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Help with Differential equation for Spherical Harmonics.

  1. Jun 23, 2010 #1
    [tex] u(r,\theta,\phi)=R(r) Y(\theta,\phi)[/tex]

    Where Y is the spherical harmonics

    [tex] \frac{\partial^2 Y}{\partial \theta^2} + cot\theta \frac{\partial Y}{\partial \theta} + csc^2 \frac{\partial^2 Y}{\partial \phi^2} + \mu Y = 0[/tex]

    The book said this equation has nontrivial solutions when

    [tex]\mu = \mu_n = n(n+1) \hbox { , for } n=0,1,2.......[/tex]

    Can anyone explain why?
     
  2. jcsd
  3. Jun 24, 2010 #2
    Are you able to find solutions for the Legendre equation?

    [tex]\left[(1-x)^2y'\right]' + \mu y = 0[/tex]

    I ask this, b/c you should be to separate the variables a second time; that is, to assume that:

    [tex]Y(\theta,\phi) = \Theta(\theta) \Phi(\phi)[/tex]

    If you plug in this for your second above, you'll generate two (independent) differential equations: one in [tex]\theta[/tex] and one in [tex]\phi[/tex]. One of these equations will be simple and straightforward, w/ a straightforward solution + boundary conditions. Something to the effect of:

    [tex]y'' = -m^2y[/tex]

    which can be solved by inspection.

    The second equation will be (roughly) in the form of the Legendre equation above.

    If you go through the math, you should be able to arrive at the result you state, namely, that:

    [tex]\mu = \mu_n = n(n+1)[/tex]

    We can go into more detail if you like.
     
  4. Jun 25, 2010 #3
    Thanks for the reply and sorry I did not respond yesterday. I was trimming trees the whole day!!! I'll look into this and respond back later.

    Thanks
    Alan
     
  5. Jun 25, 2010 #4
    I was imcomplete on my original post. This is from Poisson's equation where:

    [tex]\nabla^2u(r,\theta,\phi) = -ku(r,\theta,\phi)[/tex] where

    [tex]r^2\frac{\partial^2R}{\partial r^2} + 2r\frac{\partial R}{\partial r} + (r^2k-\mu)R = 0 \;\hbox{ and } \;\;\frac{\partial^2Y}{\partial \theta^2} + cot\theta \frac{\partial Y}{\partial \theta} + csc^2 \theta \frac{\partial^2Y}{\partial \phi^2} - \mu Y = 0[/tex]

    I have search through the book, for Legendre function, only [itex]\mu = n(n+1) \;\;\; n=0,1,2,3....[/itex] produce polynomial solution and with bounded solution on [-1,1]. Other values will produce infinite series and have unbounded solution at -1 and 1.

    Is that the only reason?
     
  6. Jun 26, 2010 #5
    What you're trying to do is solve the equation:

    [tex]\nabla^2u = -ku[/tex]

    where u is expressed in spherical coordinates, and where the Laplacian takes the form:

    [tex]\nabla^2 u = \frac{1}{r^2}\frac{\partial}{\partial r}\left(r^2 \frac{\partial u}{\partial r}\right) + \frac{1}{r^2\sin\theta}\frac{\partial }{\partial \theta}\left(\sin \theta \frac{\partial u}{\partial \theta}\right) + \frac{1}{r^2\sin^2\theta}\frac{\partial^2u}{\partial \phi^2}[/tex]

    Plugging this into the expression we wish to solve, and doing a little algebra, we arrive at:

    [tex]r^2u_{rr} + 2ru_r + u_{\theta \theta} + \cot \theta \cdot u_{\theta} + \csc^2\theta \cdot u_{\phi \phi} + kr^2u = 0[/tex]

    You start w/ the "ansatz" that the radial component of u is separable from the rest of the expression, that is:

    [tex]u(r,\theta,\phi) = R(r)Y(\theta,\phi)[/tex]

    Indeed, the equation is separable and upon substitution takes the form:

    [tex]\frac{r^2R''}{R} + \frac{2rR'}{R} + kr^2 + \frac{Y_{\theta\theta}}{Y} +
    \cot \theta \frac{Y_{\theta}}{Y} + \csc^2\theta \frac{Y_{\phi \phi}}{Y} = 0[/tex]

    Using the standard arguments that are used in separation of variables, we say that the "R" component of the equation must be constant, and equal to a constant let's call it [tex]\mu[/tex], and that the "Y" component likewise must be constant, and equal to the opposite value of "R", let's call it [tex]-\mu[/tex]. We arrive at two independent differential equations.

