Lectures Notes on:
Math 6410: Ordinary
Differential Equations
Chris Miles
miles@math.utah.edu
Last Updated: August 5, 2015
Contents
1 Initial Value Problems
1
1.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1
1.2 Existence and Uniqueness . . . . . . . . . . . . . . . . . . . . . . . . . . 6
1.3 Dependence on Initial Conditions . . . . . . . . . . . . . . . . . . . . . . 12
2 Linear Equations
2.1 Matrix Exponentials . . . . . . . .
2.2 Normal Forms . . . . . . . . . . .
2.3 Non-Autonomous Linear Equations
2.4 Inhomogeneous Linear Equations .
2.5 Stability and Boundedness . . . . .
2.6 Periodic Systems: Floquet Theory
3 Boundary Value Problems
3.1 Preliminaries . . . .
3.2 Linear Operators . .
3.3 Differential Operators
3.4 Eigenvalue Problems
3.5 Distribution Theory .
3.6 Green’s Functions .
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14
14
15
19
20
22
23
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31
31
32
35
40
46
50
4 Dynamical Systems Theory
57
4.1 Preliminaries . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 57
4.2 Planar Dynamics . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 60
4.3 Limit Cycles . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 62
i
ii
CONTENTS
4.4 Stability . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 67
4.5 Manifolds . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 70
4.6 Bifurcation Theory . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 74
5 Perturbation Theory
5.1 Preliminaries . . . . . . . .
5.2 Poincaré-Lindstedt Method
5.3 Method of Multiple Scales .
5.4 Averaging Method . . . . .
Index
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83
83
85
88
94
97
Disclaimer: These notes are for my personal use. They come with no claim of correctness or
usefulness.
Errors or typos can be reported to me via email and will be corrected as quickly as possible.
Acknowledgments: The content of these notes is based on lectures delivered (beautifully) by
Paul C. Bressloff for Math 6410 in Fall 2014.
The notes themselves (along with the figures) are derived from notes originally taken by Gregory Handy. Andrew Watson has graciously corrected a considerable number of typos in the
document.
1
Initial Value Problems
“One must not be too rigorous all the time in order to not to commit suicide."
— Paul C. Bressloff
1.1
Introduction
In this section, we discuss an overview of the topics covered in the first part of these notes,
those dealing with ordinary differential equations.
We first consider initial value problems, which are of the form
ẋ = f (x),
x ∈ Rn ,
f : Rn → Rn ,
(1.1)
that is, x = (x1 , . . . , xn )T and the notation ẋ just refers to the “time derivative”
ẋ :=
dx
.
dt
We can also consider this in coordinate form, which just restates (1.1) by explicitly considering
each component of the vector, resulting in
ẋ1 = f1 (x1 , . . . , xn ),
..
.
ẋn = fn (x1 , . . . , xn ).
We’ll typically use the form of (1.1) due to its compactness. Notice, we have not said
anything about the linearity of f (x) and in fact, most interesting cases involve a non-linear
f (x). We typically consider t to physically represent time and we’re provided some initial
condition x(0) = x0 .
Example 1.1 (Newtonian dynamics). Consider a particle of mass m and a position x ∈ R
whose motion is governed by the classical equation:
mẍ = −γ ẋ + F (x),
(1.2)
1
2
CHAPTER 1. INITIAL VALUE PROBLEMS
where γ is the drag coefficient and F (x) is an external applied force. Consider setting
x1 = x and x2 = ẋ, which results in the system
ẋ1 = f1 (x1 , x2 ) = x2 ,
γ
F (x1 )
ẋ2 = f2 (x1 , x2 ) = − x2 +
m
m
If γ = 0 we have a conservative system (energy is conserved). To see this, define the V (x),
the potential energy as
Z
x
V (x) := −
F (ξ) dξ.
0
Now, notice that
mẍ = −
dV
,
dx
multiplying both sides by ẋ and rewriting yields
dV
mẍẋ = −ẋ ,
dx
d
d m 2
ẋ = − V (x(t)),
dt 2
dt
and finally integrating both sides yields the total energy is some constant E, that is,
m 2
ẋ + V (x(t)) = E.
2
| {z } | {z }
kinetic
potential
We can think of this in a slightly different way. Define the momentum by ρ := mẋ and the
p2
energy function, or Hamiltonian by H(x, p) := 2m
+ V (x). We can now rewrite (1.2) in
the following form:
∂H
∂H
ẋ =
,
ṗ = −
.
(1.4)
∂p
∂x
Notice that
∂H
∂H
∂H
(x(t), ρ(t)) =
ẋ +
ρ̇ = 0.
(1.5)
∂t
∂x
∂ρ
Thus, when we couple the results of (1.5) with the system (1.4), we see that H is conserved
along trajectories.
Consider now N interacting particles (say, electrons) and let xi ∈ R3 and correspond to the
position of the ith particle in 3 dimensions and let ρi be the corresponding momentum vector
in R3 . Thus, the particle has 6 degrees of freedom. . Then we now can consider the new
Hamiltonian
N
X
X
1 2
1
H :=
ρ j + c2
,
2m
|x
−
x
|
j
i
j
j=1
i<j
1.1. INTRODUCTION
3
where we can think of the second term as a Coulomb-like effect. Again, we can get a system
analogous to (1.4) now for many particles
ẋi =
∂H
,
∂ρi
ρ̇i = −
∂H
.
∂xi
We note that the result is an extremely convenient and compact form to study the behavior of
this system. We introduce this example to illustrate that physics was historically a driving
force for the study of ordinary differential equations.
Example 1.2 (chemical kinetics). Let x1 , . . . , xn represent the concentration of n chemical
species evolving in time according to the law of mass action. This is a fundamentally
non-linear phenomenon that says that the rate of a 1 step chemical reaction depends on the
product of the concentrations of the reactants.
As an example, the following chemical reaction:
k1
A + B C.
k2
In other words, two reactants A, B form a complex C at rate k1 and the complex decomposes
back into its components at a rate of k2 . The differential equations that describe this process
are
dxA
dxB
=
= k2 xC − k1 xA xB
dt
dt
dxC
= k1 xA xB − k2 xC .
dt
Example 1.3 (predator-prey models). Let N (t) denote the number of bacteria in a culture,
which is so large that we can treat it as a continuous variable. Then, one classical model for
the populations dynamics is
N
Ṅ = K 1 −
N − H,
N (0) = N0 .
(1.7)
Nmax
If we consider rescaling by the fraction of the total population, x = N/Nmax and τ = kt,
then h = H/kNmax and (1.7) becomes
ẋ = (1 − x)x − h.
Notice, depending on the value of h, we have a different number of fixed points of the system,
ranging from 0 to 2. Specifically, if 0 < h < 1/4, there are two fixed points, and otherwise
none.
If we let z := x − x+ , then z = f (x+ ) + f 0 (x+ )z ≈ f 0 (x+ )z, which leads us to the notion a
bifurcation. Particularly, this is an example of a saddle-node bifurcation that we will examine
in more detail later.
4
CHAPTER 1. INITIAL VALUE PROBLEMS
Figure 1.1: Sketch of solutions of the predator-prey model and a corresponding bifurcation
diagram.
In general, we cannot usually find an explicit solution of non-linear ordinary differential
equations, so we hope to develop tools for analyzing their qualitative behavior.
In Chapter 1 , we will prove statements about existence and uniqueness of a solution. That
is, if we have ẋ = f, x(0) = x0 and f is continuously differentiable, can we expect a “nice”
solution to always exist?
After that, we’ll study linearization, in which we attempt to find a low dimensional solution
such as a fixed point x∗ where f (x∗ ) = 0 or limit cycles where x(t) = x(t + τ ) for all t.
We’ll see that the qualitative behavior of such solutions can often be determined by the
linearized equation. For instance, the fixed points of the linearized system are the same as
the full non-linear. In the case of limit cycles, the answer is a little more difficult, but this is
exactly Floquet Theory, discussed in Chapter 2. We’ll then later relate the local analysis
performed in Chapter 2 to the full non-linear system in Chapter 4.
Boundary value problems often occur in the study of partial differential equations, but
really fall under the category of ordinary differential equations. These are a departure from
initial value problems and therefore require dramatically different techniques. For instance,
consider a string fixed at ends x = 0, 1 and let u(t, x) be the displacement of the string.
These dynamics are described by the wave equation
2
∂ 2u
2∂ u
=
c
,
∂t2
∂x2
u(t, 0) = u(t, 1) = 0,
u(0, x) = u0 (x),
ut (0, x) = v(x).
(1.8)
We’ll study this equation in much more detail in the second half of these notes, but for now
suppose we have a solution of the form u(x, t) = w(t)y(x). This leads to the relation
1 ẅ
y 00
=
.
c2 w
y
Notice that w is a function of t and y is a function of x yet these two sides are equal, meaning
1.1. INTRODUCTION
5
they must be equal to some constant, λ. This results in the system of ODEs:
ẅ = −λc2 w
(initial value problem)
00
y = −λy
(boundary value problem).
The general solutions are
√ √ w(t) = a1 cos c λt + a2 sin c λt
√ √ y(x) = b1 cos
λx + b2 sin
λx .
Now, using the boundary
described in (1.8), we can solve for the coefficients to
√conditions
find that b1 = 0 and sin
λ = 0, meaning λ = (πn)2 , which describes the eigenvalues of
the solution. We say that this operator has a discrete spectrum as a result. By the principle
of superposition, we now have our full solution
u(x, t) =
∞
X
[a1,n cos(cnπt) + a2,n sin(cnπt)] sin(nπx).
n=1
We can determine the coefficients using Fourier series expansions of our initial data
u0 (x), v(x). This is just an illustration of the use of solving boundary value problems. More
complicated PDEs, coupled with separation of variables yields more complicated boundary
value problems. We can ask questions like: under what circumstances can we solve the
resulting boundary value problem in terms of a general Fourier series? This motivates the
study of Sturm-Liouiville problems, discussed in Chapter 3.
In Chapter 4, we’ll study dynamical systems and global dynamics. Particularly, we’ll see
that the existence and uniqueness theorems and linearization provide information about local
behavior, but what does this tell us about global behavior? Can we piece together the global
solution from local behaviors? Using geometric methods, such as basins of attractions,
separatrices, homoclinic and heteroclinic connections, and high dimensional attractors (fixed
points, limit cycles, strange attractors), we’ll construct together global behaviors. In planar
systems (two degrees of freedom), we’ll use the classical tool of phase-plane analysis to
study the nullclines and behavior of the system. Bifurcation theory is the study of tracking
changes in these qualitative behaviors, for which we’ll establish criteria.
Lastly, in Chapter 5, we’ll cover basic perturbation theory. Assuming a small parameter ε
in a system, can we use the information we get by studying the case where ε = 0 to obtain
information about when ε > 0? We’ll use power series in ε to answer this question.
6
1.2
CHAPTER 1. INITIAL VALUE PROBLEMS
Existence and Uniqueness
Some preliminaries
Consider an autonomous ODE of the form
ẋ = f (x),
x ∈ Rn ,
t ∈ R,
where f : U → Rn and U is an open subset of Rn , f ∈ C 0 (U ), that is, f is continuous.
Example 1.4. Continuity is not a sufficient condition to guarantee uniqueness of a solution
to an initial value problem with x(0) = x0 ∈ U . Consider the following IVP:
ẋ = 3x2/3 ,
x(0) = 0.
Notice that one solution is the trivial solution, that is x(t) ≡ 0. But notice, we can also
separate and integrate
Z
dx
1
= t =⇒ x(t) = t3 .
3
x2/3
Notice the issue: f is continuous but not differentiable at x = 0, hinting at a more reasonable
requirement for uniqueness.
Example 1.5. Another possible issue we can run into is whether or not a solution “blows
up” or becomes unbounded for a finite time. For instance, consider the following:
ẋ = x2 ,
x(0) = 1 =⇒ x(t) =
1
.
1−t
This solution is only valid for t ∈ (−∞, 1). There is another solution branch for t ∈ (1, ∞),
but we can’t get there if we start at t < 1. In general, it’s harder to eliminate cases like this.
In order to prove existence and uniqueness, we’ll require that f ∈ C 1 (U ), that is, its once
differentiable. We can weaken this requirement to only Lipschitz continuity.
Definition 1.1. The function f : Rn → Rn is differentiable at x0 ∈ Rn if there exists a
linear transformation Df (x0 ) ∈ L(Rn ) such that
|f (x0 + h) − f (x0 ) − Df (x0 )h|
= 0,
|h|→0
|h|
lim
where Df is the Jacobian, that is
∂fi
Df (x0 ) :=
∂xj
.
x=x0
The function f is continuously continuously differentiable if Df : Rn → L(Rn ) is a
continuous map of x in some open set U ⊆ Rn .
1.2. EXISTENCE AND UNIQUENESS
7
Definition 1.2. Suppose that V1 , V2 are two normed linear vector spaces. Then F : V1 → V2
is continuous at x0 ∈ V1 if ∀ε > 0, ∃δ > 0 such that for x ∈ V1 and kx − x0 k < δ, we have
kF (x) − F (x0 )k < ε. In particular, we’ll often use the Euclidean norm (or natural norm):
kDf (x)k := max |Df (x)u|.
kuk≤1
In other words, we simply require
∂fi
∂xj
to exist and be continuous.
Definition 1.3. Let U ⊂ Rn be an open set. Then, f : U → Rn satisfies a so-called
Lipschitz condition if ∃k > 0 such that ∀x, y ∈ U , we have that:
|f (x) − f (y)| ≤ k|x − y|.
The function is locally Lipschitz on U if for each point x0 ∈ U, ∃ an ε neighborhood
Nε (x0 ) ⊂ U defined to be
Nε (x0 ) := {x ∈ Rn : |x − x0 | < ε} ,
and a constant k0 such that ∀x, y ∈ Nε (x0 ), we have that:
|f (x) − f (y)| ≤ k0 |x − y|.
Theorem 1.1. If f ∈ C 1 (U ) then f is locally Lipschitz on U .
Proof. Let x0 ∈ U and suppose x, y ∈ Nε (x0 ) ⊂ U with k = maxx∈Nε (x0 ) kDf (x)k.
Notice that under the norm, Nε (x0 ) is convex, meaning that x + sw ∈ Nε (x0 ) for 0 ≤ s ≤ 1
with w = x − y.
P
∂fi
Let F (s) := f (x+sw) and F 0 (s) = D(f x+sw)·w, which we can rewrite as nj=1 ∂x
wj =
j
0
Fi (s) if we remember that these are actually vectors. We then notice that f (y) − f (x) =
F (1) − F (0), which we can rewrite as
Z 1
Z 1
0
f (y) = f (x) = F (1) − F (0) =
F (s) ds =
Df (x + sw) · w ds.
0
0
Therefore, we can take the norm of both sides and use the triangle inequality for integrals to
note
Z 1
|f (y) − f (x)| ≤
|Df (x + sw)| |w| ds
0
Z 1
≤
kDf (x + sw)k |w| ds
0
≤ k|x − y|.
Ultimately, all we need to take from this section is that Lipschitz continuity will be the
condition for our existence and uniqueness theorems later.
8
CHAPTER 1. INITIAL VALUE PROBLEMS
Banach Spaces
Definition 1.4. Let X be a real vector space. A norm on X is a map k · k : X → R with
the following three properties:
(N1) k0k = 0, kxk > 0 if x 6= 0.
(N2) kαxk = |α| kxk for α ∈ R.
(N3) kx + yk ≤ kxk + kyk, which is known as the Triangle Inequality.
Definition 1.5. Given a normed vector space (NVS) X, we say that a sequence of vectors
fn converges to a vector f if
lim kfn − f k = 0.
n→∞
Definition 1.6. A mapping F : X → Y is continuous if fn → f implies that F (fn ) →
F (f ).
Definition 1.7. A sequence {fn } is Cauchy if ∀ε > 0, ∃M such that
kfn − fm k < ε,
∀n, m > M.
Definition 1.8. A normed vector space is said to be complete if every Cauchy sequence has
a limit in the sequence.
Definition 1.9. A Banach space is a complete normed vector space.
Definition 1.10. Let C ⊆ X be a closed subset and consider a mapping K : C → C. A
fixed point of the mapping is an element x ∈ C such that K(x) = x.
K is said to be a contraction mapping if ∃θ ∈ [0, 1) such that
kK(x) − K(y)k ≤ θkx − yk,
∀x, y ∈ C.
Theorem 1.2 (contraction Principle). Let C be a non-empty closed subset of a Banach
space X and let K : C → C be a contraction mapping. Then K has a unique fixed point
x∗ ∈ C with
θn
kK n (x) − x∗ k ≤
kK(x) − x∗ k,
∀x ∈ C.
1−θ
Proof. We’ll first prove uniqueness of the fixed point. Consider otherwise. That is, if
x∗ = K(x∗ ) and y ∗ = K(y ∗ ), then
kx∗ − y ∗ | = kK(x∗ ) − K(y ∗ )k ≤ θkx∗ − y ∗ k,
1.2. EXISTENCE AND UNIQUENESS
9
which immediately suggests that x∗ = y ∗ since θ ∈ [0, 1).
