Symmetry Results for Reaction-Diffusion Equations^{∗}

Hans G. Kaper, Man Kam Kwong, Yi Li^{†}

Mathematics and Computer Science Division Argonne National Laboratory Argonne, IL 60439-4801

Abstract

This article is concerned with symmetry properties of the solutions of the reactiondiffusion equation Δu + f(u) = 0 in a bounded connected domain Ω in R^{N} (N = 2, 3, …). Of especial interest are nonlinear source terms f of the type f(u) = u^{p} − u^{q} with 0 ≤ q < p ≤ 1. Two results are presented.

The first result concerns the solution of a free boundary problem, where the domain Ω is unknown and u and its normal derivative ∂_{n}u are required to vanish on the boundary ∂Ω of Ω. It is shown that, if f is the sum of a continuous nondecreasing function and a Lipschitz continuous function on [0, ∞), then the free boundary problem does not have a positive solution unless Ω is a ball; in this case, any positive solution is radially symmetric around the center of the ball and decreasing with the radial distance from the center.

The second result concerns the solution of the Dirichlet problem on a ball in R^{N}, when the nonlinear source term f is continuous, but not necessarily Lipschitz continuous at 0. It is shown that, if f is the sum of a locally Lipschitz continuous function on (0, ∞) that is nonincreasing near 0 and a function that is Lipschitz continuous on [0, ∞), then any positive solution u is radially symmetric around the center of the ball and decreasing with the radial distance from the center.

AMS(MOS) Subject Classification. Primary 35J25. Secondary 35J65, 35R35.

Key Words. Reaction-diffusion equation, free boundary problem, Dirichlet problem, positive solution, radial symmetry, moving-plane method.

∗This work was supported by the Applied Mathematical Sciences subprogram of the Office of Energy Research, U. S. Department of Energy, under Contract W-31-109-Eng-38.

†Permanent address: Department of Mathematics, University of Rochester, Rochester, NY 14620. Supported in part by the National Science Foundation Grant DMS-8903192-A01.

1 Statement of the Problem

In this article we present two symmetry results for positive solutions of the reaction-diffusion equation

Δu + f(u) = 0, x ∈ Ω,  (1.1)

where Ω is a bounded connected domain in R^{N} (N = 2, 3, …). In the first problem, Ω is not specified, but both u and its normal derivative ∂_{n}u are required to vanish on the boundary ∂Ω of Ω. In the second, Ω is a ball in R^{N}, and u is required to vanish on the boundary of the ball.

The first problem would clearly be overdetermined if Ω were specified. However, as Ω is left unspecified, the extra degree of freedom may be enough to allow for a solution—which, in this case, consists of the pair (Ω, u). We refer to this problem as the free boundary problem. Of particular interest is the question whether a solution of the free boundary problem, if it exists, has any symmetry properties.

The second problem is a standard boundary value problem with Dirichlet data. We refer to it as the Dirichlet problem. The question here is whether a solution u is radially symmetric if f is continuous, but not necessarily Lipschitz continuous at the origin.

These problems are discussed in more detail in the following subsections. Our results are presented in Theorems 1 and 2 below. Section 2 contains some preliminary material for the proofs of these theorems; the proofs are given in Sections 3 and 4. This article extends and generalizes our earlier communication [1].

1.1 Free Boundary Problem

In a recent discussion of some computational problems in plasma physics, Miller et al. [2] proposed an interesting free boundary problem for reaction-diffusion equations. The physical problem, which concerns the existence of equilibrium configurations with magnetic islands in a Tokamak fusion device, can be posed as a free boundary problem for the partial differential equation

Δu + \sqrt{u} − 1 = 0, x ∈ Ω,  (1.2)

where Ω is a bounded connected domain in R^{N} (N = 2, 3, …). The function u must be positive throughout Ω, and u and its normal derivative ∂_{n}u must vanish on ∂Ω. The expression \sqrt{u} − 1 in the differential equation (1.2) is the simplest form the source term can take; another typical expression is u^{p} − u^{q}, where 0 ≤ q < p ≤ 1. Therefore, the general form of the free boundary problem is

\{ \begin{matrix}Δu + f(u) = 0, & x ∈ Ω, \\ u(x) > 0, & x ∈ Ω, \\ u(x) = 0, ∂_{n}u(x) = 0, & x ∈ ∂Ω, \end{matrix}  (1.3)

where the domain Ω ⊂ R^{N} is unspecified and f is a nonlinear function that is defined and continuous on [0, ∞), but generally not Lipschitz continuous at 0.

In [3], Kaper and Kwong investigated the free boundary problem for the radially symmetric case, assuming that Ω is a ball of (unknown) radius R centered at the origin, where the solution u depends only on the distance r = |x| from the origin. They proved that, for a broad class of functions f, there exists a pair (R, u) with R > 0 and u ∈ C^{2}((0, R)) ∩ C^{1}([0, R]), such that

\{ \begin{matrix}u′′ + \frac{N−1}{r}u′ + f(u) = 0, & 0 < r < R; \\ u(r) > 0, & 0 < r < R; \\ u′(0) = 0, \\ u(R) = 0, u′(R) = 0. \end{matrix}  (1.4)

Specifically, they showed that the existence of (R, u) follows if f and its integral F,

F (u) = − ∫_{0}^{u} f(s) ds,  (1.5)

satisfy the following conditions:

(F1) f is continuous on [0, ∞) and locally Lipschitz continuous on (0, ∞).

