Fourier-Bessel Series Expansion

phioder

Hello

Did an exercise and a small simulation to expand [tex]f(x)=x[/tex], defined on [tex]0<x<3[/tex] in a Fourier-Bessel series using Bessel functions of order one that satisfy the boundary condition [tex]J_1(3\lambda)=0[/tex] and I have some questions:

1.- Is there a rule to use an specific Bessel function order to do the expansion? In other words is it possible to do the expansion using Bessel functions of order 1,2,3...,n? If yes, which criteria does one uses to select the order?

2.- For this particular case, how does the boundary condition can be interpreted? Why does the boundary condition is of order 1?

The reason of my questions is that I would like to expand [tex]f(x)=x^2[/tex] and don't know if it is correct to use Bessel functions of other order than one; for this particular exercise what is the nature of expanding and setting the boundary condition with an specific order.

Thank you
Kind Regards

mda

Not something I've considered before, but if you are expanding about zero then the approximate behaviour is: [tex]J_n(x) = c x^n + ...[/tex]

So the first order is entirely appropriate for x, whereas for x^2... well you figure it out.

phioder

More concrete the question is how to solve or reduce algebraically and with pencil

[tex]\int_{0}^{R} x^3 J_1 (\lambda x) dx [/tex]

where [tex]\lambda[/tex] are the eigenvalues

I'm still not successful using the recurrence and differential relations

[tex]J_{n-1}(x) = \frac{2n}{x} J_{n}(x) - J_{n+1}(x)[/tex]

[tex] x^n J_{n-1}(x) = \frac{d}{dx}[x^n J_{n}(x)][/tex]

Maybe I'm completely wrong but what I think is that the [tex]f(x)[/tex] to expand has nothing to do with order of the bessel function. I suspect that it may be related to how the boundary condition is defined, for this specific problem the boundary condition is of order one and defined as [tex]J_1(3\lambda)[/tex]

Best Regards

mda

I'd be curious to know where you get the "boundary condition" from. Wikipedia does not mention this at all, and in fact their series assumes that f(b)->0, since usually the series is applied to bounded radial functions. However, given your boundary condition the series is most accurate using only the first term. And then my earlier observation comes into play... for a pure power term you should use the order corresponding to the power.

With respect to your question on integration, my experience is that Bessel integrals are not trivial. Having not mastered them myself I will not offer any advice... I tend to use Maple/Mathematica and that fails quite often.

phioder

Actually I'm curious too, and it is the reason of this thread.

coomast

Hello phioder,

I will give it some thought this weekend and will come back to you. I definitely need to catch up on some sleep before doing this :-)

give me till monday or tuesday

phioder

Hello Coomast

Thank you for your reply and help, have a good sleep

Best Regards

coomast

Hello phioder,

I looked into my books on Bessel functions and it seems that there is no definite rule for using a specific function. Even more, one can "choose" the order for the expansion because there is a general way of expanding a given function into an orthogonal Bessel series. I do not have the proof of this (I might do this in the future when I have a little bit more time). The simplest case is the following:
Assume [tex]\lambda_1, \lambda_2, etc.[/tex] are the positive roots of [tex]J_n(x)=0[/tex], then

[tex]f(x)=A_1 J_n(\lambda_1 x)+ A_2 J_n(\lambda_2 x)+ ...[/tex]

with:

[tex]A_k=\frac{2}{J_{n+1}(\lambda_k)} \int_0^1 x f(x) J_n(\lambda_k x) dx[/tex]

This shows that you can have any n.

This depends on the problem you are solving. The boundary conditions give you the exact number that you need to use. There is no reason why you should choose here, it depends on the condition.

This is something different. The integral can be solved fairly easy. You need to remember the following formulas:

[tex]\frac{d}{dx}(x^nJ_n(x))=x^nJ_{n-1}(x)[/tex]
[tex]\frac{d}{dx}(x^{-n}J_n(x))=-x^{-n}J_{n+1}(x)[/tex]
[tex]J_1(x)=-\frac{d J_0(x)}{dx}[/tex]

together with partial integration. It can be shown that the following integral:

[tex]\int x^p J_q(x) dx[/tex]

can be obtained in closed form if [tex]p+q \geq 0[/tex] and p+q odd, with p and q integers. On the other hand if p+q even than the result is in terms of:

[tex]\int J_0(x) dx[/tex]

Now to tackle the one you gave (without the limits of integration), make the substitution [tex]\lambda x=t[/tex], you get:

[tex]\frac{1}{\lambda^4}\int t^3 J_1 (t) dt =\frac{1}{\lambda^4} I_1[/tex]

Now this new integral can be obtained by partial integration as:

[tex]I_1 = \int t[t^2 J_1(t)] dt =t t^2 J_2(t)-\int t^2 J_2(t)dt[/tex]
[tex]I_1 = t^3 J_2(t)-\int t^3 \frac{J_2(t)}{t}dt[/tex]
[tex]I_1 = t^3 J_2(t)-\left[t^3 \frac{-J_1(t)}{t}-\int \frac{-J_1(t)}{t} 3t^2dt\right] [/tex]
[tex]I_1 = t^3 J_2(t)+t^2 J_1(t) -3 \int t J_1(t) dt [/tex]
[tex]I_1 = t^3 J_2(t)+t^2 J_1(t) -3 \left[-\int t d(J_0(t)) \right][/tex]
[tex]I_1 = t^3 J_2(t)+t^2 J_1(t) +3t J_0(t) -3 \int J_0(t) dt [/tex]

The final integral can't be expressed in terms of more elementary functions. It needs to be solved numerically. Now it is possible to change back to the original variable x and apply the limits. You end up with a numerical integral.

Hope this information helps you a bit forward. Assuming that you do not need to do exams on this, take your time in studying this. Step by step you will come to like it more and more. Bessel functions are a fascinating subject, together with the other differential equations (Legendre, Hermite, ...) you will eventually master a powerful way for solving partial differential equations.

best regards, Coomast

mda

Actually it can, but it involves Struve functions and obviously that makes life more complicated.

mda

Also, the original integral can alternatively be expressed in terms of Lommel functions, but again more complication and much less obvious from the outset.

phioder

Dear Coomast

Thank you very much for the answer and motivation. It helped much more than a bit forward. I'm currently trying to understand the idea of orthogonal bases and on the way want to try see the differences of expanding a function with different orders. Have seen some sights of Legendre, Hermit, Ince and really hope to understand as much as possible to solve partial differential equations.

Very best Regards
Phioter

rezaborhani

Hi
The order is depending on the problem you're solving, for example in partial differential equations with boundary values you must use the order of your differential equation that you have obtain to expand the boundary functions.

rezaborhani

Hi
The order is depending on the problem you're solving, for example in partial differential equations with boundary values you must use the order of your differential equation that you have obtain to expand the boundary functions.