Stochastic Solutions to Hamilton-Jacobi Equations

Abstract

In this expository paper we give an overview of the statistical properties of Hamilton-Jacobi Equations and Scalar Conservation Laws. The first part (Sects. 2–4) is devoted to the recent proof of Menon-Srinivasan Conjecture. This conjecture provides a Smoluchowski-type kinetic equation for the evolution of a Markovian solution of a scalar conservation law with convex flux. In the second part of the paper (Sects. 5 and 6) we discuss the question of homogenization for Hamilton-Jacobi PDEs and Hamiltonian ODEs with deterministic and stochastic Hamiltonian functions.

1 Introduction

The primary goal of these notes is to give an overview of the statistical properties of solutions to the Cauchy problem for the Hamilton-Jacobi Equation

$$\begin{aligned} u_t&= H(x,t,u_x) \ \ \ \ \ \ {\text {in }}\ \ {\mathbb R}^d \times (0,\infty ) \\ u&=u^0 \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ {\text {on }} \ \ {\mathbb R}^d \times \{t = 0\},\nonumber \end{aligned}$$

(1.1)

or, the scalar conservation law

$$\begin{aligned} \rho _t&= H(x,t,\rho )_x \ \ \ \ \ \ {\text {in }} \ \ {\mathbb R}\times (0,\infty ) \\ \rho&= \rho ^0 \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ {\text {on }} \ \ {\mathbb R}\times \{t = 0\},\nonumber \end{aligned}$$

(1.2)

where either H or $\rho ^0=\rho ^0(x)$ is random. Note that if u satisfies (1.1) and $d=1$, then $\rho =u_x$ satisfies (1.2). As is well-known, the PDE (1.1) or (1.2) does not possess classical solutions even when the initial data is smooth. In the case of Eq. (1.1), we may consider viscosity solutions to guarantee the uniqueness under some standard assumptions on the initial data and H. In the case of (1.2) with $d=1$, we consider the so-called entropy solutions.

We will be mostly concerned with the following two scenarios:

• (1) $d=1$, $H(x,t,p)=H(p)$ is convex in p and independent of (x, t), with initial data $\rho ^0$ that is either a white noise, or a Markov process.

• (2) $d\ge 1$, and H(x, t, p) is a stationary ergodic process in (x, t), and may not be convex in p.

Our aim is to give an overview of various classical and recent results and formulate a number of open problems. Sections 2–4 are devoted to (1), where we derive an evolution equation for the Markovian law of $\rho$ as a function of x or t. Sections 4 and 5 are devoted to (2), where we address the question of homogenization for such Hamiltonian functions.

2 Scalar Conservation Law with Random Initial Data

We first recall the following important features of the solutions to (1.2) when $d=1$, $H(x,t,p)=H(p)$ is convex in p, and independent of (x, t):

(i) :

If a discontinuity of $\rho$ occurs at $x=x(t)$, and $\rho _{\pm }=\rho (x(t)\pm ,t)$ represent the left and right limits of $\rho$ at x(t), then for a weak solution of (1.2) we must have the Rankin-Hugoniot Equation:

$$ \frac{dx}{dt}=-H[\rho _-,\rho _+]=:-\frac{H(\rho _+)-H(\rho _-)}{\rho _+-\rho _-}. $$

(ii) :

By an entropy solution, we mean a week solution for which the entropy condition is satisfied. In the case of convex H, the entropy condition is equivalent to the requirement

$$ \rho _-<\rho _+. $$

(iii) :

If $\rho ^0$ has a discontinuity with $\rho _->\rho _+$, then such a discontinuity disappears instantaneously by inserting a rarefaction wave between $\rho _-$ and $\rho _+$. That is a solution of the form

$$ G\left( \frac{x-c}{t}\right) , $$

where $G=(H')^{-1}$.

We next state three results.

(i) (Burgers Equation with Lévy Initial Data)

When $H(p)= \frac{1}{2}{p^2}$, (1.2) is the well-known inviscid Burgers’ equation, which has often been considered with random initial data. Burgers studied (1.2) in his investigation of turbulence [5]. Carraro and Duchon [6] defined a notion of statistical solution to Burgers’ equation and realized that it was natural to consider Lévy process initial data. This statistical solution approach was further developed in 1998 by the same authors [7] and by Chabanol and Duchon [8]. In fact any (random) entropy solution is also a statistical solution, but the converse is not true in general. In 1998, Bertoin [4] proved a closure theorem for Lévy initial data.

Theorem 1

Consider Burgers’ equation with initial data $\rho ^0(x)$ which is a Lévy process without negative jumps for $x \ge 0$, and $\rho ^0(x) = 0$ for $x < 0$. Assume that the expected value of $\rho ^0(1)$ is non-positive, ${\mathbb E}\rho ^0(1) \le 0$. Then, for each fixed $t > 0$, the process $x\mapsto \rho (x,t) -\rho (0,t)$ is also a Lévy process with

$$ {\mathbb E}\exp \big (-s(\rho (x,t) -\rho (0,t))\big ) = \exp (x \psi (s,t)) , $$

where the exponent $\psi$ solves the following equation:

$$\begin{aligned} \psi _t+\psi \psi _s=0. \end{aligned}$$

(2.1)

Remark 2.1(i) The requirement ${\mathbb E}\rho ^0(1) \le 0$ can be relaxed with minor modifications to the theorem, in light of the following elementary fact. Suppose that $\rho ^0(x)$ and $\hat{\rho }^0(x)$ are two different initial conditions for Burgers’ equation, which are related by $\hat{\rho }^0(x) = \rho ^0(x) + cx$. It is easy to check that the corresponding solutions $\rho (x,t)$ and $\hat{\rho }(x,t)$ are related for $t > 0$ by

$$ \hat{\rho }(x,t) = \frac{1}{1+ct} \left[ \rho \left( \frac{x}{1+ct},\frac{t}{1+ct}\right) + cx\right] . $$

Using this we can adjust a statistical description for a case where ${\mathbb E}\rho ^0(1) >0$ to cover the case of a Lévy process with general mean drift.

(ii) Sinai [26] and Aurell, Frisch, She [3] considered Burgers equation with Brownian motion initial data, relating the statistics of solutions to convex hulls and addressing pathwise properties of solutions.     $\square$

(ii) (Burgers Equation with white noise initial data)

Groeneboom [15] considers the white noise initial data. In other words, take two independent Brownian motions $B^{\pm }$, and take a two sided Brownian motion for the initial data

$$\begin{aligned} u^0(x)={\left\{ \begin{array}{ll}B^+(x)&{}\ \ \ \ \ {\text {if }}\ \ \ \ x\ge 0\\ B^-(x)&{}\ \ \ \ \ {\text {if }}\ \ \ \ x\le 0, \end{array}\right. } \end{aligned}$$

(2.2)

Theorem 2

Let $\rho =u_x$, where u is a viscosity solution of the PDE $u_t=\frac{1}{2}u_x^2$, subject to the initial condition $u(x,0)=u^0(x)$, with $u^0$ given as in (2.2). Then the process $x\mapsto \rho (x,t)$ is a Markov jump process with drift $-t^{-1}$ and a suitable jump measure $\nu (t,\rho _-,\rho _+)\ d\rho _+$.

We also refer to [13] for an explicit and simple formula expressing the one-point distribution of $\rho$ in terms of Airy functions.

(iii) A different particular case,

$$ -H(p) = {\left\{ \begin{array}{ll}0&{}\ \ \ \ \ {\text {if }}\ \ \ |p| \le 1,\\ \infty &{}\ \ \ \ \ {\text {otherwise.}} \end{array}\right. } $$

corresponds to the problem of determining Lipschitz minorants, and has been investigated by Abramson and Evans [1].

3 Menon-Srinivasan Conjecture

In 2007 Menon and Pego [19] used the Lévy-Khintchine representation for the Laplace exponent and observed that the evolution according to Burgers’ equation in (2.1) corresponds to a Smoluchowski coagulation equation [2], with additive collision kernel, for the jump measure of the Lévy process $\nu (\cdot , t)$. The jumps of $\nu (\cdot , t)$ correspond to shocks in the solution $\rho (\cdot , t)$. Regarding the sizes of the jumps as the usual masses in the Smoluchowski equation, it is plausible that Smoluchowski equation with additive kernel should be relevant.

It is natural to wonder whether this evolution through Markov processes with simple statistical descriptions is specific to the Burgers-Lévy case, or an instance of a more general phenomenon. The biggest step toward understanding the problem for a wide class of H is found in a 2010 paper of Menon and Srinivasan [20]. Here it is shown that when the initial condition $\rho ^0$ is a strong Markov process with positive jumps only, the solution $\rho (\cdot , t)$ remains Markov for fixed $t > 0$. The argument is adapted from that of [4] and both [20] and [4] use the notion of splitting times (due to Getoor [14]) to verify the Markov property according to its bare definition. In the Burgers-Lévy case, the independence and homogeneity of the increments can be shown to survive, from which additional regularity is immediate using standard results about Lévy processes. As [20] points out, without these properties it is not clear whether a Feller process initial condition leads to a Feller process in x at later times. Nonetheless, [20] presents a very interesting conjecture for the evolution of the generator of $\rho (\cdot , t)$, which has a remarkably nice form.

To prepare for the statement of Menon-Srinivasan Conjecture, we first examine the following simple scenario for the solutions of the PDE

$$\begin{aligned} \rho _t=H(\rho )_x=H'(\rho )\rho _x. \end{aligned}$$

(3.1)

Imagine that the initial data $\rho ^0$ satisfies an ODE of the form

$$\begin{aligned} \frac{d\rho ^0}{dx}(x)=b^0\big (\rho ^0(x)\big ), \end{aligned}$$

(3.2)

for some $C^1$ function $b^0:{\mathbb R}\rightarrow {\mathbb R}$. We may wonder whether or not this feature of $\rho ^0$ survives at later times. That is, for some function $b(\rho ,t)$, we also have

$$\begin{aligned} \rho _x(x,t)=b\big (\rho (x,t),t\big ), \end{aligned}$$

(3.3)

for $t>0$. For (3.3) to be consistent with (3.1), observe

$$ \rho _t=H'(\rho )\rho _x=H'(\rho )b(\rho ,t), $$

and as we calculate mixed derivatives, we arrive at

$$\begin{aligned} \rho _{xt}&=b_\rho (\rho ,t)\rho _t+b_t(\rho ,t) =b_\rho (\rho ,t)H'(\rho )b(\rho ,t)+b_t(\rho ,t),\\ \rho _{tx}&=H''(\rho ) b(\rho ,t)\rho _x+H'(\rho )b_\rho (\rho ,t)\rho _x= H''(\rho ) b^2(\rho ,t)+H'(\rho )b_\rho (\rho ,t)b(\rho ,t). \end{aligned}$$

As a result b must satisfy

$$\begin{aligned} b_t(\rho ,t)=H''(\rho ) b^2(\rho ,t). \end{aligned}$$

(3.4)

For a classical solution, all we need to do is solving the ODE (3.3) for the initial data $b(\rho ,0)=b^0(\rho )$ for each $\rho$. When H is convex, the solution may blow up in finite time. More precisely,

• If $b^0(\rho )\le 0$, then $b^0(\rho )\le b(\rho ,t)\le 0$ for all t and there would be no blow-up.

• If $b^0(\rho )>0$, then there exists some finite $T(\rho )>0$ such that $b(\rho ,t)$ is finite in the interval $[0,T(\rho ))$, and $b(\rho ,T(\rho ))=\infty$.

In fact the Eq. (3.4) is really “the method of characteristics” in disguise, and the blow-up of solutions is equivalent to the occurrence of shock discontinuity.

To go beyond what (3.4) offers, we now take a jump kernel $f^0(\rho ,d\rho _*)$ and assume that $\rho ^0(x)$ is a realization of a Markov process with infinitesimal generator

$$ {\mathcal L}^0h(\rho )=b^0(\rho )h'(\rho )+\int _{\rho }^\infty \big (h(\rho _*)-h(\rho )\big )\ f^0(\rho ,d\rho _*). $$

In words, $\rho ^0$ solves the ODE (3.3), with some occasional random jumps with rate $f^0$. We are assuming that the jumps are all positive to avoid rarefaction waves. We may wonder whether the same picture is valid at later times. That is, for fixed $t>0$, the solution $\rho (x,t)$, as a function of x is a Markov process with the generator

$$\begin{aligned} {\mathcal L}^th(\rho )=b(\rho ,t)h'(\rho )+\int _{\rho }^\infty \big (h(\rho _*)-h(\rho )\big )\ f(\rho ,d\rho _*,t). \end{aligned}$$

(3.5)

Menon-Srinivasan Conjecture roughly suggests that if H is convex, and we start with a Markov process with generator ${\mathcal L}^0$, then we have a Markov process at a later time with a generator of the form ${\mathcal L}^t$. Moreover, the drift of the generator satisfies (3.4), and the jump kernel $f(\rho ,d\rho _*,t)$ solves an integral equation. Before we derive an equation for the evolution of f, observe that when we assert that $\rho (x,t)$ is a Markov process in x, we are specifying a direction for x. More precisely, we are asserting that if $\rho (a,t)$ is known, then the law of $\rho (x,t)$ can be determined uniquely for all $x>a$. We are doing this for all $t>0$. In practice, we may try to determine $\rho (x,t)$ for $x>a(t)$, provided that $\rho (a(t),t)$ is specified. For example, we may wonder whether or not we can determine the law of $\rho (x,t)$ with the aid of the following procedure:

• The process $t\mapsto \rho (a(t),t)$ is a Markov process and its generator can be determined. Using this Markov process, we take a realization of $\rho (a(t),t)$, with some initial choice for $\rho (a(0),0)$.

• Once $\rho (a(t),t)$ is selected, we use the generator ${\mathcal L}^t$, to produce a realization of $\rho (x,t)$ for $x\ge a(t)$.

