Random Tessellations and Gibbsian Solutions of Hamilton–Jacobi Equations

Abstract

We pursue two goals in this article. As our first goal, we construct a family ${\mathcal M}_G$ of Gibbsian measures on the set of piecewise linear convex functions $g:{\mathbb R}^2\rightarrow {\mathbb R}$. It turns out that there is a one-to-one correspondence between the gradient of such convex functions and Laguerre tessellations. Each cell in a Laguerre tessellation is a convex polygon that is marked by a vector $\rho \in {\mathbb R}^2$. Each measure $\nu ^f\in {\mathcal M}_G$ in our family is uniquely characterized by a kernel $f(x,\rho ^-,\rho ^+)$, which represents the rate at which a line separating two cells associated with marks $\rho ^-$ and $\rho ^+$ passes through x. To construct our measures, we give a precise recipe for the law of the restriction of our tessellation to a box. This recipe involves a boundary condition, and a dynamical description of our random tessellation inside the box. As we enlarge the box, the consistency of these random tessellations requires that the kernel satisfies a suitable kinetic like PDE. Our interest in the family ${\mathcal M}_G$ stems from its close connection to Hamilton–Jacobi PDEs of the form $u_t=H(u_x)$, with $u:\Gamma \rightarrow {\mathbb R}$, for a convex set $\Gamma \subset {\mathbb R}^2\times [0,\infty )$, and $H:{\mathbb R}^2\rightarrow {\mathbb R}$ a strictly convex function. As our second goal, we study the invariance of the set ${\mathcal M}_G$ with respect to the dynamics of such Hamilton–Jacobi PDEs. In particular we conjecture the invariance of a suitable subfamily $\widehat{\mathcal M}_G$ of ${\mathcal M}_G$. More precisely, we expect that if the initial slope $u_x(\cdot ,0)$ is selected according to a measure $\nu ^{f}\in \widehat{{\mathcal M}}_G$, then at a later time the law of $u_x(\cdot , t)$ is given by a measure $\nu ^{\Theta _t(f)}\in \widehat{{\mathcal M}}_G$, for a suitable kernel $\Theta _t(f)$. As we vary t, the kernel $\Theta _t(f)$ must satisfy a suitable kinetic equation. We remark that the function u is also piecewise linear convex function in (x, t), and its law is an example of a Gibbsian measure on the set of Laguerre tessellations of certain convex subsets of ${\mathbb R}^3$.

Appendix A: The Kinetic Equation

The purpose of this section is to prove the existence of a solution of the kinetic equation. To have a more conventional notation, we write (x, t) for $(x_1,x_2)$ throughout this section. We start first with the following notation:

Notation A.1

(i) :

We fix $P^-<P^+$ two real numbers, such that the range of our piecewise constant function $\rho$ is in the box $[P^-,P^+]^2$.

(ii) :

For any measure space ${\mathcal E}$, let ${\mathcal F}_b({\mathcal E})$ be the space of real-valued bounded measurable functions defined on ${\mathcal E}$.

(iii) :

We introduce the function space ${\mathcal X}$ to be the set kernels $h \in {\mathcal F}_b({\mathbb R}\times ([P^-,P^+]^2)^2)$ such that $x \mapsto h(x,\rho ^-,\rho ^+)$ is $C^1$ and Lipschitz for all $\rho ^-$ and $\rho ^+$.

(iv) :

We equip ${\mathcal X}$ with the following norm

$$\begin{aligned} \Vert h\Vert _{{\mathcal X}}:= \sup _{x \in {\mathbb R}} \ \sup _{\rho ^-,\rho ^+}\ \left[ |h(x,\rho ^-,\rho ^+)| + |\partial _x h(x,\rho ^-,\rho ^+)| \right] . \end{aligned}$$

It is standard that $({\mathcal X},\Vert \cdot \Vert _{{\mathcal X}})$ is a Banach space.

(v) :

For any $v \ge 0$, let $\Gamma ^{v}$ and $\Gamma ^v_+$ be the sets

$$\begin{aligned} \Gamma ^{v}:=&\Big \{(\rho ^-,\rho ^+) \in ([P^-,P^+]^2)^2 : \ \rho ^-\prec \rho ^+, \ |[\rho ^-,\rho ^+]| \le v \Big \}\\ =&\Big \{(\rho ^-,\rho ^+) \in ([P^-,P^+]^2)^2 : \ \rho ^+-\rho ^-\in C^v {\setminus } \{0\}\Big \},\\ \Gamma ^{v}_+: =&\Big \{(\rho ^-,\rho ^+) \in ([P^-,P^+]^2)^2 : \ \rho ^+-\rho ^-\in C^v_+ {\setminus } \{0\}\Big \}, \end{aligned}$$

where $C^v$ and $C^v_+$ are the cones

$$\begin{aligned} C^v&=\Big \{m=(m_1,m_2)\in {\mathbb R}^2:\ m_1\ge 0, \ |m_2| \le vm_1\Big \},\nonumber \\ C_+^v&=\Big \{m=(m_1,m_2)\in {\mathbb R}^2:\ m_1,m_2\ge 0, \ m_2 \le vm_1\Big \}. \end{aligned}$$

(A.1)

(vi) :

Let $V_{\infty } \ge 0$, and $\delta _0>0$. We write ${\mathcal X}(V_\infty ,\delta _0)$ for the set of $h \in {\mathcal X}$ with the following properties:

(1):

The function $h(\cdot ,\rho ^-,\rho ^+)$ is zero for $(\rho ^-,\rho ^+) \notin \Gamma ^{V_{\infty }}$.

