A NNALI DELLA
S CUOLA N ORMALE S UPERIORE DI P ISA Classe di Scienze
D. G. D E F IGUEIREDO
P. L. F ELMER
A Liouville-type theorem for elliptic systems
Annali della Scuola Normale Superiore di Pisa, Classe di Scienze 4
esérie, tome 21, n
o3 (1994), p. 387-397
<http://www.numdam.org/item?id=ASNSP_1994_4_21_3_387_0>
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Elliptic Systems
D.G. DE FIGUEIREDO* - P.L. FELMER*
0. - Introduction
A
priori
estimates for solutions ofsuperlinear elliptic problems
canbe established
by
a blow uptechnique.
Such a method has been usedby Gidas-Spruck [GS1]
for the case of asingle equation.
Similar arguments can also be used in the case of systems. We refer to the work of JieQing [J]
and M.A. Souto
[S].
As in the scalar case, the treatment ofsystems
poses thequestion
of thevalidity
of a result which is referred as aLiouville-type
theoremfor solutions of systems of
elliptic equations
inR~.
Let us makeprecise
thisquestion.
We consider the
elliptic
systemin the whole of
R~,
N > 3. Thequestion
is to determine for which values of the exponents a and{3
theonly non-negative
solution(u, v)
of(0.1)
is(u, v) = (0, 0).
The notion of solution here is taken in the classical sense,
i.e.,
u, v C In the case of asingle equation
it has been
proved
in[GS2]
that theonly
solution of(0.2)
is u = 0 when* The first author was
partially
supported by CNPq and the second by Fondecyt and D.T.I.University of Chile.
Pervenuto alla Redazione il 26 Marzo 1993.
In dimension N = 2, a similar conclusion holds for 0 p oo. This is a
special
case of the result that asserts that any
superharmonic
function bounded below in the wholeplane
isnecessarily
constant, see[PW,
Theorem29,
p.130].
Itis also well known that in the critical case, p =
(N
+2)/(N - 2), problem (0.2)
has a two-parameter
family
of solutionsgiven by:
We see then that the critical
exponent,
in the scalar case,plays
animportant
role in the
validity
of theLiouville-type
theorem forequation (0.2).
It is naturalto
conjecture
that in the case of system(0.1)
the condition(0.3)
isreplaced by
The basis of this
conjecture
lies on the fact that the existence ofpositive
solutions for the Dirichlet
problem
for system(0.1)
in a bounded domain holdstrue if condition
(0.5)
issatisfied,
see[CFM, FF, HV].
If thisconjecture
weretrue we would have a
complete analogy
with the scalar case. A further argument in favor of thisconjecture
is a theorem of Mitidieri[M,
Theorem3.2]
whichstates that
(0.1)
has no nontrivial radialpositive
solutions of classN > 3,
provided
1 aQ
and(0.5)
holds. The resultproved
here is thefollowing:
THEOREM.
A)
If a > 0 and~3
> 0 are such that but not both areequal
tothen the
only non-negative C2
solution of(o.1 )
in the whole ofR:l
is the trivialone: u=0, v=0.
B)
If a= /3 = N - 2 ,
N+2 then u and v areradially symmetric
with respect N - 2to some
point
ofREMARK. In
[S]
it wasproved
that(0.1)
has no nontrivialnon-negative C2
solutions in the whole ofprovided
and
We see then that this establishes the
conjecture
in ahyperbolic region
of theplane (a,,3)
which is smaller than theregion
determinedby (0.5).
However(and
this isimportant
tostress)
such aregion
containspoints
which are notincluded in the
region
defined in(0.6).
Souto’sargument
in[S]
is based on thenon-existence of
positive
solutions in the whole ofR~’
of theinequality
when u
N/(N - 2).
This fact wasproved by
Gidas[G].
Theconjecture
would be
proved
if we had this non-existence result(for inequality (0.8))
up to(N
+2)/(N - 2).
