Electronic Journal of Qualitative Theory of Differential Equations 2009, No. 27, 1-10; http://www.math.u-szeged.hu/ejqtde/

Absence of nontrivial solutions for a class of partial differential equations and systems in

unbounded domains

AKROUT Kamel^{∗} and KHODJA Brahim^{†}

Department of Mathematics, Larbi Tebissi University, Tebessa, Algeria Department of Mathematics, Badji Mokhtar University, Annaba, Algeria

Abstract In this paper, we are interested on the study of the nonexistence of nontrivial solutions for a class of partial differential equations, in unbounded domains. This leads us to extend these results to m-equations systems. The method used is based on energy type identities.

Keywords: Differential equations, trivial solution, energy type identities.

1 Introduction

The study of the nonexistence of nontrivial solutions of partial differential equations and systems is the subject of several works of many authors, by using various methods to obtain the necessary and sufficient conditions, so the studied problems admit only the null solutions. The works of Esteban & Lions [2], Pohozaev [6] and Van Der Vorst [7], contains results concerning the semilinear elliptic equations and systems. A semilar result can be found in [4], where studied equations one of the form

\{ \begin{matrix}λ\frac{∂^{2}u}{∂t^{2}} − \frac{∂}{∂x} (p (x, y) \frac{∂u}{∂x}) − \frac{∂}{∂y} (q (x, y) \frac{∂u}{∂y}) + f (x, y, u) = 0 in ℝ × ω, \\ u + ε \frac{∂u}{∂n} = 0 on ℝ×∂ω, \end{matrix}

considered in H^{2}(ℝ × ω) ∩ L^{∞}(ℝ × ω), where ω = ]a_{1}, b_{1}[ × ]a_{2}, b_{2}[ and this equation does not admit nontrivial solutions if the following conditions holds

f(0) = 0, 2F (u) − uf(u) ≤ 0.

∗akroutkamel@gmail.com

†bmkhodja@yahoo.fr

In this work similar results for a class of the partial differential equations and systems were also obtained.

Let us consider the following problem in H^{2}(ℝ×Ω)∩L^{∞}(ℝ×Ω), Ω a bounded domain of ℝ^{n}, for a function λ : ℝ → ℝ, not changing sign and p : ℝ × Ω → ℝ also had not changing sign.

\{ \begin{matrix}−\frac{∂}{∂t} (λ (t) \frac{∂u}{∂t}) − ∑_{i=1}^{n} \frac{∂}{∂x_{i}} (p (t, x) \frac{∂u}{∂x_{i}} ) + f (x, u) = 0 in ℝ×Ω, \\ u + ε\frac{∂u}{∂ν} = 0 on ℝ × ∂Ω. \end{matrix}  (1.1)

We use the notations

H = L^{2} (Ω) ,

∥u (t, x)∥ = (∫_{Ω} |u(t, x)|^{2} dx)^{\frac{1}{2}} , the norm of u in H,

∥∇u (t, x)∥^{2} = ∫_{Ω} ∑_{i=1}^{n} |\frac{∂u}{∂x_{i}} |^{2} dx,

F (x, u) = ∫_{0}^{u} f(x, σ)dσ, ∀ x ∈ Ω, u ∈ ℝ.

Let L be the operator defined by

Lu (t, x) = − ∑_{i=1}^{n} \frac{∂}{∂x_{i}} (p (t, x) \frac{∂u}{∂x_{i}} ) , (t, x) ∈ ℝ × Ω,

and f : \overline{Ω} × ℝ → ℝ a real continuous function, locally Lipschitz in u, such that

f(x, 0) = 0, ∀ x ∈ \overline{Ω}.

We assume that

u ∈ H^{2}(ℝ; H) ∩ L^{∞}(ℝ; L^{∞}(Ω)),

satisfies the equation

−\frac{∂}{∂t} (λ (t) \frac{∂u}{∂t} (t)) + Lu (t, x) + f(x, u) = 0 , (t, x) ∈ ℝ × Ω,  (1.2)

under the boundary conditions

(u + ε\frac{∂u}{∂n}) (t, σ) = 0, (t, σ) ∈ ℝ × ∂Ω, Robin condition, (1.3)
u(t, σ) = 0, (t, σ) ∈ ℝ × ∂Ω, Dirichlet condition,  (1.4)
\frac{∂u(t,σ)}{∂n} = 0, (t, σ) ∈ ℝ × ∂Ω, Neumann condition.  (1.5)

