Mountain Pass Theorem

(1)

Solutions to an anisotropic system via sub-supersolution method and

Mountain Pass Theorem

Giovany Figueiredo

¹

and Julio R. S. Silva

^B^{1, 2}

1Departamento de Matemática, Universidade de Brasilia – UNB, CEP: 70910-900, Brasília–DF, Brazil

2Universidade Federal do Pará – Campus Universitário de Cametá, CEP: 68.400-000 , Cametá–PA, Brazil

Received 21 September 2018, appeared 28 June 2019 Communicated by Dimitri Mugnai

Abstract. We use the sub-supersolution method and the Mountain Pass Theorem in order to show existence and multiplicity of solutions for the quasilinear system given by











−

"_N

i=1

∑

∂

∂xi

∂u

∂xi

p_i−2

∂u

∂xi

#

=a1(x)u+Fu(x,u,v) inΩ,

−

"_N

i=1

∑

∂

∂xi

p_i∂v

∂xi

p_i−2

∂v

∂xi

#

=a₂(x)v+Fv(x,u,v) _in_Ω,

u,v>0 inΩ, u=v=0 on ∂Ω,

where a_j, j =1, 2 are functions in L^∞(_Ω)and andFu and Fvare continuous functions onΩ×_R².

Keywords: anisotropic operator, sub-supersolution method, Mountain Pass Theorem.

2010 Mathematics Subject Classification: 35J65, 35B45.

1 Introduction

In this paper we are concerned with existence and multiplicity of positive solutions for the following class of system nonlinear boundary value anisotropic problems given by











−^h

∑

N i=1

∂

∂x_i

∂u

∂x_i

p_i−2∂u

∂x_i i

= a₁(x)u+F_u(x,u,v) inΩ,

−^h

∑

N i=₁

∂

∂x_i

∂v

∂x_i

p_i−2∂v

∂x_i i

= a₂(x)u+Fv(x,u,v) inΩ, u,v>0 in Ω,

u,v∈W^1,

−→ p 0 (_Ω),

(1.1)

BCorresponding author. Email: giovany@unb.br

(2)

whereΩ⊂_R^Nis a bounded smooth domain with smooth boundary,N≥3 ,−→

p = (p₁, . . . ,p_N), p_i > 1, ∑i^N=1 1

p_i > 1, p₁ < p2 < · · · < pN < p^∗ := _N^{N p}₋_p. In this paper p denotes the harmonic mean

p= ^N

∑^Ni=1 1 p_i

.

For j= 1, 2, a_j ≥0 is a nontrivial mensurable function. More precisely, we will suppose that the functiona_j satisfy the following assumption:

(H) The functiona_j ∈L^∞(_Ω)witha_j(x)>0.

In this paper Fis a function onΩ×_R² _{of class}C¹ satisfying (H₁) There isδ>0 such that

F_s(x,s,t)≥(1−s)a₁(x), for every 0≤s ≤δ, a.e. inΩ, and

F_t(x,s,t)≥ (1−t)a₂(x), for every 0≤t ≤δ, a.e. inΩ. (H2) There is 1<r < p^∗ such that

Fs(x,s,t)≤ a₁(x)(s^r⁻¹+t^r⁻¹+1), for every 0 ≤s, and

F_t(x,s,t)≤ a₂(x)(s^r⁻¹+t^r⁻¹+1), for every 0≤t.

Thus, in order to show existence and multiplicity of solutions to problem (1.1), we define the Sobolev spaceE=W₀^1,⁻^→^p(_Ω)×W₀^1,⁻^→^p(_Ω)endowed with the norm

k(u,v)k= kuk_1,⁻^→_p +kvk_1,⁻^→_p, where

kuk_1,−→_p =

∑

N i=1

∂u

∂x_i pi

.

We say thatu,v ∈Eis a positive weak solution of (1.1) ifu,v>0 inΩand it verifies

∑

N i=1

Z

Ω

∂u

∂x_i

p_i−2

∂u

∂x_i

∂ϕ

∂x_i dx=

Z

Ωa₁(x)uϕdx+

Z

ΩF_u(x,u,v)ϕdx,

and N

i

∑

=₁ Z

Ω

∂u

∂x_i

pi−2

∂u

∂x_i

∂ψ

∂x_i dx =

Z

Ωa₂(x)vψdx+

Z

ΩFv(x,u,v)ψdx, for all ϕ,ψ∈W₀^1,⁻^→^p(_Ω).

