Symmetry properties for positive solutions to some

elliptic equations in sector domains with large

amplitude

Meijun Zhu

Department of Mathematics

The University of Oklahoma

Norman, Oklahoma 73019

E-mail: mzhu@math.ou.edu

Abstract

In this paper we study symmetry properties for positive solutions of semilinear elliptic equation Δu + f(u) = 0 with mixed boundary condition in a spherical sector Σ(α, R), where α, the amplitude of the sector, is between π and 2π. Under certain condition on f(u), we prove that all positive solutions are radially symmetric about the origin. Unlike well-known results of Gidas, Ni and Nirenberg [3] on Dirichlet problem, or early results of Berestycki and Pacella on the same problem for Σ(α, R) with an acute angle, extra conditions on f(u) are needed, as being pointed out early by Berestycki and Pacella in [1].

1 Introduction

In an elegant paper [3], Gidas, Ni and Nirenberg proved, among other things, that any positive C^{2} solution to

\{ \begin{matrix}Δu + f(u) = 0 & in B_{R} \\ u = 0 & on ∂B_{R} \end{matrix}

with Lipschitz f(u) and B_{R} = {x ∈ ℝ : |x| < R}, is spherically symmetric and ∂u/∂ρ < 0 for 0 < |x| = ρ < R. There are many extensions of such result into other types of symmetric domains, for example, whole space, half space, strip domains, etc. It is worthwhile to point out that symmetric property for positive solution does not hold if the domain is an annulus.

In [1], Berestycki and Pacella attempted to extend such symmetric result to positive solutions of the same equations on a “sectorial” domain, naturally with mixed boundary conditions. Let (ρ, θ_{1}, θ_{2}, ..., θ_{n−1}) be the polar coordinates in ℝ^{n}, ρ ≥ 0, θ_{i} ∈ [0, π] for 1 ≤ i ≤ n − 2 and θ_{n−1} ∈ [0, 2π]. Define the spherical sector Σ(α, R) of radius R > 0 and amplitude α ∈ (0, 2π) by

Σ(α, R) = {x ∈ ℝ^{n} : 0 < |x| = ρ < R, θ_{i} ∈ (0, π) for 1 ≤ i ≤ n − 2, θ_{n−1} ∈ (0, α)},

and denote

Γ_{0} = {x ∈ ∂Σ(α, R) : |x| = R}, Γ_{1} = {x ∈ ∂Σ(α, R) : θ_{n−1} = 0 or θ_{n−1} = α}.

They considered

\{ \begin{matrix}Δu + f(u) = 0 & in Σ(α, R) \\ u = 0 & on Γ_{0} \\ \frac{∂u}{∂ν} = 0 & on Γ_{1}. \end{matrix}  (1.1)

where ν is the outward unit normal. For f ∈ C^{1} and α ≤ π, they proved that if u ∈ C^{2}(\overline{Σ(α, R)}) is a positive solution to (1.1), then it must be radially symmetric about the origin and ∂u/∂ρ < 0 for 0 < |x| = ρ < R. However, for α > π, such result is not true in general. In the same paper, Berestycki and Pacella presented an interesting example: Let f(u) = u^{p} and α > π; There is a p̄(n) such that if p̄(n) < p < (n + 2)/(n − 2) (when n ≥ 3) or if p̄(n) < p (when n = 2), there exists positive solution to (1.1) which is not spherically symmetric.

It remains as an interesting problem: when α is greater than π, for which class of f(u), positive solutions to (1.1) must be radially symmetric? From the well-known Pohozaev identity, one can see that when n ≥ 3 and f(u) = u^{p} for p ≥ (n+2)/(n−2) the only nonnegative solution to (1.1) in C^{2}(\overline{Σ(α, R)}) is the trivial one (which of course is radially symmetric). More broadly, it is natural to consider the following equation in a punctured sector:

\{ \begin{matrix}Δu + f(u) = 0 & in Σ(α, R) \\ u = 0 & on Γ_{0} \\ \frac{∂u}{∂ν} = 0 & on Γ_{1} \ \{0\}. \end{matrix}  (1.2)

In this note, we shall confine us to the case of n ≥ 3 and prove the following:

Theorem 1.1 Let u > 0 be a C^{2} solution to (1.2). Assume that f(u) satisfies the following condition:

(i) f(t) ≥ 0 for t ≥ 0,

(ii) f(t)/t^{(n+2)/(n−2)} is nondecreasing for t > 0.

