Solving a Bessel Function DE in Electromagnetism | Cylindrical Coordinates

[Phrak]

I have a problem in electromagnetism giving a DE that looks something like a Lapacian or a Bessel function, I'm told. It derives from cylindrical coordinates.

.\ \ \ \ \ \ \ \ \left( \partial_{r} ^2 + \frac{1}{r}\partial_{r} - \frac{1}{r^2}\right)E = \frac{1}{c^2}\partial_{t}^2 E\ \ \ \ \ \ \ \ E=E(r,t) \ \ \ \ \ \ \ \ .

I don't know where to start. A series solution would be OK too.

 

[malawi_glenn]

well you can start by doing separation of variables ansatz.

 

[Mute]

Assume a solution of the form E(r,t) = R(r)T(t) and plug that into the DE. Then, separate all terms that depend on r from all terms that depend on t on opposite sides of the equal sign. The two sides have to be equal to a constant, since if you vary r on one side, then in order for it to be equal to all the terms that depend on T on the other side, t would have to change unless both sides were constant.

You thus get two sepearte ODEs, one for R(r) and one for T(t), which you can then solve. The T one should be easy, and the R one probably gives you a bessel function solution.

 

[malawi_glenn]

and from the time part of the function, you will see what sign your separation constant must be in order to fulfill the BC's of the time part.

 

[Phrak]

Thank you both.

My equation derives from

\partial_{r} \frac{1}{r} \partial_{r} \left( rE \right) = \frac{1}{c^2} \partial_{t} ^2 E

After some internet research, it looks a lot like the Laplacian in cylindrical coordinates which is

\frac{1}{r} \partial_{r} \left( r\partial_{r} E \right) = 0

The general solution to this one is a linear combination of the Bessel function of the first kind and the Bessel function of the second kind.

For E(r,t) to be separable as R(r)T(t) there can't be solutions having terms such as (kx-wt), right? However, the first approximation is separable as

E= E_{0} \frac{r_{0}}{r}exp(iwt), where I included some boundry conditions.

Any hints for what I need to do next?

 

[Phrak]

After flailing about I noticed I could do a variable replacement \ y=rE}.

Does this look at all familiar to anyone, in whole or part??

. \ \ \ \ \ \ \ \ \frac{\partial ^2 y}{\partial r^2}-\frac{1}{r}\frac{\partial y}{\partial r}= \frac{1}{c^2}\frac{\partial ^2 y}{\partial t^2}

The boundry condition is y=y_{0}\ exp(i \omega t)

 

[gilfred123]

prove
J-n(x) = (-1)nJn(x);