Nonexistence of nontrivial solutions for the p ( x )- Laplacian equations and systems in unbounded domains of ℝ n

Boundary Value Problems20112011:50

DOI: 10.1186/1687-2770-2011-50

Received: 12 January 2011

Accepted: 30 November 2011

Published: 30 November 2011

Abstract

In this paper, we are interested on the study of the nonexistence of nontrivial solutions for the p(x)-Laplacian equations, in unbounded domains of ℝ n . This leads us to extend these results to m-equations systems. The method used is based on pohozaev type identities.

1 Introduction

Several works have been reported by many authors, comprise results of nonexistence of nontrivial solutions of the semilinear elliptic equations and systems, under various situations, see [18]. The Pohozaĕv identity [1] published in 1965 for solutions of the Dirichlet problem proved absence of nontrivial solutions for some elliptic equations of the form
- Δ u + f ( u ) = 0  in  Ω , u = 0  on  Ω , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-50/MediaObjects/13661_2011_Article_95_Equa_HTML.gif
when Ω is a star shaped bounded open domain in ℝ n and f is a continuous function on ℝ satisfying
( n - 2 ) F ( u ) - 2 n u f ( u ) > 0 , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-50/MediaObjects/13661_2011_Article_95_Equb_HTML.gif
  1. A.
    Hareux and B. Khodja [2] established under the assumption
    f ( 0 ) = 0 , 2 F ( u ) - u f ( u ) 0 . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-50/MediaObjects/13661_2011_Article_95_Equc_HTML.gif
     
that the problems
- Δ u + f ( u ) = 0  in  J × ω , u  or  u n = 0  on  ( J × ω ) . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-50/MediaObjects/13661_2011_Article_95_Equd_HTML.gif
admit only the null solution in H2(J × ω) ∩ L(J × ω). where J is an interval of ℝ and ω is a connected unbounded domain of ℝ N such as
Λ N , Λ  = 1 , n ( x ) , Λ 0 on  ω , n ( x ) , Λ 0 , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-50/MediaObjects/13661_2011_Article_95_Eque_HTML.gif

(n(x) is the outward normal to ∂ω at the point x)

In this work we are interested in the study of the nonexistence of nontrivial solutions for the p(x)-laplacian problem
- Δ p ( x ) u = H ( x ) f ( u ) in  Ω B u = 0  on  Ω http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-50/MediaObjects/13661_2011_Article_95_Equ1_HTML.gif
(1.1)
with
B u = u Dirichlet condition 1 . 2 u ν Neumann condition 1 . 3 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-50/MediaObjects/13661_2011_Article_95_Equf_HTML.gif
where
Δ p x u = d i v u p ( x ) - 2 u http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-50/MediaObjects/13661_2011_Article_95_Equg_HTML.gif
Ω is bounded or unbounded domains of ℝ n , f is a locally lipshitzian function, H and p are given continuous real functions of C ( Ω ̄ ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-50/MediaObjects/13661_2011_Article_95_IEq1_HTML.gif verifying
F ( t ) = 0 t f ( σ ) d σ , f ( 0 ) = 0 , H ( x ) > 0 , ( x , H ( x ) ) 0  and  lim x  +  H ( x )  = 0 , p ( x ) > 1 , ( x , p ( x ) ) 0 , x Ω ̄ , a = sup x Ω ̄ 1 - n p ( x ) + ( x , p ( x ) ) p 2 ( x ) . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-50/MediaObjects/13661_2011_Article_95_Equ2_HTML.gif
(1.4)

(., .) is the inner product in ℝ n .

