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Local and global estimates for solutions of systems involving the p-Laplacian

in unbounded domains

A. Bechah

Abstract

In this paper, we study the local and global behavior of solutions of systems involving the p-Laplacian operator in unbounded domains. We extend some Serrin-type estimates which are known for simple equations to systems of equations.

1 Introduction

We consider the system

−∆pu=f(x, u, v) x∈Ω, (1.1)

−∆qv=g(x, u, v) x∈Ω, (1.2)

u=v= 0 x∈∂Ω. (1.3)

where Ω ⊂RN is an exterior domain, f, g are a given functions depending of the variables x, u, v and ∆p is the p-Laplacian operator; for 1 < p < +∞ ∆p

is defined by ∆pu = div |∇u|p2∇u

. Here, we study the local and global behavior of solutions of System (1.1)–(1.3). we follow the work of Serrin [4]

concerning the quasilinear equation

divA(x, u, ux) =B(x, u, ux), (1.4) where A and B are a given functions depending of the variables x, u, ux and ux= (∂x∂u

1, . . . ,∂x∂u

n). In particular, (1.4) generalizes the equation

−∆pu=f(x, u) x∈Ω. (1.5)

In [4], Serrin proves that if the function f is bounded by the terma|u|p1+g, where p >1 is a fixed exponent, ais a positive constant andg is a measurable function, then for each y∈Ω andR >0 we have the estimate

sup

BR(y)

u(x)≤cRNp

kukLp(B2R(y))+RNp(Rkgk

Lp−N (B2R(y)))p−11

(1.6)

Mathematics Subject Classifications: 35J20, 35J45, 35J50, 35J70.

Key words:quasilinear systems, p-Laplacian operator, unbounded domain, Serrin estimate.

2001 Southwest Texas State University.c

Submitted November 23, 2001. Published March 23, 2001.

1

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for all 0< ≤1.

In many cases, especially for unbounded domain, when we wish to show that the solution decay at infinity, the estimate (1.6) requires that the function f belongs toLα(Ω) withα > N/p, which is not trivial to prove in some cases. To avoid this difficulty Yu [5], Egnell [1] and others have proved that the solution of (1.5) have a regularity Lq(Ω) for each q ≥ p, and this for all function f bounded by a sublinear, superlinear or an homogeneous terms. We note that in the case of a mixed terms this last technique cannot be adapted. For the case of an homogeneous system see the paper of Fleckinger, Man`asevich, Stavrakakis and de Th´elin [2].

The first part of this paper is devoted to the local behavior of solutions of System (1.1)–(1.3). We obtain an estimate of Serrin type in the following cases:

1) f andg are bounded by a sum of homogeneous and critical terms.

2) f andg are bounded by a sum of homogeneous and constant terms.

Thus, we extend the results of [5], [1] concerning Equation and those of [2]

concerning System.

In the second part, we obtain a global estimates of solutions of System (1.1)–

(1.3) in the particular casef =A|u|α1u|v|β+1 and g=B|u|α+1|v|β1v under some conditions onα, β, pandq. Also we obtain another global estimate when f andg satisfy 2).

We recall thatD1,p(Ω) is the closure ofC0(Ω) with respect to the norm kukD1,p(Ω)=k∇ukLp(Ω).

p0=pp

1 is the conjugate ofp,p∗= NN p

p is the Sobolev exponent and we define Sp by

1 Sp

= inf

(k∇ukpLp(Ω)

kukpLp(Ω)

u∈W1,p(Ω)\{0} )

.

2 Local estimates for solutions of (1.1)–(1.3)

Theorem 2.1 Let (u, v)∈ D1,p(RN)× D1,q(RN) be a solution of(1.1)−(1.3) andτ= NN

p,τ¯= NN

q. Assume that max{p, q}< N,q≥pand

|f(x, u, v)| ≤C

|u|p1+|u|p1+|v|q/p0+|v|(τ p)τ q0

, (2.1)

and

|g(x, u, v)| ≤C

|v|q1+|v|τ q1+|u|p/q0+|u|(τ q)0τ p

, (2.2)

wherem0 is the conjugate of m andC is a constant. Then 1) For any R >0 andx∈RN satisfying

Cmaxn

2pSpτp1,22qpSq|B1|q−pN Rqpτq1o

×

kukp(τLp(B1)2R(x))+kvkq(τL(B1)2R(x))

<1

(2.3)

(3)

where Sp andSq are the Sobolev constants, we have kukL(BR

2

(x))

≤ c(1 +Rq)

N(N−p) p3 maxn

Rp−Np kukLp∗(BR(x)), Rq−Np kvkq/pLq(BR(x))

o . and

kvkL(BR 2(x))

≤ c(1 +Rq)

N(N−p) qp2 maxn

Rq−Nq kvkLq(BR(x)), Rp−Nq kuk

p q

Lp∗(BR(x))

o . witch cindependent of u, v, xandR.

