A NNALI DELLA
S CUOLA N ORMALE S UPERIORE DI P ISA Classe di Scienze
K EITH M ILLER
Exceptional boundary points for the nondivergence equation which are regular for the Laplace equation - and vice-versa
Annali della Scuola Normale Superiore di Pisa, Classe di Scienze 3
esérie, tome 22, n
o2 (1968), p. 315-330
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NONDIVERGENCE EQUATION WHICH
ARE REGULAR FOR THE LAPLACE
EQUATION - AND VICE-VERSA
by
KEITH MILLER(*) (Berkeley)
In
[9] Littman, Stampacchia,
andWeinberger investigated
the notionof
regular boundary points
for Dirichlet’sproblem
withrespect
to the uni-formly elliptic equation
indivergence
formwhen the coefficients are
only supposed
to be measurable. The coefficient matrix is assumedsymmetric
and the uniformellipticity
condition may be statedeigenvalues
ofwhere a is the
ellipticity constant,
0C
aC
1.They proved
that theregular points
forequation (1)
are the same as those forLaplace’s equation.
In this paper we prove that the
analogous
result is false for the uni-formly elliptic equation
innondivergence
formwith the same class of coefficients. In
fact,
it is false for every oc 1 when n = 3.Pervenuto alla Redazione il 16 Dicembre 1967.
(*) The researoh for this paper was partially supported by Air Force Contract num-
bei : AF-AFOSR 553-6-1 .
Our method involves consideration of
spines
ofvarying degrees
ofsharp-
ness
(see Figure 1)
such as theexponentially sharp spine
ofLebesgue [4,
p.
303].
In Theorem 5 of Section 5 we show that for n = 3 and for everyx
C
1 there exists a certainnondivergence equation
for which thetip
ofevery
algebraic spine
isexceptional, although
suchpoints
areregular
forLaplace’e equation.
The coefficients for thisequation, incidentally,
are C°°in the interior and become discontinuous
only
on the axis of the exteriorspine.
In Theorem 6 we take care of the K vice-versa ~portion
of thetitle;
we show that for n = 3 and for every « 1 there exists a certain nondi- vergence
equation
for which thetip
of everyspine (no
matter howsharp)
isregular, although
thetip
of theexponential spine
ofLebesgue
is egcep- tional forLaplace’s equation. Again
the coefficients are C°°except
on theaxis of the exterior
spine.
Only
in the case n = 3 do we haveexamples
of « both waysnonequi-
valence >> for a
arbitrarily
close to 1.However,
in Section 3 we will showthat for every n
> 2,
if a1 n- 1
, then an isolatedboundary point
is re-gular
for a certainnondivergence equation, although
it is of course excep- tional forLaplace’s equation.
The coefficients involved are discontinuousonly
at the isolatedpoint
andanalytic
elsewhere in Rn.There has been
great
interest in thequestion
ofregular boundary points
for
elliptic equations.
Severalauthors,
even before the paper ofLittman, Stampacchia,
andWeinberger,
had established the result that aboundary point
isregular
for theequations. (1)
and(3)
if andonly
if it isregular
for the
Laplace equation ;
all of theseproofs
however use smoothness of the coefficients in an essential way. Forexample,
R. Herv6[6] proved
this result for the
nondivergence equation (3) (also
with lower order termsincluded) provided
that the coefficients arelocally Lipschitz
continuous. For other references toprevious
results with smooth coefficients and to classical results forLaplace’s equation
see the introduction to[9].
On the
positive
side for thenondivergence equation (3),
thepresent
author has
recently
shown[10]
that everyboundary point
with an exteriorcone
(no
matter how small theaperature)
isregular
for(3) (also
with lower’ order terms
included).