To materialize the above procedure, we need to make sure that for some choice of a(t), the process $\rho (a(t),t)$ is Markovian with a generator that can be described. For a start, we may wonder whether or not we can even choose $a(t)=a$ a constant function. Put it differently, not only $x\mapsto \rho (x,t)$ is a Markov process for fixed $t\ge 0$, the process $t\mapsto \rho (x,t)$ is a Markov process for fixed x. As it turns out, this is the case if H is also increasing. In general, if we can find a negative constant c such that $H'(\rho )> c$, then $\hat{\rho }(x,t):=\rho (x-ct,t)$ satisfies

$$ \hat{\rho }_t=\hat{H}(\hat{\rho })_x, $$

for $\hat{H}(\rho )=H(\rho )-c\rho$, which is increasing. Hence, the process $t\mapsto \hat{\rho }(x,t)=\rho ( x-ct,t)$ is expected to be Markovian. In summary

• If H is increasing in the range of $\rho$, then $\rho$ is also Markovian on vertical lines $x=constant$.

• If $H'$ is bounded below by a negative constant c, then $\rho$ is Markovian on straight lines that are titled to the right with the slope $-c$.

To simplify the matter, from now on, we make two assumptions on H:

$$\begin{aligned} H'>0,\ \ \ \ \ H''\ge 0. \end{aligned}$$

(3.6)

The main consequences of these two assumptions are

• All the jump discontinuities are positive i.e. $\rho _-<\rho _+$.

• The speed of shocks are always negative.

We now argue that in fact the process $t\mapsto \rho (x,t)$ is a (time-inhomogeneous) Markov process with a generator ${\mathcal M}_t$ that is independent of x because the PDE (3.1) is homogeneous (i.e. H is independent of x). Indeed

$$\begin{aligned} {\mathcal M}_t h(\rho )=H'(\rho )b(\rho ,t)h'(\rho )+\int _{\rho }^\infty \big (h(\rho _*)-h(\rho )\big )H[\rho ,\rho _*]f(\rho ,d\rho _*,t). \end{aligned}$$

(3.7)

To explain the form of ${\mathcal M}_t$ heuristically, observe that the ODE $\frac{d\rho }{dx}=b(\rho ,t)$ leads to the ODE

$$ \frac{d\rho }{dt}=H'(\rho )b(\rho ,t). $$

On the other hand, if we fix x, then $\rho (x,t)$ experiences a jump discontinuity when a shock on the right of x crosses x. Given any $t>0$, a shock would occur at some $s>t$ because all shock speeds are negative; it is just a matter of time for a shock on the right of x to cross x. We can also calculate the rate at which this happens because we have the law of the first shock on the right of x, and its speed. Observe

• The process $x\mapsto \rho (x,t)$ is a homogeneous Markov process with a generator that changes with time.

• The process $t\mapsto \rho (x,t)$ is an inhomogeneous Markov process with a generator that does not depend on x. It is only the initial data $\rho (x,0)$ that is responsible for the changes of the statistics of $\rho (x,t)$, as x varies.

We are now in a position to derive formally an evolution equation for the generator ${\mathcal L}^t$, under the assumption (3.6). Indeed if we define

$$ w(x,t;\rho )={\mathbb E}^{\rho (0,t)=\rho }h(\rho (x,T)), $$

for $t<T$, then we expect

$$ w_t=-{\mathcal M}_t w,\ \ \ \ \ w_x={\mathcal L}^tw. $$

Differentiating these equations yields

$$ w_{tx}=-{\mathcal M}_t w_x=-{\mathcal M}_t{\mathcal L}^t w,\ \ \ \ \ w_{xt}=\frac{d{\mathcal L}^t}{dt}w+{\mathcal L}^t w_t =\frac{d{\mathcal L}^t}{dt}w-{\mathcal L}^t {\mathcal M}_t w. $$

As a result

$$\begin{aligned} \frac{d{\mathcal L}^t}{dt}={\mathcal L}^t {\mathcal M}_t-{\mathcal M}_t{\mathcal L}^t. \end{aligned}$$

(3.8)

As we match the drift parts of both sides of (3.8), we simply get (3.4). Matching the jump parts yields a kinetic-type equation of the form

$$\begin{aligned} f_t=Q(f,f)+Cf, \end{aligned}$$

(3.9)

for a quadratic operator Q and a linear operator C. The operator Q is independent of b and is given by

$$\begin{aligned} Q(f,f)(\rho _-,d\rho _+)=&\int _{\rho _-}^{\rho ^+}\big (H[\rho _*,\rho _+] -H[\rho _-,\rho _*]\big )f(\rho _-,d\rho _*)f(\rho _*,d\rho _+)\\&+\int _{\rho _+}^{\infty }\big (H[\rho _+,\rho _*] -H[\rho _-,\rho _+]\big )f(\rho _+,d\rho _*)f(\rho _-,d\rho _+)\\&+\int _{\rho _-}^{\infty }\big (H[\rho _-,\rho _+] -H[\rho _-,\rho _*]\big )f(\rho _-,d\rho _*)f(\rho _-,d\rho _+). \end{aligned}$$

If we set

$$\begin{aligned} \lambda (\rho _-)&=\lambda (f)(\rho _-)=\int _{\rho _-}^\infty f(\rho _-,d\rho _+),\\ A(\rho _-)&=A(f)(\rho _-)=\int _{\rho _-}^\infty H[\rho _-,\rho _+]f(\rho _-,d\rho _+), \end{aligned}$$

then $Q=Q^+-Q^-$, with

$$\begin{aligned} Q^+(f,f)(\rho _-,d\rho _+)=&\int _{\rho _-}^{\rho ^+}\big (H[\rho _*,\rho _+] -H[\rho _-,\rho _*]\big )f(\rho _-,d\rho _*)f(\rho _*,d\rho _+,t)\nonumber \\ Q^-(f,f)(\rho _-,d\rho _+)=&\left\{ A(\rho _+)-A(\rho _-)- H[\rho _-,\rho _+]\big (\lambda (\rho _+)-\lambda (\rho _-)\big )\right\} f(\rho _-,d\rho _+). \end{aligned}$$

(3.10)

To define the operator C we need to assume that $f(\rho _-,d\rho _+)=f(\rho _-,\rho _+)d\rho _+$ has a $C^1$ density. With a slight abuse of notion, we write $f(\rho _-,\rho _+)$ for the density of the measure $f(\rho _-,d\rho _+)$, and write C again for the action of the operator C on the density f:

$$\begin{aligned} (Cf)(\rho _-,\rho _+)&=b(\rho _-,t)f(\rho _-,\rho _+)\big (H[\rho _-,\rho _+]\big )_{\rho _-}\\&\ \ \ \ \ +\big [H[\rho _-,\rho _+]-H'(\rho _-)\big ]b(\rho _-,t)f_{\rho _-}(\rho _-,\rho _+)\\&\ \ \ \ \ +\left[ \big (H[\rho _-,\rho _+]-H'(\rho _+)\big )b(\rho _+,t)f(\rho _-,\rho _+)\right] _{\rho _+}. \end{aligned}$$

Menon-Srinivasan Conjecture has been established in [16] and [17]:

Theorem 3

Assume H is a $C^2$ function that satisfies (3.6). Let $\rho$ be an entropic solution of (3.1) such that $\rho (x,0)=0$, for $x\le 0$, and $\rho (x,0)$ is a Markov process with generator ${\mathcal L}^0$, for $x\ge 0$. Assume that b and f satisfy (3.4) and (3.9) respectively. Then the processes $t\mapsto \rho (0,t)$ and $x\mapsto \rho (x,t)$ are Markov processes with generators ${\mathcal M}_t$ and ${\mathcal L}^t$ respectively.

The typical situation, for Smoluchowski and other kinetic equations is that we have some (stochastic or deterministic) dynamics defined on a finite system, and these kinetic equations emerge upon passage to a scaling limit. The dynamics might not be definable for the infinite system, and the kinetic equation should describe statistics only approximately for a large but finite system. In the setting of Theorems 1, 2 and 3, the kinetic equations give statistics exactly without passage to a rescaled limit. We view this unusual circumstance as demanding an explanation. Further, our treatment in Sect. 4 below (tracking shocks as inelastically colliding particles) seems quite at home in the kinetic context.

4 Heuristics for the Proof of Theorem 3

Let us write $x_i(t)$ for the location of the i-th shock and $\rho _i(t)=\rho (x_i(t)+,t)$. We also write $\phi _x(m_0;t)$ for the flow associated with the velocity b; the function $m(x)=\phi _x(m_0;t)$ satisfies

$$ m'(x)=b(m(x),t),\ \ \ \ \ m(0)=m_0. $$

We can readily find the evolution $\mathbf q =(x_i,\rho _i:i\in {\mathbb Z})$, and $\varvec{\hat{q}}=(z_i,\rho _i:i\in {\mathbb Z})$, with $z_i=x_{i+1}-x_i$:

• $$ \dot{x}_i=-v^i:=-H[\hat{\rho }_{i-1},\rho _i],\ \ \ \ \ \ \dot{z}_i=-(v^{i+1}-v^i), $$

  where $\hat{\rho }_{i-1}(t)=\phi _{z_{i-1}}(\rho _{i-1}(t),t)$.

• $$ \dot{\rho }_i=w^i:=\big (H'(\rho _i)-H[\hat{\rho }_{i-1},\rho _i]\big )b(\rho _i,t). $$

• When $z_i$ becomes 0, the pair $(\rho _i,z_i)$ is omitted from $\varvec{\hat{q}}(t)$. The outcome after a relabeling is denoted by $\varvec{\hat{q}}^i(t)$.

Write

$$ \varDelta =\big \{(z_i,\rho _i:i\in {\mathbb Z}):\ z_i> 0,\ \rho _i\in {\mathbb R}\ \ \ {\text { for all }}\ \ i\in {\mathbb Z}\big \}. $$

We think of $\varvec{\hat{q}}(t)$ as a deterministic process that has an infinitesimal generator

in the interior of $\varDelta$. We only take those G such that on the boundary face of $\varDelta$ with $z_i=0$, we have $G\big (\varvec{\hat{q}}\big )=G\big (\varvec{\hat{q}}^i\big )$. This stems from the fact that we are interested in the function $\rho (x)=\rho (x;{\hat{\mathbf{q }}})$ associated with ${\hat{\mathbf{q }}}$ (or $\mathbf q$) that is defined by

$$ \sum _i \phi _{z_i}(x_i;x-x_i)1\!\!1\big (x\in [x_i,x_{i+1})\big ). $$

Note that $\rho (x;{\hat{\mathbf{q }}})=\rho (x;{\hat{\mathbf{q }}}^i)$ whenever $z_i=0$.

We make an ansatz that the law of $\varvec{\hat{q}}(t)$ is of the form:

$$ \mu \big (d\varvec{\hat{q}},t\big )=\prod _{i=-\infty }^\infty e^{-\int _0^{z_i}\lambda (\phi _y(\rho _i;t),t) dy} \ f\big (\phi _{z_i}(\rho _i;t),\rho _{i+1},t)\ dz_id\rho _{i+1}. $$

For this to be the case, we need to have

(4.1)

This equation should determine f and $\lambda$ if our ansatz is correct. To determine , we take a test function G and carry out the following calculation: After some integration by parts, we formally have

where

$$\begin{aligned} \varOmega _i^1&=\int _0^{z_i}\big [\lambda \big (\phi _y(\rho _i;t),t\big )\big ]_{\rho _i}dy -\frac{\big [f\big (\hat{\rho }_{i},\rho _{i+1},t\big )\big ]_{\rho _i}}{f\big (\hat{\rho }_{i},\rho _{i+1},t\big )} -\frac{f_{\rho _+}\big (\hat{\rho }_{i-1},\rho _{i},t\big )}{f\big (\hat{\rho }_{i-1},\rho _{i},t\big )},\\ \varOmega _i^2&=-\lambda \big (\hat{\rho }_{i},t\big ) +\frac{\big [f\big (\hat{\rho }_{i},\rho _{i+1},t\big )\big ]_{z_i}}{f\big (\hat{\rho }_{i},\rho _{i+1},t\big )},\\ \varOmega _i^3&=\frac{\int _{\hat{\rho }_{i-1}}^{\rho _i} H\big (\hat{\rho }_{i-1},\rho _*,\rho _i\big ) f\big (\hat{\rho }_{i-1},\rho _*,t\big )f\big (\rho _{*},\rho _i,t\big )\ d\rho _*}{f\big (\hat{\rho }_{i-1},\rho _{i},t\big )}, \end{aligned}$$

where $\varOmega _i^3$ represents the boundary contribution associated with $z_i=0$, and

$$ H(a,b,c):=H[b,c]-H[a,b]. $$

To explain the form of $\varOmega ^3_i$, observe that when $z_i=0$, we remove the ith-particle and relabel the particles to its right. The expression $f\big (\hat{\rho }_{i-1},\rho _{i},t\big ) f\big (\rho _{i},\rho _{i+1},t\big )\ d\rho _i$, that appears in $\mu$, can be rewritten as $f\big (\hat{\rho }_{i-1},\rho _{*},t\big ) f\big (\rho _{*},\rho _{i+1},t\big )\ d\rho _*$. The variable $\rho _{i+1}$ becomes $\rho _i$ after our relabeling, and its integral with respect to $\rho _*$ is a function of $\big (\hat{\rho }_{i-1},\rho _{i},t\big )$. If we replace this function with $f\big (\hat{\rho }_{i-1},\rho _{i},t\big )$, we recover the measure $\mu$.

