Ground state solutions of fractional Schrödinger equations with potentials and weak monotonicity condition on the nonlinear term
aa r X i v : . [ m a t h . A P ] J un GROUND STATE SOLUTIONS OF FRACTIONAL SCHR ¨ODINGEREQUATIONS WITH POTENTIALS AND WEAK MONOTONICITYCONDITION ON THE NONLINEAR TERM
CHAO JI
Abstract.
In this paper we are concerned with the fractional Schr¨odinger equation( − ∆) α u + V ( x ) u = f ( x, u ), x ∈ R N , where f is superlinear, subcritical growth and u f ( x,u ) | u | is nondecreasing. When V and f are periodic in x , . . . , x N , we show theexistence of ground states and the infinitely many solutions if f is odd in u . When V is coercive or V has a bounded potential well and f ( x, u ) = f ( u ), the ground statesare obtained. When V and f are asymptotically periodic in x , we also obtain theground states solutions. In the previous research, u f ( x,u ) | u | was assumed to be strictlyincreasing, due to this small change, we are forced to go beyond methods of smoothanalysis. Introduction
In this paper we consider the following fractional Schr¨odinger equation(1.1) ( − ∆) α u + V ( x ) u = f ( x, u ) , x ∈ R N , where α ∈ (0 , N > α , ( − ∆) α stands for the fractional Laplacian, f ∈ C ( R N × R , R )and the potential V ∈ C ( R N , R ).When α = 1, (1 .
1) becomes the classical Schr¨odinger equation(1.2) − ∆ u + V ( x ) u = f ( x, u ) , x ∈ R N . There has been a great deal of works dealing with the equation (1.2). In particular, Szulkinand Weth [22] studied the ground state solutions and the infinitely many solutions if f ( x, u )is odd in u for the strong indefinite case. In their paper, the nonlinear term f satisfies thefollowing assumption:( F ′ ) u f ( x,u ) | u | is strictly increasing on ( −∞ ,
0) and on (0 , ∞ ).They sought the ground states on the generalized Nehari manifold [15] M := { u ∈ E \ E − : Φ ′ ( u ) u = 0 and Φ ′ ( u ) v = 0 for all v ∈ E − } , Mathematics Subject Classification.
Key words and phrases.
Fractional logarithmic Schr¨odinger equation, Periodic potential, coercive po-tential, bounded potential, nonsmooth critical point theory.Chao Ji was supported by NSFC (grant No. 11301181), China Postdoctoral Science Foundation fundedproject. where H ( R N ) := E = E + ⊕ E − corresponds to the spectral decomposition of − △ + V with respect to the positive and negative part of the spectrum andΦ( u ) = 12 Z R N ( |∇ u | + V ( x ) u ) dx − Z R N F ( x, u ) dx,F ( x, u ) = R u f ( x, s ) ds . Because of the assumption ( F ′ ), for any u ∈ E \ E − , the set M intersects ˆ E ( u ) := E − ⊕ R + u = E − ⊕ R + u + in exactly one point ˆ m ( u ) which is the uniqueglobal maximum point of Φ | ˆ E ( u ) , the uniqueness of ˆ m ( u ) enables one to define a continu-ous map u ˆ m ( u ), which is important in the remaining proof. If ( F ′ ) is replaced by theweaker condition as follows( F ) u f ( x,u ) | u | is nondecreasing on ( −∞ ,
0) and on (0 , ∞ ),then M ∩ ˆ E ( u ) may be a finite line segment that an example can be seen in [25], so theargument in [22] collapses. To solve this problem, in [10], by applying linking methodsand showing the boundedness of all Cerami sequences for Φ, Liu obtained the groundstates. After that, via a non-smooth method, Pavia, Kryszewski and Szulkin in [16] gavethe ground state solutions and the infinitely many solutions if f ( x, u ) is odd in u , and theresult in [10] is an easy consequence of the approach in [16]. Motivated by [16], in thispaper we will generalize their results to the fractional Schr¨odinger equations when V and f are 1-periodic in x , . . . , x N . Since our problem is nonlocal, it is the more difficult andcomplicated. Moreover, for the coercive potential case, the bounded potential well case,the V and f are asymptotically periodic in x case, we also give the existence of groundstates of problem (1.2) via the variational methods [23].In recent years, the study of the various nonlinear equations or systems involving frac-tional Laplacian has received considerable attention. These problems mainly arise infractional quantum mechanics [8, 9], physics and chemistry [11], obstacle problems [20],optimization and finance [6] and so on. In the remarkable work of Caffarelli and Silvestre[2], the authors express this nonlocal operator ( − ∆) s as a Dirichlet-Neumann map for acertain elliptic boundary value problem with local differential operators defined on theupper half space. This technique is a valid tool to deal with the equations involving frac-tional operators in the respects of regularity and variational methods. For more resultson the fractional differential equations, we refer to [1, 5, 12, 13, 18, 19]. Recently, in [24],under the assumption ( F ′ ) and using Andrzej and Weth’s method [22], the authors showthe existence of infinitely many solutions of problem (1.2) when V and f are periodic in x , . . . , x N , f ( x, u ) is odd in u . Moveover, when V and f are asymptotically periodic in x , they give the ground state solutions. If ( F ′ ) is replaced by ( F ), the argument in [24]does not work, we will deal with this problem and improve their results.From now on, we always assume that inf x ∈ R N V ( x ) >
0. Besides of the assumption( F ), f also satisfies the following assumptions: RACTIONAL SCHR ¨ODINGER EQUATIONS 3 ( F ) | f ( x, u ) | ≤ C (1 + | u | p − ) for some C > < p < ∗ α = NN − α .( F ) f ( x, u ) = o ( u ) uniformly in x as u → F ) F ( x,u ) u → ∞ uniformly in x as | u | → ∞ , where F ( x, u ) = R u f ( x, s ) ds .Now let us state the main results of this paper. Theorem 1.1.
Assume that ( F ) − ( F ) hold, V and f is 1-periodic in x , . . . , x N , thenproblem (1.1) has a ground state solution. Let ∗ denote the action of Z N on H α ( R N ) given by(1.3) (cid:0) k ∗ u (cid:1) ( x ) := u ( x − k ) , k ∈ Z N . If V and f is 1-periodic in x , . . . , x N and u is a solution of problem (1 . k ∗ u for all k ∈ Z N . Set O ( u ) := { k ∗ u : k ∈ Z N } . Two solutions u and u are said to be geometrically distinct if O ( u ), O ( u ) are disjointand u = [ s u , t u ] u . Theorem 1.2.
Under the assumptions of Theorem 1.1 and f ( x, u ) is odd in u , there arethe infinitely many geometrically distinct solutions for problem (1.1). Theorem 1.3.
Assume that ( F ) − ( F ) and lim | x |→∞ V ( x ) = + ∞ hold, then problem(1.1) has a ground state solution. Theorem 1.4.
Assume that f ( x, u ) = f ( u ) and ( F ) − ( F ) hold, inf x ∈ R N V ( x ) ≤ V ( x ) < lim | x |→∞ V ( x ) = sup x ∈ R N V ( x ) < + ∞ , then problem (1.1) has a ground state solution. Let F be the class of functions h ∈ L ∞ ( R N ) such that for every ǫ >
0, the set { x ∈ R N : | h ( x ) | ≥ ǫ } has finite Lebesgue measure. Theorem 1.5.