Then u(x) is radially symmetric about the origin and ∂u/∂ρ < 0 for 0 < |x| = ρ < R.

Remark 1.1 Using the well-known Pohozaev identity, one can check that if f(t) satisfies condition (i) and (ii), the only nonnegative solution to (1.2) in C^{2}(\overline{Σ(α, R)}) is the trivial one. Quite surprisingly, from the proof of Theorem 1.1 we are able to give an alternative proof of this result, but only assume that solution is nonnegative and BOUNDED. Given that α ≥ π and the solution satisfies Neumann boundary condition, it is not obvious (but could be proved) to obtain the smoothness up to the boundary for the solution from the assuming that the solution is bounded. Our method can allow us to bypass these estimates. More details will be presented in Remark 2.1.

Remark 1.2 If lim_{u→+∞} f(u)/u^{p} = l > 0 and p ≥ (n + 2)/(n − 2), the existence of singular positive solution to (1.1) can be obtained by standard variational method since H_{r}(\overline{Σ(α, R)}\\{0\}) ⊂ L^{p+1}(\overline{Σ(α, R)}\\{0\}) is compact, where H_{r}(\overline{Σ(α, R)}\\{0\}) = \{u ∈ H^{1}(\overline{Σ(α, R)} \ \{0\}) ∩ H_{0}^{1}(B_{R}), : u is radial\}.

Our approach to above result is quite different from that in [1]. Keep in mind that there are counter-examples to the above result if f(u) does not satisfy condition (ii). It seems hard to obtain symmetry properties about solutions when amplitude α > π by using moving plane method directly. Instead, we first use the method of moving sphere as in, for example, [4], to obtain monotonicity about solutions along radial direction. Once we have such monotonicity, standard moving plane method is then applied to obtain the desired result. We want to emphasize that such approach seems not suitable for a sector with amplitude less than π (for n ≥ 3).

2 Proof

Throughout this section, we always assume that u(x) is a C^{2} positive solution to (1.2). Without loss of generality, we take R = 1.

For any λ ≥ 1, define

u_{λ}(x) = \frac{1}{|λx|^{n−2}} u(\frac{x}{|λx|^{2}} ), ∀ \frac{1}{λ} ≤ |x| ≤ 1.

Then u_{λ}(x) satisfies

−Δu_{λ} = \frac{1}{|λx|^{n+2}} f(|λx|^{n−2}u_{λ})  in Σ(α, 1) ∩ \{\frac{1}{λ} < |x|\}
u_{λ} > 0  on Γ_{0}
u_{λ} = u  on Σ(α, 1) ∩ \{|x| = \frac{1}{λ}\}
\frac{∂u_{λ}}{∂ν} = 0  on Γ_{1} ∩ \{|x| ≥ \frac{1}{λ}\}.

Let Σ_{λ} = {x ∈ Σ(α, 1) : |x| > 1/λ}, T_{λ} = {x ∈ Σ(α, 1) : |x| = 1/λ} and w_{λ} = u_{λ} − u. From Condition (ii), we know that for |λx| ≥ 1 and u_{λ} > 0,

\frac{f(|λx|^{n−2}u_{λ})}{(|λx|^{n−2}u_{λ})^{\frac{n+2}{n−2}}} ≥ \frac{f(u_{λ})}{uλ^{\frac{n+2}{n−2}}}.

Therefore, in Σ_{λ},

\frac{1}{|λx|^{n+2}} f(|λx|^{n−2}u_{λ}) − f(u) ≥ f(u_{λ}) − f(u) = c_{λ}(x)w_{λ}

where c_{λ}(x) = f′(ξ), and ξ is between u_{λ} and u. Thus c_{λ}(x) < C_{λ} (which is a constant depending on λ), and w_{λ} satisfies

\{ \begin{matrix}−Δw_{λ} ≥ c_{λ}(x)w_{λ} & in Σ_{λ} \\ w_{λ} ≥ 0 & on Γ_{0} ∪ T_{λ} \\ \frac{∂w_{λ}}{∂ν} = 0 & on Γ_{1} ∩ \{|x| ≥ \frac{1}{λ}\}. \end{matrix}  (2.1)

We need the following version of the maximum principle in a narrow domain.

