A NNALES DE L ’I. H. P., SECTION C
Y. S. C HOI
P. J. M C K ENNA
A singular Gierer-Meinhardt system of elliptic equations
Annales de l’I. H. P., section C, tome 17, n
o4 (2000), p. 503-522
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A singular Gierer2014Meinhardt system
of elliptic equations
Y.S.
CHOI1, P.J.
McKENNADepartment of Mathematics, University of Connecticut, Storrs, CT 06269, USA
Manuscript received 14 September 1999, revised 6 December 1999 Ann. Inst. Henri Poincaré, Analyse non linéaire 17, 4 (2000) 503-522
© 2000 Editions scientifiques et médicales Elsevier SAS. All rights reserved
ABSTRACT. - We establish the existence of solutions to a
singular non-quasimonotone system
ofequations.
Suchequations
are aspecial
case of the Gierer-Meinhardt
equations.
In the one dimensional case, theuniqueness
result is alsoproved.
© 2000Editions scientifiques
etmedicales Elsevier SAS
Key words: Singular, Gierer-Meinhardt, Elliptic system, Non-quasimonotone
AMS classification: Primary 35J25
RESUME. - Nous etablissons
igi
l’existence des solutions d’unsysteme d’équations singulieres non-quasimonotones.
Cesequations
forment uncas
special
desequations
de Gierer-Meinhardt. En dimension un, nous demontrons aussi l’unicité des solutions. © 2000Editions scientifiques
etmédicales Elsevier SAS
1. INTRODUCTION
Singular elliptic boundary
valueproblems
for asingle equation
havebeen
widely
studied in thepast
several decades. Whilestarting
out as the1 Research partially supported by NIH grant 1 P41-RR 13186-01.
study
ofordinary
differentialequations
in[19,20],
itrapidly progressed
to the
study
ofelliptic
nonlinearboundary
valueproblems,
see[5,9].
Presently,
many of the earlier results arebeing generalized
to differentoperators,
oftenquasilinear
andanisotropic,
as these are thetype
that often occur inapplications
such as fluiddynamics [1,3,4,2]
Although
there is a substantial literature onsystems
ofelliptic partial
differential
equations [17,6,18],
to ourknowledge,
there has been nosimilar
study
of results forsystems
ofsingular elliptic problems
eventhough they
arisenaturally
inapplications.
For
example,
thesystem
usually
with Neumannboundary conditions,
often occurs in thestudy
ofmorphogenesis
onexperiments
onhydra,
an animal of a few millimeters inlength,
wherethey
are known as the Gierer-Meinhardtequations [13].
With similar interaction terms, these
type
also occur in certain models ofpredator-prey
interactions[21].
The
steady
states of(1.1)
will be theobject
of this paper. Often thecase that is studied is when the
di
term is small and thed2
term islarge.
In this case, the
study
of solutions of thesystem
can beapproximated by
a
single elliptic equation [12,7].
Of course, with Neumann
boundary conditions,
oneusually
studiessituations in which the solutions remain
positive,
so that in some sensethe
equations
do not becomesingular.
In this paper, we
begin
thestudy
ofsteady
states of thesystem
with Dirichletboundary
conditions. This is the natural extension of the afore- mentioned work on thesingle equation. Specifically,
we shall prove twotheorems,
one an existence theorem for thesystem
for aspecial
choiceof the
exponents
p, q, r, s, and the other auniqueness
theorem under the additionalassumption
that the number ofspatial
dimensions is one,i.e.,
we have a
singular system
ofordinary
differentialequations.
While there is a
good
deal of literature on nonlinearelliptic systems,
there seems to be little work on thistype
ofsingular system, especially
ifthe
system
is notquasimonotone.
First some
preliminaries:
our notation will be standard. The norms denote the usual LP and L °° norms on function u on the505
Y S. CHOI, P.J. McKENNA / Ann. Inst. Henri Poincare 17 (2000) 503-522
region
Q inRn, respectively.
The Sobolev will denote itsW2,p
norm, = +!!~~)~~
is its~
norm.2. MAIN RESULT
Let Q C
RN
be a bounded domain withboundary,
where y E(0,1).
We considerpositive
solutions to thesingular system
ofelliptic equations:
with zero Dirichlet
boundary
conditions:where a is a
given positive
constant. We note that thesystem
is notquasi-monotone, making
direct construction ofcomparison
functionsimpossible.
