Electronic Journal of Differential Equations, Vol. 2012 (2012), No. 25, pp. 1–12. ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu ftp ejde.math.txstate.edu EXISTENCE OF SOLUTIONS FOR DISCONTINUOUS p(x)-LAPLACIAN PROBLEMS WITH CRITICAL EXPONENTS XUDONG SHANG, ZHIGANG WANG Abstract. In this article, we study the existence of solutions to the problem − div(|∇u|p(x)−2 ∇u) = λ|u|p u=0 ∗ (x)−2 u + f (u) x ∈ Ω, x ∈ ∂Ω, where Ω is a smooth bounded domain in RN , p(x) is a continuous function with N p(x) 1 < p(x) < N and p∗ (x) = N −p(x) . Applying nonsmooth critical point theory for locally Lipschitz functionals, we show that there is at least one nontrivial solution when λ less than a certain number, and f maybe discontinuous. 1. Introduction and statement of main results In recent years, the study of problems in differential equations involving variable exponents has been a topic of interest. This is due to their applications in image restoration, mathematical biology, the study of dielectric breakdown, electrical resistivity, polycrystal plasticity, the growth of heterogeneous sandpiles and fluid dynamics, etc. We refer the reader to [4, 5, 6, 12, 14, 20, 26] and references therein for more information. In this article, we discuss the existence of solutions to the problem ∗ − div(|∇u|p(x)−2 ∇u) = λ|u|p (x)−2 u + f (u) x ∈ Ω, u = 0 x ∈ ∂Ω, (1.1) where Ω is a smooth bounded domain in RN , p(x) is a continuous function defined p(x) on Ω with 1 < p(x) < N , p∗ (x) = NN−p(x) , and λ > 0. The function f (u) can have discontinuities, so that functionals associated with (1.1) may not be differentiable, and standard variational techniques can not be applied. There are many publications for the case when p(x) is a constant function; see for example [1, 2, 3, 9, 24]. For the existence of solutions for p(x)-Laplacian problems we refer the reader to [7, 11, 13, 16, 19, 22]. The existence of solutions for p(x)-Laplacian problems with critical growth is relatively new. In 2012, Bonder and Silva [8] extended the concentration-compactness 2000 Mathematics Subject Classification. 35J92, 35J70, 35R70. Key words and phrases. p(x)-Laplacian problem; critical Sobolev exponents; discontinuous nonlinearities. c 2012 Texas State University - San Marcos. Submitted November 7, 2011. Published February 7, 2012. 