We do not know if such a result is true. In[J]
thereare some
Liouville-type theorems; though
Jie’s resultsapply
to moregeneral nonlinearities, they
do not cover our result.An
important
feature of thepresent
paper is to show how the Method ofMoving Planes,
asdeveloped by Gidas-Ni-Nirenberg [GNN],
and morerecently by Berestycki-Nirenberg [BN],
can be used toproduce
rathersimple
andelegant proofs
ofLiouville-type
theorems forsystems.
The maindifficulty
stems fromthe fact that the domain where the solutions are considered is In such a case it is not clear where we can start the
procedure,
see details in Section 2. Thepresent
work has been motivatedby
recent papersby Caffarelli-Gidas-Spruck [CGS]
and Chen-Li[CL],
whereLiouville-type
theorems for asingle equation (results formerly proved by Gidas-Spruck [GS2])
have received a treatmentby
the Method of
Moving
Planes.The authors would like to thank Manuel del Pino for useful conversations.
1. - Some
general
facts aboutsuperharmonic (subharmonic)
functionsLet us recall the so called Hadamard Three
Spheres Theorem,
see[PW,
p.
131]:
"Let Q be an open set
containing
the setand U E
C2(Q)
with 0. For r r2, letThen
for any r in
[ri , r2 ] ."
The
proof
of this result is verysimple.
Let = a +br2-N.
. Choose aand b such that =
M(rl)
andy~(r2)
=M(r2).
Letu(x) -
anduse the maximum
principle
for subharmonic functions.LEMMA 1.1. Let u E
C2 (II~N B { o } )
be such that u 0 and 0.Then,
for each c > 0, one hasPROOF.
Letting
ri --i 0 in( 1.1 )
we obtainM(r) M(r2)
for all 0 r r2,which
implies (1.2) by taking
r2 = c.Letting
r2 -~ +00 in( 1.1 )
we getM(r)r N-2 M(rl)rN-2
for all r > rl, whichgives (1.3) by taking
r1 = ~.m COROLLARY 1.1. Let u E
C2 (RN)
be such that u > 0 and 0 in the whole ofe.
Let-
be its Kelvin transform.
Then,
for each - > 0, there arepositive constants b,
and c~ such thatPROOF. The function -w satisfies the
hypothesis
of Lemma 1.1.Choosing
c~ = we obtain
(1.4)
and the lower bound in(1.5). Choosing bE
=c- 1 1
we obtain the upper bound in(1.5).
D2. - Some
auxiliary
factsLet u > 0 and v > 0 be
C2
solutions ofsystem (o.1 )
with both a and{3 (N
+2)/(N - 2).
Let us introduce their Kelvin transformswhich are defined for One verifies that w and z
satisfy
the systemWe shall use the method of
moving planes.
Let us startby considering planes parallel
to x = 0,coming
from -00. For each real A let us defineand xA
as the reflection of x about theplane TÀ.
Let ex =(2A, 0,..., 0).
In what follows we consider A 0 and setix
= Infx
we define thefollowing
functions 1 1, 1 1 1 11
The first
step
in the use of the method ofmoving planes
is to show thatwe can start the process. That
is,
there exists a A 0 such that 0 and 0 for all x EUnfortunately,
this cannot be shown as we shallsee
shortly.
However this can be achieved with aslight
modification of the functions. Let usexplain
how.For x
e fx
we haveUsing
the invariance of theLaplacian
under a reflection and the fact thatI xÀ ixi,
it follows from(2.4)
Hence we get from
(2.3)
and(2.5)
which, using
the mean valuetheorem,
can be written aswhere
and
A)
is a real number betweenzx(z)
andz(x). Similarly,
we obtainwhere
and
A)
is a real number betweenwx(z)
andw(x).
We would like to infer that 0 and0,
for some A 0 and all x EEa
but we observe thatsystem (2.6)-(2.7)
is not amenable forapplication
of maximumprinciples.
In order to obtain a
system satisfying
thehypothesis
of maximumprinciples (see,
forinstance, [PW]
or[FM])
we introduce thefunction g
and show that the functions ~T, , , " ~ , , "where
g(x)
= for all x such that x 1 0,satisfy
thefollowing system
ofinequalities
The latter
system
iscooperative,
since bothc(x; A)
and6(x; A)
arepositive.
