A NNALI DELLA
S CUOLA N ORMALE S UPERIORE DI P ISA Classe di Scienze
P AUL H. R ABINOWITZ
On nontrivial solutions of a semilinear wave equation
Annali della Scuola Normale Superiore di Pisa, Classe di Scienze 4
esérie, tome 8, n
o4 (1981), p. 647-657
<http://www.numdam.org/item?id=ASNSP_1981_4_8_4_647_0>
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Equation
PAUL H. RABINOWITZ
The
question
of the existence of nontrivial timeperiodic
solutions of autonomous or forced semilinear waveequations
has been theobject
ofconsiderable recent interest
[1-12].
These papersstudy
theequation
(or
itsanalogue
wheref
alsodepends
on t in a timeperiodic fashion) together
with
boundary
conditions in x andperiodicity
condition s in t. Inparticular
the
following
result wasproved
in[11,
Theorem 3.37 andCorollary 4.14] :
THF,ORF,M 1.2. Let
f
E0([0, 1]
XR, R)
andsatisfy
(f1) f (x, 0)
= 0 andf (x, r)
isstrictly
monotoneincreasing
in r,(f2) f(0153, r) = o( Ir B)
atr = 0,
(f t)
there are constants if > 0 and it > 2 such that,
for Ir I
> r and x E[0, 1]
where(*)
This research wassponsored
inpart by
the Office of Naval Research under Contract No. N00014-76-C-0300.Pervenuto alla Redazione il 22 Settembre 1980.
Then
for
any T which is a rationalmultiple o f 1, equation (1.1)
possesses anon trivial continuous weak solution u
satisfying
.F’urthermore
f
E Ckimplies
u E Ck.As
part
of theproof
of Theorem1.2,
it was shown that the functionaldefined on the class of functions
satisfying (1.3) (and
of which(1.1)
isformally
the Eulerequation)
has apositive
critical value. Therefore(f.,)
and the form of I
imply that u, #
0 for thecorresponding
criticalpoint
U.Thus u is nonconstant in x and must
depend explicitly
on t. It was furtherobserved in
[11] (Theorem
5.24 and Remark5.25)
thatif g
satisfies(11)-(13)
the
equation
together
with(1.3)
also possesses a nontrivial weak solution. Indeed thearguments
of Theorem 1.2 gothrough
with minor modifications to establish this fact. However the functional one studies for this case iswhere G is the
primitive
of g.Again
thepositivity
ofJ(u)
for a criticalpoint u implies u
is nonconstant but we can nolonger
conclude that udepends explicitly
on t. In fact it is known[13, 14]
that as a consequence of(/2)-(/3)’
theordinary
differentialequation boundary
valueproblem
has an unbounded sequence of solutions which can be characterized
by
thenumber of zeros
they
possess in(0, 1).
Our
goal
in this paper is to show that if(f3)
isstrengthened somewhat, (1-.5), (1.3)
possessesinfinitely
many timedependent
solutions. More pre-cisely
we will prove:THEOREM 1.8. Let g E
([0, 1]x R, R)
and suppose gsatisfies (f1)-(f2)
and(ï3).
There is a constant It > 0 such thatfor all r 0
0.Then
for
anyT c- IQ
there is ako
E N such thatfor
allk> ko, (1.5), (1.3)
possesses a solution Uk uhioh is kT
periodic
in t andôuk/ôt =1=
0. Moreoverinfinitely many of the functions Uk are distinct. 3/3 0. jifo
REMARK 1.9. We have no estimate for the size of
ko
and do not knowif the result is false in
general
for k = 1. Note also that since(1.5)
is anautonomous
equation
withrespect
tot,
wheneveru(r, t)
is asolution,
so isu(x,
t-j- 0)
for any 0 E R. The above statement about theu,’k; being
distinctmeans in
particular
thatthey
do not differby merely
a translation in time.The
proof
of Theorem 1.8 draws on several results from[11]
and ideasfrom
[12].
