Find the steady-state temperature distribution in a spherical shell of inner radius 1 and outer radius 2. if the inner surface is held at ${\mathbf{0}}^{\mathbf{°}}$and the outer surface has its upper half at $\mathbf{100}\mathbf{°}$and its lower half at $\mathbf{0}\mathbf{°}$. Hint: r = 0 is not in the region of interest, so the solutions ${\mathit{r}}^{\mathbf{-}\mathbf{l}\mathbf{-}\mathbf{1}}$in (7.9) should be included. Replace ${\mathit{c}}_{\mathbf{l}}{\mathit{r}}^{\mathbf{l}}$in (7.11) by$\mathbf{(}{\mathbf{c}}_{\mathbf{l}}{\mathbf{r}}^{\mathbf{l}}\mathbf{+}{\mathbf{b}}_{\mathbf{l}}{\mathbf{r}}^{\mathbf{-}\mathbf{l}\mathbf{-}\mathbf{1}}\mathbf{)}$.

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$A\left(\theta \right)$.

$\begin{array}{c}A\left(\theta \right)=\left\{\begin{array}{l}\begin{array}{ll}100& 0<\theta <\frac{\pi }{2}\end{array}\\ \begin{array}{ll}0& \frac{\pi }{2}<\theta <\pi \end{array}\end{array}\right.\\ =\left\{\begin{array}{l}\begin{array}{ll}0& -1<\cos \left(\theta \right)<0\end{array}\\ \begin{array}{ll}100& 0<\cos \left(\theta \right)<1\end{array}\end{array}\right.\\ =100\left\{\begin{array}{l}01<x<0\\ 10<x<1\end{array}\right.\\ =100f\left(x\right)\end{array}$

Write the boundary conditions for the surface temperature distribution function.

$\begin{array}{c}{u}_{r=1}\left(x\right)=\sum _{l=0}^{\infty }(c_l{1}^l+b_l{1}^{-(l+1)})P_l\left(x\right)\\ =0\end{array}$

$\begin{array}{c}{u}_{r=2}\left(x\right)=\sum _{l=0}^{\infty }(c_l{2}^l+b_l{2}^{-(l+1)})P_l\left(x\right)\\ =100f\left(x\right)\end{array}$

The orthogonality relation of Legendre Polynomial is:

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

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

It is known that$b_l=-c_l$.

$\Rightarrow \sum _{l=0}^{\infty }(c_l+b_l)\underset{0}{\overset{1}{\int }}{P}_l\left(x\right)P_k\left(x\right)dx=0$

Solve further.

$\begin{array}{l}\sum _{l=0}^{\infty }(c_l{2}^l+b_l{2}^{-(l+1)})=\underset{0}{\overset{1}{\int }}{P}_l\left(x\right)P_k\left(x\right)dx\\ \sum _{l=0}^{\infty }(c_l{2}^l+b_l{2}^{-(l+1)})=100\underset{0}{\overset{1}{\int }}{P}_k\left(x\right)dx\\ \sum _{l=0}^{\infty }{c}_l(2^l-2^{(l+1)})\frac{1}{2}\underset{1}{\overset{1}{\int }}{P}_l\left(x\right)P_k\left(x\right)dx=\frac{100}{(2k+1)}[P_{k-1}\left(0\right)-P_{k+1}\left(0\right)]\\ \sum _{l=0}^{\infty }{c}_l(2^l-2^{-(l+1)})\frac{1}{(2l+1)}{\delta }_{l,k}=\frac{100}{(2k+1)}[P_{k-1}\left(0\right)-P_{k+1}\left(0\right)]\end{array}$

$\begin{array}{l}{c}_k=\frac{100}{(2^k-2^{-(k+1)})}[P_{k-1}\left(0\right)-P_{k+1}\left(0\right)]\\ u(r,\theta )=\sum _{k=0}^{\infty }\frac{100}{(2^k-2^{(k+1)})}[P_{k-1}\left(0\right)-P_{k+1}\left(0\right)](r^k-r^{-(k+1)})P_k\left(x\right)\end{array}$

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