Existence and concentration of positive ground state solutions for nonlinear fractional Schrödinger-Poisson system with critical growth
aa r X i v : . [ m a t h . A P ] F e b EXISTENCE AND CONCENTRATION OF POSITIVE GROUNDSTATE SOLUTIONS FOR NONLINEAR FRACTIONALSCHR ¨ODINGER-POISSON SYSTEM WITH CRITICAL GROWTH
KAIMIN TENG AND RAVI P. AGARWAL
Abstract.
In this paper, we study the following fractional Schr¨odinger-Poissonsystem involving competing potential functions (cid:26) ε s ( − ∆) s u + V ( x ) u + φu = K ( x ) f ( u ) + Q ( x ) | u | ∗ s − u, in R , ε t ( − ∆) t φ = u , in R ,where ε > f is a function of C class, superlinear andsubcritical nonlinearity, 2 ∗ s = − s , s > , t ∈ (0 , V ( x ) K ( x ) and Q ( x )are positive continuous function. Under some suitable assumptions on V , K and Q , we prove that there is a family of positive ground state solutions withpolynomial growth for sufficiently small ε >
0, of which it is concentrating onthe set of minimal points of V ( x ) and the sets of maximal points of K ( x ) and Q ( x ). The methods are based on the Nehari manifold, arguments of Brezis-Nirenberg and concentration compactness of P. L. Lions. Introduction
In this paper, we are concerned with the existence of ground state solution and itsconcentration phenomenon for the following fractional Schr¨odinger-Poisson system (cid:26) ε s ( − ∆) s u + V ( x ) u + φu = K ( x ) f ( u ) + Q ( x ) | u | ∗ s − u, in R , ε t ( − ∆) t φ = u , in R , (1.1)where s > , t ∈ (0 , ∗ s = − s , ε > V ( x ), K ( x ) and Q ( x ) satisfy the following hypotheses:( V ) V ( x ) ∈ C ( R , R ) and 0 < V = inf R V ( x ) < lim inf | x |→ + ∞ V ( x ) = V ∞ < + ∞ .( K ) K ( x ) ∈ C ( R , R ) and lim | x |→ + ∞ K ( x ) = K ∞ ∈ (0 , + ∞ ) and K ( x ) ≥ K ∞ for x ∈ R .( Q ) Q ( x ) ∈ C ( R , R ) and lim | x |→ + ∞ Q ( x ) = Q ∞ ∈ (0 , + ∞ ) and Q ( x ) ≥ Q ∞ .( Q ) There exists δ > ρ > | Q ( x ) − Q ( x ) | ≤ δ | x − x | α for | x − x | ≤ ρ , where − s ≤ α < and Q ( x ) = max x ∈ R Q ( x ).( H ) Θ := Θ V ∩ Θ K ∩ Θ Q = ∅ , where Θ V = { x ∈ R : V ( x ) = V = inf x ∈ R V ( x ) } andΘ K = { x ∈ R : K ( x ) = K = max x ∈ R K ( x ) } Θ Q = { x ∈ R : Q ( x ) = Q = max x ∈ R Q ( x ) } . Mathematics Subject Classification.
Key words and phrases.
Fractional Schr¨odinger-Poisson system; Concentration-compactnessprinciple; Ground state solution; Palais-Smale condition.
Without loss of generality, we may assume that 0 ∈ Θ. Here the non-local operator( − ∆) s ( s ∈ (0 , − ∆) s v ( x ) = C s P.V. Z R v ( x ) − v ( y ) | x − y | s d y = C s lim ε → Z R \ B ε ( x ) v ( x ) − v ( y ) | x − y | s d y for v ∈ S ( R ), where S ( R ) is the Schwartz space of rapidly decaying C ∞ function, B ε ( x ) denote an open ball of radius r centered at x and the normalization constant C s = (cid:16) R R − cos( ζ ) | ζ | s d ζ (cid:17) − . For u ∈ S ( R ), the fractional Laplace operator ( − ∆) s can be expressed as an inverse Fourier transform( − ∆) s u = F − (cid:16) (2 π | ξ | ) s F u ( ξ ) (cid:17) , where F and F − denote the Fourier transform and inverse transform, respectively.Formally, system (1.1) consists of a fractional Schr¨odinger equation coupled witha fractional Poisson equation. It can be regarded as the associated fractional caseof the following classical Schr¨odinger-Poisson system (cid:26) − ε ∆ u + V ( x ) u + φu = g ( x, u ) , in R , − ε ∆ φ = u , in R . (1.2)It is well known that system (1.2) has a strong physical meaning because it appearsin quantum mechanics models (see e.g. [12, 28]) and in semiconductor theory[34]. In particular, systems like (1.2) have been introduced in [8] as a model todescribe solitary waves. In (1.2), the first equation is a nonlinear stationary equation(where the nonlinear term simulates the interaction between many particles) thatis coupled with a Poisson equation, to be satisfied by φ , meaning that the potentialis determined by the charge of the wave function. For this reason, (1.2) is referredto as a nonlinear Schr¨odinger-Poisson system. In recent years, there has beenincreasing attention to systems like (1.2) on the existence of positive solutions,ground state solutions, multiple solutions and semiclassical states; see for examples[3, 4, 8, 23, 39, 48, 53] and the references therein.The other motivation to study the system (1.1) lies in the important roles thatfractional equations involving fractional operators play in the problems of Physics,Chemistry and Geometry, and so on. Indeed, fractional operators appear in manyproblems, such as: fractional quantum mechanics [26, 27], anomalous diffusion [33],financial [15], obstacle problems [41], conformal geometry and minimal surfaces[11]. With the help of the harmonic extension technique developed by Caffarelliand Silvestre [13], the non-local problem can be reduced to a local one, choosing aweighted Sobolev space as the work space, the usual variational methods have beensuccessfully applied to nonlinear problems involving fractional Laplacian, we referto interesting readers to see the related works [2, 6, 9, 14, 17, 18, 21, 45] and soon. Another power technique is that one directly take the usual fractional Sobolevspace as the work space so that the variational approaches can be applied, therelated work can be referred to see [9, 16, 17, 19, 20, 40, 43, 44] and so on.To the best of our knowledge, there are only few recent papers dealing witha similar system like (1.1). For example, in [46], we established the existenceof positive ground state solution for a similar system involving a critical Sobolevexponent (cid:26) ( − ∆) s u + V ( x ) u + φu = | u | p − u + | u | ∗ s − u, in R ,( − ∆) t φ = u , in R , (1.3) RACTIONAL SCHR ¨ODINGER-POISSON SYSTEM WITH CRITICAL GROWTH 3 by using the Nehari-Pohozaev manifold combing monotone trick with global com-pactness Lemma. Using the similar methods, in [47], positive ground state solu-tions for subcritical problem, i.e., | u | p − u + | u | ∗ s − u is replaced by | u | p − u with p ∈ (2 , ∗ s − s = t . In [50], the existence of infinitelymany (but possibly sign changing) solutions by means of the Fountain Theoremunder suitable assumptions on nonlinearity term. In [52], the authors studied theexistence of radial solutions for system (2.1) with replacing | u | p − u + | u | ∗ s − u by f ( u ), where the nonlinearity f ( u ) verifies the subcritical or critical assumptions ofBerestycki-Lions type. In [35], the authors studied the semiclassical state of thefollowing system (cid:26) ε s ( − ∆) s u + V ( x ) u + φu = f ( u ) , in R N , ε θ ( − ∆) α φ = γ α u , in R N ,where s ∈ (0 , α ∈ (0 , N ), θ ∈ (0 , α ), N ∈ (2 s, s + α ), γ α is a positive constant, f ( u ) satisfies the following subcritical growth assumptions: 0 < KF ( t ) ≤ f ( t ) t with some K > t ≥ f ( t ) t is strictly increasing on (0 , + ∞ ). By adapt-ing some ideas of Benci, Cerami and Passaseo [5, 7] and using the Ljusternick-Schnirelmann Theory, the authors obtained the multiplicity of positive solutionswhich concentrate on the minima of V ( x ) as ε →
0. Of course, recently, this meth-ods have been successfully applied to other many problems, such as: Schr¨odinger-Poisson system [23], fractional Schr¨odinger equations [24], p -Laplacian problem [1],Kirchhoff type problems [25], and the references therein. In [32], by using themethods mentioned before, Liu and Zhang proved the existence and concentrationof positive ground state solution for problem (1.1) when K ( x ) ≡ Q ( x ) ≡ − h ∆ u + V ( x ) u = K ( x ) | u | p − u + Q ( x ) | u | q − u x ∈ R N (1.4)where h >
0, 1 < q < p < N +2 N − , have been investigated by several scholars.Such as, Rabinowtiz [38] proved that if V is coercive and K , Q satisfy suitableassumptions, a result implies the existence of ground state solutions for problem(1.4), for any h >
0. Wang and Zeng [49] assumed that V ( x ) has a positive lowerbound, K ( x ) is bounded and positive, Q ( x ) is bounded (and allowed to changesign), they proved the existence of a ground state solution of (1.4), for any h > V ( x ), K ( x ) and Q ( x ), Cingolaniand Lazzo [10] used the Ljusternik-Schnirelman category to get the multiplicity ofpositive solutions for problem (1.4). When V ( x ), K ( x ) and Q ( x ) are all boundedand positive functions, Zhao and Zhao [53] considered the critical Schr¨odinger-Poisson system (1.2) with ε = 1 and g ( x, u ) = K ( x ) | u | p − u + Q ( x ) | u | ∗ − u ,and the existence of positive ground solutions was obtained. In [48], the authorsproved the existence and concentration of positive solutions for system (1.2) with g ( x, u ) = K ( x ) f ( u ) + Q ( x ) | u | ∗ − u , where f is some superlinear-4 growth nonlin-earity. In [31], the Kirchhoff type problem with competing potential was consideredand the existence and concentration behavior of positive solution were established.In addition, fractional Schr¨odinger equations involving the competing potentials of K. M. TENG AND R. P. AGARWAL the form ( − ∆) s u + V ( x ) u = K ( x ) | u | p − u + Q ( x ) | u | ∗ s − u x ∈ R N with 2 < p < ∗ s , have been considered in [45], the existence of ground state solutionwas obtained.Motivated by the above-cited works, the aim of this paper is to consider theexistence and concentration of positive solutions for fractional Schr¨odinger-Poissonsystem with competing potentials. As far as we know, there are few results on theexistence and concentration of positive solution for system (1.1), and even in the s = t = 1 case. There are some difficulties compared with classical Schrodinger-Poisson system. One is the L ∞ -estimate, owing to the work of Dipierro, Medinaand Valdinoci [17], similarly, we can get the L ∞ -estimate. The other is the decayestimate of solutions, with the help of works in [22], [2] and [24], we can establishthe decay estimate at infinity. Below we give some assumptions on the nonlinearity f :( f ) f ∈ C ( R ), f ( t ) = o ( t ), f ( t ) t > t > f ( t ) = 0 for t ≤ f ) f ( t ) t is strictly increasing on the interval (0 , ∞ );( f ) f ( t ) ≥ c t q − and | f ′ ( t ) | ≤ c (1 + | t | p − ) for some constants c , c >
0, where4 < q < p < ∗ s .Our main result is stated as follows. Theorem 1.1.
Suppose that ( V ) , ( Q ) , ( Q ) , ( K ) , ( f ) – ( f ) hold and let s ∈ ( , .Then ( i ) there exists ε > such that system (1.1) possesses a positive solution ( u ε , φ ε ) ∈ H s ( R ) × D t, ( R ) , where H s ( R ) and D t, ( R ) are defined in Section 2; ( ii ) u ε possesses a maximum point z ε in R such that lim ε → V ( z ε ) = V , lim ε → K ( z ε ) = K , lim ε → Q ( z ε ) = Q , and for any z ε → x ∈ Θ as ε → , set v ε ( x ) = u ε ( εx + z ε ) , the solution ( v ε , φ ε ) converges strongly in H s ( R ) to a solution ( v, φ ) of (cid:26) ( − ∆) s v + V ( x ) v + φv = K ( x ) f ( v ) + Q ( x ) | v | ∗ s − v, in R , ( − ∆) t φ = v , in R ; (1.5)( iii ) there exist two constants C > and C ∈ R such that u ε ≤ Cε s C ε s + | x − z ε | s for all x ∈ R . We also obtain a supplementary result of a nonexistence of ground state solutionfor system (1.1).
Theorem 1.2.
Assume that ( f ) – ( f ) hold and the continuous functions V ( x ) , K ( x ) , Q ( x ) satisfies ( H ) V ( x ) ≥ V ∞ = lim | x |→∞ V ( x ) = V , K ( x ) ≤ K ∞ and Q ( x ) ≤ Q ∞ , which one ofthe strictly inequality holds on a positive measure subset.Then for any ε > , system (1.1) has no ground state solution. The paper is organized as follows. In section 2, we present some preliminariesresults. In section 3, we will prove the compactness condition. In section 4, the ex-istence of positive ground state solutions of autonomous problem defined in section
RACTIONAL SCHR ¨ODINGER-POISSON SYSTEM WITH CRITICAL GROWTH 5
Variational Setting
In this section, we outline the variational framework for problem (1.1) and givesome preliminary Lemma. In the sequel, we denote by k · k p the usual norm of thespace L p ( R N ), the letters c i ( i = 1 , , . . . ), C i , C will be indiscriminately used todenote various positive constants whose exact values are irrelevant. We denote b u the Fourier transform of u for simplicity.It is easily seen that, just performing the change of variables u ( x ) → u ( x/ε ) and φ ( x ) → φ ( x/ε ), and taking z = x/ε , problem (1.1) can be rewritten as the followingequivalent form (cid:26) ( − ∆) s u + V ( εz ) u + φu = K ( εz ) f ( u ) + Q ( εz ) | u | ∗ s − u, in R ,( − ∆) t φ = u , in R . (2.1)which will be referred from now on.2.1. Work space stuff.
