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doi:10.7151/dmdico.1154

EXISTENCE OF A NONTRIVAL SOLUTION FOR DIRICHLET PROBLEM INVOLVING p(x)-LAPLACIAN

Sylwia Barna´ s Cracow University of Technology

Institute of Mathematics ul. Warszawska 24 31-155 Krak´ ow, Poland e-mail: sbarnas@pk.edu.pl

Abstract

In this paper we study the nonlinear Dirichlet problem involving p(x)- Laplacian (hemivariational inequality) with nonsmooth potential. By us- ing nonsmooth critical point theory for locally Lipschitz functionals due to Chang [6] and the properties of variational Sobolev spaces, we establish conditions which ensure the existence of solution for our problem.

Keywords: p(x)-Laplacian, hemivariational inequality, Cerami condition, mountain pass theorem, variable exponent Sobolev space.

2010 Mathematics Subject Classification: 35A15, 35D30, 35J60, 35M10, 35M87.

1. Introduction

Let Ω ⊆ R

N

(where N > 2) be a bounded domain with a C

2

-boundary ∂Ω.

In this paper we study the following nonlinear elliptic differential inclusion with p(x)-Laplacian

( −∆

p(x)

u (x) − λ|u(x)|

p(x)−2

u (x) ∈ ∂j(x, u(x)) a.e. in Ω,

u = 0 on ∂Ω,

(1)

where p : Ω → R is a continuous function satisfying 1 < p

:= ess inf

x∈Ω

p(x) 6 p(x) 6 p

+

:= ess sup

x∈Ω

p (x) < b p

,

(2)

(2)

where

b p

:=

(

N p

N−p

p(x) < N,

∞ p(x) > N.

(3)

The operator

p(x)

u (x) := div(|∇u(x)|

p(x)−2

∇u(x))

is the so-called p(x)-Laplacian, which for p(x) = const becomes a p-Laplacian.

Problems with ∆

p(x)

u are more complicated than with ∆

p

u because they are usually inhomogeneous and possesses ”more nonlinearity”. The function j(x, t) is locally Lipschitz in the t-variable and measurable in x-variable and by ∂j(x, t) we denote the subdifferential with respect to the t-variable in the sense of Clarke [8].

In problem (1) appears λ, for which we will assume that λ < p

p

+

λ

and λ < (p

− 1)p

+

(p

+

− 1)p

λ

, (4)

where λ

is defined by

λ

= inf

u∈W01,p(x)(Ω)\{0}

R

|∇u(x)|

p(x)

dx R

|u(x)|

p(x)

dx . (5)

When we consider

e

p := min n (p

− 1)p

+

(p

+

− 1)p

, p

p

+

o

, then

λ < pλ e

. (6)

It may happen that λ

= 0 (see Fan-Zhang [14]).

Recently, the study of p(x)-Laplacian problems has attracted more and more attention. The hemivariational inequalities with Dirichlet or Neumann boundary condition have been considered by many authors. In particular in Ge-Xue [21]

and Qian-Shen [27], the following differential inclusion involving p(x)-Laplacian

is studied (

−∆

p(x)

u (x) ∈ ∂j(x, u(x)) a.e. in Ω,

u = 0 on ∂Ω,

where p : Ω → R is a continuous function satisfying (2). In the last paper the

existence of two solutions of constant sign is proved. Also the study of variational

problems is an interesing topic in recent years. For example in Fan-Zhang [13]

(3)

some sufficient conditions for the existence of solutions for the Dirichlet problem with p(x)-Laplacian is presented. Also in Ji [23], the existence of three solutions for a differential equation is proved.

Finally we have papers with differential inclusions involving p(x)-Laplacian of the type (1). In Ge-Xue-Zhou [22], authors proved sufficient conditions to obtain radial solutions for differential inclusions with p(x)-Laplacian. The au- thors required that λ < 0 and used a key assumption on the exponent that p

+

< N . Differential inclusions type (1) with Dirichlet or Neumann boundary condition were also studied in Qian-Shen [27], Qian-Shen-Yang [28] and Dai [9].

The authors considered an inclusion involving a weighted function which is indefinite. In Dai [9], the existence of infinitely many nonnegative solutions is proved. They refused the assumption about negativity of λ but still needed as- sumption on the variable exponent, it means √

2p

> N or

2(pp+)2

> N . This assumptions were necessary to show compact embedding from W

1,p(x)

(Ω) into C

0

(Ω). More problems like this was considered by Fan [11, 12], Fan-Zhang [13], Dai [9].

The problem (1) was also considered in Barna´s [3, 4]. In the literature we can meet the so-called Landesman-Lazer condition, Ambrosetti-Rabinowitz or Tang conditions as well as some other. These conditions are usually not necessary. The problem of finding the possible widest class of nonlinearities for which nonlinear Dirichlet boundary value problem with p(x)-Laplacian has a solution is still open.

This paper is an extension of the theory considered in the above mentioned pa- pers. In contrast to [3, 4], we change the interval in which can be our parametr λ and we use the most general condition about the bahaviour of the potential j in the infinity - the Tange-type condition. Moreover, the existence of solution under no restriction on the exponent p(x) will be proved. Our method is more direct and is based on the critical point theory for nonsmooth Lipschitz functionals due to Chang [6].

All the above mentioned papers deal with the operator of p(x)-Laplacian.

These problems have a lot of applications in modelling electroheological fluids (see Ru˘zi˘cka [26]), elastic mechanics (see Zhikov [30]) and image restoration (see Diening-H¨ ast¨ o-Harjulehto-Ruˇziˇcka [10]). It is worth mentioning that the electroheological liquids are used in devices damping vibration and absorbers, therefore research into these materials are so important. More applications can be found in the books of Acerbi-Mingione [1], Chen-Levine-Rao [7], Mihˇ ailescu- Rˇ adulescu [25].