Notice thatboundary points
withonly slightly
worsegeometry,
i. e. with exterioralgebraic spines (no
matter howblunt),
areexceptional
for certainnondivergence equations, n ¿ 3,
as is shown in thepresent
paper. In Section 5 we mention several other results which areimmediately explained by
ourconstruction,
forexample
the results for La-.
place’s equation
thatexponential spines
areexceptional
when n = 3(Lebes- gue’s example)
and thatalgebraic spines
areexceptional
when n>
3(which
could be
computed
from Wiener’scriterion).
Our construction of a nonsolvable Dirichlet
problem
in Section 5 fol -lows the method of
Lebesgue?s
classicalexample
andrequires
that a boun-ded
discontinuity
of a solution at aboundary point
be removeable. This isalways possible only
if a ; otherwiseexamples
show that it mayn-1
be false. When a we show in Theorem 2a that a
singularity
atn-1
an isolated
point
or aboundary point
is removeableprovided
that a certaingrowth
condition is satis6ed at thepoint.
Pucci[13,
p.157]
haspreviously
derived
equivalent theorems; Gilbarg
and Serrin[5]
have used similar tech -niques,
but are concerned withequations
with continuous coefficients.When a
1’
n-1 we show in Theorem 2b that an isolatedsingularity
isremoveable
provided
that a certain order Holdercontinuity
holds at thepoint.
This result has no relevance to our maintopic,
but it follows imme-diately
and it appears to be new.1.
Pt·elimitiaries.
Let x = ., . ,
xn)
denote a real vector in2,
and is a real val-ued function defined
(and usually
of classC2)
in some bounded open setD of interest. We let
-0.
denote the class ofoperators
L of form(2), (3).
We
constantly
use the weak maximumprinciple :
thatis,
if u ECO C2 (~), Lit ? 0
in on8Q, then uc0
onD.
Given an L
E d3a
and a continuousfunction q;
on8Q;
the Dirichlet pro- blem is to find a solution ofis some function space with
generalized
second derivatives such that the maximumprinciple
still holds.Certainly
withonly
measure -able coefficients we cannot
expect
aC2 (0)
solution. It is notyet
clear whatshould be our choice of
9( (Q), although
recent results of Aleksandrov[11
and Pucci
( 14~
indicate that the Sobolev is alogi-
cal choice.
DEFINITION. A barrier
foi-
L at thepoint x°
EaQ
is a function tc, de- ufied in some relativeneighborhood
N =( U
an openneighborhood
of
x°),
which satisfiesregular boundary point for
L(or
for theequation
Lu =
0)
if there exists a barrier for L atx 0 Otherwise x°
is anexceptional point for
L.In the case of
Laplace’s equation, solvability
of the Dirichletproblem
for
arbitrary
continuousboundary
data isequivalent
to existence of a bar- rier at everyboundary point. Unfortunately
we cannot atpresent
prove thesame
equivalence
for thegeneral nondivergence equation.
Afterall,
with thepresent
state of thetheory
we cannot provesolvability
of(4)
even on thesphere
if the coefficients are not smooth. If we were to find anexample (4)
with no solution we
might
have reason tosuspect
that thedifficulty
iscaused
by
the nonsmooth coefficients in the interior and notby
thepatho- logy
of theboundary. However,
if the coefficients of L are Holder conti-nuous on
compact
subsets of,ia,
then we can prove that theequivalence holds,
as statedprecisely
in thefollowing
theorem. For this reason we res-trict our
example
to L with smooth(in
factC°°)
coefficients in the interior.THEOREM 1.
Suppose
thecoefficients of
L are Hi;lder continuous on eachcontpact
subsetof
0.Any
solutionof (4)
must then be inC2 (Q).
1.’here existia solution
of (4) for
every continuousboundary function
99ij*
andonly if
thereexists a barrier
for
L at everyboundary point.
Proof.
