Nonlinear Control Sy.. - Free

102 CHAPTER 3. LYAPUNOV STABILITY I. AUTONOMOUS SYSTEMS
This minimum exists since V(x) is a continuous function in the compact set Q. Thus, for
the trajectory starting at x0, we can write
V(x(t) = V(xo) + t V (x(r)) dr
J0 rt
V (x0) + J ry d-r = V (xo) + ryt.
0
It then follows that x(t) cannot stay forever inside the set U since V(x) is bounded in U.
Thus a trajectory x(t) initiating arbitrarily close to the origin must intersect the boundary
of U. The boundaries of this set are IIxii = E and the surface V(x) = 0. However, the
trajectory x(t) is such that V(x) > and we thus conclude that x(t) leaves the set U
through the sphere IIxii = E. Thus, x = 0 is unstable since given E > 0, we cannot find b > 0
such that
IIxoli < b = 11x(t)II < E.
This completes the proof.
Example 3.23 Consider again the system of Example 3.20
21 = x2+x1(/32-x1-x2)
-t2 = -XI + x2(N2 - x1 - x2)
We showed in Section 3.8 that the origin of this system is an unstable equilibrium point.
We now verify this result using Chetaev's result. Let V(x) = 1/2(x2 + x2). Thus we have
that V (O) = 0, and moreover V (x) > OVx E 1R2 # 0, z. e., V(-) is positive definite. Also
V = (x1,x2)f(x)
(xl + x2)(/32 - xi - x2).
Defining the set U by
U={xEIR2: IIxiI