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:

$$J_{1/2}(z)=\begin{cases} \sqrt{\frac{2}{\pi z}}\sin(z) & z>0 \\ 0 & z=0 \end{cases}$$

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.