Find the steady-state temperature distribution inside a hemisphere if the spherical surface is held at $\mathbf{100}\mathbf{°}$and the equatorial plane at $\mathbf{0}\mathbf{°}$. Hint: See the last paragraph of this section above.

The angle of the spherical surface of hemisphere is and with equatorial plane at$0°$.

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 $u(r,\theta )$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 $100°$. Because there is no $\varphi$dependence, we express $A\left(\theta \right)$in terms of standard Legendre polynomials in order to determine the associated coefficients $c_l$inside the Power series.

role="math" localid="1664358604445" $A\left(\theta \right)=\left\{\begin{array}{l}\begin{array}{ll}100& \leftarrow 0<\theta <\frac{\pi }{2}\end{array}\\ \begin{array}{ll}0& \leftarrow \frac{\pi }{2}<\theta <\pi \end{array}\end{array}\right.$

Substitute $x=\cos \left(\theta \right)$and$\theta \to x$.

$\begin{array}{c}A\left(x\right)=\left\{\begin{array}{l}\begin{array}{ll}0& \leftarrow -1<x<0\end{array}\\ \begin{array}{ll}100& \leftarrow 0<x<1\end{array}\end{array}\right.\\ =100\left\{\begin{array}{l}0\leftarrow -1<x<0\\ 1\leftarrow 0<x<1\end{array}\right.\\ =100f\left(x\right)\end{array}$

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

$u_{T=a^{\text{'}}}\left(x\right)=\sum _{l=0}^{\infty }{c}_l{a}^{ll}{P}_l\left(x\right)$

$u_{T=a^{\text{'}}}\left(x\right)=100f\left(x\right)$ ….. (1)

Calculate the value of coefficient $c_l$.

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

$\sum _{l=0}^{\infty }{c}_l{\left(a^{\text{'}}\right)}^l\underset{0}{\overset{1}{\int }}{P}_l\left(x\right)P_k\left(x\right)dx=\underset{0}{\overset{1}{\int }}100f\left(x\right)P_k\left(x\right)dx$ ….. (2)

The orthogonal relation of standard Legendre polynomials is as follows:

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

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

Substitute equation (3) and use above identity in equation (2).

$\sum _{l=0}^{\infty }{c}_l{\left(a^{\text{'}}\right)}^l\frac{1}{2}\frac{2}{(2l+1)}{\delta }_{l,k}=100\underset{0}{\overset{1}{\int }}f\left(x\right)P_k\left(x\right)dx$

$\begin{array}{c}{c}_k{\left(a^{\text{'}}\right)}^k\frac{1}{(2k+1)}=100\underset{0}{\overset{1}{\int }}{P}_k\left(x\right)dx\\ =100\frac{1}{(2k+1)}[P_{k-1}\left(0\right)-P_{k+1}\left(0\right)]\end{array}$

$c_k=\frac{100}{{\left(a^{\text{'}}\right)}^k}[P_{k-1}\left(0\right)-P_{k+1}\left(0\right)]$

$u(r,\theta )=100\sum _{k=0}^{\infty }{\left(\frac{r}{a^{\text{'}}}\right)}^k[P_{k-1}\left(0\right)-P_{k+1}\left(0\right)]P_k\left(x\right)$

Therefore, the steady-state temperature distribution inside a hemisphere is$u(r,\theta )=100\sum _{k=0}^{\infty }{\left(\frac{r}{a^{\text{'}}}\right)}^k[P_{k-1}\left(0\right)-P_{k+1}\left(0\right)]P_k\left(x\right)u(r,\theta )=100\sum _{k=0}^{\infty }{\left(\frac{r}{a^{\text{'}}}\right)}^k[P_{k-1}\left(0\right)-P_{k+1}\left(0\right)]P_k\left(x\right)$.