A NNALES DE LA FACULTÉ DES SCIENCES DE T OULOUSE
M ARTA G ARCÍA -H UIDOBRO R AÚL M ANÁSEVICH
Jumping nonlinearities for fourth order B.V.P.
Annales de la faculté des sciences de Toulouse 6
esérie, tome 2, n
o1 (1993), p. 53-72
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- 53 -
Jumping nonlinearities for fourth order B.V.P.(*)
MARTA
GARCÍA-HUIDOBRO(1)
andRAÚL MANÁSEVICH(2)
Annales de la Faculte des Sciences de Toulouse Vol. II, n° 1, 1993
RESUME. - Dans cet article, nous étudions le probleme avec conditions
aux bords
ou I =
i2 ~,
il i2 et J =, j2 ~,
jl j2 sont deux sous-ensembles arbitraires de{0,1,
2,3}
tels que la premiere valeur propre du probleme linéaire associéest strictement positive. Soit 03C61 une fonction propre non negative corres- pondant a Ai . On suppose
H(t)
=h( t) -
s03C61(t),
s E IR+. . Alors sous lacondition que f "croise" les k premieres valeurs propres de
(L),
on montrel’existence d’au moins 2k solutions du probleme
(P)
pour s suffisammentgrand et positif.
ABSTRACT. - In this paper we study the boundary value problem
where I =
~il , i2 ~,
ii i2 and J =~ jl , j~ ~,
j1 j2 are two arbitrarysets of integers from
~0,1, 2, 3~
which are such that the associated lineareigenvalue problem
satisfies al > 0. Let ~i be a nonnegative eigenfunction corresponding to 03BB1. Suppose
H(t)
=h(t) -
E IR+. Then, under the assumption that f "jumps" over the first k eigenvalues of problem(L),
we prove the existence of at least 2k solutions to problem(P)
for large positive s.( *) Recu le 25 fevrier 1993
This research was sponsored by D.I.U.C., research grant 89043, Universidad Catolica de Chile for M.G. H., and by FONDECYT research grant 1209-91 for R. M.
~ ) Facultad de Matematicas, Universidad Catolica de Chile, Casilla 6177 Correo 22, Santiago, Chile
(2) Departamento de Matematicas, F.C.F.M., Universidad de Chile, Casilla 170 Correo 3, Santiago, Chile
1. Introduction
In this paper we consider the
boundary
valueproblem
where I =
i2~, il i2
and J =~2~~ 31 j2
are twoarbitrary
but fixed sets of
integers
from~0, 1, 2, 3},
H eC[0, T],
andf
EC1(R)
satisfies that the limits
exist as finite real numbers. In
(1.1)-(1.2~
and from now on,u~~~ _
J~
EIN.
.We associate to
( 1.1 )- ( 1.2 ~
theeigenvalue problem
It is
known,
see[4],
that theeigenvalues
of thisproblem
form an infiniteincreasing sequence {03BBi}~i=1 ~ IRo
with no finite accumulationpoints.
Wewill assume the sets I and J to be such that A = 0 is not an
eigenvalue
of(1.4).
A necessary and sufficient condition for this to hold will begiven
insection 2.
Let
~i
be aneigenfunction corresponding
to the firsteigenvalue Ai
of(1.4).
We will consider the case when function Hin (1.1)
is of the formAlso,
we will assume that A- > 0 and that the interval(.1_ , a+ )
containsexactly
the first keigenvalues
ofproblem (1.4).
Thatis,
we will assumethat
for some k E IN . Under these
conditions,
we will prove in section 4 that forlarge
values of theparameter 8
in(1.5),
theboundary
valueproblem
( 1.1 ~-( 1.2 ~
has at least 2k solutions.A
nonlinearity , f satisfying (1.3)
is known in the literature as ajumping nonlinearity.
Boundary
valueproblems
withjumping
nonlinearities have been studiedmainly
for the second order case under Dirichlet and Neumannboundary conditions,
see forexample [6], ~(7~, [8]
and[9].
In[5],
some results for n-th orderequations
with ajumping nonlinearity
were obtained.In this paper we are
dealing
with animportant
extension of the resultsin
[5]
for the fourth order case. To make thispoint clear,
let us consider theboundary
valueproblem
where
H
eC[0, T] and f E C1 (IR ~
satisfies that the limitsexist and are finite.
