# Bessel function and Bessel D.E.

1. Mar 31, 2005

### quasar987

I'm trying to show that the Bessel function of the first kind satisfies the Bessel differential equation for m greater of equal to 1.

The Bessel function of the first kind of order m is defined by

$$J_m(x) = \sum_{n=0}^{\infty} \frac{(-1)^n}{2^{m+2n}n!(n+m)!}x^{m+2n} = x^m \sum_{n=0}^{\infty} \frac{(-1)^n}{2^{m+2n}n!(n+m)!}(x^2)^n$$

and suposedly it satisfies the Bessel differential equation, which would then write

$$x^2\frac{d^2 J}{dx^2} + x\frac{dJ}{dx} + (x^2-m^2)J = 0$$

I'm trying to help myself in doing that by following the steps laid down in the mathworld page on this subject. There is one detail however, it is that on the site, they proceed "backward". I.e. they start by saying that the Bessel function is defined by the solution to the Bessel D.E. and then proceed to find its form.

My main trouble with the steps they take is that in equation (30), they state that a_1 = 0. But a_n, as they will find at the end, is

$$a_n = \frac{(-1)^n}{2^{m+2n}n!(n+m)!}$$

and of course, a_1 is not 0 ! So what the greasy poop's going on here? :grumpy:

2. Mar 31, 2005

### dextercioby

I didn't see anything devious.You have to be careful with that change of summation variable from "n" to "l"...

Daniel.

P.S.If u still think there's something wrong,i'm sure u can find this matter in any ODE book treating Bessel's equation.

3. Mar 31, 2005

### quasar987

I didn't really get into any of the change of variable from n to l thing. My main concern is that they assert that a_1 is 0 but it's not.

4. Apr 1, 2005

### dextercioby

It is zero.Again,follow my advice and look into an ODE book which treats Bessel eq.(and implicitely the functions).

Daniel.

5. Apr 1, 2005

### quasar987

I'm very far from having access to books right now. What's wrong with...

$$a_1 = \frac{(-1)^1}{2^{m+2(1)}1!(1+m)!} = \frac{-1}{2^{m+2}(1+m)!}$$

?

6. Apr 1, 2005

### dextercioby

There you go.It can't get any more clear than that:

Daniel.

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7. Apr 1, 2005

### quasar987

thanks dexter but this link really doesn't say anything mathworld doesn't already say and so I'm still oblivious to how we can simultaneously have a_1 = a_3 = ... = 0 and

$$a_n = \frac{(-1)^n}{2^{m+2n}n!(n+m)!}$$

since it would imply

$$a_1= \frac{-1}{2^{m+2}(1+m)!}\neq 0$$

(Unless of course, a_n is defined in this way only for n even, but then it would be false to write the Bessel function as

$$J_m(x) = \sum_{n=0}^{\infty} \frac{(-1)^n}{2^{m+2n}n!(n+m)!}x^{m+2n}$$

with no mention of the crucial fact that n only takes even values.)

Last edited: Apr 1, 2005
8. Apr 1, 2005

### dextercioby

Well,if u didn't understand how it was done there,then it's not my fault,nor the material's.I can't do anything more...

Daniel.

9. Apr 1, 2005

### quasar987

You could say something like

We don't, because ....

10. Apr 1, 2005

### Data

$$J_m(x) = \sum_{n=0}^{\infty} \frac{(-1)^n}{2^{m+2n}n!(n+m)!}x^{m+2n}$$

$$= x^m\sum_{n=0}^{\infty} b_n x^{2n}$$

the terms with odd powers of $x$ aren't included in the series above.

And I don't see where mathworld or dexter's document says

$$a_n = \frac{(-1)^n}{2^{m+2n}n!(n+m)!}$$

Last edited: Apr 1, 2005
11. Apr 1, 2005

### shmoe

The series form you've written here does no have the same $$a_n$$ as you started with (in mathworlds (4)). Notice how the powers of this series are limited to m, m+2, m+4, ...., it's built in to this notation that 'every other' term drops out.

Look carefully at (36)-(40), especially (36) which shows how they've seperated the even/odd terms.

12. Apr 1, 2005

### quasar987

Right. The a_n of (4) is not the same as the a_n I assumed because in (4), x is to the n, wherehas in the final form of the Bessel function, x is to the 2n.

