In the physical sciences, the Airy function (or Airy function of the first kind) Ai(x) is a special function named after the British astronomer George Biddell Airy (1801–1892). The function Ai(x) and the related function Bi(x), are linearly independent solutions to the differential equation

$${\displaystyle {\frac {d^{2}y}{dx^{2))}-xy=0,}$$

Because the solution of the linear differential equation

$${\displaystyle {\frac {d^{2}y}{dx^{2))}-ky=0}$$

Definitions

For real values of x, the Airy function of the first kind can be defined by the improper Riemann integral:

$${\displaystyle \operatorname {Ai} (x)={\dfrac {1}{\pi ))\int _{0}^{\infty }\cos \left({\dfrac {t^{3)){3))+xt\right)\,dt\equiv {\dfrac {1}{\pi ))\lim _{b\to \infty }\int _{0}^{b}\cos \left({\dfrac {t^{3)){3))+xt\right)\,dt,}$$

${\textstyle {\tfrac {t^{3)){3))+xt}$

${\displaystyle [M,\infty ).}$

${\textstyle u={\tfrac {t^{3)){3))+xt.}$

y = Ai(x) satisfies the Airy equation

$${\displaystyle y''-xy=0.}$$

$${\displaystyle \operatorname {Bi} (x)={\frac {1}{\pi ))\int _{0}^{\infty }\left[\exp \left(-{\tfrac {t^{3)){3))+xt\right)+\sin \left({\tfrac {t^{3)){3))+xt\right)\,\right]dt.}$$

Properties

The values of Ai(x) and Bi(x) and their derivatives at x = 0 are given by

$${\displaystyle {\begin{aligned}\operatorname {Ai} (0)&{}={\frac {1}{3^{2/3}\,\Gamma \!\left({\frac {2}{3))\right))),&\quad \operatorname {Ai} '(0)&{}=-{\frac {1}{3^{1/3}\,\Gamma \!\left({\frac {1}{3))\right))),\\\operatorname {Bi} (0)&{}={\frac {1}{3^{1/6}\,\Gamma \!\left({\frac {2}{3))\right))),&\quad \operatorname {Bi} '(0)&{}={\frac {3^{1/6)){\Gamma \!\left({\frac {1}{3))\right))).\end{aligned))}$$

When x is positive, Ai(x) is positive, convex, and decreasing exponentially to zero, while Bi(x) is positive, convex, and increasing exponentially. When x is negative, Ai(x) and Bi(x) oscillate around zero with ever-increasing frequency and ever-decreasing amplitude. This is supported by the asymptotic formulae below for the Airy functions.

The Airy functions are orthogonal^[1] in the sense that

$${\displaystyle \int _{-\infty }^{\infty }\operatorname {Ai} (t+x)\operatorname {Ai} (t+y)dt=\delta (x-y)}$$

Real zeros of Ai(x) and its derivative Ai'(x)

Neither Ai(x) nor its derivative Ai(x) have positive real zeros. The "first" real zeros (i.e. nearest to x=0) are:^[2]

• "first" zeros of Ai(x) are at x ≈ −2.33811, −4.08795, −5.52056, −6.78671, ...
• "first" zeros of its derivative Ai'(x) are at x ≈ −1.01879, −3.24820, −4.82010, −6.16331, ...

Asymptotic formulae

As explained below, the Airy functions can be extended to the complex plane, giving entire functions. The asymptotic behaviour of the Airy functions as |z| goes to infinity at a constant value of arg(z) depends on arg(z): this is called the Stokes phenomenon. For |arg(z)| < π we have the following asymptotic formula for Ai(z):^[3]

$${\displaystyle \operatorname {Ai} (z)\sim {\dfrac {1}{2{\sqrt {\pi ))\,z^{1/4))}\exp \left(-{\frac {2}{3))z^{3/2}\right)\left[\sum _{n=0}^{\infty }{\dfrac {(-1)^{n}\,\Gamma \!\left(n+{\frac {5}{6))\right)\,\Gamma \!\left(n+{\frac {1}{6))\right)\left({\frac {3}{4))\right)^{n)){2\pi \,n!\,z^{3n/2))}\right].}$$

$${\displaystyle \operatorname {Ai} (z)\sim {\dfrac {e^{-\zeta )){4\pi ^{3/2}\,z^{1/4))}\left[\sum _{n=0}^{\infty }{\dfrac {\Gamma \!\left(n+{\frac {5}{6))\right)\,\Gamma \!\left(n+{\frac {1}{6))\right)}{n!(-2\zeta )^{n))}\right].}$$