We extend the above result of (1.1) to the system of m-equations of the form

\{ \begin{matrix}−\frac{∂}{∂t} (λ (t) \frac{∂u_{k}}{∂t} ) − ∑_{i=1}^{n} \frac{∂}{∂x_{i}} (p_{k} (t, x) \frac{∂u_{k}}{∂x_{i}} ) + f_{k} (x, u_{1}, ..., u_{m}) = 0 in ℝ×Ω, \\ u_{k} + ε\frac{∂u_{k}}{∂ν} = 0 on ℝ × ∂Ω. \end{matrix}

(1.6)

1 ≤ k ≤ m, where f_{k} : \overline{Ω} × ℝ^{m}→ ℝ,are real continuous functions, locally Lipschitz in u_{i}, verifing

f_{k}(x, u_{1}, ..., 0, ..., u_{m}) = 0, ∀ x ∈ \overline{Ω},
∃F_{m} : \overline{Ω} × ℝ^{m} → ℝ such that \frac{∂F_{m}}{∂s_{j}} = f_{j} (x, s_{1}, ..., s_{m}) , 1 ≤ j ≤ m,

Let L_{k} be the operators defined by

L_{k}u (t, x) = − ∑_{i=1}^{n} \frac{∂}{∂x_{i}} (p_{k} (t, x) \frac{∂u}{∂x_{i}} ) , (t, x) ∈ ℝ × Ω,

we assume that

u_{k} ∈ H^{2}(ℝ; H) ∩ L^{∞}(ℝ; L^{∞}(Ω)),

are solutions of the system

−\frac{∂}{∂t} (λ (t) \frac{∂u_{k}}{∂t} (t)) + ∑_{k=1}^{m} L_{k}u_{k} (t, x) + f(x, u_{1}, ..., u_{m}) = 0 , (t, x) ∈ ℝ × Ω,

(1.7)

1 ≤ k ≤ m, with boundary conditions

(u_{k} + ε\frac{∂u_{k}}{∂n}) (t, σ) = 0, (t, σ) ∈ ℝ × ∂Ω, Robin condition, (1.8)
u_{k}(t, σ) = 0, (t, σ) ∈ ℝ × ∂Ω, Dirichlet condition,  (1.9)
\frac{∂u_{k}(t,σ)}{∂n} = 0, (t, σ) ∈ ℝ × ∂Ω, Neumann condition.  (1.10)

According to the sign of λ, this type of problems comprises equations of both hyperbolic or elliptic type.

Our proof is based on energy type identities established in section 2, which make it possible to obtain the main nonexistence result in section 3. In section 4 we apply the results to some examples.

2 Identities of energy type

In this section, we give essential lemmas for showing the main result of this paper.

Lemma 1 Let λ and p satisfy

λ′ (t) ≤ 0 (resp ≥ 0) , ∀t ∈ ℝ,

\frac{∂p}{∂t}(t, x) ≤ 0 (resp ≥ 0) , ∀ (t, x) ∈ ℝ × Ω. (2.1)

Then the following energ identity,

\begin{matrix}−\frac{1}{2}λ (t) ∥\frac{∂u}{∂t} (t, x)∥^{2} + \frac{1}{2} ∫_{Ω} p(t, x) |∇u|^{2} dx \\ + ∫_{Ω} F (x, u)dx + \frac{1}{2ε} ∫_{∂Ω} p(t, s)u^{2} (t, s) ds = 0. \end{matrix}  (2.2)

holds for any solution of the Robin problem of (1.2) − (1.3).

Proof. The assumptions f ∈ W_{loc}^{1,∞} (Ω × ℝ) , p ∈ L^{∞}(ℝ × Ω) and f(x, 0) = 0, ∀x ∈ Ω, allow us to deduce the existence of two positive constants C_{1}and C_{2}, such that

|p(t, x)| ≤ C_{1}, |F (x, u)| ≤ C_{2} |u (t, x)|^{2} .

In addition, consider the functions

Ψ (t) = \frac{1}{2} ∫_{Ω} p(t, x) |∇u|^{2} dx, Φ (t) = ∫_{Ω} F (x, u)dx, t ∈ ℝ,

where Φ and Ψ are of class C^{1},and

|Ψ (t)| ≤ C_{1} ∥∇u(t, x)∥^{2} , |Φ (t)| ≤ C_{2} ∥u(t, x)∥^{2} , ∀t ∈ ℝ.

