Perturbation Expansions

3.2.4 Perturbation Expansions

In the equations of applied mathematics it is often possible to identify a small dimensionless parameter (say) ǫ, ǫ ≪ 1. The case when ǫ = 0 is called the reduced problem or the

unperturbed case, and one asks for a perturbation expansion, i.e., an expansion of the solution of the perturbed problem into powers of the perturbation parameter ǫ. In many

cases it can be proved that the expansion has the form c 0 +c 1 ǫ +c 2 ǫ 2 + · · · , but there are also important cases where the expansion contains fractional or a few negative powers. In this subsection, we consider an analytic equation φ(z, ǫ) = 0 and seek expansions for the roots z i (ǫ) in powers of ǫ. This has some practical interest in its own right, but it is mainly to be considered as a preparation for more interesting applications of perturbation methods to more complicated problems. A simple perturbation example for a differential equation is given in Problem 3.2.9.

i ( 0), 0), then a theorem of complex analysis tells us that z i (ǫ) is an analytic function in a neighborhood of the origin. Hence the expansion

If z i (

i (ǫ) −z i ( 0) = c 1 ǫ +c 2 ǫ +···

has a positive (or infinite) radius of convergence. We call this a regular perturbation problem . The techniques of power series reversion, presented in Sec. 3.1.4, can often be applied after some preparation of the equation. Computer algebra systems are also used in perturbation problems, if expansions with many terms are needed.

Example 3.2.5.

We shall expand the roots of

φ (z, ǫ)

≡ ǫz 2 −z+1=0

into powers of ǫ. The reduced problem −z + 1 = 0 has only one finite root, z 1 ( 0) = 1. Set z

3 ǫ 2 = 1 + xǫ, x = c 2 +c +c + · · · . Then φ(1 + xǫ, ǫ)/ǫ = (1 + xǫ) − x = 0, i.e., (

ǫ ǫ 2 1+c 2 1 +c 2 + · · ·) − (c 1 +c 2 ǫ +c 3 ǫ 2 + · · ·) = 0.

204 Chapter 3. Series, Operators, and Continued Fractions

Matching the coefficients of ǫ 0 ,ǫ 1 ,ǫ 2 , we obtain the system 1−c 1 =0⇒c 1 = 1,

2c 1 −c 2 =0⇒c 2 = 2, 2c 2 2 +c 1 −c 3 =0⇒c 3 = 5;

hence z 1 (ǫ) = 1 + ǫ + 2ǫ 2 + 5ǫ 3 +···.

Now, the easiest way to obtain the expansion for the second root, z 2 (ǫ) , is to use the fact that the sum of the roots of the quadratic equation equals ǫ −1 ; hence z 2 (ǫ) = ǫ −1 − 1 − ǫ − 2ǫ 2 +···. Note the appearance of the term ǫ −1 . This is due to a characteristic feature of this

one of the roots escapes to ∞ as ǫ → 0. This is an example of a singular perturbation problem, an important type of problem for differential equations; see Problem 3.2.7.

If ∂φ/∂z = 0, for some z i , the situation is more complicated; z i is a multiple root, and the expansions look different. If z i ( 0) is a k-fold root, then there may exist an expansion of

the form

z i (ǫ) =c 0 +c 1 ǫ 1/k

1/k +c 2 2 (ǫ ) +···

for each of the k roots of ǫ, but this is not always the case. See (3.2.32) below, where the expansions are of a different type. If one tries to determine the coefficients in an expansion of the wrong form, one usually runs into contradictions, but the question about the right form of the expansions still remains.

The answers are given by the classical theory of algebraic functions, where Riemann surfaces and Newton polygons are two of the key concepts; see, e.g., Bliss [35]. We shall, for several reasons, not use this theory here. One reason is that it seems hard to generalize some of the methods of algebraic function theory to more complicated equations, such as differential equations. We shall instead use a general balancing procedure, recommended in Lin and Segel [246, Sec. 9.1], where it is applied to singular perturbation problems for differential equations too.

