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

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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 −∞ −∞

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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

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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

-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

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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

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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. Regions of validity of Ai(x) and Bi(x). integration for f (x) is valid. To understand the further discussion π refer Figure 3. The first region of validity is from 0 to 3 , which gives us a sector that subtends an angle XOB. It is also evident 2π that if we rotate this entire sector by 3 we will get the second 2π region of validity which is also a sector from 3 to π subtending an angle AOX′. It is important to note that both these sectors lie in the region 0 ≤ θ ≤ π as required. Thus in these two sectors, the solution f (x) is the Ai function. To find the other independent solution of the differential equation, we rotate the second sector 2π by an angle of 3 which gives us a new region, which is also a sector subtending angle COD. Note that the third sector lies from 4π 5π π<θ< π 3 to 3 which lies in the region 2 . This is the region of validity of Bi(x). Refer Table 1; here θ′ is the angle of the sector representing the region of validity of f (x). Hence as shown in If f (z) is a function of Figure 3 we can select a contour for Ai(x) lying in the regions of the complex variable z 0 <θ< π and 2π <θ<π and for Bi(x), we can select a contour which is analytic in a 3 3 domain D (without any 4π <θ< 5π lying in the regions of 3 3 . singularities) lying between the contours C1 (the rectangular shaped contour) and C (the 3. Cauchy’s Deformation Theorem 2 circular contour), then the line integrals over To proceed with the discussion further, we need an important the- both the contours C1 and orem called Cauchy’s deformation theorem. We are stating this C2 are equivalent. theorem without proof. The interested reader can find more ma- terial in [3]. Proposition: If f (z) is a function of the complex variable z which is analytic in a domain D (without any singularities) lying between the contours C1 (the rectangular shaped contour) and C2 (the circular contour) as shown in Figure 4, then the line integrals over both the contours C1 and C2 are equivalent, i.e.,

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Figure 4. A simple di- agram to understand the Cauchy’s deformation theo- complex plane rem showing two contours C1 which is rectangular and D C2 which is circular. C2

C1 f(z)

Z f (z)dz = Z f (z)dz. (19)

C1 C2 Equivalently it amounts to saying that we can deform the rect-

angular contour C1 into a circular contour C2 and perform the integration without aﬀecting the ﬁnal answer.

4. Applying Cauchy’s Deformation Theorem to Perform Con- tour Integration for Airy Functions

Since g(k) is an analytic function (has a unique derivative at every point of the region in which it is analytic) without any singularities over the entire complex plane, we can use Cauchy’s deformation theorem to simplify the integral in (18). Considering Figure

5, C1 is any contour in the region where f (x) = Ai(x) is the solution in the valid region where Ai(x) is deﬁned, one can choose any other similar contour in that region. A similar argument is also true for the region where f (x) = Bi(x) is the solution; this is

solely controlled by contour C2 for which the deformation theorem is applicable. Since the function is analytic everywhere and

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Figure 5. Contours used Im for obtaining Ai(x) and Bi(x) as the solution to the Airy diﬀerential equation based on the valid regions of exis- C1 tence.

I1 Real

Contour for Ai(x) Im Im

I2 Real Real I3

Real

Contour for Bi(x) according to Cauchy’s deformation theorem mentioned in section 3. we can deform the contour C1 to contour I1 for Ai(x). Sim- ilarly we can deform the contours C2 to contour I2 and I3 for Bi(x). The contours I1, I2 and I3 are chosen for our conve-

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nience in performing the integration. Any other contour will give the same result, but the restrictions on the regions of validity mentioned in Table 1 have to be respected. I1 is the contour in the complex plane along the real axis from −∞ to +∞.