(2):

There exists a constant $\delta _0>0$ such that $\inf _{x \in {\mathbb R}} \inf _{(\rho ^-,\rho ^+) \in \Gamma ^{V_{\infty }}} h(x,\rho ^-,\rho ^+) \ge \delta _0$.

(3):

For all $\rho ^-,\rho ^+$, the function $x \mapsto h(x,\rho ^-,\rho ^+)$ is $C^2$ such that

$$\begin{aligned} \sup _{x \in {\mathbb R}} \sup _{\rho ^-,\rho ^+} |\partial ^2_x h(x,\rho ^-,\rho ^+)| < \infty . \end{aligned}$$

Likewise, we write ${\mathcal X}_+(V_\infty ,\delta _0)$ for the set of $h \in {\mathcal X}_+(V_\infty ,\delta _0)$ with the similar properties, except that the set $\Gamma ^{V_{\infty }}$ in (i) and (iii) is replaced with $\Gamma ^{V_{\infty }}_+$.

$\square$

The following theorem proves the existence of a local solution of the kinetic equation.

Theorem A.1

Given $h\in {\mathcal X}(V_\infty ,\delta _0)$, denote by $M_0:=\sup _{x \in {\mathbb R}} \sup _{\rho ^-,\rho ^+} h(x,\rho ^-,\rho ^+)$, and define the time

$$\begin{aligned} T^{*}:=\min \left( \frac{1}{12V_{\infty }M_0},\ \frac{\delta _0}{48V_{\infty }M_0^2}\right) . \end{aligned}$$

Then, there exists a unique solution

$$\begin{aligned} f: {\mathbb R}\times [0,T^{*}] \times ([P^-,P^+]^2)^2 \rightarrow {\mathbb R}, \end{aligned}$$

of the kinetic equation

$$\begin{aligned} f_t-[\rho ^-,\rho ^+]f_x=Q(f)=:Q^{+}(f)-Q^{-}(f), \end{aligned}$$

where

$$\begin{aligned}&Q^{+}(f)(x,t,\rho ^-,\rho ^+)\\&\quad =\int \left( [\rho ^+,\rho ^*]-[\rho ^*,\rho ^-]\right) f(x,t,\rho ^-,\rho ^*)f(x,t,\rho ^*,\rho ^+)\ \beta (d\rho ^*),\\&Q^{-}(f)(x,t,\rho ^-,\rho ^+)\\&\quad = \left( \int ([\rho ^+,\rho ^*]-[\rho ^-,\rho ^+])f(x,t,\rho ^+,\rho ^*)\ \beta (d\rho ^*) \right. \\&\left. \qquad - \int ([\rho ^-,\rho ^*]-[\rho ^-,\rho ^+]) f(x,t,\rho ^-,\rho ^*)\ \beta (d\rho ^*) \right) f(x,t,\rho ^-,\rho ^+), \end{aligned}$$

with $f(\cdot ,0,\cdot ,\cdot )=h$. The function f is $C^1$ in the variables (x, t) for all fixed $\rho ^-,\rho ^+$ and $f(\cdot ,\cdot ,\rho ^-,\rho ^+) \equiv 0$ for all $(\rho ^-,\rho ^+) \notin \Gamma ^{V_{\infty }}$. Furthermore, we have that

$$\begin{aligned} \sup _{t \in [0,T^{*}]} \Vert f(\cdot ,t,\cdot ,\cdot )\Vert _{{\mathcal X}} < \infty ,\quad \inf _{t \in [0,T^{*}]} \inf _{x \in {\mathbb R}} \inf _{(\rho ^-,\rho ^+) \in \Gamma ^{V_{\infty }}} f(x,t,\rho ^-,\rho ^+) \ge \frac{\delta _0}{2}. \end{aligned}$$

Moreover if $h\in {\mathcal X}_+(V_\infty ,\delta _0)$, then there exists a unique solution f with similar properties except that the set $\Gamma ^{V_{\infty }}$ must be replaced with $\Gamma ^{V_{\infty }}_+$.

Proof

(Step 1) We assume here without loss of generality that the measure $\beta$ has total mass 1 on the box $[P^-,P^+]^2$. By the following standard change of variables, we transform the previous PDE to an ODE. For instance, define the function g as

$$\begin{aligned} g(x,t,\rho ^-,\rho ^+)=f(x-[\rho ^-,\rho ^+]t,t,\rho ^-,\rho ^+). \end{aligned}$$

(A.2)

Then by the chain rule, g must verify the following ODE

$$\begin{aligned} g_t=\tilde{Q}^{+}(g)-\tilde{Q}^{-}(g) =\tilde{Q}^{+}(g)-\tilde{L}(g)g, \end{aligned}$$

where

$$\begin{aligned} \tilde{Q}^{+}(g)(x,t,\rho ^-,\rho ^+)=&\int [\rho ^-,\rho ^*,\rho ^+]\ g\left( x-([\rho ^-,\rho ^+]-[\rho ^-,\rho ^*])t,t,\rho ^-,\rho ^*\right) \\&\quad g(x-([\rho ^-,\rho ^+]-[\rho ^+,\rho ^*])t,t,\rho ^*,\rho ^+)\ \ \beta (d\rho ^*),\\ \tilde{L}(g)(x,t,\rho ^-,\rho ^+)=&\int ([\rho ^+,\rho ^*]-[\rho ^-,\rho ^+])\ g(x-([\rho ^-,\rho ^+]\\&\quad -[\rho ^+,\rho ^*])t,t,\rho ^+,\rho ^*)\ \beta (d\rho ^*) - \int ([\rho ^-,\rho ^*]-[\rho ^-,\rho ^+])\\&\quad g(x-([\rho ^-,\rho ^+]-[\rho ^-,\rho ^*])t,t,\rho ^-,\rho ^*)\ \beta (d\rho ^*). \end{aligned}$$

