the obstacle problem) - uzhuser.math.uzh.ch/ros-oton/caffarelli-the-obstacle...the obstacle problem)...

$: The Obstacle Problem) - UZHuser.math.uzh.ch/ros-oton/Caffarelli-The-obstacle...THE OBSTACLE PROBLEM) 9) We now use the definition of superharmonicity) f [ 1 1 1 1] 0> v\0371fr==C-v--v.IBsl$
ACCADEMIA NAZIONALEDEILINCEI

SCUOLA NORM ALE SUPERIORE)

LEZIONI FERMIANE)

Luis A. Caffarelli)

The Obstacle Problem)

.I\"..........#,

...,-.....,\"\"\"\"

:;;.::V

\\ \\0 t It

ti\037\\(.Q

Mathematiques)

PISA - 1998)))

Luis A. Caffarelli

Dep:lrtlnent of Mathematics and TICAM

1 he University of Texas at Austin

Austin, TX 78712-1082)

The obstacle Problem)))

CONTENTS)

Introduction ........................... Pag. 3)

1. An Apart on Super Harmonic Functions)\

7)

2. Regularity of Solutions . .)\

13)

3. Free Boundary Regularity)\

17)

3.1. - Free Boundary Regularity: Part I, generalities . . . . \"17

3.2.- Free Boundary Regularity: Part II. . . . . . . . . . . . . . \"20

3.3. - Application to our free boundary problem (still global solutions) . . . .\"

27

3.4. - The Structure of the set of singular points '.' . . . . \"33

3.5. - Proof of the main theorem . . . . . . . . . . . \"42)

Appendix -Boundary values of harmonic functions in Lipschitz Domains)

\ 45

48

4951)

A.I. - Construction for boundary Harnack principle

A.2. - Proofof Step2

A.3.- Step 3)

\

\

\

References .)\

53)))

Introduction)

The material presented here, corresponds to the Fermilectures,that

I was invited to deliver at the Scuola Normale di Pisa in the Spring of

I 998.

It was for me a great honor,for the prestige of the Scuola and the

lectures, and a touching experience, because of the enormous influencethat Italian mathematics, particularly the schools of Stampacchia and

De Giorgi, had in my work.

The obstable problem, in fact, was one of the maIn motivations

for the developlnent of the theory of variational inequalities (of which

Stampacchia was one of the main architects) and the problematic of free

boundary problems,in the late '60's early '70's.

On the other hand many of the themes of De Giorgi's work appear

in these lectures; Boundary Harnack inequalities, classificationof globalsolutions, flatness and regularity, etc.

The lectures are almost self-contained,exceptfor the work of De

Giorgi and Weinberger-Littman-Stampacchiaquoted in the Appendix. Fi-

nally, I would like to thank my colleagues at Pisa for their usual, warm

hospitality.)))

4) LUIS A. CAFFARELLI)

The obstacle problemconsistsin studying the properties of mini-

mizers of the Dirichlet integral)

D(u) =

tlvul2dX)

in a domain D of Rn

, among all those configurations u with prescribed

boundary values: UlaD == f, and constrained to remain, in D, above a

prescribedobstaclecpo

More precisely, we are given:)

a) A (smooth) domain D of R n.

b) A (smooth) function f on aD.

c) A (smooth) function cp on D, with CPlaD < f.

In the Hilbert space H I(D) of all those functions u with square

integrable gradient, we define K to be the closed convex set)

K == {u E HI, U I aD == f, u > cp} .)

On K, there is a unique point Uo that minimizes the Dirichlet integral)

D(u) =f

IVul2

dX ,)

Such a point Uo is called the \"solutionto the obstacleproblem.\"This problemis motivated by the description of the equilibrium

position of a membrane(the graph of u) that is \"attached\" at level f

along the boundary of D, and is restricted to remain above cp, (the

obstacle ).

Such a membrane will minimizethe area integral)

A(u) =f V I + IVul

2 dX)

that is linearized to Dirichlet integralfor smalldeflections.In any case,

the theory developed here applies to the \"minimal surface\", i.e., the non

linearized case, but for simplicity we will restricthere to the \"linear\"

case.)))

THE OBSTACLE PROBLEM) 5)

The same mathematical problem appears in many other contexts:

fluid filtration in porous media, elasto-plasticity, optimal control and

financial mathematics. See for instance the book of Friedman ([Fr])where many of these applications are described, as well as the classicalliteratureon this problem.

We start with some classical statements about Uo:)

LEMMA 1.)

a) Uo stays between AI== min f, and A2

== max(f, cp).

b) Uo is superharmonic, and support(\037uo) C {uo == cp}.

Point a) is a standard applicationof the weak maximum principle for

HI functions.

The minimum of two such functions is again in HI, and so, for

instance,u == min(uo, A2), is an admissible function and

D(uo) = D(u) + fIVuol2 dX > D(u) .

J{UQ>A2}

Therefore Uo == U, a.e. in D.

About point b) more than a proof, it is a matter of definition:

We would like to say that a function v is super harmonic if \037v < o.

Unfortunately this requires to take two derivatives of v.

Instead we use the weak definition provided by integrating by parts:

If v were still smooth, and l/f a Co function, we have the (Green)

formula))

f(liv)1fr =

Jvli1fr.

This allowsus to say, for a function v, in Lloc, that v is super harmonic

without taking any derivative.

DEFINITION. v in Lloc is super harmonic in D if, for any l/f EC\037'

I(D),

l/f non negative, we have

f vli1fr< 0,

(That is, we have said, heuristically,for any positive l/f, J (\037 v) l/f < 0, and

ifl/f \"approaches\" Dirac's 8, this seemsto imply \037v < O. Indeed, from the

theory of distributions, \037v is (as a distribution) a negative measure.))))

1. An Apart on Super Harmonic Functions)

Let us show that the previous definition coincides with the classical

definition of superharmonicity.)

LEMMA 2. If v is weakly superharmonic then itsaverageisa decreasing

function of R. More precisely, if 0 < R < S)

!v =

1r v(y) dy > ! v.

IBRI1BR(xo)B R(XO) Bs(Xo))

PROOF: To that effect we construct a test function l/f that relates both

averages. We start with the fundamental solution V =IXI\037-2

' and \"fit\"

under its graph a paraboloid, Po, tangent to V at IX I== R that can be

easily computed. We define)

{

V - PR for X < R

VR(X) ==

= 0 otherwise.)

We note that, except at the origin, VR is CI,I and tJ. VR == -C(R)X BR .

We also note that for R < S, V R <Vs.)))

8) LUIS A. CAPPARELLI)

\037PR)

Fig. I.)

Let us take as a test function 0/== Vs

-VR.)

I I)

s)

Fig. 2. Graph of 1/1.)

Notice that 0/ vanishes outside Bs, so 0/ is nowC\037'

I

(Rn

) and)

tJ.0/ == C(S)X Bs-

C(R)X BR.)

Finally, applying the formula, notice that the constants 1 and -1 are

both superharmonic, so we get)

C(S)IBsl- C(R)IBRI == o.)

That is)

/).1/1 = C[ I\037sl

XBs-

I\037RI XBR]

\302\267)))


We now use the definition of superharmonicity)

f [

11

1

1 ]

0> v\0371fr==C- v- v .-IBsl Bs IBRI BR)

REMARK. The function VR seems to be particular of the Laplacian.It is not so, one can construct such a V R for any divergence opera-

tor, D j (aij Dj), (uniformly elliptic, bounded measurable coefficients) by

solving the variational inequality:)

1Di(aijDj)(VR) =

RnX{VR>O}

-8xo)

(oxo: Diracs delta at Xo). Prom the Littman, Stampacchia Weinberger

theory, that says the fundamental solution V behaves likeIXI\037-2

' it can

be seen that)

{VR > O} rvXBR

.)

In fact, by scaling, it is enough to prove it for R == 1.

This provides a \"mean value theorem\" for general divergence equa-

tions, i.e., given Xo, there exists an increasing family of sets DR (Xo)

each one comparable to BR(XO ) such that if v is a supersolution of

Di(ajjDj)v == 0, the average, f DR(X) vdX, is decreasing with R.)

COROLLARY 1. Any super harmonic function v, has a unique pointwise

defined representative as)

v(Xo) == lim! veX) dX .

