Диссертация (1137347), страница 19
Текст из файла (страница 19)
According to the definition (4.18):H(tk − τ + s, (w, z), (x0 , y 0 )) − H(tk − τ, (w, z), (x0 , y 0 ))=+−b(w, z)Dw p̃(tk − τ + s, (w, z), (x0 , y 0 )) − b(w, z)Dw p̃(tk − τ, (w, z), (x0 , y 0 ))!10 0020 0Tra(w, z) − a(x , y − (tk − τ + s)x ) Dw p̃(tk − τ + s, (w, z), (x , y ))2!10 0020 0Tra(w, z) − a(x , y − (tk − τ )x ) Dw p̃(tk − τ, (w, z), (x , y )) .2Let us estimatea(w, z)− a(w, z)= a(x0 , y 0+ a(w, z)2−Dwp̃(tkthe most singular term. Others can be handled similarly.2− a(x0 , y 0 − (tk − τ + s)x0 ) Dwp̃(tk − τ + s, (w, z), (x0 , y 0 ))2− a(x0 , y 0 − (tk − τ )x0 ) Dwp̃(tk − τ, (w, z), (x0 , y 0 ))2− (tk − τ )x0 ) − a(x0 , y 0 − (tk − τ + s)x0 ) Dwp̃(tk − τ + s, (w, z), (x0 , y 0 ))2− a(x0 , y 0 − (tk − τ )x0 ) Dwp̃(tk − τ + s, (w, z), (x0 , y 0 ))− τ + s, (w, z), (x0 , y 0 )) := ∆H1 + ∆H2 .The term ∆H1 is easier to control using just Hölder property of a and the standard estimation2for the derivative of the frozen density Dwp̃, see (4.17).|∆H1 | = |a(x0 , y 0 − (tk − τ )x0 ) − a(x0 , y 0 − (tk − τ + s)x0 )|(4.78)0 0Cpc,K (tk − τ + s, (w, z), (x , y )2.×|Dwp̃(tk − τ + s, (w, z), (x0 , y 0 ))| ≤ |s|γ/2 |x0 |γ/2tk − τ + s81The control for |∆H2 | is more involved.
We have to derive the sensitivity of frozen density2derivatives with respect to the time-variable, actually, to bound |Dwp̃(tk − τ + s, (w, z), (x0 , y 0 )) −20 0Dw p̃(tk − τ + s, (w, z), (x , y ))|, namely:22|Dwp̃(tk − τ + s, (w, z), (x0 , y 0 )) − Dwp̃(tk − τ + s, (w, z), (x0 , y 0 ))| =Z 12≤ C|s||dλ∂v Dwp̃(τ, (w, z), (x0 , y 0 )|v=tk −τ +λs |0Z (14dλ|a(x0 , y 0 − x0 (tk − τ + λs))|Dwp̃(tk − τ + λs, (w, z), (x0 , y 0 ))|≤ C|s|0+2|Dw Dz p̃(tk − τ + λs, (w, z), (x0 , y 0 ))|)2+|w||Dwp̃(tk − τ + λs, (w, z), (x0 , y 0 ))|(4.79)using the Kolmogorov equation, applied to the frozen density for the last step.Thus, due to the control in (4.17), we get the following bounds for |∆H2 | :|w|1+(1 + |x0 |γ/2 )|∆H2 | ≤ C|s|(tk − τ )2−γ/2(tk − τ )5/2−γ/2Z 1×dλpc,K (tk − τ + λs, (w, z), (x0 , y 0 )).(4.80)0Together (4.79) and (4.80) provide the statement of the Lemma 4.5.4.Thus, from Lemma 4.5.4 it yields the final control:|J23 |!