Asymptotic Analysis Notes

MA4547 ASYMPTOTIC ANALYSIS

MIN HUANG

1

MA4547 ASYMPTOTIC ANALYSIS 2

Contents

1. Introduction 31.1. What this course is about 31.2. What this course is not about 41.3. Structure of the course 41.4. References 42. Basic concepts and notations 52.1. Order notation 52.2. Asymptotic expansions/series 63. Asymptotic techniques for calculating integrals 113.1. Integration by parts 113.2. Watsons lemma 123.3. Laplaces method 133.4. Method of stationary phase 173.5. *Method of steepest descent 194. Asymptotic solutions to algebraic equations 224.1. Inverse of analytic functions 224.2. Asymptotic solutions for large x 255. Asymptotics of ordinary differential equaitons 285.1. The method of iteration 285.2. Perturbation expansions 295.3. The WKB method 32


1. Introduction

In standard college level mathematics courses we learn various techniques of cal-culating integrals and solving differenital equations. However, these techniques onlyapply to specific types of simple integrals or equations which occur in textbooksmuch more often than in practice. In applications such as sciences and engineering,numerical methods are often preferred. However, numerical methods have a funda-mental limitation, namely it cannot provide any concrete formula, and is thereforeinsufficient when unknown parameters are present and when the variables tend toinfinity. For instance, the equation

(1.1) x sinx = 1

cannot be solved explicitly. A simple argument with the intermediate value The-orem shows that there are infinitely many solutions, as illustrated by the Figurebelow. Although it is simple to find a few of them using numerical methods, it isimpossible to find all of them in this way.

Figure 1.1. plot of x sinx 1

A natrual question to ask is, then, is there a way which allows us to find theapproximate values of all these solutions?

The answer is yes. The method is known as Asymptotic Analysis. It isa method of describing limiting behaviors of integrals and solutions to algebraicequations, differential equations, difference equations, etc.

For instance, using asymptotic anaysis one can find the approximate locationsof solutions of (1.1) to be xk pik + (1)

k

pik (k > 1). These are not exact solutionsbut they are very accurate, as shown in the figure below.

For practical purposes, approximate solutions are often good enough as long asthere is sufficient accuracy.

1.1. What this course is about. The focus of this course is asymptotic tech-niques for calculating integrals and solving equations. It will involve a lot of detailedsymbolic calculations. Loosely speaking we will calculate the limiting behavior ofcertain funcitons. It will be important to do plenty of exercises.


Figure 1.2. plot of x sinx 1 and the approximate roots xk

1.2. What this course is not about. This course is not about proofs of abstracttheorems. Although all the results in this course can be proved rigorously, the proofsare usually lengthy and may require advanced knowledge more suitable for graduatestudents. This course is not about computer programming either. Although thereare mathematical programs such as Mathematica and Maple which can performsymbolic calculations, they cannot give asymptotic formulas except in special cases.A thorough understanding of asymptotic analysis is necessary, even though theseprograms are sometimes used to help with detailed and messy calculations.

1.3. Structure of the course. In Section 2 we introduce basic concepts and no-tations of aymptotic analysis. These are intuitive and straightforward.

Section 3 is about calculating asymptotic formulas for integrals. We will focuson four techniques: integration by parts, Watsons lemma, Laplaces method, andmethod of stationary phase. We will also briefly mention the method of steepestdescent.

In Section 4 we introduce two methods for finding approximate solutions to alge-braic equations, namely the methods of undetermined coefficients and the methodof iteration.

In Section 5 we discuss asymptotic analysis for ordinary differential equations,and we will focus on two methods: the methods of perturbation expansion and theWKB method. We will also briefly mention singularly perturbed equations.

1.4. References. There are some good comprehensive textbooks (see below). How-ever, they are mainly geared towards graduate students.

(1) Carl M. Bender, Steven A. Orszag, Advanced Mathematical Methods forScientists and Engineers: Asymptotic Methods and Perturbation Theory(1999).

(2) Peter D. Miller, Applied Asymptotic Analysis (2006).


2. Basic concepts and notations

2.1. Order notation. We are already familiar with the standard notations of lim-its (lim) and definitions with language, etc. These are of course correct, butthey are too cumbersome for the purpose of calculations. We prefer notations thatare convenient and intuitive, and what we commonly use are three notations O, o,describing bounded by, of a smaller order than, and equivalent to respectively.

Suppose f(z) and g(z) are functions of the complex variable z defined on somedomain D in the complex plane, and possess limits as z z0 in D (here z0 isallowed to be infinity). We have the following basic notations:

f(z) = O(g(z)) (z z0)This means |f(z)| is less than a constant times |g(z)| if z is close to z0. In otherwords, |f(z)/g(z)| is bounded if z is close to z0.

Similarly we have

f(z) = o(g(z)) (z z0)This means

limzz0

f(z)

g(z)= 0

We also have the notationThis means

limzz0

f(z)

g(z)= 1

These notations are related to each other. For instance, f(z) = o(g(z)) impliesf(z) = O(g(z)). f(z) g(z) is the same as f(z) g(z) = o(g(z)).

In particular f(z) = O(1) means f is bounded for z close to z0. f(z) = o(1)means f tends to 0 as z approaches z0.

Example 1. If f(z) = 3z2 + z + 2, then we can write f(z) = O(z2) as z +;f(z) = o(z3) as z +; f(z) 3z2 as z +; f(z) 2 as z 0; f(z) 2 + zas z 0, etc.

One can formally treat O(g) as a function which is bounded by a constant times|g| and o(g) as a function which becomes arbitrarily smaller than |g| in the limit.In particular one can formally treat O(1) as a function which is bounded ando(1) as a function which tends to zero. Then one may add, subtract, multiply, ordivide these O and o notations carefully.