For convenience we will take I == yr and T = 2x.Choosing k E N,
we seek a solution Uk of(1.5)
which is 2nkperiodic
in t andauklat #
0.Making
thechange
of time scale T =t/k,
theperiod
becomes2n
again
and theproblem
to be solved iswith
tI(x, -r)
=u(x, t).
For the convenience of the reader and to set the
stage
for akey estimate,
the
argument
used in[11]
to establish the existence of nontrivial solutions of(1.10)
will be sketchedquickly.
Solutions are obtainedby
an approx- imationargument.
Tobegin (1.10)
is modified in two ways. The waveoperator 02/Ô7:2
-k2(02/ÔX2)
possesses an infinite dimensional null space in the class of functionssatisfying
theboundary
andperiodicity
conditionsof
(1.1C)
andgiven by
The fact that N is infinite dimensional
complicates
theanalysis
of(1.10)
and to introduce some
compactness
to theproblem
inN,
weperturb (1.10) by adding
a/?F«
term to the left hand side of theequation.
HerefJ>
0and V is the
(L2 orthogonal) projection
of U onto N. A seconddifficulty
in
treating (1.10)
arises due to the unrestricted rate ofgrowth
ofg(x, r)
asir I-*
oo. Weget
around thisby truncating
g. Moreprecisely g(x, r)
isreplaced by gR(x, r)
which coincides with g forir I K
and grows cubi-cally
at oo[11].
Thus(1.10)
isreplaced by
the modifiedproblem
where gK
satisfies(/i), ( f2), (ï3)
with a new constantil = min (p, 4).
Letting GK
denote theprimitive
of UK’ in a formal fashion(1.11)
canbe
interpreted
as the Eulerequation arising
from the functionalLet
Em
= span(sin jr sin n-r, sinjx cos br 10 j, n m}
The
strategy pursued
in[11]
was to find a criticalpoint U.k
ofJ 1.0. let
m -+ oo, and then
let p
- 0 toget
a solutionUk
of(1.10)
with greplaced by
gx . Then L°° boundsfor Uk independent
of K show if we chooseK(k) sufficiently large UK(X, Uk)
=9(XI Uk)
so(1.10)
obtains. Aseparate
com-parison argument
isrequired
to prove thatUk =1=
0.The fii st
step
incarrying
out the details of the aboveargument
involvesobtaining
an upper boundMk
for cmkJ(Uk; k, fl, K) with Mk
inde-pendent
of m,fl,
and .g. For the currentproblem
which alsodepends on k,
it is crucial to know the behavior of
Mk
as a function of k. Thus we will take a closer look at c,,,;, and use a variant of anargument
of[12]. By
Lemma 1.13 of
[11],
cmk can be characterized in a minimax fashion. We will not write down this characterizationexplicitly
but will note a consequence of it which in turnprovides
an upper bound for Cmk. Setmk = span {sinjx
sin n-r,sinjx cos n-r 10 j,
% m and%2 > j2 k2)
and
where
ak = V2/n
soII "PkJILI == 1.
SetPmk
=(/JmkEÐ
span ?Pk. Thenby
Lemma 1.3. of
[11]
Inequality (1.13)
will lead to a suitable choice forMk .
Note thatby (ta) (or
even(fa)),
there are constants «l,(X2>0
andindependent
of K such thatfor all r E
R, x
E[0, n]. Consequently
J - oo as u ---> oo in1Jfmk (under li - 11 ,)
so there isa point z - Zt
at which the maximum in(1.13)
is achieved.Writing
where E (/Jmk, liil I, = 1,
and,,2 + 52 == 1
andsubstituting (1.15)
into(1.13) gives
o- - o- -
Combining (1.14)
and(1.16)
shows thatApplying
the Holderinequality yields
where A is a constant
independent
ofm, k, P,
K. Henceby (1.13), (1.18),
and the form of
J,
for a constant M
independent
of m,k, P,
K.Letting
m - oo and thenf3
-->0,
andformalizing
what we havejust
shown
gives :
LEMMA 1.20. Tlnder the
hypotheses of
Theorem 1.8(with
l = n and T ==2:n),
,for
allk E N,
there exists a solutionITk of (1.10) satisfying
«ith M
independent of
k and K.It remains to show that for all k
sufficiently large, auk/at:t=
0 andinfinitely
many of the functionsuk(x, t)
=Uk(x, r)
are distinct. IfU
isindependent
of r for anysubsequence
of k’stending
to oo,Ul
=Uk(X)
isa classical solution of
(1.7).
