Electronic Journal of Qualitative Theory of Differential Equations 2009, No. 44, 1-14; http://www.math.u-szeged.hu/ejqtde/

SOME RESULTS OF NONTRIVIAL SOLUTIONS FOR A NONLINEAR PDE IN SOBOLEV SPACE

SAMIA BENMEHIDI AND BRAHIM KHODJA

Abstract. In this study, we investigate the question of nonexistence of nontrivial solutions of the Robin problem

(P)  | \begin{matrix}− \frac{∂^{2}u}{∂x^{2}} − ∑_{s=1}^{n} \frac{∂}{∂y_{s}} a_{s}(y, \frac{∂u}{∂y_{s}}) + f(y, u) = 0 in Ω = ℝ × D, \\ u + ε \frac{∂u}{∂n} = 0 on ℝ × ∂D. \end{matrix}

where a_{s} : D × ℝ → ℝ are H^{1}- functions with constant sign such that

(H_{1})  2 ∫_{0}^{ξ_{s}}a_{s}(y, t_{s})dt_{s} − ξ_{s}a_{s}(y, ξ_{s}) ≤ 0, s = 1, ..., n

and f : D × ℝ → ℝ is a real continuous locally Liptschitz function such that

(H_{2})  2F (y, u) − uf(y, u) ≤ 0,

We show that the function

E(x) = ∫_{D} |u(x, y)|^{2} dy

is convex on ℝ . Our proof is based on energy (integral) identities.

(D = ∏_{k=1}^{n}]α_{k}, β_{k}[ , ε > 0 and F (y, u) = ∫_{0}^{u} f(y, τ)dτ) .

1. Introduction

The problem of existence and nonexistence of nontrivial solutions of problems of the form

| \begin{matrix}−Δu + f(u) = 0 in Ω, \\ u = 0 on ∂Ω, \end{matrix}

has been investigated by many authors under various situations. Previous works have been reported by Berestycky, Gallouet & Kavian [1] , M. J. Esteban & P. L. Lions [2], Pucci & J. Serrin [9] and Pohozaev [10]. To illustrate some

2000 Mathematics Subject Classification. Primary 35J65; Secondary 35P70. Key words and phrases. Nonlinear equations, convex functions , Robin condition.

EJQTDE, 2009 No. 44, p. 1 of the typical known results, let us consider Dirichlet problem

| \begin{matrix}−Δu + f(u) = 0, u ∈ C^{2}(\overline{Ω}), \\ u = 0 on ∂Ω, \end{matrix}

Under hypothesis

\{ \begin{matrix}∇u ∈ L^{2}(Ω), \\ f(0) = 0, \\ F (u) = ∫_{0}^{u} f(s)ds ∈ L^{1}(Ω), \end{matrix}

where Ω is a connected unbounded domain of ℝ^{N} 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) Esteban & Lions [2] established that the Dirichlet problem does not have nontrivial solutions.

Berestycky, Gallouet & Kavian [1] established that the problem

| \begin{matrix}−Δu − u^{3} + u = 0, \\ u ∈ H^{2}(ℝ^{2}) \end{matrix}

admits a radial solution This same solution satisfies

| \begin{matrix}−Δu − u^{3} + u = 0, \\ u ∈ H^{2}(]0, +∞[ × ℝ) \\ \frac{∂u}{∂n} = 0 on \{0\} × ℝ, \end{matrix}

this shows that analogous Esteban-Lions result for Neumann problems is not valid.

The Pohozaev identity published in 1965 for solutions of the Dirichlet problem proved absence of nontrivial solutions for some elliptic equations when Ω is a star shaped bounded domain in ℝ^{n} and f a continuous function on ℝ satisfying:

(n − 2)F (u) − 2nuf(u) > 0,

where, n = dim ℝ^{n}.

When

Ω = J × ω,

where J ⊂ ℝ is unbounded interval and ω ⊂ ℝ^{n} domain , Haraux & Khodja [3] established under the assumption

\{ \begin{matrix}f(0) = 0, \\ 2F (u) − uf(u) ≤ 0, \end{matrix}

EJQTDE, 2009 No. 44, p. 2 if we assume that u ∈ H^{2}(J × ω) ∩ L^{∞}(J × ω) is a solution of the problems

||||||

−Δu + f(u) = 0 in Ω,
(u or \frac{∂u}{∂n}) = 0 on ∂ (J × ω) .

Then these two problems (Dirichlet and Neumann) do have only trivial solution.

When

f(u) = u (u + 1) (u + 2) ,

and

Ω = ℝ ×]0, a[ (a < π) ,

Neumann problem

| \begin{matrix}−Δu + u (u + 1) (u + 2) = 0 in Ω, \\ \frac{∂u}{∂n} = 0 on ∂Ω, \end{matrix}

is still open.

