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Convex Configurations In Free Boundary Problems

Arshak Petrosyan

DOCTORAL THESIS, 2000

DEPARTMENT OF MATHEMATICSROYAL INSTITUTE OF TECHNOLOGY

STOCKHOLM, SWEDEN

Akademisk avhandling som med tillstand av Kungl Tekniska Hogskolan framlaggestill offentlig granskning for avlaggande av filosofie doktorsexamen i matematiktorsdagen den 25 maj kl 13.00 i Kollegiesalen, Administrationsbyggnaden, KunglTekniska Hogskolan, Valhallavagen 79, Stockholm.

c© 2000 Arshak PetrosyanISBN 91-7170-573-2Hogskoletryckeriet, KTH, 2000

ABSTRACT

The thesis consists of the following three papers on free boundary problems forparabolic and elliptic partial differential equations with certain convexity assump-tions on the initial data.

Convexity and uniqueness in a free boundary problem arising in combustion theory

We consider solutions to a free boundary problem for the heat equation, describingthe propagation of flames. Suppose there is a bounded domain� ⊂ QT =

Rn× (0, T) for someT > 0 and a functionu > 0 in � such that

1u − ut = 0 in�u = 0, |∇u| = 1 on0: = ∂� ∩ QT

u(·,0) = u0 on�0,

(P)

where�0 is a given domain inRn andu0 is a positive and continuous function in�0, vanishing on∂�0. If �0 is convex andu0 is concave in�0, then we showthat (u, �) is unique and the time sections�t are convex for everyt ∈ (0, T),provided the free boundary0 is locally the graph of a Lipschitz function and thefixed gradient condition is understood in the classical sense.

On existence and uniqueness in a free boundary problem from combustionwith L. Caffarelli

We continue the study of problem (P) above under certain geometric assump-tions on the initial data. The problem arises in the limit asε → 0 of a singularperturbation problem{

1uε − uεt = βε(uε) in QT ,

uε(·,0) = uε0 onRn,(Pε)

whereβε(s) = (1/ε)β(s/ε) is a nonnegative Lipschitz function,suppβε = [0, ε]and

∫ ε0 βε(s)ds = 1/2. Generally, no uniqueness of limit solutions can be ex-

pected. However, if the initial data is starshaped, we show that the limit solution isunique and coincides with the minimal classical supersolution. In the case when�0 is convex andu0 is log-concave and satisfies the condition−M ≤ 1u0 ≤ 0,we prove that the minimal supersolution is a classical solution of the free boundaryproblem for a short time interval.

A free boundary problem for ∞-Laplace equationwith J. Manfredi and H. Shahgholian

We consider a free boundary problem for thep-Laplacian

1pu = div(|∇u|p−2

∇u),

describing the nonlinear potential flow past convex profileK with prescribed pres-sure gradient|∇u(x)| = a(x) on the free stream line. The main purpose of thispaper is to study the limit asp → ∞ of the classical solutions of the problemabove, existing under certain convexity assumptions ona(x). We show, as onecan expect, that the limit solves the corresponding problem for the∞-Laplacian

1∞u = ∇2u∇u · ∇u,

in a certain weak sense, strong however, to guarantee the uniqueness. We showalso that in the special casea(x) ≡ a0 > 0 the limit coincides with an explicitsolution, given by a distance function.

2000Mathematics Subject Classification:Primary 35R35, 35K05, 35J60Key Words: Free boundary problems, convexity, classical solutions, the heat

equation,p-Laplacian,∞-Laplacian, the propagation of flames, nonlinear poten-tial flow.

ACKNOWLEDGMENTS

I wish to express my sincere thanks to my supervisor Henrik Shahgholian, whointroduced me to the subject and guided during the work on this thesis. I alsothank Bjorn Gustafsson, who acted as my second supervisor, for a number ofuseful discussions and valuable advices.

My special thanks are to Luis Caffarelli for inviting me to the University ofTexas at Austin in April 1999 and February 2000. Without this the second paperwould not exist. I also thank Juan Manfredi for productive discussions on themysterious infinite Laplacian in Institut Mittag-Leffler, Stockholm; this resultedin a paper, included in the thesis.

I gratefully acknowledge the financial support from the Swedish Institute forthe academic year 98/99 and the Department of Mathematics of the Royal Instituteof Technology for the year 99/00. The latter was actually my second home duringthese two years in Stockholm.

As an undergraduate student I was studying at Yerevan State University andI thank all my teachers and especially my diploma advisor Norair Arakelian forteaching me mathematics.

And last but not least, I thank my family and all my friends for their constantsupport and encouragement. Your contribution cannot be overestimated.

Stockholm, April 2000

Arshak Petrosyan

CONTENTS

Introduction. Preliminaries and Overview

The propagation of flame . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .1Log-concavity in space . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3Convex classical solutions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .4A free boundary problem for∞-Laplacian . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5References . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7

Paper I. Convexity and uniqueness in a free boundary problem arising in com-bustion theory

1. Introduction and main result . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 12. Convexity of level sets . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 33. On caloric functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 54. Proof of the main theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .6

References . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9

Paper II. On existence and uniqueness in a free boundary problem from combus-tion, with L. Caffarelli

1. Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 12. Uniqueness in the starshaped case . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .33. The convex case . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .74. Lipschitz regularity in time . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 95. Some technical lemmas . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 116. The minimal element ofB . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 137. Further properties of the minimal element . . . . . . . . . . . . . . . . . . . . . . . . . . . 148. The classical solution ofP for short time . . . . . . . . . . . . . . . . . . . . . . . . . . . . 17

References . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 22

Paper III. A free boundary problem for∞-Laplace equation,with J. Manfrediand H. Shahgholian

1. Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 12. Preliminaries . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 33. Classical solutions fora(x) ≡ a0 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7

viii

4. Weak solutions for generala(x) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 105. (u∞, �∞) is a weak subsolution . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 126. Uniform gradient bound forup as p → ∞ . . . . . . . . . . . . . . . . . . . . . . . . . . 147. On∞-harmonic functions near singular convex boundary points . . . . . . . 158. C1 regularity of∂�∞ . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 199. Stable solutions of the boundary value problem . . . . . . . . . . . . . . . . . . . . . . 2010. (u∞, �∞) is a weak supersolution . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2211. The limit asp → 1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 26

Appendix: Classical solutions of (FBp) . . . . . . . . . . . . . . . . . . . . . . . . . . . . 27References . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 29

Introduction

Introduction. Preliminaries and Overview

THE PROPAGATION OF FLAME

Free boundaries, also known as interfaces and moving boundaries, are a prioriunknown sets, arising in solutions of nonlinear partial differential equations (PDE.)An example is the front of a curved premixed flame, which at timet is given as∂�(t), where�(t) = {x ∈ Rn: u(x, t) > 0} is the unburnt zone. The functionu,assumed to be nonnegative here, has the meaningu = λ(Tc − T), whereTc is thetemperature of the flame andλ is a normalization factor. We have thus

1u − ut = 0 in� = {u > 0}.(1)

Besides, the following condition should be satisfied on the free boundary

|∇u| = a on ∂�(t),(2)

for t > 0, wherea > 0 is a given constant, or in the general formulation of theproblem, a function depending onx andt . To complete the conditions, we givethe initial data

u(·,0) = u0 in �0,(3)

where�0 is the initial domain. In what follows, we will refer to the system (1)–(3)as problem (P).

The problem above is derived under certain simplifying assumptions from therather complicated system of PDE, describing the process of combustion, in thelimit of high activation energy. This method goes back to Zel’dovich and Frank-Kamenetskiı [ZF] in 1938. However, the rigorous mathematical study was post-poned until recently.

Mathematically, we consider the Cauchy problem inQT = Rn× (0, T) for a

semilinear parabolic equation

1uε − uεt = βε(uε) in QT ,

uε(·,0) = uε0 onRn,(Pε)

1

2 INTRODUCTION

�(t) = unburnt zone

u > 0

1u − ut = 0

u ≡ 0|∇u| = a

flame front

FIG. 1. The problem (P): a time section profile

whereuε0 approximate the initial datau0 as ε → 0 andβε ≥ 0 are Lipschitzfunctions such that

supportβε = [0, ε] ,∫

[0,ε]βε(s)ds = M.(4)

The functionsβε are normally constructed from a single functionβ by

βε(s) =1

εβ

(s

ε

).

The quantityε > 0 has the meaning of1/E with E the high activation energy inthe corresponding combustion model.

As proved by Caffarelli and Vazquez [CV], the family{uε} is locally uniformlybounded inQT in C1,1/2

x,t norm and converges for a subsequenceε j → 0 to afunction u, which solves (P) in a certain weak sense with the constanta in (2)given by

a =√

2M,

whereM is the constant in (4).To illustrate the passage to the limitε → 0, let us consider the one-dimensional

stationary case. This means that we have an ODE

φεxx = βε(φε) onR.

Assumingφε, φεx ≥ 0 andφε, φεx → 0 asx → −∞, we multiply both sides ofthe equations byφεx and integrate from−∞ to x. We will obtain

φεx =√

2Bε(φε),

INTRODUCTION 3

whereBε(s) =∫ s

0 βε(τ )dτ , so that

Bε(s)

= 0 s ≤ 0∈ [0,M ] , 0 ≤ s ≤ ε

= M, ε ≤ s.

Henceφε will become a linear function with the slope√

2M as it reaches thelevel ε. If we normalize nowφε by φε(0) = ε, the functionsφε(x) will convergeuniformly toφ(x) =

√2Mx+, which is a solution of (P) with a =

√2M .

LOG-CONCAVITY IN SPACE

Now we address the following question: how the geometry of�0 determinesthe geometry of�(t)? In particular we are interested in whether the convexity of�0 implies the convexity of�(t). The first thing one can notice is that the answerdepends very much on the initial datau0 as well. The main result of Paper I(Theorem 1) in this thesis says that, indeed,�(t) will stay convex provided�0is convex andu0 is log-concave in�0 and as long as the free boundary condition|∇u| = a is satisfied in the classical sense, i.e.,

lim |∇u(y, t)| = a asy → x ∈ ∂�(t), y ∈ �(t).

However, the technical assumption of Lipschitz regularity of∂� in time, corre-sponding to the finiteness of the speed of the flame propagation, is also needed.

The appearance oflog-concavity ofu0 is related to the following fact, which isemployed heavily in the proof of the above result:

Suppose we have a domain� ⊂ QT = Rn× (0, T) with convex time sections

�(t) = {x ∈ Rn: (x, t) ∈ �} andu is a solution to the Dirichlet problem

1u − ut = 0 in �; u = 0 on ∂� ∩ QT ; u(·,0) = u0 in �0 = �(0),

whereu0 > 0 in �0 and continuously vanishes on∂�0. Thenlog-concavity ofu0implies thelog-concavity ofu(·, t) for everyt ∈ [0, T ].

This statement can be proved using Korevaar’s Concavity maximum principle[Ko], see Theorem 2, Paper I. We want to sketch here an alternative proof, based onthe ideas of Brascamp and Lieb [BL]. Without loss of generality we may assumethat� has regular enough lateral boundary.

For s > 0 denote byLs = {u0 > s} the level sets of the initial functionu0.Thenlog-concavity ofu0 means that for0< λ < 1

Ls ⊃ (1 − λ)Ls0 + λLs1 whenevers = s1−λ0 sλ1 .(5)

Let nowωx,t (A) denote the caloric measure of a convex setA ⊂ �0 with respectto� at(x, t) ∈ �. Then for pointsx0, x1 ∈ �(t) andx = (1−λ)x0+λx1 ∈ �(t)

4 INTRODUCTION

set

f (s) = ωx0,t (Ls), g(s) = ωx1,t (Ls), andh(s) = ωx,t (Ls).

The key step now is to prove the inequality

h(s) ≥ f (s0)1−λg(s1)

λ, whenevers = s1−λ0 sλ1 .(6)

This follows from the Brunn-Minkowski type inequality for convex sets in theGaussian spaceC([0, t ] ,Rn) with the Wiener measure, if we use the connectionbetween Brownian motion and the caloric measureωx,t (A), which is the proba-bility that the particle, starting its motion atx, will stay in�(t − τ) at the timeτand will hit a point inA at timet .

Having (6), we conclude immediately from [Theorem 3.2, BL] that

u(x, t) ≥ u(x0, t)1−λu(x1, t)

λ,

since

u(x0, t) = || f ||L1(R+), u(x1, t) = ||g||L1(R+)

andu(x, t) = ||h||L1(R+).

Henceu(·, t) is log-concave.

CONVEX CLASSICAL SOLUTIONS

We know already that the classical solutions of (P) will stay convex in spacefor convex initial domain�0 andlog-concaveu0, but we don’t know yet whethersuch solutions exist. The limits of solutions of (Pε), also called limit solutions,give us only weak solutions of problem (P) and a priori even the uniqueness ofsuch limits is not clear. However, ifu0 is starshaped, or more precisely if all thelevel setsLs = {u0 > s} are starshaped with respect to the same point (say theorigin), then there is only one limit solution. In fact, one can show that this limitcoincides with the minimal supersolution of (P). This is Theorem 2.7 of Paper II.

The existence of classical solutions of (P), or problems similar to (P), withcertain geometric assumptions on the initial data, such as radial symmetry wasproved by several authors, using different methods, see e.g. references in [Va]. Inthe case of problem (P) with log-concave initialu0 and convex�0, perhaps thebest suited is the one due to Beurling [Be]. This was successfully applied by Henrotand Shahgholian [HS1] in the elliptic free boundary problem, closely related to(P). The idea, basicly, is to consider the intersection of all supersolutions withconvex time sections and show that it is a classical solution of the problem. Thereare complications in our case connected with the regularity in time, but puttingthe extra condition−M ≤ 1u0 ≤ 0 onu0, one can prove that a classical solutionexists for a short time intervalT = T(u0), and that time levels�(t) are convexand shrink in time. This is Theorem 8.1 of Paper II.

INTRODUCTION 5

Here we want to explain how the method of Henrot and Shahgholian works andat the same time to formulate a problem, which appears also in Paper III.

Given is a convex compact setK in Rn with nonempty interior. The problem isto find a domain� ⊃ K in Rn, and a continuous functionu > 0 in �, vanishingon ∂� such that

1pu = 0 in � \ K(7)

|∇u| = a on ∂�(8)

u = 1 on K ,(9)

wherea = a(x) > 0 is a given function offK . Here for simplicity we assumea ≡ consthere. We will refer to the system (7)–(9) as to problem (FBp). Theoperator1p for p ∈ (1,∞) is the p–Laplacian


∇u).

The reader not familiar with1p may think of the usual Laplacian, correspondingto p = 2, up to end of this section. In the next section, however, we are going tolet p → ∞ in (FBp).

We call a pair(u, �) a supersolution of (FBp) if u ∈ C(�), u > 0 in �, u = 0on ∂�, and

1pu ≤ 0 in � \ K , lim supx→∂�

|∇u| ≤ a, andu ≥ 1 on K ,

cf. (7)–(9).If we consider now the familyE∗ of all supersolutions(u, �) of (FBp) with

convex�, then one can show the existence of the minimal element(u∗, �∗) ofE∗, by taking�∗ to be the intersection of all�’s. Also u∗ is the solution of theDirichlet problem in�∗

\ K , with boundary values1 on ∂K and0 on ∂�∗. Thetrick of [HS1] is now, very roughly, that if(u∗, �∗) is not a classical solution of(FBp), and say|∇u∗

| < a in a neighborhood of a pointx ∈ ∂�∗, then by thetranslation of the supporting hyperplane to�∗ at x we can cut a small cap from�∗, thus constructing a smaller supersolution. That is why(u∗, �∗) should be aclassical solution.

Let us remark here that an essential ingredient in the method above is the con-vexity of level sets of capacitary potentialsu in convex rings� \ K , proved byLewis [Le].

A FREE BOUNDARY PROBLEM FOR ∞-LAPLACIAN

Problem (FBp), formulated in the previous section, describes the nonlinear(power-law) potential flow past the profileK with fixed pressure gradient|∇u| = a

6 INTRODUCTION

K

u ≡ 1

1pu = 0

|∇u| = a

u ≡ 0

FIG. 2. The problem (FBp)

on the free stream line. Besides the fluid mechanical application, the problemis of interest in electrochemical machining for shaping hard metals, see Laceyand Shillor [LS]. Also, as shown by Acker [Ac], (FBp) appears in the shapeoptimization problem for the minimization of the heat flow.

In the sequel we will assume thata(x) ≥ a0 > 0 and1/a(x) is concave offK and thatK is a convex compact, with regular enough boundary (specifically, itsatisfies the uniform interior ball condition.) Then (FBp) has a unique classicalsolution for everyp ∈ (1,∞), see [HS2]. We denote this solution by(up, �p) tostress the dependence onp.

An interesting thing that one can notice is that the domains�p, which are convex,increase withp and all are contained in the set{ dist(x, K ) < 1/a0}. The similarthing happens withup and the bound from above is the function1−a0 dist(x, K ).Hence, there is a limit

(u∞, �∞) = limp→∞

(up, �p).

The main result of Paper III, Theorem 4.4, shows that, as one can expect,(u∞, �∞)

solves, in a certain weak sense, a free boundary problem (FB∞), formally obtainedfrom (FBp) by pluggingp = ∞ in (7)–(9). The operator1∞ is the∞-Laplacian

1∞u = ∇2u∇u · ∇u,

which is an elliptic operator, degenerate on all directions orthogonal to∇u.By now it is very well known that the limits asp → ∞ of p-harmonic functions

(i.e. weak solutions of1pu = 0) are viscosity solutions of1∞u = 0, see e.g.[BDM]. The latters are also known as∞-harmonic functions. Generally,∞-harmonic functions are notC2 regular and, surprisingly, nobody knows if theyareC1. However, other important things such as the maximum principle, due to

INTRODUCTION 7

Jensen [Je], and the Harnack inequality, due to Lindqvist and Manfredi [LM], areproved to be true.