2) Moreover,

lim

|x|→+u(x) = lim

|x|→+v(x) = 0.

Remark 2.2 There exists an R0 such that for all R < R0, (2.3) is satisfied uniformly for all x∈Ω. This follows from the absolute continuity of the func- tionalsA7→R

A|u|pdxandA7→R

A|v|dx. To be more specific, for each >0 there exists η >0 such that for all R >0 and x∈RN satisfying|BR(x)| ≤η, we have R

BR(x)|u|pdx < andR

BR(x)|v|dx < .

Proof Letx∈ RN be fixed. For y ∈ B2R(x) and any function hdefined on B2R(x) we define

˜h(t) =h(y), t= y−x R . Since (u, v) is a solution for (1.1)–(1.3), then (˜u,˜v) satisfies

−∆pu˜=Rpf(y,u,˜ v),˜ (2.4)

−∆qv˜=Rqg(y,u,˜ v).˜ (2.5) In this proof c denotes a positive constant independent of u, v, x and R. For any ball B⊂B2(0), we have

∀w∈ W01,p(B) kwkpL(B)≤Spk∇wkpLp(B),

∀w∈ W01,q(B) kwkqL(B)≤2qp|B1(0)|q−pN Sqk∇wkqLq(B). (2.6) Sp and Sq are the Sobelev constants. Let (mn)n be a sequence of positive numbers satisfying σ < ∞ where σ is defined below and (rn)n a decreasing sequence defined by

r0= 2, rn = 2− 1 σ

n1

X

i=0

mi+p p

1/p0

,

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whereR is positive andσ=

X

i=0

mi+p p

1/p0

. We denote byBn =B(0, rn) and we define η ∈ C0(RN) so that 0 ≤η ≤1, η = 1 in Bn+1, supp(η)⊂Bn

and

|∇η| ≤c

mn+p p

1/p0

. (2.7)

We multiply (2.4) by|u˜|mn˜uηq, and integrate overBn. Using (2.1), we obtain I1+I2≤Rp(I3+I4+I5+I6), (2.8) where

I1= (1 +mn) Z

Bn

ηq|u˜|mn|∇˜u|pdx, I2=q

Z

Bn

ηq1∇η.∇u˜|∇u˜|p2|˜u|mnudx,˜ I3=C

Z

Bn

|u˜|p+mnηqdx, I4=C

Z

Bn

|u˜|p+mnηqdx, I5=C

Z

Bn

|u˜|mn˜u|v˜|q/p0ηqdx, I6=C

Z

Bn

|u˜|mn˜u|v˜|(τ p)τ q0ηqdx.

Since 1 +mn =(p1)mp n+mnp+p, we deduce from Young inequality and the facts p≤q,|η| ≤1, that for anys >0

|I2| ≤ qsp0 p0

mn+p p

Z

Bn

ηq|∇u˜|p|u˜|mndx + q

psp

mn+p p

p0p Z

Bn

|∇η|p|u˜|mn+pdx

Choosingssuch that qsp

0

p012, and using (2.7), we have

|I2| ≤ 1 2I1+c

Z

Bn

|u˜|mn+pdx. (2.9) We deduce from (2.8) and (2.9)

I1≤2Rp

6

X

i=3

Ii+c Z

Bn

|u˜|mn+pdx. (2.10)

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Using Sobolev inequality and observing that for anya≥0 andb≥0 (a+b)p≤ 2p1(ap+bp), we have

ηq/pmn+pp

p

L(Bn)≤2p1Sp(I7+I8), (2.11) where

I7= (q p)p

Z

Bn

ηqp|∇η|p|u˜|mn+pdx≤c

mn+p p

p1Z

Bn

|u˜|mn+pdx, and

I8=

mn+p p

pZ

Bn

ηq|u˜|mn|∇u˜|pdx≤

mn+p p

p1

I1, thus we deduce from (2.10) that

ηq/pmn+pp

p

L(Bn)

mn+p p

p1

c Z

Bn

|u˜|mn+pdx+ 2pSpRp

6

X

i=3

Ii

! . (2.12) First step. We construct the sequences (pn)n and (qn)n by

pn=pτn, qn=qτn, and we set

mn=p(τn−1), and ln=q(τn−1).

