A NNALI DELLA
S CUOLA N ORMALE S UPERIORE DI P ISA Classe di Scienze
C ARLOS E. K ENIG
W EI -M ING N I
On the elliptic equation Lu − k + K exp[2u] = 0
Annali della Scuola Normale Superiore di Pisa, Classe di Scienze 4
esérie, tome 12, n
o2 (1985), p. 191-224
<http://www.numdam.org/item?id=ASNSP_1985_4_12_2_191_0>
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Elliptic Equation [2u]
CARLOS E. KENIG
(*) -
WEI-MING NI(*)
1. - Introduction.
Let
(M, g)
be a Riemannian manifold of dimension 2 and .g be agiven
function on M. One may ask the
following question:
can we find a newmetric gl on .M such that K is the Gaussian curvature of gl and gl is
con f ormaZ to g (i.e.
there exists apositive function
on .M such that g, = Ifone
writes (p
= exp[2u],
then this isequivalent
to theproblem
ofsolving
the
elliptic equation
on
M,
where are theLaplace-Beltrami operator
and the Gaussian curvature of M in the metric g. Thisproblem
has been consideredby
many authors. When M iscompact,
we refer to Kazdan an Warner[KW]
for detailsand references
(see
also a recent surveyby
Kazdan[K]).
When M is~2, and g
is the usualmetric, equation (1.1)
reduces toon R2. For
equation (1.2),
there have been some non-existence results due to variousauthors, e.g. Ahlfors [A],
Wittich[W],
Osserman[Os], Sattinger [S],
Oleinik
[O],
Ni[N1],
etc. The first existence result forequation (1.2)
seemsdue to
Ni [N1]
in the case Itcomplements previous
non-existence results. Theapproach
used in[N1]
is via the barriermethod,
i.e. super- and sub-solutions.Using
anentirely
differentapproach, namely, weighted
Sobolev spaces, McOwen
[M1]
refined some of Ni’s resultsby obtaining
more
precise asymptotic
behaviour of the solutionsgiven by
his method.(*)
Both authorsare partially supported by
NSF. The first author is also par-tially supported by
the Alfred P. Sloan Foundation.Pervenuto alla Redazione il 10 Febbraio 1984 e in forma definitiva il 14 dicem- bre 1984.
More
recently,
Aviles[Av]
and Mcowen[M2]
used thisweighted
Sobolevspace
approach
to obtain some existence results when .Kchanges sign
anddecays
at 00.Returning
to the moregeneral equation (1.1),
we observethat,
undersome
appropriate hypothesis
on the metric g,equation (1.1 )
may be reduced toequation (1.1)
via the Uniformization Theorem.However,
in order toapply previous
results to this reducedequation (1.2),
it is essential to haveasymptotic
estimates for this conformaldiffeomorphism
99guaranteed by
the Uniformization Theorem. Forexample,
thefollowing
estimatesfor soine
positive
constant C and for[x[ I large,
would be sufficient. This kind of estimates have been establishedby
Ahlfors and Bers[AB]
underthe condition that the metric gii =
~,2(x) ~i~
outside acompact
set in R2 for some > 0. We would also like topoint
outthat,
in the abovereduction,
it is
important
that k is the Gaussian curvature of(11~2, g). Therefore,
froma differential
equations point
ofviaw,
a naturalquestion
arises: can we solveequation (1.1)
if k or isperturbed?
i.e. whathappens
to thesolvability
of
(1.1)
if wereplace
the curvature kby something
similar or if 4 - isreplaced by
anelliptic operator
ofdivergence
form? The Uniformization Theoremapproach
does not seem togive
an answer.The purpose of this
present
paper is tostudy
thesolvability
of equa-tion
(1.1)
as well as toinvestigate
the abovequestion.
We shall free our-selves from
geometry
and discuss theequation
on
R2,
where .L = isuniformly elliptic, k,
..g aregiven locally
bounded functions with
appropriate
conditions at oo. It turns out we canextend the results in
[N1]
to theequation (1.3).
Weagain adapt
the barriermethod used in
[N2]. However, in
thepresent
case, it is much more involved to construct barriers. The newingredient
that we need is the construction of entire solutions to the linearequation
Lu =f,
withsharp asymptotic
behavior at oo. We
accomplish
thisby adapting
atechnique
of Friedman[F].
