Existence and Multiplicity of Solutions for Nonlocal Neumann Problem with Non-Standard Growth

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1 Existence and Multiplicity of Solutions for Nonlocal Neumann Problem with Non-Standard Growth Francisco Julio S.A. Corrêa Universidade Federal de Campina Grande Centro de Ciências e Tecnologia Unidade Acadêmica de Matemática CEP: , Campina Grande - PB - Brazil fjsacorrea@gmail.com & Augusto César dos Reis Costa Universidade Federal do Pará Instituto de Ciências Exatas e Naturais Faculdade de Matemática CEP: , Belém - PA - Brazil aug@ufpa.br June 22, 205 Abstract In this paper we are concerning with questions of existence of solution for a nonlocal and non-homogeneous Neumann boundary value problems involving the p(x-laplacian in which the non-linear terms have critical growth. The main tools we will use are the generalized Sobolev spaces and the Mountain Pass Theorem. MSC: 35J60; 35J70; 58E05 Keywords: nonlocal problem; Neumann boundary conditions; trace; Sobolev spaces with variable exponent; critical exponent. Partially supported by CNPq-Brazil under Grant 30807/203-2.

2 Introduction In this paper we are going to study questions of existence of solutions for the nonlocal and non-homogeneous equation, with critical growth and Neumann boundary conditions, given by M M ( ( p(x ( u p(x + u p(x dx (div( u p(x2 u + u p(x2 u [ r = λf(x, u F (x, udx] in p(x ( u p(x + u p(x p(x2 u dx u ν [ κ = γg(x, u G(x, uds] on,, (. where IR N is a bounded smooth domain of IR N, p C(, f : IR IR, g : IR IR and M : IR + IR + are continuous functions enjoying some conditions which u u will be stated later, F (x, u = f(x, ξdξ, G(x, u = g(x, ydy, u is the outer 0 0 ν unit normal derivative, λ, r, γ, κ are real parameters, and p(x is the p(x-laplacian operator, that is, p(x u = ( N p(x2 u u, < p(x < N. x i x i i= Bonder and Silva [3], in 200, extended the concentration-compactness principle by Lions [6] for variable exponent spaces and applied the result to the Dirichlet problem involving the p(x-laplacian with subcritical and critical growth and showed the existence and multiplicity of solutions via Mountain Pass Theorem, Truncation argument and Krasnoselskii genus, inspired in Azorero and Peral [2], who solved the problem for the p-laplacian. It is important to note that there exists similar result due to Bonder and Silva [3], on concentration-compactness principle, obtained by Fu [3]. Liang and Zhang [5], in 20, showed multiplicity of solutions to the Neumann problem involving the p(x-laplacian, with critical growth, via truncation argument. Guo and Zhao [4], in 202, showed existence and multiplicity of solutions for the nonlocal Neumann problem involving the p(x-laplacian, with subcritical growth, using Mountain Pass Theorem and Fountain Theorem. Bonder, Saintier and Silva [4], in 203, presented the concentration-compactness principle for variable exponent in the trace case and in [5] showed existence of solution for the Neumann problem with critical growth on the boundary of the domain, via Mountain Pass Theorem. In this article, we discuss existence of solutions for the nonlocal problem (. with critical growth on the domain and on its boundary. We study the nonlocal condition 2

3 for the two following important classes of functions: M(τ = a + bτ η with a 0, b > 0, η and m 0 M(τ m, where m 0 and m are positive constants. Note that the original Kirchhoff term is included in our analysis. We will study the problem with the following critical Sobolev exponents p (x = Np(x N p(x and p (x = (N p(x N p(x, (.2 where p is critical exponent from the point of view of the trace. Problems in the form (. are associated with the energy functional J λ,γ (u = M ( for all u W,p(x (, where M(t = p(x ( u p(x + u p(x dx λ [ ] r+ F (x, udx r + γ [ ] κ+ (.3 G(x, uds κ + t 0 M(sdS, ds denotes the boundary measure, and W,p(x ( is the generalized Lebesgue-Sobolev space whose precise definition and properties will be established in section 2. Depending on the behavior of the functions p, q the above functional is differentiable and its Fréchet-derivative is given by J λ,γ(uv = M ( λ [ ( p(x ( u p(x + u p(x dx u p(x2 u v + u p(x2 uv dx ] r [ ] κ F (x, udx f(x, uvdx γ G(x, uds g(x, uvds (.4 for all u, v W,p(x (. So, u W,p(x ( is a weak solution of problem (. if, and only if, u is a critical point of J λ,γ. Theorem. (i Assume κ = 0, g(x, u = u q(x2 u, q : [, and A := {x : q(x = p (x} = and M(τ = a + bτ η, with a 0, b > 0, η. Moreover, assume the existence of functions p(x, q(x, β(x C + (, see section 2, positive constants A, A 2 such that A t β(x f(x, t A 2 t β(x for all t 0 and for all x, with f(x, t = 0 for all t < 0. Furthermore, (η + p + < β (r + < q and (η + (p+ η+ < (p η ( r+ A (β r+ (r +. Then there exists λ A 2 (β + r 0 > 0 such that for all λ > λ 0 and for all γ > 0 there exists a nontrivial solution to (.. 3

