Convex Configurations In Free Boundary Problems - Purdue Math

VK Convex Configurations In Free Boundary Problems

Arshak Petrosyan

DOCTORAL THESIS, 2000 DEPARTMENT OF MATHEMATICS ROYAL INSTITUTE OF TECHNOLOGY STOCKHOLM, SWEDEN

Akademisk avhandling som med tillst˚and av Kungl Tekniska Högskolan framlägges till offentlig granskning för avläggande av filosofie doktorsexamen i matematik torsdagen den 25 maj kl 13.00 i Kollegiesalen, Administrationsbyggnaden, Kungl Tekniska Högskolan, Valhallavägen 79, Stockholm.

c 2000 Arshak Petrosyan

ISBN 91-7170-573-2 Högskoletryckeriet, KTH, 2000

ABSTRACT The thesis consists of the following three papers on free boundary problems for parabolic and elliptic partial differential equations with certain convexity assumptions 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, describing the propagation of flames. Suppose there is a bounded domain ⊂ Q T = Rn × (0, T ) for some T > 0 and a function u > 0 in such that

(P)

 in  1u − u t = 0 u = 0, |∇u| = 1 on 0: = ∂ ∩ Q T  u(·, 0) = u 0 on 0 ,

where 0 is a given domain in Rn and u 0 is a positive and continuous function in 0 , vanishing on ∂0 . If 0 is convex and u 0 is concave in 0 , then we show that (u, ) is unique and the time sections t are convex for every t ∈ (0, T ), provided the free boundary 0 is locally the graph of a Lipschitz function and the fixed gradient condition is understood in the classical sense. On existence and uniqueness in a free boundary problem from combustion

with L. Caffarelli We continue the study of problem (P) above under certain geometric assumptions on the initial data. The problem arises in the limit as ε → 0 of a singular perturbation problem (Pε )

1u ε − u εt = βε (u ε ) u ε (·, 0) = u ε0

in Q T , on Rn ,

where R εβε (s) = (1/ε)β(s/ε) is a nonnegative Lipschitz function, supp βε = [0, ε] and 0 βε (s)ds = 1/2. Generally, no uniqueness of limit solutions can be expected. However, if the initial data is starshaped, we show that the limit solution is unique and coincides with the minimal classical supersolution. In the case when 0 is convex and u 0 is log-concave and satisfies the condition −M ≤ 1u 0 ≤ 0, we prove that the minimal supersolution is a classical solution of the free boundary problem for a short time interval.

A free boundary problem for ∞-Laplace equation

with J. Manfredi and H. Shahgholian We consider a free boundary problem for the p-Laplacian 1 p u = div(|∇u| p−2 ∇u), describing the nonlinear potential flow past convex profile K with prescribed pressure gradient |∇u(x)| = a(x) on the free stream line. The main purpose of this paper is to study the limit as p → ∞ of the classical solutions of the problem above, existing under certain convexity assumptions on a(x). We show, as one can expect, that the limit solves the corresponding problem for the ∞-Laplacian 1∞ u = ∇ 2 u∇u · ∇u, in a certain weak sense, strong however, to guarantee the uniqueness. We show also that in the special case a(x) ≡ a0 > 0 the limit coincides with an explicit solution, given by a distance function.

2000 Mathematics Subject Classification: Primary 35R35, 35K05, 35J60 Key Words: Free boundary problems, convexity, classical solutions, the heat equation, p-Laplacian, ∞-Laplacian, the propagation of flames, nonlinear potential flow.

ACKNOWLEDGMENTS I wish to express my sincere thanks to my supervisor Henrik Shahgholian, who introduced me to the subject and guided during the work on this thesis. I also thank Björn Gustafsson, who acted as my second supervisor, for a number of useful discussions and valuable advices. My special thanks are to Luis Caffarelli for inviting me to the University of Texas at Austin in April 1999 and February 2000. Without this the second paper would not exist. I also thank Juan Manfredi for productive discussions on the mysterious infinite Laplacian in Institut Mittag-Leffler, Stockholm; this resulted in a paper, included in the thesis. I gratefully acknowledge the financial support from the Swedish Institute for the academic year 98/99 and the Department of Mathematics of the Royal Institute of Technology for the year 99/00. The latter was actually my second home during these two years in Stockholm. As an undergraduate student I was studying at Yerevan State University and I thank all my teachers and especially my diploma advisor Norair Arakelian for teaching me mathematics. And last but not least, I thank my family and all my friends for their constant support and encouragement. Your contribution cannot be overestimated. Stockholm, April 2000 Arshak Petrosyan

CONTENTS Introduction. Preliminaries and Overview The propagation of flame . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .1 Log-concavity in space . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3 Convex classical solutions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4 A free boundary problem for ∞-Laplacian . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5 References . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7 Paper I. Convexity and uniqueness in a free boundary problem arising in combustion theory 1. 2. 3. 4.

Introduction and main result . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1 Convexity of level sets . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3 On caloric functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5 Proof of the main theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6 References . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9

Paper II. On existence and uniqueness in a free boundary problem from combustion, with L. Caffarelli 1. 2. 3. 4. 5. 6. 7. 8.

Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1 Uniqueness in the starshaped case . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3 The convex case . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7 Lipschitz regularity in time . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9 Some technical lemmas . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11 The minimal element of B . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 13 Further properties of the minimal element . . . . . . . . . . . . . . . . . . . . . . . . . . . 14 The classical solution of P for short time . . . . . . . . . . . . . . . . . . . . . . . . . . . . 17 References . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 22

Paper III. A free boundary problem for ∞-Laplace equation, with J. Manfredi and H. Shahgholian 1. Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1 2. Preliminaries . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3 3. Classical solutions for a(x) ≡ a0 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7

viii 4. Weak solutions for general a(x) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10 5. (u ∞ , ∞ ) is a weak subsolution . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 12 6. Uniform gradient bound for u p as p → ∞ . . . . . . . . . . . . . . . . . . . . . . . . . . 14 7. On ∞-harmonic functions near singular convex boundary points . . . . . . . 15 8. C 1 regularity of ∂∞ . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 19 9. Stable solutions of the boundary value problem . . . . . . . . . . . . . . . . . . . . . . 20 10. (u ∞ , ∞ ) is a weak supersolution . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 22 11. The limit as p → 1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 26 Appendix: Classical solutions of (F B p ) . . . . . . . . . . . . . . . . . . . . . . . . . . . . 27 References . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 29

Introduction

Introduction. Preliminaries and Overview

THE PROPAGATION OF FLAME Free boundaries, also known as interfaces and moving boundaries, are a priori unknown sets, arising in solutions of nonlinear partial differential equations (PDE.) An example is the front of a curved premixed flame, which at time t is given as ∂(t), where (t) = {x ∈ Rn : u(x, t) > 0} is the unburnt zone. The function u, assumed to be nonnegative here, has the meaning u = λ(Tc − T ), where Tc is the temperature of the flame and λ is a normalization factor. We have thus (1)

1u − u t = 0 in = {u > 0}.

Besides, the following condition should be satisfied on the free boundary (2)

|∇u| = a

on ∂(t),

for t > 0, where a > 0 is a given constant, or in the general formulation of the problem, a function depending on x and t. To complete the conditions, we give the initial data (3)

u(·, 0) = u 0

in 0 ,

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 the rather complicated system of PDE, describing the process of combustion, in the limit of high activation energy. This method goes back to Zel’dovich and FrankKamenetski˘ı [ZF] in 1938. However, the rigorous mathematical study was postponed until recently. Mathematically, we consider the Cauchy problem in Q T = Rn × (0, T ) for a semilinear parabolic equation (Pε )

1u ε − u εt = βε (u ε ) u ε (·, 0) = u ε0 1

in Q T , on Rn ,

2

INTRODUCTION

flame front u≡0 |∇u| = a

u>0 (t) = unburnt zone 1u − u t = 0

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

where u ε0 approximate the initial data u 0 as ε → 0 and βε ≥ 0 are Lipschitz functions such that Z (4) support βε = [0, ε] , βε (s)ds = M. [0,ε]

The functions βε are normally constructed from a single function β by βε (s) =

1 s β . ε ε

The quantity ε > 0 has the meaning of 1/E with E the high activation energy in the corresponding combustion model. As proved by Caffarelli and Vazquez [CV], the family {u ε } is locally uniformly 1,1/2 bounded in Q T in C x,t norm and converges for a subsequence ε j → 0 to a function u, which solves (P) in a certain weak sense with the constant a in (2) given by √ a = 2M, where M is the constant in (4). To illustrate the passage to the limit ε → 0, let us consider the one-dimensional stationary case. This means that we have an ODE φxε x = βε (φ ε ) on R. Assuming φ ε , φxε ≥ 0 and φ ε , φxε → 0 as x → −∞, we multiply both sides of the equations by φxε and integrate from −∞ to x. We will obtain φxε =

p 2Bε (φ ε ),

3

INTRODUCTION

where Bε (s) =

Rs 0

βε (τ )dτ , so that  s≤0 =0 Bε (s) ∈ [0, M] , 0 ≤ s ≤ ε .  = M, ε≤s

√ Hence φ ε will become a linear function with the slope 2M as it reaches the ε ε ε level ε. If we normalize √ converge √ now φ by φ (0) = ε, the functions φ (x) will uniformly to φ(x) = 2M x+ , which is a solution of (P) with a = 2M. LOG-CONCAVITY IN SPACE Now we address the following question: how the geometry of 0 determines the 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 answer depends very much on the initial data u 0 as well. The main result of Paper I (Theorem 1) in this thesis says that, indeed, (t) will stay convex provided 0 is convex and u 0 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

as y → x ∈ ∂(t), y ∈ (t).

However, the technical assumption of Lipschitz regularity of ∂ in time, corresponding to the finiteness of the speed of the flame propagation, is also needed. The appearance of log-concavity of u 0 is related to the following fact, which is employed heavily in the proof of the above result: Suppose we have a domain ⊂ Q T = Rn × (0, T ) with convex time sections (t) = {x ∈ Rn : (x, t) ∈ } and u is a solution to the Dirichlet problem 1u − u t = 0 in ;

u = 0 on ∂ ∩ Q T ;

u(·, 0) = u 0 in 0 = (0),

where u 0 > 0 in 0 and continuously vanishes on ∂0 . Then log-concavity of u 0 implies the log-concavity of u(·, t) for every t ∈ [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 on the ideas of Brascamp and Lieb [BL]. Without loss of generality we may assume that has regular enough lateral boundary. For s > 0 denote by Ls = {u 0 > s} the level sets of the initial function u 0 . Then log-concavity of u 0 means that for 0 < λ < 1 (5)

Ls ⊃ (1 − λ)Ls0 + λLs1

whenever s = s01−λ s1λ .

Let now ω x,t (A) denote the caloric measure of a convex set A ⊂ 0 with respect to at (x, t) ∈ . Then for points x0 , x1 ∈ (t) and x = (1 − λ)x0 + λx1 ∈ (t)

4

INTRODUCTION

set f (s) = ω x0 ,t (Ls ),

g(s) = ω x1 ,t (Ls ),

and h(s) = ω x,t (Ls ).

The key step now is to prove the inequality (6)

h(s) ≥ f (s0 )1−λ g(s1 )λ ,

whenever s = s01−λ s1λ .

This follows from the Brunn-Minkowski type inequality for convex sets in the Gaussian space C([0, t] , Rn ) with the Wiener measure, if we use the connection between Brownian motion and the caloric measure ω x,t (A), which is the probability that the particle, starting its motion at x, will stay in (t − τ ) at the time τ and will hit a point in A at time t. Having (6), we conclude immediately from [Theorem 3.2, BL] that u(x, t) ≥ u(x0 , t)1−λ u(x1 , t)λ , since u(x0 , t) = || f || L 1 (R+ ) ,

u(x1 , t) = ||g|| L 1 (R+ )

and u(x, t) = ||h|| L 1 (R+ ) .

Hence u(·, t) is log-concave. CONVEX CLASSICAL SOLUTIONS We know already that the classical solutions of (P) will stay convex in space for convex initial domain 0 and log-concave u 0 , but we don’t know yet whether such 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 of such limits is not clear. However, if u 0 is starshaped, or more precisely if all the level sets Ls = {u 0 > s} are starshaped with respect to the same point (say the origin), then there is only one limit solution. In fact, one can show that this limit coincides 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), with certain geometric assumptions on the initial data, such as radial symmetry was proved by several authors, using different methods, see e.g. references in [Va]. In the case of problem (P) with log-concave initial u 0 and convex 0 , perhaps the best suited is the one due to Beurling [Be]. This was successfully applied by Henrot and Shahgholian [HS1] in the elliptic free boundary problem, closely related to (P). The idea, basicly, is to consider the intersection of all supersolutions with convex time sections and show that it is a classical solution of the problem. There are complications in our case connected with the regularity in time, but putting the extra condition −M ≤ 1u 0 ≤ 0 on u 0 , one can prove that a classical solution exists for a short time interval T = T (u 0 ), and that time levels (t) are convex and shrink in time. This is Theorem 8.1 of Paper II.

5

INTRODUCTION

Here we want to explain how the method of Henrot and Shahgholian works and at the same time to formulate a problem, which appears also in Paper III. Given is a convex compact set K in Rn with nonempty interior. The problem is to find a domain ⊃ K in Rn , and a continuous function u > 0 in , vanishing on ∂ such that (7) (8) (9)

1pu = 0 |∇u| = a u=1

in \ K on ∂ on K ,

where a = a(x) > 0 is a given function off K . Here for simplicity we assume a ≡ const here. We will refer to the system (7)–(9) as to problem (F B p ). The operator 1 p for p ∈ (1, ∞) is the p–Laplacian 1 p u = div(|∇u| p−2 ∇u). The reader not familiar with 1 p may think of the usual Laplacian, corresponding to p = 2, up to end of this section. In the next section, however, we are going to let p → ∞ in (F B p ). We call a pair (u, ) a supersolution of (F B p ) if u ∈ C(), u > 0 in , u = 0 on ∂, and 1 p u ≤ 0 in \ K ,

lim sup |∇u| ≤ a,

and u ≥ 1 on K ,

x→∂

cf. (7)–(9). If we consider now the family E ∗ of all supersolutions (u, ) of (F B p ) with convex , then one can show the existence of the minimal element (u ∗ , ∗ ) of E ∗ , by taking ∗ to be the intersection of all ’s. Also u ∗ is the solution of the Dirichlet problem in ∗ \ K , with boundary values 1 on ∂ K and 0 on ∂∗ . The trick of [HS1] is now, very roughly, that if (u ∗ , ∗ ) is not a classical solution of (F B p ), and say |∇u ∗ | < a in a neighborhood of a point x ∈ ∂∗ , then by the translation 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 a classical solution. Let us remark here that an essential ingredient in the method above is the convexity of level sets of capacitary potentials u in convex rings \ K , proved by Lewis [Le]. A FREE BOUNDARY PROBLEM FOR ∞-LAPLACIAN Problem (F B p ), formulated in the previous section, describes the nonlinear (power-law) potential flow past the profile K with fixed pressure gradient |∇u| = a

6

INTRODUCTION

|∇u| = a 1pu = 0 u≡1 K u≡0

FIG. 2. The problem (F B p )

on the free stream line. Besides the fluid mechanical application, the problem is of interest in electrochemical machining for shaping hard metals, see Lacey and Shillor [LS]. Also, as shown by Acker [Ac], (F B p ) appears in the shape optimization problem for the minimization of the heat flow. In the sequel we will assume that a(x) ≥ a0 > 0 and 1/a(x) is concave off K and that K is a convex compact, with regular enough boundary (specifically, it satisfies the uniform interior ball condition.) Then (F B p ) has a unique classical solution for every p ∈ (1, ∞), see [HS2]. We denote this solution by (u p , p ) to stress the dependence on p. An interesting thing that one can notice is that the domains p , which are convex, increase with p and all are contained in the set { dist (x, K ) < 1/a0 }. The similar thing happens with u p and the bound from above is the function 1 − a0 dist (x, K ). Hence, there is a limit (u ∞ , ∞ ) = lim (u p , p ). 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 (F B∞ ), formally obtained from (F B p ) by plugging p = ∞ in (7)–(9). The operator 1∞ is the ∞-Laplacian 1∞ u = ∇ 2 u∇u · ∇u, which is an elliptic operator, degenerate on all directions orthogonal to ∇u. By now it is very well known that the limits as p → ∞ of p-harmonic functions (i.e. weak solutions of 1 p u = 0) are viscosity solutions of 1∞ u = 0, see e.g. [BDM]. The latters are also known as ∞-harmonic functions. Generally, ∞harmonic functions are not C 2 regular and, surprisingly, nobody knows if they are C 1 . 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], are proved to be true. Before we go any further, let us focus on the case a(x) ≡ a0 in problem (F B∞ ). 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 shown in Theorem 3.3 of Paper III by constructing explicit subsolutions of (F B p ) for large p. For nonconstant function a(x), the analysis of the limit is much more complicated and the best we can prove is that condition (8) satisfied in the following weak sense (80 )

u ∞ (y) → a(x) as y → x ∈ ∂∞ , y ∈ ∞ . dist (y, ∂∞ )

