The hird momen for he parabolic Anderson model Le Chen Universiy of Kansas Thursday nd Augus, 8 arxiv:69.5v mah.pr] 5 Sep 6 Absrac In his paper, we sudy he parabolic Anderson model saring from he Dirac dela iniial daa: x u, x = λu, xẇ, x, u, x = δ x, x R, where Ẇ denoes he space-ime whie noise. By evaluaing he hreefold conour inegral in he hird momen formula by Borodin and Corwin ], we obain some lici formulas for Eu, x 3 ]. One applicaion of hese formulas is given o show he exac phase ransiion for he inermiency fron of order hree. Keywords. The sochasic hea equaion; parabolic Anderson model; Dirac dela iniial condiion; space-ime whie noise; momen formula; inermiency frons; growh indices. AMS subjec classificaion. Primary 6H5; Secondary 35R6. Inroducion In his paper, we derive an lici formula for he hird momen of he following parabolic Anderson model PAM 3], u, x = λu, x Ẇ, x, x R, >, x. u, = δ, where > is he diffusion parameer and δ is he Dirac dela measure wih a uni mass a zero. The noise is assumed o be he space-ime whie noise, i.e., ] E Ẇ, x Ẇ s, y = δ sδ x y. Universiy of Kansas, Deparmen of Mahemaics. 45 Snow Hall, 46 Jayhawk Blvd. Lawrence, Kansas, 6645-7594, USA. Email: chenle@ku.edu or chenle@gmail.com.
The soluion o. is undersood in he mild form u, x = G, x + λ G s, x yus, yw ds, dy,. R d where he sochasic inegral in. is in he sense of Walsh 3] and G, x = π / x..3 Explici formulas boh for he second momen and for he wo-poin correlaion funcion were obained in 4, 7] for general iniial measure u, = µ such ha µ G, x < for all > and x R. Here, µ = µ + µ is he Jordan decomposiion and µ = µ + +µ. When he iniial daa is he dela measure wih a uni mass a zero, i.e., µ = δ, hese formulas reduce o he following simple forms see Corollary.8 of 4]: E u, xu, y] = G, xg, y + λ, x + y and in paricular, E u, x ] = G /, x 4 G / λ 4 λ x y 4 In hese formulas, we have used he following special funcions Φx = x x y λ,.4 λ + λ4 4π e λ 4 4 Φ λ..5 π / e y / dy, erfx = x e y dy, π x = erfx. Berini and Cancrini obained he wo-poin correcion funcion in an inegral form; see Corollary.5 in ]. This inegral has been evaluaed licily in 4], which gives he same form as.4. On he oher hand, for he dela iniial condiion, in a beauiful work by Borodin and Corwin ], i is showed ha for any x x k, k ] E u, x j = z A z B k πı k z A z B λ z j + x j z j dz j,.6 j= A<B k where ı = and he z j inegraion is over α j + ır wih α > α + λ > α 3 + λ > ;.7 see Appendix A. in ] where hey assume =. Thanks o condiion.7, his formula.6 can be ransformed ino he following form by inroducing anoher kk / inegrals, which resuls in a formula wih kk + / inegrals: k ] E u, x j = k πı k z j + x j z j dz j j= j= ].8 z A z B s AB z A z B λ ds AB. A<B k j=