In our first theorem we apply the sub-supersolution method to establish the existence of a weak solution for (1.1).

Theorem 1.1. Assume that conditions(H),(H₁)and(H₂)hold. Ifka_jk_∞ is small, for j=1, 2, then system(1.1)has a positive weak solution.

In order to establish the existence of two solutions for problem (1.1), we also assume

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(H₃) There ares₀,t₀>0 such that

0<F(x,s,t)≤θss Fs(x,s,t) +θtt Ft(x,s,t), a.e in Ω, for allt ≥t₀ands≥s₀ inΩ, where _p¹∗ <θ_s,θ_t < _p¹

N.

Theorem 1.2. Assume that conditions (H), (H₁)–(H₃)hold. Then, problem (1.1) has two positive weak solutions ifka_jk_∞ is small, for j=1, 2.

A considerable effort has been devoted during the last years to the study anisotropic problems. With no hope to be thorough, let us mention, for example [1,2,4–7,9–14,16,20–22] and references therein.

In some sense our paper is a natural continuation of the studies initiated in [2] and it com- pletes the results obtained there, because we study the existence and multiplicity of solutions for a system involving an anisotropic operator using subsolution & supersolution method.

This paper seems to be the first to show results on an elliptic system involving an anisotropic operator.

When p_i = 2 we have

∑^Ni=1 ∂

∂x_i

_∂x^∂u

i

p_i−2∂u

∂x_i

= _∆u and when p_i = p we have ∑i^N=1 ∂

∂xi

^∂u

∂xi

p_i−2∂u

∂xi =_∆_pu. Both cases are called isotropic cases or non-anisotropic cases and this kind of problem has been studied by many authors.

This paper is organized as follows. In the Section 2 we prove the unicity of solutions for the Linear anisotropic problem, a Comparison Principle and a regularity result for solutions to this class of problems. In the Section 3 we prove Theorem 1.1. Theorem 1.2 is proved in Section 4.

2 Technical results

We start proving a result of unicity of solution to the linear problem and a Comparison Prin- ciple of the anisotropic operator.

Lemma 2.1. There is u∈W₀^1,⁻^→^p(_Ω)the unique solution of problem







−^h

∑

N i=1

∂

∂x_i

∂w

∂x_i

pi−2∂w

∂x_i i

=a(x)inΩ, w=0on ∂Ω.

(2.1)

Proof. Consider the operatorT :W₀^1,⁻^→^p(_Ω)→(W₀^1,⁻^→^p(_Ω))⁰ such thathTu,φiis given by hTu,φi=

∑

N i=1

Z

Ω

∂u

∂x_i

pi−2

∂u

∂x_i

∂φ

∂x_i dx.

Since the inequality C_i

∂u

∂x_i − ^∂v

∂x_i

pi

≤

*

∂u

∂x_i

pi−2

∂u

∂x_i −

∂v

∂x_i

pi−2

∂v

∂x_i, ∂u

∂x_i − ^∂v

∂x_i +

(2.2) is true for some C_i >0 and for alli=_{1, . . . ,}N, we have that

hTu−Tv,u−vi>₀ _{for all}u,v∈W₀^1,⁻^→^p(_Ω)_withu6=v.

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Moreover, ifkuk →+∞, then, without loss of generality, we can assume that Z

Ω

∂u

∂x_i

pi

dx≥1, for alli=1, 2, . . . ,N.

Hence, since 1< p₁ ≤ p_i, for alli=1, 2, . . . ,N, we have

∑

N i=1

Z

Ω

∂u

∂x_i

p_i

dx≥

∑

N i=1

_Z

Ω

∂u

∂x_i

p_i

dx

p1

pi≥ ¹

N^p¹⁻¹ _N

i

∑

=1

∂u

∂x_i L^pi

p₁

, which implies

kulimk→_∞

hTu,ui

kuk = +_∞.

Thus, by Minty-Browder’s Theorem [8, Théorème 5.16], there exists a unique u ∈ W₀^1,⁻^→^p(_Ω) that satisfiesTu=a(x).