One of the steps in the proof of (80 ) is the following fact. Suppose we have a 2-dimensional cone 0(θ) = {r eiα : |α| < θ} with 0 < θ < π/2. Then there is an ∞-harmonic barrier v > 0 in 0(θ ), vanishing on the sides of 0(θ ) and having the form v(r eiα ) = r 1+ε F(α) with ε = ε(θ ) > 0. This has the following consequence: If u(x) is an ∞-harmonic function in a convex domain , vanishing on ∂, then near singular boundary points x0 ∈ ∂ the function u(x) has the behavior u(x) = o(|x − x0 |) as x → x0 , x ∈ . This is true for any dimension n ≥ 2. Indeed, a boundary point x0 is singular iff there are two supporting hyperplanes to at x0 , which form an angle 2θ < π. Then using 2-dimensional barrier v in 0(θ ) as above in combination with Jensen’s maximum principle we prove the statement. Let us make the following observation concerning the regularity of barriers v in 2-dimensional cones. All barriers, constructed in Paper III (Section 7) are only C 1,1/3 regular, although they are real analytic off the symmetry axis α = 0. In fact, if the ∞-harmonic barrier is symmetric with respect to α = 0, it cannot be C 2 regular, otherwise it will be linear on the symmetry axis and will vanish identically. As a conclusion, let us consider the limit p → 1+ in the problem (F B p ) with the same assumptions on K and a(x). As expected, p shrinks to K as p → 1+, but only if the dimension n ≥ 2. In the dimension n = 1 all the problems (F B p ) are identical and hence so are the solutions (u p , p ) for every p ∈ [1, ∞]. REFERENCES [Ac]

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

8 [Be]

INTRODUCTION

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 as p → ∞ of 1 p u p = f and related extremal problems, Rend. Sem. Mat. Univ. Politec. Torino 1989, Special Issue, 15–68. [BL]

H. J. Brascamp and E. H. Lieb, On extensions of the Brunn-Minkowski and Prékopa-Leindler theorems, including inequalities for log concave functions, and with an application to the diffusion equation, J. Functional Analysis 22 (1976), no. 4, 366–389.

[CV]

L. A. Caffarelli and J. L. Vazquez, A free-boundary problem for the heat equation arising in flame 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 for the 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 with non-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 Equations 1995 (1995), no. 04, approx. 5 pp. (electronic)

[LS]

A. A. Lacey and M. Shillor, Electrochemical and electro-discharge machining with a threshold current, 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-Kamenetski˘ı, 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 arising in combustion theory Arshak Petrosyan* Department of Mathematics, Royal Institute of Technology, 100 44, Stockholm, Sweden E-mail: [email protected]

We consider solutions to a free boundary problem for the heat equation, describing the propagation of flames. Suppose there is a bounded domain ⊂ Q T = Rn × (0, T ) for some T > 0 and a function u > 0 in such that  in  u t = 1u u = 0, |∇u| = 1 on 0: = ∂ ∩ Q T  u(·, 0) = u 0 on 0 , where 0 is a given domain in Rn and u 0 is a positive and continuous function in 0 , vanishing on ∂0 . If 0 is convex and u 0 is concave in 0 , then we show that (u, ) is unique and the time sections t are convex for every t ∈ (0, T ), provided the free boundary 0 is locally the graph of a Lipschitz function and the fixed gradient condition is understood in 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 heat equation. Suppose there is a domain ⊂ Q T : = Rn × (0, T ) for some T > 0 and a positive smooth function u in such that (1) (2) (3)

u t = 1u u = 0, |∇u| = 1 u(·, 0) = u 0

in on 0 on 0 ,

where 0: = ∂ ∩ Q T is the (free) lateral boundary of , 0 ⊂ Rn is the initial domain and u 0 is a prescribed positive continuous function in 0 , that vanishes * The author was supported by the Swedish Institute 1

2

ARSHAK PETROSYAN

continuously on 00 : = ∂0 . Then we say the pair (u, ) or, when there is no ambiguity, to be a solution to problem (P). This problem, in mathematical framework, was introduced by L.A. Caffarelli and J.L. Vazquez [CV]. It describes propagation of so-called premixed equi-diffusional flames in the limit of high activation energy. In this problem the time sections (4)

t = {x ∈ Rn : (x, t) ∈ }

represent the unburnt (fresh) zone in time t, 0t : = ∂t corresponds to the flame front, and u = c(Tc − T ) is the normalized temperature. For further details in combustion theory we refer to paper [V] of J.L. Vazquez. The existence of weak solutions to problem (P) as well as their regularity under suitable conditions on the data were established in [CV]. However, we should not expect any uniqueness result unless we impose some special geometrical restrictions. In this paper we study the case when the initial domain 0 is bounded and convex, and the initial function u 0 is concave. Throughout the paper we make the following assumptions concerning solutions (u, ) to problem (P). First, the boundary of consists of three parts: (5)

∂ = 0 ∪ 0 ∪ T ,

where T is a nonvoid open set in the plane t = T . The presence of nonempty T excludes the extinction phenomenon in time t ∈ [0, T ]. This assumption is rather of technical character, that can be avoided with the following simple procedure. Consider the extinction time (6)

T = sup{t: t 6= ∅}.

Then every domain (τ ) = ∩ {0 < t < τ }, τ ∈ (0, T ), has nonempty “upper bound” τ . Therefore we can consider first (τ ) instead of and then let τ → T . Next, we assume that for every (x0 , t0 ) ∈ 0 there exists a neighborhood V in Rn × R such that (after a suitable rotation of x-axes) (7)

∩ V = {(x, t) = (x 0 , xn , t): xn > f (x 0 , t)} ∩ V ∩ Q T

where f is a Lipschitz function, defined in V 0 = {(x 0 , t): ∃xn with (x 0 , xn , t) ∈ V }. Further, for u we assume that it is continuous up to the boundary ∂ and can be extended smoothly through T . The gradient condition in (2) is understood in the classical sense (8)

lim

t 3y→x

|∇u(y, t)| = 1

for every x ∈ ∂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 and u 0 be a concave function in 0 . Suppose that (u, ) is a solution to this problem in the sense described above. Then (u, ) is a unique solution. Moreover, the time sections t of are convex for every t ∈ (0, T ).

The plan of the paper is as follows. In Section 2 we prove a theorem on the convexity of level sets of solutions to a related Dirichlet problem. In Section 3 we recall some properties of caloric functions in Lipschitz domains. And finally in Section 4 we prove Theorem 1. 2. CONVEXITY OF LEVEL SETS In this section we establish some auxiliary results, which are, however, of independent interest. Let u 0 and 0 be as in problem (P) and a domain ⊂ Q T meets conditions (5) and (7). Then by the Petrowski criterion [P] is a regular domain for the Dirichlet problem for the heat equation (in the Perron sense), and its parabolic boundary is given by ∂ p = 0 ∪ 0.

(9)

We fix one such domain and denote by u the solution to the Dirichlet problem (10)

u t = 1u,

u = u0

on 0 ,

u = 0 on 0.

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

(11)

Ls (u(·, t)) = {x ∈ t : u(x, t) > s}

are convex for every fixed s > 0 and t ∈ (0, T ), where u is the solution to the Dirichlet problem (10). The proof is based on the Concavity maximum principle originally due to N. Korevaar [K1] and [K2]. For a function v on set v(x, t) + v(y, t) x+y C(x, y, t) = (12) −v ,t . 2 2 The function C is defined on an open subset D of the fiber product e = {(x, y, t): (x, t), (y, t) ∈ }.

4

ARSHAK PETROSYAN

e if the time sections of are convex. Note also that if v is Note that D = extended to the “upper bound” T of , then C is extended to the “upper bound” DT of D. Lemma 3 (Concavity maximum principle).

2,1 Let v ∈ C x,t () ∩ C( ∪ T )

satisfy to a parabolic equation (13)

vt = a i j (t, ∇v)vi j + b(t, x, v, ∇v)

in

with smooth coefficients, and such that b is nonincreasing in v and jointly concave in (x, v). Then the function C cannot admit its positive maximum at any point of D ∪ 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 of b 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 necessary to assume also the nonnegativeness of vt ; see A. Kennington [Ke]. Proof of Theorem 2. Assume first, that the functions f in the local representations (7) of are smooth in (x 0 , t) and strictly convex in x 0 and that u 0 is smooth. These assumptions imply the smoothness up to ∂ of the solution u to (10). Also, the positivity of u 0 implies the positivity of u. Define now v = log(u). We claim then that v(·, t) are concave functions in t for every t ∈ (0, T ]. Clearly, this will imply the statement of the theorem. For this purpose, we consider the concavity function C, defined above, and show that C ≤ 0 on D ∪ DT . Suppose the contrary. Then take a maximizing sequence (xk , yk , tk ) ∈ D ∪ DT such that (14)

lim C(xk , yk , tk ) = sup C > 0. D∪DT

Without loss of generality we may assume that there exists limit (x0 , y0 , t0 ) = lim (xk , yk , tk ). Direct calculation shows, that v satisfies (15)

vt = 1v + |∇v|2 ,

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 case t0 = 0


5

and x0 , y0 ∈ 0 is impossible, since v(·, 0) = log(u 0 ) is concave in 0 . Next, x0 ∈ 0t0 but y0 6= x0 is also excluded by the strict convexity of t ’s, since then C(xk , yk , tk ) → −∞. So, it remains to consider the last case x0 = y0 ∈ 0t0 . We observe now that by the boundary point lemma, the outward spatial normal derivatives u ν < 0 on 0 ∪ 0T . Besides, u ν < 0 also on 00 since u 0 is concave and positive in 0 and vanishes on 00 . Therefore we can carry out the same reasonings as in [CS, Proof of Lemma 3.1] (see also [GP, Lemma 3.2]) to obtain that lim inf C(xk , yk , tk ) < 0, which contradicts (14). Therefore C ≤ 0 in D ∪ DT and v(·, t) is concave in t for every t ∈ (0, T ]. This proves the theorem in the considering case. To prove the theorem in the general case, we use approximation of by domains with smooth lateral boundary and with strictly convex time sections, and a relevant smooth concave approximations of u 0 . 3. ON CALORIC FUNCTIONS In this section we recall some properties of caloric functions in Lipschitz domains. They will be used in the next section, where we prove Theorem 1. The main reference here is the paper [ACS] by I. Athanasopoulos, L. Caffarelli and S. Salsa. As in the previous section we consider a domain , satisfying conditions (5) and (7). Let also u be the solution to (10). Consider a neighborhood V of a point (x0 , t0 ) ∈ 0, where (7) holds. The function u vanishes on 0 ∩ V , is positive and satisfies the heat equation in ∩ V . In other words, u is caloric. We start with the following lemma from [ACS], which states that a caloric function u is “almost harmonic” in time sections near the lateral boundary 0. Lemma 4 ([ACS, Lemma 5]). There exist ε > 0 and a neighborhood Q of the point (x0 , t0 ) ∈ 0 such that the functions

(16)

w+ = u + u 1+ε ,

w− = u − u 1+ε

are respectively sub- and superharmonic in Q ∩ ∩ {t = t0 }. We will need also the following lemma on asymptotic development of u near the boundary point (x0 , t0 ). Lemma 5 ([ACS, Lemma 6]). Suppose there exists an n-dimensional ball B ⊂ c ∩ {t = t0 } such that B ∩ 0 = {(x0 , t0 )}. Then near x0 in t0

(17)

u(x, t0 ) = α(x − x0 , ν)+ + o(|x − x0 |)

for some α ∈ [0, ∞) and where ν denotes the outward radial direction of B at (x0 , t0 ).

6

ARSHAK PETROSYAN

In the next lemma we show that α in (17) is in fact the nontangential limit of |∇u(y, t0 )| as y → x0 . Lemma 6. Under the conditions of Lemma 5, let also K ⊂ t0 be an ndimensional truncated cone with the vertex at (x0 , t0 ) such that |x − x0 | ≤ c1 dist(x, 0t0 ) for every x ∈ K and some constant c1 . Then

(18)

lim

K 3y→x0

∇u(y, t0 ) = αν,

where α and ν are as in the asymptotic development (17). Proof. By [ACS, Corollary 4], there exists a neighborhood V of the point (x0 , t0 ) such that (19)

|u t (x, t)| ≤ c2

u(x, t) , dx,t

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

for all (x, t) ∈ V ∩ . Take an arbitrary sequence yk → x0 , yk ∈ K , and consider the functions (20)

vk (z) = u(yk + rk z, t0 )/rk ,

rk = |yk − x0 |,

defined on the ball B = B(0, ρ), ρ = 1/(2c1 ). Using (17) and (19), we obtain that for large k (21)

|vk (z)| < (α + 1)(1 + ρ)

and (22)

|1vk (z)| = rk |1u(yk + rk z, t0 )| = rk |u t (yk + rk z, t0 )| ≤ 2c1 c2

uniformly in B. Then C 1,β norms of vk are locally uniformly bounded in B for a β ∈ (0, 1); see e.g. [LU]. Therefore a subsequence of vk converges locally in C 1 norm to a function v0 in B. We may also assume that over this subsequence there exists e0 = lim ek , where ek = (yk − x0 )/|yk − x0 |. Then, using (17), we can compute that v0 (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 sequence yk → 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 Theorem 1. Denote by ∗ the spatial convex hull of , in the sense that the time


7

sections ∗t are the convex hulls of t for every t ∈ (0, T ). Since is assumed to satisfy (5) and (7), ∗ will also satisfy similar conditions. In particular, we may apply the results of two previous sections to ∗ . The lateral boundary of ∗ will be denoted by 0 ∗ 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 every x 0 ∈ 0t∗0 , 0 < t0 ≤ T ,

(23)

lim inf |∇u ∗ (y, t0 )| ≥ 1.

∗t 3y→x0 0

Proof. From Lemma 4 it follows that there are ε and s0 such that the function w+ (y) = u ∗ (y, t0 ) + u ∗(1+ε) (y, t0 ) is subharmonic in the ringshaped domain {u ∗ (·, t0 ) < s0 }. Let now y ∈ ∗t0 and u ∗ (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 in Rn to L∗s at the point y. After a suitable translation and rotation in spatial variable we may assume that y = 0, the supporting plane is x1 = 0, and L∗s ⊂ {x1 < 0}. Let x ∗ ∈ ∂∗t0 have the maximal positive x1 coordinate. Since ∗t0 is the convex hull of t0 , there must be x ∗ ∈ ∂∗t0 ∩ ∂t0 . Take now β ∈ (0, 1) and consider a function v(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. Note that the maximum can be admitted either at x ∗ or at y = 0. We show that the former case cannot occur. Indeed, the plane x1 = x1∗ is supporting to the convex set ∗t0 and therefore there exists a ball B ⊂ ∗t0 c ⊂ ct0 , “touching” both boundaries ∂∗t0 and ∂to at x ∗ and with the outward radial direction ν = −e1 = (−1, 0, . . . , 0). Therefore from Lemma 5 we will have the following asymptotic developments for u and u ∗ near x ∗ in t0 and ∗t0 respectively: (24) (25)

u(x, t0 ) = α(x1∗ − x1 )+ + o(|x − x ∗ |) u ∗ (x, t0 ) = α ∗ (x1∗ − x1 )+ + o(|x − x ∗ |).