For the convenience of the reader in the next section we briefly present the

basic notions and facts from the theory, which will be used in the study of prob-

lem (1). Moreover, we present the main properties of the general Lebesgue and

variable Sobolev spaces.

(4)

2. Mathematical preliminaries

Let X be a Banach space and X

its topological dual. By k · k we will denote the norm in X and by h·, ·i the duality brackets for the pair (X, X

). In analogy with the directional derivative of a convex function, we introduce the notion of the generalized directional derivative of a locally Lipschitz function f at x ∈ X in the direction h ∈ X by

f

0

(x; h) = lim sup

y→x,λց0

f (y + λh) − f (y)

λ .

The function h 7−→ f

0

(x, h) ∈ R is sublinear and continuous so it is the support function of a nonempty, w

-compact and convex set

∂f (x) = {x

∈ X

: hx

, h i 6 f

0

(x, h) for all h ∈ X}.

The set ∂f (x) is known as generalized or Clarke subdifferential of f at x. If f is convex, then ∂f (x) coincides with the subdifferential in the sense of convex analysis.

Let f : X → R be a locally Lipschitz function. From convex analysis it is well know that a proper, convex and lower semicontinuous function g : X → R = R ∪ {+∞} is locally Lipschitz in the interior of its effective domain dom(g) = {x ∈ X : g(x) < ∞}. A point x ∈ X is said to be a critical point of the locally Lipschitz function f : X → R, if 0 ∈ ∂f (x).

Lemma 2.1 (Qian-Shen-Zhu [29]). Let f, g : X → R be two locally Lipschitz functions. Then

(a) f

0

(x; h) = max{hξ, hi : ξ ∈ ∂f (x)};

(b) (f + g)

0

(x; h) 6 f

0

(x; h) + g

0

(x; h);

(c) (−f )

0

(x; h) = f

0

(x; −h) and f

0

(x, kh) = kf

0

(x; h) for every k > 0;

(d) the function (x; h) → f

0

(x; h) is upper semicontinuous.

For more details on the generalized subdifferentials we refer to Clarke [8] and Gasi´ nski-Papageorgiou [20].

We say that f satisfies the ”nonsmooth Cerami condition” (nonsmooth C- condition for short), if any sequence {x

n

}

n>1

⊆ X such that {f (x

n

)}

n>1

is bounded and (1 + kx

n

k)m(x

n

) → 0 as n → ∞, where m(x

n

) = min{kx

k

: x

∈ ∂f (x

n

)}, has a strongly convergent subsequence.

The first theorem is due to Chang [6] and extends to a nonsmooth settings

the well known ”mountain pass theorem” due to Ambrosetti-Rabinowitz [2].

(5)

Theorem 2.2. If X is a reflexive Banach space, R : X → R is a locally Lipschitz functional satisfying nonsmooth C-condition and for some ρ > 0 and y ∈ X such that kyk > ρ, we have

max{R(0), R(y)} < inf

kxk=ρ

{R(x)} =: η,

then R has a nontrivial critical point x ∈ X such that the critical value c = R(x) > η is characterized by the following minimax principle

c = inf

γ∈Γ

max

06τ 61

{R(γ(τ))}, where Γ = {γ ∈ C([0, 1], X) : γ(0) = 0, γ(1) = y}.

In order to discuss problem (1), we need to state some properties of the spaces L

p(x)

(Ω) and W

1,p(x)

(Ω), which we call generalized Lebesgue-Sobolev spaces (see Fan-Zhao [15, 16]).

Let

E (Ω) = {u : Ω −→ R : u is measurable}.

Two functions in E(Ω) are considered to be one element of E(Ω), when they are equal almost everywhere. Firstly, we define the variable exponent Lebesgue space by

L

p(x)

(Ω) = {u ∈ E(Ω) : Z

|u(x)|

p(x)

dx < ∞}, with the norm

kuk

p(x)

= kuk

Lp(x)(Ω)

= inf n λ > 0 :

Z

u(x)

λ

p(x)

dx 6 1 o .

Next, the generalized Lebesgue-Sobolev space W

1,p(x)

(Ω) is defined as W

1,p(x)

(Ω) = {u ∈ L

p(x)

(Ω) : ∇u ∈ L

p(x)

(Ω; R

N

)}

with the norm

kuk = kuk

W1,p(x)(Ω)

= kuk

p(x)

+ k∇uk

p(x)

.

Then (L

p(x)

(Ω), k·k

p(x)

) and (W

1,p(x)

(Ω), k·k) are separable and refelxive Banach spaces. By W

01,p(x)

(Ω) we denote the closure of C

0

(Ω) in W

1,p(x)

(Ω).

Lemma 2.3 (Fan-Zhao [15]). If Ω ⊂ R

N

is an open domain, then

(a) if q ∈ C(Ω) and 1 < p(x) 6 q(x) < p

(x) for almost all x ∈ Ω, where

(6)

p

(x) =

(

N p(x)

N−p(x)

p(x) < N

∞ p(x) > N,

then W

1,p(x)

(Ω) is embedded continuously and compactly in L

q(x)

(Ω);

(b) Poincar´e inequality in W

01,p(x)

(Ω) holds i.e., there exists a positive constant ˆ

c such that

kuk

p(x)

6 ˆ c k∇uk

p(x)

for all u ∈ W

01,p(x)

(Ω);

(c) (L

p(x)

(Ω))

= L

p(x)

(Ω), where

p(x)1

+

p1(x)

= 1 and for all u ∈ L

p(x)

(Ω) and v ∈ L

p(x)

(Ω), we have

Z

|uv|dx 6  1 p

+ 1

p

′−

 kuk

p(x)

kvk

p(x)

.

Lemma 2.4 (Fan-Zhao [15]). Let ϕ(u) = R

|u(x)|

p(x)

dx for u ∈ L

p(x)

(Ω) and let {u

n

}

n>1

⊆ L

p(x)

(Ω).

