• No results found

Airy Functions Demystified - II: The Mathematical Physics of an Innocent Looking Differential Equation – Asymptotic Forms and Some Applications

N/A
N/A
Protected

Academic year: 2022

Share "Airy Functions Demystified - II: The Mathematical Physics of an Innocent Looking Differential Equation – Asymptotic Forms and Some Applications"

Copied!
33
0
0

Loading.... (view fulltext now)

Full text

(1)

Airy Functions Demystified – II

The Mathematical Physics of an Innocent Looking Differential Equation – Asymptotic Forms and Some Applications

M S Ramkarthik and Elizabeth Louis Pereira

M S Ramkarthik (left) is an Assistant Professor at the

Department of Physics, Visvesvaraya National Institute of Technology, Nagpur. He is working in the field of quantum information theory, quantum computing,

entanglement and mathematical physics.

Elizabeth Louis Pereira (right) is a second-year M.Sc.

student working under the guidance of Dr M S

Ramkarthik at the Department of Physics,

Visvesvaraya National Institute of Technology, Nagpur and her interests are

in theoretical and mathematical physics.

Reckoning the importance of Airy functions and their over- whelming applications, we present here the derivation of their asymptotic forms for real arguments in detail as a continua- tion to the part I of the article. When needed, the associated concepts and the insights are motivated so that the reader is not left in the dark. Finally, some physical situations where this differential equation appears are explained.

1. Introduction

Considering the importance of Airy functions [1] that we have introduced in part I of the article1 with a detailed derivation, we further discuss the nature of these functions for large arguments.

As already discussed in part I of the article, these functions in real variablexare defined as follows:

Ai(x)= 1 2π

+∞

Z

−∞

cos kx+ k3 3

!

dk, (1)

and

Bi(x)= 1 2π

+∞

Z

−∞

e

kx−k33

+sin kx+ k3 3

!!

dk. (2) 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.

Vol.26, No.6, DOI: https://doi.org/10.1007/s12045-021-1179-z

(2)

Figure 1. Graph of the Airy functions Ai(x) and Bi(x) in the complete range from−∞to+∞.

-20 -15 -10 -5 0 5

x 0

1 2

solution

Ai(x) Bi(x)

d2y

dx2 = xy. (3)

1Airy Functions Demystified The

– I, Resonance, Vol.26, No.5, pp.629–647, 2021.

general solution, for (3) is obtained by doing its Fourier trans- form. This is elaborately discussed in section 2 of part I of the article. This general solution is

y= f(x) = 1 2π

+∞

Z

−∞

ei

kx+k33

dk. (4)

Here f(x) in (4) is a linear combination of Ai(x) andBi(x). The

Keywords

Airy and Bairy functions, sta- tionary phases, asymptotic forms, WKB method, bouncing ball.

behavior ofAi(x) andBi(x) forx ∈ Ris shown inFigure1.

In this article we develop the detailed derivation of asymptotic forms of Airy functions and few applications of these functions.

2. A Brief Introduction to the Methods of Stationary Phases

To understand the rest of this article, we need to do what is called asymptotic evaluation of integrals, where we analyze what hap-

(3)

pens to the integral when the argument of the function becomes large in both directions. In this section, we give a fairly detailed discussion about the Laplace method [2]. The Laplace method is used to determine the leading order behavior of the following integral forx ∈ R,

I(s)= Zb

a

g(x)es f(x)dx, (5)

ass →+∞. In this method we assume thatI(s) converges for s very large and also thatg(x) and f(x) are smooth enough (nearly constant) to be replaced by local Taylor series approximations.

Also note, f(x) should be atleast twice differentiable. Stationary point of a differentiable function of one variable is a point on the graph of a function, where the function’s first derivative is zero.

Now, suppose f(x) has the only stationary point atx0(i.e. f(x0)= 0). Now consider the functionsh(x) andl(x) such that,

l(x)= s f(x), and

h(x)=es f(x).

