Hamiltonian System and Dissipative System

Hamiltonian system and Dissipative system:

Motivating example: (energy and DE) y00 + qy = 0 system: y0 = v, v0 = −qy y00 + py0 + qy = 0 system: y0 = v, v0 = −qy − pv

1 1 Energy function: E(y, v) = v2 + qy2 2 2

1 where v2 is the kinetic energy (note that m = 1) 2 1 and qy2 is the potential energy (example: q = g) 2

1 y00 + qy = 0

d d 1 1 E(y(t), v(t)) = v2(t) + qy2(t) = v(t)v0(t) + qy(t)y0(t) dt dt 2 2

= v(t)(−qy(t)) + qy(t) · v(t) = 0 (energy is conserved) y00 + py0 + qy = 0

d d 1 1 E(y(t), v(t)) = v2(t) + qy2(t) = v(t)v0(t) + qy(t)y0(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: dx dy = 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 Hamil- tonian 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 (1857- 1918), 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 + bl + mg sin θ = 0 dt2 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, v0 = −q sin θ (v is angular velocity)

1 conserved quantity: H(θ, v) = v2 − q cos θ 2

1 √ solution orbit: v2 − 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 0 1 ! Linearization at equilibrium point: J = , − cos θ 0 √ eigenvalues: λ = ± − cos θ

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 l Period of small swinging: T = 2π . 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, v0 = −q sin θ − pv (v is angular velocity)

1 Lyapunov function: H(θ, v) = v2 − q cos θ 2

Nullcline: v = 0, v = −(q/p) sin θ = 0

Equilibrium points: (θ, v) = (kπ, 0)

10 0 1 ! Linearization at equilibrium point: J = , −q cos θ −p eigenvalues: λ2 + pλ + q cos θ = 0, q −p ± p2 − 4q cos θ λ = 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 Hamilto- nian 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 tan- gent 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 1 Gm m Kinetic energy: X m r˙ 2, Potential energy: X i j 2 i i i=1 1≤i

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 Poincar`e,(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 dx ∂H dy + = 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), y0 = 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 − y2, y0 = x(2 − y)

(2) x0 = x cos(xy), y0 = −y cos(xy) + x2.

Method of ﬁnding H(x, y): 1. Integrate: H(x, y) = R 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 ∂2H   !  ∂x∂y ∂y2  a b J =   =  ∂2H ∂2H  −c −a  − −  ∂x2 ∂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, y0 = −x − + x2, 4 y2 x2 x3 (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