Harnack Inequalities and Applications for Stochastic Equations

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p. 1/32 Harnack Inequalities and Applications for Stochastic Equations PhD Thesis Defense Shun-Xiang Ouyang Under the Supervision of Prof. Michael Röckner & Prof. Feng-Yu Wang March 6, 29

p. 2/32 Outline Introduction; Harnack inequalities for OU processes: show the main technique; Harnack inequalities for Other Cases.

p. 3/32 Part I. Introduction We introduce the four keywords in the full title Harnack Inequalities and Applications for the Transition semigroups Associated with Stochastic Equations

Transition Semigroups Let H be a real separable Hilbert space with inner product, and norm. Let X be a solution process to some stochastic equation. X : [, ) Ω H is a map. ω Ω, X.(ω), X.(ω): [, ) H is a path. t [, ), X t ( ), X t : (Ω,F t, P) (H, B(H)) is a r.v.. H X x t (ω) µ x t := P (X x t ) 1. x t P t f(x) f B b (H) = f(xt x )dp= E P f(xt x ) Ω = f(z)dµ x t. H p. 4/32

p. 5/32 Harnack Inequality & Applications The Harnack inequality we are interested in is of the following form (P t f) α (x) C(t,α,x,y)P t f α (y), for every x,y H,f C + b (H),α > 1, where C(t,α,x,y) is a constant independent of f. It was first introduced by Wang in 1997. Applications including Regularizing Property, Contractivity, Heat Kernel Estimates, Bound of the Norm of Heat Kernel etc..

p. 6/32 How to get Harnack inequalities? 1. Wang, 1997. Semigroup Calculus. For diffusions on manifolds with curvature bounded below. Calculate φ (s), where φ(s) = log P s (P t s f) α (x s ) and x s is a geodesic connecting x and y on [,t]. 2. Röckner & Wang, 23. Semigroup Calculus+Relative Density(for Lévy Case) For OU Process (Generalized Mehler Semigroup). 3. Arnaudon et al., 26. Coupling+Girsanov Transformation For diffusions on mainifolds with curvature unbounded below.

p. 7/32 Main Methods There are two levels of Measure Transformation Relative Density Coupling + Girsanov Trans. Hölder s Ineq. + Measure Trans. Trans. of µ t on (H, B(H)) Trans. of P on (Ω, F)

Part II Main Technique p. 8/32

Ornstein-Uhlenbeck Processes We shall consider OU process dx t = AX t dt + dz t to show the main idea for other cases. Single-valued SEs: dx t = AX t dt + F(t,X t )dt + dl t. SE S(O)DE with irregular drift OU process Gauss SPDE (linear) Lévy fbm BM { quasilinear Gauss OU Perturbation of process Multi-valued SEs: dx t ÃX t dt + BX t dt + σ t dw t. MSDEs (on R d ) & MSEEs (on Gelfand triple) p. 9/32

p. 1/32 Deterministic Linear Diff. Equation Recall that x t = e kt x + t r e k(t s) u s ds slolves { dxt = kx t + ru t dt, x = x R. Generalization: k A, e kt S t = e At, r R. Solve the linear control system similarly { dxt = Ax t dt + R 1/2 u t dt, x = x H. x t = S t x + t S t s R 1/2 u s ds.

p. 11/32 Null Controllable Consider { dxt = Ax t dt + R 1/2 u t dt, x = y x R. Null Controllable in time T if for each initial data, u L 2 ([,T], H) s.t. x T =. Null Controllable S T (H) Q 1/2 T (H), Q T = T S t RS t dt. Minimal Energy: Let Γ T = Q 1/2 T S T. Then { } T Γ T (x y) 2 = inf u s 2 ds: u L 2,x = y x,x T =.

p. 12/32 Gauss OU Processes Mild solution of the Langevin s Equation: dx t = AX t dt + R 1/2 dw t, X = x H. x Ornstein-Uhlenbeck process (blue curve) X t = e κt x + t e κ(t s) dw s deterministic process (black curve) e κt x o random perturbation (red curve) t e κ(t s) dw s X t = S t x + t S t s R 1/2 dw s, t.