We will prove the existence of a solution g by an approximation scheme and then recover the desired f via the equation (A.2). Define the functional ${\mathcal H}: {\mathcal X}\times {\mathbb R}\rightarrow {\mathcal X}$, ${\mathcal H}(g,t):={\mathcal H}^+(g,t)-{\mathcal K}(g,t)g$, by

$$\begin{aligned} {\mathcal H}^+(h,t)(x,\rho ^-,\rho ^+) :=&\int [\rho ^+,\rho ^*,\rho ^-]\ h(x-([\rho ^-,\rho ^+]-[\rho ^-,\rho ^*])t,\rho ^-,\rho ^*) \\&\quad h(x-([\rho ^-,\rho ^+]-[\rho ^+,\rho ^*])t,\rho ^*,\rho ^+)\ \beta (d\rho ^*),\\ {\mathcal K}^+(h,t)(x,\rho ^-,\rho ^+) :=&\int ([\rho ^+,\rho ^*]-[\rho _-,\rho ^+])\\&\quad h(x-([\rho ^-,\rho ^+]-[\rho ^+,\rho ^*])t,\rho ^+,\rho ^*)\ \beta (d\rho ^*) \\&\quad - \int ([\rho ^-,\rho ^*]-[\rho ^-,\rho ^+])\\&\quad h(x-([\rho ^-,\rho ^+]-[\rho ^-,\rho ^*])t,\rho ^-,\rho ^*)\ \beta (d\rho ^*). \end{aligned}$$

Our goal is to prove the existence of a local solution $g: [0,T^{*}] \mapsto {\mathcal X}$ to the inhomogeneous ODE

$$\begin{aligned} \dot{g}(t)={\mathcal H}(g(t),t) \end{aligned}$$

(A.3)

under the initial condition $g(0)=h$. As the function space $({\mathcal X},\Vert \cdot \Vert _{{\mathcal X}})$ is clearly Banach, we will construct a Cauchy sequence $(g_n)_{n \in {\mathbb N}}$ of elements in $C([0,T^{*}],{\mathcal X})$ that will converge to our desired solution g.

(Step 2) For any fixed $n \in {\mathbb N}$, we define the polygonal function $g_n$ such that $g_n(0)=h$ and

$$\begin{aligned} \dot{g}_n(t)={\mathcal H}\left( g_n\left( \frac{j}{n}\right) ,\frac{j}{n} \right) \text { for all } t \in \left( \frac{j}{n},\frac{j+1}{n}\right) \end{aligned}$$

for all $j \ge 0$. Let us denote $g_n^{j}=g_n \left( \frac{j}{n} \right)$, then it is clear that all $g_n^{j}$ are $C^2$ in the variable x. We have that

$$\begin{aligned} n(g_n^{j+1}-g_n^{j})={\mathcal H}\left( g_n^{j},\frac{j}{n} \right) . \end{aligned}$$

(A.4)

Let us prove first that $g_n^{j}(\cdot ,\rho ^-,\rho ^+) \equiv 0$ for all $(\rho ^-,\rho ^+) \notin \Gamma ^{V_{\infty }}$ by induction on j. Suppose this is true for j and we wish to prove it for $j+1$. Take $x \in {\mathbb R}$, $(\rho ^-,\rho ^+) \notin \Gamma ^{V_{\infty }}$, and take any $\rho ^*$ such that $\rho ^- \prec \rho ^* \prec \rho ^+$. Since $C^v$ of (A.1) is a cone, we have that either $(\rho ^-,\rho ^*) \notin \Gamma ^{V_{\infty }}$ or $(\rho ^*,\rho ^+) \notin \Gamma ^{V_{\infty }}$. In either cases

$$\begin{aligned}&g_n^{j}\left( x-\left( [\rho ^-,\rho ^+]-[\rho ^-,\rho ^*]\right) \frac{j}{n},\rho ^-,\rho ^*\right) \\&\quad g_n^{j}\left( x-([\rho ^-,\rho ^+]-[\rho ^+,\rho ^*])\frac{j}{n},\rho ^*,\rho ^+\right) = 0, \end{aligned}$$

by the induction hypothesis. As a result, $g_n^{j+1}(x,\rho ^-,\rho ^+) = 0$, as desired.

Next, let us define

$$\begin{aligned} m_j:&=\inf _{x \in {\mathbb R}} \inf _{(\rho ^-,\rho ^+) \in \Gamma ^{V_{\infty }}} g_n^{j}(x,\rho ^-,\rho ^+),\qquad M_j:=\sup _{x \in {\mathbb R}} \sup _{(\rho ^-,\rho ^+) \in \Gamma ^{V_{\infty }}} |g_n^{j}(x,\rho ^-,\rho ^+)|,\\ M'_j:&=\sup _{x \in {\mathbb R}} \sup _{(\rho ^-,\rho ^+) \in \Gamma ^{V_{\infty }}} |\partial _x g_n^{j}(x,\rho ^-,\rho ^+)|, \!\quad M''_j:=\sup _{x \in {\mathbb R}} \sup _{(\rho ^-,\rho ^+) \in \Gamma ^{V_{\infty }}} \!|\partial ^2_x g_n^{j}(x,\rho ^-,\rho ^+)|, \end{aligned}$$