Incidentally,
a refinement of theargument gives
the existence of aglobal
fundamental solution for
L,
with estimatesdepending only
on the uniformellipticity
constants of L. While this fact is not needed in the paper, we sketch itsproof
in anappendix,
since we have been unable to find aproof
of itin the
literature, and
it may prove useful in the future.Once
sharp
results for the linearequation
Lu =f
areobtained,
we usea
rearrangement inequality
of Talenti[T],
to be able to construct solutions to(1.3)
underoptimal assumptions
on K.In the case when L is
degenerate,
i.e.(aii)
isuniformly elliptic
oncompact
sets and its
eigenvalues
behave likeIxlb
at oo with0,
this barrier method also works. In these cases, weemploy
thetheory
ofdegenerate elliptic equations developed by Fabes, Jerison, Kenig
andSerapioni [FKS], [FJK].
Weaker versions of our
present
results and methods in the case when L isuniformly elliptic
were announcedby
us in the AMS SummerMeeting
of 1982 at Toronto
(in
thespecial
session (Nonlinearpartial
differentialequations
inphysics
andgeometry))) August, 1982,
and at the 803rd AMSmeeting
at New YorkCity (in
thespecial
session «Variationalproblems
in Riemannian
geometry), April,
1983(see [KN2],
The methods and the flavor of this paper arequite
close to ourprevious
work on the corres-ponding problem
forprescribing
scalar curvatures[KN1].
The
organization
of this paper is as follows. In section2,
we treat a modelcase of
(1.3)
when L _--_L1,
the standardLaplacian.
In section3,
we extendthese results to the case L is
uniformly elliptic.
In sections 4 and5,
we treatequation (1.3)
in the case when L isdegenerate. Finally, in
theappendix
we prove the existence of a
global
fundamental solution foruniformly elliptic operators,
y withprecise asymptotic
behavior.In
closing,
we should mention that L.Karp [K1)]
has some results related to ours.ACKNOWLEDGEMENT. Part of this research was carried out while the first author was
visiting
the Centre for MathematicalAnalysis
at the Austra- lian NationalUniversity,
yCanberra,
y Australia. The first author wishes to thank the Centre for its financialsupport,
and its staff for their warmhospitality during
hisstay
in Canberra.2. - The
special
case L - L1.To make our
approach
moretransparent,
we firststudy
the casei.e. the
equation
are
given,
and we areonly
interested in entire solutions on 1~2.In this
section,
we shallpresent
some existence resultsusing
the barrier method. We also include a non-existence result and anexample
which(together)
show that our existence result(Theorem 2.2)
issharp.
THEOREM 2.2.
Suppose
at co
for
some E > 0 on R2. andat oo
for
some a > thenfor
everyR2
equation (2.1)
possesses an entire solution u with(.611 CXJ.
PROOF. We use the
super-and
sub-solution method(See,
e.g.[KN1],
or
[Ng])
which isby
now well-known.First,
we construct a sub-solution. Letwhere 6 = a2. An easy
computation
shows thatLet v. solve more
precisely,
letBy dividing
R2 into the threeregions {y
E R2:lyc-R2 2[r[
>>
Iyl >
and{y
we may estimate v8. It is not hardto obtain
that,
at ooFor
simplicity,
we setand then
at oo.
Now,
let u2 = v2-~- C,
whereO2
is a constant to be chosen later. Wecompute
From the estimate of v, and the
hypothesis
on the behavior of K at oo,we have
at oo. Since
we
always have 12 -~-
IT exp[2v2]
> 0 at oo. For the finitepart,
since12
> 0 in is bounded belowby
apositive
constant. It is thuspossible
to chooseC2
to be alarge negative
constant so thaton R2 .
Therefore, u2
= V2+ O2
whereC,
is alarge negative
constant is asubsolution.
Next,
we construct asuper-solution
U, with on R2. Without loss ofgenerality,
we may assumeK(0 ) fl
0. LetBe
be a ball ofradius C)
centeredat 0 and contained in the interior of the
support
ofK,
and letfi(z)
=be a radial
function, positive
in withoutside Be
and(i.e. fi
islarge
andpositive
inBQ~~).
Let
Similarly, dv1
=h
and at o0Let ~c1 = vl
+ C1
whereCI
is alarge positive
constant to be chosen later.We
compute,
yOutside
B~ ,
7 ThusII -
k-E-
.K exp[2v,]
exp[2 C1] c 0
theresince on ll$2. In
B~, fi - 15
is(fixed
andtherefore)
bounded whilek- C 0
onB~,
y and soon B~ ,
7thus,
1 we may choose01
> 0 solarge
thaton
B~ .