In he following, we will firs show ha when k =, one can recover.4 by evaluaing he double conour inegrals in.6. Then we proceed o derive some formulas, more lici han.6, for he hird momen. As an applicaion, we esablish he hird exac phase ransiion see below. These resuls are summarized in he following hree Theorems.,. and.7. Theorem. Second momen. When k =, he double conour inegrals on he righ-hand side of.6 are equal o G, x G, x + λ 4 G /, x + x λ 4 λ x x x x λ 4, which recovers.4 wih x x. Noe ha in case of k = 3, here are six inegrals in.8. In he nex heorem, we will firs evaluae he hree conour inegrals in.8 which leads o.9. Then we proceed o evaluae wo real inegrals in.9 leading o.. Finally, by applying he mean-value heorem, we evaluae he las real inegral o obain an lici ression which is handy for applicaions; see.. Theorem. Third momen. Suppose ha he iniial daa is he Dirac dela measure δ x. The following saemens are rue: For all > and x i R, i =,, 3, E u, x u, x u, x 3 ] λ s λ s λ s 3 ds ds 3 s + s + s 3 + x 3 x s + s + s 3 + x 3 x s + s s 3 + x x = 9/ π 3/ ds x 3 + s + s 3 x + s s 3 For all x R, Eu, x 3 λ λ ] = 5/ π 3 4 3x + + π x s s 3s + λ 3ss λ 4 + 3s λ s3s λ 4 3/ λ π 3x λ 4 4 3x..9 λ + s Φ ] λ s Φ ds Φ λ.. 3 For all > and x R, here are some wo consans a, b depending on, and λ in he following range a Φ λ b. 3
such ha E u, x 3] = 3x π 3/ + λ b a λ 4 π 4 3x λ 4 + 4 3x λ π + b λ 4 λ 4 3π5 3x Φ Φ λ 3 λ.a.b.c..d Remark.3. I is known ha u, x is sricly posiive for all > and x R a.s.; see 6]. Hence, E u, x 3 ] = E u, x 3 ]. In he following, le u δ and u denoe he soluions o. saring from he dela measure δ and Lebesgue s measure i.e., u, x =, respecively. The following corollary is a consequence of.9 and he fac ha E u, x k] k ] = E u δ, x i dx... dx k..3 R R i= Corollary.4. Suppose ha he iniial daa is Lebesgue s measure, i.e., u, x =. Then E u, x 3] = 9/ π 3/ dx dx dx 3 R R R λ s λ s λ s 3 ds ds ds 3 s + s + s 3 + x 3 x s + s + s 3 + x 3 x s + s s 3 + x x x 3 + s + s 3 x + s s 3 x s s..4 Remark.5. Berini and Cancrini sudied. wih λ = and claimed in Theorem.6 of ] ha E u, x k] kk kk = Φ..5 4! X. Chen showed in 8] ha.5 is correc only for k = ; see also Remark.6 in 4]. Noe ha here are six inegrals in.4. Afer inegrals over dx dx 3, he ression becomes oo complicaed and oo long o handle. We leave i o ineresed readers o simplify his formula. 4
Remark.6 Asympoics. Noe ha he leading orders for large ime boh in.5 wih k = 3 for u and in. for u δ are he same. Acually, X. Chen 8] showed ha he asympoics of he k-h momen of u noe ha i is no u δ does saisfy, as.5, ha lim log E u, x k] = kk..6 4! As an easy consequence of., we see ha his fac is sill rue for u δ, namely, lim log E u δ, x 3] = λ4 kk 4! = λ4, for all x R..7 k=3 Of course, he formula. iself conains more informaion han hese asympoics. Now we sae one applicaion of hese momen formulas. I is known ha he soluion o. wih a general iniial condiion is inermien 3, 4, ], which means informally ha he soluion in quesion develops many all peaks. An ineresing phenomenon is ha when he iniial daa has compac suppor, hese all peaks will propagae in space wih cerain speed depending on he value of λ; see some simulaions in Figure. The spaial frons of hese all peaks are called he