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Mathematics LibreTexts

4.2: Higher Dimensions

[ "article:topic", "showtoc:no" ]
  • Page ID
    2149
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    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.

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