Qualitative properties of solutions to elliptic singular problems

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Qualitative

_Properties

_of

Solutions

to

Elliptic

Singular

Problems*

S.BERHANU

a,

F. GLADIALIb

and G. PORRUb,t

a_{MathematicsDepartment,Temple University, Philadelphia,PA19122,USA,"} b_{Dipartimentodi Matematica, Via Ospedale 72, 09124, Cagliari, Italy}

(Received23 February 1998,"Revised7August1998)

WeinvestigatethesingularboundaryvalueproblemAu/u-’ 0 inD,u 0 onOD,where

3’>0.For7>1,wefindtheestimate

lu(x)

bo62/(n+l)(x)l

<

where_b0dependson 7only,6(x)denotes thedistance from x toOD and/3is asuitable constant.For3’>0,weprovethat the functionu(+7)/2isconcavewheneverDis convex. Asimilarresultiswellknownfor the equationAu₊up 0,with0_<p_<1.Forp 0, p and_3’_> weproveconvexity sharpness results.

Keywords." Singular equations; Boundary behaviour; Convexity 1991 MathematicsSubject_{Classification:}35B45, 35E10

1.

INTRODUCTION

Let N

>

andletDC

R

Nbeabounded smoothdomain.In

[1-3,5],

the problem

Au

upinD,

u(x)

--.

+oe

asx

OD

isinvestigated.Itisproved that such a problem, for p

>

1, has a unique positive solution

u(x).

Moreover,

forp

>

3thereexists a

_constant/3

>

0 such that

.(x)

2/(1-P)(x)

</36(x)

in D,

(1.1)

*Partially supportedby MURST 40% andbyaUniversity of Cagliari Research Project. Correspondingauthor. E-mail:porru@vaxca1.unica.it.

wherea0is a constantdependingonponlyand

(x)

denotesthedistance from xto

OD [1,3,5].

In

[6,13]

it isprovedthat theproblem

Au+uP--0

inD, u-0 on0D,

(1.2)

for p

<

0 has a unique positive solution

u(x)

continuous up to the boundary

OD.

Inthesamepapersit isalsoprovedthat,for p

<

1,there

existpositiveconstants

A, A

suchthat

/62/(1-P)

_(x)

<_

(X)

<_ A62/(1-P)

_(x).

InSection2 of the present paperweshall provethat, for p

<

-1 there

exists/3

>

0such that

u(x)

82/(1-P)(x)

<

t6(P+I)/(P-1)(X)

in

D,

(1.3)

where

_b0

is a constant depending onp only. We emphasize that the

constantsa0in

(1.1)

and

b0

in

(1.3)

areindependent of the geometry ofthe domainDandevenofthedimensionN.Wealsofind aboundaryestimate

forthecase p 1.

Now,

considerproblem

(1.2)

with0

<

p

<

1.Itiswell known that this problemhasapositivesolution

u(x).

Suchasolutionisnot concave even

in the radialcase. Indeed, ifu

u(r)

thenthe corresponding equation readsas

(r

N-1

_ut)

-

rN-1up

O,

which implies that

(r

N-1

u’) <

0and

u’(r) <

0in

(0,

R].

Since

_u(R)

0,we

have

N-1

u"

(R)

+

u’ (R)

0,

(1.4)

whence,

u"(R) >

0.Thisshowsthat

u(r)

is not concave nearr R. It is known

[9,10]

that if

u(x)

is a solution of problem

(1.2)

with 0

<

p

<

in a convex domainD then, the function v u(l-p)/2 is

Ifp 1, instead of problem

(1.2)

weconsiderthe following

Au+Alu=0

inD, u=0 on0D,

where

_A1

isthe firsteigenvalue ofD.Ifuisthe (positive) correspond-ing eigenfunction then, v=logu is concave whenever D is convex

(see [9-11]).

In

[9,

p.

122]

itiswrittenthatif

u(x)

is asolutionto

(1.2)

ina convex

domainDthen thefunction

V(X)

S-(p+l)/2ds

isagoodcandidatetobeconcave.