inermiency frons. Conus and Khoshnevisan 9] inroduced he following lower and upper growh indices of order p o characerize hese inermiency frons, { } λp := sup λp := inf { α > : lim sup α > : lim sup sup x α sup x α log E u, x p > log E u, x p < },.8..9 We call he case λp = λp he p-h exac phase ransiion. Chen and Dalang 4] esablished he second exac phase ransiion, namely, if he iniial daa is a nonnegaive measure wih compac suppor, hen λ := λ = λ = λ.. This improves he resul by Conus and Khoshnevisan 9] ha π λ λ λ λ. Chen and Kunwoo 5] sudied he sochasic hea equaion he equaion wih λu in. replaced by σu on R d subjec o a Gaussian noise ha is whie in ime and colored in space. Boh nonrivial lower and upper bounds for he second growh indices were obained. More recenly, Huang, Lê and Nualar sudied he PAM on R d wih a Gaussian noise ha may have colors in boh space and ime; see ] and ]. They obained some nonrivial, someimes maching, bounds for he lower and upper growh indices of all orders p. In paricular, hey showed in he curren seing ha p λp = λp = λp = λ.. Noe ha in 8], he iniial daa is assumed o a nonnegaive funcion ha saisfies < inf u, x sup u, x <. x R x R For sudying asympoic properies, his assumpion is essenially equivalen o he case ha u, x. 5
The following heorem is an easy corollary of our momen formula, which recovers heir resul. for p = 3. a λ = 3 b λ = 4 c λ = 5 d λ = 6 e λ = 7 f λ = 8 Figure : Some simulaions of he soluion o. wih various λ ranging from 3 o 8. These simulaions are runcaed a he level 5 in order o compare he size of hese six space-ime cones. The iniial daa is πµ / x /µ wih µ = 5, which is an approximaion o he dela iniial measure. These simulaions are made up o =.. Theorem.7 The hird exac phase ransiion. If u, x = δ x, hen λ3 := λ3 = λ3 = 3 λ.. Proof. Because b in. goes o as goes o infiniy, we see ha Hence, lim sup log E u, x p = λ4 x α 3α..3 lim sup log E u, x p > α < x α 3 λ,.4 which shows ha λ3 = /3 λ. Similarly, one can show ha λ3 = /3 λ. In he following, we will prove Theorems. and. in Secions and 3, respecively. 6
Second momen: Proof of Theorem. The proof of Theorem. severs as a warm-up for he more involved proof of Theorem.. Proof of Theorem.. Suppose ha x x. Fix any α > α + λ. Then, from.8, z z λ E u, x u, x ] = πı ] s ds z z z + x z z + x z dz dz = λ ds πı s dz z + x sz dz z z z + x + sz.. The dz inegraion over α + ır is equal o dz z z z + x + sz π = x + s dz z + x ] + s z + x ] ] + s z + x +s π/ π = ı s + x z 3/ + x + s.. The dz inegraion over α + ır gives ı Hence, π z + x sz x s + x + s =ı π =ı π Eu, x u, x ] = = π dz π/ s + x x π 3/ z + x s x s + x + s s + x x s + x + s x + x + x x + λ 4 z + x + s dz ] z + x ] s + x x + s s + x x..3 7 + λ s ds
