On a p(x)-kirchhoff Equation with Critical Exponent and an Additional Nonlocal Term

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1 On a p(x-kirchhoff Equation with Critical Exponent and an Additional Nonlocal Term 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 March, 204 Abstract We study the existence of solutions for some classes of p(x-kirchhoff equation with critical exponent and an additional nonlocal term on Sobolev spaces with variable exponent. We use a version of the Concentration Compactness Principle. MSC: 35J60; 35J70; 58E05 Keywords: p(x-laplace operator; Sobolev spaces with variable exponent; critical exponent. Introduction In this paper we are going to study questions of existence of solutions of the p(x- Kirchhoff equation with critical growth, with an additional nonlocal term, Partially supported by CNPq-Brazil under Grant 30056/200-5.

2 M ( [ r p(x u p(x p(x u = λf(x, u F (x, u] + u q(x2 u in, u = 0 on, (. where IR N is a bounded smooth domain, f : IR IR and M : IR + IR + are continuous functions enjoying some conditions which will be stated later, F (x, u = u f(x, ξdξ, λ, r > 0 are real parameters and p(x is the p(x-laplacian operator, 0 that is, p(x u = ( N p(x2 u u, < p(x < N. x i x i i= We assume the following hypotheses: there are positive constants A, A 2 and a function β(x C + (, see section 2, such that A t β(x f(x, t A 2 t β(x, (.2 for all t 0 and for all x, with f(x, t = 0 for all t < 0. Furthermore, and < p = min x p(x max p(x = p + < N, (.3 x < β + (r + < q(x p = Np(x N p(x, (.4 with {x ; q(x = p (x} =. We point out that β + and β are as in the definition (.3. [ r Note that the term F (x, u] makes sense because, in view of assumption (.2, F (x, u 0 for all u IR. Problems in the form (. are associated with the energy functional J λ (u = M ( p(x u p(x λ [ ] r+ F (x, u r + t q(x u q(x (.5 for all u W,p(x 0 (, where M(t = M(sds and W,p(x 0 ( is the generalized 0 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 2

3 J λ(u.v = M ( p(x u p(x [ u p(x2 u v λ u q(x2 uv ] r F (x, u f(x, uv (.6 for all u, v W,p(x 0 (. So, u W,p(x 0 ( is a weak solution of problem (. if, and only if, u is a critical point of J λ. We use the concentration-compactness principle of Lions [9] to the variable exponent spaces, extended by Bonder and Silva [] to the generalized Lebesgue-Sobolev spaces, to prove our main result, as follows: Theorem. (i Assume (.2,(.3, (.4 and M(t = a + bt, with a > 0, b > 0. Moreover, assume 2p + < β (r + and 2(p+ 2 ( r+ A (β r+ (r + <. Then there exists λ > 0 such p A 2 (β + r that for all λ > λ there exists a nontrivial solution. (ii Assume (.2, (.3 and (.4. Moreover, assume there exists 0 < m 0 and m such that m 0 M(t m, with m p + ( r+ A (β r+ (r + < and p + < β (r +. m 0 A 2 (β + r Then there exists λ > 0 such that for all λ > λ there exists a nontrivial solution. (iii Assume (.2, (.3, (.4 and M(t = t α, with α >. Moreover, assume αp + < β (r + and α(p+ α ( r+ (p < A (β r+ (r +. Then there exists λ > 0 α A 2 (β + r such that for all λ > λ there exists a nontrivial solution. We should point out [ that the novelty in the present paper is the appearance of another nonlocal term, F (x, u] r, the critical growth and the use of the p(x-laplacian and the generalized Lebesgue-Sobolev spaces. Another point we have to emphasize is that if we consider the problem M ( [ r p(x u p(x p(x u = λf(x, u F (x, u] + (u + q(x in, u = 0 on, (.7 where u + is the positive part of u, the solutions we obtain are positive. We should point out that some ideas contained in this paper were inspired by the articles Corrêa and Figueiredo [2], [3], Fan [4], Mihailescu and Radulescu [0] and Yongqiang []. This paper is organized as follows: In section 2 we present some preliminaries on the variable exponent spaces. In section 3, we prove our main result. 3

