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

Akrout Kamel

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Department of mathematics and informatics. Tebessa university. Algeria

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Boundary Value Problems 2011, 2011:50 doi:10.1186/1687-2770-2011-50


The electronic version of this article is the complete one and can be found online at: http://www.boundaryvalueproblems.com/content/2011/1/50


Received:12 January 2011
Accepted:30 November 2011
Published:30 November 2011

© 2011 Kamel; licensee Springer.

This is an Open Access article distributed under the terms of the Creative Commons Attribution License (http://creativecommons.org/licenses/by/2.0), which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.

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 ℝn. 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 [1-8]. 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 ,

when Ω is a star shaped bounded open domain in ℝn and f is a continuous function on ℝ satisfying

( n - 2 ) F ( u ) - 2 n u f ( u ) > 0 ,

A. Hareux and B. Khodja [2] established under the assumption

f ( 0 ) = 0 , 2 F ( u ) - u f ( u ) ≤ 0 .

that the problems

- Δ u + f ( u ) = 0  in  J × ω , u  or  ∂ u ∂ n = 0  on  ∂ ( J × ω ) .

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 ℝN such as

∃ Λ ∈ ℝ N , ∥ Λ ∥  = 1 , n ( x ) , Λ ≥ 0 on  ∂ ω , n ( x ) , Λ ≠ 0 ,

(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 (1.1)

with

B u = u Dirichlet condition 1 . 2 ∂ u ∂ ν Neumann condition 1 . 3

where

Δ p x u = d i v ∣ ∇ u ∣ p ( x ) - 2 ∇ u

Ω is bounded or unbounded domains of ℝn, f is a locally lipshitzian function, H and p are given continuous real functions of C ( Ω ̄ ) 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 ) . (1.4)

(., .) is the inner product in ℝn.

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 , (1.6)

with

B u k = u k Dirichlet condition  1 . 7 ∂ u k ∂ ν Neumann condition  1 . 8

Where {fk} are locally lipshitzian functions verify

f k ( s 1 , . . . , s k - 1 , 0 , u k + 1 , . . . , s m ) = 0 , ( 0 ≤ k ≤ m ) , ∃ F m : ℝ m → ℝ : ∂ F m ∂ s k ( s 1 , . . . , s m ) = f k ( s 1 , . . . , s m ) .

H is previously defined and pk functions of C 1 ( Ω ¯ ) 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 ) (1.9)

2 Integral identities

Let

L p ( x ) ( Ω ) = u  measurable real function : ∫ Ω ∣ u ( x ) ∣ p ( x ) d x <  +  ∞ ,

with the norm

∣ u ∣ L p ( x ) ( Ω ) = ∣ u ∣ p ( x ) = inf λ > 0 : ∫ Ω ∣ u ( x ) λ ∣ p ( x ) d x ≤ 1 ,

and

W 1 , p ( x ) ( Ω ) = { u   ∈ L p x ( Ω ) : ∣ ∇ u ∣ ∈ L p x ( Ω ) } ,

with the norm

∥ u ∥ W 1 , p ( x ) ( Ω ) = ∣ u ∣ L p ( x ) ( Ω ) + ∣ ∇ u ∣ L p ( x ) ( Ω ) .

Denote W 0 1 , p ( x ) ( Ω ) the closure of C 0 ∞ ( Ω ) in W1, p(x) (Ω),

Lemma 1 Let u ∈ W 0 1 , p x Ω ∩ L ∞ Ω ̄ solution 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 (2.1)

Lemma 2 Let u ∈ W 0 1 , p ( x ) ( Ω ) ∩ L ∞ ( Ω ̄ ) solution 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 (2.2)

Proof Multiplying the equation (1.1) by ∑ j = 1 n x i ∂ u ∂ x i and integrating the new equation by parts in Ω ∩ BR, BR = 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

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 )

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

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

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 (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 , (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

On (Ω ∩ ∂BR) we have n i = x i ∣ x ∣

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

We remark that if Ω in bounded, so for R is little greater, we get Ω ∩ ∂BR = 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 , we should see

∫ 0 + ∞ d r ∫ Ω ∩ ∂ B R 1 + 1 p ( x ) ∣ ∇ u ∣ p ( x ) + ∣ H ( x ) ∥ F ( u ) ∣ d s < + ∞

consequently we can always find a sequence (Rn)n, such as

lim n → + ∞ R n → + ∞ and lim n → + ∞ M ( R n ) → 0 .

