Unbalanced (p,2) -fractional problems with critical growth
aa r X i v : . [ m a t h . A P ] J a n Unbalanced ( p, Deepak Kumar ∗ and K. Sreenadh † Department of Mathematics,Indian Institute of Technology Delhi,Hauz Khaz, New Delhi-110016, India.
Abstract
We study the existence, multiplicity and regularity results of non-negative solutionsof following doubly nonlocal problem:( P λ ) ( − ∆) s u + β ( − ∆) s p u = λa ( x ) | u | q − u + (cid:18)Z Ω | u ( y ) | r | x − y | µ dy (cid:19) | u | r − u in Ω ,u = 0 in R n \ Ω , where Ω ⊂ R n is a bounded domain with C boundary ∂ Ω, 0 < s < s < n > s ,1 < q < p <
2, 1 < r ≤ ∗ µ with 2 ∗ µ = n − µn − s , λ, β > a ∈ L dd − q (Ω), for some q < d < ∗ s := nn − s , is a sign changing function. We prove that each nonnegative weaksolution of ( P λ ) is bounded. Furthermore, we obtain some existence and multiplicityresults using Nehari manifold method. Key words:
Non-local operator, Fractional ( p,
2) Laplacian, Choquard equation, Neharimanifold.
In this article, we are concerned with the regularity, existence and multiplicity results forsolutions to ( p, P λ ) ( − ∆) s u + β ( − ∆) s p u = λa ( x ) | u | q − u + (cid:18)Z Ω | u ( y ) | r | x − y | µ dy (cid:19) | u | r − u in Ω u = 0 in R n \ Ω , where Ω is a bounded domain in R n with C boundary ∂ Ω, 0 < s < s < n > s ,1 < q < p <
2, 1 < r ≤ ∗ µ with 2 ∗ µ = n − µn − s , λ, β >
0, 0 < µ < n and a ∈ L dd − q (Ω), where ∗ email: [email protected] † e-mail: [email protected] q < d < ∗ s and 2 ∗ s = nn − s , is a sign changing function such that a + ( x ) := max { a ( x ) , } 6≡ − ∆) sm is the Fractional m -Laplace operator, defined as follows( − ∆) sm u ( x ) = − ǫ → Z R n \ B ǫ ( x ) | u ( y ) − u ( x ) | m − ( u ( x ) − u ( y )) | x − y | n + ms dy, for m > s ∈ (0 , s = s = 1, and β = 0, the following equation with Choquardtype nonlinearity − ∆ u + V ( x ) u = (cid:0) | x | − µ ∗ | u | p (cid:1) | u | p − u, in R n (1.1)has been studied extensively. In case of n = 3 , p = 2 and µ = 1, S. Pekar [34] describes (1.1) inquantum mechanics of a polaron at rest. Under the same assumptions, P. Choquard, in 1976,used (1.1), in modeling of an electron trapped in its own hole, in a certain approximation toHartree-Fock theory of one component plasma [23]. Buffoni et al. [10] proved existence of atleast one non-trivial solution of (1.1) when p = 2, n = 3 and potential V is sign changingperiodic function with the assumption that 0 lies in the gap of the spectrum of the operator − ∆ + V . In [2], Alves et al. obtained the existence of a non-trivial solution via penalizationmethod of the problem − ∆ u + V ( x ) u = (cid:0) | x | − µ ∗ F ( u ) (cid:1) f ( u ) , in R n , where 0 < µ < n , n = 3, V is a continuous real valued function and F is the primitive of f .Gao and Yang [18] studied the following Brezis-Nirenberg type problem − ∆ u = λu + Z Ω | u ( y ) | ∗ µ | x − y | µ dy ! | u ( x ) | ∗ µ − u ( x ) in Ω , u = 0 on ∂ Ω , where Ω ⊂ R n is a bounded domain, n ≥ < µ < n . They proved existence, multiplicityand non-existence results with respect to the parameter λ .In the nonlocal case D’ Avenia et al. [15] considered( − ∆) s u + ωu = (cid:0) | x | α − n ∗ | u | p (cid:1) | u | p − u, in R n , where ω > p > s ∈ (0 , p -Laplacian. For related work on thistype of problems, we refer to [22, 27, 28, 39, 41] and the references therein.Partial differential equations involving operator − ∆ p − ∆ q , known as the ( p, q )-Laplacian,arises from important applications such as biophysics, plasma physics, reaction-diffusion (see[8, 13, 17, 25, 40]). Due to this, a lot of work has been done in last decade on ( p, q )-Laplacianproblems. Among them, Papageorgiou and Rˇadulescu [30] considered the problem − ∆ p u − ∆ u = f ( x, u ) in Ω , u = 0 on ∂ Ω , where Ω is a bounded domain and f is a Carath´eodory function. Among other results, authorsproved existence of four non-trivial solutions for the case 1 < p <
2. Aizicovici et al. [1] provedexistence of constant sign and nodal solutions for the problem − ∆ p u − µ ∆ u = f ( x, u ) in Ω , u = 0 on ∂ Ω , where Ω is a bounded domain, p > µ >
0. For general p and q , Yin and Yang [42]considered − ∆ p u − ∆ q u = | u | p ∗ − u + θV ( x ) | u | r − u + λf ( x, u ) in Ω , u = 0 on ∂ Ω , where 1 < r < q < p < n and f ( x, u ) is a subcritical perturbation and they proved multiplicityof solutions using Lusternik-Schnirelman theory. Marano et. al [24] studied the problem withCarath´eodory function having critical growth. Using critical point theory with truncationarguments and comparison principle authors also proved bifurcation type result.As far as problems involving fractional ( p, q )-Laplacian is concerned, there is not much liter-ature available. Bhakta and Mukherjee [7] considered the problem( Q θ,λ ) n ( − ∆) s p u + ( − ∆) s q u = | u | p ∗ s − u + θV ( x ) | u | r − u + λf ( x, u ) in Ω , u = 0 in R n \ Ω , where 0 < s < s <
1, 1 < r < q < p < ns , and V and f are some appropriate functions. Herethey proved ( Q θ,λ ) has infinitely many weak solutions for some range of λ and θ . Moreover, for V ( x ) ≡ λ = 0, and assuming r > q and certain other conditions on n and r , they proved theexistence of cat Ω (Ω) many solutions of ( Q θ,λ ) using Lusternik-Schnirelmann category theory.Goel et al. [21] studied the following fractional ( p, q )-Laplacian problem( − ∆) s p u + β ( − ∆) s q u = λa ( x ) | u | δ − u + b ( x ) | u | r − u in Ω , u = 0 in R n \ Ω , (1.2)where Ω ⊂ R n is a bounded domain, 1 < δ ≤ q ≤ p < r ≤ p ∗ s = npn − ps , 0 < s ps , λ, β >
0, and a and b are sign changing functions. Using Nehari manifoldmethod authors proved existence of at least two non-negative and non-trivial solutions inthe subcritical case for all β > λ . For the critical case under somerestriction on δ , they obtained multiplicity results in some range of β and λ . Furthermore,they proved weak solutions of (1.2) are in the space L ∞ (Ω) ∩ C ,α loc (Ω), for some α ∈ (0 , ≤ q ≤ p < r < p ∗ s .Inspired from all these works we study regularity, existence and multiplicity results of ( p, P λ ) is bounded, in the case µ < s . Inthis regard, we prove the following theorem. Theorem 1.1
