Диссертация (1137347), страница 20

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We thus derive that for all 0 < t ≤ T, (x, y) ∈ (Rd )2 :|H (r) (t, (x, y), (x0 , y 0 ))| ≤ ((1 ∨ T (1−γ)/2 )c1 )r|Φ(t, (x, y), (x0 , y 0 ))| ≤Cpc,K (t, (x, y), (x0 , y 0 )).t1−γ/2(4.97)Then,Dxα R(t, (x, y), (x0 , y 0 ))Z=limZ tτ →0+ZduR2dτZdut/2=:t/2R2dDxα p̃(u, (x, y), (w, z))Φ(t − u, (w, z), (x0 , y 0 )dwdzDxα p̃(u, (x, y), (w, z))Φ(t − u, (w, z), (x0 , y 0 ))dwdzlim Dxα Rτ (t, (x, y), (x0 , y 0 )) + Dxα Rf (t, (x, y), (x0 , y 0 )).τ →0(4.98)The contribution Dxα Rf (t, (x, y), (x0 , y 0 )) does not exhibit time singularities in the integral, sinceon the considered integration set u ≥ 21 t.Thus, from inequalities (4.17) and (4.97):|Dxα Rf (t, (x, y), (x0 , y 0 ))| ≤Cpc,K (t, (x, y), (x0 , y 0 )).(t − s)(|α|−γ)/2(4.99)We should put more effort in the estimation of the remainder part:Dxα Rτ (t, (x, y), (x0 , y 0 )).

For|α| = 2 we apply some kind of the cancellation properties of the Gaussian kernels as in [KM17].871Introducefor an, κ2 ∈ R2d : Ĉt := arbitrary κId0dIRs =, B = d×d andsIdId0d×dRt0∗Rt−u Ba(κ1 , κ2 )B ∗ Rt−udu,!exp1p̃κ2,κ(u, (x, y), (w, z))Z :==− 12 hĈuε,−1 Z, Zi(2π)d det(Ĉuε (κ1 , κ2 ))1/2,w−x.z − y − xu(4.100)Hence, for all multi-index α, |α| = 2:Z12Dxα p̃κ ,κ (u, (x, y), (w, z))dwdz = 0.(4.101)R2dIntroducing the centering function cα (u, (x, y), (w, z)) := (Dxα p̃κwe derive that:Dxα Rτ (t, (x, y), (w, z))Z=Zt/2ZduZduτ+t/2R2d1,κ2(u, (x, y), (w, z)))|(κ1 ,κ2 )=(x,y) ,(Dxα p̃ − cα )(u, (x, y), (w, z))Φ(t − u, (w, z), (x0 , y 0 ))dwdzcα (u, (x, y), (w, z))(Φ(t − u, (w, z), (x0 , y 0 )) − Φ(t − u, (x, y + xu), (x0 , y 0 ))dwdzR2dτ:= (Rτ,1 + Rτ,2 )(t, (x, y), (x0 , y 0 )),(4.102)exploiting the centering condition (4.101) to introduce the last term of the first equality.12Since cα (u, (x, y), (w, z)) contains p̃κ ,κ (u, (x, y), (w, z)))|(κ1 ,κ2 )=(x,y) as a ‘true‘ density w.r.t.(w, z) we can cancel with Φ(t − u, (x, y + xu), (x0 , y 0 )).We recall that|cα (u, (x, y), (w, z))| ≤Cpc,K (u, (x, y), (w, z)).uOn the one hand, the terms Dxα p̃(u, (x, y), (w, z)), cα (u, (x, y), (w, z)) differ in their frozen coefficients(respectively at point w, z and x, y).

Moreover there is no back-flows w.r.t. the second variable incα (u, (x, y), (w, z)). Exploiting the Hölder property in space of the mollified coefficients, it is thenseen that:#"|w − x|γ|z − y − xu|γαα|(Dx p̃ − c )(u, (x, y), (w, z))| ≤ C+pc,K (u, (x, y), (w, z))uu≤Cpc,K (u, (x, y), (w, z)),u1−γ/2where the last inequality comes from the standard absorption of the additional time singularity withthe Gaussian density pc,K (u, (x, y), (w, z)).

Thus, from (4.97):|Rτ,1 (t, (x, y), (x0 , y 0 ))| ≤Ct|α|−γpc,K (t, (x, y), (x0 , y 0 )).(4.103)The key idea to control the contribution of the remainder part is to use the smoothing effect comesfrom the kernel Φ.88Lemma 4.6.1. For Au := {(w, z) ∈ R2d : |w − x| +one has:|z−y−xu|u≤ ct1/2 } (recall as well that u ∈ [0, 2t ])|Φ(t − u, (x, y + xu), (x0 , y 0 )) − Φ(t − u, (w, z), (x0 , y 0 ))|!≤C|x − w|γ/2 + |z − y − ux|γ/2 pc,K (t − u, (w, z), (x0 , y 0 )).(t − u)1−γ/4(4.104)Proof. From the definition of Φ and the smoothing effect of the kernel H in (4.96), it suffices to prove≤ c(u0 − u)1/2 }:that on the set Āu := {z, w ∈ R2d : |x − w| + |z−y−xu|u0 −u|H(u0 − u, (x, y + xu), (x00 , y 00 )) − H(u0 − u, (w, z), (x00 , y 00 ))|≤C|x − w|γ/2 + |z − y − xu|γ/2pc,K (u0 − u, (w, z), (x00 , y 00 )),(u0 − u)1−γ/4(4.105)for u0 ∈ (u, t], u ∈ [0, t/2].Let us first prove (4.105).

