Question

Do Problem 11 if the curved surface is held at ${\mathbf{cos}}^{\mathbf{2}}\mathbf{\theta }$and the equatorial plane at zero. Careful: The answer does not involve ${\mathbf{P}}_{\mathbf{2}}$; read the last sentence of this section.

Short answer

Therefore, the steady-state temperature distribution inside a hemisphere is$u(r,\theta )=\sum _{s=0}^{\infty }\{[P_{s-1}\left(0\right)-P_{s+1}\left(0\right)]-(2s+1)P_s\left(1\right)\}{\left(\frac{r}{a}\right)}^s{P}_s\left(x\right)$.

Step-by-step solution

Given Information:

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

Definition of steady-state temperature:

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.

Calculate the steady-state temperature distribution function:

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

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

$A\left(\theta \right)=\left\{\begin{array}{l}\begin{array}{ll}{\cos }^2\theta & \leftarrow 0<\theta <\frac{\pi }{2}\end{array}\\ \begin{array}{ll}0& \leftarrow \frac{\pi }{2}<\theta <\pi \end{array}\end{array}\right.$

Substitute $x=\cos \theta$and$\theta \to x$.

$\begin{array}{c}A\left(x\right)=\left\{\begin{array}{l}\begin{array}{ll}0& \leftarrow -1<x<0\end{array}\\ \begin{array}{ll}{x}^2& \leftarrow 0<x<1\end{array}\end{array}\right.\\ =x^2\left\{\begin{array}{l}0\leftarrow -1<x<0\\ 1\leftarrow 0<x<1\end{array}\right.\\ =x^{2}f\left(x\right)\end{array}$

Simplify Legendre polynomials:

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

$u_{r=a}\left(x\right)=\sum _{l=0}^{\infty }cl{a}^l{P}_l\left(x\right)$

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

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

$\sum _{l=0}^{\infty }{c}_l{a}^l\underset{0}{\overset{1}{\int }}{P}_l\left(x\right)P_s\left(x\right)dx=\underset{0}{\overset{1}{\int }}{x}^{2}f\left(x\right)P_s\left(x\right)dx$ ….. (2)

The orthogonal relation of standard Legendre polynomials is as follows:

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

$\underset{0}{\overset{1}{\int }}{P}_l\left(x\right)P_s\left(x\right)dx=\frac{1}{2}\frac{2}{(2l+1)}{\delta }_{l,s}$ ….. (3)

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

Use integration by part in equation (2).

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

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

Calculate the value of coefficient$c_l$.

Substitute equation (3) in equation (2).

$\begin{array}{l}\sum _{l=0}^{\infty }{c}_l{a}^l\frac{1}{2}\frac{2}{(2l+1)}{\delta }_{l,s}=-P_s\left(1\right)+\frac{2}{(2s+1)}[P_{s-1}\left(0\right)-P_{s+1}\left(0\right)]\\ c_s{a}^s\frac{1}{(2s+1)}=\frac{1}{(2s+1)}[P_{s-1}\left(0\right)-P_{s+1}\left(0\right)]-P_s\left(1\right)\\ c_s=\frac{1}{a^s}\left([P_{s-1}\left(0\right)-P_{s+1}\left(0\right)]\right)-(2s+1)P_s\left(1\right))\end{array}$

$u(r,\theta )=\sum _{s=0}^{\infty }[P_{s-1}\left(0\right)-P_{s+1}\left(0\right)]-(2s+1)P_s\left(1\right){\left(\frac{r}{a}\right)}^s{P}_s\left(x\right)$

Therefore, the steady-state temperature distribution inside a hemisphere is$u(r,\theta )=\sum _{s=0}^{\infty }[P_{s-1}\left(0\right)-P_{s+1}\left(0\right)]-(2s+1)P_s\left(1\right){\left(\frac{r}{a}\right)}^s{P}_s\left(x\right)$.