Directory UMM :Journals:Journal_of_mathematics:EJQTDE:

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

n
P

∂2u
∂
∂u
 −
as (y, ∂y
) + f (y, u) = 0 in Ω = R × D,
−
2

s

∂x
∂y
s
s=1

(P )



∂u
 u+ε
= 0 on R × ∂D.
∂n
where as : D × R → R are H 1 - functions with constant sign such that
(H1 )

2

Rξs

as (y, ts )dts − ξs as (y, ξs ) ≤ 0, s = 1, ..., n

0

and f : D × R → R is a real continuous locally Liptschitz function such
that
(H2 )

2F (y, u) − uf (y, u) ≤ 0,

We show that the function
E(x) =

Z

|u(x, y)|2 dy

D

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

is convex
n
Ru
Q
]αk , βk [ , ε > 0 and F (y, u) = f (y, τ )dτ .
D=
k=1

0

1. Introduction
The problem of existence and nonexistence of nontrivial solutions of problems of the form

 −∆u + f (u) = 0 in Ω,




 u = 0 on ∂Ω,

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

 −∆u + f (u) = 0, u ∈ C 2 (Ω),



 u = 0 on ∂Ω,

Under hypothesis


∇u ∈ L2 (Ω),



f (0) = 0,
Ru


 F (u) = f (s)ds ∈ L1 (Ω),
0

where Ω is a connected unbounded domain of RN such as
∃Λ ∈ RN , kΛk = 1, hn(x), Λi ≥ 0 on ∂Ω, hn(x), Λi =
6 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


 −∆u − u3 + u = 0,


u ∈ H 2 (R2 )

admits a radial solution
This same solution satisfies

 −∆u − u3 + u = 0,

 u ∈ H 2 (]0, +∞[ × R)

 ∂u

= 0 on {0} × R,
∂n
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 Rn and f a continuous function
on R satisfying:
(n − 2)F (u) − 2nuf (u) > 0,
where, n = dim Rn .
When
Ω = J × ω,
where J ⊂ R is unbounded interval and ω ⊂ Rn domain , Haraux & Khodja
[3] established under the assumption

f (0) = 0,
2F (u) − uf (u) ≤ 0,
EJQTDE, 2009 No. 44, p. 2

if we assume that u ∈ H 2 (J × ω) ∩ L∞ (J × ω) is a solution of the problems

 −∆u + f (u) = 0 in Ω,




 u or ∂u = 0 on ∂ (J × ω) .
∂n

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

 −∆u + u (u + 1) (u + 2) = 0 in Ω,



 ∂u

= 0 on ∂Ω,
∂n
is still open.

In this work, let ai , i = 1, ..., n be a sequence in H 1 (D × R) verifying
ai (y, 0) = 0 in D =

n
Y

]αk , βk [,

k=1

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


n
∂u
∂2u X ∂
) + f (y, u) = 0 in Ω = R × D.
ai (y,
(1.1)

− 2−
∂x
∂yi
∂yi
i=1

We assume that

u ∈ H 2 (Ω) ∩ L∞ (Ω),
and satisfies
(1.2)
(1.3)
(1.4)

u(x, s) = 0, (x, s) ∈ R × ∂D
or
∂u
(x, s) = 0, (x, s) ∈ R × ∂D
∂n
or



∂u
u+ε
(x, s) = 0, (x, s) ∈ R × ∂D
∂n

Let us denote by:

Γ = R × ∂D = Γα1 ∪ Γβ1 ∪ ... ∪ Γαn ∪ Γβn ,


Γµi = {(x, y1 , ...yi−1 , µi , yi+1 , ..., yn ) , x ∈ R, 1 ≤ i ≤ n}
EJQTDE, 2009 No. 44, p. 3

the boundary of Ω,
n (x, s) = (0, n1 (x, s) , ..., nn (x, s)) , the outward normal to ∂Ω at the point (x, s)
and

 2
∂ u (x, y)
the second derivative of u with respect to yi at point (x, y).
∂yi2
i=1,...,n
If z ∈ Ω, k = 1, 2, ...n and τ ∈ {α1 , β1 , α2 , β2 , ..., αn , βn } one writes:
z := (x, y) = (x, y1 , ..., yn )
zkτ := (y1 , ..., yk−1 , τ, yk+1 , ..., yn ) ,
dzk∗ := dy1 ...dyk−1 dyk+1 ...dyn ,
Rβ1

α1

...