Lemma 2.1 Assume that u(x) satisfies the following:

−Δu ≥ c(x)u  in Σ(α, 1) ∩ (B_{r_{1}} \ B_{r_{2}})
u ≥ 0  on Σ(α, 1) ∩ (∂B_{r_{1}} ∪ ∂B_{r_{2}})
\frac{∂u}{∂ν} = 0  on Γ_{1} ∩ \{r_{2} ≤ |x| ≤ r_{1}\}

where |c(x)| ≤ C_{λ} and 1/λ ≤ r_{2} ≤ r_{1} ≤ 1. There is a constant δ_{λ} depending on C_{λ} such that, if r_{1} − r_{2} ≤ δ_{λ}, then u(x) ≥ 0 in Σ(α, 1) ∩ (B_{r_{1}} \ B_{r_{2}}).

Proof. Choose δ_{λ} sufficiently small so that for 0 < r_{1} − r_{2} < δ_{λ}

inf_{H_{r}^{1}(B_{r_{1}+δ_{λ}}\B_{r_{2}−δ_{λ}})\\{0\}} \frac{∫_{B_{r_{1}+δ_{λ}}\B_{r_{2}−δ_{λ}}} |∇u|^{2}}{∫_{B_{r_{1}+δ_{λ}}\B_{r_{2}−δ_{λ}}} |u|^{2}} := Λ_{λ} ≥ C_{λ},

where H_{r}^{1}(B_{r_{1}+δ_{λ}} \ B_{r_{2}−δ_{λ}}) = {v ∈ H_{0}^{1}(B_{r_{1}+δ_{λ}} \ B_{r_{2}−δ_{λ}}) : v is radial}. Let u_{0} > 0 be one of the extremal functions of the above infimum. Notice that u_{0} is radially symmetric. Thus w = u/u_{0} satisfies

−Δw − 2\frac{∇u_{0}}{u_{0}} · ∇w + (Λ_{λ} − c(x))w ≥ 0  in Σ(α, 1) ∩ (B_{r_{1}} \ B_{r_{2}})
w ≥ 0  on Σ(α, 1) ∩ (∂B_{r_{1}} ∪ ∂B_{r_{2}})
\frac{∂w}{∂ν} = 0  on Γ_{1} ∩ \{r_{2} ≤ |x| ≤ r_{1}\}.

It then follows from the maximum principle and boundary condition that w ≥ 0 in Σ(α, 1) ∩ (B_{r_{1}} \ B_{r_{2}}). Therefore u(x) ≥ 0 in Σ(α, 1) ∩ (B_{r_{1}} \ B_{r_{2}}).

We are now ready to prove Theorem 1.1.

Proposition 2.1 For any λ > 1,

w_{λ}(x) ≥ 0  ∀x ∈ Σ_{λ}.  (2.2)

Proof. This can be proved through two steps.

Step 1. There is a δ_{1} > 0, such that for all 1 ≤ λ < 1 + δ_{1}, (2.2) holds in Σ_{λ}. Notice that w_{λ} satisfies (2.1). We easily verify step 1 from Lemma 2.1 by choosing

δ_{1} sufficiently small.

We thus define

λ_{0} = sup{λ > 1 : w_{μ}(x) ≥ 0 in Σ_{μ}, ∀μ ≤ λ}.

Step 2. We claim: λ_{0} = +∞.

(2.2) follows from the claim directly.

To prove above claim, we argue by contradiction. If λ_{0} < ∞, then from the strong maximum principle and boundary condition we know that

w_{λ_{0}} > 0 in Σ_{λ_{0}} and w_{λ_{0}} > 0 on Γ_{1} ∩ \overline{Σ}_{λ_{0}} \ \{|x| = \frac{1}{λ_{0}} \}.

Thus, for any 0 < δ_{2} < λ_{0}/2, there is a ϵ_{2} = ϵ_{2}(δ_{2}) > 0, such that

w_{λ_{0}} > ϵ_{2} in \overline{Σ}_{λ_{0}−δ_{2}}.

