Consider the wave equation in $\mathbb{R}^3$: $$\frac{\partial^2 u }{\partial t^2}= \nabla^2u$$

It is known that there are solutions with spherical symmetry of the form: $$u(x,t)=\frac{v(r,t)}{r} ; r^2=x^2+y^2+z^2$$ With $v(r,t)$ being a solution of the 1-dimensional wave equation in the coordinate $r$.

My question is, is there any solution with spherical symmetry of the 3-dimensional wave equation which is not of this form?


No. A solution $u(x,t)$ has spherical symmetry if and only if it can be written as $w(r,t)$ for some function $w$. This is $v(r,t)/r$, where $v(r,t) = r w(r,t)$. Substituting this function into the wave equation in $\mathbb R^3$, you get $$ \frac{1}{r} \frac{\partial^2 v}{\partial t^2} = \frac{1}{r} \frac{\partial^2 v}{\partial r^2} $$ so to be a solution, $v$ must satisfy the 1D wave equation.

  • $\begingroup$ Is there any proof or source for the implicit affirmation that such $w(r,t)$ must be of the form $w(r,t)=\frac{v(r,t)}{r}$? $\endgroup$ – Mario Jun 22 '17 at 15:14
  • $\begingroup$ Change to spherical coordinates. There is no dependence on $\phi$ or $\theta$, so ... $\endgroup$ – Robert Israel Jun 22 '17 at 16:56

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.