Asymptotic Analysis Notes - Heriot-Watt University

2.2. SINGULAR PERTURBATION PROBLEMS15Hencex( ) 1 1 1 22 18 · · · .Remark. If we were to try the iterative method, we might try to exploitthe significance of x 1, and choose the following decomposition for anappropriate iterative scheme:r x2(x 1) (x 1) x x 1 .1 xExample (non-integral powers). Find the first three terms in power seriesapproximations for the root(s) of(1 )x2 2x 1 0 ,(2.6)near x 1, where is a small parameter.Remark. Once we have realized that we need to pose a formal power11expansion in say powers of n , we could equivalently set δ n and expand1in integer powers of δ. At the very end we simply substitute back that δ n .This approach is particularly convenient when you use either Maple1 to tryto solve such problems perturbatively.Example (transcendental equation). Find the first three terms in thepower series approximation of the root ofex 1 ,(2.7)where is a small parameter.2.2. Singular perturbation problemsExample. Consider the following quadratic equation: x2 x 1 0 .(2.8)The key term in this last equation that characterizes the number of solutionsis the first quadratic term x2 . This term is ‘knocked out’ when 0. Inparticular we notice that when 0 there is only one root to the equation,namely x 1, whereas for 6 0 there are two! Such cases, where the characterof the problem changes significantly from the case when 0 1 to the casewhen 0, we call singular perturbation problems. Problems that are notsingular are regular.For the moment, consider the exact solutions to (2.8) which can be determined using the quadratic equation formula, 1 1 1 4 .x 2 1You can download a Maple worksheet from the course webpage which will help you toverify your algebra and check your homework answers.

162. PERTURBATION METHODSExpanding these two solutions (for small):(1 2 2 5 3 · · · ,x 1 1 2 2 5 3 · · · .We notice that as 0, the second singular root ‘disappears off to negativeinfinity’.Iteration method. In order to retain the second solution to (2.8) as 0and keep track of its asymptotic behaviour, we must keep the term x2 as asignificant main term in the equation. This means that x must be large. Notethat at leading order, the ‘ 1’ term in the equation will therefore be negligiblecompared to the other two terms, i.e. we have1x . This suggests the following sensible rearrangement of (2.8), x2 x 0 (2.9)11x , xand hence the iterative scheme11,xn 1 xnwithNote that in this case1x0 . 11f (x) . xHence1 x21 1 · 2 x ,f 0 (x) when x 1/ . Therefore, since is small, f 0 (x) is small when x is closeto the root, and further, we expect an order improvement in accuracy witheach iteration. The first two steps in the iterative process revealsx1 1 1, and11 1 1 1 ··· . x2

2.2. SINGULAR PERTURBATION PROBLEMS17Expansion method. To determine the asymptotic behaviour of the singular root by the expansion method, we simply pose a formal power seriesexpansion for the solution x( ) starting with an 1 term instead of the usual 0 term:1x( ) x 1 x0 x1 2 x2 · · · .(2.10) Using this ansatz2, i.e. substituting (2.10) into (2.8) and equating powers of 1 , 0 and 1 etc. generates equations which can be solved for x 1 , x0 andx1 etc. and thus we can write down the appropriate power series expansionfor the singular root.Rescaling method. There is a more elegant technique for dealing withsingular perturbation problems. This involves rescaling the variables beforeposing a formal power series expansion. For example, for the quadratic equation (2.8), setXx , and substitute this into (2.8) and multiply through by ,X2 X 0 .(2.11)This is now a regular perturbation problem. Hence the problem of findingthe appropriate starting point of a trial expansion for a singular perturbationproblem is transformed into the problem of finding the appropriate rescalingthat regularizes the singular problem. We can now apply the standard methods we have learned thusfar to (2.11), remembering to substitute back X xat the very end to get the final answer.Note that a practical way to determine the appropriate rescaling to try isto use arguments analogous to those that lead to (2.9) above.2.2.1. Example (singular perturbation problem). Use an appropriatepower series expansion to find an asymptotic approximation as 0 , correctto O( 2 ), for the two small roots of x3 x2 2x 3 0 .Then by using a suitable rescaling, find the first three terms of an asymptoticexpansion as 0 of the singular root.Example (transcendental equation). Consider the problem of finding thefirst few terms of a suitable asymptotic approximation to the real large solutionof the transcendental equation xex 1 ,(2.12) xexwhere 0 1. First we should get an idea of how the functionsand 1in (2.12) behave. In particular we should graph the function on the left-handside xex as a function of x and see where its graph crosses the graph of theconstant function 1 on the right-hand side. There is clearly only one solution,2Ansatz is German for “approach”.