For α ∈ (0 , D α, ( R ) as follows D α, ( R ) = n u ∈ L ∗ α ( R ) (cid:12)(cid:12)(cid:12) | ξ | α b u ( ξ ) ∈ L ( R ) o which is the completion of C ∞ ( R ) under the norm k u k D α, = (cid:16) Z R | ( − ∆) α u | d x (cid:17) = (cid:16) Z R | ξ | α | b u ( ξ ) | d ξ (cid:17) and for any α ∈ (0 , S α > S α = inf u ∈D α, R R | ( − ∆) α u | d x (cid:16) R R | u ( x ) | ∗ α d x (cid:17) ∗ α . (2.2)The fractional Sobolev space H α ( R ) can be described by means of the Fouriertransform, i.e. H α ( R ) = n u ∈ L ( R ) (cid:12)(cid:12)(cid:12) Z R ( | ξ | α | b u ( ξ ) | + | b u ( ξ ) | ) d ξ < + ∞ o . In this case, the inner product and the norm are defined as( u, v ) = Z R ( | ξ | α b u ( ξ ) b v ( ξ ) + b u ( ξ ) b v ( ξ )) d ξ k u k H α = (cid:18) Z R ( | ξ | α | b u ( ξ ) | + | b u ( ξ ) | ) d ξ (cid:19) . From Plancherel’s theorem we have k u k = k b u k and k| ξ | α b u k = k ( − ∆) α u k .Hence k u k H α = (cid:18) Z R ( | ( − ∆) α u ( x ) | + | u ( x ) | ) d x (cid:19) , ∀ u ∈ H α ( R ) . We denote k · k by k · k H α in the sequel for convenience. K. M. TENG AND R. P. AGARWAL
In terms of finite difference, the fractional Sobolev space H α ( R ) also can bedefined as follows H α ( R ) = n u ∈ L ( R ) (cid:12)(cid:12)(cid:12) | u ( x ) − u ( y ) || x − y | + α ∈ L ( R × R ) o endowed with the natural norm k u k = (cid:18) Z R | u | d x + Z R Z R | u ( x ) − u ( y ) | | x − y | α d x d y (cid:19) . Indeed, in view of Proposition 3.4 and Proposition 3.6 in [36], we have k ( − ∆) α u k = Z R | ξ | α | b u ( ξ ) | d ξ = 1 C ( α ) Z R Z R | u ( x ) − u ( y ) | | x − y | α d x d y. It is well known that H α ( R ) is continuously embedded into L p ( R ) for 2 ≤ r ≤ ∗ α (2 ∗ α = − α ), and is locally compactly embedded into L rloc ( R ) for 1 ≤ r < ∗ α .For any ε >
0, let H ε = { u ∈ H s ( R ) | R R V ( εx ) | u | d x < ∞} be the Sobolevspace endowed with the norm k u k ε = (cid:16) Z R ( | ( − ∆) s u | + V ( εx ) | u | ) d x (cid:17) . Clearly, by the assumption ( V ), k · k ε and k · k are equivalent norm on H ε uniformlyfor ε >
0. Moreover, H ε is continuously embedded into L r ( R ) for 2 ≤ r ≤ ∗ s andlocally compact embedded into L rloc ( R ) for 1 ≤ r < ∗ s .2.2. Formulation of Problem (1.1) and preliminaries.
By ( f ) and ( f ), forany ε >
0, there exists C ε > f ( u ) ≤ ε | u | + C ε | u | p − , F ( u ) ≤ ε | u | + C ε | u | p for all u ∈ R . (2.3)By ( f ) and ( f ), one can easily check that( 14 f ( t ) t − F ( t )) ′ > , f ′ ( u ) u − f ( u ) u > , < F ( u ) < f ( u ) u, for any u = 0 . (2.4)For problem (2.1), we first apply the usual ”reduction” argument to reduce it toa single equation involving just u .From [47], the author has proved that if 4 s + 2 t ≥
3, for each u ∈ H s ( R ), theLax-Milgram theorem implies that there exists a unique φ tu ∈ D t, ( R ) such that Z R ( − ∆) t φ tu ( − ∆) t v d x = Z R u v d x, ∀ v ∈ D t, ( R ) , that is φ tu is a weak solution of( − ∆) t φ tu = u , x ∈ R and the representation formula holds φ tu ( x ) = c t Z R u ( y ) | x − y | − t d y, x ∈ R , c t = π − − t Γ( − t )Γ( t ) , which is called t -Riesz potential.Substituting φ tu in (2.1), it reduces to a single equation, i.e., the fractionalSchr¨odinger equation with a non-local term φ tu u :( − ∆) s u + V ( εx ) u + φ tu u = K ( εx ) f ( u ) + Q ( εx ) | u | ∗ s − u, x ∈ R , (2.5) RACTIONAL SCHR ¨ODINGER-POISSON SYSTEM WITH CRITICAL GROWTH 7 whose solutions can be obtained by looking for critical points of the functional I ε : H ε → R defined by I ε ( u ) = 12 Z R (cid:16) | ( − ∆) s u | + V ( εx ) u (cid:17) d x + 14 Z R φ tu u d x − Z R K ( εx ) F ( u ) d x − ∗ s Z R Q ( εx ) | u ( x ) | ∗ s d x which is well defined in H ε and I ε ∈ C ( H ε , R ). Moreover, h I ′ ε ( u ) , v i = Z R (cid:16) ( − ∆) s u ( − ∆) s v + V ( εx ) uv + φ tu uv − K ( εx ) f ( u ) v − Q ( εx ) | u | ∗ s − uv (cid:17) d x. Definition 2.1. (1) We call ( u ε , φ ε ) ∈ H ε × D t, ( R ) is a weak solution of system (1.1) if u ε is a weak solution of problem (2.5) .(2) We call u ∈ H ε is a weak solution of (2.5) if Z R (cid:16) ( − ∆) s u ( − ∆) s v + V ( εx ) uv + φ tu uv − K ( εx ) f ( u ) v − Q ( εx ) | u | ∗ s − uv (cid:17) d x = 0 , for any v ∈ H ε . Obviously, the weak solutions of (2.5) are the critical points of I ε . Now let ussummarize some properties of φ tu . Lemma 2.2.
For every u ∈ H ε with s + 2 t ≥ , define Φ( u ) = φ tu ∈ D t, ( R ) ,where φ tu is the unique solution of equation ( − ∆) t φ = u . Then there hold: ( i ) If u n ⇀ u in H ε , then Φ( u n ) ⇀ Φ( u ) in D t, ( R ) ; ( ii ) Φ( tu ) = t Φ( u ) for any t ∈ R ; ( iii ) For u ∈ H ε , one has k Φ( u ) k D t, ≤ C k u k t ≤ C k u k ε , Z R Φ( u ) u d x ≤ C k u k t ≤ C k u k ε , where constant C is independent of u ; ( iv ) Let s + 2 t > , if u n ⇀ u in H ε and u n → u in L r ( R ) for ≤ r < ∗ s , then Z R φ tu n u n v d x → Z R φ tu uv d x for any v ∈ H ε and Z R φ tu n u n d x → Z R φ tu u d x ;( v ) For u, v ∈ H ε , there holds k Φ( u ) − Φ( v ) k D t, ≤ C ( k u k ε + k v k ε )( k u − v k ε ) . Proof.
We only need to check that ( vi ) and ( v ) hold, the proof of others can befound in [47].( iv ) By H¨older’s inequality and 4 s + 2 t > t < − s , we havethat (cid:12)(cid:12)(cid:12) Z R φ tu n u n v d x − Z R φ tu uv d x (cid:12)(cid:12)(cid:12) = (cid:12)(cid:12)(cid:12) Z R Z R u ( y ) v ( y ) | x − y | − t ( u n ( x ) − u ( x )) d x d y + Z R φ tu n ( u n − u ) v d x (cid:12)(cid:12)(cid:12) ≤ Z R ( φ tu ) ( φ tv ) | u n ( x ) − u ( x ) | d x + k φ tu n k ∗ t k u n − u k t k v k t ≤ k φ tu k ∗ t k φ tv k ∗ t ( k u n k t + k u k t ) k u n − u k t + k φ tu n k ∗ t k u n − u k t k v k t → K. M. TENG AND R. P. AGARWAL for any v ∈ H ε . For the second part, using the similar argument, we have (cid:12)(cid:12)(cid:12) Z R φ tu n u n d x − Z R φ tu u d x (cid:12)(cid:12)(cid:12) = (cid:12)(cid:12)(cid:12) Z R Z R ( u n ( x ) u n ( y ) − u ( x ) u ( y ) ) | x − y | − t d x d y (cid:12)(cid:12)(cid:12) = (cid:12)(cid:12)(cid:12) Z R φ tu n ( u n − u ) d x + Z R φ tu ( u n − u ) d x (cid:12)(cid:12)(cid:12) ≤ k φ tu n k ∗ t k u n − u k t + k φ tu k ∗ t k u n − u k t ≤ C (cid:16) k u n k ε + k u k ε (cid:17)(cid:16) k u n k t + k u k t (cid:17) k u n − u k t → . ( v ) By the definition of φ tu and φ tv , using H¨older’s inequality and 4 s + 2 t ≥ H ε ֒ → L t ( R ), we have that k Φ( u ) − Φ( v ) k D t, = Z R | ( − ∆) s ( φ tu − φ tv ) | d x = Z R ( u − v )( φ tu − φ tv ) d x ≤ C k φ tu − φ tv k D t, (cid:16) Z R | u − v | t d x (cid:17) t ≤ C k Φ( u ) − Φ( v ) k D t, (cid:16) Z R ( | u || u − v | + | v || u − v | ) t d x (cid:17) t ≤ C k Φ( u ) − Φ( v ) k D t, (cid:16) k u k t + k v k t (cid:17) k u − v k t ≤ C k Φ( u ) − Φ( v ) k D t, (cid:16) k u k ε + k v k ε (cid:17) k u − v k ε . (cid:3) The minimax level of the autonomous equation associated with equation (2.5)( − ∆) s u + νu + φ tu u = κf ( u ) + µ | u | ∗ s − u, in R , (2.6)plays an important role in the proof of compactness of ( P S ) sequence and concen-tration behavior of solutions, where µ, ν, κ > φ tu = R R u ( y ) | x − y | − t d x . For ν >
0, let E ν = { u ∈ H s ( R ) (cid:12)(cid:12)(cid:12) Z R ν | u | d x < ∞} be a Sobolev space endowed with the norm k u k E ν = (cid:16) Z R (cid:16) | ( − ∆) s u | + ν | u | (cid:17) d x (cid:17) . In fact, the Sobolev space H ε = E ν = H s ( R ) for any ε > ν > I ν,κ,µ : E ν → R is given by I ν,κ,µ ( u ) = 12 Z R ( | ( − ∆) s u | + νu ) d x + 14 Z R φ tu u d x − κ Z R F ( u ) d x − µ ∗ s Z R | u ( x ) | ∗ s d x. It is easy to see that I ν,κ,µ ∈ C ( E ν , R ) and h I ′ ν,κ,µ ( u ) , ϕ i = Z R (cid:16) ( − ∆) s u ( − ∆) s ϕ + νuϕ (cid:17) d x + Z R φ tu uϕ d x − κ Z R f ( u ) ϕ d x RACTIONAL SCHR ¨ODINGER-POISSON SYSTEM WITH CRITICAL GROWTH 9 − µ Z R | u ( x ) | ∗ s − uϕ d x, for any u, ϕ ∈ E ν . Moreover, the critical points of I ν,κ,µ in E ν are weak solutionsof equation (2.6).In section 3, we will apply the concentration-compactness principle of P. L. Lions[29, 30] and vanishing Lemma [40] to prove the compactness of ( P S ) c sequence of I ε on some low energy level. We first recall these results as follows. Proposition 2.3.
Let ρ n ( x ) ∈ L ( R N ) be a non-negative sequence satisfying Z R N ρ n ( x ) d x = l > . Then there exists a subsequence, still denoted by { ρ n ( x ) } such that one of the fol-lowing cases occurs.(i) (compactness) there exists y n ∈ R N , such that for any ε > , exists R > suchthat Z B R ( y n ) ρ n ( x ) d x ≥ l − ε, n = 1 , , · · · . (ii) (vanishing) for any fixed R > , there holds lim n →∞ sup y ∈ R N Z B R ( y ) ρ n ( x ) d x = 0 . (iii) (dichotomy) there exists α ∈ (0 , l ) such that for any ε > , there exists n ≥ , ρ (1) n , ρ (2) n ∈ L ( R N ) , for n ≥ n , there holds k ρ n − ( ρ (1) n + ρ (2) n ) k L ( R N < ε, (cid:12)(cid:12)(cid:12) Z R N ρ (1) n ( x ) d x − α (cid:12)(cid:12)(cid:12) < ε, (cid:12)(cid:12)(cid:12) Z R N ρ (2) n ( x ) d x − ( l − α ) (cid:12)(cid:12)(cid:12) < ε and dist(supp ρ (1) n , supp ρ (2) n ) → ∞ , as n → ∞ . Lemma 2.4.
Assume that { u n } is bounded in H s ( R N ) and it satisfies lim n → + ∞ sup y ∈ R N Z B R ( y ) | u n ( x ) | d x = 0 where R > . Then u n → in L r ( R N ) for every < r < ∗ s . Compactness
Define the Nehari manifold associated to the functional I ε as N ε = n u ∈ H ε \{ } (cid:12)(cid:12)(cid:12) G ε ( u ) = 0 o , where G ε ( u ) = h I ′ ε ( u ) , u i = Z R (cid:16) | ( − ∆) s u | + V ( εx ) u (cid:17) d x + Z R φ tu u d x − Z R K ( εx ) f ( u ) u d x − Z R Q ( εx ) | u ( x ) | ∗ s d x. Thus, for any u ∈ N ε , we have that I ε ( u ) = 14 Z R ( | ( − ∆) s u | + V ( εx ) u ) d x + Z R K ( εx )( 14 f ( u ) u − F ( u )) d x + 4 s − Z R Q ( εx ) | u ( x ) | ∗ s d x. Remark 3.1.
Observing that s > implies that s + 2 t > holds trivially. Also, we define the Nehari manifold associated with functional I µ,κ,ν as follows N ν,κ,µ = { u ∈ E ν \{ } (cid:12)(cid:12)(cid:12) G ν,κ,µ ( u ) = 0 } , where G ν,κ,µ ( u ) = k u k E ν + Z R φ tu u d x − κ Z R f ( u ) u d x − µ Z R | u ( x ) | ∗ s d x. Particularly, when µ = V ∞ , κ = K ∞ , µ = Q ∞ , and µ = V , κ = K , µ = Q , thefunctional I ∞ and I are defined as I ∞ ( u ) = 12 Z R ( | ( − ∆) s u | + V ∞ u ) d x + 14 Z R φ tu u d x − Z R K ∞ F ( u ) d x − ∗ s Z R Q ∞ | u ( x ) | ∗ s d x and I ( u ) = 12 Z R ( | ( − ∆) s u | + V u ) d x + 14 Z R φ tu u d x − Z R K F ( u ) d x − ∗ s Z R Q | u ( x ) | ∗ s d x. Also, we denote the Nehari manifolds by N ∞ = { u ∈ H s ( R ) \{ } (cid:12)(cid:12)(cid:12) h I ′∞ ( u ) , u i = 0 } N = { u ∈ H s ( R ) \{ } (cid:12)(cid:12)(cid:12) h I ′ ( u ) , u i = 0 } . In order to find the least energy solutions of problem (2.5) and (2.6), we define theleast energy levels as follows m ε = inf u ∈N ε I ε ( u ) , m ν,κ,µ = inf u ∈N ν,κ,µ I ν,κ,µ ( u ) , m = inf u ∈N I ( u ) , m ∞ = inf u ∈N ∞ I ∞ ( u ) . The following lemma describes some properties of the Nehari manifold N ε , N ν,κ,µ and I ε . Lemma 3.2.
