# Homogeneous systems of linear ordinary differential equation - Lecture 8 Transition matrix ∗

```Homogeneous systems of linear ordinary
differential equation - Lecture 8 ∗
September 27, 2012
Transition matrix
ẋ(t) = A(t)x(t), x(t0 ) = x0
where A(t) is a m &times; n matrix whose entries are continuous functions of time
t.
The solution is x(t) = Φ(t, t0 )x0 , where Φ(t, t0 ) is the transition matrix given
in terms of the Peano-Baker series:
Z tZ s
Z t
A(s)A(σ) dσds + . . .
A(s) ds +
Φ(t, t0 ) = I +
t0
t0
t0
Special case A(t) ≡ A is a constant n &times; n matrix:
Z tZ sZ
Z tZ s
Z t
A2 dσds +
A ds +
Φ(t, t0 ) = I +
t0
t0
Z tZ
t0
s
t0
Z
σ
A3 dτ dσds =
t0
t0
Z tZ
Z
t0
t
t0
A3 dτ dσds + . . .
t0
s
A3 (σ − t0 ) dσds
t0
(s − t0 )2
ds
2
t0
1
1
= A3 (t − t0 )3 = A3 (t − t0 )3
6
3!
=
The k-th term
Z t Z σk−1
t0
t0
σ
&middot;&middot;&middot;
t0
Z
A3
σ0
Ak dσ0 dσ1 . . . dσk−1 =
t0
1 k
A (t − t0 )k
k!
The Peano-Baker series can be written as:
∞
Φ(t, t0 ) = I+A&middot;(t−t0 )+
X (A &middot; (t − t0 ))k
1
1 2
A (t−t0 )2 + A3 (t−t0 )3 +. . . =
= eA&middot;(t−t0 )
2!
3!
k!
k=0
∗ This
work is being done by various members of the class of 2012
1
2
Dynamic Systems Theory
Scalar systems with constant coefficients
ẋ = ax, x(0) = x0 ⇔ x(t) = eat &middot; x0
Solve this by more elementary means
dx
dx
= ax ⇒ dx = axdt ⇒
dt
x
Z t
a dτ = at + C ⇒ x = eat &middot; eC
⇒ ln(x) =
ẋ = ax ⇒
0
and x(0) = x0 ⇒ eC = x0
For scalar systems with time dependent coefficients
ẋ = a(t)x(t), x(0) = x0
we can write the solution
x(t) = e
Rt
0
a(σ) dσ
x0
For a n &times; n matrix whose entries are functions of t
Z tZ s
Z t
A(s)A(σ) dσds + . . .
A(s) ds +
Φ(t, t0 ) = I +
t0
t0
6= e
Rt
t0
A(σ) dσ
t0
(in general)
There is a very special case in which the matrix case has a simplication that
is similar to the scalar case:
ẋ = f (t)Ax(t), x(0) = x0
where f (t) is a scalar cotinuous function and A is a constant n &times; n matrix.
Z
t
Z tZ
σ1
f (σ)A dσ +
f (σ1 )f (σ2 )A2 dσ2 dσ1 + . . .
0
0
0
Z t Z σ1
Z t
f (σ) dσ &middot; A +
=I+
f (σ1 )f (σ2 ) dσ2 dσ1 A2 + . . .
Φ(t, t0 ) = I +
0
Let γ(t) =
Rt
0
0
0
f (s) ds, the above series can be rewritten as
Φ(t, t0 ) = I + γ(t)A +
1
1
[γ(t)]2 A2 + [γ(t)]3 A3 + . . .
2!
3!
(Exercise: evaluate the k-th term in the series to show
that this is correct).
Rt
Seeing this pattern, we write Φ(t, t0 ) = eγ(t)A = e( 0 f (s) ds)A .
3
Dynamic Systems Theory
Specific examples
Recall the controlled pendulum system
θ̈ + cθ̇ +
g
sin(θ) = u(t)
l
θ1∗ = 0 and θ2∗ = π are the two equilibrium points.
Linearize about the stable equilibrium (θ1∗ ) – first assuming c = 0 (no friction).
g
ẍ + x = ū.
l
For today’s lecture we’re assuming ū = 0 (no forcing). Thus to put the
system into the framework under discussion we first orderize as follows:
p
p
p
x2
ẋ1 = gl ẋ = glp
x1 = gl x
⇒
x2 = ẋ
ẋ2 = ẍ = − gl x = − gl x1
ẋ1
0
=
ẋ2
−α
α
x1
&middot;
x2
0
p
where α = gl .
To solve this differential equation, we compute

0
−α

α
t
0
2
3
1 0 α
1
0 α
0 α
2
e
=I+
t+
t +
t3 + . . .
