Hamiltonian system and Dissipative system:
Motivating example: (energy and DE)
y 00 + qy = 0 system: y 0 = v, v 0 = −qy
y 00 + py 0 + qy = 0 system: y 0 = v, v 0 = −qy − pv
Energy function: E(y, v) =
1 2
1
v + qy 2
2
2
1 2
where v is the kinetic energy (note that m = 1)
2
1 2
and qy is the potential energy (example: q = g)
2
1
y 00 + qy = 0
d
d 1 2
1 2
E(y(t), v(t)) =
v (t) + qy (t) = v(t)v 0(t) + qy(t)y 0(t)
dt
dt 2
2
= v(t)(−qy(t)) + qy(t) · v(t) = 0 (energy is conserved)
y 00 + py 0 + qy = 0
d
d 1 2
1 2
E(y(t), v(t)) =
v (t) + qy (t) = v(t)v 0(t) + qy(t)y 0(t)
dt
dt 2
2
= v(t)(−qy(t) − pv(t)) + qy(t) · v(t) = −p[v(t)]2 ≤ 0
(energy is dissipated)
2
Definition:
dy
dx
= f (x, y),
= g(x, y).
dt
dt
If there is a function H(x, y) such that for each solution orbit
d
(x(t), y(t)), we have
H(x(t), y(t)) = 0, then the system is a
dt
Hamiltonian system, and H(x, y) is called conserved quantity.
(or energy function, Hamiltonian)
If there is a function H(x, y) such that for each solution orbit
d
(x(t), y(t)), we have
H(x(t), y(t)) ≤ 0, then the system is a
dt
dissipative system, and H(x, y) is called Lyapunov function.
(or energy function)
3
Example: If a satellite is circling around the earth, it is a Hamiltonian system; but if it drops to the earth, it is a dissipative
system.
Example: For linear system, center is a Hamiltonian system, and
spiral sink is a dissipative system.
Jacobian: Carl Gustav Jacob Jacobi (1804-1851), Germany
(Prussian) mathematician
Hamiltonian: William Rowan Hamilton (1805-1865), Irish mathematician
Lyapunov function: Aleksandr Mikhailovich Lyapunov (18571918), Russian mathematician
4
(1) Carl Gustav Jacob Jacobi, (2) William Rowan Hamilton,
(3) Aleksandr Mikhailovich Lyapunov
http://www-groups.dcs.st-and.ac.uk/˜history/
5
A nonlinear oscillator: pendulum
d2θ
dθ
ml 2 + bl + mg sin θ = 0
dt
dt
m: mass of the bob
l: length of rod,
g: gravity constant
b: friction constant,
θ(t): the angle between the rod and the vertical direction
simplified version: θ00 + pθ0 + q sin θ = 0
b
g
p = , and q = .
m
l
Linearization: θ00 + pθ0 + qθ = 0 (harmonic oscillation)
sin θ ≈ θ − θ3/6 + · · · ≈ θ, when θ ≈ 0
6
Ideal pendulum: (p = 0, no friction)
θ00 + q sin θ = 0 or θ0 = v, v 0 = −q sin θ
(v is angular velocity)
conserved quantity: H(θ, v) =
1 2
v − q cos θ
2
√
1 2
solution orbit: v − q cos θ = C or v = ± 2q cos θ + 2C
2
Nullcline: v = 0, sin θ = 0 (θ = kπ)
Equilibrium points: (θ, v) = (kπ, 0)
7
Linearization at equilibrium point: J =
√
eigenvalues: λ = ± − cos θ
0
1
− cos θ 0
!
,
When θ = 2kπ, cos θ = 1, λ = ±i. (center)
When θ = (2k + 1)π, cos θ = −1, λ = ±1. (saddle)
So θ = (2k + 1)π (pendulum at the top of circle) is unstable.
θ = 2kπ (pendulum at the bottom of circle) is neutrally stable.
s
Period of small swinging: T = 2π
l
.
g
8
Qualitative analysis: there are three kinds of solutions
1. Periodic solution: around θ = 2kπ; the oscillation is less than
a full circle; the oscillation keeps forever since there is no friction;
the period of the oscillation seems increasing as the amplitude
2π
2π
increases, the minimum period is √ = q
(same as linear
q
g/l
pendulum). The energy is small.
2. “Rotating solution”: velocity is never 0; θ is either increasing
(counter-clockwise rotation) or decreasing (clockwise rotation).
The energy is large.
3. Saddle connection: the unstable orbit of the saddle point
(but also the stable orbit of another saddle point); separatrix
between the periodic and rotating solutions; hard to observe in
experiment.
9
Pendulum with friction: (p > 0)
θ00 + +pθ0 + q sin θ = 0 or θ0 = v, v 0 = −q sin θ − pv
(v is angular velocity)
Lyapunov function: H(θ, v) =
1 2
v − q cos θ
2
Nullcline: v = 0, v = −(q/p) sin θ = 0
Equilibrium points: (θ, v) = (kπ, 0)
10
Linearization at equilibrium point: J =
0
1
−q cos θ −p
!
,
2 + pλ + q cos θ = 0,
eigenvalues:
λ
q
λ=
−p ±
p2 − 4q cos θ
2
When θ = 2kπ, cos θ = 1, (sink or spiral sink)
When θ = (2k + 1)π, cos θ = −1. (saddle)
So θ = (2k + 1)π (pendulum at the top of circle) is unstable.