On the other hand

To make the above formal calculation rigorous, we switch from the infinite sum to a finite sum. For this, we restrict the dynamics to an interval, say [0, L]. The configuration now belongs to

$$ \varDelta _L=\cup _{n=0}^\infty \varDelta ^L_n, $$

with $\varDelta ^L_n$ denoting the set

Again, what we have in mind is that $\rho _i(t)=\rho (x_i(t)+,t)$ with $x_1,\dots ,x_n$ denoting the location of all shocks in (0, L). For our purposes, we need to come up with a candidate for the law $\mu (t,d\mathbf q )$ of $\mathbf q (t)$ in $\varDelta _L$. The restriction of $\mu$ to $\varDelta _L^n$ is denoted by $\mu ^n$ and is given by

$$ \ell ( d\rho _0,t)\exp \left\{ -\sum _{i=0}^n\int _0^{x_{i+1}-x_i}\lambda (\phi _y(\rho _i;t),t) dy \right\} \prod _{i=0}^{n -1} \ f\big (\phi _{x_{i+1}-x_i}(\rho _i;t),\rho _{i+1},t)\ dx_{i+1}d\rho _{i+1}, $$

where f solves (3.9) and $\ell$ is the law of $\rho (0,t)$, which is a Markov process with generator ${\mathcal M}={\mathcal M}_t$:

$$\begin{aligned} \dot{\ell }={\mathcal M}^*\ell . \end{aligned}$$

(4.2)

To simplify the presentation, we assume

$$ \ell (d\rho _0,t)=\ell (\rho _0,t)\ d\rho _0. $$

As for the dynamics of q, we have the following rules:

(i) :

So long as $x_i$ remains in $(x_{i-1},x_{i+1})$, it satisfies

$$ \dot{x}_i=-v^i:=-H[\hat{\rho }_{i-1},\rho _i], $$

where $\hat{\rho }_{i-1}(t)=\phi _{z_{i-1}}(\rho _{i-1}(t),t)$.

(ii) :

We have $\dot{\rho }_0=w^0:=H'(\rho _0)b(\rho _0,t)$ and for $i>0$,

$$ \dot{\rho }_i=w^i:=\big (H'(\rho _i)-H[\hat{\rho }_{i-1},\rho _i]\big )b(\rho _i,t). $$

(iii) :

When $z_i=x_{i+1}-x_i$ becomes 0, then $\mathbf q (t)$ becomes $\mathbf q ^i(t)$, that is obtained from $\mathbf q (t)$ by omitting $(\rho _i,x_i)$.

(iv) :

With rate

$$ H[\hat{\rho }_n,\rho _{n+1}]f\big (\hat{\rho }_n,\rho _{n+1},t), $$

the configuration q gains a new particle $(x_{n+1},\rho _{n+1})$, with $x_{n+1}=L$. This new configuration is denoted by $\mathbf q (\rho _{n+1})$.

We note that since H is increasing, all velocities are negative. Moreover, when the first particle of location $x_1$ crosses the origin, a particle is lost.

We wish to establish (4.1). We write $G^n$ for the restriction of a smooth function $G:\varDelta ^L\rightarrow {\mathbb R}$ to $\varDelta ^L_n$. Recall that we only consider those test functions G that cannot differentiate between $\mathbf q$ and $\mathbf q ^i$ (respectively $\mathbf q (\rho _{n+1})$), when $x_i=x_{i+1}$ (respectively $x_{n+1}=L$). We need to verify

(4.3)

for all $n\ge 0$. Here and below, we write $\nu ^n$ for the restriction of a measure $\nu$ to $\varDelta ^L_n$. Also, given $H:\varDelta _L\rightarrow {\mathbb R}$, we write $H^n$ for the restriction of the function H to the set $\varDelta ^L_n$. To verify (4.3), we show

(4.4)

for every $C^1$ function G. It is instructive to see why (4.3) (or its integrated version (4.4)) is true when $n=0$ and 1 before treating the general case. As we will see below, the cases $n=0,1$ are already equivalent to the Eq. (3.9). As a warm-up, we first assume that $n=0$ and $b=0$. In this case the Eq. (4.3) is equivalent to the fact that the law $\ell$ of $\rho (0,\cdot )$ is governed by a Markov process with generator ${\mathcal M}_t$. The case $n=0$ and general b leads to the general form of ${\mathcal M}_t$ for the evolution of $\rho (0,\cdot )$, and an equation for $\lambda$ that is a consequence of (3.9). The full Eq. (3.9) shows up when we consider the case $n=1$.

The case $n=0$ and $b=0$ . As it turns out, the function $\lambda (\rho ,t)=\lambda (\rho )$ is independent of time when $b=0$. We simply have

$$\begin{aligned} \mu ^0(d\rho _0,t)=e^{-L\lambda (\rho _0)}\ell (d\rho _0,t), \ \ \ \ \ \mu ^0_t(d\rho _0,t)=e^{-\lambda (\rho _0)L} \ell _t(d\rho _0,t). \end{aligned}$$

(4.5)

On the other hand, the right-hand side of (4.4) is of the form $\varOmega ^1_0+\varOmega ^2_0$, where $\varOmega ^1_0$ comes from rule $\mathbf (i)$, and $\varOmega ^2_0$ comes from the stochastic boundary dynamics. Indeed

$$\begin{aligned} \varOmega ^1_0&=\int H[\rho _0,\rho _1]G^0(0,\rho _1)e^{-\lambda (\rho _1)L}f(\rho _0,\rho _1,t)\ d\rho _1\ \ell (d\rho _0,t) \\&\ \ \ \ \ \ \ \ \ \ -\int H[\rho _0,\rho _1]G^1(0,\rho _0,L,\rho _1)e^{-\lambda (\rho _0)L}f(\rho _0,\rho _1,t)\ d\rho _1\ \ell (d\rho _0,t),\nonumber \end{aligned}$$

(4.6)

which we get from the boundary terms when we apply an integration by parts to the integral

$$ -\int H[\rho _0,\rho _1] G_{x_1}^1(0,\rho _0,x_1,\rho _1)\ e^{-\lambda (\rho _0)x_1-\lambda (\rho _1)(L-x_1)} f(\rho _0,\rho _1,t)\ d\rho _1\ \ell (d\rho _0,t)\ dx_1. $$

We note that the other terms of the integration by parts formula contribute to the case $n=1$ and do not contribute to our $n=0$ case. Moreover,

$$ \varOmega ^2_0=\int H[\rho _0,\rho _1]f(\rho _0,\rho _1,t)\big (G^1(0,\rho _0,L,\rho _1)- G^0(0,\rho _0)\big )e^{-\lambda (\rho _0)L}\ d\rho _1\ \ell (d\rho _0,t). $$

From this and (4.6) we learn

$$\begin{aligned} \varOmega ^1_0+\varOmega ^2_0=&\int H[\rho _0,\rho _1]G^0(0,\rho _1)e^{-\lambda (\rho _1)L}f(\rho _0,\rho _1,t)\ d\rho _1\ \ell (d\rho _0,t)\\&\ \ \ \ \ \ \ \ \ \ -\int H[\rho _0,\rho _1]f(\rho _0,\rho _1,t) G^0(0,\rho _0)e^{-\lambda (\rho _0)L}\ d\rho _1\ \ell (d\rho _0,t)\\ =&\int H[\rho _1,\rho _0]G^0(0,\rho _0)e^{-\lambda (\rho _0)L}f(\rho _1,\rho _0,t)\ d\rho _0\ \ell (d\rho _1,t)\\&\ \ \ \ \ \ \ \ \ \ -\int H[\rho _0,\rho _1]f(\rho _0,\rho _1,t) G^0(0,\rho _0)e^{-\lambda (\rho _0)L}\ d\rho _1\ \ell (d\rho _0,t)\\ =&\int G^0(0,\rho _0)e^{-\lambda (\rho _0)L}\ \big ({\mathcal M}_t^*\ell \big )(d\rho _0,t)=\int G^0(0,\rho _0)e^{-\lambda (\rho _0)L}\ \ell _t(d\rho _0,t), \end{aligned}$$

as desired.     $\square$

The case $n=0$ and general b. To ease the notation, we write

$$ \varGamma (\rho ,x,t)=\int _0^x \lambda (\phi _y(\rho ;t),t)\ dy. $$

When $n=0$, the right-hand side of (4.4) equals

$$\begin{aligned} \int&G^0(0,\rho _0)\left[ H'(\rho _0)b(\rho _0,t)\ell (\rho _0,t)\varGamma _{\rho }(\rho _0,L,t)-\big (H'(\rho _0)b(\rho _0,t)\ell (\rho _0,t)\big )_{\rho _0} \right] \ e^{-\varGamma (\rho _0,L,t)}\ d\rho _0\\&+\int H[\rho _0,\rho _1]G^0(0,\rho _1)e^{-\varGamma (\rho _1,L,t)}f(\rho _0,\rho _1,t)\ d\rho _1\ \ell (d\rho _0,t)\\&-\int H\big [\phi _L(\rho _0;t),\rho _1\big ]f\big (\phi _L(\rho _0;t),\rho _1,t\big )G^1(0,\rho _0,L,\rho _1)e^{-\varGamma (\rho _0,L,t)}\ d\rho _1\ \ell (d\rho _0,t)\\&+\int H\big [\phi _L(\rho _0;t),\rho _1\big ]f\big (\phi _L(\rho _0;t),\rho _1,t\big ) \big (G^1(0,\rho _0,L,\rho _1)-G^0(0,\rho _0)\big )e^{-\varGamma (\rho _0,L,t)}\ d\rho _1\ \ell (d\rho _0,t) \end{aligned}$$

This simplifies to

$$\begin{aligned} \int&G^0(0,\rho _0)\left[ H'(\rho _0)b(\rho _0,t)\ell (\rho _0,t)\varGamma _{\rho }(\rho _0,L,t)-\big (H'(\rho _0)b(\rho _0,t)\ell (\rho _0,t)\big )_{\rho _0} \right] \ e^{-\varGamma (\rho _0,L,t)}\ d\rho _0\\&\ \ +\int H[\rho _*,\rho _0]G^0(0,\rho _0)e^{-\varGamma (\rho _0,L,t)}f(\rho _*,\rho _0,t)\ d\rho _0\ \ell (d\rho _*,t)\\&\ \ -\int H\big [\phi _L(\rho _0;t),\rho _1\big ]f\big (\phi _L(\rho _0;t),\rho _1,t\big ) G^0(0,\rho _0)e^{-\varGamma (\rho _0,L,t)}\ d\rho _1\ \ell (d\rho _0,t)\\&\ \ \ \ \ \ =\int G^0(0,\rho _0)\varLambda (\rho _0,t)\ e^{-\varGamma (\rho _0,L,t)}\ d\rho _0, \end{aligned}$$

where $\varLambda (\rho _0,t)$ equals

$$\begin{aligned}&H'(\rho _0)b(\rho _0,t)\ell (\rho _0,t)\varGamma _{\rho }(\rho _0,L,t)- \big (H'(\rho _0)b(\rho _0,t)\ell (\rho _0,t)\big )_{\rho _0}\\&\ \ +\int H[\rho _*,\rho _0]f(\rho _*,\rho _0,t)\ \ell (d\rho _*,t)-\int H\big [\phi _L(\rho _0;t),\rho _1\big ]f\big (\phi _L(\rho _0;t),\rho _1,t\big ) \ d\rho _1\ \ell (\rho _0,t). \end{aligned}$$

We need to match $\varLambda (\rho _0,t)$ with the corresponding term on left-hand side of (4.4), which, by (4.2) takes the form

$$\begin{aligned} -\varGamma _t&(\rho _0,L,t)\ \ell (\rho _0,t)-\big (H'(\rho _0)b(\rho _0,t)\ell (\rho _0,t)\big )_{\rho _0}\\&+\int H[\rho _*,\rho _0]f(\rho _*,\rho _0,t)\ \ell (d\rho _*,t) -A(\rho _0,t)\ \ell (\rho _0,t), \end{aligned}$$

where

$$ A(\rho _0,t)=\int H[\rho _0,\rho _*]f(\rho _0,\rho _*,t)\ d\rho _*. $$

We are done if we can verify

$$\begin{aligned} \varGamma _t(\rho _0,L,t)+H'(\rho _0)b(\rho _0,t)\varGamma _{\rho }(\rho _0,L,t)=A\big (\phi _L(\rho _0;t),t\big )-A(\rho _0,t). \end{aligned}$$

(4.7)

Equivalently

$$\begin{aligned}&\int _0^L\big [\lambda (\phi _y(\rho _0;t),t)\big ]_t\ dy+ H'(\rho _0)b(\rho _0,t)\int _0^L\big [\lambda (\phi _y(\rho _0;t),t)\big ]_{\rho _0}\ dy =\int _0^L \big [A\big (\phi _y(\rho _0;t),t\big )\big ]_y\ dy. \end{aligned}$$

For this, it suffices to check

$$ \big [\lambda (\phi _y(\rho _0;t),t)\big ]_t+ H'(\rho _0)b(\rho _0,t)\ \big [\lambda (\phi _y(\rho _0;t),t)\big ]_{\rho _0} = \big [A\big (\phi _y(\rho _0;t),t\big )\big ]_y. $$

Note that if $u(y,\rho )=A\big (\phi _y(\rho ;t),t\big )$, then

$$ u_y(y,\rho _0)=b(\rho _0,t)u_{\rho }(y,\rho _0). $$

Hence for (4.7), it suffices to show

$$\begin{aligned} \big [\lambda (\phi _y(\rho _0;t),t)\big ]_t+ H'(\rho _0)b(\rho _0,t)\big [\lambda (\phi _y(\rho _0;t),t)\big ]_{\rho _0} = b(\rho _0,t)\big [A\big (\phi _y(\rho _0;t),t\big )\big ]_{\rho _0}. \end{aligned}$$

(4.8)

To have a more tractable formula, let us write $T_yh(m)=h(\phi _y(m;t))$. The family of operators $\{T_y:\ y\in {\mathbb R}\}$, is a group in y. Moreover, if $({\mathcal B}h)(m)=b(m,t)h'(m)$, then

$$\begin{aligned} \frac{dT_y}{dy}={\mathcal B}T_y=T_y {\mathcal B}. \end{aligned}$$

(4.9)

Using this, we may rewrite (4.8) as

$$\begin{aligned} \big [\lambda (\phi _y(\rho _0;t),t)\big ]_t+ H'(\rho _0)b(\phi _y(\rho _0;t),t)\lambda _{\rho }(\phi _y(\rho _0;t),t) = b(\phi _y(\rho _0;t),t)A_{\rho }\big (\phi _y(\rho _0;t),t\big ). \end{aligned}$$

(4.10)

