Lecture 13 - Tue 2023 10 24
Continue with Hamiltonian Systems; Liapunov Functions; Limit cycles.
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                                                                  INDEX THEORY
Properties I forgot to stress last lecture:
02) Field is changed continuously *without* vanishing on curve ...... I = const.
03) If curve does not include any fixed points ...................... I = 0.
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Show, again, that for Hamiltonian systems dH/dt = 0.        HAMILTONIAN SYSTEMS.
Recall: All the equations of non-dissipative classical
        mechanics have this form.
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Finish lecture 12; from GRADIENT SYSTEMS onwards.
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Definition: Isolated closed trajectory.                            LIMIT CYCLES.
Example: van der Pol oscillator
         \ddot{x} + mu*(x^2-1)*\dot{x} + x = 0,  mu > 0.
         Decay for x large and growth for x small.
         Only one critical point.
         Later on we will see this yields a limit cycle.
For now: numerical ... assigned in problem set #05.
Draw what phase plane looks like for $mu$ small.
     Give hand-waving justification, and explain how averaging yields radius:
     Solution approximately a circle.
     Plug in approximate solution into equation  \dot{E} = mu*(1-x^2)*\dot{x}^2,
     where E = 0.5*(x^2+\dot{x}^2) = 0.5*r^2 to compute radius for which
     \dot{E} is approximately zero.
As $mu$ grows, sinusoidal character is lost, later on we will see what happens
     when $mu$ is large. For example; x = x(t) gets "jerky", and looks a bit
     like a square wave.
Explain physics behind the equation [electronic oscillator].
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One more way to rule out periodic orbits:                     DULAC's CRITERION.
\dot{x} = f(x) valid in a simply connected region of R^2
There is a scalar g(x) defined there such that div(g(x)\dot{x}) = div(g*f)
    never vanishes in the region [hence it is either positive or negative].
Then no periodic solutions can exist in the region.

Proof: Use Gauss theorem  \int_Q \div(g*\dot{x}) dA = \int_C g*\dot{x}.\hat{n} ds
where C = periodic orbit, Q = interior of C, \hat{n} = unit normal to C.
The left in this equation is not zero, while the right is zero. Contradiction.

EXAMPLE:  \dot{x} = x*(2-x-y)   and   \dot{y} = y*(4*x-x^2-3)
          Then use g = 1/(x*y) on first quadrant.

Note: Dulac's criterion can be "boosted" so as to show that a given periodic
      solution is unique [extend it to regions with one hole and apply it to
      the annulus between two periodic solutions, to reach a contradiction];
      but we will not go into this.
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