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

Authors: Qing-Qing Hu; Baoqiang Yan
Journal: Journal of Function Spaces
 (2021)

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## 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.


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## Body

## 1. Introduction

We are concerned with the following general second-order differential equation(1)−ω′′t=βgt,ωt,ω0=0,ω1=∫01ωsdγs,where β is a positive parameter, g:0,1×ℝ⟶ℝ is a continuous function, and γs is a nondecreasing function, 0<∫01sdγs<1, 0<∫01s2dγs<1.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
(2)χ′′+atgχ=0,0<t<1,χ0=0,χ1=∫ξηγzχzdz,where 0<ξ<η<1 and gχ 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
(3)χ′′+χtgχ=0,0≤t≤1,χ0=0,χ1=∫ξηχzdγzby the theory of fixed point index on a cone; Benchohra et al. [33] investigated the following second-order equation
(4)χ′′+gt,χ=0,0≤t≤1,χ0=0,χ1=∫01χzγzdzvia contraction principle and Leray-Schauder alternative theorem; Galvis [41] considered the nonlinear second-order problem
(5)χ′′+χtgt,χ=0,0≤t≤r,χ0=0,χr=β∫0aχzdzby 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:
(6)W0,1=ω∈W1,20,1:ω0=0,ω1=∫01ωsdγs,endowed with the norm ω=∫01ω′2dt1/2.The paper is organized as follows. In Section2, 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].


## 2. Preliminaries

Definition 1.
LetE¯,ς and E¯1,ς1 be metric spaces, and the operator ϕ~:E¯⟶E¯1 meets the conditions as follows:
(1)
The mappingϕ~ is surjective(2)
ςω,ν=ς1ϕ~ω,ϕ~ν, for any ω,ν∈E¯
Then,E¯,ς is isometric isomorphic to E¯1,ς1; moreover, ϕ~ is named as an isometric isomorphic mapping.Lemma 2.
LetE¯ be linear normed space. Suppose that for any ε~∈0,2, there exist δ~=δ~(ε~)>0,ω+ν=1, ω−ν≥ε satisfying
(7)ω+ν2≤1−δ~ε~;then, space E¯ is uniformly convex.Lemma 3 (Milman theorem [44]).
A uniformly convex Banach space is reflexive.Lemma 4 (Clarkson inequality [44]).
Forω,ν∈Lp((0,1)), 2≤q<+∞, one has
(8)ω+ν2p+ω−ν2p≤ωp+νp2.Lemma 5 ([45]).
LetX be a nonempty set. Y and Φ are two real functions on X. Assume that there are α>0 and ω0, ω1∈X such that
(9)Yω0=Φω0=0,Yω1>α,supω∈Y−1£¨−∞,α£©Φx<αΦω1Yω1;then, for all ρ, satisfying
(10)supω∈Y−1£¨−∞,α£©Φx<ρ<αΦω1Yω1,one has
(11)supβ≥0infω∈XYω+βρ−Φω<infω∈Xsupβ≥0Yω+βρ−Φω.Theorem 6 (three-critical-point theorem [46]).
LetX be a separable and reflexive Banach space, Y:X⟶ℝ is a continuously Ga^teaux differentiable and sequentially weakly lower semicontinuous functional whose Ga^teaux derivative generates a continuous inverse on X∗, and Φ:X⟶ℝ is a continuously Ga^teaux differentiable function whose Ga^teaux derivative is compact. Assume
(1)
for allβ∈0,+∞,
(12)lim∥ω∥→+∞Yω+βΦω=+∞,(2)
there exists a continuous concave functionH:0,+∞⟶ℝ such that
(13)supβ≥0infω∈XYω+βΦω+Hβ<infω∈Xsupβ≥0Yω+βΦω+Hβ.
Then, there exist an open intervalΩ⊆0,+∞ and a positive real number p such that, for each β∈Ω, the equation
(14)Y′ω+βΦ′ω=0
has at least three solutions inX whose norms are less than p.


