91影视

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 $u(r,胃)$inside a sphere with a radius of r = 1 in this issue. The surface temperature function is written as$A(胃,蠒)$.

Consider the equation below:

$\begin{array}{c}A(胃,蠒)=3\sin \left(胃\right)\cos \left(胃\right)\sin \left(蠒\right)\\ =3\sqrt{1鈭�{\cos }^2\left(胃\right)}\cos \left(胃\right)\sin \left(蠒\right)\end{array}$

The surface temperature function $A(胃,蠒)$, unlike previous problems, is dependent on$蠒$.

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

$u(r,胃)=\underset{l=0}{\overset{\mathrm{鈭�}}{鈭�}}{c}_l{r}^l{P}_l^m\left(\cos \left(胃\right)\right)$

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

$u_{r=1}\left(x\right)=\underset{l=0}{\overset{\mathrm{鈭�}}{鈭�}}{c}_l{P}_l^1\left(x\right)$

$u_{r=1}\left(x\right)=3\sqrt{1鈭�x^2}x\sin \left(蠒\right)\text{鈥�}$ 鈥�.. (1)

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

Write the Legendre polynomials with associated orthogonally:

$\underset{鈭�1}{\overset{1}{鈭�}}{P_l}^1\left(x\right)P_{蟿}^1\left(x\right)dx=\frac{2}{(2l+1)}\frac{(l+1)!}{(l鈭�1)!}{未}_{l,蟿}$ 鈥�.. (2)

$\begin{array}{c}\underset{鈭�1}{\overset{1}{鈭�}}{P_l}^1\left(x\right)P_{蟿}^1\left(x\right)dx=\frac{2}{(2l+1)}\frac{(l+1)!}{(l鈭�1)!}{未}_{l,蟿}\\ =\frac{2}{(2l+1)}\frac{(l+1)l(l鈭�1)!}{(l鈭�1)!}{未}_{l,蟿}\\ =\frac{2l(l+1)}{(2l+1)}{未}_{l,蟿}\end{array}$

Standard Legendre polynomials have the following identity:

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

Use integration by part in $鈭�\sqrt{1鈭�x^2}x{P}_{蟿}^1\left(x\right)dx$:

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

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

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

$\begin{array}{l}\underset{l=0}{\overset{\mathrm{鈭�}}{鈭�}}{c}_l\frac{2l(l+1)}{(2l+1)}{未}_{l,蟿}=3(鈭�12(P_{蟿}\left(1\right)鈭�P_{蟿}(鈭�1))+\frac{6}{(2蟿+1)}[P_{蟿鈭�1}(鈭�1)鈭�P_{蟿+1}(鈭�1)])\\ c_{蟿}\frac{2蟿(蟿+1)}{(2蟿+1)}=鈭�36(P_{蟿}\left(1\right)鈭�P_{蟿}(鈭�1))+\frac{18}{(2蟿+1)}[P_{蟿鈭�1}(鈭�1)鈭�P_{蟿+1}(鈭�1)])\\ c_{蟿}=\frac{鈭�18(2蟿+1)}{蟿(蟿+1)}(P_{蟿}\left(1\right)鈭�P_{蟿}(鈭�1))+\frac{9}{蟿(蟿+1)}[P_{蟿鈭�1}(鈭�1)鈭�P_{蟿+1}(鈭�1)]\end{array}$

$u(r,胃,蠒)=\underset{蟿=0}{\overset{\mathrm{鈭�}}{鈭�}}\{\frac{鈭�18(2蟿+1)}{蟿(蟿+1)}(P_{蟿}\left(1\right)鈭�P_{蟿}(鈭�1))+\frac{9}{蟿(蟿+1)}[P_{蟿鈭�1}(鈭�1)鈭�P_{蟿+1}(鈭�1)]\}{r}^{蟿}{P}_{蟿}^1\left(x\right)\sin \left(蠒\right)$

Therefore, the steady-state temperature distribution inside a sphere of radius 1 is$u(r,胃,蠒)=\underset{蟿=0}{\overset{\mathrm{鈭�}}{鈭�}}\{\frac{鈭�18(2蟿+1)}{蟿(蟿+1)}(P_{蟿}\left(1\right)鈭�P_{蟿}(鈭�1))+\frac{9}{蟿(蟿+1)}[P_{蟿鈭�1}(鈭�1)鈭�P_{蟿+1}(鈭�1)]\}{r}^{蟿}{P}_{蟿}^1\left(x\right)\sin \left(蠒\right)$.