THEOREM 1. - Let S2 C
RN
be a bounded domain with bound- ary, where y E(0, 1).
There existpositive
solutions(u, v)
inC2 (SZ )
nC 1 (SZ )
toEqs. (2.2)
withboundary
conditions(2.3).
In the one dimen-sional domain case, the solutions are
unique.
Moreover suchunique
so-lutions u and v are
symmetric
about the midpoint of
the one dimensional domain.Remark 1. - The
equations
with
positive
constantsd,
a, and~8
can be reduced to our case.By dividing
the secondequation by d,
we can, without loss ofgenerality,
assume d = 1. Now with a new variable U = we obtain
Eqs. (2.2)
inU and v.
Remark 2. - In the one dimensional case, we can
by -
translation -assume the domain is
[-L, L]
for some L > 0. Once we establishthe existence and the
uniqueness
ofsolutions,
thesymmetry
claim in the above theorem is automatic. Sinceif
not, thenU (x )
=u ( -x )
andv (x)
=v (-x)
will constitute another solutions. This contradicts theuniqueness
result.Remark 3. - In the
proof
of thistheorem,
all the constructions involv-ing
functions inC2(S2) (for example,
w inEq. (3.1))
can bereplaced by
functions in
C2(Q)
nonly.
Withdomain,
suchregularity
canbe achieved
by
the L ptheory.
Hence the theorem is still valid for 1 domain.To prove the existence of
solutions,
we have to divide into three cases:a
1,
a =1,
and a > 1. First we start with the trivial case a = 1. The other two cases will beinvestigated
in later sections.THEOREM 2. - Let Q be a bounded domain with
boundary,
where y E
(0,1). If a
=1,
then there exists aunique
solution(u, v)
in.
Proof - Simple algebraic manipulation
onEqs. (2.2) gives
theunique
solution u = v = w, where w satisfies
Ow - w -f- 1 = 0
with zero Dirichletboundary
conditions. It is noted that the Schauder’s estimateon linear
elliptic equations
ensures that w EC2(Q).
DThroughout
this paper, we let > 0and cp
> 0 be the firsteigenvalue
and the
corresponding eigenfunction satisfying
with the normalization max03A9 03C6 = 1.
3. EXISTENCE OF SOLUTIONS WHEN a 1
In this
section,
westudy
the existence of solutions as stated in Theorem 1 when a1, by using
Schauder’s fixedpoint
theorem. Wecan construct an upper bound function
(hence
apointwise bound),
butto ensure that a
positive
solutionexists,
we need to construct atricky integral
lower bound(see
the definition of the set Sbelow).
First,
we define w to be the solutionsatisfying
Hence w E
C2+Y (S2 ) .
The maximumprinciple
ensures that w > 0 in Q .With w well
defined,
let w be the solution of the linearequations:
507
Y S. CHOI, P.J. McKENNA / Ann. Inst. Henri Poincare 17 (2000) 503-522
Again
W E(Q),
and w > 0by
the maximumprinciple.
Many apriori
estimates for thistype
of linearequations
exist. The results that we will need are summarized as follows.LEMMA
then there exists a constant
k03B2
> 0 suchthat ~~C 1 f ~~.
Proof. -
This is a consequence of LPtheory
on linearelliptic equations
with p >
N,
followedby
the Sobolev’simbedding.
DWhen a
1,
it is easier tostudy
anequivalent system
to(2.2).
Letz = v - u. Then
Eqs. (2.2)
can be converted into:subject
to zero Dirichletboundary
conditions for both u and z. We note that thissystem
is still notquasi-monotone
if we look forpositive
solutions for u and z.
THEOREM 3. - Let Q be a bounded domain with
boundary. If
0 a
1,
then there existpositive
solutions(u, v)
in nC1 (S2).
Proof -
It suffices to establish existence ofpositive
solutions(u, z)
forEqs. (3.3)
with zero Dirichletboundary
conditions on both u and z. Thenv = u + z will also be
positive.
Let
ki and ka
be the constants in Lemma1,
when~8
is 1 anda,
respectively.
