Annales Academiæ Scientiarum Fennicæ Mathematica Volumen 24, 1999, 109–121

SOME QUALITATIVE ASPECTS OF A FREE BOUNDARY PROBLEM FOR THE p-LAPLACE OPERATOR

Hassan Hosseinzadeh and Henrik Shahgholian

Shahid Bahounar University, Faculty of Mathematics and Computer Science Kerman, Iran Royal Institute of Technology, Department of Mathematics S-100 44 Stockholm, Sweden; henriks@math.kth.se

Abstract. Given an L^{∞} -function μ ≥ 0, with compact support in R^{n}, and a bounded domain Ω (⊃ supp (μ)). Suppose there exists a function u, satisfying the following overdetermined boundary value problem

div(|∇u|^{p−2}∇u) = −μ in Ω,
u = 0  on ∂Ω,
−∂u/∂ν = 1  on ∂Ω,

where ν is the outward unit normal vector on ∂Ω (the boundary of Ω). We assume that ∂Ω is C^{2} , u ∈ C^{2}(\overline{Ω} \ supp (μ)) ∩ C^{1}(\overline{Ω}), and 1 < p < ∞.

In this paper we study uniqueness, uniform bounds and some geometric properties for solutions (u, Ω) of the above overdetermined (free) boundary value problem. We show, among other things, that if (u_{i}, Ω_{i}) (i = 1, 2) are two solutions to the above problem for a fixed μ, with Ω_{1} convex, then Ω_{2} ⊂ Ω_{1} . Consequently if both Ω_{1} and Ω_{2} are convex, then they coincide. Using an argument due to J. Serrin, we also prove certain types of symmetry and monotonicity, along lines, for u.

0. Introduction

The mean value property for harmonic functions

(0.1)  ∫_{∂B(x^{0},r)} h dσ = c_{n}r^{n−1}h(x^{0}),

is well known and has a central role in classical potential theory. Here σ is the surface element, and c_{n} is the area of the unit sphere in R^{n} (n ≥ 2). Now (0.1) can be written as

(0.2)  ∫_{∂B(x^{0},r)} h dσ = ∫ h dμ,

1991 Mathematics Subject Classification: Primary 35R35, 31B20, 35J60.

Henrik Shahgholian thanks Swedish Natural Science Research Council and University of Mazandaran for support. He also thanks T. Kilpeläinen for stimulating discussions on the p-Laplacian.

where μ = c_{n}r^{n−1}δ_{x^{0}} , is a constant multiple of the Dirac measure. Hence, letting

h_{y}(x) = \{ \begin{matrix}(1/c_{2}) log |x − y|^{−1} & n = 2, \\ (1/(n − 2)c_{n})|x − y|^{2−n} & n ≥ 3, \end{matrix}

where y is fixed, and defining

u(y) = ∫_{∂B(x^{0},r)} h_{y}(x) dσ − ∫ h_{y}(x) dμ in B(x^{0}, r),

we have (by (0.2)) u = 0, and −∂u/∂ν = 1 on ∂B(x^{0}, r) (see [K]). Here ν is the outward normal vector on ∂B(x^{0}, r). The function u, thus defined, satisfies (in the sense of distributions)

(0.3)  \{ \begin{matrix}Δu = −μ & in B(x^{0}, r), \\ u = 0 & on ∂B(x^{0}, r), \\ −∂u/∂ν = 1 & on ∂B(x^{0}, r). \end{matrix}

In general, one may consider the following question: For a given (positive Radon) measure (in this paper, apart from Examples 1–6, we only consider positive L^{∞} functions) with compact support, find a domain Ω (⊃ supp (μ)), and a function u satisfying (in some weak sense)

(0.4)  \{ \begin{matrix}Δ_{p}u = −μ & in Ω, \\ u = 0 & on ∂Ω, \\ −∂u/∂ν = 1 & on ∂Ω, \end{matrix}

where Δ_{p}u = div(|∇u|^{p−2}∇u), is the p-Laplace operator and 1 < p < ∞. This is an overdetermined (free) boundary value problem and hence not always wellposed (see Examples 4 and 5 below). The existence depends, strongly, on the concentration of the measure μ. For p = 2 there are some results on the existence for certain types of μ; see [GS] and [Sh1]. As for other values of p, there are no results known to the authors.