The basic idea is very simple: each term in an equation behaves like some power of ǫ . The equation cannot hold unless there is a β such that a pair of terms of the equation behave like Aǫ β (with different values of A), and the ǫ-exponents of the other terms are larger than or equal to β. (Recall that larger exponents make smaller terms.)

Let us return to the previous example. Although we have already determined the expansion for z 2 (ǫ) (by a trick that may not be useful for problems other than single analytic equations), we shall use this task to illustrate the balancing procedure. Suppose that

z 2 (ǫ)

∼ Aǫ α , (α < 0). The three terms of the equation ǫz 2 − z + 1 = 0 then get the exponents

Try the first two terms as the candidates for being the dominant pair. Then 1 + 2α = α, hence α = −1. The three exponents become −1, −1, 0. Since the third exponent is larger

3.2. More about Series 205 than the exponent of the candidates, this choice of pair seems possible, but we have not

shown that it is the only possible choice. Now try the first and the third terms as candidates. Then 1 + 2α = 0, hence α = − 1

2 . The exponent of the noncandidate is − 1

2 ≤ 0; this candidate pair is thus impossible. Finally, try the second and the third terms. Then α = 0, but we are only interested in negative values of α.

The conclusion is that we can try coefficient matching in the expansion z 2 (ǫ) =

c −1 ǫ −1 +c 0 +c 1 ǫ + · · · . We don’t need to do it, since we know the answer already, but it indicates how to proceed in more complicated cases.

Example 3.2.6.

First consider the equation z 3 −z 2 + ǫ = 0. The reduced problem z 3 −z 2 = 0 has a single root, z 1 = 1, and a double root, z 2,3 = 0. No root has escaped to ∞. By a similar coefficient matching as in the previous example we find that z 1 (ǫ) = 1 − ǫ − 2ǫ 2 + · · · . For the double root, set z = Aǫ β , β > 0. The three terms of the equation obtain the exponents 3β, 2β, 1. Since 3β is dominated by 2β we conclude that 2β = 1, i.e., β = 1/2,

2,3 (ǫ) =c 0 ǫ

+c 1 ǫ +c 2 ǫ 3/2 +···. By matching the coefficients of ǫ, ǫ 3/2 ,ǫ 2 , we obtain the system

−c 2

0 +1=0⇒c 0 = ±1,

1 +c 0 =0⇒c 1 = , 2

−2c 0 c 1 3

c 2 2 −2c 5 0 2 −c 1 + 2c 0 c 1 +c 1 c 2 0 =0⇒c 2 =± 8;

hence z 2,3 (ǫ)

There are, however, equations with a double root, where the perturbed pair of roots do not behave like ±c 1/2 0 ǫ as ǫ → 0. In such cases the balancing procedure may help.

Consider the equation

(3.2.32) The reduced problem is z 2 = 0, with a double root. Try z ∼ Aǫ α , α > 0. The exponents

2 1 + ǫ)z 2 + 4ǫz + ǫ = 0.

of the three terms become 2α, α + 1, 2. We see that α = 1 makes the three exponents all equal to 2; this is fine. So, set z = ǫy. The equation reads, after division by ǫ 2

3. Coefficient matching yields the result z

1 + ǫ)y 2 + 4y + 1 = 0, hence y(0) = a ≡ −2 ±

2 = ǫy = aǫ + (−a 2 /( 2(a + 2)))ǫ +···,

where all exponents are natural numbers. If ǫ is small enough, the last term included can serve as an error estimate. A more

reliable error estimate (or even an error bound) can be obtained by inserting the truncated expansion into the equation. It shows that the truncated expansion satisfies a modified equation exactly . The same idea can be applied to equations of many other types; see Problem 3.2.9.

206Chapter 3. Series, Operators, and Continued Fractions