Propagation of Chaos in the Nonlocal Adhesion Models for Two Cancer Cell Phenotypes

Abstract

We establish a quantitative propagation of chaos for a large stochastic systems of interacting particles. We rigorously derive a mean-field system, which is a diffusive cell-to-cell nonlocal adhesion model for two different phenotypes of tumors, from that stochastic system as the number of particles tends to infinity. We estimate the error between the solutions to a N-particle Liouville equation associated with the particle system and the limiting mean-field system by employing the relative entropy argument.

Appendices

Appendix A. Poisson Jump Process

In this appendix, we briefly introduce the Poisson jump process and its properties used in the paper. We refer to Durrett (2010), Jeanblanc (2007), Protter (2004) and (Schönbucher 2003, Chapter 4). For the basic definitions for stochastic processes we refer to (Baudoin 2014; Protter 2004).

Let $(\Omega , { {\mathcal {F}} }, P)$ be a complete probability space with a filtration $({ {\mathcal {F}} }_t)_{t\ge 0}$.

Definition 2

A nonnegative integer valued stochastic process $E=(E_t)_{t\ge 0}$ defined on $(\Omega , { {\mathcal {F}} }, P)$ is called Poisson jump process with intensity or rate $\lambda$ if the process has independent and stationary increments:

• (i) $E_0=0$ a.s.

• (ii) $E_{t_k} - E_{t_{k-1}}$ $(k=1, \dots , n)$ are independent for $0= t_0< t_1<\cdots < t_n$.

• (iii) $E_t-E_s$ is $Poisson(\lambda (t-s))$, that is,

  $$\begin{aligned} P(E_t-E_s = k)= e^{-\lambda (t-s)} \lambda ^k (t-s)^k/k! \, \text{ for } k=0, 1, 2, \ldots . \end{aligned}$$

$E_t$ is thought as a number of events occurring in [0, t] with the rate $\lambda$. It satisfies ${\mathbb {E}} E_t = \lambda t$, $\mathrm{Var} E_t = \lambda t$. From the independence increments property we find $E_t-\lambda t$ is a martingale such that

$$\begin{aligned} {\mathbb {E}} (E_t-E_s| { {\mathcal {F}} }) = \lambda (t-s). \end{aligned}$$

The Poisson process with constant intensity $\lambda$ is generalized to a counting processes $N=(N_t)_{t\ge 0}$² with a predictable compensator process $A=(A_t)_{t\ge 0}$, for which the compensated process $N_t-A_t$ is a local martingale. The compensator can give an information about the probabilities of jumps over the next time step:

$$\begin{aligned} {\mathbb {E}} (N_{t+ h} -N_t | { {\mathcal {F}} }_t) = {\mathbb {E}}(A_{t+ h}- A_t| { {\mathcal {F}} }_t) \end{aligned}$$

if we restrict to the case the compensated process is a true martingale. If A is differentiable, we extend the definition of (homogeneous) intensity $\lambda$ for $E_t$ to the intensity process $\lambda _t$ for $N_t$:

$$\begin{aligned} A_t= \int _0^t \lambda _s\, ds. \end{aligned}$$

Definition 3

Let $\lambda$ be a nonnegative $({ {\mathcal {F}} })_{t\ge 0}$ adapted process defined on $({ \Omega }, { {\mathcal {F}} }, P)$ such that $\int _0^t \lambda _s\,ds <\infty .$ A counting process N is said to be an inhomogeneous Poisson process with stochastic intensity $\lambda$ if the process

$$\begin{aligned}M_t = N_t - \int _0^t \lambda _s\,ds, \quad t\ge 0\end{aligned}$$

is a martingale.