One may also change the problem slightly by letting μ be identically zero, and instead having nonzero Dirichlet boundary conditions, working in a subdomain of R^{n}; see [AC] for p = 2, and [HS] for 1 < p < ∞, and convex subdomains.

In this paper, we will not consider the question of existence or regularity of solutions to (0.4). We only consider some qualitative aspects of the solutions, assuming they already exist. Our results concern uniqueness (Theorem 1), uniform bounds (Theorem 2), monotonicity along lines (Theorem 3), and starshapeness (Theorem 4).

Rules. Throughout this paper the following rules apply: Ω denotes a bounded domain, with C^{2} boundary; except from Examples 1–6, μ always denotes a nonnegative L^{∞} -function with

supp (μ) ⊂ Ω;

ν denotes the unit normal vector on the boundary, and pointing outward, of a domain in consideration.

Any solution u to (0.4) is extended continuously to entire R^{n}, with u ≡ 0 in R^{n} \ Ω; we also assume that if (u, Ω) is a solution to (0.4) then u ∈ C^{1}(\overline{Ω}) and it is C^{2} in an interior neighborhood of ∂Ω, and has a C^{2} extension across ∂Ω. It is also noteworthy that since u is constant on ∂Ω and positive in Ω, we will have

|∇u| = −∂u/∂ν on ∂Ω.

1. Examples

In this section we give a few explicit examples, where μ will be a non-negative measure. For this purpose we define a_{1} = (p − 1)/(p − n), a_{2} = p/(p − 1), a_{3} = −n/(p − 1), and a_{4} = a_{1} + r^{1/a_{1}}(1/a_{2} − a_{1}), for a fixed r.

Example 1. Let

u(x) = \{ \begin{matrix}a_{1}(1 − |x|^{1/a_{1}}), & p ≠ n, \\ log |x|^{−1}, & p = n. \end{matrix}

Then on ∂B(0, 1), u = 0, and |∇u| = 1, and in B(0, 1) we have Δ_{p}u = −c_{n}δ_{0} , where c_{n} is as in (0.1).

Example 2. Let p ≠ n, and set f(r) = r − r^{1−1/a_{1}} . Then it is not hard to see that for each 0 < r_{1} < 1 there is r_{2} > 1 such that f(r_{1}) = −f(r_{2}). Hence r_{1} + r_{2} = r_{1}^{1−1/a_{1}} + r_{2}^{1−1/a_{1}} . This can be used to show that the function u defined as

u(x) = \{ \begin{matrix}a_{1}r_{1}((|x|/r_{1})^{1/a_{1}} − 1), & r_{1} ≤ |x| ≤ 1, \\ a_{1}r_{2}(1 − (|x|/r_{2})^{1/a_{1}}), & 1 ≤ |x| ≤ r_{2}. \end{matrix}

is Lipschitz. Now define Ω = {x : r_{1} < |x| < r_{2}}. Then (u, Ω) solves (0.4) with dμ = (r_{1}^{n−1} + r_{2}^{n−1})χ_{∂B(0,1)}dσ.

Example 3. Let p ≠ n, 0 < r < 1, and define u as

u(x) = \{ \begin{matrix}(−r^{a_{3}}/a_{2})|x|^{a_{2}} + a_{4}, & |x| ≤ r, \\ a_{1}(1 − |x|^{1/a_{1}}), & r < |x| ≤ 1. \end{matrix}

Then Δ_{p}u = −nr^{−n}χ_{B(0,r)} and u = 0 and |∇u| = 1 on ∂B(0, 1).

For the case p = n, in Examples 2 and 3, one may define a similar function using the log-function.

Example 4 (Non-existence). Here we give a simple example of an L^{∞} - function μ, such that (0.4) has no solutions. Fix μ and let (u, Ω) solve (0.4). Then, integration by parts gives

(1.1)  ∫ dμ = − ∫_{Ω} div(|∇u|^{p−2}∇u) = − ∫_{∂Ω} |∇u|^{p−2}∇u · ν = ∫_{∂Ω} dσ.