4 (ii Assume κ = 0, g(x, u = u q(x2 u, q : [, and A := {x : q(x = p (x}. Moreover, assume the existence of functions p(x, q(x, β(x C + (, see section 2, positive constants A, A 2 such that A t β(x f(x, t A 2 t β(x for all t 0 and for all x, with f(x, t = 0 for all t < 0. Furthermore, assume there exists 0 < m 0 and m such that m 0 M(τ m, with m p + ( r+ A (β r+ (r + < m 0 A 2 (β + r and p + < β (r + < q. Then there exists λ > 0 such that for all λ > λ and for all γ > 0 there exists a nontrivial solution to (.. (iii Suppose r = 0, f(x, u = u q(x2 u, q : [, and A := {x : q(x = p (x} =. Moreover, assume the existence of functions p(x, q(x, β(x C + (, positive constants A, A 2 such that A t β(x g(x, t A 2 t β(x for all t 0 and for all x, with g(x, t = 0 for all t < 0. Furthermore, assume there exists 0 < m 0 and m such that m 0 M(τ m, with m p + ( κ+ A (β κ+ (κ + < and m 0 A 2 (β + κ p + < β (κ + < q. Then there exists γ > 0 such that for all γ > γ and for all λ > 0 there exists a nontrivial solution to (.. (iv Suppose r = 0, f(x, u = u q(x2 u, q : [,, A := {x : q(x = p (x} = and M(τ = a + bτ η, com a 0, b > 0, τ 0, and η. Moreover, assume the existence of functions p(x, q(x, β(x C + (, positive constants A, A 2 such that A t β(x g(x, t A 2 t β(x for all t 0 and for all x, with g(x, t = 0 for all t < 0. Furthermore, assume (η + p + < β (κ + < q and (η + (p + η+ ( κ+ A (β κ+ (κ + <. Then there exists γ (p η A 2 (β + κ 2 > 0 such that for all γ > γ 2 and for all λ > 0 there exists a nontrivial solution to (.. We should point [out that the ] novelty in the present paper is the appearance of r [ κ, the integral terms, F (x, udx and G(x, uds] the critical growth on Neumann boundary conditions, the use of the p(x-laplacian and the generalized Lebesgue- Sobolev spaces. We should point out that some ideas contained in this paper were inspired by the articles Corrêa and Figueiredo [[6], [7]], Fan [[8], [9]], Fu [3], Liang and Zhang [5], Mihailescu and Radulescu [7] and Yao [20]. This paper is organized as follows: In section 2 we present some preliminaries on the variable exponent spaces. In section 3, we proof of our main result. 2 Preliminaries on variable exponent spaces First of all, we set C + ( := { h; h C(, h(x > for all x } 4