1 2 X. SHANG, Z. WANG EJDE-2012/25 principle of Lions to the variable exponent spaces and proved the existence of solutions to the problem −∆p(x) u = |u|q(x)−2 u + λ(x)|u|r(x)−2 u x ∈ Ω, u = 0 x ∈ ∂Ω. Where Ω is a smooth bounded domain in RN , with q(x) ≤ p∗ (x) and the set {q(x) = p∗ (x)} 6= ∅, we can find a similar result in [15]. Fu [17] studied the existence of solutions for p(x)-Laplacian equationd involving the critical exponent and obtained a sequence of radially symmetric solutions. In the present paper, we study the discontinuous p(x)-Laplacian problems with critical growth for (1.1). To handle the gaps at the discontinuity points, our approach uses nonsmooth critical point theory for locally Lipschitz functionals, we obtain some general results for the simple case when f has only one point of discontinuity. 1,p(x) (Ω) is a solution Because f is discontinuous, we say that a function u ∈ W0 of the multivalued problem associated to (1.1) if u satisfies ∗ −∆p(x) u − λ|u|p (x)−2 u ∈ fb(u) a.e. in Ω, where fb(u) is the multivalued function fb(u) = [f (u), f (u)] with f (t) = lim inf f (s), s→t f (t) = lim sup f (s). s→t In this article, we assume f : R → R is a measurable function satisfying: (F1) f (t) = 0 if t ≤ 0 and for all t ∈ R, there exist the limits: f (t + 0) = lim+ f (t + δ); δ→0 f (t − 0) = lim+ f (t − δ). δ→0 (F2) there exist C1 , C2 > 0 such that |f (t)| ≤ C1 + C2 |t|q(x)−1 , where q(x) ∈ C(Ω) such that p(x) < q(x) < p∗ (x). (F3) f (t) = o(|t|p(x)−1 ) as t → 0. Rt (F4) f (t)t ≥ q− F (t) > 0, for all t ∈ R\{0}, where F (t) = 0 f (s)ds. Note that by hypothesis (F1), f (u) = max{f (u − 0), f (u + 0)}, f (u) = min{f (u − 0), f (u + 0)}. Theorem 1.1. Suppose f satisfies (F1)-(F4). Then there exists λ0 > 0 such that (1.1) has a nontrivial solution for every λ ∈ (0, λ0 ). One of the main motivations is to consider the particular case associated with (1.1), ∗ −∆p(x) u = λ|u|p (x)−2 u + bh(u − a)|u|q(x)−2 u x ∈ Ω, (1.2) u = 0 x ∈ ∂Ω, where h(t) = 0 if t ≤ 0 and h(t) = 1 if t > 0, a and b are positive real parameters, p(x) < q(x) < p∗ (x). As a direct consequence of Theorem 1.1, we have Theorem 1.2. For every a, b > 0, there exists λ0 > 0 such that for every λ ∈ (0, λ0 ), Equation (1.2) has a nontrivial solution satisfying meas{x ∈ Ω : u(x) > a} > 0. The rest of this article is organized as follows: In section 2 we introduce some necessary preliminary knowledge; in section 3 contains the proof of our main results. EJDE-2012/25 p(x)-LAPLACIAN PROBLEMS WITH CRITICAL EXPONENTS 3 2. Preliminaries First, we recall some definitions and properties of generalized gradient of locally Lipschitz functionals, which will be used later. Let X be a Banach space, X ∗ be its topological dual and h·, ·i be the duality. A functional I : X → R is said to be locally Lipschitz if for every u ∈ X there exists a neighborhood U of u and a constant K > 0 depending on U such that |I(u) − I(v)| ≤ Kku − vk, ∀u, v ∈ U. For a locally Lipschitz functional I, we define the generalized directional derivative at u ∈ X in the direction v ∈ X by I(u + h + δv) − I(u + h) I 0 (u; v) = lim