HANOI PEDAGOGICAL UNIVERSITY 2
DEPARTMENT OF MATHEMATICS
———o0o———

PHAN THI THUY CHINH

LINEAR DIFFERENTIAL EQUATIONS

BACHELOR THESIS

Hanoi – 2019


HANOI PEDAGOGICAL UNIVERSITY 2
DEPARTMENT OF MATHEMATICS
———o0o———

PHAN THI THUY CHINH

LINEAR DIFFERENTIAL EQUATIONS

BACHELOR THESIS
Major: Analysis

SUPERVISOR:

Dr. Tran Van Bang

Hanoi – 2019

Bachelor thesis

PHAN THI THUY CHINH

Thesis Acknowledgement
I would like to express my gratitude to the teachers of the Department
of Mathematics, Hanoi Pedagogical University 2, the teachers in the analytic group as well as the teachers involved. The lecturers have imparted
valuable knowledge and facilitated for me to complete the course and
the thesis.
In particular, I would like to express my deep respect and gratitude
to Dr. Tran Van Bang, who has direct guidance, help me complete this
thesis.
Due to time, capacity and conditions are limited, so the thesis can
not avoid errors. Then, I look forward to receiving valuable comments
from teachers and friends.
Ha Noi, May 5, 2019
Student

Phan Thi Thuy Chinh

i


Bachelor thesis

PHAN THI THUY CHINH

Thesis Assurance
I assure that the data and the results of this thesis are true and not
identical to other topics. I also assure that all the help for this thesis has

been acknowledge and that the results presented in the thesis has been
identified clearly.
Ha Noi, May 5, 2019
Student

Phan Thi Thuy Chinh

ii


Contents

Notation

iv

Preface

1

1

2

Preliminaries
1.1

Matrix theory . . . . . . . . . . . . . . . . . . . . . . . .

2


1.2

Differential equations . . . . . . . . . . . . . . . . . . . .

7

1.3

Solving differential equations . . . . . . . . . . . . . . . .

8

1.4

Existence and uniqueness theorems . . . . . . . . . . . .

10

1.5

Phase space and flows . . . . . . . . . . . . . . . . . . .

11

1.6

Limit sets and trajectories . . . . . . . . . . . . . . . . .

13


2 LINEAR DIFFERENTIAL EQUATIONS

15

2.1

Autonomous linear differential equations . . . . . . . . .

17

2.2

Normal forms . . . . . . . . . . . . . . . . . . . . . . . .

19

2.3

Invariant manifolds . . . . . . . . . . . . . . . . . . . . .

30

2.4

Geometry of phase space . . . . . . . . . . . . . . . . . .

33

2.5


Floquet Theory . . . . . . . . . . . . . . . . . . . . . . .

35

Conclusion

40

References

41

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PHAN THI THUY CHINH

Notation
To study the contents of the thesis, we need to understand the mathematical symbols. We will give some common symbols used in the thesis,
helping readers read and understand easier.
x˙
Derivates with respect to time t
γ(x)

The trajectory through x

γ + (x)


The positive semi-trajectory through x

γ − (x)

The negative semi-trajectory through x

Λ(x)

The w−limit set of x

A(x)

The α-limit set of x

T

Period

A

The square n × n matrix

A−1

The invertible matrix of A

diag (a1 , ..., an ) The diagonal matrix
ρ ± iω


A pair of complex conjugate eigenvalues

Im (z)

Imaginary part of z

Re (z)

Real part of z

E u (0)

The unstable manifold

E c (0)

The centre manifold

E s (0)

The stable manifold

ϕ (T )

The Floquet multiplier

σ

A Floquet exponent


Φ (t)

The fundamental matrix

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PHAN THI THUY CHINH

Preface
Differential equation is an important research content in Mathematics. In addition, it is widely applied in engineering, physics, economics
and some other industries. Differential equations are used throughout
the sciences to model dynamic processes.
In our university program, we have been studying how to solve some
classes of specific linear differential equations, a few qualitative properties such as existence, uniqueness of the solution. Another way is to use
the properties of matrix analysis that can help us get a good presentation of the results of linear differential equations. That is the reason why
we would like to study linear differential equations.