Thus we can say thatx0is a stationary point forh(x) andl(x) as well. Now, consider the ratios

l(x0)

l(x) = s f(x0)

s f(x) = f(x0) f(x) , and

h(x0)

h(x) = es f(x0)

es f(x) =es(f(x0)−f(x)).

As s increases, the ratio involving h will grow exponentially, while the ratio involving ldoes not change. Thus we can say that significant contributions to the integral I(s) will be from the pointsxin the neighborhood ofx0. Here we assume thatx0lies in the middle of the domain, i.e., not an endpoint of the interval of integration. Now, we write f(x) as a Taylor series aroundx= x0,

f(x)= f(x0)+ f(x0) (x−x0)+ 1

2!f′′(x0) (x−x0)2+O((x−x0)3). (6)

(4)

As we know, f(x0) = 0 and suppose f(x) is a global maximum atx = x0 then, f′′(x0) < 0 =⇒ f′′(x0) = −|f′′(x0)|.Thus our

f(x) can be approximated locally upto second order as, f(x)≈ f(x0)− 1

2!|f′′(x0)|(x−x0)2. (7) Similarly, for the functiong(x) sincex0is the stationary point we can writeg(x) asg(x)≈g(x0). Now (5) becomes

I(s) ≈ Zb

a

g(x0)es(f(x0)−2!1|f′′(x0)|(x−x0)2)dx. (8)

Sinceg(x0) andes f(x0)are just some constants, we get

I(s)≈g(x0)es f(x0)

b

Z

a

e

−s

|f′′(x0)|(x−x0)2 2!

dx. (9)

Considering the whole plane, we can replace

b

R

a

by

+∞

R

−∞

, then

I(s)≈g(x0)es f(x0)

+∞

Z

−∞

e

−s

|f′′(x0)|(x−x0)2 2

dx. (10)

Considering the celebrated Gaussian integral which is

+∞

Z

−∞

eαt2dt= rπ

α

, (11)

and using it in (10) we can see thatα = s(|f′′(x0)|/2) and t = x−x0is our new variable. Thus our integralI(s) in (10) becomes

I(s)≈g(x0)es f(x0)

+∞

Z

−∞

eαt2dt, (12)

I(s)≈g(x0)es f(x0)

r π

s (|f′′(x0)|/2), (13)

(5)

I(s)≈g(x0)es f(x0) s

s|f′′(x0)|, (14) for s → +∞. It is important to note that the Laplace method is used when g(x) and f(x) are real and the exponential is purely real. However, many situations demand that the exponential be purely complex and this comes under the ambit of what is called

‘the method of stationary phases’, described below. The method of stationary phases is similar to the Laplace method with a dif- ference of s replaced byiM, where M ∈ R. Thus the integral I(s) in (5) becomes

I(M)=

+∞

Z

−∞

g(x)eiM f(x)dx. (15) More generally (15) can be written as,

+∞

Z

−∞

g(M,z)ei f(M,z)dz, (16)

where, f(M,z) is the phase and this function does not have any poles in the complex plane. In the absence of poles the next ma- jorly contributing points to the integral are the stationary points.

The function f(M,z) has a behavior similar toM f(z). This method assumes that whenM→+∞, f(M,z) takes the value of f(M,z0) where all other points ofz ∈ Ccontribute lesser andz=z0is the stationary point. This point z0 is found by the stationary points condition f(z=z0)=0. The functions f(M,z) andg(M,z) show the same behavior. =⇒ g(z = z0) = 0. If f(M,z) has a global maxima then f′′(z0) < 0 =⇒ f′′(z0) = −|f′′(z0)|and making use of (10), we can write (15) as

I(M)≈g(M,z0)eiM f(z0)

+∞

Z

−∞

e

−iM

|f′′(z0)|(z−z0)2 2!

dz. (17)

(6)

Making use of the integral below which has been worked out in detail in the appendix,