(F2) There exists a β > 0, such that F (u) > 0 for 0 < u < β, F (β) = 0, and f(u) > 0 for u ≥ β.

(F3) With β as in (F2), ∫_{0}^{β} F^{−1/2}(u) du < ∞.

(F4) lim inf_{u→∞} f(u) > 0.

The solution of (1.4) is unique if f satisfies the additional condition

(F5) With β as in (F2), u ↦ f(u)/(u − β) is nonincreasing for u > β.

The uniqueness follows immediately from an earlier result of Kaper and Kwong [4], which is itself an extension of an earlier result of Peletier and Serrin [5]. The conditions (F1) through (F5) are satisfied for functions f of the type f(u) = u^{p} − u^{q} for most, but not all, values of p and q in the range 0 ≤ q < p ≤ 1.

In this article we turn again to the original problem (1.3) and address the issue whether a classical solution (Ω, u), if it exists, is necessarily radially symmetric. Numerical results presented for the case N = 2 in [2] seem to indicate that the implication is false if f(u) = \sqrt{u} − 1; the function u in [2, Fig. 6] is clearly not radially symmetric.

We prove the following theorem.

Theorem 1 If f is such that

f(u) = f_{1}(u) + f_{2}(u),  (1.6)

where f_{1} is continuous and nondecreasing and f_{2} is Lipschitz continuous on [0, ∞), then any classical solution (Ω, u) of the free boundary problem (1.3) that satisfies the condition u ∈ C^{2}(\overline{Ω}) is necessarily radially symmetric—that is, Ω is an open ball in R^{N}, Ω = B_{R}(x_{0}) say, and u is radially symmetric about the center x_{0} of the ball. Furthermore,

\frac{∂u}{∂r} < 0, 0 < r = |x − x_{0}| < R.  (1.7)

The conditions of the theorem, together with the conditions (F1) through (F5) above, guarantee that the free boundary problem (1.3) has one and only one classical solution (Ω, u) and that this solution is necessarily radially symmetric.

The conditions of Theorem 1 are satisfied, for example, if f(u) = u^{p}−1, where 0 < p ≤ 1. (Take f_{1}(u) = u^{p} and f_{2}(u) = −1.) They are not satisfied, however, if f(u) = u^{p} − u^{q}, where 0 < q < p ≤ 1.

The proof of Theorem 1 is given in Section 3.

1.2 Dirichlet Problem

Our second result is for the Dirichlet problem for the reaction-diffusion equation (1.1) in a ball of radius R (R > 0) in R^{N} (N = 2, 3, …). Without loss of generality, we may assume that the ball is centered at the origin, so Ω = B_{R}(0).

According to the celebrated result of Gidas, Ni, and Nirenberg [6, Section 1.1, Theorem 1, and Section 2.3, Remark 1], any classical solution of the Dirichlet problem,

\{ \begin{matrix}Δu + f(u) = 0, & x ∈ B_{R}(0), \\ u(x) > 0, & x ∈ B_{R}(0), \\ u(x) = 0, & x ∈ ∂B_{R}(0), \end{matrix}  (1.8)

is necessarily radially symmetric about the origin if f is the sum of a function that is continuous and nondecreasing and a function that is Lipschitz continuous on [0, ∞); furthermore, ∂u/∂r < 0 for 0 < r < R. If such a decomposition of f is not possible, then it is an open problem whether classical solutions of (1.8), if they exist, are indeed radially symmetric. Actually, the examples given in [6, Section 2.4] show that these cases could be very delicate.

We prove the following theorem.

Theorem 2 If f is such that

f(u) = f_{1}(u) + f_{2}(u),  (1.9)

where f_{1} is continuous on [0, ∞), locally Lipschitz continuous on (0, ∞), and nonincreasing near 0, and f_{2} is Lipschitz continuous on [0, ∞), then any classical solution u of (1.8) is necessarily radially symmetric about 0. Furthermore,

\frac{∂u}{∂r} < 0, 0 < r = |x| < R.  (1.10)

The theorem covers the case f(u) = u^{p} − u^{q}, where 0 < q < p ≤ 1, as f is decreasing on [0, (q/p)^{1/(p−q)}).

The proof of Theorem 2 is given in Section 4.

1.3 Discussion

The following discussion clarifies the relationship between Theorems 1 and 2.

The central question of interest to us was whether the existence of a positive solution (Ω, u) of the free boundary problem (1.3) implies its radial symmetry when the nonlinear source term f is continuous, but not necessarily Lipschitz continuous at 0. In particular, our interest focused on functions f of the type f(u) = u^{p} − u^{q}, where 0 ≤ q < p ≤ 1.

The case f(u) = u^{p} − 1 (0 < p ≤ 1) is indeed covered by Theorem 1, but f(u) = u^{p} − u^{q} (0 < q < p ≤ 1) is not. In fact, we were unsuccessful in our attempts at extending Theorem 1 to include functions of the latter type. However, observing that such functions are Lipschitz continuous everywhere except at 0 and decreasing near 0, we found that the moving-plane method could again be used to prove the radial symmetry of any positive solution of the Dirichlet problem for the reaction-diffusion equation (1.1) if Ω is specified to be a ball. This is the result expressed in Theorem 2.

Theorem 1 bears an interesting relationship to the celebrated result of Gidas, Ni, and Nirenberg [6] quoted in the beginning of Subsection 1.2. Gidas, Ni, and Nirenberg showed that, if Ω is a ball, then any solution of the Dirichlet problem is necessarily radially symmetric, provided f is the sum of a function that is continuous nondecreasing and a function that is Lipschitz continuous on [0, ∞). Theorem 1 shows that, by imposing the extra condition ∂_{n}u = 0 in (1.3), we have forced ourselves into a situation where there is no (positive) solution unless Ω is a ball.

