“Broadening the knowledge base and supporting the long term professional sustainability of the Research University Centre of Excellence
at the University of Szeged by ensuring the rising generation of excellent scientists.””
Doctoral School of Mathematics and Computer Science
Stochastic Days in Szeged 26.07.2012.
Relaxed Sector Condition and Random Walk in Divergence Free Random Drift Field
Bálint Tóth
(Budapest University of Technology and Economics)
B ´ ALINT T ´ OTH
(TU Budapest and U of Bristol)
Relaxed Sector Condition and
Random Walk in Divergence Free Random Drift Field
KR ´ AMLI-FEST, Szeged, 26-27 July 2013
Notation:
(Ω, π, τz : z ∈ Zd) probabability space
with ergodic Zd-action E = {e ∈ Zd : |e| = 1} possible steps of the rw ve : Ω → [−1,1], e ∈ E
◦ ve(ω) + v−e(τeω) ≡ 0 vector field
◦ X
e∈E
ve(ω) ≡ 0 divergence-free
◦
Z
Ω ve(ω)dπ(ω) = 0, no overall drift
Ve(ω, x) := ve(τxω) the vector field over Zd
Helmholtz’s Theorem:
d = 2: The height function: Z2∗ := Z2 + (1/2,1/2).
There exists H : Ω × Z2∗ → R scalar field with stationary incre- ments such that
V = curl H, Ve(x) = H(x + e + ee
2 ) − H(x + e − ee 2 )
d = 3: The stream field: Z3∗ := Z2 + (1/2,1/2,1/2).
There exists He : Ω × Z3∗ → R, e ∈ E, vector field with stationary increments such that
V = curl H, Ve(ω, x) = . . . explain in plain words
Examples and essentially different cases:
◦ H stationary + ergodic + bounded [S.M. Kozlov (1985)]:
h ∈ L∞, H(ω, x) = h(τxω) − h(ω)
◦ H stationary + ergodic + unbounded + curlH bounded:
h ∈ L2 \ L∞, H(ω, x) = h(τxω) − h(ω)
this is the case discussed today. H−1-condition
X
◦ H {stationary + ergodic} increments (but not stationary).
+ curlH bounded. No H−1-condition, superdiffusive.
◦◦ randomly oriented Manhattan-lattice
◦◦ six-vertex / square ice (d = 2)
◦◦ dimer tiling (d = 2)
The random walk in random environment: 0 < ε < 1/(2d), fixed
PωXn+1 = x + eX0n, Xn = x = 1
2d + εVe(ω, x)
The environment process:
ηn := τXnω
Stationary and ergodic Markov process on (Ω, π). (Due to div- freeness.)
Some operators on the Hilbert space L2(Ω, π):
L2(Ω, π)-gradient : ∇ef(ω) := f(τeω) − f(ω)
∇∗e = ∇−e
L2(Ω, π)-Laplacian : ∆f(ω) := 1 d
X
e∈E
(f(τeω) − f(ω))
∆∗ = ∆ ≤ 0
multiplication ops. : Mef(ω) := ve(ω)f(ω) Me∗ = Me
A commutation relation – due to div-freeness of v:
X
e∈E
Me∇e + X
e∈E
∇eM−e = 0
The transition operator / infinitesimal generator of the en- vironment process:
L = P − I = 1
2∆ + ε X
e∈E
Me∇e =: −S + A
Martingale decomposition of the displacement:
ϕ : Ω → Rd, ϕ(ω) := X
e∈E
ve(ω)e
Then:
Yn := Xn − ε
n−1 X
k=0
ϕ(ηk)
is a martingale with stationary+ergodic increments.
Xn = Yn + ε
n−1 X
k=0
ϕ(ηk)
Gaol: understand diffusive behaviour of Pn−1k=0ϕ(ηk).
Relaxed Sector Condition [I. Horv´ath, B. T´oth, B. Vet˝o (2012)]
Theorem: Efficient martingale approximation (a la Kipnis-Varadhan) holds for Pn−1
k=0 ϕ(ηk) if
(1) ” S−1/2AS−1/2 ” is skew self-adjoint
(not just skew symmetric).
(2) ϕ ∈ Ran(S−1/2) H−1-condition Remarks:
(1) Extends Varadhan et al.’s Graded Sector Condition.
(2) Proof: partly reminiscent of Trotter-Kurtz.
Two possible definitions of ” S−1/2AS−1/2 ”:
B := X
e∈E
(−∆)−1/2∇e Me (−∆)−1/2 = ” S−1/2AS−1/2 ”
on C := Dom(−∆)−1/2 = Ran(−∆)1/2 Be := (−∆)−1/2 X
e∈E
Me (−∆)−1/2∇e = ” S−1/2AS−1/2 ”
on Ce := {f ∈ L2 : X
e∈E
Me (−∆)−1/2∇ef ∈ Dom(−∆)−1/2} Facts (easy): (1) B is skew symmetric on C.
(2) C ⊂ Ce and Be
C = B.
(3) Be = Be and Be = −B∗.
Wanted: B = Be, or, equivalently B = −B∗
What is missing from skew self-adjointmess of ” S−1/2AS−1/2 ”?
von Neumann’s criterion:
B skew symmetric, and Ran(B ± I) = H
⇔
B essentially skew self-adjoint
Needed:
X
e∈E
Me (−∆)−1/2∇eψ = (−∆)1/2ψ ⇒ ψ = 0.
Warning: Formal manipulation deceives: ψ /∈ Dom(−∆)−1/2!
Raise it to the lattice Zd: change of notation: from now on:
∇,∆, . . . = lattice gradient, lattice Laplacian, . . . Wanted:
NO nontrivial scalar field Ψ : Ω × Zd → R with stationary increment, and E Ψ = 0 solves the PDE
∆Ψ = V · ∇Ψ. (1)
” Ψ = (−∆)−1/2ψ ”, ∇Ψ(ω, x) = (−∆)−1/2∇ψ(τxω)
Note similarity: No sublinearly growing harmonic function on Zd.
H−1 assumed:
The height function/stream field is stationary, L2. Hence
n→∞lim n−1E Xn2 < ∞.
Let Ψ be solution of (1). Then n 7→ Rn := Ψ(Xn) is a martingale with stationary and ergodic increments.
ρ2 := E(Rn+1 − Rn)2 ,
(1 − 2dε)kψ k2 ≤ ρ2 ≤ (1 + 2dε)kψ k2 and
n−1/2Rn ⇒ N(0, ρ2)
IF
lim
|x|→∞|x|−1 |Ψ(x) | = 0 a.s. (2) then ρ = 0,
X
.Na¨ıve guess: (2) holds for 0-mean fields with stationary incre- ments. (d = 1: ergodic theorem. Generally no true in d ≥ 2.)
Proof of (2) in d = 2:
Maximum principle: max
x∈Λ |Ψ(x) | = max
x∈∂Λ|Ψ(x) | By ergodic thm . . . : N−1 max
x∈∂[−N,N]2
|Ψ(x) | −→P 0,
Thus: N−1 max
x∈[−N,N]2
|Ψ(x) | −→P 0,
Altogether:
P |Rn | > ε√ n
= P |Rn | > ε√
n, |Xn | > K√
n + P |Rn | > ε√
n, |Xn | ≤ K√ n
≤ P |Xn | > K√
n + P max |Ψ(x) | > ε√
n → 0