Lemma 2.2. IfΩis a bounded domain and if u,v ∈W₀^1,⁻^→^p(_Ω)satisfy







−

"

∑

N i=1

∂

∂x_i

∂u

∂x_i

pi−2

∂u

∂x_i

! #

≤ −

"

∑

N i=1

∂

∂x_i

∂v

∂x_i

pi−2

∂v

∂x_i

! #

inΩ, u≤v on∂Ω,

then u≤v a.e. inΩ.

Proof. Taking 0≤φ=max{u−v, 0} ∈W^1,

−→ p

0 (_Ω)as a test function, we obtain Z

Ω^T[u>v]

∑

N i=1

∂u

∂x_i

p_i−2

∂u

∂x_i −

∂v

∂x_i

p_i−2

∂v

∂x_i, ∂u

∂x_i − ^∂v

∂x_i

dx≤0.

From inequality (2.2), we conclude that k(u−v)⁺k ≤0, this impliesu≤va.e. inΩ.

Before proving theL^∞-regularity we enunciate an iteration lemma by Stampacchia that we will use.

Lemma 2.3 (See [18]). Assume that φ : [0,∞) → [0,∞) is a nonincreasing function such that if h > k > k₀, for some α > 0,β > 1, φ(h) ≤ C(φ(k))^β/(h−k)^α. Then φ(k₀+d) = 0, where d^α = c2^β^αβ⁻¹φ(k₀)^β⁻¹.

Lemma 2.4. Let v∈W^1,

−→ p

0 (_Ω)be a solution to problem







−

"

∑

N i=1

∂

∂x_i

∂v

∂x_i

pi−2 ∂v

∂x_i

! #

= f inΩ, v=0 on∂Ω.

such that f ∈ L^r(_Ω)with r > p^∗/(p^∗−p₁).Then v∈ L^∞(_Ω).In particular, ifkfk_r is small, then alsokvk_∞ is small.

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Proof. Considerv_k =sign(u)(|u| −k)⁺, then v_k ∈W₀^1,⁻^→^p(_Ω)and _∂x^∂v

i = ^∂v_∂x^k

i in A(k) ={x ∈_Ω:

|u(x)| > k}. Let |A(k)| be the Lebesgue measure of A(k). Using v_k as test function and the Hölder inequality, we have

∑

N i=1

Z

A(k)

∂v_k

∂x_i

pi

dx=

Z

Ωf v_kdx≤ Z

Ω|v_k|^p^∗dx _p¹∗Z

Ω|f|^rdx ¹_r

|A(k)|¹⁻ ^p¹^∗⁺¹^r

. Let

0<S= _inf

u∈D^1,⁻^→^p(_R^N),kuk_p∗=1

_N

i

∑

=1

1 p_i

∂u

∂x_i

pi

, see [12].

Once thatp_i ≥ p₁ >1, we have S

_Z

Ω|u|^p^∗dx ^p_p¹∗

≤

∑

N i=1

Z

Ω

∂u

∂x_i

pi

dx, for all u∈W₀^1,⁻^→^p(_Ω). This implies

S _Z

A(k)

|v_k|^p^∗dx

p1−1 p∗

≤ _Z

Ω|f|^rdx ¹_r

|A(k)|¹⁻ ^p¹^∗⁺¹^r

. Note that if 0<k <h, A(h)⊂ A(k)and

|A(h)|^p¹^∗(h−k) = _Z

A(h)

(h−k)^p^∗dx _p¹∗

≤ _Z

A(k)

|v_k|^p^∗dx _p¹∗

, then

|A(h)| ≤ ¹ (h−k)^p^∗

1 S

p∗ p1−1

kfk

p∗ p1−1

r |A(k)| ^p

∗ p1−1

1− _p¹∗+¹_r . Sincer > _p∗^p−^∗p1, we have β:= _p^p^∗

1−1

1− _p¹∗ +¹_r>1. Therefore, if we define

φ(h) =|A(h)|, α= p^∗, β= ^p

∗

p₁−₁

1− 1

p^∗ + ¹ r

, k₀=0, we have thatφis a nonincreasing function and

φ(h)≤ ^C

(h−k)^α^φ(k)^β, for all h>k>0.

