# On the Existence of Solutions for a Class of Schrödinger–Kirchhoff-Type Equations with Sign-Changing Potential

Authors: Guocui Yang; Jianwen Zhou; Shengzhong Duan
Journal: Discrete Dynamics in Nature and Society
 (2022)

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## Abstract

In this paper, we consider the following Kirchhoff problem−a+b∫R3∇u2dxΔu+λVxu=up−2u,inR3u∈H1R3 where a,b>0 are constants, λ is a positive parameter, and 4<p<6. Under suitable assumptions on Vx, the existence of nontrivial solution is obtained via variational methods. The potential Vx is allowed to be sign-changing.


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## Body

## 1. Introduction and Main Results

In this paper, we consider the following Kirchhoff type problem:(1)−a+b∫R3∇u2dxΔu+λVxu=up−2u,inR3u∈H1R3,where a,b>0 are constants, λ is a positive parameter, 4<p<6, and the potential V satisfies the following conditions: V1V∈CR3,R and V is bounded belowV2 there exists a constant c>0 such that the set x∈R3:Vx≤c is nonempty and measx∈R3:Vx≤c<+∞, where meas denote the Lebesgue measure in R3This kind of assumptions was first introduced by Bartsch and Qiang Wang [1] in the study of the nonlinear Schrödinger equations and has attracted the attention of several researchers.In recent years, the Kirchhoff problem on a bounded domainΩ⊂RN(2)−a+b∫R3∇u2dxΔu+Vxu=up−2u,inΩ,u=0,on∂Ω,has been studied by many authors (see, for example, [2–8]). More recently, many researchers focused on the Kirchhoff problem defined on the whole space R3, i.e., the following problem:(3)−a+b∫R3∇u2dxΔu+λVxu=fx,u,inR3u∈H1R3,where V:R3⟶R is a potential function and f∈CR3×R,R. In [9], Wu studied (3) by using a symmetric Mountain Pass Theorem under the following assumptions about potential VVV∈CR3,R, infx∈R3Vx≥a0>0, where a0>0 is a constant. Moreover, for any M>0 , measx∈R3:Vx≤M<+∞, where meas denotes the Lebesgue measure in R3.Under this condition, by Lemma 3.4 in [10], the embedding H1R3↪LsR3 is compact for any s∈2,6. Hence, the corresponding results in [9] have been obtained by using the variational techniques in a standard way. In [11–13], the authors considered Kirchhoff type problem (3) with a steep potential well. Precisely, the potential function satisfies the following conditions besides V2: V3V∈CR3,R and V≥0 on R3V4Ω=intV−10 is a nonempty open set with locally Lipschitz boundary and Ω¯=V−10By using this conditions, Sun and Wu [11] considered (3) in the case where the nonlinearity fx,s is asymptotically k-linear k=1,2,4 with respect to s at infinity. Du et al. [12] studied (3) when fx,u behaves like up−2u with 4<p<6 and proved the existence and asymptotic behavior of ground state solutions. Zhang and Du [13] investigated the existence and asymptotic behavior of positive solutions for (3) by combining the truncation technique and the parameter-dependent compactness lemma for b small and λ large in the case where fx,u behave like up−2u with 2<p<4. For more results about Kirchhoff type problems, we refer the reader to [14–18] and the references therein.Under the assumption ofV1, the potential V may change sign. The purpose of this paper is to consider the multiplicity of solutions for (1) in this case. To our best knowledge, there is no existence result of solutions for (1) with sign-changing potentials. Our main result as follows.Theorem 1.
Suppose thatV1 and V2 and 4<p<6 hold. Then, system (1) possesses infinitely many distinct pairs of nontrivial solutions whenever λ>0 is sufficiently large.