We remark that new symmetry results for solutions of reaction-diffusion equations on nonsmooth domains have been obtained recently by Garofalo and Lewis [7] and Lewis and Vogel [8].

2 Maximum Principle and Boundary Lemmas

In this section, we summarize the maximum principle and two boundary lemmas for elliptic differential expressions in a form suitable for the subsequent analysis. Throughout this subsection, D is an arbitrary domain (i.e., an open connected set) in R^{N} and L denotes a uniformly elliptic differential expression on D,

Lu ≡ a^{ij}(x)∂_{ij}u + b^{i}(x)∂_{i}u + c(x)u,

where a^{ij}, b^{i}, c ∈ L^{∞}(D). (We use the notation ∂_{i} = ∂/∂x_{i} and ∂_{ij} = ∂^{2}/∂x_{i}∂x_{j}, together with the summation convention for repeated indices.)

Lemma 1 (Maximum Principle). Suppose that u ∈ C^{2}(D) satisfies the inequalities u ≥ 0 and Lu ≤ 0 on D. If u vanishes at some point in D, then u = 0 on D.

Proof. See [9, Chap. 2, Theorem 6].

Lemma 2 (Hopf Boundary Lemma). Suppose that u ∈ C^{2}(D) satisfies the inequalities u > 0 and Lu ≤ 0 on D. Suppose, furthermore, that P ∈ ∂D lies on the boundary of a ball B in D. If u is continuous in D ∪ P and u(P) = 0, then ∂_{ν}u(P) < 0 for any outward direction ν.

Proof. See [6, Section 1.3, Lemma H] and [9, Chap. 2, Theorem 8].

The derivative ∂_{ν}u is defined as the derivative in the direction ν = (ν_{1}, …, ν_{N}); that is, ∂_{ν} = ν_{i}∂_{i}. The vector ν is said to point outward from D if its scalar product with n, the unit normal vector in the outward direction, is positive (ν_{i}n_{i} > 0).

Lemma 3 (Serrin Boundary Lemma). Suppose that P ∈ ∂D and that, near P, ∂D consists of two transversally intersecting C^{2}-hypersurfaces ρ = 0 and σ = 0, where ρ < 0 and σ < 0 in D and (a^{ij}∂_{i}ρ∂_{j}σ)(P) ≥ 0.

Suppose that u ∈ C^{2}(\overline{D}), u > 0 and Lu ≤ 0 on D, and u(P) = 0.

(i) If (a^{ij}∂_{i}ρ∂_{j}σ)(P) > 0, then ∂_{ν}u(P) < 0 for any outward direction ν that is transverse to the hypersurfaces ρ = 0 and σ = 0.

(ii) If (a^{ij}∂_{i}ρ∂_{j}σ)(P) = 0, then ∂_{ν}u(P) < 0 or ∂_{ν}^{2}u(P) > 0 for any outward direction ν that is transverse to the hypersurfaces ρ = 0 and σ = 0, provided a^{ij} is twice continuously differentiable in D ∪ P and ∂_{t}(a^{ij}∂_{i}ρ∂_{j}σ)(P) ≥ 0 for any derivative in a direction t that is tangent to the submanifold {ρ = 0} ∩ {σ = 0}.

Proof. See [6, Section 1.3, Lemma S] and [10, Lemma 2].

The following lemma is a nonlinear generalization of the Hopf Boundary Lemma.

Lemma 4 Let f = f_{1} + f_{2}, where f_{1} is continuous on [0, ∞) and nondecreasing near 0 and f_{2} is Lipschitz continuous on [0, ∞), with f(0) ≥ 0. Suppose that u ∈ C^{2}(D) satisfies the inequalities u > 0 and Δu + f(u) ≤ 0 on D. Suppose, furthermore, that P ∈ ∂D lies on the boundary of a ball B in D. If u is continuous in D ∪ P and u = 0 on ∂D near P, then ∂_{ν}u(P) < 0 for any outward direction ν.

Proof. As f_{1} is nondecreasing near 0, we have f_{1}(u) ≥ f_{1}(0) on D near P. Furthermore, f_{2} is Lipschitz continuous on [0, ∞), so there exists a bounded function c on D such that f_{2}(u) − f_{2}(0) = cu on D. Therefore, u satisfies the differential inequality Δu + cu ≤ −f(0) or, since f(0) ≥ 0, Δu + cu ≤ 0 on D. The assertion of the lemma follows from the Hopf Boundary Lemma (Lemma 2).

The final lemma covers reaction-diffusion equations where f(0) is negative. It requires more smoothness of the solution u; cf. [6, Section 2.2, Lemma 2.1].

Lemma 5 Let f be continuous on [0, ∞), with f(0) < 0. Suppose that u ∈ C^{2}(D) satisfies the inequality u > 0 on D and that Δu + f(u) = 0 on D. Suppose, furthermore, that P ∈ ∂D lies on the boundary of a ball B in D. If u is twice continuously differentiable in D ∪ P and u = 0 on ∂D near P, then ∂_{ν}u(P) < 0 or ∂_{ν}^{2}u(P) > 0 for any outward direction ν.

Proof. Because u vanishes identically on ∂D near P, any directional derivative in the tangent plane vanishes at P. Furthermore, as u > 0 on D, it must be the case that ∂_{n}u(P) ≤ 0.