Since (8) is satisfied at the point x ∗ , we conclude by Lemma 6 that α = 1. Next, u ∗ ≥ u in and hence α ∗ ≥ α = 1. Observe now that w+ admits the same representation as (25). Hence for the function v(x) introduced above (26)

v(x) = (α ∗ − β)(x1∗ − x1 ) + βx1∗ + o(|x − x ∗ |).

Let now ν 0 be a spatial unit vector with (ν 0 , e1 ) < 0 such that x ∗ + hν 0 ∈ t0 for small h > 0. The existence of such a ν 0 follows from the local representation of

8

ARSHAK PETROSYAN

∂t0 as the graph of a Lipschitz function. Then v(x ∗ + hν 0 ) > v(x ∗ ) by (26) and consequently v has no maximum at x ∗ . Therefore v admits its maximum at the origin y = 0. Hence (27)

w+ (0) − w+ (he1 ) βh − 0 ≥ lim = β. h→0+ h→0+ h h

|∇w+ (0)| = lim

Letting β → 1 we obtain that |∇w+ (y)| ≥ 1, provided u ∗ (y, t0 ) < s0 . Now observe that ∇w+ = (1 + (1 + ε)u ∗ε )∇u ∗ . This proves the lemma. Proof of Theorem 1. Prove first that the domain coincides with its spatial convex hull ∗ , studied above. For this purpose we apply the Lavrentiev principle. As a reference point we take xmax ∈ 0 , a maximum point for the initial function u 0 . Without loss of generality we may assume that xmax = 0. Since u 0 is concave, u 0 (λx) ≤ u 0 (x)

(28)

for every λ ≥ 1 and x ∈ 0 (λ) = λ−1 0 . For λ ≥ 1 define u ∗λ (x, t) = u ∗ (λx, λ2 t)

(29)

in ∗ (λ) = {(x, t): (λx, λ2 t) ∈ ∗ }. 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 of u ∗λ0 and u on the parabolic boundary ∂ p ∗ (λ0 ) (see (28)), we obtain that u ∗λ0 ≤ u in ∗ (λ0 ). Let now ν be the normal vector of a supporting plane in Rn to the convex domain ∗ (λ0 )t0 at the point x0 , pointing into ∗ (λ0 )t0 . From Lemmas 5, 6 and 7 and the definition of u ∗λ we conclude that ∇u ∗λ0 (x0 + hν, t0 ) → λ0 α ∗ ν with α ∗ ≥ 1, as h → 0+. From elementary calculus there exists θ ∈ (0, 1) such that (31)

(∂/∂ν)u(x0 + θ hν, t0 ) u(x0 + hν, t0 ) = ∗ ≥1 (∂/∂ν)u ∗λ0 (x0 + θ hν, t0 ) u λ0 (x0 + hν, t0 )

and hence (32)

lim sup

t0 3y→x0

∂ ∂ ∗ u(y, t0 ) ≥ lim u λ0 (x0 + hν) = λ0 α ∗ > 1, h→0+ ∂ν ∂ν

which violates condition (8) at the point (x0 , t0 ). Therefore ∗ = , i.e. the time sections t are convex, for every t ∈ (0, T ).


9

It remains to prove the uniqueness of . For this we make the following observation. Let 0 be another solution. Then if everywhere in the proof of inclusion ∗ ⊂ above we replace ∗ by (0 )∗ , but leave unchanged, we will obtain that (0 )∗ ⊂ . Since and 0 are interchangeable, also we will have ∗ ⊂ 0 . Therefore 0 = and the proof of Theorem 1 is completed. ACKNOWLEDGMENT The 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 the regularity 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 in flame 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 variational problems, 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 boundary condition, 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, Academic Press, 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 problem from combustion Luis Caffarelli Department of Mathematics, University of Texas at Austin, Austin, Texas 78712, USA E-mail: [email protected]

and Arshak Petrosyan Department of Mathematics, Royal Institute of Technology, 100 44 Stockholm, Sweden E-mail: [email protected]

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

Key Words: parabolic free boundary problem, singular perturbation, limit solution, classical solution, uniqueness, existence.

1. INTRODUCTION In this paper we consider a free boundary problem for the heat equation, that consists of finding a nonnegative continuous function u in Q T = Rn × (0, T ), T > 0, such that

(P)

  1u − u t = 0 in = {u > 0} |∇u| = 1 on ∂ ∩ Q T , and  u(·, 0) = u 0 ,

with given nonnegative initial function u 0 ∈ C0 (Rn ). (Here 1 = 1x and ∇ = ∇x .) The problem P arises in modeling the propagation of laminar flames as the limit of the singular perturbation problem (see [CV] and, for further detail in combustion 1

2

L. CAFFARELLI AND A. PETROSYAN

theory, [Va]) (Pε )

1u ε − u εt = βε (u ε ) u ε (·, 0) = u 0,ε ,

as ε → 0+, where u 0,ε approximate u 0 in a proper way, βε ≥ 0, βε (s) = (1/ε)β(s/ε), with β a Lipschitz function, support β = [0, 1], and 1

Z (1.1) 0

β(s)ds =

1 . 2

The family of solutions {u ε } is uniformly bounded in C x,t -norm on compact subsets of Q T and every locally uniform limit u = lim j→∞ u ε j in Q T of its subsequence with ε j → 0 is a solution of P in a certain weak sense, see [CV]. We will refer to these limits as limit solutions of P. In the two-phase case, i.e. when there is no sign restriction on u, problems Pε approximate a free boundary problem, where the fixed gradient condition |∇u| = 1 is replaced by the gradient jump condition |∇u + |2 − |∇u − |2 = 1. Limit solutions of this problem were studied detailly from the local point of view in [CLW1], [CLW2]. It was shown that a limit solution u of P is a viscosity solution in a domain D if {u = 0}◦ ∩ D = ∅. For the one-phase problem, and recently for the two-phase problem as well, three concepts of solutions, limit, viscosity and classical, were shown to agree with each other to produce a unique solution under certain conditions on u 0 (see [LVW]), that guarantee the existence of the classical solutions. In this paper we provide another uniqueness theorem (Theorem 2.7) for limit solutions for starshaped u 0 , see (S) from Section 2. In fact, we prove that the unique limit solution coincides with the minimal classical supersolution of P (see Definition 2.1) in this case. The existence and analyticity of classical solutions of P, among other things, were proved in [GHV] in the case when the initial data is radially symmetric, by using the elliptic-parabolic approach. The classical solutions to free boundary problems, similar to P, were constructed in [Me] and [AG]. In this paper we prove a short time existence theorem of classical solutions of P in the so-called convex case (Theorem 8.1), when 0 = {u 0 > 0} is convex and u 0 is log-concave and superharmonic in 0 , along with other assumptions. We consider the minimal element among the classical supersolutions of P that have the following geometric property, expected from the classical solution: the time sections (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 enables us to apply a technique due to A. Henrot and H. Shahgholian [HS1], [HS2], to show that it is a classical solution of P. 1,1/2

A FREE BOUNDARY PROBLEM FROM COMBUSTION

3

2. UNIQUENESS IN THE STARSHAPED CASE Definition 2.1. A pair (u, ), where u is a continuous nonnegative function in Q T = Rn ×[0, T ], T > 0, and = {u > 0}, is called a (classical) supersolution of P if

(i) 1u − u t = 0 in ; (ii) lim sup3(y,s)→(x,t) |∇u(y, s)| ≤ 1 for every (x, t) ∈ ∂ ∩ Q T ; (iii) u(·, 0) ≥ u 0 . Respectively, a pair (u, ) is a subsolution of P if conditions (ii) and (iii) are satisfied with opposite inequality signs and lim inf instead of lim sup in (ii). A pair (u, ) is a classical solution of P if it is sub- and supersolution of P at the same time. Next, a supersolution (u, ) of P is a strict supersolution of P if there is a δ > 0 such that the stronger inequalities (ii’) lim sup3 (y,s)→(x,t) |∇u(y, s)| ≤ 1 − δ (iii’) u(·, 0) ≥ u 0 + δ on 0 = {u 0 > 0}

for every (x, t) ∈ ∂ ∩ Q T ;

hold. Analogously the strict subsolutions are defined.

Remark 2.2. According to [Theorem 6.1, CLW1], every limit solution u = lim j→∞ u ε j of P is its classical supersolution in the sense of Definition 2.1. Generally, for the two-phase problem we have that if u is a limit solution of P then (u + = max{u, 0} and u − = min{u, 0}) lim sup |∇u − (y, s)| ≤ γ

(y,s)→(x,t)

implies lim sup |∇u + (y, s)| ≤

(y,s)→(x,t)

q 1 + γ2

for every free boundary point (x, t) ∈ ∂{u > 0} ∩ Q T . When u ≥ 0, we have u − ≡ 0, hence one can take γ = 0.

Remark 2.3. Suppose that the initial function u 0 is starshaped with respect to a point x0 in the following sense: (S)

u 0 (λx + x0 ) ≥ u 0 (x + x0 )

for every λ ∈ (0, 1) and x ∈ Rn ,

or, equivalently, all the level sets Ls (u 0 ) = {u 0 > s} are starshaped with respect to the same point x0 . In the sequel, we will always assume x0 = 0.

4


Let (u, ) be a supersolution of P. Let λ and λ0 be two real numbers with 0 < λ < λ0 < 1. Define (2.1)

u λ (x, t) = (1/λ0 )u(λx, λ2 t)

in Q T /λ2 . The rescaling of variables is taken so that u λ , like u, satisfies the heat equation in its positivity set (2.2)

λ = {(x, t): (λx, λ2 t) ∈ }.

Moreover, the selection λ < λ0 < 1 makes the pair (u λ , λ ) not only a supersolution of P, but also a strict supersolution.

Lemma 2.4. Let the initial function u 0 satisfy condition (S). Then every subsolution of P is smaller than every supersolution of P.

In this lemma and further in the paper we say that a pair (u 0 , 0 ) is smaller than (u, ) if 0 ⊂ and u 0 ≤ u. Proof. Let (u, ) be a supersolution and (u 0 , 0 ) a subsolution of P in Q T . We need to prove only that 0 ⊂ ; the inequality u 0 ≤ u will follow from this inclusion by the maximum principle. In the case when u ∈ C 1 () and u 0 ∈ C 1 (0 ), the statement can be proved by the Lavrent’ev rescaling method as follows. Suppose (2.3)

λ0 = sup{λ ∈ (0, 1): 0 ⊂ λ } < 1,

where λ as in (2.2). Then 0 ⊂ λ0 and there is a common point (x0 , t0 ) ∈ ∂0 ∩ ∂λ0 ∩ Q T . Let λ0 < λ00 < 1 and u λ0 be as in (2.1). Then u 0 ≤ u λ0 in 0 . At the common point (x0 , t0 ) this inequality implies ∂ν u 0 (x0 , t0 ) ≤ ∂ν u λ0 (x0 , t0 ), where ν is the inward spatial normal vector for both ∂0 and ∂λ0 at (x0 , t0 ) (recall that we are in C 1 case.) This leads to a contradiction, since ∂ν u 0 (x0 , t0 ) = |∇u 0 (x0 , t0 )| ≥ 1 and ∂ν u λ0 (x0 , t0 ) = |∇u λ0 (x0 , t0 )| = λ0 < 1. Therefore λ0 = 1 and 0 ⊂ . The general case can be reduced to the considered regular case by the following e) be a subsolution. Choose 0 < λ < λ0 < 1 close to 1 and procedure. Let (e u, regularize e u by setting u(x, t) = (e u λ (x, t + h) − η)+ e0 ). Then we will for small h, η > 0. Analogously regularize a subsolution (ue0 , arrive in the considered regular case and can finish the proof by letting first h, η → 0+ and then λ → 1−. The following proposition opens a way to prove the uniqueness results for limit solutions of P.


5

Proposition 2.5. Let e u be a strict supersolution of P in Q T , T > 0, and u ε solutions of Pε , where u 0,ε are nonnegative uniform approximations of u 0 and support u 0,ε → support u 0 . Then

lim sup u ε (x, t) ≤ e u (x, t)

(2.4)

ε→0+

for every (x, t) ∈ Q T . Proof. Consider the ordinary differential equation ε φss (s) = γε (φ ε (s)),

(2.5)

where γε (s) = (1/ε)γ (s/ε) and γ is obtained from β in Pε by cβ(s) s ∈ [a, 1] (2.6) γ (s) = 0 s 6∈ [a, 1] . Here a ∈ (0, 1) (to be specified later) and the constant c > 1 is chosen so that Z (2.7)

1

γ (s)ds =

a

1 , 2

cf. (1.1). Let us consider the solution φ ε of (2.5), normalized by (2.8)

φ ε (s) = aε

for s ≤ 0,

and φ ε (s) > aε

for s > 0.

The family {φ ε } is recovered from a single function φ, a solution of φss = γ (φ) with the appropriate normalization, through the relation φ ε (s) = εφ(s/ε). Using this, we can find a constant C > 0 such that (2.9)

φ ε (s) = (s − Cε) + ε

for s ≥ Cε;

or in other words, that φ ε is a linear function with slope 1, for s ≥ Cε. Indeed, multiply both sides of (2.5) by φsε and integrate from −∞ to s. We will obtain (2.10)

(φsε )2 = 20ε (φ ε ),

Rs where 0ε (s) = a γε (σ )dσ . By (2.7) 0ε (s) = 1/2 for s ≥ ε and therefore φ ε will become linear function with slope 1 as it reaches ε. Now (2.9) follows with C satisfying φ(C) = 1. Moreover, (2.8)–(2.10) imply that (2.11)

(s − Cε) + ε ≤ φ ε (s) ≤ s + aε

for s ≥ 0.

6


Let now e u be a strict supersolution of P and consider the following regularization u(x, t) = (e u (x, t + h) − η)+ , for h, η > 0 small. Without loss of generality we may assume that there is δ ∈ (0, 1) such that (2.12)

|∇u| ≤ 1 − δ

if u < η

and u(·, 0) > u 0 + δ.

Consider now the compositions wε (x, t) = φ ε (u(x, t)) for sufficiently small ε > 0. Since φ ε (s) is constant for s ≤ 0, we can rewrite ˜ t + h) − η). Therefore wε is well-defined C 1,1 (Q T −h ) function, wε = φ ε (u(x, and satisfies ε 1wε − wtε = φsε (u) [1u − u t ] + φss (u)|∇u|2

in Q T −h . The first term in the right-hand side is 0 everywhere. If now the value of a in the definition (2.6) is chosen so that Z 1 1 β(s)ds = (1 − δ)2 , 2 a then the constant c in the same definition equals (1 − δ)−2 and we obtain ε 1wε − wtε = φss (u)|∇u|2 ≤ γε (φ ε (u))(1 − δ)2 ≤ βε (wε ),

if ε > 0 is so small that φ ε (s) ∈ support γε = [aε, ε] implies s < η, or explicitly ε < η/C. Besides, from (2.11), for small values of ε, wε (·, 0) = φ ε (u(·, 0)) ≥ u(·, 0) − Cε ≥ u 0,ε

2

1.5

1

0.5

-1

1

FIG. 1. Profile of φ(s)

2

3


7

and therefore wε is a C 1,1 (Q T −h ) supersolution of Pε . Using the comparison principle, we conclude that u ε (x, t) ≤ wε (x, t) for every (x, t) ∈ Q T −h . Passing to the limit as ε → 0+, we obtain lim sup u ε (x, t) ≤ lim w ε (x, t) = u(x, t) ε→0+

ε→0+

for every (x, t) ∈ Q T . Letting h, η → 0, we complete the proof of the proposition. Remark 2.6. One can formulate and prove a result analogous to Proposition 2.5 for strict subsolutions of P. Details are left to the reader. From Proposition 2.5 we derive now a uniqueness theorem in the starshaped case (S). Theorem 2.7. Let the initial function u 0 satisfy condition (S) for some x 0 ∈ Rn . If nonnegative u 0,ε approximate u 0 uniformly and support u 0,ε → support u 0 , then the limit solution of P is unique and coincides with the minimal supersolution of 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 assume x0 = 0. Suppose first that u is a classical supersolution of P. As was noted in Remark 2.3, u λ (x, t) = (1/λ0 )u(λx, λ2 t)

0 < λ < λ0 < 1

is a strict supersolution in Q T /λ2 ⊃ Q T , so that we can apply Proposition 2.5. Then, letting λ → 1−, we will arrive at (2.13)

lim sup u ε (x, t) ≤ u(x, t). ε→0+

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

3. THE CONVEX CASE The next sections of this paper will be devoted to the proof of the existence of classical solutions of P under the following convexity assumptions on data: (C1)

u 0 is superharmonic and log-concave in convex bounded 0 = {u 0 > 0}.

