R ENDICONTI
del
S EMINARIO M ATEMATICO
della
U NIVERSITÀ DI P ADOVA
G IOVANNA C ITTI
On the exterior Dirichlet problem for
∆u − u + f (x, u) = 0
Rendiconti del Seminario Matematico della Università di Padova, tome 88 (1992), p. 83-110
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for 0394u - u + f(x, u)
=0.
GIOVANNA
CITTI (*)
1. Introduction.
In this paper we
study
the existence of anonnegative
nontrivialsolution of the Dirichlet
problem
where Q is an exterior domain
(i.e.
S~ = and 0 is aregular
boun-ded open set in
R~)
andf
is a continuous function such that there existsSince the domain S~ is
unbounded,
the Sobolevembedding
c
Ho (Q), 2 ~ p 2n/(n - 2),
is notcompact,
and standard variationaltechniques
do notapply. However,
it ispossible
to usecomparison
methods between
equation (1.1)
and the«equation
atinfinity
for which existence results are well
known,
because of its radial struc- ture(see
forexample [1],
and[2]).
These methods were first introduced
by Ding
andNi [3],
andP. L. Lions
[4], [5],
whoproved
an existence result for(1.1),
when(*) Indirizzo dell’A.:
Dipartimento
di Matematica, Piazza di Porta S. Donato 5, 40100Bologna,
Italia.S~ =
]R~
andusing
asimple
variant of the concentrationcompactness
method.Then Benci and Cerami
[6]
and P. L. Lions[7] proved
arepresenta-
tion theorem for all the Palais-Smale sequences of
(1.1),
whichgives
aprecise
estimate of the energy levels where the Palais-Smale condition for the functional related toequation (1.1)
can fail.Hence, using
a minimax method Benci and Cerami[6]
obtained anexistence result for
(1.1),
when Q is «almostequal »
toR~,
and
(1.3)
has anunique positive
solution.(This
last condition isactually
a consequence of the
hypothesis f ~ (u) _ ~
because of a later re-sult of
[8]). Very recently
Bahri and Lions[9] improved
thisresult,
andshowed that
(1.1)
has a solution when S~ is anarbitrary
exteriordomain,
and as
x ~ -~ + ~ exponentially
fast. Theproof
isobtained
by studying
the energy levels found in therepresentation theorem,
with verypowerful algebraic topology
methods.Let’s
finally
recall that Coffman and Marcus[10]
obtained someexistence
theorems,
for(1.1),
when S~ is an almostspherically
symme- tric exteriordomain, by using
acompletely
differentapproach.
In this work
using
thetechnique
introducedby
Bahri andLions,
westudy problem (1.1)
forgeneral
exterior domains andgeneral
functionsf satisfying (1.2).
The paper isorganized
as follows: inparagraph
2 westate our main Theorem
(Theorem 2.1),
andgive
someexamples;
in 3we introduce some
notations,
and recall some well known results wewill
apply
in thefollowing section;
in 4 wegive
theproof
of Theorem2.1;
theproofs
of some rather technical lemma are collected in 5.Acknowledgment.
I’m verygrateful
to Prof. M.Ferri,
whokindly helped
me inunderstanding
sometricky algebraic topology arguments
in
[9].
2. Let’s consider the
following problem
where Q is an exterior domain. Assume that a is
bounded, measurable, nonnegative
and such that(2.3)
3c >0, e
>0,
R > 0 such thata(x) ~
We will also assume that
f :
~ x 1R ~ R iscontinuous, strictly positive
for t >
0,
and t--~ f(x, t)
is of classC
and odd for every x E Q. Moreover there exist p, q such that 2 p q2n/(n - 2),
andfor every fixed x E ~2 .
Finally,
we will suppose that there exists a convex functionfoo
of classC
1 such thatUnder these
hypotheses
we prove thefollowing theorem,
which is ourmain result.
THEOREM 2.1.
If
the«equation
atinfinity»
associated to(2.1 )
has a
unique- around
statesolution,
then(2.1)
has at least a solu- tion.We recall that a
ground
state solution toequation (2.7)
is a classicalstrictly positive
solution w such thatm(r)
- 0 as r - + 00. It is well kno-wn that every
ground
state of(2.7)
isradially symmetric
withrespect
to a
point
in R~. In Theorem 2.1 we meanuniqueness
of theground
sta-te
radially symmetric
withrespect
to theorigin.