Define A - + and M -(1
+~.1 ) ( I I w ( I ~
+ Let 8 >0,
to be determinedlater,
andS ==
(u, ,z)
EC1(Q)
xC1(Q): 0
u w,0
zw, u|~03A9
=z|~03A9
=0,
When 8 is
small, S
isnon-empty. Moreover,
S isclosed, bounded,
andconvex.
Define the map T such that for all
(u, z)
eS, ( T u , T z )
solve the linearsystem
ofequations:
with zero Dirichlet
boundary
conditions for both T z and T u .Using
Schauder’s estimate on
Eq. (3.4a),
we can conclude that Tz eSince u E
S,
we haveu
w. Hencetaking
the difference of theEqs. (3.2)
and
(3.4a) gives 0 (w - Tz) - a (w - Tz) X
0 with zeroboundary
condition for
( w - T z ) .
Hence the maximumprinciple gives u) ~
T z > 0in Q. The strict
inequality
is a consequence of thestrong
maximumprinciple
since 0 in(3.4a).
Inaddition,
it follows from Lemma 1that ~Tz~C1 A.
With T z > 0 in
Q, uf(u
+ 1. Hence L ptheory guarantees
a solution T u EW2,p
forEq. (3.4b)
with any p > N. Infact,
since bothu and Tz are in
u/(u
+Tz)
is in We can concludethat T u e
C2(Q) by
interior Schauder’sestimate,
and can thereforeemploy
the classical maximumprinciple
on T u .Taking
the differenceof the
Eqs. (3.1)
and(3.4b)
leadsto w >
T u > 0by using
the(classical)
maximum
principle.
Inaddition,
lemma 1gives ~Tu~C1
A.Now for any
(u, z)
ES,
we have a uniform L °° bound on( 1 - a ) u
andu / (u
+Tz).
This leads to a uniformW2,p
norm bound on T z and T u for any p > N. Since iscompactly
contained inC 1,
T is acompact
map if we can show T maps S into S.To show T : 5’ 2014~
S,
it suffices to check the twointegral inequalities
inthe definition of S.
Everything
else has been shownalready.
From(3.4a), multiply by cp
andintegrate
overSZ ,
This ensures that Tz satisfies the second
integral inequality
in thedefinition of S.
’
Finally,
we wish toverify ~, dividing
itsproof
into twocases. Define
Qs
=={x
E Q :dist(x,
>e ~ .
We will take e= ~
inthe
following,
with 6 > 0sufficiently small,
to be determined in theproof.
In the
following,
allpositive
constants mi areindependent
of .Y.S. CHOI, P.J. McKENNA / Ann. Inst. Henri Poincare 17 (2000) 503-522 509 Since w E for x E
Q B
we havew(x) ~
for some constant m ~ > 0. Hence
Since the first
eigenfunction w
has non-zeroslope
at theboundary,
thereexists a
positive
constant m2 such thatAdding
the above twoinequalities, ~u~1 m 3 ( 1
+ for someconstant m 3 >
0,
which isindependent
of .Since we have a uniform
bound
on theC1 norm
on u ES, Nirenberg- Gagliardo’s inequality (statement (3)
in Theorem2.2, [ 10])
ensures that there exists a 6~ E(0, 1),
andpositive
constants m4 and m5 such that for all uE S,
Now from
Eq. (3.4a),
Lemma 1gives
+M) 8 ~ 1-e~ ~2 .
Hence
by choosing 8
> 0sufficiently small,
Now, multiplying Eq. (3.4b) by 03C6
andintegrating
over we haveTherefore 03C6Tu
203B4. Hence T : S - S in this case.Case
II: u03C6 (1
+M)03B4.
From
Eq. (3.4b),
Hence T
maps S
into S in both cases.By
the Schauder’s fixedpoint
theorem,
T has a fixedpoint (u, z),
which satisfiesEqs. (3.3)
with zeroboundary
conditions. Sinceu ~
0 inS2,
we have z > 0 in Qby using
thestrong
maximumprinciple
onEq. (3.3a).
This in turnimplies
u > 0using Eq. (3.3b). Moreover,
interior Schauder’s estimate ensures thatthey
areD
4. EXISTENCE OF SOLUTIONS WHEN a > 1
Define g
EC2 (S2 )
to be the solution ofFirst we need an existence result for a scalar
singular equation.
LEMMA 2. - Let S2 be a bounded domain with
boundary.