(a) for a 6= 0, we have

kuk

p(x)

= a ⇐⇒ ϕ(

ua

) = 1;

(b) we have

kuk

p(x)

< 1 (respectively = 1, > 1) ⇐⇒ ϕ(u) < 1 (respectively = 1, > 1);

(c) if kuk

p(x)

> 1, then

kuk

pp(x)

6 ϕ (u) 6 kuk

pp(x)+

; (d) if kuk

p(x)

6 1, then

kuk

pp(x)+

6 ϕ (u) 6 kuk

pp(x)

. Similarly to Lemma 2.4, we have the following result.

Lemma 2.5 (Fan-Zhao [15]). Let Φ(u) = R

(|∇u(x)|

p(x)

+ |u(x)|

p(x)

)dx for u ∈ W

1,p(x)

(Ω) and let {u

n

}

n>1

⊆ W

1,p(x)

(Ω). Then

(a) for a 6= 0, we have

kuk = a ⇐⇒ Φ( u a ) = 1;

(b) we have

kuk < 1 (respectively = 1, > 1) ⇐⇒ Φ(u) < 1 (respectively = 1, > 1);

(7)

(c) if kuk > 1, then

kuk

p

6 Φ(u) 6 kuk

p+

; (d) if kuk 6 1, then

kuk

p+

6 Φ(u) 6 kuk

p

. Consider the following function

J (u) = Z

1

p(x) |∇u|

p(x)

dx, for all u ∈ W

01,p(x)

(Ω).

We know that J ∈ C

1

(W

01,p(x)

(Ω)) and −div(|∇u|

p(x)−2

∇u) is the derivative op- erator of J in the weak sense (see Chang [5]). We denote

A = J

: W

01,p(x)

(Ω) → (W

01,p(x)

(Ω))

, then

hAu, vi = Z

|∇u(x)|

p(x)−2

(∇u(x), ∇v(x))

RN

dx (7)

for all u, v ∈ W

01,p(x)

(Ω).

Lemma 2.6 (Fan-Zhang [13]). If A is the operator defined above, then A is a continuous, bounded and strictly monotone operator of type (S)

+

i.e., if u

n

→ u weakly in W

01,p(x)

(Ω) and lim sup

n→∞

hAu

n

, u

n

− ui 6 0, implies that u

n

→ u in W

01,p(x)

(Ω).

3. Existence of Solution

We start by introducing our assumptions for the nonsmooth potential j(x, t).

H (j) j : Ω × R → R is a function such that j(x, t) satisfies j(x, 0) = 0 almost everywhere on Ω and

(i) for all t ∈ R, the function Ω ∋ x → j(x, t) ∈ R is measurable;

(ii) for almost all x ∈ Ω, the function R ∋ t → j(x, t) ∈ R is locally Lipschitz;

(iii) for almost all x ∈ Ω and all v ∈ ∂j(x, t), we have

|v| ≤ a(x) + c

1

|t|

r(x)−1

,

(8)

with a ∈ L

+

(Ω), c

1

> 0 and r ∈ C(Ω) such that p

+

< r

:= min

x∈Ω

r(x) 6 r(x) 6 r

+

:= max

x∈Ω

r (x) < b p

;

(iv) there exists µ > 0 such that lim sup

t→0

j(x, t)

|t|

p(x)

6 −µ, uniformly for almost all x ∈ Ω.

As for the behaviour of j in +∞ and −∞, we will consider one of the following two different conditions:

H(j)

1

(v) we have

lim sup

|t|→∞

v

t − j(x, t)

|t|

p(x)

6 0, uniformly for almost all x ∈ Ω and all v

∈ ∂j(x, t);

(vi) there exists u ∈ W

01,p(x)

(Ω) \ {0}, such that 1

p

Z

|∇u(x)|

p(x)

dx + λ

p

Z

|u(x)|

p(x)

dx 6 Z

j(x, u(x))dx,

where λ

:= max{0, −λ}.

H(j)

2

(v) there exists constants ν > p

+

and T > 0 such that νj (x, t) 6 −j

0

(x, t; −t) and ess inf j(·, t) > 0, for almost all x ∈ Ω and all t ∈ R such that |t| > T .

Remark 3.1. Hypothesis H(j)

1

(vi) can be replaced by (vi)’ there exists u ∈ W

01,p(x)

(Ω) \ {0}, such that

c kuk

p+

6 Z

j(x, u(x))dx, if kuk > 1, or

c kuk

p

6 Z

j(x, u(x))dx, if kuk 6 1, where c := max{

p1

,

λp

}.

Condition (vi)’ is more restrective, but easier to verify then condition (vi).

(9)

Lemma 3.2. If hypotheses H (j) and H(j)

2

hold, then (a) the function

f : k → 1

k

ν

j(x, kt)

is locally Lipschitz on (0, +∞) for almost all x ∈ Ω and all t ∈ R\{0};

(b) there exist constant l > 0 such that

j (x, t) > l|t|

ν

, for almost all x ∈ Ω and all t such that |t| > T .

Proof. Suppose that U is a compact set in (0, +∞). From H(j)(ii), we know that there exists K > 0 such that

|j(x, u

1

) − j(x, u

2

)| 6 K|u

1

− u

2

| for all u

1

, u

2

∈ tU.

(8)

Now, let us fix t ∈ R\{0} and x ∈ Ω. For some k

1

, k

2

∈ U, we have

|f (k

1

) − f (k

2

)| = 1

k

1ν

j(x, k

1

t ) − 1

k

2ν

j(x, k

2

t) 6 1

k

ν1

|j(x, k

1

t ) − j(x, k

2

t )| + |k

2ν

j(x, k

2

t ) − k

1ν

j(x, k

2

t )|

k

ν1

k

2ν

= 1

k

ν1

|j(x, k

1

t ) − j(x, k

2

t )| + |j(x, k

2

t )|

k

1ν

k

2ν

|k

ν2

− k

1ν

|.

