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--- math121b:04-06 [2020/04/06 06:41]
pzhou created
+++ math121b:04-06 [2020/04/06 10:24] (current)
pzhou [Steady State temperature distribution inside a unit ball]
@@ Line 10: / Line 10: @@
 $$ u (r, \theta, \phi) = R(r) \Theta(\theta) \Phi(\phi). $$
 Then $  \Delta u = \lambda u$ is equivalent to
-$$ \frac{1}{u} \Delta u = \frac{1}{R} \frac{1}{r^2} \left( \frac{\d}{\d r} r^2  \frac{\d R}{\d r} \right) +
+$$ \lambda = \frac{1}{u} \Delta u = \frac{1}{R} \frac{1}{r^2} \left( \frac{\d}{\d r} r^2  \frac{\d R}{\d r} \right) +
 \frac{1}{\Theta} \frac{1}{r^2} \frac{1}{\sin \theta} \frac{\d }{\d \theta} ( \sin \theta \frac{\d \Theta }{\d \theta}) + \frac{1}{\Phi } \frac{1}{r^2 \sin^2 \theta} \frac{\d^2 \Phi}{\d \phi^2}. $$
 which in turns breaks into equations
@@ Line 16: / Line 16: @@
 $$ \frac{1}{\sin \theta} \frac{\d }{\d \theta} ( \sin \theta \frac{\d \Theta}{\d \theta}) +  \frac{\lambda_\phi}{\sin^2 \theta} \Theta  = \lambda_\theta \Theta  $$
 $$ \frac{1}{R} \frac{1}{r^2} \left( \frac{\d}{\d r} r^2  \frac{\d R}{\d r} \right) + \frac{\lambda_\theta}{r^2} = \lambda_r \quad \Rightarrow  \left( \frac{\d}{\d r} r^2  \frac{\d R}{\d r} \right) +  (\lambda_\theta - \lambda_r r^2) R =  0 $$
+The total eigenvalue $\lambda = \lambda_r$.
 The equation about $\Phi$ has $\sin(m\phi), \cos(m\phi)$ as eigenfunctions, with eigenvalues $\lambda_\phi = -m^2$.
@@ Line 21: / Line 22: @@
 The equation about $\Theta$ has associated Legendre polynomial as solution, recall that $y(\theta) = P_l^m (\cos \theta)$ solves equation (Problem 12.10.2)
 $$  \frac{1}{\sin \theta} \frac{\d }{\d \theta} ( \sin \theta \frac{\d y}{\d \theta}) + \left( l(l+1) -  \frac{m^2}{\sin^2 \theta} \right) y = 0 $$
-By comparison, we get eigenvalues $\lambda_\theta = l (l+1)$.
+By comparison, we get eigenvalues $\lambda_\theta = -l (l+1)$.
-The equation about $R$ is related to the spherical Bessel function's equation, $y = j_n(r), y_n(r)$ solves
+The equation about $R$ is related to the spherical Bessel function's equation. Recall  $y(r) = j_n(r), y_n(r)$ solves
 $$  \left( \frac{\d}{\d r} r^2  \frac{\d y}{\d r} \right) +   (r^2 - n(n+1) ) y = 0 $$
+Hence, for $\lambda_\theta = l(l+1)$, we can solve $R(r)$ to get
+$$ R(r) = \begin{cases}
+j_l( k r), y_l(kr) & \lambda_r = -k^2 \cr
+j_l( i k r), y_l(i kr) & \lambda_r = k^2 \cr
+r^l, r^{-l-1} & \lambda_r=0
+\end{cases}
+$$
+Usually, we require $R(0)$ to be finite, and $y_l(r) \sim r^{-l-1}$ as $r \to 0$, hence we only keep the $j_l(kr), j_l(ikr), r^l$ solution above. The case $\lambda_r$ can be obtained as the leading order behavior of $j_l(kr)$ after rescaling by a constant
+$$ r^l = \lim_{k \to 0} k^{-l} j_l( k r). $$
+In some cases, suppose we solve the equation in the exterior of a ball and require the solution to decay as $r \to \infty$, then we will not ask $R(0)$ to be finite, instead we require $R(\infty) = 0$. Then the $y_l(kr), y_l(ikr), r^{-l-1}$ solutions will come into play.
+=== Summary ===
+The general eigenfunction is
+$$u(r,\theta,\phi) =  \begin{cases} j_l(kr) \cr y_l(kr) \end{cases}
+ P_l^m(\cos \theta) \begin{cases} \cos(m \phi) \cr \sin(m \phi) \end{cases}, \quad \lambda = -k^2
+$$
+===== Steady State temperature distribution inside a unit ball =====
+We solve $ \Delta u = 0$ for $r \leq 1$ with $u(r=1, \theta, \phi) = f(\theta,\phi)$. We may first decompose $f(\theta, \phi)$ as
+$$ f(\theta, \phi) = \sum_{l=0}^\infty \sum_{m=0}^l [a_{l m} \cos(m\phi) + b_{l m} \sin(m \phi) ] P_l^m(\cos \theta)  $$
+with $b_{l 0}=0$.
+Then the solulution for $u$ is obtained by setting $\lambda = \lambda_r=0$
+$$ 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).  $$
+To obtain the coefficients $a_{lm}, b_{lm}$ from $f(\theta, \phi)$, we uses orthogonality of these functions $P_l^m(\cos \theta) \cos(m \phi), P_l^m(\cos \theta) \sin(m \phi)$ on the two sphere with volume form $\sin \theta d\theta d\phi$. For example, we claim that, if $(l, m) \neq (l', m')$, then
+$$ \int_{\phi = 0}^{2\pi} \int_{\theta=0}^\pi P_l^m(\cos \theta) \cos(m \phi) P_{l'}^{m'}(\cos \theta) \cos(m' \phi) \sin \theta d\theta d\phi = 0$$
+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.

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