182. PERTURBATION METHODSwhich will be positive, and also large when is small. In fact when 0 1,then1xex 1 x 1 , confirming that we expect the root to be large. The question is how large, ormore precisely, exactly how does the root scale with ?Given that the dominant term in (2.12) is ex , taking the logarithm of bothsides of the equation (2.12) might clarify the scaling issue. x ln x ln 0x ln ln x x ln 1 ln x , where in the last step we used that ln A ln A1 . Now we see that when 0 1 so that x 1, then x ln x and the root must lie near to ln 1 ,i.e. x ln 1 .This suggests the iterative scheme1 xn 1 ln ln xn , with x0 ln 1 . Note that in this case we can identify f (x) lnand dln 1 ln xf 0 (x) dx1 x1 x 1 , ln 1 1 ln x,when x is close to the root. Therefore f 0 (x) is small since is small. Anothergood reason for choosing the iteration method here is that a natural expansionsequence is not at all obvious. The iteration scheme gives x1 ln 1 ln ln 1 .Thenx2 ln ln ln ln 1 1 1 1 ln x11 ln ln ln ln ln ln 1 1 ln ln1 ln ln 1 1 ln 1 !!ln ln 1 ln 1 ln 1 !,

2.2. SINGULAR PERTURBATION PROBLEMS19where in the last step we used that ln AB ln A ln B. Hence, using theTaylor series expansion for ln(1 x), i.e. ln(1 x) x, we see that as 0 , 1 lnln · · · .x ln 1 ln ln 1 ln 1

CHAPTER 3Asymptotic series3.1. Asymptotic vs convergent seriesExample (the exponential integral). This nicely demonstrates the difference between convergent and asymptotic series. Consider the exponentialintegral function defined for x 0 byZ tedt .Ei(x) txLet us look for an analytical approximation to Ei(x) for x 1. Repeatedlyintegrating by parts gives t Z teeEi(x) dt t xt2xZ tee x dtxt2x t Z tee xe 2 dt2xt xt3xZ tee x e x 2 2dt xxt3x.!Z te(N 1)!11N xN 1 ( 1) N ! e 2 · · · ( 1)dt .Nx xxtN 1x{z} {z} RN (x)SN (x)Here we set SN (x) to be the partial sum of the first N terms, 12! x 1N 1 (N 1)!SN (x) e · · · ( 1)x x2 x3xNand RN (x) to be the remainder after N termsZ teNRN (x) ( 1) N !dt .tN 1xThe series for which SN (x) is the partial sum is divergent for any fixed x; noticethat for large N the magnitude of the N th term increases as N increases! Ofcourse RN (x) is also unbounded as N , since SN (x) RN (x) must bebounded because Ei(x) is defined (and bounded) for all x 0.21

223. ASYMPTOTIC SERIESSuppose we consider N fixed and let x become large:Z teNdtRN (x) ( 1) N !tN 1xZ tedt ( 1)N N !N 1txZ tedt N!tN 1xZ N! N 1e t dtxxN ! x N 1 e ,xwhich tends to zero very rapidly as x . Note that the ratio of RN (x) tothe last term in SN (x) is RN (x) RN (x) (N 1)!e x x N(N 1)!e x x NN !e x x (N 1)(N 1)!e x x NN, xwhich also tends to zero as x . Thus Ei(x) SN (x) o last term in SN (x) (3.1)as x . In particular, if x is sufficiently large and N fixed, SN (x) gives agood approximation to Ei(x); the accuracy of the approximation increases asx increases for N fixed. In fact, as we shall see, this means we can write 12! x 1 ··· ,Ei(x) ex x2 x3as x .Note that for x sufficiently large, the terms in SN (x) will successivelydecrease initially—for example 2!x 3 x 2 for x large enough. However atsome value N N (x), the terms in SN (x) for N N will start to increasesuccessively for a given x (however large) because the N th term,( 1)N 1 e x(N 1)!,xNis unbounded as N .Hence for a given x, there is an optimal value N N (x) for which thegreatest accuracy is obtained. Our estimate (3.1) suggests we should take N to be the largest integral part of the given x.In practical terms, such an asymptotic expansion can be of more value thana slowly converging expansion. Asymptotic expansions which give divergentseries, can be remarkably accurate: for Ei(x) with x 10; N 10, butS4 (10) approximates Ei(10) with an error of less than 0.003%.

