Applied Mathematics E-Notes, 13(2013), 228-233 c
Available free at mirror sites of http://www.math.nthu.edu.tw/ amen/
ISSN 1607-2510
Existence And Uniqueness Of Weak Solution For
p-Laplacian Problem In RN
Mohammed Verid Abdelkadery, Anass Ourraouiz
Received 20 July 2013
Abstract
This paper shows the existence and uniqueness of a weak solution of a problem
in RN ; which involves the p-Laplacian through the Browder Theorem.
1
Introduction
The present paper is concerned with the elliptic problem:
(P)
where 1 < p < N; N
pu
3,
p
+ m(x) j u jp
2
u = f (x; u) in RN ;
denotes the p-Laplacian de…ned by
pu
= div(j ru jp
2
ru):
We make the following assumptions.
(m0 ) m 2 C(RN ; R) and 0 < m(x) < +1:
Let = p p(q+1) and p =
L (RN ) such that
Np
N p.
There exist a 2 L(p
(f1 ) f satis…es
j f (x; s) j
)0
(RN ) and b 2 L1 (RN ) \
a(x) + b(x) j s jq ;
where
1<q
p
1:
(f2 ) f : RN R ! R be a carathéodory (CAR) function which is decreasing with
respect to the second variable, i.e.,
f (x; s1 )
for a.e. x 2
and s1 ; s2 2 R; s1
f (x; s2 )
s2 :
Mathematics Subject Classi…cations: 35J30, 35J60, 35J92.
Mohamed I, Faculty of Science, Department of Mathematics, Oujda, Morocco.
z University Mohamed I, Faculty of Science, Department of Mathematics, Oujda, Morocco.
y University
228
M. V. Abdelkader and A. Ourraoui
229
The goal of this paper is to prove the following result:
THEOREM 1. Assume that (m0 ) holds and f 2 CAR(RN
(f2 ): Then the problem (P) has a unique weak solution.
R) satis…es (f1 ) and
When p = 2, the problem (P) is a normal Schrodinger equation which has been
extensively studied. There are several studies of the existence of solutions of (P) on a
bounded domain of RN : We mention the results obtained in [1, 2] and [6] for the case of
bounded domains. In recent years, more and more attention is paid to the quasilinear
elliptic setting on RN : The main di¢ culty in the study of p-Laplacian equations in RN
arises from the lack of compactness.
In the squeal, we recall some basic de…nitions and notations which will be used
throughout the paper. Whereas, the last part of the article is dedicated to the demonstration of our main result.
DEFINITION 1.We say that u 2 W 1;p (RN ) is a weak solution of problem (P) if
Z
Z
Z
p 2
p 2
jruj
rurvdx +
m(x) juj
uvdx =
f (x; u)vdx
RN
RN
RN
for all v 2 W 1;p (RN ):
For simplicity let X = W 1;p (RN ): According to condition (m0 ); we can introduce
a new norm de…ned as follows
k u k=
Z
p
RN
jruj dx +
Z
p
RN
m(x) juj dx
1
p
:
DEFINITION 2. Let K be a Banach space. An operator A : K ! K veri…es
hAu
Av; u
vi
0
(1)
for any u; v 2 K is called a monotone operator. An operator A is called strictly
monotone if for u 6= v the strict inequality holds in (1). An operator A is called
strongly monotone if there exists C > 0 such that
hAu
Av; u
vi
Cku
v k2
for any u; v 2 K:
We recall Browder Theorem.
Theorem 3 (cf. [3]). Let A be a re‡exive real Banach space. Moreover, let A :
X ! X be an operator which is: bounded, demicontinuous, coercive, and monotone
on the space X . Then, the equation A(u) = f has at least one solution u 2 X for
each f 2 X : If moreover, A is strictly monotone operator, then the equation (P) has
precisely one solution u 2 X for every f 2 X :
230
Existence and Uniqueness of Solution for p-Laplacian Problem
We de…ne the operator A : X ! X by
A := I
F;
where the operators I and F are de…ned from X into X as
Z
Z
p 2
p
hI(u); vi =
jruj
rurvdx +
m(x) juj
RN
2
uvdx
RN
and
hF (u); vi =
Z
f (x; u)vdx;
RN
for all u; v 2 X:
By De…nition 1, the main tool in searching the weak solutions of (P) is to …nding
u 2 X which satis…es the operator equation Au = 0:
2
Proof of The Main Result
We denote by C and Ci ; i = 1; 2::: the general positive constants which are the exact
values may change from line to line.
PROOF OF THEOREM 1. In order to apply Browder Theorem, we split the proof
in several steps,
Step1. We prove that A is bounded. We know that the functional
Z
1
p
p
(jruj + m(x) juj ) dx
(u) =
p
N
R
is of class C 1 (cf. [5]) and I is the derivative operator of
in the weak sense, so it
yields I is bounded and continuous. Let u 2 X; such that k u k< K. Using Hölder’s
inequality, we obtain
kF (x; u)kX
= sup j hF (x; u); vi j
kvk=1
sup
kvk=1
sup
Z
RN
a(x) jvj dx +
8
< Z
kvk=1 :
sup
kvk=1
a(p
dx
q
RN
1
(p )0
b(x) juj jvj dx
+
RN
(" Z
C3 kak(p
)
0
Z
RN
(p )0
a
dx
1
(p )0
RN
)0
Z
+ C4 K q kbk ;
hence A is bounded.