We concentrate on the second derivatives in H which yields the mostsingular contributions:Tr((a(x, y + xu) − a(x00 , y 00 − x00 (u0 − u)))Dx2 p̃(u0 − u, (x, y + xu), (x00 , y 00 ))−Tr((a(w, z) − a(x00 , y 00 − x00 (u0 − u)))Dx2 p̃(u0 − u, (w, z), (x00 , y 00 ))= Tr((a(x, y + xu) − a(w, z))Dx2 p̃(u0 − u, (w, z), (x00 , y 00 )))−Tr((a(x, y + xu) − a(x00 , y 00 − x00 (u0 − u)))×(Dx2 p̃(u0 − u, (x, y + xu), (x00 , y 00 )) − Dx2 p̃(u0 − u, (w, z), (x00 , y 00 )))=: I + II.(4.106)Then, from (4.17),|I|≤ C≤|x − w|γ + |z − y − xu|γ/2pc,K (u0 − u, (w, z), (x00 , y 00 ))(u0 − u)C|x − w|γ/2 + |z − y − xu|γ/2pc,K (u0 − u, (w, z), (x00 , y 00 )))(u0 − u)1−γ/4(4.107)using that (w, z) ∈ Āu for the second inequality.

Now, from the explicit expression of the secondorder derivatives in (4.100), (AD2) and usual computations we also derive:!|II||x − x00 |γ + |y + xu − (y 00 − (u0 − u)x00 )|γ/2≤1(|w − x00 + λ(x − w)|2c(u0 − u)0)!|y 00 − z − (u − u0 )w + λ(y + xu − z)|2+×c(u0 − u)3Zdλexp(u0 − u)2d−89(4.108)!|w − x||z − y − xu|+.(u0 − u)3/2(u0 − u)5/2Due to Āu definition the term|w−x|(u0 −u)3/2|y+xu−z|+ (u0 −u)5/2 brings the singularity of order1u0 −u .Moreover|w − x00 + λ(x − w)|2|y 00 − z − (u − u0 )w + λ(y + xu − z)|2−c(u0 − u)c(u0 − u)3!|w − x00 |2|x − w|2|y 00 − z − (u − u0 )w|2|z − y − xu|2≤−+,−+c(u0 − u)c(u0 − u)3c(u0 − u)c(u0 − u)3!|x − w|2|z − y − xu|2≤ C for (w, z) ∈ Āu .+c(u0 − u)c(u0 − u)3−Finally,C(|w − x00 |γ + |z − (y 00 − (u0 − u)x00 )|γ/2 )(4.109) ≤exp(u0 − u)(u0 − u)2d≤(|w − x00 |2|y 00 − z − (u − u0 )w|2−−0c(u − u)c(u0 − u)3)C|x − w|γ/2 + |z − y − xu|γ/2pc,K (u0 − u, (w, z), (x00 , y 00 ))).(u0 − u)1−γ/4using the usual convexity argument and the fact that on Au for the constant C large enough!|z − y − xu||w − x|γ/2γ/2+≤ C |x − w|+ |z − y − xu|(u0 − u)1/2(u0 − u)3/2for the last inequality.Recalling that we want to establish (4.104) on Au , we consider the case: if (w, z) 6∈ Āu , we getfrom (4.96) for the tails differences:Zt0Zdu|H(u0 − u, (x, y + xu), (x00 , y 00 )) − H(u0 − u, (w, z), (x00 , y 00 ))|ĀcuuX×|(H (i) )(t − u0 , (x00 , y 00 ), (x0 , y 0 ))|dx00 dy 00i≥1Zt≤Cudu0Z(pc,K (u0 − u, (x, y + xu), (x00 , y 00 )) + pc,K (u0 − u, (w, z), (x00 , y 00 )))Ācu|z − y − xu|γ/6C|x − w|γ/6+pc,K (t − u0 , (x00 , y 00 ), (x0 , y 0 ))dx00 dy 00× 01−5γ/1201−3γ/12(u − u)(u − u)(t − u0 )1−γ/2Z tZ|x − w|γ/6 + |z − y − xu|γ/6du0≤ C101−γ/4(t − u0 )1−γ/2u Ācu (u − u)()× pc,K (t − u, (w, z), (x0 , y 0 )) + pc,K (t − u, (x, y + ux), (x0 , y 0 )) dx00 dy 00exploiting that (w, z) ∈ Ācu and the fact that on Ācu :1|z − y − xu|γ/6|x − w|γ/6+.