# SIR Epidemics on Evolving Erdős-Rényi Graphs

Wenze Chen  School of Mathematics and Statistics, Jiangsu Normal University,
Xuzhou, China, 221116<EMAIL_ADDRESS>,  Yuewen Hou  School of
Mathematics and Statistics, Jiangsu Normal University, Xuzhou, China, 221116.
<EMAIL_ADDRESS>and  Dong Yao  Research Institute of Mathematics,
Jiangsu Normal University, Xuzhou, China, 221116<EMAIL_ADDRESS>
###### Abstract.

In the standard SIR model, infected vertices infect their neighbors at rate
$\lambda$ independently across each edge. They also recover at rate $\gamma$.
In this work we consider the SIR-$\omega$ model where the graph structure
itself co-evolves with the SIR dynamics. Specifically, $S-I$ connections are
broken at rate $\omega$. Then, with probability $\alpha$, $S$ rewires this
edge to another uniformly chosen vertex; and with probability $1-\alpha$, this
edge is simply dropped. When $\alpha=1$ the SIR-$\omega$ model becomes the
evoSIR model. Jiang et al. proved in [7] that the probability of an outbreak
in the evoSIR model converges to 0 as $\lambda$ approaches the critical
infection rate $\lambda_{c}$. On the other hand, numerical experiments in [7]
revealed that, as $\lambda\to\lambda_{c}$, (conditionally on an outbreak) the
fraction of infected vertices may not converge to 0, which is referred to as a
discontinuous phase transition. In [1] Ball and Britton give two (non-
matching) conditions for continuous and discontinuous phase transitions for
the fraction of infected vertices in the SIR-$\omega$ model. In this work, we
obtain a necessary and sufficient condition for the emergence of a
discontinuous phase transition of the final epidemic size of the SIR-$\omega$
model on Erdős-Rényi graphs, thus closing the gap between these two
conditions.

##  1\. Introduction

In the SIR model, individuals are in one of three states: $S=$ susceptible,
$I=$ infected, $R=$ removed (cannot be infected). Often this epidemic takes
place in a homogeneously mixing population. However, in the real world we
often have a graph $G$ that gives the social structure of the population;
vertices represent individuals and edges connections between them. $S-I$ edges
become $I-I$ at rate $\lambda$, i.e., after a time with an
Exponential($\lambda$) distribution. An individual remains infected for an
amount of time which has an Exponential($\gamma$) distribution. Once
individuals leave the infected state, they enter the removed state and stay
removed forever.

In many real world scenarios, the network structure itself doesn’t remain
fixed but rather evolves with time. Britton et al. [2] introduced the _SIR-
$\omega$ model_ where the graph is allowed to co-evolve with the epidemic. In
addition to the dynamics in the standard SIR model, in the SIR-$\omega$ model
$S-I$ edges are broken at rate $\omega$. With probability $1-\alpha$ the
susceptible individual drops this edge, and with probability $\alpha$ $S$ re-
connects to another individual chosen uniformly at random from the other $n-2$
individuals in the graph. This model is introduced in [2]. Mathematical
analysis for this model has been done in [1, 2, 8]. When $\alpha=1$, the
SIR-$\omega$ model is also called the _evoSIR model_ , where ‘evo’ stands for
‘evolving’. The evoSIR model was studied in [7] and the authors there proved
that as the infection rate $\lambda$ approaches the critical value
$\lambda_{c}$, the probability of an outbreak converges to 0, similarly to the
case of static graphs. In other words, the transition for the probability of
an outbreak is always continuous. However, as discovered by simulations in
[7], the evoSIR model exhibits _discontinuous phase transitions_ in terms of
the final epidemic size for certain values of parameters of the model. More
precisely, when $\lambda$ approaches $\lambda_{c}$, conditionally on an
outbreak the fraction of infected individuals _doesn’t_ converge to 0. This is
in sharp contrast with the case of SIR epidemics on static Erdős-Rényi graphs
(and the configuration model), where the phase transition for the fraction of
infecteds is always continuous. To avoid potential confusions, when we say
‘phase transition’ below we always refer to the transition for the final
epidemic size conditionally on an outbreak.

For the simpler _SI_ epidemic where individuals stay infected forever (i.e.,
no recoveries), [1] and [4] give an (almost) necessary and sufficient
condition for discontinuous phase transitions of the final epidemic size in
the _SI- $\omega$ model_ on Erdős-Rényi graphs and the _evoSI_ model on the
configuration model, respectively. Note that Erdős-Rényi graphs are contiguous
to the configuration model with Poisson degree distribution. In fact, [1]
found out the limit of scaled final epidemic size, which can be expressed as a
solution to certain equation.

However, the situation becomes much more complicated for SIR epidemics on
evolving graphs. [1] give two conditions for continuous and discontinuous
phase transitions of the SIR-$\omega$ model. But there is a gap between these
two conditions. In addition, not much is known about the final epidemic size.
See Section 1.2.3 for more details.

In this work we give a necessary and sufficient condition for the continuity
of the phase transition of the SIR-$\omega$ model (Theorem 1.1), resolving a
conjecture in [1, Remark 2.4]. For the case of continuous phase transitions we
also prove the convergence of (scaled) final epidemic size (Theorem 1.4).

The rest of the introduction is organized as follows. We state our main
results in Section 1.1. Section 1.2 is devoted to a quick review of related
work. Section 1.3 contains the proof strategies as well and the organization
of this paper.

###  1.1. Main results

We start from some terminologies and definitions. The term _final epidemic
size_ , denoted by $T^{(n)}$ when the graph has size $n$, refers to the total
number of individuals that have ever been infected. For SIR epidemics it is
the number of vertices that are eventually removed. We say an _outbreak_
occurs if the epidemic infects more than $\epsilon n$ individuals ($n$ is the
size of the total population) for some $\epsilon>0$ independent of $n$, i.e.,
$T^{(n)}\geq\epsilon n$. The critical infection rate $\lambda_{c}$ is the
smallest infection rate such that an outbreak occurs with probability bounded
away from 0 as $n\to\infty$. Mathematically,

$\lambda_{c}=\inf\\{\lambda>0:\liminf_{n\to\infty}\mathbb{P}(T^{(n)}\geq\epsilon
n)>0\mbox{ for some }\epsilon>0\\}.$ (1.1)

The phase transition for the final epidemic size is said to be continuous if
for any $\epsilon>0$,

$\lim_{\lambda\to\lambda_{c}}\mathbb{P}(T^{(n)}>n\epsilon|\mbox{an outbreak
occurs})=0.$ (1.2)

If (1.2) fails for some $\epsilon_{0}>0$, then the phase transition is said to
be discontinuous.

As mentioned in the beginning of this paper, the SIR-$\omega$ model has four
parameters: $\gamma$ representing the recovery rate, $\lambda$ the infection
rate, $\omega$ the rate of dropping or rewiring and $\alpha$ the probability
of rewiring. We focus on the Erdős-Rényi($n$,$\mu/n$) graph, which is a random
graph on $n$ vertices where each pair gets connected with probability $\mu/n$.

It has been proved in [7] for the evoSIR model and [1] for the SIR-$\omega$
model that

$\lambda_{c}=\frac{\omega+\gamma}{\mu-1}.$ (1.3)

Turning to the phase transition of the final epidemic size, Ball and Britton
proved in [1, Theorem 2.4] that if

$\omega(2\alpha-1)>\gamma\mbox{ and
}\mu>\frac{2\omega\alpha}{\omega(2\alpha-1)-\gamma},$ (1.4)

then there exists $r_{0}=r_{0}(\mu,\gamma,\omega,\alpha)>0$ such that

$\lim_{n\to\infty}\mathbb{P}(T^{(n)}>nr_{0}|\mbox{an outbreak occurs})=1\mbox{
for all }\lambda>\lambda_{c}.$ (1.5)

They also showed that the phase transition is continuous when

$\omega(3\alpha-1)\leq\gamma\mbox{ or
}\mu\leq\frac{2\omega\alpha}{\omega(3\alpha-1)-\gamma}.$ (1.6)

Note that there is a gap between (1.4) and (1.6). They conjectured in [1,
Remark 2.4] that (1.4) should be both necessary and sufficient for a
discontinuous phase transition. The following theorem, which is the main
contribution of this paper, confirms this conjecture.

######  Theorem 1.1.

Suppose

$\omega(2\alpha-1)\leq\gamma\mbox{ or
}\mu\leq\frac{2\omega\alpha}{\omega(2\alpha-1)-\gamma},$ (1.7)

then for any $\epsilon>0$, there exists $\lambda_{1}>\lambda_{c}$ such that

$\lim_{n\to\infty}\mathbb{P}(T^{(n)}<\epsilon n|\mbox{an outbreak
occurs})=1\mbox{ for all }\lambda\in(\lambda_{c},\lambda_{1}).$ (1.8)

Combining (1.8) and (1.5), the phase transition for the final epidemic size of
the SIR-$\omega$ model near $\lambda_{c}$ is discontinuous if and only if
(1.4) holds.

Theorem 1.1 is proved by showing the convergence of $T^{(n)}/n$ in the case of
(1.7). Our approach is inspired by [1]. We now introduce a few more notations.
Suppose the underlying graph has $n$ vertices, then we set

  * •

$S^{(n)}(t)$: the number of susceptible vertices at time $t$,

  * •

$I^{(n)}(t)$: the number of infected vertices at time $t$,

  * •

$I^{(n)}_{E}(t)$: the number of infectious (i.e., infected–susceptible) edges
at time $t$,

  * •

$\tau^{(n)}$: the first time when $I_{E}^{(n)}$ reaches 0,

  * •

$W^{(n)}(t)$: the number of susceptible–susceptible edges at time $t$ that
were created by rewirings,

  * •

${\mathbf{X}}^{(n)}(t)=(S^{(n)}(t),I^{(n)}(t),I^{(n)}_{E}(t),W^{(n)}(t))$.

Note that $\tau^{(n)}$ can be understood as the _terminal time_ since no more
vertices can be infected after $\tau^{(n)}$. Also, recall that we use
$T^{(n)}$ to denote the final epidemic size, which is equal to
$n-S^{(n)}(\tau^{(n)})$. As will be clear soon, ${\mathbf{X}}^{(n)}$ is more
amenable to analysis after a _random time-transform_. Specifically, we
multiply all transition rates of the SIR-$\omega$ process (which is a Markov
chain) by $n/\lambda I_{E}^{(n)}(t)$. We use a hat to denote the quantities in
the time-changed process. For instance, $\widehat{\tau}^{(n)}$ is the terminal
time in the time-transformed model. Note that
${\mathbf{X}}(0)=\widehat{{\mathbf{X}}}(0)$. Also, $\widehat{T}^{(n)}$ is the
final epidemic size in the time-changed model, which is necessarily equal to
$T^{(n)}$. For the ease of notations we sometimes suppress the dependence on
$n$ when there are no ambiguities. We emphasize that _these notations are
fixed throughout the paper_.

Our goal is to give a deterministic approximation for
$\widehat{\mathbf{X}}(t)$. Consider the following system of equations

$\displaystyle\frac{d\widehat{s}}{dt}$ $\displaystyle=-1,$ (1.9)
$\displaystyle\frac{d\widehat{i}}{dt}$
$\displaystyle=-\frac{\gamma}{\lambda}\frac{\widehat{i}}{\widehat{i}_{E}}+1,$
(1.10) $\displaystyle\frac{d\widehat{i}_{E}}{dt}$
$\displaystyle=-1-\frac{\gamma}{\lambda}+\mu\widehat{s}-\frac{\widehat{i}_{E}}{\widehat{s}}+2\frac{\widehat{w}}{\widehat{s}}-\frac{\omega}{\lambda}(1-\alpha+\alpha(1-\widehat{i})),$
(1.11) $\displaystyle\frac{d\widehat{w}}{dt}$
$\displaystyle=\frac{\omega}{\lambda}\alpha\widehat{s}-2\frac{\widehat{w}}{\widehat{s}}.$
(1.12)

As will be mentioned later, these equations can be obtained by dividing the
right hand sides of equations (1.22)-(1.25) by $\lambda i_{E}(t)$. In fact,
Ball and Britton also mentioned this system of equations in [1, Section
6.5.1], but they didn’t make use of this to analyze the SIR-$\omega$ model.
Note that $(\widehat{s}(t),\widehat{w}(t))$ already forms a closed sub-system,
which can be explicitly solved for any initial condition with
$\widehat{w}(0)=0$ as follows

$\displaystyle\widehat{s}(t)$ $\displaystyle=\widehat{s}(0)-t,$ (1.13)
$\displaystyle\widehat{w}(t)$
$\displaystyle=\frac{\omega\alpha}{\lambda}(\widehat{s}(0)-t)^{2}\log\frac{\widehat{s}(0)}{\widehat{s}(0)-t}.$
(1.14)

We will show in Theorem 1.2 below that the system (1.9)-(1.12) admits a unique
solution under initial conditions that are relevant to the evolution of the
SIR-$\omega$ model.

  * •

_Positive initial condition_ :

$\widehat{i}(0)\in(0,1),\,\widehat{s}(0)\in(0,1-\widehat{i}(0)),\,\widehat{i}_{E}(0)>0,\,\widehat{w}(0)=0.$

  * •

_Zero initial condition_ :

$\widehat{s}(0)=1,,\,\widehat{i}(0)=\widehat{i}_{E}(0)=0,\,\widehat{w}(0)=0.$

The positive initial condition corresponds to a positive fraction of
individuals initially infected while the zero initial condition is useful for
analyzing the case of one initially infected node.

We denote the solution to (1.9)-(1.12) by
$\widehat{\mathbf{x}}(t):=(\widehat{s}(t),\widehat{i}(t),\widehat{i}_{E}(t),\widehat{w}(t))$.
For any solution $\widehat{{\mathbf{x}}}(t)$, we let $t_{*}$ to be the first
time after 0 when $\widehat{i}_{E}(t)$ hits 0, i.e.,

$t_{*}:=\inf\\{0<t\leq\widehat{s}(0):\widehat{i}_{E}(t)=0\\}.$ (1.15)

Here we use the convention that $\inf\emptyset=\widehat{s}(0)$. We shall show
in Section 3.2 that $t_{*}<\widehat{s}(0)$.

We now state the existence and uniqueness theorem up to time $t_{*}$.

######  Theorem 1.2.

For any positive initial condition or zero initial condition (in this case
assuming additionally that $\lambda>\lambda_{c}$), there exists a unique
solution $\widehat{\mathbf{x}}(t)$ to the system (1.9) -(1.12) up to time
$t_{*}$.

With Theorem 1.2 established, we can now state the convergence theorem for the
stochastic process $\widehat{\mathbf{X}}/n$.

######  Theorem 1.3.

Suppose
$\widehat{\mathbf{X}}^{(n)}(0)/n\xrightarrow{\mathbb{P}}\widehat{\mathbf{x}}(0)$,
then

$\sup_{t\leq\widehat{s}(0)\wedge\widehat{\tau}^{(n)}}\left(\left|\frac{1}{n}\widehat{S}^{(n)}(t)-\widehat{s}(t)\right|+\left|\frac{1}{n}\widehat{W}^{(n)}(t)-\widehat{w}(t)\right|\right)\xrightarrow{\mathbb{P}}0,$
(1.16)

and

$\sup_{t\leq
t_{*}\wedge\widehat{\tau}^{(n)}}\left\|\frac{1}{n}\widehat{X}^{(n)}(t)-\widehat{x}(t)\right\|_{2}\xrightarrow{\mathbb{P}}0.$
(1.17)

Theorem 1.3 can be used derive the convergence of the scaled final epidemic
size $T^{(n)}/n$ under (1.7).

######  Theorem 1.4.

Assume that (1.7) holds. Let $t_{*}$ be given by (1.15).

