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In 1D the wave equation $$\frac{\partial^2 u}{\partial t^2} = c^2 \frac{\partial^2 u}{\partial r^2}$$ can be satisfied with a wave $$u(r,t) = f(r-ct).$$

In 3D the wave equation $$\frac{\partial^2 u}{\partial t^2} = c^2 \frac{1}{r^2}\frac{\partial}{\partial r}\biggl(r^2 \frac{\partial u}{\partial r}\biggr)$$ can be satisfied with a wave $$u(r,t) = \frac{1}{r}\,f(r-ct)$$ whose amplitude fades to comply with conservation of energy.

One might expect that in 2D the wave equation $$\frac{\partial^2 u}{\partial t^2} = c^2 \frac{1}{r}\frac{\partial}{\partial r}\biggl(r \frac{\partial u}{\partial r}\biggr)$$ could be satisfied with a wave $$u(r,t) = \frac{1}{\sqrt r}\,f(r-ct),$$ but that's not the case. What is the reason for it?

Yrogirg
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    Shouldn't those wave equations be ${\partial\over\partial r}\left(r^k{\partial u\over\partial r}\right)$? (i.e., no $u$ in the outer partial.) – Paul Sinclair Oct 31 '15 at 03:50
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    Well, we know the actual solution. In 2D, this solution shows that if you have a compactly supported initial condition, then you do not see a disturbance at a given point for only a finite period of time. Instead you see a disturbance at a given point forever, with decaying amplitude. Evans proves this using the method of descent (solve the problem in 3 dimensions, then treat the 2 dimensional problem as a 2 dimensional problem with no dependence on $x_3$, then write the integrals in a more natural 2D form). – Ian Oct 31 '15 at 14:43
  • Typo: "treat the 2 dimensional problem as a three dimensional problem with no dependence on $x_3$, ..." – Ian Nov 01 '15 at 18:10

1 Answers1

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Let's see what happends if we consider the $n$-dimensional wave equation for radial functions, $$ \newcommand{\pl}{\partial} \pl_t^2 u = c^2 r^{-(n-1)} \pl_r ( r^{n-1} \pl_r u) $$ and try an ansatz of the form $$ u(r,t) = r^{s} f(r-ct). $$ By direct computation, we find that \begin{align} \pl_t^2 u & = c^2 r^s f''(r-ct), \\ c^2 r^{-(n-1)} \pl_r ( r^{n-1} \pl_r u) & = c^2 r^s ( f''(r-ct) \\ & \phantom{=} + (n-1+2s) f'(r-ct) r^{-1} + s(n-2+s) f(r-ct) r^{-2} ). \end{align} These two match identically if $$ \begin{cases} n-1+2s = 0 ,\\ s(n-2+s) = 0. \end{cases} $$ There are two cases.

  1. Take $s = 0$, then $n = 1$.
  2. Take $n-2+s = 0$, then $s = -1$ and $n = 3$.

These are exactly the two cases mentioned in your question. As you see, there are no other dimensions in which such an ansatz works. Of course, it's nice to have a more conceptual explanation than just arithmetic coincidence; for this, see the comments.

Michał Miśkiewicz
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