θ = 2kπ (pendulum at the bottom of circle) is stable.
11
Qualitative analysis: Most solutions will eventually tend to
an equilibrium (2kπ, 0) as the oscillation becomes smaller and
smaller. When the damping is so large, the approach to (2kπ, 0)
might be fast and direct (sink type). Otherwise, there are some
oscillation. When the energy is so large, it is possible that the
solution will go over the top for several times before it settles at
the equilibrium point.
Comparison of Hamiltonian and dissipative systems The
level curves of H(x, y) are “orbits” of a system. For a Hamiltonian system, the solution will stay on an orbit, and for a dissipative system, the solution will not stay on one orbit, but go to
orbit with smaller energy, eventually approaching an equilibrium
point.
For a Hamiltonian system, the direction of the vector field is tangent to the level curve of H(x, y), and for a dissipative system,
the direction of the vector field points to a lower energy level.
12
A famous Hamiltonian system: N -body problem:
The motions for N gravitationally interacting bodies.
N
X
X
Gmimj
1
2
Kinetic energy:
mir˙i , Potential energy:
i=1 2
1≤i<j≤N |ri − rj |
Gm m
Newton’s law of gravitational force: F = |r −ri |2j
i
j
Three body problem: (N = 3)
d2q1i
q −q
q −q
m1 2 = km1m2 2i 3 1i + km1m3 3i 3 1i
dt
r12
r13
q3i − q2i
d2q2i
q1i − q2i
+ km2m3
m2 2 = km2m1
3
3
dt
r12
r23
d2q3i
q1i − q3i
q2i − q3i
m3 2 = km3m1
+ km3m2
, (i = 1, 2, 3),
3
3
dt
r13
r23
13
Example: (a) Three bodies: Sun, Earth and Moon; (b) the
whole solar system.
Mathematical challenge: (1) Given the masses, initial positions
and initial velocities, can you describe the solution? (2) Find all
possible orbits?
Answer: General solutions cannot be found for N ≥ 3. People
are trying to find some special interesting solutions.
Mathematicians who have tried the problem: Newton, Lagrange,
Euler, Poincare, and any famous mathematician before 1900.
New solutions: http://www.ams.org/new-in-math/cover/orbits1.html
Another book: Celestial Encounters, The Origins of Chaos and
Stability, by Florin Diacu and Philip Holmes, Princeton University
Press 1996
14
(a) Jules Henri Poincarè, (b) Leonhard Euler, (c) Zhihong Jeff
Xia (a professor in Northwestern University, 1962-, the only one
who is alive, see also your textbook page 481)
Xia solved a 100-year old question about N -body problem in his
PhD thesis in 1989
15
When is the system Hamiltonian?
A solution of Hamiltonian system stays on a level curve of H(x, y),
that is, H(x(t), y(t)) = C.
∂H dy
∂H dx
+
= 0,
∂x dt
∂y dt
dx
∂H(x, y) dy
∂H(x, y)
so a form of equation is
=
,
=−
.
dt
∂y
dt
∂x
∂f
∂g
Criterion: If x0 = f (x, y), y 0 = g(x, y) and
= − , then it is a
∂x
∂y
Hamiltonian system.
16
Example: Determine whether the systems are Hamiltonian. If
they are, find the Hamiltonian functions.
(1) x0 = x − y 2, y 0 = x(2 − y)
(2) x0 = x cos(xy), y 0 = −y cos(xy) + x2.
Method of finding H(x, y):
R
1. Integrate: H(x, y) = f (x, y)dy + C(x) (use Eq.1)
2. Find C(x) by taking ∂H(x, y)/∂x (use Eq.2)
17
Form of Hamiltonian system:
dx
∂H(x, y) dy
∂H(x, y)
=
,
=−
.
dt
∂y
dt
∂x

∂ 2H

 ∂x∂y
J =
 ∂ 2H

− 2
∂x
∂ 2H



2
∂y
=
2
∂ H 

−
a
b
−c −a
!
∂y∂x
q
Eigenvalues: λ2 − a2 + bc = 0, λ = ± a2 − bc
So an equilibrium point in a Hamiltonian system is always either
a saddle of a center.
If the system has a sink, source, spiral sink or spiral source, then
it is not a Hamiltonian system.
18
y
Example: Consider the system: x0 = y, y 0 = −x − + x2,
4
2
2
3
y
x
x
(a) Verify that L(x, y) =
+
−
is a Lyapunov function for
2
2
3
the system.
(b) Sketch the level sets of L.
Problems on pendulum:
1. If the arm length of the ideal pendulum is doubled from l to
2l, what is the effect on the period of small amplitude swinging
solution?
2. Will an ideal pendulum clock that keeps perfect time on earth
run slow or fast on the moon?
3. What relationship must hold between the parameters b, m and
l for the period of a small swing back and forth of the damped
pendulum to be one second?
19