Chapter 14, Problem 1CR

To determine

The steady state temperature $u\left(r,\theta \right)$ in a circular plate of radius $c$ if the function of temperature on the circumference is $u\left(c,\theta \right)=\{\begin{array}{l}{u}_0,0<\theta <\pi \\ -u_0,\pi <\theta <2\pi \end{array}$.

Answer to Problem 1CR

The steady state temperature in a circular plate is $\underset{\_}{u\left(r,\theta \right)=\frac{2u_0}{\pi }{\displaystyle \sum _{n=1}^{\infty }\frac{1-{\left(-1\right)}^n}{n}}{\left(\frac{r}{c}\right)}^n\sin n\theta }$.

Explanation of Solution

Given:

The temperature on the circumference is $u\left(c,\theta \right)=\{\begin{array}{l}{u}_0,0<\theta <\pi \\ -u_0,\pi <\theta <2\pi \end{array}$.

Calculation:

Consider the boundary value problem is as follows:

$\frac{{\partial }^{2}u}{\partial r^2}+\frac{1}{r}\frac{\partial u}{\partial r}+\frac{1}{r^2}\frac{{\partial }^{2}u}{\partial {\theta }^2}=0$

The separation of the variables of the above equation taking separation constant $\lambda$ is as follows:

$\frac{r^2{R}^{″}+r{R}^{\prime }}{R}=\lambda \text{and}-\frac{{\Theta }^{″}}{\Theta }=\lambda$

The equation of variable $r$ is as below:

$r^2{R}^{″}+r{R}^{\prime }-\lambda R=0\mathrm{......}\left(1\right)$

The equation of variable $\theta$ is as below:

${\Theta }^{″}+\lambda \Theta =0\mathrm{......}\left(2\right)$

Consider the equation $u$ as a function of $r$ and $\theta$ as follows:

$u=R\left(r\right)\Theta \left(\theta \right)\mathrm{......}\left(3\right)$

For $\lambda =0$ the general solution of equation (2) is written as follows:

$\Theta \left(\theta \right)=c_1+c_2\theta$

At given boundary condition $c_2=0$ and substitute this value in the above equation.

$\Theta \left(\theta \right)=c_1\mathrm{......}\left(4\right)$

The general solution of the equation (1) as follows:

$R\left(r\right)=c_3{r}^{\alpha }+c_4{r}^{-\alpha }$

At given boundary condition $\alpha =0\text{and}{c}_4=0$ and substitute these values in the above equation.

$R\left(r\right)=c_3\mathrm{......}\left(5\right)$

Substitute the value of $\Theta \left(\theta \right)$ from equation (4) and the value of $R\left(r\right)$ from equation (5) in equation (3).

$u_0=c_1{c}_3$

Rewrite the term $c_1{c}_3=A_0$ in the above equation.

$u_0=A_0\mathrm{......}\left(6\right)$

For $\lambda ={\alpha }^2$ the general solution of the equation (2) is as follows:

$\Theta \left(\theta \right)=c_1\cos \alpha \theta +c_2\sin \alpha \theta$

At given boundary condition $\alpha =n$ substitute this value in the above equation.

$\Theta \left(\theta \right)=c_1\cos n\theta +c_2\sin n\theta \mathrm{......}\left(7\right)$

When $\alpha =n\text{and}\lambda =n^2$ the general solution of the equation (1) is follows:

$R\left(r\right)=c_3{r}^n+c_4{r}^{-n}$

At given boundary condition $c_4=0$ and substitute this value in the above equation.

$R\left(r\right)=c_3{r}^n\mathrm{......}\left(8\right)$

Substitute the value of $\Theta \left(\theta \right)$ from equation (7) and the value of $R\left(r\right)$ from equation (8) in equation (3).

$u_n=c_3{r}^n\left(c_1\cos n\theta +c_2\sin n\theta \right)$

Rewrite the term $c_1{c}_3=A_n\text{and}{c}_2{c}_3=B_n$ in the above equation.

$u_n=r^n\left(A_n\cos n\theta +B_n\sin n\theta \right)\mathrm{......}\left(9\right)$

Apply superposition principle on equation (6) and equation (9).

$u\left(r,\theta \right)=A_0+{\displaystyle \sum _{n=1}^{\infty }{r}^n\left(A_n\cos n\theta +B_n\sin n\theta \right)}\mathrm{......}\left(10\right)$

At given boundary condition the temperature is $u\left(c,\theta \right)=\{\begin{array}{l}{u}_0,0<\theta <\pi \\ -u_0,\pi <\theta <2\pi \end{array}$.

The value of $A_0$ is obtained as follows:

$\begin{array}{c}{A}_0=\frac{1}{2\pi }{\displaystyle \underset{0}{\overset{\pi }{\int }}{u}_{0}d\theta }+\frac{1}{2\pi }{\displaystyle \underset{\pi }{\overset{2\pi }{\int }}\left(-u_0\right)d\theta }\\ =0\end{array}$

The value of $A_n$ is obtained as follows:

$\begin{array}{c}{A}_n=\frac{1}{c^n\pi }{\displaystyle \underset{0}{\overset{\pi }{\int }}{u}_0\cos n\theta d\theta }+\frac{1}{c^n\pi }{\displaystyle \underset{\pi }{\overset{2\pi }{\int }}\left(-u_0\right)\cos n\theta d\theta }\\ =0\end{array}$

The value of $B_n$ is obtained as follows:

$\begin{array}{c}{B}_n=\frac{1}{c^n\pi }{\displaystyle \underset{0}{\overset{\pi }{\int }}{u}_0\sin n\theta d\theta }+\frac{1}{c^n\pi }{\displaystyle \underset{\pi }{\overset{2\pi }{\int }}\left(-u_0\right)\sin n\theta d\theta }\\ =\frac{2u_0}{c^{n}n\pi }\left[1-{\left(-1\right)}^n\right]\end{array}$

Substitutes the values of $A_0,A_n\text{and}{B}_n$ in equation (10).

$\begin{array}{c}u\left(r,\theta \right)=0+{\displaystyle \sum _{n=1}^{\infty }{r}^n\left(0\cos n\theta +\frac{2u_0}{c^{n}n\pi }\left[1-{\left(-1\right)}^n\right]\sin n\theta \right)}\\ =\frac{2u_0}{\pi }{\displaystyle \sum _{n=1}^{\infty }\frac{1-{\left(-1\right)}^n}{n}}{\left(\frac{r}{c}\right)}^n\sin n\theta \end{array}$

Thus, the steady state temperature in a circular plate is $\underset{\_}{u\left(r,\theta \right)=\frac{2u_0}{\pi }{\displaystyle \sum _{n=1}^{\infty }\frac{1-{\left(-1\right)}^n}{n}}{\left(\frac{r}{c}\right)}^n\sin n\theta }$.