p. 13/32 Gauss OU Semigroup Assume that Q t, t, is of trace class. (For the exixtence of the solution.) t S t s R 1/2 dw s µ t := N(,Q t ). Transition semigroup of X t = S t x + t S t s R 1/2 dw s, t. is given by P t f(x) = H f(s t x + z)µ t (dz).

p. 14/32 Gaussian Measures N(m, Q) Suppose m H (mean); Q: a trace class operator on H (variance). µ := N(m,Q) is a Gaussian measure on H if ˆµ(h) = exp(i m,h 1 2 Qh,h ), where ˆµ is the Fourier transformation of µ: ˆµ(h) := exp(i x, h )µ(dx). H

Cameron-Martin Formula: Infty Dim. Consider two Gaussian measures on H µ = N(a,Q), & ν = N(,Q) If a / Q 1/2 (H) then µ ν. If a Q 1/2 (H) then µ ν and dµ (x) = exp ( Q 1/2 x,q 1/2 a 12 ) dν Q 1/2 a 2 for every x H. Here Q 1/2 x,q 1/2 a := k=1 x,e k a,e j λ k, where Qe j = λ j e j : e j is eigenvectors of Q with eigenvalue λ j. p. 15/32

p. 16/32 Assume Q t is of trace class; S t (H) Q 1/2 (H). Harnack Inequality For Gauss OU semigroup P t : (P t f) α (x) exp ( ) β 2 Γ t(x y) 2 (P t f α (y)), t >, f C + b (H), x,y H and α,β > 1 satisfying 1 α + 1 β = 1.

P t f(x)= H Proof f(s t x + z)n(,q t )(dz) = f(s t y + z) dn(s t(x y),q t ) (z)dn(,q t )(z) H dn(,q t ) ( = f(s t y + z) exp 1 H 2 Q 1/2 t S t (x y) 2 ) + Q 1/2 t S t (x y),q 1/2 t z µ t (dz) exp ( 12 ) ( Γ t(x y) 2 f α (S t y + z)µ t (dz) H ( H exp [ β Q 1/2 t S t (x y),q 1/2 t z ( ) β 1 = exp Γ t (x y) 2 (P t f α (y)) 1 α 2 ] ) 1 β ) 1 α p. 17/32

p. 18/32 Transformation of Measures on Ω Let (Xt x Assume, Q) and (Y y t, P) be two processes on (Ω,F T). P t f(x) = E Q f(x x t ), P tf(y) = E Q f(y y t ); Q = ρ T P; X x T = Y y T. P T f(x)= E Q f(x x T )= E Qf(Y y T ) = E P ρ T f(y y T ) (E P ρ β T )1/β (E P f α (Y y T ))1/α = (E P ρ β T )1/β (P T f α (y)) 1/α, we see (P T f) α (x) P t f α (y) [E P ρ β T ] α β.

Lévy Process Lévy process is a process inhomogeneous in time and space, i.e., with independent and stationary increments. Let (Z t ) t T be a Lévy process with characteristic triple (b,r,ν) under probability measure (Ω, (F t ) t T, P). with E P exp(i z,z t ) = exp( tη(z)), z H, η(z) = i z,b + 1 2 Rz,z + H [ 1 exp(i z,x ) + i z,x ½{ x 1} (x) ] ν(dx). Lévy-Itô decomposition: Z t = bt + W t + J t, where W t is an R-Wiener process, J t is a jump process independent of W t. p. 19/32

p. 2/32 Girsanov s Theorem Let (ψ(t)) t T be an R 1/2 (H)-valued F t -predictable process, independent J t such that Eρ W (T) = 1 with ( T ρ W (T) = exp ψ(s),dw(s) 1 ) T ψ(s) 2 2 ds. Then Z(t) := Z(t) t ψ(s)ds, t T is also a (b,r,ν) Lévy process on (Ω,F, (F t ) t T ) under Q = ρ W (T)P. In the following, we take ψ(t) = R 1/2 u t, t T.

