Bessel functions.

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the tree
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Bessel functions.

Postby the tree » Thu Nov 25, 2010 5:15 pm UTC

(this is for uni, so I guess it counts as homework help)

I've only just come across these functions and suddenly I need to know how to work them, so please bear with me and my possibly kind of dumb questions.

From what I've seen (reading from Advanced Engineering Mathematics, Kreyszig) Bessel's equation is:
[math]x^2 \frac{\mbox{d}^2 y}{\mbox{d} x^2} + x \frac{\mbox{d} y}{\mbox{d} x} + (\alpha^2 - x^2 ) y =0[/math]
But in some cases (by which I mean on Wikipedia) the [imath]y[/imath] coefficient is [imath](x^2 - \alpha^2 )[/imath], I don't think the distinction is important, but I'd like to make sure, is it?
Anyway, the ODE that I've come up against is:
[math]x^2 \frac{\mbox{d}^2 y}{\mbox{d} x^2} + x \frac{\mbox{d} y}{\mbox{d} x} + (a - bx^2) y =0[/math]
Where [imath]a,b[/imath] both involve some unknowns, which I think I need to know in order to get a general solution. I can't readily see a way to get the [imath]y[/imath] co-efficient looking like [imath](\alpha^2 - x^2 )[/imath] so that I'd be able to say the finite solution is [imath]J_\alpha (x)[/imath], could anyone help me there?

I know for sure that I need the solution to be finite, hence it being the first kind of Bessel function - this condition, it was hinted at me, also required [imath]\alpha[/imath] to be an integer, but no matter how many times I read the chapter in my book, or the wikipedia article, or whatever, I kind find anything to confirm or deny that?

If I were able to know some properties of the unkowns in the ODE that I've got, from answering those questions above then that'd give me information about other ODEs that I'm looking at and help me move on with the problem that I'm trying to tackle.

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Eebster the Great
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Re: Bessel functions.

Postby Eebster the Great » Thu Nov 25, 2010 11:01 pm UTC

The Bessel function of the first kind Jn(x) and the Bessel function of the second kind Yn(x) are two linearly independent solutions to the ODE [imath]x^2 \frac{d^2 y}{dx} + x \frac{dy}{dx} + (x^2 - n^2) y = 0[/imath]. To get the factor of b, you will need to have a factor of [imath]\sqrt{b}[/imath] in the argument (simple chain rule), and you will probably need an i or -i since the signs are reversed. And of course set [imath]n = \sqrt{a}[/imath].

I don't feel like working through it but it shouldn't give you much trouble.

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