A spatial stochastic epidemic model: law of large numbers and central limit theorem

Abstract

We consider an individual-based SIR stochastic epidemic model in continuous space. The evolution of the epidemic involves the rates of infection and recovering of individuals. We assume that individuals move randomly on the two-dimensional torus according to independent Brownian motions. We define the sequences of empirical measures, which describe the evolution of the positions of the susceptible, infected and removed individuals. We prove the convergence in probability, as the size of the population tends to infinity, of those sequences of measures towards the solution of a system of parabolic PDEs. We show that appropriately centrered renormalized sequences of fluctuations around the above limit converge in law, as the size of the population tends to infinity, towards a Gaussian distribution valued process, solution of a system of linear PDEs with highly singular Gaussian driving processes. In the case where the individuals do not move we also define and study the law of large numbers and central limit theorem for the same sequence.

Appendix

We first recall that for any $s>0$, the family $(\rho ^{i,s}_{n_{1},n_{2}})_{i,n_{1},n_{2}}$ (as defined in Proposition 2.3) is an orthonormal basis of $H^{s}({\mathbb {T}}^{2})$.

In this appendix we prove the next two Lemmas.

Lemma 7.1

We have,

$$\begin{aligned}&\displaystyle \underset{x\in {\mathbb {T}}^{2}}{\sup }\sum \limits _{i,n_{1},n_{2}}(\rho _{n_{1},n_{2}}^{i,s}(x))^{2}<\infty ~\text {iff}~ s>1,\\&\quad \displaystyle \underset{x\in {\mathbb {T}}^{2}}{\sup }\sum \limits _{i,n_{1},n_{2}}(\nabla \rho _{n_{1},n_{2}}^{i,s}(x))^{2}<\infty ~\text {iff}~s>2. \end{aligned}$$

Proof

As for any $x\in {\mathbb {T}}^{2}$, $i \in [|1,8|]$, $0\le (f_{n_{1},n_{2}}^{i}(x))^{2}\le 4$,

$$\begin{aligned} \displaystyle \sum \limits _{i,n_{1},n_{2}}(\rho _{n_{1},n_{2}}^{i,s}(x))^{2}= & {} \displaystyle \sum \limits _{i,n_{1},n_{2}}\frac{(f_{n_{1},n_{2}}^{i})^{2}(x)}{(1+\gamma \pi ^{2}(n_{1}^{2}+n_{2}^{2}))^{s}} , \hbox { and }\\ \displaystyle \sum \limits _{i,n_{1},n_{2}} (\nabla \rho _{n_{1},n_{2}}^{i,s}(x))^{2}= & {} \displaystyle \pi ^{2}\sum \limits _{i=1}^{4}\sum \limits _{n_{1}>0,n_{2}>0,even}\frac{n_{1}^{2}+n_{2}^{2}}{(1+\gamma \pi ^{2}(n_{1}^{2}+n_{2}^{2}))^{s}}(f_{n_{1},n_{2}}^{i})^{2}(x)\\&+\sum \limits _{n_{1}>0,even}\frac{\pi ^{2}n_{1}^{2}[(f_{n_{1},0}^{6})^{2}(x)+(f_{n_{1},0}^{5})^{2}(x)]}{(1+\gamma \pi ^{2}n_{1}^{2})^{s}}\\&+\sum \limits _{n_{2}>0,even}\frac{\pi ^{2}n_{2}^{2}[(f_{n_{1},0}^{8})^{2}(x)+(f_{n_{1},0}^{7})^{2}(x)]}{(1+\gamma \pi ^{2}n_{2}^{2})^{s}} \\ \end{aligned}$$

So

$$\begin{aligned} \displaystyle \sum \limits _{i,n_{1},n_{2}}\{(\rho _{n_{1},n_{2}}^{i,s}(x))^{2}\le & {} 1+ 16 \sum \limits _{n_{1}>0,n_{2}>0,even}\frac{1}{(1+\gamma \pi ^{2}(n_{1}^{2}+n_{2}^{2}))^{s}}\\&+\,8\sum \limits _{i=1}^{2}\sum \limits _{n_{i}>0,even}\frac{1}{(1+\gamma \pi ^{2}n_{i}^{2})^{s}}, and\\ \displaystyle \sum \limits _{i,n_{1},n_{2}}(\nabla \rho _{n_{1},n_{2}}^{i,s}(x))^{2}\le & {} 16\pi ^{2}\sum \limits _{n_{1}>0,n_{2}>0,even}\frac{n_{1}^{2}+n_{2}^{2}}{(1+\gamma \pi ^{2}(n_{1}^{2}+n_{2}^{2}))^{s}}\\&+\,8\pi ^{2}\sum \limits _{i=1}^{2}\sum \limits _{n_{i}>0,even}\frac{n_{i}^{2}}{(1+\gamma \pi ^{2}n_{i}^{2})^{s}}\\ \end{aligned}$$