If ∂_{n}u(P) < 0, then ∂_{ν}u(P) < 0 for any outward direction ν.

If ∂_{n}u(P) = 0, then all directional derivatives at P are zero. Using a local coordinate system with the origin at P, one coordinate along the outward normal vector, and the remaining N − 1 coordinates in the tangent plane, we readily verify that Δu(P) = ∂_{n}^{2}u(P). Since u is twice continuously differentiable on D ∪ P and u(P) = 0, we have Δu(P) = −f(0) > 0; hence, ∂_{n}^{2}u(P) > 0. Then also ∂_{ν}^{2}u(P) > 0 for any outward direction ν.

3 Proof of Theorem 1

Suppose (Ω, u) is a classical solution of (1.3), where f satisfies the conditions of Theorem 1.

According to Lemma 4, we necessarily have ∂_{n}u < 0 on ∂Ω if f(0) ≥ 0, so it must be the case that f(0) < 0.

We use the moving-plane method for the proof of the theorem. This method, due to Alexandroff (see [11, Section VII.1]) was first applied by Serrin [10] to prove a symmetry result for the constant mean-curvature equation Δu = −c, c > 0, and has been used since then by several authors.

Let T_{λ} be the hyperplane

T_{λ} = {x = (x_{1}, …, x_{N}) ∈ R^{N} : x_{1} = λ},  (3.1)

and let S_{λ} be the reflection operator about T_{λ},

S_{λ} : x = (x_{1}, x_{2}, …, x_{N}) ↦ S_{λ}x = (2λ − x_{1}, x_{2}, ⋯, x_{N}), x ∈ R^{N}.  (3.2)

As Ω is bounded, T_{λ} and Ω are disjoint for all sufficiently large λ. Let λ_{1} be defined by

λ_{1} = inf{λ : T_{λ} ∩ Ω = ∅}.  (3.3)

Without loss of generality, we may assume that the set ∂Ω ∩ T_{λ_{1}} consists of a single point; if necessary, we rotate the coordinate system.

For λ < λ_{1}, λ sufficiently close to λ_{1}, T_{λ} cuts a cap Σ_{λ} off Ω,

Σ_{λ} = {x ∈ Ω : x_{1} > λ}, λ < λ_{1}.  (3.4)

The reflection of Σ_{λ} about T_{λ} is S_{λ}(Σ_{λ}),

S_{λ}(Σ_{λ}) = {S_{λ}x : x ∈ Σ_{λ}}.  (3.5)

For all λ sufficiently close to λ_{1}, S_{λ}(Σ_{λ}) is a proper subset of Ω. Let λ_{0} be defined by

λ_{0} = inf{λ : S_{λ}(Σ_{λ}) ⊂ Ω}.  (3.6)

As Ω is bounded, λ_{0} is finite. Obviously, λ_{0} < λ_{1}. For λ = λ_{0}, either the boundary of S_{λ_{0}}(Σ_{λ_{0}}) is internally tangent to the boundary of Ω at some point P not on T_{λ_{0}}, or T_{λ_{0}} is orthogonal to the boundary of Ω at some point P on ∂Ω.

For each λ ∈ (λ_{0}, λ_{1}), let u_{λ} be defined in terms of u by the expression

u_{λ}(x) = u(S_{λ}x), x ∈ Σ_{λ}.  (3.7)

The following lemma plays a crucial role.

Lemma 6 If u_{λ} ≥ u on Σ_{λ}, then either u_{λ} > u on Σ_{λ} or u_{λ} = u on Σ_{λ}. In the former case, ∂_{1}u < 0 on Ω ∩ T_{λ}.

Proof. Because u is a solution of (1.3), we have the identity Δ(u_{λ} − u) + f_{1}(u_{λ}) − f_{1}(u) + f_{2}(u_{λ}) − f_{2}(u) = 0 on Σ_{λ}. As f_{1} is nondecreasing, f_{1}(u_{λ}) ≥ f_{1}(u). Furthermore, f_{2} is Lipschitz on [0, ∞), so there exists a bounded function c_{λ} on Σ_{λ}, such that f_{2}(u_{λ})−f_{2}(u) = c_{λ}(u_{λ} − u). Hence, u_{λ} − u ≥ 0 and Δ(u_{λ} − u) + c_{λ}(u_{λ} − u) ≤ 0 on Σ_{λ}. The first part of the lemma follows from the Maximum Principle (Lemma 1).

Obviously, u_{λ} − u = 0 on Ω ∩ T_{λ}, so the Hopf Boundary Lemma (Lemma 2) implies that −∂_{1}(u_{λ} − u) < 0 on Ω ∩ T_{λ}. Since −∂_{1}u_{λ} = ∂_{1}u on Ω ∩ T_{λ}, the second part of the lemma follows.

Let Λ be the set

Λ = {λ ∈ (λ_{0}, λ_{1}) : u_{λ}(x) > u(x), x ∈ Σ_{λ}}.  (3.8)

We claim that Λ = (λ_{0}, λ_{1}). The proof is given in two steps; in the first step, we prove that Λ is nonempty and contains all λ sufficiently close to λ_{1}; in the second step, we prove that Λ actually contains the entire interval (λ_{0}, λ_{1}). The proof requires that u be twice continuously differentiable in the closure of Ω.

Lemma 7 If u ∈ C^{2}(\overline{Ω}), then there exists an ε > 0 such that (λ_{1} − ε, λ_{1}) ⊂ Λ.