We extend this technique to the system of m-equations
- Δ p k x u = H ( x ) f k ( u 1 , . . . , u m )  in  Ω , 1 k m , B u k = 0  on  Ω , 1 k m , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-50/MediaObjects/13661_2011_Article_95_Equ3_HTML.gif
(1.6)
with
B u k = u k Dirichlet condition  1 . 7 u k ν Neumann condition  1 . 8 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-50/MediaObjects/13661_2011_Article_95_Equh_HTML.gif
Where {f k } are locally lipshitzian functions verify
f k ( s 1 , . . . , s k - 1 , 0 , u k + 1 , . . . , s m ) = 0 , ( 0 k m ) , F m : m : F m s k ( s 1 , . . . , s m ) = f k ( s 1 , . . . , s m ) . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-50/MediaObjects/13661_2011_Article_95_Equi_HTML.gif
H is previously defined and p k functions of C 1 ( Ω ¯ ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-50/MediaObjects/13661_2011_Article_95_IEq2_HTML.gif class, verify
p k ( x ) > 1 , ( x , p k ( x ) ) 0 , x Ω ̄ . a k = sup x Ω ̄ 1 - n p k ( x ) + ( x , p k ( x ) ) p k 2 ( x ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-50/MediaObjects/13661_2011_Article_95_Equ4_HTML.gif
(1.9)

2 Integral identities

Let
L p ( x ) ( Ω ) = u  measurable real function : Ω u ( x ) p ( x ) d x <  +  , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-50/MediaObjects/13661_2011_Article_95_Equj_HTML.gif
with the norm
u L p ( x ) ( Ω ) = u p ( x ) = inf λ > 0 : Ω u ( x ) λ p ( x ) d x 1 , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-50/MediaObjects/13661_2011_Article_95_Equk_HTML.gif
and
W 1 , p ( x ) ( Ω ) = { u L p x ( Ω ) : u L p x ( Ω ) } , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-50/MediaObjects/13661_2011_Article_95_Equl_HTML.gif
with the norm
u W 1 , p ( x ) ( Ω ) = u L p ( x ) ( Ω ) + u L p ( x ) ( Ω ) . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-50/MediaObjects/13661_2011_Article_95_Equm_HTML.gif

Denote W 0 1 , p ( x ) ( Ω ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-50/MediaObjects/13661_2011_Article_95_IEq3_HTML.gif the closure of C 0 ( Ω ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-50/MediaObjects/13661_2011_Article_95_IEq4_HTML.gif in W1, p(x)(Ω),