Besides of ( F ) − ( F ) , V and f also satisfies the following assumptions: ( V ) There exists a functions V p ∈ C ( R N , R ) , 1-periodic in x , . . . , x N , such that ( V − V p ) ∈F , and V ( x ) ≤ V p ( x ) for all x ∈ R N . ( F ) There exists a function f p ∈ C ( R N × R , R ) , 1-periodic in x , . . . , x N , such that ( i ) | f ( x, u ) | ≥ | f p ( x, u ) | , ( x, u ) ∈ ( R N , R ) ; ( ii ) | f ( x, u ) − f p ( x, u ) | ≤ | h ( x ) | ( | u | + | u | p − ) , ( x, u ) ∈ ( R N , R ) , h ∈ F and p ∈ [2 , ∗ α ) ; ( iii ) u f p ( x,u ) | u | is nondecreasing on ( −∞ , and on (0 , ∞ ) .Then problem (1.1) has a ground state solution. The rest of the paper is organized as follows. In Section 2, we present some necessarypreliminary knowledge. In Section 3 we prove Theorem 1.1 and Theorem 1.2. In Section4 we prove Theorem 1.3. In Section 5 Theorem 1.4 is proved and in the final section weprove Theorem 1.5.
CHAO JI
Notation.
C, C , C etc. will denote positive constants whose exact values are inessential. h . , . i is the inner product in the Hilbert space E .2. Preliminaries
For any α ∈ (0 , − ∆) α u of a function u : R N → R , withsufficient decay, is defined by F (( − ∆) α u )( ξ ) = | ξ | α F ( u )( ξ ) , ξ ∈ R N , where F denotes the Fourier transform, that is, F ( φ )( ξ ) = 1(2 π ) N Z R N e − iξ · x φ ( x ) dx ≡ b φ ( ξ ) , for function φ in the Schwartz class. ( − ∆) α u can also be computed by the following sin-gular integral: ( − ∆) α u = c N,α
P.V. Z R N u ( x ) − u ( y ) | x − y | N +2 α dy, here P.V. is the principal value and c N,α ia a normalization constant.The fractional Sobolev space H α ( R N ) is defined by H α ( R N ) = n u ∈ L ( R N ) : | u ( x ) − u ( y ) || x − y | N + α ∈ L ( R N × R N ) o , endowed with the norm(2.1) k u k H α ( R N ) = (cid:16) Z R N u dx + Z Z R N × R N | u ( x ) − u ( y ) | | x − y | N +2 α dxdy (cid:17) , where the term[ u ] H α ( R N ) = k ( − ∆) α u k L ( R N ) := (cid:16) Z Z R N × R N | u ( x ) − u ( y ) | | x − y | N +2 α dxdy (cid:17) is the so-called Gagliardo semi-norm of u .For the basic properties of the fractional Sobolev space H α ( R N ), we refer to [7, 18, 19]. Proposition 2.1 ([7, 19]) . Let < α < such that α < N . Then there exists a constant C = C ( N, α ) > such that k u k L ∗ α ( R N ) ≤ C k u k H α ( R N ) for every u ∈ H α ( R N ) , where ∗ α = NN − α is the fractional critical exponent. Moreover,the embedding H α ( R N ) ⊂ L q ( R N ) is continuous for any q ∈ [2 , ∗ α ] , and is locally compactwhenever q ∈ [2 , ∗ α ) . RACTIONAL SCHR ¨ODINGER EQUATIONS 5
Proposition 2.2 ([19]) . Assume that { u n } is bounded in H α ( R N ) and it satisfies lim n →∞ sup y ∈ R N Z B R ( y ) | u n ( x ) | dx = 0 , where R > . Then u n → in L q ( R N ) for every < q < ∗ α . Let E be the Hilbert subspace of u ∈ H α ( R N ) under the norm k u k = (cid:16) Z R N V ( x ) u dx + Z Z R N × R N | u ( x ) − u ( y ) | | x − y | N +2 α dxdy (cid:17) . It is clear that this norm is equivalent to (2.1) if V ∈ L ∞ ( R N ) and E ⊂ H α ( R N ) iflim | x |→∞ V ( x ) = + ∞ . Moreover, by [17], if V is coercive, for any q ∈ [2 , ∗ α ), the embed-ding E ֒ → L q ( R N ) is compact.The energy functional J on E associated with problem (1.1) is(2.2) J ( u ) := 12 Z R N (cid:16) | ( − ∆) α u | + V ( x ) u (cid:17) dx − Z R N F ( x, u ) dx. Under the assumptions ( F ) − ( F ), J ∈ C ( E, R ) and its critical points are solutions ofproblem (1.1). Define Nehari manifold associated to the functional J ,(2.3) N = n u ∈ E \ { } : Z R N (cid:16) | ( − ∆) α u | + V ( x ) u (cid:17) dx = Z R N f ( x, u ) u dx o . It is easy to know that N is closed and if u is a nontrivial critical point of J , then u ∈ N .3. Proofs of Theorem 1.1 and Theorem 1.2
First, by ( F ) and ( F ), for any ǫ > C ǫ > | f ( x, u ) | ≤ ǫ | u | + C ǫ | u | p − for all x ∈ R N and u ∈ R . It is also easy to see from ( F ) that f ( x, tu ) = 0 if u = 0 and t > u = 0, by ( F ) and ( F ), one has(3.2) 12 f ( x, u ) u ≥ F ( x, u ) ≥ . Lemma 3.1. (i)For any u ∈ E \ { } , there exists t u = t ( u ) > such that t u u ∈ N and J ( t u u ) = max t ≥ J ( tu ) .(ii) There exists α > such that k ω k ≥ α for all ω ∈ N .(iii) If ω ∈ N , then there exist < s ω ≤ ≤ t ω such that [ s ω , t ω ] ω ⊂ N . Moreover, J ( sω ) = J ( ω ) , J ′ ( sω ) = sJ ′ ( ω ) for all s ∈ [ s ω , t ω ] and J ( z ) < J ( ω ) for all other z ∈ R + u \ [ s ω , t ω ] ω .(iv)There exists ρ > , such that c := inf N J ≥ inf S ρ J > , where S ρ := { u ∈ E : k u k = ρ } .(v)If V ⊂ E \ { } is a compact subset, then there exists R > such that J ≤ on R + u \ B R (0) for every u ∈ V .(vi) J is coercive on E , i.e., J ( u ) → ∞ , as k u k → ∞ . CHAO JI
Proof. (i) Let u ∈ E \ { } be fixed and define the function g ( t ) := J ( tu ) on [0 , ∞ ). By(3 .