With the same methods we are
going
to use in Theorem2.1,
it is also
possible
to proveTHEOREM 2.1’. Under the same
hypotheses
as in Theorem 2.1 the DirichLetproblem
has at least a solution
if
aij = aji E L 00(Q) for
everyi, j,
andwhere
aij
is the Kroneckerfunction.
We will prove
only
Theorem2.1;
with easyadaptations,
it is pos- sible to obtain also theproof
of 2.1’.Using
theuniqueness
theorems forground
state solutionsproved by Kwong
andZhang [11],
we cangive
moreexplicit hypotheses
onf~ ,
which are sufficient to
apply
Theorem 2.1(or
Theorem2.1’ ).
COROLLARY 2.2. Assume that
f
andfoo satisfy
all theprevious hypotheses,
and let 0 > 0 be such that - u +foo (u)
0 in(0, 0),
and- u > 0 in
(e, + (0). If
then
(2.1)
or(2.1)’
has at least a solution.PROOF.
By (2.4)
the function isincreasing
on(0, m ), hence,
sincefoo
isCB (p - Consequently
Besides
(( - u
+ufl (u))/( - u + f~ (u)) ~
1 in the interval(0,0).
Hen-ce,
by
Theorem 1 in[11], (2.7)
has aunique positive
solution.Significative examples coming
withinCorollary
2.2 are the fol-lowing
ones:extended as odd functions to u 0.
For somewhat different
examples
we also refer the reader to the paper on theuniqueness
of radialground
states of MacLeod and Serrin[12].
3.
Preliminary
remarks and sketch of theproof
of Theorem 2.1.Let us call
and
I is of class
CB
and its criticalpoints
are the weak solutions of(2.1).
Analogously
we define I ~ the functional associated to theequation
atinfinity:
Finally
let’s callREMARK 3.1. u is a weak solution of
(2.1)
if andonly
if u is a criti-cal
point
of(cf. [5],
page266).
Moreover11M
satisfies thefollowing properties:
such that
k(u) u E M,
and B:1m >0 ,
Let us recall the
proof
of(3.1).
Sincek(u)
u EM,
On the other
side,
Analogously,
if1),
then(3.1) immediately
follows from(3.2), (3.3)
and(3.4).
REMARK 3.2.
If problem (2.1)
has nosolutions,
thenIndeed if
(3.5)
is nottrue,
then there exists u E M such thatI(u) =
= inf
I, hence, by
the firstpart
of Remark3.1, u
is a solution of(2.1).
Moreover,
if(3.6)
does nothold,
inf I inf100,
and(2.1)
has a solutionM Moo
by
thecomparison
theorems of[3]
and[5].
In the
sequel
we willalways
assume thatequation (2.7)
has anunique positive
solution m(up
to a translation in theindependent variable).
This function is
radially symmetric
about somepoint
inradially decreasing,
and if r is the radialcoordinate,
there exists C > 0 such that((3.9)
is wellknown,
for(3.7)
and(3.8)
cf.[13]).
We also recall the
following representation
theorem for the Palais Smale sequences related to the functional I:PROPOSITION 3.3. Let be a
nonnegative
sequence inHo’(0)
such that - c and - 0 as h - + 00. Then there exist a
number
7% e N,
and m sequences in =1,
... , m, such that-~
every i~ j ~
-~ + 00 as h -~ + 00 andWe omit the
proof
of thisproposition,
which can be obtainedusing
concentration
compactness principle
as in[6], [9] (see
also[14]).
Let usonly
noteexplicitly
that the Palais Smale sequences of the functionalare also Palais Smale sequences of I.
COROLLARY 3.4.