Example 2. We have the following facts:O(f)O(g) = O(|f |+ |g|)O(f)O(g) = O(fg)O(f)o(g) = o(fg)O(f)

1 + o(g)= O(f)

f g if and only if f = g(1 + o(1))If f1 g and f2 g then f1 f2 = o(g)


A simple but important fact to remember is exponential functions increase muchfaster than powers and powers increase much faster than logs. In other wordslnx = o(xn) and xn = o(ex) as x + for all n > 0.2.2. Asymptotic expansions/series. We want to describe the limiting behaviorof complicated functions using simple functions as approximations. Obviously weoften need to use a finite or infinite sum of simple functions. One such examplethat we are already familiar with is Taylor series. If a function is smooth enoughthen the Taylor series can approximate the function arbitrarily accurately, and theremainder can be estimated using Taylors theorem. For instance, we have

ex = 1 + x+x2

2+x3

6+x4

24+ o(x4) (x 0)

Figure 2.1. The exponential function y = ex (solid red curve)and the corresponding Taylor polynomial of degree four (dashedgreen curve) around the origin.

This is a special case of an asymptotic expansion or asymptotic series. Ifa function f has an asymptotic expansion of the form

n=0

ann(z)

it means

(2.1) f(z) =Nn=0

ann(z) + o(N (z)) (z z0)

We often write


f(z) n=0

ann(z)

for short. This is a very general definition. If the infinite sum on the right handside is convergent, then this is the same as the notation introduced before, butthe sum does not have to be convergent. In other words, the formal series

n=0

ann(z)

may not be equal to any function. In this case, we must use (2.1) as the definition.The funcitons n(z) can be any type of function, but in practice we only deal withsimple functions such as powers, exponentials, and logs.

A natrual question that arises is why are we doing this? After all when weencounter a divergent power series our first reaction is to discard it as nonsense.However, in fact there are situations when a divergent power series would makemore sense than a convergent one.

Example 3. The so-called error function is defined as

erf(z) =2pi

z0

et2

dt

It is a rescaled version of the cumulative distribution function of the standardnormal distribution in probability and statistics. It is an entire function, and hasa Taylor series representation

erf(z) =2pi

n=0

(1)nz2n+1n!(2n+ 1)

This series is convergent for all z, but how useful is it really?If we take the first 20 terms (n = 0, 1, ...19) of this series we get a long Taylor

polynomial of degree 39 which looks fine for z close to 0 but fails to approximatethe function when z becomes close to 3 (relative error bigger than 20%).

The asymptotic series for the error function as z + is

(2.2) erf(z) 1 + ez2

zpi

n=0

(1)n+1(2n 1)!!(2z2)n

where (2n 1)!! = 1 3 5...(2n 1) and (1)!! = 1. The series is divergent bythe ratio test, but if we take just three terms and compare the values to the errorfunciton, the result is amazing (see figure below): the three-term truncation

ez2

(1

2piz3 1

piz

)+ 1

approximates the error function with a relative error less than 7% for all z > 1. Forz > 2 the relative error is less than 0.02%.

The divergent asymptotic series is far more useful than the convergent Taylorseries for large values of z. However, since it is divergent, the accuracy of approxi-maton near a fixed z may not become better if one takes more terms of the series.Suppose I want an approximation of the error function near z = 1 with a relative


Figure 2.2. The error function (blue curve) and the correspond-ing Taylor polynomial of degree 39 (orange curve) around the ori-gin.

Figure 2.3. The error function (blue curve) and the correspond-ing 3-term asymptotic series (orange curve) at infinity.

error less than 1%, then the asymptotic series (2.2) will be useless no matter howmany terms I take. In that case the Taylor series would be a much better choice.Taking more terms of the asymptotic series will improve the accuracy for larger zbut may actually make the approximation worse for a fixed z. In practice we rarelyneed a lot of terms of asymptotic series.

We already know that we can add, subtract, multiply, differentiate, and inte-grate power series. We can also divide Laurent series. Similarly, one can performarithmetic operations (carefully) on asymptotic series, as well as integrate them,but one in general cannot differentiate asymptotic series, unless it only containspowers of z.

Example 4. suppose


f(z) = z + sin(z2)

then we have

f(z) z (z +)but

f (z) = 1 + 2z cos(z2) 1 (z +)It is improtant to keep in mind that asymptotic series are merely approximations.

Even if an asymptotic series is convergent, it may not be equal to the function.

Example 5. The function

f(z) =1

z 1 + ez

has the convergent asymptotic expansion

f(z) n=1

zn (z +)

since

1

z 1 =1

z

1

1 1/z =1

z

n=0

(1

z)n =

n=1

zn

but of course it is not equal to the sum. Note that the sum will not be an asymptoticexpansion of f if z . This is because is an essential singularity of f .

Since asymptotic expansions are just approximations, there is no uniquenessin the sense that a function must have a fixed asymptotic series (there are manydifferent ways to approximate a function). There is only restricted uniqueness. Forexample, in the case of z , if an asymptotic series consists only of powers ofz, exponentials of z, and logs of z, then the coefficient of each term is unique. Inparticular, the coefficient of each term of (2.2) is unique. In practice this is notan important issue. Theoretical physicists use all kinds of approximate formulassince most of their equations cannot be solved explicitly. Sometimes there are a fewdifferent formulas for calculating the same physical quantity, and they are used indifferent circumstances. As for us, we are primarily interested in asmptotic seriesconsisting of powers of z only, which are called asymptotic power series. Forinstance, (2.1) is 1 plus an exponential function times an asymptotic power series.


Exercise. 1. Find the complete asymptotic power series for the following functionsas z +:

(1)z

z + 2

(2)1

(z 1)2(3) ln(z + 1) ln z(4)

ze1/tdt

2. Find the first three terms of the asymptotic power series for the following func-tions as z +:

(1)z + 1

(2)1

z2 + z + 1

(3) sin(z

1 z )


3. Asymptotic techniques for calculating integrals

3.1. Integration by parts. We already know that one can use integration byparts to evaluate certain integrals. The same technique can also be used to findasymptotic behaviors of integrals.

Example 6. The exponential integral function E1 is given by

E1(x) =

x

et

tdt

The integral cannot be evaluated explicitly, but we can perform integration byparts as follows:

x

et

tdt = e

t

t|x

x

et

t2dt =

ex

x+O(

ex

x2)

This is a simple asymptotic expansion of only one term. To obtain more termsone can continue to perform integration by parts.

x

et

tdt =

ex

x x

et

t2dt =

ex

x ex

x2+ 2

x

et

t3dt = ...