sup . δ h→0, δ↓0 It is easy to show that I 0 (u; v) is subadditive and positively homogeneus. The generalized gradient of I at u is the set ∂I(u) = {w ∈ X ∗ : I 0 (u; v) ≥ hw, vi, ∀v ∈ X}. Then, for each v ∈ X, I 0 (u; v) = max{hω, vi : ω ∈ ∂I(u)}. A point u ∈ X is a critical point of I if 0 ∈ ∂I(u). It is easy to see that if u ∈ X is a local minimum or maximum, then 0 ∈ ∂I(u). Next, we recall some definitions and basic properties of the generalized Lebesgue1,p(x) Sobolev spaces Lp(x) (Ω) and W0 (Ω), where Ω ⊂ RN is an arbitrary domain with smooth boundary. Set C+ (Ω) = {p(x) ∈ C(Ω) : p(x) > 1, ∀x ∈ Ω}, p+ = max p(x), x∈Ω p− = min p(x). x∈Ω For any p(x) ∈ C+ (Ω), we define the variable exponent Lebesgue space Z Lp(x) (Ω) = {u ∈ M (Ω) : |u(x)|p(x) dx < ∞}, Ω with the norm Z u | |p(x) dx ≤ 1}, Ω µ where M (Ω) is the set of all measurable real functions defined on Ω. Define the space |u|p(x) = inf{µ > 0 : 1,p(x) W0 (Ω) = {u ∈ Lp(x) (Ω) : |∇u| ∈ Lp(x) (Ω)}, with the norm kuk = |u|p(x) + |∇u|p(x) . Proposition 2.1 ([18, 21]). There is a constant C > 0 such that for all u ∈ 1,p(x) W0 (Ω), |u|p(x) ≤ C|∇u|p(x) . 1,p(x) So |∇u|p(x) and kuk are equivalent norms in W0 (Ω). Hence we will use the 1,p(x) norm kuk = |∇u|p(x) for all u ∈ W0 (Ω). R Proposition 2.2 ([18, 21]). Set ρ(u) = Ω |u|p(x) dx. For u, un ∈ Lp(x) (Ω), we have: (1) |u|p(x) < 1 (= 1; > 1) ⇔ ρ(u) < 1 (= 1; > 1). 4 X. SHANG, Z. WANG (2) (3) (4) (5) p EJDE-2012/25 p − + If |u|p(x) > 1, then |u|p(x) ≤ ρ(u) ≤ |u|p(x) . p+ p− If |u|p(x) < 1, then |u|p(x) ≤ ρ(u) ≤ |u|p(x) . limn→∞ un = u ⇔ limn→∞ ρ(un − u) = 0. limn→∞ |un |p(x) = ∞ ⇔ limn→∞ ρ(un ) = ∞. Proposition 2.3 ([18]). If q(x) ∈ C+ (Ω) and q(x) < p∗ (x) for any x ∈ Ω, the imbedding W 1,p(x) (Ω) → Lq(x) (Ω) is compact. Proposition 2.4 ([21]). The conjugate space of Lp(x) (Ω) is Lq(x) (Ω), where 1 q(x) p(x) = 1. For any u ∈ L Z q(x) (Ω) and v ∈ L |uv|dx ≤ ( Ω 1 p(x) + (Ω), 1 1 + )|u|p(x) |v|q(x) . p− q− s(x) Proposition 2.5 ([16]). If |u|q(x) ∈ L q(x) (Ω), where q(x), s(x) ∈ L∞ + (Ω), q(x) ≤ s(x), then u ∈ Ls(x) (Ω) and there is a number q ∈ [q− , q+ ] such that ||u|q(x) | s(x) = q(x) (|u|s(x) )q . 1,p(x) Let Iλ (u) : W0 Z Iλ (u) = Ω (Ω) → R be the energy functional defined as Z Z 1 1 p∗ (x) |∇u|p(x) dx − λ |u| dx − F (u)dx, ∗ p(x) Ω p (x) Ω (2.1) R denote Φ(u) = Ω F (u)dx. We say that Iλ (u) satisfies the nonsmooth (P S)c condition, if any sequence {un } ⊆ X such that Iλ (un ) → c and m(un ) = min{kwkX ∗ : w ∈ ∂Iλ (un )} → 0, as n → ∞, possesses a convergent subsequence. To prove our main results, we use the generalizations of the mountain pass theorem [10]. Theorem 2.6. Let X be a reflexive Banach space, I : X → R is locally Lipschitz functional which satisfies the