# Existence and Multiplicity of Nontrivial Solutions for a Class of Semilinear Fractional Schrödinger Equations

Authors: Xinsheng Du; Anmin Mao
Journal: Journal of Function Spaces
 (2017)

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## Abstract

This paper is concerned with the existence of solutions to the following fractional Schrödinger type equations:-∆su+Vxu=fx,u,x∈RN, where the primitive of the nonlinearity f is of superquadratic growth near infinity in u and the potential V is allowed to be sign-changing. By using variant Fountain theorems, a sufficient condition is obtained for the existence of infinitely many nontrivial high energy solutions.


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## Body

## 1. Introduction and the Main Result

In this work, under the assumptions thatV satisfies some weaker conditions than those in [1] and the primitive of f satisfies a more general superquadratic condition near infinity, we study the existence of infinitely many nontrivial high energy solutions to the following fractional Schrödinger equations:(1)-∆su+Vxu=fx,u,x∈RN,where s∈(0,1), N>2, and f:RN×R→R is a continuous function with some proper growth conditions. Here (-∆)s is the so-called fractional Laplacian operator of order s∈(0,1), which can be characterized as (-∆)su=F-1(ξ2sFu), with F being the usual Fourier transform in RN (-∆s is the pseudodifferential operator with symbol ξs).We need the following assumptions onV(x):(V1) V∈C(RN,R) is bounded from below.(V2) For any M>0, there exists r0>0 such that(2)limy→∞measx∈RN:x-y≤r0,Vx≤M=0,where meas denotes the Lebesgue measure in RN.More recently, Di Nezza et al. [2] have proved that (-∆)s can be reduced to the standard Laplacian -∆ as s→1. When s=1, problem (1) is the classical Schrödinger equations(3)-∆u+Vxu=fx,u,x∈RN.Over the past several decades, the existence and multiplicity of nontrivial solutions to problem (3) have been studied extensively by numerous researches using a variety of methods and techniques; see, for instance, [3–7].The nonlinear equations involving fractional Laplacian, which is a powerful tool for the descriptions of physics, probability, and finance, have attracted the attention of many researchers and have been successfully applied in various fields; see, for instance, [1, 8, 9] and the reference therein.We need the following conditions onf and its primitive F(x,u)=∫0uf(x,s)ds:(f1)  f∈C(RN×R,R) and there exist constant C>0 and υ∈(2,2s∗) such that (4)fx,u≤Cu+uυ-1,∀x,u∈RN×R,where 2s∗ denotes the critical Sobolev exponent, that is, 2s∗=2N/N-2s.(f2)  F(x,0)≡0, Fx,u≥0 for all (x,u)∈RN×R, and limu→∞F(x,u)/u2=+∞ uniformly on RN.(f3) There exists a constant θ≥1 such that(5)θF~x,u≥F~x,su,∀x,u∈RN×R,s∈0,1,where F~(x,u)≔uf(x,u)-2F(x,u).f
4
f
x
,
-
u
=
-
f
x
,
u, ∀x,u∈RN×R.Our main theorem of this work reads as follows.Theorem 1.
Suppose that(V1), (V2), and f1–f4 hold. Then problem (1) possesses infinitely many nontrivial solutions.Remark 2.
Conditions like(V1) and (V2) have been given in [1], but there infRNV(x)>0 is required. As shown in [10], the condition (f3) due to [11] is somewhat weaker than the condition that fx,u/u is nondecreasing in u for all x∈RN.Remark 3.
It follows from(V1) that there exists a constant V0≥0 such that V¯(x)≔V(x)+V0≥1 for all x∈RN. Let f¯(x,u)=f(x,u)+V0u for all (x,u)∈RN×R and consider the following new fractional Schrödinger equations(6)-∆su+V¯xu=f¯x,u,x∈RN.Then, problem (1) and problem (6) are equivalent. Evidently, the hypotheses (V1),(V2), and (f1)–(f4) still hold for V¯ and f¯ provided that those hold for V and f. Hence, we can assume without loss of generality that V(x)≥1 for all x∈RN in (V1).


