The Brownian motion as the limit of a deterministic system of hard-spheres

Abstract

We provide a rigorous derivation of the Brownian motion as the limit of a deterministic system of hard-spheres as the number of particles $N$ goes to infinity and their diameter $\varepsilon$ simultaneously goes to $0$, in the fast relaxation limit $\alpha = N\varepsilon ^{d-1}\rightarrow \infty$ (with a suitable diffusive scaling of the observation time). As suggested by Hilbert in his sixth problem, we rely on a kinetic formulation as an intermediate level of description between the microscopic and the fluid descriptions: we use indeed the linear Boltzmann equation to describe one tagged particle in a gas close to global equilibrium. Our proof is based on the fundamental ideas of Lanford. The main novelty here is the detailed study of the branching process, leading to explicit estimates on pathological collision trees.

Appendices

Appendix A: Asymptotic control of the exclusion

For the sake of completeness, we recall here the proof of Proposition 3.2. We omit all subscripts $\beta$ to simplify the presentation.

• First step: asymptotic behaviour of the partition function.

We first prove that in the scaling $N {\varepsilon }^{d-1}\ \equiv \alpha$, with $\alpha \ll 1/\varepsilon$,

$$\begin{aligned} 1 \le {\mathcal {Z}}_N^{-1} {\mathcal {Z}}_{N-s} \le \big ( 1 - {\varepsilon }\alpha {\kappa }_d \big )^{-s}, \end{aligned}$$

(7.1)

where ${\kappa }_d$ denotes the volume of the unit ball in ${\mathbf {R}}^d$. The first inequality is due to the immediate upper bound

$$\begin{aligned} {\mathcal {Z}}_N \le {\mathcal {Z}}_{N-s}. \end{aligned}$$

Let us prove the second inequality. We have by definition

$$\begin{aligned} {\mathcal {Z}}_{s+1} = \int _{ {\mathbf {T}}^{d(s+1)} } \left( \prod _{1 \le i \ne j \le s+1} \mathbf{1}_{|x_i-x_{j} |>{\varepsilon }}\right) \, dX_{s+1}. \end{aligned}$$

By Fubini’s equality, we deduce

$$\begin{aligned} {\mathcal {Z}}_{s+1} \!=\! \int _{ {\mathbf {T}}^{ds} } \!\left( \int _{ {\mathbf {T}}^{d} } \left( \prod _{1 \le i \le s} \mathbf{1}_{|x_i-x_{s+1} |>{\varepsilon }}\right) \, dx_{s+1}\right) \left( \prod _{1 \le i\ne j \le s} \mathbf{1}_{|x_i-x_{j} |>{\varepsilon }}\right) \, dX_{s}. \end{aligned}$$

Since

$$\begin{aligned} \int _{ {\mathbf {T}}^{d} } \left( \prod _{1 \le i \le s} \mathbf{1}_{|x_i-x_{s+1} |>{\varepsilon }}\right) \, dx_{s+1}\ge 1- {\kappa }_d s {\varepsilon }^d , \end{aligned}$$

we deduce the lower bound

$$\begin{aligned} {\mathcal {Z}}_{s+1} \ge {\mathcal {Z}}_{s} ( 1- {\kappa }_d s {\varepsilon }^d ) \ge {\mathcal {Z}}_{s} (1-{\kappa }_d {\varepsilon }\alpha ), \end{aligned}$$

where we used $s \le N$ and the scaling $N {\varepsilon }^{d-1} \equiv \alpha$. This implies by induction

$$\begin{aligned} {\mathcal {Z}}_N \ge {\mathcal {Z}}_{N-s} \big ( 1 - {\varepsilon }\alpha {\kappa }_d \big )^s. \end{aligned}$$

That proves (7.1).

• Second step: convergence of the marginals.

Let us introduce the short-hand notation

$$\begin{aligned} dZ_{(s+1,N)} :=dz_{s+1} \dots dz_N. \end{aligned}$$

We compute for $s \le N$

$$\begin{aligned}&M_N^{(s)} (Z_s)\\&\quad = {\mathcal {Z}}_N^{-1} \mathbf{1}_{Z_s \in {\mathcal {D}}_{\varepsilon }^s}\left( \frac{\beta }{2\pi }\right) ^{ \frac{sd}{2}}\exp \left( -\frac{\beta }{2} |V_s|^2\right) \\&\qquad \times \int _{ {\mathbf {R}}^{d(N-s)}} \left( \frac{\beta }{2\pi }\right) ^{ \frac{(N-s)d}{2}} \exp \left( -\frac{\beta }{2} \sum _{i=s+1}^N|v_i|^2\right) \, dV_{(s+1,N)}\\&\qquad \times \int _{ {\mathbf {T}}^{d(N-s)} } \left( \prod _{s+1 \le i \ne j \le N} \mathbf{1}_{|x_i - x_j| > {\varepsilon }} \right) \left( \prod _{i '\le s < j'} \mathbf{1}_{|x_{i'} - x_{j'}| > {\varepsilon }} \right) \times dX_{(s+1,N)}. \end{aligned}$$