The complete asymptotic series is

x

et

tdt = ex

n=1

(1)n1(n 1)!xn

This is obviously a divergent series, but the accuracy of approximation is quitegood.

4 5 6 7 8 9 10

0.002

0.004

0.006

0.008

0.010

0.012

0.014

Figure 3.1. The exponential integral function E1(blue curve) andthe corresponding 3-term asymptotic series (red curve) at infinity.

In the table below we compare the numerical values of E1 with its 3-term and4-term truncated asymptotic series. The 4-term series is less accurate than the3-term series for x 6 4, but more accurate for x > 5. For convergent series moreterms means better accuracy but for divergent series this is not always true.


x = 3 x = 4 x = 5 x = 6 x = 7

E1(x) 0.0130484 0.00377935 0.0011483 0.000360082 0.000115482ex

(2x3 1x2 + 1x

)0.0147517 0.00400655 0.00118588 0.000367223 0.000116976

ex( 6x4 + 2x3 1x2 + 1x) 0.0110638 0.00357727 0.00112119 0.000355747 0.000114697

Table 1. 3-term and 4-term asymptotic expansions of E1(x)

3.2. Watsons lemma. The Laplace transform is a very powerful tool in sciencesand engineering. This involves simple Laplace integrals of the type

(3.1) I(t) = b0

f(p)eptdp

where b > 0 may be taken as . In real applications t often stands for time, andpeople want to know the large time behavior of a certain (physical, chemical...)system. The result is given by

Theorem 7 (Watsons lemma). Assume the integrand in (3.1) is bounded for larget and

(3.2) f(p) n=0

anp+n

where > 1, > 0. Then

I(t) n=0

an(+ n+ 1)

t+n+1

where is the gamma function

(3.3) (x) = 0

sx1esds

Proof. The main idea is very simple: just plug (3.2) into (3.1) and integrate termby term. The key step is

b0

paeptdp = 0

paeptdp+ bpaeptdp =

(a+ 1)

ta+1+O(ebt/2)

The complete proof requires more detailed estimates and can be found in mosttextbooks.

Example 8. We can use Watsons lemma to obtain the asymptotic series for theLaplace integral

I1(t) =

0

ept

1 + p2dp

Note that

1

1 + p2=

n=0

(1)np2n


By Watsons lemma we get

I1(t) n=0

(1)n(2n+ 1)!t2n+1

since (n) = n! for every integer n.

5 6 7 8 9 10

0.12

0.14

0.16

0.18

0.20

0.22

0.24

.

Figure 3.2. The function I1(t) (blue curve) and the correspond-ing 3-term asymptotic series (red curve) at infinity.

Remark. Watsons lemma is valid not just for t +, but for all |t| , arg t (pi/2, pi/2) .

3.3. Laplaces method. We are familiar with elementary functions such as poly-nomials, exponentials, logs, and trigonometric functions. However, these functionsare often insufficient to describe complicated phenomena encountered in sciencesand engineering. For instance, many differential equations cannot be solved usingthese elementary funcitons. Instead, their solutions are the so-called special func-tions. One example is the Gamma function (3.3) we have seen before. Many specialfunctions have integral representations, the simplest among which are Laplace typeand Fourier type integrals. The method for calculating asymptotic behaviors ofthese integrals are called Laplaces method and the method of stationary phaserespectively. We will study these two methods in this and the next section.

A general Laplace integral is of the form

(3.4) I(x) = ba

f(p)exg(p)dp

where we assume f and g to be real continuous functions.At first sight, one may want to do use the substitution s = g(p) to transform

the integral into the simple type (3.1) and apply Watsons lemma. However, thisdoes not always work, since g(p) is not necessarily one-to-one and even if it is theremay not be an explicit formula for g1. Instead, we will need to examine thebehavior of g(p) carefully.

One important observation is that the value of I(x) for large x depends cruciallyon the (global) maximum value of the kernel exg(p), which depends on the maximum


value of g(p). This is because if g(p1) < g(p2) then exg(p1) is exponentially smallcompared to exg(p2). Thus the points near p2 contributes much more than thepoints near p1.

Example 9. Lets consider the Laplace integral 10

pex(pp2)dp. The function

p p2reaches its maximum value at p = 1/2. The function pex(pp2) decays veryfast if x is large and p deviates from 1/2 (see figure below).

Figure 3.3. Graph of the functionpex(pp2) for x = 30

For instance, if x = 30, then 10pex(pp

2)dp 292.514 and 0.60.4

pex(pp2)dp

164.241. Most contributions to the integral come from points near 1/2.

In general the behaviors of f and g in (3.4) may be very complicated and it is verydifficult to obtain even the leading (first) term asymptotics for I(x). Fortunately,functions encountered in practice are usually quite nice and it is often sufficient tohave the leading term asymptotics. We will consider three most important casesbelow.Case 1. The maximum of g is reach at the left endpoint a, g(a) < 0 and f(a) 6= 0.Case 2. The maximum of g is reach at the right endpoint b, g(b) > 0 and f(b) 6=

0.Case 3. The maximum of g is reach at some (unique) point c (a, b), g(c) = 0,

g(c) < 0, and f(c) 6= 0.In each case we can approximate the function g with its Taylor polynomial of degree1 or 2, and the function f with its value at the maximum point. To be specific, wehaveCase 1. g(p) g(a) + g(a)(p a), f(p) f(a)Case 2. g(p) g(b) + g(b)(p a), f(p) f(b)Case 3. g(p) g(c) + 12g(c)(p c)2, f(p) f(c)Using these approximations the integral I(x) can be calculated asymptotically.