  1. (a)

Suppose
$\widehat{\mathbf{X}}^{(n)}(0)/n\xrightarrow{\mathbb{P}}\widehat{\mathbf{x}}(0)$
where $\widehat{i}(0)>0$ and $\widehat{i}_{E}(0)>0$. Then,

$\frac{T^{(n)}}{n}\xrightarrow{\mathbb{P}}1-\widehat{s}(0)+t_{*}\mbox{ as
}n\to\infty.$ (1.18)

  2. (b)

Suppose $\lambda>\lambda_{c}$ and for each $n$ the epidemic starts from one
infected vertex with the rest being susceptible. Then conditionally on an
outbreak,

$\frac{T^{(n)}}{n}\xrightarrow{\mathbb{P}}t_{*}\mbox{ as }n\to\infty.$ (1.19)

######  Remark 1.

Note that the right hand sides of (1.18) and (1.19) can be written as
$1-\widehat{s}(t_{*})$. Theorem 1.4 verifies [1, Conjecture 2.1] under the
condition (1.7). We conjecture that this condition is not essential, i.e.,
Theorem 1.4 holds without assuming (1.7).

Theorem 1.1 follows from Theorem 1.4 via a direct computation. See Section 6.

###  1.2. Previous work

####  1.2.1. SIR on static graphs

So far the SIR model on several random graph models (including Erdős-Rényi
graphs and the configuration model) has been well understood. For the case of
Erdős-Rényi($n$,$\mu/n$) graphs, consider the function

$G_{1}(z)=\exp\left(-\mu\frac{\lambda}{\lambda+\gamma}(1-z)\right).$

The following theorem is well known. See, e.g., [9].

######  Theorem 1.5.

The critical value for SIR model on static Erdős-Rényi($n$,$\mu/n$) is given
by $\lambda_{c}=\gamma/(\mu-1)$. The probability of an outbreak converges to
$1-z_{0}$ where $z_{0}<1$ is the solution to the equation $G_{1}(z)=z$, i.e.,

$\lim_{n\to\infty}\mathbb{P}(T^{(n)}\geq\epsilon n)=1-z_{0}$

for all small enough $\epsilon>0$. In addition, conditionally on an outbreak,
the scaled final epidemic size $T^{(n)}/n$ also converges to $1-z_{0}$.

It follows from this theorem that the phase transition for final epidemic size
of the SIR model on static Erdős-Rényi graphs is continuous, since we have
$1-z_{0}\to 0$ as $\lambda\to\lambda_{c}$.

Similar results hold for the configuration model. Take any degree sequence
$(D_{1},\ldots,D_{n})$, denoted by $\mathbf{D}_{n}$, the configuration model
$\mathbb{CM}(n,\mathbf{D}_{n})$ is constructed by uniformly randomly pairing
$D_{1},\ldots,D_{n}$ half-edges attached to vertices $1,\ldots,n$,
respectively. Two paired half-edges form a single edge. If $D_{i},1\leq i\leq
n$, are i.i.d. sampled from a distribution $D$ (with minor adjustment to make
sure that the sum $\sum_{i=1}^{n}D_{i}$ is even), then we may write
$\mathbb{CM}(n,D)$ for $\mathbb{CM}(n,\mathbf{D}_{n})$. It is known that the
local neighborhood of $\mathbb{CM}(n,D)$ resembles a two-stage Galton-Watson
tree where the offspring distribution of the root is $D$ while the offspring
distribution of later generations is $D^{*}-1$ ($D^{*}$ is the size-biased
version of $D$). See, e.g., [11, Chapter 7] for a detailed introduction to the
configuration model. Let $z^{\prime}_{0}$ be the survival probability for this
two-stage Galton-Watson tree. Then for the SIR model on $\mathbb{CM}(n,D)$,
$\lambda_{c}=\gamma m_{1}/(m_{2}-2m_{1})$ where $m_{i}$ is the $i$-th moment
of $D$. In addition, both the probability of an outbreak and the scaled final
epidemic size (conditionally on an outbreak) converge to $z^{\prime}_{0}$,
which goes to 0 as $\lambda\to\lambda_{c}$. See, e.g., [6] for statements and
proofs of these results.

####  1.2.2. SI on evolving graphs

As mentioned before, in SI type models infected vertices will not recover.

For the SI-$\omega$ epidemics on Erdős-Rényi($n,\mu/n$) graphs, Ball and
Britton [1] found out the limit of $T^{(n)}/n$ (which can be expressed as the
solution to an explicit equation) and gave a necessary and sufficient
condition for the phase transition to be discontinuous. See [1, Theorem 2.5
and Theorem 2.6]. More precisely, they showed that the phase transition for
the the fraction of infecteds is discontinuous if and only if $\alpha>1/3$ and
$\mu>3\alpha/(3\alpha-1)$.

When $\alpha=1$, the SI-$\omega$ model is also called the evoSI model. Durrett
and Yao [4] analyzed the phase transition of final epidemic size in the evoSI
model with underlying graph $\mathbb{CM}(n,D)$. Let $\mu_{i}$ be the $i$-th
factorial moment of $D$, i.e.,

$\mu_{i}=\mathbb{E}[D(D-1)\cdots(D-i+1)].$

Set

$\Delta=-\frac{\mu_{3}}{\mu_{1}}+3(\mu_{2}-\mu_{1}).$ (1.20)

They showed that the phase transition is continuous if $\Delta<0$ and
discontinuous if $\Delta>0$. Note that when $D\sim\mathrm{Poisson}(\mu)$ and
$\alpha=1$, $\Delta=-\mu^{2}+3(\mu^{2}-\mu)$. Thus the condition $\Delta>0$
corresponds to $\mu>3/2$, which is the condition given by Ball and Britton by
setting $\alpha=1$.

####  1.2.3. SIR on evolving graphs

Jiang et al. [7] studied the evoSIR model (SIR-$\omega$ with $\alpha=1$) on
Erdős-Rényi($n$,$\mu/n$) graphs. They showed in [7, Theorem 1] that the
critical infection rate for evoSIR model is give by

$\lambda_{c}=\frac{\gamma+\omega}{\mu-1}.$ (1.21)

In addition, the probability of an outbreak converges to
$1-z^{\prime\prime}_{0}$ where $z^{\prime\prime}_{0}<1$ is the solution to the
equation $G_{2}(z)=z$ with $G_{2}$ defined by

$G_{2}(z)=\exp\left(-\mu\frac{\lambda}{\lambda+\omega+\gamma}(1-z)\right).$

Consequently, as $\lambda$ approaches $\lambda_{c}$, the probability of an
outbreak also converges to 0. The authors there also used numerical
experiments to discover that for certain parameters the phase transition for
final epidemic size is discontinuous. See Figure 1.

Figure 1. Simulation of the final epidemic size of the evoSIR model on an
Erdős-Rényi graph with $\mu=5$ and $\omega=4$. In this case
$\lambda_{c}=1.25$. The top curve is the simulation of final size of evoSIR.
The bottom curve is the final size of the delSIR (SIR-$\omega$ with
$\alpha=0$) epidemic with the same parameters. Other two curves are
approximations of evoSIR, which are discussed in [7].

Britton et al. introduced the SIR-$\omega$ model in [2]. Ball and Britton [1]
analyzed the SIR-$\omega$ model on an Erdős-Rényi($n$,$\mu/n$) graph using a
novel construction that couples the graph and the epidemic. We will explain
the construction in Section 2. Using this construction, they proved the
convergence of $\mathbf{X}^{(n)}(t)/n$ to a system of ODEs over any finite
time horizon. Specifically, let $\mathbf{x}(t)=(s(t),i(t),i_{E}(t),w(t))$ be
the solution of the following system of equations

$\displaystyle\frac{ds}{dt}$ $\displaystyle=-\lambda i_{E},$ (1.22)
$\displaystyle\frac{di}{dt}$ $\displaystyle=-\gamma i+\lambda i_{E},$ (1.23)
$\displaystyle\frac{di_{E}}{dt}$ $\displaystyle=-\lambda i_{E}-\gamma
i_{E}+\lambda\mu i_{E}s-\lambda\frac{i_{E}^{2}}{s}+2\lambda
i_{E}\frac{w}{s}-\omega i_{E}(1-\alpha+\alpha(1-i)),$ (1.24)
$\displaystyle\frac{dw}{dt}$ $\displaystyle=\omega\alpha i_{E}s-2\lambda
i_{E}\frac{w}{s}.$ (1.25)

Ball and Britton proved the following result in [1, Theorem 2.2].

######  Theorem 1.6.

Suppose ${\mathbf{X}}^{(n)}(0)/n\xrightarrow{\mathbb{P}}\mathbf{x}(0)$ where
$i(0)>0$ and $i_{E}(0)>0$. Then, for any $t_{0}>0$,

$\sup_{0\leq t\leq
t_{0}}\left\|\frac{{\mathbf{X}}^{(n)}(t)}{n}-\mathbf{x}(t)\right\|_{2}\xrightarrow{\mathbb{P}}0\mbox{
as }n\to\infty.$

Note that this theorem doesn’t tell us what happens when the time is $\infty$,
thus we cannot use it to deduce the final epidemic size. Ball and Britton
conjectured [1, Conjecture 2.1] that the limit of $n\to\infty$ and
$t\to\infty$ can be interchanged so that $T^{(n)}/n\to 1-s(\infty)$. Our
Theorem 1.4 confirms this conjecture under the condition (1.7).

By constructing models that are upper and lower bounds for the SIR-$\omega$
model, Ball and Britton proved in [1, Theorem 2.4] that the final epidemic
size has a discontinuous phase transition under (1.4) and a continuous
transition under (1.6). These two conditions have a gap. We will close this
gap by directly analyzing the SIR-$\omega$ model itself.

###  1.3. Proof strategies and organization of the paper

Our proof relies on the construction of the SIR-$\omega$ model given in [1].
We will give the details in the next section. Using that construction we can
derive evolution equations for quantities like $S^{(n)}(t)$. As pointed out in
[1], a common approach to analyze epidemic models is performing a suitable
time-transform so that the system becomes more solvable. In the case of the
SIR-$\omega$ model, after the time-change, the limiting system of ODEs change
from (1.22)-(1.25) to (1.9)-(1.12). Note that the equation for
$\widehat{i}^{\prime}$ involves the term $\widehat{i}/\widehat{i}_{E}$, which
is not Lipschitz for $\widehat{i}_{E}$ near 0. Thus one cannot directly apply
classical results (e.g., [3]) to obtain the convergence of the terminal time
$\widehat{\tau}^{(n)}$ and the fraction of infected vertices $T^{(n)}/n$.

We now give an quick overview of the proof of (1.17) in the case of positive
initial condition to illustrate the idea to overcome this barrier. Fix any
$\epsilon>0$, we can divide the interval $[0,t_{*}]$ into $[0,t_{*}-\epsilon]$
and $[t_{*}-\epsilon,t_{*}]$. We note that $\widehat{i}_{E}$ is bounded away
from 0 in $[0,t_{*}-\epsilon]$. Thus standard arguments (tightness plus the
uniqueness of the limit) can be used to derive (1.17) for $t\leq
t_{*}-\epsilon$. To deal with the other time interval
$[t_{*}-\epsilon,t_{*}]$, we need two observations:

  * •

$\widehat{i}(t_{*}-\epsilon)$ and $\widehat{i}_{E}(t_{*}-\epsilon)$ can be
arbitrarily small by taking $\epsilon$ to be small. It follows that
$\widehat{I}(t_{*}-\epsilon)/n$ and $\widehat{I}_{E}(t_{*}-\epsilon)/n$ must
also be small with high probability, by the convergence in the interval
$[0,t_{*}-\epsilon]$ that we just proved.

  * •

The evolution equations for $\widehat{I}$ and $\widehat{I}_{E}$, (4.32) and
(4.35), imply that the upward drift for $\widehat{I}$ and $\widehat{I}_{E}$ is
not large.

Combining these two observations, we conclude that $\widehat{I}/n$ and
$\widehat{I}_{E}/n$ remain small in the time interval
$[t_{*}-\epsilon,t_{*}]$. Since $\widehat{i}$ and $\widehat{i}_{E}$ are also
small in this interval, we finish the proof of (1.17).

In order to obtain the convergence of $\widehat{\tau}^{(n)}$ and prove Theorem
1.4, it is desirable to know whether the derivative of $\widehat{i}_{E}$ is
negative or 0 at $t_{*}$. However this seems not trivial. To overcome this
issue, we just consider the two cases $i^{\prime}_{E}(t_{*})<0$ and
$i^{\prime}_{E}(t_{*})=0$ separately. The first case is relatively easy. In
the second case we need more careful estimates and this is where we need the
condition (1.7).

To prove Theorem 1.1 we also need to separate the condition (1.7) into two
sub-cases according to whether the equality in (1.7) holds or not.

The rest of the paper is organized as follows. An alternative construction of
the SIR-$\omega$ model and several consequences of this construction are given
in Section 2. We prove the existence and uniqueness of the solution to the
system of ODEs (1.9)-(1.12) (Theorem 1.2) in Section 3. The convergence of the
time-changed SIR-$\omega$ model to differential equations (Theorem 1.3) is
proved in Section 4. We then use Theorem 1.3 to prove the limit of the scaled
final epidemic size under the condition (1.7) (Theorem 1.4) in Section 5.
Finally in Section 6, we use Theorem 1.4 to prove the continuity of the phase
transition under the condition (1.7) (Theorem 1.1).

##  2\. Construction of the SIR-$\omega$ process

The key to our proof is the following construction for the SIR-$\omega$ model
on Erdős-Rényi graphs, proposed in [1]. Initially there are $I(0)$ infected
vertices and $I_{E}(0)$ edges attached to these vertices. For each of these
edges it has a ‘free’ endpoint which is undetermined at this stage. No edges
are attached to other vertices. There are three types of Poisson process
occurring on vertices or edges. Below we use the term _infected edges_ to
denote the edges that are connected to infected vertices and have a free
endpoint.

  * •

Recovery Poisson processes for each infected vertex with rate $\gamma$. Each
infected vertex recovers with rate $\gamma$. When the recovery clock rings,
the infected vertex becomes recovered and all edges attached to this infected
vertex are no longer infected.

  * •

Rewiring/dropping Poisson processes for each infected edge with rate $\omega$.
Each infected edge has an independent Poisson process with rate $\omega$ on
it. Whenever the clock rings, this edge is dropped with probability $1-\alpha$
and rewired to a uniformly chosen vertex excluding itself with probability
$\alpha$. Suppose at time $t$ such a clock rings on an edge $e$ attached to
vertex $x$ and that edge decides to be rewired. Then vertex $x$ loses $e$ at
time $t$.

    * –

If $e$ is rewired to an infected vertex $y$, then the number of infected edges
of $y$ increases by 1. This case occurs with probability $(I(t)-1)/(n-1)$.

    * –

If $e$ is rewired to a susceptible vertex (which we don’t specify at this
moment), then $e$ is added to the pool of rewired susceptible-susceptible
edges $\mathcal{W}^{(n)}(t)$ and its cardinality $W^{(n)}(t)=W^{(n)}(t-)+1$.

    * –

If $e$ is rewired to a recovered vertex, then the effect is similar to
dropping.

  * •

Infection Poisson processes for each infected edge with rate $\lambda$. Each
infected edge is connected to a (uniformly chosen) susceptible vertex with
rate $\lambda$. Suppose infected edge $e$ is connected to susceptible vertex
$y$. Then $y$ immediately becomes infected. We now attach a random number of
edges to $y$ (which become infected edges immediately) given by

$\mbox{Poisson}\left(\frac{S^{(n)}(t)}{n}\mu\right)+\mbox{Binomial}\left(W^{(n)}(t-),\frac{2}{S^{(n)}(t-)}\right).$

We also delete all infected edges (other than those newly added to $y$) with
probability $1/S^{(n)}(t-)$, independently.

To explain the construction, note that we have required infected edges to be
paired to susceptible vertices only. The definition of Erdős-Rényi graphs
implies that the number of edges originally stemming from $y$ that connect to
susceptible vertices at time $t$ is close in distribution to
Poisson($S^{(n)}(t)\mu/n$). On the other hand, the number of rewired
susceptible edges in the pool $\mathcal{W}^{(n)}(t-)$ that have an endpoint as
$y$ has a $\mbox{Binomial}(W^{(n)}(t-),2/S^{(n)}(t-))$ distribution. Note that
all these operations have to be done immediately after the infection Poisson
clock rings.