nonsmooth (P S)c condition, I(0) = 0 and there are ρ, r > 0 and e ∈ X with kek > r, such that I(u) ≥ β if kuk = r and I(e) ≤ 0. Then there exists u ∈ X such that 0 ∈ ∂I(u) and I(u) = c. Where c = inf max I(γ(t)), γ∈Γ 0≤t≤1 Γ = {γ ∈ C([0, 1], X)|γ(0) = 0, γ(1) = e}. Recall the concentration-compactness principle for variable exponent spaces. This will be the keystone that enable us to verify that Iλ satisfies the nonsmooth (P S)c condition. 1,p(x) Proposition 2.7 ([8]). Let {un } converge weakly to u in W0 (Ω) such that ∗ |un |p (x) and |∇un |p(x) converge weakly to nonnegative measures ν and µ on RN respectively. Then, for some countable set J, we have: P ∗ (i) ν = |u|p (x) + j∈J νj δxj , P (ii) µ ≥ |∇u|p(x) + j∈J µj δxj , 1 p∗ (xj ) (iii) Sνj 1 p(xj ) ≤ µj , EJDE-2012/25 p(x)-LAPLACIAN PROBLEMS WITH CRITICAL EXPONENTS 5 where xj ∈ Ω, δxj is the Dirac measure at xj , νj and µj are constants and S is the best constant in the Gagliardo-Nirenberg-Sobolev inequality for variable exponents, namely kuk1,p(x) 1,p(x) : u ∈ W0 (Ω), u 6= 0 . S = inf |u|p∗ (x) 3. Proof of main results In this section, we denote by un * u the weak convergence of a sequence un to 1,p(x) u in W0 (Ω), and o(1) denote a real vanishing sequence, C and Ci , i = 1, 2, . . . are positive constants, |A| is the Lebesgue measure of A and p0 (x) as the conjugate 1,p(x) function of p(x). u ∈ W0 (Ω) is called a solution of (1.1) if u is a critical point of Iλ (u) and satisfies ∗ − div(|∇u|p(x)−2 ∇u) − λ|u|p (x)−2 u ∈ [f (u), f (u)] a.e. x ∈ Ω. 1,p(x) Lemma 3.1. The function Φ(u) is locally Lipschitz on W0 1,p(x) Proof. By (F2), Proposition 2.4 and 2.5, for all u, v ∈ W0 Z Z v |Φ(u) − Φ(v)| ≤ |f (t)|dtdx ZΩ Zuv ≤ |C1 + C2 |t|q(x)−1 |dtdx Ω ≤ (|C1 | (Ω). (Ω), u q(x) q(x)−1 q−1 + C3 |u|q−1 q(x) + C3 |v|q(x) )|u − v|q(x) . 1,p(x) From Proposition 2.3, we obtain that there is a neighborhood U ⊂ W0 such that |Φ(u) − Φ(v)| ≤ Kku − vk, (Ω) of u 1,p(x) where K > 0 depends on max{kuk, kvk}. So, Φ(u) is locally Lipschitz in W0 The proof is complete. (Ω). From Lemma 3.1, by Chang’s results we have that Iλ (u) is locally Lipschitz and 0 1,p(x) ω ∈ ∂Iλ (u) if and only if there is ω ∈ W −1,p (x) (Ω) such that for all ϕ ∈ W0 (Ω), Z Z Z ∗ |∇u|p(x)−2 ∇u∇ϕdx − λ |u|p (x)−2 uϕdx − ωϕdx, (3.1) hω, ϕi = Ω Ω Ω and ω(x) ∈ [f (u(x)), f (u(x))] a.e. x ∈ Ω. (3.2) 1,p(x) Lemma 3.2. Assume (F1), (F2). Let {un } be a bounded sequence in W0 (Ω) such that Iλ (un ) → c and m(un ) → 0. Then there exists a subsequence (denoted 1,p(x) again by un ) and some u ∈ W0 (Ω), such that |∇un |p(x)−2 ∇un * |∇u|p(x)−2 ∇u p(x) weakly in [L p(x)−1 (Ω)]N . Proof. The proof is similar to that of [25, Theorem 1]. Because {un } is bounded 1,p(x) 1,p(x) in W0 (Ω), there