Next, we’ll prove existence. Fix x0 and consider the iteration xn := K n (x0 ), then
kxn+1 − xn k ≤ θkxn − xn−1 k ≤ · · · ≤ θn kx1 − x0 k.
Therefore, by the triangle inequality, we have
kxn − xm k ≤
n
X
kxj − xj−1 k ≤ θ
m
n−m−1
X
θj kx1 − x0 k,
j=0
j=m+1
which we can now observe is a geometric series
θ
m
n−m−1
X
θj kx1 − x0 k =
j=0
θm (1 − θn−m )
θm
kx1 − x0 k ≤
kx1 − x0 k → 0.
1−θ
1−θ
(1.11)
Thus, we have proved that {xn } is a Cauchy sequence. From the completeness of the space,
we know that xn → x∗ ∈ C. Finally,
kK(x∗ ) − x∗ k = lim kxn+1 − xn k = 0,
n→∞
∗
therefore x is a fixed point. The error estimate is immediate by taking limn→∞ of (1.11).
Picard’s Method of Successive Approximations
Consider x(t) as a solution of the IVP ẋ = f (x), where x(0) = x0 , x ∈ Rn . Then, if and
only if x(t) is a continuous function then it satisfies the integral solution
Z t
x(t) = x0 +
f (x(s)) ds.
0
The successive approximations to the solution of this ODE are defined by the sequence
Z t
u0 (t) := x0 ,
un+1 (t) := x0 +
f (un (s)) ds,
k = 0, 1, 2, . . . .
0
We have made no comment on whether this is a good approximation or not.
Example 1.6. Consider ẋ = Ax, where x(0) = x0 . Then, applying this iteration technique,
we have that
u0 (t) = x0
Z
u1 (t) = x0 +
t
Z
f (u0 (s)) ds = x0 +
t
Ax0 = x0 + Ax0 t = x0 (1 + At)
Z t
Z t
A2
u2 (t) = x0 +
f (u1 (s)) ds = x0 +
x0 (1 + At) = x0 1 + At +
t .
2
0
0
0
0
10
CHAPTER 1. INITIAL VALUE PROBLEMS
By induction, it’s clear our result is
Ak k
uk (t) = x0 1 + A + · · ·
t = x0 eAt
k!
as k → ∞,
(1.12)
where we have used the definition of the matrix exponential in (1.12).
Fundamental Existence and Uniqueness Result
Theorem 1.3. Let U be an open subset of Rn containing x and f be differentiable, i.e.,
f ∈ C 1 (U ), then ∃a > 0 such that ẋ = f (x), x(0) = x0 has a unique solution on the interval
I = [−a, a]
Proof. This proof is so long we’ll split it into four parts.
(a) Since f ∈ C 1 (U ), this suggests that f is locally Lipschitz by a previous theorem. This
means that ∀x, y ∈ Nε (x0 ), we have that |f (x) − f (y)| ≤ k|x − y| for a given ε and
k > 0. Let b = ε/2 and set N0 = {x ∈ Rn : |x − x0 | < b} ⊆ Nε , then denote
M := maxx∈N0 |f (x)|.
Introduce the successive approximation uk (t) defined by Picard’s scheme. Assume that
∃a > 0 such that uk (t) is defined and continuous on [−a, a] and
max |uk (t) − x0 | ≤ b.
(1.13)
[−a,a]
If this is true, it follows that f (uk (t)) is defined and continuous on I = [−a, a]. Hence,
Z
uk+1 (t) = x0 +
t
f (uk (s)) ds
0
is also defined and continuous on this interval with
Z
|uk+1 (t) − x0 | ≤
t
|f (uk (s))| ds ≤ M a,
∀t ∈ [−a, a].
0
Choosing 0 < a ≤ b/M , it follows by induction that uk (t) is defined, continuous,
and satisfies the condition (1.13) for all t ∈ [−a, a] for k = 1, 2, . . .. So, in summary,
given the Lipschitz condition, we now have a neighborhood N0 where uk exists and is
continuous.
1.2. EXISTENCE AND UNIQUENESS
11
(b) By the definition of uk (t), we have
t
Z
|f (u1 (s)) − f (u0 (s)) ds
|u2 (t) − u1 (t)| ≤
0
t
Z
≤k
|u1 (s) − u0 (s)| ds
0
≤ ka max |u1 (t) − x0 |
[−a,a]
≤ kab
by (1.13).
Suppose that
max |uj (t) − uj−1 (t)| ≤ (ka)j−1 b,
j ≥ 2,
(1.14)
[−a,a]
then we have that
Z
t
|uj+1 (t) − uj (t)| ≤
|f (uj (s)) − f (uj−1 (s))| ds
0
≤ ka max |uj (t) − uj−1 (t)|
[−a,a]
≤ (ka)j b,
and therefore (1.14) holds by induction.
(c) Setting α = ka with 0 < a < 1/k, we see that for m > k ≥ N and t ∈ [−a, a], we have
m−1
X
|um (t) − uk (t)| ≤ uj+1 (t) − uj (t)
≤
≤
j=k
∞
X
|uj+1 (t) − uj (t)|
j=N
∞
X
αj b
j=N
n
=
α
b→0
1−α
as N → ∞.
In other words, {uk (t)} is a Cauchy sequence. Completeness of C[−a, a] implies that
{uk } converges to some u(t) ∈ C uniformly for all t ∈ [−a, a]. Taking the limit in
(1.13) gives us
Z t
Z t
lim uk+1 (t) = x0 + lim
f (uk (s)) ds =⇒ u(t) = x0 +
f (u(s)) ds.
k→∞
k→∞
0
0
In other words, u̇(t) = f (u) and u(0) = x0 , this we’ve found the existence of a solution.
12
CHAPTER 1. INITIAL VALUE PROBLEMS
(d) Lastly, we must prove uniqueness. Consider u(t) and v(t) are two solutions to the initial
value problem. The continuous function |u(t) − v(t)| achieves its maximum at some
point t1 ∈ [−a, a], thus
ku − vk = max |u(t) − v(t)|
t∈[−a,a]
Z t1
=
[f (u(s)) − f (v(s))] ds
0
Z t1
≤k
|u(s) − v(s)| ds
0
≤ kaku − vk,
but since k < 1, this implies that ku − vk = 0 and therefore u(t) = v(t) on t ∈ [−a, a]
due to the continuity of these solutions.
1.3
Dependence on Initial Conditions
Theorem 1.4 (Gronwall’s Lemma).
If u, v, c are positive functions on [0, t] where c is
R
t
0
differentiable and v(t) ≤ c(t) +
Z
u(s)v(s) ds, then
t
v(t) ≤ c(0) exp
Z
u(s) ds +
0
t
Z
0
c (s) exp
0
t
u(τ ) dτ
ds.
s
Proof. Let
Z
t
u(s)v(s) d(s) =⇒ Ṙ(t) = v(t)u(t) ≤ [c(t) + R(t)] u(t),
R(t) =
0
which we can summarize as Ṙ − uR ≤ Cu, which we can apply an integrating factor
Z t
Z t
d
exp −
u(s) ds R(t) = exp −
u(s) ds c(t)u(t),
dt
0
0
which then suggests that
Z t
Z s
Z t
exp −
u(τ ) dτ R(t) ≤
exp −
u(τ ) dτ u(s)c(s) ds,
0
0
0
since R(0) = 0, and therefore
Z
R(t) ≤
t
Z
exp
0
t
u(τ ) dτ
s
c(s)u(s) ds.
1.3. DEPENDENCE ON INITIAL CONDITIONS
13
Returning to the first inequality, we have that
Z t
Z t
exp
u(τ ) dτ c(s)u(s) ds
v(t) ≤ c(t) +
0
s
Z t
Z t
d
c(s) −
= c(t) +
exp
u(τ ) dτ ds
ds
0
s
Z t
t Z t
Z t
0
= c(t) − c(s) exp
u(τ ) dτ +
c (s) exp
u(τ ) dτ ds
s
0
s
0
Z t
Z t
Z t
0
u(s) ds +
c (s) exp
u(τ ) dτ ds.
≤ c(0) exp
0
0
(after IBP)
s
Theorem 1.5 (Dependence on Initial Conditions). Consider the equation ẋ = f (x), x ∈
Rn , t ∈ I where f is Lipschitz continuous with constant k. Consider the two IVPs and their
solutions
ẋ = f (x), x(0) = y =⇒ x0 (t) on I,
ẋ = f (x), x(0) = y + η =⇒ xε (t) on I.
If |η| ≤ ε, where ε > 0, then
|x0 (t) − xε (t)| ≤ εekt ,
t ∈ I.
Proof. The two IVPs are equivalent to the integral equations
Z t
Z t
x0 (t) = y +
f (x0 (s)) ds,
xε (t) = y + η +
f (xε (s)) ds.
0
0
Subtracting these two, we find
Z
t
|x0 (t) − xε (t)| ≤ |η| +
|f (x0 (s)) − f (xε (s))| ds
0
Z t
≤ε+
k|x0 (s) − xε (s)| ds,
0
and applying Gronwall’s Lemma, with the following mappings:
u(s) = k,
the desired result is immediate.
c(s) = ε,
v(s) = |x0 (s) − xε (s)|,
2
Linear Equations
“British slang lesson: ‘giving it a welly’ means pulling hard, from Wellington
boots. Use your imagination on how to use this inappropriately."
— Paul C. Bressloff
In this chapter, we’ll consider the general, autonomous, linear ODE of the form
ẋ = Ax,
2.1
x ∈ Rn .
(2.1)
Matrix Exponentials
Definition 2.1 (matrix exponential). The exponential of the n × n matrix A is defined by
the power series
∞
X
Ak
k=0
k!
:= eA .
Note, this is non-trivial, as Aj is sometimes hard to compute. We’re not even sure the series
converges.
Theorem 2.1. Let A be an n × n matrix with constant coefficients. Then the unique solution
of (2.1) with initial condition x(0) = x0 is
x(t) = etA x0 .
Proof. Note that, assuming uniform convergence,
∞
∞
X tk−1 Ak
d X (tA)k
d tA ẋ =
e x0 =
x0 =
x0 = AetA x0 = Ax(t),
dt
dt k=0 k!
(k
−
1)!
k=1
thus the solution does indeed satisfy the differential equation. We can now prove uniqueness
by supposing that y(t) is another solution. Setting
z(t) = e−tA y(t) =⇒ ż(t) = −Ae−tA y(t) + e−tA ẏ(t) = 0,
14
2.2. NORMAL FORMS
15
since y(t) is a solution. Therefore,
z(t) = x0 = e−tA y =⇒ y(t) = x0 etA ,
and therefore y(t) = x(t), meaning our solution is unique. We lastly need to prove uniform
convergence of the power series. Take the norm in this case to be the induced norm
kAk = sup
v∈Rn
v6=0
|Av|
= sup |Av|.
|v|
kvk=1
Now, we note that, under this induced norm
kABk = sup |ABv| ≤ sup kAk|Bv| = kAk kBk,
kvk=1
kvk=1
from which, we can immediately
conclude that kAk k ≤ kAkk . Recall that
P
k
absolutely convergent if k kA k/k! < ∞, and the latter holds if
∞
X
kAk k
k=0
2.2
k!
≤
∞
X
kAkk
k=0
k!
P
k
Ak /k! is
= ekAk .
Normal Forms
We have several cases for the eigenvalues of the matrix A that dictate the behavior.
Real, distinct eigenvalues
Suppose that A has n real, distinct eigenvalues λj , j = 1, . . . , n with corresponding
eigenvectors ej . That is, Aej = λj ej . Define
P := [e1 , . . . en ],
and since the eigenvectors are linearly independent, we can immediately conclude that
det P 6= 0 and therefore P is invertible. Note that
AP = [Ae1 , . . . Aen ] = [λ1 e1 , . . . , λn en ] = P diag(λ1 , . . . , λn ),
which suggests that we have the similarity transformation
A = P ΛP −1
or
P −1 AP = Λ.
Returning back to the ODE, perform the change of variables x = P y, and taking a derivative
yields
ẏ = P −1 ẋ = P −1 Ax = P −1 AP y = Λy.
16
CHAPTER 2. LINEAR EQUATIONS
In component form, this is equivalent to ẏj = λj yj for j = 1, . . . , n, which are n independent
ODEs which can be solved directly: yj (t) = exp{λi t}yj (0), which in matrix form is
equivalent to
y(t) = eΛ ty(0) =⇒ x(t) = P eΛt y(0) = P eΛt P −1 x(0).
From this, we can conclude how to evaluate the matrix exponential explicitly:
etA = P etΛ P −1 .
In other words, evaluating the matrix exponential boils down to finding the eigensystem of
the matrix.
Complex eigenvalues
First, note that complex eigenvalues come in conjugate pairs since A is a real matrix. For
simplicity, suppose A is 2 × 2 with eigenvalues λ = ρ + iω with eigenvectors e1 , e∗1 .
Define
P := [Im(e1 ), Re(e1 )],
and now note that
AP = [Im{(ρ + iω)e1 }, Re{(ρ + iω)e1 }]
= [ρ Im(e1 ) + ω Re(e1 ), ρ Re(e1 ) − ω Im(e1 )]
ρ −ω
=P
= P Λ.
ω ρ
In other words, we again have the similarity transform
Λ = P −1 AP,
but now Λ is no longer diagonal.
Returning to the ODE, again consider the change of variables x = P y, meaning that ẏ = Λy
and our solution is again
y(t) = etΛ y(0),
but is this computable? Decompose
Λ=D+C =
ρ 0
0 −ω
+
,
0 ρ
ω 0
and note since DC = CD, we can write
ρt
∞ k
1
e
0 X 0 −ω
tΛ
tD tC
e =e e =
.
0 eρt
ω 0
k!
k=0
2.2. NORMAL FORMS
17
We make the observation by computing the iterates
2n
ω
0
0
−ω 2n+1
2n
n
2n+1
C =
(−1) ,
C
=
(−1)n ,
0 ω 2n
ω 2n+1
0
that in the diagonals we get the even terms of the exponential Taylor series, and the
off-diagonals are the odd terms, meaning we just end up with
cos ωt − sin ωt
tC
e =
,
sin ωt cos ωt
and therefore
tΛ
e
cos ωt − sin ωt
=e
.
sin ωt cos ωt
ρt
Theorem 2.2. Let A be a real n × n matrix with k distinct eigenvalues λ1 , . . . , λk and
m = (n − k)/2 pairs of distinct complex conjugate eigenvalues ρj + iωj , then there exists
an invertible matrix P such that
ρj −ωj
−1
P AP = Λ = diag(λ1 , . . . , λk , B1 , . . . , Bm ),
Bj :=
.
ωj ρj
That is, Λ is block diagonal, where the blocks can be treated independently. Moreover,
etA = P etΛ P −1 .
Degenerate eigenvalues
Note that an eigenvalue λ, of a matrix A satisfies the characteristic polynomial
0 = det(A − λI) =
m
Y
(λ − λj )αj ,
j=1
where m < n (due to repeated roots), and λj 6= λk . The number αj is called the algebraic
multiplicity, where
gj := dim(kern(A − λj I)),
is called the geometric multiplicity.
Suppose that A has m distinct eigenvalues λ1 , . . . , λm , again m < n, then the generalized
eigenspace of λk is
Ek := {x ∈ Rn : (A − λk )αk x = 0} .
Associated with each degenerate eigenvalue are ak linearly independent solutions of the form
P1 (t)eλk t , P2 (t)eλk t , . . . , Pαk (t)eλk t ,
18
CHAPTER 2. LINEAR EQUATIONS
with the degree of Pj (t) < αk .
Consider the example where n = 2. Recall that a matrix A satisfies its only characteristic
equation ( by the Cayley-Hamilton theorem), if
(A − λI)2 x = 0,
∀x ∈ R2 ,
meaning that A has a repeated eigenvalue. There are two cases for such.
(1) The most simple case is described by
(A − λI)x = 0,
λ 0
∀x =⇒ A =
,
0 λ
and we are done, since in this case, the geometric and algebraic multiplicity are both 2.
(2) Consider the existence of a second eigenvector e2 , where
(A − λI)e2 6= 0,
which suggests that the algebraic multiplicity of this eigenvalue is 2, but the geometric
is 1. Define e1 = (A − λI)e2 , and note that
(A − λI)e1 = (A − λI)2 e2 = 0,
which suggests that
Ae2 = e1 + λe2 =⇒ Ae1 = λe1 ,
thus,
λ 1
A[e1 , e2 ] = [e1 , e2 ]
,
0 λ
and in summary, if we call P = [e1 , e2 ], we again have that
λ 1
−1
P AP = Λ =
.
0 λ
The transformed system we then have is
ẏ1 = λy1 + y2 ,
ẏ2 = λy2 ,
which suggests that
y2 (t) = eλt y2 (0)
and therefore
d −λt y1 e
= y2 (t)e−λt = y2 (0),
d
which suggests that, after integrating,
y1 (t)e−λt − y1 (0) = y2 (0)t =⇒ y1 (t) = eλt [y1 (0) + y2 (0)t] .
Therefore, the stability is still determined by the sign of the eigenvalue λ.
2.3. NON-AUTONOMOUS LINEAR EQUATIONS
2.3
19
Non-Autonomous Linear Equations
In this section, we consider the more general form
ẋ = A(t)x.
(2.2)
In other words, our matrix now changes with time.
Definition 2.2. Let ψj (t) be vector-valued functions, that are continuous on t ∈ R and no
ψj ≡ 0. If there exists constants αj such that
n
X
αj ψj (t) ≡ 0,
j=1
then the functions ψj are said to be linearly dependent, otherwise the functions are said to
be linearly independent.
Theorem 2.3. Any n + 1 non-zero solution of (2.2) are linearly dependent.
Proof. Let the solutions be denoted φ1 (t), . . . , φn+1 (t). For a fixed t0 , the n + 1 constant
vectors φj (t0 ) are linearly dependent. That is, there exists constants aj such that
n+1
X
αj φj (t0 ) = 0.
j=1
Consider the superposition of these solutions
x(t) :=
n+1
X
αj φj (t) =⇒ x(t0 ) = 0,
ẋ + A(t)x,
j=1
where the last part holds by the principle of superposition. By uniqueness, we know that
x(t) ≡ 0, meaning that φj are linearly dependent.
Theorem 2.4. There exists a set of n linearly independent solutions of (2.2).
Proof. The existence theorem implies that there exists a set of solutions ψ1 (t), . . . ψn (t)
corresponding to the existence conditions that [ψj (0)]m = δjm . Since
P {ψj (0)} is linearly
independent, so is the set {ψj (t)}, which suggests that φj (t) = j αj ψj (t) is non-zero
everywhere.
Theorem 2.5. Let φ1 (t), . . . , φn (t) be any set of n linearly independent solutions of (2.2),
then every solution is a linear combination of the φj (t).
20
CHAPTER 2. LINEAR EQUATIONS
Proof. Let φ(t) be any non-trivial
solution. {φ1 (t), . . . , φn (t), φ(t)} is then linearly depenP
dent, meaning that φ(t) = j αj φj (t).
Definition 2.3. Let φj (t) be n linearly independent solutions of (2.2), then the n × n matrix
Φ(t) := [φ1 (t), . . . , φn (t)],
is called a fundamental matrix.
Notice, a fundamental matrix is not unique for a particular system, and is related to others by
a non-singular constant matrix. That is,
Φ2 (t) = Φ1 (t) + C.
Theorem 2.6. The solution of ẋ = A(t)x, x(t0 ) = x0 is given by
x(t) = Φ(t)Φ−1 (t0 )x0 .
Proof. Given the definition of a fundamental matrix, we know that our solution is
x(t) = Φ(t)a
for some constant vector a. However, we utilize the initial condition
x(t0 ) = x0 = Φ(t0 )a =⇒ a = Φ−1 (t0 )x0 .
Note, in the autonomous case, Φ(t) = etA .
2.4
Inhomogeneous Linear Equations
In this section, we consider equations of the form
ẋ = A(t)x + f (t),
x(t0 ) = x0 .
For constant A, we can solve this directly by Green’s functions
Z t
0
tA
x(t) = e x(0) +
e(t−t )A f (t0 ) dt0 .
0
We’ll consider the more general case of (2.3), by trying a solution of the form
x(t) = Φ(t)Φ−1 (t0 )[x0 + φ(t)],
(2.3)
2.4. INHOMOGENEOUS LINEAR EQUATIONS
21
where the initial conditions imply that φ(t0 ) = 0. Substituting this back into the ODE yields
Φ̇(t)Φ−1 (t0 )[x0 + φ(t)] + Φ(t)Φ−1 (t0 )φ̇(t) = A(t)Φ(t)Φ−1 (t0 )[x0 + φ(t)] + f (t),
and since Φ̇(t) = A(t)Φ(t), then
Φ(t)Φ−1 (t0 )φ̇(t) = f (t) =⇒ φ̇(t) = Φ(t0 )Φ−1 (t)f (t),
which we can integrate to yield
Z
t
φ(t) = Φ(t0 )
Φ−1 (s)f (s) ds.
t0
Therefore, combining all of this, our solution is
−1
Z
t
x(t) = Φ(t)Φ (t0 )x0 + Φ(t)
Φ−1 (s)f (s) ds.
(2.4)
t0
Solutions of the form (2.4) are said to be obtained by variation of parameters.
Example 2.1. Consider the linear system
ẋ1 = x2 + et
ẋ2 = x1
ẋ3 = te−t (x1 + x2 ) + x3 + 1.
In this case, we split this into our standard notation of ẋ = A(t)x + f (t), where