Before we go any further, let us focus on the casea(x) ≡ a0 in problem (FB∞).In this case the problem has an explicit classical solution

u(x) = 1 − a0 dist(x, K ), and� = {u > 0} = {x: dist(x, K ) < 1/a0}.

Therefore we expect the limit(u∞, �∞) to coincide with this pair. This is shownin Theorem 3.3 of Paper III by constructing explicit subsolutions of (FBp) forlarge p.

For nonconstant functiona(x), the analysis of the limit is much more compli-cated and the best we can prove is that condition (8) satisfied in the following weaksense

u∞(y)

dist(y, ∂�∞)→ a(x) asy → x ∈ ∂�∞, y ∈ �∞.(8′)

One of the steps in the proof of (8′) is the following fact. Suppose we have a2-dimensional cone0(θ) = {reiα: |α| < θ} with 0 < θ < π/2. Then thereis an∞-harmonic barrierv > 0 in 0(θ), vanishing on the sides of0(θ) andhaving the formv(reiα) = r 1+εF(α) with ε = ε(θ) > 0. This has the followingconsequence:

If u(x) is an ∞-harmonic function in a convex domain�, vanishing on∂�,then near singular boundary pointsx0 ∈ ∂� the functionu(x) has the behavioru(x) = o(|x − x0|) asx → x0, x ∈ �.

This is true for any dimensionn ≥ 2. Indeed, a boundary pointx0 is singulariff there are two supporting hyperplanes to� at x0, which form an angle2θ < π .Then using2-dimensional barrierv in 0(θ) as above in combination with Jensen’smaximum principle we prove the statement.

Let us make the following observation concerning the regularity of barriersv

in 2-dimensional cones. All barriers, constructed in Paper III (Section 7) are onlyC1,1/3 regular, although they are real analytic off the symmetry axisα = 0. Infact, if the∞-harmonic barrier is symmetric with respect toα = 0, it cannot beC2

regular, otherwise it will be linear on the symmetry axis and will vanish identically.As a conclusion, let us consider the limitp → 1+ in the problem (FBp) with

the same assumptions onK anda(x). As expected,�p shrinks toK asp → 1+,but only if the dimensionn ≥ 2. In the dimensionn = 1 all the problems (FBp)are identical and hence so are the solutions(up, �p) for every p ∈ [1,∞].

REFERENCES

[Ac] A. Acker, Heat flow inequalities with applications to heat flow optimization problems, SIAMJ. Math. Anal.8 (1977), no. 4, 604–618.

8 INTRODUCTION

[Be] A. Beurling,On free-boundary problems for the Laplace equation, Sem. analytic functions 1(1958), 248–263.

[BDM] T. Bhattacharya, E. DiBenedetto and J. Manfredi,Limits asp → ∞ of1pup = f and relatedextremal problems, Rend. Sem. Mat. Univ. Politec. Torino1989, Special Issue, 15–68.

[BL] H. J. Brascamp and E. H. Lieb,On extensions of the Brunn-Minkowski and Prekopa-Leindlertheorems, including inequalities for log concave functions, and with an application to thediffusion equation, J. Functional Analysis22 (1976), no. 4, 366–389.

[CV] L. A. Caffarelli and J. L. Vazquez,A free-boundary problem for the heat equation arising inflame propagation, Trans. Amer. Math. Soc.347(1995), no. 2, 411–441.

[HS1] A. Henrot and H. Shahgholian,Existence of classical solutions to a free boundary problem forthe p-Laplace operator: (I) the exterior convex case, to appear in J. Reine Angew. Math.

[HS2] A. Henrot and H. Shahgholian,The one-phase free boundary problem for the p-Laplacian withnon-constant Bernoulli boundary condition, in preparation.

[Je] R. Jensen,Uniqueness of Lipschitz extensions: minimizing the sup norm of the gradient, Arch.Rational Mech. Anal.123(1993), no. 1, 51–74.

[Ko] N. Korevaar,Convex solutions to nonlinear elliptic and parabolic boundary value problems,Indiana Univ. Math. J.32 (1983), no.4, 603–614.

[Le] J. L. Lewis,Capacitary functions in convex rings, Arch. Rational Mech. Anal.66(1977), no. 3,201–224.

[LM] P. Lindqvist and J. J. Manfredi,The Harnack inequality for∞-harmonic functions, Electron.J. Differential Equations1995(1995), no. 04, approx. 5 pp. (electronic)

[LS] A. A. Lacey and M. Shillor,Electrochemical and electro-discharge machining with a thresholdcurrent, IMA J. Appl. Math.39 (1987), no. 2, 121–142.

[Va] J. L. Vazquez,The free boundary problem for the heat equation with fixed gradient condition,Free boundary problems, theory and applications (Zakopane, 1995), 277–302, Pitman Res.Notes Math. Ser.363, Longman, Harlow, 1996.

[ZF] Ya. B. Zel’dovich and D. A. Frank-Kamenetskiı, The theory of thermal propagation of flames,Zh. Fiz. Khim.,12(1938), 100–105 (in Russian; English translation in “Collected Works of Ya.B. Zeldovich,” vol. 1, Princeton Univ. Press 1992.)

Paper I

Convexity and uniqueness in a free boundary problem arisingin combustion theory

Arshak Petrosyan*

Department of Mathematics, Royal Institute of Technology, 100 44, Stockholm, SwedenE-mail: [email protected]

We consider solutions to a free boundary problem for the heat equation, describing thepropagation of flames. Suppose there is a bounded domain� ⊂ QT = Rn

× (0, T) forsomeT > 0 and a functionu > 0 in � such that

ut = 1u in �u = 0, |∇u| = 1 on0: = ∂� ∩ QTu(·,0) = u0 on�0,

where�0 is a given domain inRn andu0 is a positive and continuous function in�0,vanishing on∂�0. If �0 is convex andu0 is concave in�0, then we show that(u, �) isunique and the time sections�t are convex for everyt ∈ (0, T), provided the free boundary0 is locally the graph of a Lipschitz function and the fixed gradient condition is understoodin the classical sense.

Key Words: free boundary problems, heat equation, convexity of level sets, uniqueness,caloric functions in Lipschitz domains.

1. INTRODUCTION AND MAIN RESULT

In this paper we consider solutions to a free boundary problem for the heatequation. Suppose there is a domain� ⊂ QT : = Rn

× (0, T) for someT > 0and a positive smooth functionu in � such that

ut = 1u in �(1)

u = 0, |∇u| = 1 on0(2)

u(·,0) = u0 on�0,(3)

where0: = ∂� ∩ QT is the (free) lateral boundary of�, �0 ⊂ Rn is the initialdomain andu0 is a prescribed positive continuous function in�0, that vanishes

* The author was supported by the Swedish Institute

1

2 ARSHAK PETROSYAN

continuously on00: = ∂�0. Then we say the pair(u, �) or, when there is noambiguity,� to be asolution to problem(P). This problem, in mathematicalframework, was introduced by L.A. Caffarelli and J.L. Vazquez [CV]. It describespropagation of so-called premixed equi-diffusional flames in the limit of highactivation energy. In this problem the time sections

�t = {x ∈ Rn: (x, t) ∈ �}(4)

represent theunburnt (fresh) zonein time t , 0t : = ∂�t corresponds to theflamefront, andu = c(Tc − T) is thenormalized temperature. For further details incombustion theory we refer to paper [V] of J.L. Vazquez.

The existence of weak solutions to problem (P) as well as their regularity undersuitable conditions on the data were established in [CV]. However, we should notexpect any uniqueness result unless we impose some special geometrical restric-tions. In this paper we study the case when the initial domain�0 is bounded andconvex, and the initial functionu0 is concave. Throughout the paper we makethe following assumptions concerning solutions(u, �) to problem (P). First, theboundary of� consists of three parts:

∂� = �0 ∪ 0 ∪�T ,(5)

where�T is a nonvoid open set in the planet = T . The presence of nonempty�T

excludes theextinction phenomenonin time t ∈ [0, T ]. This assumption is ratherof technical character, that can be avoided with the following simple procedure.Consider theextinction time

T� = sup{t :�t 6= ∅}.(6)

Then every domain�(τ ) = � ∩ {0 < t < τ }, τ ∈ (0, T�), has nonempty “upperbound”�τ . Therefore we can consider first�(τ ) instead of�and then letτ → T�.

Next, we assume that for every(x0, t0) ∈ 0 there exists a neighborhoodV inRn

× R such that (after a suitable rotation ofx-axes)

� ∩ V = {(x, t) = (x′, xn, t): xn > f (x′, t)} ∩ V ∩ QT(7)

where f is aLipschitzfunction, defined inV ′= {(x′, t): ∃xn with (x′, xn, t) ∈ V}.

Further, foru we assume that it is continuous up to the boundary∂� and can beextended smoothly through�T . The gradient condition in (2) is understood in theclassical sense

lim�t3y→x

|∇u(y, t)| = 1(8)

for everyx ∈ ∂�t , 0< t ≤ T .The main result of this paper is as follows.

CONVEXITY AND UNIQUENESS IN A FREE BOUNDARY PROBLEM 3

Theorem 1. In problem(P) let �0 be a bounded convex domain andu0 bea concave function in�0. Suppose that(u, �) is a solution to this problem inthe sense described above. Then(u, �) is a unique solution. Moreover, the timesections�t of� are convex for everyt ∈ (0, T).

The plan of the paper is as follows. In Section 2 we prove a theorem on theconvexity of level sets of solutions to a related Dirichlet problem. In Section 3 werecall some properties of caloric functions in Lipschitz domains. And finally inSection 4 we prove Theorem 1.

2. CONVEXITY OF LEVEL SETS

In this section we establish some auxiliary results, which are, however, of inde-pendent interest.

Let u0 and�0 be as in problem (P) and a domain� ⊂ QT meets conditions (5)and (7). Then by the Petrowski criterion [P]� is a regular domain for the Dirichletproblem for the heat equation (in the Perron sense), and its parabolic boundary isgiven by

∂p� = �0 ∪ 0.(9)

We fix one such domain� and denote byu the solution to the Dirichlet problem

ut = 1u, u = u0 on�0, u = 0 on0.(10)

Theorem 2. Let the time sections�t of the domain�be convex fort ∈ [0, T ].Let alsou0 be a concave function on�0, positive in�0 and vanishing on∂�0.Then the level sets

Ls(u(·, t)) = {x ∈ �t : u(x, t) > s}(11)

are convex for every fixeds > 0 and t ∈ (0, T), whereu is the solution to theDirichlet problem (10).

The proof is based on the Concavity maximum principle originally due to N. Ko-revaar [K1] and [K2]. For a functionv on� set

C(x, y, t) =v(x, t)+ v(y, t)

2− v

(x + y

2, t

).(12)

The functionC is defined on an open subsetD of the fiber product

� = {(x, y, t): (x, t), (y, t) ∈ �}.

4 ARSHAK PETROSYAN

Note thatD = � if the time sections of� are convex. Note also that ifv isextended to the “upper bound”�T of �, thenC is extended to the “upper bound”DT of D.

Lemma 3 (Concavity maximum principle). Letv ∈ C2,1x,t (�)∩ C(�∪�T )

satisfy to a parabolic equation

vt = ai j (t,∇v)vi j + b(t, x, v,∇v) in �(13)

with smooth coefficients, and such thatb is nonincreasing inv and jointly concavein (x, v). Then the functionC cannot admit its positive maximum at any point ofD ∪ DT .

Proof. See [K2], the proof of Theorem 1.6. Though the result is proved there for

cylindrical domains, the proof is valid also in our case.

Remark. There are several formulations of this principle in the elliptic case.The strongest version states that it is sufficient to require harmonic concavity ofb in (x, v) instead of concavity; see B. Kawohl [Ka], and A. Greco and G. Porru[GP]. In the parabolic case, in order to use such an extension, it seems necessaryto assume also the nonnegativeness ofvt ; see A. Kennington [Ke].

Proof of Theorem 2.Assume first, that the functionsf in the local representa-tions (7) of� are smooth in(x′, t) and strictly convex inx′ and thatu0 is smooth.These assumptions imply the smoothness up to∂� of the solutionu to (10). Also,the positivity ofu0 implies the positivity ofu. Define nowv = log(u). We claimthen thatv(·, t) are concave functions in�t for everyt ∈ (0, T ]. Clearly, this willimply the statement of the theorem. For this purpose, we consider the concavityfunctionC, defined above, and show thatC ≤ 0 on D ∪ DT . Suppose the contrary.Then take a maximizing sequence(xk, yk, tk) ∈ D ∪ DT such that

lim C(xk, yk, tk) = supD∪DT

C > 0.(14)

Without loss of generality we may assume that there exists limit

(x0, y0, t0) = lim (xk, yk, tk).

Direct calculation shows, thatv satisfies

vt = 1v + |∇v|2,(15)

in�and hence the Concavity maximum principle is applicable. Therefore(x0, y0, t0) 6∈

D∪ DT . We want now to exclude also the other possibilities. First, the caset0 = 0


andx0, y0 ∈ �0 is impossible, sincev(·,0) = log(u0) is concave in�0. Next,x0 ∈ 0t0 but y0 6= x0 is also excluded by the strict convexity of�t ’s, since thenC(xk, yk, tk) → −∞. So, it remains to consider the last casex0 = y0 ∈ 0t0.We observe now that by the boundary point lemma, the outward spatial normalderivativesuν < 0 on0 ∪ 0T . Besides,uν < 0 also on00 sinceu0 is concaveand positive in�0 and vanishes on00. Therefore we can carry out the samereasonings as in [CS, Proof of Lemma 3.1] (see also [GP, Lemma 3.2]) to obtainthatlim inf C(xk, yk, tk) < 0, which contradicts (14). ThereforeC ≤ 0 in D ∪ DT

andv(·, t) is concave in�t for everyt ∈ (0, T ]. This proves the theorem in theconsidering case.

To prove the theorem in the general case, we use approximation of� by domainswith smooth lateral boundary and with strictly convex time sections, and a relevantsmooth concave approximations ofu0.

3. ON CALORIC FUNCTIONS

In this section we recall some properties of caloric functions in Lipschitz do-mains. They will be used in the next section, where we prove Theorem 1. Themain reference here is the paper [ACS] by I. Athanasopoulos, L. Caffarelli andS. Salsa.

As in the previous section we consider a domain�, satisfying conditions (5)and (7). Let alsou be the solution to (10). Consider a neighborhoodV of a point(x0, t0) ∈ 0, where (7) holds. The functionu vanishes on0 ∩ V , is positive andsatisfies the heat equation in� ∩ V . In other words,u is caloric.

We start with the following lemma from [ACS], which states that a caloricfunctionu is “almost harmonic” in time sections near the lateral boundary0.

Lemma 4 ([ACS, Lemma 5]). There existε > 0 and a neighborhoodQ ofthe point(x0, t0) ∈ 0 such that the functions

w+ = u + u1+ε, w− = u − u1+ε(16)

are respectively sub- and superharmonic inQ ∩� ∩ {t = t0}.

We will need also the following lemma on asymptotic development ofu nearthe boundary point(x0, t0).

Lemma 5 ([ACS, Lemma 6]). Suppose there exists ann-dimensional ballB ⊂ �c

∩ {t = t0} such thatB ∩ 0 = {(x0, t0)}. Then nearx0 in �t0

u(x, t0) = α(x − x0, ν)+

+ o(|x − x0|)(17)

for someα ∈ [0,∞) and whereν denotes the outward radial direction ofB at(x0, t0).

6 ARSHAK PETROSYAN

In the next lemma we show thatα in (17) is in fact the nontangential limit of|∇u(y, t0)| asy → x0.

Lemma 6. Under the conditions of Lemma 5, let alsoK ⊂ �t0 be ann-dimensional truncated cone with the vertex at(x0, t0) such that|x − x0| ≤

c1dist(x, 0t0) for everyx ∈ K and some constantc1. Then

limK3y→x0

∇u(y, t0) = αν,(18)

whereα andν are as in the asymptotic development (17).

Proof. By [ACS, Corollary 4], there exists a neighborhoodV of the point(x0, t0) such that

|ut (x, t)| ≤ c2u(x, t)

dx,t, dx,t = dist(x, 0t ),(19)

for all (x, t) ∈ V ∩�. Take an arbitrary sequenceyk → x0, yk ∈ K , and considerthe functions

vk(z) = u(yk + rkz, t0)/rk, rk = |yk − x0|,(20)

defined on the ballB = B(0, ρ), ρ = 1/(2c1). Using (17) and (19), we obtainthat for largek

|vk(z)| < (α + 1)(1 + ρ)(21)

and

|1vk(z)| = rk|1u(yk + rkz, t0)| = rk|ut (yk + rkz, t0)| ≤ 2c1c2(22)

uniformly in B. ThenC1,β norms ofvk are locally uniformly bounded inB fora β ∈ (0,1); see e.g. [LU]. Therefore a subsequence ofvk converges locally inC1 norm to a functionv0 in B. We may also assume that over this subsequencethere existse0 = lim ek, whereek = (yk − x0)/|yk − x0|. Then, using (17),we can compute thatv0(z) = α(z, ν) + α(e0, ν) in B, hence∇v0(0) = αν.Therefore, over a subsequence,lim ∇u(yk, t0) = lim ∇vk(0) = ∇v0(0) = αν.

Since the sequenceyk → x0, yk ∈ K was arbitrary, this proves the lemma.

4. PROOF OF THE MAIN THEOREM

In this section� will be a solution to problem (P), under conditions of The-orem 1. Denote by�∗ the spatial convex hullof �, in the sense that the time


sections�∗t are the convex hulls of�t for everyt ∈ (0, T). Since� is assumed to

satisfy (5) and (7),�∗ will also satisfy similar conditions. In particular, we mayapply the results of two previous sections to�∗. The lateral boundary of�∗ willbe denoted by0∗ and the solution to the Dirichlet problem, corresponding to (10),by u∗.

In the proof of Theorem 1 we use ideas of A. Henrot and H. Shahgholian [HS].The key step is to prove the following lemma.

