A symmetry result for fully nonlinear problems in exterior domains

Abstract

We study an overdetermined fully nonlinear problem driven by one of the Pucci’s Extremal Operators in an external domain. Under certain decay assumptions on the solution, we extend Serrin’s symmetry result, i.e, every domain where the solution exists must be radial.

Introduction

This paper aims to be a contribution on the field of symmetry of non negative solutions of fully nonlinear elliptic equations. Following the approach started by Serrin [14], employing the famous moving plane method, we show the radiality of positive solutions on an external domain setting.

We consider the problem:

$$\begin{aligned} -\mathcal {M}_{r,R} ^\pm (D^2 u) = f(u) \qquad&\text { in } {\mathbb {R}}^N \setminus G,\, N \ge 2 \end{aligned}$$

(1)

$$\begin{aligned} u > 0 \qquad&\text { in } {\mathbb {R}}^N \setminus G. \end{aligned}$$

(2)

$$\begin{aligned} \lim \limits _{|x|\rightarrow \infty }{u(x)}{|x|^{\gamma }}= c_0 > 0 \end{aligned}$$

(3)

where $\gamma$ and $c_0$ are positive constants, $\mathcal {M}_{r,R}^\pm$ denotes either one of the extremal Pucci’s operators whose definition will be recalled in Sect. 2. Throughout the text the subscript r, R will be omitted whenever confusion will not arise.

Here we assume that $f:{\mathbb {R}} \rightarrow {\mathbb {R}}$ is a locally Lipschitz function satisfying

$$\begin{aligned} \frac{f(u)-f(v)}{u-v} \le c(|u|+|v|)^\alpha \text { for |{ u}|,\,|{ v}| sufficiently small and } \alpha > \frac{2}{\gamma }, \end{aligned}$$

(4)

This is a natural assumption which paired with the growth assumption (3) guarantees the validity of a maximum principle in unbounded domains.

We also assume that $\tilde{N}= \frac{r}{R}(N-1) + 1$, is greater than 2.

In (1)–(2) G is a domain defined as:

$$\begin{aligned} G= \bigcup \limits _{i=1}^k G_i \end{aligned}$$

(5)

where $k \in {\mathbb {N}}^+$ and $G_i$ are bounded $\mathcal {C}^2$ domains such that $\bar{G_i} \cap \bar{G_j} = \emptyset$ for $i \ne j$.

We furthermore impose the following boundary conditions. For every $1\le i \le k$:

$$\begin{aligned} \frac{\partial u}{\partial \nu } = \alpha _i \le 0 \qquad&\text { on } \partial G_i. \end{aligned}$$

(6)

$$\begin{aligned} u = a_i > 0 \qquad&\text { on } \partial G_i. \end{aligned}$$

(7)

where, $\alpha _i$ and $a_i$ are constants and $\nu$ is the external normal with respect to the boundary of G.

In the particular case where the Pucci operator coincides with the Laplacian, i.e, $r=R=1$, the problem has been extensively covered in the literature.

In the early 70’s Serrin [14] proved that for $\Omega$, a bounded set with $\mathcal {C}^2$ boundary and f a continuos differentiable function, every positive solution of

$$\begin{aligned} -\Delta u = f(u,|\nabla u|) \qquad&\text { in } \Omega \\ u > 0 \qquad&\text { in } \Omega \\ u = 0 \qquad&\text { on } \partial \Omega \\ \frac{\partial u}{\partial \nu } = \text { constant } \qquad&\text { on } \partial \Omega \end{aligned}$$

is radial and furthermore $\Omega$ must be a ball.

Serrin’s result has then been extended in several directions, and would be impossible to exhaustively list all the literature, however, we present a few references for the different extensions: for the p-Laplace operator (see [4] and [9]), in the nonlocal setting ([5] and [17]) and under partially overdetermined boundary conditions see [8].

Concerning overdetermined problems for the Laplacian in external domains, Reichel (see [12] and [13]), under strong assumptions, first considered domains with one cavity and then on a follow-up article extended the result to domains with multiple cavities. Under more general assumptions, Sirakov ([16]), also obtained the result.

Both approaches are based on a topological arguments and strongly rely on the classical Serrin’s Corner Lemma (see [14]). This presents a great difficulty to study the analogous problem in the fully nonlinear setting, since Serrin’s Corner Lemma is not true in general.

In the last few years, some advances have been made regarding extending Serrin’s result for Pucci’s Extremal Operators in a bounded domain $\Omega$.

Two different kinds of assumptions have been suggested to treat the problem. In [2], I. Birindelli and F. Demengel have proved symmetry when $\mathcal {M}^\pm$ is a perturbation of the laplacian, by providing a variation of Serrin’s Corner Lemma.

Instead of assuming a constraint on the elliptic constants, Silvestre and Sirakov also obtained the same result in [15], but under geometric assumptions on the domains.

Our approach follows [2] where a smallness regime of the ratio $\frac{R}{r}$ is assumed in order to circumvent the lack of Serrin’s Corner Lemma.

Here we study the external domain problem (1)–(7) and prove:

FormalPara Theorem 1.1

Let u be a $\mathcal {C}^{2,\beta }$ solution of (1)–(7), with $\beta > 0$. There exists a positive $t_1$ which depends only on r and $\beta$ such that if $1 \le \frac{R}{r} < t_1$, then, G has only one connected component. Moreover G is a ball and the solution u is radial with respect to the center of this ball.