    The equation for R:

    [tex]\frac{r^2R''}{R} + \frac{2rR'}{R} + kr^2 = \mu[/tex]

    [tex]r^2R'' + 2rR' + \left(kr^2 - \mu\right)R = 0[/tex]

    and the equation for Y:

    [tex]\frac{Y_{\theta\theta}}{Y} + \cot\theta \frac{Y_{\theta}}{Y} + \csc^2\theta \frac{Y_{\phi\phi}}{Y} = -\mu[/tex]

    [tex]Y_{\theta\theta} + \cot \theta Y_{\theta} + \csc^2\theta Y_{\phi\phi} + \mu Y = 0[/tex]

    Now, which is these two equations do you want to work w/ first? In your initial post you mentioned only Y, but in your most recent post you brought up R as well..
     
  7. Jun 26, 2010 #6
    Let's keep working w/ the second equation, Y, which, as I indicated earlier, can be separated much like the radial component was separated:

    [tex]Y(\theta,\phi) = \Theta(\theta)\Phi(\phi)[/tex]

    Plugging this into the above expression for Y, we have:

    [tex]\sin^2\theta \cdot \mu + \frac{\sin \theta}{\theta} \frac{\partial}{\partial \theta}\left[\sin \theta \frac{\partial \Theta}{\partial \theta}\right] + \frac{1}{\Phi}\frac{\partial^2 \Phi}{\partial \phi^2} = 0[/tex]

    So indeed, the equation is separable, and again, each "component" of the equation must be equal to a constant. Let's call this constant [tex]m^2[/tex], and let's set the [tex]\Phi[/tex] component equal to the negative of this constant. We have then:

    [tex]\frac{\Phi''}{\Phi} = -m^2[/tex]

    The solution to this equation can be obtained by inspection:

    [tex]\Phi(\phi) = Ke^{im\phi}[/tex]

    The boundary condition on [tex]\Phi[/tex] specifies that [tex]\Phi(0) = \Phi(2\pi)[/tex], so we have that:

    [tex]K = K\left[\cos\left(2m\pi\right) + i \sin\left(2m\pi\right)\right][/tex]

    The boundary condition on sin imposes that [tex]2m = n[/tex], or that [tex]m=n/2[/tex], where n is an integer. The boundary condition on cos imposes that m is an integer. We choose m as an integer (0,1,2,...) to satisfy both boundary conditions.

    So we've now solved for one of the "components" of u, the azimuthal component. We have, specifically, that:

    [tex]u(r,\theta,\phi) = R(r)\Theta(\theta) e^{im\phi}[/tex]

    where m=0,1,2,3...

    The Legendre equation comes into play, in solving for the other components..
     
  8. Jun 27, 2010 #7
    Thanks for the reply. I understand all the equation you wrote, it is exactly like what I have except [itex]\Theta(\theta)\rightarrow \; P_n^m(cos\theta)[/itex]. I just stop type it all and try to put in as little as possible because my experience from the past if I type in step by step, nobody reply because it is too long and people fell asleep!!! :rofl: :rofl: :rofl: :rofl:

    [itex]\mu[/itex] can only have non trivial solution if [itex]\mu=n(n+1)[/itex]

    [tex] \frac{\partial^2Y}{\partial \theta^2} + cot\theta \frac{\partial Y}{\partial \theta} + csc^2 \theta \frac{\partial^2Y}{\partial \phi^2} + \mu Y = 0[/tex]

    [tex] Y(\theta,\phi) = \Theta(\theta)\Phi(\phi) [/tex]

    [tex] \frac{\partial^2Y}{\partial \theta^2} + cot\theta \frac{\partial Y}{\partial \theta} + csc^2 \theta \frac{\partial^2Y}{\partial \phi^2} + \mu Y = \frac{\Theta ''}{\Theta} + cot\theta \frac{\Theta '}{\Theta} + csc^2 \frac{\Phi ''}{\Phi} + \mu = 0[/tex]

    [tex]\Rightarrow\; sin^2\theta \frac{\Theta ''}{\Theta} + sin^2\theta cot\theta \frac{\Theta '}{\Theta} + sin^2\theta \mu = -\frac{\Phi ''}{\Phi} = m^2 [/tex]

    Therefore:

    [tex]\Phi''+m^2\Phi = 0 \Rightarrow \Phi(\phi) = e^{jm\phi}[/tex]

    [tex] \Theta '' + cot\theta \Theta ' + (\mu-csc^2\theta m^2) = 0[/tex]

    Let [itex] s= cos\theta [/itex]

    [tex] \Theta '' + cot\theta \Theta ' + (\mu-csc^2\theta m^2) = (1-s^2)\frac{d^2\Theta}{ds^2} - 2s\frac{d\Theta}{ds} + (\mu -\frac{1}{1-s^2}m^2)\Theta = 0[/tex]

    The Last one is Assoiate Legendre equation and the solution is [itex]P^m_n(s)=P_n^m(cos\theta)[/itex].

