The Vlasov–Fokker–Planck equation with high dimensional parametric forcing term

Abstract

We consider the Vlasov–Fokker–Planck equation with random electric field where the random field is parametrized by countably many infinite random variables due to uncertainty. At the theoretical level, with suitable assumption on the anisotropy of the randomness, adopting the technique employed in elliptic PDEs (Cohen and DeVore in Acta Numerica 24:1–159, 2015) , we prove the best N approximation in the random space enjoys a convergence rate, which depends on the summability of the coefficients of the random variable, higher than the Monte-Carlo method. For the numerical method, based on the adaptive sparse polynomial interpolation (ASPI) method introduced in Chkifa et al. (Found Comput Math 14:601–603, 2014), we develop a residual based adaptive sparse polynomial interpolation (RASPI) method which is more efficient for multi-scale linear kinetic equation, when using numerical schemes that are time dependent and implicit. Numerical experiments show that the numerical error of the RASPI decays faster than the Monte-Carlo method and is also dimension independent.

Appendices

Appendices

1.1 The proof of lemma 3.7

1.1.1 The proof of the time independent random electric field

We will first prove the case when the random electric field is time independent, that is, $E(t,x,z)\equiv E(x,z)$.

Lemma A.1

Under Condition 3.2, for all $\mathbf{z }\in U$, the following estimates hold,

$$\begin{aligned}&|\nu | = 0: \partial _t G^0_i + \frac{\eta }{\epsilon }\left\Vert h\right\Vert _V^2 \le 0; \end{aligned}$$

(A.1)

$$\begin{aligned}&|\nu | > 1: \partial _tG_i^\nu + \frac{\eta }{\epsilon }\left\Vert \partial ^{\nu }h\right\Vert _V^2\le \frac{2}{\lambda \epsilon }\sum _{\nu _j\ne 0} \nu _j^2C_j \left\Vert \partial ^{\nu -e_j}h \right\Vert _V^2, \end{aligned}$$

(A.2)

where $i = 1,2$, $\eta = \frac{C_s}{10}$, and

$$\begin{aligned}&\frac{\epsilon }{2\lambda } \left\Vert \partial ^{\nu }h \right\Vert _V^2 \le G_1^\nu \le \frac{3}{2\lambda \epsilon } \left\Vert \partial ^{\nu }h \right\Vert _V^2,\quad \frac{1}{2\lambda \epsilon } \left\Vert \partial ^{\nu }h \right\Vert _V^2 \le G_2^\nu \le \frac{3}{2\lambda \epsilon } \left\Vert \partial ^{\nu }h \right\Vert _V^2 . \end{aligned}$$

(A.3)

Proof

For $E(t,x,\mathbf{z }) \equiv E(x, \mathbf{z })$, the microscopic equation (2.13) for the perturbative solution $h(t,x,v, \mathbf{z })$ is simplified to,

$$\begin{aligned} \epsilon \partial _th + v\partial _xh - \frac{1}{\epsilon } {\mathcal {L}}h = E\left( \partial _v - \frac{v}{2}\right) h, \end{aligned}$$

(0.4)

where ${\mathcal {L}}$ is the linearized Fokker Planck operator defined in (2.14) that satisfies the local coercivity property as in (2.15). If one multiplies $\sqrt{M}$ and $v\sqrt{M}$ to (0.4) respectively, and integrates them over v, then one gets the macroscopic equations,

The first equation is the perturbative continuity equation, while the second one is the perturbative momentum equation. Notice the operators $\Pi$ and $1-\Pi$ are perpendicular to each other in $L^2_{x,v}$, that is,

$$\begin{aligned} \left\Vert h\right\Vert ^2 = \left\Vert \Pi h \right\Vert ^2 + \left\Vert (1-\Pi ) h\right\Vert ^2 = \left\Vert \sigma \right\Vert ^2 + \left\Vert (1-\Pi ) h\right\Vert ^2. \end{aligned}$$

(A.7)