Also,
let theintegers
ro and r in(1.7)
be such that0 rp r
3. This last condition ensures that A = 0 is not aneigenvalue
of the associated linear
problem
Setting H(t)
= +h(t),
it follows from~5~ that (1.71
has at least 2ksolutions for
large
values of s when the interval(A-~, a_ ),
A-0,
containsthe first k
eigenvalues
of(1.9).
The purpose of the
present
paper is to treat all theremaining
open cases for the fourth order case, for which thecorresponding
linearproblem
doesnot have A = 0 as an
eigenvalue
and thenonlinearity ,f jumps
over the firstk
eigenvalues.
These cases arerepresented by (1.1)-(1.2),
andthey
form aclass of
twenty boundary
valueproblems,
ten of which arenonselfadjoint.
Let I ~ =
~i3, i4 ~, i3 i4
and T ~ =~,~3, ,?4 ~ ~ ~3 .~4 be, respectively,
the
complementary
sets of I and J withrespect
to~0,1, 2, 3~
and setI* _
{3 - i4 ,
3 -i3~
and J* ={3 - j4 , 3 - ,j3~.
It can beeasily
verifiedthat the
adjoint problem
to(1.4)
isThis
adjoint problem
willplay
a fundamental role in theproof
of some ofour intermediate results.
For s >
0,
let us make the substitution u = s y in( 1.1 ~- ( 1.2 ~ .
For slarge,
and because of conditions
(1.3)
and(1.5),
theresulting problem
could bestudied
by considering
thelimiting equation
as s --~ o0or its linearised version
where as
usual, y+
=0~ and y = y+ - y-, together
with theboundary
conditions(1.2).
For some technical reasons that will become clearlater,
it will be convenient to consider aslight
modification ofnamely,
theboundary
valueproblem
whete zn
is a constantmultiple
of~1 ~n,
apositive eigenfunction correspond- ing
to the firsteigenvalue
of(1.4)
with Treplaced by
T-~-1 /n,
n E IN.The
problem
ofsearching
solutions to(1.12)
or to( 1.1 ~- ( 1.2 ~
can bethought
as a twoparameter dependant
initial valueproblem. Thus,
insection
2,
afterlisting
down some known factsconcerning
the linearproblem (1.4)
and itscorresponding adjoint,
we will prove some resultsconcerning (1.12)
and its related initial valueproblem.
These results will make itpossible
to work with a oneparameter boundary
valueproblem
andthus,
to use
shooting techniques
to prove ourmultiplicity
result. In section 3 wefirst prove the existence of two
specific
solutions to(1.1~-(1.2)
and then usethe results of section 2 to prove some basic lemmas that will be used in section 4 to prove our main
multiplicity
result.Finally,
we want to remark that our results will still hold true if wereplace
the differentialoperator
Lu - in(1.1) by
thegeneral
fourthorder
disconjugate
linear differentialoperator.
2.
Preliminary
resultsWe
begin
this section with a summary of theproperties
of the solutionsto the linear
problem (1.4)
that we will use.A direct
computation
shows that a necessary and sufficient condition that A = 0 is not aneigenvalue
of(1.4)
is that the elements of I ~ and Jsatisfy
Similarly,
= 0 is not aneigenvalue
of theadjoint problem (1.10)
if ando nly
ifHenceforth in this paper we assume that
(2.1)
and(2.2)
are satisfied. Infact,
it is not difficult to show that(2.1)
holds if andonly
if(2.2)
holds.It is known that the
eigenvalues
of(1.4) (resp. (1.10))
aresimple.
Mo-reover, an
eigenfunction (resp. ~z ) corresponding
to the i-theigenvalue .iZ (resp.
hasexactly
i - 1 zeros on(0, T),
all of which aresimple.
Inparticular, ~1 (resp. ~i )
can be takenstrictly positive
on(0, T). Also,
theonly
derivatives of~i (resp.
which vanish at theendpoints
0 or T arethose
appearing
in(1.2).
The
following proposition
may beproved by
induction on m.PROPOSITION 2.1.2014 The sequence
of eigenvalues ~~1?.,z ~n°-1 of ~1.l~~
and1 10 satisfy
= I~~for
ever m E ’N .The
following
three results areparticular
cases of the moregeneral
casestudied
by
Elias in[2]
and[3].
.PROPOSITION 2.2. - Let
S(u)
denote the numberof changes of sign of
u on
(0, T).
