Particle system algorithm and chaos propagation related to non-conservative McKean type stochastic differential equations

Abstract

We discuss numerical aspects related to a new class of NonLinear Stochastic Differential Equation (NLSDE) in the sense of McKean, which are supposed to represent non conservative nonlinear Partial Differential Equations (PDEs). We propose an original interacting particle system for which we discuss the propagation of chaos. We consider a time-discretized approximation of this particle system to which we associate a random function which is proved to converge to a solution of a regularized version of a nonlinear PDE.

Appendix

In this appendix, we present the proof of Lemma 4.4. We first proceed with the proof of some intermediary inequalities.

Lemma 5.1

We suppose Assumption 1. Let $N \in \mathbb N^{\star }$. Let $(\xi ^{i,N})_{i=1, \ldots ,N}$ be (a solution of) the interacting particle system (3.3); let $(\tilde{\xi }^{i,N})_{i=1, \ldots ,N}$ and $\tilde{u}$ as defined as in the discretized interacting particle system (4.1).

The random variables $V_t^i := e^{\int _0^t\Lambda \big (s,\tilde{\xi }^{i,N}_{s},u^{S^N(\tilde{\xi })}_{s}(\tilde{\xi }^{i,N}_{s})\big )ds}$ and $\tilde{V}_t^i := e^{\int _0^t\Lambda \big (r(s),\tilde{\xi }^{i,N}_{r(s)},\tilde{u}_{r(s)}(\tilde{\xi }^{i,N}_{r(s)})\big )ds}$, for all $t \in [0,T]$, $i \in \{ 1, \ldots ,N\}$ fulfill the following.

• 1. For all $t \in [0,T]$, $i \in \{1, \ldots ,N\}$

  $$\begin{aligned} \mathbb E\big [\big \vert \tilde{V}_t^i-V_t^i \big \vert ^2\big ]&\le C\delta t + C\mathbb E\left[ \int _0^t \big \vert \tilde{\xi }^{i,N}_{r(s)} - \tilde{\xi }^{i,N}_{s} \big \vert ^2 \ ds \right] \nonumber \\&\quad +\, C \mathbb E\left[ \int _0^t \big \vert \tilde{u}_{r(s)}\big (\tilde{\xi }^{i,N}_{r(s)}\big )-u^{S^N(\tilde{\xi })}_s\big (\tilde{\xi }^{i,N}_s\big )\big \vert ^2 ds \right] , \end{aligned}$$

  (6.1)

  where C is a real positive constant depending only on $M_{\Lambda }$, $L_{\Lambda }$ and T.

• 2. For all $(t,y) \in [0,T] \times \mathbb R^d$, $i \in \{1, \ldots ,N\}$

  $$\begin{aligned} \big \vert \tilde{u}_{t}(y)-u_t^{S^N(\tilde{\xi })}(y)\big \vert ^2 \le \frac{M_K}{N}\sum _{i=1}^N K\big (y-\tilde{\xi }^{i,N}_t\big ) \; \big \vert \tilde{V}_t^i-V_t^i \big \vert ^2 \ . \end{aligned}$$

  (6.2)

Proof of Lemma 5.1

Let us fix $t \in [0,T]$, $i \in \{1, \ldots ,N\}$. To prove (6.1), it is enough to recall that $\Lambda$ being Lipschitz w.r.t. the space variables and $\frac{1}{2}$-Holder continuous w.r.t. the time variable, the inequality (2.6) yields

$$\begin{aligned} \big \vert \tilde{V}_t^i - V_t^i \big \vert ^2&\le 3e^{2tM_{\Lambda }}L^2_{\Lambda } \int _0^t \Big [\vert r(s)-s\vert +\Big \vert \tilde{\xi }^{i,N}_{r(s)}-\tilde{\xi }^{i,N}_s\Big \vert ^2 \nonumber \\&\quad +\Big \vert \tilde{u}_{r(s)}\Big (\tilde{\xi }^{i,N}_{r(s)}\Big )-u_s^{S^N(\tilde{\xi })}\Big (\tilde{\xi }^{i,N}_s\Big )\Big \vert ^2 \Big ]\,ds\ , \end{aligned}$$

(6.3)

and taking the expectation in both sides of (6.3) implies (6.1) with $C := 3e^{2TM_{\Lambda }}L^2_{\Lambda }$.