This for $y=0$ takes the form

$$\begin{aligned} \lambda _t(\rho _0,t)+H'(\rho _0)b(\rho _0,t)\lambda _\rho (\rho _0,t)=b(\rho _0,t)A_\rho (\rho _0,t). \end{aligned}$$

(4.11)

Because of our choice of $\lambda$, namely

$$\begin{aligned} \lambda (t,\rho -)=\int _{\rho -}^\infty f(\rho _-,\rho _+,t)\ d\rho _+, \end{aligned}$$

(4.12)

we can deduce (4.10) from (3.9) after integrating both sides of (3.9) with respect to $\rho _-$. On account of (4.11), the claim (4.10) would follow if we can show

$$\begin{aligned} X(\rho _0,y,t):=\big [\phi _y(\rho _0;t)\big ]_t-\big [H'(\phi _y(\rho _0;t))-H'(\rho _0)\big ]b(\phi _y(\rho _0;t),t)=0. \end{aligned}$$

(4.13)

This is true for $y=0$. Differentiating with respect to y yields

$$\begin{aligned} X_y(\rho _0,y,t)&=\big [b(\phi _y(\rho _0;t),t)\big ]_t-\big [H'(\phi _y(\rho _0;t))\big ]_yb(\phi _y(\rho _0;t),t)\\&\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ -\big [H'(\phi _y(\rho _0;t))-H'(\rho _0)\big ]\big [b(\phi _y(\rho _0;t),t)\big ]_y\\&=b_t(\phi _y(\rho _0;t),t)+b_\rho (\phi _y(\rho _0;t),t)\big [\phi _y(\rho _0;t)\big ]_t -H''(\phi _y(\rho _0;t))b^2(\phi _y(\rho _0;t),t)\\&\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ -\big [H'(\phi _y(\rho _0;t))-H'(\rho _0)\big ](bb_\rho )(\phi _y(\rho _0;t),t)\\&=b_\rho (\phi _y(\rho _0;t),t)\big [\phi _y(\rho _0;t)\big ]_t-\big [H'(\phi _y(\rho _0;t))-H'(\rho _0)\big ](bb_\rho )(\phi _y(\rho _0;t),t)\\&=b_\rho (\phi _y(\rho _0;t),t)X(\rho _0,y,t), \end{aligned}$$

where we used (3.4) for the third equality. As a result.

$$ X(\rho _0,y,t)=X(\rho _0,0,t)\exp \left[ \int _0^yb_\rho (\phi _z(\rho _0;t),t)\ dz\right] =0. $$

This completes the proof of (4.2), when $n=0$.     $\square$

As we have seen so far, the case $n=0$ is valid if (4.11), a consequence of the kinetic equation (3.9), is true. On the other hand the case $n=1$ is equivalent to the kinetic equation. Before embarking on the verification of (4.3) for $n=1$, let us make some compact notions for some of the expressions that come into the proof. Given a realization $\mathbf q =\big (0,\rho _0,x_1,\rho _1,\dots ,x_n,\rho _n\big )\in \varDelta _n^L,$ we define

$$\begin{aligned} \rho \big (x,t;\mathbf q \big )&=\sum _{i=0}^{n}\phi _{x-x_i}\big (\rho _i;t\big ) 1\!\!1\big (x_i\le x<x_{i+1}\big ),\\ \varGamma (\mathbf q ,t)&=\int _0^L\lambda \big (\rho \big (y,t;\mathbf q \big )\big )\ dy=\sum _{i=0}^{n}\varGamma (\rho _{i},x_{i+1}-x_{i},t),\\ \hat{\rho }_{i-1}&=\rho (x_i-,t;\mathbf q )=\phi _{x_i-x_{i-1}}(\rho _{i-1};t), \end{aligned}$$

where $\lambda$ is defined by (4.12). Note that by (4.13),

$$\begin{aligned} \frac{d\hat{\rho }_{i}}{dt}=\big [H'(\hat{\rho }_{i})-H'(\rho _{i})\big ]b(\hat{\rho }_i,t). \end{aligned}$$

(4.14)

The case $n=1$. We have $\dot{\mu }^1=X_1\mu ^1$, where

$$\begin{aligned} X_1(\mathbf q ,t)&=-\varGamma _t(\mathbf q ,t)+\frac{ \ell _t(\rho _0,t)}{\ell (\rho _0,t)} +\frac{\big [f\big (\hat{\rho }_0,\rho _{1},t)\big ]_t}{f\big (\hat{\rho }_0,\rho _{1},t)}. \end{aligned}$$

On the other hand , with

$$ Y_1(\mathbf q ,t)=\sum _{j=1}^7 Y_{1j}(\mathbf q ,t)=\sum _{j=1}^7 Y_{1j}, $$

where

$$\begin{aligned} Y_{11}&=H'(\rho _0)b(\rho _0,t)\left[ \varGamma _{\rho }(\rho _0,x_1,t) -\frac{\big [f\big (\hat{\rho }_0,\rho _{1},t\big )\big ]_{\rho _0}}{f\big (\hat{\rho }_0,\rho _{1},t\big )}\right] -\frac{\big (H'(\rho _0)b(\rho _0,t){\ell (\rho _0,t)}\big )_{\rho _0}}{\ell (\rho _0,t)}\\ Y_{12}&=\big (H'(\rho _1)-H[\hat{\rho }_0,\rho _1]\big )b(\rho _1,t)\varGamma _{\rho }(\rho _1,L-x_1,t) \\ Y_{13}&=\frac{\big [\big (H[\hat{\rho }_0,\rho _1]-H'(\rho _1)\big )b(\rho _1,t)f(\hat{\rho }_0,\rho _1,t)\big ]_{\rho _1}}{f\big (\hat{\rho }_0,\rho _1,t\big )}\\ Y_{14}&=\frac{\big (H\big [\hat{\rho }_0,\rho _{1}\big ] f\big (\hat{\rho }_0,\rho _1,t\big )\big )_{x_1}}{f\big (\hat{\rho }_0,\rho _{1},t\big )}+ H\big [\hat{\rho }_0,\rho _{1}\big ] \left[ \lambda \big (\phi _{L-x_{1}}(\rho _{1};t),t\big )-\lambda \big (\hat{\rho }_0,t\big )\right] \\ Y_{15}&=\frac{\int H(\rho _*,\rho _0)f(\rho _*,\rho _0,t)\ \ell (d\rho _*,t)}{\ell (\rho _0,t)}\\ Y_{16}&= -\int H\big [\phi _{L-x_1}(\rho _{1};t),\rho _{*}\big ] f\big (\phi _{L-x_1}(\rho _{1};t),\rho _*,t\big )\ d\rho _*=-A\big (\phi _{L-x_1}(\rho _{1};t),t\big )\\ Y_{17}&=\frac{\int \big (H[\rho _*,\rho _1]-H\big [\hat{\rho }_0,\rho _{*}\big ]\big ) f\big (\hat{\rho }_0,\rho _*,t\big ) f(\rho _*,\rho _{1},t)\ d\rho _*}{f\big (\hat{\rho }_0,\rho _1,t\big )}. \end{aligned}$$

Here,

• The term $Y_{11}$ comes from an integration by parts with respect to the variable $\rho _0$. The dynamics of $\rho _0$ as in rule (ii) is responsible for this contribution.

• The terms $Y_{12}$ and $Y_{13}$ come from an integration by parts with respect to the variable $\rho _1$. The dynamics of $\rho _1$ as in rule (ii) is responsible for these two contributions.

• The term $Y_{14}$ comes from an integration by parts with respect to the variable $x_1$. The dynamics of $x_1$ as in rule (i) is responsible for this contribution.

• The term $Y_{15}$ comes from the boundary term $x_1=0$ in the integration by parts with respect to the variable $x_1$ when there are two particles at $x_1$ and $x_2$. This boundary condition represents the event that $x_1$ has reached the origin after which $\rho _0$ becomes $\rho _1$, and $(x_2,\rho _2)$ is relabeled $(x_1,\rho _1)$.

• The term $Y_{16}$ comes from the boundary term $x_2=L$ in the integration by parts with respect to the variable $x_2$, and the stochastic boundary dynamics as in the rule (iv). The boundary term $x_2=L$ cancels part of the contribution of the boundary dynamics as we have already seen in our calculation in the case $n=0$.

• The rule (iii) is responsible for the term $Y_{17}$. When $n=2$, the particles at $x_1$ and $x_2$ travel towards each other with speed $H[\hat{\rho }_1,\rho _2]-H[\hat{\rho }_0,\rho _1]$. As $x_2$ catches up with $x_1$, the particle $x_1$ disappears and its density $\rho _1=\hat{\rho }_1$ is renamed $\rho _*$, and is integrated out. We then relabel $(x_2,\rho _2)$ as $(x_1,\rho _1)$.

We wish to show that $X_1=Y_1$. After some cancellation, this simplifies to

$$ X'_1=Y'_1:=\hat{Y}_{11}+Y_{12}+Y_{13}+Y_{14}+Y_{16}+Y_{17}, $$

where

$$\begin{aligned} X'_1&=-\varGamma _t(\mathbf q ,t)-A(\rho _0,t)+ \frac{\big [f(\hat{\rho }_0,\rho _{1},t)\big ]_t}{f(\hat{\rho }_0,\rho _{1},t)},\\ \hat{Y} _{11}&=H'(\rho _0)b(\rho _0,t) \left[ \varGamma _{\rho }(\rho _0,x_1,t)-\frac{\big [f(\hat{\rho }_0,\rho _{1},t)\big ]_{\rho _0}}{f(\hat{\rho }_0,\rho _{1},t)}\right] . \end{aligned}$$

(The same cancellation led to the Eq. (4.7).) Observe that $\varGamma (\mathbf q ,t)=\varGamma (\rho _0,x_1,t)+\varGamma (\rho _1,L-x_1,t)$. Moreover, by (4.7),

$$\begin{aligned}&\varGamma _t(\rho _0,x_1,t)+H'(\rho _0)b(\rho _0,t)\varGamma _{\rho }(\rho _0,x_1,t)= A(\hat{\rho }_0,t)-A(\rho _0,t)\\&\varGamma _t(\rho _1,L-x_1,t)+H'(\rho _1)b(\rho _1,t)\varGamma _{\rho }(\rho _1,L-x_1,t)= A\big (\phi _{L-x_1}(\rho _1;t),t\big )-A(\rho _1,t). \end{aligned}$$

As a result,

$$\begin{aligned} -\varGamma _t(\mathbf q ,t)-A(\rho _0,t)&=H'(\rho _0)b(\rho _0,t)\varGamma _{\rho }(\rho _0,x_1,t) +H'(\rho _1)b(\rho _1,t)\varGamma _{\rho }(\rho _1,L-x_1,t)\\&\ \ \ \ -A\big (\phi _{L-x_1}(\rho _1;t),t\big )+A(\rho _1,t) -A(\hat{\rho }_0,t\big ). \end{aligned}$$

Using this, we learn that the equality $X'_1=Y'_1$ is equivalent to the identity

$$\begin{aligned} \big [f\big (\hat{\rho }_0,\rho _{1},t)\big ]_t&= H\big [\hat{\rho }_0,\rho _{1}\big ] \left[ \lambda \big (\phi _{L-x_{1}}(\rho _{1};t),t\big )-\lambda \big (\hat{\rho }_0,t\big )\right] f\big (\hat{\rho }_0,\rho _{1},t\big )\\&\ \ \ \ +\left[ A\big (\hat{\rho }_0,t\big )-A(\rho _1,t)\right] f\big (\hat{\rho }_0,\rho _{1},t\big )\\&\ \ \ \ +\int \big (H[\rho _*,\rho _1]-H\big [\hat{\rho }_0,\rho _{*}\big ]\big ) f\big (\hat{\rho }_0,\rho _*,t\big ) f(\rho _*,\rho _{1},t)\ d\rho _*\\&\ \ \ \ +{\big [\big (H[\hat{\rho }_0,\rho _1]-H'(\rho _1)\big )b(\rho _1,t)f(\hat{\rho }_0,\rho _1,t)\big ]_{\rho _1}} \\&\ \ \ \ -H'(\rho _0)b(\rho _0,t)\big [f\big (\hat{\rho }_0,\rho _{1},t)\big ]_{\rho _0} -H[\hat{\rho }_0,\rho _1]b(\rho _1,t)\varGamma _{\rho }(\rho _1,L-x_1,t)f\big (\hat{\rho }_0,\rho _{1},t\big )\\&\ \ \ \ +\big (H\big [\hat{\rho }_0,\rho _{1}\big ] f\big (\hat{\rho }_{0},\rho _1,t\big ) \big )_{x_1}. \end{aligned}$$

By the group property (4.9), we can assert that for any $C^1$ function h,

$$ \big [h(\hat{\rho }_0)\big ]_{x_1}=b(\hat{\rho }_0,t)h'(\hat{\rho }_0)=b(\rho _0,t)\big [h(\hat{\rho }_0)\big ]_{\rho _0}. $$

We use this and the definition of the quadratic operator Q in (3.10) to deduce that $X'_1=Y'_1$ is equivalent to the identity

$$\begin{aligned} \big [f(\hat{\rho }_0,\rho _{1},t)\big ]_t&=Q(f,f)(\hat{\rho }_0,\rho _{1},t)+H\big [\hat{\rho }_0,\rho _{1}\big ] \left[ \lambda \big (\phi _{L-x_{1}}(\rho _{1};t),t\big )-\lambda \big (\rho _1,t\big )\right] f\big (\hat{\rho }_0,\rho _{1},t\big )\\&\ \ \ \ +{\big [\big (H[\hat{\rho }_0,\rho _1]-H'(\rho _1)\big )b(\rho _1,t)f(\hat{\rho }_0,\rho _1,t)\big ]_{\rho _1}}\\&\ \ \ \ -H'(\rho _0)b(\hat{\rho }_0,t)f_{\rho _-}\big (\hat{\rho }_0,\rho _{1},t) -H[\hat{\rho }_0,\rho _1]b(\rho _1,t)\varGamma _{\rho }(\rho _1,L-x_1,t)f\big (\hat{\rho }_0,\rho _{1},t\big )\\&\ \ \ \ +b(\hat{\rho }_0,t)H[\hat{\rho }_0,\rho _{1}] f_{\rho _-}(\hat{\rho }_{0},\rho _1,t) +b(\hat{\rho }_0,t)H_{\rho _-}[\hat{\rho }_0,\rho _{1}] f(\hat{\rho }_{0},\rho _1,t). \end{aligned}$$