The paper is organized as follows. In Sect. 2, we state some preliminary lemmas which are used to prove the main result and introduce the necessary notation. We reserve Sect. 3 to provide the proof of Theorem 1.1 which will be divided in 10 steps. The proof closely follows [16], although some arguments are modified to deal with the fully nonlinear case. For the reader’s convenience we include full details also for the steps which are the same as in the semilinear case.

FormalPara Remark 1.2

This work has been produced during the authors PhD studies at Sapienza Universita di Roma.

1 Notation and Preliminary Results

Here we introduce some notation needed to apply the Moving Plane Method (see [10]), in order to show that, for every direction $\gamma \in \mathcal {S}^{N-1}$, our solution is symmetric in that direction. Since our operator is invariant with respect to rotations we may, without loss of generality, set $\gamma = e_1$, the first vector of a canonical base in ${\mathbb {R}}^N, \, N\ge 2$.

Hence we define, for $\lambda \in {\mathbb {R}}$

$$\begin{aligned} T_\lambda =\{ x \in {\mathbb {R}}^N \,|\, x_1 = \lambda \}\\ D_\lambda =\{ x \in {\mathbb {R}}^N \,|\, x_1 > \lambda \}\\ \end{aligned}$$

For every point $x = (x_1,x') \in {\mathbb {R}}^N$, with $x'\in {\mathbb {R}}^{N-1}$, we set $x^\lambda = (2\lambda -x_1,x')$, i.e, $x^\lambda$ is the reflection of x with respect to $T_\lambda$.

Then for every $A \subset {\mathbb {R}}^N$, we define

$$\begin{aligned} A^\lambda = \text { the reflection of A with respect to } T_{\lambda },\, i.e, A^\lambda =\{ x \in {\mathbb {R}}^N \,|\, x^\lambda \, \in A\}\end{aligned}$$

In particular we set $\Sigma _\lambda = D_\lambda {\setminus } ({\bar{G}} \cup {\bar{G}}^\lambda )$.

We will also use the notation

$$\begin{aligned}&\Gamma _t(A)= A-t e_1, \text { for } t\in {\mathbb {R}} \\&\Gamma (A) = \bigcup \limits _{t \in {\mathbb {R}}} \Gamma _t(A)\\ \end{aligned}$$

and for $z \in {\mathbb {R}}^N \setminus G$ we define,

$$\begin{aligned} \Gamma _{t}(z) = z-t e_1 \text { for } t\in {\mathbb {R}} \end{aligned}$$

For $i = 1,\ldots ,k$ we define:

$$\begin{aligned} d_i = \inf \{\lambda \in {\mathbb {R}} \,|\, T_\mu \cap \bar{G_i} = \emptyset \text { for all } \mu> \lambda \} \\ \lambda _i = \inf \{\lambda \in {\mathbb {R}} \,|\, (D_\mu \cap \bar{G_i})^\mu \subset G_i \text { and } \langle \nu (z),e_1\rangle> 0 \\ \text { for all } \mu > \lambda \text { and all } z \in T_\mu \cap \partial {G_i}\ \}\\ d= \max d_i \qquad \lambda _* = \max \lambda _i \end{aligned}$$

where, as before, $\nu (z)$ denotes the exterior normal with respect to $\partial G_i$.

Let $z \in \partial G_i ^\lambda \cap D_\lambda$, for some $i \in \{1,\ldots ,k\}$, be such that $\Gamma _{t}(z) \in G_i ^\lambda$ for small positive values of t. Then we define:

$$\begin{aligned}{\underline{t}}={\underline{t}}(z) = \min \{t > 0 \,|\, z-t e_1 \in \partial G_i \cup \partial G_i^\lambda \}\end{aligned}$$

Now we recall the definition of the Pucci’s Operators and associated quantities.

For a twice differentiable real function u defined on an open set $\Omega$ we define for $x \in \Omega$

$$\begin{aligned} \mathcal {M}_{r,R}^+ (D^2 u)(x) = R \sum _{\mu _i>0} \mu _i(x) + r \sum _{\mu _i<0} \mu _i (x) \\ \mathcal {M}_{r ,R }^- (D^2 u(x)) = r \sum _{\mu _i >0} \mu _i(x) + R \sum _{\mu _i <0} \mu _i(x) \\ \end{aligned}$$

where $0<r\le R$ are called ellipticity constants and $\mu _i=\mu _i(D^2 u(x))$, $i=1,\ldots ,N$ represent the eigenvalues associated to the hessian matrix $D^2 u (x)$. Furthermore we introduce (as in [6, 7]) the dimension like quantities $\tilde{N}_{+}$ and $\tilde{N}_{-}$ for $\mathcal {M}_{r,R}^\pm$, defined as

$$\begin{aligned} \tilde{N}_{+}= \frac{r}{R}(N-1)+1 \qquad \text { and } \tilde{N}_{-}= \frac{R}{r}(N-1)+1 \end{aligned}$$

(8)

In the scope of this work, only $\tilde{N}_{+}$ plays a role, therefore throught the article we will refer to $\tilde{N}_{+}$ as $\tilde{N}$.

We will now recall some results obtained in previous papers which will be needed later on.