In this work, let a_{i}, i = 1, ..., n be a sequence in H^{1}(D × ℝ) verifying

a_{i}(y, 0) = 0 in D = ∏_{k=1}^{n}]α_{k}, β_{k}[,

and f : D×ℝ → ℝ a locally Lipschitz continuous function such that f(y, 0) = 0 in D, so that u = 0 is a solution of the equation

(1.1)  −\frac{∂^{2}u}{∂x^{2}} − ∑_{i=1}^{n} \frac{∂}{∂y_{i}} (a_{i}(y, \frac{∂u}{∂y_{i}})) + f(y, u) = 0 in Ω = ℝ × D.

We assume that

u ∈ H^{2}(Ω) ∩ L^{∞}(Ω),

and satisfies

(1.2)  u(x, s) = 0, (x, s) ∈ ℝ × ∂D

or

(1.3)  \frac{∂u}{∂n}(x, s) = 0, (x, s) ∈ ℝ × ∂D

or

(1.4)  (u + ε\frac{∂u}{∂n}) (x, s) = 0, (x, s) ∈ ℝ × ∂D

Let us denote by:

\begin{matrix}Γ = ℝ × ∂D = Γ_{α_{1}} ∪ Γ_{β_{1}} ∪ ... ∪ Γ_{α_{n}} ∪ Γ_{β_{n}} , \\ (Γ_{μ_{i}} = \{(x, y_{1}, ...y_{i−1}, μ_{i}, y_{i+1}, ..., y_{n}) , x ∈ ℝ, 1 ≤ i ≤ n\}) \end{matrix}

the boundary of Ω,

n (x, s) = (0, n_{1} (x, s) , ..., n_{n} (x, s)) , the outward normal to ∂Ω at the point (x, s)

and

(\frac{∂^{2}u (x, y)}{∂y_{i}^{2}})_{i=1,...,n} the second derivative of u with respect to y_{i} at point (x, y).

If z ∈ Ω, k = 1, 2, ...n and τ ∈ {α_{1}, β_{1}, α_{2}, β_{2}, ..., α_{n}, β_{n}} one writes:

z := (x, y) = (x, y_{1}, ..., y_{n})

z_{k}^{τ} := (y_{1}, ..., y_{k−1}, τ, y_{k+1}, ..., y_{n}) ,

dz_{k}^{∗} := dy_{1}...dy_{k−1}dy_{k+1}...dy_{n},

∫_{α_{1}}^{β_{1}} ... ∫_{αi_{−1}}^{β_{i−1}} ∫_{α_{i+1}}^{β_{i+1}} ... ∫_{α_{n}}^{β_{n}} f(x, y)dy_{1}...dy_{i−1}dy_{i+1}...dy_{n} := ∫_{D_{i}^{∗}} f(x, y)dz_{i}^{∗}.

The objective of this paper is to extend the results of [3], [5] to problems (1.1) − (1.2), (1.1) − (1.3) and (1.1) − (1.4).

2. Integral identities

We begin this section by giving an integral identity useful in the sequel.

Lemma 1. Let

a_{i} ∈ H^{1} (D×ℝ) , i = 1, ..., n

satisfy

a_{i}(., ξ_{i}) : \overline{D} → ℝ, > 0 or < 0, ∀ξ_{i}, i ∈ \{1, ..., n\} ,

and assume f : D × ℝ → ℝ a locally Lipschitz continuous function. Then any solution u ∈ H^{2}(ℝ×D)∩L^{∞}(ℝ×D) of (1.1) satisfying (1.4), verifies for each x ∈ ℝ and ε ≠ 0 the integral identity

(2.1) \begin{matrix}∫_{D} (−\frac{1}{2} |\frac{∂u}{∂x}|^{2} + ∑_{i=1}^{n} A_{i}(y, \frac{∂u}{∂y_{i}}) + F (y, u)) (x, y)dy \\ +ε ∑_{i=1}^{n} ∫_{D_{i}^{∗}} (A_{i} (z_{i}^{α_{i}}, ε^{−1}u(x, z_{i}^{α_{i}})) + A_{i}(z_{i}^{β_{i}}, (−ε^{−1}) u(x, z_{i}^{β_{i}})))dz_{i}^{∗} = 0 \end{matrix}

Proof. Let

H : ℝ → ℝ

the function defined by

H (x) = ∫_{D} (−\frac{1}{2} |\frac{∂u}{∂x}|^{2} + ∑_{i=1}^{n} A_{i}(y, \frac{∂u}{∂y_{i}}) + F (y, u)) (x, y)dy.