1) and Proposition 2.1, for ǫ small enough we obtain that g ( t ) = t k u k − Z R N F ( x, tu ) dx ≥ t k u k − ǫt Z R N u dx − C ǫ t p Z R N | u | p dx. ≥ t k u k − C t p k u k p . Since u = 0 and 2 < p < ∗ α , g ( t ) > t > u ( x ) = 0, | tu ( x ) | → ∞ as t → + ∞ . By ( F ), we have g ( t ) = t k u k − Z R N F ( x, tu ) dx = t k u k − t Z u =0 F ( x, tu )( tu ) u dx which yields that g ( t ) → −∞ as t → ∞ . Therefore max t ≥ g ( t ) is achieved at some t u = t ( u ) >
0, and g ′ ( t u ) = J ′ ( t u u ) u = 0, t u u ∈ N .(ii) For any ω ∈ N , by (3.1), we have0 = h J ′ ( ω ) , ω i = k ω k − Z R N f ( x, ω ) ω dx ≥ k ω k − ǫ Z R N ω dx − C ǫ Z R N | ω | p dx ≥ k ω k − C k ω k p . Since 2 < p < ∗ α , there exists α > k ω k ≥ α .(iii) For any ω ∈ N , for ǫ small, we have Z R N ( | ( − ∆) α ω | + V ( x ) ω ) dx = Z R N f ( x, ω ) ω dx. (3.3)Moreover, set s > sω ∈ N , that is0 = h J ′ ( sω ) , sω i = s Z R N ( | ( − ∆) α ω | + V ( x ) ω ) dx − Z R N f ( x, sω ) sω dx, and Z R N ( | ( − ∆) α ω | + V ( x ) ω ) dx = Z R N f ( x, sω ) sω ω dx. (3.4)Combing (3.3) and (3.4), we have Z R N (cid:16) f ( x, sω ) sω − f ( x, ω ) ω (cid:17) ω dx = 0(3.5) RACTIONAL SCHR ¨ODINGER EQUATIONS 7
By ( F ), there exist s ω , t ω such that s ω ∈ (0 , t ω ≥
1, and for any s ∈ [ s ω , t ω ], f ( x, sω ) ω = f ( x, ω ) sω and sω ∈ N . Moreover, it is an immediate consequence from(i) that J ( sω ) = J ( ω ), J ′ ( sω ) = sJ ′ ( ω ) for all s ∈ [ s ω , t ω ] and J ( z ) < J ( ω ) for all other z ∈ R + u \ [ s ω , t ω ] ω .(iv) According to (3 . k u k ≤ ǫ small, we have J ( u ) = 12 k u k − Z R N F ( x, u ) dx ≥ k u k − C ǫ k u k − C C ǫ k u k p ≥ k u k − C C ǫ k u k p . Since p >
2, inf u ∈ S ρ J ( u ) > k u k = ρ and ρ small enough. Moreover, according to (i)and (ii), for every ω ∈ N , there exists t >
0, such that tω ∈ S ρ and J ( ω ) ≥ J ( sω ), so c := inf N J ≥ inf S ρ J > k u k = 1 for every u ∈ V . Arguingby contraction, suppose that there exist { u n } ⊂ V and t n u n ∈ R + u n , n ∈ N such that J ( t n u n ) ≥ n ∈ N and t n → ∞ as n → ∞ . Up to a subsequence, we may assumethat u n → u ∈ S := { u ∈ E : k u k = 1 } . It is clear that0 ≤ J ( t n u n ) t n k u n k = 12 − Z R N F ( x, t n u n ) t n u n u n dx. By ( F ) and Fatou’s lemma, one has Z R N F ( x, t n u n ) t n u n u n dx → + ∞ , this yields a contradiction.(vi) Arguing by contraction, suppose there exists a sequence { ω n } ⊂ N such that k ω n k → ∞ and J ( ω n ) ≤ d for some d >
0. Let v n = ω n k ω n k , then v n ⇀ v in E and v n ( x ) → v a.e. x ∈ R N , up to a subsequence. For some R >
0, choose y n ∈ R N satisfy Z B R ( y n ) v n dx = sup y ∈ R N Z B R ( y ) v n dx. Suppose R B R ( y n ) v n dx → n → ∞ , then v n → L p ( R N ) by Proposition 2.2. More-over, by (3.1), R R N F ( x, sv n ) dx → s ∈ R and therefore d ≥ J ( ω n ) ≥ J ( sv n ) = s − Z R N F ( x, sv n ) dx → s , CHAO JI this yields a contraction if s > √ d . So there exists δ > Z B R ( y n ) v n dx ≥ δ. (3.6)Since J is invariant under translations of the form u u ( · − k ) with k ∈ Z N , we mayassume that { y n } is bounded in R N . By (3.6) and Fatou’s lemma, we have v = 0.Moreover, by (iv) and ( F ), we obtain0 ≤ J ( ω n ) k ω n k = 12 − Z R N F ( x, ω n ) k ω n k dx = 12 − Z R N F ( x, ω n ) ω n v n dx → −∞ . This yields a contradiction. Thus, J is coercive on N . (cid:3) Remark 3.2.
In the proof of Lemma 3.1(iv), to show that v = 0, we use the assumption f and V are 1-periodic in x , . . . , x N . So for the coercive potential case, the boundedpotential well case and the V and f are asymptotically periodic in x case, we need toadapt the proof.In virtue of Lemma 3.1 (iii), for any u ∈ E \ { } , there exist ω ∈ N and 0 < s ω ≤ ≤ t ω such that m ( u ) := [ s ω , t ω ] ω = R + u ∩ N , where m : E \ { } → E is a multiplevalued map. Define b Ψ( u ) := J ( m ( u )) = max v ∈ R + u J ( v ) . This is a single-valued map since J is constant on R + u ∩ N . If ( F ′ ) holds instead of ( F ),by the same proof as in [22], for any u ∈ E \ { } , there exists a unique positive number t > J ( tu ) = max t> J ( tu ), so b Ψ ∈ C ( E \ { } , R ). But under our assumptions, u m ( u ) may not be single value, thus b Ψ may not be C in E \ { } and the criticalpoints theory for smooth functionals does not work, we need the nonsmooth methods in[4]. Proposition 3.3. b Ψ : E \ { } → R is a locally Lipschitz continuous map.Proof. If u ∈ E \ { } , then there exist a neighborhood U ⊂ E \ { } of u and R > J ( ω ) ≤ u ∈ U and ω ∈ R + u , k ω k ≥ R . Arguing by contraction, supposethat there exist sequence { u n } , { ω n } such that u n → u , ω n ∈ R + u n , J ( ω n ) > k ω n k → ∞ . Since u , u , u , . . . is a compact subset, by Lemma 3.1(v), we have J ( ω ) ≤ R > ω ∈ R + u n , n = 0 , , , . . . , k ω k ≥ R , this is a contraction.Let t u ∈ m ( u ), t u ∈ m ( u ), where u , u ∈ U , then k m ( u ) k , k m ( u ) k ≤ R . By themaximality property of m ( u ) and the mean value theorem, b Ψ( u ) − b Ψ( u ) = J ( t u ) − J ( t u ) ≤ J ( t u ) − J ( t u ) ≤ t sup s ∈ [0 , k J ′ ( t ( su + (1 − s ) u )) k u − u k≤ C k u − u k , RACTIONAL SCHR ¨ODINGER EQUATIONS 9 where C depends on R but independs on the particular choice of in m ( u ), m ( u ). Similarlythe above inequality, we also have b Ψ( u ) − b Ψ( u ) ≤ C k u − u k . This completes the proof. (cid:3)
