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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?

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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.

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  • $\begingroup$ 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

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