91Ó°ÊÓ

The surface temperature of sphere of radius 1 is$\raisebox{1ex}{$\mathrm{π}$}\!\left/ \!\raisebox{-1ex}{$2$}\right.−\mathrm{θ}$.

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 $P_l\left(\cos \left(\mathrm{θ}\right)\right)$.

For simplicity consider the equation as given below.

$\begin{array}{l}x=\cos \mathrm{θ}\\ \mathrm{θ}={\cos }^{−1}x\end{array}$

The surface temperature of sphere is$\raisebox{1ex}{$\mathrm{π}$}\!\left/ \!\raisebox{-1ex}{$2$}\right.−\mathrm{θ}$

$\begin{array}{c}\frac{\mathrm{π}}{2}−\mathrm{θ}=\frac{\mathrm{π}}{2}−{\cos }^{−1}x\\ ={\sin }^{−1}x\end{array}$

It is known that:

$u=\underset{l=0}{\overset{\mathrm{∞}}{∑}}{c}_l{r}^l{P}_l\left(\cos \mathrm{θ}\right)$

Hence,

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

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

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

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

$\begin{array}{c}{c}_1=\frac{3}{2}\underset{−1}{\overset{1}{∫}}x{\sin }^{−1}xdx\\ =\frac{3}{2}[{\left[\frac{x^2}{2}{\sin }^{−1}x\right]}_{−1}^1−{∫}_{−1}^1\frac{d}{dx}\left(x\right)\underset{−4}{\overset{1}{∫}}{\sin }^{−4}xdx]\\ =\frac{3}{2}[[\frac{1}{2}×\frac{x}{2}+\frac{1}{2}×\frac{\mathrm{π}}{2}]−\frac{1}{2}\underset{−1}{\overset{1}{∫}}\frac{x^2}{\sqrt{1−x^2}}dx]\\ =\frac{3}{2}[\frac{\mathrm{π}}{2}−\frac{1}{2}\underset{−1}{\overset{1}{∫}}\frac{x^2}{\sqrt{1−x^2}}dx]\end{array}$

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

Then$dx=\cos udu$

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

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

Simplify further.

$c_1=\frac{3\mathrm{π}}{4}−\frac{3}{4}{[\frac{1}{2}u−\frac{\sin 2u}{4}]}_{−\frac{\mathrm{π}}{2}}^{\frac{\mathrm{π}}{2}}$

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

Calculate the remaining terms.

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

Use the equation given below.

$u=\underset{l=0}{\overset{\mathrm{∞}}{∑}}{c}_l{r}^l{P}_l\left(\cos \mathrm{θ}\right)$

$u=\frac{3\mathrm{π}}{8}r{P}_1\left(\cos \mathrm{θ}\right)+\frac{7\mathrm{π}}{128}{r}^3{P}_3\left(\cos \mathrm{θ}\right)+……$

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

$\frac{3\mathrm{π}}{8}r{P}_1\left(\cos \mathrm{θ}\right)+\frac{7\mathrm{π}}{128}{r}^3{P}_3\left(\cos \mathrm{θ}\right)+……$.