= π = π s x x x + x x + x = G, x G, x + λ 4 G / s x x + λ ds + x x + λ 4 x x λ + x x + λ 4 x x + λ, x + x This proves Theorem.. 4 + λ π λ 4 λ x x 4 z + λ e z dz x x λ 3 Third momen: Proof of Theorem. x x λ..4 We firs prove wo lemmas. For all β R, > and n, denoe Λ n β, := s + βs s n ds. 3. Lemma 3.. For all > and β R, π β Λ β, = 4 Λ β, = π 3/ + β β 4 4 and for all n, Λ n β, = β Λ n β, + β In paricular, we have ha Λ β, = β π 3/ 4 + β + β 8 4 Λ 3 β, = + β 3 πβ 5/ 8 + β + 6 β 6 4, 3. β, 3.3 n Λ n β,. 3.4 β, 3.5 β Proof. We firs calculae Λ β, : Λ β, = s β + β ds 4 = β z β dz 4 8. 3.6
As for Λ β,, Λ β, = = β π 4 = = + β Λ β,, + erf s + βs β] d s + βs β. s + βs ds + β Λ β, which is equal o he righ-hand side of 3.3 afer simplificaion. For any n, Λ n β, = = = = s + βs β] s n s + βs ds s n d s + βs + β Λ n β, s n s + βs ds n n Λ n β, + β Λ n β,, which proves 3.4. In paricular, Λ β, = β 4 + + β Λ β,, 4 Λ 3 β, = + β 3 3β 8 + + β3 3 Λ β,. 4 8 + β Λ n β, Then by 3., one obains 3.5 and 3.6. This complees he proof of Lemma 3.. The nex lemma is a corollary of Lemma 3.. We neverheless sae i separaely for he convenience of our applicaion. Lemma 3.. For all a, b, d R and c >, i holds ha πad + b cd c d as + b css d ds = Φ + a c 4 c. 3.7 Now we are ready o prove Theorem.. Proof of Theorem.. Fix x, x, x 3 R wih x x x 3, >, and α, α, α 3 R wih α > α + λ > α 3 + λ. Similar o he proof of Theorem., from.8, we have ha E u, x u, x u, x 3 ] = λ ds πı 3 s λ λ ds s ds 3 s 3 9
dz z + x s s z dz z + x + s s 3 z dz 3 z z z z 3 z z 3 z 3 + x 3 + s + s 3 z 3. 3.8 Now we will calculae he riple inegral over dz 3 dz dz. Recall ha x x x 3. The dz 3 -inegral is equal o dz 3 z z z z 3 z z 3 z 3 + x 3 + s + s 3 z 3 =z z dz 3 z z 3 z z 3 z 3 + x 3 + s + s 3 z 3 =z z x 3 + s + s 3 dz 3 z 3 + x 3 + s + s 3 z 3 + x ] 3 + s + s 3 x3 + s + s 3 + z z 3 + x ] 3 + s + s 3 x3 + s + s 3 + z =ı 5/ π x 3 + s + s 3 z z + z + x 3 + s + s 3 z + x 3 + s + s 3 =:F z. 3.9 The dz -inegral can be obained in he same way as above dz z + x + s s 3 z F z =ı 5/ π x 3 + s + s 3 x + s s 3 dz z + x + s s 3 z + x + s s 3 z + x ] + s s 3 + z + x 3 + s + s 3 z + x ] ] + s s 3 s + s + s 3 x + x 3 =ı 4 π s + s + s 3 x + x 3 x 3 + s + s 3 x + s s 3 + z + x 3 + s + s 3 z + x + s s 3 ] =:F z. 3.
Similarly, one can calculae he dz -inegral, which is equal o dz z + x s s z F z Hence, =ı 4 π s + s + s 3 x + x 3 x 3 + s + s 3 x + s s 3 x s s dz z + x s s { + z + x ] s s + s + s + s 3 x + x 3 z + x ] } s s + s + s s 3 x + x =ı 3 9/ π 3/ s + s + s 3 x + x 3 s + s + s 3 x + x 3 s + s s 3 x + x x 3 + s + s 3 x + s s 3 x s s =:F s, s, s 3. E u, x u, x u, x 3 ] λ s λ s λ s 3 ds ds 3 s + s + s 3 x + x 3 s + s + s 3 x + x 3 s + s s 3 x + x = 9/ π 3/ ds x 3 + s + s 3 x + s s 3 x s s 3.. 3. Finally, for general x i R, i =,, 3, wihou ordering, one obains.9 by symmery. In his par, we assume x = x = x = x 3. Then F defined in 3. is equal o F s, s, s 3 =ı 3 9/ π 3/ s + s + s 3 s + s + s 3 s + s s 3 s + s + s 3 + s s s s 3 + s s 3 3x. 3.3 Now we are going o apply Lemma 3. o evaluae he ds 3 -inegral. Firs noice ha s + s + s 3 s + s + s 3 s + s s 3 = s + s s + s s + s + 3 s + 4s s + s s3 + 3s s s 3 s 3 3 =: a + a s 3 + a s 3 + a 3 s 3 3. 3.4