(i − 1)γ γ,pc,K (τ, (x, y), (w, z))dτCτB 1+220R2di=1()γ/2sss(1 + |x0 |1+γ/2 )++(tk − τ ) (tk − τ )5/2−γ/2(tk − τ )2−γ/2Z 1dλpc,K (tk − τ + sλ, (w, z), (x0 , y 0 ))dwdz.Z≤××tk /2Zr+1 rγ/2r−1Y(4.81)0The terms(tk −τ )2−γ/2!(i−1)γγdτC r+1 τ rγ/2B 1+,pc,K (τ, (x, y), (w, z))220R2di=1()Z1sdλpc,K (tk − τ + sλ, (w, z), (x0 , y 0 ))dwdz(tk − τ )2−γ/20!Zr1(r+1)γ Ys r+2 2(i − 1)γ γCtkB 1+,dλpc,K (tk + sλ, (x, y), (x0 , y 0 )),tk220i=1Z×≤in the convolution directly leads to prove the induction hypothesis :tk /2Zr−1Yγ/2(4.82)thus, we have to concentrate on the other two terms (tsk −τ ) + (tk −τ )s5/2−γ/2 in (4.81) and find outwhich term dominates on the current interval.γ/2Since on the interval we are considering tk − τ > tk /2 > s it is true that: (tsk −τ ) > (tk −τs)2−γ/2 .82The first term again lead us to the standard computations as in (4.83) and the second term canγ/2be bounded with (tsk −τ ) .Finally,!Z tk /2 Zr−1Y(i−1)γγdτC r+1 (1 + |x0 |1+γ/2 )τ rγ/2B 1+,pc,K (τ, (x, y), (w, z))220R2di=1!Z1sγ/2ss×dλpc,K (tk − τ + sλ, (w, z), (x0 , y 0 ))dwdz++(tk − τ ) (tk − τ )2−γ/2(tk − τ )5/2−γ/20!Z tk /2 Zr−1Y(i − 1)γ γsγ/2 (1 + |x0 |1+γ/2 )dτC r+1 τ rγ/2B 1+,pc,K (τ, (x, y), (w, z))≤22(tk − τ )0R2di=1Z 1×dλpc,K (tk − τ + sλ, (w, z), (x0 , y 0 ))(4.83)0which together with (4.83) gives us the final controlrY(i − 1)γ γsγ/2 (1 + |x0 |1+γ/2 ) r+2 (r+1)γ32B 1+|J2 | ≤Ctk,tk22i=1!Z1dλpc,K (tk + sλ, (x, y), (x0 , y 0 )),0Taking together:#"Z 1rY(i − 1)γ γs|x|sr+2 (r+1)γ/2B(1 +tk, )+ 3/2 Cdλpc,K (tk + λs, (x, y), (x0 , y 0 )),|J1 | ≤tk220tki=1!Zr1γ/20 γ/2(r+1)γ Y(i − 1)γ γs (1 + |x | ) r+2 2CtkB 1+,dλpc,K (tk + sλ, (x, y), (x0 , y 0 )),|J2 | ≤tk220i=1and equilibrating with |x| and |x0 |:s|x|(tk )3/2≤s[|x − x0 | + |x0 |]3/2tk"=|x − x0 |1/2tk#1/2tk0+ |x |s3/2tk=s2−γ/2tk+s|x0 |3/2.tkyields to the proof of the induction hypothesis (4.74) .Directly from Lemma 4.75, (4.20), (4.2.1) and the inequality (tk )−1 ≤ 2(tj )−1 for 1 ≤ k ≤ j/2,83one can get:|(4.73)| ≤ C(b, T )Eγ/2,1j−1 ZXk=11Z×tk+1tk((u − tk )γ/2(u − tk )du(1 + |x|1+γ/2 )+3/2t2dkRtkZ)pc,K (tj − u, (w, z), (x0 , y 0 ))dwdz(tj − u)1−γ/2)Z (j−1 Z tk+1X(u − tk )γ/2 (1 + |x|1+γ/2 )du= C(b, T )Eγ/2,1tktkR2ddλpc,K (tk + λ(u − tk ), (x, y), (w, z))0k=11Zpc,K (tj − u, (w, z), (x0 , y 0 ))dwdz(tj − u)1−γ/2)Z (j−1 Z tk+1X(u − tk )(1 + |x|1+γ/2 )+C(b, T )Eγ/2,1du3/2R2dtkk=1 tkdλpc,K (tk + λ(u − tk ), (x, y), (w, z))×01Z×dλpc,K (tk + λ(u − tk ), (x, y), (w, z))0pc,K (tj − u, (w, z), (x0 , y 0 ))dwdz(tj − u)1−γ/2=: ∆1 Dd,1 + ∆2 Dd,1For all η ∈ (0, γ) it holds:∆1 Dd,1≤ C(b, T )Eγ/2,1j−1 ZXk=1tk+1(Zduγ−η2dλpc,K (tk + λ(u − tk ), (x, y), (w, z))0≤ C(b, T )Eγ/2,1 (1 + |x|1+γ/2 )h(1 + |x|1+γ/2 ))1−η/2tkR2dtk1Z×(u − tk )pc,K (tj − u, (w, z), (x0 , y 0 ))dwdz(tj − u)1−γ/2γ−η2pc,K (ū, (x, y), (x0 , y 0 )),sup(4.84)ū∈[tj −h,tj ]∆1 Dd,2≤ C(b, T )Eγ/2,1j−1 ZXk=1tk+1du×dλpc,K (tk + λ(u − tk ), (x, y), (w, z))0≤ C(b, T )Eγ/2,1 (1 + |x|1+γ/2 )h(u − tk )γ−η (1 + |x|1+γ/2 )γ−η2)1/2+γ/2tkR2dtk1Z(Zpc,K (tj − u, (w, z), (x0 , y 0 ))dwdz(tj − u)1−γ/2pc,K (ū, (x, y), (x0 , y 0 )).sup(4.85)ū∈[tj −h,tj ]Thus,|Dd,1 |≤|(4.72)| + |(4.84)| + |(4.85)|≤ C(b, T )Eγ/2,1 hγ−η2(1 + |x|1+γ/2 )suppc,K (ū, (x, y), (x0 , y 0 )),(4.86)ū∈[tj −h,tj ]for η ∈ (0, γ).• Bounds for the term Dd,2 .j−1 ZXk=0tk+1tkZdup(tk , (x, y), (w, z))[H(tj − u, (w, z), (x0 , y 0 )) − H(tj − tk , (w, z), (x0 , y 0 ))]dwdz(4.87)R2d84As usual, consider the case k = 0 separately:ZZ hdup(0, (x, y), (w, z))[H(tj − u, (w, z), (x0 , y 0 )) − H(tj , (w, z), (x0 , y 0 ))]dwdzR2d0hZZ=H(tj − u, (x, y), (x0 , y 0 )) − H(tj , (x, y), (x0 , y 0 ))]dwdzduR2d0Zh≤ CZR2d0≤|H(tj − u, (x, y), (x0 , y 0 ))| + |H(tj , (x, y), (x0 , y 0 ))]|dwdzduChp (t , (x, y), (x1−γ/2 c,K jtj0, y 0 )) ≤ Chγ/2 pc,K (tj , (x, y), (x0 , y 0 )).(4.88)The result in Lemma 4.5.4 yields for k > 1:Ztk+1Zp(tk , (x, y), (w, z))[H(tj − u, (w, z), (x0 , y 0 )) − H(tj − tk , (w, z), (x0 , y 0 ))]dwdzduR2dtkZ tk+1ZC(1 + |x0 |1+γ/2 )pc,K (tk , (x, y), (w, z))t)(k(u − tk )γ/2(u − tk )(u − tk )++×(t − u)(t − u)2−γ/2(t − u)5/2−γ/2Z 1×dλpc,K (t − u + (u − tk )λ, (w, z), (x0 , y 0 ))dwdz≤duR2d(4.89)0To equilibrate with the most singular termrameter β:(u−tk )|x0 |(1+|x0 |γ/2 )(t−u)5/2−γ/2we have to balance with the pa-(u − tk )β(u − tk )(1−β)(u − tk )β≤C,5/2−γ/2−(1−β)(1−β)(t − u)(t − u)(t − u)5/2−γ/2−(1−β)To have the integrable singularity one has to impose the following conditions on γ and β: 5/2 −γ/2 − (1 − β) < 1, which is only possible if γ/2 > 1/2 + β.
