Chapter 1
Introduction
In this chapter, we describe the aims of perturbation theory in general terms, and
give some simple illustrative examples of perturbation problems. Some texts and
references on perturbation theory are [6], [7], and [10].
1.1
Perturbation theory
Consider a problem
P ε (x) = 0
(1.1)
depending on a small, real-valued parameter ε that simplifies in some way when
ε = 0 (for example, it is linear or exactly solvable). The aim of perturbation theory
is to determine the behavior of the solution x = xε of (1.1) as ε → 0. The use of a
small parameter here is simply for definiteness; for example, a problem depending
on a large parameter ω can be rewritten as one depending on a small parameter
ε = 1/ω.
The focus of these notes is on perturbation problems involving differential equations, but perturbation theory and asymptotic analysis apply to a broad class of
problems. In some cases, we may have an explicit expression for xε , such as an
integral representation, and want to obtain its behavior in the limit ε → 0.
1.1.1
Asymptotic solutions
The first goal of perturbation theory is to construct a formal asymptotic solution of
(1.1) that satisfies the equation up to a small error. For example, for each N ∈ N,
we may be able to find an asymptotic solution xεN such that
P ε (xεN ) = O(εN +1 ),
where O(εn ) denotes a term of the the order εn . This notation will be made precise
in Chapter 2.
1
2
Introduction
Once we have constructed such an asymptotic solution, we would like to know
that there is an exact solution x = xε of (1.1) that is close to the asymptotic solution
when ε is small; for example, a solution such that
xε = xεN + O(εN +1 ).
This is the case if a small error in the equation leads to a small error in the solution.
For example, we can establish such a result if we have a stability estimate of the
form
|x − y| ≤ C |P ε (x) − P ε (y)|
where C is a constant independent of ε, and |·| denotes appropriate norms. Such an
estimate depends on the properties of P ε and may be difficult to obtain, especially
for nonlinear problems. In these notes we will focus on methods for the construction
of asymptotic solutions, and we will not discuss in detail the existence of solutions
close to the asymptotic solution.
1.1.2
Regular and singular perturbation problems
It is useful to make an imprecise distinction between regular perturbation problems
and singular perturbation problems. A regular perturbation problem is one for which
the perturbed problem for small, nonzero values of ε is qualitatively the same as
the unperturbed problem for ε = 0. One typically obtains a convergent expansion
of the solution with respect to ε, consisting of the unperturbed solution and higherorder corrections. A singular perturbation problem is one for which the perturbed
problem is qualitatively different from the unperturbed problem. One typically
obtains an asymptotic, but possibly divergent, expansion of the solution, which
depends singularly on the parameter ε.
Although singular perturbation problems may appear atypical, they are the most
interesting problems to study because they allow one to understand qualitatively
new phenomena.
The solutions of singular perturbation problems involving differential equations
often depend on several widely different length or time scales. Such problems can
be divided into two broad classes: layer problems, treated using the method of
matched asymptotic expansions (MMAE); and multiple-scale problems, treated by
the method of multiple scales (MMS). Prandtl’s boundary layer theory for the highReynolds flow of a viscous fluid over a solid body is an example of a boundary layer
problem, and the semi-classical limit of quantum mechanics is an example of a
multiple-scale problem.
We will begin by illustrating some basic issues in perturbation theory with simple
algebraic equations.
Algebraic equations
1.2
3
Algebraic equations
The first two examples illustrate the distinction between regular and singular perturbation problems.
Example 1.1 Consider the cubic equation
x3 − x + ε = 0.
(1.2)
We look for a solution of the form
x = x0 + εx1 + ε2 x2 + O(ε3 ).
(1.3)
Using this expansion in the equation, expanding, and equating coefficients of εn to
zero, we get
x30 − x0 = 0,
3x20 x1 − x1 + 1 = 0,
3x0 x2 − x2 + 3x0 x21 = 0.
Note that we obtain a nonlinear equation for the leading order solution x0 , and
nonhomogeneous linearized equations for the higher order corrections x1 , x2 ,. . . .
This structure is typical of many perturbation problems.
Solving the leading-order perturbation equation, we obtain the three roots
x0 = 0, ±1.
Solving the first-order perturbation equation, we find that
x1 =
1
.
1 − 3x20
The corresponding solutions are
x = ε + O(ε2 ),
1
x = ±1 − ε + O(ε2 ).
2
Continuing in this way, we can obtain a convergent power series expansion about
ε = 0 for each of the three distinct roots of (1.2). This result is typical of regular
perturbation problems.
An alternative — but equivalent — method to obtain the perturbation series is
to use the Taylor expansion
x(ε) = x(0) + ẋ(0)ε +
1
ẍ(0)ε2 + . . . ,
2!
where the dot denotes a derivative with respect to ε. To compute the coefficients,
we repeatedly differentiate the equation with respect to ε and set ε = 0 in the result.
4
Introduction
For example, setting ε = 0 in (1.2), and solving the resulting equation for x(0), we
get x(0) = 0, ±1. Differentiating (1.2) with respect to ε, we get
3x2 ẋ − ẋ + 1 = 0.
Setting ε = 0 and solving for ẋ(0), we get the same answer as before.
Example 1.2 Consider the cubic equation
εx3 − x + 1 = 0.
(1.4)
Using (1.3) in (1.4), expanding, and equating coefficents of εn to zero, we get
−x0 + 1 = 0,
−x1 + x30 = 0,
−x2 + 3x20 x1 = 0.
Solving these equations, we find that x0 = 1, x1 = 1, . . . , and hence
x(ε) = 1 + ε + O(ε2 ).
(1.5)
We only obtain one solution because the cubic equation (1.4) degenerates to a linear
equation at ε = 0. We missed the other two solutions because they approach infinity
as ε → 0. A change in the qualitative nature of the problem at the unperturbed
value ε = 0 is typical of singular perturbation problems.
To find the other solutions, we introduce a rescaled variable y, where
x(ε) =
1
y(ε),
δ(ε)
and y = O(1) as ε → 0. The scaling factor δ is to be determined. Using this
equation in (1.4), we find that
ε 3 1
y − y + 1 = 0.
δ3
δ
(1.6)
In order to obtain a nontrivial solution, we require that at least two leading-order
terms in this equation have the same order of magnitude. This is called the principle
of dominant balance.
Balancing the first two terms, we find that∗
ε
1
= ,
δ3
δ
which implies that δ = ε1/2 . The first two terms in (1.4) are then O(ε−1/2 ), and the
third term is O(1), which is smaller. With this choice of δ, equation (1.6) becomes
y 3 − y + ε1/2 = 0.
∗ Nonzero
constant factors can be absorbed into y.
Algebraic equations
5
Solving this equation in the same way as (1.2), we get the nonzero solutions
1
y = ±1 − ε1/2 + O(ε).
2
The corresponding solutions for x are
1
1
x = ± 1/2 − + O ε1/2 .
2
ε
The dominant balance argument illustrated here is useful in many perturbation
problems. The corresponding limit, ε → 0 with x(ε) = O(ε−1/2 ), is called a distinguished limit.
There are two other two-term balances in (1.6). Balancing the second and third
terms, we find that
1
=1
δ
or δ = 1. The first term is then O(ε), so it is smaller than the other two terms. This
dominant balance gives the solution in (1.5). Balancing the first and third terms,
we find that
ε
= 1,
δ3
or δ = ε1/3 . In this case, the first and third terms are O(1), but the second term
is O(ε−1/3 ). Thus, it is larger than the terms that balance, so we do not obtain a
dominant balance or any new solutions.
In this example, no three-term dominant balance is possible as ε → 0, but this
can occur in other problems.
Example 1.3 A famous example of the effect of a perturbation on the solutions of
a polynomial is Wilkinson’s polynomial (1964),
(x − 1)(x − 2) . . . (x − 20) = εx19 .
The perturbation has a large effect on the roots even for small values of ε.
The next two examples illustrate some other features of perturbation theory.
Example 1.4 Consider the quadratic equation
(1 − ε)x2 − 2x + 1 = 0.
Suppose we look for a straightforward power series expansion of the form
x = x0 + εx1 + O(ε2 ).
We find that
x20 − 2x0 + 1 = 0,
2(x0 − 1)x1 = x20 .
6
Introduction
Solving the first equation, we get x0 = 1. The second equation then becomes 0 = 1.
It follows that there is no solution of the assumed form.
This difficulty arises because x = 1 is a repeated root of the unperturbed problem. As a result, the solution
x=
1 ± ε1/2
1−ε
does not have a power series expansion in ε, but depends on ε1/2 . An expansion
x = x0 + ε1/2 x1 + εx2 + O(ε3/2 )
leads to the equations x0 = 1, x21 = 1, or
x = 1 ± ε1/2 + O(ε)
in agreement with the exact solution.