Lemma 7. For everyx0 ∈ 0∗t0, 0< t0 ≤ T ,

lim inf�∗

t03y→x0

|∇u∗(y, t0)| ≥ 1.(23)

Proof. From Lemma 4 it follows that there areε ands0 such that the functionw+(y) = u∗(y, t0) + u∗(1+ε)(y, t0) is subharmonic in the ringshaped domain{u∗(·, t0) < s0}. Let now y ∈ �∗

t0 andu∗(y, t0) = s < s0. Then y ∈ `∗s =

∂Ls(u∗(·, t0)). By Theorem 2,L∗s = Ls(u∗(·, t0)) is convex and therefore there

exists a supporting plane inRn to L∗s at the pointy. After a suitable translation

and rotation in spatial variable we may assume thaty = 0, the supporting planeis x1 = 0, andL∗

s ⊂ {x1 < 0}. Let x∗∈ ∂�∗

t0 have the maximal positivex1-coordinate. Since�∗

t0 is the convex hull of�t0, there must bex∗∈ ∂�∗

t0 ∩ ∂�t0.Take nowβ ∈ (0,1) and consider a functionv(x) = w+(x) + βx1. Since�∗

t0 ∩ {x1 > 0} ⊂ {u∗(·, t0) < s0}, v is subharmonic in�∗t0 ∩ {x1 > 0} and

therefore it must admit its maximum value on the boundary of this domain. Notethat the maximum can be admitted either atx∗ or aty = 0. We show that the formercase cannot occur. Indeed, the planex1 = x∗

1 is supporting to the convex set�∗t0

and therefore there exists a ballB ⊂ �∗t0

c⊂ �c

t0, “touching” both boundaries∂�∗t0

and∂�to at x∗ and with the outward radial directionν = −e1 = (−1,0, . . . ,0).Therefore from Lemma 5 we will have the following asymptotic developments foru andu∗ nearx∗ in �t0 and�∗

t0 respectively:

u(x, t0) = α(x∗

1 − x1)+

+ o(|x − x∗|)(24)

u∗(x, t0) = α∗(x∗

1 − x1)+

+ o(|x − x∗|).(25)

Since (8) is satisfied at the pointx∗, we conclude by Lemma 6 thatα = 1. Next,u∗

≥ u in � and henceα∗≥ α = 1. Observe now thatw+ admits the same

representation as (25). Hence for the functionv(x) introduced above

v(x) = (α∗− β)(x∗

1 − x1)+ βx∗

1 + o(|x − x∗|).(26)

Let nowν′ be a spatial unit vector with(ν′,e1) < 0 such thatx∗+ hν′

∈ �t0 forsmallh > 0. The existence of such aν′ follows from the local representation of

8 ARSHAK PETROSYAN

∂�t0 as the graph of a Lipschitz function. Thenv(x∗+ hν′) > v(x∗) by (26) and

consequentlyv has no maximum atx∗. Thereforev admits its maximum at theorigin y = 0. Hence

|∇w+(0)| = limh→0+

w+(0)− w+(he1)

h≥ lim

h→0+

βh − 0

h= β.(27)

Lettingβ → 1 we obtain that|∇w+(y)| ≥ 1, providedu∗(y, t0) < s0. Now ob-

serve that∇w+ = (1 + (1 + ε)u∗ε)∇u∗. This proves the lemma.

Proof of Theorem 1.Prove first that the domain� coincides with its spatialconvex hull�∗, studied above. For this purpose we apply the Lavrentiev principle.As a reference point we takexmax ∈ �0, a maximum point for the initial functionu0. Without loss of generality we may assume thatxmax = 0. Sinceu0 is concave,

u0(λx) ≤ u0(x)(28)

for everyλ ≥ 1 andx ∈ �0(λ) = λ−1�0. Forλ ≥ 1 define

u∗λ(x, t) = u∗(λx, λ2t)(29)

in �∗(λ) = {(x, t): (λx, λ2t) ∈ �∗}. Suppose now that�∗

6⊂ �. Then

λ0 = inf{λ:�∗(λ) ⊂ �} > 1,(30)

�∗(λ0) ⊂ �, and there exists a common point(x0, t0) ∈ 0∗(λ0) ∩ 0 with t0 ∈

(0, T). Show that this leads to a contradiction. Indeed, by construction,u∗λ0

satisfies the heat equation in�∗(λ0). Comparing the values ofu∗λ0

andu on theparabolic boundary∂p�

∗(λ0) (see (28)), we obtain thatu∗λ0

≤ u in �∗(λ0). Letnow ν be the normal vector of a supporting plane inRn to the convex domain�∗(λ0)t0 at the pointx0, pointing into�∗(λ0)t0. From Lemmas 5, 6 and 7 and thedefinition ofu∗

λ we conclude that∇u∗λ0(x0 + hν, t0) → λ0α

∗ν with α∗≥ 1, as

h → 0+. From elementary calculus there existsθ ∈ (0,1) such that

(∂/∂ν)u(x0 + θhν, t0)

(∂/∂ν)u∗λ0(x0 + θhν, t0)

=u(x0 + hν, t0)

u∗λ0(x0 + hν, t0)

≥ 1(31)

and hence

lim sup�t03y→x0

∂

∂νu(y, t0) ≥ lim

h→0+

∂

∂νu∗λ0(x0 + hν) = λ0α

∗ > 1,(32)

which violates condition (8) at the point(x0, t0). Therefore�∗= �, i.e. the time

sections�t are convex, for everyt ∈ (0, T).


It remains to prove the uniqueness of�. For this we make the following obser-vation. Let�′ be another solution. Then if everywhere in the proof of inclusion�∗

⊂ � above we replace�∗ by (�′)∗, but leave� unchanged, we will obtainthat(�′)∗ ⊂ �. Since� and�′ are interchangeable, also we will have�∗

⊂ �′.Therefore�′

= � and the proof of Theorem 1 is completed.

ACKNOWLEDGMENTThe author thanks Henrik Shahgholian for a number of useful discussions concerning this paper.

REFERENCES

[ACS] I. Athanasopoulos, L. Caffarelli and S. Salsa,Caloric functions in Lipschitz domains and theregularity of solutions to phase transition problems, Annals of Math.143(1996), 413–434.

[CV] L.A. Caffarelli and J.L. Vazquez,A free-boundary problem for the heat equation arising inflame propagation, Trans. Amer. Math. Soc.347(1995), no.2, 411–441.

[CS] L.A. Caffarelli and J. Spruck,Convexity properties of solutions to some classical variationalproblems, Comm. Part. Diff. Eq.7(11)(1982), 1337–1379.

[GP] A. Greco and G. Porru,Convexity of solutions to some elliptic partial differential equations,SIAM J. Math. Anal.24 (1993), no. 4, 833-839.

[HS] A. Henrot and H. Shahgholian,Convexity of free boundaries with Bernoulli type boundarycondition, Nonlin. Anal. Theory Meth. Appl.28 (1997), no. 5, 815–823.

[K1] N. Korevaar,Capillary surface convexity above convex domains, Indiana Univ. Math. J.32(1983), no. 1, 73–82.

[K2] N. Korevaar,Convex solutions to nonlinear elliptic and parabolic boundary value problems,Indiana Univ. Math. J.32 (1983), no.4, 603–614.

[Ka] B. Kawohl,Rearrangements and Convexity of Level Sets in PDE, Lecture Notes in Mathematics,1150, Springer-Verlag, Berlin Heidelberg New York Tokyo, 1985.

[Ke] A.U. Kennington,Convexity of level curves for an initial value problem, J. Math. Anal. Appl.133(1988), 324-330.

[LU] O.A. Ladyzhenskaya and N.N. Uraltseva,Linear and Quasilinear Elliptic Equations, AcademicPress, New York, 1968.

[V] J.L. Vazquez,The free boundary problem for the heat equation with fixed gradient condition,Free boundary problems, theory and applications (Zakopane, 1995), 277–302, Pitman Res.Notes Math. Ser.,363, Longman, Harlow, 1996.

[P] I. G. Petrowski,Zur Ersten Randwertaufgaben der Warmeleitungsgleichung, Compositio Math.1 (1935), 383-419.

Paper II

On existence and uniqueness in a free boundary problemfrom combustion

Luis Caffarelli

Department of Mathematics, University of Texas at Austin, Austin, Texas 78712, USAE-mail: [email protected]

and

Arshak Petrosyan

Department of Mathematics, Royal Institute of Technology, 100 44 Stockholm, SwedenE-mail: [email protected]

We study a free boundary problem for the heat equation describing the propagationof laminar flames under certain geometric assumptions on the initial data. The problemarises as the limit of a singular perturbation problem, and generally no uniqueness of limitsolutions can be expected. However if the initial data is starshaped, we show that the limitsolution is unique and coincides with the minimal classical supersolution. Under certainconvexity assumption on the data, we prove that the latter is a classical solution of the freeboundary problem for a short time interval

Key Words:parabolic free boundary problem, singular perturbation, limit solution, clas-sical solution, uniqueness, existence.

1. INTRODUCTION

In this paper we consider a free boundary problem for the heat equation, thatconsists of finding a nonnegative continuous functionu in QT = Rn

× (0, T),T > 0, such that 1u − ut = 0 in � = {u > 0}

|∇u| = 1 on ∂� ∩ QT , andu(·,0) = u0,

(P)

with given nonnegative initial functionu0 ∈ C0(Rn). (Here1 = 1x and∇ = ∇x.)The problemP arises in modeling the propagation of laminar flames as the limit ofthe singular perturbation problem (see [CV] and, for further detail in combustion

1

2 L. CAFFARELLI AND A. PETROSYAN

theory, [Va]) {1uε − uεt = βε(uε)uε(·,0) = u0,ε,

(Pε)

as ε → 0+, whereu0,ε approximateu0 in a proper way,βε ≥ 0, βε(s) =

(1/ε)β(s/ε), with β a Lipschitz function,supportβ = [0,1], and∫ 1

0β(s)ds =

1

2.(1.1)

The family of solutions{uε} is uniformly bounded inC1,1/2x,t -norm on compact

subsets ofQT and every locally uniform limitu = lim j →∞ uε j in QT of itssubsequence withε j → 0 is a solution ofP in a certain weak sense, see [CV].We will refer to these limits aslimit solutions ofP.

In the two-phase case, i.e. when there is no sign restriction onu, problemsPεapproximate a free boundary problem, where the fixed gradient condition|∇u| = 1is replaced by the gradient jump condition|∇u+

|2−|∇u−

|2

= 1. Limit solutionsof this problem were studied detailly from the local point of view in [CLW1],[CLW2]. It was shown that a limit solutionu of P is a viscosity solutionin adomainD if {u = 0}

◦∩ D = ∅. For the one-phase problem, and recently for

the two-phase problem as well, three concepts of solutions, limit, viscosity andclassical, were shown to agree with each other to produce a unique solution undercertain conditions onu0 (see [LVW]), that guarantee the existence of the classicalsolutions. In this paper we provide another uniqueness theorem (Theorem 2.7) forlimit solutions for starshapedu0, see (S) from Section 2. In fact, we prove that theunique limit solution coincides with the minimal classical supersolution ofP (seeDefinition 2.1) in this case.

The existence and analyticity of classical solutions ofP, among other things,were proved in [GHV] in the case when the initial data is radially symmetric, byusing the elliptic-parabolic approach. The classical solutions to free boundaryproblems, similar toP, were constructed in [Me] and [AG].

In this paper we prove a short time existence theorem of classical solutions ofP in the so-called convex case (Theorem 8.1), when�0 = {u0 > 0} is convex andu0 is log-concave and superharmonic in�0, along with other assumptions. Weconsider the minimal element among the classical supersolutions ofP that havethe following geometric property, expected from the classical solution: the timesections�(t) = {u(·, t) > 0} are convex domains shrinking in time; see [Pe],[CV]. We prove that the minimal supersolution with this property has Lipschitz(in space and in time) lateral boundary for the short time interval, which enablesus to apply a technique due to A. Henrot and H. Shahgholian [HS1], [HS2], toshow that it is a classical solution ofP.

A FREE BOUNDARY PROBLEM FROM COMBUSTION 3

2. UNIQUENESS IN THE STARSHAPED CASE

Definition 2.1. A pair (u, �), whereu is a continuous nonnegative functionin QT = Rn

×[0, T ], T > 0, and� = {u > 0}, is called a(classical) supersolutionof P if

(i) 1u − ut = 0 in �;(ii) lim sup�3(y,s)→(x,t) |∇u(y, s)| ≤ 1 for every(x, t) ∈ ∂� ∩ QT ;(iii) u(·,0) ≥ u0.

Respectively, a pair(u, �) is a subsolutionof P if conditions (ii) and (iii) aresatisfied with opposite inequality signs andlim inf instead oflim sup in (ii).

A pair (u, �) is aclassical solutionof P if it is sub- and supersolution ofP atthe same time.

Next, a supersolution(u, �) of P is astrict supersolutionof P if there is aδ > 0such that the stronger inequalities

(ii’) lim sup�3(y,s)→(x,t) |∇u(y, s)| ≤ 1 − δ for every(x, t) ∈ ∂� ∩ QT ;(iii’) u(·,0) ≥ u0 + δ on�0 = {u0 > 0}

hold. Analogously thestrict subsolutionsare defined.

Remark 2.2.According to [Theorem 6.1, CLW1], every limit solutionu =

lim j →∞ uε j of P is its classical supersolution in the sense of Definition 2.1. Gen-erally, for the two-phase problem we have that ifu is a limit solution ofP then(u+

= max{u,0} andu−= min{u,0})

lim sup(y,s)→(x,t)

|∇u−(y, s)| ≤ γ

implies

lim sup(y,s)→(x,t)

|∇u+(y, s)| ≤

√1 + γ 2

for every free boundary point(x, t) ∈ ∂{u > 0} ∩ QT . Whenu ≥ 0, we haveu−

≡ 0, hence one can takeγ = 0.

Remark 2.3.Suppose that the initial functionu0 is starshapedwith respect toa pointx0 in the following sense:

u0(λx + x0) ≥ u0(x + x0) for everyλ ∈ (0,1) andx ∈ Rn,(S)

or, equivalently, all the level setsLs(u0) = {u0 > s} are starshaped with respectto the same pointx0. In the sequel, we will always assumex0 = 0.


Let (u, �) be a supersolution ofP. Let λ andλ′ be two real numbers with0< λ < λ′ < 1. Define

uλ(x, t) = (1/λ′)u(λx, λ2t)(2.1)

in QT/λ2. The rescaling of variables is taken so thatuλ, like u, satisfies the heatequation in its positivity set

�λ = {(x, t): (λx, λ2t) ∈ �}.(2.2)

Moreover, the selectionλ < λ′ < 1 makes the pair(uλ, �λ) not only a supersolu-tion of P, but also astrict supersolution.

Lemma 2.4. Let the initial functionu0 satisfy condition (S). Then every sub-solution ofP is smaller than every supersolution ofP.

In this lemma and further in the paper we say that a pair(u′, �′) is smaller than(u, �) if �′

⊂ � andu′≤ u.

Proof. Let (u, �) be a supersolution and(u′, �′) a subsolution ofP in QT .We need to prove only that�′

⊂ �; the inequalityu′≤ u will follow from this

inclusion by the maximum principle.In the case whenu ∈ C1(�) andu′

∈ C1(�′), the statement can be proved bythe Lavrent’ev rescaling method as follows. Suppose

λ0 = sup{λ ∈ (0,1):�′⊂ �λ} < 1,(2.3)

where�λ as in (2.2). Then�′⊂ �λ0 and there is a common point(x0, t0) ∈

∂�′∩ ∂�λ0 ∩ QT . Letλ0 < λ′

0 < 1 anduλ0 be as in (2.1). Thenu′≤ uλ0 in �′.

At the common point(x0, t0) this inequality implies∂νu′(x0, t0) ≤ ∂νuλ0(x0, t0),whereν is the inward spatial normal vector for both∂�′ and ∂�λ0 at (x0, t0)(recall that we are inC1 case.) This leads to a contradiction, since∂νu′(x0, t0) =

|∇u′(x0, t0)| ≥ 1and∂νuλ0(x0, t0) = |∇uλ0(x0, t0)| = λ0 < 1. Thereforeλ0 = 1and�′

⊂ �.The general case can be reduced to the considered regular case by the following

procedure. Let(u, �) be a subsolution. Choose0 < λ < λ′ < 1 close to1 andregularizeu by setting

u(x, t) = (uλ(x, t + h)− η)+

for smallh, η > 0. Analogously regularize a subsolution(u′, �′). Then we willarrive in the considered regular case and can finish the proof by letting firsth, η →

0+ and thenλ → 1−.

The following proposition opens a way to prove the uniqueness results for limitsolutions ofP.


Proposition 2.5. Let u be a strict supersolution ofP in QT , T > 0, anduε solutions ofPε, whereu0,ε are nonnegative uniform approximations ofu0 andsupportu0,ε → supportu0. Then

lim supε→0+

uε(x, t) ≤ u(x, t)(2.4)

for every(x, t) ∈ QT .

Proof. Consider the ordinary differential equation

φεss(s) = γε(φε(s)),(2.5)

whereγε(s) = (1/ε)γ (s/ε) andγ is obtained fromβ in Pε by

γ (s) =

{cβ(s) s ∈ [a,1]0 s 6∈ [a,1] .

