A NNALES DE LA FACULTÉ DES SCIENCES DE T OULOUSE
E RIC N ABANA
Uniqueness for positive solutions of p-Laplacian problem in an annulus
Annales de la faculté des sciences de Toulouse 6
esérie, tome 8, n
o1 (1999), p. 143-154
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- 143 -
Uniqueness for positive solutions
of p-Laplacian problem in
anannulus(*)
ERIC
NABANA(1)
Annales de la Faculte des Sciences de Toulouse Vol. VIII, nO 1, 1999
A l’aide d’une methode de tir et de quelques identites
de comparaison, nous donnons un résultat d’unicite pour les solutions radiales positives d’un probleme de Dirichlet avec le p-laplacien dans un
anneau.
ABSTRACT. 2014 By means of a shooting method and some comparison identities, we give a uniqueness result of positive radial solutions for p-
Laplacian Dirichlet problem in an annulus.
1. Introduction and Results
We
give
here auniqueness
result forpositive
radial solutions of theproblem
where S2 =
{x
E[
0 ab},
andApu
=1 p +00 .
We start
by transforming
the radialequation
of(P)
into the form :Our
uniqueness
result is then derivedby using
somecomparison
identities established here for a initial valueproblem
associated to this second order differentialequation.
~*~ ) Recu le 23 mai 1995, accepte le 10 novembre 1998
(1) L.A.M.F.A., Faculte de Mathématiques et Informatique, Université de Picardie
Jules-Verne, 33 rue Saint-Leu, F-80039 Amiens Cedex 1 (France)
E-mail : nabana@u-picardie.fr
In Nabana-de Thelin
([4], [5]),
wegive uniqueness
results ofpositive
radial solutions of
(P)
for Q a ball In this paper weessentially
followthe
approach
used in[5].
Thedifficulty
is reduced to the obtention of thecomparison
identities(I1), (12)
and(13)
in Section 2.This kind of
technique
is associated to several names in theparticular
case p = 2
(see
forexample :
Kolodner[3],
Coffman([1], [2]),
Ni[6],
Ni-Nussbaum
[7]).
It isimportant
to remark that this method is madeby
thefirst time for
p ~
2 in[5].
Let
u(x)
= a radial solution of(P).
Then u verifies thefollowing problem
By making
achange
of variables of the kindthen
problem (1.1)
is transformed intowhere a and
,Q depend
on a and b.For the remainder of this
section,
we firstgive
auniqueness
result for theproblem ( 1.2) .
THEOREM l.l. -
Suppose
p > 2. Assume that F : Il8[ a , ~ [ ~
II8 isC1
withF(0, s)
= 0for
s > a.Suppose
that there exists M > 0 suchthat, for
0 w M and a s,Q,
Fsatisfies
thefollowing
conditions :Then
problem (1.2)
has at most onepositive
solution.Theorem 1.1 is the fundamental tool in the
proofs
of thefollowing
theorems which are
generalisations
of some results in~7~.
.THEOREM 1.2.-
Suppose
that n > p > 2.Suppose
thatf
:(u, r)
E[0, ~ [
x(a, b ]
~f(u, r)
E[0, ~ [
isCl
and thatf(0 r)
= 0for
allThen
(P)
has at most onepositive
radial solutionprovided, for
u > 0 anda r
b,
thefunction f satisfies
thefollowing
conditionswhere ~C =
(n
-p)/(p
-1)
>0,
THEOREM 1.3.- Let S2 =
{x
ERp | (
0 a|x| b}
with p > 2.Suppose
thatf
:(u, r) ~ [ 0 , ~ [ x a , b ] ~ I(u,r)
E( 0 , ~ [
isCl
andthat
f(0, r)
= 0for
all r E~a, b~.
Then
(P)
has at most onepositive
radial solutionprovided, for
u > 0 anda r
b,
thefunction f satisfies (1.,~~
and thefollowing
conditionsIf
~ f (u,
- 0 in Theorem 1.2 and in Theorem1.3,
we obtain thefollowing corollary.