As

recalledabove,thisfactisknown-to betruefor 0

_<

p

<

1.InSection3of the present paperweshall prove that thestatementinaboveisalsotruefor p

<

0. Furthermore,weshall find the following sharpnessresults. Let

u(x)

be asolution of

(1.2)

and let e

>

0.Then:

(1)

ifp-0, the function V--’U1/2+e is not concave in some convex

domain;

(2)

ifp 1, the functionv u is not concave in some convexdomain;

(3)

ifp<-1, the function V"-U(1-p)/2+e is not concave even in the

radial case.

2.

BOUNDARY

BEHAVlOUR

In this section the domain D is assumed to be bounded, smooth and satisfyingauniforminterior and exteriorspherecondition.For₇

>

1,we considerthe followingproblem

Au+u-7=0

inD, u=0 on

OD.

_(2.1)

LEMMA

2.1

_If

u(x)

isapositive solutiontoproblem

(2.1)

with "7

>

1, then

lim

u(x)

[("7

+

1):

]

1/(+1)

Proof

Let R

>

0. Consider first the case of D

B(R),

a ball with radius R. The corresponding (radial) solution z

z(r)

satisfies, for 0<r<R,

N-1

z"

z’

₊

z

-’

O,

z’(O)

O,

z(R)

0.

_(2.2)

Multiplying

(2.2)

by

z’

weget

z"z

+

z

-z

<

0. After integrationover

(0,

r)

wefind

The latter inequality implies

zt

>

z(1-,)/2

(2.3)

Insertionof

(2.3)

into

Eq. (2.2)

yields

N-l(

2

11/2z(1-’)/2+z-’<O’r

whence

I

z(+’)/2

+

<

0.

r

Lete

>

0. Since

z(r)

0as r

R,

thereexistsr,

<

Rsuch that

z"+

z-(1-

e/2)

<

0 in

_(r,,R).

Since

_{f(r) <}

0,fromthisinequalitywe find

and

(z’())

(z’(r,))

2 2

e/2

[zl_.(r)

zl_7(re)]

>

0.

1-7

Forsome_r0

>

re,wealsohave

(z’(r))

2 e

Z1

2

>

_7-1

-7(r)

VrE(ro,

R)

or

z(7_l)/2(r)zt(r)_(l_6.)l/2(,

2,

1/1/2

Integrationover

_{(r, R)}

yields

z(r) > (1

e)l/(,+_)L

2(’-[(’

+

1)

211)

1/(7+1)

(R- r)

2/(’+1)

Vr

(ro, R).

(2.4)

Nowconsiderthe annulusD

B(R,

).

Letw

w(r)

beasolution to

problem

(2.1)

in

B(R,

R).

We have,forR

<

r

<

R,

N-1

w"

+

w’

₊

w

-

O,

w(R)

O,

w()

O.

(2.5)

Ifr

is apointin

_{(R, R)}

where

wt(rl)

O,from

(2.5)

wefind, forR

<

r

<

r,

(w,(r))2

fr,

2

wl-’(r)

wl-f(rl)

(N- 1)

_7(w’(t))

dt+

2

_7-1

(2.6)

By (2.5)

wealsofind

/r

tN-lw-’(t)

dt

<

w’(r)

_r--

w

-’

(t)

dt. RN-1

Using the latter inequality togetherwith de

l’pital

rulewe find lim

(’7-1)f(1/t)(w’(t))

2 dt rR

r

-1

f’

w-’(t)

dt lim

w’

!r)

<

lim

RN

Using de

l’(3pital

rule once more we get

(9’- 1)

ff(1/t)(w’(t))

2

dt

r

v-1

_w(r)

lim lim 0.

-

w-()

-<

--w’()

Recall that

w(r)

0asr Randthat, by

(2.6),

w’(r)

-+oeas r

-

R.If the integral of

(w’)

2

over

(0, rl)

is finite then we cannot apply de

l’pital

rule, but in this case, the limit is trivial.