Suppose µ < s and the function a ( x ) is bounded in Ω . Let u be a non-negative solution of problem ( P λ ) , then u ∈ L ∞ (Ω) . Regarding the existence and multiplicity results, using the method of minimization over somesuitable subset of the Nehari manifold we obtain existence of at least two non-trivial non-negative solutions for all β > λ in some range for the subcritical case. In the criticalcase, we prove the existence of solutions by identifying the first critical level (as defined inLemma 5.2), below which the Palais-Smale sequences contain a convergent subsequence. Wefirst prove multiplicity results for all β > λ but with some restrictionon q . For this we estimate the fractional p -Laplacian norm of family of minimizers of S , thebest constant of the embedding of the space X into L ∗ s (Ω) (see Lemma 5.4). Later, weremove this restriction on q and prove multiplicity result for small λ and β . We show thefollowing existence and multiplicity theorems for problem ( P λ ). Theorem 1.2
Let r < ∗ µ . Then, there exists λ > such that problem ( P λ ) has at least twonon-negative solutions for all λ ∈ (0 , λ ) and β > . Theorem 1.3
Let r = 2 ∗ µ and the function a ( x ) be continuous in Ω . Then, there existconstants Λ , Λ > such that(i) ( P λ ) admits at least one non-negative solution for all λ ∈ (0 , Λ) and β > ,(ii) ( P λ ) admits at least two non-negative solutions for all λ ∈ (0 , Λ ) , β > and(I) for all q > , provided < p < n/ ( n − s ) ,(II) for all q ∈ (cid:0) , n (2 − p ) n − s (cid:1) ∪ (cid:0) n n − np − s , p (cid:1) , provided n/ ( n − s ) ≤ p < . Next, we remove this restriction on q , to obtain the existence of second solution in the criticalcase for all 1 < q < p < λ and β . In this regard we state our theoremas follows. Theorem 1.4
Let r = 2 ∗ µ , < q < p < and the function a ( x ) be continuous in Ω .Then, there exist constants Λ , Λ , β > such that ( P λ ) has at least two solutions for all λ ∈ (0 , Λ ) and β ∈ (0 , β ) . We point out that the study of non-autonomous functionals characterized by the fact thatthe energy density changes its ellipticity and growth properties according to the point hasbeen continued by Mingione et al. [4, 5, 6], R˘adulescu et al. [3, 12, 31, 32, 37, 43], etc.Some of the abstract methods used in this paper can be found in the recent monograph byPapageorgiou, R˘adulescu and Repovˇs [33].The paper is organized as follows: In section 2, we provide variational setting and reg-ularity result. In section 3, we define Nehari manifold associated to problem ( P λ ) and givefibering map analysis and some preliminary results. In section 4, we prove the existence andmultiplicity results in the subcritical case. In section 5, we have the existence and multiplicityof solutions in the critical case. Let Ω be any open subset of R n , consider the function space, which were introduced in [38]for p i = 2 and in [20] for general p , X p i ,s i := (cid:26) u ∈ L p i ( R n ) : u = 0 a.e. in R n \ Ω , Z Q | u ( x ) − u ( y ) | p i | x − y | n + p i s i dxdy < ∞ (cid:27) , where p = 2, p = p , 0 < s i < Q = R n \ (Ω c × Ω c ), which is a reflexive Banach spacewhen endowed with the norm k u k X pi,si := (cid:18)Z Q | u ( x ) − u ( y ) | p i | x − y | n + p i s i dxdy (cid:19) pi . (2.1)Notice that the integral in (2.1) can be extended to R n as u = 0 a.e. on R n \ Ω.For simplicity, we denote X := X ,s and X := X p,s and corresponding norms by k · k X and k · k X , respectively. From [38], we have the continuous embedding of X into L m (Ω) for1 ≤ m ≤ ∗ s , therefore we define S m = inf u ∈ X \{ } k u k X k u k m . For the sake of convenience, we denote S ∗ s = S .Regarding the spaces X and X , we have the following relation. Lemma 2.1
Let < p ≤ and < s < s < , then there exists a constant C = C ( | Ω | , n, p, s , s ) > such that k u k X ≤ C k u k X , ∀ u ∈ X . Proof.
Proof follows from [21, Lemma 2.1]. (cid:3)
The nonlocal nonlinear Choquard term present in the right hand side of ( P λ ) is well defineddue to the following result. Theorem 2.2 (Hardy-Littlewood-Sobolev inequality) Let t, r > and < µ < n with /t + µ/n + 1 /r = 2 , f ∈ L t ( R n ) and h ∈ L r ( R n ) . There exists a sharp constant C ( t, r, µ, n ) > independent of f, h , such that Z R n Z R n f ( x ) h ( y ) | x − y | µ dxdy ≤ C ( t, r, µ, n ) | f | t | h | r . In general, let f = h = | u | r , then by Hardy-Littlewood-Sobolev inequality, we get Z R n Z R n | u ( x ) | r | u ( y ) | r | x − y | µ dxdy ≤ C ( n, µ, r ) k u k rtr , if | u | r ∈ L t ( R n ) for some t > t + µn = 2 . Thus, for u ∈ H s ( R n ), by Sobolevembedding theorems, we must have2 n − µn ≤ r ≤ n − µn − s . The term 2 ∗ µ := (2 n − µ ) / ( n − s ) is known as the upper critical exponent in the sense ofHardy-Littlewood-Sobolev inequality. In particular, when r = n − µn − s , for u ∈ X , we have Z R n Z R n | u ( x ) | ∗ µ | u ( y ) | ∗ µ | x − y | µ dxdy ! / ∗ µ ≤ C ( n, µ ) ∗ µ k u k ∗ s , where C ( n, µ ) is a suitable constant and 2 ∗ s = 2 n/ ( n − s ). Let us define S H := inf u ∈ H s ( R n ) \{ } R R n R R n | u ( x ) − u ( y ) | | x − y | n +2 s dxdy (cid:18)R R n R R n | u ( x ) | ∗ µ | u ( y ) | ∗ µ | x − y | µ dxdy (cid:19) ∗ µ , which is achieved if and only if u is of the form C (cid:18) tt + | x − x | (cid:19) ( n − s ) / , for x ∈ R n , for some x ∈ R n , C > t > S H = S (cid:0) C ( n, µ ) (cid:1) / ∗ µ . (2.2) Lemma 2.3 ([29, Lemma 2.2]) Define S H (Ω) := inf u ∈ X \{ } R Q | u ( x ) − u ( y ) | | x − y | n +2 s dxdy (cid:18)R Ω R Ω | u ( x ) | ∗ µ | u ( y ) | ∗ µ | x − y | µ dxdy (cid:19) / ∗ µ . Then S H = S H (Ω) and S H (Ω) is never achieved except when Ω = R n . Lemma 2.4 ([29, Lemma 3.4]) Let k u k NL = Z Ω Z Ω | u ( x ) | ∗ µ | u ( y ) | ∗ µ | x − y | µ dxdy ! / . ∗ µ , then k · k NL defines a norm on Y := { u : Ω → Ω : u is measurable, k u k NL < ∞} . Notations:
For simplicity, for p = 2 and p = p , we will use the following notations K i ( u, v ) = Z Q | u ( x ) − u ( y ) | p i − ( u ( x ) − u ( y ))( v ( x ) − v ( y )) | x − y | n + p i s i dxdy, for all u, v ∈ X i , A r ( u, v ) = Z Ω Z Ω | u ( y ) | r | u ( x ) | r − u ( x ) v ( x ) | x − y | µ dxdy for all u, v ∈ X and ˆ r = n n − µ r . Definition 2.5