8


It is easily seen that (C1) implies (S) with x0 the maximum point of u 0 . In the sequel we will always assume that x0 = 0. As it follows from Lemma 2.4 and Theorem 2.7, a classical solution of P, if exists, coincides with the minimal supersolution, and therefore with the only limit solution 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: (3.1)

(t) is convex and shrinks in time for t ∈ [0, T ] ,

cf. [Theorem 1.1, Pe] and [CV]. Definition 3.1. A supersolution (u, ) of P in Q T is said to be in class B if satisfies (3.1) and moreover ∂ ∩ Q T is Lipschitz regular in time.

The Lipschitz regularity in time is understood in the following sense: for every (x0 , t0 ) ∈ ∂ ∩ Q T there exists a neighborhood V such that (3.2)

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

for a suitable spatial coordinate system and where f is a globally defined function, uniformly Lipschitz in time. We point out that in spatial coordinates f can be chosen to be convex, if time sections (t) are convex. If the class B just defined has a minimal element, then it is a good candidate for a classical solution of P. This idea goes back to Beurling’s celebrated paper [Be]. We set \ (3.3) ∗ = (u,)∈B

and let also u ∗ be a solution to the Dirichlet problem (3.4)

1u ∗ − (u ∗ )t = 0 in ∗ ;

u ∗ = 0 on ∂∗ ∩ Q T ;

u ∗ (·, 0) = u 0 .

We can show that under some additional conditions on u 0 and for small T ≤ T (u 0 ) the pair (u ∗ , ∗ ) is the minimal element of B and indeed a classical solution of P. The conditions on u 0 are as follows. First (C2)

u 0 ∈ C 0,1 (0 ) and

lim

0 3x→∂0

|∇u 0 (x)| = 1;

and, next, there exists a constant M > 0 such that (C3)

1u 0 (x) + M(u 0 (x) − ∇u 0 (x) · x) ≥ 0 for every x ∈ 0 .

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

1u 0 (x) ≥ −M 0

for every x ∈ 0 .


9

for certain M 0 > 0. Indeed, (C3’) readily follows from (C3) in view of the boundedness of w(x) = u 0 (x) − ∇u 0 (x) · x in 0 . Suppose now that (C3’) holds. We have w(x) ≥ u 0 (x) > 0 in 0 and lim inf w(x) ≥ lim inf −∇u 0 (x) · x ≥ ρ,

0 3x→∂0

0 3x→∂0

where ρ > 0 is the diameter of the circle, centered at x0 = 0 and inscribed in 0 . Hence, there is an ε > 0 such that w(x) ≥ ε in 0 and (C3) follows with M = M 0 /ε. Remark 3.2. Before we proceed, we check that the class B is not empty and that so is ∗ . For the latter it suffices to show the existence of a subsolution of P, in view of the comparison principle (Lemma 2.4.) Indeed, let u 0 satisfies (C1)–(C3). Consider the functions (3.5)

u(x, t) = u 0 (x)

and (3.6)

√ v(x, t) = 1 − 2Mtu 0

x

√ 1 − 2Mt

in Q T . We claim that the first is a super- and the second is a subsolution of P in Q T for T ≤ 1/(2M), but in a sense a little bit different from Definition 2.1, which will not affect on the comparison principle. The matter is that (3.5) and (3.6) are not caloric in their positivity sets but super- and subcaloric respectively. Indeed, for (3.5) this follows from superharmonicity of u 0 in 0 and for (3.6) from the direct computation 1v(x, t) − vt (x, t) = (1 − 2Mt)−1/2 (1u 0 (ξ ) + M(u 0 (ξ ) − ∇u 0 (ξ ) · ξ )) ≥ 0, √ for ξ = x/ 1 − 2Mt ∈ 0 and t ∈ [0, 1/(2M)]. The gradient condition for both functions follows from (C2).

4. LIPSCHITZ REGULARITY IN TIME The following lemma plays one of the fundamental roles in our study. Lemma 4.1. Let u 0 satisfy (C1)–(C3) and let ∗ be given by (3.3). Then

∂∗ ∩ Q 1/(2M) is Lipschitz regular in time.

10


Proof. Let (u, ) ∈ B. For small ε, h > 0 let us define w(x, t) =

1 u((1 − ε)x, (1 − ε)2 (t + h)) 1−ε

in Q (1−ε)−2 T −h , that is we first stretch a little and then drop it down and thus obtain 1−ε,h = {(x, t): ((1 − ε)x, (1 − ε)2 (t + h) ∈ }. Clearly, a pair (w, 1−ε,h ) again will lie in B if the condition w(·, 0) ≥ u 0

(4.1)

is verified. In view of Remark 3.2 and Lemma 2.4 w(x, 0) =

1 1 u((1 − ε)x, (1 − ε)2 h) ≥ v((1 − ε)x, (1 − ε)2 h) 1−ε 1−ε p 1 − 2M(1 − ε)2 h 1−ε = u0 p x = u 0 (x) 1−ε 1 − 2M(1 − ε)2 h

if we choose (4.2)

h=

2ε − ε2 . 2M(1 − ε)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 (t) = 1−ε,h 1−ε

t − h . (1 − ε)2

Running over all (u, ) ∈ B, we may conclude therefore that (4.3)

1 ∗ (t) ⊃ ∗ 1−ε

t −h . (1 − ε)2

Since ∗ (t) shrinks in time, the inclusion (4.3) is not trivial if t − h < t, (1 − ε)2 for certain small ε, h > 0, given by (4.2). The latter is equivalent to to the inequality t < 1/(2M). Besides, for these t, (4.3) implies also the Lipschitz regularity of ∂∗ in time variable.


11

Remark 4.2. Let u ∗ be a solution of (3.4) for T < 1/(2M) and continue the line of reasonings from the previous lemma. For small ε, h > 0, given by (4.2), the following inequality will be satisfied 1 u ∗ ((1 − ε)x, (1 − ε)2 (t + h)) − u ∗ (x, t) ≥ 0. 1−ε Let now ε go to 0. We will obtain u ∗ (x, t) − ∇u ∗ (x, t) · x + (1/M − 2t)u ∗t (x, t) ≥ 0 in ∗ ∩ Q 1/(2M) . The reader can see the similarity of this condition with (C3). In particular, we have that u ∗t is bounded from below in every ∗ ∩ Q T with T < 1/(2M).

5. SOME TECHNICAL LEMMAS For supersolutions (u, ) ∈ B of P one can replace the gradient condition (ii) in Definition 2.1 with u(x, t) ≤1 3(x,t)→(x0 ,t0 ) d (x, t) lim sup

for every (x0 , t0 ) ∈ ∂ ∩ Q T , where d (x, t) = distance(x, ∂(t)). This is taken care of in the next lemma. Lemma 5.1. Let be a bounded domain in Q T such that (t) are convex for t ∈ (0, T ) and ∂ ∩ Q T is Lipschitz in time. Let also u be a nonnegative function, continuously vanishing on ∂ ∩ Q T , and such that 1u − u t = 0 in . Then

(5.1)

lim sup

3(x,t)→(x0 ,t0 )

|∇u(x, t)| =

u(x, t) , 3(x,t)→(x0 ,t0 ) d (x, t) lim sup

for every (x0 , t0 ) ∈ ∂ ∩ Q T . Proof. Denote by α the left lim sup in (5.1) and by β the right one. The inequality α ≥ β follows immediately from the finite-increment formula, and therefore we focus on the inequality α ≤ β. If β = ∞ there is nothing to prove. Therefore assume β is finite. Let a sequence {(xk , tk )} be such that α =

12


limk |∇u(xk , tk )|. Let also z k ∈ ∂(tk ) be chosen such that dk : = d (xk , tk ) = |xk − z k |. Set u k (y, s) =

1 u(z k + dk y, tk + dk2 s). dk

Let ek = (xk − z k )/dk and assume that {ek } converges to e = (0, . . . , 0, 1). In view of the Lipschitz regularity of ∂ ∩ Q T in t and the convexity of (t)’s, the positivity sets k = {u k > 0} will converge (over a subsequence) to a subset of 5+ = {(x, t): x · e > 0} that has a cylindrical form D × R, where D is an unbounded convex subset of the halfspace {x · e > 0}, containing a spatial ball B(e, 1). Since u k satisfy the heat equation in k and are locally uniformly bounded (since we assume β is finite), a subsequence of {u k } will converge in C 1 -norm on compact subsets of D ◦ × R to a nonnegative caloric function v, which enjoys the following properties (5.2)

|∇v(e, 0)| = α

(5.3)

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

Suppose first that D is not a halfspace. Then by a theorem of Phragmén-Lindelöf type, the condition (5.3) will imply that v ≡ 0 in D × R; in particular we will have α = 0 and hence α ≤ β. If D is a halfspace, that is D × R = 5+ , condition (5.3) is rewritten as (5.4)

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

and we observe that the only caloric functions in 5+ with property (5.4) are functions of the form v(x, t) = γ (x · e) for nonnegative γ ≤ β. We give a short proof of this statement. Extend v into a caloric function in the whole Rn × R by the odd reflection v(x 0 , xn , t) = −v(x 0 , −xn , t), where x 0 = (x1 , . . . , xn−1 ) and xn = x · e, and consider the derivative ∂v/∂ xn . It is a caloric function in Rn × R and from (5.4), by the interior gradient estimate, |∂v/∂ xn | ≤ Cβ, where C is an absolute constant. From the Liouville theorem ∂v/∂ xn is identically constant, say γ , and therefore v(x) = γ xn , since it vanishes on 5 = ∂5+ . That 0 ≤ γ ≤ β follows from (5.4). Return to our v. In this case (5.2) implies γ = α and therefore α ≤ β. The proof is complete. The following lemma is an elliptic counterpart of the lemma above and is proved in a similar way; therefore, the proof is omitted. Lemma 5.2. Let D be a bounded spatial convex domain and u a nonnegative function in D, continuously vanishing on ∂ D and such that 1u = f in D with f


13

bounded. Then (5.5)

lim sup |∇u(x)| = lim sup D3x→x0

D3x→x0

u(x) , d(x)

where d(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 in Q 1/(2M) . Then the Dirichlet problem (3.4) is solvable in the classical sense. In this section we show that (u ∗ , ∗ ) is a supersolution of P and hence the minimal element of B. Lemma 6.1. Let u 0 satisfy (C1)–(C3) and T ≤ 1/(2M). Then the pair (u ∗ , ∗ ) is the minimal element of B.

Proof. The only thing we have to verify is that (u ∗ , ∗ ) is a supersolution of P. Let (u k , k ) ∈ B be such that T (i) ∗ = k k ; (ii) the sequence {k } is decreasing; (iii) k (0) = 0 and u k (·, 0) = u 0 . We can construct such a sequence as follows. Let (u k , k ) ∈ B satisfy (i). Next, in order to have (ii) we observe the following. Denote by u k,m the solution of the Dirichlet problem in k,m = k ∩ m with the initial function u k,m (·, 0) = min{u k (·, 0), u m (·, 0)} and vanishing on the lateral boundary, then u k,m (x, t) ≤ min{u k (x, t), u m (x, t)} for every (x, t) ∈ k,m . Besides, for the distance functions we will have dk,m (x, t) = min{dk (x, t), dm (x, t)} for every (x, t) ∈ k,m . Therefore, using Lemma 5.1, we can conclude that (u k,m , k,m ) ∈ B. If now (iii) is not satisfied, we can replace k by the intersection of all m with m ≤ k and thus to make {k } decreasing. In order to have (iii), let us take as 1 in the original sequence the cylindrical domain 0 × (0, T ) (see Remark 3.2). Now, since we can assume k are decreasing, we will have k (0) = 0 . To satisfy the second condition in (iii) just replace u k with the solution of the corresponding Dirichlet problem in k .

14


Denote now by ωk the caloric measure of 0 with respect to k ; that is 1ωk − (ωk )t = 0 in k ;

ωk = 0 on ∂k ∩ Q T ;

ω(·, 0) = 1 in 0 .

Let (6.1)

C = sup |∇u 0 (x)|. x∈0

Then we can control the growth of |∇u k | in k . Namely, (6.2)

|∇u k (x, t)| ≤ 1 + (C − 1)ωk (x, t)

for every (x, t) in k . This follows from the maximum principle for subcaloric functions, since v(x, t) = |∇u k | satisfies 1v − vt ≥ 0 in k . For the next step, observe that since u k are caloric in k and uniformly bounded, aTsubsequence of {u k } will converge in C 1 norm on compact subsets of ∗ = ∗ k k to a function u . We may assume also that over this subsequence, the corresponding caloric measures ωk converge to a caloric measure ω∗ of 0 with respect 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 in B 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. Shahgholian [HS1], [HS2]. Originally it was applied to the Bernoulli type free boundary problem for p-Laplace operator, an elliptic problem, whose free boundary condition is analogous to that of P. Definition 7.1. A point (x, t) ∈ ∂ ∩ Q T , where satisfies (3.1), is said to be extreme if x ∈ ∂(t) is extreme for (t). The latter means that x is not a convex combination of points on ∂(t), other than x.

Lemma 7.2. Let u 0 satisfy (C1)–(C3) and T ≤ 1/(2M). Then the pair (u ∗ , ∗ ) satisfies

(7.1)

lim sup

∗ 3(x,t)→(x0 ,t0 )

|∇u ∗ (x, t)| = 1

for every extreme point (x0 , t0 ) ∈ ∂∗ ∩ Q T .


15

Proof. Let us point out that it is enough to prove the lemma in the case when x0 is an extremal point of ∂∗ (t0 ), which means that there is a spatial supporting hyperplane to ∗ (t0 ), touching ∂∗ (t0 ) at x0 only. This follows from the fact that the extremal points are dense among the extreme points. Suppose now x0 ∈ ∂∗ (t0 ) is extremal and that (7.1) is not true. Then, in view of Lemmas 5.1 and 4.1, there exists a (space-time) neighborhood V of (x0 , t0 ) and α > 0 such that (7.2)

u ∗ (x, t) ≤ (1 − α)d∗ (x, t)

for every (x, t) ∈ V ∩ ∗ . We may assume additionally that the intersection V ∩ ∗ is given by (3.2). Let now 5 be a spatial supporting hyperplane to ∗ (t0 ), such that 5∩∂∗ (t0 ) = {x0 }. By translation and rotation, we may assume that x0 = 0 and that 5 = {xn = 0}. Moreover let ∗ (t0 ) ⊂ {xn > 0}. Using the extremality of (x0 , t0 ), it is easy to see that there are δ0 > 0 and η0 > 0 such that {(x, t) ∈ ∗ : xn ≤ η0 and t ∈ [t0 − δ0 , t0 ]} ⊂ V. Let us consider the function h(t) = − min xn ∗ x∈ (t)

for t ∈ [t0 − δ0 , t0 ]. In view of Lipschitz regularity of ∂∗ in time, h(t) ≤ L(t0 − t) for t ∈ [t0 − δ0 , t0 ], where L is the Lipschitz constant of f in t. Let now η1 ∈ (0, η0 ) be very small and a constant C ≥ L be chosen such that h(t0 − δ0 ) ≤ Cδ0 − η1 . Further, we can find δ1 ∈ (0, δ0 ] such that h(t0 − δ1 ) = Cδ1 − η1 and h(t) ≥ C(t0 − t) − η1 for every t ∈ [t0 − δ1 , t0 ]. Define now a domain ⊂ Q T by giving its time sections as follows  ∗ t ∈ (0, t0 − δ1 )  (t), (7.3) (t) = ∗ (t) ∩ {xn > η1 − C(t0 − t)}, t ∈ [t0 − δ1 , t0 ]  (t0 ), t ∈ (t0 , T ).

