Random and Deterministic perturbations of dynamical systems. Leonid Koralov

Transcription

1 Random and Deterministic perturbations of dynamical systems Leonid Koralov

2 - M. Freidlin, L. Koralov Metastability for Nonlinear Random Perturbations of Dynamical Systems, Stochastic Processes and Applications 120 (2010), no. 7, D. Dolgopyat, L. Koralov Averaging of incompressible flows on 2-d surfaces, submitted to Journal of American Math. Society. - D. Dolgopyat, M. Freidlin, L. Koralov Deterministic and stochastic perturbations of areapreserving flows on a 2-d torus (to appear in Ergodic Theory and Dynamical Systems). 1

3 Part 1. Incompressible flow: ẋ(t) =v(x(t)), x(0) = x 0 R 2 or x 0 M. (a) Hamiltonian flows. 3

4 Perturbation: dx ε t = 1 ε v(xε t )dt + σ(x κ,ε t )dw t (random), dx ε t = 1 ε v(xε t )dt+b(xκ,ε t )dt (deterministic). The dynamics consists of the fast motion (with speed of order 1/ε) along the unperturbed trajectories together with the slow motion (with speed of order 1) in the direction transversal to the unperturbed trajectories. 4

5 Averaging - consider h : R 2 G. Then h(x ε t ) Y t as ε 0. Locally (away from the vertices of the graph): dy b(y t dt = t ), (deterministic), where T (Y t ) T (h) = γ(h) dl H, b(h) = γ(h) b, H dl, H dy t = σ(y t )dw t +b(y t )dt (random perturbations). Behavior at the vertices. Random perturbations - Freidlin and Wentzell. Deterministic perturbations - regularization required. (Brin and Freidlin). 5

6 (b) Locally Hamiltonian flows (there are regions where the unperturbed dynamimcs is ergodic). Example: H = H 0 (x 1,x 2 )+αx 1 + βx 2, α/β - irrational. 6

7 M - manifold with an area form, v - incompressible vector field, X ε t - process with generator Lε = 1 ε L v + L D. 7

8 Unperturbed dynamics: U 1,..., U m - periodic sets E 1,..., E n - ergodic components Flow on E i is isomorphic to a special flow over an interval exchange transformation. Graph: - Each edge corresponds to one of U k - Three types of vertices: (a) Those corresponding to E i, (b) Those corresponding to saddle points, (c) Those corresponding to equilibriums (but not saddles). The flow is Hamiltonian on U k with a Hamiltonian H. Denote: h k -coordinateoni k. 8

9 Theorem 1 The measure on on C([0, ), G) induced by the process Yt ε = h(xt ε) converges weakly to the measure induced by the process with the generator L with the initial distribution h(x0 ε). The limiting process is described via its generator L, which is defined as follows. Let L k f(h k )=a k (h k )f + b k (h k )f be the differential operator on the interior of the edge I k (coefficients are defined below). For f D(L), we define Lf = L k f in the interior of each edge, and as the limit of L k f at the endpoints of I k. 9

10 D(L) consists of f C(G) C 2 (I k ) such that (a) lim hk 0 L k f(h k )=q V exist and are the same for all edges entering the same vertex V. (b) At vertices corresponding to E i : n p V k k=1 lim f (h k ) = lim L h k 0 hk kf(h k ). 0 (the same with 0 instead of q V corresponding to saddles). for vertices 10

11 Coefficients: In local coordinates in U k (ω = dxdy): dx ε t = 1 ε v(xε t )dt + u(x ε t )dt + σ(x ε t )dw t. Then, a k (h k )= 1 2 T 1 (h k ) γ k (h k ) α H, H dl H and b k (h k )= 1 2 T 1 (h k ) γ k (h k ) 2 u, H + α H dl, H where α = σσ. p V k = ±1 2 γ k α H, H H dl. 11

12 Ingredients of the proof. (1) Assume (temporarily) that the area measure λ is invariant for the process for each ε. For the limit Y t of Yt ε have = h(xt ε ), we should T E[f(Y T ) f(y 0 ) Lf(Y s)ds] =0. 0 Need to prove the following lemma. Lemma1 For each function f D(L) and each T>0we have E x [f(h(xt ε T )) f(h(xε 0 )) 0 Lf(h(Xε s))ds] 0 uniformly in x T 2 as ε 0. 12

13 (2) Localization (can deal with a star-shaped graph with one accessible vertex) 13

14 (3) Need: E x [f(h(x ε T )) f(h(xε 0 )) T 0 Lf(h(Xε s ))ds] 0 Split [0,T] into intervals: [0,σ 0 ], [σ 0,τ 1 ], [τ 1,σ 1 ], [σ 1,τ 2 ],... On intervals [τ n,σ n ] (inside periodic component) - averaging (Freidlin-Wentzell) with small modifications. 14

15 On intervals [σ n,τ n+1 ] (getting from the ergodic component into the periodic component): E x [f(h(x ε τ n+1 )) f(h(x ε σ n )) τn+1 σ n Lf(h(X ε s))ds] E ν [f(h(x ε τ )) f(h(xε 0 )) τ f (0)ε α E ν τ Lf(0). 0 Lf(h(Xε s ))ds] - How can we calculate E ν τ? - Why can we assume that we start with the invariant measure ν? 15