βR
i−1 βR
i+1

αi−1 αi+1

...

β
Rn

f (x, y)dy1 ...dyi−1 dyi+1 ...dyn :=

αn

R

f (x, y)dzi∗ .

Di∗

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
ai ∈ H 1 (D×R) , i = 1, ..., n
satisfy
ai (., ξi ) : D → R, > 0 or < 0, ∀ξi , 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 6= 0 the integral identity
!
 
n
R
1  ∂u 2 P
∂u
−   +
) + F (y, u) (x, y)dy
Ai (y, ∂y
i
2 ∂x
i=1
D
(2.1)
n R




P
+ε
(Ai ziαi , ε−1 u(x, ziαi ) + Ai (ziβi , −ε−1 u(x, ziβi )))dzi∗ = 0
i=1 Di∗

Proof. Let

H:R→R
the function defined by
!
 
Z
n
∂u
1  ∂u 2 X
Ai (y,
) + F (y, u) (x, y)dy.
H (x) =
−   +
2 ∂x
∂yi
D

i=1

EJQTDE, 2009 No. 44, p. 4

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


n
R
∂2u
∂u
∂u ∂ 2 u P
d
∂u
H(x) =
+
+ f (y, u)
ai (y, ∂yi )
−
(x, y)dy
2
dx
∂yi ∂x   ∂x
 D 2 ∂xn∂x i=1

R
∂ u P ∂
∂u
∂u
)
+ f (y, u)
ai (y, ∂y
(x, y)dy
=
− 2−
i
∂x
∂y
∂x
i
i=1
D


n R
P
αi ∂u
ai ∂u
αi
βi ∂u
βi
∂u
(x,
z
))
(x,
z
))
(x,
z
)
−
a
(z
(x,
z
,
)
dzi∗ .
+
ai (ziβi , ∂y
i i
i
i
i
i
∂yi
i
∂x
∂x
∗
i=1 D
i

Indeed a simple use of Fubini’s theorem and an integration by parts yields
 2 
 2  
Z Z β i
Z
∂ u
∂ u
∂u
∂u
ai (y,
)
)
(x, y)dy =
dyi dzi∗
ai (y,
∂yi
∂yi ∂x
∂y
∂y
∂x
i
i
αi
D

Di∗

=

Z Z

Di∗

+

Z 

ai (ziβi ,

βi

∂
−
∂yi

αi

∂u
)
∂yi



Z

−

Di∗

=

∂u
∂x

∂
∂yi

D

+

Z

ai (ziβi ,

∂u(x, ziβi )
)
∂yi

Di∗









∂u
∂u
)
(x, y) dyi dzi∗
ai (y,
∂yi
∂x

(x, ziβi ) −



∂u
∂x

ai (y,



α
ai (zi i ,

∂u
)
∂yi



∂u
∂x





α
(x, zi i )

dzi∗

 
∂u
∂u
)
(x, y) dy
∂yi
∂x
α

(x, ziβi )

−

a
ai (zi i ,

∂u(x, zi i )
)
∂yi



∂u
∂x



!

α
(x, zi i )

dzi∗ .

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


Z
n
∂u
d
∂u
∂2u X ∂
H(x) =
− 2−
) + f (y, u)
ai (y,
(x, y) dy
dx
∂x
∂yi
∂yi
∂x
i=1

D

+

n Z
X

ai (ziβi ,

i=1 D ∗

!
αi
∂u(x, ziβi ) ∂u
ai ∂u(x, zi ) ∂u
αi
βi
) (x, zi ) − ai (zi ,
) (x, zi ) dzi∗ .
∂yi
∂x
∂yi
∂x

i

As u satisfies equation (1.1), the above expression reduces to
(2.3)
!
αi
n Z
βi
X
∂u(x,
z
)
∂u(x,
z
)
∂u
∂u
d
a
α
i
i
ai (ziβi ,
H(x) =
) (x, ziβi ) − ai (zi i ,
) (x, zi i ) dzi∗ .
dx
∂yi
∂x
∂yi
∂x
i=1 D ∗
i

EJQTDE, 2009 No. 44, p. 5

Now observe that

(2.4)



∂u
u+ε
∂n



(x, s) = 0 on ∂Ω, is equivalent to


∂u


) (x, y1 , .., yi−1 , αi , yi+1 , .., yn ) = 0
(u − ε


∂yi


∂u

 (u + ε
) (x, y1 , .., yi−1 , βi , yi+1 , .., yn ) = 0
∂yi

x ∈ R, αi < yi < βi .