4 2 Preliminaries on variable exponent spaces First of all, we set C + ( = { h; h C(, h(x > for all x } 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 <. We consider L p(x ( equipped with the Luxemburg norm u p(x = inf µ > 0; u(x p(x µ. The generalized Lebesgue - Sobolev space W,p(x ( is defined by with the norm W,p(x ( = { u L p(x (; u L p(x ( } u,p(x = u p(x + u p(x. We define W,p(x 0 ( as being the closure of Cc ( in W,p(x ( with respect the norm u,p(x. Accordingly to Fan - Zhao [7], the spaces L p(x (, W,p(x ( and W,p(x 0 ( are separable and reflexive Banach spaces. Furthermore, if the Lebesgue measure of is finite, p, p 2 C( and p (x p 2 (x, for all x, then we have the continuous embedding L p2(x ( L p(x (. The proof of the following propositions may be found in Fan-Shen-Zhao [5], Fan- Zhang [6] and [7]. Proposition 2. 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. Proposition 2.2 Set ρ(u = u(x p(x. For all u L p(x (, we have: 4

5 . For u 0, u p(x = λ ρ( u λ = ; 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 + lim ρ(u k = +. k + Proposition 2.3 (Poincaré Inequality If u W,p(x 0 (, then where C is a constant independent of u. u p(x C u p(x, Note that, in view of Poincaré inequality, the norms,p(x and u = u p(x are equivalent in W,p(x 0 (. From now on we work on W,p(x 0 ( with the norm u = u p(x. We denote by L p (x ( the conjugate space of L p(x (, where p(x + p (x =, for all x. Proposition 2.4 (Hölder Inequality If u L p(x ( and v L p (x (, then ( uv p + u p p(x v p (x. Proposition 2.5 (Fan-Zhang [6] Let L p(x that L p(x (u(v = u p(x2 u v, then : W,p(x 0 ( (W,p(x 0 ( be such u, v W,p(x 0 (, (i L p(x : W,p(x 0 ( (W,p(x 0 ( 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 0 ( and lim sup(l p(x (u n L p(x (u, u n u 0, then u n u in W,p(x 0 (; (iii L p(x : W,p(x 0 ( (W,p(x 0 ( is a homeomorphism. 5

6 Proposition 2.6 (Bonder-Silva [] 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 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. Where {x i } i I A and S is the best constant in the Gagliardo-Nirenberg-Sobolev inequality for variable exponents, namely S = 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 0 ( is a Palais-Smale sequence at level c for the functional J λ if J λ (u j c and J λ(u j 0 in (W,p(x 0 (. (2.8 The number c is called the energy level c. The functional J λ satisfies the Palais-Smale condition if any Palais-Smale sequence at level c has a convergent subsequence. 3 Proof of Theorem. Proof. (i Here we are assuming (.2, (.3, (.4 and M(t = a + bt. Moreover, 2p + < β (r + and 2(p+ 2 ( r+ A (β r+ (r + <. p A 2 (β + r The proof follows from several lemmas. Lemma 3. If (u j W,p(x 0 ( is a Palais-Smale sequence, with energy level c, then (u j is bounded in W,p(x 0 (. 6

7 Proof. Since (u j is a Palais-Smale sequence with energy level c, we have J λ (u j c and J λ(u j 0. But ( J λ (u = a p(x u p(x + b ( 2 2 p(x u p(x λ [ ] r+ F (x, u r + q(x u q(x, and ( ( ( J λ(u.v = a u p(x2 u v + b p(x u p(x u p(x2 u v [ ] r λ F (x, u f(x, uv u q(x2 uv. with Then 2(p + 2 C + u j J λ (u j θ J λ(u j u j, p < θ < ( A A 2 r+ (β r+ (r + (β + r. (3.9 Now suppose that (u j is unbounded in W,p(x 0 (. Thus, passing to a subsequence if necessary, we get u j > and C + u j ( a p a u + j p, θ which is a contradiction because p >. Hence (u j is bounded in W,p(x 0 (. Lemma 3.2 Let (u j W,p(x 0 ( be a Palais-Smale sequence, with ( energy level c and = min q(x where A is the set given on Proposition 2.6. If c < x A θ (as N, where a = min{a /p+, a /p }, then index set I = and u j u strongly in L q(x (. Proof. By Proposition 2.6 and Lemma 3., 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 (. To prove this, let us consider the extension of Lions concentration-compactness principle. We can see the functions u W,p(x 0 ( 7

