Heat flow, heat content and the isoparametric property

Abstract

Let M be a Riemannian manifold and $\Omega$ a compact domain of M with smooth boundary. We study the solution of the heat equation on $\Omega$ having constant unit initial conditions and Dirichlet boundary conditions. The purpose of this paper is to study the geometry of domains for which, at any fixed value of time, the normal derivative of the solution (heat flow) is a constant function on the boundary. We express this fact by saying that such domains have the constant flow property. In constant curvature spaces known examples of such domains are given by geodesic balls and, more generally, by domains whose boundary is connected and isoparametric. The question is: are they all like that? This problem is the analogous (for the heat equation) of the classical Serrin’s problem for harmonic domains. In this paper we give an affirmative answer to the above question: in fact we prove more generally that, if a domain in an analytic Riemannian manifold has the constant flow property, then every component of its boundary is an isoparametric hypersurface. For space forms, we also relate the order of vanishing of the heat content with fixed boundary data with the constancy of the r-mean curvatures of the boundary and with the isoparametric property. Finally, we discuss the constant flow property in relation to other well-known overdetermined problems involving the Laplace operator, like the Serrin problem or the Schiffer problem.

Appendix: Proof of Lemma 27

This section is based on the paper by Ulrich Tamm [26], and we will follow closely the notation there.

Given a sequence $\{c_0,c_1,c_2,\dots \}$ form the infinite matrix

$$\begin{aligned} A=\begin{pmatrix}c_0&{}c_1&{}c_2&{}\cdots \\ c_1&{}c_2&{}c_3&{}\cdots \\ c_2&{}c_3&{}c_4&{}\cdots \\ \vdots &{}\vdots &{}\vdots &{}\ddots \\ \end{pmatrix}\end{aligned}$$

(68)

and let $A_n$ be the $n\times n$ sub-matrix in the upper left corner:

$$\begin{aligned} A_1=(c_0), \quad A_2=\begin{pmatrix} c_0&{}c_1\\ c_1&{}c_2\end{pmatrix}, \quad A_3=\begin{pmatrix} c_0&{}c_1&{}c_2\\ c_1&{}c_2&{}c_3\\ c_2&{}c_3&{}c_4\end{pmatrix}, \dots \end{aligned}$$

In this way we obtain a sequence of Hankel matrices. We will also consider shifted sequences, namely for all $k\ge 0$ we set

$$\begin{aligned} A^{(k)}=\begin{pmatrix}c_k&{}c_{k+1}&{}c_{k+2}&{}\cdots \\ c_{k+1}&{}c_{k+2}&{}c_{k+3}&{}\cdots \\ c_{k+2}&{}c_{k+3}&{}c_{k+4}&{}\cdots \\ \vdots &{}\vdots &{}\vdots &{}\ddots \\ \end{pmatrix}\end{aligned}$$

and let $A^{(k)}_n$ be the $n\times n$ sub-matrix in the upper left corner of $A^{(k)}$; by convention $A^{(0)}=A$. For all $k=0,1,\dots$ we set

$$\begin{aligned} d_n^{(k)}=\det A^{(k)}_n. \end{aligned}$$

By definition, the sequence $\{d_1^{(k)}, d_2^{(k)}, d_3^{(k)}, \dots \}$ is called the Hankel transform of the sequence $\{c_k,c_{k+1}, c_{k+2}\dots \}$.

Our interest is in the sequence

$$\begin{aligned} c_m=\left( {\begin{array}{c}2m+1\\ m\end{array}}\right) , \quad m\ge 0. \end{aligned}$$

(69)

It is just the sequence defined in (54) with offset at $m=0$:

$$\begin{aligned} c_m=a_{m+1}=\{1,3,10,35,126,462,1716, 6735, \dots \}. \end{aligned}$$

Then

$$\begin{aligned} A^{(0)}=\begin{pmatrix}1&{}3&{}10&{}35&{}\cdots \\ 3&{}10&{}35&{}126&{}\cdots \\ 10&{}35&{}126&{}462&{}\cdots \\ \vdots &{}\vdots &{}\vdots &{}\vdots &{}\ddots \\ \end{pmatrix}, \quad \text {and}\quad A^{(1)}=\begin{pmatrix}3&{}10&{}35&{}126&{}\cdots \\ 10&{}35&{}126&{}462&{}\cdots \\ 35&{}126&{}462&{}1716&{}\cdots \\ \vdots &{}\vdots &{}\vdots &{}\vdots &{}\ddots \\ \end{pmatrix}. \end{aligned}$$