If one takes $\partial ^\nu$ and $\partial ^\nu \partial _x$ to (0.4), and multiplies $\partial ^\nu _\mathbf{z }h$ and $\partial ^\nu \partial _xh$ respectively, then integrates them over x, v and adds the two equations together, one has

$$\begin{aligned}&\frac{\epsilon }{2}\partial _t \left\Vert \partial ^{\nu }h \right\Vert _V^2 +\frac{\lambda }{\epsilon } \left\Vert (1-\Pi )\partial ^{\nu }h\right\Vert _{V,\omega }^2 \nonumber \\ \le&\underbrace{\left\langle \partial ^\nu \left( E\left( \partial _v-\frac{v}{2}\right) h \right) , \partial ^\nu h\right\rangle }_{I} +\underbrace{\left\langle \partial ^\nu \partial _x\left( E\left( \partial _v-\frac{v}{2}\right) h \right) , \partial ^\nu \partial _xh\right\rangle }_{II}, \end{aligned}$$

(A.8)

where the second term II comes from the hypocoercivity of ${\mathcal {L}}$ in (2.15). If one takes $\partial ^\nu$ to (A.6) , and multiplies $\partial ^\nu \partial _x\sigma$, then integrates it over x, v, one has

(A.9)

Here the first term on the LHS and the first term on the RHS come from

$$\begin{aligned}&\left\langle \partial _t \partial ^{\nu }u, \partial ^{\nu }\partial _x\sigma \right\rangle = \partial _t\left\langle \partial ^{\nu }u, \partial ^{\nu }\partial _x\sigma \right\rangle - \left\langle \partial ^{\nu }u, \partial _t \partial ^{\nu }\partial _x\sigma \right\rangle \nonumber \\ =&\partial _t\left\langle \partial ^{\nu }u, \partial ^{\nu }\partial _x\sigma \right\rangle + \frac{1}{\epsilon }\left\langle \partial ^{\nu }u, \partial ^{\nu }\partial _x^2u \right\rangle = \partial _t\left\langle \partial ^{\nu }u, \partial ^{\nu }\partial _x\sigma \right\rangle - \frac{1}{\epsilon }\left\Vert \partial ^{\nu }\partial _xu\right\Vert ^2, \end{aligned}$$

(A.10)

where the second equality is because of the continuity equation (A.5). The second term on the LHS of (A.9) is because of

$$\begin{aligned}&\frac{1}{\epsilon }\left\langle \partial ^{\nu }u, \partial ^{\nu }\partial _x\sigma \right\rangle = \left\langle -\frac{1}{\epsilon } \partial ^{\nu }\partial _xu, \partial ^{\nu }\sigma \right\rangle = \left\langle \partial _t \partial ^{\nu }\sigma , \partial ^{\nu }\sigma \right\rangle = \frac{1}{2}\partial _t\left\Vert \partial ^{\nu }\sigma \right\Vert ^2. \end{aligned}$$

(A.11)

Furthermore since,

$$\begin{aligned}&\left\langle \partial ^{\nu }\partial _x\sigma , \partial ^{\nu }\partial _x\sigma \right\rangle + \left\langle \int v^2(1-\Pi )\partial _x \partial ^{\nu }h\sqrt{M}\,dv , \partial ^{\nu }\partial _x\sigma \right\rangle \nonumber \\&\quad \ge \frac{1}{2}\left\Vert \partial ^{\nu }\partial _x\sigma \right\Vert ^2-\frac{1}{2}\left\Vert (1-\Pi )\partial ^{\nu }\partial _xh \right\Vert ^2_\omega , \end{aligned}$$

(A.12)

this gives the third term on the LHS and the second term on the RHS of (A.9).

Furthermore, integrate (A.5) over x, by the periodic boundary condition, one has,

$$\begin{aligned} \partial _t \int \sigma dx = 0. \end{aligned}$$

Since from (2.12), we know that

$$\begin{aligned} \int \sigma (0,x,z) dx = \int h\sqrt{M}dxdv = 0, \end{aligned}$$

so

$$\begin{aligned} \int \sigma (t,x,z) dx = \int \sigma (0,x,z) dx = 0. \end{aligned}$$

Similar equality can be obtained for $\partial ^\nu \sigma$. Therefore, one can apply the Poincare inequality (1.1) to $\left\Vert \partial _x\sigma \right\Vert ^2$ to get,

$$\begin{aligned} \left\Vert \partial ^\nu \partial _x\sigma \right\Vert ^2 \ge C_s\left\Vert \partial ^\nu \sigma \right\Vert _V^2. \end{aligned}$$