Hence we see that:

• $\displaystyle \sum \limits _{i,n_{1},n_{2}}(\rho _{n_{1},n_{2}}^{i,s}(x))^{2}<\infty$ provided the series $\sum \limits _{n_{1}>0,n_{2}>0}\frac{1}{(1+\gamma \pi ^{2}(n_{1}^{2}+n_{2}^{2}))^{s}}$; $\sum \limits _{n_{1}>0}\frac{1}{(1+\gamma \pi ^{2}n_{1}^{2})^{s}}$ and $\sum \limits _{n_{2}>0}\frac{1}{(1+\gamma \pi ^{2}n_{2}^{2})^{s}}$ converge.

• $\displaystyle \sum \limits _{i,n_{1},n_{2}}(\nabla \rho _{n_{1},n_{2}}^{i,s}(x))^{2}<\infty$ provided the series $\sum \limits _{n_{1}>0,n_{2}>0}\frac{n_{1}^{2}+n_{2}^{2}}{(1+\gamma \pi ^{2}(n_{1}^{2}+n_{2}^{2}))^{s}}$; $\sum \limits _{n_{1}>0}\frac{n_{1}^{2}}{(1+\gamma \pi ^{2}n_{1}^{2})^{s}}$ and $\sum \limits _{n_{2}>0}\frac{n_{2}^{2}}{(1+\gamma \pi ^{2}n_{2}^{2})^{s}}$ converge.

- Convergence of the series

• (1) Convergence of $\sum \limits _{n_{1}>0,n_{2}>0}\frac{n_{1}^{2}+n_{2}^{2}}{(1+\gamma \pi ^{2}(n_{1}^{2}+n_{2}^{2}))^{s}}$

  It is so easy to see that $\sum \limits _{n_{1}\ge 1,n_{2}\ge 1}\frac{n_{1}^{2}+n_{2}^{2}}{(1+\gamma \pi ^{2}(n_{1}^{2}+n_{2}^{2}))^{s}}$ and $\displaystyle \int _{1}^{+\infty }\int _{1}^{+\infty }\frac{x^{2}+y^{2}}{(1+\gamma \pi ^{2}(x^{2}+y^{2}))^{s}}dxdy$ are of the same type (either convergent or divergent), and the latter is of the same type as

  $$\begin{aligned} \displaystyle \int _{1}^{+\infty }\frac{r^{3}}{(1+\gamma \pi ^{2}r^{2})^{s}}dr \le \frac{1}{\gamma ^{s}\pi ^{2s}}\int _{1}^{+\infty }r^{3-2s}dr \end{aligned}$$

  and $\int _{1}^{+\infty }r^{3-2s}dr<\infty$ iff $\hbox {s}>2$.

  Thus $\sum \limits _{n_{1}\ge 1,n_{2}\ge 1}\frac{n_{1}^{2}+n_{2}^{2}}{(1+\gamma \pi ^{2}(n_{1}^{2}+n_{2}^{2}))^{s}}$ converges iff $\hbox {s}>2$.

• (2) By the same argument as previously $\sum \limits _{n_{1}>0,n_{2}>0}\frac{1}{(1+\gamma \pi ^{2}(n_{1}^{2}+n_{2}^{2}))^{s}}$ converges for s>1.

• (3) By the comparison criterion the series $\sum \limits _{n_{1}>0}\frac{1}{(1+\gamma \pi ^{2}n_{1}^{2})^{s}}$ and $\sum \limits _{n_{2}>0}\frac{1}{(1+\gamma \pi ^{2}n_{2}^{2})^{s}}$ converge for $s>\frac{1}{2}$.

