Asymptotic s

8/11/2019 Asymptotic s

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INTEGRAL ASYMPTOTICS AND THE WKB APPROXIMATION

EMILY MEISSENADVISOR: KEN MCLAUGHLIN

Abstract. In this paper, we investigate integral asymptotics the asymptotic behavior as x of somefunction I (x ) with an integral representation. Using this theory, we study solutions to Airys equation. Thenwe discuss the WKB Approximation method, which approximates the solution to a specic type of secondorder differential equation.

Contents

1. Introduction 2

2. Integral Asymptotics 22.1. Laplaces Method 2

2.2. Method of Steepest Descent 8

3. The WKB Approximation 17

3.1. Formulation 18

3.2. Turning Points and the Connection Formula 19

3.3. Application: The Eigenvalue Problem 22

4. Extensions 23

References 23

Date : February 23, 2013.1


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1. Introduction

This paper is an exposition on asymptotic methods. In general, asymptotic methods aim to describe thebehavior of solutions or functions in a limiting sense as some parameter changes in value. In this paper,we look at specic cases of integrals and differential equations. Solutions of such equations commonly lackclosed-forms, and so asymptotic methods are often useful for determining their leading-order behavior.

We will begin by developing two common methods to study integral asymptotics. The rst, LaplacesMethod, was presented by Pierre-Simon Laplace in 1774 and handles integrals of real-valued functions. Thesecond extends this method to handle to complex integrals and was used rst by Riemann in 1863; it isknown as the Method of Steepest Descent.

We then look at a specic differential equation containing a small parameter. The WKB Method will beused to nd approximate solutions. The credit for derivation of the WKB Method is usually given to Wentzel,Kramers, and Brillouin, who developed it in 1926 [4]. Although it is sometimes referred to by other nameswhen credit is given to those who formulated the method previously, we refer to it as the WKB Methodin this paper. Its primary usefulness is seen in physics to derive approximate solutions to Schr odingersEquation. We formulate an approximate solution valid on R using results derived in the section on Methodof Steepest Descent. For a more physical approach to the WKB Method, see Griffiths [3].

2. Integral Asymptotics

The general study of asymptotics looks at the behavior of a function I (x) as x or x 0. Integralasymptotics focuses on functions which are represented in integral form, for example the Gamma function:I (x) =

0e t tx 1dt.

Later, we will look at the behavior of the Gamma function as x . In general, the integrals studied canbe complex- or real-valued. We look at two methods of nding the asymptotics of such integrals, one whichapplies to real-valued functions and another which extends to complex-valued functions.

2.1. Laplaces Method.Suppose the function I (x) has the form

I (x) = b

af (t)exg ( t ) dt

where all numbers are real-valued. As x , one would expect the largest contribution of the integral tocome from wherever g(t) is largest. If g(t) has a maximum at c [a, b] and f (c) = 0, then we can start byapproximatingI (x)

c+

cf (t)exg ( t ) dt

if c = a, b or

I (x) c+

cf (t)exg ( t ) dt

if c = a (with a similar form if c = b). As x , the rest of the original integral is exponentially smallcomparatively. Then, because our interval is restricted to near t = c, we can replace f (t) and g(t) with theirTaylor expansions. The only step remaining is to remove the dependence on in the nal expansion.

2.1.1. A Simple Example.For a rst example, let us consider the integral

I (x) = 1

0sin( t)e xt dt

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5 10 15 20

5 10 2

0.1

0.15

x

y

I (x)Approximation 5 10 15 20

0.40.3

0.2

0.1x

y

Relative Error

Figure 1. A plot of I (x) = 1

0 sin(t)e xt dt and its leading order approximation, x 2 . As

x , the relative error, 1 x 2 /I (x), goes to 0.

(Section 6.2, Exercise 4 in [2]). We see that g(t) =

t attains a maximum at t = 0 over the range of

integration, so as x , the leading order behavior will be dominated by the integral near t = 0. Further,in this region sin( t) t . Thus, we can approximateI (x)

0te xt dt.

We can now integrate by parts:

I (x) te xt

x 0+

1x

0e xt dt

= e x

x e x

x2 +

1x2

.

For small , we see that I (x) x 2 to leading order as x . In our next example, we will give a detailedanalysis of the error incurred in each approxiation. This example is meant as an illustration, although onecould control the error. See Figure 1 for a comparison to the exact value. To obtain more terms for anasymptotic expansion, we would leave more terms in the sin( t) expansion before evaluating any integrals.

Because the leading order is dominated by the region near t = 0, we could have just as easily approximated

I (x)

0te xt dt,

which would have provided the same answer because the integral on t[ ,) would be exponentially smallcompared the integral on t[0, ).

2.1.2. An Extended Example: The Gamma Function.The Gamma function is an extension of the factorial such that ( n + 1) = n! and is dened as

(x + 1) =

0e t tx dt.

Although this can be extended to the complex plane by letting x be complex, we consider only x > 0 realand look for an approximation as x .

As written, the Gamma Function does not t our general Laplace integral form. Using tx = exp ( x log t),we can rewrite our integral as

(x + 1) =

0exp(t + x log t) dt.

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0.5 1 1.5 2 2.5 3

2

4

6

x

y

I (x)Approximation

2 4 6 8 10

2 10 2

4 10 2

6 10 2

8 10 2

x

y

Relative Error

Figure 2. (z) versus Stirlings approximation G(x). As x , 1 G(x)/ (x) 0.

One might already guess, due to the integrand not being in the form f (t)exp( xg(t)), that we might have aproblem. Let us dene

h(t) = t + x log tand investigate where our integrand is maximal. We see from

h (t) = 1 + xt

= 0

that the critical point depends on x; specically it is at t = x. To avoid such complications, we make thechange of variables t = xu into the original integral. This gives us

(x + 1) =

0exp(xu ) (xu )x x du

= xx +1

0exp(xu + x log u) du

= xx +1 e x

0exp( x xu + x log u) du.