By adding $\theta _i$(A.8) $+ \frac{1}{2\epsilon }$ (A.9), and the fact that

$$\begin{aligned} \left\Vert \partial ^{\nu }\partial _xu\right\Vert ^2 \le \left\Vert (1-\Pi )\partial ^{\nu }\partial _xh \right\Vert ^2\le \left\Vert (1-\Pi )\partial ^{\nu }h \right\Vert _{V,\omega }^2, \end{aligned}$$

one has,

$$\begin{aligned}&\partial _t G_i^{\nu } +\frac{\lambda \theta _i}{\epsilon } \left\Vert (1-\Pi )\partial ^{\nu }h\right\Vert _{V,\omega }^2 + \frac{ C_s}{4\epsilon }\left\Vert \partial ^{\nu }\sigma \right\Vert _V^2 \nonumber \\&\quad \le \frac{3}{4\epsilon }\left\Vert (1-\Pi )\partial ^{\nu }h \right\Vert _{V,\omega }^2 + \theta _i\left( I+II\right) -\frac{1}{2\epsilon }III, \end{aligned}$$

(A.13)

where $C_s$ comes from the Poincare inequality (1.1). In order to find an estimate for $G^\nu _i$, one needs to estimate terms $\theta _i\left( I+II\right) -\frac{1}{2\epsilon }III$. First notice that

$$\begin{aligned}&\left( \partial _v + \frac{v}{2}\right) \partial ^\nu \partial _x^i (\Pi h) = -\frac{v}{2}\left( \partial ^\nu \partial _x^i \sigma \right) \sqrt{M}+ \frac{v}{2}\left( \partial ^\nu \partial _x^i \sigma \right) \sqrt{M}= 0, \quad \text {for }i = 0, 1, \end{aligned}$$

(A.14)

which implies that terms I and II can be simplified to, for $i = 0,1$,

$$\begin{aligned} I, II&= -\left\langle \partial ^\nu \partial _x^i(Eh), \left( \partial _v + \frac{v}{2}\right) \partial ^\nu \partial _x^ih\right\rangle \nonumber \\&= -\left\langle \partial ^\nu \partial _x^i(Eh), \left( \partial _v + \frac{v}{2}\right) \left( \partial ^\nu \partial _x^i( \Pi h) + (1-\Pi )\partial ^\nu \partial _x^ih\right) \right\rangle \nonumber \\&= -\left\langle \partial ^\nu \partial _x^i(Eh), \left( \partial _v + \frac{v}{2}\right) (1-\Pi )\partial ^\nu \partial _x^i h\right\rangle . \end{aligned}$$

(A.15)

Another inequality that will be used frequently later is that for $i = 0,1$,

$$\begin{aligned}&\left\Vert \left( \partial _v + \frac{v}{2}\right) \left( 1-\Pi \right) \partial ^\nu \partial _x^i h\right\Vert ^2 \nonumber \\&\quad = \left\Vert \partial _v(1-\Pi )\partial ^\nu \partial _x^i h\right\Vert ^2 + \frac{1}{4}\left\Vert v (1-\Pi )\partial ^\nu \partial _x^i h \right\Vert ^2 + \int _\Omega \frac{v}{2}\partial _v\left( (1-\Pi )\partial ^\nu \partial _x^i h\right) ^2dxdv \nonumber \\&\quad = \left\Vert \partial _v(1-\Pi )\partial ^\nu \partial _x^i h\right\Vert ^2 + \frac{1}{4}\left\Vert v (1-\Pi )\partial ^\nu \partial _x^i h \right\Vert ^2 \nonumber \\&\qquad - \frac{1}{2} \left\Vert (1-\Pi )\partial ^\nu \partial _x^i h \right\Vert ^2 \le \left\Vert (1-\Pi )\partial ^\nu \partial _x^i h \right\Vert ^2_\omega . \end{aligned}$$

(A.16)

