Диссертация (1137347), страница 18
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Such sensitivities lead to highertime singularities and make the unbounded transport term appear. The higher time-singularitiesyield the stated restriction on the Hölder index γ. The unbounded transport gives the term |x| ∧ |x0 |in the above bound. We finally can reach a global error of order hβ , β < γ − 1/2 which is close tothe expected one in hγ/2 when γ goes to 1.4.5.1Proof of Theorem 4.5.1The basic idea to prove Theorem 4.5.1 consists in applying parametrix expansion to both densitiesp(t, (x, y), (x0 , y 0 )), ph (t, (x, y), (x0 , y 0 ).
The convergence of the parametrix series expansion for thesolution of (4.9) and the scheme (4.23) follows from Section 4.2 above.In order to derive bounds for the difference of densities in (4.60), let us introduce for 0 ≤ j < j 0 ≤N ∀(x, y), (x0 , y 0 ) ∈ R2d × R2d,∗ ,Xpd (tj , (x, y), (x0 , y 0 )) :=p̃ ⊗h H (r) (tj , (x, y), (x0 , y 0 )).(4.61)r∈NFrom (4.20) and the semigroup property (4.16) it follows thatγ/2p̃ ⊗h H(tj , (x, y), (x0 , y 0 )) ≤ C(b, T, γ)tj B(1, γ2 )pc,K (tj , (x, y), (x0 , y 0 )) which is by induction yieldsthat for all r ≥ 1, ∀(x, y), (x0 , y 0 ) ∈ R2d × R2d,∗ ,|p̃ ⊗h H (r) (tj , (x, y), (x0 , y 0 ))|rY(i − 1)γ γr rγ/2,pc,K (tj , (x, y), (x0 , y 0 ))B 1+≤ C tj22i=11−γwith C := C(λ, γ)(|b|∞ T 2 + 1).From the last inequality we readily get that the series in (4.61) converges absolutely and uniformlyon R2d × R2d,∗ and that ∀(x, y), (x0 , y 0 ) ∈ R2d × R2d,∗ ,pd (tj , (x, y), (x0 , y 0 )) ≤ Eγ/2,1 (C(|b|∞ T 1/2 + T γ/2 )pc,K (tj , (x, y), (x0 , y 0 ))).(4.62)As the result we decompose the total global error into two terms:|(p − ph )(ti , (x, y), (x0 , y 0 ))| ≤ |(p − pd )(ti , (x, y), (x0 , y 0 ))| + |(pd − ph )(ti , (x, y), (x0 , y 0 ))|.Error bound on pd − ph (same discrete convolution)Remark that for r ≥ 1 as it can be decomposed with the classical approach from [KM02] (see also[Fri18] and [KM17] for connections with the current Hölder settings).(r)p̃ ⊗h H (r) − p̃ ⊗h Hh = (r−1)p̃ ⊗N H (r−1) ⊗h (H − Hh ) + p̃ ⊗N H (r−1) − p̃ ⊗h Hh⊗h Hh .For the sum from r = 1 to r = ∞ it yields:pd − ph = pd ⊗h (H − Hh ) + (pd − ph ) ⊗h Hh .By induction, for 0 ≤ j < j 0 ≤ N one gets for all (x, y), (x0 , y 0 ) ∈ R2d × R2d,∗ ,74(pd − ph )(tj , (x, y), (x0 , y 0 )) =X{pd ⊗ (H − Hh )} ⊗h H (r) (tj , (x, y), (x0 , y 0 )).(4.63)r≥0As the result, it is sufficient for us to establish the right control for each term in the sum (4.63).Lemma 4.5.2.
