Applied Mathematics E-Notes, 16(2016), 11-20 c
Available free at mirror sites of http://www.math.nthu.edu.tw/ amen/
ISSN 1607-2510
On The Spectrum Of (p(x); q(x))-Laplacian In RN
Abdelrachid El Amroussy, Anass Ourraouiz, Mostafa Allaouix
Received 27 October 2014
Abstract
In this work, we study the existence of a family of eigenvalues for a nonhomogeneous problem involving variable exponents in RN by using the variational
approach.
1
Introduction
Large number of papers was devoted to study elliptic equations and variational problems with variable exponent. They were of considerable importance in the theory of
partial di¤erential equations. Some of these problems come from di¤erent areas of
applied mathematics and physics such as Micro Electro-Mechanical systems, surface
di¤usion on solids or image processing and restoration. For more inquiries on modeling
physical phenomena involving p(x)-growth condition, we refer to [1–10].
Motivated by the works [11, 12], we consider the following problem
p(x) u
=
r(x) 2
q(x) u
+ m(x) juj
p(x) u
= div(jujp(x)
u in RN ;
(1)
where
2
u)
N
is the p(x)-Laplacian operator and p; q : R
! R are Lipschitz continuous functions
with
1 < p := inf p(x) p(x) sup p(x) := p+ < N; N 3;
RN
RN
where m is a positive weight for a.e x 2 R
N
such that
m 2 L1 (RN ) \ L
with
(x) =
(x)
(RN )
q (x)
:
q (x) r(x)
Suppose that
1 < q(x) < r = inf r(x)
RN
r+ = sup r(x) < p(x)
RN
Mathematics Subject Classi…cations: 35J60, 35D05, 35J20, 35J40.
Mohamed I, Faculty of sciences, Department of Mathematics, Oujda, Morocco
z National School of Applied Sciences of Al-Hoceima, Morocco
x University Mohamed I, Faculty of Sciences, Department of Mathematics, Oujda, Morocco
y University
11
(2)
12
The Spectrum of Laplacian with Variable Expoment
and
N q(x)
:
N q(x)
r+ < q (x) with q (x) =
(3)
De…ne
1
=
inf
u2W 1;p(x) (RN )\W 1;q(x) (RN );u6=0
R
RN
and
=
h
R
1
p(x)
RN
inf
u2W 1;p(x) (RN )\W 1;q(x) (RN );u6=0
We state our main result.
p(x)
1
jruj
+ q(x)
jruj
R
r(x)
m(x)
juj
dx
RN r(x)
q(x)
i
dx
;
h
i
p(x)
q(x)
jruj
+ jruj
dx
:
R
r(x)
m(x)
juj
dx
N
R
THEOREM 1. Assume that conditions (2) and (3) hold. Then we have:
(i)
1
> 0 and
(ii) There exists
2[
1 ; +1[
2]0;
1]
is an eigenvalue of the problem (1).
such that
2]0; [ is not an eigenvalue of problem (1).
DEFINITION 1. We say that 2 R is an eigenvalue of the problem (1) if there
exists u 2 W 1;p(x) (RN ) \ W 1;q(x) (RN ) n f0g such that
Z
Z
Z
p(x) 2
q(x) 2
r(x) 2
jruj
rurvdx +
jruj
rurvdx =
m(x) juj
uvdx;
RN
RN
RN
for all v 2 W 1;p(x) (RN ) \ W 1;q(x) (RN ) n f0g:
When p(x) = q(x) = 2 and m(x) = 1, the problem (1) is a normal Schrodinger
equation (see [13, 14]). The case m = 1 and p(x) = q(x) in a bounded domain has been
studied by Fan, Zhang and Zhao in [15, 16]. The case of an inde…nite weight m 6= 1 in
RN with p(x) = q(x) was considered by [11].
This article is organized as follows. In section 2, we give the necessary notations
and preliminaries. We include some useful results involving the variable exponents
Lebesgue and Sobolev spaces in order to facilitate the reading of the paper. Finally, in
section 3, we prove the main result.