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

n Z 
X


d
ai −1
αi
αi ∂u
βi
βi ∂u
βi
−1
ai (zi , −ε
u(x, zi )) (x, zi ) − ai (zi , ε u(x, zi )) (x, zi ) dzi∗ .
H(x) =
dx
∂x
∂x
i=1 D ∗
i

= −ε

n Z
X

i=1 D ∗
i



∂ 
Ai (ziβi , −ε−1 u(x, ziβi )) + Ai ziαi , ε−1 u(x, ziαi ) dzi∗ ,
∂x

i.e


n Z 

X




d 
H(x) + ε
Ai ziαi , ε−1 u(x, ziαi ) + Ai (ziβi , −ε−1 u(x, ziβi )) dzi∗  = 0.
dx
i=1 D

i

Integrating this expression,with respect to x one obtains
n Z 

X




Ai ziαi , ε−1 u(x, ziαi ) + Ai (ziβi , −ε−1 u(x, ziβi )) dzi∗ = const.
H(x)+ε
i=1 D ∗
i

Since
u (x, y) ∈ H 2 (R × D),
one must get


+∞
Z
Z
n


X






Ai ziαi , ε−1 u(x, ziαi ) + Ai (ziβi , −ε−1 u(x, ziβi ) )dzi∗  dx < ∞.
H(x) + ε
i=1 D ∗

−∞

i

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
!
 
Z
n
∂u
1  ∂u 2 X
Ai (y,
) + F (y, u) (x, y)dy = 0.
−   +
(2.5)
2 ∂x
∂yi
D

i=1

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
Z 





Ai ziαi , ε−1 u(x, ziαi ) + Ai (ziβi , −ε−1 u(x, ziβi )) dzi∗ = 0.
Di∗

If one supposes that u(x, s) = 0 for (x, s) ∈ R × ∂D, it is immediate, that
Ai (ziαi , 0) = Ai (ziβi , 0), ∀i = 1, ..., n.
Now if the boundary condition is

∂u
(x, s) = 0 for (x, s) ∈ R × ∂D, then
∂n

∂u
(x, s) = h∇u.ni (x, s) = 0 (x, s) ∈ R × ∂D,
∂n
i.e
∂u
(x, ziαi ) =
∂x
∂u
(x, z1α1 ) =
∂y1

∂u
(x, ziβi ) = 0
∂x
∂u
(x, z1β1 ) = 0
∂y1
.
.
.
∂u
∂u
(x, ziαi ) =
(x, ziβi ) = 0
∂yi
∂yi
.
.
.
∂u
∂u
(x, znαn ) =
(x, znβn ) = 0
∂yn
∂yn
consequently
a

ai (zi i ,














 , x ∈ R, αi ≤ yi ≤ βi ,













∂u
∂u
α
(x, zi i )) = ai (ziβi ,
(x, ziβi )) = 0, ∀i = 1, ..., n.
∂yi
∂yi

because
ai (x, 0) = 0, ∀x ∈ D, ∀i = 1, ..., n..
Finally one gets
Ai (ziαi , 0) = Ai (ziβi , 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 ai , i = 1, ..., n and f satisfying respectively
ai (., ξi ) : D → R, > 0 or < 0,
(3.1)

∀ξi , ∀i ∈ {1, ..., n}
2Ai (y, ξi ) − ai (y, ξi ) ξi ≤ 0,

(3.2)

2F (y, u) − uf (y, u) ≤ 0 ,

and assume
u ∈ H 2 (Ω) ∩ L∞ (Ω)
to be a solution of (1.1) − (1.4). Then the function
Z
x 7→ E(x) = |u(x, y)|2 dy is convex on R.
D

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

 ∂u 2
∂2u
1 ∂2
2
(u 2 ) (x, y) = (
u −   )(x, y).
∂x
2 ∂x2
∂x
u
In fact by multiplying equation (1.1) by and integrating the new equation
2
over D, we obtain
!