A function u ∈ X is said to be a solution of problem ( P λ ) , if for all v ∈ X K ( u, v ) + β K ( u, v ) − λ Z Ω a ( x ) | u | q − uv dx − A r ( u, v ) = 0 . The Euler functional I λ : X → R associated to the problem ( P λ ) is defined as I λ ( u ) = 12 k u k X + βp k u k pX − λq Z Ω a ( x ) | u | q dx − r Z Ω Z Ω | u ( x ) | r | u ( y ) | r | x − y | µ dxdy. Now we present a regularity result, namely the L ∞ bound, for weak solutions of problem( P λ ). Proof of Theorem 1.1 : Let u be a nonnegative solution of ( P λ ). For ϑ > T > φ ( t ) = φ T,ϑ ( t ) = , if t ≤ t ϑ , if 0 < t < TϑT ϑ − t − ( ϑ − T ϑ , if t ≥ T. Then, φ ′ ( t ) = , if t ≤ ϑt ϑ − , if 0 < t < TϑT ϑ − , if t ≥ T. As u ( x ) ≥
0, it is easy to observe that φ ( u ) ≥ , φ ′ ( u ) ≥ uφ ′ ( u ) ≤ ϑφ ( u ) . Since φ is Lipschitz, we have φ ( u ) ∈ X and k φ ( u ) k X = (cid:18)Z Q | φ ( u ( x )) − φ ( u ( y )) | | x − y | n +2 s dxdy (cid:19) / ≤ K k u k X . Moreover, we have S k φ ( u ) k ∗ s ≤ k φ ( u ) k X = Z Ω φ ( u )( − ∆) s φ ( u ) . (2.3)Since φ is convex, we observe that φ ( u ) φ ′ ( u ) ∈ X and Z Ω φ ( u )( − ∆) s φ ( u ) ≤ Z Ω φ ( u ) φ ′ ( u )( − ∆) s u. (2.4)Define g : R → R as g ( t ) = φ ( t ) φ ′ ( t ), then g is increasing, and set G ( t ) = R t g ′ ( s ) p ds for t ∈ R . Using the inequality (see [9, Lemma A.2]) | a − b | p − ( a − b ) (cid:0) g ( a ) − g ( b ) (cid:1) ≥ | G ( a ) − G ( b ) | p , for a, b ∈ R , we obtain Z Q | u ( x ) − u ( y ) | p − (cid:0) u ( x ) − u ( y ) (cid:1)(cid:0) φ ( u ( x )) φ ′ ( u ( x )) − φ ( u ( y )) φ ′ ( u ( y )) (cid:1) | x − y | n + ps dxdy ≥ Z Q | G ( u ( x )) − G ( u ( y )) | p | x − y | n + ps dxdy ≥ . (2.5)Noting the fact that 1 < q <
2, that is, 0 < q − <
1, there exists a constant
C > λ | a ( x ) | u | q − u | ≤ C (1 + | u | ). Thus, from (2.3), (2.4) and (2.5), we deduce that S k φ ( u ) k ∗ s ≤ k φ ( u ) k X = Z Ω φ ( u )( − ∆) s φ ( u ) ≤ Z Ω φ ( u ) φ ′ ( u )( − ∆) s u + Z Q | u ( x ) − u ( y ) | p − (cid:0) u ( x ) − u ( y ) (cid:1)(cid:0) φ ( u ( x )) φ ′ ( u ( x )) − φ ( u ( y )) φ ′ ( u ( y )) (cid:1) | x − y | n + ps dxdy = λ Z Ω a ( x ) | u | q − uφ ( u ) φ ′ ( u ) dx + Z Ω Z Ω | u ( y ) | r | u ( x ) | r − u ( x ) φ ( u ) φ ′ ( u ) | x − y | µ dxdy ≤ C Z Ω (1 + u ) φ ( u ) φ ′ ( u ) + Z Ω Z Ω | u ( y ) | r | u ( x ) | r − u ( x ) φ ( u ) φ ′ ( u ) | x − y | µ dxdy. Using the relation uφ ′ ( u ) ≤ ϑφ ( u ) and φ ′ ( u ) ≤ ϑ (1 + φ ( u )), we get S k φ ( u ) k ∗ s ≤ Cϑ Z Ω (cid:0) φ ( u ) + φ ( u ) (cid:1) + ϑ Z Ω Z Ω | u ( y ) | r | u ( x ) | r − φ ( u ) | x − y | µ dxdy. (2.6)Now, using H¨older inequality, we see that Z Ω φ ( u ) ≤ (cid:18)Z Ω φ ( u ) (cid:19) / | Ω | / ≤ Z Ω φ ( u ) + 12 | Ω | . (2.7)Next, to estimate the second term in (2.6), we follow the approach similar to [39, Proof ofTheorem 1(2)]. Using Theorem 2.2, we have Z Ω Z Ω | u ( y ) | r | u ( x ) | r − φ ( u ) | x − y | µ dxdy ≤ C ( n, µ ) k| u | r k n n − µ k| u | r − φ ( u ) k n n − µ . (2.8)By the embedding results of X , we get u ∈ L ∗ s , so we can assume k u k ∗ s ≤ C . EmployingH¨older inequality, we get k| u | r k n n − µ ≤ C . Next, for m >
0, we deduce that (cid:18)Z Ω (cid:0) | u | r − φ ( u ) (cid:1) n n − µ (cid:19) n − µ n = (cid:20) Z { u
0. Bymeans of (2.10), it is easy to notice that k u k ∗ s ϑ < ∞ . Thus, from (2.11), we get k u k ∗ s ϑ ≤ ( Cϑ ) ϑ k u k ϑ ∗ s ∗ µ = C ϑ ϑ ϑ k u k ∗ s ϑ < ∞ , and k u k ∗ s ϑ ≤ C ϑ + ϑ ϑ ϑ ϑ ϑ k u k ∗ s ϑ , similarly we obtain k u k ∗ s ϑ k ≤ C P km =1 12 ϑm k Y m =1 ( ϑ ϑm m ) k u k ∗ s ϑ . (2.12)By ratio test we can see that X m ∈ N ϑ m < ∞ . Let z k = Q km =1 ϑ ϑm m . Again by using ratiotest, we get ln z k = P km =1 ln ϑ m ϑ m is convergent. Hence there exists a positive constant A suchthat C P km =1 12 ϑm Q km =1 ( ϑ ϑm m ) ≤ A for all k ≥
0. Therefore, (2.12) implies k u k ∗ s ϑ k ≤ A k u k ∗ s ϑ < ∞ . (2.13)We claim that u ∈ L ∞ (Ω). Suppose not, then there exists ε > M ⊂ Ω with | M | > u ( x ) ≥ A k u k ∗ s ϑ + ε a.e. x ∈ M. k u k ∗ s ϑ k ≥ (cid:18)Z M | u ( x ) | ∗ s ϑ k (cid:19) ∗ s ϑk ≥ ( A k u k ∗ s ϑ + ε ) | M | ∗ s ϑk , this implies that lim inf k →∞ k u k ∗ s ϑ k ≥ ( A k u k ∗ s ϑ + ε ) , which is a contradiction to (2.13). Hence, u ∈ L ∞ (Ω). (cid:3) Due to the fact that 1 < r , we see that I λ ( tu ) → −∞ , as t → ∞ for u ( ∈ X . Hence,the functional I λ is not bounded below on X . Therefore, it is necessary to restrict I λ to aproper subset of X on which it is bounded below. For this reason, we consider the Nehariset M λ associated to ( P λ ), which is defined as M λ = { u ∈ X \ { } : hI ′ λ ( u ) , u i = 0 } , where h , i is the duality between X and its dual space. Obviously, M λ contains all thesolution of ( P λ ). To study the critical points of the functional I λ , we define the fibering mapsassociated to it. For u ∈ X , define ϕ u : R + → R as ϕ u ( t ) = I λ ( tu ), that is ϕ u ( t ) = t k u k X + β t p p k u k pX − λt q q Z Ω a ( x ) | u | q dx − t r r Z Ω Z Ω | u ( x ) | r | u ( y ) | r | x − y | µ dxdy,ϕ ′ u ( t ) = t k u k X + βt p − k u k pX − λt q − Z Ω a ( x ) | u | q dx − t r − Z Ω Z Ω | u ( x ) | r | u ( y ) | r | x − y | µ dxdy, (3.1) ϕ ′′ u ( t ) = k u k X + β ( p − t p − k u k pX − ( q − λt q − Z Ω a ( x ) | u | q dx − (2 r − t r − Z Ω Z Ω | u ( x ) | r | u ( y ) | r | x − y | µ dxdy. (3.2)It is clear that tu ∈ M λ if and only if ϕ ′ u ( t ) = 0 and in particular, u ∈ M λ if and only if ϕ ′ u (1) = 0. Hence, it is natural to split M λ into three parts corresponding to local minima,local maxima and points of inflection, namely M λ := (cid:8) u ∈ M λ : ϕ ′′ u (1) = 0 (cid:9) , and M ± λ := (cid:8) u ∈ M λ : ϕ ′′ u (1) ≷ (cid:9) . Define σ λ := inf { I λ ( u ) | u ∈ M λ } and σ ± λ := inf { I λ ( u ) | u ∈ M ± λ } . Lemma 3.1 I λ is coercive and bounded below on M λ . Proof.
Proof follows using H¨older inequality and Sobolev embedding results. (cid:3) Lemma 3.2
There exists λ > such that for all λ ∈ (0 , λ ) , we have M λ = ∅ . Proof.
