Skip to main content
\(\newcommand{\id}{\mathrm{id}}\) \( \newcommand{\Span}{\mathrm{span}}\) \( \newcommand{\kernel}{\mathrm{null}\,}\) \( \newcommand{\range}{\mathrm{range}\,}\) \( \newcommand{\RealPart}{\mathrm{Re}}\) \( \newcommand{\ImaginaryPart}{\mathrm{Im}}\) \( \newcommand{\Argument}{\mathrm{Arg}}\) \( \newcommand{\norm}[1]{\| #1 \|}\) \( \newcommand{\inner}[2]{\langle #1, #2 \rangle}\) \( \newcommand{\Span}{\mathrm{span}}\)
Mathematics LibreTexts

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