Z
n
X
∂2u
∂
u
∂u
− 2 −
0=
(x, y)dy
) + f (y, u)
ai (y,
∂x
∂yi
∂yi
2
D

i=1

(3.3)
!
 


Z
n

1  ∂u 2 1 X
1 ∂2
u
∂u
∂
2
ai (y,
=
−
u +   −
) u + f (y, u)(x, y) dy.
4 ∂x2
2 ∂x
2
∂yi
∂yi
2
i=1

D

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

∂
∂yi

D

=−

Z

D






Z Zβi
∂u
∂u
∂
) (u) (x, y) dy = (
) udyi )dzi∗ =
ai (y,
ai (y,
∂yi
∂yi
∂yi
Di∗ αi

∂u ∂u
)
(x, y) dy+
ai (y,
∂yi ∂yi

Z

(ai (ziβi ,

αi
∂u(x, ziβi )
βi
αi ∂u(x, zi )
)u(x, zi )−ai (zi ,
)u(x, ziαi ))dzi∗
∂yi
∂yi

Di∗

Instead of (3.3), we obtain
!
 
Z
n

1  ∂u 2 1 X
∂u
∂u
1
1 ∂2
2
ai (y,
u +   +
)
+ uf (y, u) (x, y)dy
−
4 ∂x2
2 ∂x
2
∂yi ∂yi 2
D

i=1

EJQTDE, 2009 No. 44, p. 8

n

1X
=
2

Z

(ai (ziβi ,

i=1 D ∗

∂u(x, ziβi )
∂u(x, ziαi )
)u(x, ziβi ) − ai (ziαi ,
)u(x, ziαi ))dzi∗
∂yi
∂yi

i

From (2.4) it follows that
n Z
X
∂u(x, ziβi )
∂u(x, ziαi )
)u(x, ziβi ) − ai (ziαi ,
)u(x, ziαi ))dzi∗
(ai (ziβi ,
∂yi
∂yi
i=1 D ∗
i

n Z
X

=

i=1 D ∗
i

n Z
X

=



(ai (ziβi , −ε−1 u(x, ziβi ))u(x, ziβi )−ai (ziαi , ε−1 u(x, ziαi ))u(x, ziαi ))dzi∗
(

i=1 D ∗
i

i.e
Z



1
1
ai (ziβi , −ε−1 u(x, ziβi )) −ε−1 u(x, ziβi )− −1 ai (ziαi , ε−1 u(x, ziαi ))ε−1 u(x, ziαi ))dzi∗
−1
−ε
ε

!
 2
n
X




∂u
∂u
1
1
∂u
1
1 ∂2
ai (y,
u2 +   +
)
+ uf (y, u) (x, y)dy
−
4 ∂x2
2 ∂x
2
∂yi ∂yi 2
i=1

D
n

εX
+
2

Z

i=1 D ∗
i





(ai (ziβi , −ε−1 u(x, ziβi )) −ε−1 u(x, ziβi )+ai (ziαi , ε−1 u(x, ziαi ))ε−1 u(x, ziαi ))dzi∗ = 0,

Combining this formula and (2.1) we obtain
!
 2
Z
n
X




1
∂u
1
∂u
∂u
1
1 ∂2
u2 +   +
ai (y,
)
+ uf (y, u) (x, y)dy
−
4 ∂x2
2 ∂x
2
∂yi ∂yi 2
i=1

D

n

+

εX
2

Z

i=1 D ∗
i





(ai (ziβi , −ε−1 u(x, ziβi )) −ε−1 u(x, ziβi )+ai (ziαi , ε−1 u(x, ziαi ))ε−1 u(x, ziαi ))dzi∗
=

Z

D

+ε

n Z
X

i=1 D ∗
i

i.e

!
 
n
∂u
1  ∂u 2 X
Ai (y,
) + F (y, u) (x, y)dy
−   +
2 ∂x
∂yi
i=1





(Ai ziαi , ε−1 u(x, ziαi ) + Ai (ziβi , −ε−1 u(x, ziβi )))dzi∗
Z

D

=

Z

D

n
X
i=1

  !