8 as functions in W,p(x (IR N, with u = 0 in IR N \. By identifying the functions u j q(x and u q(x as functions of L (IR N, we obtain u j q(x ϕ = u j q(x ϕ u q(x ϕ = IR N IR N u q(x ϕ, for all ϕ C 0 (IR N. Taking ϕ C 0 (IR N such that ϕ(x = in with compact support containing the convergence follows. ( Now if (u j is Palais-Smale sequence, with energy level c, and c < θ (as N, where a = min{a /p+, a /p }, then I =. Indeed, suppose that I. Let ϕ C0 (IR N be such that 0 ϕ, ϕ(0 ( =, with support in the unit ball of IR N. Let us consider x xi the functions ϕ i,ε (x = ϕ, with ε > 0. ε As J λ(u j 0 in ( W,p(x 0 we obtain that But, J λ(u j (ϕ i,ε u j = lim J λ(u j (ϕ i,ε u j = 0. [ ( ] a + b p(x u j p(x u j p(x2 u j (ϕ i,ε u j [ ] r λ F (x, u j f(x, u j (ϕ i,ε u j u j q(x2 u j (ϕ i,ε u j. When j([ we ( get ] ( 0 = lim a + b p(x u j p(x u j p(x2 u j (ϕ i,ε u j + [ ] r (a + bt 0 ϕ i,ε dµ ϕ i,ε dν λ F (x, u f(x, u(ϕ i,ε u, where t 0 = lim j p(x u j p(x. We may show that, lim ε 0 ( lim j [ a + b ( ] ( p(x u j p(x see Shang-Wang [2]. On the other hand, lim (a + bt 0 ϕ i,ε dµ = (a + bt 0 µ i ϕ(0, ε 0 and [ ] r λ F (x, u f(x, u(ϕ i,ε u 0, as ε 0. u j p(x2 u j (ϕ i,ε u j 0, lim ϕ i,ε dν = ν i ϕ(0 ϵ 0 Then, (a + bt 0 µ i ϕ(0 = ν i ϕ(0 implies that aµ i ν i. By Sν /p (x i i µ /p(x i i we obtain, 8

9 S p(xi ν p(x i/p (x i i µ i. So, as p(xi ν p(x i/p (x i i aµ i ν i. Thus, as p(xi ν p(x i/p (x i ν (p p(x i /p (x i i and a /p(xi S ν (p p(x i /p (x i p(x i i = ν /N i. Therefore (as N ν i, where a = min{a /p+, a /p }. On the other hand, using θ satisfying (3.9 c = lim J λ (u j = lim (J λ (u j θ J λ(u j u j, i = i.e., ( c = lim a p(x u j p(x + b ( 2 2 p(x u j p(x λ [ ] r+ F (x, u j r + q(x u j q(x a u j p(x b ( θ θ p(x u j p(x u j p(x + λ [ ] r F (x, u j f(x, u j u j + u j q(x, ( θ θ a ( c lim u p + j p(x + b ( 2 [ ] u 2(p + 2 j p(x λ r+ F (x, u j r + q(x u j q(x a u j p(x b ( 2 u θ θp j p(x + λ [ ] r F (x, u j f(x, u j u j + u j q(x, (( θ θ a c lim p a ( b u + j p(x + θ 2(p + b ( 2 u 2 θp j p(x [ ] λ Ar+ 2 r+ u r + (β r+ j β(x q(x u j q(x ] + λar+ r+ θ (β + [ u r j β(x + u j q(x, θ (( a c lim p a ( b u + j p(x + θ 2(p + b ( 2 u 2 θp j p(x + λar+ θ (β + [ ] r+ u r j β(x A r+ [ ] λ 2 r+ u r + (β r+ j β(x + u j q(x θ q(x u j q(x. So, we have( c lim u j q(x θ q(x u j q(x. Now setting A δ = (B δ (x = {x : dist(x, A < δ} and δ = min q(x, x A x A δ we obtain ( c lim A δ θ ( u j q(x = δ θ ( u q(x + ν i δ A δ i I 9