+∞

Z

−∞

eiαt2dt= r π

|α|ei sgn(α)π4, (18)

where

sgn(α)= α

|α|, we can write (17) as

I(M)≈g(M,z0)eiM f(z0) s

|M f′′(z0)|e−i

π

4. (19)

If f(M,z) has a global minima then f′′(z0) > 0 =⇒ f′′(z0) =

|f′′(z0)|and making use of (10), we can write (15) as

I(M)≈g(M,z0)

+∞

Z

−∞

e

iM

f(z0)+|f′′(z0)|(z−z0)2 2!

dz, (20)

I(M)≈g(M,z0)eiM f(z0)

+∞

Z

−∞

e

iM

|f′′(z0)|(z−z0)2 2!

dz. (21) Making use of the integral in (18), we can easily get

I(M)≈g(M,z0)eiM f(z0) s

|M f′′(z0)|ei

π

4. (22)

A noteworthy point is that if the integral has more than one max- ima/minima, the value of the final integral under asymptotic ap- proximations will be the sum of all the contributions at those in- dividual maximas/minimas.

3. Asymptotic Forms of Airy Functions for Real Argument

It is a natural question to ask what happens to the Airy and Bairy functions asx→+∞and x→ −∞. These are called the asymp- totic forms [3], which are useful in studying physical phenomena

(7)

like WKB approximation, quantum bouncing ball, Airy’s disc in diffraction, etc. FromFigure 1 it is clear thatAi(x) → 0 expo- nentially for x > 0 andAi(x) → 0 in an oscillatory fashion as x < 0. However, Bi(x) → 0 in an oscillatory fashion as x < 0 but the function Bi(x) is quite notorious as x → +∞. InFigure 1 for x > 0, Bi(x) increases very rapidly. The question is how to capture the analytic expressions for Ai(x) and Bi(x) in these asymptotic limits. In this article, the asymptotic forms given are of first-order, and thus the error between these asymptotic forms and original Airy functions is larger at xcloser to 0. If we con- sider higher-order terms of this asymptotic expansion, the errors will considerably decrease. The asymptotic forms of Airy func- tions given here are the solutions of Airy’s differential equations at a larger limit (x→+∞and x → −∞, forx ∈ R). We obtain all the asymptotic forms of Ai(x) andBi(x) in this article, and to obtain the asymptotic form of Bi(x) for x > 0, we follow a dif- ferent but interesting procedure. The derivations which we give in this article will be useful to those students who would like to delve deep. Consider the solution obtained by Fourier transform of (3),

f(x)= 1 2π

+∞

Z

−∞

ei

tx+t33

dt. (23)

Consider the functionS(t,x) given by, S(t,x)=tx+ t3

3, (24)

thus (23) becomes

f(x)= 1 2π

+∞

Z

−∞

eiS(t,x)dt. (25)

In order to use the techniques given in section 2, expand S(t,x) in the form of a Taylor series around the stationary point t = t0. As x → ±∞value of the integral is majorly contributed around

(8)

t=t0, thus

S(t,x) = S(t0,x)+ d dtS(t,x)

t=t

0

(t−t0) +1

2!

d2 dt2S(t,x)

t=t

0

(t−t0)2+O(t−t0)3. (26) According to the condition of stationary phases,

d dtS(t,x)

t=t

0

=0, x+t20 =0, t0= +√

−x, (27)

t0=−√

−x. (28)

We now make a detailed study of what happens when x → −∞

andx→+∞using the above.

Case1: WhenAi(x) andBi(x) are oscillatory. Now we consider for x → −∞, here x < 0. We will write x as−|x|. This will reduce the stationary points tot0 = √

|x| and t0 = −√

|x|. Thus forx→ −∞the contribution to our integral f(x) as given by (25) will have major contribution due to these two stationary points.