By Lemma2.3, we haveφ(d) =0 ford=ckfk

1 p1−1

r |_Ω|^β⁻^α¹/S^p¹¹⁻¹, then kuk_∞ ≤ ^ckfk

1 p1−1

r |_Ω|^β⁻^α¹ S^p¹¹⁻¹

.

3 Proof of Theorem 1.1

We say that[(u,v),(u,u)]is a pair of sub and supersolution for the problem (1.1), respectively, ifu,v∈E∩L^∞(_Ω)_,u,u ∈E∩L^∞(_Ω)

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a) u≤u,v≤ v in Ωandu=0≤u,v=0≤v on ∂Ω, b) Givenϕ,ψ, with ϕ,ψ≥0, we have











∑

N i=₁ Z

Ω

∂u

∂x_i

p_i−2∂u

∂x_i

∂ϕ

∂x_i ≤

Z

Ωa₁(x)uϕ+

Z

ΩFu(x,u,w)ϕdx for all w∈[v,v]

∑

N i=1

Z

Ω

∂v

∂x_i

pi−2∂v

∂x_i

∂ψ

∂x_i ≤

Z

Ωa2(x)vψ+

Z

ΩFv(x,w,v)ψdx for all w∈[u,u]

(3.1)











∑

N i=1

Z

Ω

∂u

∂x_i

p_i−2∂u

∂x_i

∂ϕ

∂x_i ≥

Z

Ωa₁(x)uϕ+

Z

ΩF_u(x,u,w)ϕdx for all w∈[v,v]

∑

N i=1

Z

Ω

∂v

∂x_i

p_i−2∂v

∂x_i

∂ψ

∂x_i ≥

Z

Ωa₂(x)vψ+

Z

ΩF_v(x,w,v)ψdx for all w∈[u,u]

(3.2)

Lemma 3.1. Assume that(H), (H₁)and(H2)hold. Ifka_jk_∞ is small, for j = 1, 2, then there exist u,v,u,v∈ E^TL^∞(_Ω)such that

i) k(u,v)k_∞ ≤ δ, whereδ is the constant that appeared in the hypothesis(H₁). ii) 0< u(x)≤u(x)a.e inΩand0<v(x)≤ v(x)a.e inΩ.

iii) (u,v)is a subsolution and(u,v)is a supersolution of (1.1).

Proof. By Lemma2.1, there is a unique positive solutionu ∈W₀^1,⁻^→^p(_Ω)satisfying the problem below







−

"

∑

N i=1

∂

∂x_i

∂u

∂x_i

p_i−2 ∂u

∂x_i

! #

=a₁(x) inΩ, u=0 on∂Ω.

Similary, there exists a unique positive solutionv∈W₀^1,⁻^→^p(_Ω)satisfying







−

"

∑

N i=1

∂

∂x_i

∂v

∂x_i

p_i−2

∂v

∂x_i

! #

=a₂(x) inΩ, v =₀ _on ∂Ω.

By Lemma 2.4, u,v ∈ L^∞(_Ω) and there exist C₁,C2 > 0 such that kuk_∞ ≤ C₁kak_∞ and kvk_∞ ≤C₂kak_∞. Now we fixka_jk_∞, withj=1, 2 so that

kuk_∞ ≤ ^δ

2 and kvk_∞ ≤ ^δ 2, which ends the proof of the condition (i).

In order to prove ii), we invoke Lemma 2.1 one more time to show that there exists a unique positive solutionu∈W^1,

−

→p

0 (_Ω)^TL^∞(_Ω)







−

"

∑

N i=1

∂

∂x_i

∂u

∂x_i

pi−2

∂u

∂x_i

! #

=1+a₁(x) in Ω, u=0 on ∂Ω

(3.3)

(7)

and there exists a unique positive solutionv∈W₀^1,⁻^→^p(_Ω)







−

"

∑

N i=1

∂

∂x_i

∂v

∂x_i

p_i−2

∂v

∂x_i

! #

=1+a₂(x) inΩ, v=0 on∂Ω.