In Sect. 2.1, we used that $\alpha _k \chi _1(\Xi ^i_{s^-})$ ($k=1,2$) are intensities of the time-changed Poisson processes ${\tilde{E}}_k^i(t) = E^i(\alpha _k \int _0^t \chi _k(\Xi ^i_{s^-}) \,ds$ at time t (Oelschlager 1989) to see the expectation of the stochastic integral with respect to ${\tilde{E}}_k^i(t)$ is as same as that with respect to $A_k^i(t):=\alpha _k \int _0^t \chi _1(\Xi ^i_{s^-})\,ds$ ($k=1,2$).

Appendix B. Proof of (3.5)

In this appendix, we prove (3.5) following the similar argument in (Lim et al. 2020, Lemma 4.1). Although the overall idea is similar, we add more details in this probabilistic argument for reader’s convenience.

We note that the process $(\mathbf {X}_{t}^{N}, \mathbf {\Xi }_{t}^{N})$ given by (1.2)–(1.3) is a Markov process since it is a solution to a SDE driven by Markov processes $\{B_t^i\}_{t\ge 0}, \{E_t^i\}_{t\ge 0}$ for $i=1, \dots , N$. We refer to (Protter 2004, Theorem 32, Chapter 5) for the Markov nature of the process defined by stochastic integral against Lévy processes. Fix $t_0>0$ and we will show that (3.5) holds at $t=t_0$. Let u and ${\tilde{u}}$ be the solutions of the following equations:

$$\begin{aligned} \begin{aligned} \partial _t u&= { {\mathcal {L}} }^*_{r,N} u \quad \text{ with } u(t_0) = \rho _N(t_0), \\ \partial _t {\tilde{u}}&= { {\mathcal {L}} }^*_{r,N} {\tilde{u}} \quad \text{ with } {\tilde{u}}(t_0) = {\bar{\rho }}_N(t_0). \end{aligned} \end{aligned}$$

(B.1)

By Theorem 2 in Sect. 4, u and ${\tilde{u}}$ are in $C([t_0, \infty ); W^{1, p}(\Pi ^N))$ ($F=0$ case) with initial data $\rho _N(t_0), {\bar{\rho }}_N(t_0) \in W^{1, p}(\Pi ^N)$. We find that

$$\begin{aligned} \frac{d}{dt}{ {\mathcal {H}} }(u|{\tilde{u}})(t) = \int _{\Pi ^N} \left( \log \left( \frac{u(t)}{{\tilde{u}}(t) }\right) { {\mathcal {L}} }^*_{r,N} u(t)- u(t) \left( \frac{{ {\mathcal {L}} }^*_{r,N} u(t)}{u(t)} - \frac{{ {\mathcal {L}} }^*_{r,N} \tilde{u}(t)}{{\tilde{u}}(t)}\right) \right) dm^N\nonumber \\ \end{aligned}$$

(B.2)

for $t> t_0$. Let us assume the non-positivity of (B.2) for the moment. Note that the LHS of (3.5) is formally obtained if we evaluate $\frac{d}{dt}{ {\mathcal {H}} }(u|{\tilde{u}})(t)$ at $t=t_0$. Indeed, we see that (B.2) converges to

$$\begin{aligned} \int _{\Pi ^N} \left( \log \left( \frac{\rho _N}{{\bar{\rho }}_N} \right) { {\mathcal {L}} }^*_{r,N} \rho _N - \rho _N \left( \frac{{ {\mathcal {L}} }^*_{r,N} \rho _N}{\rho _N} - \frac{{ {\mathcal {L}} }^*_{r,N} {\bar{\rho }}_N}{{\bar{\rho }}_N}\right) \right) dm^N \end{aligned}$$

(B.3)

as $t \searrow t_0$; the difference of two integrals (B.2) and (B.3) is

$$\begin{aligned}&\int _{\Pi ^N} \left( ({ {\mathcal {L}} }^*_{r,N} u - { {\mathcal {L}} }^*_{r,N} \rho _N) \log \frac{u}{{\tilde{u}}} + { {\mathcal {L}} }^*_{r,N}\rho _N (\log u-\log \rho _N - (\log {\tilde{u}} - \log {\bar{\rho }}_N))\right) dm^N\nonumber \\&\quad + \int _{\Pi ^N} \frac{u(t)}{{\tilde{u}}(t)} ( { {\mathcal {L}} }^*_{r,N} {\tilde{u}} - { {\mathcal {L}} }^*_{r,N} {\bar{\rho }}_N) + { {\mathcal {L}} }^*_{r,N}\bar{\rho }_N\left( \frac{u}{{\tilde{u}}}- \frac{\rho _N}{\bar{\rho }_N}\right) dm^N. \end{aligned}$$