Based on (A.15) and (A.16), we will bound the term $\theta _i\left( I+II\right) -\frac{1}{2\epsilon }III$ for the cases $|\nu | = 0, |\nu |>1$ respectively. Firstly, for the case $|\nu | = 0$,

$$\begin{aligned}&\theta _i\left( I+II\right) -\frac{1}{2\epsilon }III \nonumber \\&\quad = \theta _i\left\langle Eh, \left( \partial _v + \frac{v}{2}\right) (1-\Pi ) h\right\rangle + \theta _i\left\langle \partial _x\left( Eh\right) , \left( \partial _v + \frac{v}{2}\right) (1-\Pi ) h\right\rangle \nonumber \\&\qquad - \frac{1}{2\epsilon }\left\langle E\sigma , \partial _x\sigma \right\rangle \nonumber \\&\quad \le \frac{\theta _i}{2}\left\Vert E \right\Vert _{L^\infty _x}\left( \epsilon \left\Vert h \right\Vert _V^2 + \frac{1}{\epsilon }\left\Vert (1-\Pi ) h\right\Vert _{V,\omega }^2\right) \nonumber \\&\quad + \frac{\theta _i}{2}\left\Vert \partial _xE \right\Vert _{L^\infty _x}\left( \epsilon \left\Vert h \right\Vert ^2 + \frac{1}{\epsilon }\left\Vert (1-\Pi ) \partial _xh\right\Vert _\omega ^2\right) \nonumber \\&+ \frac{1}{4\epsilon } \left\Vert E \right\Vert _{L^\infty _x}\left( \left\Vert \sigma \right\Vert ^2 + \left\Vert \partial _x\sigma \right\Vert ^2\right) . \end{aligned}$$

(A.17)

Since $\left\Vert E \right\Vert _{L^\infty (U,W^\infty _x)} \le C_E$, and $\left\Vert h \right\Vert ^2_V = \left\Vert \sigma \right\Vert _V^2 + \left\Vert (1-\Pi )h \right\Vert _V^2$, one can further simplify (A.17) to

$$\begin{aligned}&(A.17) \le C_E\left( \left( \frac{\epsilon \theta _i}{2} + \frac{1}{4\epsilon }\right) \left\Vert \sigma \right\Vert _V^2 + \frac{\theta _i}{\epsilon }\left\Vert (1-\Pi ) h\right\Vert _{V,\omega }^2\right) . \end{aligned}$$

(A.18)

Plug the above estimate into (A.13), one has,

$$\begin{aligned}&\partial _t G_i^0 +\frac{\lambda \theta _i}{\epsilon } \left\Vert (1-\Pi )h\right\Vert _{V,\omega }^2 + \frac{ C_s}{4\epsilon }\left\Vert \sigma \right\Vert _V^2 \nonumber \\ \le&\frac{3}{4\epsilon }\left\Vert (1-\Pi )h \right\Vert _{V,\omega }^2 + C_E\left( \left( \frac{\epsilon \theta _i}{2} + \frac{1}{4\epsilon }\right) \left\Vert \sigma \right\Vert _V^2 + \frac{\theta _i}{\epsilon }\left\Vert (1-\Pi ) h\right\Vert _{V,\omega }^2\right) , \end{aligned}$$

(A.19)

which implies,

$$\begin{aligned}&\partial _t G_i^{0} + C^0_h \left\Vert (1-\Pi )h\right\Vert _{V,\omega }^2 + C^0_\sigma \left\Vert \sigma \right\Vert _V^2 \le 0, \end{aligned}$$

(A.20)

where

$$\begin{aligned}&C^0_h = \frac{\lambda \theta _i}{\epsilon } -\frac{3}{4\epsilon } - \frac{C_E\theta _i}{\epsilon },\quad C^0_\sigma = \frac{ C_s}{4\epsilon } - C_E\left( \frac{\epsilon \theta _i}{2} + \frac{1}{4\epsilon }\right) . \end{aligned}$$

(A.21)

Since $C_E\le \frac{\lambda C_s}{8} \le \frac{\lambda }{16}$, so $\lambda - C_E \ge \frac{15}{16}\lambda$, which implies $\left( \lambda - C_E\right) \frac{\theta _i}{\epsilon } \ge \frac{4}{\lambda }$ for both $i = 1,2$. Therefore $C_h^0 \ge \frac{3}{\epsilon }$. Since $C_s - C_E \ge \frac{7}{8}C_s$, and $\epsilon C_E\theta _i\le \frac{\epsilon C_s}{7}$ for both $i = 1,2$, therefore $C_\sigma ^0 \ge \frac{7C_s}{32\epsilon } - \frac{\epsilon C_s}{14} \ge \frac{C_s}{10\epsilon }$. Hence plug back $C^0_h, C^0_\sigma \ge \frac{C_s}{10\epsilon }$ to (A.20), one has,

(A.22)

which is exactly (A.1) in Lemma A.1.