From part b) of Proposition 2.1 in [26] we see that, for all n:

$$\begin{aligned} d_n^{(0)}=1, \quad d_n^{(1)}=2n+1, \end{aligned}$$

(70)

and the following expression holds for $k\ge 2$:

$$\begin{aligned} d_n^{(k)}=\Pi _{1\le i\le j\le k}\dfrac{i+j-1+2n}{i+j-1}. \end{aligned}$$

Now introduce an indeterminate x and consider the infinite matrices

$$\begin{aligned} \mathcal{A}(x)=\begin{pmatrix}1&{}x&{}x^2&{}\cdots \\ c_0&{}c_1&{}c_2&{}\cdots \\ c_1&{}c_2&{}c_3&{}\cdots \\ c_2&{}c_3&{}c_4&{}\cdots \\ \vdots &{}\vdots &{}\vdots &{}\ddots \\ \end{pmatrix}, \quad \mathcal{A}^{(1)}(x)=\begin{pmatrix}1&{}x&{}x^2&{}\cdots \\ c_1&{}c_2&{}c_3&{}\cdots \\ c_2&{}c_3&{}c_4&{}\cdots \\ c_3&{}c_4&{}c_5&{}\cdots \\ \vdots &{}\vdots &{}\vdots &{}\ddots \\ \end{pmatrix}\end{aligned}$$

and the sequences of polynomials $\{P_n(x)\}, \{P_n^{(1)}(x)\}$ defined by

$$\begin{aligned} \left\{ \begin{aligned}&P_0(x)=1, \quad P_1(x)=\begin{vmatrix} 1&x\\c_0&c_1\end{vmatrix}, \quad P_2(x)=\begin{vmatrix} 1&x&x^2\\c_0&c_1&c_2\\c_1&c_2&c_3\end{vmatrix}, \quad ...\\&P_0^{(1)}(x)=1, \quad P_1^{(1)}(x)=\begin{vmatrix} 1&x\\c_1&c_2\end{vmatrix}, \quad P_2^{(1)}(x)=\begin{vmatrix} 1&x&x^2\\c_1&c_2&c_3\\c_2&c_3&c_4\end{vmatrix}, \quad ...\end{aligned}\right. \end{aligned}$$

Recalling that $c_m=a_{m+1}$ we see that proving Lemma 27 amounts to prove that, for all n:

$$\begin{aligned} P_n(4)=(-1)^n, \quad P_n^{(1)}(4)=(-1)^n(n+1). \end{aligned}$$

(71)

Note that, by (70), the leading coefficient of $P_n(x)$ (resp. $P_n^{(1)}(x)$) is $(-1)^{n}d_n=(-1)^n$ (resp. $(-1)^nd_n^{(1)}=(-1)^n(2n+1)$). Then, the polynomials

$$\begin{aligned} t_n(x)=(-1)^{n}P_n(x), \quad t_n^{(1)}(x)=\dfrac{(-1)^n}{2n+1}P_n^{(1)}(x) \end{aligned}$$

(72)

are of degree n and monic. Taking into account (71) and (72), to prove Lemma 27 it is then sufficient to show the following fact.

Lemma 29

For the polynomials defined in (72) one has, for all n:

$$\begin{aligned} t_n(4)=1,\quad t_n^{(1)}(4)=\dfrac{n+1}{2n+1}. \end{aligned}$$

To prove Lemma 29, we use the theory of orthogonal polynomials. We first observe:

Lemma 30

The polynomials $\{t_n(x)\}$ defined in (72) coincide with the polynomials defined in equations (1.7) and (1.8) in [26]. They are orthogonal with respect to the linear operator T mapping $x^k$ to its moment $c_k$ for all k: i.e.

$$\begin{aligned} T(t_m(x)\cdot t_n(x))=0\quad \text {for all } m\ne n. \end{aligned}$$

Similarly, the polynomials $\{t_n^{(1)}(x)\}$ are orthogonal with respect to the shifted linear operator mapping $x^k$ to $c_{k+1}$ for all k.