The hypotheses on u, a_{i}, i = 1, ..., n and f imply that H is absolutely continuous and thus differentiable almost everywhere on ℝ, we have (2.2)

\frac{d}{dx}H(x) = ∫_{D} (−\frac{∂u}{∂x} \frac{∂^{2}u}{∂x^{2}} + ∑_{i=1}^{n} a_{i}(y, \frac{∂u}{∂y_{i}})\frac{∂^{2}u}{∂y_{i}∂x} + f(y, u)\frac{∂u}{∂x}) (x, y)dy

= ∫_{D} (−\frac{∂^{2}u}{∂x^{2}} − ∑_{i=1}^{n} \frac{∂}{∂y_{i}} (a_{i}(y, \frac{∂u}{∂y_{i}})) + f(y, u)) (\frac{∂u}{∂x}) (x, y)dy

+ ∑_{i=1}^{n} ∫_{D_{i}^{∗}} (a_{i}(z_{i}^{β_{i}}, \frac{∂u}{∂y_{i}} (x, z_{i}^{β_{i}}))\frac{∂u}{∂x}(x, z_{i}^{β_{i}}) − a_{i}(z_{i}^{a_{i}} , \frac{∂u}{∂y_{i}} (x, z_{i}^{α_{i}}))\frac{∂u}{∂x}(x, z_{i}^{α_{i}})) dz_{i}^{∗}.

Indeed a simple use of Fubini’s theorem and an integration by parts yields

∫_{D} a_{i}(y, \frac{∂u}{∂y_{i}}) (\frac{∂^{2}u}{∂y_{i}∂x}) (x, y)dy = ∫_{D_{i}^{∗}} (∫_{α_{i}}^{β_{i}} a_{i}(y, \frac{∂u}{∂y_{i}}) (\frac{∂^{2}u}{∂y_{i}∂x}) dy_{i}) dz_{i}^{∗}

= ∫_{D_{i}^{∗}} (∫_{α_{i}}^{β_{i}} −\frac{∂}{∂y_{i}} (a_{i}(y, \frac{∂u}{∂y_{i}})) \frac{∂u}{∂x} (x, y) dy_{i}) dz_{i}^{∗}

+ ∫_{D_{i}^{∗}} (a_{i}(z_{i}^{β_{i}}, \frac{∂u}{∂y_{i}}) (\frac{∂u}{∂x}) (x, z_{i}^{β_{i}}) − a_{i}(z_{i}^{α_{i}} , \frac{∂u}{∂y_{i}}) (\frac{∂u}{∂x}) (x, z_{i}^{α_{i}})) dz_{i}^{∗}

= ∫_{D} −\frac{∂}{∂y_{i}} (a_{i}(y, \frac{∂u}{∂y_{i}})) (\frac{∂u}{∂x}) (x, y) dy

+ ∫_{D_{i}^{∗}} (a_{i}(z_{i}^{β_{i}}, \frac{∂u(x, z_{i}^{β_{i}})}{∂y_{i}}) (\frac{∂u}{∂x}) (x, z_{i}^{β_{i}}) − a_{i}(z_{i}^{a_{i}} , \frac{∂u(x, z_{i}^{α_{i}})}{∂y_{i}}) (\frac{∂u}{∂x}) (x, z_{i}^{α_{i}})) dz_{i}^{∗}.

By summing up these formulas with respect to i and substituting them in (2.2), one obtains

\frac{d}{dx}H(x) = ∫_{D} (−\frac{∂^{2}u}{∂x^{2}} − ∑_{i=1}^{n} \frac{∂}{∂y_{i}} (a_{i}(y, \frac{∂u}{∂y_{i}})) + f(y, u)) (\frac{∂u}{∂x}) (x, y) dy

+ ∑_{i=1}^{n} ∫_{D_{i}^{∗}} (a_{i}(z_{i}^{β_{i}}, \frac{∂u(x, z_{i}^{β_{i}})}{∂y_{i}})\frac{∂u}{∂x}(x, z_{i}^{β_{i}}) − a_{i}(z_{i}^{a_{i}} , \frac{∂u(x, z_{i}^{α_{i}})}{∂y_{i}})\frac{∂u}{∂x}(x, z_{i}^{α_{i}})) dz_{i}^{∗}.

As u satisfies equation (1.1), the above expression reduces to (2.3)

\frac{d}{dx}H(x) = ∑_{i=1}^{n} ∫_{D_{i}^{∗}} (a_{i}(z_{i}^{β_{i}}, \frac{∂u(x, z_{i}^{β_{i}})}{∂y_{i}})\frac{∂u}{∂x}(x, z_{i}^{β_{i}}) − a_{i}(z_{i}^{a_{i}} , \frac{∂u(x, z_{i}^{α_{i}})}{∂y_{i}})\frac{∂u}{∂x}(x, z_{i}^{α_{i}})) dz_{i}^{∗}.

