# Overdetermined problems in annular domains with a spherical-boundary component in space forms

## Abstract

We obtain a Serrin-type symmetry of the solutions to various overdetermined boundary value problems in annular domains with a spherical-boundary component in space forms by using the maximum principle for suitable subharmonic functions and integral identities.

## 1 Introduction

In his pioneering work , Serrin proved that if there exists a solution of the following overdetermined boundary value problem for a smooth bounded open connected domain $$\Omega \subset \mathbb{R}^{n}$$

\begin{aligned} \textstyle\begin{cases} \Delta u =-1 & \text{in } \Omega , \\ u=0 & \text{on } \partial \Omega , \\ \frac{\partial u}{\partial \nu} = \mathrm{const}=c & \text{on } \partial \Omega , \end{cases}\displaystyle \end{aligned}

then the solution u is radially symmetric and Ω is a ball. Here, ν denotes the outward unit normal to Ω. His proof is based on the moving-plane method, which was initiated by Alexandrov . Immediately, Weinberger  gave an alternative simple proof of Serrin’s symmetry result, which was based on the maximum principle for a suitable subharmonic function and some integral identities. Thereafter, overdetermined boundary value problems have been actively studied. For instance, Serrin’s symmetry result has been generalized into space forms (see [46, 9, 11, 15, 17] for example and references therein).

On the other hand, one may still expect the radial symmetry of the solutions to overdetermined problems in annular domains. To describe this precisely, let $$\Omega _{0}$$ and $$\Omega _{1}$$ be simply connected bounded $$C^{2}$$ domains in $$\mathbb{R}^{n}$$ ($$n \geq 2$$) such that $$\overline{\Omega}_{1} \subset \Omega _{0}$$. For the annular domain $$\Omega :=\Omega _{0} \setminus \overline{\Omega}_{1}$$, consider the following overdetermined boundary value problem

\begin{aligned} \textstyle\begin{cases} \Delta u = -1 & \text{in } \overline{\Omega}, \\ u = 0, \qquad \frac{\partial u}{\partial \nu} = c_{0}& \text{on } \partial \Omega _{0}, \\ u = a > 0, \qquad \frac{\partial u}{\partial \nu} = c_{1} & \text{on } \partial \Omega _{1}, \end{cases}\displaystyle \end{aligned}
(1.1)

where ν is the outward unit normal to Ω and $$c_{0}$$, $$c_{1}$$, and a are real constants. In 1990, Philippin  proved that if each domain $$\Omega _{i}$$ ($$i=0, 1$$) is star shaped then the solution to the overdetermined problem (1.1) is radially symmetric and the domain Ω is a standard annulus (see also [1, 12] for more general related results). Under the additional condition that $$0\leq u \leq a$$ in Ω̅, Reichel  obtained the same result. Later, Sirakov  removed the extra condition (see [8, 21] for $$n=2$$). Recently, Kamburov-Sciaraffia  constructed a bounded real analytic annular domain $$\Omega \subset \mathbb{R}^{n}$$, which is different from a standard annulus, satisfying that the overdetermined problem (1.1) admits a solution $$u\in C^{\infty }(\overline{\Omega})$$ with $$a>0$$ and $$c_{0}=c_{1}<0$$.

In this paper, we generalize the above Serrin-type result for annular domains in $$\mathbb{R}^{n}$$ to space forms by using the maximum principle for suitable subharmonic functions and some integral identities. This approach was also used to obtain the radial symmetry of solutions to partially overdetermined problems in domains inside a convex cone in space forms by the authors of . In Sect. 2, we study overdetermined problems in annular domains with inner spherical boundary in the unit sphere $$\mathbb{S}^{n}$$. For the equations $$\Delta u = -n\cos r$$ and $$\Delta u +nu=-n$$, we obtain radial-symmetry results (see Theorem 2.1 and Theorem 2.2). Moreover, we prove rigidity theorems (Theorem 2.4 and Theorem 2.5) for annular domains in $$\mathbb{S}^{n}$$ under suitable conditions on the inner spherical boundary. In Sect. 3, we consider the equation $$\Delta u -nu=-n$$ on an annular domain with outer spherical boundary in the hyperbolic space $$\mathbb{H}^{n}$$. In Theorem 3.2, we are able to prove a Serrin-type symmetry result for such domains. Furthermore, under the additional assumption that the annular domain is weakly star shaped (see Definition 2.3), we obtain the same result with a weakened Dirichlet condition on the outer spherical boundary (see Theorem 3.3).

## 2 Annular domains with inner spherical boundary

In this section, we study overdetermined boundary value problems in an annular domain whose inner boundary is spherical. Before we state our results, we start with some notations. Let $$M^{n}$$ be an n-dimensional space form of constant sectional curvature $$K=0, 1$$, and −1: the corresponding spaces are the Euclidean space $$\mathbb{R}^{n}$$, the unit sphere $$\mathbb{S}^{n}$$, and the hyperbolic space $$\mathbb{H}^{n}$$, respectively. These spaces can be regarded as the warped product space $$M=I\times \mathbb{S}^{n-1}$$ with the metric $$g=dr^{2} +h(r)^{2} g_{\mathbb{S}^{n-1}}$$, where r denotes the distance from the pole p of the model space and $$g_{\mathbb{S}^{n-1}}$$ denotes the round metric on $$\mathbb{S}^{n-1}$$. Moreover, the warping function $$h(r)$$ is given by

• $$h(r) = r$$ on $$I = [0,\infty )$$ in $$\mathbb{R}^{n}$$;

• $$h(r) = \sin r$$ on $$I = [0,\pi )$$ in $$\mathbb{S}^{n}$$;

• $$h(r) = \sinh r$$ on $$I = [0,\infty )$$ in $$\mathbb{H}^{n}$$.

Now we prove the radial symmetry of the solution to an overdetermined boundary value problem on annular domains in $$\mathbb{S}^{n}$$ with inner spherical boundary (see Fig. 1).

### Theorem 2.1

Let Ω be an annular domain in $${\mathbb{S}}^{n} \setminus B_{R}(N)$$ such that $$\partial B_{R} (N) \subset \partial \Omega$$, where $$B_{R} (N) \subset {\mathbb{S}}^{n}$$ denotes the closed geodesic ball of radius $$0< R< \pi$$ centered at the north pole $$N \in {\mathbb{S}}^{n}$$. Suppose there is a solution $$u\in C^{2}(\overline{\Omega})$$ satisfying that

$$\textstyle\begin{cases} \Delta u = -n \cos r = -nh' & \textit{in } \Omega , \\ u=0, \qquad \frac{\partial u}{\partial \nu} = \mathrm{const} = c_{1} & \textit{on } \partial \Omega \setminus \partial B_{R} (N), \\ u=\mathrm{const}=a>0,\qquad \frac{\partial u}{\partial \nu} = \sin R & \textit{on } \partial B_{R} (N), \end{cases}$$

where ν is the outward unit normal to Ω and $$r(x) = \operatorname{dist}(N,x)$$. Assume that either Ω is contained in the upper hemisphere $${\mathbb{S}}_{+}^{n}$$ or u is positive. Then, Ω is the standard annulus $$\{x\in {\mathbb{S}}^{n} : R < r(x) < R_{1}\}$$ and the solution u is radial and is given by

$$u(x)= \cos r(x) - \cos R_{1},$$

where $$R_{1} = \sin ^{-1}(-c_{1})$$.

### Proof

In the case where Ω is contained in $$\mathbb{S}^{n}_{+}$$, we see that u is positive in Ω by the maximum principle. Thus, we may assume that u is positive in Ω.