Lemma 1 Let u W 0 1 , p x Ω L Ω ̄ http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-50/MediaObjects/13661_2011_Article_95_IEq5_HTML.gifsolution of the equation (1.1) - (1.2), we have
Ω 1 - n p ( x ) + ( x , p ( x ) ) p 2 ( x ) 1 - ln u p ( x ) - a u p ( x ) d x + H ( x ) ( n F ( u ) - a u f ( u ) ) + ( x , H ( x ) ) F ( u ) d x = Ω 1 - 1 p x u p ( x ) ( x , ν ) d s http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-50/MediaObjects/13661_2011_Article_95_Equ5_HTML.gif
(2.1)
Lemma 2 Let u W 0 1 , p ( x ) ( Ω ) L ( Ω ̄ ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-50/MediaObjects/13661_2011_Article_95_IEq6_HTML.gifsolution of the equation (1.1) - (1.3), we have
Ω 1 - n p ( x ) + ( x , p ( x ) ) p 2 ( x ) 1 - ln u p ( x ) - a u p ( x ) d x + H ( x ) ( n F ( u ) - a u f ( u ) ) + ( x , H ( x ) ) F ( u ) d x = Ω 1 - 1 p x u p ( x ) + H ( x ) F ( u ) ( x , ν ) d s http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-50/MediaObjects/13661_2011_Article_95_Equ6_HTML.gif
(2.2)
Proof Multiplying the equation (1.1) by j = 1 n x i u x i http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-50/MediaObjects/13661_2011_Article_95_IEq7_HTML.gif and integrating the new equation by parts in Ω ∩ B R , B R = B (0, R)
- Ω B R d i v u p ( x ) - 2 u j = 1 n x j u x j d x = - i , j = 1 n Ω B R x i u p ( x ) - 2 u x i x j u x j d x = Ω B R u p ( x ) + u p ( x ) - 2 i , j = 1 n x j u x i 2 u x i x j d x - i , j = 1 n ( Ω B R ) u p ( x ) - 2 u x i u x j x j ν i d s http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-50/MediaObjects/13661_2011_Article_95_Equn_HTML.gif
Introducing the following result
u p ( x ) - 2 i = 1 n u x i 2 u x i x j = 1 p ( x ) x j u p ( x ) - p x j p 2 ( x ) u p ( x ) ln u p ( x ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-50/MediaObjects/13661_2011_Article_95_Equo_HTML.gif
we have
Ω B R u p ( x ) + j = 1 n x j p ( x ) x j u p ( x ) - j = 1 n ( x , p ( x ) ) p 2 ( x ) u p ( x ) ln u p ( x ) d x - ( Ω B R ) i , j = 1 n ( Ω B R ) u p ( x ) - 2 u x i u x j x j ν i d s = Ω B R 1 - n p ( x ) + ( x , p ( x ) ) p 2 ( x ) 1 - ln u p ( x ) u p ( x ) d x - ( Ω B R ) i , j = 1 n u p ( x ) - 2 u x i u x j x j ν i - j = 1 n 1 p ( x ) u p ( x ) x j ν j d s http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-50/MediaObjects/13661_2011_Article_95_Equp_HTML.gif
On the other hand
Ω B R H ( x ) f ( u ) j = 1 n x j u x j d x = j = 1 n Ω B R x j H ( x ) x j ( F ( u ) ) d x = - Ω B R ( n H ( x ) + ( x , H ( x ) ) ) F ( u ) d x + j = 1 n ( Ω B R ) H ( x ) F ( u ) x j ν j d s http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-50/MediaObjects/13661_2011_Article_95_Equq_HTML.gif
these results conduct to the following formula
Ω B R 1 - n p ( x ) + ( x , p ( x ) ) p 2 ( x ) 1 - ln u p ( x ) u p x d x + ( n H ( x ) + ( x , H ( x ) ) ) F ( u ) d x = ( Ω B R ) i , j = 1 n u p ( x ) - 2 u x i u x j x j ν i - j = 1 n 1 p ( x ) u p ( x ) - H ( x ) F ( u ) x j ν j d s http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-50/MediaObjects/13661_2011_Article_95_Equ7_HTML.gif
(2.3)
Multiplying the present equation (1.1) by au and integrating the obtained equation by parts in Ω, we obtain
( Ω B R ) a u p ( x ) - a u H ( x ) f ( u ) d x = ( Ω B R ) a u p ( x ) u ν u d s = 0 , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-50/MediaObjects/13661_2011_Article_95_Equ8_HTML.gif
(2.4)
Combining (2.3) and (2.4) we obtain
Ω B R 1 - n p ( x ) + ( x , p ( x ) ) p 2 ( x ) 1 - ln u p ( x ) - a u p ( x ) d x + H ( x ) ( n F ( u ) - a u f ( u ) ) + ( x , H ( x ) ) F ( u ) d x = ( Ω B R ) i , j = 1 n u p ( x ) - 2 u x i u x j x j ν i - j = 1 n 1 p ( x ) u p ( x ) - H ( x ) F ( u ) x j ν j d s = Ω B R i , j = 1 n u p ( x ) - 2 u x i u x j x j ν i - j = 1 n 1 p ( x ) u p ( x ) - H ( x ) F ( u ) x j ν j d s + Ω B R i , j = 1 n u p ( x ) - 2 u x i u x j x j ν i - j = 1 n 1 p ( x ) u p ( x ) - H ( x ) F ( u ) x j ν j d s http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-50/MediaObjects/13661_2011_Article_95_Equr_HTML.gif

On (Ω ∩ ∂B R ) we have n i = x i x http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-50/MediaObjects/13661_2011_Article_95_IEq8_HTML.gif

so the last integral is major by
M ( R ) = R Ω B R 1 + 1 p ( x ) u p ( x ) + H ( x ) F ( u ) d s http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-50/MediaObjects/13661_2011_Article_95_Equs_HTML.gif

We remark that if Ω in bounded, so for R is little greater, we get Ω ∩ ∂B R = ϕ, then M (R) = 0.