Now observe that (u + ε\frac{∂u}{∂n}) (x, s) = 0 on ∂Ω, is equivalent to

(2.4) \{ \begin{matrix}(u − ε\frac{∂u}{∂y_{i}}) (x, y_{1}, .., y_{i−1}, α_{i}, y_{i+1}, .., y_{n}) = 0 \\ (u + ε\frac{∂u}{∂y_{i}}) (x, y_{1}, .., y_{i−1}, β_{i}, y_{i+1}, .., y_{n}) = 0 \end{matrix} x ∈ ℝ, α_{i} < y_{i} < β_{i}.

This allows to write formula (2.3) in the following form

\frac{d}{dx}H(x) = ∑_{i=1}^{n} ∫_{D_{i}^{∗}} (a_{i}(z_{i}^{β_{i}}, (−ε^{−1}) u(x, z_{i}^{β_{i}}))\frac{∂u}{∂x}(x, z_{i}^{β_{i}}) − a_{i}(z_{i}^{a_{i}} , ε^{−1}u(x, z_{i}^{α_{i}}))\frac{∂u}{∂x}(x, z_{i}^{α_{i}})) dz_{i}^{∗}.

= −ε ∑_{i=1}^{n} ∫_{D_{i}^{∗}} \frac{∂}{∂x} (A_{i}(z_{i}^{β_{i}}, −ε^{−1}u(x, z_{i}^{β_{i}})) + A_{i} (z_{i}^{α_{i}}, ε^{−1}u(x, z_{i}^{α_{i}}))) dz_{i}^{∗},

i.e

\frac{d}{dx} (H(x) + ε ∑_{i=1}^{n} ∫_{𝔻_{i}} (A_{i} (z_{i}^{α_{i}}, ε^{−1}u(x, z_{i}^{α_{i}})) + A_{i}(z_{i}^{β_{i}}, (−ε^{−1}) u(x, z_{i}^{β_{i}}))) dz_{i}^{∗}) = 0.

Integrating this expression,with respect to x one obtains

H(x)+ε ∑_{i=1}^{n} ∫_{D_{i}^{∗}} (A_{i} (z_{i}^{α_{i}}, ε^{−1}u(x, z_{i}^{α_{i}})) + A_{i}(z_{i}^{β_{i}}, (−ε^{−1}) u(x, z_{i}^{β_{i}}))) dz_{i}^{∗} = const.

Since

u (x, y) ∈ H^{2}(ℝ × D),

one must get

∫_{−∞}^{+∞}(H(x) + ε ∑_{i=1}^{n} ∫_{D_{i}^{∗}} (A_{i} (z_{i}^{α_{i}}, ε^{−1}u(x, z_{i}^{α_{i}})) + A_{i}(z_{i}^{β_{i}}, (−ε^{−1}) u(x, z_{i}^{β_{i}})))dz_{i}^{∗}) dx < ∞.

We conclude that the constant is null which is the desired result. Lemma 2. Let u be chosen as in Lemma1.1. If one assumes u to be solution of problems (1.1) − (1.3) or (1.1) − (1.4), then for each x ∈ ℝ, the solution u verifies

(2.5) ∫_{D} (−\frac{1}{2} |\frac{∂u}{∂x}|^{2} + ∑_{i=1}^{n} A_{i}(y, \frac{∂u}{∂y_{i}}) + F (y, u)) (x, y)dy = 0.

EJQTDE, 2009 No. 44, p. 6

Proof. To prove (2.5) it suffices to show that the second term of (2.1) vanishes if u verifies (1.2) or (1.3), i.e

∫_{D_{i}^{∗}} (A_{i} (z_{i}^{α_{i}}, ε^{−1}u(x, z_{i}^{α_{i}})) + A_{i}(z_{i}^{β_{i}}, (−ε^{−1}) u(x, z_{i}^{β_{i}}))) dz_{i}^{∗} = 0.

If one supposes that u(x, s) = 0 for (x, s) ∈ ℝ × ∂D, it is immediate, that

A_{i} (z_{i}^{α_{i}}, 0) = A_{i}(z_{i}^{β_{i}}, 0), ∀i = 1, ..., n.

Now if the boundary condition is \frac{∂u}{∂n}(x, s) = 0 for (x, s) ∈ ℝ × ∂D, then

\frac{∂u}{∂n}(x, s) = ⟨∇u.n⟩ (x, s) = 0 (x, s) ∈ ℝ × ∂D,

i.e

\begin{matrix}\frac{∂u}{∂x}(x, z_{i}^{α_{i}}) = \frac{∂u}{∂x}(x, z_{i}^{β_{i}}) = 0 \\ \frac{∂u}{∂y_{1}} (x, z_{1}^{α_{1}}) = \frac{∂u}{∂y_{1}} (x, z_{1}^{β_{1}}) = 0 \\ . \\ . \\ . \\ \frac{∂u}{∂y_{i}} (x, z_{i}^{α_{i}}) = \frac{∂u}{∂y_{i}} (x, z_{i}^{β_{i}}) = 0 \\ . \\ . \\ . \\ \frac{∂u}{∂y_{n}} (x, z_{n}^{α_{n}}) = \frac{∂u}{∂y_{n}} (x, z_{n}^{β_{n}}) = 0 \end{matrix} | , x ∈ ℝ, α_{i} ≤ y_{i} ≤ β_{i},

consequently

a_{i}(z_{i}^{a_{i}} , \frac{∂u}{∂y_{i}} (x, z_{i}^{α_{i}})) = a_{i}(z_{i}^{β_{i}}, \frac{∂u}{∂y_{i}} (x, z_{i}^{β_{i}})) = 0, ∀i = 1, ..., n.

because

a_{i}(x, 0) = 0, ∀x ∈ \overline{D}, ∀i = 1, ..., n..

