Massachusetts Institute of Technology Department of Electrical Engineering and Computer Science 6.243j (Fall 2003): DYNAMICS OF NONLINEAR SYSTEMS by A. Megretski Lecture 11: Volume Evolution And System Analysis 1 Lyapunov analysis, which uses monotonicity of a given function of system state along trajectories of a given dynamical system, is a major tool of nonlinear system analysis. It is possible, however, to use monotonicity of volumes of subsets of the state space to predict certain properties of system behavior. This lecture gives an introduction to such methods. 11.1 Formulae for volume evolution This section presents the standard formulae for evolution of volumes. 11.1.1 Weighted volume Let U be an open subset of R n , and ‰ : U ∞? R be a measureable function which is bounded on every compact subset of U. For every hypercube x,r) = {x = [x 1 ; x 2 ; ...; x n ] : |x k ? ?Q(? x k |? r} contained in U, its weighted volume with respect to ‰ is deflned by x n?1 +r ? ? ? ? ?? x 1 +r ? x 2 +r ? ? ? x n +r? ? V ‰ (Q(?x,r)) = ... ‰(x 1 ,x 2 ,...,x n )dx n dx n?1 ... dx 2 dx 1 . ? ? ? ?x 1 ?r x 2 ?r x n?1 ?r x n ?r Without going into the flne details of the measure theory, let us say that the weighted volume of a subset X ‰ U with respect to ‰ is well deflned if there exists M > 0 such that for every ? > 0 there exist (countable) families of cubes {Q 1 k } (all contained in k } and {Q 2 1 Version of October 31, 2003 ? ? ? ? ‰ 2 U ) such that X is contained in the union of Q i k , the union of all Q 2 is contained in the k union of X and Q 1 k , and k ) < ?, V |‰| (Q 2 V |‰| (Q 1 k ) < M, k k in which case the volume V ‰ (X) is (uniquely) deflned as the limit of V ‰ (Q 2 k ) k as ? ? 0 and Q 2 k are required to have empty pair-wise intersections. A common alternative notation for V ‰ (X) is V ‰ (X) = ‰(x)dx. x?X When ‰ · 1, we get a deflnition of the usual (Lebesque) volume. It can be shown that the weighted volume is well deflned for every compact subset of U , and also for every open subset of U for which the closure is contained in U . It is important to remember that not every bounded subset of U has a volume, even when ‰ · 1. 11.1.2 Volume change under a smooth map The rules for variable change in integration allow one to trace the change of weighted volume under a smooth transformation. Theorem 11.1 Let U be an open subset of R n . Let F : U ∞? U be an injective Lipschitz function which is difierentiable on an open subset U 0 of U such that the complement of U 0 in U has zero Lebesque volume. Let ‰ : U ∞? R be a given measureable function which is bounded on every compact subset of U . Then, if ‰-weighted volume is deflned for a subset X ‰ U , ‰-weighted volume is also deflned for F (X), ‰ F -weighted volume is deflned for X, where x))|, dF/dx deflned for ?‰(F (x))| det(dF/dx(? x, ‰ F (?x) = 0, otherwise, and V ‰ (F (X)) = V ‰ F (X). Note that the formula is not always valid for non-injective functions (because of possi- ble “folding”). It is also useful to remember that image of a line segment (zero Lebesque volume when n > 1) under a continuous map could cover a cube (positive Lebesque volume). ? fl fl fl fl fl fl ? 3 11.1.3 Volume change under a difierential ow Let consider the case when the map F = S t is deflned by a smooth difierential ow. Remember that, for a difierential function g : R n ∞? R n , div(g) is the trace of the Jacobian of g. Theorem 11.2 Let U be an open subset of R n . Let f : U ∞? R n and ‰ : U ∞? R be ?continuously difierentiable functions. For T > 0 let U T be the set of vectors x ≤ U such that the ODE x˙(t) = f (x(t)), has a solution x : [0, T ] ∞? U such that x(0) = ?x. Let S T : U T ∞? U be the map deflned by S T (x(0)) = x(T ). Then, if X is contained in a compact subset of U T and has a ‰-weighted volume, the map t ∞? V ‰ (S t (X)) is well deflned, difierentiable, and its derivative at t = 0 is given by dV ‰ (S t (X)) = V div(‰f ) (X). dt Proof According to Theorem 11.2, V ‰ (S t (X)) = ‰(S t (? x))|d?x))| det(dS t (x)/dx(? x. X Note that fl dS t (?x) fl fl = f (?x), dt t=0 and dS t (x)/dx(? x), where x) = ￠(t, ? d￠(t, ?x) df fl = fl ￠(t, ? x) = I. x), ￠(0, ? dt dx x=S t (?x) Hence det(dS t (x)/dx) > 0, and, at t = 0, d d x))| = det(￠(t, ?