We start with a version of the Hopf Lemma for non-proper operators

Lemma 2.1

[1] Let $\Omega \subset {\mathbb {R}}^N$ be a smooth domain and let $b, c \in {L}^\infty (\Omega )$. Suppose u $\in \mathcal {C}(\bar{\Omega })$ is a viscosity solution of

$$\begin{aligned} \mathcal {M}^-(D^2 u) -b(x)|Du|+c(x)u\le 0 \qquad&\text { in } \Omega . \end{aligned}$$

(9)

$$\begin{aligned} u \ge 0 \qquad&\text { in } \Omega . \end{aligned}$$

(10)

Then either $u \equiv 0$ or $u>0$ in $\Omega$. Furthermore, at any point $x_0 \in \partial \Omega$ where $u(x_0)=0$, we have

$$\begin{aligned} \liminf \limits _{t \rightarrow 0} \frac{u(x_0+tv)-u(x_0)}{t} < 0 \end{aligned}$$

(11)

for every $v \in {\mathbb {R}}^N$ such that $\langle v,\nu (x_0) \rangle > 0$, where $\nu (x_0)$ is the external normal at $x_0$ with respect to the boundary of $\Omega$.

Lemma 2.2

[16] Let $i \in \{ 1,\ldots , k\}$, if $\lambda \ge \lambda _*$, then any $z \in D_\lambda \cap \partial {G_i}^\lambda$ has one and only one of the following properties,( see Fig. 1):

• I) $\Gamma _t(z) \in \Sigma _\lambda$ for small positive values of t or there exists a sequence

  $t_m$ $\searrow 0$ such that $\Gamma _{t_m}(z) \in D_\lambda \cap \partial {G_i}^\lambda$

• II) ${\underline{t}} \le d(z,T_\lambda )$, the open segment $(\Gamma _{{\underline{t}}}(z),z)$ belongs to $G_i ^\lambda$, and

  $\Gamma _{{\underline{t}}}(z) \in \partial G_i ^\lambda$.

• III) ${\underline{t}} < d(z,T_\lambda )$, the open segment $(\Gamma _{{\underline{t}}}(z),z)$ belongs to $G_i ^\lambda$, and $\Gamma _{{\underline{t}}}(z) \in \partial G_i$.

• IV) $\lambda = \lambda _*$ and $z\in \partial G_i ^\lambda \cap \partial G_i$

Fig. 1

The four different types of points of $\partial G_{i}^\lambda$ $\cap$ $D_\lambda$: The arcs (A, B], [H, I), (J, K) are of type (I). (B, C], [E, F], [G, H) are of type (II), (C, D), (D, E), (F, G) are of type (III) and D is of type (IV)

Lemma 2.3

[2] Let f be a locally Lipschitz function, suppose that $\Omega$ is a bounded $\mathcal {C}^{2,\beta }$ domain and suppose that $H_0$ is an hyperplane such that

• there is $P \in H_0 \cap \partial \Omega$ such that $\nu (P) \in H_0$;

• $\Omega ^-$ is the intersection of $\Omega$ with one of the half spaces bounded by $H_0$ and $\Omega ^+$, its reflection with respect to $H_{0}$ is contained in $\Omega$.

Let $u\ge 0$ be a viscosity solution of

$$\begin{aligned} \mathcal {M}_{r,R}^\pm (D^2 u) + f(u) = 0 \qquad \text { in } \Omega \end{aligned}$$

Let $u_0$ be the reflection of in $\Omega ^+$. If $u_0 > u$ in a neighborhood of P in $\Omega$, $u(P)=u_0(P)$ and $\nabla u_0(P) = \nabla u(P) \ne 0$, then there exists a $t_1 > 1$, such that if $\frac{R}{r} < t_1$ and for any direction v pointing inside $\Omega ^+$

$$\begin{aligned} \partial ^2 _v u_0 (P) > \partial ^2 _v u (P) \end{aligned}$$

(12)

Lemma 2.4

[2] Let u be a $\mathcal {C}^1$ solution of (1)–(7), if $\partial G$ is the graph of a $\mathcal {C}^2$ function $\psi$, then for every $x_0 \in \partial G,$ $D^2u(x_0)$ only depends on $\psi (x_0), \nabla \psi (x_0) \text { and } D^2\psi (x_0)$.

Lemma 2.5

[11] Let $\Omega \subset {\mathbb {R}}^N$ be an unbounded smooth domain. Suppose $u \in \mathcal {C}^2(\Omega )$ satisfies for $x \in \Omega$

$$\begin{aligned} \mathcal {M}_{r,R}^-(D^2 u)) +c(x) u \le 0 \end{aligned}$$

where, c(x) is a locally bounded real function such that

$$\begin{aligned} c(x) < \frac{-q (R(q+1)-r(N-1))}{|x|^2} \end{aligned}$$

for some $q \in (0, \tilde{N}-2)$. If, $\liminf \limits _{|x|\rightarrow \infty } u(x) |x|^q \ge 0$ and $u \ge 0 \text { on } \partial \Omega$ then $u\ge 0$ in $\Omega$.

2 Proof of Theorem 1

Using all the notations introduced in Sect. 2, we consider, for $x \in \Sigma _\lambda$, the function

$$\begin{aligned} w_\lambda (x) = u_\lambda (x)-u(x)= u(x^\lambda ) - u(x). \end{aligned}$$

where u is a solution of (1)–(7) for either one of the operators $\mathcal {M}_{r,R} ^\pm$.

The proof of Theorem 1 will be obtained through several steps.

