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.

$\mathbf{cos\theta }\mathbf{-}{\mathbf{cos}}^{\mathbf{3}}\mathbf{\theta }$.

Short answer

The steady-state temperature distribution$u(r,\theta )=\frac{2}{5}r{P}_1\left(\cos \theta \right)-\frac{2}{5}{r}^3{P}_3\left(\cos \theta \right)$.

Step-by-step solution

Given Information.

An expression has been given as$\cos \theta -{\cos }^3\theta$.

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,

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

And to separate the variable the solution assumed is of the form$\mathit{u}\mathbf{=}\mathit{R}\mathbf{\left(}\mathbf{r}\mathbf{\right)}\mathit{\Theta }\mathbf{\left(}\mathbf{\theta }\mathbf{\right)}\mathit{Z}\mathbf{\left(}\mathbf{z}\mathbf{\right)}$.

Solve the equation.

Start with the relation mentioned below.

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

At the boundary r = a = 1 , you have

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

Rewrite the equation.

$\begin{array}{c}u(a,\theta )=\cos \left(\theta \right)-{\cos }^3\left(\theta \right)\\ =\sum _{l=0}^{\infty }{c}_l{P}_l\left(\cos \right(\theta \left)\right)\end{array}$

It is known that a leading term in every Legendre polynomial is of the same order as the polynomial.

$P_l\left(x\right)=x^l+\text{lowerorderterms}$

Express $x^4$as mentioned below

$\begin{array}{l}x=P_1\left(x\right)+\dots \\ x^3=P_3\left(x\right)+\dots \end{array}$

Write the first few Legendre polynomials.

$\begin{array}{l}{P}_0\left(x\right)=1\\ P_1\left(x\right)=x\\ P_2\left(x\right)=\frac{1}{2}(3x^2-1)\\ P_3\left(x\right)=\frac{1}{2}(5x^3-3x)\end{array}$

$\begin{array}{l}{P}_4\left(x\right)=\frac{1}{8}(35x^4-30x^2+3)\\ P_5\left(x\right)=\frac{1}{8}(63x^5-70x^3+15x)\\ P_6\left(x\right)=\frac{1}{16}(231x^6-315x^4+105x^2-5)\end{array}$

Rewrite the expression and solve further.

Rewrite the Expression.

$x=P_1\left(x\right)$

$\begin{array}{c}{x}^3=\frac{1}{5}(2P_3\left(x\right)+3x)\\ =\frac{1}{5}(2P_3\left(x\right)+3P_1\left(x\right))\end{array}$

Solve further.

$\begin{array}{c}x-x^3=P_1\left(x\right)-\frac{1}{5}(2P_3\left(x\right)+3P_1\left(x\right))\\ =\frac{2}{5}{P}_1\left(x\right)-\frac{2}{5}{P}_3\left(x\right)\end{array}$

The boundary condition expressed in terms of Legendre polynomials.

$\begin{array}{c}u(a,\theta )=x-x^3\\ =\frac{2}{5}{P}_1\left(x\right)-\frac{2}{5}{P}_3\left(x\right)\\ =\sum _{l=0}^{\infty }{c}_l{P}_l\cos \left(x\right)\end{array}$

See the coefficients where only $c_1,c_3$survive.

$\begin{array}{l}{\text{c}}_1\text{=}\frac{\text{2}}{5}\\ {\text{c}}_3\text{=-}\frac{\text{2}}{5}\end{array}$

Hence, the solution is $u(a,\theta )=\frac{2}{5}r{P}_1\left(\cos \theta \right)-\frac{2}{5}{r}^3{P}_3\left(\cos \theta \right)$.