As

aconsequence of

this estimate,

(2.6)

yields, for agiven e

>

0,

(w’(r))

<

w-(r)

Vr

(,r).

2

_7-1

Integratingover

(R, r)

with r

<

&, one finds

7+1

w(7+1)/2<(2)

1/27-1

(1

+

e)l/2(r-

R),

or

w(r) <

(1-I-)1/(7+1)

F2(

’c--ir(’r

+

1):1

1/(7+1)

(r-R)2/(+1).

(2.7)

Recall thatDhas auniform interiorand exteriorsphere condition.

Take apoint PE

OD. We

may assumethat P

(R,

0,...,

0),

thatDis

contained in the annulus

B(R, R)

with center in

(2R,

0,...,

0)

and R large, and thatDcontainsthe ball

B(R)

with center in

(0,...,

0).

Note

that

B(R, R)

and

B(R)

are tangentto

OD

in P. If

u(x)

is the solution

of problem

(2.1)

inD,if

_w(x)

isthesolution intheannulus

B(R,[)

and if

z(x)

isthe solutioninB then, bythe comparison principle,wehave

Let 0

<

r

<

R. Ifwe take a point x

(r,

0,...,

0)

then, by using

(2.4)

and

(2.7)

itfollows that

(1

)

1/(7+1)

[(")/+

1):]

1/(3’+1)

1/(7+1)

<

u(x) < (1

+

)2/(7+1)l(")

Sincee is arbitrary, the lemma follows.

THEOREM 2.2

_If

U(X)

isapositive solutiontoproblem

(2.1)

with "7

>

1, thenthereexists

>

0such that

+

1):.]

’/(7+)

(2/(7+1)(X)

i

<

fl(7-1)/(7+l)(x)

VX

E

D,

where

_6(x)

denotes thedistance

_from

xto

OD.

Proof

Following

[5],

put

W(x)

bo52/(7+l)(x)

+

fitS(x),

with

and 0

</3

willbefixedlater. Writing instead of

8(x),

onefinds

2

_{62/(9,+1)_}

flSi

Wi

bo

18i

q-7+1

Since

_6i6i:

and

-AS= (N- 1)K=

H, Kbeing themeancurvatureof the levelsurfaces of

6(x),

wefind

A

W

_b

-7₆-27/(7+1) _-Jr-

_b0

2 _H8(1-7)/@+1)

₊

_fill.

_(2.8)

Onthe otherhand, byapplyingTaylorexpansiontothefunction

f(t)

(bo5

/(+)

+

t)

-onefinds

Weclaimthat,for

5(x) _<

5o

small

and/3

large,wehave

_q,b--l/35-

+

3’(7

1)

₂

+

b--r-z/32ts-z/(.r+

2

< b0

Hi5(1-3’)/(3’+1)

_+/3H.

7+1

(2.10)

Indeed,letus rewrite

_(2.10)

as

’T

_{I_bo}

_{_+_/(,.y}

_{_+_}

i)5(,-I)/(,+I)]

b-’r-2

-y 2

</5b

-’-

q-

_bo

H(52/(’+1)q--

_H5.

"7+1

The left handside canbe made negative for

5(x)

<

5o

bychoosing

/55.r_)/(.+1)

b____o

(2 11)

7+1"

Now,

decrease

_5o

andincrease/3untilthe right handside ispositive.This ispossiblebecause, bythesmoothness of

D,

HisboundednearOD.The

claim isproved.

By (2.8)-(2.10)

itfollows that

A

_W+W

-’r<0

on{xD:5(x)<50}.

Obviously,

W(x)

u(x)

on

OD. Increase/3

and decrease

_5o

accordingto

(2.11)

until we have

W(x)> u(x)

for

5(x)=5o.

Observe that this is

Hence,

bythe comparison principle for elliptic equations[7,Theorem

9.2]

wefind

u(x) <

bo62/(7+l)(x)

-+-

(2.12)

for all xEDwith

_{6(x) <_}

_60.