1


Chapter 1
Preliminaries
1.1

Matrix theory

We consider A is the n × n matrix, we have some basic definitions,

operations and its properties.
Definition 1.1.1. In linear algebra, a square matrix A is called invertible, if existing a square matrix B such that:
AB = BA = In .
where In = diag (1, 1, ..., 1) is denoted the n × n identity matrix. In this
case, the matrix B is uniquely determined by A and is called the inverse
of A, denoted by A−1 .
Proposition 1.1.2. i) A is invertible ⇔ det A = 0;
ii) If A and B are invertible then AB is also invertible and (AB)−1 =
B −1 A−1 ;
iii) At

−1

t

= A−1 .

In linear algebra, we had learnt how to find the invertible matrix of
A (n × n matrix). We can list three method as follow:
Method 1.1.3. (Using determinant method)

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To start, we recall the complement of a element. Let A is the n×n matrix. If we omit the ith row and the j th column of A, we will get a submatrix (n − 1)×(n − 1) of A, denoted by Mij . Then, A = (−1)i+j det Mij ×
Mij is said to be the complement algebra of the element set on the ith

row and the j th column of A. Hence, we get the formula to find out the
invertible matrix of A:
• If det A = 0, then A is not invertible.
• If det A = 0, then A is invertible and
A−1 =

1
PA .
det A

Example 1.1.4. Let the matrix A


1 2 1


0 1 1


1 2 3
Find the invertible matrix of A.
We get,
1 2 1
det A = 0 1 1 = 2 = 0.
1 2 3
Thus, A is invertible.
Applying determinant method above, we get:
A11 = (−1)1+1

1 1


A12 = (−1)1+2

0 1

3

2 3

1 3

= 1,

= 1,


Bachelor thesis

PHAN THI THUY CHINH

0 1

A13 = (−1)1+3

1 2

= −1,

Similarly,
A22 = 2,

A23 = 0,
A31 = 1,
A32 = −1,
A32 = 1.
Thus, the invertible matrix of A is:


1 −4 1


 1 2 −1


−1 0 1
Example 1.1.5. Let the matrix
1 −2
3

4

Therefore, we apply the formula
−1

a b
c d

−1

=


a b

=

c d

1
ad − bc

d

−b

−c

a

We get the invertible matrix is:
1 −2
3

4

−1

=

1
10


Method 1.1.6. (Using Gauss method)

4

4

2

−3 1

.

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PHAN THI THUY CHINH

To find out the invertible matrix of A, we build the n × 2n matrix
[A | In ] .
where A is an n × n matrix and In is the identity matrix. Then, using
primary transformations according to row to change [A | In ] into [In | B].
Thus, B is the invertible matrix of A, B = A−1 .
Method 1.1.7. (Solving the system of equations to identify the invertible matrix)
Let an n × n matrix





a a . . . a1n
 11 12

a a . . . a 
2n 
 21 22
 .

.
.
..
. . ... 
 ..


an1 an2 . . . ann
To detect the invertible matrix of A, we set the system of equations:


a11 x1 + a12 x2 + . . . + a1n xn = y1




a x + a x + ... + a x = y
21 1
22 2
2n n
2
(1.1)

.
.

.




a x + a x + . . . + a x = y
n1 1
n2 2
nn n
n
where x1 , x2 , . . . , xn are variables and y1 , y2 , . . . , yn are parameters.
We consider two cases:
• If all y1 , y2 , . . . , yn the system of equations has only unique root:


x1 = b11 y1 + b12 y2 + . . . + b1n yn




x = b y + b y + . . . + b y
1
21 1
22 2
2n n
..


.




x = b y + b y + . . . + b y
n
n1 1
12 2
nn n

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PHAN THI THUY CHINH

then





A−1

b b . . . b1n

 11 12
b b . . . b 

2n 
 21 22
= .

.
.
..
. . ... 
 ..


bn1 bn2 . . . bnn

• If existing y1 , y2 , . . . , yn such that the system of equations has no
root or countless roots, A is not invertible.
Definition 1.1.8.

a
 11
a
 21
 .
 ..

an1

Let A is the n × m matrix and B


b

b
...
a12 . . . a1m
 11 12

b
a22 . . . a2m 
 21 b22 . . .

,
 .