Proof

We closely follow the exposition in [26], page 3, with a slight change of notation to suit our needs. For all $n\ge 1$, consider the following equation in the unknowns $\alpha _{n,0},\dots ,\alpha _{n,n-1}$:

$$\begin{aligned} \begin{pmatrix}c_0&{}c_1&{}c_2&{}\cdots &{}c_{n-1}\\ c_1&{}c_2&{}c_3&{}\cdots &{}c_{n}\\ c_2&{}c_3&{}c_4&{}\cdots &{}c_{n+1}\\ \ \vdots &{}\vdots &{}\vdots &{}\ddots &{}\vdots \\ c_{n-1}&{}c_n&{}c_{n+1}&{}\cdots &{}c_{2n-2}\\ \end{pmatrix}\cdot \begin{pmatrix} \alpha _{n,0}\\ \alpha _{n,1}\\ \alpha _{n,2}\\ \vdots \\ \alpha _{n,n-1}\end{pmatrix}=\begin{pmatrix} -c_n\\ -c_{n+1}\\ -c_{n+2}\\ \vdots \\ -c_{2n-1}\end{pmatrix}. \end{aligned}$$

(73)

In our case, we know that $\det A_n=1$ for all n, hence $A_n$ is non-singular and (73) has a unique solution. As in (1.8) of [26], define, for $n\ge 1$, the monic polynomials

$$\begin{aligned} \tau _n(x)\doteq x^n+\alpha _{n,n-1}x^{n-1}+\alpha _{n,n-2}x^{n-2}+\dots +\alpha _{n,1}x+\alpha _{n,0}. \end{aligned}$$

(74)

Now, by Cramer’s rule, $\alpha _{n,k}$ is the determinant of the matrix obtained by replacing the $(k+1)$-st column of $A_n$ by the column in the right-hand side of (73); in turn, this determinant is also equal to $(-1)^n$ times the coefficient of $x^k$ in $P_n(x)$. This means that

$$\begin{aligned} \tau _n(x)=t_n(x) \end{aligned}$$

as asserted. The orthogonality of the sequence $\{t_n(x)\}$ with respect to the linear mapping T has been proved in a book by Bressoud (as cited in [26]); however it is not difficult to verify that, from definition (72) of $t_n(x)$, one has $T(x^m\cdot t_n(x))=0$ for all $m=0,\dots ,n-1$, which easily implies this fact. A similar argument applies to the polynomials $t_n^{(1)}(x)$ associated to the shifted sequence. $\square$

Proof of Lemma 29 Orthogonal polynomials satisfy a three-term recursive relation: following [26], pages 5–6 and equation (1.19), one can write:

$$\begin{aligned} \left\{ \begin{aligned}&t_n(x)=(x-\alpha _n)t_{n-1}(x)-\beta _{n-1}t_{n-2}(x), \quad t_0(x)=1, t_1(x)=x-\alpha _1\\&t_n^{(1)}(x)=(x-\alpha _n^{(1)})t_{n-1}^{(1)}(x)-\beta _{n-1}^{(1)}t_{n-2}^{(1)}(x) \quad t_0^{(1)}(x)=1, t_1^{(1)}(x)=x-\alpha _1^{(1)}\end{aligned}\right. \end{aligned}$$

(75)

for suitable numerical sequences $\alpha _n\doteq \alpha _n^{(0)},\beta _n\doteq \beta _n^{(0)}, \alpha _n^{(1)}, \beta _{n-1}^{(1)}$. In fact, for arbitrary k, the coefficients $\alpha _n^{(k)}, \beta _{n-1}^{(k)}$ can be computed in terms of the coefficients $q_n^{(k)},e_n^{(k)}$ in the continued fraction expansion of $1-xF(x)$, where $F(x)=\sum _{m=0}^{\infty }c_{m+k}x^m$ (see (1.14), (1.19) and (1.20) and (1.21) in [26]):

$$\begin{aligned} \alpha _1=q_1\quad \text {and, for } n\ge 1: \quad \alpha _{n+1}^{(k)}=q_{n+1}^{(k)}+e_{n}^{(k)},\quad \beta _n^{(k)}=q_n^{(k)}\cdot e_n^{(k)}, \end{aligned}$$

(76)

For our sequence $c_m=\left( {\begin{array}{c}2m+1\\ m\end{array}}\right)$ the corresponding coefficients have been computed in the equation (2.6) of Corollary 2.1 of [26].

$$\begin{aligned} q_n^{(k)}=\dfrac{(2n+2k)(2n+2k+1)}{(2n+k-1)(2n+k)},\quad e_n^{(k)}=\dfrac{(2n-1)(2n)}{(2n+k)(2n+k+1)}. \end{aligned}$$