 ∂u 2
1 ∂2
2
−
u +   (x, y)dy
4 ∂x2
∂x

!
1
∂u ∂u
1
∂u
) − ai (y,
)
) + F (y, u) − uf (y, u) (x, y)dy
(Ai (y,
∂yi
2
∂yi ∂yi
2
EJQTDE, 2009 No. 44, p. 9

+ε

n Z 
X
i=1 D ∗
i

+ε

n Z 
X
i=1 D ∗
i



Ai ziαi , ε−1 u(x, ziαi )


1
αi −1
αi
αi
−1
− ai (zi , ε u(x, zi ))ε u(x, zi ) dzi∗
2








1
Ai (ziβi , −ε−1 u(x, ziβi )) − ai (ziβi , −ε−1 u(x, ziβi )) −ε−1 u(x, ziβi ) dzi∗ .
2

Hypotheses (3.1) and (3.2) imply that


Z
Z
d2 
2
|u (x, y)| dy  ≥ 4
dx2
D

D

This completes the proof.

2


 ∂u


 ∂x (x, y) dy ≥ 0, ∀x ∈ R.

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 ai , 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
!
 
Z
n
∂u ∂u
1
1 ∂ 2 2
1  ∂u 2 1 X
u +   +
ai (y,
)
+ uf (y, u) (x, y)dy
−
4 ∂x2
2 ∂x
2
∂yi ∂yi 2
i=1

D

=

1
2

n Z
X
i=1 D

(ai (ziβi ,

∂u(x, ziβi )
∂u(x, ziαi )
)u(x, ziβi ) − ai (ziαi ,
)u(x, ziαi ))dzi∗
∂yi
∂yi

i

∂u
(x, s) = 0, for (x, s) ∈ R × ∂D this formula reduces
Now if u (x, s) = 0 or ∂n
to
!
 2
Z
n
X




1
1 ∂2
∂u
∂u
∂u
1
1
−
ai (y,
u2 −   +
)
+ uf (y, u) (x, y)dy = 0
4 ∂x2
2 ∂x
2
∂yi ∂yi 2
i=1

D

We can now employ (2.5) to transform this identity into the following form
!
 2
Z
n
X




1 ∂2
∂u
∂u
1
∂u
1
1
−
ai (y,
u2 +   +
)
+ uf (y, u) (x, y)dy
4 ∂x2
2 ∂x
2
∂yi ∂yi 2
i=1
D
!


Z
n
∂u(x, y)
1  ∂u(x, y) 2 X
Ai (y,
+
) + F (y, u(x, y)) dy
− 
=
2
∂x 
∂yi
i=1

D

i.e

Z

D

  !

 ∂u 2
1 ∂2
2
u +   (x, y)dy
−
4 ∂x2
∂x

EJQTDE, 2009 No. 44, p. 10

=

Z

D

n 
X
i=1

1
∂u ∂u
∂u(x, y)
) − ai (y,
)
Ai (y,
∂yi
2
∂yi ∂yi



!
1
+ F (y, u) − uf (y, u) (x, y)dy.
2

Our assumptions on ai , 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
ai (y,

∂u(x, y)
∂u(x, y)
)=
∂yi
∂yi

then the equation (1.1) becomes
(4.1)

−∆u + f (y, u) = 0 in Ω
Application 4.2 : We can put
ai (y,

∂u(x, y)
∂u
(x, y)) = ci
∂yi
∂yi

with ci are reals constants. In this case (1.1) can be rewritten as
n

(4.2)

∂2u X ∂2u
− 2−
ci 2 + f (y, u) = 0 in Ω
∂x
∂yi
i=1

Application 4.3 : We can also put
ai (y,

(4.3)

∂u(x, y)
∂u(x, y)
) = pi (y)
∂yi
∂yi

with pi (y) < 0 or > 0 in D, it follows that the equation(1.1) is
equivalent to


n
∂2u X ∂
∂u
− 2−
pi (y)
+ f (y, u) = 0 in Ω
∂x
∂yi
∂yi
i=1

We observe that in this three applications, we have

2Ai (y, ξi ) − ai (y, ξi ) ξi ≡ 0, ∀ξi , 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


n ∂ 
 ∂2u P
∂u
 −
)
+ θ (y) |u|p−1 u = 0 in Ω = R × D
−
a
(y,
i
∂yi
 ∂x2 i=1 ∂yi

(5.1) 


