Three Solutions For A Quasi-Linear Elliptic Problem

Three Solutions For A Quasi-Linear Elliptic Problem

Mostafa Allaoui

, Abdelrachid El Amrouss

, Anass Ourraoui

Received 10 July 2012

Abstract

This paper shows the existence of at least three solutions for Navier problem involving the p(x)-biharmonic operator. Our technical approach is based on a theorem obtained by B. Ricceri.

1 Introduction

Analysis of solutions of speci…c boundary value problems is of considerable importance in the theory of partial di¤erential equations, especially for equations of fourth order.

Its interest is widely justi…ed with many physical examples and arises from a variety of nonlinear phenomena. It is used in non-Newtonian ‡uids, in some reaction-di¤usion problems, as well as in ‡ow through porous media. It also appears in nonlinear elasticity petroleum extraction and in the theory of quasi-regular and quasi-conformal mappings.

For more detailed references on physical and mathematical background, we refer to [1, 2, 3, 8].

The present work is concerned with the following p(x)-biharmonic problem with Navier boundary condition,

(P)

2

p(x)u= f(x; u) + g(x; u) in ; u= u= 0 on @ ;

where is a bounded open domain in R^N with smooth boundary @ , ²_p(x)u = (j j^{p(x) 2} u) is the p(x)-biharmonic with p 2 C( ); p(x) > 1 for every x 2 and ; 2R+. We de…neF(x; t) =Rt

0f(x; s)ds,G(x; t) =Rt

0g(x; s)dsand we denote byp := inf_x2 p(x)andp⁺:= sup_x2 p(x):

Throughout this paper, we suppose the following assumptions.

There exist two positive constantsC; and 2C( ) with := inf

x2

(x); ⁺:= sup

x2

(x)and1< ⁺< p ; such that

(F₁)F(x; t) 0for a.e. x2 and x2[0; ]:

Mathematics Sub ject Classi…cations: 35J60, 35D05, 35J20, 35J40.

yUniversity Mohamed I, Faculty of Science, Department of Mathematics, Oujda, Morocco

zUniversity Mohamed I, Faculty of Science, Department of Mathematics, Oujda, Morocco

xUniversity Mohamed I, Faculty of Science, Department of Mathematics, Oujda, Morocco

51

(F₂)There existsq₁(x)2C( )withp⁺< q₁ q₁(x)< p₂(x)such that lim sup

t!0

F(x; t)

jtj^q1(x) <+1; uniformly for a.e. x2 with

p₂(x) =

( N p(x)

N 2p(x) if p(x)<^N₂; +1 if p(x) ^N₂: (F3)jF(x; t)j C(1 +jtj ^(x))forx2 and for allt2R: (F4)F(x;0) = 0for a.e. x2 :

(G) sup_{(x;t)2 R1+t}^G(x;t)q2 (x) <+1, whereq2(x)2C( )andq2(x)< p₂(x)forx2 : The goal of this paper is to prove the following result.

THEOREM 1. Assume that (F1)to (F4)and (G) are satis…ed. Then there exist an open interval [0;+1[and a positive real numbere such that for every 2 , there exists >0 such that for each 2[0; ], problem (P)has at least three weak solutions whose norms inW^2;p(x)( )\W₀^1;p(x)( )are less thane.

Many authors consider the existence of nontrivial solutions for some fourth order problems such as [2, 3]. This is a generalization of the classical p-biharmonic operator (j uj^{p 2}) obtained in the case when p is a positive constant. Here we point out that the p(x)-biharmonic operator possesses more complicated nonlinearities than p- biharmonic, for example, it is inhomogeneous and usually it does not have the so-called

…rst eigenvalue, since the in…mum of its principle eigenvalue is zero. This study is inspired by the results of [6] and [7], we are to prove the existence of three solutions of problem (P), and the technical approach is based on the three-critical-points theorem of Ricceri [11, 12].

This paper is divided into three sections organized as follows: in section 2 we start with some preliminary basic results on the theory of Lesbegue-Sobolev spaces with variables exponent (we refer to the book of Musielak [10], Mih¼ailescu and R¼adulescu [9]), we recall the three-critical-points theorem of Ricceri with some required results.

In section 3, we give the proof of the main result.

2 Preliminaries

In order to deal with the problem (P), we need some theory of variable exponent Sobolev space. For convenience, we only recall some basic facts which will be used later. Suppose that R^N is a bounded domain with smooth boundary @ : Let C₊( ) = fp 2 C( ) and essinf_x2 p(x) > 1g for any p(x) 2 C₊( ): Set p = min_x2 p(x),p⁺= max_x2 p(x)and

p_k(x) = N p(x)

N kp(x) ifkp(x)< N andp_k(x) = +1ifkp(x) N:

De…ne the variable exponent Lebesgue space by L^p(x)( ) = u: !Rmeasurable:

Z

juj^p(x)dx <1 :

ThenL^p(x)( )endowed with the norm kukp(x)= inf >0 :

Z u ^p(x)

dx 1 ;

becomes a separable and re‡exive Banach space.

