The Bessel functions are solutions to Bessel's equation,
$$
x^2y''+xy'+(x^2 - \nu^2)y=0
$$
The parameter $\nu$ is called the order of the equation.
This equation has a regular singular point at $x_0=0.$ Since that is the
only singular point, the series solution about $x_0=0$ encounters no other
singular point so the series solution about $x_0=0$ will converge in the
entire punctured plane, $0 < |x| < \infty$ (by a "punctured" plane
we mean the plane which has a single point cut out, in this case the point
$x_0=0$).
- (6 points) Find the series solution about $x_0=0$ of Bessel's equation
of order
$\nu=1/2$ corresponding to the larger root of the indicial equation.
- (2 points) Find the pattern for the coefficients of the series
solution constructed in the previous problem. Show that the series is equal to
$\displaystyle \frac{\sin(x)}{\sqrt{x}}.$
In problem 1, following the pattern in the text, you picked
$a_0=1,$ but you could have picked any other value for $a_0$ and gotten
another solution. This just multiplies the solution you found by a
constant. For example, if you had picked $a_0=2,$ then every coefficient
would be multiplied by 2, so the overall function would be multiplied by
2. That's not anything new, anytime you find a solution
to a homogeneous linear differential
equation, any other multiple of the solution solves the same equation.
But this can cause confusion if we want to talk about "the" Bessel
function. In order to avoid confusion, we pick a
particular solution and call it $J_{\nu}(x),$ the Bessel function of order
$\nu.$ This is similar to choosing units in a physical problem. The agreed
convention is to define
$$
J_{\nu}(x)=\sum_{k=0}^{\infty} (-1)^k
\frac{x^{2k+\nu}}{2^{2k+\nu}k!\Gamma(k+\nu+1)}
$$
This is not the series you found in problem 1. Since
$\Gamma(3/2)=\sqrt{\pi}/2$ as was worked out in a previous extra credit
assignment, you get
$\displaystyle J_{1/2}(x)=\sqrt{\frac{2}{\pi x}}\sin(x).$ You may be
wondering why anyone would pick to multiply the series you
found by $\sqrt{2/\pi}$ and make that "the" Bessel function. The reason is
that part of the power of Bessel functions lies in the identities that
relate Bessel functions of different orders, as in problem 3 below. The
form of those identities is made simpler by choosing the $a_0$ terms for
different orders in a clever pattern. So you use $a_0=1$ for $J_0(x)$
(which was worked out in the text) but $a_0=\sqrt{2/\pi}$ for
$J_{1/2}(x)$ and so on. Feel free to stop by my office if you want to see
more about this normalization and why it is convenient.
$J_{1/2}(x)$ is one solution of the Bessel equation of order $1/2.$
Since the Bessel equation of order $1/2$ is a second-order linear
homogeneous equation, we need to find a second linearly independent
solution in order to build the general solution to the equation. We will
now prove one identity for Bessel functions and use that to find that
second solution.
- (2 points) Show $\displaystyle
\frac{d}{dx}\left(x^{\nu}J_{\nu}(x)\right)
= x^{\nu}J_{\nu-1}(x).$
- (2 points) Using the identity in the previous problem, show
$\displaystyle J_{-1/2}(x)=\sqrt{\frac{2}{\pi x}}\cos(x).$
Since the parameter $\nu$ is squared in Bessel's equation,
$x^2y''+xy'+(x^2-\nu^2)y=0,$ you get the same equation with $\nu=1/2$ and
$\nu=-1/2.$ So both $J_{1/2}(x)$ and $J_{-1/2}(x)$ are solutions, and
since they are linearly independent (as you can check by computing the
Wronskian), that makes the general solution
$y=c_1J_{1/2}(x)+c_2J_{-1/2}(x).$
It will always be the case that $J_{\nu}(x)$ and $J_{-\nu}(x)$ both
satisfy Bessel's equation of order $\nu,$ though they usually won't have a
simple representation in terms of $\sin(x)$ and $\cos(x)$ (or what would
be the point of defining a new type of functions). As long as $\nu$ is not
an integer, $J_{\nu}(x)$ and $J_{-\nu}(x)$ will also be linearly
independent and so you build the general solution of Bessel's equation of
order $\nu$ from these two functions. If $\nu$ is an integer, then they
are not linearly independent, in fact $J_n(x)=(-1)^nJ_{-n}(x)$ for
any integer $n.$ In this case you need the Bessel functions of the second
kind, $Y_n(x),$ which add a logarithmic singularity to the mix. And there
are many more identities and tricks involving both kinds of Bessel
functions. As I noted in class, you can fill a whole book about the Bessel
functions. I hope the few problems you've done here provide a small taste
of this area of mathematics.
If you have any problems with this page, please contact
bennett@math.ksu.edu.
©2010 Andrew G. Bennett