Using (8), we obtain

|f (k

1

) − f (k

2

)|

6 K

k

ν1

|k

1

− k

2

| |t| + |j(x, k

2

t )|

k

1ν

k

2ν

|k

1

− k

2

| |k

1ν−1

k

2

+ k

1ν−2

k

22

+ . . . + k

1

k

ν−12

|.

Since U is a compact set, we have

|f (k

1

) − f (k

2

)| 6 s|k

1

− k

2

|,

for some s = s(x, t, U ) > 0. This implies that the function f is locally Lipschitz on (0, +∞).

Moreover, when we consider the subdifferential in the sense of Clarke of the function f , we obtain

∂f (k) = ∂  1

k

ν

j(x, kt) 

⊆ −ν

k

ν+1

j(x, kt) + 1

k

ν

∂j(x, kt)t,

(10)

for all k ∈ (0, +∞). By virtue of the Lebourg mean value theorem for locally Lipschitz functions, for k > 1 we can choose ξ ∈ (1, k), such that

f (k) − f (1) ∈ h −ν

ξ

ν+1

j(x, ξt) + 1

ξ

ν

∂j(x, ξt)t i (k − 1)

= k − 1

ξ

ν+1

(−νj(x, ξt) + ∂j(x, ξt)ξt), (9)

for almost all x ∈ Ω and all t ∈ R\{0}. From definition of subdifferential in the sense of Clarke, we have

hη, −ξti 6 j

0

(x, ξt; −ξt), for all η ∈ ∂j(x, ξt).

Combinig this with (9) and using H(j)

2

(v), we obtain k − 1

ξ

ν+1

(−νj(x, ξt) + ∂j(x, ξt)ξt) > k − 1

ξ

ν+1

(−νj(x, ξt) − j

0

(x, ξt; −ξt)) > 0 for almost all x ∈ Ω and all t such that |t| > T . Thus from (9), it follows that f (k) − f (1) =

k1ν

j (x, kt) − j(x, t) > 0, and so

k

ν

j(x, t) 6 j(x, kt),

for almost all x ∈ Ω, all k > 1 and t such that |t| > T. Therefore, j(x, t) = j(x, t

T T ) > t

ν

T

ν

j(x, T ), for t > T, and

j(x, t) = j(x, t

T (−T )) > t

ν

T

ν

j (x, −T ), for t 6 −T.

It follows that

j (x, t) > l|t|

ν

,

where l =

T1ν

min{ess inf j(·, T ), ess inf j(·, −T )} > 0 for almost all x ∈ Ω and all t such that |t| > T .

Remark 3.3. (i) If H(j) and H(j)

1

hold, then exist T > 0 such that for almost all x ∈ Ω, all v

∈ ∂j(x, t) and some ε > 0, we have that v

t − j(x, t) 6 ε for

|t| > T . On the other hand, from the definition of subdifferential in the sense

(11)

of Clarke, we have −v

t 6 j

0

(x, t; −t). Hence

j (x, t) + ε > −j

0

(x, t; −t) (10)

for all t such that |t| > T .

(ii) If H(j) and H(j)

2

hold, then for almost all x ∈ Ω and all v

∈ ∂j(x, t), we have

j (x, t) 6 νj(x, t) 6 −j

0

(x, t; −t) 6 v

t

for all t such that |t| > T where ν > p

+

. So from Lemma 3.2, we know that lim sup

|t|→∞

v

t − j(x, t)

|t|

p(x)

> lim sup

|t|→∞

(ν − 1)j(x, t)

|t|

p(x)

> lim sup

|t|→∞

(ν − 1)l|t|

ν

|t|

p(x)

= +∞, uniformly for almost all x ∈ Ω and all v

∈ ∂j(x, t).

(iii) Hypotheses H(j)

1

and H(j)

2

exclude each other.

We introduce locally Lipschitz functional R : W

01,p(x)

(Ω) → R defined by R(u) =

Z

1

p(x) |∇u(x)|

p(x)

dx − Z

λ

p(x) |u(x)|

p(x)

dx − Z

j(x, u(x))dx,

for all u ∈ W

01,p(x)

(Ω).

Lemma 3.4. If hypotheses H(j) and H(j)

1

hold and λ ∈ (−∞,

(p(p+−1)p−1)p+

λ

), then R satisfies the nonsmooth C-condition.

Proof. Let {u

n

}

n>1

⊆ W

01,p(x)

(Ω) be a sequence such that {R(u

n

)}

n>1

is bounded and (1 + ku

n

k)m(u

n

) → 0 as n → ∞, where m(u

n

) = min{ku

k

: u

∈ ∂R(u

n

)}.

We will show that {u

n

}

n>1

⊆ W

01,p(x)

(Ω) is bounded.

Because |R(u

n

)| 6 M for all n > 1, we have

−M 6 Z

1

p(x) |∇u

n

(x)|

p(x)

dx − Z

λ

p(x) |u

n

(x)|

p(x)

dx − Z

j(x, u

n

(x))dx.

(11)

Since ∂R(u

n

) ⊆ (W

01,p(x)

(Ω))

is weakly compact, nonempty and the norm func- tional is weakly lower semicontinuous in a Banach space, then we can find u

n

∂R(u

n

) such that ku

n

k

= m(u

n

), for n > 1.

(12)

Consider the operator A : W

01,p(x)

(Ω) → (W

01,p(x)

(Ω))

defined by (7). Then, for every n > 1, we have

u

n

= Au

n

− λ|u

n

|

p(x)−2

u

n

− v

n

, (12)

where v

n

∈ ∂ψ(u

n

) ⊆ L

p(x)

(Ω), for n > 1, with

p(x)1

+

p(x)1

= 1 and ψ : W

01,p(x)

(Ω) → R is defined by ψ(u

n

) = R

j(x, u

n

(x))dx. We know that, if v

n

∈ ∂ψ(u

n

), then v

n

(x) ∈ ∂j(x, u

n

(x)) (see Clarke [8]).

