Bessel's Differential Equation

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Bessel's equation can be solved by a Frobenius Ansatz introduced in Series Solution of ODEs.

\begin{align} z^2 \, y'' + z \, y' + (z^2-\nu^2) \, y =0 \end{align} were $y(z)$. To bring the equation in stardard form we devide by $z^2$. \begin{align} y'' + \frac{1}{z} y' + \left( 1-\frac{\nu^2}{z^2} \right) \, y =0 \end{align} Now we plug in our ansatz $y=z^{\sigma} \sum_{n=0}^{\infty} a_n \, z^n$. Recall the derivatives \begin{align} y'=\sum_{n=0}^{\infty} (n+ \sigma) a_n \, z^{n + \sigma -1} \end{align} and \begin{align} y''=\sum_{n=0}^{\infty} (n+ \sigma)(n + \sigma -1) a_n \, z^{n + \sigma -2} \end{align} and plug in \begin{align} \sum_{n=0}^{\infty} (n+ \sigma)(n + \sigma -1) a_n \, z^{n + \sigma -2} + \frac{1}{z} \sum_{n=0}^{\infty} (n+ \sigma) a_n \, z^{n + \sigma -1} + \left( 1-\frac{\nu^2}{z^2} \right) \, z^{\sigma} \sum_{n=0}^{\infty} a_n \, z^n =0 \end{align} \begin{align} \sum_{n=0}^{\infty} \left[(n+ \sigma)(n + \sigma -1)+(n+ \sigma)-\nu^2\right] \, a_n \, z^{n + \sigma -2} + z^{\sigma} \sum_{n=0}^{\infty} a_n \, z^n =0 \end{align} or by shifting indices and deviding by $z^{\sigma}$ \begin{align} \sum_{n=0}^{\infty} \left[(n+ \sigma)(n + \sigma -1)+(n+ \sigma)-\nu^2\right] \, a_n \, z^{n} + \sum_{n=0}^{\infty} a_n \, z^{n+2} =0 \tag{1} \end{align} Simlifying further we have \begin{align} \sum_{n=0}^{\infty} \left[(n+ \sigma)(n + \sigma)-\nu^2 \, a_n + z^2 \right]a_n \,z^{n} =0 \end{align} which gives for $n=0$ and $z=0$ the equation \begin{align} \sigma^2-\nu^2 =0 \end{align} that we could have obtained directly by the formula \begin{align} (\sigma)(\sigma -1) + s(z) \sigma + t(z) =0 \\ \end{align} with $s=1$ and $t=z^2-\nu^2$ at $z=0$. The quadradic equation has the simple roots $\sigma=\pm \nu$.

To calulate the coefficients $a_n$ plug in the obtained values for $\sigma$ in equation (1), that can be written as \begin{align} \sum_{n=0}^{\infty} \left[(n+ \sigma)(n + \sigma)-\nu^2 \, a_n + a_{n-2} \right]z^{n} =0 \end{align} For $\sigma=+\nu$ we find \begin{align} [(n+ \nu)^2-\nu^2] \, a_n + a_{n-2} &=0\\ n(n+2\nu) \, a_n + a_{n-2}&=0 \\ a_n&=- \frac{a_{n-2}}{n(n+2\nu)} \end{align} similarly for $\sigma=-\nu$ \begin{align} [(n- \nu)^2-\nu^2] \, a_n + a_{n-2} &=0\\ n(n-2\nu) \, a_n + a_{n-2}&=0 \\ a_n&=- \frac{a_{n-2}}{n(n-2\nu)} \end{align}

$\sigma=+\nu$ $\sigma=-\nu$
$a_0=1$ $a_0=1$
$a_2=\frac{1}{2(2+2\nu)}$ $a_2=\frac{1}{2(2-2\nu)}$
$a_4=-\frac{a_2}{4(4+2\nu)}=-\frac{1}{4(4+2\nu)(2+2\nu)}$ $a_4=\frac{a_2}{4(4-2\nu)}=\frac{1}{4(4-2\nu)(2-2\nu)}$
$a_6=\frac{a_4}{6(6+2\nu)}=\frac{1}{6(6+2\nu)4(4+2\nu)2(2+2\nu)}$ $a_6=\frac{a_4}{6(6-2\nu)}=\frac{1}{6(6-2\nu)4(4-2\nu)2(2-2\nu)}$
$\vdots$ $\vdots$
\begin{align}a_{2k}&=\frac{(-1)^k a_0}{2k \dots 4 \cdot 2 \cdot (2k+\nu) \dots (4+\nu)(2+\nu)}\\ &=\frac{(-1)^k a_0}{2^{2k}k! (k+2\nu) \dots (2+2\nu)(1+2\nu)} \end{align} \begin{align}a_{2k}&=\frac{(-1)^k a_0}{2k \dots 4 \cdot 2 \cdot (2k-\nu) \dots (4-\nu)(2-\nu)}\\ &=\frac{(-1)^k a_0}{2^{2k}k! (k-2\nu) \dots (2-2\nu)(1-2\nu)} \end{align}