Thuswe can
apply
the maximumprinciple (Thm.
1.1 in[FM]), provided
Our next purpose is to establish.
PROPOSITION
2.1. There exists A* 0 such that for all AA*,
0 and forall XEIA-
We shall use in the
proof
ofthe above proposition
someproperties
offl7x (and
a similar set ofproperties
ofZx),
which are collected in the lemmas below. We consider A 0.LEMMA 2.1.
(i)
If > 0 for all x in somepunctured
ball cEa,
and if
then
Wa
attains its infimum infx.
(ii)
There exists a 0, such that if A~
andc(A)
0, thenWa
attains its infimum in~a .
PROOF.
(i)
Since 0as Ix I
- oo, we can find ri 1 > 0 such thatWA (x) - c(A) a( e)>
forx I
> ri . ThusW
attains its infimum on thecompact
set2
( ) |x >r
1 a pBr, (o)
n SinceWa
vanishes onTA,
the result follows.(ii) Using Corollary
1.1 we findpositive
numbers c 1 and r such thatw(x) >
ci, for 01,
andw(x) c1
forIxl
> r. Choose A =min{-r, -1}.
Then for A
A .
LEMMA 2.2.
(i)
There is apositive
constantdx, depending only
on w andA,
such thatfor all x E
fx - B1(eÀ),
where 0.(ii)
There is anRa
> 0,depending only
on A and w, such thatfor all x E where 0.
PROOF.
(i)
If~3
> 1, it follows from 0 thatwa (x) ~(x; A) w(x).
We then estimate
w(x) using inequality (1.5)
ofCorollary
1.1. We see thatd
a can be taken as{ (y) |y| |y|_1] Then 1j;(x; À) dA . Using
gthe definition of
6(x;A)
we obtain(2.11 ) readily. Actually, inequality (2.11 )
holds for all x E
If 0
~3
1 then we estimateusing
the otherpart
ofinequality (1.5)
to obtainwhere dA),
== 1}.
From here we obtain(2.11)
sinceQ -
1 0.(ii)
Observe thatSo
(2.12)
followsimmediately
from the aboveinequality together
with(2.11).
D LEMMA 2.3. There exists
A* A
suchthat,
for all AA*, inequality
(2.12)
holds for all x EiA,
whereWA(x)
0.PROOF. We first observe that
da
if So for all estimatefor all x E
ix,
where 0. Hence theright
side of(2.13)
isnegative
iflxl
>R~ (as
in Lemma2.2(ii)).
Now take A* such that ERN : Ixj
>0
LEMMA 2.4. Assume that there is XO C
Ea
such thatThen 0.
Ifmoreover |x0 |
>Rx (as
in the version of Lemma2.2(ii)
for
Za ),
thenPROOF. Since = 0 and 0, it follows from
(2.8)
thatwhich
implies
that 0. Next we can write(2.14)
asAs we said before an
analogue
to Lemma 2.2 holds forZx.
So the first bracketis
negative,
and the conclusion follows. DPROOF OF PROPOSITION 2.1. Let us assume,
by contradiction,
that forsome A A*
(A*
= least of the twoA*’s, the
one defined in Lemma 2.3 and the one from the version of Lemma 2.3 forZa ),
we haveWithout loss of
generality,
we may assume that 0 for some x Efx.
The
argument
below showsthat
theconjunction
or in(2.16)
can bereplaced by
and.Indeed,
since A*A,
theassumption
0, for some x EEa, implies
via Lemma2.1(ii)
thatthe
infimum ofW a
is attained at apoint
xo inP7x;
and Lemma 2.4 states that 0.We next use Lemma 2.3 to see that the first bracket in
(2.15)
isnegative,
and then
Now we do a similar
argument
withZa
and conclude thatwhere xl E
Ea
is thepoint
whereZx
attains its infimum.Finally
we use(2.17), (2.18)
and the fact that2A(xi) :5
and to come to acontradiction. D
We next define
and
PROPOSITION 2.2. If
Ao
0, then 0 and 0 for all REMARK. It followsreadily
fromProposition
2.2 that both w and z aresymmetric
with respect to thehyperplane Tao .