## 2. Preliminaries

As a matter of convenience, without loss of generality, we may assume thata=1 and b=1. Consequently, we are dealing with the Kirchhoff type problem as(4)−1+∫R3∇u2dxΔu+λVxu=up−2u,inR3u∈H1R3..Let(5)H1R3=u∈L2R3:∇u∈L2R3,be the usual Sobolev space with the standard inner product and norm as follows:(6)u,v=∫R3∇u∇v+uvdx,u=∫R3∇u2+u2dx1/2.In our problem, we work in the space defined by(7)Eλ=u∈H1R3:∫R3∇u2+λV+xu2dx<+∞,with the inner product and the norm as follows:(8)u,vEλ=∫R3∇u⋅∇v+λV+xuvdx,uEλ=u,uEλ1/2,where V±x=max±Vx,0 and Vx=V+x−V−x. It follows from the conditions V1andV2 and the Hölder and Sobolev inequalities that(9)∫R3∇u2+u2dx≤∫R3∇u2dx+Vc2/3∫Vcu6dx1/3+1c∫R3/VcVxu2dx≤max1+Vc2/3S−1,c−1∫R3∇u2+Vxu2dx,which implies that the embedding Eλ↪H1R3 is continuous. Here, S is the best constant for the embedding of D1,2R3 in L6R3. Combine with the continuity of the following embedding:(10)H1R3↪LsR3,2≤s≤6.There is a constantas>0 such that(11)us≤asuEλ,∀u∈Eλ.As a consequence, the functionalIλ:Eλ⟶R given by(12)Iλu=12∫R3∇u2+λVxu2dx+14∫R3∇u22−1p∫R3updx.is well defined, and it is of class C1 with derivative(13)<Iλ′u,v>=∫R3∇u∇vdx+∫R3λVxuvdx+∫R3∇u2dx∫R3∇u∇vdx−∫R3up−2uvdx.for all u,v∈Eλ. As in [19], let(14)Fλ=u∈Eλ:suppu⊂V−10,∞,and denote the orthogonal complement of Fλ in Eλ by Fλ⊥. Consider the eigenvalue problem(15)−Δu+V+xu=μV−xu,u∈Fλ⊥.In view ofV1 and V2, the quadratic form u↦∫R3V−xu2dx is weakly continuous. We have the following proposition.Proposition 1 (see Lemma 2.1 in [19]).
SupposeV1 and V2 and V−≠0 hold. Then, for each fixed j,iμjλ⟶0 as λ⟶+∞iiμjλ is a non-increasing continuous function of λwhereμjλ=infdimM≥j,M⊂F⊥/λsupuEλ2:u∈M,∫R3λV−xu2dx=1j=1,2,3,… is sequence of positive eigenvalues of problem P satisfying μ1λ≤μ2λ≤⋯≤μjλ⟶∞ as j⟶∞ and the corresponding eigenfunctions ejλj=1∞.
Let(16)Eλ−=spanejλ:μjλ≤1,Eλ+=spanejλ:μjλ>1.
Then,(17)Eλ=Eλ−⊕Fλ⊕Eλ−.Moreover, dimEλ−<+∞ for every fixed λ>0.To complete the proof of our theorem, we need the following results.Theorem 2 (see Theorem 9.12 in [20]).
LetE be an infinite dimensional Banach space, and let I∈C1E,R be even, satisfying (PS) condition and I0=0. If E=V⊕X, where V is finite dimensional and I satisfies the following:I1 there are constants ρ,α>0 such that I|∂Bρ∩X≥αI2 for each finite dimensional subspace E˜⊂E, there is an R=RE˜ such that I≤0 on E˜/BRE˜thenI possesses an unbounded sequence of critical values.