It is clear from the expression of ${\mathcal H}$ that we have for all $j \ge 0$,

$$\begin{aligned} n(m_{j+1}-m_j) \ge -6V_{\infty }M_j^2,\quad n(M_{j+1}-M_j) \le 6V_{\infty } M_j^2. \end{aligned}$$

(A.5)

Let us prove first by induction on j the following inequality,

$$\begin{aligned} rj<1\quad \implies \quad M_j \le M_0 (1-rj)^{-1}. \end{aligned}$$

(A.6)

where $r=\frac{6V_{\infty }M_0}{n}$. The verification for $j=0$ is trivial. Assume it is true for j, then from the second inequality in (A.5), it suffices to prove that

$$\begin{aligned} (1-rj)^{-1} \left( 1+r(1-rj)^{-1}\right) \le \left( 1-r(j+1) \right) ^{-1}. \end{aligned}$$

This inequality is equivalent to

$$\begin{aligned} (1-(j-1)r)(1-(j+1)r) \le (1-jr)^2, \end{aligned}$$

which is clearly true. As an immediate consequence of (A.6) we have that

$$\begin{aligned} \sup _{t \in [0,T^{*}]} \Vert g_n(t)\Vert _{L^{\infty }} \le M_0 \sup _{t \in [0,T^{*}]} \left( 1-\frac{6 \left\lfloor nt \right\rfloor V_{\infty } M_0}{n}\right) ^{-1} \le 2M_0. \end{aligned}$$

By differentiating the identity (A.4), we also have that

$$\begin{aligned} n(M'_{j+1}-M'_j) \le 12V_{\infty } M'_j M_j, \end{aligned}$$

So for all j such that $\frac{j+1}{n} \le T^*$, we have that

$$\begin{aligned} M'_{j+1} \le M'_j\left( 1+\frac{24V_{\infty }M_0}{n}\right) , \end{aligned}$$

from which it follows that

$$\begin{aligned} M'_j \le M'_0\left( 1+\frac{24V_{\infty }M_0}{n}\right) ^j\le M'_0 e^{\frac{2j}{nT^*}}\le M'_0 e^2, \end{aligned}$$

and hence

$$\begin{aligned} \sup _{t \in [0,T^{*}]} \Vert \partial _x g_n(t)\Vert _{L^{\infty }} \le M'_0e^2. \end{aligned}$$

Likewise, by differentiating twice the identity (A.4), we get that

$$\begin{aligned} n(M''_{j+1}-M''_j) \le 12V_{\infty }M_jM''_j+12V_{\infty }(M'_j)^2\le 24V_{\infty }M_0 M''_j+12V_{\infty }(M'_0)^2e^4. \end{aligned}$$

From this, it follows by similar arguments as before that

$$\begin{aligned} M''_j \le \left( M''_0+\frac{(M'_0)^2e^4}{2M_0}\right) e^{\frac{24V_{\infty }M_0j}{n}}, \end{aligned}$$

and hence

$$\begin{aligned} \sup _{t \in [0,T^{*}]} \Vert \partial ^2_x g_n(t)\Vert _{L^{\infty }} \le \left( M''_0+\frac{(M'_0)^2e^4}{2M_0}\right) e^{2}. \end{aligned}$$

Now since we have for every j such that $\frac{j+1}{n} \le T_{*}$,

$$\begin{aligned} m_{j+1} \ge m_j -\frac{24V_{\infty }M_0^2}{n}, \end{aligned}$$

it follows easily that

$$\begin{aligned} \inf _{t \in [0,T_{*}]} \inf _{x \in {\mathbb R}} \inf _{(\rho ^-,\rho ^+) \in \Gamma ^{V_{\infty }}} g_n(t)(x,\rho ^-,\rho ^+) \ge \frac{\delta _0}{2} >0. \end{aligned}$$

We have hence proved that all the approximating functions $(g_n)_{n \in {\mathbb N}} \in {\mathcal C}([0,T^{*}],{\mathcal X})$ are supported on $\Gamma ^{V_{\infty }}$ in the $(\rho ^-,\rho ^+)$ variables, and are uniformly bounded from above and below by positive constants in their supports.

(Step 3) To finish the proof, we shall show that the sequence $\{g_n\}$ is Cauchy. This is achieved by obtaining Lipschitz estimates on $g_n$. Observe that for any $s<t$, and $k_1,k_2 \in {\mathcal X}$ that are $C^2$ in the x-variable and supported on $\Gamma ^{V_{\infty }}$ such that

$$\begin{aligned} \max (\Vert \partial ^2_x k_1\Vert _{L^{\infty }},\Vert \partial ^2_x k_2\Vert _{L^{\infty }})<\infty , \end{aligned}$$

it is straightforward to show

$$\begin{aligned} \Vert {\mathcal H}(k_1,t)-{\mathcal H}(k_2,t)\Vert _{{\mathcal X}} \le&6V_{\infty }\left( \Vert k_1\Vert _{{\mathcal X}} +\Vert k_2\Vert _{{\mathcal X}}\right) \Vert k_1-k_2\Vert _{{\mathcal X}},\\ \Vert {\mathcal H}(k_1,t)-{\mathcal H}(k_1,s)\Vert _{{\mathcal X}} \le&72V_{\infty }^2\Vert k_1\Vert _{{\mathcal X}} \left( \Vert k_1\Vert _{{\mathcal X}}+\Vert \partial ^2_x k_1\Vert _{L^{\infty }}\right) (t-s). \end{aligned}$$