## 2. The Proof of Main Result

In this section,s∈(0,1) is a fixed number. We denote by ·p the usual norm of the space LP(RN). cii=1,2,… or C denotes some positive constants.In the light of finite differences, the nonhomogeneous Sobolev spaceHs(RN) is defined by (7)HsRN=u∈L2RN:ux-uyx-yN/2+s∈L2RN×RN.It is a Hilbert space, when endowed with the scalar product given by (8)u,v=∫RNuvdx+∫RN∫RNux-uyvx-vyx-yN+2sdxdy.The corresponding norm is therefore (9)uHs=∫RNu2dx+∫RN∫RNux-uy2x-yN+2sdxdy1/2.The spaceHsRN is also denoted by the Fourier transform. Indeed, it is defined as follows: (10)HsRN=u∈L2RN:∫RN1+ξ2sFuξ2dξ<+∞.This space has a Hilbert structure when endowed with the scalar product (11)u,v=∫RN1+ξ2sFuξFvξdξ,so that the corresponding norm is (12)uHs=∫RN1+ξ2sFuξ2dξ1/2.To illustrate the relationship of the above two norms, let us start from the concept of Schwartz functionS (is dense in Hs(RN)), that is, the rapidly decreasing C∞ function on RN, which will be used later. If u∈S, the fractional Laplacian (-∆)s acts on u as (13)-∆sux=CN,sP.V.∫RNux-uyx-yN+2sdy=CN,slimϵ→0+∫RN∖B0,ϵux-uyx-yN+2sdy,where the symbol P.V. represents the principle value of the integral and the constant C(N,s) depends only on the space dimension N and on the order s. We can write an integral expression for C(N,s) in the form (14)CN,s=∫RN1-cos⁡ζζN+2sdζ.In [2], the authors have proved (15)-∆su=Fξ2sFu,uHs2=2CN,s∫RNξ2sFu2dξ,where (16)uHs=∫RN∫RNux-uy2x-yN+2sdxdy1/2is the Gagliardo (semi)norm. Moreover, by the Plancherel formula in Fourier analysis, it is easy to show that (17)uHs2=2CN,s-∆s/2uL22.Hence, the norms on Hs(RN) which was defined above are all equivalent.For our problem (1), the Hilbert space H is defined by(18)H=u∈HsRN:∫RNξ2sFu2dξ+∫RNVxu2dx<+∞.The inner product and the norm are defined as (19)u,v=∫RNVxuvdx+∫RNξ2sFuξFvξdξ,u=∫RNξ2sFu2dξ+∫RNVxu2dx1/2.Under the assumptions(V1) and (V2), we have the following lemma due to [10].Lemma 4.
The Hilbert spaceH is compactly embedded in Lt(RN) for 2≤t<2s∗=2N/(N-2s).Definition 5.
A weak solution to problem (1) is a function u∈H such that (20)∫RNVxuvdx+∫RNξ2sFuξFvξdξ=∫RNfx,uvdx,∀u∈H.The energy functional associated with problem (1) is defined by (21)Φu=12∫RNξ2sFuξ2dξ+12∫RNVxu2dx-∫RNFx,udx,u∈H.By Lemma 4 and conditions (f1) and f2, we can prove that Φ is well defined and Φ∈C1H with(22)Φ′u,v=∫RNVxuvdx+∫RNξ2sFuξFvξdξ-∫RNfx,uvdx,∀u,v∈H.Therefore weak solutions of (1) correspond to critical points of Φ.LetX be a Banach space with the norm · and let {Xj} be a sequence of subspace of X with dim⁡Xj<+∞ for each j∈N. Further, X=⨁j∈NXj¯, the closure of the direct sum of all Xj. Set (23)Wk=⨁j=0kXj,Zk=⨁j=k∞Xj¯,Bk=u∈Wk:u≤ρk,Sk=u∈Zk:u=rkfor ρk>rk>0. Consider a family of C1 functionals Φλ:X→R defined by (24)Φλu=Au-λBu,λ∈1,2.The following variant Fountain theorem comes from Zou [12].Theorem 6.
Assume that the functionalΦλ defined above satisfies the following:
(
A
1
)  Φλ maps bounded sets to bounded sets uniformly for λ∈[1,2], and Φλ(-u)=Φλ(u) for all (λ,u)∈[1,2]×X.
(
A
2
)  B(u)≥0 for all u∈X,A(u)→+∞ or B(u)→+∞ as u→+∞.
(
A
3
