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Elementary Differential Equations and Boundary Value Problems - Boyce W.E.

Boyce W.E. Elementary Differential Equations and Boundary Value Problems - John Wiley & Sons, 2001. - 1310 p.
Download (direct link): elementarydifferentialequations2001.pdf
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Damped Pendulum. We continue the discussion of the damped pendulum begun in Examples 2 and 3. Near the origin the nonlinear equations (8) are approximated by the linear system (16), whose eigenvalues are
- y \/Y2 - 4a>2 2
(18)
The nature of the solutions of Eqs. (8) and (16) depends on the sign of y2 - 4a>2 as
follows.
1. If y2 - 4a>2 > 0, then the eigenvalues are real, unequal, and negative. The critical point (0, 0) is an asymptotically stable node of the linear system (16) and of the almost linear system (8).
2. If y2 - 4a>2 = 0, then the eigenvalues are real, equal, and negative. The critical point (0, 0) is an asymptotically stable (proper or improper) node of the linear system (16). It may be either an asymptotically stable node or spiral point of the almost linear system (8).
3. If y 2 - 42 < 0, then the eigenvalues are complex with negative real part. The critical point (0, 0) is an asymptotically stable spiral point of the linear system (16) and of the almost linear system (8).
r1, r2 =
Thus the critical point (0, 0) is a spiral point of the system (8) if the damping is small and a node if the damping is large enough. In either case, the origin is asymptotically stable.
9.3 Almost Linear Systems 485
Let us now consider the case y 2 4a>2 < 0, corresponding to small damping, in more detail. The direction of motion on the spirals near (0, 0) can be obtained directly from Eqs. (8). Consider the point at which a spiral intersects the positive /-axis (x = 0 and > 0). At such a point it follows from Eqs. (8) that dx/dt > 0. Thus the point (x, ) on the trajectory is moving to the right, so the direction of motion on the spirals is clockwise.
The behavior of the pendulum near the critical points (nn, 0), with n even, is the same as its behavior near the origin. We expect this on physical grounds since all these critical points correspond to the downward equilibrium position of the pendulum. The conclusion can be confirmed by repeating the analysis carried out above for the origin. Figure 9.3.3 shows the clockwise spirals at a few of these critical points.
Now let us consider the critical point (, 0). Here the nonlinear equations (8) are approximated by the linear system (17), whose eigenvalues are
y \f Y 2 + 4m2 2 '
(19)
One eigenvalue (r1) is positive and the other (r2) is negative. Therefore, regardless of the amount of damping, the critical point x = , = 0isan unstable saddle point of both the linear system (17) and of the almost linear system (8).
To examine the behavior of trajectories near the saddle point (, 0) in more detail we write down the general solution of Eqs. (17), namely,
^ + CJl) &, (20)
U/ 2 \r2
where C1 and C2 are arbitrary constants. Since r1 > 0 and r2 < 0, it follows that the solution that approaches zero as t corresponds to C1 = 0. For this solution v/u = r2, so the slope of the entering trajectories is negative; one lies in the second quadrant (C2 < 0) and the other lies in the fourth quadrant (C2 > 0). For C2 = 0 we obtain the pair of trajectories exiting from the saddle point. These trajectories have slope r1 > 0; one lies in the first quadrant (C1 > 0) and the other lies in the third quadrant (C1 < 0).
The situation is the same at other critical points (nn, 0) with n odd. These all correspond to the upward equilibrium position of the pendulum, so we expect them to be unstable. The analysis at (, 0) can be repeated to show that they are saddle points oriented in the same way as the one at (, 0). Diagrams of the trajectories in the neighborhood of two saddle points are shown in Figure 9.3.4.
r1
r2 =
FIGURE 9.3.3 Asymptotically stable spiral points for the damped pendulum.
486
Chapter 9. Nonlinear Differential Equations and Stability
EXAMPLE
4
The equations of motion of a certain pendulum are
dx/dt = j, dj/dt = -9 sinx - 1 j, (21)
where x = 6 and j = d6/ dt. Draw a phase portrait for this system and explain how it shows the possible motions of the pendulum.
By plotting the trajectories starting at various initial points in the phase plane we obtain the phase portrait shown in Figure 9.3.5. As we have seen, the critical points (equilibrium solutions) are the points (nn, 0), where n = 0, 1, 2,.... Even values of n, including zero, correspond to the downward position of the pendulum, while odd values of n correspond to the upward position. Near each of the asymptotically stable critical points the trajectories are clockwise spirals that represent a decaying oscillation about the downward equilibrium position. The wavy horizontal portions of the trajectories that occur for larger values of | j| represent whirling motions of the pendulum. Note that a whirling motion cannot continue indefinitely, no matter how large | j| is; eventually the angular velocity is so much reduced by the damping term that the pendulum can no longer go over the top, and instead begins to oscillate about its downward position.
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