∂u
 (u + ε )(x, s) = 0, (x, s) ∈ R × ∂D
∂n
where
θ : D → R,
is a nonnegative continuous real function, p ≥ 1 does not have nontrivial
solutions.
Indeed,
2F (y, u) − uf (y, u) = θ (y) (

2
− 1) |u|p+1 ≤ 0.
p+1

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

n ∂ 
 ∂2u P
∂u
 −
)
+ ρ (y) u = 0 in R × D
−
a
(y,
i
∂yi
 ∂x2
∂yi
i=1

(5.2)


 u + ε ∂u = 0 on R × ∂D
∂n

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
d2



1

4 dx2

Z

D



(|u(x, y)|2 dx =

Z

D

!
 2
 ∂u 
1
  + F (y, u) − uf (y, u) dx
 ∂x 
2

Z  2
 ∂u 
=   (x, y)dx ≥ 0
∂x
D

Example 3. Let

θ1 , θ2 : D → R,
be two continuous nonnegative functions, p, q ≥ 1 and
f (y, u) = mu + θ1 (y) |u|p−1 u + θ2 (y) |u|q−1 u), m ∈ R.
EJQTDE, 2009 No. 44, p. 12

The problem


n ∂ 
P
 ∂2u
∂u
 −
)
+ f (y, u) = 0 in R × D
−
a
(y,
i
∂yi

∂x2
i=1 ∂yi

(5.3)



∂u
 u+ε
= 0 on R × ∂D
∂n

does not have nontrivial solutions.
It suffices to remark that,

2F (y, u) − uf (y, u) =
θ1 (y) (

2
2
− 1) |u|p+1 + θ2 (y) (
− 1) |u|q+1 ≤ 0
p+1
q+1

and then apply theorem 3.1.
References
[1] H. BERESTYCKY, T. GALLOUET AND O. KAVIAN, Equations de
champs scalaires non-lin´eaires dans R2 . C. R. Acad. Sc. Paris, S´erie 1, 297, (1983),
307-310.
[2] M. J. ESTEBAN AND P. L. LIONS, Existence and nonexistence results for
semi linear elliptic problems in unbounded domains. Proc. Roy. Soc. Edimburgh 93A(1982), 1-14.
[3] A. HARAUX AND B. KHODJA, Caract`ere trivial de la solution de certaines
´equations aux d´eriv´ees partielles non lin´eaires dans des ouverts cylindriques de
RN . Portugaliae Mathematica. Vol. 42, Fasc. 2, (1982), 209-219.
[4] B. KHODJA, Nonexistence of solutions for semilinear equations and systems in
cylindrical domains. Comm. Appl. Nonlinear Anal. (2000), 19-30.
[5] B. KHODJA, A nonexistence result for a nonlinear PDE with Robin condition, International Journal of Mathemlatics and Mathematical Sciences, volume 2006, Article
ID 62601, 1-12
[6] B.KHODJA AND A. MOUSSAOUI, Nonexistence results for semilinear systems in unbounded domains,Electron. J. Diff. Eqns., Vol. 2009 (2009), No. 02, 1-11.
[7] E. MITIDIERI, Nonexistence of positive solutions of semilinear elliptic systems in
RN . Diff. and Int. Equations, 9 (1996), 465-479.
[8] M. H. PROTTER AND H. F. WEINBERGER, Maximum principle in differential equations, Prentice-Hall, Englewood Cliffs, New Jersey, (1967).
[9] PUCCI & J. SERRIN, A general variational identity, Ind. Univers. Mat.
Journal. Vol. 35, N3 (1986), 681-703.
[10] S. I. POHOZAEV, Eigenfunctions of the equation ∆u + 1f (u) = 0. Soviet.
Math. Dokl. 6 (1965), 1408-1411.
[11] D. DE FIGUEIREDO, Semilinear elliptic systems . Nonlinear Functional Analysis
and Application to differential Equations (1998), 122-152, ICTP Trieste ITALY, 21
April-9 May 1997. World Scientific.

EJQTDE, 2009 No. 44, p. 13

(Received June 1, 2009)

Badji Mokhtar University, department of mathematics P.O.12 Annaba, ALGERIA
E-mail address: samiabenmehidi@yahoo.fr
Badji Mokhtar University, department of mathematics P.O.12 Annaba, ALGERIA
E-mail address: bmkhodja@yahoo.fr

EJQTDE, 2009 No. 44, p. 14