If (2.1)
has nosolutions,
andsatisfies
thesame
assumptions
as inProposition 3.3,
thenThese results
point
out the crucial roleplayed by
the level sets ofI,
which
corresponds
to aninteger multiple
of I~(c~),
andby
the sets of fi-nite sums of translations of m. Hence the idea of
[9]
is to defineand
and to
study
theirtopological properties
with a deformation argu- ment.For every uo E
Wm
we consider theCauchy problem
This has a
unique
solution t -u(t, uo)
defined on all ofR,
such thatu(t, uo)
E M + forevery t
E R. Let’s denoteT~(uo)
is continuous as a function of uo , for every 6 ~ 0and,
if(2.1)
hasno
solutions, T~(uo)
+ 00 for every d~ > 0. Hence we can defineand
Then the
following
lemma holds:LEMMA 3.5.
(2.1)
has not asolution,
then(Wm, retracts by deformation through
r on(W m -1, Wm-i),
andfor
everys > 0 there exist 8 >
0,
ël e(0, ~)
such thatand
PROOF. This lemma can be
proved
with the sametechniques
as(4.7) in [9].
For the sake ofbrevity
we omit theproof.
Now,
tocomplete
the sketch of theproof
of our Theorem2.1,
we in- troduce somealgebraic topology arguments.
Toexplain
the mainidea,
we first describe it in a
particular
case which is howeveralready
con-tained in
[9].
We assume that a isconstant, f (x, u)
= u, and Q isan exterior domain. Let’s denote
where k is defined in Remark 3.1
and ?
is a C °° function suchthat 9
= 0in
Il~n ~~’, ~
= 1 in aneighborhood
of + w , and 1. In order tostudy
thehomology
groups ofBm
Bahri and Lions also need to intro- duce suitable manifoldsTm
such that B:1m E N(3.15) (Tm ,
has the samehomology
groups as and callh*
theisomorphism
between these groups.With these notations
they
prove that there exists a natural number~ such that
Then
they
assumeby
contradiction that(2.1)
has nosolutions,
andthey
built a suitable commutative
diagram involving
the groups(which
are not trivial because=
Z2
for every m ~li).
This is done in twosteps.
The first is based on a
simple property of Bm.
Asc
Bm , [9]
can consider thefollowing
commutativediagram,
where a is the
connecting homomorphism
and i is theembedding
(3.16). By (3.15)
aconnecting homomorphism
is also induced on(Tm , aTm),
and thefollowing diagram
commutesIn the second
step
Bahri and Lions use a deformation lemma(which
isanalogous
to our Lemma3.5),
the cupproduct
and someproperties
ofTm
to defme twohomomorphisms da and da
such that thefollowing
dia-gram is commutative
-
and sends the
generator
of to thegenerator
of·
In this way
they
build this chain ofhomomorphisms:
Since is zero
(by (3.16))
and issurjective,
then also~
1 ( Wl )
has to be zero. However a direct compu- tation shows thatZ2
and is anisomorphism.
This con-tradiction proves that
(2.1)
has a solution.In the
general
case, whenf
is not a power, but itmerely
is a func-tion
verifying
the behaviorhypotheses (2.4)
withf~
convex, we can notprove
(3.16),
and we can not work as before. However we show that there exist twoneighborhoods
ofrespectively
calledBm-i
1 andsuch that
has the same
homology
groups as(B~ ,
and
for a
suitable g
e N. Hence with more technicalarguments,
we built acommutative
diagram analogous
to thepreceding
one and we concludethe
proof
of Theorem 2.1.4. Proof of Theorem 2.1.
We argue
by contradiction,
and we assume that(2.1)
has no solu- tion. Hence inparticular (3.5)
and(3.6)
in Remark 3.2hold,
andWo=0.
STEP 0.
Definition of
themanifold Tm
=Sf)
1 andof
anhomomorphism
between a suitable subsetsof
it and(this argument
isanalogous
to(3.15)).
,S is a
(n - 1)-dimensional sphere
embedded in Q such that hS c Q forevery h a
1(it
is not restrictive to assume that ,S is asphere
of radius1,
because this may be achievedby
asimple scaling).