We show that if the condition Cmaxn

2pSpRpτn(p1),22qp|B1|q−pN SqRqτn(q1)o

×

ku˜kp(τLp(B1)2)+k˜vkq(τL(B1)2)

< 1, is satisfied, the solution (˜u,v) belongs to˜ Lpn+1(Bn+1)×Lqn+1(Bn+1).

First, we start by estimating the integrals (Ii), i= 3, . . . ,6. We have I3=C

Z

Bn

|u˜|p+mnηqdx≤cku˜kpLnpn(Bn). (2.13) Remarking that mpn+1

n +

q p0

qn = 1, we deduce from H¨older inequality that I5=C

Z

Bn

|u˜|mnu˜|˜v|q/p0ηqdx≤cku˜kmLpnn+1(Bn)kv˜kq/pLqn0(Bn). (2.14) We write mn+p = p(τ −1) +mn +p , q = τ q(mpn+p

n+1). Observing that

mn+p

pn+1 +p(τp1) = 1,we deduce from H¨older inequality I4=C

Z

Bn

|u˜|p+mnηqdx≤C Z

Bn

|u˜|p(τ1)|u˜|p+mnητ q(

mn+p pn+1)

dx

≤Ck˜ukp(τLp(B1)n)q/pτnkpL(Bn).

(2.15)

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Remark that τ q (τ p)0 − q

p0 =q(τ−1), q(τ−1)

τ q +mn+ 1 pn+1 +

q p0

qn+1 = 1, (2.16) and

τmn+ 1 pn+1

q p0

qn+1

= 1, then from H¨older inequality, we have

I6=C Z

Bn

|u˜|mnu˜|v˜|(τ p)τ q0ηqdx

≤C Z

Bn

|v˜|q(τ1)ητ q(mn+1pn+1)|u˜|mn+1ητ q(

q p0 qn+1)

|˜v|q/p0dx

≤Ckv˜kq(τLτ q(B1)n)q/pτnkp(

1+mn pn )

L(Bn)kη˜vτnkq(

p0q qn) L(Bn).

(2.17)

Substitutingmn byp(τn−1) in (2.12), we obtain

q/pτnkpL(Bn)−τn(p1)2pSpRp(I4+I6)

≤τn(p1)

c Z

Bn

|u˜|pndx+ 2pSpRp(I3+I5)

.

(2.18)

It follows from (2.13) - (2.17) and the factp≤qthat kηq/pτnkpL(Bn)−C2pSpRpτn(p1)

ku˜kp(τLp∗(B1)n)q/pτnkpL(Bn)

+kv˜kq(τLτ q(B1)n)q/pτnkp

(1+mn) pn

L(Bn)kηv˜τnkq(

q p0 qn) L(Bn)

!

≤c(1 +Rqn(q1)

ku˜kpLnpn(Bn)+ku˜kmLpnn+1(Bn)kv˜kq/pLqn0(Bn)

.

(2.19)

Similarly, we have

kηv˜τnkqL(Bn)−C22qpSq|B1|q−pN Rqτn(q1)

k˜vkq(τL(B1)n)kηv˜τnkqL(Bn)

+ku˜kp(τLτ p(B1)n)kη˜vτnkq

(1+ln) qn

L(Bn)q/pτnkp(

p q0 pn) L(Bn)

≤c(1 +Rqn(q1)k˜vkqLnqn(Bn)+cRqτn(q1)k˜vklLnqn+1(Bn)ku˜kp/qLpn0(Bn).

(2.20)

Next, we defineθn+1= max{kηq/pτnkpL(Bn),kηv˜τnkqL(Bn)}, and

En = max{ku˜kpLnpn(Bn),kv˜kqLnqn(Bn)}1/pn. Simple computations using H¨older inequality and the definition ofEn andθn, show that

θn+1−Cmaxn

2pSpRpτn(p1),22qp|B1|q−pN SqRqτn(q1)o

×

ku˜kp(τLp(B1)n)+kv˜kq(τL(B1)n)

θn+1≤c(1 +Rqn(q1)Enpn.

(2.21)

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We know that there existsR0>0 such that for anyR < R0

Cmaxn

2pSpRpτn(p1),22qp|B2|q−pN SqRqτn(q1)o

×

ku˜kp(τLp∗(B1)2)+kv˜kq(τL(B1)2)

<1.