Fort0= √

|x|,

S(t0,x)=t0x+ t03

3 =|x|12(−|x|)+|x|32 3

, (29)

S(t0,x)=−2

3|x|32, (30)

d2S dt2 t=t

0

=2t0 =2|x|12. (31) Fort0=−√

|x|,

S(t0,x)=t0x+ t03

3 =−|x|12(−|x|)+(−|x|32

3 ), (32) S(t0,x)= 2

3|x|32, (33)

d2S dt2 t=t

0

=2t0 =−2|x|12. (34)

(9)

Now plugging (26) into (25), we get

f(x)= 1 2π

+∞

Z

−∞

e

i

S(t0,x)+d dtS(t

,x) t=t

0

(t−t0)+1 2!

d2 dt2S(t

,x) t=t

0

(t−t0)2

dt. (35) In case of x < 0, to obtain Ai(x), we would consider a contour for integrating f(x) in the complex plane passing through both the stationary points. Here, we are integrating forAi(x) along the ℜ(t) from−∞to+∞and thus the two stationary points contribute as they are purely real. This contour of−∞to+∞along the real axis is in the region of validity of Ai(x) as mentioned inTable1 of section 2 in part I of the article. Taking into account the contri- butions from both the stationary points, the resultant integration as given by (35) will be

Ai(x) = 1 2π

+∞

Z

−∞

e

i

23|x|32+2|x|12

t−|x|

1 2

2

2!

dt

+ 1 2π

+∞

Z

−∞

e

i

2

3|x|32−2|x|12

t−

−|x|

1 2

2

2!

dt. (36) The first integral in (36) is an integral withS(t,x) expanded as a Taylor series around t=t0 =√

|x|and the second integral is with S(t,x) aroundt=t0=−√

|x|. Thus we have

Ai(x) = 1 2πe−i 23|x|

3 2

+∞

Z

−∞

ei|x|

1 2

t−|x|12

2

dt

+ 1 2πei 23|x|

3 2

+∞

Z

−∞

e−i|x|

1 2

t+|x|12

2

dt. (37) Using the integral given in (18) the first integral of (37) boils down to

+∞

Z

−∞

ei|x|

1 2

t−|x|12

2

dt= s

π

|x|12

ei(+1)π4, (38)

(10)

Figure 2. Asymptotic form of Ai(x) for x < 0 (x

−∞) showing oscillatory be- havior.

-15 -10 -5 0

x -0.5

0 0.5 1 1.5

Ai(x)

Asymptotic form Actual function

-5 0

0 1

x<0

The second integral of (37) becomes,

+∞

Z

−∞

e−i|x|

1 2

t+|x|12

2

dt= s

π

|x|12

ei(−1)π4. (39) Thus (37) becomes

Ai(x)= 1 2π

e−i 23|x|

3 2

s π

|x|12

eiπ4 + 1 2π

ei 23|x|

3 2

s π

|x|12

e−iπ4, (40) collecting the terms,

Ai(x)= |x|14 2√

π e−i

2 3|x|32π

4

+ei

2 3|x|32π

4

!

. (41)

Using Euler’s formula which ise±iθ =cos(θ)±isin(θ), Ai(x)= |x|14

2√ π

2 cos 2

3|x|32 − π 4

!!

, (42)

= |x|14

√ π

sin 2

3|x|32 −π 4+ π

2

!!

, (43)

(11)

Ai(x)= |x|14

√ π

sin 2 3|x|32 +

π 4

!

. (44)

(44) is the final asymptotic form ofAi(x) when x→ −∞. Figure 2 shows the comparative plot between Ai(x) (actual functions in blue) and Ai(x) (asymptotic in red). We closely see that the red and blue curves do not match near x = 0 as the case should be.

The inset clearly shows the behavior near x=0. Now in order to find the asymptotic expansion for Bi(x) whenx → −∞ (x < 0), we consider an integralL(x) as follows from (36).