A straightforward computation yields

$$\operatorname{Hess}h'= - h'g \quad \text{and} \quad \Delta h'= -nh',$$

where $$\operatorname{Hess}h'$$ denotes the Hessian of $$h'$$ and g denotes the metric of $${\mathbb{S}}^{n}$$. Note that

\begin{aligned} (\Delta u)^{2} \leq n \operatorname{tr}\bigl(\operatorname{Hess} ^{2}u\bigr), \end{aligned}
(2.1)

where $$\operatorname{Hess}^{2} = \operatorname{Hess}\circ \operatorname{Hess}$$. Moreover, equality holds if and only if Hessu is proportional to the metric g. By the polarized Bochner formula,

\begin{aligned} \Delta \bigl\langle \nabla \bigl(u-h'\bigr), \nabla u \bigr\rangle & = \bigl\langle \nabla \bigl(\Delta \bigl(u-h' \bigr)\bigr),\nabla u \bigr\rangle + \bigl\langle \nabla \bigl(u-h' \bigr), \nabla (\Delta u) \bigr\rangle \\ &\quad {} + 2 \operatorname{tr}\bigl(\operatorname{Hess}\bigl(u-h' \bigr) \circ \operatorname{Hess}u\bigr) + 2 \operatorname{Ric}\bigl(\nabla \bigl(u-h'\bigr), \nabla u\bigr). \end{aligned}
(2.2)

From (2.1), it follows that

$$\operatorname{tr}\bigl(\operatorname{Hess}\bigl(u-h'\bigr) \circ \operatorname{Hess}u\bigr) = \operatorname{tr}\bigl(\operatorname{Hess} ^{2}u\bigr)+ h' \Delta u = \operatorname{tr}\bigl( \operatorname{Hess} ^{2}u\bigr)- nh^{\prime 2} \geq 0.$$

Thus, (2.2) becomes

$$\Delta \bigl\langle \nabla \bigl(u-h'\bigr), \nabla u \bigr\rangle \geq -n \bigl\langle \nabla \bigl(u-h'\bigr), \nabla h' \bigr\rangle + 2(n-1) \bigl\langle \nabla \bigl(u-h' \bigr), \nabla u \bigr\rangle .$$

Since $$u>0$$ in Ω, we obtain

\begin{aligned} &\int _{\Omega } u \Delta \bigl\langle \nabla \bigl(u-h'\bigr), \nabla u \bigr\rangle \,dV \\ &\quad \geq -n \int _{\Omega } u \bigl\langle \nabla \bigl(u-h' \bigr), \nabla h' \bigr\rangle \,dV + 2 (n-1) \int _{\Omega } u \bigl\langle \nabla \bigl(u-h' \bigr), \nabla u \bigr\rangle \,dV . \end{aligned}
(2.3)

Since

$$\frac{\partial h'}{\partial \nu} = \sin R = \frac{\partial u}{\partial \nu} \quad \text{on } \partial B_{R}(N),$$

using the divergence theorem, we obtain

\begin{aligned} \int _{\Omega } \bigl\langle \nabla \bigl(u-h' \bigr), \nabla \bigl(u^{2}\bigr) \bigr\rangle \,dV & = \int _{\Omega } \operatorname{div}\bigl( u^{2} \nabla \bigl(u-h'\bigr) \bigr)\,dV - \int _{\Omega } u^{2} \Delta \bigl(u-h' \bigr) \,dV \\ &= \int _{\partial \Omega } u^{2} \frac{\partial }{\partial \nu} \bigl(u-h'\bigr) \,d\sigma = 0, \end{aligned}

which yields

$$\int _{\Omega } u \bigl\langle \nabla \bigl(u-h' \bigr) ,\nabla u \bigr\rangle \,dV = 0.$$
(2.4)

From (2.4),

\begin{aligned} \int _{\Omega } u \bigl\langle \nabla \bigl(u-h' \bigr), \nabla h' \bigr\rangle \,dV &= - \int _{\Omega } u \bigl\langle \nabla \bigl(u-h' \bigr), -\nabla h' \bigr\rangle \,dV \\ &= - \int _{\Omega } u \bigl\vert \nabla \bigl(u-h' \bigr) \bigr\vert ^{2} \,dV . \end{aligned}
(2.5)

Combining (2.3), (2.4), and (2.5),

\begin{aligned} \int _{\Omega } u \Delta \bigl\langle \nabla \bigl(u-h' \bigr), \nabla u \bigr\rangle \,dV & \geq n \int _{\Omega } u \bigl\vert \nabla \bigl(u-h' \bigr) \bigr\vert ^{2}\,dV \geq 0. \end{aligned}
(2.6)

On the other hand, from Green’s identity,

\begin{aligned} \int _{\Omega } u \Delta \bigl\langle \nabla \bigl(u-h' \bigr), \nabla u \bigr\rangle \,dV & = \int _{\Omega } \bigl\langle \nabla \bigl(u-h' \bigr), \nabla u \bigr\rangle \Delta u \,dV \\ &\quad{} + \int _{\partial \Omega } u \frac{\partial }{\partial \nu} \bigl\langle \nabla \bigl(u-h'\bigr), \nabla u \bigr\rangle \,d\sigma \\ &\quad{} - \int _{\partial \Omega } \bigl\langle \nabla \bigl(u-h' \bigr), \nabla u \bigr\rangle \frac{\partial u}{\partial \nu} \,d\sigma . \end{aligned}
(2.7)

Using the divergence theorem and the boundary conditions, we obtain

\begin{aligned} \int _{\Omega } \bigl\langle \nabla \bigl(u-h' \bigr) , \nabla \bigl(uh'\bigr) \bigr\rangle \,dV & = \int _{\Omega } \operatorname{div}\bigl(uh'\nabla \bigl(u-h'\bigr)\bigr) \,dV - \int _{\Omega } uh'\Delta \bigl(u-h' \bigr) \,dV \\ &= \int _{\partial \Omega } uh'\frac{\partial }{\partial \nu} \bigl(u-h'\bigr) \,d\sigma =0, \end{aligned}

which implies that

\begin{aligned} \int _{\Omega } \bigl\langle \nabla \bigl(u-h' \bigr), \nabla u \bigr\rangle \Delta u \,dV & = -n \int _{\Omega } h' \bigl\langle \nabla \bigl(u-h'\bigr) , \nabla u \bigr\rangle \,dV \\ & = n \int _{\Omega } u \bigl\langle \nabla \bigl(u-h' \bigr), \nabla h' \bigr\rangle \,dV \\ & = -n \int _{\Omega } u \bigl\vert \nabla \bigl(u-h' \bigr) \bigr\vert ^{2} \,dV \leq 0. \end{aligned}
(2.8)

Since Ω is a level set of u, u is parallel to ν on Ω. Thus,

\begin{aligned} \int _{\partial \Omega } \bigl\langle \nabla \bigl(u-h' \bigr), \nabla u \bigr\rangle \frac{\partial u}{\partial \nu} \,d\sigma & = {c_{1}}^{2} \int _{\partial \Omega \setminus \partial B} \bigl\langle \nabla \bigl(u-h' \bigr), \nu \bigr\rangle \,d\sigma \\ & = {c_{1}}^{2} \int _{\partial \Omega } \bigl\langle \nabla \bigl(u-h' \bigr), \nu \bigr\rangle \,d\sigma \\ & = {c_{1}}^{2} \int _{\Omega } \Delta \bigl(u-h'\bigr) \,dV =0. \end{aligned}
(2.9)

Substituting (2.8) and (2.9) into (2.7),

$$\int _{\Omega } u \Delta \bigl\langle \nabla \bigl(u-h' \bigr), \nabla u \bigr\rangle \,dV \leq a \int _{\partial B_{R}(N)} \frac{\partial }{\partial \nu} \bigl\langle \nabla \bigl(u-h'\bigr), \nabla u \bigr\rangle \,d\sigma .$$
(2.10)

To compute the right-hand side of (2.10), we choose a local orthonormal frame $$\{e_{i}\}_{i=1}^{n}$$ such that

$$e_{n} = \frac{\partial }{\partial r}.$$

Since $$u_{r}$$ is constant on $$\partial B_{R} (N)$$ and each $$e_{i}$$ is tangent to $$\partial B_{R}(N)$$ for all $$i=1, \ldots , n-1$$, we have

$$u_{ri} = 0 \quad \text{on } \partial B_{R} (N)$$
(2.11)

for all $$i =1, \ldots , n-1$$. Since $$\partial B_{R} (N)$$ is a level set of u,

$$u_{i}=0 \quad \text{and} \quad u_{ij}=0\quad \text{on } \partial B_{R} (N)$$
(2.12)

for all $$i,j=1,\ldots , n-1$$. From (2.11) and (2.12),

$$\Delta u = (n-1)\frac{\cos r}{\sin r}u_{r} + u_{rr} \quad \text{on } \partial B_{R} (N),$$

which yields that

$$-nh' = (n-1)\frac{\cos R}{\sin R} (-\sin R ) + u_{rr} \quad \text{on } \partial B_{R} (N).$$