With this choice ofCi ,
we haveon R2 and u, is a
super-solution. Now, comparing
c1andU2
we see that atinfinity,
yIt is therefore
possible
to chooseCI
> 0 evenlarger
to assure that ~c1 > Uzon R2
(the
finitepart
does not create any difficulties atall).
Now,
our assertion followsfrom,
e.g. Theorem 5 in or Theorem 2.10 in[N2] .
REMARK 2.3:
(i)
Note that there is nosign
conditionimposed on k,
and the constant acould be
large negative (it depends only
on In this caseis actuaily
R2
allowed to have
polynomial growth
at oo.(ii)
The bound on oc isactually
bestpossible by
the nonexistence result below and theexample
at the end of this section.(iii)
LetIt is easy to see that
by setting
w = u - v,equation (2.1)
reduces to equa- tion(1.2)
and therefore Theorem 2.2 follows from theprevious
work in[N1].
We
prefer
our firstproof
because itgeneralizes
to the moregeneral equation (1.3)
in itspresent
form andthereby
ourapproach
is madetransparent.
Furthermore,
the idea in our firstproof
may beapplied
to handle the situa- tion where k is notintegrable (see
the remark below and Theorem2.4).
(iv)
Our method issimple, elementary
andyet
very flexible. Fork,
~ with otherasymptotic behavoirs,
it is stillpossible
to use this barrier method to obtain someresults;
we shallonly
include one of those resultshere, just
to illustrate thispoint.
THEOREM 2.4.
Suppose k,
K are bothnegative
on 1~2.If
there exist constantsC2 , C4 , ~1, 12, 13
allpositive,
such that,and
at oo,
then, equaction (2.1) possesses act
least one entire solution on R2.PROOF. The
proof
goesalong
the same line as that of Theorem 2.2.First,
let11
be a functionpositive
in the unit ballB1, identically
zero outsideBi
and withwhere a is an
arbitrary
number which isbigger than -!
max(I + l2, 2 + Z2.
Let
then
4wi
=f 1
and at oo,Set ul
= V,-~-- C5, where 05>
0 will be chosen later.Now, compute
At
infinity,
we haveThus,
at oo,since 2a -
Z2 > i .
For the finiteregion,
sincef 1-
k isfixed,
andK 0,
we can choose 0
so large
that the aboveinequality
also holds on, say, With this choice of
C~,
we see that0,
is a
super-solution.
Now,
we areready
to construct a sub-solution ~2 with on R2. Let~2(x) c 0
on l~~ andfor every with
(Note
that this ispossible
sincek(x) c-
at00).
Set U2 = v2 -06,
where
06
0 will be chosen later andThen, similarly, Llv2 = f 2
and at co,Therefore,
at o0since
06
0 and 2a -1:,
> 2. For the finiteregion,
say~x~
we can chooseC6 large negative
so thatsince k 0.
Thus,
u, is a subsolution ifCg
islarge negative.
Since a >0,
U2 at oo. Thus on R2
(by varying 06 again
ifnecessary)
andour result follows.
The
argument
we use inproving
thefollowing
non-existence result goes back to Wittich[W]
and has been usedby
various authors(see,
e.g.[S], [Ni]).
Wepresent
here a variant of thistechnique.
THEOREM 2.5.
I f k > 0
on R2for
where00, e
arepositive constants,
theneq2cation (2.1)
possessesno entire solution on l~2
for .K c 0
on ~2 withat oo,
f or
some aREMARK 2.6.
Following
thepointwise
conditions on k and K may bereplaced by
someappropriate integral
conditions. To thisend,
we
define,
as in[N,],
where Sr
is thesphere
of radius r centered at theorigin,
and dS denotesthe volume element of Note that
k -
.K if ~ isradially symmetric
and use the convention that
.9(r)
= 0 ifNow,
the conclusion of Theorem 2.5 still holds if wereplace k by k
andK
by k.
PROOF OF THEOREM 2.5.
Suppose u
is a solution of(2.1)
on R2. Setv = - ~,
equation (2.1)
becomesUsing
Green’sidentity,
Jensen’s and Schwarz’sinequalities
we arrive at(see
p. 350[Ni]
fordetails)
Relabel
this, writing
v,k,
.g instead ofV, k, IK.