γ ≤ c1γ5γ(u0 − u)1− 2(u0 − u)1− 4(u0 − u)1− 12Let us consider precisely the compatibility of pc,K (t−u, (x, y+xu), (x0 , y 0 )) and pc,K (t−s, (x, y), (x0 , y 0 ).90Observe that,pc,K (t − u, (x, y + xu), (x0 , y 0 ))Cexp(t − u)2d≤("|x − x0 |2|y + ux + x(t − u) − y 0 |2−+t−u(t − u)3≤ Cpc,K (t − u, (x, y), (x0 , y 0 )),#)(4.109)recalling that t − u is of order t for the last inequality.To sum up, for (w, z) ∈ Ācu ∩ Au it holds:t/2|x − w|γ/6 + |y − z − xu|γ/6pc,K (u, (x, y), (w, z))uτAu ∩Ācu1pc,K (t − u, (w, z), (x0 , y 0 ))dwdz×(t − u)1−3γ/4!γ/6!γ/6#"Z t/2 Z11|z − y − xu||x − w|√=Cdu+pc,K (u, (x, y), (w, z))uu1−γ/12u3/2u1−3γ/12τAu ∩Ācu|Rτ,2 (t, (x, y), (x0 , y 0 ))| ≤ C 2×ZZdu1Cpc,K (t − u, (w, z), (x0 , y 0 ))dwdz ≤ 1−γ/2 pc,K (t, (x, y), (x0 , y 0 )).(t − u)1−3γ/4tAs the result, we completed the proof of the Lemma.Then, we can derive from (4.17), (4.102) and (4.104) that on Au :t/2|x − w|γ/2 + |y − z − xu|γ/2pc,K (u, (x, y), (w, z))uτAu1which now compatible to absorb the singularity ×pc,K (t − u, (w, z), (x0 , y 0 ))dwdz(t − u)1−γ/4Z t/2 ZC|x − w|γ/2 + |y − z − xu|γ/2+ γ/4dupc,K (u, (x, y), (w, z))utτACu|Rτ,2 (t, (x, y), (x0 , y 0 ))| ≤ C 2ZZdu×{|Φ(t − u, (w, z), (x0 , y 0 ))| + |Φ(t − u, (x, y − xu), (x0 , y 0 ))|}dwdz.(4.110)On the complementary set Acu it holds:Z|t/2Zduτcα (u, (x, y), (w, z))(Φ(t − u, (w, z), (x0 , y 0 )) − Φ(t − u, (x, y + xu), (x0 , y 0 ))dwdz|AcuZZC t/2|z − (y + ux)|γ/2≤ γ/4 pc,K (u, (x, y), (w, z))|w − x|γ/2 +t τuγ/2Acu"#× |Φ(t − u, (w, z), (x0 , y 0 ))| + |Φ(t − u, (x, y + xu), (x0 , y 0 ))| dwdz ZZC t/211≤ γ/4 dup (u, (x, y), (w, z))1−γ/4 (t − u)1−γ/2 c,Kt τAcu u"#× pc,K (t − u, (w, z), (x0 , y 0 )) + pc,K (t − u, (x, y + xu), (x0 , y 0 )) dwdz (4.111)91Plugging (4.109) into (4.111) one can get the bound on Acu :Z|t/2Zduτcα (u, (x, y), (w, z))(Φ(t − u, (w, z), (x0 , y 0 )) − Φ(t − u, (x, y + xu), (x0 , y 0 ))dwdz|AcuZ t/2 duC1≤ γ/4 pc,K (t, (x, y), (x0 , y 0 )) ≤ 1−γ/2 pc,K (t, (x, y), (x0 , y 0 )).1−γ/41−γ/2tu(t − u)tτC(4.112)thus, taking Lemma 4.6.1 into account, we have:t/2|x − w|γ/2 + |y − z − xu|γ/2pc,K (u, (x, y), (w, z))uτAuC1pc,K (t − u, (w, z), (x0 , y 0 ))dwdz + 1−γ/2 pc,K (t, (x, y), (x0 , y 0 ))×1−γ/4(t − u)tC≤ 1−γ/2 pc,K (t, (x, y), (x0 , y 0 )),t τ,2R (t, (x, y), (x0 , y 0 )) ≤ CZZduwhich together with (4.103), (4.102), (4.99) and (4.98) give the statement of the Section.92Bibliography[AKH17]P.

Andersson and A. Kohatsu-Higa. Unbiased simulation of stochastic differential equations using parametrix expansions. Bernoulli, 23:2028–2057, 2017.[Aro59]D. G Aronson. The fundamental solution of a linear parabolic equation containing a smallparameter. Illinois Journal of Mathematics, 3:580–619, 1959.[Bai17]N.T.G. Bailey. The mathematical theory of infectious diseases and its applications.

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