I need to verify Bessel function expension.

[yungman]

I am almost certain I understand the Bessel function expension correctly, but I just want to verify with you guys to be sure:

1) $$J_{p}(\alpha_{j}x)=\sum_{n=0}^{\infty}\frac{(-1)^{n}\alpha_{j}^{2n+p}x^{2n+p}}{n!\Gamma(n+p+1)2^{2n+p}}$$

2) $$f(x)=\sum_{j=1}^{\infty}A_{j}J_{p}(\alpha_{j}x)=\sum_{j=1}^{\infty}[A_{j}\sum_{n=0}^{\infty}\frac{(-1)^{n}\alpha_{j}^{2n+p}x^{2n+p}}{n!\Gamma(n+p+1)2^{2n+p}}]$$

3) $$\int_{0}^{R}xJ_{p}(\alpha_{j}x)J_{p}(\alpha_{k}x)dx=\int_{0}^{R}x[\sum_{n=0}^{\infty}\frac{(-1)^{n}\alpha_{j}^{2n+p}x^{2n+p}}{n!\Gamma(n+p+1)2^{2n+p}}][\sum_{n=0}^{\infty}\frac{(-1)^{n}\alpha_{k}^{2n+p}x^{2n+p}}{n!\Gamma(n+p+1)2^{2n+p}}]dx$$

Please take a look and let me know if I am correct or not from studying the books.

Thanks

Alan

 

[yungman]

Any body please!

 

[Astronuc]

Equations 1 and 3 look correct.

I'm not sure what one is trying to with equation 2. One is simply defining f(x) as a weighted sum of J_p(a_jx), but it looks OK.

Curious about using $$\Gamma(n+p+1)$$ as opposed to (n+p)!

One can also check - http://mathworld.wolfram.com/BesselFunctionoftheFirstKind.html

 

[yungman]

Equations 1 and 3 look correct.

I'm not sure what one is trying to with equation 2. One is simply defining f(x) as a weighted sum of J_p(a_jx), but it looks OK.

Curious about using $$\Gamma(n+p+1)$$ as opposed to (n+p)!

One can also check - http://mathworld.wolfram.com/BesselFunctionoftheFirstKind.html

Thanks for your help. 2) is just expand the $$J_{p}(\alpha_{j}x) and J_{p}(\alpha_{k}x)$$ out only. The reason I put this in because I have problem doing integration following this logic to prove the identity of integral on [0,a] of $$J^{2}_{p}(\alpha_{j}x)$$

I want to verify 2) so I can continue to figure out what I do wrong. With 2), I can pull out all the constant term, gamma function and all, then I only have to integrate$$x^{4n+2p+1}$$ on [0,a].


Regarding to the gamma function $$\Gamma(n+p+1)=(n+p)!$$ only if p is an integer. I have to read more on the link that you provide, I cannot pull it out of my head right at the moment, I have to do a little reading before I can answer that. All I know if p=1/2, your will have some number times $$\sqrt{\pi}$$. I don't think you can get the answer using (n+p)!.

 

[RedX]

I'm not sure what one is trying to with equation 2. One is simply defining f(x) as a weighted sum of J_p(a_jx), but it looks OK.

Equation 2 is taking an arbitrary function f(x) and writing it as an expansion of Bessel functions of order p.

In order to do this, f(x) can't blow up at the origin (or else you would have to include Bessel functions of the 2nd kind in your expansion). The Bessel functions of any order 'p' form a complete orthogonal set with weight x (and if your function doesn't blow up at the origin, then your expansion will only involve Bessel functions of the 1st kind of order 'p'), where the orthogonal set is $$J_p (\alpha_i x)$$ where $$\alpha_i=\frac{r_i}{R}$$ where the r_i are the roots of the bessel function of order p, and x=R is a boundary where the function vanishes.

The orthogonality follows from the fact that $$J_m(kx)$$ is a solution of the Bessel equation (written here in Sturm-Lioville form):

$$(xy')'+k^2xy-\frac{m^2}{x}y=0$$

which is a Hermitian linear differential operator with eigenvalue k^2 of weight x:

$$\mathcal Ly=[x\frac{d^2}{dx^2}+\frac{d}{dx}-\frac{m^2}{x}]y=-k^2xy$$

when appropriate boundary conditions are applied, which is done by discretizing or quantizing the allowed eigenvalues k to $$\frac{r_i}{R}$$