.Then, for
i =1, 2, 3,
where denotes the number
of boundary
conditionsimposed
to~?.,.z
in(1.2)
among~,..,z, ~~,
. . ., ~~m 1~
.Moreover,
thesechanges of sign
are theonly
zerosof ~{~~
andthey
are allsimple.
PROPOSITION 2.3. - The
eigenvalues Am,
m e INof (I./ ), regarded
asfunctions of
theendpoint T,
are continuous andstrictly decreasing.
PROPOSITION 2.4. -
Ifi
=(11 , 12), J
=(ji , j2)
are two setsof different integers
such thatik ik, and ik jk for
k =1,
2 and at least oneof
theseinequalities
isstrict,
thenAm
>Am for
all m e INand if only
oneof
theinequalities
isstrict,
thenAm+i
>Am
>Am where Am
, m e IN denotes the m-theigenvalue of (I./)
with I and Jreplaced by I
andJ.
Let us consider now the semilineat
boundary
valueproblem
where
y+
=max~y, 0~, y-
=y+
- y andfor some k E IN.
Then, problem (2.4~-(2.5)
has the two trivial solutionsIt is clear that yi is
strictly positive
on(0, T),
y2 isstrictly negative
on(0, T)
and that these are the
only
solutions of(2.4)-(2.5)
which do notchange sign
on
(0, T).
By
means of the substitutionwe see
that y
is a solution to(2.5)
if andonly
if w is a solution towhere
g(~~ _ ~1+~+ - ~_ ~- .
.Next,
let us consider the Initial Value Problemwhere x E
C(IR).
. We recall that I U{i3, i4}
={0,1, 2, 3}.
Letw(
. , c,d)
bethe solution to this
problem.
PROPOSITION 2.5. Given any
to
> 0 and any c EIR,
there ezists aunique
d E IR such thatZ.cn~~ ~
c,d)
= 0 .Proof.
- Let the real numbers cl, c2,dl, d~
be such that cl > c2 anddl
>d2,
with at least one of theinequalities being
strict. Onsetting wi(t)
:=w(t,
cZ,dZ),
i =1, 2,
we see that w = w2 satisfies the initial valueproblem
From our
assumptions
on cj anddi,
there exists 5 > 0 such thatw(t)
> 0for t E
(0, Suppose
there existstl
> 0 such thatw(t)
> 0 for t E(0, tl ~
and = o. Since g is an
increasing function,
we see from the first of(2.12)
that > 0 for t E(o, tl )
and thus anapplication of Taylor’s
Theorem
yields
implying
the contradiction > 0.Hence, w(t)
> 0 for every t > 0.Similarly,
Now,
for fixedto
> 0 and c EIR,
consider the sequences~cvn } °° 1
anddefined
by
w~ =n)
and vn = c,-n). By
setting
i =j2
in(2.14),
we obtain thatimplying
thatHence,
from thecontinuity
of solutions to(2.11)
on the initial conditionsand from
(2.13)-(2.14),
there exists aunique
d E IR such thatHence the
proposition.
DThe
following corollary
is a direct consequence of(2.14).
COROLLARY 2.6. - Let
to
> 0 befixed. Then,
theunique
real number dgiven by proposition
2.5 is auniformly
continuousdecreasing function of
c.In what follows we will denote
by w ( ~ ,
x,c)
the solution of(2.11)
whichsatisfies the conditions =
0, i
EI, ~,ut23 ~ ~0)
= c andw~~2 ~ (T)
= 0.Let denote a
positive eigenfunction corresponding
to the firsteigen-
value
03BB1,n
of(1.4) with 0 , T] replaced by [0, T ] and
setwhere n is chosen
large enough,
say n > nl(see proposition 2.3),
to haveLet
ae(t)
= in(2.11).
From now on, we will refer toproblem (2.11)
with xreplaced by zn
as(2.11)n.
Setc)
-iv( ~
zn,c)
andw( . ,c)
=w ( ~ c).
.Clearly,
oo,iv~2~ ( ~ c)
converges,uniformly
onbounded
intervals,
toiv~2~ ( . , c)
for i =0, 1, 2,
3.We will now construct two different solutions to
(2.11)~.
To thiseffect,
we first prove the
following
lemma.LEMMA 2.7. - For any m G IN and ~1 G every nontrivial solution
y(t, ~) of
is such that
y~~1 ~
hasezactly
m zeros on(0, T),
allof
which aresimple.