For each v ∈ E , the generalized directional derivative b Ψ ◦ ( u ; v ) in the direction v isdefined by b Ψ ◦ ( u ; v ) = lim sup h → t ↓ b Ψ( u + h + tv ) − b Ψ( u + h ) t . The function v b Ψ ◦ ( u ; v ) is subadditive and positively homogeneous, and then is convex.The generalized gradient of b Ψ at u , denoted ∂ b Ψ( u ), is defined to be subdifferential of theconvex function b Ψ ◦ ( u ; v ) at θ , that is, ω ∈ ∂ b Ψ( u ) ⊂ E if and only if for all v ∈ E , b Ψ ◦ ( u ; v ) ≥ h ω, v i . If 0 ∈ ∂ b Ψ( u ), i.e. b Ψ ◦ ( u ; v ) ≥
0, for all v ∈ E , we call u is a critical point of b Ψ. Wecall a sequence { u n } is a Palais-Smale sequence for b Ψ(PS-sequence for short) if b Ψ( u n ) isbounded and there exist ω n ∈ ∂ b Ψ( u n ) such that ω n →
0. The functional b Ψ satisfies thePS-condition if each PS-sequence has a convergent subsequence.We shall use some notations S := { u ∈ E : k u k = 1 } , T u S := { v ∈ E : h u, v i = 0 } , Ψ = b Ψ | S , Ψ d := { u ∈ S : Ψ( u ) ≤ d } , Ψ c := { u ∈ S : Ψ( u ) ≥ c } , Ψ dc := Ψ c ∩ Ψ d ,K := { u ∈ S : 0 ∈ ∂ b Ψ( u ) } , K c := Ψ cc ∩ K, ∂ Ψ( u ) := ∂ b Ψ( u ) , if u ∈ S. Proposition 3.4. (i) u ∈ S is a critical point of b Ψ if and only if m ( u ) consists of criticalpoints of J . The corresponding critical values coincide.(ii) { u n } ⊂ S is a PS-sequence for b Ψ if and only if there exist ω n ∈ m ( u n ) such that { ω n } is a PS-sequence for J .Proof. (i) In fact, we need to show that for u ∈ S , b Ψ ◦ ( u ; v ) ≥ v ∈ E if andonly if m ( u ) consists of critical points of J . It is clear that E = R u ⊕ T u S , and by themaximizing property of m ( u ), J ′ ( ω ) v = 0 for all ω ∈ m ( u ) and v ∈ R u . Fixed s ∈ R , since b Ψ is locally Lipschitz continuous and b Ψ( u ) = b Ψ( σu ) for all σ >
0, we have | b Ψ( u + h + t ( su )) − b Ψ( u + h ) | = | b Ψ((1 + ts ) u + h ) − b Ψ((1 + ts )( u + h )) | ≤ Ct | s |k h k for t > k h k small enough. Thus b Ψ ◦ ( u ; su ) = 0 for all s ∈ R .Let s u > s u u ∈ m ( u ), by the maximizing property of m ( u ) and the mean value theorem, b Ψ( u + h + tv ) − b Ψ( u + h ) t = J ( s u + h + tv ( u + h + tv )) − J ( s u + h ( u + h )) t ≤ J ( s u + h + tv ( u + h + tv )) − J ( s u + h + tv ( u + h )) t = s u + h + tv J ′ ( s u + h + tv ( u + h + θtv )) v where t > θ ∈ (0 , N is bounded away from 0 and J is coercive on N , thus s u + h + tv is bounded. Letting h → t ↓ b Ψ ◦ ( u ; v ) ≤ sJ ′ ( su ) v, (3.7)where s n := s u + h n + t n v → s >
0. Moreover, since N is closed and b Ψ( u ) = lim n →∞ b Ψ( u + h n + t n v ) = lim n →∞ J ( s n ( u + h n + t n v ))= lim n →∞ (cid:16) s n k u + h n + t n v k − Z R N F ( x, s n ( u + h n + t n v )) dx (cid:17) = s k u k − Z R N F ( x, su ) dx, so su ∈ N . From Lemma 3.1(iii), it is possible that R + u ∩ N is a line segment, not apoint, hence s may be different for different v . We set s , s correspond to v and v , byLemma 3.1(iii), we have s u = τ ( s u ) and J ′ ( s u ) v = τ J ′ ( s u ) v for some τ >
0. Fromthis and (3.7), for y ∈ ∂ Ψ( u ), one has(3.8) h y, v i ≤ b Ψ ◦ ( u ; v ) ≤ τ ( v ) J ′ ( su ) v, where τ is bounded and bounded away from 0(by constants independent of v ). It followsthat u is a critical point of b Ψ if and only if m ( u ) consists of critical points of J .(ii)We take y n ∈ ∂ Ψ( u n ) and ω n ∈ m ( u n ). Since J is coercive on N and J ( m ( u n )) isbounded, the sequence { m ( u n ) } is bounded. As in (3.8), we have(3.9) h y n , v i ≤ b Ψ ◦ ( u n ; v ) ≤ τ n ( v ) J ′ ( ω n ) v, where τ n is bounded and bounded away from 0 because so is m ( u n ). We complete theproof. (cid:3) Remark 3.5. (1) Because of b Ψ ◦ ( u ; su ) = 0 for all s ∈ R , so ∂ Ψ( u ) ⊂ T u S .(2) If ω n ∈ m ( u n ) is a PS-sequence of J , then so is any sequence ω ′ n ∈ m ( u n ).The pseudo-gradient vector field H : S \ K → T S for Ψ be very important. For u ∈ S ,we define(3.10) ∂ − Ψ( u ) := { γ ∈ ∂ Ψ( u ) : k γ k = min a ∈ ∂ Ψ( u ) k a k} RACTIONAL SCHR ¨ODINGER EQUATIONS 11 and µ ( u ) := inf β ∈ S {k ∂ − Ψ( β ) k + k u − β k} . Because ∂ Ψ( u ) is a closed and convex set, from [16], it follows that γ in (2.1) exists andis unique, so we have K = { u ∈ S : ∂ − Ψ( u ) = 0 } . By [3], the map u
7→ k ∂ − Ψ( u ) k is lower semicontinuous but not continuous in general. Toregularize k ∂ − Ψ( u ) k , the function µ be introduced. Lemma 3.6.
The function µ is continuous and u ∈ K if and only if µ ( u ) = 0 .Proof. Let u, v, β ∈ S , by the definition of µ , we have µ ( u ) ≤ k ∂ − Ψ( β ) k + k u − β k ≤ k ∂ − Ψ( β ) k + k v − β k + k u − v k . So µ ( u ) ≤ inf β ∈ S {k ∂ − Ψ( β ) k + k v − β k} + k u − v k = µ ( v ) + k u − v k . Similarity, we have µ ( v ) − µ ( v ) ≤ k u − v k , Hence µ is Lipschitz continuous and is also continuous.Since 0 ≤ µ ( u ) ≤ k ∂ − Ψ( β ) k , it is easy to see that µ ( u ) = 0 if u ∈ K . Now suppose µ ( u ) = 0. In virtue of the definition of µ ( u ), there exist β n ⊂ S such that ∂ − Ψ( β n ) → β n → u . Moreover, by the map u
7→ k ∂ − Ψ( u ) k is lower semicontinuous, so u ∈ K . (cid:3) Proposition 3.7.
