Old symmetry problem revisited

1. Introduction

Let \(D\) be bounded smooth connected domain in \(\mathbb{R}^3\), \(S\) be its boundary, \(N\) is the outer unit normal to \(S\), \(u_N\) is the normal derivative of \(u\) on \(S\), \(|D|\) is the volume of \(D\) and \(|S|\) is the suraface area of \(S\). Various symmetry problems were considered in [1, 2]. Consider the problem Our result is the following:

Ttheorem 1.1. If problem (1) is solvable then \(D\) is a ball.

This result was proved by different methods in [3] and in [4]. The proof, given in the next section, is novel, short and is based on a new idea. We assume that \(D\subset \mathbb{R}^2\) so that \(S\) is a curve. Then the ballis a disc.

2. Proof of Theorem 1.1

Proof. Let \(s\) be the curve length, \(\bf{s}\) be the point on \(S\) corresponding to the parameter \(s\), \(\{x(s), y(s)\}\) be the parametric representation of \(S\), \({\bf{s}}=x(s)e_1+y(s)e_2\), where \(\{e_j\}|_{j=1,2}\) is a Cartesian basis in \(\mathbb{R}^2\). It is known that \(\frac{d{\bf{s}}}{ds}={\bf{t}}(s)\) is the tangent unit vector to \(S\) at the point \(\bf{s}\) and

where \(k(s)\ge 0\) is the curvature of \(S\) and \(\bf{\nu}(s)\) is the normal to \(S\). Since \(u_N=\nabla u\cdot N=m>0\) on \(S\) the convexity of \(S\) does not change sign, so \(\nu\) does not change sign, \(k(s)>0\) \(\forall s\in S\) and \(N(s)=-\bf{\nu}(s)\) \(\forall s\in S\). Differentiate the identity \(u(x(s),y(s))=0\) with respect to \(s\) and get \(\nabla u \cdot \bf{t}=0\). Differentiate this identity and use (1)-(2) to get where \({\bf t} =t_1e_1+t_2e_2\). Rewrite (3) as Equation (4) holds in every coordinate system obtained from \(\{x,y\}\) by rotations. Clearly \(u_{xx}(s), u_{yy}(s), u_{xy}(s)\) cannot vanish simultaneously due to (4). Also \(u_{xx}(s), u_{yy}(s)\) cannot vanish simultaneously due to the first equation in (1). Equation (4) holds in any coordinate system obtained from a fixed Cartesian system by rotations. Equation (1) on the boundary yields: We prove that (4) and (5) are not compatible (lead to a contradiction) except when \(S\) is a circle. Let \(u_{xx}:=p\), \(u_{xy}:=q\). Denote by \(A\) the \(2\times 2\) matrix with the elements \(A_{11}=p\),\(A_{22}=1-p\), where (5) was used, \(A_{12}=A_{21}=q\). Let \(I\) be the identity matrix. The equation \(\det (A-\lambda I)=\lambda^2 -\lambda -p^2-q^2+p=0\) has two solutions, so the eigenvalues of \(A\) are: The corresponding eigenvectors are Note that \(\lambda_{+}+\lambda_{-}=1\), \(\lambda_{+} \lambda_{-}=-p^2-q^2+p.\) Thus, \(\lambda_{+}>0\). The eigenvectors are orthogonal: \(e_1\cdot e_2=0\) but not normalized: \(\|e_1\|^2=\|e_2\|^2=1+\gamma^2\). Since \(\|e_1\|^2\) is invariant under rotations of a Cartesian coordinate system, so is \(\gamma^2\). Let \(w:=\{t_1,t_2\}\). Then (4) implies Since \(e_1\) and \(e_2\) form an orthogonal basis in \(\mathbb{R}^2\) one can find unique constants \(c_1,\, c_2\) such that Solving this linear algebraic system for \(c_1,\, c_2\) one gets: where \(\Delta=1+\gamma^2\) is the determinant of the matrix of the system (9). Substitute \(w\) from (9) into (8) and get: where we have used the relations: \(Ae_j=\lambda_je_j\), \(\lambda_1:=\lambda_+\), \(\lambda_2:=\lambda_-\), \((e_1,e_2)=0\), \(\|e_j\|^2=1+\gamma^2\), \((Ae_j,e_j)=\lambda_j(1+\gamma^2)\), \(j=1,2\). Using (10) one gets from (11): We prove that (12) leads to a contradiction unless \(S\) is a circle. Assume first that \(\lambda_{-}< 0\) and recall that \(\lambda_{+}>0\). Choose a point \(s\in S\) and the Cartesian coordinate system such that \(t_1(s)+\gamma(s) t_2(s)=0\). This is possible since \(\gamma^2\) is invariant under rotations and the only restriction on the real-valued \(t_1,\,t_2\) is the relation \(t_1^2+t_2^2=1\). Since \(\lambda_{-}< 0\) and \(t_2-\gamma t_1\neq 0\), we have a contradiction with inequality (12).

Assume now that \(\lambda_{-}\ge 0\) and \(\lambda_{-}\neq \lambda_{+}\). Then the left side of (12) is not a constant as a function of \(\{t_1,t_2\}\), that is, not a constant with respect to rotations of the coordinate system, while its right side is a constant. Thus, we have a contradiction.

Suppose finally that \(\lambda_{-}= \lambda_{+}\). Then \(\lambda_{-}= \lambda_{+}=\frac 1 2\) at any \(s\in S\). This implies by formula (6) that \(p=\frac 1 2\), \(u_{yy}=\frac 1 2\) and \(q=0\) on \(S\) for all \(s\in S\). By formula (7) one gets \(\gamma=0\), \(\|e_j\|=1\). Consequently, by formula (4), it follows that

\(\kappa(s)=\frac 1 {2m}\). Thus, the curvature of \(S\) is a constant, so \(S\) is a circle of a radius \(a\). Thus, \(m=\frac {\pi a^2}{2\pi a}=\frac a 2\), \(k(s)=\frac 1 a\) and the solution to problem (1) is \(u=\frac {|x|^2-a^2}{4}\). Obviously this \(u\) solves equation (1) and satisfies the first boundary condition in (1). The second boundary condition is also satisfied: \(u_N|_{S}=a/2\).

Theorem 1.1 is proved in the two-dimensional case. We leave to the reader to consider the three-dimensional case, see [5]. Theorem 1.1 is proved.

Competing Interests

The author declares that he has no competing interests.