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

$\mathbf{cos\theta }\mathbf{-}\mathbf{3}{\mathbf{sin}}^{\mathbf{2}}\mathbf{\theta }$.

An expression has been given as $\cos \theta -3{\sin }^2\theta$.

The total of the second-order partial derivatives of , the unknown function, with respect to the Cartesian coordinates equals , according to Laplace's equation.

Laplace’s equation in cylindrical coordinates is${\mathbf{\nabla }}^{\mathbf{2}}\mathit{u}\mathbf{=}\frac{\mathbf{1}}{\mathbf{r}}\frac{\mathbf{\partial }}{\mathbf{\partial }\mathbf{r}}\mathbf{\left(}\mathbf{r}\frac{\mathbf{\partial }\mathbf{u}}{\mathbf{\partial }\mathbf{r}}\mathbf{\right)}\mathbf{+}\frac{\mathbf{1}}{{\mathbf{r}}^{\mathbf{2}}}\frac{{\mathbf{\partial }}^{\mathbf{2}}\mathbf{u}}{\mathbf{\partial }{\mathbf{\theta }}^{\mathbf{2}}}\mathbf{+}\frac{{\mathbf{\partial }}^{\mathbf{2}}\mathbf{u}}{\mathbf{\partial }{\mathbf{z}}^{\mathbf{2}}}\mathbf{=}\mathbf{0}$

And to separate the variable the solution assumed is of the form$\mathit{u}\mathbf{=}\mathit{R}\mathbf{\left(}\mathbf{r}\mathbf{\right)}\mathit{\Theta }\mathbf{\left(}\mathbf{\theta }\mathbf{\right)}\mathit{Z}\mathbf{\left(}\mathbf{z}\mathbf{\right)}$.

Rewrite the given question.

$\begin{array}{c}A\left(\theta \right)=\cos \left(\theta \right)-3{\sin }^2\left(\theta \right)\\ =\cos \left(\theta \right)+3{\cos }^2\left(\theta \right)-3\end{array}$

Due to the fact that $A\left(\theta \right)$is independent of $\varphi ,$the associated Legendre polynomials $P_l^m\left(\cos \left(\theta \right)\right)$reduce to the standard Legendre polynomials $P_l\left(\cos \left(\theta \right)\right)$.

In order to find the steady-state temperature distribution function. Express $A\left(\theta \right)$in terms of the standard Legendre polynomials which in turn leads us to determine the corresponding coefficients $c_I.$

Substitute$x=\cos \left(\theta \right)$

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

$u_{r=1}\left(x\right)=3x^2+x-3$ ….. (1)

Multiply the above equation by $P_{\nu }$and integrate.

$\sum _{l=0}^{\infty }{c}_l\underset{-1}{\overset{1}{\int }}{P}_l\left(x\right)P_{\nu }\left(x\right)dx=\underset{-1}{\overset{1}{\int }}(3x^2+x-3)P_{\nu }\left(x\right)dx$ ….. (2)

Legendre orthogonally relation:

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

Insert equation (3) into equation (2):

$\sum _{l=0}^{\infty }{c}_l\frac{2}{(2l+1)}{\delta }_{l,\nu }=\underset{-1}{\overset{1}{\int }}(3x^2+x)P_{\nu }\left(x\right)dx-3\underset{-1}{\overset{1}{\int }}{P}_{\nu }\left(x\right)$ ….. (4)

Legendre identity:

$\underset{a}{\overset{1}{\int }}{P}_n\left(x\right)dx=\frac{1}{(2n+1)}[P_{n-1}\left(a\right)-P_{n+1}\left(a\right)]$ ….. (5)

Combine equation (4) and (5).

$c_{\nu }=\frac{2\nu +1}{2}\underset{-1}{\overset{1}{\int }}(3x^2+x)P_{\nu }\left(x\right)dx-\frac{3}{2}[P_{\nu -1}(-1)-P_{\nu +1}(-1)]$

Hence the final answer is as given below.

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