2.1 Step 1: $\exists \bar{\lambda } \in {\mathbb {R}}$ such that $\forall \lambda \ge \bar{\lambda }, w_\lambda \ge 0 ,$

Proof

Since $u \rightarrow 0$ as $|x|\rightarrow \infty$, we can take $\bar{\lambda }$ such that $\bar{\lambda } > d$, and, for $i \in \{1,\ldots ,k\}$

$$\begin{aligned} u(x) \le \frac{1}{2} \min \limits _{i} a_i \qquad \text { for } \vert x \vert > \bar{\lambda }. \end{aligned}$$

Hence

since $z \in \partial G_i ^\lambda$ implies $|z| > |\lambda |.$

The above proves that $w_\lambda$ is positive on $\partial G_i^\lambda$. The proof on the rest of the domain is as follows:

Note that regardless of the operator, $\mathcal {M}_{r,R}^+$ or $\mathcal {M}_{r,R}^-$, $w_\lambda$ satisfies

$$\begin{aligned} \mathcal {M}_{r,R}^- (D^2 u) + \frac{f(u_\lambda )-f(u)}{w_\lambda } w_\lambda \le 0 \qquad&\text { in } \Sigma _\lambda . \end{aligned}$$

(13)

$$\begin{aligned} w_\lambda \ge 0 \qquad&\text { on } \partial \Sigma _\lambda . \end{aligned}$$

(14)

Since by hypothesis $u=O(|x|^{-\gamma })$ and (4) holds, we have that,

$\frac{f(u_\lambda )-f(u)}{w_\lambda } = O(|x|^{-2})$ and $\liminf \limits _{|x|\rightarrow \infty } w_\lambda |x|^q = 0$, for some positive $q < \min \{\gamma ,\tilde{N} -2\}$. Thus, Lemma 2.5 applies and we get $w_\lambda \ge 0$. $\square$

Now we define

$$\begin{aligned} \lambda _0 = \inf \{ \lambda \in {\mathbb {R}} \,\vert \, w_\mu \ge 0 \text { in } \Sigma _\lambda , \; \forall \mu > \lambda \}. \end{aligned}$$

The above set is nonempty due to the previous step and clearly $\lambda _0$ is finite. Also it follows from the continuity of u that $w_{\lambda _0} \ge 0$.

2.2 Step 2: $\frac{\partial u}{\partial x_1} < 0$ in $\{ x=(x_1,x')\in {\mathbb {R}}^N \,|\, x_1 > \max \{\lambda _0, d\}\}$

Proof

Let $\mu > \max \{d,\lambda _0\}$. By Lemma 2.1 either $w_\mu \equiv 0$ or $w_\mu > 0$ in $\Sigma _\mu$ with $\frac{\partial w_\mu }{\partial \nu } < 0$ on the points of $\partial \Sigma _\mu$ such that $w_\mu = 0$.

If $w_\mu > 0$ in $\Sigma _\mu$, we get on $T_\mu$

$$\begin{aligned} w_\mu = 0, \qquad 0>\frac{\partial w_\mu }{\partial \nu } =\frac{\partial w_\mu }{\partial (-x_1)}=- \frac{\partial w_\mu }{\partial x_1}=2\frac{\partial u}{\partial x_1} \end{aligned}$$

(15)

and so the assertion holds. Let us prove that $w_\mu \equiv 0$ in $\Sigma _\mu$ cannot occur.

Indeed, if $w_\mu \equiv 0$ we have two cases:

• 1. $w_\lambda >0$ in $\Sigma _\lambda$ for all $\lambda > \mu$.

• 2. $w_{\tilde{\mu }} \equiv 0$ in $\Sigma _{\tilde{\mu }}$ for some $\tilde{\mu } > \mu$.

Case 1: We just repeat the above argument for every $\lambda > \mu$ and obtain

$$\begin{aligned} \frac{\partial u}{\partial x_1}< 0 \text { in } T_\lambda \text { so that, } \frac{\partial u}{\partial x_1}< 0 \text { in } \Sigma _\mu = \bigcup \limits _{\lambda >\mu }T_\lambda . \end{aligned}$$

(16)

Then take $x \in \Sigma _\mu$ such that $\{ x + t e_1\,,\, t\in {\mathbb {R}}\} \cap G = \emptyset$ (such x always exists since G is bounded) and notice that:

$$\begin{aligned} u(x) \le u(x^{\lambda _0}) =u( (x^{{\lambda _0}} )^\mu )< u(x) \end{aligned}$$

(17)

where the last inequality follows from $\frac{\partial u}{\partial x_1}< 0 \text { in } \Sigma _\mu$. The contradiction in (17) shows that Case 1 does not hold.

