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

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.

The first author was partially supported by NSF Grants DMS-1522184, DMS-1819012, DMS-1107291: RNMS KI- Net, and NSFC Grant No. 31571071. The work of the third author has been funded by the Alexander von Humboldt-Professorship program, the European Union’s Horizon 2020 research and innovation programme under the Marie Sklodowska-Curie grant agreement No.765579-ConFlex, Grant MTM2017-92996-C2-1-R COSNET of MINECO (Spain), ELKARTEK project KK-2018/00083 ROAD2DC of the Basque Government, ICON of the French ANR and Nonlocal PDEs: Analysis, Control and Beyond, AFOSR Grant FA9550-18-1-0242, and Transregio 154 Project *Mathematical Modelling, Simulation and Optimization using the Example of Gas Networks* of the German DFG.

Authors and Affiliations

1. School of Mathematical Sciences, Institute of Natural Sciences, MOE-LSEC and SHL-MAC, Shanghai Jiao Tong University, Shanghai, 200240, China

   Shi Jin

2. Department of Mathematics, University of Wisconsin-Madison, Madison, WI, 53706, USA

   Yuhua Zhu

3. Chair in Applied Analysis, Alexander von Humboldt-Professorship, Department of Mathematics Friedrich- Alexander-Universität Erlangen-Nürnberg, 91058, Erlangen, Germany

   Enrique Zuazua

4. Chair of Computational Mathematics, Fundación Deusto, Av. de las Universidades 24, 48007, Bilbao, Basque Country, Spain

   Enrique Zuazua

5. Departamento de Matemáticas, Universidad Autónoma de Madrid, 28049, Madrid, Spain

   Enrique Zuazua

Corresponding author

Correspondence to Shi Jin.

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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,

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

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,

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,

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

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

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

Furthermore since,

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,

Since from (2.12), we know that

so

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,

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

one has,

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

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

Another inequality that will be used frequently later is that for \(i = 0,1\),

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 \(|\nu | = 0\),

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

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

which implies,

where

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

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

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,

where

and

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

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.

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