10 ( θ ( ν i δ θ (as N. δ Since δ > 0 is arbitrary and q is continuous, we get c Therefore, the index set I is empty if, c < ( θ (as N. ( θ (as N. Lemma 3.3 Let (u j W,p(x ( 0 ( be a Palais-Smale sequence with energy level c. If c < θ (as N, there exist u W,p(x 0 ( and a subsequence, still denoted by (u j, such that u j u in W,p(x 0. Proof. From J λ(u j 0, we have ( J λ(u j.(u j u = a u j p(x2 u j (u j u + ( ( b p(x u j p(x u j p(x2 u j (u j u [ ] r λ F (x, u j f(x, u j (u j u u j q(x2 u j (u j u and and 0. Note that there exists nonnegative constants c, c 2, c 3 and c 4 such that c ( [ c 3 p(x u j p(x c 2 F (x, u j ] r c4. By Lemma 3.2, u j u in L q(x ( and using Hölder inequality we obtain f(x, u j (u j u 0 0

11 If we take u j q(x2 u j (u j u 0. 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 2.5 we have u j u in W,p(x 0 (. Lemma 3.4 (i For all λ > 0, there are α > 0, ρ > 0, such that J λ (u α, u = ρ. (ii There exists an element w 0 W,p(x 0 ( with w 0 > ρ and J λ (w 0 < α. Proof. (i We have, ( J λ (u a p(x u p(x λ [ r + ] r+ A 2 β(x uβ(x u q(x. q So, J λ (u a ( u p(x λ ( r+ [ ] A2 r+ u β(x u q(x. p + r + β q If u is small enough, from Proposition 2.2 we obtain u p(x u p+ and from the immersions W,p(x 0 ( L β(x ( and W,p(x 0 ( L q(x (, we get u β(x M β u β and u q(x dx M q 2 u q. Hence J λ (u a p + u p+ λ ( r+ A2 M β (r+ r + β u β (r+ q M q 2 u q. Using the assumption 2p + < β (r + q, J λ (u a p + u p+ λ ( r+ A2 M β (r+ r + β u β (r+ β (r + M q 2 u β (r+, can be written as J λ (u a p + u p+ λ ( r+ A2 M β (r+ r + β + β (r + M q 2 u β (r+. Taking ρ = u, ( J λ (u ρ p+ a λ ( r+ A2 M β (r+ p + r + β + q M q 2 ρ β (r+p +,

12 and the result follows. (ii Take 0 < w W,p(x 0 (. For t >, ( a J λ (tw p tp+ w p(x + b ( 2 2(p 2 t2p+ w p(x + λ ( ( A r+ t β (r+ w β(x r + β + q + tq w q(x. Then we have lim J λ(tw =, t and the proof is over. By Lemma 3.4, we may use the Mountain Pass Theorem [3], which guarantees the existence of a sequence (u j W,p(x 0 ( such that where a candidate for critical value is J λ (u j c and J λ(u j 0 in (W,p(x 0 (, c = inf sup J λ (h(t h C t [0,] and C = {h : [0, ] W,p(x 0 ( : h continuous and h(0 = 0, h( = w 0 }. For 0 < t < and fixing w W,p(x 0 (, ( a J λ (tw p tp w p(x + b ( 2 2(p 2 t2p w p(x λ ( ( A r+ t β (r+ w β(x r + β + q + tq w q(x J λ (tw a ( p tp w p(x + b ( 2 2(p 2 t2p w p(x λ ( ( A r+ t β (r+ w β(x. r + β + ( aã Setting g(t = p + bã2 t p λ ( A β bt (r+, 2(p 2 r + β + where ã = w p(x and b ( r+, = w β(x we obtain sup Jλ (tw g(t. Note that ( aã + bã 2 β (r+p g(t have a critical point of maximum t λ = 2p λ (, and that t A λ 0 β bβ + when λ. By the continuity of J 2

13 Then exists λ such that λ λ lim λ ( sup J λ (tw < t 0 This completes the proof of part (i. sup J λ (tw t 0 = 0. ( θ (as N. Proof. (ii Here we follow the same ideas of part (i and we assume (.2, (.3 and (.4. Furthermore, let us suppose there exists 0 < m 0 and m such that m 0 M(t m, with m p + ( r+ A (β r+ (r + < and p + < β (r +. m 0 (β + r A 2 Lemma 3.5 If (u j W,p(x 0 ( is a Palais-Smale sequence, with energy level c, then (u j is bounded in W,p(x 0 (. Proof. Since (u j is a Palais-Smale sequence with energy level c, we have J λ (u j c and J λ(u j 0. But ( J λ (u = M p(x u p(x λ [ ] r+ F (x, u r + q(x u q(x, and with J λ(u.v = M Then ( p(x u p(x [ u p(x2 u v λ u q(x2 uv. C + u j J λ (u j θ J λ(u j u j, m p + m 0 < θ < ( A A 2 If we assume (u j is unbounded, we obtain C + u j ] r F (x, u f(x, uv r+ (β r+ (r + (β + r. (3.0 ( m0 p m u + j p, θ which is a contradiction because p >. Hence (u j is bounded in W,p(x 0 (. 3