.. . . .
.. . . . .. 
 ..
.
.
. 

bm1 bm2 . . .
an2 . . . anm

We get a n × n matrix C being

c
 11
c
 21
 .
 ..


cn1

is an m × p matrix

b1p

b2p 

.. 
. 

bmp

the matrix production, C = AB,

c12 . . . c1p

c22 . . . c2p 

.. . . . .. 
. 
.

cn2 . . . cnp

such that

n


Cij = ai1 b1j + ... + ain bnj =

aik bkj ,
k=1

for i = 1, n and j = 1, p.
Definition 1.1.9. (Power of a matrix)
We can raise the n × n matrix to nonnegative integer power by multiplying it by itself repeatedly in the same way as for ordinary number.
That is
A0 = I, A1 = A, Ak = AA . . . A.
Computing the k th power of a matrix, we need carry out k − 1 times
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PHAN THI THUY CHINH

the time of a single matrix multiplication, which is done with the trivial
algorithm (repeated multiplication).
An easy case for exponentiation is that of a diagonal matrix. Since
the product of diagonal matrices amounts to simply multiplying corresponding with diagonal elements together, the k − 1 power of a diagonal
matrix is obtained by raising the entries to the power k.

1.2

Differential equations

In this section, we see that solutions to differential equations can be
represented as curves in some appropriate spaces. Let’s the autonomous

equation
x˙ = f (x), x ∈ Rn ,

(1.2)

Definition 1.2.1. The curve (x1 (t), ..., xn (t)) in Rn is an integral curve
of the equation (1.2) if and only if
(x˙ 1 (t), ..., x˙ n (t)) = f (x1 (t), ..., xn (t)),
for all t ∈ I.
In other words, (x1 (t), ..., xn (t)) is the solution of (1.2) on I . Thus, the
tangent to the integral curve at (x1 (t0 ), ..., xn (t0 )) is f (x1 (t0 ), ..., xn (t0 )).
Example 1.2.2. Consider the differential equation
x˙ = x; x : I → R.
We have x˙ + x = 0 then the integral curve of this equation is x = ce−t .
Definition 1.2.3. Consider x˙ = f (x). The solution of this differential
equation define a flow, ϕ(x, t) satisfied ϕ(x, t) is solution of the equation
(1.2) with the initial condition x(0) = x.
Hence
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d
ϕ(x, t) = f (ϕ(x, t)),
dt
for all t and ϕ(x, 0) = x.
Then, the solution x(t) with x(0) = x0 is ϕ(x0 , t).


1.3

Solving differential equations

To solve differential equations, we will start with a simple example. Consider the equation
x¨ + x = 0,

(1.3)

with the initial conditions x (0) = 0 and x˙ (0) = b. We can write down
the solution
x (t) = a cos t + b sin t.

(1.4)

The solution is meaningless unless the properties of the functions sine
and cosine of t are well known. We should make a question in this circumstance: how did we solve this equation? We will give three different
methods to solve it.
Method 1.3.1. If we set y = x,
˙ then the equation (1.3) becomes
x˙
y˙

=

0

1


x

−1 0

y

with the initial conditions x (0) = 0 and y (0) = b.
0 1
Put A =
and ω = (x, y)T . The equation becomes
−1 0
ω˙ = Aω,
with ω (0) = (a, b)T .
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PHAN THI THUY CHINH

This equation has solutions etA ω (0). Therefore, we need to calculate
the matrix etA and defined by
∞

e

tA

=
k=0


tn An
.
n!

We easily show that
2n

A

=

(−1)n

0

0

(−1)n

and
2n+1

A

=

0

(−1)n


(−1)n+1

0

and so

etA

2n

n

t (−1) / (2n)!


=


t

n
2n+1

t

2n+1

n


(−1) / (2n + 1)!

n

(−1)n+1 / (2n + 1)!

t2n (−1)n / (2n)!

n






n

which we recognize as being series solutions for sine and cosine of t to
show
etA =

cos t

sin t

− sin t cos t

Hence, x (t) = a cos t + b sin t and x (t) = −a sin t + b cos t.
Method 1.3.2. Try a trial solution of the form x = ect and find c since
the equation (1.3) is linear. If x1 and x2 are independent solutions, then

the general solution is a sum of x1 and x2 . Substituting x = ect into the
differential equation, this equation becomes
e¨ct + ect = 0
⇔ c2 + 1 ect = 0

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PHAN THI THUY CHINH

⇔ c = ±i.
Therefore, the solution is x (t) = c1 eit + c2 e−it where c1 and c2 are the
coefficients and determined from the initial conditions.
Method 1.3.3. The differential equation can be written as follow:
d
+i
dt
Put v (t) =

d
− i x = 0.
dt

d
− i x so
dt
d
d

+ i v = 0 or e−it
veit = 0.
dt
dt

Hence veit = c1 or v = c1 e−it . Now replace v by the definition of v in
terms of x and solve another linear first order differential equation to
obtain x the sum of e±it as in method (1.3.2).