If Ω is not bounded, such as |∇u| ∈ W1, p(x)(Ω), F(u) ∈ L1 (Ω) and lim x + H ( x ) 0 , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-50/MediaObjects/13661_2011_Article_95_IEq9_HTML.gif we should see
0 + d r Ω B R 1 + 1 p ( x ) u p ( x ) + H ( x ) F ( u ) d s < + http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-50/MediaObjects/13661_2011_Article_95_Equt_HTML.gif
consequently we can always find a sequence (R n ) n , such as
lim n + R n + and lim n + M ( R n ) 0 . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-50/MediaObjects/13661_2011_Article_95_Equu_HTML.gif

In the problem (1.1) - (1.2), u| Ω= 0. Then, u = u ν n http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-50/MediaObjects/13661_2011_Article_95_IEq10_HTML.gif, we obtain the identity (2.1).

In the problem (1.1) - (1.3), u ν Ω = 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-50/MediaObjects/13661_2011_Article_95_IEq11_HTML.gif, we obtain the identity (2.2). ■

Lemma 3 Let u k W 0 1 , p k ( x ) ( Ω ) L ( Ω ̄ ) ( 1 k m ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-50/MediaObjects/13661_2011_Article_95_IEq12_HTML.gif, solution of the system (1.6) - (1.7). Then for the constants a k of ℝ, we have
Ω k = 1 m 1 - n p k ( x ) + ( x , p k ( x ) ) p k 2 ( x ) 1 - ln u k p k ( x ) - a k u k p k ( x ) + H ( x ) n F m ( u 1 , . . . , u m ) - k = 1 m a k u k f k ( u 1 , . . . , u m ) + + ( x , H ( x ) ) F m ( u 1 , . . . , u m ) d x = Ω k = 1 m 1 - 1 p k ( x ) u k p k ( x ) ( x , ν ) d s http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-50/MediaObjects/13661_2011_Article_95_Equ9_HTML.gif
(2.5)
Lemma 4 Let u k W 0 1 , p ( Ω ) L ( Ω ̄ ) ( 1 k m ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-50/MediaObjects/13661_2011_Article_95_IEq13_HTML.gif, solutions of the system (1.6) - (1.8). Then for the constants a k of ℝ, we have
Ω k = 1 m 1 - n p k ( x ) + ( x , p k ( x ) ) p k 2 ( x ) 1 - ln u k p k ( x ) - a k u k p k ( x ) + H ( x ) n F m ( u 1 , . . . , u m ) - k = 1 m a k u k f k ( u 1 , . . . , u m ) + + ( x , H ( x ) ) F m ( u 1 , . . . , u m ) ] d x = Ω k = 1 m 1 - 1 p k ( x ) u k p k ( x ) + H ( x ) F m ( u 1 , . . . , u m ) ( x , ν ) d s http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-50/MediaObjects/13661_2011_Article_95_Equ10_HTML.gif
(2.6)
Proof Multiplying the equation (1.6) by j = 1 n x i u k x i http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-50/MediaObjects/13661_2011_Article_95_IEq14_HTML.gif and integrating the new equation by part in Ω ∩ B R , B R = B (0, R), we get
Ω B R 1 - n p k ( x ) + ( x , p k ( x ) ) p k 2 ( x ) 1 - ln u k p k ( x ) u k p k ( x ) d x = ( Ω B R ) i , j = 1 n u k p k ( x ) - 2 u k x i u k x j x j ν i - j = 1 n 1 p k ( x ) u k p k ( x ) x j ν j d s http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-50/MediaObjects/13661_2011_Article_95_Equv_HTML.gif
On the other hand
Ω B R H ( x ) f k ( u 1 , . . . , u m ) j = 1 n x j u k x j d x = j = 1 n Ω B R x j H ( x ) u k x j u k ( F m ( u 1 , . . . , u m ) ) d x http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-50/MediaObjects/13661_2011_Article_95_Equw_HTML.gif
These results conduct to the following formula
Ω B R 1 - n p k ( x ) + ( x , p k ( x ) ) p k 2 ( x ) 1 - ln u k p k ( x ) u k p k ( x ) + j = 1 n x j H ( x ) u k x j u k ( F m ( u 1 , . . . , u m ) ) d x = ( Ω B R ) i , j = 1 n u k p k ( x ) - 2 u k x i u k x j x j ν i - j = 1 n 1 p k ( x ) u k p k ( x ) x j ν j d s http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-50/MediaObjects/13661_2011_Article_95_Equx_HTML.gif
Doing the sum on k of 1 to m, we obtain
Ω B R k = 1 m 1 - n p k ( x ) + ( x , p k ( x ) ) p k 2 ( x ) 1 - ln u k p k ( x ) u k p k ( x ) + j = 1 n x j H ( x ) x j F m ( u 1 , . . . , u m ) d x = ( Ω B R ) k = 1 m i , j = 1 n u k p k ( x ) - 2 u k x i u k x j x j ν i + k = 1 m j = 1 n 1 p k ( x ) u k p k ( x ) x j ν j d s http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-50/MediaObjects/13661_2011_Article_95_Equy_HTML.gif
which leads to the following identity
Ω B R k = 1 m 1 - n p k ( x ) + ( x , p k ( x ) ) p k 2 ( x ) 1 - ln u k p k ( x ) u k p k ( x ) - ( n H ( x ) + ( x , H ( x ) ) ) F m ( u 1 , . . . , u m ) d x = ( Ω B R ) k = 1 m i , j = 1 n u k p k ( x ) - 2 u k x i u k x j x j ν i + k = 1 m 1 p k x u k p k ( x ) + H ( x ) F m ( u 1 , . . . , u m ) ( x , ν ) d s http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-50/MediaObjects/13661_2011_Article_95_Equ11_HTML.gif
(2.7)
Now, multiply the equation (1.1) by au and integrating the obtained equation by parts in Ω ∩ B R
( Ω B R ) a k u p k ( x ) - a k u k H ( x ) f k ( u 1 , . . . , u m ) d x = 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-50/MediaObjects/13661_2011_Article_95_Equ12_HTML.gif
(2.8)