Let crm be the group ofpermutation of {1,.... m},
9Dm = ~ (xi , ... , xm)
E= xj 1
9Dm
andDm
be aa-m-invariant
tubularneighborhoods
ofDm ,
such thatDm cc Dm
andSo
= Next we denoteThe group om acts on S m x
1 and So
x we will denoterespect- ively
S m 1 and,So
1 thequotient
under the action of om.Let us
finally
introduce the continuous mapwhere m is the
unique
solution of(2.7), l~
is defined in Remark 3.1 and pis a fixed function of class
C °° ,
which isidentically
zero in 1 in aneighborhood
of + 00, and 0 ~p K
1 inREMARK 4.1.
If
we setthen
Bm c
M + Vm E I~. Moreover hdefines
anhomomeorphism
between
PROOF. We recall the
proof
of thisremark, already
contained in[9]
(Lemma III.1),
because we areusing
a different notation. We willonly
prove
that,
if(xl , ... , xN)
are distinctpoints
inR~.
yl , ... , YN E II~n andthen Yi = ...
= YN = 0.
Integrating
theprevious expression
on weget
Let’s make the
change
of variable y - xi = zDifferentiating
withrespect to ~,
for everypolynomial
p we haveConsequently, (taking
into account that w isalways strictly posi- tive)
and we
conclude,
since thepoints xj
are distinct and1I(z
+ = 1 ifz ~ I
is
big.
STEP 1.
Proposition 4.2,
and thesubsequent
assertion(4.4)
areanalogous
to(3.16),
and allows us todefine
adiagram analogous
to(3.17).
PROPOSITION 4.2. Let s, El, 6 be the real value
defined
in the de-formation
Lemma 3.5. Then there exists ,u EN,
Y) E[0,1]
and ~ > 0 suchthat
where h is
defined
in(4.1 ), Wm
in(3.10)
andV(m, ë)
in(3.11 ).
h also sends anysufficiently
smallneighborhood of 3(Sm
0 XDnm-1)
to5m
(We’ll
prove thisproposition
in Section5.)
Let i and be
neighborhood
of which retractby
de-formation of and such that the closure of
4n -I
1 si embedded inthe interior Then also =
h(Dm
U s mX 3m-I)
and=
h(Dm
areneighborhoods
of i whichretract
by
deformation on If we call r r thedilatation,
from Remark 4.1 and
Proposition 4.2,
it follows thatHence we can consider the
following diagram,
which is commutative:where
a *
are theconnecting homomorphisms, j * :
-
H* B Bm - 2 )
is inducedby
the naturalembedding, r*
is inducedby
the retraction ofB,,,-,
on p* and q* are theprojection
on thequotient.
Let us recall that~’3~
isexactly
the natural connectionhomomorphism
of thetriple (Wm, Wm-1, Wm-2).
On the otherside,
since is an
homomorphism,
Hence the function
naturally
induces anhomomorphism,
which we call
again a * ,
such that the
following diagram
is commutativeSince we have assumed that
(2.1 )
has nosolutions,
from the deform-ation Lemma 3.5 it follows that
The natural
embedding
is an
isomorphism,
and induces anhomomorphism
such that the
following diagram
is commutative:vious
diagram simply
reduces toSTEP 2.
Definition of
twohomomorphisms da
andda analogous
to(3.18).
PROPOSITION 4.3. There exists a continuous
function
such that the
following diagram
is commutativewhere t is the
projection
on thefirst
coordinate(cf. [9], Proposition iii.1).
Let’s denote
the
homomorphism
induced on thecohomology
groups. Then for everywe can defme
where n is the cap
product,
and thefollowing diagram
is commuta-tive :
Moreover
(cf. [9], (5.9)),
we haveand there exists such that
sends the
generator
in thegenerator.
In this way we have defined a commutative
diagram
where
(8 . d~ ) ... (8 . d~ )
is anisomorphism.
STEP 3. Conclusion
of
theproof.
Since
Wo
=0,
andSo
=,S,
thisdiagram
becomes:By (4.2)
= 0.Hence, by
thecommutativity
of thisdiagram,
also
’
is
identically
zero.On the other side
by Proposition 4.3,
onV( 1, ë)
isdefined,
afunction
which is a left inverse of
h,
hence 1 can not be zero. This contradic- tion proves that(2.1)
has a solution.5. Proof of
Proposition
4.2.Let’s
begin
withproving
some remarks.PROPOSITION 5.1. From the
hypotheses
we have made onf, if foLLo-
ws that
for
every(aI,
... ,am)
suchthat ai ~
0for
every i =1, ...,
0. Letus note
explicitly
that(2P - 2)/(p
+ 2P -2)
>1 /2;
this will be crucial in theproof of Proposition
4.2.PROOF.
