A NNALES DE L ’I. H. P., SECTION C
P. N. S RIKANTH
Symmetry breaking for a class of semilinear elliptic problems
Annales de l’I. H. P., section C, tome 7, n
o2 (1990), p. 107-112
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Symmetry breaking
for
aclass of semilinear elliptic problems
P. N. SRIKANTH
Tata Institute of Fundamental Research, P.O. Box 1234, Bangalore-560 012, India
Vol. 7, n° 2, 1990, p. 107-112. Analyse non linéaire
- - - - --~
ABSTRACT. - In this article we consider the
problem
where Q is an annulus and p > 1. We prove there is
always
a break of symmetryalong
the branch of radial solutions if the thickness of Q issufficiently
small.Key words : Symmetry Breaking, Critical Exponents, Annulus, Morse Index.
RESUME. - Dans cet
article,
nous considerons leprobleme
ou Q est un anneau et p > 1. Nous montrons
qu’il
y atoujours
une cassede
symetrie
dans la branche des solutions radiales, sil’épaisseur
de Q estsuffisamment
petite.
Classifcation A.M.S. : Primary 35 J 25; Secondary 35 P 30.
Annales de l’Institut Henri Poincaré - Analyse non linéaire - 0294-1449 Vol. 7/90/02/107/06/$2.60/0 Gauthier-Villars
108 P. N. SRIKANTH
1. INTRODUCTION
In this article we consider the parameter
dependent problem
where Q c (~n is an annulus and prove that
along
the branch of radialsolutions
(ux, À)
of(Px)
the symmetry breaks. That is, we show there existsa critical
Ào
such that everyneighbourhood
of(u~o, Ào)
contains nonradial solutions ofIn
[1] using
variational arguments Brezis andNirenberg proved
that for/? close to - including
the casen + 2
there exists both radial 1 and pn-2 g p
n - 2
a a d nonradial solutions for when ~, is close to zero. Further
they
alsoposed
thequestion
whether these nonradial solutions come from a break of symmetryalong
the branch of radial solutions of The main result of this article answers thisquestion partially,
in the sense that we showsymmetry
always
breaks if the annulus is thin. Moreover our argumentsonly ask p
> 1(p can
be greaterthan-~2014 ),
but how thin the annulusn-2
should be
depends
on p. We refer the reader to[4]
and to the referencesquoted
there for other results on symmetrybreaking.
2. MAIN RESULT
Before we state and prove the main result we will
specify
thefollowing.
Throughout
this section we shall assume Q={x E
1I x ~
1 +E~
andconsider the
problem (Px)
asposed
on this Q. Also we will use a and b inplace
of 1 and 1 + Erespectively.
We will useCo + °‘ (Q)
to denote the set ofcontinuously
differentiable functions on Q which vanish on aS2 and whose first order derivatives are Holder continuous with exponent a. We will use(Q)
to denote thesubspace
of radial functions. We shall alsouse X to denote another closed
subspace
ofCo° °‘ (Q)
which we definebelow.
x2, ..., ..., -x"_1,
It is well known
(see [5])
that(Px)
admits anunique
continuous branch of radial solutionsX)
inCo~ °‘ (Q)
x R and theprojection
on the ~, axis is( -
oo , whereÀ1
denotes the firsteigenvalue
of(-A)
on the domainQ. Moreover the solutions ux are
radially nondegenerate
if E is small, seeAnnales de l’Institut Henri Poincaré - Analyse non linéaire
Theorem 1.7 of
[5]. Throughout
what follows we shall assume our E is such that radialnondegeneracy
holds.THEOREM. -
Along
the branchof
radial solutions~,)
the symmetry breaks at some point(u~o, Xo)
with 0~,o if
E is smallenough.
We break up the
proof
of this Theorem into several steps.Step
1. - The Morse-Index of the solution(uo, 0) (i.
e. thepoint
onthe branch
(ux, À) corresponding
to~, = o)
isbigger
orequal
to(n + 1 )
inthe space
Co+°‘ (Q).
Proof of Step
1. - Notice thatby
the definition of Morse Index it isenough
toproduce orthogonal
functions whichsatisfy
to conclude that the Morse Index of uo is greater or
equal
to(n + 1).
It iseasy to see v --_ uo satisfies
(1)
because uo is a solution of(Po).
Now we construct with w
orthogonal
to uo and satis-fying (1).
n
Define w
(x)
= uo(x) .
wo(x)
where wo(jc)=03A3
xi.Clearly
w(x)
EC5’ ex (Q)
and is
orthogonal
to uo since uo is radial. Asimple computation yields
where u0x:~ =
~u0. Clearly (1)
is satisfiedby
w ifXi
the firsteigenvalue
of~~
( - A)
on Q isbigger
than201420142014
as can be seenby
thefollowing.
Multiply (2) by w
on both sides andintegrate keeping
in mind Mo is radial and areorthogonal directions,
theintegration
of terms onright
of
(2) yield
Vol. 7, n° 2-1990.