Here we are acting the quadratic operator Q on functions because we are assuming that $f(\rho _,d\rho _+,t) =f(\rho _,\rho _+,t) \ d\rho _+$, is absolutely continuous with respect to the Lebesgue measure. We now use (4.13) to assert that $X'_1=Y'_1$ is equivalent to the identity

$$\begin{aligned} f_t(\hat{\rho }_0,\rho _{1},t)&= Q(f,f)(\hat{\rho }_0,\rho _{1},t) +b(\hat{\rho }_0,t)H_{\rho _-}[\hat{\rho }_0,\rho _{1}] f(\hat{\rho }_{0},\rho _1,t)\\&\ \ \ \ + H\big [\hat{\rho }_0,\rho _{1}\big ] \left[ \lambda \big (\phi _{L-x_{1}}(\rho _{1};t),t\big )-\lambda \big (\rho _1,t\big )\right] f\big (\hat{\rho }_0,\rho _{1},t\big )\\&\ \ \ \ + \big [H[\hat{\rho }_0,\rho _{1}]-H'(\hat{\rho }_0)\big ]b(\hat{\rho }_0,t)f_{\rho _-}\big (\hat{\rho }_0,\rho _{1},t)\\&\ \ \ \ +{\big [\big (H[\hat{\rho }_0,\rho _1]-H'(\rho _1)\big )b(\rho _1,t)f(\hat{\rho }_0,\rho _1,t)\big ]_{\rho _1}}\\&\ \ \ \ -H[\hat{\rho }_0,\rho _{1}] b(\rho _1,t)\varGamma _{\rho }(\rho _1,L-x_1,t)f\big (\hat{\rho }_0,\rho _{1},t\big ). \end{aligned}$$

On the other hand, by the definition of $\varGamma$,

$$\begin{aligned} \nonumber b(\rho _1,t)\varGamma _{\rho }(\rho _1,L-x_1,t)&=\int _0^{L-x_1}b(\rho _1,t) \left[ \lambda \big (\phi _y(\rho _1;t),t\big )\right] _{\rho _1}\ dy\\&=\int _0^{L-x_1} \big [\lambda \big (\phi _y(\rho _1;t),t\big )\big ]_y\ dy\\&=\lambda \big (\phi _{L-x_{1}}(\rho _{1};t),t\big )-\lambda (\rho _1,t),\nonumber \end{aligned}$$

(4.15)

where we used the group property (4.9) for the second equality. This leads to

$$\begin{aligned} f_t(\hat{\rho }_0,\rho _1,t)\,=\,&Q(f,f)(\hat{\rho }_0,\rho _{1},t) +b(\hat{\rho }_0,t)f(\hat{\rho }_0,\rho _1,t)H_{\rho _-}[\hat{\rho }_0,\rho _1]\\&\ +\big [H[\hat{\rho }_0,\rho _1]-H'(\hat{\rho }_0)\big ]b(\hat{\rho }_0,t) f_{\rho _-}(\hat{\rho }_0,\rho _1,t)\\&\ +\left[ \big (H[\hat{\rho }_0,\rho _1]-H'(\rho _1)\big )b(\rho _1,t)f(\hat{\rho }_0,\rho _1,t)\right] _{\rho _1}, \end{aligned}$$

which is exactly our kinetic equation!     $\square$

General n. We write $\dot{\mu }^n=X_n\mu ^n$. We have,

$$\begin{aligned} X_n=-\varGamma _t(\mathbf q ,t)+\frac{ \ell _t(\rho _0,t)}{\ell (\rho _0,t)} +\sum _{i=1}^{n}\frac{\big [f(\hat{\rho }_{i-1},\rho _{i},t)]_t}{f(\hat{\rho }_{i-1},\rho _{i},t)}. \end{aligned}$$

(4.16)

By (4.7), and (4.15),

$$ \varGamma _t(\mathbf q ,t)=\sum _{i=0}^{n}\left\{ \big (A(\hat{\rho }_i,t)-A(\rho _i,t)\big )-H'(\rho _i) \big (\lambda (\hat{\rho }_i,t)-\lambda (\rho _i,t)\big ) \right\} . $$

From this, (4.14) and (3.10) we deduce

$$\begin{aligned} X_n&=\frac{ \ell _t(\rho _0,t)}{\ell (\rho _0,t)}+ \sum _{i=1}^{n}\frac{Q^+(f,f)(\hat{\rho }_{i-1},\rho _{i},t)}{f(\hat{\rho }_{i-1},\rho _{i},t)}+ \sum _{i=1}^{n}\frac{(Cf)(\hat{\rho }_{i-1},\rho _{i},t)}{f(\hat{\rho }_{i-1},\rho _{i},t)}\\&\ \ \ \ \ \ +\sum _{i=0}^{n}\left\{ H'(\rho _i) \big (\lambda (\hat{\rho }_i,t)-\lambda (\rho _i,t)\big )+A(\rho _i,t)-A(\hat{\rho }_i,t)\right\} \\&\ \ \ \ \ \ -\sum _{i=1}^{n}\left\{ A(\rho _i,t)-A(\hat{\rho }_{i-1},t) -H[\hat{\rho }_{i-1},\rho _i]\big (\lambda (\rho _{i},t)-\lambda (\hat{\rho }_{i-1},t)\big ) \right\} \\&\ \ \ \ \ \ +\sum _{i=1}^{n}\big [H'(\hat{\rho }_{i-1})-H'(\rho _{i-1})\big ]b(\hat{\rho }_{i-1},t)\frac{f_{\rho _-}(\hat{\rho }_{i-1},\rho _{i},t)}{f(\hat{\rho }_{i-1},\rho _{i},t)}\\&=\frac{ \ell _t(\rho _0,t)}{\ell (\rho _0,t)}+ \sum _{i=1}^{n}\frac{Q^+(f,f)(\hat{\rho }_{i-1},\rho _{i},t)}{f(\hat{\rho }_{i-1},\rho _{i},t)}+ \sum _{i=1}^{n}\frac{(Cf)(\hat{\rho }_{i-1},\rho _{i},t)}{f(\hat{\rho }_{i-1},\rho _{i},t)}\\&\ \ \ \ \ \ + H'(\rho _0)\big (\lambda (\hat{\rho }_0,t)-\lambda (\rho _0,t)\big )+A(\rho _0,t)-A(\hat{\rho }_n,t)\\&\ \ \ \ \ \ +\sum _{i=1}^{n}\left\{ H'(\rho _i) \big (\lambda (\hat{\rho }_i,t)-\lambda (\rho _i,t)\big )+H[\hat{\rho }_{i-1},\rho _i]\big (\lambda (\rho _{i},t)-\lambda (\hat{\rho }_{i-1},t)\big )\right\} \\&\ \ \ \ \ \ +\sum _{i=1}^{n}\big [H'(\hat{\rho }_{i-1})-H'(\rho _{i-1})\big ]b(\hat{\rho }_{i-1},t)\frac{f_{\rho _-}(\hat{\rho }_{i-1},\rho _{i},t)}{f(\hat{\rho }_{i-1},\rho _{i},t)}. \end{aligned}$$

For the right-hand side of (4.3) we write , where

$$\begin{aligned} Y_n=Y'_{11}+Y''_{11}+Y'''_{11}+Y_{n2}+Y_{n3}+Y'_{n4}+Y''_{n4} +Y_{n5}+Y_{n6}+Y_{n7}, \end{aligned}$$

(4.17)

with $Y_{15}$ independent of n and defined before, $Y_{n6}=-A(\hat{\rho }_n,t)$, and

$$\begin{aligned} Y'_{11}&=H'(\rho _0)b(\rho _0,t)\varGamma _{\rho }(\rho _0,x_1,t) = H'(\rho _0)\big (\lambda (\hat{\rho }_0,t)-\lambda (\rho _0,t)\big )\\ Y''_{11}&=-H'(\rho _0)b(\rho _0,t)\frac{\big [f\big (\hat{\rho }_0,\rho _{1},t\big )\big ]_{\rho _0}}{f\big (\hat{\rho }_0,\rho _{1},t\big )}\\ Y'''_{11}&=-\frac{\big (H'(\rho _0)b(\rho _0,t){\ell (\rho _0,t)}\big )_{\rho _0}}{\ell (\rho _0,t )}\\ Y_{n2}&=\sum _{i=1}^{n} \big (H'(\rho _i)-H[\hat{\rho }_{i-1},\rho _i]\big )b(\rho _i,t) \varGamma _{\rho }(\rho _i,x_{i+1}-x_i,t)\\&=\sum _{i=1}^{n} \big (H'(\rho _i)-H[\hat{\rho }_{i-1},\rho _i]\big )\big (\lambda (\hat{\rho }_i,t)-\lambda (\rho _i,t)\big ), \end{aligned}$$

$$\begin{aligned} Y_{n3}&=\sum _{i=1}^{n-1}\frac{\big [\big (H[\hat{\rho }_{i-1},\rho _i]-H'(\rho _i)\big )b(\rho _i,t)f(\hat{\rho }_{i-1},\rho _i,t) f(\hat{\rho }_{i},\rho _{i+1},t)\big ]_{\rho _i}}{f(\hat{\rho }_{i-1},\rho _i,t\big )f(\hat{\rho }_{i},\rho _{i+1},t)}\\&\ \ \ \ \ \ \ \ \ \ \ +\frac{\big [\big (H[\hat{\rho }_{n-1},\rho _n] -H'(\rho _n)\big )b(\rho _n,t) f(\hat{\rho }_{n-1},\rho _n,t)\big ]_{\rho _n}}{f(\hat{\rho }_{n-1},\rho _n,t)},\\ Y'_{n4}&=\sum _{i=1}^{n-1} \frac{\big [H\big [\hat{\rho }_{i-1},\rho _{i}\big ]f\big (\hat{\rho }_{i-1},\rho _i,t\big ) f\big (\hat{\rho }_{i},\rho _{i+1},t\big )\big ]_{x_i}}{f\big (\hat{\rho }_{i-1},\rho _i,t\big ) f\big (\hat{\rho }_{i},\rho _{i+1},t\big )}+ \frac{\big [H\big [\hat{\rho }_{n-1},\rho _{n}\big ]f\big (\hat{\rho }_{n-1},\rho _n,t\big )\big ]_{x_n}}{f\big (\hat{\rho }_{n-1},\rho _n,t\big )},\\ Y''_{n4}&=\sum _{i=1}^n H\big [\hat{\rho }_{i-1},\rho _i\big ] \left( \lambda \big (\hat{\rho }_{i},t\big )- \lambda \big (\hat{\rho }_{i-1},t\big )\right) ,\\ Y_{n7}&=\sum _{i=1}^{n}\frac{Q^+(f,f)\big (\hat{\rho }_{i-1},\rho _{i},t)}{f\big (\hat{\rho }_{i-1},\rho _i,t\big )}, \end{aligned}$$

where we used (4.15) for the first and fifth equality. We wish to show that $X_n=Y_n$. From

$$ \frac{ \ell _t(\rho _0,t)}{\ell (\rho _0,t)}=Y_{15}+Y'''_{11}-A(\rho _0,t), $$

and some cancellation, the equality $X_n=Y_n$ simplifies to $X'_n=Y'_n$, where

$$\begin{aligned} X'_n(\mathbf q ,t)&=\sum _{i=1}^{n}\frac{(Cf)(\hat{\rho }_{i-1},\rho _{i},t)}{f(\hat{\rho }_{i-1},\rho _{i},t)} +\sum _{i=1}^{n}\big [H'(\hat{\rho }_{i-1})-H'(\rho _{i-1})\big ]b(\hat{\rho }_{i-1},t)\frac{f_{\rho _-}(\hat{\rho }_{i-1},\rho _{i},t)}{f(\hat{\rho }_{i-1},\rho _{i},t)}\\&=\sum _{i=1}^{n}\frac{\big [\big (H[\hat{\rho }_{i-1},\rho _i]-H'(\rho _i)\big )b(\rho _i,t)f(\hat{\rho }_{i-1},\rho _i,t) \big ]_{\rho _i}}{f(\hat{\rho }_{i-1},\rho _i,t\big )}\\&\ \ \ \ \ \ +\sum _{i=1}^{n}\big [H[\hat{\rho }_{i-1},\rho _i]-H'(\rho _{i-1})\big ]b(\hat{\rho }_{i-1},t)\frac{f_{\rho _-}(\hat{\rho }_{i-1},\rho _{i},t)}{f(\hat{\rho }_{i-1},\rho _{i},t)} +b(\hat{\rho }_{i-1},t)H_{\rho _-}[\hat{\rho }_{i-1},\rho _i], \end{aligned}$$

and $Y'_n=Y''_{11}+Y_{n3}+Y'_{n4}\$. Observe that $Y'_{n4}=Y'_{n41}+Y'_{n42}+Y'_{n43}$, and $Y_{n3}=Y_{n31}+Y_{n32}$, where

$$\begin{aligned} Y'_{n41}&=\sum _{i=1}^n H_{\rho _-}\big [\hat{\rho }_{i-1},\rho _{i}\big ]b(\hat{\rho }_{i-1},t),\\ Y'_{n42}&=\sum _{i=1}^n H\big [\hat{\rho }_{i-1},\rho _{i}\big ] \frac{\big [f\big (\hat{\rho }_{i-1},\rho _i,t\big ) \big ]_{x_i}}{f\big (\hat{\rho }_{i-1},\rho _i,t\big )},\\ Y'_{n43}&=\sum _{i=1}^{n-1} H\big [\hat{\rho }_{i-1},\rho _{i}\big ] \frac{\big [f\big (\hat{\rho }_{i},\rho _{i+1},t\big )\big ]_{x_i}}{f\big (\hat{\rho }_{i},\rho _{i+1},t\big )},\\ Y_{n31}&=\sum _{i=1}^{n}\frac{\big [\big (H[\hat{\rho }_{i-1},\rho _i]-H'(\rho _i)\big )b(\rho _i,t)f(\hat{\rho }_{i-1},\rho _i,t) \big ]_{\rho _i}}{f(\hat{\rho }_{i-1},\rho _i,t\big )},\\ Y_{n32}&=\sum _{i=1}^{n-1}\big (H[\hat{\rho }_{i-1},\rho _i]-H'(\rho _i)\big )b(\rho _i,t) \frac{\big [f(\hat{\rho }_{i},\rho _{i+1},t)\big ]_{\rho _i}}{f(\hat{\rho }_{i},\rho _{i+1},t)}. \end{aligned}$$

