Massachusetts Institute of Technology Department of Electrical Engineering and Computer Science 6.243j (Fall 2003): DYNAMICS OF NONLINEAR SYSTEMS by A. Megretski Lecture 8: Local Behavior at Eqilibria 1 This lecture presents results which describe local behavior of autonomous systems in terms of Taylor series expansions of system equations in a neigborhood of an equilibrium. 8.1 First order conditions This section describes the relation between eigenvalues of a Jacobian a ? (?x 0 ) and behavior of ODE x˙(t) = a(x(t)) (8.1) or a difierence equation x(t + 1) = a(x(t)) (8.2) in a neigborhood of equilibrium ?x 0 . In the statements below, it is assumed that a : X ≥? R n is a continuous function deflned on an open subset X ‰ R n . It is further assumed that ?x 0 ? X, and there exists an n-by-n matrix A such that |a(? x 0 ) ? A–|x 0 + –) ? a(? ? 0 as |–| ? 0. (8.3) |–| If derivatives da k /dx i of each component a k of a with respect to each cpomponent x i of x exist at ?x 0 , A is the matrix with coe–cients da k /dx i , i.e. the Jacobian of the system. However, difierentiability at a single point ?x 0 does not guarantee that (8.3) holds. On the other hand, (8.3) follows from continuous difierentiability of a in a neigborhood of ?x 0 . 1 Version of October 3, 2003 ? ?? 2 Example 8.1 Function a : R 2 ≥? R 2 , deflned by ? ? 2 ? 2 ? (? 2 x 2 ) 2 ? ?x 1 x 1 x 2 x 1 ? ? 2 x 1 ? a = ? 2 ? 2 x 2 ??x 2 x 1 x 2 + (? 2 ? ? 2 ) 2 x 2x 1 for x ≤ x 1 and ?? = 0, and by a(0) = 0, is difierentiable with respect to ? x 2 at every point ?x ? R 2 , and its Jacobian a ? (0) = A equals minus identity matrix. However, condition (8.3) is not satisfled (note that a ? (? x = 0). x) is not continuous at ? 8.1.1 The continuous time case Let us call an equilibrium ?x 0 of (8.1) exponentially stable if there exist positive real num- bers ?,r,C such that every solution x : [0,T] ≥? X with |x(0) ? ?x 0 | < ? satisfles x 0 |? Ce ?rt |x(0) ? ?|x(t) ? ? x 0 | ? t ? 0. The following theorem can be attributed directly to Lyapunov. Theorem 8.1 Assume that a(?x 0 ) = 0 and condition (8.3) is satisfled. Then (a) if A = a ? (?x 0 ) is a Hurwitz matrix (i.e. if all eigenvalues of A have negative real part) then ?x 0 is a (locally) exponentially stable equilibrium of (8.1); x 0 ) has an eigenvalue with a non-negative real part then ?(b) if A = a ? (? x 0 is not an exponentially stable equilibrium of (8.1); x 0 ) has an eigenvalue with a positive real part then ?(c) if A = a ? (? x 0 is not a stable equilibrium of (8.1). Note that Theorem 8.1 does not cover all possible cases: if A is not a Hurwitz matrix and does not have eigenvalues with positive real part then the statement says very little, and for a good reason: the equilibrium may turn out to be asymptotically stable or unstable. Note also that the equilibrium ?x = 0 from Example 8.1 (where a is difierentiable but does not satisfy (8.3)) is not stable, despite the fact that A = ?I has all eigenvalues at ?1. Example 8.2 The equilibrium ?x = 0 of the ODE x˙(t) = fix(t) + flx(t) 3 is asympotically stable when fi < 0 (this is due to Theorem 8.1), but also when fi = 0 and fl < 0. The equilibrium is not stable when fi > 0 (due to Theorem 8.1), but also when fi = 0 and fl > 0. In addition, the equilibrium is stable but not asymptotically stable when fi = fl = 0. 