Find the steady-state temperature distribution inside a sphere of radius 1 when the surface temperatures are as given in Problems 1 to 10.

$\raisebox{1ex}{$\mathbf{\pi }$}\!\left/ \!\raisebox{-1ex}{$\mathbf{2}$}\right.\mathbf{-}\mathbf{\theta }$

The surface temperature of sphere of radius 1 is$\raisebox{1ex}{$\pi $}\!\left/ \!\raisebox{-1ex}{$2$}\right.-\theta$.

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

The standard Legendre polynomials is $P_l\left(\cos \left(\theta \right)\right)$.

For simplicity consider the equation as given below.

$\begin{array}{l}x=\cos \theta \\ \theta ={\cos }^{-1}x\end{array}$

The surface temperature of sphere is$\raisebox{1ex}{$\pi $}\!\left/ \!\raisebox{-1ex}{$2$}\right.-\theta$

$\begin{array}{c}\frac{\pi }{2}-\theta =\frac{\pi }{2}-{\cos }^{-1}x\\ ={\sin }^{-1}x\end{array}$

It is known that:

$u=\sum _{l=0}^{\infty }{c}_l{r}^l{P}_l\left(\cos \theta \right)$

Hence,

$c_m=\frac{2m+1}{2}\underset{-1}{\overset{1}{\int }}\left|x\right|P_m\left(x\right)dx$ ….. (1)

Take m = 0 and put in equation (1).

$\begin{array}{c}{c}_0=\frac{1}{2}\underset{-1}{\overset{1}{\int }}{\sin }^{-1}xdx\\ =0\end{array}$

Take m = 1 and put in equation (1).

$\begin{array}{c}{c}_1=\frac{3}{2}\underset{-1}{\overset{1}{\int }}x{\sin }^{-1}xdx\\ =\frac{3}{2}[{\left[\frac{x^2}{2}{\sin }^{-1}x\right]}_{-1}^1-{\int }_{-1}^1\frac{d}{dx}\left(x\right)\underset{-4}{\overset{1}{\int }}{\sin }^{-4}xdx]\\ =\frac{3}{2}[[\frac{1}{2}\times \frac{x}{2}+\frac{1}{2}\times \frac{\pi }{2}]-\frac{1}{2}\underset{-1}{\overset{1}{\int }}\frac{x^2}{\sqrt{1-x^2}}dx]\\ =\frac{3}{2}[\frac{\pi }{2}-\frac{1}{2}\underset{-1}{\overset{1}{\int }}\frac{x^2}{\sqrt{1-x^2}}dx]\end{array}$

Put $x=\sin u$and change the limits accordingly.

Then$dx=\cos udu$

$\begin{array}{c}{c}_1=\frac{3}{2}[\frac{\pi }{2}-\frac{1}{2}\underset{\raisebox{1ex}{$-\pi $}\!\left/ \!\raisebox{-1ex}{$2$}\right.}{\overset{\raisebox{1ex}{$\pi $}\!\left/ \!\raisebox{-1ex}{$2$}\right.}{\int }}\frac{{\sin }^{2}u}{\sqrt{1-{\sin }^{2}u}}\cos udu]\\ =\frac{3}{2}[\frac{\pi }{2}-\frac{1}{2}\underset{\raisebox{1ex}{$-\pi $}\!\left/ \!\raisebox{-1ex}{$2$}\right.}{\overset{\raisebox{1ex}{$\pi $}\!\left/ \!\raisebox{-1ex}{$2$}\right.}{\int }}\frac{{\sin }^{2}u}{\cos u}\cos udu]\\ =\frac{3}{2}[\frac{\pi }{2}-\frac{1}{2}\underset{\raisebox{1ex}{$-\pi $}\!\left/ \!\raisebox{-1ex}{$2$}\right.}{\overset{\raisebox{1ex}{$\pi $}\!\left/ \!\raisebox{-1ex}{$2$}\right.}{\int }}{\sin }^{2}udu]\end{array}$

$c_1=\frac{3}{2}[\frac{\pi }{2}-\frac{1}{2}\underset{\raisebox{1ex}{$-\pi $}\!\left/ \!\raisebox{-1ex}{$2$}\right.}{\overset{\raisebox{1ex}{$\pi $}\!\left/ \!\raisebox{-1ex}{$2$}\right.}{\int }}\left(\frac{1-\cos 2u}{2}\right)du]$

Simplify further.

$c_1=\frac{3\pi }{4}-\frac{3}{4}{[\frac{1}{2}u-\frac{\sin 2u}{4}]}_{-\frac{\pi }{2}}^{\frac{\pi }{2}}$

$\begin{array}{c}{c}_1=\frac{3\pi }{4}-\frac{3}{4}[\frac{1}{2}(\frac{\pi }{2}+\frac{\pi }{2})-\frac{\sin 2\left(\frac{\pi }{2}\right)}{4}+\frac{\sin 2(-\frac{\pi }{2})}{4}]\\ =\frac{3\pi }{4}-\frac{3\pi }{8}\\ =\frac{3\pi }{8}\end{array}$

Calculate the remaining terms.

$\begin{array}{c}{c}_2=0\\ c_3=\frac{7\pi }{128}\mathrm{.......}\end{array}$

Use the equation given below.

$u=\sum _{l=0}^{\infty }{c}_l{r}^l{P}_l\left(\cos \theta \right)$

$u=\frac{3\pi }{8}r{P}_1\left(\cos \theta \right)+\frac{7\pi }{128}{r}^3{P}_3\left(\cos \theta \right)+\dots \dots$

Hence the steady-state temperature distribution inside a sphere of radius 1:

$\frac{3\pi }{8}r{P}_1\left(\cos \theta \right)+\frac{7\pi }{128}{r}^3{P}_3\left(\cos \theta \right)+\dots \dots$.