From the choice of the sequence {u

n

}

n>1

⊆ W

01,p(x)

(Ω), at least for a subse- quence, we have

|hu

n

, w i| 6 ε

n

kwk

1 + ku

n

k for all w ∈ W

01,p(x)

(Ω), (13)

with ε

n

ց 0. Putting w = u

n

in (13) and using (12), we obtain

−ε

n

6 − Z

|∇u

n

(x)|

p(x)

dx + λ Z

|u

n

(x)|

p(x)

dx + Z

v

n

(x)u

n

(x)dx.

(14)

Now, let us consider two cases.

Case 1. Let λ < 0. We define λ

:= max{0, −λ}.

From (11) and (14), we have

−M − ε

n

6  1

p

− 1  Z

|∇u

n

(x)|

p(x)

dx + λ

 1

p

− 1  Z

|u

n

(x)|

p(x)

dx +

Z

v

n

(x)u

n

(x)dx − Z

j(x, u

n

(x))dx.

(15)

By virtue of hypotheses H(j)

1

(v), we know that for some sufficiently small con- stant c > 0 there exist constant L

1

> 0, such that

v

t − j(x, t)

|t|

p(x)

6 c,

uniformly for almost all x ∈ Ω and all t such that t > L

1

. It immediately follows that

v

t − j(x, t) 6 c|t|

p(x)

. (16)

On the other hand, from the Lebourg mean value theorem (see Clarke [8]), for almost all x ∈ Ω and all t ∈ R, we can find v ∈ ∂j(x, kt) with 0 < k < 1, such that

|j(x, t) − j(x, 0)| 6 |v||t|.

(13)

So from hypothesis H(j)(iii), for almost all x ∈ Ω, we have

|j(x, t)| 6 a(x)|t| + c

1

|t|

r(x)

6 a (x)|t| + c

1

|t|

r+

+ c

2

,

for some c

2

> 0. Then for almost all x ∈ Ω and all t such that |t| < L

1

, it follows that

|j(x, t)| 6 c

3

, (17)

for some c

3

> 0. Therefore, from (16) and (17) it follows that for almost all x ∈ Ω and all t ∈ R, we have

v

t − j(x, t) 6 c|t|

p(x)

+ c

4

, (18)

for some c

4

> 0. From (15), we obtain λ



1 − 1 p

 Z

|u

n

(x)|

p(x)

dx 6 M + ε

n

+ Z

v

n

(x)u

n

(x)dx − Z

j(x, u

n

(x))dx.

If we use (18), we get λ



1 − 1 p

 Z

|u

n

(x)|

p(x)

dx 6 M + ε

n

+ c Z

|u

n

(x)|

p(x)

dx + Z

c

4

dx, for all n > 1, which leads to

h λ



1 − 1 p

 − c i Z

|u

n

(x)|

p(x)

dx 6 c

5

,

for some c

5

:= M + ε

1

+ c

4

|Ω| > 0. Because c is some sufficiently small positive constant, so we know that λ



1 −

p1

 − c > 0 and

the sequence {u

n

}

n>1

⊆ L

p(x)

(Ω) is bounded (19)

(see Lemma 2.4 (c) and (d)).

Now, consider again (15). We obtain

 1 − 1 p

 Z

|∇u

n

(x)|

p(x)

dx 6 M + ε

n

+ Z

v

n

(x)u

n

(x)dx − Z

j(x, u

n

(x))dx.

In a similar way, by using (18) we have

 1 − 1

p

 Z

|∇u

n

(x)|

p(x)

dx 6 M + ε

n

+ c Z

|u

n

(x)|

p(x)

dx + Z

c

4

dx,

(14)

for all n > 1. From (19) we know that {u

n

}

n>1

⊆ L

p(x)

(Ω) is bounded, so

 1 − 1 p

 Z

|∇u

n

(x)|

p(x)

dx 6 c

6

, where c

6

> 0. Because 

1 −

p1



> 0, we have that

the sequence {∇u

n

}

n>1

⊆ L

p(x)

(Ω; R

N

) is bounded (20)

(see Lemma 2.4 (c) and (d)).

From (19) and (20), we deduce that

the sequence {u

n

}

n>1

⊆ W

01,p(x)

(Ω) is bounded (see Lemma 2.5 (c) and (d)).

Case 2. Now, let λ > 0.

Again from (11) and (14), we have

−M − ε

n

6  1

p

− 1  Z

|∇u

n

(x)|

p(x)

dx + λ  1 − 1

p

+

 Z

|u

n

(x)|

p(x)

dx +

Z

v

n

(x)u

n

(x)dx − Z

j(x, u

n

(x))dx.

(21)

From the definition of λ

(in Case 2 we have that λ

> 0), we get λ

Z

|u

n

(x)|

p(x)

dx 6 Z

|∇u

n

(x)|

p(x)

dx, (22)

for all n > 1. Using this fact in (21), we have h

λ

 1 − 1

p

 + λ  1

p

+

− 1 i Z

|u

n

(x)|

p(x)

dx 6 M + ε

n

+

Z

v

n

(x)u

n

(x)dx − Z

j(x, u

n

(x))dx.

(23)

In a similar way like in Case 1, by using (18) in (23), we obtain h λ



1 − 1 p

 + λ  1 p

+

− 1 

− c i Z

|u

n

(x)|

p(x)

dx 6 c

7

,

(15)

for some c

7

:= M + ε

1

+ c

4

|Ω| > 0. From sufficiently small c > 0, we know that λ

(1 −

p1

) + λ(

p1+

− 1) − c > 0, hence

the sequence {u

n

}

n>1

⊆ L

p(x)

(Ω) is bounded (24)

(see Lemma 2.4 (c) and (d)).