Case 2: We take $y \in \Sigma _\mu$ such that $\{ y + t e_1\,,\, t\in {\mathbb {R}}\} \cap G = \emptyset$ and notice that

$$\begin{aligned} 0 < u(y) =u(y^{\tilde{\mu }})= u((y^{\tilde{\mu }})^\mu ) \end{aligned}$$

(18)

Since $\tilde{\mu } - \mu > 0$, continuing to reflect with respect to $T_{\tilde{\mu }}$ and $T_\mu$ alternatively, we can construct a sequence of points $y_n$ such that $u(y_n)=u(y)$ and $|y_n| \rightarrow \infty$. This is clearly a contradiction with the hypothesis that $u(x) \rightarrow 0$ as $|x| \rightarrow \infty$. So also Case 2 cannot hold. $\square$

2.3 Step 3: $\lambda _0 \le d$

Proof

Assume not, then, $\lambda _0 > d$. Since u is a continuous function, $w_{\lambda _0} \ge 0$ in $\Sigma _{\lambda _{0}}$. Moreover $\frac{\partial u}{\partial x_1} < 0$ in $\{ x_1 > \lambda _0\}$ by Step 2. By the strong maximum principle, either $w_{\lambda _{0}}>0$ or $w_{\lambda _{0}}\equiv 0$. Assume $w_{\lambda _{0}}\equiv 0$. Take $y=(y_1,y'),z=(z_1,z') \in \partial G_1^{\lambda _0}$ such that ${\lambda _{0}}<y_1<z_1$ and $y'=z'$. Then $u(z) < u(y)$. On the other hand we also have

$$\begin{aligned} u(y)= u(y^{\lambda _{0}})=a_1=u(z^{\lambda _{0}})=u(z) \end{aligned}$$

(19)

Thus we get a contradiction which shows that $w_{\lambda _{0}}\not \equiv 0$.

Now consider the case $w_{\lambda _{0}}>0$. By definition of $\lambda _0$ there is a sequence {$\lambda _m$} such that $\lambda _m \nearrow {\lambda _{0}}$ and for every m there exists a minimizer $x_m \in \Sigma _{\lambda _m}$ such that $w_{\lambda _m}(x_m) < 0$ and $\nabla w_{\lambda _m} (x_m) = 0$.

Since ${\lambda _{0}}> d$ we fix $m_0$ such that

$$\begin{aligned} dist(\bar{G^{\lambda _m}},T_{{\lambda _{0}}}) \ge \frac{1}{2} dist(\bar{G^{\lambda _0}},T_{{\lambda _{0}}})>0 \end{aligned}$$

(20)

for $m \ge m_0$. We will now break in two cases

• 1. $x_m \in Int(\Sigma _{\lambda _m})$ for every $m \ge m_0$.

• 2. $x_m \in \partial \Sigma _{\lambda _m}$ for some $m \ge m_0$

Case 1: If $x_m \rightarrow x_0$, then passing to the limit in the definition of $x_m$ we obtain that $w_{\lambda _0}(x_0)=0$ and $\nabla u(x_0) =0$. Therefore we must have, by the strong maximum principle, $x_0 \in \partial \Sigma _{\lambda _{0}}$, but that is a clear contradiction since Hopf Lemma implies $\nabla u(x_0) \ne 0$.

If $x_m$ diverges, then we consider the sequence of points $\{y_m\}$ that minimize $w_{\lambda _m} |x|^q$ in $\Sigma _{\lambda _m}$, for $q > 0$ to be chosen. Clearly if $x_m$ diverges so does $y_m$. The function $w_{\lambda _m}|x|^q$ satisfies $\text { in } \Sigma _{\lambda _m} \cap \{ w_{\lambda _m} < 0\}:$

$$\begin{aligned} \mathcal {M}_{r,R}^-(D^2 w_{\lambda _m}|x|^q) - b(x) |\nabla (w_{\lambda _m}|x|^q) | + c(x) w_{\lambda _m}|x|^q \le 0 \end{aligned}$$

(21)

where,

$$\begin{aligned}b(x)= \frac{2\sqrt{N} R |\nabla (|x|^{-q}|)}{|x|^{-q}}\end{aligned}$$

and,

$$\begin{aligned} c_m(x)= \frac{f(u_{\lambda _m})-f(u)}{w_\lambda } + \frac{q(R(q+1))-r(N-1)) }{|x|^2} < 0 \end{aligned}$$

(22)

To see the above take $\phi$ as a test function for $w_{\lambda _m}|x|^q$. Thus, $\phi |x|^{-q}$ is a test function for $w_{\lambda _m}$. Since $w_{\lambda _m}$ is a subsolution, we obtain by using $\phi |x|^{-q}$ as a test function:

$$\begin{aligned} 0 \ge \frac{f(u_{\lambda _m})-f(u)}{w_\lambda } \phi |x|^{-q} +\mathcal {M}_{r,R}^-(\phi |x|^{-q}) \end{aligned}$$

(23)

Using

$$\begin{aligned} D^2(\phi |x|^q)= |x|^q D^2\phi + D^2(|x|^q) \phi + 2 \nabla (|x|^q) \otimes \nabla \phi \end{aligned}$$

and that for every $A \in \mathcal {M}_{NxN}$, whose eigenvalues belong to (r, R),

$$\begin{aligned} Tr(A(p\otimes q)) \le \sqrt{N} R|p||q| \end{aligned}$$

(24)

we obtain (21).

If q belongs to $(0,\tilde{N}-2)$ and $|x| > L$, for L sufficiently large, by (3) and (4) we get $c_m < 0$. Now we fix m such that $|y_m| > L+2$ and consider the domain $\Sigma _{\lambda _m} \setminus B_{L+1}$. By the weak maximum principle, which we can apply since $c_m <0$, inequality (21) implies that $w_{\lambda _m}|x|^q$ cannot achieve its minimum at $y_m$ in $\Sigma _{\lambda _m} {\setminus } B_{L+1}$ unless it is constant. However, if q is chosen to be smaller than $\gamma$, by 3, we obtain $\lim \limits _{|x|\rightarrow \infty } w_{\lambda _m}|x|^q = 0$. Hence a contradiction.

