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Ch 9'2: Autonomous Systems and Stability

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Title: Ch 9'2: Autonomous Systems and Stability


1
Ch 9.2 Autonomous Systems and Stability
1/ Autonomous Systems
  • We consider systems of 2 simultaneous
    differential equations of the form
  • With the initial conditions
  • in vector form
  • or
  • where x xi yj, f(x)F(x,y)i G(x,y)j,
    x0 x0i y0j, and
  • In vector form, the solution x ?(t) ?(t)i
    ?(t)j is a curve traced out by a moving point in
    the xy-plane (phase plane).

2
  • Note that the functions F and G depend on x and
    y, but not t.
  • Such a system is said to be autonomous.
  • The system x' Ax, where A is a constant matrix,
    is an example of an autonomous sys. However, if
    one or more of the elements of the coef matrix A
    is a function of t, then the system is
    nonautonomous.
  • In particular our autonomous system (
    )
  • has a direction field that is independent of
    time.
  • It follows that only one trajectory passes
    through each point (x0, y0) in the phase plane.
  • Thus all solutions to an initial value problem of
    the form
  • lie on the same trajectory, regardless of the
    time t0 at which they pass through (x0, y0).

3
2/ Stability and Instability
  • We consider the autonomous system x' f(x) and
    denote the magnitude of x by x.
  • The points where f(x) 0 are the critical
    points and correspond to constant, or
    equilibrium, solutions of the system of
    equations.
  • A critical point x0 is said to be stable if, for
    all ? gt 0 there is a ? gt 0 such that every
    solution x ?(t) satisfying ?(0) - x0 lt ?
    exists for all positive t and satisfies ?(t) -
    x0 lt ? for all t ? 0.
  • Otherwise, x0 is unstable.
  • A critical point x0 is said to be asymptotically
    stable if it is stable and if there exists a ?0 gt
    0 such that if a solution x ?(t) satisfies
  • ?(0) - x0 lt ?0, then ?(t) ? x0 as t ? ?
  • (Thus trajectories that start sufficiently
  • close to x0 not only stay close to x0
  • but must eventually approach x0 as
  • t ? 8.)

4
3/ The Oscillating Pendulum
  • Suppose a mass m is attached to one end of a
    weightless rigid rod of length L, with the other
    end of the rod supported at the origin O.
  • The position of the pendulum is described by the
    angle ? (counterclockwise direction taken to be
    positive).
  • Gravitational force mg acts downward,
  • Damping force cd? /dt , c gt 0, always opposite
    to the direction of motion.
  • The governing equation is
  • Rewriting our equation in standard form

5
  • To convert this eq. into a system of two first
    order equations, we let
  • x ?
  • y d? /dt
  • To find the critical points of this autonomous
    system, solve
  • These points correspond to 2 physical equilibrium
    positions, one with the mass directly below point
    of support (? 0), and the other with the mass
    directly above point of support (? ?).
  • (Intuitively, the first is stable and the second
    is unstable.)

6
  • If a mass is slightly displaced from a lower
    equilibrium position, it will oscillate with
    decreasing amplitude, and slowly approach an
    equilibrium position as the damping force
    dissipates initial energy. This type of motion
    illustrates asymptotic stability.
  • If a mass is slightly displaced from an upper
    equilibrium position, it will rapidly fall, and
    then approach a lower equilibrium position. This
    type of motion illustrates instability
  • For c (or ? ) 0, if the mass is displaced
    slightly from the lower equilibrium position, it
    will oscillate indefinitely with constant
    amplitude about the equilibrium position. This
    motion is stable but not asymptotically stable.

7
4/ Determination of Trajectories
  • Consider the autonomous system
  • It follows that
  • which is a first order equation in the variables
    x and y.
  • If we can solve this equation using methods from
    Chapter 2, then the implicit expression for the
    solution, H(x,y) c, gives an equation for the
    trajectories of
  • (Note this approach is applicable only in
    special cases.)

8
Example 3
  • Consider the system
  • It follows that
  • The solution of this separable equation is
  • Thus the trajectories are hyperbolas, as shown
    below.
  • The direction of motion can by inferred
  • from the signs of dx/dt and dy/dt in the
  • four quadrants.

9
Example 4 Separable Equation
  • Consider the system
  • It follows that
  • The solution of this separable equation is
  • Note that the equilibrium points are found by
    solving
  • hence (-2, 2) and (2, 2) are the equilibrium
    points.
  • Note
  • that (-2, 2) is a center and (2, 2) is a saddle
    pt.
  • Also, one trajectory leaves the saddle point
  • (at t -?), loops around the center, and
  • returns to the saddle point (at t ?).
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