Theproof
of the first assertion is trivial. Let it be a solution of(4)
and consider anysnbsphere 8
with S c Q.By
the well known exis- tencetheory
forequations
with Holder continuous coePficients[4,
p.339]
there exists a
C2 solution it,
on S which agrees with 2c on We laverequired
that the maximumprinciple
also hold for the class9f; hence,
u = Ul on S. See
[10,
p.98]
forproof
of the second assertion.We now define certain
types
ofspines,
in order ofincreasing sharp-
ness. Let D be the
body
of revolution obtainedby rotating (for n > 2)
the shaded area shown in
Figure
1 about the xn axis. We may assume thataD
isexcept
at0,
thetip
of theintruding spine.
Let(r,0)
be.
polar
coordinates with thepositive
Xn axis aspolar
axis. Thatis, r (x) = I x I
and We say that
aD
contains a conicalspine
it’there exists a
positive
constant k such that(~2013 8) ~ k
as x --~ 0 on¿Q;
3~
contains analgebraic spine
if there existpositive
constants k and p such that(n
-9) =
krP as x -+ 0 on8Q
contains anexponeittial spine
if there exist
positive
constantsk, C,
and A such that(71
-9) ~
Ic exp(- Cr-1)
Fig. 1
as x -+ 0 on
00.
As adegenerate
case we say that contains a linesegment spine
if(~z
-0)
= 0 for all xsufficiently
near 0 on for exam-ple,
S~ =;x : 0 ~ x ( 1
and .x is not on thenegative xn axisl.
Finally,
we have the case of an isolatedboundary point,
forexample,
S~ = yx : 0 ~ x ~ 1;.
ThisaD
may be considered ’ to have acompletely degenerate spine ;
we have made thebody
of thespine
so thin that wehave
wiped i~
outcompletely, leaving only
itstip,
0.2.
Removeabte singularities.
In this section we very
simply
andprecisely
characterize thegrowth condition,
independence
upon a, which is necessary to insure that a sin-gularity
at an interior orboundary point
be removeable. What weactually
show is that the maximum
principle
continues to hold as if thesingularity
were not there. It therefore
might
be moreprecise
to call such a resultan «extended maximum
principle »,
inkeeping
with theterminology
ofGilbarg
and Serrin[5],
rather than a theorem on « removeablesingularities ».
Consider
radially symmetric
functions of the formw (x) = g (r).
Theclass
Eo.
is invariant under translation and rotation of the coordinate axes.At each fixed
point x° ~
0 we choose coordinate axes withorigin at x°
such that
the yi
axis is in the radial direction and axes arein the
tangential
directions. Withrespect
to these axes, = grr , and the cross derivatives are all zero. There-fore,
theoperators
Lapplied
to radialfunctions,
areexactly
of theform
where the functions
oci (x)
are measureable andsatisfy a ~ ai (x) ~ 1.
We consider theparticular operator Lo
which at eachpoint
has coefficient 1 for the second derivative in the radial direction and coefficient a for the second derivatives in thetangential
directions. Theequation
then has solutions w
(x)
= 1 andThus for the solution .w in
(7)
ispositive
and tends to-~
oo atthe
origin. Moreover,
it is a solution of Pucci’smaximizing equation [13],
as is
easily
seen from therepresentation (5),
so it is asupersolution (i.
e., L1C ~0)
for all L Hence it can be used in the standard fashion to removesingularities
which grow moreslowly
than itself.THEOREM 2a.
Suppose
Q is a bounded open set and xO E Q(either
XO E Qor x° E
aS2
isallowed). Suppose
2b.
Suppose
Q is a bounded openset,
’
Proof.
Asmentioned,
theproof
of(2a)
is standard. We consider theproof
of(2b). Suppose,
for the sake ofcontradiction,
that u has apositive
maximum value M on
S~. By
thestrong
maximumprinciple,
this maximumcan be assumed
only
at x°. Let ~S be asphere
centered aboutx°,
ofradius R. some N
3f
on aS and u = M atx°. Now,
the func- tion - w(x) = - B1-(l(n-l)
is a solution of themaximizing equation,
as is seen from the
representation (5). Hence,
thefunction v (x)
= M -is a
supersolution
which is h u on theboundary
of the deleted Thus u(x)
~ v(x) by
the maxi-mum
principle,
which contradicts thehypothesized growth
condition on-
3.
