Ordinary Differential Equations
Lecture Notes
Victor Ivrii
Department of Mathematics,
University of Toronto
© by Victor Ivrii, 2021,
Toronto, Ontario, Canada
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Contents
Contents
i
1 Introduction to ODE
1.1 Basic Mathematical Models and Direction Fields . . . . . . .
1.2 Solutions of Some Differential Equations . . . . . . . . . . .
1.3 Classification of Differential Equations . . . . . . . . . . . .
2
3
7
10
2 First-Order Differential Equations
2.2 Separable Differential Equations . . . . . . . . . . . .
2.1 Linear Differential Equations . . . . . . . . . . . . . .
Miscellaneous Equations . . . . . . . . . . . . . . . .
2.4 Linear and Nonlinear Differential Equations . . . . .
Lagrange and Clairaut equations . . . . . . . . . . .
Singular Solutions to First Order Equations . . . . .
2.6 Exact Differential Equations and Integrating Factors
2.8 Existence and Uniqueness Theorem (optional) . . . .
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36
38
40
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3 Second-Order Linear Differential Equations
60
3.1 Differential Equations with Constant Coefficients . . . . . . 61
3.2 Solutions of Linear Homogeneous Equations; the Wronskian
65
3.3 Complex Roots of the Characteristic Equation . . . . . . . . 71
3.4 Repeated Roots; Reduction of Order . . . . . . . . . . . . . 77
3.5 Method of Undetermined Coefficients . . . . . . . . . . . . . 82
3.6 Variation of Parameters . . . . . . . . . . . . . . . . . . . . 88
4 Higher Order Linear Differential Equations
95
4.1 Solutions of Linear Homogeneous Equations . . . . . . . . . 96
4.2 Homogeneous Equations with Constant Coefficients . . . . . 102
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Contents
4.3
4.4
ii
The Method of Undetermined Coefficients . . . . . . . . . . 110
The Method of Variation of Parameters . . . . . . . . . . . . 114
7 Systems of First-Order Linear Equations
4.4 Introduction . . . . . . . . . . . . . . . . . . . . . . . . .
4.4 Matrices . . . . . . . . . . . . . . . . . . . . . . . . . . .
4.4 Linear Algebra . . . . . . . . . . . . . . . . . . . . . . .
4.4 Basic Theory of Systems of First-Order Linear Equations
4.4 Homogeneous Linear Systems with Constant Coefficients
4.4 Complex-Valued Eigenvalues . . . . . . . . . . . . . . . .
7.8 Repeated Roots . . . . . . . . . . . . . . . . . . . . . . .
7.7 Fundamental Matrices . . . . . . . . . . . . . . . . . . .
7.9 Nonhomogeneous Linear Systems . . . . . . . . . . . . .
7.A. Examples: n = 3 . . . . . . . . . . . . . . . . . . . . . . .
9 Nonlinear Differential Equations and Stability
9.1 The Phase Plane: Linear Systems . . . . . . . . . . .
9.2 Autonomous Systems and Stability . . . . . . . . . .
9.3 Locally Linear Systems . . . . . . . . . . . . . . . . .
9.4 Competing Species and Other Examples . . . . . . .
9.5 Integrable Systems and Predator–Prey Equations . .
9.6 Lyapunov’s Second Method . . . . . . . . . . . . . .
9.7 Periodic Solutions and Limit Cycles . . . . . . . . . .
9.8 Chaos and Strange Attractors: the Lorenz Equations
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Preface
These are Lecture Notes for MAT 244 “Introduction to Ordinary Differential
Equations” at Faculty of Arts and Science, University of Toronto. This is a
sophomore class for all but Math Specialist students.
I was teaching it for several years and the last time it at Fall of 2020.
This time the class was taught online due to COVID-19 pandemic and I
made beamer slides for lectures. These slides were reformatted to a book
format.
These Lecture Notes are addition rather than substitution for our standard textbook Elementary Differential Equations and Boundary Value Problems, 11th Edition, by William E. Boyce, Richard C. DiPrima and Douglas
B. Meade (referred as Textbook). Earlier editions, especially 10th, are also
admissible.
We cover Chapters 1–4, 7 and 9 from Textbook, however some sections
are permuted, some material removed and other material added, exposition
is different.
We used online plotter (which I also recommend to students):
https://aeb019.hosted.uark.edu/pplane.html
c b a This work is licensed under a Creative Commons Attribution-ShareAlike 4.0
International License.
1
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Chapter 1
Introduction to ODE
What are ODEs?
Ordinary Differential Equation (ODE) of the first order is an equation
F (x, y, y ′ ) = 0
(1)
dy
where x is an argument, y = y(x) is an unknown function, y ′ = dx
is its
derivative, F (x, y, z) is a given function of 3 variables.
Solution to (1) is a function y(x) satisfying this equation. y(x) is a
solution on interval I (finite or infinite) if it is defined on I and satisfy (1)
there.
We call (1) differential equation because it contains not only y(x) but
also its derivative. We call it ordinary differential equation because y(x) is
a function of 1 variable and y ′ is ordinary derivative.
Equations, invoking partial derivatives like
∂z ∂z
F (x, y, z, , ) = 0,
∂x ∂y
are called Partial Differential Equations (PDE) and studied in the different
class (f.e. APM346 “Partial Differential Equations”).
Similarly, equations, containing second derivatives
F (x, y, y ′ , y ′′ ) = 0
(2)
are called ODEs of the second order and containing derivatives up to order
m
F (x, y, y ′ , y ′′ , . . . , y (m) ) = 0
2
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(3)
Chapter 1. Introduction to ODE
3
are called ODEs of order m.
We often consider ODEs resolved with respect to the highest order derivative
y (m) = f (x, y, y ′ , y ′′ , . . . , y (m−1) )
(4)
Students, taking this class sometimes wonder “Where are all theorems?
We just solve equations!” It is a nature of the beast. The proof of the most
important Existence and Uniqueness Theorem is only sketched. The reason
is simple: the rigorous proof requires Real Analysis which some of you take
next year.
With APM346 it is even worse: many of Existence and Uniqueness
Theorems are not even formulated.
But mathematicians began to solve ODEs in late 17-th century and the
proofs came only 150 years later. During these 150 years mathematicians
did pretty amazing things!
1.1
1.1.1
Basic Mathematical Models and
Direction Fields
Some Basic Mathematical Models
Example 1.1.1 (Radioactive decay). Consider a radioactive material. Let
x(t) be a quantity of this material at time t. From time t to time t + dt
some part of this material decays and this quantity is proportional to the
total quantity of the material at the given moment and dt (dt ≪ 1–this
means that interval is very short). Therefore
dx := x(t + dt) − x(t) = −kx(t)dt
or
x′ = −kx
where k is the coefficient of proportionality. One can guess the solution
x(t) = Ce−kt with an arbitrary constant C. Check that this is a solution! To
find C one needs to know how much material was there at time 0: C = x(0).
Check it!
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Chapter 1. Introduction to ODE
4
Example 1.1.2 (Amoebae). Let x(t) be a number of amoebaes at time t.
From time t to time t + dt some of these amoebae divide into two; their
number is proportional to the total quantity of the amoebae at the given
moment and dt (dt ≪ 1–this means that interval is very short). Therefore
dx := x(t + dt) − x(t) = kx(t)dt
or
x′ = kx
where k is the coefficient of proportionality. One can guess the solution
x(t) = Cekt with an arbitrary constant C. Check that this is a solution! To
find C one needs to know how many was there at time 0: C = x(0). Check
it!
Example 1.1.3 (Falling object). Let an object of mass m falls from some
height. Let v(t) be its velocity (directed up) at time t. Then there are two
forces: a gravity mg and air resistance −kv(t) (it is proportional to v(t) but
has an opposite direction) and coefficient k depends on the size and the
shape of the object:
m
k
dv
= −mg − kv =⇒ v ′ = −g − v.
dt
m
Indeed, according to Newton’s law the force equal mw, where w =
an acceleration.
dv
dt
is
Example 1.1.4 (The spring). Consider a mass m on the spring:
Let x(t) be a deviation of the center of the ball from the equilibrium.
Then according to Hooke’s law the force is proportional to x(t) and have an
opposite direction. Therefore
m
Indeed, w =
stiffness.
d2 x
dt2
d2 x(t)
= −kx(t).
dt2
is an acceleration and k is the coefficient of spring
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Chapter 1. Introduction to ODE
5
x
Example 1.1.5 (Mathematical pendulum). Consider a mass on the stick:
Let x = x(t) be a deviation of the pendulum from the vertical at the
moment t. Then
g
d2 x
= − sin(x)
2
dt
ℓ
where ℓ is the length of pendulum (more precisely: the distance from the
anchor to the center of the mass). We will show it in Chapter 9.
Example 1.1.6 (Celestial mechanics). Let x(t) (it is a vector!) be a position
of the planet (with a negligibly small mass) at the moment t and the Sun is
at 0.
Then
d2 x
x
= −Gm0 3
2
dt
|x|
where m0 is the mass of the Sun and G is the gravitational constant.
Indeed, acceleration is on the left, and the gravity force divided by the
mass of the planet is on the right.
I. Newton actually solved it analytically and proved that Kepler’s laws of
celestial mechanics follow from it but it would not be the case if the gravity
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Chapter 1. Introduction to ODE
6
pull was not proportional to the inverse square from the distance!! It is how
Newton’s gravity law was discovered.
Example 1.1.7 (Celestial mechanics. II). However, there are more than one
planet and their masses are small (but not negligible) in comparison with
the mass of the Sun and all of them pull one another. So the more precise
system is
n
X
d2 xj
xk − xj
=
−G
,
m
k
dt2
|xk − xj |3
k=0
j = 0, . . . , n.
This is n-body problem; it has no analytical solution and is very hard.
But mathematicians of the 18-th and 19-th centuries (hundred years after I.
Newton), using that masses of planets are much smaller than the mass of
the Sun, solved it approximately.
Comparing their calculations and the astronomical observations they
discovered the relative masses of the planets. Finally, they even calculated
where an unknown planet should be–and astronomers looked in the place
prescribed by mathematician and here it (Neptune) was!
Example 1.1.8 (Brachistochrone). I. Newton solved the problem from the
Calculus of Variations (which did not exist then): find the fastest slide
between two fixed points A and B (in the initial point A the velocity is 0).
One may think that this is a straight line, but it is the shortest, not the
fastest. The fastest is called a brachistochrone, which is described by the
second-order ODE
y ′′ = −
1 + y ′2
⇐⇒ y 1 + y ′2 = C
2y
with a constant C and is a part of the cycloid :
A
x
y
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Chapter 1. Introduction to ODE
1.1.2
7
Direction Fields and Integral Curves
Definition 1.1.1. Assume that at each point (x, y) of the plane a direction
is given. Direction is defined by its slope k = k(x, y). Then we say that
direction field is given.
In particular, k(x, y) = 0 if the direction here is a horizontal and k(x, y) =
∞ if the direction is vertical.
Definition 1.1.2. Integral curve of the direction field is a curve which is
tangent to this field at each point. It means that it has the same slope:
dy
= k(x, y).
dx
(1.1.1)
It means exactly that y = y(x) is a solution to equation (1.1.1).
Figure 1.1: Direction field and several different integral curves
We will deal a lot with such fields in Chapters 7 and 9.
1.2
Solutions of Some Differential Equations
We solve some equations graphically.
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Chapter 1. Introduction to ODE
8
Example 1.2.1 (Radioactive decay). Recall that equation is: x′ = −kx.
Draw a direction field (for k = 1/2) and integral curves
Example 1.2.2 (Falling object). Recall that equation is: v ′ = −kv − g.
Draw a direction field (for k = 1/2, g = 1) and integral curves
We see that at v = −g/k direction field becomes horizontal, so v = −g/k
(constant) is a solution. It is called terminal velocity. For a person without
a parachute it is ≈ −60m/s and with pa arachute ≈ −6m/s.
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Chapter 1. Introduction to ODE
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Example 1.2.3 (Amoebae). Recall that equation is: x′ = kx.
Draw a direction field (for k = 1/2) and integral curves
Example 1.2.4 (Amoebae. II). Assume that we harvest amoebae with a
constant speed (in the textbook it field mice and owl).Then equation is:
x′ = kx − g. Draw a direction field (for k = 1/2, g = 1) and integral curves
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Chapter 1. Introduction to ODE
10
As x = g/k the direction field is horizontal, so x = g/k is a solution.
As x(0) > g/k we see that x(t) is increasing (so amoebae number grows).
As x(0) < g/k we see that x(t) is decreasing and the colony becomes extinct.
Example 1.2.5 (Amoebae. III). Assume that the colony has a limited
resources of the food. The popular model is x′ = (k − αx)x.
Draw a direction field (for k = 1/2, α = 1/4) and integral curves
As x = 0, and as x = k/α the direction field is horizontal, so x = k/α is
a solution. As 0 < x(0) < k/α we see that x(t) is increasing (so amoebae
number grows but never reaches the limit k/α). As x(0) > k/α we see that
x(t) is decreasing (so amoebae number decays but never reaches the limit
k/α).
More complicated biological models (namely, Lotka—Volterra predatorprey model and Lotka—Volterra two competing species model) will be
considered in Chapter 9.
1.3
Classification of Differential Equations
We actually covered them but please, read this section in the Textbook.
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Chapter 2
First-Order Differential
Equations
2.2
Separable Differential Equations
We start from the simplest equations, which are separable equations because
we can separate variables (and then integrate equations), leaving “Section
2.1. Linear Differential Equations; Method of Integrating Factors” for the
next lecture.
So we want to separate variables x and y in the equation
dy
= f (x, y);
dx
(2.2.1)
it is possible if and only if f (x, y) is a product (or ratio) of two functions,
one of them depending on x and another on y:
dy
M (x)
=
.
dx
N (y)
(2.2.2)
dy
M (x)
=
.
dx
N (y)
(2.2.2)
In this equation
we want y on the left and x on the right and blue are those factors which
are on their places while red are factors which are on the wrong places.
11
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Chapter 2. First-Order Differential Equations
12
So we need to multiply by dx thus moving it to the right, and by N (y)
thus moving it to the left:
N (y) dy = M (x) dx.
(2.2.3)
Now both x and y are on the right places and we can integrate:
Z
Z
N (y) dy = M (x) dx,
(2.2.4)
resulting in
H(y) = G(x) + C
(2.2.5)
with H ′ (y) = N (y) and G′ (x) = M (x).
So, we got a general solution in the implicit form
F (x, y) := H(y) − G(x) = C.
(2.2.6)
Remark 2.2.1. (a) We got an arbitrary constant C here! So it is a general
solution because particular solutions would be obtained by choosing different
values of C and it is implicit because usually it is not resolved with respect
to y.
(b) Why just one constant? Should not it be two constants in (2.2.5)
H(y) + C1 := H(y) − G(x) = C2 ?
No, because in fact we would get the same (2.2.6)
F (x, y) = H(y) − G(x) = C := C2 − C1 .
Example 2.2.1.
dy
= y.
dx
Solution. Multiplying by dx and dividing by y we get
dy
= dx
y
and integrating we get
ln(y) = x + C =⇒ y = ex+C .
Right?
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(2.2.6)
Chapter 2. First-Order Differential Equations
13
There are some problems with this solution: we missed solution y = 0
because we divided by y and we also missed y < 0 because it should be
ln |y| on the left,
ln |y| = x + C =⇒ |y| = ex+C =⇒ y = ±eC ex = C1 ex
with C1 = ±eC . Now C1 is an arbitrary constant, and we can add value
C1 = 0 if not in the process but in the final expression, and now we fixed all
glitches, but it is boring!!!
So, there is a shortcut which should be used in all assignments (but you
need to understand what is behind this shortcut):
dy
= dx;
y
integrating we get
ln(y) = x + ln(C) =⇒ y = Cex .
Alternatively, we could do it like this:
dy
= dx;
y
integrating we get
ln
y
= x =⇒ y = Cex .
C
Example 2.2.2.
dy
y
= .
dx
x
Solution. Multiplying by dx and dividing by y we get
dy
dx
=
y
x
and integrating we get
ln(y) = ln(x) + ln(C) =⇒ y = Cx.
Now it is merrier!
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Chapter 2. First-Order Differential Equations
14
Example 2.2.3.
y dy
= (y 2 + 1)x.
dx
Solution. Multiplying by 2dx and dividing by y 2 + 1 we get
2y dy
= 2x dx
y2 + 1
and integrating we get
p
2
ln(y 2 + 1) = x2 + ln(C) =⇒ y 2 + 1 = Cex =⇒ y = ± Cex2 − 1.
Example 2.2.4. Consider now Cauchy’s problem consisting of equation and
initial condition
(
y ′ = ay + b,
y(0) = 0
with constants a, b.
Solution. (a) We start from the equation:
dy
a dy
= ay + b =⇒
= a dx =⇒ ln(ay + b) = ax + ln(C)
dx
ay + b
b
=⇒ ay + b = Ceax =⇒ y = C1 eax − .
a
(b) Plugging into initial condition we get
0 = C1 −
b
b
=⇒ C1 = .
a
a
(c) Plugging C1 into general solution we get finally
y=
b ax
e −1 .
a
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Chapter 2. First-Order Differential Equations
15
Example 2.2.5. Yet another Cauchy’s problem

p
y ′ = 1 − y 2 ,
y(0) = 1 .
2
Solution. (a) We start from the equation:
dy p
dy
= 1 − y 2 =⇒ p
= dx =⇒ arcsin(y) = x + C.
dx
1 − y2
π
(b) Plugging into initial condition we get = C. Plugging it into general
6
solution we get solution to the Cauchy’s problem
π
π
arcsin(y) = x +
=⇒ y = sin x + .
6
6
Example 2.2.6. Find the general solution and solution to the Cauchy’s
problem

2
y ′ = 1 + y ,
1 + x2

y(0) = 1.
Solution. (a) We start from the equation:
dy
1 + y2
dy
dx
=
=⇒
=
dx
1 + x2
1 + y2
1 + x2
=⇒ arctan(y) = arctan(x) + arctan(C).
(b) Then the general solution is
y = tan(arctan(x) + arctan(C)) =
x+C
.
1 − Cx
(c) Plugging into initial condition we get C = 1. Plugging it into general
solution we get solution to the Cauchy’s problem
y=
1+x
.
1−x
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Chapter 2. First-Order Differential Equations
16
Example 2.2.7 (Population dynamics. I).
dx
= kx(a − x),
dt
k > 0, a > 0,
where x(t) is a number of rabbits which multiply according to the law
dx
= kax =⇒ x = Cekat when there are few of them, but they suffer due
dt
to overcrowding when there are too many.
Solution. Separating variables we get
Z
Z h
dx
dx
dx
dx i
= k dt =⇒ akt = a
=
+
x(a − x)
x(a − x)
x
a−x
x
= ln(x) − ln(a − x) + ln(C) =⇒
= Cekat
a−x
Finally, we get
x=
a
1 − Ce−kat
We see that x(t) tends to a sustainable level a, from below, if 0 < x(0) < a
and from above if x(0) > a.
Figure 2.1: Population dynamics. I
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Chapter 2. First-Order Differential Equations
17
Example 2.2.8 (Population dynamics. II).
(
x′ = kx(b − y),
y ′ = ℓy(x − a)
k, ℓ > 0, a, b > 0,
where x(t) is a number of rabbits and y(t) is a number of foxes.
Solution. We cannot solve this problem, but since equations do not include
t explicitly, we have an autonomous system and can exclude t, dividing
equations and then separate vriables
dx
kx(b − y)
(x − a) dx
(y − b) dy
=
=⇒
=−
.
dy
ℓy(x − a)
kx
ℓy
Then
Z
(y − b) dy
(x − a) dx
=−
kx
ℓy
y b
x a
=⇒ − ln(x) + − ln(y) = C.
k k
ℓ ℓ
Z
So, we have lines which are level lines of
F (x, y) =
x a
y b
− ln(x) + − ln(y).
k k
ℓ ℓ
Observe that f (x) = xk − ka ln(x) has the plot with a single minimum at x = a
and level lines are like on the next page with equilibrium at x = a, y = b
and oscillations around it.
x=a
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Chapter 2. First-Order Differential Equations
18
Figure 2.2: Population dynamics. II
Remark 2.2.2. We return to this example in Chapter 9, Predator-Pray model
when you learn more from Calculus II.
2.1
2.1.1
Linear Differential Equations
Linear Differential Equations: Definitions
Definition 2.1.1. 1a
(a) First order linear differential equation is equation of the form
A(t)y ′ + B(t)y = G(t),
(2.1.1)
or equivalently
y ′ + p(t)y = g(t)
p(t) =
B(t)
G(t)
, g(t) =
.
A(t)
A(t)
(2.1.2)
(b) g(t) is the right-hand side (expression); if g(t) = 0 equation is called
homogeneous; otherwise it is called inhomogeneous.
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Chapter 2. First-Order Differential Equations
2.1.2
19
Linear Homogeneous Equations
Linear homogeneous equation play a very special role in what follows. They
are also separable equations and therefore could be soved instantly:
Z
dy
dy
= −p(t)y =⇒
= −p(t) dt =⇒ ln(y) = − p(t) dt + ln(C)
dt
y
and we proved
Lemma 2.1.1. General solution of the linear homogeneous equation
y ′ + p(t)y = 0
(2.1.3)
is given by
Z
y = exp(−P (t))
2.1.3
with P (t) :=
p(t) dt.
(2.1.4)
Linear Inhomogeneous Equations: Method of
Integrating Factors
Consider now the general linear equation
y ′ + p(t)y = g(t)
(2.1.2)
and multiply it by a factor µ(t) (to be determined):
µ(t)y ′ + µ(t)p(t)y = µ(t)g(t).
(2.1.5)
Definition 2.1.2. µ(t) is an integrating factor if equation (2.8.5) is of the
form
′
µ(t)y = µ(t)g(t).
(2.1.6)
Why so? – Because equation (2.1.6) could be easily integrated:
Z
µ(t)y =
t
µ(τ )g(τ ) dτ + C
−1
=⇒ y = µ(t)
Z
t
µ(τ )g(τ ) dτ + Cµ(t)−1 . (2.1.7)
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Chapter 2. First-Order Differential Equations
20
However (2.1.6) could be rewritten as
µ(t)y ′ + µ′ (t)y = µ(t)g(t)
and it is equivalent to (2.1.5) if and only if
µ′ = p(t)µ
(2.1.8)
which is a linear homogeneous equation with p(t) replaced by −p(t), and
therefore
Z
µ(t) = exp(P (t))
with P (t) := p(t) dt
(2.1.9)
is an integrating factor for (2.1.2).
So we have proven
Theorem 2.1.2. The general solution to linear inhomogeneous equation
y ′ + p(t)y(t) = g(t)
(2.1.2)
is given by
y(t) = e
−P (t)
Z
t
eP (τ ) g(τ ) dτ + Ce−P (t) .
(2.1.10)
Remark 2.1.1. (a) We need to find just one integrating factor rather than
all of them.
(b) In Section 2.6 we consider integrating factors for more general nonlinear
equations.
(c) It is much more important to remember the method than formula
(2.1.10).
Example 2.1.1. Find the general solution to the equation and solution to
the Cauchy’s problem
(
y ′ + tan(t)y = cos(t),
y(0) = 1.
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Chapter 2. First-Order Differential Equations
21
Solution. (a) Finding integrating factor:
dµ
= tan(t) dt
µ
Z
=⇒ ln(µ) = tan(t) dt = − ln(cos(t)) =⇒ µ(t) =
µ′ (t) = µ(t) tan(t) =⇒
1
.
cos(t)
(remember, we need just one integrating factor, so no constant is needed
there).
(b) Multiplying equation by the integrating factor we get
sin(t)y
y
y′
+
=
1
=⇒
= t + C =⇒ y = (t + C) cos(t)
cos(t) cos2 (t)
cos(t)
′
y(t)
=
cos(t)
is the general solution to the equation y ′ + tan(t)y = cos(t). Here we need a
constant C.
(c) Plugging into initial condition we get C = 1 and therefore
y = (t + 1) cos(t)
is the solution to the Cauchy’s problem.
2.1.4
Linear Inhomogeneous Equations: Method of
Variation of Parameters
Linear Inhomogeneous Equations: Method of Variation of Parameters Now let us consider another method, namely Method of Variation of
Parameters which is equivalent to the Method of Integrating Factors for
first-order linear equations but has a different idea behind it and generalizes
to a completely different class of the equations (see Chapter 3).
So we consider inhomogeneous equation
y ′ + p(t)y = g(t)
(2.1.2)
and corresponding homogeneous equation
y ′ + p(t)y = 0
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(2.1.3)
Chapter 2. First-Order Differential Equations
22
According to (2.1.4) the general solution to (2.1.3) is
Z
−P (t)
y = Cy1
with y1 = e
, P (t) = p(t) dt.
(2.1.11)
We will look at the solution of the inhomogeneous equation (2.1.2) in the
form (2.1.11) but with C1 which is not a constant but an unknown function
u:
y = uy1
with y1 : y1′ + p(t)y1 = 0.
(2.1.12)
Then
(uy1 )′ + p(t)uy1
= g(t),
′
′
=⇒ u y1 + uy1 + puy1 = g(t)
=⇒ u′ y1 + u y1′ + py1 = g(t)
with brackets equal 0 due to (2.1.12). Then
u′ y1 (t) = g(t) =⇒ u′ = y1 (t)−1 g(t)
Z t
y1 (τ )−1 g(τ ) dτ + C
=⇒ u(t) =
Z t
=⇒ y(t) = uy1 (t) = y1 (t)
y1 (τ )−1 g(τ ) dτ + Cy1 (t).
So,
Z
y(t) = uy1 (t) = y1 (t)
t
y1 (τ )−1 g(τ ) dτ + Cy1 (t).
Recall that according to (2.1.11) y1 (t) = e−P (t) with P (t) =
got again
Z t
−P (t)
y(t) = e
eP (τ ) g(τ ) dτ + Ce−P (t) .
(2.1.13)
R
p(t) dt; so we
(2.1.10)
The last term in (2.1.10) is C1 y1 (t) which is the general solution of the
homogeneous equation. This is a reflection of much more general Theorem
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Chapter 2. First-Order Differential Equations
23
Theorem 2.1.3. The general solution to the inhomogeneous equation equals
to the particular solution to this equation plus the general solution to the
corresponding homogeneous equation.
This Theorem holds for all linear equations, including higher-order ODEs,
PDEs, integral equations and so on.
Remark 2.1.2. (a) We replaced an arbitrary constant parameter C by an
unknown function u. This is why the method is called Method of Variation
of Parameters.
(b) For m-th order linear inhomogeneous ODEs there will be not 1 but m
parameters and this method has been invented for 2-nd order ODEs first.
(c) Compare Method of Integrating Factor and Method of Variation of
Parameters. For first-order linear ODE it is the same since µ(t) = y1 (t)−1 .
One could notice this even without explicit formulas for µ(t) and y1 (t).
Indeed, in the Method of Integrating Factor the last term is Cµ−1 (t) and in
the Method of Variation of Parameters it is Cy1 (t). therefore y1 (t) = Cµ(t)−1
(all these constants may differ).
Example 2.1.2.
ty
= t.
t2 + 1
Solution. (a) Solving the corresponding homogeneous equation:
y′ −
ty
dy
t dt
1
=⇒
=
=⇒
ln(y)
=
ln(t2 + 1) + ln(C)
t2 + 1
y
t2 + 1
2
√
=⇒ y = C t2 + 1.
√
(b) Plugging y = u t2 + 1 into inhomogeneous equation we get
y′ =
√
t
u′ t2 + 1 = t =⇒ u′ = √
2
t +1
Z
=⇒ u =
(c) Finally,
y=
√
t2 + 1 + C
√
√
√
t dt
= t2 + 1 + C.
t2 + 1
√
t2 + 1 = t2 + 1 + C t2 + 1.
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Chapter 2. First-Order Differential Equations
2.1.5
24
Solution to the Cauchy’s Problem
Consider Cauchy’s problem
(
y ′ + p(t)y(t) = g(t),
y(t0 ) = y0 .
Let us make all integrals with the lower limit t0 :
Z t
p(τ ) dτ,
P (t) =
t0
Z t
−P (t)
eP (τ ) g(τ ) dτ + Ce−P (t) .
y(t) = e
(2.1.14)
(2.1.15)
t0
Plugging into initial condition we get C = y(t0 ) (since P (t0 ) = 0).
We arrive to
Theorem 2.1.4. Solution to the Cauchy’s problem (2.1.14) is given by
Z t
−P (t)
y(t) = e
eP (τ ) g(τ ) dτ + y0 e−P (t)
(2.1.16)
t0
with P (t) defined by (2.1.15).
Miscellaneous Equations
2.1.6
Bernoulli Equations
In this lecture we’ll cover miscellaneous types of the first-order ODES. We
start from
Definition 2.1.3. Bernoulli’s equation is equation of the form
y ′ + p(t)y = g(t)y n
with n ∈ R, n ̸= 0, 1.
(2.1.17)
This restriction n ̸= 0, 1 is because for n = 0 we get a linear equation,
and for n = 1 we get even linear homogeneous equation
y ′ + (p(t) − g(t))y = 0.
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Chapter 2. First-Order Differential Equations
25
Bernoulli’s equations are closely related to linear equations. Indeed, we
can rewrite it in the form
y ′ y −n + p(t)y 1−n = g(t)
and observing that due to the chain rule from Calculus I
′
1
y 1−n
y ′ y −n =
1−n
we conclude that the substitution
z = y 1−n
(2.1.18)
z ′ + (1 − n)p(t)z = (1 − n)g(t).
(2.1.19)
reduces it to the linear equation
Example 2.1.3. Find the general solution and the solution to the Cauchy’s
problem
(
y′ + y = y2,
y(0) = −1.
Solution. It is a separable equation y ′ = y 2 − y but it is also a Bernoulli’s
equation and we solve it as such.
(a) Here n = 2 and we need to make a substitution z = y 1−n = y −1 and we
get
z ′ − z = −1.
If you forgot (2.1.19), note that z = y −1 =⇒ y = z −1 and then
(z −1 )′ + z −1 = (z −1 )2 =⇒ −z −2 z ′ + z −1 = z −2 =⇒ z ′ − z = −1.
(b) So, z ′ − z = −1. Solving corresponding homogeneous linear equation
z ′ − z = 0 we get z = Cet and using substitution z = uet we get
Z
′ t
′
−t
u e = −1 =⇒ u = −e =⇒ u = − e−t dt = e−t + C
=⇒ z = e−t + C et = 1 + Cet .
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Chapter 2. First-Order Differential Equations
26
(c) Therefore
y = z −1 =
1
1 + Cet
is the general solution.
(d) Plugging to initial condition we get −1 =
y = z −1 =
1
,
1 − 2et
1
1+C
=⇒ C = −2. Then
t > − ln(2)
is solution to the Cauchy’s problem. Here t > − ln(2) because as t = − ln(2)
solution blows up.
Remark 2.1.3. The fact, that Bernoulli’s equation could be reduced to linear
equation does not mean that we should do it. Rather we should apply the
same method as for linear equations.
So, we consider again
y ′ + p(t)y = g(t)y n
with n ∈ R, n ̸= 1.
(2.1.17)
Solving corresponding linear homogeneous equation y ′ + p(t)y = 0 we find
that
Z t
−P (t)
y = Ce
,
P (t) =
p(t) dt.
So we are looking to
y = ue−P (t) .
Plugging it to (2.1.17) we get
′
n
ue−P (t) + p(t)ue−P (t) = g(t) ue−P (t)
=⇒ u′ e−P (t) = g(t)un e−nP (t)
=⇒ u′ u−n = g(t)e−(n−1)P (t)
Z t
1
1−n
=⇒
u
=
g(τ )e−(n−1)P (τ ) dτ
1−n
Z t
1
1−n
−(n−1)P (τ )
=⇒ u = (1 − n)
g(τ )e
dτ
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(2.1.20)
Chapter 2. First-Order Differential Equations
27
and plugging it into (2.1.20): y = ue−P (t) wwe get
Z t
1
1−n
−(n−1)P (τ )
g(τ )e
dτ
y = (1 − n)
e−P (t) .
(2.1.21)
Remark 2.1.4. No need to remember the formula. Remember the method!
Using ready formula will not bring you a full mark in assessments.
2.1.7
Homogeneous Equations
Definition 2.1.4. The equation is said to be homogeneous if the right-hand
side of the equation
dy
= f (x, y)
dx
y
x
can be expressed as a function of the ratio
(2.1.22)
only, i.e. f (x, y) = φ
y
x
.
Remark 2.1.5. (a) “Homogeneous equations” and “Linear Homogeneous
Equations” are completely different animals. Do not confuse them!
(b) Equation is called homogeneous, because function f is homogeneous of
order 0
f (λx, λy) = f (x, y)
∀x, y, λ ̸= 0.
(2.1.23)
So, we consider a homogeneous equation
dy
y
=φ
.
dx
x
(2.1.24)
For homogeneous equations you need to remember the following substitution
y = ux.
(2.1.25)
Then
(ux)′ = φ(u) =⇒ u′ x + u = φ(u) =⇒ u′ x = φ(u) − u,
which is a separable equation
dx
du
=
=⇒ ln(x) =
x
φ(u) − u
Z
du
.
φ(u) − u
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Chapter 2. First-Order Differential Equations
We got a solution in the parametric form

Z
du x = exp
,
φ(u) − u

y = ux.
28
(2.1.26)
Remark 2.1.6. We say that this is a solution in the parametric form because
both x and y are obtained as functions of a parameter u.
Example 2.1.4. Find the general solution and the solution to the Cauchy’s
problem

2
2
y ′ = x + y ,
x2 + xy

y(1) = 0.
Solution. Let us check that it is a homogeneous equation: if we plug y = ux
we get on the right
1 + u2
x2 + x2 u2
=
.
x2 + x2 u
1+u
Then equation becomes
1 + u2
1 + u2
1−u
dx
(1 + u) du
=⇒ u′ x =
−u=
=⇒
=
1+u
1+u
1+u
x
1−u
Z
Z 2 (1 + u) du
=⇒ ln(x) =
=
−1 −
du = −u − 2 ln(u − 1) + ln C
1−u
u−1
=⇒ x = exp −u − 2 ln(u − 1) = C(u − 1)−2 e−u
u′ x + u =
So we got a general solution in the parametric form
(
x = C(u − 1)−2 e−u ,
y = Cu(u − 1)−2 e−u .
To solve the Cauchy’s problem we plug x = 1 and u = 0 (think why). Then
1 = C and
(
x = (u − 1)−2 e−u ,
y = u(u − 1)−2 e−u .
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Chapter 2. First-Order Differential Equations
2.1.8
29
Riccati’s equations
Riccati’s equation
dy
= q1 (t) + q2 (t)y + q3 (t)y 2
dt
(2.1.27)
usually cannot be solved in the quadratures, which means by taking integrals
and solving functional equations, like we did before.
However, if some particular solution is known, we can find the general
solution (see Problems to Chapter 2).
2.4
2.4.1
Linear and Nonlinear Differential
Equations
Notion of Solution
There are profound differences between linear and non-linear equations,
including, but not exclusively, ODEs. In this lecture we discuss Existence
and Uniqueness of Solutions (which should be a name of the section). In
Section 2.8 we sketch proofs. Honestly, rigorous proofs require a some
knowledge of Real Analysis, very few of you will take in the future.
First, however, we will define what solution is. Let I ⊂ R be an interval
(finite, or infinite, or semi-infinite) (α, β), −∞ ≤ α < β ≤ ∞).
Definition 2.4.1. Let f (x, y) be a function of two variables, continuous in
a domain (open set) D ⊂ R2 (you should already learn what does it mean
in Calculus II, or will learn shortly). Then we call y = y(x) a solution to
ODE on interval I
y ′ = f (x, y)
(2.4.1)
if
(a) y(x) is a continuously differentiable on I,
(b) and for each x ∈ I (and for each x ∈ I (2.4.1) is fulfilled (including
(x, y(x)) ∈ D).
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Chapter 2. First-Order Differential Equations
2.4.2
30
Existence and Uniqueness for Linear Equations
Theorem 2.4.1. Consider a Cauchy’s problem for a linear first-order ODE
y ′ + p(t)y = g(t),
y(t0 ) = y0 ,
(2.4.2)
(2.4.3)
with p(t) and g(t) continuous on I ∋ t0 .
Then this problem (2.4.2)–(2.4.3) has a solution on I and it is unique.
Proof. We already have a formula:
Z t
−P (t)
eP (τ ) g(τ ) dτ + y0 e−P (t) ,
y(t) = e
Z
t0
2.4.3
t
P (t) =
p(τ ) dτ.
t0
Existence and Uniqueness for General
Equations
Theorem 2.4.2. Consider a Cauchy’s problem for a first-order ODE
y ′ = f (x, y),
y(x0 ) = y0 ,
(2.4.4)
(2.4.5)
with (x0 , y0 ) ∈ D. Assume that
(a) f is a continuous function in domain D ⊂ R2 ;
(b) f satisfies Lipschitz’s condition with respect to y
|f (x, y1 ) − f (x, y2 )| ≤ L|y1 − y2 |
∀x, y1 , y2 : (x, y1 ) ∈ D, (x, y2 ) ∈ D. (2.4.6)
Then
(i) There exist an interval I := (x0 − δ, x0 + δ), δ > 0 and a solution y(x)
on I to (2.4.4)–(2.4.5);
(ii) This solution is unique: if y1 (x) and y2 (x) two solutions on I = (x0 −
δ, x0 + δ) (δ > 0 is arbitrary), both satisfying y1 (x0 ) = y2 (x0 ) = y0 , then
y1 (x) = y2 (x).
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Chapter 2. First-Order Differential Equations
31
Remark 2.4.1. (a) Let f and ∂f
be continuous in D. Then assumption
∂y
(2.4.6) is fulfilled for any closed rectangle
R = [x0 − δ, x0 + δ] × [y0 − N, y0 + N ] ⊂ D.
(b) Theorem 2.4.1 claims a global existence while Theorem 2.4.2 claims only
a local existence.
Why Theorem 2.4.2 does not claim a global existence even if D is a strip
D = (x0 − δ, x0 + δ) × R?
Example 2.4.1.
(
y ′ = y 1+α , α > 0
y(0) = 1.
This equation is separable, solving it we get −αy −1−α dy = −αdx =⇒
y = C − αx; it satisfies initial condition as 1 = C. So we get
−α
y = 1 − αx
− α1
which blows-up (goes to infinity as x ↗ x∗ := α1 . Thus it is a solution only
on (−∞, x∗ ).
(a) y ′ = y 2
(b) y ′ = y|y|
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Chapter 2. First-Order Differential Equations
32
One can think that the reason of the blow-up is the superlinear growth
of f (x, y) with respect to y. Indeed, with no more then linear growth we
can get a global existence
Theorem 2.4.3. Assume that conditions of Theorem 2.4.2 are fulfilled in
D = I × R, I = (α, β) ∋ x0 .
Further, assume that
|f (x, y)| ≤ M (|y| + 1)
∀x ∈ I, y.
(2.4.7)
Then there exists a (unique) solution to (2.4.4)–(2.4.5) on I.
Remark 2.4.2. Theorem 2.4.1 follows from Theorem 2.4.3. Check it!
There is a more general theorem, than Theorem 2.4.3, without Lipschitz’s
condition:
Theorem 2.4.4. Consider a Cauchy’s problem for a first-order ODE
y ′ = f (x, y),
y(x0 ) = y0 ,
(4)
(2.4.5)
with (x0 , y0 ) ∈ D. Assume that f is a continuous function in domain
D ⊂ R2 .
Then there exist an interval I := (x0 − δ, x0 + δ), δ > 0 and a solution
y(x) on I to (2.4.4)–(2.4.5).
Remark 2.4.3. Theorem 2.4.4 claims only existence but not uniqueness of
the solution.
Example 2.4.2.