Then,

Φ′ (t) = ∫_{Ω} f(x, u)\frac{∂u}{∂t}dx, ∀t ∈ ℝ,

and

\begin{matrix}Ψ′ (t) = ∫_{Ω} (∑_{i=1}^{n} p(t, x)\frac{∂u}{∂x_{i}} \frac{∂^{2}u}{∂x_{i}∂t} + \frac{1}{2} \frac{∂p}{∂t}(t, x) |∇u|^{2}) dx \\ = − ∫_{Ω} ∑_{i=1}^{n} \frac{∂}{∂x_{i}} (p(t, x)\frac{∂u}{∂x_{i}} ) \frac{∂u}{∂t}dx + \frac{1}{2} ∫_{Ω} \frac{∂p}{∂t}(t, x) |∇u|^{2} dx + ∫_{∂Ω} p(t, s)\frac{∂u}{∂ν} \frac{∂u}{∂t} (t, s) ds. \end{matrix}

Define the function K : ℝ → ℝ by

K (t) = −\frac{1}{2}λ (t) ∥\frac{∂u}{∂t} (t, x)∥^{2} + Ψ(t) + Φ(t).

The function K is absolutely continuous and differentiable in ℝ, and

K′(t) = −\frac{1}{2}λ′ (t) ∥\frac{∂u}{∂t} (t, x)∥^{2} − λ (t) ∫_{Ω} \frac{∂u}{∂t} \frac{∂^{2}u}{∂t^{2}}dx + Ψ′(t) + Φ′(t)

= \frac{1}{2}λ′ (t) ∥\frac{∂u}{∂t} (t)∥^{2} + \frac{1}{2} ∫_{Ω} \frac{∂p}{∂t}(t, x) |∇u|^{2} dx + ∫_{∂Ω} p(t, s)\frac{∂u}{∂ν} \frac{∂u}{∂t} (t, s) ds

+ ∫_{Ω} (−\frac{∂}{∂t} (λ (t) \frac{∂u}{∂t}) − ∑_{i=1}^{n} \frac{∂}{∂x_{i}} (p(t, x)\frac{∂u}{∂x_{i}} ) + f(x, u)) \frac{∂u}{∂t}dx.

Because u is solution of (1.2) − (1.3), we deduce that

K′(x) = \frac{1}{2}λ′ (t) ∥\frac{∂u}{∂t} (t, x)∥^{2} + \frac{1}{2} ∫_{Ω} \frac{∂p}{∂t} |∇u|^{2} dx + ∫_{∂Ω} p(t, x)\frac{∂u}{∂ν} \frac{∂u}{∂t} (t, s) ds,

while on the boundary,

∫_{∂Ω} p(t, x)\frac{∂u}{∂ν} \frac{∂u}{∂t} (t, s) ds = −\frac{1}{ε} ∫_{∂Ω} p(t, s)\frac{∂u}{∂t}u (t, s) ds.

Also

K′(t) = \frac{1}{2}λ′ (t) ∥\frac{∂u}{∂t} (t, x)∥^{2} + \frac{1}{2} ∫_{Ω} \frac{∂p}{∂t}(t, x) |∇u|^{2} dx − \frac{1}{ε} ∫_{∂Ω} p(t, s)\frac{∂u}{∂t}u (t, s) ds

= \frac{1}{2}λ′ (t) ∥\frac{∂u}{∂t} (t, x)∥^{2} + \frac{1}{2} ∫_{Ω} \frac{∂p}{∂t}(t, x) |∇u|^{2} dx

−\frac{1}{2ε} \frac{∂}{∂t} ( ∫_{∂Ω} p(t, s)u^{2} (t, s) ds) + \frac{1}{2ε} ∫_{∂Ω} \frac{∂p}{∂t}(t, s)u^{2} (t, s) ds,

i.e

\begin{matrix}\frac{d}{dt} (K(t) + \frac{1}{2ε} ∫_{∂Ω} p(t, s)u^{2} (t, s) ds) = \frac{1}{2}λ′ (t) ∥\frac{∂u}{∂t} (t, x)∥^{2} \\ +\frac{1}{2} ∫_{Ω} \frac{∂p}{∂t}(t, x) |∇u|^{2} dx + \frac{1}{2ε} ∫_{∂Ω} \frac{∂p}{∂t}(t, s)u^{2} (t, s) ds. \end{matrix}

We set

M (t) = K(t) + \frac{1}{2ε} ∫_{∂Ω} p(t, s)u^{2} (t, s) ds.