Case 1. I(x) ba

f(a)ex(g(a)+g(a)(pa))dp f(a)exg(a)

0

exg(a)sds = f(a)e

xg(a)

xg(a)


Case 2. I(x) ba

f(b)ex(g(b)+g(b)(pb))dp f(b)exg(b)

0

exg(b)sds =

f(b)exg(b)

xg(b)

Case 3. I(x) ba

f(c)ex(g(c)+g(c)(pc)2/2)dp f(c)exg(c)

exg(c)s2/2ds =

2pif(c)exg(c)xg(c)

The last integral in Case 3 follows from

exg(c)s2/2ds =

2xg(c)

eu2

du =

2pixg(c)

The reason we can change the integral bounds to infinity is because the additionalerrors introduced by this approximation are exponentially small.

Now lets summarize our results.

Proposition 10 (Laplaces method). Consider the leading order large x asymp-totics of the general Laplace integral

I(x) =

ba

f(p)exg(p)dp

(1) If the maximum of g is reached at the left endpoint a, g(a) < 0, andf(a) 6= 0, then

I(x) f(a)exg(a)

xg(a)(2) If the maximum of g is reached at the right endpoint b, g(b) > 0, and

f(b) 6= 0, then

I(x) f(b)exg(b)

xg(b)(3) If the maximum of g is reached at some (unique) point c (a, b), g(c) = 0,

g(c) < 0, and f(c) 6= 0, then

I(x)

2pif(c)exg(c)xg(c)Example 11. Lets consider the Laplace integral

10

pex(pp2)dp. Since c = 1/2 is

the unique maximum point for p p2, and

p p2 = 14(p 1

2

)2+O

((p 1

2)4)

we havef(c) =

1

2, g(c) =

1

4, g(c) = 2

using laplaces method we get

10

pex(pp2)dp

piex/4

2x

As shown in the table below the formula is quite accurate.


x=5 x=10 x=15 x=20 10pex(pp

2)dp 1.22585 3.3276 9.66976 29.3645piex/4

2x

1.38334 3.41414 9.72979 29.4105Table 2. Asymptotics of a Laplace integral

n=1 n=2 n=3 n=4 n=5n! 1 2 6 24 120

2pinnnen 0.922137 1.919 5.83621 23.5062 118.019Table 3. Stirlings formula

The next example is a bit advanced but the conclusion is very useful. We knowthat factorials increase faster than even exponentials, but how fast does a factorialincrease? The answer is given by the famous Stirlings formula.

Example 12 (Stirlings formula). Recall that the Gamma function satisfies

(x+ 1) =

0

sxesds

To find out the large x asymptotics we first perform the (somewhat mysterious)transforms

(x+ 1) =

0

es+x ln sds = x 0

exr+x ln(xr)dr = xx+1 0

ex(r+ln r)dr

Note that r + ln r reaches its maximum at c = 1 and

r + ln r = 1 12

(r 1)2 +O ((r 1)3)This means

f(c) = 1, g(c) = 1, g(c) = 1By Laplaces method we get

0

ex(r+ln r)dr

2piexx

Therefore we obtain Stirlings formula

(x+ 1)

2pixxxex

This means

n!

2pinnnen

As shown in the table below the formula is very accurate even for smaller vauesof n.


3.4. Method of stationary phase. In this section we study large x asymptoticsof general Fourier integrals of the type

(3.5) I(x) = ba

f(p)eixh(p)dp

If a = b = and h(p) = p this would be the usual Fourier transform, which canbe calculated using integration by parts. In general, the function eixh(p) oscillatesrapidly for large x. This means there will be a lot of cancellations in the integral.Thus we naturally expect to have the following

Theorem 13 (Riemann-Lebesgue Lemma). If f(p) is integrable and h(p) is contin-uously dierentiable on [a, b], but h(p) is not constant over any subinterval of [a, b],then as x we have

ba

f(p)eixh(p)dp = o(1)

The Riemann-Lebesgue lemma itself does not give any asymptotic information,but it is essential for proving the asymptotic methods below. Due to the massivecancellations of Fourier integrals, the main contribution should come from pointswhere the cancellations are the weakest. Possible candidates are endpoints (sinceonly one side of the interval is counted for the integral and there is no cancellationfrom the other side) as well as points where h(c) = 0 (since h(p) h(c)+(pc)2/2,two sides of the interval centered at c do not cancel each other). In other words,the main contribution to the integral come from the critical points. As in the caseof Laplace integrals, the behaviors of f and h in (3.5) may be very complicated anddifficult to anlayze, but in practice it is often sufficient to consider the followingtwo important cases:

Case 1. h(p) 6= 0 for all p [a, b].Case 2. There is a unique point c (a, b) such that h(c) = 0. This point c is

called a stationary point.

The first case can be handled by integration by parts.

I(x) =

ba

f(p)eixh(p)dp =f(p)eixh(p)

ixh(p)|ba

1

ix

ba

(f(p)

h(p)

)eixh(p)dp

The second integral is of order o(1/x) by the Riemann-Lebesgue lemma, whichmeans

I(x) f(b)eixh(b)

ixh(b) f(a)e

ixh(a)

ixh(a)

In the second case the main contribution comes from the stationary point c,which (I believe) is why the method is called stationary phase.

We use the approximations h(p) h(c) + 12h(c)(p c)2, f(p) f(c) to obtain


Figure 3.4. Graph of p cos(x(p p2)) with x = 200. There arerapid oscillations except near the stationary point 1/2.

I(x) =

ba

f(p)eixh(p)dp ba

f(c)eix(h(c)+h(c)(pc)2/2)dp

f(c)eixh(c)

eix(h(c)(pc)2/2)dp f(c)eixh(c)

2

x|h(c)|

eisgn(h(c))s2ds

where sgn(y) = 1 for y > 0 and sgn(y) = 1 for y < 0. The reason we can changethe integral bounds to infinity is because the additional errors introduced by thisapproximation are of a smaller order than the main integral. Now we use the specialvalue of the integral

eis2

ds =pieipi/4

to obtain the final formula

I(x) f(c)eixh(c)+isgn(h(c))pi/4

2pi

x|h(c)|Note that the decay rate 1/

x is weaker than the endpoint case (1/x). This

means the contribution from the stationary point is stronger than contributionsfrom endpoints. Thus there is no need to consider endpoints as long as a stationarypoint exists.

