1. ## PDE Separation Problem

Hey I have been given the below question but am really struggling to get even the first part:

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

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

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

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

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

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

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

(a) State why the differential equation with the given boundary conditions can't be solved by the method of separating variables.

The problem I am having is that I can separate the equation into to parts so I am assuming it is to do with the boundary conditions why it cannot be solved, but I don't know what aspect of the conditions effects this?

2. Originally Posted by jezzyjez
Hey I have been given the below question but am really struggling to get even the first part:

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

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

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

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

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

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

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

(a) State why the differential equation with the given boundary conditions can't be solved by the method of separating variables.

The problem I am having is that I can separate the equation into to parts so I am assuming it is to do with the boundary conditions why it cannot be solved, but I don't know what aspect of the conditions effects this?
Let's suppose it can. So the solution would be of the form

$\displaystyle \phi = R(r) T(\theta)$.

The boundary conditions become

$\displaystyle \phi(r,0) = R(r) T(0) = 0\;\;\; \Rightarrow\;\;\; T(0) = 0$

and

$\displaystyle \phi(r,\frac{\pi}{2}) = R(r) T(\frac{\pi}{2}) = 1$

but this can't happen since $\displaystyle R(r)$ varies.

3. Originally Posted by Danny
Let's suppose it can. So the solution would be of the form

$\displaystyle \phi = R(r) T(\theta)$.

The boundary conditions become

$\displaystyle \phi(r,0) = R(r) T(0) = 0\;\;\; \Rightarrow\;\;\; T(0) = 0$

and

$\displaystyle \phi(r,\frac{\pi}{2}) = R(r) T(\frac{\pi}{2}) = 1$

but this can't happen since $\displaystyle R(r)$ varies.
I still dont understand the boundary conditions you have mentioned, dont they just mean if $\displaystyle \theta = 0$ the potential is 0 and if $\displaystyle \theta = \frac{\pi}{2}$ potential is 1. surely the other boundary cond is the one I need to look at???

**************EDIT 15:24*********************

4. Originally Posted by jezzyjez
I still dont understand the boundary conditions you have mentioned, dont they just mean if $\displaystyle \theta = 0$ the potential is 0 and if $\displaystyle \theta = \frac{\pi}{2}$ potential is 0. surely the other boundary cond is the one I need to look at???
But you said that $\displaystyle \phi = 1$ at $\displaystyle \theta = \frac{\pi}{2}$ not $\displaystyle \phi = 0.$

5. Sorry thats what i meant but i still dont fully understand

6. $\displaystyle \phi(r,\frac{\pi}{2}) = R(r) T(\frac{\pi}{2}) = 1$
$\displaystyle R(r) = \frac{1}{T(\frac{\pi}{2})} = const$
and
$\displaystyle R(r) = const$ this is not possible