(5.1) immediately
follows from(2.4).
Let’s prove(5.2).
Since
f ~
is convex, for everyfixed j E ~ 1, m I,
we haveMultiplying by
aj weget
Hence, summing on j,
weget (5.2).
Inparticular
Then, dividing by
, we haveLet’s now prove
(5.3)
when rrz =2;
in other words we will prove that for every a, b >0,
(since f’
isincreasing)
(if
we make thechange
of variable s = r + 0 in the secondinteg- ral)
(5.3)
can now beproved by
induction on m,using
thisinequality
and(5.4).
PROPOSITION 5.2. E be
radically
sym-metric,
and assume that3a, f3
> 0 3C E R such thatthen
for
anarbitrary
VI v I
= 1(the left
hand side isindependent of
v,since ~
is a radialfunction).
Now the result follows
by
dominated convergenceTheorem,
as in[9]
Lemma 11.2.
Let us now prove
Proposition
4.2.By
the definition of(,So -1 )m
for every there exists ym > 0 such thatWe can also assume that
where e is defined in
(2.6)
and aoo in(2.2).
Because of the definition of h(see (4.1)),
we haveonly
to prove thatsuch that
Indeed
(4.3)
is very easy if m = 1. If m ~ 2 and x ==
(x1,
... , Xm,tl,
... , E(,So -1 )m
x from(5.5)
and(5.7)
it followsthat if
then, by (5.6)
and(5.8)
x e
if x
E xthen x
Eby (5.9).
Finally (5.10), together
with(5.8)
proves(4.2).
We first prove
(5.10).
Ifby
contradictionOn the other side the
(n - I)-measure independent
of xi. Hence if we callit 1,
=
C(8),
thenand this is a contradiction.
Let’s now prove
(5.7).
/ m BWe’ll first show that
B
~=1ti wi /
2013~~ as1 /m
where k is the function defined in Remark 3.1
(since
w is a solution of(2.7))
uniformly
withrespect
to(xl ,
... ,Xm)
E(So -1 )m. Analogously
and
mti
- 1. Hencekti
- 1 as A -~ + oo,and ti
-~1 /m,
andm /
we conclude that
V(m, ~)
here and in thesequel
we write ki=l
instead
B
~=1 1//
.Let’s prove
(5.8).
We will denoteFirst recall the
following
estimates(see [9], 6.20)
Now,
since (JJ is a solution of(2.7),
we haveand,
if we assume that ad£,
fromProposition
5.2 we haveHence we
get
Let us estimate the second therm in the functional I
(by (2.4)
and(2.6))
(by Proposition 5.2)
(by 5.3))
(by 2.4))
Hence
By
theproof
of(5.7),
forevery i
=1,
..., m,kti
- 1 as À - + 00 andY) ~
0,
henceOn the other
side, a’/(2
hence fromProposition 5.2,
it follows that there existI, j e (I, ..., 7%)
such thatConsequently,
if ~ isbig enough
weget:
(since
the functionhas a maximum at the
point (1, ...,1)~
(since
is a solution of(2.7))
Let’s now prove
(5.9).
We’ll
begin
withestimating
Arguing
as before weget
Then, by
Remark 3.1It is not difficult to prove that the function
m
constrained on the
manifold Eti = 1,
has a strict maximum in thepoint (1/m, ...,1/m),
then317
> 0 such thatVY) 17 30(Y) 3À(y): V(ti ,
... ,tm)
E~~_i,VA>~
Consequently
there exists).(r¡)
and such that k m -Now
The function
has a strict maximum at the
point ( 1...1 ).
Since Ym , we can assume thatbelongs
to anarbitrary
smallneighborhood
of 1 forevery i =
1,
..., m. On the otherside,
from the estimate we havejust obtained,
weget
Hence 30 > 0 such that
if ~ is
big enough.
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