It has been proved by Ball and Britton in [1, Section 2.3] that the process
just constructed (which we temporarily call the coupled SIR-$\omega$ model)
has the same law as the original SIR-$\omega$ model on Erdős-
Rényi($n$,$\mu/n$) graphs, after conditioning on the event

$\begin{split}\Omega_{\ref{simple}}:=\\{&\mbox{The graph induced from the
construction}\\\ &\mbox{of the couple SIR-$\omega$ model is a simple
graph}\\}.\end{split}$ (2.1)

Then Ball and Britton showed that

$\lim_{n\to\infty}\mathbb{P}(\Omega_{\ref{simple}})>0.$ (2.2)

Set

$\Omega_{\ref{simple0}}:=\\{\mbox{An outbreak occurs in the coupled
SIR-$\omega$ model}\\}.$ (2.3)

Using the techniques in the proof of [6, Theorems 2.7 and 2.10], we can show
that

$\liminf_{n\to\infty}\mathbb{P}(\\{\Omega_{\ref{simple0}}\cap\Omega_{\ref{simple}})>0.$
(2.4)

Equation (2.4) implies that it suffices to prove Theorem 1.1 for the coupled
SIR-$\omega$ model. To see this, we use a subscript ‘old’ to denote the
original SIR-$\omega$ model and ‘new’ to denote the couple SIR-$\omega$ model.
Suppose (1.8) holds for the coupled model, then for $\lambda<\lambda_{1}$,

$\lim_{n\to\infty}\mathbb{P}(T^{(n)}_{\textrm{new}}>\epsilon n)=0.$ (2.5)

By (2.4) and (2.5),

$\limsup_{n\to\infty}\mathbb{P}(T^{(n)}_{\textrm{new}}>\epsilon
n|\Omega_{\ref{simple0}}\cap\Omega_{\ref{simple}})\leq\limsup_{n\to\infty}\frac{\mathbb{P}(T^{(n)}_{\textrm{new}}>\epsilon
n)}{\mathbb{P}(\Omega_{\ref{simple0}}\cap\Omega_{\ref{simple}})}=0.$ (2.6)

Since the original model and the couple model are equivalent upon conditioning
on $\Omega_{\ref{simple}}$,

$\lim_{n\to\infty}\mathbb{P}(T^{(n)}_{\textrm{old}}>\epsilon
n|\Omega_{\ref{simple0}})=\lim_{n\to\infty}\mathbb{P}(T^{(n)}_{\textrm{new}}>\epsilon
n|\Omega_{\ref{simple0}}\cap\Omega_{\ref{simple}})=0.$

This proves Theorem 1.1 for the original model. Hence in this paper we will
not distinguish between the original SIR-$\omega$ model and the coupled
SIR-$\omega$ model.

In the rest of this section we will state several implications of the
construction, which will be used in later proofs.

######  Lemma 2.1.

Let $A_{e}$ be the total number of edges appearing in the SIR-$\omega$
process. Then exists a constant $C>0$ such that

$\mathbb{P}(A_{e}\geq 2\mu n)\leq\frac{C}{n}.$ (2.7)

###### Proof.

From the construction of the SIR-$\omega$ model, $A_{e}$ is bounded from above
the sum of two random variables

$I_{E}(0)+\sum_{i=1}^{n-I(0)}F_{i}\mbox{ where }F_{i}\mbox{ i.i.d.
}\sim\mbox{Poisson}(\mu).$

Clearly $I_{E}(0)$ is bounded from above by the sum of $I(0)n$ independent
Bernoulli($\mu/n$) variables. Let $U_{1}$ have the distribution
Binomial($n^{2}$, $\mu/n$) and $U_{2}$ the distribution Binomial($n^{2}/2$,
$\mu/n$). For the case $I(0)>n/2$,

$A_{e}\mbox{ is stochastically dominated by }U_{1}+\sum_{i=1}^{n/2}F_{i}.$

For the case $I(0)\leq n/2$,

$A_{e}\mbox{ is stochastically dominated by }U_{2}+\sum_{i=1}^{n}F_{i}.$

We have

$\mathbb{E}(U_{1})=\mu n,\,\textrm{Var}(U_{1})\leq\mu n.$ (2.8)

Thus

$\mathbb{P}(U_{1}\geq 5\mu n/4)=\mathbb{P}(U_{1}-\mathbb{E}(U_{1})\geq\mu
n/4)\leq\frac{\textrm{Var}(U_{1})}{(\mu n/4)^{2}}\leq\frac{C}{n}.$ (2.9)

We also have

$\mathbb{P}\left(\sum_{i=1}^{n/2}F_{i}\geq
3n\mu/4\right)=\mathbb{P}\left(\sum_{i=1}^{n/2}(F_{i}-\mu)\geq
n\mu/4\right)\leq\frac{\mathrm{Var}(Y_{1})n/2}{(n\mu/4)^{2}}\leq\frac{C}{n}.$
(2.10)

Combining (2.9) and (2.10),

$\begin{split}\mathbb{P}\left(A_{e}\geq 2\mu n,I(0)\geq
n/2\right)&\leq\mathbb{P}\left(U_{1}+\sum_{i=1}^{n/2}F_{i}\geq 2\mu
n\right)\\\ &\leq\mathbb{P}(U_{1}\geq 5\mu
n/4)+\mathbb{P}\left(\sum_{i=1}^{n/2}F_{i}\geq 3n\mu/4\right)\\\
&\leq\frac{C}{n}+\frac{C}{n}\leq\frac{2C}{n}.\end{split}$ (2.11)

This proves Lemma 2.1 for the case $I(0)\geq n/2$. The case $I(0)\leq n/2$ can
be proved similarly. Combining these two cases we deduce Lemma 2.1. ∎

By Lemma 2.1, $\mathbb{P}(A_{e}\geq 2\mu n)\to 0$. From now on we work on the
event $\\{A_{e}\leq 2\mu n\\}$.

######  Lemma 2.2.

The number of times that any given edge is rewired is stochastically dominated
by Geometric($\alpha\omega/(\omega+\lambda)$). Consequently, the number of
rewired edges that any given vertex receives is stochastically dominated by
Binomial($2\mu n$,$2\omega/(\lambda n)$).

###### Proof.

The rewiring/dropping Poisson processes and infection Poisson processes occur
independently for each edge in the construction of the SIR-$\omega$ model. By
properties of Poisson processes, the probability that rewiring/dropping occurs
before infection is $\omega/(\omega+\lambda)$. Also note that the probability
of rewiring is $\alpha$ in the rewiring/dropping Poisson processes. As a
result, for any given edge $e$,

$\mathbb{P}(\mbox{edge }e\mbox{ is rewired when one of the two clocks
rings})\leq\alpha\omega/(\omega+\lambda).$ (2.12)

Since whether each attempt to rewire succeeds or not is independent, the
number of times that any given edge is rewired is stochastically dominated by
a Geometric distribution with success probability
$\alpha\omega/(\omega+\lambda)$. On the other hand, when a rewiring occurs,
the probability of any given vertex $x$ being selected to receive the rewired
edge is at most $1/(n-1)$. As a result, the probability that any given edge
$e$ has been rewired to any given vertex $x$ is bounded from above by

$1-\mathbb{E}\left[\left(1-\frac{1}{n-1}\right)^{Z}\right]\mbox{ where
}Z\sim\mbox{Geometric}(\alpha\omega/(\omega+\lambda)),$ (2.13)

which is further bounded by (for large $n$)

$\mathbb{E}\left(\frac{Z}{n-1}\right)=\frac{1}{n-1}\left(\left(1-\frac{\alpha\omega}{\omega+\lambda}\right)^{-1}-1\right)\leq\frac{\omega}{(n-1)\lambda}\leq\frac{2\omega}{\lambda
n}.$ (2.14)

Since we are working on the event $\\{A_{e}\leq 2\mu n\\}$, we know that the
number of edges that have the potential to be rewired is less than or equal to
$2\mu n$. Since the infection Poisson processes and rewiring/dropping Poisson
processes are independent for different edges, the total number of rewired
edges that any given vertex receives is stochastically dominated by

$\mbox{Binomial}\left(2\mu n,\frac{2\omega}{\lambda n}\right).$

∎

######  Lemma 2.3.

Let $N_{i}$ be the number of edges added to vertex $i$ when it first becomes
infected. Then $N_{i}$ is stochasticallly dominated by the independent sum

$\mbox{Poisson}(\mu)+\mbox{Binomial}(2\mu n,\omega/(\lambda n)).$

Consequently, for some constant $C_{\ref{ni4}}$ we have

$\mathbb{E}(N_{i}^{4})\leq C_{\ref{ni4}}.$ (2.15)

Similarly, setting $\widehat{N}_{i}$ to be the number of edges of vertex $i$
when it becomes recovered, then

$\mathbb{E}(\widehat{N}_{i}^{4})\leq C_{\ref{ni4}}.$ (2.16)

###### Proof.

We know that $N_{i}$ is equal to the independent sum

$\mbox{Poisson}\left(\frac{S^{(n)}(t)}{n}\mu\right)+\mbox{Binomial}\left(W^{(n)}(t-),\frac{2}{S^{(n)}(t-)}\right)$

where the second part in the sum stands for the number of edges rewired to
vertex $i$ when $i$ first becomes infected. By Lemma 2.2 we see $N_{i}$ is
stochasticallly dominated by the independent sum Poisson($\mu$)+Binomial($2\mu
n$,$2\omega/(\lambda n)$). Set $V_{i}\sim\mbox{Poisson}(\mu)$,
$Z_{i}\sim\mbox{Binomial}(2\mu n$,$2\omega/(\lambda n))$. Then we have

$\mathbb{E}(N_{i}^{4})\leq\mathbb{E}((V_{i}+Z_{i})^{4}).$ (2.17)

Note that the fourth moment of Poisson($\mu$) is clearly finite, and

$\mathbb{E}(Z_{i}^{4})\leq\sum_{j=1}^{4}(2\mu
n)^{j}\left(\frac{2\omega}{\lambda n}\right)^{j}\leq C.$ (2.18)

Hence for some sufficiently large constant $C_{\ref{ni4}}$,

$\mathbb{E}((V_{i}+Z_{i})^{4})\leq\mathbb{E}(8(V_{i}^{4}+Z_{i}^{4}))\leq
C_{\ref{ni4}},$ (2.19)

which then implies $\mathbb{E}(N_{i}^{4})\leq C_{\ref{ni4}}$ and proves
(2.15). Equation (2.16) can be proved in exactly the same way as (2.15) since
$\widehat{N}_{i}$ is also dominated by the independent sum of Poisson($\mu$)
and Binomial($2\mu n$,$\omega/(\lambda n)$). ∎

##  3\. Existence and uniqueness of the solution to (1.11)-(1.11)

In this section we prove the existence and uniqueness of the solution to the
system (1.9)-(1.12). Note that we have already given closed form expressions
for $\widehat{s}(t)$ and $\widehat{w}(t)$ in (1.14) and (1.14) (which are
necessarily unique). Hence it remains to deal with $\widehat{i}$ and
$\widehat{i}_{E}$. We treat the case of positive initial condition in Section
3.1, and the case of zero initial condition in Section 3.2.

We now show that, the $t_{*}$ defined in (1.15) is strictly smaller than
$\widehat{s}(0)$. To see this, we use an argument by contradiction. Assume
that $t_{*}=\widehat{s}(0)$. Then $\widehat{i}_{E}(t)>0$ for all
$0<t<\widehat{s}(0)$. Equations (1.11), (1.13) and (1.14) together imply the
existence of an $\epsilon>0$ so that

$\frac{d\widehat{i}_{E}(t)}{dt}\leq-1-\frac{\widehat{i}_{E}(t)}{\widehat{s}(t)}$
(3.1)

holds true for all $t\in(\widehat{s}(0)-\epsilon,\widehat{s}(0))$. It follows
that

$\frac{d}{dt}\left(\frac{\widehat{i}_{E}(t)}{\widehat{s}(t)}\right)=\frac{1}{\widehat{s}(t)}\left(\frac{d\widehat{i}_{E}(t)}{dt}+\frac{\widehat{i}_{E}(t)}{\widehat{s}(t)}\right)\leq-\frac{1}{\widehat{s}(t)}.$
(3.2)

Integrating (3.2) from $\widehat{s}(0)-\epsilon$ to $t$, we see that

$\frac{\widehat{i}_{E}(t)}{\widehat{s}(t)}-\frac{\widehat{i}_{E}(\widehat{s}(0)-\epsilon)}{\widehat{s}(\widehat{s}(0)-\epsilon)}\leq\int^{t}_{\widehat{s}(0)-\epsilon}\frac{-1}{\widehat{s}(u)}du=\log\left(\frac{\widehat{s}(0)-t}{\epsilon}\right),$
(3.3)

which goes to $-\infty$ as $t\to\widehat{s}(0)$. We have thus arrived at a
contradiction. Therefore we must have $t_{*}<\widehat{s}(0)$.

###  3.1. Existence and uniqueness for positive initial condition

In this case the existence and uniqueness for the pair
$(\widehat{i},\widehat{i}_{E})$ follows from standard results in ODE theory.
Indeed, the right hand sides of (1.10) and (1.11) are Lipschitz functions of
$\widehat{i}$ and $\widehat{i}_{E}$ when $\widehat{i}_{E}$ is bounded away
from 0. Existence theory for ODE implies that, given any solution
$(\widehat{i},\widehat{i}_{E})$ defined for small $t>0$, one can extend its
domain to $t_{*}$ where $\lim_{t\to t_{*}}\widehat{i}_{E}(t)=0$. We set
$\widehat{i}_{E}(t_{*})=0$.

As for the uniqueness, assume that there are two solution
$(\widehat{i}_{1},\widehat{i}_{E,1})$ and
$(\widehat{i}_{2},\widehat{i}_{E,2})$. Given any $\delta>0$,

$\begin{split}b_{\delta,1}&=\inf\\{t\geq 0:\widehat{i}_{E,1}\leq\delta\\},\\\
b_{\delta,2}&=\inf\\{t\geq 0:\widehat{i}_{E,2}\leq\delta\\},\\\
b_{\delta}&=b_{\delta,1}\wedge b_{\delta,2}.\end{split}$ (3.4)

We define $t_{*,1}$ and $t_{*,2}$ to be the first time after 0 that
$\widehat{i}_{E,1}$ and $\widehat{i}_{E,2}$ hit 0, respectively. Then
$b_{\delta,1}\leq t_{*,1}<\widehat{s}(0)$. Hence $b_{\delta}<\widehat{s}(0)$.
By (1.10) and (1.11), for $t\leq b_{\delta}$,

$\begin{split}\left|\widehat{i}_{E,1}(t)-\widehat{i}_{E,2}(t)\right|&\leq\int_{0}^{t}\frac{\left|\widehat{i}_{E,1}(u)-\widehat{i}_{E,2}(u)\right|}{\widehat{s}(u)}du+\frac{\omega\alpha}{\lambda}\int_{0}^{t}\left|\widehat{i}_{1}(u)-\widehat{i}_{2}(u)\right|du\\\
&\leq\left(\frac{1}{\widehat{s}(0)-b_{\delta}}+\frac{\omega\alpha}{\lambda}\right)\int_{0}^{t}\left(\left|\widehat{i}_{E,1}(u)-\widehat{i}_{E,2}(u)\right|+\left|\widehat{i}_{1}(u)-\widehat{i}_{2}(u)\right|\right)du,\end{split}$
(3.5)

and

$\begin{split}\left|\widehat{i}_{1}(t)-\widehat{i}_{2}(t)\right|&\leq\frac{\gamma}{\lambda}\int_{0}^{t}\left|\frac{\widehat{i}_{1}(u)}{\widehat{i}_{E,1}(u)}-\frac{\widehat{i}_{2}(u)}{\widehat{i}_{E,2}(u)}\right|du\\\
&\leq\frac{\gamma}{\lambda}\
\int_{0}^{t}\frac{\left|\widehat{i}_{E,2}\widehat{i}_{1}(u)-\widehat{i}_{2}(u)\widehat{i}_{E,1}(u)\right|}{\widehat{i}_{E,1}(u)\widehat{i}_{E,2}(u)}du\\\
&\leq\frac{\gamma}{\lambda}\
\int_{0}^{t}\left(\frac{\left|\widehat{i}_{1}(u)-\widehat{i}_{2}(u)\right|}{\widehat{i}_{E,1}(u)}+\frac{\widehat{i}_{2}(u)\left|\widehat{i}_{E,1}(u)-\widehat{i}_{E,2}(u)\right|}{\widehat{i}_{E,1}(u)\widehat{i}_{E,2}(u)}\right)du\\\
&\leq\frac{\gamma}{\lambda}\left(\frac{1}{\delta}+\frac{1}{\delta^{2}}\right)\int_{0}^{t}\left(\left|\widehat{i}_{E,1}(u)-\widehat{i}_{E,2}(u)\right|+\left|\widehat{i}_{1}(u)-\widehat{i}_{2}(u)\right|\right)du.\end{split}$
(3.6)

Combining (3.5) and (3.6), we see that for all $t\leq b_{\delta}$ and some
constant $C>0$,

$\left|\widehat{i}_{E,1}(t)-\widehat{i}_{E,2}(t)\right|+\left|\widehat{i}_{1}(t)-\widehat{i}_{2}(t)\right|\leq
C\int_{0}^{t}\left(\left|\widehat{i}_{E,1}(u)-\widehat{i}_{E,2}(u)\right|+\left|\widehat{i}_{1}(u)-\widehat{i}_{2}(u)\right|\right)du.$

By Gronwall’s inequality we deduce that, for all $t\leq b_{\delta}$

$\widehat{i}_{E,1}(t)=\widehat{i}_{E,2}(t)\mbox{ and
}\widehat{i}_{1}(t)=\widehat{i}_{2}(t).$ (3.7)

Since $\delta$ can be taken to be arbitrarily small, we conclude that
$t_{*,1}=t_{*,2}$, and for all $t\leq t_{*,1}$,

$\widehat{i}_{E,1}(t)=\widehat{i}_{E,2}(t)\mbox{ and
}\widehat{i}_{1}(t)=\widehat{i}_{2}(t).$ (3.8)

This proves the uniqueness part.