 t
 
0
1 0
e
0





0
1
0
0
1 .
,
x0 =
A(t) =
,
f (t) =
−t
−t
1
te
te
1
−1
We consider the homogeneous equation (where f ≡ 0), and we see that ẍ1 = x1 , which
leads to
ẋ1 = x2 , ẋ2 = x1 , ẋ3 − x3 = te−t (x1 + x2 ),
and solving for x1 suggests that x1 = e±t , meaning that we now have three possible cases
x1
1
0
t 1
−t
=e
, e
,
.
x2
1
−1
0
Consider the first case, where
ẋ3 − x3 = 2t =⇒ x3 = −2(1 + t) + cet ,
22
CHAPTER 2. LINEAR EQUATIONS
and in the other two cases, x3 = cet . Recall that we need three linearly independent solutions
to construct a fumdamental matrix, which we can do by choosing c appropriately in each
case. One such fundamental matrix is




et
e−t 0
e−t
e−t
0
1
et
−e−t 0  ,
et
−e−t
0 ,
Φ(t) = 
Φ−1 (t) = 
2
t
−2t
−2t
−2(1 + t)
0
e
2(1 + t)e
2(1 + t)e
2e−t
and after some pretty intense algebra, we can conclude
3 1
3 −t
1 1
3
t
x1 =
+ t e − e ,
x2 =
+ t et + e−t ,
4 2
4
4 2
4
2.5
x3 = 3et − t2 − et − 4.
Stability and Boundedness
Theorem 2.7. For the regular linear system (2.1), the zero solution is Lyapunov stable on
t ≥ t0 for an arbitrary t0 if and every if every solution is bounded as t → ∞
Proof. First suppose that the zero-solution x∗ (t) = 0 is stable. This means that ∀ε >
0, ∃δ > 0 such that
|x(t0 )| < δ =⇒ |x(t)| < ε.
Let Ψ(t) = [ψ1 (t), . . . , ψn (t)] be the fundamental matrix with Ψ(t0 ) = δI/2, or [ψj (t0 )]m =
δδj,m /2.
Any solution can be written as x(t) = Ψ(t)c for some constant c, and stability show thats
|ψj (t) | =
δ
< δ =⇒ |ψj (t)| < ε,
2
∀j.
Therefore,
n
n
n
X
X
X
|x(t)| = |
cj ψj (t) ≤
|cj ||ψj (t)| ≤ ε
|cj | < ∞,
j=1
j=1
j=1
thus, the solution is bounded.
In the other direction, suppose that every solution is bounded. Let Φ(t) be any fundamental
matrix. Boundedness implies ∃M > 0 such that kΦ(t)k < M for t > t0 . Given any ε > 0,
let
ε
δ=
.
M kΦ−1 (t0 )k
We can then write any solution as
x(t) = Φ(t)Φ−1 (t0 )x(t0 ),
2.6. PERIODIC SYSTEMS: FLOQUET THEORY
23
and it follows that if |x(t0 )| < δ, then
kx(t)k ≤ kΦ(t)kkΦ−1 (t0 )k|x(t0 )| ≤ M
ε
= ε.
Mδ
Theorem 2.8. All solutions for the inhomogeneous system (2.3) have the same stability
property as the zero solution of the homogeneous system (2.1).
Proof. Let x∗ (t) be a solution to the inhomogeneous equation whose stability we wish to
determine. Let x(t) be any other solution. It follows that if y(t) := x(t) − x∗ (t), then
ẏ(t) = A(t)y(t).
Lyapunov stability of x∗ implies that ∀ε > 0, ∃δ > 0 such that |x(t0 ) − x∗ (t0 )| < δ implies
that |x(t) − x∗ (t)| < ε.
In terms of y, this is equivalent to |y(t0 )| < δ =⇒ |y(t)| < ε. This is exactly the condition
for Lyapnuov stability of the zero solution determined by the previous theorem.
2.6
Periodic Systems: Floquet Theory
Consider a non-autonomous linear system
x ∈ Rn ,
ẋ = P (t)x,
P (t + T ) = P (t),
∀t,
(2.5)
that is, T is the minimal period of P . Such an equation may arise from linearizing about a
limit cycle solution of a standard homogeneous equation.
Theorem 2.9 (Floquet). The regular system (2.5), where P is an n × n has at least one
non-trivial solution χ(t) such that
χ(t + T ) = µχ(t),
where µ is a constant called a Floquet multiplier.
Proof. Let Φ(t) = [φ1 (t), . . . , φn (t)] be a fundamental matrix. Then,
Φ̇(t) = P (t)Φ(t),
but since P (t + T ) = P (t), then we also have that
Φ̇(t + T ) = P (t)Φ(t + T )
24
CHAPTER 2. LINEAR EQUATIONS
Denote Ψ(t) = Φ(t + T ), which is also a fundamental matrix. It follows from the linear
independence of the solutions that ψi (t) = φi (t + T ) are linear combinations of φk (t),
meaning that
φi (t + T ) = Φ(t)ei =⇒ Φ(t + T ) = Φ(t)E,
E := [e1 , . . . , en ],
where det E 6= 0. Let µ be an eigenvalue of E and let v be the corresponding eigenvector.
That is, (E − µI)v = 0, and consider the solution χ(t) = Φ(t)v, which suggests
χ(t + T ) = Φ(t + T )v = Φ(t)Ev = µΦ(t)v = µχ(t).
The eigenvalues of E are called the Floquet multipliers or characteristic numbers, where
|µ| < 1 implies convergence in some sense.
Theorem 2.10. The constants µ are independent of the choice of Φ.
Proof. Let Φ(t) and Φ∗ (t) be two fundamental matrices. Then, we know these differ by a
constant, Φ∗ (t) = Φ(t)A, where det A 6= 0, and therefore
Φ∗ (t + T ) = Φ(t + T )A
= Φ(t)EA
= Φ∗ (t)A−1 EA
= Φ∗ (t)E ∗ ,
E ∗ := A−1 EA.
Clearly E, E ∗ have the same eigenvalues, but more explicitly
det (E ∗ − µI) = det A−1 (E − µI)A = det A−1 A det(E − µI) = det(E − µI).
If we choose Φ(t) such that Φ(0) = I, then E = Φ(T ).
Definition 2.4. Let µ be a Floquet multiplier. Set µ = eρT . We call ρ a characteristic or
Floquet exponent. Note that ρ is only defined up to an additive multiple of 2πi/T . Make
the restriction that −π < Im(ρT ) < π.
Theorem 2.11. Suppose that the matrix E for which Φ(t + T ) = Φ(t)E has n distinct
eigenvalues µi , then (2.5) has n linearly independent solutions of the form
χi (t) = pi (t)eρi t ,
pi (t + T ) = pi (t).
Proof. For each µi = eρiT , there exists a solution
χi (t + T ) = µi χi (t) = eρi T χi (t),
which suggests that
χi (t + T )e−ρi (t+T ) = χi (t)e−ρi t .
|
{z
} | {z }
pi (t+T )
pi (t)
2.6. PERIODIC SYSTEMS: FLOQUET THEORY
25
The linear independence follows from the construction that χi (t) = Φ(t)vi . For distinct
eigenvalues µi , we know that the eigenvectors vi are linearly independent. Since Φ(t) is
non-singular, we see that hte solutions χi (t) are also linearly-independent.
Stability of a Limit Cycle
Suppose that ẋ = f (x) has a limit cycle solution u(t) with period T . Linearizing around the
limit cycle, taking v := x − u, we see that
v̇ = ẋ − u̇ = f (u + v) − f (u(t)) = Df (u(t))v(t)
(by Taylor expansion),
which allows us to conclude that
v̇ = P (t)v(t).
Note that
u̇ = f (u(t)) =⇒ ü = Df (u(t))u̇(t),
d
(u̇) = Df (u(t))u̇(t),
dt
which suggests that one solution of v̇ = P (t)v(t) is v = u̇.
Since u̇ is a T -periodic solution, it must have a Floquet multiplier of µ = 1, which means
that phase shifts on the limit cycle persist.
Also, u̇(t) is the vector tangent to the limit cycle at any given time t. The existence of a
multiplier with µ = 1 reflects the phase-shift invariance of an autonomous system (with
regard to drifts around the limit cycle). The limit cycle is locally stable provided the other
n − 1 Floquet multipliers lie within the unit circle.
Example 2.2. Consider the example where
ẋ1
1 1
x1
=
,
ẋ2
0 h(t)
x2
h(t) = (2 + sin t − cos t)−1 (cos t + sin t),
and note that
d
ẋ2 h(t)x2 =⇒
dt
x2
2 + sin t − cos t
=0
=⇒
x2 (t) = b(2t sin t − cos t).
Then, we examine the next equation
ẋ1 −x1 = b(2t sin t−cos t) =⇒
d −t e x1 = e−t b(2+sin t−cos t) =⇒ x1 (t) = aet −b(2t sin t),
dt
where we have made use of an integrating factor to solve for x1 . Note that we can construct
two linearly independent solutions by setting either a = 1, b = 0 or a = 0, b = 1, yielding
−2 − sin t
et
Φ(t) =
,
2 + sin t − cos t 0
26
CHAPTER 2. LINEAR EQUATIONS
and introduce the non-singular matrix E such that Φ(t + 2π) = Φ(t)E, since it’s clear our
system is 2π periodic. We compute E directly
−1 −2 1
−2 e2π
1 0
−1
E = Φ (0)Φ(2π) =
=
.
1 0
1
0
0 e2π
Thus, our eigenvalues here are µ1 = 1, µ2 = e2π which correspond to ρ1 = 0, ρ2 = 1. The
general solution is then
−2 − sin t
1 t
0t
1t
x(t) = p1 (t)e + p2 (t)e = b
+a
e,
2 + sin t − cos t
0
Note that p1 , p2 are each 2π periodic.
Definition 2.5. Let Φ(t) = [φ1 , . . . , φn ] be a fundamental solution with φ̇j = P (t)φj , then
W (t) := det Φ(t),
is called the Wronskian of the system.
Theorem 2.12 (Liouville’s (or Abel’s) formula). Let Φ be any fundamental matrix of (2.5),
then for any arbitrary t0 ,
Z
t
W (t) = W (t0 ) exp
tr{P (s)} ds .
t0
Proof. Using Liebnitz’s rule, we have that
n
X
dW
=
∆k ,
where
dt
k=1


φ̇11 · · · φ̇n1

.. 
..
∆1 := det  ...
.
. 
φ̇1n · · · φ̇nn
P

P
m p1m φ1m · · ·
m p1m φnm


..
..
..
= det 
.

.
.
φ̇1n

···
φnn

φ11 · · · φn1
 ..
..  ,
..
=
p1m det  .
.
. 
m=1
φ1n · · · φnn
n
X
and therefore,
n
X
dW
=
pkk W (t) = tr{P }W (t) =⇒ W (t) = W (t0 ) exp
dt
k=1
Z
t
tr{P (s)} ds .
t0
2.6. PERIODIC SYSTEMS: FLOQUET THEORY
27
Theorem 2.13. For the system (2.5) where P (t) is T -periodic, then the Floquet multipliers
satisfy
n
Y
T
Z
µj = exp
tr{P (s)} ds .
0
j=1
Proof. Let Ψ(t) be the fundamental matrix with Ψ(0) = I =⇒ E = Ψ(T ). Our multipliers
satisfy the characteristic equation
0 = det{E − µI} =
n
Y
(µj − µ) =⇒ det E =
j=1
n
Y
µj ,
j=1
and therefore, by the previous theorem
n
Y
Z
T
µj = det Ψ(T ) = W (T ) = exp
tr{P (s)} ds .
0
j=1
The usefulness of this theorem is: if we know some information about our Floquet multipliers,
we can use this relation to establish information about the others. Next is an example of
doing so.
Example 2.3 (Lienard equation). Consider the generalized Lienard equation
ẍ + f (x)ẋ + g(x) = 0,
and suppose that there exists a T periodic solution x(t) = φ(t). Linearizing about this
solution and rewriting as a first order system where ẋ = y and ẏ = −f (x)y − g(x) yields
ẋ
0
1
x
x
=
= P (t)
.
ẏ
−Df (φ)y − Dg(φ) −f (φ)
y
y
Note that tr P (t) = −f (φ(t)), and we know that since there is a periodic solution, µ1 = 1,
and therefore, µ2 must satisfy
Z
µ2 = exp −
T
f (φ(s)) ds ,
0
and it follows that the limit cycle is locally stable provided that
Z
T
f (φ(s)) ds ≥ 0 =⇒ µ2 ≤ 0.
0
28
CHAPTER 2. LINEAR EQUATIONS
Example 2.4 (Mathieu equation). Consider the equation
ẍ + (α + β cos t)x = 0,
(2.6)
which can be rewritten as the first order system
ẋ = P (t)x,
P (t) =
0
1
.
−α − β cos t 0
Note that tr P = 0, which suggests that µ1 · µ2 = 1. Hence, µ1 , µ2 satisfy a quadratic
characteristic equation
p
1
µ2 − φ(α, β)µ + 1 = 0 =⇒ µ1,2 = φ ± φ2 − 4.
2
The behavior of solutions therefore depends on the unknown function φ(α, β). There are
several cases:
(i) If φ > 2, then µ1,2 are both real and positive. Setting µ1,2 = exp{±σT }, where
T = 2π, we can write the general solution as
χ(t) = c1 p1 (t)eσt + c2 p2 (t)e−σ t,
where p1,2 (t) are each 2π periodic. Note that this is unbounded.
(ii) If φ = 2, then µ1 = µ2 = 1, meaning that ρ1 = ρ2 = 0. There is one solution of
period 2π, a limit cycle.
(iii) If −2 < φ < 2, then µ1,2 are of the form of a complex conjugate pair. Since |µ1,2 | = 1,
it follows that ρ1,2 = ±iν. Therefore, the general solution is
χ(T ) = c1 eiνt p1 (t) + c2 e−iνt p2 (t),
which is a quasi-periodic solution for ν/2π irrational. Also, this solution is bounded.
(iv) If φ = −2, this suggests that µ1 = µ2 = −1 and ρ1,2 = i/2. There exists a solution
satisfying χ(t + 2π) = −χ(t), which means our solution is now 4π periodic. The
other solution is unbounded.
(v) If φ < −2, then µ1,2 are real, distinct, and negative, and our solution is therefore
χ(t) = c1 e(σ+i/2)t p1 (t) + c2 e(−σ+i/2)t p2 (t),
where σ > 0, and p1 , p2 are 2π periodic. Yet again, this solution is unbounded.
2.6. PERIODIC SYSTEMS: FLOQUET THEORY
29
We can now consider the transition curves separating these cases. Specifically, the curves
where φ(α, β) = ±2 partition the parameter space where all solutions are bounded (|φ| < 2)
from regions where all solutions are unbounded (|φ| > 2).
Although φ is not known explicitly, we do know that along a transition curve, there are
solutions with period 2π when φ = 2 or 4π, when φ = −2. We can determine the parameter
values at which these periodic solutions exist.
A 2π periodic solution can be represented by the Fourier series
∞
X
χ(t) =
cn eint ,
n=−∞
which when we substitute into the Mathieu equation, yields
X
X
β it
2
int
−it
cn eint = 0,
−
n cn e + α + (e + e )
2
n
n
and after collecting coefficients, we have the relation
1
1
βcn+1 + (α − n2 )cn + βcn−1 = 0.
2
2
Assuming α 6= n2 , this can be thought of as a tridiagonal “infinite matrix” equation
Γc = 0,
. .
.. . . . . .
.
.. γ1 1


Γ = ... 0 γ0

..
. 0 0
.. . . . .
.
.
.
.
. ..
.
γ1 0 0 ..

.. ,
1 γ0 0 .