y ′ = 1 y 1−α , 1 > α > 0
α

y(0) = 0.
This equation is separable, solving it αy −1+α dy = dx =⇒ y α = x − C,
satisfying initial condition 0 = C we get
1
y = xα .
For some α ∈ (0, 1) it is(defined only for x ≥ 0, but it does not matter: we
0
x < 0,
can always set y(x) =
. But y(x) = 0 is another solution
1/α
(αx)
x > 0.
to the same problem!
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Chapter 2. First-Order Differential Equations
2.4.4
33
Consequences of Existence and Uniqueness
Theorem
Consequences of Existence and Uniqueness Theorem Let us discuss
some corollaries of Theorem 2.4.3.
Corollary 2.4.5. In the framework of Theorem 2.4.2 there exists a general
solution is y = y(x, C) and all solutions could be obtained from it.
Proof. Indeed, we can define y(x, C) as a solution to equation (2.4.4) with
initial condition y(x0 ) = C.
But what happens if we are only in the framework of Theorem 2.4.4?
Look for the previous example y ′ = 31 y 2/3 .
Solution y = 0 is really “a bad boy”: it is not a part of the general
solution and at each point of it another solution (which is a part of the
general solution) branches out. Such solutions are called singular solutions.
Here α = 13 and the general solution is y(x, C) = (x − C)3 .
2
Figure 2.3: y ′ = 13 y 3
We also can construct a “composite solution”:

3

(x − C1 ) x < C1 ,
C 1 ≤ x ≤ C2 ,
y(x) = 0

(x − C )3
x > C2 ,
2
1
Example 2.4.3. y ′ = y|y|− 3 which by continuity means that y ′ = 0 for y = 0.
The general solution here is y = ±(x − C)3 , x > C.
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Chapter 2. First-Order Differential Equations
34
2
Figure 2.4: y ′ = 31 y 3 –composite solution
1
Figure 2.5: y ′ = y|y|− 3
However for x < C we are forced to set y = 0 and the composite solution
looks like one of these:
1
Figure 2.6: y ′ = y|y|− 3 –composite solutions
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Chapter 2. First-Order Differential Equations
35
Remark 2.4.4. If the general solution is defined by F (x, y, C) = 0 then the
singular solution could be obtained by solving

F (x, y, C) = 0
(2.4.8)
 ∂F (x, y, C) = 0;
∂C
C = C(x, y) should be excluded from these equations.
Consider again directional fields. Assume first that the directions
⃗ℓ(x, y) = (α(x, y), β(x, y)) are not vertical: α(x, y) ̸= 0.
Then the equation of integral curves (reminder: integral curve of the
vector field ⃗ℓ(x, y) is a curve, tangent to it at each point).
dy
dx
=
α(x, y)
β(x, y)
(2.4.9)
is equivalent to
y ′ = f (x, y) :=
β(x, y)
α(x, y)
(2.4.10)
and if f satisfies assumptions of Theorem 2.4.2, then through each point
(x0 , y0 ) passes exactly one integral curve.
In particular, integral curves do not intersect.
On the other hand, as we have seen, in more general settings integral
curves may intersect.
Remark 2.4.5. (a) So where α(x, y) does not vanish we are fine.
(b) On the other hand, if β(x, y) does not vanish, we simply consider y as
an argument and x as a function.
(c) From this point of view, unpleasant are points in which both α(x, y)
and β(x, y) vanish. Such points are called stationary points of vector field
⃗ℓ(x, y) and they will be studied in great details in Chapters 7 and 9. This is
forbidden for directional fields.
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Chapter 2. First-Order Differential Equations
36
Optional reading:
Lagrange and Clairaut equations
Lagrange equation
Lagrange equation is of the form
y = xφ(y ′ ) + ψ(y ′ )
(2.A.1)
with φ(p) − p ̸= 0. To solve it we plug p = y ′ and differentiate equation:
pdx = φ(p)dx + xφ′ (p) + ψ ′ (p) dp.
(2.A.2)
This is a linear ODE with respect to x. We find the ¡em¿general solution¡/em¿
x = f (p, C) and then y = f (p, C)φ(p) + ψ(p):
(
x = f (p, C)
(2.A.3)
y = f (p, C)φ(p) + ψ(p)
gives us a general solution in the parametric form, provided
φ(p) − p ̸= 0.
(2.A.4)
Further, equation (2.A.1) can have a singular solution (or solutions; see
Singular Solutions to First Order Equations)
y = xφ(p) + ψ(p),
(2.A.5)
where p is a root of equation φ(p) − p = 0.
Clairaut equation
Clairaut’s equation is a particular case of Lagrange equation with φ(p) = p.
In this case (2.A.3)1 becomes just x + ψ ′ (p) = 0 and (2.C.3)2 becomes
y = Cx + ψ(C)
which is a general solution, and (2.A.5) becomes
(
x = −ψ ′ (p)
y = xp + ψ(p)
which is a singular solution in the parametric form.
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(2.A.6)
(2.A.7)
Chapter 2. First-Order Differential Equations
37
Riccati’s equation
General. General Riccati equation is
y ′ = a(x)y + b(x)y 2 + c(x).
(2.B.1)
It is closely connected with 2nd order linear homogeneous equations which
we will study in Chapter 3:
u′′ + p(x)u′ + q(x)u = 0.
(2.B.2)
Indeed, plugging into (2.B.2) u = eϕ we get
h
i
′′
′2
′
ϕ + ϕ + p(x)ϕ + q(x) eϕ = 0
which is (2.B.1) with y = ϕ′ , b = −1, a = −p, q = −c.
On the other hand, plugging into (2.B.1) y = zb(x)−1 we get with respect
to z again Riccati equation with the coefficient at z 2 equal 1.
While there is no general method to solve Riccati’s equation, there is a
remarkable fact: if a particular solution y1 (x) is known, the general solution
is
y(x) = y1 (x) + u(x),
(2.B.3)
where u satisfies Bernoulli’s equation
u′ = b(x)u2 + [2b(x)y1 + a(x)]u.
(2.B.4)
Indeed, one needs just to plug (2.B.3) into (2.B.1).
Special Case 1: a, b, c are constants. Then we can separate variables.
Special Case 2: y ′ = by 2 + cxn with constants b, c.
(a) If n = 0 we get Special Case 1.
1
(b) If n = −2 we have a homogeneous equation after substution y = .
z
(c) If n ̸= 2 see More reading http://eqworld.ipmnet.ru/en/solutions/
ode/ode0106.pdf.
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Chapter 2. First-Order Differential Equations
38
Other special cases. More reading
http://eqworld.ipmnet.ru/en/solutions/ode/ode0123.pdf
Singular Solutions to First Order Equations
Definition 2.C.1. Let us consider equation
F (x, y, y ′ ) = 0
(2.C.1)
with continuously differentiable function F (x, y, p).
We call y = y(x) a singular solution if it differentiable, satisfies (2.C.1)
and at each point (x0 , y0 ), y0 = y(x0 ) it touches another solution z(x), which
differs from it as x ̸= x0 :
y(x0 ) = z(x0 ),
(2.C.2)
y ′ (x0 ) = z ′ (x0 ),
(2.C.3)
y(x) ̸= z(x)
for x ̸= x0 .
(2.C.4)
Finding Singular Solution
First it means that equation (2.C.1) violates conditions of Existence and
Uniqueness Theorem (Theorem 2.8.4), which means
∂F (x, y, y ′ )
= 0.
∂y ′
(2.C.5)
Indeed, if this partial derivative is not 0 then by Implicit Function Theorem
of Calculus II in the vicinity of (x0 , y0 , p0 ), y0 = y(x0 ), p0 = y ′ (x0 ) equation
(2.C.1) could be resolved with respect to y ′ :
y ′ = f (x, y)
with a continuously differentiable f (x, y) and conditions of this Theorem
are fulfilled.
So we solve the system


F (x, y, p) = 0,
(2.C.6)
∂F (x, y, y ′ )

=
0,

∂y ′
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Chapter 2. First-Order Differential Equations
39
find y = y(x), p = p(x), check that we got y = y(x) which is a solution,
which means that p(x) = y ′ (x). To do so, we differentiate the first equation
in (2.C.6) by x and use chain rule:
∂F
∂F ′ ∂F ′
+
y +
p =0
∂x
∂x
∂p
and using the second equation in (2.C.6) we get
∂F
∂F
(x, y, p) +
(x, y, p)p = 0.
(2.C.7)
∂x
∂x
We also need to check that t each point (x0 , y0 ), y0 = y(x0 ) it touches
another solution z(x), which differs from it as x ̸= x0 .
Example 2.C.4. (a) Check that for Lagrange and Clairaut equations equations (see Lagrange equation)
xφ(y ′ ) + ψ(y ′ ) − y = 0
(with φ(p) ̸= p and φ(p) = p identically) equations (2.C.6) define y(x), p(x)
(defined there) and (2.C.7) is satisfied as well.
(b) Check that for equation
y ′3 − y − kx = 0
equations (2.C.6) define p = 0, y = kx but (2.C.7) is satisfied if and only if
k = 0.
General Solution and Singular Solution. If we found the general
solution to equation (2.C.1)
G(x, y, C) = 0
(2.C.8)
with an arbitrary constant C, then the singular solution could be found as