From these decompositions, we learn that $X'_n=Y'_n$ is equivalent to $X''_n=Y''_n$, where

$$\begin{aligned} X''_n&=\sum _{i=1}^{n}\big [H[\hat{\rho }_{i-1},\rho _i]-H'(\rho _{i-1})\big ]b(\hat{\rho }_{i-1},t)\frac{f_{\rho _-}(\hat{\rho }_{i-1},\rho _{i},t)}{f(\hat{\rho }_{i-1},\rho _{i},t)}, \end{aligned}$$

and $Y''_n=Y''_{11}+Y_{n32}+Y'_{n42}+Y'_{n43}.$ By the group property (4.9),

$$ \big (h(\hat{\rho }_{i-1})\big )_{x_i}=b(\rho _{i-1},t) \big (h(\hat{\rho }_{i-1})\big )_{\rho _{i-1}},\ \ \ \ \ \ \big (h(\hat{\rho }_{i})\big )_{x_i}=-b(\rho _{i},t) \big (h(\hat{\rho }_{i})\big )_{\rho _{i}}, $$

This allows us to write

$$\begin{aligned} Y'_{n42}+Y'_{n43}=\sum _{i=1}^{n-1}&H\big [\hat{\rho }_{i-1},\rho _{i}\big ] \left\{ b(\rho _{i-1},t)\frac{\big [f\big (\hat{\rho }_{i-1},\rho _i,t\big ) \big ]_{\rho _{i-1}}}{f\big (\hat{\rho }_{i-1},\rho _i,t\big )}-b(\rho _{i},t) \frac{\big [f\big (\hat{\rho }_{i},\rho _{i+1},t\big )\big ]_{\rho _i}}{f\big (\hat{\rho }_{i},\rho _{i+1},t\big )}\right\} \\&+H\big [\hat{\rho }_{n-1},\rho _{n}\big ] b(\rho _{n-1},t)\frac{\big [f\big (\hat{\rho }_{n-1},\rho _i,t\big ) \big ]_{\rho _{n-1}}}{f\big (\hat{\rho }_{n-1},\rho _n,t\big )}. \end{aligned}$$

Hence

$$\begin{aligned} Y''_{n}&=-\sum _{i=0}^{n-1} H'(\rho _i)b(\rho _i,t) \frac{\big [f(\hat{\rho }_{i},\rho _{i+1},t)\big ]_{\rho _i}}{f(\hat{\rho }_{i},\rho _{i+1},t)} +\sum _{i=1}^n H\big [\hat{\rho }_{i-1},\rho _{i}\big ] b(\rho _{i-1},t)\frac{\big [f\big (\hat{\rho }_{i-1},\rho _i,t\big ) \big ]_{\rho _{i-1}}}{f\big (\hat{\rho }_{i-1},\rho _i,t\big )}\\&=\sum _{i=1}^n \left[ H\big [\hat{\rho }_{i-1},\rho _{i}\big ]-H'(\rho _{i-1})\right] b(\rho _{i-1},t)\frac{\big [f\big (\hat{\rho }_{i-1},\rho _i,t\big ) \big ]_{\rho _{i-1}}}{f\big (\hat{\rho }_{i-1},\rho _i,t\big )}\\&=\sum _{i=1}^n \left[ H\big [\hat{\rho }_{i-1},\rho _{i}\big ]-H'(\rho _{i-1}\right] b(\hat{\rho }_{i-1},t)\frac{f_{\rho _-} \big (\hat{\rho }_{i-1},\rho _i,t\big )}{f\big (\hat{\rho }_{i-1},\rho _i,t\big )}=X''_n, \end{aligned}$$

as desired. For the third equality, we have used (4.9). This completes the proof.     $\square$

So far, we have been able to formally show that the law of $\mathbf q (t)$ is $\mu (d\mathbf q ,t)$, by verifying the forward Eq. (4.3) for every n (recall that n represents the number of particles/shock discontinuities in the interval (0, L)). Our verification of (4.3) is rather tedious but elementary. Our verification is formal at this point because the evolution of $\mathbf q$ is governed by a discontinuous deterministic dynamics that is interrupted by stochastic Markovian entrance of new particles at the boundary point L. By selecting a pair $\ell$ and f that are differentiable with respect to time, it is not hard to justify our calculation for the left-hand side of (4.3), as it appeared in (4.16). It is the justification of the right-hand side as in (4.17) that requires additional work.

Writing $\varPhi _s^t(\mathbf q )$ for $\mathbf q (t)$ with initial condition $\mathbf q (s)=\mathbf q$, it suffices to show that for every nice function $G:\varDelta _L\rightarrow {\mathbb R}$,

$$\begin{aligned} \frac{d}{ds}{\mathbb E}\ G\big (\varPhi _s^t(\mathbf q )\big )=\frac{d}{ds}\int G\big (\varPhi _s^t(\mathbf q )\big )\ \mu (d\mathbf q ,s)=0. \end{aligned}$$

(4.18)

Clearly (4.18) implies

$$ {\mathbb E}G(\mathbf q (t))={\mathbb E}\int G\big (\varPhi _0^t(\mathbf q )\big )\ \mu (d\mathbf q ,0)=\int G(\mathbf q )\ \mu (d\mathbf q ,t), $$

which means that the law of $\mathbf q (t)$ is given by our candidate $\mu (d\mathbf q ,t)$. For a rigorous proof of this, we calculate the left time-derivative of ${\mathbb E}\ G\big (\varPhi _s^t(\mathbf q )\big )$ by hand and show that this left-derivative equals to

$$ \sum _{n=0}^\infty \int G (X_n-Y_n)\ d\mu ^n. $$

The details can be found in [16] and [17].

5 Homogenizations for Hamiltonian ODEs

The Hamilton-Jacobi PDE may be used to model the growth of an interface that is described as a graph of a height function. More precisely, the graph of a solution

$$ u:{\mathbb R}^d\times [0,\infty )\rightarrow {\mathbb R}, $$

of the Hamilton-Jacobi equation

$$\begin{aligned} u_t+H(x,u_x)=0, \end{aligned}$$

(5.1)

describes an interface at time t in microscopic coordinates. If the ratio of micro to macro scale is a large number n, then

$$ u^n(x,t)=\frac{1}{n} u(nx,nt), $$

is the corresponding macroscopic height function. In practice n is large and we may obtain a simpler description of our model if the large n limit of $u^n$ exists and satisfies a simple equation. Indeed $u^n$ satisfies

$$ u^n_t+H\big (nx,u^n_x\big )=0, $$

and this equation must be solved for an initial condition of the form $u^n(x,0)=g(x)$, where g represents the initial macroscopic height function. Let us define

$$ (\varGamma _ng) (x)=ng\left( \frac{x}{n}\right) ; $$

the job of the operator $\varGamma _n$ is to turn a macroscopic height function to its associated microscopic height function. We also write $T_t=T_t^H$ for the semigroup associated with the PDE (5.1). More precisely, $T_tu^0(x)=u(x,t)$ means

$$\begin{aligned} {\left\{ \begin{array}{ll}&{}u_t+H(x,u_x)=0,\ \ \ \ t>0,\\ &{}u(x,t)=u^0(x), \end{array}\right. } \end{aligned}$$

(5.2)

In terms of the operators $T_t$ and $\varGamma _n$, we simply have $u^n=\big (\varGamma _n^{-1}\circ T_{nt}\circ \varGamma _n\big )(g).$ Put it differently,

$$\begin{aligned} T^{H\circ \varGamma _n}_t=\varGamma _n^{-1}\circ T^H_{nt}\circ \varGamma _n, \end{aligned}$$

(5.3)

where $\gamma _n(x,p)=(nx,p)$. If we write T(H) for $T^H_1$, then in particular we have

$$ T({H\circ \gamma _n})=\varGamma _n^{-1}\circ T(H)^n\circ \varGamma _n. $$

The hope is that under some assumptions on H, the large n-limit of $u^n$ exists and the limit $\bar{u}$ provides a reduced and simpler description of the growth model under study. For example, when H is 1-periodic in x-variable, the high oscillations of $H\circ \gamma _n$, may result in the convergence of $u^n$ to a function $\bar{u}$, that solves the homogenized equation

$$\begin{aligned} \bar{u}_t+\bar{H}(\bar{u}_x)=0. \end{aligned}$$

(5.4)

When this happens, we write .

More generally, write ${\mathcal H}$ for the space of all $C^1$ Hamiltonian functions and define the natural translation operator

$$ \tau _a H(x,p)=H(x+a,p), $$

for every $a\in {\mathbb R}^d$. We then take a probability measure ${\mathbb P}$ on ${\mathcal H}$ that is translation invariant and ergodic. We wish to take advantage of the ergodicity to assert that $T^{H\circ \gamma _n}_t\rightarrow T^{\bar{H}}_t$, ${\mathbb P}$-almost surely, as $n\rightarrow \infty$. If this happens for a deterministic function $\bar{H}$, then we write . We note

• If ${\mathbb P}$ is supported on the set

  $$ A:=\big \{\tau _aH^0:\ a\in {\mathbb R}^d\big \}, $$

  for some 1-periodic Hamiltonian function $H^0$, then A is isomorphic to the d-dimensional torus and we are back to the periodic scenario.

• If ${\mathbb P}$ is supported on the topological closure (with respect to the uniform norm), of the set

  $$ A:=\big \{\tau _aH^0:\ a\in {\mathbb R}^d\big \}, $$

  for some Hamiltonian function $H^0$, and this closure is a compact set, then $H^0$ is almost periodic and the homogenization would allow us to find the large n-limit of $T^{H\circ \gamma _n}_t\rightarrow T^{\bar{H}}_t$, for almost all choices of H in the compact support of ${\mathbb P}$. In this case $\bar{A}$ has the structure of an Abelian Lie group and ${\mathbb P}$ is the corresponding Haar measure.

To explore the homogenization question further, we discuss the connection between Hamiltonian ODE and Hamilton-Jacobi PDE. For a classical solution, the method of characteristics suggests that at least for short times, we can solve (5.2) in terms of the flow of the Hamiltonian ODE

$$\begin{aligned} \dot{x}&=H_p(x,p),\\ \dot{p}&=-H_x(x,p).\nonumber \end{aligned}$$

(5.5)

Equivalently we write $\dot{z}=J\nabla H(z)$, where $z=(x,p)$, and

$$ J = \left[ {\begin{array}{*{20}l} 0 &{} I \\ { - I} &{} 0 \\ \end{array} } \right] , $$

with I denoting the $d\times d$ identity matrix. Writing $\phi _t=\phi ^H_t$ for the flow of (5.5), we have

$$\begin{aligned} \phi _t^H\left\{ \big (x,\nabla u^0(x)\big ):\ x\in {\mathbb R}^d\right\} = \left\{ \big (x, u_x(x,t)\big ):\ x\in {\mathbb R}^d\right\} , \end{aligned}$$

(5.6)

provided that the left-hand side remains a graph of a function. As we mentioned earlier, the Eq. (5.2) does not possess $C^1$ solutions in general. This has to do with the fact that if $\phi _t$ folds the graph of $\nabla u^0$, then the left-hand side of (5.6) is no longer a graph of a function and (5.6) has no chance to be true. One possibility is that we trim the left-hand side (5.6) and hope for

$$\begin{aligned} \phi _t^H\left\{ \big (x,\nabla u^0(x)\big ):\ x\in {\mathbb R}^d\right\} \supseteq \left\{ \big (x,u_x(x,t)\big ):\ x\in {\mathbb R}^d\right\} , \end{aligned}$$

(5.7)

For this to work, we have to give-up the differentiability of u. This geometric and rather naive idea does not suggest how the trimming should be carried out.

Alternatively, we may add a small viscosity term of the form $\varepsilon \varDelta u$ to the right-hand side of (5.1) to guarantee the existence of a unique classical solution, and pass to the limit $\varepsilon \rightarrow 0$. The outcome is known as a viscosity solution (see [12]). As it turns out, under some coercivity assumption on H, we can guarantee the existence of a solution that is differentiable almost everywhere. We can now modify the right-hand side of (5.7) accordingly and wonder whether or not

$$\begin{aligned} \phi _t^H\left\{ \big (x,\nabla u^0(x)\big ):\ x\in {\mathbb R}^d\right\} \supseteq \left\{ \big (x, u_x(x,t)\big ):\ x\in {\mathbb R}^d, \ \ u_x(x,t){\text { exists}}\right\} , \end{aligned}$$

(5.8)

is true. The answer is affirmative if H is convex in p. However (5.8) may fail if we drop the convexity assumption. To explain this in the case of piecewise smooth solutions, we recall that if H is convex in p, the only discontinuity we can have is a shock discontinuity. In this case, at every point (a, t), with $t>0$, we can find a solution $(x(s),p(s):s\in [0,t])$ (the so-called backward characteristic) such that $x(t)=a$. If $\rho =u_x$ is continuous at a, this backward characteristic is unique and $p(t)=\rho (a,t)$. If $\rho$ is discontinuous at (a, t), then $\rho (a,t)$ is multi-valued and for each possible value p of $\rho (a,t)$, there will be a solution to the Hamiltonian ODE with $(x(t),p(t))=(a,p)$. In both cases, we still have (5.8).

