# Nontrivial Solutions for 4-Superlinear Schrödinger–Kirchhoff Equations with Indefinite Potentials

Authors: Wei Chen; Yue Wu
Journal: Journal of Function Spaces
 (2021)

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## Abstract

This paper is devoted to the 4-superlinear Schrödinger–Kirchhoff equation−a+b∫ℝ3∇u2dxΔu+Vxu=fx,u,inℝ3, where a>0, b≥0. The potential V here is indefinite so that the Schrödinger operator −Δ+V possesses a finite-dimensional negative space. By using the Morse theory, we obtain nontrivial solutions for this problem.


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## Body

## 1. Introduction and Main Results

In this work, we consider the Schrödinger–Kirchhoff type equation of the form(1)−a+b∫ℝ3∇u2dxΔu+Vxu=fx,u,x∈ℝ3,where a>0, b≥0 are constants. This equation arises when we look for stationary solutions of the equation(2)ρ∂2u∂t2−P0h+E2L∫0L∂u∂x2dx∂2u∂x2=0proposed by Kirchhoff [1] as an extension of the classical D’Alembert’s wave equation for free vibrations of elastic strings, where ρ, h, P0, and L are positive constants. In [2], J.L. Lions introduced an abstract functional analysis framework to the equation(3)utt−a+b∫Ω∇u2dxΔu=fx,u,where Ω⊂ℝ3 is a bounded domain. After that, (3) received much attention. For more details on the physical and mathematical background of this problem, we refer to [1–3] for references.Problem (1) has been studied extensively by many researchers. Some interesting studies by variational methods can be found in, for example, [3–13] and references therein. Under suitable conditions, it is well known that weak solutions to (1) correspond to critical points of the energy functional Φ:H1ℝ3→ℝ,(4)Φu=12∫ℝ3a∇u2+Vxu2dx+b4∫ℝ3∇u2dx2−∫ℝ3Fx,udx,where Fx,t=∫0tfx,sds. We emphasize that in the papers mentioned above, the authors only considered the case that the Schrödinger operator −Δ+V is positively definite. In this situation, a typical way to deal with (1) is to use the mountain pass theorem (cf. [14]). However, when the potential V is negative somewhere, the zero function u=0 is no longer a local minimizer of Φ. In this case, the functional Φ would not enjoy the general linking geometry due to the nonlocal term ∫ℝ3∇u2dx. Hence, the classical linking theorem (see, e.g., ([15], Theorem 2.12)) is also not applicable.Such an indefinite situation was studied in [16]. To overcome these difficulties and the difficulty that the Sobolev embedding H1ℝ3°L2ℝ3 is not compact, it is assumed in [16] that(5)μV−1−∞,M<∞forallM>0,V0,so that the related weighted Sobolev space is compactly embedded into L2ℝ3. Then, via Morse theory, they obtained nontrivial critical points of Φ.In this paper, we will consider the case of more generalV such that the abovementioned compact embedding may not be true. We assume the potential V satisfies.(V)V∈Cℝ3 is bounded such that the quadratic form(6)Qu≔∫ℝ3a∇u2+Vxu2dx,which is nondegenerate and the negative space of Q is finite-dimensional.For the nonlinearityf, we make the following assumptions:(f1) f∈Cℝ3×ℝ there exist C>0 and p∈4,6 such that(7)fx,t≤C1+tp−1forallx∈ℝ3;(f2) fx,t=ot as t→0 uniformly in x∈ℝ3 and(8)limt→∞Fx,tt4=+∞forallx∈ℝ3;(f3) 0<4Fx,t≤tfx,t for x∈ℝ3 and t≠0;(f4) for any r>0, there holds(9)limx→∞sup0<t≤rfx,tt=0.Remark 1.
(i)
(V) is a generic assumption onV. For example, if V:ℝ3→ℝ such that(10)limx→∞Vx=V∞∈0,∞,and 0 does not happen to be a spectrum point of the Schrödinger operator −Δ+V, then such V satisfies our assumption (V).(ii)
Note that in (f2), we have not required that the limit (8) holds uniformly(iii)
In order to produce critical points ofΦ, eventually, we will encounter the compactness problem. For this issue, we assume assumption (f4). It is easily see that let a:ℝ3→0,∞ be continuous, limx→∞ax=0 and p∈4,6, then(11)fx,t=axtp−2tsatisfies (f1)–(f4).Now, we are ready to state our main results.Theorem 2.
Assume that (V) and (f1)–(f4) are satisfied. Then, problem ((1)) has a nontrivial solution.Theorem 3.
Assume that (V) and (f1)–(f4) are satisfied. Moreover, iffx,·is odd, then problem ((1)) has a sequence of solutionsunsuch that the energyΦun→+∞.It is known that if the quadratic formQ is indefinite, usually, it is more difficult to verify the boundedness of the (PS) sequence. In [16], this is done by taking advantage of the compact embedding mentioned before. Under our present setting, the related Sobolev embedding H1ℝ3°L2ℝ3 is not compact. We will illustrate a general technique to establish the boundedness of (PS) sequences. Moreover, it is also worth to point out that the weak limit of the bounded (PS) sequence is not obviously a critical point of Φ. Since we cannot easily see that Φ′ is weakly sequentially continuous in H1ℝ3 by direct calculations due to the existence of nonlocal term ∫ℝ3∇u2dx. Indeed, in general, we do not know ∫ℝ3∇un2dx→∫ℝ3∇u2dx from un⇀u in H1ℝ3.The paper is organized as follows. In the next section, we prove that the (PS) sequences ofΦ are bounded and Φ satisfies the (PS) condition. In Section 3, we recall some concepts and results in infinite-dimensional Morse theory; then, we analyze the critical groups of Φ at infinity; finally, we give the proof of Theorem 11. Having established the (PS) condition, the proof of Theorem 3 is quite similar to that of ([16], Theorem 3); therefore, we omit it here.


