Question

Find the interior temperature in a hemisphere if the curved surface is held at $\mathit{u}\mathbf{=}\mathit{c}\mathit{o}\mathit{s}\mathit{\theta }$and the equatorial plane at $\mathit{u}\mathbf{=}\mathbf{1}$.

Short answer

The interior temperature in a hemisphere is given by,

$u=1-\frac{1}{2}r{P}_1\left(\cos \theta \right)+\frac{7}{8}{r}^3{P}_3\left(\cos \theta \right)-\frac{11}{16}{r}^5{P}_5\left(\cos \theta \right)\dots$

Step-by-step solution

Given information:

The curved surface is$u=\cos \theta$.

Definition of Laplace’s equation:

The total of the second-order partial derivatives of R, the unknown function, in Cartesian coordinates equals 0, according to Laplace's equation.

Use Laplace’s equation:

Write the Laplace equation in the spherical coordinates.

$$\begin{gathered} {\nabla }^2=\frac{1}{r^2}\frac{\partial }{\partial r}\left(r^2\frac{\partial u}{\partial r}\right)+\frac{1}{r^2\sin \theta }\frac{\partial }{\partial \theta }\left(\sin \theta \frac{\partial u}{\partial \theta }\right)+\frac{1}{r^2{\sin }^2\theta }\frac{{\partial }^{2}u}{\partial {\Phi }^2} \\ =0 \end{gathered}$$

Use separation of variables to separate the equation.

$u=R\left(r\right)\Theta \left(\theta \right)\Psi \left(\Phi \right)$

Write the general solution as the series of basic functions.

$u=\sum _{l=0}^{\infty }{c}_l{r}^{l}Pl\left(\cos \theta \right)$

Apply Boundary conditions:

First check the odd-even nature of the function.

$$\begin{gathered} f\left(u\right)=\cos \theta \\ f\left(-u\right)=\cos \left(-\theta \right) \\ f\left(-u\right)=\cos \theta \\ f\left(-u\right)=f\left(u\right) \end{gathered}$$

Hence, the function is even.

The curved surface is at $u=\cos \theta$and the equatorial plane is:

$$\begin{gathered} c_l=\underset{0}{\overset{1}{\int }}\cos \theta P_l\left(\cos \theta \right)d\theta \\ =\frac{l+1}{2l+1}\underset{0}{\overset{1}{\int }}{P}_{l+1}\left(\cos \theta \right)d\theta +\frac{l}{2l+1}\underset{0}{\overset{1}{\int }}{P}_{l-1}\left(\cos \theta \right)d\theta \end{gathered}$$

Here,

$$\begin{gathered} \underset{0}{\overset{1}{\int }}{P}_{l+1}d\theta =\left\{\begin{array}{l}0,l\text{odd}\\ {\left(-1\right)}^{(l+4)/2}\frac{\left(l+2\right)!}{2^{l+2}{\left[\frac{l+2}{2}!\right]}^2(l+1)}, \text{if}l\text{is even}\end{array}\right. \\ \underset{0}{\overset{1}{\int }}{P}_{l-1}d\theta =\left\{\begin{array}{l}{\delta }_{1,l}, l\text{odd}\\ {\left(-1\right)}^{(l+2)/2}\frac{\left(l\right)!}{2^l{\left[\frac{l}{2}!\right]}^2\left(l-1\right)}, \text{if}l\text{is even}\end{array}\right. \end{gathered}$$

Hence,the interior temperature in a hemisphere is given by,

.$u=1-\frac{1}{2}r{P}_1\left(\cos \theta \right)+\frac{7}{8}{r}^3{P}_3\left(\cos \theta \right)-\frac{11}{16}{r}^5{P}_5\left(\cos \theta \right)\dots$