For the case where $|\nu | > 0$, based on (A.15), (A.16) and the definition of $E = {\bar{E}}(x) + \sum _{j\ge 1}E_j(x)z_j$, one can bound $\theta _i\left( I+II\right) -\frac{1}{2\epsilon }III$ by

$$\begin{aligned}&\theta _i\left( I+II\right) -\frac{1}{2\epsilon }III\nonumber \\&\quad \le \theta _i\left\langle \partial _x\left( E\partial ^{\nu }h\right) , \left( \partial _v+\frac{v}{2}\right) (1-\Pi )\partial ^{\nu }\partial _xh\right\rangle - \frac{1}{2\epsilon } \left\langle E\partial ^{\nu }\sigma , \partial ^{\nu }\partial _x\sigma \right\rangle \nonumber \\&\qquad +\sum _{\nu _j\ne 0}\left( \theta _i\left\langle \nu _jE_j\partial ^{\nu -e_j}h, (\partial _v+\frac{v}{2})(1-\Pi )\partial ^{\nu }h\right\rangle \right. \nonumber \\&\qquad \left. + \theta _i\left\langle \partial _x\left( \nu _jE_j\partial ^{\nu -e_j}h\right) , (\partial _v+\frac{v}{2})(1-\Pi )\partial ^{\nu }\partial _xh\right\rangle \right. \nonumber \\&\qquad \left. -\frac{1}{2\epsilon } \left\langle \nu _jE_j\partial ^{\nu -e_j}\sigma , \partial ^{\nu }\partial _x\sigma \right\rangle \right) . \end{aligned}$$

(A.23)

Since the three terms in the second line of the above equation is similar to the case where $|\nu | = 0$, hence one can get similar estimates as in (A.18). In addition, by assumption $\left\Vert E_j \right\Vert _{L^\infty (U,W^{1,\infty }_x)} \le C_j$, one has

$$\begin{aligned}&(A.23)\nonumber \\&\quad \le C_E\left( \left( \frac{\epsilon \theta _i}{2}+\frac{1}{4\epsilon }\right) \left\Vert \partial ^{\nu }\sigma \right\Vert _V^2+ \frac{\theta _i}{\epsilon }\left\Vert (1-\Pi )\partial ^{\nu }h \right\Vert _{V,\omega } ^2\right) \nonumber \\&\qquad +\frac{\theta _i}{2}\sum _{\nu _j\ne 0} C_j\left( \epsilon \nu _j^2\left\Vert \partial ^{\nu -e_j}h \right\Vert _V^2 + \frac{1}{\epsilon }\left\Vert (1-\Pi )\partial ^{\nu }h \right\Vert _{V,\omega }^2\right) \nonumber \\&\qquad +\frac{1}{4\epsilon }\sum _{\nu _j\ne 0}C_j\left( \nu _j^2\left\Vert \partial ^{\nu -e_j}\sigma \right\Vert ^2 + \left\Vert \partial ^{\nu }\partial _x\sigma \right\Vert ^2 \right) \nonumber \\&\quad \le C_E\left( \frac{1}{2}\left( \epsilon \theta _i+\frac{1}{\epsilon }\right) \left\Vert \partial ^{\nu }\sigma \right\Vert _V^2+ \frac{3\theta _i}{2\epsilon }\left\Vert (1-\Pi )\partial ^{\nu }h \right\Vert _{V,\omega } ^2\right) \nonumber \\&\qquad +\sum _{\nu _j\ne 0} \nu _j^2C_j\left( \frac{\epsilon \theta _i}{2} + \frac{1}{4\epsilon }\right) \left\Vert \partial ^{\nu -e_j}h \right\Vert _V^2 , \end{aligned}$$

(A.24)

where the second inequality is because that $\sum _{j\ge 1}C_j \le C_E$. Hence plug (A.24) into (A.13), one has,

$$\begin{aligned}&\partial _t G^\nu _i + C^\nu _h \left\Vert (1-\Pi )\partial ^\nu h\right\Vert _{V,\omega }^2 + C^\nu _\sigma \left\Vert \partial ^\nu \sigma \right\Vert _V^2 \le \sum _{\nu _j\ne 0} \nu _j^2C_j\left( \frac{\epsilon \theta _i}{2} + \frac{1}{4\epsilon }\right) \left\Vert \partial ^{\nu -e_j}h \right\Vert _V^2, \end{aligned}$$