R\037O

B R(XO))

Also, v is lower semicontinuous, that is)

v(Xo) < lim v(X).x\037xo)))


We now go back to our solution Uo of the obstacle problem:

About b) Uo is superharmonic. Indeed any positive 1fr E CJ is an

admissible perturbation, i.e.,)

f i Vu ol2

dX <f

(Vuo + 8cp)2

=

fi Vu ol

2+ 28

fVuoVcp + 8

2

f(Vcp)2,)

Therefore)

o <f

V Uo V cp ,

COROLLARY 2. Uo is pointwise defined,lowersemicontinuous, and the

set {uo > cp}is open. (Moreprecisely,ifuo(Xo)> cp(Xo) + 8, there exists

a neighborhoodofXo where uo(X) > cp(X) + 8/2.))

COROLLARY 3. \037uo is supported in the closed set {uo==cp}.)

COROLLARY 4. Uo is continuous.)

The proof of Corollary4 is a consequenceof Evans theorem, that)

says:)

THEOREM 1. Let v be a superharmonic function, and suppose that

vlsupport(\037v) is continuous. Then v is continuous.)

Let me sketch the proof of the Theorem 1. Suppose that there exists

a sequence Xk \037 Xo, such that lim V(Xk) # v(Xo).Then)

a) Xo E support(\037v), if not v would be harmonic in a neighborhood of

Xo and thus continuous. (Remember that by definition, support(\037v)

is closed.)

b) We can assume that X k \037 support(\037v), since v is continuous there.

c) lim v(Xk ) == a > v(Xo) == 0 (no loss of generality) by seminconti-

nuity.

d) Further, by semicontinuity, given 8 > 0, we may assume veX) > -8,

for I X - X 0 I< 8.)))


B28k

(Yk))

Fig. 3.)

Let Yk be the closest point to Xk In support(\037v) (in particular

Ok== IYk

-Xkl

< IXo-

Xkl \037 0.)

Then v(Y k ) \037 v(Xo) == o. Let us find a contradiction: by superhar-

monicity)

1

1v(Yk ) > v(Y) dY .-

I B2Dk (Y k ) IB28k (Yk))

But we evaluate)

r v(Y) dY = rv + r v = It + h .

}B28k (Yk) }

B28k (YkhB8k (Xk) )B8k (Xk))

For II, we use that v > -s, once our configuration is close to Xo.

Therefore II > -sIB2Dk(Yk)l.

For 12, we use that v is harmonic inBDk (X k ). Thus the mean value

theorem holds and)

1 2 ==IBDklv(X k ))

Therefore)

1

v(Y k ) > -8 + 2nv(Xd \302\267

For s < a /2n

we obtain a contradiction.)))


Before we go on, let me list a few properties of harmonic and

superharmonicfunctions that are an easy consequence of the mean value

theorem:)

a) Harnack inequality

If v is harmonic and non-negative in B I, then for R < 1)

sup v < C(R)infv.BR BR)

(C(R) goes to infinity when R goes to one as a (negative) powerof(1-

R)).)

b) Derivative estimates

If v is harmonic in BI(0))

lV'v(O)1 < C osc v.Bl (0))

Note that harmonicity is a linear, translation invariant property, thus

derivatives of harmonic functions are again harmonic, and by iterating

b), and scaling, we get1

b/) ID(k)v(O)1 <

C(k)k osc v. (Check it!).r B r (0))

c) Strict maximum principle

Given a superharmonic function v In D, v cannot have a localIninimum in D, unless v is identically constant.

Finally, we will also need the)

d) Solvability of Laplace's equation in a ballGiven f continuous In a B1, then there existsa unIque harmonic

function v, such that)

VlaBl== f \302\267)))

2. Regularity of Solutions)

We are now ready to launch into the study of the regularity propertiesof u and the set u ==

cpo We start with the regularity of u. The CI,I

regularity of u, when cp is CI,l is due to Frehse [F].The proof below

is in [C-K]. We will always stay away from aD.

THEOREM 2. HUp to CI,I, U is as regular as cp\". More precisely)

a) Assume that cp has a modulus of continuity a (r), then u has modulus of

continuity Ca (2r)b) Assume now that Vcp has modulus of continuity a(r), then Vu has

Inodulus Ca(2r).)

To prove Theorem2, we start with)

LEMMA 3. Let u(X o ) == cp(Xo). Then HU separates from cp with the

speed dictated by a(r) \", more precisely in case a))

sup (u -cp)

< Ca(2r) .

Br (Xo))

In case b))

sup (u-

cp)< Cra(2r).

Br (Xo))))


PROOF: We prove case b). Let L == cp(Xo) + (Vcp(X o), X -Xo) be the

linear part of cp at Xo. Then, by definition of a(r), on Br(Xo))

L - ra(r) < cp(X)< u(X).)

Let us show that on Br /2,)

u(X) < L + Cra(r) .)

Consider)

W == u - (L - ra(r)) .)

This is a non-negative superharmonic function in Br.Let us split it in W == WI + W2, with WI harmonic and equal to W

in a Br.

Thus, since W is superharmonic and non negative)

o <WI

< W ,)

and hence also)

o <W2

< W .)

We have that)

WI (Xo) < u(Xo) - (L - ra(r)) ==cp(Xo)

- (L - ra(r)) == ra(r) .)

By Harnack inequality)

Wt!B r /2< Cra(r) \302\267)

About W2, it is superharmonic and vanishes on aBr. Thus, it attains

its maximum in the support of its Laplacian.But)

\037W2 == \037u .)

So it attains its maximum at a point X I, where u ==cp.)))


But remember that W2 < W == U - (L - ra(r)). Thus)

W(XI)< <P(XI)

- (L - r(a(r)) < Ca(r).)

The proof of the lemma is complete.

In particularif for instance <p is C I,I, u \"lifts away\" from <p in a

quadratic fashion, that is (u- <p)(X) < CIX-

Xo12.

From part b) of the Theorem, let me just show that if <p is C I, I,

then u is C1,I (away from aD). This just follows by scaling: Let

XI E Q == {u > <p}, d(X I , A) == d(X, Xo) == p (A == {u == <p}, Xo E A).

Then on Bp(XI ), u is harmonic. But on B4p (X O ), u has quadratic bounds

away from Lxo. (Now, <p being CI,l, we can take ra(r) == Cr2

.) Thus,)

1 1IID

2u(XI)11

<2\"

osc (u - Lxo) < C2 r2< C \302\267

P Bp (Xl) r)

We have now completed the local regularity theory of u, i.e., u E C1,I

is as good an estimate as we may hope for, since \037u jumps from zero

to \037<p across aQ.)))

3. Free Boundary Regularity)

3.1.- FreeBoundary Regularity: Part I, generalities)

We now begin to study the regularity of aQ.

The material in this part can be found in [C-1] and [C-3].

To study the free boundary regularity, we reduce it to a local prob-lem, and consider the new variable w == u - cp. Then, we have the

following local problem.

DEFINITION (NORMALIZED SOLUTIONS). In the unit ball of Rn

we are

given afunction w with the following properties:

a) w > 0, w is CI,I.

b) On the set r2 == {w > OJ, \037w = 1.

c) The point 0 belongsto aQ(i.e.,isafreeboundary point).

QUESTION. What can we say about the geometry of aQ?

SOME REMARKS.. a) On Q == {w > O} we really have)

\037w == \037(u-

cp)== -\037cp == g(X).)

Since u is superharmonic, it cannot touch cp at a point where \037cp> 0,

so near the free boundary we should expect g(X)> O. In fact, if \037cp)))


and V' \037cp do not vanish simultaneously (a necessary non-degeneracycondition),a variation of Hopf's principle shows that g(X) > 0 near the

free boundary.

REMARK. b) We have made g(X) = 1. All it is necessary is g(X) of

class Ca for the general theory and g(X) of class cI,a, to show that

singular points lay in smooth manifolds, but these assumptions would

fill the proofs of little technicalities.

REMARK c): IMPORTANT RESCALING OBSERVATIONS. The function W A (X) ==

>:'zw()..X)satisfies the same conditions, in B)..instead of BI'

We start with the following)

LEMMA 4: OPTIMAL REGULARITY. w restricted to BI/2 is boundedby a

universal constant and its C I,Inorm is also bounded by a universal constant.)