(B.4)

For the moment we assume the positivity of solutions $\rho _N$ and ${\bar{\rho }}_N$ to (3.1) and (3.2) respectively, which will be justified in Lemma 3 below. By integrating by parts, we have

$$\begin{aligned}(B.4)&=-\frac{1}{2}\int _{\Pi ^N} (\nabla _N u - \nabla _N \rho _N) \cdot \left( \frac{\nabla _N u }{u} - \frac{\nabla _N {\tilde{u}}}{{\tilde{u}}} \right) dm^N \\&\quad +\int _{\Pi ^N} (S_N^* u-S_N^* \rho _N) (\log u - \log {\tilde{u}})\,dm^N \\&\quad - \frac{1}{2} \int _{\Pi ^N} \nabla _N \rho _N\cdot \left( \frac{\nabla _N u}{u} - \frac{\nabla _N \rho _N}{\rho _N}- \left( \frac{\nabla _N {\tilde{u}}}{{\tilde{u}}} - \frac{\nabla _N {\bar{\rho }}_N}{{\bar{\rho }}_N} \right) \right) dm^N \\&\quad + \int _{\Pi ^N} S_N^* \rho _N ( \log u - \log \rho _N - (\log {\tilde{u}} - \log {\bar{\rho }}_N))\,dm^N \\&\quad -\frac{1}{2} \int _{\Pi ^N} \left( \nabla _N \left( \frac{u}{\tilde{u}}\right) \cdot (\nabla _N {\tilde{u}} - \nabla _N {\bar{\rho }}_N) - 2\frac{u}{{\tilde{u}}} ( S_N^* {\tilde{u}} - S_N^* {\bar{\rho }}_N) \right) dm^N \\&\quad -\frac{1}{2} \int _{\Pi ^N} \left( \nabla _N {\bar{\rho }}_N \cdot \nabla _N \left( \frac{ u}{ {\tilde{u}} }- \frac{\rho _N}{ {\bar{\rho }}_N}\right) - 2 S_N^* {\bar{\rho }}_N \left( \frac{ u}{ {\tilde{u}} }- \frac{\rho _N}{ {\bar{\rho }}_N} \right) \right) dm^N . \end{aligned}$$

Using $u, {\tilde{u}}, \rho _N, {\bar{\rho }}_N >0$ and $u, {\tilde{u}} \in C([t_0, \infty ); W^{1, p}(\Pi ^N))$ for $p>dN$, we see that the difference is bounded above by $C ( \Vert u(t) - \rho _N\Vert _{W^{1, p}(\Pi ^N)} + \Vert {\tilde{u}}(t) - {\bar{\rho }}_N\Vert _{W^{1, p}(\Pi ^N)} )$, which is vanishing as $t \searrow t_0$. It concludes (3.5).

Proof of the non-positivity of (B.2). In what follows, we suppress the superscript N in $\mathbf {Y_t^N}=(\mathbf {X}_{t}^{N}, \mathbf {\Xi }_{t}^{N})$. Let $\mathbf {Y}_t$, ${\tilde{\mathbf {Y}}}_t$ be the Markov process defined by (1.2)–(1.3) with initial distribution $\rho _N$, ${\bar{\rho }}_N$ at $t=t_0$ respectively. We denote the solutions of (B.1) at time t by $u_t$ and ${\tilde{u}}_t$, which are distribution of $\mathbf {Y}_t$, ${\tilde{\mathbf {Y}}}_t$. Let $\kappa$ be the Borel measure on $\Pi \times \Pi$ induced by the joint distribution of $(\mathbf {Y}_t, \mathbf {Y}_{t+h})$ for $t>t_0, h>0$. That is

