundary Condition

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 e ﬀ ectiveness of our problem better, the example is listed.

For the integral boundary value problems such as (1), many researchers studied mainly by the method of topologi-cal degree. For examples, using the fixed point theorem, Ma [39] studied ordinary second-order equation as below χ ′ ′ + a t ð Þg χ ð Þ = 0, 0 < t < 1, where 0 < ξ < η < 1 and gðχÞ 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 kωk = ½ . 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 g is neither superlinear nor sublinear in our paper. Some ideas of our proof come from [42,43].

Separability and Reflexivity for Space W
In this section, we illustrate that W is a separable and reflexive real Banach space to guarantee our main results. The following theorem is given. Theorem 7. W is a separable and reflexive real Banach space.
Proof. The proof is divided into three parts.
It is well known that W is a normed linear space. Let {ω n ðxÞ} be an arbitrary cauchy sequence of space W . According to Morrey's inequality [47], one has where M 0 is a constant, r ∈ ð0, 1Þ. So, for any n, ω n ðxÞ is uniform convergence in space W , which means By Lebesgue control convergence theorem, we have From the second of (16) and (17), we obtain ωð1Þ = Ð 1 0 ω ðsÞdγðsÞ. Therefore, Part 1. holds.
Define mapping T : where w ∈ Λ. ConstantsÂ, M 1 , M 2 , and M 3 in (3) are denote bŷ Let Θ = T ðwÞ, and it is easy to obtain Θ ∈ W . Consider set b Λ = fM = T ðwÞ: w ∈ Λg; so, b Λ ⊂ W is an enumerable subset. Next, let us demonstrate that there exists M n ⊂ b Λ with lim n→+∞ M n = w, for any w ∈ W . Owing to w ∈ W , there exists sequence fw n g ⊂ Λ such that lim n→+∞ w n = w. Let M n = T ðw n Þ and write M n − w n k k= Hence, Part 2. holds. Part 3. W is reflexive. Define operator P : That is to say, P : ω ⟶ ω′ ∈ L 2 ð0, 1Þ, for all ω ∈ W : Let set Consequently, Next, we will prove operator P is an isometric isomorphic mapping. Evidently, mapping P is surjective. By Definition 1, we only illustrate P as isometric. According to mapping P : W ⟶ Y, we deduce Hence, operator P is an isometric isomorphic mapping. Then, we demonstrate Y that is uniformly convex. For By Lemma 2, Y is uniformly convex. Owing to P is an isometric isomorphic mapping, and W is also uniformly convex. Part 3. holds by Lemma 3.
We complete the proof of the theorem.

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 ðt, ηÞ ∈ ½0, 1 × ℝ, and g : ½0, 1 × ℝ ⟶ ℝ is a continuous function. For our main results, the following lemma is first given.

Journal of Function Spaces
Proof. Consider the following functions. 2α for any ω ∈ W . Notice that the critical points of Γ are the generalized solutions of problem (1). Therefore, we just validate that Y and Φ accord with the conditions of Theorem 6. It is obvious that Y is a continuously Gâteaux differentiable and sequentially weakly lower semicontinuous functional whose Gâ teaux derivative yields a continuous inverse on W * , and Φ is a cntinuously Gâteaux differentiable functional whose Gâ teaux derivative is compact.
By Lemma 5, selecting HðβÞ = ρβ, 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.

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.