We distinguish the following cases:
Case 1: u ∈ M λ such that Z Ω a ( x ) | u | q dx = 0 . From (3.1), we have k u k X + β k u k pX − Z Ω Z Ω | u ( x ) | r | u ( y ) | r | x − y | µ dxdy = 0 . Therefore, φ ′′ u (1) = k u k X + β ( p − k u k pX − (2 r − Z Ω Z Ω | u ( x ) | r | u ( y ) | r | x − y | µ dxdy = (2 − r ) k u k X + β ( p − r ) k u k pX < , which implies u / ∈ M λ . Case 2: u ∈ M λ such that Z Ω a ( x ) | u | q dx = 0 . If u ∈ M λ , then from (3.1) and (3.2), we have(2 − q ) k u k X + β ( p − q ) k u k pX = (2 r − q ) Z Ω Z Ω | u ( x ) | r | u ( y ) | r | x − y | µ dxdy and (3.3)(2 r − k u k X + β (2 r − p ) k u k pX = λ (2 r − q ) Z Ω a ( x ) | u | q dx. (3.4)Define E λ : M λ → R as E λ ( u ) = (2 r − k u k X + β (2 r − p ) k u k pX (2 r − q ) − λ Z Ω a ( x ) | u | q dx, then from (3.4), E λ ( u ) = 0 for all u ∈ M λ . Additionally, using H¨older inequality, we have E λ ( u ) ≥ k u k qX (cid:20)(cid:18) r − r − q (cid:19) k u k (2 − q ) X − λ k a k dd − q S − q d (cid:21) , (3.5)Now from (3.3) and Theorem 2.2, we get k u k X ≥ (cid:18) (2 − q ) S r ˆ r (2 r − q ) C ( n, µ ) (cid:19) r − . Using this in (3.5), we obtain E λ ( u ) ≥ k u k qX (cid:18) r − r − q (cid:19) (cid:18) (2 − q ) S r ˆ r (2 r − q ) C ( n, µ ) (cid:19) − q r − − λ k a k dd − q S − q d ! . Set λ := (2 r − S q d (2 r − q ) k a k dd − q (cid:18) (2 − q ) S r ˆ r (2 r − q ) C ( n, µ ) (cid:19) − q r − > , (3.6)then from (3.5), for λ ∈ (0 , λ ) we get E λ ( u ) > , for all u ∈ M λ , which is a contradiction.Therefore, M λ = ∅ for all λ ∈ (0 , λ ). (cid:3) ψ u : R + −→ R by ψ u ( t ) = t − q k u k X + βt p − q k u k pX − t r − q Z Ω Z Ω | u ( x ) | r | u ( y ) | r | x − y | µ dxdy, then ψ ′ u ( t ) = (2 − q ) t − q k u k X + β ( p − q ) t p − q − k u k pX − (2 r − q ) t r − q − Z Ω Z Ω | u ( x ) | r | u ( y ) | r | x − y | µ dxdy. Then trivially, tu ∈ M λ if and only if t is a solution of ψ u ( t ) = λ R Ω a ( x ) | u | q dx and if tu ∈ M λ ,then ϕ ′′ tu (1) = t q − ψ ′ u ( t ). Moreover, we see ψ u ( t ) → −∞ as t → ∞ , ψ u ( t ) > t smallenough and ψ ′ u ( t ) < t large enough. Now based on the sign of Z Ω a ( x ) | u | q dx , we willstudy the fibering map ϕ u . Lemma 3.3
Let u ( ∈ X and λ ∈ (0 , λ ) .(i) If R Ω a ( x ) | u | q dx > , then there exist unique t < t max < t such that t u ∈ M + λ and t u ∈ M − λ . Moreover, I λ ( t u ) = min ≤ t ≤ t I λ ( tu ) and I λ ( t u ) = max t ≥ t max I λ ( tu ) .(ii) If R Ω a ( x ) | u | q dx < , then there exists unique t > such that t u ∈ M − λ . Proof. ( i ) Let u ∈ X such that R Ω a ( x ) | u | q dx >
0. We claim that there exists unique t max > ψ ′ u ( t max ) = 0. To prove this, it is sufficient to show the existence of unique t max such that F u ( t max ) = β ( p − q ) k u k pX , where F u ( t ) := (2 r − q ) t r − p Z Ω Z Ω | u ( x ) | r | u ( y ) | r | x − y | µ dxdy − (2 − q ) t − p k u k X . By the fact that p < < r , we see that F u ( t ) < t small enough, F u ( t ) → ∞ as t → ∞ . Hence, there exists unique ˆ t > F u (ˆ t ) = 0. Moreover, thereexists unique ˜ t > F ′ u (˜ t ) = 0. Therefore, there exists unique t max > ˆ t > F u ( t max ) = β ( p − q ) k u k pX p . Using these, we conclude that ψ u is increasing in (0 , t max ),decreasing in ( t max , ∞ ). As a consequence,(2 − q ) t max k u k X ≤ (2 − q ) t max k u k X + β ( p − q ) t pmax k u k pX = t rmax (2 r − q ) Z Ω Z Ω | u ( x ) | r | u ( y ) | r | x − y | µ dxdy ≤ (2 r − q ) t rmax C ( n, µ ) S − r ˆ r k u k rX . Define T := 1 k u k X (cid:18) (2 − q ) S r ˆ r (2 r − q ) C ( n, µ ) (cid:19) / (2 r − ≤ t max then, ψ u ( t max ) ≥ ψ u ( T ) ≥ T − q k u k X − T r − q C ( n, µ ) S − r ˆ r k u k rX = k u k qX (cid:18) r − r − q (cid:19) (cid:18) (2 − q ) S r ˆ r (2 r − q ) C ( n, µ ) (cid:19) − q r − ≥ . Since λ < λ , then there exist t < t max and t > t max such that ψ u ( t ) = ψ u ( t ) = λ Z Ω a ( x ) | u | q dx . That is, t u, t u ∈ M λ . Also ψ ′ u ( t ) > ψ ′ u ( t ) > t u ∈ M + λ and4 t u ∈ M − λ . Since ϕ ′ u ( t ) = t q ( ψ u ( t ) − λ R Ω a ( x ) | u | q dx ), ϕ ′ u ( t ) < t ∈ [0 , t ) and ϕ ′ u ( t ) > t ∈ ( t , t ). So, I λ ( t u ) = min ≤ t ≤ t I λ ( tu ) . Also, ϕ ′ u ( t ) > t ∈ [ t , t ) , ϕ ′ u ( t ) = 0and ϕ ′ u ( t ) < t ∈ ( t , ∞ ) implies I λ ( t u ) = max t ≥ t max I λ ( tu ) . ( ii ) Let u ∈ X be such that R Ω a ( x ) | u | q dx <
0. From ( i ), we have ψ u is increasing in (0 , t max ),decreasing in ( t max , ∞ ) and ψ ′ u ( t max ) = 0. Since λ R Ω a ( x ) | u | q dx < ψ u ( t max ) >
0, thereexists unique t > ψ u ( t ) = λ R Ω a ( x ) | u | q dx and ψ ′ u ( t ) <
0, which implies t u ∈ M − λ , that is t u is a local maximum. (cid:3) Lemma 3.4
Let λ be defined as in (3.6) , then the following holds.(i) There exists a constant C > such that σ λ ≤ σ + λ ≤ − (2 − q )( r − qr C < .(ii) inf {k u k : u ∈ M − λ } > . Proof.
To prove ( i ), let u ∈ X such that R Ω a ( x ) | u | q dx >
0. It implies there exists t = t ( u ) > t u ∈ M + λ that is, ϕ ′′ t u (1) >
0. Therefore, we have I λ ( t u ) ≤ (cid:18) − qq (cid:19) (cid:18) r − (cid:19) k t u k X + β (cid:18) p − qq (cid:19) (cid:18) r − p (cid:19) k t u k pX ≤ (cid:18) − qq (cid:19) (cid:18) r − (cid:19) k t u k X ≤ − (2 − q )( r − qr C , where C = k t u k X . This implies σ + λ ≤ − (2 − q )( r − qr C < ii ), let u ∈ M − λ , then ϕ ′′ u (1) < − q ) k u k X ≤ (2 − q ) k u k X + β ( p − q ) k u k pX < (2 r − q ) Z Ω Z Ω | u ( x ) | r | u ( y ) | r | x − y | µ dxdy ≤ (2 r − q ) C ( n, µ ) S − r ˆ r k u k rX , that is, k u k X > (cid:18) (2 − q ) S r ˆ r (2 r − q ) C ( n, µ ) (cid:19) / (2 r − . Hence the result follows. (cid:3)
Lemma 3.5
Let λ ∈ (0 , λ ) , and z ∈ M λ , then there exists ǫ > and a differentiable function ξ : B (0 , ǫ ) ⊆ X → R + such that ξ (0) = 1 , the function ξ ( w )( z − w ) ∈ M λ and h ξ ′ (0) , w i = 2 K ( z, w ) + pβ K ( z, w ) − qλ R Ω a ( x ) | z | q − zwdx − r A r ( z, w )(2 − q ) k z k X + ( p − q ) k z k pX − (2 r − q ) Z Ω Z Ω | z ( x ) | r | z ( y ) | r | x − y | µ dxdy , for all w ∈ X . Proof.