Under the assumptions ( V ) , ( Q ) , ( K ) and ( f ) − ( f ) , the the fol-lowing statements hold: ( i ) N ε is a manifold of C -class diffeomorphic to the unite sphere of H ε ; ( ii ) For every u ∈ H ε \{ } , there exists a unique t u > such that t u u ∈ N ε and I ε ( t u u ) = max t ≥ I ε ( tu );( iii ) For every u ∈ N ε , there exists C > such that k u k ε ≥ C > ; ( iv ) If { u n } is a ( P S ) c sequence in H ε , i.e., I ε ( u n ) → c and I ′ ε ( u n ) → in ( H ε ) ′ as n → ∞ , then there exists u ∈ H ε such that u n ⇀ u in H ε and I ′ ε ( u ) = 0 if s + 2 t > ; ( v ) Let { u n } ⊂ H ε be a sequence satisfying h I ′ ε ( u n ) , u n i → and Z R (cid:16) K ( εx ) f ( u n ) u n + Q ( εx ) | u n | ∗ s (cid:17) d x → a RACTIONAL SCHR ¨ODINGER-POISSON SYSTEM WITH CRITICAL GROWTH 11 as n → ∞ , where a is a positive constant. Then, up to a subsequence, there exists t n > such that h I ′ ε ( t n u n ) , t n u n i = 0 and t n → as n → ∞ ;( vi ) Let { u n } be a sequence such that u n ∈ N ε and I ε ( u n ) → m ε , then we mayassume that { u n } is a ( P S ) m ε sequence in H ε .Proof. The proof of ( ii ) and ( iii ) is standard, we only to verify the remain conclu-sions.( i ) Let u ∈ N ε , by computation, using (2.4), we get h G ′ ε ( u ) , u i = 2 Z R (cid:16) | ( − ∆) s u | + V ( εx ) u (cid:17) d x + 4 Z R φ tu u d x − Z R K ( εx ) (cid:16) f ′ ( u ) u + f ( u ) u (cid:17) d x − ∗ s Z R Q ( εx ) | u ( x ) | ∗ s d x = − Z R (cid:16) | ( − ∆) s u | + V ( εx ) u (cid:17) d x − Z R K ( εx ) (cid:16) f ′ ( u ) u − f ( u ) u (cid:17) d x − (2 ∗ s − Z R Q ( εx ) | u ( x ) | ∗ s d x ≤ − k u k ε < . ( iv ) Let { u n } ⊂ H ε be a ( P S ) c sequence, it is easy to show that { u n } is boundedin H ε . By the reflexivity of H ε and using the Sobolev embedding property, upto a subsequence, still denoted by { u n } , we may assume that there exists u ∈ H ε such that u n ⇀ u in H ε , u n → u in L rloc ( R ) with 1 ≤ r < ∗ s and u n → u a.e.in R . By (2.3), we see that f ( u ) ∈ L ∗ s − p − ( R ) because of 2 < ∗ s − p − < ∗ s − Q ( εx ) ∗ s − ∗ s | u n | ∗ s − u n and K ( εx ) f ( u n ) are bounded in L ∗ s ∗ s − ( R ) and L θ ∗ s − p − ( R ) with θ ∈ ( p − ∗ s − ∗ s ∗ s − , ∗ s ∗ s − ), respectively. Here using s > , we canchoose θ ∈ ( p − ∗ s − ∗ s ∗ s − , ∗ s ∗ s − ) such that 2 < θ ∗ s − p − < ∗ s , 2 < θ (2 ∗ s − < ∗ s and2 < θ (2 ∗ s − θ (2 ∗ s − − ( p − < ∗ s . Using the fact u n → u a..e in R , we obtain that Q ( εx ) ∗ s − ∗ s | u n | ∗ s − u n ⇀ Q ( εx ) ∗ s − ∗ s | u | ∗ s − u in L ∗ s ∗ s − ( R ) (3.1)and K ( εx ) f ( u n ) ⇀ K ( εx ) f ( u ) in L θ ∗ s − p − ( R ) . (3.2)Next, we show that φ tu n → φ tu a.e. in R . In fact, using 2 s + 2 t > − t < p < s , q > s > − t so that 2 p ′ , q ′ ∈ (2 , ∗ s ), using H¨older’s inequality,we deduce that | φ tu n ( x ) − φ tu ( x ) | ≤ c t Z R | u n ( y ) − u ( y ) || x − y | − t d y ≤ c t (cid:16) Z | x − y |≤ R | x − y | p (3 − t ) d y (cid:17) p (cid:16) Z | x − y |≤ R | u n − u | p ′ d y (cid:17) p ′ + (cid:16) Z | x − y | >R | x − y | q (3 − t ) d y (cid:17) q (cid:16) Z | x − y | >R | u n − u | q ′ d y (cid:17) q ′ concluding the pointwise convergence. Owing to φ tu n u n is bounded in L γ ( R ),where γ satisfies 2 ≤ γ ∗ t ∗ t − γ ≤ ∗ s and 2 ≤ γ ′ ≤ ∗ s (using 4 s + 2 t ≥ φ tu n u n → φ tu u a.e. in R , we have that φ tu n u n ⇀ φ tu u in L γ ( R ) . (3.3)Therefore, combining with (3.1), (3.2), (3.3) and using the weakness convergencein H ε , we infer that h I ′ ε ( u n ) , ϕ i − h I ′ ε ( u ) , ϕ i → , for any ϕ ∈ H ε . which leads to I ′ ε ( u ) = 0.( v ) By the assumptions, it is easy to see that k u n k ε = 0 for large n . Using theconclusion ( ii ), there exists t n > t n u n ∈ N ε i.e., h I ε ( t n u n ) , t n u n i = 0.Now we prove that t n → n → ∞ . Set a n = k u n k ε , b n = Z R φ tu n u n d x, c n = Z R K ( εx ) f ( u n ) u n d x, d n = Z R Q ( εx ) | u n | ∗ s d x. By assumption, we have that c n + d n → a > n → ∞ . Passing to a subsequence,we may assume that a n → a , b n → b , c n → c , d n → d . Thus, a + b = c + d = a and a > a > h I ε ( t n u n ) , t n u n i = 0, by (2.3) and ( f ), we have that t n a n + t n b n ≥ t qn K ∞ Z R | u n | q d x + t ∗ s n d n and t n a n + t n b n ≤ C (cid:16) εt n Z R | u n | d x + t pn Z R | u n | p d x (cid:17) + t ∗ s n d n which imply that there exist T , T > < T ≤ t n ≤ T . Hence, up to asubsequence, still denoted by { t n } , we may assume that t n → T as n → ∞ . Since h I ′ ε ( t n u n ) , t n u n i = 0 and h I ′ ε ( u n ) , u n i →
0, we get T a n + T b n = Z R K ( εx ) f ( T u n ) T u n d x + T ∗ s d n + o n (1) , a n + b n = c n + d n + o n (1)which leads to Z R K ( εx ) (cid:16) f ( T u n )( T u n ) − f ( u n )( u n ) (cid:17) u n d x = ( 1 T − a n + (1 − T ∗ s − ) d n + o n (1)= ( 1 T − a + (1 − T ∗ s − ) d + o n (1) . Observing that ( T − a + (1 − T ∗ s − ) d < T >
1, but it follows fromFatou’s Lemma and u n → u a.e. in R that Z R K ( εx ) (cid:16) f ( T u )( T u ) − f ( u ) u (cid:17) u d x ≥ T <
1. Hence, it is only true that T = 1.( vi ) Suppose { u n } be a minimizing sequence of I ε constrained in N ε . By the Eke-land’s variational principle in [51] (Theorem 8.5, Page 122), there exists a sequence { v n } ⊂ N ε such that I ε ( v n ) → m ε , k v n − u n k ε → I ′ ε ( v n ) − λ n G ′ ε ( v n ) → n → ∞ . Thus, in view of v n ∈ N ε , we have that h I ′ ε ( v n ) , v n i = λ n h G ε ( v n ) , v n i + RACTIONAL SCHR ¨ODINGER-POISSON SYSTEM WITH CRITICAL GROWTH 13 o n (1) = o n (1). We assume that lim n →∞ h G ε ( v n ) , v n i = γ ≤
0, if γ = 0, owing to theproof of ( i ), we see that k v n k ε → n → ∞ . This yields a contradiction with m ε >
0. Therefore, λ n → n → ∞ and so I ′ ε ( v n ) → n → ∞ . Hence,without loss of generalization, we may assume that I ε ( u n ) → m ε and I ′ ε ( u n ) → n → ∞ , i.e., { u n } is a ( P S ) m ε sequence for I ε . (cid:3) The functional I ε satisfies the mountain pass geometry. Lemma 3.3.
Suppose ( V ) , ( K ) , ( Q ) and ( f ) − ( f ) hold, then the functional I ε has the following properties: ( i ) there exists α, ρ > such that I ε ( u ) ≥ α for k u k ε = ρ ; ( ii ) there exists e ∈ H ε satisfying k e k ε > ρ such that I ε ( e ) < . The proof of Lemma 3.3 is standard and hence is omitted. By Lemma 3.3 andTheorem 1.15 in [51] (Mountain pass theorem without Palais-Smale condition), itfollows that there exists a (
P S ) c ε sequence { u n } ⊂ H ε such that I ε ( u n ) → c ε and I ′ ε ( u n ) → n → ∞ , where c ε = inf γ ∈ Γ ε max t ∈ [0 , I ε ( γ ( t )), where Γ ε = { γ ∈ C ([0 , , H ε ) | γ (0) =0 , I ε ( γ (1)) < } . Similarly to the arguments in [38] or [23], by (2.4), the equiv-alent characterization of c ε is given by c ε = m ε = inf u ∈ H ε \{ } max t ≥ I ε ( tu ) . The following Lemma gives the estimate of the critical value c ε . Lemma 3.4.
Suppose that ( V ) , ( K ) , ( Q ) , ( Q ) and ( f ) − ( f ) hold, then theinfinimum c ε satisfies < c ε < s Q ( x ) − s s S s s , for ε small enough, where S s is the best Sobolev constant defined by (2.2) .Proof. We define u ε ( x ) = ψ ( x − x /ε ) U ε ( x ) , x ∈ R , where U ε ( x ) = ε − s u ∗ ( x ), u ∗ ( x ) = κ ( ε + | x − x /ε | ) − s (see Appendix), and ψ ∈ C ∞ ( R ) such that 0 ≤ ψ ≤ R , ψ ( x ) ≡ B R (0) and ψ ≡ R \ B R (0).From Lemma A.2 and Lemma A.3 in Appendix, we know that Z R | ( − ∆) s u ε ( x ) | d x ≤ k ( − ∆) s ¯ u k + O ( ε − s ) , (3.4) Z R | u ε ( x ) | ∗ s d x = k ¯ u k ∗ s ∗ s + O ( ε ) , (3.5)where ¯ u = κ (1+ | x | ) − s is such that S s = | ( − ∆) s ¯ u k k ¯ u k ∗ s , and Z R | u ε ( x ) | p d x = O ( ε (2 − p )3+2 sp ) , p > − s , O ( ε (2 − p )3+2 sp | log ε | ) , p = − s , O ( ε − s p ) , < p < − s . (3.6) By ( ii ) of Lemma 3.2, there exists t ε > t ≥ I ε ( tu ε ) = I ε ( t ε u ε ). Hence d I ε ( tu ε )d t (cid:12)(cid:12)(cid:12) t = t ε = 0, that is t ε Z R ( | ( − ∆) s u ε | + V ( εx ) u ε ) d x + t ε Z R φ tu ε u ε d x = Z R K ( εx ) f ( t ε u ε ) t ε u ε d x + t ∗ s ε Z R Q ( εx ) | u ε | ∗ s d x. By ( f ), we have that t ε Z R ( | ( − ∆) s u ε | + V ( εx ) u ε ) d x + t ε Z R φ tu ε u ε d x ≥ t ∗ s ε Z R Q ( εx ) | u ε | ∗ s d x. It follows from ( iii ) of Lemma 2.2 that t ∗ s ε ≤ k Q ( εx ) ∗ s u ε k ∗ s ∗ s (cid:16) t ε Z R ( | ( − ∆) s u ε | + V ( εx ) u ε ) d x + Ct ε k u ε k t (cid:17) . (3.7)(3.4), (3.5), (3.6) and (3.7) imply that | t ε | ≤ C , where C is independent of ε > C > t ε ≥ C > ε > ε n → n → ∞ such that t ε n → n → ∞ . Therefore0 < c ε ≤ sup t ≥ I ε ( tu ε n ) = I ε ( t ε n u ε n ) → , which is a contradiction.Denote g ( t ) = t R R | ( − ∆) s u ε | d x − t ∗ s ∗ s R R Q ( x ) | u ε | ∗ s d x , by (3.4) and (3.5),it is easy to check thatsup t ≥ g ( t ) = s (cid:16) R R | ( − ∆) s u ε | d x (cid:17) s (cid:16) Q ( x ) R R | u ε | ∗ s d x (cid:17) − s s = s Q ( x ) − s s (cid:16) k ( − ∆) s ¯ u k + O ( ε − s ) (cid:17) s (cid:16) k ¯ u k ∗ s ∗ s + O ( ε ) (cid:17) − s s ≤ s Q ( x ) − s s (cid:16) k ( − ∆) s ¯ u k k ¯ u k ∗ s (cid:17) s + O ( ε − s )= s Q ( x ) − s s S s s + O ( ε − s ) . Thus I ε ( t ε u ε ) ≤ sup t ≥ g ( t ) + C Z R V ( εx ) | u ε | d x + C Z R φ tu ε u ε d x − C Z R | u ε | q +1 d x + Z R (cid:16) Q ( x ) − Q ( εx ) (cid:17) | u ε | ∗ s d x ≤ s Q ( x ) − s s S s s + O ( ε − s ) + C Z R | u ε | d x + C (cid:16) Z R | u ε | t d x (cid:17) t − C Z R | u ε | q +1 d x + Z R (cid:16) Q ( x ) − Q ( εx ) (cid:17) | u ε | ∗ s d x ≤ s Q ( x ) − s s S s s + O ( ε − s ) + C (cid:16) Z R | u ε | t d x (cid:17) t − C Z R | u ε | q +1 d x RACTIONAL SCHR ¨ODINGER-POISSON SYSTEM WITH CRITICAL GROWTH 15 + Z R (cid:16) Q ( x ) − Q ( εx ) (cid:17) | u ε | ∗ s d x. (3.8)where we have used (3.6) and s > which implies 2 < − s .By (3.6), we have thatlim ε → + (cid:16) R R | u ε | t d x (cid:17) t ε − s ≤ lim ε → + O ( ε t +4 s − ) ε − s = 0 , t > − s ,lim ε → + O ( ε t +4 s − | log ε | ) ε − s = 0 , t = − s ,lim ε → + O ( ε − s ) | log ε | ) ε − s = 0 , t < − s .(3.9)Since s > and q > s − s , then q + 1 > − s , 2 s − − s ( q + 1) <
0. Thuslim ε → + R R | u ε | q +1 d xε − s = lim ε → + O ( ε − − s ( q +1) ) ε − s = + ∞ . (3.10)By the hypothesis ( Q ), we deduce that Z R (cid:16) Q ( x ) − Q ( εx ) (cid:17) | u ε | ∗ s d x = Cε (cid:16) Z | εx − x |≤ ρ | εx − x | α ψ ( x − x /ε ) ∗ s ( ε + | x − x /ε | ) d x + Z | εx − x | >ρ ψ ( x − x /ε ) ∗ s ( ε + | x − x /ε | ) N − s d x (cid:17) ≤ Cε (cid:16) ε α − Z ρ ε r α (1 + r ) d r + 1 ε Z + ∞ ρ ε r (1 + r ) d r (cid:17) ≤ Cε (cid:16) ε α − Z r α (1 + r ) d r + ε α − Z ρ ε r α (1 + r ) d r + 1 (cid:17) ≤ C ( ε α + ε + ε ) . Since − s ≤ α < , then Z R (cid:16) Q ( x ) − Q ( εx ) (cid:17) | u ε | ∗ s d x ≤ O ( ε − s ) . (3.11)Therefore, combining with (3.8), (3.9), (3.10) and (3.11), we conclude that I ε ( t ε u ε ) < s Q ( x ) − s s S s s for ε small enough and thus the proof is completed. (cid:3) Lemma 3.5.