Let us denote

$$\begin{aligned} M:=\max \left( 2M_0,M'_0e^2, \left( M''_0+\frac{(M'_0)^2e^4}{2M_0}\right) e^{2}\right) \end{aligned}$$

The constant M is a uniform upper bound on the supremum norm of $g_n(t),\partial _x g_n(t),\partial ^2_x g_n(t)$ for all $t \in [0,T^{*}]$ and $n \in {\mathbb N}$. We have that

$$\begin{aligned} \Vert \dot{g}_n(t)-\dot{g}_m(t)\Vert _{{\mathcal X}} =&\left\| {\mathcal H}\left( g_n \left( \frac{\left\lfloor nt \right\rfloor }{n}\right) ,\frac{\left\lfloor nt \right\rfloor }{n}\right) -{\mathcal H}\left( g_m \left( \frac{\left\lfloor mt \right\rfloor }{m}\right) ,\frac{\left\lfloor mt \right\rfloor }{m}\right) \right\| _{{\mathcal X}}\\ \le&\left\| {\mathcal H}\left( g_n \left( \frac{\left\lfloor nt \right\rfloor }{n}\right) ,\frac{\left\lfloor nt \right\rfloor }{n}\right) -{\mathcal H}\left( g_n \left( \frac{\left\lfloor nt \right\rfloor }{n}\right) ,t\right) \right\| _{{\mathcal X}}\\&+ \left\| {\mathcal H}\left( g_n \left( \frac{\left\lfloor nt \right\rfloor }{n}\right) ,t\right) -{\mathcal H}\left( g_n(t),t\right) \right\| _{{\mathcal X}}\\&+ \left\| {\mathcal H}(g_n(t),t)-{\mathcal H}(g_m(t),t) \right\| _{{\mathcal X}}\\&+ \left\| {\mathcal H}(g_m(t),t)-{\mathcal H}\left( g_m \left( \frac{\left\lfloor mt \right\rfloor }{m}\right) ,t\right) \right\| _{{\mathcal X}}\\&+ \left\| {\mathcal H}\left( g_m \left( \frac{\left\lfloor mt \right\rfloor }{m}\right) ,t\right) -{\mathcal H}\left( g_m \left( \frac{\left\lfloor mt \right\rfloor }{m}\right) ,\frac{\left\lfloor mt \right\rfloor }{m}\right) |\right\| _{{\mathcal X}} \\ \le&144V_{\infty }^2M^2 \left( \frac{1}{n}+\frac{1}{m}\right) +12V_{\infty }M \left( \left\| g_n\left( \frac{\left\lfloor nt \right\rfloor }{n}\right) -g_n(t)\right\| _{{\mathcal X}}\right. \\&+ \left. \left\| g_m\left( \frac{\left\lfloor mt \right\rfloor }{m}\right) -g_m(t)\right\| _{{\mathcal X}}\right) +12V_{\infty }M\Vert g_n(t)-g_m(t)\Vert _{{\mathcal X}}. \end{aligned}$$

On the other hand,

$$\begin{aligned} \left\| g_n \left( \frac{\left\lfloor nt \right\rfloor }{n} \right) -g_n(t)\right\| _{{\mathcal X}} \le \frac{1}{n}\left\| {\mathcal H}\left( g_n \left( \frac{ \left\lfloor nt \right\rfloor }{n} \right) \right) \right\| _{{\mathcal X}} \le \frac{6V_{\infty } M^2}{n} \end{aligned}$$

and similarly for the term concerning m. Hence there exist two positive constants $C_1,C_2$ that only depend on $V_{\infty }$ and M, such that for all $t \in [0,T^{*}]$,

$$\begin{aligned} \Vert \dot{g}_n(t)-\dot{g}_m(t)\Vert _{{\mathcal X}} \le C_1\left( \frac{1}{n}+\frac{1}{m}\right) +C_2\Vert g_n(t)-g_m(t)\Vert _{{\mathcal X}}, \end{aligned}$$

which implies

$$\begin{aligned} \left| \left| g_n(t)-g_m(t)\right| \right| _{{\mathcal X}} \le C_1 \left( \frac{1}{n}+\frac{1}{m}\right) t+C_2\int _{0}^{t}\Vert g_n(s)-g_m(s)\Vert _{{\mathcal X}}\ ds. \end{aligned}$$

This, and the Gronwall’s inequality give

$$\begin{aligned} \sup _{t \in [0,T^{*}]} \Vert g_n(t)-g_m(t)\Vert _{{\mathcal X}} \le C_1\left( \frac{1}{n}+\frac{1}{m} \right) T^{*}\left( 1+c_2T^{*}e^{C_2T^{*}}\right) \end{aligned}$$

which implies that $(g_n)_{n \in {\mathbb N}}$ is a Cauchy sequence and therefore admits a limit $g_{\infty } \in {\mathcal C}([0,T^{*}],{\mathcal X})$. The function $g_{\infty }$ (that we now regard as a function of the four variables $(x,t,\rho ^-,\rho ^+)$) is $C^1$ in the variables x and t, and verify the inhomogeneous ODE (A.3) and is bounded uniformly from below by $\frac{\delta _0}{2}$ and is such that

$$\begin{aligned} \sup _{t \in [0,T^*]} \sup _{x \in {\mathbb R}} \sup _{\rho ^-,\rho ^+} g_{\infty }(x,t,\rho ^-,\rho ^+) \le M. \end{aligned}$$

Moreover, for any fixed x and t in its domain of definition, the function $(\rho ^-,\rho ^+) \mapsto g_{\infty }(x,t,\rho ^-,\rho ^+)$ is supported on $\Gamma ^{V_{\infty }}$. Now defining

$$\begin{aligned} f(x,t,\rho ^-,\rho ^+)=g_{\infty }(x+[\rho ^-,\rho ^+]t,t,\rho ^-,\rho ^+) \end{aligned}$$

(A.7)

f is again $C^1$ in x and t, verify the same properties as $g_{\infty }$ and verifies the desired kinetic equation.