For z ∈ M λ , define a function H z : R × X → R given by H z ( t, w ) := hI ′ λ ( t ( z − w )) , ( t ( z − w )) i = t k z − w k X + βt p k z − w k pX − t q λ Z Ω a ( x ) | z − w | q dx − t r Z Ω Z Ω | ( z − w )( x ) | r | ( z − w )( y ) | r | x − y | µ dxdy. Then H z (1 ,
0) = hI ′ λ ( z ) , z i = 0 and by Lemma 3.2, we have ∂∂t H z (1 , = 0 . Therefore, byimplicit function theorem result follows (for details see [21, Lemma 3.5]). (cid:3)
Proposition 3.6
Let λ ∈ (0 , λ ) , then there exists a minimizing sequence { u k } ⊂ M λ suchthat I λ ( u k ) = σ λ + o k (1) and I ′ λ ( u k ) = o k (1) . Proof.
Using Lemma 3.1 and Ekeland variational principle [16], there exists a minimizingsequence { u k } ⊂ N λ such that I λ ( u k ) < σ λ + 1 k , and (3.7) I λ ( u k ) < I λ ( v ) + 1 k k v − u k k X for each v ∈ M λ . By taking k large, using equation (3.7) and Lemma 3.4, we deduce that I λ ( u k ) = (cid:18) − r (cid:19) k u k k X + β (cid:18) p − r (cid:19) k u k k pX − λ (cid:18) q − r (cid:19) Z Ω a ( x ) | u k | q dx< σ λ + 1 k < σ + λ < , which gives us u k k large enough. Now, using H¨older’s inequality, we get qr ( − σ + λ ) S q d λ (2 r − q ) k a k dd − q /q ≤ k u k k X ≤ λ (2 r − q ) k a k dd − q q ( r − S q d / (2 − q ) . Then, proof of the result I ′ λ ( u k ) →
0, as k → ∞ follows exactly on the same line of [21,Proposition 4.1]. (cid:3) r < ∗ µ ) In this section, we prove the existence of at least two non-negative solutions of problem ( P λ ). Lemma 4.1
The functional I λ satisfies ( P S ) c condition for all c ∈ R . That is, if { u k } ⊂ X satisfies I λ ( u k ) = c + o k (1) and I ′ λ ( u k ) = o k (1) in X ′ , (4.1) then { u k } has a convergent subsequence in X . Proof.
Let { u k } ⊂ X satisfies (4.1). Then it is easy to verify that sequence { u k } is boundedin X . So up to subsequence u k ⇀ u weakly in X , u k → u strongly in L ν (Ω) , ≤ ν < ∗ s and u k ( x ) → u ( x ) a.e. in Ω. Since hI ′ λ ( u k ) − I ′ λ ( u ) , ( u k − u ) i → k → ∞ , we have o k (1) = hI ′ λ ( u k ) − I ′ λ ( u ) , u k − u i = K ( u k , u k − u ) − K ( u , u k − u ) + β ( K ( u k , u k − u ) − K ( u , u k − u )) − λ Z Ω a ( x ) (cid:0) | u k ( x ) | q − u k ( x ) − | u ( x ) | q − u ( x ) (cid:1) ( u k ( x ) − u ( x )) dx − (cid:0) A r ( u k , u k − u ) − A r ( u , u k − u ) (cid:1) . (4.2)Using H¨older inequality and the fact that d < ∗ s , we obtain Z Ω a ( x ) | u k | q − u k ( u k − u ) dx ≤ k a k dd − q k u k k q − d k u k − u k d → k → ∞ . Again, using H¨older inequality, Hardy-Littlewood-Sobolev inequality with the fact ˆ r = nr n − µ < ∗ s , we deduce that A r ( u k , u k − u ) = Z Ω Z Ω | u k ( x ) | r | u k ( y ) | r − u k ( y )( u k ( y ) − u ( y )) | x − y | µ dxdy ≤ C ( n, µ ) k u k k r − r k u k − u k ˆ r → , as k → ∞ . Now, we claim that the sequence { u k } has a convergent subsequence.Using the definition of K , it is easy to see that K ( u k , u k − u ) −K ( u , u k − u ) = k u k − u k X .Furthermore, since we know that | a − b | η ≤ C η (cid:0) ( | a | η − a − | b | η − b )( a − b ) (cid:1) η ( | a | η + | b | η ) − η for a, b ∈ R n , < η ≤ , where C η is some positive constant depending on η . Set a = u k ( x ) − u k ( y ), b = u ( x ) − u ( y )and then using H¨older inequality, we deduce that k u k − u λ k pX ≤ C ( K ( u k , u k − u ) − K ( u , u k − u )) p (cid:18)Z Q | u k ( x ) − u k ( y ) | p + | u ( x ) − u ( y ) | p | x − y | n + ps (cid:19) − p and boundedness of { u k } in X (follows from boundedness in X and Lemma 2.1), implies k u k − u k pX ≤ C ( K ( u k , u k − u ) − K ( u , u k − u )) p . Collecting all these informations in (4.2), we obtain o k (1) = hI ′ λ ( u k ) − I ′ λ ( u ) , u k − u i ≥ C (cid:0) k u k − u k X + β k u k − u k X (cid:1) . Hence, it concludes proof of the claim. (cid:3) Proof of Theorem 1.2 :
Using Proposition 3.6 and Lemma 4.1, there exist minimizingsequences { u k } ∈ M + λ , { v k } ∈ M − λ and u and v ∈ X such that u k → u and v k → v strongly in X for λ ∈ (0 , λ ). Therefore, for λ ∈ (0 , λ ), u , v are weak solutions of problem( P λ ). By means of Lemma 3.4, we conclude that u , v
0, hence u ∈ M + λ and v ∈ M − λ .Moreover, I λ ( u ) = σ + λ and I λ ( v ) = σ − λ . Since M + λ ∩ M − λ = ∅ , therefore u and v aredistinct solutions.Now we prove non-negativity of u . If u ≥
0, then we have a non negative solution of problem( P λ ), which is also a minimizer for I λ in M + λ , otherwise we have | u | 6≡
0, hence by fiberingmap analysis we get unique t > t u ∈ M + λ . We note that ψ | u | (1) ≤ ψ u (1) = λ Z Ω a ( x ) | u | q = ψ | u | ( t ) ≤ ψ u ( t ) and 0 < ψ ′ u (1), because of the fact u ∈ M + λ , whichimplies t ≥
1. Thus, σ + λ ≤ ϕ | u | ( t ) ≤ ϕ | u | (1) ≤ ϕ u (1) = σ + λ . Hence, I λ ( t | u | ) = ϕ | u | ( t ) = σ + λ and t | u | ∈ M + λ that is, t | u | is a nonnegative solutionof problem ( P λ ) in M + λ . (cid:3) r = 2 ∗ µ ) In this section we assume the function a ( x ) is continuous in Ω and a + ( x ) = max { a ( x ) , } 6≡
0. Then without loss of generality we may assume there exists δ > m a :=inf x ∈ B δ a ( x ) > Theorem 5.1
Let { u k } ⊂ M λ be a ( P S ) c sequence for I λ such that u k ⇀ u weakly in X ,then I ′ λ ( u ) = 0 . Moreover, there exists a positive constant C = C ( q, s , n, S, | Ω | ) such that I ′ λ ( u ) ≥ − C λ − q , where C = (cid:18) (2 . ∗ µ − q )(2 − q )2 . . ∗ µ q (cid:19) (cid:18) . ∗ µ − q . ∗ µ − (cid:19) q − q S − q − q k a k − q ∞ | Ω | ∗ s − q )2 ∗ s − q ) . (5.1) Proof.