We haveMaking
thechange
of variables r = exp[t], m(t)
=v(egp [t]),
weget
(2.7)
m+
exp[2t] k(exp [t] )
exp[2t] K(exp [t] )
exp[- 2m],
where the dot ~ means differentiation with
respect
to t. Observe that0, (since K 0).
Let to
=log Ro.
Then--
ofr
t ~ to.
From(2.7),
weconclude,
fort ~ to
Integrating,
for
Integrating again,
we obtainfor Denote the constants
by
(2.8)
reduces tofor all
By (2.7)
and ourassumptions on k,
where ð>
0 is a small constant to be chosen later.By (2.9)
and ourhypo-
thesis on
K,
we derive thefollowing
estimate fort > to
if we choose 6 = 2 -
oc/JL (Note
that ð> 0by
theassumption
ona).
Thus(2.10) becomes,
fort>to,
We now
proceed
toget
a contradiction.Multiplying (2.11) by
m(which
reverses the
inequality)
andintegrating,
with D =
(2 C16)
exp[-
Notice that exp[- + D
is
nondecreasing
and ism2(to) (which
may be taken to bepositive)
at t =to . Integrating (2.12)
fromto
tot,
we obtainfor all
t ~ to . Letting
t we obtain a contradiction.EXAMPLE 2.13. Define
where 8 > 0 is an
arbitrary
constant. Then the constant in Theorem 2.2 iswhich is
precisely
the constant thatappeared
in Theorem 2.5. Thus we learn that theexponent
is the
dividing point
for existence and non-existence. We should also remark thatalthough
the function k in(2.14)
is notcontinuous, it
is clear that we can smooth out thediscontinuity
of k andyet
maintain itsintegral
as closeas we want to the
original
one.3. - The
uniformly elliptic
case.The main
goal
of this section is to extend the resultin §
2 to theequation
on
R2,
where(we
use the summationconvention),
and thematrix is
symmetric, bounded, measurable,
anduniformly elliptic
on R2 . The first
difficulty
one encounters intrying
togeneralize
theproofs
in section 2 is the existence of
global
solutions to Lu =f in R2,
withprecise asymptotic
behavior at oo. In order to do so weadapt
atechnique
due toFriedman
[F]. A
refinement of theargument
alsoyields
the existence ofa
global
fundamental solution forL,
with estimates whichdepend only
on the
ellipticity constants,
for L. This is notactually
needed in our paper, ybut,
since we have not been able to find aproof
of this statement in the litera-ture,
we will include it in anappendix,
for future reference.Using
our results on the linearequation
Lu =f,
we thenproceed
to solve(3.1), obtaining
ageneralization
of Theorem 2.2. When L =LJ,
our theoremreduces to the
sharp
result in Theorem 2.2.We assume
throughout
thatand
that,
if = then Sometimes we will referto I and !~. as the
ellipticity
constants of Z.DEFINITION 3.2. Consider the reflection
(Kelvin transformation) given by
Solutions u to.Lu(x)
=f (x)
inIxl
> 1correspond,
underthis reflection to solutions to
Lv(y)
= in 0jyj 1,
where= =
ôz,
and(see [KN1],
Lemma2).
Since the matrix(bi,-
isunitary, (akZ)
satisfies the sameellipticity
estimates asLet g
be the Green’s functionfor
L,
withpole
at0,
in the ball~y~ 1.
Letg(x)
= -g(x/lxI2). g(x)
willbe called the Green’s function for .L in
Ixl
>1,
withpole
at oo. It satisfies thefollowing properties:
Moreover,
there exist constants01
andO2,
whichdepend only
on A andA (and
which can be estimatedexplicitly
in terms of A andA),
such thatfor For this last
property,
see[LSW].
THEOREM 3.3.
Suppose
thatf (x)
islocally
bounded onR2, f
isintegrable
on
R2, II(X) IV
oo,for
some p > 1.Then,
there exists a solu- 1.1>1tion u to Lu =
f on R2..M’oreover,
the solution u has theasymptotic
behavioru(x)
= =0(l),
as 0153 -+ oo, whereg(x)
is thefunction for
Lin >
1,
withpole
at oo.PROOF. As mentioned
before,
weadapt
atechnique
due to Friedman( [F~ ).
We
proceed by steps.
Step
1. SolveIn order to do so, we solve
Since =
for some p
>1, by our assump-
tion on
f
the existence and boundedness of follows from[LSW].
Wethen set
wi(z)
=Step
2. SolveThis
step
is clearby [LSW].