Proof. Let P be the point where ∂Ω intersects the hyperplane T_{λ_{1}}. At P, the outward normal vector is oriented in the positive x_{1}-direction.

Because u vanishes on ∂Ω near P and ∂_{1}u(P) = 0, it follows from Lemma 5 that ∂_{1}^{2}u(P) > 0. This inequality extends by continuity to a small arc Γ_{ε} = ∂Ω ∩ B_{ε}(P). Also, ∂_{1}u = 0 on Γ_{ε}, so integrating over x_{1} from Γ_{ε} into Ω and reducing ε if necessary, we find that ∂_{1}u < 0 in a small domain Ω_{ε} = Ω ∩ B_{ε}(P) near P. By choosing λ in (λ_{0}, λ_{1}) sufficiently close to λ_{1} we can ensure that Σ_{λ} and its reflection S_{λ}(Σ_{λ}) are entirely contained in Ω_{ε}. For such λ we certainly have u_{λ} > u on Σ_{λ}, so λ ∈ Λ.

Lemma 8 If u ∈ C^{2}(\overline{Ω}), then Λ = (λ_{0}, λ_{1}).

Proof. Suppose that there exists a monotonically decreasing sequence {λ^{i}} of values λ^{i} ∈ (λ_{0}, λ_{1}) that belong to Λ and converge to some λ ∈ (λ_{0}, λ_{1}). As λ^{i} ∈ Λ, we have u_{λ^{i}} > u on Σ_{λ^{i}}, so u_{λ} ≥ u on Σ_{λ}, by continuity. Since λ ∈ (λ_{0}, λ_{1}), we can apply Lemma 6 and conclude that u_{λ} > u on Σ_{λ}. This result shows that the set Λ is left-closed in (λ_{0}, λ_{1}).

Next we prove that the set Λ is left-open in (λ_{0}, λ_{1}). The proof is by contradiction, where we assume that there exists a λ ∈ (λ_{0}, λ_{1}) that belongs to Λ and a monotonically decreasing sequence {λ^{i}} of values λ^{i} ∈ (λ, λ_{1}) that do not belong to Λ, with λ^{i} → λ as i → ∞.

As λ^{i} ∉ Λ, we can find a point x^{i} ∈ Σ_{λ^{i}}, such that u_{λ^{i}}(x^{i}) ≤ u(x^{i}). Because Ω is bounded, we can extract a convergent subsequence of {x^{i}}, say {x^{i}} itself, whose limit point, P say, is in the closure of Σ_{λ}. At P, we have u_{λ}(P) ≤ u(P). We claim that no such point P exists.

Since λ ∈ Λ, we know that u_{λ} > u on Σ_{λ}, so P ∉ Σ_{λ}.

Suppose that P ∈ ∂Σ_{λ}\(\overline{Ω} ∩ T_{λ}). Then P ∈ ∂Ω, so u(P) = 0. On the other hand, u_{λ}(P) = u(S_{λ}P) > 0, because λ_{0} < λ < λ_{1} and, therefore, S_{λ}P ∈ Ω. Thus we find that u_{λ}(P) > u(P). But we have already shown that u_{λ}(P) ≤ u(P), so we must conclude that P ∉ ∂Σ_{λ}\(\overline{Ω} ∩ T_{λ}).

Suppose that P ∈ \overline{Ω} ∩ T_{λ}. The line segment ℓ_{i} joining x^{i} and S_{λ^{i}}x^{i} lies entirely in Ω. Therefore, the inequality u_{λ^{i}}(x^{i}) ≤ u(x^{i}) implies that ∂_{1}u(y^{i}) ≥ 0 at some point y^{i} ∈ ℓ_{i}. Now, x^{i} → P and S_{λ^{i}}x^{i} → S_{λ}P as i → ∞, and if P ∈ \overline{Ω} ∩ T_{λ}, as supposed, then S_{λ}P = P. Thus, ℓ_{i} shrinks to the single point P and we find that ∂_{1}u(P) ≥ 0. According to Lemma 6, ∂_{1}u < 0 on Ω ∩ T_{λ}, so we must conclude that P ∉ Ω ∩ T_{λ}.

The only remaining possibility is P ∈ ∂Ω ∩ T_{λ}. At such a point P, we have ∂_{1}^{2}u(P) > 0, according to Lemma 5. This inequality extends by continuity to a small domain inside Ω near P, so ∂_{1}u is a (strictly) increasing function of x_{1} there. But then it follows that ∂_{1}u < 0 on a small domain inside Ω near P, since ∂_{1}u = 0 on ∂Ω. However, this conclusion contradicts the assumption that there exists a sequence {(λ^{i}, x^{i})} with λ^{i} ∈ (λ, λ_{1}), x^{i} ∈ Σ_{λ^{i}} ⊂ Σ_{λ}, and converging to (λ, P), where u_{λ^{i}}(x^{i}) = u(S_{λ^{i}}x^{i}) ≤ u(x^{i}). We must therefore conclude that P ∉ ∂Ω ∩ T_{λ}.

Thus, we have exhausted all the possibilities and conclude that no such point P exists. The assumption that there exists a λ ∈ (λ_{0}, λ_{1}) that belongs to Λ and a monotonically decreasing sequence {λ^{i}} of values λ^{i} ∈ (λ, λ_{1}) that do not belong to Λ, with λ^{i} → λ as i → ∞, is untenable. Hence, Λ is left-open in (λ_{0}, λ_{1}).