Assume p E
p > 0,
and = 0. Then there exists aunique positive
solution u EC2 (SZ )
nC 1 (S2 )
to thesingular equation
Moreover, u >
g.Proof -
We follow theproof
in[9]. Essentially, although
theequation
is
singular
at theboundary,
if we can construct apositive
upper solution andpositive
lower solution with the upper solutionlarger
than the lowersolution,
and both vanish at theboundary,
then a classical solution in nC(Q)
exists.Define M =
A
for some constant A > 0. ThenRecall
{x
e Q :dist(x, > ~ { . If E > 0
issufficiently small,
thenthere exists a constant m > 0 such
that
for all x e Itis clear then .
if we choose A to be
sufficiently large.
Y.S. CHOI, P.J. McKENNA / Ann. Inst. Henri Poincare 17 (2000) 503-522 511
For x E since there exists a ~ > 0 such that
minj
~.By
furtherincreasing
A if necessary,Thus 0 u - a u + + 1
~
0 for all x e Q. Hence u is an an upper solution forEq. (4.2)
when A issufficiently large.
Next we define u =
By taking
eo > 0sufficiently small, u >
u .Reducing
eo further if necessary, we haveHence _u is a lower
solution,
and there exists a classical solution u eC2(Q)
nC (Q )
forEqs. (4.2). Uniqueness
of such solution followseasily by using
the maximumprinciple.
Now since p E and =
0,
we have/? ~ Mcp
for someconstant M > 0.
Therefore,
This norm bound ensures that u e
by
Lemma 1.Finally
theclaim
u >
g is a consequence of the maximumprinciple.
Thiscompletes
the
proof
of this lemma. DAnalogous
to the case when a1,
we willstudy
anequivalent system
to
Eqs. (2.2)
when a > 1. This time wedefine z
--_ u - v, andstudy
with zero
boundary
conditions for both z and v.Again
we willemploy
Schauder’s fixed
point
theorem inproving
thefollowing
theorem.THEOREM 4. - Let Q be a bounded domain with
boundary. If
a >
1,
then there existpositive
solutions(u, v)
inC2 (S2 )
nC 1 (SZ ).
Proof -
It suffices to establish the existence ofpositive
solutions(v, z)
for
Eqs. (4.3)
with zero Dirichletboundary
conditions on both v and z.Then u = v + z will also be
positive.
Let
ki
be the constant in Lemma 1when ~ = 1,
and B == xmax{a - 1, 1}.
Let A >1,
to be determinedlater,
andA1 1)A.
Now define
For
large A,
S isnon-empty. Moreover, S
isclosed, bounded,
and convex.Define the map T such that for all
( v , z)
ES, ( T v , T z )
solve thesystem
ofequations :
with zero Dirichlet
boundary
conditions for both Tz and T v . Note that the firstequation
is linear.Using
Schauder’s estimate onEq. (4.4a),
wecan conclude that Tz e
C2+Y (S2 ) .
The maximumprinciple
alsogives
Tz > 0.
Moreover,
it follows from Lemma 1 thatThe second
Eq. (4.4b)
is asingular equation
as in Lemma2,
and has asolution T v E
C2(S2)
n andT v >
g.If we can
show B,
andA,
then T : S - S. To prove the firstinequality, multiplying Eq. (4.4a) by Tz,
andintegrating
over
Q ,
weget [ [ T z [ ( H 1 (a -1 ) [ [ v [ [ 2
afterusing
theCauchy-S chwarz’s inequality.
Then fromEq. (4.4b),
we haveHence,
This
quadratic inequality
leads to:Y.S. CHOI, P.J. McKENNA / Ann. Inst. Henri Poincare 17 (2000) 503-522 513
This
jH~ norm
bound willhelp
us to obtain theC~ norm
bound.Let mi
be
somegeneric positive
constants. Take any x E Q . We haveg (x ) > m1dist(x,
for some m 1 >0,
andT .z (x )
by
the mean value theorem forsome §
EQ ,
which lies in thestraight
line
connecting
x and its closestpoint
to Hence fix some p >N,
by using
the Sobolev’simbedding
and LP estimate onEq. (4.4a).
Define8 --_
(p - 2) / p .
Hence 0 8 1. Wehave ~v~p ~v~03B8~~v~1-03B82.