Step
3. Let 81 ={0153
E ll~2 :Ixl = 1 ~.
Define anoperator
W inR) by
thefollowing procedure.
Fix an
hEO(81;R),
(To
do so, it isenough
to solve(This step
is clearby [LSW]).
(iii)
DefineW (h)
=z yx~ =1.
SinceW(h)
is a Holder continuous func-tion,
it is easy to see that W is acompact
linearoperator.
Alsoand
W(1)
= 1.Thus, 11 W ~~
- 1.Olearly I
= 1 is aneigenvalue. Suppose
that h is a
corresponding eigenfunction,
i.e.W (h)
=h,
and suppose that h is non-constant. Then w and z are bothnon-constant,
and= h,
whilezlx =1 -
ffiax x ax Ih(x)f.
Thisclearly
violates thestrong
maximumprinciple. Thus, h
must be aconstant,
and theeigenspace
is1-dimensional.
By
the Fredholmtheory,
the corange of I-W has dimension1,
the kernel of I-W* also has dimension
1,
and if Eker(I-W*), C
is inthe range of I-W if and
only
if = 0.Step
4.ft(Z2)
~ 0. Ifnot,
Z2 is in the range of I-W. Thus we can find hso that z,
+ W(h)
= h. Define nowu(x)
on R2by
Then, u
is welldefined,
since z,+ z
= w+ g
for 1Ixi
2.Moreover,
Lu = 0 in R2.
But, w
isbounded,
andg(x)
behaves likeloglxl
at oo-whichcontradicts the minimum
principle.
Step
5. Let A _ -Then,
there exists h so that Zl+ + ~W (h)
= h. Let w, z be thecorresponding
solutions in(i), (ii)
ofstep
3.Define
Then,
it is easy to check that u is an entire solution of Lu =f
in R2.Step
6. ~, =If.
If we show thisequality,
since wi and ware boundedin
Ixl
>1,
the theorem will follow. LetThen,
Let
§5(z)
= andchange
variables.Then,
Set now =
q3(x) - 1,
so that1J’(x)
E ThenThus,
and our assertion follows.
Before we
proceed
toapply
Theorem 3.3 toequation (3.1),
we need ashapr
estimate on the function
g(x)
defined in 3.2.Assume that
Then,
we haveLEMMA 3.4. For with 0
~ 4n,
we have thatPROOF. We first note that the result is
sharp,
since when(aii(x»)
=I,
g(x) = 1/2n log Ixl. By reflection,
we have to show thatand
L
isgiven by
a matrixakl,
with Pickradial,
and = 1. Let = Let Ve solveThe results of
[LSW]
show that vE converges to(- g)
as E -> 0uniformly
on
compact
subsets{0}. Therefore,
wehave, by
Fatou’s lemmaWe are therefore reduced to
studying f
exp as s- 0. We willpi
appeal
to the results and notations of[T].
Notethat, by
the minimumprinciple,
Let nowv~ (y)
be thespherically symmetric decreasing rearrangement
of vE, i.e. the function from into[0, + oo),
whoselevel sets
y ~ 1, va ( y) > t~
are concentric discs with the same measure as the level sets~y : v~( y) > t~,
and such thatv~ (y)
= 0 for=
1.Clearly,
we haveThe main result in
[T]
shows thatv:(y) pointwise,
where solvesTherefore,
since #
> 0. We claim thatIn
fact,
and qg is radial.
Therefore, by
thesuperharmonicity
of(l/2jr) log zi)
as a function
of z,
this is smaller thanlog
The lemma nowsince follows
COROLLARY 3.5.
If ~, ~~ ~ 2 c ai j (x) ~i ~; ,
thenfor
everyfl,
with 0A4n,
rPROOF. Let
LA
=8;,
and g~ the Green’s function forL).
with
pole
at oo theng(x)
= Thecorollary
now follows from 3.4.We now
proceed
toapply
the results in Theorem 3.3 andCorollary
3.5to
study equation (3.1).
THEOREM 3.6.
If
at 00,
for
somepositive
constants s,C,
andK
0 onR2,
withat 00
for
some A > 0 andthen
for
every a, 0 C a~4y~ ~ C 2 [4~~, - 2 f 1~ - a], equation (3.1 )
possessesan entire solution u, with
PROOF. We follow the outline of the
proof
of Theorem 2.2. First weconstruct a subsolution. Set
By Corollary 3.5,
dz oo.Moreover,
y there exists p > 1 such thatChoose now 6 =
6(a)
> 0 soa/2.