14 Lemma 3.6 Let (u j W,p(x ( 0 ( be a Palais-Smale sequence, with energy level c. If c < θ (m 0 S N /p, where m 0 = min{m + /p 0, m 0 }, then index set I = and u j u strongly in L q(x (. Proof. Following the same steps as in Lemma ( 3.2, we show that the sequence (u j converges strongly in L q(x (, and that if c < θ (m 0 S N /p, where m 0 = min{m + 0, /p m 0 }, then index set I =. As J λ(u j 0 in ( W,p(x 0 we obtain But, J λ(u j (ϕ i,ε u j = lim J λ(u j (ϕ i,ε u j = 0. [ ( ] M p(x u j p(x u j p(x2 u j (ϕ i,ε u j [ ] r λ F (x, u j f(x, u j (ϕ i,ε u j u j q(x2 u j (ϕ i,ε u j. When j([ ( we have ] ( 0 = lim M p(x u j p(x u j p(x2 u j (ϕ i,ε u j + [ ] r M(t 0 ϕ i,ε dµ ϕ i,ε dν λ F (x, u f(x, u(ϕ i,ε u, where t 0 = lim t We may show that, p(x u j p(x. ( [ ( ] ( lim lim M ε 0 j p(x u j p(x you can see in Shang-Wang [2]. On the other hand, lim M(t 0 ϕ i,ε dµ = M(t 0 µ i ϕ(0, ε 0 and [ ] r λ F (x, u f(x, u(ϕ i,ε u 0, as ε 0. u j p(x2 u j (ϕ i,ε u j 0, lim ϕ i,ε dν = ν i ϕ(0 ϵ 0 Then, M(t 0 µ i ϕ(0 = ν i ϕ(0 implies that m 0 µ i ν i. By Sν /p (x i i µ /p(x i i and proceeding as in item (i we obtain, (m 0 S N ν i, where m 0 = min{m 0 /p +, m 0 /p }. If c is the energy level, using θ satisfying (3.0 and considering I, we have 4

15 (( m0 c lim p m ] u + j p(x + λar+ r+ θ θ (β + [ u r j β(x [ ] λ Ar+ 2 r+ u r + (β r+ j β(x + u j q(x u θ q j q(x. ( c lim u j q(x θ q(x u j q(x. Now setting A δ = (B δ (x = {x : dist(x, A < δ}, we obtain c lim A ( δ θ ( ( x A θ ν i δ q A δ u j q(x = θ (m 0 S N. δ Since δ > 0 is arbitrary and q is continuous, we get ( c θ (m 0 S N. Therefore, if c < in L q(x (. ( θ ( u q(x dx + ν i δ A δ i I ( θ (m 0 S N, the index set I is empty and u j u strongly Lemma 3.7 Let (u j W,p(x ( 0 ( be a Palais-Smale sequence, with energy level c. If c < θ (m 0 S N, there exist u W,p(x 0 ( and a subsequence, still denoted by (u j, such that u j u in W,p(x 0. Proof. From we have J λ(u j.(u j u = M Note that λ ( 0. [ m 0 J λ (u j 0, p(x u j p(x u j p(x2 u j (u j u ] r F (x, u j f(x, u j (u j u u j q(x2 u j (u j u ( p(x u j p(x m 5