(2.6)

Herea ∈ (0,1) (to be specified later) and the constantc > 1 is chosen so that∫ 1

aγ (s)ds =

1

2,(2.7)

cf. (1.1). Let us consider the solutionφε of (2.5), normalized by

φε(s) = aε for s ≤ 0, and φε(s) > aε for s> 0.(2.8)

The family{φε} is recovered from a single functionφ, a solution of

φss = γ (φ)

with the appropriate normalization, through the relationφε(s) = εφ(s/ε). Usingthis, we can find a constantC > 0 such that

φε(s) = (s − Cε)+ ε for s ≥ Cε;(2.9)

or in other words, thatφε is a linear function with slope 1, fors ≥ Cε. Indeed,multiply both sides of (2.5) byφεs and integrate from−∞ to s. We will obtain

(φεs)2

= 20ε(φε),(2.10)

where0ε(s) =∫ s

a γε(σ )dσ . By (2.7)0ε(s) = 1/2 for s ≥ ε and thereforeφε

will become linear function with slope1 as it reachesε. Now (2.9) follows withC satisfyingφ(C) = 1. Moreover, (2.8)–(2.10) imply that

(s − Cε)+ ε ≤ φε(s) ≤ s + aε for s ≥ 0.(2.11)


Let nowu be a strict supersolution ofP and consider the following regularization

u(x, t) = (u(x, t + h)− η)+,

for h, η > 0 small. Without loss of generality we may assume that there isδ ∈ (0,1) such that

|∇u| ≤ 1 − δ if u < η andu(·,0) > u0 + δ.(2.12)

Consider now the compositions

wε(x, t) = φε(u(x, t))

for sufficiently smallε > 0. Sinceφε(s) is constant fors ≤ 0, we can rewritewε = φε(u(x, t + h) − η). Thereforewε is well-definedC1,1(QT−h) function,and satisfies

1wε − wεt = φεs(u) [1u − ut ] + φεss(u)|∇u|2

in QT−h. The first term in the right-hand side is0 everywhere. If now the valueof a in the definition (2.6) is chosen so that∫ 1

aβ(s)ds =

1

2(1 − δ)2,

then the constantc in the same definition equals(1 − δ)−2 and we obtain

1wε − wεt = φεss(u)|∇u|2

≤ γε(φε(u))(1 − δ)2 ≤ βε(w

ε),

if ε > 0 is so small thatφε(s) ∈ supportγε = [aε, ε] impliess < η, or explicitlyε < η/C. Besides, from (2.11), for small values ofε,

wε(·,0) = φε(u(·,0)) ≥ u(·,0)− Cε ≥ u0,ε

-1 1 2 3

0.5

1

1.5

2

FIG. 1. Profile ofφ(s)


and thereforewε is a C1,1(QT−h) supersolution ofPε. Using the comparisonprinciple, we conclude thatuε(x, t) ≤ wε(x, t) for every(x, t) ∈ QT−h. Passingto the limit asε → 0+, we obtain

lim supε→0+

uε(x, t) ≤ limε→0+

wε(x, t) = u(x, t)

for every(x, t) ∈ QT . Letting h, η → 0, we complete the proof of the proposi-tion.

Remark 2.6. One can formulate and prove a result analogous to Proposition 2.5for strict subsolutions ofP. Details are left to the reader.

From Proposition 2.5 we derive now a uniqueness theorem in the starshapedcase (S).

Theorem 2.7. Let the initial functionu0 satisfy condition (S) for somex0 ∈

Rn. If nonnegativeu0,ε approximateu0 uniformly andsupportu0,ε → supportu0,then the limit solution ofP is unique and coincides with the minimal supersolutionof P.

For the existence of limit solutions and their local properties we refer to [CLW1],[CLW2] and [CV].

Proof. Without loss of generality we may assumex0 = 0. Suppose first thatuis a classical supersolution ofP. As was noted in Remark 2.3,

uλ(x, t) = (1/λ′)u(λx, λ2t) 0< λ < λ′ < 1

is a strict supersolution inQT/λ2 ⊃ QT , so that we can apply Proposition 2.5.Then, lettingλ → 1−, we will arrive at

lim supε→0+

uε(x, t) ≤ u(x, t).(2.13)

Let nowu be a limit solution ofP. By Remark 2.2u is a classical supersolution ofP. Therefore (2.13) holds again. It is not difficult to understand that this completesthe proof of the theorem.

3. THE CONVEX CASE

The next sections of this paper will be devoted to the proof of the existence ofclassical solutions ofP under the followingconvexityassumptions on data:

u0 is superharmonic andlog-concavein convex bounded�0 = {u0 > 0}.

(C1)


It is easily seen that (C1) implies (S) withx0 the maximum point ofu0. In thesequel we will always assume thatx0 = 0.

As it follows from Lemma 2.4 and Theorem 2.7, a classical solution ofP, ifexists, coincides with the minimal supersolution, and therefore with the only limitsolution of P. Next, from condition (C1) we may expect that the time sections�(t) = {x: (x, t) ∈ �} of a classical solution(u, �)enjoys the following property:

�(t) is convex and shrinks in time fort ∈ [0, T ] ,(3.1)

cf. [Theorem 1.1, Pe] and [CV].

Definition 3.1. A supersolution(u, �) of P in QT is said to be in classB if� satisfies (3.1) and moreover∂� ∩ QT is Lipschitz regular in time.

The Lipschitz regularity in time is understood in the following sense: for every(x0, t0) ∈ ∂� ∩ QT there exists a neighborhoodV such that

V ∩� = {xn > f (x1, . . . , xn−1, t)} ∩ V,(3.2)

for a suitable spatial coordinate system and wheref is a globally defined function,uniformly Lipschitz in time. We point out that in spatial coordinatesf can bechosen to be convex, if time sections�(t) are convex.

If the classB just defined has a minimal element, then it is a good candidate fora classical solution ofP. This idea goes back to Beurling’s celebrated paper [Be].We set

�∗=

⋂(u,�)∈B

�(3.3)

and let alsou∗ be a solution to the Dirichlet problem

1u∗− (u∗)t = 0 in �∗

; u∗= 0 on ∂�∗

∩ QT ; u∗(·,0) = u0.(3.4)

We can show that under some additional conditions onu0 and for smallT ≤ T(u0)

the pair(u∗, �∗) is the minimal element ofB and indeed a classical solution ofP.The conditions onu0 are as follows. First

u0 ∈ C0,1(�0) and lim�03x→∂�0

|∇u0(x)| = 1;(C2)

and, next, there exists a constantM > 0 such that

1u0(x)+ M(u0(x)− ∇u0(x) · x) ≥ 0 for everyx ∈ �0.(C3)

Let us point out that if (C2) holds, then (C3) can be replaced by

1u0(x) ≥ −M ′ for everyx ∈ �0.(C3’)


for certain M ′ > 0. Indeed, (C3’) readily follows from (C3) in view of theboundedness ofw(x) = u0(x)−∇u0(x) · x in�0. Suppose now that (C3’) holds.We havew(x) ≥ u0(x) > 0 in �0 and

lim inf�03x→∂�0

w(x) ≥ lim inf�03x→∂�0

−∇u0(x) · x ≥ ρ,

whereρ > 0 is the diameter of the circle, centered atx0 = 0 and inscribed in�0. Hence, there is anε > 0 such thatw(x) ≥ ε in �0 and (C3) follows withM = M ′/ε.

Remark 3.2. Before we proceed, we check that the classB is not empty andthat so is�∗. For the latter it suffices to show the existence of a subsolution ofP,in view of the comparison principle (Lemma 2.4.)

Indeed, letu0 satisfies (C1)–(C3). Consider the functions

u(x, t) = u0(x)(3.5)

and

v(x, t) =√

1 − 2Mtu0

(x

√1 − 2Mt

)(3.6)

in QT . We claim that the first is a super- and the second is a subsolution ofP inQT for T ≤ 1/(2M), but in a sense a little bit different from Definition 2.1, whichwill not affect on the comparison principle. The matter is that (3.5) and (3.6) arenot caloric in their positivity sets but super- and subcaloric respectively. Indeed,for (3.5) this follows from superharmonicity ofu0 in �0 and for (3.6) from thedirect computation

1v(x, t)− vt (x, t) = (1 − 2Mt)−1/2(1u0(ξ)+ M(u0(ξ)− ∇u0(ξ) · ξ)) ≥ 0,

for ξ = x/√

1 − 2Mt ∈ �0 andt ∈ [0,1/(2M)]. The gradient condition for bothfunctions follows from (C2).

4. LIPSCHITZ REGULARITY IN TIME

The following lemma plays one of the fundamental roles in our study.

Lemma 4.1. Let u0 satisfy (C1)–(C3) and let�∗ be given by (3.3). Then∂�∗

∩ Q1/(2M) is Lipschitz regular in time.


Proof. Let (u, �) ∈ B. For smallε, h > 0 let us define

w(x, t) =1

1 − εu((1 − ε)x, (1 − ε)2(t + h))

in Q(1−ε)−2T−h, that is we first stretch� a little and then drop it down and thusobtain

�1−ε,h = {(x, t): ((1 − ε)x, (1 − ε)2(t + h) ∈ �}.

Clearly, a pair(w,�1−ε,h) again will lie inB if the condition

w(·,0) ≥ u0(4.1)

is verified. In view of Remark 3.2 and Lemma 2.4

w(x,0) =1

1 − εu((1 − ε)x, (1 − ε)2h) ≥

1

1 − εv((1 − ε)x, (1 − ε)2h)

=

√1 − 2M(1 − ε)2h

1 − εu0

(1 − ε√

1 − 2M(1 − ε)2hx

)= u0(x)

if we choose

h =2ε − ε2

2M(1 − ε)2.(4.2)

Therefore(w,�1−ε,h) ∈ B if h is given by (4.2). Note now that the time levels of�1−ε,h are given by the identity

1

1 − ε�(t) = �1−ε,h

(t

(1 − ε)2− h

).

Running over all(u, �) ∈ B, we may conclude therefore that

1

1 − ε�∗(t) ⊃ �∗

(t

(1 − ε)2− h

).(4.3)

Since�∗(t) shrinks in time, the inclusion (4.3) is not trivial if

t

(1 − ε)2− h < t,

for certain smallε, h > 0, given by (4.2). The latter is equivalent to to the inequalityt < 1/(2M). Besides, for theset , (4.3) implies also the Lipschitz regularity of∂�∗

in time variable.


Remark 4.2. Let u∗ be a solution of (3.4) forT < 1/(2M) and continue theline of reasonings from the previous lemma. For smallε, h > 0, given by (4.2),the following inequality will be satisfied

1

1 − εu∗((1 − ε)x, (1 − ε)2(t + h))− u∗(x, t) ≥ 0.

Let nowε go to0. We will obtain

u∗(x, t)− ∇u∗(x, t) · x + (1/M − 2t)u∗t (x, t) ≥ 0

in �∗∩ Q1/(2M). The reader can see the similarity of this condition with (C3).

In particular, we have thatu∗t is bounded from below in every�∗

∩ QT withT < 1/(2M).

5. SOME TECHNICAL LEMMAS

For supersolutions(u, �) ∈ B of P one can replace the gradient condition (ii)in Definition 2.1 with

lim sup�3(x,t)→(x0,t0)

u(x, t)

d�(x, t)≤ 1

for every(x0, t0) ∈ ∂� ∩ QT , where

d�(x, t) = distance(x, ∂�(t)).

This is taken care of in the next lemma.

Lemma 5.1. Let� be a bounded domain inQT such that�(t) are convex fort ∈ (0, T) and∂�∩ QT is Lipschitz in time. Let alsou be a nonnegative function,continuously vanishing on∂� ∩ QT , and such that1u − ut = 0 in �. Then

lim sup�3(x,t)→(x0,t0)

|∇u(x, t)| = lim sup�3(x,t)→(x0,t0)

u(x, t)

d�(x, t),(5.1)

for every(x0, t0) ∈ ∂� ∩ QT .

Proof. Denote byα the left lim sup in (5.1) and byβ the right one. Theinequalityα ≥ β follows immediately from the finite-increment formula, andtherefore we focus on the inequalityα ≤ β. If β = ∞ there is nothing toprove. Therefore assumeβ is finite. Let a sequence{(xk, tk)} be such thatα =


limk |∇u(xk, tk)|. Let alsozk ∈ ∂�(tk) be chosen such thatdk: = d�(xk, tk) =

|xk − zk|. Set

uk(y, s) =1

dku(zk + dky, tk + d2

k s).

Let ek = (xk − zk)/dk and assume that{ek} converges toe = (0, . . . ,0,1). Inview of the Lipschitz regularity of∂� ∩ QT in t and the convexity of�(t)’s,the positivity sets�k = {uk > 0} will converge (over a subsequence) to a subsetof 5+

= {(x, t): x · e > 0} that has a cylindrical formD × R, whereD is anunbounded convex subset of the halfspace{x · e > 0}, containing a spatial ballB(e,1). Sinceuk satisfy the heat equation in�k and are locally uniformly bounded(since we assumeβ is finite), a subsequence of{uk} will converge inC1-norm oncompact subsets ofD◦

× R to a nonnegative caloric functionv, which enjoys thefollowing properties

|∇v(e,0)| = α(5.2)

0 ≤ v(x, t) ≤ β distance(x, ∂D) for every(x, t) ∈ D × R.(5.3)

Suppose first thatD is not a halfspace. Then by a theorem of Phragmen-Lindeloftype, the condition (5.3) will imply thatv ≡ 0 in D ×R; in particular we will haveα = 0 and henceα ≤ β.

If D is a halfspace, that isD × R = 5+, condition (5.3) is rewritten as

0 ≤ v(x, t) ≤ β(x · e) for every(x, t) ∈ 5+(5.4)

and we observe that the only caloric functions in5+ with property (5.4) arefunctions of the formv(x, t) = γ (x · e) for nonnegativeγ ≤ β. We give a shortproof of this statement. Extendv into a caloric function in the wholeRn

× R bythe odd reflectionv(x′, xn, t) = −v(x′,−xn, t), wherex′

= (x1, . . . , xn−1) andxn = x · e, and consider the derivative∂v/∂xn. It is a caloric function inRn

× Rand from (5.4), by the interior gradient estimate,|∂v/∂xn| ≤ Cβ, whereC is anabsolute constant. From the Liouville theorem∂v/∂xn is identically constant, sayγ , and thereforev(x) = γ xn, since it vanishes on5 = ∂5+. That0 ≤ γ ≤ β

follows from (5.4).Return to ourv. In this case (5.2) impliesγ = α and thereforeα ≤ β. The proof

is complete.

The following lemma is an elliptic counterpart of the lemma above and is provedin a similar way; therefore, the proof is omitted.

Lemma 5.2. Let D be a bounded spatial convex domain andu a nonnegativefunction inD, continuously vanishing on∂D and such that1u = f in D with f


bounded. Then

lim supD3x→x0

|∇u(x)| = lim supD3x→x0

u(x)

d(x),(5.5)

whered(x) = distance(x, ∂D).

6. THE MINIMAL ELEMENT OF B

From Lemma 4.1 we know that if (C1)–(C3) are satisfied then�∗ given by(3.3) will have a Lipschitz in time lateral boundary inQ1/(2M). Then the Dirichletproblem (3.4) is solvable in the classical sense. In this section we show that(u∗, �∗) is a supersolution ofP and hence the minimal element ofB.

Lemma 6.1. Let u0 satisfy (C1)–(C3) andT ≤ 1/(2M). Then the pair(u∗, �∗) is the minimal element ofB.

Proof. The only thing we have to verify is that(u∗, �∗) is a supersolution ofP. Let (uk, �k) ∈ B be such that

(i) �∗=

⋂k�k;

(ii) the sequence{�k} is decreasing;(iii) �k(0) = �0 anduk(·,0) = u0.

We can construct such a sequence as follows. Let(uk, �k) ∈ B satisfy (i). Next,in order to have (ii) we observe the following. Denote byuk,m the solution ofthe Dirichlet problem in�k,m = �k ∩ �m with the initial functionuk,m(·,0) =

min{uk(·,0),um(·,0)} and vanishing on the lateral boundary, then

uk,m(x, t) ≤ min{uk(x, t),um(x, t)} for every(x, t) ∈ �k,m.

Besides, for the distance functions we will have

d�k,m(x, t) = min{d�k(x, t),d�m(x, t)} for every(x, t) ∈ �k,m.

Therefore, using Lemma 5.1, we can conclude that(uk,m, �k,m) ∈ B. If now (iii)is not satisfied, we can replace�k by the intersection of all�m with m ≤ k andthus to make{�k} decreasing.

In order to have (iii), let us take as�1 in the original sequence the cylindricaldomain�0 × (0, T) (see Remark 3.2). Now, since we can assume�k are decreas-ing, we will have�k(0) = �0. To satisfy the second condition in (iii) just replaceuk with the solution of the corresponding Dirichlet problem in�k.


Denote now byωk the caloric measure of�0 with respect to�k; that is

1ωk − (ωk)t = 0 in �k; ωk = 0 on ∂�k ∩ QT ; ω(·,0) = 1 in �0.

Let

C = supx∈�0

|∇u0(x)|.(6.1)

Then we can control the growth of|∇uk| in �k. Namely,

|∇uk(x, t)| ≤ 1 + (C − 1)ωk(x, t)(6.2)

for every(x, t) in �k. This follows from the maximum principle for subcaloricfunctions, sincev(x, t) = |∇uk| satisfies1v − vt ≥ 0 in �k.

For the next step, observe that sinceuk are caloric in�k and uniformly bounded,a subsequence of{uk} will converge inC1 norm on compact subsets of�∗

=⋂k�k to a functionu∗. We may assume also that over this subsequence, the

corresponding caloric measuresωk converge to a caloric measureω∗ of �0 withrespect to�∗. Then in the limit we will obtain from (6.2)

|∇u∗(x, t)| ≤ 1 + (C − 1)ω∗(x, t) for every(x, t) ∈ �∗.

As a consequence,(u∗, �∗) is inB and therefore is its minimal element.

7. FURTHER PROPERTIES OF THE MINIMAL ELEMENT

The method used in this and the next section is due to A. Henrot and H. Shahgho-lian [HS1], [HS2]. Originally it was applied to the Bernoulli type free boundaryproblem forp-Laplace operator, an elliptic problem, whose free boundary condi-tion is analogous to that ofP.

Definition 7.1. A point (x, t) ∈ ∂� ∩ QT , where� satisfies (3.1), is saidto beextremeif x ∈ ∂�(t) is extreme for�(t). The latter means thatx is not aconvex combination of points on∂�(t), other thanx.