Now let μ be defined as dμ = Cχ_{D}dx, where

C = n^{(n−1)/n}(\frac{c_{n}}{volume (D)})^{1/n},

and c_{n} is as in (0.1). Then (1.1) reduces to

n^{(n−1)/n} · (\frac{c_{n}}{volume (D)})^{1/n} · volume (D) = area (∂Ω),

and hence, by the (strict) inclusion D ⊂ Ω,

c_{n}^{1/n}(n volume (Ω))^{(n−1)/n} > c_{n}^{1/n}(n volume (D))^{(n−1)/n} = area (∂Ω).

This obviously contradicts the well-known isoperimetric inequality [B]. Therefore for μ as above there cannot exist a solution to (0.4).

Example 5 (Non-existence). Let dμ = f(x) dx, and suppose (u, Ω) solves (0.4). Set M = sup f, and let B(x^{0}, r_{Ω}) be the smallest ball containing Ω. Then we claim

(1.2)  n < Mr_{Ω}.

This provides us with a test for non-existence.

To prove (1.2), we define

v(x) = (\frac{p − 1}{p})\frac{r_{Ω}^{p/(p−1)} − |x − x^{0}|^{p/(p−1)}}{r_{Ω}^{1/(p−1)}} .

Then

Δ_{p}v = \frac{−n}{r_{Ω}}.

Now if (1.2) fails, then

Δ_{p}u = −f ≥ −M ≥ \frac{−n}{r_{Ω}} = Δ_{p}v in Ω.

Since also u = 0 ≤ v on ∂Ω, we may apply Lemma 1 (a) and (c) (Section 3) to deduce that u < v in Ω. (Or at least in some interior neighborhood of ∂Ω; see the remark following Lemma 1.) Now let y ∈ ∂Ω correspond to the largest distance to x^{0} , i.e. |y − x^{0}| = r_{Ω} , and observe that the unit outward normal vector ν at y equals (y − x^{0})/|y − x^{0}| and that u(y) = v(y) = 0. Invoking part (b) of Lemma 1 we conclude

(1.3)  −1 = \frac{∂u}{∂ν}(y) > \frac{∂v}{∂ν} (y) = −1,

which is a contradiction. Hence (1.2) holds.

Example 6 (Perimeter of Ω). Let (u, Ω) solve (0.4), and let us drop the assumption that ∂Ω is C^{2} . The boundary condition ∂u/∂ν = −1 is to be interpreted as (replaced by) |∇u| = 1, in the sense that for any small ε > 0 there is a small neighborhood N_{ε} of ∂Ω such that

(1.4)  | |∇u| − 1| < ε,  in Ω ∩ N_{ε}.

We will show that Ω has finite perimeter; see [Z] for a definition. Let us define v = u^{p/(p−1)} (observe that u ≥ 0). Then one may easily verify that

Δ_{p}v = C_{p}(uΔ_{p}u + |∇u|^{p}), C_{p} = (\frac{p}{p − 1})^{p−1}.

Since Δ_{p}u = 0 in {0 < u < ε}, for ε small, we have

Δ_{p}v = C_{p}|∇u|^{p} ≥ \frac{1}{2}C_{p}.

Using this and that |∇v| = 0 on {u = 0}, we have (by (1.4))

\begin{aligned}\frac{1}{2}C_{p} ∫_{\{0<u<ε\}} dx & ≤ ∫_{\{0<u<ε\}} Δ_{p}v dx = ∫_{\{u=ε\}} |∇v|^{p−2}∇v · ν dσ \\ & = C_{p} ∫_{\{u=ε\}} |∇u|^{p−1}u dσ = C_{p}ε ∫_{\{u=ε\}} |∇u|^{p−1} dσ = C_{p}ε ∫ dμ. \end{aligned}

This shows that

(1.5)  χ_{{u>ε}} → χ_{{u>0}}  in L^{1}(R^{n}).