Inparticular,
itfollows
that any nontrivial solutionof (2.20)
with~_
=~+
issuch that
~
hasezactly
k zeros on(0, T)
and that these zeros aresimple.
Proof.
-Suppose
on thecontrary
that there exist mo E IN and.1o
Esuch that
y~~l ~ ( ~ , a~
has less than mo zeros on(0, T),
andconsider the set
A
_ ~ ~ ~ , a~
has at leastmo zeros
on (o, T) , a ,
From
proposition
2.2above,
and sincei3
andji j2,
we obtain that~~~1 ~ o+ 1
hasexactly
mosimple
zeros on(0,T)
and therefore E A.Also,
from ourassumption, A
is bounded belowby Ao
and inf A = A*.We claim that
y{~1 ~ ( ~ , a* ) ,
hasexactly
mo - 1 zeros on(0, T)
and~l* ~
= 0.Indeed,
ify~~l ~ ( ~ , .~* ~
has less than mo zeros on(0, T),
then
yt~l ~ ( ~ , .1*
+E)
would have less than mo zeros on(0,T)
for all smallenough
E > 0.Also,
ify{~1 ~ ( ~ , A*)
has more than mo - 1 zeros on(0,T),
then for
sufficiently
small E >0, ~(~)(., , a* - E~
would have at least mo zeros on(0, T).
Since both alternatives contradict the definition ofA*,
our claimfollows and A* =
Àmo.
SinceAo
E(03BBm0, 03BBm0 +1 )
this is also a contradiction.Therefore, y~~l ~ ( ~ , ao )
has at least mo zeros on(0,T).
A similarreasoning
shows that
y{~1 ~ ( . ,
can have at most mo zeros on(0,T)
and the lemmafollows. 0
PROPOSITION 2.8.2014 For each n E
IN,
there exist two real numbers 0 and0,
such that the solutionof equation
~Z.11~~, is
such thativ~~l ~(
hasezactly
k zeros on( 0, T)
and allof
these zeros are
simple.
Proof.
- Let En > 0 be such that the solutiony( ~ , a+)
of(2.20)
forwhich
y~~3 ~ ( 0 ~
= 1 satisfiesThen,
the functionssatisfy equation (2.11)~
on[0, T],
= 0 for t 67,
= e~,w(i3)-k,n(0) =
-~n andw(j2)±k,n(T)
= 0. Thusw±k,n(t) = wn(t,c±k,n) for
W -k,n 0 - -En an
w:l:k,n
- o. us w:l:k,n t - Wnt,
c:l:k,n lor i 6[0, T],
where = QConsider now the
problem
The conditions on
A+
and A-imply
that the solutionw( - , c)
of thisproblem
satisfies
w{~1 ~ (T, c~ ~
0.Actually,
under the additionalassumption
thatA+
>A 2,
we have thefollowing
result.THEOREM 2.9. - Let A-
Ai
anda+
>h2
. Then the solutionw(
-, c~,
c E
~-1,1~ of problem ~,~.2,~~
is such that:i) zv{~1 ~( - 1) changes sign exactly
once on(0, T);
ii) } ( - , -1)
does notchange sign
on(0, T~
.Remark. - From the
point
of view of ourmultiplicity result,
the as-sumption A+
>A2
is not a restrictive one, as it will be seen later.Proof.
- Let~i
be thepositive eigenfunction corresponding
to the firsteigenvalue
=Ai
of(1.10),
andmultiply equation (2.22) by ~1.
We seethen that w satisfies the
equation
Since A-
Ai ~+, G(t)
> 0 forevery t
> 0 andtherefore, evaluating
at t = T and
observing
that the left hand side of(2.23)
can be written inthe two ways
we obtain that
A direct
computation
shows that(-l)~ ~~ ~(T)
0 for allpossible
choices
of j1,
andtherefore,
we have that0,
and we conclude that the number ofsign changes
oftt/~)(’ c)
is odd when c = 1 and it iseven when c = 20141.
We will now show the same holds true for the number of
sign changes
ofw( ,c).