.
γ1 1 γ1 ..
. . . . . . ..
.
.
. .
..
.
..
.
..
γn :=
1 β
.
2 α − n2
The curve φ(α, β) = 2 corresponds to values of α for which det Γ = 0. The 4π periodic
solution can similarly be expanded as
X
1
1 2
1
int/2
χ(t) =
dn e
=⇒ βdn+2 + α − n dn + dn−2 = 0.
2
4
2
n
This equation splits into odd n and even n, and again leads to a similar “infinite matrix”
equation. The even set leads to det Γ = 0, which is because a 2π periodic solution is also
4π periodic. For the odd coefficients, we obtain a new condition det Γ̂ = 0.
The resulting transition curves are a complex tongue structure that get smaller from left to
right.
30
CHAPTER 2. LINEAR EQUATIONS
Figure 2.1: Transition curves of the Mathieu equation.
We can perform a perturbation analysis on this equation to get the leading order behavior of
the transition curves, as well. Again, consider the Mathieu equation (2.6), and suppose that
|β| is small. Then, we take the power expansion of α
α = α(β) = α0 + α1 β + α2 β 2 + · · · ,
and we also take the series expansion of either the 2π or 4π periodic
x(t) = x0 (t) + βx1 (t) + β 2 x2 (t) + · · · .
Substituting into the Mathieu equation and collecting powers of β, we find the first few
powers
ẍ0 + α0 x0 = 0
ẍ1 + α0 x1 = −(α1 + cos t)x0
ẍ2 + α0 x2 = −α2 x0 − (α1 cos t)x1 .
Since we desire a period of 2π, we take α0 = n2 , where n = 0, 1, . . . and period 4π if
α0 = (n + 1/2)2 . Both solutions can be considered together by writing α0 = n2 /4, again,
where n is an integer. For the sake of illustration, take n = 0 or n = 1.
First, considering n = 1, we have α0 = 1/4, and so
t
t
+ b0 sin ,
2
2
which, when plugging into the second term in our expansion yields
1
t
1
t a0
3t b0
3t
ẍ1 + x1 = −a0 α1 +
cos − b0 (α1 − ) sin − cos − sin ,
2
2
2
2
2
2
2
2
x0 = a0 cos
where we clearly have the freedom to choose a0 and b0 , but we want our solution to be
bounded, so we must avoid resonance. That is, we want to avoid solutions that match our input
frequencies. We can do this either by taking b0 = 0, α1 = −1/2 or by a0 = 0, α1 = 1/2.
Hence, α = 1/4 ± β/2 + O(β 2 ), which is the shape of a tongue starting at 1/4.
3
Boundary Value Problems
“I have to drink for medical reasons. Well, I guess that’s a lie. I just drink for
regular reasons."
— Paul C. Bressloff
3.1
Preliminaries
Definition 3.1. Let V be a complex vector space. An inner product is the mapping
h·, ·, i : V ⊕ V → C, with the following properties:
(1) hα1 f1 + α2 f2 , gi = α1∗ hf1 , gi + α2∗ hf2 , gi.
(2) hf, f i > 0f for f 6≡ 0.
(3) hf, gi = hg, f i∗ .
Associated with every inner product is an associated norm kf k =
p
hf, f i.
Definition 3.2. If V is complete with respect to its natural norm, then it is a Hilbert space.
Theorem 3.1 (Cauchy-Schwarz inequality). If f, g ∈ V , where V is an inner product
space, then
| hf, gi | ≤ kf k kgk.
Proof. The proof is obviously true for g or f identically equal to 0, so take g 6= 0 without
loss of generality, then take
h := f −
hf, gi g
=⇒ hh, gi = 0,
kgk2
and therefore
2
2
2
hf,
gi
g
= hh, hi + | hf, gi | hg, gi ≥ | hf, gi | ,
kf k = h
+
kgk2 kgk2
kgk2
2
31
32
CHAPTER 3. BOUNDARY VALUE PROBLEMS
and multiplying by kgk2 and taking square roots, we are done. If h = 0, this implies that
f, g are linearly independent.
Suppose that {uj }nj=1 is an orthonormal set. Then for every f ∈ V , we can write
f=
n
X
huj , f i uj + f⊥ ,
j=0
since
*
uk , f −
n
X
+
huj , f i uj
= huk , f i − huk , f i = 0,
j=0
since huj , uj ki = δjk , so that f⊥ is orthogonal to the orthonormal set. Also note that
+
*
n
n
X
X
huk , f i uk
kf k2 = f⊥ +
huj , f i uj , f⊥ +
j=0
n
X
k=0
= kf⊥ k2 +
= kf⊥ k2 +
huj , f i huk , f i huj , uk i
j,k=0
n
X
| huj , f i k2 .
j=0
This leads us directly to the following theorem:
Theorem 3.2 (Bessel’s inequality). Let f ∈ V and {uj }nj=0 be an orthonormal set. Then
n
X
| huj , f i |2 ≤ kf k2 .
j=0
3.2
Linear Operators
Definition 3.3. A linear operator is a linear mapping A : D(A) → H, where D(A)
denotes the domain of A which is a linear subspace of H, some Hilbert space.
Definition 3.4. A linear operator A is symmetric if its domain is dense (in H)and if
hg, Af i = hAg, f i ,
∀f, g ∈ D(A).
Definition 3.5. A number z ∈ C is called an eigenvalue of A if there exists a non-zero
vector u ∈ D(A) such that
Au = zu.
3.2. LINEAR OPERATORS
33
The eigenspace is
kern{A − zI} = {u ∈ D(A) : (A − z)u = 0}.
An eigenvalue is said to be simple if
dim(kern{A − zI}) = 1.
Theorem 3.3. Let A be a symmetric linear operator, then all the eigenvalues are real and
the eigenvectors of distinct eigenvalues are orthogonal.
Proof. Suppose λ is an eigenvalue with normalized eigenvector u, then
λ = hu, Aui = hAu, ui = λ∗ =⇒ λ ∈ R.
Consider Au1 = λ1 u1 and Au2 = λ2 u2 , then
(λ1 − λ2 ) hu1 , u2 i = hAu1 , u2 i − hu1 , Au2 i = 0 =⇒ u1 ⊥ u2 if λ1 6= λ2 .
Note that the theorem says nothing about the existence of eigenvalues.
Definition 3.6. The linear operator A for D(A) = H is bounded if
kAk := sup kAf k < ∞,
kf k=1
and by construction
kAf k ≤ kAkkf k,
meaning that, boundedness implies continuous for a linear operator.
Definition 3.7. A linear operator with D(A) = H is compact if every sequence {Afn } has
a convergent subsequence whenever {fn } is bounded. Note that by definition, every compact
operator is bounded.
Theorem 3.4. A compact symmetric operator has an eigenvalue λ0 for which
|λ0 | = kAk.
Proof. Set a = kAk with a 6= 0. Since
kAk2 = sup kAf k2 = sup hAf, Af i = sup f, A2 f ,
kf k=1
kf k=1
kf k=1
we can conclude there is a normalized sequence {un } such that
a2 = lim un A2 un .
n→∞
34
CHAPTER 3. BOUNDARY VALUE PROBLEMS
Since A is compact, we can choose {un } such that {A2 un } converges, which implies that
there exists some u
lim A2 un = au.
n→∞
Now, note that
k(A2 − a2 )un k2 = kA2 un k2 − 2a2 un , A2 un + a4 ≤ 2a2 (a2 − un , A2 un ).
Hence, we see that
lim (A2 − a2 )un = 0 =⇒ lim un = u,
n→∞
n→∞
which implies that
(A2 − a2 )u = 0 = (A + a)(A − a)u = 0.
Thus, we have either that
(A − a)u = 0 =⇒ λ0 = a or v = (A − a)u 6= 0 =⇒ (A + a)v = 0 =⇒ λ0 = −a.
Note that, for a bounded operator, there does not exist an eigenvalue for which |λ| > kAk.
We can iterate the previous theorem to establish the existence of an orthonormal basis of
eigenvalues. To do so, set
H (1) = {f ∈ H : hu0 , f i = 0}, where Au0 = λ0 u0 .
If f ∈ H (1) , then we have that
hAf, u0 i = λ0 hf, u0 i = 0 =⇒ Af ∈ H (1) .
Denote the restriction of A to H (1) by A1 . This is also a compact symmetric operator.
Applying the above theorem, there exists some eigenvalue λ1 and a normalized eigenvector
u1 such that
A1 u1 = λ1 u1 , |λ1 | = kA1 k < kAk,
hu1 , u0 i = 0.
Hence, we generate a sequence of eigenvalues {λj } and normalized eigenvectors {uj } such
that uj huik for k < j. This procedure will not stop unless H is finite dimensional, however
λj = 0,
∀j ≥ n if An = 0.
Theorem 3.5 (Spectral theorem for compact operators). Suppose H is a an inner product
space and A : H → H is a compact operator.
Then there exists a sequence of real eigenvalues {λj } such that λj → 0. The corresponding
normalized eigenvectors form an orthonormal set and every element f ∈ R(A), where R(A)
is the range of A, can be written as
f=
N
X
huj , f i uj
j=0
If R(A) is dense, then the eigenvectors form an orthonormal basis.
(3.1)
3.3. DIFFERENTIAL OPERATORS
35
Proof. Without loss of generality, take H to be infinite dimensional. The existence of the
eigenvalues λj , which we know are real, and eigenvectors uj has been established by the
previous theory.
If λj 6→ 0, then there exists a bounded subsequence vk = uk /λk for which
kAvk − Avi k2 = kuk − ui k2 = 2,
which violates the assumption of compactness. Next, let f = Ag ∈ R(A) and set
fn =
n
X
huj , f i uj ,
gn =
j=0
Observe that
fn =
n
X
n
X
huj , gi uj .
j=0
huj , Agi uj =
j=0
n
X
λk huj , gi uj = Agn .
j=0
Thus, we see that
kf − fn k = kA(g − g − n)k = kAn+1 (g − gn )k ≤ |λn+1 |kg − gn k ≤ |λn+1 |kgk,
since g − gn ∈ H (n+1) . Letting n → ∞ shows that fn → f , which proves (3.1).
Finally, let f ∈ H and suppose R(A) is dense. For a fixed ε > 0, there exists an fε ∈ R(A)
such that kf − fε k < ε/2. Moreover, from the previous step, there exists ∃fˆε in the span of
{uj } for sufficiently large n such that
kfε − fˆε k < ε/2 < ε,
and it follows that, from Bessel’s inequality,
kf − fn k ≤ kf − fˆε k < ε, for sufficiently large n.
We’d like to use the Spectral theorem for differential operators, but unfortunately, differential
operators are not compact. In fact, they’re often not even bounded. In a few sections, we’ll
study Green’s functions which remedy this problem, as they are effectively the inverse of
differential operators, and are compact.
3.3
Differential Operators
For this section, define the formal differential operator on [a, b] to be
L := p0 (x)
dn
dn−1
+
p
(x)
+ · · · + pn (x),
1
dxn
dxn−1
36
CHAPTER 3. BOUNDARY VALUE PROBLEMS
and is defined without worrying about the domain D(L) of functions it is applied to.
A general linear differential operator is defined to be a formal linear differential operator
together with boundary conditions.
We also require that D(L) to contain only functions that are sufficiently differentiable so
that if f ∈ D(L) then f ∈ L2 [a, b], say, and Lf ∈ L2 [a, b], which is then further restricted
by the boundary conditions.
Adjoint Operators
Given a formal LDO and a weight function w(x) ∈ R such that w(x) > 0 for x ∈ [a, b], we
construct antoher operator L† such that for any sufficiently differentiable u(x) and v(x), we
have
h
∗ i
d
Q[u, v],
w u∗ Lv − v L† u
=
dx
for some function Q, which is bilinear in u, v and their first n − 1 derivatives. In this case,
L† is called the formal adjoint of L.
If we now define an inner product,
b
Z
wu∗ v d,
hu, wiw :=
a
b
and if u, v have boundary conditions such that Q[u, v] = 0 then
a
†
hu, Lviw = L u, v .
Example 3.1 (Sturm-Liouville Operator). The general second order operator is defined by
d2
∂
+
p
(x)
+ p2 (x),
1
dx2
∂x
where pi (x) is real and w(x) = 1. This is a general second order operator. Now, we use
integration by parts to find the adjoint.
Z b
hu, Lvi =
u∗ [p0 v̈ + p1 v̇ + p2 ] dx
a
b Z b
b Z b
Z b
d
d
∗
∗
∗
∗
= u p0 v̇ −
{p1 u } v̇ dx + u p1 v −
{p1 u } v dx +
u∗ p2 dx
dx
dx
a
a
a
a
a
L := p0 (x)
Z
=
a
b
d2
d
v
{p0 u} − {p1 u} + p2 u
2
dx
dx
∗
h
ib
d
dx + u∗ p0 v̇ − v {p0 u∗ } + u∗ p1 v .
dx
a
3.3. DIFFERENTIAL OPERATORS
37
From this, we can conclude two things. The adjoint operator:
d2
d2
d
d
0
p
+
p
=
p
+ p000 − p01 + p2 ,
p
−
+
(2p
−
p
)
1
2
0
0
1
0
dx2
dx
dx2
dx
and the boundary conditions
L† =
Q[u, v] = u∗ p0 v̇ − vp0 u̇∗ + (p1 − p00 )u∗ v.
Ideally, we want a formally self-adjoint operator, that is L = L† . In this case, we require that
2p00 − p1 = p1 =⇒ p00 = p1 ,
and also
p000 − p01 + p2 = p2 =⇒ p000 = p01 =⇒ p1 = p00 ,
and the resulting operator is
d n do
p0
+ p2 ,
dx
dx
which is called the Sturm-Liouville operator, and is by construction, self-adjoint.
L=
One way to make a general second order operator formally self adjoint is to introduce a
suitable weight function w(x). Consider
L := p0
d2
d
+ p1
+ p2 ,
2
dx
dx
and suppose that p0 is positive definite on [a, b] and pj ∈ R. Take w to be
Z y
p1 (y)
1
w=
exp
dy ,
p2
a p0 (y)
and then
Lv =
1
0
(wp0 v 0 ) + p2 v,
w
and therefore
Z
b
nd
o
{wp0 v̇} + p2 vw dx
dx
a
b Z b
= u∗ wp0 v 0 −
w {p0 v̇ u̇∗ − p2 vu∗ } dx
a
a
Z bn
o
d
=
v {wp0 u̇∗ } + vp2 u∗ dx + [wp0 (u∗ v̇ − u̇∗ v)]ba
dx
a
b
Z b
∗
=
v(x) (Lu) dx + Q(u, v) .
hu, Lviw =
a
u∗
a
Recall that a finite self-adjoint (Hermitian) matrix has a complete set of orthonormal
eigenvectors, which extend to the range of a compact symmetric operator.
38
CHAPTER 3. BOUNDARY VALUE PROBLEMS
Example 3.2 (A simple eigenvalue problem). Consider the linear differential operator
L := −
d2
,
dx2
D(L) := f, Lf ∈ L2 [0, 1], f (0) = f (1) = 0 .
We take the standard inner product with w = 1, which means that
1
hf1 , Lf2 i − hLf1 , f2 i = [f10∗ f2 − f1∗ f20 ]0 = 0,
by the boundary conditions. Therefore, this operator is not only formally self adjoint, but
also self adjoint since the domains of each are teh same.
The eigenvalue equation Lψ = λψ, or equivalently
ψ̈ + λψ = 0,
ψ(1) = ψ(0) = 0,
has solutions ψn (x) = sin nπx with λn = n2 π 2 , which tells us that the eigenvalues are
real, and the eigenfunctions are orthogonal. We know from Fourier analysis that (once
normalized), these eigenfunctions are complete, that is any f ∈ L2 [0, 1] has a convergent
expansion
∞
X
√
f (x) =
an 2 sin nπx.
n=1
Thus, this example behaves in the nicest way. Consider now, a different operator:
L := −i
d
,
dx
with the same boundary conditions. Note now,
Z 1
∗
hf1 , Lf2 i − h(Lf1 ) , f2 i =
[f1∗ (−if20 ) − (−if10 )∗ f2 ] dx = −i[f1∗ f2 ]10 = 0.
0
Thus, L is self-adjoint, but the eigenvalue equation becomes
−i
d
ψ = λψ =⇒ ψ(x) = eiλx .
dx
Clearly, this cannot satisfy the boundary conditions, and therefore there are no eigenvalues.
Our ODE is overdetermined. Thus, this illustrates that differential operators that are similar
can behave very differently.
Adjoint Boundary Conditions
We can consider removing the requirement that D(L) = D(L† ), that is, we choose the
domain of L† such that for any u ∈ D(L† ) and v ∈ D(L), then we have that Q[u, v] = 0 by
the adjoint boundary condition.
3.3. DIFFERENTIAL OPERATORS
39
Example 3.3. Consider again the operator described by
L := −i
Now, note that
Z 1
0
∗
d
,
dx
D(L) := {f : Lf ∈ L2 [0, 1], f (1) = 0}.
0
∗
∗
u (−iv ) dx = −i{ u (1)v(1) −u (0)v(0)} +
| {z }
Z
1
{−iu0 }∗ v dx,
0
=0 since v(1)=0
and then we choose the boundary condition for u such that the other term vanishes, for
instance, u(0) = 0, meaning that our adjoint is
d
D(L† ) = {f : L† f ∈ L2 [0, 1], f (0) = 0}.
L† = −i ,
dx
Note that since the form of the operator is the same, L and L† are formally self adjoint, but
since D(L) 6= D(L⊥ ), these operators are not self-adjoint (or symmetric).
Self-Adjoint Boundary Conditions
A formally self-adjoint operator L is truly self adjoint if and only if L = L† and D(L) =
D(L† ). We’ll consider an example of such similar to the one above.
Example 3.4. Consider the same formal operator, with no specifications on boundary
conditions
L := −i
d
,
dx
which, when checking the inner product, yields
Z 1
Z 1
∗
0
u (−iv ) dx −
(−iu0 )∗ v dx = −i {u∗ (1)v(1) − u∗ (0)v(0)} .
0
0
For this to be truly self adjoint, we need the right hand side to vanish, or, in other words,
u∗ (1)
v(0)
=
=k
∗
u (0)
v(1)
(constant).
However, we need D(L) = D(L† ), so we can interchange the requirement of u, v and also
produce the requirement
u(1)
v(1)
=
,
u(0)
v(0)
which implies that k ∗ = k −1 or that k = eiθ for θ ∈ R. Therefore, we have twisted
boundary conditions, which require that
u(1)
v(1)
=
= eiθ =⇒ D(L) = D(L† ) = {f : Lf ∈ L2 [0, 1], f (1) = eiθ f (0)}.
u(0)
v(0)
The eigenfunctions of this operator are then un = exp{(2πn + θ)ix}, which form an
orthonormal set, with eigenvalues 2πn + θ, which are all real.
40
CHAPTER 3. BOUNDARY VALUE PROBLEMS
Example 3.5 (Sturm-Liouiville operator). Recall the Sturm-Liouville operator
d n
do
p(x)
+ q(x),
dx
dx
For this to be self-adjoint, we require that
x ∈ [a, b].
L :=
b
hu, Lvi − hLu, vi = [p(u∗ v 0 − u0∗ v)]a = 0,
which is equivalent to
u0∗ (a)
v 0 (a)
=
u∗ (a)
v(a)
and
v 0 (b)
u0 (a)
=
,
v(a)
u(a)
and therefore,
αy(a) + βy 0 (a) = 0,
3.4
α̂y(b) + β̂y 0 (b) = 0, for y = u, v.
Eigenvalue Problems
It can be proven that a truly self-adjoint linear differential operator with respect to the L2 [a, b]
inner product possesses a complete set of mutually orthogonal eigenfunctions. The set of
eigenvalues belongs to the spectrum σ(L) , and forms the point spectrum when an eigenvalue
is associated with an eigenfunction in L2 [a, b].
When an operator acts on a function over an infinite interval, then the eigenfunctions may
fail to be normalizable (recall that we had e−iλψ as eigenfunctions to −id /dx ), and the
associated eigenvalues are then said to belong to the continuous spectrum. The spectrum
may be partially complete and partially continuous. There is also a third component called
the residual spectrum, but this is empty for self-adjoint operators. We will look at these sets
more thoroughly next.
Discrete Spectrum
Let L be a self-adjoint operator on [a, b], and consider
Z b
Lφn (x) = λn φn (x),
n ∈ Z,
φ∗n (x)φm (x) dx = δnm .
a
Completeness of these eigenfunctions can be expressed by the condition
X
φn (x)φ∗n (y) = δ(x − y).
n∈Z
Multiplying both sides by f (y) and integrating with respect to f (y) yields
Z b
X
f (x) =
φn (x)
φ∗n (y)f (y) dy =⇒ hφn , f i = f (x).
n∈Z
a
3.4. EIGENVALUE PROBLEMS
41
Continuous Spectrum
Consider the linear differential operator on R:
H := −
d2
,
dx2
and consider only the interval [−L/2, L, 2] and impose periodic boundary conditions
φ(−L/2) = φ(L/2) on the eigenvalue equation
−φ̈ = λφ =⇒ φ(x) = eikx , λ = k 2 .
By the boundary conditions, eikL = 1 and therefore we have discrete eigenvalues kn = 2πn/L,
so our eigenfunctions are
1
φn (x) = √ exp{ikn x}.
L
The completeness condition is then
X1
exp{ikn (x − x0 )} = δ(x − x0 ),
L
n∈Z
x, x0 ∈ [−L/2, L/2].
As L → ∞, the eigenvalues become so close together that they’re indistinguishable and the
spectrum becomes R itself. The sum over n now becomes an integral
Z
dk
1X
f (kn ) →
f (k).
L n∈Z
R 2π
In the limit, the completeness condition still holds
Z
0 dk
eik(x−x )
= δ(x − x0 ),
2π
R
which we can think of as the Fourier transform limit of Fourier series. When L = ∞,
φk (x) = eikx is no loner normalizable:
Z
Z
2
|φk (x)| dx =
1 dx = ∞ =⇒ φk 6∈ L2 ,
R
R
so strictly speaking, φk are no longer eigenfunctions. More rigorously, a point λ ∈ σC (L),
the continuous spectrum, if ∀ε > 0, we can construct an approximate eigenfunction φε
where kφε k = 1 but
kHφε − λφε k < ε.
We clearly want this to be 0 but we cannot due to the non-normalizability of the eigenfunctions.
In other words, the inverse operator (H − λI)−1 is unbounded.
42
CHAPTER 3. BOUNDARY VALUE PROBLEMS
Mixed Spectrum
Consider the eigenvalue equation
d2
2
Hψ := − 2 − 2 sech x ψ = λψ,
dx
x ∈ R.
We’ll try a solution of the form
ψk (x) = (a+b tanh x)e
ikx
=⇒
d2
2
− 2 − 2 sech x ψk = k 2 ψk (x)−2(ikb+a) sech2 xeikx .
dx
The second term vanishes by our choice of a, b, and therefore, we’re left with the equation
Hψk = k 2 ψk
provided that ψk = b[−ik + tanh x]eikx .
We can attempt to normalize this in some sense, that is,
ψk (x) = √
1
0
eikx [−ik + tanh x] =⇒ ψk∗ (x)ψk (x0 ) = e−ik(x−x ) as |x| → ∞.
2
1+k
In this case, the spectrum σ(H) contains a continuum of states with λ = k 2 . Consider the
special case of λ = 0, in which we have
∂ 2 ψ0
1
= −2 sech2 xψ0 (x) =⇒ ψ0 (x) = √ sech x,
2
∂x
2
thus, since we’ve found a normalizable eigenfunction, λ = 0 is actually a discrete eigenvalue.
Consider now the completeness condition, or rather, the deviation from the completeness
condition:
Z ∞