## 3. Separability and Reflexivity for SpaceW

In this section, we illustrate thatW 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.
Part 1. Space W is a Banach space.
It is well known thatW is a normed linear space. Let {ωnx} be an arbitrary cauchy sequence of space W. According to Morrey’s inequality [47], one has
(15)ωnx−ωxC0,r≤M0ωnx−ωxW,where M0 is a constant, r∈0,1. So, for any n, ωnx is uniform convergence in space W, which means
(16)ωn0=ω0=0,limn→+∞ωn1=ω1.
By Lebesgue control convergence theorem, we have(17)limn→+∞ωn1=limn→+∞∫01ωnsdγs=∫01limn→+∞ωnsdγs=∫01ωsdγs.
From the second of (16) and (17), we obtain ω1=∫01ωsdγs. Therefore, Part 1. holds.
Part 2. Space W is separable.
LetΛ⊂W1,20,1 be an enumerable subset. Due to the separability of space W1,20,1, there exists wn⊂Λ with limn→+∞wn=w, for any w∈W1,20,1.
Define mappingT:Λ⟶W satisfying
(18)Twt=wt+1M1−A^t2+t∫01wsdγs−y1−M2t2+M3t+1w0,where w∈Λ. Constants A^, M1, M2, and M3 in (3) are denote by
(19)A^=1−∫01tdγt1−∫01t2dγt,M1=∫01tdγt−∫01t2dγt1−∫01t2dγt,M2=∫01dγt−∫01tdγt∫01tdγt−∫01t2dγt,M3=∫01t2dγt−∫01dγt∫01tdγt−∫01t2dγt.
LetΘ=Tw, and it is easy to obtain Θ∈W. Consider set Λ^=M=Tw:w∈Λ; so, Λ^⊂W is an enumerable subset.
Next, let us demonstrate that there existsMn⊂Λ^ with limn→+∞Mn=w, for any w∈W. Owing to w∈W, there exists sequence wn⊂Λ such that limn→+∞wn=w. Let Mn=Twn and write
(20)Mn−wn=∫01Mn−wn′2dt1/2=∫011M1−2A^t+1∫01ωnsdγs−ωn1−2M2t+M3ωn02dt1/2=∫011M12−2A^t+12∫01ωnsdγs−ωn12+2M2t+M32ωn20−21M1−2A^t+12M2t+M3∫01ωnsdγs−ωn1ωn0dt1/2=4A^23+1−2A^∫01ωnsdγs−ωn12+4M223+M32+2M2M3ωn20−2M1−4A^M23−A^M3+M2+M3∫01ωnsdγs−ωn1ωn01/2→4A^23+1−2A^∫01ωsdγs−ω12+4M223+M32+2M2M3ω20−2M1−4A^M23−A^M3+M2+M3∫01ωsdγs−ω1ω01/2=0,asn→+∞.
Hence,Part 2. holds.
Part 3.W is reflexive.
Define operatorP:
(21)W⟶L20,1.
That is to say,P:ω⟶ω′∈L20,1, for all ω∈W. Let set
(22)Y=ω′:ω∈Wendowed with the norm
(23)v=∫01v2dt,we obtain
(24)Y⊂L20,1.
Consequently,(25)Y⊂L20,1,ω′=v.
Next, we will prove operatorP 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
(26)ω=v=Pω.
Hence, operatorP is an isometric isomorphic mapping.
Then, we demonstrateY that is uniformly convex. For ∀ε~>0, ω,v∈Y, such that ω=v=1,ω−v≥ε~, choosing δ~ε~=1−1−ε~2/4>0. On account of Lemma 4, particularly p=2, we deduce
(27)ω+v2≤12ω2+v2−ω−v22≤1−ε~24=1−δ~ε~.
By Lemma2, 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.


