# Nontrivial Solution for the Fractionalp-Laplacian Equations via Perturbation Methods

Authors: Huxiao Luo; Shengjun Li; Xianhua Tang
Journal: Advances in Mathematical Physics
 (2017)

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## Abstract

We study the existence of nontrivial solution of the following equation without compactness:(-Δ)pαu+up-2u=f(x,u),  x∈RN, where N,p≥2,  α∈(0,1),  (-Δ)pα is the fractional p-Laplacian, and the subcritical p-superlinear term f∈C(RN×R) is 1-periodic in xi for i=1,2,…,N. Our main difficulty is that the weak limit of (PS) sequence is not always the weak solution of fractional p-Laplacian type equation. To overcome this difficulty, by adding coercive potential term and using mountain pass theorem, we get the weak solution uλ of perturbation equations. And we prove that uλ→u as λ→0. Finally, by using vanishing lemma and periodic condition, we get that u is a nontrivial solution of fractional p-Laplacian equation.


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## Body

## 1. Introduction

This article is concerned with the fractionalp-Laplacian equations(1)-Δpαu+up-2u=fx,u,x∈RN,where N,p≥2,  α∈(0,1), and f satisfies the following conditions. (f1)
f∈C(RN×R), f is 1-periodic in xi for i=1,2,…,N, and (2)limt→∞fx,ttq-1=0,limt→∞fx,tttp=+∞uniformly inx∈RN for some q∈(p,pα∗), where pα∗=Np/N-αp;(f2)
f(x,t)=o(tp-2t) as t→0 uniformly for x∈RN;(f3)
F~(x,u)≔1/pf(x,u)u-F(x,u)>0 if u≠0, and there exists c0≥0 and σ>max⁡{N/pα,1} such that (3)fx,uσ≤c0uσp-1F~x,u,∀x∈RN,u≥r0,whereF(x,t)=∫0tf(x,s)ds.The fractionalp-Laplacian is defined on smooth functions by (4)-Δpαux=2limϵ→0∫RN∖Bϵxux-uyp-2ux-uyx-yN+αpdy,x∈RN. This definition is consistent, up to a normalization constant depending on N and α, with the usual definition of the linear fractional Laplacian operator (-Δ)α when p=2. There is, currently, a rapidly growing literature on problems involving these nonlocal operators. This type of problem arises in many different applications, such as continuum mechanics, phase transition phenomena, population dynamics, and game theory, as they are the typical outcome of stochastically stabilization of Lévy processes; see [1–9] and the references therein. The literature on nonlocal operators and their applications is very interesting and quite large; we refer the interested reader to [4, 10–21] and the references therein. For the basic properties of fractional Sobolev spaces, we refer the interested reader to [22, 23].The main purpose of this paper is to consider the existence of nontrivial solutions for equation (1). Our main difficulty is that the weak limit of (PS) sequence is not always the weak solution of (1). To overcome this problem, we apply the perturbation method [22, 24–26]. First, we consider the perturbation equation by adding coercive potential term(5)-Δpαu+λVxup-2u+up-2u=fx,u,x∈RN,where λ∈(0,1] is a parameter and V(x) satisfies the following conditions:(V1)
V∈C(RN), infRNV(x)≥V0>0.(V2)
measx∈RN∣Vx≤M<+∞,  ∀M>0.And we prove that the energy functional of (5) has the geometry of the mountain pass theorem that it satisfies the Cerami condition and finally that the obtained solutions {uλ} have the uniform bounds. Finally, we verify that uλ⇀u as λ→0 and u is the nontrivial solution of (1).Now, we give the main result of this article.Theorem 1.
Suppose that(f1)–(f3) hold. Then (1) possesses at least a nontrivial solution.Remark 2.
In order to get our result, there are mainly three difficulties.(i)
The working space has not compactness.(ii)
The classical AR condition for the nonlinearity is not satisfied.(iii)
If{un} is a Palais Smale sequence of Φ (see Section 2) and un converges weakly to u0, one can not obtain that u0 is a weak solution of the fractional p-Laplacian type equation (1).Notation 1.
In this paper we make use of the following notation:(i)
·p is the usual norm of the space Lp(RN).(ii)
c, C and ci, Ci denote positive (possibly different) constants.(iii)
We denote the weak convergence inX and its X∗ by “⇀” and the strong convergence by “→”.(iv)
o(1) denote being infinitely small (possibly different) when n→∞.


