Airy Functions Demystified – I∗

GENERAL ARTICLE Airy Functions Demystified – I∗ The Mathematical Physics of an Innocent Looking Differential Equation – Basics and Numerical Solutions M S Ramkarthik and Elizabeth Louis Pereira Airy functions have been long important in various areas of mathematics, physics, and astronomy. Even though the Airy differential equation looks extremely simple, its solutions are generally ‘notorious’ in nature. In this expository article, we have given a thorough description of the Airy functions with real arguments. Majority of the books (with the exception of specialized books on the subject) do not pay attention in detail to these functions. Taking this as a motivating factor, we M S Ramkarthik (left) is an have written this article for postgraduates and students be- Assistant Professor at the Department of Physics, ginning research to help them understand the nitty-gritty of Visvesvaraya National these beautiful functions. The solutions of the Airy differen- Institute of Technology, tial equation called the Airy and Bairy functions are explained Nagpur. He is working in the in detail. Finally, we have given some ideas to numerically ob- field of quantum information theory, quantum computing, tain these functions. entanglement and mathematical physics. Elizabeth Louis Pereira 1. Introduction (right) is a second-year M.Sc. student working under the Sir George Biddell Airy was an erudite English mathematician guidance of Dr M S and Astronomer Royal from 1835 to 1881. He greatly contributed Ramkarthik at the Department of Physics, to the study of planetary orbits, measuring the mean density of Visvesvaraya National Earth, solutions of certain two-dimensional problems in solid me- Institute of Technology, chanics, and fixing the location of the prime meridian (Green- Nagpur and her interests are wich). Here, we emphasize his mathematical theory developed in theoretical and mathematical physics. for explaining the rainbow. Earlier to the famous Airy functions of Sir G B Airy, Newton and Descartes used ray optics to trace the path of the rainbow which could explain the geometry of rain- Keywords Airy and Bairy functions, Fourier transform, contour integration, numerical techniques. ∗Vol.26, No.5, DOI: https://doi.org/10.1007/s12045-021-1166-4 RESONANCE | May 2021 629 GENERAL ARTICLE bows to certain extent [1]. Airy used wave theory and could suc- cessfully explain supernumerary rainbows and their intensity pro- files also. He introduced the following function for the cumula- tive amplitude for the whole wavefront: +∞ π A(m) = Z cos mζ − ζ3 dζ. (1) 2 −∞ Supernumerary bows are By doing some tedious calculations mentioned later in this ar- formed due to ticle, we arrive at the following two linearly independent Airy interference of functions of the real variable x as it is known in its present form wavefronts. These bows are formed in the [2]. +∞ primary rainbow. 1 k3 Ai(x) = Z cos kx + ! dk, (2) π 3 0 or +∞ 1 k3 Ai(x) = Z cos kx + ! dk, (3) 2π 3 −∞ +∞ 3 3 1 kx− k k Bi(x) = Z e 3 + sin kx + !! dk, (4) π 3 0 or +∞ 3 3 1 kx− k k Bi(x) = Z e 3 + sin kx + !! dk. (5) 2π 3 −∞ Where Ai(x) and Bi(x) are known as Airy and Bairy functions respectively, and they are the two independent solutions of the following differential equation. d2y = xy. (6) dx2 2. Evaluation of Airy Functions From Differential Equations In this section, we use simple results from Fourier transforms to evaluate the solutions of the differential equation given by (6), and 630 RESONANCE | May 2021 GENERAL ARTICLE the calculations are shown below. Any general piecewise continuous function f (x) has a Fourier transform F , which is given as (with k being complex): +∞ 1 F ( f (x)) = Z f (x)e−ikxdx, (7) 2π −∞ taking the Fourier transform on both sides of (6) we get, d2y F ! = F (xy), (8) dx2 considering the L.H.S. of (8), +∞ d2y 1 d2y F ! = Z e−ikxdx, (9) dx2 2π dx2 −∞ integrating by parts on the R.H.S. of (9) we get +∞ +∞ d2y 1 d2y 1 d d2y F ! = e−ikx Z dx! − Z e−ikx Z dx! dx, 2 2 2 dx 2π dx −∞ 2π dx dx −∞ (10) d2y F ! = −k2F (y). (11) dx2 For evaluating the R.H.S. of (8) we use a small trick based on the Integration by parts derivative of the Fourier transform, define F (y) as, R (uv)dx = u R vdx − du vdx dx +∞ R dx R 1 F (y) = Z ye−ikxdx, (12) 2π −∞ differentiating (12) with respect to k we get, +∞ d d 1 F (y) = Z ye−ikxdx , (13) dk dk 2π −∞ as e−i kx is a continuous function and its first derivative exists, therefore, +∞ +∞ d 1 d 1 F (y) = Z ye−ikx dx = Z (−ix)ye−ikxdx, (14) dk 2π dk 2π −∞ −∞ RESONANCE | May 2021 631 GENERAL ARTICLE dF (y) F (xy) = i , (15) dk using (8), (11), and (15), it is easy to see that dF (y) − k2F (y) = i , (16) dk the only solution to (16) is 3 i k F (y) = e 3 , (17) taking the inverse Fourier transform of (17), we get a solution satisfying the differential (6). Hence we have the final solution as +∞ 3 1 i kx+ k y = f (x) = Z e 3 dk. (18) 2π −∞ Since the Airy’s Since the Airy’s differential equation is a second order differen- differential equation is a tial equation, it will have two linearly independent simultaneous second order differential solutions which are denoted by Ai(x) and Bi(x). Any linear com- equation, it will have two linearly independent bination of Ai(x) and Bi(x) in the form aAi(x) + bBi(x) where a, b simultaneous solutions are real constants is also solution of (6). which are denoted by Ai(x) and Bi(x). 2.1 A Look into the Complex Plane It is useful to go to the complex plane in order to understand the solutions of the Airy differential equation as given by (18). Let us 3 i kx+ k consider the function g(k) = e 3 of a complex variable k in order to obtain the present forms of Ai(x) and Bi(x) [(3) and (5)] by doing integration of the general function given in (18) along different contours. Considering the graphs of Ai(x) and Bi(x) in Figure 1 we can see that they mimic each other’s behavior closely as x → −∞ but along the positive real axis they have different behaviors (Ai(x) → 0 as x → +∞; Bi(x) → +∞ as x → +∞). It is also worth mentioning that both Ai(x) and Bi(x) don’t have extremum at the point x = 0. However, if we are a little careless, we can be easily misled to believe that these functions attain maximum at x = 0. Ai can be a complex function but Bi we define it to be a real 632 RESONANCE | May 2021 GENERAL ARTICLE 2 Figure 1. Graph of the Ai(x) Airy functions Ai(x) and Bi(x) Bi(x) in the complete range from −∞ to +∞. 1 solution 0 -20 -15 -10 -5 0 5 x Figure 2. Behavior of the imaginary axis z ∈ C imaginary part ℑ(k) of the z(r, θ) argument k in the complex ℑ(k) > 0 plane. 0 < θ < π real axis ℑ(k) < 0 π < θ < 2π function for any of its argument. By looking at the graph for Ai(x) in Figure 1, we can see that for x → +∞, Ai(x) → 0 is only possible when ℑ(exponent) > 0 of (18). This implies that the contours which can satisfy this condition and give us the solution RESONANCE | May 2021 633 GENERAL ARTICLE Figure 3. Regions show- A B ing the valid range of angles Y Imaginary where Ai(x) and Bi(x) live in the complex plane. Ai(x) Ai(x) π π X′ 3 3 X O Real =2π XOA 3 =2π AOC 3 Bi(x) Bi(x) 2π π π COX = 3 6 6 C D Y′ Ai(x) lie in the first and second quadrants of the complex plane. Similarly considering for Bi(x), we can conclude that for x → +∞, Bi(x) → +∞ we can say that it is only possible only when ℑ(exponent) < 0. This implies that the contours which can satisfy to give us the second linearly independent solution Bi(x) lie in the third and fourth quadrants of the complex plane. This can be visualized from Figure 2. The leading term of the exponential in (18) is k3, thus for getting the individual behaviors of Ai(x) and Bi(x) from the general solution, as given by (18), we use the famous Eulerian representation of complex numbers k = reiθ, this implies k3 = r3e3iθ. Since ℑ(k3) > 0 for Ai(x), we should have 0 < 3θ < π which means that π 0 < θ < 3 should be the region where any contour should give Ai(x). But this is not the complete story, we know a rotation of the plane by an angle of 2π, or an integral multiple of 2π will also give the same result, i.e., a rotation should give a region in which a contour along which when we integrate, we will get f (x) as the θ 2π solution. This implies that a rotation of with an angle 3 or an 2π integral multiple of 3 will also be the region where the contour 634 RESONANCE | May 2021 GENERAL ARTICLE Sr.no. θ θ′ f (x) π ′ π 1 0 <θ< 3 0 <θ < 3 Ai(x) 2π <θ< π 2π 2π <θ′ <π 2 0 + 3 3 + 3 3 Ai(x) + 2π <θ< π + 2π 4π <θ′ < 5π 3 0 2 3 3 2 3 3 3 Bi(x) Table 1.

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