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

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y = 0, -a2 sin x — y Ó = 0.

We obtain y = 0 and x = ±nn, where n is an integer. These points correspond to two physical equilibrium positions, one with the mass directly below the point of support (â = 0) and the other with the mass directly above the point of support (â = n). Our intuition suggests that the first is stable and the second is unstable.

9.2 Autonomous Systems and Stability

475

More precisely, if the mass is slightly displaced from the lower equilibrium position, it will oscillate back and forth with gradually decreasing amplitude, eventually approaching the equilibrium position as the initial potential energy is dissipated by the damping force. This type of motion illustrates asymptotic stability.

On the other hand, if the mass is slightly displaced from the upper equilibrium position, it will rapidly fall, under the influence of gravity, and will ultimately approach the lower equilibrium position in this case also. This type of motion illustrates instability. In practice, it is impossible to maintain the pendulum in its upward equilibrium position for any length of time without an external constraint mechanism since the slightest perturbation will cause the mass to fall.

Finally, consider the ideal situation in which the damping coefficient c (or ó) is zero. In this case, if the mass is displaced slightly from its lower equilibrium position, it will oscillate indefinitely with constant amplitude about the equilibrium position. Since there is no dissipation in the system, the mass will remain near the equilibrium position, but will not approach it asymptotically. This type of motion is stable, but not asymptotically stable. In general, this motion is impossible to achieve experimentally because the slightest degree of air resistance or friction at the point of support will eventually cause the pendulum to approach its rest position.

These three types of motion are illustrated schematically in Figure 9.2.3. Solutions of the pendulum equations are discussed in more detail in the next section.

(a) (b) (c)

FIGURE 9.2.3 Qualitative motion of a pendulum. (a) With air resistance. (b) With or without air resistance. (c) Without air resistance.

Determination of Trajectories The trajectories of a two-dimensional autonomous system can sometimes be found by solving a related first order differential equation. From Eqs. (1) we have

dy = dy/dt G (x, y)

dx dx/dt F(x, y)’ ()

which is a first order equation in the variables x and y. Observe that such a reduction

is not usually possible if F and G depend also on t. If Eq. (14) can be solved by any

of the methods of Chapter 2, and if we write solutions (implicitly) in the form

H(x, y) = c, (15)

then Eq. (15) is an equation for the trajectories of the system (14). In other words, the trajectories lie on the level curves of H(x, y). Keep in mind that there is no general

476

Chapter 9. Nonlinear Differential Equations and Stability

EXAMPLE

1

way of solving Eq. (14) to obtain the function H, so this approach is applicable only in special cases.

Find the trajectories of the system

dx/dt = y, dy/dt = x. (16)

In this case, Eq. (14) becomes

dy x

-f = -. (17)

dx y

This equation is separable since it can be written as

ydy = x dx,

and its solutions are given by

H(x, y) = y2 - x2 = c, (18)

where c is arbitrary. Therefore the trajectories of the system (16) are the hyperbolas

shown in Figure 9.2.4. The direction of motion on the trajectories can be inferred from

the fact that both dx/ dt and dy/ dt are positive in the first quadrant. The only critical point is the saddle point at the origin.

Another way to obtain the trajectories is to solve the system (16) by the methods of Section 7.5. We omit the details, but the result is

x = c1et + c2e-t, y = c1et — c2e-t.

Eliminating t between these two equations again leads to Eq. (18).

FIGURE 9.2.4 Trajectories of the system (16).

9.2 Autonomous Systems and Stability

477

EXAMPLE

2

PROBLEMS

Find the trajectories of the system

dx dy

— = 4 — 2y, — = 12 — 3x . (19)

dt y' dt v '

From the equations

4 — 2 y = 0, 12 — 3x2 = 0

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