We then have an integrand of the form f (t)exp( xg(t)) with f (t) = 1 and g(u) = 1 u + log u. Taylorexpanding around u = 1, the critical point of g(u), gives g(u) = (u 1)2/ 2 + O (u 1)3 (used in step(2) below). Then applying Laplaces Method,(x + 1) xx +1 e

x 1+

1exp( xg(u)) du(1)

xx +1 e x

1+

1exp x(u 1)2 / 2 du(2)

xx +1 e x

exp x(u 1)2/ 2 du(3)

= xx +1 e x 2x

exp s2 ds= 2x x

e

x

= G(x).

This result is known as Stirlings Approximation (see Figure 2). We made three approximations: (1)restricting the range of integration, (2) truncating the Taylor series of g(t), and (3) expanding the range of

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integration. In this extended example, we will now look at the error introduced in each of these approxima-tions.

Interval restriction. The error introduced in (1) is

E 1(x, ) = xx +1 e x

1+exp( xg(u)) du + xx +1 e x

1

0exp( xg(u)) du,

which we want to show is small compared to our nal approximation. Let E +1 (x, ) denote the rst termand E 1 (x, ) the second. We begin by looking at E

+1 (x, ). Because g(u) is concave on (0 , ), as seen inFigure 3, we see that g(u) is less than its tangent line at u = 1 + , y(u) = a + b(u 1 ), on our intervalof interest, (1 + ,). Here a = log(1 + ) and b = /(1 + ). Thus,

E +1 (x, ) xx +1 e x

1+exp( xy(u)) du

= xx +1 exp x 1 + a b(1 + )

1+exp( xbu) du

= xx +1exp(x + xa )

xb .

Note that a = log(1 + ) < 0. In particular, by Taylor series expansiona = log(1 + ) = 2 / 2 + 3 / 3 4 / 4 + 2 / 2 + 3 / 3 = 2 / 6.

Then we have

E +1 (x, ) xx

b exp(x)exp x 2 / 6 .

Thus E +1 (x, ) is exponentially small compared to G(x) = 2x (x/e )x , i.e.E +1 (x, )

G(x) 1 + 2x exp x

2 / 6 .

We now show that E 1 (x, ) is similarly small. We use the tangent line at u = 1

, y(u) = a + b(u

1+ ),

which again dominates g(u). Here a = log(1 ) + and b = /(1 ). ThenE 1 (x, ) xx +1 e

x 1

0exp( xy(u)) du

= xx +1 exp x 1 + a b(1 ) 1

0exp( xbu) du

xx +1 exp x 1 + a b(1 ) 1

exp( xbu) du

= xx +1exp(x + xa)

xb

< xx

b exp(x)exp x 2 / 2 .

Here we used that a = log(1 ) + < 2 / 2 < 0 which we getby Taylor expanding,

g(u)

1 1 1 +

Figure 3. g(u) and tangent lines at1 + and 1 . Notice that g(u) isconcave on (0 ,

) and so is less than

its tangent line taken at any point from(0, ).

a = log(1 ) + = 2 / 2 3 / 3 4 / 4 < 2 / 2.Then

E 1 (x, )G(x)

< 1 2x exp x

2 / 2 .

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Taylor series truncation. The error introduced in (2) is

E 2(x, ) = xx +1 e x 1+

1exp( xg(u)) exp x(u 1)2 / 2 du.

Note that g(u) > (u 1)2 / 2 for u[1 ,1+ ], evident due to its Taylor series being an alternating series,and so this expression is positive. The difficulty in calculating this quantity is that we cannot explicitlycalculate the integral of the rst term. To remedy this, we look for an exact transformation f (u) = z suchthat g(u) = 1 u + log( u) = z2 / 2. We can then express

1+

1exp( xg(u)) du =

f (1+ )

f (1 )exp xz 2 / 2 (z)dz

where (z) = f 1(z) = u. We would like to show that (z) exists and is analytic near f (1) = 0. If this isthe case, then we can use Taylors Remainder Theorem to get

(4) (z) = (0) + (0)z + (c(z))z2 / 2

where f (1 ) < c < f (1 + ). We can then bound E 2(x, ).We start by dening f (u) =

2g(u), taking branch cuts for the square root and logarithm along the

negative real axis. Although g(1) = 0, we shall prove that f (u) is analytic near u = 1 and does not have abranch point.

First, g(u) has Taylor series

g(u) =

k =2

(1)k (u 1)kk

= 12

(u 1)2 + 13

(u 1)3 14

(u 1)4 + . . . .Then

2g(u) = ( u 1)2 1

23

(u 1) + 12

(u 1)2 + . . . = ( u 1)2 (u).

Letting u 1 = r exp( i) with (, ], we see (u 1)2 = r 2 exp(2 i) = r exp( i) is analytic. Also,we see (u) = 1 + k=1 2(1)k (u 1)k / (k + 2) is analytic in u and (u) = 0 or near u = 1, so thereis no branch point at u = 1. Because we do not need to worry about the branch cuts of the square root orlogarithm along the negative real axis, we see f (u) is analytic near u = 1 and can be written as

(5) f (u) = ( u 1) 1/ 2(u).

We now show that (z) = f 1(z) exists and is analytic near z = 0. Because f (u) has a simple zero atu = 1 and f (1) = 0, there exists a contour C in the u-plane around u = 1 inside which f (u) = 0 except atu = 1 and f (u) = 0. Because f (u) is analytic near u = 1, the Argument Principle gives us

12i C

f (u)f (u)

du = 1 .

Let s be a point in the z-plane near z = 0. Letting gs (u) = f (u) s, there exists a contour C within C suchthat(6) I (s) = 1

2i C gs (u)gs (u)

du = 12i C

f (u)f (u) s

du = 1

is constant for all s inside C due to Hurwitzs Theorem. Then, because u is analytic in the u-plane,uf (u)/ (f (u) s) has the same poles as f (u)/ (f (u) s), of which there is only one at u = (s). Theresidue at this location given in (6) now gets multiplied by u evaluated at (s). Thus, by the ResidueTheorem we have

12i C

uf (u)f (u) s

du = (s).