G(x, y, p) = 0,
(2.C.9)
∂G

(x, y, p) = 0.
∂p
One need to check that p(x) is not a constant but really depends on x.
More details and examples could be found in
https://www.math24.net/singular-solutions-differential-equations/
We skip Section 2.5 since main content has been covered already and we
will cover more advanced material in Chapter 9.
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Chapter 2. First-Order Differential Equations
2.6
2.6.1
40
Exact Differential Equations and
Integrating Factors
Exact Differential Equations
We consider a differential equation in the form
M (x, y) dx + N (x, y) dy = 0
(2.6.1)
which is clearly equivalent to
M (x, y) + N (x, y)y ′ = 0.
(2.6.2)
Remark 2.6.1. Surely, (2.6.2) means that x can be chosen as an independent
variable, which includes assumption that N (x, y) does not vanish, but (2.6.1)
does not require it. In this section we deal only with equation in the form
(2.6.1).
Definition 2.6.1. Equation (2.6.1) is called exact if there exists function
U (x, y), which is differentiable as a function of two variables, such that its
full differential is equal to the left-hand side:
dU = M (x, y) dx + N (x, y) dy.
(2.6.3)
Remark 2.6.2. (a) The notion of the full differential of the function is given
in the beginning of Calculus II. In particular, U (x, y) is differentiable,
provided it is continuously differentiable, that means partial derivatives ∂U
∂x
and ∂U
are
continuous.
Then
∂y
dU =
∂U
∂U
dx +
dy.
∂x
∂y
(b) Often the following notation is used Ux =
as Ux′ ).
∂U
∂x
(2.6.4)
(note we do not write it
(c) If you take Analysis II (a.k.a. Calculus II on steroids) you will learn
differential forms. In particular, M (x, y) dx + N (x, y) dy is 1-form. Then it
is called exact if it is equal dU for some U .
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Chapter 2. First-Order Differential Equations
41
Why we would want to have an equation which is exact? – Because we
integrate it immediately:
dU (x, y) = 0 =⇒ U (x, y) = C
(2.6.5)
and y is an implicit function of x.
Remark 2.6.3. From Calculus II: equation (2.6.5) defines level lines of
function U (x, y). Properties of level lines are studied in Calculus II and we
return to this topic in Chapter 9.
Level lines are very useful: contour lines in topographical maps, isotherms
and isobars in the weather forecast, equipotential lines in electrostatics and
so on.
Task
(a) Determine if equation is exact.
(b) Find this function U (x, y).
2.6.2
When Equation is Exact?
So, we want to find U (x, y) such that
∂U
= M (x, y),
∂x
∂U
= N (x, y).
∂y
(2.6.6)
Two equations and just one function! Similar systems are called overdetermined and there is no surprise that we must impose some conditions for
them to have a solution.
Let us differentiate the first equation by y and the second by x:
∂ 2U
∂M
=
,
∂y∂x
∂y
∂ 2U
∂N
=
.
∂x∂y
∂x
Those are second order derivatives and in the first equality we differentiate
first by x and then by y, while in the second equality we differentiate first
by y and then by x.
∂ 2U
∂M
=
,
∂y∂x
∂y
∂ 2U
∂N
=
.
∂x∂y
∂x
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Chapter 2. First-Order Differential Equations
42
However, we know from Calculus II that second and higher order derivatives do not depend in which order they are taken:
∂ 2U
∂ 2U
=
.
∂y∂x
∂x∂y
(2.6.7)
Therefore we get a necessary condition for equation to be exact:
Theorem 2.6.1. If equation M (x, y) dx + N (x, y) dy is exact then
∂N
∂M
=
.
∂y
∂x
(2.6.8)
Remark 2.6.4. In Calculus II equality
∂ 2U
∂ 2U
=
∂y∂x
∂x∂y
2
2
is conditioned: all second order partial derivatives including ∂∂xU2 and ∂∂yU2
must be continuous, that is M (x, y) and N (x, y) must be continuously
differentiable. But we do not care!
(a) In all examples M (x, y) and N (x, y) are continuously differentiable.
(b) In more advanced mathematics all functions can be differentiated as
many times as we want (albeit derivatives will be not a functions but so
called distributions) and second and higher order derivatives always do not
depend in which order they are taken.
Question
(a) Is condition
∂M
∂y
=
∂N
∂x
not only necessary, but also sufficient?
(b) If so, how to find function U (x, y)?
From
∂U
= M (x, y),
∂x
∂U
= N (x, y).
∂y
we get, integrating the first equation by x
Z x
U (x, y) =
M (s, y) ds + ϕ(y)
x0
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(2.6.6)1,2
(2.6.9)
Chapter 2. First-Order Differential Equations
43
with some fixed x0 and an arbitrary
function ϕ(y).
Rx
Indeed, let V (x, y) = x0 M (s, y) ds; then ∂V
= M and therefore
∂x
∂(U −V )
= 0, which means exactly that U − V does not depend on x, it
∂x
is a function (any function) of y alone.
So, we have solved (2.6.6)1 by (2.6.9). Let us plug it into (2.6.6)2 :
Z
∂ x
M (s, y) ds + ϕ(y) = N (x, y)
∂y x0
Z x
∂M
⇐⇒
(s, y) ds + ϕ′ (y) = N (x, y)
∂y
x0
and using equality My = Nx
Z x
∂N
⇐⇒
(s, y) ds + ϕ′ (y) = N (x, y)
∂s
x0
⇐⇒ N (x, y) − N (x0 , y) + ϕ′ (y) = N (x, y)
and this means exactly that ϕ′ (y) = N (x0 , y).
Remark 2.6.5. ϕ is a function of y alone and ϕ′ is an ordinary derivative.
So, we have ϕ′ (y) = N (x0 , y) and since N (x0 , y) is also a function of y
alone, we can integrate it:
Z y
ϕ(y) =
N (x0 , t) dt + C0
y0
with some fixed y0 and an arbitrary constant C0 . Plugging it into (2.6.9) we
arrive to
Z y
Z x
U (x, y) =
N (x0 , t) dt +
M (s, y) ds + C0 .
(2.6.10)
y0
x0
Remark 2.6.6. (a) This formula is not symmetric with respect to x and y.
Surely the alternative formula also works
Z x
Z y
U (x, y) =
M (s, y0 ) ds +
N (x, t) dt + C0 .
(2.6.11)
x0
y0
(b) I strongly encourage not to use any of these formulas in the assignments
but to repeat the arguments leading to them; see examples below.
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Chapter 2. First-Order Differential Equations
44
The above arguments really work only in the rectangular domain. We
proved
Theorem 2.6.2. Let D = {(x, y) : a < x < b, c < y < d}. Then equation
M (x, y) dx + N (x, y) dy = 0
(2.6.1)
is exact if and only if M and N satisfy
∂M
∂N
=
.
∂y
∂x
(2.6.8)
Then there exist a function U , given by (2.6.10) or (2.6.11) such that Ux = M ,
Uy = N and equation (2.6.1) integrates to
U (x, y) = C
(2.6.5)
with an arbitrary constant C.
Remark 2.6.7. We do not need C0 in (2.6.10) or (2.6.11): it is absorbed in
C in the right of (2.6.5).
Example 2.6.1.
ex cos(2y) dx + 9y 2 − 2ex sin(2y) dy = 0.
Solution. (a) Checking that equation is exact:
M (x, y) = ex cos(2y) =⇒ My = −2ex sin(2y);
N (x, y) = 9y 2 − 2ex sin(2y) =⇒ Nx = −2ex sin(2y);
?
My = Nx
Yes, hooray! Equation is exact.
(b) Finding U :
Ux = M (x, y) = ex cos(2y)
Z
=⇒ U = ex cos(2y) dx = ex cos(2y) + ϕ(y)
remember, that x and y are independent variables when finding U ; plugging
it to the second equation
Uy = −2ex sin(2y) + ϕ′ (y) = 9y 2 − 2ex sin(2y),
=⇒ ϕ′ (y) = 9y 2 =⇒ ϕ(y) = 3y 3
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Chapter 2. First-Order Differential Equations
45
(skipping +C0 which will be absorbed into C)
=⇒ U = ex cos(2y) + 3y 3 = C
is a general solution.
Example 2.6.2. In the previous Example 2.6.1 find solution such that
y(0) = 0.
Solution. Plugging x = 0, y = 0 into ex cos(2y) + 3y 3 = C we get C = 1
and therefore
ex cos(2y) + 3y 3 = 1
is the required solution.
Example 2.6.3.
cos(x) cos(y) dx + sin(x) sin(y) dy = 0.
Solution.
Ux = cos(x) cos(y) =⇒ U = sin(x) cos(y) + ϕ(y)
=⇒ Uy = − sin(x) sin(y) + ϕ′ (y) = sin(x) sin(y)
ϕ′ (y) = 2 sin(x) sin(y) =⇒ ϕ(y) = −2 sin(x) cos(y)
=⇒ U (x, y) = sin(x) cos(y) − 2 sin(x) cos(y) = − sin(x) cos(y)
− sin(x) cos(y) = C.
This was an example of a very wrong solution.
(a) We have not checked, if equation is exact. – It is not!
(b) But so far we were just careless. The realy grave error was done when
we got ϕ′ (y) = 2 sin(x) sin(y). Since ϕ(y) should be a function of y alone, it
is impossible to satisfy, and we should abort solution and write “Rats! It is
impossible”. Instead we moved ahead.
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Chapter 2. First-Order Differential Equations
2.6.3
46
General Domains (optional)
Definition 2.6.2. This was an example of a very wrong solution.
(a) An open set M is connected if each pair of z ∈ M , w ∈ M may be
joined by curve lying entirely M .
(b) Domain is called simply-connected if it is connected and does not
have “holes inside”
(a) This domain is simplyconnecded
(b) This domain is not simplyconnecded
Theorem 2.6.3. Let D be a simply-connected domain. Then equation
M (x, y) dx + N (x, y) dy = 0
(2.6.1)
is exact if and only if M and N satisfy
∂M
∂N
=
.
∂y
∂x
Then there exist a function U , given
Z
U (x, y) =
M (x, y) dx + N (x, y) dy + C0
(2.6.8)
(2.6.12)
γ
where (x0 , y0 ) is some initial point and γ is any curve from (x0 , y0 ) to (x, y),
entirely inside D (the result does not depend on the choice of it).
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Chapter 2. First-Order Differential Equations
47
Then equation (2.6.1) integrates to
U (x, y) = C
(2.6.5)
with an arbitrary constant C.
Remark 2.6.8. (a) Integral in (2.6.12) is a line integral which you (will)
study in the end of Calculus II. Formulas (2.6.10) and (2.6.11) are simply
examples of such integrals:
(a) In formula (10)
(b) In formula (11)
(b) If domain is not simply continued, then condition (2.6.8) is only necessary, but not sufficient. For example, equation
x dy − y dx
=0
x2 + y 2
with
M (x, y) =
−y
,
+ y2
x2
N (x, y) =
x2
x
+ y2
satisfies this condition, but U (x, y) in this case is a polar angle, which cannot
be a continuous single-valued function in the domain {(x, y) : (x, y) ̸= (0, 0)}.
punctured point (0, 0)
Going once counter-clockwise around 0, we return to the same point but
the polar angle increases by 2π.
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Chapter 2. First-Order Differential Equations
2.6.4
48
Integrating Factors
Let us recall what we learned on the previous Lecture W3L2. So, we consider
a differential equation in the form
M (x, y) dx + N (x, y) dy = 0.
(2.6.1)
Recall that this equation is called exact if there exists a continuously differentiable function U (x, y) such that
dU (x, y) = M (x, y) dx + N (x, y) dy =⇒ U (x, y) = C.
(2.6.5)
So, in this case equation (2.6.1) can be integrated immediately.
Recall that (2.6.5) is equivalent to
∂U
= M (x, y),
∂x
∂U
= N (x, y).
∂y
(2.6.6)
The main result of the previous lecture was
Theorem 2.6.1. (i) If equation
M (x, y) dx + N (x, y) dy = 0
(2.6.1)
is exact then
∂N
∂M
=
.
∂y
∂x
(eqn-2.64.8)
(ii) Conversely, if a domain is simply-connected and (2.6.8) holds, then
equation (2.6.1) is exact.
Definition 2.6.3. Consider equation
M (x, y) dx + N (x, y) dy = 0
(2.6.1)
and multiply it by µ(x, y). If the resulting equation
µ(x, y)M (x, y) dx + µ(x, y)N (x, y) dy = 0
is exact, we call µ(x, y) integrationg factor.
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(2.6.13)
Chapter 2. First-Order Differential Equations
49
Plugging µM and µN instead of M and N into
∂M
∂N
=
,
∂y
∂x
(2.6.8)
we get
∂(µM )
∂(µN )
=
∂y
∂x
∂µ
∂N
∂µ
∂M
+M
=µ
+N
⇐⇒ µ
∂y
∂y
∂x
∂x
which is equivalent to
M
∂N
∂µ
∂µ
∂M −N
=µ
−
.
∂y
∂x
∂x
∂y
(2.6.14)
Remark 2.6.9. Equation (2.6.14) is a first order linear partial differential
equation. Such equations are studied in the beginning of any PDE class, f.
e. APM346, and they are no more simple than the original ODE (2.6.1).
However, there are cases when integrating factor could be found in the
specific form, and in this class we consider three following cases:
1. µ(x, y) = µ(x);
2. µ(x, y) = µ(y);
3. µ(x, y) = µ(xy).
In all these cases we will get a very simple first order ODE for a function
of one variable µ and we will be able to solve it.
2.6.5
Case 1. µ(x, y) = µ(x)
Plugging µ = µ(x) into (2.6.14) we get
∂M −N µ (x) = µ
−
∂x
∂y
′
∂N
− ∂N
+
µ′
∂x
⇐⇒
=
µ
N
∂M
∂y
.
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(2.6.15)
Chapter 2. First-Order Differential Equations
50
Since µ = µ(x), this works if and only if
−Nx + My
=: f (x)
N
Integrating we get ln(µ(x)) =
is a function of x alone.
R
(2.6.16)
f (x) dx.
Example 2.6.4. Consider linear ODE
y ′ + p(x)y = g(x) ⇐⇒
p(x)y − g(x) dx + dy = 0.
(2.6.17)
−Nx + My
= p(x) and
N
R
with P (x) = p(x) dx as we did
Then M (x, y) = p(x)y − g(x), N (x, y) = 1 and
µ′
= p(x) =⇒ µ(x) = eP (x)
µ
studying linear ODEs.
we get
Example 2.6.5. (a) Find integrating factor and then a general solution of
ODE
y + 3y 2 e2x dx + 1 + 2ye2x dy = 0
(2.6.18)
(b) Also, find a solution satisfying y(0) = 1.
Solution. (a) Since M (x, y) = y + 3y 2 e2x and N (x, y) = 1 + 2ye2x , we get
My − Nx
My − Nx = 1 + 2ye2x and
= 1.
N
µ′
Therefore we are looking for µ = µ(x),
= 1 =⇒ ln(µ) = x (we do
µ
not need a constant here) and µ = ex .
Multiplying (2.6.18) by µ we get
yex + 3y 2 e3x dx + ex + 2ye3x dy = 0.
(b) Then
Ux = yex + 3y 2 e3x
Z
=⇒ U =
yex + 3y 2 e3x dx = yex + y 2 e3x + φ(y)
?
=⇒ Uy = ex + 2ye3x + φ′ (y) = N
=⇒ φ(y) = 0 =⇒ U (x, y) = yex + y 2 e3x
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Chapter 2. First-Order Differential Equations
51
where we do not need a constant. Finally
U (x, y) = yex + y 2 e3x = C
is a general solution.
(c) Plugging x = 0, y = 0 we get C = 2 and
U (x, y) = yex + y 2 e3x = 2
satisfies y(0) = 1.
2.6.6
Case 2. µ(x, y) = µ(y)
This case is very similar to the previous one. Plugging µ = µ(y) into (2.6.14)
we get
∂N
∂M M µ (y) = µ
−
∂x
∂y
∂N
∂M
′
− ∂y
µ
∂x
=
.
⇐⇒
µ
N
′
(2.6.19)
Since µ = µ(x), this works if and only if
Nx − My
=: g(y)
M
Integrating we get
ln(µ(y)) =
is a function of y alone.
R
(2.6.20)
g(y) dy.
Example 2.6.6. (a) Find integrating factor and then a general solution of
ODE
−y sin(x) + y 3 cos(x) dx + 3 cos(x) + 5y 2 sin(x) dy = 0
(2.6.21)
√
π
(b) Also, find a solution satisfying y( ) = 2.
4
Solution. (a) Since M (x, y) = −y sin(x) + y 3 cos(x) and
N (x, y) = 3 cos(x) + 5y 2 sin(x) we get My − Nx = 2 sin(x) − 2y 2 cos(x) and
My − Nx
= − y2 .
M
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Chapter 2. First-Order Differential Equations
Therefore we are looking for µ = µ(y),
52
µ′
=
µ
2
y
=⇒ ln(µ) = 2 ln(y) and
µ = y2.
Multiplying (2.6.21) by µ we get
−y 3 sin(x) + y 5 cos(x) dx + 3y 2 cos(x) + 5y 4 sin(x) dy = 0.
(b) Then
Ux = −y 3 sin(x) + y 5 cos(x)
Z
=⇒ U =
−y 3 sin(x) + y 5 cos(x) dx = y 3 cos(x) + y 5 sin(x) + φ(y)
?
=⇒ Uy = y 3 cos(x) + y 5 + φ′ (y) = N =⇒ φ(y) = 0
=⇒ U (x, y) = y 3 cos(x) + y 5 sin(x)
where we do not need a constant. Finally
U (x, y) = y 3 cos(x) + y 5 sin(x) = C
is a general solution.
(c) Plugging x =
√
π
, y = 2 we get C = 6 and
4
√
π
satisfies y( ) = 2.
4
U (x, y) = y 3 cos(x) + y 5 sin(x) = 6
2.6.7
Case 3. µ(x, y) = µ(xy)
Plugging µ = µ(y) into (2.6.14) and taking into account that
∂µ
= µ′ y we get
∂x
(M x − N y)µ′ (xy) = µ
⇐⇒
∂N
− ∂N
∂x
−
∂x
+ ∂M
∂y
∂µ
∂y
= µ′ x and
∂M ∂y
µ′
=
.
µ
Mx − Ny
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(2.6.22)
Chapter 2. First-Order Differential Equations
53
Since µ = µ(xy), this works if and only if
Nx − My
=: h(xy)
Mx − Ny
Integrating we get ln(µ(t)) =
is a function of xy alone.
R
(2.6.23)
h(t) dt.
Example 2.6.7. (a) Find integrating factor and then a general solution of
ODE
3y cos(x + y) − xy sin(x + y) dx
+ 3x cos(x + y) − xy sin(x + y) dy = 0. (2.6.24)
(b) Also, find a solution satisfying y
π π
= .
2
2
Solution. (a) Since
M (x, y) = 3y cos(x + y) − xy sin(x + y),
N (x, y) = 3x cos(x + y) − xy sin(x + y)
we get
My − Nx = (y − x) sin(x + y) =⇒
My − Nx
1
=−
xM − yN
xy
Therefore we are looking for µ = µ(xy).
µ′ (t)
1
Then
=
=⇒ ln(µ) = ln(t) and µ = t = xy.
µ(t)
t
Multiplying (2.6.24) by µ we get
3xy 2 cos(x + y) − x2 y 2 sin(x + y) dx
+ 3x2 y cos(x + y) − x2 y 2 sin(x + y) dy = 0.
(b) Then
Ux = 3xy 2 cos(x + y) − x2 y 2 sin(x + y)
Z
=⇒ U =
3xy 2 cos(x + y) − x2 y 2 sin(x + y) dx
= x2 y 2 cos(x + y) + φ(y)
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Chapter 2. First-Order Differential Equations
54
where we integrated by parts.
Then
?
Uy = 3x2 y cos(x + y) − x2 y 2 sin(x + y) + φ′ (y) = N
=⇒ φ′ (y) = 0 =⇒ φ(y) = 0.
Finally
U = x2 y 2 cos(x + y) = C
is a general solution.
(c) Plugging x = y =
π
π4
, we get C = −
and
2
16
U = x2 y 2 cos(x + y) = −
satisfies y
2.6.8
π4
16
π π
= .
2
2
Final Remarks
Remark 2.6.10. Let µ(x, y) be an integrating factor
µM dx + µN dy = dU.
(a) Let F be an arbitrary function of one variable. Then ν(x, y) = F ′ (U (x, y))µ(x, y)
is also an integrating factor:
νM dx + νN dy = dV
(2.6.25)
with V (x, y) = F (U (x, y)).
(b) Conversely, if ν(x, y) is also an integrating factor, that is (2.6.25) holds,
then V = F (U (x, y)) for some function F of one variable, and ν(x, y) =
F ′ (U (x, y))µ(x, y).
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Chapter 2. First-Order Differential Equations
2.8
2.8.1
55
Existence and Uniqueness Theorem
(optional)
Elements of the Real Analysis
Definition 2.8.1. Let
(a) C 0 ([a, b]) := {f : [a, b] → R : f is continuous} be a space of continuous
functions on [a, b] with a norm
∥f ∥ ≡ max |f (x)|
[a,b]
and a distance
dist(f, g) ≡ ∥f − g∥ = max |f (x) − g(x)|, f, g ∈ C 0 ([a, b]);
[a,b]
(b) C 1 ([a, b]) := {f : [a, b] → R : df /dx exists and it is continuous} be a
space of continuously differentiable functions on [a, b];
(c) A Cauchy sequence in a metric space (i.e. a set with a distance satisfying
triangle inequality and such that dist(f, g) = dist(g, f ) and dist(f, g) =
0 ⇐⇒ f = g) is a sequence {fn }n≥1 such that
lim dist(fn , gm ) = 0.
n,m→∞
(C 0 ([a, b]) is an example of a metric space);
(d) A complete metric space is a metric space such that for every Cauchy
sequence {fn }n≥1 , there exists a point f := limn→∞ fn , in that space such
that
lim dist(fn , g) = 0.
n→∞
Cauchy theorem from 1st year Calculus says that the real numbers form
a complete metric space.
Theorem 2.8.1. C 0 ([a, b]) is complete (with respect to dist(f, g)).
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Chapter 2. First-Order Differential Equations
56
Proof. Assume {gn }n≥1 is a Cauchy sequence in C 0 ([a, b]). This implies
that {gn (x)}n≥1 is a Cauchy sequence in R, for any x ∈ [a, b]. By Cauchy’s
theorem, this last sequence converges to a number, which we denote with
g(x). We obtain in this way a function g defined on [a, b]. Moreover,
lim dist(gn ; g) = 0
n→∞
since for any ϵ > 0, there is N such that if m, n ≥ N ,
|gn (x) − gm (x)| < ϵ/2 for all x ∈ [a, b].
Taking the limit for m → ∞ one obtains
|gn (x) − g(x)| ≤ ϵ/2 < ϵ for all x ∈ [a, b].
Finally, g is continuous: Given ϵ > 0, take N for which
|gn (x) − g(x)| < ϵ/3,
n ≥ N, for all x ∈ [a, b].
Select n ≥ N as above. Since gn (x) is continuous, one has that for any
x, y ∈ [a, b], 0 < |x − y| < δ,
|gn (x) − gn (y)| < ϵ/3
for an appropriate δ = δ(ϵ/3) > 0. It follows immediately that
|g(x) − g(y)| < ϵ
for 0 < |x − y| < δ, x, y ∈ [a, b], and hence g is continuous.
Lemma 2.8.2. Let f (x, y) be a function with ∂f
continuous. Put
∂y
Z 1
f (x, y1 ) − f (x, y2 )
∂f
∇f (x, y1 , y2 ) :=
=
(x, (1 − s)y1 + sy2 )ds.
y1 − y2
0 ∂y
Ry
(It follows from h(y1 ) − h(y2 ) = y12 h′ (t)dt by a change of variable t =
(1 − s)y1 + sy2 .)
Denote
B = max
|x|,|y|≤b
∂f
(x, y) .
∂y
Then
∇f (x, y1 , y2 ) ≤ B for all |x|, |y1 |, |y2 | ≤ b.
Proof. Show this yourself! It is easy!
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Chapter 2. First-Order Differential Equations
2.8.2
57
Existence and Uniqueness Theorem
Theorem 2.8.3. Let f (x, y) be continuous and ∂f
exist and be bounded in
∂y
the “box” |x − x̄| ≤ b, |y − ȳ| ≤ b. Then Cauchy’s problem
y ′ = f (x, y),
y(x̄) = ȳ
(2.8.1)
(2.8.2)
has a unique solution y = y(x) on interval (x̄ − a′ , x̄ + a′ ) with sufficiently
small a′ > 0.
Proof. Denote
A=
max
|x−x̄|≤a,|y−ȳ|≤b
|f (x, y)|,
B=
∂f
(x, y)
∂y
max
|x−x̄|≤a,|y−ȳ|≤b
′
and let us redefine a := a = min{b/A, 1/2B} (so that a · A ≤ b and
a · B ≤ 1/2).
(a) First of all we claim that (2.8.1)–(2.8.2) is equivalent to a single integral
equation
Z x
y(x) = I(y)(x) := ȳ +
f (s, y(s))ds.
(2.8.3)
x̄
- Indeed, if y satisfies (2.8.1)–(2.8.2) then integrating (2.8.1) from x̄ to
x we arrive to y(x) − y(x̄) = I(y)(x) and using (2.8.2) we arrive to
(2.8.3).
- Conversely if y satisfies (2.8.3) then y ∈ C 1 (x̄ − a, x̄ + a) (because I(y)
is a continuously differentiable) and differentiating (2.8.3) we arrive to
(2.8.1); plugging x = x̄ into (2.8.3) we arrive to (2.8.2).
(b) Note that for any y, z ∈ C 0 ([x̄ − a, x̄ + a]) such that |y(x) − ȳ| ≤ b,
|z(x) − ȳ| ≤ b we have dist(y, z) ≤ 2b.
(c) I(y) defined above is a contraction, that is
dist(I(y), I(z)) ≤ q dist(y, z)
(2.8.4)
for some q < 1. Indeed, due to the Lemma 2.8.2:
Z x
|I(y)(x) − I(z)(x)| = |
∇f (s, y(s), z(s))(y(s) − z(s))ds|
x̄
≤ aB · dist(y, z)
and, since aB ≤ 1/2, we can take q = 1/2.
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Chapter 2. First-Order Differential Equations
58
Remark 2.8.1. It follows from (2.8.3) that I(gi ) = gi for i = 1, 2 implies
g1 = g2 . Show this yourself. This proves uniqueness.
(d) Any sequence composed of y0 ∈ C 0 ([x̄ − a, x̄ + a]) with ∥y(x) − ȳ∥ ≤ b
(for instance y0 ≡ 0), yn := I(yn−1 ), n ≥ 1, is a Cauchy sequence: indeed,
because q = 1/2,
lim q n = 0
n→∞
Take n(ϵ) such that q n < ϵ/2b for all n ≥ n(ϵ). Let m ≥ n ≥ n(ϵ). Then
dist(gm , gn ) = ∥I n (ym−n − y0 )∥ ≤ q n ∥ym−n − y0 ∥ ≤ q n 2b
< (ϵ/2b) · 2b = ϵ.
(e) By making use of Theorem 2.8.1, there exists y ∈ C 0 ([−a, a]) such that
limn→∞ dist(yn , y) = 0, and hence |y(x) − ȳ| ≤ b for |x − x0 | ≤ a.
Since
dist(y, I(y)) ≤ dist(y, yn ) + dist(yn , I(y))
1
= dist(y, yn ) + dist(I(yn−1 ), I(y)) ≤ dist(y, yn ) + dist(yn−1 , y)
2
and
lim dist(y, yn ) = 0 = lim dist(yn−1 , y),
n→∞
n→∞
it follows dist(y, I(y)) = 0, i.e. y = I(y).
2.8.3
Existence theorem
One can prove
Theorem 2.8.4. Let f (x, y) be continuos in the “box” |x− x̄| ≤ b, |y− ȳ| ≤ b.
Then Cauchy problem (2.8.1)–(2.8.2) has a solution y = y(x) on interval
(x̄ − a′ , x̄ + a′ ) with sufficiently small a′ > 0.
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Chapter 2. First-Order Differential Equations
59
Sketch of the proof. Consider Euler approximations with the step h:
yh,n+1 = yh,n + f (xn , yh,n )h,
xn = x̄ + nh,
yh,0 = ȳ
(2.8.5)
and on “step” intervals (xn , xn+1 ) we apply a linear approximation yh (x) =
yh,n + f (xn , yh,n )(x − xn ).
Here we take n = 0, 1, 2, . . . but we can go also in the opposite direction
(replacing h by −h). So, we get a piecewise linear function yh (x).
One can prove that
(a) Functions yh (x) are defined on interval [x̄ − a, x̄ + a] with a redefined
as a′ = min(a, b/A) (see proof of Theorem 2.8.3) and are uniformly bounded
there: |yh (x) − ȳ| ≤ b;
(b) Functions yh (x) are uniformly continuous which means that for each
ϵ > 0 there exists δ > 0 such that |x − x′ | < δ =⇒ |yh (x) − yh (x′ )| < ϵ;.
Indeed, δ = ϵ/A works. Uniformly here and above means that bound b
and δ do not depend on h;
(c) |yh (x) − I(yh )(x)| ≤ εh for all x ∈ [x̄ − a, x̄ + a] with εh → 0 as h → 0.
Let us take hm = 2−m → +0 as m → ∞.
Now we apply Arzelá–Ascoli theorem from Real Analysis:
Theorem 2.8.5. From sequence of functions yhm (x) satisfying 1–2 one can
select a subsequence yhmk (x) converging in C([x̄ − a, x̄ + a]): yhmk (x) → y(x).
Since step hmk → 0 the limit is by no means piecewise linear!
Then obviously I(yhmk ) → I(y). Further, (c) implies that y = I(y) and
therefore y satisfies (2.8.3) and thus it satisfies (2.8.1)–(2.8.2).
Remark 2.8.2. (a) In contrast to Theorem 2.8.3 Theorem 2.8.4 does not
1
imply uniqueness of solution; indeed, example y ′ = y 3 analyzed before shows
the lack of uniqueness;
(b) Both Theorems 2.8.3 and 2.8.5 are based on fixed point equation y = I(y)
but existence of the fixed point y is due to different ideas: in Theorem 2.8.3 it
exists and is unique because map y → I(y) is contractive; in Theorem 2.8.5
it exists (but is not necessarily unique) because map y → I(y) is compact
(we did not define this notion).
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Chapter 3
Second-Order Linear
Differential Equations
Introduction
In this Chapter we consider second-order ODEs. The general form of such
equations (resolved with respect to the highest-order derivative) is
y ′′ = f (t, y, y ′ ).
(3.1.1)
Such equations often appear in physics, in particularly mechanics: for
example, y is the position of the point (of the mass m) on the line, t time,
and equation is
y ′′ =
1
F (t, y, y ′ ).
m
where y ′ is the velocity, y ′′ is the acceleration, and F is the force, which can
depend on time, position and velocity. This is the Second Newton’s Law of
Motion.
No surprise that a general solution depends on two arbitrary parameters,
y = φ(t, C1 , C2 )
(3.1.2)
and to determine them (and thus to find a unique solution) we need two
additional conditions. In this class we consider only Cauchy’s promlem
(a.k.a initial problem, or initial value problem)
y ′ (t0 ) = y1
y(t0 ) = y0 ,
60
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(3.1.3)
Chapter 3. Second-Order Linear Differential Equations
61
(in the mechanical model we define both initial position and initial velocity).
It does not mean that the other problems are not reasonable, but we do
not cover them in this class.
We mainly concentrate on Second-Order Linear Differential Equations
y ′′ + p(t)y ′ + q(t)y = g(t).
3.1
(3.1.4)
Differential Equations with Constant
Coefficients
Definition 3.1.1. (a) If both p(t) and q(t) are constant, we say that (3.3.4)
is a Linear ODE with constant coefficients, otherwise we say that it is a
Linear ODE with variable coefficients.
(b) If g(t) = 0 say that (3.3.4) is a Linear Homogeneous ODE, otherwise
we say that it is a Linear Inhomogeneous ODE.
So, consider linear homogeneous ODE with constant coefficients:
ay ′′ + by ′ + cy = 0,
a ̸= 0,
(3.1.5)
Let us try to guess the solution: y = ekt (we have a good reason for this:
differentiation does not change it, just adds coefficient). Then y ′ = kekt ,
y ′′ = k 2 ekt (and so on) and plugging to (3.1.5) we get
(ak 2 + bk + c)ekt = 0,
⇐⇒ L(k) := ak 2 + bk + c = 0.
(3.1.6)
Definition 3.1.2. L(k) = ak 2 + bk + c is called a characteristic polynomial,
(3.1.6) is called a characteristic equation, and its roots are characteristic
roots.
Theorem 3.1.1. (i) y = ekt with real k is a solution to a linear homogeneous ODE with constant coefficients
ay ′′ + by ′ + cy = 0,
a ̸= 0,
(3.1.5)
if and only if k is a characteristic root, that means L(k) = ak 2 + bk + c = 0.
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Chapter 3. Second-Order Linear Differential Equations
62
(ii) If characteristic roots k1,2 are real and distinct, k1 =
̸ k2 then the general
solution to (3.1.5) is
y = C1 ek1 t + C2 ek2 t .
(3.1.7)
Proof. (i) The proof of Statement (i) is obvious.
(ii) We postpone a real proof of Statement (ii), now just observing that
we need to have two constants C1 and C2 to satisfy initial conditions and
that if k1 = k2 we get the same solution but we need two non-proportional
solutions.
Remark 3.1.1. (a) There are two distinct characteristic roots k1 and k2 if
and only if the the discriminant ∆ = b2 − 4ac is positive; then
√
−b ± b2 − 4ac
.
(3.1.8)
k1,2 =
2a
(b) Assuming that a = 1 we can graphically show the allowed (so far) range
of b, c:
c
b
(c) The case of repeated roots ∆ = 0, k1 = k2 will be covered by Section 3.3,
and the case of complex roots ∆ < 0 will be covered in Section 3.4.
Example 3.1.1. Find the general solution to
y ′′ − y = 0.
Solution. Characteristic equation is k 2 − 1 = 0, characteristic roots are
k1,2 = ±1 and the general solution is y = C1 et + C2 e−t . We plot some
solutions:
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Chapter 3. Second-Order Linear Differential Equations
y
63
y = et
y = (3et − 3e−t )/4
y = (et + e−t )/2
t
y = e−t
Example 3.1.2. Find the general solution to
y ′′ − 5y ′ + 6y = 0.
Solution. Characteristic equation is k 2 − 5k + 6 = 0, characteristic roots are
k1 = 1 and k2 = 2 and the general solution is y = C1 e2t + C2 e3t .
We plot some solutions:
y
y = 32 e3t
y = e3t + 12 e2t
y = 23 e2t
y = e3 t − 21 e2t
t
Remark 3.1.2. Recall Vieta’s Theorem from algebra: roots of k 2 + pk + q = 0
satisfy
k1 + k2 = −p,
k1 k2 = q
There is a generalization to higher-order equations.
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Chapter 3. Second-Order Linear Differential Equations
3.1.1
64
Cauchy’s Problem
There is a fast method to solve Cauchy’s problem
a0 y ′′ + a1 y ′ + a2 y = 0,
y(t0 ) = y0 ,
y ′ (t0 ) = y0′ .
(3.1.9)
(3.1.10)
Observe that
(a) We can plug t − t0 instead of t, so solution would be
y(t) = C1 ek1 (t−t0 ) + C2 ek2 (t−t0 ) ;
(b) We can use eα(t−t0 ) cosh(β(t − t0 )) and eα(t−t0 ) sinh(β(t − t0 )) instead of
ek1 t and ek2 t if k1,2 = α ± β; so solution would be
y(t) = D1 eα(t−t0 ) cosh(β(t − t0 )) + D2 eα(t−t0 ) sinh(β(t − t0 )).
(3.1.11)
with unknown constants D1 and D2 .
Then plugging (3.1.11) to (3.1.10) we get
D1 = y0 ,
βD2 + αD1 = y0′ =⇒ D2 =
1 ′
y0 − αy0 .
β
Remark 3.1.3. Hyperbolic functions a
cosh(x) =
ex + e−x
2
sinh(x) =
ex − e−x
2
and
are very useful. In particular
(cosh(x))′ = sinh(x),
cosh2 (x) − sinh2 (x) = 1,
(sinh(x))′ = cosh(x),
sinh(x)
tanh(x) =
.
cosh(x)
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(3.1.12)
Chapter 3. Second-Order Linear Differential Equations
3.2
3.2.1
65
Solutions of Linear Homogeneous
Equations; the Wronskian
Solutions of Linear Homogeneous Equations
In this section we deal with general 2-nd order linear homogeneous ODEs,
which are not necessarily with constant coefficients:
L[y] := a0 (t)y ′′ + a1 (t)y ′ + a2 (t)y = 0.
(3.2.1)
Assuming that a0 (t) does not vanish we can divide by it, so we can
always assume that the leading coefficient a0 (t) = 1.
We need the following
Theorem 3.2.1 (Existence and Uniqueness Theorem). Assume that a0 (t),
a1 (t), a2 (t) and g(t) are continuous functions on interval I = (α, β) with
−∞ ≤ α < β ≤ ∞, which could be open, closed, or open on one end and
closed on another. Assume also that a0 (t) does not vanish anywhere on I.
Then for t0 ∈ I and y0 , y0′ ∈ C the Cauchy’s problem
L[y] :=a0 (t)y ′′ + a1 (t)y ′ + a2 (t)y = g(t),
y(t0 ) = y0 ,
y ′ (t0 ) = y0′
(3.2.2)
(3.2.3)
has a unique twice continuously differentiable on I solution y(t).
We postpone the proof of the Existence and Uniqueness Theorem until
Chapter 7.
Theorem 3.2.2. Assume that a0 (t), a1 (t), a2 (t) are continuous functions
on interval I = (α, β) with −∞ ≤ α < β ≤ ∞, which could be open, closed,
or open on one end and closed on another. Assume also that a0 (t) does not
vanish anywhere on I.
Then solutions of the linear homogeneous equation
L[y] :=a0 (t)y ′′ + a1 (t)y ′ + a2 (t)y = 0
(3.2.4)
form a 2-dimensional linear space which means that there two linearly
independent solutions y1 (t) and y2 (t) such that any other solution can be
represented as their superposition
y(y) = C1 y1 (t) + C2 y2 (t)
in the unique way.
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(3.2.5)
Chapter 3. Second-Order Linear Differential Equations
66
Proof. (a) Observe first that because equation is linear and homogeneous,
a superposition (linear combination) of solutions is also a solution.
(b) Let us define solutions y1 (t) and y2 (t) from Cauchy’s problems
(
(
L[y] = 0,
L[y] = 0,
and
(3.2.6)
′
y(t0 ) = 1, y (t0 ) = 0
y(t0 ) = 0, y ′ (t0 ) = 1
They exist due to the Existence and Uniqueness Theorem. Then if
y(t) = C1 y1 (t) + C2 y2 (t), then
y(t0 ) = C1
and
y ′ (t0 ) = C2 ,
(3.2.7)
so decomposition, if exists, is unique. In particular, if C1 y1 (t) + C2 y2 (t) = 0
then C1 = C2 = 0 (so y1 (t) and y2 (t) are linearly independent).
(c) Let y(t) be any solution. Define z(t) := C1 y1 (t) + C2 y2 (t) with C1 and
C2 defined by (3.2.7) .
Then L[z] = 0, z(t0 ) = C1 = y(t0 ) and z ′ (t0 ) = C2 = y ′ (t0 ), so z(t)
satisfies exactly the same problem as y(t). But solution is unique and
therefore y(t) = z(t) = C1 y1 (t) + C2 y2 (t).
Thus y1 (t) and y2 (t) form basis in the space of solutions.
Definition 3.2.1. The basis in {y1 (t), y2 (t)} in the space of solutions is
called a fundamental system of solutions.
Example 3.2.1. For equation y ′′ − y = 0
(a) {et , e−t } is a fundamental system of solutions.
(b) {cosh(t), sinh(t)} is a fundamental system of solutions (cosh(t) =
t
−t
sinh(t) = e −e
).
2
(c) {2et , 3e−t − et } is a fundamental system of solutions.
(d) {2et , 3et } is not a fundamental system of solutions.
(e) {2et , 3e2t } is not a fundamental system of solutions.
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et +e−t
,
2
Chapter 3. Second-Order Linear Differential Equations
67
Example 3.2.2. For equation y ′′ − y ′ = 0
(a) {et , 1} is a fundamental system of solutions.
(b) Suggest any other fundamental system of solutions (to this equation).
Example 3.2.3. For equation y ′′ = 0
(a) {1, t} is a fundamental system of solutions.
(b) Suggest any other fundamental system of solutions (to this equation).
3.2.2
Wronskian
Definition 3.2.2. Let y1 , y2 , . . . , yn be functions defined and (n − 1)-times
differentiable on interval I. Then
y1
y2
. . . yn
y1′
y2′
. . . yn′
. . . ..
.
W [y1 , y2 , . . . , yn ] := ...
..
.
(n−2)
y2
(n−1)
y2
y1
y1
(n−2)
. . . yn
(n−1)
. . . yn
(3.2.8)
(n−2)
(n−1)
is a Wronskian of y1 , y2 , . . . , yn .
Example 3.2.4. (a) For n = 2
W [y1 , y2 ] :=
y1 y2
y1′
y2′
= y1 y2′ − y1′ y2 .
(3.2.9)
(b) For n = 3
y1 y2 y3
W [y1 , y2 , y3 ] := y1′ y2′ y3′ .
y1′′ y2′′ y3′′
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(3.2.10)
Chapter 3. Second-Order Linear Differential Equations
68
Theorem 3.2.3. Let y1 , y2 be solutions to the second order equation
L[y] :=a0 (t)y ′′ + a1 (t)y ′ + a2 (t)y = 0.
(3.2.4)
d
a1
W [y1 , y2 ] = − W [y1 , y2 ].
dt
a0
(3.2.11)
Then
Proof. Expressing from equation yj′′ = − aa01 yj′ −
a2
y
a0 j
and plugging it into
d
W [y1 , y2 ] = (y2′′ y1 − y2 y1′′ )
dt
we see that the right-hand expression equals to − aa10 (y2′ y1 −y2 y1′ ) = − aa10 W [y1 , y2 ].
Example 3.2.5. Let y1 = ek1 t , y2 = ek2 t are solutions to linear constant
coefficients equation. Then one can see easily that
W [y1 , y2 ] = (k2 − k1 )e(k2 +k1 )t
and
d
W [y1 , y2 ] = (k1 + k2 )W [y1 , y2 ].
dt
By Vieta’s formula k1 + k2 = − aa01 .
Corollary 3.2.4. In the framework of Theorem 3.2.3
(i) The following equality holds:
Z a (t) 1
W [y1 , y2 ](t) = C exp −
dt .
a0 (t)
(3.2.12)
(ii) If a0 , a1 , a2 are continuous functions and a0 does not vanish on I then
either W [y1 , y2 ](t) does not vanish anywhere on I or is 0 identically.
Proof. Proof follows from (3.2.11).
Theorem 3.2.5. (i) Let y1 and y2 be linearly dependent on I.
W [y1 , y2 ] = 0 identically.
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Then
Chapter 3. Second-Order Linear Differential Equations
69
(ii) Let y1 and y2 be two solution of the linear homogeneous equation, with
a0 , a1 , a2 continuous on I and a0 not vanishing there. If W [y1 , y2 ] = 0 then
y1 , y2 are linearly dependent.
Proof. (i) If y1 and y2 are linearly dependent, then either y1 = 0 or y2 = Cy1 .
One can easily check that in both cases W [y1 , y2 ] = 0.
(ii) Let W [y1 , y2 ](t0 ) = 0. Then either y1 (t0 ) = y1′ (t0 ) = 0 or y2 (t0 ) =
Cy1 (t0 ) and y2′ (t0 ) = Cy1′ (t0 ).
- In the former case y1 satisfies L[y1 ] = 0, y1 (t0 ) = y1′ (t0 ) = 0 and
therefore y1 (t) ≡ 0.
- In the latter case z = y2 − Cy1 =⇒ L[z] = 0, z(t0 ) = z ′ (t0 ) = 0 and
therefore z(t) ≡ 0 and y2 (t) ≡ Cy1 (t).
Remark 3.2.1. In the general case W [y1 , y2 ] ≡ 0 does not imply that y1 , y2
are linearly dependent. For example,
(
(
0 t ≤ 0,
t2 t < 0,
y1 (t) = 2
and
y2 (t) =
t t > 0,
0 t ≥ 0,
are linearly independent but W [y1 , y2 ] ≡ 0.
y2
y1
Corollary 3.2.6. Assume that we know solution y1 (t) of linear homogeneous
second order equation. Then to findanother solution (not proportional to
y1 ) we need to find W = exp − aa10 dt and then
Z
y2 (t) = y1 (t)
W (t)
dt.
y12 (t)
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(3.2.13)
Chapter 3. Second-Order Linear Differential Equations
70
Proof. Note that (up to constant factor) y1 y2′ − y2 y1′ = W . Dividing by y12
we see that
Z
W (t)
W
y2
y2 ′ y1 y2′ − y2 y1′
dt.
=
= 2 =⇒
=
2
y1
y1
y1
y1
y12 (t)
which implies (3.2.13).
Example 3.2.6. One can see easily that y = et satisfyes
y ′′ (t − 1) − y ′ t + y = 0.
Then
t
W′
=
=⇒ ln(W ) =
W
t−1
Z
t
dt =
t−1
Z
1+
1 dt = t + ln(t − 1)
t−1
(we are losing constant but we do not care since we need to find just one
y2 ). Then W = (t − 1)et and
Z
t
y2 (t) = e (t − 1)e−t dt = et × −te−t = −t,
where we integrated by parts and lost another constant, we also don’t care
about. One can check that y = −t satisfies equation.
If we want to find a second order equation with the fundamental system
of solutions {y1 (t), y2 (t)} such that W [y1 , y2 ] ̸= 0, we write the third order
Wronskian invoking y1 , y2 and y
y
W [y, y1 , y2 ] := y ′
y 1 y2
y1′ y2′ = 0.
(3.2.14)
y ′′ y1′′ y2′′
We will prove it later.
Example 3.2.7. Let y1 (t) = t2 , y2 (t) = t + 1. Then
y
t2 t + 1
W [y, y1 , y2 ] = y ′ 2t
y ′′
2
1
0
= −(t2 + 2t)y ′′ + (2t + 2)y ′ − 2y = 0 .
required equation
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Chapter 3. Second-Order Linear Differential Equations
3.3
3.3.1
71
Complex Roots of the Characteristic
Equation
Complex Numbers
Recall that to solve linear homogeneous equation with constant coefficients
ay ′′ + by ′ + cy = 0, a ̸= 0, one needs to solve a characteristic equation
L(k) := ak 2 + bk + c = 0 and if it has two distinct real roots k1 and k2 , then
the general solution is y = C1 ek1 t + C2 ek2 t .
What should we do if characteristic roots k1,2 are complex and conjugate?
—We need to define, what does it mean ekt if k is a complex number, so that
(ekt )′ = kekt .
At this moment a very brief theory of complex numbers would suffice,
more detailed exposition will be required in Chapter 4. A complex number
is k = α + iβ, with real α = Re(k) and β = Im(k), called a real part of k
and an imaginary part of k correspondingly.
Complex numbers could be added and multiplied as
(α′ + iβ ′ ) + (α′′ + iβ ′′ ) = (α′ + α′′ ) + i(β ′ + β ′′ ),
(α′ + iβ ′ )(α′′ + iβ ′′ ) = (α′ α′′ − β ′ β ′′ ) + i(α′ β ′′ − β ′ α′′ )
(so i2 = −1) and all the usual number laws work.
Every complex number k has a conjugate k:
α + iβ = α − iβ
and absolute value |k|:
|α + iβ| =
p
α2 + β 2 .
Then
kk = |k|2 .
Finally, we can delete complex numbers if the denominator is not 0:
k1
k1 k2
=
.
k2
|k2 |2
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Chapter 3. Second-Order Linear Differential Equations
3.3.2
72
Complex Exponents
Recall from Calculus I that for real t the Taylor decomposition holds
t
e =
∞
X
tn
n=0
n!
.
The great thing with the converging power series is that we can plug here
complex t. This leads to the Theory of Functions of Complex Variables,
some of you may take later. So, let us plug it instead of t, remembering
that i2 = −1 :
eit =
∞ n n
X
i t
n=0
n!
=
∞
∞
X
i2m t2m X i2m+1 t2m+1
+
(2m)! m=0 (2m + 1)!
m=0
(we consider even and odd powers separately)
∞
∞
X
X
(−1)m t2m
(−1)m t2m+1
=
+i
= cos(t) + i sin(t)
(2m)!
(2m
+
1)!
m=0
m=0
because from Calculus I we know that
∞
X
(−1)m t2m
= cos(t),
(2m)!
m=0
∞
X
(−1)m t2m+1
= sin(t).
(2m
+
1)!
m=0
Thus we arrive to
Definition 3.3.1 (Euler’s formula).
eit = cos(t) + i sin(t).
(3.3.1)
Corollary 3.3.1.
eit + e−it
= Re(eit ),
2
eit − e−it
sin(t) =
= Im(eit ).
2i
cos(t) =
(3.3.2)
(3.3.3)
Now we make a final
Definition 3.3.2.
e(α+iβ)t := eαt cos(βt) + i sin(βt) ,
α, β ∈ R.
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(3.3.4)
Chapter 3. Second-Order Linear Differential Equations
73
Lemma 3.3.2. Consider complex-valued function ekt of real variable t
(k = α + iβ with α, β ∈ R), defined by (3.3.4). Then
′
ekt = kekt ,
(3.3.5)
and
e(k1 +k2 )t = ek1 t ek2 t .
(3.3.6)
Proof. Check it–it is easy, just use the definition! For the second equality
also use trigonometric formulas cos(x + y) = cos(x) cos(y) − sin(x) sin(y)
and sin(x + y) = sin(x) cos(y) + cos(x) sin(y).
3.3.3
Complex Roots of the Characteristic Equation
All heavy lifting has been done:
Theorem 3.3.3. (i) y = ekt with complex k is a solution to a linear homogeneous ODE with constant coefficients
ay ′′ + by ′ + cy = 0,
a ̸= 0,
(3.3.7)
if and only if k is a characteristic root, that means L(k) = ak 2 + bk + c = 0.
(ii) If characteristic roots k1,2 are distinct, k1 ̸= k2 then the general solution
to (3.3.7) is
y = C1 ek1 t + C2 ek2 t .
(3.3.8)
(iii) If characteristic roots k1,2 = α ± iβ (β > 0) are complex, distinct and
conjugate then the general solution to (3.3.7) is also
y = eαt A cos(βt) + B sin(βt)
(3.3.9)
with arbitrary constants A and B.
Proof. Proof of Statements (i) and (ii) follows from the rule of the differentiation of the complex exponent.
To prove Statement (iii) we just rewrite e(α±iβ)t = eαt cos(βt) + i sin(βt)
and denote A = 12 (C1 + C2 ), B = 2i1 (C1 − C2 ).
We will call (3.3.7) solution in the real form while (3.3.8) is called solution
in the complex form.
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Chapter 3. Second-Order Linear Differential Equations
3.3.4
74
Oscillations
Consider cases of complex conjugate roots α ± iβ. If α = 0 then
y = A cos(βt) + B sin(βt) = C
A
C
cos(βt) +
B
sin(βt)
C
√
with C = A2 + B 2 .
Since A2 + B 2 = C 2 one can find φ such that A = C cos(φ) and
B = −C sin(φ). Then
y = C cos(φ) cos(βt) − sin(φ) sin(βt)
and finally
y = C cos(βt + φ).
(3.3.10)
y(t)
φ
β
C
T
t
−C
So, we have oscillations, with amplitude C, angular frequency β, period
T = 2π
and phase φ.
β
Assume now that α < 0. Then we have damped oscillations.
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Chapter 3. Second-Order Linear Differential Equations
75
y(t)
Ceαt
t
−Ceαt
Remark 3.3.1. Maximas and minimas of y(t) us where y ′ (t) = 0, that
is tn : tan(βtn + φ) = αβ , which for α ̸= 0 do not coincide with where
cos(βtn + φ) = ±1.
For physical processes which lead to such solutions, read Section 3.7
“Mechanical and Electrical Vibrations” of the Textbook.
3.3.5
Cauchy’s Problem
There is a fast method to solve Cauchy’s problem
a0 y ′′ + a1 y ′ + a2 y = 0,
y(t0 ) = y0 ,
y ′ (t0 ) = y0′ .
(3.3.11)
(3.3.12)
Observe that
(a) We can plug t − t0 instead of t, so solution would be
y(t) = D1 eα(t−t0 ) cos(β(t − t0 )) + D2 eα(t−t0 ) sin(β(t − t0 )).
(3.3.13)
with unknown constants D1 and D2 .
Then plugging (3.3.11) to (3.3.8) we get
D1 = y0 ,
βD2 + αD1 = y0′ =⇒ D2 =
1 ′
y0 − αy0 .
β
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(3.3.14)
Chapter 3. Second-Order Linear Differential Equations
76
Example 3.3.1. (a) Find the general solution of y ′′ + 2y ′ + 2y = 0.
(b) Solve Cauchy’s problem y(0) = 2, y ′ (0) = 0.
(c) Find local minima and maxima of the solution to the Cauchy’s problem.
Solution. (a) Characteristic equation k 2 + 2k + 2 = 0 has roots −1 ± i and
therefore the general solution is
y = e−t A cos(t) + B sin(t) .
(b) Plugging into initial data we get A = 2, −A + B = 0 =⇒ B = 2 and
y = e−t 2 cos(t) + 2 sin(t)
1
√
1
= 2 2e−t √ cos(t) + √ sin(t))
2
2
√ −t
π
= 2 2e cos(t − )
4
(c) with extremums as y ′ (t) = −4e−t sin(t) = 0 =⇒ tn = πn, y(tn ) =
2(−1)n e−πn ; for odd n we have local maxima and for odd n local minima.
y(t)
t
So far we have not considered case k1 = k2 .
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Chapter 3. Second-Order Linear Differential Equations
77
c
k1,2 = ±βi
k1,2 = α ± βi, α > 0 k1,2 = α ± βi, α < 0
k1 > k2 > 0
k2 < k1 < 0
b
k1 > 0 > k2
3.4
3.4.1
Repeated Roots; Reduction of Order
Repeated Roots
Recall that to solve linear homogeneous equation with constant coefficients
ay ′′ + by ′ + cy = 0, a ̸= 0, one needs to solve a characteristic equation
L(k) := ak 2 + bk + c = 0.
If equation has two equal roots k1 = k2 = k, then we know so far only
one linearly independent solution y1 = ekt . How we get the second one?
We leave a more developed theory for Section 3.5 and especially for
Chapter 4, and just use the formula from Section 3.2,
Z
W (t)
y2 (t) = y1 (t)
dt,
y12 (t)
Z a
1
dt .
W (t) = exp −
a0
For constant coefficient equation
and therefore W = e(k1 +k2 )t and
k1 t
y2 (t) = e
Z
(k2 −k1 )t
e
a1
a0
= −(k1 + k2 ) due to Vieta’s formula,
1
e(k2 −k1 t)t k2 ̸= k1 ,
k2 − k1
dt = e

