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DOI: 10.24193/subbmath.2020.2.05

Existence and multiplicity of solutions to the Navier boundary value problem for a class of (p(x), q(x))-biharmonic systems

Hassan Belaouidel, Anass Ourraoui and Najib Tsouli

Abstract. In this article, we study the following problem with Navier boundary conditions.





∆(a(x,∆u)) =Fu(x, u, v), in Ω

∆(a(x,∆v)) =Fv(x, u, v), in Ω, u=v= ∆u= ∆v= 0 on∂Ω.

By using the Mountain Pass Theorem and the Fountain Theorem, we establish the existence of weak solutions of this problem.

Mathematics Subject Classification (2010):35J30, 35J60, 35J92.

Keywords: Fourth-order, variable exponent, Palais Smale condition, mountain pass theorem.

1. Introduction

In recent years, the study of differential equations and variational problems with p(x)-growth conditions was an interesting topic, which arises from nonlinear elec- trorheological fluids and elastic mechanics. In that context we refer the reader to Ruzicka [15], Zhikov [20] and the reference therein; see also [4, 7, 8, 5].

Fourth-order equations appears in many context. Some of theses problems come from different areas of applied mathematics and physics such as Micro Electro- Mechanical systems, surface diffusion on solids, flow in Hele-Shaw cells (see [10]).

In addition, this type of equations can describe the static from change of beam or the sport of rigid body.

In [1] the authors studied a class ofp(x)-biharmonic of the form

∆(|∆u|p(x)−2∆u) =λ|u|q(x)−2u in Ω, u= ∆u= 0 on∂Ω,

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where Ω is a bounded domain inRN, with smooth boundary∂Ω,N ≥1,λ≥0 . In [3], A. El Amrouss and A. Ourraoui considered the below problem and us- ing variational methods, by the assumptions on the Carath´eodory function f, they establish the existence of Three solutions the problem of the form

∆(|∆u|p(x)−2∆u) +a(x)|u|p(x)−2u=f(x, u) +λg(x, u) in Ω, Bu=T u= 0 on∂Ω.

Inspired by the above references, the work of L. Li [11]and [14], the aim of this article is to study the existence and multiplicity of weak solutions for (p(x), q(x))−biharmonic type system





∆(a(x,∆u)) =Fu(x, u, v), in Ω

∆(a(x,∆v)) =Fv(x, u, v), in Ω, u= ∆u= 0, v= ∆v= 0 on∂Ω,

(1.1)

where Ω is a bounded domain inRN with smooth boundary∂Ω,N ≥1,

2p(x)u:= ∆(|∆u|p(x)−2∆u),

is thep(x)-biharmonic operator,p,qare continuous functions on Ω with inf

x∈Ω

p(x)>max

1,N 2

, inf

x∈Ω

q(x)>max

1,N 2

and F : Ω×R2 → R is a function such that F(., s, t) is continuous in Ω, for all (s, t) ∈ R2, F(x, ., .) is C1 in R2 for every x ∈ Ω, and Fu, Fv denote the partial derivative ofF, with respect tou, v respectively such that

(F1) For all (x, s, t)∈Ω×R2, we assume lim

|s|→0

Fs(x, s, t)

|s|p(x)−1 = 0, lim

|t|→0

Ft(x, s, t)

|s|q(x)−1 = 0.

(F2) For all (x, s, t)∈Ω×R2, we assume

F(x, s, t) =o(|s|p(x)−1+|t|q(x)−1)as|(s, t)| → ∞.

(F3) There existsu >0,v >0 such thatF(x, u, v)>0 for a.e x∈Ω

(F4) There exist λ > 0 such thatF(x, s, t) ≥λ(|s|α(x)− |t|β(x)) for all (s, t)∈ R2, with

α > r+, 1< β≤β+< r. (F5) For all (x, s, t)∈Ω×R2F(x,−s,−t) =−F(x, s, t).

Let a: Ω×RN → RN to be a continuous potential derivative with respect to ξ of the mapping A: Ω×RN → RN where a=DA=A0, with the assumption as below

(A1) A(x,0) = 0 , for all x∈Ω.

(A2) a(x, ξ)≤C1(1 +|ξ|r(x)−1),C1>0 andr > p+, r> q+.

