# PDE Separating Variables for 3d spherical wave equation

1. Nov 14, 2011

### King Tony

1. The problem statement, all variables and given/known data
Just need someone else to double check more work. I just want to know if I'm separating these variables correctly.

2. Relevant equations

$$\frac{\partial^2u}{\partial t^2} = c^2\nabla^2u$$

3. The attempt at a solution

Allow $u(\rho, \theta, \phi, t) = T(t)\omega(\rho, \theta, \phi)$

where $\rho$ is the radius, $\theta$ is the cylindrical angle and $\phi$ is the azimuthal angle. Then, separating the time dependent variable is as follows,

$$\frac{T''(t)}{c^2T(t)} = \frac{\nabla^2\omega}{\omega} = -\lambda$$

From this, I know the time dependent ODE. The problem I'm having is with the spacial variable separation. Now we have,

$$\nabla^2\omega = -\lambda\omega$$

Allow $\omega(\rho, \theta, \phi) = P(\rho)\Theta(\theta)\Phi(\phi)$, then we get

$$\frac{\Theta\Phi}{\rho^2}\frac{d}{d\rho}(\rho^2 \frac{dP}{d\rho}) + \frac{P\Theta}{\rho^2sin\phi}\frac{d}{d\phi}(sin \phi \frac{d\Phi}{d\phi}) + \frac{P\Phi}{\rho^2sin^2\phi}\frac{d^2\Theta}{d \theta^2} + \lambda P\Theta\Phi = 0$$

By dividing by $\frac{P\Theta\Phi}{\rho^2sin^2 \phi}$, we can isolate the theta variable and move it to the other side of the equation, this introduces a seperation constant, $\mu$. We get:

$$\frac{sin^2 \phi}{P}\frac{d}{d\rho}(\rho^2 \frac{dP}{d\rho}) + \frac{sin \phi}{\Phi}\frac{d}{d\phi}(sin \phi \frac{d\Phi}{d\phi}) + \lambda\rho^2 sin^2 \phi = -\frac{d^2\Theta}{d \theta^2} = \mu$$

Solving the theta ODE (with periodic BC) gives $\mu = m^2, m = 0, 1, 2, ...$

and we can move on to the next step, namely, finding our ODEs for rho and phi.

$$\frac{sin^2 \phi}{P}\frac{d}{d\rho}(\rho^2 \frac{dP}{d\rho}) + \frac{sin \phi}{\Phi}\frac{d}{d\phi}(sin \phi \frac{d\Phi}{d\phi}) + \lambda\rho^2 sin^2 \phi - m^2 = 0$$

Divide by $sin^2 \phi$ and shuffle equations to get the rho and phi dependent ODEs with seperation constant $\nu$

$$\frac{1}{P}\frac{d}{d\rho}(\rho^2 \frac{dP}{d\rho}) + \lambda\rho^2 = -\frac{1}{sin \phi \Phi}\frac{d}{d\phi}(sin \phi \frac{d\Phi}{d\phi}) + \frac{m^2}{sin^2 \phi} = \nu$$

Finally, we end up with our ODEs for rho and phi,

$$\frac{d}{d\rho}(\rho^2 \frac{dP}{d\rho}) + (\lambda\rho^2 - \nu)P = 0$$

$$\frac{d}{d\phi}(sin \phi \frac{d\Phi}{d\phi}) + (-\nu sin \phi - \frac{m^2}{sin \phi})\Phi = 0$$

I have a couple questions about this, it seems that I have a couple signs mixed up (compared to my book (Haberman)) and I don't know if I have done this entirely correctly. I greatly value your responses. Thank you!

- Tony