Английская Википедия:Airy function
Шаблон:Short description Шаблон:About
In the physical sciences, the Airy function (or Airy function of the first kind) Шаблон:Math is a special function named after the British astronomer George Biddell Airy (1801–1892). The function Шаблон:Math and the related function Шаблон:Math, are linearly independent solutions to the differential equation <math display="block">\frac{d^2y}{dx^2} - xy = 0 , </math> known as the Airy equation or the Stokes equation.
Because the solution of the linear differential equation <math display="block">\frac{d^2y}{dx^2} - ky = 0</math> is oscillatory for Шаблон:Math and exponential for Шаблон:Math, the Airy functions are oscillatory for Шаблон:Math and exponential for Шаблон:Math. In fact, the Airy equation is the simplest second-order linear differential equation with a turning point (a point where the character of the solutions changes from oscillatory to exponential).
Definitions
For real values of Шаблон:Mvar, the Airy function of the first kind can be defined by the improper Riemann integral: <math display="block">\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,</math> which converges by Dirichlet's test. For any real number Шаблон:Mvar there is a positive real number Шаблон:Mvar such that function <math display="inline">\tfrac{t^3}3 + xt</math> is increasing, unbounded and convex with continuous and unbounded derivative on interval <math>[M,\infty).</math> The convergence of the integral on this interval can be proven by Dirichlet's test after substitution <math display="inline">u=\tfrac{t^3}3 + xt.</math>
Шаблон:Math satisfies the Airy equation <math display="block">y - xy = 0.</math> This equation has two linearly independent solutions. Up to scalar multiplication, Шаблон:Math is the solution subject to the condition Шаблон:Math as Шаблон:Math. The standard choice for the other solution is the Airy function of the second kind, denoted Bi(x). It is defined as the solution with the same amplitude of oscillation as Шаблон:Math as Шаблон:Math which differs in phase by Шаблон:Math:
<math display="block">\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.</math>
Properties
The values of Шаблон:Math and Шаблон:Math and their derivatives at Шаблон:Math are given by <math display="block">\begin{align}
\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{align}</math> Here, Шаблон:Math denotes the Gamma function. It follows that the Wronskian of Шаблон:Math and Шаблон:Math is Шаблон:Math.
When Шаблон:Mvar is positive, Шаблон:Math is positive, convex, and decreasing exponentially to zero, while Шаблон:Math is positive, convex, and increasing exponentially. When Шаблон:Mvar is negative, Шаблон:Math and Шаблон:Math 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 <math display="block"> \int_{-\infty}^\infty \operatorname{Ai}(t+x) \operatorname{Ai}(t+y) dt = \delta(x-y)</math> again using an improper Riemann integral.
- Real zeros of Шаблон:Math and its derivative Шаблон:Math
Neither Шаблон:Math nor its derivative Шаблон:Math have positive real zeros. The "first" real zeros (i.e. nearest to x=0) are:[2]
- "first" zeros of Шаблон:Math are at Шаблон:Math
- "first" zeros of its derivative Шаблон:Math are at Шаблон:Math
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 Шаблон:Mvar goes to infinity at a constant value of Шаблон:Math depends on Шаблон:Math: this is called the Stokes phenomenon. For Шаблон:Math we have the following asymptotic formula for Шаблон:Math:[3]
<math display="block"> \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].</math> or<math display="block"> \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].</math> where <math>\zeta = \tfrac 23 z^{3/2}.</math> In particular, the first few terms are[4]<math display="block">\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) </math> There is a similar one for Шаблон:Math, but only applicable when Шаблон:Math:
<math display="block"> \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].</math> A more accurate formula for Шаблон:Math and a formula for Шаблон:Math when Шаблон:Math or, equivalently, for Шаблон:Math and Шаблон:Math when Шаблон:Math but not zero, are:[3][5]<math display="block">\begin{align}
\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{align}</math>
When Шаблон:Math these are good approximations but are not asymptotic because the ratio between Шаблон:Math or Шаблон:Math 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 Шаблон:Math and Шаблон:Math. Similarly to before, when Шаблон:Math:[5]
<math display="block"> \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].</math>
When Шаблон:Math we have:[5]
<math display="block"> \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].</math>
Similarly, an expression for Шаблон:Math and Шаблон:Math when Шаблон:Math but not zero, are[5]
<math display="block">\begin{align}
\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{align}</math>
Complex arguments
We can extend the definition of the Airy function to the complex plane by <math display="block">\operatorname{Ai}(z) = \frac{1}{2\pi i} \int_{C} \exp\left(\tfrac{t^3}{3} - zt\right)\, dt,</math> where the integral is over a path C starting at the point at infinity with argument Шаблон:Math and ending at the point at infinity with argument π/3. Alternatively, we can use the differential equation Шаблон:Math to extend Шаблон:Math and Шаблон:Math to entire functions on the complex plane.