Example 1.5 Consider the transcendental equation
xe−x = ε.
(1.7)
As ε → 0+ , there are two possibilities:
(a) x → 0, which implies that x = ε + ε2 + O(ε2 );
(b) e−x → 0, when x → ∞.
In the second case, x must be close to log 1/ε.
To obtain an asymptotic expansion for the solution, we solve the equation iteratively using the idea that e−x varies much more rapidly than x as x → 0. Rewriting
(1.7) as e−x = ε/x and taking logarithms, we get the equivalent equation
1
x = log x + log .
ε
Thus solutions are fixed points of the function
1
f (x) = log x + log .
ε
We then define iterates xn , n ∈ N, by
1
xn+1 = log xn + log ,
ε
1
x1 = log .
ε
Defining
1
L1 = log ,
ε
1
L2 = log log
ε
,
Eigenvalue problems
7
we find that
x2
= L1 + L2 ,
x3
= L1 + log(L1 + L2 )
L2
+O
= L1 + L2 +
L1
L2
L1
2 !
.
At higher orders, terms involving
1
L3 = log log log
ε
,
and so on, appear.
The form of this expansion would be difficult to guess without using an iterative
method. Note, however, that the successive terms in this asymptotic expansion
converge very slowly as ε → 0. For example, although L2 /L1 → 0 as ε → 0, when
ε = 0.1, L1 ≈ 36, L2 ≈ 12; and when ε = 10−5 , L1 ≈ 19, L2 ≈ 1.
1.3
Eigenvalue problems
Spectral perturbation theory studies how the spectrum of an operator is perturbed
when the operator is perturbed. In general, this question is a difficult one, and
subtle phenomena may occur, especially in connection with the behavior of the
continuous spectrum of the operators. Here, we consider the simplest case of the
perturbation in an eigenvalue.
Let H be a Hilbert space with inner product h·, ·i, and Aε : D(Aε ) ⊂ H → H a
linear operator in H, with domain D(Aε ), depending smoothly on a real parameter
ε. We assume that:
(a) Aε is self-adjoint, so that
hx, Aε yi = hAε x, yi
for all x, y ∈ D(Aε );
(b) Aε has a smooth branch of simple eigenvalues λε ∈ R with eigenvectors
xε ∈ H, meaning that
Aε xε = λε xε .
(1.8)
We will compute the perturbation in the eigenvalue from its value at ε = 0 when ε
is small but nonzero.
A concrete example is the perturbation in the eigenvalues of a symmetric matrix.
In that case, we have H = Rn with the Euclidean inner product
hx, yi = xT y,
8
Introduction
and Aε : Rn → Rn is a linear transformation with an n × n symmetric matrix (aεij ).
The perturbation in the eigenvalues of a Hermitian matrix corresponds to H = Cn
with inner product hx, yi = xT y. As we illustrate below with the Schrödinger
equation of quantum mechanics, spectral problems for differential equations can be
formulated in terms of unbounded operators acting in infinite-dimensional Hilbert
spaces.
We use the expansions
Aε
= A0 + εA1 + . . . + εn An + . . . ,
xε
= x0 + εx1 + . . . + εn xn + . . . ,
λε
= λ0 + ελ1 + . . . + εn λn + . . .
in the eigenvalue problem (1.8), equate coefficients of εn , and rearrange the result.
We find that
(A0 − λ0 I) x0 = 0,
(A0 − λ0 I) x1 = −A1 x0 + λ1 x0 ,
n
X
{−Ai xn−i + λi xn−i } .
(A0 − λ0 I) xn =
(1.9)
(1.10)
(1.11)
i=1
Assuming that x0 6= 0, equation (1.9) implies that λ0 is an eigenvalue of A0
and x0 is an eigenvector. Equation (1.10) is then a singular equation for x1 . The
following proposition gives a simple, but fundamental, solvability condition for this
equation.
Proposition 1.6 Suppose that A is a self-adjoint operator acting in a Hilbert space
H and λ ∈ R. If z ∈ H, a necessary condition for the existence of a solution y ∈ H
of the equation
(A − λI) y = z
(1.12)
is that
hx, zi = 0,
for every eigenvector x of A with eigenvalue λ.
Proof. Suppose z ∈ H and y is a solution of (1.12). If x is an eigenvector of A,
then using (1.12) and the self-adjointness of A − λI, we find that
hx, zi = hx, (A − λI) yi
= h(A − λI) x, yi
=
0.
Eigenvalue problems
9
In many cases, the necesary solvability condition in this proposition is also sufficient, and then we say that A − λI satisfies the Fredholm alternative; for example,
this is true in the finite-dimensional case, or when A is an elliptic partial differential
operator.
Since A0 is self-adjoint and λ0 is a simple eigenvalue with eigenvector x0 , equation
(1.12) it is solvable for x1 only if the right hand side is orthogonal to x0 , which implies that
λ1 =
hx0 , A1 x0 i
.
hx0 , x0 i
This equation gives the leading order perturbation in the eigenvalue, and is the
most important result of the expansion.
Assuming that the necessary solvability condition in the proposition is sufficient,
we can then solve (1.10) for x1 . A solution for x1 is not unique, since we can add
to it an arbitrary scalar multiple of x0 . This nonuniqueness is a consequence of the
fact that if xε is an eigenvector of Aε , then cε xε is also a solution for any scalar cε .
If
cε = 1 + εc1 + O(ε2 )
then
cε xε = x0 + ε (x1 + c1 x0 ) + O(ε2 ).
Thus, the addition of c1 x0 to x1 corresponds to a rescaling of the eigenvector by a
factor that is close to one.
This expansion can be continued to any order. The solvability condition for
(1.11) determines λn , and the equation may then be solved for xn , up to an arbitrary
vector cn x0 . The appearance of singular problems, and the need to impose solvabilty
conditions at each order which determine parameters in the expansion and allow for
the solution of higher order corrections, is a typical structure of many pertubation
problems.
1.3.1
Quantum mechanics
One application of this expansion is in quantum mechanics, where it can be used
to compute the change in the energy levels of a system caused by a perturbation in
its Hamiltonian.
The Schrödinger equation of quantum mechanics is
i~ψt = Hψ.
Here t denotes time and ~ is Planck’s constant. The wavefunction ψ(t) takes values
in a Hilbert space H, and H is a self-adjoint linear operator acting in H with the
dimensions of energy, called the Hamiltonian.
10
Introduction
Energy eigenstates are wavefunctions of the form
ψ(t) = e−iEt/~ ϕ,
where ϕ ∈ H and E ∈ R. It follows from the Schrödinger equation that
Hϕ = Eϕ.
Hence E is an eigenvalue of H and ϕ is an eigenvector. One of Schrödinger’s
motivations for introducing his equation was that eigenvalue problems led to the
experimentally observed discrete energy levels of atoms.
Now suppose that the Hamiltonian
H ε = H0 + εH1 + O(ε2 )
depends smoothly on a parameter ε. Then, rewriting the previous result, we find
that the corresponding simple energy eigenvalues (assuming they exist) have the
expansion
E ε = E0 + ε
hϕ0 , H1 ϕ0 i
+ O(ε2 )
hϕ0 , ϕ0 i
where ϕ0 is an eigenvector of H0 .
For example, the Schrödinger equation that describes a particle of mass m moving in Rd under the influence of a conservative force field with potential V : Rd → R
is
i~ψt = −
~2
∆ψ + V ψ.
2m
Here, the wavefunction ψ(x, t) is a function of a space variable x ∈ Rd and time
t ∈ R. At fixed time t, we have ψ(·, t) ∈ L2 (Rd ), where
R
L2 (Rd ) = u : Rd → C | u is measurable and Rd |u|2 dx < ∞
is the Hilbert space of square-integrable functions with inner-product
Z
u(x)v(x) dx.
hu, vi =
Rd
The Hamiltonian operator H : D(H) ⊂ H → H, with domain D(H), is given by
H=−
~2
∆ + V.
2m
If u, v are smooth functions that decay sufficiently rapidly at infinity, then Green’s
theorem implies that
Z
~2
hu, Hvi =
u −
∆v + V v dx
2m
Rd
Z
~2
~2
∇ · (v∇u − u∇v) −
(∆u)v + V uv dx
=
2m
2m
Rd
Eigenvalue problems
11
~2
−
∆u + V u v dx
=
2m
Rd
= hHu, vi.
Z
Thus, this operator is formally self-adjoint. Under suitable conditions on the potential V , the operator can be shown to be self-adjoint with respect to an appropriately
chosen domain.
Now suppose that the potential V ε depends on a parameter ε, and has the
expansion
V ε (x) = V0 (x) + εV1 (x) + O(ε2 ).