There exists a locally Lipschitz continuous vector field H : S \ K → T S with k H ( u ) k ≤ and inf {h γ, H ( u ) i : γ ∈ ∂ Ψ( u ) } > µ ( u ) for all u ∈ S \ K . If J is even,then H may be chosen to be odd. This follows by an easy inspection of the proof of Proposition 2.10 in [16].
Proof of Theorem 1.1 . Since c := inf u ∈ S Ψ( u ) = inf ω ∈N J ( ω ) > { u n } ⊂ S such that Ψ( u n ) → c and(3.11) Ψ( v ) ≥ Ψ( u n ) − n k v − u n k for all v ∈ S. For a given v ∈ T u n S , let z n ( t ) = u n + tv k u n + tv k . It is clear that k u n + tv k − O ( t ) as t → b Ψ( u n + tv ) = Ψ( z n ( t )). From (3.11), we have b Ψ ◦ ( u n ; v ) ≥ lim sup t ↓ b Ψ( u n + tv ) − b Ψ( u n ) t = lim sup t ↓ Ψ( z n ( t )) − Ψ( u n ) t ≥ − n k v k . Since J is coercive on N , { m ( u n ) } is bounded. Moreover, by (3.9), one has − n k v k ≤ b Ψ ◦ ( u ; v ) ≤ τ n ( v ) J ′ ( ω n ) v, where ω n ∈ m ( u n ) ⊂ N and τ n is bounded and bounded away from 0. Since for any v ∈ R ω n , J ′ ( ω n ) v = 0, { ω n } is a bounded PS-sequence of J . Up to a subsequence, ω n ⇀ u , ω n → u in L loc ( R N ) and ω n → u in a.e. x ∈ R N . If ω n → L p ( R N ), then R R N F ( x, ω n ) dx → R R N f ( x, ω n ) ω n dx → n → ∞ , this implies that k ω n k → n → ∞ and this contradicts with Lemma 3.1(ii), so ω n L p ( R N ), by Proposition2.2, for some R > δ >
0, there exist { y n } such that Z B R ( y n ) ω n dx ≥ δ. Since J and N are invariant under translations of the form ω ω ( · − k ) with k ∈ Z N , wemay assume that { y n } is bounded in R N . By Fatou’s lemma, we know that u = 0. Nowwe show that u is a ground state solution. By (3.2) and Fatou’s lemma c := lim n →∞ J ( ω n ) = lim n →∞ (cid:16) J ( ω n ) − J ′ ( ω n ) ω n (cid:17) = lim n →∞ Z R N (cid:16) f ( x, ω n ) ω n − F ( x, ω n ) (cid:17) dx ≥ Z R N (cid:16) f ( x, u ) u − F ( x, u ) (cid:17) dx = J ( u ) − J ′ ( u ) u = J ( u ) ≥ c. The proof is completed. (cid:3)
Remark 3.8.
Under the assumptions of Theorem 1.1, if f ( x, u ) ≥ u ≥ f ( x, u ) = 0, u ≤
0, then we may obtain a nonnegative ground state solution. Put u + := max { u, } .Noting that the conclusion of Theorem 1.1 holds for the functional J + ( u ) := 12 Z R N (cid:16) | ( − ∆) α u | + V ( x ) u (cid:17) dx − Z R N F ( x, u + ) dx. So we get a ground state solution u of the equation( − ∆) α u + V ( x ) u = f ( x, u + ) , x ∈ R N . Using u − := min { u, } as a test function in above equation, and integrating by parts, weobtain Z R N ( − ∆) α u · u − dx = − Z R N V ( x )( u − ) dx ≤ . RACTIONAL SCHR ¨ODINGER EQUATIONS 13
But we know that Z R N ( − ∆) α u · u − dx = Z Z R N × R N ( u ( x ) − u ( y ))( u − ( x ) − u − ( y )) | x − y | N +2 α dxdy ≥ Z Z { u> }×{ u< } ( u ( x ) − u ( y ))( − u − ( y )) | x − y | N +2 α dxdy + Z Z { u< }×{ u< } ( u − ( x ) − u − ( y )) | x − y | N +2 α dxdy + Z Z { u< }×{ u> } ( u ( x ) − u ( y )) u − ( x ) | x − y | N +2 α dxdy ≥ . Thus u − = 0 and u ≥ f ( x, u ) is odd in u . To prove the existence of infinitely manygeometrically distinct solutions, we assume the contrary. Since for each [ s ω , t ω ] ω ⊂ N there corresponds a unique point u ∈ S . Assume that F is a finite set and choose a subset F of K such that −F = F and each orbit O ( ω ) has a unique representative in F . Lemma 3.9.
The mapping m − : N → S is Lipschitz continuous. Lemma 3.10. κ := inf {k v − w k : v, w ∈ K, v = w } > . The proofs of the above two lemmas are similar with Lemma 2.11 and Lemma 2.13 in[22], so we omit it here.
Lemma 3.11 ([22]) . Let d ≥ c . If { ( v n } , { v n } ⊂ Ψ d are two Palais-Smale sequences for Ψ , then either k v n − v n k → as n → ∞ or lim sup n →∞ k v n − v n k ≥ ρ ( d ) > , where ρ ( d ) depends on d but not on the particular choice of PS-sequences in Ψ d . Let H be the pseudo-gradient vector field in Proposition and η : → S \ K be the flowdefined by ddt η ( t, w ) = − H ( η ( t, w )) ,η (0 , w ) = w, (3.12)where G := { ( t, w ) : w ∈ S \ K, T − ( w ) < t < T + ( w ) } and ( T − ( w ) , T + ( w )) is the maximalexistence time for the trajectory t η ( t, w ) which passing through ω at t = 0. Note that η is odd in w because H is and t Ψ( η ( t, w )) is strictly decreasing by the properties ofa pseudogradient. Lemma 3.12.