L(x) = 1 2π

−∞

Z

+∞

e

i

23|x|32+2|x|12

t−|x|

1 2

2

2!

dt

+ 1 2π

+∞

Z

−∞

e

i

2

3|x|32−2|x|12

t−

−|x|

1 2

2

2!

dt, (45) where the first part of integral L(x) is along a contour from +∞

to−∞withS(t,x) expanded in Taylor series around the pointt= t0 =√

|x|(which is opposite for the case we considered inAi(x)) and the second part is similar toAi(x) (refer to the explanation in section 4 of part I of the article).

L(x) = − 1 2π

+∞

Z

−∞

e

i

23|x|32+2|x|12

t−|x|

1 2

2

2!

dt

+ 1 2π

+∞

Z

−∞

e

i

2

3|x|32−2|x|12

t−

−|x|

12

2

2!

dt, (46)

= − 1 2πe−i23|x|

3 2

+∞

Z

−∞

ei|x|

1 2

t−|x|12

2

dt

+ 1 2πei23|x|

3 2

+∞

Z

−∞

e−i|x|

1 2

t+|x|12

2

dt. (47)

(12)

Figure 3. Asymptotic form of Bi(x) for x < 0 (x

−∞) showing oscillatory be- havior.

-20 -15 -10 -5 0

x 0

1 2 3

Bi(x)

Actual function Asymptotic form

-5 0

0 1 2 3

x<0

Using a similar approach as we adopted forAi(x), and using (18), we get

L(x)=− 1 2πe−i23|x|

3 2

s π

|x|12ei

π

4 + 1

2πei23|x|

3 2

s π

|x|12e−i

π

4, (48)

L(x)= |x|14 2√

π

−e−i

2 3|x|32π4

+ei

2 3|x|32π4

!

, (49)

L(x)= |x|14 2√

π

2isin 2 3|x|32

π 4

!!

, (50)

L(x)= |x|14

√ π

sin 2 3|x|32

π 4

!

i. (51)

Thus in order to obtainBi(x), we useL(x) to writeBi(x)=iL(x).

This implies the change of phase of the contour for L(x) by π2. This is done in order to getBi(x) in the region of f(x) =Bi(x) as discussed in section 2 of part I of the article. Thus we finally get Bi(x) as,

Bi(x)=−|x|14

√ π

sin 2

3|x|32 −π 4

!

. (52)

(13)

Now, cos

θ+ π 4

= cos

θ− π 4 + π

2

=cos

θ− π 4

cos π

2

−sin

θ− π 4

sin π

2

=−sin

θ− π 4

. (53)

(52) can be written in terms of cosine form using (53) Bi(x)= |x|14

√ π

cos 2

3|x|32 + π 4

!

. (54)

(54) is the final asymptotic form ofBi(x) when x→ −∞. Figure 3 shows the comparative plot between Bi(x) (actual functions in blue) and Bi(x) (asymptotic in red). We closely see that the red and blue curves do not match near x = 0 as the case should be.

The inset clearly shows the behavior nearx=0.

Case2: Now we consider for x→+∞, here x>0. Considering (27) and (28) stationary points reduce tot0 =i√

xandt0=−i√ x.

Thus for x →+∞the contribution to our integral f(x) will have significant contributions from these two stationary points. Con- sider the argument of the exponential as given by (4)

Fort0 =i √ x,

S(t0,x)= t30

3 +t0x= −ix32

3 +(ix12)x, (55) S(t0,x)= 2i

3x32, (56)

=⇒ d2S dt2 t=t

0

=2t0=2ix12. (57) Fort0 =−i √

x,

S(t0,x)= t30

3 +t0x= ix32

3 +(−ix12)x, (58) S(t0,x)=−2i

3x32, (59)

=⇒ d2S dt2 t=t

0

=2t0=−2ix12. (60)

(14)

Considering the above values of S(t0,x) and d2S dt2 t=t

0

for t0 = i√

xandt0=−i√ x.