###  3.2. Existence and uniqueness for zero initial condition

We start from the comparison principle for the system (1.10)-(1.11) under
positive initial condition.

######  Lemma 3.1.

Suppose that there are two solutions $(\widehat{i}_{1},\widehat{i}_{E,1})$ and
$(\widehat{i}_{2},\widehat{i}_{E,2})$ such that
$\widehat{i}_{1}(0)\leq\widehat{i}_{2}(0),\widehat{i}_{E,1}(0)\leq\widehat{i}_{E,2}(0)$
and $\widehat{i}_{E,2}(0)>0$. Then for all $t\leq t_{*,2}$,

$\widehat{i}_{1}(t)\leq\widehat{i}_{2}(t)\mbox{ and
}\widehat{i}_{E,1}(t)\leq\widehat{i}_{E,2}(t).$ (3.9)

###### Proof.

Denote the right hand of (1.10) and (1.11) by
$F_{1}(\widehat{i},\widehat{i}_{E})$ and $F_{2}(\widehat{i},\widehat{i}_{E})$,
respectively. It is clear that for any $(x_{1},y_{1})$ and $(x_{2},y_{2})$
such that $y_{1},y_{2}>0$, we have

$F_{1}(x_{1},y_{1})\leq F_{1}(x_{2},y_{2})\mbox{ if }x_{1}=x_{2}\mbox{ and
}y_{1}\leq y_{2},$

and

$F_{2}(x_{1},y_{1})\leq F_{2}(x_{2},y_{2})\mbox{ if }y_{1}=y_{2}\mbox{ and
}x_{1}\leq x_{2}.$

Thus the system (1.10)-(1.11) satisfies the Kamker-Müller conditions, which in
turn implies that this system is monotone with respect to the initial
condition. See [10] for general theories of monotone dynamical systems. ∎

######  Remark 2.

In fact, the comparison principle holds true for the time-reversed version of
(1.10) and (1.11) as well. In other words, if

$\widehat{i}_{1}(t)\leq\widehat{i}_{2}(t),\,\widehat{i}_{E,1}(t)\leq\widehat{i}_{E,2}(t)\mbox{
and }\widehat{i}_{E,1}(u)>0\mbox{ for all }u\in(0,t),$

then for all $u\leq t$,

$\widehat{i}_{1}(u)\leq\widehat{i}_{2}(u)\mbox{ and
}\widehat{i}_{E,1}(u)\leq\widehat{i}_{E,2}(u).$

######  Remark 3.

Suppose that there are two solutions $(\widehat{i}_{1},\widehat{i}_{E,1})$ and
$(\widehat{i}_{2},\widehat{i}_{E,2})$ such that
$\widehat{i}_{1}(t)\leq\widehat{i}_{2}(t)$ for all $t\leq t_{*,1}\wedge
t_{*,2}$. Then one can show that
$\widehat{i}_{E,1}(t)\leq\widehat{i}_{E,2}(t)$ for all $t\leq t_{*,1}\wedge
t_{*,2}$.

Recall that for zero initial condition $\widehat{i}(0)=\widehat{i}_{E}(0)=0$
and $\widehat{s}(0)=1$. Using Lemma 3.1 we can deduce the existence of a
solution to (1.10) and (1.11). Indeed, consider the pair of functions
$(\widehat{i}_{\epsilon},\widehat{i}_{E,\epsilon})$ given by the solution to
(1.10)-(1.11) with the initial condition
$\widehat{i}_{\epsilon}(0)=\widehat{i}_{E,\epsilon}(0)=\epsilon$. (The
existence of $(\widehat{i}_{\epsilon},\widehat{i}_{E,\epsilon})$ is guaranteed
by the results proved in Section (3.1).) Set
$t^{\epsilon}_{*}:=\inf\\{t>0:\widehat{i}_{E,\epsilon}(t)=0\\}$. We claim that

$\liminf_{\epsilon\to 0}t^{\epsilon}_{*}>0.$ (3.10)

###### Proof of (3.10).

Using (1.11), we see that for some $C_{\ref{hati'}}>0$,

$\left|\widehat{i}^{\prime}_{E,\epsilon}\right|\leq
C_{\ref{hati'}},\forall\,\epsilon\leq 1,\,t\leq 1/2.$ (3.11)

Hence $\widehat{i}_{E,\epsilon}\leq\epsilon+C_{\ref{hati'}}t$. Hence

$\begin{split}\frac{d\widehat{i}_{E,\epsilon}}{dt}&=-1-\frac{\gamma}{\lambda}-\frac{\omega}{\lambda}+\mu(1-t)-\frac{\widehat{i}_{E,\epsilon}}{1-t}+\frac{2\widehat{w}}{\widehat{s}}+\frac{\omega\alpha}{\lambda}\widehat{i}_{\epsilon}\\\
&\geq\mu-1-\frac{\gamma+\omega}{\lambda}-\mu
t-\frac{\epsilon+C_{\ref{hati'}}t}{1-t}.\end{split}$ (3.12)

Thanks to the assumption $\lambda<\lambda_{c}=(\gamma+\omega)/(\mu-1)$ for
zero initial condition,

$\mu-1-\frac{\gamma+\omega}{\lambda}>\mu-1-\frac{\gamma+\omega}{\lambda_{c}}=\mu-1-\frac{\gamma+\omega}{(\omega+\gamma)/(\mu-1)}=0.$
(3.13)

By (3.11), (3.12) and (3.13),

$\widehat{i}^{\prime}_{E,\epsilon}\geq 0\mbox{ for
}t\leq\left(\mu-1-\frac{\gamma+\omega}{\lambda}-2\epsilon\right)\min\left\\{\frac{1}{2\mu},\left(\left(\mu-1-\frac{\gamma+\omega}{\lambda}\right)+2C_{\ref{hati'}}\right)^{-1}\right\\}.$

This implies that $t^{\epsilon}_{*}$ is uniformly bounded from below for all
small $\epsilon$. Hence (3.10) follows. ∎

Define $t^{0}_{*}=\liminf_{\epsilon\to 0}t^{\epsilon}_{*}>0$. By Lemma 3.1,
for each $t<t^{0}_{*}$, $\widehat{i}_{\epsilon}(t)$ and
$\widehat{i}_{E,\epsilon}(t)$ are both decreasing in $\epsilon$. By Lemma 3.1,
$t^{\epsilon}_{*}$ actually monotonically decreases to $t^{0}_{*}$. Thus we
can define

$\begin{split}\widehat{i}(t)&=\lim_{\epsilon\to
0}\widehat{i}_{\epsilon}(t),\\\ \widehat{i}_{E}(t)&=\lim_{\epsilon\to
0}\widehat{i}_{E,\epsilon}(t).\end{split}$ (3.14)

Clearly $(\widehat{i}(t),\widehat{i}_{E}(t))$ satisfies (1.10) and (1.11) with
$\widehat{i}(0)=\widehat{i}_{E}(0)=0$. Thus we obtain a nontrivial solution of
(1.10)-(1.11) since $t^{0}_{*}>0$. To prepare for the proof of uniqueness we
need one more ingredient. Suppose that $(\widehat{i}(t),\widehat{i}_{E}(t))$
is a solution to (1.10)-(1.11). We claim that

$\limsup_{t\to 0}\frac{\widehat{i}(t)}{\widehat{i}_{E}(t)}<\infty.$ (3.15)

To see this, note that by (1.10),

$\widehat{i}(t)\leq t,\,\forall t\geq 0.$ (3.16)

Repeating the calculation in (3.12),

$\widehat{i}^{\prime}_{E}(0)=\lim_{t\to
0}\frac{\widehat{i}_{E}(t)}{t}=-1-\frac{\gamma}{\lambda}+\mu-\frac{\omega}{\lambda}>-\frac{\gamma+\omega}{\lambda_{c}}+\mu-1=-\frac{\gamma+\omega}{(\mu-1)/(\gamma+\omega)}+\mu-1=0.$
(3.17)

Equation (3.15) follows from (3.16) and (3.17).

We now prove the uniqueness part. Suppose that
$(\widehat{i}_{1}(t),\widehat{i}_{E,1}(t))$ and
$(\widehat{i}_{2}(t),\widehat{i}_{E,2}(t))$ are two solutions that satisfy
(1.10)-(1.11). We divide the proof into two steps.

Step 1: We first show that either

$\widehat{i}_{1}(t)\leq\widehat{i}_{2}(t),\,\widehat{i}_{E,1}(t)\leq\widehat{i}_{E,2}(t),\,\forall
t\leq t_{*,1}\wedge t_{*,2},$ (3.18)

or

$\widehat{i}_{1}(t)\geq\widehat{i}_{2}(t),\,\widehat{i}_{E,1}(t)\geq\widehat{i}_{E,2}(t),\,\forall
t\leq t_{*,1}\wedge t_{*,2},$ (3.19)

To see this, we consider two cases.

  * •

Case ①: there exists $t_{1}\in(0,t_{*,1}\wedge t_{*,2})$ such that
$\widehat{i}_{1}(t_{1})=\widehat{i}_{2}(t_{1})$. Without loss of generality
let us assume that $\widehat{i}_{1}(t_{1})=\widehat{i}_{2}(t_{1})$ and
$\widehat{i}_{E,1}(t_{1})\leq\widehat{i}_{E,2}(t_{1})$. Then by the comparison
principle for the system (1.10)-(1.11) and its backward-in-time version
(Remark 2),

$\widehat{i}_{1}(t)\leq\widehat{i}_{2}(t),\,\widehat{i}_{E,1}(t)\leq\widehat{i}_{E,2}(t),\,\forall
t\leq t_{*,1}\wedge t_{*,2}.$

This proves (3.18).

  * •

Case ②: the opposite of Case ①. In this case, we must have either

$\widehat{i}_{1}(t)\leq\widehat{i}_{2}(t),\,\forall t\in(0,t_{*,1}\wedge
t_{*,2}),$

or

$\widehat{i}_{1}(t)\geq\widehat{i}_{2}(t),\,\forall t\in(0,t_{*,1}\wedge
t_{*,2}).$

Using Remark 3 we see that either (3.18) or (3.19) holds true.

Step 2: Without loss of generality we assume that (3.18) holds true. Using
(1.11),

$\begin{split}\widehat{i}_{E,2}(t)-\widehat{i}_{E,1}(t)&\leq\int_{0}^{t}\frac{\omega\alpha}{\lambda}(\widehat{i}_{2}(u)-\widehat{i}_{1}(u))du.\end{split}$
(3.20)

Define the function

$H(t):=\sup_{0\leq u\leq
t}\left|\widehat{i}_{1}(u)-\widehat{i}_{2}(u)\right|.$

For $t$ small, using (1.10), (3.17), (3.15) and (3.20),

$\begin{split}\widehat{i}_{2}(t)-\widehat{i}_{1}(t)&=\frac{\gamma}{\lambda}\int_{0}^{t}\left(\frac{\widehat{i}_{1}(u)}{\widehat{i}_{E,1}(u)}-\frac{\widehat{i}_{2}(u)}{\widehat{i}_{E,2}(u)}\right)du\\\
&\leq\frac{\gamma}{\lambda}\int_{0}^{t}\left(\frac{\widehat{i}_{1}(u)}{\widehat{i}_{E,1}(u)}-\frac{\widehat{i}_{1}(u)}{\widehat{i}_{E,2}(u)}\right)du\\\
&=\frac{\gamma}{\lambda}\int_{0}^{t}\frac{\widehat{i}_{1}(u)(\widehat{i}_{E,2}(u)-\widehat{i}_{E,1}(u))}{\widehat{i}_{E,1}(u)\widehat{i}_{E,2}(u)}du\\\
&\leq
C\int_{0}^{t}\frac{1}{u}\left(\int_{0}^{u}\left(\widehat{i}_{2}(r)-\widehat{i}_{1}(r)\right)dr\right)du.\end{split}$
(3.21)

It follows that

$H(t)\leq C\int_{0}^{t}\frac{1}{u}uH(u)du\leq C\int_{0}^{t}H(u)du.$ (3.22)

Since $H(0)=0$, applying the Gronwall’s ineqaulity we deduce that $H(t)=0$ for
all $t$ small. This implies that $\widehat{i}_{E,1}(t)=\widehat{i}_{E,2}(t)$
for all $t$ small enough (say, for $t\leq\epsilon_{0}$). From the proof of the
results in Section 3.1 and the fact that
$\widehat{i}_{E,1}(\epsilon_{0})=\widehat{i}_{E,2}(\epsilon_{0})>0$, we see
that $\widehat{i}_{E,1}(t)=\widehat{i}_{E,2}(t)$ for all $\epsilon_{0}\leq
t\leq t_{*,1}\wedge t_{*,2}$. Hence $t_{*,1}=t_{*,2}$ and
$(\widehat{i}_{1}(t),\widehat{i}_{E,1}(t))=(\widehat{i}_{2}(t),\widehat{i}_{E,2}(t))$
for all $t\leq t_{*,1}$.

##  4\. Approximations of the SIR-$\omega$ process by differential equations

The proof of Theorem 1.3 is based on Dykin’s formula applied to the
construction of the SIR-$\omega$ model using Poisson processes in Section 2.
Dykin’s formula says that, if $V(t)$ is a Markov chain, then for any function
$f$ in the domain of $L$,

$f(V(t))-\int_{0}^{t}Lf(V(s))ds$

is a martingale, where $L$ is the Markov generator for $V(t)$.

This section is divided into three subsections. We first prove (1.16). Then we
establish (1.17) for the case of positive initial condition where
$\widehat{i}_{E}(0)$ and $\widehat{i}(0)$ are positive. Finally, we prove
(1.17) for zero initial condition case where
$\widehat{i}(0)=\widehat{i}_{E}(0)=0$.

###  4.1. Proof of (1.16)

We start from the analysis of $\widehat{S}(t)$. Corresponding to the three
types of Poisson processes, we can divide the change of $\widehat{S}(t)$ into
three categories.

  * •

Change of $\widehat{S}(t)$ due to recovery Poisson processes: 0.

  * •

Change of $\widehat{S}(t)$ due to rewiring/dropping Poisson processes: 0.

