PDE Separating Variables for 3d spherical wave equation

[King Tony]

Homework Statement

Just need someone else to double check more work. I just want to know if I'm separating these variables correctly.

Homework Equations

[tex]\frac{\partial^2u}{\partial t^2} = c^2\nabla^2u[/tex]

The Attempt at a Solution

Allow [itex]u(\rho, \theta, \phi, t) = T(t)\omega(\rho, \theta, \phi)[/itex]

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

[tex]\frac{T''(t)}{c^2T(t)} = \frac{\nabla^2\omega}{\omega} = -\lambda[/tex]

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

[tex]\nabla^2\omega = -\lambda\omega[/tex]

Allow [itex]\omega(\rho, \theta, \phi) = P(\rho)\Theta(\theta)\Phi(\phi)[/itex], then we get

[tex]\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[/tex]

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

[tex]\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[/tex]

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

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

[tex]\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[/tex]

Divide by [itex]sin^2 \phi[/itex] and shuffle equations to get the rho and phi dependent ODEs with separation constant [itex]\nu[/itex]

[tex]\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[/tex]

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

[tex]\frac{d}{d\rho}(\rho^2 \frac{dP}{d\rho}) + (\lambda\rho^2 - \nu)P = 0[/tex]

[tex]\frac{d}{d\phi}(sin \phi \frac{d\Phi}{d\phi}) + (-\nu sin \phi - \frac{m^2}{sin \phi})\Phi = 0[/tex]

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

 

[mighty2000]

Dear Tony,

Thank you for sharing your work with us. It looks like you have separated the variables correctly. However, it is always a good idea to double check your work and make sure you have the correct signs and constants. I would recommend going through your calculations again and carefully checking each step. It may also be helpful to compare your work with the solutions in your textbook or with other resources online.

Additionally, I noticed that you have not included the boundary conditions for your problem. It is important to consider these when solving the ODEs for rho and phi, as they will affect the solutions. Make sure to include these in your calculations and check that your final solutions satisfy the boundary conditions.

Overall, your work looks good and it seems that you have a good understanding of separating variables and solving ODEs. Keep up the good work and don't hesitate to reach out if you have any further questions.