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Math Help - Legendre Polynomial problem

  1. #1
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    Legendre Polynomial problem

    Find  \theta \ \ s.t. \ \ \Delta^{2}\phi=0, \ \ a<r<b \ \ with \ \ \phi=\sin^{2}\theta, \ \ both \ \ when \ \ r=a \ \ and \ \ r=b
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  2. #2
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    Quote Originally Posted by silversand View Post
    Find  \theta \ \ s.t. \ \ \Delta^{2}\phi=0, \ \ a<r<b \ \ with \ \ \phi=\sin^{2}\theta, \ \ both \ \ when \ \ r=a \ \ and \ \ r=b
    Not clear, provide context and explain notation please.

    CB
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  3. #3
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    Quote Originally Posted by silversand View Post
    Find  \theta \ \ s.t. \ \ \Delta^{2}\phi=0, \ \ a<r<b \ \ with \ \ \phi=\sin^{2}\theta, \ \ both \ \ when \ \ r=a \ \ and \ \ r=b
    I'm guessing what you mean is find \phi not \theta . From your BC, it's desirable to switch to polar coords, so your PDE becomes

     <br />
\phi_{rr} + \frac{\phi_r}{r} + \frac{\phi_{\theta \theta}}{r^2} = 0<br />
with your BC \phi(a,\theta) = \phi(b,\theta) = \frac{1}{2} - \frac{\cos 2 \theta}{2}

    We'll shift the BC a bit by introducing \psi = \phi - \frac{1}{2} so the PDE is the same but the new BCs is \psi(a,\theta) = \psi(b,\theta) = - \frac{\cos 2 \theta}{2}. Fortunately for us, this problem admits separable solutions so

     <br />
\psi = \left( c_1 r^2 + \frac{c_2}{r^2}\right) \left( A \sin 2 \theta + B \cos 2 \theta \right)<br />

    Applying the boundary condtions gives us (we can set B = 1 wlog)

     <br />
c_1 = - \frac{1}{2(a^2+b^2)},\;\;\; c_2 = -\frac{1}{2} \frac{a^2b^2}{a^2+b^2},<br />

    which leads to the final solution

     <br />
\phi = \frac{1}{2} - \frac{1}{2(a^2+b^2)} \left(r^2 + \frac{a^2b^2}{r^2} \right)\cos 2 \theta.<br />

    Side note: Title says Legendre polynomial - not sure why? They typically arise in 3D geometries but nothing in this post suggest that the problem is in \mathbb R^3.
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