Combining (2.7) and (2.8), we get the identities (2.5) and (2.6).

The rest of the proof is similar to the that of lemma 1. ■

3 Principal Result

theorem 3.1 If u W 0 1 , p ( x ) ( Ω ) L ( Ω ̄ ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-50/MediaObjects/13661_2011_Article_95_IEq15_HTML.gifbe a solution of the problem (1.1) - (1.2), Ω is star shaped and that a, H, f and F verify the following assumptions
n F ( u ) - a u f ( u ) 0 , x Ω , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-50/MediaObjects/13661_2011_Article_95_Equ13_HTML.gif
(3.1)
( x , H ( x ) ) F ( u ) 0 , x Ω . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-50/MediaObjects/13661_2011_Article_95_Equ14_HTML.gif
(3.2)

Then, the problem admits only the null solution.

Proof Ω is star shaped, imply that
Ω 1 - 1 p ( x ) u p ( x ) ( x , ν ) d s 0 . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-50/MediaObjects/13661_2011_Article_95_Equ15_HTML.gif
(3.3)
On the other hand, the condition (3.1) give
Ω 1 - n p ( x ) + ( x , p ( x ) ) p 2 ( x ) 1 - ln u p ( x ) - a u p ( x ) d x + H ( x ) ( n F ( u ) - a u f ( u ) ) + ( x , H ( x ) ) F ( u ) d x 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-50/MediaObjects/13661_2011_Article_95_Equ16_HTML.gif
(3.4)
(1.4), (3.3) and (3.4), allow to get
F ( u ) = 0  in  Ω . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-50/MediaObjects/13661_2011_Article_95_Equz_HTML.gif
So, the problem (1.1) - (1.2) becomes
- d i v u p ( x ) - 2 u = 0  in  Ω , u = 0  on  Ω . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-50/MediaObjects/13661_2011_Article_95_Equ17_HTML.gif
(3.5)
Multiplying the equation (3.5) by u and integrating over Ω, we get
Ω u p ( x ) d x = 0 . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-50/MediaObjects/13661_2011_Article_95_Equaa_HTML.gif
So
u = 0 , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-50/MediaObjects/13661_2011_Article_95_Equab_HTML.gif

