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

$\mathbf{3}\mathbf{sin\theta cos\theta sin\varphi }$Hint: See equation (7.10) and Chapter 12, equation (10.6).

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:

$\begin{array}{c}A(\theta ,\varphi )=3\sin \left(\theta \right)\cos \left(\theta \right)\sin \left(\varphi \right)\\ =3\sqrt{1-{\cos }^2\left(\theta \right)}\cos \left(\theta \right)\sin \left(\varphi \right)\end{array}$

The surface temperature function $A(\theta ,\varphi )$, unlike previous problems, is dependent on$\varphi$.

The steady-state temperature distribution function will be written in a Power series that explicitly includes the corresponding Legendre polynomials $P_l^m\left(\cos \left(\theta \right)\right)$due to the $\varphi$dependence. From an algebraic standpoint, the analogous Power series is of the type .

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

Its coefficients are calculated by expressing it as $P_l^m\left(\cos \left(\theta \right)\right)$in terms of the corresponding Legendre polynomials. Let $x=\cos \left(\theta \right)$be the chosen value.

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

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

The term $\sin \left(m\varphi \right)$as one functional portion of the spherical harmonics $Y_l^m(\theta ,\varphi )$determines the of the corresponding Legendre polynomials $P_l^m\left(\cos \left(\theta \right)\right)$. As can be observed from the surface temperature function $A(x,\varphi )=3\sqrt{1-x^2}x\sin \left(\varphi \right)$, where $m=1$in this case.

Write the Legendre polynomials with associated orthogonally:

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

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

Standard Legendre polynomials have the following identity:

$\underset{a}{\overset{1}{\int }}{P}_n\left(x\right)dx=\frac{1}{(2n+1)}[P_{n-1}\left(a\right)-P_{n+1}\left(a\right)]$ ….. (3)

Use integration by part in $\int \sqrt{1-x^2}x{P}_{\tau }^1\left(x\right)dx$:

$\begin{array}{c}\int \sqrt{1-x^2}x{P}_{\tau }^1\left(x\right)dx=\int (1-x^2)x{P}_{\tau }\left(x\right)dx\\ =\int (x-x^3)P_{\tau }\left(x\right)dx\\ =(x-x^3)P_{\tau }\left(x\right)-\int (1-3x^2)P_{\tau }\left(x\right)dx\\ =(x-x^3)P_{\tau }\left(x\right)-[(1-3x^2)P_{\tau }\left(x\right)-\int (-6x)P_{\tau }\left(x\right)dx]\end{array}$

$\begin{array}{c}\int \sqrt{1-x^2}x{P}_{\tau }^1\left(x\right)dx=(x-x^3)P_{\tau }\left(x\right)-(1-3x^2)P_{\tau }\left(x\right)+[(-6x)P_{\tau }\left(x\right)-\int (-6)P_{\tau }\left(x\right)dx]\\ =(x-x^3)P_{\tau }\left(x\right)-(1-3x^2)P_{\tau }\left(x\right)-\left(6x\right)P_{\tau }\left(x\right)+6\int P_{\tau }\left(x\right)dx\end{array}$

Equations (2) and (3), as well as the multiple integration section, should be inserted into equation (1).

$\begin{array}{l}\sum _{l=0}^{\infty }{c}_l\frac{2l(l+1)}{(2l+1)}{\delta }_{l,\tau }=3(-12(P_{\tau }\left(1\right)-P_{\tau }(-1))+\frac{6}{(2\tau +1)}[P_{\tau -1}(-1)-P_{\tau +1}(-1)])\\ c_{\tau }\frac{2\tau (\tau +1)}{(2\tau +1)}=-36(P_{\tau }\left(1\right)-P_{\tau }(-1))+\frac{18}{(2\tau +1)}[P_{\tau -1}(-1)-P_{\tau +1}(-1)])\\ c_{\tau }=\frac{-18(2\tau +1)}{\tau (\tau +1)}(P_{\tau }\left(1\right)-P_{\tau }(-1))+\frac{9}{\tau (\tau +1)}[P_{\tau -1}(-1)-P_{\tau +1}(-1)]\end{array}$

$u(r,\theta ,\varphi )=\sum _{\tau =0}^{\infty }\{\frac{-18(2\tau +1)}{\tau (\tau +1)}(P_{\tau }\left(1\right)-P_{\tau }(-1))+\frac{9}{\tau (\tau +1)}[P_{\tau -1}(-1)-P_{\tau +1}(-1)]\}{r}^{\tau }{P}_{\tau }^1\left(x\right)\sin \left(\varphi \right)$

Therefore, the steady-state temperature distribution inside a sphere of radius 1 is$u(r,\theta ,\varphi )=\sum _{\tau =0}^{\infty }\{\frac{-18(2\tau +1)}{\tau (\tau +1)}(P_{\tau }\left(1\right)-P_{\tau }(-1))+\frac{9}{\tau (\tau +1)}[P_{\tau -1}(-1)-P_{\tau +1}(-1)]\}{r}^{\tau }{P}_{\tau }^1\left(x\right)\sin \left(\varphi \right)$.