  * •

Change of $\widehat{S}(t)$ due to infection Poisson Processes: decreases by 1.
The rate for this to occur is

$\lambda\widehat{I}_{E}(t)\times\frac{n}{\lambda\widehat{I}_{E}(t)}=n.$

Here the factor $\frac{n}{\lambda\widehat{I}_{E}(t)}$ comes from the fact that
we are considering the time-changed dynamics.

Collecting these three types of changes and applying Dykin’s formula, we see
that

$\widehat{S}(t)=\widehat{S}(0)+\int_{0}^{t}(-n)du+M_{1}(t),$ (4.1)

where $M_{1}(t)$ is a martingale. Note that we can also define $M_{1}(t)$ for
$t\geq\widehat{\tau}$ ($\widehat{\tau}$ is the first time that
$\widehat{I}_{E}$ reaches 0) by simply setting
$M_{1}(t)=M_{1}(\widehat{\tau})$. Then $M_{1}(t)$ is a martingale defined for
all $t\geq 0$.

By (4.1), the quadratic variation process of $M_{1}(t)$, $\langle
M_{1}(t),M_{1}(t)\rangle$, is equal to the quadratic variation process of
$\widehat{S}(t)$. The quadratic variation process of $\widehat{S}(t)$ can be
bounded using previous analysis for the change of $\widehat{S}(t)$. In
particular, since the clocks in infection Poisson processes can effectively
ring at most $n$ times, we have

$\langle\widehat{S}(t),\widehat{S}(t)\rangle\leq n\times 1^{2}=n,\forall
t\leq\widehat{\tau}.$

Consequently,

$\sup_{t\geq 0}\mathbb{E}\left(M_{1}(t)^{2}\right)\leq n.$ (4.2)

By $L^{2}$ maximal inequality applied to the martingale $M_{1}(t)$, we have

$\mathbb{E}\left(\sup_{0\leq t\leq\widehat{\tau}}M_{1}(t)^{2}\right)\leq 4n.$
(4.3)

Markov inequality then implies that

$\mathbb{P}\left(\sup_{t\leq\widehat{\tau}}\left|M_{1}(t)\right|>n^{2/3}\right)\leq\frac{4n}{n^{4/3}}\leq
4n^{-1/3}.$ (4.4)

Dividing both sides of (4.1), we get

$\frac{\widehat{S}(t)}{n}=\frac{\widehat{S}(0)}{n}-t+\frac{M_{1}(t)}{n}.$
(4.5)

Recall that we have assumed that
$\widehat{S}(0)/n\xrightarrow{\mathbb{P}}\widehat{s}(0)$. Combining (4.4) and
(4.5),

$\sup_{t\leq\widehat{\tau}}\left|\widehat{S}(t)/n-(\widehat{s}(0)-t)\right|\xrightarrow{\mathbb{P}}0,$
(4.6)

as desired.

Now we analyze $\widehat{W}(t)$.

  * •

Change of $\widehat{W}(t)$ due to recovery Poisson processes: 0.

  * •

Change of $\widehat{W}(t)$ due to rewiring/dropping Poisson processes:
increases by 1 with probability $\alpha\widehat{S}(t)/n$, otherwise stays the
same. Thus the mean of the change is $\alpha\widehat{S}(t)/n$. The rate for
such an event to occur is

$\omega\widehat{I}_{E}(t)\times\frac{n}{\lambda\widehat{I}_{E}(t)}=\frac{n\omega}{\lambda}.$

  * •

Change of $\widehat{W}(t)$ due to infection Poisson processes: decreases by

$\mbox{Binomial}(\widehat{W}(t),2/\widehat{S}(t)).$

The mean of the change is $2\widehat{W}(t)/\widehat{S}(t)$. The rate for such
an event to occur is

$\lambda\widehat{I}_{E}(t)\times\frac{n}{\lambda\widehat{I}_{E}(t)}=n.$

Hence we can write

$\widehat{W}(t)=\widehat{W}(0)+\int_{0}^{t}\left(\frac{\alpha\widehat{S}(u)}{n}\frac{n\omega}{\lambda}-n\frac{2\widehat{W}(u)}{\widehat{S}(u)}\right)du+M_{2}(t).$
(4.7)

We first control the martingale $M_{2}(t)$. Using our analysis for the change
of $\widehat{W}(t)$,

$\langle
M_{2}(t),M_{2}(t)\rangle=\langle\widehat{W}(t),\widehat{W}(t)\rangle\leq\sum_{i=1}^{n}N_{i}^{2}+Q(t),$
(4.8)

where $Q(t)$ is the number of arrivals of rewiring/dropping Poisson processes
by time $t$ and $N_{i}$ is the number of edges added to vertex $i$ when vertex
$i$ first becomes infected. Since $Q(t)$ has total rate $\omega n/\lambda$,

$\mathbb{E}(Q(t))\leq\frac{\omega nt}{\lambda}.$ (4.9)

By Lemma 2.3 we see

$\mathbb{E}\left(\sum_{i=1}^{n}N_{i}^{2}\right)\leq Cn.$ (4.10)

Combining (4.9) and (4.10) we see that there exists some constant $C$ such
that for all $t\leq 2$,

$\mathbb{E}\langle M_{2}(t),M_{2}(t)\rangle\leq Cn.$ (4.11)

Similarly to (4.4) we deduce that

$\mathbb{P}\left(\sup_{t\leq\widehat{\tau}\wedge
2}\left|M_{2}(t)\right|>n^{2/3}\right)\leq\frac{Cn}{n^{4/3}}\leq Cn^{-1/3}.$
(4.12)

To finish the proof of (1.16), we need to show

$\sup_{t\leq\widehat{s}(0)\wedge\widehat{\tau}}\left|\widehat{W}(t)/n-\widehat{w}(t)\right|\xrightarrow{\mathbb{P}}0.$
(4.13)

We first prove a weaker version of (4.13). Fix any $\epsilon>0$. We now prove

$\sup_{0\leq
t\leq(\widehat{s}(0)-\epsilon)\wedge\widehat{\tau}}\left|\widehat{W}(t)/n-\widehat{w}(t)\right|\xrightarrow{\mathbb{P}}0.$
(4.14)

###### Proof of (4.14).

We prove (4.14) in two steps. First we show that $\widehat{W}(t)/n$ is a tight
sequence in D[0,$\widehat{s}(0)-\epsilon$], the space of RCLL functions on the
interval [0,$\widehat{s}(0)-\epsilon$]. Then we show the uniqueness of
possible sequential limit.

Since $\widehat{W}(0)\to\widehat{w}(0)$, we know
$\\{\widehat{W}(0)/n\\}_{n\geq 1}$ is a tight sequence of random variables. To
establish tightness of $\\{\widehat{W}(t)/n,0\leq t\leq 1-\epsilon\\}_{n\geq
1}$, we need to show that for any fixed $\epsilon^{\prime},\delta>0$, there
exist $\theta>0$ and an integer $n_{0}$ so that for all $n\geq n_{0}$,

$\mathbb{P}\left(\sup_{\left|t_{1}-t_{2}\right|\leq\theta,t_{1}<t_{2}\leq\widehat{s}(0)-\epsilon}\left|\frac{\widehat{W}(t_{1})}{n}-\frac{\widehat{W}(t_{2})}{n}\right|\geq\delta\right)\leq\epsilon^{\prime}.$
(4.15)

Assuming (4.15) for the moment, we see that $\\{\widehat{W}(t)/n,0\leq
t\leq\widehat{s}(0)-\epsilon\\}$, as an element of
D[0,$\widehat{s}(0)-\epsilon$], satisfies condition (ii) of Proposition 3.26
in [5]. Consequently, $\\{\widehat{W}(t)/n,0\leq
t\leq\widehat{s}(0)-\epsilon\\}_{n\geq 1}$ is a tight sequence.

Now we show that any sequential limit of $\widehat{W}(t)/n$ coincides with
$\widehat{w}(t)$. By the tightness of $\\{\widehat{W}(t)/n,t\geq 0\\}_{n\geq
1}$, we see for any subsequence of $\widehat{W}(t)/n$ we can extract a further
subsequence that converges in distribution to a process $\check{w}(t)$ with
continuous sample path. By the Skorokhod representation theorem we can assume
the convergence is actually in the almost sure sense, and we can also assume
that $\widehat{S}(t)/n$ converges a.s. to $\widehat{s}(t)$ in the interval
[0,$\widehat{s}(0)-\epsilon$] and $M_{2}(t)/n$ converges to 0 a.s.. It remains
to prove that the limit point $\check{w}(t)$ coincides with $\widehat{w}(t)$,
which is then necessarily independent of the subsequence. But this is clearly
the case. Indeed, by dividing both sides of (4.7) by $n$ and then sending $n$
to $\infty$, we obtain that

$\check{w}(t)=\widehat{w}(0)+\int_{0}^{t}\frac{\omega}{\lambda}\alpha\widehat{s}(u)du-\int_{0}^{t}2\frac{\check{w}(u)}{\widehat{s}(u)}du,$
(4.16)

for $t\leq\widehat{s}(0)-\epsilon$. This equation has a unique solution
$\check{w}(t)=\widehat{w}(t)$.

It remains to prove (4.15). We have

$\frac{\widehat{W}(t_{2})}{n}-\frac{\widehat{W}(t_{1})}{n}=\int_{t_{1}}^{t_{2}}\left(\frac{\alpha\omega\widehat{S}(u)}{\lambda
n}-\frac{2\widehat{W}(u)}{\widehat{S}(u)}\right)du+\frac{M_{2}(t_{2})-M_{2}(t_{1})}{n}.$
(4.17)

We define the event

$\Omega_{\ref{s>ep}}:=\\{\widehat{S}^{n}(t)\geq\epsilon n/2,\forall
t\leq(\widehat{s}(0)-\epsilon)\wedge\widehat{\tau}\\}\cap\\{\sup_{t\leq\widehat{\tau}\wedge
2}\left|M_{2}(t)\right|\leq n^{2/3}\\}.$ (4.18)

By (4.6) and (4.12), $\mathbb{P}(\Omega_{\ref{s>ep}})\to 1$ as $n\to\infty$.
On $\Omega_{\ref{s>ep}}$, for $t_{1}<t_{2}<\widehat{s}(0)-\epsilon$, using
(4.17),

$\begin{split}&\left|\frac{\widehat{W}(t_{2})}{n}-\frac{\widehat{W}(t_{1})}{n}\right|\\\
\leq&(t_{2}-t_{1})\left(\frac{\alpha\omega}{\lambda}+\frac{2\mu n}{\epsilon
n/2}\right)+\frac{2}{n}\sup_{t\leq\widehat{\tau}\wedge
1}\left|M_{2}(t)\right|\\\
\leq&C\left((t_{2}-t_{1})+n^{-1/3}\right).\end{split}$ (4.19)

This proves (4.15) and thus also completes the proof of (4.14). . ∎

Having proved (4.14), we now use it to show the original statement (1.16). In
other words, given any $\epsilon^{\prime}>0$, we need to show

$\lim_{n\to\infty}\mathbb{P}\left(\sup_{0\leq
t\leq\widehat{s}(0)\wedge\widehat{\tau}}\left|\frac{\widehat{W}(t)}{n}-\widehat{w}(t)\right|>\epsilon^{\prime}\right)=0.$
(4.20)

Denote the oscillation of a function $h(t)$ in any given interval $[a,b]$ by

$\mbox{Osc}\left[h(t),[a,b]\right]:=\sup_{a\leq u\leq v\leq
b}\left|h(u)-h(v)\right|.$ (4.21)

Observe that

$\begin{split}&\sup_{0\leq
t\leq\widehat{s}(0)\wedge\widehat{\tau}}\left|\frac{\widehat{W}(t)}{n}-\widehat{w}(t)\right|\\\
\leq&\sup_{0\leq
t\leq(\widehat{s}(0)-\epsilon)\wedge\widehat{\tau}}\left|\frac{\widehat{W}(t)}{n}-\widehat{w}(t)\right|+\mbox{Osc}\left[\frac{\widehat{W}(t)}{n},[(\widehat{s}(0)-\epsilon)\wedge\widehat{\tau},\widehat{s}(0)\wedge\widehat{\tau}]\right]\\\
&+\mbox{Osc}[\widehat{w}(t),[\widehat{s}(0)-\epsilon,\widehat{s}(0)]]\\\
:=&I_{1}+I_{2}+I_{3}.\end{split}$ (4.22)

Fix any $\epsilon^{\prime}>0$, we need to find $\epsilon$ so that

$\lim_{n\to\infty}\mathbb{P}(I_{1}>\epsilon^{\prime}/3)=\lim_{n\to\infty}\mathbb{P}(I_{2}>\epsilon^{\prime}/3)=0\mbox{
and }I_{3}\leq\epsilon^{\prime}/3.$ (4.23)

Once (4.23) is proved, (4.20) follows from the union bound.

Equation (4.14) already implies that $\mathbb{P}(I_{1}>\epsilon^{\prime}/3)\to
0$ for any fixed $\epsilon$ and $\epsilon^{\prime}$, so we only need to
estimate $I_{2}$ and $I_{3}$. From (1.14) we have the explicit expression for
$\widehat{w}(t)$:

$\widehat{w}(t)=\frac{\omega\alpha}{\lambda}(\widehat{s}(0)-t)^{2}\log\frac{\widehat{s}(0)}{\widehat{s}(0)-t}.$

We can compute its derivative

$\widehat{w}^{\prime}(t)=\frac{\omega\alpha}{\lambda}\left(2(t-\widehat{s}(0))\log\frac{\widehat{s}(0)}{\widehat{s}(0)-t}+(\widehat{s}(0)-t)\right).$
(4.24)

From this we see that the derivative of $\widehat{w}(t)$ is uniformly bounded.

$\sup_{0\leq t\leq\widehat{s}(0)}\left|\widehat{w}^{\prime}(t)\right|\leq C.$
(4.25)

It follows that

$I_{3}=\mbox{Osc}[\widehat{w},[\widehat{s}(0)-\epsilon,\widehat{s}(0)]]\leq
C\epsilon.$ (4.26)

Also note that

$\widehat{w}(\widehat{s}(0)-\epsilon)=\frac{\omega\alpha}{\lambda}\epsilon^{2}\log(\widehat{s}(0)/\epsilon)\leq
C\epsilon.$

Hence by choosing $\epsilon$ small enough we can make

$I_{3}\leq\epsilon^{\prime}/3\mbox{ and
}\widehat{w}(\widehat{s}(0)-\epsilon)\leq\epsilon^{\prime}/3.$ (4.27)

It remains to control $I_{2}$. Note that

$\displaystyle I_{2}\left\\{\begin{array}[]{l}=0,\ \text{ if }\
\widehat{\tau}<\widehat{s}(0)-\epsilon;\\\
\leq\sup_{\widehat{s}(0)-\epsilon\leq
t\leq\widehat{s}(0)\wedge\widehat{\tau}}\widehat{W}(t)/n,\ \text{ if }\
\widehat{\tau}\geq\widehat{s}(0)-\epsilon.\end{array}\right.$

Define the event

$\Omega_{\ref{ws0-ep}}=\\{\widehat{\tau}<\widehat{s}(0)-\epsilon\\}\cup\\{\widehat{\tau}\geq\widehat{s}(0)-\epsilon,\widehat{W}(\widehat{s}(0)-\epsilon)>\epsilon^{\prime}n/12\\}.$
(4.28)

By (4.14),

$\lim_{n\to\infty}\mathbb{P}\left(\Omega_{\ref{ws0-ep}}\right)=1.$ (4.29)

By (4.17), on
$\Omega_{\ref{s>ep}}\cap\Omega_{\ref{ws0-ep}}\cap\\{\widehat{\tau}\geq\widehat{s}(0)-\epsilon\\}$,
for all $\widehat{s}(0)-\epsilon\leq t\leq\widehat{s}(0)\wedge\widehat{\tau}$,

$\begin{split}\frac{\widehat{W}(t)}{n}&\leq\frac{\widehat{W}(\widehat{s}(0)-\epsilon)}{n}+\frac{\alpha\omega}{\lambda}\epsilon+\frac{2\sup_{t}\left|M_{2}(t)\right|}{n}\\\
&\leq\frac{\epsilon^{\prime}}{12}+\frac{\alpha\omega}{\lambda}\epsilon+\frac{2n^{2/3}}{n},\end{split}$
(4.30)

which is smaller than $\epsilon^{\prime}/3$ for $\epsilon$ small enough and
$n$ large enough.