Next, by (1.4), for any ε > 0 there is δ_{ε} ↘ 0 such that |∇u| > 1 − δ_{ε} in {u ≤ ε}. Hence

∫ dμ = − ∫_{Ω} Δ_{p}u dx = ∫_{∂{u>ε}} |∇u|^{p−1} dσ ≥ (1 − δ_{ε})^{p−1} Perimeter ({u > ε}).

Letting ε → 0 and using (1.5) and the lower-semicontinuity of the variation measure (perimeter) we obtain

Perimeter (Ω) = Perimeter ({u > 0}) ≤ ∫ dμ.

2. Main results

In this section we state our main results, which will be proven in Section 4. These results are of qualitative nature (uniqueness and geometric features) and for p = 2 they where proven in [GS] and [Sh2]. To avoid technical difficulties we assume, throughout this section, that u is C^{1} on \overline{Ω}, it is C^{2} in an interior neighbourhood of ∂Ω and it has a C^{2} extension across ∂Ω. Also once again we stress that supp (μ) ⊂ Ω, and μ is a bounded L^{∞} -function with compact support.

Theorem 1. Let (u_{j}, Ω_{j}) (j = 1, 2) be solutions to (0.4), with supp (μ) ⊂ Ω_{1} ∩ Ω_{2} . Then the following hold.

(1) If Ω_{1} is convex then Ω_{2} ⊂ Ω_{1} .

(2) If Ω_{1} ∩ Ω_{2} is convex, then Ω_{1} ≡ Ω_{2} .

(3) If both Ω_{1} and Ω_{2} are convex, then Ω_{1} ≡ Ω_{2} and u_{1} ≡ u_{2} .

Theorem 2. Let 1 < p ≤ n and (u, Ω) be a solution to (0.4), with Ω bounded. Then there exists R = R(μ, p, n) such that

Ω ⊂ B(0, R).

For the next result we need some new definitions and notations.

For a fixed unit vector a ∈ R^{n} and for λ ∈ R set

T_{λ} = T_{a,λ} := {x · a = λ},  Tλ^{−} := {x · a < λ},  Tλ^{+} := {x · a > λ}.

For x ∈ R^{n} let x^{λ} denote the reflected point with respect to T_{λ} and set u^{λ}(x) = u(x^{λ}). Also let

Ω_{λ} = Ω ∩ Tλ^{+} = the cap cut off by T_{λ},
Ω̃_{λ} = {x^{λ} : x ∈ Ω_{λ}} = the reflection of Ω_{λ} in T_{λ}.

Theorem 3. Assume that for some unit vector a ∈ R^{n} and some λ_{0} ∈ R we have

supp (μ) ∩ Tλ^{+} = ∅,

for all λ ≥ λ_{0} . Then for any solution u to (0.4) the following hold.

(2.1)  Ω̃_{λ} ⊂ Ω  for all λ ≥ λ_{0},
(2.2)  u ≤ u^{λ}  in Ω_{λ} for all λ > λ_{0},
(2.3)  a · ∇u ≤ 0  in Ω_{λ_{0}}.

Corollary 1. Let (u, Ω) be a solution to (0.4) and suppose supp (μ) ⊂ K, a compact convex set. Then for any x ∈ ∂Ω \ K the inward normal ray L_{x}, of ∂Ω at x intersects K. Moreover, ∂Ω \ K is Lipschitz.

Proof. If for x ∈ ∂Ω \ K we have L_{x} ∩ K = ∅ then one can find a ∈ R^{n} and λ_{0} ∈ R such that K ⊂ Ta,λ_{0}^{−}, L_{x} ⊂ Ta,λ_{0}^{+} . The first inclusion implies that the assumptions of Theorem 3 are satisfied and hence (2.1) holds. This in particular means that ∂Ω ∩ Ta,λ_{0}^{+} is a graph, seen from T_{a,λ_{0}} , and therefore the outward normal vector ν_{y} at every point y ∈ ∂Ω ∩ Ta,λ_{0}^{+} makes an angle 0 ≤ θ < \frac{1}{2}π with a, i.e.