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Thus (z) exists and is analytic near z = 0 and the following formula is revealing:

12i C

(w)w s

dw = 12i C

(f (u)) f (u)f (u) s

du = 12i C

uf (u)f (u) s

du = (s).

Because we now know that (z) is analytic near z = 0, we know it has a Taylor series around z = 0. We

solve for the Taylor coefficients (n )

(z) in (4). First, we have f (1) = 0, f (1) = 1, f (1) = 2. Next, throughdifferentiation and the chain rule of f ( (z)) = z, we nd (0) = 1, (0) = 1, (0) = 2, so (z) = 1 2z + (c(z))z2 / 2.

Using this expansion for (z) and letting M = supc ( f (1 ) ,f (1+ ))

(c),

E 2(x, ) = xx +1 e x f (1+ )

f (1 )exp xz 2 / 2 1 2z + (c(z))z2 / 2 dz

exp xz 2 / 2 dz

= xx +1 e x

f (1 )exp xz 2 / 2 dz +

f (1+ )

exp xz 2 / 2 dz

+ f (1+ )

f (1 )exp(xz 2 / 2) 2z + (c(z))z2 / 2 dz

xx +1 e x

exp xz 2 / 2 dz +

exp xz 2 / 2 dz + xx +1 e x

exp xz 2 / 2 2z + Mz2 / 2 dz

= E 3(x, ) + xx +1 e x M 2x3where E 3(x, ) is the same as the error introduced in (3) and will be bounded next.

Thus,E 2(x, )

G(x) | (c)|

2x +

E 3(x, )G(x)

.

E 2(x, ) is the dominating error but is still small compared to our nal approximation G(x).

Interval expansion. Finally, we would like to show that the error introduced in (3),

E 3(x, ) = xx +1 e x

1+exp x(u 1)2 / 2 du + xx +1 e

x 1

exp x(u 1)2 / 2 du,is small compared to our nal approximation. We label the rst term E +3 (x, ) and the second E

3 (x, ).

We bound E +3 (x, ). First,

I (y) =

yexp s2 ds

=

0exp (s + y)2 ds

= exp y2

0exp u2 2uy du

exp y2

0exp(2uy) du

= exp y2

2y .

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Then, by a change of variables s = x/ 2(u 1),E +3 (x, ) = x

x +1 e x x2 I x2 12 xx +1 exp(x)exp x 2 / 2 ,which is exponentially small compared to our nal approximation, i.e.

E +3 (x, )

G(x) x8 exp x 2 / 2 .

Similarly, by the change of variables s = x/ 2(1 u),E 3 (x, ) = x

x +1 e x x2 I x2 12 xx +1 exp(x)exp x 2 / 2and we again see

E 3 (x, )G(x) x8 exp x 2 / 2 .

2.2. Method of Steepest Descent.In general, the integral under consideration may be complex-valued. The Method of Steepest Descent, alsoknown as the Saddle-Point Method, generalizes the idea of Laplaces Method to complex integrals of theform

I (x) =

f (z)exp( xg(z)) dz

=

f (z)exp x (z) + i(z) dz

where is a contour in the complex-plane, and (z), (z), and x are real-valued. Unfortunately, we canno longer claim that the main contribution comes from around wherever g(z) attains a real or absolutemaximum along now, rapid oscillations in the imaginary component (z) can lead to cancellation.

Instead, this method usually prescribes deforming into a contour along which (z) is constant. It canbe shown these paths of constant phase are also the paths of steepest ascent or decent of (z) [1]. From thisfamily of contours, we choose one on which (z) attains a maximum on the interior, which requires g (z) = 0at some point on the contour. We then linearly approximate the contour at the saddle point and apply theideas of Laplaces Method on the remaining integral.

2.2.1. Airys Equation.As an example, we will look at deriving the asymptotic behavior of solutions to the equation

y (x) xy(x) = 0 ,also known as the Airy Equation. To nd the solutions, we start by supposing they take the form

y(x) =

f (z) exp( xz )dz

where is some contour in the complex plane.

Assuming we can differentiate under the integral and then integrating by parts, we obtain

y (x) xy = z2f (z) exp( xz )dz xf (z) exp( xz )dz= z2f (z) exp( xz )dz f (z) exp( xz )| + f (z) exp( xz )dz.

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(a) Allowable contour regions(shaded).

C 1

(b) The contour for Ai( x).

C 2

(c) Bi(x) uses this contour as wellas the ( b) contour.

Figure 4. Contours of solutions of the Airy equation.

Let us require that f (z)exp( xz )

= 0. We are then left with z2f (z) + f (z) exz dz = 0. This reducesto the condition z2f (z) + f (z) = 0, or equivalently f (z) = c exp(z3 / 3).Returning to the boundary conditions, f (z)exp( xz )

= c exp(z3 / 3 + xz ) = 0. If we pick to extend

to innity in the complex plane, the direction is restricted by Re z3 > 0 as . If z = r exp( i), thisgives us (/ 6, / 6)(/ 2, 5/ 6)(7/ 6, 3/ 2). See Figure 4a .Hence, solutions to Airys Equation may be represented in integral form as

y(x) = c exp z3 / 3 + xz dzwith appropriately chosen .

Two of such solutions, Ai( x) and Bi( x), are commonly chosen as a basis for all solutions of the Airyequation. These are known as the Airy and Bairy functions. Ai( x) uses the contour = C 1 shown in Figure4b and c = i/ (2) and can be simplied to

Ai(x) = 1

0cos(t3 / 3 + xt )dt.

Bi(x) adds solutions using = C 1 , c = 1/ (2), and = C 2 , c = 1 / , and simplies toBi(x) =

1

0exp t3 / 3 + xt + sin t3 / 3 + xt dt.