Combining
with theinequality
that wejust obtained,
weget
Employing
Lemma 1 onEq. (4.4b)
now with(4.6),
and A >1,
itfollows that
if we take A
sufficiently large
in the lastinequality.
Hence T : 5’ 2014~ S.Take p > N. For all
( v , .z )
ES,
because of thebound (4.6),
we have aW2,p
norm bound on both v and z. SinceW2,p
iscompactly
contained inT is a
compact
map.By
the Schauder’s fixedpoint theorem,
we have a fixedpoint (v, z)
in S which satisfies
Eqs. (4.3).
The solution v and z arepositive,
and inC 2 ( S2 )
nC 1 ( S2 ) by
similararguments
as in theproof
of Theorem 3. 0 Theorems 2 to 4 establish the existence of solution as summarized in Theorem 1. We now move on tostudy uniqueness
of solution for one dimensional domains in the next section.5.
UNIQUENESS
OF THE SOLUTION IN ONE DIMENSIONWithout loss of
generality,
we let the domain Q be(0,1),
andstudy
theuniqueness
ofpositive
solutions(u, v)
inC2 (O, 1 )
nC 1 [O, 1] satisfying Eqs. (2.2).
There is no distinction between the a 1 case and the a > 1case.
Because
u / v
> 0 inQ ,
then fromEq. (2.2a),
we have u" - u 0 in(0,1). Application
of theHopf’s
lemma(p. 5,
Theorem2, [14]) gives
u’(0)
> 0.Similarly, Eq. (2.2b) yields v’ (0)
> 0. Similaranalysis
can bemade at x = 1. Now as x -
0,
we haveThus all solutions
(u, v)
inC2 (o, 1)
n1]
are in factC2 [o, 1].
Weneed an extension of the result in Lemma 2.
LEMMA 3. - Let the same
assumptions
in Lemma 2 hold. Let ui i be thepositive
solutionof Eqs. (4.2)
with zeroboundary
conditions whenp(x)
ispi (x ),
i =1, 2.
Assumep2 (x ) >
pi(x)
in[o, 1 ]
andp2 ~
p 1, then u 2 > uin [0, 1 ] .
Proof -
Define u w u 2 - u 1. Then fromEq. (4.2)
for i =1, 2,
weobtain
Since =
0,
andu (0)
> 0 fori = 1, 2,
hence thepositive
coefficientp2 (x) /Cu 1 u2)
behaves like1 /x
as x -~ 0. Now animproved
versionof the maximum
principle (p. 6,
Theorem3, [14])
allows for such unbounded coefficient(so long
as we have asign restriction),
we cantherefore conclude û > 0 in
[0, 1]
unless û isidentically
zero. Sincep2 ~
p~ in[0, 1],
we can exclude this case. Theproof
of this lemma is nowcomplete.
DLEMMA 4. - Let
(u i , vi )
EC2 [o, 1 ]
bepositive
solutionsof (2.2)
withzero
boundary conditions,
i =1, 2.
Assume(u2, v2) ~ (u 1, v1)
in[o, 1].
Then u
and v2 Q
v 1.Proof. -
Note that if u 2 _ ui, then v2 = vi 1 from thegoverning Eq. (2.2a). Similarly
when v2 --_ then u2 = ui 1by using Eq. (2.2b).
Since
(u2, v2) ~ (Mi,
in[0, 1],
we have uand v2 Q
DTHEOREM 5. - Let
(u i ,
E1]
bepositive
solutionsof (2.2)
with zero
boundary conditions,
i =1, 2.
Assume(u2, v2) ~ (u 1, v1)
in[0, 1].
Thenu2 ~
uand v2 ~
vi,i. e.,
neither componentofeach
solutioncan be ordered.
Proof -
We will assumev2 >
vi in[0, 1],
and arrive at a contradiction.First we note that
Y.S. CHOI, P.J. McKENNA / Ann. Inst. Henri Poincare 17 (2000) 503-522 515
Multiplying
the firstequation by
u 2, the secondequation by
ui, andsubtracting
oneequation
fromanother,
we haveSince vi
by
Lemma4,
thisgives
a contradiction. Hencev2 ~
vl .Next we assume
u 2 >
u 1. From Lemmas 3 and4,
we obtain v2 > v 1.We have
just proved
that this isimpossible.
