91影视

The angle of the curved surface of hemisphere is ${\cos }^2胃$and with the equatorial plane at$0掳$.

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 $u(r,胃)$inside a hemisphere with a radius of a' in this problem. The equatorial plane is maintained at a temperature of $0掳$, whereas the spherical surface $A\left(胃\right)$is maintained at a temperature of${\cos }^2胃$.

Because there is no $蠒$dependence, express $A\left(胃\right)$in terms of standard Legendre polynomials in order to determine the associated coefficients $c_l$inside the Power series.

$A\left(胃\right)=\left\{\begin{array}{l}\begin{array}{ll}{\cos }^2胃& 鈫�0<胃<\frac{蟺}{2}\end{array}\\ \begin{array}{ll}0& 鈫�\frac{蟺}{2}<胃<蟺\end{array}\end{array}\right.$

Substitute $x=\cos 胃$and$胃鈫�x$.

$\begin{array}{c}A\left(x\right)=\left\{\begin{array}{l}\begin{array}{ll}0& 鈫�鈭�1<x<0\end{array}\\ \begin{array}{ll}{x}^2& 鈫�0<x<1\end{array}\end{array}\right.\\ =x^2\left\{\begin{array}{l}0鈫�鈭�1<x<0\\ 1\text{鈥夆赌夆赌�}鈫�0<x<1\end{array}\right.\\ =x^{2}f\left(x\right)\end{array}$

Write $A\left(x\right)$standard Legendre polynomials $P_l\left(x\right)$for radius r = a.

$u_{r=a}\left(x\right)=\underset{l=0}{\overset{\mathrm{鈭�}}{鈭�}}cl{a}^l{P}_l\left(x\right)$

$u_{r=a}\left(x\right)=x^{2}f\left(x\right)$ 鈥�.. (1)

Multiply $P_s\left(x\right)$in equation (1) and integrate.

$\underset{l=0}{\overset{\mathrm{鈭�}}{鈭�}}{c}_l{a}^l\underset{0}{\overset{1}{鈭�}}{P}_l\left(x\right)P_s\left(x\right)dx=\underset{0}{\overset{1}{鈭�}}{x}^{2}f\left(x\right)P_s\left(x\right)dx$ 鈥�.. (2)

The orthogonal relation of standard Legendre polynomials is as follows:

$\underset{0}{\overset{1}{鈭�}}{P}_l\left(x\right)P_s\left(x\right)dx=\frac{1}{2}{鈭�}_{鈭�1}^1{P}_l\left(x\right)P_s\left(x\right)dx$

$\underset{0}{\overset{1}{鈭�}}{P}_l\left(x\right)P_s\left(x\right)dx=\frac{1}{2}\frac{2}{(2l+1)}{未}_{l,s}$ 鈥�.. (3)

Use identity of Legendre polynomial. i.e.$\underset{b}{\overset{1}{鈭�}}{P}_n\left(x\right)dx=\frac{1}{(2n+1)}[P_{n鈭�1}\left(b\right)鈭�P_{n+1}\left(b\right)]$

Use integration by part in equation (2).

$\begin{array}{c}\underset{0}{\overset{1}{鈭�}}{x}^{2}f\left(x\right)P_s\left(x\right)dx=\underset{0}{\overset{1}{鈭�}}{x}^2{P}_s\left(x\right)dx\\ ={x^2{P}_s\left(x\right)|}_0^1鈭�\underset{0}{\overset{1}{鈭�}}2x{P}_s\left(x\right)dx\\ =P_s\left(1\right)鈭�{2x{P}_s\left(x\right)|}_0^1+2\underset{0}{\overset{1}{鈭�}}{P}_s\left(x\right)dx\\ =P_s\left(1\right)鈭�2P_s\left(1\right)+\frac{2}{(2s+1)}[P_{s鈭�1}\left(0\right)鈭�P_{s+1}\left(0\right)]\end{array}$

$\underset{0}{\overset{1}{鈭�}}{x}^{2}f\left(x\right)P_s\left(x\right)dx=鈭�P_s\left(1\right)+\frac{2}{(2s+1)}[P_{s鈭�1}\left(0\right)鈭�P_{s+1}\left(0\right)]$

Calculate the value of coefficient$c_l$.

Substitute equation (3) in equation (2).

$\begin{array}{l}\underset{l=0}{\overset{\mathrm{鈭�}}{鈭�}}{c}_l{a}^l\frac{1}{2}\frac{2}{(2l+1)}{未}_{l,s}=鈭�P_s\left(1\right)+\frac{2}{(2s+1)}[P_{s鈭�1}\left(0\right)鈭�P_{s+1}\left(0\right)]\\ c_s{a}^s\frac{1}{(2s+1)}=\frac{1}{(2s+1)}[P_{s鈭�1}\left(0\right)鈭�P_{s+1}\left(0\right)]鈭�P_s\left(1\right)\\ c_s=\frac{1}{a^s}\left([P_{s鈭�1}\left(0\right)鈭�P_{s+1}\left(0\right)]\right)鈭�(2s+1)P_s\left(1\right))\end{array}$

$u(r,胃)=\underset{s=0}{\overset{\mathrm{鈭�}}{鈭�}}[P_{s鈭�1}\left(0\right)鈭�P_{s+1}\left(0\right)]鈭�(2s+1)P_s\left(1\right){\left(\frac{r}{a}\right)}^s{P}_s\left(x\right)$

Therefore, the steady-state temperature distribution inside a hemisphere is$u(r,胃)=\underset{s=0}{\overset{\mathrm{鈭�}}{鈭�}}[P_{s鈭�1}\left(0\right)鈭�P_{s+1}\left(0\right)]鈭�(2s+1)P_s\left(1\right){\left(\frac{r}{a}\right)}^s{P}_s\left(x\right)$.