Applied Mathematics E-Notes, 13(2013), 51-59 c
Available free at mirror sites of http://www.math.nthu.edu.tw/ amen/
ISSN 1607-2510
Three Solutions For A Quasi-Linear Elliptic Problem
Mostafa Allaouiy, Abdelrachid El Amroussz, Anass Ourraouix
Received 10 July 2012
Abstract
This paper shows the existence of at least three solutions for Navier problem
involving the p(x)-biharmonic operator. Our technical approach is based on a
theorem obtained by B. Ricceri.
1
Introduction
Analysis of solutions of speci…c boundary value problems is of considerable importance
in the theory of partial di¤erential equations, especially for equations of fourth order.
Its interest is widely justi…ed with many physical examples and arises from a variety
of nonlinear phenomena. It is used in non-Newtonian ‡uids, in some reaction-di¤usion
problems, as well as in ‡ow through porous media. It also appears in nonlinear elasticity
petroleum extraction and in the theory of quasi-regular and quasi-conformal mappings.
For more detailed references on physical and mathematical background, we refer to
[1, 2, 3, 8].
The present work is concerned with the following p(x)-biharmonic problem with
Navier boundary condition,
(P)
where
2
p(x) u
u=
= f (x; u) + g(x; u) in
u = 0 on @ ;
;
is a bounded open domain in RN with smooth boundary @ ,
2
p(x) u
p(x) 2
=
(j j
u) is the p(x)-biharmonic with p 2 C( ); p(x) > 1 for every x 2
Rt
Rt
and ; 2 R+ . We de…ne F (x; t) = 0 f (x; s)ds, G(x; t) = 0 g(x; s)ds and we denote
by p := inf x2 p(x) and p+ := supx2 p(x):
Throughout this paper, we suppose the following assumptions.
There exist two positive constants C; and 2 C( ) with
:= inf
(x);
x2
such that
(F1 ) F (x; t)
0 for a.e. x 2
+
:= sup (x) and 1 <
+
<p ;
x2
and x 2 [0; ]:
Mathematics Subject Classi…cations: 35J60, 35D05, 35J20, 35J40.
Mohamed I, Faculty of Science, Department of Mathematics, Oujda, Morocco
z University Mohamed I, Faculty of Science, Department of Mathematics, Oujda, Morocco
x University Mohamed I, Faculty of Science, Department of Mathematics, Oujda, Morocco
y University
51
52
Three Solutions for a Quasi-Linear Elliptic Problem
(F2 ) There exists q1 (x) 2 C( ) with p+ < q1
lim sup
t!0
uniformly for a.e. x 2
F (x; t)
jtj
q1 (x)
q1 (x) < p2 (x) such that
< +1;
with
p2 (x) =
(
N p(x)
N 2p(x)
+1
if p(x) <
if p(x)
N
2;
N
2:
(x)
(F3 ) jF (x; t)j C(1 + jtj
) for x 2 and for all t 2 R:
(F4 ) F (x; 0) = 0 for a.e. x 2 :
G(x;t)
:
(G) sup(x;t)2 R 1+t
q2 (x) < +1, where q2 (x) 2 C( ) and q2 (x) < p2 (x) for x 2
The goal of this paper is to prove the following result.
THEOREM 1. Assume that (F1 ) to (F4 ) and (G) are satis…ed. Then there exist
an open interval
[0; +1[ and a positive real number e such that for every 2 ,
there exists > 0 such that for each 2 [0; ], problem (P) has at least three weak
1;p(x)
solutions whose norms in W 2;p(x) ( ) \ W0
( ) are less than e.
Many authors consider the existence of nontrivial solutions for some fourth order
problems such as [2, 3]. This is a generalization of the classical p-biharmonic operator
(j ujp 2 ) obtained in the case when p is a positive constant. Here we point out
that the p(x)-biharmonic operator possesses more complicated nonlinearities than pbiharmonic, for example, it is inhomogeneous and usually it does not have the so-called
…rst eigenvalue, since the in…mum of its principle eigenvalue is zero. This study is
inspired by the results of [6] and [7], we are to prove the existence of three solutions of
problem (P), and the technical approach is based on the three-critical-points theorem
of Ricceri [11, 12].