t
k2 = k1 ,

 1 ek2 t k ̸= k ,
2
1
= k2 − k1
 k1 t
te
k2 = k1 .
k1 t


As expected, we got nothing new for k2 ̸= k1 , but for k2 = k1 we got
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Chapter 3. Second-Order Linear Differential Equations
78
Theorem 3.4.1. If characteristic roots k1,2 are equal, k1 = k2 then the
general solution to ay ′′ + by ′ + cy = 0 is
y = (C1 + C2 t)ek1 t .
(3.4.1)
If C2 = 0 we know the plot; if C2 ̸= 0 plot always change the sign and
looks like
y(t)
t
Another proof. (a) Consider k2 ̸= k1 , then ek2 t − ek1 t is a solution. But if
k2 → k1 it tends to 0. To prevent it, we divide by k2 − k1 :
ek2 t − ek1 t
k2 − k1
and calculate the limit as k2 → k1 . By the L’Hôpital’s rule, we need to
differentiate by k2 both the numerator and denominator getting tek2 t and 1
correspondingly; so the limit is tek1 t .
(b) Another way would be to consider the case of roots k ± iβ (remember,
repeated root happens on the border between two distinct real roots, and
two complex conjugate roots) and divide ekt sin(βt) by β and calculate the
limit as β → 0. Again by the L’Hôpital’s rule we get tekt .
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Chapter 3. Second-Order Linear Differential Equations
3.4.2
79
Cauchy’s Problem
There is a fast method to solve Cauchy’s problem
a0 y ′′ + a1 y ′ + a2 y = 0,
y(t0 ) = y0 ,
y ′ (t0 ) = y0′ .
(3.4.2)
(3.4.3)
Observe that
(a) We can plug t − t0 instead of t, so solution would be
y(t) = D1 + D2 (t − t0 ) ek1 (t−t0 )
(3.4.4)
with unknown constants D1 and D2 .
Then plugging (3.4.11) to (3.4.10) we get
D1 = y 0 ,
D2 + k1 D1 = y0′ =⇒ D2 = y0′ − k1 y0 .
(3.4.5)
Example 3.4.1. (a) Find the general solution to y ′′ − 6y ′ + 9y = 0.
(b) Solve Cauchy’s problem y(0) = 1, y ′ (0) = 0.
Solution. (a) Characteristic equation k 2 − 6k + 9 = 0 has roots k1 = k2 = 3
and therefore general solution is
y(t) = C1 e3t + C2 te3t .
(b) Plugging into initial conditions we get C1 = 1, 3C1 + C2 = 0 and
therefore
y(t) = e3t − 3te3t
solves Cauchy’s problem.
Now we covered also k1 = k2 .
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Chapter 3. Second-Order Linear Differential Equations
80
c
k1,2 = ±βi
k1,2 = α ± βi, α > 0 k1,2 = α ± βi, α < 0
k1 > k2 > 0
k2 < k1 < 0
b
k1 > 0 > k2
3.4.3
Euler’s Equations
Second order Euler’s equations are equations of the form
a0 x2 y ′′ + a1 xy ′ + a2 y = 0,
x > 0,
(3.4.6)
with constant a0 ̸= 0, a1 , a2 .
They play important role in different applications and they are closely
related to equations with constant coefficients. Rewriting (3.4.6) as
a0 x(xy ′ )′ + (a1 − a0 )xy ′ + a2 y = 0,
x > 0,
(3.4.7)
and observing that
xy ′ = x
dy
dy
=
dx
d ln(x)
(3.4.8)
we see that making a change of variables t = ln(x) with −∞ < t < ∞ we
reduce Euler’s equation to equation with constant coefficients
a0 yt′′ + (a1 − a0 )yt′ + a2 y = 0.
(3.4.9)
Recall that for such equations exponential solutions y(t) = ekt play
important role. Therefore for Euler’s equations power solutions y(x) = xk
play the same role. Indeed, plugging directly to (3.4.6) y(x) = xk we get
a0 k(k − 1) + a1 k + a2 xk = 0,
⇐⇒ L(k) := a0 k(k − 1) + a1 k + a2 = 0
(3.4.10)
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Chapter 3. Second-Order Linear Differential Equations
81
which is called indicial equation and coincides with characteristic equation
for (3.4.9).
(a) If indicial equation
L(k) := a0 k(k − 1) + a1 k + a2 = 0
(3.4.9)
equation has two real distinct roots k1 and k2 then the general solution for
Euler’s equation is
y(x) = C1 xk1 + C2 xk2 .
(3.4.11)
(b) If indicial equation has a real repeating root k1 = k2 then the general
solution for Euler’s equation is
y(x) = C1 + C2 ln(x) xk1 .
(3.4.12)
(c) If indicial equation has a two complex conjugate roots k1,2 = α ± iβ
then the general solution for Euler’s equation is
y(x) = C1 cos(β ln(x)) + C2 sin(β ln(x)))xα .
(3.4.13)
Example 3.4.2. (a) x2 y ′′ − 2y = 0. Indical equation k 2 − k − 2 = 0 has roots
k1 = −1 and k2 = 2 and the general solution is
y(x) = C1 x−1 + C2 x2 .
(b) x2 y ′′ − 3xy ′ + 4y = 0. Indical equation k 2 − 4k + 4 = 0 has roots
k1 = k2 = 2 and the general solution is
y(x) = C1 + C2 ln(x) x2 .
(c) x2 y ′′ + xy ′ + 4y = 0. Indical equation k 2 + 4 = 0 has roots k1,2 = ±i
and the general solution is
y(x) = C1 cos(ln(x)) + C2 sin(ln(x)).
(d) x2 y ′′ − 3xy ′ + 5y = 0. Indical equation k 2 − 4k + 5 = 0 has roots
k1,2 = 2 ± i and the general solution is
y(x) = C1 cos(ln(x)) + C2 sin(ln(x)) x2 .
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Chapter 3. Second-Order Linear Differential Equations
3.5
82
Method of Undetermined Coefficients
Now we turn our attention to inhomogeneous equations
L[y] := a0 (t)y ′′ + a1 (t)y ′ + a2 (t)y = g(t).
(3.5.1)
In this lecture we start from some rather trivial general observations, and
then consider constant coefficients equations with the exponential or similr
right-hand expression.
3.5.1
General Observations
Theorem 3.5.1. (i) Let zj be a solution to L[zj ] = gj (t); then z(t) =
c1 z1 (t) + c2 z2 (t) is a solution to L[z] = g(t) := c1 g1 (t) + c2 g2 (t).
(ii) Let ȳ(t) be some solution to L[y] = g(t). Then the general solution to
L[y] = g(t) is given by
y(t) = ȳ(t) + y ∗ (t),
(3.5.2)
where y ∗ (t) is the general solution to homogeneous equation L[y ∗ ] = 0.
Proof. (i) Statement (i) follows from linearity of L: L[c1 z1 +c2 z2 ] = c1 L[z1 ]+
c2 L[z2 ].
(ii) Linearity of L implies Statement (ii) as well: if we found some solution
ȳ(t) to L[y] = g, then L[y] = g ⇐⇒ L[y − ȳ] = 0.
Remark 3.5.1. We call ȳ(t) a particular solution.
Therefore, to find a general solution to inhomogeneous equation we need
to find
- a particular solution to it and
- a general solution to the corresponding homogeneous equation.
3.5.2
Method of Undetermined Coefficients: k is not
a Root
We consider constant coefficient equations with exponential (or similar)
right-hand expression. We start from the simplest
L[y] = a0 y ′′ + a1 y ′ + a2 y = cekt .
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(3.5.3)
Chapter 3. Second-Order Linear Differential Equations
83
Let us try y(t) = Aekt with undetermined (so far) coefficient A. Then
want
L[Aekt ] = L(k)Aekt = cekt ,
with a characteristic polynomial
L(k) = a0 k 2 + a1 k + a2 .
(3.5.4)
(3.5.5)
Then
Lemma 3.5.2. If k is not a characteristic root, then (3.5.3) has a particular
c
.
solution ȳ(t) = Aekt with A = L(k)
Example 3.5.1. Solve y ′′ − 4y ′ + 3y = 24e−t .
Solution. Since L(k) = k 2 − 4k + 3 and L(−1) = 8 ̸= 0 we have a particular
solution ȳ(t) = 24
e−t . On the other hand, characteristic roots are 1, 3, so
8
the general solution is
y = 3e−t + C1 et + C2 e3t
with arbitrary constants C1 , C2 .
Example 3.5.2. Find a particular solution to y ′′ − 4y ′ + 3y = 8 cos(t).
Solution. Now k = ±i because cos(t) = 12 (eit + e−it ) and ±i are not characteristic roots.
Since 8 cos(t) = 8 Re(eit ) and the coefficients of L are real, it is sufficient
to find a particular solution z(t) to L[z] = 8eit and then to calculate its real
part. Since L(i) = 2 − 4i we have
z(t) =
4(1 + 2i)
4 + 8i it
8
eit =
eit =
e
2 − 4i
(1 − 2i)(1 + 2i)
5
and
4 8
4
8
y(t) = Re ( + i)(cos(t) + i sin(t) = cos(t) − sin(t).
5 5
5
5
Alternative solution. Due to the same reason we are looking for y(t) =
A cos(t) + B sin(t) where now instead of one complex uncertain coefficient we
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Chapter 3. Second-Order Linear Differential Equations
84
have two real uncertain coefficients A and B. Plugging it to y ′′ − 4y ′ + 3y = 0
we get
[−A cos(t) − B sin(t)] − 4[−A sin(t) + B cos(t)] + 3[A cos(t) + B sin(t)]
= 8 cos(t)
or
(
2A − 4B = 8,
8
4
8
4
=⇒ A = , B = − =⇒ y(t) = cos(t) − sin(t).
5
5
5
5
4A + 2B = 0
These approaches are equivalent, use whatever you like!
Example 3.5.3 (Forced Oscillations).
y ′′ + ω 2 y = c cos(ω0 t),
ω > 0, ω0 > 0.
(3.5.6)
Solution. Plugging ȳ(t) = A cos(ω0 t) + B sin(ω0 t) to equation we get
A(ω 2 − ω02 ) cos(ω0 t) + B(ω 2 − ω02 ) sin(ω0 t) = c cos(ω0 t)
c
c
=⇒ A = 2
, B = 0 =⇒ y(t) = 2
cos(ω0 t)
2
ω − ω0
ω − ω02
and therefore
y(t) =
c
cos(ω0 t) + C1 cos(ωt) + C2 sin(ωt)
ω 2 − ω02
(3.5.7)
because characteristic roots are ±iω. This works only if ω ̸= ω0 (no
resonance).
y ′′ + 4y = 5 cos(3t), y(0) = y ′ (0) = 0 =⇒ y(t) = − cos(3t) + cos(2t)
Remark 3.5.2. Is the sum y(t) = y1 (t) + y2 (t) of two periodic functions
periodic?
(a) If periods T1 and T2 are equal, T1 = T2 then y(t) is periodic with the
same period T = T1 = T2 .
(b) If periods T1 and T2 are commensurable (T1 : T2 = n1 : n2 with integers
n1 , n2 with greatest common divisor 1), then y(t) is periodic with period
T = n2 T1 = n1 T2 (the least common multiple) as seen on previous slide.
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Chapter 3. Second-Order Linear Differential Equations
85
y(t)
t
(c) If periods T1 and T2 are not commensurable, T1 : T2 is irrational then y(t)
is (usually) non-periodic, but there are almost exact repetitions separated
by large intervals. Such functions are called quasiperiodic.
y(t)
t
Example 3.5.4 (Forced Damped Oscillations).
y ′′ + 2γy ′ + ω 2 y = c cos(ω0 t),
ω > 0, ω0 > 0, ω0 > γ > 0.
(3.5.8)
Solution. Here complex approach rules: Plugging z(t) = Aeiω0 t to equation
and using L(iω0 ) = (ω 2 − ω02 ) + 2iγω0 we get
z(t) =
c
(ω 2
−
ω02 )
+ 2iγω0
eiω0 t
and
c
iω0 t
y(t) = Re
e
(ω 2 − ω02 ) + 2iγω0
+ e−γt C1 cos(ω ∗ t) + C2 sin(ω ∗ t) ,
ω∗ =
q
ω02 − γ 2 (3.5.9)
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Chapter 3. Second-Order Linear Differential Equations
86
because characteristic roots are −γ ± iω ∗ .
Observe that since γ > 0 for large positive t solution approximately
equal to the first term,
c
iω0 t
ȳ(t) = Re
e
(ω 2 − ω02 ) + 2iγω0
which could be rewritten in real form.
Observe, that the amplitude of ȳ(t)
|c|
p
2
(ω − ω02 )2 + γ 2 ω02
reaches its maximum,
when the angular frequency of the driving force ω0
p
∗
2
equals = ω = ω0 − γ 2 .
y(t)
t
y ′′ + 2y ′ + 2y = 5 cos(t), y(0) = y ′ (0) = 0
=⇒ y(t) = cos(t) + 2 sin(t) − e−t cos(t) + 3 sin(t)
√
and the blue line plots y(t) while the red line plots ȳ(t); ω = 2, ω0 = ω ∗ = 1.
Example 3.5.5 (Resonance).
y ′′ + ω 2 y = c cos(ω0 t),
ω = ω0 > 0.
(3.5.10)
Solution. We do not have a theory so far but let us apply Example 3.5.3
with ω ̸= ω0 and solution
c
y(t) = 2
cos(ω0 t) + C1 cos(ωt) + C2 sin(ωt).
(7)
ω − ω02
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Chapter 3. Second-Order Linear Differential Equations
87
We want ω0 → ω and it looks reasonable to set
y(t) = c
cos(ω0 t) − cos(ωt)
ω 2 − ω02
and to use L’Hp̂ital’s rule as ω0 → ω, resulting in
ȳ(t) =
c
t sin(ωt).
2ω
(3.5.11)
y(t)
t
y ′′ + y = cos(t), y(0) = y ′ (0) = 0
z ′′ + z = cos(1.1t), z(0) = z ′ (0) = 0
and the blue line plots y(t) while the red line plots z(t) .
However, if we fix ω0 and ω solution on large time intervals
2 sin ω02−ω t sin ω02+ω t
cos(ω0 t) − cos(ωt)
=
c
ω 2 − ω02
ω 2 − ω02
looks differently (oscillations with slowly changing amplitude)
3.5.3
Method of Undetermined Coefficients: k is a
Root
The previous example suggests that if k is a (simple) characteristic root
then a particular solution should be Atekt . Indeed
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Chapter 3. Second-Order Linear Differential Equations
88
y(t)
t
Lemma 3.5.3. (i) If k is a simple characteristic root, then
L[y] = a0 y ′′ + a1 y ′ + a2 y = cekt .
has a particular solution ȳ(t) = Atekt with A =
(3)
c
;
L′ (k)
(ii) If k is a double characteristic root, then (3) has a particular solution
¯ = At2 ekt with A = ′′c .
y(t)
L (k)
Here L′ (k) and L′′ (k) are derivatives of L(k)
.
Proof. Proof (in more general settings) follows from formula
L[u(t)ekt ] =
X L(j) (k)
j
j!
u(j) (t)ekt
(3.5.12)
which will be proven in Chapter 4. All details there too.
Remark 3.5.3. Here L(j) (k) denotes j-th derivative of L(k) with respect to k
while u(j) (t) denotes j-th derivative of u(t) with respect to t.
3.6
3.6.1
Variation of Parameters
Variation of Parameters: Method
Now we consider a universal method to solve Linear Inhomogeneous ODEs,
as long as we can find a fundamental system of solutions of the corresponding
homogeneous ODE.
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Chapter 3. Second-Order Linear Differential Equations
89
So let us consider equation
L[y] := a0 (t)y ′′ + a1 (t)y ′ + a2 (t)y = g(t).
(3.6.1)
We know that the solutions to the corresponding homogeneous equation
L[y] = 0 are
y = C1 y1 (t) + C2 y2 (t),
L[yj ] = 0.
(3.6.2)
We will look for solutions to (3.6.1) in the form
y = u1 (t)y1 (t) + u2 (t)y2 (t),
(3.6.3)
with unknown functions u1 and u2 .
What? We had one unknown function and now we have two? That’s
the idea: having extra unknown functions we can impose an extra equation!
So, we have
y =u1 y1 + u2 y2 .
(3.6.3)
Differentiating we get
y ′ =u1 y1′ + u2 y2′ + u′1 y1 + u′2 y2
but we want to get rid off selected terms, so we impose an extra condition
u′1 y1 + u′2 y2 = 0
(3.6.4)
arriving to
y ′ =u1 y1′ + u2 y2′ .
(3.6.5)
Differentiating again we get
y ′′ =u1 y1′′ + u2 y2′′ + u′1 y1′ + u′2 y2′ ,
(3.6.6)
but we could impose only one condition–because we had only one extra
function.
Therefore:
y =u1 y1 + u2 y2 ;
y ′ =u1 y1′ + u2 y2′ ,
y ′′ =u1 y1′′ + u2 y2′′ + u′1 y1′ + u′2 y2′ .
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(3.6.3)
(3.6.5)
(3.6.6)
Chapter 3. Second-Order Linear Differential Equations
90
and therefore, multiplying by a2 , a1 and a0 respectively and adding we get
L[y] =u1 L[y1 ] + u2 L[y2 ] + a0 (u′1 y1′ + u′2 y2′ ).
Since L[yj ] = 0 we get
want
L[y] = a0 (u′1 y1′ + u′2 y2′ ) = g(t)
and therefore we get (3.6.7) but recall (4):
u′1 y1 + u′2 y2 =0,
g(t)
=: f (t).
u′1 y1′ + u′2 y2′ =
a0 (t)
(3.6.4)
(3.6.7)
We got a system of linear algebraic equations with respect to u′1 , u′2 :
(
u′1 y1 + u′2 y2 =0,
u′1 y1′ + u′2 y2′ =f (t).
The determinant of this system is
W (t) =
y1 y2
(3.6.8)
y1′ y2′
which is Wronskian W [y1 , y2 ]. Since y1 , y2 is a fundamental system, W [y1 , y2 ]
does not vanish and we can solve the system
0 y2
u′1 =
f y2′
W1
f y2
=
,
=−
W
W
y1 y2
y1′ y2′
y1 0
u′2 =
y2 f
W2
f y1
=
.
=
W
W
y1 y2
(3.6.9)
y1′ y2′
We derived these formulas (3.6.9) using Kramer’s rule from Linear
Algebra–but you could do it by hand.
Now we need to integrate and we are done:
Z t
Z t
f (s)y2 (s)
f (s)y1 (s)
ds, u2 (t) =
ds
(3.6.10)
u1 (t) = −
W (s)
W (s)
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Chapter 3. Second-Order Linear Differential Equations
and since y(t) = u1 (t)y1 (t) + u2 (t)y2 (t)
Z
Z t
f (s)y2 (s)
ds + y2 (t)
y(t) = −y1 (t)
W (s)
Z
=
t
t
91
f (s)y1 (s)
ds
W (s)
y2 (t)y1 (s) − y1 (t)y2 (s)
f (s) ds.
W (s)
(3.6.11)
So we proved the following
Theorem 3.6.1. The general solution to
L[y] := a0 (t)y ′′ + a1 (t)y ′ + a2 (t)y = g(t).
(3.6.1)
is given by
Z
y(t) =
t
y2 (t)y1 (s) − y1 (t)y2 (s)
f (s) ds
W (s)
(3.6.11)
where {y1 , y2 } is the fundamental system of solutions to the corresponding
homogeneous equation, W = W [y1 , y2 ] is their Wronskian, and f = a−1
0 g.
Remark 3.6.1. (a) Numerator in the integrand is skew-symmetric (a.k.a.
anti-symmetric) with respect to y1 and y2 (which means that it changes the
sign when we permute y1 and y2 ) but the denominator is also skew-symmetric
and therefore the whole thing is symmetric.
(b) Integrals
Z
u1 = −
t
f (s)y2 (s)
ds and u2 (t) =
W (s)
Z
t
f (s)y1 (s)
ds
W (s)
(10)
are defined up to additive constants, C1 and C2 and therefore y(t) is defined
up to C1 y1 (t) + C2 y2 (t) which is the general soluton to the corresponding
homogeneous equation–as it should.
3.6.2
Cauchy’s problem
Theorem 3.6.2. The solution to the Cauchy’s problem
L[y] :=a0 (t)y ′′ + a1 (t)y ′ + a2 (t)y = g(t).
y(t0 ) = y0
y ′ (t0 ) = y0′
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(3.6.1)
(3.6.12)
Chapter 3. Second-Order Linear Differential Equations
92
is given by
t
y2 (t)y1 (s) − y1 (t)y2 (s)
f (s) ds
W (s)
t0
y1 (t)y2′ (t0 ) − y2 (t)y2′ (t0 )
y2 (t)y1 (t0 ) − y1 (t)y2 (t0 ) ′
+
y0 +
y0 . (3.6.13)
W (t0 )
W (t0 )
Z
y(t) =
where {y1 , y2 } is the fundamental system of solutions to the corresponding
homogeneous equation, W = W [y1 , y2 ] is their Wronskian, and f = a−1
0 g.
Proof. (a) Consider first the case y0 = y0′ = 0. Observe that expression
(3.6.13) is obtained from (3.6.11) when we take integrals in (3.6.10) from t0
to t which corresponds to u1 (t0 ) = u2 (t0 ) = 0.
Since
y(t) = u1 (t)y1 (t) + u2 (t)y2 (t)
y ′ (t) = u1 (t)y1′ (t) + u2 (t)y2′ (t)
we get y(t0 ) = y ′ (t0 ) = 0.
(b) Then due to the linearity we need to add the solution z(t) to homogeneous equation with z(t0 ) = y0 , z ′ (t0 ) = y0′ (linearity allows us to scatter
data between different problems). Then z(t) = C1 y1 (t) + C2 y2 (t) with
C1 =
y0 y2′ (t0 ) − y0′ y2 (t0 )
,
W (t0 )
C2 =
−y0 y2′ (t0 ) + y0′ y2 (t0 )
W (t0 )
which implies that z(t) is equal to the second line in (3.6.13).
Definition 3.6.1. Integral kernel in the first line of (3.6.13)
C(t, s) =
y2 (t)y1 (s) − y1 (t)y2 (s)
W (s)
(3.6.14)
is called Cauchy’s function.
Then (3.6.13) becomes
Z t
∂C(t, t0 )
y(t) =
C(t, s)f (s) ds + C(t, t0 )y0′ −
y0 .
∂t0
t0
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(3.6.15)
Chapter 3. Second-Order Linear Differential Equations
3.6.3
93
Examples
Example 3.6.1. Find the general solution of y ′′ + y =
1
and solution,
cos3 (t)
satisfying y(0) = y ′ (0) = 0.
Solution. Since y1 (t) = cos(t), y2 (t) = sin(t) we get

 ′
sin(t)

′
′

,
u1 =
 u1 cos(t) + u2 sin(t) = 0,
cos3 (t)
=⇒
1
1

−u′1 sin(t) + u′2 cos(t) =

3
 u′2 =
cos (t)
cos2 (t)

1
u1 =
+ C1 ,
2 cos2 (t)
=⇒

u2 = tan(t) + C2 .
Then
y(t) =
1
sin2 (t)
+
+ C1 cos(t) + C2 sin(t)
2 cos(t)
cos(t)
is the general solution.
To solve this Cauchy’s problem we need to take u1 (0) = u2 (0) = 0, thus
C1 = − 12 , C2 = 0 and
y(t) =
sin2 (t) 1
sin2 (t)
1
+
− cos(t) =
.
2 cos(t)
cos(t)
2
2 cos(t)
We can rewrite a general solution as
y(t) =
sin2 (t)
+ C1 cos(t) + C2 sin(t)
2 cos(t)
(with different coefficients).
Example 3.6.2. Find the general solution of y ′′ − y =
satisfying y(0) = y ′ (0) = 0.
et
2
and solution,
+1
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Chapter 3. Second-Order Linear Differential Equations
94
Solution. Since y1 (t) = et , y2 (t) = e−t we get

 ′ t
e−t
e−t

′ −t

= e−t −
,
u1 e + u2 e = 0,
u′1 = t
−t
e
+
1
1
+
e
=⇒
2
u′1 et − u′2 e−t =

et

t
u′2 =
e +1
et + 1

Z −t e


e−t −
dt = −e−t + ln(1 + e−t ) + C1 ,
u 1 = −
1 + e−t
=⇒
Z

et

u 2 =
dt = ln(et + 1) + C2 .
et + 1
Then
y(t) = −e−t + ln(1 + e−t ) + C1 et + ln(et + 1)e−t + C2 e−t
−t
t
t
t
−t
= −e − t + ln(1 + e ) + C1 e + ln(e + 1)e + C2 e−t .
is the general solution.
To solve this Cauchy’s problem we need to take u1 (0) = u2 (0) = 0, thus
C1 = 1 − ln(2), C2 = − ln(2) and
y = −e−t + ln(1 + e−t ) + 1 − ln(2) et + ln(et + 1) − ln(2) e−t .
Example 3.6.3. Find the general solution of t2 y ′′ + ty ′ − y = 4t2 et .
Solution. Since indicial equation is k 2 − 1 = 0, we have y1 (t) = t, y2 (t) = t−1
and
 ′
(
′ −1
u1 t + u2 t = 0,
u′1 = 2et ,
2 t =⇒
u′ − u′ t−2 = 4t e
u′2 = −2t2 et
1
2
2
t

t
u1 = 2e + C1 ,
Z
=⇒
u2 = −2 t2 et dt = (−2t2 + 4t − 4)et + C2
Then
y(t) = 2et + C1 t + (−2t2 + 4t − 4)et + C2 t−1
=4(1 − t−1 )et + C1 t + C2 t−1
is the general solution.
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Chapter 4
Higher Order Linear
Differential Equations
4.1.1
Introduction
Introduction In this Chapter we consider higher-order ODEs. The general
form of such equations (resolved with respect to the highest-order derivative)
is
y (n) = f (t, y, y ′ , . . . , y (n−1) ).
(4.1.1)
We say that n is an order of equation (4.1.1).
The theory of such equations is very similar to the theory of second order
equations (n = 2) but there are some complications. No surprise that a
general solution depends on n arbitrary parameters,
y = φ(t, C1 , C2 , . . . , Cn )
(4.1.2)
and to determine them (and thus to find a unique solution) we need n
additional conditions. In this class we consider only Cauchy’s problem (a.k.a
initial problem, or initial value problem)
(n−1)
y(t0 ) = y0 , y ′ (t0 ) = y1 , . . . , y (n−1) (t0 ) = y0
95
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.
(4.1.3)
Chapter 4. Higher Order Linear Differential Equations
4.1
4.1.1
96
Solutions of Linear Homogeneous
Equations
General results
We mainly concentrate on Linear Differential Equations
L[y] := a0 (t)y (n) + a1 (t)y (n−1) + . . . + an (t)y = g(t).
(4.1.4)
Assuming that a0 (t) does not vanish we can divide by it, so we can always
assume that the leading coefficient a0 (t) = 1.
Definition 4.1.1. (a) If all coefficients a0 , . . . , an are constant, we say that
(4.1.4) is a Linear ODE with constant coefficients, otherwise we say that it
is a Linear ODE with variable coefficients.
(b) If g(t) = 0, we say that (4.1.4) is a Linear Homogeneous ODE, otherwise
we say that it is a Linear Inhomogeneous ODE.
We need the following
Theorem 4.1.1 (Existence and Uniqueness Theorem). Assume that
a0 (t), a1 (t), . . . an (t) and g(t) are continuous functions on interval I = (α, β)
with −∞ ≤ α < β ≤ ∞, which could be open, closed, or open on one end
and closed on another. Assume also that a0 (t) does not vanish anywhere on
(n−1)
interval I. Then for t0 ∈ I and y0 , y0′ , . . . , y0
∈ C the Cauchy’s problem
L[y] :=a0 (t)y (n) + a1 (t)y (n−1) + . . . + an (t)y = g(t),
y(t0 ) = y0 , y ′ (t0 ) = y0′ , . . . , y (n−1) (t0 ) =
(n−1)
y0
(4.1.4)
(4.1.3)
has a unique n times continuously differentiable on I solution y(t).
We postpone the proof of the Existence and Uniqueness Theorem until
Chapter 7.
Theorem 4.1.2. Assume that a0 (t), a1 (t), . . . an (t) are continuous functions
on interval I = (α, β) with −∞ ≤ α < β ≤ ∞, which could be open, closed,
or open on one end and closed on another. Assume also that a0 (t) does not
vanish anywhere on I. Then solutions of the linear homogeneous equation
L[y] :=a0 (t)y (n) + a1 (t)y (n−1) + . . . + an (t)y = 0
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(4.1.5)
Chapter 4. Higher Order Linear Differential Equations
97
form a n-dimensional linear space which means that there n linearly independent solutions y1 (t), y2 , . . . , yn (t) such that any other solution can be
represented as their superposition
y(t) = C1 y1 (t) + C2 y2 (t) + . . . + Cn yn (t)
(4.1.6)
in the unique way.
Proof. (a) Observe first that because equation is linear and homogeneous,
a superposition (linear combination) of solutions is also a solution.
(b) Let us define solutions yj (t) from Cauchy’s problems
(
L[y] = 0,
(k)
yj (t0 ) = δj−1,k
where δpq
(
1 p = q,
=
0 p=
̸ q
k = 0, . . . , n − 1
(4.1.7)
is a Kronecker’s symbol.
These solutions exist due to the Existence and Uniqueness Theorem.
Then if y(t) = C1 y1 (t) + C2 y2 (t) + . . . + Cn yn (t), then
Ck = y (k−1) (t0 )
(4.1.8)
so decomposition, if exists, is unique. In particular,
y(t) = C1 y1 (t) + C2 y2 (t) + . . . + Cn yn (t) ≡ 0
=⇒ C1 = C2 = . . . = Cn = 0
(so y1 (t), . . . , yn (t) are linearly independent).
(c) Let y(t) be any solution. Define z(t) := C1 y1 (t) + . . . + Cn yn (t) with
C1 , . . . , Cn defined by (4.1.8).
Then L[z] = 0, and
z (j) (t0 ) = Cj+1 = y (j) (t0 )
j = 0, 1, . . . , n − 1
so z(t) satisfies exactly the same problem as y(t). But solution is unique
and therefore y(t) = z(t) = C1 y1 (t) + . . . + Cn yn (t).
Thus {y1 (t), . . . , yn (t)} form basis in the space of solutions.
Definition 4.1.2. The basis in {y1 (t), y2 (t), . . . , yn (t)} in the space of
solutions is called a fundamental system of solutions.
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Chapter 4. Higher Order Linear Differential Equations
4.1.2
98
Wronskian
Definition 4.1.3. Let y1 , y2 , . . . , yn be functions defined and (n − 1)-times
differentiable on interval I. Then
y1
y2
. . . yn
y1′
W [y1 , y2 , . . . , yn ] := ...
y2′
..
.
...
..
.
(n−2)
y2
(n−1)
y2
y1
y1
yn′
..
.
(n−2)
. . . yn
(n−1)
. . . yn
(4.1.9)
(n−2)
(n−1)
is a Wronskian of y1 , y2 , . . . , yn .
The following property of Wronskian could be handy:
Theorem 4.1.3. Let y1 , . . . , yn and α be functions. Then
W [αy1 , . . . , αyn ] = αW [y1 , . . . , yn ].
(4.1.10)
Proof. Indeed, the first row in W [αy1 , . . . , αyn ] is just αy1 , . . . , αyn , that
means the first row in W [y1 , . . . , yn ], multiplied by α; so we can move factor
α outside.
The second row in W [αy1 , . . . , αyn ] is αy1′ + α′ y1 , . . . , αyn′ + α′ yn , that
means the second row in W [y1 , . . . , yn ] multiplied by α plus the first row,
multiplied by α′ ; we can ignore this addition and move factor α outside.
And so on: each row in W [αy1 , . . . , αyn ] is the corresponding row in
W [y1 , . . . , yn ] multiplied by α plus a linear combination of the previous rows.
We can ignore this linear combination and factor α outside.
Example 4.1.1.
W [eat , ebt cos(ct), ebt sin(ct)] = e3bt W [e(a−b)t , cos(ct), sin(ct)]
=e
3bt
e(a−b)t ,
cos(ct),
sin(ct)
(a − b)e(a−b)t ,
−c sin(ct),
c cos(ct)
2 (a−b)t
(a − b) e
=e
(a+2b)t
2
, −c cos(ct), −c2 sin(ct)
1,
cos(ct),
sin(ct)
(a − b)
−c sin(ct),
c cos(ct)
(a − b)2 , −c2 cos(ct), −c2 sin(ct)
= c[(a − b)2 + c2 ]e(a+2b)t
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Chapter 4. Higher Order Linear Differential Equations
99
where we multiplied the first row by c2 and added to the third; the rest is
obvious.
Theorem 4.1.4. Let y1 , y2 , . . . , yn be solutions to the n-th order equation
L[y] :=L[y] := a0 (t)y (n) + a1 (t)y (n−1) + . . . + an (t)y = 0.
(4.1.5)
Then
d
a1
W [y1 , y2 , . . . , yn ] = − W [y1 , y2 , . . . , yn ].
dt
a0
(4.1.11)
Proof. We need to differentiate determinant. Recall that the determinant
is a sum (with some signs) of the products, such that from each row (and
each column) exactly one element is present. Recall that to differentiate a
product
(u1 u2 · · · un )′ = u′1 u2 · · · un + u1 u′2 u3 · · · un + . . . + u1 · · · un−1 u′n
(Leibniz’s rule). Therefore
Lemma 4.1.5. The derivative of the n-determinant equals to the sum
of n determinants, obtained from the original one, in which one row is
differentiated.
Let us apply this rule to W [y1 , y2 , . . . , yn ]: if we differentiate the first
row, we get the second row, so we get determinant with two equal rows, and
this is 0. The same is true if we differentiate any other row except the last
one, and therefore only this term would not vanish.
So we get
y1
y1′
d
W [y1 , y2 , . . . , yn ] = ...
dt
(n−2)
y1
(n)
y1
y2
. . . yn
y2′
..
.
. . . yn′
. . . ..
.
(n−2)
. . . yn
(n)
. . . yn
y2
y2
(n−2)
(n)
Let us multiply the first row by an , the second row by an−1 , . . . , (n − 1)th row by a2 and add to the last row multiplied by a0 (and we multiply
determinant by a−1
0 to compensate). We get the last row with elements
(n)
(n−2)
a0 y k + a2 y k
(n−1)
+ . . . + an yk = L[yk ] − a1 yk
(n−1)
= −a1 yk
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Chapter 4. Higher Order Linear Differential Equations
100
due to L[yk ] = 0. So we get
y1
y1′
d
..
W [y1 , y2 , . . . , yn ] = a−1
.
0
dt
(n−2)
y1
(n−1)
−a1 y1
y2
. . . yn
y2′
..
.
. . . yn′
. . . ..
.
(n−2)
(n−2)
y2
. . . yn
(n−1)
−a1 y2
(n−1)
. . . −a1 yn
= −a−1
0 a1 W [y1 , y2 , . . . , yn ]
where we moved factor −a1 from the last line. Done!
Corollary 4.1.6. In the framework of Theorem 4.1.4
(i) The following equality holds
Z a (t) 1
W [y1 , y2 , . . . , yn ](t) = C exp −
dt .
a0 (t)
(4.1.12)
(ii) If a0 , a1 , . . . , an are continuous functions and a0 does not vanish on
I then either W [y1 , y2 , . . . , yn ](t) does not vanish anywhere on I or is 0
identically.
Proof. Proof follows from (4.1.11).
Theorem 4.1.7. (i) Let y1 , y2 , . . . , yn be linearly dependent on I. Then
W [y1 , y2 , . . . , yn ] = 0 identically.
(ii) Let y1 , y2 , . . . , yn be solution of the linear homogeneous equation of order
n, with a0 , a1 , . . . , an continuous on I and a0 not vanishing there. If
W [y1 , y2 , . . . , yn ] = 0 then y1 , y2 , . . . , yn are linearly dependent.
Proof. (i) If y1 , y2 , . . . , yn are linearly dependent, then nontrivial linear
combination of y1 , y2 , . . . , yn is 0: C1 y1 + C2 y2 + . . . + Cn yn ≡ 0. Then
the linear combination of the columns of W [y1 , y2 , . . . , yn ] with the same
coefficients is 0, and then W [y1 , y2 , . . . , yn ] = 0.
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Chapter 4. Higher Order Linear Differential Equations
101
(ii) Let W [y1 , y2 , . . . , yn ](t0 ) = 0. Then some nontrivial linear combination
of the columns of W [y1 , y2 , . . . , yn ](t0 ) is 0:
n
X
(k)
Cj yj (t0 ) = 0
k = 0, . . . , n − 1.
j=1
Then z(t) = C1 y1 (t) + C2 y2 (t) + . . . + Cn yn (t) satisfies L[z] = 0 and also
z (k) (t0 ) = 0, and therefore it is identically 0 and therefore y1 , y2 , . . . , yn are
linearly dependent.
If we want to find a n-th order equations with the fundamental system
of solutions {y1 (t), y2 (t), . . . , yn } such that W [y1 , y2 , . . . , yn ] ̸= 0, we write
the (n + 1)-th order Wronskian invoking y1 , y2 , . . . , yn and y
W [y, y1 , y2 , . . . , yn ] = 0.
(4.1.13)
Indeed, if y is a linear combination of y1 , . . . , yn then W [y, y1 , y2 , . . . , yn ] =
0. On the other hand, decomposing W [y, y1 , y2 , . . . , yn ] by the first column
we get
W [y, y1 , y2 , . . . , yn ] = a0 y (n) + a1 (t)y (n−1) + . . . an (t)y
with
a0 = (−1)n W [y1 , y2 , . . . , yn ].
This is how instructors write equations so they know their solutions in
advance!
4.1.3
Order Reduction
(a) Assume that we know some non-trivial solution y1 of the linear homogeneous equation L[y] = 0. Then, plugging y = zy1 we get
L[zy1 ] = Ln [y1 ]z (n) + Ln−1 [y1 ]z (n−1) + . . . + L1 [y1 ]z ′ + L0 [y1 ]z = 0
(4.1.14)
with L0 = L. Indeed, to get z we need to treat it as a constant factor, and
then L[zy1 ] = zL[y1 ], so L0 = L.
Since L[y1 ] = 0, we get that (4.1.14) does not include z without derivatives:
L[zy1 ] = Ln [y1 ]z (n) + Ln−1 [y1 ]z (n−1) + . . . + L1 [y1 ]z ′ = 0
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Chapter 4. Higher Order Linear Differential Equations
102
Therefore denoting z ′ by v, we will get (n − 1)-th order equations for it:
Ln [y1 ]v (n−1) + Ln−1 [y1 ]z (n−2) + . . . + L1 [y1 ]v = 0.
(b) Due to Theorem 4.1.4
y2
yn = y1n W [v2 , . . . , vn ]
W [y1 , y2 , . . . , yn ] = y1n W 1, , . . . ,
y1
y1
with vj = (yj /y1 )′ , j = 2, . . . , n.
(c) If we know m < n linearly independent solutions y1 , . . . , ym (m < n)
then applying these arguments we will get for v = (y/y1 )′ (n − 1)-th order
equation, with known (m − 1) linearly independent solutions vj = (yj /y1 )′ ,
j = 2, . . . , m.
Continuing, we will be able to get (n − m)-th order equation.
4.2
4.2.1
Homogeneous Equations with Constant
Coefficients
Introduction
So, consider linear homogeneous ODE with constant coefficients:
L[y] := a0 y (n) + a1 y (n−1) + . . . + an y = 0.
(4.2.1)
As in the case n = 2, plugging y = ekt , k ∈ C, we arrive to equation
L(k) := a0 k n + a1 k n−1 + . . . + an = 0
(4.2.2)
Definition 4.2.1. L(k) is called a characteristic polynomial, (4.2.2) is called
a characteristic equation, and its roots are characteristic roots.
Theorem 4.2.1. y = ekt with complex k is a solution to a linear homogeneous ODE with constant coefficients if and only if k is a characteristic root,
that means L(k) = 0.
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Chapter 4. Higher Order Linear Differential Equations
4.2.2
103
Algebra of Polynomials
Now we need to provide some facts from the algebra of polynomials. First
we need
Theorem 4.2.2 (Principal Theorem of the Algebra of Polynomials). Let
L(k) := a0 k n + a1 k n−1 + . . . + an
(4.2.3)
be a polynomial of degree n (a0 ̸= 0) with complex coefficients. Then L(k)
has exactly n roots k1 , k2 , . . . , kn ∈ C, that is
L(k) = a0 (k − k1 )(k − k2 ) · · · (k − kn ).
(4.2.4)
Optional proof is provided in the “Appendix. Fundamental Theorem of
Algebra”. If you take MAT334 “Complex Variables” there will be a proof
(actually, several of them) as well.
Theorem 4.2.3 (Vieta’s Theorem). Let L(k) defined by (4.2.3) be a polynomial of degree n (a0 =
̸ 0) with complex coefficients and k1 , k2 , . . . , kn be
it’s roots. Then

a1

k1 + k2 + . . . + kn = − ,


a0



a2



k1 k2 + k1 k3 + . . . + kn−1 kn =
,


a0

a3
(4.2.5)
k
k
k
+
.
.
.
+
k
k
k
=
−
1
2
3
n−2
n−1
n

a0





...