COROLLARY 1.4. -
Suppose
that n > p > 2 and thatThen,
theproblem
has at most one
positive
radial solutionprovided
2. Proof of Results
We
begin
this sectionby giving
a theorem on existence anduniqueness
of a
positive
solution of thefollowing
initial valueproblem :
THEOREM 2.1.-
Suppose
that n > p and that thefunction f : (u, r)
E1I8 x
lI8+
~f (u, r)
~ R isCl.
Thenfor
each d >0,
theproblem (2.1)
has a
unique positive
solutionu(r)
=u(r, d) defined
in a maximal intervalJd
C~ a , ( .
. Thefunction
d -u(r, d)
isCl
and r -u(r, d)
isC2 if
1 p
2,
and at allpoints
r > a such thatu’(r) #
0if
p > 2.We prove this theorem in the
Appendix.
2.1 Proof of Theorem 1.1
Let
w(s, d)
be a solution of thefollowing
initial valueproblem
Consider the set
and the map
The domain D is open in
(a, co)
and sincew’ (T(d), d~ ~ 0,
it arises from theimplicit
function theorem that T isCl
at theneighbourhood
of D’spoints
and that ...Put
p(s, d)
=8w(s,
Then p satisfies thefollowing problem
As in the
proof
of Theorem 2.2 in[6],
we aregoing
to show that T isstrictly decreasing
on D. Sinceit is sufficient to prove
d)
0.Put
Then, ’Ø
satisfies theequation
Note
that,
from thehypothesis F(0, s)
= 0 forall s >
a and theuniqueness
of
w(s, d),
we deduce thatv(s, d) ~
0at s >
a for whichw(s, d)
= 0. In thesame way, if
p(so)
=0,
thennecessarily ~(sp) ~
0.Now,
for agiven d
>0,
put/~
=T(d).
LEMMA 2.1. - Let X =
(p -
Thenand the
function
vanishes at least once in the interval(a, ,Q).
Proof.
- Theidentity (I1)
isstraightforward.
Suppose
that p does not vanish in(a, ,Q).
Thenp(s)
> 0 in(a, ~i).
Integrating (I1)
from a to,0,
we obtainSince X is continuous
on [a, /3]
withthen,
the left-hand of(2.4)
would benegative
while theright-hand
remainspositive
because of the condition(Cl)
of Theorem1.1;
this is a contradic- tion. DSince
F(w, s)
> 0 for 0w M,
and a s,~,
there exists aunique point
03B30 ~(03B1, 03B2)
such that =0,
v > 0 on( a ,
and v 0 on~’Yo ~ a~~
.LEMMA 2.2.- Let Y =
(s - a)zu~~ - ((s - Then, for
alla , ,0~ B {yp},
we havewith
p’
=p/(p - 1).
Letço
=min{s
>a | p(s)
=0}.
. Then yp03BE0.
Proof.
- Astraightforward computation gives
thecomparison identity
(12).
Let e > 0 small
enough
and consider the function H definedby
Then, (12)
can beput
into thefollowing
formThe existence of
ço
isgiven by
Lemma2.1,
and we havecp(s)
> 0 fora s
ço
and 0.Suppose
thatço
7; Thenv(s)
> 0 for all s such thatço .
Integrating (2.5)
from a + e togo
andby using
condition(C2)
in Theorem1.1,
we obtainTherefore
thus
w’ (o) 0 ;
this is a contradiction. DLEMMA 2.3. - Let Z = - Then :
and
~p
is theunique
zeroof
p in]
a[.
Proof.
- Theidentity (13)
isstraightforward.
Suppose
that p has others zeros in(a , ,0 ] and
letThen, p
0 in the interval and > 0.By integrating (13)
frowço
to andby considering
the condition(C3),
we havethus >
0;
a contradiction with the Lemma 2.2. DIt follows from Lemma 2.3 that
p(s)
0 onço
s/3.