p. 21/32 OU Process Driven by Lévy Process Y y t driven by Z t on (Ω,F t, P). Y y = y, dy y t = AY y t dt + dz t. X x t driven by Z t on (Ω,F t, Q). (Q = ρ W P) X x = x, dx x t = AX x t dt + d Z t, = AX x t dt + dz t R 1/2 u t dt. P t f(y) = E P f(y y t ) P t f(x) = E Q f(x x t ) Hence (note x T = ) X x t = Y y t x t X T = Y T

p. 22/32 Drift Transformation for Coupling H x t Yt x := S t x + S t s dz s X x t := S t x + t t S t s d Z s = Yt x S t s R 1/2 u s ds = Y y t x t y o t Y y t := S t y + S t s dz s T t

p. 23/32 Harnack Inequlity: Lévy OU By the procedure introduced before, for Lévy OU semigroup P t, we have (P t f) α (x) exp ( β 2 t ) u s 2 ds (P t f α (y)), t >,α,β > 1 satisfying α 1 + β 1 = 1, f C + b (H), x,y H. Taking infimum over all null control u, we get (P t f) α (x) exp ( ) β 2 Γ t(x y) 2 (P t f α (y)),

p. 24/32 Remarks There are two essential points in the methods Successful coupling; Absolute Continuity of Measures. We study these two topics in two Chapters in the thesis: Gluing of martingales and its applications on coupling; Absolute continuity of Lévy processes in infinite dimensional spaces.

p. 25/32 Part III. Other Cases We shall use the coupling methods to establish Harnack inequalities for other cases: SDE with classical monotone condition; SDE with irregular drift MSEE The key is to choose proper drifts. Fisrst consider Harnack inequalities for the transition semigroup of SDE dy t = dw t + b(x t )dt. Assume b(x): R d R d satisfying linear growth condition.

p. 26/32 Assume there exist K R s.t. SDEs (1) b(x) b(y),x y K x y 2, x,y R d. Consider dx t = dw t + b(x t )dt ξ t x y X t Y t {t<τ} dt, X = x X t Y t dy ½ t = dw t + b(y t )dt, Y = y, where τ is the coupling time of X t and Y t and x X t Coupling Time ξ t = 2K e Kt 1 e 2Kt, y Y t τ X t = Y t T

p. 27/32 Harnack inequalities for SDEs (1) For every x,y R d, f C + b (Rd ) and T >, ( βk x y (P T f) α 2) (x) exp 1 e 2KT P T f α (y).

Harnack inequalities for SDE (2) Assume that b(x) is of linear growth, the solution to equation dy t = dw t + b(x t )dt is weakly unique and there exists a nonnegative increasing function g on [, ) such that sup x y =r 1 b(x) b(y),x y g(r). r Then (P T f) α (x) exp β 2 T [ g( x y ) + ξ t x y T ξ u du ] 2 dt P T f α (y), where ξ t is any positive continuous function on [,T]. p. 28/32

p. 29/32 Choose a Drift SDE (2) Consider ( ) ( ) ( ) d X t Y t = 1 2 σε dw 1,t dw 2,t + b(x t ) γ t b(y t ) dt, where ε = 1/n 4, ( σ ε ( 2 ε + ε)i ( 2 ε ) ε)i = ( 2 ε ε)i ( 2 ε + ε)i, γ t = [ + Xt Y t X t Y t,b(t,x t) b(t,y t ) ε(d 1) { X X t Y t t Y t 1 n 2 } ξ s x y T ξ s ds ½ ] X t Y t X t Y t ½ { X t Y t >}.

p. 3/32 Choose a Drift for MSEEs Consider the following coupled multivalued stochastic evolution equation dx t AX t dt + BX t dt + σ(t)dw t U t dt, X = x D(A) dy t AY t dt + BY t dt + σ(t)dw t Y = y D(A) where the drift U t is of the following form U t = η t(x t Y t ) X t Y t δ ½ {t<τ}. H

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p. 32/32 Thanks Thanks for your Attention.