Lemma 7.2. Let u0 satisfy (C1)–(C3) andT ≤ 1/(2M). Then the pair(u∗, �∗) satisfies

lim sup�∗3(x,t)→(x0,t0)

|∇u∗(x, t)| = 1(7.1)

for every extreme point(x0, t0) ∈ ∂�∗∩ QT .


Proof. Let us point out that it is enough to prove the lemma in the case whenx0 is anextremalpoint of ∂�∗(t0), which means that there is a spatial supportinghyperplane to�∗(t0), touching∂�∗(t0) atx0 only. This follows from the fact thatthe extremal points are dense among the extreme points.

Suppose nowx0 ∈ ∂�∗(t0) is extremal and that (7.1) is not true. Then, in viewof Lemmas 5.1 and 4.1, there exists a (space-time) neighborhoodV of (x0, t0) andα > 0 such that

u∗(x, t) ≤ (1 − α)d�∗(x, t)(7.2)

for every (x, t) ∈ V ∩ �∗. We may assume additionally that the intersectionV ∩�∗ is given by (3.2).

Let now5be a spatial supporting hyperplane to�∗(t0), such that5∩∂�∗(t0) =

{x0}. By translation and rotation, we may assume thatx0 = 0 and that5 = {xn =

0}. Moreover let�∗(t0) ⊂ {xn > 0}. Using the extremality of(x0, t0), it is easyto see that there areδ0 > 0 andη0 > 0 such that

{(x, t) ∈ �∗: xn ≤ η0 andt ∈ [t0 − δ0, t0]} ⊂ V.

Let us consider the function

h(t) = − minx∈�∗(t)

xn

for t ∈ [t0 − δ0, t0]. In view of Lipschitz regularity of∂�∗ in time,

h(t) ≤ L(t0 − t)

for t ∈ [t0 − δ0, t0], whereL is the Lipschitz constant off in t . Let nowη1 ∈

(0, η0) be very small and a constantC ≥ L be chosen such that

h(t0 − δ0) ≤ Cδ0 − η1.

Further, we can findδ1 ∈ (0, δ0] such that

h(t0 − δ1) = Cδ1 − η1

and

h(t) ≥ C(t0 − t)− η1

for everyt ∈ [t0 − δ1, t0].Define now a domain� ⊂ QT by giving its time sections as follows

�(t) =

�∗(t), t ∈ (0, t0 − δ1)

�∗(t) ∩ {xn > η1 − C(t0 − t)}, t ∈ [t0 − δ1, t0]�(t0), t ∈ (t0, T).

(7.3)


Let alsou be a solution to the Dirichlet problem in� for the heat equation, zero onthe lateral boundary and equalu0 in�0. We claim that ifη1 is small enough, then(u, �) is inB. This will lead to a contradiction since� 6⊂ �∗ and the lemma willfollow. Since� satisfies (3.1), we need to verify only that(u, �) is a supersolutionof P

By [Lemma 5, ACS], there existsε = ε(n, L) > 0 such that in a neighborhoodof (x0, t0), the function

w∗(x) = w∗(x; t) = u∗(x, t)+ u∗(x, t)1+ε

is subharmonic inx. Moreover the size of the neighborhood depends only onn andL. We may supposeV has this property. Next, note that we can takeC ≤ L + 1if η1 is sufficiently small andδ0 is fixed. This will make the boundary of newconstructed� (L + 1)-Lipschitz in time. Therefore we may assume also that

w(x) = w(x; t) = u(x, t)+ u(x, t)1+ε

is subharmonic in the neighborhoodV .We are ready to prove now that(u, �) is indeed a supersolution.Case 1. t ∈ (0, t0 − δ1). This is the simplest case, sinceu = u∗ in � ∩

0< t < t0 − δ1: all the properties ofu there follows from those ofu∗.Case 2.t ∈ [t0 − δ1, t0], the most interesting case. First of all, since�(t) ⊂

�∗(t) for theset , we have alsou ≤ u∗ there. Let now consider a partD(t) of�∗(t)between to planes:51 = {xn = η1 −C(t0 − t)} and50 = {xn = η0 −C(t0 − t)}.Compare there two functionsw(x) = w(x; t) and`(x) = xn − η1 + C(t0 − t).On51 both functions are0. Next

`(x) = η0 − η1 on50.

t0

t0 − δ1

t0 − δ0

�

FIG. 2. Construction of� from�∗ in profile


To estimatew on50, let us first estimateu on50. Thus

u(x, t) ≤ u∗(x, t) ≤ (1 − α)d�∗(x, t) ≤ (1 − α)(xn − L(t0 − t)) ≤ (1 − α)η0

and therefore, ifη0 is small enough, we will obtain

w(x) ≤ (1 − α/2)η0 on50.

Choose nowη1 so small that(1 − α/2)η0 ≤ η0 − η1. Thenw ≤ ` on∂D(t) and,sincew is subharmonic and is harmonic (linear), we conclude thatw ≤ ` inD(t). Along with u ≤ u∗ this gives

lim sup�3(x,t)→∂�

u(x, t)

d�(x, t)≤ 1,(7.4)

wheret is free to vary within[t0 − δ1, t0].Case 3.t ∈ (t0, T). Since�(t) shrink in time andu0 is superharmonic in�0,

considering the time derivativeut in�, we can infer from the maximum principlefor the heat equation thatut ≤ 0 in�. In particular, we will have in the cylindricalportion of� with t > t0 that u(x, t) ≤ u(x, t0) in �(t) = �(t0) and applyingLemma 5.1 we conclude that (7.4) is valid also in this case.

Summing up, we see that (7.4) holds for allt ∈ (0, T), and by Lemma (5.1) this

implies(u, �) ∈ B, which is a contradiction.

8. THE CLASSICAL SOLUTION OF P FOR SHORT TIME

In this section we prove

Theorem 8.1. Letu0 satisfy (C1)–(C3). Then the minimal element(u∗, �∗)

of B is a classical solution ofP in Q1/(2M). Moreover this classical solution isunique, the time sections�∗(t) are convex and shrinking in time andu∗(·, t) arelog-concave in�∗(t), 0< t < 1/(2M).

Up to now we have not used fully thelog-concavity condition onu0. Thefollowing lemma will exploit this property.

Lemma 8.2. Let u be a solution of a Dirichlet problem for the heat equationin a bounded domain� in QT , zero on∂� ∩ QT andu(·,0) = u0. If the timesections�(t) are convex for everyt ∈ [0, T ] andu0 is log-concave in�0 then sois u(·, t) in �(t) for everyt ∈ (0, T).

Proof. The proof, using Korevaar’s Concavity maximum principle [Ko] canbe found in [Pe]. We point out that an alternative proof can be given based on Bras-

camp and Leib’s paper [BL].


Remark 8.3. However, we can replace the condition oflog-concavity ofu0 byanother one that guarantees the convexity of level sets{u(·, t) > s} for everyu asin Lemma 8.2. In fact only this property will be used.

The convexity of level sets is used in the following lemma, which is mainly dueto [HS1], [HS2].

Lemma 8.4. Let D be a bounded spatial convex domain withC1 regularboundary,V a neighborhood of∂D andw a smooth positive subharmonic functionin D ∩ V , continuously vanishing on∂D. If the level lines{w = s} are strictlyconvex surfaces for0< s< s0 then the condition

lim supD∩V3x→x0

|∇w(x)| ≥ 1

for every extreme pointx0 ∈ ∂D implies that

|∇w(x)| ≥ 1

for everyx with 0< w(x) < s0.

Proof. Let y0 ∈ D ∩ V be such thatw(y0) = s ∈ (0, s0), so`s = {w(x) = s}is a strictly convex surface. Denote by5 a tangent hyperplane tos at y0. Bytranslation and rotation we may assume thaty0 = 0 and5 = {xn = 0} andthat`s lies in the lower halfspace{xn ≤ 0}. Choose now an extreme pointx0 ∈

∂D ∩ {xn ≥ 0} such that it has the maximalxn-coordinate among the points of∂D. Although this point can be not uniquely defined, we will denote itx0(y0) toindicate the way it was constructed fromy0.

Suppose now, that along withC1 regularity of∂D, w is C1 regular up to∂D.The core inequality is then

|∇w(y0)| ≥ |∇w(x0(y0))|.(8.1)

To prove, letα = |∇w(x0(y0))| andβ ∈ (0, α). Consider a functionf (x) =

w(x) + βxn in D+= D ∩ {xn > 0}. The function f is subharmonic inD+ and

therefore admits its maximum on∂D+. A simple analysis shows that the maximumof f is admitted either at the originy0 or atx0. Let us exclude the latter possibility.We have that∂xnw(x0) = −|∇w(x0)| = −α and hence∂xn f (x0) = β − α < 0,which is impossible ifx0 is a maximum point. Therefore the maximum off isadmitted aty0 = 0 and as a consequence we have

|∇w(0)| = −∂xnw(0) = − limh→0+

w(hen)− w(0)

h≥ lim

h→0+

βh

h= β.


Lettingβ → α−, we obtain (8.1). This, of course, proves the lemma in the caseconsidered.

Consider now the general case. Choose a sequence of points{x j ∈ D ∩ V}

converging tox0 so that

limj →∞

|∇w(x j )| = lim supx→x0

|∇w(x)| ≥ 1.

Let sj = w(x j ) and a domainD j be bounded by the level surface`sj . Constructpointsy j ∈ `s on the same level surface asy0 so thatx j = x0(y j ) for the domainD j . It can be done as follows. Take the tangent hyperplane5 j to `sj at x j andmove it down towardss until the plane touchess and definey j to be the touchingpoint. Now, the functionw is C1 regular up to the boundary ofD j and thereforewe may apply (8.1) to obtain

|∇w(y j )| ≥ |∇w(x j )|.

It is clear that the proof will be completed as soon as we show thaty j → y0. Due tostrict convexity of s, this is indeed so, if the outer normalsν j of the tangent planes5 j to `sj atx j converge to the unit vectoren = (0,0, · · · ,1). In its turn, the latterstatement is a consequence of theC1 regularity of∂D and the proof of the lemma is

complete.

For the proof of the following lemma we thank Vladimir Maz’ya.

Lemma 8.5. Let D be a bounded spatial convex domain andx0 a singularpoint on∂D, such that there are more than one supporting hyperplanes toD atx0. Let alsoV be a neighborhood ofx0 andw a subharmonic function inD ∩ V ,continuously vanishing on∂D ∩ V . Then

limD∩V3x→x0

w(x)

d(x)= 0,(8.2)

whered(x) = distance(x, ∂D).

Proof. As a first step, we note, that placingD between two supporting hyper-planes atx0 which form an angle of openingα < π , one can easily construct abarrier function and prove that for smallr

S(r ): = maxBr (x0)∩D

u(x) ≤ C1r π/α(8.3)

whereC1 is some constant, andBr (x0) = {x: |x − x0| < r }.Next, we claim that

w(x) ≤ C2d(x)S(4r )/r for x ∈ ∂Br (x0) ∩ D(8.4)


with a constantC2 > 0.Indeed, letx1 be a point on∂D with |x − x1| = d(x). Note thatd(x) ≤ r ,

hence|x0 − x1| ≤ 2r and in particularB2r (x1) ⊂ B4r (x0). Let5 be a supportinghyperplane toD at x1 and5+ be the halfspace, which containsD. Let ω bea harmonic measure of∂B2r (x1) ∩ 5+ with respect toB2r (x1) ∩ 5+ andv: =S(4r )ω. Then1v = 0 ≤ 1w in B2r (x1) ∩ D andv ≥ w on∂(B2r (x1) ∩ D) andthereforev(y) ≥ w(y) in B2r (x1) ∩ D by the maximum principle. In particular,at x we have

w(x) ≤ v(x) ≤ S(4r )ω(x) ≤ S(4r )C3

rd(x)

with C3 a universal constant, such that

ω(y) ≤C3

r|y − x1|

for y ∈ Br (x1) ∩5+, and (8.4) follows.

Now, the estimates (8.3) and (8.4) terminate the proof.

Proof of Theorem 8.1. First observe that we need only to show that(u∗, �∗)

is a subsolution. The properties of(u∗, �∗) in the second part of the theoremfollow from inclusion (u∗, �∗) ∈ B and Lemma 8.2. The uniqueness followsfrom Lemma 2.4.

Recall that from Lemma 7.2 we know that

lim sup�∗3(x,t)→(x0,t0)

|∇u∗(x, t)| = 1

for every extreme point(x0, t0) ∈ ∂�∗∩ Q1/(2M). Denote byR the set of all

t0 ∈ (0,1/(2M)) such that

lim sup�∗(t0)3x→x0

|∇u∗(x, t0)| = 1

for every extreme pointx0 ∈ ∂�∗(t0).The reader can easily see the difference between these two properties: if in the

former case(x, t) goes to(x0, t0) by varying in space and time, in the latter casethe timet0 is fixed.

Let us prove that the complementJ = (0,1/(2M))\R is a union of a countablefamily of nowhere dense subsets of(0,1/(2M)). This follows from the continuityof |∇u∗

| in�∗. Indeed, let{Uk} be an open countable basis for topology inRn. Ift0 ∈ J then there exist an extreme pointx0 ∈ ∂�∗(t0), and natural numbersk andm such thatx0 ∈ Uk and|∇u∗(x, t0)| ≤ 1 − (1/m) for everyx ∈ Uk ∩ �∗(t0).Denote now the set of allt0 ∈ J with suchk andmbyJk,m. ThenJ =

⋃k,mJk,m.


Besides, as easy to see,Jk,m are nowhere dense in(0,1/(2M)) and our assertionfollows.

Consider now, as in the proof of Lemma 7.2, the function

w∗(x) = w∗(x; t) = u∗(x, t)+ u∗(x, t)1+ε.

Let t0 ∈ (0,1/(2M)). By [Lemma 5, ACS], there existε > 0, δ > 0 ands0 > 0such thatw∗

= w∗(·; t) is subharmonic in a convex ringD(t) = {0< w∗(x, t) <s0} whenevert ∈ (t0 − δ, t0 + δ). Besides, from Lemma 8.2, the level surfaces ofu∗ and therefore those ofw∗ are convex. Moreover, they are strictly convex dueto real analyticity ofu∗(·, t) in �∗(t).

Now, we point out that ift ∈ R, then∂�∗(t) is C1 regular. Otherwise therewould exist a singular extreme pointx0 ∈ ∂�∗(t) with

lim�∗(t)3x→x0

|∇u∗(x, t)| = 0,(8.5)

which contradicts to the definition ofR. Indeed, ifx0 ∈ ∂�∗(t) is singular thenby Lemma 8.5w∗(x; t)/d�(x, t) → 0, or, equivalently,u∗(x, t)/d�(x, t) → 0 asx → x0, and (8.5) will follow from Lemma 5.2. Note that Lemma 5.2 is applicablehere sincef (x): = u∗

t (x, t) = 1u∗(x, t) is bounded in�∗(t) (see Remark 4.2).Let nowt ∈ (t0 − δ, t0 + δ) ∩R. Then Lemma 8.4 implies that

|∇w∗(x; t)| ≥ 1,

if 0 < w∗(x; t) < s0 and t is as above. This inequality is extended for allt ∈ (t0−δ, t0+δ)because of everywhere density ofRand continuity of|∇w∗(x; t)|in �∗. Since|∇w∗

| and|∇u∗| are asymptotically equivalent whenu∗

→ 0, weobtain immediately that

lim inf�∗3(x,t)→(x0,t0)

|∇u∗(x, t)| ≥ 1

wheneverx0 ∈ ∂�∗(t0). Sincet0 ∈ (0,1/(2M)) was arbitrary, we conclude that(u∗, �∗) is indeed a classical solution ofP in Q1/(2M). The theorem is proved.

ACKNOWLEDGMENTThe second author thanks the Department of Mathematics of the University of Texas at Austin for

the visiting appointment and the hospitality during the visit. He also thanks the Swedish Institute andGustaf Sigurd Magnuson’s foundation of the Royal Swedish Academy of Sciences for partial support.


REFERENCES

[ACS] I. Athanasopoulos, L. Caffarelli and S. Salsa,Caloric functions in Lipschitz domains and theregularity of solutions to phase transition problems, Annals of Math.143(1996), 413–434.

[AG] D. Andreucci and R. Gianni,Classical solutions to a multidimensional free boundary problemarising in combustion theory, Comm. Partial Differential Equations19(1994), no. 5-6, 803–826.

[Be] A. Beurling,On free-boundary problems for the Laplace equation, Sem. analytic functions 1(1958), 248–263.

[BL] H. J. Brascamp and E. H. Lieb,On extensions of the Brunn-Minkowski and Prekopa-Leindlertheorems, including inequalities for log concave functions, and with an application to thediffusion equation, J. Functional Analysis22 (1976), no. 4, 366–389.

[CLW1] L. A. Caffarelli, C. Lederman and N. Wolanski,Uniform estimates and limits for a two phaseparabolic singular perturbation problem, Indiana Univ. Math. J.46 (1997), no. 2, 453–489.

[CLW2] L. A. Caffarelli, C. Lederman and N. Wolanski,Pointwise and Viscosity solutions for the limitof a two phase parabolic singular perturbation problem, Indiana Univ. Math. J.46 (1997),no. 3, 719–740.

[CV] L. A. Caffarelli and J. L. Vazquez,A free-boundary problem for the heat equation arising inflame propagation, Trans. Amer. Math. Soc.347(1995), no. 2, 411–441.

[GHV] V. A. Galaktionov, J. Hulshof and J. L. Vazquez,Extinction and focusing behaviour of sphericaland annular flames described by a free boundary problem, J. Math. Pures Appl. (9)76 (1997),no. 7, 563–608.


[HS2] A. Henrot and H. Shahgholian,Existence of classical solutions to a free boundary problem forthe p-Laplace operator: (II) the interior convex case, to appear in Indiana Univ. Math. J.

[Ko] N. Korevaar,Convex solutions to nonlinear elliptic and parabolic boundary value problems,Indiana Univ. Math. J.32 (1983), no. 4, 603–614.

