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PDE Separating Variables for 3d spherical wave equation

  1. Nov 14, 2011 #1
    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

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

    3. 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 spacial 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 seperation 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 seperation 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
  2. jcsd
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