an


k1 k2 · · · kn = (−1)n .
a0
Proof. One needs to use a decomposition (4.2.4), open all parentheses and
observe that the coefficient at k n−j should be equal to the sum of all possible
of j-products (which means j of factors) of non-repeating numbers k1 , . . . , kn ,
multiplied by (−1)j a0 .
Remark 4.2.1. Vieta’s theorem is rather useful when trying to guess roots
of the polynomials in the assignments.
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Chapter 4. Higher Order Linear Differential Equations
104
Roots k1 , k2 , . . . , km are not necessarily distinct. Collecting equal root,
we rewrite (4.2.4) as
L(k) = a0 (k − k1 )m1 (k − k2 )m2 · · · (k − ks )ms ,
m1 + . . . + ms = n.
(4.2.6)
where now k1 , . . . , ks are distinct.
Definition 4.2.2. (a) We say that mj is a multiplicity of root kj .
(b) Roots of multiplicity 1 are called simple roots, roots of multiplicity 1
are called double roots, roots of multiplicity 3 are called triple roots, . . .
(c) So we can reformulate Theorem 4.2.1: A polynomial of degree n has
exactly n complex roots, counting their multiplicities.
(d) Sometimes we say that k is root of multiplicity 0 if k is not a root at all!
Theorem 4.2.4. k∗ is a root of multiplicity m if and only if
L(k∗ ) = 0, L′ (k∗ ) = 0, . . . , L(m−1) (k∗ ) = 0, L(m) (k∗ ) ̸= 0,
(4.2.7)
where L(j) denotes j-th derivative of L(k) .
Proof. Rewriting L(k) = T (k)(k − k∗ )m where P (k∗ ) ̸= 0, we see that
L(k) = T (k)(k − k∗ )m ,
L′ (k) = mT (k)(k − k∗ )m−1 + Q1 (k)(k − k∗ )m ,
L′′ (k) = m(m − 1)T (k)(k − k∗ )m−2 + Q2 (k)(k − k∗ )m−1 ,
...
L(j) (k) = m(m − 1) · · · (m − j + 1)(k − k∗ )m−j T (k) + Qj (k)(k − k∗ )m−j+1 ,
j = 0, . . . , m
with polynomials Qj . then we get (4.2.7) and also useful formula
L(m) (k∗ )
T (k∗ ) =
.
m!
(4.2.8)
Now to finish we consider roots of a complex number. Note that a
complex number z could be written as z = reiφ with r = |z| and φ (it is
called an argument of the complex number z and denoted arg(z), it is a polar
angle on the complex plane C:
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Chapter 4. Higher Order Linear Differential Equations
105
Im(z)
z
φ
Re(z)
φ is defined up to 2πm, m ∈ CZ. Then equation wn = z has n simple roots
wm = r1/n ei(φ+2πm)/n ,
m = 0, 1, . . . , n − 1
(4.2.9)
(the rest of them just repeats).
Example 4.2.1.
1. z 2 = 1 has two roots z1,2 = ±1.
2. z 3 = 1 has three roots z1 = 1, z2,3 = − 21 ±
√
3
i.
2
3. z 4 = 1 has four roots z1,2 = ±1, z3,4 = ±i.
4.2.3
Solutions to Homogeneous Equations with
Constant Coefficients
So, consider such equation
L[y] :=a0 y (n) + a1 y (n−1) + . . . + an y = 0
(4.2.1)
and the corresponding characteristic polynomial
L(k) :=a0 k n + a1 k n−1 + . . . + an .
(4.2.3)
If all characteristic roots k1 , k2 , . . . , kn are distinct, then we have n
solutions y1 (t) = ek1 t , y2 (t) = ek2 t , . . . , yn (t) = ekn t and the general solution
is
y(t) = C1 ek1 t + C2 ek2 t + . . . + Cn ekn t
(4.2.10)
exactly like in the case n = 2.
Now we cover the case when some roots are not simple.
Theorem 4.2.5. Let k1 , k2 , . . . , ks be characteristic roots of multiplicities
m1 , m2 , . . . , ms respectively, m1 + m2 + . . . + ms = n. Then
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Chapter 4. Higher Order Linear Differential Equations
106
(i) Functions
yj,p (t) = tp ekj t ,
j = 1, . . . , s; p = 0, . . . , mj − 1
(4.2.11)
form a fundamental system of solutions.
(ii) The general solution to (4.2.1) is
y(t) =
j −1
s m
X
X
Cj,p tp ekj t .
(4.2.12)
j=1 p=0
Proof. (i) First prove that yj,p (t) are solutions. To do so observe that L is
a polynomial of ddt :
d d n−1
n
L[y] := a0
+ a1
+ . . . + an [y]
dt
dt
m
d d
− kj j [y]
=a0 Tj
dt dt
and that
d kt
d
(e z) = ekt
+k z
dt
dt
and therefore
m
d d
d kt d mj z =0
L[ekj z] = a0 Tj
( − kj j (ekt z) = a0 Tj
e (
dt dt
dt
dt
provided z(t) is a polynomial of degree < mj .
(ii) We skip the proof that these functions are linearly independent. It looks
trivial, but the proof is rather technical.
Example 4.2.2. y ′′′ − 3y ′ + 2y = 0
Solution. Writing characteristic equation k 3 − 3k + 2 = 0 we see that k1 = 1
is a characteristic root. To find two other roots, observe that due to Vieta’s
theorem k2 + k3 = −1, k2 k3 = −2 and therefore k2 = 1, k3 = −2.
So k1 = 1, m1 = 2 and k2 = −2, m2 = 1 and
y(t) = (C1 + C2 t)et + C3 e−2t
is a general solution (we number coefficients to our pleasure).
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Example 4.2.3. y (4) − 16y = 0
Solution. Writing characteristic equation k 4 − 16 = 0 we see that k1,2 = ±2,
k3,4 = ±2i and
y(t) =C1 e2t + C2 e−2t + C3 e2it + C4 e−2it
=C1 e2t + C2 e−2t + D1 cos(2t) + D2 sin(2t).
is a general solution.
Example 4.2.4. y (4) + 2y ′′ + y = 0.
Solution. Writing characteristic equation k 4 + 2k 2 + 1 = 0 we see that it is
(k 2 + 1)2 = 0 and therefore k1,2 = i, k3,4 = −i and
y(t) =(C1 + C2 t)eit + (C3 + C4 t)e−it
=(D1 + D2 t) cos(t) + (D1 + D2 t) sin(t).
is a general solution.
Example 4.2.5. y ′′′ − 6y ′′ + 12y ′ − 8y = 0.
Solution. Writing characteristic equation k 3 − 6k 2 + 12k − 8 = 0 we see that
it is (k − 1)3 = 0 and therefore k1,2,3 = 1 and
y(t) =(C1 + C2 t + C3 t2 )e2t
is a general solution.
4.2.4
Euler’s Equations
Like for n = 2 Euler’s equation is
a0 xn y (n) + a1 xn−1 y (n−1) + . . . + an−1 xy ′ + an y = 0,
x > 0.
(4.2.13)
Using substitution t = ln(x) it is reduced to a linear equations with the
constant coefficients
(n)
b0 y t
(n−1)
+ b1 y t
+ . . . + bn−1 yt′ + bn y = 0,
−∞ < t < ∞.
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(4.2.14)
Chapter 4. Higher Order Linear Differential Equations
108
Again, the indicial equation for (4.2.13)
a0 k(k − 1) · · · (k − n + 1) + a1 k(k − 1) · · · (k − n + 2) + . . .
+ an−2 k(k − 1) + an−1 k + an = 0 (4.2.15)
coincides with characteristic equation for (4.2.14)
b0 k n + b1 k n−1 + . . . + bn−1 k + bn = 0.
(4.2.16)
(a) To real root kj of multiplicity mj correspond solutions
yj,p (x) = (ln(x))p xkj ,
p = 0, . . . , mj − 1.
(4.2.17)
(b) To complex root kj = αj ± iβj (βj > 0) of multiplicity mj correspond
solutions
(
yj,p (x) = (ln(x))p xαj cos(βj ln(x)),
p = 0, . . . , mj − 1.
(4.2.18)
zj,p (x) = (ln(x))p xαj sin(βj ln(x)),
Appendix. Fundamental Theorem of Algebra
You need to know the statement of this theorem, albeit proof is optional:
Theorem 4.A.1. Consider polynomial
P = a0 z n + a1 z n−1 + . . . + an−1 z + an
(4.A.1)
where n ≥ 1, aj ∈ C and a0 =
̸ 0. Then there exists z ∈ C such that
P (z) = 0.
Proof. We provide one proof among of many known, based on completely
different ideas.
Step 1. Let
R = max
1≤k≤n
ak
a0
1/k
.
Assume |z| ≥ 2R; then |P (z)| ≥ |an | = |P (0)|. Indeed, using
|z| k
|a0 | ≥ Rk |a0 | ≥ |ak |
2
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Chapter 4. Higher Order Linear Differential Equations
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for k = 1, 2, . . . , n we conclude that
|
n
X
ak z
n−k
|≤
k=1
n
X
n−k
|ak | · |z|
≤
k=1
n
X
|z|n
k=1
1 n n
1 k
|a0 | = |a0 | 1 −
|z| .
2
2
Therefore
n
|P (z)| = |a0 z +
n
X
ak z
n−k
n
| ≥ |a0 | · |z| − |
k=1
n
X
ak z n−k |
k=1
≥
1 n
a0 | ≥ |an | = |P (0)|
2
as required.
Step 2. Since P (x + yi) = A(x, y) + iB(x, y) where A(x, y) and B(x, y) are
polynomials
in two variables with real coefficients, it implies |P (x + yi)| =
p
2
|A(x, y)| + |B(x, y)|2 . Therefore map C ∋ z → |P (z)| ∈ R+ is continuous
and hence (Calculus II) attains a minimal value on any closed bounded set
(such sets are called compacts), in particular on D = {z ∈ C : |z| ≤ 2R}.
Due to Step 1 such value is also a minimal value on all complex plane C.
Thus, there is z0 ∈ C with |z0 | ≤ 2R such that |P (z0 )| = minz∈C |P (z)|.
Step 3. We may assume that z0 = 0 by considering a polynomial Q(z) =
P (z + z0 ) instead of P (z). Given a local minima at 0 of z → |P (z)| we claim
that P (0) = 0 (which suffices).
Indeed, assume that P (0) = an ̸= 0. Let us rewrite P (z) in inverse order
and skip terms with coefficients equal 0: P (z) = an + an−k z k + . . . where
an ̸= 0, an−k ̸= 0, and k ≥ 1; recall that a0 ̸= 0. Then
P (z) = an + an−k z k + an−k z k Q(z)
with
Q(z) =
a
n−k+1
an−k
+
an−k+2 2
a0 n−k z + ...
z
.
an−k
an−k
By Moivre formula there exists z∗ ∈ C such that an + an−k z∗k = 0. Also
there is δ0 ∈ (0, 1) such that if |z| < δ0 |z∗ | then |Q(z)| ≤ 41 .
Let zδ = δ · z∗ and 0 < δ ≤ δ0 . Then
1
|P (zδ )| = |an + δ k an−k z∗k + δ k an−k z∗k Q(δz∗ )| ≤ |an − δ k ak | + δ k |an |
4
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Chapter 4. Higher Order Linear Differential Equations
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where we use an−k z∗k = −an . Hence
1
3
|P (zδ )| ≤ |an |(1 − δ k ) + δ k |an | = |an |(1 − δ k ) < |an | = |P (0)|.
4
4
This contradicts to the assumption that z0 = 0 is a minima. Done!
4.3
The Method of Undetermined
Coefficients
So, consider a linear inhomogeneous ODE with constant coefficients:
L[y] := a0 y (n) + a1 y (n−1) + . . . + an y = g(t),
a0 ̸= 0,
(4.3.1)
and assume that the right-hand expression is an exponent g(t) = cekt (real
or complex). Let us write a characteristic polynomial
L(k) := a0 k n + a1 k n−1 + . . . + an .
(4.3.2)
If k is not a characteristic root, then, exactly as in the case n = 2 y = Aekt
c
is a particular solution.
with A =
L(k)
What happens if k a characteristic root of multiplicity m?
Theorem 4.3.1. Let kj be a characteristic root of multiplicity mj . Then
equation (4.3.1) with g(t) = cekj t has a particular solution y(t) = Atmj ekj t
c
with A = (mj )
where L(p) (k) is a p-th derivative of L(k).
L
(kj )
Proof. Recall that in this case L(k) = T (k)(k − kj )mj with
T (kj ) =
Then using notation D :=
d
dt
1 (mj )
L
(kj ) ̸= 0.
mj !
(4.3.3)
we have a formula
L[z(t)ekt ] = L(D)[z(t)ekt ] = T (D)(D − kj )mj [z(t)ekj t ]
= T (D)[z (mj ) (t)ekj t ] (4.3.4)
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Chapter 4. Higher Order Linear Differential Equations
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and if we plug here z = Atmj we get z (mj ) = mj !A and
want
L[Atmj ekj t] = mj !T (kj )Aekj t = L(mj ) (kj )Aekj t = cekj t .
Recall that L(p) (k) means p-th derivative by k but z (p) (t) means p-th derivative by t.
There is a more general theorem:
Theorem 4.3.2. Let kj be a characteristic root of multiplicity mj . Then
equation
L[y] := a0 y (n) + a1 y (n−1) + . . . + an y = g(t)
, a0 ̸= 0,
(4.3.1)
with g(t) = Pq (t)ekj t , with
Pq (t) := b0 tq + b1 tq−1 + . . . + bq ,
(4.3.5)
a polynomial of degree q, has a particular solution y(t) = Qq (t)tmj ekj t with
Qq (t) := B0 tq + B1 tq−1 + . . . + Bq ,
(4.3.6)
a polynomial of degree q with uncertain coefficients B0 , . . . , Bq .
Proof. (a) Assume first that mj = 0 so kj is not a root at all. Then we use
the following formula
n
X
1 (p)
L[z(t)e ] =
L (k)z (p) (t)ekt
p!
p=0
kt
(4.3.7)
which we prove later.
Then taking z(t) = Qq (t) we reduce our equation to
n
X
1 (p)
L (kj )Q(p)
q (t) = Pq (t).
p!
p=0
(p)
Note that Qq (t) is a polynomial of degree q − p and if Pq (t) = b0 tq + . . .,
and Qq (t) = B0 tq + . . ., where . . . denote polynomials of degree ≤ q − 1,
b0
then taking B0 = L(k
makes leading coefficients on the left and on the right
j)
equal; then problem has been reduced to polynomials of degree ≤ q − 1.
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Chapter 4. Higher Order Linear Differential Equations
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(b) Let now mj > 0. We use again
L[z(t)ekt ] = L(D)[z(t)ekt ] = T (D)[z (mj ) (t)ekj t ].
(4)
Then using T (D) instead of L(D) (recall that T (kj ) ̸= 0) we can construct
w(t) = z (mj ) (t), a polynomial of degree q, such that the right-hand expression
here is equal to Pq (t).
Then integrating mj times w(t) by t , we get z(t), a polynomial of degree
q + mj . However in this polynomial we can ignore terms of degree ≤ mj − 1
(=they do not affect w) and take z(t) = Qq (t)tmj .
Proof of (4.3.7). To prove it recall that L = L(D), a polynomial with
respect to D = ddt and
D[z(t)ekj t ] = ekj t (D + kj )[z(t)]
and therefore
L(D)[z(t)ekj t ] = ekj t L(D + kj )[z(t)]
and decomposing
L(D + kj ) =
X 1
L(p) (kj )Dp
p!
p≥0
(4.3.8)
we arrive to (4.3.7).
Remark 4.3.1. Application of the Taylor decomposition (8) with operator D
is justified because L(k) and T (k) are polynomials.
Definition 4.3.1. Expression Pp (t)ekt where P (t) is a polynomial of degree
p and k ∈ C is called a quasipolynomial of degree p.
Example 4.3.1. Find particular solutions to
(a) y ′′′ − 3y ′ + 2y = 58 cos(t);
(b) y ′′′ − 3y ′ + 2y = 8 cosh(2t);
(c) y ′′′ − 3y ′ + 2y = 9et .
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Chapter 4. Higher Order Linear Differential Equations
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Solution. (a) Since cos(t) = Re(eit ) and ±i are not characteristic roots
(L(k) = k 3 − 3k + 2), we get
y(t) = Re
58
58 it e = Re
eit = Re 2(2 + 5i)eit
L(i)
−5i + 2
= 4 cos(t) − 20 sin(t).
(b) Since 8 cosh(2t) = 4e2t +4e−2t , and k = 2 is not a characteristic root, but
k = −2 is a simple characteristic root, (L(k) = k 3 − 3k + 2, L′ (k) = 3k 2 − 3)
we get a solution as a sum of two particular solutions y(t) = y1 (t) + y2 (t),
with
4 2t
e = e2t ,
L(2)
4
4
te−2t = te−2t
y2 (t) = ′
L (−2)
9
y1 (t) =
and
4
y(t) = 2e2t + te−2t .
9
(c) Since 1 is a double characteristic root and L′′ (1) = 6 we get
y(t) =
9
L′′ (1)
3
t2 et = t2 et .
2
Example 4.3.2. Find a particular solution to y ′′′ − 3y ′ + 2y = 36t sinh(t).
Solution. Since 36 sinh(t) = 18et + 18e−t and L(D) = (D + 2)(D − 1)2 we
have y(t) = y1 (t) + y2 (t) with y1 (t) = (At + B)t2 et and y2 (t) = (Et + F )e−t .
(a) Recall that to find (At + B) we should first solve (D + 3)w = 18t and
taking w = at + b we get 3at + 3b + a = 18t =⇒ a = 6, b = −2 =⇒ w(t) =
6t − 2.
Integrating twice we get z(t) = t3 − t2 and y1 (t) = (−t3 − t2 )et .
(b) To find Et+F we need to solve L(−1)(Et+F )+L′ (−1)(Et+F )′ = −18t
(because higher derivatives of Et + F vanish). So 4(Et + F ) = −18t =⇒
E = − 92 , F = 0 and y2 (t) = − 92 te−t .
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Chapter 4. Higher Order Linear Differential Equations
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Finally, y(t) = (t3 − t2 )et + 92 te−t .
Example 4.3.3. Find a particular solution to y (4) + 2y ′′ + y = 36t sin(t).
Solution. Since L(k) = (k 2 + 1)2 and ±i are double roots, to solve u(4) +
2u′′ + u = 36teit we need to take u = (At + B)t2 .
Then T (D) = (D + i)2 and we need to solve T (i)(at + b) + T ′ (i)(at + b)′ =
36t =⇒ −4(at + b) + 4ia = 36t =⇒ a = −9, b = 9i and w(t) = −9t + 9i.
Integrating twice we get z(t) = − 32 t3 + 9i2 t2 and
3
9i u(t) = − t3 + t2 eit .
2
2
Finally,
3
9i 2 it 9
3
3
y(t) = Im(u(t)) = Im − t + t e = − t3 sin(t) − t2 cos(t).
2
2
2
2
4.4
The Method of Variation of Parameters
Now we consider a generalization of the Method of Variation of Parameters
to arbitrary n.
4.4.1
General solution
So let us consider equation
L[y] := a0 (t)y (n) + a1 (t)y (n−1) + . . . + an (t)y = g(t).
(4.4.1)
We know that the solutions to the corresponding homogeneous equation
L[y] = 0 are
y = C1 y1 (t) + C2 y2 (t) + . . . + Cn yn (t),
L[yj ] = 0.
(4.4.2)
We will look for solutions to (4.4.1) in the form
y = u1 (t)y1 (t) + u2 (t)y2 (t) + . . . + un (t)yn (t),
(4.4.3)
with unknown functions u1 , . . . , un .
Since we have now not 2 but n unknown functions we will be able to
impose not one, but (n − 1) conditions.
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Chapter 4. Higher Order Linear Differential Equations
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So, we have
y =u1 y1 + u2 y2 + . . . + un yn .
(4.4.4)
Differentiating we get
y ′ =u1 y1′ + u2 y2′ + . . . + un yn′ + u′1 y1 + u′2 y2 + . . . + u′n yn
(4.4.5)
but we want to get rid off selected terms, so we impose an extra condition
u′1 y1 + u′2 y2 + . . . + u′n yn = 0
(4.4.6)
arriving to
y ′ =u1 y1′ + u2 y2′ + . . . + un yn′ .
(4.4.7)
Differentiating again we get
y ′′ =u1 y1′′ + u2 y2′′ + . . . un yn′′ + u′1 y1′ + u′2 y2′ + . . . + u′n yn′ ,
(4.4.8)
but we request to be selected terms equal 0.
Continuing this we get
(k)
(k)
y (k) =u1 y1 + u2 y2 + . . . + un yn(k) ,
(k−1)
u′1 y1
(k)
u′2 y2
+
+ ... +
...............
(n)
k = 0, . . . , n − 1
(k−1)
u′1 y1
= 0,
k = 0, . . . , n − 2,
(n)
y (n) =u1 y1 + u2 y2 + . . . + un yn(n)
(n−1)
+u′1 y1
(n−1)
+ u′2 y2
+ . . . + u′n yn(n−1)
and therefore, multiplying by an , an−1 , . . . , a0 respectively and adding we
get
L[y] =u1 L[y1 ] + u2 L[y2 ] + . . . un [Lyn ]
(n−1)
+a0 (u′1 y1
(n−1)
+ u′2 y2
+ . . . + u′n yn(n−1) ).
Since L[yj ] = 0 we get
(n−1)
L[y] = a0 (u′1 y1
(n−1)
+ u′2 y2
want
+ . . . + u′n yn(n−1) ) = g(t).
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Chapter 4. Higher Order Linear Differential Equations
116
We got a system of linear algebraic equations with respect to u′1 , u′2 , . . . , u′n :
 ′
uy
+ u′2 y2
+ . . . + u′n yn
= 0,

 1 1

′ ′
′ ′
′ ′


u1 y1
+ u2 y 2
+ . . . + un yn
= 0,



..............................

(n−2)
(n−2)

u′1 y1
+ u′2 y2
+ . . . + u′n yn(n−2) = 0,





u′1 y1(n−1) + u′2 y2(n−1) + . . . + u′n yn(n−1) = f (t) = g .
a0
The determinant of this system is W [y1 , y2 , . . . , yn ](t). Since y1 , y2 , . . . , yn
is a fundamental system, W [y1 , y2 , . . . , yn ] does not vanish and we can solve
the system.
According to Kramer’s rule u′k = W [y1 ,yW2k,...,yn ] where Wk is the same
determinant with k-th column replaced by a column all 0 but the last one is
f ; f.e.
0 y2
. . . yn
y2′
. . . yn′
. . . ..
.
0
W1 = ... ...
(n−2)
. . . yn
(n−1)
. . . yn
0 y2
f y2
(n−2)
(n−1)
Decomposing by the k-th column we get Wk = (−1)n+k W [y1 , . . . , yk−1 , yk+1 , . . . , yn ]f
and finally
u′k (t) = (−1)n+k
W [y1 , . . . , yk−1 , yk+1 , . . . , yn ](t)
f (t).
W [y1 , y2 , . . . , yn ](t)
Now we need to integrate and we have uk (t):
Z t
W [y1 , . . . , yk−1 , yk+1 , . . . , yn ](s)
n+k
uk (t) = (−1)
f (s) ds;
W [y1 , y2 , . . . , yn ](s)
(4.4.9)
(4.4.10)
multiplying by yk (t) and adding we arrive to the following formula:
Z t
n
X
W [y1 , . . . , yk−1 , yk+1 , . . . , yn ](s)yk (t)
n+k
y(t) =
(−1)
f (s) ds.
W
[y
1 , y2 , . . . , yn ](s)
k=1
(4.4.11)
So we proved the following
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Chapter 4. Higher Order Linear Differential Equations
117
Theorem 4.4.1. The general solution to
L[y] := a0 (t)y (n) + a1 (t)y (n−1) + . . . + an (t)y = g(t)
is given by
Z t
C(t, s)f (s) ds
y(t) =
(4.4.1)
(4.4.12)
where {y1 , y2 , . . . , yn } is the fundamental system of solutions to the corresponding homogeneous equation, W = W [y1 , y2 , . . . , Wn ] is their Wronskian,
f = a−1
0 g and
n
X
W [y1 , . . . , yk−1 , yk+1 , . . . , yn ](s)yk (t)
.
C(t, s) =
(−1)n+k
W [y1 , y2 , . . . , yn ](s)
k=1
(4.4.13)
Remark 4.4.1. Numerator in the integrand is skew-symmetric (a.k.a. antisymmetric) with respect to all functions (which means that it changes
the sign when we permute any two of them) but the denominator is also
skew-symmetric and therefore the whole thing is symmetric.
4.4.2
Cauchy’s Problem
Theorem 4.4.2. The solution to the Cauchy’s problem
L[y] := a0 (t)y (n) + a1 (t)y (n−1) + . . . + an (t)y = g(t)
y(t0 ) = y0 , y ′ (t0 ) = y0′ , . . . , y (n−1) (t0 ) =
(n−1)
y0
(4.4.1)
(4.4.14)
is given by
Z
t
y(t) =
C(t, s)f (s) ds +
t0
n−1
X
∂ n−1−k
(k)
(−1)k
C(t, t0 )y0
∂t0
k=0
(4.4.15)
where Cauchy function C(t, s) is defined by (4.4.13), and f = a−1
0 g.
(n−1)
Proof. For y0 = y0′ = . . . = y0
uk (t0 ) = 0 and formulas
(k)
(k)
= 0 this follows immediately from
y (k) = u1 y1 + u2 y2 + . . . + un yn(k) ,
k = 0, . . . , n − 1.
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Chapter 4. Higher Order Linear Differential Equations
118
In the general case proof is technical, based on the fact that
y(t) = ȳ(t) +
n
X
ck yk (t)
k=1
where ȳ(t) is satisfy Cauchy’s problem with the right-hand expression g(t)
and the rest of y(t) must satisfy
n
X
(k)
cj yj
(k)
= y0 .
j=1
This is a linear algebraic system, which should be solved and ck plugged
into y(t). You are strongly encouraged to use method rather than the ready
formula.
4.4.3
Examples
Example 4.4.1. Find the general solution of
y ′′′ − 7y ′ + 6y =
40
.
+1
e2t
Solution. Characteristic equation is L(k) = k 3 −7k+6 = (k−1)(k 2 +k−6) =
(k − 1)(k − 12)(k + 3) and roots asre k1 = 1, k2 = 2, k3 = −3 and
y = C1 et + C2 e2t + C3 e−3t
is a solution to homogeneous equation. We look for solution to our equation
in the same form, albeit C1 , C2 , C3 replaced by u1 , u2 , u3 .
Then
 ′ t
u1 e + u′2 e2t + u′3 e−3t = 0,


 ′ t
u1 e + 2u′2 e2t − 3u′3 e−3t = 0,


u′ et + 4u′ e2t + 9u′ e−3t = 40 .
1
2
3
e2t + 1
Subtracting the first from the second equations, we see that u′2 e2t = 4u′3 e−3t
and then u′1 et = −5u′3 e−2t and then
C3′ e−3t =
e2t
2
,
+1
C1′ et = −
10
,
+1
e2t
C2′ e2t =
e2t
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8
.
+1
Chapter 4. Higher Order Linear Differential Equations
Z
u1 = −
Z
8dt
=8
2t
e (e2t + 1)
u2 =
Z
u3 =
10dt
= −10
t
e (e2t + 1)
2e3t dt
= 2
e2t + 1
Z Z Z −2t
e
et −
e−t −
119
et dt
e2t + 1
= 10e−t + 5 arctan(et ) + c1 ,
e2t − 2t
dt
e +1
= −4e−2t − 4 ln(e2t + 1) + c2 ,
et dt = 2et − 2 arctan(et ) + c3 .
e2t + 1
Finally we get
y = 10e−t + 5 arctan(et ) + c1 et + −4e−2t − 4 ln(e2t + 1) + c2 e2t
t
t
+ 2e − 2 arctan(e ) + c3 e−3t .
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Chapter 7
Systems of First-Order Linear
Equations
4.4
4.4.1
Introduction
General
Now we begin to study systems of ODEs:

(m)
(m−1)

x1 = f1 (x1 , . . . , x′n ; x1 , . . . , x′n ; . . . ; x1
, . . . , x(m−1)
; t),
n




(m−1)
 x(m)
= f2 (x1 , . . . , x′n ; x1 , . . . , x′n ; . . . ; x1
, . . . , x(m−1)
; t),
2
n
..


.



 (m)
(m−1)
xn =f1 (xn , . . . , x′n ; x1 , . . . , x′n ; . . . ; x1
, . . . , x(m−1)
; t),
n
(4.4.1)
is a system of n equations of order m.
We can rewrite it in vector form
x(m) = f (x, x′ , . . . , x(m−1) ; t)
f 
!
x1
x2
with vector-columns x =
..
.
xn
(4.4.2)
1
f2
and f =  .. .
.
fn
If equations describe, for example, 1-dimensional motion, system describe
multidimensional motion, or systems of points. We had examples at the
very beginning of class (celestial mechanics); see also examples in Section
7.1 of the Textbook.
120
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Chapter 7. Systems of First-Order Linear Equations
121
Observe that m-th order equation can be reduced to the first order system
of m equations. Indeed, consider the such equation:
x(m) = f (x, . . . , x(m−1) ; t)
(4.4.3)
and denote x1 := x, x2 := x′ , . . . , xm := x(m−1) . Then equation (4.4.3) is
reduced to

x′1 = x2 ,





x′2 = x3 ,


..
(4.4.4)
.




x′m−1 = xm ,


 ′
xm = f (x1 , . . . , xm ; t).
Remark 4.4.1. (a) Then x is m times continuously differentiable solution
to (4.4.3) if and only if x is once differentiable solution to (4.4.4);
(b) The same procedure works for m-th order n-system which can be reduced
to te first order mn-system (instead of x we use x and make mn-vector
consisting of m n-blocks).
(c) However, the converse reduction (first order n-system to a single n-th
order equation) is possible only in some cases.
The Existence and Uniqueness Theorem, Existence Theorem, and Global
Existence and Uniqueness Theorem (see below) for the first order systems
are by no means different from the same theorems for first order equations
(see Sections 2.4 and also Section 2.8 for ideas of the proofs).
Due to reduction these theorems work for m-th order systems and, in
particular, for m-th order equations.
Theorem 4.4.1. Consider a Cauchy’s problem for a first-order system
x = f (x, t),
x(t0 ) = x0 ,
with (x0 , t0 ) ∈ D ⊂ Rnx × Rt . Assume that
(a) f is a continuous function in domain D ⊂ Rn+1 ;
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(4.4.5)
(4.4.6)
Chapter 7. Systems of First-Order Linear Equations
122
(b) f satisfies Lipschitz’s condition with respect to x
|f (x; t) − f (y; t)| ≤ L|x − y|
∀t, x, y : (x; t) ∈ D, (y; t) ∈ D. (4.4.7)
Then
(i) There exist an interval I := (t0 − δ, t0 + δ), δ > 0 and a solution y(x)
on I to (4.4.5)–(4.4.7);
(ii) This solution is unique: if x(t) and y(t) two solutions on I = (t0 −
δ, t0 + δ) (δ > 0 is arbitrary), both satisfying x(t0 ) = y(t0 ) = x0 , then
x(t) = y(t).
Theorem 4.4.2. Assume that conditions of Theorem 4.4.1 are fulfilled in
D = Rnx × Rt , I = (α, β) ∋ t0 .
Further, assume that
|f (x; t)| ≤ M (|x| + 1)
∀t ∈ I, x
(4.4.8)
Then there exists a (unique) solution to (4.4.5)–(4.4.6) on I.
There is a more general theorem, than Theorem 4.4.1, without Lipschitz’s
condition and uniqueness:
Theorem 4.4.3. Consider a Cauchy’s problem for a first-order system
x = f (x, t),
x(t0 ) = x0 ,
(7.1.5)
(7.1.6)
with (x0 , t0 ) ∈ D ⊂ Rnx × Rt .
Assume that f is a continuous function in domain D.
Then there exist an interval I := (t0 − δ, t0 + δ), δ > 0 and a solution
y(x) on I to (4.4.5)–(4.4.6).
4.4.2
Linear First Order Systems of ODEs
The main subject of this Chapter are Linear First Order Systems of ODEs
 ′
x1 = a11 (t)x1 + a12 (t)x2 + . . . a1n (t)xn + f1 (t),




 x′2 = a21 (t)x1 + a22 (t)x2 + . . . a2n (t)xn + f2 (t),
(4.4.9)
..


.


 ′
xn = an1 (t)x1 + an2 (t)x2 + . . . ann (t)xn + fn (t),
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Chapter 7. Systems of First-Order Linear Equations
123
homogeneous (when f1 = f2 = . . . = fn = 0) and inhomogeneous.
Such systems could be written using matrix notations
x′ = A(t)x + f
(4.4.10)
where


 a11 (t) a12 (t) . . . a1n (t) 


 a (t) a22 (t) . . . a2n (t) 
A(t) =  21.
..
.. 
...
 ..
.
. 


an1 (t) an2 (t) . . . ann (t)
(4.4.11)
is the matrix of the system (4.4.9).
We are mainly interested in constant coefficients systems when matrix
A(t) is constant (which means that all elements ajk are constant).
4.4
Matrices
4.4.1
Definitions
You are supposed to know matrices from Linear Algebra class (which is
prerequisite). To refresh, read Section 7.2 of the Textbook. Reminders
- m × n-matrix is

 a11 a12

a22
a
A =  .21
..
 ..
.

am1 am2

. . . a1n 

. . . a2n 
.. 
...
. 

. . . amn
with real or complex elements (sometimes we mention it explicitely).
- In particular m × 1 matrices are called vector-columns and 1 × n
matrices are called vector-rows.
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Chapter 7. Systems of First-Order Linear Equations
4.4.2
124
Operations
(a) Addition of matrices. What is the sum of two matrices A + B (including: when we can add two matrices?) Properties (commutativity and
associativity) What is zero matrix 0 ?
(b) Multiplication of matrix by a number (real or complex). What is the
product λA? Properties: (distributivity and associativity)
(c) Multiplication of matrices. What is the product AB (including: when
we can multiply two matrices?) Properties: (distributivity and associativity).
Is there commutativity? What is square matrix? What is identity matrix
I?
(d) Powers of matrices. What is the product An (including: when it is
defdined?) What is inverse matrix A−1 ? When it exists? – non-zero
determinant. In particular: (AB)−1 = B −1 A−1 .
(e) Transposition. What is transposed matrix AT ? What is relations
between transposition and operations (a) – (d)? In particular: (AB)T =
B T AT , (AT )T = A.
(f) Complex-conjugate matrix. What is complex conjugate matrix A? What
is relations between complex-conjugation and operations (a)–(e)?
T
(g) Adjoint matrix (also Hermitian conjugate matrix ) is A∗ := A . What
is relations between finding adjoint and operations and operations (a)–(f)?
4.4.3
Vectors
(a) Vectors are vector-columns, sometimes written as x = (x1 , . . . , xn )T .
(b) Dot-product of vectors x and y is x · y := xT y.
(c) Inner product (or scalar product) of vectors x and y is (x, y) := xT y.
It coincides with dot-product for real vectors; for complex vectors
dot-product does not make much sense.
p
(d) The length of the vector x is |x| = (x, x).
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Chapter 7. Systems of First-Order Linear Equations
4.4.4
125
Determinant
(a) What is det(A)? Properties, in particular: When det(A) = 0? det(AB) =
det(A) det(B), det(A)T = det(A) and det(A−1 ) = (det(A))−1 .
(b) Finding inverse matrix: Gaussian elimination method.
4.4
Linear Algebra
This is material you also supposed to know from Linear Algebra.
Please refresh from Section 7.3 of the Textbook – topics
(a) Systems of Linear Algebraic Equations;
(b) Linear Independence
Various aspects of the Topic
(a) Eigenvalues and Eigenvectors
will be covered in this class in details, as required by the main material.
4.4
4.4.1
Basic Theory of Systems of First-Order
Linear Equations
General
Theorem 4.4.1. Assume that A(t) is continuous matrix-function (that
means that all its elements ajk (t) are continuous functions) on interval
I = (α, β) with −∞ ≤ α < β ≤ ∞, which could be open, closed, or open on
one end and closed on another.
Then solutions of the linear homogeneous equation
x′ (t) = A(t)x(t)
(4.4.1)
form a n-dimensional linear space which means that there n linearly independent solutions x(1) (t), x(2) , . . . , x(n) (t) such that any other solution can
be represented as their superposition
x(t) = C1 x(1) (t) + C2 x(2) (t) + . . . + Cn x(n) (t)(t)
in the unique way.
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(4.4.2)
Chapter 7. Systems of First-Order Linear Equations
126
Proof. (a) Observe first that because system is linear and homogeneous, a
superposition (linear combination) of solutions is also a solution.
(b) Let us define solutions x(j) (t) from Cauchy’s problems
( ′
x (t) = A(t)x(t),
x(j) (t0 ) = ξ (j)
(j)
where ξ (1) , . . . , ξ (n) form a basis in Rn , f. e. ξk = δjk
(
1 j = k,
=
0 j=
̸ k
(4.4.3)
is
a Kronecker’s symbol.
These solutions exist due to the Existence and Uniqueness Theorem.
Then if x(t) = C1 x(1) (t) + C2 x(2) (t) + . . . + Cn x(n) (t), then
x(t0 ) = C1 ξ (1) + C2 ξ (2) + . . . + Cn ξ (n)
(4.4.4)
so decomposition, if exists, is unique. In particular,
x(t) = C1 x(1) (t) + C2 x(2) (t) + . . . + Cn x(n) (t) ≡ 0
=⇒ C1 = C2 = . . . = Cn = 0
(1)
(n)
(so x (t), . . . , x (t) are linearly independent).
(c) Let x(t) be any solution. Define y(t) := C1 x(1) (t) + . . . + Cn x(n) (t) with
C1 , . . . , Cn uniquely defined by (4.4.4).
Then y(t) satisfies the same system (4.4.1) as well and y(t0 ) = x(t0 ); so
y(t) satisfies exactly the same problem as x(t). But solution is unique and
therefore x(t) = y(t) := C1 x(1) (t) + . . . + Cn x(n) (t).
Thus {x(1) (t), . . . , x(n) (t)} form basis in the space of solutions.
Definition 4.4.1. The basis in {x(1) (t), . . . , x(n) (t)} in the space of solutions
is called a fundamental system of solutions.
Definition 4.4.2.


(1)
(2)
(n)
x
(t)
x
(t)
.
.
.
x
(t)
1
1
 1

 (1)

(2)
(1)
x2 (t) x2 (t) . . . x2 (t) 
X(t) :=  .
..
.. 
..
 ..
.
.
. 