In particular,

$$\begin{aligned} \left\{ \begin{aligned}&q_n=q_n^{(0)}=\dfrac{2n+1}{2n-1}\\&e_n=e_n^{(0)}=\dfrac{2n-1}{2n+1}\end{aligned}\right. \quad \text {and}\quad \left\{ \begin{aligned}&q_n^{(1)}=\dfrac{(2n+2)(2n+3)}{(2n)(2n+1)}\\&e_n^{(1)}=\dfrac{(2n-1)(2n)}{(2n+1)(2n+2)}.\end{aligned}\right. \end{aligned}$$

From (76) we obtain $\alpha _1=q_1=3$, $\alpha _1^{(1)}= q_1^{(1)}=\frac{10}{3}$ and for $n\ge 2$ (after some calculations):

$$\begin{aligned} \left\{ \begin{aligned}&\alpha _n=2\\&\beta _{n-1}=1\end{aligned}\right. , \quad \text {and}\quad \left\{ \begin{aligned}&\alpha _n^{(1)}=2+\frac{4}{(2n+1)(2n-1)}\\&\beta _{n-1}^{(1)}=\dfrac{(2n-3)(2n+1)}{(2n-1)^2}\end{aligned}\right. \end{aligned}$$

(77)

From (75) and (77) one has the following recursive scheme for the sequence $\{t_n(x)\}$:

$$\begin{aligned} \left\{ \begin{aligned}&t_n(x)=(x-2)t_{n-1}(x)-t_{n-2}(x)\\&t_0(x)=1\\ {}&t_1(x)=x-3\end{aligned}\right. \end{aligned}$$

Setting $X_n=t_n(4)$ we obtain

$$\begin{aligned} \left\{ \begin{aligned}&X_n=2X_{n-1}-X_{n-2}\\&X_0=1\\ {}&X_1=1\end{aligned}\right. \end{aligned}$$

hence $X_n=1$ for all n. That is:

$$\begin{aligned} t_n(4)=1 \end{aligned}$$

for all n. This shows the first relation in the Lemma.

About the sequence $\{t_n^{(1)}(x)\}$ we know:

$$\begin{aligned} \left\{ \begin{aligned}&t_n^{(1)}(x)=(x-\alpha _n^{(1)})t_{n-1}^{(1)}(x)-\beta _{n-1}^{(1)}t_{n-2}^{(1)}(x)\\&t_0^{(1)}(x)=1\\ {}&t_1^{(1)}(x)=x-\frac{10}{3}\end{aligned}\right. \end{aligned}$$

where $\alpha _n^{(1)}$ and $\beta _{n-1}^{(1)}$ are as in (77). We are interested in the sequence $X_n=t_n^{(1)}(4),$ which obeys the recursive law:

$$\begin{aligned} \left\{ \begin{aligned}&X_n=(4-\alpha _n^{(1)})X_{n-1}-\beta _{n-1}^{(1)}X_{n-2}\\&X_0=1\\ {}&X_1=\frac{2}{3}\end{aligned}\right. \end{aligned}$$

By induction on n, let us show that $X_n=\frac{n+1}{2n+1}$. In fact this holds for $n=1$. Assume it to be true for all $k\le n-1$. By (77):

$$\begin{aligned} 4-\alpha _n^{(1)}=\dfrac{8n^2-6}{(2n-1)(2n+1)}, \quad \beta _{n-1}^{(1)}=\dfrac{(2n-3)(2n+1)}{(2n-1)^2} \end{aligned}$$

hence:

$$\begin{aligned} X_n&=(4-\alpha _n^{(1)})X_{n-1}-\beta _{n-1}^{(1)}X_{n-2}\\&=\dfrac{8n^2-6}{(2n-1)(2n+1)}\cdot \frac{n}{2n-1}-\dfrac{(2n-3)(2n+1)}{(2n-1)^2}\cdot \frac{n-1}{2n-3}\\&=\frac{n+1}{2n+1} \end{aligned}$$

which shows the claim. With this, the proof of Lemma 27 is complete.

Remark We point out that the expression of $\alpha _{n+1}^{(k)}$ in the statement of Corollary 2.3 in [26] is wrong; this is due to an incorrect algebraic manipulation of $q_{n+1}^{(k)}+e_n^{(k)}$ in the proof of Corollary 2.3 there. The correct value of $\alpha _{n+1}^{(1)}$ shown in (77) is directly computed from the expressions of $q_n^{(1)}$ and $e_n^{(1)}$ taken from Corollary 2.1, equation (2.6) of [26].