The isolated boundary point
as aregular point.
When a
1
theexponent
of the solution zr in(7)
ispositive.
n-1
Hence w itself is a solution with a non-removeable bounded isolated sin-
gularity. Moreover, w
is a barrier at theorigin
for theoperator Lo ;
hencethe
following
result:THEOREM 3. The
origin is
an isolatedregular boundary point for
theoperator L 0
° described in(6)
when x . Thisoperator
hascoefficients
n - 1 ‘
analytic
in Rn ’nlinu8 theorigin.,
as claimed in the introduction.Incidentally, to
is a solution of Pucci’sminimizing equation
hence,
within the class ofequations 0,
L E minimizes the so-lution
(when
itexists)
of the Dirichletproblem
on~x :
0C I x 1~ t
forboundary values 99
= 0 at 0 1on ~ =
1.4.
The special solutioiis.
We now
develop
someparticular equations
which have solutions of the formu (x) = ra f (8),
0),
~1,
where u is aC2
functionexcept
on thenegative Xn axis,
butwhere f’
has a certainasymptotic growth
as 0 --~ n-.Let Po
be apositive
constant andlet fl
be a function of 0 defined asfollows :
fl
is monotone and C °° on[0, ~c].
At each fixed
point x° ~
0 we find it convenient to choose the follow-ing
Euclidean coordinatesystem
withorigin
atx° :
let thepositive y1
1axis lie in the direction of
increasing r,
let thepositive yz
axis lie in the direction ofincreasing 0.
and let the axes be chosenarbitrarily
indirections
orthogonal
to the y,, y2plane.
We then defineL1
to be theoperator
whose value at each x0 isgiven by
Clearly
the coefficientsof Li
withrespect
to theoriginal
axes are C°°except
on the
negative
xnaxis,
as claimed in the introduction. For functions of the formt-If (0), A
any realnumber,
we haveWe
normalize, setting J* (0)
= 1. The condition that)-If (0)
be aC2
solutionof
L1 (r~ f’ (8)) = 0
for0 c 9 C n, r * 0,
is then thatj8
be a solution ofthe initial value
problem
The final condition insures that is
C2
even on thepositive
rn axis.For our purposes we will
always
useIn the first case aL is in in the second L is the
Laplacian,
and inthe third L is in
d2a . Actually,
thedevelopment (a)-(12)
isjust
aspeciali-
zation of a
general representation
forL (i-If (0)), L E Eo.,
to be found in[10],
but we have
thought
it moreenlightening
to include thecomplete descrip-
tion of
.L1
here.Now comes the crucial observation
concerning
theproblable
behaviorof the solution of
(11)
as 02013>-h. If welook,.
as in the method of Frobe- nious[3,
pp.132-135],
for solutions of(ltlt)
of the form(n
- g (n -9), where u
is a constantand g
isregular
and non zero at theorigin,
y then the third term in because it has no factor of ln -0)"~ ~
does notenter into the indicial
equation
for p. We obtain roots = 0and p 2
== 1 -
(it
-2) independently
of A.(In
case ,u2 --- 0 weexpect
a solutionof the form
log (~c
-0) g (~z
-0)).
NA’e thereforesuspect
that the solutionof
(11 )
will have thefollowing asymptotic
behavior as(vr20130)20130+:
ne-gative
powergrowth
for(n20132)B0>1, logarithmic growth
for(n20132)B0=1,
and bounded
growth
for(n
-2) ~io
1. Thefollowing
theorem confirms thisexpectation.
Wedelay
itsproof
to Section 6.THEOREM 4.
(a)
For every real A andpositive Po
there exists aunique
solution(call
it
F) of problem ( 11 ).
