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Extra Credit 4
Bessel Functions

    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 $x_0=0$).

  1. (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. (2 points) Find the pattern for the coefficients of the series solution constructed in the previous problem. Show that the series is equal to $\displaystyle \frac{\sin(x)}{\sqrt{x}}.$

    3. 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.

  3. (2 points) Show $\displaystyle \frac{d}{dx}\left(x^{\nu}J_{\nu}(x)\right) = x^{\nu}J_{\nu-1}(x).$

  4. (2 points) Using the identity in the previous problem, show $\displaystyle J_{-1/2}(x)=\sqrt{\frac{2}{\pi x}}\cos(x).$

    5. 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