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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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known as the gradient of V, is normal to the curve V(x, y) = c and points in the direction of increasing V. In the present case, V increases outward from the origin, so V V points away from the origin, as indicated in Figure 9.6.16. Next, consider a trajectory x = (), y = (t) ofthe system (6), and recall that the vector T(t) = '() + ;(t)j is tangent to the trajectory at each point; see Figure 9.6.1b. Let x1 = (^), y1 = (^) be a point of intersection of the trajectory and a closed curve V (x, y) = c. At this point '(t1) = F(xv y1), r(t1) = G(x1, y1), so from Eq. (7) we obtain
V(xv y{) = Vx(xv {)'(t^ + Vy(x1, ^'(^
= [ Vx(xv 1) + Vy(x1, y1)j] (t1)i + 7(t1>j]
= VV(x1, 1) T(t) (10)
Thus V(xv y,1) is the scalar product of the vector VV (x1; y1) and the vector T(^). Since (x1, y1) < 0, it follows that the cosine of the angle between VV (x1; y1) and T(t1) is also less than or equal to zero; hence the angle itself is in the range [/2, 3/2]. Thus the direction ofmotion on the trajectory is inward with respect to V(xv y,1) = c or, at worst, tangent to this curve. Trajectories starting inside a closed curve V(x1, y1) = c (no matter how small c is) cannot escape, so the origin is a stable point. If V(^1, y^) < 0, then the trajectories passing through points on the curve are actually pointed inward. As a consequence, it can be shown that trajectories starting sufficiently close to the origin must approach the origin; hence the origin is asymptotically stable.
A geometric argument for Theorem 9.6.2 follows in a somewhat similar manner. Briefly, suppose that is positive definite, and suppose that given any circle about the origin there is an interior point (x1; y) at which V(x1; y1) > 0. Consider a trajectory that starts at (^1, y1). Along this trajectory it follows from Eq. (8) that V must increase, since V(x1, 1) > 0; furthermore, since V(x1; y) > 0, the trajectory cannot approach the origin because V(0, 0) = 0. This shows that the origin cannot be asymptotically
516
Chapter 9. Nonlinear Differential Equations and Stability
EXAMPLE
2
stable. By further exploiting the fact that V(x, y) > 0, it is possible to show that the origin is an unstable point; however, we will not pursue this argument.
Use Theorem 9.6.1 to show that (0, 0) is a stable critical point for the undamped pendulum equations (2). Also use Theorem 9.6.2 to show that (, 0) is an unstable critical point.
Let V be the total energy given by Eq. (4):
V(x, y) = mgL(1 - cos x) + 2mL2y2. (4)
If we take D to be the domain -/2 < x < /2, -to < y < to, then V is positive
there except at the origin, where it is zero. Thus V is positive definite on D. Further,
as we have already seen,
V = (mgL sin x)(y) + (mL2y)(-gsin x)/L = 0
for all x and y. Thus Vis negative semidefinite on D. Consequently, by the last statement in Theorem 9.6.1, the origin is a stable critical point for the undamped pendulum. Observe that this conclusion cannot be obtained from Theorem 9.3.2 because (0, 0) is a center for the corresponding linear system.
Now consider the critical point (, 0). The Liapunov function given by Eq. (4) is no longer suitable because Theorem 9.6.2 calls for a function V for which Vis either positive or negative definite. To analyze the point (, 0) it is convenient to move this point to the origin by the change of variables x = + u, y = v. Then the differential equations (2) become
g
du/dt = v, dv/dt = sin u, (11)
and the critical point is (0, 0) in the uv-plane. Consider the function
V(u, v) = v sin u (12)
and let Dbe the domain -/4 < u < /4, -to < v < to. Then
V = (v cos u)(v) + (sin u)[(g/L) sin u] = v2cos u + (g/L) sin2 u (13)
is positive definite in D. The only remaining question is whether there are points in
every neighborhood of the origin where V itself is positive. From Eq. (12) we see that V(u, v) > 0 in the first quadrant (where both sin u and v are positive) and in the third quadrant (where both are negative). Thus the conditions of Theorem 9.6.2 are satisfied and the point (0, 0) in the uv-plane, or the point (, 0) in the xy-plane, is unstable. The damped pendulum equations are discussed in Problem 7.
From a practical point of view we are often more interested in the basin of attraction. The following theorem provides some information on this subject.
Theorem 9.6.3 Let the origin be an isolated critical point of the autonomous system (6). Let the function V be continuous and have continuous first partial derivatives. If there is a bounded domain DK containing the origin where V(x, y) < K, V is positive definite,
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