(b) If
1 ~(n
-2) Po
and 0[ ~
1 then the solution F hasexactly
one zero occurs in
(~/2, on),
is monotonedecreasing
on[0, on),
and isasymptotic
to anegative
constant times(n
-8)1~~’~-2»~ (or
-log (n - 9)
when1 )
as 9-+n-.,
(c) If (n
-2) ~Bo [
1 thenfor
att real 1 tlae solution F iscontinuous
onalt
[0, n] ;
moreover,for
all8ufficiently
, F remainspositive
on[0, n].
5.
Exceptional points and vice-versa.
Lebesque
mentions in[8,
p.353]
that thetip
of everyalgebraic spine
is
regular
forLaplace’s equation
when n =3.
Hisproof
seemsmistaken ;
heintroduced the barrier
+ x2 + xa , perhaps
under the mistaken notionthat its zero level surface x3
= - (xi -~-
forms anarbitrarily sharp algebraic spine
asActually
one can show that no barrier 2c forLaplace’s equation
at thetip
of analgebraic spine
can be Holdercontinuous;
for,
suppose it wereHolder
continuous of orderÀ, 1 ~
0. Then let 0[ ~,1 [ ~,.
The function of the form u =
rÅlf(9)
which is harmonic on0 ~ 8 ~/(0)
==I I fc, (0) = 0,
has a zero91 n,
as is seen in theorem 4b. Since thespine
issharper
than any cone, we could dominate uby
amultiple
of wnear the vertex on the cone 0 c 0
01 ,
and conclude that u must be Holder continuous oforder A,
a contradiction.However, Lebesgue’s
result still holds. It appears as aproblem
inKellog [7,
p.334].
The author has checked it outusing
Weiner’s criterion.One can underestimate the
capacity
of acylinder
oflength h
and radiusp 2013 by
a constant timeshlllog (J 1.
One then can underestimate the ca- 2pacity ri
of the ithsegment
of the exteriorspine (see Kellog’s notation)
in terms of the
capacity
of an enclosedcylinder,
thusestablishing
thatthe
tip
of anyalgebraic spine
isregular.
We now establish that the
tip
of everyalgebraic spine
isexceptional
for the
operator L1
when we letPo =1,
with a any constantsatisfying
a
1
C a1,
and n = 3. Then forevery 1, 0 .1. [ 1,
the solution u(x)
==="
1
I’A F (0)
described in Theorem 4b isasymptotic
to- crA (71 - c
apositive
constant. be a domain whoseboundary
contains thealgebraic
J 1. della .Scuola Sup. - Pisa.
Act
spine (n - 6) ~r1-á.
Thus u(x)
--~ - c as x -~ 0 on8Q ;
but u(x)
-+ 0 asx --~ 0 on
straight
line rays. Hence u restricted toS~
has continuousboundary
values and u assumes theseboundary
valuescontinuously except
for a bounded
discontinuity
at 0.However,
we can show that no solu-tion v of the same
equation
can assume theseboundary
valuescontinuously;
for if so would
satisfy E, (u
- v) = 0 inQ, u -
v Ec2 (0) n co (T2 - - 0}),
and(u
-v)
= 0 on8Q - (0)
with at most a boundeddiscontinuity
at 0. Since
L1
hasellipticity
constant a> 1 ,
2 we may inferby
Theorem2a that this bounded
singularity
isremoveable,
and(u
-v)
= 0 on Q-(0j.
Therefore,
the Dirichletproblem
for theseboundary
values onaD
isnonsolvable. All
points
of other than 0 areregular (there
exists anexterior
sphere
at each of thesepoints). This,
with Theorem1, implies
that 0 is not a
regular point
for this domain and thisequation.
Of course 0 is also an
exceptional point
for thisequation
for any do- main with asharper spine. Letting A-+
0+ we see that thetip
of anyalgebraic spine,
no matter howblunt,
is alsoexceptional.
1 ’
THEOREM 5. Let n = 3 2 a 1. Let
Lie
be theoperator
de-fined by (9) with flo = 1.