The perturbation in a simple energy eigenvalue
E ε = E0 + εE1 + O(ε2 ),
assuming one exists, is given by
R
E1 =
V (x)|ϕ0 (x)|2 dx
RdR 1
,
|ϕ0 (x)|2 dx
Rd
where ϕ0 ∈ L2 (Rd ) is an unperturbed energy eigenfunction that satisfies
−
~2
∆ϕ0 + V0 ϕ0 = E0 ϕ0 .
2m
Example 1.7 The one-dimensional simple harmonic oscillator has potential
V0 (x) =
1 2
kx .
2
The eigenvalue problem
~2 00 1 2
ϕ + kx ϕ = Eϕ,
ϕ ∈ L2 (R)
2m
2
is exactly solvable. The energy eigenvalues are
1
En = ~ω n +
n = 0, 1, 2, . . . ,
2
−
where
r
ω=
k
m
is the frequency of the corresponding classical oscillator. The eigenfunctions are
2
ϕn (x) = Hn (αx)e−α
x2 /2
,
where Hn is the nth Hermite polynomial,
Hn (ξ) = (−1)n eξ
2
dn −ξ2
e ,
dξ n
12
Introduction
and the constant α, with dimensions of 1/length, is given by
√
mk
2
α =
.
~
The energy levels Enε of a slightly anharmonic oscillator with potential
V ε (x) =
1 2
k
kx + ε 2 W (αx) + O(ε2 )
2
α
where ε > 0 have the asymptotic behavior
1
Enε = ~ω n + + ε∆n + O(ε2 )
2
as ε → 0+
as ε → 0+ ,
where
R
∆n =
2
W (ξ)Hn2 (ξ)e−ξ dξ
R
.
Hn2 (ξ)e−ξ2 dξ
For an extensive and rigorous discussion of spectral perturbation theory for
linear operators, see [9].
1.4
Nondimensionalization
The numerical value of any quantity in a mathematical model is measured with
respect to a system of units (for example, meters in a mechanical model, or dollars
in a financial model). The units used to measure a quantity are arbitrary, and a
change in the system of units (for example, to feet or yen, at a fixed exchange rate)
cannot change the predictions of the model. A change in units leads to a rescaling
of the quantities. Thus, the independence of the model from the system of units
corresponds to a scaling invariance of the model. In cases when the zero point of a
unit is arbitrary, we also obtain a translational invariance, but we will not consider
translational invariances here.
Suppose that a model involves quantities (a1 , a2 , . . . , an ), which may include dependent and independent variables as well as parameters. We denote the dimension
of a quantity a by [a]. A fundamental system of units is a minimal set of independent units, which we denote symbolically by (d1 , d2 , . . . , dr ). Different fundamental
system of units can be used, but given a fundamental system of units any other derived unit may be constructed uniquely as a product of powers of the fundamental
units, so that
αr
1 α2
[a] = dα
1 d2 . . . d r
(1.13)
for suitable exponents (α1 , α2 , . . . , αr ).
Example 1.8 In mechanical problems, a fundamental set of units is d1 = mass,
d2 = length, d3 = time, or d1 = M , d2 = L, d3 = T , for short. Then velocity
Nondimensionalization
13
V = L/T and momentum P = M L/T are derived units. We could use instead
momentum P , velocity V , and time T as a fundamental system of units, when mass
M = P/V and length L = V T are derived units. In problems involving heat flow, we
may introduce temperature (measured, for example, in degrees Kelvin) as another
fundamental unit, and in problems involving electromagnetism, we may introduce
current (measured, for example, in Ampères) as another fundamental unit.
The invariance of a model under the change in units dj 7→ λj dj implies that it
is invariant under the scaling transformation
α
α
r,i
ai → λ1 1,i λ2 2,i . . . λα
ai
r
i = 1, . . . , n
for any λ1 , . . . λr > 0, where
α
α
r,i
[ai ] = d1 1,i d2 2,i . . . dα
.
r
(1.14)
Thus, if
a = f (a1 , . . . , an )
is any relation between quantities in the model with the dimensions in (1.13) and
(1.14), then f has the scaling property that
α
α
α1,1 α2,1
αr
1 α2
r,n
r,1
an .
a1 , . . . , λ1 1,n λ2 2,n . . . λα
λ2 . . . λ α
λα
r
r
1 λ2 . . . λr f (a1 , . . . , an ) = f λ1
A particular consequence of the invariance of a model under a change of units is
that any two quantities which are equal must have the same dimensions. This fact
is often useful in finding the dimension of some quantity.
Example 1.9 According to Newton’s second law,
force = rate of change of momentum with respect to time.
Thus, if F denotes the dimension of force and P the dimension of momentum,
then F = P/T . Since P = M V = M L/T , we conclude that F = M L/T 2 (or
mass × acceleration).
Example 1.10 In fluid mechanics, the shear viscosity µ of a Newtonian fluid is the
constant of proportionality that relates the viscous stress tensor T to the velocity
gradient ∇u:
T =
1
µ ∇u + ∇uT .
2
Stress has dimensions of force/area, so
[T ] =
ML 1
M
=
.
T 2 L2
LT 2
14
Introduction
The velocity gradient ∇u has dimensions of velocity/length, so
[∇u] =
L1
1
= .
TL
T
Equating dimensions, we find that
[µ] =
M
.
LT
We can also write [µ] = (M/L3 )(L2 /T ). It follows that if ρ0 is the density of the
fluid, and µ = ρ0 ν, then
[ν] =
L2
.
T
Thus ν, which is called the kinematical viscosity, has the dimensions of diffusivity.
Physically it is the diffusivity of momentum. For example, in time T√
, viscous effects
lead to the diffusion of momentum over a length scale of the order νT .
Scaling invariance implies that we can reduce the number of quantities appearing in the problem by the introduction of dimensionless variables. Suppose that
(a1 , . . . , ar ) are a set of (nonzero) quantities whose dimensions form a fundamental
system of units. We denote the remaining quantities in the model by (b1 , . . . , bm ),
where r + m = n. Then for suitable exponents (β1,i , . . . , βr,i ), the quantity
Πi =
bi
β
a1 1,i
β
. . . ar r,i
is dimensionless, meaning that it is invariant under the scaling transformations
induced by changes in units. Such dimensionless quantities can often be interpreted
as the ratio of two quantities of the same dimension appearing in the problem (such
as a ratio of lengths, times, diffusivities, and so on). Perturbation methods are
typically applicable when one or more of these dimensionless quantities is small or
large.
Any relationship of the form
b = f (a1 , . . . , ar , b1 , . . . , bm )
is equivalent to a relation
Π = f (1, . . . , 1, Π1 , . . . , Πm ).
Thus, the introduction of dimensionless quantities reduces the number of variables
in the problem by the number of fundamental units involved in the problem. In
many cases, nondimensionalization leads to a reduction in the number of parameters
in the problem to a minimal number of dimensionless parameters. In some cases,
one may be able to use dimensional arguments to obtain the form of self-similar
solutions.
Nondimensionalization
15
Example 1.11 Consider the following IVP for the Green’s function of the heat
equation in Rd :
ut = ν∆u,
u(x, 0) = Eδ(x).
Here δ is the delta-function. The dimensioned parameters in this problem are the
diffusivity ν and the energy E of the point source. The only length and times
scales are those that come from the independent variables (x, t), so the solution is
self-similar.
We have [u] = θ, where θ denotes temperature, and, since
Z
u(x, 0) dx = E,
Rd
we have [E] = θLd . Dimensional analysis and the rotational invariance of the
Laplacian ∆ imply that
E
|x|
.
u(x, t) =
f √
(νt)d/2
νt
Using this expression for u(x, t) in the PDE, we get an ODE for f (ξ),
ξ
d−1
d
00
f +
+
f 0 + f = 0.
2
ξ
2
We can rewrite this equation as a first-order ODE for f 0 + 2ξ f ,
0
d−1
ξ
ξ
0
0
f + f = 0.
f + f +
2
ξ
2
Solving this equation, we get
ξ
b
f 0 + f = d−1 ,
2
ξ
where b is a constant of integration. Solving for f , we get
Z −ξ2
2
2
e
f (ξ) = ae−ξ /4 + be−ξ /4
dξ,
ξ d−1
where a s another constant of integration. In order for f to be integrable, we must
set b = 0. Then
aE
|x|2
u(x, t) =
exp −
.
4νt
(νt)d/2
Imposing the requirement that
Z
u(x, t) dx = E,
Rd
16
Introduction
and using the standard integral
Z
|x|2
d/2
dx = (2πc) ,
exp −
2c
Rd
we find that a = (4π)−d/2 , and
u(x, t) =
E
|x|2
exp
−
.