(1)
(2)
(n)
xn (t) xn (t) . . . xn (t)
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(4.4.5)
Chapter 7. Systems of First-Order Linear Equations
127
is a fundamental matrix for system x′ (t) = A(t)x(t).
It is n × n-matrix, consisting of vector-columns x(1) (t), . . . , x(n) (t) and
because those satisfy x′ (t) = A(t)x(t), the fundamental matrix satisfies
X ′ (t) = A(t)X(t).
(4.4.6)
Indeed, matrix A applies to each column of X(t).
Remark 4.4.1. One can rewrite
x(t) = C1 x(1) (t) + C2 x(2) (t) + . . . + Cn x(n) (t)(t)
(7.4.2)
x(t) = X(t)c
(4.4.7)
as
where c = (C1 , . . . , Cn )T is a vector-column.
4.4.2
Wronskian
Definition 4.4.3. Let x(1) (t), x(2) (t), . . . , x(n) (t) be functions defined and
differentiable on interval I. Then
(1)
x1
(1)
x2
..
.
x1
(1)
(2)
W [x (t), x (t), . . . , x
(n)
x
(t)] := . 2
..
(1)
xn
(2)
. . . x1
(2)
...
..
.
(2)
xn
(n)
(1)
x2
..
.
(4.4.8)
(n)
. . . xn
is a Wronskian of x(1) (t), x(2) (t), . . . , x(n) (t).
Remark 4.4.2. So Wronskian is a determinant of the matrix (4.4.5) except
here we do not assume that x(1) (t), x(2) (t), . . . , x(n) (t) are solutions.
Theorem 4.4.2. Let x(1) (t), x(2) (t), . . . , x(n) (t) be solutions to x′ (t) =
A(t)x(t). Then
d
W [x(1) , x(2) , . . . , x(n) ](t) = tr[A(t)] W [x(1) , x(2) , . . . , x(n) ](t)
dt
where
tr[A] = a11 + a22 + . . . + ann
is the trace of matrix A.
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(4.4.9)
(4.4.10)
Chapter 7. Systems of First-Order Linear Equations
128
Remark 4.4.3. Trace is one of the matrix invariants. Determinant is another
(but there are n of them. Details–when we consider matrix eigenvalues).
Proof of Theorem 4.4.2. We need to differentiate determinant. Recall that
the derivative of the determinant is a sum of n determinants, each has the
same elements, but in j-th term the j-th row is differentiated. So
d
W [x(1) , x(2) , . . . , x(n) ](t) = V1 + V2 + . . . + Vn
dt
with
(1)′
x1
(1)
x2
..
.
(1)
xn
x1
x
V1 = . 2
..
xn
(2)′
. . . x1
(n)′
(2)
...
..
.
(2)
. . . xn
(1)
x2
..
.
(n)
and so on.
Recall that for solutions we have
(j)′
xk
(j)
(j)
= ak1 x1 + ak2 x2 + . . . akn x(j)
n
and decomposing Vk by k-th row we see that
Vk = ak1 Wk1 + ak2 Wk2 + . . . + akn Wkn
where Wkj is the same determinant W but with k-th row replaced by j-th
row.
If k =
̸ j we have a determinant with two identical rows and it is 0. For
j = k we have exactly W . Therefore Wkj = δjk W and Vk = akk W and,
finally
W ′ = (a11 + a22 + . . . + ann )W.
Remark 4.4.4. How is it connected to the corresponding results from Chapter 4? Recall that when we reduce equation
b0 y (n) + b1 y (n−1) + . . . + bn y = 0
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Chapter 7. Systems of First-Order Linear Equations
129
to the first-order system, we set x1 = y, x2 = y ′ , . . . , xn = y (n−1) so the
Wronskian of n scalar functions y (1) , . . . y (n) is exactly a Wronskian of n
vector-functions x(1) , . . . , x(n) .
One can see easily that the matrix of x′ = Ax in this reduction is


0
1
0
... 0


 0
0
1
... 0 




 0
0
0
... 0 


A= .
..
..
.. 
..
.
.
 .
.
.
. 


 0

0
0
.
.
.
1


−an −an−1 −an−2 . . . −a1
with ak =
bk
b0
and therefore tr[A] = − bb10 .
Corollary 4.4.3. In the framework of Theorem 4.4.2
(i)
Z
W [y1 , y2 , . . . , yn ](t) = C exp − tr[A] dt .
(4.4.11)
(ii) If A is continuous functions then
- either W [x(1) , . . . , x(n) ](t) does not vanish and X(t) is non-degenerate
anywhere on I
- or W [x(1) , . . . , x(n) ](t) is 0 and X(t) is degenerate everywhere on I .
Proof. Proof follows from (4.4.11).
4.4
4.4.1
Homogeneous Linear Systems with
Constant Coefficients
General
Let us consider a system with constant coefficients
x′ = Ax
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Chapter 7. Systems of First-Order Linear Equations
130
For n = 1 we had a single equation x′ = Ax and solution was x(t) = CeAt .
Let us try, like in the case of linear equations (of any order) with constant
coefficients x(t) = ekt ξ. Here we need to put a constant vector ξ to have a
vector-valued solution x. Then
kξekt = Aξekt
which is equivalent to
(A − kI)ξ = 0.
4.4.2
(4.4.2)
Linear Algebra: Eigenvalues and Eigenvectors
Now we need to discuss some material from Linear Algebra class (see Section
7.3 of the Textbook); more to follow later.
Definition 4.4.1. Let
(A − kI)ξ = 0
(7.5.2)
hold with ξ ̸= 0. Then k is an eigenvalue of mathrix A and ξ a corresponding
eigenvector.
Remark 4.4.1. (a) These notions eigenvalue and eigenvector are extremely
important mathematical notions with enormous value to applications. For
linear operators which are (kind of) generalizations of matrices, instead of
eigenvector often is used eigenfunction (if operators act in functional spaces)
and those are very important for different fields, for example, Quantum
Mechanics.
(b) The set of eigenvalues of A is called the spectrum of A (for operators it
is more complicated).
Since ξ =
̸ 0, then matrix (A − kI) should have non-trivial kernel (nullspace) Ker(A − kI) ̸= {0}, so it is degenerate which happens if and only if
its determinant is 0:
P (k) := det(A − kI) = 0,
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(4.4.3)
Chapter 7. Systems of First-Order Linear Equations
131
which means exactly that
a11 − k
P (k) =
a21
..
.
an1
a12
a13 . . .
a1n
a22 − k a23 . . .
..
.. . .
.
.
.
a2n
..
.
an2
= 0.
(4.4.4)
an3 . . . ann − k
Definition 4.4.2. P (k) = det(A − kI) is a characteristic polynomial and
P (k) = det(A − kI) = 0 is called a characteristic equation of matrix A.
So we just proved the following
Theorem 4.4.1. k is an eigenvalue of matrix A if and only if it is a
characteristic root, that means a root of the characteristic equation of A.
Theorem 4.4.2. (i) Characteristic polynomial is a polynomial of degree n,
its leading coefficient is (−1)n . Therefore it has exactly n roots (counting
multiplicities).
(ii) Coefficient at k n−1 is equal to (−1)n−1 tr(A), and coefficient at k 0 is
det(A).
(iii) On the other hand, coefficient at k n−1 is equal to (−1)n−1 (k1 + . . . + kn );
coefficient at k n−2 is equal to (−1)n (k1 k2 +k1 k3 +. . .+kn−1 kn ); . . . ; coefficient
at k 0 is equal to k1 k2 · · · kn .
(iv) Characteristic polynomials of matrices A and Q−1 AQ coincide for any
non-degenerate matrix Q.
Proof. Statements (a)–(c) are trivial, Statement (d) follows from
det(Q−1 (A − kI)Q) = det(Q−1 ) det(A − kI) det(Q) = det(A − kI)
because det(Q−1 ) det(Q) = 1.
Remark 4.4.2. Recall that matrices Q−1 AQ and A are called similar ; so
coefficients of the characteristic polynomials for similar matrices coincide
and we refer to them as spectral invariants; there are n of them, the first one
is tr(A) and the last one is det(A); others have neither names nor special
notations.
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Chapter 7. Systems of First-Order Linear Equations
132
Let us talk about eigenvectors; this is trivial:
Theorem 4.4.3. (i) The linear combinations of eigenvectors corresponding
to the same eigenvalue k is again an eigenvector, corresponding to this
eigenvalue k.
(ii) ξ is an eigenvector of A, corresponding to eigenvalue k if and only if
Q−1 ξ is is an eigenvector of Q−1 AQ, corresponding to this eigenvalue k.
Definition 4.4.3. The set of eigenvectors, corresponding to the same
eigenvalue k is the eigenspace of A corresponding to eigenvalue k. Eigenspace
is a linear vector space (we include 0 ).
Theorem 4.4.4. (i) Assume that all characteristic roots are simple. Then
to each root corresponds exactly one eigenvector (up to a constant factor).
Select the basis consisting of these eigenvectors ξ (1) , . . . , ξ (n) . Then the
matrix of A in this basis is diagonal:


k1 0 . . . 0 


 0 k2 . . . 0 
A=.

 .. . . . . . . ... 


0 0 . . . kn
(4.4.5)
(ii) Conversely, if matrix is A is diagonal, then eigenvalues are diagonal
elements, and eigenvectors are basis vectors.
Proof. Trivial. The case of non-simple characteristic roots is way more
complicate. We discuss it later.
Remark 4.4.3 (Useful observation). Eigenvalues of upper triangular matrices
(with all elements below the (main) diagonal equal to 0) and of lower
triangular matrices (with all elements above the (main) diagonal equal to 0)
are exactly diagonal elements.
4.4.3
General Solution: Distinct Eigenvalues
Theorem 4.4.5. Assume that all characteristic roots are simple. Then
to each root corresponds exactly one eigenvector (up to a constant factor).
Then the general solution to equation is
x(t) = C1 ek1 t ξ (1) + C2 ek2 t ξ (2) + . . . + Cn ekn t ξ (n)
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(4.4.6)
Chapter 7. Systems of First-Order Linear Equations
133
with arbitrary constants C1 , . . . , Cn .
Proof. Obviously, (4.4.6) defines a solution. As t = 0 we have
x(0) = C1 ξ (1) + C2 ek2 t ξ (2) + . . . + Cn ekn t ξ (n)
and since ξ (1) , . . . , ξ (n) is a basis, any vector x0 could be decomposed over
it. Therefore Cauchy’s problem x(0) = x0 has a solution for any x0 in the
form (4.4.6).
4.4.4
n = 2, Eigenvalues Real and Distinct
We consider several examples in which matrix A is real and characteristic
roots are distinct and real.
Case Ia: 0 < k1 < k2 . Assume that n = 2 and eigenvalues are real,
distinct and positive: 0 < k1 < k2 . Let ξ (1) and ξ (2) be corresponding
eigenvectors. Then in virtue of Theorem 4.4.5 the general solution is on the
form (4.4.6):
x(t) = C1 ek1 t ξ (1) + C2 ek2 t ξ (2) .
In particular, solutions with C2 = 0 or C1 = 0 are x(t) = Cj ekj t ξ j with
j = 1, 2 respectively and therefore they are movements along vectors ξ (1)
and ξ (2) “out”.
Any other solution is also a movement “out” but as t → +∞ it is almost
in the direction ξ (2) but as t → −∞ it is much closer to ξ (1) .
In the basis {ξ (1) , ξ (2) } the curves a parabolas y = C|x|k2 /k1 . As t → −∞
all solutions tend to equilibrium 0 and as t → +∞ all solutions (except
x(t) = 0 escape to infinity. It is an unstable node.
!
3 1
Example 4.4.1. (a) Consider A =
. Then characteristic equation
1 3
!
α
(1)
is (k − 3)2 − 1 = 0 =⇒ k1 = 2, k2 = 4 and to find ξ =
we have
β
!
!
!
1 1
α
1
= 0 =⇒ α+β = 0 and we can select ξ (1) =
. Similarly
1 1
β
−1
!
1
ξ (2) =
. These eigenvectors are orthogonal because A is symmetric.
1
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Chapter 7. Systems of First-Order Linear Equations
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ξ (2)
ξ (1)
Figure 7.1: Unstable node
(b) Consider A =
6 1
!
. Then characteristic equation is (k − 6)2 − 4 = 0
4 6
=⇒ k1 = 4, k2 = 8 and to find ξ (1) =
α
!
we have
β
=⇒ 2α + β = 0 and we can select ξ (1) =
2 1
4 2
!
1
!
α
!
=0
β
. Similarly ξ (2) =
−2
These eigenvectors are not orthogonal because A is not symmetric.
!
1
.
2
Case Ib: 0 > k1 > k2 . Assume that n = 2 and eigenvalues are real, distinct
and negative: 0 > k1 > k2 . Let ξ (1) and ξ (2) be corresponding eigenvectors.
Then in virtue of Theorem 4.4.5 the general solution is on the form (4.4.6).
In particular, solutions with C2 = 0 or C1 = 0 are x(t) = Cj ekj t ξ j with
j = 1, 2 respectively and therefore they are movements along vectors ξ (1)
and ξ (2) “in”.
Any other solution is also a movement “in” but as t → −∞ it is almost
in the direction ξ (2) but as t → ∞ it is much closer to ξ (1) .
In the basis {ξ (1) , ξ (2) } the curves a parabolas y = C|x|k2 /k1 . As t → ∞
all solutions tend to equilibrium 0 and as t → −∞ all solutions (except
x(t) = 0 escape to infinity. It is a stable node.
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Chapter 7. Systems of First-Order Linear Equations
135
ξ (2)
ξ (1)
Figure 7.2: Unstable node
!
−3 −1
Example 4.4.2. (a) Consider A =
. Then characteristic equa−1 −3
!
α
tion is (k + 3)2 − 1 = 0 =⇒ k1 = −2, k2 = −4 and to find ξ (1) =
we
β
!
!
!
−1 −1
α
1
have
= 0 =⇒ α + β = 0 and we can select ξ (1) =
.
−1 −1
β
−1
!
1
Similarly ξ (2) =
.
1
!
−6 −1
(b) Consider A =
. Then characteristic equation is (k+6)2 −4 =
−4 −6
!
!
!
α
2
1
α
0 =⇒ k1 = 4, k2 = 8 and to find ξ (1) =
we have
=0
β
4 2
β
!
!
1
1
=⇒ 2α + β = 0 and we can select ξ (1) =
. Similarly ξ (2) =
.
−2
2
These eigenvectors are not orthogonal because A is not symmetric.
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Chapter 7. Systems of First-Order Linear Equations
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ξ (2)
ξ (1)
Figure 7.3: Stable node
ξ (2)
ξ (1)
Figure 7.4: Stable node
Case II: k1 < 0 < k2 . Assume that n = 2 and eigenvalues are real, distinct
and of different signs: k1 < 0 < k2 . Let ξ (1) and ξ (2) be corresponding
eigenvectors. Then in virtue of Theorem 4.4.5 the general solution is on
the form (4.4.6). In particular, solutions with C2 = 0 or C1 = 0 are
x(t) = Cj ekj t ξ j with j = 1, 2 respectively and therefore they are movements
along vectors ξ (1) are “in” and along ξ (2) are“out”.
Any other solution is bypassing 0 and t → +∞ it is almost in the
direction ξ (2) but as t → −∞ it is much closer to ξ (1) .
In the basis {ξ (1) , ξ (2) } the curves a hyperbolas y = C|x|k2 /k1 . It is a
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Chapter 7. Systems of First-Order Linear Equations
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saddle.
Example 4.4.3. (a) Consider A =
1 3
!
. Then characteristic equation is
3 1
(k − 1)2 − 9 = 0 =⇒ k1 = −2, k2 = 4 and to find ξ (1) =
α
!
we have
β
3 3
3 3
!
α
!
!
1
= 0 =⇒ α + β = 0 and we can select ξ (1) =
(stable
−1
β
direction, corresponding to k1 < 0). Similarly ξ (2) =
!
1
(stable direction,
1
corresponding to k2 > 0).
Again, ξ (1) and ξ (2) are orthogonal because A is symmetric.
ξ (2)
ξ (1)
Figure 7.5: Saddle
(b) Consider A =
2
3
!
. Then characteristic equation is (k−2)2 −36 = 0
12 2
=⇒ k1 = −4, k2 = 8 and we can select ξ (1) =
!
1
. Similarly ξ (2) =
−2
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!
1
2
.
Chapter 7. Systems of First-Order Linear Equations
138
ξ (2)
ξ (1)
Figure 7.6: Saddle
Online plotter:
https://aeb019.hosted.uark.edu/pplane.html
4.4
Complex-Valued Eigenvalues
If eigenvalues are distinct then the same formula for n = 2
x(t) = C1 ek1 t ξ (1) + C2 ek2 t ξ (2)
(4.4.1)
holds but we want to understand how real solutions (with A also real) look
like. Since A is real, then for complex-conjugate k1,2 = α ± iβ, β > 0
eigenvectors are complex-conjugate as well ξ (1,2) = f (1) ± if (2) and for real
solutions we need to take coefficients also complex-conjugate.
Let us start from the beginning, considering system
(
x′ = ax + by,
(4.4.2)
y ′ = cx + dy
with A =
a b
!
and real a, b, c, d.
c d
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Chapter 7. Systems of First-Order Linear Equations
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Then
1
p
p
1
k1,2 =
a + d ± (a + d)2 − 4(ad − bc) =
a + d ± (a − d)2 + 4bc
2
2
and eigenvalues are complex conjugate if
−4bc > (a − d)2 .
(4.4.3)
Case III: a + d = 0. Assume first that α = 0 ⇐⇒ a + d = 0. Then one
can prove easily that
′
−cx2 + by 2 + 2axy = 0.
(4.4.4)
So, along trajectories
−cx2 + by 2 + 2axy = C
(4.4.5)
with arbitrary constant C. It follows from (4.4.3) that cb + a2 < 0 and these
curves are ellipses with axis along eigenvectors of the matrix
!
−c a
B=
.
(4.4.6)
a b
We need to figure out the orientation: clock-wise or counter-clock-wise.
It depends on the sign of
y ′ x − x′ y = (cx + dy)x − (ax + by)y = cx2 − by 2 + (d − a)xy
(4.4.7)
and quadratic form here is positive definite if c > 0 (and then b < 0) and
negative definite if c < 0 (and then b > 0).
Therefore if a = −d and (4.4.3) holds then trajectory are ellipses with
counter-clock-wise orintation as c > 0 and clock-wise orientation if c < 0.
Example 4.4.1.
(
x′ = 3x − 4y,
y ′ = 4x − 3y
2
Solution. So,
! a = 3, b = 4, c = −4 and d = −3, then bc + a < 0 and
!
4 3
1
B=
with eigenvalues λ1 = 7 and λ2 = 1 and eigenvectors
3 4
1
!
√
1
and
. So ratio of semiaxis is 1 : 7.
−1
Since c < 0 it is clock-wise oriented.
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Chapter 7. Systems of First-Order Linear Equations
140
Figure 7.7: Center
Case IVa: a + d > 0. Assume now that α > 0. Then it moves along
ellipse but also away from the origin.
Example 4.4.2.
(
x′ = 4x + 4y,
y ′ = −4x − 2y
Solution. It differs from the previous example by adding 1 to each diagonal
element, which means exactly adding factor et to x(t).
Case IVb: a + d < 0. Assume now that α < 0. Then it moves along
ellipse but also towards the origin the origin.
Example 4.4.3.
(
x′ = 2x + 4y,
y ′ = −4x − 4y
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Chapter 7. Systems of First-Order Linear Equations
141
Figure 7.8: Unstable Focus (Unstable Spiral Point; Source)
Solution. It differs from the Example 1 by adding −1 to each diagonal
element, which means exactly adding factor e−t to x(t).
Figure 7.9: Stable Focus (Unstable Spiral Point; Sink)
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Chapter 7. Systems of First-Order Linear Equations
142
In your assignments useful is the following
Rotation rule. If bottom-left element in A is > 0, rotation counterclockwise; if bottom-left element in A is < 0, rotation is counter-clockwise.
It is justified exactly as in the case of center (α = 0):
y ′ x − x′ y = (cx + dy)x − (ax + by)y = cx2 − by 2 + (d − a)xy > 0
∀(x, y) ̸= (0, 0) (4.4.8)
if c > 0 and < 0 if c < 0.
Integrability. If (anbd only if) a + d = 0 the system is integrable:
H(x, y) := −cx2 + by 2 + 2axy = C.
(4.4.9)
From Calculus II. Function H(x, y) = px2 + 2qxy + ry 2 has ay point
(0, 0)
(a) Minimum, if pr > q 2 and p > 0; then level lines are ellipses.
(b) Maximum, if pr > q 2 and p < 0; then level lines are ellipses.
(c) A saddle point (minimax) if pr < q 2 ; then level lines are hyperbolas.
Here p = −c, r = b and q = a. Case a = −d and bc + a2 < 0 is in
the framework of the previous lecture. In this special case hyperbolas are
described by k12 x2 + k2 y 2 = C in the basis of eigenvectors of mathrix
!
−c a
B=
.
a b
with k1 < 0 < k2 .
7.8
7.8.1
Repeated Roots
General Theory
Now consider the most complicated case of the constant coefficients system
x′ = Ax
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(7.8.1)
Chapter 7. Systems of First-Order Linear Equations
143
when characteristic polynomial
P (k) = det(A − k I)
(7.8.2)
P (k) = (−1)k (k − k1 )n1 (k − k2 )n2 · · · (k − ks )ns
(7.8.3)
has repeated roots
with disjoint k1 , . . . , ks ., n1 + . . . + ns = n.
Recall that for a characteristic root kj which also is an eigenvalue we
could write a corresponding solution
x(j) (t) = ξ (j) ekj t
(7.8.4)
where ξ (j) is a corresponding eigenvector.
If to the same eigenvalue kj corresponds nj linearly independent eigenvectors ξ (j,1) ,. . . , ξ (j,nj ) , we have nj linearly independent solutions
x(j,p) (t) = ξ (j,p) ekj t ,
p = 1, . . . , nj ,
(7.8.5)
and there is no difference with what we had before.
There are important cases when it is so, the matrix A is diagonalizable
and the dimension of eigenspace, corresponding to eigenvalue kj , is equal to
nj .
Recall that matrix A is diagonalizable, when there exists a non-degenerate
matrix Q such that Λ = Q−1 AQ is a diagonal matrix (with eigenvalues
on the main diagonal). Also recall that the eigenspace, corresponding to
eigenvalue kj is a subspace, consisting of all eigenvectors, corresponding to
eigenvalue kj (and it also includes 0 ).
Remark 7.8.1.
ΛQ−1
Λ = Q−1 AQ ⇐⇒ A = QΛ
but each form has its advantages. Here vector-columns of Q are eigenvectors
of A.
Reminder (if you ever learned this in your Linear Algebra class):
Theorem 7.8.1. Matrix A is diagonalizable in the following cases (not
exclusively):
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Chapter 7. Systems of First-Order Linear Equations
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(a) All characteristic roots are simple (we had this case already),
(b) A is self-adjoint, that means A∗ = A where A is an adjoint (Hermitian conjugate) matrix. In this case eigenvalues are real and eigenvectors
corresponding to different eigenvalues are orthogonal, and matrix Qcould be
taken unitary.
(c) A is skew-self-adjoint, that means A∗ = −A. In this case eigenvalues are purely imaginary and eigenvectors corresponding to different
eigenvalues are orthogonal, and matrix Q could be taken unitary.
(d) A is unitary, that means A∗ = A−1 . In this case eigenvalues are
on the unit circle {z : |z|} ⊂ C, eigenvectors corresponding to different
eigenvalues are orthogonal, and matrix Q could be taken unitary.
(e) A is normal, that means A∗ A = AA∗ . In this case eigenvectors
corresponding to different eigenvalues are orthogonal, and matrix Q could
be taken unitary.
What should we do? However, what should we do if the dimension of
eigenspace mj is less than the multiplicity of the root nj : mj < nj ? (we’ll
see that mj ≤ nj ).
Let us try x(t) = ekj t (tξ + η). Then x′ (t) = ekj t (kj tξ + kj η + ξ) and
x′ (t) = Ax(t) iff
kj tξ + kj η + ξ = tAξ + Aη,
which means that
(A − kj I)ξ = 0,
(A − kj I)η = ξ.
(7.8.6)
(7.8.7)
Here (7.8.6) means that ξ is an eigenvector, but (7.8.6) and (7.8.7)
together mean that
(A − kj I)2 η = 0.
(7.8.8)
If dimension of null-space of (A − kj I)2 is equal to nj then we are done.
This is the case when n = 2. So we consider this as an example, and continue
the general theory later.
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Chapter 7. Systems of First-Order Linear Equations
7.8.2
145
Repeated roots: n = 2
Consider real A. Then as n = 2 we can have the following cases: first
k1 = k2 = k, and
(a) m = 2 (there are two eigenvectors).
(b) m = 1 (there is just one eigenvector).
!
k 0
Case (a) is easy; then A =
,k=
̸ 0 (case k = 0 is trivial) and
0 k
x = ekt ξ are rays.
(a) k > 0, Unstable
straight node
(b) k < 0, Stable
straight node
Case (b) m = 1 is more interesting.
kt
Again, let k > 0 and first we have solutions
! x = ±e ξ where ξ is the
only eigenvector: let, for example, A =
0 −1
1
η=
!
1
0
we get ξ =
! !
−1 −1
1
1
1
0
=
. Then k = 1 and selecting
2
!
−1
: The general solution is
1
x(t) = (C1 tξ + C1 η + C2 ξ)ekt
(7.8.9)
and if C1 > 0 as t → +∞ it escapes to infinity and it is directed mainly
along ξ (because of extra factor t) while as t → −∞ it tends to the origin,
mainly along −ξ (also because of extra factor t); for C1 < 0 it directions
are opposite.
As k < 0 we have a very similar arguments except as t → +∞ it goes to
origin mainly in the direction ξ and as t → −∞ it escapes to infinity mainly
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Chapter 7. Systems of First-Order Linear Equations
146
−ξ
Figure 7.10: k > 0, Unstable improper node
in the direction −ξ; consider, for example, A =
!
−2 −1
0
. Then k = −1
2
but both ξ and η are the same.
ξ
Figure 7.11: k < 0, Stable improper node
Remark 7.8.2. Doing your assignments, you do not need to pick η and then
calculate ξ, you may just find eigenvector ξ and use the following
Rotation rule. If bottom-left element in A is > 0, rotation mainly
counter-clockwise; if bottom-left element in A is < 0, rotation mainly
clockwise;
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Chapter 7. Systems of First-Order Linear Equations
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It is justified exactly as in the case of complex roots
y ′ x − x′ y = (cx + dy)x − (ax + by)y = cx2 − by 2 + (d − a)xy > 0
∀(x, y) ̸ ∥ ξ (7.8.10)
if c > 0 and < 0 if c < 0. Or you can think about repeated roots case as
a borderline of complex roots case and it inherits this rule from the latter.
However, instead of the infinite number of rotations in the case of spiral
point, we have now just half-rotation!
In both examples bottom-left element in A is 1 > 0.
7.8.3
General Theory (continued)
Return now to the general theory. So if the dimension of Ker (A − kj I)2
null-space of (A − kI)2 equals nj we are done. But what if it is still less?
2
Then we will try ekj t ( t2 ξ + tη + ζ) and get
(A − kI)ξ = 0,
(A − kI)η = ξ,
(A − kI)ζ = η
(7.8.6)
(7.8.7)
(7.8.11)
which means exactly that ζ ∈ Ker (A − kj I)3 , and if dimension of the
latter = nj we are done; otherwise we continue with increasing powers until
we reach nj .
Theorem 7.8.2. Dimension of Ker (A − kI)nj = nj .
It follows from Theorem 7.8.3 below. Thus we will stop our ascent here
or before.
Therefore if


(A − kI)ξ (1) = 0,




(A − kI)ξ (2) = ξ (1) ,
(7.8.12)
..


.



(A − kI)ξ (m) = ξ (m−1) ,
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Chapter 7. Systems of First-Order Linear Equations
148
then
x(t) = ekt
m
X
j=1
tm−j−1
ξ (j)
(m − j − 1)!
(7.8.13)
is a solution.
Definition 7.8.1. (a) Vectors belonging to Ker (A − kI)nj are called
generalized eigenvectors. So, ξ (2) , . . . , ξ (m) are generalized eigenvectors.
(b) Ker (A − kI)nj is a root space, corresponding to eigenvalue k.
7.8.4
Linear Algebra: Jordan Normal Form
Linear Algebra: Jordan Normal Form To justify the previous we need
the following statement from Linear Algebra:
Theorem 7.8.3 (Jordan Normal Form Theorem). Let A be a n × n matrix.
Then there exists a non-degenerate matrix Q, such that Q−1 AQ is a Jordan
normal form matrix


J 1 0 0 . . . 0 


 0 J2 0 . . . 0 
−1
Q AQ =  .
(7.8.14)
.. .. . .
.. 
 ..

.
.
.
.


0 0 0 . . . Jq
where J 1 , . . . J q are Jordan blocks with kp on the main diagonal and 1 on
the diagonal above it:


kp 1 0 . . . 0 0


 0 kp 1 . . . 0 0 


 .. .. .. . .

.
.
.
.
Jp =  . . .
(7.8.15)
. . .




 0 0 0 . . . kp 1 
0
0
0 ...
0
kp
where among kp could be equal, but the total dimension of Jordan blocks with
the same kp = kj is equal to nj .
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Chapter 7. Systems of First-Order Linear Equations
149
Found over internet
https://www.math.upenn.edu/∼moose/240S2015/slides7-23.pdf
Example 7.8.1. Consider a linear homogeneous equation of order n and
reduce it to the first order system in the canonical way. Then there is only
one eigenvector no matter how large multiplicity nj of the root is, so in this
case we ascent up to nj .
We can prove it considering matrix A, but we can justify it from the
fact, that we in Chapter 4 always had a solution containing tnj −1 .
7.7
7.7.1
Fundamental Matrices
General
Recall that the fundamental matrix of the system
x′ (t) = A(t)x(t)
(7.7.1)
is a matrix X(t) with a columns x(1) (t), . . . , x(n) (t)


(1)
(2)
(n)
x1 x1 . . . x1 
 (1) (2)
(n) 
x2 x2 . . . x2 
X(t) =  .
.
..
.. 
...
 ..
. 
.


(1)
(2)
(n)
xn xn . . . xn
(7.7.2)
It satisfies matrix equation
X ′ (t) = A(t)X(t).
(7.7.3)
Using fundamental matrix one can solve inhomogeneous system
x′ (t) = A(t)x(t) + g(t).
(7.7.4)
Indeed, let us apply the method of variations of the parameters. So we look
for a solution in the form
x(t) = u1 (t)x(1) (t) + u2 (t)x(2) (t) + . . . + un (t)x(n) (t)
u
1 (t)
= X(t)U (t),
u (t)

2
U (t) =  ..  (7.7.5)
.
un (t)
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with unknown functions u1 (t), . . . , un (t). Then
x′ (t) = X ′ (t)U (t) + X(t)U ′ (t) = A(t)X(t)U (t) + X(t)U ′ (t)
want
= A(t)x(t) + g(t) = X(t)U (t) + g(t)
which is equivalent to
U ′ (t) = X −1 (t)g(t).
Then
Z
U (t) =
t
X
−1
Z
(s)g(s) ds =⇒ x(t) =
t
X(t)X −1 (s)g(s) ds.
(7.7.6)
Further, there is a formula for a solution of the Cauchy’s problem
x′ (t) = A(t)x(t) + g(t)
x(t0 ) = x0 .
(7.7.4)
(7.7.7)
Namely,
Z
t
x(t) =
X(t)X −1 (s)g(s) ds + X(t)X −1 (t0 )x0 .
(7.7.8)
t0
Indeed, the first term in the right-hand expression solves the Cauchy’s
problem with x(t0 ) = 0, and the last term solves Cauchy’s problem with
g(t) = 0 and x(t0 ) = x0 .
Remark 7.7.1. One can check easily, that if we consider linear inhomogeneous
higher order equation, reduce it to a first order system in the standard way
and solve it like this, it would be exactly the same process, as if we solved
as in Chapter 4.
But this is so much more transparent!
7.7.2
Constant Coefficients Systems and Matrix
Exponents
If A(t) = A is constant, we can construct X(t) using matrix exponent
∞ r r
X
tA
X(t) = etA :=
.
(7.7.9)
r!
r=0
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Chapter 7. Systems of First-Order Linear Equations
151
Indeed,
etA
′
=
∞
X
tr−1 Ar
(r − 1)!
r=1
because the derivative of the term with r = 0 is 0]
=
∞ r r+1
X
tA
r!
r=0
=A
∞ r r
X
tA
r!
r=0
= AetA .
Furthermore, for t = 0 we have etA = A0 = I.
We claim that the “normal” exponent rule
e(s+t)A = etA esA
(7.7.10)
applies. Indeed,
e(s+t)A =
∞
r
∞
X
r!
Ar
Ar X X
=
tp sr−p
(t + s)r
r!
p!(r − p)!
r!
r=0 p=0
r=0
where we used binomial formula
=
∞ X
∞
X
p=0 q=0
tp sq
Ap+q
= etA esA .
p!q!
Therefore, plugging it into formula formula
Z t
x(t) =
X(t)X −1 (s)g(s) ds + X(t)X −1 (t0 )x0 .
(7.7.8)
t0
we arrive to
Z
t
x(t) =
e(t−s)A g(s) ds + e(t−t0 )A x0 .
(7.7.11)
t0
How to calculate etA ? In virtue to Jordan Normal Form Theorem
A = QJ Q−1 and then Ar = QJ r Q−1 and therefore etA = QetJ Q−1 .
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Chapter 7. Systems of First-Order Linear Equations
152
Obviously etJ is block-diagonal with diagonal blocks etJ p . Since each block
is kp I p + N p , where I p and N p are the blocks of the same size as J p ,


0 1 0 ... 0 0


0 0 1 . . . 0 0 


 .. .. .. . . .. .. 
Np = . . .
. . .




0 0 0 . . . 0 1 
0 0 0 ... 0 0
we see that N rp has 1 on the diagonal which is r units higher than the main
one and 0 everywhere else, and then using formula
et(kp I p +N p ) = etkp I p etN p = etkp etN p
because I p and N p commute and finally

2
3
1 t t2! t3! . . .


0 1 t t2 . . .

2!

.
etJ p = ekp t  .. .. .. . .
. ..
. . .

0 0 0 . . . 0

0 0 0 ... 0
tm−1
(m−1)!
tm−2
(m−2)!
..
.
tm
m!



tm−1 
(m−1)! 
1
t
0
1






with described above m × m-matrix N p (calculate N 2p , N 3p , . . . , N m−1
and
p
m
N p ).
7.9
Nonhomogeneous Linear Systems
Basically we covered it already. See also the Textbook.
7.A. Examples: n = 3
Example 7.A.1.


4 −2 −4


′

x = −3
5
8
x
2 −2 −3
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Chapter 7. Systems of First-Order Linear Equations
153
Solution. Characteristic equation:
4−k
−2
−4
−3
5−k
8
2
−2
−3 − k
−k
5
= −k 3 + k 2 4 + 5 − 3
8
−2 −3
+
4 −4
2 −3
+
4
−3
−2 5
4 −2 −4
+ −3
5
8
2 −2 −3
= −k 3 + 6k 2 − 11k + 6 = 0
with characteristic roots k1 = 1, k2 = 2, k3 = 3 we guess k1 = 1 and
exploiting Vieta’s theorem we have k2 + k + 3 = 5, k2 k3 = 6).
Looking for eigenvectors: k = 1

   

   
3 −2 −4
α
0
3 −2 −4
α
0

   

   
−3
   

   
4
8
2
4

 β  = 0 =⇒ 0
 β  = 0
2 −2 −4
γ
0
2 −2 −4
γ
=⇒ β = −2γ, α = 0 =⇒ ξ (1)
Similarly ξ (2)
 
 
2
1
 
 
(3)



= 1 and ξ = −1
 and
1
0
x(t) = C1 et ξ (1) + C1 e2t ξ (2) + C3 e3t ξ (3) .
All directions are unstable. We call it unstable node.
Example 7.A.2.


−8 10 20


′

x =  3 −1 −4
x
−4
4
9
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0


0
 

= −2
.
1
Chapter 7. Systems of First-Order Linear Equations
154
Solution. Characteristic equation (like in Example 7.A.1):
−8 − k
10
20
3
−1 − k
−4
−4
4
9−k
= −k 3 + 7k − 6 = 0
and characteristic roots k1 = 1, k2 = 2, k3 = −3.

Looking for eigenvectors (like in Example 7.A.1) we get ξ (1)
ξ (2)

0
 

=
−2,
1
 
 
1
2
 
 
(3)



= 1 and ξ = −1
 and
0
1
x(t) = C1 et ξ (1) + C2 e2t ξ (2) + C3 e−3t ξ (3) .
ξ (1) and ξ (2) are unstable, ξ (3) is stable. It is a saddle. Another kind of a
saddle would be when two directions are stable and one unstable.
Example 7.A.3.


7 −10 −20


 x.
x′ = 
0
3
4


2 −4 −7
Solution. Characteristic equation (like in Example 7.A.1):
−8 − k
10
20
3
−1 − k
−4
−4
4
9−k
= −k 3 + 3k 2 − 7k + 5
and one characteristic root is k1 = 1. Then k1 + k2 + k3 = 3, k1 k2 k3 = 5 and
we have k 2 − 2k + 5 = 0 =⇒ k2,3 = 1 ± 2i (exploiting Vieta’s theorem).
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Chapter 7. Systems of First-Order Linear Equations
155

0
 

=
−2,
1

Looking for eigenvectors (like in Example 7.A.1) we get ξ (1)

2−i


2+i





 and ξ (3) = −1 + i and
ξ (2) = 
−1
−
i




1
1
(2)
(2)
t (1)
t
x(t) = C1 e ξ +C2 e cos(2t) Re ξ
− sin(2t) Im ξ
+C3 et cos(2t) Im ξ (2) + sin(2t) Re ξ (2) .
All directions are unstable. Think: what we would have if
1. k1 = −1, k2,3 = 1 ± 2i,
2. k1 = 1, k2,3 = −1 ± 2i,
3. k1 = −1, k2,3 = −1 ± 2i.
Example 7.A.4.


2 −1 −2


′

x = −5
6 16
 x.
3 −3 −8
Solution. Characteristic equation (like in Example 7.A.1):
2−k
−1
−2
−5
6−k
16
3
−3
−8 − k
= −k 3 + 3k − 2
and one characteristic root is k1 = k2 = 1, k3 = −2.
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Chapter 7. Systems of First-Order Linear Equations
Looking for eigenvectors (like in Example 7.A.1) we get ξ (1)

0
 

and ξ (3) = 
−2. We do
1
(2)
(A − kI)ξ = ξ (1) :

1 −1

−5
5

3 −3
156
 
1
 

=
1
0

not have an eigenvector ξ (2) ; instead we look for
   
−2
α
1
   




16 β  = 1
 =⇒ γ = 1.
−9
γ
0
 
3
 
(2)

Selecting β = 0 we have α = 3 and ξ = 
0.
1
Finally
x(t) = C1 et ξ (1) + C2 et ξ (1) t + ξ (2) + C3 e−2t ξ (3)
where ξ (2) is a generalized eigenvector.
ξ (1) and ξ (2) are unstable directions, ξ (3) is stable direction.
Example 7.A.5.