(2.4)  ν_{y} · a > 0  for all y ∈ ∂Ω ∩ Ta,λ_{0}^{+}  (scalar product).

Now the second inclusion (L_{x} ⊂ Ta,λ_{0}^{+} ) implies that ν_{x} (outward normal at x) points into Ta,λ_{0}^{−}. Hence the angle between a and ν_{x} is greater than \frac{1}{2}π, i.e. ν_{x} · a < 0. Since ∂Ω is C^{1} we may choose y ∈ ∂Ω ∩ Ta,λ_{0}^{+} near x with ν_{y} pointing into Ta,λ^{−}, i.e. ν_{y} ·a < 0. This contradicts (2.4). Hence the first statement of the corollary follows. The second statement follows easily by varying a and λ_{0} such that K ⊂ Ta,λ_{0}^{−}.

For the proof of Theorem 3 we shall use some reflection methods related to the “moving plane method” which has previously been used in similar problems in [Se], [GNN], [BN], [Sh2], [GS].

Theorem 4. Let (u, Ω) solve (0.4) and suppose there is α ≤ −1 such that

(2.5)  t^{α}μ(x/t) ≤ μ(x),

for all 0 < t ≤ 1 and x ∈ R^{n}. Then Ω is starshaped with respect to the origin, and moreover

(2.6)  t^{β}u(x/t) ≤ u(x)

for β = (p + α)/(p − 1) and all 0 < t ≤ 1 and x ∈ Ω.

3. Auxiliary lemmas

This section is devoted to some technical lemmas.

Lemma 1 (Hopf’s comparison principle). Let D ⊂ R^{n} be bounded, and v_{1}, v_{2} ∈ C^{1}(\overline{D}), with

Δ_{p}v_{1} ≤ Δ_{p}v_{2}.

Then the following hold.

(a) If v_{1} ≥ v_{2} on ∂D, then v_{1} ≥ v_{2} in D.

(b) Suppose v_{1} > v_{2} in D, v_{1}(x^{0}) = v_{2}(x^{0}) for some x^{0} ∈ ∂D, |∇v_{2}| ≥ γ > 0 in D (for some γ > 0), and D satisfies the interior sphere condition. Then ∂v_{1}/∂ν(x^{0}) < ∂v_{2}/∂ν(x^{0}), where ν is the unit outward normal vector on ∂D, at x^{0} .

(c) If v_{1} ≥ v_{2} and v_{1} ≢ v_{2} in D, |∇v_{2}| ≥ γ > 0 (for some γ > 0), then v_{1} > v_{2} in D.

Remark. Throughout this paper we will, repeatedly, use Lemma 1. It is thus crucial that the relation |∇v_{2}| > γ, for some γ > 0, holds for the function v_{2} in part (b) and (c) of the lemma. This, however, is not needed for part (a) of the lemma. Now in our applications of this lemma we will always consider part (a) in the entire domain in consideration and part (b) and (c) in a small subdomain with x^{0} on its boundary. Since the magnitude of the gradient of any solution to (0.4) approaches one, continuously, the required condition in the lemma is fulfilled, near the boundary for solutions to (0.4). In the sequel we will omit mentioning this argument.

Lemma 2 (Serrin’s boundary point lemma). Let D be a domain with C^{2} boundary and let T be a plane containing the normal to ∂D at some point x^{0} ∈ ∂D. Denote by D′ the portion of D that lies on some particular side of T. Suppose that w_{1} and w_{2} are of class C^{2}(\overline{D}), w_{1} ≢ w_{2} in D, w_{1}(x^{0}) = w_{2}(x^{0}) and satisfy

Δ_{p}w_{1} ≤ Δ_{p}w_{2},  w_{1} ≥ w_{2}  in D′.

Then either

\frac{∂w_{1}}{∂τ} > \frac{∂w_{2}}{∂τ}  or  \frac{∂^{2}w_{1}}{∂τ^{2}} > \frac{∂^{2}w_{2}}{∂τ^{2}}  at x^{0},

where τ is any direction entering D′ non-tangentially.