Finally we remark that in the above proof, we may replace the set $\Gamma ^{V_{\infty }}$ with $\Gamma ^{V_{\infty }}_+$. $\square$

For the second part of this section, we will prove the existence of the solution to the Kolmogorov forward equation both in space x and time t. More precisely, we wish to address the existence of a unique uniformly positive solution $\ell$ of the equations (2.3) and (2.4), provided that the kernel f is uniformly positive. We remark that these equations are consistant by Proposition 4.1. Because of this, we only need to solve (2.3) in $[a^-,a^+]$ for an initial condition $\ell (a^-,t,\cdot )$ that solves (2.4). The existence of a solution to (2.3) can be carried out by standard arguments. However, we need to ensure the constructed solution is uniformly positive in $\Lambda$, if the initial $\ell ^0(\rho )=\ell (a^-,t_0,\rho )$ is uniformly positive. Observe that if $\ell$ solves (2.3), then

$$\begin{aligned} \frac{d}{dx}\int \ell (x,t,\rho )\ \beta (d\rho )=0, \end{aligned}$$

because the $\beta$-integral of the right-hand side of (2.3) is 0. This means

$$\begin{aligned} \int \ell (x,t)\ \beta (d\rho )=1, \end{aligned}$$

(A.8)

if this is the case for $x=a^-$. On the other hand, if the total integral of $\beta$ is one, $f\ge \delta _1$ for some positive constant $\delta _1$, and $\ell$ is a solution of (2.3) satisfying (A.8), then

$$\begin{aligned} \ell _x(x,t,\rho )\ge \delta _1-\lambda (x,t,\rho )\ell (x,t,\rho ), \end{aligned}$$

which leads to the lower bound

$$\begin{aligned} \ell (x,t,\rho )\ge \ell (a^-,t,\rho )e^{-\int _{a^-}^x \lambda (\theta ,t,\rho )\ d\theta } +\delta _1\int _{a^-}^xe^{-\int _{y}^x \lambda (\theta ,t,\rho )\ d\theta }\ dy. \end{aligned}$$

From this we learn that $\ell$ is uniformly positive in $\Lambda$ if this is the case on the left boundary side of $\Lambda$. By assumption, $\ell$ is uniformly positive at $(a^-,t_0)$, and as t varies, the function $t\mapsto \ell (a^-,t,\rho )$ satisfies (2.4). If the kernel f is supported in $\Gamma ^{V_{\infty }}_+$, then $[\rho ^-,\rho ^+]f\ge 0$, and a repetition of the above reasoning guarantees

$$\begin{aligned} \ell (a^-,t,\rho )\ge \ell (a^-,t_0,\rho )e^{-\int _{t_0}^x A(a^-,\theta ,\rho )\ d\theta }. \end{aligned}$$

In summary, when the kernel f is supported in $\Gamma ^{V_{\infty }}_+$, and is uniformly positive on its support, we can construct a unique uniformly positive solution $\ell$ to forward equations (2.3) and (2.4) by standard arguments. However some care is needed if $[\rho ^-,\rho ^+]$ can change sign in the support of our kernel f. In this case, we can guarantee the existence of a uniformly positive solution to (2.3) and (2.4) if we either replace the time interval $[0,T^*]$ with a shorter interval, or assume that the initial $\ell (a^-,t_0,\rho )$ is sufficiently positive. As an example, we demonstrate how a lower bound of 1/6 on the initial $\ell$ can guarantee the positivity of the solution.

Theorem A.2

Fix $a^{-}<a^{+}$. Let $\ell ^{0} : [P^-,P^+]^2 \rightarrow [0,+\infty )$ be a measurable function such that there exists two constants $c,C>0$ with

$$\begin{aligned} c \le \ell ^{0}(\rho ) \le C \text { for all } \rho \end{aligned}$$

and $\int \ell ^{0}(\rho )\ \beta (d\rho )=1$. Moreover, assume that $c \ge \frac{1}{6}$. Then there exists a $C^1$ solution $\ell : [a^{-},a^{+}] \times [0,T^{*}] \times [P^-,P^+]^2 \rightarrow [0,+\infty )$ to the equations (2.3) and (2.4) such that $\ell (a^{-},0,\cdot )=\ell ^{0}$, and such that $\ell$ is uniformly bounded below by a positive constant and

$$\begin{aligned} \int \ell (x,t,\rho )\ \beta (d\rho )=1, \end{aligned}$$

for all $(x,t) \in [a^{-},a^{+}] \times [0,T^{*}]$.

