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

$\mathbf{\left|}\mathbf{cos\theta }\mathbf{\right|}$.

The surface temperature of sphere of radius 1 is$\left|\cos \left(\theta \right)\right|$.

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.

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

For simplicity, consider$x=\cos \theta$.

$\begin{array}{c}{u}_{r=1}=\left|\cos \left(\theta \right)\right|\\ =\left|x\right|\end{array}$

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).

$\begin{array}{c}{c}_0=\frac{1}{2}\underset{-1}{\overset{1}{\int }}\left|x\right|dx\\ =\frac{1}{2}{\left[\frac{x^2}{2}\right]}_{-1}^1\\ =0\end{array}$

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

$\begin{array}{c}{c}_1=\frac{3}{2}\underset{-1}{\overset{1}{\int }}\left|x\right|dx\\ =\frac{3}{2}{\left[\frac{x^2}{2}\right]}_{-1}^1\\ =0\end{array}$

Take m = 2 and put in equation (1).

$\begin{array}{c}{c}_2=\frac{5}{4}\underset{-1}{\overset{1}{\int }}\left|x\right|(3x^2-1)dx\\ =\frac{5}{4}\underset{-1}{\overset{1}{\int }}(3\left|x\right|x^2-\left|x\right|)dx\\ =\frac{5}{4}{(3\frac{x^4}{4}-\frac{x^2}{2})}_{-1}^1\\ =\frac{5}{4}(\frac{3}{4}-\frac{1}{2}-\frac{3}{4}+\frac{1}{2})\end{array}$

$c_2=\frac{5}{8}$

Take m = 3 and put in equation (1).

$c_3=0$

Take m = 4 and put in equation (1).

$\begin{array}{c}{c}_4=\frac{9}{16}\underset{-1}{\overset{1}{\int }}\left|x\right|(35x^4-30x^2+3)dx\\ =2\times \frac{9}{16}\underset{0}{\overset{1}{\int }}(35x^5-30x^3+3x)dx\end{array}$

$\begin{array}{c}{c}_4=\frac{9}{8}{(35\frac{x^6}{6}-30\frac{x^4}{4}+3\frac{x^2}{2})}_0^{1}dx\\ =\frac{9}{8}(\frac{35}{6}-\frac{30}{4}+\frac{3}{2})\\ =\frac{9}{8}\left(\frac{140-180+36}{24}\right)\end{array}$

$\begin{array}{c}{c}_4=\frac{9}{8}(\frac{35}{6}-\frac{30}{4}+\frac{3}{2})\\ =\frac{9}{8}\left(\frac{140-180+36}{24}\right)\\ =\frac{9}{8}\left(\frac{-4}{24}\right)\\ =\frac{-3}{16}\end{array}$

Use the equation s given below.

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

$u=\frac{1}{2}{P}_0\left(\cos \theta \right)+\frac{5}{8}{r}^2{P}_2\left(\cos \theta \right)-\frac{3}{16}{r}^4{P}_4\left(\cos \theta \right)+\dots$

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

$\frac{1}{2}{P}_0\left(\cos \theta \right)+\frac{5}{8}{r}^2{P}_2\left(\cos \theta \right)-\frac{3}{16}{r}^4{P}_4\left(\cos \theta \right)+\dots$.