That is,

$$u_{rr} = -nh' + (n-1) \cos R =-h'\quad \text{on } \partial B_{R} (N).$$

Note that

\begin{aligned} \frac{\partial }{\partial \nu} \bigl\langle \nabla \bigl(u-h'\bigr),\nabla u \bigr\rangle & = \bigl\langle \nabla _{\nu }\nabla \bigl(u-h'\bigr) , \nabla u \bigr\rangle + \bigl\langle \nabla \bigl(u-h'\bigr), \nabla _{\nu } \nabla u \bigr\rangle \\ & = 2 \operatorname{Hess}u (\nabla u,\nu ) - \operatorname{Hess}h'( \nabla u, \nu ) - \operatorname{Hess}u \bigl( \nabla h', \nu \bigr) \\ & = 2 \operatorname{Hess}u (\nabla u,\nu ) + h' \frac{\partial u}{\partial \nu} - \operatorname{Hess}u \bigl( \nabla h', \nu \bigr) \\ & = 2 u_{rr}\frac{\partial u}{\partial \nu} +h' \frac{\partial u}{\partial \nu} - u_{rr} \frac{\partial h'}{\partial \nu} \\ & = \sin R\bigl(u_{rr} + h'\bigr)=0 \quad \text{on } \partial B_{R} (N). \end{aligned}

Thus,

$$\int _{\Omega } u \Delta \bigl\langle \nabla \bigl(u-h' \bigr), \nabla u \bigr\rangle \,dV \leq 0.$$
(2.13)

Combining (2.6) and (2.13), we conclude that

$$\int _{\Omega } u \Delta \bigl\langle \nabla \bigl(u-h' \bigr), \nabla u \bigr\rangle \,dV = 0.$$

Note that the equality in (2.6) holds when $$\nabla (u-h') \equiv 0$$ in Ω. Thus,

$$u(x) = h'+ c = \cos r + c$$

for some constant c. Since u vanishes on $$\partial \Omega \setminus \partial B_{R}(N)$$ and Ω is connected, the set $$\partial \Omega \setminus \partial B_{R}(N)$$ is the boundary of the geodesic ball $$B_{R_{1}} (N)$$ centered at N with radius $$R_{1} = \cos ^{-1}(-c)$$. Hence, Ω must be the standard annulus $$\{x\in \mathbb{S}^{n}: R < r(x) < R_{1}\}$$. Furthermore, the constant c can be expressed in terms of the constant $$c_{1}$$. To see this, observe that on $$\partial \Omega \setminus \partial B_{R}(N)$$

$$\frac{\partial u}{\partial \nu} = \langle -\sin r \nabla r, \nabla r \rangle = -\sin R_{1} = c_{1},$$

which implies that $$R_{1} = \sin ^{-1}(-c_{1})$$. Therefore, the solution is given by

$$u(x) = \cos r + c = \cos r - \cos R_{1},$$

where

$$R_{1} = \sin ^{-1}(-c_{1}).$$

□

We remark that the center N of the geodesic ball $$B_{R}(N)$$ can be replaced by any point $$p \in \mathbb{S}^{n}$$ in Theorem 2.1. In this case, the solution u is radially symmetric with respect to the point p. Note that consistency requires $$-1 \leq c_{1} < 0$$ and $$R < R_{1}$$.

### Theorem 2.2

Let Ω be an annular domain in $${\mathbb{S}}_{+}^{n} \setminus B_{R}(N)$$ such that $$\partial B_{R}(N) \subset \partial \Omega$$, where $$B_{R}(N) \subset {\mathbb{S}}_{+}^{n}$$ denotes the closed geodesic ball of radius $$0< R<\frac{\pi}{2}$$ centered at the north pole $$N \in {\mathbb{S}}^{n}$$. Given $$R< R_{1} <\frac{\pi}{2}$$, suppose there is a solution $$u\in C^{2}(\overline{\Omega})$$ such that

$$\textstyle\begin{cases} \Delta u +nu = -n & \textit{in } \Omega , \\ u=0, \qquad \frac{\partial u}{\partial \nu} = - \frac{\sin R_{1}}{\cos R_{1}} & \textit{on } \partial \Omega \setminus \partial B_{R}(N), \\ u=\mathrm{const}=a > 0,\qquad \frac{\partial u}{\partial \nu} = \frac{\sin R}{\cos R_{1}} & \textit{on } \partial B_{R}(N), \end{cases}$$

where ν is the outward unit normal to Ω. Then, Ω is the standard annulus $$\{x\in {\mathbb{S}}^{n} : R < r(x) < R_{1}\}$$ and the solution u is radial and is given by

$$u(x)= \frac{1}{\cos R_{1}}\bigl(\cos r(x) -\cos R_{1}\bigr),$$

where $$r(x) = \operatorname{dist}(N,x)$$.

### Proof

Using the Bochner formula and (2.1), we have

\begin{aligned} \Delta \vert \nabla u \vert ^{2} & = 2 \bigl\langle \nabla ( \Delta u), \nabla u \bigr\rangle + 2 \operatorname{tr}\bigl(\operatorname{Hess} ^{2}u\bigr)+ 2 \operatorname{Ric}(\nabla u, \nabla u) \\ & \geq -2n \vert \nabla u \vert ^{2} + \frac{2}{n} ( \Delta u)^{2} + 2(n-1) \vert \nabla u \vert ^{2} \\ & = \frac{2}{n} (-n-nu)\Delta u - 2 \vert \nabla u \vert ^{2} \\ & = -2 \Delta u - \Delta u^{2}. \end{aligned}
(2.14)

Define

$$P(u) := \vert \nabla u \vert ^{2} + 2u + u^{2}.$$

Then, (2.14) shows that P is a subharmonic function in Ω. We claim that the function P is constant in Ω and the constant a satisfies that

$$a+1=\frac{\cos R}{\cos R_{1}}.$$

To see this, we first note that, on the boundary Ω, the function P is given by

\begin{aligned} P(u) = \textstyle\begin{cases} \frac{\sin ^{2}R_{1}}{\cos ^{2} R_{1}} & \text{on } \partial \Omega \setminus \partial B_{R}(N), \\ \frac{\sin ^{2} R}{\cos ^{2} R_{1}} + 2a + a^{2} & \text{on } \partial B_{R}(N). \end{cases}\displaystyle \end{aligned}

Suppose that

$$\frac{\sin ^{2} R}{\cos ^{2} R_{1}} + 2a + a^{2} > \frac{\sin ^{2}R_{1}}{\cos ^{2} R_{1}}.$$

Then,

\begin{aligned} (a+1)^{2} > \frac{\sin ^{2} R_{1} - \sin ^{2} R}{\cos ^{2} R_{1}} +1 & = \frac{1-\cos ^{2}R_{1} - (1-\cos ^{2} R)}{\cos ^{2} R_{1}} +1 = \frac{\cos ^{2} R}{\cos ^{2} R_{1}}. \end{aligned}

Since a is positive,

$$a+1 > \frac{\cos R}{\cos R_{1}}.$$
(2.15)

By the maximum principle,

$$\max_{\Omega } P(u) = \max_{\partial \Omega } P(u) = \frac{\sin ^{2} R}{\cos ^{2} R_{1}} + 2a + a^{2}.$$