Now lets summarize our results.

Proposition 14 (Method of stationary phase). Consider the leading order large xasymptotics of the general Fourier integral

(3.6) I(x) = ba

f(p)eixh(p)dp


x=10 x=12 x=14 x=16 x=18 10p cos(x(p p2))dp -0.0551303 -0.164609 -0.224068 -0.227672 -0.18095212 cos(

xpi4 )

pix -0.0401627 -0.153561 -0.215588 -0.220967 -0.175521Table 4. Asymptotics of a Fourier integral

(1) If h(p) 6= 0 for all p [a, b] then

I(x) f(b)eixh(b)

ixh(b) f(a)e

ixh(a)

ixh(a)(2) If there is a unique point c (a, b) such that h(c) = 0, then

I(x) f(c)eixh(c)+isgn(h(c))pi/4

2pi

x|h(c)|Example 15. Consider the Fourier integral

I2(x) =

10

p cos(x(p p2))dpWe first write the cosine function as the real part of a complex exponential:

10

p cos(x(p p2))dp =


part of g(p) is a constant near the saddle point. This new contour is called a steepestdescent contour. Then one can apply Laplaces method to obtain the asymptoticbehavior of I(x) for large x. We will not go into the details here since this methodis quite complicated.


Exercise. 1. Find the complete asymptotic series for large positive x of the fol-lowing functions:

(1) I(x) = x

et

t2dt

(2) The cosine integral function I(x) = x

cos t

tdt

(3) The error function erf(x) =2pi

x0

et2

dt

(4) I(x) = 0

exp

2 + pdp

(5) I(x) = 10

ln(1 + p)exp2

dp

(6) I(x) = 10

exp

1 + pdp

2. Find the first three terms of the asymptotic series for large positive x of thefollowing functions:

(1) I(x) = x

et

1 tdt

(2) I(x) = x

ett2

dt

(3) I(x) = 10

ex(1x)p

1 + pdp

(4) I(x) = 11ex(pp

2)dp

3. Find the leading term asymptotics for large positive x of the following functions:

(1) I(x) = 10

pex(p+p3)dp

(2) I(x) = 20

p2ex(3pp3)dp

(3) I(x) = 21

ln(1 + p)ex(2pp3)dp

(4) I(x) = 10

(1 + 2p)ex(p2p)dp

(5) I(x) = 10

ep+ix(p+p3)dp

(6) I(x) = 20

p2eix(3pp3)dp

(7) I(x) = 10

(1 + 2p)eix(p2p)dp

(8) The Bessel function J1(x) =1

pi

pi0

cos(t x sin t)dt


4. Asymptotic solutions to algebraic equations

In this section we will introduce methods for finding asymptotic solutions tocertain algebraic equations which cannot be solved explicitly.

4.1. Inverse of analytic functions. We know from complex analysis that if afunction f(z) is analytic and f (a) 6= 0, then f(z) is invertible near z = a, and itsinverse f1 is analytic near f(a). This means f1 has a Taylor series expansion.But how can we calculate this series? There is a formula called Lagrange-Brmannformula, which is extremely complicated. Fortunately there are much easier ways.Since we can adopt the new variable x = za and new function g(z) = f(z)f(a),we can assume a = 0 and f(a) = 0 for simplicity.

Fact 16 (Method of undetermined coefficients). If f(z) has the Taylor series f(z) =n=1

bnzn, then one can solve the equation f(z) = w by taking z =

n=1

anwn and

solve for the coefficients of an using the equation

k=1

bk(

n=1

anwn)k = w

Example 17. Lets find the first three terms of the Taylor series of the inversefunction of z + z2 2 sin z at 0. This means we want to solve the equation

z + z2 2 sin z = wThis means

w = z + z2 + z3

3+ o(z3)

By taking z =n=1

anwn we get the equation (note that z w)

w = n=1

anwn +

( n=1

anwn

)2+

1

3

( n=1

anwn

)3+ o(w3)

Lets write out the terms up to degree three

w = a1w a2w2 a3w3 + a21w2 + 2a1a2w3 +a313w3 + o(w3)

This implies all the coefficients of w,w2, w3 must be zero, that is

1 = a1

a2 + a21 = 0

a3 + 2a1a2 + a31

3= 0

So we get a1 = 1, a2 = 1, a3 = 5/3. The first three terms of the inverseseries is therefore


z = w + w2 43w3 + o(w3)

In general it is not possible to find a formula for all terms.

Example 18. If we want to find the inverse function of 1 + z2 + z5 near z = 1, wecan take

z 1 =n=1

an(w 3)n

since if z = 1then w = 1 + z2 + z5 = 3. This means

w 3 = (n=1

an(w 3)n + 1)2 1 + (n=1

an(w 3)n + 1)5 1

Then we can proceed as before (but using w1 = w 3 as a new variable) to findthe first few terms of the series.

Fact 19 (Method of iteration). If f(z) has the Taylor series f(z) =n=1

bnzn, then

the equation f(z) = w is the equivalent to

(4.1) z =

w n=2

bnzn

b1We can solve this using iteration. In the first step we set z = 0 and compute

the right hand side of (4.1), which yields z = w/b1. In the second step we plug thisvalue into the right hand side of (4.1) again, which gives

w

b1n=2

bnb1

(w

b1

)n=w

b1 b2b31w2 + o(w2)

In the third step we plug the valuew

b1 b2b31w2 into the right hand side of (4.1) again

to get

w

b1n=2

bnb1

(w

b1 b2b31w2)n

=w

b1 b2b31w2 +

2b2 + b3b41

w3 + o(w3)

We may continue this procedure to obtain the desired inverse power series.

Example 20. Lets find the first three terms of the Taylor series of the inversefunction of z z2 + 2 sin z using the iteration method. Since

w = z + z2 + z3

3+ o(z3)

we have

z = w + z2 + z3

3+ o(z3)

The first iteration is just w. The second iteration is


w + w2and the third iteration is

w + (w + w2)2 + (w + w2)3

3= w + w2 5

3w3 + o(w3)

just as before.

