LECTURE 33: APPLICATIONS OF NONLINEAR SYSTEMS
MINGFENG ZHAO
November 24, 2014
Conservative equations
For a conservative equation x00 + f (x) = 0, let y = x0 , then we have the system:
x0 = y,
and y 0 = −f (x).
Then there are never any asymptotically stable points for the system x0 = y, y 0 = −f (x), and the trajectories
satisfy
1 2
y +
2
Z
f (x) dx = C,
which is called the Hamiltonian or energy of the system.
Pendulum
Suppose we have a mass m (in kilograms) on a pendulum of length L (in meters).
Figure 1. Pendulum Problem
Let θ(t) be the angle between the vertical line and the pendulum at time t (in seconds), then
θ00 +
g
sin(θ) = 0 ,
L
1
2
MINGFENG ZHAO
which is a conservative equation. In Chapter 2, we studied that when |θ| is very small, we can think sin(θ) ≈ θ, recall
sin(x)
lim
= 1, then
x→0
x
g
θ00 + θ = 0 .
L
Let ω = θ0 , then we can study the following two dimensional system:

θ

0

 =
ω

ω
− Lg sin(θ)

Find trajectories: Let’s solve
g sin(θ)
dω
=− ·
.
dθ
L
ω
Then the trajectories satisfy:
1 2 g
ω − cos(θ) = C.
2
L
Find critical points: Let’s solve
w = 0,
and
−
g
sin(θ) = 0.
L
Then all critical points are:

Notice that the Jacobian matrix of 
(kπ, 0),

ω
− Lg
 is:
sin(θ)



Since det 
0
− Lg
cos(kπ)
(kπ, 0), linearization is:
1
0
1
− Lg cos(θ)
0

.

 = g cos(kπ) 6= 0, then the system is almost linear at (kπ, 0). Hence we know that at
L
0

u

v

 1
Since cos(kπ) =
 −1
k ∈ Z.
if k is even
if k is odd
0

 =
0
1
− Lg cos(kπ)
0
, then we know that

u

v

.
LECTURE 33: APPLICATIONS OF NONLINEAR SYSTEMS
3
I. At (2kπ, 0), the linearization is:

u

0

 =
v
0
1
− Lg
0


u
.

v


r
r
0
1
g
g
 has eigenvalues λ1 = i
 has eigenvalues
and λ2 = −i
. Since 
Since 
L
L
− Lg cos(θ) 0
− Lg 0
r
r
g
g
λ1 = i
cos(θ) and λ2 = −i
cos(θ) when θ is close to (2kπ, 0). So we know that the system has a
L
L
stable center at critical point (2kπ, 0).

0
1

II. At ((2k + 1)π, 0), the linearization is:

u

0

 =
v

Since 
0
g
L
1

r
 has eigenvalues λ1 =
0
critical point ((2k + 1)π, 0).
0
g
L
1

u

0

.
v
g
and λ2 = −
L
r
g
. Then the system has an unstable saddle at
L
Figure 2. Various possibilities for the motion of the pendulum
Since the trajectories satisfy:
1 2 g
ω − cos(θ) = C.
2
L
4
MINGFENG ZHAO
Then
r
w=±
2g
cos(θ) + C.
L
Figure 3. Phase plane diagram and some trajectories of the nonlinear pendulum equation
Now for any initial condition (θ(0), ω(0)) = (θ0 , 0), we have C = −
r
0
θ =ω=±
2g
cos(θ0 ), that is,
L
r
2g
2g
2g p
cos(θ) −
cos(θ0 ) = ±
· cos(θ) − cos(θ0 ).
L
L
L
Let’s compute the period T of θ(t). It’s easy to see that T equals 4 times of the time from θ0 to 0. So we only
consider θ0 = ω is positive case, that is,
dθ
= θ0 = ω =
dt
r
2g p
· cos(θ) − cos(θ0 ).
L
Then
dt
=
dθ
s
L
1
·p
.
2g
cos(θ) − cos(θ0 )
So we know that
T
=
4
Z
θ0
s
0
L
1
·p
dθ.
2g
cos(θ) − cos(θ0 )
That is, the period is:
s
T =4
L
2g
Z
θ0
0
1
·p
dθ .
cos(θ) − cos(θ0 )
From the above formula, we know that
lim T = ∞ .
θ0 →π
LECTURE 33: APPLICATIONS OF NONLINEAR SYSTEMS
5
In fact,
Z
0
π
1
p
cos(θ) + 1
Z
dθ
π
=
0
Z
1 − cos(u)
π
=
0
=
1
p
1
√
2
1
q
Z
0
2 sin2
π
u
2
1
sin
u
2
du Let u = π − θ
du
du
Z π
1
1
√
u du Since sin(x) ≤ x for all x ≥ 0
2 0 2
√ Z π1
2
du
=
0 u
≥
=
∞.
Recall the general solution to the linearized pendulum equation θ00 +
g
√
√
θ = 0 is: θ(t) = C1 cos ( gLt) + C2 sin ( gLt).
L
So the period Tlinear for the the linearized pendulum equation is:
Tlinear
2π
= p g = 2π
L
s
L
.
g
Let’s compare T and Tlinear :
Figure 4. Plot of T and Tlinear with
g
T − Tlinear
= 1 (left), and
(right)
L
T
Department of Mathematics, The University of British Columbia, Room 121, 1984 Mathematics Road, Vancouver, B.C.
Canada V6T 1Z2
E-mail address: mingfeng@math.ubc.ca