3 8.1.2 Proof of Theorem 8.1 The proof of (a) can be viewed as an excercise in “storage function construction” outlined in the previous lecture. Indeed, assuming, for simplicity, that ?x 0 = 0, (8.1) can be re- written as x˙(t) = Ax(t) + w(t), w(t) = a(x(t)) ?Ax(t). Here the linear part has standard storage functions V LTI (? x ? P ?x) = ? x, P = P ? with supply rates LTI (? w) = 2? ? P(A? ?x, ? x x + w). In addition, due to (8.3), for every – > 0 there exists ? > 0 such that the nonlinear component w(t) satisfles the sector constraint 2 NL (x(t), w(t)) = –|x(t)| 2 ?|w(t)| ? 0, as long as |x(t)| < ?. Since A is a Hurwitx matrix, P = P ? can be chosen positive deflnite and such that PA + A ? P = ?I. Then (? w) = LTI (? w) + ? NL (? w)x, ? x, ? x, ? x| 2 + 2? ? w ??|? ? ? w|??|w| 2 ,= (?– ?1)|? x ? P ? w| 2 ? (?– ?1)|x| 2 ?2∈P∈￠|x|￠|? ? where ∈P∈ is the largest singular value of P, is a supply rate for the storage function V = V LTI for every constant ? ? 0. When ? = 16∈P∈ and – = 0.25/?, we have x, ? x (? w) ??0.5|?| 2 , which proves that, for |x(t)| < ?, the inequality 1 V (x(t)) ??0.5|x(t)| 2 ?? V (x(t)). 2∈P ?1 ∈ Hence V (x(t)) ? e ?dt V (x(0)) ? t ? 0, where d = 1/2∈P ?1 ∈, as long as |x(t)| < ?. Since ?1 ∈P∈￠|x(t)|? V (x(t)) ?∈P ?1 ∈ ￠|x(t)|, this implies (a). The proofs of (b) and (c) are more involved, based on showing that solutions which start at ?x 0 + –v, where v is an eigenvector of A corresponding to an eigenvalue with a non- negative (strictly positive) real part, cannot converge to ?x 0 quickly enough (respectively, diverge from ?x 0 ). ? 4 To prove (b), take a real number d ? (0,r/2) such that no two eigenvalues of A sum up to ?2d. Then P = P ? be the unique solution of the Lyapunov equation P(A + dI) + (A ? + dI)P = ?I. Note that P is non-singular: otherwise, if Pv = 0 for some v ≤= 0, it follows that ? ? (A ? ?|v| 2 = v (P(A + dI) + (A ? + dI)P)v = (Pv) ? (A + dI)v + v + dI)(Pv) = 0. In addition, P = P ? is not positive semideflnite: since, by assumption, A + dI has an eigenvector u ≤= 0 which corresponds to an eigenvalue ? with a positive real part, we have ?|u| 2 = ?2Re(?)u Pu, hence u ? Pu < 0. Let ? > 0 be small enough so that 2? ? 2 ?x ? Pw ? 0.5|x| for |w|? ?|x|. By assumption, there exists – > 0 such that x) ?A? ? ?|a(? x|? ?|x| for |x|? –. Then d (e 2dt 2dt x(t) ? Px(t)) = e (2dx(t) ? Px(t) + 2x(t) ? PAx(t) + 2x(t) ? P(a(x(t)) ?Ax(t))) dt 2 ??0.5e 2dt |x(t)| as long as x(t) is a solution of (8.1) and |x(t)| ? –. In particular, this means that if x(0) ? Px(0) ??R < 0 and |x(0)|? – then e 2dt x(t) ? Px(t) ??R for as long as |x(t)|? –, which contradicts exponential stability with rate r > 2d. The proof of (c) is similar to that of (a). 8.1.3 The discrete time case The results for the discrete time case are similar to Theorem 8.1, with the real parts of the eigenvalues being replaced by the difierence between their absolute values and 1. Let us call an equilibrium ?x 0 of (8.2) exponentially stable if there exist positive real numbers ?,r,C such that every solution x : [0,T] ≥? X with |x(0) ? ?x 0 | < ? satisfles x 0 |? Ce ?rt |x(0) ? ?|x(t) ? ? x 0 | ? t = 0,1,2,.... Theorem 8.2 Assume that a(?x 0 ) = 0 and condition (8.3) is satisfled. Then ? ? ? ? ? 5 (a) if A = a ? (? 0 ) is a Schur matrix (i.e. if all eigenvalues of A have absolute value less x than one) then ? 0 is a (locally) exponentially stable equilibrium of (8.2); x (b) if A = a ? (? 0 ) has an eigenvalue with absolute value greater than 1 then ? 0 is not anx x exponentially stable equilibrium of (8.2); (c) if A = a ? (? 0 ) has an eigenvalue with absolute value strictly larger than 1 then ?x x 0 is not a stable equilibrium of (8.2). 