Under assumptions (ÂD), for all 0 ≤ j < j 0 ≤ N, for all (x, y), (x0 , y 0 ) ∈ R2d ×R2d,∗ ,one has|{pd ⊗ (H − Hh )} ⊗h H (r) (tj , (x, y), (x0 , y 0 ))| ≤ Chγ/2 pc,K (ti , (x, y)(4.64)for some constant c := c(λ, γ) ≥ 1 and a non decreasing positive fuction T → C := C(T, b, σ).Proof. First, let us consider the first step separately. For j = 1 directly from the kernel functiondefinition it follows that:(H − Hh )(tj , (w, z), (ŵ, ẑ)) =1 2p̃(tj , (w, z)(ŵ, ẑ))hb(ŵ, ẑ)Dx p̃(tj , (w, z), (ŵ, ẑ))i + Tr (a(w, z) − a(ŵ, ẑ − ŵtj ))Dw2−h−1 (ph − p̃h )(tj , (w, z), (ŵ, ẑ)).From (4.20) it follows that:1 2|hb(ŵ, ẑ)Dx p̃(tj , (w, z), (ŵ, ẑ))i + Tr (a(w, z) − a(ŵ, ẑ − ŵtj ))Dwp̃(tj , (w, z)(ŵ, ẑ)) |212−γ/2≤ C(|b|∞ tj+ 1) 1−γ/2 pc,K (tj , (w, z), (ŵ, ẑ)).tjIn [LM10] authors achieved the following control to prove Lemma 4.1( see [LM10], Appendix, A1,proof of Lemma 4.1, the case (b) which absolutely covers our model and assumptions):h−1 (ph − p̃h )(tj , (w, z), (ŵ, ẑ)) ≤C(T, b, σ)1−γ/2tjpc,K (tj , (w, z), (ŵ, ẑ)) for tj = h.Combining the last two estimates together one cat get for tj = h, ∀(w, z), (ŵ, ẑ) ∈ R2d × R2d,∗ :|(H − Hh )(tj , (w, z), (ŵ, ẑ))| ≤ (|H| + |Hh |)(tj , (w, z), (ŵ, ẑ))C≤ 1−γ/2 pc,K (tj , (w, z), (ŵ, ẑ)),tj(4.65)where T → C := C(T, b, σ) is a non-decreasing positive function.Now we make use of the decomposition: ∀i ∈ [2, N ]pd ⊗ (H − Hh )(ti , (x, y), (w, z))=i−2 ZXhk=0dudvpd (tk , (x, y), (u, v))(H − Hh )(ti − tk , (u, v), (w, z))R2dZ+hpd (ti−1 , (x, y), (u, v))(H − Hh )(h, (u, v), (z, w))dudv.R2d(4.66)75From (4.62), (4.65) and the semigroup property (4.16), we derive:Zdp (ti−1 , (x, y), (u, v))(H − Hh )(h, (u, v), (w, z))dudv hR2d≤Chpc,K (ti , (x, y), (w, z)),h1−γ/2(4.67)where T → C := C(T, b, σ) is a positive non-decreasing function.Let us again mention the paper [LM10].
We would like to emphasize that under previous assumptions for coefficients in our model, according to the paper, there exist a constant c := c(λ, γ) > 1 suchthat for all 1 < j < j 0 ≤ N :|H(tj 0 − tj , (x, y), (x0 y 0 )) − Hh (tj 0 − tj , (x, y), (x0 y 0 ))|Chpc,K (tj , (x, y), (x0 , y 0 ))≤(tj 0 − tj )−2+γ/2(4.68)where T → C = C(T, b, σ) is a positive non-decreasing function.
The case j = 1 has been alreadyproved in (4.65).From (4.62), (4.68) and the semigroup property (4.16) one gets: i−2 ZXdhdudvp (tk , (x, y), (u, v))(H − Hh )(ti − tk , (u, v), (w, z))R2dk=0≤ Chγ/2 pc,K (ti , (x, y), (w, z)).Due to all the previous estimates, we derive∀i ∈ [2, N ], |pd ⊗h (H − Hh )(ti , (x, y), (w, z))| ≤ Chγ/2 pc,K (ti , (x, y), (w, z)),where T → C := C(T, b, σ) is a non-decreasing positive function and (4.64) follows by induction.From Lemma 4.5.2, we obtain ∀(x, y), (x0 , y 0 ) ∈ R2d × R2d,∗ :|(pd − ph )(ti , (x, y), (x0 , y 0 ))| ≤ C(T, b, σ)hγ/2 pc,K (ti , (x, y), (x0 , y 0 )).(4.69)Remark 4.5.1. We would like to emphasize that up to now the bound we have in (4.69) is muchbetter than the one which has been stated in the Theorem 4.5.1.