Consequently, on $\Omega_{\ref{s>ep}}\cap\Omega_{\ref{ws0-ep}}$,
$I_{2}\leq\epsilon^{\prime}/3$. It follows that

$\lim_{n\to\infty}\mathbb{P}(I_{2}>\epsilon^{\prime}/3)\leq\lim_{n\to\infty}\mathbb{P}\left(\left(\Omega_{\ref{s>ep}}\cap\Omega_{\ref{ws0-ep}}\right)^{c}\right)=0,$
(4.31)

since both $\mathbb{P}(\Omega_{\ref{s>ep}})$ and
$\mathbb{P}(\Omega_{\ref{ws0-ep}})$ tend to 1 as $n\to\infty$. Equation (4.23)
now follows from (4.27) and (4.31). This completes the proof of (1.16).

###  4.2. Proof of (1.17) for positive initial condition

We first analyze $\widehat{I}(t)$.

  * •

Change of $\widehat{I}(t)$ due to recovery Poisson processes: decreases by 1.
The rate for such an event to occur is

$\gamma\widehat{I}(t)\times\frac{n}{\lambda\widehat{I}_{E}(t)}=\frac{\gamma\widehat{I}(t)n}{\lambda\widehat{I}_{E}(t)}.$

  * •

Change of $\widehat{I}(t)$ due to rewiring/dropping Poisson processes: 0.

  * •

Change of $\widehat{I}(t)$ due to infection Poisson processes: increases by 1.
The rate for such an event to occur is

$\lambda\widehat{I}_{E}(t)\times\frac{n}{\lambda\widehat{I}_{E}(t)}=n.$

Thus we can decompose $\widehat{I}(t)$ as follows:

$\widehat{I}(t)=\widehat{I}(0)+\int_{0}^{t}\left(-\frac{\gamma\widehat{I}(t)n}{\lambda\widehat{I}_{E}(t)}+n\right)du+M_{3}(t).$
(4.32)

The quadratic variation of the martingale $M_{3}(t)$ can be bounded in the
similar way to $M_{1}(t)$:

$\langle
M_{3}(t),M_{3}(t)\rangle=\langle\widehat{I}(t),\widehat{I}(t)\rangle\leq\sum_{i=1}^{n}1^{2}+\sum_{i=1}^{n}1^{2}=2n.$
(4.33)

Similarly to (4.4),

$\mathbb{P}\left(\sup_{t\leq\widehat{\tau}}\left|M_{3}(t)\right|>n^{2/3}\right)\leq\frac{8n}{n^{4/3}}\leq
8n^{-1/3}.$ (4.34)

Now we turn to $\widehat{I}_{E}(t)$. Let $\widehat{I}_{E}(i,t)$ be the
infected edges of vertex $i$ at time $t$. If $i$ is not infected at time $t$
then $\widehat{I}_{E}(i,t)=0$. We can classify the change of
$\widehat{I}_{E}(t)$ as follows.

  * •

Change of $\widehat{I}_{E}(t)$ due to recovery Poisson processes: decreases by
$\widehat{I}_{E}(i,t)$ if vertex $i$ recovers. The total rate of change is
equal to

$-\frac{n}{\lambda\widehat{I}_{E}(t)}\sum_{i=1}^{n}\gamma\widehat{I}_{E}(i,t)=-\frac{n\gamma}{\lambda}.$

(The minus sign means that the the change is negative.)

  * •

Change of $\widehat{I}_{E}(t)$ due to rewiring/dropping Poisson processes:
decreases by 1 with probability $1-\alpha+\alpha(1-\widehat{I}(t)/n)$,
otherwise stays the same. Thus the mean of the change is
$-(1-\alpha+\alpha(1-\widehat{I}(t)/n))$. The rate for such an event to occur
is

$\omega\widehat{I}_{E}(t)\times\frac{n}{\lambda\widehat{I}_{E}(t)}=\frac{n\omega}{\lambda}.$

  * •

Change of $\widehat{I}_{E}(t)$ due to Infection Poisson processes: increases
by

$-1+\mbox{Poisson}\left(\frac{(\widehat{S}(t)-1)\mu}{n}\right)+\mbox{Binomial}\left(\widehat{W}(t),2/\widehat{S}(t)\right)-\mbox{Binomial}\left(\widehat{I}_{E}(t),\frac{1}{\widehat{S}(t)}\right).$

The mean of the change is

$-1+\frac{(\widehat{S}(t)-1)\mu}{n}+\frac{2\widehat{W}(t)}{\widehat{S}(t)}-\frac{\widehat{I}_{E}(t)}{\widehat{S}(t)}.$

The rate for such an event to occur is

$\lambda\widehat{I}_{E}(t)\times\frac{n}{\lambda\widehat{I}_{E}(t)}=n.$

We now decompose $\widehat{I}_{E}(t)$ into the drift part and martingale part.

$\begin{split}\widehat{I}_{E}(t)=&\widehat{I}_{E}(0)+\int_{0}^{t}\left(-\frac{n\gamma}{\lambda}\right)du+\int_{0}^{t}\frac{n\omega}{\lambda}\left(-\left(1-\alpha+\alpha\left(1-\frac{\widehat{I}(t)}{n}\right)\right)\right)du\\\
&+\int_{0}^{t}n\left(-1+\frac{(\widehat{S}(u)-1)\mu}{n}+\frac{2\widehat{W}(u)}{\widehat{S}(u)}-\frac{\widehat{I}_{E}(u)}{\widehat{S}(u)}\right)du+M_{4}(t).\end{split}$
(4.35)

Let $N_{i}$ be the number of edges added to vertex $i$ when $i$ first becomes
infected and $\widehat{N}_{i}$ be the number of infected edges of $i$ just
before it becomes recovered. The quadratic variation of $M_{4}(t)$ can be
bounded by

$\langle
M_{4}(t),M_{4}(t)\rangle=\langle\widehat{I}_{E}(t),\widehat{I}_{E}(t)\rangle\leq\sum_{i=1}^{n}(\widehat{N}_{i}^{2}+(N_{i}+1)^{2})+Q(t),$
(4.36)

where $Q(t)$ is the number of arrivals of rewiring/dropping Poisson processes
by time $t$. By (4.9),

$\mathbb{E}(Q(t))\leq\frac{\gamma nt}{\lambda}.$ (4.37)

By Lemma 2.3, for some constant $C>0$,

$\mathbb{E}\left(\sum_{i=1}^{n}(\widehat{N}_{i}^{2}+({N}_{i}+1)^{2})\right)\leq
Cn.$ (4.38)

Similarly to the proof of (4.4), using (4.36), (4.37) and (4.36), there exists
some constant $C^{\prime}$ such that

$\mathbb{P}\left(\sup_{t\leq\widehat{\tau}\wedge
2}\left|M_{4}(t)\right|>n^{2/3}\right)\leq\frac{C^{\prime}n}{n^{4/3}}=C^{\prime}n^{-1/3}.$
(4.39)

For any $\delta>0$, we can define $\widehat{\tau}_{\delta}$ by

$\widehat{\tau}_{\delta}=\inf\\{t\geq 0:\widehat{I}_{E}(t)\leq\delta n\\}.$
(4.40)

Let

$\mathbf{\widehat{Y}}(t):=(\widehat{I}(t),\widehat{I}_{E}(t));$ (4.41)
$\mathbf{\widehat{y}}(t):=(\widehat{i}(t),\widehat{i}_{E}(t)).$ (4.42)

In order to prove (1.17), we first prove the following weaker version of it:

$\sup_{0\leq t\leq
t_{*}\wedge\widehat{\tau}_{\delta}}\left|\mathbf{\widehat{Y}}(t)/n-\mathbf{\widehat{y}}(t)\right|\xrightarrow{\mathbb{P}}0.$
(4.43)

###### Proof of (4.43).

Similarly to the proof of (1.16), we divide the proof of (4.43) into two
steps. For $t\geq\widehat{\tau}_{\delta}$ we set $\mathbf{\widehat{Y}}(t)$ to
be $\mathbf{\widehat{Y}}(\widehat{\tau}_{\delta})$. We first show that
$\mathbf{\widehat{Y}}(t)/n$ is a tight sequence in D[0,$t_{*}$]. Since
$\mathbf{\widehat{Y}}(0)/n\to\mathbf{\widehat{y}}(0)/n$, we know
$\\{\mathbf{\widehat{Y}}(0)/n\\}_{n\geq 1}$ is a tight sequence of random
variables. To establish tightness of
$\\{\mathbf{\mathbf{\widehat{Y}}}(t)/n,0\leq t\leq t_{*}\\}_{n\geq 1}$ we need
to show that for any fixed $\epsilon_{1},\epsilon_{2}>0$, there exist
$\theta>0$ and $n_{0}>0$ so that for $n\geq n_{0}$,

$\mathbb{P}\left(\sup_{\left|t_{1}-t_{2}\right|\leq\theta,t_{1}<t_{2}\leq
t_{*}\wedge\widehat{\tau}_{\delta}}\left\|\frac{\mathbf{\widehat{Y}}(t_{1})}{n}-\frac{\mathbf{\widehat{Y}}(t_{2})}{n}\right\|_{2}\geq\epsilon_{2}\right)\leq\epsilon_{1}.$
(4.44)

Note that $\mathbf{\widehat{Y}}(t)/n$ is tight if and only if both $I(t)/n$
and $I_{E}(t)/n$ are tight. This follows from the inequality

$\begin{split}\left\|\widehat{Y}(t_{2})-\widehat{Y}(t_{1})\right\|_{2}=&\left\|(\widehat{I}(t_{2})-\widehat{I}(t_{1}),\widehat{I}_{E}(t_{2})-\widehat{I}_{E}(t_{1}))\right\|_{2}\\\
=&\sqrt{\left|\widehat{I}(t_{2})-\widehat{I}(t_{1})\right|^{2}+\left|\widehat{I}_{E}(t_{2})-\widehat{I}_{E}(t_{1})\right|^{2}}\\\
\leq&\left|\widehat{I}(t_{2})-\widehat{I}(t_{1})\right|+\left|\widehat{I}_{E}(t_{2})-\widehat{I}_{E}(t_{1})\right|.\end{split}$
(4.45)

Assuming (4.44) for the moment, we show that the limit of
$\mathbf{\widehat{Y}}(t)/n$ coincides with $\mathbf{\widehat{y}}(t)$.
Similarly to the proof that the limit of $\widehat{W}(t)/n$ coincides with
$\widehat{w}(t)$, we would like to show that any sequential limit point
$\mathbf{\check{y}}(t)$ coincides with $\mathbf{\widehat{y}}(t)$, which is
then necessarily unique. This is clearly the case since, after dividing both
sides of (4.32), (4.35) by $n$ and then sending $n$ to $\infty$, we obtain
that

$\displaystyle\check{i}(t)$
$\displaystyle=\widehat{i}(0)-\int_{0}^{t}\frac{\gamma\check{i}(u)}{\lambda\widehat{i}_{E}(u)}+t,$
(4.46) $\displaystyle\check{i}_{E}(t)$
$\displaystyle=\widehat{i}_{E}(0)+\int_{0}^{t}\left(\mu\widehat{s}(u)+\frac{2\widehat{w}(u)}{\widehat{s}(u)}-\frac{\check{i}_{E}(u)}{\widehat{s}(u)}+\frac{\omega\alpha}{\lambda}\check{i}(u)\right)du-\frac{\lambda+\omega+\gamma}{\lambda}t.$
(4.47)

For $t\leq t_{*}$, this equation has a unique solution
$\mathrm{\check{y}}(t)=\mathbf{\widehat{y}}(t)$, as proved in Theorem 1.2.

It remains to establish (4.44). Using (4.32) and (4.35),

$\frac{\widehat{I}(t_{2})}{n}-\frac{\widehat{I}(t_{1})}{n}=\int_{t_{1}}^{t_{2}}\left(1-\frac{\gamma\widehat{I}(u)}{\lambda\widehat{I}_{E}(u)}\right)du+\frac{M_{3}(t_{2})-M_{3}(t_{1})}{n},$
(4.48)

and

$\begin{split}\frac{\widehat{I}_{E}(t_{2})}{n}-\frac{\widehat{I}_{E}(t_{1})}{n}=&\int_{t_{1}}^{t_{2}}\left(\frac{\widehat{S}(u)-1}{n}\mu+\frac{2\widehat{W}(u)}{\widehat{S}(u)}-\frac{\widehat{I}_{E}(u)}{\widehat{S}(u)}+\frac{\omega\alpha\widehat{I}(u)}{\lambda
n}-\frac{\lambda+\omega+\gamma}{\lambda}\right)du\\\
&+\frac{M_{4}(t_{2})-M_{4}(t_{1})}{n}.\end{split}$ (4.49)

We define the event

$\Omega_{\ref{iie}}=\\{\widehat{S}(t)\geq\frac{\widehat{s}(0)-t_{*}}{2}n,\,\forall
t\leq t_{*}\wedge\widehat{\tau}\\}\cap\\{\sup_{t\leq\tau\wedge
2}\left|M_{3}(t)\right|\leq n^{2/3}\\}\cap\\{\sup_{t\leq\tau\wedge
2}\left|M_{4}(t)\right|\leq n^{2/3}\\}.$ (4.50)

By (4.6), (4.34) and (4.39),

$\lim_{n\to\infty}\mathbb{P}(\Omega_{\ref{iie}})=1.$ (4.51)

On $\Omega_{\ref{iie}}$, for $t_{1}<t_{2}\leq\widehat{\tau}_{\delta}\wedge
t_{*}$,

$\begin{split}\left|\frac{\widehat{I}(t_{2})}{n}-\frac{\widehat{I}(t_{1})}{n}\right|\leq(t_{2}-t_{1})\left(1+\frac{2\gamma}{\lambda\delta}\right)+\frac{2n^{2/3}}{n}\leq
C(t_{2}-t_{1})+2n^{-1/3},\end{split}$ (4.52)

and

$\begin{split}&\frac{\widehat{I}_{E}(t_{2})}{n}-\frac{\widehat{I}_{E}(t_{1})}{n}\\\
\leq&(t_{2}-t_{1})\left(\mu+\frac{4\mu}{\widehat{s}(0)-t_{*}}+\frac{4\mu}{\widehat{s}(0)-t_{*}}+\frac{\omega\alpha}{\lambda}+\frac{\lambda+\omega+\lambda}{\lambda}\right)+\frac{2n^{2/3}}{n}\\\
\leq&C(t_{2}-t_{1})+2n^{-1/3}.\end{split}$ (4.53)

Equation (4.44) now follows from (4.45), (4.52) and (4.53). This also
completes the proof of (4.43). ∎

Having proved the weaker version (4.43), we now turn to the original version
(1.17). We need the following two lemmas to prepare for its proof.