5 and for each h C + ( we define h + := max h(x and h := min h(x. We denote by M( the set of real measurable functions defined on. For each p C + (, we define the generalized Lebesgue space by L p(x ( := { } u M(; u(x p(x dx <. We consider L p(x ( equipped with the Luxemburg norm u p(x := inf µ > 0; u(x p(x dx µ. We denote by L p (x ( the conjugate space of L p(x (, where p(x + p (x =, for all x. The proof of the following propositions may be found in Bonder and Silva [3], Bonder, Saintier and Silva [[4], [5]], Fan, Shen and Zhao [0], Fan and Zhang [] and Fan and Zhao [2]. Proposition 2. (Hölder Inequality If u L p(x ( and v L p (x (, then Proposition 2.2 Set ρ(u :=. For u 0, u p(x = λ ρ( u λ = ; ( uvdx p + u p p(x v p (x. 2. u p(x < (= ; > ρ(u < (= ; > ; 3. If u p(x >, then u p p(x ρ(u u p+ p(x ; 4. If u p(x <, then u p+ p(x ρ(u u p p(x ; 5. lim k + u k p(x = 0 6. lim k + u k p(x = + lim ρ(u k = 0; k + u(x p(x dx. For all u L p(x (, we have: lim ρ(u k = +. k + 5

6 The generalized Lebesgue - Sobolev space W,p(x ( is defined by W,p(x ( := { u L p(x (; u L p(x ( } with the norm Denoting ρ,p(x := proposition: u := u p(x + u p(x. ( u p(x + u p(x dx u W,p(x (, we have the following Proposition 2.3 For all u, u j W,p(x (, we have:. u < (= ; > ρ,p(x (u < (= ; > ; 2. If u >, then u p ρ,p(x (u u p+ ; 3. If u <, then u p+ ρ,p(x (u u p ; 4. lim j + u j = 0 5. lim j + u j = + lim ρ,p(x(u j = 0; j + lim ρ,p(x(u j = +. j + Proposition 2.4 Suppose that is a bounded smooth domain in IR N and p C( with p(x < N for all x. If p C( and p (x p (x ( p (x < p (x for x, then there is a continuous (compact embedding W,p(x ( L p(x (, where p (x = Np(x N p(x. The Lebesgue spaces on are defined as L q(x ( := {u u : IR is measurable and u q(x ds < }, and the corresponding (Luxemburg norm is given by u L q(x ( := u q(x, := inf{λ > 0 : u(x λ q(x ds }. Proposition 2.5 Suppose that is a bounded smooth domain in IR N and p, q C( with p(x < N for all x. Then there is a continuous and compact embedding W,p(x ( L q(x (N p(x (, where q(x < p (x = N p(x. 6

7 Proposition 2.6 (Fan and Zhang [] Let L p(x : W,p(x ( (W,p(x ( be such that L p(x (u(v = u p(x2 u vdx, u, v W,p(x (, then (i L p(x : W,p(x ( (W,p(x ( is a continuous, bounded and strictly monotone operator; (ii L p(x is a mapping of type S +, i.e., if u n u in W,p(x ( and lim sup(l p(x (u n L p(x (u, u n u 0, then u n u in W,p(x (; (iii L p(x : W,p(x ( (W,p(x ( is a homeomorphism. Proposition 2.7 (Bonder Saintier and Silva [4]Let (u j j IN W,p(x ( be a sequence such that u j u weakly in W,p(x (. Then there exists a countable set I positive numbers (µ i I and (ν i I and points {x i } A := {x : q(x = p (x} such that u j q(x ds ν = u q(x ds + ν i δ xi, weakly in the sense of measures. (2.5 i I u j p(x dx µ u p(x dx + i I µ i δ xi, weakly in the sense of measures. (2.6 p(x i q(x T xi ν i i µ i, (2.7 where T xi = sup T (p(., q(., ɛ,i, Γ ɛ,i is the localized Sobolev trace constant where ɛ>0 ɛ,i = B ɛ (x i and Γ ɛ,i = B ɛ (x i. Proposition 2.8 (Bonder and Silva [3] Let q(x and p(x be two continuous functions such that < inf p(x sup p(x < N and q(x p (x in. x x Let (u j be a weakly convergent sequence in W,p(x ( with weak limit u and such that: u j p(x µ weakly-* in the sense of measures. u j q(x ν weakly-* in the sense of measures. In addition we assume that A := {x : q(x = p (x} is nonempty. Then, for some countable index set I, we have: ν = u q(x + ν i δ xi, ν i > 0 i I 7