ThenL1
has C°°coefficients
on Iln -(the
closeda
.
negative
x3axis), aLl
E and 0 is anexceptional boundary point for
theequation L,u
= 0 on any domain S2 whoseboundary
contains analgebraic spine
with axis on thenegative
X3 axis. 0 is ai-egular boundary point f’or Laplace’s equation
on such domains.Similarly, using
Theorem4b,
we see that thetip
of everyexponential spine
isexceptional
forLaplace’s equation
when n = 3.If we instead let
~Bo
= a, 0 a1,
then Theorem 4c tells us that the solution F of(11) stays positive
on[0, ~~
for allsufficiently
small A.Thus r~ F (0)
is a barrier at theorigin
forL,
when~S~
contains a line.
segment spine. Actually,
a much morestraighforward
construction of sucha barrier is
given
in Lemma7 ;
the barrier there constructed is then used in Lemma 8 to prove Theorem 4c.THEOREM 6. Let n =
3,
0~
a1,
and letL1
be theoperator defined by (9;
withPo
= a. ThenL1
EEa
and 0 is aregular boundary point for L1
on the
domain Q
=(the
unit batt minus the closednegative
X3axi8l.
Hou,e-0 is an
exceptional boundary point for Laplace’s equation
on this doniain.The two theorems above were stated
only
in the case n = 3 becauseonly
in this dimension do we obtainexamples
of « both waysnonequiva-
lence of
regular boundary points »
for aarbitrarily
close to 1.However,
theasymptotic growth
of ~’ revealed in Theorem4, depending
as it doesonly
on -
2)
alsogives
information forhigher
dimensions.When n ~ 4,
and then thetip
of everyalgebraic spine
is ex-ceptional
for theoperator L1. (Since
this includes the case a =1,
for which.L1 equals
theLaplacian,
we have here noexample of nonequivalence)
Whenand
= a then thetip
of the linesegment spine
is re-gular
forL1.
In summary, when n = 3 Theorems 5
and
6give examples
of excep- tionalpoints
aad vice-versa » for aarbitrarily
close to 1. When n =27
Theorem 3
gives
anexample
of « vice-versa » for aarbitrarily
close to 1.When it ~
4,
theparagraph
above and Theorem 3give examples
of« vice-versa » for a
sufficiently
distant from 1.6.
Proof of Theorem
4.LEMMA 1. For every real ~, and
Inositive (n
-2) flo
there exists aunique
solution
(call
itF) of
the initial valueproblem (11).
We need
only require
thatfl
be apositive
and continuous function on[0, n].
Existence is then asimple
case of a theorem of the au-thor’s
[11]
which states that there exists aunique
solution of theproblem
provided
that (Y is continuous in its threevariables, Lipschitz
continuousin its second and third
variables,
and monotonenonincreasing
in its thirdvariable,
and that the function p is continuous for t > 0 ando (t-2)
ast -+ 0+ .
LEMMA 2a. Tlte14e exists a solution H
of (lla)
on(0, n)
EC«2
(0, n], H~ (n) = 0, H (n) = 1.
2b. 1’he14e exists acnother solution K
of
on(0, n)
which is asymp-2c:
F,
the solution the initial valueproblem (11),
is a linear com-bination
of
H andK,
F = aH-~-
bK.Proof : Letting
t = ~ -0,
we see that Lemma 2 a also follows imme-diately
from the existencetheory
forproblem (13),
Now consider Lemma 2b.
Letting f (t)
=(or
g(t) log
t ifI
- (n
-2) ~80
=0)
we see thatf
satisfies( l la) iffi g
satisfies a certain equa- tion of the formThus, provided (n - 2, (14)
isexactly
in the form(13a)
and thereexists a solution
with g’ (0) = 0, g (0)
= 1.The above method suffices for the
example
used in Theorem5,
for theren = 3, ~B - 1 ,
1. We have included it for itssimplicity.
However,
for thegeneral theory
with(n
-2) ~8o
anarbitrary positive
con-stant we shall have to go to the method of Frobenius and use the
analy- ticity
ofequation
at0 =
n.We
try
for solutions of the formand
get
pi = 0and P2
= 1 -(n
-2) ~8a
as solutions of the indicialequation.
