91Ó°ÊÓ

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 $u(r,\mathrm{θ})$inside a sphere with a radius of r = 1 in this issue. The surface temperature function is written as$A(\mathrm{θ},\mathrm{ϕ})$.

Consider the equation below:

$\begin{array}{c}A(\mathrm{θ},\mathrm{ϕ})={\sin }^2\left(\mathrm{θ}\right)\cos \left(\mathrm{θ}\right)\cos \left(2\mathrm{ϕ}\right)−\cos \left(\mathrm{θ}\right)\\ =(1−{\cos }^2\left(\mathrm{θ}\right))\cos \left(\mathrm{θ}\right)(\cos \left(2\mathrm{ϕ}\right)−1)\end{array}$

The steady-state temperature distribution function $u(r,\mathrm{θ},\mathrm{ϕ})$is directly dependent on the related Legendre polynomials $P_l^m\left(\cos \left(\mathrm{θ}\right)\right)$due to the $\mathrm{ϕ}$dependency in the surface temperature function $A(\mathrm{θ},\mathrm{ϕ})$. As a result, the $u(r,\mathrm{θ},\mathrm{ϕ})$Power series has the form as below.

$u(r,\mathrm{θ},\mathrm{ϕ})=\underset{l=0}{\overset{\mathrm{∞}}{∑}}{c}_l\left(\mathrm{ϕ}\right)r^l{P}_l^m\left(\cos \left(\mathrm{θ}\right)\right)$

Now, Let x be$\cos \mathrm{θ}$.

$u_{r=1}\left(x\right)=\underset{l=0}{\overset{\mathrm{∞}}{∑}}{c}_l{P}_l^2\left(x\right)$

$u_{r=1}\left(x\right)=(1−x^2)x(\cos \left(2\mathrm{ϕ}\right)−1)$ ….. (1)

Multiply equation (1) by $P_{\mathrm{ν}}\left(x\right)$and integrate.

$\underset{l=0}{\overset{\mathrm{∞}}{∑}}{c}_l\underset{−1}{\overset{1}{∫}}{P_l}^2\left(x\right)P_{\mathrm{ν}}^2\left(x\right)dx=\underset{−1}{\overset{1}{∫}}(1−x^2)x{P}_{\mathrm{ν}}^2\left(x\right)dx(\cos \left(2\mathrm{ϕ}\right)−1)\text{ }$ ….. (2)

Write the Legendre polynomials with associated orthogonality:

$\underset{−1}{\overset{1}{∫}}{P_l}^2\left(x\right)P_{\mathrm{ν}}^2\left(x\right)dx=\frac{2}{(2l+1)}\frac{(l+2)!}{(l−2)!}{\mathrm{δ}}_{l,\mathrm{ν}}$ ….. (3)

$\begin{array}{c}\underset{−1}{\overset{1}{∫}}{P_l}^2\left(x\right)P_{\mathrm{ν}}^2\left(x\right)dx=\frac{2}{(2l+1)}\frac{(l+2)(l+1)(l−1)(l−2)!}{(l−2)!}{\mathrm{δ}}_{l,\mathrm{ν}}\\ =\frac{2(l+2)(l+1)(l−1)}{(2l+1)}{\mathrm{δ}}_{l,\mathrm{ν}}\end{array}$

Substitute equation (3) into equation (2).

$\begin{array}{l}\underset{l=0}{\overset{\mathrm{∞}}{∑}}{c}_l\frac{2(l+2)(l+1)(l−1)}{(2l+1)}{\mathrm{δ}}_{l,\mathrm{ν}}=\underset{−1}{\overset{1}{∫}}(x−x^3)P_{\mathrm{ν}}\left(x\right)dx(\cos \left(2\mathrm{ϕ}\right)−1)\\ c_{\mathrm{ν}}\frac{2(\mathrm{ν}+2)(\mathrm{ν}+1)(\mathrm{ν}−1)}{(2\mathrm{ν}+1)}=\underset{−1}{\overset{1}{∫}}(x−x^3)P_{\mathrm{ν}}\left(x\right)dx(\cos \left(2\mathrm{ϕ}\right)−1)\\ c_{\mathrm{ν}}=\frac{(2\mathrm{ν}+1)}{2(\mathrm{ν}+2)(\mathrm{ν}+1)(\mathrm{ν}−1)}\underset{−1}{\overset{1}{∫}}(x−x^3){P_{\mathrm{ν}}}^2\left(x\right)dx(\cos \left(2\mathrm{ϕ}\right)−1)\end{array}$

$u(r,\mathrm{θ},\mathrm{ϕ})=\underset{\mathrm{ν}=0}{\overset{\mathrm{∞}}{∑}}\left\{\frac{(2\mathrm{ν}+1)}{2(\mathrm{ν}+2)(\mathrm{ν}+1)(\mathrm{ν}−1)}\underset{−1}{\overset{1}{∫}}(x−x^3){P_{\mathrm{ν}}}^2\left(x\right)dx\right\}{r}^{\mathrm{ν}}{P}_{\mathrm{ν}}^2\left(x\right)(\cos \left(2\mathrm{ϕ}\right)−1)$

Therefore, the steady-state temperature distribution inside a sphere of radius 1 is

$u(r,\mathrm{θ},\mathrm{ϕ})=\underset{\mathrm{ν}=0}{\overset{\mathrm{∞}}{∑}}\left\{\frac{(2\mathrm{ν}+1)}{2(\mathrm{ν}+2)(\mathrm{ν}+1)(\mathrm{ν}−1)}\underset{−1}{\overset{1}{∫}}(x−x^3){P_{\mathrm{ν}}}^2\left(x\right)dx\right\}{r}^{\mathrm{ν}}{P}_{\mathrm{ν}}^2\left(x\right)(\cos \left(2\mathrm{ϕ}\right)−1)$.