Assume that ( V ) , ( Q ) , ( Q ) , ( K ) and ( f ) − ( f ) hold. If c ε < min { m ∞ , s Q ( x ) − s s S s s } , then I ε satisfies the ( P S ) condition for c ε .Proof. Let { u n } be a ( P S ) sequence of I ε at the level c ε , i.e., I ε ( u n ) → c ε , I ′ ε ( u n ) → H ε ) ′ . (3.12)It is easy to check that { u n } is bounded in H ε . Thus, up to a subsequence, stilldenoted by { u n } , we may assume that there exists u ∈ H ε such that u n ⇀ u in H ε , u n → u in L rloc ( R ) for 2 ≤ r < ∗ s .Next, we aim to show that u n → u in H ε . For this purpose, set ρ n ( x ) = 14 | ( − ∆) s u n | + 14 V ( εx ) | u n | + 14 K ( εx )( f ( u n ) u n − F ( u n ))+ 4 s − Q ( εx ) | u n | ∗ s . Clearly, { ρ n } is bounded in L ( R ). Hence, up to a subsequence, still denoted by { ρ n } , we may assume that Ψ( u n ) := k ρ n k → l as n → ∞ . Obviously, l > c ε >
0. In fact, l = c ε .Next, we apply Proposition 2.3 to { ρ n } .If { ρ n } vanishing, then { u n } also vanishing, i.e., there exists R > n →∞ sup y ∈ R Z B R ( y ) | u n | d x = 0 . By Lemma 2.4, one has u n → L r ( R ), 2 < r < ∗ s . Thus, by (2.3) and ( iii ) ofLemma 2.2, we can deduce that Z R K ( εx ) F ( u n ) d x → Z R K ( εx ) f ( u n ) u n d x → , Z R φ tu n u n d x → . From (3.12), we get I ε ( u n ) = 12 k u n k ε − ∗ s Z R Q ( εx ) | u n | ∗ s d x + o n (1) (3.13)and o n (1) = h I ′ ε ( u n ) , u n i = k u n k ε − Z R Q ( εx ) | u n | ∗ s d x + o n (1) . (3.14)We may assume that there exist L ≥ k u n k ε → L , Z R Q ( εx ) | u n | ∗ s d x → L . Obviously, L >
0, otherwise, a contradiction with c ε >
0. By (2.2), we have that Z R Q ( εx ) | u n | ∗ s d x ≤ Q x Z R | u n | ∗ s d x ≤ Q ( x ) (cid:16) S − s Z R | ( − ∆) s u n | d x (cid:17) ∗ s ≤ Q ( x ) S − − s s k u n k ∗ s ε which implies that L ≥ Q ( x ) − s s S s s . Combing (3.13), we can deduce that c ε = s L ≥ s Q ( x ) − s s S s s , this contradictswith the assumption. Hence, vanishing does not occur.Next, we show the dichotomy does not occur. Suppose by contradiction thatthere exist α ∈ (0 , l ) and { y n } ⊂ R such that for every ε n →
0, we can choose { R n } ⊂ R + ( R n > R /ε + R ′ , for any fixed ε > R , R ′ are positive constantsdefined later) with R n → ∞ satisfyinglim sup n →∞ (cid:12)(cid:12)(cid:12) α − Z B Rn ( y n ) ρ n ( x ) d x (cid:12)(cid:12)(cid:12) + (cid:12)(cid:12)(cid:12) ( l − α ) − Z R \ B Rn ( y n ) ρ n ( x ) d x (cid:12)(cid:12)(cid:12) < ε n . (3.15)Let ξ : R + ∪ { } → R + be a cut-off function such that 0 ≤ ξ ≤ ξ ( t ) = 1 for t ≤ ξ ( t ) = 0 for t ≥ | ξ ′ ( t ) | ≤
2. Set v n ( x ) = ξ (cid:16) | x − y n | R n (cid:17) u n ( x ) , w n ( x ) = (cid:16) − ξ (cid:16) | x − y n | R n (cid:17)(cid:17) u n ( x ) , Then by (3.15), we see thatlim inf n →∞ Ψ( v n ) ≥ α, lim inf n →∞ Ψ( w n ) ≥ l − α. RACTIONAL SCHR ¨ODINGER-POISSON SYSTEM WITH CRITICAL GROWTH 17
Denote Ω n = B R n ( y n ) \ B R n ( y n ), by (3.15), then R Ω n ρ n ( x ) d x → n → ∞ ,which leads to Z Ω n (cid:16) | ( − ∆) s u n | + V ( εx ) | u n | (cid:17) d x → , Z Ω n | u n | d x → , Z Ω n | u n | ∗ s d x → R Ω n φ tu n u n d x →
0. There-fore, by similar arguments as Lemma 3.4 in [47], we have that Z R | ( − ∆) s u n | d x = Z R | ( − ∆) s v n | d x + Z R | ( − ∆) s w n | d x + o n (1) , (3.16) Z R V ( εx ) u n ( x ) d x = Z R V ( εx ) v n ( x ) d x + Z R V ( εx ) w n ( x ) d x + o n (1) , (3.17) Z R K ( εx ) F ( u n ) d x = Z R K ( εx ) F ( v n ) d x + Z R K ( εx ) F ( w n ) d x + o n (1) , (3.18) Z R K ( εx ) f ( u n ) u n d x = Z R K ( εx ) f ( v n ) v n d x + Z R K ( εx ) f ( w n ) w n d x + o n (1) , (3.19) Z R Q ( εx ) | u n | ∗ s d x = Z R Q ( εx ) | v n | ∗ s d x + Z R Q ( εx ) | w n | ∗ s d x + o n (1) , (3.20) Z R φ tu n u n d x ≥ Z R φ tv n v n d x + Z R φ tw n w n d x + o n (1) . (3.21)We need to check that (3.18) and (3.19) hold. Indeed, by ( K ), (2.3) and H¨older’sinequality, we infer that (cid:12)(cid:12)(cid:12) Z R K ( εx ) (cid:16) F ( u n ) − F ( v n ) − F ( w n ) (cid:17) d x (cid:12)(cid:12)(cid:12) = (cid:12)(cid:12)(cid:12) Z Ω n K ( εx ) (cid:16) F ( u n ) − F ( ξu n ) − F ((1 − ξ ) u n ) (cid:17) d x (cid:12)(cid:12)(cid:12) ≤ C (cid:16) ε Z Ω n | u n | d x + Z Ω n | u n | p d x (cid:17) ≤ C h ε (cid:16) Z Ω n | u n | d x (cid:17) θ (cid:16) Z Ω n | u n | ∗ s d x (cid:17) − θ )2 ∗ s + (cid:16) Z Ω n | u n | d x (cid:17) pθ (cid:16) Z Ω n | u n | ∗ s d x (cid:17) p (1 − θ ∗ s i → n → ∞ , where θ ∈ (0 ,
1) and θ ∈ (0 ,
1) satisfy = θ + − θ ∗ s , p = θ + − θ ∗ s .Similarly, we can verify that (3.19) hold.Hence, by (3.16)–(3.21), we getΨ( u n ) ≥ Ψ( v n ) + Ψ( w n ) + o n (1) . Then l = lim n →∞ Ψ( u n ) ≥ lim inf n →∞ Ψ( v n ) + lim inf n →∞ Ψ( w n ) ≥ α + l − α = l, hence lim n →∞ Ψ( v n ) = α, lim n →∞ Ψ( w n ) = l − α. (3.22)By (3.12), (3.16)-(3.21), we have o n (1) = h I ′ ε ( u n ) , u n i ≥ h I ′ ε ( v n ) , v n i + h I ′ ε ( w n ) , w n i + o n (1) . (3.23) We distinguish the following two cases:Case 1. Up to a subsequence, we may assume that h I ′ ε ( v n ) , v n i ≤ h I ′ ε ( w n ) , w n i ≤
0. Without loss of generality, we suppose that h I ′ ε ( v n ) , v n i ≤
0, then k v n k ε + Z R φ tv n v n d x − Z R K ( εx ) f ( v n ) v n d x − Z R Q ( εx ) | v n | ∗ s d x ≤ . (3.24)By ( ii ) of Lemma 3.2, for any n , there exists t n > t n v n ∈ N ε and then h I ′ ε ( t n v n ) , t n v n i = 0, i.e., t n k v n k ε + t n Z R φ tv n v n d x − Z R K ( εx ) f ( t n v n ) t n v n d x − t ∗ s n Z R Q ( εx ) | v n | ∗ s d x = 0 . (3.25)Combining (3.24) and (3.25), we have( 1 t n − k v n k ε − Z R K ( εx ) (cid:16) f ( t n v n )( t n v n ) − f ( v n ) v n (cid:17) v n d x − ( t ∗− s n − Z R Q ( εx ) | v n | ∗ s d x ≥ t n ≤ f ). Then, by t n v n ∈ N ε and (2.4) (impliesthat f ( su ) su − F ( su ) is nondecreasing in s ∈ (0 , + ∞ )), we have that c ε ≤ I ε ( t n v n ) = I ε ( t n v n ) − h I ′ ε ( t n v n ) , t n v n i = 14 t n k v n k ε + 14 Z R K ( εx ) (cid:16) f ( t n v n ) t n v n − F ( t n v n ) (cid:17) d x + 4 s − t ∗ s n Z R Q ( εx ) | v n | ∗ s d x ≤ Ψ( v n ) → α < l = c ε which is a contradiction.Case 2. Up to a subsequence, we may assume that h I ′ ε ( v n ) , v n i > h I ′ ε ( w n ) , w n i >
0. By (3.23), we see that h I ′ ε ( v n ) , v n i = o n (1) and h I ′ ε ( w n ) , w n i = o n (1). In view of(3.16)–(3.21), we have that I ε ( u n ) ≥ I ε ( v n ) + I ε ( w n ) + o n (1) . (3.26)If the sequence { y n } ⊂ R is bounded, we will deduce a contradiction by comparing I ε ( w n ) and m ∞ . In fact, by the assumptions ( V ), ( K ) and ( Q ), for any δ >
0, thereexists R >
0, such that | K ( εx ) − K ∞ | ≤ δ, | Q ( εx ) − Q ∞ | ≤ δ, V ( εx ) − V ∞ > − δ, ∀| x | ≥ R /ε. (3.27)By the boundedness of { y n } ⊂ R , there exists R ′ > | y n | ≤ R ′ . Thus, R \ B R n ( y n ) ⊂ R \ B R n − R ′ (0) ⊂ R \ B R /ε (0) for n large enough. From (3.27), wecan deduce that Z R (cid:16) V ( εx ) − V ∞ (cid:17) | w n | d x = Z | x − y n | >R n (cid:16) V ( εx ) − V ∞ (cid:17) | w n | d x ≥ − δ Z | x − y n | >R n | w n | d x ≥ − δC, and by the arbitrariness of δ , this leads to Z R (cid:16) V ( εx ) − V ∞ (cid:17) | w n | d x ≥ o n (1) . (3.28) RACTIONAL SCHR ¨ODINGER-POISSON SYSTEM WITH CRITICAL GROWTH 19
Moreover, it is easy to check that Z R (cid:16) K ( εx ) − K ∞ (cid:17) F ( w n ) d x = o (1) , Z R (cid:16) K ( εx ) − K ∞ (cid:17) f ( w n ) w n d x = o n (1)(3.29)and Z R (cid:16) Q ( εx ) − Q ∞ (cid:17) | w n | ∗ s d x = o n (1) . (3.30)It follows from (3.28)–(3.30) that I ε ( w n ) ≥ I ∞ ( w n ) + o n (1) and o n (1) = h I ′ ε ( w n ) , w n i ≥ h I ′∞ ( w n ) , w n i + o n (1) . (3.31)If h I ′∞ ( w n ) , w n i ≤ n large enough, similar to the proof of Case 1, we get thatthere exists t n ≤ t n w n ∈ N ∞ . Thus, by (3.28)–(3.30) m ∞ ≤ I ∞ ( t n w n ) = I ∞ ( t n w n ) − h I ′∞ ( t n w n ) , t n w n i = 14 t n k w n k + 14 Z R K ∞ (cid:16) f ( t n w n ) t n w n − F ( t n w n ) (cid:17) d x + 4 s − t ∗ s n Z R Q ∞ | w n | ∗ s d x ≤ k w n k ε + 14 Z R K ( εx ) (cid:16) f ( w n ) w n − F ( w n ) (cid:17) d x + 4 s − Z R Q ( εx ) | w n | ∗ s d x + o n (1)= Ψ( w n ) + o n (1) → l − α = c ε − α < c ε which contradicts with the assumption c ε < m ∞ .Observing that h I ′∞ ( w n ) , w n i → h I ′ ε ( v n ) , v n i → n → ∞ (indeed, R R (cid:16) K ∞ f ( w n ) w n + Q ∞ | w n | ∗ s (cid:17) d x → A , R R (cid:16) K ( εx ) f ( v n ) v n + Q ( εx ) | v n | ∗ s (cid:17) d x → B , where A, B >
0, otherwise, contradicts with (3.22)), by ( v ) of Lemma 3.2, thereexist two sequences { t n } ⊂ R + and { s n } ⊂ R + satisfying t n → s n → n → ∞ , respectively, such that t n w n ∈ N ∞ , s n v n ∈ N ε . Hence, by (3.31), we get I ε ( w n ) ≥ I ∞ ( w n ) + o n (1) = I ∞ ( t n w n ) + o n (1) ≥ m ∞ + o n (1)and I ε ( v n ) = I ε ( s n v n ) + o n (1) ≥ m ε + o n (1) = c ∞ + o n (1) . Therefore, by (3.26), we have c ε ≥ m ∞ + c ε > m ∞ , a contradiction.If { y n } ⊂ R is unbounded, we choose a subsequence, stilled denoted by { y n } ,such that | y n | ≥ R n . Then B R n ( y n ) ⊂ R \ B R n (0) ⊂ R \ B R /ε (0). Similarly tothe proof (3.28)–(3.30), we can infer that Z R (cid:16) V ( εx ) − V ∞ (cid:17) | v n | d x ≥ o n (1) Z R (cid:16) Q ( εx ) − Q ∞ (cid:17) | v n | ∗ s d x = o n (1)and Z R (cid:16) K ( εx ) − K ∞ (cid:17) F ( v n ) d x = o (1) , Z R (cid:16) K ( εx ) − K ∞ (cid:17) f ( v n ) v n d x = o n (1)Similarly as to the case that { y n } is bounded, we can obtain a contradiction bycomparing I ε ( v n ) with m ∞ . Thus, dichotomy does not happen. According to the above arguments, the sequence { ρ n } must be compactness,i.e., there exists { y n } ⊂ R such that for every ˆ δ >