(2.22)

Also, remark that

θn+1≥max{ku˜kpLnpn+1(Bn+1),k˜vkqLnqn+1(Bn+1)}

≥max{ku˜kpLn+1pn+1(Bn+1),k˜vkqLn+1qn+1(Bn+1)}1/τ

=En+1pn .

(2.23)

Therefore, from (2.21) - (2.23), and the factp≤q En+1pn ≤c(1 +Rqn(q1)Enpn. So

En+1≤(c(1 +Rq))1/pnτn(q−1)pn En. This implies that

ku˜kLpn+1(Bn+1)≤En+1≤(c(1 +Rq))

P i=0

1 pτ i τ

P i=0

i(q−1) pτ i E0. Since P

i=0 1

i = Np2 andP

i=0 i(q1)

i <∞, we deduce that ˜u∈Lpn+1(Bn+1).

Similarly, we have k˜vkq/pLqn+1(Bn+1)≤ k˜vk

qn+1 pn+1

Lqn+1(Bn+1)≤En+1≤(c(1 +Rq))

P i=0 1

pτ i τ

P i=0

i(q−1) pτ i E0, thereforev∈Lqn+1(Bn+1).

Second step We remark that hypothesis (2.3) is equivalent to Cmaxn

2pSpRpτp1,22qp|B1|q−pN SqRqτq1o

ku˜kp(τLp(B1)2)+kv˜kq(τLq(B1)2)

<1.

We assume that R, u and v satisfy (2.3), which by the first step implies that (˜u,v)˜ ∈L2(B1)×L2(B1). We let δ= τ2τ2τ+1 andχ = τδ. It is clear that 1< δ < τ, and soχ >1. We construct a sequences (sn)n and (tn)n by

sn=pχn, tn=qχn. In this step mn andrn are defined by

mn=p χn

δ −1

, and

r0= 1, rn= 1− 1 2σ

n1

X

i=0

mi+p p

1/p0

,

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which implies mn+p=sn/δ. Now, we estimate the integrals (Ii)i=3,...,6. We have

I3≤cku˜ksn

Lsnδ (Bn)≤cku˜ksLnsn(Bn). (2.24) Remarking that msn+1

n +tq/p0

n = 1, it follows from H¨older inequality that I5≤cku˜kmn+1

Lsnδ (Bn)kv˜kq/p0

Ltnδ (Bn)≤ ku˜kmLsnn+1(Bn)kv˜kq/pLsn0(Bn). (2.25) We have p(τ21)+mns+p

n = 1, thus from H¨older inequality we have

I4≤cku˜kp(τL21)(Bn)ku˜ksLnsn(Bn)≤cku˜ksLnsn(Bn). (2.26) Observing that q(τ21)+msn+1

n +q/pt 0

n = 1, it follows from H¨older inequality that I6≤c

Z

Bn

|v˜|q(τ1)|u˜|mn+1|˜v|q/p0dx

≤ckv˜kq(τL2(B1)n)ku˜kmLsnn+1(Bn)kv˜kq/pLtn0(Bn)

≤cku˜kmLsnn+1(Bn)kv˜kq/pLtn0(Bn).

(2.27)

We deduce from (2.12), (2.24)–(2.27) and the factp≤qthat

ηq/pχn

p

L(Bn)≤cχn(q1)(1 +Rq)

ku˜ksLnsn(Bn)+ku˜kmLsnn+1(Bn)kv˜kq/pLtn0(Bn)

(2.28) Similarly, we have

ηv˜χn

q

L(Bn)≤cχn(q1)(1 +Rq)

kv˜ktLntn(Bn)+kv˜klLntn+1(Bn)ku˜kp/qLsn0(Bn)

(2.29) As in the first step, we let Λn= maxn

ku˜ksLnsn(Bn),k˜vktLntn(Bn)

o1/sn

Γn = maxn

q/pχnkpL(Bn),kηv˜χnkqL(Bn)

o and Υn= maxn

ku˜ksLnsn(Bn),kv˜ktLntn(Bn)

otn1

. Simple computations show that ku˜kmLsnn+1(Bn)k˜vkq/pLtn0(Bn)≤minn

Λsnntnno

, (2.30)

and

kv˜klLntn+1(Bn)ku˜kp/qLsn0(Bn)≤minn

Λsnntnno

. (2.31)

Also, remark that Γn≥maxn

ku˜ksLnsn+1 (Bn+1),kv˜ktLntn+1 (Bn+1)

o

= Λsn+1n= Υtnn. (2.32)

