Solving the inhomogeneous wave equation in polar coordinates?

Question

Imagine you have the following wave equation, with an excitation term $\mu/r^2$: $$ \frac{1}{c^2}\frac{\partial^2 \psi}{\partial t^2} - \Delta \psi = \frac{\mu}{r^2} $$ $$ \frac{1}{c^2}\frac{\partial^2 \psi}{\partial t^2} - \frac{\partial^2 \psi}{\partial r^2} - \frac{1}{r}\frac{\partial \psi}{\partial r} - \frac{1}{r^2} \frac{\partial^2 \psi}{\partial \theta^2} = \frac{\mu}{r^2} $$ Using separation of variables: $\psi(r, \theta, t) = \phi(r, \theta) T(r)$: $$ \frac{d^2T}{dt^2} = - k^2 c^2 T\\ \; \\ \frac{\partial^2 \phi}{\partial r^2} + \frac{1}{r}\frac{\partial \phi}{\partial r} + \frac{1}{r^2} \frac{\partial^2 \phi}{\partial \theta^2} - \frac{\mu}{r^2} =-k^2\phi $$ The first equation is simply harmonic motion: $$ T(t) \propto e^{ikc t} $$ The second one is a little bit trickier. Because of the linearity of the equation I started solving the homogeneous equation: $$ \frac{\partial^2 \phi}{\partial r^2} + \frac{1}{r}\frac{\partial \phi}{\partial r} + \frac{1}{r^2} \frac{\partial^2 \phi}{\partial \theta^2} = -k^2\phi \\ \; \\ \phi(r, \theta) \propto J_m(kr)e^{im\theta} $$ Where $J_m(x)$, is the Bessel Function of first kind. Then I assumed a particular solution would look like: $$ \phi(r, \theta) = u(r, \theta) J_m(kr)e^{im\theta} $$ And my problem would be reduced to find $u(r, \theta)$. But, I cannot continue. Could someone help? What I'm doing wrong? Any hint?

Answer 1

If you have in inhomogenous term only dependent on $r$, you should definitly do a seperation of variables for $r$ and not $t$. In fact, as @naturallyInconsistent has already pointed out, you can even do a seperation of variables for all three of them. I don't know if this is what you want, or if you want to keep it more general by using $\phi(r,\vartheta)$, but just in case I have written down the calculations for seperating all three variables:

With $\psi(t,r,\vartheta)=T(t)R(r)\Theta(\vartheta)$ you get: $$\frac{\mathrm{d}^2T}{\mathrm{d}t^2} =-k^2c^2T$$ $$\frac{\mathrm{d}^2R}{\mathrm{d}r^2} +\frac{1}{r}\frac{\mathrm{d}R}{\mathrm{d}r} =\frac{1}{r}\frac{\mathrm{d}}{\mathrm{d}r}\left(r\frac{\mathrm{d}R}{\mathrm{d}r}\right) =-\left(\frac{C}{r^2}+k^2\right)R$$ $$\frac{\mathrm{d}^2\Theta}{\mathrm{d}\vartheta^2}+\mu =C\Theta$$ with a constant $C\in\mathbb{R}$. (Keep in mind here, that from your second equation to your forth equation, the sign in front of $\mu$ has switched. I kept the upper sign.) If you now multiply the first equation with $R\Theta$ and divide by $c^2$, multiply the second equation with $T\Theta$ and multiply the third equation with $TR$ and divide by $r^2$ and then substract the second and the third equation from the first, you get exactly your wave equation.

• The equation for $T$ can be solved quite easily with the exponential function as you have already explained. The solution is $T(t)=e^{ikct}$.
• The equation for $\Theta$ can be simplified, when replacing $\Theta$ with $\widetilde\Theta=\Theta-\frac{\mu}{C}$ to get $\frac{\mathrm{d}^2\widetilde\Theta}{\mathrm{d}\vartheta^2} =C\widetilde\Theta$. Since $\Theta$ (and therefore $\widetilde\Theta$) is periodic with $\Theta(\vartheta+2\pi)=\Theta(\vartheta)$ (or $\widetilde\Theta(\vartheta+2\pi)=\widetilde\Theta(\vartheta)$), a suitable choice for $C$ is $-n^2$ with a natural $n$, so that $\Theta(\vartheta)=a\sin(n\vartheta)+b\cos(n\vartheta)+\frac{\mu}{n^2}$ with constants $a,b\in\mathbb{R}$.

Answer 2

You have quite a few mistakes. You have sign mistakes.

And then it is weird that you knew how to do separation of variables to get the time dependence out. Why do you not see that angle dependence is also the same? Then you would have an equation purely for the radial dependence, and the solution should look like Bessel too, just with some alteration. You have to be consistent in how to apply the mathematical machinery.