Method 1 : The Ai(x) integral using (25) and integrating from

−∞to+∞along contours which contain the two stationary points on the imaginary axis gives us

Ai(x) = 1 2π

+∞

Z

−∞

ei

2i 3x32+ix12

t−ix12

2!

dt

+ 1 2π

+∞

Z

−∞

ei

2i 3x32−ix12

t+ix12

2!

dt, (61)

Jordan’s Lemma:

Suppose a functionf(z) is analytic everywhere in the upper half of the complex plane (ℑ(z)>0) outside the circle|z|>R0and

|f(z)| ≤MRwhereMRis a positive constant and

R→+∞lim MR =0, then we can shift the contour fromCtoCR(whereCR

is a semicircle of radius Rgiven byz=Re where 0θπand R>R0) then

R→+∞lim R

CR

f(z)eiαzdz=0 whereα >0

= 1 2πe23x

3 2

+∞

Z

−∞

e−x

1 2

t−ix12

2

dt

+ 1 2π

e23x

3 2

+∞

Z

−∞

ex

1 2

t+ix12

2

dt. (62)

As mentioned earlier in section 2 of part I of the article, as x → +∞ we see that Ai(x) → 0. This implies that Ai(x) is expo- nentially decreasing, thereby, its magnitude has an upper bound.

Thus, according to Jordan’s lemma, we can move our contour, which is originally along the realX axis from−∞ to+∞into a contour to include the stationary pointt0=i√

x. This is only pos- sible when we neglect the second term of (62), as the first term only converges whenx→+∞and not the second. Note that, the stationary pointt0 = −i√

x will not contribute toAi(x). Finally (62) becomes

Ai(x)= 1 2πe23x

3 2

+∞

Z

−∞

e−x

1 2

t−ix12

2

dt. (63)

The above integral is of the Gaussian form (easily derivable using

(15)

Figure 4.Asymptotic form ofAi(x) forx>0 (x+∞) showing exponential decay.

2 3 4 5

x

-0.01 0 0.01 0.02 0.03 0.04

Ai(x)

Actual function Asymptotic form

2 2.25 2.5

0.02 0.025 0.03 0.035

x>0

the gamma function) as,

+∞

Z

−∞

eαt2dt= r π

|α|. (64) We simplifyAi(x) as

Ai(x) = 1 2π

e23x

3 2

√ π x14

, (65)

Ai(x)= x14 2√

π e23x

3

2. (66)

(66) is the final asymptotic form ofAi(x) whenx→+∞.

Figure4 shows the comparative plot betweenAi(x) (actual func- tions in blue) and Ai(x) (asymptotic in red). We closely see that the red and blue curves do not match nearx=0 as the case should be. The inset clearly shows the behavior near x=0.

Method 2 : We give an alternative method for obtaining (66).

From (4), we have

(16)

Ai(x)= 1 2π

+∞

Z

−∞

ei

k3 3+kx

dk, (67)

herex > 0. In this letk = k˜+i√

x =⇒ dk = dk˜,where ˜k is a new variable. Whenk=−∞we have ˜k =−∞and whenk= +∞

we have ˜k= +∞. Thus ourAi(x) in (67) becomes Ai(x)= 1

+∞

Z

−∞

ei

(ix+k˜)3

3 +x(i

x+k)˜

!

dk˜, (68)

= 1 2π

+∞

Z

−∞

e

i

−ix 3 2−3xk+3ix˜

1 2 ˜k2+˜k3

3 +x

ix12k

dk˜, (69)

= 1 2π

+∞

Z

−∞

ei

2

3ix32+ix12k˜2+k˜33

dk˜, (70)

= 1 2π

+∞

Z

−∞

e

23x32−x12k˜2+ik˜33

dk˜, (71)

= 1 2πe23x

3 2

+∞

Z

−∞

e

xk˜2+ik˜33dk˜. (72)

Asx→+∞; √

x→+∞, thus the leading order term of the above integral will bee

xk˜2, which beautifully reduces to the Gaus- sian integral. As √

xgoes on increasing the graph of the Gaus- sian function (e

xk˜2) will get narrower along the independent variable ˜k axis and the value of function will go on increasing at k˜ =0.