According to Lemma 7, the set Λ contains an interval (λ_{1} − ε, λ_{1}) with ε > 0. Since Λ is both left-open and left-closed in (λ_{0}, λ_{1}), it must be the case that Λ = (λ_{0}, λ_{1}).

Next, we investigate what happens at λ = λ_{0}. We recall that, at λ = λ_{0}, either the boundary of S_{λ_{0}}(Σ_{λ_{0}}) is internally tangent to ∂Ω at some point not on T_{λ_{0}}, or T_{λ_{0}} is orthogonal to ∂Ω.

Lemma 9 If u ∈ C^{2}(\overline{Ω}), then Ω is symmetric about T_{λ_{0}} and u_{λ_{0}} = u on Σ_{λ_{0}}.

Proof. By Lemma 8, Λ = (λ_{0}, λ_{1}), so u_{λ} > u on Σ_{λ} for all λ ∈ (λ_{0}, λ_{1}). Then u_{λ_{0}} ≥ u on Σ_{λ_{0}}, by continuity. The same arguments as in Lemma 6 yield the inequality Δ(u_{λ_{0}} − u) + c_{λ_{0}}(u_{λ_{0}} − u) ≤ 0 on Σ_{λ_{0}}, where c_{λ_{0}} is bounded, so it follows from the Maximum Principle (Lemma 1) that either u_{λ_{0}} > u on Σ_{λ_{0}} or u_{λ_{0}} = u on Σ_{λ_{0}}. We claim that the former case is impossible.

If the boundary of S_{λ_{0}}(Σ_{λ_{0}}) is internally tangent to ∂Ω at some point S_{λ_{0}}P, then u_{λ_{0}}(P) = u(P). The boundary of Σ_{λ_{0}} is smooth near P, so if u_{λ_{0}} > u on Σ_{λ_{0}}, we can apply the Hopf Boundary Lemma (Lemma 2) and conclude that ∂_{1}(u_{λ_{0}} − u)(P) < 0. But ∂_{1}u_{λ_{0}}(P) and ∂_{1}u(P) are both zero, because the normal derivative of u vanishes on ∂Ω, so we have a contradiction.

On the other hand, if T_{λ_{0}} is orthogonal to ∂Ω and P is a point of ∂Ω ∩ T_{λ_{0}}, we can apply the Serrin Boundary Lemma (Lemma 3) at P. Near such a point P, ∂Σ_{λ_{0}} consists of two orthogonally intersecting hypersurfaces ρ(x) = λ_{0} − x_{1} = 0 and σ(x) = 0. A simple computation shows that we are in case (ii) of Lemma 3. Because u and u_{λ_{0}} coincide on T_{λ_{0}}, any directional derivative of u_{λ_{0}} − u in the hyperplane T_{λ_{0}} is zero. Moreover, u = 0 on ∂Ω, so any directional derivative in the tangent plane to ∂Ω at P is also zero. Consequently, ∂_{ν}(u_{λ_{0}} − u)(P) = 0 for any outward direction ν. If u_{λ_{0}} > u on Σ_{λ_{0}}, then it follows from the Serrin Boundary Lemma (Lemma 3) that ∂_{ν}^{2}(u_{λ_{0}} − u)(P) > 0 for any outward direction ν that is transverse to ∂Ω and T_{λ_{0}}. In particular, taking ν = (−1/\sqrt{2}, 0, …, 0, 1/\sqrt{2}), we find (−∂_{1} + ∂_{N})^{2}(u_{λ_{0}} − u)(P) > 0. But ∂_{1}^{2}u_{λ_{0}}(P) = ∂_{1}^{2}u(P) and ∂_{N}^{2}u_{λ_{0}}(P) = ∂_{N}^{2}u(P), while ∂_{1}∂_{N}u_{λ_{0}}(P) = −∂_{1}∂_{N}u(P), so the inequality reduces to ∂_{1}∂_{N}u(P) > 0. However, since both u and ∂_{n}u are identically zero on ∂Ω, the only second-order derivative that does not vanish at P is ∂_{n}^{2}u; in fact, ∂_{n}^{2}u(P) = −f(0). At P, the normal vector is in the linear manifold spanned by x_{2}, …, x_{N}, so it must be the case that ∂_{1}∂_{N}u(P) = 0. Again, we have arrived at a contradiction.

We therefore conclude that u_{λ_{0}} = u on Σ_{λ_{0}}.

Now, if the union of Σ_{λ_{0}}, S_{λ}(Σ_{λ_{0}}), and Ω ∩ T_{λ_{0}} were a proper subset of Ω, then a part of ∂S_{λ_{0}}(Σ_{λ_{0}})\(Ω ∩ T_{λ_{0}}) would be in Ω. But then we would have a contradiction, as u vanishes on ∂S_{λ_{0}}(Σ_{λ_{0}})\(Ω ∩ T_{λ_{0}}). Hence, Ω is the union of Σ_{λ_{0}}, S_{λ}(Σ_{λ_{0}}), and Ω ∩ T_{λ_{0}}.

Conclusion of the Proof. If (Ω, u) is a classical solution of (1.3) and u ∈ C^{2}(\overline{Ω}), then Lemma 9 applies, so Ω is symmetric about the hyperplane T_{λ_{0}} and u_{λ_{0}} = u on Σ_{λ_{0}}. Moreover, u_{λ} > u and ∂_{1}u < 0 on Σ_{λ} for all λ ∈ (λ_{0}, λ_{1}).