Ok, so this whole page from mathworld won't help me at all to show that the Bessel function of the first kind satisfies the Bessel differential equation for m greater of equal to 1, will it?

Anyone sees how this can be done?

I've tried playing with the following recurrence relations (about halfway on the page), but without success :grumpy:.

13. Apr 1, 2005

### dextercioby

Solving the equation and finding the initial solution (given by the problem) is an equivalent approach.

Daniel.

14. Apr 1, 2005

### Data

Why don't you just substitute it into the equation and show that it satisfies it? There's a bit of algebra, but it's straightforward and not that long anyhow.

15. Apr 1, 2005

### quasar987

It doesn't add up for me. What I did is set

$$J_p(x) = x^p f(x)$$

Where

$$f(x) = \sum_{n=0}^{\infty}a_n X^n$$

where

$$a_n = \frac{(-1)^n}{2^{p+2n}n!(n+p)!}$$

and

$$X = X(x) = x^2$$

That way, f(x) is a power serie with a radius of convergence $R \neq 0$, so according to a theorem seen in class,

$$f ' (x) = \sum_{n=0}^{\infty} \frac{d}{dx}a_n x^{2n} = \sum_{n=0}^{\infty}2n a_n x^{2n-1}$$

and

$$f '' (x) = \sum_{n=0}^{\infty}2n(2n-1) a_n x^{2n-2}$$

With this I construct $x^2 J''(x)$, $x J'(x)$ and $(x^2-p^2)J(x)$, and add them up. However, after double checking all my calculation, I am left with

$$\sum_{n=0}^{\infty}(4np+4n^2+x^2) a_n x^{2n+p}$$

Is there anything wrong with what I did above?

How did you do it Data?

16. Apr 1, 2005

### Data

Well, if you're going to pull the $x^p$ out of the sum, then you'll need to use product rule later to evaluate $J''$. Anyways, here's how I'd go about it:

We have

$$J_m(x) = \sum_{n=0}^{\infty} \frac{(-1)^n}{2^{m+2n}n!(n+m)!}x^{m+2n}, \; m \geq 0$$

which is a power series with radius of convergence $\infty$, so differentiating termwise is justified. Then

$$J^\prime_m(x) = \sum_{n=0}^{\infty} \frac{(-1)^n(m+2n)}{2^{m+2n}n!(n+m)!}x^{m+2n-1}$$

$$J^{\prime\prime}_m(x) = \sum_{n=0}^{\infty} \frac{(-1)^n(m+2n)(m+2n-1)}{2^{m+2n}n!(n+m)!}x^{m+2n-2}$$

Our differential equation is

$$x^2\frac{d^2\psi}{dx^2} + x\frac{d\psi}{dx} + (x^2-m^2)\psi = 0$$

so substituting $J$ for $\psi$ on the left side gives

$$x^2\frac{d^2J}{dx^2} + x\frac{dJ}{dx} + (x^2-m^2)J$$

$$= x^2\sum_{n=0}^{\infty} \left[\frac{(-1)^n(m+2n)(m+2n-1)x^{m+2n-2}}{2^{m+2n}n!(n+m)!}\right] + x\sum_{n=0}^{\infty} \left[\frac{(-1)^n(m+2n)x^{m+2n-1}}{2^{m+2n}n!(n+m)!}\right] + (x^2 - m^2)\sum_{n=0}^{\infty} \left[\frac{(-1)^nx^{m+2n}}{2^{m+2n}n!(n+m)!}\right]$$

$$= \sum_{n=0}^{\infty} \left[\frac{(-1)^n(m+2n)(m+2n-1)x^{m+2n}}{2^{m+2n}n!(n+m)!}\right] + \sum_{n=0}^{\infty} \left[\frac{(-1)^n(m+2n)x^{m+2n}}{2^{m+2n}n!(n+m)!}\right] + \sum_{n=0}^{\infty} \left[\frac{(-1)^nx^{m+2n+2}}{2^{m+2n}n!(n+m)!}\right] - \sum_{n=0}^{\infty} \left[\frac{(-1)^nm^2x^{m+2n}}{2^{m+2n}n!(n+m)!}\right]$$