16 and there exists nonnegative constants c and c 2 such that and [ c F (x, u j ] r c2. By Lemma 3.6, u j u in L q(x (, and using Hölder inequality we obtain If we take f(x, u j (u j u 0 u j q(x2 u j (u j u 0. 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 2.5 we have u j u in W,p(x( 0. Lemma 3.8 (i For all λ > 0, there are α >, ρ > 0, such that J λ (u α, u = ρ. (ii There exists an element w 0 W,p(x 0 ( with w 0 > ρ and J(w 0 < α. Proof. (i We have, ( J λ (u m 0 p(x u p(x λ [ r + Let ρ = u be, where u is small enough, m0 J λ (u ρ ( p+ λ ( A2 p + r + β and the result follows. (ii Take 0 < w W,p(x 0 (. For t >, J λ (tw m ( p tp+ w p(x + λ r + q + tq ] r+ A 2 β(x uβ(x u q(x. q r+ M β (r+ + ( ( A t β (r+ β + w q(x. q M q 2 ρ β (r+p +, w β(x r+ 6

17 Then we have and the proof is over. lim J λ(tw =, t By Lemma 3.8, we can use Mountain Pass Theorem [3], which guarantees the existence of a sequence (u j W,p(x 0 ( such that where a candidate for critical value is J λ (u j c and J λ(u j 0 in (W,p(x 0 (, c = inf sup J λ (h(t h C t [0,] and C = {h : [0, ] W,p(x 0 ( : h continuous and h(0 = 0, h( = w 0 }. For 0 < t < and fixing w W,p(x 0 (, J λ (tw m ( p tp w p(x λ ( ( A r+ t β (r+ w β(x r + β + q + tq w q(x and J λ (tw m ( p tp w p(x λ ( ( A r+ t β (r+ w β(x. r + β + ( m ã Defining g(t = t p λ ( A β bt (r+, p r + β + where ã = w p(x and b ( r+, = w β(x we obtain sup Jλ (tw g(t. Note that β m g(t have a critical point of maximum t λ = ã (r+p λ (, and that t A λ 0 β bβ + when λ. By continuity of J Then exists λ such that λ λ lim λ ( sup J λ (tw < t 0 This completes the proof of part (ii. sup J λ (tw t 0 = 0. ( θ (m 0 S N. 7

18 Proof. (iii Here we follow the same ideas of part (i and (ii and also assumes (.2, (.3 and (.4. Moreover, in this item, we assume M(t = t α α(p + α with α >, (p < α ( r+ A (β r+ (r + and αp + < β (r +. (β + r A 2 Lemma 3.9 If (u j W,p(x 0 ( is a Palais-Smale sequence, with energy level c, then (u j is bounded in W,p(x 0 (. Proof. Since (u j is a Palais-Smale sequence with energy level c, we have J λ (u j c and J λ(u j 0. But J λ (u = ( α α p(x u p(x λ [ ] r+ F (x, u r + q(x u q(x, and with J λ(u.v = Then ( α p(x u p(x [ u p(x2 u v λ u q(x2 uv. C + u j J λ (u j θ J λ(u j u j, ] r F (x, u f(x, uv α(p + α ( r+ (p < θ < A (β r+ (r +. (3. α A 2 (β + r If we assume (u j unbounded, we obtain ( C + u j α(p + u α θ(p α j p, which is a contradiction because p >. Hence (u j is bounded in W,p(x 0 (. Lemma 3.0 Let (u j W,p(x ( 0 ( be a Palais-Smale sequence, with energy level c. If c < θ (t S N /p, where t = min{t + /p, t } with t > 0, then index set I = and u j u strongly in L q(x (. Proof. Following the same steps as in Lemma ( 3.2, we show that the sequence (u j converges strongly in L q(x (, and that if c < θ (t S N /p, where t = min{t +, /p t } with t > 0, then index set I =. 8

19 As J λ(u j 0 in ( W,p(x 0 we obtain that But, J λ(u j (ϕ i,ε u j = lim J λ(u j (ϕ i,ε u j = 0. ( α p(x u j p(x u j p(x2 u j (ϕ i,ε u j [ ] r λ F (x, u j f(x, u j (ϕ i,ε u j u j q(x2 u j (ϕ i,ε u j. When j we have ( α ( 0 = lim p(x u j p(x u j p(x2 u j (ϕ i,ε u j + [ ] r t α 0 ϕ i,ε dµ ϕ i,ε dν λ F (x, u f(x, u(ϕ i,ε u, where t 0 = lim t We may show that, p(x u j p(x. ( α ( lim lim ε 0 j p(x u j p(x see Shang-Wang [2]. On the other hand, lim ε 0 tα 0 ϕ i,ε dµ = t α 0 µ i ϕ(0, lim ϵ 0 and [ ] r λ F (x, u u j p(x2 u j (ϕ i,ε u j 0, ϕ i,ε dν = ν i ϕ(0 f(x, u(ϕ i,ε u 0, as ε 0. Then, t α 0 µ i ϕ(0 = ν i ϕ(0 implies that t µ i ν i, with 0 < t t α 0. By Sν /p (x i i µ /p(x i i and proceeding as in item (i, we obtain, (t S N ν i, /p where t = min{t + /p, t }. If c is the energy ( ( level, using θ satisfying (3. and considering I, we have α c = lim α p(x u j p(x λ [ ] r+ F (x, u j r + q(x u j q(x u θ(p α j p(x + λ [ ] r F (x, u j f(x, u j u j + u j q(x, θ θ and 9