• (4) By the comparison criterion the series $\sum \limits _{n_{1}>0}\frac{n_{1}^{2}}{(1+\gamma \pi ^{2}n_{1}^{2})^{s}}$ and $\sum \limits _{n_{2}>0}\frac{n_{2}^{2}}{(1+\gamma \pi ^{2}n_{2}^{2})^{s}}$ converge for $s>\frac{3}{2}$ $\square$

Lemma 7.2

Under the assumption (H2), for any $t\ge 0$, we have

$$\begin{aligned} \underset{x}{\sup }\Big \Vert \displaystyle \frac{K(x,.)}{\int _{{\mathbb {T}}^{2}}K(x',.)\mu _{t}(dx')}\Big \Vert ^{2}_{H^{3}}<\infty \end{aligned}$$

Proof

We have

$$\begin{aligned} \Big \Vert \displaystyle \frac{K(x,.)}{\int _{{\mathbb {T}}^{2}}K(x',.)\mu _{t}(dx')}\Big \Vert ^{2}_{H^{3}}&=\displaystyle \sum \limits _{|\eta |\le 3}\int _{{\mathbb {T}}^{2}}\Big \vert D^{\eta } \frac{K(x,y)}{\int _{{\mathbb {T}}^{2}}K(x',y)\mu _{t}(dx')}\Big |^{2} dy, \end{aligned}$$

Now if we let $\displaystyle w_{t}(x,y)=\frac{K(x,y)}{\int _{{\mathbb {T}}^{2}}K(x',y)\mu _{t}(dx')}$, for any $y\in {\mathbb {T}}^{2}$, one has