This term can be controlled bettersince we do not feel the explosion which comes for p − pd when we basically have to investigate thedifference between the time integral and the Riemann sums.Error bound on p − pdIt still remains to control the difference p − pd . For r ≥ 1, we write the decomposition according tothe same iteration procedure as in [KM02],p̃ ⊗ H(r)− p̃ ⊗h H(r)=(r−1)(r−1)p̃ ⊗ H⊗ H − p̃ ⊗ H⊗h H + p̃ ⊗ H (r−1) − p̃ ⊗ H (r−1)⊗h H.76Summing up from r = 1 to ∞ we getp − pd = p ⊗ H − p ⊗h H + (p − pd ) ⊗h H.As in the paper [KM17]:(p − pd )(tj , (x, y), (x0 , y 0 )) = (p ⊗ H − p ⊗h H)(tj , (x, y), (x0 , y 0 ))+(p − pd ) ⊗h H(tj , (x, y), (x0 , y 0 ))=X(p ⊗ H − p ⊗h H) ⊗h H (r) (tj , (x, y), (x0 , y 0 )),(4.70)r≥0The key point is thus to control |p ⊗ H − p ⊗h H|.For that purpose let us write:=(p ⊗ H − p ⊗h H)(tj , (x, y), (x0 , y 0 ))Zj−1 Z tk+1Xdu{p(u, (x, y), (w, z))H(tj − u, (w, z), (x0 , y 0 ))k=0−=p(tk , (x, y), (w, z))H(tj − tk , (w, z), (x0 , y 0 ))}dwdzZj−1 n Z tk+1Xdu{[p(u, (x, y), (w, z)) − p(tk , (x, y), (w, z))]k=0×+R2dtkR2dtkH(tj − u, (w, z), (x0 , y 0 ))}dwdzj−1 n ZXk=0tk+1tkoZ{p(tk , (x, y), (w, z))duR2d×[H(tj − u, (w, z), (x0 , y 0 )) − H(tj − tk , (w, z), (x0 , y 0 ))]}dwdz=:(Dd,1 + Dd,2 )(tj , (x, y), (x0 , y 0 )).o(4.71)• Bounds for the term Dd,1 .- For k = 0, one readily gets:Z h Z0 0du{[p(u, (x, y), (w, z)) − p(0, (x, y), (w, z))]H(tj − u, (w, z), (x , y ))}dwdz 0R2dZ hChdu≤pc,K (tj , (x, y), (x0 , y 0 ))≤ Cpc,K (tj , (x, y), (x0 , y 0 ))1−γ/2(tj )1−γ/20 (tj − u)Chγ/2 pc,K (tj , (x, y), (x0 , y 0 )).≤(4.72)- For k ∈ [1, j − 1] we are interested to control the sum:j−1 ZXk=1tk+1tkZdu{p(u, (x, y), (w, z)) − p(tk , (x, y), (w, z))}H(tj − u, (w, z), (x0 , y 0 ))dwdz.R2d(4.73)To proceed with the case for k ≥ 1 one needs the following result:77Lemma 4.5.3.
Under (ÂD) there exist constants C(λ, γ), c := c(λ, γ) ≥ 1 such that for all r ≥ 0,for all (x, y), (x0 , y 0 ) ∈ R2d × R2d,∗ one has|p̃ ⊗ H (r) (tk + (u − tk ), (x, y), (x0 , y 0 )) − p̃ ⊗ H (r) (tk , (x, y), (x0 , y 0 ))|()γ/2(u − tk )1+γ/2r+1 rγ/2 (u − tk )≤ Eγ/2,1 (1 + |x|)Ctk+3/2tktkZ 1r−1Y(i − 1)γ γ×dλpc,K (tk + λ(u − tk ), (x, y), (x0 , y 0 ))B(1 +, )220i=1(4.74)from which it follows that|p(tk + (u − tk ), (x, y), (x0 , y 0 )) − p(tk , (x, y), (x0 , y 0 ))|()(u − tk )γ/2(u − tk )1+γ/2≤ C(b, T )Eγ/2,1 (1 + |x|)+3/2tktkZ 1dλpc,K (tk + λ(u − tk ), (x, y), (x0 , y 0 )).×(4.75)0Proof.