Conditions (2.1) imply that

M′ (t) ≤ 0 (resp ≥ 0) , ∀t ∈ ℝ,

i.e H is monotonous. But also this function verifies

lim_{|t|→+∞}M (t) = 0,

because M ∈ L^{2} (ℝ) . Hence M (t) = 0, ∀t ∈ ℝ, and this gives the desired result.

Lemma 2 Let λ and p verify (2.1). The solution of the Dirichlet problem (1.2) , (1.4) or the Neumann problem (1.2) , (1.5) , satisfies the following energ identity

−\frac{1}{2}λ (t) ∥\frac{∂u}{∂t} (t, x)∥^{2} + \frac{1}{2} ∫_{Ω} p(t, x) |∇u|^{2} dx + ∫_{Ω} F (x, u)dx = 0.  (2.3)

Proof. For the problem (1.2) , (1.4), the fact that u = 0 on the boundary implies that

\begin{matrix}∫_{∂Ω} p(t, x)\frac{∂u}{∂ν} \frac{∂u}{∂t} (t, s) ds = \frac{d}{dt} ( ∫_{∂Ω} p(t, x)\frac{∂u}{∂ν}u (t, s) ds) \\ − ∫_{∂Ω} \frac{∂p}{∂t} \frac{∂u}{∂ν}u (t, s) ds − ∫_{∂Ω} p(t, x)\frac{∂^{2}u}{∂t∂ν} u (t, s) ds = 0. \end{matrix}

For the problem (1.2) , (1.5), the fact that \frac{∂u}{∂ν} = 0 on the boundary implies that

∫_{∂Ω} p(t, x)\frac{∂u}{∂ν} \frac{∂u}{∂t} (t, s) ds = 0.

The remainder of the proof is similar to that of Lemma 1.

Lemma 3 Let λ and p_{k} satisfy

λ′ (t) ≤ 0 (resp ≥ 0) , ∀t ∈ ℝ,

\frac{∂p_{k}}{∂t} (t, x) ≤ 0 (resp ≥ 0) , 1 ≤ k ≤ m, ∀ (t, x) ∈ ℝ × Ω. (2.4)

Then any solutions of the system (1.7)−(1.8) satisfies for all t ∈ ℝ, the following energetic identity

\begin{matrix}−\frac{1}{2} ∑_{k=1}^{m} λ (t) ∥\frac{∂u_{k}}{∂t} (t, x)∥^{2} + \frac{1}{2} ∑_{k=1}^{m} ∫_{Ω} p_{k}(t, x) |∇u_{k}|^{2} dx \\ + ∫_{Ω} F_{m}(x, u_{1}, ..., u_{m})dx + \frac{1}{2ε} ∑_{k=1}^{m} ∫_{∂Ω} p_{k}(t, s)u_{k}^{2} (t, s) ds = 0. \end{matrix}  (2.5)

Lemma 4 Let λ and p_{k} verify (2.5). Then the solutions of the systems (1.7) − (1.9) or (1.7) − (1.10) , satisfies for all t ∈ ℝ, the following estimate

−\frac{1}{2} ∑_{k=1}^{m} λ (t) ∥\frac{∂u_{k}}{∂t} (t, x)∥^{2} + \frac{1}{2} ∑_{k=1}^{m} ∫_{Ω} p_{k}(t, x) |∇u_{k}|^{2} dx + ∫_{Ω} F_{m}(x, u_{1}, ..., u_{m})dx = 0.

(2.6)

Proof. Let us define the function K_{m} : ℝ → ℝ by

K_{m} (t) = −\frac{1}{2} ∑_{k=1}^{m} λ (t) ∥\frac{∂u_{k}}{∂t} (t, x)∥^{2} + Ψ_{m}(t) + Φ_{m}(t),

where the functions Ψ_{m} and Φ_{m} are defined as follows

\begin{matrix}Ψ_{m} (t) = \frac{1}{2} ∫_{Ω} ∑_{k=1}^{m} p_{k}(t, x) |∇u_{k}|^{2} dx, t ∈ ℝ, \\ Φ_{m} (t) = ∫_{Ω} F_{m}(x, u_{1}, ..., u_{m})dx, t ∈ ℝ, \end{matrix}

the rest of the proof is similar to the proofs of the preceding lemmas.