PROOF: Apply Theorem 2 to u == W -2\037IXI

2, cp

== -2\037IXI

2.

LEMMA 5: OPTIMALGRADIENT BOUND.)

lV'w(Xo)1 < C(w(X O ))1/2 .)

PROOF: Let w(X o) == h > O. Then, from the second derivative bounds,

B(Ch)J/2(XO)C Q. (If 3 XI E B Ch l/2(X O ) n A, find a contradiction!). In

BCCh )J/2(XO),

\037w == 1 and)

Ch -IX

- Xol2v==w+

2n)

is harmonic and non-negative.

From Harnack and Interiorestimates)

C CIVw(Xo)1 = IVv(Xo)1 <

h 1/2osc V <

1/2v(X o)

(C)B(Ch)1/2

(C h)

C< -h == Ch

1 / 2-

h 1/2.)

LEMMA 6: (MAXIMUM GROWTH). Let Xo E Q, thenSUPBr(Xo)

W > Cr 2.)))


PROOF: It is enough, by continuity, to prove it for Xo E Q. Let v ==

W -2\037IX

- Xo12. Then v is harmonic in Q n Br(X o ), and positive at

Xo. It should take a positive maximumon a(Q n Br(Xo)). But on

(aQ) n Br(Xo ), W = 0, so v is negative.

Thus the maximum takes place at a point X 1 E a Br. There

1 2o < V == w(X I )- -r .

2n)

LEMMA 7. The \"level strip\", Sh =={O < w < h2} C Q, satisfies

meas(Sh n Br) == ISh n Brl < Ch r n - l.)

We split the proof into two steps:)

LEMMA 7 A. Let We == De W be the directional derivative of w in the

directione. Then)

rIVw e l

2 < Ch r n - l,

l{O\037we\037h}nBr)

PROOF: By the rescaling properties of w (i.e., by looking at Wr ==

\037w(rX)) it is enough to look at r == 1 (Check this fact!!). We truncater

We at levels sand h: W e == min[(w e - s)+, h], and write the usual

formula)

rV W e VW e + we[}.we = r

w eDvwe,

lB} laB}

Since \037we = 0 in Q (in particular for We > s) and IDvwel < ID2

wl< C,

we immediately get)

rIVw e l

2 < Ch ,l{\302\243<we<h})

PROOF OF LEMMA 6: Again we prove it for r == 1)

Sh C {1V'wl < h} (from Lemma 4) c n{W:f:en< h}.)

Thus

ISh n Btl = r [}.w< C rID

2wI2 < b r IVw el

2dX ,

JShnBI JShnBl J(w\037en <hI)))


COROLLARY 5. The neighborhood N8 of the free boundary)

(N8(S)) == {X : d(X, S) <8})

has measure)

IN 8 n Brl < 8r n - l.)

In particular, thefree boundary has locallyfinite n -1 dimensionalHausdorff

measure, and)

Hn-l(aQ n Br) < C r n - I.)

PROOF: (N8 n Q) c S8, and thus)

IN8nQnBri < 8mr n - l.)

But Q has uniform positive density along the free boundary, I.e., for

Xo E aQ, we have)

I Br(Xo) n QI > JL>OIBrl)

for some universal constant JL.

Indeed, from Lemma 6 (maximum growth), W(XI) ==sUPBr(Xo)

W >

Cr 2, and from CI,I estimates,IVwIB2r(Xo>

< Cr, Thus w must remain

positive in BCr (X I) for C small enough.This completesthe \"generalities\" part of our discussion, that is those

properties and techniques common to many free boundary problems:

optimal regularity, to allow for rescalings,maximum possible growth

to provide stability of the free boundariesand coincidence sets under

rescalings and measure theoretical propertiesof the free boundaries.)

3.2. - Free Boundary Regularity: Part II)

We now go to a special issue in each freebo?ndary problem, that

IS the classification of global solutions to our problem. Namely if we)))


plan, as in the theory of minimal surfaces to prove local regularity by

a blow up argument we want to see the special form of the \"blowuplimits\" of our problem.

The main theorem in this regard is that global solutions are convex.

More precisely)

THEOREM 3. ([C1]) Let, as before, W be a normalized solution of our

problem in BI (0). Then, there exists a universal modulusof continuity a (r)

(a (0+) == 0), such that second pure derivatives of w)

Dee w)

satisfy)

Deew(X) > -a(IXI) .)

We prove the theorem through the following lemmaappliedinductively.)

LEMMA 8. Assume that Br(Xo) C Q(w), and tangent to the free bound-

ary, a Q, at a point Yo.

Let -a ==infBr(xo) Deew .

Then Deew(Xo) > -a + CaMfor some C, M depending only on

dimension.)

PROOF OF LEMMA 8: By rescaling w to w ==-h-w(rx), we may assumer

r == 1. (Note that this rescaling does not change bounds on second

derivatives.)

We do the following construction: in the ray (Xo, Yo) choose a point,

Y I== Xo + (1 - h) (yo

- X o), at distance h from Yo. Since Yo belongs

to aQ)

w(YI ) < Ch2 and V'w(Y I ) < Ch .)

Starting at YI, the direction +e or -e points \"inwards\" to the ball

(i.e., (+ or - e, Yo- Xo) < 0). Say +e. (Note that Dee ==

D-e,-e.)

Therefore if Y2== YI + *h 1/2e, the segment from YI to Y2, remains at

distance at least h/4 from aB I (for h small).)))

22) LUIS A. CAPFARELLI)

Iy; - \037I rv h 1/ 22 1)

IY 1-

YOI==h)

Y1)

Bl (X O))

Fig. 4.)

At Y2, w is still positive. Let us show that this means that Dee W

must be \"almost positive\" at some point in the segment I == (Y I , Y2 ).

Indeed we represent W(Y2)- w(Y I ) as a double integral of Dee

along I)

o < w(Y 2) == w(Y I ) + (V'w(Y I ). Y2

- Y I ))

h 1/ 2

+ff

Deew < h2+h. \037 +

ffDee.

I I)

That is)

f fDeew > -Ch

2 - Ch3 / 2,

I)

But I has length h 1/2. Therefore)

(h 1/2)2supDeeW > -C h 2 - Ch3 / 2

,I)

or)

sup Dee W > - Chi/2.)))


We have, therefore at least, a point Y3 E I, with Dee W (Y3) > -Ch l / 2.

We want now to choose h so that we have an actual gain over the

previous bound:)

Dee W + a > o.)

Thus we choose-Ch l / 2 = -aI2, i.e., h =U\037J

2. We have that)

DeeW(Y3) + a > a12.)

We now apply Harnack inequality to the non-negativeharmonicfunction)

veX) == Deew + a .)

It says that)

Deew(Xo) + a = v(Xo) >CV(Y3)

\302\267(1

- IY31)M= C\037

(h)M = Ca1+2M ,)

The lemma is complete.)

COROLLARY 6. Let W be a normalized solution.

If DeewlQ> -a, then DeewlQnBl/2

> -a + CaM.

PROOF: Let Xo E Q n B I / 2 , and Br (X o) be the largest ball in Q con-

taining Xo. Since 0 E aQ, Br(Xo) C BI and the lemmaapplies.PROOF OF THEOREM: By induction we have that, for w a normalized

solution, since w1BI/2 is CI,1)

Dee w lB I /2> -C = -ao.)

We apply the lemma inductively and we get)

D ee w lB2_k> -ak)

with)

-ak+1 > -ak + Car \302\267)

This implies ak > -k- e for some small \302\243,or a (r) == -I log r I-e.)))


COROLLARY 7. Let w be a solution in BM, i.e., w(X) =\037w(MX)

is

a normalizedsolution.Then)

DaawlBJ> a

( \037 )

\302\267

COROLLARY 8. Let w be a solutionin Rn

, then w is convex, and A(w) ==

{w== O} is convex.)

At this point let us pause for a moment and study what are the

possible candidates for global solutionsw:

a) w ==1(x+)2 (in some systems of coordinates).

These are the \"blow up\" limits that we expect to get if we start

from a point where aQ is a smooth surfaceseparating Q from A.