Since P is not a constant function under our assumption, we have

$$\frac{\partial P}{\partial \nu}> 0 \quad \text{on } \partial B_{R}(N)$$
(2.16)

by the Hopf lemma. Choose a local orthonormal frame $$\{e_{i}\}_{i=1}^{n}$$ such that $$e_{n} = \frac{\partial}{\partial r}$$. Since $$u_{r}$$ is constant on $$\partial B_{R}(N)$$ and each $$e_{i}$$ is tangent to $$\partial B_{R}(N)$$ for $$i=1, \ldots , n-1$$,

$$u_{ri} = 0 \quad \text{on }\partial B_{R}(N)$$
(2.17)

for all $$i=1, \ldots , n-1$$. Since $$\partial B_{R}(N)$$ is a level set of u,

$$u_{i}=0 \quad \text{and}\quad u_{ij}=0 \quad \text{on } \partial B_{R}(N)$$
(2.18)

for all $$i,j=1, \ldots , n-1$$. From (2.17) and (2.18),

$$\Delta u = (n-1)\frac{\cos r}{\sin r}u_{r} + u_{rr} \quad \text{on } \partial B_{R}(N),$$

which yields that

$$-n-na = (n-1)\frac{\cos R}{\sin R} \biggl(-\frac{\sin R}{\cos R_{1}} \biggr) + u_{rr} \quad \text{on } \partial B_{R}(N).$$

That is,

$$u_{rr} = -n -na + (n-1) \frac{\cos R}{\cos R_{1}} \quad \text{on } \partial B_{R}(N).$$

Thus,

\begin{aligned} \frac{\partial P}{\partial \nu} & =2 \operatorname{Hess}u(\nabla u, \nu ) + 2 \frac{\partial u}{\partial \nu} + 2u \frac{\partial u}{\partial \nu} \\ & = 2u_{rr} \frac{\partial u}{\partial \nu} + 2 \frac{\partial u}{\partial \nu}+2u \frac{\partial u}{\partial \nu} \\ & = 2 (n-1) \frac{\sin R}{\cos R_{1}} \biggl(\frac{\cos R}{\cos R_{1}} -1 -a \biggr) \quad \text{on } \partial B_{R}(N). \end{aligned}

By (2.15),

$$\frac{\partial P}{\partial \nu} < 0 \quad \text{on } \partial B_{R}(N),$$

which is a contradiction to (2.16). Therefore, we see that

$$\frac{\sin ^{2} R}{\cos ^{2} R_{1}} + 2a +a^{2} \leq \frac{\sin ^{2} R_{1}}{\cos ^{2} R_{1}}.$$
(2.19)

Then,

$$-\frac{\cos R}{\cos R_{1}}\leq a+1 \leq \frac{\cos R}{\cos R_{1}}.$$
(2.20)

Suppose that P is not a constant function. Then, we obtain

$$P< \frac{\sin ^{2} R_{1}}{\cos ^{2} R_{1}} \quad \text{in }\Omega$$
(2.21)

by the maximum principle. Let

$$X = h\frac{\partial }{\partial r},$$

where $$h(r)=\sin r$$. Clearly,

$$h'(r) = \cos r>0, \qquad h''(r) = -h(r),\quad \text{and}\quad \operatorname{div}X = nh' \quad \text{on } {\mathbb{S}}_{+}^{n}.$$
(2.22)

From (2.21),

$$\frac{\sin ^{2} R_{1}}{\cos ^{2} R_{1}} \int _{\Omega } h' \,dV > \int _{ \Omega } P h' \,dV = \int _{\Omega } \bigl( \vert \nabla u \vert ^{2} h' + u^{2} h' + 2uh' \bigr)\,dV .$$
(2.23)

Note that

$$\bigl\langle \nabla h', \nabla u \bigr\rangle = \bigl\langle h'' \nabla r, \nabla u \bigr\rangle = h''u_{r}.$$

Applying the divergence theorem, we obtain

\begin{aligned} \int _{\Omega } \vert \nabla u \vert ^{2} h' \,dV & = \int _{\Omega } \bigl\langle \nabla \bigl(uh' \bigr),\nabla u \bigr\rangle \,dV - \int _{\Omega } u \bigl\langle \nabla h' , \nabla u \bigr\rangle \,dV \\ & = \int _{\Omega } \operatorname{div} \bigl(uh'\nabla u \bigr) \,dV - \int _{\Omega } uh' \Delta u \,dV - \int _{\Omega } uh'' u_{r} \,dV \\ & = \int _{\partial \Omega }uh'\frac{\partial u}{\partial \nu} \,d\sigma - \int _{\Omega } uh'(-n-nu)\,dV + \int _{\Omega }uhu_{r} \,dV \\ & = \frac{a \cos R\sin R}{\cos R_{1}} \bigl\vert \partial B_{R}(N) \bigr\vert + n \int _{ \Omega } uh' \,dV \\ &\quad {}+ n \int _{\Omega } u^{2} h' \,dV + \int _{\Omega } uhu_{r} \,dV . \end{aligned}
(2.24)

Using (2.22), we see that

\begin{aligned} \int _{\Omega } uhu_{r} \,dV & = \int _{\Omega } u \biggl\langle \nabla u, h \frac{\partial }{\partial r} \biggr\rangle \,dV = \frac{1}{2} \int _{\Omega } \bigl\langle \nabla u^{2} , X \bigr\rangle \,dV \\ & = \frac{1}{2} \int _{\Omega } \operatorname{div} \bigl(u^{2} X \bigr) \,dV - \frac{1}{2} \int _{\Omega } u^{2} \operatorname{div} (X ) \,dV \\ & = \frac{1}{2} \int _{\partial \Omega } u^{2} \langle X , \nu \rangle \,d\sigma - \frac{n}{2} \int _{\Omega } u^{2} h' \,dV \\ & = -\frac{a^{2}}{2} \sin R \bigl\vert \partial B_{R}(N) \bigr\vert - \frac{n}{2} \int _{ \Omega } u^{2} h' \,dV , \end{aligned}

which is equivalent to

$$\int _{\Omega } u^{2} h' \,dV = - \frac{a^{2}}{n} \sin R \bigl\vert \partial B_{R}(N) \bigr\vert - \frac{2}{n} \int _{\Omega } uhu_{r} \,dV .$$
(2.25)

Substituting (2.24) and (2.25) into (2.23), we obtain

\begin{aligned} \frac{\sin ^{2} R_{1}}{\cos ^{2} R_{1}} \int _{\Omega } h' \,dV &> a \sin R \bigl\vert \partial B_{R}(N) \bigr\vert \biggl(\frac{\cos R}{\cos R_{1}} - \frac{a(n+1)}{n} \biggr) \\ &\quad + (n+2) \int _{\Omega } uh' \,dV - \biggl(1+ \frac{2}{n} \biggr) \int _{\Omega } uhu_{r} \,dV . \end{aligned}
(2.26)

Using the following Pohozaev identity [35, 14, 20]

$$\operatorname{div} \biggl(\frac{ \vert \nabla u \vert ^{2}}{2}X-hu_{r} \nabla u \biggr) = \frac{n-2}{2} \vert \nabla u \vert ^{2} h' - hu_{r} \Delta u,$$

we have

\begin{aligned} &\int _{\Omega }\operatorname{div} \bigl( \vert \nabla u \vert ^{2} X - 2hu_{r} \nabla u \bigr)\,dV \\ &\quad = (n-2) \int _{\Omega } \vert \nabla u \vert ^{2} h'\,dV - 2 \int _{\Omega } hu_{r}(-n-nu)\,dV . \end{aligned}

From (2.24) and (2.25),

\begin{aligned} &\int _{\Omega } \operatorname{div} \bigl( \vert \nabla u \vert ^{2} X - 2hu_{r} \nabla u\bigr) \,dV \\ &\quad = (n-2) \int _{\Omega } \vert \nabla u \vert ^{2} h' \,dV + 2n \int _{\Omega } hu_{r} \,dV + 2n \int _{\Omega } uhu_{r} \,dV \\ &\quad = (n-2) \biggl( \frac{ a \cos R\sin R}{\cos R_{1}} \bigl\vert \partial B_{R}(N) \bigr\vert + n \int _{\Omega } uh'\,dV + n \int _{\Omega } u^{2} h' \,dV + \int _{ \Omega } uhu_{r} \,dV \biggr) \\ &\quad \quad {}+2n \int _{\Omega } \langle X , \nabla u \rangle \,dV + 2n \int _{\Omega } uhu_{r} \,dV \\ &\quad = (n-2) \biggl( \frac{ a \cos R\sin R}{\cos R_{1}} \bigl\vert \partial B_{R}(N) \bigr\vert + n \int _{\Omega } uh'\,dV + n \int _{\Omega } u^{2} h' \,dV \biggr) \\ &\quad \quad {}+ (3n-2) \int _{\Omega } uhu_{r} \,dV + 2n \int _{\Omega } \operatorname{div} (uX ) \,dV - 2n \int _{\Omega } u \operatorname{div} (X )\,dV \\ &\quad = (n-2) a \sin R \bigl\vert \partial B_{R}(N) \bigr\vert \biggl( \frac{\cos R}{\cos R_{1}} -a \biggr) -2an \sin R \bigl\vert \partial B_{R}(N) \bigr\vert \\ &\quad \quad {}+(n+2) \biggl( \int _{\Omega } uhu_{r} \,dV -n \int _{\Omega }uh' \,dV \biggr). \end{aligned}
(2.27)