3.1. ASYMPTOTIC VS CONVERGENT SERIES23Magnitude of last term in SN (x)0.10.090.08 (N 1)! e x/xN 0.070.060.050.040.030.02x 2x 2.5x 30.01012345N678910Figure 1. The behaviour of the magnitude of the last termin SN (x), i.e. ( 1)N 1 (N 1)!e x x N , as a function of Nfor different values of x.Basic idea. Consider the following power series expansion about z z0 : Xn 0an (z z0 )n .(3.2) Such a power series is convergent for z z0 r, for some r 0,provided (see the Remainder Theorem)RN (x) Xn N 1an (z z0 )n 0 ,as N for each fixed z satisfying z z0 r. A function f (z) has an asymptotic series expansion of the form (3.2)as z z0 , i.e.f (z) Xn 0an (z z0 )n ,provided RN (x) o (z z0 )N ,as z z0 , for each fixed N .(3.3)

243. ASYMPTOTIC SERIESExponential integral .060.040.020 0.0211.522.53x3.544.55Logarithm of error in the asymptotic approximations1log10 S1(x) Ei(x) log10 S2(x) Ei(x) 0log10 S3(x) Ei(x) 1 2log10N S (x) Ei(x) log10 S4(x) Ei(x) 3 4 511.522.53x3.544.55Figure 2. In the top figure we show the behaviour of the Exponential Integral function Ei(x) and four successive asymptotic approximations. The lower figure shows how the magnitude of the difference between the four asymptotic approximations and Ei(x) varies with x—i.e. the error in the approximations is shown (on a semi-log scale).

3.2. ASYMPTOTIC EXPANSIONS253.2. Asymptotic expansionsDefinition. A sequence of gauge functions {φn (x)}, n 1, 2, . . . is saidto form an asymptotic sequence as x x0 if, for all n,as x x0 . φn 1 (x) o φn (x) ,Examples. (x x0 )n as x x0 ; x n as x ; (sin x)n as x 0.Definition. If {φn (x)} is an asymptotic sequence of functions as x x0 ,we say that Xan φn (x)n 1where the an are constants, is an asymptotic expansion (or asymptotic approximation) of f (x) as x x0 if for each Nf (x) NXn 1 an φn (x) o φN (x) ,(3.4)as x x0 , i.e. the error is asymptotically smaller than the last term in theexpansion.Remark. An equivalently property to (3.4) isf (x) N 1Xn 1 an φn (x) O φN (x) ,as x x0 (which can be seen by using that the guage functions φn (x) forman asymptotic sequence).Notation. We denote an asymptotic expansion byf (x) Xan φn (x) ,n 1as x x0 .Definition. If the gauge functions form a power sequence, then the asymptotic expansion is called an asymptotic power series.Examples. xn as x 0; (x x0 )n as x x0 ; x n as x .

263. ASYMPTOTIC SERIES3.3. Properties of asymptotic expansionsUniqueness. For a given asymptotic sequence {φn (x)}, the asymptoticexpansion of f (x) is unique; i.e. the an are uniquely determined as followsf (x);x x0 φ1 (x)f (x) a1 φ1 (x)a2 lim;x x0φ2 (x).PN 1f (x) n 1an φn (x)aN lim;x x0φN (x)a1 limand so forth.Non-uniqueness (for a given function). A given function f (x) may havemany different asymptotic expansions. For example as x 0,tan x x 31 x3 2 515 x ··· sin x 12 (sin x)3 38 (sin x)5 · · · .Subdominance. An asymptotic expansion may be the asymptotic expansion of more than one function. For example, if as x x0f (x) Xn 0an (x x0 )n ,then alsof (x) eas x x0 because eIn fact,1 (x x0 )2 1(x x0 )2 Xn 0an (x x0 )n , o (x x0 )n as x x0 for all n. Xn 0an (x x0 )nis asymptotic as x x0 to any function which differs from f (x) by a functiong(x) so long as g(x) 0 as x x0 faster than all powers of x x0 . Such afunction g(x) is said to be subdominant to the asymptotic power series; theasymptotic power series of g(x) would beg(x) Xn 00 · (x x0 )n .Hence an asymptotic expansion is asymptotic to a whole class of functionsthat differ from each other by subdominant functions.