######  Lemma 4.1.

Consider the case of positive initial condition. For any $\epsilon>0$, there
exists an $\delta_{*}\in(0,\widehat{i}_{E}(0))$, s.t. for any
$\delta<\delta_{*}$, we have

$\lim_{n\to\infty}\mathbb{P}(\widehat{\tau}_{\delta}\geq t_{*}-\epsilon)=1.$
(4.54)

###### Proof of Lemma 4.58.

Observe that for any $\epsilon>0$, we have

$a:=\inf_{0\leq t\leq t_{*}-\epsilon}\widehat{i}_{E}(t)>0.$ (4.55)

We now set

$\delta<\delta_{*}:=\min\left\\{\frac{a}{2},\frac{\widehat{i}_{E}(0)}{2}\right\\}.$
(4.56)

Note that, if $\widehat{\tau}_{\delta}\leq t_{*}-\epsilon$, then

$\widehat{\tau}_{\delta}\wedge t_{*}=\widehat{\tau}_{\delta}.$

Hence by (4.43),

$\begin{split}\lim_{n\to\infty}\mathbb{P}\left(\widehat{\tau}_{\delta}\leq
t_{*}-\epsilon,\left|\frac{\widehat{I}_{E}(\widehat{\tau}_{\delta})}{n}-{\widehat{i}_{E}(\widehat{\tau}_{\delta})}\right|\geq\frac{a}{2}\right)=0.\end{split}$
(4.57)

On the other hand, we claim that the event

$\Omega_{\ref{taundelta}}=\left\\{\widehat{\tau}_{\delta}\leq
t_{*}-\epsilon,\left|\frac{\widehat{I}_{E}(\widehat{\tau}_{\delta})}{n}-{\widehat{i}_{E}(\widehat{\tau}_{\delta}}\right|<\frac{a}{2}\right\\}$
(4.58)

is empty. Indeed, by (4.55) and the definition of $\widehat{\tau}_{\delta}$
and $\delta_{*}$, on $\Omega_{\ref{taundelta}}$,

$\frac{\widehat{I}_{E}(\widehat{\tau}_{\delta})}{n}\leq\delta\leq\frac{a}{2}\mbox{
and }\widehat{i}_{E}(\widehat{\tau}_{\delta})\geq a.$ (4.59)

Consequently,

$\left|\frac{\widehat{I}_{E}(\widehat{\tau}_{\delta})}{n}-{\widehat{i}_{E}(\widehat{\tau}_{\delta})}\right|\geq
a-\frac{a}{2}=\frac{a}{2}.$

Hence

$\Omega_{\ref{taundelta}}=\emptyset.$ (4.60)

Combining (4.57) and (4.60),

$\begin{split}\lim_{n\to\infty}\mathbb{P}\left(\widehat{\tau}_{\delta}\leq
t_{*}-\epsilon\right)=\lim_{n\to\infty}\mathbb{P}\left(\widehat{\tau}_{\delta}\leq
t_{*}-\epsilon,\left|\frac{\widehat{I}_{E}(\widehat{\tau}_{\delta})}{n}-{\widehat{i}_{E}(\widehat{\tau}_{\delta})}\right|\geq\frac{a}{2}\right)=0,\end{split}$
(4.61)

as desired. ∎

######  Lemma 4.2.

We have the convergence

$\lim_{t\to t_{*}}\widehat{i}(t)=0.$ (4.62)

###### Proof of Lemma 4.2.

By (1.11), we see that $\left|\widehat{i}^{\prime}_{E}(t)\right|$ is uniformly
bounded. Hence there exists a constant $C$ such that

$\widehat{i}_{E}(t)\leq C(t_{*}-t)$ (4.63)

for all $t\leq t_{*}$. Assuming $\limsup_{t\to t_{*}}\widehat{i}(t)=a>0$, then
by the fact that $\widehat{i}^{\prime}(t)\leq 1$, there exists an $\epsilon>0$
such that $\widehat{i}(t)>a/2$ for $t\in(t_{*}-\epsilon,t_{*})$. By (1.10) and
(4.63),

$\widehat{i}(t)-\widehat{i}(t_{*}-\epsilon)=-\int_{t_{*}-\epsilon}^{t}\frac{\widehat{i}(u)}{\widehat{i}_{E}(u)}du+(t-t_{*}+\epsilon)\leq-\int_{t_{*}-\epsilon}^{t}\frac{a/2}{C(t_{*}-u)}du+(t-t_{*}+\epsilon),$

which goes to $-\infty$ as $t\to t_{*}$. This is a contradiction. Hence we
must have $\lim_{t\to t_{*}}\widehat{i}(t)=0$, proving Lemma 4.2. ∎

We now come back to the proof of (1.17). Given any $\epsilon^{\prime}>0$, we
need to prove

$\lim_{n\to\infty}\mathbb{P}\left(\sup_{0\leq t\leq
t_{*}\wedge\widehat{\tau}}\left|\frac{\widehat{I}(t)}{n}-\widehat{i}(t)\right|>\epsilon^{\prime}\right)=0,$
(4.64)

and

$\lim_{n\to\infty}\mathbb{P}\left(\sup_{0\leq t\leq
t_{*}\wedge\widehat{\tau}}\left|\frac{\widehat{I}_{E}(t)}{n}-\widehat{i}_{E}(t)\right|>\epsilon^{\prime}\right)=0.$
(4.65)

###### Proof.

Given the $\epsilon^{\prime}>0$, we let $\epsilon>0$ be some number to be
determined. By Lemma 4.1, there exists a $\delta>0$ such that the event
$\\{\widehat{\tau}_{\delta}\geq t_{*}-\epsilon\\}$ holds with high
probability. On this event,

$[0,t_{*}\wedge\widehat{\tau}]=[0,t_{*}\wedge\widehat{\tau}_{\delta}]\cup[t_{*}-\epsilon,t_{*}\wedge\widehat{\tau}].$
(4.66)

Consequently,

$\begin{split}&\sup_{0\leq t\leq
t_{*}\wedge\widehat{\tau}}\left|\frac{\widehat{I}(t)}{n}-\widehat{i}(t)\right|\\\
\leq&\sup_{0\leq t\leq
t_{*}\wedge\widehat{\tau}_{\delta}}\left|\frac{\widehat{I}(t)}{n}-\widehat{i}(t)\right|+\mbox{Osc}\left[\frac{\widehat{I}(t)}{n},[t_{*}-\epsilon,t_{*}\wedge\widehat{\tau}]\right]\\\
&+\mbox{Osc}[\widehat{i}(t),[t_{*}-\epsilon,t_{*}]]\\\
:=&J_{1}+J_{2}+J_{3}.\end{split}$ (4.67)

Here Osc stands for the oscillation of a function, as defined in (4.21).

Our goal is to find an $\epsilon>0$ so that

$\lim_{n\to\infty}\mathbb{P}(J_{1}>\epsilon^{\prime}/3)=\lim_{n\to\infty}\mathbb{P}(J_{2}>\epsilon^{\prime}/3)=0\mbox{
and }J_{3}\leq\epsilon^{\prime}/3,$ (4.68)

Once (4.68) is proved, (4.64) follows by the union bound.

Equation (4.44) already implies $\mathbb{P}(J_{1}>\epsilon^{\prime}/3)\to 0$
for any fixed $\epsilon,\epsilon^{\prime}$. By Lemma 4.2,
$J_{3}\leq\epsilon^{\prime}/3$ provided that $\epsilon$ is small enough.
Therefore, to prove (4.68), it remains need to estimate $J_{2}$. For
$t_{1}<t_{2}\leq t_{*}\wedge\widehat{\tau}$,

$\begin{split}\frac{\widehat{I}(t_{2})}{n}-\frac{\widehat{I}(t_{1})}{n}\leq\int_{t_{1}}^{t_{2}}\left(1-\frac{\gamma\widehat{I}(u)}{\lambda\widehat{I}_{E}(u)}\right)du+\frac{2\sup_{t\leq\widehat{\tau}\wedge
2}\left|M_{3}(t)\right|}{n}.\end{split}$ (4.69)

Thus for any $t\in[t_{*}-\epsilon,t_{*}\wedge\widehat{\tau}]$,

$\frac{\widehat{I}(t)}{n}\leq\epsilon+2n^{-1}\sup_{t\leq\widehat{\tau}\wedge
2}\left|M_{3}(t)\right|+\left|\frac{\widehat{I}(t_{*}-\epsilon)}{n}\right|.$
(4.70)

Repeating the derivation of (4.30), one can show that the right hand side of
(4.70) is smaller than $\epsilon^{\prime}/3$ with high probability for large
$n$ and small $\epsilon$. This implies that
$\lim_{n\to\infty}\mathbb{P}(J_{2}>\epsilon^{\prime}/3)=0$ for $\epsilon$
small and completes the proof of (4.64). Equation (4.65) can be proved
similarly. Equation (1.17) now follows from (4.64) and (4.65).

∎

For future references we record the following corollary of Lemma 4.1.

######  Corollary 1.

Suppose $\widehat{i}(0)>0$ and $\widehat{i}_{E}(0)>0$. For any $\epsilon>0$,

$\lim_{n\to\infty}\mathbb{P}(\widehat{\tau}\geq t_{*}-\epsilon)=1.$ (4.71)

Consequently, for any $\epsilon>0$, we have

$\mathbb{P}\left(T^{(n)}\geq(1-\widehat{s}(0)+t_{*}-\epsilon)n\right)=1.$
(4.72)

###### Proof of Corollary 1.

From the definition of $\widehat{\tau}$, we see that

$\widehat{\tau}\geq\widehat{\tau}_{\delta},\,\forall\delta>0.$

By Lemma 4.1, for any $\epsilon>0$, we can find a $\delta>0$, so that

$\lim_{n\to\infty}\mathbb{P}(\widehat{\tau}_{\delta}\geq t_{*}-\epsilon)=1.$

It follows that

$\lim_{n\to\infty}\mathbb{P}(\widehat{\tau}\geq
t_{*}-\epsilon)\geq\lim_{n\to\infty}\mathbb{P}(\widehat{\tau}_{\delta}\geq
t_{*}-\epsilon)=1.$

Equation (4.72) follows from the convergence of $\widehat{S}(t)/n$ to $\\\
hat{s}(t)(=hat{s}(0)-t)$. Indeed, by (4.71) and (1.9) we have

$\mathbb{P}(\widehat{S}(t_{*}-\epsilon)\leq\widehat{s}(t_{*}-\epsilon)+\epsilon,\widehat{\tau}>t_{*}-\epsilon)=1.$
(4.73)

Since

$\widehat{s}(t_{*}-\epsilon)+\epsilon=\widehat{s}(0)-(t_{*}-\epsilon)+\epsilon=\widehat{s}(0)-t_{*}+2\epsilon,\mbox{and
}T^{(n)}=n-\widehat{S}(\widehat{\tau})\geq n-\widehat{S}(t_{*}-\epsilon),$

we obtain that

$\lim_{n\to\infty}\mathbb{P}\left(T^{(n)}\geq(1-\widehat{s}(0)+t_{*}-2\epsilon)n\right)=1.$
(4.74)

Equation (4.72) then follows since $\epsilon>0$ is arbitrary. ∎

###  4.3. Proof of (1.17) for zero initial condition

Conditionally on the event that an outbreak occurs, we may assume that for
some $\epsilon_{0}>0$, $T^{(n)}>2\epsilon_{0}n$. We claim that, with high
probability, $\widehat{\tau}\geq\epsilon_{0}$. Indeed, by (1.9),

$\begin{split}\lim_{n\to\infty}\mathbb{P}\left(T^{(n)}>2\epsilon_{0}n,\widehat{\tau}<\epsilon_{0}\right)\leq&\lim_{n\to\infty}\mathbb{P}(\widehat{S}(\widehat{\tau})<(1-2\epsilon)n,\widehat{\tau}<\epsilon_{0})\\\
\leq&\lim_{n\to\infty}\mathbb{P}\left(\left|\widehat{S}(\widehat{\tau})-(1-\widehat{\tau})n\right|\geq\epsilon
n\right)=0.\end{split}$ (4.75)

We need the following rough bounds for $\widehat{I}(t)$ and
$\widehat{I}_{E}(t)$.

######  Lemma 4.3.

By lowering the value of $\epsilon_{0}$ if needed, we can find two constants
$C_{\ref{rub}}$ and $C_{\ref{rlb}}$ such that the following two equations hold
for any $\epsilon>0$.

$\lim_{n\to\infty}\mathbb{P}\left(\widehat{I}(t)+\widehat{I}_{E}(t)\leq(C_{\ref{rub}}t+\epsilon)n,\,\forall
t\leq\epsilon_{0}\wedge\widehat{\tau}\right)=1.$ (4.76)
$\lim_{n\to\infty}\mathbb{P}\left(\widehat{I}_{E}(t)\geq(C_{\ref{rlb}}t-\epsilon)n,\,\forall
t\leq\epsilon_{0}\wedge\widehat{\tau}\right)=1.$ (4.77)

###### Proof.

Define the events

$\Omega_{\ref{rul1}}=\\{\sup_{0\leq t\leq
2\wedge\widehat{\tau}}\left(\left|M_{3}(t)\right|+\left|M_{4}(t)\right|\right)\leq\epsilon
n/4\\},$ (4.78)

and

$\Omega_{\ref{rul}}=\\{\widehat{S}(t)\geq(1-\epsilon-t)n,\forall t\leq
1\wedge\widehat{\tau}\\}.$ (4.79)

Using (4.34), (4.39) and the assumption

$\frac{\widehat{I}(0)}{n}\to\widehat{i}(0)=0,\,\frac{\widehat{I}_{E}(0)}{n}\to\widehat{i}_{E}(0)=0,$

we see that

$\lim_{n\to\infty}\mathbb{P}(\Omega_{\ref{rul1}})=\lim_{n\to\infty}\mathbb{P}(\Omega_{\ref{rul2}})=1.$

Define

$\Omega_{\ref{omeganew}}=\Omega_{\ref{rul1}}\cap\Omega_{\ref{rul2}}.$ (4.80)

Then we have $\mathbb{P}(\Omega_{\ref{omeganew}})\to 1$. On
$\Omega_{\ref{omeganew}}$, using (4.32), we get

$\widehat{I}(t)\leq\widehat{I}(0)+t+M_{3}(t)\leq t+\epsilon n,$

and

$\begin{split}\widehat{I}_{E}(t)&\leq\widehat{I}_{E}(0)+n\int_{0}^{t}\left(\mu+\frac{2\widehat{W}(u)}{\widehat{S}(u)}\right)du+M_{4}(t)\\\
&\leq\epsilon n+n\int_{0}^{t}\left(\mu+\frac{4\mu}{1-u-\epsilon}\right)du\\\
&\leq\epsilon n+9\mu nt,\end{split}$

provided that $\epsilon$ and $t$ are both smaller than $1/4$. This proves
(4.76). To show (4.77), note that

$\widehat{I}_{E}(t)\geq
n\int_{0}^{t}\left(-\frac{\gamma}{\lambda}-\frac{\omega}{\lambda}-1+\mu\frac{\widehat{S}(u)-1}{n}-\frac{\widehat{I}_{E}(u)}{\widehat{S}(u)}\right)du+M_{4}(t).$
(4.81)

On $\Omega_{\ref{omeganew}}$, $\widehat{S}(u)\geq(1-\epsilon-t)n$. In
addition, $\widehat{I}_{E}(t)\leq(9\mu t+\epsilon)n$, as we have just shown.
Hence, for sufficiently small $\epsilon$ and $t$,

$\begin{split}\widehat{I}_{E}(t)&\geq
n\left(\left(-\frac{\gamma+\omega+\lambda}{\lambda}+\mu-\mu\epsilon\right)t-\frac{\mu
t^{2}}{2}-\int_{0}^{t}\frac{9\mu t+\epsilon}{1/2}du\right)-\epsilon n\\\ &\geq
n\left(-\frac{\gamma+\omega+\lambda}{\lambda}+\mu\right)t/2-2\epsilon
n,\end{split}$ (4.82)

which proves (4.77). ∎

Let $\epsilon_{0}$ be small enough so that Lemma 4.3 holds. Define

$\widehat{\tau}_{\delta,\epsilon_{0}}:=\inf\\{t\geq\epsilon_{0}:\widehat{I}_{E}(t)\leq\delta
n\\}.$

We now prove the tightness of
$\\{(\widehat{I}(t)/n,\widehat{I}_{E}(t)/n),\,t\leq\widehat{\tau}_{\delta,\epsilon_{0}}\\}$.
To this end, we need to show that, for any fixed
$\epsilon^{\prime},\delta_{1}>0$, there exist $\theta>0$ and $n_{0}>0$ so that
for $n\geq n_{0}$,

$\mathbb{P}\left(\sup_{\left|t_{1}-t_{2}\right|\leq\theta,t_{1}<t_{2}\leq\widehat{\tau}_{\delta,\epsilon_{0}}}\left|\frac{\widehat{I}(t_{1})}{n}-\frac{\widehat{I}(t_{2})}{n}\right|\geq\delta_{1}\right)\leq\epsilon^{\prime},$
(4.83)

and

$\mathbb{P}\left(\sup_{\left|t_{1}-t_{2}\right|\leq\theta,t_{1}<t_{2}\leq\widehat{\tau}_{\delta,\epsilon_{0}}}\left|\frac{\widehat{I}_{E}(t_{1})}{n}-\frac{\widehat{I}_{E}(t_{2})}{n}\right|\geq\delta_{1}\right)\leq\epsilon^{\prime}.$
(4.84)