Now, again from (21), we have

 1 − 1 p

 Z

|∇u

n

(x)|

p(x)

dx 6 M + ε

n

+ λ  1 − 1

p

+

 Z

|u

n

(x)|

p(x)

dx +

Z

v

n

(x)u

n

(x)dx − Z

j(x, u

n

(x))dx.

(25)

Using (18) and (24) in (25), we obtain

 1 − 1 p

 Z

|∇u

n

(x)|

p(x)

dx 6 c

8

, for some c

8

> 0. Since 1 −

p1

> 0, then we have that

the sequence {∇u

n

}

n>1

⊆ L

p(x)

(Ω; R

N

) is bounded (26)

(see Lemma 2.4 (c) and (d)).

From (24) and (26), we have that

the sequence {u

n

}

n>1

⊆ W

01,p(x)

(Ω) is bounded (see Lemma 2.5 (c) and (d)).

From Cases 1 and 2, we have that

the sequence {u

n

}

n≥1

⊆ W

01,p(x)

(Ω) is bounded.

Hence, by passing to a subsequence if necessary, we may assume that u

n

→ u weakly in W

01,p(x)

(Ω),

u

n

→ u in L

p(x)

(Ω), (27)

for some u ∈ W

1,p(x)

(Ω). Putting w = u

n

− u in (13) and using (12), we obtain

hAu

n

, u

n

−ui−λ Z

|u

n

(x)|

p(x)−2

u

n

(x)(u

n

−u)(x)dx−

Z

v

n

(x)(u

n

−u)(x)dx 6 ε

n

,

(16)

with ε

n

ց 0. If we pass to the limit as n → ∞, we get lim sup

n→∞

hAu

n

, u

n

− ui 6 0

(see Barna´s [3, 4]). So from Lemma 2.6, we have that u

n

→ u in W

01,p(x)

(Ω) as n → ∞. Thus R satisfies the (C)-condition.

Lemma 3.5. If hypotheses H(j) and H(j)

2

hold, then R satisfies the nonsmooth C-condition.

Proof. Let {u

n

}

n>1

⊆ W

01,p(x)

(Ω) be a sequence such that {R(u

n

)}

n>1

is bounded and (1 + ku

n

k)m(u

n

) → 0 as n → ∞. We will show that {u

n

}

n>1

⊆ W

01,p(x)

(Ω) is bounded.

Since |R(u

n

)| 6 M for all n > 1, then we have Z

1

p(x) |∇u

n

(x)|

p(x)

dx − Z

λ

p(x) |u

n

(x)|

p(x)

dx − Z

j(x, u

n

(x))dx 6 M, (28)

so Z

|∇u

n

(x)|

p(x)

dx − λ

+

p

+

p

Z

|u

n

(x)|

p(x)

dx − p

+

Z

j(x, u

n

(x))dx 6 p

+

M, (29)

where λ

+

:= max{λ, 0}. From the choice of the sequence {u

n

}

n>1

⊆ W

01,p(x)

(Ω), at least for a subsequence, we have

|hu

n

, w i| 6 ε

n

kwk

1 + ku

n

k for all w ∈ W

01,p(x)

(Ω), (30)

with ε

n

ց 0 and u

n

is like (12) in Lemma 3.4. Taking w = u

n

in (30) and using (12), we obtain

− Z

|∇u

n

(x)|

p(x)

dx + λ Z

|u

n

(x)|

p(x)

dx + Z

v

n

(x)u

n

(x)dx 6 ε

n

. (31)

From the definition of subdifferential in the sense of Clarke, we have hv

n

, −u

n

i 6 j

0

(x, u

n

; −u

n

), for some v

n

∈ ∂j(x, u

n

(x)).

It follows that

hv

n

, u

n

i > −j

0

(x, u

n

; −u

n

),

(17)

for all n > 1. Using this fact in (31), we obtain

− Z

|∇u

n

(x)|

p(x)

dx − λ

Z

|u

n

(x)|

p(x)

dx

− Z

j

0

(x, u

n

(x); −u

n

(x))dx 6 ε

n

, (32)

where λ

:= max{−λ, 0}. Adding (29) and (32), we have

− 

λ

+ λ

+

p

+

p

 Z

|u

n

(x)|

p(x)

dx − Z

p

+

j(x, u

n

(x)) + j

0

(x, u

n

(x); −u

n

(x))dx 6 ε

n

+ M p

+

. Thus,

− Z

νj(x, u

n

(x)) + j

0

(x, u

n

(x); −u

n

(x))dx + (ν − p

+

) Z

j(x, u

n

(x))dx 6 ε

n

+ M p

+

+ 

λ

+ λ

+

p

+

p

 Z

|u

n

(x)|

p(x)

dx, (33)

for almost all x ∈ Ω and all n > 1, where ν > p

+

. From H(j)(iii) and the Lebourg mean value theorem, similarly as in Lemma 3.4, we can show that for almost all x ∈ Ω and all |t| < T , there exists constants c

9

, c

10

> 0 such that

|j(x, t)| 6 c

9

and |j

0

(x, t; −t)| 6 c

10

. So

− Z

{|un|<T }

(νj(x, u

n

(x)) + j

0

(x, u

n

(x); −u

n

(x)))dx > −c

11

, (34)

for some c

11

> 0. From H(j)

2

(v), we know that there exists constants T, c

12

> 0 such that

− Z

{|un|>T }

(νj(x, u

n

(x)) + j

0

(x, u

n

(x); −u

n

(x)))dx > −c

12

. (35)

Using (34) and (35), we obtain

− Z

(νj(x, u

n

(x)) + j

0

(x, u

n

(x); −u

n

(x)))dx > −c

13

, (36)

where c

13

:= c

11

+ c

12

> 0.