| det(dS t (x)/dx(? x)) dt dt d￠(t, ?x) fl = trace dt t=0 = div(f )(?x), where the equality d det(A(? )) = det(A(? ))trace A(? ) ?1 dA(? ) d? d? was used. Finally, at t = 0, d ‰(S t (? x))| = (∈‰)(? x) + ‰(? x) = div(‰f )(x).x))| det(dS t (x)/dx(? x)f (? x)div(f )(? dt 4 11.2 Using volume monotonicity in system analysis Results from the previous section allow one to establish invariance (monotonicity) of weighted volumes of sets evolving according to dynamical system equations. This section discusses application of such invariance in stability analysis. 11.2.1 Volume monotonicity and stability Given an ODE model x˙(t) = f(x(t)), (11.1) where f : R n ∞? R n is a continuously difierentiable function, condition div(f) < 0, if satisfled everywhere except possibly a set of zero volume, guarantees strictly monotonic decrease of Lebesque volume of sets of positive volume. This, however, does not guarantee stability. For example, the ODE x˙ 1 = ?2x 1 , x˙ 2 = x 2 does not have a stable equilibrium, while volumes of sets are strictly decreasing with its ow. However, it is possible to make an opposite statement that a system for which a positively weighted volume is strictly monotonically increasing cannot have a stable equi- librium. Theorem 11.3 Let f : R n ∞? R n and ‰ : R n ∞? R be continuously difierentiable functions such that ‰-weighted volume of every ball in R n is positive, and div(f‰)(?x) ? 0 for all ?x ≤ R n . Then system (11.1) has no asymptotically stable equilibria and no asymptotically stable limit cycles. Proof Assume to the contrary that x 0 : R ∞? R n is a stable equilibrium or a stable limit cycle solution of (11.1). Then there exists ? > 0 such that lim min |x(t) ? x 0 (?)| = 0 t?? ? for every solution x = x(￠) of (11.1) such that x(0) belongs to the ball B 0 = {? ?x : |x ? x 0 (0)| ? ?. Let v(t) = V ‰ (S t (B 0 )), where S t is the system ow. By assumption, v is monotonically non-increasing, v(0) = 0, and v(t) ? 0 as t ? ?. The contradiction proves the theorem. 5 11.2.2 Volume monotonicity and strictly invariant sets Let us call a set X ‰ R n strictly invariant for system (11.1) if every maximal solution x = x(t) of (11.1) with x(0) ≤ X is deflned for all t ≤ R and stays in X for all t ≤ R. Obviously, if X is a strictly invariant set then, for every weight ‰, V ‰ (S t (X)) does not change as t changes. Therefore, if one can flnd a ‰ for which div(‰f) > 0 almost everywhere, the strict invariance of X should imply that X is a set of a zero Lebesque volume, i.e. the following theorem is true. Theorem 11.4 Let U be an open subset of R n . Let f : U ∞? R n and ‰ : U ∞? R be continuously difierentiable functions. Assume that div(f‰) > 0 for almost all points of U. Then, if X is a bounded closed subset of U which is strictly invariant for system (11.1), the Lebesque volume of X equals zero. As a special case, when n = 2 and ‰ · 1, we get the Bendixon theorem, which claims that if, in a simply connected region U, ￥(f) > 0 almost everywhere, there exist no non-equilibrium periodic trajectories of (11.1) in U. Indeed, a non-equilibrium periodic trajectory on a plane bounds a strictly invariant set. 11.2.3 Monotonicity of singularly weighted volumes So far, we considered weights which were bounded in the regions of interest. A recent ob- servation by A. Rantzer shows that, when studying asymptotic stability of an equilibrium, it is most beneflcial to consider weights which are singular at the equilibrium. In particular, he has proven the following stability criterion. Theorem 11.5 Let f : R n ∞? R n and ‰ : R n /{0} ∞? R be continuously difierentiable functions such that f(0) = 0, ‰(x)f(x)/|x| is integrable over the set |x|? 1, and div(f‰) > 0 for almost all x ≤ R n . If either ‰ ? 0 or 0 is a locally stable equilibrium of (11.1) then for almost all initial states x(0) the corresponding solution x = x(t) of (11.1) converges to zero as t ? ?. To prove the statement for the case when x = 0 is a stable equilibrium, for every r > 0 consider the set X r of initial conditions x(0) for which sup |x(t)| > r ? T > 0. t?[T,?) The set X is strictly invariant with respect to the ow of (11.1), and has well deflned ‰-weighted volume. Hence, by the strict ‰-weighted volume monotonicity, the Lebesque measure of X r equals zero. Since this is true for all r > 0, almost every solution of (11.1) converges to the origin. 6 Example 11.1 (Rantzer) The system ?2x 1 + x 1 ? x 2 x˙ 1 = 2 2 x˙ 2 = ?6x 2 + 2x 1 x 2 satisfles conditions of Theorem 11.5 with ‰(x) = |x| ?4 .