$$\begin{aligned} \kappa (B) = P( (\mathbf {Y}_t, \mathbf {Y}_{t+h}) \in B )\quad \text{ for } \text{ any } B\in { {\mathcal {B}} }(\Pi \times \Pi ), \end{aligned}$$

where ${ {\mathcal {B}} }(\Pi \times \Pi )$ is the collection of Borel sets in $\Pi \times \Pi$. We denote the induced Borel measure on $\Pi$ by $\mathbf {Y}_t$ by $\lambda$. The induced measures ${\tilde{\kappa }}$, $\tilde{\lambda }$ are similarly defined.

Since the space $\Pi$ is sufficiently nice, it is possible to disintegrate $\kappa$ and ${\tilde{\kappa }}$. More precisely, we can find a kernel $\lambda$ (${\tilde{\lambda }}$) -a.e. uniquely determined kernel $\mu _{\cdot }$ (${\tilde{\mu }}_{\cdot }$) that maps $\Pi$ to the space of Borel measures on $\Pi$ such that

• (i) For $x \in \Pi$, $\mu _x( \cdot )$ is a Borel measure on $\Pi$.

• (ii) $\mu _{\cdot }(B)$ is $\lambda$-measurable for any fixed $B \in { {\mathcal {B}} }(\Pi )$.

• (iii) $\int _{\Pi \times \Pi } f(x, y) d\kappa (x, y) = \int _{\Pi }\int _{\Pi } f(x,y) \mu _x (dy) \lambda (dx)$ for any bounded Borel function on $\Pi \times \Pi$.

The similar assertion holds for ${\tilde{\mu }}_{\cdot }$. The disintegration theorem holds in more general circumstance, but we won’t go further, see for instance (Dellacherie and Meyer 1978, III-70). By (i)–(iii), we have $\{\mu _x (dy)\}_{x\in \Pi }$ and $\{{\tilde{\mu }}_x (dy)\}_{x\in \Pi }$ such that

$$\begin{aligned} P( \mathbf {Y}_t \in A, \mathbf {Y}_{t+h} \in B)&= \int _A\int _B \mu _x(dy) d\lambda (x) \\ P({\tilde{\mathbf {Y}}}_t \in A, {\tilde{\mathbf {Y}}}_{t+h} \in B)&= \int _A\int _B {\tilde{\mu }}_x(dy) d{\tilde{\lambda }}(x) \end{aligned}$$

for $A, B \in { {\mathcal {B}} }(\Pi )$. On the other hands, since $\{ \mathbf {Y}_t\}$, $\{{\tilde{\mathbf {Y}}}_t\}$ are the same Markov processes, there exists the transition kernel $P_h(x, dy)$ such that

$$\begin{aligned} P( \mathbf {Y}_t \in A, \mathbf {Y}_{t+h} \in B)&= \int _A\int _B P_h(x, dy) d\lambda (x), \\ P({\tilde{\mathbf {Y}}}_t \in A, {\tilde{\mathbf {Y}}}_{t+h} \in B)&= \int _A\int _B P_h(x, dy) d{\tilde{\lambda }}(x). \end{aligned}$$

Thus we conclude

$$\begin{aligned} \mu _x (dy) = P_h(x, dy) = {\tilde{\mu }}_x (dy). \end{aligned}$$

We use $u_{t+h| t}(x, y)$ to denote the density of $\mu _x (dy)$ with respect to dy, where we abuse the notation dy to stand for $dm=dy \otimes d \sharp$ the canonical measure on $\Pi = {\mathbb {T}}^d \times \{ 1, 2\}$. ${\tilde{u}}_{t+h| t}(x, y)$ is defined in the same way. $u_{t+h| t}(x, y)$ is so called a conditional density, which is a density of conditional probability

$$\begin{aligned} P(u_{t+h} \in B|u_t = x) = \int _B \mu _x (dy). \end{aligned}$$