Let us start with the base for the induction. To control the difference between frozen densitiesat the step r = 0 one can write, applying the mean-value theorem and the Kolmogorov equation,takeing s = u − tk for a moment:p̃(tk + s, (x, y), (x0 , y 0 )) − p̃(tk , (x, y), (x0 , y 0 ))Z 1= s∂τ p̃(tk + λs, (x, y), (x0 , y 0 ))|τ =tk +λs dλ0Z11Tr(a(x0 , y 0 − x0 (tk + λs))Dx2 p̃(tk + λs, (x, y), (x0 , y 0 )))0 2+ hx, ∇y p̃(tk + λs, (x, y), (x0 , y 0 ))iZ 11|x|≤ Cs+p̃(tk + λs, (x, y), (x0 , y 0 ))(tk + λs) (tk + λs)3/20!Z11|x|≤ Cs+ 3/2dλp̃(tk + λs, (x, y), (x0 , y 0 )).tk0t= s(4.76)kSo (4.74) is valid for r = 0.
Now proceeding by induction we assume that (4.74) is valid for r ≥ 0.By a change of variables one has:p̃ ⊗ H (r+1) (tk + s, (x, y), (x0 , y 0 )) − p̃ ⊗ H (r+1) (tk , (x, y), (x0 , y 0 ))Ztk +s=dτp̃ ⊗ H (r) (τ, (x, y), (w, z))H(t + s − τ, (w, z), (x0 , y 0 ))dwdz0R2dZ tkZ−dτp̃ ⊗ H (r) (τ, (x, y), (w, z))H(tk − τ, (w, z), (x0 , y 0 ))dwdz0R2dZ tk +s Z=dτp̃ ⊗ H (r) (tk + s − τ, (x, y), (w, z))H(τ, (w, z), (x0 , y 0 ))dwdzZtkR2dZ+tkZ{p̃ ⊗ H (r) (tk + s − τ, (x, y), (w, z))dτ0R2d−p̃ ⊗ H (r) (tk − τ, (x, y), (w, z))}H(τ, (w, z), (x0 , y 0 ))dwdz= I + J.78From (4.20) and Lemma 4.2.1 one can get|I| ≤C r+2tk +sZ1−γ/2tkYr(i − 1)γ γ(tk + s − τ )dτ,pc,K (tk , (x, y), (x0 , y 0 ))B 1+22i=1rYs r+2 (r+1)γ/2(i − 1)γ η2≤ C,)pc,K (tk , (x, y), (x0 , y 0 )),tkB 1+tk22i=1rγ/2tkwhere we used that s = u − tk ∈ [0, h] for the last inequality.For the second term to full fill the induction assumption in time let us decompose J = J1 + J2 ,where:Z tk /2 ZJ1 =dτ{p̃ ⊗ H (r) (tk + s − τ, (x, y), (w, z))R2d0−p̃ ⊗ H (tk − τ, (x, y), (w, z))}H(τ, (w, z), (x0 , y 0 ))dwdzZZ tkdτ{p̃ ⊗ H (r) (tk + s − τ, (x, y), (w, z))(r)J2=R2dtk /2−p̃ ⊗ H (r) (tk − τ, (x, y), (w, z))}H(τ, (w, z), (x0 , y 0 ))dwdz.First, assume that s = u − tk ∈ [0, tk /2].