8 µ u p(x + i I µ i δ i, µ i > 0 S q ν /p (x i i µ /p(x i i, i I. Where {x i } i I A and S q is the best constant in the Gagliardo-Nirenberg-Sobolev inequality for variable exponents, namely S q = S q ( = inf φ C 0 φ Lp(x (. ( φ L q(x ( Definition 2. We say that a sequence (u j W,p(x ( is a Palais-Smale sequence for the functional J λ,γ if J λ,γ (u j c and J λ,γ(u j 0 in (W,p(x (, (2.8 where and c = inf sup J λ,γ (h(t > 0 (2.9 h C t [0,] C = {h : [0, ] W,p(x ( : h continuous and h(0 = 0, J λ,γ (h( < 0}. The number c is called the energy level c. When (2.8 implies the existence of a subsequence of (u j, still denoted by (u j, which converges in W,p(x (, we say that J λ,γ satisfies the Palais-Smale condition. 3 Proof of Theorem. Proof. (i The proof follows from several lemmas. Lemma 3. If (u j W,p(x ( is a Palais-Smale sequence, with energy level c, then (u j is bounded in W,p(x (. Proof. Since (u j is a Palais-Smale sequence with energy level c, we have J λ,γ (u j c and J λ,γ(u j 0. Then taking θ such that (η + (p + η+ < θ < (p η C + u j (J λ,γ (u j θ J λ,γ(u j u j ( ( a uj p(x + u j p(x dx p(x ( A A 2 + b η + 8 r+ (β r+ (r + (β + r (3.0 ( ( η+ uj p(x + u j dx p(x p(x

9 λ [ ] r+ F (x, u j dx γ r + q(x u q(x ds a u j p(x2 u j u j dx θ a u j p(x2 u j u j dx + λ [ ] r F (x, u j f(x, u j u j + γ u j q(x2 u j u j ds θ θ θ b ( ( η ( uj p(x + u j p(x ( dx uj p(x2 u j u j + u j p(x2 u j u j dx θ p(x Thus, C + u j ( a ( uj p(x + u p + j p(x ( b ( dx + uj p(x + u (η + (p + η+ j p(x η+ dx λa r+ [ ] r+ 2 u (r + (β r+ j β γ dx u q(x ds a u q j p(x dx θ a [ ] u j p(x dx + λar+ r+ u β(x γ + u θ θ(β + r j q(x ds θ b ( ( uj p(x + u θ(p η j p(x η+ dx, and ( a C + u j p a ( ( uj p(x + u + j p(x dx θ ( b + (η + (p + b ( ( uj p(x + u η+ θ(p η j p(x η+ dx ( λa r+ + θ(β + λa r+ [ ] ( r+ 2 γ u β(x + r (r + (β r+ θ γ u q j q(x ds. Now suppose that (u j is unbounded in W,p(x (. Thus, passing to a subsequence if necessary, we ( get u j > and we obtain a C + u j p a ( b u p + + θ (η + (p + b u (η+p η+ θ(p ( η A r+ +λ θ (β + Ar+ 2 u β± (r+, r r + (β r+ which is a contradiction because β ± (r + > p >. Hence (u j is bounded in W,p(x (. Lemma ( 3.2 Let (u j W,p(x ( be a Palais-Smale sequence, with energy level c and t 0 = lim j p(x ( u j p(x + u j p(x dx. If c < ( θ ( p(x inf γ /p(x i a /p(xi i p (x i p (x T i p(x i q xi, i I where a = t with 0 < t < bt η 0, then index set I = and u j u strongly in L q(x (. 9

10 Proof. Assume that u j u weakly in W,p(x (. By Proposition 2.7 and Lemma 3., we have u j q(x ds ν = u q(x ds + ν i δ xi, i I weakly in the sense of measures. u j p(x dx µ u p(x dx + i I µ i δ xi, weakly in the sense of measures. p(x i q(x T xi ν i i µ i, i I. If I = then u j u strongly in L q(x (. Suppose that I. Let x i be a singular point of the measures µ and ν. We consider φ C0 (IR N, such that 0 φ(x, ( φ(0 = 0 and support in the unit ball of IR N. Consider the functions x xi φ i,ε (x = φ for all x IR N and ε > 0. ε As J λ,γ(u j 0 in ( W,p(x we obtain that J λ,γ(uφ i,ε u j = i.e., γ J λ,γ(uφ i,ε u j = γ lim J λ,γ(u j (φ i,ε u j = 0. ( ( η a + b p(x ( u j p(x + u j p(x dx ( uj p(x2 u j φ i,ε u j + u j p(x2 u j φ i,ε u j dx [ ] r u j q(x2 u j φ i,ε u j ds λ F (x, u j dx f(x, u j (φ i,ε u j dx 0, ( ( η ( a + b p(x ( u j p(x + u j p(x dx ( uj p(x2 u j φ i,ε u j + u j p(x φ i,ε + u j p(x φ i,ε dx [ ] r F (x, u j dx f(x, u j (φ i,ε u j dx 0. u j q(x φ i,ε ds λ When j([ we get η ] 0 = lim a + b p(x ( u j p(x + u j p(x dx [ ( η ] a + b p(x ( u j p(x + u j p(x dx u p(x φ i,ε dx ( u j p(x2 u j φ i,ε u j dx [ ] r (a + bt η 0 φ i,ε dµ γ φ i,ε dν λ F (x, udx f(x, u(φ i,ε udx. + 0

