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! :)