This paper is divided into three sections organized as follows: in section 2 we start
with some preliminary basic results on the theory of Lesbegue-Sobolev spaces with
variables exponent (we refer to the book of Musielak [10], Mih¼
ailescu and R¼
adulescu
[9]), we recall the three-critical-points theorem of Ricceri with some required results.
In section 3, we give the proof of the main result.
2
Preliminaries
In order to deal with the problem (P), we need some theory of variable exponent
Sobolev space. For convenience, we only recall some basic facts which will be used
later. Suppose that
RN is a bounded domain with smooth boundary @ : Let
C+ ( ) = fp 2 C( ) and ess inf x2 p(x) > 1g for any p(x) 2 C+ ( ): Set p =
minx2 p(x), p+ = maxx2 p(x) and
pk (x) =
N p(x)
if kp(x) < N and pk (x) = +1 if kp(x)
N kp(x)
N:
Allaoui et. al
53
De…ne the variable exponent Lebesgue space by
Lp(x) ( ) =
u:
! R measurable :
Z
p(x)
juj
dx < 1 :
Then Lp(x) ( ) endowed with the norm
kukp(x) = inf
Z
>0:
u
p(x)
dx
1 ;
becomes a separable and re‡exive Banach space.
PROPOSITION 1 (cf. [5]). Set (u) =
p
(1) kukp(x)
1 ) kukp(x)
(2) kukp(x)
1 ) kukp(x)
p+
R
juj
p(x)
dx: If u 2 Lp(x) ( ), we have
p+
(u)
kukp(x) :
(u)
kukp(x) :
p
De…ne the variable exponent Sobolev space W k;p(x) ( ) by
W k;p(x) ( ) = fu 2 Lp(x) ( ) : D u 2 Lp(x) ( ) and j j
where
D u=
with = ( 1 ; 2 ; :::;
the norm kuk = j
@j j
@ 1 x1 :::@
N
kg
xN
k;p(x)
a multi-index and j j = N
( ) with
i=1 i : The space W
kj kD ukp(x) is a separable and re‡exive Banach space.
N)
PROPOSITION 2 (cf. [5]). For p; r 2 C+ ( ) such that r(x)
there is a continuous and compact embedding
pk (x) for all x 2
,
W k;p(x) ( ) ,! Lr(x) ( ):
k;p(x)
We denote W0
( ) by the closure of C01 ( ) in W k;p(x) ( ):
1;p(x)
REMARK 1 (cf. [3]). (W 2;p(x) ( ) \ W0
( ); k : k) is a separable and re‡exive
Banach space. By the above remark and proposition 2.2 there is a continuous and
1;p(x)
compact embedding of W 2;p(x) ( ) \ W0
( ) into Lr(x) ( ) where r(x) < p2 for all
x2 :
PROPOSITION 3 (cf. [5]). Set %(u) =
have
p+
p
(1) kuk 1 ) kuk
%(u) kuk :
p
p+
(2) kuk 1 ) kuk
%(u) kuk :
(3) k un k! 0 , %(un ) ! 0:
R
j uj
p(x)
dx: For u; un 2 W 2;p(x) ( ), we
54
Three Solutions for a Quasi-Linear Elliptic Problem
(4) k un k! +1 ) %(un ) ! +1:
The proof is similar to proof in ([5], Theorem 3.1).
PROPOSITION 4 (cf. [5]). For any u 2 Lp(x) ( ) and v 2 Lq(x) ( ), we have
Z
1
1
+
kukp(x) kvkq(x)
uvdx
p
q
where
1
1
+
= 1:
p(x) q(x)
DEFINITION 1. We say that u 2 X is a weak solution of problem (P) if
Z
Z
Z
p(x) 2
j uj
u vdx =
f (x; u)vdx +
g(x; u)vdx
for all v 2 X:
We de…ne
I(u) =
Z
Z
1
p(x)
j uj
dx; J(u) =
p(x)
and
(u) =
where
F (x; t) =
Z
Z
Set
hL(u); vi =
G(x; u)dx
t
f (x; s)ds; G(x; t) =
0
Z
F (x; u)dx
Z
t
g(x; s)ds:
0
j uj
p(x) 2
u vdx for u; v 2 X:
PROPOSITION 5 (cf. [2, 3]).