Bi(x) is chosen so that its phase differs by that of Ai( x) by / 2 for negative x and has the same amplitudeof oscillation as x .

Because these solutions form a basis for the solutions of Airys equation, any solution can be written in the

form c1Ai(x) + c2Bi(x). Note that Ai( x) 0 as x , which can be shown using the Reimann-LebesgueLemma, and Bi( x) as x , so any solution with c2 = 0 will behave like c2Bi(x) for large x 0.

2.2.2. Airy Function Asymptotics.Airy Function, > 0. Starting with the function

Ai() = i2 C 1 exp s3 / 3 + s ds

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14 12 10 8 6 4 2 2 4

0.5

1

1.5

x

yAi(x)Bi(x)

Figure 5. Ai(x) and Bi( x).

where C 1 is the imaginary axis as shown in Figure 4b , we look for the leading order behavior as + .We rst make the change of variables z = s :

Ai() = i 2 C 1 exp 3/ 2 z3 / 3 + z dz.

We look to deform C 1 into a contour which passes through a saddle point of g(z) = z3 / 3 + z and onwhich g(z) has constant imaginary part. The saddle points of g(z) are at z = 1. Letting z = x + iy, wehaveg(z) = x3 / 3 + xy2 + x + i y3 / 3 x2y + y .

In order for Im[ g(z)] to be constant, we have y(y2 / 3x2 + 1) = C . Since Im[g(z)] = 0 at the saddle points,we see C = 0. One possible curve which satises Im[ g(z)] = 0 and passes through a saddle point is y = 0,or z = x, shown as D 3 in Figure 6a ; however, this line cannot be deformed into C 1 . The other two possiblechoices are the hyperbolas from y2/ 3 x2 + 1 = 0, shown in Figure 6a as D 1 and D 2 . Using either of thesehyperbolas gives Re [ g(z)] = 8x3 / 32x; thus, D1 will attain a real maximum at z = 1 and Re[g(z)] as x along the contour, which is what we want, whereas D 2 will attain a real minimum at z = 1.

We deform C 1 into D1 as shown in Figure 6b using L1 and L2 and Cauchys Integral Theorem, since ourintegrand is analytic, by showing that L 1 and L 2 0 as R .

Dene

I 1(, R ) = i 2 L 1 exp 3/ 2 z3 / 3 + z dz.

Along L1 , z = x + iR . Let R = 1 + R2 / 3 and x = R x.I 1(, R ) =

i 2

R

0exp

3/ 2

3x3 3xR 2 + iR 3 3ix 2R 3x + 3 iR dx

= i

2 1

0 R exp

3/ 2

3 3R x

3 3 R xR 2 R x expi 3/ 2

3R3 3 2R x2R + R dx.

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D 3

D 2

D 1

(a) Contours for Im z 3 / 3 + z = 0 where arrows indicatedirection of increasing real part.

iR

iRD 1

C 1

L 1

L 2(b) Steepest descent contour D1 for Ai(), > 0. C 1 isdeformed into D 1 by L1 and L2 . Arrows indicate directionof traversal.

Figure 6. Contours for Ai( ), > 0. Shading indicates where Re z3 / 3 + z < 0.

Then

I 1(, R ) 2

1

0 R exp

3/ 2

3 3R x

3 3 R xR 2 R x dx

2

1

0 R exp

x 3/ 2

3 3R 3 R R2 R dx

=

2

3 R3/ 2 ( 3R 3 R R2 R ) exp

3/ 2

3 3R 3 R R

2

R 1

2

33/ 2 (8R2 / 3 1)

exp3/ 2

3 3 8R3 / 3 R 1 using R R/ 3 as R

0 as R .An almost identical argument holds for L2 where z = x iR .Thus, we have deformed the contour C 1 into D1 along whichIm [g(z)] = 0, so

Ai() = i 2 D 1 exp 3/ 2 z3 / 3 + z dz

= i 2 D 1

exp 3/ 2Re

z3 / 3 + z dz.

Along D1 , we have a maximum at z = 1 and can apply the ideasbehind Laplaces Method. First, everything away from z = 1on D 1 is negligible. We approximate D 1 near z = 1 by the linez = 1 + i where is real-valued, shown as D1 in Figure 7.Observe that z = 1 corresponds to = 0, and so everythingaway from = 0 is negligible. Then

z3 / 3 + z = 2/ 3 2 + i 3 / 3,

D 1

D 1

Figure 7. We approximate the steep-est descent contour D1 near z = 1 byD1 .

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and so Re [g(z)] = 2/ 3 2 on D1 . Applying Laplaces Method, for large > 0 we haveAi() = i 2 D 1 exp 3/ 2Re z3 / 3 + z dz

i 2 D 1 near

z = 1

exp 3/ 2Re z3 / 3 + z dz

2 exp

23

3/ 2

exp 3/ 2 2 d

2

exp 23

3/ 2

exp 3/ 2 2 d

= 1

2 1/ 4 exp 23

3/ 2 .

This is the leading order approximation for Ai( ) as .Airy Function, < 0. In this case, we let z = s . Then

Ai() = i 2

C 1

exp ()3/ 2 z3 / 3 z dz.

We again look to deform C 1 . The saddle points of g(z) = z3 / 3 z are at z = i. Letting z = x + iy,g(z) = x3 / 3 + xy2 x + i y3 / 3 x2y y .

Requiring Im[ g(z)] to be constant gives us y3 / 3 x2y y = C . Because Im[ g(z)] = 2/ 3 at the saddlepoints, we require y3/ 3 x2y y = 2/ 3, which leads to the solutions shown in Figure 8a and 8b . In eachgure the solutions share a constant imaginary part and, noting the arrows which indicate the direction of increasing Re[ g(z)], we build a steepest descent curve by taking the solid paths. We then deform the contourC 1 into E 1 + E 2 , as shown in Figure 9, where Re [g(z)] attains a maximum at z = i along E 1 and at z = ialong E 2 .