Since u k ⇀ u in X , it implies { u k } is a bounded sequence in X , and up tosubsequence, u k → u in L ν (Ω) , ≤ ν < ∗ s and u k → u a.e. in Ω. Now from the proof of[21, Theorem 4.3] it follows that K i ( u k , v ) → K i ( u, v ) for i = 1 , , and Z Ω a ( x ) (cid:0) | u k ( x ) | q − u k ( x ) − | u ( x ) | q − u ( x ) (cid:1) v ( x ) dx → k → ∞ , for all v ∈ X . From the continuous embedding of X into L ∗ s , we get u k ⇀u weakly in L ∗ s , as k → ∞ . Therefore, | u k | ∗ µ ⇀ | u | ∗ µ in L ∗ s / ∗ µ (Ω) and we know that Rieszpotential defines a continuous linear map from L ∗ s / ∗ µ (Ω) to L nµ (Ω), thus we have | x | − µ ∗ | u k | ∗ µ ⇀ | x | − µ ∗ | u | ∗ µ in L nµ (Ω) . | u k | ∗ µ − u k ⇀ | u | ∗ µ − u in L ∗ s / (2 ∗ µ − (Ω). Combining all these facts, weobtain Z Ω Z Ω | u k ( x ) | ∗ µ | u k ( y ) | ∗ µ − u k ( y ) φ ( y ) | x − y | µ dxdy → Z Ω Z Ω | u ( x ) | ∗ µ | u ( y ) | ∗ µ − u ( y ) φ ( y ) | x − y | µ dxdy Therefore, we have hI ′ λ ( u k ) − I ′ λ ( u ) , φ i = K ( u k , φ ) − K ( u, φ ) + β ( K ( u k , φ ) − K ( u, φ )) − Z Ω a ( x ) (cid:0) | u k ( x ) | q − u k ( x ) − | u ( x ) | q − u ( x ) (cid:1) φ ( x ) dx − (cid:0) A ∗ µ ( u k , φ ) − A ∗ µ ( u, φ ) (cid:1) → φ ∈ X . This implies I ′ λ ( u ) = 0. In particular, hI ′ λ ( u ) , u i = 0, that is I λ ( u ) = (cid:18) − . ∗ µ (cid:19) k u k X + β (cid:18) p − . ∗ µ (cid:19) k u k pX − λ (cid:18) q − . ∗ µ (cid:19) Z Ω a ( x ) | u | q dx ≥ (cid:18) − . ∗ µ (cid:19) k u k X − λ (cid:18) q − . ∗ µ (cid:19) Z Ω a ( x ) | u | q dx. (5.2)By H¨older inequality, Sobolev embeddings and Young inequality, we obtain λ Z Ω a ( x ) | u | q dx ≤ λ k a k ∞ S − q | Ω | ∗ s − q ∗ s k u k qX = q (cid:18) − . ∗ µ (cid:19) (cid:18) q − . ∗ µ (cid:19) − ! q k u k qX λ q (cid:18) − . ∗ µ (cid:19) (cid:18) q − . ∗ µ (cid:19) − ! − q k a k ∞ | Ω | ∗ s − q ∗ s S − q ≤ (cid:18) − . ∗ µ (cid:19) (cid:18) q − . ∗ µ (cid:19) − k u k X + (cid:18) − q (cid:19) (cid:18) . ∗ µ − q . ∗ µ − (cid:19) q − q S − q − q k a k − q ∞ | Ω | ∗ s − q )(2 − q )2 ∗ s λ − q , (5.3)Therefore, result follows from equations (5.2) and (5.3) with C = (cid:18) q − . ∗ µ (cid:19) (cid:18) − q (cid:19) (cid:18) . ∗ µ − q . ∗ µ − (cid:19) q − q S − q − q k a k − q ∞ | Ω | ∗ s − q )(2 − q )2 ∗ s . (cid:3) Lemma 5.2 (Palais-Smale range). I λ satisfies the ( P S ) c condition with c ∈ ( −∞ , c ∞ ) ,where c ∞ := (cid:18) n − µ + 2 s n − µ ) (cid:19) S n − µn − µ +2 s ( C ( n, µ )) n − s n − µ +2 s − C λ − q and C is the positive constant defined in (5.1) . Proof.
Let { u k } be a ( P S ) c sequence of I λ in X . Then we have12 k u k k X + βp k u k k pX − λq Z Ω a ( x ) | u k | q dx − . ∗ µ k u k k . ∗ µ NL = c + o k (1) (5.4)and k u k k X + β k u k k pX − λ Z Ω a ( x ) | u k | q dx − k u k − u k . ∗ µ NL = o k (1) . (5.5)As an easy consequence of this, we get { u k } is a bounded sequence in X . Therefore, up toa subsequence, u k ⇀ u in X , for some u ∈ X and by Lemma 5.1, we see that u is a criticalpoint of I λ . Claim: u k → u strongly in X .Since, u k → u strongly in L ν (Ω) for 1 ≤ ν < ∗ s , it implies R Ω a ( x ) | u k | q dx → R Ω a ( x ) | u | q dx .Also, by Brezis -Leib Lemma, we have k u k k p i X i = k u k − u k p i X i + k u k p i X i + o k (1) , ≤ i ≤ , p = 2 , p = p and k u k k . ∗ µ NL = k u k . ∗ µ NL + k u k − u k . ∗ µ NL + o k (1) . (5.6)Therefore, by using equations (5.4), (5.5) and (5.6), we get12 k u k − u k X + βp k u k − u k pX − . ∗ µ k u k − u k . ∗ µ NL = c − I λ ( u ) + o k (1) k u k − u k X + β k u k − u k pX − k u k − u k . ∗ µ NL = o k (1) . (5.7)Hence, let k u k − u k X + β k u k − u k pX → l and k u k − u k . ∗ µ NL → l, as k → ∞ . If l = 0, thenclaim is proved. So, we assume l >
0, then l / ∗ µ = (cid:0) lim k →∞ k u k − u k . ∗ µ NL (cid:1) / ∗ µ ≤ ( C ( n, µ )) / ∗ µ lim k →∞ (cid:0) S − k u k − u k X (cid:1) ≤ ( C ( n, µ )) ∗ µ S − l. This implies l ≥ C ( n, µ ) − ∗ µ − S ∗ µ ∗ µ − , that is, l ≥ S n − µn − µ +2 s ( C ( n,µ )) n − s n − µ +2 s . Now, from (5.7), we have c − I λ ( u ) ≥ (cid:0) k u k − u k X + β k u k − u k pX (cid:1) − . ∗ µ k u k − u k . ∗ µ NL = (cid:18) n − µ + 2 s n − µ ) (cid:19) l. Therefore, with the help of Theorem 5.1, we get c ≥ (cid:18) n − µ + 2 s n − µ ) (cid:19) l + I λ ( u ) ≥ (cid:18) n − µ + 2 s n − µ ) (cid:19) S n − µn − µ +2 s ( C ( n, µ )) n − s n − µ +2 s − C λ − q = c ∞ , which is a contradiction. (cid:3) Proof of Theorem 1.3 (i): Let γ > λ ∈ (0 , γ ), c ∞ > { γ , λ } >
0. By Proposition 3.6, for all λ ∈ (0 , Λ), there exists a minimizingsequence u k ∈ M λ such that I λ ( u k ) = σ λ + o k (1) and I ′ λ ( u k ) = o k (1). Now using the fact σ λ ≤ σ + λ < u λ ∈ X such that u k → u λ in X .By Theorem 5.1 and due to the fact σ λ <
0, we get I ′ λ ( u λ ) = 0 and R Ω a ( x ) | u λ | q > u λ
0, thus u λ ∈ M λ . Next, we will prove that u λ ∈ M + λ . Suppose on the contrary u λ ∈ M − λ , then by the fibering map analysis there exist t < t = 1 such that t u λ ∈ M + λ and t u λ ∈ M − λ . Since φ u λ is increasing in [ t , t ), it implies σ λ ≤ I λ ( t u λ ) < I λ ( tu λ ) ≤ I λ ( u λ ) = σ λ for t ∈ ( t , u λ ∈ M + λ and σ λ = I λ ( u λ ) = σ + λ . Moreover,by using same assertions and arguments as in proof of Theorem 1.2, we obtain that u λ isnonnegative. (cid:3) Consider the family of functions U ǫ ( x ) = ǫ − ( n − s u ∗ ( xǫ ), where u ∗ ( x ) = ¯ u (cid:16) xS s (cid:17) , ¯ u ( x ) = ˜ u ( x ) k ˜ u k ∗ s and ˜ u ( x ) = d ( b + | x | ) − n − s , d ∈ R \ { } and b >
0. ([38]) U ǫ satisfies( − ∆) s U ǫ = | U ǫ | ∗ s − U ǫ , in R n , and k U ǫ k X = k U ǫ k ∗ s ∗ s = S n s . (5.8)Let η ∈ C ∞ c (Ω) be such that η = 1 in B δ (0), η = 0 in B c δ (0), and 0 ≤ η ≤ u ǫ = ηU ǫ , then we have the following estimates Lemma 5.3 (see [19, 38]) The following hold true(i) k u ǫ k X ≤ S n s + O ( ǫ n − s ) (ii) k u ǫ k ∗ s ∗ s = S n s + O ( ǫ n ) (iii) k u ǫ k NL ≤ C ( n, µ ) n . ∗ µs S n − s s H + O ( ǫ n ) (iv) k u ǫ k NL ≥ (cid:0) C ( n, µ ) n s S n − µ s H − O ( ǫ n ) (cid:1) ∗ µ . Now, we will estimate the fractional p -Laplacian norm of the family of functions { u ǫ } . Lemma 5.4
Let < p < , then there exists C > such that k u ǫ k pX ≤ C ǫ n − s p if < p < n/ ( n − s ) ǫ − p n if n/ ( n − s ) ≤ p < . (5.9) Proof.