2
Preliminary Notes
In order to deal with the problem (1), we need some theory of variable exponent Sobolev
Space. For convenience, we only recall some basic facts which will be used later.
De…ne the variable exponent Lebesgue space Lp(x) (RN ) by
Z
p(x)
Lp(x) (RN ) = u : RN ! R measurable :
juj
dx < 1 :
RN
El Amrouss et al.
13
Then Lp(x) ( ) endowed with the norm
jujp(x) = inf
>0:
Z
p(x)
u
dx
1
RN
becomes a separable and re‡exive Banach space.
De…ne the variable exponent Sobolev space W 1;p(x) (RN ) by
W 1;p(x) (RN ) = fu 2 Lp(x) (RN ) : ru 2 Lp(x) (RN )g
equipped with the norm
kukp(x) = inf
(
>0:
Z
ru
p(x)
dx
RN
)
1 ;
which is also a separable and re‡exive Banach space.
PROPOSITION 1 (cf. [17]). Let (u) =
Then the following statements hold.
(1) If kukp(x)
1, then kukp(x) p
(2) If kukp(x)
1, then kukp(x) p
+
R
RN
p(x)
jruj
dx and u 2 W 1;p(x) (RN ):
+
(u)
kukp(x) p :
(u)
kukp(x) p :
(3) limn!1 kun kp(x) = 0 (resp +1) if, and only if, limn!1 (un ) = 0 (resp +1).
R
REMARK 1. We have similar results (1) and (2) of Propositiion 1 for
p(x)
juj
dx.
RN
1 (u)
=
0
PROPOSITION 2 (cf. [17]). For any u 2 Lp(x) (RN ) and v 2 Lp (x) (RN ); we have
Z
2 jujp(x) jvjp0 (x)
uvdx
RN
with
1
1
+
= 1:
p(x) p0 (x)
PROPOSITION 3 (cf. [17]). Suppose that p is Lipschitz continuous, q : RN ! R is
a measurable function and p(x) q(x) p (x); 8x 2 RN : Then there is a continuous
embedding
W 1;p(x) (RN ) ,! Lq(x) (RN ):
If
is a bounded open subset of RN with cone property, then the embedding
W 1;p(x) ( ) ,!,! Lq(x) ( )
14
The Spectrum of Laplacian with Variable Expoment
is compact.
In what follows, for simplicity let
W = W 1;p(x) (RN ) \ W 1;q(x) (RN );
which will be endowed with the following norm
kuk = kukp(x) + kvkq(x) ;
R
which makes W a Banach space separable and re‡exive and denote by
assertions of Proposition 1 remain valid with the norm k k :
=
R
RN
: The
PROPOSITION 4 (cf. [5]). Suppose that p; q are are Lipschitz continuous functions
and r is a measurable function. If q(x) r(x) p(x); then Lp(x) (RN ) \ Lq(x) (RN ) ,!
Lr(x) (RN ) with a continuous embedding.
Analogously, we have the following interesting result (see [5]).
PROPOSITION 5. Under the assumptions of Proposition 4, we have a continuous
embedding from W into Lr(x) (RN ):
3
Proof of the Main Result
First, by the assumption (2) we can see that
p(x)
jruj
r+
q(x)
+ jruj
and
jruj
r+
juj
, jruj
r
+ juj
i
p(x)
q(x)
+ jruj
r(x)
juj
r
jruj
:
i
The continuous embedding from W 1;r (RN ) into Lr (RN ); i = ; implies that there
exist two positive constants C1 ; C2 such that
Z
Z
Z
Z
r+
r+
r
r
jruj dx C1 juj dx and
jruj dx C1 juj dx:
Then there exists C > 0 such that
Z h
Z
i
p(x)
q(x)
2
jruj
+ jruj
dx C
and thus
Z
So
1
p+
juj
r+
+ juj
1
1
p(x)
q(x)
jruj
+
jruj
dx
p(x)
q(x)
> 0: Similarly, we get
> 0:
r
dx
Cr
2jmj1
C
jmj1
Z
Z
m(x) juj
r(x)
m(x) r(x)
juj
dx:
r(x)
dx
El Amrouss et al.
