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Boundary Value Problems

, 2011:50 | Cite as

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

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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.

Keywords

Bounded Domain Elliptic Equation Dirichlet Problem Nontrivial Solution Laplacian Equation 
These keywords were added by machine and not by the authors. This process is experimental and the keywords may be updated as the learning algorithm improves.

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, 2, 3, 4, 5, 6, 7, 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 , Open image in new window
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 , Open image in new window
  1. A.
    Hareux and B. Khodja [2] established under the assumption
    f ( 0 ) = 0 , 2 F ( u ) - u f ( u ) 0 . Open image in new window
     
that the problems
- Δ u + f ( u ) = 0  in  J × ω , u  or  u n = 0  on  ( J × ω ) . Open image in new window
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 , Open image in new window

(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 Open image in new window
(1.1)
with
B u = u Dirichlet condition 1 . 2 u ν Neumann condition 1 . 3 Open image in new window
where
Δ p x u = d i v u p ( x ) - 2 u Open image in new window
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 ̄ ) Open image in new window 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 ) . Open image in new window
(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 , Open image in new window
(1.6)
with
B u k = u k Dirichlet condition  1 . 7 u k ν Neumann condition  1 . 8 Open image in new window
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 ) . Open image in new window
H is previously defined and p k functions of C 1 ( Ω ¯ ) Open image in new window 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 ) Open image in new window
(1.9)

2 Integral identities

Let
L p ( x ) ( Ω ) = u  measurable real function : Ω u ( x ) p ( x ) d x <  +  , Open image in new window
with the norm
u L p ( x ) ( Ω ) = u p ( x ) = inf λ > 0 : Ω u ( x ) λ p ( x ) d x 1 , Open image in new window
and
W 1 , p ( x ) ( Ω ) = { u L p x ( Ω ) : u L p x ( Ω ) } , Open image in new window
with the norm
u W 1 , p ( x ) ( Ω ) = u L p ( x ) ( Ω ) + u L p ( x ) ( Ω ) . Open image in new window

Denote W 0 1 , p ( x ) ( Ω ) Open image in new window the closure of C 0 ( Ω ) Open image in new window in W1, p(x)(Ω),

Lemma 1 Let u W 0 1 , p x Ω L Ω ̄ Open image in new windowsolution 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 Open image in new window
(2.1)
Lemma 2 Let u W 0 1 , p ( x ) ( Ω ) L ( Ω ̄ ) Open image in new windowsolution 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 Open image in new window
(2.2)
Proof Multiplying the equation (1.1) by j = 1 n x i u x i Open image in new window 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 Open image in new window
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 ) Open image in new window
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 Open image in new window
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 Open image in new window
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 Open image in new window
(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 , Open image in new window
(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 Open image in new window

On (Ω ∩ ∂B R ) we have n i = x i x Open image in new window

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 Open image in new window

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 , Open image in new window we should see
0 + d r Ω B R 1 + 1 p ( x ) u p ( x ) + H ( x ) F ( u ) d s < + Open image in new window
consequently we can always find a sequence (R n ) n , such as
lim n + R n + and lim n + M ( R n ) 0 . Open image in new window

In the problem (1.1) - (1.2), u| Ω= 0. Then, u = u ν n Open image in new window, we obtain the identity (2.1).

In the problem (1.1) - (1.3), u ν Ω = 0 Open image in new window, we obtain the identity (2.2). ■

Lemma 3 Let u k W 0 1 , p k ( x ) ( Ω ) L ( Ω ̄ ) ( 1 k m ) Open image in new window, 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 Open image in new window
(2.5)
Lemma 4 Let u k W 0 1 , p ( Ω ) L ( Ω ̄ ) ( 1 k m ) Open image in new window, 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 Open image in new window
(2.6)
Proof Multiplying the equation (1.6) by j = 1 n x i u k x i Open image in new window 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 Open image in new window
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 Open image in new window
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 Open image in new window
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 Open image in new window
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 Open image in new window
(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 Open image in new window
(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 ̄ ) Open image in new windowbe 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 , Open image in new window
(3.1)
( x , H ( x ) ) F ( u ) 0 , x Ω . Open image in new window
(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 . Open image in new window
(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 Open image in new window
(3.4)
(1.4), (3.3) and (3.4), allow to get
F ( u ) = 0  in  Ω . Open image in new window
So, the problem (1.1) - (1.2) becomes
- d i v u p ( x ) - 2 u = 0  in  Ω , u = 0  on  Ω . Open image in new window
(3.5)
Multiplying the equation (3.5) by u and integrating over Ω, we get
Ω u p ( x ) d x = 0 . Open image in new window
So
u = 0 , Open image in new window