exist a subsequence and u ∈ W0 (Ω) such that un * u in 1,p(x) p(x) W0 (Ω) and un → u in L (Ω) as n → ∞. We claim that the set J given by Proposition 2.7 is finite. Choose a function ϕ(x) ∈ C0∞ (RN ) such that 0 ≤ ϕ(x) ≤ 1, ϕ(x) ≡ 1 on B(0, 1) and ϕ(x) ≡ 0 on 6 X. SHANG, Z. WANG EJDE-2012/25 x−x RN \ B(0, 2). Let ϕj,ε (x) = ϕ( ε j ), for any x ∈ RN , > 0 and j ∈ J. It is clear 1,p(x) 1,p(x) that {ϕj,ε un } ⊂ W0 (Ω) for any j ∈ J, and is bounded in W0 (Ω). Take ϕ = ϕj,ε un in hωn , ϕi, we obtain Z Z p(x)−2 |∇un | ∇un · un ∇ϕj,ε dx + |∇un |p(x) ϕj,ε dx Ω Ω Z Z (3.3) ∗ ω n ϕj,ε un dx = o(1). −λ |un |p (x) ϕj,ε dx − Ω Ω 1,p(x) Taking into account that ωn ∈ ∂Iλ (un ), by (F2) and un * u in W0 0 −1,p (x) 0 infer that ω n is bounded in W (Ω), and so there exists ω 0 ∈ W such that 0 ω n * ω 0 in W −1,p (x) (Ω) and ω 0 ∈ [f (u), f (u)]. Let n → ∞ in (3.3), by Proposition 2.7, we have Z lim |∇un |p(x)−2 ∇un · un ∇ϕj,ε dx n→∞ Ω Z Z Z =λ ϕj,ε dν − ϕj,ε dµ + ω 0 ϕj,ε u dx. Ω Ω (Ω), we −1,p (x) (Ω) (3.4) (3.5) Ω By Hölder inequality it is easy to check that Z 0 ≤ | lim |∇un |p(x)−2 ∇un ∇ϕj,ε · un dx| n→∞ ≤ Z Ω p+ |∇un | dx p+p −1 Z Z |∇un |p− dx p−p −1 Z + |un |p− |∇ϕj,ε |p− dx − p1 − Ω Ω Z Z p+ −1 ≤ C4 ( |∇un |p+ dx) p+ ∗ |un |(p+ ) dx (p 1 )∗ + B(xj ,2ε) Ω + C5 p1 Ω Ω + |un |p+ |∇ϕj,ε |p+ dx + Z |∇un |p− dx p−p −1 Z ∗ |un |(p− ) dx − (p 1 )∗ − → 0, as → 0. B(xj ,2ε) Ω From (3.5), as → 0, we obtain λνj = µj . From Proposition 2.7, we conclude that νj = 0 or νj ≥ S N max{λ − pN + ,λ − pN − }. (3.6) It implies that J is a finite set. Without loss of generality, let J = {1, 2, . . . , m}. For any δ > 0, we denote Ωδ = {x ∈ Ω|dist(x, xj ) > δ}. Choose R large enough such that Ω ⊂ {x ∈ RN ||x| < R}, ψ(x) ∈ C ∞ (RN ), 0 ≤ ψ(x) ≤ 1, ψ(x) ≡ 0 on B(0, 2R) and ψ(x) ≡ 1 on RN \ B(0, 3R). Take > 0 small enough such that B(xi , ) ∩ B(xj , ) = ∅, ∀i, j ∈ J, i 6= j and ∪m j=1 B(xj , ) ⊂ B(0, 2R). We take ψ (x) = 1 − m X ϕj,ε − ψ(x), x ∈ RN . j=1 Then ψ (x) ∈ C ∞ (RN ), supp ψ ⊂ B(0, 3R) with ψ (x) = 0 on ∪m j=1 B(xj , /2) and ψ (x) = 1 on (RN \ B(xj , )) ∩ B(0, 2R). EJDE-2012/25 p(x)-LAPLACIAN PROBLEMS WITH CRITICAL EXPONENTS 7 1,p(x) As {ψ un } is bounded in W0 (Ω), let ϕ = ψ un in hωn , ϕi, we obtain Z Z p(x)−2 |∇un | ∇un · un ∇ψ dx + |∇un |p(x) ψ dx Ω Ω Z Z ∗ ω n ψ un dx = o(1). −λ |un |p (x) ψ dx − Ω By (3.4) and un * u in Ω 1,p(x) W0 (Ω), Z lim n→∞ we can easily obtain Z ω n ψ un dx = ω 0 ψ udx. Ω Ω P p∗ (x) + j∈J νj δxj , we obtain Since ψ (x) = 0 on ∪m j=1 B(xj , 2 ) and ν = |u| Z Z Z ∗ ∗ lim |un |p (x) ψ dx = ψ dν = |u|p (x) ψ dx. n→∞ Ω Ω Ω Hence Z p(x) |∇un | lim n→∞ Ω Z ψ dx = lim (− |∇un |p(x)−2 ∇un · un ∇ψ dx) n→∞ Ω Z Z ∗ ω 0 ψ udx. +λ |u|p (x) ψ dx + Ω (3.7) Ω In the same way, taking