(3.4)

Note that, for all 0≤ ϕ,ψ∈W^1,

−→ p

0 (_Ω), we have

∑

N i=1

Z

Ω

∂u

∂x_i

pi−2

∂u

∂x_i

∂ϕ

∂x_i dx =

Z

Ω[a₁(x) +1]ϕdx≥

Z

Ωa₁(x)ϕdx=

∑

N i=1

Z

Ω

∂u

∂x_i

pi−2

∂u

∂x_i

∂ϕ

∂x_i dx and

∑

N i=1

Z

Ω

∂v

∂x_i

pi−2

∂v

∂x_i

∂ψ

∂x_i dx=

Z

Ω[a₂(x) +1]ψdx≥

Z

Ωa₂(x)ψdx=

∑

N i=1

Z

Ω

∂v

∂x_i

pi−2

∂v

∂x_i

∂ψ

∂x_i dx.

Then, from Lemma2.2 we conclude thatu(x)≤ u(x)a.e. inΩandv(x)≤ v(x)a.e. inΩ, which proves the conditionii).

Our final task is to check that the conditioniii)holds. First, we use the maximum principle in [9, Corollary 4.4] and conclude thatu,v>0. Now using the definition ofu, vand(H₁), we obtain, for each ϕ,ψ≥0,

∑

N i=1

Z

Ω

∂u

∂x_i

pi−2 ∂u

∂x_i

∂ϕ

∂x_i −

Z

Ωa₁(x)uϕdx−

Z

ΩF_u(x,u,v)ϕdx

≤

Z

Ωa₁(x)ϕdx−

Z

Ωa₁(x)uϕdx−

Z

Ω(1−u)a₁(x)ϕdx

=0 and

∑

N i=1

Z

Ω

∂v

∂x_i

pi−2

∂v

∂x_i

∂ψ

∂x_i −

Z

Ωa₂(x)vψdx−

Z

ΩFv(x,u,v)ψdx

≤

Z

Ωa₂(x)ϕdx−

Z

Ωa₂(x)vψdx−

Z

Ω(₁−v)a₁(x)ψdx

=0

Then,(u,v)is a subsolution for problem (1.1).

Now, we use(H2), (3.3) and (3.4) we have forka_jk_∞sufficiently small such that

∑

N i=1

Z

Ω

∂u

∂x_i

pi−2 ∂u

∂x_i

∂ϕ

∂x_i −

Z

Ωa₁(x)uϕ−

Z

ΩFu(x,u,v)ϕdx

≥1− ka₁k_∞kuk_∞− ka₁k_∞− ka₁k_∞kuk^r_∞⁻¹− ka₁k_∞kvk^r_∞⁻¹

Z

Ωϕdx>0 and

∑

N i=1

Z

Ω

∂v

∂x_i

pi−2

∂u

∂x_i

∂ψ

∂x_i −

Z

Ωa₂(x)vψ−

Z

ΩF_v(x,u,v)ψdx

≥1− ka2k_∞kuk_∞− ka2k_∞− ka2k_∞kuk^r_∞⁻¹− ka2k_∞kvk^r_∞⁻¹

Z

Ωψdx>0 Thenu,vis a supersolution of (1.1).

(8)

Consider the functions

G_s(x,s,t) =











a₁(x)u(x) +F_s(x,u(x),t), s>u(x)

a₁(x)s+F_s(x,s,t), u(x)≤s≤ u(x) a₁(x)u(x) +Fs(x,u(x),t), s<u(x),

(3.5)

and

G_t(x,s,t) =











a₂(x)v(x) +F_t(x,s,v(x)), t> v(x)

a₂(x)t+F_t(x,s,t), v(x)≤t ≤v(x) a2(x)v(x) +Ft(x,s,v(x)), t< v(x),

(3.6)

and the auxiliary problem











−

"

∑

N i=1

∂

∂x_i

∂u

∂x_i

pi−2 ∂u

∂x_i

! #

=_G_u(_x,_u,_v)_in_Ω, u>0 inΩ,

−

"

∑

N i=1

∂

∂x_i

∂v

∂x_i

pi−2

∂v

∂x_i

! #

=G_v(x,u,v)in Ω, u,v >0 inΩ,

u,v ∈W₀^1,⁻^→^p(_Ω).

