A symmetry theorem in two-phase heat conductors

1.   Introduction

In the previous paper ^[7], we considered the Cauchy problem for the heat diffusion equation in the whole Euclidean space consisting of two media with different constant conductivities, where initially one medium has temperature 0 and the other has temperature 1. There, the large time behavior, either stabilization to a constant or oscillation, of temperature was studied. The present paper deals with the case where one medium is bounded and the interface is of class $C^{2, \alpha}$, and introduces an overdetermined problem with the condition that the interface is stationary isothermic.

To be precise, let $\Omega$ consist of a finite number, say $m$, of bounded domains $\{ \Omega_j\}$ in $\mathbb R^N$ with $N \ge 2$, where each $\partial\Omega_j$ is of class $C^{2, \alpha}$ for some $0 < \alpha < 1$ and $\overline{\Omega}_i\cap \overline{\Omega}_j = \emptyset$ if $i\not = j$. Denote by $\sigma = \sigma(x)\ (x \in \mathbb R^N)$ the conductivity distribution of the whole medium given by

where $\sigma_-, \sigma_+$ are positive constants with $\sigma_- \not = \sigma_+$. The diffusion over such multiphase heat conductors has been dealt with also in ^{[3,4,9,10,11]}.

We consider the unique bounded solution $u = u(x, t)$ of the Cauchy problem for the heat diffusion equation:

where ${\mathcal X}_{\Omega}$ denotes the characteristic function of the set $\Omega$. The maximum principle gives

Our symmetry theorem is stated as follows.

Theorem 1.1. If there exists a function $a:(0, +\infty)\to(0, +\infty)$ satisfying

then $\Omega$ must be a ball.

If $\partial\Omega$ is of class $C^6$, then Theorem 1.1 can be proved by the method employed in [3, Theorem 1.5 with the proof,pp. 335–341], where concentric balls are characterized. The proof there consists of four steps summarized as follows: (i) reduction of (1.2) to elliptic problems by the Laplace-Stieltjes transform $\lambda\int_0^\infty e^{-\lambda t}u(x, t)dt$ for all sufficiently large $\lambda > 0$, (ii) construction of precise barriers based on the formal WKB approximation where the fourth derivatives of the distance function to $\partial\Omega$ together with the assumption (1.4) are used, (iii) showing that the mean curvature of $\partial\Omega$ is constant with the aid of the precise asymptotics as $\lambda \to \infty$ and the transmission conditions on the interface $\partial\Omega$, (iv) Alexandrov's soap bubble theorem ^[1] from which we conclude that $\partial\Omega$ must be a sphere.

The approach of the present paper is different from that in ^[3] and only requires $\partial\Omega$ to be of class $C^{2, \alpha}$ for some $\alpha > 0$. Here the proof consists of two ingredients: (i) reduction to elliptic problems by the Laplace-Stieltjes transform $\lambda\int_0^\infty e^{-\lambda t}u(x, t)dt$ for some $\lambda$, for instance $\lambda = 1$, (ii) the method of moving planes due to Serrin ^[6,8,12,13] with the aid of the transmission conditions on $\partial\Omega$. To apply the method of moving planes, the solutions need to be of class $C^2$ up to the interface $\partial\Omega$ from each side, which is guaranteed if $\partial\Omega$ is of class $C^{2, \alpha}$.

2.   Introducing a Laplace-Stieltjes transform

Let $u = u(x, t)$ be the unique bounded solution of (1.2) satisfying (1.4). We use the Gaussian bounds for the fundamental solutions of diffusion equations due to Aronson [2, Theorem 1,p. 891] (see also [5, p. 328]). Let $g = g(x, \xi, t)$ be the fundamental solution of $u_t = \mbox{ div}(\sigma\nabla u)$. Then there exist two positive constants $\lambda < \Lambda$ such that

for all $(x, t), (\xi, t) \in \mathbb R^N\times(0, +\infty)$. Note that $u$ is represented as

Define the function $v = v(x)$ by

With the function $a$ in (1.4), we set $a^* = \int_0^\infty e^{-t}a(t) dt$. Then, (1.3) yields that $0 < a^* < 1$. Set

Then we observe that

Here, $n$ denotes the outward unit normal vector to $\partial\Omega$, the inequalities in (2.5) and (2.6) follow from the maximum principle, (2.7) expresses the transmission conditions on the interface $\partial\Omega$, and (2.8) follows from (2.1) and (2.2).

3.   Proof of Theorem 1.1

Let us apply directly the method of moving planes due to Serrin ^[6,8,12,13] to our problem in order to show that $\Omega$ must be a ball. The point is to apply the method to both the interior $\Omega$ and the exterior $\mathbb R^N\setminus\overline{\Omega}$ at the same time. For the method of moving planes for $\mathbb R^N\setminus\overline{\Omega}$, we refer to ^[8,12]. In this procedure, the supposition that $\Omega$ is not symmetric will lead us to the contradiction that the transmission conditions (2.7) do not hold.

Let $\gamma$ be a unit vector in $\mathbb R^N, \; \lambda\in\mathbb R,$ and let $\pi_\lambda$ be the hyperplane $x\cdot\gamma = \lambda.$ For large $\lambda, \; \pi_\lambda$ is disjoint from $\overline{\Omega}$; as $\lambda$ decreases, $\pi_\lambda$ intersects $\overline{\Omega}$ and cuts off from $\Omega$ an open cap $\Omega_\lambda = \Omega \cap \{ x \in \mathbb R^N : x\cdot\gamma > \lambda\}$.