ϕ = ψ u in hωn , ϕi, we obtain Z Z p(x)−2 |∇un | ∇un · u∇ψ dx + |∇un |p(x)−2 ∇un ∇u · ψ dx Ω Ω Z Z ∗ −λ |un |p (x)−2 un uψ dx − ω n ψ udx = o(1). Ω Ω Thus Z |∇un |p(x)−2 ∇un ∇u · ψ dx Z Z Z (3.8) ∗ = lim (− |∇un |p(x)−2 ∇un · u∇ψ dx) + λ |u|p (x) ψ dx + ω 0 ψ udx. lim n→∞ Ω n→∞ Ω Ω Ω So, from (3.7) and (3.8), as n → ∞, we have Z 0≤ (|∇un |p(x)−2 ∇un − |∇u|p(x)−2 ∇u)(∇un − ∇u)dx Ωδ Z ≤ ψ (|∇un |p(x)−2 ∇un − |∇u|p(x)−2 ∇u)(∇un − ∇u)dx Ω Z |∇un |p(x)−2 ∇un ∇ψ · (u − un )dx = Ω Z + ψ |∇u|p(x)−2 ∇u · (∇u − ∇un )dx + o(1) Ω ≤ k∇ψ k∞ · ||∇un |p(x)−1 |p0 (x) · |u − un |p(x) Z + ψ |∇u|p(x)−2 ∇u · (∇un − ∇u)dx + o(1). Ω Thus we have Z lim n→∞ Ωδ (|∇un |p(x)−2 ∇un − |∇u|p(x)−2 ∇u)(∇un − ∇u)dx = 0. (3.9) 8 X. SHANG, Z. WANG EJDE-2012/25 Denote gn (x) = (|∇un |p(x)−2 ∇un − |∇u|p(x)−2 ∇u)(∇un − ∇u), then gn (x) ≥ 0, and by (3.9), gn (x) → 0 a.e. on Ωδ . Let E be a compact subset of Ωδ , suppose gn (x) → 0 a.e. on E. If ∇un were not convergence to ∇u everywhere on E, there would at least exist x0 ∈ E such that lim ∇un (x0 ) 6= ∇u(x0 ). n→∞ Then |∇un (x0 )|p(x0 ) = |∇un (x0 )|p(x0 )−2 ∇un (x0 )∇u(x0 ) + |∇u(x0 )|p(x0 )−2 ∇un (x0 )∇u(x0 ) − |∇u(x0 )|p(x0 ) + gn (x0 ). By the interpolation inequality, ||∇un (x0 )|p(x0 )−2 ∇un (x0 )∇u(x0 )| ≤ |∇un (x0 )|p(x0 )−1 · |∇u(x0 )| ≤ 1 |∇un (x0 )|p(x0 ) + C1 |∇u(x0 )|p(x0 ) , and ||∇u(x0 )|p(x0 )−2 ∇un (x0 )∇u(x0 )| ≤ |∇u(x0 )|p(x0 )−1 · |∇un (x0 )| ≤ 2 |∇u(x0 )|p(x0 ) + C2 |∇un (x0 )|p(x0 ) . We choose 1 , 2 properly, because gn (x0 ) is bounded, then |∇u(x0 )| ≤ C. Let ∇u(x0 ) = η, so we can assume ∇un (x0 ) → η 6= η. Thus gn (x0 ) → (|η|p(x0 )−2 η − |η|p(x0 )−2 η)(η − η) > 0. This contradicts gn (x0 ) → 0. Hence, ∇un (x0 ) → ∇u(x0 ) everywhere on E. So ∇un (x0 ) → ∇u(x0 ) a.e. on Ωδ . Since δ is arbitrary, we obtain ∇un (x0 ) → ∇u(x0 ) a.e. on Ω. Since {|∇un |p(x)−2 ∇un } is integrable in L1 (Ω), we obtain that as n → ∞, |∇un |p(x)−2 ∇un * |∇u|p(x)−2 ∇u p(x) weakly in [L p(x)−1 (Ω)]N . The proof is complete Lemma 3.3. Suppose f satisfies (F2), (F4). Then Iλ satisfies the nonsmooth (P S)c condition provided c < ( p1+ − 1 N q− )S max{λ 1− pN + ,λ 1− pN − }. 1,p(x) Proof. Let {un } ⊂ W0 (Ω) be such that Iλ (un ) → c and m(un ) → 0 as n → ∞. We must show the existence of a subsequence of {un } which converges strongly in 1,p(x) W0 (Ω). First, we show that {un } is bounded. We know that Z Z Z 1 1 p(x) p∗ (x) |∇un | dx − λ |un | dx − F (un )dx Iλ (un ) = p∗ (x) Ω Ω p(x) Z ZΩ Z (3.10) 1 λ p(x) p∗ (x) ≥ |∇un | dx − ∗ |un | dx − F (un )dx. p+ Ω p− Ω Ω Let ωn ∈ ∂Iλ (un ) such that kωn k = m(un ) = o(1). From (3.1) we have Z Z Z p(x) p∗ (x) hωn , un i = |∇un | dx − λ |un | dx − ω n un dx, Ω Ω Ω (3.11) EJDE-2012/25 p(x)-LAPLACIAN PROBLEMS WITH CRITICAL