This isclearly
true if~i
= 0.0,
let7= I, J = (0, j2 ),
anddenote
by Ai
the firsteigenvalue
of(1.4)
withI,
Jreplaced by 7,
J. Fromproposition 2.4,
we have thatAi Ai A2
andtherefore,
onmultiplying equation (2.22) by ~i,
thepositive eigenfunction corresponding
to the firsteigenvalue
of(1.10)
withI*,
J*replaced by I*, J*,
we obtain thatSince
~1"‘(T)
>0, (2.26) implies
thatw(T)
0.Now,
for t E(0, T~,
let w = on(2.23~.
Then v satisfies the differentialequation
-
Since from
proposition
2.2 the function(~i~~ 2 - 2~i ~i~~
isstrictly positive
on
(0, T),
we conclude that ifv~(t) >
0 for t E(a,,C~~,
thenNow,
let c = 1 and suppose thatt~(’, 1) changes sign
at three differentpoints
on(0, T). Then,
since~i(t)
> 0 for every t E(0, T),
so does v atthe same three
points.
Let us denote thesepoints by ti, i
=1, 2,
3 with0
t1 t2 t3 T,
and assume thatv(t)
0 for t E(tl, t2)
andv(t)
> 0for t E
(~2, Then,
it follows the existence of an interval(a, !Q)
C(tl, t2)
such that
v"(a) > 0, v"(,Ci)
0 andv’(t) >
0 for t E(a, ~Q) contradicting (2.28).
.Thus, w( ,1 ) changes sign exactly
once on( 0, T) .
.For the case c =
-1,
the sameargument
as above shows thatw(t, -1 )
0for t E
(0, T)
when theintegers it, i2
are nonconsecutive.Indeed,
in thiscase, either
v(0)
= 0 orv’(0)
= 0 and it follows from(2.28)
thatz.v( ~ , -1) changes sign
at most once on(0,T). Thus,
we conclude from(2.25)
and(2.26)
that in factw(t, -1)
0 forevery t
E(0, T).
Wheni2
=i1
+1,
this result followsby considering
the solutionWj, j
E~0, 1, 2, 3}
of the linearproblem
It can be
easily
verified that thefunction zj,k
:= wk satisfies the first of(2.29)
and does notchange sign
on(o, Also,
0and
z{~} (T) _
for every choice of theintegers j,
1~ E0 1, 2, 3 .
Therefore,
sincewo(t)
0 for every t E( 0, T ~
andwo (T )
>0, by using
aboot
strap argument,
we conclude thatw3 (t)
0 for t E(0, T). Finally,
anapplication
of Rolle’s Theorem and theboundary
conditions =0,
i E
, i2 ~
andz.cJ{~2 ~ (T ~
= 0yield
the result of the theorem. DCOROLLARY 2.10. - Under the
assumptions of
theorem2.9,
there exists two real numbers cl > 0 and co 0 such that the solutionsw~( ~
i =
0,
1of ~~.11~n satisfy:
i) (j1)n(., c1) changes sign exactly
once on(0, T);
it) )(., c0)
does notchange sign
on(0, T)
.Proof.
- Thecorollary
followseasily
from the fact that the solutionwn~~ ( - , c)
ofconverges in
C4 ~ 0 , T
+(1/n)~
to the solutionw(
. ,c)
of(2.22)
as b goes too+. a
We finish this section with a result
concerning
themultiplicity
of thezeros of a solution to
(2.11)n.
Since itsproof
is very similar to that of Lemma 3.3 in~5~,
we will omit it.PROPOSITION 2.11. -
If
c~ 0,
then t~e solutionwn( ~ , c) of (~.~l~n
andits derivatives may have
only simple
zeros on(0, T).
.3. Basic lemmas
Let n E
IN,
n > nl and consider thefamily
ofboundary
valueproblems
Herein we will refer to
(3.1)oo-(3.2)oo
whenspeaking
of thelimiting problem (1.1)-(1.2)
wit h Hsatisfying
condition(1.5).
Let
Wn
be the Banach space of functionsf
EC3 ~ 4 , T+ (1/~)]
( f
EC3~ 0 , T])
whichsatisfy (3.2)n ((3.2)oo) provided
with the usual norm, and letOur first result in this section is similar to the
corresponding
one in~6~,
andthus we will
only
sketch theproof.
LEMMA 3.1. There exists ~0 > 0 such that
for
any E E(0, co],
there exists an so > 0 such that
for
s > so, theboundary
valueproblem
~3.1~~-~3.~~n
possesses aunique nonnegative
solution and aunique nonpositive
solutionsatisfying
Moreover,
there exists no E IN such thatfor
n > no, neither eo nor sodepend
on n, andfor
s > so,where y1 is the
function defined
in~~. ?’~.