dk ∗
0
I = δ(x − x ) −
ψk (x)ψk (x0 ) 6= 0 (because we have a discrete eigenvalue)
2π
−∞
Z ∞
o
dk n −ik(x−x0 )
=
e
− ψk∗ (x)ψk (x0 )
2π
Z−∞
∞
dk −ik(x−x0 ) 1
=
e
{1 + ik(tanh x − tanh x0 ) − tanh x tanh x0 −} . (3.2)
2
2π
1
+
k
−∞
To evaluate this integral, we cite a contour integration, which results in
Z ∞
0
dk e−ik(x−x )
1
0
= e−|x−x |
2
2
−∞ 2π 1 + k
as well as the result that
Z ∞
−∞
(3.3)
0
dk e−πk(x−x ) ik
i
0
= sgn(x − x0 )e−|x−x | .
2
2π 1 + k
2
(3.4)
3.4. EIGENVALUE PROBLEMS
43
Using the two integrals (3.3) and (3.4) to evaluate (3.2), we see that
I=
1
0
{1 + sgn(x − x0 )(tanh x − tanh x0 ) − tanh x tanh x0 } e−|x−x | ,
2
and assuming x > x0 , this reduces to
1
1
0
I = (1 + tanh x)(1 − tanh x0 )e−(x−x ) = sech x sech x0 = ψ0 (x)ψ0 (x0 ).
2
2
Thus, this results in the completeness relation
Z ∞
dk ∗
0
ψ0 (x)ψ0 (x ) +
ψk (x)ψk (x0 ) = δ(x − x0 ).
2π
−∞
If this were a sum of discrete eigenvalues, we would have to deliberately neglect k = 0, but
since k = 0 has measure zero in the integral, it won’t be captured even if we still integrate
over R.
Rayleigh-Ritz Variational Principle
The eigenvalues of the Sturm-Liouville operator satisfy the conditions
Lφ := −(pφ0 )0 +qφ = λφ,
x ∈ [a, b],
αφ(a)+βφ0 (a) = 0,
α̂φ(b)+ β̂φ0 (b) = 0.
Assume that there exists countably infinite eigenvalues that are real valued and non-negative
which correspond to a set of eigenfunctions that form a complete orthonormal set. That is,
b
Z
0 ≤ λ1 ≤ λ2 ≤ · · · ,
φm (x)φ∗n (x) dx = δmn .
hφm , φn i =
a
Consider a smooth bounded function u with eigenfunction expansion
X
u(x) =
uk φk (x),
uk = hu, φk i ,
hu, ui = 1.
k∈Z
Introduce the energy integral (for a R-valued u):
Z
E[u] := hu, Lui =
b
Z
u(x)Lu(x) dx =
a
a
b
0 2
2
p(u ) + qu
b
dx − puu ,
0
a
after integrating by parts. Using the boundary conditions at x = a, b, this reduces to
Z
E[u] =
a
b
α̂
α
p(u0 )2 + qu2 dx + p(b)u2 (b) − p(a)u2 (a).
β
β̂
44
CHAPTER 3. BOUNDARY VALUE PROBLEMS
We can also calculate the energy integral another way, substituting into the generalized
Fourier series:
(
)
Z bX
X
E[u] = hu, Lui =
uk φk (x)L
uj φj (x) dx
a k∈Z
=
XZ
j,k∈Z
=
X
j∈Z
b
{φk (x)Lφj (x)} uk uj dx
a
uk uj λj hφk , φj i
j,k∈Z
=
X
λk u2k .
k∈Z
By the ordering of the eigenvalues, we know that
X
X
E[u] =
λk u2k ≥ λ1
u2k = λ1 ,
k∈Z
k∈Z
2
since kuk2 = 1 =⇒
k uk = 1 by Parseval’s identity. Thus, since E[φ1 ] = λ1 , we deduce
that the smallest eigenvalue is obtained by minimizing E[u] with respect to all admissible
functions u such that kuk= 1 and the minimum occurs when u = φ1 . In reality, we don’t
know the eigenfunctions and need to find them, which we now know we can do by using
P
hu, Lui
,
u∈D(L) hu, ui
such that kuk = 1.
λ1 = inf
This is sometimes called the Raleigh quotient. Similarly, if we restrict the class of admissible
functions by requiring u satisfy uk = hu, φk i = 0 for k = 1, 2, . . . , n − 1, then we can say
that
∞
X
hu, Lui
,
E[u] =
λk u2k ≥ λn =⇒ λn = inf
u∈D(L) hu, ui
k=n
⊥
u∈vn
where vn is spanned by {φ1 , . . . , φn−1 }, resulting in a slightly different Rayleigh quotient.
Proof. We’ll now prove that λn → ∞ as n → ∞. For simplicity, consider Dirichlet
boundary conditions so that
Rb
{p(u0 )2 + qu2 } dx
E[u]
a
=
,
Rb
kuk2
u2 dx
a
and let pM := maxx∈(a,b) {p(x)}, pN = minx∈(a,b) {p(x)} and define qN , qM identically. We
can now define a new Rayleigh quotient
EM [u]
E[u]
)
≥
=⇒ λ(M
≥ λn ,
n
2
2
kuk
kuk
3.4. EIGENVALUE PROBLEMS
45
(N )
(N )
(M )
and similarly, we get that λn ≤ λn . Therefore, λn ≤ λn ≤ λn . Now, consider two
intervals IN , IM such that IN ⊂ [a, b] ⊂ IM . Consider the new variational problem for
EM [u]/kuk2 with [a, b] replaced by IN .
The admissible functions must now vanish at ∂In , which places additional constraints on the
(M )
(M )
functions and therefore produces a smaller search space. This implies that λn ≤ λ̂n ,
and therefore we miss the minimum.
Similarly, consider EN (u)/kuk2 on IM . The admissible functions don’t have to vanish at
(N )
(N )
x = a, b which suggests that our search space is larger and therefore λ̂n ≤ λn , and
piecing this all together, we have
)
)
)
)
λ̂(N
≤ λ(N
≤ λn ≤ λ(M
≤ λ̂(M
n
n
n
n .
(N )
(M )
Thus, the eigenvalues λ̂n and λn can be determined by explicitly solving a constant
(M )
(N )
coefficient boundary value problem. We always find that λ̂n , λ̂N → ∞ as n → ∞.
Hence, λn → ∞ as well, and reaches ∞ at a non-finite n.
Proof. We can also prove completeness of the eigenfunctions of the Sturm-Liouville operator.
Consider the orthonormal set of eigenfunctions φk , where again,
Lφk := −(pφ0k )0 + qφk = λk φk ,
x ∈ [a, b].
Let Rn represent the error, that is,
Z
n
Rn (x) := u(x) −
uk φk (x).
k=1
We wish to show that Rn → 0, which would suggest that the eigenfunction expansion is
valid, an equivalent condition of completeness. Since u and φk are admissible, so is Rn as it
is just a linear combination of these. Also, we know that
*
+
n
X
hRn , φj i = u −
φk , φj = 0.
k=1
Therefore, we consider again the Rayleigh quotient,
E[Rn ]
E[Rn ]
≥ λn+1 =⇒ kRn k ≤
.
hRn , Rn i
λn+1
We have already established that λn → ∞, meaning that kRn k → 0 so long as E[Rn ] is
bounded. Note that
"
#
" n
#
Z b
n
n
X
X
X
uk
{pRn0 φ0k + qRn φk } dx.
E[u] = E Rn +
φk uk = E[Rn ]+E
uk φk +2
k=1
k=1
k=1
a
46
CHAPTER 3. BOUNDARY VALUE PROBLEMS
The last term can be written as
b
Z b
Z b
0
0
Rn {−(pφk ) + qφk } dx + pRn φk = λk
Rn φk dx = λk hRn , φk i = 0,
a
a
a
since Rn = 0 on the boundaries and k ≤ n. Also, note that
" n
# Z ( n
) ( n
)
n
n
b
X
X
X
X
X
E
uk φk =
uk φk L
uk φk dx =
uj uk hφj , φk i λk =
λk u2k .
k=1
a
k=1
k=1
j,k=1
k=1
Therefore, combining these two results, we have that
E[Rn ] = E[u] −
n
X
λk u2k ≤ E[u],
k=1
and is therefore bounded since E[u] is bounded (which is true since u is bounded), and we
know that λk ≥ 0. We can conclude then that kRn k → 0 as n → ∞.
3.5
Distribution Theory
Recall that a linear operator L : V → V , where V is some function space, satisfies
L[αf1 + βf2 ] = αLf1 + βLf2 .
For a finite dimensional space, we can represent a linear operator as a matrix:
X
y = Ax,
yi =
Aij xj ,
j
which, for infinite dimensional cases, we can think of as integrals as natural analogs. That is,
Z b
g = Af,
g(x) =
A(x, y)f (y) dy.
a
The most commonly used space is L2 [a, b] for our purposes. If A(x, y) is an ordinary
function it is called an integral kernel. In function spaces, the Dirac delta function plays
the role of the identity matrix. That is,
Z b
f (x) =
δ(x − y)f (y) dy.
a
We can informally think of δ(x) as the limit of a sequence of rectangles with with ε and
height 1/ε and therefore area 1. In this case, call the approximation δε (x − a). We see then
that
Z ∞
Z
1 a+ε/2
1
2
2
kδε k =
|δε (x)| dx = 2
dx = → ∞.
ε a−ε/2
ε
−∞
3.5. DISTRIBUTION THEORY
47
Figure 3.1: Approximation of δ(x − a).
Therefore, this definition is slightly problematic. We can also think of this arising from
Fourier series, noting that
X
einπx/L e−inπx/L = δ(x − x0 ).
n
If we then define
Z
n
eikx
δn (x) :=
−n
In this case,
2
Z
∞
kδn k =
−∞
dk
1 sin nx
=
.
2π
π x
sin2 nx 2
|n|
x dx =
.
2
π
π
To make sense of δ(x) mathematically, we exploit the notion of a dual space.
Definition 3.8. The dual space V ∗ of a vector space V is the space of all linear functionals
f : V → R (or C).
Test Functions
We treat δ(x) as an element of the dual space of a vector space T of test functions. Test
functions are typically analytic (infinitely differentiable) that tend rapidly to 0 at ±∞. The
precise definition of T depends on the problem.
(i) If one is making extensive use of the Fourier transform, one common choice is the
Schwartz space S(R). This is the space of analytic functions φ such that
m n d φ
|φ|min = sup |x| m < ∞,
∀m, n ≥ 0.
dx
x∈R
(ii) C ∞ of compact support are also common. That is, functions that are identically 0
outside of some finite interval.
In a sense, the “nice” behavior of test functions compensates for the “naughty” behavior of
things like δ(x). The latter are referred to as generalized functions or distributions.
48
CHAPTER 3. BOUNDARY VALUE PROBLEMS
Not every linear map is included in these function spaces. We require that the map be
continuous. That is, if φn → φ, then the distribution u must satisfy u(φn ) → u(φ).
In terms of the dual space formalization, we define δ as the map
hδ, φi = φ(0),
∀φ ∈ T .
Note, this notation does not mean an inner product, as δ, φ are in different function spaces.
It is simply shorthand for the test function, distribution pair.
Similarly, we can define (δ 0 , φ) = −φ0 (0), which makes sense if we consider our informal
definition
∞
Z ∞
Z ∞
0
δ (x − y)f (y) dy = δ(x − y)f (y)
−
δ(x − y)f 0 (y) dy = −f 0 (x).
−∞
−∞
−∞
In general, we then have that
δ (n) , φ = (−1)n φ(n) (0).
Note, L2 functions are not “nice” enough for their dual space to accommodate δ function.
The Hilbert space L2 is its own dual space (called reflexive). The Riesz representation
theorem asserts that any linear map F : L2 → R can be written as F [f ] = h`, f i for ` ∈ L2 ,
however δ 6∈ L2 , as it’s not even normalizable.
Weak Derivatives
Define the weak derivative (or distributional derivative) v(x) or u0 (x) of a distribution u by
requiring its evaluation on a test function φ ∈ T to be
Z ∞
Z ∞
0
0
hu , φi =
u φ(x) dx := −
u(x)φ0 (x) dx = − hu, φ0 i .
−∞
−∞
Note that this interchange, which looks like integration by parts actually is only formal. It is
taken to be the definition of the derivative.
Example 3.6.
Z
∞
−∞
d
|x|φ(x) dx = −
dx
Z
∞
|x|φ0 (x) dx
−∞
∞
Z
Z
0
=−
xφ (x) dx +
xφ0 (x) dx
0
−∞
Z ∞
Z 0
=
φ(x) dx −
φ(x) dx
0
−∞
Z ∞
=
sgn(x)φ(x) dx.
−∞
Thus, in the weak sense,
d
|x|
dx
= sgn(x).
0
3.5. DISTRIBUTION THEORY
49
Example 3.7.
Z
∞
−∞
d
sgn(x)φ(x) dx = −
dx
Z
∞
sgn(x)φ0 (x) dx
−∞
Z 0
Z ∞
0
φ (x) +
φ0 (x) dx
=−
0
Z ∞ −∞
δ(x)φ(x) dx.
= 2φ(0) = 2
−∞
Therefore,
d
dx
sgn(x) = 2δ(x).
Example 3.8. We can also solve differential equations in the weak sense. Consider
u0 = 0,
which, in the weak sense, is equivalent to
0 = hu0 , φi = − hu, φ0 i ,
∀φ ∈ T .
We need to determine the action of u on all test functions ψ, that is hu, ψi = 0, whenever
ψ(x) = φ0 (x) for some ψ ∈ T .
First, we’ll prove that ψ(x) = φ0 (x) for some ψ ∈ T if and only if
Z
∞
ψ(x) dx = 0.
(3.5)
−∞
If ψ(x) = φ0 (x). Conversely, if (3.5) holds, then we can define a test function by
Z
x
φ(x) :=
ψ(x0 ) dx0 .
−∞
Consider φ0 ∈ T such that
R
φ0 (x) dx = 1 and write
Z
∞
φ(x) = φ0 (x)
Z
φ(s) ds + φ(x) − φ0 (x)
−∞
It follows immediately that hu, ψi = 0 and therefore
hu, φi = hhu, φ0 i , φi = hc, φi .
Thus, u = constant in the weak sense.
∞
−∞
φ(s) ds .
50
3.6
CHAPTER 3. BOUNDARY VALUE PROBLEMS
Green’s Functions
Green’s functions are a method for solving inhomogeneous equation
Lf = h, where L is an LDO.
Roughly speaking, the Green’s function is an integral kernel that represents the “inverse” of
L.
Theorem 3.6 (Fredholm Alternative). Let V be a finite dimensional vector space with an
inner product and let A be a linear operator A : V → V . Then either
(i) Ax = b has a unique solution (that is, A−1 exists).
(ii) Ax = 0 has a non-trivial solution.
In the latter case, Ax = b has no solution unless b is orthogonal to all solutions v of A† v = 0,
that is, all null vectors of the adjoint.
This result continues to hold for linear differential operators on function spaces provided
that L† is defined with appropriate adjoint boundary conditions where kern(L) 6= kern(L† )
unless the derivative of boundary conditions is equal to the order of the equation.
Example 3.9. Consider the following boundary value problem
Ly =
dy
,
dx
y(0) = y(1) = 0,
x ∈ [0, 1].
We check the null space Ly = 0, which only has the trivial solution y = 0. Hence, if a
solution of Ly = f exists, it will be unique. The adjoint operator is then
L† = −
d
with no boundary conditions.
dx
In this case, L† y = 0 has a non-trivial solution y = 1. Therefore, a solution to Ly = f only
exists when
Z
1
h1, f i =
f (x) dx = 0.
0
If this condition is satisfied, then
Z
y(x) =
x
f (s) ds.
0
This actually follows from our naive attempt at this solution despite the fact that we had two
boundary conditions for a first order system.
3.6. GREEN’S FUNCTIONS
51
Constructing Green’s functions
We consider the construction, in general, for homogeneous boundary conditions. We’ll
particularly consider the differential operator L where kern(L) = kern(L† ) = {0}. We
represent the inverse operator L−1 by the integral kernel
(L−1 )x,ξ = G(x, ξ) such that Lx G(x, ξ) = δ(x − ξ),
which can be thought of in the way that AA−1 = I for matrices. Given this definition, the
solution to Ly = f can be written as
Z
y(x) = G(x, ξ)f (ξ) dξ,
since
Z
Lx y =
Z
Lx G(x, ξ)f (ξ) dξ =
δ(x − ξ)f (ξ) dξ = f (x).
Three conditions we require of G:
(i) The function χ(x) = G(x, ξ) for a fixed ξ must have some discontinuity at x = ξ in
order to generate a δ function.
(ii) χ(x) must obey Lχ = 0 for x 6= ξ.
(iii) χ(x) also obeys the same homogeneous boundary conditions as y(x).
Consider the Sturm-Liouville equation
(p(x)y 0 )0 + q(x)y(x) = f (x),
y 0 (a)
β
= ,
y(a)
α
y 0 (b)
β̂
= ,
y(b)
α̂
x ∈ [a, b].
For a fixed ξ, LG = (pG0 )0 + qG = δ(x − ξ). We require G(x, ξ) to be continuous at x = ξ,
otherwise we generate a term δ 0 . Therefore, let us write
(
AyL (x)yR (ξ), x < ξ
G(x, ξ) =
AyL (ξ)yR (x), x > ξ.
Note that this is continuous at x = ξ by construction. Specifically, take yL to satisfy Ly = 0
and the left boundary conditions and yR to satisfy the right boundary conditions. In order to
determine how to satisfy the equation for G at x = ξ, we integrate LG = δ from ξ − ε to
ξ + ε to obtain
ξ+
dG0 1
0
0
p(ξ) {G (ξ + ε, ξ) − G (ξ − ε, ξ)} = 1,
ε → 0 =⇒
=
.
dx ξ−
p(ξ)
52
CHAPTER 3. BOUNDARY VALUE PROBLEMS
Using the continuous solution, we now have
Ap(ξ) {yL (ξ)yR0 (ξ) − yL0 (ξ)yR (ξ)} = 1,
{z
}
|
:=W (ξ)
but note that this is exactly Ap(ξ)W (ξ), where W is the Wronskian. We can now use this to
solve for A explicitly, and we have our Green’s function
(
1
y (x)yR (ξ), x < ξ
p(ξ)W (ξ) L
G(x, ξ) =
1
y (ξ)yR (x), x > ξ.
p(ξ)W (ξ) L
For the Sturm-Liouville equation, the product p(ξ)W (ξ) is actually a constant. Using the
form of the operator, we can show that
dW
p1 (ξ)
=−
W (ξ),
dξ
p(ξ)
where p1 (ξ) = p0 (ξ).
Therefore, using our form of the Wronskian:
Z ξ 0
p (x)
W (0)
W (ξ) = W (0) exp −
dx = W (0) exp {− ln p(ξ)/p(0)} =
p(0),
p(ξ)
0 p(x)
which suggests that
W (ξ)p(ξ) = W (0)p(0) = constant.
The fact that this product is constant implies that we have the symmetry G(x, ξ) = G(ξ, x).
The above work also requires that W 6= 0. If W = 0, then yL ∼ yR , that is, they are
linearly dependent and the only solution yR would satisfy both boundary conditions where
kern L 6= ∅.
Example 3.10. Consider the boundary value problem
−
d2 y
= f (x),
dx2
y(0) = y(1) = 0.
If we look at the homogeneous equation, d2 y/dx2 = 0, we see that solutions are of the form
ax + b. Coupling this with the boundary conditions, it’s clear that
yL (x) = x,
yR (x) = 1 − x =⇒ W = yL0 yR − yL yR0 = 1.
Also note that here, p(ξ) = −1. Piecing this all together, we find that our Green’s function
is then
(
x(1 − ξ), x < ξ
G(x, ξ) =
.
ξ(1 − x), x > ξ
We then can use this as an effective inverse to solve our original problem:
Z 1
Z x
Z 1
y(x) =
G(x, ξ)f (ξ) dξ = (1 − x)
ξf (ξ) dξ + x
(1 − ξ)f (ξ) dξ.
0
0
x
3.6. GREEN’S FUNCTIONS
53
Initial Value Problems
We can think of initial value problems are boundary value problems where all the boundary
conditions are imposed at one end of the interval. Consider the general form
dy
− Q(t)y = F (t),
dt
y(0) = 0.
We seek a Green’s function such that
nd
o
Lt G(t, t0 ) :=
− Q(t) G(t, t0 ) = δ(t − t0 ),
dt
G(0, t0 ) = 0.
The unique solution of Lt G = 0 and G(0, t0 ) = 0 for t < t0 is G ≡ 0. Around t = t0 , we
have the following jump condition
Z t
0
0
0
0
0
0
G(t + ε, t ) − G(t − ε, t ) = 1 =⇒ G(t, t ) = H(t − t ) exp
Q(s) ds ,
t0
where H(t) is the Heaviside function, which imposes causality on the system.
Example 3.11 (Forced, damped, harmonic oscillator). The corresponding equation for
this physical system is
ẍ + 2γ ẋ Ω2 + γ 2 x = F (t),
x(0) = ẋ(0) = 0.
In this case, following this procedure, one finds that the Green’s function is
1
0
G(t, t0 ) = H(t − t0 ) e−γ(t−t ) sin Ω(t − t0 ).
Ω
This form is somewhat convenient because it identifies the damping component and the
oscillatory component.
Modified Green’s Functions
When the equation Ly = 0 has a non-trivial solution, the structure of the Green’s function
and the method for obtaining it changes. In fact, the standard Green’s function does not even
exist.
Example 3.12.
Ly := −y 00 = f (x),
y 0 (0) = y 0 (1) = 0.
The equation Ly = 0 has a non-trivial solution y(x) = constant, so dim kern L = 1. The
operator L is self-adjoint, that is, L† = L and the Fredholm Alternative then requires that
Z 1
h1, f (x)i =
f (x) dx = 0, since kern L† = kern L.
0
54
CHAPTER 3. BOUNDARY VALUE PROBLEMS
We can not define the Green’s function as a solution to
−
∂
G(x, ξ) = δ(x − ξ),
∂x2
G0 (0, ξ) = G0 (1, ξ) = 0,
because we make the observation that, integrating both sides,
Z
1
Z
δ(x − ξ) dx = 1 6=
0
0
1
∂
G(x, ξ) dx = 0.
∂x2
This suggests solving the modified equation
−
∂
G(x, ξ) = δ(x − ξ) − 1.
∂x2
A general solution to −y 00 = −1 is y = A + Bx + x2 /2, and therefore the functions
1
yL = A + x2 ,
2
1
yR = c − x + x2 ,
2
satisfy the boundary conditions and we have our modified Green’s function
(
c − ξ + 12 x2 , 0 < x < ξ
G(x, ξ) =
c − x + 21 x2 , ξ < x < 1.
One finds that, around x = ξ,
lim {G0 (ξ − ε, ξ) − G0 (ξ + ε, ξ)} = 1.
ε→0
Note that c is an arbitrary constant of ξ, so choose c = ξ 2 /2 + 1/3, so G is symmetric and
therefore we have
(
1
− ξ + 12 (x2 + ξ 2 ), 0 < x < ξ
G(x, ξ) = 13
− x + 21 (x2 + ξ 2 ), ξ < x < 1.
3
The solution to our original problem is then
Z
y(x) =
1
G(ξ, x) df (ξ)dξ + A,
0
noting that y(x) does not depend on the choice of c.
3.6. GREEN’S FUNCTIONS
55
Inhomogeneous Boundary Conditions
Suppose that we wish to solve
d2 y
= f (x),
dx2
y(0) = a, y(1) = b.
In the corresponding case of homogeneous boundary conditions where a = b = 0, the
Green’s function is
(
x(1 − ξ), x < ξ
G(x, ξ) =
(3.6)
ξ(1 − x), x > ξ.
From Green’s theorem, we see that
2 Z 1
Z 1
∂2
dy
1
y(x) − 2 G(x, ξ) dx = {G0 (x, ξ)y(x) − G(x, ξ)y 0 (x)}0 .
G(x, ξ) − 2 dx−
dx
∂x
0
0
The right hand side is then
G0 (1, ξ)y(1) − G0 (0, ξ)y(0) = −(1 − ξ)a − ξb,
where we have used the definition of G described in (3.6). Therefore, our full solution is
Z
1
G(x, ξ)f (x) dx + (1 − ξ)a + ξb.
y(ξ) =
0
We only needed the Green’s functions for the homogeneous boundary case.
Eigenfunction Expansions
Suppose there exists a complete set of eigenfunctions {φn } for some operator L. That is,
Lφn = λn φn ,
λn 6= 0.
Then, L is invertible and has a Green’s function. We take the Green’s function to be
G(x, ξ) :=
X φn (x)φ∗ (ξ)
n
n
λn
.