## 2. Variational Framework

Before stating this section, we define the Gagliardo seminorm by(6)uα,p=∫2RNux-uypx-yN+αpdxdy1/p, where u:RN→R is a measurable function. On one hand, we define fractional Sobolev space by (7)Wα,pRN=u∈LpRN:uis measurable,uα,p<∞ endowed with the norm (8)uα,p=uα,pp+upp1/p, where (9)up=∫RNuxpdx1/p. Moreover, (1) is variational and its solutions are the critical points of the functional defined in Wα,p(RN) by (10)Φu=1puα,pp+1pupp-∫RNFx,udx,∀u∈Wα,p. From (f1), it is easy to check that Φ is well defined on Wα,p(RN) and Φ∈C1(Wα,p(RN),R), and (11)Φ′u,v=∫2RNux-uyp-2ux-uyvx-vyx-yN+αpdxdy+∫RNup-2uvdx-∫RNfx,uvdx.On the other hand, we consider the fractional Sobolev space(12)Xα≔u∈Wα,p:∫RNVxupdx<∞ endowed with the norm (13)u≔uXα=uα,pp+∫RNVxupdx1/p. We also need the following inner norm: (14)uλ=uα,pp+λ∫RNVxupdx1/p and let Xλα=(Xα,·λ). Obviously, we have (15)λup≤uλp≤up;the two norms · and ·λ are equivalent. Next, the following lemma discusses the continuous and compact embedding for Xα↪Lq(RN) for all q∈[p,pα∗]. For the proof of the lemma, it was proved in [27] in the case p=2. For the general case, the proof is similar. We give it here for readers convenience.Lemma 3.
Assume that(V1) and (V2) hold. Then Xα is continuously embedded in Ls(RN) for all s∈[p,pα∗]. Moreover, Xα can be compactly embedded into Ls(RN) for all s∈[p,pα∗).Proof.
Let{un}⊂Xα be a bounded sequence of Xα such that un⇀0 in Xα. Then, by Theorem 2.1 in [22], un→0 in Llocq(RN) for p≤q<pα∗. We claim that(16)un⟶0strongly inLpRN.To prove (16), we only need to show that, for any ε>0, there exists R>0 such that (17)∫RN∖BRunxpdx<ε.Set (18)BR=x∈RN∣x<R,AR,M=x∈RN∖BR∣Vx≥M,BR,M=x∈RN∖BR∣Vx<M; then (19)∫AR,Munxpdx≤∫RNVxMunxpdx≤unXαpM.Now choose σ∈(1,pα∗/p) such that 1/σ+1/σ′=1; then we have (20)∫BR,Munxpdx≤∫BR,Munxpσdx1/σmeasBR,M1/σ′≤CunXαpmeas⁡BR,M1/σ′. Since unXαp is bounded and condition (V2) holds, we may choose R, M large enough such that unXαp/M and meas(B(R,M)) are small enough. Hence, ∀ε>0, we have (21)∫RN∖BRunxpdx=∫AR,Munxpdx+∫BR,Munxpdx<ε from which (16) follows.
To prove the lemma for general exponentq, we use an interpolation argument. Let un⇀0 in Xα, we have just proved that un→0 in Lp(RN). That is, (22)∫RNunxpdx⟶0, as n→∞. Moreover, because the embedding Xα⇀Lpα∗(RN) is continuous and {un} is bounded in Xα, we also have (23)supn∫RNunxpα∗dx<∞. Since q∈(p,pα∗), there is a number λ∈(0,1) such that 1/q=λ/p+1-λ/pα∗. Then by Hölder inequality (24)∫RNunxqdx=∫RNunxλqunx1-λqdx≤unpλqunpα∗1-λq⟶0. This implies un→0 in Lq(RN).From Lemma3, there exists γs>0 such that (25)us≤γsu,∀u∈Xα, where us denotes the usual norm in Ls(RN) for all p≤s≤pα∗.Next, we define the energy functionalΦλ on Xα by (26)Φλu=1puλp+1pupp-∫RNFx,udx=Φu+λp∫RNVxupdx,∀u∈Xα. We also need the following inner norm: (27)uλ,p=uλp+upp1/p; by Lemma 3, we have that the norms · and ·λ,p are also equivalent. By (V1), (V2), the energy functional Φλ:Xα→R is well defined and of class C1(Xα,R). Moreover, the derivative of Φλ is (28)Φλ′u,v=∫2RNux-uyp-2ux-uyvx-vyx-yN+αpdxdy+λ∫RNVxup-2uvdx+∫RNup-2uvdx-∫RNfx,uvdx=Φ′u,v+λ∫RNVxup-2uvdx for all u,v∈Xα.In what follows, we give the vanishing lemma which is introduced by Lion.Lemma 4 (see [28]).
Assume{uk} is a bounded sequence in Wα,p which satisfies (29)limk→+∞supy∈RN∫BRyukxpdx=0 for some R>0. Then (30)uk⟶0inLq,∀p<q<NpN-αp.