Hence, λ ds 3 s 3 F s, s, s 3 =ı 3 9/ π 3/ 3x s + s + s s 3 a k s k 3 s 3 λ k= + + s s s 3 ds 3 =ı 3 9/ π 3/ 3x s + s + s s 3 λ a k Λ k + s s,, 3.5 k= where he funcions Λ k, are defined in 3.. Afer some edious ansion and simplificaion, we see ha 3 λ a k Λ k k= + s s, = b + b s + b s πλ 3/ s s + λ 6 + λ 4 9 s + s 4 4 πλ 3/ + 6 + λ 4 9 s + s s s + λ s s + λ 8 4 =:A s + A s + A 3 s, 3.6 where b = s + λ 4 s λ4, b = 5s λ 4 4 and b =. 3.7 Therefore, he ds -inegral is equal o λ ds s k= 3x s λ ds 3 s 3 F s, s, s 3 = ı 3 9/ π 3/ 3 k= By Lemma 3., ds s λ + s s A s λ = b k Λ k s, = 8 6s + 3λ ds s λ + s s A k s. 3.8
+ 6 πλ 3/ 5/ 5s + λ s λ λ s 4. 3.9 Similarly, for he A erm, we have ha ds s λ + s s A s = 4 πλ 3/ s + λ 4 ds 3s λ 4 + 3s 9 s s + s 6 + λ 4 = πλ 3/ 5/ 3s + λ s + λ, 3. 4 where we have applied Lemma 3. in he las sep. The inegraion wih respec o A 3 is much more complicaed. Insead, we claim ha for r >, r Ir := ds s λ + s s A 3 s = r πλ 3/ ds s λ 8 + s s 6 + λ 4 9 s + s s s + λ s s + λ 4 { = π 6 λ λ s + λ s + λ 4 4 } s λ r + s λ λ + 5 4 s + + 4 3r + λ 3r 4 + r ] λ 6s + λ s + s + λ 4 r + s + λ 4 4 + 3 s + λ s + λ πs 4 4 s λ λ s + 4 s λ r + s λ + 4 + 4 3r 4 + r λ 6s + λ s + s + λ 4 r + s + λ 4 4 ] s λ 4 5 π λ s λ + 3 3 r 4 r s ] λ, 3. 3
which can be verified direcly by differeniaing boh sides. Because x x π / e x as x and lim x x =, we see ha { lim Ir = π r 6 λ λ s + λ s + λ 4 4 } s λ λ s + 5 4 + ] + 3 s + λ s + λ πs 4 4 s λ λ s + 4 ] s λ + + 4 5 ] π λ s λ + 3. 3. 4 Therefore, ds s λ + s s A 3 s { = π 6 λ λ 3/ s + λ s + λ 4 4 } s λ λ s + 5 4 + 3 s + λ s + λ πs 4 4 ] s λ λ s + 4 s λ 4 5 ] π λ s λ + 3. 3.3 4 Finally, combining all hese hree inegrals, we see ha λ λ λ s ds s ds 3 s 3 =ı 3 3x π λ 3s + λ 3 F s, s, s 3 λ 4 3 s λ 4 4 s + λ
+ π λ 3s λ λ 4 3 3 3 s λ /3 λ s 4 ] + 5/ 5/ π 3/ λ 4 s 4 s λ / =:Θs. Therefore, he hird momen is equal o E u, x 3] = πı 3 3.4 Θs ds. 3.5 Because x = Φ x and hanks o Lemma 3., π 3x 5/ 5/ π 3/ λ 4 s 3 4 s λ / ds = π 3/ 3x s s s λ ds 5/ = 3x λ λ 4 + π 3/ π 4 3x Φ λ, 3.6 one can obain. afer some simplificaion. 3 By he mean-value heorem, here exi wo consans a, b in he range given by. such ha Eu, x 3 bλ λ ] = 5/ π 3 4 3x 3s + λ 3s s λ ds 4 aλ λ + 5/ π 3 4 3x 3s λ s 3s λ ds 4 + 3x π 3/ λ λ 4 + π 4 3x Φ λ. 3.7 Then one can apply Lemma 3. o evaluae hese wo inegrals. This complees he whole proof of Theorem.. Acknowledgemens L.C. would like o hank Rober Dalang and Davar Khoshnevisan for many helpful commens. Khoshnevisan asked L.C. in 4 wheher one could obain an lici formula for he hird momen, which moivaed he curren sudy. There was no much progress on his problem unil when L.C. was vising he Simons Cener for Geomery and Physics for he conference Sochasic Parial Differenial Equaions May 6-, 6, Ivan Corwin poined ou o L.C. he ransform from.6 o.8. This became he saring poin of he whole calculaion in his paper. Here L.C. would like o ress his sincere graiude o him. Finally, L.C. would also like o hank he organizer Marin Hairer for he wonderful conference. 5
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