# Thread: PDE non seperables

1. ## PDE non seperables

Hey I have been given the below question but am really struggling to get it still I understand how to solve problems using seperation of variables but this has got me stumped

We want to find the value of the electrostatic potential $\phi(r,\theta)$ in the quadrant.

$0 \leq r \leq b$ , $0 \leq \theta \leq \frac{\pi}{2}$

where $r$ and $\theta$ are the standard polar co-ordinates and $b$ is a given positive real number. Assuming electrostatic potential is the solution of Laplaces equation in polar co-ordinates

$\frac{\delta^2 \phi}{\delta r^2} +\frac{1}{r}\frac{\delta\phi}{\delta r} + \frac{1}{r^2} \frac{\delta^2\phi}{\delta\theta^2} = 0$

The boundary conditions on $\phi(r,\theta)$ are

$\phi(b,\theta) = \frac{4}{\pi^2} \theta(\pi - \theta),$

$\phi(r,0) = 0, \phi(r,\frac{\pi}{2}) = 1$

(a) show that the function $\phi(\theta) = \frac{2\theta}{\pi}$ is a solution of the initial equation that satisfies the first set of boundary conditions

Any help would be great

Thanks

2. Originally Posted by jezzyjez
Hey I have been given the below question but am really struggling to get it still I understand how to solve problems using seperation of variables but this has got me stumped

We want to find the value of the electrostatic potential $\phi(r,\theta)$ in the quadrant.

$0 \leq r \leq b$ , $0 \leq \theta \leq \frac{\pi}{2}$

where $r$ and $\theta$ are the standard polar co-ordinates and $b$ is a given positive real number. Assuming electrostatic potential is the solution of Laplaces equation in polar co-ordinates

$\frac{\delta^2 \phi}{\delta r^2} +\frac{1}{r}\frac{\delta\phi}{\delta r} + \frac{1}{r^2} \frac{\delta^2\phi}{\delta\theta^2} = 0$

The boundary conditions on $\phi(r,\theta)$ are

$\phi(b,\theta) = \frac{4}{\pi^2} \theta(\pi - \theta),$

$\phi(r,0) = 0, \phi(r,\frac{\pi}{2}) = 1$

(a) show that the function $\phi(\theta) = \frac{2\theta}{\pi}$ is a solution of the initial equation that satisfies the first set of boundary conditions

Any help would be great

Thanks
Actually, it's the second BC.

if $\phi(r,\theta) = \frac{2 \theta}{\pi}$ then

$
\phi(r,0) = \frac{2 0}{\pi} = 0
$

$
\phi(r,\frac{\pi}{2}) = \frac{2}{\pi} \cdot \frac{\pi}{2} = 1.
$

3. Ok yeh I understand this next part askes to write a PDE and BC's for the function

$\psi(r,\theta) = \phi(r,\theta) - \phi_0(\theta)$

Then using seperation of variables solve $\psi (r,\theta)$ and hence write the solution $\phi (r,\theta)$ of the original eqn/BC.

Any help would be great.

4. Originally Posted by jezzyjez
Ok yeh I understand this next part askes to write a PDE and BC's for the function

$\psi(r,\theta) = \phi(r,\theta) - \phi_0(\theta)$

Then using seperation of variables solve $\psi (r,\theta)$ and hence write the solution $\phi (r,\theta)$ of the original eqn/BC.

Any help would be great.
THe idea here is to transform the two BC to have them both zero so a separation of variables will work. IF we let

$
\phi = \psi + \frac{2}{\pi} \theta
$

the PDE becomes

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

and the BCs

$
\psi(r,0) = 0,\;\; \psi\left(r ,\frac{\pi}{2}\right) = 0
$

and the IC

$
\psi(b,\theta) = \frac{2 \theta}{\pi} \left(1 - 2 \frac{\theta}{\pi}\right).$

Now separation of variables will work.