The situation is far more complex when H is not convex. What may cause the violation of (5.8) is the occurrence of a rarefaction type solutions. To explain this, let us assume that $d=1$, and H depends on p only. There are three momenta (or densities) $a_1<a_2<a_3$ such that

• The graph of H is convex and below its cord in $[a_1,a_2]$.

• The graph of H is concave and above its cord in $[a_2,a_3]$.

• The graph of H is below its cord in the interval $[a_1,a_3]$.

Now imagine that we have two discontinuities at x(t) and y(t) with $x(t)<y(t)$, and both are shock discontinuities. Assume

• The left and right values of $\rho$ at x(t) are $a'_2(t)<a'_3(t)$.

• The left and right values of $\rho$ at y(t) are $a'_3(t)>a'_1(t)$.

• These two shock discontinuities meet at some instant $t_0$ with $a_i'(t_0)=a_i$.

As a result, at the moment $t_0$ the two shock discontinuities are replaced with a rarefaction wave. Now if we take a point (x, t) inside the fan of this rarefaction wave (for which necessarily $t>t_0$), then at such (x, t) the connection with the initial data is lost and $(x,u_x(x,t))$ does not belong to the left-hand side of (5.8).

Motivated by the failure of (5.8) for viscosity solutions, we formulate a question.

Question 5.1:

Is there a notion of generalized solution for (5.1) for which (5.8) is always true?

Using some ideas from topology and symplectic geometry the notion of geometric solution has been developed by Chaperon [9,10,11], Sikorav [25] and Viterbo [27]. The main features of this solution is as follows:

• (i) The geometric solution satisfies (5.8) always.

• (ii) The geometric solution satisfies (5.2) at every differentiability point of u.

• (iii) The geometric solution coincides with the viscosity solution when H is convex in p.

• (iv) Writing $\hat{T}_tu^0$ for the geometric solution of (5.2) with the initial condition $u^0$, we do not in general have $\hat{T}_t\circ \hat{T}_s=\hat{T}_{t+s}$ (except when H is convex in p).

Needless to say the last feature of the geometric solution is a serious flaw and does not provide a satisfactory answer for Question 5.1. Nonetheless the geometric solution provides a useful notion that helps us to connect the Eq. (5.2) to the Hamiltonian ODEs.

Because of the intimate relation between the Hamilton-Jacobi Equation and the Hamiltonian ODE, we may wonder whether a homogenization phenomenon occurs for the latter. More precisely, does the high-n limit of

$$ \phi ^{H\circ \gamma _n}_t=\gamma _n^{-1}\circ \phi _{nt}^H\circ \gamma _n, $$

exist in a suitable sense? Note that $H\circ \gamma _n$ has no pointwise limit and the existence of pointwise limit of $\phi ^{H\circ \gamma _n}_t$ is not expected either. Writing $\phi _H$ for $\phi ^H_1$, we may wonder in what sense, if any, the sequence $\phi _{H\circ \gamma _n}$ has a limit. We note

$$ \phi _{H\circ \gamma _n}= \gamma _n^{-1}\circ \phi ^{n}_H\circ \gamma _n=:S_n(\phi _H). $$

We now discuss the existence of some interesting metric on the space ${\mathcal H}$ that is weaker than uniform norm and is closely related to the flow properties of the Hamiltonian ODEs. More importantly, there is a chance that $H\circ \gamma _n$ converges with respect to such metrics.

There are two metrics on ${\mathcal H}$ that are well-suited for our purposes. These metrics were defined by Hofer and Viterbo; the proofs of non-triviality of these metrics are highly non-trivial. Let us write down a wish-list for what our metric should satisfy.

Let us write ${\mathcal D}$ for the space of maps $\varphi$ such that $\varphi =\phi _H$ for some smooth Hamiltonian function $H: {\mathbb R}^{2d}\times [0,1]\rightarrow {\mathbb R}$. (Any such map is symplectic as we will see later.) Assume that there exists a function $E:{\mathcal D}\rightarrow [0,\infty )$ with the following properties: For $\varphi ,\psi ,\tau \in {\mathcal D}$,

(i) :

$E(\varphi )=E(\varphi ^{-1})$.

(ii) :

$E(\varphi )=E\big (\tau ^{-1}\varphi \tau )$.

(iii) :

$E(\varphi \psi )\le E(\varphi )+E(\psi )$.

(iv) :

$E(\varphi )=0$ if and only if $\varphi = id$.

(v) :

$E\big (\gamma _{\ell }^{-1}\varphi \gamma _\ell \big )=\ell ^{-1}E(\varphi )$, where $\gamma _\ell (x,p)=(\ell x,p)$ and $\ell \in (0,\infty )$.

Here and below we simply write $\varphi \psi$ for $\varphi \circ \psi$ and think of ${\mathcal D}$ as a group with multiplication given by the map composition.

From E, we build a metric D on ${\mathcal D}$ by $D(\varphi ,\psi )=E\big (\varphi \psi ^{-1}\big )$. This metric has the following properties:

Proposition 1

(i) $D(\varphi \tau ,\psi \tau )=D(\tau \varphi ,\tau \psi )=D(\varphi ,\psi )$ for $\varphi ,\psi ,\tau \in {\mathcal D}$.

(ii) For $\varphi _1,\psi _1\dots ,\varphi _k,\psi _k$, we have

$$ D(\varphi _1\dots \varphi _k,\psi _1\dots \psi _k)\le \sum _{i=1}^kD(\varphi _i,\psi _i). $$

(iii) For $S_n(\varphi )=\gamma _{n}^{-1}\circ \varphi ^n\circ \gamma _n,$ we have

$$ D\big (S_n(\varphi ),S_n(\psi )\big )\le D(\varphi ,\psi ). $$

In the case of a homogenization, we expect $S_n(\varphi )\rightarrow \bar{\varphi }$, where $\bar{\varphi }=\phi _{\bar{H}}$, for a Hamiltonian function $\bar{H}$ that is independent of x. Write ${\mathcal D}_0$ for the space of such $\bar{\varphi }$. We note that $S_n(\bar{\varphi })=\bar{\varphi }$. As a result, for any $\bar{\varphi } \in {\mathcal D}_0$,

$$\begin{aligned} D\big (S_n(\varphi ), \bar{\varphi }\big )= D\big (S_n(\varphi ),S_n(\bar{\varphi })\big )\le D\big (\varphi , \bar{\varphi }\big ), \end{aligned}$$

(5.9)

by Proposition 1(iii). As was noted by Viterbo [28], (5.9) implies that the set of limit points of the sequence $(S_n(\varphi ):\ n\in {\mathbb N})$ is a singleton: If $\bar{\varphi }$ and $\bar{\psi }$ are two limit points, then given $\delta >0$, we find $n,m\in {\mathbb N}$ such that

$$ D(S_n(\varphi ),\bar{\varphi })\le \delta ,\ \ \ \ D(S_{m}(\varphi ),\bar{\psi })\le \delta . $$

From this and (5.9) we learn,

$$ D(S_{nm}(\varphi ),\bar{\varphi })\le \delta ,\ \ \ \ D(S_{nm}(\varphi ),\bar{\psi })\le \delta , $$

because $S_{nm}=S_n\circ S_m$. Hence $D(\bar{\varphi },\bar{\psi })\le 2\delta$. By sending $\delta \rightarrow 0$ we deduce that $\bar{\varphi }=\bar{\psi }$.

A natural question is whether we have homogenization with respect to such a metric.

Question 5.2:

Given $\varphi \in {\mathcal D}$, does the large n limit of the sequence $\big \{S_n(\varphi )\big \}$ exist with respect to a metric D as above?     $\square$

6 Lagrangian Manifolds and Viterbo’s Metric

The Question 5.2 has been answered affirmatively by Viterbo [28] when the Hamiltonian H is periodic in x and the metric D is the Viterbo’s metric. We continue with a brief discussion of Viterbo’s metric.

To simplify our presentation, let us assume that H is 1-periodic in x. We may also regard $u(\cdot ,t)$ as a function on the d-dimensional torus ${\mathbb T}^d$.

To examine the left-hand side of (5.8), assume that the initially the solution of the ODE (5.5) satisfies the relationship $p=\nabla u^0(x)$, for some smooth function $u^0$. Whenever (5.6) is true, then at time t we have a similar relationship between the components of $\phi _t(x,p)$. Let us write $M^t:=\phi _t(M^0)$, where

$$ M^0=\big \{(x,\nabla u^0(x)):\ x\in {\mathbb T}^d\big \}. $$

To get a feel for $M^t=\phi _t^H(M^0)$, observe that $M^0$ is a graph of a an exact derivative. Let us refer to such manifolds as an exact Lagrangian graph. In general if

$$ M=\big \{(x,X(x)):\ x\in {\mathbb T}^d\big \}, $$

then vectors of the form

$$ \hat{a}:=\begin{bmatrix}a\\(DX)(x)a\end{bmatrix}, $$

are tangents to M at x. What makes M exact is that if $X=\nabla u$, then the matrix $A=DX=D^2u$ is symmetric. To state this directly in terms of the tangent vectors, observe

Hence the symmetry of A is equivalent to identically. (Here $\bar{\omega }$ is the standard symplectic 2-form of ${\mathbb R}^{2d}$.) Motivated by this we call a manifold M Lagrangian if the restriction of $\bar{\omega }$ to M is identically 0. The point of this definition is that if $M^0$ is the graph of an exact derivative, then $\varphi (M^0)$ may not be a graph of a function. However, when $\varphi$ preserves the form $\bar{\omega }$, then $\varphi (M^0)$ is always a Lagrangian. We say a map $\varphi$ is symplectic if it preserves $\bar{\omega }$ in the following sense:

$$ \bar{\omega }\big ((D\varphi )(x)a,(D\varphi )(x)b\big )=\bar{\omega }(a,b), $$

for every $x\in {\mathbb T}^d$ and every pair of vectors $a,b\in {\mathbb R}^{2d}$.

It is well-known that the correct topology for the viscosity solution comes from the uniform norm; this has to do with the fact the viscous approximation of Hamilton-Jacobi Equation satisfies a maximum principle that survives as we send the viscous term to 0. Since we are now interested in Hamiltonian ODE, we may try to define some kind of metrics on Lagrangian manifolds of the form $\phi _t^H\big (M^0\big )$, where $M^0$ is an exact Lagrangian. Let us write ${\mathcal L}_0$ for the set of exact Lagrangian graphs, and define

$$\begin{aligned} {\mathcal H}_0&=\big \{H:{\mathbb T}^d\times {\mathbb R}^d\times [0,1]\rightarrow {\mathbb R}:\ H\ {\text { is }} C^1 \ \ {\text { and 1-periodic in }}\ x\big \}\\ {\mathcal L}&=\big \{\phi _H(M):\ \ H\in {\mathcal H}_0, \ \ M \in {\mathcal L}_0\big \} \end{aligned}$$

When M is the graph of $\nabla u$, for some $C^1$ function $u:{\mathbb T}^d\rightarrow {\mathbb R}$, we refer to u as the generating function of M. When this is the case, we write ${\mathcal G}(M)=u$. We also write

$$ L(u):=\big \{(x,\nabla u(x)):\ x\in {\mathbb T}^d\big \}. $$

Viterbo defines a metric on ${\mathcal L}$ that is a generalization of the $L^\infty$-metric on its generating function. In other words, the metric D is defined in such way that if $M^0$ and $M^1$ are two exact Lagrangian graphs, then

$$ D\big (M,M')=\Vert {\mathcal G}(M)-{\mathcal G}(M')\Vert _{\infty }, $$

where by $\Vert \cdot \Vert _\infty$ we really mean the total oscillation:

$$ \Vert u\Vert _{\infty }=\max u-\min u. $$

This definition is quite natural because ${\mathcal L}(u)={\mathcal L}(u+c)$, for any constant c.

To guess how to extend the definition of this metric to ${\mathcal L}$, we need to develop a better understanding of the Hamiltonian ODEs. First, we claim that there exists a functional ${\mathcal I}={\mathcal I}^H$ on the space of the paths $z(\cdot )=(x,p)(\cdot )$, such that $\dot{z}=J\nabla H(z,t)$ if and only if $z(\cdot )$ is a critical point of ${\mathcal I}$. Writing the Hamiltonian ODE as $J\dot{z}+\nabla H(z,t)=0$, it is not hard to come up with an example for ${\mathcal I}$; we use a quadratic term to produce the linear part $J\dot{z}$, and H to produce $\nabla H$. The following function ${\mathcal I}:C^1([0, 1];{\mathbb T}^d\times {\mathbb R}^d)\rightarrow {\mathbb R}$, is the integral of the celebrated Cartan-Poincaré form:

$$ {\mathcal I}(z)=\int _0^1\left[ p(t)\cdot \dot{x}(t)-H(z(t),t)\right] \ dt. $$

Formally, $\partial {\mathcal I}(z)=-J\dot{z}-\nabla H(z,t)$. More precisely, if $\eta :[0,1]\rightarrow {\mathbb T}^d\times {\mathbb R}^d$, satisfies $\eta (0)=\eta (1)=0$, then $\psi (\delta )={\mathcal I}(z+\delta \eta )$ satisfies

$$ \dot{\psi }(0)=-\int _0^1\big (J\dot{z}(t)+\nabla H(z(t),t)\big )\cdot \eta (t)\ dt. $$

We now use this to come up with a generating-like function for $M^1=\phi _H\big (M^0\big )$, where $M^0={\mathcal G}(u^0)$. To this end, let us define

$$ \varGamma :=\big \{z:[0,1]\rightarrow {\mathbb T}^d\times {\mathbb R}^d:\ z\in C^1\big \},\ \ \ \ \varGamma (a)=\big \{z=(x,p)\in \varGamma \ : \ x(1)=a\big \}. $$

In words, $\varGamma (a)$ consists of position/momentum paths with the position component reaching a at time 1. We note that if $z\in \varGamma (a)$ and $\eta \in \varGamma (0)$, then $z+\delta \eta \in \varGamma (a)$ for all $\delta \in {\mathbb R}$. We then define $\hat{\mathcal I}:\varGamma (a)\rightarrow {\mathbb R}$ by

$$ \hat{\mathcal I}(z)=u^0(x(0))+{\mathcal I}(z)=u^0(x(0))+\int _0^1\left[ p(t)\cdot \dot{x}(t)-H(z(t),t)\right] \ dt. $$