4νt
(4πνt)d/2
Example 1.12 Consider a sphere of radius L moving through a fluid with constant
speed U . A primary quantity of interest is the total drag force D exerted by the
fluid on the sphere. We assume that the fluid is incompressible, which is a good
approximation if the flow speed U is much less than the speed of sound in the
fluid. The fluid properties are then determined by the density ρ0 and the kinematic
viscosity ν. Hence,
D = f (U, L, ρ0 , ν).
Since the drag D has the dimensions of force (M L/T 2 ), dimensional analysis implies
that
UL
2 2
.
D = ρ0 U L F
ν
Thus, the dimensionless drag
D
= F (Re)
ρ0 U 2 L2
is a function of the Reynold’s number
Re =
UL
.
ν
The function F has a complicated dependence on Re that is difficult to compute
explicitly. For example, F changes rapidly near Reynolds numbers for which the
flow past the sphere becomes turbulent. Nevertheless, experimental measurements
agree very well with the result of this dimensionless analysis (see Figure 1.9 in [1],
for example).
Dimensional analysis leads to continuous scaling symmetries. These scaling symmetries are not the only continuous symmetries possessed by differential equations.
The theory of Lie groups and Lie algebras provides a systematic method for computing all continuous symmetries of a given differential equation [13]. Lie originally
introduced the notions of Lie groups and Lie algebras precisely for this purpose.
Example 1.13 The full group of symmetries of the one-dimensional heat equation
ut = uxx
Nondimensionalization
17
is generated by the following transformations [13]:
u(x, t) 7→ u(x − α, t),
u(x, t) 7→ u(x, t − β),
u(x, t) 7→ γu(x, t),
u(x, t) 7→ u(δx, δ 2 t),
2
u(x, t) 7→ e−εx+ε t u(x − 2εt, t),
x
t
−ηx2
1
u
,
,
exp
u(x, t) 7→ √
1 + 4ηt
1 + 4ηt 1 + 4ηt
1 + 4ηt
u(x, t) 7→ u(x, t) + v(x, t),
where (α, . . . , η) are constants, and v(x, t) is an arbitrary solution of the heat
equation. The scaling symmetries involving γ and δ can be deduced by dimensional arguments, but the symmetries involving ε and η cannot.
For further discussion of dimensional analysis and self-similar solutions, see [1].
18
Chapter 2
Asymptotic Expansions
In this chapter, we define the order notation and asymptotic expansions. For additional discussion, see [3], and [12].
2.1
Order notation
The O and o order notation provides a precise mathematical formulation of ideas
that correspond — roughly — to the ‘same order of magnitude’ and ‘smaller order
of magnitude.’ We state the definitions for the asymptotic behavior of a real-valued
function f (x) as x → 0, where x is a real parameter. With obvious modifications,
similar definitions apply to asymptotic behavior in the limits x → 0+ , x → x0 ,
x → ∞, to complex or integer parameters, and other cases. Also, by replacing | · |
with a norm, we can define similiar concepts for functions taking values in a normed
linear space.
Definition 2.1 Let f, g : R \ 0 → R be real functions. We say f = O(g) as x → 0
if there are constants C and r > 0 such that
|f (x)| ≤ C|g(x)| whenever 0 < |x| < r.
We say f = o(g) as x → 0 if for every δ > 0 there is an r > 0 such that
|f (x)| ≤ δ|g(x)| whenever 0 < |x| < r.
If g 6= 0, then f = O(g) as x → 0 if and only if f /g is bounded in a (punctured)
neighborhood of 0, and f = o(g) if and only if f /g → 0 as x → 0.
We also write f g, or f is ‘much less than’ g, if f = o(g), and f ∼ g, or f is
asymptotic to g, if f /g → 1.
Example 2.2 A few simple examples are:
(a) sin 1/x = O(1) as x → 0
19
20
Asymptotic Expansions
(b) it is not true that 1 = O(sin 1/x) as x → 0, because sin 1/x vanishes is
every neighborhood of x = 0;
(c) x3 = o(x2 ) as x → 0, and x2 = o(x3 ) as x → ∞;
(d) x = o(log x) as x → 0+ , and log x = o(x) as x → ∞;
(e) sin x ∼ x as x → 0;
(f) e−1/x = o(xn ) as x → 0+ for any n ∈ N.
The o and O notations are not quantitative without estimates for the constants
C, δ, and r appearing in the definitions.
2.2
Asymptotic expansions
An asymptotic expansion describes the asymptotic behavior of a function in terms
of a sequence of gauge functions. The definition was introduced by Poincaré (1886),
and it provides a solid mathematical foundation for the use of many divergent series.
Definition 2.3 A sequence of functions ϕn : R \ 0 → R, where n = 0, 1, 2, . . ., is
an asymptotic sequence as x → 0 if for each n = 0, 1, 2, . . . we have
ϕn+1 = o (ϕn )
as x → 0.
We call ϕn a gauge function. If {ϕn } is an asymptotic sequence and f : R \ 0 → R
is a function, we write
f (x) ∼
∞
X
an ϕn (x)
as x → 0
(2.1)
n=0
if for each N = 0, 1, 2, . . . we have
f (x) −
N
X
an ϕn (x) = o(ϕN )
as x → 0.
n=0
We call (2.1) the asymptotic expansion of f with respect to {ϕn } as x → 0.
Example 2.4 The functions ϕn (x) = xn form an asymptotic sequence as x → 0+ .
Asymptotic expansions with respect to this sequence are called asymptotic power
series, and they are discussed further below. The functions ϕn (x) = x−n form an
asymptotic sequence as x → ∞.
Example 2.5 The function log sin x has an asymptotic expansion as x → 0+ with
respect to the asymptotic sequence {log x, x2 , x4 , . . .}:
1
log sin x ∼ log x + x2 + . . .
6
as x → 0+ .
Asymptotic expansions
21
If, as is usually the case, the gauge functions ϕn do not vanish in a punctured
neighborhood of 0, then it follows from Definition 2.1 that
PN
f (x) − n=0 an ϕn (x)
aN +1 = lim
.
x→0
ϕN +1
Thus, if a function has an expansion with respect to a given sequence of gauge functions, the expansion is unique. Different functions may have the same asymptotic
expansion.
Example 2.6 For any constant c ∈ R, we have
1
+ ce−1/x ∼ 1 + x + x2 + . . . + xn + . . .
1−x
as x → 0+ ,
since e−1/x = o(xn ) as x → 0+ for every n ∈ N.
Asymptotic expansions can be added, and — under natural conditions on the
gauge functions — multiplied. The term-by-term integration of asymptotic expansions is valid, but differentiation may not be, because small, highly-oscillatory terms
can become large when they are differentiated.
Example 2.7 Let
f (x) =
1
+ e−1/x sin e1/x .
1−x
Then
f (x) ∼ 1 + x + x2 + x3 + . . .
as x → 0+ ,
but
f 0 (x) ∼ −
cos e1/x
+ 1 + 2x + 3x2 + . . .
x2
as x → 0+ .
Term-by-term differentiation is valid under suitable assumptions that rule out
the presence of small, highly oscillatory terms. For example, a convergent power
series expansion of an analytic function can be differentiated term-by-term
2.2.1
Asymptotic power series
Asymptotic power series,
f (x) ∼
∞
X
an xn
as x → 0,
n=0
are among the most common and useful asymptotic expansions.
22
Asymptotic Expansions
If f is a smooth (C ∞ ) function in a neighborhood of the origin, then Taylor’s
theorem implies that
N
X
f (n) (0) n when |x| ≤ r,
x ≤ CN +1 xN +1
f (x) −
n!
n=0
where
CN +1 = sup
|x|≤r
(N +1) f
(x)
(N + 1)!
.
It follows that f has the asymptotic power series expansion
f (x) ∼
∞
X
f (n) (0) n
x
n!
n=0
as x → 0.
(2.2)
The asymptotic power series in (2.2) converges to f in a neighborhood of the origin
if and only if f is analytic at x = 0. If f is C ∞ but not analytic, the series may
converge to a function different from f (see Example 2.6 with c 6= 0) or it may
diverge (see (2.4) or (3.3) below).
The Taylor series of f (x) at x = x0 ,
∞
X
f (n) (x0 )
(x − x0 )n ,
n!
n=0
does not provide an asymptotic expansion of f (x) as x → x1 when x1 6= x0 even if
it converges. The partial sums therefore do not generally provide a good approximation of f (x) as x → x1 . (See Example 2.9, where a partial sum of the Taylor
series of the error function at x = 0 provides a poor approximation of the function
when x is large.)
The following (rather surprising) theorem shows that there are no restrictions
on the growth rate of the coefficients in an asymptotic power series, unlike the case
of convergent power series.
Theorem 2.8 (Borel-Ritt) Given any sequence {an } of real (or complex) coefficients, there exists a C ∞ -function f : R → R (or f : R → C) such that
X
f (x) ∼
an xn
as x → 0.
Proof.
