Download (direct link):
(b) Asymptotic stability of the critical point (0, 0) can be shown by constructing a better Liapunov function as was done in part (d) of Problem 7. However, the analysis for a general function g is somewhat sophisticated and we only mention that an appropriate form for V is
V(x, y) = 2 y2 + Ayg(x) + j g(s) ds,
where A is a positive constant to be chosen so that V is positive definite and V is negative definite. For the pendulum problem [g(x) = sin x] use V as given by the preceding equation with A = ³ to show that the origin is asymptotically stable.
Hint: Use sin x = x — ax3/3! and cos x = 1 — âx2/2! where a and â depend on x, but 0 <a< 1 and 0
< â < 1 for —n/2 < x < n/2; let x = r cos9, y = r sin9, and showthat V(r cos9, r sin9) = — 1 r2[1 + 2 sin29 + h(r, 9)], where lh(r, 9)| < 2 if r is sufficiently small. To showthat V is positive definite use cos x = 1 — x2/2 + yx4/4!, where ó depends on x, but 0 < ó < 1 for —æ/2 < x < n/2.
In Problems 10 and 11 we will prove part of Theorem 9.3.2: If the critical point (0, 0) of the almost linear system
dx/dt = anx + a12y + F1(x, y), dy/dt = a21 x + a22y + Gl(x, y) (i)
is an asymptotically stable critical point of the corresponding linear system
dx/dt = an x + a12 y, dy/dt = a21 x + a22 y, (ii)
then it is an asymptotically stable critical point of the almost linear system (i). Problem 12 deals with the corresponding result for instability.
9.7 Periodic Solutions and Limit Cycles
10. Consider the linear system (ii).
(a) Since (0, 0) is an asymptotically stable critical point, show that a11 + a22 < 0 and an a22 - a12a21 > 0. (See Problem 21 of Section 9.1.)
(b) Construct a Liapunov function V(x, y) = Ax2 + Bxy + Cy2 such that V is positive definite and Vis negative definite. One way to ensure that Vis negative definite is to choose A, B, and C so that V(x, y) = -x2 - y2. Show that this leads to the result
A_ a21 + a22 + (a11 a22 - a12a2j) â _ a12a22 + a11a21
= 2Ä ’ = Ä '
+ a12 + (a11 a22 - a12a21)
where Ä = (a11 + a22)(a11a22 - a12a21).
(c) Using the result of part (a) show that A > 0 and then show (several steps of algebra are required) that
4AC - B2 = (a21 + a22 + a21 + a22)(a11a22 - a12a2j) + 2(a11 a22 - a12a21)2 > q
Thus by Theorem 9.6.4, V is positive definite.
11. In this problem we show that the Liapunov function constructed in the preceding problem is also a Liapunov function for the almost linear system (i). We must show that there is some region containing the origin for which Vis negative definite.
(a) Show that
V(x, y) = -(x2 + y2) + (2Ax + By)F1(x, y) + (Bx + 2Cy)G1(x, y).
(b) Recall that F1(x, y)/r — Q and G1(x, y)/r — Q as r = (x2 + y2)1/2 — Q. This means that given any e > Q there exists a circle r = R about the origin such that for Q < r < R, | F1(x, y)| < er,and | G1 (x, y)| <er.Letting Mbe the maximum of |2 A|, | B|, and |2C|, show by introducing polar coordinates that R can be chosen so that V(x, y) < Q for r < R.
Hint: Choose e sufficiently small in terms of M.
12. In this problem we prove a part of Theorem 9.3.2 relating to instability.
(a) Show that if a11 + a22 > Q and a11a22 - a12 a21 > Q, then the critical point (Q, Q) of the linear system (ii) is unstable.
(b) The same result holds for the almost linear system (i). As in Problems 10 and 11 construct a positive definite function V such that V(x, y) = x2 + y2 and hence is positive definite, and then invoke Theorem 9.6.2.
9.7 Periodic Solutions and Limit Cycles
In this section we discuss further the possible existence of periodic solutions of second order autonomous systems
x' = f(x). (1)
Such solutions satisfy the relation
x(t + T) = x(t) (2)
Chapter 9. Nonlinear Differential Equations and Stability
for all t and for some nonnegative constant T called the period. The corresponding trajectories are closed curves in the phase plane. Periodic solutions often play an important role in physical problems because they represent phenomena that occur repeatedly. In many situations a periodic solution represents a “final state” toward which all “neighboring” solutions tend as the transients due to the initial conditions die out.
A special case of aperiodic solution is a constant solution x = x0, which corresponds to a critical point of the autonomous system. Such a solution is clearly periodic with any period. In this section, when we speak of a periodic solution, we mean a nonconstant periodic solution. In this case the period T is positive and is usually chosen as the smallest positive number for which Eq. (2) is valid.