Proof

Without loss of generality let us assume that $a^{-}=0$ and denote $a^{+}=a$. We will construct a two-parameter function $\ell :[-V_{\infty }T^{*},a] \times [0,T^{*}] \rightarrow {\mathcal F}_b([P^-,P^+]^2)$. The reason why we extend the space domain to $[-V_{\infty }T^{*},a]$ instead of [0, a] will be made clear later. Let us define first $\ell (\cdot ,0)$ on $[-V_{\infty }T^{*},a]$ using the first ODE in the x-direction. The utility of the condition $c \ge \frac{1}{6}$ is to ensure the non-negativity of $\ell$ as we run the ODE backwards from $0 \rightarrow -V_{\infty }T^{*}$. Our strategy for proving the existence of the solution of the ODE at $t=0$ is done in a similar fashion as the kinetic equation via an approximation scheme. In other words, we construct a polygonal approximating $\ell _n : [-V_{\infty }T^{*},a] \rightarrow {\mathcal F}_b([P^-,P^+]^2)$ by putting $\ell _n(0)=\ell ^{0}$, and for any $k \in {\mathbb Z}$ by the inductive relation. More precisely, we put $f^k(\rho ,\rho _*):=f(k/n,0,\rho ,\rho _*)$, and require that functions $\ell _n^{k}:=\ell _n \left( \frac{k}{n} \right) \in {\mathcal F}_b([P^-,P^+]^2)$ to satisfy

$$\begin{aligned} n \left( \ell ^{k+1}_n(\rho )-\ell ^k_n(\rho )\right) =&\int f^k \left( \rho ^*,\rho \right) \ell _n^k(\rho ^*)\ \beta (d\rho ^*) -\left( \int f^k\left( \rho ,\rho ^+\right) \ \beta (d\rho ^+) \right) \ell _n^k(\rho ), \end{aligned}$$

for $k\ge 0$, and

$$\begin{aligned} -n \left( \ell ^{k-1}_n(\rho )-\ell ^k_n(\rho )\right) =&\int f^k \left( \rho ^*,\rho \right) \ell _n^k(\rho ^*)\ \beta (d\rho ^*) \\&-\left( \int f^k\left( \rho ,\rho ^+\right) \ \beta (d\rho ^+) \right) \ell _n^k(\rho ), \end{aligned}$$

for $k \le 0$. The intermediate values $\ell _n(x)$ for $x \in (\frac{k}{n},\frac{k+1}{n})$ are obtained by linear interpolation. As an initial observation, remark that

$$\begin{aligned} \int \ell _n^{k}(\rho )\ \beta (d\rho )=\int \ell _n^{k \pm 1}(\rho )\ \beta (d\rho ), \end{aligned}$$

and hence

$$\begin{aligned} \int \ell _n^{k}(\rho )\ \beta (d\rho )=1,\quad {\text { for all }} \; k \in {\mathbb Z}. \end{aligned}$$

Now, if we take $n \ge M_0$ where $M_0=\Vert f(\cdot ,0,\cdot ,\cdot )\Vert _{L^{\infty }}$, then by induction it follows that $\ell _n^{k} \ge 0$ for all $k \ge 0$, as we have that

$$\begin{aligned} \ell _n^{k+1}(\rho )&=\ell _n^{k}(\rho )+\frac{1}{n} \left( \int f^k \left( \rho ^*,\rho \right) \ell _n^k(\rho ^*)\ \beta (d\rho ^*) - \left( \int f^k\left( \rho ,\rho ^+\right) \ \beta (d\rho ^+) \right) \ell _n^k(\rho ) \right) \\&\ge \ell _n^{k}(\rho ) - \frac{M_0}{n} \ell _n^{k}(\rho ), \end{aligned}$$

which in turn implies the following lower bound

$$\begin{aligned} \ell _n^{k}(\rho ) \ge \ell ^0(\rho )\left( 1 -\frac{M_0}{n} \right) ^k \ge \ell ^{0}(\rho ) e^{-\frac{M_0 k}{n}}\ \ \text { for all } \ k \ge 0. \end{aligned}$$

On the other hand, for $k \le 0$ we have

$$\begin{aligned} \ell _n^{k-1}(\rho )&=\ell _n^{k}(\rho )-\frac{1}{n} \int f^k (\rho ^*,\rho ) \ell _n^k(\rho ^*)\ \beta (d\rho ^*) +\frac{1}{n} \left( \int f^k\left( \rho ,\rho ^+\right) \ \beta (d\rho ^+) \right) \ell _n^k(\rho ), \end{aligned}$$

which leads to

$$\begin{aligned} \ell _n^{k}(\rho ) \ge \ell ^{0}(\rho )-\frac{M_0k}{n}, \end{aligned}$$

because

$$\begin{aligned} \int f^k \left( \rho ^*,\rho \right) \ell _n^k(\rho ^*)\ \beta (d\rho ^*) \le M_0 \int \ell _n^{k}(\rho ^*)\ \beta (d\rho _*) =M_0. \end{aligned}$$

In particular, if $\frac{k}{n} \ge -V_{\infty }T^{*}$, then $\frac{M_0k}{n}\ge -\frac{1}{2}$, and

$$\begin{aligned} \inf _{\rho } \ell _n^{k}(\rho ) \ge c -\frac{1}{12} \ge \frac{1}{12}. \end{aligned}$$

We have therefore constructed the polygonal approximating function $\ell _n : [-V_{\infty }T^{*},a] \rightarrow {\mathcal F}_b([P^-,P^+]^2)$ such that it is uniformly bounded from below by $\min ({1}/{12},c e^{-M_0 a })$. The sequence $(\ell _n)_{n \in {\mathbb N}}$ is a Cauchy sequence in the space $C([-V_{\infty }T^{*},a],{\mathcal F}_b([P^-,P^+]^2)$ where ${\mathcal F}_b([P^-,P^+]^2)$ is viewed as a Banach space equipped with the uniform norm. We obtain that the limit $\ell _{\infty }:=\lim _{n \rightarrow \infty } \ell _n$ is a solution to the ODE

$$\begin{aligned} (\ell _{\infty })_x(x,\rho )= & {} \int f(x,0,\rho ^*,\rho )\ell _{\infty }(x,\rho ^*)\ \beta (d\rho ^*) \\&- \left( \int f(x,0,\rho ,\rho ^+)\ \beta (d\rho ^+) \right) \ell _{\infty }(x,\rho ) \end{aligned}$$