Let η : R → R be a C ∞ -function such that
1 if |x| ≤ 1,
η(x) =
0 if |x| ≥ 2.
We choose a sequence of positive numbers {δn } such that δn → 0 as n → ∞ and
n
x ≤ 1,
(2.3)
|an | x η
δn C n
2n
Asymptotic expansions
23
where
kf kC n = sup
x∈R
n X
(k) f (x)
k=0
n
denotes the C -norm. We define
f (x) =
∞
X
an xn η
n=0
x
δn
.
This series converges pointwise, since when x = 0 it is equal to a0 , and when x 6= 0
it consists of only finitely many terms. The condition in (2.3) implies that the
sequence converges in C n for every n ∈ N. Hence, the function f has continuous
derivatives of all orders.
2.2.2
Asymptotic versus convergent series
We have already observed that an asymptotic series need not be convergent, and a
convergent series need not be asymptotic. To explain the difference in more detail,
we consider a formal series
∞
X
an ϕn (x),
n=0
where {an } is a sequence of coefficients and {ϕn (x)} is an asymptotic sequence as
x → 0. We denote the partial sums by
SN (x) =
N
X
an ϕn (x).
n=0
Then convergence is concerned with the behavior of SN (x) as N → ∞ with x fixed,
whereas asymptoticity (at x = 0) is concerned with the behavior of SN (x) as x → 0
with N fixed.
A convergent series define a unique limiting sum, but convergence does not give
any indication of how rapidly the series it converges, nor of how well the sum of
a fixed number of terms approximates the limit. An asymptotic series does not
define a unique sum, nor does it provide an arbitrarily accurate approximation of
the value of a function it represents at any x 6= 0, but its partial sums provide good
approximations of these functions that when x is sufficiently small.
The following example illustrates the contrast between convergent and asymptotic series. We will examine another example of a divergent asymptotic series in
Section 3.1.
Example 2.9 The error function erf : R → R is defined by
Z x
2
2
e−t dt.
erf x = √
π 0
24
Asymptotic Expansions
2
Integrating the power series expansion of e−t term by term, we obtain the power
series expansion of erf x,
2
(−1)n
1 3
2n+1
erf x = √
x
+ ... ,
x − x + ... +
3
(2n + 1)n!
π
which is convergent for every x ∈ R. For large values of x, however, the convergence
is very slow. the Taylor series of the error function at x = 0. Instead, we can use
the following divergent asymptotic expansion, proved below, to obtain accurate
approximations of erf x for large x:
2 ∞
e−x X
(2n − 1)!! 1
erf x ∼ 1 − √
(−1)n+1
2n
xn+1
π n=0
as x → ∞,
(2.4)
where (2n − 1)!! = 1 · 3 · . . . · (2n − 1). For example, when x = 3, we need 31 terms
in the Taylor series at x = 0 to approximate erf 3 to an accuracy of 10−5 , whereas
we only need 2 terms in the asymptotic expansion.
Proposition 2.10 The expansion (2.4) is an asymptotic expansion of erf x.
Proof.
We write
2
erf x = 1 − √
π
Z
∞
2
e−t dt,
x
2
and make the change of variables s = t ,
1
erf x = 1 − √
π
Z
∞
s−1/2 e−s ds.
x2
For n = 0, 1, 2, . . ., we define
Z
∞
Fn (x) =
s−n−1/2 e−s ds.
x2
Then an integration by parts implies that
2
e−x
1
Fn (x) = 2n+1 − n +
Fn+1 (x).
x
2
By repeated use of this recursion relation, we find that
1
1 − √ F0 (x)
π
"
#
2
1 e−x
1
= 1− √
− F1 (x)
x
2
π
1
1
1
1·3
−x2
−
= 1− √ e
+ 2 F2 (x)
x 2x3
2
π
erf x =
Asymptotic expansions
=
25
2
1 · 3 · . . . · (2N − 1)
1
1
1
1 − √ e−x
− 3 + . . . (−1)N
x 2x
2N x2N +1
π
1 · 3 · . . . · (2N + 1)
+ (−1)N +1
F
(x)
.
N +1
2N +1
It follows that
2
N
e−x X
1 · 3 · . . . · (2n − 1)
erf x = 1 − √
(−1)n
+ RN +1 (x)
2n x2n+1
π n=0
where
1 1 · 3 · . . . · (2N + 1)
RN +1 (x) = (−1)N +1 √
FN +1 (x).
2N +1
π
Since
≤
∞
s−n−1/2 e−s ds
x2
Z ∞
1
e−s ds
x2n+1 x2
Z
|Fn (x)| = 2
≤
e−x
,
x2n+1
we have
2
|RN +1 (x)| ≤ CN +1
e−x
,
x2N +3
where
CN =
1 · 3 · . . . · (2N + 1)
√
.
2N +1 π
This proves the result.
2.2.3
Generalized asymptotic expansions
Sometimes it is useful to consider more general asymptotic expansions with respect
to a sequence of gauge functions {ϕn } of the form
f (x) ∼
∞
X
fn (x),
n=0
where for each N = 0, 1, 2, . . .
f (x) −
N
X
n=0
fn (x) = o(ϕN +1 ).
26
Asymptotic Expansions
For example, an expansion of the form
f (x) ∼
∞
X
an (x)ϕn (x)
n=0
in which the coefficients an are bounded functions of x is a generalized asymptotic
expansion. Such expansions provide additional flexibility, but they are not unique
and have to be used with care in many cases.
Example 2.11 We have
∞
X
1
∼
xn
1 − x n=0
as x → 0+ .
We also have
∞
X
1
∼
(1 + x)x2n
1 − x n=0
as x → 0+ .
This is a generalized asymptotic expansion with respect to {xn | n = 0, 1, 2, . . .}
that differs from the first one.
Example 2.12 According to [12], the following generalized asymptotic expansion
with respect to the asymptotic sequence {(log x)−n }
∞
sin x X n!e−(n+1)x/(2n)
∼
x
(log x)n
n=1
as x → ∞
is an example showing that “the definition admits expansions that have no conceivable value.”
Example 2.13 A physical example of a generalized asymptotic expansion arises in
the derivation of the Navier-Stokes equations of fluid mechanics from the Boltzmann
equations of kinetic theory by means of the Chapman-Enskog expansion. If λ is the
mean free path of the fluid molecules and L is a macroscopic length-scale of the
fluid flow, then the relevant small parameter is
ε=
λ
1.
L
The leading-order term in the Chapman-Enskog expansion satisfies the NavierStokes equations in which the fluid viscosity is of the order ε when nondimensionalized by the length and time scales characteristic of the fluid flow. Thus the leading
order solution depends on the perturbation parameter ε, and this expansion is a
generalized asymptotic expansion.
Stokes phenomenon
2.2.4
27
Nonuniform asymptotic expansions
In many problems, we seek an asymptotic expansion as ε → 0 of a function u(x, ε),
where x is an independent variable.∗ The asymptotic behavior of the function
with respect to ε may depend upon x, in which case we say that the expansion is
nonuniform.
Example 2.14 Consider the function u : [0, ∞) × (0, ∞) → R defined by
u(x, ε) =
1
.
x+ε
If x > 0, then
ε
ε2
1
1 − + 2 + ...
u(x, ε) ∼
x
x x
as ε → 0+ .
On the other hand, if x = 0, then
u(0, ε) ∼
1
ε
as ε → 0+ .
The transition between these two different expansions occurs when x = O(ε). In
the limit ε → 0+ with y = x/ε fixed, we have
1
1
u(εy, ε) ∼
as ε → 0+ .
ε y+1
This expansion matches with the other two expansions in the limits y → ∞ and
y → 0+ .
Nonuniform asymptotic expansions are not a mathematical pathology, and they
are often the crucial feature of singular perturbation problems. We will encounter
many such problems below; for example, in boundary layer problems the solution
has different asymptotic expansions inside and outside the boundary layer, and in
various problems involving oscillators nonuniformities aries for long times.
2.3
Stokes phenomenon
An asymptotic expansion as z → ∞ of a complex function f : C → C with an
essential singularity at z = ∞ is typically valid only in a wedge-shaped region
α < arg z < β, and the function has different asymptotic expansions in different
wedges.† The change in the form of the asymptotic expansion across the boundaries
of the wedges is called the Stokes phenomenon.
∗ We
† We
consider asymptotic expansions with respect to ε, not x.
consider a function with an essential singularity at z = ∞ for definiteness; the same phenomenon occurs for functions with an essential singularity at any z0 ∈ C.
28
Asymptotic Expansions
Example 2.15 Consider the function f : C → C defined by
2
f (z) = sinh z 2 =
2
ez − e−z
.
2
Let |z| → ∞ with arg z fixed, and define
Ω1 = {z ∈ C | −π/4 < arg z < π/4} ,
Ω2 = {z ∈ C | π/4 < arg z < 3π/4 or −3π/4 < arg z < −π/4} .