Let us fix $y \in \mathbb R^d$. Concerning (6.2), by recalling the third line equation of (4.1) and the equation (1.6) (with $m = S^N(\tilde{\xi })$), we have

$$\begin{aligned} \Big \vert \tilde{u}_{t}(y)-u_t^{S^N(\tilde{\xi })}(y)\Big \vert ^2= & {} \left| \frac{1}{N} \sum _{i=1}^N K\Big (y- \tilde{\xi }_t^{i,N}\Big ) \tilde{V}^i_t - \frac{1}{N} \sum _{i=1}^N K\Big (y- \tilde{\xi }_t^{i,N}\Big ) V^i_t \right| ^2 \nonumber \\= & {} \left| \frac{1}{N} \sum _{i=1}^N K\Big (y- \tilde{\xi }_t^{i,N}\Big ) \big ( \tilde{V}^i_t - V^i_t \big ) \right| ^2 \nonumber \\\le & {} \frac{1}{N}\sum _{i=1}^N K^2\Big (y-\tilde{\xi }^{i,N}_t\Big ) \big \vert \tilde{V}_t^i - V_t^i \big \vert ^2 \nonumber \\\le & {} \frac{M_K}{N}\sum _{i=1}^N K\Big (y-\tilde{\xi }^{i,N}_t\Big ) \; \big \vert \tilde{V}_t^i-V_t^i \big \vert ^2 \ , \end{aligned}$$

(6.4)

which concludes the proof of (6.2) and therefore of Lemma 5.1.$\square$

Proof of Lemma 4.4

All along this proof, C will denote a positive constant that only depends

$T,M_K,m_{\Phi },m_g,L_K,L_\Phi ,L_g$ and $M_{\Lambda },L_{\Lambda }$ and that can change from line to line.

Let us fix $t \in [0,T]$.

• Inequality (4.8) of Lemma 4.4 is simply a consequence of the following computation:

  $$\begin{aligned} \mathbb E\Big [\Big \vert \tilde{\xi }^{i,N}_{r(t)}-\tilde{\xi }^{i,N}_{t}\Big \vert ^2\Big ]= & {} \mathbb E\left[ \left| \int _{r(t)}^{t}\Phi \Big (r(s),\tilde{\xi }^{i,N}_{r(s)},\tilde{u}_{r(s)}\Big (\tilde{\xi }^{i,N}_{r(s)}\Big )\Big )\,dW_s\right. \right. \\&\left. \left. \; +\int _{r(t)}^{t}g\Big (r(s),\tilde{\xi }^{i,N}_{r(s)},\tilde{u}_{r(s)}\Big (\tilde{\xi }^{i,N}_{r(s)}\Big )\Big )\,ds\right| ^2 \right] \\\le & {} 4\mathbb E\left[ \int _{r(t)}^{t}\Big \vert \Phi \Big (r(s),\tilde{\xi }^{i,N}_{r(s)},\tilde{u}_{r(s)}(\tilde{\xi }^{i,N}_{r(s)}\Big )\Big ) - \Phi (r(s),0,0) \Big \vert ^2\,ds\right] \\&\;+\, 4\mathbb E\left[ \int _{r(t)}^{t} \vert \Phi (r(s),0,0) \vert ^2\,ds\right] \\&\; +\, 4(t-r(t))\mathbb E\left[ \int _{r(t)}^{t}\Big \vert g\Big (r(s),\tilde{\xi }^{i,N}_{r(s)},\tilde{u}_{r(s)}\Big (\tilde{\xi }^{i,N}_{r(s)}\Big )\Big ) \right. \\&\left. \;-\, g(r(s),0,0)\vert ^2\,ds \right] \\&\; +\,4(t-r(t))\mathbb E\left[ \int _{r(t)}^{t} \vert g(r(s),0,0) \vert ^2\,ds\right] \\\le & {} 8\big (L_{\Phi }^2+(t-r(t))L_{g}^2\big )\int _{r(t)}^t \mathbb E\Big [\Big \vert \tilde{\xi }^{i,N}_{r(s)} \Big \vert ^2 \Big ] + \mathbb E\Big [\Big \vert \tilde{u}_{r(s)}\Big (\tilde{\xi }^{i,N}_{r(s)}\Big ) \Big \vert ^2 \Big ] ds \\&+ \; 4(t-r(t)) \Big ( \sup _{s \in [0,T]} \vert \Phi (s,0,0) \vert ^2 + (t-r(t)) \sup _{s \in [0,T]} \vert g(s,0,0) \vert ^2 \Big )\\\le & {} C\delta t \ ,\quad \text {as soon as }\quad \delta t \in \, ]0, 1[ \ , \end{aligned}$$

  where we have used the fact, under items 1. and 6. of Assumption 2, that the second order moment of $\tilde{\xi }^{i,N}_s$ is uniformly bounded. $\Lambda$ being uniformly bounded (item 3. of Assumption 2), the function $\tilde{u}$ as well. We have finally invoked item 6. of Assumption 2.