Finally one gets

A_{i} (z_{i}^{α_{i}}, 0) = A_{i}(z_{i}^{β_{i}}, 0) = 0, ∀i = 1, ..., n.

3. Main results

The goal of this section is to establish the nonexistence of nontrivial solutions to Robin problem.

EJQTDE, 2009 No. 44, p. 7 Theorem 1. Let a_{i}, i = 1, ..., n and f satisfying respectively

\begin{matrix}a_{i}(., ξ_{i}) : \overline{D} → ℝ, > 0 or < 0, \\ 2A_{i} (y, ξ_{i}) − a_{i} (y, ξ_{i}) ξ_{i} ≤ 0, \end{matrix} ∀ξ_{i}, ∀i ∈ \{1, ..., n\} (3.1)

2F (y, u) − uf(y, u) ≤ 0 , (3.2) and assume

u ∈ H^{2}(Ω) ∩ L^{∞}(Ω)

to be a solution of (1.1) − (1.4). Then the function

x ↦ E(x) = ∫_{D} |u(x, y)|^{2} dy is convex on ℝ.

Proof. To begin the proof, we see that almost everywhere in Ω = ℝ × D, we have

(u\frac{∂^{2}u}{∂x^{2}}) (x, y) = (\frac{1}{2} \frac{∂^{2}}{∂x^{2}} (u^{2}) − |\frac{∂u}{∂x}|^{2})(x, y).

In fact by multiplying equation (1.1) by \frac{u}{2} and integrating the new equation over D, we obtain

0 = ∫_{D} (−\frac{∂^{2}u}{∂x^{2}} − ∑_{i=1}^{n} \frac{∂}{∂y_{i}} (a_{i}(y, \frac{∂u}{∂y_{i}})) + f(y, u)) \frac{u}{2}(x, y)dy

(3.3)

= ∫_{D} (−\frac{1}{4} \frac{∂^{2}}{∂x^{2}} (u^{2}) + \frac{1}{2} |\frac{∂u}{∂x}|^{2} − \frac{1}{2} ∑_{i=1}^{n} \frac{∂}{∂y_{i}} (a_{i}(y, \frac{∂u}{∂y_{i}})) u + \frac{u}{2}f(y, u)(x, y)) dy.

A simple use of Fubini’s theorem and an integration by parts yields,

∫_{D} \frac{∂}{∂y_{i}} (a_{i}(y, \frac{∂u}{∂y_{i}})) (u) (x, y) dy = ∫_{D_{i}^{∗}} (∫_{α_{i}}^{β_{i}} \frac{∂}{∂y_{i}} (a_{i}(y, \frac{∂u}{∂y_{i}})) udy_{i})dz_{i}^{∗} =

= − ∫_{D} a_{i}(y, \frac{∂u}{∂y_{i}})\frac{∂u}{∂y_{i}} (x, y) dy+∫_{D_{i}^{∗}} (a_{i}(z_{i}^{β_{i}}, \frac{∂u(x, z_{i}^{β_{i}})}{∂y_{i}})u(x, z_{i}^{β_{i}})−a_{i}(z_{i}^{α_{i}}, \frac{∂u(x, z_{i}^{α_{i}})}{∂y_{i}})u(x, z_{i}^{α_{i}}))dz_{i}^{∗}

Instead of (3.3), we obtain

∫_{D} (−\frac{1}{4} \frac{∂^{2}}{∂x^{2}} (u^{2}) + \frac{1}{2} |\frac{∂u}{∂x}|^{2} + \frac{1}{2} ∑_{i=1}^{n} a_{i}(y, \frac{∂u}{∂y_{i}})\frac{∂u}{∂y_{i}} + \frac{1}{2}uf(y, u)) (x, y)dy

EJQTDE, 2009 No. 44, p. 8

= \frac{1}{2} ∑_{i=1}^{n} ∫_{D_{i}^{∗}} (a_{i}(z_{i}^{β_{i}}, \frac{∂u(x, z_{i}^{β_{i}})}{∂y_{i}})u(x, z_{i}^{β_{i}}) − a_{i}(z_{i}^{α_{i}}, \frac{∂u(x, z_{i}^{α_{i}})}{∂y_{i}})u(x, z_{i}^{α_{i}}))dz_{i}^{∗}