2
2
If z ∈ Ω1 , then Re z 2 > 0 and ez e−z , whereas if z ∈ Ω2 , then Re z 2 < 0 and
2
2
ez e−z . Hence
(
1 z2
e
as |z| → ∞ in Ω1 ,
f (z) ∼ 21 −z2
e
as
|z| → ∞ in Ω2 .
2
The lines arg z = ±π/4, ±3π/4 where the asymptotic expansion changes form are
2
2
called anti-Stokes lines. The terms ez and e−z switch from being dominant to
subdominant as z crosses an anti-Stokes lines. The lines arg z = 0, π, ±π/2 where
the subdominant term is as small as possible relative to the dominant term are
called Stokes lines.
This example concerns a simple explicit function, but a similar behavior occurs
for solutions of ODEs with essential singularities in the complex plane, such as error
functions, Airy functions, and Bessel functions.
Example 2.16 The error function can be extended to an entire function erf : C →
C with an essential singularity at z = ∞. It has the following asymptotic expansions
in different wedges:

√
π)
as z → ∞ with z ∈ Ω1 ,
 1 − exp(−z 2 )/(z √
erf z ∼ −1 − exp(−z 2 )/(z π) as z → ∞ with z ∈ Ω2 ,
√

as z → ∞ with z ∈ Ω3 .
− exp(−z 2 )/(z π)
where
Ω1 = {z ∈ C | −π/4 < arg z < π/4} ,
Ω2 = {z ∈ C | 3π/4 < arg z < 5π/4} ,
Ω3 = {z ∈ C | π/4 < arg z < 3π/4 or 5π/4 < arg z < 7π/4} .
Often one wants to determine the asymptotic behavior of such a function in one
wedge given its behavior in another wedge. This is called a connection problem
(see Section 3.5.1 for the case of the Airy function). The apparently discontinuous
change in the form of the asymptotic expansion of the solutions of an ODE across
an anti-Stokes line can be understood using exponential asymptotics as the result of
a continuous, but rapid, change in the coefficient of the subdominant terms across
the Stokes line [2].
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72
Chapter 5
Method of Multiple Scales: ODEs
The method of multiple scales is needed for problems in which the solutions depend
simultaneously on widely different scales. A typical example is the modulation of
an oscillatory solution over time-scales that are much greater than the period of
the oscillations. We will begin by describing the Poincaré-Lindstedt method, which
uses a ‘strained’ time coordinate to construct periodic solutions. We then describe
the method of multiple scales.
5.1
Periodic solutions and the Poincaré-Lindstedt expansion
We begin by constructing asymptotic expansions of periodic solutions of ODEs. The
first example, Duffing’s equation, is a Hamiltonian system with a family of periodic
solutions. The second example, van der Pol’s equation, has an isolated limit cycle.
5.1.1
Duffing’s equation
Consider an undamped nonlinear oscillator described by Duffing’s equation
y 00 + y + εy 3 = 0,
where the prime denotes a derivative with respect to time t. We look for solutions
y(t, ε) that satisfy the initial conditions
y 0 (0, ε) = 0.
y(0, ε) = 1,
We look for straightforward expansion of an asymptotic solution as ε → 0,
y(t, ε) = y0 (t) + εy1 (t) + O(ε2 ).
The leading-order perturbation equations are
y000 + y0 = 0,
y0 (0) = 1,
73
y00 (0) = 0,
74
Method of Multiple Scales: ODEs
with the solution
y0 (t) = cos t.
The next-order perturbation equations are
y100 + y1 + y03 = 0,
y1 (0) = 0,
y10 (0) = 0,
with the solution
y1 (t) =
3
1
[cos 3t − cos t] − t sin t.
32
8
This solution contains a secular term that grows linearly in t. As a result, the
expansion is not uniformly valid in t, and breaks down when t = O(ε) and εy1 is
no longer a small correction to y0 .
The solution is, in fact, a periodic function of t. The straightforward expansion
breaks down because it does not account for the dependence of the period of the
solution on ε. The following example illustrates the difficulty.
Example 5.1 We have the following Taylor expansion as ε → 0:
cos [(1 + ε)t] = cos t − εt sin t + O(ε2 ).
This asymptotic expansion is valid only when t ¿ 1/ε.
To construct a uniformly valid solution, we introduced a stretched time variable
τ = ω(ε)t,
and write y = y(τ, ε). We require that y is a 2π-periodic function of τ . The choice
of 2π here is for convenience; any other constant period — for example 1 — would
lead to the same asymptotic solution. The crucial point is that the period of y in τ
is independent of ε (unlike the period of y in t).
Since d/dt = ωd/dτ , the function y(τ, ε) satisfies
ω 2 y 00 + y + εy 3 = 0,
y(0, ε) = 1,
y 0 (0, ε) = 0,
y(τ + 2π, ε) = y(τ, ε),
where the prime denotes a derivative with respect to τ .
We look for an asymptotic expansion of the form
y(τ, ε) = y0 (τ ) + εy1 (τ ) + O(ε2 ),
ω(ε) = ω0 + εω1 + O(ε2 ).
Periodic solutions and the Poincaré-Lindstedt expansion
75
Using this expansion in the equation and equating coefficients of ε0 , we find that
ω02 y000 + y0 = 0,
y00 (0) = 0,
y0 (0) = 1,
y0 (τ + 2π) = y0 (τ ).
The solution is
y0 (τ ) = cos τ,
ω0 = 1.
After setting ω0 = 1, we find that the next order perturbation equations are
y100 + y1 + 2ω1 y000 + y03 = 0,
y1 (0) = 0,
y10 (0) = 0,
y1 (τ + 2π) = y1 (τ ).
Using the solution for y0 in the ODE for y1 , we get
y100 + y1
= 2ω1 cos τ − cos3 τ
µ
¶
3
1
=
2ω1 −
cos τ − cos 3τ.
4
4
We only have a periodic solution if
ω1 =
3
,
8
and then
y1 (t) =
1
[cos 3τ − cos τ ] .
32
It follows that
y = cos ωt +
1
ε [cos 3ωt − cos ωt] + O(ε2 ),
32
3
ω = 1 + ε + O(ε2 ).
8
This expansion can be continued to arbitrary orders in ε.
The appearance of secular terms in the expansion is a consequence of the nonsolvability of the perturbation equations for periodic solutions.
Proposition 5.2 Suppose that f : T → R is a smooth 2π-periodic function, where
T is the circle of length 2π. The ODE
y 00 + y = f
76
Method of Multiple Scales: ODEs
has a 2π-periodic solution if and only if
Z
f (t) cos t dt = 0,
Z
f (t) sin t dt = 0.
T
T
Proof. Let L2 (T) be the Hilbert space of 2π-periodic, real-valued functions with
inner product
Z
hy, zi =
y(t)z(t) dt.
T
We write the ODE as
Ay = f,
where
A=
Two integration by parts imply that
d2
+ 1.
dt2
Z
y (z 00 + z) dt
hy, Azi =
ZT
(y 00 + y) z dt
=
T
=
hAy, zi,
meaning that operator A is formally self-adjoint in L2 (T). Hence, it follows that if
Ay = f and Az = 0, then
hf, zi =
hAy, zi
=
hy, Azi
=
0.
The null-space of A is spanned by cos t and sin t. Thus, the stated condition is
necessary for the existence of a solution.
When these solvability conditions hold, the method of variation of parameters
can be used to construct a periodic solution
y(t) =
Thus, the conditions are also sufficient.
In the equation for y1 , after replacing τ by t, we had
f (t) = 2ω1 cos t − cos 3t.
This function is orthogonal to sin t, and
o
n
hf, cos ti = 2π 2ω1 cos2 t − cos4 t ,
¤
Periodic solutions and the Poincaré-Lindstedt expansion
77
where the overline denotes the average value,
Z
1
f=
f (t) dt.
2π T
Since
cos2 t =
1
,
2
cos4 t =
3
,
8
the solvability condition implies that ω1 = 3/8.
5.1.2
Van der Pol oscillator
We will compute the amplitude of the limit cycle of the van der Pol equation with
small damping,
¡
¢
y 00 + ε y 2 − 1 y 0 + y = 0.
This ODE describes a self-excited oscillator, whose energy increases when |y| < 1
and decreases when |y| > 1. It was proposed by van der Pol as a simple model of a
beating heart. The ODE has a single stable periodic orbit, or limit cycle.
We have to determine both the period T (ε) and the amplitude a(ε) of the limit
cycle. Since the ODE is autonomous, we can make a time-shift so that y 0 (0) = 0.
Thus, we want to solve the ODE subject to the conditions that
y(t + T, ε) = y(t, ε),
y(0, ε) = a(ε),
y 0 (0, ε) = 0.