## 2. Palais-Smale Condition

Throughout of this paper, we always denoteE=H1ℝ3. In view of assumption (V), we may choose an equivalent norm · on E such that(12)Qu=u+2−u−2,u=u++u−,u±∈E±,where E+ and E− are positive and negative spaces of Q, respectively, and E=E+⊕E−. Here and in what follows, u± denotes the orthogonal projection of u on E±. Then, the functional Φ can be rewritten as(13)Φu=12u+2−u−2+b4∫ℝ3∇u2dx2−∫ℝ3Fx,udx.By (f1), Φ is of class C1 on E with the derivative given by(14)Φ′u,v=∫ℝ3a∇u·∇v+Vxuvdx+b∫ℝ3∇u2dx∫ℝ3∇u·∇vdx−∫ℝ3fx,uvdx,for all u,v∈E. Then, u solves (1) if and only if it is a critical point of Φ.Lemma 4.
Letunbe a (PS) sequence ofΦ, that is,(15)supnΦun<∞,Φ′un→0.
Then,un is bounded in E.Proof.
Suppose by contradiction thatun→∞. Let vn=un−1un. Then, passing to a subsequence, there exists v∈E such that(16)vn=vn++vn−⇀v=v++v−inE.Ifv=0, then vn−→v− since dimE−<∞. Noting that vn+2+vn−2=1, we have(17)vn+2−vn−2≥12.By (f3) we deduce that, for n large enough,(18)1+supnΦun+un≥Φun−14Φ′un,un=14un+2−un−2+∫ℝ314fx,unun−Fx,undx≥14un2vn+2−vn−2≥18un2,contradicting un→∞.Now suppose thatv≠0. Then, the set Θ=x∈ℝ3:vx≠0 has positive Lebesgue measure. For x∈Θ, we have unx→∞, and hence (8) implies(19)Fx,unxun4xvn4x→+∞.Then, the Fatou lemma yields(20)∫ΘFx,unun4vn4dx→+∞.On the other hand, for largen,(21)∫ΘFx,unun4vn4dx=1un4∫ΘFx,undx≤1un4∫ℝ3Fx,undx=1un412un+2−un−2+b4∫ℝ3∇un2dx2≤1+b4∇un24un4≤1+M,where M>0 is a constant and ·s denotes the standard norm in Lsℝ3. This is also a contradiction.In conclusion, we deduce that the (PS) sequenceun is bounded.To get a convergent subsequence of the (PS) sequence, we need the following lemma.Lemma 5.
Letun⇀uinE. Then,(22)lim¯n→∞∫ℝ3∇u·∇un−udx≥0.Proof.
Sinceun⇀u in E, we have ∇un⇀∇u in L2ℝ3. Hence,(23)∫ℝ3∇un·∇udx→∫ℝ3∇u2dx.Defineψu=∫ℝ3∇u2dx in E. It is easy to see that ψ is continuous and convex. Hence, ψ is weakly lower semicontinuous in E, so that(24)lim¯n→∞∫ℝ3∇un2dx≥∫ℝ3∇u2dx.Consequently,(25)lim¯n→∞∫ℝ3∇u·∇un−udx=lim¯n→∞∫ℝ3∇un2dx−limn→∞∫ℝ3∇un·∇udx≥∫ℝ3∇u2dx−∫ℝ3∇u2dx=0.The lemma is proved.Lemma 6.