${\displaystyle \zeta ={\tfrac {2}{3))z^{3/2}.}$

$${\displaystyle \operatorname {Ai} (z)={\frac {e^{-\zeta )){2\pi ^{1/2}z^{1/4))}\left(1-{\frac {5}{72\zeta ))+{\frac {385}{10368\zeta ^{2))}+O(\zeta ^{-3})\right)}$$

$${\displaystyle \operatorname {Bi} (z)\sim {\frac {1}((\sqrt {\pi ))\,z^{1/4))}\exp \left({\frac {2}{3))z^{3/2}\right)\left[\sum _{n=0}^{\infty }{\dfrac {\Gamma \!\left(n+{\frac {5}{6))\right)\,\Gamma \!\left(n+{\frac {1}{6))\right)\left({\frac {3}{4))\right)^{n)){2\pi \,n!\,z^{3n/2))}\right].}$$

$${\displaystyle {\begin{aligned}\operatorname {Ai} (-z)\sim &{}\ {\frac {1}((\sqrt {\pi ))\,z^{1/4))}\sin \left({\frac {2}{3))z^{3/2}+{\frac {\pi }{4))\right)\left[\sum _{n=0}^{\infty }{\dfrac {(-1)^{n}\,\Gamma \!\left(2n+{\frac {5}{6))\right)\,\Gamma \!\left(2n+{\frac {1}{6))\right)\left({\frac {3}{4))\right)^{2n)){2\pi \,(2n)!\,z^{3n))}\right]\\[6pt]&{}-{\frac {1}((\sqrt {\pi ))\,z^{1/4))}\cos \left({\frac {2}{3))z^{3/2}+{\frac {\pi }{4))\right)\left[\sum _{n=0}^{\infty }{\dfrac {(-1)^{n}\,\Gamma \!\left(2n+{\frac {11}{6))\right)\,\Gamma \!\left(2n+{\frac {7}{6))\right)\left({\frac {3}{4))\right)^{2n+1)){2\pi \,(2n+1)!\,z^{3n\,+\,3/2))}\right]\\[6pt]\operatorname {Bi} (-z)\sim &{}{\frac {1}((\sqrt {\pi ))\,z^{1/4))}\cos \left({\frac {2}{3))z^{3/2}+{\frac {\pi }{4))\right)\left[\sum _{n=0}^{\infty }{\dfrac {(-1)^{n}\,\Gamma \!\left(2n+{\frac {5}{6))\right)\,\Gamma \!\left(2n+{\frac {1}{6))\right)\left({\frac {3}{4))\right)^{2n)){2\pi \,(2n)!\,z^{3n))}\right]\\[6pt]&{}+{\frac {1}((\sqrt {\pi ))\,z^{\frac {1}{4))))\sin \left({\frac {2}{3))z^{3/2}+{\frac {\pi }{4))\right)\left[\sum _{n=0}^{\infty }{\dfrac {(-1)^{n}\,\Gamma \!\left(2n+{\frac {11}{6))\right)\,\Gamma \!\left(2n+{\frac {7}{6))\right)\left({\frac {3}{4))\right)^{2n+1)){2\pi \,(2n+1)!\,z^{3n\,+\,3/2))}\right].\end{aligned))}$$

When |arg(z)| = 0 these are good approximations but are not asymptotic because the ratio between Ai(−z) or Bi(−z) and the above approximation goes to infinity whenever the sine or cosine goes to zero. Asymptotic expansions for these limits are also available. These are listed in (Abramowitz and Stegun, 1983) and (Olver, 1974).

One is also able to obtain asymptotic expressions for the derivatives Ai'(z) and Bi'(z). Similarly to before, when |arg(z)| < π:^[5]

$${\displaystyle \operatorname {Ai} '(z)\sim -{\dfrac {z^{1/4)){2{\sqrt {\pi ))\,))\exp \left(-{\frac {2}{3))z^{3/2}\right)\left[\sum _{n=0}^{\infty }{\frac {1+6n}{1-6n)){\dfrac {(-1)^{n}\,\Gamma \!\left(n+{\frac {5}{6))\right)\,\Gamma \!\left(n+{\frac {1}{6))\right)\left({\frac {3}{4))\right)^{n)){2\pi \,n!\,z^{3n/2))}\right].}$$

When |arg(z)| < π/3 we have:^[5]

$${\displaystyle \operatorname {Bi} '(z)\sim {\frac {z^{1/4))((\sqrt {\pi ))\,))\exp \left({\frac {2}{3))z^{3/2}\right)\left[\sum _{n=0}^{\infty }{\frac {1+6n}{1-6n)){\dfrac {\Gamma \!\left(n+{\frac {5}{6))\right)\,\Gamma \!\left(n+{\frac {1}{6))\right)\left({\frac {3}{4))\right)^{n)){2\pi \,n!\,z^{3n/2))}\right].}$$