11 We may show that, lim (a + ε 0 btη 0 ( u p(x2 u φ i,ε udx 0, see Shang-Wang [8]. On the other hand, lim (a + ε 0 btη 0 φ i,ε dµ = (a + bt η 0µφ(0, γ lim φ i,ε dν = γνφ(0 ɛ 0 and [ ] r λ F (x, udx f(x, u(φ i,ε udx 0, (a+bt 0 u p(x φ i,ε dx 0 as ε 0. Then, (a + bt η 0µ i φ(0 = γν i φ(0 implies that γ aµ i ν i. By T xi ν /p (x i i µ /p(x i i we obtain γ a ( p(xi p(x T xi ν i /p (x i i γ aµ i ν i. Thus γ a ( p(xi p(x T xi ν i /p (x i i = ν (p (x ip(x i /p (x i i and γ /p(xi a /p(xi T xi ν (p (x ip(x i /p(x i p (x i i. Therefore ν i ( γ /p(x i a /p(x i T xi p(x ip (x i p (x i p(x i. On the other hand, using θ satisfying (3.0 c = lim J λ,γ (u j = lim (J λ,γ (u j θ J λ,γ(u j u j. Thus, ( γ c lim θ γ u q j q(x ds ( γ c θ γ ( u q(x ds + ν q i δ xi dν ( i I c θ ( p(x inf γ /p(x i a /p(xi i p (x i p (x T i p(x i q xi. i I Therefore, the index set I is empty if, c < ( γ θ γ ν q i ( θ ( p(x inf γ /p(x i a /p(xi i p (x i p (x T i p(x i q xi. i I Lemma ( 3.3 Let (u j W,p(x ( be a Palais-Smale sequence with energy level c. If c < θ γ ( p(x inf γ /p(x i a /p(xi i p (x i p (x T i p(x i q xi, there exist u W,p(x ( and a i I subsequence, still denoted by (u j, such that u j u in W,p(x (. Proof. From we have J λ,γ(u j 0,

12 and J λ,γ(u j φu j = ( ( η a + b p(x ( u j p(x + u j p(x dx ( uj p(x2 u j (u j u + u j p(x2 u j (u j u dx [ λ γ u j q(x2 u j (u j uds ] r F (x, u j dx f(x, u j (u j udx 0, Note that there exists nonnegative constants c, c 2, c 3 and c 4 such that c a + b ( η p(x ( u j p(x + u j p(x dx c 2 [ c 3 F (x, u j dx] r c4. But using Hölder inequality, we have u j p(x2 u j (u j udx u j p(x u j u dx C u p(x/p(x u j u p(x, and f(x, u j (u j udx A 2 u j β(x u j u dx C 2 u β(x/β(x u j u β(x, where C and C 2 are positive constants. Thus and u j p(x2 u j (u j udx 0 f(x, u j (u j udx 0. By Lemma 3.2, u j u in L q(x ( and using Hölder inequality we obtain If we take u j q(x2 u j (u j u dx 0. L p(x (u j (u j u = u j p(x2 u j (u j udx, we obtain L p(x (u j (u j u 0. We also have L p(x (u(u j u 0. So (L p(x (u j L p(x (u, u j u 0. From Proposition 2.6 we have u j u in W,p(x (. 2

