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Laplace equation polar coordinates

dwsmith

Well-known member
Feb 1, 2012
1,673
I have never solved an equation in polar form. I am not sure with how to start.

Solve Laplace's equation on a circular disk of radius a subject to the piecewise boundary condition
$$
u(a,\theta) = \begin{cases}
1, & \frac{\pi}{2} - \epsilon < \theta < \frac{\pi}{2} + \epsilon\\
0, & \text{otherwise}
\end{cases}
$$
where $\epsilon \ll 1$. Physically, this would reflect the electric potential distribution on a conducting disk whose edge is almost completely grounded except a small portion of angular extent $\Delta\theta = 2\epsilon$ around the location $\theta = \frac{\pi}{2}$. Obtain the solution to this problem and plot the solution for the case of $a = 1$ and $\epsilon = 0.05$.
 

Sudharaka

Well-known member
MHB Math Helper
Feb 5, 2012
1,621
I have never solved an equation in polar form. I am not sure with how to start.

Solve Laplace's equation on a circular disk of radius a subject to the piecewise boundary condition
$$
u(a,\theta) = \begin{cases}
1, & \frac{\pi}{2} - \epsilon < \theta < \frac{\pi}{2} + \epsilon\\
0, & \text{otherwise}
\end{cases}
$$
where $\epsilon \ll 1$. Physically, this would reflect the electric potential distribution on a conducting disk whose edge is almost completely grounded except a small portion of angular extent $\Delta\theta = 2\epsilon$ around the location $\theta = \frac{\pi}{2}$. Obtain the solution to this problem and plot the solution for the case of $a = 1$ and $\epsilon = 0.05$.
The method of solving a Laplace's equation in polar coordinates can be found >>here<< (Refer Example 3).
 

dwsmith

Well-known member
Feb 1, 2012
1,673
The method of solving a Laplace's equation in polar coordinates can be found >>here<< (Refer Example 3).
I figured out how to solve in polar. However, this problem is still posing a lot of difficulty.
 

dwsmith

Well-known member
Feb 1, 2012
1,673
By separation of variables, we have
$$
\begin{cases}
\Theta(\theta) = A\cos\lambda\theta + B\sin\lambda\theta\\
R(r) = r^{\pm\lambda}
\end{cases}
$$
So how do I use the conditions now?
 

dwsmith

Well-known member
Feb 1, 2012
1,673
So I have that
$$
u(r,\theta) = \alpha_0 + \sum_{n = 1}^{\infty}r^{n}(A_n\cos n\theta + B_n\sin n\theta).
$$
Using the final condition, we have
$$
u(a,\theta) = \alpha_0 + \sum_{n = 1}^{\infty}a^{n}(A_n\cos n\theta + B_n\sin n\theta) = \begin{cases}
1, & \frac{\pi}{2} - \epsilon < \theta < \frac{\pi}{2} + \epsilon\\
0, & \text{otherwise}
\end{cases}
$$
Correct?
$$
A_na^n\int_{-\pi}^{\pi}\cos^2 n\theta d\theta = \left[\int_{-\pi}^{\frac{\pi}{2}-\epsilon}0\cos n\theta d\theta + \int_{\frac{\pi}{2}-\epsilon}^{\frac{\pi}{2}+\epsilon}\cos n\theta d\theta + \int_{\frac{\pi}{2}+\epsilon}^{\pi}0\cos n\theta d\theta\right]
$$
 
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dwsmith

Well-known member
Feb 1, 2012
1,673
So I have that
$$
u(r,\theta) = \alpha_0 + \sum_{n = 1}^{\infty}r^{n}(A_n\cos n\theta + B_n\sin n\theta).
$$
Using the final condition, we have
$$
u(a,\theta) = \alpha_0 + \sum_{n = 1}^{\infty}a^{n}(A_n\cos n\theta + B_n\sin n\theta) = \begin{cases}
1, & \frac{\pi}{2} - \epsilon < \theta < \frac{\pi}{2} + \epsilon\\
0, & \text{otherwise}
\end{cases}
$$
Correct?
$$
A_na^n\int_{-\pi}^{\pi}\cos^2 n\theta d\theta = \left[\int_{-\pi}^{\frac{\pi}{2}-\epsilon}0\cos n\theta d\theta + \int_{\frac{\pi}{2}-\epsilon}^{\frac{\pi}{2}+\epsilon}\cos n\theta d\theta + \int_{\frac{\pi}{2}+\epsilon}^{\pi}0\cos n\theta d\theta\right]
$$
\begin{alignat*}{3}
A_n & = & \frac{1}{\pi a^n}\left[\int_{-\pi}^{\frac{\pi}{2} - \epsilon}0\cos n\theta d\theta + \int_{\frac{\pi}{2} - \epsilon}^{\frac{\pi}{2} + \epsilon}\cos n\theta d\theta + \int_{\frac{\pi}{2} + \epsilon}^{\pi}0\cos n\theta d\theta\right]\\
& = & \frac{1}{\pi a^n}\int_{\frac{\pi}{2} - \epsilon}^{\frac{\pi}{2} + \epsilon}\cos n\theta d\theta\\
& = & \left.\frac{1}{n\pi a^n}\sin n\theta\right|_{\frac{\pi}{2} - \epsilon}^{\frac{\pi}{2} + \epsilon}\\
& = & \frac{1}{n\pi a^n}\left[\sin n\left(\frac{\pi}{2} + \epsilon\right) - \sin n\left(\frac{\pi}{2} - \epsilon\right)\right]
\end{alignat*}

Can I get $A_n$ in a more usable form?
Here is what I got for $B_n$ and $A_0$.

\begin{alignat*}{3}
B_n & = & \frac{1}{\pi a^n}\left[\int_{-\pi}^{\frac{\pi}{2} - \epsilon}0\sin n\theta d\theta + \int_{\frac{\pi}{2} - \epsilon}^{\frac{\pi}{2} + \epsilon}\sin n\theta d\theta + \int_{\frac{\pi}{2} + \epsilon}^{\pi}0\sin n\theta d\theta\right]\\
& = & \frac{1}{\pi a^n}\int_{\frac{\pi}{2} - \epsilon}^{\frac{\pi}{2} + \epsilon}\sin n\theta d\theta\\
& = & -\left.\frac{1}{n\pi a^n}\cos n\theta\right|_{\frac{\pi}{2} - \epsilon}^{\frac{\pi}{2} + \epsilon}\\
& = & -\frac{1}{n\pi a^n}\left[\cos n\left(\frac{\pi}{2} + \epsilon\right) - \cos n\left(\frac{\pi}{2} - \epsilon\right)\right]
\end{alignat*}
\begin{alignat*}{3}
A_0 & = & \frac{1}{\pi}\left[\int_{-\pi}^{\frac{\pi}{2} - \epsilon}0 + \int_{\frac{\pi}{2} - \epsilon}^{\frac{\pi}{2} + \epsilon}d\theta + \int_{\frac{\pi}{2} + \epsilon}^{\pi}0d\theta\right]\\
& = & \frac{1}{\pi}\int_{\frac{\pi}{2} - \epsilon}^{\frac{\pi}{2} + \epsilon}d\theta\\
& = & \left.\frac{\theta}{\pi}\right|_{\frac{\pi}{2} - \epsilon}^{\frac{\pi}{2} + \epsilon}\\
& = & \frac{1}{\pi}\left[\frac{\pi}{2} + \epsilon - \frac{\pi}{2} + \epsilon\right]\\
& = & \frac{2\epsilon}{\pi}
\end{alignat*}
 
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