0, there exists e R >
0, we have R R \ B e R ( y n ) ρ n ( x ) d x < ˆ δ . By the interpolation inequality, we have that Z R \ B e R ( y n ) | u n | m d x ≤ (cid:16) Z R \ B e R ( y n ) | u n | d x (cid:17) mθ (cid:16) Z R \ B e R ( y n ) | u n | ∗ s d x (cid:17) m (1 − θ ∗ s < C ˆ δ (3.32)where m ∈ [2 , ∗ s ], θ ∈ [0 ,
1] satisfies m = θ + − θ ∗ s . This means that the sequence {| u n | m } with 2 ≤ m ≤ ∗ s is also compactness.We claim that the sequence { y n } is bounded. If not, up to a subsequence, wecan choose r n such that | y n | ≥ r n ≥ e R + R /ε with r n → + ∞ . For n large enough, B e R ( y n ) ⊂ R \ B r n − e R (0) ⊂ R \ B R /ε (0). By (3.32), we can infer that Z R (cid:16) V ( εx ) − V ∞ (cid:17) | u n | d x = Z B e R ( y n ) + Z R \ B e R ( y n ) (cid:16) V ( εx ) − V ∞ (cid:17) | u n | d x ≥ − δC + o n (1) ≥ o n (1)Similarly, we can obtain that Z R (cid:16) K ( εx ) − K ∞ (cid:17) F ( u n ) d x = o n (1) , Z R (cid:16) K ( εx ) − K ∞ (cid:17) f ( u n ) u n d x = o n (1) , and Z R (cid:16) Q ( εx ) − Q ∞ (cid:17) | u n | ∗ s d x = o n (1) . Thus, I ε ( u n ) ≥ I ∞ ( u n ) + o n (1) and o n (1) = h I ′ ε ( u n ) , u n i ≥ h I ′∞ ( u n ) , u n i + o n (1).If h I ′∞ ( u n ) , u n i ≤
0, by the similar arguments as Case 1, there exists t n ≤ n such that t n u n ∈ N ∞ . Hence, m ∞ ≤ I ∞ ( t n u n ) = I ∞ ( t n u n ) − h I ′∞ ( t n u n ) , t n u n i = 14 t n k u n k + 14 Z R K ∞ (cid:16) f ( t n u n ) t n u n − F ( t n u n ) (cid:17) d x + 4 s − t ∗ s n Z R Q ∞ | u n | ∗ s d x ≤ k u n k ε + 14 Z R K ( εx ) (cid:16) f ( u n ) w n − F ( u n ) (cid:17) d x + 4 s − Z R Q ( εx ) | u n | ∗ s d x + o n (1)= I ε ( u n ) − h I ′ ε ( u n ) , u n i + o n (1) → c ε which is a contradiction.If h I ′∞ ( u n ) , u n i = o n (1), by ( v ) of Lemma 3.2, there exists t n → t n u n ∈ N ∞ . Hence, for n large enough, we have that c ε = I ε ( u n ) + o n (1) ≥ I ∞ ( u n ) + o n (1) = I ∞ ( t n u n ) + o n (1) ≥ m ∞ + o n (1) ≥ m ∞ which is a contradiction. Therefore, the claim is true.In view of the boundedness of { y n } and u n → u in L rloc ( R ) for 2 ≤ r < ∗ s , using(3.32), it is easy to check that u n → u in L r ( R ) for 2 ≤ r < ∗ s . Set f u n = u n − u , RACTIONAL SCHR ¨ODINGER-POISSON SYSTEM WITH CRITICAL GROWTH 21 by the weakness convergence and Brezis-Lieb Lemma, one has k u n k ε = k u k ε + k f u n k ε + o n (1) (3.33)and Z R Q ( εx ) | u n | ∗ s d x = Z R Q ( εx ) | u | ∗ s d x + Z R Q ( εx ) | f u n | ∗ s d x + o n (1) . (3.34)By H¨older’s inequality and using u n → u in L r ( R ) for 2 ≤ r < ∗ s , it is easy toverify that Z R K ( εx ) F ( u n ) d x = Z R K ( εx ) F ( u ) d x + o n (1) (3.35) Z R K ( εx ) f ( u n ) u n d x = Z R K ( εx ) f ( u ) u d x + o n (1) (3.36) Z R K ( εx ) f ( u n ) v d x = Z R K ( εx ) f ( u ) v d x + o n (1) (3.37)and Z R Q ( εx ) | u n | ∗ s − u n v d x = Z R Q ( εx ) | u | ∗ s − uv d x + o n (1) . (3.38)Therefore, by ( iv ) of Lemma 2.2 and (3.37)–(3.38), it is easy to see that h I ′ ε ( u n ) , v i →h I ′ ε ( u ) , v i for any v ∈ H ε , i.e., I ′ ε ( u ) = 0 and then u ∈ N ε , so that I ε ( u ) ≥
0. Con-sequently, by ( iv ) of Lemma 2.2 and (3.33)–(3.38), we have that c ε + o n (1) = I ε ( u n ) ≥ I ε ( u n ) − I ε ( u ) = 12 k f u n k ε − ∗ s Z R Q ( εx ) | f u n | ∗ s d x + o n (1)and o n (1) = h I ′ ε ( u n ) , u n i − h I ′ ε ( u ) , u i = k f u n k ε − Z R Q ( εx ) | f u n | ∗ s d x + o n (1)By similar arguments as the proof of vanishing (see (3.13), (3.14)), it is easy to geta contradiction with the limit k f u n k ε → L >
0. Thus, L = 0 and hence u n → u in H ε . (cid:3) Relation between c ε and m , m ∞ In this section, we shall compare the energy level c ε of problem (2.3) with theenergy level m and m ∞ of limit equation (2.6). For this purpose, we should provethe existence of positive ground state solutions for the autonomous problem (2.6).For reader’s convenience, we rewrite it as follows( − ∆) s u + νu + φ tu u = κf ( u ) + µ | u | ∗ s − u, in R (4.1)where µ, ν, κ > φ tu = R R u ( y ) | x − y | − t d x .Similar to Section 3, we can easily prove that the functional I ν,κ,µ verifies themountain pass geometry. Then, applying Theorem 1.15 in [51], there exists a( P S ) c ν,κ,µ sequence { u n } ⊂ E ν such that I ν,κ,µ ( u n ) → c ν,κ,µ and I ′ ν,κ,µ ( u n ) → n → ∞ , where c ν,κ,µ can be characterized by the following relations c ν,κ,µ = inf γ ∈ Γ ν,κ,µ max t ∈ [0 , I ν,κ,µ ( γ ( t )) = inf u ∈N ν,κ,µ I ν,κ,µ ( u ) = inf u ∈ E ν \{ } max t ≥ I ν,κ,µ ( tu ) , where Γ ν,κ,µ = { γ ∈ C ([0 , , E ν ) | γ (0) = 0 , I ν,κ,µ ( γ (1)) < } . For obtaining acompactness of the above sequence, we give the following estimate for c ν,κ,µ . Lemma 4.1.
For any ν, κ, µ > , the infinimum c ν,κ,µ satisfies < c ν,κ,µ < s µ − s s S s s , where S s is the best Sobolev constant defined by (2.2) .Proof. We define u ε ( x ) = ψ ( x ) U ε ( x ) , x ∈ R , where U ε ( x ) = ε − − s u ∗ ( x/ε ), u ∗ ( x ) = e u ( x/ S ss ) k e u k ∗ s , κ ∈ R \{ } , µ > x ∈ R are fixed constants, e u ( x ) = κ ( µ + | x − x | ) − − s , and ψ ∈ C ∞ ( R ) such that0 ≤ ψ ≤ R , ψ ( x ) ≡ B δ and ψ ≡ R \ B δ . From Proposition 21 andProposition 22 in [42], Lemma 3.3 in [46], we know that Z R | ( − ∆) s u ε ( x ) | d x ≤ S s s + O ( ε − s ) , (4.2) Z R | u ε ( x ) | ∗ s d x = S s s + O ( ε ) , (4.3)and Z R | u ε ( x ) | p d x = O ( ε (2 − p )3+2 sp ) , p > − s , O ( ε (2 − p )3+2 sp | log ε | ) , p = − s , O ( ε − s p ) , < p < − s . (4.4)Similar to the proof of ( ii ) of Lemma 3.2, there exists t ε > t ≥ I ν,κ,µ ( tu ε ) = I ν,κ,µ ( t ε u ε ). Hence d I ν,κ,µ ( tu ε )d t (cid:12)(cid:12)(cid:12) t = t ε = 0, that is t ε Z R ( | ( − ∆) s u ε | + νu ε ) d x + t ε Z R φ tu ε u ε d x = κ Z R f ( t ε u ε ) t ε u ε d x + µt ∗ s ε Z R | u ε | ∗ s d x. By ( f ), we have that t ε Z R ( | ( − ∆) s u ε | + νu ε ) d x + t ε Z R φ tu ε u ε d x ≥ µt ∗ s ε Z R | u ε | ∗ s d x. It follows from ( iii ) of Lemma 2.2 that t ∗ s ε ≤ µ k u ε k ∗ s ∗ s (cid:16) t ε Z R ( | ( − ∆) s u ε | + νu ε ) d x + Ct ε k u ε k t (cid:17) . (4.5)Therefore, (4.2), (4.3), (4.4) and (4.5) imply that | t ε | ≤ C , where C is independentof ε > C > t ε ≥ C > ε > ε n → n → ∞ such that t ε n → n → ∞ . Therefore0 < c ν,κ,µ ≤ sup t ≥ I ν,κ,µ ( tu ε n ) = I ν,κ,µ ( t ε n u ε n ) → , which is a contradiction. RACTIONAL SCHR ¨ODINGER-POISSON SYSTEM WITH CRITICAL GROWTH 23
Denote g ( t ) = t R R | ( − ∆) s u ε | d x − µt ∗ s ∗ s R R | u ε | ∗ s d x , by (4.2) and (4.3), it iseasy to check thatsup t ≥ g ( t ) = s (cid:16) R R | ( − ∆) s u ε | d x (cid:17) s (cid:16) µ R R | u ε | ∗ s d x (cid:17) − s s = s µ − s s (cid:16) S s s + O ( ε − s ) (cid:17) s (cid:16) S s s + O ( ε ) (cid:17) − s s = s µ − s s (cid:16) S s s + O ( ε ) (cid:17)(cid:16) S s s + O ( ε − s ) S s s + O ( ε ) (cid:17) s ≤ s µ − s s S s s + O ( ε − s ) . Thus I ν,κ,µ ( t ε u ε ) ≤ sup t ≥ g ( t ) + C Z R | u ε | d x + C Z R φ tu ε u ε d x − C Z R | u ε | q +1 d x ≤ s µ − s s S s s + O ( ε − s ) + C Z R | u ε | d x + C (cid:16) Z R | u ε | t d x (cid:17) t − C Z R | u ε | q +1 d x ≤ s µ − s s S s s + O ( ε − s ) + C (cid:16) Z R | u ε | t d x (cid:17) t − C Z R | u ε | q +1 d x. where we have used (4.4) and s > which implies 2 < − s .Observing thatlim ε → + (cid:16) R R | u ε | t d x (cid:17) t ε − s ≤ lim ε → + O ( ε t +4 s − ) ε − s = 0 , t > − s ,lim ε → + O ( ε t +4 s − | log ε | ) ε − s = 0 , t = − s ,lim ε → + O ( ε − s ) | log ε | ) ε − s = 0 , t < − s .Since s > and q >
3, then q + 1 > − s , 2 s − − s ( q + 1) <
0. Thuslim ε → + R R | u ε | q +1 d xε − s = lim ε → + O ( ε − − s ( q +1) ) ε − s = + ∞ . Therefore, we have proved that for ε small enough, there holds I ν,κ,µ ( t ε u ε ) < s µ − s s S s s and thus the proof is completed. (cid:3) Similar arguments to Lemma 3.5, we can obtain the compactness of the (
P S ) c ν,κ,µ sequence. It is stated as follows. Lemma 4.2.
Assume c ν,κ,µ < s µ − s s S s s and let { u n } be the ( P S ) sequence atthe level c ν,κ,µ . Then there exists { y n } ⊂ R such that for every ξ > , there exists b R > such that Z R \ B b R ( y n ) (cid:16) | ( − ∆) s u n | + | u n | + | u n | ∗ s (cid:17) d x < ξ. Proof.
It is easy to verify that the (
P S ) c ν,κ,µ sequence { u n } which satisfying I ν,κ,µ ( u n ) → c ν,κ,µ and I ′ ν,κ,µ ( u n ) → n → ∞ , is bounded in E ν .Let ρ n ( x ) = | (∆) s u n | + ν | u n | + κ ( f ( u n ) u n − F ( u n )) + s − µ | u n | ∗ s , then ρ n ∈ L ( R ) is also bounded.Checking the proof of Lemma 3.5 line by line, we find that it is only to provethe Case 2.Case 2. Up to a subsequence, we may assume that h I ′ ν,κ,µ ( v n ) , v n i > h I ′ ν,κ,µ ( w n ) , w n i >
0. Thus, h I ′ ν,κ,µ ( v n ) , v n i = o n (1) and h I ′ ν,κ,µ ( w n ) , w n i = o n (1).Similar to the proof of ( v ) of Lemma 3.2, there exist two sequences { t n } ⊂ R + and { s n } ⊂ R + satisfying t n → s n → n → ∞ , respectively, such that t n w n ∈ N ν,κ,µ , s n v n ∈ N ν,κ,µ . Therefore, I ν,κ,µ ( v n ) = I ν,κ,µ ( s n v n ) + o n (1) , I ν,κ,µ ( w n ) = I ν,κ,µ ( t n w n ) + o n (1)which leads to a contradiction that c ν,κ,µ = I ν,κ,µ ( u n ) + o n (1) ≥ I ν,κ,µ ( v n ) + I ν,κ,µ ( w n ) + o n (1)= I ν,κ,µ ( s n v n ) + I ν,κ,µ ( t n w n ) + o n (1) ≥ c ν,κ,µ + o n (1) . Hence, dichotomy does not happen. The sequence { ρ n } must be compactness, i.e.,there exists { y n } ⊂ R such that for every ξ >
0, there exists b R >
0, we have R R \ B b R ( y n ) ρ n ( x ) d x < ξ . The proof is completed. (cid:3) Remark 4.3.
By the interpolation inequality, we could show that the sequence {| u n | m } with ≤ m ≤ ∗ s is also compactness. Proposition 4.4.