Lemma 3. Let (u, Ω) solve (0.4). Let T_{λ} be a plane containing the normal to ∂Ω at some point x^{0} ∈ ∂Ω. Suppose, moreover, Ω̃_{λ} ⊂ Ω. Then the first and second derivatives of u and ũ coincide at x^{0} .

The proofs of Lemmas 1, 2 and 3 are much the same as that of the case p = 2. Indeed, since |∇u| > 0 near the boundary, in all three lemmas, we may rewrite the p-Laplacian as

h_{1}(|∇u|)Δu + h_{2}(|∇u|) ∑_{i,j} u_{i}u_{j}u_{ij},

where h_{1} and h_{2} are positive C^{1} functions in their argument. Hence we will have a uniformly elliptic operator, which is translation and rotation invariant. However, for the reader’s convenience we give exact references for the proofs of the lemmas. Lemma 1 is proven in [T, Lemma 3.2, Propositions 3.4.1, 3.4.2]. A proof for Lemma 2 can be found in ([Se, Lemmas 1 and 2]). The proof of Lemma 3 is also the same as in ([Se, p. 307]).

Lemma 4. Let (u, Ω) be a solution to (0.4). Consider a hyperplane T, which cuts off Ω a cap Ω′, and such that supp (μ) ∩ \overline{Ω′} = ∅. Then

(3.1)  d := sup_{x∈∂Ω′} dist (x, T) < sup_{T} u.

Moreover if x^{0} ∈ T is such that u(x^{0}) = sup_{T} u, then

(3.2)  −\frac{∂u}{∂a}(x^{0}) > 1,

where a is the unit normal vector to T pointing inward Ω′.

Proof. Since the problem is rotation and translation invariant we may assume that T = {x_{1} = 0}, and Ω′ = {x_{1} > 0} ∩ Ω. Now let y ∈ ∂Ω′, be such that d = dist (y, T) and observe that (∂u/∂x_{1})(y) = −1. Then define

h(x) = s(d − x_{1}),

where

s := \frac{sup_{T} u}{d}.

Now observing that

(3.3)  Δ_{p}h = Δ_{p}u = 0 in Ω′,  h ≥ u on ∂Ω′,  u(y) = h(y) = 0

we may invoke Lemma 1 (a)–(c) to deduce

−s = \frac{∂h}{∂x_{1}} (y) < \frac{∂u}{∂x_{1}} (y) = −1,

i.e.

(3.4)  \frac{sup_{T} u}{d} = s > 1,

which proves (3.1). Next using (3.3)–(3.4) and the fact that u(x^{0}) = h(x^{0}) we obtain

1 < s = −\frac{∂h}{∂x_{1}} (x^{0}) ≤ −\frac{∂u}{∂x_{1}} (x^{0}),

i.e. (3.2) holds.

Lemma 5 (Pasting lemma, [HKM, 7.9]). Let u be any solution to (0.4) and extend it to the entire R^{n} by defining it to be zero in the exterior of Ω. Then

Δ_{p}(u − c) ≥ −μ,

for any constant c.

The proof of this lemma follows easily from the pasting lemma [HKM, 7.9]. However, for the reader’s convenience we give the proof.

Proof. Take a small neighborhood N of ∂Ω such that supp (μ) ∩ N = ∅, and define

v = \{ \begin{matrix}max(u, 0) & in N ∩ Ω, \\ 0 & in R^{n} \ Ω. \end{matrix}

Then, by [HKM, 7.9], −v is p-superharmonic in N. Hence v−c, is p-subharmonic in N which is the desired result.

4. Proofs of the main theorems

Proof of Theorem 1. Since the proofs of (1) and (2) are similar, we only prove (1); (3) follows from (1) or (2). Suppose Ω_{2} \ Ω_{1} ≠ ∅, then we reach a contradiction. Extend u_{j} by zero to R^{n} \ Ω_{j}, for j = 1, 2, and let x^{0} ∈ ∂Ω_{1} be such that u_{2}(x^{0}) = sup_{∂Ω1} u_{2} . Define now w(x) = u_{2} − u_{2}(x^{0}) in Ω_{1} . Then by Lemma 5 Δ_{p}w ≥ −μ. Hence

Δ_{p}w ≥ Δ_{p}u_{1} in Ω_{1},  w ≤ u_{1} on ∂Ω_{1},  w(x^{0}) = u_{1}(x^{0}) = 0.