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(A3) Aisr(x)-uniformly convex: there exists a constantk >0 such that A

x,ξ+η

2

≤ 1

2A(x, ξ) +1

2A(x, η)−k|ξ−η|r(x), for allx∈Ω, ξ, η∈RN.

(A4) Aisr(x)-subhomogenuous, for all (x, ξ)∈Ω×RN,

|ξ|r(x)≤a(x, ξ)≤r(x)A(x, ξ).

(A5) For all (x, s)∈Ω×RN a(x,−s) =−a(x, s).

The main results of this paper are the following theorems.

Theorem 1.1. Assume that(A1)−(A4)and(F1)−(F3)hold. Then the problem (1.1) has two weak solutions.

Theorem 1.2. Assume that(A1)−(A5)and(F1)−(F5)hold. Then the problem (1.1) has a sequence of weak solutions such that φ(±(uk, vk))→+∞, as k→+∞ with φ is a energy associated of the problem (1.1)defined in (2.2).

This paper is organized as three sections. In Section 2, we recall some basic properties of the variable exponent Lebegue-Sobolev spaces. In Section3 we give the proof of main results.

2. Preliminaries

To studyp(x))-Laplacian problems, we need some results on the spacesLp(x))(Ω) and Wk,p(x))(Ω), and properties of p(x))-Laplacian, which we use later. Let Ω be a bounded domain ofRN, denote

C+(Ω) ={h(x);h(x)∈C(Ω), h(x)>1,∀x∈Ω}.

For anyh∈C+(Ω), we define

h+= max{h(x); x∈Ω}, h= min{h(x); x∈Ω};

For anyp∈C+(Ω), we define the variable exponent Lebesgue space Lp(x))(Ω) =n

u;uis a measurable real-valued function such that Z

|u(x)|p(x))dx <∞o ,

endowed with the so-calledLuxemburg norm

|u|p(x))= inf (

µ >0;

Z

u(x) µ

p(x))

dx≤1 )

.

Then (Lp(x))(Ω),| · |p(x))) becomes a Banach space.

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Proposition 2.1 ([9]). The space (Lp(x))(Ω),| · |p(x))) is separable, uniformly convex, reflexive and its conjugate space is Lq(x)(Ω) whereq(x) is the conjugate function of p(x)), i.e.,

1

p(x))+ 1 q(x) = 1, for allx∈Ω. For u∈Lp(x))(Ω) andv∈Lq(x)(Ω), we have

Z

uvdx

≤ 1 p + 1

q

|u|p(x))|v|q(x)≤2|u|p(x))|v|q(x). The Sobolev space with variable exponentWk,p(x))(Ω) is defined as

Wk,p(x))(Ω) ={u∈Lp(x))(Ω) :Dαu∈Lp(x))(Ω),|α| ≤k}, where

Dαu= ∂|α|

∂xα11∂xα22. . . ∂xαNNu,

with α = (α1, . . . , αN) is a multi-index and |α| =

N

X

i=1

αi. The space Wk,p(x))(Ω) equipped with the norm

kukk,p(x)) = X

|α|≤k

|Dαu|p(x)),

also becomes a separable and reflexive Banach space. For more details, we refer the reader to [6, 9, 13]. Denote

pk(x) =

( N p(x)

N−kp(x) ifkp(x)< N, +∞ ifkp(x)≥N for anyx∈Ω,k≥1.

Proposition 2.2 ([9]). Forp, r∈C+(Ω) such that r(x)≤pk(x)for allx∈Ω, there is a continuous embedding

Wk,p(x))(Ω),→Lr(x)(Ω).

If we replace ≤with<, the embedding is compact.

We denote byW0k,p(x)(Ω) the closure ofC0(Ω) in Wk,p(x)(Ω). Then the func- tion space

W2,p(x)(Ω)∩W01,p(x)(Ω)

,kukp(x)

is a separable and reflexive Banach space, where

kukp(x)= infn µ >0 :

Z

∆u(x) µ

p(x)

≤1o .

Remark 2.3. According to [[18] Theorem 4.4. ], the norm k · k2,p(x) is equivalent to the normk · kp(x)in the spaceX. Consequently, the normsk · k2,p(x),k · kandk · kp(x) are equivalent.