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Thus, we deduce from (2.28)–(2.32) that

Λsn+1n≤cχn(q1)(1 +Rq) Λsnn, and so

Λn+1≤cδ/snχn(q−1)δsn (1 +Rq)δ/snΛn. Which implies that

ku˜kLsn(Bn)≤Λn≤c

P

i=0 δ siχ

P

i=0 i(q−1)δ

si (1 +Rq)

P

i=0 δ si Λ0. SinceP

i=0 δ

si =p(τδτδ),andP

i=0 i(q1)δ

si <∞,then ku˜kL(B1

2

)≤ lim

n+supku˜kLsn(Bn)

≤c(1 +Rq)p(τ−δ)δτ maxn

ku˜kLp(B1),k˜vkq/pLq(B1)

o . Similarly, we have

Υn+1≤ctnδ χn(q−1)δtn (1 +Rq)tnδ Υn

Asntends to infinity, we obtain kv˜kL(B1

2

)≤ lim

n+supk˜vkLtn(Bn)

≤c(1 +Rq)q(τ−δ)δτ maxn

k˜vkLp(B1),ku˜k

p q

Lq(B1)

o . By the imbeddings

Lp(B1)⊂Lp(B1) and Lq(B1)⊂Lq(B1), and the fact

δτ

τ−δ = τ

(τ−1)2 =N(N−p) p2 , we have

ku˜kL(B1

2

)≤c(1 +Rq)

N(N−p) p3 maxn

ku˜kLp∗(B1),kv˜kq/pLq(B1)

o , and

k˜vkL(B1

2

)≤c(1 +Rq)N(N−p)qp2 maxn

k˜vkLp∗(B1),ku˜k

p q

Lq(B1)

o. Coming back to (u, v) by a simple change of variables, we find

kukL(BR

2(x))

≤ c(1 +Rq)

N(N−p) p3 maxn

Rp−Np kukLp∗(BR(x)), Rq−Np kvkq/pLq(BR(x))

o .

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and

kvkL(BR

2(x))

≤ c(1 +Rq)

N(N−p) qp2 max

Rq−Nq kvkLq∗(BR(x)), Rp−Nq kuk

p q

Lp(BR(x))

o . The proof of 2) follows from 1) and Remark 2.2 ♦ Proposition 2.3 Let (u, v)∈ D1,p(RN)× D1,q(RN)a solution of (1.1)–(1.3).

We assumeq≥p,

|f(x, u, v)| ≤C

|u|p1+|v|q/p0+ 1

, (2.33)

and

|g(x, u, v)| ≤C

|v|q1+|u|p/q0+ 1

, (2.34)

wherem0 is the conjugate ofm. Then kukL(B1)≤c(1 +Rq)pN2 maxn

1, Rp−Np kukLp(B2), Rq−Np kvkq/pLq∗(B2)

o

, (2.35) and

kvkL(B1)≤c(1 +Rq)pqN maxn

1, Rp−Nq kuk

p q

Lp(B2), Rq−Nq kvkLq∗(B2)

o. (2.36) Proof We use the same change of variables as in the proof of Theorem 2.1.

Thus, we obtain that (˜u,˜v) satisfies (2.4) and (2.5). Also we keep the same sequences (mn)n, (rn)n, (Bn)nand the same functionη. We multiply Equation (2.4) by|u˜|mnuη˜ q, and integrate overBn. Using (2.33), we have

I1+I2≤Rp(I3+I4+I5), (2.37) where

I1= (1 +mn) Z

Bn

ηq|u˜|mn|∇˜u|pdx, I2=q

Z

Bn

ηq1∇η.∇u˜|∇u˜|p2|˜u|mnudx,˜ I3=C

Z

Bn

|u˜|p+mnηqdx, I4=C

Z

Bn

|u˜|mn˜u|v˜|q/p0ηqdx, I5=C

Z

Bn

|u˜|mn˜uηqdx.

The integralsI1, I2, I3 and I4 are the same to those obtained in Theorem 2.1.