We may thus apply Lemma 1 to deduce that

∂u_{2}/∂ν(x^{0}) = ∂w/∂ν(x^{0}) > ∂u_{1}/∂ν(x^{0}) = −1

i.e.

(4.1)  −∂u_{2}/∂ν(x^{0}) < 1.

Now, using the convexity of ∂Ω_{1} and that supp (μ) ⊂ Ω_{1} ∩ Ω_{2} , we can take a supporting plane T at x^{0} such that the assumptions of Lemma 4 are fulfilled. But then, (4.1) contradicts (3.2). This completes the proof of the theorem.

Proof of Theorem 2. By scaling, we may assume supp (μ) ⊂ B(0, 1) and μ ≤ 1. We also consider the case 1 < p < n; since for p = n we need only to replace the fundamental solution |x|^{(p−n)/(p−1)} , in the definition below, with the logarithmic kernel. Define

v = \{ \begin{matrix}a|x|^{p/(p−1)} + b, & |x| ≤ 1, \\ c|x|^{(p−n)/(p−1)}, & |x| ≥ 1, \end{matrix}

where

a = −n^{1/(1−p)}(\frac{p − 1}{p}),  c = (\frac{p − 1}{n − p})n^{1/(1−p)},  b = c − a.

Then v satisfies

Δ_{p}v = −χ_{B(0,1)} ≤ −μ ≤ Δ_{p}u in Ω, v > 0 in R^{n}.

Hence Lemma 1 can be applied to obtain v ≥ u. Define now

d := sup_{x∈∂Ω} dist (x, B(0, 1)).

Then by Lemma 4, and the above

d ≤ sup_{∂B(0,1)} u ≤ sup_{∂B(0,1)} v ≤ c.

This proves the theorem.

Proof of Theorem 3. Obviously Ω_{λ} = ∅, for large values of λ. We start moving T_{λ}, by decreasing λ, towards Ω and until it intersects Ω and produces the cap Ω_{λ}. Then, at the beginning of this intersection the reflected cap Ω̃_{λ} remains inside Ω, since ∂Ω is C^{2} . Now if the theorem fails, then for some λ > λ_{0} one of the following can occur.

(1) Ω̃_{λ} becomes internally tangent to ∂Ω at some point not on T_{λ}.

(2) T_{λ} reaches a position where it is orthogonal to ∂Ω at some point x^{0} , on ∂Ω. Suppose situation (1) occurs. Then u^{λ}, the reflection of u in T_{λ}, is p-harmonic in Ω̃_{λ}, and satisfies u^{λ} ≤ u on ∂Ω̃_{λ}. Therefore we can apply part (a) of Lemma 1 to conclude that u^{λ} ≤ u. Next we observe that near x^{0} , the touching point of ∂Ω̃_{λ} and ∂Ω, both |∇u^{λ}| and |∇u| are > 0. Hence part (b) and (c) of Lemma 1 can be applied to deduce

−1 = ∂u^{λ}/∂ν(x^{0}) > ∂u/∂ν(x^{0}) = −1,

which is a contradiction, and implies that situation (1) above cannot occur.

Now if situation (2) arises, then one may use Lemmas 2 and 3 to reach a contradiction. This completes the proof of (2.1).

Next (2.3) follows from (2.2). Now by (2.1) u^{λ} ≥ u = 0 on ∂Ω ∩ Tλ^{+} , and u^{λ} = u on T_{λ}. Since Δ_{p}u^{λ} ≤ Δ_{p}u in Ω_{λ} we can apply Lemma 1 (a) to deduce (2.2).