Case 2: Since we have that $x_m \in \partial G^{\lambda _m}$.

Since $dist(G^{\lambda _m},T_{{\lambda _{0}}}) > 0$, we get that $\{x_m +t e_1\} \subset D_{\lambda _0}$ for small positive values of t.

Now we will use Lemma 2.2 to obtain a contradiction. Clearly $x_m$ is not a point of type (III) or (IV). Let us prove it cannot be of types (I) and (II).

Type (I): $\Gamma _t(x) \in \Sigma _{\lambda _m}.$ Then, by Step 2,

$$\begin{aligned} 0 \le \frac{\partial w_{\lambda _m}}{\partial (-x_1)} (x_m)= \frac{\partial u}{\partial (x_1)}(x_m) + \frac{\partial u}{\partial (x_1)}(x_m ^\lambda ) <\frac{\partial u}{\partial (x_1)}(x_m ^\lambda ) \end{aligned}$$

(25)

We will now show that $\frac{\partial u}{\partial (x_1)}(x_m ^\lambda ) \le 0$ to obtain a contradiction

Since $\Gamma _{t}(x_m) \in \Sigma _{\lambda _m}$, this implies that $\langle \nu (x_m ^\lambda ,e_1) \ge 0$. Since $u \equiv a_i$ on $\partial G_i$ we obtain

$$\begin{aligned} \frac{\partial u}{\partial \xi } = 0 \qquad \forall \xi \text { in the tangent space to }\partial G_i \end{aligned}$$

Therefore

$$\begin{aligned} \frac{\partial u}{\partial x_1} (x_m^\lambda )=\frac{\partial u}{\partial \nu } (x_m^\lambda ) \langle \nu (x_m^\lambda ,e_1)\rangle \le 0 \end{aligned}$$

(26)

Let us now see that it is not possible to have a sequence $t_m \searrow 0$ such that $\Gamma _{t_m} \in \partial G_i^{\lambda _m}$. If that was the case we would have, by taking $y \in \Gamma _{t_m}(x_m)$, that

$$\begin{aligned} w_{\lambda _m}(x_m) \le w_{\lambda _m}(y) = a_i -u(y) < a_i - u(x_m) = w_{\lambda _m}(x_m). \end{aligned}$$

(27)

which is not possible. Type (II): Clearly $\Gamma _{{\underline{t}}} \in \partial G_i ^{\lambda _m} \cap D_{\lambda _0}$ hence as before

$$\begin{aligned} w_{\lambda _m}(x_m) \le w_{\lambda _m}(\Gamma _{{\underline{t}}}) = a_i -u(\Gamma _{{\underline{t}}}) < a_i - u(x_m) = w_{\lambda _m}(x_m). \end{aligned}$$

(28)

$\square$

2.4 Step 4: For any $z \in \partial G$ and any unit vector $\eta$ for which$\langle \eta , \nu (z) \rangle > 0$, we can find a small enough ball $B_\delta (z)$ such that $\frac{\partial u}{\partial {\eta }} < 0$ in $B_\delta (z){\setminus } {\bar{G}}$

The proof of Step 4 is done by induction and will use some of the following steps. For clarity’s sake we will delay the proof until all the remaining steps have been proved.

2.5 Step 5: $w_\lambda >0$ for any $\lambda \in [\lambda _0,\infty ) \cap (\lambda _*,\infty )$

Proof

Based on Steps 2 and 3, we may assume $\lambda \le d$. By the strong maximum principle we just have to see that $w_\lambda$ is not identically zero in a connected component Z of $\Sigma _\lambda$. We will proceed arguing by contradiction, i.e, we assume that such a connected component exists.

First we observe that for every Y connected component of $\Sigma _\lambda$ we have

$$\begin{aligned} dist(\partial Y \cap D_\lambda , T_\lambda ) = 0. \end{aligned}$$

(29)

In fact, Y is connected, hence $Y^\lambda$ is a connected component of $({\mathbb {R}}^N {\setminus } {\bar{G}}) {\setminus } \bar{D_\lambda }$. From that it follows that either $Y^\lambda$ contains a left neighborhood of $T_\lambda$ or

$$\begin{aligned} dist(\partial Y^\lambda \setminus \bar{D_\lambda }, T_\lambda ) = 0. \end{aligned}$$

(30)

However, the former case is not possible since $\lambda \le d$ means that G $\cap T_\lambda$ is non empty. Thus, (29) follows by reflecting (30). Therefore take $\{z_m\}$ a sequence in $\partial Z \cap D_\lambda$ such that $z_m \rightarrow z_0 \in T_\lambda \cap \partial Z \cap D_\lambda$.

Since $\lambda > \lambda _*$, $\langle \nu (z_m ^\lambda ),e_1\rangle <0$, for m sufficiently large, and the open segment $(z_m,z_m ^\lambda )$ is contained in the ball $B_\delta (z_0) {\setminus } {\bar{G}}$, where delta is the one given by Step 4. Therefore by Step 4, u strictly decreases in $(z_m ^\lambda ,z_m)$, which means $w_\lambda (z_m) >0$. This is a contradiction with the fact that w is identically zero in Z. $\square$

2.6 Step 6:$\frac{\partial u}{\partial x_1} < 0$ in $D_{\tilde{\lambda }} \setminus {\bar{G}}$ for $\tilde{\lambda }=\max \{\lambda _0,\lambda _*\}$

Proof

The proof follows in the same way as in the proof of Step 2, using Step 5 and Hopf Lemma. $\square$

2.7 Step 7:$\lambda _0 \le \lambda _*$

Proof

We will argue by contradiction, suppose $\lambda _0 > \lambda _*$. By Step 5, we know that $w_{\lambda _0}$ is positive in $\Sigma _{\lambda _0}$, therefore we may obtain as in Step 3, a sequence of $\lambda _m \nearrow \lambda _0$, such that $w_{\lambda _m}(x_m) <0$, for some $x_m \in \Sigma _{\lambda _m}$. We will break into the following cases:

Case 1: There is a subsequence of $\{x_m\}$ such that $x_m \in$ Int$\Sigma _{\lambda _m}$.

