91Ó°ÊÓ

The surface temperature of sphere of radius 1 is$5{\cos }^3\mathrm{θ}−3{\sin }^2\mathrm{θ}$.

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.

Take the steady-state temperature distribution function as $u(r,\mathrm{θ})$.

Consider the given surface temperature as$A\left(\mathrm{θ}\right)$.

$\begin{array}{c}A\left(\mathrm{θ}\right)=5{\cos }^3\mathrm{θ}−3{\sin }^2\mathrm{θ}\\ =5{\cos }^3\mathrm{θ}+3{\cos }^2\mathrm{θ}−3\end{array}$

The standard Legendre polynomials is $P_l\left(\cos \left(\mathrm{θ}\right)\right)$.

For simplicity consider $x=\cos \mathrm{θ}$

$\underset{l=0}{\overset{\mathrm{∞}}{∑}}{c}_l{1}^l{P}_l\left(x\right)=5x^3+3x^2−3$ ….. (1)

Multiply equation (1) by $P_{\mathrm{Î\pm }}$and integrate the function.

$\underset{l=0}{\overset{\mathrm{∞}}{∑}}{c}_l\underset{−1}{\overset{1}{∫}}{P}_l\left(x\right)P_{\mathrm{Î\pm }}\left(x\right)=\underset{−1}{\overset{1}{∫}}(5x^3+3x^2−3)P_{\mathrm{Î\pm }}\left(x\right)dx\text{ }$ ….. (2)

The Orthogonal relation of Legendre polynomials:

$\underset{−1}{\overset{1}{∫}}{P}_l\left(x\right)P_{\mathrm{Î\pm }}\left(x\right)=\frac{2}{(2l+1)}{\mathrm{δ}}_{l,\mathrm{Î\pm }}$ ….. (3)

The Identity of Legendre polynomials

$\underset{a}{\overset{1}{∫}}{P}_{n}dx=\frac{1}{(2n+1)}[P_{n−1}\left(a\right)−P_{n+1}\left(a\right)]$ ….. (4)

Put equation (3) and (4) in (2) and simplify further.

$\underset{l=0}{\overset{\mathrm{∞}}{∑}}{c}_l\underset{−1}{\overset{1}{∫}}{P}_l\left(x\right)P_{\mathrm{Î\pm }}\left(x\right)=\underset{−1}{\overset{1}{∫}}(5x^3+3x^2−3)P_{\mathrm{Î\pm }}\left(x\right)dx\text{ }$

$\begin{array}{l}\underset{l=0}{\overset{\mathrm{∞}}{∑}}{c}_l\frac{2}{(2l+1)}{\mathrm{δ}}_{l,\mathrm{Î\pm }}=\underset{−1}{\overset{1}{∫}}(5x^3+3x^2)dx−3\underset{−1}{\overset{1}{∫}}{P}_{\mathrm{Î\pm }}\left(x\right)dx\\ c_{\mathrm{Î\pm }}\frac{2}{(2\mathrm{Î\pm }+1)}=\underset{−1}{\overset{1}{∫}}(5x^3+3x^2)dx−\frac{3}{(2\mathrm{Î\pm }+1)}[P_{\mathrm{Î\pm }−1}(−1)−P_{\mathrm{Î\pm }+1}(−1)]\\ c_{\mathrm{Î\pm }}=\frac{(2\mathrm{Î\pm }+1)}{2}\underset{−1}{\overset{1}{∫}}(5x^3+3x^2)dx−\frac{3}{2}[P_{\mathrm{Î\pm }−1}(−1)−P_{\mathrm{Î\pm }+1}(−1)]\end{array}$

$u(r,\mathrm{θ})=\underset{\mathrm{Î\pm }=0}{\overset{\mathrm{∞}}{∑}}\{\frac{(2\mathrm{Î\pm }+1)}{2}\underset{−1}{\overset{1}{∫}}(5x^3+3x^2)dx−\frac{3}{2}[P_{\mathrm{Î\pm }−1}(−1)−P_{\mathrm{Î\pm }+1}(−1)]\}{r}^{\mathrm{Î\pm }}{P}_{\mathrm{Î\pm }}\left(x\right)$

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

$u(r,\mathrm{θ})=\underset{\mathrm{Î\pm }=0}{\overset{\mathrm{∞}}{∑}}\{\frac{(2\mathrm{Î\pm }+1)}{2}\underset{−1}{\overset{1}{∫}}(5x^3+3x^2)dx−\frac{3}{2}[P_{\mathrm{Î\pm }−1}(−1)−P_{\mathrm{Î\pm }+1}(−1)]\}{r}^{\mathrm{Î\pm }}{P}_{\mathrm{Î\pm }}\left(x\right)$