13 Lemma 3.4 (i For all λ > 0, there are α > 0, ρ > 0, such that J λ,γ (u α, u = ρ. (ii There exists an element w 0 W,p(x ( with w 0 > ρ and J λ,γ (w 0 < α. Proof. (i We have, J λ,γ (u a ( p(x ( u p(x + u p(x dx η+ p(x ( u p(x + u p(x dx ] r+ A 2 β(x uβ(x dx γ u q(x ds. q ( + b η + λ [ r + So, a J λ,γ (u ( u p(x + u p(x dx p + ( b η+ + ( u p(x + u p(x dx (η + (p+ η+ λ ( r+ [ ] A2 r+ u β(x γ dx u q(x ds. r + β q If u < is small enough, from Proposition 2.3 we obtain ( u p(x + u p(x u p+ and from the immersions W,p(x ( L β(x ( and W,p(x ( L q(x ( (continuous, we get u β(x dx M β u β and u q(x dx M q 2 u q. Hence a b (η+ J λ,γ (u p + u p+ + (η + (p + η+ u p+ λ ( r+ A2 M β (r+ r + β u β (r+ γ q M q 2 u q. Using the assumption (η + p + < β (r + < q, a b (η+ J λ,γ (u p + u (η+p+ + (η + (p + η+ u p+ λ ( r+ A2 M β (r+ r + β u β (r+ can be written as ( a J λ,γ (u p + b u p+ (η+ + (η + (p + η+ λ ( r+ A2 M β (r+ γ r + β + β (r + M q 2 u β (r+. Taking ρ = u, γ β (r + M q 2 u β (r+, 3

14 J λ,γ (u ( ρ (η+p+ a p + b λ + (η + (p + η+ r + γ + β (r + M q 2 and the result follows. (ii Take 0 < w W,p(x (. For t > (, a J λ,γ (tw p tp+ Then we have and the proof is over. + λ r + ( w p(x + w p(x dx ( (η+ + btp+ (η + (p η+ r+ ( t β (r+ ( A2 β η+ ( w p(x + w p(x dx w β(x dx lim J λ,γ(tw =, t r+ γ ( r+ A2 M β (r+ β ] ρ β (r+(η+p + q + tq w q(x ds., By Lemma 3.4, we may use the Mountain Pass Theorem [9], which guarantees the existence of a sequence (u j W,p(x ( such that J λ,γ (u j c and J λ,γ(u j 0 in (W,p(x (, where a candidate for critical value is c = inf sup J λ,γ (h(t h C t [0,] and C = {h : [0, ] W,p(x ( : h continuous and h(0 = 0, h( = w 0 }. For 0 < t < and fixing w W,p(x ( (, a J λ,γ (tw p tp ( w p(x + w p(x dx (η+ ( η+ + btp ( w p(x + w p(x dx (η + (p η+ λ ( ( A r+ t β+ (r+ w β(x γ w q(x ds. J λ,γ (tw r + β + a p tp ( ( (η+ + btp (η + (p η+ λ ( A r + β + ( w p(x + w p(x dx q + tq+ η+ ( w p(x + w p(x dx ( r+ w β(x. t β+ (r+ 4

15 ( aã Setting g(t = p + bã η+ t p λ ( A β bt (r+, (η + (p η+ r + β + where ã = ( w p(x + w p(x dx and b = w β(x dx, we obtain sup J λ (tw g(t. Note that g(t has a critical point of maximum t λ = [ β + ((η + (p η aã + b(ã η+ λa (η + (p η bβ and that t λ 0 when λ. By the continuity of J λ,γ lim λ Then exists λ 0 such that λ λ 0 sup J λ,γ (tw < t 0 This completes the proof of part (i. ( sup J λ,γ (tw t 0 = 0. ] β + (r+p, ( θ γ ( p(x inf γ /p(x i a /p(xi i p (x i p (x T i p(x i q xi i I Proof.(ii Like the item (i. Proof. (iii The proof follows from several lemmas. Lemma 3.5 Let (u j W,p(x ( be a Palais-Smale sequence for J λ,γ bounded. then (u j is Proof. Let (u j be a Palais-Smale sequence with energy level c, i.e., J λ,γ (u j c and J λ,γ(u j 0. Then taking θ such that we have So C + u j m p + m 0 < θ < C + u j ( A A 2 κ+ (β κ+ (κ + (β + κ, (3. (J λ,γ (u j θ J λ,γ(u j u j. ( m0 p m ( ( uj p(x + u + j p(x dx θ ( γa κ+ θ(β + γa κ+ [ 2 κ (κ + (β κ+ 5 + ( λ θ λ u q j q(x dx κ+ u ds] β(x.

