Annealed Brownian motion in a heavy tailed Poissonian potential

Transcription

1 Annealed Brownian motion in a heavy tailed Poissonian potential Ryoki Fukushima Tokyo Institute of Technology Workshop on Random Polymer Models and Related Problems, National University of Singapore, May 21-25, / 25

2 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 2 / 25

3 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.) 2 / 25

4 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 3 / 25

5 0 B t exp{ t 0 V ω(b s )ds} : large, P : large, P 0 : small 4 / 25

6 0 B t exp{ t 0 V ω(b s )ds} : large, P : small, P 0 : large 4 / 25

7 2. Light tailed case Donsker and Varadhan (1975) When v(x) = o( x d 2 ) as x, { E E 0 [exp as t. t 0 }] V ω (B s ) ds } = exp { c(d, κ)t d d+2 (1 + o(1)) ( ) ) = P 0 B [0,t] B(x, t 1 d+2 R0 ) P (ω(b(x, t 1 d+2 R0 )) = 0, Remark c(d, κ) = inf U {κλd (U) + U }. 5 / 25

8 R 0 t 1 d+2 B t 0 6 / 25

9 One specific strategy gives dominant contribution to the partition function. It occurs with high probability under the annealed path measure. 7 / 25

10 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. 8 / 25

11 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. 9 / 25

12 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 }. 10 / 25

13 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 lower order terms. 10 / 25

14 O(t 1 α ) B t 0 1 o(t α ) 11 / 25

15 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 / 25

16 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. 13 / 25

17 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. 13 / 25

18 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. In addition, the term C(d, α) x 2 φ(x) 2 dx says that V ω (locally) looks like a quadratic function. 13 / 25

19 O(t 1 α ) B t 0 α d+2 O(t 4α ) 14 / 25

20 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 15 / 25

21 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 15 / 25

22 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). 16 / 25

23 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 Remark: X s := t α d+2 4α B t α d+2 2α log t). B t = t α d+2 4α X α+d 2 s t 2α 16 / 25

24 4. Outline of the proof (of localization) Observation: The 1st statement implies the 2nd one and vice versa. 17 / 25

25 4. Outline of the proof (of localization) Observation: The 1st statement implies the 2nd one and vice versa. Indeed, it is easy to believe 2nd 1st. 17 / 25

26 4. Outline of the proof (of localization) Observation: The 1st statement implies the 2nd one and vice versa. Indeed, it is easy to believe 2nd 1st. To see 1st 2nd, let L t := 1 t 0 δ B s ds and rewrite Q t (dω) = 1 [ E 0 [E e t Lt,Vω ] ] P Lt t (dω), Z t t where Z t is the normalizing constant and P Lt t (dω) := exp{ t L t, V ω }P(dω). E [exp{ t L t, V ω }] 17 / 25

27 Assuming the 1st statement, we may replace L t by δ ml with t m Lt = xl t (dx) in the following: P Lt t (V ω (x) V ω (m t (ω)) C(d, α)t α d+2 α x 2) 1? 18 / 25

28 Assuming the 1st statement, we may replace L t by δ ml with t m Lt = xl t (dx) in the following: Let m Lt = 0 for simplicity. P δ 0 t (V ω (x) V ω (0) C(d, α)t α d+2 α x 2) 1? This can be easily proved since (ω, P δ 0 t ) is nothing but the Poisson point process with intensity e t( x α 1) dx. 18 / 25

29 Assuming the 1st statement, we may replace L t by δ ml with t m Lt = xl t (dx) in the following: Let m Lt = 0 for simplicity. P δ 0 t (V ω (x) V ω (0) C(d, α)t α d+2 α x 2) 1? This can be easily proved since (ω, P δ 0 t ) is nothing but the Poisson point process with intensity e t( x α 1) dx. Remark In fact, a slightly weaker localization bound is enough to do the above replacement. 18 / 25

30 This observation is useless (circular argument) as it is. But due to the last remark, there is a chance to go as follows: crude control on the potential, crude control on the trajectory, fine control on the potential, fine control on the trajectory. 19 / 25

31 This observation is useless (circular argument) as it is. But due to the last remark, there is a chance to go as follows: crude control on the potential, crude control on the trajectory, fine control on the potential, fine control on the trajectory. Assume V ω attains its local minimum at 0 for simplicity. 19 / 25

32 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 ) / 25

33 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) ]. 20 / 25

34 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 ). 20 / 25

35 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 ) / 25

36 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 21 / 25

37 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 21 / 25

38 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 22 / 25

39 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 / 25

40 4.3 Fine control on the potential The crude control on the trajectory is good enough to yield ( Q t V ω (x) V ω (0) C(d, α)t α d+2 α x 2 ) in B(0, t α d+2 4α +ϵ ) / 25

41 4.4 Fine control on the trajectory V ω (x) V ω (0) t α d+2 x 4α (log t) 1 2 +ϵ 0 t α d+2 4α (log t) 1 2 +ϵ 24 / 25

42 4.4 Fine control on the trajectory V ω (x) V ω (0) t α d+2 x 4α (log t) 1 2 +ϵ 0 t α d+2 4α (log t) 1 2 +ϵ By penalizing a crossing, Q t (B [0,t] B (0, t α d+2 1 4α (log t) +ϵ)) / 25

43 Thank you! 25 / 25