16


Let also u be a solution to the Dirichlet problem in for the heat equation, zero on the lateral boundary and equal u 0 in 0 . We claim that if η1 is small enough, then (u, ) is in B. This will lead to a contradiction since 6⊂ ∗ and the lemma will follow. Since satisfies (3.1), we need to verify only that (u, ) is a supersolution of P By [Lemma 5, ACS], there exists ε = ε(n, L) > 0 such that in a neighborhood of (x0 , t0 ), the function w ∗ (x) = w ∗ (x; t) = u ∗ (x, t) + u ∗ (x, t)1+ε is subharmonic in x. Moreover the size of the neighborhood depends only on n and L. We may suppose V has this property. Next, note that we can take C ≤ L + 1 if η1 is sufficiently small and δ0 is fixed. This will make the boundary of new constructed (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 neighborhood V . We are ready to prove now that (u, ) is indeed a supersolution. Case 1. t ∈ (0, t0 − δ1 ). This is the simplest case, since u = u ∗ in ∩ 0 < t < t0 − δ1 : all the properties of u there follows from those of u ∗ . Case 2. t ∈ [t0 − δ1 , t0 ], the most interesting case. First of all, since (t) ⊂ ∗ (t) for these t, we have also u ≤ u ∗ there. Let now consider a part D(t) of ∗ (t) between to planes: 51 = {xn = η1 − C(t0 − t)} and 50 = {xn = η0 − C(t0 − t)}. Compare there two functions w(x) = w(x; t) and `(x) = xn − η1 + C(t0 − t). On 51 both functions are 0. Next `(x) = η0 − η1

on 50 .

t0 t0 − δ1 t0 − δ0

FIG. 2. Construction of from ∗ in profile


17

To estimate w on 50 , let us first estimate u on 50 . Thus u(x, t) ≤ u ∗ (x, t) ≤ (1 − α)d∗ (x, t) ≤ (1 − α)(xn − L(t0 − t)) ≤ (1 − α)η0 and therefore, if η0 is small enough, we will obtain w(x) ≤ (1 − α/2)η0

on 50 .

Choose now η1 so small that (1 − α/2)η0 ≤ η0 − η1 . Then w ≤ ` on ∂ D(t) and, since w is subharmonic and ` is harmonic (linear), we conclude that w ≤ ` in D(t). Along with u ≤ u ∗ this gives u(x, t) ≤ 1, 3(x,t)→∂ d (x, t) lim sup

(7.4)

where t is free to vary within [t0 − δ1 , t0 ]. Case 3. t ∈ (t0 , T ). Since (t) shrink in time and u 0 is superharmonic in 0 , considering the time derivative u t in , we can infer from the maximum principle for the heat equation that u t ≤ 0 in . In particular, we will have in the cylindrical portion of with t > t0 that u(x, t) ≤ u(x, t0 ) in (t) = (t0 ) and applying Lemma 5.1 we conclude that (7.4) is valid also in this case. Summing up, we see that (7.4) holds for all t ∈ (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. Let u 0 satisfy (C1)–(C3). Then the minimal element (u ∗ , ∗ )

of B is a classical solution of P in Q 1/(2M) . Moreover this classical solution is unique, the time sections ∗ (t) are convex and shrinking in time and u ∗ (·, t) are log-concave in ∗ (t), 0 < t < 1/(2M). Up to now we have not used fully the log-concavity condition on u 0 . The following lemma will exploit this property. Lemma 8.2. Let u be a solution of a Dirichlet problem for the heat equation in a bounded domain in Q T , zero on ∂ ∩ Q T and u(·, 0) = u 0 . If the time sections (t) are convex for every t ∈ [0, T ] and u 0 is log-concave in 0 then so is u(·, t) in (t) for every t ∈ (0, T ).

Proof. The proof, using Korevaar’s Concavity maximum principle [Ko] can be found in [Pe]. We point out that an alternative proof can be given based on Brascamp and Leib’s paper [BL].

18


Remark 8.3. However, we can replace the condition of log-concavity of u 0 by another one that guarantees the convexity of level sets {u(·, t) > s} for every u as in 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 due to [HS1], [HS2]. Lemma 8.4. Let D be a bounded spatial convex domain with C 1 regular

boundary, V a neighborhood of ∂ D and w a smooth positive subharmonic function in D ∩ V , continuously vanishing on ∂ D. If the level lines {w = s} are strictly convex surfaces for 0 < s < s0 then the condition lim sup |∇w(x)| ≥ 1 D∩V 3x→x0

for every extreme point x0 ∈ ∂ D implies that |∇w(x)| ≥ 1 for every x with 0 < w(x) < s0 . Proof. Let y0 ∈ D ∩ V be such that w(y0 ) = s ∈ (0, s0 ), so `s = {w(x) = s} is a strictly convex surface. Denote by 5 a tangent hyperplane to `s at y0 . By translation and rotation we may assume that y0 = 0 and 5 = {xn = 0} and that `s lies in the lower halfspace {xn ≤ 0}. Choose now an extreme point x0 ∈ ∂ D ∩ {xn ≥ 0} such that it has the maximal xn -coordinate among the points of ∂ D. Although this point can be not uniquely defined, we will denote it x0 (y0 ) to indicate the way it was constructed from y0 . Suppose now, that along with C 1 regularity of ∂ D, w is C 1 regular up to ∂ D. The core inequality is then (8.1)

|∇w(y0 )| ≥ |∇w(x0 (y0 ))|.

To prove, let α = |∇w(x0 (y0 ))| and β ∈ (0, α). Consider a function f (x) = w(x) + βxn in D + = D ∩ {xn > 0}. The function f is subharmonic in D + and therefore admits its maximum on ∂ D + . A simple analysis shows that the maximum of f is admitted either at the origin y0 or at x0 . Let us exclude the latter possibility. We have that ∂xn w(x0 ) = −|∇w(x0 )| = −α and hence ∂xn f (x0 ) = β − α < 0, which is impossible if x0 is a maximum point. Therefore the maximum of f is admitted at y0 = 0 and as a consequence we have |∇w(0)| = −∂xn w(0) = − lim

h→0+

w(hen ) − w(0) βh ≥ lim = β. h→0+ h h


19

Letting β → α−, we obtain (8.1). This, of course, proves the lemma in the case considered. Consider now the general case. Choose a sequence of points {x j ∈ D ∩ V } converging to x0 so that lim |∇w(x j )| = lim sup |∇w(x)| ≥ 1.

j→∞

x→x0

Let s j = w(x j ) and a domain D j be bounded by the level surface `s j . Construct points y j ∈ `s on the same level surface as y0 so that x j = x0 (y j ) for the domain D j . It can be done as follows. Take the tangent hyperplane 5 j to `s j at x j and move it down towards `s until the plane touches `s and define y j to be the touching point. Now, the function w is C 1 regular up to the boundary of D j and therefore we 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 that y j → y0 . Due to strict convexity of `s , this is indeed so, if the outer normals ν j of the tangent planes 5 j to `s j at x j converge to the unit vector en = (0, 0, · · · , 1). In its turn, the latter statement is a consequence of the C 1 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 and x 0 a singular point on ∂ D, such that there are more than one supporting hyperplanes to D at x0 . Let also V be a neighborhood of x0 and w a subharmonic function in D ∩ V , continuously vanishing on ∂ D ∩ V . Then

(8.2)

lim

D∩V 3x→x0

w(x) = 0, d(x)

where d(x) = distance(x, ∂ D). Proof. As a first step, we note, that placing D between two supporting hyperplanes at x0 which form an angle of opening α < π , one can easily construct a barrier function and prove that for small r (8.3)

S(r ): =

max u(x) ≤ C1r π/α

Br (x0 )∩D

where C1 is some constant, and Br (x0 ) = {x: |x − x0 | < r }. Next, we claim that (8.4)

w(x) ≤ C2 d(x)S(4r )/r

for x ∈ ∂ Br (x0 ) ∩ D

20


with a constant C2 > 0. Indeed, let x1 be a point on ∂ D with |x − x1 | = d(x). Note that d(x) ≤ r , hence |x0 − x1 | ≤ 2r and in particular B2r (x1 ) ⊂ B4r (x0 ). Let 5 be a supporting hyperplane to D at x1 and 5+ be the halfspace, which contains D. Let ω be a harmonic measure of ∂ B2r (x1 ) ∩ 5+ with respect to B2r (x1 ) ∩ 5+ and v: = S(4r )ω. Then 1v = 0 ≤ 1w in B2r (x1 ) ∩ D and v ≥ w on ∂(B2r (x1 ) ∩ D) and therefore v(y) ≥ w(y) in B2r (x1 ) ∩ D by the maximum principle. In particular, at x we have C3 w(x) ≤ v(x) ≤ S(4r )ω(x) ≤ S(4r ) d(x) r with C3 a universal constant, such that ω(y) ≤

C3 |y − x1 | r

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 theorem follow from inclusion (u ∗ , ∗ ) ∈ B and Lemma 8.2. The uniqueness follows from 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 ) ∈ ∂∗ ∩ Q 1/(2M) . Denote by R the set of all t0 ∈ (0, 1/(2M)) such that lim sup |∇u ∗ (x, t0 )| = 1

∗ (t0 )3x→x0

for every extreme point x0 ∈ ∂∗ (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 case the time t0 is fixed. Let us prove that the complement J = (0, 1/(2M))\R is a union of a countable family of nowhere dense subsets of (0, 1/(2M)). This follows from the continuity of |∇u ∗ | in ∗ . Indeed, let {Uk } be an open countable basis for topology in Rn . If t0 ∈ J then there exist an extreme point x0 ∈ ∂∗ (t0 ), and natural numbers k and ∗ m such that x0 ∈ Uk and |∇u ∗ (x, t0 )| ≤ 1 − (1/m) for every x ∈ US k ∩ (t0 ). Denote now the set of all t0 ∈ J with such k and m by Jk,m . Then J = k,m Jk,m .


21

Besides, as easy to see, Jk,m are nowhere dense in (0, 1/(2M)) and our assertion follows. 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 exist ε > 0, δ > 0 and s0 > 0 such that w ∗ = w ∗ (·; t) is subharmonic in a convex ring D(t) = {0 < w∗ (x, t) < s0 } whenever t ∈ (t0 − δ, t0 + δ). Besides, from Lemma 8.2, the level surfaces of u ∗ and therefore those of w∗ are convex. Moreover, they are strictly convex due to real analyticity of u ∗ (·, t) in ∗ (t). Now, we point out that if t ∈ R, then ∂∗ (t) is C 1 regular. Otherwise there would exist a singular extreme point x0 ∈ ∂∗ (t) with (8.5)

lim

∗ (t)3x→x0

|∇u ∗ (x, t)| = 0,

which contradicts to the definition of R. Indeed, if x0 ∈ ∂∗ (t) is singular then by Lemma 8.5 w ∗ (x; t)/d (x, t) → 0, or, equivalently, u ∗ (x, t)/d (x, t) → 0 as x → x0 , and (8.5) will follow from Lemma 5.2. Note that Lemma 5.2 is applicable here since f (x): = u ∗t (x, t) = 1u ∗ (x, t) is bounded in ∗ (t) (see Remark 4.2). Let now t ∈ (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 all t ∈ (t0 −δ, t0 +δ) because of everywhere density of R and continuity of |∇w ∗ (x; t)| in ∗ . Since |∇w ∗ | and |∇u ∗ | are asymptotically equivalent when u ∗ → 0, we obtain immediately that lim inf

∗ 3(x,t)→(x0 ,t0 )

|∇u ∗ (x, t)| ≥ 1

whenever x0 ∈ ∂∗ (t0 ). Since t0 ∈ (0, 1/(2M)) was arbitrary, we conclude that (u ∗ , ∗ ) is indeed a classical solution of P in Q 1/(2M) . The theorem is proved. ACKNOWLEDGMENT The 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 and Gustaf Sigurd Magnuson’s foundation of the Royal Swedish Academy of Sciences for partial support.

22


REFERENCES [ACS] I. Athanasopoulos, L. Caffarelli and S. Salsa, Caloric functions in Lipschitz domains and the regularity 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 problem arising in combustion theory, Comm. Partial Differential Equations 19 (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 Prékopa-Leindler theorems, including inequalities for log concave functions, and with an application to the diffusion equation, J. Functional Analysis 22 (1976), no. 4, 366–389.

[CLW1] L. A. Caffarelli, C. Lederman and N. Wolanski, Uniform estimates and limits for a two phase parabolic 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 limit of 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 in flame 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 spherical and annular flames described by a free boundary problem, J. Math. Pures Appl. (9) 76 (1997), no. 7, 563–608. [HS1]


[HS2]

A. Henrot and H. Shahgholian, Existence of classical solutions to a free boundary problem for the 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 boundary problem from combustion, preprint. [Me]

A. M. Me˘ı 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 combustion theory, 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, USA E-mail: [email protected]

and Arshak Petrosyan and Henrik Shahgholian Department of Mathematics, Royal Institute of Technology, 100 44 Stockholm, Sweden E-mail: {arshak,henriks}@math.kth.se

We consider a free boundary problem for the p-Laplacian 1 p u = div(|∇u| p−2 ∇u), describing nonlinear potential flow past convex profile K with prescribed pressure gradient |∇u(x)| = a(x) on the free stream line. The main purpose of this paper is to study the limit as p → ∞ of the classical solutions of the problem above, existing under certain convexity assumptions on a(x). We show, as one can expect, that the limit solves the corresponding problem for the ∞-Laplacian 1∞ u = ∇ 2 u∇u · ∇u, in a certain weak sense, strong however, to guarantee the uniqueness. We show also that in the special case a(x) ≡ a0 > 0 the limit coincides with an explicit solution, given by a distance 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 configurations in potential flow in fluid mechanics. This time, however, the focus is on nonlinear flows with power law generalization (see e.g. [AM], [HS1–3].) The latter refers to the p–Laplace operator

1 p u = div(|∇u| p−2 ∇u), 1

1 < p < ∞.

2

J. MANFREDI, A. PETROSYAN AND H. SHAHGHOLIAN

The peculiar nonlinearity and degeneracy of the operator make the problems of potential flow both more realistic and also much harder to analyze. Our concern, in this paper, will be the limit case p → ∞ of the work of A. Henrot and H. Shahgholian [HS1–3]. Let us first formulate the problem. Let K be a compact convex subset in Rn , and a(x) a positive continuous function in K c = Rn \ K . For p ∈ (1, ∞) consider the following Bernoulli-type free boundary problem: find a pair (u, ) where ⊃ K is a domain in Rn and u is a nonnegative continuous function in such that  in \ K  1pu = 0 u = 0 and |∇u| = a on ∂ (F B p )  u=1 on K . Problem (F B p ), as mentioned earlier, describes nonlinear potential flow past the profile K and ∂ describes the stream line with prescribed pressure |∇u| = a. This problem for p = 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 further references. 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 on a(x) and regularity assumptions on K to be specified later, problem (F B p ) admits a unique classical solution, which we denote by (u p , p ); see [HS3]. The main purpose of this paper is to study the behavior of the pair (u p , p ) as p → ∞. For reader’s convenience we outline some steps of the proof for 1 < p < ∞ in the appendix. It it known that the limits of p-harmonic functions (weak solutions of 1 p u = 0) as p → ∞ are viscosity solutions of the equation 1∞ u = 0 (in other words, ∞– harmonic function), where 1∞ u = ∇ 2 u∇u · ∇u, see [BDM] (see also [Ar].) The operator 1∞ is known as the ∞–Laplace operator. Since the free boundary condition is the same for all problems (F B p ), we expect that the limit (u ∞ , ∞ ) = lim (u p , p ), p→∞

if it exists, is a solution of the limiting problem   1∞ u = 0 u = 0 and |∇u| = a (F B∞ )  u=1

in \ K on ∂ on K

in a certain sense. Another point of view is to treat problem (F B∞ ) independently and interpret the limit process above as one of the ways of solving the problem. This approach gives 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 (F B∞ ) explicitly. For instance consider the case a(x) ≡ a0 . Then one can easily see that a pair (u, ) with (1.1)

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

solves problem (F B∞ ) in the classical sense (see Section 3). Now one can ask whether the explicit solution (1.1) is a limit of solutions of (F B p ). We show that this is indeed the case and moreover (1.1) is the unique classical solution of (F B∞ ) with a(x) ≡ a0 (see Theorem 3.3.) The complications in the case of more general a(x) are connected with the fact that generally it is not known whether ∞–harmonic functions are C 1 regular or not, and the problem is how to interpret the free boundary condition. We overcome this difficulty by introducing the notion of the weak solution of (F B∞ ). Then we show that the only weak solution of (F B∞ ) is nothing but the limit (u ∞ , ∞ ) (see Theorem 4.4.) We conclude the paper with the brief analysis of the limit p → 1+. The following phenomenon takes place: the domains p shrink to K (in dimension n ≥ 2.) 2. PRELIMINARIES Throughout the paper we assume that K is a convex compact in Rn , n ≥ 2, which satisfies the uniform interior ball condition, i.e. there exists δ > 0 such that for every x ∈ ∂ K there is a ball B of radius δ with the property B ⊂ K and ∂ B ∩ ∂ K = {x}. Next the function a(x) will be positive and continuous in K c = Rn \ K with inf a(x) = a0 > 0.