On the other hand, since

$$\nabla u = \frac{\partial u}{\partial \nu} \nu \quad \text{on } \partial \Omega ,$$

applying the divergence theorem, we have

\begin{aligned} &\int _{\Omega } \operatorname{div}\bigl( \vert \nabla u \vert ^{2} X - 2hu_{r} \nabla u\bigr) \,dV \\ &\quad = \int _{\partial \Omega } \vert \nabla u \vert ^{2} \langle X ,\nu \rangle \,d\sigma - 2 \int _{\partial \Omega } \langle X , \nabla u \rangle \langle \nabla u, \nu \rangle \,d\sigma \\ &\quad = - \int _{\partial \Omega } \vert \nabla u \vert ^{2} \langle X,\nu \rangle \,d\sigma \\ &\quad = - \frac{\sin ^{2} R_{1}}{\cos ^{2} R_{1}} \int _{\partial \Omega \setminus \partial B_{R}(N)} \langle X,\nu \rangle \,d\sigma - \frac{\sin ^{2} R}{\cos ^{2} R_{1}} \int _{\partial B_{R}(N)} \langle X,\nu \rangle \,d\sigma \\ &\quad = - \frac{\sin ^{2} R_{1}}{\cos ^{2} R_{1}} \int _{\Omega } \operatorname{div} (X )\,dV + \frac{\sin ^{2} R_{1} -\sin ^{2} R}{\cos ^{2} R_{1}} \bigl(-\sin R \bigl\vert \partial B_{R}(N) \bigr\vert \bigr) \\ &\quad = -n \frac{\sin ^{2} R_{1}}{\cos ^{2} R_{1}} \int _{\Omega } h' \,dV - \frac{\sin R(\sin ^{2} R_{1} -\sin ^{2} R)}{\cos ^{2} R_{1}} \bigl\vert \partial B_{R}(N) \bigr\vert . \end{aligned}
(2.28)

Combining (2.27) and (2.28),

\begin{aligned} n \frac{\sin ^{2} R_{1}}{\cos ^{2} R_{1}} \int _{\Omega } h' \,dV & = - \frac{\sin R(\sin ^{2} R_{1} - \sin ^{2} R)}{\cos ^{2} R_{1}} \bigl\vert \partial B_{R}(N) \bigr\vert \\ &\quad{} - (n-2) a \sin R \bigl\vert \partial B_{R}(N) \bigr\vert \biggl( \frac{\cos R}{\cos R_{1}} -a \biggr) \\ &\quad{} + 2na \sin R \bigl\vert \partial B_{R}(N) \bigr\vert \\ &\quad{} + (n+2) \biggl(n \int _{\Omega }uh' \,dV - \int _{\Omega } uhu_{r} \,dV \biggr). \end{aligned}
(2.29)

Combining (2.26) and (2.29),

$$0 < \frac{\sin ^{2} R - \sin ^{2} R_{1}}{\cos ^{2} R_{1}} - 2a(n-1) \frac{\cos R}{\cos R_{1}} + 2an +(2n-1)a^{2}.$$

Using (2.19), we can rewrite the above inequality as

\begin{aligned} 0 &< -a^{2} -2a -2a(n-1) \frac{\cos R}{\cos R_{1}} + 2na + (2n-1) a^{2} \\ &=2a(n-1) \biggl(a + 1 - \frac{\cos R}{\cos R_{1}} \biggr). \end{aligned}

However, this is a contradiction to (2.20). Therefore, we see that P is a constant function. It follows that the two boundary values of P must be equal:

$$\frac{\sin ^{2} R_{1}}{\cos ^{2} R_{1}} = \frac{\sin ^{2} R}{\cos ^{2} R_{1}} +2a +a^{2},$$

which gives

$$a+1=\frac{\cos R}{\cos R_{1}}.$$
(2.30)

Since $$\Delta P = 0$$ in Ω, the equality in (2.14) holds. This implies that Hessu must be proportional to the metric g by (2.1). Thus,

\begin{aligned} \operatorname{Hess}u = \frac{\Delta u}{n}g = (-1-u)g. \end{aligned}
(2.31)

Let $$\gamma : J \rightarrow \mathbb{S}^{n}_{+}$$ be a unit-speed maximal geodesic satisfying that

$$\gamma (0)=N, \qquad \nabla _{\gamma '(s)}\gamma '(s) = 0 \quad \text{and} \quad \bigl\vert \gamma '(s) \bigr\vert ^{2} =1.$$

Define $$f(s):=u(\gamma (s))$$ on J. Then,

$$f'(s) = \bigl\langle \nabla u, \gamma '(s) \bigr\rangle$$

and by (2.31)

\begin{aligned} f''(s) & = \bigl\langle \nabla _{\gamma '(s)} \nabla u,\gamma '(s) \bigr\rangle + \bigl\langle \nabla u, \nabla _{\gamma '(s)}\gamma '(s) \bigr\rangle \\ & = \operatorname{Hess}u \bigl(\gamma '(s),\gamma '(s)\bigr) \\ & = \bigl(-1-u\bigl(\gamma (s)\bigr)\bigr) \bigl\vert \gamma '(s) \bigr\vert ^{2} \\ & = -1-f(s). \end{aligned}

By the boundary conditions along $$\partial B_{R}(N)$$,

$$f(R)=a \quad \text{and} \quad f'(R) = - \frac{\sin R}{\cos R_{1}}.$$

Thus, we obtain the following initial value problem:

$$\textstyle\begin{cases} f''(s) +f(s) = -1 & \text{on } J, \\ f'(R) = - \frac{\sin R}{\cos R_{1}}, \\ f(R) = a. \end{cases}$$

The general solution is given by

$$f(s) = c_{1} \cos s +c_{2} \sin s -1$$

for some constants $$c_{1}$$ and $$c_{2}$$. From the initial conditions,

\begin{aligned}& -\frac{\sin R}{\cos R_{1}} = f'(R) = -c_{1} \sin R +c_{2} \cos R, \\& a =f(R) = c_{1} \cos R + c_{2} \sin R -1. \end{aligned}

Using (2.30), we obtain

\begin{aligned}& c_{1} = (a+1) \cos R + \frac{\sin ^{2} R}{\cos R_{1}} = \frac{1}{\cos R_{1}}, \\& c_{2} = (a+1) \sin R - \frac{\sin R}{\cos R_{1}} \cos R = 0. \end{aligned}

Thus, the solution to the initial value problem is given by

$$f(s) = \frac{\cos s}{\cos R_{1}}-1.$$

Since $$f(s)$$ depends only on the distance, we conclude that

$$u(x) = \frac{1}{\cos R_{1}}\bigl(\cos r(x) -\cos R_{1}\bigr),$$

where $$r(x) = \operatorname{dist}(x,N)$$. Since u vanishes on $$\partial \Omega \setminus \partial B_{R}(N)$$ by the boundary condition, $$\partial \Omega \setminus \partial B_{R}(N)$$ is the boundary of the geodesic ball centered at N with radius $$R_{1}$$. Therefore, Ω is the standard annulus $$\{x\in {\mathbb{S}}^{n} : R < r(x) < R_{1}\}$$. □

In Theorem 2.2, we assumed that the annular domain is contained in the upper hemisphere $$\mathbb{S}^{n}_{+}$$ and assumed that $$u=a>0$$ and $$\frac{\partial u}{\partial \nu}>0$$ on the inner spherical boundary. Instead, assuming each boundary component of the annular domain is geometrically rather simple, we are able to prove the same radial symmetry of the domain in the case where $$u=a<-1$$ and $$\frac{\partial u}{\partial \nu}<0$$ on the spherical boundary.