Proof
. Theproof
is based on the Banach Fixed Point Theoremapplied
to each of the
mappings
where
Gn (t, T~
is the Green’s functioncorresponding
to theproblem
In a first
step,
thepositive
numbers and are determinedby
the
requirement
that the restriction be a contraction for~ E
(0,
and s > and in a secondstep,
so,n is determinedby
therequirement
that forS 2: SO,n
and for all e E(0, Fn,s
C(Here,
forf
EWn,
=9
EWn II 9 - r
We will show that there exists
no 2:
nl such thatfor n >
no, eo,n and so,n may be takenindependent
of n.Indeed,
LetMi
be a uniform upper bound forI
on[0, M2
be such thati f j(~) - a+ I M2 for £ E ~ 0 , oo).
Also,
let 03B4 > 0 be such that 5T/2,
and 51/9 Ml M2.
Now setmo = and choose > 0 such that if
u E
(yl ),
thenu(t)
> 0 on(0, T)
andu(t)
>mo/2
on~ b/2 T - b/2 ~.
Let no E IN be such that
Then, if n 2:
no, and w EWn
is such thatwe
have, by setting
that un E
Woo
andfor i =
0, 1, 2,
3.Therefore,
which
implies
thatun(t)
> 0 for all t E(0,T)
and >mo/2
fort E ~ b~2 ~ T _ b~2 ~.
Hence, w(t)
> 0 for every t E~0 , T
+(1/n~~
andw(t)
>mo/2
for allLet us set ~o = Since
f ~(~~
=a+,
there exists R > 0 such thatNow,
for 6- E(0,
choose so > 0 such thatand such that
for s >
so,Then ~~
depends only
on e andh,
and it can beeasily
verifiedthat for
each n 2:
noand S 2:
so, is a contraction and CThus,
the existence of aunique
solution un,~to
(3.1)n-(3.2)n satisfying
the first of(3.4)
followsby setting
=where zn,s is the
unique
fixedpoint
of inThe existence of is
proved similarly.
0Let v = su - in
~3.1)n,
go =f
- g, and consider theboundary
valueproblem
For d E
IR,
letwn( ~
c,d)
be the solution to(2.11)n,
and denoteby v~~~( ~
c,d)
the solution to(3.14)n
whichsatisfy
-0,
i E1,
v~23 ~ (o)
= c,vt24 ~(o)
= d.It can be
easily
verified that there exists some constantMo
> 0independent
of sand n, such that for i =0, 1, 2,
3 and t E[ 0 ,
T -~(1/n) ] ,
,Now let Co and cl be as in
corollary 2.10,
seteg
=max{-c0 , c1}
and letdo
> 0 be such that forevery n ~
no and(c) ~ cQ,
Let also the constant
Mi
be such thatand choose so such that for
and n > no, it holds that
Then,
for s> s~, n >
no,~c) c~
anddo,
i --_1, 2,
we haveOn
setting dl
=d2,
= =1, 2 in (3.21),
we conclude that for each cE ~ [2014c~ , c~], (3.14)n-(3.15~
has at least one solutionAlso,
from(2.13)
and theidentity
we obtain that for
every n ~
noand s ~
sl ,Let now be as in
proposition
2.8.LEMMA 3.2. There exists
s2 >
sl, such thatfor
all n > no ands > s2
any solutionof (3.14)n-(3.15)
with c =satisfies
that itsj1-th
order derivative has
exactly
k zeros on~ 0 , T ~,
allof
which aresimple
andcontained on
(0, T )
.Proof
. - Let ~ E(0 , ] be
such that if v EC3[0 , T
+(1/n) ]
satisfiesp~
i E~~ v~~a)(T~
- 0 andIlv
- ~, cs [o,fi+(1/~)l I E,
’ thenhas exactly
k zeroson [0 , T ],
all of themsimple.
Let s2 > sl be chosen so that any solution to(3.14~~
with c = satisfies(3.23)
forand
(3.19)-(3.20)
are satisfied for e = ~. Let vn,s denote any such asolution,
and
for s ~
s2 define the functionThen,
it can beeasily
verifiedby using
Gronwall’sLemma,
thatand the result follows from the choice
For co and cl as in
corollary 2.10,
thefollowing
results can beproved similarly.