(18)

Now, let us consider (ν − p

+

)

Z

j(x, u

n

(x))dx = (ν − p

+

)  Z

{|un|>T }

j(x, u

n

(x))dx

+ Z

{|un|<T }

j(x, u

n

(x))dx  . (37)

Again by the use of the Lebourg mean value theorem, inequality (37) and Lemma 3.2, we have

(ν − p

+

) Z

j(x, u

n

(x))dx > (ν − p

+

)  l

Z

|u

n

(x)|

ν

dx − c

9

|Ω| 

> c

14

ku

n

k

νν

− c

15

, (38)

for some c

14

, c

15

> 0 and all n > 1.

Using (36) and (38) in (33), we have c

14

ku

n

k

νν

6 ε

n

+ M p

+

+ c

13

+ c

15

+ 

λ

+ λ

+

p

+

p

 Z

|u

n

(x)|

p(x)

dx.

Since ν > p

+

for almost all x ∈ Ω, so it follows that

the sequence {u

n

}

n>1

⊆ L

ν

(Ω) is bounded.

For any n > 1 such that ku

n

k

p(x)

6 1 we have ku

n

k

pp(x)+

<

Z

|u

n

(x)|

p(x)

dx <

Z

(1 + |u

n

(x)|

p

)dx < |Ω| + ku

n

k

νν

6 c

16

, for some c

16

> 0 (see Lemma 2.4). On the other hand, for any n > 1 such that ku

n

k

p(x)

> 1, we have

ku

n

k

pp(x)

<

Z

|u

n

(x)|

p(x)

dx <

Z

(1 + |u

n

(x)|

p+

)dx < |Ω| + ku

n

k

νν

6 c

17

. Thus

the sequence {u

n

}

n>1

⊆ L

p(x)

(Ω) is bounded.

(39)

Now, consider again (28) and multiply it by ν > p

+

to obtain Z

ν

p

+

|∇u

n

(x)|

p(x)

dx − Z

λ

+

ν

p

|u

n

(x)|

p(x)

dx − ν Z

j(x, u

n

(x))dx 6 νM,

(40)

(19)

for all n > 1. Adding (32) and (40), we get Z

 ν p

+

− 1 

|∇u

n

(x)|

p(x)

dx − 

λ

+ λ

+

ν p

 Z

|u

n

(x)|

p(x)

dx

− Z

νj(x, u

n

(x) + j

0

(x, u

n

(x); −u

n

(x))dx 6 ε

n

+ νM.

From (39) we know that the sequence {u

n

}

n≥1

⊆ L

p(x)

(Ω) is bounded and using the inequality (36), we obtain

 ν

p

+

− 1  Z

|∇u

n

(x)|

p(x)

dx < c

18

, for some c

18

> 0 and all n > 1. Since

pν+

− 1 > 0, then

the sequence {∇u

n

}

n>1

⊆ L

p(x)

(Ω, R

N

) is bounded.

(41)

From (39) and (41), we have that

the sequence {u

n

}

n>1

⊆ W

1,p(x)

(Ω) is bounded.

The rest of proof is similar as the proof of Lemma 3.4.

Lemma 3.6. If hypothesis H (j) hold, λ <

pp+

λ

and θ ∈ (r

+

, p b

), then there exist β

1

, β

2

> 0 such that for all u ∈ W

01,p(x)

(Ω) with kuk < 1, we have

R(u) > β

1

kuk

p+

− β

2

kuk

θ

.

Proof. From hypothesis H(j)(iv), we can find δ > 0, such that for almost all x ∈ Ω and all t satisfying |t| 6 δ, we have

j (x, t) 6 − µ 2 |t|

p(x)

.

On the other hand, from hypothesis H(j)(iii), we know that for almost all x ∈ Ω and all t such that |t| > δ, we have

|j(x, t)| 6 a

1

|t| + c

1

|t|

r(x)

,

for some a

1

, c

1

> 0. Thus for almost all x ∈ Ω and all t ∈ R we get j (x, t) 6 − µ

2 |t|

p(x)

+ γ|t|

θ

, (42)

with some γ > 0 and r

+

< θ < p b

.

(20)

Let us consider two cases.

Case 1. Let λ 6 0.

By using (42), we obtain that R(u) =

Z

1

p(x) |∇u(x)|

p(x)

dx − Z

λ

p(x) |u(x)|

p(x)

dx − Z

j(x, u(x))dx

>

Z

1

p

+

|∇u(x)|

p(x)

dx + Z

µ

2 |u(x)|

p(x)

dx − γ Z

|u(x)|

θ

dx.

So, we have

R(u) > β

1

h Z

|∇u(x)|

p(x)

dx + Z

|u(x)|

p(x)

dx i

− γkuk

θθ

,

where β

1

:= min{

p1+

,

µ2

}.

Case 2. Let λ > 0.

Using (42) and (22), we obtain that R(u) =

Z

1

p(x) |∇u(x)|

p(x)

dx − Z

λ

p(x) |u(x)|

p(x)

dx − Z

j(x, u(x))dx

>

Z

1

p

+

|∇u(x)|

p(x)

dx − Z

|λ|

p

|u(x)|

p(x)

dx +

Z

µ

2 |u(x)|

p(x)

dx − γ Z

|u(x)|

θ

dx

> 1 p

+

Z

|∇u(x)|

p(x)

dx + µ 2

Z

|u(x)|

p(x)

dx

− |λ|

λ

p

Z

|∇u(x)|

p(x)

dx − γkuk

θθ

=  1

p

+

− |λ|

λ

p

 Z

|∇u(x)|

p(x)

dx + µ 2

Z

|u(x)|

p(x)

dx − γkuk

θθ

. We have that

p1+

λ|λ|p

> 0, so

R(u) > β

1

h Z

|∇u(x)|

p(x)

dx + Z

|u(x)|

p(x)

dx i

− γkuk

θθ

,

where β

1

:= min{

p1+

λ|λ|p

,

µ2

}.

