4.2: Higher Dimensions

Set

$$
\Box u=u_{tt}-c^2\triangle u,\ \ \triangle\equiv\triangle_x=\partial^2/\partial x_1^2+\ldots+
\partial^2/\partial x_n^2,
\]

and consider the initial value problem

$$\begin{eqnarray} \label{wavehigher1} \Box u&=&0\ \ \ \mbox{in} \mathbb{R}^n\times\mathbb{R}^1\\ \label{wavehigher2} u(x,0)&=&f(x)\\ \label{wavehigher3} u_t(x,0)&=&g(x), \end{eqnarray}$$

where $f$ and $g$ are given $C^2(\mathbb{R}^2)$-functions.

By using spherical means and the above d'Alembert formula we will derive a formula for the solution of this initial value problem.

Method of Spherical means

Define the spherical mean for a $C^2$-solution $u(x,t)$ of the initial value problem by

$$\begin{equation} \label{mean1} M(r,t)=\frac{1}{\omega_n r^{n-1}}\int_{\partial B_r(x)}\ u(y,t)\ dS_y, \end{equation}$$

where

$$
\omega_n=(2\pi)^{n/2}/\Gamma(n/2)
\]

is the area of the n-dimensional sphere, $\omega_n r^{n-1}$ is the area of a sphere with radius $r$.

From the mean value theorem of the integral calculus we obtain the function $u(x,t)$ for which we are looking at by
$$\begin{equation} \label{uM} u(x,t)=\lim_{r\to0} M(r,t). \end{equation}$$
Using the initial data, we have
$$\begin{eqnarray} \label{mean2} M(r,0)&=&\frac{1}{\omega_n r^{n-1}}\int_{\partial B_r(x)}\ f(y)\ dS_y=:F(r)\\ \label{mean3} M_t(r,0)&=&\frac{1}{\omega_n r^{n-1}}\int_{\partial B_r(x)}\ g(y)\ dS_y=:G(r), \end{eqnarray}$$
which are the spherical means of $f$ and $g$.

The next step is to derive a partial differential equation for the spherical mean. From definition ($\ref{mean1}$) of the spherical mean we obtain, after the mapping $\xi=(y-x)/r$, where $x$ and $r$ are fixed,
$$M(r,t)=\frac{1}{\omega_n }\int_{\partial B_1(0)}\ u(x+r\xi,t)\ dS_\xi.$$
It follows
$$\begin{eqnarray*} M_r(r,t)&=&\frac{1}{\omega_n }\int_{\partial B_1(0)}\ \sum_{i=1}^n u_{y_i}(x+r\xi,t)\xi_i\ dS_\xi\\ &=&\frac{1}{\omega_n r^{n-1}}\int_{\partial B_r(x)}\ \sum_{i=1}^n u_{y_i}(y,t)\xi_i\ dS_y. \end{eqnarray*}$$
Integration by parts yields
$$\frac{1}{\omega_n r^{n-1}}\int_{B_r(x)}\ \sum_{i=1}^n u_{y_iy_i}(y,t)\ dy$$
since $\xi\equiv (y-x)/r$ is the exterior normal at $\partial B_r(x)$. Assume $u$ is a solution of the wave equation, then
$$\begin{eqnarray*} r^{n-1}M_r&=&\frac{1}{c^2\omega_n}\int_{B_r(x)}\ u_{tt}(y,t)\ dy\\ &=&\frac{1}{c^2\omega_n }\int_0^r\ \int_{\partial B_c(x)}\ u_{tt}(y,t)\ dS_ydc. \end{eqnarray*}$$
The previous equation follows by using spherical coordinates. Consequently
$$\begin{eqnarray*} (r^{n-1}M_r)_r&=&\frac{1}{c^2\omega_n}\int_{\partial B_r(x)}\ u_{tt}(y,t)\ dS_y\\ &=&\frac{r^{n-1}}{c^2}\frac{\partial^2}{\partial t^2}\left(\frac{1}{\omega_n r^{n-1}} \int_{\partial B_r(x)}\ u(y,t)\ dS_y\right)\\ &=&\frac{r^{n-1}}{c^2}M_{tt}. \end{eqnarray*}$$
Thus we arrive at the differential equation
$$(r^{n-1}M_r)_r=c^{-2}r^{n-1}M_{tt},$$
which can be written as
$$\begin{equation} \label{EPD} M_{rr}+\frac{n-1}{r}M_r=c^{-2}M_{tt}. \end{equation}$$
This equation ($\ref{EPD}$) is called Euler-Poisson-Darboux equation.

Contributors and Attributions

• Prof. Dr. Erich Miersemann (Universität Leipzig)

• Integrated by Justin Marshall.