1.4

Existence and uniqueness theorems

Consider the differential equation in the form
x˙ = f (x, t), x ∈ Rn , f : Rn × R → Rn
where the dot is denoted by the differentiation with respect to time t.
The simplest example of differential equation is the linear differential
equation
x˙ = Ax.

(1.5)

where A is an n × n matrix with constant coefficients. If the initial
condition at t = 0 is x0 , the equation has solutions x = etA x0 .
∞
(tA)k
(tA)2
(tA)k
tA
= I + tA +

+ ... +
+ ...
Where e =
k!
2!
k!
k=0

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Theorem 1.4.1. (Local existence and uniqueness)
Suppose that x˙ = f (x, t) and f : Rn × R −→ Rn is continuously
differentiation. Then existing maximal t1 > 0, t2 > 0 such that a solution
x(t) with x(t0 ) = x0 exists and is unique for all t ∈ (t0 − t1 , t0 + t1 ).
Theorem 1.4.2. (Continuity of solutions)
Suppose that f is C r (r times continuously differentiation) and r ≥ 1,
in some neighbourhood of (x0 , t0 ). Then existing

> 0 and δ > 0 such

that if |x−x0 | < there is a unique solution x(t) defined on [t0 −δ, t0 +δ]
with x(t0 ) = x . The solutions depend on continuity x and t.

1.5


Phase space and flows

Lemma 1.5.1. (Properties of the flow)
(i) ϕ(x, 0) = x;
(ii) ϕ(x, t + s) = ϕ(ϕ(x, t), s) = ϕ(ϕ(x, s), t) = ϕ(x, s + t).
Example 1.5.2. Consider the equation
x˙ = Ax with x(0) = x0 .
The solution of equation is x = x0 etA .Then the flow ϕ(x0 , t) = x0 etA .
⇒ The flow ϕ(x, t) = xetA .
We will check properties of the flow, we have:
i, ϕ(x, 0) = xe0 = x.
ii, We have ϕ(x, t) = etA x,
We infer, ϕ(x, t + s) = xe(t+s)A .
ϕ(ϕ(x, t), s) = ϕ(x, t)esA = xetA esA = xe(t+s)A ,
ϕ(ϕ(x, s), t) = ϕ(x, s)etA = xesA etA = xe(t+s)A ,
and ϕ(x, s + t) = xe(s+t)A .
We infer that, (x, t + s) = ϕ(ϕ(x, t), s) = ϕ(ϕ(x, s), t) = ϕ(x, s + t).
Definition 1.5.3. A point x is stationary point of the flow if and only
if ϕ(x, t) = x, ∀t. Thus, at a stationary point f (x) = 0.
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PHAN THI THUY CHINH

Example 1.5.4. Consider the equation

x˙ = −x
x(0) = x

0

x is stationary point ⇔ x = ϕ(x, t)
⇔ x = xe−t , ∀t
⇔ x(e−t − 1) = 0, ∀t
⇔ x = 0.
Hence, the flow has an unique stationary point, that is x = 0.
Example 1.5.5. Finding the stationary points of x˙ = x(1 − x).
Stationary points exist where x˙ = 0 given function the stationary
points are the solutions to x(1 − x) = 0.
A solution of this equation is x = 0 then there is a stationary point
at x = 0.
A solution also exists where 1 − x = 0. Hence, there is another stationary point at x = 1.
Definition 1.5.6. A point x is periodic of (minimal) period T if and
only if

ϕ(x, t + T ) = ϕ(x, t) ∀t
ϕ(x, t + s) = ϕ(x, t) f or all 0 ≤ s < T
The curve Γ = {y|y = ϕ(x, t), 0 ≤ t < T } is called a periodic orbit of
the differential equation and is a closed curve in phase space.

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Example 1.5.7. Consider the differential equations


x˙1 = x2
x˙ = −x
2

with the matrix A =

0

1

−1 0

(1.6)

1

and the initial condition is x(0) =

a
b

.

We have
(1.6) ⇒

x¨1 = x˙ 2
x˙ 2 = −x1

⇒ x¨1 + x1 = 0.