(21)

As θ < p

(x), then W

01,p(x)

(Ω) is embedded continuously in L

θ

(Ω) (see Lemma 2.3(c)). So there exists ̺ > 0 such that

kuk

θ

6 ̺ kuk for all u ∈ W

01,p(x)

(Ω).

(43)

Using (43) and Lemma 2.5(d), for all u ∈ W

01,p(x)

(Ω) with kuk < 1, we have R(u) > β

1

kuk

p+

− β

2

kuk

θ

,

where β

2

= γ̺

θ

.

Using Lemmata 3.4, 3.5 and 3.6, we can prove the following existence theorem for problem (1).

Theorem 3.7. If hypotheses H(j) and H(j)

1

hold, λ < pλ e

(see (6)), then problem (1) has a nontrival solution.

Proof. From Lemma 3.6 we know that there exist β

1

, β

2

> 0, such that for all u ∈ W

01,p(x)

(Ω) with kuk < 1, we have

R(u) > β

1

kuk

p+

− β

2

kuk

θ

= β

1

kuk

p+

 1 − β

2

β

1

kuk

θ−p+

 .

Since p

+

< θ , if we choose ρ ∈ (0, 1) small enough, we will deduce that R(u) > L, for all u ∈ W

01,p(x)

(Ω), with kuk = ρ and some L > 0.

Now, let u ∈ W

01,p(x)

(Ω) be as in hypothesis H(j)

1

(vi). We have R(u) =

Z

1

p(x) |∇u(x)|

p(x)

dx − Z

λ

p(x) |u(x)|

p(x)

dx − Z

j(x, u(x))dx

6 1

p

Z

|∇u(x)|

p(x)

dx + λ

p

Z

|u(x)|

p(x)

dx − Z

j(x, u(x))dx.

From hyphothesis H(j)

1

(vi), we get R(u) 6 0. This permits the use of Theorem 2.2 which gives us u ∈ W

01,p(x)

(Ω) such that R(u) > 0 = R(0) and 0 ∈ ∂R(u).

From the last inclusion we obtain

0 = Au − λ|u|

p(x)−2

u − v

, where v

∈ ∂ψ(u). Hence

Au = λ|u|

p(x)−2

u + v

,

so for all v ∈ C

0

(Ω), we have hAu, vi = λh|u|

p(x)−2

u, v i + hv

, v i.

(22)

So we have

Z

|∇u(x)|

p(x)−2

(∇u(x), ∇v(x))

RN

dx

= Z

λ |u(x)|

p(x)−2

u(x)v(x)dx + Z

v

(x)v(x)dx, for all v ∈ C

0

(Ω).

From the definition of the distributional derivative we have

( −div(|∇u(x)|

p(x)−2

∇u(x)) = λ|u(x)|

p(x)−2

u(x) + v(x) in Ω,

u = 0 on ∂Ω,

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so (

−∆

p(x)

u (x) − λ|u(x)|

p(x)−2

u (x) ∈ ∂j(x, u(x)) in Ω,

u = 0 on ∂Ω.

(45)

Therefore u ∈ W

01,p(x)

(Ω) is a nontrivial solution of (1).

Theorem 3.8. If hypotheses H(j) and H(j)

2

hold, λ <

pp+

λ

, then problem (1) has a nontrival solution.

Proof. From Lemma 3.6 we know that for all u ∈ W

01,p(x)

(Ω) with kuk < 1, we have that R(u) > L with some L > 0.

Using Lemma 3.2(b), for any u ∈ W

01,p(x)

(Ω)\{0} we have R(tu) =

Z

1

p(x) |∇tu(x)|

p(x)

dx − Z

λ

p(x) |tu(x)|

p(x)

dx − Z

j(x, tu(x))dx

6 t

p+

 1 p

Z

|∇u(x)|

p(x)

dx + λ

p

Z

|u(x)|

p(x)

dx 

− Z

j(x, tu(x))dx 6 c · t

p+

 Z

(|∇u(x)|

p(x)

+ |u(x)|

p(x)

)dx 

− l · t

ν

Z

|u(x)|

ν

dx + L

2

, where λ

:= max{0, −λ}, c = max{

p1

,

λp

}.

Since ν > p

+

, then we get that R(tu) → −∞ when t → ∞. This permits the use of Theorem 2.2 which gives us u ∈ W

01,p(x)

(Ω) such that R(u) > 0 = R(0) and 0 ∈ ∂R(u).

Therefore u ∈ W

01,p(x)

(Ω) is a nontrival solution of (1).

Remark 3.9. A nonsmooth potential satisfying hypotheses H(j) and H(j)

1

is

for example the one given by the following function (see Rys 1):

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j

1

(x, t) =

 

 

−µ|t|

p(x)

if |t| 6 1,

(µ + σ − |2|

p(x)

)|t| − 2µ − σ + |2|

p(x)

if 1 < |t| 6 2, ln |t| + σ − |2|

p(x)

− ln |2| if |t| > 2,

with µ, σ > 0 and continuous function p : Ω → R which satisfies 1 < p

6 p(x) 6 p

+

< p b

.

- 6

Rys. 1

t j1(x, t)

A A A A A A

     

1 2

−1

−2

−µ σ− 2p(x)

A nonsmooth potential satisfying hypotheses H(j) and H(j)

2

is for example the one given by the following function (see Rys 2):

- 6

Rys. 2

t j2(x, t)

−1 1

−µ

j

2

(x, t) =

( −µ|t|

p(x)

if |t| 6 1,

|t|

q(x)

− µ − 1 if |t| > 1,

with µ > 0 and continuous functions p, q : Ω → R which satisfies 1 < p

6 p(x) 6 p

+

< q

6 q(x) 6 q

+

< p b

.

Of course j

1

does not satisfy H(j)

2

and j

2

does not satisfy H(j)

1

(see Remark

3.3).

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Received 21 June 2013

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