choosing $a_1=0$ all odd coefficients are zero \begin{align} a_{2k+1}=0 \end{align} From the coefficients in the table above we can immediately see, that we run into truble if the $\nu$ is an integer, since $a_n$ diverges for some $n_0 \pm 2\nu=0$. All coefficients above this $n=n_0$ are not defined and we are unable to find two independent solutions as we will see.

  • If $\sigma=-nu \leq 0$, $\nu$ is not allowed to be a positive integer
  • If $\sigma=\nu \geq 0$, $\nu$ is not allowed to be a negative integer

Our Frobenius ansatz for $\sigma=+ \nu$ now takes the form \begin{align} y_{+v}(z)=\sum_{n=0}^{\infty} a_n \, z^{n+\nu} =\sum_{k=0}^{\infty} \frac{(-1)^k a_0}{2^{2k}k! (k+\nu) \dots (2+\nu)(1+\nu)} z^{2k+\nu} \end{align} This looks already pretty good, but we can make this look even better by choosing a more clever $a_0$. To make the clever choice more obvious let us recall the basic property of the Gamma Function, which is nothing but the generalisation of the factorial. \begin{align} \Gamma(x+1)=x \Gamma(x) \end{align} so we have \begin{align} \Gamma(k+\nu +1)=k+\nu \Gamma(k+\nu)=(k+\nu)(k-1+\nu) \Gamma(k-1+\nu)= \dots \end{align} applying this rule $k$ times we have \begin{align} \Gamma(k+\nu +1)=(k+\nu)(k-1+\nu) \dots (2+\nu)(1+\nu) \Gamma(1+\nu) \end{align} Perfect! The factors in front of $\Gamma(1+\nu)$ are just the same we encounter in the expression for $a_n$. So it seems resonalbe, to make the clever choice. \begin{align} a_0=\frac{1}{2^{\nu} \, \Gamma(1+\nu)} \end{align}

If so, the solution to Bessel's equation can be written in the neat form \begin{align} J_{v}(z) =\sum_{k=0}^{\infty} \frac{(-1)^k }{k! \, \Gamma(k+\nu +1)} \left( \frac{z}{2} \right)^{2k+\nu} \end{align} This is one Bessel function of $\nu$-th order. In the same way we can find the second solution for $-\nu$ \begin{align} J_{-v}(z) =\sum_{k=0}^{\infty} \frac{(-1)^k }{k! \, \Gamma(k-\nu +1)} \left( \frac{z}{2} \right)^{2k-\nu} \end{align} This equations are called bessel equations of the first kind, they build the general solution of Bessel's equation for non integer $\nu$ \begin{align} y(z)=c_1 \cdot J_v(z) + c_2 \cdot J_{-v}(z) \end{align}

If $\nu$ is an integer we can use the simple factorial notation. \begin{align} J_{v}(z) =\sum_{k=0}^{\infty} \frac{(-1)^k }{k! \, (k+\nu)!} \left( \frac{z}{2} \right)^{2k+\nu} \end{align} and \begin{align} J_{-v}(z) =\sum_{k=0}^{\infty} \frac{(-1)^k }{k! \, (k-\nu)!} \left( \frac{z}{2} \right)^{2k-\nu} \end{align} If we shift variables $k=j+n$ the last expression gives \begin{align} J_{-v}(z) =\sum_{k=0}^{\infty} \frac{(-1)^{j+n} }{(j+n)! \, j!} \left( \frac{z}{2} \right)^{2k-\nu} \end{align} This solutions are dependent \begin{align} J_{v}(z) =(-1)^n \, J_{-v}(z) \end{align}