2

b) w = 'L7 Ai

x

4with Ai > 0, 'L Ai

= 1, always in some system of

coordinates.)

These are the solutions we expect if we started from a point where

A had very little density or further an isolated point of A.

In fact:

REMARK. If w is global and A a half space, w is as in a) from Cauchy-

Kovalevski theorem.

If w is global and A has empty interior then w IS as in b) from

Liouville theorem.)

c) We can construct a radial solutionwith A a ball:)

.l... I X 12

/2n)

\037

subtract)

fundamental

solution)

Fig. 5.)))


So life is not so simple.But at least we can say that if W IS a

global solution, 0 E A and the trace of A in say a BI is not \"so thin\"

that is \"almost\" lower dimensional, then near the originall levelsurfacesof ware smooth Lipschitz surfaces.

Let us write, from now on X == (X', x n ) with x' == (XI, . .. , Xn-I).)

LEMMA 9. Let w be a global solution(with 0 E aQ) and assume that

Bp(-te n ) C A,lorsome 0 < t < 1/2.

Then a) for Ix'i <\037,

-t < X n < 1, for any unit vector a, with

an > 0, la' I< p/8, we have)

al) D(1w > 0

a2) All level surfaces, w == A are Lipschitz graphs.)

CX n == I(x', A) with

II/IILip< - \302\267

p)

a3) Den w(X) > C(p)d(X, aA).

a4) For la' I< p/16 we have D(1w > CDenw.)

NOTE. a2), a3) and \037) follow from al).)

PROOF: Since w is convex, the directionalderivative, D(1 w is monotone

along any line in the direction a, and thus, it becomes positive once

such a line intersectsA.

This proves al) and thus a2). This also proves that w grows

quadratically in the vertical direction that is, if X == (x', x n ), then

w(X) > Cp21x n

- I(x', 0)12. Indeed, consider the ball)

B == B 1I fl

(x', I(x', 0)).Tb p Xn-)

From maximal growth)

w(Y) == sup w > Cp2

1xn- 11

2

B)

but w is monotone increasing along the segment that joins Y to X, from)))


Finally, this implies a3: Indeed

j(X' ,Xn)

Den dYn > Cp2

1xn- 11

2,

(x',f(x'\302\273

therefore, sup] Dell > Cp2

1xn- fl. But this can only happen at a point

along I, at distance Cp2

1x n- fl from the free boundary.

From Harnack inequality, this also holds for all of the segment

between say C Ip 2/xn- fl and /xn

- fl with some large constant C(p).

(Notice that the factor IXn- f I scales out of the computation.)

Finally, a4) follows by expressing any such direction a, as)

a == aa + be n)

with a a direction for which al) applies.Then a, b can be chosen positive and b > bo > 0, b o a universal

constant (1/8?). (The reader can check this out.)

We will now invoke the theory of harmonic functions in Lipschitz

domains to deduce that all these level surfaces are uniformly C I,a all

the way to the free boundary. We will develop this theory as soon as

we completethe free boundary regularity theory.

Let me state the main theorems(see[CFMS],[K-J]and [A-C]).

THEOREM 4. Consider the do.main Bt == BI n {x n > OJ. Let VI, V2 be

two non-negative solutions of a divergenceoperator(aij bounded measur-

able and uniformly elliptic))

Di(aijDj)v == 0)

and)

VilXll=O = 0 \302\267

Let us normalize them so VI (1en)==

V2(\037en)== 1. Then, the quotient

u(X) = VI (X)

V2 (X)

is-bounded and Holder continuous up to Xn == 0 in B02' with lIuliLoo,

II u II c a < C depending only on the ellipticityof aij.)))


REMARK. The hypotheses on Vi are invariant under bilipschitz transforma-

tions of Bt. Indeed,we transfer the weak formulation of DiaijDjv == O.

That is)

jDj1/faijDj v = 0

for any HJ test function 1/1' or)

j(V1/flAVvdX = 0,)

Change variables Y == Y(X). Then)

V'x 1/1'==

TyX V' 1/1' (Y))

and)

dX ==detT[ dY.)

So)

j(Vy1/f)TA*VyvdY = 0)

for a bounded measurable A*

.)

COROLLARY 10. Same result holds if, instead of Bt, we consider the

domain)

D == {IX'I < 1, f(X') < X <M})

for some f (X') Lipschitz, I f (X') I< M /2.)

3.3. - Application to our free boundary problem (still global solutions))

THEOREM 5. Under the hypothesis of Theorem 4, the levelsurfaces{x n == f(x', A)} are uniformly cI,a up to A == O.)))


PROOF: We show thatWa are uniformly Holder continuous in {Ix'i <Wn

\037, Ixlll < I} for any \"horizontal\" a.

To apply the theorem we miss the pOSItIVIty of W a . Write a =

f6a +e n . Then a4) applies to a (although it is not unitary) and according

to Corollary 9, the quotient

Wa is C aup to aQ.

Wen

That is [6 ( :::\037 )+ 1 is C cx

up to the free boundary and hence)

W aVa == -

Wen)

is Ca

up to the free boundary.

But Va is simply, Da f(x', A). Therefore the graphs X n == I(x', A)

are uniformly c1,a all the way up to A == 0, i.e., up to thefr\037e boundary.

We next show, in order to complete the theory, that it is not nec-

essary to pass to the limit, i.e., requireW to be a global solution to

reproduce the geometry above. This is a rather unusual fact, i.e., that

in a finite approximation we can directly get regularity, characteristic of

this particular problem. First an approximation lemma that says that we

can almost reproduce the geometry of the global solutionsif our W is

a solution on a large enoughball.)

LEMMA 10. Fix P > 0, S > O. assume that we are given a solution W

in a ball B M (0), with M == M (p, s) \"very\" large.

Assume also that wi Bl (0) has the followingproperty:)

\"The set Aw n Bl cannot be enclosedin any strip {a < X n < fJ} of

width, (fJ - a) < 4np\

Then, for M > M(p, s) large there existsa globalsolution Woo (X),

that satisfies the hypothesis of Theorem4, and such that)

II W -Woo II LOO(Bl)) lIV'w

-V'WooIILOO(Bl)

< s.)

PROOF: The proof is by compactness: assume that such an M(p, s)

does not exist. That means that if we fix (p, s), we can find a sequence)))


of solutions UJk defined in balls Bk (0) that satisfy the hypothesis of the

theorem and not the conclusion.The Wk are a compact family in cI,a in compact sets (they all vanish

with their gradient at the origin and are universally C I, Iin B k / 2). Thus

a subsequence convergesuniformly in compact sets to some function

Woo. To get a contradiction, it suffices to show that Woo is a global

solution satisfying the hypothesis of Theorem4. Indeed

a) Woo > 0, Woo E CI,I, and \037woo(x) == 1, whenever w(X) > 0

b) woo(O) == 0 and 0 E aQ becausesUPBr(O) Wk > Cr

2independently

of k and thereforesUPBr(O) Woo > Cr

2 .

Also, Woo and A(w) are convex and further:

\"A (woo) n B I cannot be enclosed in any strip of width\037np\".

Indeed of if A (woo) is enclosed in such a strip, say)

ex < Xn < f3 ,)

7

f3- ex == - np

2)

then Woo> 8 > 0 outside {ex

- s < X n <f3 + s} n BI. But Wk is

converging uniformly to W in compact sets, thus Wk>

\037> 0 outside)

{ex- s < X n <

f3 + s} n BI)

contradicting one of the hypothesis. To completethe proof of the theo-

rem, we need to show that the \"strip property\" implies that A (woo) n B I

contains a ball of radius p.

For that we invoke a lemma of F. John that says that if E is

the ellipsoid of largest volume contained in a convex set, in our case,

(A(woo ) n B I ) then nE ::) (A (woo) n BI ).

With this lemma at hand, if E has one of its diameterssmaller

than 2p, nE has one of its diameters smallerthan 2np and we can trap

A (woo) n B 1 in a 2np strip, a contradiction.

In order to be ableto jump now from the limiting configuration to

an approximatingone, we need the following curious property of the set

Q(w) for a normalizedsolution w.)))


LEMMA 11. Let h be a harmonic function in Q (w). Assume)

a) h > 0 on aQ (forinstancelimh > 0).

b) If Na denotes the a neighborhoodofaQ,bl ) hl Na

> -a

b2 ) hlQ,N a > 1.)