The general case is very similar.

Example 21. If we want to find the inverse function of 1 + z2 + z5 near z = 1, wenote that

1 + z2 + z5 = 3 + 7(z 1) + o(z 1)Thus we can take

z 1 = w 37

+ (1 + z2 + z5 3

7 (z 1))

and iterate as before, treating z1 = z 1 and w1 = w 3 as new variables. If onewishes one may write

z1 =w17

+ (1 + (z1 + 1)

2 + (z1 + 1)5 3

7 z1)

which may be easier to iterate.

Fact 22. The method of iteration can also be used to calculate asymptotic serieswhich are not power series, as long as the equation can be written in the form

(4.2) z = g(w) + h(z)

where h(z) = o(z). The first iteration is g(w). The second one is g(w) + h(g(w))(truncated to the correct order), and so on.

Example 23. If we want to find the inverse function of ez(1 + 1/z) near +, weset

w = ez(1 +1

z)

and take logs on both sides, which yields

lnw = z + ln(1 +1

z)

This means we can iterate

z = lnw ln(1 + 1z

) = lnw 1z

+1

2z2 1

3z3+ ...

The first iteration gives lnw. The second iteration gives

lnw 1lnw

The third iteration gives


lnw 1lnw 1lnw

+1

2(lnw 1lnw )2+o((lnw)2) = lnw 1

lnw+

1

2(lnw)2+o((lnw)2)

and so on.

4.2. Asymptotic solutions for large x. We can also use the method of iterationto find asymptotic solutions to algebraic equations. In particular if there are in-finitely many solutions going to infinity we can find the asymptotic values of thesesolutions. The major task is to write the equation in a proper form which allowsfor iterations.

Example 24. Lets go back to our very first example x sinx = 1. We know thereare infinitely many solutions, but how can we find them asymptotically? The firstobservation is that since x goes to infinity, sinx must small in order for the productx sinx = 1 to be true. In other words sinx = 1/x = o(1). We know that sinxis close to 0 if and only if x is close to kpi for integers k. Thus we may writex = kpi + t (0 < t < pi) and the equation becomes

(4.3) (1)k(kpi + t) sin t = 1which is equivalent to

(4.4) t = arcsin(

(1)kkpi + t

)= (1)k arcsin

(1

kpi + t

)Since t is of order 1/k and

arcsin

(1

kpi + t

) arcsin

(1

kpi

)= O(

1

k2) = o(t)

the right hand side of (4.4) is of the correct form as in (4.2), meaning we caniterate the right hand side of (4.4) after taking the large t power series. That is,we can iterate

(4.5) t = (1)k(

1

pik tpi2k2

+1 + 6t2

6pi3k3+O(

1

k)4)

The first iteration gives

t =(1)kpik

The second iteration gives

t =(1)kpik

1 (1)k/6

pi3k3

and so on.Alternatively, since t = o(1) we can expand the sin t in (4.3) and get

(1)k(kpi + t)(t t3

6+O(t5)) = 1

which implies

16pikt3 + pikt+ t2 = (1)k + o(t3)


which as the iterative form

(4.6) t =(1)kpik

t2

pik+

1

6t3 + o(t3)

The first iteration gives

t =(1)kpik

The second iteration gives

t =(1)kpik

1 (1)k/6

pi3k3

as before. Although the iterative equations (4.5) and (4.6) are different, they givethe same result. As long as we require the series to consist of powers of 1/k, theseries is unique.


Exercise. 1. Find the first three terms of the inverse power series at z = 0 of thefollowing functions:

(1) z3 z2 + sin z(2) z2 + ln(1 + z)(3) ez cos z(4)

sin z1 + z2

2. Find the first three terms of the inverse power series of the following functions:

(1)2 z31 + z

at z = 1

(2) z5 + 2z2 at z = 13. Find the first three terms of the asymptotic series of the solutions to the followingequations (for large integer k):

(1) x cosx = 2(2) sinx+

cosx

x= 0


5. Asymptotics of ordinary differential equaitons

Ordinary differential equations (ODEs) are very important in applications in-cluding sciences and engineering, but most ODEs cannot be solved explicitly. Some-times one can use power series expansions or numerical analysis to obtain satisfac-tory approximations, but in many cases we need to use asymptotic methods. Inthis section we will introduce three methods: iteration, perturbation expansion,and WKB.

5.1. The method of iteration. Suppose we ant to find out the behavior of asolution to some ODE for large x, then neither power series nor numerical simulationwould be an ideal choice since they do not have sufficient accuracy for large x.Instead, we may use iteration methods just as we did for algebraic equations. Theprocedure is as follows:

Fact 25 (Method of iteration). Suppose we have an ODE of the form

y = f(x) + g(y, y)where f(x) = o(1) can be written as a power series in 1/x, and g(y, y) = o(y),then we may iterate the right hand side to obtain an asymptotic series. The firstiteration gives y = f(x) (truncated to leading order). The second iteration givesy = f(x) + g(f(x), f (x)) (truncated to second order) and so on. y will also have apower series in 1/x.

Example 26. The ODE y = y + 1/x can be solved in terms of the exponentialintegral function we have studies before. The general solution is

y(x) = Cex exE1(x) = Cex + ex x

et

tdt

There is a unique solution of order o(1) for large x, which corresponds to C = 0.We already know how to find the asymptotic series of the solution using integrationby parts, but in fact we can find the asymptotic series without solving the equationexplicitly. We simply write

y =1

x y

and iterate the right hand side. The first iteration gives 1/x. the second gives1

x+

1

x2.The third gives

1

x+

1

x2+

2

x3, and so on. This works because the derivative

of 1/xkis of order 1/xk+1, so y belongs to the right hand side.

If it is clear that y has an power series in 1/x, one can also use the method of

undetermined coefficients by setting y(x) =k=1

xk can plug it into the ODE. For

ODEs the series are typically divergent (unlike for algebraixc equations).