The asymptotic formula for Шаблон:Math is still valid in the complex plane if the principal value of Шаблон:Math is taken and Шаблон:Mvar is bounded away from the negative real axis. The formula for Шаблон:Math is valid provided Шаблон:Mvar is in the sector <math>x\in\C : \left|\arg(x)\right| < \tfrac{\pi}{3} - \delta</math> for some positive δ. Finally, the formulae for Шаблон:Math and Шаблон:Math are valid if Шаблон:Math is in the sector <math>x\in\C : \left|\arg(x)\right| < \tfrac{2\pi}{3} - \delta.</math>
It follows from the asymptotic behaviour of the Airy functions that both Шаблон:Math and Шаблон:Math have an infinity of zeros on the negative real axis. The function Шаблон:Math has no other zeros in the complex plane, while the function Шаблон:Math also has infinitely many zeros in the sector <math>z\in\C : \tfrac{\pi}{3} < \left|\arg(z)\right| < \tfrac{\pi}{2}.</math>
Plots
| <math>\Re \left[ \operatorname{Ai} ( x + iy) \right] </math> | <math>\Im \left[ \operatorname{Ai} ( x + iy) \right] </math> | \operatorname{Ai} ( x + iy) \right| \, </math> | <math>\operatorname{arg} \left[ \operatorname{Ai} ( x + iy) \right] \, </math> |
|---|---|---|---|
| Файл:AiryAi Real Surface.png | Файл:AiryAi Imag Surface.png | Файл:AiryAi Abs Surface.png | Файл:AiryAi Arg Surface.png |
| Файл:AiryAi Real Contour.svg | Файл:AiryAi Imag Contour.svg | Файл:AiryAi Abs Contour.svg | Файл:AiryAi Arg Contour.svg |
| <math>\Re \left[ \operatorname{Bi} ( x + iy) \right] </math> | <math>\Im \left[ \operatorname{Bi} ( x + iy) \right] </math> | \operatorname{Bi} ( x + iy) \right| \, </math> | <math>\operatorname{arg} \left[ \operatorname{Bi} ( x + iy) \right] \, </math> |
|---|---|---|---|
| Файл:AiryBi Real Surface.png | Файл:AiryBi Imag Surface.png | Файл:AiryBi Abs Surface.png | Файл:AiryBi Arg Surface.png |
| Файл:AiryBi Real Contour.svg | Файл:AiryBi Imag Contour.svg | Файл:AiryBi Abs Contour.svg | Файл:AiryBi Arg Contour.svg |
Relation to other special functions
For positive arguments, the Airy functions are related to the modified Bessel functions: <math display="block">\begin{align}
\operatorname{Ai}(x) &{}= \frac1\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{align}</math> Here, Шаблон:Math and Шаблон:Math are solutions of <math display="block">x^2y + xy' - \left (x^2 + \tfrac{1}{9} \right )y = 0.</math>
The first derivative of the Airy function is <math display="block"> \operatorname{Ai'}(x) = - \frac{x} {\pi \sqrt{3}} \, K_{2/3}\!\left(\frac{2}{3} x^{3/2}\right) .</math>
Functions Шаблон:Math and Шаблон:Math 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: <math display="block">\begin{align}
\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(\frac23 x^{3/2}\right)\right].
\end{align}</math> Here, Шаблон:Math are solutions of <math display="block">x^2y + xy' + \left (x^2 - \frac{1}{9} \right )y = 0.</math>
The Scorer's functions Шаблон:Math and Шаблон:Math solve the equation Шаблон:Math. They can also be expressed in terms of the Airy functions: <math display="block">\begin{align}
\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{align}</math>
Fourier transform
Using the definition of the Airy function Ai(x), it is straightforward to show its Fourier transform is given by <math display="block">\mathcal{F}(\operatorname{Ai})(k) := \int_{-\infty}^{\infty} \operatorname{Ai}(x)\ e^{- 2\pi i k x}\,dx = e^{\frac{i}{3} (2\pi k)^3}.</math>This can be obtained by taking the Fourier transform of the Airy equation. Let <math display=inline>\hat y = \frac{1}{2\pi i}\int y e^{-ikx}dx</math>, then <math>i\hat y' + k^2 \hat y = 0</math>, which then has solutions <math>\hat y = C e^{ik^3/3}.</math> Why is there only one dimension of solutions? Because the Fourier transform requires Шаблон:Mvar to decay to zero fast enough, and unfortunately Шаблон:Math exponentially fast, so it cannot be found by Fourier transform.
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.
See also
Notes
References
- Шаблон:AS ref
- Шаблон:Citation
- Frank William John Olver (1974). Asymptotics and Special Functions, Chapter 11. Academic Press, New York.
- Шаблон:Citation
- Шаблон:Citation
External links
- Шаблон:Springer
- Шаблон:MathWorld
- Wolfram function pages for Ai and Bi functions. Includes formulas, function evaluator, and plotting calculator.
- Шаблон:Dlmf
- ↑ David E. Aspnes, Physical Review, 147, 554 (1966)
- ↑ Шаблон:Cite web
- ↑ 3,0 3,1 Шаблон:Harvtxt, Eqns 10.4.59, 10.4.61
- ↑ Шаблон:Cite web
- ↑ 5,0 5,1 5,2 5,3 Шаблон:Harvtxt, Eqns 10.4.60 and 10.4.64
- ↑ M.Kh.Khokonov. Cascade Processes of Energy Loss by Emission of Hard Photons // JETP, V.99, No.4, pp. 690-707 \ (2004).
- ↑ Miller, William Hallowes. "On spurious rainbows." Transactions of the Cambridge Philosophical Society 7 (1848): 277.
- ↑ Шаблон:Cite journal
- ↑ Шаблон:Cite book
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