In the problem (1.1) - (1.2), u|∂Ω = 0. Then, ∇ u = ∂ u ∂ ν n , we obtain the identity (2.1).

In the problem (1.1) - (1.3), ∂ u ∂ ν ∂ Ω = 0 , we obtain the identity (2.2). ■

Lemma 3 Let u k ∈ W 0 1 , p k ( x ) ( Ω ) ∩ L ∞ ( Ω ̄ ) ( 1 ≤ k ≤ m ) , solution of the system (1.6) - (1.7). Then for the constants ak 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 (2.5)

Lemma 4 Let u k ∈ W 0 1 , p ( Ω ) ∩ L ∞ ( Ω ̄ ) ( 1 ≤ k ≤ m ) , solutions of the system (1.6) - (1.8). Then for the constants ak 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 (2.6)

Proof Multiplying the equation (1.6) by ∑ j = 1 n x i ∂ u k ∂ x i and integrating the new equation by part in Ω ∩ BR, BR = 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

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

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

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

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 (2.7)

Now, multiply the equation (1.1) by au and integrating the obtained equation by parts in Ω ∩ BR

∫ ( Ω ∩ B R ) a k ∣ ∇ u ∣ p k ( x ) - a k u k H ( x ) f k ( u 1 , . . . , u m ) d x = 0 (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 ̄ ) be 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 , (3.1)

( x , ∇ H ( x ) ) F ( u ) ≤ 0 , ∀ x ∈ Ω . (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 . (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 (3.4)

(1.4), (3.3) and (3.4), allow to get

F ( u ) = 0  in  Ω .

So, the problem (1.1) - (1.2) becomes

- d i v ∣ ∇ u ∣ p ( x ) - 2 ∇ u = 0  in  Ω , u = 0  on  ∂ Ω . (3.5)

Multiplying the equation (3.5) by u and integrating over Ω, we get

∫ Ω ∣ ∇ u ∣ p ( x ) d x = 0 .

So

∣ ∇ u ∣ = 0 ,

Hence u = cte = 0, because u|∂Ω = 0. ■

theorem 3.2 If u ∈ W 0 1 , p ( x ) ( Ω ) ∩ L ∞ ( Ω ̄ ) solution 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 , (3.6)

( x , ∇ H ( x ) ) F ( u ) ≤ 0 , ∀ x ∈ Ω . (3.7)

H ( x ) F ( u ) ≥ 0 , ∀ x ∈ ∂ Ω . (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 ̄ ) solution of the system (1.6) - (1.7), Ω is a star shaped and that ak, H, fk and Fm 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 , (3.9)

( x , ∇ H ( x ) ) F m ( u 1 , . . . , u m ) ≤ 0 , ∀ x ∈ Ω . (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   . (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 . (3.12)

(1.4), (3.11) and (3.12), allow to have

F m ( u 1 , . . . , u m ) = 0  in  Ω .

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 . (3.13)

Multiplying (3.13) by uk and integrating on Ω, we have

∫ Ω ∣ ∇ u k ∣ p k ( x ) d x = 0

So

∣ ∇ u k ∣ = 0

Therefore uk = cte = 0, ∀1 ≤ k ≤ m, because uk|∂Ω = 0. ■

theorem 3.4 If u k ∈ W 0 1 , p k ( x ) ( Ω ) ∩ L ∞ ( Ω ̄ ) solution of the system (1.6) - (1.8), Ω is a star shaped and that ak, H, fk and Fm 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 , (3.14)

( x , ∇ H ( x ) ) F m ( u 1 , . . . , u m ) ≤ 0 , ∀ x ∈ Ω , (3.15)

H ( x ) F m ( u 1 , . . . , u m ) ≥ 0 , ∀ x ∈ ∂ Ω . (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 ̄ ) 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 , (4.1)

where Ω is a bounded domain of ℝn, c, μ > 0, q > 1 and p x = 1 + ∣ x ∣ 2 > 1 .

By choosing

a = sup Ω 1 - n + ( n - 1 ) ∣ x ∣ 2 1 + ∣ x ∣ 2 1 + ∣ x ∣ 2 ,

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 .

So, the problem (4.1) doesn't admit non trivial solutions if

q ≥ n - a a .

Example 2 Considering in W 0 1 , p ( x ) ( Ω ) ∩ W 0 1 , γ ( Ω ̄ ) , 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 (4.2)

where Ω is a bounded domain of ℝn, 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 )

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

So, the system (4.2) doesn't admit non trivial solutions if

γ a 1 + δ a 2 ≥ n

Competing interests

The author declares that they have no competing interests.

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