From the continuity of w_{λ} with respect to λ, we know that there is a positive constant δ_{3} < δ_{2}, such that for any λ > λ_{0} and λ − λ_{0} < δ_{3},

w_{λ} > ϵ_{2}/2 in \overline{Σ}_{λ_{0}−δ_{2}}.  (2.3)

Therefore, for λ_{0} < λ < λ_{0} + δ_{3}, w_{λ} satisfies

−Δw_{λ} ≥ c_{λ}(x)w_{λ}  in Σ_{λ} ∩ B_{1/(λ_{0}−δ_{2})}
w_{λ} ≥ 0  0n (∂B_{1/(λ_{0}−δ_{2})} ∪ ∂B_{1/λ}) ∩ \overline{Σ}_{λ}
\frac{∂w_{λ}}{∂ν} = 0  on Γ_{1} ∩ \overline{Σ_{λ} ∩ B_{1/(λ_{0}−δ_{2})}},

where |c_{λ}(x)| ≤ C_{0} = C(λ_{0}). If we choose δ_{2} sufficiently small so that

\frac{1}{λ_{0} − δ_{2}} − \frac{1}{λ} < C_{0} ∀λ_{0} < λ < λ_{0} + δ_{3},

we obtain from Lemma 2.1 that w_{λ} ≥ 0 in Σ_{λ} ∩ B_{1/(λ_{0}−δ_{2})}. Thus, combining with (2.3), we know that for any 0 ≤ λ − λ_{0} ≤ δ_{3},

w_{λ} ≥ 0 in Σ_{λ}.

This contradicts the definition of λ_{0}. We hereby prove the claim, and thus complete the proof of Proposition 2.1.

Since w_{λ} is not identically equal to zero for any λ > 1, we have that w_{λ} = 0 on T_{λ} and w_{λ} > 0 in Σ_{λ}. Denote |x| = ρ. It follows from the Hopf Lemma that

\frac{∂w_{λ}}{∂ρ} |_{T_{λ}}> 0,

that is

(\frac{∂u_{λ}}{∂ρ} − \frac{∂u}{∂ρ}) |_{T_{λ}}> 0.

On T_{λ}, we have

\begin{aligned}\frac{∂u_{λ}}{∂ρ} (x) & = \frac{∂}{∂ρ}(\frac{1}{|λx|^{n−2}} ) · u(\frac{x}{|λx|^{2}} ) + \frac{∂}{∂ρ}u(\frac{x}{|λx|^{2}} ) \\ & = −(n − 2)λ · u(\frac{x}{|λx|^{2}} ) − \frac{∂u}{∂ρ}(x). \end{aligned}  (2.4)

Notice that u(x) > 0 in Σ_{λ}. It follows that

\frac{∂u}{∂ρ}(x) < 0  ∀x ∈ Σ(α, 1).  (2.5)

To proceed, we need the following lemma. It can be proved exactly in the same way as that of Lemma 2.1 in [2], and we shall omit the details.

Lemma 2.2 Let u ≥ 0 be a C^{2} solution of (1.2). Assume that f(u) satisfies (1) f(t) ≥ 0 for t ≥ 0,

(2) \underline{lim}_{t→∞}f(t)/t^{p} > 0 for some p ≥ n/(n − 2).

Then u is a distribution solution to

Δu + f(u) = 0  in Σ(α, R)
u = 0  on Γ_{0}
\frac{∂u}{∂ν} = 0  on Γ_{1}.

Let P be a tangent plane of Γ_{1} passing through one point p ∈ Γ_{1}. Denote H_{p} = \overline{Σ}(α, 1) ∩ P. Since α ≥ π, we know that H_{p} and Γ_{0} bound a half ball in Σ(α, 1), which we denoted as Ω_{+}. Let L be the axis which passes through the origin and is orthogonal to H_{p}.

Proposition 2.2 In Ω_{+}, u is axially symmetric with respect to L-axis.

Proof. Without loss of generality, we may assume that Ω_{+} is B_{1}^{+} = {x : |x| = 1, x_{n} > 0}. Since u(x) is a distribution solution by Lemma 2.2, we can prove the above proposition via standard moving plane method.

We first move planes along positive x_{1}-direction. For any −1 ≤ λ < 0, define

B_{1,λ}^{+} = {x ∈ B_{1}^{+} : x_{1} < λ}, T_{λ} = {x ∈ B_{1}^{+} : x_{1} = λ},
x^{λ} is the reflection of x about T_{λ},
u_{λ}(x) = u(x^{λ}), w_{λ}(x) = u_{λ}(x) − u(x).