Let nowf 2(x)
=Let
v2(x)
be the solutiongiven
in Theorem 3.3 toLV2
= k+ /~. By
Theo-rem
3.3,
and our choice off~ ,
we havev2(x)
=(f k + a/2 ) g(x) + ~(1)
at o.For
simplicity,
yset k
=f k.
Let~c2(x)
=v2(x) + A2,
where the constantA2
0 is to be chosen later. Wecompute,
y at oo,Since
g(x)
- oo as -> oo, this lastexpression
willbe > 0
at oo,provided
4nh - a -
2k -
2ac >0,
or acC 2 [4n2 - 2k - a].
This holdsby
our assump- tions on a and a. For the finitepart, f 2
is bounded from belowby
apositive
constant. It is thus
possible
to chooseA2
to be alarge negative
constantso that
on R2 .
Therefore,
u, is a subsolution.Next,
we construct asuper-solution
ui with on ll~2. Without loss ofgenerality,
we may assume(otherwise (3.1)
reduces to a linearequation,
which can be solvedusing
Theorem3.3),
and that.K(o)
~ p.Sup-
pose that Let = be a radial
function, positive
inB (!/2’ with
outside 9 and
Let
v1(x)
be the solution toLVI
=f 1
onR2, given by
Theorem 3.3. At oo=
(k + + 0(1 ).
Let ui = wi+ where A1
is alarge positive
constant to be chosen later. We
compute,
yOutside
Be, fi - Thus, f, -
k-~-
.K exp[2v,]
exp[2A,] 0 there,
since
.K c 0
on R2. InBe, tl - k
isbounded,
while.K c -
C 0 onB~,
Iand so
Thus,
we may chooseA,
> 0 solarge
thatWith this choice of we have
and ui is a
supersolution. Now, comparing
u, and u2, we see that atinfinity,
yIt is therefore
possible
to chooseA,
> 0 evenlarger
to assure that ul > U2on R2
(the
finitepart
does not create any difficulties atall).
The Theoremnow follows from Theorem 5 in
[KNI].
REMARK 3.9. When
Let f3 ==
2 -(aj2n) ,
if andonly
if a 4n -Thus,
in thiscase,
(3.8)
reduces to the condition in Theorem 2.2.Hence,
Theorem 3.6is
sharp.
REMARK 3.10. Recall that
I
for whereCi
andO2 depend only on A,
and can beexplicitly
estimated in termsof A,
A.Then,
if 4~~, -2 f k ~ 0,
a sufficient condition for(3.7)
to hold isthat at 00, for some On the other
hand,
if 4nh -2 f k 0,
a sufficient condition for(3.7)
to hold is thatI
at oo, for some
REMARK 3.11. Remark 2.3 and Theorem 2.4 can be carried out in this
more
general setting
in an almost wordby
word fashion. The details areomitted.
REMARK 3.12. It is easy to
apply
Theorem 3.6 to thegeometric equation (1.1)
to obtain existence results. The details are left to the interested readers.4. - The
degenierate
case b 0.We now consider
operators
L =aj)
where is auniformly
elliptic
matrix oncompact sets,
and whoseeigenvalues
behave like at o0 for some b 0.We shall follow the
approach
usedin §
3. Theonly
real modification is theappeal
to the results in[FKS], [FJK]
ondegenerate elliptic equations.
Because of the
similarity
of thearguments
with those usedin § 3,
we shallbe
sketchy, concentrating
on the main differences.Following
theapproach in § 3,
we first consider theproblem
offinding
solutions to
where
We start out
constructing
a Green’s functiong(x),
withpole
at o.LEMMA 4.1. There exists a Green’s
function gtx) for
L on[x [ > 1,
withpole
at 00. gsatisfies
and
for lxl
>2,
whereC,, O2 depend only
on theeigen-
values
of
theuniformty elliptic
matrixfor >
2.PROOF. g
will be constructedby
reflection. LetL
be theoperator
with matrixwhere
The size of the
eigenvalues
islyl-b,
with - b >0,
as y - 0. There-fore,
the results in[FKS], [FJK] apply
toL.
Letg(y)
be the Green’s functionfor
L
inIyl C 1,
withpole
at theorigin.
The existenceof g
isguaranteed by
the results in[FJK]. Also, by [FJK],
forIyl ~
where
w(B(o, s)) = f Izl-bdz
~ S2-b.Hence, - g(y)
~Iylb
forIyl I -1.