For each ω ∈ S \ K , lim t → T + ( ω ) η ( t, ω ) exists and is a critical point of J .Proof. If T + ( ω ) < ∞ and let 0 ≤ s < t < T + ( ω ). Then k η ( t, ω ) − η ( s, w ) k ≤ Z ts k H ( η ( τ, w )) k dτ ≤ t − s. Hence the limit exits and if it is not a critical point, then η ( · , w ) can be continued for t > T + ( ω ).Assume T + ( ω ) = ∞ . It suffices to prove that for each ǫ > t ǫ > k η ( t ǫ , ω ) − η ( t, w ) k < ǫ for any t ≥ t ǫ . Argument by contradiction, we can find ǫ ∈ (0 , ρ ( d )2 ) and { t n } ⊂ R + with t n → + ∞ and k η ( t n , ω ) − η ( t n +1 , w ) k = ǫ for all n ≥
1. Choose the smallest t n ∈ ( t n , t n +1 ) such that k η ( t n , ω ) − η ( t n , w ) k = ǫ andlet κ n := min { z ( η ( s, ω )) : s ∈ [ t n , t n ] } . By the continuity of µ , Proposition 3.6 andProposition 7.1.1(viii) in [3], we have ǫ k η ( t n , ω ) − η ( t n , w ) k ≤ Z t n t n k H ( η ( s, ω )) k ds ≤ t n − t n ≤ κ n Z t n t n inf γ ∈ ∂ Ψ( η ( s,u )) h γ, H ( η ( s, ω )) i ds = − κ n Z t n t n sup γ ∈ ∂ Ψ( η ( s,u )) h γ, − H ( η ( s, ω )) i ds ≤ − κ n Z t n t n dds Ψ( η ( s, ω )) ds = 2 κ n (Ψ( η ( t n , ω ) − Ψ( η ( t n , ω )) . Since Ψ is bounded below, Ψ( η ( t n , ω ) − Ψ( η ( t n , ω ) →
0, it follow that κ n →
0. Hence wecan find s n ∈ [ t n , t n ] such that z ( η ( s n , ω )) → n → ∞ . By the definition µ there exit ω n such that ω n − η ( s n , ω ) → ∂ − Ψ( ω n ) →
0. So lim sup n →∞ k ω n − η ( t n , ω ) k ≤ ǫ .Similarly, there exists a largest t n ∈ [ t n , t n +1 ] such that k η ( t n , ω ) − η ( t n +1 , w ) k = ǫ andwe can find ω n such that ∂ − Ψ( ω n ) → n →∞ k ω n − η ( t n +1 , ω ) k ≤ ǫ . So ǫ ≤ lim sup n →∞ k ω n − ω n k ≤ ǫ < ρ ( d ), it contradicts with Lemma 3.11. The proof iscompleted. (cid:3) Let P ⊂ S , δ > U δ ( P ) := { w ∈ S : dist( w, P ) < δ } . Lemma 3.13.
Let d ≥ c . Then for every δ > there exists ǫ = ǫ ( δ ) > such that (1) Ψ d + ǫd − ǫ ∩ K = K d (2) lim t → T + ( w ) Ψ( η ( t, w )) < d − ǫ for all w ∈ Ψ d + ǫ \ U δ ( K d ) .Proof. Since we assume that F is a finite set, so (1) holds for ǫ > U δ ( K d ) ⊂ J d +1 and δ < ρ ( d + 1). In order to find ǫ > τ := inf n µ ( ω ) : ω ∈ U δ ( K d ) \ U δ ( K d ) o and claim that τ >
0. Argument by contradiction. Assume that there exits a sequence { v n } ⊂ U δ ( K d ) \ U δ ( K d ) such that µ ( v n ) →
0. According to the definition of µ , there existsa PS-sequence { ω n } of Ψ such that k ω n − v n k → n → ∞ . Using this limit, the finitenessassumption of F and the Z N -invariance of Ψ, we may assume that ω n ∈ U δ ( ω ) \ U δ ( ω ) RACTIONAL SCHR ¨ODINGER EQUATIONS 15 for some ω ∈ K d . Let v n → ω . Since ω ∈ K d and µ is continuous, µ ( v n ) →
0. Asbefore, there exists a PS-sequence { ω n } of Ψ such that k ω n − v n k → n → ∞ , moreover, k ω n − ω k → n → ∞ . So, we have δ ≤ lim sup n →∞ k ω n − ω n k ≤ δ < ρ ( d + 1) , this contradicts with Lemma 3.11.Hence τ is positive. Choose ǫ < δτ such that (1) holds.By Lemma 3.12 and (1), the only way (2) can fail is that η ( t, w ) → ˜ ω ∈ K d as t → T + ( ω )for some w ∈ Ψ d + ǫ \ U δ ( K d ). In this case we let t := sup { t ∈ [0 , T + ( ω )) : η ( t, w ) U δ (˜ ω ) } and t := sup { t ∈ ( t , T + ( ω )) : η ( t, w ) ∈ U δ (˜ ω ) } . Then δ k η ( t , w ) − η ( t , w ) k ≤ Z t t k H ( η ( s, ω )) k ds ≤ t − t , and Ψ( η ( t , ω ) − Ψ( η ( t , ω ) = Z t t dds Ψ( η ( s, ω )) ds ≤ Z t t sup γ ∈ ∂ Ψ( η ( s,u )) h γ, − H ( η ( s, ω )) i ds ≤ − Z t t inf γ ∈ ∂ Ψ( η ( s,u )) h γ, H ( η ( s, ω )) i ds ≤ − µ ( η ( s, u ))( t − t ) ≤ − τ ( t − t ) ≤ − δτ . Hence Ψ( η ( t , w )) ≤ d + ǫ − δτ < d and therefore η ( t, w ) ˜ ω , it contradicts with ourassumption. This completes the proof. (cid:3) Proof of Theorem 1.2 . LetΣ := { A ⊂ S : A = A, A = − A } . Recall that the definition of the Krasnoselskii genus γ ( A ), for A ⊂ Σ in [21]. Define c k := inf { d ∈ R : γ (Ψ d ) ≥ k } , k ≥ . Thus c k are those numbers at which the set Ψ d change genus and it is easy to see that c k ≤ c k +1 . We claim: K c k = ∅ and c k < c k +1 for all k ∈ N . To prove this, let k ≥ d := c k . By Lemma 3.10, K d is either empty or a discreteset, hence γ ( K d ) = 0 or 1. By the continuity property of the genus, there exists δ > γ ( U ) = γ ( K d ), where U := U δ ( K d ) and δ < κ . For such δ , choose ǫ > so that the conclusions of Lemma 3.13 hold. Then for each w ∈ Ψ d + ǫ \ U there exists t ∈ [0 , T + ( w )) such that Ψ( η ( t, w )) ≤ d − ǫ . Let e = e ( w ) be the infimum of the timefor which Ψ( η ( t, w )) ≤ d − ǫ . Since d − ǫ is not a critical value of Ψ, it is easy to seeby the Implicit Function Theorem that e is a continuous mapping and since Ψ is even, e ( − w ) = e ( w ). Define a mapping h : Ψ d + ǫ \ U → Ψ d + ǫ by setting h ( w ) := η ( e ( w ) , w ).Then h is odd and continuous, so it follows from the properties of the genus and thedefinition of c k that γ (Ψ d + ǫ ) ≤ γ ( U ) + γ (Ψ d − ǫ ) ≤ γ ( U ) + k − γ ( K d ) + k − . If γ ( K d ) = 0, then γ (Ψ d + ǫ ) ≤ k −
1, it contradicts the definition of c k . So γ ( K d ) = 1and K d = ∅ . If c k = c k +1 = d , then γ ( K d ) >
1. Since this is impossible, we must have c k ≤ c k +1 and K c k = ∅ for all k ≥
1. Hence, the proof is finished. (cid:3) Proof of Theorem 1.3
In this section, we assume that V ( x ) is coercive, that is, V ( x ) → + ∞ as | x | → ∞ .To prove Theorem 1.3, we need to adapt the proof of Theorem 1.1. The main differencebetween them is how to show that the solution is nontrivial. From section 3, we know thatLemma 3.1 is very important. By a simple observation, in addition to Lemma 3.1(vi), theproof of other results in Lemma 3.1 are the same as the coercive potential case. Lemma 4.1.