To do this, one can show, similar to before, that the integral along L1 and L2 goes to zero as R .We again apply the ideas of Laplaces Method. Along E 2 , we have a maximum at z = i and Im[g(z)] =

2/ 3. Again, away from z = i the contribution from the integral along E 2 is negligible, and near z = i weapproximate E 2 by the line z = + i( + 1) with real. Then on E 2 ,z3 / 3 z = 2 2 2 3 / 3 + i 2/ 3 2 3 / 3 .Thus, Re[ g(z)] 2 2 near z = i or = 0. Following the same procedure as before,

I 2() = i 2 E 2 exp ()3/ 2 z3 / 3 z dz= i

2 exp i()3/ 2Im z3 / 3 + z E 2 exp ()3/ 2Re z3 / 3 z dz

i

2 exp

i2

3(

)3/ 2

E 2 nearz = i

exp (

)3/ 2Re

z3 / 3

z dz

i 2

exp i23

()3/ 2

exp 2()3/ 2 2 (1 + i)d

(1 + i)

2 exp i

23

()3/ 2

exp 2()3/ 2 2 d

= 1

2 ()1/ 4 exp i

23

()3/ 2 + i4

.

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E 1

(a) Solutions of Im z 3 / 3 z = 2 / 3.

E 4

(b) Solutions of Im z 3 / 3 z = 2/ 3.

Figure 8. Solutions of Im z3 / 3 z = 2/ 3 with arrows indicating direction of increas-ing real part. Shading indicates where Re z3 / 3 z < 0. To build a path of steepestdescent for Ai( ), < 0, we take the solid paths. See Figure 9. Dashing indicates paths nottaken.

iR

iR

E 2

E 1

C 1

L 1

L 2

Figure 9. Steepest descent contours E 1 and E 2 for Ai(), < 0. Arrows indicate directionof traversal. Shading indicates where Re z3 / 3 z < 0. C 1 is deformed into E 2 + E 3 byL1 and L2 .

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10 8 6 4 2 2 4 6

0.40.2

0.2

0.4

0.6

0.8

x

yAiryApproximations

10 5 5 10

5

510 2

x

yRelativeError

Figure 10. A plot of Ai( x) and its leading order approximations for x < 0 and x > 0.The relative error is also plotted. The zeros of the approximation for x < 0 do not exactlycoincide with the zeros of Ai( x), hence the relative error blows up at the zeros of Ai( x).

Along E 1 , we have a maximum at z = i and have Im[ g(z)] = 2 / 3. Near z = i, we approximate E 1 byz = + i( 1) with real. Then along E 1 ,z3 / 3 z = 2 2 + 2 3 / 3 + i 2/ 3 2 3 / 3 .Thus, we approximate Re [ g(z)] 2 2 .

I 1() = i 2 E 1 exp ()3/ 2 z3 / 3 z dz= i

2 exp i()3/ 2Im z3 / 3 + z E 1 exp ()3/ 2Re z3 / 3 z dz

i

2 exp i

2

3(

)3/ 2

E 1 nearz = i

exp (

)3/ 2Re

z3 / 3

z dz

i 2

exp i23

()3/ 2

exp 2()3/ 2 2 (1 + i)d

(1 i)

2 exp i

23

()3/ 2

exp 2()3/ 2 2 d

= 1

2 ()1/ 4 exp i

23

()3/ 2 i4

.

Thus, as ,Ai() = I 1() + I 2()

12 ()1/ 4 exp i 2

3()3/ 2 i 4 + 12 ()1/ 4

exp i 23 ()3/ 2 + i

4.

A Note on NotationAlthough it is frequently written as such, it is technically incorrect to write

Ai() 1

()1/ 4 cos

23

()3/ 2 + 4

= 1

()1/ 4 sin

23

()3/ 2 + 4

14


15/23

as because this would implyAi() =

1 ()1/ 4

sin23

()3/ 2 + 4

(1 + o(1)) .

This is false, because the zeros of Ai( ) do not exactly coincide with those of sin 23 ()3/ 2 + 4 . Instead,Ai() 12 ()1/ 4

exp i 23()3/ 2 + i 4 + 12 ()1/ 4 exp i 2

3()3/ 2 i 4

because this statement allows exibility in the location of zeros:

Ai() = 1

2 ()1/ 4 exp i

23

()3/ 2 + i4

(1 + o(1)) + 1

2 ()1/ 4 exp i

23

()3/ 2 i4

(1 + o(1)) .

Similarly, it is often falsely stated that as ,Bi()

1 ()1/ 4

sin 23

()3/ 2 + i4

= 1

()1/ 4 cos

23

()3/ 2 + i4

.

Bairy Function, > 0. We start with

Bi() = 12 C 1 exp s3 / 3 + s ds +

1 C 2 exp s3 / 3 + s ds.

Since weve already analyzed C 1 , from which we get

12 C 1 exp s3 / 3 + s ds

i2 1/ 4 exp

23

3/ 2 ,

we look at deforming C 2 .

After a change of variables z = s for > 0, we recall the saddle points of g(z) = z3 / 3 + z are atz = 1. This time, we see that we can use D 3 to help build a contour along which Im [ g(z)] is constant, asshown in Figure 11a . After noting that the integral along L1 goes to 0 as R

, we have

1 C 2 exp s3/ 3 + s ds =

D 1 exp 3/ 2 z3 / 3 + z dz +

D 3 exp 3/ 2 z3 / 3 + z dz.

Similar to the Airy Function, > 0 case, we approximate D1 by the line z = 1 + i where ranges from to 0 this is the bottom half of D1 from Figure 7. Repeating the calculation process from before,

Bi() = D 1 exp 3/ 2Re z3 / 3 + z dz

i

exp

23

3/ 2 0

exp 3/ 2 2 d

i exp 233/ 2 0

exp 3/ 2 2 d

= i

2 1/ 4 exp 23

3/ 2 .

