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I'm solving a scalar Wave Equation

$$\left(\frac{1}{c^2}\frac{\partial^2}{\partial t^2}-\nabla^2\right)u(\vec{r},t)=0$$

under the assumption that $u$ only depends on the magnitude of the position vector: $u(\vec{r},t)=u(r,t)$.

Assuming that $u(r,t)=T(t)\Psi(r)$, we have

$$\frac{\Psi(r)}{c^2}T''(t)=T(t)\nabla^2\Psi(r) \implies\frac{T''(t)}{c^2T(t)}=\frac{1}{\Psi(r)}\nabla^2\Psi(r)=-k^2$$

for some (possibly complex) constant $k$.

The differential equation for $T(t)$ is easily solved, we have

$$T(t) = Ae^{i\omega t}+Be^{-i\omega t}$$

for arbitrary values of $A,B$ and $\omega=ck$.

The spatial equation can be rewritten as

$$\left(\nabla^2+k^2\right)\Psi(r) = 0$$

Now, since $\Psi(r)$ only depends on the radial distance, we can plug in the Laplacian in spherical coordinates to obtain an ODE for $\Psi(r)$.

$$ \frac{1}{r^2}\frac{d}{dr}\left(r^2\frac{d}{dr}\Psi(r)\right)+k^2\Psi(r) = 0 $$

Calculating the derivatives and rearranging, we have

$$r^2\Psi''(r)+2r\Psi'(r)+k^2r^2\Psi(r) = 0$$

Initially, I thought that this would require Bessel Functions, as it looks very similar to the Bessel Differential Equation.

However, I was told that this can be solved using only elementary functions and I was given the hint to write $\Psi(r) = r^af(r)$ for some other function of $f(r)$ and an unknown power $a$.

With this, we have the following expressions for the derivatives that appear in our ODE:

\begin{align} \Psi'(r) &=r^af'(r)+ar^{a-1}f(r)\\ \Psi''(r) &= r^af''(r)+2ar^{a-1}f'(r)+a(a-1)r^{a-2}f(r) \end{align}

Plugging this into our ODE becomes quite messy, but we have

\begin{align} r^{a+2}f''(r)+2ar^{a+1}f'(r)+a(a-1)r^af(r)+2r^{a+1}f'(r)+2ar^af(r)+k^2r^{a+2}f(r) = 0 \end{align}

However, this is still a second order differential equation that seems much more complicated than the one we had before, so I'm not sure what this accomplished. Could anybody give me a hint on how to proceed here?

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Simplifying,

$$r^{a}\{r^2f''(r)+2(a+1)rf'(r)+[a(a+1)+k^2r^2]f(r)\}=0$$

If we put $a=-1$, then

\begin{align*} f''(r)+k^2f(r) &=0 \\ f(r) &= Ae^{ikx}+Be^{-ikx} \\ \psi(r) &= \frac{Ae^{ikx}+Be^{-ikx}}{r} \end{align*}

where the amplitude decays as $\dfrac{1}{r}$ that is consistent with spherical waves.

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  • $\begingroup$ Of course, thanks. I tried to group by order of derivatives and that got me nowhere, I should have thought of grouping by powers of $r$ - that makes the whole thing rather obvious, once you do that. Thanks again. $\endgroup$ – Tom Jan 15 '17 at 13:09
  • $\begingroup$ I'm curious, though: while it's pretty clear that $a=-1$ works in this way and $a=-1$ is the natural choice, are there solutions for other values of $a$, and if yes, how would one go about finding them? $\endgroup$ – Tom Jan 15 '17 at 13:19

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