Thusby (1.21)
with -B" =K(k) suitably large,
By (1.7),
Combining (1.21)-(1.23) yields
as k - oo
along
thissubsequence.
Moreoverby ( f 3),
Thus
Ukg(0153, Uk)
-+ 0 in Ll. From(1.23) again
we conclude thatdUk/dx
-+ 0in -L2 which
easily implies Uk
-* 0 in Loo.By (f,),
for any 8 >0,
there is a ð > 0 such thatIr I 6 implies lg(x, r) /er. Choosing 8 1/n
and klarge enough
so thatIlUklil-bl (1.23)
then showsa contradiction.
Consequently Uk depends
on z for alllarge
k.To prove the second assertion of Theorem
1.8,
suppose two functionsUk(x, z ), Uj(X, r) correspond
to the same functionof x,
t modulo a translation in time(keeping
Remark 1.9 inmind).
ThusUk(x, T)
=Uk(x, tlk)
-U(x, t)
and
1I;(r, -r)
=U;(r, tlj)
=U(x,
t-f- 0)
for some 0 E R orU,(X, r)
=U(x, kT),
Uj(x, -r)
=U(X,jí + 0).
Since U must be both 2nk and2nj periodic
int,
letting
a denote thegreatest
common divisorof j and k,
wehave j
=oy2
Ic =
crk
and U hasperiod
2na in t. Furthermoreand
similarly
Consequently
if there were a sequence of solutionsUk,
of(1.10) correspond- ing
to the same function U(up
to a translation int), by (1.27)-(1.28)
we haveand
Ck,-¿’oo
likek’ along
this sequencecontrary
to(1.19).
Thus at mostfinitely
many functionsUk(X7 T) correspond
to the same solutionui,(x, t)
of
(1.5)y (1.3)
andinfinitely
many of the functions uk must be timedependent
solutions of
(1.5), (1.3).
Theproof
of Theorem 1.8 iscomplete.
REMARK 1.30. Both the existence assertions from
[11]
and thearguments given
above usehypothesis (12)
whichrequires that g
vanish morerapidly
than
linearly
at 0. However this condition can be weakened. Thesimplest
such
generalization
would be toreplace g(x, r) by
ar+ g(x, r)
with a aconstant and for this case we have:
THEOREM
1.31. Let g satisfy (f,), (f2), (/g)
and let oc>0. Thenfor
allT c IQ,
there exists ako E N
such thatfor all k > ko,
theproblem
has a continuous weak solution Uk which is kT
periodic
in t andallklat
0.Moreover
in f initely
manyof
thesefu%ctions
are distinct.PROOF. For convenience we
again
take I == 7&, T = 27&. It was shown in[11]
that Theorem 1.2 carries over to(1.32)
for Lx > 0. It is also easy to see that theargument
of Lemma 1.20 willgive (1.21)
for thissetting.
Likewise
(1.27)-(1.29)
are unaffectedby
the a term. Thus weget
Theo-rem 1.31
provided
that we can showUk(x, í) depends
on r for alllarge
k.If
not,
theanalogues
of(1.22)-(1.23)
here areand
Thus
(1.19), (1.33)-(1.34),
and(1,)
show thatUkg(x, Uk.)
-* 0 in Ell as k - ooas in
(1.24)-(1.25).