Proposition 2.4 ([2]). If we denoteρ(u) =R

|∆u|p(x)dx, then foru, un ∈X, we have (1) kukp<1 (respectively=1;>1) ⇐⇒ ρ(u)<1 (respectively= 1;>1);

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(2) kukp≤1⇒ kukpp+≤ρ(u)≤ kukpp; (3) kukp≥1⇒ kukpp ≤ρ(u)≤ kukpp+;

(4) kunkp→0 (respectively→ ∞)⇐⇒ ρ(un)→0 (respectively→ ∞).

Note that the weak solutions of problem (1.1) are considered in the generalized Sobolev space

X =

W2,p(x)(Ω)∩W01,p(x)(Ω)

×

W2,q(x)(Ω)∩W01,q(x)(Ω) equipped with the norm

k(u, v)k= max{kukp(x),kukq(x)}.

Remark 2.5 (see [19]). As the Sobolev space X is a reflexive and separable Banach space, there exist (en)n∈N ⊆X and (fn)n∈N ⊆X such that fn(el) =δnl for any n, l∈N and

X = span{en:n∈N}, X= span{fn:n∈N}w

. Fork∈N, denote by

Xk= span{ek}, Yk =⊕kj=1Xj, Zk =⊕k Xj. For everym >1 ,u, v∈Lm(Ω), we define

|(u, v)|m:= max{|u|m,|v|m}.

Lemma 2.6 (See[8]). Define

βk:= sup{|(u, v)|m;k(u, v)k= 1,(u, v)∈Zk}, wherem:= max

x∈Ω

(p(x), q(x)). Then, we have lim

k→∞βk= 0.

2.1. Existence and multiplicity of weak solutions

Definition 2.7. We say that (u, v)∈X is weak solution of (1.1) if Z

a(x,∆u)∆ϕdx+

Z

a(x,∆v)∆ϕdx= Z

Fu(x, u, v)ϕdx+ Z

Fv(x, u, v)ϕdx, (2.1) for allϕ∈X.

The functional associated to (1.1) is given by φ(u, v) =

Z

A(x,∆u)dx+ Z

A(x,∆v)dx− Z

F(x, u, v)dx. (2.2) It should be noticed that under the condition (F1)−(F2) the functionalφis of class C1(X,R) and

φ0(u, v)(ψ, ϕ) = Z

a(x,∆u)∆ψdx+ Z

a(x,∆v)∆ϕdx (2.3)

− Z

Fu(x, u, v)ψdx− Z

Fv(x, u, v)ϕdx, ∀(ψ, ϕ)∈X.

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Then, we know that the weak solution of (1.1) corresponds to critical point of the functionalφ.

Definition 2.8. We say that

(1) The C1-functional φsatisfies the Palais-Smale condition (in short (P S) condi- tion) if any sequence (un)n∈N⊆X for which, (φ(un))n∈N⊆R is bounded and φ0(un)→0 asn→ ∞, has a convergent subsequence.

(2) The C1-functionalφsatisfies the Palais-Smale condition at the level c(in short (P S)c condition) forc∈Rif any sequence (un)n∈N⊆X for which,φ(un)→c andφ0(un)→0 asn→ ∞, has a convergent subsequence.

(3) The C1-functional φ satisfies the (P S)c condition for c ∈ R if any sequence (un)n∈N⊆Xfor which,un∈Ynfor eachn∈N,φ(un)→candφ0|Y

n)(un)→0 as n→ ∞withYn, n∈Nas defined in Remark 2.5, has a subsequence convergent to a critical point ofφ.

Remark 2.9. It is easy to see that if φ satisfies the (P S) condition, then φ satisfies the (P S)c condition for everyc∈R.

Proof of Theorem1.1.To prove Theorem 1.1, we shall use the Mountain Pass theorem [16]. We first start with the following lemmas.

Lemma 2.10. Under the assumptions(F1)-(F3)and(A1)-(A3)φis sequentially weakly lower semi continuous and coercive .

Proof.By (F1)-(F2), we see that

|F(x, s, t)| ≤C3(1 +|s|p(x)+|t|q(x)), ∀(s, t)∈R2. (2.4) By the compact embeddings

X ,→Lp(x)(Ω), X ,→Lq(x)(Ω), we deduce thatw7→R

F(x, w)dxis sequentially lower semi continuous∀w∈R2. Since

w7→

Z

A(x,∆u)dx+ Z

A(x,∆v)dx is convex uniformly, so it is sequentially lower semi continuous.