110 P. N. SRIKANTH
Hence
Step
1 follows.Step
2. - In this step we draw some conclusions based onStep
1. FromStep
1 it follows that as we movealong
the continuous branch(ux, À)
asÀ varies over
(0,
there must be apoint Ào
such that(u~o, Ào)
lies onthe branch and uÀo is a
degenerate
solution of(P~o),
i. e.has a nontrivial solution v. It is classical that we can write v in the form
where cpo is a constant and for
k >-_ l,
cpk is aneigenfunction
of theLaplacian
on the(n -1 ) sphere corresponding
to the k-th nonradialeigenvalue.
Alsoak’s
are solutions of theequation
where
k+k(k+n-2) for k >_-
1. Also from the remark on radial nonde- generacy made at thebeginning
of this section ao(r)
= 0. We now makeanother assertation in
Step
3keeping
in mind the arguments above and alsousing
the same notations.Step
3. - We claim there exists~, E (o,
which we shall still denoteby Ào (which
may be different fromÀo
ofStep 2)
such thathas a nontrivial solution and this v has
exactly
the formProof of Step
3. - To prove our assertion we make a carefulstudy
ofthe set of
equations
we have in(4).
Webegin by considering
theweighted eigenvalue problem
where yi
(~,)
is the firsteigenvalue
associated with the +~,).
It isclear y 1
(A)
is a continuous function of A as(ux, A)
lie on a continuousbranch. Now as
along
the branch it is clear > 1,since the operator on the left side of
(7)
is morepositive
than(-A).
.4nnalo.s de l’Institut Henri Poincaré - Analyse non linéaire
Hence it is clear that if we have to meet any situation as described in
Step
2 then yl(~,)
must beexactly
1 for some~,o.
This is because for all k > 1, the operator on the left hand side in(4)
is morepositive
than whenk =1. Hence it is clear that there exists a
~,o
such that(5)
has a solution vwith
with a 1 (r)
> 0[since
itcorresponds
to firsteigenvalue
y 1(~,o)] .
We claimthat in this
situation,
that iswhen al
> 0, all theak’s
--_ 0for k >__
2. Thisfollows from
(4)
since ak(r) satisfy
with j~ > V k > 1, then
ak’s
shouldchange sign
forall k ?_
2. But thenStrum’s
comparison
theorem would force al tochange sign.
Hence sinceal > 0 all
ak’s ==
0for k >_
2(we
havegiven
this argumentonly
because itis very
general, however,
in the situation we are in it follows in astraight
forward way since the operator in the left in
(8)
is morepositive
for anyk >_ 2 than when
k =1 ).
Thiscompletes
theproof
ofStep
3.Proof of
the Theorem. - Now we restrict our consideration ofproblem (Px)
to the space X defined at thebeginning
of this section. Due to this restriction the n-dimensionaldegeneracy proved
inStep
3 reduces to thecase of an 1-dimensional
degeneracy.
From all theprevious
arguments itnow follows that as we move
along
the branch(ux, ~,)
thetopological degree
has tochange
at thepoint
Thischange
indegree
leads toa
secondary
bifurcation.By
theuniqueness
of radial solutions which wehave assumed, it now follows that the the new solutions
arising
from thesecondary
bifurcation are nonradial and hence our claim of break of symmetry.Remark 1. - In the case when 1
n + 2
a combination ofStep
1n-2 p
and a result due to H. Hofer
[2]
leads to the conclusion that theequation (Po) (i. e.
when~, = o)
has atleast one nonradialsolution,
since the radial solution cannot be the Mountain PassSolution,
for such a solution must have Morse Index 1.Remark 2. -
Using completely
different arguments Li Yan-Yan has shown in[3],
that the number of solutions of(Po)
tends toinfinity
as thethickness of the annulus goes to zero. We believe that this increase in the number of solutions occurs because as the thickness of the annulus goes to zero the Morse Index of the radial solution of
(Po)
goes toinfinity
asVol. 7, n° 2-1990.
112 P. N. SRIKANTH
was
pointed
out to the authorby
Prof. A. Bahri(To
see thisreplace
thefunction wo of
Step
1by
the functions cpk referred to inStep 2).
We believethis
produces
more and moresecondary
bifurcationpoints
on our branch(ux, ~,)
and the branches due to thesesecondary
bifurcation lead to moreand more solutions. However so far we are not able to prove this. Infact it seems very hard to use the results of Crandall-Rabinowitz on bifurcation from
simple eigenvalues
due to the difficulties involved inproducing
agood
transvers direction.REFERENCES
[1] H. BREZIS and L. NIRENBERG, Positive Solutions of Nonlinear Elliptic Problems Involving
Critical Soboler Exponents, C.P.A.M., Vol. 36, 1983, pp. 437-477.
[2] H. HOFER, The Topological Degree at a Critical Point of Mountain Pass Type, A.M.S.
Summer Institute, Berkeley, 1985.
[3] YAN-YAN LI, Ph. D. Thesis, Courant Institute, 1988.
[4] M. RAMASWAMY and P. N. SRIKANTH, Symmetry Breaking for a Class of Semilinear
Elliptic Problems, T.A.M.S., 304, NO2, 1987, pp. 839-845.
[5] W.-M. NI and R. D. NUSSBAUM, Uniqueness and Nonuniqueness for Positive Radial Solutions of -0394u+f(u, r)=0, C.P.A.M., Vol. 38, 1985, pp. 67-108.
(Manuscript received October 3rd,1988.)
Annales de l’Institut Henri Poincaré - Analyse non linéaire