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

${\mathbf{sin}}^{\mathbf{2}}\mathbf{\theta cos\theta cos}\mathbf{2}\mathbf{\varphi }\mathbf{-}\mathbf{cos\theta }$(See problem 9).

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 a steady-state temperature.

Compute the steady-state temperature distribution function $u(r,\theta )$inside a sphere with a radius of r = 1 in this issue. The surface temperature function is written as$A(\theta ,\varphi )$.

Consider the equation below:

$\begin{array}{c}A(\theta ,\varphi )={\sin }^2\left(\theta \right)\cos \left(\theta \right)\cos \left(2\varphi \right)-\cos \left(\theta \right)\\ =(1-{\cos }^2\left(\theta \right))\cos \left(\theta \right)(\cos \left(2\varphi \right)-1)\end{array}$

The steady-state temperature distribution function $u(r,\theta ,\varphi )$is directly dependent on the related Legendre polynomials $P_l^m\left(\cos \left(\theta \right)\right)$due to the $\varphi$dependency in the surface temperature function $A(\theta ,\varphi )$. As a result, the $u(r,\theta ,\varphi )$Power series has the form as below.

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

Now, Let x be$\cos \theta$.

$u_{r=1}\left(x\right)=\sum _{l=0}^{\infty }{c}_l{P}_l^2\left(x\right)$

$u_{r=1}\left(x\right)=(1-x^2)x(\cos \left(2\varphi \right)-1)$ ….. (1)

Multiply equation (1) by $P_{\nu }\left(x\right)$and integrate.

$\sum _{l=0}^{\infty }{c}_l\underset{-1}{\overset{1}{\int }}{P_l}^2\left(x\right)P_{\nu }^2\left(x\right)dx=\underset{-1}{\overset{1}{\int }}(1-x^2)x{P}_{\nu }^2\left(x\right)dx(\cos \left(2\varphi \right)-1)$ ….. (2)

Write the Legendre polynomials with associated orthogonality:

$\underset{-1}{\overset{1}{\int }}{P_l}^2\left(x\right)P_{\nu }^2\left(x\right)dx=\frac{2}{(2l+1)}\frac{(l+2)!}{(l-2)!}{\delta }_{l,\nu }$ ….. (3)

$\begin{array}{c}\underset{-1}{\overset{1}{\int }}{P_l}^2\left(x\right)P_{\nu }^2\left(x\right)dx=\frac{2}{(2l+1)}\frac{(l+2)(l+1)(l-1)(l-2)!}{(l-2)!}{\delta }_{l,\nu }\\ =\frac{2(l+2)(l+1)(l-1)}{(2l+1)}{\delta }_{l,\nu }\end{array}$

Substitute equation (3) into equation (2).

$\begin{array}{l}\sum _{l=0}^{\infty }{c}_l\frac{2(l+2)(l+1)(l-1)}{(2l+1)}{\delta }_{l,\nu }=\underset{-1}{\overset{1}{\int }}(x-x^3)P_{\nu }\left(x\right)dx(\cos \left(2\varphi \right)-1)\\ c_{\nu }\frac{2(\nu +2)(\nu +1)(\nu -1)}{(2\nu +1)}=\underset{-1}{\overset{1}{\int }}(x-x^3)P_{\nu }\left(x\right)dx(\cos \left(2\varphi \right)-1)\\ c_{\nu }=\frac{(2\nu +1)}{2(\nu +2)(\nu +1)(\nu -1)}\underset{-1}{\overset{1}{\int }}(x-x^3){P_{\nu }}^2\left(x\right)dx(\cos \left(2\varphi \right)-1)\end{array}$

$u(r,\theta ,\varphi )=\sum _{\nu =0}^{\infty }\left\{\frac{(2\nu +1)}{2(\nu +2)(\nu +1)(\nu -1)}\underset{-1}{\overset{1}{\int }}(x-x^3){P_{\nu }}^2\left(x\right)dx\right\}{r}^{\nu }{P}_{\nu }^2\left(x\right)(\cos \left(2\varphi \right)-1)$

Therefore, the steady-state temperature distribution inside a sphere of radius 1 is

$u(r,\theta ,\varphi )=\sum _{\nu =0}^{\infty }\left\{\frac{(2\nu +1)}{2(\nu +2)(\nu +1)(\nu -1)}\underset{-1}{\overset{1}{\int }}(x-x^3){P_{\nu }}^2\left(x\right)dx\right\}{r}^{\nu }{P}_{\nu }^2\left(x\right)(\cos \left(2\varphi \right)-1)$.