This is a key point - from now we haveto assume that the Hölder index γ is at least bigger than 1/2. And the parameter beta have to bechosen as follows: 0 < β < γ − 1/2. Basically, we can just rewrite γ := 1/2 + β for some β ∈ (0, 1/2].Under the mentioned assumptions we can achieve the total rate of convergence hβ , β ∈ (0, 1/2]which, according to the restrictions we imposed before, means in case γ is close to 1 (getting closerto Lipschitz assumptions on coefficients) one can get the "standard" convergence rate of hγ−1/2 .As the result, for any β ∈ (0, 1/2], summing (4.89), one has:j−1 ZXk=1tk+1tkZdup(tk , (x, y), (w, z))R2d×[H(tj − u, (w, z), (x0 , y 0 )) − H(tj − tk , (w, z), (x0 , y 0 ))]dwdzZj−1 Z tk+1X(u − tk )β (1 + |x0 |1+γ/2 )Cdupc,K (tk , (x, y), (w, z))(t − u)5/2−γ/2−(1−β)R2dk=1 tkZ 1dλpc,K (t − u + (u − tk )λ, (w, z), (x0 , y 0 ))dwdz≤C(b, T )(1 + |x0 |1+γ/2 )hβ×≤0supū∈[tj −h,tj ]85pc,K (ū, (x, y), (x0 , y 0 ))(4.90)|Dd,2 | ≤ (4.88) + (4.90) ≤ C(b, T )(1 + |x0 |1+γ/2 )hβpc,K (ū, (x, y), (x0 , y 0 )).sup(4.91)ū∈[tj −h,tj ]From (4.86) and (4.91) we finally get:(p ⊗ H − p ⊗h H)(tj , (x, y), (x0 , y 0 )) = (Dd,1 + Dd,2 )(tj , (x, y), (x0 , y 0 ))≤ C(b, T )Eγ/2,1 hγ−η2|x|(1 + |x|γ/2 )pc,K (ū, (x, y), (x0 , y 0 ))supū∈[tj −h,tj ]+C(b, T )(1 + |x0 |1+γ/2 )hβpc,K (ū, (x, y), (x0 , y 0 ))sup(4.92)ū∈[tj −h,tj ]Consequently, we also obtain|(p ⊗ H − p ⊗h H) ⊗h H(tj , (x, y), (x0 , y 0 ))|≤ C 2 ((1 + |x|1+γ/2 ) ∨ (1 +γ/2|x0 |1+γ/2 ))tj hβ B(1,(4.93)γ) sup pc,K (ū, (x, y), (x0 , y 0 )),2 ū∈[tj −h,tj ]where T → C(T, b, σ) is non decreasing function.and by induction on r ≥ 0, we get:|(p ⊗ H − p ⊗h H) ⊗h H (r) (tj , (x, y), (x0 , y 0 ))|r γ2≤ C r+1 ((1 + |x|1+γ/2 ) ∨ (1 + |x0 |1+γ/2 ))hβ tj×sup(4.94)rYγ γB 1 + (i − 1) ,2 2i=1pc,K (ū, (x, y), (x0 , y 0 )).(4.95)ū∈[tj −h,tj ]Plugging this in (4.70), due to the asymptotic of the Gamma function, one gets:|(p − pd )(ti , (x, y), (x0 , y 0 ))|≤ C(T, b, σ, γ, β)((1 + |x|1+γ/2 ) ∨ (1 + |x0 |1+γ/2 ))hβsupū∈[tj −h,tj ]Combining with (4.69) we complete the proof of the Theorem 4.5.1.86pc,K (ū, (x, y), (x0 , y 0 )).4.6AppendixProof of Theorem 4.4.2.Proof.
For the sake of simplicity, we consider the most singular case of ε = 1.We start from the parametrix representation as usual. The basic strategy is to consider derivativesfor the main part at first and then - for the reminder term. Having bounds for the main term from(4.17) we turn to the rest of the parametrix sum.Dxα p(t, (x, y), (x0 , y 0 )) = Dxα p̃(t, (x, y), (x0 , y 0 )) +∞XhiDxα p̃ ⊗ H (r) (t, (x, y), (x0 , y 0 )) ,r=1for |α| = 1, 2.Since the first order derivative gives an integrable time singularity. For the case |α| = 2 we haveto discuss precisely. Let us denote∞Xp̃ ⊗ H (r) (t, (x, y), (x0 , y 0 )) = p̃ ⊗ Φ(t, (x, y), (x0 , y 0 )),R(t, (x, y), (x0 , y 0 )):=Φ(t, (x, y), (x0 , y 0 ))∞X:=H (r) (t, (x, y), (x0 , y 0 )).r=1r=1Inequality (4.20) for H then yields for all r ∈ N∗ , 0 < t ≤ T, (x, y), (x0 , y 0 ) ∈ (R2d )2 :with the conventionQ0i=1r−1YrγγγB( , 1 + (i − 1) )pc,K (t, (x, y), (x0 , y 0 ))t−1+ 2 ,22i=1(4.96)= 1.
![](https://s.studizba.com/z.php?f=/templates/si/i/noavatar.png&w=100&h=100&t=1"){kind=avatar}