From (2.4) it follows that

∑_{i=1}^{n} ∫_{D_{i}^{∗}} (a_{i}(z_{i}^{β_{i}}, \frac{∂u(x, z_{i}^{β_{i}})}{∂y_{i}})u(x, z_{i}^{β_{i}}) − a_{i}(z_{i}^{α_{i}}, \frac{∂u(x, z_{i}^{α_{i}})}{∂y_{i}})u(x, z_{i}^{α_{i}}))dz_{i}^{∗}

= ∑_{i=1}^{n} ∫_{D_{i}^{∗}} (a_{i}(z_{i}^{β_{i}}, (−ε^{−1}) u(x, z_{i}^{β_{i}}))u(x, z_{i}^{β_{i}})−a_{i}(z_{i}^{α_{i}}, ε^{−1}u(x, z_{i}^{α_{i}}))u(x, z_{i}^{α_{i}}))dz_{i}^{∗}

= ∑_{i=1}^{n} ∫_{D_{i}^{∗}} (\frac{1}{−ε^{−1}} a_{i}(z_{i}^{β_{i}}, −ε^{−1}u(x, z_{i}^{β_{i}})) (−ε^{−1}) u(x, z_{i}^{β_{i}})−\frac{1}{ε^{−1}} a_{i}(z_{i}^{α_{i}}, ε^{−1}u(x, z_{i}^{α_{i}}))ε^{−1}u(x, z_{i}^{α_{i}}))dz_{i}^{∗}

i.e

∫_{D} (−\frac{1}{4} \frac{∂^{2}}{∂x^{2}} (u^{2}) + \frac{1}{2} |\frac{∂u}{∂x}|^{2} + \frac{1}{2} ∑_{i=1}^{n} a_{i}(y, \frac{∂u}{∂y_{i}})\frac{∂u}{∂y_{i}} + \frac{1}{2}uf(y, u)) (x, y)dy

+\frac{ε}{2} ∑_{i=1}^{n} ∫_{D_{i}^{∗}} (a_{i}(z_{i}^{β_{i}}, (−ε^{−1}) u(x, z_{i}^{β_{i}})) (−ε^{−1}) u(x, z_{i}^{β_{i}})+a_{i}(z_{i}^{α_{i}}, ε^{−1}u(x, z_{i}^{α_{i}}))ε^{−1}u(x, z_{i}^{α_{i}}))dz_{i}^{∗} = 0,

Combining this formula and (2.1) we obtain

∫_{D} (−\frac{1}{4} \frac{∂^{2}}{∂x^{2}} (u^{2}) + \frac{1}{2} |\frac{∂u}{∂x}|^{2} + \frac{1}{2} ∑_{i=1}^{n} a_{i}(y, \frac{∂u}{∂y_{i}})\frac{∂u}{∂y_{i}} + \frac{1}{2}uf(y, u)) (x, y)dy

+\frac{ε}{2} ∑_{i=1}^{n} ∫_{D_{i}^{∗}} (a_{i}(z_{i}^{β_{i}}, (−ε^{−1}) u(x, z_{i}^{β_{i}})) (−ε^{−1}) u(x, z_{i}^{β_{i}})+a_{i}(z_{i}^{α_{i}}, ε^{−1}u(x, z_{i}^{α_{i}}))ε^{−1}u(x, z_{i}^{α_{i}}))dz_{i}^{∗}

= ∫_{D} (−\frac{1}{2} |\frac{∂u}{∂x}|^{2} + ∑_{i=1}^{n} A_{i}(y, \frac{∂u}{∂y_{i}}) + F (y, u)) (x, y)dy

+ε ∑_{i=1}^{n} ∫_{D_{i}^{∗}} (A_{i} (z_{i}^{α_{i}}, ε^{−1}u(x, z_{i}^{α_{i}})) + A_{i}(z_{i}^{β_{i}}, (−ε^{−1}) u(x, z_{i}^{β_{i}})))dz_{i}^{∗}

i.e

∫_{D} (−\frac{1}{4} \frac{∂^{2}}{∂x^{2}} (u^{2}) + |\frac{∂u}{∂x}|^{2}) (x, y)dy

= ∫_{D} (∑_{i=1}^{n}(A_{i}(y, \frac{∂u}{∂y_{i}}) − \frac{1}{2}a_{i}(y, \frac{∂u}{∂y_{i}})\frac{∂u}{∂y_{i}}) + F (y, u) − \frac{1}{2}uf(y, u)) (x, y)dy

EJQTDE, 2009 No. 44, p. 9

+ε ∑_{i=1}^{n} ∫_{D_{i}^{∗}} (A_{i} (z_{i}^{α_{i}}, ε^{−1}u(x, z_{i}^{α_{i}})) − \frac{1}{2}a_{i}(z_{i}^{α_{i}}, ε^{−1}u(x, z_{i}^{α_{i}}))ε^{−1}u(x, z_{i}^{α_{i}})) dz_{i}^{∗}

+ε ∑_{i=1}^{n} ∫_{D_{i}^{∗}} (A_{i}(z_{i}^{β_{i}}, (−ε^{−1}) u(x, z_{i}^{β_{i}})) − \frac{1}{2}a_{i}(z_{i}^{β_{i}}, (−ε^{−1}) u(x, z_{i}^{β_{i}})) (−ε^{−1}) u(x, z_{i}^{β_{i}})) dz_{i}^{∗}.