Since we want to use $\hat{\mathcal I}$ to build a generating function for $M^1$, observe that $\varGamma (0)$ is an infinite dimensional vector space and any $z\in \varGamma (a)$ can be written as

$$ z(t)=(a,0)+\xi (t), $$

with $\xi \in \varGamma (0)$. If $M^1$ is still a graph of function and has a generating function $u^1$, then what is happening is that we have a solution z satisfying $\dot{z}=J\nabla H(z,t)$ with

$$ z(0)=\big (x(0),\nabla u^0(x(0))\big ),\ \ \ \ z(1)=\big (x(1),\nabla u^1(x(1))\big ). $$

Moreover, if u solves (1.1), then $u^1(x)=u(x,1)$. Note that if $w(t)=u(x(t),t)$, then $\dot{w}=p\cdot \dot{x}-H(z,t)$, or

$$ u^1(x(1))=u^0(x(0))+\int _0^1\left[ p(t)\cdot \dot{x}(t)-H(z(t),t)\right] \ dt. $$

To separate x(1) from the rest of information in the path $z(\cdot )$, we define ${\mathcal J}:{\mathbb T}^d\times \varGamma (0)\rightarrow {\mathbb R}$, by

$$ {\mathcal J}(a;\xi )=\hat{\mathcal I}\big ((a,0)+\xi \big ). $$

In other words, if $z=(a,0)+\xi =(x,p)$, and $\xi =(x',p)$, then $x'(t)=x(t)-x(1)=x(t)-a$. Now, if we set

$$ \hat{\psi }(\delta )=\hat{\mathcal I}(z+\delta \eta )={\mathcal J}(a;\xi +\delta \eta ), $$

for $z\in \varGamma (a)$, and $\eta =(\hat{x},\hat{p})\in \varGamma (0)$, then

$$ \frac{d\hat{\psi }}{d\delta }(0)=\big (\nabla u^0(x(0))-p(0)\big )\cdot \hat{x}(0) -\int _0^1\big (J\dot{z}(t)+\nabla H(z(t),t)\big )\cdot \eta (t)\ dt. $$

We can now assert

$$ \partial _{\xi }{\mathcal J}(a;\xi )=0\ \ \ \ \iff \ \ \ \ p(0)=\nabla u^0(x(0)),\ \ {\text { and }}\ \ z=(a,0)+\xi \ \ \ {\text { satisfies }} \dot{z}=J\nabla H(z,t). $$

On the other hand, if we set $\bar{\psi }(\delta )=\hat{\mathcal I}\big (z+(\delta b,0)\big ) ={\mathcal J}(a+\delta b;\xi )$, then

$$ \partial _a{\mathcal J}(a;\xi )\cdot b=\frac{d\bar{\psi }}{d\delta }(0)=\nabla u^0(x(0))\cdot b -\int _0^1 H_x(z(t),t)\cdot b\ dt. $$

As a result, if $\partial _{\xi }{\mathcal J}(a;\xi )=0$, then

$$ \partial _a{\mathcal J}(a;\xi )= \nabla u^0(x(0)) -\int _0^1 H_x(z(t),t)\ dt=\nabla u^0(x(0))+\int _0^1 \dot{p}(t)\ dt=p(1). $$

From this we deduce

$$ \phi ^H\big (M^0\big )=\left\{ \big (a, {\partial _a {\mathcal J}}(a,\xi )\big ):\ a\in {\mathbb T}^d,\ \ {\partial _ { \xi }{\mathcal J}}(a,\xi )=0\right\} , $$

where $a=x(1)$ represents the position at time 1. We think of ${\mathcal J}(a;\xi )$ as a generalized generating function (or in short GG function) of $M=M^1$. The Lagrangian $M^1$ is exact if for every $(a,p)\in {\mathbb T}^d\times {\mathbb R}^d$, there is at most one solution z to the Hamiltonian ODE with $x(1)=a, p(1)=p.$ Our aim is to associate a nonnegative number E(M) to $M\in {\mathcal L}$ that in the case of an exact Lagrangian $M={\mathcal G}(u)$,

$$ E(M)=E^+(M)-E^-(M), $$

where $E^\pm (M)$ are two critical values of u, namely the maximum and minimum of u. In the case of a non-exact M, we may use the functional ${\mathcal J}={\mathcal J}_M$ to select two critical points $z^\pm =(a^\pm ,\xi ^\pm )$ of the functional ${\mathcal J}_M$ to define

$$ E^\pm (M)={\mathcal J}_M(a^\pm ,\xi ^\pm )=\hat{\mathcal I}(z^\pm ). $$

The main question now is how to select the critical paths $z^\pm$. The classical theories of Morse and Lusternik-Schnirelman would provide us with systematic ways of selecting critical values of a scalar-valued function on a manifold. (See for example Appendix E of [21] for an introduction on LS Theory.) These theories are applicable if the underlying manifold is finite-dimensional and their generalizations to infinite dimensional setting are highly nontrivial. (Floer Theory is a prime example of such generalization.) However in our setting it is possible to approximate the functional ${\mathcal I}$ or ${\mathcal J}$ with a function that is defined on ${\mathbb T}^d\times {\mathbb R}^N$ for a suitable N that depends on H and $u^0$ and could be large. More precisely, we may try to find a generating function $S:{\mathbb T}^d\times {\mathbb R}^N\rightarrow {\mathbb R}$ such that

$$ M=\big \{(x, S_x(x,\xi )):\ x\in {\mathbb T}^d,\xi \in {\mathbb R}^N, S_\xi (x,\xi )=0\big \}, $$

In fact any manifold of this form is automatically a Lagrangian manifold, simply because the tangent vectors at a point of the form $(x, S_x(x,\xi ))$ are still of the form $\big (v, A(x,\xi )v\big ); v\in {\mathbb R}^d$, where $A=S_{xx}$ is a symmetric matrix.

To explain the existence of such finite dimensional generating functions, we need to make another observation about the flows of Hamiltonian ODEs.

We may regard the symplectic property of $\varphi =\phi _1^H$, as saying that its graph

$$ Gr(\varphi ):\big \{(x,\varphi (x)):\ x\in {\mathbb T}^d\times {\mathbb R}^d\big \}, $$

is Lagrangian with respect to the 2-form $\omega \oplus (-\omega )$ in ${\mathbb R}^{4d}$. This Lagrangian manifold is an exact graph when the set $Gr(\varphi )$ can be expressed as a graph of the gradient of a scalar-valued function. But now because of the form of the symplectic form $\omega \oplus (-\omega )$, this must be done in a twisted way. More precisely, if $\varphi (x,p)=(X,P)$, then the generating function would depend for example on (X, p). In the case of an exact symplectic map, we may find a scalar-valued function S(X, p) such that

$$ \varphi \big (S_p(X,p),p\big )=\big (X,S_X(X,p)\big ). $$

The identity map has the generating function $p\cdot X$. This suggests writing $S(X,p)=X\cdot p-w(X,p)$ with w periodic in X. In terms of w,

$$ \varphi \big (X-w_p(X,p),p\big )=\big (X,p-w_X(X,p)\big ). $$

Now imagine that $M=\varphi (M^0)$, where both $M^0$ and $\varphi$ are exact with generating functions $u^0$ and $S(X,p)=X\cdot p-w(X,p)$. Then

$$ \hat{S}(X;x,p)=u^0(x)+p\cdot (X-x)-w(X,p)=:p\cdot (X-x)-\hat{w}(X;x,p), $$

is a GG function for $M^1$: If $\xi =(x,p)$, then

$$ \hat{S}_{\xi }(X;\xi )=0\ \ \ \iff \ \ \ p=\nabla u^0(x),\ \ \ x=X-w_p(X,p). $$

As a result

$$ \hat{S}_{\xi }(X;\xi )=0\ \ \ \implies \ \ \ \varphi (x,p)=\big (X, \hat{S}_X(X;\xi )\big ), $$

because $\hat{S}_X=p-w_X(X,p)=P$.

As we mentioned earlier, the identity map has a generating function. Using Implicit Function Theorem, it is not hard to show that any symplectic map that is $C^1$-close to the identity also possesses a generating function. Now if $\varphi =\phi _H$ is the time-one map associated with a smooth Hamiltonian, then we can find $\delta >0$ sufficiently small, such that the map $\varphi =\phi ^H_\delta$ is sufficiently close to the identity map and possesses a generating function. In general, each $\phi ^H$ can be expressed as $\varphi ^1\circ \dots \circ \varphi ^N$ with each $\varphi ^i$ possessing a generating function as above. If each $\varphi ^i$ has a generating function of the form $X\cdot p-w^i(X,p)$, then $M=\varphi (M^0)$ has a generating function of the form

$$ \hat{S}\big (x_N;\xi \big )=\hat{S}\big (x_N;x_0,p_0,\dots ,x_{N-1},p_{N-1}\big ): =u^0(x_0)+\sum _{i=0}^{N-1}\big [p_i\cdot (x_{i+1}-x_i)-w^i(x_{i+1},p_i)\big ]. $$

We refer to [22] and Chapter 9 of [21] for more details on generating functions.

So far we know that our Lagrangian manifolds possess finite-dimensional generating functions. The next question to address is that how we can select appropriate critical values $E^\pm (M)$ for $\hat{S}(X;\xi )$.

For the rest of this section, we assume that M is a Lagrangian manifold with a generating function $S(x,\xi )$. More precisely,

$$\begin{aligned} M=\big \{(x, S_x(x,\xi )):\ x\in {\mathbb T}^d,\xi \in {\mathbb R}^N, S_\xi (x,\xi )=0\big \}, \end{aligned}$$

(6.1)

and $S(x,\xi )$ is a nice perturbation of a quadratic function in $\xi$. By this we mean that there exists a quadratic function $B(\xi )= A\xi \cdot \xi$ such that A is an invertible symmetric matrix, and

$$ \sup _{x,\xi }|S(x,\xi )-B(\xi )|,\ \ \ \ \sup _{x,\xi }|S_{\xi }(x ,\xi )-\nabla B(\xi )|<\infty . $$

We wish to put a metric on the space ${\mathcal L}$ of such Lagrangians. For this, we first wish to define the size E(M) of a Lagrangian manifold M. If M is an exact Lagrangian graph with generating function u, we simply set

$$ E(M)=\max u-\min u. $$

If M can be represented as in (6.1), then E(M) is defined by

$$ E(M)=E^+(M)-E^-(M), $$

where $E^-(M)$ and $E^+(M)$ are two critical values of the generating function S that are the analog of $\min u$ and $\max u$. To explain our strategy for selecting $E^{\pm }(M)$, first imagine that $S(x,\xi )=u(x)+B(\xi )$. Then we still have $E^-(M)=\min u=u(x_-)$ and $E^+(M)=\max u=u(x_+)$, because both $(x_{\pm },0)$ are critical points of S. After all 0 is a critical value for B. We may apply Lusternik-Schnirelman (LS) Theory, to assert that the function S also has two critical points that are very much the analogs of $(x_\pm ,0)$. (See [28] and Appendix E of [21].) We are now ready to define a metric on the space Lagrangian manifolds who possess generating function as in (6.1). If M and $M'$ are two Lagrangian manifolds with generating functions S and $S'$ respectively, then we define a new generating function

$$ \big (S\ominus S'\big )(x,\xi _1,\xi _2)=S(x,\xi _1)-S'(x,\xi _2). $$

This new generating function produces a new Lagrangian manifold

$$\begin{aligned} M\ominus M'&=\big \{(x, S_x(x,\xi _1)-S'_x(x,\xi _2)):\ x\in {\mathbb T}^d,\xi \in {\mathbb R}^N,\xi \in {\mathbb R}^{N'},S_\xi (x,\xi )=0, S'_\xi (x,\xi _2)=0\big \}\\&=\big \{(x,p-p'):\ (x,p)\in M,\ (x,p')\in M'\big \}. \end{aligned}$$

This generating function is a bounded perturbation of $\big (B\ominus B'\big )(\xi _1,\xi _2)=B(\xi _1)-B(\xi _2)$. We set

$$ D(M,M')=E\big (S\ominus S'\big ). $$

We now would like to use the above metric to define a metric for Hamiltonian functions or their corresponding flows that was defined by Viterbo:

$$ {\mathcal D}(H,H')=\sup \left\{ D\big (\phi _H(M),\phi _{H'}(M)\big ): \ M\in {\mathcal L}\right\} . $$

Theorem 4

(Viterbo [28]) The large n-limit of $H\circ \gamma _n$ exists with respect to the Viterbo Metric D. Moreover, if the limit is denoted by ${\mathcal B}(H)$, then ${\mathcal B}$ satisfies the following properties

• (i) For every symplectic $\varphi \in {\mathcal D}$, we have ${\mathcal B}(H\circ \varphi )={\mathcal B}(H)$.

• (ii) If $\{H,K\}:=J\nabla H\cdot K=0$, then ${\mathcal B}(H+K)={\mathcal B}(H)+{\mathcal B}(K)$.

This should be compared with the Lions-Papanicolaou-Varadhan [18] homogenization result.

Theorem 5

Assume that H(x, p) is a $C^1$, x-periodic Hamiltonian function with

$$ \lim _{|p|\rightarrow \infty }\inf _x H(x,p)=\infty . $$

Then the large n limit of $T^{H\circ \gamma _n}$ exists. The limit is of the form $T^{\bar{H}}$, for a Hamiltonian function that is independent of x.

In fact when H is convex in p; otherwise they could be different. Moreover, Theorem 6.2 has been extended to the random ergodic setting when H is convex in p in Rezakhanlou-Tarver [23] and Souganidis [24]. A natural question is whether or not Theorem 6.1 can be extended to the random setting.

Question 6.1:

Can we extend Viterbo’s metric (or Hofer’s metric) to the random setting and does the large n limit of $H\circ \gamma _n$ exist for a stationary ergodic Hamiltonian H?     $\square$