3 1 i kx+ k Ai(x) = Z e 3 dk, (20) 2π I1 +∞ 3 1 i kx+ k Ai(x) = Z e 3 dk, (21) 2π −∞ +∞ 1 k3 k3 Ai(x) = Z cos kx + ! + i sin kx + !! dk, (22) 2π 3 3 −∞ since sin(θ) is an odd function and cos(θ) is an even function,

+∞ 1 k3 Ai(x) = Z cos kx + ! dk. (23) π 3 0 Now, we will show that the integral of Ai(x) as mentioned in (23) is actually one of the linearly independent solutions of the diﬀer- ential (6). We can write Ai(x) as,

+∞ 1 k3 Ai(x) = Z cos kx + ! dk, (24) 2π 3 −∞

3 1 i kx+ k Ai(x) = f (x) = Z e 3 dk. (25) 2π L Now look at (24) in the complex plane and consider the function f (x) in the complex plane along a general contour L. Replace x by z, where z ∈ C.

3 1 i kz+ k f (z) = Z e 3 dk, (26) 2π L

2 k3 ′′ d  1 i kz+  f (z) =  Z e 3 dk . (27) dz2 2π     L 

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3 i kz+ k Since e 3 is a continuous function in the complex plane, dif- ferentiation under the integral sign can be done as follows;

2 3 1 d i kz+ k f ′′(z) = Z e 3 dk. (28) 2π dz2 L

3 1 i kz+ k f ′′(z) = Z −k2e 3 dk. (29) 2π L Now consider, f ′′(z) − z f (z), (30)

3 3 1 i kz+ k i kz+ k f ′′(z) − z f (z) = Z −k2e 3 − ze 3 ! dk, (31) 2π L

3 1 i i kz+ k = Z −k2 − z e 3 dk, (32) 2π i L

3 −i i kz+ k = Z −i k2 + z e 3 dk, (33) 2π L

3 i i kz+ k = Z i k2 + z e 3 dk, (34) 2π L

3 i d i kz+ k f ′′(z) − z f (z) = Z e 3 dk. (35) 2π dk L

3 i kz+ k The right hand side of (35) is 0 as the function e 3 is 0 at both the limits and the derivative is with respect to k. Therefore,

f ′′(z) − z f (z) = 0. (36)

Thus we can say that the general solution in (18) is the solution of Airy’s diﬀerential equation. However, for Bi(x) we consider two integrals along the contours I2 and I3. I2 is the contour going from −∞ to 0 along the negative imaginary axis and the negative real axis from 0 to −∞. I3 is the contour going from the positive real axis from +∞ to 0 and along the negative imaginary axis from 0 to −∞. We rather use an interesting notation based

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on the Eulerian representation of complex numbers, −∞ and +∞ along the imaginary axis in the below section are represented as −i π i π +∞e 2 and as +∞e 2 respectively. Also, −∞ along the real axis iπ is denoted by +∞e . From (18), we get by using the contour I2, the integral, 3 1 i kx+ k L = Z e 3 dk, (37) 1 2π I2 0 +∞eiπ 3 3 1 i kx+ k 1 i kx+ k L = Z e 3 dk + Z e 3 dk. (38) 1 2π 2π −i π +∞e 2 0 Consider 0 3 1 i kx+ k Z e 3 dk, (39) 2π −i π +∞e 2 let k = −it , k3 = it3, dk = −idt, −i π then the new limits are k = 0 =⇒ t = 0 ; k = +∞e 2 =⇒ t = +∞. 0 0 3 3 1 i kx+ k 1 tx− t Z e 3 dk = Z e 3 (−idt), (40) 2π 2π −i π +∞ +∞e 2 +∞ 3 i tx− t = Z e 3 (dt), (41) 2π 0 now consider +∞eiπ 3 1 i kx+ k Z e 3 dk, (42) 2π 0 let k = −t , k3 = −t3, dk = −dt then the new limits are k = 0 =⇒ t = 0 ; k = +∞eiπ =⇒ t = +∞. +∞eiπ +∞ 3 3 1 i kx+ k 1 −i tx+ t Z e 3 dk = Z e 3 (−dt), (43) 2π 2π 0 0 using Euler’s formula for eiθ as cos(θ) + i sin(θ), +∞ +∞ 1  t3 t3  = − Z cos tx + ! dt − i Z sin tx + ! dt , (44) 2π  3 3     0 0 

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from (41) and (44), we get L1 ﬁnally as,

+∞ 3 3 i tx− t t L = Z e 3 + sin tx + !! dt 1 2π 3 0 +∞ 1 t3 − Z cos tx + ! dt. (45) 2π 3 0 Now we consider the integral as given in (18) but now along the contours I3 as 3 1 i kx+ k L = Z e 3 dk. (46) 2 2π I3