If we choose $\nu$ an integer, one branch of the solution (belonging eather to $\sigma=\nu>0$ or $\sigma=\nu < 0$) can not be computed, thus we can not find two independent solutions although we have two different $\sigma$'s. In this case we can find the second independent solution by the formula \begin{align} Y_{v}(z) =\frac{J_v(z) cos (\nu \pi) - J_{-v}(z)}{sin (v \pi)} \end{align} called bessel function of the second kind or webers function. For integer $\nu$ the general solution to Bessel's equation is \begin{align} y(z)=c_1 \cdot J_v(z) + c_2 \cdot Y_{v}(z) \end{align}


Find the general solution of \begin{align} z^2 \, y'' + z \, y' + \left[z^2- \left( \frac{1}{2} \right) ^2 \right] \, y =0 \end{align} This is Bessel's equation for $\nu=1/2$, which is not an integer so the general solution is. \begin{align} y(z)=c_1 \cdot J_v(z) + c_2 \cdot J_{-v}(z) \end{align} The Bessel functions turn out just nicely! \begin{align} J_{\pm v}(z) =\sum_{k=0}^{\infty} \frac{(-1)^k }{k! \, \Gamma(k \pm \nu +1)} \left( \frac{z}{2} \right)^{2k \pm \nu}\\ J_{\pm 1/2}(z) =\sum_{k=0}^{\infty} \frac{(-1)^k }{k! \, \Gamma(k \pm \left( \frac{1}{2} \right) +1)} \left( \frac{z}{2} \right)^{2k \pm \left( \frac{1}{2} \right)}\\ \end{align} We use the property of the Gamma Function $\Gamma(x+1)=x \Gamma(x)$ and $\Gamma\left( \frac{1}{2} \right)=\sqrt{\pi}$. We find for positive $\nu$ \begin{align} J_{ 1/2}(z) &=\frac{1}{\Gamma(3/2)}\left( \frac{z}{2} \right)^{1/2}-\frac{1}{\Gamma(5/2)}\left( \frac{z}{2} \right)^{5/2}+\frac{1}{2!\Gamma(7/2)}\left( \frac{z}{2} \right)^{9/2} - \dots \\ J_{1/2}(z) &=\frac{1}{\left( \frac{1}{2} \right) \sqrt{\pi}}\left( \frac{z}{2} \right)^{1/2}-\frac{1}{\left( \frac{3}{2} \right) \left( \frac{1}{2} \right)\sqrt{\pi} }\left( \frac{z}{2} \right)^{5/2}+\frac{1}{2! \left( \frac{5}{2} \right) \left( \frac{3}{2} \right) \left( \frac{1}{2} \right)\sqrt{\pi} }\left( \frac{z}{2} \right)^{9/2} - \dots \\ J_{ 1/2}(z) &=\frac{\left( \frac{z}{2} \right)^{1/2}}{\left( \frac{1}{2} \right) \sqrt{\pi}} \left[1- \frac{z^2}{3!}+ \frac{z^5}{5!} - \dots \right] = \frac{\left( \frac{z}{2} \right)^{1/2}}{\left( \frac{1}{2} \right) \sqrt{\pi}} \frac{sin (z)}{z}= \sqrt{\frac{2}{\pi z}} sin(z)\\ \end{align} and for negative $\nu$ \begin{align} J_{- 1/2}(z) &=\frac{1}{\Gamma(3/2)}\left( \frac{z}{2} \right)^{-1/2}-\frac{1}{\Gamma(5/2)}\left( \frac{z}{2} \right)^{3/2}+\frac{1}{2!\Gamma(7/2)}\left( \frac{z}{2} \right)^{7/2} - \dots \\ J_{- 1/2}(z) &=\frac{\left( \frac{z}{2} \right)^{-1/2}}{\left( \frac{1}{2} \right) \sqrt{\pi}} \left[1- \frac{z^2}{3!}+ \frac{z^5}{5!} - \dots \right] = \sqrt{\frac{2}{\pi z}} cos(z)\\ \end{align} Ultimately the general solution is

\begin{align} y(z)=c_1 \cdot J_v(z) + c_2 \cdot J_{-v}(z)=c_1 \cdot \sqrt{\frac{2}{\pi z}} sin(z) + c_2 \sqrt{\frac{2}{\pi z}} cos(z) \end{align}

Further Reading:

  • Riley, Hobson, Bence - Mathematical Methods for Physics and Engineering
  • G. B. Folland - Fourier Analysis and its Applications
  • Schaums Outlines - Fourier Analysis