2 −1 −2


′

x = 1
0 −2
 x.
0
0
1
Solution. Characteristic equation (like in Example 7.A.1):
2 − k −1
−2
1
−k
−2
0
0
1−k
= −k 3 + 3k 2 − 3k + 1 = 0
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Chapter 7. Systems of First-Order Linear Equations
157
and one characteristic root is k1 = k2 = k3 = 1.
Looking for eigenvectors (like in Example 7.A.1) we get

   
1 −1 −2
α
0
   

1 −1 −2 β  = 0

   
0
0
0
γ

0
 
(1)

= −2
. We selected ξ this way because the
0

 
1
 
(1)
(3)

and ξ = 1
 and ξ
0
1
left-hand expression above is always proportional to this vector, so we can
solve (A − kI)ξ (2) = ξ (1)

   
 
1
1
1 −1 −2
α
   
 

(2)
1 −1 −2 β  = 1 =⇒ ξ = 0
 

   
0
0
0
0
γ
0
where β = γ = 0 was selected arbitrarily.
Finally
x(t) = C1 et ξ (1) + C2 et tξ (1) + ξ (2) + C3 et ξ (3)
where ξ (2) is a generalized eigenvector.
All directions are unstable.
Example 7.A.6.


2 −1 −2


 x.
x′ = 
−1
4
6


1 −2 −3
Solution. Characteristic equation (like in Example 7.A.1):
2−k
−1
−2
−1
4−k
6
1
−2
−3 − k
= −k 3 + 3k 2 − 3k + 1 = 0
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Chapter 7. Systems of First-Order Linear Equations
158
and one characteristic root is k1 = k2 = k3 = 1.
Looking for eigenvectors (like in Example 7.A.1) we get

   
1 −1 −2
α
0

   
−1




3
6 β  = 0

.
1 −2 −4
γ
0
Here we would getjust
 one eigenvector (think why)
 so
we do things differently: we take ξ
(3)
1
1
  (2)
 
(3)
(1)
(2)



= 0, ξ = (A − kI)ξ = −1
 and ξ = (A − kI)ξ
0
1
is an eigenvector.
Finally
t (1)
x(t) = C1 e ξ
t
+ C2 e tξ
(1)
+ξ
(2)
t
+ C3 e
t2
2
ξ
(1)
+ tξ
where ξ (2) and ξ (3) are generalized eigenvectors.
All directions are unstable.
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(2)
+ξ
(3)


0
 

=  2

−1
Chapter 9
Nonlinear Differential
Equations and Stability
Introduction
In this Chapter we will deal almost exclusively with non-linear first order
differential 2 × 2-systems
x′ (t) = f (x(t); t))
(9.1.1)
that is
(
x′ = f (x, y; t),
y ′ = g(x, y; t)
with x =
x
!
,
f=
y
f
!
(9.1.2)
g
but sometimes we consider arbitrary dimension n.
Definition 9.1.1. Such system is called autonomous if f does not depend
on t.
In this study important role plays the study of local behaviour of solutions
near points of equilibrium.
Definition 9.1.2. Consider autonomous system (9.1.1). Point x0 is point
of equilibrium (or stationary point) if x(t) ≡ x0 is a solution to (9.1.1).
Since x(t) ≡ x0 is constant, plugging it into (9.1.1) we get
159
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Chapter 9. Nonlinear Differential Equations and Stability
160
Theorem 9.1.1. x0 is a stationary point if and only if
f (x0 ) = 0.
(9.1.3)
In the next Lecture we consider a linearization when nonlinear system
(9.1.1) is approximated near equilibrium point by a linear system with
constant coefficients (like one we studied in Chapter 7). So it this lecture
we recall results of that chapter specifically for 2 × 2-systems and discuss
them in more depth.
9.1
The Phase Plane: Linear Systems
So, consider a linear 2 × 2 system
x′ (t) = Ax(t).
(9.1.4)
What are stationary points here? They must satisfy a linear system
Ax0 = 0 and if matrix A is non-degenerate (det A ̸= 0) the only stationary
point is x0 = 0 .
Recall that we need to consider eigenvalues k1,2 ̸= 0 of A which we must
find from the characteristic equation
det(A − kI) = 0
(9.1.5)
and then corresponding eigenvectors of A, ξ (1) and ξ (2)
(A − kj I) = 0
(9.1.6)
and then the general solution is
x(t) = c1 ek1 t ξ (1) + c2 ek2 t ξ (2) .
(9.1.7)
It is not necessarily correct when k1 = k2 .
9.1.1
Real, Unequal Eigenvalues of the Same Sign
This happens when the roots of the characteristic equation
k 2 − (a + d)k + (ad − bc) = 0
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Chapter 9. Nonlinear Differential Equations and Stability
161
are real (that is the discriminant is positive)
(a − d)2 + 4bc > 0
(9.1.8)
ad − bc > 0.
(9.1.9)
and of the same sign
If both roots are negative 0 > k2 > k1 , then we get a stable node (or
nodal sink ): all solutions tend to 0 as t → +∞ mainly along ξ (2) , and all
solutios (except x(t) ≡ 0 ) escape to infinity as t → −∞, mainly along ξ (1) .
But in applications to non-linear systems we are interested in behaviour
near 0 , that is as t → +∞.
In the basis {ξ (1) , ξ (2) } those are x = C|y|α with α = k1 /k2 .
!
−8
2
Example 9.1.1. A =
. Eigenvalues are k1 = −9, k2 = −6, α = 32
1 −7
!
!
−2
1
(1)
(2)
and eigenvectors are ξ =
and ξ =
.
1
1
If both roots are positive 0 < k2 < k1 , then we get an unstable node (or
nodal source): all solutions tend to 0 as t → −∞ mainly along ξ (2) , and all
solutios (except x(t) ≡ 0) escape to infinity as t → ∞, mainly along ξ (1) .
But in applications to non-linear systems we are interested in behaviour
near 0 , that is as t → −∞.
!
8 −2
Example 9.1.2. A =
. It is from Example 9.1.1 with the opposite
−1
7
sign, so , eigenvalues are opposite:
k1 =!6, k2 = 3, α = 2 and eigenvectors
!
are the same: ξ (1) =
−2
, ξ (2) =
1
the movement is opposite.
9.1.2
1
. Therefore the same picture but
1
Real Eigenvalues of the Opposite Sign
In this case discriminant is positive
(a − d)2 + 4bc > 0
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(9.1.8)
Chapter 9. Nonlinear Differential Equations and Stability
ξ (1)
162
ξ (2)
Figure 9.1: Stable node
but the determinant is negative
ad − bc < 0.
(9.1.10)
While analytic formulas are the same, conclusion is different: assuming
that k1 > 0 > k2 the only trajectories tending to 0 as t → −∞ are straight
rays along ξ (1) , and the only trajectories tending to 0 as t → +∞ are straight
rays along ξ (2) , and other trajectories bypass 0 .
!
−1 10
Example 9.1.3. A =
. Eigenvalues are k1 = 9, k2 = −6, α = − 32
5 4
!
!
1
−2
and eigenvectors are ξ (1) =
and ξ (2) =
.
1
1
9.1.3
Complex Eigenvalues with Nonzero Real Part
In this case discriminant must be negative
(a − d)2 + 4bc < 0
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(9.1.11)
Chapter 9. Nonlinear Differential Equations and Stability
ξ (1)
163
ξ (2)
Figure 9.2: Unstable node
but the trace nonzero
a + d ̸= 0
(9.1.12)
We get a a stable spiral point, if a + d < 0, and an unstable spiral point, if
a + d > 0, counter-clockwise if c > 0, and clockwise if c < 0.
!
−3 −13
Example 9.1.4. A =
. Eigenvalues are k1,2 = −2 ± 18i. So
25 −1
stable spiral point (−2 < 0), counter-clockwise (25 > 0).
!
3 13
Example 9.1.5. A =
. This is Example 9.1.4 with the opposite
−25 1
sign. Eigenvalues are k1,2 = 2 ± 18i. So unstable spiral point (2 > 0),
clockwise (−25 < 0).
9.1.4
Robust and Fragile Cases
These cases
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Chapter 9. Nonlinear Differential Equations and Stability
ξ (2)
164
ξ (1)
Figure 9.3: Saddle
(a) Node:
a) stable;
b) unstable;
(b) Saddle;
(c) Spiral:
(a) stable counter-clockwise;
(c) stable clockwise;
(b) unstable counter-clockwise;
(d) unstable clockwise;
I would refer as robust (not a standard terminology, though): they are
defined exclusively by inequalities and when we change a, b, c, d a little, these
inequalities would not break.
The rest I would call fragile (again not a standard terminology, though):
they are defined by at least one equality, and it may not survive no matter
how little we change a, b, c, d. Consider interesting cases, when there is just
one equality.
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165
Figure 9.4: Stable spiral point, counter-clockwise
9.1.5
Fragile Cases
Let discriminant equal 0:
(a − d)2 + 4bc = 0.
(9.1.13)
a + d ̸= 0.
(9.1.14)
but trace is not:
Then eigenvalues coincide but they are not 0. Assume that
bc ̸= 0.
(9.1.15)
Then there is just one eigenvector, and we have improper node.
What can happen to it, when we change coefficients just a bit? It can
become either node or spiral.
!
!
−3
1
1
Example 9.1.6. A =
. Then k1 = k2 = −2 and ξ (1) =
. It
−1 −1
1
is stable improper node (−4 < 0), clockwise (−1 < 0).
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Chapter 9. Nonlinear Differential Equations and Stability
Figure 9.5: Unstable spiral point, clockwise
ξ (1)
Figure 9.6: Stable improper node, clockwise
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166
Chapter 9. Nonlinear Differential Equations and Stability
167
Let us change matrix elements
ξ (2) ξ (1)
(a) Stable node
Stable node: A =
(b) Stable spiral
−3.1
!
1
−3
1
!
.
Stable spiral A =
.
−1 −.9
−1.1 −1
Look the same? On the left eigenvectors do not differ much. On the
right, imaginary part is 0.3 so you do not see that there is a lot of rotations.
9.1.6
Pure Imaginary Eigenvalues
Then discriminant is negative
(a − d)2 + 4bc < 0
(9.1.11)
a + d = 0.
(9.1.16)
but the trace zero
It is a center. It is counter-clockwise if c > 0, and clockwise, if c < 0.
!
−1 −13
Example 9.1.7. A =
.
25
1
Eigenvalues are k1,2 = ±18i. So stable spiral point (0 = 0), counterclockwise (25 > 0).
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168
Figure 9.7: Center, counter-clockwise
What can happen to it, when we change coefficients just a bit? It can
become a spiral either stable or unstable (but orientation will be preserved).
(a) Stable Spiral
(b) Unstable Spiral
!
!
−1.3 −13
−.7 −13
Stable spiral: A =
. Unstable spiral A =
.
25
.7
25 1.3
On the left it collapses, on the right it expand–but slow (in comparison
to rotation)
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Chapter 9. Nonlinear Differential Equations and Stability
9.1.7
169
One of Eigenvalues is Zero
This happens when determinant is 0 but trace is not. Then equilibrium
points occupy the whole line (along ξ (2) , corresponding to eigenvalue k2 = 0).
The other eigenvalue is not, k1 ̸= 0 and the movement is going along
lines, parallel to ξ (1) , towards equilibrium line if k1 < 0 and from it if k1 > 0.
What can happen to it, when we change coefficients just a bit? It can
become either a node or a saddle.
!
!
1 1
1
Example 9.1.8. A =
. Then k1 = 2, ξ (1) =
and k1 = 0,
1 1
1
!
−1
ξ (2) =
.
1
0
0
(a) Unstable Node
Unstable node: A =
1.1 1
(b) Saddle
!
.
Saddle A =
1 1
They look the same, put look more carefully at origin!
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.9 1
1 1
!
.
Chapter 9. Nonlinear Differential Equations and Stability
9.1.8
170
More Degenerate Cases
Other cases are described by more than one equality:
(a) k1 = k2 = 0 by two equalities;
(b) k1 = k2 and there are two eigenvectors ξ (1) and ξ (2) by three equalities;
(c) k1 = k2 = 0 and there are two eigenvectors ξ (1) and ξ (2) by four
equalities;
and we don’t need them!
9.2
9.2.1
Autonomous Systems and Stability
Definitions
In this lecture we discuss the notions of stability, asymptotic stability
and instability. There will be definitions, examples and discussions, but
no theorems or methods (beside of computer simulation); however it is
important for all future expositions.
So, we consider autonomous system
x′ = f (x);
(9.2.1)
so f does not depend on t explicitly.
Definition 9.2.1. (a) Solution x(t) is stable if for each ε > 0 there exists
δ = δ(ε) > 0 such that if y(t) is another solution and
∥x(0) − y(0)∥ < δ
(9.2.2)
then
∥x(t) − y(t)∥ < ε
∀t > 0.
(9.2.3)
(b) Solution x(t) is asymptotically stable if
- it is stable and also
- there exists δ > 0 such that if (9.2.3) holds then
∥x(t) − y(t)∥ → 0
as t → +∞.
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(9.2.4)
Chapter 9. Nonlinear Differential Equations and Stability
171
(c) Solution x(t) is unstable if it is not stable: there exists ε > 0 such that
for any δ > 0 there exists solution y(t) with ∥x(0) − y(0)∥ < δ and there
also exists t > 0 with ∥x(t) − y(t)∥ ≥ ε.
- In other words: solution x(t) is stable, if any other solution y(t)
which starts sufficiently close to it at moment t = 0 (or any other
moment–for autonomous systems the choice of of the initial moment
does not matter) will stay sufficiently close to it in the future–note
t > 0; solution x(t) is asymptotically stable if this solution y(t) will
also move closer and closer to it in the future, and solution is unstable
if no matter how close to it y(t) starts, in some remote future they
may be not too close enough.
- Why “remote future”? Because if we limit time by an arbitrarily
large but fixed time T then there is a kind of stability (provided f
satisfies Lipschitz condition): for any ε > 0 and any T > 0 there exists
δ = δ(ε, T) > 0 such that
∥x(0) − y(0)∥ < δ =⇒ ∥x(t) − y(t)∥ < ε
for all t ∈ [−T, T ].
Remark 9.2.1. (a) For linear homogeneous systems with constant coefficients x′ = Ax all solutions are stable, asymptotically stable or unstable in
the same time (think why) and thus we can talk about those for x(t) ≡ 0 .
(b) In particular, for linear homogeneous 2 × 2 systems with constant
coefficients recall pictures and you’ll see instantly that
- stable nodes and stable spirals are asymptotically stable;
- centers are stable but not asymptotically stable;
- unstable nodes, unstable spirals and saddles are unstable.
(c) For linear homogeneous n × n systems with constant coefficients the
rule is also simple:
- If all eigenvalues (real or complex) have their real parts < 0, then
asymptotic stability–think about exponents.
- If there is at least one eigenvalue with real part > 0, then instability–
think about exponents.
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172
- Assume that all eigenvalues have their real parts ≤ 0. Consider eigenvalues with real parts = 0. Then if all of them have complete sets of
eigenvectors then stability, but not asymptotic stability–think about
exponents. But if there is at least one of them with not enough
eigenvectors, then instability (because factor tk appears).
(d) We talk about “stability into future”, but we can talk about “stability
into past” either by reversing time t 7→ −t or by simply taking t < 0 instead
of t > 0 and t → −∞ instead of t → +∞.
Definition 9.2.2. Let x(t) be asymptotically stable. Then the basin of
attraction of x(t) is a set Ω ∈ Rn such consisting of points y 0 , such that the
solution of our system y(t) with y(0) = y 0 satisfies ∥x(t) − y(t)∥ → 0 as
t → +∞.
Why this terminology? – Think about river basin or lake basin in
geography.
Remark 9.2.2. For linear homogeneous systems with constant coefficients,
if 0 is asymptotically stable then the basin of attraction of x(t) = 0 is the
whole space, but for non-linear systems it is much more complicated.
9.2.2
Examples
We consider several examples from mechanics: one-dimensional movement
with the force depending only on position: x′′ = f (x) (here f (x) is the force
and we assume that mass m = 1). Denote y = x′ the velocity, then
(
x′ = y,
(9.2.5)
y ′ = f (x).
Multiplying the second equation by y we get
y(t)2 ′
= −(V (x(t)))′ =⇒
2
y2
H(x, y) :=
+ V (x) = E
2
yy ′ = f (x)x′ =⇒
(9.2.6)
with potential
Z
V (x) = −
x
f (x) dx
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(9.2.7)
Chapter 9. Nonlinear Differential Equations and Stability
173
and constant E.
2
In Physics potential = potential energy, my2 is kinetic energy (remember,
mass m = 1) so E is an energy and (9.2.7) is an energy conservation law.
Consider stationary points f = (y, f (x))T = 0 ⇐⇒ y = 0, f (x) = 0. So
stationary points (equilibria) are those points on the phase plane where
velocity y = 0 and force f (x) = 0.
So, those are critical points of potential V (x)–maxima and minima (we
so far ignore critical points which are also inflection points).
One can guess that minima of potential correspond to stable equilibrium
points and maxima correspond to unstable equilibtrium points.
This is correct because the trajectories are (parts of) level lines of the
2
energy function H(x, y) := y2 +V (x) and minima V (x) correspond to minima
of H(x, y) and maxima to saddles of it (indeed, max by x meets min by y!).
Example 9.2.1 (Harmonic oscillations). f (x) = −kx (this is Hooke’s law)
and the system is x′ = y, y ′√= −kx and solutions are x = C cos(ωt + φ),
y = −Cω sin(ωt + φ), ω = k, which are harmonic oscillations with the
angular frequency ω and amplitude C.
2
2
Meanwhile V (x) = kx2 and E = kC2 . Recall the well-known picture:
Figure 9.8: Harmonic oscillations
Example 9.2.2 (Anharmonic oscillations). f (x) = −x3 − x. Then V (x) =
2
x4
+ x2 :
4
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Chapter 9. Nonlinear Differential Equations and Stability
174
V (x)
x
It has a single minimum, no maxima, and thus there is a single equilibrium
point (0, 0) which is stable.
Figure 9.9: Unharmonic oscillations
Equilibrium (0, 0) is stable. But are oscillations stable? They are unstable
because in nonlinear oscillations (also called anharmonic oscillations) periods
depend on amplitude and in this example oscillations with larger amplitudes
have smaller periods. So, changing a bit amplitude we change a bit a period
but in the very long run this error accumulates and eventually becomes pretty
large.
This is a reason why Textbook does not even discuss stability of solutions
which are not stationary, and we will not do it in the future (except if
specifically mentioned).
Example 9.2.3 (Anharmonic Oscillations with Liquid Friction). Let us add
a friction (a liquid one, with the force, proportional to the velocity and
directed against it): x′′ = −x3 − x − x′ .
Then equilibrium will be asymptotically stable, and the basin of attraction the whole plane.
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Chapter 9. Nonlinear Differential Equations and Stability
3
2
175
Example 9.2.4. f (x) = −x2 − x. Then V (x) = x3 + x2 :
It has a single minimum and a single maximum, and thus there are two
equilibrium points: (0, 0) which is stable and (−1, 0) which is unstable.
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Chapter 9. Nonlinear Differential Equations and Stability
old
176
Equilibrium (0, 0) is stable, and movement with the energy below thresh1
oscillates, while above escapes to x = −∞.
6
Example 9.2.5. Let us add a friction x′′ = −x2 − x − x′ .
Then equilibrium at (0, 0) will be asymptotically stable. Can you see
the basin of attraction of (0, 0)? It is bounded by two trajectories entering
into unstable equilibrium at (−1, 0).
2
4
Example 9.2.6 (Double Well). f (x) = x − x3 . Then V (x) = − x2 + x2 :
It has two minima and a single maximum, and thus there are three
equilibrium points: (−1, 0) an (1, 0) which are stable and (0, 0) which is
unstable.
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Chapter 9. Nonlinear Differential Equations and Stability
177
Equilibriums (±1, 0) are stable, and movement with the energy below
threshold 0 oscillates in it’s own small well, while above oscillates in the
large common well
Example 9.2.7. Let us add a friction x′′ = x − x3 − x′ .
Then equilibriums at (±1, 0) will be asymptotically stable. Can you
see the basin of attraction of each of them? They are separated by two
trajectories entering into unstable equilibrium at (0, 0).
Example 9.2.8.
In this case f (x) = −gl sin(x) where x is an angle and
V (x) = gl(1 − cos(x)) where we count potential relatively to
the bottom. Minima of V (x) are at
x = 2πn, and maxima at 2πn + π, n ∈ CZ, which correspond
to different angles but the same physical positions.
x
(a) So we see that really minimas of V (x) corresponds to centers and
maximas to saddles. Centers are stable equilibrium points and saddles
are unstable.
(b) But what are those lines which are going above and below and are
not closed? – Those correspond to enegies so high, that pendulum
goes through top point and not oscillates but rotates clockwise or
counter-clockwise.
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Chapter 9. Nonlinear Differential Equations and Stability
178
Example 9.2.9 (Mathematical Pendulum with Liquid Friction). Let us add
a liquid friction: x′′ = − sin(x) − x′ . Find the basin of attraction for each
stable equilibrium (physically they are the same, but the number of rotations
the pendulum does before entering the damped oscillations mode matters).
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Chapter 9. Nonlinear Differential Equations and Stability
9.3
179
Locally Linear Systems
9.3.1
Linearization
In this section we discuss a linearization or a linear approximation to
nonlinear systems, which is a very common and powerful tool. We start
from the general idea.
Let us consider a nonlinear system
x′ = f (x; t)
(9.3.1)
and let us know one solution x0 (t). Let us consider solutions x(t), close to
x0 (t): ∥x(t) − x0 (t)∥ ≤ ε ≪ 1. Then, assuming that f is twice continuously
differentiable,
x′ (t) = f (x0 (t); t) + Df (x0 (t); t)(x(t) − x0 (t)) + O(ε2 )
where Df (x0 (t); t) is a Jacobi matrix, that is

∂f1
∂f1
...
 ∂x1 ∂x2
 ∂f2 ∂f2
...

Df :=  ∂x. 1 ∂x. 2
 ..
.. . . .

∂fn
∂fn
...
∂x1
∂x2
(9.3.2)
a matrix of first derivatives:

∂f1
∂xn 
∂f2 
∂xn 
.
.. 
. 
(9.3.3)
∂fn
∂xn
Recall that O(ε2 ) means not exceeding M ε2 where M is some fixed constant
while ε → 0. Rewriting (9.3.2)
x′ (t) = f (x0 (t); t) + Df (x0 (t); t)(x(t) − x0 (t)) + O(ε2 )
(2)
and remembering that x0 (t) is a solution
x′0 (t) = f (x0 (t); t)
and subtracting we get
y ′ (t) = Df (x0 (t); t)y(t) + O(ε2 )
(9.3.4)
with y(t) := (x(t) − x0 (t)).
Finally, dropping O(ε2 ) we get a linearization of our nonlinear system
on solution x0 (t).
y ′ (t) = Df (x0 (t); t)y(t).
So, we got a linear homogeneous system (9.3.5).
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(9.3.5)
Chapter 9. Nonlinear Differential Equations and Stability
180
Main assumption. From now on we assume that the original system is
autonomous and x0 (t) = x0 is a stationary point.
Then (9.3.5) is a linear homogeneous system with constant coefficients.
Main question. What can we say about stationary point x0 from the
properties of matrix Df (x0 )?
9.3.2
Stability
Here we will give the answer for arbitrary n.
Theorem 9.3.1. Let x0 be a stationary point of the autonomous n×n-system
x′ (t) = f (x(t)),
f (x0 ) = 0.
(9.3.6)
Then
(i) If all eigenvalues of Df (x0 ) have negative real parts then x0 is asymptotically stable.
(ii) If at least one of eigenvalues of Df (x0 ) have a positive real part then
x0 is unstable.
We are not prove this (as any other) theorem of this section. At the
moment we consider only simple examples that if all eigenvalues have nonpositive real parts but at least one has a real part 0 then linearization does
not provide an answer.
Example 9.3.1. Consider n = 1 and equation x′ = f (x) := ax3 . Then x0 = 0
is a stationary point, f ′ (0) = 0.
(a) If a < 0 then x0 = 0 is asymptotically stable.
(b) If a > 0 then x0 = 0 is unstable.
(a) a < 0
(b) a > 0
Example 9.3.2. Consider n = 1 and equation x′ = f (x) := ax2 . Then x0 = 0
is a stationary point, f ′ (0) = 0.
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Chapter 9. Nonlinear Differential Equations and Stability
181
(a) If a < 0 then x0 = 0 is asymptotically stable from the right, unstable
from the left, and unstable in total.
(b) If a > 0 then then x0 = 0 is asymptotically stable from the left,
unstable from the right and unstable in total.
(a) a < 0
(b) a > 0
In some cases one-sided stability makes sense.
9.3.3
Classification: Robust Cases
Here we will give the answer for n = 2. No surprise that if the stationary
point 0 of the linear homogeneous system is of the robust type, then we can
give a definitive answer:
Theorem 9.3.2. Let x0 be a stationary point of the autonomous 2×2-system
x′ (t) = f (x(t)),
f (x0 ) = 0.
(9.3.6)
Let k1 and k2 be eigenvalues of Df (x0 ). Then
(i) If k1 < k2 < 0 then x0 is a stable proper node: all trajectories near
x0 tend to x0 as t → +∞, all trajectories save two enter into x0 along
directions ±ξ (1) and these two enter along directions ξ (2) and −ξ (2) , where
here and below ξ (1) and ξ (2) are corresponding eigenvectors.
(ii) If k1 > k2 > 0 then x0 is a unstable proper node: all trajectories near
x0 tend to x0 as t → −∞, all trajectories save two enter into x0 along
directions ±ξ (1) and these two enter along directions ξ (2) and −ξ (2) .
(iii) If k1 > 0 > k2 then x0 is a saddle: all trajectories near x0 save four
bypass x0 , two of remaining trajectories enter into x0 along directions ξ (2)
and −ξ (2) , and other two exit x0 along directions ξ (1) and −ξ (1) . It is also
unstable.
(iv) If k1,2 are complex conjugate eigenvalues, and Re(k1,2 ) < 0 then x0 is
a stable spiral point: all trajectories near x0 tend to x0 as t → +∞, and
make the infinite number of rotations.
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Chapter 9. Nonlinear Differential Equations and Stability
182
(v) If k1,2 are complex conjugate eigenvalues, and Re(k1,2 ) > 0 then x0 is a
unstable spiral point: all trajectories near x0 tend to x0 as t → +∞, and
make the infinite number of rotations.
Example 9.3.3.
(
x′ = −x + y 2 ,
y ′ = −2y + x2 .
Then for x0 = 0 we have Df (x0 ) =
k2 = −1, ξ (2) =
!
1
−1
0
!
0 −2
, k1 = −2, ξ (1) =
1
.
0
(a) Nonlinear system
!
0
(b) Linearization
Example 9.3.4.
(
x′ = x + y 2 ,
y ′ = −y + x2 .
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,
Chapter 9. Nonlinear Differential Equations and Stability
Then for x0 = 0 we have Df (x0 ) =
k2 = −1, ξ (2) =
!
0
1
0
!
0 −1
, k1 = 1, ξ (1) =
183
!
1
,
0
.
1
(a) Nonlinear system
(b) Linearization
Example 9.3.5.
(
x′ = −x + 2y + 20y 3
y ′ = −2x − y.
!
−1
2
Then for x0 = 0 we have Df (x0 ) =
, k1,2 = −1 ± 2i.
−2 −1
9.3.4
Classification: Fragile Cases
But what happens in fragile cases? We will consider
(a) Centers (k1,2 are purely imaginary)
(b) Straight nodes (k1 = k2 ̸= 0, and there are two linearly independent
eigenvectors).
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Chapter 9. Nonlinear Differential Equations and Stability
(a) Nonlinear system
184
(b) Linearization
(c) Improper nodes (k1 = k2 ̸= 0, and there is only one linearly independent eigenvector).
Centers can remain centers, or become spiral points (stable or unstable);
orientation, however, remains the same.
Example 9.3.6.
(
x′ = y + ax(x2 + y 2 ),
y ′ = −x + ay(x2 + y 2 ).
0 1
!
, k1,2 = ±i. For linearization
−1 0
we have just circles with clockwise orientation.
In the polar coordinates (r, θ) equations become r′ = ar3 , θ′ = −1 and we
see that there is a rotation with a speed −1 (thus clockwise) and movement
along radius; as a < 0 it is toward 0 , and as a > 0 it is away from 0 . Thus
we get spiral points but they are not like ones we had: before each rotation
added a constant factor to r, but now near 0 movement along r slows down,
so trajectories become more compressed.
Then for x0 = 0 we have Df (x0 ) =
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Chapter 9. Nonlinear Differential Equations and Stability
(a) a < 0
185
(b) a > 0
Example 9.3.7.
(
x′ = sin(y),
y ′ = − sin(x).
At 0 linearization is the same, center remains center but away from 0 the
lines become less and less circle-like.
But what about straight or improper nodes? Textbook (Table 9.3.1, on
page 412) says that they can remain nodes or become spiral points. However,
if f (x) is twice continuously differentiable (in fact much weaker assumption
is required) it is incorrect: node remains a node! –proper or improper.
Theorem 9.3.3. Let x0 be a stationary point of the autonomous 2×2-system
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x′ (t) = f (x(t)),
f (x0 ) = 0.
(9.3.6)
Chapter 9. Nonlinear Differential Equations and Stability
186
Assume that f (x) is twice continuously differentiable. Let k1 = k2 =
̸ 0 be
eigenvalues of Df (x0 ). Then
(i) If k1 = k2 < 0 and there is just one linearly independent eigenvector ξ (1) ,
then x0 is a stable improper node: all trajectories near x0 tend to x0 as
t → +∞, all trajectories enter into x0 along directions ±ξ (1) .
(ii) If k1 = k2 > 0 and there is just one linearly independent eigenvector
ξ (1) , then x0 is a unstable improper node: all trajectories near x0 tend to
x0 as t → −∞, all trajectories exit from x0 along directions ±ξ (1) .
(iii) If k1 = k2 < 0 and there are two linearly independent eigenvectors,
then x0 is a stable straight node: all trajectories near x0 tend to x0 as
t → +∞, and for each ξ ̸= 0 there is exactly one trajectory entering x0 in
the direction ξ.
(iv) If k1 = k2 > 0 and there are two linearly independent eigenvectors,
then x0 is a unstable straight node: all trajectories near x0 tend to x0 as
t → −∞, and for each ξ ̸= 0 there is exactly one trajectory exiting x0 in
the direction ξ.
Only for less smooth f (x) node can become a spiral:
Example 9.3.8.
(
x′ = −x − yσ(r)
y ′ = −y + σ(r)
with r =
Then for x0 = 0 we have Df (x0 ) =
−1
0
p
x2 + y 2 .
!
0 −1
, k1,2 = 1. For linearization
we have a stable straight node.
In the polar coordinates (r, θ) equations become r′ = r, θ′ = σ(r) and
the question is how many rotations will be made.
Since dθ
= r−1 σ(r) we need to calculate
dr
Z
I := r−1 σ(r) dr
0
and for σ(r) = | log(r)|−1 we get I = ∞. However function f (x) is only
once continuously differentiable!
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9.4
187
Competing Species and Other Examples
9.4.1
Competing Species
In this lecture we consider Competing Species Model (one of Volterra-Lotka
models) describing two species competing for food.
(
x′ = x(a1 − σ1 x − α1 y),
(9.4.1)
y ′ = y(a2 − σ2 y − α2 x).
Without “α”-terms it would be just two equations, describing two species
without competition. Obviously, we are interested only in x ≥ 0, y ≥ 0, and
all positive constants a1 , a2 , σ1 , σ2 , α1 , α2 .
Finding stationary points from
(
0 = x(a1 − σ1 x − α1 y),
0 = y(a2 − σ2 y − α2 x)
we have
a1 − σ1 x = 0, a1 − σ1 x − α1 y = 0,
x = 0,
x = 0,
y = 0,
a2 − σ2 y = 0,
y = 0,
a2 − σ2 y − α2 x = 0,
A0 (0, 0),
A2 (0, σ2−1 a2 )
A1 (σ1−1 a1 , 0);
A3 (x3 , y3 ).
where A0 , A1 , A2 belong to the border of the quadrant {x > 0, y > 0} and
A3 (x3 , y3 ),
x3 =
σ 2 a1 − α 1 a2
,
σ1 σ2 − α1 α2
y3 =
σ1 a2 − α2 a1
,
σ1 σ2 − α1 α2
(9.4.2)
may belong to this quadrant {x > 0, y > 0} or not.
We see that x = 0 and y = 0 are solutions; so we have the following
picture:
!
a1 − 2σ1 x − α1 y
−α1 x
Finding Jacobi matrix J =
and
−α2 y
a2 − 2σ2 y − α2 x
calculating it in stationary points
A0 (0, 0)
A2 (0, σ2−1 a2 )
A1 (σ1−1 a1 , 0)
A3 (x3 , y3 )


 
 
 
−1
−1
a
0
a − σ2 α2 a2
0
−a
−σ1 α1 a1
−σ x −α1 x3
 1
  1
  1
  1 3

−1
−1
0 a2
−σ2 α2 a2
−a2
0
a2 − σ1 α1 a1
−α2 y3 −σ2 y3
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Chapter 9. Nonlinear Differential Equations and Stability
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A2
A3 where are you?
A0
A1
Figure 9.10: A3 where are you?
and therefore eigenvalues are
A0 (0, 0)
a1 , a 2
A2 (0, σ2−1 a2 )
A1 (σ1−1 a1 , 0)
a1 − σ2−1 α2 a2 , −a2 −a1 , a2 − σ1−1 α1 a1 .
Therefore
(a) A0 is unstable node (good news: no total extinction)
(b) A1 may be either stable node or a saddle;
(c) A2 may be either stable node or a saddle.
Theorem 9.4.1. (i) A3 belongs to {x > 0, y > 0} if and only if both A1
and A2 are of the same type: stable nodes or saddles.
(ii) If both A1 and A2 are stable nodes, then A3 is a saddle.
(iii) If both A1 and A2 are saddles, then A3 is a stable node.
Proof. (i) Note that A3 is an intersection of two straight lines
a1 − σ1 x − α1 y = 0
a2 − σ2 y − α2 x = 0.
The first line connects (a1 /σ1 , 0) and (0, a1 /α1 ), while the second one
(a2 /α2 , 0) and (0, a2 /σ2 ).
The first two pictures correspond to a saddle and a stable node, the third
to two saddles and fourth to two nodes.
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Chapter 9. Nonlinear Differential Equations and Stability
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a1 /α1
a2 /σ2
a2 /σ2
a1 /α1
a2 /α2
a1 /σ1
a1 /α1
a2 /σ2
a2 /α2 a1 /σ1
a2 /σ2
a1 /α1
a1 /σ1 a2 /α2
a2 /α2 a1 /σ1
(
A2 : a1 − σ2−1 α2 a2 and − a2
Indeed, recall eigenvalues:
A1 : a2 − σ1−1 α1 a1 and a1 .


