Annealed Brownian motion in a heavy tailed Poissonian potential

Annealed Brownian motion in a heavy tailed Poissonian potential Ryoki Fukushima Research Institute of Mathematical Sciences Stochastic Analysis and Applications, Okayama University, September 26, 2012 1 / 27

Introduction Anderson (1958): low lying spectra of + V ω are pure point and corresponding eigenfunctions are strongly localized. 2 / 27

Introduction Anderson (1958): low lying spectra of + V ω are pure point and corresponding eigenfunctions are strongly localized. Consider the parabolic counterpart (Gärtner-Molchanov 1990, later many others): t u = ( V ω )u, u(0, x) = δ 0 (x). The eigenfunction expansion u(t, x) = k=1 e tλ k φ k (0)φ k (x) tells us that u(t, ) localizes as well. 2 / 27

1. Setting ) ({B t } t 0, P x : κ -Brownian motion on R d ( ω = ) δ ωi, P : Poisson point process on R d i with unit intensity 3 / 27

1. Setting ) ({B t } t 0, P x : κ -Brownian motion on R d ( ω = ) δ ωi, P : Poisson point process on R d i with unit intensity Potential For a non-negative and integrable function v, V ω (x) := i v(x ω i ). (Typically v(x) = 1 B(0,1) (x) or x α 1 with α > d.) 3 / 27

Annealed measure We are interested in the behavior of Brownian motion under the measure { t } exp V ω (B s )ds P P 0 ( ) 0 Q t ( ) = { t }]. E E 0 [exp V ω (B s )ds The configuration is not fixed and hence Brownian motion and ω i s tend to avoid each other. 0 4 / 27

0 B t exp{ t 0 V ω(b s )ds} : large, P : large, P 0 : small 5 / 27

0 B t exp{ t 0 V ω(b s )ds} : large, P : small, P 0 : large 5 / 27

2. Light tailed case Donsker and Varadhan (1975) When v(x) = o( x d 2 ) as x, { t }] } E E 0 [exp V ω (B s ) ds = exp { c(d, κ)t d d+2 (1 + o(1)) 0 as t. Remark c(d, κ) = inf U {κλd (U) + U }. 6 / 27

Suppose v is compactly supported for simplicity. Then { t }] E E 0 [exp V ω (B s ) ds 0 ( P 0 B[0,t] U) ) P (ω(u) = 0) exp{ κλ D (U)t U }. Optimizing over U, we get the correct lower bound. 7 / 27

Suppose v is compactly supported for simplicity. Then { t }] E E 0 [exp V ω (B s ) ds 0 ( P 0 B[0,t] U) ) P (ω(u) = 0) exp{ κλ D (U)t U }. Optimizing over U, we get the correct lower bound. maximizer = B(x, t 1 d+2 R0 ) 7 / 27

R 0 t 1 d+2 B t 0 8 / 27

One specific strategy gives dominant contribution to the partition function. It occurs with high probability under the annealed path measure. 9 / 27

Sznitman (1991, d = 2) and Povel (1999, d 3) When v has a compact support, there exists such that D t (ω) B ( 0, t 1 d+2 (R0 + o(1)) ) Q t ( B [0,t] B ( D t (ω), t 1 d+2 (R0 + o(1)) )) t 1. Remark Bolthausen (1994) proved the corresponding result for two-dimensional random walk model. 10 / 27

3. Heavy tailed case Pastur (1977) When v(x) x α (α (d, d + 2)) as x, { t }] } E E 0 [exp V ω (B s )ds = exp { a 1 t d α (1 + o(1)), 0 where a 1 = B(0, 1) Γ ( ) α d α. 11 / 27

3. Heavy tailed case Pastur (1977) When v(x) x α (α (d, d + 2)) as x, { t }] } E E 0 [exp V ω (B s )ds = exp { a 1 t d α (1 + o(1)), 0 where a 1 = B(0, 1) Γ ( ) α d α. Unfortunately, this first order asymptotics tells us little about the Brownian motion. 11 / 27

In fact, Pastur s proof goes as follows: { t }] E E 0 [exp V ω (B s )ds 0 E[exp{ tv ω (0)}] exp{ a 1 t d α }. 12 / 27

In fact, Pastur s proof goes as follows: { t }] E E 0 [exp V ω (B s )ds 0 E[exp{ tv ω (0)}] exp{ a 1 t d α }. The effort of the Brownian motion is hidden in the higher order terms. A bit more careful inspection of the proof shows... 12 / 27

O(t 1 α ) B t 0 1 o(t α ) 13 / 27

F. (2011) When v(x)= x α 1 (d < α < d + 2), { E E 0 [exp = exp t 0 }] V ω (B s )ds { a 1 t d α (a2 + o(1))t α+d 2 2α }, where a 2 := { inf φ 2 =1 } κ φ(x) 2 + C(d, α) x 2 φ(x) 2 dx. Remark The proof is an application of the general machinery developed by Gärtner-König 2000. 14 / 27

Recalling the Donsker-Varadhan LDP ( 1 t ) { P 0 δ Bs ds φ 2 (x)dx exp t t 0 } κ φ(x) 2 dx, we expect the second term explains the behavior of the Brownian motion. 15 / 27

Recalling the Donsker-Varadhan LDP ( 1 t ) { P 0 δ Bs ds φ 2 (x)dx exp t t 0 } κ φ(x) 2 dx, we expect the second term explains the behavior of the Brownian motion. In particular, since P 0 ( B[0,t] B(x, R) ) exp{ tr 2 }, the localization scale should be tr 2 = t α+d 2 2α R = t α d+2 4α. 15 / 27