(A.25)

where

$$\begin{aligned}&C^\nu _h = \frac{\lambda \theta _i}{\epsilon } -\frac{3}{4\epsilon } - \frac{3C_E\theta _i}{2\epsilon } \ge \frac{\eta }{\epsilon },\quad C^\nu _\sigma = \frac{ C_s}{4\epsilon } - \frac{C_E}{2}\left( \epsilon \theta _i+ \frac{1}{\epsilon }\right) \ge \frac{\eta }{\epsilon } \end{aligned}$$

(A.26)

and

$$\begin{aligned}&\frac{\epsilon \theta _i}{2} + \frac{1}{4\epsilon } \le \frac{2}{\lambda \epsilon }, \end{aligned}$$

(A.27)

for both $i = 1,2$, which gives (A.2) in Lemma A.1. $\square$

1.1.2 The proof of lemma 3.7

If one takes $\partial ^\nu$ and $\partial ^\nu \partial _x$ to (2.13), and multiplies $\partial ^\nu _\mathbf{z }h$ and $\partial ^\nu \partial _xh$ respectively, then integrates them over x, v and adds the two equations together, one has

$$\begin{aligned}&\frac{\epsilon }{2}\partial _t \left\Vert \partial ^{\nu }h \right\Vert _V^2 +\frac{\lambda }{\epsilon } \left\Vert (1-\Pi )\partial ^{\nu }h\right\Vert _{V,\omega }^2 \nonumber \\ \le&\left\langle \partial ^\nu \left( E\left( \partial _v-\frac{v}{2}\right) h \right) , \partial ^\nu h\right\rangle +\left\langle \partial ^\nu \partial _x\left( E\left( \partial _v-\frac{v}{2}\right) h \right) , \partial ^\nu \partial _xh\right\rangle , \nonumber \\&-\underbrace{\left\langle v\sqrt{M}\partial ^\nu \left( \left( E - E^\infty \right) e^{-\phi ^\infty }\right) , \partial ^\nu h\right\rangle }_{IV} -\underbrace{\left\langle v\sqrt{M}\partial ^\nu \partial _x\left( \left( E - E^\infty \right) e^{-\phi ^\infty } \right) , \partial ^\nu \partial _xh\right\rangle }_{V}. \end{aligned}$$

(A.28)

If one multiplies $\sqrt{M}$ and $v\sqrt{M}$ to (2.13), and integrates it over v, then one gets

Then if one takes $\partial ^\nu$ to (A.30), and multiplies $\partial ^\nu \partial _x\sigma$, then integrates it over x, v, one has

(A.31)

By comparing (A.28) and (A.31) with (A.8) and (A.9), we actually only need to bound terms IV, V, VI. Similar to (A.13), one has the following estimates for $G_i^\nu$,

(A.32)

First notice, for $i = 0, 1$,

(A.33)

So one can break

in $(IV+ V)$, therefore

(A.34)

For the term VI, one can bound it by

(A.35)

For $|\nu | = 0$, plug (A.34) and (A.35) into (A.32), and based on the estimates (A.19) we have already got in Appendices 0.1, one has

(A.36)

which implies,

(A.37)

where

(A.38)

Since $C_E\le \frac{\lambda C_s}{8} \le \frac{\lambda }{16}$, so $\lambda - 2C_E \ge \frac{7}{8}\lambda$, which implies $\left( \lambda - C_E\right) \frac{\theta _i}{\epsilon } \ge 1$ for both $i = 1,2$. Therefore $C_h^0 \ge \frac{1}{4\epsilon }$. Since $\frac{C_s}{4} - \frac{3C_E}{8} \ge \frac{11}{64}C_s$, and $\epsilon C_E\theta _i\le \frac{\epsilon C_s}{7}$ for both $i = 1,2$, therefore $C_\sigma ^0 \ge \frac{11C_s}{64\epsilon } - \frac{\epsilon C_s}{14} \ge \frac{C_s}{10\epsilon }$. Hence plug back $C^0_h, C^0_\sigma \ge \frac{C_s}{10\epsilon }$ to (A.37), one has,

(A.39)

For $|\nu | > 0$, plug (A.34) and (A.35) into (A.32), and based on the estimates (A.24), one has

(A.40)

which implies

(A.41)

for the same $C^\nu _h, C^\nu _h$ defined in (A.38). This completes the proof.