Note that $u_t$ and $u_{t+h}$ are the x- and y- marginal densities of the joint distribution of $(Y_t, Y_{t+h})$, while the density of $\kappa$ is denoted by $u_{t, t+h}$. The notations with tilde are defined accordingly. As in (Lim et al. 2020, Lemma 4.1), we define the average conditional relative entropy between $u_{t+h|t}$ and ${\tilde{u}}_{t+h|t}$ by

$$\begin{aligned} { {\mathcal {H}} }( u_{t+h|t} | {\tilde{u}}_{t+h|t}): = \int _{\Pi } u_t(x) \int _{\Pi } u_{t+h|t}(x,y) \log \frac{ u_{t+h|t}}{{\tilde{u}}_{t+h|t}} \,dydx. \end{aligned}$$

Applying the chain rule (Dupuis and Ellis 1997, Theorem C.3.1) of the relative entropy, we have

$$\begin{aligned} { {\mathcal {H}} }( u_{t, t+h} | {\tilde{u}}_{t, t+h}) = { {\mathcal {H}} }( u_{t} | \tilde{u}_{t})+{ {\mathcal {H}} }( u_{t+h|t} | {\tilde{u}}_{t+h|t}). \end{aligned}$$

The last quantity is zero since $u_{t+h|t} = {\tilde{u}}_{t+h|t}$. If we proceed the above discussion to the joint distribution of $(\mathbf {Y}_{t+h}, \mathbf {Y}_t)$, we similarly arrive at

$$\begin{aligned} { {\mathcal {H}} }( u_{ t+h,t} | {\tilde{u}}_{t+h,t}) = { {\mathcal {H}} }( u_{t+h} | \tilde{u}_{t+h})+{ {\mathcal {H}} }( u_{t|t+h} | {\tilde{u}}_{t|t+h}). \end{aligned}$$

Finally using ${ {\mathcal {H}} }( u_{ t+h,t} | {\tilde{u}}_{t+h,t}) ={ {\mathcal {H}} }( u_{t, t+h} | {\tilde{u}}_{t, t+h})$ and the non-negativity of the relative entropy ${ {\mathcal {H}} }( u_{t|t+h} | {\tilde{u}}_{t|t+h})$, we have

$$\begin{aligned} { {\mathcal {H}} }( u_{t} | {\tilde{u}}_{t}) \ge { {\mathcal {H}} }( u_{t+h} | \tilde{u}_{t+h}), \end{aligned}$$

thus (B.2) $\le 0$.

In the lemma below we provide the positivity of $\rho _N$ and ${\bar{\rho }}_N$ to (3.1) and (3.2) respectively, which was assumed in showing (B.4) to vanish.

Lemma 3

Let $\rho _N$ and ${\bar{\rho }}_N$ be the solutions of (3.1) and (3.2) respectively, which well-posedness are given in Theorem 2 in Sect. 4 with initial data $\rho _N(0), {\bar{\rho }}_N(0) \in W^{1, p}(\Pi ^N)$ for $p>dN$. If the initial data $\rho _N(0)$ and ${\bar{\rho }}_N(0)$ satisfy

$$\begin{aligned} \inf _{(\mathbf {x},\varvec{\xi }) \in \Pi ^N}\rho _N(0)> 0 \quad \text{ and } \quad \inf _{(\mathbf {x},\varvec{\xi }) \in \Pi ^N}{\bar{\rho }}_N(0) > 0, \end{aligned}$$

then the solutions $\rho _N$ and ${\bar{\rho }}_N$ remain positive.

