Question

Question:Find the steady-state temperature distribution inside a sphere of radius 1 when the surface temperatures are as given in Problems 1 to 10.
$\mathbf{35}{\mathbf{cos}}^{\mathbf{4}}\mathbf{\theta }$.

Short answer

The steady-state temperature distribution.

$\mathrm{u}\left(\mathrm{r},\mathrm{\theta }\right)=8{\mathrm{r}}^4{\mathrm{P}}_4\left(\mathrm{cos\theta }\right)+20{\mathrm{r}}^2{\mathrm{P}}_2\left(\mathrm{cos\theta }\right)+7{\mathrm{P}}_0\left(\mathrm{cos\theta }\right)$.

Step-by-step solution

Given Information.

An expression has been given as $35{\cos }^4\theta$.

Definition of Laplace’s equation.

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

Laplace’s equation in cylindrical coordinates is

${\mathbf{\nabla }}^{\mathbf{2}}\mathit{u}\mathbf{=}\frac{\mathbf{1}}{\mathbf{r}}\frac{\mathbf{\partial }}{\mathbf{\partial }\mathbf{r}}\left(\mathit{r}\frac{\partial \mathit{u}}{\partial \mathit{r}}\right)\mathbf{+}\frac{\mathbf{1}}{{\mathbf{r}}^{\mathbf{2}}}\frac{{\mathbf{\partial }}^{\mathbf{2}}\mathbf{u}}{\mathbf{\partial }{\mathbf{\theta }}^{\mathbf{2}}}\mathbf{+}\frac{{\mathbf{\partial }}^{\mathbf{2}}\mathbf{u}}{\mathbf{\partial }{\mathbf{z}}^{\mathbf{2}}}\mathbf{=}\mathbf{0}$

And to separate the variable the solution assumed is of the form .

$\mathbf{u}\mathbf{=}\mathbf{R}\left(r\right)\mathbf{\Theta }\left(\theta \right)\mathbf{Z}\left(z\right)$

Solve the equation.

Start with the relation mentioned below.

At the boundary , you have

$$\begin{gathered} u\left(a,\theta \right)=35{\cos }^4\left(\theta \right) \\ =\sum _{l=0}^{\infty }{c}_l{P}_l\left(\cos \left(\theta \right)\right) \end{gathered}$$

Rewrite the equation by taking,

$x=\cos \left(\theta \right)$

$$\begin{gathered} \mathrm{u}\left(\mathrm{a},\mathrm{\theta }\right)=35{\cos }^4\left(\mathrm{x}\right) \\ =\sum _{\mathrm{l}=0}^{\infty }{\mathrm{c}}_{\mathrm{l}}{\mathrm{P}}_{\mathrm{l}}\left(\mathrm{x}\right) \end{gathered}$$
It is known that a leading term in every Legendre polynomial is of the same order as the polynomial.

$P_l\left(x\right)=x^l+\text{lower order terms}$

Now, express x⁴ as belox

$x^4=P_4+\dots$

Write the first few Legendre polynomials
$$\begin{gathered} P_0\left(x\right)=1 \\ {P}_1\left(x\right)=x \\ {P}_2\left(x\right)=\frac{1}{2}\left(3x^2-1\right) \\ {P}_3\left(x\right)=\frac{1}{2}\left(5x^3-3x\right) \end{gathered}$$
$$\begin{gathered} P_4\left(x\right)=\frac{1}{8}\left(35x^4-30x^2+3\right) \\ {P}_5\left(x\right)=\frac{1}{8}\left(63x^5-70x^3+15x\right) \\ {P}_6\left(x\right)=\frac{1}{16}\left(231x^6-315x^4+105x^2-5\right) \end{gathered}$$

Rewrite the expression and solve further.

Rewrite the Expression as given by,
$$\begin{gathered} 35x^4=8P_4\left(x\right)+30x^2-3 \\ =8P_4\left(x\right)+30x^2-3P_0\left(x\right) \end{gathered}$$
Express in terms of Legendre polynomials
$x^2=\frac{1}{3}\left(2P_2\left(x\right)+P_0\left(x\right)\right.$

Therefore, you have,
$$\begin{gathered} 35x^4=8P_4\left(x\right)+30\frac{1}{3}\left(2P_2\left(x\right)+P_0\left(x\right)-3P_0\left(x\right)\right. \\ =8P_4\left(x\right)+20P_2\left(x\right)+7P_0\left(x\right) \end{gathered}$$
The boundary condition expressed in terms of Legendre polynomials.
$$\begin{gathered} u\left(a,\theta \right)=35x^4 \\ =8P_4\left(x\right)+20P_2\left(x\right)+7P_0\left(x\right) \\ =\sum _{l=0}^{\infty }{c}_l{P}_l\cos \left(x\right) \end{gathered}$$
See the coefficients where only survive.


Hence, the solution is $\mathrm{u}\left(\mathrm{r},\mathrm{\theta }\right)=8{\mathrm{r}}^4{\mathrm{P}}_4\left(\mathrm{cos\theta }\right)+20{\mathrm{r}}^2{\mathrm{P}}_2\left(\mathrm{cos\theta }\right)+7{\mathrm{P}}_0\left(\mathrm{cos\theta }\right)$.