(3.7)

We define the functionalΦ:E→_Rby Φ(u,v) =

∑

N i=1

Z

Ω

1 p_i

∂u

∂x_i

pi

dx+

∑

N i=1

Z

Ω

1 p_i

∂v

∂x_i

pi

dx−

Z

Ω

G(x,u,v)dx. (3.8) We haveΦ∈ C¹ E,R

with Φ⁰(u,v)(ϕ,ψ) =

∑

N i=1

Z

Ω

∂u

∂x_i

p_i−2

∂u

∂x_i

∂ψ

∂x_i dx+

∑

N i=1

Z

Ω

∂v

∂x_i

p_i−2

∂v

∂x_i

∂ϕ

∂x_i dx

−

Z

ΩGu(x,u,v)ψdx−

Z

ΩGv(x,u,v)ϕdx, for allu,v,ψ,ϕ∈E.

From(H₂)and definition of GsandGt, we have that

|Gs(x,s,t)| ≤K₁, for some K₁>0, a.e. inΩ (3.9) and

|G_t(x,s,t)| ≤K₂, for some K₂ >0, a.e. inΩ. (3.10) From(3.9)and (3.10) , we have that Φis coercive. Then, we can obtain that (u_n,v_n) is a bounded sequence in E such that

Φ(u_n,v_n)→c=inf

M Φ, where

M =(u,v)∈ E:u≤u≤u a.e. inΩ andv≤v ≤va.e. in Ω .

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Hence, up to subsequence, we have











(u_n,v_n)*(u,v)_inE,

(u_n,v_n)→v inL^s(_Ω)×L^s(_Ω), 1≤s< p^∗, (un(x),vn(x))→(u(x),v(x))a.e inΩ.

(3.11)

Now, note that M is closed and convex in E. By [19, Therem 1.2], the restriction Φ M

attains its infimum at a point (u,v) in M. Using the same argument as in the proof of [19, Therem 2.4], we see that (u,v) weakly solves (3.7). Since G_s(x,s,t) = a₁(x)s+F_s(x,s,t) for s ∈ [u,u] and Gt(x,s,t) = a₂(x)t+Ft(x,s,t) for t ∈ [v,v] then (u,v) is a positive weak solution of (1.1).

4 Proof of Theorem 1.2

Let(u,v)∈E∩L^∞(_Ω)the subsolution of Problema (1.1). In our next result we prove that the functionalΦsatisfies the geometric hypotheses of the Mountain Pass Theorem (to see [3]).

Consider the functions Gbs(x,s,t) =

(a₁(x)s+F_s(x,s,t), s >u(x)

a₁(x)u(x) +F_s(x,u(x)_,t)_, s ≤u(x)_, ^(4.1) Gbt(x,s,t) =

(a₂(x)t+F_t(x,s,t), t>v(x)

a2(x)v(x) +Ft(x,v(x),t), t≤v(x) ^(4.2) and define the functional Φb :W^1,

−

→p

0 (_Ω)→_Rby Φb(u,v) =

∑

N i=1

Z

Ω

1 p_i

∂u

∂x_i

pi

dx+

∑

N i=1

Z

Ω

1 p_i

∂v

∂x_i

pi

dx−

Z

ΩGb(x,u,v)dx. (4.3) Note that by(H₂),(4.1) and (4.2) , we have

Gb_s(x,s,t)≤fC₁|t|+a₁(x)|s|^r+a₁(x)s|t|^r, for all s≥0, (4.4) and

Gb_t(x,s,t)≤Cf₂|t|+a₂(x)|t|^r+a₁(x)t|s|^r, for allt ≥0, (4.5) for some constantsfC₁,Cf₂>0.

Lemma 4.1. The functionalΦb satisfies the(PS)-condition for every c∈_R. Proof. Let(u_n,v_n)⊂Ebe a sequence such that

Φb(u_n,v_n)→c and Φb⁰(u_n,v_n)→0. (4.6) Using(H₃)and Sobolev’s embedding, there areC₁,C₂ >0 such that

C1+k(un,vn)k ≥Φb(_u_n_,_v_n)−^hθ_u_nΦˆ⁰(_u_n_,_v_n)(_u_n_{, 0}) +θ_v_nΦb⁰(_u_n_,_v_n)(_0,_v_n)ⁱ

≥C₂k(u_n,v_n)k^p¹, (4.7)

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where get that (u_n,v_n) is a bounded sequence inEand hence, up to subsequence, we have











(un,vn)*(u,v) inE,

(u_n,v_n)→(u,v) in L^s(_Ω)×L^s(_Ω), 1≤s< p^∗, (u_n(x),v_n(x))→(u(x),v(x)) a.e. in Ω.