Similarly, an expression for Ai'(−z) and Bi'(−z) when |arg(z)| < 2π/3 but not zero, are^[5]

$${\displaystyle {\begin{aligned}\operatorname {Ai} '(-z)\sim &{}-{\frac {z^{1/4))((\sqrt {\pi ))\,))\cos \left({\frac {2}{3))z^{3/2}+{\frac {\pi }{4))\right)\left[\sum _{n=0}^{\infty }{\frac {1+12n}{1-12n)){\dfrac {(-1)^{n}\,\Gamma \!\left(2n+{\frac {5}{6))\right)\,\Gamma \!\left(2n+{\frac {1}{6))\right)\left({\frac {3}{4))\right)^{2n)){2\pi \,(2n)!\,z^{3n))}\right]\\[6pt]&{}-{\frac {z^{1/4))((\sqrt {\pi ))\,))\sin \left({\frac {2}{3))z^{3/2}+{\frac {\pi }{4))\right)\left[\sum _{n=0}^{\infty }{\frac {7+12n}{-5-12n)){\dfrac {(-1)^{n}\,\Gamma \!\left(2n+{\frac {11}{6))\right)\,\Gamma \!\left(2n+{\frac {7}{6))\right)\left({\frac {3}{4))\right)^{2n+1)){2\pi \,(2n+1)!\,z^{3n\,+\,3/2))}\right]\\[6pt]\operatorname {Bi} '(-z)\sim &{}\ {\frac {z^{1/4))((\sqrt {\pi ))\,))\sin \left({\frac {2}{3))z^{3/2}+{\frac {\pi }{4))\right)\left[\sum _{n=0}^{\infty }{\frac {1+12n}{1-12n)){\dfrac {(-1)^{n}\,\Gamma \!\left(2n+{\frac {5}{6))\right)\,\Gamma \!\left(2n+{\frac {1}{6))\right)\left({\frac {3}{4))\right)^{2n)){2\pi \,(2n)!\,z^{3n))}\right]\\[6pt]&{}-{\frac {z^{1/4))((\sqrt {\pi ))\,))\cos \left({\frac {2}{3))z^{3/2}+{\frac {\pi }{4))\right)\left[\sum _{n=0}^{\infty }{\frac {7+12n}{-5-12n)){\dfrac {(-1)^{n}\,\Gamma \!\left(2n+{\frac {11}{6))\right)\,\Gamma \!\left(2n+{\frac {7}{6))\right)\left({\frac {3}{4))\right)^{2n+1)){2\pi \,(2n+1)!\,z^{3n\,+\,3/2))}\right]\\\end{aligned))}$$

Complex arguments

We can extend the definition of the Airy function to the complex plane by

$${\displaystyle \operatorname {Ai} (z)={\frac {1}{2\pi i))\int _{C}\exp \left({\tfrac {t^{3)){3))-zt\right)\,dt,}$$

The asymptotic formula for Ai(x) is still valid in the complex plane if the principal value of x^2/3 is taken and x is bounded away from the negative real axis. The formula for Bi(x) is valid provided x is in the sector ${\displaystyle x\in \mathbb {C} :\left|\arg(x)\right|<{\tfrac {\pi }{3))-\delta }$ for some positive δ. Finally, the formulae for Ai(−x) and Bi(−x) are valid if x is in the sector ${\displaystyle x\in \mathbb {C} :\left|\arg(x)\right|<{\tfrac {2\pi }{3))-\delta .}$

It follows from the asymptotic behaviour of the Airy functions that both Ai(x) and Bi(x) have an infinity of zeros on the negative real axis. The function Ai(x) has no other zeros in the complex plane, while the function Bi(x) also has infinitely many zeros in the sector ${\displaystyle z\in \mathbb {C} :{\tfrac {\pi }{3))<\left|\arg(z)\right|<{\tfrac {\pi }{2)).}$

Plots

${\displaystyle \Re \left[\operatorname {Ai} (x+iy)\right]}$	${\displaystyle \Im \left[\operatorname {Ai} (x+iy)\right]}$	${\displaystyle \left|\operatorname {Ai} (x+iy)\right|\,}$	${\displaystyle \operatorname {arg} \left[\operatorname {Ai} (x+iy)\right]\,}$

${\displaystyle \Re \left[\operatorname {Bi} (x+iy)\right]}$	${\displaystyle \Im \left[\operatorname {Bi} (x+iy)\right]}$	${\displaystyle \left|\operatorname {Bi} (x+iy)\right|\,}$	${\displaystyle \operatorname {arg} \left[\operatorname {Bi} (x+iy)\right]\,}$