## 4. Main Results

In this section, we will show there are at least three solutions for problem (1) mainly by Theorem 6. Define function
(28)ρt,η=∫0ηgt,sdsfor any t,η∈0,1×ℝ, and g:0,1×ℝ⟶ℝ is a continuous function. For our main results, the following lemma is first given.Lemma 8.
Suppose that there exist six positive constantsK, P, L, A1, A2andA^ with P<LK/8A^ and 0<A1<A2<1, such that
(1)
ρt,η≥0, t,η∈0,A1∪A2,1×0,L/4A^;(2)
maxt,η∈0,1×−P,Pρt,η<64A^2P2/KL2∫A1A2ρt,L/4A^dt,then, there exist α>0, μ∈W such that 2α<μ2 and
(29)maxt,η∈0,1×−P,Pρt,η<2α∫01ρt,μtdtμ2.Proof.
Define(30)μt=Lt−1−∫01tdγt1−∫01t2dγtt2.
Notice that spaceW is normed μ=∫01μ′2dt1/2. It is readily found that μt∈W. In a simple calculation, we have
(31)μ2=∫01μ′2dt=L21−10A^32,where
(32)A^=1−∫01tdλt1−∫01t2dλt.
By the hypothesesP<LK/8A^, choosing α=2P2, 0<A1<1/2A^<A2<1, any number and K=163−10A^2A^2/9, we have 2α<∥μ∥2. According to assumption (a), one has
(33)2α∫01ρt,μtdtω2≥4P2μ2∫A1A2ρt,L4A^dt=4P2L21−10A^/32∫A1A2ρt,L4A^dt=64A^2P2KL2∫A1A2ρt,L4A^dt>maxt,η∈0,1×−P,Pρt,η.
Thus, inequality (29) holds.Theorem 9.
Assume that there exist eight positive constantsK, P, L, A1, A2, c, κ, and A^ with P<LK/8A^, 0<A1<A2<1 and κ<2, such that
(1)
ρt,η≥0, t,η∈0,A1∪A2,1×0,L/4A^;(2)
maxt,η∈0,1×−P,Pρt,η<64A^2P2/KL2∫A1A2ρt,L/4A^dt;(3)
ρt,η≤c1+ηκ, for t∈0,1 and η∈ℝ
Then, there exists an open intervalΩ⊆0,+∞ and a positive real number p such that for each β∈Λ, problem (1) has at least three solutions belonging to W whose norms are less than p.Proof.
Consider the following functions.(34)2α∫01ρt,μtdt∥ω∥2≥4P2∥μ∥2∫A1A2ρt,L4A^dt=4P2L21−10A∧/32∫A1A2ρt,L4A^dt=64A∧2P2KL2∫A1A2ρt,L4A^dt>maxt,η∈0,1×−P,Pρt,η.for any ω∈W.
Notice that the critical points ofΓ 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 Ga^teaux differentiable and sequentially weakly lower semicontinuous functional whose Ga^teaux derivative yields a continuous inverse on W∗, and Φ is a cntinuously Ga^teaux differentiable functional whose Ga^teaux derivative is compact.
In addition, by conditionc and Poincare′ inequality, one readily has
(35)lim∥ω∥→+∞Γω=lim∥ω∥→+∞Yω+βΦω=+∞,
for anyβ≥0.
Next, we prove that there existα>0,ω∈W satisfying
(36)supω∈Φ−1£¨−∞,α£©−Φω<−αΦωYω.
From Lemma8, ωt=μt, one has maxt∈0,1ωt=L/4A^; on the other hand, ω=L1−10A^/3. Thus, maxt∈0,1∣ω∣≤ω, for all ω∈W. Thus,
(37)Y−1−∞,α⊆ω∈W:ω≤2α,0≤t≤1,for any α>0. Moreover,
(38)supω∈Y−1−∞,α−Φω≤sup∥ω∥2≤2α∫01ρt,ωdt≤maxt,η∈0,1×−α2,α2ρt,η.
From Lemma8, one has
(39)maxt,η∈0,1×−α2,α2ρt,η<2α∫01ρt,ωtdtω2=−αΦωYω.
So, inequality (36) holds.
By Lemma5, selecting Hβ=ρβ, we have
(40)supβ≥0infω∈WYx+βΦx+ρ<infω∈Wsupβ≥0Yx+βΦx+ρ.
Therefore, the proof is complete by Theorem6.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.(41)−ω″t=βtfω,ω0=0,ω1=∫01ωsds,where gt,ω=tfω, γs=s. Choosing L=30, fω defined by
(42)fω=eω,ifω≤L3,ω+eL/3−L3,ifω>L3.Proof.
According to the integrability offω and fω≥0 for ω∈0,η, where η∈0,L/4A^, we have
(43)∫0ηfydy≥0,which means t∫0ηfydy≥0, for t∈0,A1∪A2,1. So, the assumption (a) of Theorem 9 is satisfied.
Define(44)Fη=∫0ηfydy.
Thus,(45)Fη=eη−1,ifη≤L3,ηeL/3−L3+eL/31−L3+L218+η22−1,ifη>L3.
Owing togt,ω=tfω, one has
(46)maxt,η∈0,1×−P,Pρt,η≤maxt,η∈0,1×−P,P∫0ηgt,ydy=maxt∈0,1tFP.
Choosingκ=3/2, c=eL/3, P=1, K=9, A2=1/2, and A2=3/4, we guarantee that conditions (b) and (c) of Theorem 9 are satisfied.



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