The cases where $x_m$ diverges or converges to a point on the regular part of $\partial \Sigma _{\lambda _m}$ are analogous to what have been done in Step 3. Furthermore the proof of Step 5 shows that $w_{\lambda _m} >0$ for m sufficiently large.

Case 2: There is a subsequence of $\{x_m\}$ such that $x_m \in$ $\partial \Sigma _{\lambda _m}$.

Clearly $x_m$ is not in $T_{\lambda _m}$, therefore $\{x_m\}$ is a bounded sequence and passing to a subsequence we may assume $x_m \in$ $\partial G_{i}^{\lambda _m}$, for a fixed $i \in \{1,\ldots ,k\}$. We will now proceed by using Lemma 2.2 to reach a contradiction.

First let us show that $x_m$ is not contained in $\partial \Sigma _{\lambda _m} \cap \{x_1 \le \lambda _0\}$. If that was the case then $x_m \rightarrow x_0 \in T_{\lambda _0}\cap \partial G$, therefore the same argument as in Step 5 would imply a contradiction. Henceforth we will assume that $x_m \in \partial G_{i ^\lambda _m} \cap \{x_1 > \lambda _0\}$. It is clear that $x_m$ is not of type (IV) since $\lambda _m > \lambda _*$. Arguing as in Step 3 we also can exclude the case where $x_m$ is of type (I). We will now consider two cases in order to finish the proof. Let $y_m = \Gamma _{t_m}(x_m)$.

Case (a): $y_m \in D_{\lambda _0}$ for some $m \in {\mathbb {N}}$.

If $x_m$ is of type (II), then $y_m \in \partial G_i ^{\lambda _m}\cap D_{\lambda _0}$. Thus we may obtain a contradiction as in Step 3. If $x_m$ is of type (III), then $y_m \in \partial G_i \cap D_{\lambda _0}$, by Steps 4 and 6 we obtain that u is strictly decreasing from $y_m$ to $x_m$. However this contradicts the fact $w_{m(x_m)} < 0$ since

$$\begin{aligned} w_{\lambda _m}(x_m) = a_i - u(x_m) = u(y_m) - u(x_m) > 0 \end{aligned}$$

(31)

Case 2: $y_m \in {\bar{D}}_{\lambda _m} \setminus {D}_{\lambda _0}$ for every $m \in {\mathbb {N}}$.

Since $y_m$ $\in$ $\partial G \cup \partial G ^{\lambda _m}$, $\{y_m\}$ is bounded. Thus we may obtain that up to a subsequence $y_m$ converges to $y_0 \in \partial G_i ^{\lambda _0} \cap T_{\lambda _0}$. Therefore for m sufficiently large, $y_m$ and its projection on $T_{\lambda _0}$ belong to $B_\delta (y_0)$, where $\delta$ is given by Step 4. Again combining Steps 4 and 6 we obtain that u decreases on the whole segment $(y_m,x_m)$ and get a contradiction as in the previous case. $\square$

2.8 Step 8: $w_{\lambda _*} \equiv 0$ in at least one connected component of $\Sigma _{\lambda _*}$

Proof

If there exists $z_0 \in \partial G \cap T_{\lambda _*}$ such that $\langle \nu (z_0), e_1 \rangle = 0$, we will show that $w_{\lambda _*}$ is identically zero in the connected component that contains $z_0$ on its boundary. For that, assume $u > u_{\lambda _*}$ in a certain ball around $z_0$. We take $t_1$ as in the assumption of Theorem 1.1, then, since we assume $\frac{R}{r} < t_1$, Lemma 2.3 implies that for every direction v pointing inside $\Sigma _{\lambda _*}$ either $\partial _v u_{\lambda _*}(z_0) > \partial _v u(z_0)$ or $\partial _v ^2 u_{\lambda _*}(z_0) > \partial _v u^2 u(z_0)$.

The first inequality cannot happen, since on $\partial G_i$, $\frac{\partial u}{\partial _v}$ is constant. The second inequality also cannot happen since $D^2 u_{\lambda _*}(z_0)=D^2u(z_0)$ because by Lemma 2.4 the hessian only depends on the shape of the boundary. Therefore we obtain $u\equiv u_{\lambda _*}$ in $B(z_0,R)$, by the strong maximum principle, and also $u\equiv u_{\lambda _*}$ in the whole connected component which contains $z_0$.

If there is not such $z_0$ then, by the definition of $\lambda _*$ we can find a point $z_1 \in \partial G \cap D_{\lambda _*}$ such that $z_1^{\lambda _*} \in \partial G$ is a point of internal tangency. Due to the boundary conditions, $w_{\lambda _*} (z_1)=0=\frac{\partial w_{\lambda _*}}{\nu }(z_1)$. That contradicts Hopf Lemma.