The free boundary problem (1.3) is rotationally invariant, so the positive x_{1}-direction is not privileged in any way. Therefore, Ω is symmetric in every direction—that is, Ω is a ball, because it is connected—and the gradient of u in any radial direction is negative, as claimed.

4 Proof of Theorem 2

Throughout this section Ω denotes the ball of radius R (R fixed) centered at the origin,

Ω = B_{R}(0), R > 0.  (4.1)

The proof of Theorem 2 is also based on the moving-plane method; in the notation of Eqs. (3.3) and (3.6), we have

λ_{0} = 0, λ_{1} = R.  (4.2)

Suppose u is a classical solution of the Dirichlet problem (1.8), where f satisfies the conditions of Theorem 2. For each λ ∈ (λ_{0}, λ_{1}), u defines a function u_{λ} on Σ_{λ} as in (3.7),

u_{λ}(x) = u(S_{λ}x), x ∈ Σ_{λ}.  (4.3)

Lemma 10 If u_{λ} ≥ u on Σ_{λ}, then either u_{λ} > u on Σ_{λ} or u_{λ} = u on Σ_{λ}. In the former case, ∂_{1}u < 0 on Ω ∩ T_{λ}.

Proof. The proof is similar to the proof of Lemma 6. Because u is a solution of (1.8), we have the identity Δ(u_{λ} − u) + f(u_{λ}) − f(u) = 0 on Σ_{λ}. As u > 0 on Σ_{λ} and f is locally Lipschitz on (0, ∞), there exists a locally bounded function c_{λ} on Σ_{λ}, such that f(u_{λ}) − f(u) = c_{λ}(u_{λ} − u). Hence, u_{λ} − u ≥ 0 and Δ(u_{λ} − u) + c_{λ}(u_{λ} − u) = 0 on Σ_{λ}. The first part of the lemma follows from the Maximum Principle (Lemma 1). The second part of the lemma follows from the Hopf Boundary Lemma (Lemma 2).

We define the set Λ as in (3.8),

Λ = {λ ∈ (λ_{0}, λ_{1}) : u_{λ}(x) > u(x), x ∈ Σ_{λ}} .  (4.4)

As before, our objective is to show that Λ = (λ_{0}, λ_{1}).

Lemma 11 There exists an ε > 0 such that (λ_{1} − ε, λ_{1}) ⊂ Λ.

Proof. We prove the lemma by contradiction, assuming that in every domain Σ_{λ} there is a point y where u_{λ}(y) ≤ u(y), no matter how close λ is to λ_{1}.

Let P be the point where ∂Ω intersects the hyperplane T_{λ_{1}}. Let c = sup{|f_{2}(u) − f_{2}(v)|/|u − v|} : u, v ∈ [0, ∞)}. We choose η > 0 sufficiently small that the first (positive) eigenvalue μ_{1}(Ω_{η}) of the Dirichlet problem for −Δ on Ω_{η} = Ω∩B_{η}(P) satisfies the inequality μ_{1}(Ω_{η}) > c. Such a choice is always possible, as μ_{1} is a decreasing function of |Ω_{η}| and μ_{1}(Ω_{η}) → ∞ as |Ω_{η}| → 0. We restrict the discussion to values of λ sufficiently close to λ_{1} that Σ_{λ} and its reflection S_{λ}(Σ_{λ}) are both entirely contained in Ω_{η}.

First, we assume that in every domain Σ_{λ} there is a point y where u_{λ}(y) < u(y). Because u = 0 on ∂Ω and u > 0 on Ω, it is certainly true that u_{λ} ≥ u on ∂Σ_{λ}\(Ω_{η}∩T_{λ}). Furthermore, u_{λ} = u on Ω_{η} ∩ T_{λ}, so u_{λ} ≥ u on ∂Σ_{λ}. But then there exists a neighborhood N of y, which is entirely contained in Σ_{λ}, such that u_{λ} < u on N and u_{λ} = u on ∂N.

Now, f_{1} is nonincreasing near 0 and u vanishes on ∂Ω, so for λ sufficiently close to λ_{1} we certainly have f_{1}(u_{λ}) ≥ f_{1}(u) on N. Furthermore, f_{2}(u) − f_{2}(u_{λ}) ≤ c(u − u_{λ}) on N, so Δ(u−u_{λ})+c(u−u_{λ}) ≥ 0 on N. Multiplying both sides of this inequality by u−u_{λ}, which is positive on N, integrating over N, and using the inequality (Δv, v) ≤ −μ_{1}(N)(v, v), where μ_{1}(N) is the first (positive) eigenvalue for the Dirichlet problem for −Δ on N, we find that μ_{1}(N) ≤ c. But now we have a contradiction, as μ_{1}(N) > μ_{1}(Ω_{η}) and η was chosen in such a way as to ensure that μ_{1}(Ω_{η}) > c.

On Σ_{λ}, both u and u_{λ} are positive. Since f is locally Lipschitz continuous on (0, ∞), we can invoke the Maximum Principle (Lemma 1) to rule out the possibility that u_{λ}(y) = u(y). Thus we conclude that u_{λ} > u on Σ_{λ} for all λ sufficiently close to λ_{1}.

Lemma 12 Λ = (λ_{0}, λ_{1}).

Proof. As in the proof of Lemma 8, we show that Λ is left-closed in (λ_{0}, λ_{1}). The next task is to show that Λ is left-open in (λ_{0}, λ_{1}). The proof is similar to the proof in Lemma 8 and is based on an argument by contradiction, but it differs in a critical detail. The proof of Lemma 8 depends on Lemma 5, which holds only if f(0) < 0, and in the current situation the existence of a classical solution of (1.8) does not imply anything about the sign of f(0).