Assume that ( f ) – ( f ) hold. Then problem (2.6) has at least apositive ground state solution in E ν satisfying lim | x |→∞ u ( x ) = 0 .Proof. From the above arguments, we see that there exists a (
P S ) c ν,κ,µ sequence for I ν,κ,µ . From Lemma 4.2, the sequence { u n } is bounded and verifies the compactnessin the sense of Proposition 2.3. Set c u n ( · ) = u n ( · + y n ). Using the invariance of R by translation, we see that { v n } is a bounded ( P S ) c ν,κ,µ sequence and Z R \ B b R (0) (cid:16) | ( − ∆) s c u n | + | c u n | + | c u n | ∗ s (cid:17) d x < ξ. Since { c u n } is bounded in E ν , up to a subsequence, still denoted by { c u n } , thereexists b u ∈ E ν such that c u n ⇀ b u in E ν , c u n → b u in L rloc ( R ) for 2 ≤ r < ∗ s . FromRemark 4.3, we see that R R \ B b R (0) | c u n | r d x ≤ ξ and hence c u n → b u in L r ( R ) for2 ≤ r < ∗ s . Similar arguments to the proof of compactness in Lemma 3.5, we canconclude that c u n → b u in E ν . Hence, I ν,κ,µ ( b u ) = c ν,κ,µ and I ′ ν,κ,µ ( b u ) = 0, that is, b u is a nontrivial critical point of I ν,κ,µ . From the equivalent characterize of mountainvalue, we conclude that b u is a nontrivial ground state solution of problem (2.6).Finally, we only need to show that b u is positive. For simplicity, we replace b u by u in the following discussion. If we replace I ν,κ,µ by the following functional I + ν,κ,µ ( u ) = 12 k u k E ν + 14 Z R φ tu u d x − Z R κF ( u ) d x − ∗ s Z R µ | u + | ∗ s d x, RACTIONAL SCHR ¨ODINGER-POISSON SYSTEM WITH CRITICAL GROWTH 25 where u ± = max {± u, } , then we see that all the calculations above can be repeatedword by word. So, there exists a nontrivial ground state critical point u ∈ E ν of I + ν,κ,µ . Hence, 0 = h ( I + ν,κ,µ ) ′ ( u ) , u − i = h u, u − i − Z R φ tu ( u − ) d x which implies that Z R Z R ( u ( x ) − u ( y ))( u − ( x ) − u − ( y )) | x − y | s d x d y ≥ . However, by computation, we have that Z R Z R ( u ( x ) − u ( y ))( u − ( x ) − u − ( y )) | x − y | s d x d y = Z { u ( x ) ≥ }×{ u ( y ) < } ( u ( x ) − u ( y )) u ( y ) | x − y | s d x d y + Z { u ( x ) < }×{ u ( y ) ≥ } ( u ( y ) − u ( x )) u ( x ) | x − y | s d x d y − Z { u ( x ) < }×{ u ( y ) < } ( u ( x ) − u ( y )) | x − y | s d x d y ≤ . Hence, Z R Z R ( u ( x ) − u ( y ))( u − ( x ) − u − ( y )) | x − y | s d x d y = 0which leads to u − = 0, and thus u ≥ u f ( x, u ) = µ | u | ∗ s − + κf ( u ) − νu − φ tu u , by ( f ) and ( f ), it is easy to checkthat f ( x, u ) ≤ µu ∗ s − + 1, for any u ≥
0. Using Proposition 4.1.1 in [17], we seethat u ∈ L ∞ ( R ) and check the proof of Proposition 4.1.1 word by word, using4.1.6 and 4.1.7, we can obtain that k u k ∞ ≤ C ( Z R | u | ∗ s d x + Z R | u | ∗ s β d x ) ∗ s ( β − ≤ C h(cid:16) Z R | u | ∗ s d x (cid:17) ∗ s ( β − + (cid:16) Z R | u | ∗ s d x (cid:17) β − i (4.6)where C > u , β = ∗ s +12 . This yields u ∈ L r ( R ) for all r ∈ [2 , + ∞ ]. Moreover, φ tu ∈ L ∞ ( R ). Therefore, according to Proposition 2.9 in [41],and s > , we see that u ∈ C ,α ( R ) for any 0 < α < s −
1. Thus, by Lemma 3.2in [20], we have that( − ∆) s u ( x ) = − C (3 , s ) Z R u ( x + y ) + u ( x − y ) − u ( x ) | x − y | s d x d y, ∀ x ∈ R . Assume that there exists x ∈ R such that u ( x ) = 0, then from u ≥ u − ∆) s u ( x ) = − C (3 , s ) Z R u ( x + y ) + u ( x − y ) | x − y | s d x d y < . However, observe that ( − ∆) s u ( x ) = − νu ( x ) − ( φ tu u )( x )+ κf ( u ( x ))+ µu ( x ) ∗ s − =0, a contradiction. Hence, u ( x ) >
0, for every x ∈ R . Finally, the fact u ∈ L r ( R ) ∩ C ,α ( R ) for 2 ≤ r ≤ ∞ implies that lim | x |→∞ u ( x ) = 0. The proof iscompleted. (cid:3) Lemma 4.5.
There exists ε ∗ > such that c ε < m ∞ for all ε ∈ (0 , ε ∗ ) . Proof.
By the assumption ( V ), there must exist a w ∈ R + such that V < w < V ∞ .Hence, m < m w,K ,Q ≤ m w,K ∞ ,Q ∞ < m ∞ owing to K ≥ K ∞ and Q ≥ Q ∞ .Indeed, using Proposition 4.4, choose u be a ground state solution of problem (4.1)with ν = V ∞ , κ = K ∞ and µ = Q ∞ , such that I ∞ ( u ) = m ∞ . Then there holds I ∞ ( u ) = max t ≥ I ∞ ( tu ) and there exists t > t u ∈ N w,K ∞ ,Q ∞ and I w,K ∞ ,Q ∞ ( t u ) = max t ≥ I w,K ∞ ,Q ∞ ( tu ). Hence m ∞ = max t ≥ I ∞ ( tu ) ≥ max t ≥ I w,K ∞ ,Q ∞ ( tu ) = I w,K ∞ ,Q ∞ ( t u )= I ∞ ( t u ) = I w,K ∞ ,Q ∞ ( t u ) + 12 ( V ∞ − w ) Z R | u | d x> I w,K ∞ ,Q ∞ ( t u ) ≥ m w,K ∞ ,Q ∞ . Similarly, we can show that m < m w,K ∞ ,Q ∞ .Taking ν = w , κ = K ∞ and µ = Q ∞ , in view of Proposition 4.4, we knowthat there exists v ∈ N w,K ∞ ,Q ∞ such that I w,K ∞ ,Q ∞ ( v ) = m w,K ∞ ,Q ∞ . Let η ∈ C ∞ ( R , [0 , η ( x ) = 1 if | x | ≤
1, and η ( x ) = 0 if | x | ≥
2. For θ >
0, set u θ ( x ) = η ( x/θ ) v ( x ). Similar to the proof of ( ii ) of Lemma 3.2, thereexists t θ > t θ u θ ∈ N w,K ∞ ,Q ∞ . We claim that there exists θ > I w,K ∞ ,Q ∞ ( t θ u θ ) < m ∞ . For convenience, we denote u = t θ u θ . In fact, if I w,K ∞ ,Q ∞ ( t θ u θ ) ≥ m ∞ , for all θ >
0. From the definition of η ( x ), using Lemma5 in [37], u θ → v in H s ( R ) as θ → ∞ . Recall that v ∈ N w,K ∞ ,Q ∞ , N w,K ∞ ,Q ∞ isclosed in H s ( R ), thus we obtain that t θ →
1. Thus m ∞ ≤ lim inf θ →∞ I w,K ∞ ,Q ∞ ( t θ u θ ) = I w,K ∞ ,Q ∞ ( v ) = m w,K ∞ ,Q ∞ < m ∞ which is impossible. Hence our claim is true. Since the compact support set of u denoted by supp u is compact and V (0) = V , we can choose ε ∗ > V ( εx ) ≤ w for all x ∈ supp u and ε ∈ (0 , ε ∗ ). Thus, we havemax t ≥ I ε ( tu ) ≤ max t ≥ I w,K ∞ ,Q ∞ ( tu ) = I w,K ∞ ,Q ∞ ( u ) < m ∞ for any ε ∈ (0 , ε ∗ )which implies that c ε < m ∞ for ε ∈ (0 , ε ∗ ). (cid:3) Proposition 4.6.
Suppose that ( V ) , ( Q ) , ( Q ) , ( K ) , ( f ) – ( f ) hold and s ∈ ( , . Then there exists e ε > small, such that for each ε ∈ (0 , e ε ) , problem (2.5) has at least a positive ground state solution u ε satisfying lim | x |→∞ u ε ( x ) = 0 .Proof. From Lemma 3.3, Lemma 3.4, Lemma 4.5 and Lemma 3.5, there exists asmall e ε >
0, such that I ε has a nontrivial critical point u ε ∈ H ε for ε ∈ (0 , e ε ).Hence, u ε is a nontrivial ground state solution of problem (2.5). Similar argumentsas Proposition 4.4, we can prove that for ε ∈ (0 , e ε ), u ε is a positive ground statesolution for problem (2.5) with lim | x |→∞ u ε ( x ) = 0. (cid:3) In the end of this section, we will establish the relation between lim ε → c ε and m .We state the relation as the following Lemma. Lemma 4.7. lim ε → c ε = m . RACTIONAL SCHR ¨ODINGER-POISSON SYSTEM WITH CRITICAL GROWTH 27
Proof.
First, we show that there exists ε > c ε ≥ m for all ε ∈ (0 , ε ).We suppose by contradiction that for any given ε >
0, there exists some ε ∈ (0 , ε )such that c ε < m . By Proposition 4.6, we can choose u be a ground state solutionof problem (2.5) such that I ε ( u ) = c ε = max t ≥ I ε ( tu ) < m . Similar to theproof of ( ii ) of Lemma 3.2, there exists t > t u ∈ N such that I ( t u ) = max t ≥ I ( tu ). Hence, by V ( ε x ) ≥ V , K ( ε x ) ≤ K and Q ( ε x ) ≤ Q ,we get that m > I ε ( u ) = max t ≥ I ε ( tu ) ≥ max t ≥ I ( tu ) = I ( t u ) ≥ m which is a contradiction.Next, we will prove that lim sup ε → c ε ≤ m . Let u be a nontrivial ground statesolution of equation (4.1) with ν = V , κ = K and µ = Q , that is I ( u ) = m .For each θ >
0, set u θ ( x ) = η ( x/θ ) u ( x ), where η ∈ C ∞ ( R , [0 , η ( x ) = 1 if | x | ≤
1, and η ( x ) = 0 if | x | ≥
2. From the definition of η ( x ), usingLemma 5 in [37], u θ → u in H s ( R ) as θ → ∞ .For each ε > θ >
0, there exists t ε,θ > I ε ( t ε,θ u θ ) = max t ≥ I ε ( tu θ ).Thus, h I ′ ε ( t ε,θ u θ ) , t ε,θ u θ i = 0, that is, t ε,θ Z R (cid:16) | ( − ∆) s u θ | + V ( εx ) | u θ | (cid:17) d x + t ε,θ Z R φ tu θ u θ d x = Z R K ( εx ) f ( t ε,θ u θ ) t ε,θ u θ d x + t ∗ s ε,θ Z R Q ( εx ) | u θ | ∗ s d x. (4.7)Similar to Lemma 3.5 or Lemma 4.2, we can deduce that for each θ >
0, 0 < lim ε → t ε,θ = t θ < ∞ . Taking the limit as ε → t θ Z R (cid:16) | ( − ∆) s u θ | + V | u θ | (cid:17) d x + t θ Z R φ tu θ u θ d x = Z R K f ( t θ u θ ) t θ u θ d x + t ∗ s θ Z R Q | u θ | ∗ s d x. This yields to t θ u θ ∈ N , i.e., I ( t θ u θ ) = max t ≥ I ( tu θ ). Recall that u θ → u in H s ( R ) as θ → ∞ and u ∈ N , thus, using ( ii ) of Lemma 3.2, we have t θ → θ → ∞ . By the definition of c ε , we have thatlim sup ε → c ε ≤ lim sup ε → max t ≥ I ε ( tu θ ) = lim sup ε → I ε ( t ε,θ u θ ) = I ( t θ u θ ) . Let θ → ∞ , we get that lim sup ε → c ε ≤ I ( u ) = m . The proof is completed. (cid:3) Concentration behavior
In this section, we study the concentration behavior of ground state solutions ofsystem (1.1). In this section, we choose H s ( R ) as our work space since H s ( R ) = E ν = H ε for any ε > ν > e ε >
0, such that for each ε ∈ (0 , e ε ),problem (2.5) possesses a positive ground state solution v ε ∈ H s ( R ) satisfying I ε ( v ε ) = c ε and I ′ ε ( v ε ) = 0. Now, we study the behavior of the family { v ε } . Lemma 5.1.
For the family { v ε } satisfying I ε ( v ε ) = c ε and I ′ ε ( v ε ) = 0 , there exist b ε > , such that for all ε ∈ (0 , b ε ) , there exist a family { y ε } ⊂ R , and constants R, σ > such that Z B R ( y ε ) | v ε | d x ≥ σ. (5.1) Proof.
Suppose by contradiction that (5.1) does not happen. Then there exists asequence ε n → n → ∞ such thatlim n →∞ sup y ∈ R Z B R ( y ) | v ε n | d x = 0 . Taking a similar discussion as that in the proof of Lemma 3.5, we can easily obtaina contradiction. Hence, (5.1) holds. (cid:3)
For simplicity, we denote w ε ( x ) := v ε ( x + y ε )(= u ε ( εx + εy ε )) . By the fact that I ( v ε ) = c ε and I ′ ε ( v ε ) = 0, so w ε is a positive ground state solutionto the following equation( − ∆) s w + V ( εx + εy ε ) w + φ tw w = K ( εx + εy ε ) f ( w ) + Q ( εx + εy ε ) | w | ∗ s − w. (5.2) Lemma 5.2.