Now we prove thatφis coercive. From (F2) forεsmall enough, there existδ >0 such that

|F(x, s, t)|≤ε(|s|p(x)+|t|q(x)), for|(s, t)|> δ, and thus we have

|F(x, s, t)|≤ε(|s|p(x)+|t|q(x)) + max

|(s,t)|≤δ|F(x, s, t)| ||(s, t)|,∀(s, t)∈R2,

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for a.ex∈Ω. Consequently, fork(u, v)k>1 we obtain φ(u, v) ≥

Z

A(x,∆u)dx+ Z

A(x,∆v)dx

− ε Z

|u|p(x)dx−ε Z

|v|q(x)dx− max

|(u,v)|≤δ|F(x, u, v)| Z

|(u, v)|dx

≥ Z

1

r(x)|∆u|r(x)dx+ Z

1

r(x)|∆v|r(x)dx

− Cε Z

|u|p(x)dx−Cε Z

|v|q(x))dx− max

|(u,v)|≤δ|F(x, u, v)| Z

|(u, v)|dx

≥ 1 r+max

kukrr(x) ,kvkrr(x)

−2Cεmax

kukpp(x)+ ,kvkqq(x)+

−Cε|Ω| max

|(u,v)|≤δ|F(x, u, v)|max

kukpp(x)+ ,kvkqq(x)+ .

Therefore, φ is coercive and has a global minimizer (u1, v1) which is a nontrivial because by (F3)

φ(u1, v1)≤φ(u, v)<0.

Lemma 2.11. Under the assumptions (F1)-(F3) and (A1)-(A4). Then φsatisfies the Palais-smale condition.

Proof.Letwn= (un, vn)⊂X be a Palais-smale sequence, then φ0(wn)→0 inX, φ(wn)→l∈R. We show that (wn) is bounded. By (A5) we have

φ(wn) = Z

A(x,∆un)dx+ Z

A(x,∆vn)dx− Z

F(x, un, vn)dx

≥ Z

1

r(x)|∆un|r(x)dx+ Z

1

r(x)|∆vn|r(x)dx− Z

F(x, un, vn)dx, and we get

φ0(un, vn)(un, vn)

= Z

a(x,∆un)∆undx+ Z

a(x,∆vn)∆vndx

− Z

Fun(x, un, vn)undx− Z

Fvn(x, un, vn)vndx

≤ Z

r(x)A(x,∆un)dx+ Z

r(x)A(x,∆vn)dx

− Z

Fun(x, un, vn)undx− Z

Fvn(x, un, vn)vndx.

Using the fact thatFs, Ft∈C(Ω×R2,R) and with (F1)−(F2), forε >0 there exists δ >0 andη >0 such that

|Fs(x, s, t)| ≤ε|s|p(x)−1, |Ft(x, s, t)| ≤ε|t|q(x)−1,

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and

|F(x, s, t)| ≤ε(|s|p(x)+|t|q(x)), for all|s, t)| ≤δ, and for all|s, t)| ≥η.

Then we have

|Fs(x, s, t)s| ≤ε|s|p(x), |Ft(x, s, t)t| ≤ε|t|q(x), (2.5) and

|F(x, s, t)| ≤ε(|s|p(x)+|t|q(x)), for all|s, t)| ≤δ, and for all|s, t)| ≥η.

It yields,

− 1

2r+φ0(un, vn)(un, vn)

≥ − 1 2r+

Z

r(x)A(x,∆un)dx− 1 2r+

Z

r(x)A(x,∆vn)dx

+ 1

2r+ Z

Fun(x, un, vn)undx+ Z

Fvn(x, un, vn)vndx

≥ − 1 2r+

Z

r(x)A(x,∆un)dx− 1 2r+

Z

r(x)A(x,∆vn)dx

+ 1

2r+ Z

Fun(x, un, vn)undx+ Z

Fvn(x, un, vn)vndx

.