) There exists ρk>rk>0 such that(25)bkλ=infu∈Zk,u=rkΦλu>akλ=maxu∈Wk,u=ρkΦλu,∀λ∈1,2.Then(26)bkλ≤ckλ=infγ∈Γkmaxu∈BkΦλγu,∀λ∈1,2,where Γk=γ∈CBk,X:γis odd,γ∂Bk=id. Moreover, for almost every λ∈[1,2], there exists a sequence {unk(λ)} such that (27)supnunkλ<∞,Φλ′unkλ⟶0,Φλunkλ⟶ckλasn⟶∞.In order to use Theorem6 to prove the main result, we define the functionals A, B, and Φλ on the working space H by (28)Au=12∫RNξ2sFuξ2dξ+12∫RNVxu2dx,Bu=∫RNFx,udx,Φλu=Au-λBu=12u2-λ∫RNFx,udxfor all u∈H and λ∈[1,2]. We choose an orthonormal basis ej:j∈N of H and let Xj=span{ej} for all j∈N. Obviously Φ1=Φ. In order to complete the proof of our theorem, we need the following lemmas.Lemma 7.
Assume that(V1), (V2), and (f1) hold. Then there exists a positive integer k1 and a sequence rk→∞ as k→∞ such that(29)bkλ≔infu∈Zk,u=rkΦλu>0,∀k≥k1,where Zk=⨁j=k∞Xj¯=span{ek,…}¯ for all k∈N.Proof.
From the definition ofΦλ(u) and (f1), there holds(30)Φλu≥12u2-2∫RNFx,udx≥12u2-2c1u22+uυυ,∀λ,u∈1,2×H,where c1 is a constant. Let(31)l2k=supu∈Zk,u=1u2,lυk=supu∈Zk,u=1uυ,∀k∈N.Since H is compactly embedded into both L2(RN) and Lυ(RN), we have(32)l2k⟶0,lυk⟶0ask⟶∞.From (30) and (31), it follows that(33)Φλu≥12u2-2c1l22ku2-2c1lυυkuυ,∀λ,u∈1,2×Zk.From (32), there exists a positive integer k1 such that (34)2c1l22k≤18,∀k≥k1.For each k≥k1, let rk=16c1lυυ(k)1/(2-υ). Then rk→+∞ as k→∞, since υ>2. It is immediate to check that(35)bkλ≔infu∈Zk,u=rkΦλu≥rk24>0,∀k≥k1.Hence, the proof of Lemma 7 is complete.Lemma 8.
Assume that(V1), (V2), (f1), and (f2) hold. Then for the positive integer k1 and the sequence {rk} obtained in Lemma 7, there exist ρk>rk for each k≥k1, such that(36)akλ=maxu∈Wk,u=ρkΦλu<0,∀k≥k1,where Wk=⨁j=1kXj=span{e1,…,ek} for all k∈N.Proof.
To begin with, we carry out that for any finite dimensional subspaceF∈H there exists a constant ε>0 such that(37)measx∈RN:ux≥εu≥ε,∀u∈F∖0.Assume for contradiction that, for any n∈N, there exists un∈F∖{0} such that(38)measx∈RN:ux≥unn<1n.Let vn=un/un∈F for all n∈N. Then vn=1 for all n∈N, and(39)measx∈RN:vnx≥1n<1n,∀n∈N.Up to a subsequence, if necessary, we can say that vn→v0 in H for some v0∈F since F is of finite dimension. Evidently, v0=1. By Lemma 4 and the equivalence of any two norms on F, we conclude that(40)∫RNvn-v02dx⟶0asn⟶∞.Since v0≠0, we see that there exists a constant δ0>0 satisfying(41)measx∈RN:v0x≥δ0≥δ0.For any n∈N, we define the sets(42)Λn=x∈RN:vnx≤1n,Λnc=RN∖Λn=x∈RN:vnx≥1n.Let Λ0=x∈RN:v0x≥δ0. Then for sufficiently large n, from (39) and (41), one can easily see that(43)meas⁡Λn∩Λ0≥meas⁡Λ0-measΛnc≥δ0-1n≥δ02.So that, for n large enough, we obtain (44)∫RNvn-v02dx≥∫Λn∩Λ0vn-v02dx≥∫Λn∩Λ0vn-v02dx≥δ0-1n2·meas⁡Λn∩Λ0≥δ038>0.This is impossible, and (37) holds.
SinceWk is finite dimensional for each k∈N, we deduce from this and (37) that(45)meas⁡Λuk≥εk,∀u∈Wk∖0,where Λuk=x∈RN:ux≥εku for all k∈N and u∈Wk∖0. By (f2), for each k∈N, there exists a constant bk>0 such that(46)Fx,u≥u2εk3,∀x∈RN,u≥bk.So, by (45), (46), and (f2), for any k∈N and λ∈[1,2], we have(47)Φλu≤12u2-∫RNFx,udx≤12u2-∫Λuku2εk3dx≤12u2-εk2u2·measΛukεk3≤12u2-u2=-12u2for all u∈Wk with u≥bk/εk. Let (48)ρk>max⁡rk,bkεk,∀K≥k1.We see that (49)akλ≔akλ=maxu∈Wk,u=ρkΦλu≤-ρk22<0,∀k≥k1.The proof of Lemma 8 is complete.Proof of Theorem1.