Proof

We only show this for $\rho _N$ since almost the same argument can be applied for ${\bar{\rho }}_N$. Noticing $\xi _i \in \{1,2\}$ for all $i=1,\dots ,N$, for the proof, we use the induction on m, based on the number of cases $\xi _i = 2$. Let us start with $m=0$, i.e., $\xi _i = 1$ for all $i=1,\dots ,N$. In this case, we observe (2.5) to read

$$\begin{aligned} \partial _t \rho _N&= (\Delta _N + F^*_N)\rho _N \\&\quad + \alpha _1 \sum _{i=1}^N \left( \chi _2(\xi _i)\rho _N(\mathbf {x}, \tilde{\Theta }_2^i(\varvec{\xi }))- \chi _1(\xi _i)\rho _N(\mathbf {x}, \varvec{\xi }) \right) \\&\quad +\alpha _2 \sum _{i=1}^N\left( \chi _1(\xi _i)\rho _N(\mathbf {x}, \tilde{\Theta }_1^i(\varvec{\xi }))- \chi _2(\xi _i)\rho _N(\mathbf {x}, \varvec{\xi }) \right) \\&=(\Delta _N + F^*_N)\rho _N - \alpha _1 N \rho _N + \alpha _2 \sum _{i=1}^N \rho _N(\mathbf {x}, \tilde{\Theta }_1^i(\varvec{\xi })). \end{aligned}$$

Thus, by applying Feynman–Kac’s formula and continuity argument, we get

$$\begin{aligned} \inf _{(\mathbf {x},\varvec{\xi }) \in \Pi ^N}\rho _N(t) > 0. \end{aligned}$$

Indeed, let us consider a d-dimensional process ${\bar{X}}^i_t$ satisfying

$$\begin{aligned} {\bar{X}}^i_t = {\bar{X}}^i_0 +\sigma (1)\int _0^t \,dB^i_s + \frac{1}{N} \sum _{j=1}^N \int _0^t F({\bar{X}}^i_s-{\bar{X}}^j_s, 1)\,ds, \quad i=1,\dots , N, \end{aligned}$$

where $\bar{\mathbf {X}}_0^N = ({\bar{X}}^1_0, \dots , {\bar{X}}^N_0) = \mathbf {x}$. Then, the generator of the process $\bar{\mathbf {X}}_t^N = ({\bar{X}}^1_t, \dots , {\bar{X}}^N_t)$ is the following differential operator:

$$\begin{aligned} {\mathcal {G}}_N := \frac{\sigma (1)^2}{2} \sum _{i=1}^N \Delta _{x_i} -\frac{1}{N} \sum _{i,j=1}^N F(x_i-x_j, 1) \cdot \nabla _{x_i}. \end{aligned}$$

Since $x \mapsto F(x,1)$ is globally Lipschitz and $F \in L^\infty (\Pi )$, we find

$$\begin{aligned} \begin{aligned}&\rho _N(\mathbf {x},1,\dots , 1,t) \\&\quad = \mathbb {E} \exp \left( -\alpha _1 Nt -\int _0^t f(\bar{X}_s)\,ds \right) \rho _N(\bar{\mathbf {X}}_t^N, 1, \dots , 1, 0) \\&\qquad + \alpha _2 \sum _{k=1}^N \mathbb {E} \int _0^t \exp \left( -\alpha _1 N(t-s)-\int _s^t \sum _{i,j=1}^N f({\bar{X}}_\tau )\,d\tau \right) \rho _N(\bar{\mathbf {X}}_s^N, {\tilde{\Theta }}_1^k(\varvec{\xi }),s)\,ds \end{aligned} \end{aligned}$$

(B.5)

satisfies

$$\begin{aligned} \partial _t \rho _N(\mathbf {x},\varvec{\xi }) = {\mathcal {G}}_N \rho _N(\mathbf {x},\varvec{\xi }) - \left( N + f(x) \right) \rho _N(\mathbf {x},\varvec{\xi }) + \alpha _2 \sum _{i=1}^N \rho _N(\mathbf {x}, {\tilde{\Theta }}_1^i(\varvec{\xi })). \end{aligned}$$