Thus,

φ(un, vn)− 1

2r+φ0(un, vn)(un, vn)

≥ Z

A(x,∆un)dx+ Z

A(x,∆vn)dx− Z

F(x, un, vn)dx

− 1

2r+ Z

r(x)A(x,∆un)dx− 1 2r+

Z

r(x)A(x,∆vn)dx

− Z

F(x, un, vn)dx+ 1 2r+

Z

Fun(x, un, vn)undx+ Z

Fvn(x, un, vn)vndx

≥ 1 2

Z

|∆un|r(x)dx+ Z

|∆vn|r(x)dx

− Z

F(x, un, vn)dx

+ 1

2r+ Z

Fun(x, un, vn)undx+ Z

Fvn(x, un, vn)vndx

≥ 1 2max

kunkrr(x)+ ,kvnkrr(x)+

−(Cε+ε) Z

|un|p(x)dx−(Cε+ε) Z

|vn|q(x))dx.

Sincer> p+>1, r> q+>1, by the compact embeddings X ,→Lp(x)(Ω), X ,→Lq(x)(Ω),

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we deduce

φ(un, vn)− 1

2r+φ0(un, vn)(un, vn)

≥ 1 2max

kunkrr(x)+ ,kvnkrr(x)+

−2(C0ε+ε)k(un, vn)k

≥ 1

2 −2(C0ε+ε)

k(un, vn)k, whereC0 is positive constant.

Forεsmall enough withR= 12−2(C0ε+ε)>0, we get k(un, vn)k ≤ 1

R

φ(un, vn)− 1

2r+φ0(un, vn)(un, vn)

.

Since φ(un, vn) is bounded andφ0(un, vn)(un, vn) → 0 as n → ∞, then (un, vn) is bounded in X, passing to a subsequence, so (un, vn) *(u, v) in X and (un, vn)→ Lp(x)(Ω)×Lq(x)(Ω). We show that (un, vn)→(u, v) inX.

φ0(un, vn) ((un, vn)−(u, v))

= Z

a(x,∆un)∆(un−u)dx+ Z

a(x,∆vn)∆(vn−v)dx

− Z

Fun(x, un, vn)(un−u)dx− Z

Fvn(x, un, vn)(vn−v)dx.

Since

Z

a(x,∆un)∆(un−u)dx+ Z

a(x,∆vn)∆(vn−v)dx

= |φ0(un, vn) ((un, vn)−(u, v)) + Z

Fun(x, un, vn)(un−u)dx +

Z

Fvn(x, un, vn)(vn−v)dx|

≤ kφ0(un, vn)kX?k(un, vn)−(u, v)k +

Z

|Fun(x, un, vn)||(un−u)|dx+ Z

|Fvn(x, un, vn)||(vn−v)|dx.

By (2.5), we have Z

|Fun(x, un, vn)||(un−u)|dx+ Z

|Fvn(x, un, vn)||(vn−v)|dx

≤ ε Z

|un−u|p(x)+|vn−v|q(x) dx,

we get

lim sup

n→+∞

Z

a(x,∆un)∆(un−u)dx+ Z

a(x,∆vn)∆(vn−v)dx

≤0.

Sincea(x, ξ) is of (S+) type, we see that (un, vn)→(u, v) in X.

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Now, we verified the conditions of Mountain Pass Theorem. By H¨older’s inequality, from (F1) there exists δ >0 such that

|F(x, u, v)| ≤

Z u 0

Fs(x, s, v)dx+ Z v

0

Ft(x,0, t)dx+F(x,0,0)

≤ ε

Z u 0

|s|p(x)−1dx+ Z v

0

|t|q(x)−1dx

+|F(x,0,0)|

≤ ε(|u|p(x)+|v|q(x)) +M, for all|u, v)| ≤δ, withM := max

x∈Ω

F(x,0,0) and by (F2), there existsM(δ)>0 such that

|F(x, u, v)|≤M(δ)(|u|p(x)+|v|q(x)), for|(u, v)|> δ.

Therefore, fork(u, v)k=%small enough, we have φ(u, v) ≥

Z

A(x,∆u)dx+ Z

A(x,∆v)dx−ε Z

|(u,v)|<δ

|u|p(x)+|v|q(x) dx

− M(δ) Z

|(u,v)|>δ

(|u|p(x)+|v|q(x))−M meas{|(u, v)|< δ}

≥ 1

r+max

kukrr(x)+ ,kvkrr(x)+

− min(εC, M(δ)C0) max

kukpp(x) ,kvkqq(x)

−M meas{|(u, v)|< δ}

= g(%).