### Definition 2.3

An annular domain Ω is called weakly star shaped with respect to p if each component of the boundary Ω can be written as a graph over a geodesic sphere with center p.

### Theorem 2.4

Let Ω be an annular domain in $${\mathbb{S}}^{n} \setminus B_{R}(N)$$ such that $$\partial B_{R}(N) \subset \partial \Omega$$, where $$B_{R}(N)\subset {\mathbb{S}}_{+}^{n}$$ denotes the closed geodesic ball of radius R centered at the north pole $$N \in {\mathbb{S}}^{n}$$. Assume that Ω is weakly star shaped with respect to N. Given $$R< R_{1}<\pi$$, suppose there is a solution $$u\in C^{2}(\overline{\Omega})$$ such that

$$\textstyle\begin{cases} \Delta u +nu = -n & \textit{in } \Omega \\ u=0, \qquad \frac{\partial u}{\partial \nu} = - \frac{\sin R_{1}}{\cos R_{1}} & \textit{on } \partial \Omega \setminus \partial B_{R}(N) \\ u=a < -1,\qquad \frac{\partial u}{\partial \nu} = \frac{\sin R}{\cos R_{1}}< 0 & \textit{on } \partial B_{R}(N), \end{cases}$$

where ν is the outward unit normal to Ω. Then, Ω is the standard annulus $$\{x\in {\mathbb{S}}^{n} : R < r(x) < R_{1}\}$$ and the solution u is radial and is given by

$$u(x)=\frac{1}{\cos R_{1}}\bigl(\cos r(x) -\cos R_{1}\bigr),$$

where $$r(x) = \operatorname{dist}(N,x)$$.

### Proof

Define two P-functions as follows:

$$P(u)= \vert \nabla u \vert ^{2}+ 2u + u^{2} \quad \text{and} \quad \widetilde{P}(u) = \bigl\langle \nabla u, \nabla h' \bigr\rangle + uh'+h',$$

where $$h(r)=\sin r$$ and $$r(x)=\operatorname{dist}(N, x)$$. Then,

$$\Delta P \geq 0 \quad \text{and} \quad \Delta \widetilde{P}=0.$$

The function P on the boundary is given by

$$P(u) = \textstyle\begin{cases} \frac{\sin ^{2}R_{1}}{\cos ^{2} R_{1}} & \text{on } \partial \Omega \setminus \partial B_{R}(N), \\ \frac{\sin ^{2} R}{\cos ^{2} R_{1}} + 2a + a^{2} & \text{on } \partial B_{R}(N). \end{cases}$$

Suppose that

$$\frac{\sin ^{2} R}{\cos ^{2} R_{1}} + 2a + a^{2} > \frac{\sin ^{2}R_{1}}{\cos ^{2} R_{1}}.$$

In this case, we use the same argument as in the proof of Theorem 2.2 to obtain a contradiction. Therefore, we see that

$$\frac{\sin ^{2} R}{\cos ^{2} R_{1}} + 2a +a^{2} \leq \frac{\sin ^{2} R_{1}}{\cos ^{2} R_{1}},$$

which gives

$$\frac{\cos R}{\cos R_{1}}\leq a+1 \leq -\frac{\cos R}{\cos R_{1}}.$$
(2.32)

Now suppose that neither P nor is a constant function. Choose an orthonormal frame $$\{e_{1}, \ldots , e_{n}\}$$ on $$\partial \Omega \setminus \partial B_{R}(N)$$ such that $$e_{n} = \nu$$. Using the same argument as above, we obtain

$$u_{i} = 0, \qquad u_{ \nu i}=0, \qquad u_{ij}=0 \quad \text{for all } i,j =1,\ldots , n-1.$$

By the Hopf lemma,

\begin{aligned} 0&< \frac{\partial P}{\partial \nu} = 2 \operatorname{Hess}u(\nabla u, \nu ) + 2 \frac{\partial u}{\partial \nu} + 2 u \frac{\partial u}{\partial \nu} \\ & = 2u_{\nu \nu} \frac{\partial u}{\partial \nu} + 2 \frac{\partial u}{\partial \nu} \\ & = - \frac{2\sin R_{1}}{\cos R_{1}}( u_{\nu \nu} + 1) \quad \text{on } \partial \Omega \setminus \partial B_{R}(N). \end{aligned}

Since $$\frac{\sin R}{\cos R_{1}}<0$$ by our assumption,

$$u_{\nu \nu} +1 > 0 \quad \text{on } \partial \Omega \setminus \partial B_{R}(N).$$
(2.33)

Note that both the maximum and minimum values of the function are attained on the boundary of Ω because is a harmonic function. We have the following two possibilities:

1. (a)

has the maximum value at $$y_{1} \in \partial \Omega \setminus \partial B_{R}(N)$$;

2. (b)

has the maximum value at $$y_{2} \in \partial B_{R}(N)$$.

In case (a), by the Hopf lemma,

\begin{aligned} 0&< \frac{\partial \widetilde{P}}{\partial \nu}(y_{1}) = \operatorname{Hess}u \bigl( \nabla h', \nu \bigr) + \operatorname{Hess}h'( \nabla u, \nu ) + u \frac{\partial h'}{\partial \nu} + h' \frac{\partial u}{\partial \nu} + \frac{\partial h'}{\partial \nu} \\ & = u_{\nu \nu} \frac{\partial h'}{\partial \nu} - h' \frac{\partial u}{\partial \nu} + u \frac{\partial h'}{\partial \nu} + h' \frac{\partial u}{\partial \nu} + \frac{\partial h'}{\partial \nu} \\ & = (u_{\nu \nu} +1) \frac{\partial h'}{\partial \nu}. \end{aligned}

Note that

$$\frac{\partial h'}{\partial \nu} = - \sin r \langle \nabla r, \nu \rangle < 0 \quad \text{on } \partial \Omega \setminus \partial B_{R}(N),$$

since $$\Omega \cup B_{R} (N)$$ is star shaped with respect to N. Thus,

$$u_{\nu \nu}(y_{1}) + 1 < 0,$$

which is a contradiction to (2.33). In case (b),

\begin{aligned} 0< \frac{\partial \widetilde{P}}{\partial \nu}(y_{2}) = \frac{\partial h'}{\partial \nu} (n-1) \biggl( \frac{\cos R}{\cos R_{1}} -a-1 \biggr). \end{aligned}

Note that

$$\frac{\partial h'}{\partial \nu} = - \sin r \langle \nabla r , \nu \rangle = \sin R >0 \quad \text{on } \partial B_{R}(N).$$

However, (2.32) shows that

$$\frac{\partial \widetilde{P}}{\partial \nu}(y_{2})\leq 0,$$

which is a contradiction. Therefore, either P or is a constant function.

Suppose is a constant function. This implies that

$$\frac{\partial \widetilde{P}}{\partial \nu} = 0 \quad \text{on } \partial \Omega$$

and

$$u_{\nu \nu} + 1 =0 \quad \text{on }\partial \Omega \setminus \partial B_{R}(N).$$

Then,

$$\frac{\partial P}{\partial \nu} = 0\quad \text{on } \partial \Omega \setminus \partial B_{R}(N).$$

Since P has the maximum value on $$\partial \Omega \setminus \partial B_{R}(N)$$, P is a constant function in Ω by the Hopf lemma. Thus, we may assume that P is a constant function. Therefore, the two boundary values are equal, which implies that

$$\frac{\sin ^{2} R_{1}}{\cos ^{2} R_{1}} = \frac{\sin ^{2} R}{\cos ^{2} R_{1}} + 2a +a^{2}.$$

The remaining part of the proof is exactly the same as that of Theorem 2.2. Finally, we conclude that

$$u(x) = \frac{1}{\cos R_{1}} \bigl(\cos r(x)- \cos R_{1}\bigr),$$

where $$r(x) = \operatorname{dist}(N,x)$$. Moreover, Ω is the standard annulus $$\{x\in {\mathbb{S}}^{n} : R < r(x) < R_{1}\}$$. □

In Theorem 2.4, changing the boundary conditions on $$\partial B_{R}(N)$$ into

$$u=a>-1 \quad \text{and}\quad \frac{\partial u}{\partial \nu} = \frac{\sin R}{\cos R_{1}}>0\quad \text{on } \partial B_{R}(N)$$

gives the same conclusion. More precisely, applying the same argument as in the proof of Theorem 2.4, we have the following.