From (2.5) we know that ∂u/∂ρ ≤ 0 for |x| = ρ < 1 on hyperplane H_{p}. Thus, w_{λ} satisfies

\{ \begin{matrix}−Δw_{λ} = f′(ξ)w_{λ} & in B_{1,λ}^{+} \\ w_{λ} ≥ 0 & on ∂B_{1,λ}^{+} \ \{x ∈ B_{1,λ} : x^{λ} = 0\}, \end{matrix}  (2.6)

where ξ is a number between u(x) and u_{λ}(x).

It is clear that for λ close to −1, w_{λ} ≥ 0 in B_{1,λ}^{+}. We can define

λ_{0} = sup{λ < 0 : w_{μ} ≥ 0 in B_{1,μ}^{+}, for any − 1 ≤ μ ≤ λ}.

Claim: λ_{0} = 0. This can be proved through a contradiction argument. Suppose that λ_{0} < 0. Let x̃ be the point so that x̃^{λ_{0}} = 0. Since u(x) is a distribution solution, we know that for any ϵ_{∗} > 0, there is a δ_{∗} > 0 such that w_{λ_{0}} > δ_{∗} in \overline{B}_{ϵ_{∗}}^{+}(x̃). Therefore w_{λ_{0}} > 0 in B_{1,λ_{0}}^{+}. From continuity of w_{λ}, we know that there is a 0 < ϵ_{1} such that for any λ_{0} < λ < λ_{0} + ϵ_{1}, w_{λ} ≥ 0 in \overline{B}_{1,ϵ_{0}}^{+}. In B_{1,λ_{0}+ϵ_{1}}^{+} \ B_{1,ϵ_{0}}^{+}, f′(ξ) ≤ C. From the narrow domain maximum principle, we know that when ϵ_{1} is sufficiently small, for any λ_{0} < λ < λ_{0} + ϵ_{1}, w_{λ} ≥ 0 in B_{1,λ}^{+} \ B_{1,ϵ_{0}}^{+}. Thus w_{λ} ≥ 0 in B_{1,λ}^{+} for any λ_{0} < λ < λ_{0} + ϵ_{1}. This contradicts the definition of λ_{0}. The claim is proved.

It follows that for any −1 ≤ λ ≤ 0, w_{λ} ≥ 0. Similarly, we can move planes from x_{1} negative direction and any x_{i} direction for i = 2, ..., n − 1, thus obtain the proposition.

Due to Proposition 2.2, we then can extend u(x) into B_{1} ball as the following:

ũ(x) = \{ \begin{matrix}u(x) & ∀x ∈ \overline{Σ}(α, 1) \\ u(−x) & ∀x ∈ B_{1} \ \overline{Σ}(α, 1). \end{matrix}

It is easy to check that ũ(x) is a C^{2} positive solution to

\{ \begin{matrix}Δũ + f(ũ) & in B_{1} \ \{0\} \\ ũ = 0 & on ∂B_{1}. \end{matrix}  (2.7)

From Lemma 2.1 in [2] we know that ũ is a distribution solution in B_{1}. Therefore we can use the moving plane method as above to show that ũ(x) = ũ(|x|), hence u(x) = u(|x|).

Remark 2.1 We now give some details about remark 1.1. Let us assume that f(u) satisfies condition (i) and (ii). If u ≥ 0 is a bounded solution to (1.2), we know from (2.7) that ũ ∈ C^{1}(B_{1}(0)) by elliptic estimates. From the strong maximum principle, we know that u = 0 or u > 0 in Σ(α, 1). If u > 0, then u(0) > 0. It then follows from (2.4) that ∂u/∂ρ(x) → −∞ as |x| = 1/λ → 0. This contradicts the fact that u ∈ C^{2}(\overline{Σ(α, 1)}).

References

[1] H. Berestycki, and F. Pacella, Symmetry properties for positive solutions of elliptic equations with mixed boundary conditions, J. Funct. Anal. 87(1989), 177-211.

[2] L. Caffarelli, B. Gidas and J. Spruck, Asymptotic symmetry and local behavior of semilinear elliptic equations with critical Sobolev growth, Comm. Pure Appl. Math. 42 (1989), 271-297.

[3] B. Gidas, W.M. Ni and L. Nirenberg, Symmetry and related properties via the maximum principle, Comm. Math. Phys. 68(1979), 209-243.

[4] Y.Y. Li, and M. Zhu, Uniqueness theorems through the method of moving spheres, Duke Math J. vol., 80(1995) 383-417.