Using the Poincaré-Lindstedt method, we introduce a strained variable
τ = ωt,
and look for a 2π-periodic solution y(τ, ε), where ω = 2π/T . Since d/dt = ωd/dτ ,
we have
¡
¢
ω 2 y 00 + εω y 2 − 1 y 0 + y = 0,
y(τ + 2π, ε) = y(τ, ε),
y(0, ε) = a,
y 0 (0, ε) = 0,
where the prime denotes a derivative with respect to τ We look for asymptotic
expansions,
y(τ, ε) = y0 (τ ) + εy1 (τ ) + O(ε2 ),
ω(ε) = ω0 + εω1 + O(ε2 ),
a(ε) = a0 + εa1 + O(ε2 ).
78
Method of Multiple Scales: ODEs
Using these expansions in the equation and equating coefficients of ε0 , we find that
ω02 y000 + y0 = 0,
y0 (τ + 2π) = y0 (τ ),
y0 (0) = a0 ,
y00 (0) = 0.
The solution is
y0 (τ ) = a0 cos τ,
ω0 = 1.
The next order perturbation equations are
¡
¢
y100 + y1 + 2ω1 y000 + y02 − 1 y00 = 0,
y1 (τ + 2π) = y1 (τ ),
y1 (0) = a1 ,
y10 (0) = 0.
Using the solution for y0 in the ODE for y1 , we find that
¢
¡
y100 + y1 = 2ω1 cos τ + a0 a20 cos2 τ − 1 sin τ.
The solvability conditions, that the right and side is orthogonal to sin τ and cos τ
imply that
1 3 1
a − a0 = 0,
8 0 2
ω1 = 0.
We take a0 = 2; the solution a0 = −2 corresponds to a phase shift in the limit cycle
by π, and a0 = 0 corresponds to the unstable steady solution y = 0. Then
1
3
y1 (τ ) = − sin 3τ + sin τ + α1 cos τ.
4
4
At the next order, in the equation for y2 , there are two free parameters, (a1 , ω2 ),
which can be chosen to satisfy the two solvability conditions. The expansion can
be continued in the same way to all orders in ε.
5.2
The method of multiple scales
The Poincaré-Linstedt method provides a way to construct asymptotic approximations of periodic solutions, but it cannot be used to obtain solutions that evolve
aperiodically on a slow time-scale. The method of multiple scales (MMS) is a more
general approach in which we introduce one or more new ‘slow’ time variables for
each time scale of interest in the problem. It does not require that the solution
depends periodically on the ‘slow’ time variables.
The method of multiple scales
79
We will illustrate the method by applying it to Mathieu’s equation for y(t, ε),
y 00 + (1 + δ + ε cos kt) y = 0,
where (δ, ε, k) are constant parameters, and the prime denotes a derivative with
respect to t. This equation describes a parametrically forced linear oscillator whose
frequency is changed sinusoidally in time (e.g. small amplitude oscillations of a
swing). The equilibrium y = 0 is unstable when the √
frequency k of the parametric
forcing is sufficiently close to the resonant frequency 1 + δ of the unforced (ε = 0)
oscillator.
We suppose that ε ¿ 1 and
δ = εδ1 ,
where δ1 = O(1) as ε → 0. We will consider the case k = 2, which corresponds to
the strongest instability, when y(t, ε) satisfies
y 00 + (1 + εδ1 + ε cos 2t) y = 0.
The idea of the MMS is to describe the evolution of the solution over long timescales of the order ε−1 by the introduction of an additional ‘slow’ time variable
τ = εt.
We then look for a solution of the form
y(t, ε) = ỹ(t, εt, ε),
where ỹ(t, τ, ε) is a function of two time variables (t, τ ) that gives y when τ is
evaluated at εt.
Applying the chain rule, we find that
y 0 = ỹt + εỹτ ,
y 00 = ỹtt + 2εỹtτ + ε2 ỹτ τ ,
where the subscripts denote partial derivatives. Using this result in the original
equation, and denoting partial derivatives by subscripts, we find that ỹ(t, τ, ε) satisfies
ỹtt + 2εỹtτ + ε2 ỹτ τ + (1 + εδ1 + ε cos 2t) ỹ = 0.
In fact, ỹ(t, τ, ε) only has to satisfy this equation when τ = εt, but we will require
that it satisfies it for all (t, τ ). This requirement implies that y satisfies the original
ODE. We have therefore replaced an ODE for y by a PDE for ỹ. At first sight,
this may not appear to be an improvement, but as we shall see we can use the
extra flexibility provided by the dependence of ỹ on two variables to obtain an
asymptotic solution for y that is valid for long times of the order ε−1 . Specifically,
80
Method of Multiple Scales: ODEs
we will require that y(t, τ, ε) is a periodic function of the ‘fast’ variable t. Moreover,
we only need to solve ODEs in t to construct this asymptotic solution.
We expand
ỹ(t, τ, ε) = y0 (t, τ ) + εy1 (t, τ ) + O(ε2 ).
We use this expansion in the equation for ỹ, and equate coefficients of ε0 and ε to
zero. We find that
y0tt + y0 = 0,
y1tt + y1 + 2y0tτ + (δ1 + cos 2t) y0 = 0.
The solution of the first equation is
y0 (t, τ ) = A(τ )eit + c.c.
Here, it is convenient to use complex notation. The amplitude A(τ ) is an arbitrary
complex valued function of the ‘slow’ time, and c.c. denotes the complex conjugate
of the preceding terms.
Using this solution in the second equation, and writing the cosine in terms of
exponentials, we find that y1 satisfies
y1tt + y1
= −2iAτ eit − A (δ1 + cos 2t) eit + c.c.
µ
¶
1
1
= − Ae3it − 2iAτ + δ1 A + A∗ eit + c.c.
2
2
Here, the star denotes a complex conjugate. The solution for y1 is periodic in t,
and does not contain secular terms in t, if and only if the coefficient of the resonant
term eit is zero, which implies that A(τ ) satisfies the ODE
1
2iAτ + δ1 A + A∗ = 0.
2
Writing A = u + iv in terms of its real and imaginary parts, we find that
µ
¶
µ
¶µ
¶
u
0
δ1 /2 − 1/4
u
=
v τ
−δ1 /2 − 1/4
0
v
The solutions of this equation are proportional to e±λτ where
r
1 1
− δ12 .
λ=
2 4
Thus, in the limit ε → 0, the equilibrium y = 0 is unstable when |δ1 | < 1/2, or
|δ| <
1
|ε|.
2
The method of averaging
5.3
81
The method of averaging
Consider a system of ODEs for x(t) ∈ Rn which can be written in the following
standard form
x0 = εf (x, t, ε).
(5.1)
Here, f : Rn × R × R → Rn is a smooth function that is periodic in t. We assume
the period is 2π for definiteness, so that
f (x, t + 2π, ε) = f (x, t, ε).
Many problems can be reduced to this standard form by an appropriate change of
variables.
Example 5.3 Consider a perturbed simple harmonic oscillator
y 00 + y = εh(y, y 0 , ε).
We rewrite this equation as a first-order system and remove the unperturbed dynamics by introducing new dependent variables x = (x1 , x2 ) defined by
µ
¶ µ
¶µ
¶
y
cos t sin t
x1
=
.
y0
− sin t cos t
x2
We find, after some calculations, that (x1 , x2 ) satisfy the system
x01 = −εh (x1 cos t + x2 sin t, −x1 sin t + x2 cos t, ε) sin t,
x02 = εh (x1 cos t + x2 sin t, −x1 sin t + x2 cos t, ε) cos t,
which is in standard periodic form.
Using the method of multiple scales, we seek an asymptotic solution of (5.1)
depending on a ‘fast’ time variable t and a ‘slow’ time variable τ = εt:
x = x(t, εt, ε).
We require that x(t, τ, ε) is a 2π-periodic function of t:
x(t + 2π, τ, ε) = x(t, τ, ε).
Then x(t, τ, ε) satisfies the PDE
xt + εxτ = f (x, t, ε).
We expand
x(t, τ, ε) = x0 (t, τ ) + εx1 (t, τ ) + O(ε2 ).
At leading order, we find that
x0t = 0.
82
Method of Multiple Scales: ODEs
It follows that x0 = x0 (τ ) is independent of t, which is trivially a 2π-periodic
function of t. At the next order, we find that x1 satisfies
x1t + x0τ = f (x0 , t, 0) ,
(5.2)
x1 (t + 2π, τ ) = x1 (t, τ ).
The following solvability condition is immediate.
Proposition 5.4 Suppose f : R → Rn is a smooth, 2π-periodic function. Then
the n × n system of ODEs for x(t) ∈ Rn ,
x0 = f (t),
has a 2π-periodc solution if and only if
Z 2π
1
f (t) dt = 0.