Φsatisfies the (PS) condition.Proof.
Letun be a (PS) sequence. From Lemma 4, we know that un is bounded in E. Up to a subsequence, we may assume that un⇀u in E. Hence,(26)∫ℝ3a∇un·∇u+Vxunudx→∫ℝ3a∇u2+Vxu2dx=u+2−u−2.Consequently,(27)o1=Φ′un,un−u=∫ℝ3a∇un·∇un−u+Vxunun−udx+b∫ℝ3∇un2dx∫ℝ3∇un·∇un−udx−∫ℝ3fx,unun−udx=∫ℝ3a∇un2+Vxun2dx−∫ℝ3a∇un·∇u+Vxunudx+b∫ℝ3∇un2dx∫ℝ3∇un·∇un−udx−∫ℝ3fx,unun−udx=o1+un+2−u+2−un−2−u−2+b∫ℝ3∇un2dx∫ℝ3∇un·∇un−udx−∫ℝ3fx,unun−udx.BecausedimE−<∞, we have un−→u− and thus un−→u−. Collecting all infinitesimal terms, we obtain(28)un+2−u+2=o1+∫ℝ3fx,unun−udx−b∫ℝ3∇un2dx∫ℝ3∇un·∇un−udx.We claim that(29)lim¯n→∞∫ℝ3fx,unun−udx≤0.Givenε>0. For r≥1, by (f1) and Hölder’s inequality, we have(30)∫un≥rfx,unun+−u+dx≤2C∫un≥runp−1un+−u+dx≤2Crp−6∫un≥run5un+−u+dx≤2Crp−6un65un+−u+6.Sincep<6, we may fix r large enough such that,(31)∫un≥rfx,unun+−u+dx≤ε3,for all n. Moreover, it follows from (f4) that there exists R>0 such that(32)∫x≥Run≤rfx,unun+−u+dx≤un2un+−u+2supt≤r,x≥Rfx,tt≤ε3,for all n. Finally, from (f1) and (f2) we deduce that, for any ε>0, there exists cε>0 such that(33)Fx,t≤εt2+cεtp,fx,t≤εt+cεtp−1.Note thatun+→u+ in LsBR0 for every s∈2,2∗. Consequently, for n large enough,(34)∫x≤Run≤rfx,unun+−u+dx≤∫x≤Run≤runun+−u+dx+c1∫x≤Run≤runp−1un+−u+dx≤un2un+−u+L2BR0+c1unpp−1un+−u+LpBR0≤ε3.Combining (31), (32), and (34), we obtain that(35)∫ℝ3fx,unun+−u+dx≤ε,for n large enough. Since dimE−<∞, we obtain that (29) holds by the arbitrariness of ε.Now using Lemma5, we deduce from (28) that(36)lim¯n→∞un+2−u+2=lim¯n→∞∫ℝ3fx,unun−udx−b∫ℝ3∇un2dx∫ℝ3∇un·∇un−udx=lim¯n→∞∫ℝ3fx,unun−udx−lim¯n→∞b∫ℝ3∇un2dx∫ℝ3∇un·∇un−udx≤lim¯n→∞∫ℝ3fx,unun−udx≤0.Combining this with the weakly lower semicontinuous of the norm, we have(37)u+2≤lim¯n→∞un+2≤lim¯n→∞un+2≤u+2.That isun+→u+. Reminding un−→u−, we get un→u. Thus, un→u in E.