LEMMA 3.3. - There exists s3 > s2 such that
for
s > s3 any solutionof problem ~3.Il~~n-~3.5~
is such that: ’i) for
c = co, itsj1-th
order derivative isstrictly negative
in(0 , T
+(1/n~~ ~
it) for
c = cl, its order derivative hasexactly
one zero on(0 , T
+(1/n~~
and this zero issimple.
LEMMA 3.4. - There exists s4 > s3 such that
for
s > s4 and c incompact
intervals notcontaining 0,
any solutionof ~3.ll~~n-~3.5~
and its derivatives may haveonly simple
zeros on(0
T +(1/n~~ .
4. The
multiplicity
resultWe are now in a
position
to prove ourmultiplicity
result. As wementioned
before,
this will be doneby using shooting techniques.
To thisend, for s >
s4and j
=1, 2,
...,~,
we define the setsPj,n
={x
>0 any
solution to(3.14)n-(3.15)
is suchthat its
j1-th
order derivative has at(4.1) least j
zeros on(0, T)
for each c E, x ~ ~
and
Nj,n
=~ x 0 any
solution to(3.14)n-(3.15)
is such’ that its
ji-th
order derivative has at(4.2) least j
zeros on(0, T)
for each c ~~ x ; c._~~n ~ ~.
From lemma
3.2,
both sets arenonempty
for allj. Also,
from lemma3.3,
is bounded
for j
=2,
... , J~ and is boundedfor j
=1, 2,
... , k so that we can setWe will first prove the
following
result.LEMMA 4.1.2014
Any
solution to(3.14)n-(3.15)
with c =satisfies
theboundary
conditions(1.2).
Proof
. We will prove that any such a solution has atmost ~ 2014 1 simple
zeros on
(0, T)
and atleast j
zeros on(0, T]. Indeed,
suppose forexample
that has more
than j -1
zeros on(0, T). Then,
due to the fact that thezeros are
simple
as well as to the continuousdependence
ofon t and on c
(the
latter onebeing
uniform withrespect
to t E[0, T]),
we conclude that any solution of
(3.14)n-(3.15)
with c = + g willsatisfy
that itsj1-th
order derivative has atleast j
zeros on(0 , T)
for everysufficiently
small g > 0 in contradiction with the definition of A similarargument
shows that has atleast j
zeros on(0, T]
and we concludethat
v~~l ~ (T)
= 0 andvt~~~
hasexactly j -
1 zeros on(0, T).
DNow,
since 0 cifor j
=2, ... , ~
and 0 > c_~!~ > co for,j
=1, 2,
... ,~,
we may assume,by considering subsequences
if necessary, that the sequences and~ C_~~n } ~ no
areconvergent.
Let us setWe will now state and prove our main result.
THEOREM 4.2. Let
f
EC1 (R~ satisfy
condition~1.3~
and suppose thatfor
some k EIN,
,Then, for
any h EC~
0, T ~,
there exists s* > 0 such thatfor
s >s*,
theboundary
valueproblem
has at least 21~ distinct solutions.
Proof
. Let us first consider the case k = 1. . In this case, we set s* = so and we have that the functions andgiven by
lemma 3.1 are twodifferent solutions to
(B.V.P.)
and our assertion follows.Let now k > 1.
Letting
s* = s4, it is clear that the solutionsj
=2, 3,
..., ~
andj
=1, 2,
... , k of(3.14)~-(3.15)
with c =c~
andc =
respectively, satisfy
theboundary
conditions(1.2),
and that theirj1-th
order derivative can have atmost j - 1
zeros on(0, T).
We will showthat in fact
they
haveexactly j -
1 zeros on(0, T). Suppose
forexample
that
~"~ has
lessthan j-1
zeros on(0, T).
Then at least one of the zeros ofthe
jl-th
order derivative of a solution of(3.14)n-(3.15)
with c =is
leaving
the interval(0, T) through
one of itsendpoints
as n - ~.Thus,
the number of derivatives of which vanish at that
endpoint
is at least3,
in contradiction withproposition
2.10.Similarly, v~31 ~
hasexactly j -
1zeros on
(0, T)
and the theorem followsby setting
and = ~
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CODDINGTON(E.)
and LEVINSON(N.)
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ELÍAS(U.) .
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ELÍAS(U.) .
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