Hypotheses (3.1) and (3.2) imply that

\frac{d^{2}}{dx^{2}} (∫_{D} |u (x, y)|^{2} dy) ≥ 4 ∫_{D} |\frac{∂u}{∂x} (x, y)|^{2} dy ≥ 0, ∀x ∈ ℝ.

This completes the proof.

Remark 1. The convexity of the function E(x) on ℝ implies the triviality of the solution u(x, y) of the problem (1.1) − (1.4).

Theorem 2. Let the function a_{i}, i = 1, ..., n and f be as described as in Theorem 3.1. We assume u ∈ H^{2}(Ω) ∩ L^{∞}(Ω) is a solution of (1.1) − (1.2) or (1.1) − (1.3), then the function E(x) defined above is convex on ℝ.

Proof. By similar arguments as in the proof of Theorem 3.1, we obtain

∫_{D} (−\frac{1}{4} \frac{∂}{∂x^{2}}^{2} (u^{2}) + \frac{1}{2} |\frac{∂u}{∂x}|^{2} + \frac{1}{2} ∑_{i=1}^{n} a_{i}(y, \frac{∂u}{∂y_{i}})\frac{∂u}{∂y_{i}} + \frac{1}{2}uf(y, u)) (x, y)dy

= \frac{1}{2} ∑_{i=1}^{n} ∫_{D_{i}} (a_{i}(z_{i}^{β_{i}}, \frac{∂u(x, z_{i}^{β_{i}})}{∂y_{i}})u(x, z_{i}^{β_{i}}) − a_{i}(z_{i}^{α_{i}}, \frac{∂u(x, z_{i}^{α_{i}})}{∂y_{i}})u(x, z_{i}^{α_{i}}))dz_{i}^{∗}

Now if u (x, s) = 0 or \frac{∂u}{∂n} (x, s) = 0, for (x, s) ∈ ℝ × ∂D this formula reduces to

∫_{D} (−\frac{1}{4} \frac{∂^{2}}{∂x^{2}} (u^{2}) − \frac{1}{2} |\frac{∂u}{∂x}|^{2} + \frac{1}{2} ∑_{i=1}^{n} a_{i}(y, \frac{∂u}{∂y_{i}})\frac{∂u}{∂y_{i}} + \frac{1}{2}uf(y, u)) (x, y)dy = 0

We can now employ (2.5) to transform this identity into the following form

∫_{D} (−\frac{1}{4} \frac{∂^{2}}{∂x^{2}} (u^{2}) + \frac{1}{2} |\frac{∂u}{∂x}|^{2} + \frac{1}{2} ∑_{i=1}^{n} a_{i}(y, \frac{∂u}{∂y_{i}})\frac{∂u}{∂y_{i}} + \frac{1}{2}uf(y, u)) (x, y)dy

= ∫_{D} (−\frac{1}{2} |\frac{∂u(x, y)}{∂x}|^{2} + ∑_{i=1}^{n} A_{i}(y, \frac{∂u(x, y)}{∂y_{i}}) + F (y, u(x, y))) dy

i.e

∫_{D} (−\frac{1}{4} \frac{∂^{2}}{∂x^{2}} (u^{2}) + |\frac{∂u}{∂x}|^{2}) (x, y)dy

= ∫_{D} (∑_{i=1}^{n} (A_{i}(y, \frac{∂u(x, y)}{∂y_{i}}) − \frac{1}{2}a_{i}(y, \frac{∂u}{∂y_{i}})\frac{∂u}{∂y_{i}} ) + F (y, u) − \frac{1}{2}uf(y, u)) (x, y)dy.

Our assumptions on a_{i}, and f imply the desired result.

4. Applications

A practical tool for characterizing the assumption (3.1) or (3.2) of Theorem 3.1 is the following Proposition.

Proposition 1. Let

f : ℝ → ℝ

a Lipschitzian real function such that

f (0) = 0.

We suppose that f is concave on ] − ∞, 0[ and convex on ]0, +∞[. Then the function f satisfies the assumption (3.1) or (3.2) of Theorem 3.1.