Then there exists a universal ao such that for a < ao, the hypothesis

above imply that h > 0 in Q n BI / 2 .)

PROOF: Suppose not, then there exists an Xo, in Na nB I / 2 , where h < 0

(since outsideof Na, h > 1). Consider, in B I / 4 (X O ), the function)

v = h(X) - /)

[W(X)

-2\037

IX- X O/

2

]

,)

Then (this must sound familiar!))

a) v is harmonic in B I /4(XO)n Q.

b) v(Xo) < O.)

Thus)

c) v must have a negative minimum in a(B I / 4 n Q).)

But along aQ, v > 0, thus the minimum occurs along aBI/4.Let us see that this is not possible. On aBI /4

n N a ,)

8

(

1

)

2V > -a - C8a 2

+ - - > 0-2n 4)

if 8 > 200na, and a small. OnaBI/4

n (Q\" N a ))

v > 1 - C82> 0)

if 8 (universally) small.

As a corollarywe get the following.)

D)))


THEOREM 6. There exists an s ==s(p) such that

If w satisfies the conditionsofLemma9, that is: w is a solution in a

ball BM, M(p, s(p)) == M(p) and A(w) n BI cannotbe trapped in a

strip of width 2np.

Then, in an appropriatesystem of coordinates,)

a) conclusions of Lemma 8 holdfor w (with Ix'i <\037

substituted by Ix'i <

f6and la'i < 16 substituted by la'i <

f2in \037);

b) conclusions of Theorem 5 hold, that is, all level surfaces f(x', A) are

uniforlnly C l,aUp to A == O.)

PROOF: From Lemma 10, we have that there exists a Woo, satisfying

the hypothesis of Lemma8, such that Iw-

WooiLoo, lV'w- V'WooiLoo< s.

Therefore,the crucial properties a3), a4) that make f (x', A) uniformly

Lipschitz graphs, are \"almost\" satisfied. We will now use Lemma 11, to

show that these properties are \"fully\" satisfied. We start by organizing

the information we have by putting Lemma 9 and Lemma 10 together.)

LEMMA 12. Let w satisfy the hypothesisofLemlna 9, and let Woo be its.s globalapproximation. Then, in the domain Ix'i <

\037,-t < X n < 1, we

have (Nt; (S) is the s-neighborhood of S))

a) aQ (w) CNC(p)v'i(aQ (Woo)).

b) Den w(X) > C(p)[d(X, aA) - C,jS].c) Forla'i< R)

Daw(X) > C(p)[d(X, aA) - C,jS].)

d) w(X) > C(p)(d(X, aA) - C,jS)2.)

PROOF: All we have to prove is a) and then we just put the estimates

in Lemmas 8 and 9 together.

To prove a) we note that if Xo is in Q(w oo) and d(X o , aQ(w oo )) >

C(p),jS then woo(Xo)> C(p)(C(p),jS)2> 2s. Thus, w(Xo) > 0 and

Xo cannot belong to aQ.)))


-On the other hand if Xo E Q(w) n A(w oo), from non-degeneracy,)

sup w > C(C(p).J8)2 > 2s ,

BC(p)v'i)

thusBC(p)v'\302\243(Xo)

cannot be contained in A(w oo). 0

Now, to complete the proof of Theorem 6, all we have to prove is

that w satisfies aI) of Lemma 8. (Now with la'i <16). Since a2), a3)

and a4) follow from it.

From c) of Lemma 12,)

Daw(X) > [C(p)d(X,aA)- c.J8].)

We apply Lemma 11 to Daw in B p / 8 . Let h =\037f/\037

.

Then if d(X, aA) >\037:\037;)

heX) > 2 - Cs 1 / 4 > 1 .)

28 1/ 4If d (X, a A) < (;(p)

,)

heX)> -Cs

l / 4.)

If we choose s small enough so that C (p)s1/4 < (p j8)ao, we

have heX) > 0 in Bp/I6(0). Since outsideof Bp / 16, Da is already non-

negative, due to the fact that we are away from A, the proof is complete.)-

By inverting the relation M == M(p) into p ==p(M), we have

proven the following theorem.)

THEOREM 7. Let w be a normalized solution.

Then there is a universal modulus of continuity a(r) (more precisely

a(r) ==p(\037))

such that iffor one value ofr, say ro, A(w) n Bra cannot

be enclosed in a strip ofwidth roa (ra), then, in an rJ neighborhood of the

origin,the free boundary is a C I,a surface X n == f (X') with)

CIIfll c l,a < -.

ro)))


PROOF: Let us renormalizew by ro, i.e., consider)

1W ==

2\"w(ro, X) .ro)

Then w is defined in a ball BM of radius M == ..1 and A(w) n BI cannotro

be enclosed in any strip of width a(ro) ==p(M). Thus, Lemma 10

applies.)

3.4. - The Structureofthe set of singular points)

Now, it only remains the study of the structure of the set Sing(w) of

singular points of N, that is, those Xo in aQ for which IA n Br(Xo)1 C

Sra(r) (a strip of width ra(r)) for every positive r.

Our main objectiveis to prove the following)

THEOREM 8. Given a singular point, say Xo == 0 E aQ)

a) There existsa unique non-negative quadratic polynomial)

1 TQxo == -(X MX)

2)

}vith \037Qxo

== trace M == 1, such that)

I(w - Qxo)(X)1< IXI2

a(IXI))

for some (universal) modulus of continuity a(IXI).

b) M(Xo) is continuous on Xo(for Xo in Sing(w))

c) If dim ker M == k, the singular set Singe w), lays, in a neighborhoodofXO, in a k-dimensional C I

manifold.)

The size of the neighborhood dependson the smallestnon zero eigen-

value of M.)))


Before entering into the proof of this theorem, let us point out that,

because of compactness, if Xo is a singular point, thenWIBr(Xo)

looks

\"more and more\" as a quadratic polynomial:)

LEMMA 11. Given s, there exists an M(s), such that, if W is a solution in

B M and if 0 is a singularpointofa Q, then, for some non-negative quadratic

polynomial Q == XT

AX, with \037 Q== 1, we have that)

Iw-

QIBI < S .)

PROOF: Suppose not, then there is a sequence of solutionsWk(X) de-

fined in Bk (0), that have zero as a singularpoint, and for which there

is no such polynomial.Let us take a subsequence, Wk, that converges uniformly in compact

sets to a global solution,Woo. We prove: Woo is a quadratic polynomial

Q as above.

REMARK 1. Aoo has empty interior. If not, Aoo being convex, Aoo n B I

will also have nonempty interior, i.e., will contain a ballBro (Xo), where

Woo = O. Prom nondegeneracy Wk = 0 on Bro /2(XO)

for k large. If not)

SUPWk> Cr5 for any Y k E B ro/2 (X O ) n Q(Wk) ,

B I (Yk)4 rO)

contradicting the uniform convergence of Wk to WOe

But, since 0 is a singular point of Wk, A(Wk)nB I must be contained

in a strip of width a(\037) according to Theorem 7, a contradiction as soon

asa(\037)

becomes smaller than roo

REMARK 2. Woo is a quadratic polynomial. Indeed \037Woo = 1 and has

quadratic growth, so)

1 2h == Woo

- -IXI2n

is globally harmonicwith quadratic growth, thus h and hence Woo is a

quadratic polynomial.)))


COROLLARY 11. Let w be a normalizedsolution,and0a singular point.

Then given s, there exists a roes)so thatfor any r < ro, we have a quadraticpolynolnial Qr

== XT

AX, with)

Iw - QrlBr< sr

2.)

As before, we invert the relation s (r) == a (r) and say: if w is a normalizedsolution and zero is a singular point, there existsa Qr such that)

Iw- QrlBr

< r2

a(r).)

The problem with Lemma 11, is the standard problem in singularity

theory, for instance in minimal surface theory: take a sequenceof blow

ups of a minimal surfaces and you get a minimal cone.

The problemis that different sequences may give different cones,

that is the cone (or in our case the quadratic polynomial) may slowly

rotate. The question is thus how to \"glue\" the polynomials Qr that

approximate a normalized solutionat all different levels Br.