$$\begin{aligned}&\displaystyle \frac{\partial w_{t}}{\partial {y_{1}}}(x,y)=\frac{\frac{\partial K}{\partial {y{1}}}(x,y)}{\int _{{\mathbb {T}}_{2}}K(u,y)\mu _{t}(du)}-\frac{K(x,y)\int _{{\mathbb {T}}^{2}}\frac{\partial K}{\partial {y_{1}}}(u,y)\mu _{t}(du)}{(\int _{{\mathbb {T}}^{2}}K(u,y)\mu _{t}(du))^{2}}.\\&\displaystyle \frac{\partial ^{2} w_{t}}{\partial {y_{2}y_{1}}}(x,y)=\frac{\frac{\partial ^{2} K}{\partial {y_{2}y_{1}}}(x,y)}{\int _{{\mathbb {T}}^{2}}K(u,y)\mu _{t}(du)}-\frac{\frac{\partial K}{\partial {y_{1}}}(x,y)\int _{{\mathbb {T}}^{2}}\frac{\partial K}{\partial {y_{2}}}(u,y)\mu _{t}(du)}{(\int _{{\mathbb {T}}^{2}}K(u,y)\mu _{t}(du))^{2}}\\&\quad -\displaystyle \frac{\frac{\partial K}{\partial {y_{2}}}(x,y)\int _{{\mathbb {T}}^{2}}\frac{\partial K}{\partial {y_{1}}}(u,y)\mu _{t}(du)+K(x,y)\int _{{\mathbb {T}}^{2}}\frac{\partial ^{2} K}{\partial {y_{2}y_{1}}}(u,y)\mu _{t}(du)}{(\int _{{\mathbb {T}}^{2}}K(u,y)\mu _{t}(du))^{2}}\\&\quad +2\frac{K(x,y)\int _{{\mathbb {T}}^{2}}\frac{\partial K}{\partial {y_{1}}}(u,y)\mu _{t}(du)\int _{{\mathbb {T}}^{2}}\frac{\partial K}{\partial {y_{2}}}(u,y)\mu _{t}(du)}{(\int _{{\mathbb {T}}^{2}}K(u,y)\mu _{t}(du))^{3}}. \\&\displaystyle \frac{\partial ^{3} w_{t}}{\partial {y_{1}y_{2}y_{1}}}(x,y)=\frac{\frac{\partial ^{3} K}{\partial {y_{1}y_{2}y_{1}}}(x,y)}{\int _{{\mathbb {T}}^{2}}K(u,y)\mu _{t}(du)}-2\frac{\frac{\partial ^{2} K}{\partial {y_{2}y_{1}}}(x,y)\int _{{\mathbb {T}}^{2}}\frac{\partial K}{\partial {y_{1}}}(u,y)\mu _{t}(du)}{(\int _{{\mathbb {T}}_{2}}K(u,y)\mu _{t}(du))^{2}}\\&\quad -\displaystyle \frac{\frac{\partial ^{2} K}{\partial {y_{1}y_{1}}}(x,y)\int _{{\mathbb {T}}^{2}}\frac{\partial K}{\partial {y_{2}}}(u,y)\mu _{t}(du)+2\frac{\partial K(x,y)}{\partial y_{1}}\int _{{\mathbb {T}}^{2}}\frac{\partial ^{2} K}{\partial {y_{1}y_{2}}}(u,y)\mu _{t}(du)}{(\int _{{\mathbb {T}}^{2}}K(u,y)\mu _{t}(du))^{2}}\\&\quad +4\frac{\frac{\partial K}{\partial y_{1}}(x,y)\int _{{\mathbb {T}}^{2}}\frac{\partial K}{\partial {y_{2}}}(u,y)\mu _{t}(du)\int _{{\mathbb {T}}^{2}}\frac{\partial K}{\partial {y_{1}}}(z,y)\mu _{t}(du)}{(\int _{{\mathbb {T}}^{2}}K(u,y)\mu _{t}(du))^{3}}\\&\quad -\displaystyle \frac{\frac{\partial K}{\partial y_{2}}(x,y)\int _{{\mathbb {T}}^{2}}\frac{\partial ^{2} K}{\partial {y_{2}y_{1}}}(u,y)\mu _{t}(du)+K(x,y)\int _{{\mathbb {T}}^{2}}\frac{\partial ^{3} K}{\partial {y_{1}y_{2}y_{1}}}(u,y)\mu _{t}(du)}{(\int _{{\mathbb {T}}^{2}}K(u,y)\mu _{t}(du))^{2}}\\&\quad \displaystyle +2\frac{\int _{{\mathbb {T}}^{2}}\frac{\partial K}{\partial {y_{1}}}(u,y)\mu _{t}(du)\Big [\frac{\partial K}{\partial {y_{2}}}(x,y)\int _{{\mathbb {T}}_{2}}\frac{\partial K}{\partial {y_{1}}}(u,y)\mu _{t}(du)+2K(x,y)\int _{{\mathbb {T}}^{2}}\frac{\partial ^{2} K}{\partial {y_{2}y_{1}}}(u,y)\mu _{t}(du)\Big ]}{(\int _{{\mathbb {T}}^{2}}K(u,y)\mu _{t}(du))^{3}}\\&\quad +2\frac{K(x,y)\int _{{\mathbb {T}}^{2}}\frac{\partial ^{2} K}{\partial {y_{1}y_{1}}}(u,y)\mu _{t}(du)\int _{{\mathbb {T}}^{2}}\frac{\partial K}{\partial {y_{2}}}(u,y)\mu _{t}(du)}{(\int _{{\mathbb {T}}^{2}}K(u,y)\mu _{t}(du))^{3}}\\&\quad +6\frac{K(x,y)\Big (\int _{{\mathbb {T}}^{2}}\frac{\partial K}{\partial {y_{1}}}(u,y)\mu _{t}(du)\Big )^{2}\int _{{\mathbb {T}}^{2}}\frac{\partial K}{\partial {y_{2}}}(u,y)\mu _{t}(du)}{(\int _{{\mathbb {T}}^{2}}K(u,y)\mu _{t}(du))^{4}}. \end{aligned}$$

Furthermore from Lemma 5.1, $\forall |\eta |\le 3$, $x\in {\mathbb {T}}^{2}$, $D^{\eta }K(x,y)$ is bounded by a constant independent of x. Thus since $\forall y\in {\mathbb {T}}^{2}$, $\int _{{\mathbb {T}}^{2}}K(u,y)\mu _{t}^{N}(du)=\int _{{\mathbb {T}}^{2}} K(u,y)f(t,u)(du)$ is lower bounded by a constant independent of y and $f(t,.)\le \delta _{2}$ then we deduce from the above calculations that $\displaystyle \sum \limits _{|\eta |\le 3}\int _{{\mathbb {T}}^{2}}\Big \vert D^{\eta } \frac{K(x,y)}{\int _{{\mathbb {T}}^{2}}K(x',y)\mu _{r}(dx')}\Big |^{2} dy$ is bounded by a constant independent of x. Hence the result.$\square$