We deduce, by symmetry,

$$\begin{aligned} M_N^{(s)} = {\mathcal {Z}}_N^{-1} \mathbf{1}_{Z_s \in {\mathcal {D}}_{\varepsilon }^s}M^{\otimes s} \Big ( {\mathcal {Z}}_{N-s} - {\mathcal {Z}}^\flat _{(s+1,N)} \Big ) \end{aligned}$$

(7.2)

with the notation

$$\begin{aligned} {\mathcal {Z}}^\flat _{(s+1,N)} := \int _{{\mathbf {T}}^{d(N-s) } } \left( 1 - \prod _{i \le s < j}\mathbf{1}_{|x_i - x_j| > {\varepsilon }}\right) \prod _{s+1 \le k \ne \ell \le N} \mathbf{1}_{|x_k- x_\ell | > {\varepsilon }}\, dX_{(s+1,N)}. \end{aligned}$$

From there, the difference $\mathbf{1}_{Z_s \in {\mathcal {D}}_{\varepsilon }^s}M^{\otimes s} - M_{N}^{(s)}$ decomposes as a sum

$$\begin{aligned} \mathbf{1}_{Z_s \in {\mathcal {D}}_{\varepsilon }^s}M^{\otimes s} - M_{N}^{(s)}= & {} \left( 1 - {\mathcal {Z}}_{N}^{-1} {\mathcal {Z}}_{N-s}\right) \mathbf{1}_{Z_s \in {\mathcal {D}}_{\varepsilon }^s}M^{\otimes s} \nonumber \\&+{\mathcal {Z}}_N^{-1} {\mathcal {Z}}^\flat _{(s+1,N)} \mathbf{1}_{Z_s \in {\mathcal {D}}_{\varepsilon }^s}M^{\otimes s} . \end{aligned}$$

(7.3)

By (7.1), there holds $1- {\mathcal {Z}}_{N}^{-1} {\mathcal {Z}}_{N-s} \rightarrow 0$ as $N \rightarrow \infty$, for fixed $s$. Since $M^{\otimes s}$ is uniformly bounded, this implies that the first term in the right-hand side of (7.3) tends to 0 as $N$ goes to $\infty$. Besides, by

$$\begin{aligned} 0 \le 1 - \prod _{i \le s < j}\mathbf{1}_{|x_i - x_j| > {\varepsilon }} \le \sum _{i \le s < j} \mathbf{1}_{|x_i-x_j|<{\varepsilon }}, \end{aligned}$$

we bound

$$\begin{aligned}&{\mathcal {Z}}^\flat _{(s+1,N)}\\&\quad \le \sum _{1 \le i \le s}\int _{{\mathbf {T}}^{d(N-s)}} \left( \sum _{s+1 \le j \le N} \mathbf{1}_{|x_i-x_j|<{\varepsilon }}\right) \prod _{s+1 \le k \ne \ell \le N} \mathbf{1}_{|x_k - x_ \ell | > {\varepsilon }} \, dX_{(s+1,N)}. \end{aligned}$$

Given $1 \le i\le s$, there holds by symmetry and Fubini’s equality,

$$\begin{aligned}&\int _{{\mathbf {T}}^{d(N-s)} }\left( \sum _{s+1 \le j \le N} \mathbf{1}_{|x_i-x_j|<{\varepsilon }}\right) \prod _{s+1 \le k \ne l \le N} \mathbf{1}_{|x_k - x_l| > {\varepsilon }}\ dX_{(s+1,N)}\\&\quad \le (N-s) \int _{{\mathbf {T}}^{d } } \mathbf{1}_{|x_i - x_{s+1}| < {\varepsilon }} \, dx_{s+1} \int _{{\mathbf {T}}^{d(N-s-1)} } \prod _{s+2\le k \ne l \le N} \mathbf{1}_{|x_k - x_l| > {\varepsilon }} \, dX_{(s+2,N)} \\&\quad = (N-s) \int _{{\mathbf {T}}^{d } } \mathbf{1}_{|x_i - x_{s+1}| < {\varepsilon }} \, dx_{s+1} \, \times \, {\mathcal {Z}}_{N-s-1}, \end{aligned}$$

so that

$$\begin{aligned} {\mathcal {Z}}^\flat _{(s+1,N)} \le s (N-s) {\varepsilon }^d {\kappa }_d {\mathcal {Z}}_{N-s-1}. \end{aligned}$$

(7.4)

By (7.1), we obtain

$$\begin{aligned} {\mathcal {Z}}_N^{-1} {\mathcal {Z}}^\flat _{(s+1,N)} \le {\varepsilon }\alpha s {\kappa }_d \big (1 - {\varepsilon }\alpha {\kappa }_d \big )^{-(s+1)}, \end{aligned}$$

and the upper bound tends to 0 as $N \rightarrow \infty$, for fixed $s$. This implies convergence to 0 of the second term in the right-hand side of (7.3).

