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
\begin{equation}\label{e1}
\nabla^2 u=1 \quad in \quad D, \quad u|_S=0,
\quad u_N|_S=m=|D|/|S|.
\end{equation}
(1)
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
\begin{equation}\label{e2}
\frac{d\bf{t}}{ds}=k(s)\bf{\nu} (s),
\end{equation}
(2)
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
\begin{equation}\label{e3}
u_{xx}t_1^2(s) +2u_{xy}t_1(s)t_2(s) +u_{yy}t_2^2(s)+\nabla u \cdot k(s)\nu (s)=0,
\end{equation}
(3)
where \({\bf t}
=t_1e_1+t_2e_2\). Rewrite (3) as
\begin{equation}\label{e4}
u_{xx}t_1^2(s) +2u_{xy}t_1(s)t_2(s) +u_{yy}t_2^2(s)=m k(s).
\end{equation}
(4)
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:
\begin{equation}\label{e5}
u_{xx}+u_{yy}=1.
\end{equation}
(5)
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:
\begin{equation}\label{e6}
\lambda_{\pm}=\frac 1 2 \pm (\frac 14+p^2+q^2-p)^{1/2}=\frac 1 2 \pm [(\frac 1 2-p)^2+q^2]^{1/2}.
\end{equation}
(6)
The corresponding eigenvectors are
\begin{equation}\label{e7}
e_1=\{1, \gamma\}, \quad e_2=\{-\gamma, 1\},\quad \gamma:=\frac q{p+\lambda_{+}-1}.
\end{equation}
(7)
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
\begin{equation}\label{e8}
(Aw,w)=mk(s)>0.
\end{equation}
(8)
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
\begin{equation}\label{e9}
c_1e_1+c_2e_2=w.
\end{equation}
(9)
Solving this linear algebraic system for \(c_1,\, c_2\) one gets:
\begin{equation}\label{e10}
c_1=\frac{t_1+\gamma t_2}{\Delta},\quad c_2=
\frac{t_2-\gamma t_1}{\Delta},
\end{equation}
(10)
where \(\Delta=1+\gamma^2\) is the determinant of the matrix of the system (9).
Substitute \(w\) from (9) into (8)
and get:
\begin{equation}\label{e11}
[c_1^2\lambda_{+}+c_2^2\lambda_{-}](1+\gamma^2)=mk(s)>0,
\end{equation}
(11)
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):
\begin{equation}\label{e12}
(t_1+\gamma t_2)^
2 \lambda_{+}+(t_2-\gamma t_1)^2\lambda_{-}=mk(s)(1+\gamma^2)>0.
\end{equation}
(12)
We prove that (12) leads to a contradiction unless \(S\) is a circle.
Assume first 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.
