Solving Second order non-Homogeneous PDE

[femiadeyemi]

Hi Everyone,

I was reading a paper and I found it hard to comprehend how some of the equations were arrived at, probably because my math rottenness. Anyway I need your help on understanding how these equations were arrived at. The problem goes like this:

We have this PDE in cylindrical co-ordinate:

\frac{\partial^{2} }{\partial t^{2}} P - c_{0}^{2} \Delta P = \frac{\partial}{\partial t} q(\underline{r},t)

and q(\underline{r},t) = δ(t) Q(\underline{r})

Here is a quote from the paper:

In axially symmetric case P(\underline{r},t) = P(r,z,t), the radial co-ordinate is expressed through a superposition of Bessel function J_{0} (λ r) with continuous (P(r,z,t) = ∫^{∞}_{0} P_{λ}(z,t) J_{0} (λ r) λ dλ) or discrete series of λ values.

In both cases for spectrum amplitude P_{λ}(z,t) one get equation:

\frac{\partial^{2} }{\partial t^{2}} P_{λ} - c_{0}^{2} (\frac{\partial^{2} P_{λ}}{\partial z^{2}} - λ^{2} P_{λ} ) = 0

Basically my question is, how does the author arrived at equation P(r,z,t) = ∫^{∞}_{0} P_{λ}(z,t) J_{0} (λ r) λ dλ and \frac{\partial^{2} }{\partial t^{2}} P_{λ} - c_{0}^{2} (\frac{\partial^{2} P_{λ}}{\partial z^{2}} - λ^{2} P_{λ} ) = 0

And secondly, it is possible to solve the above PDE without using any boundary conditions because I didn't see any BC or IC used in arriving at the two equations (that is, P(r,z,t) = ∫^{∞}_{0} P_{λ}(z,t) J_{0} (λ r) λ dλ and \frac{\partial^{2} }{\partial t^{2}} P_{λ} - c_{0}^{2} (\frac{\partial^{2} P_{λ}}{\partial z^{2}} - λ^{2} P_{λ} ) = 0)?

Looking forward for your assistance and thank you in advance.

 

[Simon Bridge]

Look up "Bessel function" ... he gets there because he recognized the type of DE and looked it up. This happens a lot. All the lambdas is because of the substitution he needed to make the Bessel function solution work.

Notice that it's a definite integral and P_\lambda will need boundary conditions which will be provided by the particular system.

I don't know what that big delta is doing there though. Is it supposed to be \nabla ... or maybe the Laplace operator \nabla^2?
If you expand the operator you should see the starting point.

Aside: In physics papers the BCs can often be implied by the type of problem - remember they are written for other professional academics working in, and familiar with, the field.

 

[femiadeyemi]

Look up "Bessel function" ... he gets there because he recognized the type of DE and looked it up. This happens a lot. All the lambdas is because of the substitution he needed to make the Bessel function solution work.

Notice that it's a definite integral and P_\lambda will need boundary conditions which will be provided by the particular system.

I don't know what that big delta is doing there though. Is it supposed to be \nabla ... or maybe the Laplace operator \nabla^2?
If you expand the operator you should see the starting point.

Aside: In physics papers the BCs can often be implied by the type of problem - remember they are written for other professional academics working in, and familiar with, the field.

Thank you very your reply. I've been going through my old mathematical physics text, from what I've read so far it seem the author is using Hankel transformation but I'm not totally sure though, but it worth giving it a shot!

To answer your question. Yes, the operator in the pde is a Laplace operator.

On the BCs of the problem, is it possible to deduce the BCs' without been a professional in the field? And if the answer is yes, can you tell me how, I can have a feeling of what the BCs' and ic's are?

 

[Simon Bridge]

To answer your question. Yes, the operator in the pde is a Laplace operator.

Yes that makes sense - it's just that, around here, it's sometimes an error: someone sees a nabla and they think it's supposed to be a delta...

If you expand out the laplassian in cylindrical coords, simply for no angular dependence, you'll find you can make it look like the Bessel equation.

On the BCs of the problem, is it possible to deduce the BCs' without been a professional in the field?

Yes you can, one of the beauties of science is that you shouldn't need to be an "insider" to be able to understand it. You need to look at the specific physics being investigated and ask yourself what the limits are on that system.

One of the things that can help is to work with someone who can be in the room with you. There is a limit to what I will do - I can help you steer but you have to do the actual driving. Fun... isn't it ;)

 

[femiadeyemi]

There is a limit to what I will do - I can help you steer but you have to do the actual driving. Fun... isn't it ;)

Thanks! I will definitely love to drive with your help :)
Let me give it a shot and I will let you know how it goes. Once again, thank you!