To see why this works, note that
(
)
X φn (x)φ∗ (ξ)
X Lφn (x)φ∗ (ξ) X
n
n
LG(x, ξ) = L
=
=
φn (x)φ∗n (ξ) = δ(x − ξ),
λ
λ
n
n
n
n
n
by the completeness of the eigenvalues.
56
CHAPTER 3. BOUNDARY VALUE PROBLEMS
Example 3.13. Consider the operator
L := −
d2
,
dx2 2
D(L) = {y : Ly ∈ L2 [0, 1], y(0) = y(1) = 0}.
We’ve established the Green’s function for this operator is
(
x(1 − ξ), x < ξ
G(x, ξ) =
ξ(1 − x), x > ξ.
Note that Lφn = λn φn has solution of the form φn (x) = sin nπx, where λn = n2 π 2 /2.
Therefore, we can write the Green’s function as the eigenfunction expansion
∞ X
2
G(x, ξ) =
sin nπx sin nπξ.
n2 π 2
n=1
When one or more of the eigenvalues is/are zero, a modified Green’s function is necessary
and obtained by simply omitting the corresponding terms in the series. That is,
Gmod (x, ξ) =
X φn (x)φ∗ (ξ)
n
.
λn
n
λn 6=0
In this case,
LGmod (x, ξ) = δ(x − ξ) −
X
n
λn =0
and again the Fredholm Alternative applies.
φn (x)φ∗n (ξ),
4
Dynamical Systems Theory
“One can never actually prove all Russians are awful lecturers. There might
still be one counterexample out there. It’s the classic black swan."
— Paul C. Bressloff
4.1
Preliminaries
Figure 4.1: Trajectory of a particle and the flow map on a cloud of points.
Definition 4.1. Let U ⊂ Rn and let f ∈ C 1 (U ). For x0 ∈ U , let φ(t, x−) be the solution
to the initial value problem
ẋ = f (x),
x(0) = x0 ,
t ∈ I(x0 ).
Then, the set of mappings φt defined by
φt (x0 ) := φ(t, x0 )
is called the flow of the ODE or vector field of f (x).
Here, I(x0 ) is the maximal time interval over which a solution exists. If x0 is fixed, then
φ(·, x0 ) : I → U defines a solution curve or trajectory of the system through x0 ∈ U . On
the other hand, if x0 varies over some set M ⊂ U , then the flow φt : M → U defines the
motion of the cloud of points in M .
57
58
CHAPTER 4. DYNAMICAL SYSTEMS THEORY
Theorem 4.1. For all x0 ∈ U , if t ∈ I(x0 ), and s ∈ I(φt (x0 )), then s + t ∈ I(x0 ) and
φs+t = φs (φt (x0 )).
Proof. Let I(x0 ) = (α, β) and define x : (α, s + t) → U to be
(
φ(r, x0 ),
α<r≤t
x(r) :=
φ(r − t, φt (x0 )), t ≤ r ≤ s + t.
Then x(r) is a solution of the IVP on (α, s + t) which implies that s + t ∈ I(x0 ) and by
uniqueness,
φs+t (x0 ) = x(s + t) = φ(s, φt (x0 )) = φs (φt (x0 )).
Note, for all x0 ∈ U, φ−t (φt (x0 )) = x0 and φt (φ−t (x0 )) = x0 . In other words, flows form
an Abelian group.
Example 4.1. Consider the IVP
ẋ = f (x) :=
1
.
x
(4.1)
Here, f ∈ C 1 (U ) for U = {x ∈ R : x > 0}. The solution to the IVP for x(0) = x0 ∈ U is
q
x
0
I(x0 ) = − , ∞ .
φt (x0 ) = 2t + x20 ,
2
Figure 4.2: Sketch of where flows are defined for (4.1).
Definition 4.2. Let U ⊂ Rn and f ∈ C 1 (U ). Let φt (U ) be the flow of ẋ = f (x) for all
t ∈ R. Then a set S ⊂ U is called invariant with respect to the flow φt if
φt (S) ⊂ S,
∀t ∈ R.
Similarly, S is positively (or negatively) invariant with respect to φt if φt (S) ⊂ S for all
t ≥ 0 or( t ≤ 0).
4.1. PRELIMINARIES
59
Definition 4.3. The linear system
ẋ = Ax,
A := Df (x0 )
is called the linearization of f (x) around x = x0 .
It turns out that for any hyperbolic system, the behavior of the linearizion in a neighborhood
of x0 is topologically equivalent to the non-linear system. We’ll investigate this more later.
Definition 4.4. A point x is periodic of a minimal period T if and only if
φ(t + T, x) = φ(t, x),
∀t ∈ R,
φ(t + s, x) 6= φ(t, x)∀0 < s < T.
The trajectory
Γ := {y : y = φ(t, x), 0 ≤ t ≤ T }
is called a periodic orbit.
Definition 4.5. Suppose that φt (x) is defined for x ∈ Rn and t ∈ R. The trajectory through
x is the set
Γ(x) :=
[
φt (x),
t∈R
and the positive and negative semi-trajectories, Γ+ , Γ− , are the same except limited to t > 0
and t < 0 respectively.
Definition 4.6. The ω-limit set of x is defined to be
ω(x) := {y ∈ Rn : ∃ a sequence {tn }∞
n=0 : tn → ∞, φ(tn , x) → y}.
Example 4.2. Consider a stable periodic orbit Γ. Then ω(x) = Γ if x lies in the basin of
attraction of Γ.
Figure 4.3: The ω-limit set of a stable periodic orbit.
60
CHAPTER 4. DYNAMICAL SYSTEMS THEORY
Similarly, we can define the α-limit set as the same except tn → −∞ but this is used far less
often.
Theorem 4.2. The set ω(x) is closed and positive invariant. Furthermore, if Γ+ (x) is
bounded, then ω(x) is compact and non-empty.
Proof. The definition of ω(x) implies that ω(x) is closed, as it is effectively a set of “limit
points”. Suppose that ρ ∈ ω(x) so that ∃{tn } with tn → ∞ and φ(tn , x) → ρ. Since
φ(t + tn , x) = φ(t, φ(tn , x)), we can take the limit tn → ∞ for a fixed t to find that
that φ(t + tn , x) → φ(t, ρ) as n → ∞. That is, φ(t, ρ) ∈ ω(x). Hence, the positive
semi-trajectory through ρ lies in ω(x), which suggests that ω(x) is positive invariant.
If Γ+ (x) is bounded, then clearly so is ω(x). A basic result from real analysis: a bounded set
in Rn with an infinite number of points has at least one point of accumulation, and therefore
ω(x) is non-empty. A bounded, closed, non-empty set in Rn is compact.
We could also prove analogous results for α(x) and Γ− (x).
Definition 4.7. A closed invariant set A ⊂ U ⊂ Rn is called an attracting set of ẋ = f (x)
if there exists some neighborhood M ⊂ A such that
∀x ∈ M : φt (x) ∈ M,
∀t ≥ 0, φt (x) → A,
as t → ∞.
An attractor is an attracting set which contains a dense orbit.
Examples of attractors are stable equilibria, stable periodic orbits, and strange attractors
(n ≥ 3, chaos theory). Some interesting properties of strange attractors are, they they
contain:
(i) a countable set of periodic orbits of arbitrarily large orbits.
(ii) an uncountable set of non-periodic motions
(iii) a dense orbit.
We won’t investigate these any further.
4.2
Planar Dynamics
Consider the general, autonomous, second order system
ẋ = X(x, y),
ẏ = Y (x, y).
4.2. PLANAR DYNAMICS
61
We can assume that the conditions of existence hold, then we can consider sketching phase
diagrams. Particularly, it is useful to draw nullclines , which are defined to be the curves
where
Y (x, y) = 0 (zero slope) ,
X(x, y) = 0 (infinite slope).
Points where nullclines meet determine the fixed points of the system. Regions in the (x, y)
plane where X, Y have the same sign have phase paths with positive slope and regions where
the signs differ have phase paths with negative slope. The direction of adjacent phase paths
varies continuously.
Example 4.3. Consider the second order system
ẋ = y(1 − x2 ),
ẏ = −x(1 − y 2 ).
(4.2)
Fixed points can be found by setting both of these equations equal to 0, which we find yields
the set of points (0, 0), (±1, ±1), (±1, ∓1), so 5 points total. The phase paths satisfy
x(1 − y 2 )
dy
=−
=⇒ −
dx
y(1 − x2 )
Z
x
dx =
1 − x2
Z
y
dy =⇒ (1−x2 )(1−y 2 ) = constant.
1 − y2
Figure 4.4: Nullclines and example paths for (4.2).
We can plot example paths, for instance for x = ±1, y = ±1. The nullclines ẋ = 0 are
y = 0 and x = ±1. For ẏ = 0, we have x = 0 and y = ±1. Particularly, for c = 1 − ε
with 0 < ε < 1, we see that x2 , y 2 1 and therefore 1 − x2 − y 2 + x2 /y 2 = 1 − ε, which
suggests that 1 − x2 − y 2 ≈ 1 − ε. Therefore, we conclude flows are circular around the
origin.
62
CHAPTER 4. DYNAMICAL SYSTEMS THEORY
Example 4.4 (Predator-Prey). Let x(t) be a population of prey at time t and let y(t) be a
population of predators. Consider the predator-prey system
ẋ = ax − cxy,
ẏ = −by + dxy.
(4.3)
The fixed points of this system are easily found to be at (0, 0) and (b/d, a/c). The phase
paths can be described by
Z
Z
a − cy
dx
dy
−by + dxy
=
=⇒
dy =
(dx − b),
dx
ax − cxy
y
x
which mean that solutions are of the form
a log y + b log x − cy − xd = C,
x, y > 0.
This can be rewritten as f (x)+g(y) = C, where f (x) = b log x−xd and g(y) = a log y −cy.
Figure 4.5: Sketch of solutions of (4.3)
The function f (x) has a global maximum at x = b/d since f 0 (x) = b/x − d, f 00 (x) =
−b/x2 < 0. Similarly, g(y) has a maximum at y = a/c. Thus, we can conclude that
F (x, y) = f (x) + g(y) has a global max at (b/d, a/c), the fixed point.
4.3
Limit Cycles
Theorem 4.3. Bendixson’s Negative Criterion There are no closed paths in a simply
connected region of the phase plane on which
Γ(x, y) :=
∂X ∂Y
+
is of one sign.
∂x
∂y
Proof. Impose the usual smoothness conditions so that we can apply the divergence theorem.
Suppose there exists a closed path C in a region D where Γ(x, y) has one sign. Then, by the
4.3. LIMIT CYCLES
63
Figure 4.6: Sketch for the proof of Bendixson’s negative criterion.
divergence theorem, we have that
ZZ Z
∂X ∂Y
+
dxdy = (X, Y ) · n ds.
∂x
∂y
M
C
Since the integrand on the left hand side is of one sign, it follows that it cannot vanish.
however, we know that (X, Y ) · n ≡ 0 so we have a contradiction.
Figure 4.7: A local traversal.
Definition 4.8. A local traversal L is a line segment that all trajectories of the ODE cross
from the same side. If x0 is not a fixed point, then it is always possible to construct a traversal
in a neighborhood of x0 by continuity.
Lemma 4.1. If a trajectory γ(x) intersects a local traversal L several times, the successive
crossing points move monotonically along L.
Proof. This is a consequence of the Jordan Curve Lemma, which states that: a closed
curve in the plane separates into two parts, an interior and an exterior. The closed curve
containing the orbit from P1 to P2 defines an interior (shaded in Figure: 4.8), within which
the orbit cannot reenter, and therefore, P3 must move beyond P2 and so on.
Corollary 4.1. If x ∈ ω(x0 ) is not a fixed point and x ∈ γ(x0 ), then γ(x) is a closed curve.
Proof. Since x ∈ γ(x0 ), it follows that ω(x) = ω(x0 ). Thus, x ∈ ω(x0 ) =⇒ x ∈ ω(x).
Choose L to be a local traversal through x. Since x ∈ ω(x), there exists an increasing
sequence of times {tn } such that φ(x, tn ) → x with φ(x, tn ) ∈ L and φ(x, 0) = x. Unless
φ(x, tn ) = x for all n, that is, we have a periodic orbit, we have a sequence of points on L
that move close to x, contradicting our lemma.
64
CHAPTER 4. DYNAMICAL SYSTEMS THEORY
Figure 4.8: Sketches for the lemma. Leftmost figure describes the basic case. Right two
sketches show “cheating” contradictions.
Figure 4.9: Sketches for the proof of the corollary. The leftmost figure comes from
application of the lemma, but note that φ(x, tn ) → x.
Theorem 4.4 (Poincaré-Bendixson theorem). For the ODE ẋ = f (x), suppose that the
trajectory γ(x0 ) enters and does not leave some closed bounded domain D and that there are
no fixed points in D. Then, there is at least one periodic orbit in D and this orbit is in the
ω-limit set of x0 .
Proof. Since γ(x0 ) enters the domain D, ω(x0 ) is non-empty and is contained in D. Let
x ∈ ω(x0 ) and note that x is not a limit point (by assumption). If x ∈ γ(x0 ), then γ(x)
is periodic by the previous corollary. Suppose x 6∈ γ(x0 ). Since x ∈ ω(x0 ), we have that
γ + (x) ⊂ ω(x0 ), since the ω-limit set is positive invariant, and therefore γ + (x) ⊂ D. Since
D is compact, γ + (x) has a limit point x∗ ∈ D so that x∗ ∈ ω(x), but if x∗ ∈ γ + (x), then
γ(x∗ ) is closed by the previous corollary.
Therefore, the final step is to establish that x∗ 6∈ γ + (x), which would lead to a contradiction.
Choose a local traversal L through x∗ . Since x∗ ∈ ω(x), the trajectory γ + (x) must intersect
L at points P1 , P2 that accumulate monotonically to x∗ . However, γ + (x) ⊂ ω(x0 ), which
suggests that Pi ∈ ω(x0 ), hence γ(x0 ) passes arbitrarily close to Pi and then Pj where j > i,
which is an infinite number of times. If γ(x0 ) intersections with L are not monotonic, this
would contradict the lemma about traversals.
4.3. LIMIT CYCLES
65
Figure 4.10: Sketch of the traversals in the proof of the Poincaré-Bendixon theorem.
In practice, we are typically looking for an annular region D with a source in the hole (so
trajectories enter D across the inner boundary) and the outer boundary is chosen so that the
radial component of the vector field is always pointing inward.
Example 4.5. Consider the system
1
ẋ = y + x(1 − 2r2 ),
4
1
ẏ = −x + y(1 − r2 ,
2
r := x2 + y 2 .
For a fixed point, we have
1
x = − x(1−2r2 )(1−r2 ) =⇒ x = y = 0 or (1−2r2 )(1−r2 ) = 8 =⇒ 2r4 −3r2 +9 = 0,
8
which has no real solution. Thus, the origin is our only fixed point. Also note that
1
1
1
1
rṙ = xẋ + y ẏ = x2 (1 − 2r2 ) + y 2 (1 − r2 ) = r2 (1 + sin2 θ) − r4 ,
4
2
4
2
from which, we can conclude that ṙ > 0 for all θ provided that r/4−r3 /2 > 0 =⇒ r2 < 1/2
and ṙ < 0 provided that r − r3 > 0 =⇒ r > 1. Thus, we can immediately apply the
Poincaré-Bendixon theorem in the annular region
1
D = (r, θ) : < r < 1
2
Example 4.6 (Lienard equation). Consider the system corresponding to a damped non-linear
oscillator:
ẍ + f (x)ẋ + x = 0.
Note that this is a generalization of the Van der Pol equation:
ẍ + x = µ(1 − x2 )ẋ,
m > 0.
Assume that f (x) is Lipschitz continuous and that it satisfies the following properties:
66
CHAPTER 4. DYNAMICAL SYSTEMS THEORY
(i) F (x) :=
Rx
0
f (s) ds is an odd function.
(ii) F (x) → ∞ as x → ∞ and ∃β > 0 such that if x > β then F (x) > 0.
(iii) There exists α > 0 such that for 0 < x < α, then F (x) < 0.
Figure 4.11: Sketch F (x) corresponding to the Lienard equation.
We can convert this to a first order system
ẋ = y − F 0 (x),
ẏ = −x.
We will show that there exists at least one periodic solution to this equation. If α = β, then
there exists a unique periodic orbit that is globally attracting (except from the origin).
First, note that there exists a unique fixed point at the origin. Linearizing about (0, 0) yields
ẋ = y − F 0 (0)x,
ẏ = x,
which has eigenvalues
1p 2
1
λ1,2 = − f (0) ±
f (0) − 4,
2
2
and since f (0) < 0, we see that this fixed pont repels. Introduce the radial coordinates
1
1
R := r2 = (x2 + y 2 ),
2
2
in which, our system becomes
rṙ = xẋ = y ẏ = x(y − F (x)) − yx = −xF (x).
For −α < x < α, we have that F (x) < 0 so that Ṙ > 0. Consider a solution that starts at
(0, y0 ), where y0 > 0. We see that
(
0,
x=0
dy
x
dy
=−
=⇒
=
.
dx
y − F (x)
dx
∞, y = F (x)
4.4. STABILITY
67
Figure 4.12: Behavior of the Lienard system. Note that |y1 | < y0 .
By choosing y0 large enough, we will show that the behavior is described by Figure: 4.12.
If the conditions of this figure hold, then the reflection of the orbit starting from y0 under
x → −x and y → −y generates an invariant set bounded by the paths of orbits and the
segments [−y1 , y0 ] and [−y0 , y1 ].
For this to hold, we need to prove the condition that |y1 | < y0 for some y0 > 0. Consider
R(x, y) to be defined the same as above, and note that
Z
Z
Z Z
xF (x)
dR,
R(0, y1 ) − R(0, y0 ) =
dR =
+
−
dx +
y − F (x)
BEC
ABECD
AB
CD
where,
dR =
dR dt
xF (x)
dx = −
dx.
dt dx
y − F (x)
Since F (x) is bounded for 0 ≤ x ≤ β, the first expression vanishes as y0 → ∞. Assume
that y1 → −∞ as y0 → ∞, otherwise the proof is satisfied trivially. Write the second
expression as
Z
Z
Z
Ṙ
dR =
F (x) dy =
dy.
BEC
BEC
BEC ẏ
Since x > β along BEC, we have that F (x) > 0. The integration is carried out from the
positive to negative of y, meaning the integral is therefore negative for sufficiently large y0
and therefore R(0, y1 ) − R(0, y0 ) < 0 =⇒ |y1 | < y0 .
4.4
Stability
The notion of stability is not as straightforward as other aspects we’ve covered thus far. Just
saying a solution is “stable” is not specific enough, so we’ll have to define several notions of
stability.
Definition 4.9. Consider the ODE ẋ = f (x, t), x ∈ Rn . Let φ(x, t) be the flow corre-
sponding to this ODE, then φ(x, t) is said to be Lyapnuov stable for t ≥ t0 if and only
68
CHAPTER 4. DYNAMICAL SYSTEMS THEORY
if
∀ε > 0, ∃δ(ε, t0 ) > 0 : |x − y| < δ =⇒ |φ(x, t) − φ(y, t)| < ε for t ≥ t0 ,
otherwise the solution is said to be Lyapunov unstable.
Figure 4.13: Figure illustrating asymptotic stability of a fixed point.
Definition 4.10. Let φ(x, t) be Lyapnuov stable for t ≥ t0 . If additionally ∃η(t0 ) > 0 such
that
|x − y| < η =⇒ lim |φ(x, t) − φ(y, t)| = 0,
t→∞
then the solution is said to be asymptotically stable.
We can also consider the stability of periodic solutions, which are not generally stable in the
sense of Lyapunov stability. Orbits starting out in a neighborhood of a periodic solution will
tend to have slightly different frequencies and thus a different phase shift. This motivations
us to define a Poincaré section or map.
Consider the general, autonomous system ẋ = f (x), x ∈ Rn , where we have some periodic
solution φ(t). We now construct an n − 1 dimensional traversal V to the closed orbit, that is,
a manifold that is punctured by the closed orbit and nowhere tangent to it. We can think of
this as a “strobing” of the system in some sense. We only see x0 , x1 and so on.
Figure 4.14: Manifold V involved in the construction of a Poincaré map.
4.4. STABILITY
69
The closed orbit intersects V at a point a. Consider an orbit γ(x0 ) starting at x0 ∈ V and
following the orbit until it returns to V . The point x0 is therefore mapped V → V by the
flow and the return map, or Poincaré map P is a map x0 → P (x0 ) → P 2 (x0 ). .
Definition 4.11. Given an autonomous system with a periodic solution φ(t), transversal
manifold V and Poincaré map P with a fixed point a, we say that φ(t) is stable if, for each
ε > 0, we can find δ(ε) such that
|x0 − a| ≤ δ, x0 ∈ V =⇒ |P n (x0 ) − a| ≤ ε,
n = 1, 2, . . . .
The above solution is also asymptotically stable if it is stable and there exists a delta such
that
|x0 − a| < δ, x0 ∈ V =⇒ lim P n (x0 ) = a.
n→∞
Non-autonomous systems
Consider the non-autonomous equation ẋ = f (x, t), where x ∈ Rn and f (x, t+T ) = f (x, t).
We can apply the previous definition of stability by rewriting the equation as an n + 1
dimensional autonomous system
ẋ = f (x, θ),
θ̇ = 1,
θ(0) = 0,
where we note that θ is defined θ = θ mod T , so (θ, x) ∈ S 1 ⊕ Rn . The transversal is
now n dimensional, which leads us to a natural choice for a Poincaré section, which is the
mapping Rn → Rn at times 0, T, T 2 and so on.
Example 4.7. Consider the system
ẍ + 2µẋ + ω02 x = h cos ωt,
0 < µ < ω0 ,
h, ω > 0.
We rewrite this as the first order system in the manner described previously:
ẋ = y,
ẏ = −ω02 x − 2µy + h cos ωθ,
θ̇ = 1,
θ(0) = 0.
The solution to this system can be found, and is
q
q
−µt
−µt
2
2
ω0 − µ2 t + c2 e sin
ω0 − µ2 t + α cos ωt + β sin ωt,
x(t) = c1 e cos
where
α=
(ω02 − w) h
(ω02
2,
ω2)
β=
2µωh
2.
+
−
+ (ω02 − ω 2 )
Note that µ = 0 corresponds to a singularity due to resonance, but µ > 0 by assumption so
we’re in the clear. The constants c1 , c2 are determined by the initial conditions x(0), ẋ(0),
such that
ẋ(0) + µx(0) − µα − ωβ
p
c1 = x(0) − α,
c2 =
.
ω02 − µ2
4µ2 ω 2
4µ2 ω 2
70
CHAPTER 4. DYNAMICAL SYSTEMS THEORY
It is clear from the explicit form that we have a periodic solution of period 2π/ω when
c1 = c2 = 0, that is, x(0) = α, ẋ(0) = ωβ. We therefore construct the Poincaré mapping
P by strobing at times 0t = 0, 2π/ωpand so on. The point (α, ωβ) is the fixed point of P .