(2.1)

x∈K c

We will assume also that 1/a(x) is locally concave in K c . (Sometimes we will call such functions harmonic concave.) Moreover, assuming w.l.o.g. that the origin is an interior point of K , we will require (2.2)

λa(λx) > a(x) for every x ∈ K c and λ > 1.

We start with the definition. Definition 2.1. A pair (u, ), where is a domain in Rn that contains K

and u is a nonnegative continuous function on , is called a classical supersolution of (F B p ) for 1 < p < ∞ if (i) u ∈ C 1 ( \ K ) and 1 p u ≤ 0 in \ K in the sense of distributions; (ii) lim sup y→x |∇u(y)| ≤ a(x) for every x ∈ ∂;

4


(iii) u ≥ 1 on K and u = 0 on ∂. Analogously, we define classical subsolutions of (F B p ), by reversing the inequality signs in (i), (ii) and (iii) and replacing lim sup by lim inf in (ii). A classical solution of (F B p ) is a pair (u, ) which is a classical sub- and supersolution at the same time. As a straightforward consequence of condition (2.2), we have the following comparison principle. Lemma 2.2 (Comparison principle). For problem (F B p ), every classical subsolution is smaller than every classical supersolution.

More precisely, the lemma should be read: If (u 0 , 0 ) is a classical sub- and (u, ) is a classical supersolution of (F B p ), then 0 ⊂ and u 0 ≤ u in 0 . Proof. We will consider first the case when u 0 and u are C 1 up to ∂0 and ∂ respectively. Suppose that 0 6⊂ . Then we have λ0 = min{λ ≥ 1: λ−1 0 ⊂ } > 1. Here λ−1 0 = {λ−1 x: x ∈ } =: 0λ . Then 0λ0 ⊂ and there is a common point x0 ∈ ∂(0λ0 ) ∩ ∂. Consider the rescaled function u 0λ0 (x) = u 0 (λ0 x) in 0λ0 . Then usual comparison principle for p–harmonic functions says that u 0λ0 ≤ u in 0λ0 . But u 0λ0 (x0 ) = u(x0 ) = 0, hence there must be λ0 a(λ0 x0 ) ≤ |∇u λ0 (x0 )| ≤ |∇u(x0 )| ≤ a(x0 ), which contradicts to (2.2). Therefore 0 ⊂ and moreover u 0 ≤ u in 0 by the comparison principle for p-harmonic functions. The reasonings above are known as the Lavrent’ev principle (cf. [La]). The case when u 0 and u are not C 1 up to the boundary can be handled in various ways, reducing it to the case above. One of the ways is to consider u 0 − ε and u − ε on the level sets 0ε = {u 0 ≥ ε} and ε = {u ≥ ε} respectively and deduce that the constant λ0 = λ0 (ε) as above but for 0ε and ε tends to 1 as ε goes to 0+. The comparison lemma above gives us the uniqueness of the classical solution for (F B p ) if it exists. The following existence theorem is due to A. Henrot and H. Shahgholian. For reader’s convenience we give a brief sketch of the proof of this theorem at the end of this paper.


5

Theorem 2.3 ([HS3]). There exists a unique classical solution (u p , p ) of the free boundary problem (F B p ). Moreover p is convex and ∂ p is C 1 regular.

Next we will use the favorable geometric situation to compare solutions of (F B p ) for different p. This lemma has appeared already in [Ja]. Lemma 2.4. Let ⊃ K be a convex domain and u be the p–capacitary potential (1 < p < ∞) of the convex ring \ K , i.e. a continuous function on \ K such that

1 p u = 0 in \ K ; u = 0 on ∂; u = 1 on ∂ K . Then 1q u ≤ 0, 1q u ≥ 0,

if 1 < q ≤ p and if p ≤ q ≤ ∞.

Proof. By a result of J. Lewis [Le], |∇u| > 0 in \ K and therefore u is a real analytic in \ K . Then one can write 1 p u in a nondivergence form (2.3)

1 p u = |∇u| p−2 (1u + |∇u|−2 ( p − 2)1∞ u).

Let us introduce operators (2.4)

L p u: = |∇u|2− p 1 p u = 1u + |∇u|−2 ( p − 2)1∞ u.

Then L q u has always the same sign as 1q u for every 1 < q < ∞. Using now assumption that 1 p u = 0 in \ K and hence L p u = 0, we deduce that (2.5)

L q u = |∇u|−2 (q − p)1∞ u

in \ K .

Hence the lemma will follow as soon as we prove (2.6)

1∞ u ≥ 0.

To this end, take a point x ∈ \ K and choose a coordinate system centered at x such that the n–axis is directed as ∇u. Let u(x) = s. By [Le] the level set Ls (u) = {y: u(y) > s} is convex and therefore its boundary `s (u) = ∂Ls (u) can be given near the point x as a graph {y: yn = f (y 0 )} of a convex C ∞ function f in

6


∞

p

K

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

the coordinate system chosen above (here y 0 = (y1 , . . . , yn−1 ).) Differentiating twice the identity u(y 0 , f (y 0 )) = s and using that f i (0) = 0 we obtain (2.7)

u ii + u n f ii = 0 at 0 for i = 1, . . . , n − 1.

The indices mean differentiation with respect to the corresponding y-coordinates. In particular u n = |∇u| at 0. Observe now that in these new coordinates the condition 1 p u = 0 at x means precisely (2.8)

n−1 X

u ii + ( p − 1)u nn = 0 at 0.

i=1

From (2.7) and (2.8) we obtain (2.9)

1∞ u(x) = u nn u 2n =

n−1 3 κu ≥ 0, p−1 n

Pn−1 where κ = 1/(n − 1) i=1 f ii is the mean curvature of `s (u) at the point x, which is nonnegative. Since x was an arbitrary point in \ K , the proof of the lemma is complete. Corollary 2.5 (Variational solution). The classical solutions (u p , p ) of (F B p ) are nondecreasing in p ∈ (1, ∞) and are uniformly bounded. More precisely, p0 ⊂ p00 and u p0 ≤ u p00 in p0 for every 1 < p 0 ≤ p 00 < ∞ and there exists a bounded domain 0 and a function u 0 on 0 such that p ⊂ 0 and u p ≤ u 0 for all 1 < p < ∞. Hence there exists

(2.10)

(u ∞ , ∞ ) = lim (u p , p ) p→∞

which we will call a variational solution of (F B∞ ).


7

Proof. The monotonicity part is an immediate corollary of Theorem 2.3, Lemma 2.4 and Lemma 2.2. As the uniform bound (u 0 , 0 ) we can use u 0 (x) = 1 − a0 dist (x, K ),

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

where, recall, a0 = infx∈K c a(x) > 0. Indeed, the pair (u 0 , 0 ) is a classical supersolution of (F B p ) for all p ∈ (1, ∞) and hence the statement follows. (See also Step 1 in the proof of Theorem 3.3.)

3. CLASSICAL SOLUTIONS FOR a(x) ≡ a0 Precisely as for (F B p ), we define the classical (sub- and super-) solutions of (F B∞ ). Definition 3.1. A pair (u, ), where is a domain in Rn that contains K

and u is a nonnegative continuous function on , is called a classical supersolution of (F B∞ ) if (i) u ∈ C 1 ( \ K ) and 1∞ u ≤ 0 in \ K in the viscosity sense; (ii) lim sup y→x |∇u(y)| ≤ a(x) for every x ∈ ∂; (iii) u ≥ 1 on K and u = 0 on ∂. Classical subsolutions and classical solutions of (F B∞ ) are defined by the respective modifications of (i)–(iii) as in Definition 2.1.

Remark 3.2. Lemma 2.2 remains valid for p = ∞ with the same proof. The only properties of the operator needed is the comparison principle and the possibility to rescale sub- and supersolutions. Observe, however, that the comparison principle for 1∞ is nontrivial and was proved by R. Jensen in [Je]. As we have mentioned in the introduction, problem (F B∞ ) has a simple explicit solution (u, ) given by (1.1) when a(x) ≡ a0 > 0. From the other hand we know from Corollary 2.5 that there is always a variational solution. The next theorem says that these two are the same. Theorem 3.3 (Classical solution). In the case a(x) ≡ a0 > 0 the variational solution (u ∞ , ∞ ) of (F B∞ ) as defined in Corollary 2.5 is given by

(3.1)

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

Moreover, (u ∞ , ∞ ) is the unique classical solution of (F B∞ ).

8


Proof. Set u(x) = 1 − a0 dist (x, K ) and = { dist (x, K ) < 1/a0 }. Step 1. Show that for every p ∈ (1, ∞) there holds p ⊂ and u p ≤ u in p .

(3.2)

By Lemma 2.2 it is sufficient to show that (u, ) is a classical supersolution of (F B p ). Since the gradient and boundary value conditions (ii) and (iii) in Definition 2.1 are readily satisfied the only condition to be verified is 1 p u ≤ 0. The latter holds due to concavity of u. Hence (3.2) is proved. Passing to the limit in (3.2) as p → ∞ we obtain ∞ ⊂ and u ∞ ≤ u in ∞ .

(3.3)

Step 2. In order to reverse inequalities in (3.3), we construct explicit subsolutions of (F B p ) from (u, ) as follows. Assume for the moment that K has C 2 regular boundary; then the distance function dist (x, K ) is C 2 in K c as well. For ε > 0 let φε (s) =

(3.4)

eεs − 1 ε

and δε ∈ (0, 1) be such that φε (1 − δε ) = 1.

(3.5)

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

(3.6) in

ε = {x: u(x) > δε } = {x: dist (x, K ) < 1/a0 − δε }. We claim that (3.7)

(u ε , ε ) is a classical subsolution of (F B p ) for p > p(ε).

By construction, the gradient and boundary value conditions are satisfied and we have to verify only that 1 p u ε ≥ 0 in ε \ K for p large. For this purpose, take a point x ∈ ε \ K and choose a coordinate system (y1 , . . . , yn ) centered at x with n axis directed as ∇u ε (x) (or, equivalently, as ∇u(x).) These are the same


9

coordinates used in the proof of Lemma 2.4. Observe that the level sets of u ε are those of u, but parameterized in another way. Assume u(x) = s. Then the level line `s (u) = `φε (s−δε ) (u ε ) can be represented near x as a graph {y: yn = f (y 0 )} of a convex C 2 function. In new coordinates we can calculate that at x u εj j = φε00 (u − δε )u 2j + φε0 (u − δε )u j j

(3.8)

for j = 1, . . . , n. Using that u i = 0 for i = 1, . . . , n − 1, u n = a0 and u nn = 0 at x as well as a formula (2.7) (which is still valid) we obtain (3.9)

u iiε = −φε0 (u − δε )a0 f ii (x) for i = 1, . . . , n − 1

and u εnn = φε00 (u − δε )a02

(3.10)

at point x. Observe now that φε satisfies the differential equation φε00 = εφε0 . Employing this fact we find that |∇u ε |2− p 1 p u ε =

n−1 X

u iiε + ( p − 1)u εnn = φε0 (u − δε )a0 (( p − 1)a0 ε − (n − 1)κ),

i=1

(3.11) Pn−1 where κ = (1/(n − 1)) i=1 f ii is the mean curvature of the level line `s (u) = `(1−s)/a0 (dist (·, K )) at x. Recall now that we assume K has a uniform interior ball property hence the mean curvature of ∂ K at every point is bounded by a finite constant, say κ0 . But then the mean curvature of every level line of the distance function dist (x, K ) at every point is not greater than κ0 . Hence we have κ ≤ κ0 for the κ in (3.11). If now p is so large that ( p − 1)a0 ε − (n − 1)κ0 ≥ 0, then from (3.11) we will have immediately 1 p u ε ≥ 0 at x. This means that (3.7) is true with n−1 p(ε) = 1 + κ0 . a0 ε Letting ε → 0 we conclude that (3.12)

⊂ ∞

and u ≤ u ∞ in

10


in the case when K has C 2 regular boundary. If not, let K η ⊂ K be a strictly convex subset with C 2 boundary approximating K as η → 0. Then proceeding as above we prove first that {x: dist (x, K η ) < 1/a0 } ⊂ ∞ and 1−a0 dist (x, K η ) ≤ u ∞ (x). Letting η → 0 we prove (3.12) in the general case. Step 3. The uniqueness of (u ∞ , ∞ ) follows readily from Lemma 2.2, see Remark 3.2. The proof of the theorem is complete.

4. WEAK SOLUTIONS FOR GENERAL a(x) Now we return to the general case, when a(x) is a positive harmonic concave function for x ∈ K c , satisfying (2.1) and (2.2). In this case we don’t have an explicit solution as for the case a(x) ≡ a0 , but we can still hope that the variational solution is a classical solution of (F B∞ ). The first difficulty we face is that u ∞ may fail to be C 1 . However, as we will see later, the family {u p } of classical solutions of (F B p ) is uniformly bounded in the Lipschitz norm, and hence u ∞ is at least Lipschitz. The main purpose of this section is to introduce a notion of a weak solution of (F B∞ ) that doesn’t use the differentiability properties of u. Definition 4.1. A pair (u, ), where is a domain in Rn that contains K

and u is a nonnegative continuous function on , is called a weak supersolution of (F B∞ ) if (i) 1∞ u ≤ 0 in \ K in the viscosity sense; (ii) lim sup y→x u(y)/ dist (y, ∂) ≤ a(x) for every x ∈ ∂; (iii) u ≥ 1 on K and u = 0 on ∂. Analogously, we define weak subsolutions of (F B∞ ), by reversing the inequality signs in (i), (ii) and (iii) and replacing lim sup by lim inf in (ii). A weak solution of (F B∞ ) is a pair (u, ) which is a weak sub- and supersolution at the same time.

Remark 4.2. The comparison principle, Lemma 2.2, still works for the weak solutions defined above, with a minor modification in the proof. This implies in particular the uniqueness of the weak solution. The following lemma provides compatibility of the above definition with the definition of classical solutions. Lemma 4.3. A classical (super-, sub-) solution of (F B∞ ) is a weak (super-, sub-) solution of (F B∞ ).


11

Proof. 1) Supersolutions. This case is simple. For y ∈ \ K we have u(y) = u(y) − u(z) = (y − z) · ∇u(ξ ) ≤ dist (y, ∂)|∇u(ξ )|, where z = z(y) ∈ ∂ is chosen such that dist (y, ∂) = |y − z| and ξ is a certain point on the line segment, joining y and z. Since y → x ∈ ∂ implies ξ → x, we obtain u(y) lim sup ≤ lim sup |∇u(y)| ≤ a(x) dist (y, ∂) as \ K 3 y → x ∈ ∂. 2) Subsolutions. This case needs more attention. Let (u, ) be a classical subsolution of (F B∞ ), and consider the function v(x) = u(x)/a(x). Assume first a(x) is C 1 regular. Then v is C 1 as well and its gradient is given by ∇v =

∇a ∇u −u 2 . a a

In particular lim inf |∇v(y)| ≥ 1

\K 3y→x

for every x ∈ ∂. Then we can proceed as in the proof of Claim 1 in [Vo]. Let y ∈ \ K and v(y) = s. Suppose 0 < s < sε , where sε (for small ε > 0) is chosen such that for z ∈ \ K v(z) < sε

(4.1)

implies 1 − ε < |∇v(z)|.