Proof of Theorem 4. For β = (p + α)/(p − 1), define

u_{t}(x) = t^{β}u(x/t), Ω_{t} = {x : x/t ∈ Ω},

and observe that by (2.5), the origin is in the interior of the support of μ and thus inside Ω. Hence for small values of t, Ω_{t} ⊂ Ω. Now define t_{0} to be the largest number such that Ω_{t} ⊂ Ω for all 0 < t < t_{0} , and let y ∈ ∂Ω ∩ ∂Ω_{t_{0}} . Since μ is of relation (2.5), we have

Δ_{p}u_{t_{0}} ≥ Δ_{p}u, in Ω_{t_{0}},  0 = u_{t_{0}} ≤ u on ∂Ω_{t_{0}},  u_{t_{0}}(y) = u(y) = 0.

If the conclusion in the theorem fails, then

(4.2)  t_{0} < 1,

and u_{t_{0}} ≢ u. We may thus invoke Lemma 1 (a)–(c) to conclude

(4.3)  −t_{0}^{β−1} = \frac{∂u_{t_{0}}}{∂ν} (y) > \frac{∂u}{∂ν}(y) = −1.

The assumption α ≤ −1 implies β − 1 ≤ 0, which in turn gives that (4.3) is in direct contradiction with (4.2). We thus conclude that t_{0} = 1 and Ω is starshaped. To show (2.6), we consider once again u_{t}(x) as above and observe, by the starshapeness, Ω_{t} ⊂ Ω (for all 0 < t ≤ 1). Hence by (2.5) Δ_{p}u_{t} ≥ Δ_{p}u in Ω_{t}, and u_{t} ≤ u on ∂Ω_{t}. Now Lemma 1 (a) gives the desired result.

5. Concluding remarks

Most of the results of this paper can easily be generalized to hold for the following situations:

(1) Degenerate operators of the form div(φ_{1}(u, |∇u|)∇u) + φ_{2}(u, |∇u|), with regular ingredients;

(2) μ a positive measure;

(3) ∂Ω has finite perimeter with no a priori smoothness assumptions;

(4) relaxation of ∂u/∂ν = −1 on ∂Ω to \overline{lim}_{Ω∋x→∂Ω}|∇u| = 1. The reason that we work with L^{∞} -functions rather than measures is the belonging of the solution to the space W^{1,p}(R^{n}). However, outside the support of μ, the solution is as good as Lipschitz, if e.g. we have the boundary behavior for the gradient as in (0.4). Therefore if we assume the solution is in the space W^{1,p}(R^{n}), then we can repeat the above arguments, without any changes, to obtain Theorems 1–4, for arbitrary nonnegative measures.

The proof of Theorem 2, given here, does not apply when p > n. The reason is that in this case, we cannot find nonnegative supersolutions in R^{n}, merely because of the behavior of the function |x|^{(p−n)/(p−1)} , near the infinity. We believe, however, that the theorem should hold for all 1 < p < ∞.

As mentioned earlier, most of the existing literature is concerned with the case p = 2. It is known that measures, which are enough concentrated on their support, do admit solutions; e.g., for finite positive combinations of Dirac masses there always exists a solution for (0.4), when p = 2. For other values of p, this is an open question. A natural way of finding solutions is through minimization of the functional

∫_{R^{n}} \frac{1}{p}|∇v|^{p} − μv + (\frac{p − 1}{p})χ_{\{v>0\}},

over the set

{v : v ∈ W_{0}^{1,p}(R^{n}), v ≥ 0}.

A minimizer, or stationary point, u of this functional expects to satisfy

(5.1)  Δ_{p}u = −μ + ℋ ^{n−1}⌊∂Ω,

where Ω = {u > 0}, and ℋ ^{n−1}⌊∂Ω is the n-dimensional Hausdorff measure supported on ∂Ω. Of course if ∂Ω is already known to be “nice” then it is not hard to prove that (5.1) holds. Also, for smooth boundaries, (5.1) expresses that |∇u| = 1 on ∂Ω. Thus the main problem, in using minimization, consists of showing the smoothness of the free boundary. This question (for p = 2) has been treated by H.W. Alt and L.A. Caffarelli in their pioneering paper [AC]. It is most likely that the same technique works for the p-Laplacian.

Qualitative aspects of a free boundary problem 121

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Received 14 May 1997