Hence u = cte = 0, because u| Ω= 0. ■

theorem 3.2 If u W 0 1 , p ( x ) ( Ω ) L ( Ω ̄ ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-50/MediaObjects/13661_2011_Article_95_IEq16_HTML.gifsolution of the problem (1.1) - (1.3), Ω is a star shaped and that a, H, F and F verify the following conditions
n F ( u ) - a u f ( u ) 0 , x Ω , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-50/MediaObjects/13661_2011_Article_95_Equ18_HTML.gif
(3.6)
( x , H ( x ) ) F ( u ) 0 , x Ω . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-50/MediaObjects/13661_2011_Article_95_Equ19_HTML.gif
(3.7)
H ( x ) F ( u ) 0 , x Ω . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-50/MediaObjects/13661_2011_Article_95_Equ20_HTML.gif
(3.8)

Therefore, the problem admits only the null solution.

Proof Similar to the proof of theorem 1. ■

theorem 3.3 If u k W 0 1 , p k ( x ) ( Ω ) L ( Ω ̄ ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-50/MediaObjects/13661_2011_Article_95_IEq17_HTML.gifsolution of the system (1.6) - (1.7), Ω is a star shaped and that a k , H, f k and F m verify the following conditions
n F m ( u 1 , . . . , u m ) - k = 1 m a k u k f k ( u 1 , . . . , u m ) 0 , x Ω , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-50/MediaObjects/13661_2011_Article_95_Equ21_HTML.gif
(3.9)
( x , H ( x ) ) F m ( u 1 , . . . , u m ) 0 , x Ω . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-50/MediaObjects/13661_2011_Article_95_Equ22_HTML.gif
(3.10)

So, the system admits only the null solutions.

Proof Ω is a star shaped, implies that
Ω k = 1 m 1 - 1 p k ( x ) u k p k ( x ) ( x , ν ) d s 0 . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-50/MediaObjects/13661_2011_Article_95_Equ23_HTML.gif
(3.11)
On the other hand, the conditions (3.9) and (3.10), give
Ω k = 1 m 1 - n p k ( x ) + ( x , p k ( x ) ) p k 2 ( x ) 1 - ln u k p k ( x ) - a k u k p k ( x ) + H ( x ) n F m ( u 1 , . . . , u m ) - k = 1 m a k u k f k ( u 1 , . . . , u m ) + + ( x , H ( x ) ) F m ( u 1 , . . . , u m ) d x 0 . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-50/MediaObjects/13661_2011_Article_95_Equ24_HTML.gif
(3.12)
(1.4), (3.11) and (3.12), allow to have
F m ( u 1 , . . . , u m ) = 0  in  Ω . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-50/MediaObjects/13661_2011_Article_95_Equac_HTML.gif
So the system (1.6) - (1.7) becomes
- d i v u k p k ( x ) - 2 u k = 0  in  Ω , 1 k m , u k = 0  on  Ω , 1 k m . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-50/MediaObjects/13661_2011_Article_95_Equ25_HTML.gif
(3.13)
Multiplying (3.13) by u k and integrating on Ω, we have
Ω u k p k ( x ) d x = 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-50/MediaObjects/13661_2011_Article_95_Equad_HTML.gif
So
u k = 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-50/MediaObjects/13661_2011_Article_95_Equae_HTML.gif