This is solved by the use of a monotonicity formula:)

THEOREM 9 ([ACF]). In B I (0), let u I, U2 be two continuous functions

such that)

a) Have disjoint supports: u I.

u2 == o.

b) u I (0)== u2 (0) == O.

C)Ui\037Ui > O.)

Then)

T(R) ==

(

\037 ( I VU l12

dX

) (

\037 ( I Vu 212

dX

)

== 7i .72R2 } BR

rn - 2 R2 }BR

r n - 2)

is monotone increasing in R.)))


Fig. 6.)

Let me make several remarks about this theorem whoseproofwe

postpone to the end of the discussion. The standard picture to understand

this theorem, is to have in Bl (0) a (relatively nice) surface S, through

the origin, separating B I in two domains D I and D 2 .

In each of them we have a harmonic function Vi that vanishes along

S. Then

a) Each of the termsIi can be understood as an average of I V'Vi 12

,

i.e., we are dividing the volume integral in a domain of size rv Rn

by

a factor R2r n - 2 rv R n.

In fact if VI is the positive part of a linear function, VI (X) ==axi)

1i (R) = C(n)a2

.)

with C (n) a precise constant related to the volume of the unit ball of

R n.)

b) Ii has linear scaling, i.e., if u(X) ==tU(AX),

7i( \037 ,u)

=7i(R,u).

c) If S is smooth at zero, i.e., Dv Vi exists, then)

lim 7i == C(n)(D v Vj)2 .R\037O)))


In particular)

C2(n)(DvVI)2(DvV2)2< 7(1/2) .)

d) On the other hand one may suspect that the quantity Ti (R) could

be uncontrolled for some \"not too bad\" Vi. That is not so: considerthe

function n\0372in B I / 2 , and extended to a function V on B I in a smooth,r

non-negative way, so that V = 0 near Bl.

Then)

Bl/2)

Fig. 7. Graph of V.)

Ti(1f2)= 2

2 r

(Vn\037;2

)

dX} B'/2 r

< 22

l,

11

( \037)

\302\267V dX

1

22 V. 2

== 2 (\037V)\037dX <CliviIlL2(B)

Bl -.....Bi/22 1)

e) Finally, a remark on homogeneous harmonic functions in cones. Letr be a cone with vertex at the origin, i.e., given a subset bO C SI, set)

r =={

X : \037 E \037o C S I

}

.

IXI)))


Then we can look for homogeneousharmonic functions on r, heX) ==

ra

I(a), vanishing on ar by finding eigenfunctions on bO:)

(n-- 1) 1

\037h == Drr h + Dr h +2\037ahr r

= r a - 2[(a(a

- 1) + (n - l)a)/(a) + \037a I(a)])

So if I is an eigenfunctionof A/(a) + \037a I(a) == 0, in bO, we

can associateto I the homogeneous harmonic function heX) == r a I(a),wherea is related to A by)

a(a + n - 2) == A.)

For any positive A, we have two roots, one bigger than zero and one less

than -en - 2) (the extremal case being a constant and the fundamental

solution).

We sketch now the proof of the theorem.

PROOF: By scaling, it is enough to prove it for r == 1.)

T' (1) ==71'72 + 71 Ti - 47172,)

so we want to prove)T/ 7,'--.l+\037-4>0.71 72

-

We now reduce the formula to integrals on bi == Di n S I)

J

(V'Ui)2

J

!:1(ut1 (

n - 2 2

)Ti

==2

dX >2 dX == Ui DrUi + U

ida

r n - r n -1:' 2

I)

(We use f U\037V - V\037U ==f1: UVv

- vU v ) while)

71' =\037

( VU i)2 dX ,)

So)

71' f (u r )2 + (ua)2 da

Ti

-

fUU r + n22u2da\302\267)))


Note at this point thatJ

Ju:tis minimized by AI, the first eigenfunction

of the Spherical Laplacian t::\".a, in \037l, so we want to split UU r in an

optimal fashion to spread its controlbetween J (u r )2 and J u\037, i.e.,)

JUU r =

\037[A Ju

2+

\037 J ui],)

That will leave us with)

2J u; + J(u a )2

\037J u; + [A + (n - 2)]J (u)2

\302\267)

To perfectly balance both terms, we want)

1 [A+(n-2)]

A Al)

or A[A + n - 2] == Al .)

If we choose A in such a way we can say that)

T/\037 > 2A. .

T;- I)

Thus)T/ T,'-1.. + -1. - 4 >

2(A I + A 2 - 2) .'Ii T2

But note that Ai, from the formula above, is precisely the homogeneity

in r of the function hi, harmonic in the cone ri, generated by \037i,

vanishing on a r i .

That is, the monotonicity formula has been reduced to the following

question:)

Given two disjoint cones r i in Rn

, let hi be the corresponding

homogeneous,non-negative harmonic functions, is it true that the

sum of their homogeneities,Ai, is always bigger or equal than two?)

Note that for a linear function we have equality (i.e., we have

homogeneity one on both sides) and it is natural to guess that this is

the extremal configuration.We sketch the main ideas of the proof.)))


Step a) (Sperner [S])By symmetrization, among all domains h in SI of prescribed mea-

sure (I h I), the one that minimizes A, and thus A, is the spherical cup)

hI == SI n {XI> a}) h2 == S2 n {XI < a}.)

There)J

a(u )

2( 1 - (X )

2)

(n-2)/2 dx'\\

.f

-I Xl I I

A I == In

Ja-I u

2(1

- (XI)2)n/2dXI

(by changingvariables XI == COS () in polar coordinates).)

Step b) (Friedlandand Hayman [F-H])

The minimum decreases with dimension, SInce an n I dimensional

configuration, can be extended (without changing the homogeneities)toa higher, n2 == n I + k dimension, so we may try to scale properly and

study the problem for large dimension.Note that since A(A + (n - 2)) == A,)

1

[ ]

A

(

0(1)

)A == - V (n - 2)2- 4A - (n - 2) == +

2 (n - 2) (n- 2)2)

,.., ,..,

for n large, so it is enough to study Al + A2 for n going to infinity,

with)

A_IJ\037I (U XI )2(1

-(XI )2)(n-2)/2 dXI

I-

nJ\037I

u2

(1- (XI)2)n/2dXI

\302\267

This suggests the change of variable)

y == ,JnXI .)

Then u y=

)nU

XI ' and)

A- _ J\037III/2(Uy)2(1 -y2 /n)(n-2)/2 dy

I--

S\302\267

J- 0iu 2(1 - y2/n)n/2dy)))


But then (1 -y2jn)(n-2)/2 converges in compact sets to e- y2 / 2 and Al

converges to the eigenvalue of the Gaussian

J.\037oo(uy)2e-y2/2 dy

J\037oou

2 e- y2 / 2dy

.

Our problem thus becomes: Given s E R, let Al be the Gaussian

eigenvalue for (-00, s), A2 for (s, (0), is it true that)

Al +A2 > 2A(-00,0) ?)

(Remember that A(-oo, 0) corresponds to the plane solution,for which

the desired inequality holds.))

Step c) ([B-K-P])For the Gaussian, A is a convex function of s.

IDEA OF THE PROOF: The eigenfunction u).. satisfies an equation

x 2 x2Dxe- Dxu == e- u.)

Such an equation)

Dxf Dx u == -Afu.

Can always be reduced, by taking a new variable v == fI/2u, to

(f 1/2)\"Dxx v -

fl/2V = AV \302\267)

In our case)

Dxx v -( IJ:t

-\037

)

V=AV.

A theorem of Brascamp and Lieb, then says that, if UI, U2 are convex

sets of R Il, V is a convex potential and A the first Dirichlet eigenvalue

in U of)

LlV-VV==AV,)

then)

A(tUI + (1 - t)U2)<tA(UI) + (1 - t)A(U2).)

We apply this theorem for A(a) above.

We now go back to the classification theorem for singularpoints.)))


3.5. - Proof ofthe main theorem)

SOME PREVIOUS REMARKS. Let 0 be a singularpoint of aQ. Recall that)

a) (A n Br) C Sra(r) (a strip of width ra(r)), and that

b) 3 Qr ==1(X

T M rX), with M non-negative, trace of M == 1, such

that)

Iw - Qrl < r2a(r)

IVw - Qrl < ra 1/2(r).)