−α 0
2 −α 0
3! −α 0
3
2
1 0 −α3
1 −α2 0
1 0
0 α
2
t3
t
+
=
+
t+
0
0
α2
0 1
−α 0
2
3! α3
6
4
5
1 α4 0
1 −α6 0
1
0
α5
5
4
+
t +
t +
t6 + . . .
0
α6
4! 0 α4
5! −α5 0
6!

1 − 12 α2 t2 +
=
=
αt −
cos(αt)
− sin(αt)
1 4 4
1 6 6
4! α t − 6! α t +
1 5 5
1 3 3
3! α t + 5! α t − . . .
sin(αt)
cos(αt)
...
1 3 3
1 5 5
3! α t − 5! α t + . . .
1 2 2
1 4 4
1 6 6
2 α t + 4! α t − 6! α t + . . .
−αt +
1−
Check that this satisfies the differential equation
d
−α sin(αt) α cos(αt)
cos(αt) sin(αt)
=
−α cos(αt) −α sin(αt)
dt − sin(αt) cos(αt)
0 α
cos(αt) sin(αt)
=
&middot;
−α 0
− sin(αt) cos(αt)
and
!
4
Dynamic Systems Theory
cos(α &middot; 0)
− sin(α &middot; 0)
sin(α &middot; 0)
cos(α &middot; 0)
=
1
0
0
=I
1
The solution to the origina problem is
x1 (0)
cos(αt) sin(αt)
ẋ1 (t)
&middot;
=
x2 (0)
− sin(αt) cos(αt)
ẋ2 (t)
s
s
l
(cos(αt)x1 (0) + sin(αt)x2 (0))
g
s
l
= cos(αt)x(0) +
sin(αt)ẋ(0)
g
s
r r g
l
g
= cos
t x(0) +
sin
t ẋ(0)
l
g
l
x(t) =
l
x1 (t) =
g
Example 2
ẋ = Ax
where A is a 3 &times; 3 matrix

0
A =  kz
−ky
−kz
0
kx

ky
−kx 
0
where kx2 + ky2 + kz2 = 1. A is a skew symetric matrix (aij = −aji ).
d
kx(t)k =
Note: The norm of the x(t) satisfying this equation is constant ( dt
0). Thus the solution “lives” on the surface of the unit sphere.


 2

 
ky kx
kx kz
−kz − ky2
1 0 0
0
−kz ky
ky kz  t2
−kz2 − kx2
0
−kx  t +  ky kx
eAt = 0 1 0 +  kz
kx kz
ky kz
−ky2 − kx2
0 0 1
−ky kx
0


−ky
kz
}|
{ z
}|
{
z
 0
2
2
2
2
2
2 
−k
(−k
−
k
)
−
k
k
−k
k
−
k
(k
+
k
)  t3 + . . .
z
z
y
y

z
x
y
z
y
x
+

−kz
0
−ky
ky
−kx
0
1 2 2
1 3
1 2 4
1 5
= I + At + A t − At − A t + At
3!
4!
2
5!
1 3
1 5
1
1
1 2
2
= I + A t − t + t − ... + A
t − t4 + t6 − . . .
3!
5!
2
4!
6!
= I + A sin(t) + A2 (1 − cos(t))
5
Dynamic Systems Theory
Simplications that always work

Compute e

−3/2
1/2

1/2 
t
−3/2
. Directly
1 5/2
−3/2 1/2
0
t+
+
1/2 −3/2
1
2 −3/2
1
0
−3/2 2
t + ...
5/2
(I can’t see a pattern.)
Suppose I express A w.r.t.another basis {~
p1 , p~2 } and suppose that in this
λ1 0
basis A has the form
.
0 λ2
A~
p1 = λ1 p~1
A~
p2 = λ2 p~2
A p~1
p~2 = λ1 p~1 λ2 p~2 = p~1
λ1 0
AP = P
0 λ2
λ1 0
−1
P AP =
=Λ
0 λ2
λ1
p~2
0
0
λ2
1
= I + P ΛP −1 t + (P ΛP −1 )(P ΛP −1 )t2 + . . .
2
1
1
1
= I + P ΛP −1 t + (P Λ2 P −1 )t2 + (P Λ3 P −1 )t3 + (P Λ4 P −1 )t4 + . . .
2
3!
4!
1 2 2
1 3 3
1 4 4
= P I + Λt + Λ t + Λ t + Λ t + . . . P −1
2
3!
4!
λt
1
e
0
=P
P −1
0
eλ 2 t
eAt = eP ΛP
−1
t
```