[LVW] C. Lederman, J. L. Vazquez and N. Wolanski,Uniqueness of solutions to a free boundaryproblem from combustion, preprint.

[Me] A. M. Meı rmanov,On a problem with a free boundary for parabolic equations, (Russian) Mat.Sb. (N.S.)115(157) (1981), no. 4, 532–543.

[Pe] A. Petrosyan,Convexity and uniqueness in a free boundary problem arising in combustiontheory, preprint.

[Va] J. L. Vazquez,The free boundary problem for the heat equation with fixed gradient condition,Free boundary problems, theory and applications (Zakopane, 1995), 277–302, Pitman Res.Notes Math. Ser.363, Longman, Harlow, 1996.

Paper III

A free boundary problem for ∞–Laplace equation

Juan Manfredi

Department of Mathematics, University of Pittsburgh, Pittsburgh, PA 15260, USAE-mail: [email protected]

and

Arshak Petrosyan and Henrik Shahgholian

Department of Mathematics, Royal Institute of Technology, 100 44 Stockholm, SwedenE-mail: {arshak,henriks}@math.kth.se

We consider a free boundary problem for thep-Laplacian


∇u),

describing nonlinear potential flow past convex profileK with prescribed pressure gradient|∇u(x)| = a(x) on the free stream line. The main purpose of this paper is to study the limitasp → ∞ of the classical solutions of the problem above, existing under certain convexityassumptions ona(x). We show, as one can expect, that the limit solves the correspondingproblem for the∞-Laplacian

1∞u = ∇2u∇u · ∇u,

in a certain weak sense, strong however, to guarantee the uniqueness. We show also thatin the special casea(x) ≡ a0 > 0 the limit coincides with an explicit solution, given by adistance function.

Key Words:free boundary problems; classical solutions; weak solutions;∞–Laplacian;p–Laplacian.

1. INTRODUCTION

In the past few years there has been a renewed interest in geometric configu-rations in potential flow in fluid mechanics. This time, however, the focus is onnonlinear flows with power law generalization (see e.g. [AM], [HS1–3].) Thelatter refers to thep–Laplace operator


∇u), 1< p < ∞.

1

2 J. MANFREDI, A. PETROSYAN AND H. SHAHGHOLIAN

The peculiar nonlinearity and degeneracy of the operator make the problems ofpotential flow both more realistic and also much harder to analyze.

Our concern, in this paper, will be the limit casep → ∞of the work of A. Henrotand H. Shahgholian [HS1–3]. Let us first formulate the problem.

Let K be a compact convex subset inRn, anda(x)a positive continuous functionin K c

= Rn\ K . For p ∈ (1,∞) consider the following Bernoulli-type free

boundary problem: find a pair(u, �) where� ⊃ K is a domain inRn andu is anonnegative continuous function in� such that 1pu = 0 in � \ K

u = 0 and |∇u| = a on ∂�u = 1 on K .

(FBp)

Problem (FBp), as mentioned earlier, describes nonlinear potential flow past theprofile K and∂� describes the stream line with prescribed pressure|∇u| = a.This problem forp = 2 is well studied and there is an extensive list of literature.We refer to the paper of A. Acker and R. Meyer [AM] for background and furtherreferences. We also mention the pioneering work of H. W. Alt and L. A. Caffarelli[AC] in this connection.

For p ∈ (1,∞), under suitable convexity assumptions ona(x) and regularityassumptions onK to be specified later, problem (FBp) admits a unique classicalsolution, which we denote by(up, �p); see [HS3]. The main purpose of this paperis to study the behavior of the pair(up, �p) asp → ∞. For reader’s conveniencewe outline some steps of the proof for1< p < ∞ in the appendix.

It it known that the limits ofp-harmonic functions (weak solutions of1pu = 0)as p → ∞ are viscosity solutions of the equation1∞u = 0 (in other words,∞–harmonic function), where

1∞u = ∇2u∇u · ∇u,

see [BDM] (see also [Ar].) The operator1∞ is known as the∞–Laplace operator.Since the free boundary condition is the same for all problems (FBp), we expectthat the limit

(u∞, �∞) = limp→∞

(up, �p),

if it exists, is a solution of the limiting problem 1∞u = 0 in � \ Ku = 0 and |∇u| = a on ∂�u = 1 on K

(FB∞)

in a certain sense.Another point of view is to treat problem (FB∞) independently and interpret

the limit process above as one of the ways of solving the problem. This approachgives rise to new kinds of questions such as uniqueness and regularity of solutions.

A FREE BOUNDARY PROBLEM FOR∞-LAPLACE EQUATION 3

It is worth noting that in certain special cases one can write down a solution of(FB∞) explicitly. For instance consider the casea(x) ≡ a0. Then one can easilysee that a pair(u, �) with

u(x) = 1 − a0 dist(x, K ) and � = { dist(x, K ) < 1/a0}(1.1)

solves problem (FB∞) in the classical sense (see Section 3). Now one can askwhether the explicit solution (1.1) is a limit of solutions of (FBp). We show thatthis is indeed the case and moreover (1.1) is the unique classical solution of (FB∞)with a(x) ≡ a0 (see Theorem 3.3.)

The complications in the case of more generala(x) are connected with the factthat generally it is not known whether∞–harmonic functions areC1 regular ornot, and the problem is how to interpret the free boundary condition. We overcomethis difficulty by introducing the notion of the weak solution of (FB∞). Then weshow that the only weak solution of (FB∞) is nothing but the limit(u∞, �∞) (seeTheorem 4.4.)

We conclude the paper with the brief analysis of the limitp → 1+. Thefollowing phenomenon takes place: the domains�p shrink to K (in dimensionn ≥ 2.)

2. PRELIMINARIES

Throughout the paper we assume thatK is a convex compact inRn, n ≥ 2,which satisfies theuniform interior ball condition, i.e. there existsδ > 0 suchthat for everyx ∈ ∂K there is a ballB of radiusδ with the propertyB ⊂ K and∂B ∩ ∂K = {x}.

Next the functiona(x) will be positive and continuous inK c= Rn

\ K with

infx∈K c

a(x) = a0 > 0.(2.1)

We will assume also that1/a(x) is locally concave inK c. (Sometimes we will callsuch functionsharmonic concave.) Moreover, assuming w.l.o.g. that the origin isan interior point ofK , we will require

λa(λx) > a(x) for everyx ∈ K c andλ > 1.(2.2)

We start with the definition.

Definition 2.1. A pair (u, �), where� is a domain inRn that containsKandu is a nonnegative continuous function on�, is called aclassical supersolutionof (FBp) for 1< p < ∞ if

(i) u ∈ C1(� \ K ) and1pu ≤ 0 in � \ K in the sense of distributions;(ii) lim supy→x |∇u(y)| ≤ a(x) for everyx ∈ ∂�;


(iii) u ≥ 1 on K andu = 0 on ∂�.

Analogously, we defineclassical subsolutionsof (FBp), by reversing the in-equality signs in (i), (ii) and (iii) and replacinglim supby lim inf in (ii).

A classical solutionof (FBp) is a pair(u, �) which is a classical sub- andsupersolution at the same time.

As a straightforward consequence of condition (2.2), we have the followingcomparison principle.

Lemma 2.2 (Comparison principle). For problem (FBp), every classicalsubsolution is smaller than every classical supersolution.

More precisely, the lemma should be read:If (u′, �′) is a classical sub- and(u, �) is a classical supersolution of (FBp), then�′

⊂ � andu′≤ u in �′.

Proof. We will consider first the case whenu′ andu areC1 up to∂�′ and∂�respectively. Suppose that�′

6⊂ �. Then we have

λ0 = min{λ ≥ 1:λ−1�′⊂ �} > 1.

Hereλ−1�′= {λ−1x: x ∈ �} =:�′

λ. Then�′λ0

⊂ � and there is a commonpoint x0 ∈ ∂(�′

λ0) ∩ ∂�. Consider the rescaled function

u′λ0(x) = u′(λ0x) in �′

λ0.

Then usual comparison principle forp–harmonic functions says thatu′λ0

≤ u in�′λ0

. But u′λ0(x0) = u(x0) = 0, hence there must be

λ0a(λ0x0) ≤ |∇uλ0(x0)| ≤ |∇u(x0)| ≤ a(x0),

which contradicts to (2.2). Therefore�′⊂ � and moreoveru′

≤ u in �′ by thecomparison principle forp-harmonic functions.

The reasonings above are known as theLavrent’ev principle(cf. [La]).The case whenu′ andu are notC1 up to the boundary can be handled in various

ways, reducing it to the case above. One of the ways is to consideru′−ε andu−ε

on the level sets�′ε = {u′

≥ ε} and�ε = {u ≥ ε} respectively and deduce thatthe constantλ0 = λ0(ε) as above but for�′

ε and�ε tends to1 asε goes to0+.

The comparison lemma above gives us the uniqueness of the classical solutionfor (FBp) if it exists. The following existence theorem is due to A. Henrot andH. Shahgholian. For reader’s convenience we give a brief sketch of the proof ofthis theorem at the end of this paper.


Theorem 2.3 ([HS3]). There exists a unique classical solution(up, �p)

of the free boundary problem (FBp). Moreover�p is convex and∂�p is C1

regular.

Next we will use the favorable geometric situation to compare solutions of(FBp) for different p. This lemma has appeared already in [Ja].

Lemma 2.4. Let � ⊃ K be a convex domain andu be the p–capacitarypotential (1 < p < ∞) of the convex ring� \ K , i.e. a continuous function on� \ K such that

1pu = 0 in � \ K ; u = 0 on ∂�; u = 1 on ∂K .

Then

1qu ≤ 0, if 1< q ≤ p and

1qu ≥ 0, if p ≤ q ≤ ∞.

Proof. By a result of J. Lewis [Le],|∇u| > 0 in� \ K and thereforeu is a realanalytic in� \ K . Then one can write1pu in a nondivergence form

1pu = |∇u|p−2(1u + |∇u|

−2(p − 2)1∞u).(2.3)

Let us introduce operators

L pu: = |∇u|2−p1pu = 1u + |∇u|

−2(p − 2)1∞u.(2.4)

ThenLqu has always the same sign as1qu for every1 < q < ∞. Using nowassumption that1pu = 0 in � \ K and henceL pu = 0, we deduce that

Lqu = |∇u|−2(q − p)1∞u in � \ K .(2.5)

Hence the lemma will follow as soon as we prove

1∞u ≥ 0.(2.6)

To this end, take a pointx ∈ � \ K and choose a coordinate system centered atx such that then–axis is directed as∇u. Let u(x) = s. By [Le] the level setLs(u) = {y: u(y) > s} is convex and therefore its boundary`s(u) = ∂Ls(u) canbe given near the pointx as a graph{y: yn = f (y′)} of a convexC∞ function f in


�∞ K�p

FIG. 1. The variational solution:�∞ = lim p→∞�p

the coordinate system chosen above (herey′= (y1, . . . , yn−1).) Differentiating

twice the identityu(y′, f (y′)) = s and using thatfi (0) = 0 we obtain

ui i + un fi i = 0 at0 for i = 1, . . . ,n − 1.(2.7)

The indices mean differentiation with respect to the correspondingy-coordinates.In particularun = |∇u| at 0. Observe now that in these new coordinates thecondition1pu = 0 at x means precisely

n−1∑i =1

ui i + (p − 1)unn = 0 at0.(2.8)

From (2.7) and (2.8) we obtain

1∞u(x) = unnu2n =

n − 1

p − 1κu3

n ≥ 0,(2.9)

whereκ = 1/(n−1)∑n−1

i =1 fi i is the mean curvature of`s(u) at the pointx, whichis nonnegative. Sincex was an arbitrary point in� \ K , the proof of the lemma is

complete.

Corollary 2.5 (Variational solution). The classical solutions(up, �p) of(FBp) are nondecreasing inp ∈ (1,∞) and are uniformly bounded. More pre-cisely,�p′ ⊂ �p′′ andup′ ≤ up′′ in �p′ for every1 < p′

≤ p′′ < ∞ and thereexists a bounded domain�0 and a functionu0 on�0 such that�p ⊂ �0 andup ≤ u0 for all 1< p < ∞. Hence there exists

(u∞, �∞) = limp→∞

(up, �p)(2.10)

which we will call avariational solutionof (FB∞).


Proof. The monotonicity part is an immediate corollary of Theorem 2.3,Lemma 2.4 and Lemma 2.2. As the uniform bound(u0, �0) we can use

u0(x) = 1 − a0 dist(x, K ), �0 = {u0 > 0} = {x: dist(x, K ) < 1/a0},

where, recall,a0 = infx∈K c a(x) > 0. Indeed, the pair(u0, �0) is a classical su-persolution of (FBp) for all p ∈ (1,∞) and hence the statement follows. (See also

Step 1in the proof of Theorem 3.3.)

3. CLASSICAL SOLUTIONS FOR a(x) ≡ a0

Precisely as for (FBp), we define the classical (sub- and super-) solutions of(FB∞).

Definition 3.1. A pair (u, �), where� is a domain inRn that containsKandu is a nonnegative continuous function on�, is called aclassical supersolutionof (FB∞) if

(i) u ∈ C1(� \ K ) and1∞u ≤ 0 in � \ K in the viscosity sense;(ii) lim supy→x |∇u(y)| ≤ a(x) for everyx ∈ ∂�;(iii) u ≥ 1 on K andu = 0 on ∂�.

Classical subsolutionsandclassical solutionsof (FB∞) are defined by the respec-tive modifications of (i)–(iii) as in Definition 2.1.

Remark 3.2. Lemma 2.2 remains valid forp = ∞ with the same proof. Theonly properties of the operator needed is the comparison principle and the possi-bility to rescale sub- and supersolutions. Observe, however, that the comparisonprinciple for1∞ is nontrivial and was proved by R. Jensen in [Je].

As we have mentioned in the introduction, problem (FB∞) has a simple explicitsolution(u, �) given by (1.1) whena(x) ≡ a0 > 0. From the other hand we knowfrom Corollary 2.5 that there is always a variational solution. The next theoremsays that these two are the same.

Theorem 3.3 (Classical solution). In the casea(x) ≡ a0 > 0 the varia-tional solution(u∞, �∞) of (FB∞) as defined in Corollary 2.5 is given by

u∞(x) = 1 − a0 dist(x, K ) and �∞ = { dist(x, K ) < 1/a0}.(3.1)

Moreover,(u∞, �∞) is the unique classical solution of (FB∞).


Proof. Set

u(x) = 1 − a0 dist(x, K ) and � = { dist(x, K ) < 1/a0}.

Step 1.Show that for everyp ∈ (1,∞) there holds

�p ⊂ � and up ≤ u in �p.(3.2)

By Lemma 2.2 it is sufficient to show that(u, �) is a classical supersolution of(FBp). Since the gradient and boundary value conditions (ii) and (iii) in Defini-tion 2.1 are readily satisfied the only condition to be verified is1pu ≤ 0. Thelatter holds due to concavity ofu. Hence (3.2) is proved. Passing to the limit in(3.2) asp → ∞ we obtain

�∞ ⊂ � and u∞ ≤ u in �∞.(3.3)

Step 2.In order to reverse inequalities in (3.3), we construct explicit subsolutionsof (FBp) from (u, �) as follows. Assume for the moment thatK hasC2 regularboundary; then the distance functiondist(x, K ) is C2 in K c as well. Forε > 0let

φε(s) =eεs − 1

ε(3.4)

andδε ∈ (0,1) be such that

φε(1 − δε) = 1.(3.5)

(More explicitly δε = 1 − (1/ε) log(1 + ε).) Define

uε(x) = φε(u − δε)(3.6)

in

�ε = {x: u(x) > δε} = {x: dist(x, K ) < 1/a0 − δε}.

We claim that

(uε, �ε) is a classical subsolution of (FBp) for p > p(ε).(3.7)

By construction, the gradient and boundary value conditions are satisfied and wehave to verify only that1puε ≥ 0 in �ε \ K for p large. For this purpose, takea pointx ∈ �ε \ K and choose a coordinate system(y1, . . . , yn) centered atxwith n axis directed as∇uε(x) (or, equivalently, as∇u(x).) These are the same


coordinates used in the proof of Lemma 2.4. Observe that the level sets ofuε arethose ofu, but parameterized in another way. Assumeu(x) = s. Then the levelline `s(u) = `φε(s−δε)(u

ε) can be represented nearx as a graph{y: yn = f (y′)}

of a convexC2 function. In new coordinates we can calculate that atx

uεj j = φ′′ε (u − δε)u

2j + φ′

ε(u − δε)u j j(3.8)

for j = 1, . . . ,n. Using thatui = 0 for i = 1, . . . ,n − 1, un = a0 andunn = 0at x as well as a formula (2.7) (which is still valid) we obtain

uεi i = −φ′ε(u − δε)a0 fi i (x) for i = 1, . . . ,n − 1(3.9)

and

uεnn = φ′′ε (u − δε)a

20(3.10)

at pointx. Observe now thatφε satisfies the differential equation

φ′′ε = εφ′

ε.

Employing this fact we find that

|∇uε|2−p1puε =

n−1∑i =1

uεi i + (p − 1)uεnn = φ′ε(u − δε)a0((p − 1)a0ε − (n − 1)κ),

(3.11)

whereκ = (1/(n − 1))∑n−1

i =1 fi i is the mean curvature of the level line`s(u) =

`(1−s)/a0(dist(·, K )) at x. Recall now that we assumeK has a uniform interiorball property hence the mean curvature of∂K at every point is bounded by a finiteconstant, sayκ0. But then the mean curvature of every level line of the distancefunctiondist(x, K ) at every point is not greater thanκ0. Hence we haveκ ≤ κ0for theκ in (3.11). If nowp is so large that

(p − 1)a0ε − (n − 1)κ0 ≥ 0,

then from (3.11) we will have immediately1puε ≥ 0 at x. This means that (3.7)is true with

p(ε) = 1 +n − 1

a0εκ0.