+
Z
juj
RN
q
p
p
dx
Z
p
(bv) p
q
p
p
q
dx
RN
p
juj
q
p
dx
Z
RN
b
1 #
Z
RN
9
=
;
p
jvj
1
p
dx
)
M. V. Abdelkader and A. Ourraoui
231
Step 2. We prove that A is demicontinuous. It is well known that the functional
Z
1
p
p
(jruj + m(x) juj dx
(u) =
p
RN
is of class C 1 : Since I is the Fréchet derivative of hence I is continuous. Now we
check that F is completely continuous that is, if un * u then F (un ) ! F (u) and it is
well be done. Let un is weakly convergent to u in X so un is bounded in X. Set
Bk = x 2 RN : jxj < k ;
so we have jbjL
Z
(RN nBk )
RN nBk
converges to zero as n ! +1. For all v 2 X we have
Z
a(x) jvj dx
jvj
RN nBk
C kvk
p
dx
Z
! p1
Z
RN nBk
0
(p )
RN nBk
(p )0
jaj
dx
!
jaj
dx
!
1
(p )0
1
(p )0
:
Similarly,
Z
Z
q
RN nBk
b(x) juj jvj dx
RN nBk
Z
RN nBk
juj
p
juj
p
q
C kuk kvk
! pq
Z
dx
! pq
Z
RN nBk
Z
(b(x) jvj) p
RN nBk
b(x) dx
RN nBk
! p1
p
jvj
!1
p
q
Z
dx
! pp
b
RN nBk
q
!1
:
According to previous inequalities we have,
Z
(f (x; un )
f (x; u))vdx
RN nBk
C kvk
Z
b(x) dx
RN nBk
!1
+kvk
Z
RN nBk
ja(x)j
0
(p )
which yields that
Z
(f (x; un )
RN nBk
f (x; u))vdx ! 0
for k su¢ ciently large. From the compact embedding W 1;p (Bk ) ,! Lq (Bk ), we can
infer that
Z
Z
f (x; un )vdx !
f (x; u)vdx
Bk
Bk
!
dx
1
(p )0
;
232
Existence and Uniqueness of Solution for p-Laplacian Problem
and then we have
Z
(f (x; un )
RN
Z
+
(f (x; un )
Bk
f (x; u))vdx
Z
=
(f (x; un )
f (x; u))vdx
RN nBk
f (x; u))vdx !
0:
So F is completely continuous and then F is continuous.
Step 3. We prove that A is monotone. We recall the following inequality for p
x; y 2 RN (see [4])
jyj
p
p
p 2
jxj + p jxj
x(y
x) +
2;
p
jy
2p
xj
:
1
1
p 2
rv (ru
Let
hI(u)
I(v); u
vi =
Z
RN
+
Z
jruj
p 2
rudx
p 2
m(x)(juj
RN
u
jrvj
p 2
jvj
v)(u
rv)dx
v)dx:
We obtain that
hI(u)
I(v); u
vi
2
p2p 1
= Cp ku
Z
1 RN
p
vk :
jru
p
rvj dx +
Z
RN
m(x) ju
p
vj dx
(2)
Therefore, A is strongly monotone. ( see e.g. [7]). Further, since f is decreasing with
respect to the second variable,
Z
hF (u) F (v); u vi =
(f (x; u) f (x; v))(u v)dx 0:
RN
It follows that A is strongly monotone.
Step 4. We prove that A is a coercive operator. We have
Z
Z
1
1
p
p
hAu; ui =
(jruj + m(x) juj ) dx
f (x; u)udx
kuk
kuk RN
RN
Z
1
p
q
kuk
(a(x) juj + b(x) juj juj)dx
kuk
RN
1
p
kuk
C1 kak(p )0 kuk C2q+1 kuk kuk ;
kuk
which yields the coercivity of A for 1 < q < p 1: In the case when q = p 1; since
X ,! Lp (RN ) with continuous embedding, then by a similar argument to that used in
[1], A is coercive.
Step 5. From the previous steps, the assumptions of Theorem 3 are ful…lled.
Therefore, problem (P) has a weak solution. For the uniqueness of weak solution for
M. V. Abdelkader and A. Ourraoui
233
problem (P); suppose that u and v be a weak solutions of (P) such that u 6= v. By (2)
it follows that
0 = hAu
Av; u
vi
Cp ku
p
vk
0:
Then u = v and the proof now is completed.
This solution cannot be trivial provided that we suppose f (x; 0) 6= 0; because in
this case A0 6= 0:
Acknowledgment. The authors would like to thank the anonymous referee for
the valuable comments.
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