Using the same notations which were used for evaluating L1, we get,

0 3 1  i kx+ k  L = Z e 3  dk 2 2π     +∞  −i π +∞e 2 3 1  i kx+ k  +  Z e 3 dk , (47) π   2    0    again using the Euler’s formula,

eiθ = cos θ + i sin θ, (48) we get

+∞ 1  k3 k3  L2 = − Z cos kx + ! + i sin kx + !! dk 2π  3 3     0  −i π +∞e 2 3 1  i kx+ k  +  Z e 3 dk . (49) π   2    0    Considering the second integral in the above expression,

− π +∞e i 2 3 i kx+ k Z e 3 dk. (50) 0

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Let ik = t , −ik3 = t3, dk = −idt, −i π then the new limits are k = 0 =⇒ t = 0 ;k = +∞e 2 =⇒ t = +∞. −i π +∞e 2 +∞ 3 3 i kx+ k tx− t Z e 3 dk = Z e 3 (−idt). (51) 0 0

Thus L2 ﬁnally becomes

+∞ 1  k3 k3  L = − Z cos kx + ! + i sin kx + !! dk 2 2π  3 3     0  +∞ 3 1 tx− t + Z e 3 (−idt), (52) 2π 0

+∞ 1  t3  L = − Z cos tx + !! dt 2 2π  3     0  +∞ 3 3 i  k tx− t  + − Z sin tx + ! + e 3 ! dt . (53) 2π  3     0 

As the two contours I2 and I3 lie in the region of validity of both Ai(x) and Bi(x) (refer Table 1), from (45) and (53) we can see that the integrals along the two contours I2 and I3 give a linear combination of Ai(x) and Bi(x). Thus Ai(x) will be given by

Ai(x) = − (L1 + L2) .

The negative sign in above expression of Ai(x) is a natural adjust- ment to the limits of integral considered for Ai(x) using contour I1 and the contour made by adding I2 and I3. Similarly to obtain the trigonometric expression for Bi(x), we choose the contour

including I3 and I2 with direction of I2 reversed and changing π the phase by 2 (Bi(x) is always real) i.e., Bi(x) is deﬁned as

Bi(x) = i (L2 − L1) . (54)

This diﬀerence between the expressions of Ai(x) and Bi(x) using the same two contours is due to their contrasting nature for x > 0.

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Considering (54), we get

+∞ 1  t3  Bi(x) = i − Z cos tx + !! dt 2π  3     0  +∞ 3 3 i  k tx− t  +i − Z sin tx + ! + e 3 ! dt 2π  3     0  +∞ 3 3 i tx− t t −i Z e 3 + sin tx + !! dt 2π 3 0 +∞ 1 t3 +i Z cos tx + ! dt. (55) 2π 3 0

Thus (55) is simpliﬁed as,

+∞ 3 3 1 tx− t t Bi(x) = Z e 3 + sin tx + !! dt. (56) π 3 0

5. Numerical Methods for the Solution of Airy’s Diﬀerential Equation

An effective way of solving Airy’s differential equation is the Nu- An effective way of merov’s method [4] and the Runge–Kutta method [5]. In this ar- solving Airy’s ticle, we present a brief view of these methods for the sake of differential equation is the Numerov’s method completeness. Numerov’s method was deduced to solve certain and the Runge–Kutta differential equations in astronomy of the form, method.

d2y = U(x) + W(x)y = F(x, y). (57) dx2 It is clear from (57) that the method works for the case when the differential equation is devoid of the first derivative of y w.r.t x, also note that F(x, y) is a function, which is linear in y. Numerov’s method involves a series of algorithms and iterations, also some initial conditions to achieve a general solution, which depends on the physical situation under consideration. We can apply Nu- merov’s method to integrate a differential equation from an initial

