Bessel function series expansion

[hasan_researc]

Homework Statement

This is the how the question begins.

1. Bessel's equation is z^{2}\frac{d^{2}y}{dz^{2}} + z\frac{dy}{dz} + \left(z^{2}- p^{2}\right)y = 0.

For the case p^{2} = \frac{1}{4}, the equation has two series solutions which (unusually) may be expressed in terms of elementary functions:

J_{1/2} = \left(\frac{2}{\pi z}\right)^{1/2} sin z
J_{-1/2} = \left(\frac{2}{\pi z}\right)^{1/2} cos z

[ The factors \left(\frac{2}{\pi z}\right)^{1/2} are supefluous, but are included by convention, for reasons that are not relevant to the present purposes.]

Clearly J_{-1/2} is singular at z = 0. Show that J_{1/2}(0) = 0.

2......(for later)

Homework Equations

The Attempt at a Solution

I am going to assume the solutions J_{1/2} and J_{-1/2} without worrying about why/how they come about.

Obviously, when z = 0, cos z \neq 0. Therefore, \left(\frac{2}{\pi z}\right)^{1/2} blows up and J_{-1/2} is singular at z = 0.

On the other hand, if we draw separately the graphs of \left(\frac{2}{\pi z}\right)^{1/2} and sin z and then combine the two in a single graph of J_{1/2}, we find that it is sinusoidal with an amplitude given by \left(\frac{2}{\pi z}\right)^{1/2}. This means that the curve oscillates as it moves towards z = 0 with an amplitude that tends to infinity as z tends to 0. How do I conclude from this that J_{1/2}(0) = 0.

 

[ystael]

Compare the order to which \sin z vanishes at 0 to the order of growth of \sqrt{2/(\pi z)} at 0. From that you should be able to prove an estimate of the form J_{1/2}(z) = O(f(z)) as z \to 0 for an appropriately chosen bound f.

 

[hasan_researc]

Thanks for your help!

I don't understand what 'order' means in this context.
How do we compare the orders?

"an estimate of the form J_{1/2}(z) = O(f(z)) as z \to 0 for an appropriately chosen bound f.": I don't understand!

 

[ystael]

"Order" is short for "order of growth", which means a simple function (usually a power, log, or exponential function or simple combination of those functions) which you can use to "measure" the speed with which a given function vanishes or grows near a point. These orders are measured up to a constant factor.

For example, \csc x grows like 1/x as x \to 0, that is, you can choose positive constants c, C so that c|x^{-1}| < |\csc x| < C|x^{-1}| for x close enough to zero. One common way to try to find an order of growth for a function is to expand it in a Taylor or Laurent series.

For the meaning of O(f(z)) look here: Wikipedia article on Big-O notation Using this notation you could express the example above as \csc x = \Theta(1/x) as x \to 0.

The point of my hint is that, to prove that J_{1/2}(0) = 0, you should examine the orders of growth near zero of the functions whose product is J_{1/2}, to prove that the product can be estimated by a function you know vanishes at 0.

 

[hasan_researc]

I will have a think about this method while we move on to the next part of the question.

2. By means of the substitution z = kx, show that J_{1/2}(kx) and J_{-1/2}(kx) are solutions of the following equations:

x^{2}\frac{d^{2}y}{dx^{2}} + x\frac{dy}{dx} + \left(k^{2}x^{2} - p^{2}\right)y = 0.

3. Show that this equation, subject to boundary conditions y(0) = 0, y(1) = 0, has eigenvalues k = n\pi, n = 1,2,3... and eigenfunctions y_{n} = J_{1/2}(n\pi x).

I am fine with 2, but din't know how to begin 3.

 

[benorin]

Try performing the limit calculation for a nore solid answer: show that \lim_{z\rightarrow 0} \left(\frac{2}{\pi z}\right)^{1/2} \sin z=1

 

[hasan_researc]

\lim_{z\rightarrow 0} \left(\frac{\sin z}{z}\right)=1, so
\lim_{z\rightarrow 0} \left(\frac{2}{\pi z}\right)^{1/2} (\left \sin z \right)
= \lim_{z\rightarrow 0} \left(\frac{2z}{\pi}\right)^{1/2}\left(\frac{\sin z}{z}\right)
= \lim_{z\rightarrow 0} \left(\frac{2z}{\pi}\right)^{1/2} \lim_{z\rightarrow 0} \left(\frac{\sin z}{z}\right)

The first limit is 0 and the second is 1, so the final answer is 0: a contradiction!

Thoughts?

 

[jackmell]

Show that J_{1/2}(0) = 0.

2......(for later)

Homework Equations

The Attempt at a Solution

I am going to assume the solutions J_{1/2} and J_{-1/2} without worrying about why/how they come about.

Obviously, when z = 0, cos z \neq 0. Therefore, \left(\frac{2}{\pi z}\right)^{1/2} blows up and J_{-1/2} is singular at z = 0.

On the other hand, if we draw separately the graphs of \left(\frac{2}{\pi z}\right)^{1/2} and sin z and then combine the two in a single graph of J_{1/2}, we find that it is sinusoidal with an amplitude given by \left(\frac{2}{\pi z}\right)^{1/2}. This means that the curve oscillates as it moves towards z = 0 with an amplitude that tends to infinity as z tends to 0. How do I conclude from this that J_{1/2}(0) = 0.

The function:

J_{1/2}(z)=\sqrt{\frac{2}{\pi z}}\sin(z)

is undefined at z=0. However, it's limit there is zero by L'Hospital's rule so that if we explicitly define:

J_{1/2}(z)=\begin{cases} \sqrt{\frac{2}{\pi z}}\sin(z) &amp; z&gt;0 \\<br /> 0 &amp; z=0<br /> \end{cases}<br />

then J_{1/2}(z) becomes continuous for real z greater than or equal to zero.

and that thing about drawing separate graphs is not correct. If you plot the function, you should get a sinusoidal wave for positive real z.