Following the ideas of [38, Proposition 21], we define the following sets D := { ( x, y ) ∈ R n : x ∈ B δ , y ∈ B cδ , | x − y | > δ/ } and E := { ( x, y ) ∈ R n : x ∈ B δ , y ∈ B cδ , | x − y | ≤ δ/ } . u ǫ , it is clear that k u ǫ k pX = Z R n | u ǫ ( x ) − u ǫ ( y ) | p | x − y | n + ps dxdy = Z B δ Z B δ | U ǫ ( x ) − U ǫ ( y ) | p | x − y | n + ps dxdy + Z D +2 Z E + Z B cδ Z B cδ ! | u ǫ ( x ) − u ǫ ( y ) | p | x − y | n + ps dxdy. (5.10)By [38, Claim 10 of Proposition 21], for x, y ∈ B cδ , we have | u ǫ ( x ) − u ǫ ( y ) | ≤ Cǫ ( n − s ) / min { , | x − y |} . Therefore, I = Z B cδ Z B cδ | u ǫ ( x ) − u ǫ ( y ) | p | x − y | n + ps dxdy ≤ Cǫ n − s p Z B δ Z R n min { , | x − y | p }| x − y | n + ps dxdy = Cǫ n − s p . Next, by [38, (4.21)], for x ∈ B δ , y ∈ B cδ with | x − y | ≤ δ/
2, we have | u ǫ ( x ) − u ǫ ( y ) | ≤ Cǫ ( n − s ) / | x − y | . Then, proceeding similarly, we get I = Z E | u ǫ ( x ) − u ǫ ( y ) | p | x − y | n + ps dxdy ≤ Cǫ n − s p . To evaluate I := R D | u ǫ ( x ) − u ǫ ( y ) | p | x − y | n + ps dxdy , we first note that due to the fact 1 < p <
2, thereexists A p > | u ǫ ( x ) − u ǫ ( y ) | p ≤ | U ǫ ( x ) − U ǫ ( y ) | p + | U ǫ ( y ) − u ǫ ( y ) | p + A p | U ǫ ( x ) − U ǫ ( y ) | p − | U ǫ ( y ) − u ǫ ( y ) | . Now, using | u ǫ ( y ) | ≤ | U ǫ ( y ) | ≤ Cǫ ( n − s ) / for all y ∈ B cδ ([38, (4.17)]), we obtain Z D | U ǫ ( y ) − u ǫ ( y ) | p | x − y | n + ps dxdy ≤ p Z D | U ǫ ( y ) | p | x − y | n + ps dxdy ≤ Cǫ n − s p Z D | x − y | n + ps dxdy = Cǫ n − s p . (5.11)Furthermore, Z D | U ǫ ( x ) − U ǫ ( y ) | p − | U ǫ ( y ) − u ǫ ( y ) || x − y | n + ps dxdy ≤ Z D (cid:18) | U ǫ ( x ) | p − | U ǫ ( y ) || x − y | n + ps + | U ǫ ( y ) | p | x − y | n + ps (cid:19) dxdy. (5.12)By [38, (4.17)] and definition of U ǫ , we have | U ǫ ( x ) | p − | U ǫ ( y ) | ≤ Cǫ n − s (cid:16) b + (cid:12)(cid:12) xǫS / (2 s ) (cid:12)(cid:12) (cid:17) − n − s ( p − . δ e = δ/ǫ , ξ = xǫS / (2 s and ζ = | x − y | , we deduce that Z D | U ǫ ( x ) | p − | U ǫ ( y ) || x − y | n + ps dxdy ≤ Cǫ n − s ǫ n Z ξ ∈ B δǫ Z | ζ | >δ/ (cid:0) b + | ξ | (cid:1) − n − s ( p − | ζ | − ( n + ps ) dξdζ ≤ Cǫ n + n − s " Z ξ ∈ B δǫ \ B (cid:0) b + | ξ | (cid:1) − n − s ( p − ≤ Cǫ n + n − s (cid:0) ǫ ( n − s )( p − − n + 1 (cid:1) ≤ Cǫ ( n − s )( p − / . Using this together with (5.11) and the fact that ( n − s ) / ≤ ( n − s )( p − / Z D | U ǫ ( x ) − U ǫ ( y ) | p − | U ǫ ( y ) − u ǫ ( y ) || x − y | n + ps dxdy ≤ Cǫ n − s p . Thus, from (5.10), we obtain k u ǫ k pX = Z R n | u ǫ ( x ) − u ǫ ( y ) | p | x − y | n + ps dxdy ≤ Z R n | U ǫ ( x ) − U ǫ ( y ) | p | x − y | n + ps dxdy + Cǫ n − s p . (5.13)Now, it remains to evaluate only k U ǫ k pX . We recall U ǫ ( x ) = ǫ − n − s u ∗ (cid:0) xǫ (cid:1) and U ( x ) = u ∗ ( x ).Using suitable change of variables, we deduce that k U ǫ k pX = ǫ − n − s p Z R n | u ∗ ( xǫ ) − u ∗ ( yǫ ) | p | x − y | n + ps dxdy = ǫ − n − s p Z R n | u ∗ ( z ) − u ∗ ( w ) | p | z − w | n + ps dzdw, then, using Lemma 2.1 and (5.8), we get k U ǫ k pX ≤ Cǫ − n − s p k U k pX ≤ Cǫ n (1 − p ) . (5.14)Hence, (5.13) and (5.14) give the required result of the Lemma. (cid:3) Lemma 5.5
There exists Λ > such that for every λ ∈ (0 , Λ ) and β > , there holds σ − λ < c ∞ , where c ∞ is as defined in Lemma 5.2, in each of the following cases(I) for all q > , if < p < n/ ( n − s ) ,(II) for all q ∈ (cid:0) , n (2 − p ) n − s (cid:1) ∪ (cid:0) n n − np − s , p (cid:1) , if n/ ( n − s ) ≤ p < . Proof.