To completetheproofofthetheorem,let

W(x)

with

_bo

_and/3

asbefore.Wefind

-AW

_bff’6

-2’/(7+1)

₊

_b0

2

_{H6(1-’)/(’7+1)}

"7+1

(2.13)

By

applyingTaylorexpansiontothefunction

f(t)

(b0t52/(’+1)

t)

-onefinds

Weclaimthat,for

6(x) <_

60

small

and/3

largewehave

2

b-’)’-lfl6

-1

> b0

Ht

(1-’y)/(’+I)

3H.

(2.15)

Indeed,inequality

(2.15)

canbewritten as

Theclaimfollows easily.

By (2.13)-(2.15)

itfollows that

If necessary,

increase/3

and decrease

₆₀

untilwehave

W(x)< u(x)

for

6(x)

0.

Againbythe comparison principlewe find

>

(2.16)

for all x Dwith

_(x)

<

_0.

The

constants/3

and

₀

in

(2.12)and (2.16)can

be takenwiththesame

values.Therefore,

u(x)

2/(7+1)

_(X)

<

3

(’-1/(’+1

_(x),

for allx EDsuch that

6(x)

<

60.

Increasing/3 again, the theorem follows.

By

using Theorem 2.2 one finds easily that the gradient of

u(x)

is

unboundedneartheboundaryofD.For_3’ 1, Theorem 2.2fails.Now

wegiveadirectestimateof thegradientinthiscase, improvingaresult

of[13].

Considerthe followingproblem:

Au+p(x)u

-1=0

inD, u=0 on0D,

(2.17)

whereDis a domainsatisfyinganinteriorspherecondition andp(x)is a

smooth function satisfying

0

<

pl

<_ p(x).

(2.18)

PROPOSITION 2.3

_If

u(x)

is a solution to problem

(2.17)

with p(x) satisfying

(2.18)

then the gradient

_of

u(x)

isunboundedineachpoint

_of

OD.

Proof

Letz

_z(r)

bethesolutionofthe followingproblem: N-

_lzt

z"

₊

pl

z-

=0, O

<

r

<

R,

z’(O)

=O,

z(R)

=O.

(2.19)

By (2.19)

we find

(rN-lzt)

<

0, whence

z’(r) <

0 in

(0, R).

Multiplyingby

f, Eq. (2.19)

implies

Integratingon

(0,

r)

we find

(z’)

+

Pllog

z-

<

0, whence,

(zz’)

2

<

2plz2(r)log

z(0)

The latterinequalityimplies that

limzz O. r-+R

Let

_ro

E

(R/2, R)

suchthat

_zzl

_<

pR

_forr0<r<R.

4(N-1)

Using

Eq. (2.19)

once more wefind,on

(ro, R)

zz p

<-

p

<

z r z

\--47r

2z

Since z

<

0,from the latter inequalitywefind

Ztt

Z Pl Z

2 Z

Integrationover

(ro,

r)

yields

(z’(r))

2

(z’(ro))

:z

>

Pllog

Finally,wefindthat

z(ro)

in

(r0,

R)

-z’(r)

>

log

Now consider problem

(2.17)

in D. We claim that the interior derivative ofthe solution u approaches

+ec

as x OD. Indeed, for

Po

E

OD,

consider aball BCD and tangent to

OD

at

_Po.

Let zbe the

solutionofproblem

(2.19)

in suchaball._{Since Pl}

<_

p(x),onefinds

Az

₊

p(x)z

-1

_{>_}

₀

inB.

By

the comparison principle[7,Theorem

9.2]

between the last inequality and

Eq. (2.17)

one findsthat

u(x)

>

z(x)

inB.

As

aconsequence,ifPis a

pointinBcloseto

_P0

wehave

u(P)

u( 0)

_>

z( 0)

Using the last inequality togetherwith

(2.20),

the proposition follows.

3.

CONVEXITY

Inthissection,DC]Nisassumedtobebounded,smooth andconvex.