Hence this contradiction enables us to concludeu 2 ~
u 1. DSuppose
there are two distinctpositive
solutions =1, 2.
Define û = u2 - u
and
v = v2 - In the abovetheorem,
we havejust
established that neither û nor v can be
single-signed.
We nowproceed
toexclude the
sign changing
cases for û and v .We will need the
following lemma,
theproof
of which wepostpone
tillthe next section.
LEMMA 5. - Both û and v have
finite
numberof zeros
in[o, 1 ],
unlessthey
areidentically
zero.We now prove the
uniqueness
of solution under this additionalassumption.
DefineBecause of the Lemma
5, I+
consists offinitely
manydisjoint
closedintervals,
say(I+)i,
i =l,
... , ml,i.e., I+
=Ui(I+)i.
It is noted that in each(I+)i ,
there can be at mostfinitely
manypoints
at which û = 0. Tosimplify
ournotation,
weemploy I+
to denote any interval(1+) i.
Similarremarks
apply
forI_, J+,
and J-(see Fig. 1).
LEMMA 6. - The
following four
cases hold:(a) J+~
(b) J_,
-
(c) I+~
(d)
/..Proof. -
Examine case(a)
first. AssumeI+
CJ+,
and letI+ = [a, b]
for some
0 a b
1. From bothEqs. (5.1)
for i =1, 2, simple algebraic manipulation yields
Since =
u2 (a)
andu 1 (b)
=u2 (b),
the aboveequation
issimplified
to
Since
u2 {b) u 1 (b)
andu;(a),
the first two terms on the lefthand side are
non-positive.
Theintegral
term isstrictly negative,
since[a, b]
cJ+,
andv
is zeroonly
onfinitely
manypoints
on(a, b).
Hencewe arrive at a contradiction. The
proof
of case(a)
iscomplete.
Theproof
of case
(b)
is similarby studying
the sameEqs. (5.1).
Next we look at case
(c).
Assume J- C1+
and let J- =[a, b]
forsome clear that
u 2 / v2
> on(a, b).
Hence fromEqs. (2.2)
for i =1, 2,
weget
with the
boundary
conditionsv (a )
=v(b)
= 0. The maximumprinciple
gives v
> 0 on(a, b).
This contradicts the interval[a, b]
= J-. Theproof
Y.S. CHOI, P.J. McKENNA / Ann. Inst. Henri Poincare 17 (2000) 503-522 517
of case
(c)
iscomplete.
Theproof
of case(d)
is similar to that for case(c).
DTHEOREM 6. - Let Q be a bounded interval in the one dimensional
case. Then
positive
solutions(u, v)
inC2 (0, 1)
nC 1 [O, 1]
areunique.
Proof -
Without loss ofgenerality, by picking
which solution weassign
as i =1,
we can assume û > 0 in aneighborhood
of x = 0. Letû > 0 on
[0, al], [a2, a3], ... , [a2n, a2n-~-1]~
0 on[a, a2], [a3, a4],
.... Thus a2n+1 =
1,
or a2n+2 =1, depending
on whether the last intervalthat û does not vanish is
positive
ornegative.
We divide into two cases in
studying
the interval[0,
Case I: v > 0 near x = 0.
On the interval
(0, al ),
V has tochange sign exactly
once. For ifv
doesnot
change sign,
then it contradicts case(a)
in Lemma 6. If vchanges sign
more than once, then it contradicts case(c)
in Lemma 6.Case II: v 0 near x = 0.
On the interval
(0, al ),
v cannotchange sign.
For if vchanges sign,
then it contradicts case
(c)
in Lemma 6.Hence in both cases, v is
(strictly) negative
at x = al . It has tochange sign exactly
once inside the interval(ai , a2).
For if v does notchange sign
in(a I , a2),
it contradicts case(b)
in Lemma 6. If vchanges sign
more than once, then it contradicts case
(d)
in Lemma 6. Thus v > 0 atx = a2.
A
repetition
of sucharguments gives
v 0 at x = a3, t) > 0 at x = a4,and so on. If the last interval is
[a2n, 1 ],
then we know v > 0 at x = a2n . . Because of case(a)
in Lemma6,
v has tochange sign
at some interiorpoint
in(a2n, a2n+I).
