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# 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

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

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

\label{uM}
u(x,t)=\lim_{r\to0} M(r,t).

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

\label{EPD}
M_{rr}+\frac{n-1}{r}M_r=c^{-2}M_{tt}.

This equation (\ref{EPD}) is called Euler-Poisson-Darboux equation.

### Contributors

• Integrated by Justin Marshall.