8.2 Higher order conditions When the Jacobian A = a ? (? x 0 has no eigenvalues x 0 ) of (8.1) evaluated at the equilibrium ? with positive real part, but has some eigenvalues on the imaginary axis, local stability analysis becomes much more complicated. Based on the proof of Theorem 8.1, it is natural to expect that system states corresponding to strictly stable eigenvalues will behave in a predictably stable fashion, and hence the behavior of system states corresponding to the eigenvalues on the imaginary axis will determine local stability or instability of the equilibrium. 8.2.1 A Center Manifold Theorem In this subsection we assume for simplicity that ?x 0 = 0 is the studied equilibrium of (8.1), i.e. a(0) = 0. Assume also that a is k times continuously difierentiable in a neigborhood of ?x 0 = 0, where k ? 1, and that A = a ? (0) has no eigenvalues with positive real part, but has eigenvalues on the imaginary axis, as well as in the open left half plane Re(s) < 0. Then a linear change of coordinates brings A into a block-diagonal form A c 0 A = 0 A s , where A s is a Hurwitz matrix, and all eigenvalues of A c have zero real part. Theorem 8.3 Let a : R n ≥? R n be k ? 2 times continuously difierentiable in a neigbor- hood of ?x 0 = 0. Assume that a(0) = 0 and A c 0 a (0) = A = 0 A s , where A s is a Hurwitz p-by-p matrix, and all eigenvalues of the q-by-q matrix A c have zero real part. Then (a) there exists ? > 0 and a function h : R q ≥? R p , k ? 1 times continuously difier- entiable in a neigborhood of the origin, such that h(0) = 0, h ? (0) = 0, and every solution x(t) = [x c (t); x s (t)] of (8.1) with x s (0) = h(x c (0)) and with |x c (0)| < ? satisfles x s (t) = h(x 0 (t)) for as long as |x c (t)| < ?; 6 (b) for every function h from (a), the equilibrium ?x 0 = 0 of (8.1) is locally stable (asymptotically stable) [unstable] if and only if the equilibrium ?x c = 0 of the ODE dotx c (t) = a([x c (t); h(x c (t))]) (8.4) is locally stable (asymptotically stable) [unstable]; (c) if the equilibrium ?x c = 0 of (8.4) is stable then there exist constants r > 0, > 0 such that for every solution x = x(t) of (8.1) with |x(0)| < r there exists a solution x c = x c (t) of (8.4) such that lim e t |x(t) ? [x c (t); h(x c (t))]| = 0. t?? The set of points x = [x c ; h(? x M c = {? ? x c )] : |? c | < ?}, where ? > 0 is small enough, is called the central manifold of (8.1). Theorem 8.3, called frequently the center manifold theorem, allows one to reduce the dimension of the system to be analyzed from n to q, as long as the function h deflning the central manifold can be calculated exactly or to a su–cient degree of accuracy to judge local stability of (8.4). Example 8.3 This example is taken from Sastry, p. 312. Consider system x˙ 1 (t) = ?x 1 (t) + kx 2 (t) 2 , x˙ 2 (t) = x 1 (t)x 2 (t), where k is a real parameter. In this case n = 2, p = q = 1, A c = 0, A s = ?1, and k can be arbitrarily large. According to Theorem 8.3, there exists a k times difierentiable function h : R ≥? R such that x 1 = h(x 2 ) is an invariant manifold of the ODE (at least, in a neigborhood of the origin). Hence ky 2 = h(y) + h ˙ (y)h(y)y for all su–ciently small y. For the 4th order Taylor series expansion 3 4 ˙ 4 h(y) = h 2 y 2 + h 3 y + h 4 y 4 + o(y ), h(y) = 2h 2 y + 3h 3 y 2 + 4h 4 y + o(y 3 ), comparing the coe–cients on both sides of the ODE for h yields h 2 = k, h 3 = 0, h 4 = ?2k 2 . Hence the center manifols ODE has the form x˙ c (t) = kx c (t) 3 + o(x c (t) 3 ), which means stability for k < 0 and instability for k > 0.