This completes the proof of Proposition 3.2.

Appendix B: Recollisions in the torus

We show here how to adapt the arguments of [21] to prove Lemma 5.2.

• To build the set of “bad velocities”, we use the correspondence between the torus and the whole space with periodic structure. Asking that there exists $u \in [0,t]$ such that

  $$\begin{aligned} d\big ((x_1-v_1 u ),( x _2-v_2 u )\big ) \le {\varepsilon }, \end{aligned}$$

  boils down to having

  $$\begin{aligned} (x_1-v_1 u )-( x _2-v_2 u ) \in \bigcup _{k\in {\mathbf {Z}}^d} B_{{\varepsilon }}(k). \end{aligned}$$

Then, by the triangular inequality and provided that ${\varepsilon }<\bar{a}$,

$$\begin{aligned} ( x^0_1-v_1 u )-( x^0_2-v_2 u ) \in \bigcup _{k\in {\mathbf {Z}}^d} B_{3\bar{a}}(k ). \end{aligned}$$

Now, since $|v_1-v_2|\le 2E$ and $u \in [0,t]$, this implies that

$$\begin{aligned} s (v_1-v_2) \in \left( \bigcup _{k\in {\mathbf {Z}}^d} B_{3 \bar{a}}( x^0_1- x^0_2 + k)\right) \cap B_{0}(2Et). \end{aligned}$$

In other words, $v_1-v_2$ has to belong to a finite union of cones of vertex 0

• At most one of which is of solid angle $( \bar{a} /{\varepsilon }_0)^{d-1}$;

• The other ones (at most $(4E t)^d$) are of solid angle $c \, \bar{a}^{d-1}$.

The intersection $K(\bar{x}_1-\bar{x}_2, {\varepsilon }_0, \bar{a})$ of these cones and of the sphere of radius $2E$ is of size

$$\begin{aligned} |K(\bar{x}_1-\bar{x}_2, {\varepsilon }_0, \bar{a}) | \le CE^d \left( \Big ( \frac{\bar{a}}{{\varepsilon }_0}\Big )^{d-1} + (E t )^d {\bar{a}}^{d-1}\right) . \end{aligned}$$

• In order to prove the second estimate, we need to refine a little bit the previous argument. Asking that there exists $u \in [\delta ,t]$ such that

  $$\begin{aligned} d((x_1-v_1 u ),( x _2-v_2 u) )\le {\varepsilon }_0, \end{aligned}$$

  boils down to having

  $$\begin{aligned} u (v_1-v_2) \in B_{3{\varepsilon }_0}( x^0_1- x^0_2+ k), \end{aligned}$$

  (8.1)

  for some $k \in {\mathbf {Z}}^d \cap B_{2Et }( x^0_2 - x^0_1)$.

  • If $| x^0_1- x^0_2 + k| \ge 1/4$, condition (8.1) implies that $v_1-v_2$ belongs to the intersection of $B_{2E}(0)$ and some cone of vertex 0 and solid angle ${\varepsilon }_0^{d-1}$.

  • If $| x^0_1- x^0_2 + k| \le 1/4$ (which can happen only for one value of $k$), denoting by $n$ any unit vector normal to $\bar{x}_1-\bar{x}_2 + k$, we deduce from (8.1) that

    $$\begin{aligned} u |(v_1-v_2) \cdot n | \le 3{\varepsilon }_0 \end{aligned}$$

    from which we deduce that $v_1-v_2$ belongs to the intersection of $B_{2E}(0)$ and some cylinder of radius ${\varepsilon }_0/\delta$.

  The union $K_\delta ( x^0_1- x^0_2, {\varepsilon }_0, \bar{a})$ of these “bad” sets is therefore of size

  $$\begin{aligned} |K_\delta ( x^0_1- x^0_2, {\varepsilon }_0, \bar{a}) | \le CE \left( \left( \frac{{\varepsilon }_0}{\delta } \right) ^{d-1} + E^{d-1} \big (E t \big )^d {{\varepsilon }_0}^{d-1}\right) . \end{aligned}$$

The lemma is proved. $\square$