• Now, let us focus on the second inequality (4.9) of Lemma 4.4. Note that for any $y\in \mathbb R^d$, the following inequality holds:

  $$\begin{aligned}&\vert \tilde{u}_{r(t)}(y)-\tilde{u}_t(y)\vert \nonumber \\&\quad \le \frac{1}{N} \sum _{i=1}^N \left[ \left| K\Big (y-\tilde{\xi }^{i,N}_{r(t)}\Big ) -K\Big (y-\tilde{\xi }^{i,N}_t\Big )\right| \, e^{\int _0^{r(t)}\Lambda \big (r(s),\tilde{\xi }^{i,N}_{r(s)},\tilde{u}_{r(s)}(\tilde{\xi }^{i,N}_{r(s)})\big )ds} \right. \nonumber \\&\qquad + \left. K\Big (y-\tilde{\xi }^{i,N}_t\Big ) \,\left| e^{\int _0^{r(t)}\Lambda \big (r(s),\tilde{\xi }^{i,N}_{r(s)},\tilde{u}_{r(s)}(\tilde{\xi }^{i,N}_{r(s)})\big )ds}- e^{\int _0^t\Lambda \big (r(s),\tilde{\xi }^{i,N}_{r(s)},\tilde{u}_{r(s)}(\tilde{\xi }^{i,N}_{r(s)})\big )ds} \right| \right] \ . \nonumber \\ \end{aligned}$$

  (6.5)

  Using the fact that K and $\Lambda$ are bounded, one can apply (2.6) to bound the second term of the sum on the r.h.s. of the above inequality as follows:

  $$\begin{aligned}&K\Big (y-\tilde{\xi }^{i,N}_t\Big ) \,\left| e^{\int _0^{r(t)}\Lambda \big (r(s),\tilde{\xi }^{i,N}_{r(s)},\tilde{u}_{r(s)}\big (\tilde{\xi }^{i,N}_{r(s)}\Big )\big )ds}- e^{\int _0^t\Lambda \big (r(s),\tilde{\xi }^{i,N}_{r(s)},\tilde{u}_{r(s)}(\tilde{\xi }^{i,N}_{r(s)})\big )ds}\right| \nonumber \\&\quad \le M_Ke^{tM_{\Lambda }}(t-r(t))M_{\Lambda } \le C\delta t \ . \end{aligned}$$

  (6.6)

  The first term of the sum on the r.h.s. of (6.5) is bounded using the Lipschitz property of K and the fact that $\Lambda$ is bounded.

  $$\begin{aligned}&\left| K\Big (y-\tilde{\xi }^{i,N}_{r(t)}\Big ) -K\Big (y-\tilde{\xi }^{i,N}_t\Big )\right| \, e^{\int _0^{r(t)}\Lambda \big (r(s),\tilde{\xi }^{i,N}_{r(s)},\tilde{u}_{r(s)}\big (\tilde{\xi }^{i,N}_{r(s)}\big )\big )ds}\nonumber \\&\quad \le L_Ke^{tM_{\Lambda }} \Big \vert \tilde{\xi }^{i,N}_{r(t)}-\tilde{\xi }^{i,N}_t\vert \ . \end{aligned}$$

  (6.7)

  Injecting (6.6) and (6.7) in (6.5), for all $y \in \mathbb R^d$, we obtain

  $$\begin{aligned} \vert \tilde{u}_{r(t)}(y) - \tilde{u}_{t}(y) \vert \le C \delta t + \frac{L_Ke^{tM_{\Lambda }}}{N} \sum _{i=1}^N \Big \vert \tilde{\xi }^{i,N}_{r(t)}-\tilde{\xi }^{i,N}_{t} \Big \vert , \end{aligned}$$

  which finally implies that

  $$\begin{aligned} \Vert \tilde{u}_{r(t)}-\tilde{u}_t \Vert _{\infty }^2\le C\delta t ^2+\frac{C}{N}\sum _{i=1}^N \Big \vert \tilde{\xi }^{i,N}_{r(t)}-\tilde{\xi }^{i,N}_t\Big \vert ^2 \ . \end{aligned}$$

  We conclude by using inequality (4.8) of Lemma 4.4 after taking the expectation of the r.h.s. of the above inequality.

• Finally, we deal with inequality (4.10) of Lemma 4.4. Observe that the error on the left-hand side can be decomposed as

  $$\begin{aligned} \mathbb E\Big [\Big \Vert \tilde{u}_{r(t)}-u_t^{S^N(\tilde{\xi })}\Big \Vert ^2_{\infty }\Big ]\le & {} 2\mathbb E\big [\big \Vert \tilde{u}_{r(t)}-\tilde{u}_t\big \Vert ^2_{\infty }\big ]+2\mathbb E\Big [\Big \Vert \tilde{u}_{t}-u_t^{S^N(\tilde{\xi })}\Big \Vert ^2_{\infty }\Big ] \nonumber \\\le & {} C\delta t +2\mathbb E\Big [\Big \Vert \tilde{u}_{t}-u_t^{S^N(\tilde{\xi })}\Big \Vert ^2_{\infty }\Big ]\ , \end{aligned}$$

  (6.8)

  where we have used inequality (4.9) of Lemma 4.4.