Now the assumption(f1) and the definition of Φλ(u) imply that Φλ maps bounded sets to bounded sets uniformly for λ∈1,2. From (f4), we obtain Φλ(u)=Φλ(-u) for all (λ,u)∈[1,2]×H. Therefore, the condition (A1) of Theorem 6 holds. Clearly, the condition (A2) holds by the definitions of A(u)B(u) and (f2). Moreover, Lemmas 7 and 8 imply that the condition (A3) holds for all k≥k1, where k1 is given in Lemma 7. Thus, it follows from Theorem 6 that, for each k≥k1, a.e. λ∈[1,2], there exists a sequence {umk(λ)}⊂H such that(50)supmumkλ<∞,Φλ′umkλ⟶0Φλumkλ⟶ckλasm⟶∞,where (51)ckλ=infγ∈Γkmaxu∈BkΦλγu,∀λ∈1,2,with Bk=u∈Wk:u≤ρk and Γk=γ∈CBk,X:γis odd,γ∂Bk=id.Furthermore, the proof of Lemma7 shows that(52)ckλ∈bk¯,ck¯,∀k≥k1,λ∈1,2,where ck¯=maxu∈BkΦ1(u) and bk¯≔rk2/4→∞ as k→∞.For eachk≥k1, (50) shows that we can choose λn→1 (depending on k) and can get the corresponding sequences satisfying(53)supmumkλn<∞,Φλn′umkλn⟶0asm⟶∞.We prove the following 2 claims.Claim 1.
For eachλn given above, the sequence {umk(λn)}m∞ has a convergent subsequence.
Throughout this paragraph, for the simplicity of notation, we denoteum=umk(λn) by m∈N throughout this paragraph. Indeed, by (53), we may let(54)um⇀u0asm⟶∞,for some u0∈H. With the aid of Riesz Representation Theorem, Φλn′:H→H∗ and B′:H→H∗ can be written as Φλn′:H→H and B′:H→H, respectively. Hence(55)um=Φλn′um+λnB′um,∀m∈N.From Lemma 4, (53), (54), and (55), we see immediately that the right-hand side of (55) converges strongly in H, so that um→u0 in H, and then the proof of Claim 1 is complete.
Therefore, by Claim1, up to subsequences, we may assume(56)limm→∞umkλn=unk,∀n∈N,k≥k1,and hence from (50) and (52) we deduce that(57)Φλn′unk=0,Φλnunk∈bk¯,ck¯,∀n∈N,k≥k1.Claim 2.
For eachk≥k1, the sequence {unk}n=1∞ in (56) is bounded.
Throughout this paragraph, for notational simplicity, we will writeun=unk for all n∈N.
Otherwise, up to subsequences, we may assume(58)un⟶∞,wn≔unun⇀w0∈Hasn⟶∞.By Lemma 4 and (58), we can prove that(59)wn⟶w0inLpRNfor2≤p<2s∗.Up to subsequences, if necessary, we have(60)wnx⟶w0xa.e.x∈RN.If w0=0, then there exists a sequence {tn}⊂[0,1] such that(61)Φλntnun=maxt∈0,1Φλntun.For any D>0, setting wn¯=4Dwn, by (59) and (f2), we obtain that(62)0≤∫RNFx,wn¯dx≤C∫RNwn¯2+wn¯υdx⟶0asn⟶∞.Thus, for n large enough, there holds (63)Φλntnun≥Φλnwn¯=12wn¯2-λn∫RNFx,wn¯dx=2D-λn∫RNFx,wn¯dx≥D.These yield immediately limn→∞Φλn(tnun)=+∞. Since Φλn(0)=0 and Φλn(un)∈[bk¯,ck¯] in (57), we obtain tn∈(0,1) in (61) for n large enough, which shows that(64)0=ddtt=tnΦλn′tun=Φλn′tun,tnun.Hence, from (64) and (f3), we deduce that (65)Φλnun=Φλnun-12Φλn′un,un=λn2∫RNF~x,undx≥λn2θF~x,tnundx=1θΦλntnun-12θΦλn′tnun,tnun=1θΦλntnun⟶+∞asn⟶∞,where θ is given in (f3). By (57), this is impossible.
Sincew0≠0, of course Ω≔{x∈RN:w0(x)≠0} has positive measure. Equation (58) shows that(66)unx⟶∞,x∈Ω.From (58), (66), (f2), and Fatou’s lemma, it is easy to see that(67)12-Φλnunun2=λn∫RNFx,unun2dx≥λn∫Ωwn2Fx,unun2dx⟶+∞asn⟶∞.This contradiction shows that w0≠0 cannot hold and concludes the proof of Claim 2.For eachk≥k1, using the same arguments in the proof of Claim 1, one can also prove that the sequence unkn=1∞ has a strong convergent subsequence with the limit uk being just the critical point of Φ=Φ1 by Claim 2 and (57). Clearly, Φuk∈bk¯,ck¯ for all k≥k1. These imply that Φ have infinitely many nontrivial critical points of since bk¯→+∞ as k→∞. Therefore, problem (1) admits infinitely many nontrivial solutions and the proof of Theorem 1 is complete.



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