But then v has to become zero at or before x = 1because of its
boundary
condition. It will contradict case(c)
in Lemma 6.The case that the last interval is
[a2n+l, 1 ]
where u 0 can be treatedsimilarly.
Thiscomplete
theuniqueness proof
if Lemma 5 holds. aThe
proof
of Lemma 5 is the contents of the next section.6. PROOF OF LEMMA 5
In this
section,
we will establish Lemma 5. This will finish theproof
of Theorem 1 as well.
LEMMA 7. - Let w =
(w 1, w2)
EC2 [a, b]
be a nontrivial solutionof
a linear system
of
twoequations
on the interval[a, b] :
where A E
C [a, b].
0 andA2{ (x) ~ 0 for all
x E[a, b],
thenneither w nor w2 can have
infinite
numberofzeros
in[a, b].
~
Proof -
First we assume that there are infinite number of zeros{xn }, n = 1, 2, ... ,
for w 1 in[a, b].
Take asubsequence
of{xn }
if necessary,we can assume xn --~ xo for some xo E
[a, b].
Without loss ofgenerality,
assume that
{xn }
is a monotonedecreasing
sequence.(The
other case thatit is a monotone
increasing
sequence can be treated in the sameway. ) By
Rolle’stheorem,
we canreadily
conclude that =0,
=
0,
and = 0. EvaluateEq. (6. la)
at x = xo, we haveAI2(XO)W2 (xo)
= 0. Sinceby assumption 0,
we obtainW2(XO)
= 0. If=
0,
thenuniqueness
of initial valueproblem
forEqs. (6.1 )
forcesw = 0 and w2 == 0 for all x e
[a, b].
Since w is not a trivialsolution,
wecan assume that 0.
Let 8 > 0 be
sufficiently
small and x E(xo,
xo+ 8].
Thenbecause behaves like
o ( (x - w2 (x ) ~ > m (x - xo )
for somem >
0, and IA12(x)1
has apositive
minimum on the interval[a, b].
HenceEq. (6.1 a) gives w 1
is non-zero and issingle-signed
for x E(xo,
xo +8].
This contradicts that there are infinite zeros of w on
(xo,
xo +8].
Hencew
has
to have finite number of zeros in[a, b].
The
proof
of w2having
finite number of zeros in[a, b]
is similar. It invokes theassumption A21
is nonzero for all x e[a, b].
DFrom
Eqs. (2.2),
asimple
calculation shows that if w ~(M, v),
then wsatisfies
Eqs. (6.1 )
with matrixThis matrix A is
singular
near theboundary
x = 0 and x = 1. However for any subinterval[a, b]
c(0,1),
Lemma 7 allows us to conclude that there are finite number of zeros in[a, b]
for û and v. It remains to showY.S. CHOI, P.J. McKENNA / Ann. Inst. Henri Poincare 17 (2000) 503-522 519
that there are no infinite number of zeros of û and
v
near x = 0 and x = 1.First we recall the
following simple
theorem.THEOREM
7. - If f
EC[0, 1],
and u EC2 [o, 1] satisfies u" = f(x)
with
u (0)
=u’ (0)
=0,
thenfor
any x E[0, 1],
We need this formula to establish the
uniqueness
theorem of initial valueproblem
for asingular,
linearsystem
ofequations
below.LEMMA 8. - Let
Eqs. (6.1)
hold with w EC2 [o, 1 /2].
Moreoverw(0)
=0, w’(0)
=0,
and A EC(o, 1/2]. Furthermore,
letlimx~oxA(x)
exist. Then w n 0 on
[o, 1 /2].
Proof -
Define the matrixB(x)
=x A (x ) .
Hence B EC [o, 1 / 2]
dueto the
assumption
on the existence of the above limit. Let M ==(where
theinfinity
norm here is theinfinity
matrixnorm on a 2 x 2
matrix),
and k ==(where
theinfinity
norm here is theinfinity
vector norm on a 2 x 1vector).
It isnoted that k is well defined because
w(0)
=0,
and w is inC~ [0,1/2].
Now formula
(6.3)
can beapplied
toyield
which in turns lead to:
Hence,
Putting Eq. (6.5)
back intoEq. (6.4),
we obtain animproved
estimate:Using induction,
it is easy to check that for anypositive integer n,
+
1 ) ! ) .