There exists θ > 0 such that g(%) > θ > 0. Since φ(0,0) = 0, we conclude that φ satisfies the conditions of Mountain Pass Theorem. Then there exists (u2, v2) such thatφ0(u2, v2) = 0.

Proof of Theorem 1.2. To prove Theorem 1.2, above, will be based on a variational approach, using the critical points theory, we shall prove that theC1-functionalφhas a sequence of critical values. The main tools for this end are “Fountain theorem” (see Willem [16, Theorem 6.5]) which we give below.

Theorem 2.12 (“Fountain theorem”,[16]). LetX be a reflexive and separable Banach space,φ∈C1(X,R)be an even functional and the subspacesXk, Yk, Zk as defined in remark 2.5. If for eachk∈N there existρk > rk >0such that

(1) infx∈Zk,kxk=rkφ(x)→ ∞as k→ ∞, (2) maxx∈Yk,kxk=ρkφ(x)≤0,

(3) I satisfies the(P S)c condition for every c >0.

ThenI has a sequence of critical values tending to+∞.

According to Lemma 2.6, (F5) and (A5), Φ∈ C1(X,R) is an even functional. We will prove that ifkis large enough, then there existρk > νk >0 such that

bk:= inf{Φ(u)/u∈Zk,kuk=νk} →+∞ ask→+∞; (2.6) ak:= max{Φ(u)/u∈Yk,kuk=ρk} →0 as k→+∞. (2.7)

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For any (u, v) ∈ Zk, kvkq(x) > 1, kukp(x) > 1 and k(u, v)k = ηk, (ηk will be specified later), by (2.4) we have

φ(u, v) = Z

A(x,∆u)dx+ Z

A(x,∆v)dx− Z

F(x, u, v)dx

≥ 1 r+max

kukrr(x) ,kvkrr(x))

− Z

C3(1 +|u|p(x)+|v|q(x))dx

≥ 1 r+max

kukrr(x) ,kvkrr(x))

−C3 Z

dx−C3 Z

|u|p(x)dx−C3 Z

|v|q(x)dx

≥ 1

r+k(u, v)kr−C3kk(u, v)k)p+−C3kk(u, v)k)q+−C3|Ω|

≥ 1

r+k(u, v)kr−C4βkk(u, v)km−C3|Ω|, wheremis defined in Lemma 2.6. We fix

ηk = 1

r+C4βkb m−r1

→+∞ask→+∞.

Consequently

φ(u, v)≥ηk 1

r+ηrk−1−C4βkbηm−1k

−C3|Ω|.

Then,

φ(u, v)→+∞as k→+∞.

Proof of (2.7).From (F4), there existsλ >0 such that F(x, s, t)≥λ(|s|α(x)− |t|β(x)), withα> r+, β+< r.

Therefore, by Lemma 2.1 [12] and Lemma 3.1 [17], for anyω:= (u, v)∈Yk with kωk= 1 and 1< t=ρk, we have

φ(tω) = Z

A(x, t∆u)dx+ Z

A(x, t∆v)dx− Z

F(x, tω)dx

≤ Z

tr(x)A(x,∆u)dx+ Z

tr(x)A(x,∆v)dx

−λ Z

|tu|α(x)dx+λ Z

|tv|β(x)dx

≤tr+ Z

A(x,∆u)dx+ Z

A(x,∆v)dx

−λtα Z

|u|α(x)dx+λtβ Z

|v|β(x)dx.

Byα> r+> βanddimYk<∞, we conclude thatφ(tu, tv)→ −∞asktωk →+∞

forω∈Yk. By applying the fountain Theorem, we achieved the proof of Theorem 1.2.

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Hassan Belaouidel

Department of Mathematics, Faculty of Sciences of Oujda University Mohamed I, Oujda, Morocco

e-mail:[email protected] Anass Ourraoui

Department of Mathematics, Faculty of Sciences of Oujda University Mohamed I, Oujda, Morocco

e-mail:[email protected] Najib Tsouli

Department of Mathematics, Faculty of Sciences of Oujda University Mohamed I, Oujda, Morocco

e-mail:[email protected].

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