11.2.2 Volume monotonicity and strictly invariant sets Let us call a set X C R strictly invariant for system(11.1)if every maximal solution x=x(t)of(1) with a(0)∈ x is defined for all t∈ R and stays in X for all t∈ R. Obviously, if X is a strictly invariant set then, for every weight p, Ve(St(X)) does not change as t changes. Therefore, if one can find a p for which div(pf)>0 almo 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 Letu be an open subset of R". Let f: U HR andP: UHR be continuously differentiable functions. Assume that div(p)>0 for almost all points ofU Then, if X is a bounded closed subset of U which is strictly invariant for system(11.1), the Lebesque volame of X equals zero As a special case, when n=2 and p= l, we get the Bendiron theorem, which claims that if, in a simply connected region U, >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 ol servation by A. Rantzer shows that, when studying asymptotic stability of an equilibrium, it is most beneficial to consider weights which are singular at the equilibrium In particular, he has proven the following stability criterion Theorem 11.5 Let f: R"HR" and P: R /10 bR be continuously differentiable functions such that f(0)=0, P(a)f(a)/l is integrable over the set r|> 1, and div(p)> 0 for almost all E R. If either p20 or0 is a locally stable equilibrium of(11.1) then for almost all initial states a(0) the corresponding solution x=a(t)of(11.1) converges to zero as t→∞ To prove the statement for the case when a =0 is a stable equilibrium >0 consider the set Xr of initial conditions (0) for which sup r(t>r VT> The set X is strictly invariant with respect to the How of (11.1), and has well defined e-weighted volume. Hence, by the strict weighted volume monotonicity, the Lebesque measure of X equals zero. Since this is true for all r>0, almost every solution of (11.1) converges to the origin5 11.2.2 Volume monotonicity and strictly invariant sets Let us call a set X � Rn strictly invariant for system (11.1) if every maximal solution x = x(t) of (11.1) with x(0) ≤ X is defined 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�(St(X)) does not change as t changes. Therefore, if one can find 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 Rn. Let f : U ∞� Rn and � : U ∞� R be continuously differentiable 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 beneficial to consider weights which are singular at the equilibrium. In particular, he has proven the following stability criterion. Theorem 11.5 Let f : Rn ∞� Rn and � : Rn/{0} ∞� R be continuously differentiable 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 ≤ Rn. 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 Xr 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 flow of (11.1), and has well defined �-weighted volume. Hence, by the strict �-weighted volume monotonicity, the Lebesque measure of Xr equals zero. Since this is true for all r > 0, almost every solution of (11.1) converges to the origin