Spherically symmetric waves¶

Spherically symmetric three-dimensional waves propagate in the radial direction $r$ only so that $u = u(r,t)$. The fully three-dimensional wave equation

$$ \frac{\partial^2u}{\partial t^2}=\nabla\cdot (c^2\nabla u) + f $$

then reduces to the spherically symmetric wave equation

$$ \begin{equation} \frac{\partial^2u}{\partial t^2}=\frac{1}{r^2}\frac{\partial}{\partial r} \left(c^2(r)r^2\frac{\partial u}{\partial t}\right) + f(r),\quad r\in (0,R),\ t>0 \thinspace . \label{_auto1} \tag{1} \end{equation} $$

Assume that the wave velocity $c$ is constant. One can easily show that the function $v(r,t) = ru(r,t)$ fulfills a standard wave equation in Cartesian coordinates. To this end, insert $u=v/r$ in

$$ \frac{1}{r^2}\frac{\partial}{\partial r} \left(c^2(r)r^2\frac{\partial u}{\partial t}\right) $$

to obtain

$$ r\left(\frac{d c^2}{dr}\frac{\partial v}{\partial r} + c^2\frac{\partial^2 v}{\partial r^2}\right) - \frac{d c^2}{dr}v \thinspace . $$

The two terms in the parenthesis can be combined to

$$ r\frac{\partial}{\partial r}\left( c^2\frac{\partial v}{\partial r}\right), $$

which is recognized as the variable-coefficient Laplace operator in one Cartesian coordinate. The spherically symmetric wave equation in terms of $v(r,t)$ now becomes

$$ \begin{equation} \frac{\partial^2u}{\partial t^2}= \frac{\partial}{\partial r} \left(c^2(r)\frac{\partial v}{\partial t} -\frac{1}{r}\frac{d c^2}{dr}v\right) + rf(r),\quad r\in (0,R),\ t>0 \thinspace . \label{_auto2} \tag{2} \end{equation} $$

In the case of constant wave velocity $c$, this equation reduces to the wave equation in a single Cartesian coordinate:

$$ \begin{equation} \frac{\partial^2u}{\partial t^2}= \frac{\partial}{\partial r} \left(c^2(r)\frac{\partial v}{\partial t}\right) + rf(r),\quad r\in (0,R),\ t>0 \thinspace . \label{wave:app:rsymm:Cart} \tag{3} \end{equation} $$

That is, any program for solving the one-dimensional wave equation in a Cartesian coordinate system can be used to solve (3), provided the source term is multiplied by the coordinate. Moreover, if $r=0$ is included in the domain, spherical symmetry demands that $\partial u/\partial r=0$ at $r=0$, which means that

$$ \frac{\partial u}{\partial r} = \frac{1}{r^2}\left( r\frac{\partial v}{\partial r} - v\right) = 0,\quad r=0, $$

implying $v(0,t)=0$ as a necessary condition. For practical applications, we exclude $r=0$ from the domain and assume that some boundary condition is assigned at $r=\epsilon$, for some $\epsilon >0$.