16 Now suppose that (u j is unbounded in W,p(x (. Thus, passing to a subsequence if necessary, we get u j > and we obtain C + u j ( p u + j p, θ which is a contradiction because p >. Hence (u j is bounded in W,p(x (. Lemma ( 3.6 Let (u j W,p(x ( be a Palais-Smale sequence, with energy level c. If c < θ qa λ(m 0 S N, where m 0 = min{(λ m 0 /p+, (λ m 0 /p }, then index set I = and u j u strongly in L q(x (. Proof. Assume that u j u weakly in W,p(x (. By Proposition 2.8 and Lemma 3.5, we have u j q(x ν = u q(x + ν i δ xi, ν i > 0 i I u j p(x µ u p(x + i I µ i δ xi, µ i > 0 Sν /p (x i i µ /p(x i i, i I. If I = then u j u strongly in L q(x (. Suppose that I. Let x i be a singular point of the measures µ and ν. We consider φ C0 (IR N, such that 0 φ(x, φ(0 ( = 0 and and support in the unit ball of IR N. Consider the functions φ i,ε (x = x xi φ for all x IR N and ε > 0. ε As J λ,γ(u j 0 in ( W,p(x we obtain that lim J λ,γ(u j (φ i,ε u j = 0. ( J λ,γ(u(φ i,ε u j = M p(x ( u j p(x + u j p(x dx ( uj p(x2 u j φ i,ε u j + u j p(x2 u j φ i,ε u j dx [ ] κ λ u j q(x2 u j φ i,ε u j dx γ G(x, u j ds g(x, u j (φ i,ε u j ds 0. When j( ( we get 0 = lim M p(x ( u j p(x + u j p(x dx ( u j p(x2 u j φ i,ε u j dx ( M p(x ( u j p(x + u j p(x dx u j p(x φ i,ε dx + 6

17 [ ] κ M(t 0 φ i,ε dµ λ φ i,ε dν γ G(x, uds g(x, u(φ i,ε uds, ( where t 0 = lim j p(x ( u j p(x + u j p(x dx. We may show that, lim M(t 0 ( u p(x2 u φ i,ε udx 0, see Shang-Wang [8]. ε 0 and γ 0. On the other hand, λ lim φ i,ε dν = νφ(0 ɛ 0 lim M(t 0 φ i,ε dµ = M(t 0 µφ(0, ε 0 [ ] κ G(x, uds g(x, u(φ i,ε uds 0, M(t 0 u p(x φ i,ε dx 0 quando ε Then, M(t 0 µ i φ(0 = λν i φ(0 implies that λ m 0 µ i ν i. By Sν /p (x i i µ /p(x i i we get (m 0 S N ν i, where m 0 = min{(λ m 0 /p+, (λ m 0 /p }. On the other hand, using θ satisfying (3. we have c = lim (J λ (u j θ J λ,γ(u j u j. Note that, ( λ c lim θ λ u j q(x dx. q(x Considering A δ = (B δ (x = {x : dist(x, A < δ}, Therefore, c x A ( λ θ λ ( qa u q(x dx + ( λ ν i δ A δ i I θ λ qa ν i. δ c Therefore, the index set I is empty if, c < ( λ θ λ qa (m 0 S N. ( θ qa λ(m 0 S N. 7

18 Lemma ( 3.7 Let (u j W,p(x ( be a Palais-Smale sequence with energy level c. If c < θ qa λ(m 0 S N, there exist u W,p(x ( and a subsequence, still denoted by (u j, such that u j u in W,p(x (. Proof. From J λ,γ(u j 0, we have ( J λ,γ(u j (u j u = M p(x ( u j p(x + u j p(x dx ( uj p(x2 u j (u j u + u j p(x2 u j (u j u dx λ u j q(x2 u j (u j udx [ ] κ γ G(x, u j ds g(x, u j (u j uds 0. Note that there exists nonnegative constants c and c 2 such that [ c Using Hölder inequality, one has G(x, u j ds] r c2. u j p(x2 u j (u j udx u j p(x u j u dx C u p(x p(x/p(x u j u p(x, and g(x, u j (u j uds A 2 u j β(x u j u ds C 2 u β(x β(x/β(x u j u β(x, where C and C 2 are positive constants. Thus and u j p(x2 u j (u j udx 0 g(x, u j (u j uds 0. By Lemma 3.6, u j u in L q(x ( and using Hölder inequality we obtain If we take u j q(x2 u j (u j udx 0. 8