Example 27. Lets consider the ODE

(5.1) y y + y + y2 = 2 + 1x

The general solution cannot be found explicitly. However, we can use the methodof iteration to find certain special asymptotic solutions. There is no simple solution


of order o(1) due to the 2 on the right hand side. However, ther are two solutionsbehaving like constants. They satisfy

y + y2 = 2

That is, we have solutions y1 = 1 + o(1) and y2 = 2 + o(1). Lets focus on y1and y2 can be similarly studied (this is an exercise). By setting y = 1 + u we canrewrite (5.1) as

u u + 3u+ u2 = 1x

This gives the iterative equation

u =1

3

(1

x u2 + u u

)The first iteration gives u =

1

3x. The second gives u =

1

3x 4

27x2. The third

gives u = 22243x3

427x2

+1

3x, and so on.

It is possible to prove the existence real solutions which the asymptotic solutionsapproximate, but the proof is very advanced.

5.2. Perturbation expansions. ODEs arising in practice often contain unknownparameters and people often want to find out the limiting behavior when the pa-rameter is small.

Fact 28. If an ODE contains a small parameter which is not multiplying the

term of the highest derivative, we may write the solution as y =k=0

kyk, plug it

into the ODE, collect terms according to powers of , and solve the yks one by one.

Example 29. Lets consider the initial value problem

(5.2) y(x) = (1 + x)y(x); y(0) = 1; y(0) = 1

and we are interested in the behavior of the solution for small and boundedx. The solution can be expressed in terms of Airy functions, which are fairly

complicated. Instead, we write y =k=0

kyk and plug this into the ODE which

gives

k=0

kyk = (1 + x)k=0

kyk

Collecting power of we get

y0 = y0; y0(0) = 1; y0(0) = 1

y1 y1 = xy0; y1(0) = 0; y1(0) = 0and in general


Figure 5.1. Solutions to the ODE (5.2) (blue curve) and its 3-term perturbation expansion for = 0.2 (orange curve)

yn+1 yn+1 = xyn; yn+1(0) = 0; yn+1(0) = 0We can solve them term by term. This yields

y0(x) = ex

y1(x) =1

8ex

(e2x(2x2 2x+ 1) 1)

y2(x) =1

192ex

(6x2 + e2x

(6x4 20x3 + 36x2 36x+ 15)+ 6x 15)

and so on.The solution is therefore written as

y(x) = ex+

8ex

(e2x(2x2 2x+ 1) 1)+ 2

192ex

(6x2 + e2x

(6x4 20x3 + 36x2 36x+ 15)+ 6x 15)+O(3)

As shown in the figure below, the approximation is quite accurate.

Another example is

Example 30. Lets consider the initial value problem

(5.3) y = y + y2; y(0) = 1; y(0) = 0

By setting y =k=0

kyk we get the equation

k=0

kyk =k=0

kyk +

( k=0

kyk

)2Collecting power of we get

y0 = y0; y0(0) = 1; y

0(0) = 0


Figure 5.2. Solutions to the ODE (5.3) (blue curve) and its 3-term perturbation expansion for = 0.2 (orange curve)

y1 y1 = y20 ; y1(0) = 0; y1(0) = 0

y2 y2 = 2y0y1; y2(0) = 0; y2(0) = 0and so on. This means

y0(x) = 1

y1(x) = x+ ex 1

y2(x) = x2 + 2exx+ 4x 6ex + 6

and so on. So

y(x) = 1 + (x+ ex 1) + 2(x2 + 2exx+ 4x 6ex + 6) +O(3)The plots are shown below.

5.2.1. Singular perturbations. If the small parameter appears in front of the termwith the highest derivative, the regular perturbation expansion method may fail.

Example 31. The initial value problem

y y = x; y(0) = 1has a solution

y(x) = x + (+ 1)ex/There is no expansion in powers . If one attempts such an expansion, the first

term should satisfy y0(x) = x; y0(0) = 1 which has no solution.


We say this problem is singularly (instead of regularly) perturbed. To solve theproblem, we need to perform the substitution x = t which transforms the ODEinto

y(t) y(t) = t; y(0) = 1The new ODE is regularly perturbed and can be solved using the perturbation

expansion method. However, if we want the result to be valid for say x [0, 1], wewill need t [0, 1/]. The new expansion may not be valid for t so large. In thatcase we need to use the new ODE for small t, the orignal ODE for x of order 1,and perhaps also use a formula for the intermediate region (for instance t 1/)to get a uniform description for the solution. The method of singular perturbationis quite complicated.

5.3. The WKB method. The WKB method is a popular asymptotic methodused in physics, and it is named after three physicists. There are several differentsenarios where it can be applied, and we will focus on a particularly important one,namely the large x behavior of solution to ODEs. We have seen previously thatthere may be solution which have asymptotic power series in 1/x, but these arejust special cases. General solutions will not behave like that and simple equationslike y = y has no such solution. However, for a fairly large class of linear ODEsthere are solutions of the type

(5.4) y xecxk=0

akxk

where > 0 and c may be complex.

Fact 32 (WKB method). To find a solution of the type (5.4), one can simply plugit into the ODE and determine the unknown coefficients , , c, ak, ... However, itis usually simpler if one first plug y = ecx

into the ODE and determine c, , thenplug y = xecx

into the ODE to determine , and finally plug in the full expansion(5.4).

Example 33. Lets begin with the ODE

y(x) = (4 +1

x)y(x)

The general solutions can be represented using hypergeometric functions, and theyare quite complicated. The WKB solution is relatively simple. First we attemptthe approximate solution y = ecx

b

and plug it into the ODE. This gives

bcxb2ecxb (bcxb + b 1) = (2 + 1

x)ecx

b

To leading order we haveb2c2x2b2 = 4

This means b = 1, c = 2 since there are two linearly independent solutions.Lets focus on the solution with c = 2. The other one is left as an exercise. Thenwe attempt the approximate solution y = xae2x and plug it into the ODE. Thisgives


e2xxa2(a2 + a(4x 1) x) = 0

which means a = 1/4. Finally we let

y(x) = x1/4e2x(1 +

k=1

akxk)

and plug this into the ODE. Here we fix a0 = 1 since the solution to the homoge-neous ODE has an arbitrary multiplicative constant. We get

(64a1 + 3)e2x

16x7/4 (128a2 21a1)e

2x

16x11/4 (192a3 77a2)e

2x

16x15/4+ ... = 0

This means a1 = 364, a2 = 63

8192, a3 = 1617

524288, and so on. The WKB solu-

tion is

y(x) x1/4e2x(1 364x 63

8192x2 1617

524288x3+O(x4))

Our next example is the Bessel equation whose solutions are called Bessel func-tions.