## 3. Proofs of the Main Result

The proof of Theorem1 is divided into several lemmas. We show that the functional Φλ has the geometry of the mountain pass theorem that it satisfies the Cerami condition and finally that the obtained solutions have the uniform bounds.Lemma 5.
Suppose that(V1), (V2), (f1)–(f3) are satisfied. Then there exists ρ>0, η>0, such that infΦλu∣u∈Xα,u=ρ>η for fixed λ∈(0,1], where ρ and η are independent of λ.Proof.
For anyε>0, it follows from (f1) and (f2) that there exists Cε such that(31)fx,s≤εsp-1+Cεsq-1,s∈R,x∈RN, where p<q<pα∗, and then (32)Fx,s≤εpsp+Cεqsq,s∈R,x∈RN. For ρ>0, let (33)Σρ=u∈Xα∣u≤ρ.So, from the Sobolev inequality, one has (34)∫RNFx,udx≤εp∫RNupdx+Cεq∫RNuqdx≤γpεpup+γqCεquq. So one has, for u∈∂Σρ, (35)Φλu=1puα,pp+1pupp+λp∫RNVxupdx-∫RNFx,udx≥λpuα,pp+λp∫RNVxupdx-γpεpup-γqCεquq=λ-γpεpup-γqCεquq, since 0<λ≤1. Hence, by fixing ε∈(0,1/λγp) and letting ρ>0 be small enough, it is easy to see that there is η>0 such that this lemma holds.Lemma 6.
Suppose that(V1), (V2), (f1)–(f3) are satisfied. then there exists e∈Xα with e>ρ such that Φλ(e)<0 for fixed λ∈(0,1], where ρ is given by Lemma 5.Proof.
Using(f1), we obtain there exists T>0 such that (36)Fx,t>tpε-Tpε,t∈R,x∈RN. Next, for φ∈C0∞(RN) we have (37)∫RNFx,tφtpdx≥1ε∫RNφpdx-Tpεtp∫suppφdx. This implies (38)limt→∞∫RNFx,tφtpdx≥1ε∫RNφpdx, for all ε>0. Since ε is arbitrary, by the above inequality, we get (39)limt→∞∫RNFx,tφtpdx=+∞. Consequently, (40)Φλtφtp=1pφλp+1pφpp-∫RNFx,tφtpdx≤1pφp+1pφpp-∫RNFx,tφtpdx⟶-∞ as t→+∞. Hence, let t0 be big enough and e=t0φ; then we have Φλ(e)<0; we complete the proof.Definition 7.
We say thatJ satisfies Cerami condition in E, if, for any sequence {un}⊂E such that (41)Jun⟶c,1+unEJ′un⟶0,as n→∞, there exists a convergent subsequence of {un}.Lemma 8.
Suppose that(V1), (V2) and (f1)–(f3) are satisfied. Then the functional Φλ(un) satisfies Cerami condition.Proof.
Let{un} be a sequence in Xα so that(42)Φλun⟶cλ,1+unΦλ′un⟶0.We shall prove that {un} contains a convergent subsequence.