Ai(x)= 1 2πe23x

3 2

+∞

Z

−∞

e

x˜k2dk˜, (73)

= 1 2π

e23x

3 2

r π

√x

. (74)

(17)

Figure 5. Plots showing the degree of resemblance betweenAi(x),Bi(x) and the Gaussiane−x2.

-2 0 2 4

x 0

1 2

functions

Ai(x) Bi(x) e-x

2

Thus we get our resultant asymptotic form ofAi(x) forx>0 as

Ai(x)= x14 2√

π e23x

3

2. (75)

The above mentioned method gaveAi(x), as substitutingk with k+i˜ √

xmeans that the contour along the real axis is shifted to pass through i√

x. This change of variable lies in the complex plane with ℑ(k) > 0. But a similar approach will not work for Bi(x).

In case of obtaining Bi(x) as the solution, we would naturally be tempted to substitute k with ˜k−i√

x. The whole idea of the above method is based on the fact thatAi(x) forx >0 resembles the behavior of the Gaussian function e−x2, which is not same for Bi(x), as you can see inFigure 5. These are systematically tabulated inTable1.

In order to derive the asymptotic expression for Bi(x) for x > 0, we use a different approach as follows. As we know that Airy’s differential equation is a second order linear differential equation, we have three pairs of linearly independent solutions of Ai(z) in

(18)

Table 1. Regions of resem- blance of Ai(x) and Bi(x)

withe−x2. Sr. No. x Ai(x) Bi(x)

1 x→ −∞ yes yes

2 x→+∞ yes no

Figure 6. The complex plane and the angles which we use to getBi(z) in terms ofAi(z) using the cube roots of unity.

I

2

I

3

Imaginary

Real I

1

− 2π 3 2π 3

the complex plane as mentioned below.

Ai(z) andBi(z),

Ai(z) andAi

ze3i

,

Ai(z) andAi

ze−2

π 3 i

.

ThusBi(z) can be written in terms ofAi(z) andAi ze23πi

orAi(z) andAi

ze23πi orAi

ze23πi

and Ai ze23πi

. From the above, it is clear that Bi(z) can be written as a linear combination of Ai(z).

InFigure6 we have shown the regions or angles of definition for the three integrals I1, I2 and I3 respectively. Dictated by (4) integralI1is along the real axis from−∞to+∞as given by,

(19)

I1(z) = 1 2π

+∞

Z

−∞

ei

tz+t33

dt, (76)

wheretis a dummy variable. IntegralI2is also from−∞to+∞

but at an angle of 23π with respect to the positive real axis, i.e., from−∞e3ito+∞e3i. Similarly, integralI3is from−∞to+∞

at an angle of−23π, i.e., from−∞e23πi to+∞e23πi. Consider the following integralI2,

I2(z)= 1 2π

+∞e2

π 3i

Z

−∞e3i

ei

tz+t33

dt. (77)

In (77) we substitute t = e23πik. This substitution is useful as it reduces the limits of integration to −∞ and +∞, i.e., when t =

−∞e23πi, k = −∞ and when t = +∞e23πi, k = +∞. Also dt = e3idk. ThusI2(z) finally becomes,

I2(z) = 1 2π

+∞

Z

−∞

ei

ke2

π 3i

z+k3e32πi

e2

π 3idk

=e23πi 1 2π

+∞

Z

−∞

ei

ke3iz+k33

dk. (78)

Using (76) in (78), we can write integralI2(z) as, I2(z)=e23πiI1

ze23πi

. (79)

A similar approach forI3(z) gives us I3(z) =e23πiI1

ze23πi

. (80)

As we know thatg(k) =ei

kz+k33

is an entire function, by Cauchy’s integral theorem [4], we have,I1(z)+I2(z)+I3(z)= 0 and ex- pressing these integrals in terms ofI1(z) we get,

I1(z)+e2

π 3iI1

ze2

π 3i

+e2

π 3iI1

ze2

π 3i

=0. (81)

(20)

According to section 2 of part I of the paper, our Ai(z) is along the real axis, we can say thatI1(z) = Ai(z).(81) gives us the ex- pression relating Ai(z), Ai

ze23πi

and Ai ze23πi

with each other as follows.