Example 34. The Bessel equation of order 1 is

x2y + xy + (x2 1)y = 0To find a WKB solution, we first attempt y = ecx

b

and plug it into the ODEwhich gives

ecxb (b2c2x2b + b2cxb + x2 1) = 0

To leading order this means

b2c2x2b = x2which implies b = 1, c = i. We focus on the solution with c = i as the other

one is just the complex conjugate of the first one. We let y = xaeix and plug it intothe ODE which gives

eixxa(a2 + 2iax+ ix 1) = 0

To leading order this means a = 1/2. Then we let

y(x) = x1/2eix(1 +k=1

akxk)

and plug this into the ODE. This gives

1

4x9/2eix((3 8ia1)x4 + (5a1 16ia2)x3 + 3(7a2 8ia3)x2 + ...) = 0

This means

3 8ia1 = 0

5a1 16ia2 = 0


7a2 8ia3 = 0and so on. Consequently a1 =

3i

8, a2 =

15

128, a3 = 105i

1024, ... So the solution is

y(x) x1/2eix(1 + 3i8x

+15

128x2 105i

1024x3+O(x4))

The other solution is just the complex conjugate

y(x) x1/2eix(1 3i8x

+15

128x2+

105i

1024x3+O(x4))

Our next example is the Airy equation whose solutions are called Airy functions.

Example 35. The Airy equation is

(5.5) y = xy

To find a WKB solution, we first attempt y = ecxb

and plug it into the ODEwhich gives

ecxb (b2c2x2b + (b 1)bcxb x3)

x2= 0

To leading order this means

b2c2x2b x3 = 0which implies b = 3/2, c = 2/3. Lets first look at the solution with c = 2/3.

We let y = xae(2/3)x3/2

and plug it into the ODE which gives

1

2e2x3/2

3 xa2(

2a2 + a(

4x3/2 2)

+ x3/2)

= 0

To leading order this means a = 1/4. Then we let

y(x) = x1/4e(2/3)x3/2

(1 +

k=1

akx3k/2)


e2x3/2

3

((96a2 77a1)x9/2 + (48a1 5)x6 + (144a3 221a2)x3 + ...

)16x33/4

= 0

which means a1 =5

48, a2 =

385

46088, a3 =

85085

663552, ...

So the solution is

y(x) x1/4e(2/3)x3/2(1 + 548x3/2

+385

46088x3+

85085

663552x9/2+O(x6))

Alternatively, one may do the change of variable t = 23x3/2. This means

dy

dx=dy

dt

dt

dx= x1/2

dy

dt=

(3t

2

)1/3dy

dt

and therefore


d2y

dx2=dt

dx

d

dt(

(3t

2

)1/3dy

dt) =

(3t

2

)1/3(

(3

2

)1/3t2/3

3

dy

dt+

(3t

2

)1/3d2y

dt2) =

t1/3

22/331/3dy

dt+(

3t

2)2/3

d2y

dt2

In the new variable the Airys ODE becomes

(5.6) y(t) +1

3ty(t) = y(t)

We let y = taet and plug it into the ODE which gives

1

3etta2

(3a2 + a(6t 2) + t) = 0

This means a = 1/6. Then we let

y(t) = t1/6et(1 +k=1

aktk)


et((72a1 5)t4 + (144a2 77a1)t3 + (216a3 221a2)t2 + ...

)36t37/6

= 0

which means a1 =5

72, a2 =

385

10368, a3 =

85085

2239488, ... One can check that the so-

lution is the same as before.However, since (5.6) has a symmetry, namely if y(t) is a solution then y(t)

is also a solution, we can immediately get the other solution by changing x3/2 tox3/2 in the first solution:

y(x) x1/4e(2/3)x3/2(1 548x3/2

+385

46088x3 85085

663552x9/2+O(x6))

The accuracy of this approximation is very good.

1.5 2.0 2.5 3.0 3.5 4.0

0.1

0.2

0.3

0.4

Figure 5.3. Airy function y (blue curve) and its 4-term asymp-totics (red curve)


Exercise. 1. Find the first three terms of the asymptotic power series of solutionsto the following ODEs with the given behavior:

(1) y 2y + y2 = 1x

; y = o(1)

(2) y + yy + y =1

x2; y = o(1)

(3) y + sin y =2

x; y = o(1)

(4) y + y2 = 1 +1

x; y = 1 + o(1)

(5) y (1 + 1x

)y =x

1 + x; y = 1 + o(1)

(6) xy y + y = 1x

; y = o(1)

(7) y + (y)2 + xy = x 1; y = 1 + o(1)2. Find the first three terms of the small perturbation expansions of solutions tothe following ODEs:

(1) y +2

xy + y2 = 0; y(1) = 1

(2) y + y + y2 = 0; y(0) = 1(3) y y + xy = 0; y(0) = 1, y(0) = 0(4) y + y e2xy = 1; y(0) = 0, y(0) = 1(5) y + y 2

x2y = x; y(0) = 1

(6) y + y xy = 1; y(0) = 1, y(0) = 0(7) (1 + x)y 6y = 2; y(1) = 0, y() = o(1)

3. Find the first three terms of the WKB solutions to the following ODEs (eachequation has two linearly independent solutions):

(1) y(x) = (4 +1

x)y(x)

(2) y(x) = (1 1x

+1

x2)y(x)

(3) The linearized Riccati equation x2y + (x2 2)y = 0(4) The Bessel equation of order 2: x2y + xy + (x2 4)y = 0(5) The Weber differential equation y + (1 x

2

4)y = 0

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