To prove the lemma, we will show that there exists ǫ > t ≥ I λ ( tu ǫ ) < c ∞ .Let γ > λ ∈ (0 , γ ), c ∞ > I λ ( tu ǫ ) ≤ t k u ǫ k X + β t p p k u ǫ k pX ≤ t k u ǫ k X + Cβ t p p k u ǫ k X ≤ C ( t + t p ) . t > ≤ t ≤ t I λ ( tu ǫ ) < c ∞ . Let Θ ( t ) = t k u ǫ k X + β t p p k u ǫ k pX − t . ∗ µ . ∗ µ k u ǫ k . ∗ µ NL , then we note that Θ (0) = 0, Θ ( t ) > t small enough, Θ ( t ) < t large enough, and there exists t ǫ > t ≥ Θ ( t ) = Θ ( t ǫ ) , that is 0 = Θ ′ ( t ǫ ) = t ǫ k u ǫ k X + βt p − ǫ k u ǫ k pX − t . ∗ µ − ǫ k u ǫ k . ∗ µ NL which gives us t . ∗ µ − pǫ = 1 k u ǫ k . ∗ µ NL (cid:0) t − pǫ k u ǫ k X + β k u ǫ k pX (cid:1) < C (1 + t − pǫ ) . Since 2 . ∗ µ >
2, there exists t > t ǫ ≤ t for all ǫ >
0. Also, for δ > δ < δ , we have Z Ω a ( x ) | u ǫ | q dx = Z B δ (0) a ( x ) | u ǫ | q dx ≥ m a Z B δ (0) | u ǫ | q dx ≥ m a Z B δ (0) | U ǫ | q dx. Therefore, we obtainsup t ≥ t I λ ( tu ǫ ) ≤ sup t ≥ Θ ( t ) − t q m a q λ Z B δ (0) | U ǫ | q = t ǫ k u ǫ k X + β t pǫ p k u ǫ k pX − t . ∗ µ ǫ . ∗ µ k u ǫ k . ∗ µ NL − t q m a q λ Z B δ (0) | U ǫ | q ≤ sup t ≥ t k u ǫ k X − t . ∗ µ . ∗ µ k u ǫ k . ∗ µ NL ! + β t p p k u ǫ k pX − t q m a q λ Z B δ (0) | U ǫ | q . (5.15)Let Υ( t ) = t k u ǫ k X − t . ∗ µ . ∗ µ k u ǫ k . ∗ µ NL . It is easy to verify that Υ attains its maximum atˆ t = (cid:18) k u ǫ k X k u ǫ k . ∗ µNL (cid:19) / (2 ∗ µ − . Therefore,sup t ≥ Υ( t ) = Υ(ˆ t ) = n − µ + 2 s n − µ ) k u ǫ k X k u ǫ k . ∗ µ NL ! ∗ µ / (2 ∗ µ − , using the estimates of Lemma 5.3 and (2.2), we havesup t ≥ Υ( t ) ≤ n − µ + 2 s n − µ ) ( C ( n, µ ) ∗ µ S H ) n s + O ( ǫ n − s ) (cid:0) C ( n, µ ) n s S n − µ s H − O ( ǫ n ) (cid:1) ∗ µ ∗ µ / (2 ∗ µ − ≤ n − µ + 2 s n − µ ) S n − µn − µ +2 s H + O ( ǫ n − s ) . (5.16)4For δ >
0, sufficiently small such that B δ (0) ⋐ Ω, 2 δ < δ and 0 < ǫ < δ/
2, we have thefollowing estimate ([19]) Z B δ (0) | U ǫ | q ≥ C ǫ n − ( n − s q , if n < ( n − s ) qǫ n | ln ǫ | , if n = ( n − s ) qǫ ( n − s q , if n > ( n − s ) q. (5.17)Thus, using (5.9), (5.16) and (5.17) in (5.15), we getsup t ≥ t I λ ( tu ǫ ) ≤ n − µ + 2 s n − µ ) S n − µn − µ +2 s H + C ǫ n − s + C ǫ n − s p if 1 < p < n/ ( n − s ) ǫ − p n if n/ ( n − s ) ≤ p < . − λC ǫ n − ( n − s q , if n < ( n − s ) qǫ n | ln ǫ | , if n = ( n − s ) qǫ ( n − s q , if n > ( n − s ) q. (5.18)Now, we consider the following cases. Case (I): If 1 < p < n/ ( n − s ).In this case ( n − s ) p/ < n (1 − p/
2) and 1 < q < p < n/ ( n − s ) < n/ ( n − s ), therefore(5.18) implies sup t ≥ t I λ ( tu ǫ ) ≤ n − µ + 2 s n − µ S n − µn − µ +2 s H + C ǫ n − s p − λC ǫ ( n − s q . And, therefore there exists γ > λ ∈ (0 , γ ) C ǫ n − s p − λC ǫ n − s q < − C λ − q . Case (II): If n/ ( n − s ) ≤ p < n − s ) p/ ≥ n (1 − p/ t ≥ t I λ ( tu ǫ ) ≤ n − µ + 2 s n − µ ) S n − µn − µ +2 s H + C ǫ − p n − λC ǫ n − ( n − s q , if n < ( n − s ) qǫ n | ln ǫ | , if n = ( n − s ) qǫ ( n − s q , if n > ( n − s ) q. (5.19) Subcase (a): If n > ( n − s ) q .In this case, we see that ( n − s )2 q < n (1 − p ), if q < n (2 − p ) n − s . Then, proceeding similar toCase(I), there exists γ > λ ∈ (0 , γ ) C ǫ n − s p − λC ǫ n − s q < − C λ − q . Subcase (b): If n < ( n − s ) q .Choosing ǫ = ( λ − q ) n (2 − p ) ≤ δ , (5.19) yieldssup t ≥ t I λ ( tu ǫ ) ≤ n − µ + 2 s n − µ ) S n − µn − µ +2 s H + C λ − q − λC λ n (2 − q )(2 − p ) (cid:0) n − ( n − s q (cid:1) . Clearly, 1 + 4 n (2 − q )(2 − p ) (cid:18) n − n − s q (cid:19) < − q if q > n/ (4 n − np − s ). Then, in this situation, there exists γ > λ ∈ (0 , γ ), C λ − q − C λ λ n (2 − q )(2 − p ) (cid:16) n − n − s q (cid:17) < − C λ − q . Let Λ = min { γ , γ , γ , Λ , ( δ ) n (2 − p ) / } >
0, then for all λ ∈ (0 , Λ ) and sufficiently small ǫ >
0, we obtain sup t ≥ I λ ( tu ǫ ) < c ∞ . Now choosing δ > u ǫ ∈ X and using Lemma 3.3, there exists˜ t > tu ǫ ∈ M − λ . Hence, σ − λ ≤ I λ (˜ tu ǫ ) ≤ sup t ≥ I λ ( tu ǫ ) < c ∞ . This completes proof of the Lemma. (cid:3)
Proof of Theorem 1.3 (ii): With the help of Lemma 3.5 and Proposition 3.6, we get aminimizing sequence { v k } ⊂ M − λ , which is also ( P S ) σ − λ sequence. By Lemmas 5.5 and 5.2,there exists v λ ∈ X such that v k → v λ in X . Using Theorem 5.1 and strong convergence, v λ ∈ M − λ and I λ ( v λ ) = σ − λ . Thus, v λ is a weak solution of problem ( P λ ) in M − λ and since u λ ∈ M + λ , u λ = v λ . (cid:3) Lemma 5.6
There exists Λ , β > such that for every λ ∈ (0 , Λ ) and β ∈ (0 , β ) , thereholds σ − λ < c ∞ . Proof.
Here we only give outline of the proof because arguments are similar to the previouslemma and [21, Lemma 4.7]. Let β = ǫ α , where α > ( n − s ) and taking into account (5.16)and (5.17), from (5.15), it follows thatsup t ≥ t I λ ( tu ǫ ) ≤ n − µ + 2 s n − µ ) S n − µn − µ +2 s H + C ǫ n − s − λC ǫ n − ( n − s q , if n < ( n − s ) qǫ n | ln ǫ | , if n = ( n − s ) qǫ ( n − s q , if n > ( n − s ) q. The case, when n > ( n − s ) q , follows exactly on the same lines of Case(I) of Lemma 5.5.For the other case, we set ǫ = ( λ − q ) n − s ≤ δ and proceed similarly to [21, Lemma 4.7], toget the required result of the lemma for some Λ , β > (cid:3) Proof of Theorem 1.4 : Proof follows exactly on the same lines of Proof of Theorem 1.3(ii) by using Lemma 5.6 instead of Lemma 5.5. (cid:3)
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