### Theorem 2.5

Let Ω be an annular domain in $${\mathbb{S}}^{n} \setminus B_{R}(N)$$ such that $$\partial B_{R}(N) \subset \partial \Omega$$, where $$B_{R}(N)\subset {\mathbb{S}}_{+}^{n}$$ denotes the geodesic ball of radius R centered at the north pole $$N \in {\mathbb{S}}^{n}$$. Assume that Ω is a weakly star-shaped domain with respect to N. Given $$R < R_{1}< \pi$$, suppose there is a solution $$u\in C^{2}(\overline{\Omega})$$ such that

$$\textstyle\begin{cases} \Delta u +nu = -n & \textit{in } \Omega \\ u=0, \qquad \frac{\partial u}{\partial \nu} = - \frac{\sin R_{1}}{\cos R_{1}} & \textit{on } \partial \Omega \setminus \partial B_{R}(N) \\ u=a >-1, \qquad \frac{\partial u}{\partial \nu} = \frac{\sin R}{\cos R_{1}}>0 & \textit{on } \partial B_{R}(N), \end{cases}$$

where ν is the outward unit normal to Ω. Then, Ω is the standard annulus $$\{x\in {\mathbb{S}}^{n} : R < r(x) < R_{1}\}$$ and the solution u is radial and is given by

$$u(x)= \frac{1}{\cos R_{1}}\bigl(\cos r(x) -\cos R_{1}\bigr),$$

where $$r = \operatorname{dist}(N,x)$$.

## 3 Annular domains with outer spherical boundary

In the unit sphere $$\mathbb{S}^{n}$$, an overdetermined boundary value problem for an annular domain Ω with outer spherical boundary can be regarded as the problem for the domain with an inner spherical boundary. To be precise, for $$p\in \mathbb{S}^{n}$$, let Ω be a domain in $$\mathbb{S}^{n} \setminus B_{R}(p)$$ such that $$\partial B_{R}(p) \subset \partial \Omega$$, where $$B_{R}(p)$$ denotes the geodesic ball of radius R centered at p. Then, Ω is also a domain in $$B_{\pi -R} (-p)$$ such that $$\partial B_{\pi -R} (-p) \subset \partial \Omega$$. This observation gives the following result, which is basically the same as Theorem 2.1.

### Theorem 3.1

Let Ω be an annular domain in $$B_{R} (N)$$ such that $$\partial B_{R} (N) \subset \partial \Omega$$, where $$B_{R} (N)\subset {\mathbb{S}}^{n}$$ denotes the geodesic ball of radius $$0< R<\pi$$ centered at the north pole $$N \in {\mathbb{S}}^{n}$$. Suppose there is a solution $$u\in C^{2}(\overline{\Omega})$$ such that

$$\textstyle\begin{cases} \Delta u = -n\cos r = -nh' & \textit{in } \Omega , \\ u=0, \qquad \frac{\partial u}{\partial \nu} = \mathrm{const} = c_{1} & \textit{on } \partial \Omega \setminus \partial B_{R}(N), \\ u=\mathrm{const}=a< 0, \qquad \frac{\partial u}{\partial \nu} = - \sin R & \textit{on } \partial B_{R}(N), \end{cases}$$

where ν is the outward unit normal to Ω and the function $$h(r)$$ is defined as before with $$r(x) = \operatorname{dist}(N,x)$$. Assume that either Ω is contained in $${\mathbb{S}}_{+}^{n}$$ or u is negative. Then, Ω is an annulus $$\{x\in {\mathbb{S}}^{n} : R_{1} < r(x) < R\}$$ and the radial solution u is given by

$$u(x)= \cos r(x) - \cos R_{1}.$$

In the following, we prove the radial symmetry of the solution to an overdetermined boundary value problem on annular domains in $$\mathbb{H}^{n}$$ with outer spherical boundary (see Fig. 2).

### Theorem 3.2

Let Ω be an annular domain in $$B_{R}(p)$$ such that $$\partial B_{R}(p) \subset \partial \Omega$$, where $$B_{R}(p)\subset {\mathbb{H}}^{n}$$ denotes the geodesic ball of radius R centered at $$p \in {\mathbb{H}}^{n}$$. Given $$0< R_{1}< R$$, suppose there is a solution $$u\in C^{2}(\overline{\Omega})$$ such that

$$\textstyle\begin{cases} \Delta u -nu = -n & \textit{in } \Omega , \\ u=0,\qquad \frac{\partial u}{\partial \nu} = \frac{\sinh R_{1}}{\cosh R_{1}} & \textit{on } \partial \Omega \setminus \partial B_{R}(p), \\ u=\mathrm{const}=a ,\qquad \frac{\partial u}{\partial \nu} = - \frac{\sinh R}{\cosh R_{1}} & \textit{on } \partial B_{R}(p), \end{cases}$$

where ν is the outward unit normal to Ω. If $$1-\frac{\cosh R}{\cosh R_{1}} \leq a <0$$, then Ω is the standard annulus $$\{x\in {\mathbb{H}}^{n} : R_{1} < r(x) < R\}$$. Moreover, the solution u is radial and is given by

$$u(x)= \frac{1}{\cosh R_{1}} \bigl(\cosh R_{1} - \cosh r(x) \bigr) ,$$

where $$r(x) = \operatorname{dist} (p,x)$$.

### Proof

Let $$P(u):= |\nabla u|^{2} + 2u - u^{2}$$. Then, from the Bochner formula,

$$\Delta P \geq 0.$$

Note that the maximum value of P is attained on the boundary of Ω by the maximum principle. Using the boundary conditions,

$$P(u) = \textstyle\begin{cases} \frac{\sinh ^{2}R_{1}}{\cosh ^{2} R_{1}} & \text{on } \partial \Omega \setminus \partial B_{R}(p), \\ \frac{\sinh ^{2} R}{\cosh ^{2} R_{1}} + 2a - a^{2} & \text{on } \partial B_{R}(p). \end{cases}$$

From the assumption that $$1-\frac{\cosh R}{\cosh R_{1}} \leq a <0$$, we have

$$(a-1)^{2} \leq \frac{\cosh ^{2} R}{\cosh ^{2} R_{1}}.$$

Then,

\begin{aligned} a^{2} -2a & \leq \frac{\cosh ^{2} R}{\cosh ^{2} R_{1}}-1 = \frac{(\sinh ^{2} R +1) - (\sinh ^{2} R_{1} +1)}{\cosh ^{2} R_{1}} = \frac{\sinh ^{2} R}{\cosh ^{2} R_{1}} - \frac{\sinh ^{2} R_{1}}{\cosh ^{2} R_{1}}. \end{aligned}

Thus,

$$\frac{\sinh ^{2} R}{\cosh ^{2} R_{1}} +2a -a^{2} \geq \frac{\sinh ^{2} R_{1}}{\cosh ^{2} R_{1}},$$

which means that the function P defined on Ω attains its maximum value on $$\partial B_{R}(p)$$. Suppose P is not a constant function. From the Hopf lemma, it follows that

$$\frac{\partial P}{\partial \nu} >0 \quad \text{on }\partial B_{R}(p).$$
(3.1)