Simple computations used before show that

ηq/pmn+pp

p

L(Bn)

mn+p p

p1

c Z

Bn

|u˜|mn+pdx+ 2pSpRp

5

X

i=3

Ii

! . (2.38)

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Now, we define (pn)n and (qn)n by

pn=pτn, qn=qτn,

and let mn =p(τn−1),and , ln =q(τn−1). Then we estimate the integrals Ii, i= 3, . . . ,5. It is clear from (2.13) and (2.14) that

I3≤cku˜kpLnpn(Bn) and I4≤cku˜kmLpnn+1(Bn)k˜vkq/pLqn0(Bn). (2.39) On the other hand

I5≤C Z

Bn

|u˜|mn+1dx=cku˜kmLmn+1n+1(Bn)≤c|Bn|(mn1 pn1 )(mn+1)ku˜kmLpnn+1(Bn)

≤c|B2|p−1pτ nku˜kmLpnn+1(Bn)

≤cku˜kmLpnn+1(Bn).

(2.40) We deduce from (2.38)–(2.40) that

ku˜kpLnpn+1(Bn+1)≤ kηq/pτnkpL(Bn)

≤cτn(p1)

ku˜kpLnpn(Bn)

+Rp

ku˜kpLnpn(Bn)+ku˜kmLpnn+1(Bn)k˜vkq/pLqn0(Bn)+ku˜kmLpnn+1(Bn) . (2.41) Similarly, we have

k˜vkqLnqn+1(Bn+1)≤ kη˜vτnkqL(Bn)

≤cτn(q1)

k˜vkqLnqn(Bn)

+Rq

kv˜kqLnqn(Bn)+kv˜klLnqn+1(Bn)ku˜kp/qLpn0(Bn)+kv˜klLnqn+1(Bn) . (2.42) Following the proof of Theorem 2.1 we let

En = maxn

1,ku˜nkpLnpn(Bn),k˜vnkqLnqn(Bn)

o1/pn

and Fn=n

1,k˜unkpLnpn(Bn),kv˜nkqLnqn(Bn)

oqn1

. We obtain ku˜kL(B1)≤ lim

n+supku˜kLpn(Bn)≤En

≤c(1 +Rq)pN2 E0

=c(1 +Rq)

N p2 maxn

1,ku˜kLp(B2),kv˜kq/pLq(B2)

o .

(2.43)

(12)

k˜vkL(B1)≤ lim

n+supkv˜kLqn(Bn)≤Fn

≤c(1 +Rq)pqN F0

=c(1 +Rq)pqN maxn 1,ku˜k

p q

Lp(B2),kv˜kLq(B2)

o .

(2.44)

Using a simple change of variables in (2.43) and (2.44) we obtain (2.35) and (2.36).

3 Global estimates for solutions of (1.1)–(1.3)

Proposition 3.1 Let (u, v)∈ D1,p(Ω)× D1,q(Ω)a solution of (1.1)–(1.3). We assume that there exist a functionsa, b∈L1(Ω)∩L(Ω)and a constantC such that

|f(x, u, v)| ≤a(x) +C(|u|p1+|v|q/p0), (3.1)

|g(x, u, v)| ≤b(x) +C(|v|q1+|v|p/q0), (3.2) wherep >1,q >1. Then

1)(u, v)∈Lσ(Ω)×Lη(Ω) for all(σ, η)∈[p,+∞)×[q,+∞).

2) lim

|x|→+u(x) = lim

|x|→+v(x) = 0.

Proof 1) Let pn = pτn, qn = qτn, mn = τn −1, tn = τn −1, Tk(u) = max{−k,min{k, u}} and w = |Tk(u)|pmnTk(u), with k > 0. Multiplying the equation (1.1) byw and integrating over Ω, we obtain

(pmn+ 1) Z

|∇Tk(u)|p|Tk(u)|pmndx= Z

f(x, u, v)w dx.

Observing that ( 1

mn+ 1)p|∇(Tk(u))mn+1|p=Tk(u)pmn|∇Tk(u)|p, (3.3) we deduce from H¨older and Sobolev inequalities that for any 0 < γ < 1, we have

Z

|Tk(u)|τ(pmn+p))1/τ

≤c

kak1γkakγL1(Ω)kukpmLpnn(Ω)+1+kukpLnpn(Ω)+kvkq/pLqn0(Ω)kukmLpnn+1(Ω)

.

(3.4)

withcdepending from n. Lettingktend to infinity in (3.4), we obtain kukpLnpn+1(Ω)≤c

kukpmLpnn(Ω)+1+kukpLnpn(Ω)+kvkq/pLqn0(Ω)kukmLpnn+1(Ω)

. (3.5) We derive from (3.5) that u∈Lpn(Ω) for all n∈N. Similarly, we prove that v ∈ Lqn(Ω) for all n∈ N. By interpolation inequality (see [3]) we prove that

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