PROPOSITION 1 (cf. [5]). Set (u) =R

juj^p(x)dx:Ifu2L^p(x)( ), we have (1) kukp(x) 1) kuk^pp(x) (u) kuk^pp(x)⁺ :

(2) kukp(x) 1) kuk^pp(x)⁺ (u) kuk^pp(x):

De…ne the variable exponent Sobolev spaceW^k;p(x)( )by

W^k;p(x)( ) =fu2L^p(x)( ) :D u2L^p(x)( )and j j kg where

D u= @^{j j}

@ ¹x1:::@ ^NxN

with = ( ₁; ₂; :::; _N)a multi-index and j j= ^N_{i=1 i}:The space W^k;p(x)( ) with the normkuk= _{j kj}kD ukp(x)is a separable and re‡exive Banach space.

PROPOSITION 2 (cf. [5]). Forp; r2C+( ) such thatr(x) p_k(x)for allx2 , there is a continuous and compact embedding

W^k;p(x)( ),!L^r(x)( ):

We denoteW₀^k;p(x)( )by the closure ofC₀¹( )in W^k;p(x)( ):

REMARK 1 (cf. [3]). (W^2;p(x)( )\W₀^1;p(x)( );k :k)is a separable and re‡exive Banach space. By the above remark and proposition 2.2 there is a continuous and compact embedding of W^2;p(x)( )\W₀^1;p(x)( ) into L^r(x)( ) wherer(x)< p₂ for all x2 :

PROPOSITION 3 (cf. [5]). Set%(u) =R

j uj^p(x)dx:For u; un 2W^2;p(x)( ), we have

(1)kuk 1) kuk^p+ %(u) kuk^p : (2)kuk 1) kuk^p %(u) kuk^p+: (3)kunk!0,%(un)!0:

(4)ku_nk!+1 )%(u_n)!+1:

The proof is similar to proof in ([5], Theorem 3.1).

PROPOSITION 4 (cf. [5]). For anyu2L^p(x)( ) andv2L^q(x)( ), we have Z

uvdx 1

p + 1

q kukp(x)kvkq(x)

where

1 p(x)+ 1

q(x) = 1:

DEFINITION 1. We say thatu2X is aweak solutionof problem (P)if Z

j uj^{p(x) 2} u vdx= Z

f(x; u)vdx+ Z

g(x; u)vdx

for allv2X:

We de…ne

I(u) = Z 1

p(x)j uj^p(x)dx; J(u) = Z

F(x; u)dx

and

(u) = Z

G(x; u)dx where

F(x; t) = Z t

0

f(x; s)ds; G(x; t) = Z t

0

g(x; s)ds:

Set

hL(u); vi= Z

j uj^{p(x) 2} u vdxforu; v2X:

PROPOSITION 5 (cf. [2, 3]).

(i) L:X !X is a continuous, bounded and strictly monotone operator.

(ii) Lis a mapping of type(S₊), i.e. ifu_n* uinXandlim sup_n!1hL(u_n) L(u); u_n ui 0;thenu_n!uinX:

(iii) L:X !X is a homeomorphism.

PROPOSITION 6 (cf. Theorem 1 in [11]). LetX be a real re‡exive Banach space, K R an interval, I : X ! R be a sequentially weakly lower semi-continuous C¹ function whose derivative admits a continuous inverse onX andJ :X !Rbe a C¹

functional with compact derivative. In addition,Iis bounded on each bounded subset ofX. Assume that

lim

kxk!1I(x) + J(x) = +1 (1)

for 2K;and that there exists 2Rsuch that sup

2K

xinf2X(I(x) + (J(x) + ))< inf

x2Xsup

2K

(I(x) + (J(x) + )):

Then there exist a nonempty set A K and a positive number e with the following property: for every 2Aand everyC¹functional :X!Rwith compact derivative, there exists >0such that for each 2[0; ], the equation

I⁰(u) + J⁰(u) + ⁰(u) = 0

has at least three solutions inX whose norms are less thane.

PROPOSITION 7 (cf.[12]). LetX be a nonempty set, andI andJ are two real functions on X. Suppose there are >0 andu0; u12X such that

I(u₀) =J(u₀) = 0; I(u₁)> and sup

u2I ¹(] 1; ])

J(u)< J(u₁) I(u1):

Then for each satisfying

I(u0) =J(u0) = 0; I(u1)> and sup

u2I ¹(] 1; ])

J(u)< < J(u1) I(u1);

we have

sup

0

uinf2X(I(u) + ( J(u)))< inf

u2Xsup

0

(I(u) + ( J(u))):

3 Proof of the Main Result

We now turn to the proof of Theorem 1. First, we check the conditions of proposition 6.

According to proposition 5, it is clear thatIis continuously Gâteaux di¤erentiable, whose Gâteaux derivative admits a continuous inverse onX :Notice thatIis a convex and continuous functional, and then it is a weakly lower semi-continuous function.