Relation to other special functions

For positive arguments, the Airy functions are related to the modified Bessel functions:

$${\displaystyle {\begin{aligned}\operatorname {Ai} (x)&{}={\frac {1}{\pi )){\sqrt {\frac {x}{3))}\,K_{1/3}\!\left({\frac {2}{3))x^{3/2}\right),\\\operatorname {Bi} (x)&{}={\sqrt {\frac {x}{3))}\left[I_{1/3}\!\left({\frac {2}{3))x^{3/2}\right)+I_{-1/3}\!\left({\frac {2}{3))x^{3/2}\right)\right].\end{aligned))}$$

$${\displaystyle x^{2}y''+xy'-\left(x^{2}+{\tfrac {1}{9))\right)y=0.}$$

The first derivative of the Airy function is

$${\displaystyle \operatorname {Ai'} (x)=-{\frac {x}{\pi {\sqrt {3))))\,K_{2/3}\!\left({\frac {2}{3))x^{3/2}\right).}$$

Functions K_1/3 and K_2/3 can be represented in terms of rapidly convergent integrals^[6] (see also modified Bessel functions)

For negative arguments, the Airy function are related to the Bessel functions:

$${\displaystyle {\begin{aligned}\operatorname {Ai} (-x)&{}={\sqrt {\frac {x}{9))}\left[J_{1/3}\!\left({\frac {2}{3))x^{3/2}\right)+J_{-1/3}\!\left({\frac {2}{3))x^{3/2}\right)\right],\\\operatorname {Bi} (-x)&{}={\sqrt {\frac {x}{3))}\left[J_{-1/3}\!\left({\frac {2}{3))x^{3/2}\right)-J_{1/3}\!\left({\frac {2}{3))x^{3/2}\right)\right].\end{aligned))}$$

$${\displaystyle x^{2}y''+xy'+\left(x^{2}-{\frac {1}{9))\right)y=0.}$$

The Scorer's functions Hi(x) and -Gi(x) solve the equation y′′ − xy = 1/π. They can also be expressed in terms of the Airy functions:

$${\displaystyle {\begin{aligned}\operatorname {Gi} (x)&{}=\operatorname {Bi} (x)\int _{x}^{\infty }\operatorname {Ai} (t)\,dt+\operatorname {Ai} (x)\int _{0}^{x}\operatorname {Bi} (t)\,dt,\\\operatorname {Hi} (x)&{}=\operatorname {Bi} (x)\int _{-\infty }^{x}\operatorname {Ai} (t)\,dt-\operatorname {Ai} (x)\int _{-\infty }^{x}\operatorname {Bi} (t)\,dt.\end{aligned))}$$

Fourier transform

Using the definition of the Airy function Ai(x), it is straightforward to show its Fourier transform is given by

$${\displaystyle {\mathcal {F))(\operatorname {Ai} )(k):=\int _{-\infty }^{\infty }\operatorname {Ai} (x)\ e^{-2\pi ikx}\,dx=e^((\frac {i}{3))(2\pi k)^{3)).}$$

${\textstyle {\hat {y))={\frac {1}{2\pi i))\int ye^{-ikx}dx}$

${\displaystyle i{\hat {y))'+k^{2}{\hat {y))=0}$

${\displaystyle {\hat {y))=Ce^{ik^{3}/3}.}$

Applications

Quantum mechanics

The Airy function is the solution to the time-independent Schrödinger equation for a particle confined within a triangular potential well and for a particle in a one-dimensional constant force field. For the same reason, it also serves to provide uniform semiclassical approximations near a turning point in the WKB approximation, when the potential may be locally approximated by a linear function of position. The triangular potential well solution is directly relevant for the understanding of electrons trapped in semiconductor heterojunctions.

Optics

A transversally asymmetric optical beam, where the electric field profile is given by the Airy function, has the interesting property that its maximum intensity accelerates towards one side instead of propagating in a straight line as is the case in symmetric beams. This is at expense of the low-intensity tail being spread in the opposite direction, so the overall momentum of the beam is of course conserved.

Caustics

The Airy function underlies the form of the intensity near an optical directional caustic, such as that of the rainbow (called supernumerary rainbow). Historically, this was the mathematical problem that led Airy to develop this special function. In 1841, William Hallowes Miller experimentally measured the analog to supernumerary rainbow by shining light through a thin cylinder of water, then observing through a telescope. He observed up to 30 bands.^[7]

Probability

In the mid-1980s, the Airy function was found to be intimately connected to Chernoff's distribution.^[8]

The Airy function also appears in the definition of Tracy–Widom distribution which describes the law of largest eigenvalues in Random matrix. Due to the intimate connection of random matrix theory with the Kardar–Parisi–Zhang equation, there are central processes constructed in KPZ such as the Airy process. ^[9]

History

The Airy function is named after the British astronomer and physicist George Biddell Airy (1801–1892), who encountered it in his early study of optics in physics (Airy 1838). The notation Ai(x) was introduced by Harold Jeffreys. Airy had become the British Astronomer Royal in 1835, and he held that post until his retirement in 1881.