SetIzls
now
g(x) _ - g(x/~x~2). Then, g(x)
satisfies all therequired properties
THEOREM 4.2.
Suppose
thati
islocally
bounded onR2,
and thatIf(x) I
a = 2
+
e, 8 > 0 at 00.Then,
there exists a solution u to .Lu= f
on
R2. Moreover,
the solution u has theasymptotic behavior
=( f f ) g(x) + o(g(x) ~
as x -~ c>o, whereg(x)
is thefunction
constructed in Lemma 4.1.PROOF. We
proceed by steps, paralleling
Theorem 3.3.Step
1. Solvewith at 00.
We do this once more
by
reflection. Letl1h
solveSince
Lemma 2.7 in
[FJK] implies
thatWI(Y)
exists and isgiven by
where
h(y)
= As in theproof
of Theorem 1 in weestimate as follows. Write
the second term is
bounded,
and weonly
have to bound the first one for°
We first
give pointwise
estimates forg(z, y)
forIyl 2, Izl i.
Let~(~) == s)) === f
It is easy to check thatMoreover, by
Theorem 3.3 of[FJK],
whenever
d/2, d
=I1 -
We then see thatWe can now estimate the first term in
(4.3).
WriteIn the
region where [y] ~/2, ~z- y~ ~
so weget
In the
region
where]z] ~ Iyl I
andiz -
thus.In the
region where ly I > 2 z , ly 1,
and henceprovided
E C - b. We now setZul(x)
=Step
2. SolveStep
3. Define anoperator
W onR) by
thefollowing procedure:
Fix an h E
R) (i)
Solve(This
can be doneby reflection, using
the results in[FJK]).
(ii)
Solve(iii)
Define-W(h)
=W is
compact, 1, W(I)
=1,
1 is theonly eigenvalue
ofW,
thecorresponding eigenspace
is one dimensional and consistsonly
of constants.If E
ker(I - W*), ~
is in the range of I - W if andonly
ifh(C)
= 0.Step
4.A(z,)
~ 0. Theproof
is the same asstep
4 in Theorem3.3,
sinceg(x)
- 00 at 00.Step
5. Let Pick h so thatw, z be the
corresponding
solutions in(i), (ii)
ofstep
3. DefineThen,
Lit =f.
’ ’ ’
Step
6. A =If.
If we show thisequality,
since w, =0(g(x) ) by
Lemma4.1,
and
step 1,
and w isbounded, the
theorem will follow. Theequality 2
=f f
is
proved
in the same way as inStep
6 of Theorem 3.3.We can now solve the nonlinear
equation
where E = and is
uniformly elliptic
oncompact sets,
and itseigenvalues
behave like at oo, b 0.THEOREM 4.5.
If
KO
on R2 andthen, for
every a >0,
with a Cmin ~- b, s~,
a C(lj4n) (0153 - 2 f k), equation (4.4)
possesses an entire solution u on
R2,
withPROOF. We follow the
proof
of Theorem 3.6. First we construct a sub- solution. Letwhere a min
{- b, ~~.
Since =2na,
if v,, is the solution ofLv2
= k+ f2
constructed in Theorem
4.2,
then at oo,v2(x)
=+ +
Set U2 = V2+ A.2,
y where the constantA2
0 is to be chosen later. Wesee that
At oo, we have
Since a
(rt- 2 f k ),
the number - a+ 2fk +
4na+ o (1 )
remainssmaller than a fixed
negative
constant for xlarge. Thus,
asg(x) ,.
for x
large, f2(x) +
K exp[2v2]
exp[2A2] 0
for xlarge.
On the finitepart,
since C >
0,
wejust
have to chooseA2 0,
withIA21large,
toguarantee
this
inequality.
We have thus constructed a subsolution u2, withNext we shall construct a
super-solution
U.1 > U2 - Without loss ofgenerality,
we can assume
(Otherwise
we are doneby
Theorem4.2).
Asbefore,
let int
(supp K),
and letf1(X) satisfy
Let now v, be the solution to
Ev,
=fi given by
Theorem 4.2. At oo,vl(x)
=
( f k + 4na) g(x) + o(g(x) ).
Set u:, = VI+ Au A.1 > 0,
a constant to bechosen later.
In so In
Be ,
7is
bounded, .K c-
C 0. It is thereforepossible
to choose 0so
large
that+
g exp[2wi]
exp[2A,l
0.Thus,
ul is asupersolution.