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value x0 to a ﬁnal xn+1 by implementing the following equations, (58), (59), and (60) in a program. The iterative formula for the solution y is

h2 6 2yn − yn−1 + 12 (10Fn + Fn−1 + Un+1) + O(h ) yn+1 = , (58) − h2 1 12 Wn+1

where we have the notations as yn+1 = y(xn+1), Fn = F(xn, yn), etc. For n = 1 in (58), we get

h2 6 2y1 − y0 + 12 (10F1 + F0 + U2) + O(h ) y2 = . (59) − h2 1 12 W2

In order to estimate the value of y2, we need to know y1 and y0. y0 is already known from the initial conditions and y1 is given by (60)

W h2 W h2 h2 y = y 1 − 2 ! + hy′ 1 − 2 ! + (7F + 6U − U ) 1 0 24 0 12 24 0 1 2 h4W 1 − 2 (F + U ) + O h5 !   . 0 2 1 ( )  2 4  (60) 36 1 − W1h + W1W2h   4 18  Thus using (60) we ﬁnd the value of y1 and substituting it in (59) we can get y2. The values of F(x, y), W(x, y), U(x, y) are given by the function on R.H.S. of (57). Then using (58), the solution of y can be found for a large number of iterations. Note that (57) is an ubiquitous equation in physics. Some important equations can be solved by this method, including the Schr¨odinger equation for a particle moving with energy E in potential V(x) which is,

d2ψ 2m = − (E − V(x)) ψ = F(x, y), dx2 ~2 and the equation of an oscillating particle of mass m attached to spring of constant k:

d2y m = −ky = F(x, y). dt2

It is very obvious that Numerov’s method can be applied to Airy’s diﬀerential equation, as substituting U(x) = 0 and W(x) = x

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Figure 6. Ai(x) and Bi(x) 0.4 Ai(x) as obtained by Numerov’s method.

y 0

-0.4 -10 -5 0 5 x 4 Bi(x) 2 y

0 -10 -5 0 Actual functions Generated using Numerov’s method

Figure 7. Solutions ob- 0.8 Ai(x) tained for Airy diﬀerential equation using Runge–Kutta 0.4 method. y 0

-0.4 -10 -5 0 5

0.8 Bi(x) 0.4 y 0 -0.4 -10 -5 x 0 Actual functions Generated using Runge-Kutta method in (57) will give us Airy’s diﬀerential equation, and the solution obtained numerically by successive iterations of Numerov’s

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method will be a linear combination of Ai(x) and Bi(x). Figure 6 shows the comparison between the plots obtained using Nu- merov’s method (corresponding to numerically generated, colored in blue) and those obtained using Wolfram Mathematica (corresponding to the actual function, colored in red). The initial conditions to obtain solutions from Numerov’s method from

x0 are y0 = Ai(x0) in the ﬁrst and y0 = Bi(x0) in the second graph. Thus we can see an appreciable match between the actual functions: Ai(x), Bi(x), and those generated by Numerov’s method. Also there is some divergence in behavior of Numerov generated Ai(x) and actual Ai(x) for x > 0. On reducing the scale of the x-axis of these graphs, divergence in nature of Ai(x) and Bi(x) for x > 0 and x < 0 between actual and numerically generated solutions will be visible more clearly. This discrepancy can be reduced by taking smaller values of h. Figure 7 shows the comparison between Ai(x) and Bi(x) obtained from Runge–Kutta method and the values of Ai(x) and Bi(x) obtained from the actual functions using Wolfram Mathematica. The solutions of (57) obtained by numerical methods solely depend on the initial conditions. The reader interested in knowing more about Numerov’s algorithm can refer [4].

6. Summary and Conclusions

In this article, we have dealt with Airy differential equation and its solutions, namely, Airy function (Ai(x)) and the Bairy function (Bi(x)) in detail. We restrict our discussions to the case where the arguments of these functions are real. After a brief historical introduction to the topic, in sections 2 and 3, we derive the explicit integral forms of Airy functions using Fourier transform methods. After this, we add a different flavor to these functions using complex contour integration techniques like Cauchy’s deformation theorem to gain deeper insight into the subject by finding the regions of validity of these functions in the complex plane. The article will not be complete without giving at least an elementary technique to evaluate the Airy functions numerically, and to this end, we have given a flavour of the Numerov and the Runge–

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Kutta methods in section 6.