−σ1 x3 −α1 x3
, so characteristic equation is
Further, J (A3 ) = 
−α2 y3 −σ2 y3
k 2 + (a1 + a3 )k + (σ1 σ2 − α1 α2 )x3 y3 = 0.
(9.4.3)
because (σ1 x3 + σ2 y3 ) = a1 + a2 . It follows from
x3 =
σ2 a1 − α1 a2
,
σ1 σ2 − α1 α2
y3 =
σ1 a2 − α2 a1
.
σ1 σ2 − α1 α2
(ii) If both A1 and A2 are stable nodes then (σ1 σ2 − α1 α2 ) < 0 and the
roots of (9.4.3) are real of opposite signs. So, A3 is a saddle.
(iii) Finally, if both A1 and A2 are saddles then (σ1 σ2 − α1 α2 ) > 0, and the
eigenvalues are real of the same sign: indeed discriminant
(σ1 x3 + σ2 y3 )2 − 4(σ1 σ2 − α1 α2 )x3 y3
= (σ1 x3 − σ2 y3 )2 + 4α1 α2 x3 y3 > 0
Observe that solution (x(t), y(t) cannot escape to infinity because as
one can see easily x′ < 0, y ′ < 0 as x + y ≥ M with sufficiently large M .
Therefore there are four cases:
(a) A1 is a stable node and A2 is a saddle. Therefore all solutions tend to
A1 and species y become extinct.
(b) Symmetrically A2 is a stable node and A1 is a saddle. Therefore all
solutions tend to A2 and species x become extinct.
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(c) Both A1 and A2 are saddles and all solutions tend to A3 . Coexistence.
(d) Both A1 and A2 are stable nodes and all solutions tend to either of
them, one of species becomes extinct (depending on initial condition) and
we need to find basins of attractions A1 and A2 .
A2
A0
A1
Figure 9.11: Case 1: A1 is a stable node and A2 is a saddle
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A2
A0
A1
Figure 9.12: Case 2: A2 is a stable node and A1 is a saddle
A2
A3
A0
A1
Figure 9.13: Case 3: A1 and A2 are saddles, A3 is a stable node
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A2
Basin of A2
A3
Basin of A1
A0
A1
Figure 9.14: Case 4: A1 and A2 are stable nodes and A3 is a saddle
9.4.2
Examples
Consider examples, how they should be solved in the Final Assessment.
Example 9.4.1. For the system of ODEs
(
x′ = x(10 − 2x − 3y) ,
y ′ = y(3x + 2y − 12)
(a) Describe the locations of all critical points.
(b) Classify their types (including whatever relevant: stability, orientation,
etc.).
(c) Sketch the phase portraits near the critical points.
(d) Sketch the full phase portrait of this system of ODEs.
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Chapter 9. Nonlinear Differential Equations and Stability
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Solution. (a) Solving x(10 − 2x − 3y) = 0, y(3x + 2y − 12) = 0 we get cases
x = 0, y = 0
x = 0, 3x + 2y − 12 = 0
y = 0, 10 − 2x − 3y = 0
10 − 2x − 3y = 0, 3x + 2y − 12 = 0
=⇒ A1 = (0, 0),
=⇒ A2 = (0, 6)
=⇒ A3 = (5, 0),
16 6
=⇒ A4 = ( , ).
5 5
(b) Linearizations at these points have matrices
A1 = (0, 0) A2 = (0, 6)
A3 = (5, 0)
A4 = ( 16
, 6)
5 5
 
 
 

10
0
−8 0
−10 −15
− 32 − 48
5 
 
 
  5

18
12
0 −12
18 12
0
3
5 q 5
{−8, 12}
{−10, 3}
{−2 ± 76
{10, −12}
i}
5
(below are eigenvalues), the last one is from characteristic equation k 2 + 4k +
96
= 0.
5
(c) Therefore A1 , A2 and A3 are saddles, A4 is a focal stable point and
since the bottom-left number is positive, it is counter-clockwise oriented.
Directions are
A1 : ξ (1) = (1, 0)T –unstable, ξ (2) = (0, 1)T –stable (k1 > 0 > k2 );
9 T
A2 : ξ (1) = (1, − 10
) –stable, ξ (2) = (0, 1)T –unstable (k1 < 0 < k2 ).
A3 : ξ (1) = (1, 0)T –stable, ξ (2) = (1, − 13
)T (k1 < 0 < k2 )–untsable.
15
(d) One should observe that either x = 0 in every point of the trajectory, or
in no point; and that y = 0 in every point of the trajectory, or in no point.
It allows us to make a “skeleton” of the phase portrait.
Also, draw trajectories near saddles (do it by yourself!)
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A2
A4
A1
A3
A2
A4
A1
9.5
9.5.1
A3
Integrable Systems and Predator–Prey
Equations
Integrable Systems: General
In this section we discuss integrable 2 × 2 systems, their connection to what
we considered before, predator–prey equations and mechanical oscillations.
So, let us consider
(
x′ =f (x, y),
(9.5.1)
y ′ =g(x, y)
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(it is more convenient to use coordinate form here). Let us rewrite it
(excluding t) as
−g(x, y) dx + f (x, y) dy = 0.
(9.5.2)
Definition 9.5.1. System (9.5.1) is integrable if there exists an integrating
factor µ(x, y) such that after multiplication by it (9.5.2) becomes exact, that
is there exists H(x, y) such that
−µg = Hx ,
µf = Hy .
(9.5.3)
Let us consider stationary point of (9.5.1), that is (x0 , y0 ) such that
f (x0 , y0 ) = g(x0 , y0 ) = 0
(9.5.4)
µ(x0 , y0 ) ̸= 0.
(9.5.5)
assuming that
Then (x0 , y0 ) is a a critical point of H(x, y) (and under assumption (9.5.5)
conversely) and Jacoby matrix is
!
!
fx fy
Hxy
Hxx
(x0 , y0 ) = µ(x0 , y0 )
(x0 , y0 ).
(9.5.6)
gx gy
−Hyy −Hxy
Therefore for integrable systems eigenvalues are k1 and k2 = −k1 , either
real or purely imaginary, which excludes nodes and “normal” spiral points
and leaves us with saddles and centers (assuming that k1,2 ̸= 0).
Remark 9.5.1. For general systems linearization does not give a definitive
answer, if it is a center, or a “strange” spiral point, but integrable systems
are different!
Indeed, if Hessian matrix
Hess(H) :=
Hxx Hxy
!
(9.5.7)
Hxy Hyy
at (x0 , y0 ) is either positive or negative definite, that is
2
−∆ := Hxx Hyy − Hxy
>0
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(9.5.8)
Chapter 9. Nonlinear Differential Equations and Stability
196
then H(x, y) has either minimum or maximum at (x0 , y0 ) and the level lines
of it are those of the center, and if Hessian matrix is indefinite, that is
2
−∆ = Hxx Hyy − Hxy
<0
(9.5.9)
it is a saddle. One can see easily, that
(a)√in the case (9.5.8) eigenvalues of Df (x0 , y0 ) are purely imaginary k1,2 =
±i −∆, and
√
(b) in the case (9.5.9) eigenvalues of Df (x0 , y0 ) are real k1,2 = ± ∆.
9.5.2
Predator–Prey Model
Predator–prey mode is another Volterra-Lotka models
(
x′ = x(a − αy),
y ′ = y(βx − b)
(9.5.10)
where constants a, b, α, β are positive and we consider only x ≥ 0, y ≥ 0.
Let us find stationary points: x = 0 or a − αy = 0 and y=0 and
βx − y = 0. One can see easily that only two cases are possible: x = y = 0
and a − αy = βx − b = 0 which leaves us with two points: x0 = y0 = 0
(extinction) and x0 = β −1 b, y0 = α−1 a (equilibrium).
However, let us first check that system (9.5.10) is integrable. Rewriting
it as
−y(βx − b) dx + x(a − αy) dy = 0
we observe that we can separate variables
−
(βx − b)
(a − αx)
dx +
dy = 0.
x
y
so it is integrable. We actually integrated it in W2L1.
Here, however,!we simply observe that at stationary point (0, 0) Jacobi
matrix is
a
0
and it is a saddle, with two integral lines passing through
0 −b
it x = 0 and y = 0, and in (β −1 b, α−1 a) Jacobi matrix is
0
−αβ −1 b
βα−1 a
0
and it is a center with counter-clockwise rotation.
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Chapter 9. Nonlinear Differential Equations and Stability
197
(a) Note that small oscillations near equilibrium look like ellipses with
axes parallel to coordinate axes; explain why, calculate the ratio of axes of
these ellipses and find the angular frequency.
(b) Read more in the Textbook.
9.5.3
Mechanical Oscillations
Recall that 1-dimensional mechanical oscillations without friction are described by
x′′ = f (x)
(we assume that m = 1) which can be reduced to the system
(
x′ = y,
y ′ = −f (x)
(9.5.11)
(9.5.12)
and integrated to
my 2
+ V (x) = E
2
(9.5.13)
with potential
Z
V (x) = −
x
f (x) dx
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(9.5.14)
Chapter 9. Nonlinear Differential Equations and Stability
198
and full energy E.
Then stationary points are (x0 , 0) with V ′ (x0 ) = f (x0 ) = 0 and x0 is a
stable equilibrium points if x0 is minimum for V (x), that means V ′′ (x0 ) > 0,
and x0 is unstable equilibrium points if x0 is a maximum for V (x), that means
V ′′ (x0 ) < 0; (we assume that critical points of V (x) are non-degenerate).
Small oscillations near stable equilibrium points are “almost harmonic”
and described
by (x − x0 )′′ = −V ′′ (x0 )(x − x0 ) with an angular frequency
p
′′
ω ≈ V (x0 ) and a period T ≈ 2π
= √ 2π
.
ω
′′
V (x0 )
However, larger oscillations are anharmonic and they have periods
Z x2 (E)
dx
p
T = T (E) = 2
(9.5.15)
2(E − V (x))
x1 (E)
where x1 (E) and x2 (E) are the left and right end of interval where a particle
is oscillated:
(
V (x1 (E)) = V (x2 (E)) = E
and
(9.5.16)
V (x) < E ∀x : x1 (E) < x < x2 (E).
Indeed,
p
x′2
dx
+ V (x) = E =⇒ x′ = ± 2(E − V (x)) =⇒ dt = p
2
2(E − V (x))
which implies (9.5.15) because particle runs from x1 (E) to x2 (E) and back.
Remark 9.5.2. (a) This is improper integral because integrand is ∞ on both
ends. However it converges if V ′ (x1 (E)) < 0 and V ′ (x2 (E)) > 0.
Indeed, in this case |E−V (x)| ≥ ϵ|x−x1 (E)| near x1 (E) and |E−V (x)| ≥
ϵ|x − x2 (E)| near x2 (E) with ϵ > 0.
(b) On the other hand, if V (x) is twice continuously differentiable and
either V ′ (x1 (E)) = 0 or V ′ (x2 (E)) = 0 this integral diverges: particle moves
infinitely long toward unstable equilibrium.
Indeed, in this case |E −V (x)| ≤ C|x−x1 (E)| near x1 (E) or |E −V (x)| ≤
C|x − x2 (E)| near x2 (E).
2
Example 9.5.1 (Double Well). f (x) = x − x3 and V (x) = − x2 +
Here we consider the cases when particle is confined to
(a) the large common well or
(b) to the small left or right well.
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x4
:
4
Chapter 9. Nonlinear Differential Equations and Stability
199
V (x)
E
x1 (E) x2 (E)
x1 (E)
E
x1 (E)
x2 (E)
E
x
x2 (E)
Figure 9.15: Double well
9.5.4
Examples
Consider examples, how they should be solved in the Final Assessment.
Example 9.5.2. For system of ODEs
(
x′ = x2 y − 3y 2 + 2x ,
y ′ = −xy 2 + 3x2 − 2y .
(a) Find stationary points.
(b) Linearize the system at stationary points and sketch the phase portrait
of this linear system.
(c) Find the equation of the form H(x, y) = C, satisfied by the trajectories
of the nonlinear system.
(d) Sketch the full phase portrait.
Solution. (a) Solving
x2 y − 3y 2 + 2x = 0,
−xy 2 + 3x2 − 2y = 0
we multiply the these equations by y, x respectively and add: 3x3 − 3y 3 =
0 =⇒ x = y. Then x3 − 3x2 + 2 = 0 results in
x = 0, x = 1, x = 2 and A1 (0, 0), A2 (1, 1), A3 (2, 2) correspondingly.
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(b) Linearizations at these points have matrices
A1 = (0, 0)
!
2
0
A2 = (1, 1)
!
4 −5
0 −2
5 −4
A3 = (2, 2)
!
10 −2
2
correspondingly with eigenvalues
√
√
{2, −2}
{− 41i, 41i}
−10
√ √
{− 96, 96}.
Therefore A1 and A3 are saddles, A2 is either center or focus and since
bottom left 5 > 0 it is counter-clockwise oriented.
(c) Directions are
A1 : ξ (1) = (1, 0)T –unstable direction, ξ (2) = (0, 1)T –stable direction
(since k1 > 0 > k2 ).
√
√
A3 : ξ (1) = (1, 5 − 24)T –stable direction, ξ (2) = 5 − 24, 1)T –unstable
direction.
(d) It allows us to make a “skeleton” of the phase portrait (thick black
lines) on the figure; red lines are very approximate.
A3
A2
A1
(e) Rewriting equation as
(−xy 2 + 3x2 − 2y) dx − (x2 y − 3y 2 + 2x) dy = 0
⇐⇒ (−xy 2 + 3x2 − 2y) dx + (−x2 y + 3y 2 − 2x) dy
one can check that it is exact. Then
1
Hx = −xy 2 + 3x2 − 2y =⇒ H(x, y) = − x2 y 2 + x3 − 2xy + h(y)
2
?
=⇒ Hy = −x2 y − 2x + h′ (y) = −x2 y + 3y 2 − 2x =⇒ h′ (y) = 3y 2
=⇒ h(y) = y 3
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(we do not need the constant here) and solution is
1
H(x, y) = − x2 y 2 + x3 + y 3 − 2xy = C
2
Therefore (1, 1) must be center. One can check easily that it is a local minimum
of H(x, y) but we do not need it.
A3
A2
A1
9.6
9.6.1
Lyapunov’s Second Method
General
In this section we cover Lyapunov’s Second Method of establishing stability
or instability and using it we justify Lyapunov’s First Method (based on
eigenvalues of the Jacobi matrix). While more general and more powerful,
the Second Method is also more difficult: it requires a construction of some
auxiliary function V (x) which reaches a strict minimum in x0 and decreases
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Chapter 9. Nonlinear Differential Equations and Stability
202
(or not increases) along trajectories–in the proof of asymptotic stability and
stability.
On the other hand, this function should have a some kind of the the
opposite properties in the proof of instability.
In many physical problems this function is just an energy but there are
plenty of problems in which energy is not defined, or even if it is V (x) differs
from it.
Theorem 9.6.1 (Stability Theorem). Consider autonomous system
x′ (t) = f (x(t))
(9.6.1)
with f (x) continuous in domain D ⊂ Rn . Let x0 ∈ D be a stationary point:
f (x0 ) = 0 .
(i) Assume that there exists function V (x), continuously differentiable in
D, having a strict minimum at x0 and such that
f (x) · ∇V (x) ≤ 0
∀x ∈ D.
(9.6.2)
Then x0 is a stable stationary point.
(ii) Assume further, that inequality is strict except x0 :
f (x) · ∇V (x) < 0
∀x ∈ D, x ̸= x0 .
(9.6.3)
Then x0 is an asymptotically stable stationary point.
Remark 9.6.1. (a) Conditions are very weak: f is only continuous which is
insufficient for unicity.
(b) Since x0 is minimum of V (x) then ∇V (x0 ) = 0 , so f (x0 ) · ∇V (x0 ) = 0.
Proof. Consider V (x(t)). Then according to chain rule
d
V (x(t))) = ∇V (x(t)) · x′ (t) = f (x) · ∇V (x).
dt
(9.6.4)
Therefore conditions (9.6.2) and (9.6.3) mean correspondingly exactly
that V (x(t)) does not increase and strongly decays (except x0 ) along trajectories (we are considering here and below the small vicinity of x0 ).
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203
Observe that for any ε > 0 there exists κ > 0 such that
{x : V (x) ≤ V (x0 ) + κ} ⊂ {x : ∥x − x0 ∥ < ε}
and for any κ > 0 there exists δ > 0 such that
{x : ∥x − x0 ∥ < δ} ⊂ {x : V (x) ≤ V (x0 ) + κ}.
Indeed, it due to the fact that x0 is a point of the strict minimum:
V (x) < V (x0 + κ)
x0
Due to (9.6.2) V (x(t)) does not increase and once
in {x : V (x) < V (x0 ) + κ} is trapped there in the
future. Therefore, for any ε > 0 there exists δ > 0
such that ∥x(t0 ) − x0 ∥ < δ =⇒ ∥x(t) − x0 ∥ < ε
for all t > t0 . Stability is proven!
To prove asymptotic stability observe that x(t) cannot be too long in
{x : V (x0 ) + κ/2 < V (x)}.
Indeed in
Ω := {x : V (x0 ) + κ/2 ≤ V (x) ≤ V (x0 ) + κ}
function f (x) · ∇V (x) ≤ −σ for some σ > 0 because as continuous function
it is reaches maximum here and due to (9.6.3) it is negative (here we need a
strict inequality).
Therefore if x(t) is in Ω for t : t0 ≤ t ≤ t0 + T with T = κ/σ, then
(V (x(t)))′ ≤ −σ and V (x(t0 +T )) ≤ V (x(t0 ))−σT ≤ V (x0 ) –contradiction!
Therefore x(t) will be eventually falling into {x : V (x) ≤ V (x0 )) + κ′ }
with arbitrarily small κ′ > 0 and thus into {x : ∥x−x0 ∥ < ε′ } with arbitrarily
small ε′ > 0, and this is asymptotic stability!
Theorem 9.6.2 (Instability Theorem). Consider autonomous system
x′ (t) = f (x(t))
(9.6.1)
with f (x) continuous in domain D ⊂ Rn . Let x0 ∈ D be a stationary point:
f (x0 ) = 0 .
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(i) Assume that there exists function V (x), continuously differentiable in
D, and such that
f (x) · ∇V (x) ≥ 0
∀x ∈ D.
(9.6.5)
Assume that in any vicinity of x0 there exists x such that V (x) > V (x0 )
then x0 cannot be asymptotically stable.
(ii) Assume further, that inequality is strict except x0 :
f (x) · ∇V (x) > 0
∀x ∈ D, x ̸= x0 .
(9.6.6)
Then x0 is an unstable stationary point.
Proof. Proof is based on the same idea as the proof of Theorem 9.6.1, but
now condition (9.6.5) means that V (x(t)) does not decay and condition
(9.6.5) means that V (x(t)) strictly increases.
Theorem 9.6.3. Consider autonomous system (9.6.1) and assume as before
that f is continuous in D. Assume that there exists V (x), continuously
differentiable, and such that {x : V (x) ≤ K} is a closed subset D.
(i) Assume further, that (9.6.2) holds: f (x) · ∇V (x) ≤ 0 for all x ∈ D.
Then once captioned to {x : V (x) ≤ K} , x(t) remains here in the future.
(ii) Furthermore, assume that x0 is a stationary point, V (x) reaches a
strong minimum in x0 and that (9.6.3) holds: f (x) · ∇V (x) < 0 for all
x ∈ D, except x0 .
Then once captioned to {x : V (x) ≤ K} , x(t) → x0 as t → +∞ (that
means that {x : V (x) ≤ K}) belongs to the basin of attraction of x0 ).
Proof. Proof just repeats the proof of Theorem 9.6.1 (but now we are not
confined to a small vicinity of x0 ).
9.6.2
Justification of Lyapunov’s First Method
So, assume that f is continuously differentiable and f (x0 ) = 0 . Then
f (x) = J (x − x0 ) + o(∥x − x0 ∥)
with J = Df (x0 ). Recall that g = o(h) means that
g
h
(9.6.7)
→ 0 as x → x0 .
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Assume for simplicity that there is a complete set of eigenvectors of J .
Then in the coordinate system, consisting of eigenvectors ξ (j) corresponding
to real eigenvalues αj and Re(ξ (j) ) and Im(ξ (j) ) for eigenvectors ξ (j) corresponding to complex eigenvalues αj ± iβj matrix J has a block-diagonal
form:, consisting of 1 × 1 blocks αj corresponding to real eigenvalues and
!
αj βj
2 × 2 blocks
corresponding to complex eigenvalues.
−βj αj
(a) Assume that αj < 0 for all j. Consider
i
X
Xh
V (x) =
(x(j) − x(j) )2 +
(x(j) − x(j) )2 + (y (j) − y (j) )2
j
j
where the first sum corresponds to real eigenvalues, and the second to
complex ones.
Then
J (x − x0 ) · (x − x0 )
i
X
X h
(j)
=
αj |x(j) − x0 |2 +
αj (x(j) − x(j) )2 + (y (j) − y (j) )2
j
j
≤ −σ∥x − x0 ∥2
with −σ = maxj αj and therefore
σ
f (x) · ∇V (x) ≤ −σ∥x − x0 ∥2 + o(∥x − x0 ∥2 ) ≤ − ∥x − x0 ∥2
2
in the small vicinity of x0 .
Statement (ii) of Theorem 9.6.1 implies asymptotic stability.
(b) Assume that among αj are positive. Assume for simplicity that αj ̸= 0.
Consider
i
X
X h
(j)
(j)
(j)
V (x) =
ϵj (x(j) − x0 )2 +
ϵj (x(j) − x0 )2 + (y (j) − y0 )2
j
j
where the first sum corresponds to real eigenvalues, and the second to
complex ones and ϵj = ±1 if αj ≷ 0 respectively. Then
J (x − x0 ) · (x − x0 )
h
i
X
X
(j)
(j)
(j)
=
ϵj αj |x(j) − x0 |2 +
ϵj αj (x(j) − x0 )2 + (y (j) − y0 )2
j
j
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≥ σ∥x − x0 ∥2
Chapter 9. Nonlinear Differential Equations and Stability
206
with σ = minj |αj | and therefore
f (x) · ∇V (x) ≥ σ∥x − x0 ∥2 − o(∥x − x0 ∥2 ) ≥
σ
∥x − x0 ∥2
2
in the small vicinity of x0 .
Statement 2 of Theorem 9.6.2 implies instability.
9.6.3
Examples
Example 9.6.1. Consider mechanical oscillation x′′ = f (x) with potential
W (x) : f (x) = −W ′ (x) having isolated minimum at x0 . Reduce it to the
first order system setting y = x′ .
2
Note that V (x, y) := y2 + W (x) is preserved along trajectories, Statement 1 of Theorem 9.6.1 proves that (x0 , 0) is stable, and Statement 1 of
Theorem 9.6.2 implies that (x0 , 0) is not asymptotically stable.
Example 9.6.2.
(
x′ = −y + ax3 ,
y ′ = x + by 3
Then V (x, y) = x2 + y 2 and multiplying equations by 2x and 2y respectively
and adding, we get (V (x, y))′ = 2ax4 + 2by 4 .
Then
(a) Statement (ii) of Theorem 9.6.1 implies asymptotic stability of (0, 0) if
a < 0, b < 0.
(b) Statement (ii) of Theorem 9.6.2 implies asymptotic instability of (0, 0)
if a > 0, b > 0.
To investigate when a and b have different signs, let us take V (x, t) =
x2 + y 2 + cx3 y + dxy 3 . Then
U := (V (x, y))′ =2ax4 + 2by 4 − cx4 + 3(c − d)x2 y 2 + dy 4 + O(x6 + y 6 )
=(2a − c)x4 + 3(c − d)x2 y 2 + (2b + d)y 4 + O(x6 + y 6 ).
(c) If a + b < 0 let us take c = 2a + ϵ, d = −2b − ϵ with ϵ > 0 and then
U = −ϵx4 + 6(a + b − ϵ)x2 y 2 − ϵy 4 + O(x6 + y 6 ) ≤ −ϵ(x4 + y 4 )
and Statement (ii) of Theorem 9.6.1 implies asymptotic stability of (0, 0).
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(d) If a + b > 0 let us take c = 2a + ϵ, d = −2b − ϵ with ϵ < 0 and then
U ≥ −ϵ(x4 + y 4 ) and Statement (ii) of Theorem 9.6.2 implies instability of
(0, 0).
(e) Finally, as b = −a equation
(x − ay 3 ) dx + (y − ax3 ) dy = 0
x2 + y 2
2(1 + axy)2
and therefore (0, 0) is a center which is stable but not asymptotically stable.
is integrable with integrating factor (1 + axy)−3 and H(x, y) =
Figure 9.16: a = 1, b = −2
Example 9.6.3 (complicated, optional). Consider mechanical oscillation
x′′ = f (x) − Φ(x′ )x′ with potential W (x) : f (x) = −W ′ (x) having isolated
minimum at x0 and friction Φ(x′ )x′ where Φ(y) is continuous and nonnegative. Reduce it to the first order system setting y = x′ .
2
Note that V (x, y) := y2 + W (x) does not increase along trajectories
d
V (x, y) = −Φ(y)y 2 ≤ 0.
dt
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208
Figure 9.17: a = 2, b = −1
Figure 9.18: a = 1, b = −1
Theorem 9.6.1 implies that (x0 , 0) is stable. It does not imply “out of the
box” that (x0 , 0) is asymptotically stable even if Φ(y) is strictly positive but
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209
there is a trick.
Assume that Φ(y) > 0 as y ̸= 0 and that V ′ (x) ≷ 0 as x ≶ x0 .
To prove asymptotic stability we need to show that W (x(t), y(t)) →
V (x0 ) as t → +∞. If it is not so, then there exists κ > 0 such that
W (x(t), y(t)) ≥ V (x0 ) + κ for all t > 0 (because W (x(t), y(t)) does not
increase).
Observe that (x(t), y(t)) goes from (x(tk ), 0) to x(tk+1 , 0) where x(tk ) > 0
for odd k and x(tk ) < 0 for even k and since V (x(tk )) = W (x(tk ), 0) ≥
V (x0 ) + κ, there exists δ > 0 such that |x(tk ) − x0 | ≥ δ for all k. Therefore,
x(t) crosses I := (x0 − δ/2, x0 + δ/2) infinite number of times and spend an
infinite time (in total) there.
However, it is impossible: since W (x(t), y(t)) ≥ V (x0 ) + κ, x(t) ∈ I =⇒
|y| ≥ σ for some σ > 0 and then ddt W (x(t), y(t)) ≤ −ε for some ε > 0 and
for large time W (x(t), y(t) would drop below V (x0 ).
9.7
9.7.1
Periodic Solutions and Limit Cycles
Examples
In this section we discuss periodic solutions of non-linear autonomous 2 × 2systems
x′ (t) = f (x(t)).
(9.7.1)
Solution x(t) of (9.7.1) is periodic with period T ̸= 0 if
x(t + T ) = x(T )
∀t.
(9.7.2)
Remark 9.7.1. Sure, if T is period then any multiple of T is also a period.
We are interested in the minimal period, that is the smallest of all (positive)
periods.
So far we have seen periodic trajectories around a center, in which case
all trajectories close to it were periodic; for linear systems they had the
same period while for nonlinear system it was not usually the case. However,
there could exist isolated periodic trajectories.
Example 9.7.1.
(
x′ = εx(1 − x2 − y 2 ) − y,
y ′ = εy(1 − x2 − y 2 ) + x,
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(9.7.3)
Chapter 9. Nonlinear Differential Equations and Stability
which in polar coordinates is really simple
(
r′ = εr(1 − r2 ),
θ′ = 1.
210
(9.7.4)
Therefore it rotates with the constant speed around 0 and dynamic by r is
described by the first equation r′ = r(1 − r2 ):
(a) ε = 1
(b) ε = −1
In this case r = 0 and r = 1 are stationary points (but only for equation
r′ = εr(1 − r2 )), respectively stable and unstable. For ε = −1 situation is
reversed.
(a) ε = 1
(b) ε = −1
In both case we see a single periodic trajectory (also called a cycle) and
a single stationary point inside of it; as ε = 1 the cycle is stable (that means
that all other close trajectories approach it, while winding up around it)
and the stationary point inside is an unstable spiral point; as ε = 1 the
cycle is unstable (that means that all other close trajectories go away from
it, while winding up around it) and the stationary point inside is a stable
spiral point.
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211
Remark 9.7.2. (a) They do not look like spirals because they expand or
collapse too fast in comparison with rotation, but try ε = ±.2!
(b) Such cycles may be also semi-stable, more precisely, either stable from
inside, unstable from outside or ustable from inside, stable from outside.
Example 9.7.2. Let us try the system
(
(
p
x′ = xf ( x2 + y 2 ) − y,
r′ = f (r),
p
⇐⇒
θ′ = 1.
x′ = yf ( x2 + y 2 ) + x,
(9.7.5)
with f (r) having r0 ̸= 0 as a semi-stable stationary point: either f (r) > 0
for both r ≶ r0 or f (r) < 0 for both r ≶ r0 . For example f (r) = εr|1 − r2 |.
(a) ε = 1
9.7.2
(b) ε = −1
Theorems
We call such isolated cycles, which are approached by trajectories from
inside and outside as t goes to either +∞, or to −∞ limit cycles.
Theorem 9.7.1. Let f (x) be continuously differentiable in the simplyconnected domain D in R2 . Then
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212
(i) A closed trajectory of the system
x′ (t) = f (x(t))
(9.7.1)
must necessarily enclose at least one stationary point.
(ii) If it encloses only one stationary point, this stationary point cannot be
a saddle point.
Remark 9.7.3. We do not provide proof which is is from domain of smooth
2-dimensional topology.
For example, the proof of Statement 1 It is based on the fact that one
cannot comb the disc so that the “hair” would be tangent to its boundary:
if you have a smooth vector field f (x) in the disc, and it is tangent to its
boundary, it must vanish somewhere in the disc.
Theorem 9.7.2. Let f (x) = (f (x, y), g(x, y))T be continuously differentiable
in the simply-connected domain D in R2 .
If fx + gy has the same sign throughout D, then there is no closed
trajectory of the system (9.7.1) lying entirely in D.
Proof. The proof is based on Green’s formula so it is accessible only to those
who already finished Calculus II.
Assume that there is a closed trajectory γ. Denote its interior by Ω.
Applying Green’s formula
I
ZZ
f (x(s)) · n(x(s)) ds =
(fx + gy ) dxdy
γ
Ω
where we assume that γ is counter-clockwise oriented, ds is the element of
the length, and n is a unit exterior normal to γ.
Since γ is an integral curve of f , f is tangent to it in every point and
f (x(s)) · n(x(s)) = 0 and integral on the left is 0.
However assumption that either fx + gy > 0 everywhere in Ω or fx +
gy < 0 everywhere in Ω implies that integral on the right is > 0 or < 0
correspondingly. Contradiction.
Theorem 9.7.3 (Poincaré-Bendixon Theorem). Let f (x) be continuously
differentiable in the domain D in R2 . Let D′ be a bounded subdomain in D,
and let R be the region that consists of D′ plus its boundary. Assume that
R ⊂ D.
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213
Suppose that R contains no stationary point of the system (9.7.1). If
there is a solution of the system (9.7.1) that exists and stays in R for all
t ≥ t0 (for some constant t0 ), then either this solution itself is a periodic
solution (closed trajectory), or this solution spirals toward a closed trajectory
as t → +∞.
In either case, the system (9.7.1) has a periodic solution in R.
Again, the proof which uses topological properties of the plane, is beyond
our reach.
9.7.3
van der Pol’s Equation
As an example we consider van der Pol’s equation
u′′ − µ(1 − u2 )u′ + u = 0.
(9.7.6)
Remark 9.7.4. This equation describes a triode, an electronic amplifying
vacuum tube (or valve in British English) consisting of three electrodes
inside an evacuated glass envelope: a heated filament or cathode, a grid, and
a plate (anode). They were widely used in electronics (together with less
complicated diode (without grid) and more complicated tubes, with more
grids, but were replaced by semiconductors except where the high power is
needed. Those electronic tubes are remote relative of incandescent electric
lamps.
Here µ is a parameter. As µ = 0 we get a harmonic oscillator u′′ + u = 0.
For µ > 0 the second term on the left-hand side of equation (9.7.6) must
also be considered. This is the resistance term, proportional to u′ , with a
coefficient −µ(1 − u2 ) that depends on u:
(a) For large u, this term is positive and acts as usual to reduce the amplitude
of the response (negative feedback).
(b) However, for small u, the resistance term is negative and so causes the
response to grow (positive feedback).
This suggests that perhaps there is a solution of intermediate size that other
solutions approach as t increases. To analyze equation (9.7.6), we write it
as a system of two equations by introducing the variables x = u and y = u′ :
(
x′ = y,
(9.7.7)
y ′ = −x + µ(1 − x2 )y.
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Chapter 9. Nonlinear Differential Equations and Stability
The only stationary point is (0, 0) with Jacobi matrix
characteristic equation k 2 −µk+1 = 0 and eigenvalues k1,2
Then
214
!
0 1
with
−1 µ
p
= 12 (µ± µ2 − 4).
(a)t As 0 < µ < 2 it is unstable spiral point;
(b)t As µ = 2 it is unstable improper node;
(c)t As µ > 2 it is unstable proper node.
Such change of the type is usually called a bifurcation.
(a) Theorem 9.7.1 implies that if there is a closed trajectory, it must
enclose the origin.
(b) Since f = y and g = −x + µ(1 − x2 )y we have fx + gy = µ(1 − x2 )
and it is > 0 as |x| < 1, Theorem 9.7.2 implies that the closed trajectory
cannot be entirely in the vertical strip {(x, y) : |x| < 1}.
(c) Finally, Poincaré-Bendixon Theorem shows that there is a closed
trajectory (however analysis is complicated).
We do a numerical simulation:
(a) µ = .25
(b) µ = .5
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Chapter 9. Nonlinear Differential Equations and Stability
(a) µ = 1
(b) µ = 1.5
(c) µ = 2
(d) µ = 2.5
(e) µ = 5
(f) µ = 10
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215
Chapter 9. Nonlinear Differential Equations and Stability
9.7.4
216
More Examples
Example 9.7.3 (Nested Limit Cycles).
(
r′ = εr cos(r),
θ′ = 1.
There are limit cycles r =
unstable.
(a) ε = 1
π
2
+ πn. Find which are stable, and which are
(b) ε = −1
Example 9.7.4 (Several stationary points inside a limit cycle). We see a limit
cycle, and two attractive and one saddle point inside
(
x′ = y − x(9 − x2 − y 2 ) + 20x(1 + x2 + y 2 )−2 ,
y ′ = −x − y(9 − x2 − y 2 )
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217
9.8 Chaos and Strange Attractors:
the Lorenz Equations
9.8.1
The Lorenz Equations
In this section we discuss what may happen in dimension greater than
2. In early 1960-th E. N. Lorenz, American meteorologist, was computer
simulating what was supposed to be a toy-model of the Physics of Atmosphere. He considered a nonlinear autonomous 3 × 3 system with positive
coefficients σ, r, b
 ′
=: f (x, y, z),

x = −σx + σy
′
y = rx − y − xz =: g(x, y, z),
(9.8.1)

 ′
z = −bz + xy
=: h(x, y, z),
now called Lorenz equations.
Let us find stationary points x = (x, y, z)T and calculate Jacobi matrix
J in them. Stationary points are defined by

= 0,

−σx + σy
rx − y − xz = 0,
(9.8.2)

−bz + xy
= 0.
Then x = y and plugging y = x into remaining two we get
(
x(r − 1 − z) = 0 =⇒ either x = 0 or z = r − 1,
−bz + x2 = 0.
(9.8.3)
- x = 0 =⇒ y = 0, z = 0 and P 0 (0, 0, 0).
- z = r − 1 =⇒ x2 = b(r − 1) which has
(a) no solutions as r < 1 or
p
1),
(b) solutions xp= ± b(r −p
P 1,2 = (± b(r − 1), ± b(r − 1), r − 1) as r ≥ 1. For r = 1 they
coincide.
Such change is called bifurcation and it happens rather often.
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Chapter 9. Nonlinear Differential Equations and Stability
Next calculate Jacobi matrix at these points. It is equal

 

f f f
−σ
σ
0
 x y z 




J :=  gx gy gz  = r − z −1 −x
.
hx hy hz
y
x −b
Then


−σ σ
0



J (P 0 ) = 
r
−1
0


0
0 −b
218
(9.8.4)
(9.8.5)
with eigenvalues k3 = −b and eigenvalues of 2 × 2 block, found from
k 2 + (σ + 1)k + σ(1 − r) = 0
and since discriminant (1 + σ)2 − 4σ(1 − r) = (1 − σ)2 + 4σr > 0
(a) As r < 1 they are real and negative,
(b) As r > 1 they are real and opposite signs.
And here we have bifurcation as well: for r < 1 point P 0 is a stable
3D-node, and as r > 1 it is an unstable saddle with two negative and one
positive eigenvalues.
p
p
Further consider P 1,2 = (± b(r − 1), ± b(r − 1), r − 1) with r > 1:
J (P 1,2 )



−σ
σ
0
−σ
σ
0



p

=
1
−1
∓ b(r − 1)
r − z −1 −x 
.
p
p
y
x −b
± b(r − 1) ± b(r − 1)
−b
(9.8.6)
So, we need to calculate eigenvalues, and to write a characteristic polynomial (ouch!): but anyway:
k 3 + (b + σ + 1)k 2 + b(σ + r)k − 2σb(r − 1) = 0.
(9.8.7)
One of the roots is real for sure but more detailed investigation shows
that there are 2 other critical values for r:
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Chapter 9. Nonlinear Differential Equations and Stability
219
(b) As 1 < r < r1 there are three distinct real eigenvalues.
(c) As r1 < r < r2 there are one negative real eigenvalue and two complexconjugate eigenvalues with negative real parts.
(d) As r > r2 there are one negative real eigenvalue and two complexconjugate with positive real parts.
So let us consider different cases:
(a) Case r < 1 is easy: there is just one stationary point P 0 and it is
attractive, its basin of attraction is R3 and it is a kind of 3D-node.
(b) Case 1 < r < r1 is also easy: P 0 is unstable but P 1,2 are both
attractive, they are kind of 3D-nodes, and one needs just to learn
where the basin of attractions of each is.
(c) Case r1 < r < r2 is also easy: P 0 is unstable but P 1,2 are both
attractive, they are kind of 3D-spiral points, and one needs just to
learn where the basin of attractions of each is.
(d) But case r > r2 is really interesting: P 0 is unstable and P 1,2 are also
unstable and a typical trajectory is loitering between them, approaching certain subset in R3 .
9.8.2
Strange Attractor
This set is called attractor. However it is a really strange set and it is called
(officially!) strange attractor. But why this set deserves this name? In 2D
attractors were rather orderly: points (stable nodes or stable spiral points)
or limit cycles, sometimes other lines. For 3 × 3 linear systems they could
be a point, a straight line or a plane. So one could expect attractor to be
either a point, or a line, or a surface.
But this one is neither: it is a bit fatter than the plane, and for such
sets there are several (not coinciding) definition of dimension. For Lorenz
attractor it is a bit larger than 2!
Lorenz Attractor in Wikipedia
Lorenz Attractor Video
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