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1. Chapter 3: Lyapunov Stability I: Autonomous Systems 1 Denitions Consider the autonomous system x = f(x) f : D Rn (1) Denition 1 : xe is an equilibrium point of (1) if f(xe) = 0. We want to know whether or not the trajecories near an equilibrium point are well behaved. Denition 2 : xe is said to be stable if for each > 0, = () > 0 x(0) xe < x(t) xe < t t0 otherwise, the equilibrium point is said to be unstable. IMPORTANT: This notion applies to the equilibrium, not the system. A dynamical system can have several equilibrium points. 1 2. E T &% '$ sxe r x0 dds r rr rrrj Figure 1: Stable equilibrium point. E T &% '$ s xe r x0 rr rr rrj Figure 2: Asymptotically stable equilibrium point. Denition 3 : xe of the system (1) is said to be convergent if there exists 1 > 0 : x(0) xe < 1 lim t x(t) = xe. Equivalently, xe is convergent if for any given 1 > 0, T such that x(0) xe < 1 x(t) xe < 1 t t0 + T. Denition 4 : xe is said to be asymptotically stable if it is both stable and con- vergent. 2 3. Denition 5 : xe is said to be (locally) exponentially stable if there exist two real constants , > 0 such that x(t) xe x(0) xe et t > 0 (2) whenever x(0) xe < . It is said to be globally exponentially stable if (2) holds for any x Rn . Note: Clearly, exponential stability implies asymptotic stability. The con- verse is, however, not true. Remarks: We can always assume that xe = 0. Given any other equilibrium point we can make a change of variables and dene a new system with an equilibrium point at x = 0. Dene: y = x xe y = x = f(x) f(x) = f(y + xe) g(y). Thus, the equilibrium point ye of the new systems y = g(y) is ye = 0, since g(0) = f(0 + xe) = f(xe) = 0. Thus, xe is stable for the system x = f(x) if and only if y = 0 is stable for the system y = g(y). 3 4. 2 Positive Denite Functions Denition 6 : V : D R is positive semi denite in D if (i) 0 D and V (0) = 0. (ii) V (x) 0, x in D {0}. V : D R is positive denite in D if (ii) V (x) > 0 in D {0}. V : D R is negative denite (semi denite)in D if V is positive denite (semi denite). 4 5. Example 1 : (Quadratic form) V (x) : Rn R = xT Qx , Q Rnn , Q = QT . Since Q = QT , its eigenvalues i, i = 1, n, are all real. Thus V () positive denite i > 0, i = 1, , n V () positive semidenite i 0, i = 1, , n V () negative denite i < 0, i = 1, , n V () negative semidenite i 0, i = 1, , n 2 Notation: Given a dynamical system and a function V we will denote V (x) = dV dt = V x dx dt = V f(x) = V x1 , V x2 , , V xn f1(x) ... fn(x) . Example 2 : Let x = ax1 bx2 + cos x1 and dene V = x2 1 + x2 2. Thus, we have V (x) = V x f(x) = [2x1, 2x2] ax1 bx2 + cos x1 = 2ax2 1 + 2bx2 2 + 2x2 cos x1. 2 Notice that the V (x) depends on the systems equation f(x) and thus it will be dierent for dierent systems. 5 6. Note: V = xT Qx, where Q = QT if Q = Qsym + Qasym Qsym = Q + QT 2 Qasym = Q QT 2 Then, V = xT Qsymx + xT Qasymx where, Qasym = 0 a b a ... ... b 0 i.e. skew symmetric and xT Qasymx = 0 V = xT Qsymx Q is always symmetric Theorem: Rayleigh-Ritz theorem min {Q} x 2 xT Qx max {Q} x 2 6 7. 3 Stability Theorems Theorem 1 : (Lyapunov Stability Theorem) Let x = 0 be an equilibrium point of x = f(x), and let V : D R be a continuously dierentiable function such that (i) V (0) = 0, (ii) V (x) > 0 in D {0}, (iii) V (x) 0 in D {0}, thus x = 0 is stable. Theorem 2 : (Asymptotic Stability Theorem) Under the conditions of Theorem 1, if V () is such that (i) V (0) = 0, (ii) V (x) > 0 in D {0}, (iii) V (x) < 0 in D {0}, thus x = 0 is asymptotically stable. 7 8. Proof of theorem 1: Choose r > 0 and dene Br = {x Rn : x r} D is contained in D. Br, so dened is a closed and bounded (compact) set (a sphere). We now construct a Lyapunov surface inside Br and show that all trajec- tories starting near x = 0 remain inside the surface. Let = min x =r V (x) (thus > 0) Now choose (0, ) and denote = {x Br : V (x) }. Thus, by construction, Br. Assume now that x(0) . V (x) 0 V (x) V (x(0)) t 0. Trajectories starting in at t = 0 stays inside for all t 0. By the continuity of V (x), > 0: x < V (x) < (B Br). It then follows that x(0) < x(t) Br t > 0 and then x(0) < x(t) < r t 0 which means that the equilibrium x = 0 is stable. 2 Proof of theorem 2: Similar, only that V (x) < 0 in D implies that is shrinking until eventually it becomes the single point x = 0. 2 8