Assume that ( F ) − ( F ) hold and V ( x ) → + ∞ as | x | → ∞ , then J iscoercive on E ,i.e., J ( u ) → ∞ , as k u k → ∞ .Proof. Arguing by contraction, suppose there exists a sequence { ω n } ⊂ N such that k ω n k → ∞ and J ( ω n ) ≤ d for some d >
0. Let v n = ω n k ω n k . Then v n ⇀ v in E , v n → v in L p ( R N ) and v n ( x ) → v a.e. x ∈ R N , up to a subsequence. If v = 0, then by (3.1), R R N F ( x, sv n ) dx → s ∈ R and therefore d ≥ J ( ω n ) ≥ J ( sv n ) = s − Z R N F ( x, sv n ) dx → s , this yields a contraction if s > √ d . So v n ( x ) → v = 0 a.e. x ∈ R N . Moreover, by ( F )one has 0 ≤ J ( ω n ) k ω n k = 12 − Z R N F ( x, ω n ) k ω n k dx = 12 − Z R N F ( x, ω n ) ω n v n dx → −∞ . This yields a contradiction. (cid:3)
Proof of Theorem 1.3 . Similar with the proof of Theorem 1.1, there exists a boundedPS-sequence { ω n } of J . Up to a subsequence, ω n ⇀ u , ω n → u in L p ( R N ) and ω n → u ina.e. x ∈ R N . If u = 0, then R R N F ( x, ω n ) dx → R R N f ( x, ω n ) ω n dx → n → ∞ ,this implies that k ω n k → n → ∞ and this contradicts with k ω k bounded away from0, for any ω ∈ N , so u = 0 is a nontrivial solution. Moreover, by Fatou’s lemma and (3.2),it is easy to know that u is a ground state solution. (cid:3) RACTIONAL SCHR ¨ODINGER EQUATIONS 17 Proof of Theorem 1.4
We assume that inf x ∈ R N V ( x ) ≤ V ( x ) < lim | x |→∞ V ( x ) = sup x ∈ R N V ( x ) < + ∞ and f ( x, u ) = f ( u ) hold in this section. As in Section 4, except for the proof of Lemma 3.1(vi),others are the same. Now we give its proof in the bounded potential well case. Lemma 5.1.
Assume that f ( x, u ) = f ( u ) , ( F ) − ( F ) and inf x ∈ R N V ( x ) ≤ V ( x ) < lim | x |→∞ V ( x ) = sup x ∈ R N V ( x ) < + ∞ hold, then J is coercive on E ,i.e., J ( u ) → ∞ , as k u k → ∞ .Proof. Arguing by contraction, suppose there exists a sequence { ω n } ⊂ N such that k ω n k → ∞ and J ( ω n ) ≤ d for some d >
0. Let v n = ω n k ω n k . Then v n ⇀ v in E and v n ( x ) → v a.e. x ∈ R N , up to a subsequence. For some R >
0, choose y n ∈ R N satisfy Z B R ( y n ) v n dx = sup y ∈ R N Z B R ( y ) v n dx. Similar with the proof of Lemma 3.1(v), we may prove that there exists δ > Z B R ( y n ) v n dx ≥ δ. (5.1)Set e v n ( · ) = v n ( · − y n ), then we have e v n ⇀ e v in E , e v n → e v in L loc ( R N ), and e v n → e v in a.e. x ∈ R N . In virtue of (5.1), we have e v = 0. By ( F ) we obtain that0 ≤ J ( u n ) k u n k = 12 − Z R N F ( u n ) k u n k dx = 12 − Z R N F ( u n ( x − y n )) | u n ( x − y n ) | e v n dx → −∞ . This is a contradiction. we complete the proof. (cid:3)
We shall need a limiting problem(5.2) ( − ∆) α u + V ∞ u = f ( u ) , x ∈ R N . The energy functional corresponding to it is J ∞ ( u ) := 12 Z R N ( | ( − ∆) α u | + V ∞ u ) dx − Z R N F ( u ) dx. Let N ∞ = { u ∈ E \ { } : Z R N | ( − ∆) α u | + V ∞ u dx = Z R N f ( u ) u dx } be the Nehari manifold for J ∞ . Since V ∞ be constant and f independs on x , there existsa solution u ∞ = 0 for minimizes J ∞ on N ∞ by Theorem 1.1. Lemma 5.2. (i) If
V < V ∞ , then < c < c ∞ , where c ∞ := inf u ∈N ∞ J ∞ ( u ) .(ii) For { u n } ⊂ N , if J ( u n ) → d ∈ (0 , c ∞ ) and J ′ ( u n ) → , then u n ⇀ u = 0 after passingto a subsequence, u is a critical point of J and J ( u ) ≤ d . Proof. (i) Let s > s u ∞ ∈ N . Since V ( x ) < V ∞ in x ∈ R N , we have c ≤ J ( s u ∞ ) < J ∞ ( s u ∞ ) ≤ J ∞ ( u ∞ ) = c ∞ . (ii) Because J is coercive on N , { u n } is bounded. Up to a subsequence, u n ⇀ u in E , u n ( x ) → u ( x ) a.e. x ∈ R N . By Fatou’s lemma, d = J ( u n ) − h J ′ ( u n ) , u n i + o (1) = Z R N (cid:16) f ( u n ) u n − F ( u n ) (cid:17) dx + o (1) ≥ Z R N (cid:16) f ( u ) u − F ( u ) (cid:17) dx + o (1) = J ( u ) − h J ′ ( u ) , u i + o (1) = J ( u ) + o (1) . So J ( u ) ≤ d and it remains to show that u = 0. Arguing indirectly, suppose u = 0. Since u n → L loc ( R N ) and V ( x ) → V ∞ as | x | → ∞ , J ( u n ) − J ∞ ( u n ) = 12 Z R N ( V ( x ) − V ∞ ) u n dx → J ∞ ( u n ) → d . Using the H¨older and the Sobolev inequalities and taking v with k v k = 1, we obtain (cid:12)(cid:12) h J ′ ( u n ) − J ′∞ ( u n ) , v i (cid:12)(cid:12) ≤ Z R N ( V ∞ − V ( x ) | u n | | v | dx ≤ C (cid:18)Z R N ( V ∞ − V ( x )) u n dx (cid:19) / . As the right-hand side tends to 0 uniformly in k v k = 1, J ′ ( u n ) − J ′∞ ( u n ) → J ′∞ ( u n ) →
0. So 0 = h J ′ ( u n ) , u n i ≥ k u n k − C Z {| u n |≥ } | u n | p dx and if k u n k p →
0, then u n → E which is impossible because J ( u n ) → d >
0. Henceby Proposition 2.2, for some
R >
0, there are ( y n ) ⊂ R N and δ > Z B R ( y n ) u n dx ≥ δ. Let v n ( x ) := u n ( x + y n ). Since J ∞ is invariant with respect to translations by elements of R N , J ∞ ( v n ) → d and J ′∞ ( v n ) →
0. Moreover, Z B (0) v n dx = Z B ( y n ) u n dx ≥ δ and therefore v n ⇀ v = 0 after passing to a subsequence. It follows that v is a nontrivialcritical point of J ∞ and J ∞ ( v ) ≤ d < J ∞ ( v ∞ ) which is the desired contradiction. (cid:3) Proof of Theorem 1.4 . Similar with the proof of Theorem 1.1, there exists a boundedsequence { ω n } such that J ( ω n ) → c and J ′ ( ω n ) →
0. Using Lemma 5.2 we obtain a criticalpoint u = 0 of J such that J ( u ) ≤ c . So J ( u ) = c and u is a ground state solution ofproblem (1.1). The proof is completed. (cid:3) RACTIONAL SCHR ¨ODINGER EQUATIONS 19 Proof of Theorem 1.5
Now we seek the ground state solutions of problem (1.1) when V and f are asymptoti-cally periodic in x . Firstly, by a simple observation, Lemma 3.1 holds under assumptions( V ) and ( F ) − ( F ). Moreover, to prove Theorem 1.5, we need some lemmas. Lemma 6.1.