Here

$$\begin{aligned} f(x) := \frac{1}{N} \sum _{i,j=1}^N (\nabla _{x_i} \cdot F)(x_i-x_j, 1) \end{aligned}$$

and

$$\begin{aligned} f({\bar{X}}_t) := \frac{1}{N} \sum _{i,j=1}^N (\nabla _{x_i} \cdot F)({\bar{X}}^i_t-{\bar{X}}^j_t, 1). \end{aligned}$$

On the other hand, for fixed $j \in \{1,\dots , N\}$, noticing

$$\begin{aligned} \chi _1({\tilde{\Theta }}_1^i(\varvec{\xi })_j) = 1 - \delta _{i}, \quad \chi _2({\tilde{\Theta }}_1^i(\varvec{\xi })_j) = \delta _{ij}, \end{aligned}$$

and

$$\begin{aligned} {\tilde{\Theta }}_j^i({\tilde{\Theta }}_1^i(\varvec{\xi })) = {\tilde{\Theta }}_j^i(\varvec{\xi }) \quad \text{ for } i=1,2, \end{aligned}$$

we obtain

$$\begin{aligned}&\partial _t \rho _N(\mathbf {x}, {\tilde{\Theta }}_1^i(\varvec{\xi })) \\&\quad = (\Delta _N + F^*_N)\rho _N(\mathbf {x}, {\tilde{\Theta }}_1^i(\varvec{\xi }))\\&\qquad + \alpha _1 \sum _{j=1}^N \left( \delta _{ij}\rho _N(\mathbf {x}, \tilde{\Theta }_2^i(\varvec{\xi })) - (1-\delta _{ij})\rho _N(\mathbf {x}, \tilde{\Theta }_1^i(\varvec{\xi })) \right) \\&\quad + \alpha _2 \sum _{j=1}^N \left( (1 - \delta _{ij}) \rho _N(\mathbf {x}, {\tilde{\Theta }}_1^i(\varvec{\xi })) - \delta _{ij} \rho _N(\mathbf {x}, \tilde{\Theta }_1^i(\varvec{\xi })) \right) . \end{aligned}$$

Since $\rho _N(\mathbf {x}, {\tilde{\Theta }}_2^i(\varvec{\xi })) = \rho _N(\mathbf {x}, 1,\dots , 1) = \rho _N(\mathbf {x}, \varvec{\xi })$, we find

$$\begin{aligned} \begin{aligned}&\partial _t \rho _N(\mathbf {x}, {\tilde{\Theta }}_1^i(\varvec{\xi })) \\&\quad = (\Delta _N + F^*_N)\rho _N(\mathbf {x}, {\tilde{\Theta }}_1^i(\varvec{\xi })) \\&\qquad + \alpha _1\rho _N(\mathbf {x}, \varvec{\xi }) - \left( \alpha _1(N-1) - \alpha _2 (N-2) \right) \rho _N(\mathbf {x}, {\tilde{\Theta }}_1^i(\varvec{\xi })). \end{aligned} \end{aligned}$$

(B.6)

We then define

$$\begin{aligned} { {\mathcal {T}} }:= \left\{ t> 0 : \inf _{\mathbf {x}\in \mathbb {T}^{dN}} \rho _N(\mathbf {x}, 1, \dots , 1, s) > 0 \quad \text{ for } \text{ all } \quad s \in [0,t)\right\} . \end{aligned}$$

By the continuity of $\rho _N$, it is clear that ${ {\mathcal {T}} }\ne \emptyset$. Set $t_* := \sup { {\mathcal {T}} }$ and suppose $t_* < \infty$. Then by definition, $\lim _{t \rightarrow t_*-}\inf _{\mathbf {x}\in \mathbb {T}^{dN}}\rho _N(\mathbf {x}, 1, \dots , 1, t) =0$. On the other hand, applying Feynman–Kac’s formula to (B.6) and using the positivity assumption on $\inf _{(\mathbf {x},\varvec{\xi }) \in \Pi ^N}\rho _N(0)$ yield