Let

u(x)

beapositivesolution totheproblem

Au+u-’-0

inD, u-0 on0D,

(3.1)

with0

<

7.Wewant toprove that thefunction

v-/,/(1+,,/)/2

is concave.

Using

Eq. (3.1),

wefind

[11-

1+.7]

v(x)=0

onOD.

(3.2)

Av-

71Vv12

₊

inD,

v

₊₇

2

Wefirstshow that thefunction

Q(x)

7

iVvl

2

+

1+7

(3.3)

LEMMA3.1 Ifu(x)isapositivesolutiontoproblem(3.1)with’y>1,andif

v u

(1+’)/2,

then the

_function

Q(x)

_defined

asin

(3.3)

ispositiveinD.

Proof

Wehave

,),2

Q(x)=

2 with 2

Itsufficestoprovethat,when

<

7,

P(x)

<

O.

Fore

>

0,consider the solution

u(x)

totheproblem

Au+u

-’=0 inD, u=e

onOD.

The correspondingfunction

P(x)

IVul2

u’-2

satisfies

P Ukbl_k l’l-’Y1"l

Note

that the summation conventionoverrepeatedindicesisused.This

equationtogetherwithSchwarz inequality yield

(Pi-

b/-TUi)

2 (Zgkblki)2

IVUl2UkiUki,

fromwhich itfollowsthat

Pi- 2u-’rui

Pi

-+-

u

-2"r.

UkiUki

iX7ul

2

Usingthis estimate aswellas

Eq. (3.1)

wefind

AP

+

2u-Tui- Pi

17ul2

ei

>_

O.

By

the classical maximum principle, P attains its maximum value

eitherwhen

Vu

0or ontheboundaryofD.SinceDisconvex,Hopf’s boundary lemma prevents Pfrom having its maximum value on

OD

(see [14]). Hence,

ul-7

_M-7

4-< <0,

-7-

1-7

where

M,

denotes themaximumvalue

ofu(x)

u,(x).

The lemma follows

aseO.

Fordiscussingtheconcavity ofsolutionsto

Eq. (3.2),

weuseKorevaar function

[9,11]

C(x,y)

2v((x

+

y)/2)

v(x)

v(y),

x,y

e

D.

(3.4)

Thefunction

v(x)

is concave inDifandonlyif

C(x,

y)

>

0inD D.

PROPOSITION 3.2

_If

v(x)

isa positive solutiontoproblem

(3.2)

thenthe

function

C(x,

y)

defined

as in

(3.4),

cannothaveanegativeminimuminD.

Proof

By

Lemma 3.1,thefunctionatthe right handsideof

Eq. (3.2)

is

negative.

Moreover,

suchafunction isincreasingwithrespectto v.The prooffollows easily by

[9,

Theorem3.13,p.

116].

Seealso

[8,11].

To

get thepositiveness of

C(x,

y)ontheboundaryofD

D,

weprove the following

LEMMA

3.3

_If

V(X)

&apositive solutiontoproblem

(3.2)

and

_if

yE

OD,

then the

_function

(x)

2v(z)

v(x),

with z

(x

+

y)/2

Proof

If xE

OD,

wehave

b(x)

2v(z)

>

0.If xE

D,

bycomputationwe

find

v

V(z)

V(x),

x

Xv(z)

Xv(x).

Using

(3.2),

the latter equation yields

2v(z)

+

+

v-(-

(Alvv(x)

+

withA

(1

"y)/(1

₊

"y)andB

(1

₊

,)/2.

Using

(3.5),

this equationcan

berewritten as

(A]7v(z)]2 B)

/k/

+

ai)i

2v(z)v(x)

+

(3.6)

with suitable smoothfunctions a

i.

By

Lemma 3.1,the coefficient of

_b

in

(3.6)

ispositive.

Hence,

bythe classicalmaximumprinciple,we inferthat

b(x)

attains its minimum value on the boundaryofD. The lemmais

proved.

COROLLARY 3.4

_If

v(x)

isa positive solutiontoproblem

(3.2),

then the

function

C(x,

y)

_defined

asin

(3.4)

isnon-negative on

D

;.