Ai(z)+e2

π 3iAi

ze2

π 3i

+e2

π 3iAi

ze2

π 3i

=0. (82) We can writeBi(z) in terms ofAi

ze3i

andAi ze3i

as follows.

Bi(z)=i

−ei23πAi

ze23πi

+e−i23πAi

ze23πi

, (83)

(83) is a linear combination of Bi(z), Ai(ze23πi) and Ai(ze23π), where contour ofAi(ze23πi) is in the opposite direction and mul- tiplied byiasBi(z) is real.

Bi(z)=eiπ6Ai

ze23πi

+e−iπ6Ai

ze23πi

. (84)

Forz ∈ Rand z > 0, using asymptotic forms ofAi(x) as given by (66) and using modified argumentsze3iandze−2π3 i,(84) can be written as,

Bi(z) = eiπ6











zei23π14

2√ π

e

2 3z32

ei3 32











+e−i

π 6











ze−i23π14

2√ π

e

2 3z32

e−i2

π 3 3

2











, (85)

Bi(z)= z14 2√

π e23z

3

2

eiπ6e−iπ6 +e−iπ6eiπ6

, (86)

Bi(z)= z14 2√

π e23z

3

2 ×2, (87)

Bi(z)= z14

√ π

e23z

3

2. (88)

(88) is the final first order asymptotic form ofBi(z), wherez ∈ R and z > 0. (88) was verified using mathematica. For further

(21)

Figure 7.Asymptotic form ofBi(x) for x > 0 showing exponential increase.

0 2 4 6

x 0

2000 4000 6000 8000

Bi(x)

Actual function Asymptotic form

0 1 2

1

2 x>0

understanding, the reader can refer [3]. Figure7 shows the com- parative plot between Bi(x) (actual functions in blue) and Bi(x) (asymptotic in red). We closely see that the red and blue curves do not match near x= 0 as the case should be. The inset clearly shows the behavior near x = 0 with an extra cusp on taking a closer look. This cusp is due to presence of x14 being the domi- nant factor in the range 0<x<1 overe23x

3 2.

Table 2. Asymptotic forms ofAi(x) andBi(x).

Sr. no. x Ai(x) Bi(x)

1 x<0 |x|

1

4

π sin2

3|x|32 + π4 |x|14

π cos2

3|x|32 + π4 2 x>0 x

1 4

2 πe23x

3

2 x14

π e23x

3 2

4. Applications of Airy Functions

Airy functions, like other special functions, are used in studying many physical situations. These functions originated from the studies by Airy and Stokes pertaining to the origin of rainbows in

References

Related documents

The purpose of this paper is to provide a measure and a description of intra-household inequality in the case of Senegal using a novel survey in which household consumption data

The Congo has ratified CITES and other international conventions relevant to shark conservation and management, notably the Convention on the Conservation of Migratory

 In Asymptotic Analysis, we evaluate the performance of an algorithm in terms of input size (we don’t measure the actual running time).  In mathematical analysis, asymptotic

The Ashtamudi lake (Fig.l) with a water spread of 32 sq.km area h a s extensive natural oyster beds of C. The presence of oyster beds and the fairly calm nature of the

Bamber (1917) recorded a singje specimen with secondary sex characters of male, testis on the left side, ovo-testis on the right side, right and left oviducts and male ducts,

Women and Trade: The Role of Trade in Promoting Gender Equality is a joint report by the World Bank and the World Trade Organization (WTO). Maria Liungman and Nadia Rocha 

Harmonization of requirements of national legislation on international road transport, including requirements for vehicles and road infrastructure ..... Promoting the implementation

China loses 0.4 percent of its income in 2021 because of the inefficient diversion of trade away from other more efficient sources, even though there is also significant trade