x(t) = x + f (t) b(f(t)) 2 dt + o( 1 2ǫ 0 f (t) b(f(t)) 2 dt Let G be an open set containing a unique stable equlibrium point x 0 for the ODE

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1 Eit Problem. Consider ǫ (t) = + t 0 b( ǫ (s))ds + ǫβ(t) let Q,ǫ be the distribution of the solution ǫ. As ǫ 0 the measure Q,ǫ concentrates on the trajectory which is the solution of (t) = + t 0 b((s))ds There is a large deviation principle for {Q,ǫ } on C[[0, T]; R d ]. Q,ǫ (A) = ep[ More precisely for closed sets C for open sets G, f( ) A lim supǫlog Q y,ǫ (C) y lim y ǫlog Q y,ǫ (G) T + o( 2ǫ 0 ǫ )] f( ) C f( ) G T T Let G be an open set containing a unique stable equlibrium point 0 for the ODE ẋ(t) = b((t)) i.e. any solution of the ODE starting from any point in the closure Ḡ tends to 0 as t, remaining in G for all t > 0. For instance assume that G is smooth b 0 on the boundary δg points inward at every point. For any G z δg let U(T,, z) = f:,f(t)=z f(t) G for t<t T U(, z) = U(T, z) T>0 Let z 0 δg be such that U( 0, z 0 ) < U( 0, z) for all z δg, z z 0. If τ is the eit time (τ) is the eit place from G, then for any G any neighborhood N of z 0, Theorem: lim Q,ǫ[(τ) / N] 0

2 Remark. No matter where the process starts inside G intially it will follow the ODE, be driven towards 0, slow down as it reaches 0 hang around there for a very long time. Let us take two neighborhoods S, S 2 around 0, with 0 S S S 2. It is not hard to see that U(, z) is a continuous function of z, given N, we can pick S, S 2 such that U(, z) sup U(, z δs 2 z Nc 0 ) + η δs 2 We will estimate the following probabilities: if τ be the eit time from G S c lim sup ǫ sup log Q,ǫ [(τ ) N c ] δs 2 δs 2 z N c U(, z) lim ǫ δs 2 log Q,ǫ [(τ ) N] sup δs 2 U(, z 0 ) This will do it. The picture is the process will sooner or later eit from S c. But most of the time it will be pulled back to 0. There is a very small chance that it will eit in N even smaller chance of eiting from N c. So it is most likely to eit from N. First we estimate the probability that eit from S c takes too long. lim sup T lim supǫlog sup Q,ǫ [τ T] = S c Otherwise there will be paths with T 0 bounded T large. This means there will be paths with T 0 small T large. This in turn means solutions of ODE remaining in S c for too long. If the paths do not hang around for too long, the large deviation estimate applies it is much more likely to eit from N, than from N c. A special case is the gradient flow, where b() = ( V )(). 0 is a local minimum of V. Then it is not hard to see that U( 0, z) = 2[V (z) V ( 0 )]. Invariant distributions. L = a i,j ()D i,j + 2 j i,j b j ()D j µ is probability measure on R d such that (Lu)()dµ() = 0 for all smooth u with compact support. Suppose there is a unique process corresponding to L, is µ an invariant distribution fro the process? Proof dpends on duality consequently finding enough classical solutions for the equation u t = Lu 2

3 or the resolvent equation λu Lu = f which require ellipticity Hölder continuity. Assume only that the coefficients are continuous, but the process is unique. If we know that dµ = φd with φ L q we can use the L p theory in the elliptic case. To prove it in general requires several steps. Invariance Principle. Theorem: Suppose π h (, dy) is a Markov Chain such that, for every smooth u with compact support [u(y) u()]π h (, dy) (Lu)() h uniformy on compact sets, there eists a unique process with out eplosion for L, then the interpolated Markov Chain converges to the process. In particular lim h 0 nh t f(y)πh n (, dy) (T tf)() = f(y)p(t,, dy) where p is the transition probability of the process corresponding to L. Proof: Step. Let us interpolate the Markov chain call the process P h. Let us take smooth cut off function φ R () define It is easy to see that h π R h (, y) = φr ()π h (, dy) + ( φ R ())δ (dy) [u(y) u()]π R h (, dy) (LR u)() = φ R ()(Lu)() uniformly in. We will prove that the processes P R h, are tight. Let τ ǫ be the eit time from the ball of radius ǫ for the process starting from. We want to estimate sup sup Ph,[τ R ǫ δ] = F(ǫ, δ) h If u ǫ is a smooth function that is in a ball of radius ǫ 2 0 outside a ball of radius ǫ, L R u ǫ () C ǫ [u(y) u()]πh R (, dy) C ǫ h In particular is a super-martingale under P R,h u(x(nh)) u() nhc ǫ P R,h [τ ǫ δ] E[u(τ ǫ δ)] δc ǫ 3

4 Let τ, τ 2,..., τ N be the successive times at which X(nh) gets away a distance ǫ from the previous (τ i ). We proceed till τ N > T. We estimate the following. sup E[e τ i+ F τi ] ρ < ω,h P[N k] P[τ + + τ k T] e T E[e (τ + +τ k ) ] e T ρ k From the locality of L, it follows that Therefore P[min(τ,..., τ k ) δ)] kδc ǫ π R h (, B(, ǫ) c ) = o(h) [ ] sup P,h R sup X((j + )h X(jh) ǫ 0 j n as h 0. This is enough to control the oscillations. We can use the control on the modulus of continuity to prove tightness. If P R is any limit it is a solution to the martingale problem for L R. This agrees with L until th eit itme from B R the ball of radius R. Since there is no eplotion if R is large τ R is large, is bigger than T, with probability nearly one so P R,h P,h are close the limit of P,h as h 0 is P. Finally to prove that µ is the invariant measure we will construct a Markov Chain {π h (, dy)}, for which µ is inavariant which converges to {P }. Given L, we construct the resolvent Π h = (I hl) on the range of D h of bounded functions with two bounded derivatives under (I hl). The maimum principle guarantees that Π h is well defined is positivity preserving. We define a linear functional Λ on functions of two variables of the form 0 g(, y) = v 0 (y) + i u i ()w i (y) with u, w being bounded continuous functions w i = v i hlv i D h, by n Λ(g) = v 0 (y)dµ(y) + u i ()v i ()dµ() Suppose Λ is nonnegative we etend it as a non negative linear functional. Then both the marginals of Λ are µ. [Note that we can take v = the remaining v as 0. Then g(, y) = u ()]. If we take the r.c.p.d π h (, dy), µπ h = µ i= π h (v hlv) = v for smooth v we have for v with compact support. h (π hv v) = π h Lv Lv 4

5 Suppose g(, y) 0. Then consider the function n u i ()v i = Φ(v) i= defined for v = (v,..., v n ) R n. Φ is concave Φ(v() t(lv)()) is a conve function of t for all. So is the integral Therefore for h 0, i ψ(t) = Φ(v() t(lv)())dµ() ψ (0) = Φ ui (v())(lv i )()dµ() [LΦ(v)]()dµ() = 0 ψ(h) = Φ(v() h(lv)())dµ() Φ(v())dµ() We can approimate Φ by smooth conve functions. Denote v i hlv i = w i. Then [v 0 () + u i ()v i ()]dµ() [v 0 () + Φ(v())]dµ() [v 0 () + Φ(w())]dµ() But [v 0 (y) + Φ(w(y)] = [v 0(y) + i u i ()w i (y)] = g(, y) 0 5