Application 4.1 : Taking

a_{i}(y, \frac{∂u(x, y)}{∂y_{i}}) = \frac{∂u(x, y)}{∂y_{i}}

then the equation (1.1) becomes

(4.1)  −Δu + f(y, u) = 0 in Ω

Application 4.2 : We can put

a_{i}(y, \frac{∂u}{∂y_{i}} (x, y)) = c_{i} \frac{∂u(x, y)}{∂y_{i}}

with c_{i} are reals constants. In this case (1.1) can be rewritten as

(4.2)  −\frac{∂^{2}u}{∂x^{2}} − ∑_{i=1}^{n} c_{i} \frac{∂^{2}u}{∂y_{i}^{2}} + f(y, u) = 0 in Ω

Application 4.3 : We can also put

a_{i}(y, \frac{∂u(x, y)}{∂y_{i}}) = p_{i}(y)\frac{∂u(x, y)}{∂y_{i}}

with p_{i}(y) < 0 or > 0 in \overline{D}, it follows that the equation(1.1) is equivalent to

(4.3)  −\frac{∂^{2}u}{∂x^{2}} − ∑_{i=1}^{n} \frac{∂}{∂y_{i}} (p_{i}(y)\frac{∂u}{∂y_{i}} ) + f(y, u) = 0 in Ω

We observe that in this three applications, we have

2A_{i} (y, ξ_{i}) − a_{i} (y, ξ_{i}) ξ_{i} ≡ 0, ∀ξ_{i}, i = 1, ..., n.

EJQTDE, 2009 No. 44, p. 11

5. Examples

To conclude this work, let us give a few simple examples illustrating the use of Theorem 3.1. Example 1. The problem

(5.1) | \begin{matrix}−\frac{∂^{2}u}{∂x^{2}} − ∑_{i=1}^{n} \frac{∂}{∂y_{i}} (a_{i}(y, \frac{∂u}{∂y_{i}})) + θ (y) |u|^{p−1} u = 0 in Ω = ℝ × D \\ (u + ε\frac{∂u}{∂n})(x, s) = 0, (x, s) ∈ ℝ × ∂D \end{matrix}

where

θ : \overline{D} → ℝ,

is a nonnegative continuous real function, p ≥ 1 does not have nontrivial solutions.

Indeed,

2F (y, u) − uf(y, u) = θ (y) (\frac{2}{p + 1} − 1) |u|^{p+1} ≤ 0.

Theorem 3.1 give the desired result. Example 2. Let ρ : \overline{D} → ℝ, be a continuous function . The problem

(5.2)  | \begin{matrix}−\frac{∂^{2}u}{∂x^{2}} − ∑_{i=1}^{n} \frac{∂}{∂y_{i}} (a_{i}(y, \frac{∂u}{∂y_{i}})) + ρ (y) u = 0 in ℝ × D \\ u + ε\frac{∂u}{∂n} = 0 on ℝ × ∂D \end{matrix}

considered in H^{2}(ℝ × D) ∩ L^{∞}(ℝ × D) does not have nontrivial solutions.

A simple calculation gives

2F (y, u) − uf(y, u) ≡ 0.

and

\frac{1}{4} \frac{d^{2}}{dx^{2}} (∫_{D} (|u(x, y)|^{2} dx) = ∫_{D} (|\frac{∂u}{∂x}|^{2} + F (y, u) − \frac{1}{2}uf(y, u)) dx

= ∫_{D} |\frac{∂u}{∂x}|^{2} (x, y)dx ≥ 0

Example 3. Let

θ_{1}, θ_{2} : \overline{D} → ℝ,

be two continuous nonnegative functions, p, q ≥ 1 and

f(y, u) = mu + θ_{1} (y) |u|^{p−1} u + θ_{2} (y) |u|^{q−1} u), m ∈ ℝ. EJQTDE, 2009 No. 44, p. 12 The problem

(5.3)  | \begin{matrix}−\frac{∂^{2}u}{∂x^{2}} − ∑_{i=1}^{n} \frac{∂}{∂y_{i}} (a_{i}(y, \frac{∂u}{∂y_{i}})) + f(y, u) = 0 in ℝ × D \\ u + ε\frac{∂u}{∂n} = 0 on ℝ × ∂D \end{matrix}

does not have nontrivial solutions.

It suffices to remark that,

2F (y, u) − uf(y, u) =

θ_{1} (y) (\frac{2}{p + 1} − 1) |u|^{p+1} + θ_{2} (y) (\frac{2}{q + 1} − 1) |u|^{q+1} ≤ 0

and then apply theorem 3.1.

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(Received June 1, 2009)

Badji Mokhtar University, department of mathematics P.O.12 Annaba, AL-GERIA

E-mail address: samiabenmehidi@yahoo.fr

Badji Mokhtar University, department of mathematics P.O.12 Annaba, AL-GERIA

E-mail address: bmkhodja@yahoo.fr