(i) We claim that{un} is bounded in Xα. Observe that for n large(43)cλ+1≥Φλun-1pΦλ′unun=∫RNF~x,undx.Arguing indirectly, assume by contradiction that un→∞; then unλ,p→∞. Set vn=un/unλ,p; then vnλ,p=1. By Lemma 3, one has vns≤γsvnλ,p=γs for s∈[p,pα∗). Observe that from (42) and (44)Φλ′unun=unλ,pp1-∫RNfx,unvnunλ,pp-1dx it follows that(45)∫RNfx,unvnunλ,pp-1dx=∫RNfx,ununp-1vnpdx⟶1. Set for r≥0(46)hr≔infF~x,u∣x∈RN,u∈Rwithu≥r. By (f3), h(r)>0 for all r>0, and h(r)→+∞ as r→+∞. For 0≤a<b let (47)Ωna,b=x∈RN∣a≤unx<b,cab≔infF~x,uup∣x∈RN,u∈Rwitha≤u≤b. Since F~(x,u)>0 if u≠0, one has cab>0 and (48)F~x,unx≥cabunxp,∀x∈Ωna,b. It follows from (43) that(49)cλ+1≥∫Ωn0,aF~x,undx+∫Ωna,bF~x,undx+∫Ωnb,∞F~x,undx≥∫Ωn0,aF~x,undx+cab∫Ωna,bunpdx+hbΩnb,+∞. Invoking (f3), set τ≔pσ/(σ-1) and σ∗=σ/(σ-1)=τ/p. Since σ>max{1,N/pα} one sees τ∈(p,pα∗). Fix arbitrarily s∈(τ,pα∗). Using (49), (50)Ωnb,+∞≤c+1hb⟶0 as b→+∞ uniformly in n, which implies by Hölder inequality that(51)∫Ωnb,+∞vnτdx≤∫Ωnb,+∞vnsdxτ/sΩnb,+∞1-τ/s≤γsτΩnb,+∞1-τ/s⟶0 as b→+∞ uniformly in n. Using (49) again, for any fix 0<a<b,(52)∫Ωna,bvnpdx=1unλ,pp∫Ωna,bunpdx≤cλ+1cabunλ,pp⟶0 as n→+∞.
Let0<ε<1/3. By (f2) there is aε>0 such that fx,u<ε/γpup-1 for all u≤aε; consequently,(53)∫Ωn0,aεfx,ununp-1vnpdx≤∫Ωn0,aεεγpvnpdx≤εγpvnpp≤ε for all n. By (f3), (43), (51), and Hölder inequality we can take bε≥r0 large so that(54)∫Ωnbε,+∞fx,ununp-1vnpdx≤∫Ωnbε,+∞fx,unσunσp-1dx1/σ∫Ωnbε,+∞vnτdx1/σ∗≤∫RNc0F~x,undx1/σ∫RNvnτdxp-1/τ∫Ωnbε,+∞vnτdx1/τ<ε for all n. Note that there is γ=γ(ε)>0 independent of n such that fx,un≤γunp-1 for x∈Ωn(aε,bε). By (52) there is n0 such that(55)∫Ωnaε,bεfx,ununp-1vnpdx≤γ∫Ωnaε,bεvnpdx<ε for all n≥n0. Now the combination of (53), (55), and (62) implies that for n≥n0(56)∫RNfx,ununp-1vnpdx<3ε<1 which contradicts (45). Hence {un} is bounded in Xα.
(ii) By (i), we can conclude that{un} is bounded in Xα. Going if necessary to a subsequence, we can assume that un⇀u in Xα. From Lemma 3, we have un→u in Ls(RN) for all p≤s<pα∗. By the boundedness of {un} in Lp(RN), we have (57)Λ1=supn∫RNunpdx<∞. By Hölder inequality and the above inequality we also have (58)Λ2=supn∫RNunp-1udx≤supn∫RNunpdxp-1/pup<∞. Similarly, (59)Λ3=supn∫RNup-1undx<∞. By (f1), (f2), for ε>0, there exists Cε>0 such that(60)fx,t≤εtp-1+Cεtq-1,∀x,t∈RN×R. Then using Hölder inequality we have (61)∫RNfx,un-fx,uun-udx≤∫RNεunp-1+up-1+Cεunq-1+uq-1un-udx≤ε∫RNunp+up+unp-1u+up-1undx+Cε∫RNunq-1un-u+uq-1un-udx≤εΛ1+Λ2+Λ3+upp+2Cεsupnunqq-1+uqq-1un-uq. Since {un} is bounded in Lq(RN) and ε is arbitrarily small, we have(62)∫RNfx,un-fx,uun-udx⟶0,asn⟶∞.By (62) and(63)un-up=∫unx-uny-ux-uypx-yN+αpdxdy+∫Vxunx-uxpdxwe have(64)Φλ′un-Φλ′u,un-u=∫unx-unyp-2unx-uny-ux-uyp-2ux-uyx-yN+αpunx-uny-ux+uydxdy+λ∫Vxunp-2un-up-2uun-udx-∫fx,un-fx,uun-udx≥c1∫unx-uny-ux-uypx-yN+αpdxdy+λc2∫Vxun-updx-o1≥minc1,λc2un-up-o1,where we have used the following elementary inequality: (65)ap-2a-bp-2ba-b≥ca-bp,where the constant c is independent from the variable a and b. Recall that un⇀u, Φλ′(un)→0 as n→∞; it is clear that(66)Φλ′un-Φλ′u,un-u⟶0,asn⟶∞.From (64), (66), we have un-u→0 as n→∞. Therefore, Φλ(u) satisfies Cerami condition.Since Lemmas5–8 hold, the Mountain Pass Theorem [28] gives that (5) has a nontrivial solution uλ satisfying (67)Φλ′uλ=0,cλ=Φλuλ=infγ∈Γmaxt∈0,1Φλγt>Φλ0=0, where Γ=Υ∈C0,1,Xα∣Φλ0=0,Φλ1=e and e from Lemma 6.Lemma 9.
Letλk→0 and {uk}⊂E be a sequence of critical points of Φλk satisfying Φλk′(uk)=0 and Φλk(uk)≤C for some C independent of k. Then, up to a subsequence uk⇀u in Xα as k→∞ and u is a critical point of Φ.Proof.
We first claim that{uk} is bounded in Xα. Observe that(68)cλk=Φλkuk-1pΦλk′ukuk=∫RNF~x,ukdxArguing indirectly, assume by contradiction that uk→∞; then ukλk,p→∞. Set vk=uk/ukλk,p; then vkλk,p=1. By Lemma 3, one has vks≤γsvkλ,p=γs for s∈[p,pα∗). Observe that (69)Φλk′ukuk=ukλk,pp1-∫RNfx,ukvkukλk,pp-1dx=0; it follows that(70)∫RNfx,ukvkukλk,pp-1dx=∫RNfx,ukukp-1vkpdx=1.Set for r≥0(71)hr≔infF~x,u∣x∈RN,u∈Rwithu≥r. By (f3), h(r)>0 for all r>0, and h(r)→+∞ as r→+∞. For 0≤a<b let (72)Ωka,b=x∈RN∣a≤ukx<b,cab≔infF~x,uup∣x∈RN,u∈Rwitha≤u≤b.Since F~(x,u)>0 if u≠0, one has cab>0 and (73)F~x,ukx≥cabukxp,∀x∈Ωka,b. It follows from (68) that(74)cλk=∫Ωk0,aF~x,ukdx+∫Ωka,bF~x,ukdx+∫Ωkb,∞F~x,ukdx≥∫Ωk0,aF~x,ukdx+cab∫Ωka,bukpdx+hbΩkb,+∞.Invoking (f3), set τ≔pσ/(σ-1) and σ∗=σ/(σ-1)=τ/p. Since σ>max{1,N/pα} one sees τ∈(p,pα∗). Fix arbitrarily s∈(τ,pα∗). Using (74), (75)Ωkb,+∞≤cλkhb⟶0 as b→+∞ uniformly in k, which implies by Hölder inequality that(76)∫Ωkb,+∞vkτdx≤∫Ωkb,+∞vksdxτ/sΩkb,+∞1-τ/s≤γsτΩkb,+∞1-τ/s⟶0as b→+∞ uniformly in k. Using (74) again, for any fix 0<a<b,(77)∫Ωka,bvkpdx=1ukλk,pp∫Ωka,bukpdx≤cλkcabukλk,pp⟶0as k→+∞.