Moreover, andJ are continuously Gâteaux di¤erentiable functions and its Gâteaux derivatives are compact. By a similar analysis to that in Fan and Zhang (cf. [4]), by (F3) and (G), we know thatJ; 2C¹(X;R)such that

J⁰(u)v= Z

f(x; u(x))vdxand ⁰(u)v= Z

g(x; u(x))vdx

foru; v2X:Since the identity operator from X toL ^(x)is compact, so the operators J⁰ and ⁰ are compact. Obviously,Iis bounded on each bounded subset ofX.

Forkuk<1;

1

p⁺kuk^p+ I(u) 1

p kuk^p :

LetC₀>0 such thatC_{0 p}¹+kuk^p _p¹ kuk^p+:Then I(u) 1

p⁺kuk^p C₀:

Sincekuk 1, we haveI(u) _p¹+kuk^p ;and thus for anyu2X we infer that J(u) =

Z

F(x; u)dx Z

C(1 +juj ^(x)dx)

C(j j+kuk ⁺(x)+kuk (x)) C₁(1 +kuk ⁺(x))

C₂(1 +kuk ⁺);

withC₁ 0andC₂ 0:Consequently, we obtain I(u) + J(u) 1

p⁺kuk^p C₂(1 +kuk ⁺) C₀: Therefore, foru2X and 0, since ⁺< p ;we get

lim

kuk!+1(I(u) + J(u)) = +1:

Then the assumption (1) is satis…ed. In order to prove the assumption (2), we need to verify the conditions of proposition 7.

Let u0 = 0: Then I(u0) = J(u0) = 0: We show that the assumption (3) of proposition 7 holds. Let x⁰ 2 (because is a nonempty bounded open set) and r2 > r1 > 0: Take !(x) 2 C₀¹( ) with !(x) = 0 for x 2 nB(x⁰; r2); !(x) =

r2 r1(r2 kxi x⁰_i k²)whenx2B(x⁰; r2)nB(x⁰; r1)and!(x) = whenx2B(x⁰; r1) withkxk2= ( ^N_i=1(x_i)²)¹²:

Hereu₁(x) =!(x). Then we can get J(u1) = J(!) =

Z

F(x; !)dx >0:

By (F2), there exist 2[0;1]andC1>0such that

F(x; t) C1jtj^q1(x) for jtj< and a.e. x2 : Putting

K₁= sup

jtj<

C[1 +jtj ⁺] jtj^q1

; K₂= sup

jtj>

C[1 +jtj ⁺] jtj^q1

; K₃= sup

jtj<1

C[1 +jtj ⁺] jtj^q1

;

K₄= sup

jtj>1

C[1 +jtj ⁺] jtj^q1

andM = maxfC₁; K_i; i= 1; :::;4g: Thus

F(x; t)< M jtj^q1 fort2Rand a.e. x2 :

Now, …x such that 0 < < 1: If we have _p¹+kuk^p+ < 1: By the Sobolev embedding Theorem, for suitable positive constantsC₂ andC₃;we entail that

J(u) = Z

F(x; u)dx < M Z

juj^q1 C2kuk^q1 C3

q1 p+:

It follows fromq₁ > p⁺ that

lim

!0⁺

sup 1

p+kuk^p+ f J(u)g

= 0: (2)

Let! 2X as previously mentioned with the fact J(!)>0: Fix ₀ where < ₀<

1

p⁺minfk!k^p+;k!k^p ;1g 1:Now, there are two cases to be considered.

Ifk!k<1 , we have

I(u₁) = I(!) = Z 1

p(x)j !j^p(x)dx 1 p⁺

Z

j !j^p(x)dx 1

p⁺k!k^p+ 0> :

By (2), it yields sup

1 p+kkuk^p+

J(u) 2

J(u1 1

p k!k^p 2

J(u1)

I(u1) < J(u1) I(u1) :

Else ifk!k 1we obtain I(u₁) = I(!) =

Z 1

p(x)j !j^p(x)dx 1 p⁺

Z

j !j^p(x)dx 1

p⁺k!k^p 0> :

From (2), since >0, we get sup

1 p+kuk^p+

J(u) 2

J(u1)

1

p k!k^p+ 2

J(u1)

I(u1) < J(u1) I(u1) :

Thereby,

sup

1 p+kuk^p+

J(u)< J(u₁)

I(u1) : (3)

For anyu2I ¹(] 1; ]), we have Z 1

p(x)j uj^p(x)dx :

Then Z

1

p⁺j uj^p(x)dx :

Hence, Z

j uj^p(x)dx 1

1 p⁺

< ₀ 1

1 p⁺

<1:

The last inequality implies that kuk<1 and 1

p⁺kuk^p+ <

Z 1

p(x)j uj^p(x)dx ; so we conclude

I ¹(] 1; ]) fu2X : 1

p⁺ kuk^p+ < g: We deduce from the relation (3) that

sup

u2I ¹(] 1; ])

J(u)< J(u₁) I(u1) :

We can …nd such that

sup

u2I ¹(] 1; ])

J(u)< < J(u₁) I(u₁) :

Taking K = [0;+1[, the assumptions of proposition 7 are satis…ed. Then, we may easily obtain the condition (2) of proposition 6. Consequently, I; J and verify the conditions of proposition 6. So the proof is complete.

Acknowledgment. The authors would like to thank the anonymous referee for the valuable comments.

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