$$= \sum_{n=0}^\infty \left[\frac{(-1)^nx^{m+2n}}{2^{m+2n}n!(n+m)!}\left((m+2n)(m+2n-1)+(m+2n)-m^2\right)\right] + \sum_{n=0}^{\infty} \left[\frac{(-1)^nx^{m+2n+2}}{2^{m+2n}n!(n+m)!}\right]$$

$$= \sum_{n=0}^\infty \left[\frac{(-1)^nx^{m+2n}}{2^{m+2n}n!(n+m)!}\left(4n^2 + 4nm)\right)\right] + \sum_{n=0}^{\infty} \left[\frac{(-1)^nx^{m+2n+2}}{2^{m+2n}n!(n+m)!}\right]$$

by theorems, this is equal to zero if and only if the coefficient of each power of $x$ is zero. The smallest power of $x$ on the left side is $m$ when n=0 in the first series and its coefficient is (from the first, second, and last series) just

$$\frac{(-1)^0(4(0)^2+4(0)m)}{2^m0!m!} = 0$$

as desired. We can thus ignore the first term in the first series and make a change of index, $n \rightarrow n+1$. This gives

$$\sum_{n=0}^\infty \left[\frac{(-1)^{n+1}x^{m+2n+2}}{2^{m+2n+2}(n+1)!(n+m+1)!}\left(4(n+1)^2 + 4(n+1)m)\right)\right] + \sum_{n=0}^{\infty} \left[\frac{(-1)^nx^{m+2n+2}}{2^{m+2n}n!(n+m)!}\right]$$

$$=\sum_{n=0}^\infty \left(\frac{(-1)^{n+1}}{2^{m+2n+2}(n+1)!(n+m+1)!}\left(4(n+1)^2 + 4(n+1)m\right) + \frac{(-1)^n}{2^{m+2n}n!(n+m)!}\right)x^{m+2n+2}$$

$$=\sum_{n=0}^\infty \left(\frac{(-1)^{n+1}}{2^{m+2n+2}(n+1)!(n+m+1)!}\left(4n^2 + 4nm + 8n + 4m + 4\right) + \frac{(-1)^n}{2^{m+2n}n!(n+m)!}\right)x^{m+2n+2}$$

and again we look at coefficients. This time let's set

$$k_n = \frac{(-1)^{n+1}}{2^{m+2n+2}(n+1)!(n+m+1)!}$$

so our sum reduces to just

$$\sum_{n=0}^\infty k_n\left(\left(4n^2+4nm+8n+4m+4\right) + \left((-1)2^2(n+1)(n+m+1)\right)\right)x^{m+2n+2}$$

and thus the coefficient of $x^{m+2n+2}$ is

$$k_n\left(\left( 4n^2 + 4nm + 8n + 4\right) + \left((-1)2^2(n+1)(n+m+1)\right)\right)$$

$$= k_n \left(4n^2 + 4nm + 8n +4m+ 4 - 4(n^2 + 2n + nm + m + 1)\right)$$

$$= k_n\left( (4n^2 - 4n^2) + (4nm-4nm) + (8n-8n) + (4m-4m) + (4-4) \right)$$

$$= 0$$

and thus every coefficient in the series is $0$, and we have found

$$x^2\frac{d^2J}{dx^2} + x\frac{dJ}{dx} + (x^2-m^2)J = 0$$

as we wanted.

Last edited: Apr 1, 2005
17. Apr 1, 2005

### quasar987

Phew! Thanks for the long and most complete reply ever :)

18. Apr 1, 2005

### Data

Like I said, just a little arithmetic

19. Apr 1, 2005

### saltydog

Well, I worked it through also but I left the coefficients as $a_n$ and ended up with an expression:

$$\sum_{n=1}^{\infty}[(m+2n)(m+2n-1)a_n+(m+2n)a_n+a_{n-1}-m^2a_n]x^{m+2n}$$

This reduces to:

$$\sum_{n=1}^{\infty}[4n(m+n)a_n+a_{n-1}]x^{m+2n}$$

Now, substitute the expression for $a_n$ and $a_{n-1}$ and the results come out to zero.

20. Apr 1, 2005

### Data

Yeah. Essentially the same thing I did, except without writing everything out each time! Of course I just copied and pasted anyways, so it doesn't matter too much~