## 3. Critical Groups and the proof of Theorem11

Before giving the proof of Theorem11, we recall some concepts and results of infinite dimensional Morse theory [17].LetX be a Banach space, φ:X→ℝ be a C1 functional, u be an isolated critical point of φ and φu=c. Then,(38)Cqφ,u≔Hqφc,φc\0,q=0,1,2,⋯,is called the qth critical group of φ at u, where φc≔φ−1−∞,c and H∗ stands for the singular homology with coefficients in ℤ.Ifφ satisfies the (PS) condition and the critical values of φ are bounded from below by α, then following Bartsch and Li [18], we call(39)Cqφ,∞≔HqX,φα,q=0,1,2,⋯the qth critical group of φ at infinity. It is well known that the homology on the right hand does not depend on the choice of α.Proposition 7 (see [18]).
Ifφ∈C1X,ℝsatisfies the (PS) condition andCkφ,0≠Ckφ,∞for somek∈ℕ, thenφhas a nonzero critical point.Proposition 8 (see [19]).
Supposeφ∈C1X,ℝhas a local linking at0, i.e.,X=Y⊕Zand(40)φu≤0foru∈Y∩Bρ,φu≥0foru∈Z\0∩Bρ,for some ρ>0, where Bρ=u∈X:u≤ρ. If k=dimY<∞, then Ckφ,0≠0.For the proof of Theorem11, we may assume the Φ has only finitely many critical points. Since Φ satisfies the (PS) condition, the critical group C∗Φ,∞ of Φ at infinity makes sense. To study C∗Φ,∞, we need the following lemma.Before state it, we point out that the proof of the following lemma is quite different and more general ([16], Lemma 2.4), because in our case, the working Sobolev space is H1ℝ3, which can not compactly embedded into L2ℝ3.Lemma 9.
There existsA>0such that, ifΦu≤−A, then(41)ddtt=1Φtu<0.Proof.
Otherwise, there exists a sequenceun⊂E such that Φun≤−n but(42)Φ′un,un=ddtt=1Φtu≥0.Consequently,(43)un+2−un−2≤un+2−un−2+∫ℝ3fx,unun−4Fx,undx=4Φun−Φ′un,un≤−4n.Letvn=un−1un and vn± be the orthogonal projection of vn on E±. Then, vn−→v− for some v−∈E−, because dimE−<∞.Ifv−≠0, then for some v∈E\0 we have vn⇀v in E. Similar to (20), we obtain(44)∫ℝ3fx,ununun4dx≥4∫ℝ3Fx,unun4dx≥4∫v≠0Fx,unun4vn4dx→+∞.Hence, by (42), we get(45)0≤Φ′un,unun4=un+2−un−2un4+b∇un24un4−∫ℝ3fx,ununun4dx≤1+M−∫ℝ3fx,ununun4dx→−∞,a contradiction. Therefore, v−=0. From(46)vn+2+vn−2=1,we see that vn+→1. Consequently, for n large enough,(47)un+=unvn+≥uvn−=un−,violating (43). Hence, the desired result is proved.Lemma 10.
CqΦ,∞=0for allq=0,1,2,⋯.Proof.
LetB=v∈E:v≤1, S=∂B be the unit sphere in E, and A>0 be the number given in Lemma 9. Without loss of generality, we may assume that(48)−A<infu≤2Φu.Using (8), it is clear that for any v∈S,(49)Φsv→−∞,ass→+∞.So there issv>0 such that Φsvv=−A. Set u=svv, then a direct computation and Lemma 9 gives(50)ddss=svΦsv=1svddtt=1Φtu<0.By the implicit function theorem,T:v↦sv is a continuous function on S. Using the function T, as in [20–22], we can construct a strong deformation retract η:E\B→Φ−A,(51)ηu=u,ifΦu≤−A,Tuuuu,ifΦu>−A,and deduce(52)CqΦ,∞=HqE,Φ−A≅HqE,E\B=0,q=0,1,2,⋯.Now, we are ready to prove our main result.Proof of Theorem 11.
It follows from (V) and (f2) that as u→0,(53)Φu=12u+2−u−2+ou2.Hence, there existsρ>0 such that Φ is positive in E+\0∩Bρ and negative in E−\0∩Bρ, that is, Φ has a local linking with respect to the decomposition E=E+⊕E−. Therefore, Proposition 8 yields(54)CkΦ,0≠0,where k=dimE−. By Lemma 10, CkΦ,∞=0. Applying Proposition 7, we obtain that Φ has a nonzero critical point. The proof of Theorem 11 is completed.



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