Take n -~ 0. Hence w --_ 0 on[o, 1 /2] .
aWe now finish the
proof
of Lemma 5by showing
that no infinitenumber of zeros of û and v near x = 0 and x = 1. We focus on x = 0.
The other end x = 1 is similar. We first consider three cases.
Case I: û has infinite zeros in a
neighborhood of x = 0,
andv’ (o) ~
0.Because of the existence of infinite zeros of û near x =
0,
we haveu’ (o)
= 0. Thusu 1 (0)
=u 2 (o) .
Withv’ (o) ~ 0,
we knowv 1 (o) ~ v2 (0) . (Recall
that bothvl (0)
andv2 (0)
arepositive.)
Now from
Eq. (2.2a)
for i =1, 2,
we can derive:Since
hence |u2/v2 - u1/v1| is
bounded away from zero in asufficiently
smallneighborhood [0, 8]
for some 8 > 0. Reduce 8 further if necessary, theright-hand
side ofEq. (6.7)
issingle-signed
and non-zero in[0, 8],
becauseM(0)
= 0. Sinceu (o)
= 0 andu’ (o)
=0,
this will contradict that û has infinite number of zero in[0, ~ ] .
Hence case I cannot exist.Case II: û has infinite zeros in a
neighborhood of x = 0,
andv’ (0)
= 0.In this case,
M(0)
=u’ (o)
= 0 andv (o)
=v’(O)
= 0. Since w -(u, v)
satisfies
Eqs. (6.1 )
with the matrix Agiven by (6.2),
Lemma 8gives
u - 0and v - 0 on the interval
[o, 1 /2] .
Now we can use Lemma 7 to conclude thatthey
areidentically
zero on[0, 1].
Hence case II isequivalent
toidentically
zero û and v .Case III: v has infinite zeros in a
neighborhood of x = 0,
andu’ (o) ~
0.Using
similarproof
as in case I onEq. (2.2b),
we haveThe same argument shows that this case does not exist.
521
Y.S. CHOI, P.J. McKENNA / Ann. Inst. Henri Poincare 17 (2000) 503-522
When there are infinite number of zeros for û near x =
0,
andonly
finite number of zeros for
t),
this is either case I or case II. But case I doesnot exist. Thus from the conclusion of case
II,
û == 0 and v = 0 on[0, 1 ] .
Similar remark
applies
when there are infinite number of zeros for vnear x = 0 and
only
finite number zeros for M. Since case III does notexist,
we end up with case IIagain.
When there are infinite number of zeros for both û and v near x =
0,
this is case II. Hence we draw the same conclusion that both û and v are
identically
zero on[o, 1 ] .
Theproof
of Lemma 5 is nowcomplete.
7. CONCLUDING REMARKS... WHAT’S LEFT
We conclude this paper with a short outline of the main open
questions
left
by
this paper.To some extent, in
taking
all theexponents
in thecoupling
termsof
(1.1)
to be one, we have made theequation
lesssingular
than itotherwise would be. The most obvious open
question
is whathappens
for different
exponents. Presumably,
forexistence,
we need a criticalexponent
limitation onthe p
and r, and(if
thesingle-equation
literatureis any
guide)
no restriction onthe q
and s. To ourknowledge,
there areno additional results even in the one-dimensional case or when radial
symmetry
isimposed.
In
addition,
we have no information on whether the solutions we findare
radially symmetry
if theregion
is a ball. It is natural to ask if allpositive
solutions of(2.2)
areradially symmetric
or if somesymmetry-
breaking
can result.A
stronger
version of thisquestion
is whether thepositive-positive
solution is
always unique.
Note thatuniqueness
is still open even in onedimension for the more
general exponent
case. For other values of p, q, r, s, formalarguments
onEqs. (1.1)
leads to asingle elliptic equation,
which has
multiple
solutions for Neumannboundary
conditions.And
finally,
one can wonder whether these methods can begeneralized
to a wider class of nonlinearities than the pure
exponents.
REFERENCES
[1] 010Cani0107 S., Lee Keyfitz B., An elliptic problem arising form the unsteady transonic
small disturbance equation, J. Differential Equations 125 (1996) 548-574.
[2] Choi Y.S., Kim E.H., On the existence of positive solutions of quasilinear elliptic boundary value problems, J. Differential Equations 155 (1999) 423-442.