Along D3 , we have saddle points at z = 1 and z = 1. Note g(1) = 2/ 3 is a local minimum andg(1) = 2 / 3 is a global maximum of Re [ g(z)] along D 3 , so the contribution from z = 1 is dominant. Aroundz = 1, we make the exact change of variables z = 1+ with real. Then Re [ g(z)] = 2 / 3 2 3 / 3 2/ 3 2

15


16/23

iRC 2

D 3

D 1

L 1(a) Steepest descent contour built from D1 and D3 forthe C 2 part of Bi( ), > 0. Shading indicates whereRe z 3 / 3 + z < 0.

iRC 2

E 1

L 1(b) Steepest descent contour E 1 for the C 2 part of Bi( ), < 0. Shading indicates where Re z 3 / 3 z < 0.

Figure 11. Steepest descent contours for C 2 part of Bi( ), < 0, where arrows indicatedirection of traversal.

is maximal around = 0 and recall Im [ g(z)] = 0. Then D 3 exp 3/ 2 z3 / 3 + z dz =

exp i 3/ 2Im z3 / 3 + z D 3 exp 3/ 2Re z3 / 3 + z dz

D 3 near

z =1

exp 3/ 2Re z3 / 3 + z dz

exp23

3/ 2

exp 3/ 2 2 d

exp23

3/ 2

exp 3/ 2 2 d

= 1 1/ 4 exp

23

3/ 2 .

Adding these three approximations, we see as + ,Bi()

1 1/ 4 exp

23

3/ 2 .

Bairy Function, < 0. We have again already analyzed C 1 in the case of < 0 and get

12 C 1 exp s3 / 3 + s ds

i2 ()1/ 4

exp i23

()3/ 2 + i4

i2 ()1/ 4

exp i23

()3/ 2 i4

.

Looking at Figure 11b , we see that we can deform C 2 into E 1 . Using our previous analysis,1 C 2 exp s3/ 3 + s ds

i 1/ 4 exp i

23

()3/ 2 i/ 4 .16


17/23

10 8 6 4 2

0.5

1

1.5

2

x

yBi(x)Approximations

10 5 5 10

105

5

10

1510 2

x

yRelativeError

Figure 12. A plot of Bi( x) and its leading order approximations for x < 0 and x > 0. Therelative error is also plotted. Again, the zeros of the approximation for x < 0 do not exactlycoincide with the zeros of Bi( x), hence the relative error blows up at the zeros of Bi( x).

Thus,

Bi() = 12 C 1 exp s3 / 3 + s ds +

1 C 2 exp s3 / 3 + s ds

i

2 ()1/ 4 exp i

23

()3/ 2 + i4

+ i

2 ()1/ 4 exp i

23

()3/ 2 i4

.

3. The WKB Approximation

The WKB Method nds approximate solutions to the second-order differential equation

2y (x) V (x)y(x) = 0where is a small parameter. For a general potential, V (x), this equation does not have an exact solution.For example, V (x) = x2 + x3 . Even when the equation does have an exact solution, it may not be manageable,such as how using V (x) = x2 gives a solution in terms of the parabolic cylinder function. Hence, we look foran approximate solution in the limit of small .

Extension to a more general equation.Note that any second order equation of the form

z (x) + p(x)z (x) + q (x)z(x) = 0

with p(x) = 0 can be converted into WKB form2y (x)

V (x)y(x) = 0

through the change of variables z = y exp 12 p(x)dx and V (x) = 2 p (x)/ 2 + p2(x)/ 4 q (x) .

This method is most often used to localize approximate solutions to the time-independent, one-dimensionalSchr odinger equation:

2

2md2dx2

+ V (x) = E.17


18/23

Without exactly solving the equation, one can nd the allowable energy levels for a given potential. Griffiths[3] has a thorough presentation of this method in the context of the Schr odinger equation. In this paper, wereturn to our more general equation.

3.1. Formulation.For intuitions sake, let us start by looking at the case where V (x) = V 0 is a constant. Then our equationbecomes 2y (x) V 0y(x) = 0. For solutions, we have three separate cases:

y(x) =

a1 exp x V 0 / + a2 exp x V 0 / if V 0 > 0,c1 exp ix V 0 / + c2 exp ix V 0 / if V 0 < 0,c1x + c2 if V 0 = 0.This suggests that for a general V (x) we use the ansatz y(x) = A(x)exp( (x)/ ) when V (x) > 0 and

y(x) = A(x)exp( i(x)/ ) when V (x) < 0. Here A(x) is the amplitude function and (x) is the phasefunction.

We already know that we will not be able to nd an exact solution for general V (x), so we expand A(x)in a power series in powers of :

A(x) = a0(x) + a1(x) + 2a2(x) + . . . .

For shorthand, we will abbreviate A = A (x). When V (x) > 0, we compute

y (x) = [A + A / ]exp(/ ) ,

y (x) = A + 2 A / + A / + A( )2 / 2 exp( / ) .

Plugging in the expansion for A(x) and grouping by orders of , our equation becomes:

( )2a0 + 2 a0 + a0 +( )2a1 + 2 a0 +2 a1 + a1 +( )

2a2 + O( 3) V (x) a0 + a1 + 2a2 + O( 3) = 0 .

Because our equation holds in the limit as 0, it is valid to equate orders of , from which we get:O(1) : ( )2a0 = V (x)a0

= = V (x)dx.O( ) : 2 a0 + a0 + ( )2a1 = V (x)a1

= 2 a0 + a0 = 0

= a0 = c/ .The rst of these conditions, = V (x)dx, is known as the Eikonal equation. The second condition,2 a0 + a0 = 0, is known as the transport equation.We could continue this process to solve for each a i (x), but stopping here gives us a leading-order approx-

imation of y(x).