PROOF OF PROPOSITION 2.2.
By continuity
we see that 0 and0 for x E
Now
observe that0, it
follows from(2.9)
that 0, which willimply
0. So if we assume,by contradiction,
that
the
conclusion ofProposition
2.2 is not true, we conclude that both and are notidentically
zero. It follows from(2.8)
thatfor all x E
I:.xo.
Soby
the maximumprinciple (recall 0)
we seethat > 0 for x E
Similarly
> 0 for x ENext
using
theHopf
maximumprinciple
we obtain thaton the
boundary Tao. (Here a/av
isa/axl).
We shall see now that this isimpossible.
From the definition ofAo,
we conclude that there exist a sequence of realnumbers Ak
-;Ao,
withAk
>Ao,
and a sequence ofpoints
inEak,
where or
Z Àk
isnegative.
Since satisfies(2.6)
and > 0 forx E
Eao,
we conclude that issuperharmonic
inÏ:Ào.
Inparticular WÀo
issuperharmonic
in where 2R=
Theargument
in theproof
of Lemma 1.1 shows that there exists co > 0 such that
A similar
argument
shows that(We
may choose the same c, or decreaseit).
SoBy continuity
we havefor l~
sufficiently large.
It follows from Lemma
2.1(i) (using (2.20))
and Lemma 2.4 that for ksufficiently large
bothW Àk
andZ Àk
attain theirnegative
infima inÏ:Àk. [Recall
that we are
using
the contradictionhypothesis
thatW ak
or isnegative
some-where in Let us denote
by Xk
and Yk,respectively,
thepoints
of minimaof
Wak
and If follows from Lemma 2.4 that at least one of the sequences,or
lykl
is bounded. Assume thatfxkl
is bounded andpassing
to asubsequence
assume that Xk - x.By continuity
we have that 0 andW a° (x)
0.Since xf=eAo
andW a° (x)
> 0 in we conclude that x ET
, which contradicts the fact observed before thatav
av 0 onT
A similar
argument applies
if we assume thatlykl
is bounded. D3. - Proof of the theorem
We first prove Part A.
Performing
themoving plane procedure
we havetwo
possibilities:
(i)
IfAo
0, it follows fromProposition
2.2 that w and z aresymmetric
will respect to the
plane
Butlooking
atequations (2.1)
and(2.2)
we realizethat this is
impossible.
(ii)
IfAo =
0, we conclude thatw(x)
andzo(x) ~! z(x)
for allx E
10.
We canperform
theprocedure
from theright
and we will reach aao
= 0,(’B0
cannot bepositive,
in virtue of anargument
as in(i) above),
fromwhich we get
w(x) ~~- wo(x)
andz(x) > zo(x)
for all x E Eo. So w and z aresymmetric
withrespect
to theplane
x 1 = 0.(iii)
Thisreasoning
can be made from any direction. And so theonly possibility
would be that both w and z areradially symmetric
with respectto the
origin
0. But 0 was chosenarbitrarily
when weperform
the Kelvin transform. Thus u and v areradially symmetric
with respect to anypoint.
Thenthey
would be constant and from theequations
wefinally
obtain u = v = 0.Part B. We show that w and z are
symmetric
with respect to someplane parallel
to x = 0.Indeed,
ifAo
0, this follows fromProposition 2.2,
and inthis case the
plane
isTAO.
IfAo
= 0, weperform
themoving plane procedure
from the
right
and find acorresponding ’B0
> 0. Ifao
> 0, ananalogue
toProposition
2.2 shows that w and z aresymmetric
withrespect
toTAO.
IfAr0
= 0we
proceed
as in(ii)
above to conclude that both w and z aresymmetric
with
respect
to x 1 = 0. Weperform
thismoving plane procedure taking planes perpendicular
to anydirection,
and for eachdirection 1
E we finda
plane T,~
with theproperty
that both w and z aresymmetric
with respect toT1.
Asimple argument
shows that all theseplanes
intersect at asingle point,
or w = z = 0. D
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IMECC - UNICAMP Caixa Postal 6065 13081 Campinas, S.P.
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