Let0<ε<1/3. By (f2) there is aε>0 such that fx,u<ε/γpup-1 for all u≤aε; consequently,(78)∫Ωk0,aεfx,ukukp-1vkpdx≤∫Ωk0,aεεγpvkpdx≤εγpvkpp≤εfor all k. Obviously, by (31), there exists a constant C~(k)>0 such that(79)∫RNc0F~x,ukdx≤2c0ϵp∫RNukpdx+c01p+1qCϵ∫RNukqdx≤2c0ϵγpppukp+c01p+1qγqqukq<C~kfor all uk∈Xα. By (f3), (68), (79), and Hölder inequality we can take bε≥r0 large so that(80)∫Ωkbε,+∞fx,ukukp-1vkpdx≤∫Ωkbε,+∞fx,ukσukσp-1dx1/σ∫Ωkbε,+∞vkτdx1/σ∗≤∫RNc0F~x,ukdx1/σ∫RNvkτdxp-1/τ∫Ωkbε,+∞vkτdx1/τ<εfor all k. Note that there is γ=γ(ε)>0 independent of k such that fx,uk≤γukp-1 for x∈Ωk(aε,bε). By (77) there is k0 such that(81)∫Ωkaε,bεfx,ukukp-1vkpdx≤γ∫Ωkaε,bεvkpdx<εfor all k≥k0. Now the combination of (78), (80), and (81) implies that for k≥k0(82)∫RNfx,ukukp-1vkpdx<3ε<1 which contradicts with (70). Hence {uk} is bounded in Xα. We may assume up to a subsequence {uk} converges to u weakly in Xα. By Hölder inequality, we have(83)λk∫RNVxukp-2ukvdx≤λk∫RNVxp-1/pukp-1Vx1/pvdx≤λk∫RNVxukpdxp-1/p∫RNVxvpdx1/p⟶0,ask⟶∞.By the weakly continuity of Φ′ and (83), we have (84)Φ′u,v=limk→∞Φ′uk,v=limk→∞Φλk′uk,v-λk∫RNVxukp-2ukvdx=-limk→∞λk∫RNVxukp-2ukvdx=0for any v∈Xα. Hence, u is critical point of Φ(u).Proof of Theorem1.
By Lemma6, there exists a constant C>0, independent of λ, such that (85)cλ=infγ∈Γmaxt∈0,1Φλγt≤supt≥0Φ1tTu≤C. Then, we can choose a sequence λk→0. Assume that {uk}⊂Wα,p is a sequence of critical points of Φλk. According to Lemma 9, u is a critical point of Φ(u) on Wα,p; it is suffice to show that u≠0. Indeed, if u is vanishing, then (86)δ≔limk→+∞supy∈RN∫B1yukxpdx=0.By Lemma 4, we have(87)uk⟶0inLqRN,forp<q<pα∗.By (f1),(f2), for ε>0, there exists Cε>0 such that(88)fx,uu≤εup+Cεuq,0=Φλk′uk,uk=ukα,pp+λk∫RNVxukpdx-∫RNfx,ukukdx≥ukα,pp-εukpp-Cεukqq≥Cukqp-Cεukqqsince ε is arbitrary small. Hence, we have ukq≥C/Cε1/q-p>0. This conflicts with (87); thus u is nonvanishing, δ>0. Therefore, by standard method, we can obtain u≠0. The proof is completed.



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*Source: 5317213-2017-11-21.xml*