2π 0
Proof.
The solution is
Z
t
x(t) = x(0) +
f (s) ds.
0
We have
Z
t+2π
x(t + 2π) − x(t) =
f (s) ds,
t
which is zero if and only if f has zero mean over a period.
¤
If this condition does not hold, then the solution of the ODE grows linearly in
time at a rate equal to the mean on f over a period.
An application of this proposition to (5.2) shows that we have a periodic solution
for x1 if and only if x0 satisfies the averaged ODEs
x0τ = f (x0 ) ,
where
f (x) =
1
2π
Z
2π
f (x, t, 0) dt.
0
First we state the basic existence theorem for ODEs, which implies that the
solution of (5.1) exists on a time imterval ofthe order ε−1 .
Theorem 5.5 Consider the IVP
x0 = εf (x, t),
x(0) = x0 ,
The method of averaging
83
where f : Rn ×T → Rn is a Lipschitz continuous function of x ∈ Rn and a continuous
function of t ∈ T. For R > 0, let
BR (x0 ) = {x ∈ Rn | |x − x0 | < R} ,
where | · | denotes the Euclidean norm,
|x| =
n
X
2
|xi | .
i=1
Let
M=
sup
|f (x, t)|.
x∈BR (x0 ),t∈T
Then there is a unique solution of the IVP,
x : (−T /ε, T /ε) → BR (x0 ) ⊂ Rn
that exists for the time interval |t| < T /ε, where
T =
R
.
M
Theorem 5.6 (Krylov-Bogoliubov-Mitropolski) With the same notation as
the previous theorem, there exists a unique solution
x : (−T /ε, T /ε) → BR (x0 ) ⊂ Rn
of the averaged equation
x0 = εf (x),
x(0) = x0 ,
where
f (x) =
1
2π
Z
f (x, t) dt.
T
e < R, and
Assume that f : Rn × T → Rn is continuously differentiable. Let 0 < R
define
e
R
Te =
,
f
M
f=
M
sup
|f (x, t)|.
x∈BR
e (x0 ),t∈T
Then there exist constants ε0 > 0 and C > 0 such that for all 0 ≤ ε ≤ ε0
|x(t) − x(t)| ≤ Cε
for |t| ≤ Te/ε.
84
Method of Multiple Scales: ODEs
A more geometrical way to view these results is in terms of Poincaré return
maps. We define the Poincaré map P ε (t0 ) : Rn → Rn for (5.1) as the 2π-solution
map. That is, if x(t) is the solution of (5.1) with the initial condition x(t0 ) = x0 ,
then
P ε (t0 )x0 = x(t0 + 2π).
The choice of t0 is not essential here, since different choices of t0 lead to equivalent
Poincaré maps when f is a 2π-periodic function of t. Orbits of the Poincaré map
consist of closely spaced points when ε is small, and they are approximated by the
trajectories of the averaged equations for times t = O(1/ε).
5.4
The WKB method for ODEs
Suppose that the frequency of a simple harmonic oscillator is changing slowly compared with a typical period of the oscillation. For example, consider small-amplitude
oscillations of a pendulum with a slowly varying length. How does the amplitude
of the oscillations change?
The ODE describing the oscillator is
y 00 + ω 2 (εt)y = 0,
where y(t, ε) is the amplitude of the oscillator, and ω(εt) > 0 is the slowly varying
frequency.
Following the method of multiple scales, we might try to introduce a slow time
variable τ = εt, and seek an asymptotic solutions
y = y0 (t, τ ) + εy1 (t, τ ) + O(ε2 ).
Then we find that
y0tt + ω 2 (τ )y0 = 0,
y0 (0) = a,
y00 (0) = 0,
with solution
y0 (t, τ ) = a cos [ω(τ )t] .
At next order, we find that
y1tt + ω 2 y1 + 2y0tτ = 0,
or
y1tt + ω 2 y1 = 2aωωτ t cos ωt.
The WKB method for ODEs
85
We cannot avoid secular terms that invalidate the expansion when t = O(1/ε). The
defect of this solution is that its period as a function of the ‘fast’ variable t depends
on the ‘slow’ variable τ .
Instead, we look for a solution of the form
y = y(θ, τ, ε),
1
θ = ϕ(εt),
ε
τ = εt,
where we require y to be 2π-periodic function of the ‘fast’ variable θ,
y(θ + 2π, τ, ε) = y(θ, τ, ε).
The choice of 2π for the period is not essential; the important requirement is that
the period is a constant that does not depend upon τ .
By the chain rule, we have
d
= ϕτ ∂θ + ε∂τ ,
dt
and
2
y 00 = (ϕτ ) yθθ + ε {2ϕτ yθτ + ϕτ τ yθ } + ε2 yτ τ .
It follows that y satisfies the PDE
2
(ϕτ ) yθθ + ω 2 y + ε {2ϕτ yθτ + ϕτ τ yθ } + ε2 yτ τ = 0.
We seek an expansion
y(θ, τ, ε) = y0 (θ, τ ) + εy1 (θ, τ ) + O(ε2 ).
Then
2
(ϕτ ) y0θθ + ω 2 y0 = 0.
Imposing the requirement that y0 is a 2π-periodic function of θ, we find that
2
(ϕτ ) = ω 2 ,
which is satisfied if
Z
ϕ(τ ) =
τ
ω(σ) dσ.
0
The solution for y0 is then
y0 (θ, τ ) = A(τ )eiθ + c.c.,
where it is convenient to use complex exponentials, A(τ ) is an arbitrary complexvalued scalar, and c.c. denotes the complex conjugate of the preceding term.
86
Method of Multiple Scales: ODEs
At the next order, we find that
ω 2 (y1θθ + y1 ) + 2ωy0θτ + ωτ y0θ = 0.
Using the solution for y0 is this equation, we find that
ω 2 (y1θθ + y1 ) + i (2ωAτ + ωτ A) eiθ + c.c. = 0.
The solution for y1 is periodic in θ if and only if A satisfies
2ωAτ + ωτ A = 0.
It follows that
¡
¢
ω|A|2 τ = 0,
so that
ω|A|2 = constant.
Thus, the amplitude of the oscillator is proportional to ω −1/2 as its frequency
changes.
The energy E ofthe oscillator is given by
E
=
=
1 0 2
1
(y ) + 2 y 2
2
ω
1 2 2
ω |A| .
2
Thus, E/ω is constant. The quantity E/ω is called the action. It is an example of
an adiabatic invariant.
The WKB method can also be used to obtain asymptotic approximations as
ε → 0 of ODEs of the form
ε2 y 00 + V (x)y = 0.
This form corresponds to a change of variables t 7→ x/ε in the previous equations.
The corresponding WKB expansion is is
y(x, ε) = A(x, ε)eiS(x)/ε ,
A(x, ε) = A0 (x) + εA1 (x) + . . . .
This expansion breaks down at turning points where V (x) = 0, and then one must
use Airy functions (or other functions at degenerate turning points) to describe the
asymptotic behavior of the solution.
Perturbations of completely integrable Hamiltonian systems
5.5
87
Perturbations of completely integrable Hamiltonian systems
Consider a Hamiltonian system whose configuration is described by n angles x ∈ Tn ,
where Tn is the n-dimensional torus, with corresponding momenta p ∈ Rn . The
Hamiltonian H : Tn × Rn → R gives the energy of the system. The motion is
described by Hamilton’s equations
dx
∂H
=
,
dt
∂p
dp
∂H
=−
.
dt
∂x
This is a 2n × 2n system of ODEs for x(t), p(t).
Example 5.7 The simple pendulum has Hamiltonian
H(x, p) =
1 2
p + 1 − cos x.
2
A change of coordinates (x, p) 7→ (x̃, p̃) that preserves the form of Hamilton’s
equations (for any Hamiltonian function) is called a canonical change of coordinates.
A Hamiltonian system is completely integrable if there exists a canonical change
of coordinates (x, p) 7→ (ϕ, I) such that H = H(I) is independent of the angles
ϕ ∈ Tn . In these action-angle coordinates, Hamilton’s equations become
dϕ
∂H
=
,
dt
∂I
Hence, the solutions are I = constant and
dI
= 0.
dt
ϕ(t) = ω(I)t + ϕ0 ,
where
ω(I) =
∂H
.
∂I
If
H ε (ϕ, I) = H0 (I) + εH1 (ϕ, I)
is a perturbation of a completely integrable Hamiltonian, then Hamilton’s equations
have the form
dϕ
= ω0 (I) + εf (ϕ, I),
dt
dI
= εg(ϕ, I),
dt
where
∂H1
∂H1
∂H0
,
f=
,
g=−
.
ω0 =
∂I
∂I
∂ϕ
The study of these problems using multi-phase averaging methods is very subtle
(e.g. KAM theory).