Proof

Itfollows from Proposition 3.2 andLemma3.3.

THEOREM 3.5

_If

V(X)

&a positive solutiontoproblem

(3.2)

&a convex domainDthenitisstrictlyconcaveinD.

Proof

By (3.2)

andLemma3.1, thefunction w -v satisfies

Xw-

%

(AlVwl

+

B) >

0,

w

withA

(1

7)/(1

+

7)

andB

(1

+

"),)/2. By

Corollary 3.4,w is convex.

Moreover,

w

_w(Al7wl

2

+

B)

-1 is convex.

By

atheorem ofKorevaar and Lewis

[12],

weconcludethat theHessianofvhasaconstantrankin

D.

Now,

vattainsits maximum inDon acompact subsetK

c

D.Iris well knownin this case

[4]

that in anyneighborhood Uof

K,

thereis apoint

PEUwheretheHessian ofvisstrictly negative.Itfollows thatvisstrictly

concaveinD.

COROLLARY 3.6

_If

U(X)

is a positive solution toproblem

(3.1)

then it attains its maximumvalueinDatasinglepoint.

Nowweprovesomesharpnessresultsonconvexity.

1.Let D

c ]N

bea convexdomain, and let

u(x)

beapositivesolution to

theproblem

Au+l=0 inD, u=0 on0D.

(3.7)

Werecall that thefunction

v(x)

u

is concave inD

[9,10].

Weprove that the exponent

1/2

issharp.

Let N= 2,and letDbe the trianglewith vertex

_{(- / x/-, 0),}

₍₁

_{/ x/,}

₀₎

and

(0, 1).

Thefunction

u(x,

y)

y

rl(1

y)2

3x2|

=+

solvesproblem

(3.7)

in this domain.Letb(y)

4u(0,

y). Wehave

qS(y)

y(1

y).

Of course, thefunction is concavein

(0, 1).

Ifa

_>

1/2,

(b(y)) is not concave neary 1.

IfN

>

2,one canobtainthe result in abovebyusing the function

hi(X)

_---(1

XN

[

XN)

2

N-

il

2.Letu

>

0beasolutionoftheproblem

Suchafunction isthefirsteigenfunction ofD.Weknow thatv logu is concavewheneverDis convex.

Isthere anye

>

0 such that v u is concavefor allconvex domains?

The answer is negative, as one can seeby the followingexample. Let N--2, and let

D-{0<r<a,

0<0<

Hererand 0arepolarcoordinates,m is aninteger andaisthefirst zeroof the mth Besselfunction.Thefirsteigenfunction ofDis

u(x,

y)

Jm(r)

sin(toO),

Jm(r)

being the mth Besselfunction.Itisknown that

Jm(r)

behaveslike r asr 0.

Hence,

we musttake e

_<

_1/m

ifwewantthefunction u tobe

concave. Since m canbe choosen arbitrarilylarge,the resultfollows. 3.Let

z(x)

beasolutionof theproblem

Az+z-l=0

inB, z=0 on0B,

(3.8)

whereBis aball.

By

Theorem 3.5,

z(x)

is concave.Letusshowthat,given e

>

0, thefunction v z1/(1-)is not concave near

OB.

Indeed,

Z--1

l-e,

VZ

(1

/Xz

₍₁

_)v-/Xv

₍₁

_)v

--I

_IX7vl

2,

Substitutinginto

Eq. (3.8),

wefind

vl-ZZmv_

_ev-2lVvl

2

(3.9)

and since, by

(2.20),

]Vv]

as x

OB,

the right handsideof

(3.9)

becomes positiveneartheboudaryof B.

As

aconsequence,A v

>

0near

OBandtherefore,vcannotbeconcave onB.

Similarly,onecanshowthat,if

_z(x)

is asolutionof theproblem

Az+z

-’=0 inB, z=0 on0B,

with 7

>

then,for anye

>

0, the function v- z(l+’)/2+eis not concave nearOB.

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