Find now a curve γ y (t) which solves ∇v(γ y (t)) d γ y (t) = dt |∇v(γ y (t))|2

(4.2)

and γ y (s) = y.

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

for t ∈ [0, s] .

Let ` y denote the length of the curve γ y [0, s]. Then from (4.1) v(y) ≥ (1 − ε)` y ≥ (1 − ε) dist (y, ∂). Consequently lim inf

u(y) v(y) = a(x) lim inf ≥ a(x) dist (y, ∂) dist (y, ∂)

12


as \ K 3 y → x ∈ ∂ and the lemma follows in the case a(x) is C 1 regular. For continuous a(x), take a C 1 regularization aε (x) ≤ a(x), converging pointwise to a(x) as ε → 0, and use it in the reasonings above to obtain lim inf u(y)/ dist (y, ∂) ≥ aε (x). Then we let ε → 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 (F B∞ ) as defined in Corollary 2.5 is the only weak solution of (F B∞ ).

5. (u ∞ , ∞ ) IS A WEAK SUBSOLUTION The statement in the title of this section constitutes the relatively easy part in the proof of Theorem 4.4. Lemma 5.1. The variational solution (u ∞ , ∞ ) is a weak subsolution of (F B∞ ).

Proof. We use similar reasonings as those in the proof of Lemma 4.3. Let (u p , p ) be the classical solution of (F B p ) for p > 1, y ∈ p \ K and u p (y) = s. Consider the curve γ y (t) which solves (5.1)

∇u p (γ y (t)) d γ y (t) = dt |∇u p (γ y (t))|2

and γ y (s) = y.

On this curve the following relation is satisfied u p (γ y (t)) = t and one can infer from this that γ y (t) can be defined on the interval (0, s) with a continuous extension to [0, s]. The curve γ y has the following nice property. (5.2)

|∇u p (γ y (t))| % as t % .

To prove this statement, fix z 0 = γ y (t0 ) for some t0 ∈ (0, s). Choose a new coordinate system in Rn with the origin at z 0 and with the n–axis directed as ∇u p (z 0 ). Then γ y0 (t0 ) is directed as the n–axis as well. Therefore (5.3)

d |∇u p (γ y (t))| t=t = |γ 0 (t0 )|(|∇u p |)n (z 0 ) = |γ 0 (t0 )|(u p )nn (z 0 ), 0 dt


13

where the subscript n means the differentiation along the n–axis. On the other hand we have (u p )nn =

(5.4)

n−1 κ(u p )n ≥ 0 p−1

at z 0 , where κ ≥ 0 is the mean curvature of the level line {u p = t0 } at z 0 , as it follows 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 curve γ y (t) as t → 0+. Then in the limit |∇u p (γ y (0)| = a(γ y (0)). Together with (5.2) this implies |∇u p (y)| ≥ a(γ y (0)) for y ∈ p \ K .

(5.5)

More generally, we have |∇u p (γ y (t))| ≥ a(γ y (0)) for t ∈ [0, s] .

(5.6)

To proceed, we suppose p ≥ p0 > 1 and 0 < s ≤ σ < s0 < 1. Set c0 = inf a(y) and

L 0 = sup

|a(y) − a(z)| |y − z|

for y, z ∈ ∞ \ {u p0 > s0 }. Then c0 > 0 and L 0 < ∞. Using (5.6), we can infer very rough estimate |∇u p (γ y (t))| ≥ c0

for t ∈ [0, s] .

If we denote by ` y the length of the curve γ y [0, s], then the estimate above will imply u p (y) ≥ c0 ` y and consequently |y − γ y (0)| ≤ ` y ≤

u p (y) . c0

Then by the definition of L 0 and the above estimate a(γ y (0)) ≥ a(y) − L 0 |y − γ y (0)| ≥ a(y) −

L0 L0 u p (y) ≥ a(y) − σ. c0 c0

14


Going back again to (5.6), integrating along the curve γ y [0, s] and applying the previous estimate, we obtain L0 u p (y) ≥ a(γ y (0))` y ≥ a(y) − σ dist (y, ∂ p ) c0 for every p ≥ p0 and y ∈ p \ K with u p (y) ≤ σ < s0 . Letting p → ∞, we obtain L0 u ∞ (y) ≥ a(y) − σ dist (y, ∂∞ ) c0 for every y ∈ ∞ \ K with u ∞ (y) ≤ σ < s0 . It is clear now that the lemma follows.

6. UNIFORM GRADIENT BOUND FOR u p AS p → ∞ In this section we make the first step in proving that (u ∞ , ∞ ) is a weak supersolution of (F B∞ ). We remind that we assume K to satisfy the uniform interior ball condition. That is, there exists δ > 0 such that for every x ∈ ∂ K one can find a ball B of radius δ such that B ⊂ K and ∂ B ∩ ∂ K = {x}. As a consequence we have the following lemma. Lemma 6.1. There exists a constant M > 0 such that for the classical solution (u p , p ) of (F B p ) we have

|∇u p | ≤ M

in p \ K

for all p ≥ p0 > 1. The constant M here depends only on δ in the uniform interior ball condition for K , on p0 , on d0 = inf y∈∂ p0 dist (y, K ), and on the space dimension n. In particular, u ∞ is Lipschitz continuous. Proof. We mostly follow the proof of Lemma 2.3 in [HS1]. Observe first that |∇u p | satisfies the maximum principle in p \ K , namely max |∇u p | ≤ max |∇u p |.

p \K

∂ K ∪∂ p

This follows from Payne-Philippin’s inequality L u p (|∇u p | p ) ≥ 0,


15

with the elliptic operator L u p (v): = |∇u p | p−2 1v + ( p − 2)|∇u p | p−4 ∇ 2 v∇u p · ∇u p . For the proof apply Lemma 1 in [PP] with the particular choices α = −1, β = 0, γ and v = 0 to obtain L u p (|∇u p | p ) ≥

p( p − 1)2 |∇u p | p−6 (∇u p · ∇(|∇u p |2 ))2 . 4

Now, having the gradient maximum principle it suffices to prove that ∇u p is uniformly bounded on ∂ p ∪ ∂ K . Since |∇u p | = 1 on ∂ p we need only a uniform gradient bound on ∂ K . Fix now p0 > 1 and consider the function u p0 near ∂ K . For x ∈ ∂ K let Bδ (z x ) be a touching ball at x from inside of K . By the definition of d0 we have also Bδ+d0 (z x ) ⊂ p0 . If one denote by v p0 the p0 –capacitary potential of the ring Bδ+d0 (z x ) \ Bδ (z x ) then by the comparison principle we will have v p0 ≤ u p0 . Since also v p0 (x) = u p0 (x) we may conclude |∇u p0 (x)| ≤ |∇v p0 (x)| = M(δ, d0 , p0 , n). Take now p > p0 . From Corollary 2.5 we know that u p0 < u p < 1. Also u p = u p0 = 1 on ∂ K and consequently |∇u p (x)| ≤ |∇u p0 (x)| ≤ M

for x ∈ ∂ K .

The proof is complete.

7. ON ∞-HARMONIC FUNCTIONS NEAR SINGULAR CONVEX BOUNDARY POINTS In this section we study the behavior of ∞–harmonic functions near a singular convex boundary point. The results are more or less independent from the rest of the paper, and have the interest of their own. More specifically we prove the following result. Theorem 7.1. Let D be a convex open set with x 0 a singular boundary point and u a viscosity solution of 1∞ u = 0 in D ∩ B R (x0 ) for some R > 0. Suppose moreover u vanishes on ∂ D ∩ B R (x0 ). Then u(x) = O(|x − x0 |1+ε ) as x → x0 for some ε > 0.

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

16


Proof. The convexity of D helps us to reduce the problem to the 2-dimensional case. Namely, that the point x0 is a singular boundary point of D means precisely that there are at least 2 supporting hyperplanes to D at x0 . Therefore after a necessary translation and rotation we may assume x0 = 0 and that there is a cone 0 in R2 of aperture less than π and with vertex at the origin, such that D ⊂ 0 × Rn−2 . Suppose at the moment that we know the existence of a barrier v in the 2dimensional cone 0 that has the following properties: (i) v ∈ C(0) and 1∞ v ≤ 0 in the viscosity sense; (ii) v > 0 in 0 \ {0} and v(0) = 0; (iii) v(x) = O(|x|1+ε ) for some ε > 0. Then using standard arguments together with Jensen’s maximum principle we obtain u(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 be C 2 regular, the reason being that ∞–superharmonicity implies concavity of the barrier on the symmetry axis. The barriers that we will construct next are only C 1,1/3 regular. Lemma 7.2. For every θ ∈ (0, π/2) there is ε > 0 and a C 1,1/3 regular

function F(α) for α ∈ (−θ, θ) such that (using complex notations) v, given by v(r eiα ) = r 1+ε F(α),

(7.1)

satisfies (i)–(iii) above in 0 = {r eiα : α ∈ (−θ, θ )}. Proof. obtain

Let us compute formally the ∞–Laplacian for v as in (7.1). We will

1∞ v(r eiα ) = r −(1−3ε)/2 (ε(1+ε)3 F(α)3 +F 0 (α)2 ((1+ε)(1+2ε)F(α)+F 00 (α))). This can be rewritten as 1∞ v(r eiα ) = r −(1−3ε)/2 L ε (F)(α), where L ε (F) = ε(1 + ε)3 F 3 + (F 0 )2 ((1 + ε)(1 + 2ε)F + F 00 ). Substitute F = e f . This will give L ε (e f ) = e3 f (ε(1 + ε)3 + ( f 0 )4 + ( f 0 )2 ((1 + ε)(1 + 2ε) + f 00 )), or L ε (e f ) = e3 f ( f 0 )2 `ε ( f 0 ),


17

1.5 1 0.5 -40

-20

20

40

-0.5 -1 -1.5

FIG. 2. Graphs of φε (y) as ε → 0.

where `ε (y) = (1 + ε)(1 + 2ε) +

ε(1 + ε)3 + y2 + y0 y2

a first order differential operator. Let us now consider the ODE `ε (y) = 0. After integration we will get

y(α) − arctan 1+ε

r

+

y(α) ε arctan √ = α + C, 1+ε ε(1 + ε)

where C is a constant. Choose C = 0 and consider the function r y y ε φε (y) = − arctan + arctan √ 1+ε 1+ε ε(1 + ε) on the whole real axis. Then φε is a smooth strictly decreasing function with φ(0) = 0. Using that the image of arctan is the interval (−π/2, π/2) we easily obtain that the image of φε is the interval (−θε , θε ) with π θε = 2

r ε 1− . 1+ε

Then the inverse function ψε = φε−1 is well defined on the interval (−θε , θε ). Moreover, since the derivative φε0 (y) = −

y2 (y 2

+ε

+ ε2 )(y 2

+ (1 + ε)2 )

18


0.04 0.02

-0.3

-0.2

-0.1

0.1

0.2

0.3

-0.02 -0.04

FIG. 3. φε (y) near 0 as ε → 0.

is strictly negative for y 6= 0, ψε (α) is C ∞ everywhere on (−θε , θε ) except α = 0. To examine the behavior of ψε (α) near α = 0 we look at the asymptotic development of φε (y) near y = 0:

φε (y) = −

y3 + O(y 5 ). 3ε(1 + ε)3

This implies that ψε is only C 1/3 regular at the origin. Let us choose now R αε > 0 so that θ < θε and go backwards in the constructions above. Set f (α) = 0 ψε (t)dt and F(α) = e f (α). Then F is C ∞ everywhere in the cone 0ε = {r eiα : α ∈ (−θε , θε )} ⊃ 0 except the symmetry axis α = 0 where it is only C 1,1/3 regular. By construction, v(r eiα ) = r 1+ε F(α) will satisfy 1∞ v = 0 in the classical sense off the symmetry axis α = 0. We claim that in the viscosity sense 1∞ v = 0 in the whole 0ε . First observe that there are no paraboloids touching v by below at the points on the symmetry axis α = 0, otherwise F would be C 1,1 regular. This implies immediately that 1∞ v ≤ 0 in 0ε in the viscosity sense. Let now P be a paraboloid touching v by above at a point on the axis α = 0. Since v is C 1 regular, at the touching point ∇ P = ∇v, which is directed along the axis. Observe now that on the axis α = 0 v is a convex function r 1+ε F(0) and hence we must have ∇ 2 P∇ P · ∇ P ≥ 0. This implies that 1∞ v ≥ 0 in the viscosity sense in 0ε . Hence the claim follows. The rest of properties (i)-(iii) are easily verified for v and the proof of the lemma is complete. Theorem 7.1 is proved.


Remark 7.3. function

19

Observe that in the case θ < π/4 one can take as a barrier the

v(x, y) = x 4/3 − |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 that appears in the proof for ε = 1/3.

8. C 1 REGULARITY OF ∂∞ The results of this section are straightforward consequences of Theorem 7.1. The level sets {u ∞ > s} are convex sets with regular boundary {u ∞ = s} for every s ∈ [0, 1). In particular, ∂∞ is C 1 regular.

Lemma 8.1 (C 1 regularity).

C1

Proof. Let p0 > 1 and s0 ∈ (0, 1). Then from the proof of Lemma 5.1 we know that for c0 = min{a(x): x ∈ ∞ \ {u p0 > s0 }} > 0 we have |∇u p (y)| ≥ c0

whenever p ≥ p0 and 0 < u p (y) < s0 .

This implies in particular, that u p (y) − s ≥ c0 dist (y, {u p = s}) whenever p ≥ p0 and s < u p (y) < s0 . Letting p → ∞ we obtain (8.1)

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

Now, Theorem 7.1 says that (8.1) would be impossible if {u ∞ = s} were not C 1 regular. Indeed, suppose there is a singular point y0 on {u ∞ = s}. Since {u ∞ > s} is a convex open set, we can find a unit vector e and a small δ > 0 such that the truncated cone C = C(y0 , e, δ) = {y ∈ Rn : |y − y0 | < δ, angle(y − y0 , e) < δ} is contained in {u ∞ > s}. Then along the symmetry axis of C, for y = y0 + ηe with 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 |,

20


contradicting Theorem 7.1. The lemma is proved. The following corollary is a kind of compensation for the missing C 1 regularity of u ∞ . Corollary 8.2. There is a unit-vector field ν ∈ C(∞ \ K ) such that

(8.2)

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

Proof. For every y ∈ ∞ \ K there is a unique supporting hyperplane to {u ∞ < u ∞ (y)} at y. Denote by ν(y) the unit normal vector of this hyperplane, pointing into {u ∞ < u ∞ (y)}. Then Lemma 8.1 implies the continuity of ν up to ∂∞ . To see the identity (8.2), recall that u ∞ is Lipschitz continuous and therefore differentiable a.e. in ∞ \ K . At the differentiability points y where |∇u ∞ | 6= 0, evidently ∇u ∞ /|∇u ∞ | is normal to the level set and hence coincides with ν(y). The proof is complete.