  Let us consider the second term on the r.h.s. of the above inequality. To simplify the notations, we introduce the real valued random variables

  $$\begin{aligned} V_t^i := e^{\int _0^t\Lambda \big (s,\tilde{\xi }^{i,N}_{s},u^{S^N(\tilde{\xi })}_{s}\big (\tilde{\xi }^{i,N}_{s}\big )\big )ds}\quad \text {and}\quad \tilde{V}_t^i := e^{\int _0^t\Lambda \big (r(s),\tilde{\xi }^{i,N}_{r(s)},\tilde{u}_{r(s)}\big (\tilde{\xi }^{i,N}_{r(s)}\big )\big )ds}\ ,\nonumber \\ \end{aligned}$$

  (6.9)

  defined for any $i=1, \ldots N$ and $t\in [0,T]$.

  Using successively inequalities (6.1) of Lemma 5.1, (4.8) of Lemma 4.4 and (2.9) of Proposition 2.5, we have for all $i \in \{1, \ldots ,N\}$,

  $$\begin{aligned} \mathbb E\big [\big \vert \tilde{V}_t^i-V_t^i \big \vert ^2\big ]\le & {} C\delta t + C \mathbb E\left[ \int _0^t \Big \vert \tilde{u}_{r(s)}\Big (\tilde{\xi }^{i,N}_{r(s)}\Big )-u^{S^N(\tilde{\xi })}_s\Big (\tilde{\xi }^{i,N}_s\Big )\Big \vert ^2 ds \right] \nonumber \\\le & {} C\delta t + C \mathbb E\left[ \int _0^t \Big \vert \tilde{u}_{r(s)}\Big (\tilde{\xi }^{i,N}_{r(s)}\Big )-u^{S^N(\tilde{\xi })}_s\Big (\tilde{\xi }^{i,N}_{r(s)}\Big )\Big \vert ^2 ds \right] \nonumber \\&+ \; C \mathbb E\left[ \int _0^t \Big \vert u_{s}^{S^N(\tilde{\xi })}\Big (\tilde{\xi }^{i,N}_{r(s)}\Big )-u^{S^N(\tilde{\xi })}_s\Big (\tilde{\xi }^{i,N}_{s}\Big )\Big \vert ^2 ds \right] \nonumber \\\le & {} C\delta t +C\int _0^t \left[ \mathbb E\Big [\Big \Vert \tilde{u}_{r(s)}-u^{S^N(\tilde{\xi })}_s\Big \Vert _\infty ^2\Big ] + \mathbb E\Big [\Big \vert \tilde{\xi }^{i,N}_{r(s)}-\tilde{\xi }^{i,N}_s\Big \vert ^2\Big ] \right] \, ds \nonumber \\\le & {} C\delta t +C\int _0^t \mathbb E\Big [\Big \Vert \tilde{u}_{r(s)}-u^{S^N(\tilde{\xi })}_s\Big \Vert _\infty ^2\Big ] \, ds \ . \end{aligned}$$

  (6.10)

  On the other hand, inequality (6.2) of Lemma 5.1 implies

  $$\begin{aligned} \Big \Vert \tilde{u}_{t}-u_t^{S^N(\tilde{\xi })} \Big \Vert _{\infty }^2 \le \frac{M_K^2}{N}\sum _{i=1}^N \big \vert \tilde{V}_t^i-V_t^i \big \vert ^2 \ . \end{aligned}$$

  (6.11)

  Taking the expectation in both sides of (6.11) and using (6.10) give

  $$\begin{aligned} \mathbb E\Big [\Big \Vert \tilde{u}_{t}-u_t^{S^N(\tilde{\xi })}\Big \Vert ^2_{\infty }\Big ]\le & {} \frac{M_K^2}{N}\sum _{i=1}^N \mathbb E\left[ \big \vert \tilde{V}_t^i-V_t^i \big \vert ^2 \right] \le C\delta t\nonumber \\&+\,C\int _0^t \mathbb E\Big [\Big \Vert \tilde{u}_{r(s)}-u^{S^N(\tilde{\xi })}_s\Big \Vert _\infty ^2\Big ] \, ds\ . \end{aligned}$$

  (6.12)

  We end the proof by injecting this last inequality in (6.8) and by applying Gronwall’s lemma.$\square$