That means s ∈ [0, tk − u] for all u ∈ [0, tk /2] so that wecan apply the induction hypothesis and get:"# Z!tk /211(tk − τ )rγ/2 τ −(1−γ/2) dτ+ 3/2|J1 | ≤ C r+2 s(1 + |x|1+γ/2 )tk0tk!! Zr−11Y(i − 1)γ γ×B 1+,dλpc,K (tk + λs, (x, y), (x0 , y 0 ))220i=1"#rY(i − 1)γ γ11(r+1)γ/2B(1 +≤ C r+2 s(1 + |x|1+γ/2 ), )+ 3/2 tktk22tki=1Z 1×dλpc,K (tk + λs, (x, y), (x0 , y 0 )).0Now, if s ∈ (tk /2, tk ] one writes J1 = J11 + J12 with:ZZ (tk /2J11=p̃ ⊗ H (r) (tk − τ +dτR2d0− p̃ ⊗ H (r) (t − τ +J12Ztk /2=Zdτtktk+ (s − ), (x, y), (w, z))22)tk, (x, y), (w, z)) H(τ, (w, z), (x0 , y 0 ))dwdz,2(p̃ ⊗ H (r) (tk − τ +R2d0tk, (x, y), (w, z))2)− p̃ ⊗ H(r)(tk − τ, (x, y), (w, z)) H(τ, (w, z), (x0 , y 0 ))dwdz.79From the induction hypothesis and (4.20) using that tk /2 ≤ tk − τ for τ ∈ [0, tk /2] one has:Z tk /2tk11+ tk|J11 | ≤ C r+2 (1 + |x|1+γ/2 )(s − ) tk2( 2 + (tk − τ ))3/202 + (tk − τ )!rYtk rγ/2 −(1−γ/2)(i − 1)γ γ×(tk − τ + )τdτB(1 +, )222i=1Z 1×dλpc,K (tk + λs, (x, y), (x0 , y 0 ))0#"rY11(i − 1)γ γ(r+1)γ/2r+21+γ/2+ 3/2 tk, )≤ C(1 + |x|)sB(1 +tk22tki=1Z 1dλpc,K (tk + λs, (x, y), (x0 , y 0 )).×0Using the similar argument with s ≥ tk /2:"#rY1(i − 1)γ γ1(r+1)γ/221+γ/2|J1 | ≤ (1 + |x|)sB 1+,+ 3/2 C r+2 tktk22tki=1Z 1×dλpc,K (tk + λs, (x, y), (x0 , y 0 )),0which yields#rY(i − 1)γ γ11r+2 (r+1)γ/2B 1+tk,)s+ 3/2 Ctk22tki=1"|J1 |1+γ/2≤ (1 + |x|1Zdλpc,K (tk + λs, (x, y), (x0 , y 0 ))×0The last term J2 is given by the sum of three terms:Z s Z1J2 = −dτp̃ ⊗ H (r) (τ, (x, y), (w, z))H(tk − τ + s, (w, z), (x0 , y 0 ))dwdz,R2d0J22Ztk /2+s=Zp̃ ⊗ H (r) (τ, (x, y), (w, z))H(tk − τ + s, (w, z), (x0 , y 0 ))dwdz,duR2dtk /2J23tk /2Z=Zdτp̃ ⊗ H (r) (τ, (x, y), (w, z))R2d0{H(tk − τ + s, (w, z), (x0 , y 0 )) − H(tk − τ, (w, z), (x0 , y 0 ))}dwdz.×Using (4.20) and (4.2.1) one can get as usual:|J21 |≤ Cr+2Zτ0≤srγ/21γ dτ(tk + s − τ )1− 2! r−1Y!(i − 1)γ γB 1+,pc,K (tk , (x, y), (x0 , y 0 ))22i=1!rYs r+2 (r+1)γ(i − 1)γ γCtk 2B 1+,pc,K (tk , (x, y), (x0 , y 0 ))tk22i=180and similarly|J22 |≤C1r+21−γ/2tk≤s r+2 (r+1)γCtk 2tk!!(i − 1)γ γτdτB 1+pc,K (tk , (x, y), (x0 , y 0 )),22tk /2i=1!rY(i − 1)γ γ,pc,K (tk , (x, y), (x0 , y 0 )).B 1+22i=1Ztk /2+srγ/2rYTo control |J23 | we need to derive bounds for the kernel time sensitivityH(t − τ + s, (w, z), (x0 , y 0 )) − H(t − τ, (w, z), (x0 , y 0 )).Lemma 4.5.4.|H(tk − τ + s, (w, z), (x0 , y 0 )) − H(tk − τ, (w, z), (x0 , y 0 ))|!)(sγ/2 |x0 |γ/2s|w|(1 + |x0 |γ/2 )ss|x0 |γ/2++≤C+(tk − τ )(t − τ )2−γ/2(tk − τ )2−γ/2(tk − τ )5/2−γ/2Z 1dλpc,K (tk − τ + sλ, (w, z), (x0 , y 0 )).×(4.77)0Proof.
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