EXPONENTS 9 where ω n (x) ∈ [f (un ), f (un )] for a.e. x ∈ Ω. By (F4) we obtain 1 1 ω n un ≥ f (un )un ≥ F (un ). q− q− (3.12) From (3.10), (3.11) and (3.12), we obtain 1 1 1 hωn , un i ≥ ( − ) C6 (1 + kun k) ≥ Iλ (un ) − q− p+ q− Z |∇un |p(x) dx. (3.13) Ω If kun k > 1, by Proposition 2.2, we obtain ( 1 1 − )kun kp− ≤ C6 (1 + kun k). p+ q− 1,p(x) Thus {un } is bounded in W0 (Ω). Then there exist a subsequence and u ∈ ∗ 1,p(x) 1,p(x) (Ω) such that un * u in W0 (Ω), so we know that {|un |p (x)−2 un ϕ} is W0 uniformly integrable in L1 (Ω). By this fact, Lemma 3.2 and m(un ) → 0, taking n → ∞ in hωn , ϕi, we have Z Z Z ∗ 0= |∇u|p(x)−2 ∇u∇ϕdx − λ |u|p (x)−2 uϕdx − ωϕdx, ∀ ϕ ∈ C0∞ (RN ). Ω Ω Ω So we derive that ∗ − ∆p(x) u − λ|u|p (x)−2 u ∈ [f (u), f (u)]. (3.14) Now we applying Proposition 2.7 to prove that νj = 0 in (3.6). Assume νj 6= 0 for some j ∈ J. From (3.13), we have Z 1 1 1 hωn , un i ≥ ( − ) |∇un |p(x) dx. Iλ (un ) − q− p+ q− Ω Since Iλ (un ) → c and m(un ) → 0, using Proposition 2.7, taking n → ∞, we obtain Z 1 1 1 1 X c≥( − ) |∇u|p(x) dx + ( − ) µj p+ q− Ω p+ q− j∈J 1 1 N 1− N 1− N ≥( − )S max{λ p+ , λ p− }. p+ q− Since c < ( p1+ − have 1 N q− )S max{λ 1− pN Z + ,λ p∗ (x) |un | Ω 1− pN − }, then νj = 0 for all j ∈ J. Hence we Z dx → ∗ |u|p (x) dx. (3.15) Ω So we can use [15, Lemma 2.1]. Set vn = un − u and we have Z Z Z ∗ ∗ ∗ |un |p (x) dx = |vn |p (x) dx + |u|p (x) dx + o(1), Z Ω ZΩ ZΩ |∇un |p(x) dx = |∇vn |p(x) dx + |∇u|p(x) dx + o(1). Ω Ω (3.16) (3.17) Ω ∗ Thus, by (3.15) and (3.16), un → u strongly in Lp (x) (Ω). From (3.11), using (3.14) and (3.17), we obtain Z Z Z Z p(x) p(x) p∗ (x) hωn , un i = |∇vn | dx + |∇u| dx − λ |un | dx − ω n un dx + o(1). Ω Ω Ω Ω 10 X. SHANG, Z. WANG EJDE-2012/25 By (3.4) and (3.15), letting n → ∞, we conclude that Z |∇vn |p(x) dx → 0. Ω 1,p(x) This fact and Proposition 2.2 imply that un → u strongly in W0 is complete. (Ω). The proof Lemma 3.4. Suppose f satisfies (F2), (F3). Then, for every λ > 0, there are α, ρ > 0, such that Iλ (u) ≥ α, kuk = ρ. Proof. By (F2) and (F3), we have ∗ |f (t)| ≤ |t|p(x)−1 + C|t|q(x) ≤ |t|p(x)−1 + C()|t|p (x)−1 . Therefore, Z Z Z 1 1 p(x) p∗ (x) Iλ (u) = |∇u| dx − λ |u| dx − F (u)dx p∗ (x) Ω p(x) Ω Z ZΩ Z ∗ λ + C() 1 p(x) p(x) |∇u| dx − |u| dx − |u|p (x) dx. ≥ p+ Ω p− Ω p∗− Ω (3.18) we can take kuk < 1 sufficiently small such that |u|p(x) < 1 and |u|p∗ (x) < 1. From (3.18), Propositions 2.1 and 2.4, and the definition of S, using the usual arguments, we obtain 1 λ + C() p∗− p− Iλ (u) ≥ kukp+ − |u|p(x) − |u|p∗ (x) p+ p− p∗− ∗ λ + C() −p∗− 1 kukp+ − ≥ S kukp− ∗ 2p+ p− ∗ λ + C() −p∗− 1 − =( S kukp− −p+ )kukp+ . ∗ 2p+ p− Considering λ + C() −p∗− p∗− −p+ 1 S ktk − , 2p+ p∗− since p+ < p∗− , we have g(t) → 2p1+ as t → 0. Hence, there exists ρ > 0 such that g(ρ) > 0. So, we obtain α and ρ > 0, such that g(t) = Iλ (u) ≥ α, kuk = ρ. The proof is complete. 