info prev up next book cdrom email home

Airy Differential Equation

Some authors define a general Airy differential equation as

\begin{displaymath}
y''\pm k^2xy = 0.
\end{displaymath} (1)

This equation can be solved by series solution using the expansions
$\displaystyle y$ $\textstyle =$ $\displaystyle \sum_{n=0}^\infty a_n x^n$ (2)
$\displaystyle y'$ $\textstyle =$ $\displaystyle \sum_{n=0}^\infty n a_n x^{n-1} = \sum_{n=1}^\infty n a_n x^{n-1}$  
  $\textstyle =$ $\displaystyle \sum_{n=0}^\infty (n+1) a_{n+1} x^n$ (3)
$\displaystyle y''$ $\textstyle =$ $\displaystyle \sum_{n=0}^\infty (n+1)n a_{n+1} x^{n-1} = \sum_{n=1}^\infty (n+1)n a_{n+1} x^{n-1}$  
  $\textstyle =$ $\displaystyle \sum_{n=0}^\infty (n+2)(n+1) a_{n+2} x^n.$ (4)

Specializing to the ``conventional'' Airy differential equation occurs by taking the Minus Sign and setting $k^2=1$. Then plug (4) into
\begin{displaymath}
y''-xy = 0
\end{displaymath} (5)

to obtain
\begin{displaymath}
\sum_{n=0}^\infty (n+2)(n+1) a_{n+2} x^n - x \sum_{n=0}^\infty a_n x^n = 0
\end{displaymath} (6)


\begin{displaymath}
\sum_{n=0}^\infty (n+2)(n+1) a_{n+2} x^n - \sum_{n=0}^\infty a_n x^{n+1} = 0
\end{displaymath} (7)


\begin{displaymath}
2a_2 + \sum_{n=1}^\infty (n+2)(n+1) a_{n+2} x^n - \sum_{n=1}^\infty a_{n-1} x^n = 0
\end{displaymath} (8)


\begin{displaymath}
2a_2 + \sum_{n=1}^\infty [(n+2)(n+1) a_{n+2}-a_{n-1}]x^n = 0.
\end{displaymath} (9)

In order for this equality to hold for all $x$, each term must separately be 0. Therefore,
$\displaystyle a_2$ $\textstyle =$ $\displaystyle 0$ (10)
$\displaystyle (n+2)(n+1)a_{n+2}$ $\textstyle =$ $\displaystyle a_{n-1}.$ (11)

Starting with the $n=3$ term and using the above Recurrence Relation, we obtain
\begin{displaymath}
5 \cdot 4 a_5 = 20 a_5 = a_2 = 0.
\end{displaymath} (12)

Continuing, it follows by Induction that
\begin{displaymath}
a_2 = a_5 = a_8 = a_{11} = \ldots a_{3n-1} = 0
\end{displaymath} (13)

for $n=1$, 2, .... Now examine terms of the form $a_{3n}$.
$\displaystyle a_3$ $\textstyle =$ $\displaystyle {{a_0} \over {3 \cdot 2}}$ (14)
$\displaystyle a_6$ $\textstyle =$ $\displaystyle {{a_3} \over {6 \cdot 5}} = {{a_0} \over {(6 \cdot 5)(3 \cdot 2)}}$ (15)
$\displaystyle a_9$ $\textstyle =$ $\displaystyle {{a_6} \over {9 \cdot 8}} = {{a_0} \over {(9 \cdot 8)(6 \cdot 5)(3 \cdot 2)}}.$ (16)

Again by Induction,
\begin{displaymath}
a_{3n} = {{a_0} \over {[(3n)(3n-1)][(3n-3)(3n-4)]\cdots [6 \cdot 5][3 \cdot 2]}}
\end{displaymath} (17)

for $n=1$, 2, .... Finally, look at terms of the form $a_{3n+1}$,
$\displaystyle a_4$ $\textstyle =$ $\displaystyle {{a_1} \over {4 \cdot 3}}$ (18)
$\displaystyle a_7$ $\textstyle =$ $\displaystyle {{a_4} \over {7 \cdot 6}} = {{a_1} \over {(7 \cdot 6)(4 \cdot 3)}}$ (19)
$\displaystyle a_{10}$ $\textstyle =$ $\displaystyle {{a_7} \over {10 \cdot 9}} = {{a_1} \over {(10 \cdot 9)(7 \cdot 6)(4 \cdot 3)}}.$ (20)

By Induction,
\begin{displaymath}
a_{3n+1} = {{a_1} \over {[(3n+1)(3n)][(3n-2)(3n-3)] \cdots [7 \cdot 6][4 \cdot 3]}}
\end{displaymath} (21)

for $n=1$, 2, .... The general solution is therefore
$y = a_0\left[{1+\sum_{n=1}^\infty{{x^{3n}}\over{(3n)(3n-1)(3n-3)(3n-4)\cdots3\cdot2}}}\right]$
${}+a_1\left[{x+\sum_{n=1}^\infty{{x^{3n+1}}\over{(3n+1)(3n)(3n-2)(3n-3)\cdots4\cdot3}}}\right].$

(22)
For a general $k^2$ with a Minus Sign, equation (1) is
\begin{displaymath}
y''-k^2xy = 0,
\end{displaymath} (23)

and the solution is
\begin{displaymath}
y(x)= {\textstyle{1\over 3}}\sqrt{x}\left[{AI_{-1/3}\left({{...
...I_{1/3}\left({{\textstyle{2\over 3}} kx^{3/2}}\right)}\right],
\end{displaymath} (24)

where $I$ is a Modified Bessel Function of the First Kind. This is usually expressed in terms of the Airy Functions $\mathop{\rm Ai}\nolimits (x)$ and $\mathop{\rm Bi}\nolimits (x)$
\begin{displaymath}
y(x) = A' \mathop{\rm Ai}\nolimits (k^{2/3}x)+B' \mathop{\rm Bi}\nolimits (k^{2/3}x).
\end{displaymath} (25)

If the Plus Sign is present instead, then
\begin{displaymath}
y''+k^2xy = 0
\end{displaymath} (26)

and the solutions are
\begin{displaymath}
y(x)= {\textstyle{1\over 3}}\sqrt{x}\left[{AJ_{-1/3}\left({{...
...J_{1/3}\left({{\textstyle{2\over 3}} kx^{3/2}}\right)}\right],
\end{displaymath} (27)

where $J(z)$ is a Bessel Function of the First Kind.

See also Airy-Fock Functions, Airy Functions, Bessel Function of the First Kind, Modified Bessel Function of the First Kind



info prev up next book cdrom email home

© 1996-9 Eric W. Weisstein
1999-05-25