91影视

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$螖u=0$.

Write Laplace operator in spherical coordinate.

$螖=\frac{1}{r^2}\frac{鈭�}{鈭�r}\left(r^2\frac{鈭�}{鈭�r}\right)+\frac{1}{r^2\sin 胃}\frac{鈭�}{鈭�胃}\left(\sin 胃\frac{鈭�}{鈭�胃}\right)+\frac{1}{r^2{\sin }^2胃}\frac{{鈭�}^2}{鈭�{蠒}^2}$

Solve the equation

$u(r,胃,蠒)=R\left(r\right)W\left(胃\right)Q\left(蠒\right)$

$\begin{array}{c}\frac{WQ}{r^2}\frac{鈭�}{鈭�r}\left(r^2\frac{鈭�R}{鈭�r}\right)+\frac{RQ}{r^2\sin 胃}\frac{鈭�}{鈭�胃}\left(\sin 胃\frac{鈭�W}{鈭�胃}\right)+\frac{RW}{r^2{\sin }^2胃}\frac{{鈭�}^{2}Q}{鈭�{蠒}^2}=0\\ \frac{1}{R}\frac{鈭�}{鈭�r}\left(r^2\frac{鈭�R}{鈭�r}\right)+\frac{1}{W\sin 胃}\frac{鈭�}{鈭�胃}\left(\sin \left(胃\right)\frac{鈭�W}{鈭�胃}\right)+\frac{1}{Q{\sin }^2胃}\frac{{鈭�}^{2}Q}{鈭�{蠒}^2}=0\end{array}$

Write equation for $胃$dependence as below.

$\begin{array}{l}\frac{1}{R}\frac{鈭�}{鈭�r}\left(r^2\frac{鈭�R}{鈭�r}\right)+\frac{1}{W\sin 胃}\frac{鈭�}{鈭�胃}\left(\sin 胃\frac{鈭�W}{鈭�胃}\right)鈭�\frac{m^2}{{\sin }^2胃}=0\\ 鈭�\frac{1}{W\sin 胃}\frac{鈭�}{鈭�胃}\left(\sin 胃\frac{鈭�W}{鈭�胃}\right)+\frac{m^2}{{\sin }^2胃}=伪\\ \sin 胃\frac{鈭�}{鈭�胃}\left(\sin 胃\frac{鈭�W}{鈭�胃}\right)鈭�m^{2}W+伪{\sin }^2\left(胃\right)W=0\end{array}$

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

Write equation for $蠒$-dependence.

$\begin{array}{l}鈭�\frac{1}{Q}\frac{{鈭�}^{2}Q}{鈭�{蠒}^2}=m^2\\ \frac{{鈭�}^{2}Q}{鈭�{蠒}^2}+m^{2}Q=0\end{array}$

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

Write equation for r-dependence.

$\begin{array}{l}\frac{1}{R}\frac{鈭�}{鈭�r}\left(r^2\frac{鈭�R}{鈭�r}\right)=伪\\ \frac{鈭�}{鈭�r}\left(r^2\frac{鈭�R}{鈭�r}\right)鈭�伪R=0\\ \frac{鈭�}{鈭�r}\left(r^2\frac{鈭�R}{鈭�r}\right)鈭�伪R=0\\ r^2\frac{{鈭�}^{2}R}{鈭�r^2}+2r\frac{鈭�R}{鈭�r}鈭�l(l+1)R=0\end{array}$

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

Therefore, the steady-state temperature distribution in a spherical shell is $\begin{array}{c}u(r,胃,蠒)=u^{\left(i\right)}(r,胃,蠒)+u^{\left(e\right)}(r,胃,蠒)\\ =\underset{l,m=0}{\overset{\mathrm{鈭�}}{鈭�}}{A}_l^{\left(i\right)}{r}^l{P}_l^m\left(\cos 胃\right)e^{卤im蠒}+\underset{l,m=0}{\overset{\mathrm{鈭�}}{鈭�}}{B}_l^{\left(e\right)}{r}^{鈭�(l+1)}{P}_l^m\left(\cos 胃\right)e^{卤im蠒}\end{array}$