1,p(x) Next, we choose ϕ(x) ∈ W0 (Ω), such that kϕk = 1. Lemma 3.5. Suppose f satisfies (F4). Then, there exists λ0 > 0, t0 > 0 such that Iλ (t0 ϕ) < 0, and for all λ ∈ (0, λ0 ), 1 N 1 1− N 1− N − )S max{λ p+ , λ p− }. sup Iλ (tϕ) < ( p+ q− t≥0 Proof. By (F4), we have f (u)u ≥ f (u)u ≥ q− F (u), ∀u 6= 0. This implies F (tu) ≥ tq− F (u), for all t ≥ 1. Then, for any t > 1, ∗ Z Z Z tp ∗ tp+ λtp+ − ∗ |ϕ|p (x) dx − F (tϕ)dx ≤ + − tq− F (ϕ)dx = J1 (tϕ). Iλ (tϕ) ≤ p− p+ Ω p− Ω Ω EJDE-2012/25 p(x)-LAPLACIAN PROBLEMS WITH CRITICAL EXPONENTS 11 Since q− > p+ and F (ϕ) > 0, there exists t0 > 0 sufficiently large such that Iλ (t0 ϕ) < 0 and kt0 ϕk > ρ with ρ given by Lemma 3.4. If 0 ≤ t < 1, then Z tp− Iλ (tϕ) ≤ − F (tϕ)dx = J2 (tϕ). p− Ω Let J(tϕ) = max{J1 (tϕ), J2 (tϕ)}, so we have sup Iλ (tϕ) ≤ sup J(tϕ). t≥0 t≥0 Hence, we can find λ0 > 0 such that sup Jλ (tϕ) < ( t≥0 1 N 1 1− N 1− N − )S max{λ p+ , λ p− }. p+ q− So, for all λ ∈ (0, λ0 ), we have sup Iλ (tϕ) < ( t≥0 1 N 1 1− N 1− N − )S max{λ p+ , λ p− }. p+ q− The proof is complete. Proof of Theorem 1.1. It is obvious that Iλ (0) = 0. By Lemmas 3.1, 3.3–3.5, according to Theorem 2.6, there exist λ0 > 0, and for all λ ∈ (0, λ0 ), we can find 1,p(x) an u ∈ W0 (Ω) such that Iλ (u) > 0 and 0 ∈ ∂Iλ (u). Hence, u is a nontrivial solution of (1.1). The proof is complete. Proof of Theorem 1.2. In (1.2), f (u) = bh(u−a)|u|q(x)−2 u has only one discontinu1,p(x) (Ω) ity point a, so by the consequence of Theorem 1.1, we obtain that an u ∈ W0 is a nontrivial nonnegative solution of (1.2). That is, ∗ − ∆p(x) u − λ|u|p (x)−2 u ∈ fb(u) a.e. in Ω where fb(u) is the multivalued function given by ( {f (s)} s 6= a, fb(s) = [0, bh(x)uq(x)−1 ] s = a. (3.19) (3.20) If V = {x ∈ Ω : u(x) = a} exists, by (3.19) and (3.20), we have ∗ −∆p(x) u − λ|u|p (x)−2 u ∈ [0, bh(x)uq(x)−1 ] a.e. in V. Using the Morrey-Stampacchia’s theorem [23], we have −∆p(x) u = 0 a.e. x ∈ V . So ∗ −λap (x)−1 ≥0 a.e. in V. This is a contradiction. Thus |V | = 0. The proof is complete. Acknowledgments. The author wants to thank the anonymous referees for their carefully reading this paper and their useful comments. 12 X. SHANG, Z. WANG EJDE-2012/25 References [1] C. O. Alves, A. M. Bertone; A discontinuous problem involving the p-Laplacian operator and critical exponent in RN , Electron. J. Differential Equations. Vol. 2003(2003), no. 42, 1–10. [2] C. O. Alves, A. M. Bertone, J. V. Goncalves; A variational approach to discontinuous problems with critical Sobolev exponents, J. Math. Anal. 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