3 The main Result

Theorem 1 Let us suppose that λ, F and f verify

λ (t) > 0 (resp < 0) , ∀t ∈ ℝ,

2F (x, u) − uf (x, u) ≤ 0 (resp ≥ 0) , (3.1)

and (2.1) holds. Then the problem (1.2)−(1.3) admit only the null solution. Proof. Let us define the function E by

E (t) = ∥u (x, t)∥^{2} .

Multiplying equation (1.1) by u and integrating the new equation on Ω, we obtains

∫_{Ω} [−\frac{∂}{∂t} (λ (t) \frac{∂u}{∂t}) − ∑_{i=1}^{n} \frac{∂}{∂x_{i}} (p (t, x) \frac{∂u}{∂x_{i}} ) − f (x, u)] udx

= ∫_{Ω} [−\frac{1}{2} (λ′ (t) \frac{∂(u^{2})}{∂t} + λ (t) \frac{∂^{2}(u^{2})}{∂t^{2}}) + λ (t) (\frac{∂u}{∂t})^{2}

+p (t, x) |∇u|^{2} + uf (x, u)] dx − ∑_{i=1}^{n} ∫_{∂Ω} p (t, s) \frac{∂u}{∂x_{i}} u (t, s) ν_{i}ds

= −\frac{1}{2} (λ (t) E′′ (t) + λ′ (t) E′ (t)) + λ (t) ∥\frac{∂u}{∂t} (t, x)∥^{2} + ∫_{Ω} p (t, x) |∇u|^{2} dx

+ ∫_{Ω} uf (x, u) dx − ∫_{∂Ω} p (t, s) \frac{∂u}{∂ν}u (t, s) ds = 0.

Using identity (2.2), we have

\frac{d}{dt} (λ (t) E′ (t)) = λ (t) E′′ (t) + λ′ (t) E′ (t)

= 2λ (t) ∥\frac{∂u}{∂t} (t)∥^{2} + 2 ∫_{Ω} p (t, x) |∇u (t, x)|^{2} dx

+2 ∫_{Ω} u (t, x) f (x, u (t, x)) dx + \frac{2}{ε} ∫_{∂Ω} p (t, s) u^{2} (t, s) ds

= 4λ (t) ∥\frac{∂u}{∂t} (t)∥^{2} − 2 ∫_{Ω} (2F (x, u) − uf (x, u)) dx.

If λ (t) > 0, the assumption (3.1) implies that

\frac{d}{dt} (λ (t) E′ (t)) = λ (t) E′′ (t) + λ′ (t) E′ (t) ≥ 0 , ∀t ∈ ℝ.  (3.2)

We conclude that

E′ (t) ≤ 0,

otherwise,

∃t_{1} ≥ 0, E′ (t_{1}) ≥ 0.  (3.3)

Equation (3.2) implies that λ (t) E′ (t) is an increasing function

λ (t_{1}) E′ (t_{1}) ≤ λ (t) E′ (t) , ∀t ≥ t_{1},

but, one has

lim_{|t|→+∞}E′ (t) = 0,

because E′ ∈ L^{2} (ℝ), then

λ (t_{1}) E′ (t_{1}) ≤ 0 and λ (t_{1}) > 0 ⇒ E′ (t_{1}) ≤ 0,

which contradicts relation (3.3). Hence,

E′ (t) ≤ 0, ∀t ∈ ℝ

i.e E is monotonous. But, this function verifies

lim_{|t|→+∞}E (t) = 0,

witch implies that

E (t) = 0, ∀t ∈ ℝ,

Thus u = 0 in ℝ × ⩽̸.

If λ (t) < 0, we deduce by the same manner that u = 0 in ℝ × ⩽̸.

Theorem 2 Let λ, F and f verify (3.1) and (2.1) holds. Then the only solution of the problems (1.2) − (1.4) or (1.2) − (1.5) is the null solution.

Proof. Identical to that of Theorem 1.