In rescaled terms, if wr==

\037w(rX) and A r == A(wr

),r)

Iwr - Qrl Bl

< a(r), 1V'(wr - Qr)1< a I/ 2

(r)

(Ar n B I ) C Sa(r) .)

In particular, outside of Sa(r)I/3, from a priori estimates:)

1IDij(w

r - Qr)1 <a(r)2/3

osc(wr - Qr)<

a(r)I/3)

We are now ready for Step 1: Let U e == Dew, be a directional

derivative of w. Then \037ue== 0 in Q, and U e = 0 on A, so we can

apply the monotonicity formula tou\037, u-;: The function)

T -

(\037

J

lV'u\03712

) (

\037

J

lV'u-;12

)(R) -

2 2dX

2 2dx

R r n -R r n -

== T+ (R, e) T- (R, e))

is monotone in R.

In Step 1, we will evaluate r:f:.(R, e) in terms of M R(i.e., D

2QR))

T+(R, e) =\037

{

IDje\03712dX

RJ{(Dew\302\273O}nBR

r n -

jI D. wR I

2Je

dX .={DewR>O}nBl

r n - 2)))


We want to substituteDjeWR by DjeQR

==Mj\037

==eJ

MRe. From the

general remarks above

r+ (R, e) ==

j

II MR e 11

2

d X + a (r) 1/3D R 0 rn-2e W >

(we estimate the integral splittings in Sa(R)I/3 and CSa(R)1/3).Next we want to substitute the domain)

D == {De wR > O}, by DO == {DeQR== X

TMRe > O}.

If we denote by D-l:. == {De QR == XT MRe > ::l:a 1/2(R)} we have, from

the estimatesabove)

D+ C D C D- .)

Thus)

IIMR

ell2

(r 1

dX

)

_ a(r)I/3} D+ r n - 2

< T+(R, e) < IIMR

el12

(L- rn\0372

dX

)+ a(r)1/3 \302\267

But D- \" D+ consists of a strip of widthalf\037:\0372

, hence)

I j j I(

a(r)I/2

)

- < Cmin , 1 .

DX DO-

II M e II)

Therefore, with)

C*(n) = r dX2

} B+ r n -

1

we have the final estimate

r+(R, e) == IIMR

ell2

(

c* + 0 (a(r)I/2

))+ 0(a(r)I/3))

liMe II

== C*IIMRell + 0(a(r)I/3) and

r(R, e) == (C*)2I1MR

eIl2

+ 0(a(r)1/3).)

(Remember that II MR

II is uni versall y bounded since 0 < M R, and trace

MR == 1..)

This completes Step 1.

In Step 2, we \"glue\" all of the M Rby the monotonicity formula.)))


COROLLARY OF STEP 1. If RI <R2, for any unit vector e,)

IIMR1ell

2 < IIM R2el/

2+ 0(a

l / 3(R2\302\273.)

PROOF: It follows from the fact that T(R, e) is monotone in R.

LEMMA 12. IIMR I - M R

21I < 0(aI / 6

(R 2\302\273).)

PROOF: Let N == MR 2 - M R

I. Then N is symmetric, and trace N == o.

We have)

II (MR 2 - N)e1l

2< IIM

R2el1

2+ a 1/3(R2))

or)

-2{MR2 e, Ne) + IINell

2< a

l / 3(R 2).)

Choose e the eigenvector, e).., corresponding to A < 0, the smallest (more

negative) eigenvalue of N. Then)

-2AeT

M R2e + A2 < 01/3(R 2).)

Since MR 2 is non-negative, we have A2 < a l / 3

(R 2) or IAI< a

l / 6(R 2).

Since tr N == 0, II N II< a 1/6 (R 2).)

COROLLARY 12.)

a) As R goes to zero MRhas a unique limit MO.

b) IIMR - Moll < a I / 6(R).

c) IIw- X T

MoXlIlBR<

IIw- QRII + IIQR-

QOII< R

2 a l / 6(R),

d) LetQ\037o

denote the polynomial corresponding to center Xo , thenfor)

IIQ\037o

-Q\037IIIBR(XO)nBR(XI)

<IIQ\037o

-wll +

IIQ\037I

-wll

< R 2a(R) .)

In particular if II X I- X o II

< 8 R for 8 small)

IIM\037o

-M\037III

< a(r) \302\267)

The rest of the theoreln now follows.)))

APPENDIX)

Boundary values of harmonic functionsin Lipschitz Domains)

In this section we discuss the boundary regularity properties of

harmonicfunctions in Lipschitz domains, or more generally of solutionsof elliptic equations with bounded measurable coefficients.

Since the family of solutions u to an equation)

Lu ==Di(aijDju)

== 0)

is invariant under bilipschitz transformations(just write the weak formu-

lation))

lV<pT AVudX = 0

for any cp in HI (Q), we may choose as our basicdomain)

Q == Bt == BI (0) n {xn > O} .)

The main theorem we want to prove is that:)

THEOREM A. Let u, v, be two positive solutions to)

a) Lu == Lv == 0 in Q.

b) u, v take continuously the value zero on {xn==

OJ.)))


Normalize them so)

c) u(\037en)==

v(1en)== 1.)

Then)

u- is of class Cel

in(Bt;2)

v)

(all the way to X n = 0) and)

u) u)

<CV c a -),

V LOO)

where C is a universal constant dependingonly on the ellipticity of A ==[a;j].)

In order to prove this theorem we will need several classical results

on the theory of solutions to Lu == o.)

THEOREM A-I (DEGIORGI OSCILLATIONLEMMA). Let v be a subsolution

of Lv == 0 in BI, satisfying)

a) v < 1

b) I{v < O}I== M > 0)

ThenSUPBl/2

v < A(M) < 1.

THEOREM A-2 (DEGIORGI-NASH-MoSER INTERIOR HARNACK INEQUAL-

ITV). Let v be a non-negative solution in BI (0), then for r < 1)

sup u < (1 - r)-P inf u

Br(O) Br(O))

(for r close to one, we may choose constant one by making p large).)

THEOREM A-3 (LITTMAN STAMPACCHIA WEINBERGER - BEHAVIOR OF

THE FUNDAMENTAL SOLUTION). The fundamental solutionofL behaves like

that of the Laplacian, more precisely: Let BI C Q, and V satisfy)

a) L(V) == -80 (Dirac's).

b) VlaQ = o.)))


Then onBI/2)

C I C 2<V< .n2- - n2r - r -

With these tools at hand, we can prove the theorem.The proofisdivided into two main steps:

STEP 1: BOUNDARY HARNACK PRINCIPLE: If u is as in Theorem A,

above. Then)

uI B+

< M , a universal constant.1/2)

STEP 2: Show that\037

remains bounded in B02 all the way to Xn== O.

STEP 3: Iterate steps 1 and 2.

PROOF OF STEP 1: We start by noticing

a) If Yo E {x n == OJ, then sUPBr(Yo)U decreases polynomially, I.e.,

for r < R)

sup u < C(

!:.-)

asup U

Br(YO)R BR

indeed, when extended by u = 0 for X n < 0, u is a subsolution of

Lu == 0 and I{u < O}n Brl ==\037IBrl.

From DeGiorgi oscillation Lemma,

with A = A(lj2), we get:)

sup uI B

< A sup uI B

\302\267

r/2 r)

b) From the interior Harnack inequality

sup u < s-Pu

(\037en

)

== s-P .

B3/4n {Xn >s} 2)

c) Since u takes continuously the value zero at {xn} == 0, the sup

of u in B02

is attained, i.e.,)

sup u == u(Xo) == M .

Bt2)

We will now show that if M > Mo large, we can construct a

sequenceof points,X k all contained inB\"t;4'

X k ---+ {x n== OJ, and such

that U(Xk) goes to +00, a contradiction to the continuity of u up to

{xn==

OJ.)))


A.I. - Constructionfor boundary Harnack principle)

We will denote by Y k the projection of X k in the {x n== O} axis.

Prom interior Harnack (remark b) above),

M == u(Xo) < IX o -Yol-p .)

Thus, with \302\243== 1/ p)

do == IXo-

Yol< M-

s

that is Xo is very close to the {xn== O} plane.