(4.8)

Using (4.6), (4.8), (2.2), the Lebesgue dominated convergence theorem and standard argu- ments, up to subsequence, we obtain

∑

N i=1

Z

Ω

∂u_n

∂x_i − ^∂u

∂x_i

p_i

≤o_n(1),

and N

i

∑

=1

Z

Ω

∂v_n

∂x_i − ^∂v

∂x_i

pi

≤on(1), which implies(u_n,v_n)→(u,v)inE.

Lemma 4.2. Assume that(H),(H₁)–(H₃)hold. Then forka_jk_L∞ small, for j=1, 2,Φb satisfies:

i) There are R >k(u,v)kandβ>0,such that

Φb(u,v)<0< β≤ inf

(u,v)∈∂BR(0)

Φb(u,v). ii) There are e∈W₀^1,⁻^→^p(_Ω)\B_2R(0)such thatΦb(e)<β.

Proof. Since(u,v)is a subsolution of (1.1),Gc_s(x,u,t) = a₁(x)u+F_s(x,u,t)uandcG_t(x,s,v) = a₂(x)v+Ft(x,s,v)v, with p_i >1, fori=1, 2, ...N, we have

Φb(u,v) =

∑

N i=1

Z

Ω

∂u

∂x_i

p_i

dx+

∑

N i=1

Z

Ω

∂v

∂x_i

p_i

dx

−

Z

Ω(a₁(x)u+Fs(x,u,t))udx−

Z

Ω(a2(x)v+Fs(x,s,v))v dx. (4.9) Now, letk(u,v)k= R>1, without loss of generality, we can assume that

Z

Ω

∂u

∂x_i

p_i

dx≥1, for alli=1, 2, . . . ,N, and

Z

Ω

∂v

∂x_i

pi

dx≥1, for alli=1, 2, . . . ,N.

Hence, using this inequality, (4.4) and (4.5) with the Sobolev Embedding Theorem, we find positive constants, such that

Φb(u,v)≥Kk(u,v)k −c3ka₁k_L∞(_Ω)kuk_L∞(_Ω)k(u,v)k −c₄ka₁k_L∞(_Ω)k(u,v)k

−c₅ka₁k_L∞(_Ω)k(u,v)k^r−c₆ka₂k_L∞(_Ω)kvk_L∞(_Ω)k(u,v)k

−c₇ka₂k_L∞(_Ω)k(u,v)k −c₈ka₂k_L∞(_Ω)k(u,v)k^r−c₉ka₁k_L∞(_Ω)k(u,v)k^r

−c₁₀ka₁k_L∞(_Ω)k(u,v)k^r−c₁₁ka₂k_L∞(_Ω)k(u,v)k^r

−c₁₂ka2k_L∞(_Ω)k(u,v)k^r, for all k(u,v)k=R, (4.10)

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where K = min_k₁

pN, _p^k²

N . Note that, if (u,v) ∈ ∂B_R(0) with R > 1 and for ka_jk_L∞(_Ω) sufficiently small, with j= 1, 2, there existsβ ∈ _R such that Φb(u,v) ≥ β, for all(u,v) ∈ ∂B_R(0). Hence, the choices of β, R andka_jk_L∞(_Ω)combined with inequalities (_4.9)_and(_4.10)_{result in}

Φb(u,v)<0<β≤ inf

(u,v)∈∂BR(0)

Φb(u,v), which shows the conditioni).

Now, by definition ofGbswe have

Gbsu(x,su, 0)≥ F(x,su, 0) for all s≥1, a.e. inΩ.

We invoke(H₁)and(4.3)to obtain

Φb(su, 0)≤

∑

N i=1

s^p^N p₁

Z

Ω

∂u

∂x_i

p_N

−

Z

ΩF(x,su, 0)dx.