Substitution of t = 2π/ω with γ = ω02 − µ2 2π/ω yields the system
c
+
α
x(0)
x(2π/ω)
1
p
P
=P
=
ẋ(0)
ẋ(2π/ω)
c2 ω02 − µ2 + ωβ − µc
c1 e−2πµω cos γ + c2 e−2πµ/ω sin γ+α
p
p
−2πµ/ω
−2πµω
2
2
2
2
e
cos γ −c1 µ + c2 ω0 − µ − e
sin γ c1 ω0 − µ + µc2 + ωβ
=
=
c01 + α
p
.
c02 ω02 − µ2 + ωβ − µc01
(n)
(n)
It can be shown that c1 , c2 → 0 as n → ∞, and therefore,
α
n x(0)
lim P
=
.
ẋ(0)
ωβ
n→∞
4.5
Manifolds
Theorem 4.5 (Stable Manifold theorem). Near a hyperbolic fixed point x0 , the non-linear
system ẋ = f (x), x ∈ Rn has a stable and unstable, differentiable manifolds, S and U ,
tangent at x0 to the stable and unstable subspaces E S , E U (eigenvectors) of the linearized
system ẋ = Ax, A = Df (x0 ).
Furthermore, S, U have the same dimensions as E S , E U and if φt is the flow of the
system, then S, U are positively and negatively invariant (respectively) under φt such that
limt→∞ φt (x) = x0 for all x ∈ S and the same (t → −∞) applies for x ∈ U .
Before we prove this theorem, we will look at an example.
Example 4.8. Consider the system
ẋ1 = −x1 ,
ẋ2 = −x2 + x21 ,
ẋ3 = x3 + x21 .
Clearly, the origin is a fixed point, and the Jacobian at the fixed point is then Df (0, 0, 0) =
diag(−1, −1, 1). From this, we can conclude that E S is the (x1 , x2 ) plane and E U is the x3
axis. The explicit solution to this system is
x1 (t) = c1 e−t ,
x2 (t) = c2 e−t +
c21 t
e − e−2t ,
3
x3 (t) = c3 et +
c21 t
e − e−2t .
3
!
4.5. MANIFOLDS
71
Thus, the initial condition, x(0) = c := (c1 , c2 , c3 ) determines our stability. Specifically, we
can clearly see that limt→∞ φt (c) = 0 if and only if c3 + c21 /3 = 0. Hence, we can conclude
that the stable manifold is where this occurs:
S = {c ∈ R3 : c3 = −c21 /3}.
Similarly, we note that limt→−∞ φt (0) = 0 if and only if c1 = c2 = 0, which provides us our
unstable manifold:
U = {c ∈ R3 : c1 = c2 = 0}.
To prove the theorem itself, we need a few more pieces of machinery.
Definition 4.12. Let X be a metric space and let A, B ⊂ X. A homeomorphism h : A → B
is a continuous, injective map such that h−1 : B → A is continuous. The sets A, B are
called homeomorphic or topologically equivalent if there exists a homeomorphism from
A to B.
Definition 4.13. An n dimensional differentiable manifold M is a connected metric space
with an open covering {Uα }, that is, M = ∪α Uα such that
(i) For all α, Uα is homeomorphic to an open ball in Rn .
(ii) If Uα ∩ Uβ 6= ∅ and for each hα : Uα → B, hβ : Uβ → B are homeomorphisms,
then hα (Uα ∩ Uβ ) and hβ (Uα ∩ Uβ ) are subsets of of Rn and the map h : hα ◦ h−1
β :
hβ (Uα ∩ Uβ ) → hα (Uα ∩ Uβ ) is differentiable for all x ∈ hβ (Uα ∩ Uβ ) and the the det
of the Jacobian satisfies det Dhβ (x) 6= 0.
The pair (Uα , hα ) is called a chart for the manifold M and the set of charts is called an atlas.
Figure 4.15: Figure describing the basic definition of a manifold.
In general, we think of this as a map to Rn , which is differentiable, and then map back to the
curved space. We just need the same answer in the overlaps of the regions.
With these tools, we can now prove the stable manifold theorem.
72
CHAPTER 4. DYNAMICAL SYSTEMS THEORY
Proof. Consider writing the system as
A := DF (0),
ẋ = Ax + F (x),
F (x) := f (x) − Ax,
F ∈ C 1 (U ).
Also, note that F (0) = 0 and DF (0) = 0. Since the fixed pint is hyperbolic, there exists an
n × n transformation (that is invertible) C such that
P 0
−1
B = C AC =
,
0 Q
where λ1 , . . . , λk are the eigenvalues of P and have Re{λj } < 0 and λk+1 , . . . , λn are the
eigenvalues of Q with Re{λj } > 0.
Consider the change of coordinates y = C −1 x, under which, our system becomes
ẏ = By + G(y),
and define
U (t) =
eP t 0
,
0 0
G(y) := C −1 F (Cy),
0 0
.
V (t) =
0 eQt
Then, our system is effectively
u̇ = Bu,
v̇ = Bv,
kuk → 0 as t → ∞ and kvk → 0 as t → −∞.
We consider the integral equation
Z t
Z
u(t, a) = U (t)a +
U (t − s)G(u(s, a)) ds −
0
∞
V (t − s)G(u(s, a)) ds.
t
Direct differentiation shows that this satisfies the ODE, so it is an equivalent problem. Set
aj = 0 for j = k + 1, . . . , n. Initial conditions are (a k-dimensional manifold) described by
(
aj ,
j = 1, . . . , k
R ∞
uj (0, a) =
.
− 0 v(−s)G(u(s, a, . . . , ak , 0)) ds j = k + 1, . . . , n
Setting yj = uj (0, a), we have the following conditions:
yj = ψj (y, . . . , yn ),
j = k + 1, . . . , n.
This defines a stable manifold S̃ for (y1 , . . . , yn ). If y(0) ∈ S̃, then y(t) → 0 as t → ∞,
whereas if y(0) 6∈ S̃, then y(t) does not approach 0 as t → ∞. Hence, S̃ is positively
invariant. We can translate this stable manifold into x coordinates by the transformation
x = Cy.
We can also construct the unstable manifold in a similar manner.
4.5. MANIFOLDS
73
Example 4.9. Consider the system
ẋ2 = x2 + x21 .
ẋ1 = −x1 − x22 ,
(4.4)
Note that x1 is linearly stable, whereas x2 is linearly unstable. We then set u(0) (t, a) = 0 and
Z t
Z ∞
(j+1)
(j)
u
(t, a) = u(t)a +
U (t − s)G(u (s, a)) ds −
V (t − s)G(u(j) (s, a)) ds,
0
t
and we use the iterative method suggested by the proof of the stable manifold theorem. Here,
we see that
2
−1 0
−x2
A=B=
,
F (x) = G(x) =
.
0 1
x21
From this, we can start our iterative procedure
−t e
0
0 0
U (t) =
, V (t) =
,
0 0
0 et
a1
a=
.
0
The first step is easy to compute:
U
(1)
(t, a) = U (t)a =
e−t a1
,
0
and following the iterative procedure,
−t Z ∞ −t e
e a1
0
(2)
.
U (t, a) =
−
ds =
e−2t a21
t−s −2s 2
0
e e a1
− 3
t
Performing yet another step in the same manner yields
−t Z ∞
Z
1 t −(t−s) −4s 4 0
e a1
(3)
ds
U (t, a) =
−
e
e a1 ds −
et−s e−2s a21
0
a 0
t
−t
1
e a1 + 27
(e−4t − e−t ) a41
=
.
1 −2t 2
e a1
3
Setting t = 0, we see that
U
(3)
(0, a) =
a1
.
− 13 a31
Thus, we can conclude that we have an approximation to the stable manifold described by
1
1
S : x2 = ψ2 (x1 ) =⇒ x2 = − x21 =⇒ x21 + x2 = 0.
3
3
The stable and unstable manifolds are inherently local manifolds, defined in a small
neighborhood of a fixed point. We can also introduce definitions of more global objects.
74
CHAPTER 4. DYNAMICAL SYSTEMS THEORY
Figure 4.16: Sketch of the approximation of the stable and unstable manifold for (4.4).
Definition 4.14. Let φt be the flow of ẋ = f (x). The global stable and unstable manifolds
are then
W S (0) :=
[
φt (s),
W U (0) :=
t≤0
[
φt (u).
t≥0
In general to find the unstable manifold, take the process for finding the stable, except
t → −t.
4.6
Bifurcation Theory
Bifurcation theory concerns the change in qualitative behavior of a dynamical system as one
or more of the parameters are varied. If a fixed point of a flow is hyperbolic (Re λ 6= 0),
then the local behavior of the flow is completely determined by the linearized flow (up to
homeomorphism). Small perturbations of a model system will also have a hyperbolic fixed
point of the same stability type, which is called structural stability. A bifurcation of a fixed
point can only occur at parameter values for which a fixed point is non-hyperbolic. For this
reason, we have a general rule:
Criterion for detecting a bifurcation: find parameter values for which the linearized flow
has a zero (or purely imaginary) eigenvalue.
Consider the dynamical system
ẋ = F (x),
x ∈ Rm ,
where x = 0 is a non-hyperbolic fixed point. Let L be the Jacobian:
∂Fi Lij :=
.
∂xj x=0
m
Suppose that v1 , . . . , vm ⊂ R are the set of (generalized) eigenvectors of L such that
(a) The set v1 , . . . , vm1 correspond to the stable directions. That is, the linear stable manifold
is generated by these directions:
E S = span{v1 , . . . , vm1 }.
4.6. BIFURCATION THEORY
75
(b) Similarly, the unstable directions lead to the linear unstable manifold:
E U = span{vm1 +1 , . . . , vm2 }.
(c) Lastly, the remaining eigenvectors vm2 +1 , . . . , vm correspond to the eigenvectors for
which Re λ = 0, which we’ll call the linear center manifold:
E C = span{vm2 +1 , . . . , vm }.
Hence, Rm = E S ⊕ E U ⊕ E C .
Theorem 4.6 (Center Manifold Theorem). There exists a non-linear mapping h : E C →
E S ⊕ E U with h(0) = Dh(0) = 0 (that is, derivative with respect to the parameter of
interest) and a neighborhood U ⊂ Rm of x = 0 such that the center manifold
M := {(x, h(x)) : x ∈ E C }
has the following properties:
(a) invariance: the center manifold M is locally invariant with respect to ẋ = F (x). That
is, if an initial state x(0) ∈ M ∩ U , then x(t) ∈ M as long as x(t) ∈ U . In other words,
x(t) can only leave M when it leaves the neighborhood U .
(b) attractivity: If E U = {0}, that is, no eigenvalues have Re λ > 0, then the center
manifold is locally attractive. That is, all solutions stay within U tend exponentially to
some solution of ẋ = F (x) in M .
Figure 4.17: Sketch of the center manifold M and the linearized manifolds E S , E U , E C .
Note, these properties imply that M has the same dimension as E C and passes through the
fixed point at x = 0 and is tangential to E C at the origin.
For completeness, suppose that E U = {0} so that ẋ = F (x) can be written as
ẋ = Ax + f1 (x, y),
ẏ = −By + f2 (x, y),
76
CHAPTER 4. DYNAMICAL SYSTEMS THEORY
where A has Re λ = 0 and B has Re λ > 0, so x ∈ E C , y ∈ E S . Note that also,
fi (0, 0) = 0, Dfi (0, 0) = 0. From the center manifold theorem, we can approximate the
flow on M by
ẋ = Ax + f1 (x, h(x)) := G(x),
since M is defined to be
M := {(x, h(x)) : h(0) = 0, Dh(0) = 0, h : U → E S }.
To determine h, we note that on M
y = h(x) =⇒ ẏ = Dh(x)ẋ = Dh(x) {Ax + f1 (x, h(x))} ,
but also ẏ = −Bh(x) + f2 (x, h(x)), so we must match these two conditions, yielding
Dh(x) {Ax + f1 (x, h(x))} = −Bh(x) + f2 (x, h(x)).
We can solve this, at least, in principle, by Taylor expanding h and equating coefficients at
each order of h.
Example 4.10. Consider the system
ẋ = xy,
ẏ = −y − x2 ,
(4.5)
so that x is the neutral direction, with no linear terms, and y is the stable direction, meaning
that linearization yields that x, y correspond to E C , E S respectively. We try a general
solution of the form y = h(x) = ax2 + bx3 + cx4 + dx5 + O(x6 ), yielding,
ẏ = h0 (x)ẋ = h0 (x)xh(x) = (2ax+3bx2 +· · · )x(ax2 +bx3 +· · · ) = 2a2 x4 +5abx5 +O(x6 ).
On the other hand, we see that
ẏ = −y − x2 = −h(x) − x2 = −(a + 1)x2 − bx3 − cx4 − dx5 + O(x6 ).
Equating coefficients yields a = −1, b = 0, c = −2, d = 0, hence our approximation to the
center manifold is
y = h(x) = −x2 − 2x4 + O(x6 ),
ẋ = xh(x) = −x x2 + 2x4 + O(x6 ) .
From this, we deduce that the origin is a non-linear sink.
Local Bifurcations
Here, we consider bifurcations where dim E C = 1 and codimension 1 bifurcations, which
we can think of this as varying one parameter. So far, we’ve shown that the nature of a
non-hyperbolic fixed point can be analyzed in terms of the center manifold.
4.6. BIFURCATION THEORY
77
Figure 4.18: Sketch of the center manifold M and the linearized manifolds E S , E U , E C
for the example described by (4.5).
Consider the general ODE
ẋ = f (x, µ),
x ∈ Rm , µ ∈ R,
where µ is some parameter of interest, and that f (0, 0) = 0. Suppose that when µ = 0, the
linearized flow ẋ = Lx near x = 0 has some eigenvalues with real part. We then decompose
the space into Rm = E S ⊕ E C ⊕ E U and reduce the dynamics onto the center manifold.
Suppose that E U = {0}, and that we can rewrite this as the system
ẋ = A(µ)x + f1 (x, y, µ),
ẏ = −B(µ)y + f2 (x, y, µ).
All eigenvalues of A(0) therefore have 0 real part and all of the eigenvalues of B(µ) have
positive real parts in a neighborhood of µ = 0. In order to include µ on the center manifold,
we must introduce the trivial (but very important) equation
µ̇ = 0.
We then find an approximation to the extended center manifold y = h(x, µ) to be
ẋ = A(µ)x + f1 (x, h(x, µ), µ) := G(x, µ),
µ̇ = 0.
Expanding G(x, µ) as a Taylor series in a neighborhood of (x, µ) = (0, 0), we have that
1
1
1
2
2
ẋ = Gµ µ + Gµµ µ + · · · + Gµx µ + Gµµx µ + · · · x+ Gxx + Gµxx µ + · · · x2 +· · · .
2
2
2
The number of solutions of ẋ = 0 and their stability varies whether certain partial derivatives
vanish and/or their sign if they do not vanish. We will consider the canonical bifurcations
now.
Saddle-Node Bifurcation
Consider the equation
ẋ = µ − x2 .
78
CHAPTER 4. DYNAMICAL SYSTEMS THEORY
Figure 4.19: A saddle node bifurcation.
For µ < 0, there are no equilibria and for µ > 0, there is a stable/unstable pair. This is the
most simple bifurcation and is known as the saddle-node bifurcation.
In general, a 1 dimensional dynamical system ẋ = f (x, µ) has a saddle-node bifurcation at
(x, µ) = (0, 0) if the following conditions hold:
(1) f (0, 0), fx (0, 0) = 0, that is the Jacobian vanishes.
(2) fxx (0, 0) 6= 0.
(3) fµ (0, 0) 6= 0.
If fµ fxx < 0, the stable unstable pair exists for λ > 0, whereas if fλ fxx > 0, they exist for
λ < 0, which is the case illustrated in Figure: 4.19.
Transcritical Bifuraction
Consider the system
ẋ = µx − x2 ,
which clearly has fixed points at (x, µ) = (0, µ), which merge as µ → 0 and exchange
Figure 4.20: A transcritical bifurcation.
4.6. BIFURCATION THEORY
79
stability. Note that condition (3) of the saddle-node bifurcation is violated here.
In general, for ẋ = f (x, µ), there is a transcritical bifurcation at (x, µ) = (0, 0) if the
following conditions hold:
(1) f (0, 0) = 0, fx (0, 0) = 0.
(2) fxx (0, 0) 6= 0.
2
(3) fµ (0, 0) = 0, fµx
− fxx fµµ > 0.
Pitchfork Bifurcation
Consider the system
ẋ = µx − x3 .
The conditions for a pitchfork bifurcation are:
Figure 4.21: A pitchfork bifurcation.
(1) f (0, 0) = 0, fx (0, 0) = 0.
(2) fx (0, 0) = 0, fxxx (0, 0) 6= 0.
(3) fµ (0, 0) = 0, fµx (0, 0) 6= 0.
A subcritical pitchfork is when the non-zero fixed points are unstable. A supercritical
pitchfork is when the non-zero fixed points are stable.
Example 4.11. Consider the system
ẋ = x3 + x2 − (2 + µ)x + µ.
(4.6)
80
CHAPTER 4. DYNAMICAL SYSTEMS THEORY
Figure 4.22: Bifurcation diagram of the example described by (4.6).
2
The fixed point of the system occurs
√ at (x − 1)(x + 2x − µ) = 0, meaning x = 1 is a fixed
point for all µ and x = −1 ± 1 + µ exists for all µ > −1. Thus, we generate two fixed
points at µ = −1, meaning we have a saddle node bifurcation. When this branch hits the
x = 1 fixed point, it must exchange stability, meaning we have a transcritical bifurcation as
well. This occurs when
√
√
−1 ± 1 + 3 = −1 ± 4 = −1 + 2 = 1 =⇒ µ = 3.
Hopf Bifurcation
In this case, dim E C = 2, but it is still codimension one, since we’re only varying one
parameter. So far, we’ve have only considered bifurcations on a 1 dimensional center
manifold with fixed points being created or destroyed or stability exchanged.
The Hopf bifurcation involves a non-hyperbolic fixed point with a pair of eigenvalues ±iω
and therefore requires at least a 2 dimensional center manifold and the bifurcating solutions
are periodic rather than stationary. Consider the system
ẋ = µx − ωy − (x2 + y 2 )x,
ẏ = ωx + µy − (x2 + y 2 )y,
which is the “normal form” for a Hopf bifurcation. The Jacobian clearly has eigenvalues
µ ± iω. Hence, at the linear level x = 0 is a stable focus for µ < 0 and an unstable focus for
µ > 0. We expect some sort of bifurcation at µ = 0. In order to understand what happens,
change to polar coordinates, resulting in
ṙ = µr − r3 ,
θ̇ = ω.
We see clearly now that we simply have a pitchfork bifurcation in the radius.
4.6. BIFURCATION THEORY
81
Figure 4.23: A Hopf bifurcation.
Theorem 4.7 (Hopf Bifurcation theorem). Consider a dynamical system of the form
ẋ = f (x, y, µ),
ẏ = g(x, y, µ),
f (0, 0, µ) = g(0, 0, µ) = 0.
Suppose that the following conditions hold:
(1) The Jacobian matrix L, evaluated at the fixed point x = y = 0 has a pair of purely
imaginary eigenvalues. This is equivalent to the condition tr L = 0, det L > 0.
(2) If α(µ)±iω(µ) denote the pair of eigenvalues of L(µ) at the fixed points, then ∂α
(0) 6= 0.
∂µ
This is the transversality condition. In other words, we cross the imaginary axis if
fµx + gµy 6= 0 at (x, y, µ) = (0, 0, 0).
Then, a limit cycle bifurcates from the origin with an amplitude that grows with |µ|1/2 and
its period tends to 2π/|ω| as |µ| → 0.
The bifurcation is supercritical if σ < 0 and subcritical if σ > 0, where
σ :==
1
1
{fxxx + gxxy + fxyy + gyyy }+
{fxy (fxx + fyy ) − gxy (gxx + gyy − fxx gxx + fyy gyy )} .
16
16ω
The origin is stable for µ > 0 if fµx + gµy < 0 and stable for µ < 0 if fµx + gµy > 0.
Figure 4.24: Different cases describing the Hopf bifurcation. Note, (c) and (d) can be
obtained by just reversing the inequalities and exchanging the stability of the drawn diagrams.
82
CHAPTER 4. DYNAMICAL SYSTEMS THEORY
Cusp Bifurcation
Suppose that we have a dynamical system with a non-hyperbolic fixed point at x = 0 such
that
ẋ = xn + O |x|n+1 ,
n ≥ 2.
We can think of this as a degenerate version of a more general class of ODEs:
ẋ =
n−1
X
µk xk + xn + O |x|n+1 .
k=0
If µ0 = 0, this implies the origin is a fixed point. Also, if µ1 = 0, this implies that the
fixed point at the origin is non-hyperbolic. The above equation is called an unfolding of the
degenerate ODE xn . By understanding the behavior of the unfolded system near µk = 0, we
can understand the full bifurcation system.
The simplest unfolding is for n = 2, in which case, we have
ẋ = µ0 + µ1 x + x2 + O |x|3 .
We can eliminate one of the remaining parameters by completing the square with y = x−µ1 /2,
and therefore
µ2
ẏ = ν + y 2 ,
ν = µ0 − 1 .
4
Now, suppose we consider the unfolding of n = 3, described by
ẋ = µ0 + µ1 x + µ2 x2 − x3 + O |x|4 .
Completing the square here yields
ẋ = ν0 + ν1 x − x3 ,
which is now our first encounter with codimension 2. Fixed points are the solutions of the
algebraic equation ν0 + ν1 x − x3 = 0.
Figure 4.25: A cusp bifurcation.
5
Perturbation Theory
“Let’s get down and dirty, like Baywatch."
— Paul C. Bressloff
5.1
Preliminaries
We’ll motivate the basic ideas of an asymptotic expansion by an example.
Example 5.1. Consider the integral f (ε), where ε > 0 is some small parameter:
Z
f (ε) :=
0
∞
e−t
dt.
1 + εt
If we integrate by parts, this yields,
Z ∞
e−t
f (ε) = 1 − ε
dt
(1 + εt)2
0
2
3
N
N
N +1
= 1 − ε + 2ε − 3!ε + · · · + (−1) N !ε + (−1)
|
Note that
∞
Z
0
e−t
dt ≤
(1 + εt)N +2
Z
(N + 1)!ε
N +1
Z
∞
0
{z
:=RN (ε)
e−t
dt .
(1 + εt)N +2
}
∞
e−t dt = 1,
0
we can conclude that
|RN (ε)| ≤ (N + 1)!εN +1 N !εN ,
and hence, for a fixed N ,
f (ε) − PN a εk k=0 k
lim = 0,
N
ε→0 ε
where ak are the coefficients derived from integration by parts. In other words,
f (ε) =
N
X
ak εk + O εN +1 .
k=0
83
84
CHAPTER 5. PERTURBATION THEORY
This formal series is said to be an asymptotic expansion of f (ε) for a fixed N because it
provides a good approximation as ε → 0.
Important note: the above expansion is not convergent for a fixed ε since N !εN → ∞ as
N → ∞, so this is subtlety of asymptotic.
Definition 5.1. If f (ε) = O(g(ε)) as ε → 0, said “big oh”, this means that ∃M < ∞ such
that |f | ≤ M |g| as ε → 0.
Definition 5.2. We can say that f (ε) = o(g(ε)), said “little oh”, if
