Question

The surface temperature of a sphere of radius 1 is held at $\mathit{u}\mathbf{=}\mathit{s}\mathit{i}{\mathit{n}}^{\mathbf{2}}\mathit{\theta }\mathbf{+}\mathit{c}\mathit{o}{\mathit{s}}^{\mathbf{3}}\mathit{\theta }$. Find the interior temperature $\mathit{u}\left(r,\theta ,\varphi \right)$.

Short answer

The inside temperature of the sphere with radius 1 is:

$u\left(r,\theta ,\Phi \right)=1-\frac{1}{2}r{P}_1\left(\cos \theta \right)+\frac{3}{5}r{P}_1\left(\cos \theta \right)-\frac{2}{3}{r}^2{P}_2\left(\cos \theta \right)+\frac{2}{5}{r}^3{P}_3\left(\cos \theta \right)$

Step-by-step solution

Given information:

The radius of the sphere is 1.

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.

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

Step 4:Apply Boundary conditions:

First check the odd-even nature of the function.

$$\begin{gathered} f\left(x\right)={\sin }^2\theta +{\cos }^3\theta \\ ={\left(\sin \left(-\theta \right)\right)}^2+{\left(\cos \left(-\theta \right)\right)}^3 \\ =-\left({\sin }^2\theta +{\cos }^3\theta \right) \end{gathered}$$

As, $f\left(-x\right)=-f\left(x\right)$the function is odd.

Write the boundary condition as a function of Legendre polynomials.

${\sin }^2\theta +{\cos }^3\theta =\sum _{l=0}^{\infty }{c}_l{r}^{l}Pl\left(\cos \theta \right)$

Hence the inside temperature of the sphere with radius 1 is:

$u\left(r,\theta ,\Phi \right)=1-\frac{1}{2}r{P}_1\left(\cos \theta \right)+\frac{3}{5}r{P}_1\left(\cos \theta \right)-\frac{2}{3}{r}^2{P}_2\left(\cos \theta \right)+\frac{2}{5}{r}^3{P}_3\left(\cos \theta \right)$