15
Before embarking in the proof of Theorem 1, we need some auxiliar lemmas. Let
Z
Z
1
1
p(x)
q(x)
I(u) =
jruj
dx +
jruj
dx
p(x)
q(x)
and
J(u) =
Z
1
r(x)
m(x) juj
dx:
r(x)
The following lemma plays a crucial role in our di¤erent lines.
LEMMA 1.
(i) The functional I is weakly lower semi-continuous, that is, un * u implies that
I(u) lim inf n I(un ):
(ii) The functional J is weakly-strongly continuous, that is, un * u implies that
J(un ) ! J(u):
PROOF. (i) Using the convexity of the functional I, we see that the assertion is
immediate. (ii) Assume that un * u in W; which is re‡exive. Then fun g is a bounded
sequence. By Proposition 3, we have
W ,! Lq (x) (RN ):
o
n
So we obtain a boundedness of jun jq (x) : So, there is a positive constant M > 0
such that
max
n
Taking
k
jun jr(x)
q (x)
r(x)
; jujr(x)
= fx 2 RN : jxj < kg: Further, m 2 L
jmjL
(x) (RN n
k)
q (x)
r(x)
(x)
M:
(RN ) implies that
! 0 as k ! +1:
Giving " > 0; we may …nd k1 > 0 large enough such that
jmjL
(x) (RN n
k1 )
<
"
:
8M
It follows from the compact embedding W 1;r(x) ( k1 ) ,!,! Lr(x) (
Z
Z
r(x)
m(x)jun j
)dx !
m(x)jujr(x) dx;
k1
k1 );
k1
because m 2 L1 (RN ): Hence, there exists n1 > 0 such that for n
Z
k1
m(x)jun jr(x) dx
that
Z
m(x)jujr(x) dx <
k1
"
:
2
n1 ;
16
The Spectrum of Laplacian with Variable Expoment
In view of Proposition 2, we get
jJ(un )
Z
J(u)j
+
k1
Z
h
RN n
"
+
2
Z
i
m(x)jujr(x) dx
m(x)jun jr(x)
k1
RN n
[m(x)jun jr(x)
k1
m(x)jujr(x) ]dx
h
i
r(x)
r(x)
m(x) jun j
+ juj
dx
"
+ 2jmjL (x) (RN n
2
" "
+ = ":
2 2
k1 )
jun jr(x)
q (x)
r(x)
+ jujr(x)
q (x)
r(x)
Consequently, J(un ) ! J(u):
LEMMA 2. We have the following limits,
I(u)
= +1
kuk!+1 J(u)
lim
and
I(u)
= +1:
kuk!0 J(u)
(4)
lim
PROOF. When kuk ! 0; we have kukq(x) ! 0: The preceding Proposition 1 together with Proposition 4 and 5 imply
I(u)
=
J(u)
R
1
p(x)
R 1
p(x)
q(x)
jruj
dx + q(x)
jruj
dx
R
r(x)
1
m(x) r(x) juj
dx
1
q+
q+
kukq(x)
r
;
c jmj1 kukq(x)
where c > 0: Since r > q + ; then the …rst relation hold. For the seconde assertion, let
Z
1
p(x)
K(u) =
jruj
dx:
p(x)
If kukp(x)
1; then we have
1
p
kukp(x)
p+
If kukp(x)
K(u)
1
p+
kukp(x) :
p
K(u)
1
p
kukp(x) :
p
1; then we have
1
p+
kukp(x)
+
p
El Amrouss et al.