Theorem 3 Let us suppose that λ, F_{m} and f_{k}, 1 ≤ k ≤ m, satisfy

λ (t) > 0 (resp < 0) , ∀t ∈ ℝ,

F_{m} (x, u_{1}, ..., u_{m}) − ∑_{k=1}^{m} u_{k}f_{k} (x, u_{1}, ..., u_{m}) ≤ 0 (resp ≥ 0) , (3.4)

and (2.5) holds. Then the system (1.7) − (1.8) admit only the null solutions. Proof. Multiplying equation (1.6) by u_{k} and integrating the new equation on Ω, one obtains

\begin{matrix}−\frac{1}{2} (λ (t) \frac{d^{2}}{dt^{2}} ∥u_{k} (t, x)∥^{2} + λ′ (t) \frac{d}{dt} ∥u_{k} (t, x)∥^{2}) + λ (t) ∥\frac{∂u_{k}}{∂t} (t, x)∥^{2} + ∫_{Ω} p_{k} (t, x) |∇u_{k}|^{2} dx \\ + ∫_{Ω} u_{k}f_{k} (x, u_{1}, ..., u_{m}) dx − ∫_{∂Ω} p_{k} (t, s) \frac{∂u_{k}}{∂ν} u_{k} (t, s) ds = 0. \end{matrix}

The sum on k from 1 to m gives

\begin{matrix}−\frac{1}{2} (λ (t) E′′_{m} (t) + λ′ (t) E′_{m} (t)) + λ (t) ∑_{k=1}^{m} ∥\frac{∂u_{k}}{∂t} (t, x)∥^{2} + ∑_{k=1}^{m} ∫_{Ω} p_{k} (t, x) |∇u_{k}|^{2} dx \\ + ∑_{k=1}^{m} ∫_{Ω} u_{k} (t, x) f_{k} (x, u_{1}, ..., u_{m}) dx − ∑_{k=1}^{m} ∫_{∂Ω} p_{k} (t, s) \frac{∂u_{k}}{∂ν} u_{k} (t, s) ds = 0. \end{matrix}

By using identity (2.6), we deduce that

\begin{matrix}\frac{d}{dt} (λ (t) E′_{m} (t)) = λ (t) E′′_{m} (t) + λ′ (t) E′_{m} (t) \\ = 4λ (t) ∑_{k=1}^{m} ∥\frac{∂u_{k}}{∂t} (t)∥^{2} − 2 ∫_{Ω} (2F_{m} (x, u_{1}, ..., u_{m}) − ∑_{k=1}^{m} u_{k} (t, x) f_{k} (x, u_{1}, ..., u_{m})) dx. \end{matrix}

Then the assumption (3.4) gives the result.

Theorem 4 Let λ, p and F verify

λ (t) > 0, p (t, x) < 0 and F (x, u) ≤ 0, ∀ (t, x) ∈ ℝ × Ω, or λ (t) < 0, p (t, x) > 0 and F (x, u) ≥ 0, ∀ (t, x) ∈ ℝ × Ω,

(3.5)

and (2.1) holds. Then the problems (1.2) − (1.3), (1.2) − (1.4) and (1.2) − (1.5) admit only the null solution.

Proof. Assumptions (3.4) and equality (2.3) allow is

−\frac{1}{2}λ (t) ∥\frac{∂u}{∂t} (t, x)∥^{2} + \frac{1}{2} ∫_{Ω} p(t, x) |∇u|^{2} dx + ∫_{Ω} F (x, u)dx = 0,

which implies that

\frac{∂u}{∂t} (t, x) = 0, ∀ (t, x) ∈ ℝ × Ω,

i.e.

u (t, x) = u (x) .

But the following condition is necessary

∫_{ℝ×Ω} |u (t, x)|^{2} dtdx = ∫_{ℝ×Ω} |u (x)|^{2} dtdx < +∞,

then u ≡ 0.

Theorem 5 Let λ, p_{k} (1 ≤ k ≤ m)and f satisfy

λ (t) > 0, p_{k} (t, x) < 0 and F_{m} (x, u_{1}, ..., u_{m}) ≤ 0, ∀ (t, x) ∈ ℝ × Ω, or λ (t) < 0, p_{k} (t, x) > 0 and F_{m} (x, u_{1}, ..., u_{m}) ≥ 0, ∀ (t, x) ∈ ℝ × Ω.

(3.6)

and (2.4) holds. Then the systems (1.7) − (1.8) , (1.7) − (1.9) and (1.7) − (1.10) admit only the null solutions.

Proof. Similar to that of Theorem 2.