Using(H₃), there existsKe₁ >0 such that

F(x,s, 0)≥ Ke₁s^θs¹, for all s ≥max{1,s₀}, wheres0 are the constants that appear in(H3). Then,

Φb(su, 0)≤s^p^N

∑

N i=1

Z

Ω

1 p₁

∂u

∂x_i

pi

−Ke₁s^θs¹ Z

Ω|u|^θs¹ dx.

Since _p¹∗ < θ_s < _p¹

N, we conclude that Φb(su, 0) → −_∞ as s → +∞. So, we may find e=s₀(u, 0)∈Esuch thatkek> RandΦb(e)<β, which satisfies the condition ii).

Proof of Theorem1.2. Let(u,v),(u,v)be the subsolution and the supersolution of (1.1) given in Lemma (3.1) and(u₁,v₁)the solution of (1.1) obtained in Theorem1.1.

Using the Lemma4.2, we conclude, with the Mountain Pass Theorem (see [3]), that bc= _inf

γ∈_Γmax

t∈[0,1]

Φb(γ(t))_, _where_Γ={γ∈ C [_{0, 1}]_,W₀^1,⁻^→^p(_Ω)_:γ(₀) = (u,v)_, γ(₁) =e}, is critical value ofΦ.b

By (3.5),(3.6),(4.1) and (4.2),Gs(x,s,t) = Gbs(x,s,t)fors ∈ [0,u]andGt(x,s,t) = Gbt(x,s,t) fort ∈[0,v], thusΦ(u,v) =Φb(u,v)withu∈[0,u]andv∈ [0,v], whereΦandΦb are given in (3.8) and (4.3), respectively. Then,

Φb(u₁,v₁) =inf

MΦ(u,v), where

M=(u,v)∈E:u≤u≤ua.e. in Ωandv≤v≤va.e. in Ω . was given in the proof of in Theorem 1.1.

Therefore, the problem (1.1) has two weak solutions v₁,v₂ ∈W^1,

−→ p

0 (_Ω), such that Φb(u₁,v₁)≤Φb(v)<₀< β≤_bc=Φb(u₂,v₂)_.

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Recall thatu ≤ u₁ ≤u a.e. in Ωandv≤ v₁ ≤ v a.e. in Ω, thus(u₁,v₁)>0. Now, we will show that(u2,v2)>0.

Taking ((u,v)−(u₂,v₂))⁺, as test function and defining {(u₂,v₂) < (u,v)} := {x ∈ _Ω : u₂(x)<u(x) andv₂(x)<v(x)}, we have

∑

N i=1

Z

Ω

∂u2

∂x_i

pi−2

∂u2

∂x_i

∂(u−u2)⁺

∂x_i dx+

∑

N i=1

Z

Ω

∂v2

∂x_i

pi−2

∂v2

∂x_i

∂(v−v2)⁺

∂x_i dx

=

Z {u₂<u}

(a₁u+F_s(x,u,t))(u−u₂)⁺dx+

Z {v₂<v}

(a₂v+F_t(x,s,v))(v−v₂)⁺dx. (4.11) Since(u,v)is a subsolution of (1.1), using (4.11) we obtain

∑

N i=1

Z

Ω

∂u

∂x_i

pi−2

∂u

∂x_i

∂(u−u2)⁺

∂x_i dx−

∑

N i=1

Z

Ω

∂u2

∂x_i

pi−2

∂u2

∂x_i

∂(u−u2)⁺

∂x_i dx≤0

and N

i

∑

=1

Z

Ω

∂v

∂x_i

pi−2 ∂v

∂x_i

∂(v−v₂)⁺

∂x_i dx−

∑

N i=1

Z

Ω

∂v₂

∂x_i

pi−2 ∂v₂

∂x_i

∂(v−v₂)⁺

∂x_i dx≤0.

From inequality (2.2), we conclude thatk(u−u₂)⁺k_1,⁻^→_p ≤ 0 and k(v−v₂)⁺k_1,⁻^→_p ≤ 0, this implies 0<u≤u₂ a.e. inΩand 0<v≤ v₂ a.e. inΩ. We concluded that(u₂,v₂)>0.

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