3.3. PROPERTIES OF ASYMPTOTIC EXPANSIONS27Example (subdominance: exponentially small errors). The function e xis subdominant to an asymptotic power series of the form Xan x nn 0as x ; and so if a function f (x) has such a asymptotic expansion, sodoes f (x) e x , i.e. f (x) has such an asymptotic power series expansion upto exponentially small errors.Equating coefficients. If we write Xn 0 Xan · (x x0 )n n 0bn · (x x0 )n(3.5)we mean that the class of functions to which Xn 0an · (x x0 )nand Xn 0bn · (x x0 )nare asymptotic as x x0 , are the same. Further, uniqueness of asymptoticexpansions means that an bn for all n, i.e. we may equate coefficients oflike powers of x x0 in (3.5).Arithmetical operations. Suppose as x x0 ,f (x) Xan φn (x)andn 0g(x) Xbn φn (x)n 0then as x x0 ,αf (x) βg(x) Xn 0(an bn ) · φn (x) ,where α and β are constants. Asymptotic expansions can also be multipliedand divided—perhaps based on an enlarged asymptotic sequence (which wewill need to be able to order). In particular for asymptotic power series, whenφn (x) (x x0 )n , these operations are straightforward:f (x) · g(x) where cn Pnm 0 an bn m Xn 0and if b0 6 0, d0 a0 /b0 , then f (x) X dn (x x0 )n ,g(x)n 0 where dn an cn (x x0 )n ,Pn 1m 0 dm bn m /b0 .

283. ASYMPTOTIC SERIESIntegration. An asymptotic power series can be integrated term by term(if f (x) is integrable near x x0 ) resulting in the correct asymptotic expansions for the integral. Hence, if Xf (x) as x x0 , thenZn 0 Xan(x x0 )n 1 .n 1xx0an (x x0 )nf (t) dt n 0Differentiation. Asymptotic expansions cannot in general be differentiated term by term. The problem with differentiation is connected with subdominance. For instance, the two functions 11 22(x x)(x x)00f (x)andg(x) f (x) esin ediffer by a subdominant function and thus have the same asymptotic powerseries expansion as x x0 . However f 0 (x) and 111 222 300 3(x x)(x x)(x x)000 2(x x0 ) esin eg (x) f (x) 2(x x0 ) cos edo not have the same asymptotic power series expansion as x x0 .However if f 0 (x) exists, is integrable, and as x x0 ,f (x) Xn 0an (x x0 )n ,then0f (x) Xn 0n · an · (x x0 )n 1 .In particular, if f (x) is analytic in some domain, then one can differentiate anasymptotic expansion for f (x) term by term—recall that a real function f (x)is said to be analytic at a point x x0 if it can be represented by a powerseries in powers of x x0 with a non-zero radius of convergence. For example,111 2 ··· ,x 1x xas x implies, since the power series shown is in fact convergent for1is analytic for all x 1, thatevery x 1 and therefore x 1112 2 3 ··· ,2(x 1)xxas x . (Both of the power series are the Taylor series expansions forthe respective functions shown.)

3.4. ASYMPTOTIC EXPANSIONS OF INTEGRALS293.4. Asymptotic expansions of integralsIntegral representations. When modelling many physical phenomena, itis often useful to know the asymptotic behaviour of integrals of the formZ b(x)I(x) f (x, t) dt ,(3.6)a(x)as x x0 (or more generally, the integral of a complex function along acontour). For example, many functions have integral representations: the error function,2Erf(x) πZx2e t dt;0 the incomplete gamma function (x 0, a 0),Z xγ(a, x) e t ta 1 dt .0Many other special functions such as the Bessel, Airy and hypergeometricfunctions have integral representations as they are solutions of various classesof differential equations. Also, if we use Laplace, Fourier or Hankel transformations to solve differential equations, we are often left with an integralrepresentation of the solution (eg. to determine an inverse Laplace transform we must evaluate a Bromwich contour integral ). Two simple techniquesfor obtaining asymptotic expansions of integrals like (3.6) are demonstratedthrough the following two examples.Example (Inserting an asymptotic expansion of the integrand). We canobtain an asymptotic expansion of the incomplete gamma function γ(a, x) asx 0, by expanding the integrand in powers of t and integrating term byterm. Z x t2 t3γ(a, x) 1 t · · · ta 1 dt2! 3!0a 1axxa 2xa 3x ··· a(a 1) (a 2)2! (

and so on. Note that the rst equation is trivial since we actually expanded about the 0 solution, namely x0 1. Hence we see that x( ) 1 1 2 1 8 2 O( 4): For small, this expansion truncated after the third term is a good approx-imation to the act