f (ε) = 0.
lim ε→0 g(ε) Definition 5.3. The ordered sequence of functions {δj (ε)} for j = 0, 1, 2, . . . is called an
asymptotic sequence as ε → 0 if δj+1 (ε) = o(δj (ε)) as ε → 0.
Definition 5.4. Let f (ε) be a continuous function of ε and let {δj (ε)} be an asymptotic
P
sequence. The formal series N
j=0 aj δj (ε) is called an asymptotic expansion of f (ε) valid
to order δN (ε) if for all N ≥ 0,
f (ε) − PN aj δj (ε) j=0
lim = 0.
ε→0 δN (ε)
We usually write that f ∼
PN
k=0
ak δk (ε).
Note that we are making no claim of the “goodness” of each choice of N . In some cases,
N = 5 may provide a better approximation than N = 100.
Definition 5.5. In the case of a perturbation solution to an ODE, we consider asymptotic
expansions of the form
x(t, ε) =
N
X
ak (t)δk (ε),
k=0
which will typically be valid over some range of times. It is often useful to characterize the
time interval over which the above asymptotic expansion is valid, that is,
x(t, ε) − PN a (t)δ (ε) k
k=0 k
lim = 0 for 0 ≤ δ̂(ε)t ≤ c,
ε→0 δN (ε)
where c is a constant independent of ε.
Example 5.2. If we consider x(t, ε) = εt sin t for x ∈ R, t ≥ 0, we see that x(t, ε) = O(ε)
for t = O(1) but x(x, ε) = O(1) for t = O(1/ε).
5.2. POINCARÉ-LINDSTEDT METHOD
85
Naive Attempt
Consider the general ODE
ẋ = f (x, t; ε),
x ∈ Rn ,
x(0) = x̃0 .
If we Taylor expand the series in ε, we find that
f (x, t; ε) = f0 (x, t) + εf1 (x, t) + · · · ,
and then one might expect that
x(t) = x0 (t) + εx1 (t) + · · · .
If we substitute this x expansion into the expanded f and Taylor expand each fi , we can
equate powers of ε and impose the initial conditions that x0 (0) = x̃0 and xj (0) = 0 for all
other j.
Example 5.3. Consider the linear oscillator with weak damping
ẍ + 2εẋ + x = 0,
x(0) = a,
ẋ(0) = 0.
(5.1)
This has an explicit solution of
x(t) = ae
−εt
√
√
εa
−εt
2
2
√
e sin t 1 − ε .
cos t 1 − ε +
1 − ε2
Alternatively, we introduce the expansion
x(t) = x0 (t) + εx1 (t) + · · · ,
and plug this into (5.1), which yields, after equating coefficients of εn :
ẍ0 + x0 = 0,
ẍn + xn = −2ẋn−1 ,
with the initial conditions x0 (0) = a, ẋ0 (0) = 0 and xj (0) = ẋj (0) = 0 for all other j. This
leads to
x(t) = a cos t + aε {sin t − t cos t} + · · · ,
but note that, we have an εt term, meaning this expansion is only valid for t < 1/ε, which is
particularly bad as ε → 0. We can do better with more sophisticated techniques.
5.2
Poincaré-Lindstedt Method
Consider the initial value problem
ẋ = f0 (x, t) + εf1 (x, t) + · · · + εm fm (x, t) + εm+1 Rm (x, t; ε),
86
CHAPTER 5. PERTURBATION THEORY
with x(0) = µ, t ≤ h, x ∈ D ⊂ Rn , and 0 ≤ ε < ε0 . Assume that fi (x, t) is continuously
differentiable in both x, t and (m + 1 − i) times continuously differentiable in x, that
is, Lipschitz. Also suppose that Rm (x, t, ε) is continuous in all of its inputs, Lipschitz
continuous in x and bounded. Substitution into the equation for x of the formal expansion
x(t; ε) = x0 (t) + εx1 (t) + · · · εm xm (t),
generates an approximation of x(t; ε) such that
kx(t) − {x0 (t) + · · · + εm xm (t)}k = O εm+1 .
In some sense, periodic solutions are determined by their behavior over a time scale of O(1).
If we combine naive perturbation theory with the periodicity of the solution, we can generate
a convergent solution over a whole time scale of O(1). We will illustrate this method for the
differential equation
ẍ + x = εf (x, ẋ; ε).
If ε = 0, we have a degenerate case in which all solutions are periodic with period 2π (with
a center at the origin). Suppose that for ε > 0, at least one periodic solution branches off
with period T (ε), where we already know T (0) = 2π. Furthermore, assume that an initial
Figure 5.1: A figure describing the preliminaries of the Poincaré-Lindstedt method.
condition on the periodic orbit is x(0) = a(ε), ẋ(0) = 0. The expansion theorem we have
implies that on a time-scale of O(1), we have that
lim x(t, ε) = a(0) cos t.
ε→0
It is convenient to fix the period by the frequency,
ω −2 = 1 − εη(ε),
ωt = θ,
where ε = 0 when ω = 1, and η(ε) can be expanded as a Taylor series in ε. Let x0 denote
dx/dθ, in which case, we have
p
x00 + x = εη(ε)x + ε(1 − εη(ε))f (x, 1 − εη(ε))x0 , ε) = εF (x, x0 , ε, η(ε)).
This has the formal solution
Z
x(θ) = a(ε)cosθ + ε
0
θ
F (x(s), x0 (s), ε, η(ε)) sin(θ − s) ds,
5.2. POINCARÉ-LINDSTEDT METHOD
87
with x(θ + 2π) = x(θ) for all θ. Applying the periodicity condition, we require that
Z θ+2π
F (x(s), x0 (s), ε, η(ε)) sin(θ − s) ds = 0,
0
and expanding sin(θ − s) = sin θ cos s − sin s cos θ and using the fact that sin θ, cos θ are
linearly independent, we end up with the two conditions
Z 2π
I1 (a(ε), η(ε), ε) :=
F (x(s), x0 (s), ε, η(ε))) sin s ds = 0,
Z0 2π
I2 (a(ε), η(ε), ε) :=
F (x(s), x0 (s), ε, η(ε))) cos s ds = 0.
0
For each η, this is a system of two equations with two unknowns, a, η. By the implicit
function theorem, this has a unique solution in the neighborhood of ε = 0 if the Jacobian
doesn’t vanish, that is,
∂(I1 , I2 ) J =
6= 0.
∂(a, η) ε=0
So, setting ε = 0 for I1 , I2 yields
Z 2π
sin sf (a(0) cos s, −a(0) sin s, 0) ds = 0
I1 =
0
Z 2π
cos sf (a(0) cos s, −a(0) sin s, 0) ds,
I2 = πη(0)a(0) +
0
in which case,
∂I1 ∂I2
∂I1 ∂I2
∂I1
J=
−
=
πa(0) = πa(0)
∂a(0) ∂η(0) ∂η(0) ∂a(0)
∂a(0)
Z
0
2π
∂f
∂f
sin s cos s
− sin s
∂x
∂ ẋ
This provides us a concrete condition for the existence of a limit cycle for ε 6= 0.
Example 5.4 (Van der Pol equation). We return to the Van der Pol equation, described by
ẍ + x = ε(1 − x2 )ẋ,
where the right hand side is of the form εf (x, ẋ; ε) as required for the Poincaré-Lindstedt
method. As the method prescribes, set ωt = θ and ω −2 = 1 − εη(ε). We apply the
periodicity condition
Z 2π
1
1
2
2
2
2
a(0)
sin τ (1 − a(0) cos τ ) dτ = 0 =⇒ a(0) 1 − a(0) = 0 =⇒ a(0) = 2.
2
4
0
We also check the second periodicity condition
Z 2π
πη(0)a(0) −
a(0) sin τ cos τ 1 − a(0)2 cos2 τ dτ = 0,
0
ds 6= 0.
88
CHAPTER 5. PERTURBATION THEORY
which suggests that η(0) = 0 since we know that a(0) 6= 0. Lastly, we check the Jacobian
condition
Z 2π
J = πa(0)
sin τ cos τ 2a(0)2 cos τ sin τ − sin τ 1 − a(0)2 cos2 τ dτ 6= 0,
0
and therefore we can conclude a limit cycle does indeed exist. Introduce the power series
expansions
x(θ) = a0 cos θ+εγ1 (θ)+ε2 γ2 (θ)+· · · =⇒ x0 (θ) = −a0 sin θ+εγ10 (θ)+ε2 γ20 (θ)+· · · ,
which we can substitute and collect powers of ε to yield
1
1 2
00
γ1 + γ1 = a0 η0 cos θ + a0
a0 − 1 sin θ + a30 sin 3θ.
4
4
To avoid resonance, we must set a0 = 2 and η0 = 0, which is actually the result we got when
we looked at this equation in Chapter 4 with dynamical systems techniques. Our general
solution is then
a2
γ1 (θ) = a1 cos θ + b1 sin θ − 0 sin 3θ.
32
0
3
The initial condition x (0) = 0 implies that b1 = 3a0 /32. When we set a0 = 2, we finally
find that
1
3
γ1 (θ) = a1 cos θ + sin θ − sin 3θ.
4
4
2
At O(ε ), we now have
1
5
3
00
γ2 + γ2 = 2η1 +
cos θ + 2a1 sin θ + 3a1 sin 3θ − cos 3θ + cos 5θ.
4
2
4
Again, to eliminate resonance,, we set η1 = −1/8 and a1 = 0, meaning our next terms are
3
1
x(θ) = 2 cos θ + ε
sin θ − sin 3θ + O(ε2 ),
4
4
which we can use in our original relationships to find
θ = ωt,
5.3
1
1
1
ω=p
≈q
= 1 − ε2 + O(ε4 ).
2
16
1 − εη(ε)
1 + ε8 + · · ·
Method of Multiple Scales
In practice, one often finds that solutions oscillate on a time scale of O(t) with an amplitude
and phase that drift on a time scale of O(εt). This suggests looking for solutions of the form
R(εt) sin[t + O(εt)].
5.3. METHOD OF MULTIPLE SCALES
89
Hence, there are two timescales: a fast time t and a slow time τ = εt. In order to separate
the two distinct time scales, we treat t and τ as independent and introduce the asymptotic
expansion
x(t) → x(t, τ ; ε) ∼ x0 (t, τ ) + εx1 (t, τ ) + · · · .
We also split the total time derivative as
d
∂
∂
=
+ε ,
dt
∂t
∂τ
which leads to
∂x0
ẋ ∼
+ε
∂t
∂x1 ∂x0
+
∂t
∂τ
+ ··· ,
∂ 2 x0
ẍ ∼
+ε
∂t2
∂ 2 x1
∂ 2 x0
+
2
∂t2
∂t∂τ
+ ··· .
Example 5.5. Consider the system
ẍ + εẋ(x2 − 1) + x = 0,
x(0) = 1,
ẋ(0) = 0,
and we need not assume the solution is a limit cycle. Consider then, the asymptotic expansion
for t = O(1/ε):
x(t) ∼ x0 (t, τ ) + εx1 (t, τ ) + · · · ,
for which, the O(1) coefficient equation is
∂ 2 x0
+ x0 = 0,
∂t2
x0 (0, 0) = 1,
∂x0
(0, 0) = 0 =⇒ x0 = R(τ ) cos[t + O(τ )].
∂t
We next examine the O(ε) terms, for which, the equation is
∂ 2 x0 ∂x0 2
∂ 2 x1
+
x
=
−2
−
(x − 1),
1
∂t∂τ
∂τ 0
∂t2 2
x1 (0, 0) = 0,
∂x1
∂x1
(0, 0) =
(0, 0) = 0.
∂τ
∂t
Evaluating the right hand side for our known solution to x0 yields
−2
∂ 2 x0
∂x0 2
−
(x − 1) = −2 [Rτ sin {t + θ(τ )} + Rθτ cos{t + θ(τ )}]
∂t∂τ
∂τ 0
+ R sin{t + θ(τ )} R2 cos2 {t + θ(τ )} − 1
1
1
2
= 2Rτ − (4 − R ) sin{t + θ} + Rθτ cos{t + θ} + R3 sin 3{t + θ}.
4
4
In order to eliminate the secular (resonant) terms, we must set
1
2Rτ − R(4 − R2 ) = 0,
4
90
CHAPTER 5. PERTURBATION THEORY
which provides us with an ODE for R, where R(0) = 1. We also know that θτ = 0, θ(0) =
0 =⇒ θ(τ ) = 0. We can solve the R ode by setting y := R2 and therefore dy = 2RdR,
which yields,
1
dR
dy
dy
1
1
dτ = −
=−
=−
−
,
8
R(R2 − 4)
2y(y − 4)
8 y−4 y
which provides us the solution
y − 4
= −τ + c =⇒ y − 4 = Ae−τ .
ln y y
Coupled with the fact that y = 1 when τ = 0, we conclude that A = −3. Thus, piecing this
all together, we have
R(τ ) = √
2
=⇒ lim R(τ ) = 2 =⇒ x(t) ∼ 2 cos t + O(ε),
τ →∞
1 + 3e−τ
and is a stable limit cycle.
Example 5.6 (Duffing equation). Consider the system
ẍ + x = −εx3 ,
0 < ε 1,
which we can think of arising from the small angle approximation sin x ≈ x − x3 /3. Again,
take the asymptotic expansion
x(t) ∼ x0 (t, τ ) + εx1 (t, τ ) + · · · ,
using the method of multiple scales, splitting the time derivative. On O(1), we find
∂ 2 x0
+ x0 = 0 =⇒ x0 = R(τ ) cos{t + θ(τ )}.
∂τ 2
Continuing on to the O(ε) terms, we have
∂ 2 x1
∂ 2 x0
∂
3
+
x
=
−
2
−
x
=
2
[R(τ ) sin{t + θ(τ )] − R3 cos3 (t + θ),
1
0
2
∂τ
∂t∂τ
∂τ
and using trig reductions, this turns into
2
∂
[R(τ ) sin{t + θ(τ )] − R3 cos3 (t + θ) = 2Rτ sin(t + θ) + 2Rθτ cos(t + θ)
∂τ
1
− R3 [3 cos(t + θ) + cos 3(t + θ)] .
4
In order to eliminate resonance, we must take Rτ = 0 and θτ = 3R2 /8, which means that
3 2
3 2
R(τ ) = R0 , θ(τ ) = R0 τ + θ0 =⇒ R0 cos t + θ0 + R0 εt + O(ε).
8
8
5.3. METHOD OF MULTIPLE SCALES
91
Example 5.7 (Forced Duffing equation). Consider the system
∂u
∂ 2u
= Du 2 + f (u, v),
∂t
∂x
∂ 2v
∂ 2v
=
D
+ g(u, v).
v
∂t2
∂x2
We assume that u(x) = u0 , v(x) = v0 and we linearized, and taking a Fourier transform
yields the dispersion curve
ẍ + x = ε(γ cos t − κẋ − βx − x3 ).
We will perform an asymptotic expansion on this equation, again using the method of
multiple scales. The O(1) terms yield
x0 = A(τ )eit + A∗ (τ )e−it ,
and the O(ε) terms are
∂x0
∂ 2 x0
∂ 2 x1
3
+
x
=
γ
cos
t
−
κ
−
βx
−
x
−
2
1
0
0
∂t
∂t∂τo
∂τ 2 2
i
hn γ
it
2
=e
− iκA(τ ) − βA(τ ) − 3|A| − 2iAτ + other terms,
2
where the other terms contain higher orders of eint and complex conjugates. We eliminate
the resonant term by setting
−2iAτ − iκA − 3|A|2 A +
γ
= 0,
2
and in polar coordinates A = reiθ = u + iv yields
2rτ = −κr +
γ
sin θ,
2
2θτ = β + 3r2 −
We can consider the nullcline for rτ , which is described by
Figure 5.2: Nullcline for rτ .
γ
cos θ
2r
92
CHAPTER 5. PERTURBATION THEORY
r=−
γ
sin θ.
2κ
The null cline for θτ is given by
cos θ =
2r
(β + 3r2 ) = F (r, β).
γ
There are several scenarios for the nullcline depending on the value of β.
Figure 5.3: Plot of F (r, β).
If β > 0, then there exists a unique r for each θ. If 0 > β > βc , where F (rmin , βc ) = −1,
there exists two values of r for each value of cos θ in some interval (−z, 0) for 0 < z < 1.
We can calculate this explicitly, noting that
Figure 5.4: Nullclines of θτ for the different cases of β.
∂F
2
|β|
2
= (β + 9r2 ) = 0 =⇒ rmin
=
.
∂r
γ
9
5.3. METHOD OF MULTIPLE SCALES
93
In this case,
2 |β|1/2
F (rmin , β) =
3 γ
81
3|β|
−|β| +
=⇒ βc3 = − γ 2 .
9
16
The intersection of rτ = 0 and θτ = 0 nullclines determine the fixed points of the amplitude
equation, which correspond to periodic solutions of the forced Duffing equation at the
resonant frequency. Stability can be determined by the directions of the vector field.
Figure 5.5: Intersections of the nullclines for the forced Duffing equation. A-E are the
intersections of rτ , θτ nullclines for increasing κ and fixed β, where 0 > β > βc .
We discuss the intersections illustrated in Figure: 5.5:
A: for small κ, there is only one fixed point.
B-C: At κ = κ1c , there is a saddle node bifurcation leading to the formation of a saddle and
a stable fixed point.
D-E: At κ = κ2c , there is a second saddle node bifurcation, in which the saddle and other
stable fixed points annihilate.
Suppose that we increase κ from small values. Initially, we’re on the upper branch, where it
remains until κ = κ2c , when it jumps to the lower branch. Starting from high values of κ,
we jump from the lower to the upper branch at κ1c < κ2c . This is an example of hysteresis.
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CHAPTER 5. PERTURBATION THEORY
Figure 5.6: Hysteresis effect in the forced Duffing equation.
5.4
Averaging Method
Consider the system
ẍ + x = εf (x, ẋ).
If ε = 0, we know the solution to be x(t) = r0 cos(ψ0 + t). Assume for ε 6= 0, the solution
is of the form
x(t) = r(t) cos{t + ψ(t)} =⇒ ẋ = −r(t) sin{t + ψ(t)}.
If we substitute this into the ODE, we get
−ṙ sin(t + ψ) − r cos(t + ψ)ψ̇ = εf (r cos(t + ψ), −r sin(t + ψ)),
(5.2)
and we also require that the form for ẋ is consistent with the form for x(t):
ṙ cos(ψ + t) − ψ̇r sin(ψ + t) = 0.
(5.3)
We can combine these in the following two ways:
(5.2) · sin(t + ψ) − (5.3) · cos(t + ψ),
(5.2) cos(t + ψ) + (5.3) · sin(ψ + t),
to yield
ṙ = −ε sin(t+ψ)f (r cos(t+ψ), −r sin(t+ψ)),
ε
ψ̇ = − cos(t+ψ)f (r cos(t+ψ), −r sin(t+ψ)),
r
which we can rewrite as
ṙ = εR(, r, ψ, t),
ψ̇ = εΨ(r, ψ, t),
where both R and Ψ are periodic in t. Since these two functions change slowly on a time
scale of O(1), we might expect to generate an approximate solution (r0 , ψ0 ) by averaging
the right hand side over a period of 2π with r, psi fixed, yielding
ṙa = εR0 (ra , ψa ),
ψ̇a = εΨ0 (ra , ψa ),
5.4. AVERAGING METHOD
95
where these right hand sides are, as previously mentioned, the averaged functions over a
period:
Z 2π
1
R0 (r, ψ) := −
sin(t + ψ)f (r cos(t + ψ), −r sin(t + ψ)) dt = R0 (r),
2π 0
and
1
Ψ0 (r, ψ) := −
2π
2π
Z
cos θf (r cos θ, −r sin θ) dθ = Ψ0 (r).
0
Example 5.8 (Van der Pol equation). Consider, yet again, the system
ẍ + x = ε(1 − x2 )ẋ.
In this case, performing this technique yields
1
r2
1
1 2
R0 (r) = r 1 −
,
Ψ0 (r) = 0 =⇒ ṙa = εra 1 − ra ,
2
4
2
4
Ψ̇ = 0.
Theorem 5.1 (Averaging Theorem). Consider a system of the form
x ∈ U ⊂ Rn ,
ẋ = ε(x, t; ε),
0 < ε 1,
(5.4)
and f ∈ C r (r ≥ 2), defined on the domain f : Rn ⊕ R ⊕ R → Rn , is bounded on bounded
sets as well as periodic with period T > 0.
We define the average system to be
1
ẏ = ε
T
T
Z
f (y, t, 0) dt := εf¯(y).
(5.5)
0
Then, we have the following results:
(i) There exists a C r change of variables
x = y + εw(y, t; ε)
under which, (5.4) becomes
ẏ = εf (y) + ε2 f1 (y, t; ε).
(ii) If x(t), y(t) are solutions of (5.4) and (5.5) with initial conditions x0 , y0 and |x0 − y0 | =
O(ε), then
|x(t) − y(t)| = O(ε) on a time scale t = O(1/ε).
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CHAPTER 5. PERTURBATION THEORY
(iii) If y ∗ is a hyperbolic fixed point of (5.5), then ∃ε0 > 0 such that for all 0 < ε ≤ ε0 ,
(5.4) has a unique hyperbolic periodic orbit such that
γε (t) = y ∗ + O(ε) with the same stability as y ∗ ,
and γε (t) could itself also be a fixed point.
Example 5.9. Consider the general system
ẍ + x = εf (x, ẋ, t),
and take
x(t) = a(t) cos ωt +
b(t)
sin(ωt) =⇒ ẋ(t) = −ωa(t) sin ωt + b(t) cos(ωt).
ω
Plugging this ansatz into the original differential equation yields
ȧ = −
ε
sin(ωt)f (x, ẋ, t)
ω
ḃ = ε cos(ωt)f (x, ẋ, t).
If we consider the Mathieu equation again, where
ẍ = {1 + 2ε cos(2t)} x = 0 with ω = 1,
then we have that
ȧ = 2ε {a cos t + b sin t} sin t cos 2t,
ḃ = −2ε{a cos t + b sin t} cos t cos 2t.
Both of these amplitude equations are varying slowly, which is the appropriate timescale to
apply the averaging theorem, yielding
1
ȧa = − εba ,
2
1
ḃa = − εaa ,
2
and applying the initial conditions x(0) = x0 , ẋ(0) = 0, we find that
1 aa (t) = x0 e−εt/2 + eεt/2 ,
2
1 ba (t) = x0 e−εt/2 − eεt/2 .
2
Index
Abel’s formula, 26
asymptotic
expansion, 84
sequence, 84
attractor, 60
averaging
method, 94
theorem, 95
Banach space, 8
Bendixson’s negative criterion, 62
Bessel’s inequality, 32
bifurcation
cusp, 82
Hopf, 80
pitchfork, 79
saddle-node, 77
transcritical, 78
bifurcation theory, 74
boundary conditions
adjoint operator, 39
inhomogeneous, 55
twisted, 39
boundary value problems, 31
boundedness
of linear systems, 22
Cauchy sequence, 8
Cauchy-Schwarz inequality, 31
center manifold, 75
characteristic number, see Floquet multiplier
chemical kinetics, 3
complete space, 8
contraction mapping, 8
differentiability, 6
distributions, 46
dual space, 47
Duffing equation, 90
dynamical systems, 57
eigenfunctions
completeness, 40
eigenspace
generalized, 17
of an operator, 33
eigenvalue
of an operator, 32
energy integral, 43
Euclidean norm, 7
existence
of solutions to ODEs, 10
fixed point, 8
Floquet
exponent, 24
multiplier, 23
flow, 57
forced Duffing equation, 91
Fredholm alternative, 50
fundamental matrix, 20
Green’s functions, 50
eigenfunction expansion, 55
modified, 53
of initial value problems, 53
of the Sturm-Liouville equation, 51
Gronwall’s lemma, 12
Hamiltonian, 2
Hilbert space, 31
homeomorphism, 71
hysteresis, 93
97
98
initial value problems, 1
inner product, 31
weighted, 36
integral kernel, 46
invariant set, 58
Lienard equation, 27, 65
limit cycles, 62
linear dependence, 19
linear equations, 14
linearization, 59
Liouville’s formula, 26
Lipschitz condition, 7
local traversal, 63
manifold
definition, 71
global, 74
stable, 70
Mathieu equation, 28, 96
perturbation analysis, 30
transition curves, 30
matrix exponential, 14
multiple scales, 89
Newtonian dynamics, 1
norm, 8
normal forms
of linear systems, 15
nullclines, 61
ω-limit set, 59
operator
adjoint, 36
bounded, 33
compact, 33
differential, 35
linear, 32
symmetric, 32
periodic orbit, 59
perturbation theory, 83
Picard’s iteration, 9
planar dynamics, 60
INDEX
Poincaré map, 68
Poincaré-Bendixson theorem, 64
Poincaré-Lindstedt method, 85
predator-prey, 3, 62
Rayleigh quotient, 44
Rayleigh-Ritz variational principle, 43
spectral theorem, 34
spectrum, 40
continuous, 41
discrete, 40
mixed, 42
stability
asymptotic, 68
Lyapunov, 67
of a dynamical system, 67
of a limit cycle, 25
of linear systems, 22
stable manifold theorem, 70
Sturm-Liouville operator, 37
boundary conditions, 40
test functions, 47
uniqueness
of solutions to ODEs, 10
Van der Pol equation, 65, 87, 95
variation of parameters, 21
weak derivatives, 48
Wronskian, 26