(i) L : X ! X is a continuous, bounded and strictly monotone operator.
(ii) L is a mapping of type (S+ ), i.e. if un * u in X and lim supn!1 hL(un )
0; then un ! u in X:
L(u); un
(iii) L : X ! X is a homeomorphism.
PROPOSITION 6 (cf. Theorem 1 in [11]). Let X be a real re‡exive Banach space,
K
R an interval, I : X ! R be a sequentially weakly lower semi-continuous C 1
function whose derivative admits a continuous inverse on X and J : X ! R be a C 1
ui
Allaoui et. al
55
functional with compact derivative. In addition, I is bounded on each bounded subset
of X. Assume that
lim I(x) + J(x) = +1
(1)
kxk!1
for
2 K; and that there exists
2 R such that
sup inf (I(x) + (J(x) + )) < inf sup (I(x) + (J(x) + )):
2K x2X
x2X 2K
Then there exist a nonempty set A K and a positive number e with the following
property: for every 2 A and every C 1 functional : X ! R with compact derivative,
there exists > 0 such that for each 2 [0; ], the equation
0
I 0 (u) + J 0 (u) +
(u) = 0
has at least three solutions in X whose norms are less than e.
PROPOSITION 7 (cf.[12]). Let X be a nonempty set, and I and J are two real
functions on X. Suppose there are > 0 and u0 ; u1 2 X such that
I(u0 ) = J(u0 ) = 0; I(u1 ) >
and
sup
u2I
Then for each
1 (]
J(u1 )
:
I(u1 )
J(u) <
1; ])
satisfying
I(u0 ) = J(u0 ) = 0; I(u1 ) >
and
sup
u2I
1 (]
J(u) <
1; ])
<
J(u1 )
;
I(u1 )
we have
sup inf (I(u) + (
0 u2X
3
J(u))) < inf sup(I(u) + (
u2X
J(u))):
0
Proof of the Main Result
We now turn to the proof of Theorem 1. First, we check the conditions of proposition
6.
According to proposition 5, it is clear that I is continuously Gâteaux di¤erentiable,
whose Gâteaux derivative admits a continuous inverse on X : Notice that I is a convex
and continuous functional, and then it is a weakly lower semi-continuous function.
Moreover, and J are continuously Gâteaux di¤erentiable functions and its Gâteaux
derivatives are compact. By a similar analysis to that in Fan and Zhang (cf. [4]), by
(F3) and (G), we know that J; 2 C 1 (X; R) such that
Z
Z
0
0
J (u)v =
f (x; u(x))vdx and (u)v =
g(x; u(x))vdx
for u; v 2 X: Since the identity operator from X to L (x) is compact, so the operators
J 0 and 0 are compact. Obviously, I is bounded on each bounded subset of X.
56
Three Solutions for a Quasi-Linear Elliptic Problem
For kuk < 1;
1
p+
kuk
p+
1
p+
Let C0 > 0 such that C0
p
1
p+
1, we have I(u)
J(u)
=
p+
1
p
kuk
kuk
: Then
1
p
kuk
p+
I(u)
Since kuk
1
p
kuk :
p
I(u)
C0 :
p
kuk ; and thus for any u 2 X we infer that
Z
F (x; u)dx
Z
(x)
C(1 + juj
dx)
+
C(j j + kuk
C1 (1 + kuk
0 and C2
+ kuk
(x) )
(x) )
+
C2 (1 + kuk
with C1
(x)
+
);
0: Consequently, we obtain
I(u) + J(u)
Therefore, for u 2 X and
1
p
kuk
p+
0, since
lim
+
C2 (1 + kuk
+
)
C0 :
< p ; we get
(I(u) + J(u)) = +1:
kuk!+1
Then the assumption (1) is satis…ed. In order to prove the assumption (2), we need to
verify the conditions of proposition 7.