We define $\ell (\cdot ,0)=\ell _{\infty }$. We will move on now to prove the existence of the solution $\ell$ as an ODE in the time variable t. In order to preserve the non-negativity of $\ell$, we have taken advantage in the ODE in the x-direction of the positivity of the kernel $f(x,t,\rho ^-,\rho ^+)$, however in the t-direction the kernel is equal to $[\rho ^-,\rho ^+]f(x,t,\rho ^-,\rho ^+)$. To circumvent this difficulty, we take advantage of the finite speed propagation (this also explains why we have constructed $\ell (0,\cdot )$ on $[-V_{\infty }T^{*},a]$ instead of just [0, a]). For any $x \in {\mathbb R}$ and $t \in [0,T^{*}]$ we define

$$\begin{aligned} \tilde{f}(x,t,\rho ^-,\rho ^+)=f(x+V_{\infty }t,t,\rho ^-,\rho ^+) \text { for all } \rho ^-,\rho ^+. \end{aligned}$$

We define a function $\tilde{\ell } :[-V_{\infty }T^{*},a] \times [0,T^{*}] \rightarrow {\mathcal F}_b([P^-,P^+]^2)$ that satisfies the initial condition $\tilde{\ell }(x,0)=\ell _{\infty }(x)=\ell (x,0)$. Now, for $x=-V_{\infty }T^{*}$ we define $\tilde{\ell }(-V_{\infty }T^{*},\cdot ) : [0,T^{*}] \rightarrow {\mathcal F}_b([P^-,P^+]^2)$ by solving the ODE

$$\begin{aligned} \tilde{\ell }_t(-V_{\infty }T^{*},t,\rho )=&\int \left( [\rho ^*,\rho ]+V_{\infty } \right) \tilde{f}(-V_{\infty }T^{*},t,\rho ^*,\rho )\tilde{\ell }(-V_{\infty }T^{*},t,\rho ^*)\ \beta (d\rho ^*) \\&-\left( \int \left( [\rho ,\rho ^+]+V_{\infty }\right) \tilde{f}(-V_{\infty }T^{*},t,\rho ,\rho ^+)\ \beta (d\rho ^+) \right) \\&\qquad \tilde{\ell }(-V_{\infty }T^{*},t,\rho ), \end{aligned}$$

with initial condition $\tilde{\ell }(-V_{\infty }T^{*},0)=\ell _{\infty }(-V_{\infty }T^{*})$. Now, for any fixed $t \in (0,T^{*}]$ we define $\tilde{\ell } (\cdot ,t) : [-V_{\infty }T^{*},a] \rightarrow {\mathcal F}_b([P^-,P^+]^2)$ by solving the ODE on $[-V_{\infty }T^{*},a]$

$$\begin{aligned} \tilde{\ell }_x(x,t,\rho )= & {} \int \tilde{f}(x,t,\rho ^*,\rho )\tilde{\ell }(x,t,\rho ^*)\ \beta (d\rho ^*)\\&- \left( \int \tilde{f}(x,t,\rho ,\rho ^+)\ \beta (d\rho ^+) \right) \tilde{\ell }(x,t,\rho ) \end{aligned}$$

with initial condition determined by $\tilde{\ell }(-V_{\infty }T^{*},t)$. The existence of these solutions is done by exactly the same approximation scheme than before, and the function $\tilde{\ell }$ is bounded uniformly from below on the box $[-V_{\infty }T^{*},a]\times [0,T^{*}]$ due to the non-negativity of the kernels $([\rho ^-,\rho ^+]+V_{\infty })f(x,t,\rho ^-,\rho ^+)$ and $f(x,t,\rho ^-,\rho ^+)$. Moreover, if we assume that initially $f(0,\cdot )$ is $C^3$ then we get that f is $C^2$ in the variables (x, t), it follows that $\ell$ is also $C^2$ and thus from Proposition 5.1, the ODE in t is verified for all $x \in [-V_{\infty }T^{*},a]$, i.e

$$\begin{aligned} \tilde{\ell }_t(x,t,\rho )=&\int \left( [\rho ^*,\rho ]+V_{\infty } \right) \tilde{f}(x,t,\rho ^*,\rho )\tilde{\ell }(x,t,\rho ^*)\ \beta (d\rho ^*) \\&-\left( \int \left( [\rho ,\rho ^+]+V_{\infty }\right) \tilde{f}(x,t,\rho ,\rho ^+)\ \beta (d\rho ^+) \right) \tilde{\ell }(x,t,\rho ) \end{aligned}$$

Now, it suffices to define

$$\begin{aligned} \ell (x,t,\rho )=\tilde{\ell }(x-V_{\infty }t,t,\rho ) \text { for all } (x,t) \in [0,a] \times [0,T^{*}] \text{ and } \rho \in [P^-,P^+]^2 \end{aligned}$$

then $\ell$ is $C^1$ in (x, t) and verify the desired ODEs. Moreover, the total of mass of $\ell$ is conserved through space and time. $\square$