- #1

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[tex].\ \ \ \ \ \ \ \ \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) \ \ \ \ \ \ \ \ .[/tex]

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

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- Thread starter Phrak
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- #1

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[tex].\ \ \ \ \ \ \ \ \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) \ \ \ \ \ \ \ \ .[/tex]

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

- #2

malawi_glenn

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well you can start by doing separation of variables ansatz.

- #3

Mute

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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.

- #4

malawi_glenn

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- #5

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Thank you both.

My equation derives from

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

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

[tex] \frac{1}{r} \partial_{r} \left( r\partial_{r} E \right) = 0 [/tex]

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

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

Any hints for what I need to do next?

My equation derives from

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

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

[tex] \frac{1}{r} \partial_{r} \left( r\partial_{r} E \right) = 0 [/tex]

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

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

Any hints for what I need to do next?

Last edited:

- #6

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

[tex]. \ \ \ \ \ \ \ \ \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}[/tex]

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

- #7

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prove

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

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

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