9. STABLE SOLUTIONS OF THE BOUNDARY VALUE PROBLEM The disadvantage of the classical and weak solutions defined in the previous sections is that we don’t know generally their stability under the limit. In this section we intend to give a notion of a stable solution, in this sense. As the source of the definitions and notations we use User’s guide to viscosity solutions [CIL] of M. G. Crandall, H. Ishii and P. L. Lions. Suppose we have a family F p : Rn × R × Rn × S(n) → R, of upper semicontinuous proper operators, where 1 < p < ∞ and S(n) is the set of symmetric n × n matrices. The operator F is proper if F(x, r, η, X ) ≤ F(x, s, η, Y ) whenever r ≤ s and Y ≤ X , where r, s ∈ R, x, η ∈ Rn , X, Y ∈ S(n) and S(n) is equipped with its usual order. Observe, the operator −1 p and not 1 p is proper! Suppose next, we have a family of locally compact subsets O p ⊂ Rn and functions u p ∈ LSC(O p ) (lower semicontinuous functions on O p ), which solve F p (x, u p , ∇u p , ∇ 2 u p ) ≥ 0 on O p in the viscosity sense. Denote O∞ = lim sup O p p→∞


21

= {x: ∃x pk ∈ O pk such that x pk → x} and u ∞ (x) = lim inf∗ u p (x) on O∞ p→∞

= inf{lim inf u pk (x pk ): x pk ∈ O pk , x pk → x}. k→∞

Then the following theorem holds. Theorem 9.1. If the operators F p converge locally uniformly to an operator F∞ , then

F∞ (x, u ∞ , ∇u ∞ , ∇ 2 u ∞ ) ≥ 0 on O∞ in the viscosity sense. The theorem is an immediate corollary of the following lemma, which is a variation of the Proposition 4.3 in [CIL]. 2,−

Lemma 9.2. For every xˆ ∈ O∞ and (η, X ) ∈ JO u ∞ (x) ˆ there exists a ∞

2,− sequence xˆ pk ∈ pk and (ηk , X k ) ∈ JO u pk (xˆ pk ) with pk → ∞ such that p k

(xˆ pk , u pk (xˆ pk ), ηk , X k ) → (x, ˆ u ∞ (x), ˆ η, X ).

Recall that given a function u on a subset O of Rn and a point xˆ ∈ O, the 2,− second-order subjet JO u(x) ˆ is the set of pairs (η, X ) ∈ Rn × S(n) such that u(x) ≥ u(x) ˆ + η(x − x) ˆ +

(9.1)

1 X (x − x) ˆ · (x − x) ˆ + o(|x − x| ˆ 2) 2

as O 3 x → x. ˆ In addition to subjets we will also need their closures 2,−

2,− J O u(x) ˆ = {(η, X ) ∈ Rn × S(n): ∃xk ∈ O and (ηk , X k ) ∈ JO u(xk ) such that (xk , u(xk ), ηk , X k ) → (x, ˆ u(x), ˆ η, X )}.

Proof of Lemma 9.2. We refer to the proof of Proposition 4.3 in [CIL]. The only essential difference is that the locally compact sets O p do not vary with p there, 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 of Rn and 0 a relatively open portion of ∂ D. Then a lower semicontinuous function u on D ∪ 0 is a viscosity

22


solution of (BV )

F(x, u, ∇u, ∇ 2 u) ≥ 0 in D B(x, u, ∇u, ∇ 2 u) ≥ 0 on 0

with F and B proper operators, if u is a viscosity solution of G + (x, u, ∇u, ∇ 2 u) ≥ 0 on D ∪ 0, where G + (x, r, η, X ) =

F(x, r, η, X ) if x ∈ D max(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 ) ∈ J D2,− u(x), x ∈ D 2,− max(F(x, u(x), η, X ), B(x, u(x), η, X )) ≥ 0 for (η, X ) ∈ J D∪0 u(x), x ∈ 0.

Suppose now we have a sequence of continuous functions u p that solve F p ≥ 0 in D p

and

B p ≥ 0 on 0 p

in the viscosity sense, for continuous proper operators F p and B p , open subsets D p of Rn and relatively open 0 p ⊂ ∂ D p . Suppose also that D p and 0 p converges in the Hausdorff metric to D∞ and 0∞ ⊂ ∂ D∞ and that u p converges locally uniformly to a continuous function u ∞ . Then the following assertion holds. Theorem 9.4. If the operators F p and B p converges locally uniformly to operators F∞ and B∞ respectively as p → ∞ and u p are as above, then the limit u ∞ solves

F∞ ≥ 0 in D∞

and

B∞ ≥ 0 on 0∞

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 (F B∞ ).


23

Proof. Take a sequence of points yk ∈ ∞ \ K , converging to x0 ∈ ∂∞ . Let dk = dist (yk , ∂∞ ) and xk ∈ ∂∞ be such that |yk −xk | = dk . Then without loss of generality we may assume ek = (yk − xk )/dk converge to e0 = (1, 0, . . . , 0). We want to prove that lim sup k→∞

u ∞ (yk ) ≤ a(x0 ). dk

Define u ∞ to be identically zero off ∞ and consider the functions vk (x) =

u ∞ (xk + dk x) dk

x ∈ Rn .

Since u ∞ is Lipschitz continuous, the family {vk } is uniformly Lipschitz. Hence we may extract a subsequence converging uniformly on compacts to a Lipschitz function v0 . Assume that this is the whole sequence. Then u ∞ (yk ) = vk (ek ) → v0 (e0 ) dk and we will be done if we show v0 (e0 ) ≤ a(x0 ). In fact, the following identity holds:

(10.1)

v0 (x) = a(x0 )(x · e0 )+

x ∈ Rn .

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 fix s0 ∈ (0, 1), we can find a constant c0 such that (10.2)

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

see the proof of Lemmas 5.1 and 8.1. Observe next, that Dk ⊂ 5k = {x: x · ek > 0} by definition and that 0 ∈ ∂ Dk ∩ ∂5k . Since ∞ has C 1 regular boundary and Dk are dilations of ∞ with a coefficient 1/dk → ∞, we will have that Dk converge to 50 and (10.2) will give v0 (x) ≥ c0 dist (x, 5c0 ).

24


Hence the positivity set of v0 contains 50 . Since vk are 0 off 5k , v0 is 0 off 50 . This proves the statement. Step 2. There is α ≥ a(x0 ) such that v0 (x) = α(x · e0 )+ . Let ν be the vector field in Corollary 8.2 and set νk (x) = ν(xk + dk x). Then ∇vk (x) = |∇vk (x)|νk (x) a.e. in Dk . Observe now, that νk (x) → ν(x0 ) = e0 uniformly on compacts subsets of 50 . Suppose φ ∈ L 2loc (50 , Rn ) is a weak limit of a subsequence of ∇vk (note, ∇vk are uniformly bounded in L ∞ norm). Then e · e0 = 0 implies φ · e0 = 0 a.e. in 50 . This means φ = |φ|e0 a.e. in 50 . Besides, we will have ∇v0 = φ in the distributional sense and therefore ∂e v0 = 0 for any direction e orthogonal to e0 . Therefore v0 (x) = f (x · e0 ) in 50 for a certain function f , positive and continuous. Besides, f will be nondecreasing. Next, observe that v0 satisfies 1∞ v0 = 0 in the viscosity sense, since vk satisfy 1∞ vk = 0 in Dk \ {vk ≥ 1/dk }. Then we leave to the reader as a good exercise to show that for v0 of the special form f (x ·e0 ) as above this means f 00 (t)( f 0 (t))2 = 0 in the viscosity sense. This, in turn, implies that f is a linear function, i.e. f (t) = αt +β. Since f (0) = 0 we obtain f (t) = αt and consequently v0 (x) = α(x ·e0 )+ everywhere in Rn . That α ≥ a(x0 ) follows simply from the observation that α = 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 (10.3) −∇v0 · e0 + a(x0 ) ≥ 0 on ∂50 . To prove this statement, we are going to apply Theorem 9.4. We first refine our constructions. Let dk, p = dist (yk , ∂ p ) and xk, p ∈ ∂ p be such that dk, p = |yk − xk, p |. Let also ek, p = (yk − xk, p )/dk, p . Define next vk, p (x) =

u p (xk, p + dk, p x) dk, p

with the convention that u p = 0 off p .

for x ∈ Rn ,


25

Now, choose pk so large that e˜k : = ek, pk → e0 and moreover v˜k : = vk, pk → ek = {v˜k > 0} and R > 0 is fixed, then v0 uniformly on compacts. If now D ek ∩ B R (0) and ∂ D ek ∩ B R (0) converge in the Hausdorff distance to 50 ∩ B R (0) D and ∂50 ∩ B R (0) respectively. Moreover, if ν˜ k (x) = ∇ v˜k (x)/|∇ v˜k (x)| and a˜ k (x) = a(xk, pk + dk, pk x) then ν˜ k (x) and a˜ k (x) converge uniformly in B R (0) to e0 and a(x0 ). Set next for p > 1 F p (x, u, ∇u, ∇ 2 u) = −

|∇u|4− p |∇u|2 1pu = − 1u − 1∞ u. p−2 p−2

and observe that F p converges locally uniformly on compacts to F∞ = −1∞ as p → ∞. Fix now large R > 0. Then v˜k is a classical and therefore viscosity solution of the following boundary value problem (10.4)

ek ∩ B R (0) F p (x, u, ∇u, ∇ 2 u) ≥ 0 in D ek ∩ B R (0) −∇u · ν˜ k (x) + a˜ k (x) ≥ 0 on ∂ D

Applying now Theorem 9.4, we conclude immediately that v0 is a viscosity solution of −1∞ v ≥ 0 in 50 ∩ B R (0) (10.5) −∇v · e0 + a(x0 ) ≥ 0 on ∂50 ∩ B R (0) for every R. Since R is arbitrary, (10.3) follows. Step 4. In the representation v0 (x) = α(x · e0 )+ , we have α ≤ a(x0 ). This follows from (10.3). Indeed, for β < α, we have v0 (x) = α(x · e0 ) ≥ β(x · e0 ) + (x · e0 )2

for x ∈ 50 ,

as x → 0, which means that (βe0 , e0t e0 ) ∈ J52,− v0 (0). Hence we must have by 0 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{−2β 2 , −β + a(x0 )} ≥ 0. This implies −β + a(x0 ) ≥ 0

26


and since β < 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 the classical solutions (u p , p ) of (F B p ) as p → 1+. The answer is extremely simple. Theorem 11.1. As p → 1+, the domains p collapse to K , provided n ≥ 2.

Remark 11.2. We stress here that the space dimension n ≥ 2. For n = 1 the statement of Theorem 11.1 fails, since all the problems (F B p ) are the same for all p ∈ (1, ∞).

Proof of Theorem 11.1. We follow a scheme, similar to that in the proof of Theorem 3.3. For large E denote φ E (s) =

e Es − 1 E

and let δ E > 0 be such that φ E (δ E ) = 1, or explicitly δE =

log(1 + E) . E

Note that δ E → 0 as E → ∞. Let u E (x) = φ E (δ E − a0 dist (x, K )) in E = {x: dist (x, K ) < δ E /a0 }. Assume for a moment that K has C 2 regular strictly convex boundary, such that the mean curvature κ(x) ≥ κ1 > 0 on ∂ K . Then we claim that (11.1) (u E , E ) is a classical supersolution of (F B p ) for 1 < p < p(E).


27

The free boundary conditions are easily verified, hence we need to prove only that 1 p u E ≤ 0. Computing as in the proof of Theorem 3.3, we obtain |∇u E |2− p 1 p u E = φ 0E a0 (( p − 1)a0 E − (n − 1)κ), where κ is the mean curvature at a point on the level line { dist (x, K ) = s} with 0 < s < δ E /a0 . Then κ ≥ κ1 /(1 + sκ1 ) and consequently, if E is so large that δ E /a0 ≤ 1, κ ≥ κ2 = κ1 /(1 + κ1 ). Hence 1 p u E ≤ 0 as soon as ( p − 1)a0 E − (n − 1)κ2 ≤ 0 or equivalently (n − 1)κ2 . a0 E Thus (11.1) and consequently the theorem follow in the case when K has a strictly convex C 2 regular boundary. For general K we can find K η ⊃ K that converges to K as η → 0, with the properties as above, so that we can conclude lim p→1+ p ⊂ K η . Then letting η → 0 we complete the proof of the theorem. p 0 in K c and the uniform interior ball condition for K , we can construct explicit (radially symmetric) supersolutions and subsolutions to guarantee (∗). Then the following assertion holds. Proposition A.1. (u ∗ , ∗ ) is the classical solution of (F B p ).

Here is the sketch of the proof. Step 1. Prove that (u ∗ , ∗ ) is a classical supersolution of (F B P ). This is relatively easy to show, using that v = |∇u| p is a subsolution of the linearized p-Laplacian L u (v) = |∇u| p−2 1v + ( p − 2)|∇u| p−4 ∇ 2 v∇u · ∇u ≥ 0, provided u is p-harmonic. Step 2. Prove that lim sup |∇u ∗ (y)| = a(x) as y → x, y ∈ ∗ , for every extreme point x ∈ ∂∗ . The latter means that x is not a convex combination of points on ∂∗ other than x. The points x ∈ ∂∗ with a hyperplane 5 touching ∂∗ at x only are dense among all extreme points, so it is enough to prove the statement only for them. If in an arbitrary small neighborhood of such x we have |∇u ∗ | < a, then we can cut from ∗ a small cap by a parallel translation of the supporting hyperplane 5, thus constructing a new supersolution; see [HS2, Lemma 3.4]. This, however, will contradict the minimality of ∗ . Step 3. Observe that Step 2 implies C 1 regularity of ∂∗ . Indeed, let x ∈ ∂∗ with two supporting hyperplanes. Then by barrier arguments, we will have |∇u ∗ (y)| → 0 as y → x, a contradiction. Step 4. Having C 1 regularity of ∂∗ , one can show that in fact lim inf |∇u ∗ (y)| = a(x) as y → x, y ∈ ∗ , for every extreme point x ∈ ∂∗ . 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 use the assumption that 1/a(x) is concave along with the following nice observation due to P. Laurence and E. Stredulinsky [LS]. Lemma A.2. Let D be a convex domain with C 1 regular boundary ∂ D and

` ⊂ ∂ D a line segment. Let also u be a nonnegative C 1 function with convex level sets, defined in D and continuously vanishing on ∂ D. Then 1/|∇u| is convex on `. This completes the proof of the proposition and hence Theorem 2.3 follows. ACKNOWLEDGMENT Authors 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.


29

REFERENCES [AM]

A. Acker and R. Meyer, A free boundary problem for the p-Laplacian: uniqueness, convexity, and successive approximation of solutions Electron. J. Differential Equations 1995, No. 08, approx. 20 pp.

[AC]

H. W. Alt and L. A. Caffarelli,Existence and regularity for a minimum problem with free boundary, J. Reine Angew. Math. 325 (1981), 105–144.

[Ar]

G. Aronsson, Extension of functions satisfying Lipschitz conditions, Ark. Mat. 6 (1967) 551– 561.

[BDM] T. Bhattacharya, E. DiBenedetto and J. Manfredi, Limits as p → ∞ of 1 p u p = f and related extremal problems, Rend. Sem. Mat. Univ. Politec. Torino 1989, Special Issue, 15–68. [CIL]

M. G. Crandall, H. Ishii and P.-L. Lions, User’s guide to viscosity solutions of second order partial differential equations, Bull. Amer. Math. Soc. (N.S.) 27 (1992), no. 1, 1–67.

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M. Dobrowolski, On quasilinear elliptic equations in domains with conical boundary points, J. Reine Angew. Math. 394 (1989),186–195.

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[HS2]

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

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A. Henrot and H. Shahgholian, The one-phase free boundary problem for the p-Laplacian with non-constant Bernoulli boundary condition, in preparation.

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U. Janfalk, Behaviour in the limit, as p → ∞, of minimizers of functionals involving p-Dirichlet integrals, SIAM J. Math. Anal. 27 (1996), no. 2, 341–360.

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

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I. N. Krol’, The solutions of the equation Dxi (Du p−2 Dxi u) = 0 with a singularity at a boundary point (Russian) Boundary value problems of mathematical physics, 8. Trudy Mat. Inst. Steklov. 125 (1973), 127–139.

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M. A. Lavrent’ev, Variational methods for boundary value problems for systems of elliptic equations, Translated from the Russian by J. R. M. Radok, Second edition, Dover, New York, 1989.

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

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P. Laurence and E. Stredulinsky, Existence of regular solutions with convex levels for semilinear elliptic equations with nonmonotone L 1 nonlinearities. I. An approximating free boundary problem, Indiana Univ. Math. J. 39 (1990), no. 4, 1081–1114.

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L. E. Payne and G. A. Philippin, On gradient maximum principles for quasilinear elliptic equations, Nonlinear Anal. 23 (1994), no. 3, 387–398.

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A. L. Vogel, Symmetry and regularity for general regions having a solution to certain overdetermined boundary value problems, Atti Sem. Mat. Fis. Univ. Modena 40 (1992), no. 2, 443–484.

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