Let u0 = 0: Then I(u0 ) = J(u0 ) = 0: We show that the assumption (3) of
proposition 7 holds. Let x0 2
(because
is a nonempty bounded open set) and
r2 > r1 > 0: Take !(x) 2 C01 ( ) with !(x) = 0 for x 2
n B(x0 ; r2 ); !(x) =
0
0
0
k xi xi k2 ) when x 2 B(x ; r2 )nB(x ; r1 ) and !(x) = when x 2 B(x0 ; r1 )
r2 r1 (r2
2 21
with k x k2 = ( N
i=1 (xi ) ) :
Here u1 (x) = !(x). Then we can get
Z
J(u1 ) = J(!) =
F (x; !)dx > 0:
By (F2 ), there exist
2 [0; 1] and C1 > 0 such that
F (x; t)
C1 jtj
q1 (x)
for jtj <
and a.e. x 2
:
Putting
K1 = sup
jtj<
C[1 + jtj
jtj
q1
+
]
; K2 = sup
jtj>
C[1 + jtj
q1
jtj
+
]
; K3 = sup
jtj<1
C[1 + jtj
q1
jtj
+
]
;
Allaoui et. al
57
K4 = sup
C[1 + jtj
+
]
q1
and M = maxfC1 ; Ki ; i = 1; :::; 4g:
jtj
jtj>1
Thus
F (x; t) < M jtj
q1
for t 2 R and a.e. x 2 :
p+
< 1: By the Sobolev
Now, …x
such that 0 <
< 1: If we have p1+ kuk
embedding Theorem, for suitable positive constants C2 and C3 ; we entail that
J(u) =
Z
Z
F (x; u)dx < M
q1
q1
juj
C2 kuk
C3
q1
p+
:
It follows from q1 > p+ that
sup
lim
1
p+
kukp
f J(u)g
+
!0+
= 0:
(2)
Let ! 2 X as previously mentioned with the fact J(!) > 0: Fix 0 where
p+
p
1
; k!k ; 1g 1: Now, there are two cases to be considered.
p+ minfk!k
If k!k < 1 , we have
Z
Z
1
1
p(x)
p(x)
j !j
dx
j !j
dx
I(u1 ) = I(!) =
p(x)
p+
1
p+
k!k
0 > :
+
p
<
0
<
By (2), it yields
1
p+
Else if k!k
kkukp
J(u1
J(u)
sup
2
+
1
p
k!k
p
J(u1 )
<
2 I(u1 )
J(u1 )
:
I(u1 )
1 we obtain
I(u1 )
=
I(!) =
Z
1
p(x)
j !j
dx
p(x)
1
p
k!k
p+
From (2), since
Z
p(x)
j !j
dx
> :
> 0, we get
sup
1
p+
0
1
p+
kuk
J(u1 )
J(u)
2
p+
1
p
p+
k!k
J(u1 )
<
2 I(u1 )
J(u1 )
:
I(u1 )
Thereby,
sup
1
p+
kukp
+
J(u) <
J(u1 )
:
I(u1 )
(3)
58
Three Solutions for a Quasi-Linear Elliptic Problem
1
For any u 2 I
(]
1; ]), we have
Z
1
p(x)
j uj
dx
p(x)
Then
Z
Hence,
Z
1
p(x)
j uj
dx
p+
p(x)
j uj
dx
1
1
p+
<
:
:
1
0 1
p+
< 1:
The last inequality implies that kuk < 1 and
Z
1
1
p+
p(x)
kuk
<
j uj
dx
p+
p(x)
so we conclude
I
1
(]
1; ])
fu 2 X :
;
1
p+
kuk < g:
+
p
We deduce from the relation (3) that
sup
u2I
We can …nd
1 (]
J(u1 )
:
I(u1 )
J(u) <
1; ])
such that
sup
u2I
1 (]
1; ])
J(u) <
<
J(u1 )
:
I(u1 )
Taking K = [0; +1[, the assumptions of proposition 7 are satis…ed. Then, we may
easily obtain the condition (2) of proposition 6. Consequently, I; J and verify the
conditions of proposition 6. So the proof is complete.
Acknowledgment. The authors would like to thank the anonymous referee for
the valuable comments.
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