Now we use the oscillation lemmabackwards:SInce,)

sup u > u (X o) > MBdo (yo))

this implies that)

sup u == u(X I ) > T M

Bdo (Yo)

(for T =Ad/2) ' a universal constant bigger than one),

Again, by Harnack, as with do, we obtain that

d l== IX I

-YII < (TM)-S

and, by oscillation backwards,as with u (x I), we obtain that

u(X 2 ) == sup U > Tu(X I ) > T2M.

B4dl)

Once more, by Harnack,

d2 == IX2-

Y21< (T

2M)-S .)

We repeat inductively the process, and we get a sequenceof points X k ,

satisfying

a) u(X k ) > T k M

b) IXk-

Ykl< (TkM)-S

c) IX k- Xk-II < 4(T k - 1

M)-S)

All we have to make sure is that in this construction we always

stayed inside, say B 9 / 16. But T > 1 is universal, \302\243is universal and M can

be chosen as largeas we please, so we can make L IXk- X k-II < 1/16,

and get, for M >Mo, a contradiction. This proves Step 1.)))


A.2. - ProofofStep2)

We want to show now that\037

remains bounded. Sincev(\037en)

==

u(1ell)== 1, we have, from Step 1, that

vIB+< M, and from interior

1/2

Harnack, that uI B+ n { > 1/ 8 }

>\037

. So our Step 2 reduces to the following1/2

Xn_

question: take in R n, the cube Q2 (en) == {O < Xn < 2, IXj I < 1 for j <

n} andQI/2(\037en)

== {O < X n < 1, IXjl< ! for j < n}. Let FI , F2 be

two faces of Q2(en ), different from {xn== OJ. Let Vi be the function

satisfying)

a) LVi== 0 in Q2

b)V;liJQ2

= X F;,

Show that in Q 1/2,)

V 1< C V2 .)

(This is called the doubling property of harmonic measure.It states that

two adjacent \"balls\" along the boundary of a Lipschitz domain have

comparable L-(harmonic) measure. L-harmonic measure may be abso-

lutely singular with respect to Lebesgue measure so this is a nontrivial

result.))

PROOF.We will show that both Vi are comparable to the Green func-

tion G(x, et), inQI/2(\037en)'

From the oscillation lemma, by extending

(1 -Vi) identically zero across F i we get that (1 -

Vi)< A < 1, near

Fi , say on the cube Q Fiof sides one with one face lying on Fi (see

figure ).

Thus Vi (en) > (1-

A) > 0 and thus Vi is strictly positive inside Q2, sayin QI(en), the cube of sides one centered on en.

Let G(X, Y) denote the (L) Green function in the unit cube. From

[L-S-W], G(X, eI) is bounded for X on the boundary qf Ql (en), vanishes

on a Q2 and hence)

G(X,el) <(VI (X)))

In Q2(e n )-Ql(e n ).)))


en

------:rF. .J

QPj \037

I)

X n== 0)

Ql/2(1/4e n ))

Fig. 8.)

We now show that for X in RI/2 ==Ql/2(!e n ), also VI (X)

<

CG(X, eI). For that, we \"freeze\" X in QI/2(*en ) and recall that

LyG(X, Y) == 0, i.e., G is a solution in Y for X fixed, as long as

X # Y, in particular for Y \037 QI/2(*e n ).

Therefore, Step 1 applies and with X always frozen in RI/2)

G(X, Y) < CG(X, en))

for say, Y in Q3/4(ien).Since G vanishes in a Q2(en), the standard energy estimate says that)

JIVyG(X, Y)1

2 dX < CJ

G2< CG

2(X, en) \302\267

Q2(en),Ql (ien) Q2(en)'Q3/4(\037en))

We now take a Coo function 1}, vanishing in a * neighborhood of F I ,

and 1} = 1 on QI(1en) and represent VI (X), for X in QI/4(!e n ), by the

formulas (no boundary terms left))

J(17V I)(Y)L y G(X, Y) +

JV

T(17V I)A VyG(X, Y) = 0)

and)

( 17(Y)G(X, Y)LVI(Y) + ( VT

(17G)A VVI = 0)))


That give us, after subtracting

VI (X) =J

VT

17 A [VVyG- GVv] \302\267

But on the support of V'1},)

II v II H 1 < C, and II G II HI< C G (X, en) \302\267

SO VI (X) < CG(X, en). (Note that J(V'l}VI)2 is bounded from the

standard energy inequality since 1} vanishes near Fl.)

Step 2 is complete, since, going back to our u and v, we can say

that)

V <

M(LVi)F.I)

and)

1U >

MVI ,

where FI is the face opposite to X n == O.)

A.3. - Step 3)

Consists In showing a Celestimate by iteration, in the following)

way.)

LEMMA. There are constants ak, bk , \037< ak < b k

< M, and a constant

A < 1, such that:

On B:[-k')

akU < v < bk U and (b k- ak) <

A(b k - I- ak-I) .)

PROOF: By induction: Renormalize B;--k to Bt by the transformation

ii == it (2-k

x), define the positive functions

(v- aku)(X)

WI (X)== ,

b k- ak

W2(X)=

(bku - ii)(X) .bk

-ak)))


and look at the positive numbers WI (\037en)' w2(\037en).One of them IS

bigger than 1u(1en)since WI + W2

==U(1

e n).

Say WI (\037en).Then by inductive hypothesis 2wI (X) is a non-

negative solution of L == 0, vanishes on {xn== O} an 2WI (\037en)

>u(\037en).

Hence)1

>-U Bl/2

-M

or, renormalizing back, inB;--(k+1)')

2wI)

v -aku 1>-

(bk- ak)u

- 2M.)

That is, in B 2-(k+1),)

[ak+

2\037(bk

- a k

)]

U < v < bk.)

So bk+1== bk and ak+1 == ak +2\037 (b k

-ak).)))

References)

[ACF] ALT, H.W., CAFFARELLI, L.A., FRIEDMAN, A., Variational problems with

two phases and their free boundaries,Trans. Amer. Math. Soc. 282 (1984),431-461.

[A-C] ATHANASOPOULOS, I., CAFFARELLI, L.A., A theorem of real analysis and its

application to free boundary problems, Comm.Pure Appl. Math. 38 (1985),499-502.

[B-K-P] W. BECKNER, C.E. KENIG AND 1. PIPER, A convexity property of eigenvalues,with applications, to appear.

[C 1] CAFFARELLI, L.A., The regularity of free boundariesin higher dimensions,

Acta Math. 139 (1977),155-184.[C2] CAFFARELLI, L.A., Compactness methods infree boundary problems, Comm.

Partial Differential Equations 5 (1980), 427-448.

[C3] CAFFARELLI, L.A., A remark on the Hausdorff measure of a free boundary,

and the convergence of coincidencesets,Boll.Un. Mat. Ital. A (5) 18 (1981),109-113.

[CK] CAFFARELLI, L.A., KINDERLEHRER, D., Potential methods in variational in-

equalities, 1. Analyse Math. 37 (1980), 285-295.

[CFMS] CAFFARELLI, L.A., FABEs, E.\037 MORTOLA, S., SALSA, S., Boundary behavior

of nonnegative solutions ofellipticoperatorsin divergence form, Indiana Univ.

Math. 1. 30 (1981),621-640.

[F] FREHSE, 1., On the regularity of solutions of a secondorder variational in-

equality, Boll. Un. Mat. Ital. (4)6 (1972).)))

54 LUIS A. CAFPARELLI)

[Fr] FRIEDMAN, A., Variational Principles and free boundary problems, Wiley,

1982.

[F-H] S. FRIEDLAND AND W.K. HAYMAN, Eigenvalue Inequalitiesfor the Dirichlet

problem on spheres and the growth of subharmonic functions, Comment.Math. Helv. 51 (1979), 133-161.

[J-K] JERISON, D., KENIG, C., Boundary behaviour of harmonic functions in non-

tangentially accessible domains, Adv. in Math. 46 (1982), 80-147.

[S] E. SPERNER, Fur symmetrisierung von Functionen auf Sphiiren, Math. Z.

(1973),317-327.)))

the obstacle problem) - uzhuser.math.uzh.ch/ros-oton/caffarelli-the-obstacle...the obstacle problem)...

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