The characteristics equation is λ2 + 1 = 0 ⇒ λ = ±i. Hence, the solution
of equations is
x1 = a cos t + b sin t
x2 = a sin t − b cos t
We get flow ϕ(x, t) =

.

a cos t + b sin t

.
a sin t − b cos t
For all point x is periodic of period 2π of flow ϕ as
ϕ(x, t + 2π) =

a cos(t + 2π

b sin(t + 2π)

a sin(t + 2π −b cos(t + 2π)

=

a cos t

b sin t

a sin t −b cos t.


= ϕ(x, t).

and ϕ(x, t + s) = ϕ(x, t), ∀s (0 < s < 2π).

1.6

Limit sets and trajectories

Consider x˙ = f (x) with x(0) = x0 , or equivalently the flow ϕ(x0 , t).
Definition 1.6.1. The trajectory through x is the set γ(x) =

ϕ(x, t)
t∈R

+

and the positive semi-trajectory, γ (x), and the negative semi-trajectory,
γ − (x) are defined as
ϕ(x, t) and γ − (x) =

γ + (x) =
t≥0

ϕ(x, t).
t≤0

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PHAN THI THUY CHINH

Definition 1.6.2. The w−limit set of x, Λ(x), and the α−limit set of
x, A(x), are the sets
Λ(x) = {y ∈ Rn |∃tn with tn −→ ∞ and ϕ(x, tn ) −→ y as n −→ ∞}
and A(x) = {y ∈ Rn |∃sn with sn −→ −∞ and ϕ(x, sn ) −→ y as
n −→ ∞}.
With the w−limit set, Λ(x), which is the set of points which x tend to
(i.e the limit points of γ + (x)).
The α−limit set, A(x), which is the set of points that trajectory,
through x tends to in backward time.
Example 1.6.3. Consider B(0, x ).
Suppose ϕ(x, t0 ) = y.
We choose tn = t0 + 2π → +∞ as n → ∞.
⇒ ϕ(x, tn ) = ϕ(x, t0 + 2π) = ϕ(x, t0 ) = y, ∀n
⇒ ϕ(x, tn ) → y as n → ∞.
Similarly, we choose tn = t0 − 2π → −∞ as n → ∞.
⇒ ϕ(x, tn ) = ϕ(x, t0 − 2π) = ϕ(x, t0 ) = y, ∀n
⇒ ϕ(x, tn ) → y as n → ∞.
Hence, Λ(x) = B(0, x ) = A(x).

14


Chapter 2
LINEAR DIFFERENTIAL
EQUATIONS
It is really easy to solve linear differential equations. We can write down
solutions in terms of the exponential function.

Example 2.0.1. Let the autonomous differential equation:
x˙ = Ax, x ∈ Rn .

(2.1)

where A is a constant n × n matrix and the initial condition x (0) = x0 ,
then the solution is:
x (t) = etA x0 .

(2.2)

If we can exponentiate constant matrices, solving linear differential
equations is simple. This solution also help us solve equations such as:
x˙ = Ax + g (t) .

(2.3)

Putting y = e−tA x with the initial condition y (0) = x (0) = x0 . Differentiating y with respect to t:
y˙ = e−tA x˙ − Ae−tA x = e−tA g (t) .

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Hence

t


e−sA g (s) ds

y − x0 =
0

or

t
tA

e−sA g (s) ds.

tA

x (t) = e x0 + e

0

These above results depend on the standard properties of the exponential
of a matrix. Equation (2.1) is a particular case of the more general linear
differential equation is
x˙ = A (t) x, x ∈ Rn .

(2.4)

Linear equations have a simple property is that if x1 (t) and x2 (t) are
solutions, then λx1 (t) + µx2 (t) is also a solution. This superposition
principle holds in general for nonlinear systems, but it results to an
important consequence for linear systems. If we can find n independent

solutions to a linear differential equation in Rn , then any other solution
can be written as a sum of these solutions. We can represent x1 t, . . . , xn t
are independent solutions, the general solution can be written as:
x (t) = Φ (t) c,

(2.5)

where c ∈ Rn is constant and Φ (t) is the n × n matrix whose columns
are the solutions xi (t) i.e. Φ (t) = [x1 (t) , . . . , xn (t)].
Φ (t) is called a fundamental matrix and etA is a simple natural choice
of fundamental matrix for the problem x˙ = Ax. From the solution in
terms of a fundamental matrix, x (0) = x0 = Φ (0) c, so c = Φ (0)−1 x0
and x (t) = Φ (t) Φ (0)−1 x0 . Comparing to the solution (2.2), if we see
that Φ (t) is any fundamental matrix of the autonomous equation (2.1),
then
etA = Φ (t) Φ (0)−1 .
Hence, we can solve linear differential equations by the idea of a fun16


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PHAN THI THUY CHINH

damental matrix. In particular, it is useful to consider problems with
periodic coefficients, x˙ = A (t) x, with A (t) = A (t + T ) for some T > 0.
Thus, studying linear systems is very useful.

