Eigenvalue problem using Bessel Functions

[kidmode01]

Homework Statement

Bessels equation of order n is given as the following:

$$y'' + \frac{1}{x}y' + (1 - \frac{n^2}{x^2})y = 0$$

In a previous question I proved that Bessels equation of order n=0 has the following property:

$$J_0'(x) = -J_1(x)$$

Where J(x) are Bessel functions of order n=0 and order n=1 respectively.

Use this property to solve the eigenvalue problem:

$$\frac{1}{r}\frac{d}{dr}(r(\frac{d\phi}{dr}))+ \lambda^2\phi = 0$$

where we have:

$$0 < r < a$$

$$\frac{d\phi}{dr}(a) = 0$$

$$\phi(0)$$ bounded

Homework Equations

The question about the special property:
The power series for the nth Bessel function is

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

Use this to show that:

$$J_{n+1}(x) = -x^n\frac{d}{dx}(x^{-n}J_n(x))$$

The Attempt at a Solution

*Phew, that was a lot of latex!*

I've tried expanding:

$$\frac{1}{r}\frac{d}{dr}(r(\frac{d\phi}{dr}))+ \lambda^2\phi = 0$$

using the product rule, into something more familiar to see if I can make some sort of relation to the property I am supposed to use. Any help is appreciated, I've spent a lot of time rearranging terms and trying to relate things but can't quite put my finger on it.

 

[kidmode01]

In case anyone is interested, I think this is right:

Like I was trying before we expand:

$$\frac{1}{r}\frac{d}{dr}(r(\frac{d\phi}{dr}))+ \lambda^2\phi = 0$$

we get:

$$\phi''+\frac{1}{r}\phi'+\lambda^2\phi = 0$$

To get this into a more "familiar" form of bessels equation, multiply everything by $$r^2$$:

$$r^2\phi"" + r\phi'+(r^2\lambda^2-0^2)\phi = 0$$

Which is exactly Bessel's equation of order n=0.

Digging into a textbook we can learn that the following is a solution to this equation:

$$\phi(r) = C_1J_0(\lambda r) + C_2Y_0(\lambda r)$$

Since in our problem we are told that $$\phi(0)$$ is bounded. But looking into our text we find that $$Y_0(x)$$ is some ghastly function with $$ln(x)$$ inside of it. Since $$ln(x)$$ is not bounded at 0, we can conclude that $$C_2 = 0$$.

Our second condition says:

$$\frac{d\phi}{dr}(a) = 0$$

then we must have

$$\phi'(a) = -\lambda C_1J_0(\lambda a) = 0$$

So we'll throw out the case that $$C_1$$ is zero because it isn't interesting. But if $$C_1$$ is not equal to zero then we either require $$\lambda = 0$$ or $$J_0(\lambda a) = 0$$

Then if $$x_1, x_2, x_3, . . .$$ are roots of $$J_0(\lambda a)$$ we get $$\lambda = \frac{x_1}{a}, \frac{x_2}{a}, \frac{x_3}{a} . . .$$

If I'm missing something or this is completely wrong, someone let me know! :)