Furthermore,
at oo,vl(x)
=( f k + + o(g(x) ),
while =( f k
+ + o(g(x) ),
thus at 00. It is thuspossible
to choose0,
evenlarger
so that on R 2.Now,
standard results(see
Theorem 5 in
[KN1]) imply
the existence of a solution u ofequation (4.4),
with
REMARK 4.7. We have
actually
shown that(4.4)
hasinfinitely
many solutions.REMARK 4.8. Since f or
Ix > 2,
if a suff i-cient condition for
(4.6)
to hold is thatfor some a >
2 f k,
while if0,
a sufficient condition for(4.6)
to holdis that
for some
5. - The
degenerate
case b > 0.In this
section,
we consideroperators
L =a~)
with theopposite asymptotic
behavior as that in theprevious section, namely, (au)
isuniformly elliptic
oncompact
sets and itseigenvalues
behave like at oo for someb > 0. This case turns out to be similar to our
previous
paperalthough
it
actually
has a mixed flavour of two dimensional andhigher
dimensional results.We shall follow
closely
the treatment inLEMMA 5.1. There exists a
f unction
with EG = 0in
G(x) ,.
at 00.PROOF. Extend
aii(x)
to all of R2 in such a way that theeigenvalues
of are - in all of R2. Let the results of
[FJK]
and[FKS]
apply.
LetGk(x)
be the Green function for LonIxl C
k withpole
at theorigin,
an easy
rescaling argument
shows thetGk(ky)
= whereGk(y)
is theGreen
function,
withpole
at 0 in1,
ofLet = The
eigenvalues
ofLk
are --By
Theorem 3.3in [FJK],
for
C 2 ,
whereand hence
for
I -1.
Butthen,
for 0
Ixl C k/2, independently
of k. The lemma follows as in Lemma 1 in[KN1].
LEMMA 5.2.
If
w is apositive
solutionof
= 0 onIxi > 1,
then u solvesiff v(y)
=u(x(y))/w(x(y))
solveswhere y =
y(x)
=XIBXB2,
X =x(y)
=YllyB2, J(x) = IXJ-4
andThe
proof
is asimple change
of variables.THEOREM 5.6. Assume that
at c>o, then there exists a
unique
bounded solutionof (5.3)
withPROOF. Let =
G(x),
where isgiven by
Lemma 5.1. Setthen Let
L =
theeigenvalues
of(akz(Y»)
areof the order
Let
g(z, y)
be the Green functionfor Z
in|y|
1 andthen
v(z)
solves(5.4) by [FJK],
and the lastequality
definesV1(Z)
andv2(z).
As in
step
1 of theprevious section,
we can estimate vl v2 since the behavior ofg(z, y)
is similarto g there,
yImitating
theproof
inStep
1in § 4,
we obtain that V2 is bounded andthus
THEOREM 5.9. Given a continuous
function
h onlxl
=1,
there exists aunique bounded
solution uof
with
at 00.
PROOF. The
proof
is almost identical with that of Theorem5.6,
thuswe omit it here.
THEOREM 5.12. Assume that
f satisfies (5.7)
at oo with b > 8 > 0. Then there exists a bounded solutionof
.Lu =f
with theasymptotic
behavior(5.8)
at oo.
PROOF. The
proof
is almost identical with that of Theorem 3 in the condition b > 8 > 0implies
that(5.11) decays
faster than(5.8).
We now can use Theorem 5.12 to show that the nonlinear
equation
has
infinitely
many solutions on R2. Our main result is thefollowing
THEOREM 5.14.
I f
there exist an 8 such that b > 8 > 0 and suchthat, f or Ixl large
for
sorrae constant C >0,
and K c 0 onR2, equations (5.13)
possessesin f initety
many bounded solutions on R 2.
PROOF. Let
By
Theorem5.12,
there exist VI’ v,solving
= 1~ -f, Lv2
= k+ f respecti- vely
and withat oo. Set ul = v,
+ Cl ;
thenon
R2,
for any constantCi .
Set U2 = V2+ C2,
thenSince k satisfies
(5.15)
and v, is bounded(and fixed),
we can chooseO2
0with C2 ( large
to ensureThus,
with this choice ofC2,
u2 is a subsolutions. Set nowCi
= 0 and chooseO2 0, [ C2 I large
sothat U2
= V2+ C2
is a subsolution and(since
both vl, v. are