Airy Differential Equation

Some authors define a general Airy differential equation as

$$y''\pm k^2xy = 0.$$ (1)

This equation can be solved by series solution using the expansions

$$y = \sum_{n=0}^\infty a_n x^n$$ (2)

$$y' = \sum_{n=0}^\infty n a_n x^{n-1} = \sum_{n=1}^\infty n a_n x^{n-1}$$

$$= \sum_{n=0}^\infty (n+1) a_{n+1} x^n$$ (3)

$$y'' = \sum_{n=0}^\infty (n+1)n a_{n+1} x^{n-1} = \sum_{n=1}^\infty (n+1)n a_{n+1} x^{n-1}$$

$$= \sum_{n=0}^\infty (n+2)(n+1) a_{n+2} x^n.$$ (4)

Specializing to the ``conventional'' Airy differential equation occurs by taking the Minus Sign and setting $k^2=1$. Then plug (4) into

$$y''-xy = 0$$ (5)

to obtain

$$\sum_{n=0}^\infty (n+2)(n+1) a_{n+2} x^n - x \sum_{n=0}^\infty a_n x^n = 0$$ (6)

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

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

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

In order for this equality to hold for all $x$, each term must separately be 0. Therefore,

$$a_2 = 0$$ (10)

$$(n+2)(n+1)a_{n+2} = a_{n-1}.$$ (11)

Starting with the $n=3$ term and using the above Recurrence Relation, we obtain

$$5 \cdot 4 a_5 = 20 a_5 = a_2 = 0.$$ (12)

Continuing, it follows by Induction that

$$a_2 = a_5 = a_8 = a_{11} = \ldots a_{3n-1} = 0$$ (13)

for $n=1$, 2, .... Now examine terms of the form $a_{3n}$.

$$a_3 = {{a_0} \over {3 \cdot 2}}$$ (14)

$$a_6 = {{a_3} \over {6 \cdot 5}} = {{a_0} \over {(6 \cdot 5)(3 \cdot 2)}}$$ (15)

$$a_9 = {{a_6} \over {9 \cdot 8}} = {{a_0} \over {(9 \cdot 8)(6 \cdot 5)(3 \cdot 2)}}.$$ (16)

Again by Induction,

$$a_{3n} = {{a_0} \over {[(3n)(3n-1)][(3n-3)(3n-4)]\cdots [6 \cdot 5][3 \cdot 2]}}$$ (17)

for $n=1$, 2, .... Finally, look at terms of the form $a_{3n+1}$,

$$a_4 = {{a_1} \over {4 \cdot 3}}$$ (18)

$$a_7 = {{a_4} \over {7 \cdot 6}} = {{a_1} \over {(7 \cdot 6)(4 \cdot 3)}}$$ (19)

$$a_{10} = {{a_7} \over {10 \cdot 9}} = {{a_1} \over {(10 \cdot 9)(7 \cdot 6)(4 \cdot 3)}}.$$ (20)

By Induction,

$$a_{3n+1} = {{a_1} \over {[(3n+1)(3n)][(3n-2)(3n-3)] \cdots [7 \cdot 6][4 \cdot 3]}}$$ (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

$$y''-k^2xy = 0,$$ (23)

and the solution is

$$y(x)= {\textstyle{1\over 3}}\sqrt{x}\left[{AI_{-1/3}\left({{... ...I_{1/3}\left({{\textstyle{2\over 3}} kx^{3/2}}\right)}\right],$$ (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 $\mathop{\rm Bi}\nolimits (x)$

$$y(x) = A' \mathop{\rm Ai}\nolimits (k^{2/3}x)+B' \mathop{\rm Bi}\nolimits (k^{2/3}x).$$ (25)

If the Plus Sign is present instead, then

$$y''+k^2xy = 0$$ (26)

and the solutions are

$$y(x)= {\textstyle{1\over 3}}\sqrt{x}\left[{AJ_{-1/3}\left({{... ...J_{1/3}\left({{\textstyle{2\over 3}} kx^{3/2}}\right)}\right],$$ (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