We summarize some results found in[3,
p.133].
If fl2 is not an
integer
then there exist twoindependent
solutionsIf P2 =
0,
then there exist twoindependent solutions, g (t)
asabove,
andIf ,u2 is an
integer
- k(necessarily negative
since~o
ispositive)
then thereexist two
independent solutions, H (t)
as above andbut where the coefficients co , c1, ... are not
necessarily
constants but may include a factor oflog
t.tn
any case we see that Lemma 2b is valid. Fi-nally,
Hand g areindependent,
so every solution of(lla), F included,
isa linear combination of the two.
LEMMA 3. For A
) 0,
the solution F must start outstrictly decreasing
and continue
strictly decreasing
solong
as F remainsPositive.
Proof :
Theexpansion Fe (0)
=BPee (o) -~-
o(0)
inserted in(lla) yields Fee (0)
0. Hence isinitially strictly decreasing. However,
were F to ceasesirictly decreasing
while stillpositive
it would either have an interval ofconstancy (clearly impossible)
or apositive
relative minimum within(0, ~c) (also clearly impossible since 12 + (n
-2)
ispositive
and 0 at a mi-nimum).
Thiscompletes
theproof
of the lemma.To prove Theorem 4b it is sufficient to establish that the
coefficient b,
where F = aH
+ bg,
ispositive. However,
for certain values of 1(eigenva- lues)
b is zero. Forexample, A
=1 is aneigenvalue
for everyfunction fl
since the linear
function Xn
= r1 cos 0 is a solution of everyelliptic equa-
tion under consideration.
However, by comparison
of F with cos 0 wewill establish that there are no
eigenvalues satisfying
01
1.LEMMA 4. When 0
~,
1 no solutionf of (lla)
can have both a zero02
and an earlier zero9. of fe
in[0, n/2]. (By synunetry
the sameapplies
2n
[~/2, n]
2b2th02 ol)’
Proof :
Notice thatf
satisfiesequation ( l l a.)
with coefficientA2 + fl (n- 2) A
for the zero order
term,
but cosine satisfies the sameequation
with a lar-ger coefficient
12 + fl (n - £h) I.
Achange
of theindependent
variableputs
the
equation
inself adjoint
form and one then introduces the Prüfer substitution. Now at01
the Prüferphase
functionsw (9)
andw (9),
of thesolntions f’
and cosinerespectively, satisfy w (0~)
>n/2
= w(0~). Hence,
the
proof
of the SturmComparison
Theorem[2,
p.259]
establishes thatw
(0) >
w(0)
for all0 > 91; thus, f
cannot have a zero(i.e.,
co cannotreach
:1:)
untilstrictly
after the first zero of cosine.An alternate
proof
would involvecomparing
the solution(notice L1 (r~ f (8)) = 0)
with the super solution r~ cos 0(notice .L1 (1’,1.
cos0) 0)
onthe
region 8,
8(x) 82~ after
firstusing
the linear solution r1 cos 0as a « barrier at
infinity >>
toget
aPhragmen-Lindelof
theorem(see proof
of Lemma 8 for
example)
for allregions
contained in the upper half space~>0.
LEMMA 5. Fo~~
O ~ 1,
the solution -I’ )ii itst bestrictly decreasing
onall
[0, ?).
Proof :
We have seen, Lemma3,
that F continuesstrictly decrea,sing
so
long
as itstays positive.
Now ~’ cannot have a zero relative minimum in(0, ~)~
for then we would have T" - F = 0 at an interiorpoint,
and theunique
solution of this initial valueproblem
would be If’ = 0.Also,
F cannot have a
negative
relative minimum in(0, ~),
for then we would have~/2 C 82 C ol C ~
withF (82)
= 0 andFe (0,)
=0,
which contradicts Lemma 4applied
to the secondquadrant.