Remark 1 Note that one can apply these results in the field ℝ^{+} × Ω, with the condition

u (0, x) = 0, ∀x ∈ Ω.

4 Applications

Example 1 Let

θ, θ_{1}, θ_{2} : \overline{Ω} → ℝ,

be a nonnegative functions of class C (ℝ) , p, q ≥ 1 and m ∈ ℝ, such that

f(x, u) = mu + θ_{1} (x) |u|^{p−1} u + θ_{2} (x) |u|^{q−1} u.

Then the problem defined by

\{ \begin{matrix}\frac{∂^{2}u}{∂t^{2}} − ∑_{i=1}^{n} \frac{∂}{∂x_{i}} (θ (x) \frac{∂u}{∂x_{i}} ) + f(x, u) = 0 in ℝ × Ω, \\ (u + ε\frac{∂u}{∂n}) (x, σ) = 0 on ℝ × ∂Ω, \end{matrix}  (4.1)

admits anly the null solution.

In this case it suffice to check that

\begin{matrix}2F (x, u) − uf(x, u) = \\ θ_{1} (x) (\frac{2}{p+1} − 1) |u|^{p+1} + θ_{2} (x) (\frac{2}{q+1} − 1) |u|^{q+1} ≤ 0, \end{matrix}

and apply Theorem 1. Example 2 Let Ω be a bounded open of set ℝ^{n}. Then, problem

\{ \begin{matrix}\frac{∂}{∂t} (e^{−t^{2}} \frac{∂u}{∂t}) − Δu = θ (x) |u|^{p−1} u in ℝ^{+} × Ω, \\ (u + ε\frac{∂u}{∂n}) (t, σ) = 0 on ℝ^{+} × ∂Ω, \\ u (0, x) = 0, ∀x ∈ Ω, \end{matrix}  (4.2)

where

p ≥ 1, θ : \overline{Ω} → ℝ, is nonnegative,

admits only the trivial solution, u ≡ 0.

Indeed,

\begin{matrix}λ (t) = −e^{−t^{2}} < 0, λ′ (t) = 2te^{−t^{2}} ≥ 0, ∀t ≥ 0, \\ 2F (x, u) − uf(x, u) = θ (x) (\frac{2}{p+1} − 1) |u|^{p+1} ≥ 0. \end{matrix}

Theorem 1 gives the result. Example 3 Let Ω be a bounded open of set ℝ^{n}, p, q ≥ 1,Then, the system

\{ \begin{matrix}−\frac{∂}{∂t} (λ (t) \frac{∂u}{∂t}) − Δu + (p + 1) θ (x) u |u|^{p−1} |v|^{q+1} = 0 in ℝ × Ω, \\ −\frac{∂}{∂t} (λ (t) \frac{∂v}{∂t} ) − Δv + (q + 1) θ (x) v |v|^{q−1} |u|^{p+1} = 0 in ℝ × Ω, \\ (u + ε\frac{∂u}{∂n}) (t, σ) = (v + ε\frac{∂v}{∂n}) (t, σ) = 0 on ℝ × ∂Ω, \end{matrix} (4.3)

where

θ : \overline{Ω} → ℝ, is nonnegative,
λ (t) > 0 of a class L^{∞} (ℝ) ,

admit only the trivial solutions, u ≡ v ≡ 0.

Indeed, there exist a function F defined as follows

F (x, u, v) = θ (x) |u|^{p+1} |v|^{q+1} ,

witch satisfies

\frac{∂F}{∂u} = f_{1}(x, u, v) = (p + 1) θ (x) u |u|^{p−1} |v|^{q+1} , \frac{∂F}{∂v} = f_{2}(x, u, v) = (q + 1) θ (x) v |v|^{q−1} |u|^{p+1} , F (x, u, v) − uf_{1}(x, u, v) − vf_{2} (x, u, v) = −θ (x) (p + q + 1) |u|^{p+1} |v|^{q+1} ≤ 0.

Theorem 3 gives the result.

References

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[4] B. Khodja, Nonexistence of solutions for semilinear equations and systems in cylindrical domains. Comm. Appl. Nonlinear Anal. (2000), 19-30.

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[8] C. Yarur, Nonexistence of positive singular solutions for a class of semilinear elliptic systems. Electronic Journal of Diff. Equations, 8 (1996), 1-22.

(Received January 6, 2009)