The family { εy ε } which obtained in Lemma 5.1 is bounded in R forany ε ∈ (0 , b ε ) .Proof. Suppose by contradiction that { εy ε } is unbounded, then there exist twosequences ε n and { ε n y ε n } such that lim n →∞ ε n → n →∞ | ε n y ε n | = + ∞ . In thesequel, for simplicity, we denote y n = y ε n and v n = v ε n . Set w n ( · ) = v n ( · + y n ),then for each n ∈ N , w n ≥ I ε n ( w n ) = c ε n and I ′ ε n ( w n ) = 0, andfrom (5.1), we have that Z B R (0) | w n | d x ≥ σ > n ∈ N . (5.3)By Lemma 4.7, it is easy to check that { w n } is bounded in H s ( R ). Thus, up to asubsequence, still denoted by { w n } , we assume that there exists w ∈ H s ( R ) suchthat w n ⇀ w in H s ( R ), w n → w in L rloc ( R ) for 2 ≤ r < ∗ s and w n → w a.e. in R . Obviously w ≥
0. Moreover, from (5.3), we see that w n ∈ N , there exists t n > t n w n ∈ N . We claim thatlim n →∞ I ( t n w n ) = m . Since t n w n ∈ N , then clearly I ( t n w n ) ≥ m . On the otherhand, by Lemma 4.5, we have that I ( t n w n ) ≤ t n Z R (cid:16) | ( − ∆) s w n | + V ( ε n x + ε n y n ) | w n | (cid:17) d x + t n Z R φ tw n w n d x − Z R K ( ε n x + ε n y n ) F ( t n w n ) d x − Z R Q ( ε n x + ε n y n ) | w n | ∗ s d x = I ε n ( t n w n ) ≤ max t ≥ I ε n ( tw n ) = I ε n ( w n ) = c ε n = m + o n (1) (5.4)which yields to lim sup n →∞ I ( t n w n ) ≤ m and hence the claim is true.Since { w n } is bounded in H s ( R ), by (5.4), it is easy to get that { t n } is bounded.Thus, up to a subsequence, still denoted by { t n } , we may assume that lim n →∞ t n = t ≥ RACTIONAL SCHR ¨ODINGER-POISSON SYSTEM WITH CRITICAL GROWTH 29
0. If t = 0, in view of the boundedness of { w n } in H s ( R ), we see that t n w n → H s ( R ), and thus lim n →∞ I ( t n w n ) = 0, a contradiction. Hence, t > { t n w n } is bounded in H s ( R ), up to a subsequence, still denotedby { t n w n } , we may assume that t n w n ⇀ b w in H s ( R ). Since w n ⇀ w in H s ( R )and t n → t as n → ∞ , then t n w n ⇀ tw in H s ( R ) as n → ∞ . By the uniquenessof weak limit, it yields to b w = tw . Therefore, we obtain a bounded minimizingsequence { t n w n } ⊂ N as n → ∞ and h I ′ ( t n w n ) , t n w n i = 0 for any n ∈ N . Similarproof as that done in the proof of ( vi ) of Lemma 3.2, we may assume that { t n w n } is a ( P S ) m sequence for I . By Lemma 4.1, using similar argument as the proof ofProposition 4.4, we can conclude that t n w n → tw in H s ( R ). Moreover, tw ∈ N .Thus, 0 ≤ t k w n − w k ≤ | t n − t |k w n k + k t n w n − tw k → n → ∞ , that is, w n → w in H s ( R ) as n → ∞ . Therefore, by Fatou’s Lemma and t n w n ∈N , recalling that ε n → | ε n y n | → ∞ as n → ∞ , we have that m ≤ I ( tw ) < I ∞ ( tw ) = I ∞ ( tw ) − h I ′ ( tw ) , tw i = 14 Z R (cid:16) | ( − ∆) s tw | d x + Z R ( V ∞ − V | tw | (cid:17) d x + Z R ( Q − Q ∞ ∗ s ) | tw | ∗ s d x + Z R ( K f ( tw ) tw − K ∞ F ( tw )) d x ≤ lim inf n →∞ Z R (cid:16) | ( − ∆) s t n w n | d x + lim inf n →∞ Z R ( V ( ε n x + ε n y n )2 − V | t n w n | (cid:17) d x + lim inf n →∞ Z R ( K f ( t n w n ) t n w n − K ( ε n x + ε n y n ) F ( t n w n )) d x + lim inf n →∞ Z R ( Q − Q ( ε n x + ε n y n )2 ∗ s ) | t n w n | ∗ s d x ≤ lim inf n →∞ (cid:16) I ε n ( t n w n ) − h I ′ ( t n w n ) , t n w n i (cid:17) = lim inf n →∞ I ε n ( t n w n ) ≤ lim inf n →∞ max t ≥ I ε n ( tw n ) = lim inf n →∞ I ε n ( w n ) = lim inf n →∞ c ε n = m (5.5)which yields a contradiction. Thus, { εy ε } is bounded in R . (cid:3) For any ε n →
0, the subsequence { ε n y ε n } of the family { εy ε } is such that ε n y ε n → x ∗ in R , we will prove that x ∗ ∈ Θ. Lemma 5.3. x ∗ ∈ Θ .Proof. Set I x ∗ ( v ) = 12 Z R ( | ( − ∆) s v | + V ( x ∗ ) v ) d x + 14 Z R φ tv v d x − Z R K ( x ∗ ) F ( v ) d x − ∗ s Z R Q ( x ∗ ) | v ( x ) | ∗ s d x. Suppose that V ( x ∗ ) > V . Taking the similar arguments of Lemma 5.2 and replac-ing I ∞ by I x ∗ in (5.5), we can obtain a contradiction. Hence, x ∗ ∈ Θ V . By similardiscussion, we can obtain a contradiction in the case x ∗ ∈ Θ K ∩ Θ Q . Therefore, x ∗ ∈ Θ = Θ V ∩ Θ K ∩ Θ Q and the proof is completed. (cid:3) Since w ε is a positive ground state solution of problem (5.2) and I ε ( w ε ) = c ε (using the invariance of translation), by Lemma 4.7, it is easy to check that thereexists e ε > ε ∈ (0 , e ε ), w ε is bounded in H s ( R ) by a constantwhich is independent of ε . Hence, for any ε n →
0, the subsequence { w ε n } is boundedin H s ( R ), we may assume that up to a subsequence, w ε n ⇀ w in H s ( R ) andby Lemma 5.2, up to a subsequence, we also may assume that ε n y ε n → x ∈ Θ as n → ∞ . Lemma 5.4. w ε n → w in H s ( R ) and w is a positive ground state solution ofthe following problem ( − ∆) s u + V ( x ) u + φ tu u = K ( x ) f ( u ) + Q ( x ) | u | ∗ s − u. (5.6) Proof.
Similar proof of Lemma 5.2, it is easy to check that w ε n → w in H s ( R ).Therefore, h I ′ ε n ( w ε n ) , ϕ i = h ( I x ) ′ ( w ) , ϕ i , for any ϕ ∈ H s ( R ). This means that w is a nontrivial ground state solution of problem (5.6). By the similar argumentof Proposition 4.4, we can complete the proof. (cid:3) Moreover, we have the following vanishing estimate of { w ε } at infinity. Lemma 5.5. lim | x |→∞ w ε ( x ) = 0 uniformly in ε ∈ (0 , e ε ) .Proof. For any ε n → w n := w ε n is a positive ground state solution of problem(5.2), then f ( x, w n ) := K ( ε n x + ε n y ε n ) f ( w n )+ Q ( ε n x + ε n y ε n ) | w n | ∗ s − w n − V ( ε n x + ε n y ε n ) w ε n + φ tw n w n ≤ C (1 + | w n | ∗ s − ), where C is independent of n and w n , bythe estimate (4.6), we have that k w n k ∞ ≤ C k w n k α ≤ C , where C > n .Now we borrow the ideas in [2] to complete the proof. For this purpose, werewrite problem (5.2) as follows( − ∆) s w n + w n = g n ( x )where g n ( x ) := w n + f ( x, w n ). Clearly, g n ∈ L ∞ ( R ) and is uniformly bounded.From Lemma 5.4, for n → ∞ , we have that g n → g in L r ( R ) for 2 ≤ r ≤ ∗ s ,where g = w + K ( x ) f ( w ) + Q ( x ) | w | ∗ s − w − V ( x ) w − φ w w . Using someresults found in [22], we see that w n ( x ) = Z R K ( x − y ) g n ( y ) d y where K is a Bessel potential, which possesses the following properties:( K ) K is positive, radially symmetric and smooth in R \{ } ;( K ) there exists a constant C > K ( x ) ≤ C | x | s for all x ∈ R \{ } ;( K ) K ∈ L τ ( R ) for τ ∈ [1 , − s ).We define two sets A δ = { y ∈ R | | x − y | ≥ δ } and B δ = { y ∈ R | | x − y | < δ } .Hence,0 ≤ w n ( x ) ≤ Z R K ( x − y ) | g n ( y ) | d y = Z A δ K ( x − y ) | g n ( y ) | d y + Z B δ K ( x − y ) | g n ( y ) | d y. From the definition of A δ and ( K ), we have that for all n ∈ N , Z A δ K ( x − y ) | g n ( y ) | d y ≤ Cδ s k g n k ∞ Z A δ | x − y | s d y ≤ Cδ s Z A δ | x − y | s d y := Cδ s . RACTIONAL SCHR ¨ODINGER-POISSON SYSTEM WITH CRITICAL GROWTH 31
On the other hand, by H¨older’s inequality and ( K ), we deduce that Z B δ K ( x − y ) | g n ( y ) | d y ≤ Z B δ K ( x − y ) | g n − g | d y + Z B δ K ( x − y ) | g | d y ≤ (cid:16) Z B δ K s d y (cid:17) s (cid:16) Z B δ | g n − g | − s d y (cid:17) − s + (cid:16) Z B δ K d y (cid:17) (cid:16) Z B δ | g | d y (cid:17) ≤ (cid:16) Z R K s d y (cid:17) s (cid:16) Z R | g n − g | − s d y (cid:17) − s + (cid:16) Z R K d y (cid:17) (cid:16) Z B δ | g | d y (cid:17) where we have used the fact that s > so that s < − s and 2 < − s .Since (cid:16) R B δ | g | d y (cid:17) → | x | → + ∞ , thus, we deduce that there exist n ∈ N and R > ε > Z B δ K ( x − y ) | g n ( y ) | d y ≤ δ, ∀ n ≥ n and | x | ≥ R . Hence, Z R K ( x − y ) | g n ( y ) | d y ≤ Cδ s + δ, ∀ n ≥ n and | x | ≥ R . For each n ∈ { , , · · · , n − } , there exists R n > (cid:16) R B δ | g n | d y (cid:17) < δ as | x | ≥ R n . Thus, for | x | ≥ R n , we have that Z R K ( x − y ) | g n ( y ) | d y ≤ Cδ s + Z B δ K ( x − y ) | g n ( y ) | d y ≤ Cδ s + kKk (cid:16) Z B δ | g n | d y (cid:17) ≤ C ( δ s + δ )for each n ∈ { , , · · · , n − } . Therefore, taking R = max { R , R , · · · , R n − } ,we infer that for any n ∈ N , there holds0 ≤ w n ( x ) ≤ Z R K ( x − y ) | g n ( y ) | d y ≤ Cδ s + δ, for all | x | ≥ R implies that lim | x |→∞ w n ( x ) = 0 uniformly in n ∈ N . As a result, the conclusionfollows from the arbitrariness of ε n . (cid:3) Now, we give the estimate of decay properties of solutions u ε . Lemma 5.6.
There exists a constant
C > such that w ε ( x ) ≤ C | x | s , for all x ∈ R and ε ∈ (0 , e ε ) . Proof.
We borrow some ideas of the proof of Theorem 1.1 in [24] to give the proofof Lemma 5.6. By Lemma 4.2 and Lemma 4.3 in [22], by scaling, there exists acontinuous function W such that0 < W ( x ) ≤ C | x | s (5.7)and ( − ∆) s W + V W = 0 on R \ B R (0) . for some suitable R >
0. By Lemma 5.5, there exists R > R > R ) large enough such that( − ∆) s w ε + V w ε = ( − ∆) s w ε + V ( εx ) w ε + ( V − V ( εx )) w ε = K ( εx ) f ( w ε ) + Q ( ε ) w ∗ s − ε − φ tw ε w ε + ( V − V ( εx )) w ε ≤ K f ( w ε ) + Q w ∗ s − ε − V w ε ≤ x ∈ R \ B R (0). Therefore, we have obtained that( − ∆) s W ( x ) + V W ≥ ( − ∆) s w ε + V w ε on R \ B R (0) . (5.8)Let A = inf B R (0) W > Z ε ( x ) = ( B + 1) W − A w ε , where B = sup <ε< e ε k w ε k ∞ ≤ C < ∞ , where C is some positive constant independent of ε . We claim that Z ε ( x ) ≥ x ∈ R and ε ∈ (0 , e ε ). If the claim is true, we have that w ε ( x ) ≤ B + 1 A W ≤ C | x | s for all x ∈ R and ε ∈ (0 , e ε )and the conclusion is proved.Suppose by contradiction that there exist ε ∈ (0 , e ε ) and x nε ∈ R such thatinf x ∈ R Z ε ( x ) = lim n →∞ Z ε ( x nε ) < . (5.9)Since lim | x |→∞ W ( x ) = lim | x |→∞ w ε ( x ) = 0 uniformly for ε ∈ (0 , e ε ), then lim | x |→∞ Z ε ( x ) =0. Hence, the sequence { x nε } is bounded and then up to a subsequence, we mayassume that x nε → e x ε . From (5.9), we have thatinf x ∈ R Z ε ( x ) = Z ε ( e x ε ) < . (5.10)By (5.10), we get( − ∆) s Z ε ( e x ε ) + V Z ε ( e x ε ) = V Z ε ( e x ε ) − C (3 , s ) Z R Z ε ( e x ε + y ) + Z ε ( e x ε − y ) − Z ε ( e x ε ) | x − y | s d y < . Note that Z ε ( x ) ≥ AB + W − AB > B R (0), hence e x ε ∈ R \ B R (0). From(5.8), by computation, we have that( − ∆) s Z ε ( e x ε ) + V Z ε ( e x ε ) = h ( B + 1) (cid:16) ( − ∆) s W + V W (cid:17) − A (cid:16) ( − ∆) s w ε + V w ε (cid:17)i(cid:12)(cid:12)(cid:12) x = e x ε ≥ (cid:3) Proof of Theorem 1.1. ( i ) Taking ε = min { e ε , e ε, b ε } . By Proposition 4.6,for each ε ∈ (0 , ε ), problem (2.5) has at least a positive ground state solution v ε . Hence, let φ ε = φ tv ε , then ( v ε , φ ε ) is a positive solution for system (1.1) for ε ∈ (0 , ε ).( ii ) From Proposition 4.2 and Lemma 5.5, there exists R > w ε ( x ) = v ε ( x + y ε ), denoted by e x ε , is located in B R (0). RACTIONAL SCHR ¨ODINGER-POISSON SYSTEM WITH CRITICAL GROWTH 33
Thus, the global maximum point of v ε is given by x ε = e x ε + y ε . Observing that u ε ( x ) = v ε ( x/ε ), then we have that ( u ε ( x ) , φ tu ε ( x )) is a positive ground state solu-tion of system (1.1) and u ε has a global maximum point z ε = εx ε . It follows fromLemma 5.2, Lemma 5.3 and e x ε ∈ B R (0) that lim ε → V ( z ε ) = V , lim ε → K ( z ε ) = K andlim ε → Q ( z ε ) = Q . Moreover, in view of Lemma 5.4, we see that z ε → x if ε → u ε ( εx + z ε ) converges to u and u is a solution for problem (5.6). As a result,( u, φ ) is a solution of system (1.5).( iii ) By Lemma 5.6, we have that u ε ( x ) = v ε ( x/ε ) = w ε ( x/ε − y ε ) = w ε ( x + ε e x ε − z ε ε ) ≤ C | x + ε e x ε − z ε ε | s ≤ Cε s ε s + | x − z ε | s − ε s R s := Cε s C ε s + | x − z ε | s , where C = 1 − R s .6. Nonexistence of ground states
In this section, our goal is to show the nonexistence of ground state solution tosystem (1.1), that is, for each ε >
0, the ground energy c ε is not attained. Lemma 6.1.
