Question

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

$\mathbf{5}{\mathbf{cos}}^{\mathbf{3}}\mathbf{\theta }\mathbf{-}\mathbf{3}{\mathbf{sin}}^{\mathbf{2}}\mathbf{\theta }$.

Short answer

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

$u(r,\theta )=\sum _{\alpha =0}^{\infty }\{\frac{(2\alpha +1)}{2}\underset{-1}{\overset{1}{\int }}(5x^3+3x^2)dx-\frac{3}{2}[P_{\alpha -1}(-1)-P_{\alpha +1}(-1)]\}{r}^{\alpha }{P}_{\alpha }\left(x\right)$

Step-by-step solution

Given Information

The surface temperature of sphere of radius 1 is$5{\cos }^3\theta -3{\sin }^2\theta$.

Definition of steady-state temperature

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.

Calculate the steady-state temperature distribution function

Take the steady-state temperature distribution function as $u(r,\theta )$.

Consider the given surface temperature as$A\left(\theta \right)$.

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

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

For simplicity consider $x=\cos \theta$

$\sum _{l=0}^{\infty }{c}_l{1}^l{P}_l\left(x\right)=5x^3+3x^2-3$ ….. (1)

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

$\sum _{l=0}^{\infty }{c}_l\underset{-1}{\overset{1}{\int }}{P}_l\left(x\right)P_{\alpha }\left(x\right)=\underset{-1}{\overset{1}{\int }}(5x^3+3x^2-3)P_{\alpha }\left(x\right)dx$ ….. (2)

The Orthogonal relation of Legendre polynomials:

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

The Identity of Legendre polynomials

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

Use equation (3) and (4) in equation (2):

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

$\sum _{l=0}^{\infty }{c}_l\underset{-1}{\overset{1}{\int }}{P}_l\left(x\right)P_{\alpha }\left(x\right)=\underset{-1}{\overset{1}{\int }}(5x^3+3x^2-3)P_{\alpha }\left(x\right)dx$

$\begin{array}{l}\sum _{l=0}^{\infty }{c}_l\frac{2}{(2l+1)}{\delta }_{l,\alpha }=\underset{-1}{\overset{1}{\int }}(5x^3+3x^2)dx-3\underset{-1}{\overset{1}{\int }}{P}_{\alpha }\left(x\right)dx\\ c_{\alpha }\frac{2}{(2\alpha +1)}=\underset{-1}{\overset{1}{\int }}(5x^3+3x^2)dx-\frac{3}{(2\alpha +1)}[P_{\alpha -1}(-1)-P_{\alpha +1}(-1)]\\ c_{\alpha }=\frac{(2\alpha +1)}{2}\underset{-1}{\overset{1}{\int }}(5x^3+3x^2)dx-\frac{3}{2}[P_{\alpha -1}(-1)-P_{\alpha +1}(-1)]\end{array}$

$u(r,\theta )=\sum _{\alpha =0}^{\infty }\{\frac{(2\alpha +1)}{2}\underset{-1}{\overset{1}{\int }}(5x^3+3x^2)dx-\frac{3}{2}[P_{\alpha -1}(-1)-P_{\alpha +1}(-1)]\}{r}^{\alpha }{P}_{\alpha }\left(x\right)$

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

$u(r,\theta )=\sum _{\alpha =0}^{\infty }\{\frac{(2\alpha +1)}{2}\underset{-1}{\overset{1}{\int }}(5x^3+3x^2)dx-\frac{3}{2}[P_{\alpha -1}(-1)-P_{\alpha +1}(-1)]\}{r}^{\alpha }{P}_{\alpha }\left(x\right)$