Hence, when V (x) > 0 our approximate solution is

y+ (x) = 1 (V (x)) 1/ 4 exp

1 x

0 V (x) dx + 2(V (x)) 1/ 4 exp 1 x

0 V (x) dx .Doing an almost identical analysis, we nd that when V (x) < 0 our approximate solution is

y (x) = 1 (V (x)) 1/ 4 exp

i x

0 V (x) dx + 2(V (x)) 1/ 4 exp i x

0 V (x) dx .18


19/23

10 5 5

25

20

15

10

5

WKB

Exact

(a) = 0 .1

10 5 5

150

100

50

50

100

WKB

Exact

(b) = 0 .001

Figure 13. Example plot of exact solution, which is in terms of Bessel functions of the rstand second kind, along with the WKB approximation where V (x) = e2x . As V (x) 0, orx , the WKB approximation worsens. As 0, the approximation improves. Notethat the exact solution also changes with . Plotted via Mathematica.

For a purely positive or negative V (x), we can use these as our approximate solutions.

To get an idea of the error incurred in leaving out the rest of the series in our approximation, we couldset a1(x) = a0(x)w(x). Solving for w(x), we nd that w(x) gets large as V (x) 0:

w(x) = k + V (x)

8(V (x))3/ 2 +

132 (V

(x))2

(V (x))5/ 2dx.

The details of this derivation can be found in Holmes [4]. For this paper, it suffices to understand that theerror of our approximation grows as V (x) approaches 0. This is to be expected because our approximatesolutions blow up as V (x) 0.

3.1.1. A Simple Example.

For an example, we look at V (x) = e2x

, which is purely negative. Figures 13a and 13b show the exactsolutions along with the approximate solutions for two different values of . We can see that as 0,our approximate solution approaches the exact solution, although the solution differs for each . Also, asx for any value of , V (x) 0 and our approximate solution diverges from the exact.

3.2. Turning Points and the Connection Formula.Suppose we have a true solution, y(x), to the equation 2y (x) V (x)y(x) = 0 for x R . Our goal is tobuild an approximation to y(x) valid for all xR . Because we will have different approximations in differentregions for a general V (x), we denote the approximation of y(x) when V (x) > 0 as y+ (x) and when V (x) < 0as y (x). When V (x) is near 0, we will approximate y(x) by y0(x). Then, combining these approximations,we will have a global approximation to y(x). To approximate y(x) when V (x) is near 0, we will approximateV (x) by a straight line.

3.2.1. Connection Formula.Assume that V (x) has a simple zero and positive slope at x = 0. Then near zero we nd an approximatey0(x) using V (x) = ax where a = V (0) > 0. Our equation becomes

(7) 2y0 (x) axy 0(x) = 0 .Let us consider a rescaling of this equation with x = x . Then we have

2 2d2y0dx2

a

xy0 = 0 .19


20/23

I II III IV V

V (x)

y (x) valid y+ (x) valid

y0(x) valid

Figure 14. Connection regions.

Letting = a1/ 3 / 2/ 3 , we arrive at Airys Equation

d2y0dx2 xy0 = 0

with solution y0(x) = c1Ai(x) + c2Bi(x). Hence, (7) has solutions

y0(x) = c1Aia1/ 3

2/ 3 x + c2Bia1/ 3

2/ 3 x .

There exists an overlap in regions of validity for approximations y0(x) and y+ (x), and similarly an overlapfor y0(x) and y (x). Hence, in order to have a consistent global approximation, we match the solutions

by making sure they agree asymptotically in these regions. We will refer to the regions as shown in Figure14. Note the regions of overlap are regions II and IV. In the limit of small , it becomes valid to use theasymptotics of Ai and Bi for our solution y0(x) for any xed x. We start by assuming we known 1 and 2and work to determine c1 , c2 , 1 , 2 .

In region IV, x > 0 and so

Ai(x ) 1/ 6 exp 23 a1/ 2 1x3/ 2

2 a 1/ 12 x1/ 4 , Bi(x) 1/ 6 exp 23 a

1/ 2 1x3/ 2

a 1/ 12 x1/ 4 .Substituting V (x) = ax into y+ (x), we have

y+ (x) = 1 (ax ) 1/ 4 exp

23

a1/ 2 1x3/ 2 + 2(ax ) 1/ 4 exp23

a1/ 2 1x3/ 2 .

Assuming y0(x) and y+ (x) are both valid approximations of y(x) in this region, it must be true that 2a 1/ 4

=c2 1/ 2 1/ 6a 1/ 12 . For all physically-relevant problems 2 = 0, which we assume for the rest of this paper .It is then visible that 1a 1/ 4 = c12 1 1/ 2 1/ 6a 1/ 12 .

A Note on CoefficientsSuppose 2 = 0. Our argument for determining c1 , c2 from 1 , 2 relies on the approximations having thesame asymptotic behavior in region IV. Unfortunately, 1Ai(x) + 2Bi(x) 2Bi(x) asymptotically andso for any choice of c1 , our approximations will be asymptotically equivalent. Thus, we will be unable to

20


21/23

determine the relation of c1 to 1 and 2 . After nding the relation of c1 , c2 to 1 , 2 , we will be left withthree coefficients: 1 , 2 , c1 .

Similarly, if we were to start with knowing 1 , 2 , we could nd their relation to c1 , c2 ; we could thendetermine 2 but then fail to nd 1s relation to anything.

In region II, x < 0 and so

Ai(x ) 1/ 6

2 a 1/ 12 (x)1/ 4exp i

23

a1/ 2(x)3/ 2 1 + i

4

+ exp i23

a1/ 2(x)3/ 2 1 i

4

,

Bi(x ) 1/ 6

2i a 1/ 12 (x)1/ 4exp i

23

a1/ 2(x)3/ 2 1 + i

4 exp i

23

a1/ 2(x)3/ 2 1 i

4

.