2.1

Autonomous linear differential equations


In this section, we will verify that the exponential etA is a fundamental matrix for the differential equation x˙ = Ax where x ∈ Rn and A
is a linear map of Rn to itself (A is the n × n matrix with constant
coefficients).
Firstly, we need to state the definition of the exponential of a matrix
and form some its properties.
Definition 2.1.1. Let A be the n × n matrix with constant coefficients,
then the exponential of A is defined by the power series, denoted by eA :
∞
A

e =
k=0

Ak
.
k!

Similarly with the exponential of a real number, the exponential of a
matrix has same properties. The following exercises establish some these
properties.
Proposition 2.1.2. Given a matrix A( i.e. a linear map of Rn to itself)
define the norm of A, A by
A =

|Av|
.
v∈Rn ,v=0 |v|
sup


then:
i) 0 ≤ A < ∞
ii)
λA = |λ| · A , f orλ ∈ R
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Bachelor thesis

PHAN THI THUY CHINH

A+B ≤ A + B
AB ≤ A · B
Ak ≤ A

k

Based on these basic properties of the exponential we can prove the
existence and uniqueness of solutions to autonomous linear differential
equations.
Theorem 2.1.3. Let A be the n×n matrix with coefficients and x ∈ Rn .
Then the unique solution x (t) of x˙ = Ax with x (0) is x (t) = etA x0 .
Proof. We need to note that
d tA
e =
dt

∞

k=0


d
dt

tk Ak
k!

∞

=
k=1

tk−1 Ak
(k − 1)!

= AetA ,

where we have used the result quote in the second exercise above to
differentiate the sum term by term. Hence
d
d
x (t) = etA x0 = AetA x0 = Ax (t) ,
dt
dt
and so etA x0 is a root of the equation x˙ = Ax with x (0) = x0 .
To prove uniqueness, suppose that y (t) is another root to the differential equation with the same initial value, y (0) = x0 .
Set z (t) = e−tA y (t), then
z˙ = −Ae−tA y (t) + e−tA y˙ = −Ae−tA y (t) + Ae−tA y (t) = 0,
i.e., z (t) í constant. But z (0) = y (0) = x0 and so z (t) = x0 . Now, from
the definition of z (t), this implies that y (t) = etA x0 = x (t), and hence

solution is unique.
The above result shows that we can find out the solutions of linear
differential equations by exponentiating matrices, but it does not tell
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Bachelor thesis

PHAN THI THUY CHINH

us how to calculate with exponential of matrices. If we want to write
down solutions explicitly, we must learn how to take the exponential of
a matrix. If A is diagonal, A = diag (a1 , ..., an ), it should be obvious that
etA = diag ea1 t , . . . , ean t and solutions in component form are xi (t) =
eai t x0i , but for more general matrices, the solutions are not so obvious. To
deal with this problem, we need understand the idea of Jordan normal
form in linear algebra.

2.2

Normal forms

Consider a simple change of coordinates x = P y where P is the n × n
invertible matrix (det P = 0). Then x˙ = Ax implies that
y˙ = P −1 x˙ = P −1 Ax = AP y = Λx,

(2.6)

where Λ = P −1 AP and the initial value x (0) = x0 is transformed to
y (0) = P −1 x0 = y0 . In these new coordinates, the solution is y (t) =

etΛ y0 , and so transforming back to the x coordinates
x (t) = P y (t) = P etΛ y0 = P etΛ x0 .

(2.7)

Comparing to the known solution etA x0 , we see that:
etA = P etΛ P −1 .

(2.8)

Choosing P such that λ is taken a simple form which will allow us to
calculate etΛ easier.
As you know, it is very easy to exponentiate diagonal matrices. Thus,
suppose that A has n distinct real eigenvalues, λ1 , . . . , λn , with associated
eigenvectors ei , so Aei = λi ei , 1 ≤ i ≤ n. Let P = [e1 , . . . , en ], the matrix
with the eigenvectors of A as columns.
Then, since the eigenvectors of distinct real eigenvalues are real and
19