    [tex] Y(\theta,\phi) = \Theta(\theta)\Phi(\phi) = e^{jm\phi}P_n^m(cos\theta)[/tex]

    Still that does not answer my question why [itex] \mu=n(n+1)[/itex] except the solution is bounded if [itex] \mu=n(n+1)[/itex] like the book said.
     
    Last edited: Jun 27, 2010
  9. Jun 30, 2010 #8
    True, but the reason I'm writing it all out is to make sure I've got all the equations right.. :)

    Easy to get jumbled up in all this..

    I'm not sure I'm getting the same expression you are for the [tex]\Theta[/tex] expression.

    Recapping where we are, we've gotten the expression separated out as:

    [tex]u(r,\theta,\phi) = R(r)\Theta(\theta)\Phi(\phi)[/tex]

    and we've been able to establish that:

    [tex]\Phi(\phi) = Ke^{im\phi}[/tex]

    where m=1,2,3,...

    Going back to the expression for [tex]\Theta[/tex], what I get is:

    [tex]\mu \sin^2\theta + \frac{\sin \theta}{\Theta}\left[\Theta' \sin \theta\right]' = m^2[/tex]

    [tex]\mu \sin^2\theta + \frac{\sin \theta}{\Theta}\left(\Theta'' \sin \theta + \Theta' \cos \theta\right) = m^2[/tex]

    [tex]\mu \sin^2\theta + \sin^2\theta \frac{\Theta''}{\Theta} + \sin\theta \cos\theta \frac{\Theta'}{\Theta} = m^2[/tex]

    [tex]\frac{\Theta''}{\Theta} + \cot \theta \frac{\Theta'}{\Theta} + \left(\mu - m^2\csc^2\theta\right) = 0[/tex]

    [tex]\Theta'' + \cot\theta \cdot \Theta' + \left(\mu -m^2\csc^2\theta\right)\cdot \Theta = 0[/tex]

    This seems to be slightly different than the expression you arrived at (off by a factor of [tex]\Theta[/tex] in the last term)..??
     
  10. Jun 30, 2010 #9
    You're correct about deriving the associated Legendre equation.

    I'll step through the math (for my own edification, if nothing else, and for the sake of completion), and then try to answer your question in the next post.

    First, starting with:

    [tex]\Theta'' + \cot \theta \cdot \Theta' + \left(\mu - m^2\csc^2\theta\right) \cdot \Theta = 0[/tex]

    Making the subtitution:

    [tex]s = \cos \theta[/tex]

    [tex]\frac{d\Theta}{d\theta} = \frac{d\Theta}{ds} \cdot \frac{ds}{d\theta} = -\sin\theta \cdot \frac{d\Theta}{ds}[/tex]

    [tex]\frac{d^2\Theta}{d\theta^2} = -\cos\theta \cdot \frac{d\Theta}{ds} + \sin^2\theta \cdot \frac{d^2\Theta}{ds^2}[/tex]

    and so:

    [tex]\sin^2\theta \cdot \frac{d^2\Theta}{ds^2} - 2\cos\theta \cdot \frac{d\Theta}{ds} + \left(\mu - m^2\csc^2\theta\right) \cdot \Theta = 0 [/tex]

    [tex]\left(1-s^2\right)\cdot \Theta'' - 2s \cdot \Theta' + \left(\mu - \frac{m^2}{1-s^2}\right) \cdot \Theta = 0[/tex]

    which, you're correct, is the associated Legendre equation of order m.

    I'll try to answer your question in the next post..
     
  11. Jul 3, 2010 #10
    I really think [itex]\mu=n(n+1)[/itex] is the only one that give bounded solution on [-1,1]. All other values will let the solution tend to infinity at -1 or 1. It can only satisfy on (-1,1) not [-1,1]. In expansion of a function, it is required for [itex]0 < \theta < 2\pi[/itex] which translate to [-1,1] when substitude [itex]s=cos\theta[/itex].
     
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