19 L p(x (u j (u j u = u j p(x2 u j (u j u, we obtain L p(x (u j (u j u 0. We also have L p(x (u(u j u 0. So (L p(x (u j L p(x (u, u j u 0. From Proposition 2.6 we have u j u in W,p(x (. Lemma 3.8 (i For all λ > 0, there are α > 0, ρ > 0, such that J λ,γ (u α, u = ρ. (ii There exists an element w 0 W,p(x ( with w 0 > ρ and J λ,γ (w 0 < α. Proof. (i We have, ( J λ,γ (u m 0 p(x ( u p(x + u p(x dx λ u q(x dx. q So, J λ,γ (u m 0 p + ( u p(x + u p(x dx λ q γ κ + γ [ κ + ( κ+ [ A2 u q(x dx. If u < is small enough, from Proposition 2.3 we obtain β ] κ+ A 2 β(x uβ(x ds u β(x ds ] κ+ ( u p(x + u p(x u p+ and from the immersions W,p(x ( L β(x ( and W,p(x ( L q(x ( (continuous, we get u β(x dx M β u β and u q(x dx M q 2 u q. Hence J λ,γ (u m 0 p + u p+ γ ( κ+ A2 M β (κ+ κ + β u β (κ+ λ q M q 2 u q. Using the assumption p + < β (κ + < q, J λ,γ (u m 0 p + u p+ γ ( κ+ A2 M β (κ+ κ + β u β (κ+ λ β (κ + M q 2 u β (κ+, can be written as J λ,γ (u m 0 p + u p+ γ κ + Taking ρ = u, ( A2 β κ+ M β (κ+ + λ β (κ + M q 2 u β (κ+. 9

20 J λ,γ (u ρ p+ and the result follows. m0 ( p + γ r + ( A2 β κ+ M β (κ+ + (ii Take 0 < w W,p(x (. For t >, J λ,γ (tw m ( p tp+ ( w p(x + w p(x dx + γ ( κ+ ( A2 t β (κ+ κ + β Then we have and the proof is over. lim J λ,γ(tw =, t λ β (κ + M q 2 ρ β (κ+p +, λ q + tq w q(x dx κ+ w ds β(x. By Lemma 3.4, we may use the Mountain Pass Theorem [9], which guarantees the existence of a sequence (u j W,p(x ( such that J λ,γ (u j c and J λ,γ(u j 0 in (W,p(x (, where a candidate for critical value is c = inf sup J λ,γ (h(t h C t [0,] and C = {h : [0, ] W,p(x ( : h continuous and h(0 = 0, h( = w 0 }. For 0 < t < and fixing w W,p(x (, J λ,γ (tw m ( p tp ( w p(x + w p(x dx λ q + tq+ w q(x dx γ ( κ+ ( A κ+ t β+ (κ+ w ds β(x. κ + β + ( m J λ,γ (tw p tp ( w p(x + w p(x dx γ ( κ+ ( A κ+ t β+ (κ+ w ds β(x. κ + β + Settingg(t = m ã p tp γ ( κ+ A bt β + (κ+, κ + β + where ã = ( w p(x + w p(x dx and b = w β(x dx, we obtain sup J λ,γ (tw g(t. Note that g(t has a critical point of maximum t γ = [ m (β + κ ã γa κ b 20 ] β + (κ+p,

21 and that t γ 0 when γ. By the continuity of J λ,γ Then exists γ such that γ γ lim γ ( sup J λ,γ (tw < t 0 This completes the proof of part (iii. sup J λ,γ (tw t 0 = 0. ( θ qa λ(m 0 S N. Proof.(iv Similar to the proof of the item (iii. Remark 3. In a forthcoming article we will present results concerning multiplicity of solutions for the problem (.. 2

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