Find the steady-state temperature distribution in a spherical shell of inner radius 1 and outer radius 2 if the inner surface is held at $\mathbf{0}\mathbf{°}$and the outer surface has its upper half at $\mathbf{100}\mathbf{°}$and its lower half at role="math" localid="1664359640240" $\mathbf{0}\mathbf{°}$. Hint: r = 0 is not in the region of interest, so the solutions ${\mathit{r}}^{\mathbf{-}\mathbf{l}\mathbf{-}\mathbf{1}}$in (7.9) should be included. Replace ${\mathit{c}}_{\mathbf{l}}{\mathit{r}}^{\mathbf{l}}$in (7.11) by$\mathbf{(}\mathbf{c}\mathbf{l}{\mathbf{r}}^{\mathbf{l}}\mathbf{+}\mathbf{b}\mathbf{l}{\mathbf{r}}^{\mathbf{-}\mathbf{l}\mathbf{-}\mathbf{1}}\mathbf{)}$.

The inner radius of spherical shell is 1 and outer radius of spherical shell is 2.

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.

Solve Laplace equation for a sphere having radius r = a where$\Delta u=0$.

Write Laplace operator in spherical coordinate.

$\Delta =\frac{1}{r^2}\frac{\partial }{\partial r}\left(r^2\frac{\partial }{\partial r}\right)+\frac{1}{r^2\sin \theta }\frac{\partial }{\partial \theta }\left(\sin \theta \frac{\partial }{\partial \theta }\right)+\frac{1}{r^2{\sin }^2\theta }\frac{{\partial }^2}{\partial {\varphi }^2}$

Solve the equation

$u(r,\theta ,\varphi )=R\left(r\right)W\left(\theta \right)Q\left(\varphi \right)$

$\begin{array}{c}\frac{WQ}{r^2}\frac{\partial }{\partial r}\left(r^2\frac{\partial R}{\partial r}\right)+\frac{RQ}{r^2\sin \theta }\frac{\partial }{\partial \theta }\left(\sin \theta \frac{\partial W}{\partial \theta }\right)+\frac{RW}{r^2{\sin }^2\theta }\frac{{\partial }^{2}Q}{\partial {\varphi }^2}=0\\ \frac{1}{R}\frac{\partial }{\partial r}\left(r^2\frac{\partial R}{\partial r}\right)+\frac{1}{W\sin \theta }\frac{\partial }{\partial \theta }\left(\sin \left(\theta \right)\frac{\partial W}{\partial \theta }\right)+\frac{1}{Q{\sin }^2\theta }\frac{{\partial }^{2}Q}{\partial {\varphi }^2}=0\end{array}$

Write equation for $\theta$dependence as below.

$\begin{array}{l}\frac{1}{R}\frac{\partial }{\partial r}\left(r^2\frac{\partial R}{\partial r}\right)+\frac{1}{W\sin \theta }\frac{\partial }{\partial \theta }\left(\sin \theta \frac{\partial W}{\partial \theta }\right)-\frac{m^2}{{\sin }^2\theta }=0\\ -\frac{1}{W\sin \theta }\frac{\partial }{\partial \theta }\left(\sin \theta \frac{\partial W}{\partial \theta }\right)+\frac{m^2}{{\sin }^2\theta }=\alpha \\ \sin \theta \frac{\partial }{\partial \theta }\left(\sin \theta \frac{\partial W}{\partial \theta }\right)-m^{2}W+\alpha {\sin }^2\left(\theta \right)W=0\end{array}$

$W\left(\theta \right)=P_l^m\left(\cos \theta \right)$

Write equation for $\varphi$-dependence.

$\begin{array}{l}-\frac{1}{Q}\frac{{\partial }^{2}Q}{\partial {\varphi }^2}=m^2\\ \frac{{\partial }^{2}Q}{\partial {\varphi }^2}+m^{2}Q=0\end{array}$

$\begin{array}{c}Q\left(\varphi \right)=e^{\pm im\varphi }\\ =\{\begin{array}{l}\mathrm{Re}\left(Q\right)=\cos \left(m\varphi \right)\\ \mathrm{Im}\left(Q\right)=\sin \left(m\varphi \right)\end{array}\end{array}$

Write equation for r-dependence.

$\begin{array}{l}\frac{1}{R}\frac{\partial }{\partial r}\left(r^2\frac{\partial R}{\partial r}\right)=\alpha \\ \frac{\partial }{\partial r}\left(r^2\frac{\partial R}{\partial r}\right)-\alpha R=0\\ \frac{\partial }{\partial r}\left(r^2\frac{\partial R}{\partial r}\right)-\alpha R=0\\ r^2\frac{{\partial }^{2}R}{\partial r^2}+2r\frac{\partial R}{\partial r}-l(l+1)R=0\end{array}$

The general solution is $R_l\left(r\right)=A_l{r}^l+B_l{r}^{-(l+1)}$

Therefore, the steady-state temperature distribution in a spherical shell is $\begin{array}{c}u(r,\theta ,\varphi )=u^{\left(i\right)}(r,\theta ,\varphi )+u^{\left(e\right)}(r,\theta ,\varphi )\\ =\sum _{l,m=0}^{\infty }{A}_l^{\left(i\right)}{r}^l{P}_l^m\left(\cos \theta \right)e^{\pm im\varphi }+\sum _{l,m=0}^{\infty }{B}_l^{\left(e\right)}{r}^{-(l+1)}{P}_l^m\left(\cos \theta \right)e^{\pm im\varphi }\end{array}$