O(t 1 α ) B t 0 α d+2 O(t 4α ) 16 / 27

0 the below is the section along this line V ω (x) Heavy tailed case h(t) r(t) th(t) t/r(t) 2 17 / 27

0 the below is the section along this line V ω (x) Light tailed case h(t) r(t) th(t) t/r(t) 2 17 / 27

Main Theorem (F. 2012) Q t (B [0,t] B (0, t α d+2 1 4α (log t) +ϵ)) t 2 1, Q t ( V ω (x) V ω (m t (ω)) t α d+2 α C(d, α) x m t (ω) 2 {t α d+2 4α Bt α d+2 2α s } s 0 in B(0, t α d+2 4α +ϵ ) ) t 1, in law OU-process with random center, where m t (ω) is the minimizer of V ω in B(0, t α d+2 4α log t). 18 / 27

4. Outline of the proof An important feature of this model is that there are two scales in the optimal strategy. ω lives in the scale t 1/α, {B s } s t lives in the scale o(t 1/α ). 19 / 27

4. Outline of the proof An important feature of this model is that there are two scales in the optimal strategy. ω lives in the scale t 1/α, {B s } s t lives in the scale o(t 1/α ). This, for instace, prevents us from using the compactification by projecting on a torus. 19 / 27

Suppose we have a crude estimate sup B s = o(t 1/α ). 0 s t Then B s 0 viewed from ω. Thus it is plausible that Q t (dω) E[exp{ tv ω(0)} : dω]. E[exp{ tv ω (0)}] 20 / 27

Suppose we have a crude estimate sup B s = o(t 1/α ). 0 s t Then B s 0 viewed from ω. Thus it is plausible that Q t (dω) E[exp{ tv ω(0)} : dω]. E[exp{ tv ω (0)}] The right hand side is nothing but the Poisson point process with intensity e t( x α 1) dx. potential locally looks like parabola, localization, weak convergence. 20 / 27

It remains to verify sup B s = o(t 1/α ). This is a localization but 0 s t weaker than the main theorem. To this end, we need some control on V ω... circular argument? 21 / 27

It remains to verify sup B s = o(t 1/α ). This is a localization but 0 s t weaker than the main theorem. To this end, we need some control on V ω... circular argument? No! What we need is crude control which can be established with bare hands. 21 / 27

4.1 Crude control on the potential Lemma 1 ( Q t V ω (0) d α a 1t α d α ) + t 3α 3d+2 4α ( M 1, M 1 ) 1. 22 / 27

4.1 Crude control on the potential Lemma 1 Idea ( Q t V ω (0) d α a 1t α d α Z t E[exp{ tv ω (0)}] ) + t 3α 3d+2 4α ( M 1, M 1 ) 1. { { } = exp a 1 t d α, sup h>0 [ e th P(V ω (0) h) ]. 22 / 27

4.1 Crude control on the potential Lemma 1 Idea ( Q t V ω (0) d α a 1t α d α Z t E[exp{ tv ω (0)}] ) + t 3α 3d+2 4α ( M 1, M 1 ) 1. { { } = exp a 1 t d α, sup h>0 [ e th P(V ω (0) h) ]. d α a 1t α d α = h(t) is the maximizer. { t } ] E E 0 [exp V ω (B s )ds : V ω (0) is far from h(t) 0 E [exp{ tv ω (0)} : V ω (0) is far from h(t)] = o(z t ). 22 / 27

Lemma 2 Q t ( V ω (0) + V ω (x) 2h(t) + c 1 t α d+2 α x 2 for t α d+6 8α < x < M 2 t α d+6 8α ) 1. 23 / 27

Lemma 2 Q t ( V ω (0) + V ω (x) 2h(t) + c 1 t α d+2 α x 2 Idea By Lemma 1, { t } exp V ω (B s )ds 0 for t α d+6 8α < x < M 2 t α d+6 8α ) 1. { exp { th(t)} = exp d } α a 1t d α 23 / 27

Lemma 2 Q t ( V ω (0) + V ω (x) 2h(t) + c 1 t α d+2 α x 2 Idea By Lemma 1, { t } exp V ω (B s )ds 0 for t α d+6 8α Then, use [ { E exp t }] 2 (V ω(0) + V ω (x)) and Chebyshev s inequality. < x < M 2 t α d+6 8α ) 1. { exp { th(t)} = exp d } α a 1t d α exp { a } 1 t d α c2 t d 2 α x 2 23 / 27

4.2 Crude control on the trajectory V ω (x) V ω (0) M 2 t α d+6 8α t α d+6 8α 0 t α d+6 8α x M 2 t α d+6 8α 24 / 27

4.2 Crude control on the trajectory V ω (x) V ω (0) M 2 t α d+6 8α t α d+6 8α 0 t α d+6 8α x M 2 t α d+6 8α By penalizing a crossing, Q t (B [0,t] B )) (0, M 2 t α d+6 8α 1. 24 / 27

How about the critical case α = d + 2? Ôkura (1981) proved { t }] } E E 0 [exp V ω (B s ) ds = exp { c(d, κ)t d d+2 (1 + o(1)) 0 as t, where { c(d, κ) = inf φ 2 =1 { κ φ(x) 2 + 1 exp φ(y) 2 } } dy dx. x y d+2 25 / 27

Thank you! 26 / 27

How about the critical case α = d + 2? Ôkura (1981) proved { t }] } E E 0 [exp V ω (B s ) ds = exp { c(d, κ)t d d+2 (1 + o(1)) 0 as t, where { c(d, κ) = inf φ 2 =1 { κ φ(x) 2 + 1 exp φ(y) 2 } } dy dx. x y d+2 27 / 27