Inparticular
d~ 0,
so Theorem 1.1 isproved. 0
2.2 Proof of Theorem 1.2
Let r and
u(x)
=u(r),
apositive
radial solution of(P). Then,
usatisfies the
problem (1.1).
Consider thefollowing change
of variablesThen
problem (1.1)
is transformed intowhere
the function F is
given by
We are
going
toapply
Theorem 1.1 to theproblem (2.6)
in order to obtainthe
uniqueness
of u,positive
radial solution of(P).
It suffices to show thatthe conditions
(Cl), (C2)
and(C3)
of Theorem 1.1 are satisfiedby
thefunction F defined at
(2.7).
By
astraightforward computation,
we obtain anequivalence
between(Cl)
and the condition(1.3)
of Theorem 1.2 on the one hand and between(C3)
and(1.5)
on the other hand.Otherwise, (C2)
isequivalent
toThis
inequality
is satisfied iff
verifies the relation(1.4).
Theorem 1.2 is thusproved. 0
2.3 Proof of Theorem 1.3
Let r and
u(x)
=u(r) ,
apositive
radial solution of(P).
Thechange
of variables
transforms the
problem (1.1)
intowhere a
= - In(b),
and{3 = - In(a).
Put
It is easy to see that F
given
at(2.9)
satisfies the conditions(Cl),
and(C3)
of Theorem 1.1
if f
verifiesrespectively
the relations(1.3)
and(1.7).
Thecondition
(C2)
is verifiedby
F if andonly
ifWe observe that this last
inequality
is valid iff
satisfies(1.6).
Then,
the Theorem 1.3 followsimmediately
from Theorem 1.1applied
tothe
problem (2.8). ~
3.
Appendix -
Proof of Theorem 2.1Let K be a
compact in R2
definedby
where
60
> 0 and 6~0 > 0 are fixed. PutConsider E the Banach space defined
by
with the norm
where
Select
do
>0,
suchthat Id -
E £0. .Existence and
uniqueness
for(2.1)
result from thestudy
of fixedpoint
ofthe
application .~’d : (u, v)
E E -->(u, v)
E E where u and v are definedby
Let B =
{(v,, v)
EE
and~~v - dp-l~~p closed,
convexbounded subset in E.
Then
60
smallenough and |d 2014d0| ~ ~, od
is aLipschitz
map of B into itself withLipschitz
constant less than one. In fact for all(u, v)
E Bwe have
and then
Therefore C B for 6 small
enough.
Note
that,
for 6 smallenough,
there exists two constants ci > 0 and c2 > 0 such that for all v EC~([a ,
a +6 ])
withv(a)
>0,
we haveThen,
for v~ EC~([a, , a -I- b ~~
with = >0, i
=1, 2,
we obtainwhere
c(p)
=(p’ -
if 1p 2,
andc(p)
=(p’ - 1)~~-2 if p > 2.
For all
Xi
=(ui, vi)
EB,
i =1, 2,
we haveThe contraction
mapping principle implies
that has aunique
fixedpoint
X =
(u, v)
in B. The functions u and v are in, a + b ~, IIB~
and wehave ,
and
thus, u(r)
=u(r, d)
is theunique
solution of(2.1).
By using
the samearguments
of classicaltheory
on existence andunique-
ness for initial value
problem
forordinary
differentialequations,
we showthat u is
uniquely
defined on a maximal intervalJd
in[ a ,
+00[.
It is immediate to see that r ->
u(r)
isC2
onJd
if 1p 2,
and thatr --~
u(r)
isC2 only
at thepoints
r such thatu~(r) ~
0.To prove that d -~
u(r, d)
isC1,
we consider thefollowing
mapFor all X E B and 6 small
enough,
we have =Therefore, by
theimplicit
function theorem for Banach spaces, we conclude that the functions d -.u(r, d)
and d --~u’(r, d)
areC1. 0
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