Hence u = cte = 0, because u| Ω= 0. ■

theorem 3.2 If u W 0 1 , p ( x ) ( Ω ) L ( Ω ̄ ) Open image in new windowsolution 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 , Open image in new window
(3.6)
( x , H ( x ) ) F ( u ) 0 , x Ω . Open image in new window
(3.7)
H ( x ) F ( u ) 0 , x Ω . Open image in new window
(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 ̄ ) Open image in new windowsolution 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 , Open image in new window
(3.9)
( x , H ( x ) ) F m ( u 1 , . . . , u m ) 0 , x Ω . Open image in new window
(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 . Open image in new window
(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 . Open image in new window
(3.12)
(1.4), (3.11) and (3.12), allow to have
F m ( u 1 , . . . , u m ) = 0  in  Ω . Open image in new window
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 . Open image in new window
(3.13)
Multiplying (3.13) by u k and integrating on Ω, we have
Ω u k p k ( x ) d x = 0 Open image in new window
So
u k = 0 Open image in new window

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 ̄ ) Open image in new windowsolution 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 , Open image in new window
(3.14)
( x , H ( x ) ) F m ( u 1 , . . . , u m ) 0 , x Ω , Open image in new window
(3.15)
H ( x ) F m ( u 1 , . . . , u m ) 0 , x Ω . Open image in new window
(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 ̄ ) Open image in new window 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 , Open image in new window
(4.1)

where Ω is a bounded domain of n , c, μ > 0, q > 1 and p x = 1 + x 2 > 1 . Open image in new window

By choosing
a = sup Ω 1 - n + ( n - 1 ) x 2 1 + x 2 1 + x 2 , Open image in new window
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 . Open image in new window
So, the problem (4.1) doesn't admit non trivial solutions if
q n - a a . Open image in new window
Example 2 Considering in W 0 1 , p ( x ) ( Ω ) W 0 1 , γ ( Ω ̄ ) , Open image in new window 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 Open image in new window
(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 ) Open image in new window
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 Open image in new window
So, the system (4.2) doesn't admit non trivial solutions if
γ a 1 + δ a 2 n Open image in new window

Notes

References

  1. 1.
    Pohozaev SI: Eeigenfunctions of the equation Δu + λf (u) = 0. Soviet Math Dokl 1965, 1408-1411.Google Scholar
  2. 2.
    Haraux A, Khodja B: Caractère triviale de la solution de certaines équations aux dérivées partielles non linéaires dans des ouverts cylindriques de ℝN. Portugaliae Mathematica 1982, 42(Fasc 2):1-9.MathSciNetGoogle Scholar
  3. 3.
    Esteban MJ, Lions P: Existence and non-existence results for semi linear elliptic problems in unbounded domains. Proc Roy Soc Edimburgh 1982, 93-A: 1-14.CrossRefMathSciNetGoogle Scholar
  4. 4.
    Kawarno N, NI W, Syotsutani : Generalised Pohozaev identity and its applications. J Math Soc Japan 1990, 42(3):541-563. 10.2969/jmsj/04230541CrossRefMathSciNetGoogle Scholar
  5. 5.
    Khodja B: Nonexistence of solutions for semilinear equations and systems in cylindrical domains. Comm Appl Nonlinear Anal 2000, 19-30.Google Scholar
  6. 6.
    NI W, Serrin J: Nonexistence thms for quasilinear partial differential equations. Red Circ Mat Palermo, suppl Math 1985, 8: 171-185.MathSciNetGoogle Scholar
  7. 7.
    Van Der Vorst RCAM: Variational identities and applications to differential systems. Arch Rational; Mech Anal 1991, 116: 375-398.CrossRefMathSciNetGoogle Scholar
  8. 8.
    Yarur C: Nonexistence of positive singular solutions for a class of semilinear elliptic systems. Electronic Journal of Diff Equations 1996, 8: 1-22.CrossRefGoogle Scholar

Copyright information

© Kamel; licensee Springer. 2011

This article is published under license to BioMed Central Ltd. 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.

Authors and Affiliations

  1. 1.Department of mathematics and informaticsTebessa universityAlgeria

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