Denote by $\Omega^\lambda$ the reflection of $\Omega_\lambda$ with respect to the plane $\pi_\lambda$. Then, $\Omega^\lambda$ is contained in $\Omega$ at the beginning, and remains in $\Omega$ until one of the following events occurs:

(i) $\Omega^\lambda$ becomes internally tangent to $\partial \Omega$ at some point $p\in\partial\Omega\setminus \pi_\lambda;$

(ii) $\pi_\lambda$ reaches a position where it is orthogonal to $\partial \Omega$ at some point $q\in\partial\Omega\cap\pi_\lambda$ and the direction $\gamma$ is not tangential to $\partial\Omega$ at every point on $\partial \Omega \cap \{ x \in \mathbb R^N : x\cdot\gamma > \lambda\}$.

Let $\lambda_*$ denote the value of $\lambda$ at which either (i) or (ii) occurs. We claim that $\Omega$ is symmetric with respect to $\pi_{\lambda_*}$.

Suppose that $\Omega$ is not symmetric with respect to $\pi_{\lambda_*}$. Denote by $D$ the reflection of $\left(\mathbb R^N\setminus\overline{\Omega}\right)\cap\{ x \in \mathbb R^N : x\cdot\gamma > \lambda_*\}$ with respect to $\pi_{\lambda_*}$. Let $\Sigma$ be the connected component of $\left(\mathbb R^N\setminus\overline{\Omega}\right)\cap\{ x \in \mathbb R^N : x\cdot\gamma < \lambda_*\}$ whose boundary contains the points $p$ and $q$ in the respective cases (i) and (ii). Since $\Omega^{\lambda_*}\subset\Omega$, we notice that

Let $x^{\lambda_*}$ denote the reflection of a point $x\in\mathbb R^N$ with respect to $\pi_{\lambda_*}$, namely,

Using the functions $v^\pm$ defined in (2.4), we introduce the functions $w^\pm = w^\pm(x)$ by

It then follows from (2.5)–(2.8) that

and hence by the maximum principle

Note that $w^+$ can be zero in $\Omega^{\lambda_*}$ since some connected component $\Omega_j$ of $\Omega$ can be symmetric with respect to $\pi_{\lambda_*}$ and, in such a case, $w^+ \equiv 0$ in $\Omega_j$. But $w^-$ is strictly positive in $\Sigma$ since $\Omega$ is not symmetric with respect to $\pi_{\lambda_*}$.

Let us first consider the case (i). The first equality in (2.7) yields that $w^+(p) = w^-(p) = 0$. Then, it follows from (3.5) and Hopf's boundary point lemma that

where we used the fact that $n$ is the outward unit normal vector to $\partial\Omega$ as well as the inward unit normal vector to $\partial\Sigma$. It thus follows from the definition (3.2) of $w^\pm$ that

Reflection symmetry with respect to the plane $\pi_{\lambda_*}$ yields that

Indeed, we observe that

and by using (3.1), we see that

Then, combing these equalities yields (3.7). It thus follows that

On the other hand, the second equality in (2.7) shows that

which contradict (3.8).

Let us proceed to the case (ii). As in ^[13], by a translation and a rotation of coordinates, we may assume:

Since $\partial\Omega$ is of class $C^2$, there exists a $C^2$ function $\varphi : \mathbb R^{N-1}\to\mathbb R$ such that in a neighborhood of $q = 0$, $\partial\Omega$ is represented as a graph $x_N = \varphi(\hat{x})$ where $\hat{x} = (x_1, \dots, x_{N-1})\in\mathbb R^{N-1}$, where

Since the event (ii) occurs at $\lambda = 0$, we observe that the function $\frac {\partial\varphi}{\partial x_1}(0, x_2, \dots, x_{N-1})$ achieves its local maximum $0$ at $(x_2, \dots, x_{N-1}) = 0\in \mathbb R^{N-2}$, and hence

Notice that

since $x^{\lambda_*} = (-x_1, x_2, \dots, x_N)$.

The equalities (2.7) at $(\hat{x}, \varphi(\hat{x}))$ in a neighborhood of $q = 0$ are read as

Differentiating (3.11) in $x_i$ for $i = 1, \dots, N-1$ yields that at $(\hat{x}, \varphi(\hat{x}))$

Then, differentiating (3.13) in $x_j$ for $j = 1, \dots, N-1$ yields that at $(\hat{x}, \varphi(\hat{x}))$

By letting $\hat{x} = 0$ in these equalities, we obtain from (3.9) that

Next, differentiating (3.12) in $x_i$ for $i = 1, \dots, N-1$ and letting $\hat{x} = 0$ give

Since the functions $w^\pm$ are expressed as (3.10), with the aid of $\eqref{values of v+- at 0}$ we have that

The relations (3.3)–(3.5) enable us to apply Serrin's corner point lemma (see [6, Lemma S,p. 214] or [8, Serrin's Corner Lemma,p. 393]) to show that

where $\frac {\partial^2 w^\pm}{\partial s_{\pm}^2}$ denotes the second derivative of $w^\pm$ in the direction of $s_{\pm}$. Note that each of the directions $s_\pm$ respectively enters $\Omega^{\lambda_*}, \Sigma$, transversally to both of the hypersurfaces $\partial\Omega$ and $\pi_{\lambda_*}$. Thus, we have from (3.10) and (3.17) that

It then follows from (3.18) that

which contradicts (3.16) with $i = 1$. Thus $\Omega$ is symmetric with respect to $\pi_{\lambda_*}$. Since the unit vector $\gamma$ is arbitrary, $\Omega$ must be a ball and Theorem 1.1 is proved.

Acknowledgments

This research was partially supported by the Grants-in-Aid for Scientific Research (B) and (C) ($\sharp$ 18H01126 and $\sharp$ 22K03381) of Japan Society for the Promotion of Science and National Research Foundation of S. Korea grant 2022R1A2B5B01001445.

Conflict of interest

The authors declare no conflict of interest.