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,
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+.
(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
Ω
|u(x)|p(x))dx <∞o ,
endowed with the so-calledLuxemburg norm
|u|p(x))= inf (
µ >0;
Z
Ω
u(x) µ
p(x))
dx≤1 )
.
Then (Lp(x))(Ω),| · |p(x))) becomes a Banach space.
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
Ω
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
p∗k(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)≤p∗k(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
Ω
∆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
Ω|∆u|p(x)dx, then foru, un ∈X, we have (1) kukp<1 (respectively=1;>1) ⇐⇒ ρ(u)<1 (respectively= 1;>1);
(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
Ω
a(x,∆u)∆ϕdx+
Z
Ω
a(x,∆v)∆ϕdx= Z
Ω
Fu(x, u, v)ϕdx+ Z
Ω
Fv(x, u, v)ϕdx, (2.1) for allϕ∈X.
The functional associated to (1.1) is given by φ(u, v) =
Z
Ω
A(x,∆u)dx+ Z
Ω
A(x,∆v)dx− Z
Ω
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
Ω
a(x,∆u)∆ψdx+ Z
Ω
a(x,∆v)∆ϕdx (2.3)
− Z
Ω
Fu(x, u, v)ψdx− Z
Ω
Fv(x, u, v)ϕdx, ∀(ψ, ϕ)∈X.
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
ΩF(x, w)dxis sequentially lower semi continuous∀w∈R2. Since
w7→
Z
Ω
A(x,∆u)dx+ Z
Ω
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,
for a.ex∈Ω. Consequently, fork(u, v)k>1 we obtain φ(u, v) ≥
Z
Ω
A(x,∆u)dx+ Z
Ω
A(x,∆v)dx
− ε Z
Ω
|u|p(x)dx−ε Z
Ω
|v|q(x)dx− max
|(u,v)|≤δ|F(x, u, v)| Z
Ω
|(u, v)|dx
≥ Z
Ω
1
r(x)|∆u|r(x)dx+ Z
Ω
1
r(x)|∆v|r(x)dx
− Cε Z
Ω
|u|p(x)dx−Cε Z
Ω
|v|q(x))dx− max
|(u,v)|≤δ|F(x, u, v)| Z
Ω
|(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
Ω
A(x,∆un)dx+ Z
Ω
A(x,∆vn)dx− Z
Ω
F(x, un, vn)dx
≥ Z
Ω
1
r(x)|∆un|r(x)dx+ Z
Ω
1
r(x)|∆vn|r(x)dx− Z
Ω
F(x, un, vn)dx, and we get
φ0(un, vn)(un, vn)
= Z
Ω
a(x,∆un)∆undx+ Z
Ω
a(x,∆vn)∆vndx
− Z
Ω
Fun(x, un, vn)undx− Z
Ω
Fvn(x, un, vn)vndx
≤ Z
Ω
r(x)A(x,∆un)dx+ Z
Ω
r(x)A(x,∆vn)dx
− Z
Ω
Fun(x, un, vn)undx− Z
Ω
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,
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
Ω
r(x)A(x,∆un)dx− 1 2r+
Z
Ω
r(x)A(x,∆vn)dx
+ 1
2r+ Z
Ω
Fun(x, un, vn)undx+ Z
Ω
Fvn(x, un, vn)vndx
≥ − 1 2r+
Z
Ω
r(x)A(x,∆un)dx− 1 2r+
Z
Ω
r(x)A(x,∆vn)dx
+ 1
2r+ Z
Ω
Fun(x, un, vn)undx+ Z
Ω
Fvn(x, un, vn)vndx
.
Thus,
φ(un, vn)− 1
2r+φ0(un, vn)(un, vn)
≥ Z
Ω
A(x,∆un)dx+ Z
Ω
A(x,∆vn)dx− Z
Ω
F(x, un, vn)dx
− 1
2r+ Z
Ω
r(x)A(x,∆un)dx− 1 2r+
Z
Ω
r(x)A(x,∆vn)dx
− Z
Ω
F(x, un, vn)dx+ 1 2r+
Z
Ω
Fun(x, un, vn)undx+ Z
Ω
Fvn(x, un, vn)vndx
≥ 1 2
Z
Ω
|∆un|r(x)dx+ Z
Ω
|∆vn|r(x)dx
− Z
Ω
F(x, un, vn)dx
+ 1
2r+ Z
Ω
Fun(x, un, vn)undx+ Z
Ω
Fvn(x, un, vn)vndx
≥ 1 2max
kunkrr(x)+ ,kvnkrr(x)+
−(Cε+ε) Z
Ω
|un|p(x)dx−(Cε+ε) Z
Ω
|vn|q(x))dx.
Sincer−> p+>1, r−> q+>1, by the compact embeddings X ,→Lp(x)(Ω), X ,→Lq(x)(Ω),
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
Ω
a(x,∆un)∆(un−u)dx+ Z
Ω
a(x,∆vn)∆(vn−v)dx
− Z
Ω
Fun(x, un, vn)(un−u)dx− Z
Ω
Fvn(x, un, vn)(vn−v)dx.
Since
Z
Ω
a(x,∆un)∆(un−u)dx+ Z
Ω
a(x,∆vn)∆(vn−v)dx
= |φ0(un, vn) ((un, vn)−(u, v)) + Z
Ω
Fun(x, un, vn)(un−u)dx +
Z
Ω
Fvn(x, un, vn)(vn−v)dx|
≤ kφ0(un, vn)kX?k(un, vn)−(u, v)k +
Z
Ω
|Fun(x, un, vn)||(un−u)|dx+ Z
Ω
|Fvn(x, un, vn)||(vn−v)|dx.
By (2.5), we have Z
Ω
|Fun(x, un, vn)||(un−u)|dx+ Z
Ω
|Fvn(x, un, vn)||(vn−v)|dx
≤ ε Z
Ω
|un−u|p(x)+|vn−v|q(x) dx,
we get
lim sup
n→+∞
Z
Ω
a(x,∆un)∆(un−u)dx+ Z
Ω
a(x,∆vn)∆(vn−v)dx
≤0.
Sincea(x, ξ) is of (S+) type, we see that (un, vn)→(u, v) in X.
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
Ω
A(x,∆u)dx+ Z
Ω
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)
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
Ω
A(x,∆u)dx+ Z
Ω
A(x,∆v)dx− Z
Ω
F(x, u, v)dx
≥ 1 r+max
kukrr(x)− ,kvkrr(x))−
− Z
Ω
C3(1 +|u|p(x)+|v|q(x))dx
≥ 1 r+max
kukrr(x)− ,kvkrr(x))−
−C3 Z
Ω
dx−C3 Z
Ω
|u|p(x)dx−C3 Z
Ω
|v|q(x)dx
≥ 1
r+k(u, v)kr−−C3(βkk(u, v)k)p+−C3(βkk(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
Ω
A(x, t∆u)dx+ Z
Ω
A(x, t∆v)dx− Z
Ω
F(x, tω)dx
≤ Z
Ω
tr(x)A(x,∆u)dx+ Z
Ω
tr(x)A(x,∆v)dx
−λ Z
Ω
|tu|α(x)dx+λ Z
Ω
|tv|β(x)dx
≤tr+ Z
Ω
A(x,∆u)dx+ Z
Ω
A(x,∆v)dx
−λtα− Z
Ω
|u|α(x)dx+λtβ− Z
Ω
|v|β(x)dx.
Byα−> r+> β−anddimYk<∞, 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].