Therefore u k = cte = 0, ∀1 ≤ km, because u k | Ω= 0. ■

theorem 3.4 If u k W 0 1 , p k ( x ) ( Ω ) L ( Ω ̄ ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-50/MediaObjects/13661_2011_Article_95_IEq18_HTML.gifsolution of the system (1.6) - (1.8), Ω is a star shaped and that a k , H, f k and F m verify the following conditions
n F m ( u 1 , . . . , u m ) - k = 1 m a k u k f k ( u 1 , . . . , u m ) 0 , x Ω , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-50/MediaObjects/13661_2011_Article_95_Equ26_HTML.gif
(3.14)
( x , H ( x ) ) F m ( u 1 , . . . , u m ) 0 , x Ω , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-50/MediaObjects/13661_2011_Article_95_Equ27_HTML.gif
(3.15)
H ( x ) F m ( u 1 , . . . , u m ) 0 , x Ω . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-50/MediaObjects/13661_2011_Article_95_Equ28_HTML.gif
(3.16)

So, the problem admit only the null solution.

Proof Similar to the that of theorem 3. ■

4 Examples

Example 1 Considering in W 0 1 , p ( x ) ( Ω ) W 0 1 , q ( Ω ̄ ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-50/MediaObjects/13661_2011_Article_95_IEq19_HTML.gif the following problem
- d i v u p ( x ) - 2 u = c ( 1 + x ) μ u u q - 1 i n Ω , u = 0 o n Ω , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-50/MediaObjects/13661_2011_Article_95_Equ29_HTML.gif
(4.1)

where Ω is a bounded domain of n , c, μ > 0, q > 1 and p x = 1 + x 2 > 1 . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-50/MediaObjects/13661_2011_Article_95_IEq20_HTML.gif

By choosing
a = sup Ω 1 - n + ( n - 1 ) x 2 1 + x 2 1 + x 2 , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-50/MediaObjects/13661_2011_Article_95_Equaf_HTML.gif
we obtain
( x , H ( x ) ) F ( u ) = - c μ x q ( 1 + x ) μ + 1 u q + 1 < 0 , ( x , p ( x ) ) = x 2 1 + x 2 0 , n F ( u ) - a u f ( u ) = n q + 1 - a u q + 1 0 i f q n - a a . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-50/MediaObjects/13661_2011_Article_95_Equag_HTML.gif
So, the problem (4.1) doesn't admit non trivial solutions if
q n - a a . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-50/MediaObjects/13661_2011_Article_95_Equah_HTML.gif
Example 2 Considering in W 0 1 , p ( x ) ( Ω ) W 0 1 , γ ( Ω ̄ ) , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-50/MediaObjects/13661_2011_Article_95_IEq21_HTML.gif the following elliptic system
- Δ p ( x ) u = c γ ( 1 + x ) μ u u γ - 1 v δ i n Ω , - Δ q ( x ) v = c δ ( 1 + x ) μ v v δ - 1 u γ i n Ω , u = 0 o n Ω http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-50/MediaObjects/13661_2011_Article_95_Equ30_HTML.gif
(4.2)

where Ω is a bounded domain of n , c, μ, γ, δ > 0 and p, q > 1.

By choosing
a 1 = sup x Ω ̄ 1 - n p ( x ) + ( x , p ( x ) ) p 2 ( x ) a n d a 2 = sup x Ω ̄ 1 - n p ( x ) + ( x , q ( x ) ) q 2 ( x ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-50/MediaObjects/13661_2011_Article_95_Equai_HTML.gif
we obtain
( x , H ( x ) ) F ( u , v ) = - c μ 1 + x μ + 1 u γ v δ < 0 , n F ( u , v ) - a 1 u f 1 ( u , v ) - a 2 v f 2 ( u , v ) = ( n - γ a 1 - δ a 2 ) u γ v δ http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-50/MediaObjects/13661_2011_Article_95_Equaj_HTML.gif
So, the system (4.2) doesn't admit non trivial solutions if
γ a 1 + δ a 2 n http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2011-50/MediaObjects/13661_2011_Article_95_Equak_HTML.gif

Declarations

Authors’ Affiliations

(1)
Department of mathematics and informatics, Tebessa university

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Copyright

© Kamel; licensee Springer. 2011

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