20 (( c lim α(p + u α θ(p α j p(x + λar+ θ (β + [ ] r+ u r j β(x A r+ [ ] λ 2 r+ u r + (β r+ j β(x + u j q(x θ q(x u j q(x ( c lim u j q(x θ q(x u j q(x. Considering again A δ = (B δ (x = {x : dist(x, A < δ}, we obtain x A ( c lim A δ θ ( u j q(x = δ θ ( u q(x dx + ν i δ A ( δ i I θ ν i ( δ θ (t S N. δ Since δ > 0 is arbitrary and q is continuous, we get Therefore, if c < in L q(x (. c ( θ (t S N. ( θ (t S N, the index set I is empty and u j u strongly Lemma 3. Let (u j W,p(x ( 0 ( be a Palais-Smale sequence, with energy level c. If c < θ (t S N, there exist u W,p(x 0 ( and a subsequence (u j such that u j u in W,p(x 0. Proof. From we have J λ(u j.(u j u = ( λ [ J λ(u j 0, α p(x u j p(x u j p(x2 u j (u j u ] r F (x, u j f(x, u j (u j u u j q(x2 u j (u j u. Note there exists nonnegative constants c, c 2, c 3 and c 4 such that 20

21 and and c ( [ c 3 α p(x u j p(x c 2 F (x, u j ] r c4. By Lemma 3.0, u j u in L q(x (, and using the Hölder inequality we obtain, If we take f(x, u j (u j u 0 u j q(x2 u j (u j u 0. 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 2.5 we have u j u in W,p(x( 0. Lemma 3.2 (i For all λ > 0, there are α >, ρ > 0, such that J λ (u α, u = ρ. (ii There exists an element w 0 W,p(x 0 ( with w 0 > ρ and J(w 0 < α. Proof. (i We have, ( [ α J λ (u u p(x λ α(p + α r + Let ρ = u be, where u is small enough, ( J λ (u ρ αp+ λ ( A2 α(p + α r + β and the result follows. (ii Take 0 < w W,p(x 0 (. For t >, ] r+ A 2 β(x uβ(x u q(x. q r+ M β (r+ + q M q 2 ρ β (r+αp +, 2

22 J λ (tw Then we have and the proof is complete. ( ( α α(p α tαp+ w p(x λ A + r + β + q + tq w q(x. lim J λ(tw =, t ( t β (r+ w β(x r+ By Lemma 3.2, we can use Mountain Pass Theorem [3], which guarantees the existence of a sequence (u j W,p(x 0 ( such that where a candidate for critical value is J λ (u j c and J λ(u j 0 in (W,p(x 0 (, c = inf sup J λ (h(t h C t [0,] and C = {h : [0, ] W,p(x 0 ( : h continuous and h(0 = 0, h( = w 0 }. For 0 < t < and fixing w W,p(x 0 (, ( ( α ( J λ (tw α(p α tαp w p(x λ A r+ t β (r+ w β(x r + β + q + tq w q(x and ( ( α ( J λ (tw α(p α tαp w p(x λ A r+ t β (r+ w β(x. r + β + ( ã Defining g(t = t αp λ ( A β bt (r+, α(p α r + β + ( α ( r+, where ã = w p(x and b = w β(x we obtain sup Jλ (tw g(t. Note ã that g(t have a critical point of maximum t λ = that t λ 0 when λ. By continuity of J Then exists λ such that λ λ lim λ ( sup J λ (tw < t 0 sup J λ (tw t 0 λ ( A β + bβ (p α = 0. ( θ (t S N. 22 β (r+αp, and

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