91影视

The radius of the sphere is 1.

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$P_l\left(\cos \left(胃\right)\right)$.

Consider the equation

$u_{r=1}=\left\{\begin{array}{l}\begin{array}{ll}0,& 鈭�1<x<0\end{array}\\ \begin{array}{ll}x,& 0<x<1\end{array}\end{array}\right.$

Take the equation

$c_m=\frac{2m+1}{2}\underset{鈭�1}{\overset{1}{鈭�}}(5x^3+3x^2鈭�3)p_m\left(x\right)dx$ 鈥�.. (1)

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

$\begin{array}{c}{c}_0=\frac{1}{2}\underset{0}{\overset{1}{鈭�}}x{P}_0\left(x\right)dx\\ =\frac{1}{2}\underset{0}{\overset{1}{鈭�}}xdx\\ =\frac{1}{2}{\left[\frac{x^2}{2}\right]}_0^1\end{array}$

$c_0=\frac{1}{4}$

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

$\begin{array}{c}{c}_1=\frac{3}{2}\underset{0}{\overset{1}{鈭�}}(x脳x)dx\\ =\frac{3}{2}\underset{0}{\overset{1}{鈭�}}{\text{x}}^{2}dx\\ =\frac{3}{2}{\left[\frac{x^3}{3}\right]}_0^1\\ =\frac{1}{2}\end{array}$

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

$\begin{array}{c}{c}_2=\frac{5}{4}{鈭�}_0^{1}x(3x^2鈭�1)dx\\ =\frac{5}{4}{鈭�}_0^1(3x^3鈭�x)dx\\ =\frac{5}{4}{[\frac{3x^4}{4}鈭�\frac{x^2}{2}]}_0^1\\ =\frac{5}{16}\end{array}$

Use the equation

$u=\underset{l=0}{\overset{\mathrm{鈭�}}{鈭�}}{c}_l{r}^l{P}_l\left(\cos 胃\right)$

$u=\underset{l=0}{\overset{\mathrm{鈭�}}{鈭�}}{c}_l{r}^l{P}_l\left(\cos 胃\right)=\frac{1}{4}{P}_0\left(\cos 胃\right)+\frac{9}{16}r{P}_1\left(\cos 胃\right)+\frac{15}{32}{r}^2{P}_2\left(\cos 胃\right)+鈥�$

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

$\frac{1}{4}{P}_0\left(\cos 胃\right)+\frac{9}{16}r{P}_1\left(\cos 胃\right)+\frac{15}{32}{r}^2{P}_2\left(\cos 胃\right)+鈥�$.