17
In this case, we set
Then
1
p
kukp(x)
p+
K(u)
We entail that
1
p
kuk
p+
K(u)
1
p+
kukp(x)
p+
1
p
kukp(x)
p+
C0
0:
C0 :
C0 ; 8u 2 W 1;p(x) (RN ):
Thereby, there exists C1 > 0 such that
Z
Z
1
1
p(x)
q(x)
I(u) =
jruj
dx +
jruj
dx
p(x)
q(x)
1
1
p
q
kukp(x) + + kukq(x) C1 ;
p+
q
whenever u 2 W:
On the other hand, according to Proposition 4 there exists C3 > 0 such that
Z
1
r(x)
r+
r
J(u) =
m(x) juj
dx C3 (kuk + kuk ):
r(x)
Afterwards, we have
I(u)
J(u)
R
=
1
p(x)
jruj
R
p(x)
kukp(x) +
1
q+
r+
C3 (kuk
since r+ < p ; we infer that
of
I(u)
J(u)
R
1
q(x)
r(x)
1
r(x) m(x) juj
p
1
p+
dx +
q
kukq(x)
r
+ kuk
q(x)
jruj
dx
dx
C1
;
)
! +1 when kuk ! +1:
Proof of Theorem 1. We show that 1 is an eigenvalue of (1). By the de…nition
W n f0g such that
1 ; there exists fun g
1
= lim
n
I(un )
:
J(un )
Here fun g is bounded in W; because the coercivity of
I(un )
, and thus there exists
J(un )
u 2 W satisfying un * u; so
I(u)
Whether u 6= 0; then
1
lim inf I(un ) and J(un ) ! J(u):
n
=
I(u)
J(u)
and it is done. Otherwise, assume that u = 0;
so I(un ) ! 0 as n ! +1: We have I(un ) =
I(un )
J(un ) J(un );
passing to limit we get
18
The Spectrum of Laplacian with Variable Expoment
I(un ) ! 0 as n ! +1; along with kun k ! 0; hence
We conclude that for any v 2 W we have
@ I(u + "v)
@" J(u + "v)
I(un )
J(un )
! +1 which is paradoxical.
= 0:
"=0
By a straightforward computation we have
Z
jrujp(x)
2
+ jrujq(x)
2
rurvdx J(u) =
Z
m(x)jujr(x) uvdx I(u); 8v 2 W:
Consequently, 1 is an eigenvalue of problem (1).
Now, we are to prove that any > 1 is an eigenvalue of (1). Set
L(u) = I(u)
J(u):
From Lemma 1, we know that I is weakly lower semi-continuous and J is weakly-strongly
continuous, which yields L is weakly lower semicontinuous.
If we return to relation (4), it is clear that L is of class C 1 and coercive, therefore,
L admits a global minimum u in W; which is a critical point for L. We claim that u
is nontrivial. Indeed, by the characterization of 1 ; there exists v 2 W n f0g such that
I(v)
> 1 ; accordingly L(v) < 0, thence u 6= 0:
1 = J(v) ; since
Hereinafter, we check that any 2]0; [ is not an eigenvalue of problem (1). First,
r+
+
we may observe that
1 and r < p :
1 ; because
p
Next, let 2]0; [, by contradiction we assume that is an eigenvalue, then there
exists v 2 W n f0g which satis…es
Z
Z
p(x)
q(x)
r(x)
jrvj
+ jrvj
dx =
m(x) juj
dx:
On the other hand, according to de…nition of
R
p(x)
we obtain
q(x)
jrvj
+ jrvj
R
r(x)
m(x) jvj
dx
dx
= ;
which is contradictory and then the proof is achieved.
REMARK 2. When
0; for all u 2 W n f0g we have
Z
Z
Z
p(x)
q(x)
r(x)
I 0 (u)u = jruj
dx + jruj
dx
m(x) juj
dx > 0;
and then the problem (1) has no solution.
Whether m < 0; we may consider that the eigenvalue
argument works.
< 0; and the similar
El Amrouss et al.
19
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