## 3. Proof of Main Results

Lemma 1.
Suppose that4<p<6 and V1 and V2 hold. Then, there exist α,ρ>0 such that Iλu≥α for all u∈Eλ with uλ=ρ.Proof.
By Proposition 1, for each fixedλ>Λ, there exists a positive integer nλ such that μjλ≤1 for j<nλ and μjλ>1 for j≥nλ. Thus, for any u=u1+u2∈Eλ+⊕Fλ, we have(18)Iλu=12uEλ2−12∫R3λV−xu2dx+14∫R3∇u2dx2−1p∫R3updx≥12∫R3∇u12+λVxu12dx+12u2Eλ2−1p∫R3updx≥121−1μnλλuEλ2−1pupp≥121−1μnλλuEλ2−CuEλp≥121−1μnλλ−CuEλp−2uEλ2,for all u∈Bρ0¯, where Bρ0=u∈Eλ+⊕Fλ:uEλ<ρ. Since p>2, the conclusion follows by choosing ρ sufficiently small.Lemma 2.
Suppose that4<p<6 and V1andV2 hold. Then, there is a large r>0 such that Iu<0 on E˜/Br0.Proof.
Since all norms are equivalent in a finite dimensional space, there are constantsCp>0 and C>0 such that(19)uD1,2R3≤CuEλ,up≥CpuEλ,∀u∈E˜⊂Eλ.where uD1,2R32=∫R3∇u2dx. Hence, for all u∈E˜,(20)Iλu=12uEλ2−12∫R3λV−xu2dx+14∫R3∇u2dx2−1p∫R3updx≤12uEλ2+C4uEλ4−1pupp≤12uEλ2+C4uEλ4−CppuEλp.
Sincep>4, consequently, there is a large r>0 such that Iu<0 on E˜/Br0.Lemma 3.
Let4<p<6 and V1andV2 be satisfied. Then, there exists Λ>0 such that, for each c∈R, Iλ satisfies the PSc condition for all λ≥Λ.Proof.
Letun be a PSc sequence, that is, Iλun⟶c and Iλ′un⟶0. If un is unbounded in Eλ, up to a subsequence, we can assume that(21)unEλ⟶+∞,Iλun⟶c,Iλ′un⟶0,as n⟶∞, after passing to a subsequence. Set wn=un/unEλ, we can assume that wn⇀w in Eλ and wnx⟶wx a.e. x∈R3.
Ifw=0, since u↦∫R3V−xu2dx∈ is weakly continuous, we have(22)o1=1unEλ2Iλun−1pIλ′un,un=12−1p−12−1p∫R3λV−xwn2dx+14−1p1unEλ2∫R3∇un2dx2≥12−1p+o1,a contradiction. If w≠0, then the set Ω=x∈R3:ωx≠0 has positive Lebesgue measure. For x∈Ω, one has unx⟶∞ as n⟶∞; Fatou’s lemma shows that ∫Ωunp−4wn4dx⟶∞ as n⟶∞. Thus, by (9), we obtain(23)1/p∫R3unp−4wn4dx=1/2unEλ2−1/2unEλ2∫R3λV−xun4dx+1/4unEλ4∫R3∇un2dx2+o1≤1/2unEλ2+1/4unEλ4∫R3∇un2dx2+o1≤C/4+o1.
This is a contradiction. This impliesun is bounded in Eλ. We assume that unEλ≤T. Passing to a subsequence if necessary, we can assume that there exists u∈Eλ and A∈R such that(24)un⇀uinEλ,∫R3∇u2dx⟶A2,∫R3∇u2dx≤A2.
Then,Iλ′un⟶0 implies that(25)∫R3∇u∇vdx+∫R3λVxuvdx+A2∫R3∇u∇vdx−∫R3up−2uvdx=0,∀v∈Eλ.
Takingv=u in (25), we obtain(26)∫R3∇u2dx+∫R3λVxu2dx+A2∫R3∇u2dx−∫R3updx=0,∀v∈Eλ.
Letvn≔un−u. It follows from V1andV2 that(27)vn22=∫Vx≥cvn2dx+∫Vx<cvn2dx≤1λcvnEλ2+o1.
Moreover, Let0<α<min6−p/2,1,2<p<6. Then, 2<2p−α/2−α<6. By Sobolev inequalities and Hölder inequality, one has(28)vnpp=∫R3vnαvnp−αdx≤∫R3vn2dxα/2∫R3vn2p−α/2−αdx2−α/2=vn2αvn2p−α/2−αp−α≤Cλc−α/2vnEλp+o1,we know(29)o1=Iλ′un,un=unEλ2−∫R3λV−xun2dx+∇un24−unpp−uEλ2+∫R3λV−xu2dx−A2∇un22+upp=vnEλ2−vnpp+A4−A2∇u22+o1≥vnEλ2−vnpp−2vnp2+o1≥1−2apTp−2Cp−2/pλc−αp−2/2pvnEλ2+o1.
LettingΛ>0 be so large that the term in the brackets above is positive when λ≥Λ, we get vn⟶0 in Eλ. Since vn=un−u and vn⟶0, it follows that un⟶u in Eλ. This completes the proof.Proof of Theorem 1.
Obviously,I0=0. Furthermore, I is even. The conclusion follows from Lemmas 1–3 and Theorem 1.



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