Sinceand the
right
hand side of(1.35)
isuniformly
boundedin k,
it follows from(1.7)
that the functionsd2Ukjdx2
areuniformly
bounded in Ll. The boun-dary conditions Uk(O)
= 0 =Uk(n) imply
that there is Xk E(0, n)
such that(dUkjd0153)(Xk)
+ 0. Hencewhich
implies
thatThus the functions
P dUk/dx
are bounded in Loo andby (1.7) again,
soare
d2Uk/dx2.
It follows that asubsequence
ofUk
converges(in 11 - ifC’)
to asolution U of
(1.7)
as k --->- oo. But(/i)
andIIUkg(x, Uk)IJLl
- 0 as k - ooimply
U == o.Next observe that
(1.7)
can be written aswhere H is the Green’s function for -
d2/dx2
under theboundary
conditionU(O)
= 0 =U(n). Dividing (1.37) by I/Ukllol gives
By (/2),
thearguments
of theintegral operator
areuniformly
bounded in 01.Hence since this
operator
iscompact
from 01 toC2, by (t2) again
a sub-sequence of
Uk/I/Ukllol
converge to Vsatisfying /IV II 01 == 1
andor
equivalently
If a is not an
eigenvalue
of- d2/dx2
under theseboundary
conditionswe have a contradiction and the
proof
iscomplete.
Thus suppose a is aneigenvalue.
Consider theeigenvalue problems:
where q;
is C’ on[0, n].
Let2j (resp. lzj((p))
denote thej-th eigenvalue
of
(1.41) (resp. (1.42)),
theeigenvalues being
orderedaccording
to increas-ing magnitude.
As is well known anyeigenfunction corresponding
toÂm
or
flm( q;) belongs
to8m
-199
ECl([O, n], R) 1q;(0)
= 0 =q;(:n;), q;
hasexactly
m - Izeros in
(0, n),
and99’ 0 0
at all zerosof q;
in[0, n]} . (Indeed
theeigenvalues
of(1.41)
are2.
= ne2a-1 andcorresponding eigen-
functions are
multiples
ofsin mx).
Sinceg(x, q;)q;-I> 0
via(11),
we haveÅi>q;i(q;)
forall j
E N and 99 E01,
99 # 0 via a standardcomparison
theorem[15, Chapter 6]. By (1.40),
1 is aneigenvalue
of(1.41),
say 1== 2.
andV c S-.
Thusp,,,(99) 1
and since8m
open(in
the C’topology)
andUkll/U1cl/cl ---+ V
in C1along
somesubsequence,
it follows thatU,l 11 U,, 11,0,,
and therefore
U, belongs
to8_
for alllarge k
in thissubsequence.
Writ-ing (1.7)
aswe see
!lm(Uk)
= 1.By (/i) again, 9(XI Uk) Uk I
> 0except
at the m-f-1
zeros of
Uk.
An examination of theproof
of the SturmComparison
Theo-rem
[16,
pp.208-209]
thenshows Uk
has a zero between eachpair
of suc-cessive zeros of V.
Consequently Uk C-
a contradiction. Thus Theo-rem 1.31 is established.
REMARK 1.44. In
[5], Blezis,
Coron andNirenberg study (1.1), (1.3) replacing (/3) by
’
and
(and
with noanalogue
of(/,)).
If we use(/’)-(/5)
with .rdependent f
inplace
of
(f,),
it is not difficult to see that theproof
of[11]
carries over for thiscase as does Lemma 1.20 and
(1.27)-(1.29).
Thus we obtain a variant of Theorem 1.8 for this case once it is established thatUk(ae, r) depends
on 7:for
large
k. To dothis,
we argue as in theproof
of Theorem 1.8. Assume( f,)
holds
with y
= 0. Thenby (1.25)
and(flc) , tJg(0153, Uk)IILl
-* 0 as k - oo.This in turn
implies IlUk[IL’
-)- 0 via(1.7)
and(1.36).
Hence(1.26) again provides
a contradiction.It is also
possible
for us todrop (fi)
and even therequirement
that(x, 0)
= 0 in(f,)
but then a new existence mechanism isrequired
and weshall not carry out the details here.
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