91Ó°ÊÓ

The inner and the outer radius of the sphere is 1, and 2 respectively.

When a conductor reaches a point where no more heat can be absorbed by the rod, it is said to be at steady-state temperature.

Write the definition of the surface temperature distribution function$A\left(\mathrm{θ}\right)$.

$\begin{array}{c}A\left(\mathrm{θ}\right)=\left\{\begin{array}{l}\begin{array}{ll}100& \text{ â¶Ä‰â¶Ä‰â¶Ä‰â¶Ä‰â¶Ä‰â¶Ä‰â¶Ä‰}0<\mathrm{θ}<\frac{\mathrm{Ï€}}{2}\end{array}\\ \begin{array}{ll}0& \text{ â¶Ä‰â¶Ä‰â¶Ä‰â¶Ä‰â¶Ä‰â¶Ä‰â¶Ä‰}\frac{\mathrm{Ï€}}{2}<\mathrm{θ}<\mathrm{Ï€}\end{array}\end{array}\right.\\ =\left\{\begin{array}{l}\begin{array}{ll}0& \text{ â¶Ä‰â¶Ä‰â¶Ä‰â¶Ä‰â¶Ä‰â¶Ä‰}−1<\cos \left(\mathrm{θ}\right)<0\end{array}\\ \begin{array}{ll}100& \text{ â¶Ä‰â¶Ä‰â¶Ä‰â¶Ä‰â¶Ä‰â¶Ä‰â¶Ä‰â€‰â€‰}0<\cos \left(\mathrm{θ}\right)<1\end{array}\end{array}\right.\\ =100\left\{\begin{array}{l}0\text{ â¶Ä‰â¶Ä‰â¶Ä‰â¶Ä‰â¶Ä‰â¶Ä‰â¶Ä‰â€‰â€‰â€‰â€‰â€‰}1<x<0\\ 1\text{ â¶Ä‰â¶Ä‰â¶Ä‰â¶Ä‰â¶Ä‰â¶Ä‰â¶Ä‰â€‰â€‰â€‰â€‰â€‰â€‰}0<x<1\end{array}\right.\\ =100f\left(x\right)\end{array}$

Write the boundary conditions for the surface temperature distribution function.

$\begin{array}{c}{u}_{r=1}\left(x\right)=\underset{l=0}{\overset{\mathrm{∞}}{∑}}(c_l{1}^l+b_l{1}^{−(l+1)})P_l\left(x\right)\\ =0\end{array}$

$\begin{array}{c}{u}_{r=2}\left(x\right)=\underset{l=0}{\overset{\mathrm{∞}}{∑}}(c_l{2}^l+b_l{2}^{−(l+1)})P_l\left(x\right)\\ =100f\left(x\right)\end{array}$

The orthogonality relation of Legendre Polynomial is:

$\begin{array}{c}\underset{0}{\overset{1}{∫}}{P}_l\left(x\right)P_k\left(x\right)dx=\frac{1}{2}\underset{1}{\overset{1}{∫}}{P}_l\left(x\right)P_k\left(x\right)dx\\ =\frac{1}{2}\frac{2}{(2l+1)}{\mathrm{δ}}_{l,k}\end{array}$

The identity of Legendre Polynomial is$\underset{a}{\overset{1}{∫}}{P}_k\left(x\right)dx=\frac{1}{(2k+1)}[P_{k−1}\left(a\right)−P_{k+1}\left(a\right)]$.

It is known that$b_l=−c_l$.

$⇒\underset{l=0}{\overset{\mathrm{∞}}{∑}}(c_l+b_l)\underset{0}{\overset{1}{∫}}{P}_l\left(x\right)P_k\left(x\right)dx=0$

Solve further.

$\begin{array}{l}\underset{l=0}{\overset{\mathrm{∞}}{∑}}(c_l{2}^l+b_l{2}^{−(l+1)})=\underset{0}{\overset{1}{∫}}{P}_l\left(x\right)P_k\left(x\right)dx\\ \underset{l=0}{\overset{\mathrm{∞}}{∑}}(c_l{2}^l+b_l{2}^{−(l+1)})=100\underset{0}{\overset{1}{∫}}{P}_k\left(x\right)dx\\ \underset{l=0}{\overset{\mathrm{∞}}{∑}}{c}_l(2^l−2^{(l+1)})\frac{1}{2}\underset{1}{\overset{1}{∫}}{P}_l\left(x\right)P_k\left(x\right)dx=\frac{100}{(2k+1)}[P_{k−1}\left(0\right)−P_{k+1}\left(0\right)]\\ \underset{l=0}{\overset{\mathrm{∞}}{∑}}{c}_l(2^l−2^{−(l+1)})\frac{1}{(2l+1)}{\mathrm{δ}}_{l,k}=\frac{100}{(2k+1)}[P_{k−1}\left(0\right)−P_{k+1}\left(0\right)]\end{array}$

$\begin{array}{l}{c}_k=\frac{100}{(2^k−2^{−(k+1)})}[P_{k−1}\left(0\right)−P_{k+1}\left(0\right)]\\ u(r,\mathrm{θ})=\underset{k=0}{\overset{\mathrm{∞}}{∑}}\frac{100}{(2^k−2^{(k+1)})}[P_{k−1}\left(0\right)−P_{k+1}\left(0\right)](r^k−r^{−(k+1)})P_k\left(x\right)\end{array}$

Hence this is the steady state temperature inside and outside the spherical shell.