Assume ( V ) and ( F ) − ( F ) hold. Then J ( u ) ≤ J p ( u ) , for all u ∈ E . It follows an easy inspection.
Lemma 6.2.
Assume ( V ) and ( F )( ii ) hold. Assume that { u n } ⊂ E satisfies u n ⇀ and ϕ n ∈ E is bounded. Then Z R N ( V ( x ) − V p ( x )) u n ϕ n dx → , (6.1) Z R N ( f ( x, u n ) − f p ( x, u n )) ϕ n dx → , (6.2) Z R N ( F ( x, u n ) − F p ( x, u n )) dx → . (6.3)For the proof of this lemma one may refer to [24], so we omit it. Proof of Theorem 1.5 . As the same in the proof of Theorem 1.1, there exists a boundedPS-sequence { ω n } ⊂ m ( u n ) satisfies J ( ω n ) → c and J ′ ( ω n ) →
0. Up to a subsequence, ω n ⇀ u in E , ω n → u in L loc ( R N ), and ω n → u a.e. on x ∈ R N . If u = 0, u is a ground statesolution of problem (1.1) and the proof is completed. Now we show that u = 0. Arguingby contradiction, if ω n → L p ( R N ), then R R N F ( x, ω n ) dx → and R R N f ( x, ω n ) ω n dx → n → ∞ , this implies that k ω n k → n → ∞ and this contradicts with Lemma. If ω n L p ( R N ), by Proposition 2.2, for some R > δ >
0, there exist y n such that Z B r ( y n ) ω n dx ≥ δ. (6.4)Without loss of generality, we assume that y n ∈ Z N . Setting ¯ ω n ( x ) = ω n ( x − y n ), up to asubsequence, we have ¯ ω n ⇀ ¯ u in E , ¯ ω n → ¯ u in L loc ( R N ), and ¯ ω n → ¯ u a.e. on x ∈ R N . ByFatou’s lemma and (6.4), ¯ u = 0.For any ϕ ∈ E , set ϕ n ( · ) = ϕ ( · − y n ), by (6.1) and (6.2) in Lemma 6.2 , we may obtain h J ′ ( ω n ) , ϕ n i − h J ′ p ( ω n ) , ϕ n i → . Since J ′ ( ω n ) → k ϕ n k = k ϕ k , h J ′ ( ω n ) , ϕ n i →
0, so we have h J ′ p ( ω n ) , ϕ n i →
0. Be-cause V p and f p are 1-periodic in and y n ∈ Z N , one has h J ′ p ( ω n ) , ϕ n i = h J ′ p ( ¯ ω n ) , ϕ i . Since ϕ is arbitrary, J ′ p ( ¯ ω n ) → E as n → ∞ . Since J ′ p is weakly sequently continuous by,we have J ′ p (¯ u ) = 0. Now we show that J p (¯ u ) ≤ c . Replacing ϕ n by ω n in Lemma 6.2, we have Z R N ( f ( x, ω n ) − f p ( x, ω n )) ω n dx → . (6.5)Combine with (6.5) and (6.3), we have Z R N (cid:16) f p ( x, ω n ) ω n − F p ( x, ω n ) (cid:17) dx = Z R N (cid:16) f ( x, ω n ) ω n − F ( x, ω n ) (cid:17) dx + o n (1) . (6.6)Since f p ( x, ω n ) ω n − F p ( x, ω n ) is 1-periodic in x , . . . , x N , so we have Z R N (cid:16) f p ( x, ω n ) ω n − F p ( x, ω n ) (cid:17) dx = Z R N (cid:16) f p ( x, ¯ ω n )¯ ω n − F p ( x, ¯ ω n ) (cid:17) dx. (6.7)By (6.6) and (6.7), one has Z R N (cid:16) f p ( x, ¯ ω n )¯ ω n − F p ( x, ¯ ω n ) (cid:17) dx = Z R N (cid:16) f ( x, ω n ) ω n − F ( x, ω n ) (cid:17) dx + o n (1) . By and Fatou’s lemma, it follows thatlim n →∞ Z R N (cid:16) f p ( x, ¯ ω n )¯ ω n − F p ( x, ¯ ω n ) (cid:17) dx ≥ Z R N (cid:16) f p ( x, ¯ u )¯ u − F p ( x, ¯ u ) (cid:17) dx, so lim n →∞ Z R N (cid:16) f ( x, ω n ) ω n − F ( x, ω n ) (cid:17) dx ≥ Z R N (cid:16) f p ( x, ¯ u )¯ u − F p ( x, ¯ u ) (cid:17) dx. Moreover, we have c := lim n →∞ J ( ω n ) = lim n →∞ (cid:16) J ( ω n ) − J ′ ( ω n ) ω n (cid:17) = lim n →∞ Z R N (cid:16) f ( x, ω n ) ω n − F ( x, ω n ) (cid:17) dx ≥ (cid:16) f p ( x, ¯ u )¯ u − F p ( x, ¯ u ) (cid:17) dx. = J p (¯ u ) − J ′ p (¯ u )¯ u = J p (¯ u ) . Since J ′ p (¯ u ) = 0 and ¯ u = 0, by Lemma 3.1(iii), max t ≥ J p ( t ¯ u ) = J p (¯ u ) and there exists t ¯ u > t ¯ u ¯ u ∈ N . Then J ( t ¯ u ¯ u ) ≤ J p ( t ¯ u ¯ u ) ≤ max t ≥ J p ( t ¯ u ) = J p (¯ u )In virtue of J p (¯ u ) ≤ c , J ( t ¯ u ¯ u ) ≤ c . So J ( t ¯ u ¯ u ) = c . The proof is completed. (cid:3) References [1] G. Autuori, P. Pucci,
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Center for Applied Mathematics, Tianjin University, 300072 Tianjin, ChinaDepartment of Mathematics, East China University of Science and Technology, 200237Shanghai, China
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