$$\begin{aligned} \min _{i=1,\dots ,N}\inf _{\mathbf {x}\in \mathbb {T}^{dN}}\rho _N(\mathbf {x}, \tilde{\Theta }_1^i(\varvec{\xi }),t) \ge 0 \end{aligned}$$

for $t \in [0,t_*)$ due to $\rho _N(\mathbf {x}, 1,\dots ,1,t) > 0$ for $t \in [0,t_*)$. We then combine this with (B.5) to get

$$\begin{aligned} \rho _N(\mathbf {x},1,\dots , 1,t) \ge \mathbb {E} \exp \left( -\alpha _1 Nt -\int _0^t f({\bar{X}}_s)\,ds \right) \rho _N(\bar{\mathbf {X}}_t^N, 1, \dots , 1, 0). \end{aligned}$$

Taking the infimum over $x \in \mathbb {T}^{dN}$ and limit $t \rightarrow t_*-$ on both sides of the above gives

$$\begin{aligned} 0 = \lim _{t \rightarrow t_*-}\inf _{\mathbf {x}\in \mathbb {T}^{dN}}\rho _N(\mathbf {x}, 1, \dots , 1, t) > 0, \end{aligned}$$

and this is a contradiction. Hence, $t_* = \infty$.

Now we assume that $\rho _N$ is positive for $m < N$ and consider the case $m+1$. Note that the particles are indistinguishable (see Remark 3). Thus, without loss of generality, we may assume that $\xi _i = 2$ for $i=1,\dots ,m+1$, and $\xi _i =1$ for $i=m+2,\dots ,N$. We notice that the following argument also holds when $m+1 = N$. Denoting by

$$\begin{aligned} u_N^{m+1} := \rho _N(\mathbf {x}, \underbrace{2,\dots ,2}_{m+1},\underbrace{1,\dots ,1}_{N-m-1}) =: \rho _N(\mathbf {x}, \varvec{\xi }_0), \end{aligned}$$

we get

$$\begin{aligned} \partial _t u_N^{m+1}&= (\Delta _N + F^*_N)u_N^{m+1} + \alpha _1 \sum _{i=1}^{m+1} \rho _N(\mathbf {x}, {\tilde{\Theta }}^i(\varvec{\xi }_0)) \\&\quad - \left( \alpha _1(N-m+1) + \alpha _2 (m+1)\right) u_N^{m+1}+ \alpha _2 \sum _{m+2}^N \rho _N(\mathbf {x}, {\tilde{\Theta }}^i_1(\varvec{\xi }_0)). \end{aligned}$$

On the other hand, ${\tilde{\Theta }}^i(\varvec{\xi }_0) = (2,\dots , 2,1,2,\dots ,2,1,\dots ,1)$, i.e., it has only m 2’s, thus by the inductive hypothesis, $\rho _N(\mathbf {x}, {\tilde{\Theta }}^i(\varvec{\xi }_0))$ is positive for $i=1,\dots , m+1$. Hence,

$$\begin{aligned} \partial _t u_N^{m+1}&\ge (\Delta _N + F^*_N)u_N^{m+1} - \left( \alpha _1(N-m+1) + \alpha _2 (m+1)\right) u_N^{m+1}\\&\quad + \alpha _2 \sum _{m+2}^N \rho _N(\mathbf {x}, \tilde{\Theta }^i_1(\varvec{\xi }_0)) . \end{aligned}$$

Similarly as before, we next observe the equation for $\rho _N(\mathbf {x}, {\tilde{\Theta }}^i_1(\varvec{\xi }_0))$, and then again apply Feynman–Kac’s formula and continuity argument to the above to have $u_N^{m+1} (t) > 0$ under the assumption that $\inf _{(\mathbf {x},\varvec{\xi }) \in \Pi ^N}\rho _N(0) > 0$. Then by the inductive reasoning, we conclude $\rho _N(t) > 0$. $\square$

We have finished the proof of (3.5).