To prove by contradiction that Λ is left-open in (λ_{0}, λ_{1}), we assume that there exists a point λ ∈ (λ_{0}, λ_{1}) belonging to Λ and a monotonically decreasing sequence {λ^{i}} of points λ^{i} not belonging to Λ, such that λ^{i} → λ as i → ∞. The assumption implies that there exists a sequence of points {x^{i}}, x^{i} ∈ Σ_{λ^{i}}, where u_{λ^{i}}(x^{i}) ≤ u(x^{i}). The sequence {x^{i}} converges to a point P in the closure of Σ_{λ}, where u_{λ}(P) ≤ u(P).

The same arguments as in the proof of Lemma 8 can be used to show that P does not belong to Σ_{λ} or to ∂Σ_{λ}\(\overline{Ω}∩T_{λ}) or to Ω∩T_{λ}. To rule out the remaining possibility, namely, that P belongs to ∂Ω ∩ T_{λ}, we need a different argument, as the proof of Lemma 8 relies on Lemma 5.

The argument is partially similar to the argument used in the proof of Lemma 11. First, we restrict the discussion to a small subdomain D_{η} of Σ_{λ} near P. Specifically, we take D_{η} to be the intersection of Σ_{λ} with a cylinder C_{η}, which extends to the right of T_{λ} (i.e., in the half-space x_{1} ≥ λ), whose axis passes through P and runs parallel to the x_{1}-axis, and whose radius η is fixed such that the first (positive) eigenvalue of the Dirichlet problem for −Δ on D_{η} satisfies the inequality μ_{1}(D_{η}) > c.

Suppose that, in every neighborhood of P inside Σ_{λ}, there is a point y where u_{λ}(y) < u(y). Then there certainly is such a point y in D_{η}. Now, consider the values of u_{λ} and u on the boundary ∂D_{η} of D_{η}.

The boundary ∂D_{η} consists of three hypersurfaces: one (labeled S_{1}) on the boundary ∂Ω of Ω, one (labeled S_{2}) on the surface of the circular cylinder C_{η}, and one (labeled S_{3}) on the hyperplane T_{λ}. Because u = 0 on ∂Ω and u > 0 on Ω, it is certainly the case that u_{λ} ≥ u on S_{1}. Also, because ∂_{1}u < 0 on Ω ∩ T_{λ}, we have u_{λ} > u on S_{2}; if necessary, we reduce η, so C_{η} intersects ∂Ω sufficiently close to T_{λ}. Finally, u_{λ} = u on T_{λ} and therefore on S_{3}. Thus, we see that u_{λ} ≥ u on ∂D_{η}.

But then there is a neighborhood N of y, which is entirely contained in D_{η}, such that u_{λ} < u on N and u_{λ} = u on ∂N.

Now, we complete the proof as in Lemma 11 by showing that μ_{1}(N) ≤ c, and thus arrive at a contradiction. Using the Maximum Principle (Lemma 1), we then rule out the possibility that u_{λ}(y) = u(y). Hence, we conclude that u_{λ} > u on D_{η}.

However, this conclusion contradicts the assumption that there exists a sequence {(λ^{i}, x^{i})} with λ^{i} ∈ (λ, λ_{1}), x^{i} ∈ Σ_{λ^{i}} ⊂ Σ_{λ}, and converging to (λ, P), where u_{λ^{i}}(x^{i}) ≤ u(x^{i}). We must therefore conclude that P ∉ ∂Ω ∩ T_{λ}.

Again, we have exhausted all the possibilities and conclude that no point P with the stated properties exists. Therefore, Λ must be left-open in (λ_{0}, λ_{1}).

Being simultaneously nonempty near λ_{1}, left-open and left-closed in (λ_{0}, λ_{1}), Λ must be the entire interval (λ_{0}, λ_{1}).

Conclusion of the Proof. According to Lemma 12, Λ = (λ_{0}, λ_{1}). That is, u_{λ} > u and ∂_{1}u < 0 on Σ_{λ} for every λ ∈ (λ_{0}, λ_{1}). Letting λ tend to λ_{0}, we find u(S_{λ_{0}}x) ≥ u(x) for all x ∈ Σ_{λ_{0}} and ∂_{1}u ≤ 0 on Ω ∩ T_{λ_{0}}.

But the boundary value problem (1.8) is rotationally invariant, so the positive x_{1}- direction is not privileged in any way. Therefore, u is radially symmetric and ∂u/∂r < 0 for λ_{0} < r < λ_{1}.

Remark. At this point, it is clear why, in Theorem 2, we needed to assume from the start that Ω is a ball. The difficulty is in the proof of the symmetry of Ω about T_{λ_{0}}. Here, the proof of Lemma 9 relies on the fact that f is Lipschitz continuous at 0. In the case of Theorem 2, an identity of the type f(u_{λ}) − f(u) = c_{λ}(u_{λ} − u) ceases to hold when both u_{λ} and u vanish. Consequently, we cannot apply the Hopf Boundary Lemma and thus rule out the possibility that the boundary of S_{λ_{0}}(Σ_{λ_{0}}) is internally tangent to ∂Ω. This problem is, in fact, related to the lack of uniqueness of the solution of an initial value problem of the type u′ = f(u), u(0) = 0, where the forcing term f is not Lipschitz at 0.

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