Assume the continuous functions V ( x ) , K ( x ) , Q ( x ) satisfies ( H ) and ( f ) – ( f ) hold. Then for each ε > , c ε = m ∞ .Proof. Noting that H ε = H s ( R ), for any ε >
0. By ( H ), we have I ∞ ( u ) ≤ I ε ( u ),for all u ∈ H s ( R ). By ( ii ) of Lemma 3.2, for each u ∈ N ∞ , there exists t u > t u u ∈ N ε . Hence, for each u ∈ N ∞ , we have that0 < m ∞ = inf u ∈N ∞ I ∞ ( u ) ≤ max t ≥ I ∞ ( tu ) ≤ max t ≥ I ε ( tu ) = I ε ( t u u ) . By ( i ) of Lemma 3.2, one has0 < m ∞ ≤ inf u ∈N ∞ I ε ( t u u ) = inf v ∈N ε I ε ( v ) = c ε . So, it suffices to show that c ε ≤ m ∞ .By Proposition 4.4, there exists u ∞ ∈ N ∞ is a ground state solution of (2.6)with ν = V ∞ , κ = K ∞ and µ = Q ∞ . Set e n ( x ) = u ∞ ( x − y n ) where y n ∈ R and | y n | → ∞ as n → ∞ . Then, there exists t n ( e n ) > t n e n ∈ N ε , that is, t n Z R (cid:16) | ( − ∆) s u ∞ | + V ( εx + εy n ) | u ∞ | (cid:17) d x + t n Z R φ tu ∞ u ∞ d x = Z R K ( εx + εy n ) f ( t n u ∞ ) t n u ∞ d x + t ∗ s n Z R Q ( εx + εy n ) | u ∞ | ∗ s d x. (6.1)This implies that t n cannot converge to zero and infinity. Suppose that t n → t as n → ∞ . Letting n → ∞ in (6.1), we have that Z R (cid:16) | ( − ∆) s u ∞ | + V ∞ | u ∞ | (cid:17) d x + t Z R φ tu ∞ u ∞ d x = Z R K ∞ f ( tu ∞ ) u ∞ t d x + t ∗ s − Z R Q ∞ | u ∞ | ∗ s d x. In view of u ∞ ∈ N ∞ , we conclude that t = 1. Since c ∞ ≤ I ε ( t n e n ) = I ∞ ( t n u ∞ ) + t n Z R ( V ( εx + εy n ) − V ∞ ) | u ∞ | d x − Z R ( K ( εx + εy n ) − K ∞ ) F ( t n u ∞ ) d x − t ∗ s n ∗ s Z R ( Q ( εx + εy n ) − Q ∞ ) | u ∞ | ∗ s d x. (6.2)By the assumption on V , for any δ >
0, there exists
R > Z | x |≥ Rε | V ( εx + εy n ) − V ∞ || u ∞ | d x ≤ δ. By | y n | → ∞ , according to Lebesgues theorem, we havelim n →∞ Z | x | < Rε | V ( εx + εy n ) − V ∞ || u ∞ | d x = 0 . Thus, lim n →∞ Z R | V ( εx + εy n ) − V ∞ || u ∞ | d x = 0 . Similarly, we deduce thatlim n →∞ Z R | K ( εx + εy n ) − K ∞ F ( u ∞ ) d x = 0 , lim n →∞ Z R | Q ( εx + εy n ) − Q ∞ | u ∞ | ∗ s d x = 0 . Therefore, using t n → n → ∞ in (6.2), we infer that c ε ≤ m ∞ andthe proof is completed. (cid:3) Proof of Theorem 1.2.
Suppose by contradiction that there exist some ε > u ∈ N ε such that I ε ( u ) = c ε . From Lemma 6.1, c ε = m ∞ . In view ofLemma 3.2, there exists t > t u ∈ N ∞ . Thus, using the fact u ∈ N ε ,we have that m ∞ ≤ I ∞ ( t u ) ≤ I ε ( t u ) ≤ max t ≥ I ε ( tu ) = I ε ( u ) = c ε = m ∞ . Thus, m ∞ = I ∞ ( t u ) = I ε ( t u ). However, I ∞ ( t u ) = I ε ( t u ) + t Z R ( V ∞ − V ( ε x )) | u | d x + Z R ( K ( ε x ) − K ∞ ) F ( t u ) d x + t ∗ s ∗ s Z R ( Q ( ε x ) − Q ∞ ) | u | ∗ s d x< I ε ( t u )which is a contradiction. The proof is completed.7. Appendix
In this section, By a similar argument of Section 4 in [42], we give the estimateof (3.4), (3.5) and (3.6). We prove them in the general case.Let S s = S s ( e u ), e u = κ ( µ + | x − x | ) − N − s , x ∈ R N with κ ∈ R \{ } , µ > x ∈ R N are fixed constant. Using scaling and translation transform, we see that S s = S s ( e u ) = S s (¯ u ) , ¯ u ( x ) = κ (1 + | x | ) N − s , x ∈ R N . RACTIONAL SCHR ¨ODINGER-POISSON SYSTEM WITH CRITICAL GROWTH 35
Set u ∗ ( x ) = κ ( ε + | x − x /ε | ) N − s , U ε ( x ) = ε N − s u ∗ ( x ) , x ∈ R N , ε > . It is easy to check that k ( − ∆) s U ε k = k ( − ∆) s ¯ u k = S s k ¯ u k ∗ s = S s k U ε k ∗ s . Let η ∈ C ∞ ( R N ) be such that 0 ≤ η ≤ η (cid:12)(cid:12)(cid:12) B R (0) = 1, supp η ⊂ B R (0). Set u ε ( x ) = η ( x − x /ε ) U ε ( x ), for any ε > x ∈ R N . By computation, we havethe following estimate for u ε . Lemma A.1 . ( i ) Let ρ >
0. If x ∈ R N \ B ρ ( x /ε ), then | u ε ( x ) | ≤ | U ε ( x ) | ≤ C ε N − s , |∇ u ε ( x ) | ≤ C ε N − s for any ε > C and C , possibly depending on N, s, ρ .( ii ) For any x ∈ R N and y ∈ R N \ B R ( x /ε ), with | x − y | ≤ R , | u ε ( x ) − u ε ( y ) | ≤ C ε N − s | x − y | . ( iii ) For any x, y ∈ R N \ B R ( x /ε ), | u ε ( x ) − u ε ( y ) | ≤ C ε N − s min { , | x − y |} . for any ε > C and C , possibly depending on N, s, ρ . Proof. ( i ) By the definition of u ε , for any x ∈ R N \ B ρ ( x /ε ), we have | u ε ( x ) | ≤ | U ε ( x ) | ≤ Cε N − s ε + | x − x /ε | ) N − s ≤ Cε N − s ε + ρ ) N − s ≤ C ε N − s and (cid:12)(cid:12)(cid:12) ∇ u ε ( x ) (cid:12)(cid:12)(cid:12) = (cid:12)(cid:12)(cid:12) ∇ ( η ( x − x /ε )) U ε ( x ) + η ( x − x /ε ) ∇ U ε ( x ) (cid:12)(cid:12)(cid:12) ≤ C (cid:16) | U ε ( x ) | + ε N − s | x − x /ε | ( ε + | x − x /ε | ) N +2 − s (cid:17) ≤ C (cid:16) | U ε ( x ) | + ε N − s | x − x /ε | ρ ( ε + | x − x /ε | ) N +2 − s (cid:17) ≤ C ε N − s . ( ii ) Let x ∈ R N , y ∈ R N \ B R ( x /ε ), with | x − y | ≤ R , and let ξ be any point onthe segment joining x and y . Then ξ = tx + (1 − t ) y , t ∈ [0 , | ξ − x /ε | ≥ | x − x /ε | − t | x − y | ≥ R − R R . Taking ρ = R in ( i ), implies that |∇ u ε ( ξ ) | ≤ Cε N − s and so by Taylor expansion, | u ε ( x ) − u ε ( y ) | ≤ Cε N − s | x − y | , for any ε > . (7.1) ( iii ) Let x, y ∈ R N \ B R ( x /ε ). If | x − y | ≤ R , then the second conclusion of ( ii )follows from (7.2). If | x − y | > R , using the conclusion ( i ), we have that | u ε ( x ) − u ε ( y ) | ≤ | u ε ( x ) | + | u ε ( y ) | ≤ Cε N − s . The proof is completed. (cid:3)
Lemma A.2 . Let s ∈ (0 ,
1) and n > s . Then the following estimate holds true Z R N | ( − ∆) s u ε | d x ≤ S N s s + O ( ε N − s )as ε → Proof.
By the definition of u ε , we can rewrite k ( − ∆) s u ε k as follows k ( − ∆) s u ε k = 1 C ( N, s ) Z R N | u ε ( x ) − u ε ( y ) | | x − y | N =2 s d x d y = 1 C ( N, s ) (cid:16) Z B R ( x /ε ) × B R ( x /ε ) | U ε ( x ) − U ε ( y ) | | x − y | N +2 s d x d y + Z R N \ B R ( x /ε ) × R N \ B R ( x /ε ) | u ε ( x ) − u ε ( y ) | | x − y | N +2 s d x d y + 2 Z D | u ε ( x ) − u ε ( y ) | | x − y | N +2 s d x d y + 2 Z E | u ε ( x ) − u ε ( y ) | | x − y | N +2 s d x d y (cid:17) where the sets D and E are given by D = n ( x, y ) ∈ R N (cid:12)(cid:12)(cid:12) x ∈ B R ( x /ε ) , y ∈ R N \ B R ( x /ε ) and | x − y | > R o and E = n ( x, y ) ∈ R N (cid:12)(cid:12)(cid:12) x ∈ B R ( x /ε ) , y ∈ R N \ B R ( x /ε ) and | x − y | ≤ R o . Similar to the proof of Proposition 21 in [42], using Lemma A.1, it is easy to showthat, as ε →
0, there hold Z E | u ε ( x ) − u ε ( y ) | | x − y | N +2 s d x d y ≤ O ( ε N − s ) (7.2)and Z R N \ B R ( x /ε ) × R N \ B R ( x /ε ) | u ε ( x ) − u ε ( y ) | | x − y | N +2 s d x d y ≤ O ( ε N − s ) . (7.3)To complete the proof, checking the proof of Proposition 21 in [42], it is sufficientto verify the following estimate holds Z D | U ε ( x ) || U ε ( y ) || x − y | N +2 s d x d y ≤ O ( ε N − s )as ε → i ) of Lemma A.1, making the change of variables ξ = x − y , wededuce that Z D U ε ( x ) || U ε ( y ) || x − y | N +2 s d x d y ≤ Cε N − s Z D ε + | x − x /ε | ) N − s | x − y | N +2 s d x d y = Cε N − s Z D ε | x | ) N − s | x − y | N +2 s d x d y RACTIONAL SCHR ¨ODINGER-POISSON SYSTEM WITH CRITICAL GROWTH 37 ≤ Cε N − s Z B R/ε (0) | x | ) N − s d x Z | ξ | > R ε | ξ | N +2 s d ξ ≤ Cε N − s ε − s ε s = O ( ε N − s )where D ε is defined as D ε = n ( x, y ) ∈ R N (cid:12)(cid:12)(cid:12) x ∈ B R/ε (0) , y ∈ R N \ B R/ε (0) , | x − y | > R ε o . Finally, combing with (7.2), (7.3), we conclude that k ( − ∆) s u ε k ≤ C ( N, s ) Z B R ( x /ε ) × B R ( x /ε ) | U ε ( x ) − U ε ( y ) | | x − y | N +2 s d x d y + O ( ε N − s ) ≤ C ( N, s ) Z R N | U ε ( x ) − U ε ( y ) | | x − y | N +2 s d x d y + O ( ε N − s )= k ( − ∆) s ¯ u k + O ( ε N − s )as ε → (cid:3) Lemma A.3.
Let s ∈ (0 ,
1) and
N > s , p ∈ (1 , ∗ s ). Then the followingestimates hold. Z R N | u ε ( x ) | p d x = O ( ε N − N − s p ) , p > NN − s , O ( ε N − N − s p | log ε | ) , p = NN − s , O ( ε N − s p ) , < p < NN − s . (7.4)and Z R N | u ε ( x ) | ∗ s d x = Z R N | ¯ u ( x ) | ∗ s d x + O ( ε N ) (7.5)as ε → Proof.
By the definition of u ε , we have Z R N | u ε ( x ) | p d x = Z B R ( x /ε ) | U ε ( x ) | p d x + Z B R ( x /ε ) \ B R ( x /ε ) | ψ ( x − x /ε ) U ε ( x ) | p d x ≥ Cε N − s p Z B R ( x /ε ) ε + | x − x /ε | ) ( N − s ) p d x ≥ Cε N − N − s p Z R/εδ r ) ( N − s ) p r N − d r for any 0 < δ < R/ε . Therefore, Z R | u ε ( x ) | p d x ≥ Cε N − N − s p Z R/εδ r ) ( N − s ) p r N − d r ≥ Cε N − N − s p Z R/εδ r N ( p − − sp +1 dr = − C ,s ε Np − sp + C ,s ε N − ( N − s ) p , p − p N > s , C ,s ε N − ( N − s ) p | log ε | + C ,s ε N − N − s p , p − p N = 2 s , C ,s ε Np − sp − C ,s ε N − ( N − s ) p , p − p N < s = O ( ε N − N − s p ) , p > NN − s , O ( ε N − N − s p | log ε | ) , p = NN − s , O ( ε N − s p ) , < p < NN − s .On the other hand, arguing in the same way, we infer that Z R N | u ε ( x ) | p d x ≤ Z B R ( x /ε ) | U ε ( x ) | p d x ≤ Cε N − N − s p Z R/ε r N − (1 + r ) N − s p dr = Cε N − N − s p (cid:16) Z δ r N − (1 + r ) N − s p dr + Z R/εδ r N ( p − − sp +1 dr (cid:17) = O ( ε N − N − s p ) , p > NN − s , O ( ε N − N − s p | log ε | ) , p = NN − s , O ( ε N − s p ) , < p < NN − s ,where 0 < δ < R/ε . Hence, (7.4) holds true.( ii ) Z R N | u ε ( x ) | ∗ s d x = Z R N | U ε ( x ) | ∗ s d x + Z R N (cid:16) | η ( x − x /ε ) | ∗ s − (cid:17) | U ε ( x ) | ∗ s d x = k ¯ u k ∗ s ∗ s + Z R N \ B R ( x /ε ) (cid:16) | η ( x − x /ε ) | ∗ s − (cid:17) | U ε ( x ) | ∗ s d x = k ¯ u k ∗ s ∗ s + Z R N \ B R/ε (0) (cid:16) | η ( x ) | ∗ s − (cid:17) | x | ) N d x ≈ k ¯ u k ∗ s ∗ s + Z R N \ B R/ε (0) | x | N d x = k ¯ u k ∗ s ∗ s + O ( ε N )as ε → (cid:3) Acknowledgements.
This work was done when the first author visited Depart-ment of Mathematics, Texas A&M University-Kingsville under the support of ChinaScholarship Council (201508140053), and he thanks Department of Mathematics,Texas A&M University-Kingsville for their kind hospitality. The first author is alsosupported by NSFC grant 11501403.
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E-mail address : [email protected] Ravi P. AgarwalDepartment of Mathematics, Texas A & M University-Kingsville, Texas, Kingsville 78363,USA
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