16 If λ is invariant: E ν τ λ(e) E μσ λ(u),so E ν τ λ(e) λ(u) E μσ const ε α. If λ is not invariant: consider d X ε t = 1 ε v( X ε t )dt + ũ( X ε t )dt + σ( X ε t )dw t, (replace u by some ũ so that λ is invariant for the new process). By the Girsanov Theorem: ν ν, E ν τ E ν τ. So, E ν τ λ(e) λ(u) E μ σ (the gluing conditions are the same as for the measure-preserving process). 16

17 (4) Why does E x σ k 0asε 0? (time to reach U k ) Let u ε (t, y), y M \U k, be the probability that the process starting at y does not reach U k before time t. u ε ( (t, y) = L D + 1 ) t ε L v u ε u ε (0,y)=1, y M \ U k, u ε (t, y) =0, t > 0. 17

18 (a) Lemma (Zlatos): All H 1 0 (M\U k)-eigenvalues for v are zero on E implies that the L 2 (E)- norm (and so L 1 (E)-norm) of u ε (t, ) tends to zero as ε 0 for each t>0. (b) A uniform bound on fundamental solution doesn t get affected by adding an incompressible drift term. (a) and (b) imply that E x σ 0. With some effort possible to show that E x σ k 0. 18

19 Part 2: Averaging of deterministic perturbations Recall dx κ,ε t = 1 ε v(xκ,ε t )dt + b(x κ,ε )dt+ t κu(xt κ,ε )dt + κσ(xt κ,ε )dw t. Let Yt κ,ε = h(xt κ,ε ) be the corresponding process on the graph G. We demonstrated that the distribution of Yt κ,ε converges, as ε 0, to the distribution of a limiting process, which will be denoted by Zt κ. Zκ t, in turn, converges to the distribution of a limiting Markov process on G when κ 0. 19

20 The limiting process Z t can be described as follows. It is a Markov process with continuous trajectories which moves deterministically along an edge I k of the graph with the speed b k (h k )= 1 2 (T k(h k )) 1 2 b, H dl. γ k (h k ) H If the process reaches V corresponding to an ergodic component, then it either remains at V forever or spends exponential time in V and then continues with deterministic motion away from V along a randomly selected edge (with probabilities which can be specified). The same if V corresponds to a saddle point, but no exponential delay. 20

21 Theorem 2 The measure on on C([0, ), G) induced by the process Zt κ converges weakly to the measure induced by the process Z t with the initial distribution h(x0 ε). The process Z t is defined by the deterministic system. The stochastic perturbations are used just for regularization purposes. 21

22 Part 3. Ẋ x t = b(xx t ), Xx 0 = x Rd ; dx x,ε t = b(xt x,ε )dt + εσ(xt x,ε )dw t, X x,ε 0 = x. in terms of PDEs: u ε t = ε2 2 d i,j=1 a ij (x) 2 u ε x i x j + b(x) x u ε, u ε (0,x)=g(x), x R d. u ε (t, x) =Eg(X x,ε t ).

23 Action functional: S 0,T (ϕ) = 1 2 T 0 d i,j=1 if ϕ - absolutely continuous, a ij (ϕ t )( ϕ i t b i (ϕ t ))( ϕ j t b j(ϕ t ))dt, T S 0,T (ϕ) =+, otherwise. a ij =(a 1 ) ij. Quasi-potential: V mn = V (O m,o n )= inf T {S 0,T (ϕ) :ϕ(0) = O m,ϕ(t )=O n }. 22

24 τ ε mn - the time it takes the process to go from O m to a small neighborhood of O n. τ ε mn exp(v mn /ε 2 ), Consider the process (and solution of PDE) at times t(ε) withln(t(ε)) λ/ε 2. Suppose, for example, that x D 1 and V 12 <V 21. If λ<v 12,thenu ε (t(ε),x) g(o 1 ). If λ>v 12,thenu ε (t(ε),x) g(o 2 ). 23

25 Nonlinear problem: u ε t = ε2 2 d i,j=1 a ij (x, u ε ) 2 u ε x i x j + b(x) x u ε, u ε (0,x)=g. equivalent to the system dx t,x,ε s = b(x t,x,ε s )dt+εσ(xs t,x,ε,u ε (t s, Xs t,x,ε ))dw s, u ε (t, x) =Eg(X t,x,ε t ). Construct Ṽ12 using a(t, g(x 0 )). If λ<ṽ12, then still u ε (t(ε),x) g(o 1 ). If λ>ṽ12, then new effects appear for u ε and the processes. 24

26 Result in the non-linear case (2 equilibriums): Theorem: lim u ε (exp(λ/ε 2 ),x)=c n (λ), x D n. ε 0 Corollary: If x D 1, then the distribution of X exp(λ/ε2 ),x,ε exp(λ/ε 2 ) converges to the measure μ λ 1 = a 1 δ O1 + a 2 δ O2, where the coefficients a 1 and a 2 can be found from the equations c 1 (λ) = a 1 g(o 1 )+a 2 g(o 2 ), a 1 + a 2 =1. 25

27 Multiple equilibriums. - Need to look at V a(x,c) mn = V a(x,c) (O m,o n ), which determine the hierarchy of cycles; - The hierarchy of cycles may evolve in time (i.e., depends on λ). 26