LEMMA 6. For 0
C
11,
tlae solution F must beof
the aH+
bKivith b
>
0.Proof:
We show first thatb ~ 0. Suppose,
for the sake ofcontradiction,
that F = aH.
Certainly a #
0. If a> 0,
thenby
Lemma 5 F hasstayed positive
andstrictly decreasing
on[0, ~z~ ~
butapplying
the same Lemma 5to H in the
opposite
directionyelds
that H must be astrictly decreasing
function of (n -
0),
a contradiction.If a
0 then thegraph
of F wouldcross the axis
(in
the secondquadrant by
Lemma4)
and 1~’ would have both a zero and a later zero of its derivative(at
0 in the secondquadrant, contradicting
Lemma 4.Finally,
since b=1= 0,
F isstrictly decreasing,
and K(0)
--> - oo as9 --~ n-
(or Ko (0)
-+ oo in case (n -2 j flo 1),
it is clear that b>
0.This
completes
theproof
of Theorem 4b. We continue to theproof
ofTheorem 4c
by
firstconstructing
a barrier at theorigin
on the domain Q = R’l-(negative
xnaxis{
for theoperator L1
LEMMA 7. Let
(n
-2) flo
1. There exists afunction f
and numbers~c >
0 such thatProof : Let f3
be a constantsatisfying
Then letand extend it to
[0, a/2]
in such a way thatNormalization later will take care of the
"~(o)
= 1requirement.
We duenote
11 f (0)
«.~1 (rl f (0))/1.1-2
and consider the case 1 = 0. On[;r/2, n)
wehave ,
which is
uniformly negative
on[a/2, ~c~
since(~
-0)
cot 0 h - 1 and===
~o
there. Likewise on[0, ~z/2)
wehave 10 ( f )
0uniformly.
Hencethere exists an q
>
0 such thatBut,
since thisf
is bounded and Imultiplies only f
in1;L,
see(10), 1A ( f (0)) depends continuously on I, uniformly
on[0, hence lx ( f (8)) ~ - r~
on[0, 11:)
for
all |A
somepositive
IA.LEMMA 8. For all ,1
satislying
0A, u
the solution Fof
Theorem 4is
~ f’
on~U~ ~j,
wheref
and IA aredefined
in Lemma 7.Proof.
We first establish thefollowing Phragmen-Lindelöf type
resultfor the
operator 2~ :
if D is an open set contained in the open cone[x :
x is not on theclosed negative xn, axis), w
EG’2 (Q)
n co(Q), L1 w >
0 inon
8Q,
and w(x)=o
as x-~-oo inQ,
then onS~.
Ourproof
usesthe
positive supersolution
as a « barrier ati,nfinity ». By
thegrowth
condition on w, for every E
>
0 there exists an .Rsufficiently large
thati~E ~ ~v - is c 0 on the
boundary
ofDR
== ~fi ~ Bl.
HenceV, 0 on
12R by
the maximumprinciple
for bounded open sets.This,
withthe arbitrariness of ~, establishes the desired result. More
general Phragmen-
Lindelof theorems on cones may be found in
[12j.
We know that both
f
and F are continuous on[0, ~j
andequal
1 at0,
that
rlf (0)
is asupersolution
forL,
onQ,
and thatr~ F (9)
is asolution.
Suppose
now for the sake of contradiction that thereexists V
E(0, ~]
where.~’ (y)
=lcf (~)
with 0 k 1.A.hplication
of the abovePhragmen-Lindeliif
result to the subsolution on the open cone
S,~ = yx :
;x~
#
0 and 0c 8 (x) y;
would thenimply
that F(0)
Clcf’(8)
for 0-- 0 --- 11’, yielding
a contradiction when 0=0. Thiscompletes
theproof
of the lemmaand of Theorem 4c,
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