$\square$

2.9 Step 9: Let Z be a connected component of $\Sigma _{\lambda _*}$ such that $w_{\lambda _*} = 0$ in Z. Then $\partial Z \setminus T_{\lambda _*} \subset \partial G$

Proof

We shall use Lemma 2.2 in order to show that all points on $\partial Z \setminus T_{\lambda _*}$ are of symmetry type (IV). Suppose by contradiction that there is a point z that is not of symmetry type (IV). The point z cannot be of type (I), otherwise we could argue as in Step 3 and obtain a contradiction. Also it is not of type (II) or type(III). To see that just set $y=\Gamma _{{\underline{t}}}(z) \in \Sigma _{\lambda _*}$, then, by Step 6, u is strictly decreasing on the segment (y, z). If z was of type (II) we would have:

$$\begin{aligned} a_i = u(y^{\lambda _*}) \ge u(y) > u(z) = u(z^{\lambda _*}) =a_i. \end{aligned}$$

(32)

which is a contradiction. Type(III) can be excluded in an analogous fashion. $\square$

2.10 Step 10: End of the proof of Theorem 1.1

Let us denote by $G^C$ the complement of G in ${\mathbb {R}}^N$. Then, the set X,

$$\begin{aligned} X = Z \cup Z^{\lambda _*}\cup (\partial Z \cap G^C) \cup (\partial Z^{\lambda _*} \cap G^C) \end{aligned}$$

(33)

is symmetric with respect to $T_{\lambda _*}$. Note that by Step 9:

$$\begin{aligned} \partial Z\cap G^C \subset T_{\lambda _*} \cap G^C \end{aligned}$$

(34)

One may check that X is an open subset in $G^C$, therefore $G^C {\setminus } X = G^C {\setminus } {\bar{X}}$ which implies that $X =G^C$ since $G^C$ is connected. This implies that u and G are symmetric in the $x_1$ direction Therefore applying the result for every direction we obtain that G is radially symmetric and the solution u is radial.

2.11 Proof of Step 4:

Now we give the proof of Step 4 that was previously omitted.

Proof

We will prove the result by induction on k. First assume that $k=1$ in (5). If $\alpha _1 < 0$, the result is trivial by continuity, therefore lets assume $\alpha _1=0$, which implies $\nabla u \equiv 0$ and $|D^2 u|=|\partial _\nu ^2 u|$ on $\partial G$.

Choose $z_0 \in T_d \cap \partial G$, then

$$\begin{aligned} \frac{\partial u}{\partial x_1}(z_0)=\frac{\partial u}{\partial \nu }(z_0)=0 \end{aligned}$$

(35)

Since $\lambda _0 \le d$ by Step 3, then Steps 2 implies

$$\begin{aligned} \frac{\partial u}{\partial x_1}(\Gamma _{t}(z_0))<0 \end{aligned}$$

for negative t. This implies

$$\begin{aligned} 0 \ge \frac{\partial ^2 u}{\partial x_1^2}(z_0) =\frac{\partial ^2 u}{\partial \nu ^2}(z_0) \end{aligned}$$

(36)

On the other hand, the eigenvalues of $D^2 u$ for $x \in \partial G$ are

• $\frac{\partial ^2 u}{\partial \nu ^2}(x)$ with multiplicity one, (associated to the normal direction at x).

• 0 with multiplicity $N-1,$ (associated to the tangent space to $\partial G$ at x).

Therefore by (1) and (7), we obtain for x in $\partial G$

(37)

Analogously one would obtain for $\mathcal {M}_{r,R}^-$:

(38)

In either case, by (36) we get that $f(a_1)$ must be a non-negative value. If $f(a_1)$ is strictly positive, we are done since $u \in \mathcal {C}^2(\mathbb {R^N} \setminus G)$ we would have

$$\begin{aligned} \frac{\partial ^2 u}{\partial \eta ^2}=\frac{\partial ^2 u}{\partial \nu ^2} \langle \eta ,\nu \rangle ^2 < 0 \text { on } \partial G. \end{aligned}$$

(39)

This together with (35) and (36) imply the assertion.

If $f(a_1) = 0$ then all the first and second derivatives of u vanish on $\partial G$, therefore extending u as $a_1$ on G would provide a solution on the whole ${\mathbb {R}}^N$. However that is not possible for such a solution in ${\mathbb {R}}^N$, either by known results of symmetry and monotonicity (see [11]) or by Steps 1–3, the solution cannot be flat in an open set. Suppose now that the result is valid for $k-1$. Therefore, by steps 5 to 10, Theorem 1.1 is proved for $k-1$ domains $G_i$.

Set

$$\begin{aligned} I=\{i \,|\, \alpha _i< 0 \text { or } \frac{\partial ^2 u}{\partial \nu ^2} < 0 \text { on } \partial G_i \} \end{aligned}$$

(40)

and J being $\{1...k\} \setminus I$. We claim that J is empty. Define D as

$$\begin{aligned} D= \max \limits _{j \in J} d_j \end{aligned}$$

(41)

If $D< \lambda _*$, we could complete steps 5 to 10 and obtain a contradiction. If $D \ge \lambda _*$ the moving plane reaches at least one domain $G_j$ with $j \in J$, therefore arguing as in the case $k=1$ we obtain a contradiction. $\square$