Letting ε → 0 we conclude that

� ⊂ �∞ and u ≤ u∞ in �(3.12)


in the case whenK hasC2 regular boundary. If not, letKη ⊂ K be a strictlyconvex subset withC2 boundary approximatingK asη → 0. Then proceeding asabove we prove first that{x: dist(x, Kη) < 1/a0} ⊂ �∞ and1−a0 dist(x, Kη) ≤

u∞(x). Lettingη → 0 we prove (3.12) in the general case.

Step 3.The uniqueness of(u∞, �∞) follows readily from Lemma 2.2, see Re-mark 3.2.

The proof of the theorem is complete.

4. WEAK SOLUTIONS FOR GENERAL a(x)

Now we return to the general case, whena(x) is a positive harmonic concavefunction for x ∈ K c, satisfying (2.1) and (2.2). In this case we don’t have anexplicit solution as for the casea(x) ≡ a0, but we can still hope that the variationalsolution is a classical solution of (FB∞). The first difficulty we face is thatu∞

may fail to beC1. However, as we will see later, the family{up} of classicalsolutions of (FBp) is uniformly bounded in the Lipschitz norm, and henceu∞ isat least Lipschitz.

The main purpose of this section is to introduce a notion of a weak solution of(FB∞) that doesn’t use the differentiability properties ofu.

Definition 4.1. A pair (u, �), where� is a domain inRn that containsKandu is a nonnegative continuous function on�, is called aweak supersolutionof (FB∞) if

(i) 1∞u ≤ 0 in � \ K in the viscosity sense;(ii) lim supy→x u(y)/dist(y, ∂�) ≤ a(x) for everyx ∈ ∂�;(iii) u ≥ 1 on K andu = 0 on ∂�.

Analogously, we defineweak subsolutionsof (FB∞), by reversing the inequalitysigns in (i), (ii) and (iii) and replacinglim supby lim inf in (ii).

A weak solutionof (FB∞) is a pair(u, �)which is a weak sub- and supersolutionat the same time.

Remark 4.2. The comparison principle, Lemma 2.2, still works for the weaksolutions defined above, with a minor modification in the proof. This implies inparticular the uniqueness of the weak solution.

The following lemma provides compatibility of the above definition with thedefinition of classical solutions.

Lemma 4.3. A classical (super-, sub-) solution of (FB∞) is a weak (super-,sub-) solution of (FB∞).


Proof. 1) Supersolutions.This case is simple. Fory ∈ � \ K we have

u(y) = u(y)− u(z) = (y − z) · ∇u(ξ) ≤ dist(y, ∂�)|∇u(ξ)|,

wherez = z(y) ∈ ∂� is chosen such thatdist(y, ∂�) = |y − z| andξ is a certainpoint on the line segment, joiningy andz. Sincey → x ∈ ∂� impliesξ → x,we obtain

lim supu(y)

dist(y, ∂�)≤ lim sup|∇u(y)| ≤ a(x)

as� \ K 3 y → x ∈ ∂�.2) Subsolutions.This case needs more attention. Let(u, �) be a classical

subsolution of (FB∞), and consider the functionv(x) = u(x)/a(x). Assume firsta(x) is C1 regular. Thenv is C1 as well and its gradient is given by

∇v =∇u

a− u

∇a

a2.

In particular

lim inf�\K3y→x

|∇v(y)| ≥ 1

for everyx ∈ ∂�. Then we can proceed as in the proof of Claim 1 in [Vo]. Lety ∈ � \ K andv(y) = s. Suppose0 < s < sε, wheresε (for small ε > 0) ischosen such that forz ∈ � \ K

v(z) < sε implies 1 − ε < |∇v(z)|.(4.1)

Find now a curveγy(t) which solves

d

dtγy(t) =

∇v(γy(t))

|∇v(γy(t))|2and γy(s) = y.(4.2)

There exist at least one solutionγy(t) of (4.2) defined on the whole interval(0, s)with a continuous extension to[0, s] and which satisfies

v(γy(t)) = t for t ∈ [0, s] .

Let `y denote the length of the curveγy [0, s]. Then from (4.1)

v(y) ≥ (1 − ε)`y ≥ (1 − ε)dist(y, ∂�).

Consequently

lim infu(y)

dist(y, ∂�)= a(x) lim inf

v(y)

dist(y, ∂�)≥ a(x)


as� \ K 3 y → x ∈ ∂� and the lemma follows in the casea(x) is C1 regular.For continuousa(x), take aC1 regularizationaε(x) ≤ a(x), converging point-

wise toa(x) asε → 0, and use it in the reasonings above to obtain

lim inf u(y)/dist(y, ∂�) ≥ aε(x).

Then we letε → 0 to complete the proof.

One of the main purposes of this paper is to prove the following theorem.

Theorem 4.4 (Weak solution). The variational solution(u∞, �∞)of (FB∞)as defined in Corollary 2.5 is the only weak solution of (FB∞).

5. (u∞, �∞) IS A WEAK SUBSOLUTION

The statement in the title of this section constitutes the relatively easy part inthe proof of Theorem 4.4.

Lemma 5.1. The variational solution(u∞, �∞) is a weak subsolution of(FB∞).

Proof. We use similar reasonings as those in the proof of Lemma 4.3.Let (up, �p) be the classical solution of (FBp) for p > 1, y ∈ �p \ K and

up(y) = s. Consider the curveγy(t) which solves

d

dtγy(t) =

∇up(γy(t))

|∇up(γy(t))|2and γy(s) = y.(5.1)

On this curve the following relation is satisfied

up(γy(t)) = t

and one can infer from this thatγy(t) can be defined on the interval(0, s) with acontinuous extension to[0, s]. The curveγy has the following nice property.

|∇up(γy(t))| ↗ ast ↗ .(5.2)

To prove this statement, fixz0 = γy(t0) for somet0 ∈ (0, s). Choose a newcoordinate system inRn with the origin atz0 and with then–axis directed as∇up(z0). Thenγ ′

y(t0) is directed as then–axis as well. Therefore

d

dt|∇up(γy(t))|

∣∣t=t0

= |γ ′(t0)|(|∇up|)n(z0) = |γ ′(t0)|(up)nn(z0),(5.3)


where the subscriptn means the differentiation along then–axis. On the otherhand we have

(up)nn =n − 1

p − 1κ(up)n ≥ 0(5.4)

at z0, whereκ ≥ 0 is the mean curvature of the level line{up = t0} at z0, as itfollows from the computations in the proof of Lemma 2.4, cf. (2.9). Combining(5.3) and (5.4), we prove (5.2).

Let us now approach the free boundary∂�p along the curveγy(t) ast → 0+.Then in the limit

|∇up(γy(0)| = a(γy(0)).

Together with (5.2) this implies

|∇up(y)| ≥ a(γy(0)) for y ∈ �p \ K .(5.5)

More generally, we have

|∇up(γy(t))| ≥ a(γy(0)) for t ∈ [0, s] .(5.6)

To proceed, we supposep ≥ p0 > 1 and0< s ≤ σ < s0 < 1. Set

c0 = inf a(y) and L0 = sup|a(y)− a(z)|

|y − z|

for y, z ∈ �∞ \ {up0 > s0}. Thenc0 > 0 andL0 < ∞. Using (5.6), we can infervery rough estimate

|∇up(γy(t))| ≥ c0 for t ∈ [0, s] .

If we denote by y the length of the curveγy [0, s], then the estimate above willimply

up(y) ≥ c0`y

and consequently

|y − γy(0)| ≤ `y ≤up(y)

c0.

Then by the definition ofL0 and the above estimate

a(γy(0)) ≥ a(y)− L0|y − γy(0)| ≥ a(y)−L0

c0up(y) ≥ a(y)−

L0

c0σ.


Going back again to (5.6), integrating along the curveγy [0, s] and applying theprevious estimate, we obtain

up(y) ≥ a(γy(0))`y ≥

(a(y)−

L0

c0σ

)dist(y, ∂�p)

for every p ≥ p0 andy ∈ �p \ K with up(y) ≤ σ < s0.Letting p → ∞, we obtain

u∞(y) ≥

(a(y)−

L0

c0σ

)dist(y, ∂�∞)

for every y ∈ �∞ \ K with u∞(y) ≤ σ < s0. It is clear now that the lemmafollows.

6. UNIFORM GRADIENT BOUND FOR up AS p → ∞

In this section we make the first step in proving that(u∞, �∞) is a weak super-solution of (FB∞).

We remind that we assumeK to satisfy the uniform interior ball condition. Thatis, there existsδ > 0 such that for everyx ∈ ∂K one can find a ballB of radiusδsuch thatB ⊂ K and∂B ∩ ∂K = {x}. As a consequence we have the followinglemma.

Lemma 6.1. There exists a constantM > 0such that for the classical solution(up, �p) of (FBp) we have

|∇up| ≤ M in �p \ K

for all p ≥ p0 > 1. The constantM here depends only onδ in the uniforminterior ball condition for K , on p0, on d0 = infy∈∂�p0

dist(y, K ), and on thespace dimensionn.

In particular, u∞ is Lipschitz continuous.

Proof. We mostly follow the proof of Lemma 2.3 in [HS1]. Observe first that|∇up| satisfies the maximum principle in�p \ K , namely

max�p\K

|∇up| ≤ max∂K∪∂�p

|∇up|.

This follows from Payne-Philippin’s inequality

Lup(|∇up|p) ≥ 0,


with the elliptic operator

Lup(v): = |∇up|p−21v + (p − 2)|∇up|

p−4∇

2v∇up · ∇up.

For the proof apply Lemma 1 in [PP] with the particular choicesα = −1, β = 0,γ andv = 0 to obtain

Lup(|∇up|p) ≥

p(p − 1)2

4|∇up|

p−6(∇up · ∇(|∇up|2))2.

Now, having the gradient maximum principle it suffices to prove that∇up isuniformly bounded on∂�p ∪ ∂K . Since|∇up| = 1 on ∂�p we need only auniform gradient bound on∂K .

Fix now p0 > 1 and consider the functionup0 near∂K . Forx ∈ ∂K let Bδ(zx)

be a touching ball atx from inside ofK . By the definition ofd0 we have alsoBδ+d0(zx) ⊂ �p0. If one denote byvp0 the p0–capacitary potential of the ringBδ+d0(zx) \ Bδ(zx) then by the comparison principle we will havevp0 ≤ up0.Since alsovp0(x) = up0(x) we may conclude

|∇up0(x)| ≤ |∇vp0(x)| = M(δ,d0, p0,n).

Take now p > p0. From Corollary 2.5 we know thatup0 < up < 1. Alsoup = up0 = 1 on ∂K and consequently

|∇up(x)| ≤ |∇up0(x)| ≤ M for x ∈ ∂K .

The proof is complete.

7. ON ∞-HARMONIC FUNCTIONS NEAR SINGULAR CONVEXBOUNDARY POINTS

In this section we study the behavior of∞–harmonic functions near a singularconvex boundary point. The results are more or less independent from the restof the paper, and have the interest of their own. More specifically we prove thefollowing result.

Theorem 7.1. Let D be a convex open set withx0 a singular boundary pointandu a viscosity solution of1∞u = 0 in D ∩ BR(x0) for someR> 0. Supposemoreoveru vanishes on∂D ∩ BR(x0). Thenu(x) = O(|x − x0|

1+ε) asx → x0for someε > 0.

Such a result forp–harmonic functions is very well known, see e.g. in [Kr],[Do].


Proof. The convexity ofD helps us to reduce the problem to the2-dimensionalcase. Namely, that the pointx0 is a singular boundary point ofD means preciselythat there are at least2 supporting hyperplanes toD at x0. Therefore after anecessary translation and rotation we may assumex0 = 0 and that there is acone0 in R2 of aperture less thanπ and with vertex at the origin, such thatD ⊂ 0 × Rn−2.

Suppose at the moment that we know the existence of a barrierv in the 2-dimensional cone0 that has the following properties:

(i) v ∈ C(0) and1∞v ≤ 0 in the viscosity sense;(ii) v > 0 in 0 \ {0} andv(0) = 0;(iii) v(x) = O(|x|

1+ε) for someε > 0.

Then using standard arguments together with Jensen’s maximum principle weobtainu(x) = O(|x|

1+ε), as needed.Let us point out that a barrier as above, symmetric with respect to the cone’s

axis, cannot beC2 regular, the reason being that∞–superharmonicity impliesconcavity of the barrier on the symmetry axis. The barriers that we will constructnext are onlyC1,1/3 regular.

Lemma 7.2. For everyθ ∈ (0, π/2) there isε > 0 and a C1,1/3 regularfunctionF(α) for α ∈ (−θ, θ) such that (using complex notations)v, given by

v(reiα) = r 1+εF(α),(7.1)

satisfies(i)–(iii) above in0 = {reiα:α ∈ (−θ, θ)}.

Proof. Let us compute formally the∞–Laplacian forv as in (7.1). We willobtain

1∞v(reiα) = r −(1−3ε)/2(ε(1+ε)3F(α)3+F ′(α)2((1+ε)(1+2ε)F(α)+F ′′(α))).

This can be rewritten as

1∞v(reiα) = r −(1−3ε)/2Lε(F)(α),

where

Lε(F) = ε(1 + ε)3F3+ (F ′)2((1 + ε)(1 + 2ε)F + F ′′).

SubstituteF = e f . This will give

Lε(ef ) = e3 f (ε(1 + ε)3 + ( f ′)4 + ( f ′)2((1 + ε)(1 + 2ε)+ f ′′)),

or

Lε(ef ) = e3 f ( f ′)2`ε( f ′),


-40 -20 20 40

-1.5

-1

-0.5

0.5

1

1.5

FIG. 2. Graphs ofφε(y) asε → 0.

where

`ε(y) = (1 + ε)(1 + 2ε)+ε(1 + ε)3

y2+ y2

+ y′

a first order differential operator. Let us now consider the ODE

`ε(y) = 0.

After integration we will get

− arctan

(y(α)

1 + ε

)+

√ε

1 + εarctan

(y(α)

√ε(1 + ε)

)= α + C,

whereC is a constant. ChooseC = 0 and consider the function

φε(y) = − arctan

(y

1 + ε

)+

√ε

1 + εarctan

(y

√ε(1 + ε)

)on the whole real axis. Thenφε is a smooth strictly decreasing function withφ(0) = 0. Using that the image ofarctanis the interval(−π/2, π/2) we easilyobtain that the image ofφε is the interval(−θε, θε) with

θε =π

2

(1 −

√ε

1 + ε

).

Then the inverse functionψε = φ−1ε is well defined on the interval(−θε, θε).

Moreover, since the derivative

φ′ε(y) = −

y2

(y2 + ε + ε2)(y2 + (1 + ε)2)


-0.3 -0.2 -0.1 0.1 0.2 0.3

-0.04

-0.02

0.02

0.04

FIG. 3. φε(y) near0 asε → 0.

is strictly negative fory 6= 0, ψε(α) is C∞ everywhere on(−θε, θε) exceptα =

0. To examine the behavior ofψε(α) nearα = 0 we look at the asymptoticdevelopment ofφε(y) neary = 0:

φε(y) = −y3

3ε(1 + ε)3+ O(y5).

This implies thatψε is onlyC1/3 regular at the origin.Let us choose nowε > 0 so thatθ < θε and go backwards in the constructions

above. Setf (α) =∫ α

0 ψε(t)dt andF(α) = e f (α). ThenF is C∞ everywhere inthe cone0ε = {reiα:α ∈ (−θε, θε)} ⊃ 0 except the symmetry axisα = 0 whereit is only C1,1/3 regular.

By construction,v(reiα) = r 1+εF(α) will satisfy 1∞v = 0 in the classicalsense off the symmetry axisα = 0. We claim that in the viscosity sense1∞v = 0in the whole0ε. First observe that there areno paraboloids touchingv by belowat the points on the symmetry axisα = 0, otherwiseF would beC1,1 regular.This implies immediately that1∞v ≤ 0 in 0ε in the viscosity sense. Let nowPbe a paraboloid touchingv by above at a point on the axisα = 0. Sincev is C1

regular, at the touching point∇ P = ∇v, which is directed along the axis. Observenow that on the axisα = 0 v is a convex functionr 1+εF(0) and hence we musthave∇

2P∇ P · ∇ P ≥ 0. This implies that1∞v ≥ 0 in the viscosity sense in0ε.Hence the claim follows.

The rest of properties (i)-(iii) are easily verified forv and the proof of the lemma

is complete.

Theorem 7.1 is proved.


Remark 7.3. Observe that in the caseθ < π/4 one can take as a barrier thefunction

v(x, y) = x4/3− |y|

4/3= r 4/3(cos(α)4/3 − | sin(α)|4/3)

in the lemma above. One can even check that it is precisely the function thatappears in the proof forε = 1/3.

8. C1 REGULARITY OF ∂�∞

The results of this section are straightforward consequences of Theorem 7.1.

Lemma 8.1 (C1 regularity). The level sets{u∞ > s} are convex sets withC1 regular boundary{u∞ = s} for everys ∈ [0,1).

In particular, ∂�∞ is C1 regular.

Proof. Let p0 > 1 ands0 ∈ (0,1). Then from the proof of Lemma 5.1 weknow that forc0 = min{a(x): x ∈ �∞ \ {up0 > s0}} > 0 we have

|∇up(y)| ≥ c0 wheneverp ≥ p0 and0< up(y) < s0.

This implies in particular, that

up(y)− s ≥ c0 dist(y, {up = s}) wheneverp ≥ p0 ands< up(y) < s0.

Letting p → ∞ we obtain

u∞(y)− s ≥ c0 dist(y, {u∞ = s}) whenevers< u∞(y) < s0.(8.1)

Now, Theorem 7.1 says that (8.1) would be impossible if{u∞ = s} were notC1

regular. Indeed, suppose there is a singular pointy0 on{u∞ = s}. Since{u∞ > s}is a convex open set, we can find a unit vectore and a smallδ > 0 such that thetruncated coneC = C(y0,e, δ) = {y ∈ Rn: |y − y0| < δ, angle(y − y0,e) < δ}

is contained in{u∞ > s}. Then along the symmetry axis ofC, for y = y0 + ηewith 0< η < δ we will have

dist(y, {u∞ = s}) ≥ tan(δ)|y − y0|.

