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math121b:04-06 [2020/04/06 07:28] pzhou |
math121b:04-06 [2020/04/06 10:24] (current) pzhou [Steady State temperature distribution inside a unit ball] |
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Then the solulution for $u$ is obtained by setting $\lambda = \lambda_r=0$ | Then the solulution for $u$ is obtained by setting $\lambda = \lambda_r=0$ | ||
$$ u(r, \theta, \phi) = \sum_{l=0}^\infty \sum_{m=0}^l r^l [a_{l m} \cos(m\phi) + b_{l m} \sin(m \phi) ] P_l^m(\cos \theta). | $$ u(r, \theta, \phi) = \sum_{l=0}^\infty \sum_{m=0}^l r^l [a_{l m} \cos(m\phi) + b_{l m} \sin(m \phi) ] P_l^m(\cos \theta). | ||
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+ | To obtain the coefficients $a_{lm}, b_{lm}$ from $f(\theta, \phi)$, we uses orthogonality of these functions $P_l^m(\cos \theta) \cos(m \phi), P_l^m(\cos \theta) \sin(m \phi)$ on the two sphere with volume form $\sin \theta d\theta d\phi$. For example, we claim that, if $(l, m) \neq (l', m')$, then | ||
+ | $$ \int_{\phi = 0}^{2\pi} \int_{\theta=0}^\pi P_l^m(\cos \theta) \cos(m \phi) P_{l' | ||
+ | Indeed, integrating $d\phi$, we see that if $m \neq m'$ the result is zero; if $m=m'$ but $l \neq l'$, then we use the orthogonality of associated Legendre functions (see section 12.10 of Boas), to show that the integral is zero. | ||