Using 2 = 0 and our relation of c1 to 1 , y0(x) has the approximation

y0(x) 1(ax ) 1/ 4 exp i

23

a1/ 2(x)3/ 2 1 + i

4

+ exp i23

a1/ 2(x)3/ 2 1 i

4

.

For x near 0, y (x) has the approximation

y (x) 1 (ax ) 1/ 4 exp i

23

a1/ 2 1(x)3/ 2 + 2(ax ) 1/ 4 exp

23

a1/ 2 1(x)3/ 2 .We can see that for both approximations to be valid we need 1 = 1 exp (i/ 4) and 2 = 1 exp (i/ 4).

Thus we have the full approximation

y(x)

1 (V (x)) 1/ 4 exp

1 x

0 V (x) dx if x 0,2 ( a) 1/ 6 1Ai 2/ 3a1/ 3(x A) if x 0, 1 (V (x))

1/ 4 exp i

x

0 V (x) dx + i 4+ 1(V (x))

1/ 4exp

i

x

0 V (x) dx i4 if x 0.

3.2.2. Turning Point Specications.Suppose V (x) crosses 0 at A with non-zero slope a. We arrive at the similar form:

y(x)

1 (V (x)) 1/ 4 exp

1 x

A V (x) dx if x 0,2 ( a) 1/ 6 1Ai 2/ 3a1/ 3x if x 0, 1 (V (x))

1/ 4 exp i

A

x V (x) dx + i 4+ 1(V (x))

1/ 4 exp i A

x V (x) dx i 4 if x 0.A

Figure 15.General turning point,V (x) = a(x A).

3.3. Application: The Eigenvalue Problem.Consider the corresponding eigenvalue problem associated with each V (x):

(8) 2y (x) V (x)y(x) = y(x)nding such that an eigenfunction y(x) exists which is bounded as x . This is equivalent to havingV (x) = V (x) in our original problem.

21


22/23

V (x)

A B

yA + (x) valid yB + (x) valid

yA (x) and yB valid

I II III

Figure 16. Plot of example V (x) = V (x) .If V (x) > 0 does not cross the x-axis, then there is no eigenfunction corresponding to because ourboundedness requirement would give 1 = 2 = 0 on our y+ (x) approximation.Now suppose we have

V (x) which crosses the axis at multiple points. For simplication, assume

V (x) = 0

at only x = A, B and

V (x) < 0 on (A, B ), as in Figure 16. Suppose for our given , there exists a true

eigenfunction y(x) to (8). Further, suppose around A we have approximations to y(x) of yA+ (x) in regionI and yA (x) in region II, and around B we have approximations yB + (x) in region III and yB in regionII. We denote the coefficients for yA+ (x) by 1 and 2 and the coefficients for yB + (x) by 1 and 2 . Forboundedness, we see 2 = 2 = 0. In order to have a valid global approximation, we need yA (x) andyB (x) to agree in region II.

yA + (x) = 1 V (x) 1 / 4

exp 1 A

x V (x) dx , x < A,yA (x) = 1 V (x)

1 / 4exp i

x

A V (x) dx + i 4 + 1 V (x) 1 / 4 exp i x

A V (x) dx i 4 , A < x < ByB (x) = 1 V (x)

1 / 4exp i

B

x V (x) dx + i 4 + 1 V (x) 1 / 4 exp i B

x V (x) dx i 4 , A < x < ByB + (x) = 1 V (x)

1 / 4exp 1

x

B V (x) dx , B < x.We now nd conditions for yA (x) = yB (x) for x(A, B ). We begin by rewriting

yA (x) = 1 V (x) 1/ 4

exp i

B

A V (x) dx + i Bx V (x) dx + i 4+ 1(V (x))

1/ 4 expi

B

A V (x) dx i Bx V (x) dx i 4 .22


23/23

Then in order for yA (x) and yB (x) to be functionally the same, we need

1 exp i

B

A V (x) dx + i 4 = 1 exp i 4and

1 exp

i

B

A

V (x) dx

i

4=

1 exp i

4.

This gives us

exp 2i

B

A V (x) dx = exp ( i ) ,so

B

A V (x)/ dx = n + / 2, nZ . For n odd we see 1 = 1 , and for n even 1 = 1 . Thus we need 1 = ( 1)n +1 1 and

1 B

A V (x) dx = n / 2.This last condition restricts our choice of for which such a global approximation, our eigenfunction, will

exist. Recall that V (x) < 0 on (A, B ) and hence

BA

V (x) dx > 0, so our possible n values are

restricted to n = 1 , 2, 3, . . . .

3.3.1. Particle in a Potential Well.Solving this eigenvalue problem can be used to nd allowable energy levels of a particle in a potential wellalong with their corresponding wavefunctions:

2

2md2dx2

+ V (x) = E.

As stated by Griffiths (pg. 333), the result is extraordinarily powerful, for it enables us to calculate(approximate) allowed energies without ever solving the Schr odinger equation, by simply evaluating oneintegral.

4. Extensions

We could further analyze the spectral properties of the WKB eigenvalue problem, such as nding themaximum or minimum eigenvalue, the spacing between eigenvalues for large x, or the different types of spectra. Though our eigenvalue restrictions are only accurate to order , it is possible to nd higher ordercorrection terms. Also, our eigenvalue analysis can be easily extended to other forms of V (x), such as thosewith more turning points like x4 x2 or sin(x) using the same analysis as above at each turning point.

References

[1] Bender, C., & Orszag, S. (1978). Advanced Mathematical Methods for Scientists and Engineers . Inter-national series in pure and applied mathematics. McGraw-Hill.

[2] Carrier, G., Krook, M., & Pearson, C. (1983). Functions of a Complex Variable: Theory and Technique .Classics in Applied Mathematics. Society for Industrial and Applied Mathematics.

[3] Griffiths, D. (2005). Introduction to Quantum Mechanics . Pearson Prentice Hall.[4] Holmes, M. (1995). Introduction to Perturbation Methods . Texts in Applied Mathematics. Springer-

Verlag.

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