This implies by (8.1)

u∞(y)− s ≥ c0 tan(δ)|y − y0|,


contradicting Theorem 7.1. The lemma is proved.

The following corollary is a kind of compensation for the missingC1 regularityof u∞.

Corollary 8.2. There is a unit-vector fieldν ∈ C(�∞ \ K ) such that

∇u∞(y) = ν(y)|∇u∞(y)| a.e. in�∞ \ K .(8.2)

Proof. For everyy ∈ �∞ \ K there is a unique supporting hyperplane to{u∞ < u∞(y)} at y. Denote byν(y) the unit normal vector of this hyperplane,pointing into{u∞ < u∞(y)}. Then Lemma 8.1 implies the continuity ofν up to∂�∞.

To see the identity (8.2), recall thatu∞ is Lipschitz continuous and therefore dif-ferentiable a.e. in�∞ \ K . At the differentiability pointsy where|∇u∞| 6= 0, ev-idently∇u∞/|∇u∞| is normal to the level set and hence coincides withν(y). The

proof is complete.

9. STABLE SOLUTIONS OF THE BOUNDARY VALUE PROBLEM

The disadvantage of the classical and weak solutions defined in the previoussections is that we don’t know generally their stability under the limit. In thissection we intend to give a notion of a stable solution, in this sense. As the sourceof the definitions and notations we useUser’s guide to viscosity solutions[CIL]of M. G. Crandall, H. Ishii and P. L. Lions.

Suppose we have a familyFp: Rn×R×Rn

×S(n) → R, of upper semicontin-uousproperoperators, where1< p < ∞ andS(n) is the set of symmetricn × nmatrices. The operatorF is proper if

F(x, r, η, X) ≤ F(x, s, η,Y) wheneverr ≤ s andY ≤ X,

wherer, s ∈ R, x, η ∈ Rn, X,Y ∈ S(n) andS(n) is equipped with its usual order.Observe, the operator−1p and not1p is proper!

Suppose next, we have a family of locally compact subsetsOp ⊂ Rn andfunctionsup ∈ LSC(Op) (lower semicontinuous functions onOp), which solve

Fp(x,up,∇up,∇2up) ≥ 0 onOp

in the viscosity sense. Denote

O∞ = lim supp→∞

Op


= {x: ∃xpk ∈ Opk such thatxpk → x} and

u∞(x) = lim infp→∞ ∗

up(x) onO∞

= inf{lim infk→∞

upk(xpk): xpk ∈ Opk , xpk → x}.

Then the following theorem holds.

Theorem 9.1. If the operatorsFp converge locally uniformly to an operatorF∞, then

F∞(x,u∞,∇u∞,∇2u∞) ≥ 0 onO∞

in the viscosity sense.

The theorem is an immediate corollary of the following lemma, which is avariation of the Proposition 4.3 in [CIL].

Lemma 9.2. For every x ∈ O∞ and (η, X) ∈ J2,−O∞

u∞(x) there exists a

sequencexpk ∈ �pk and(ηk, Xk) ∈ J2,−Opk

upk(xpk) with pk → ∞ such that

(xpk ,upk(xpk), ηk, Xk) → (x,u∞(x), η, X).

Recall that given a functionu on a subsetO of Rn and a pointx ∈ O, thesecond-ordersubjetJ2,−

O u(x) is the set of pairs(η, X) ∈ Rn× S(n) such that

u(x) ≥ u(x)+ η(x − x)+1

2X(x − x) · (x − x)+ o(|x − x|

2)(9.1)

asO 3 x → x. In addition to subjets we will also need their closures

J2,−O u(x) = {(η, X) ∈ Rn

× S(n): ∃xk ∈ O and(ηk, Xk) ∈ J2,−O u(xk)

such that(xk,u(xk), ηk, Xk) → (x,u(x), η, X)}.

Proof of Lemma 9.2. We refer to the proof of Proposition 4.3 in [CIL]. Theonly essential difference is that the locally compact setsOp do not vary withpthere, but this does not affect on the proof.

Having Theorem 9.1 in mind, we give the following definition.

Definition 9.3. Let D be an open subset ofRn and0 a relatively openportion of∂D. Then a lower semicontinuous functionu on D ∪ 0 is aviscosity


solutionof {F(x,u,∇u,∇2u) ≥ 0 in DB(x,u,∇u,∇2u) ≥ 0 on0

(BV)

with F andB proper operators, ifu is a viscosity solution of

G+(x,u,∇u,∇2u) ≥ 0 on D ∪ 0,

where

G+(x, r, η, X) =

{F(x, r, η, X) if x ∈ Dmax(F(x, r, η, X), B(x, r, η, X)) if x ∈ 0.

Explicitly, this means{F(x,u(x), η, X) ≥ 0 for (η, X) ∈ J2,−

D u(x), x ∈ D

max(F(x,u(x), η, X), B(x,u(x), η, X)) ≥ 0 for (η, X) ∈ J2,−D∪0u(x), x ∈ 0.

Suppose now we have a sequence of continuous functionsup that solve

Fp ≥ 0 in Dp and Bp ≥ 0 on0p

in the viscosity sense, for continuous proper operatorsFp and Bp, open subsetsDp of Rn and relatively open0p ⊂ ∂Dp. Suppose also thatDp and0p convergesin the Hausdorff metric toD∞ and0∞ ⊂ ∂D∞ and thatup converges locallyuniformly to a continuous functionu∞. Then the following assertion holds.

Theorem 9.4. If the operatorsFp and Bp converges locally uniformly tooperatorsF∞ andB∞ respectively asp → ∞ andup are as above, then the limitu∞ solves

F∞ ≥ 0 in D∞ and B∞ ≥ 0 on0∞

in the viscosity sense.

10. (u∞, �∞) IS A WEAK SUPERSOLUTION

In this section we prove the second counterpart of Theorem 4.4.

Lemma 10.1. The variational solution(u∞, �∞) is a weak supersolution of(FB∞).


Proof. Take a sequence of pointsyk ∈ �∞ \ K , converging tox0 ∈ ∂�∞. Letdk = dist(yk, ∂�∞)andxk ∈ ∂�∞ be such that|yk−xk| = dk. Then without lossof generality we may assumeek = (yk − xk)/dk converge toe0 = (1,0, . . . ,0).We want to prove that

lim supk→∞

u∞(yk)

dk≤ a(x0).

Defineu∞ to be identically zero off�∞ and consider the functions

vk(x) =u∞(xk + dkx)

dkx ∈ Rn.

Sinceu∞ is Lipschitz continuous, the family{vk} is uniformly Lipschitz. Hencewe may extract a subsequence converging uniformly on compacts to a Lipschitzfunctionv0. Assume that this is the whole sequence. Then

u∞(yk)

dk= vk(ek) → v0(e0)

and we will be done if we show

v0(e0) ≤ a(x0).

In fact, the following identity holds:

v0(x) = a(x0)(x · e0)+ x ∈ Rn.(10.1)

We prove this important fact in several steps.

Step 1.Show that{v0 > 0} = 50 = {x: x · e0 > 0}.

Let Dk = {vk > 0}. Then if we fixs0 ∈ (0,1), we can find a constantc0 suchthat

vk(x) ≥ min{c0 dist(x, Dck), s0/dk};(10.2)

see the proof of Lemmas 5.1 and 8.1.Observe next, thatDk ⊂ 5k = {x: x · ek > 0} by definition and that0 ∈

∂Dk ∩ ∂5k. Since�∞ hasC1 regular boundary andDk are dilations of�∞ witha coefficient1/dk → ∞, we will have thatDk converge to50 and (10.2) will give

v0(x) ≥ c0 dist(x,5c0).


Hence the positivity set ofv0 contains50. Sincevk are0 off 5k, v0 is 0 off 50.This proves the statement.

Step 2.There isα ≥ a(x0) such thatv0(x) = α(x · e0)+.

Let ν be the vector field in Corollary 8.2 and set

νk(x) = ν(xk + dkx).

Then

∇vk(x) = |∇vk(x)|νk(x) a.e. inDk.

Observe now, thatνk(x) → ν(x0) = e0 uniformly on compacts subsets of50.Supposeφ ∈ L2

loc(50,Rn) is a weak limit of a subsequence of∇vk (note,∇vk

are uniformly bounded inL∞ norm). Thene · e0 = 0 impliesφ · e0 = 0 a.e.in 50. This meansφ = |φ|e0 a.e. in50. Besides, we will have∇v0 = φ inthe distributional sense and therefore∂ev0 = 0 for any directione orthogonal toe0. Thereforev0(x) = f (x · e0) in 50 for a certain functionf , positive andcontinuous. Besides,f will be nondecreasing.

Next, observe thatv0 satisfies1∞v0 = 0 in the viscosity sense, sincevk satisfy1∞vk = 0 in Dk \ {vk ≥ 1/dk}. Then we leave to the reader as a good exercise toshow that forv0 of the special formf (x·e0)as above this meansf ′′(t)( f ′(t))2 = 0in the viscosity sense. This, in turn, implies thatf is a linear function, i.e.f (t) =

αt +β. Since f (0) = 0we obtainf (t) = αt and consequentlyv0(x) = α(x ·e0)+

everywhere inRn.Thatα ≥ a(x0) follows simply from the observation thatα = v0(e0) and

v0(e0) = lim vk(ek) = lim(u∞(yk)/dk) ≥ a(x0),

by Lemma 5.1.

Step 3.v0 is a viscosity solution (see Section 9) of a boundary value problem{−1∞v0 ≥ 0 in 50−∇v0 · e0 + a(x0) ≥ 0 on ∂50.

(10.3)

To prove this statement, we are going to apply Theorem 9.4. We first refine ourconstructions.

Let dk,p = dist(yk, ∂�p) andxk,p ∈ ∂�p be such thatdk,p = |yk − xk,p|. Letalsoek,p = (yk − xk,p)/dk,p. Define next

vk,p(x) =up(xk,p + dk,px)

dk,pfor x ∈ Rn,

with the convention thatup = 0 off �p.


Now, choosepk so large thatek: = ek,pk → e0 and moreovervk: = vk,pk →

v0 uniformly on compacts. If nowDk = {vk > 0} and R > 0 is fixed, thenDk ∩ BR(0) and∂ Dk ∩ BR(0) converge in the Hausdorff distance to50 ∩ BR(0)and∂50 ∩ BR(0) respectively. Moreover, if

νk(x) = ∇vk(x)/|∇vk(x)| and ak(x) = a(xk,pk + dk,pk x)

thenνk(x) andak(x) converge uniformly inBR(0) to e0 anda(x0).Set next forp > 1

Fp(x,u,∇u,∇2u) = −|∇u|

4−p

p − 21pu = −

|∇u|2

p − 21u −1∞u.

and observe thatFp converges locally uniformly on compacts toF∞ = −1∞ asp → ∞.

Fix now largeR> 0. Thenvk is a classical and therefore viscosity solution ofthe following boundary value problem{

Fp(x,u,∇u,∇2u) ≥ 0 in Dk ∩ BR(0)−∇u · νk(x)+ ak(x) ≥ 0 on ∂ Dk ∩ BR(0)

(10.4)

Applying now Theorem 9.4, we conclude immediately thatv0 is a viscositysolution of {

−1∞v ≥ 0 in 50 ∩ BR(0)−∇v · e0 + a(x0) ≥ 0 on ∂50 ∩ BR(0)

(10.5)

for everyR. SinceR is arbitrary, (10.3) follows.

Step 4.In the representationv0(x) = α(x · e0)+, we haveα ≤ a(x0).

This follows from (10.3). Indeed, forβ < α, we have

v0(x) = α(x · e0) ≥ β(x · e0)+ (x · e0)2 for x ∈ 50,

asx → 0, which means that(βe0,et0e0) ∈ J2,−

50v0(0). Hence we must have by

the definition of the viscosity supersolution that

max{−1∞(β(x · e0)+ (x · e0)2),−∇(β(x · e0)+ (x · e0)

2)+ a(x0)} ≥ 0

at x = 0, or

max{−2β2,−β + a(x0)} ≥ 0.

This implies

−β + a(x0) ≥ 0


and sinceβ < α was arbitrary,

α ≤ a(x0).

The identity (10.1), Lemma 10.1 and consequently Theorem 4.4 are proved.

11. THE LIMIT AS p → 1

We would like to conclude with a brief discussion of what happens with theclassical solutions(up, �p) of (FBp) as p → 1+. The answer is extremelysimple.

Theorem 11.1. As p → 1+, the domains�p collapse toK , providedn ≥ 2.

Remark 11.2. We stress here that the space dimensionn ≥ 2. Forn = 1 thestatement of Theorem 11.1 fails, since all the problems (FBp) are the same for allp ∈ (1,∞).

Proof of Theorem 11.1. We follow a scheme, similar to that in the proof ofTheorem 3.3. For largeE denote

φE(s) =eEs

− 1

E

and letδE > 0 be such thatφE(δE) = 1, or explicitly

δE =log(1 + E)

E.

Note thatδE → 0 asE → ∞. Let

uE(x) = φE(δE − a0 dist(x, K ))

in

�E= {x: dist(x, K ) < δE/a0}.

Assume for a moment thatK hasC2 regular strictly convex boundary, such thatthe mean curvatureκ(x) ≥ κ1 > 0 on ∂K . Then we claim that

(uE, �E) is a classical supersolution of (FBp) for 1< p < p(E).(11.1)


The free boundary conditions are easily verified, hence we need to prove only that1puE

≤ 0. Computing as in the proof of Theorem 3.3, we obtain

|∇uE|2−p1puE

= φ′

Ea0((p − 1)a0E − (n − 1)κ),

whereκ is the mean curvature at a point on the level line{ dist(x, K ) = s} with0 < s < δE/a0. Thenκ ≥ κ1/(1 + sκ1) and consequently, ifE is so large thatδE/a0 ≤ 1,

κ ≥ κ2 = κ1/(1 + κ1).

Hence1puE≤ 0 as soon as

(p − 1)a0E − (n − 1)κ2 ≤ 0

or equivalently

p < 1 +(n − 1)κ2

a0E.

Thus (11.1) and consequently the theorem follow in the case whenK has a strictlyconvexC2 regular boundary. For generalK we can findKη ⊃ K that converges toK asη → 0, with the properties as above, so that we can concludelim p→1+�p ⊂

Kη. Then lettingη → 0 we complete the proof of the theorem.

APPENDIX: CLASSICAL SOLUTIONS OF ( FBp)

The main objective of this appendix is to outline the main steps in the proofof Theorem 2.3 due to A. Henrot and H. Shahgholian [HS3]. We will make thesame assumptions on the compactK and functiona(x) in K c

= Rn\ K as in the

beginning of Section 2.Let E∗

= E∗(K ,a(x)) be the subclass of all classical supersolutions(u, �) of(FBp) (see Definition 2.1) with convex support�.

Consider the intersection

�∗=

⋂(u,�)∈E∗

�

and setu∗ to be thep-capacitary potential of the convex ring�∗\ K

1pu∗= 0 in �∗

\ K , u∗= 0 on ∂�, u∗

= 1 on K .

In order for this definition to have a meaning one needs to have

E∗6= ∅ and �∗

⊃⊃ K .(∗)


Using the assumptionsa(x) ≥ a0 > 0 in K c and the uniform interior ball con-dition for K , we can construct explicit (radially symmetric) supersolutions andsubsolutions to guarantee (∗).

Then the following assertion holds.

Proposition A.1. (u∗, �∗) is the classical solution of (FBp).

Here is the sketch of the proof.

Step 1.Prove that(u∗, �∗) is a classical supersolution of (FBP). This is relativelyeasy to show, using thatv = |∇u|

p is a subsolution of the linearizedp-Laplacian

Lu(v) = |∇u|p−21v + (p − 2)|∇u|

p−4∇

2v∇u · ∇u ≥ 0,

providedu is p-harmonic.

Step 2.Prove thatlim sup|∇u∗(y)| = a(x) asy → x, y ∈ �∗, for everyextremepoint x ∈ ∂�∗. The latter means thatx is not a convex combination of points on∂�∗ other thanx. The pointsx ∈ ∂�∗ with a hyperplane5 touching∂�∗ atx onlyare dense among all extreme points, so it is enough to prove the statement only forthem. If in an arbitrary small neighborhood of suchx we have|∇u∗

| < a, then wecan cut from�∗ a small cap by a parallel translation of the supporting hyperplane5, thus constructing a new supersolution; see [HS2, Lemma 3.4]. This, however,will contradict the minimality of�∗.

Step 3.Observe thatStep 2impliesC1 regularity of∂�∗. Indeed, letx ∈ ∂�∗ withtwo supporting hyperplanes. Then by barrier arguments, we will have|∇u∗(y)| →

0 asy → x, a contradiction.

Step 4.HavingC1 regularity of∂�∗, one can show that in factlim inf |∇u∗(y)| =

a(x) asy → x, y ∈ �∗, for every extreme pointx ∈ ∂�∗. This can be shown,e.g. by the technique used in Section 5 of this paper.

Step 5.Now we have that the free boundary condition is satisfied for all points on∂�∗ except those contained in line segments` ⊂ ∂�∗. To fill these points we usethe assumption that1/a(x) is concave along with the following nice observationdue to P. Laurence and E. Stredulinsky [LS].

Lemma A.2. Let D be a convex domain withC1 regular boundary∂D and` ⊂ ∂D a line segment. Let alsou be a nonnegativeC1 function with convex levelsets, defined inD and continuously vanishing on∂D. Then1/|∇u| is convex on`.

This completes the proof of the proposition and hence Theorem 2.3 follows.

ACKNOWLEDGMENTAuthors are grateful to Institut Mittag-Leffler for its hospitality during the preparation of this paper.

The research of the third author was supported by the Swedish Natural Science Research Council.


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