1
$\begingroup$

I'm going through Griffiths EM and I've come across an example on how to solve Laplace's equaton through separation of variables, trying to convert a partial differential equation into an ODE, which is much simpler to solve. Nonetheless, in one of the steps in his reasoning, there might be a division by zero which I'd like to have checked. Here is the textbook's example:

Say you have 2 infinite planes connected to earth, as well as to each other by an infinite strip with some potential $V_0(y)$, as shown in the picture.

System of 2 infinite planes connected by an infinite strip

The boundary conditions are as follows:

  • $V=0$ when $y=0$
  • $V=0$ when $y=a$
  • $V=V_0(y)$ when $x=0$
  • $V\longrightarrow 0$ when $x\longrightarrow \infty$

We are asked to find the potential in the region inbetween the infinite planes. In order to achieve this, we first assume the potential function is of the form $V(x,y)=f(x)\cdot g(y)$ as to be able to separate variables later. Now we can substitute this expression in Laplace's equation and write:

$$g(y)\cdot \frac{d^2 f}{d x^2} + f(x)\cdot \frac{d^2 g}{d y^2}=0$$

Now, the textbook claims we can easily separate variables by dividing by $V$, in which case we'd obtain:

$$\frac{1}{f(x)}\cdot \frac{d^2 f}{dx^2} + \frac{1}{g(y)}\cdot \frac{d^2 f}{dx^2}=0$$

Nonetheless, I don't understand why we can divide by $V$ if there are regions in space in which $V=0$ as stated in the boundary conditions. If our solution is to account for the potential in the region inbetween the infinite planes, then how come we are allowed to divide by $V$? My best guess is the regions in which $V=0$ are not contained in the region inbetween the planes but rather delimit it, so we don't need them to be included in our solution as long as the boundary conditions are consistent with what we obtain near said limits.

$\endgroup$
1

2 Answers 2

0
$\begingroup$

The separation of variables itself is only a necessary condition. Regarding its sufficiency, I think the reason why you can divide the Laplace equation by $f(x)g(y)$ even if it is zero at certain points is because the zeros of a harmonic function being either the real or imaginary part of a holomorphic function are isolated, said differently, the zeros of a harmonic function cannot have an accumulation point within its domain. And when you create the linear combination of the solutions of the ODEs you will not end up getting any "funny" business from them.

$\endgroup$
0
$\begingroup$

My thoughts on this: I guess it has to do with the fact that the potential function is continuous throughout... Think about it... Had it not been continuous, the electric field in the region/point of discontinuity won't be defined... Also see that the laplacian of V being zero strictly applies to the region between the plates and not the plates themselves, you don't have any info regarding charge densities there... Now combining these two facts it's clear that any solution to the equation would tend towards zero as y tends to 0 or a... It doesn't have to be zero in the region concerned...

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.