Choose a local frame $$\{e_{i}\}_{i=1}^{n}$$ such that $$e_{n} = \frac{\partial}{\partial r}$$. Note that $$u_{r}$$ is constant on $$\partial B_{R}(p)$$. Since each $$e_{i}$$ is tangent to $$\partial B_{R}(p)$$ for $$i = 1, \ldots , n-1$$, it follows that

$$u_{ri} = 0 \quad \text{on }\partial B_{R}(p)$$
(3.2)

for $$i =1 , \ldots , n-1$$. Moreover, since $$\partial B_{R}(p)$$ is a level set of u,

$$u_{i}=0\quad \text{and}\quad u_{ij}=0\quad \text{on } \partial B_{R}(p)$$
(3.3)

for $$i,j = 1 , \ldots , n-1$$. From (3.2) and (3.3),

$$-n+na = \Delta u = (n-1)\frac{\cosh r}{\sinh r}u_{r} + u_{rr}= (n-1) \frac{\cosh R}{\sinh R} \biggl(-\frac{\sinh R}{\cosh R_{1}} \biggr) + u_{rr} \quad \text{on } \partial B_{R}(p),$$

which yields that

$$u_{rr} = -n +na + (n-1) \frac{\cosh R}{\cosh R_{1}} \quad \text{on } \partial B_{R}(p).$$

It follows that

\begin{aligned} \frac{\partial P}{\partial \nu} = -2 (n-1) \frac{\sinh R}{\cosh R_{1}} \biggl( \frac{\cosh R}{\cosh R_{1}} -1 +a \biggr) \leq 0 \quad \text{on } \partial B_{R}(p), \end{aligned}

which is a contradiction to (3.1). Hence, P is a constant function and $$\Delta P = 0$$. As before, equality holds in (2.1), which implies that

$$\operatorname{Hess}u = \frac{\Delta u}{n} g = (u-1)g,$$

where g denotes the metric of $${\mathbb{H}}^{n}$$. The same argument as in Theorem 2.2 shows that Ω is the standard annulus $$\{x\in {\mathbb{H}}^{n} : R_{1} < r(x) < R\}$$ and

$$u(x) = \frac{1}{\cosh R_{1}} \bigl(\cosh R_{1} - \cosh r(x)\bigr),$$

where $$r(x) = \operatorname{dist}(p,x)$$. □

In the proof of Theorem 3.2, the assumption that $$1-\frac{\cosh R}{\cosh R_{1}} \leq a <0$$ is necessary. Instead, under the condition that the annular domain is weakly star shaped, the assumption on the constant a can be weakened as follows.

### Theorem 3.3

Let Ω be an annular domain in $$B_{R}(p)$$ such that $$\partial B_{R}(p) \subset \partial \Omega$$, where $$B_{R}(p)\subset {\mathbb{H}}^{n}$$ denotes the geodesic ball of radius R centered at $$p \in {\mathbb{H}}^{n}$$. Assume that Ω is weakly star shaped with respect to p. Given $$0< R_{1}< R$$, suppose there is a solution $$u\in C^{2}(\overline{\Omega})$$ such that

$$\textstyle\begin{cases} \Delta u -nu = -n & \textit{in } \Omega , \\ u=0, \qquad \frac{\partial u}{\partial \nu} = \frac{\sinh R_{1}}{\cosh R_{1}} & \textit{on } \partial \Omega \setminus \partial B_{R}(p), \\ u=\mathrm{const}=a < 0, \qquad \frac{\partial u}{\partial \nu} = - \frac{\sinh R}{\cosh R_{1}} & \textit{on } \partial B_{R}(p), \end{cases}$$

where ν is the outward unit normal to Ω. Then, Ω is the standard annulus $$\{x\in {\mathbb{H}}^{n} : R_{1} < r(x) < R\}$$. Moreover, the solution u is radial and is given by

$$u(x)= \frac{1}{\cosh R_{1}} \bigl(\cosh R_{1} -\cosh r(x)\bigr),$$

where $$r(x) = \operatorname{dist} (p, x)$$.

### Proof

One can prove Theorem 3.3 by using the same argument as in the proof of Theorem 2.4. Here, we give the sketch of the proof. Define

$$P(u):= \vert \nabla u \vert ^{2} + 2u - u^{2} \quad \text{and} \quad \widetilde{P}:= \bigl\langle \nabla u, \nabla h' \bigr\rangle - uh'+h',$$

where $$h(r)=\sinh r$$ and $$r(x)=\operatorname{dist}(p, x)$$. Then,

$$\Delta P \geq 0 \quad \text{and} \quad \Delta \widetilde{P}=0.$$

The boundary conditions show that

$$P(u) = \textstyle\begin{cases} \frac{\sinh ^{2}R_{1}}{\cosh ^{2} R_{1}} & \text{on } \partial \Omega \setminus \partial B_{R}(p), \\ \frac{\sinh ^{2} R}{\cosh ^{2} R_{1}} + 2a - a^{2} & \text{on } \partial B_{R}(p). \end{cases}$$

Suppose that

$$\frac{\sinh ^{2} R}{\cosh ^{2} R_{1}} + 2a - a^{2} > \frac{\sinh ^{2}R_{1}}{\cosh ^{2} R_{1}}.$$

Then,

$$a-1 > -\frac{\cosh R}{\cosh R_{1}}.$$

Applying the maximum principle,

$$\max_{\Omega } P(u) = \max_{\partial \Omega } P(u) = \frac{\sinh ^{2} R}{\cosh ^{2} R_{1}} + 2a - a^{2}.$$

By the Hopf lemma and the assumption,

\begin{aligned} 0< \frac{\partial P}{\partial \nu} = -2 (n-1) \frac{\sinh R}{\cosh R_{1}} \biggl( \frac{\cosh R}{\cosh R_{1}} -1 +a \biggr) < 0 \quad \text{on } \partial B_{R}(p), \end{aligned}

which is a contradiction. Therefore, we see that

$$\frac{\sinh ^{2} R}{\cosh ^{2} R_{1}} + 2a -a^{2} \leq \frac{\sinh ^{2} R_{1}}{\cosh ^{2} R_{1}}.$$
(3.4)

Then,

$$a-1 \leq - \frac{\cosh R}{\cosh R_{1}}.$$
(3.5)

Suppose that neither P nor is a constant function. Then,

\begin{aligned} 0< \frac{\partial P}{\partial \nu} = 2 \frac{\sinh R_{1}}{\cosh R_{1}}( u_{\nu \nu} + 1) \quad \text{on } \partial \Omega \setminus \partial B_{R}(p), \end{aligned}

which implies that

$$u_{\nu \nu} +1 > 0\quad \text{on } \partial \Omega \setminus \partial B_{R}(p).$$
(3.6)

Since is harmonic, the maximum value of is attained on Ω. Suppose that has the maximum value at $$y_{1} \in \partial \Omega \setminus \partial B_{R}(p)$$. By using (3.6), the Hopf lemma and the assumption that $$B_{R}(p) \setminus \Omega$$ is star shaped with respect to p,

\begin{aligned} 0< \frac{\partial \widetilde{P}}{\partial \nu}(y_{1}) = (u_{\nu \nu} +1) \frac{\partial h'}{\partial \nu} = (u_{\nu \nu} +1) \sinh r \langle \nabla r,\nu \rangle < 0, \end{aligned}

which gives a contradiction. Now, suppose that has the maximum value at $$y_{2} \in \partial B$$. Similarly, by the Hopf lemma and (3.5),

\begin{aligned} \begin{aligned} 0&< \frac{\partial \widetilde{P}}{\partial \nu}(y_{2}) = \frac{\partial h'}{\partial \nu} (n-1) \biggl( \frac{\cosh R}{\cosh R_{1}} +a-1 \biggr) \\ &= \sinh r \langle \nabla r ,\nu \rangle (n-1) \biggl( \frac{\cosh R}{\cosh R_{1}} +a-1 \biggr) \\ &\leq 0, \end{aligned} \end{aligned}

which again gives a contradiction. Thus, either P or is a constant function. Suppose is a constant function. Then,

$$u_{\nu \nu} + 1 =0 \quad \text{on } \partial \Omega \setminus \partial B_{R}(p),$$

which shows that

$$\frac{\partial P}{\partial \nu} = 0 \quad \text{on } \partial \Omega \setminus \partial B_{R}(p).$$

By (3.4), P attains its maximum value on $$\partial \Omega \setminus \partial B_{R}(p)$$. However, from the Hopf lemma, it follows that P is a constant function in Ω. Thus, we have $$\Delta P =0$$ in Ω. Now, we can apply the same argument as in the proof of Theorem 2.2 to finish the proof. □

Not applicable.

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The authors would like to thank the referee for his/her useful comments and suggestions.

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