To prove (4.83), we assume that
$\delta_{1}<\min\\{\epsilon_{0},C_{\ref{rub}}\epsilon_{0}\\}$ and take

$\epsilon_{1}=\theta_{1}=\frac{\delta_{1}}{4C_{\ref{rub}}}.$ (4.85)

We divide $[0,\widehat{\tau}_{\delta,\epsilon_{0}}]$ into two sub-intervals:
$B_{1}=[0,\epsilon_{1}]$ and
$B_{2}=[\epsilon_{1}.\widehat{\tau}_{\delta,\epsilon_{0}}]$. Observe that if
$t_{1}\in B_{1}$ and $\left|t_{1}-t_{2}\right|\leq\theta$, then both $t_{1}$
and $t_{2}$ lie in the interval $[0,\epsilon_{0}]$. For
$\epsilon=\delta_{1}/4$, we have

$C_{\ref{rub}}(\epsilon_{1}+\theta_{1})+\epsilon=\frac{3\delta_{1}}{4}.$
(4.86)

Using (4.76) (with $\epsilon=\delta_{1}/4$) and (4.86), we see that, for some
$n_{1}$ large enough and all $n\geq n_{1}$,

$\mathbb{P}\left(\sup_{0<t_{2}-t_{1}\leq\theta_{1},t_{1}\in
B_{1}}\left|\frac{\widehat{I}(t_{1})}{n}-\frac{\widehat{I}(t_{2})}{n}\right|\geq\delta_{1}\right)\leq\mathbb{P}\left(\sup_{t\leq\epsilon_{1}+\theta_{1}}\widehat{I}(t)\geq\frac{3\delta_{1}}{4}n\right)\leq\frac{\epsilon^{\prime}}{2}.$
(4.87)

On the other hand, if both $t_{1}$ and $t_{2}$ are in the interval $B_{2}$,
then

$\lim_{n\to\infty}\mathbb{P}\left(\widehat{I}_{E}(t)\geq\min\left\\{\frac{C_{\ref{rlb}}\epsilon_{1}}{2},\delta\right\\}n,\,\forall
t\in(\epsilon_{1},\widehat{\tau}_{\delta,\epsilon_{0}})|\widehat{\tau}>\epsilon_{0}\right)=1.$
(4.88)

Indeed, by (4.77) with $\epsilon=C_{\ref{rlb}}\epsilon_{1}/2$,

$\lim_{n\to\infty}\mathbb{P}\left(\widehat{I}_{E}(t)\geq(C_{\ref{rlb}}\epsilon_{1}-C_{\ref{rlb}}\epsilon_{1}/2)n,\,\forall\epsilon_{1}\leq
t\leq\epsilon_{0}|\widehat{\tau}>\epsilon_{0}\right)=1.$ (4.89)

Thus (4.88) follows from (4.89) and the definition of
$\widehat{\tau}_{\delta,\epsilon_{0}}$.

By repeating the proof of (4.44), one can show that (4.88) implies that the
sequence

$\\{(\widehat{I}/n,\widehat{I}_{E}(t)/n),\epsilon_{1}\leq
t\leq\widehat{\tau}_{\delta,\epsilon_{0}}\\}_{n\geq 1}$

is a tight. Thus we can find some $\theta_{2}$ and $n_{2}>0$ such that for all
$n\geq n_{2}$,

$\mathbb{P}\left(\sup_{0<t_{2}-t_{1}\leq\theta_{2},t_{1},t_{2}\in
B_{2}}\left|\frac{\widehat{I}(t_{1})}{n}-\frac{\widehat{I}(t_{2})}{n}\right|\geq\delta_{1}\right)\leq\frac{\epsilon^{\prime}}{2}.$
(4.90)

Equation (4.83) now follows from (4.87), (4.90) and the union bound by setting

$\theta=\min\\{\theta_{1},\theta_{2}\\}\mbox{ and
}n_{0}=\max\\{n_{1},n_{2}\\}.$

Equation (4.84) can be proved in the same way. Hence we have shown the
tightness of the sequence $\\{(\widehat{I}(t)/n,\widehat{I}_{E}(t)/n),\,0\leq
t\leq\widehat{\tau}_{\delta,\epsilon_{0}}\\}$.

With the tightness established, we can proceed in the same way as in Section
4.2 to prove the uniform convergence of
$(\widehat{I}(t)/n,\widehat{I}_{E}(t)/n)$ to
$(\widehat{i}(t),\widehat{i}_{E}(t))$ for $t\leq t_{*}\wedge\widehat{\tau}$.
This completes the proof of (1.17) for the case of
$\widehat{i}(0)=\widehat{i}_{E}(0)=0$.

######  Remark 4.

Using the same proofs one can show that Lemma 4.2 and Lemma 1 also hold for
the case of zero initial condition (for Corollary 1 we need to condition on a
large outbreak).

##  5\. Final epidemic size of the SIR-$\omega$ model

In this section we prove Theorem 1.4. We only give a detailed proof for part
(a), equation (1.18), since the proof for part (b) is identical. By Corollary
1, for any $\epsilon>0$, with high probability
$T^{(n)}\geq(1-\widehat{s}(0)+t_{*}-\epsilon)n$. In order to prove Theorem 1.4
it remains to prove the other direction:

$\lim_{n\to\infty}\mathbb{P}(T^{(n)}>(1-\widehat{s}(0)+t_{*}+\epsilon)n)=0.$
(5.1)

Define the event

$\Omega_{\ref{taut000}}:=\\{\widehat{\tau}>t_{*}+\epsilon\\}.$

Using (1.9) and equality
$T^{(n)}=\widehat{T}^{(n)}=n-\widehat{S}(\widehat{\tau})$, in order to prove
(5.1) it suffices to prove

$\lim_{n\to\infty}\mathbb{P}(\Omega_{\ref{taut000}})=0.$ (5.2)

Indeed, (5.1) follows from (5.2), (1.13) and (1.16). For the rest of this
section we fix a small $\epsilon>0$. Our goal is to prove (5.2).

Using equation (4.35) of $\widehat{I}_{E}$, for $t\geq t_{*}$,

$\begin{split}\widehat{I}_{E}(t)=&\widehat{I}_{E}(t_{*})+\int_{t_{*}}^{t}n\left(-1-\frac{\gamma}{\lambda}-\frac{\omega}{\lambda}+\mu\frac{\widehat{S}(u)-1}{n}+\frac{2\widehat{W}(u)}{\widehat{S}(u)}\right)du\\\
&+\int_{t_{*}}^{t}n\left(\frac{\omega\alpha}{\lambda}\frac{\widehat{I}(u)}{n}-\frac{\widehat{I}_{E}(u)}{\widehat{S}(u)}\right)du+M_{4}(t)-M_{4}(t_{*}).\end{split}$
(5.3)

We also have, by (4.32), for $t\geq t_{*}$,

$\widehat{I}(t)=\widehat{I}(t_{*})+\int_{t_{*}}^{t}n\left(-\frac{\gamma}{\lambda}\frac{\widehat{I}(u)}{\widehat{I}_{E}(u)}+1\right)du+M_{3}(t)-M_{3}(t_{*}).$
(5.4)

Define

$\begin{split}F(t)&=-1-\frac{\gamma}{\lambda}-\frac{\omega}{\lambda}+\mu\widehat{s}(t)+2\frac{\widehat{w}(t)}{\widehat{s}(t)},\\\
F_{n}(t)&=-1-\frac{\gamma}{\lambda}-\frac{\omega}{\lambda}+\mu\frac{\widehat{S}(t)-1}{n}+\frac{2\widehat{W}(t)}{\widehat{S}(t)}.\end{split}$
(5.5)

Clearly $F$ is a continuous function of $t$. Given any $\delta>0$, we can
define an event

$\Omega_{\ref{fn-f}}(\delta):=\\{\left|F_{n}(t)-F(t)\right|\leq\delta,\forall
t\leq\widehat{\tau}^{(n)}\wedge(t_{*}+\epsilon)\\}.$

Using (1.16), for any $\delta>0$,

$\lim_{n\to\infty}\mathbb{P}\left(\Omega_{\ref{fn-f}}(\delta)\right)=1.$ (5.6)

We define a random function $E(t)$ by setting $E(t)=0$ for $t\leq t_{*}$, and
for $t\geq t_{*}$,

$E(t)=\int_{t_{*}}^{t}n\left(F_{n}(u)-F(u)\right)du+\widehat{I}_{E}(t_{*})+\int_{t_{*}}^{t}n\left(\frac{\omega\alpha}{\lambda}\frac{\widehat{I}(u)}{n}-\frac{\widehat{I}_{E}(u)}{\widehat{S}(u)}\right)du+M_{4}(t)-M_{4}(t_{*}).$
(5.7)

Then on the event $\Omega_{\ref{taut000}}$, for $t\geq t_{*}$,

$\widehat{I}_{E}(t)=\int_{t_{*}}^{t}F(u)du+E(t).$ (5.8)

We claim that

$F(t_{*})\leq 0.$ (5.9)

To see this, recall that that we have shown $\widehat{i}(t_{*})=0$ in Lemma
4.2. We also have $\widehat{i}_{E}(t_{*})=0$ by the definition of $t_{*}$.
Thus

$\widehat{i}^{\prime}_{E}(t_{*})=-1-\frac{\gamma}{\lambda}+\mu\widehat{s}(t_{*})+2\frac{\widehat{w}(t_{*})}{\widehat{s}(t_{*})}-\frac{\omega}{\lambda}=F(t_{*}).$
(5.10)

On the other hand

$\widehat{i}^{\prime}_{E}(t_{*})=\lim_{t\to
t_{*}-}\frac{\widehat{i}(t_{*})-\widehat{i}(t)}{t_{*}-t}\leq 0.$ (5.11)

Equation (5.9) thus follows from (5.10) and (5.11).

We now divide the proof of (5.2) into two subsections, according to
$F(t_{*})<0$ or $F(t_{*})=0$.

###  5.1. Proof of (5.2) for $F(t_{*})<0$

By the continuity of $F$ around $t_{*}$,

$F(t)\leq-\frac{3F(t_{*})}{4},\,\forall t\in[t_{*},t_{*}+\epsilon].$ (5.12)

provided $\epsilon$ is small. Define the event

$\Omega_{\ref{610}}=\\{\widehat{\tau}\leq
t_{*}+\epsilon\\}\cup\\{\widehat{\tau}>t_{*}+\epsilon,\widehat{I}(t)\leq\left(t-t_{*}-\frac{F(t_{*})\lambda}{8\omega\alpha}\right)n,\,\forall
t_{*}\leq t\leq\widehat{\tau}\\}.$ (5.13)

By (4.34) and (4.32),

$\lim_{n\to\infty}\mathbb{P}(\Omega_{\ref{610}})=1.$ (5.14)

On $\Omega_{\ref{610}}\cap\\{\widehat{\tau}\geq t_{*}+\epsilon\\}$, we have

$\int_{t_{*}}^{t_{*}+\epsilon}\widehat{I}(t)dt\leq\frac{\epsilon^{2}}{2}-\frac{F(t_{*})\lambda}{8\omega\alpha}\epsilon.$
(5.15)

We set

$\Omega_{\ref{611}}=\left\\{\widehat{\tau}\leq
t_{*}+\epsilon\right\\}\cup\left\\{\widehat{\tau}>t_{*}+\epsilon,\left|\widehat{I}_{E}(t_{*})\right|\leq\frac{-F(t_{*})\epsilon
n}{8},\sup_{t\leq
2\wedge\widehat{\tau}}\left|M_{4}(t)\right|\leq\frac{-F(t_{*})\epsilon
n}{8}\right\\}.$ (5.16)

By (1.17) (applied to $t=t_{*}$), the fact that $\widehat{i}_{E}(t_{*})=0$ and
(4.39),

$\lim_{n\to\infty}\mathbb{P}(\Omega_{\ref{611}})=1.$ (5.17)

On the event
$\Omega_{\ref{610}}\cap\Omega_{\ref{611}}\cap\Omega_{\ref{fn-f}}(-F(t_{*})/8)\cap\\{\widehat{\tau}>t_{*}+\epsilon\\}$,
by (5.12) and (5.15),

$\begin{split}\widehat{I}_{E}(t+\epsilon)&\leq(t_{*}+\epsilon-
t_{*})\frac{3F(t_{*})n}{4}+E(t_{*}+\epsilon)\\\ &\leq\frac{3\epsilon
F(t_{*})n}{4}-\frac{F(t_{*})\epsilon n}{8}-\frac{F(t_{*})\epsilon
n}{8}+\frac{n\omega\alpha}{\lambda}\left(\frac{\epsilon^{2}}{2}-\frac{F(t_{*})\lambda}{8\omega\alpha}\epsilon\right)-\frac{\epsilon
F(t_{*})}{8}-\frac{\epsilon F(t_{*})}{8}\\\
&=n\left(\frac{F(t_{*})}{8}\epsilon+\frac{\omega\alpha}{2\lambda}\epsilon^{2}\right),\end{split}$
(5.18)

which is smaller than $0$ if $\epsilon$ is small. Thus on
$\Omega_{\ref{610}}\cap\Omega_{\ref{611}}\cap\Omega_{\ref{fn-f}}(-F(t_{*})/8)\cap\\{\widehat{\tau}>t_{*}+\epsilon\\}$,
we necessarily have $\widehat{\tau}^{(n)}\leq t_{*}+\epsilon$, which is a
contradiction. Hence

$\Omega_{\ref{610}}\cap\Omega_{\ref{611}}\cap\Omega_{\ref{fn-f}}(-F(t_{*})/8)\cap\\{\widehat{\tau}>t_{*}+\epsilon\\}=\emptyset.$
(5.19)

Combining (5.6), (5.19), (5.14) and (5.17), we get

$\lim_{n\to\infty}\mathbb{P}(\widehat{\tau}>t_{*}+\epsilon)=0,$ (5.20)

as desired.

###  5.2. Proof of (5.2) for $F(t_{*})=0$

By (5.5) and the assumption $\lambda>\lambda_{c}$,

$F^{\prime}(t)=-\mu+\frac{2\omega\alpha}{\lambda}\left(1-\log\frac{\widehat{s}(0)}{\widehat{s}(t)}\right)\leq-\mu+\frac{2\omega\alpha}{\lambda}<-\mu+\frac{2\omega\alpha}{\lambda_{c}}.$
(5.21)

Recall that $\lambda_{c}=(\gamma+\omega)/(\mu-1)$. Thus we have

$\begin{split}-\mu+\frac{2\omega\alpha}{\lambda_{c}}&=-\mu+\frac{2\omega\alpha(\mu-1)}{\gamma+\omega}\\\
&=\left(\frac{2\omega\alpha}{\gamma+\omega}-1\right)\mu-\frac{2\omega\alpha}{\gamma+\omega}\\\
&=\left(\frac{2\omega\alpha}{\gamma+\omega}-1\right)\left(\mu-\frac{2\omega\alpha}{2\omega\alpha-\omega-\gamma}\right).\end{split}$
(5.22)

If the condition (1.7) is satisfied, then either

$\frac{2\omega\alpha}{\gamma+\omega}-1\leq\frac{2\omega\alpha}{\omega(2\alpha-1)+\omega}-1=0,$

or

$\frac{2\omega\alpha}{\gamma+\omega}-1>0\mbox{ and
}\mu\leq\frac{2\omega\alpha}{2\omega\alpha-\omega-\gamma}.$

By (5.22), in both cases

$-\mu+\frac{2\omega\alpha}{\lambda_{c}}\leq 0.$ (5.23)

Hence we can set

$C_{\ref{f't4}}:=-\left(-\mu+\frac{2\omega\alpha}{\lambda}\right)>0.$ (5.24)

It follows from (5.21) that, for $t\geq t_{*}$,

$F(t)\leq-C_{\ref{f't4}}(t-t_{*}).$

Consequently, we have

$\int_{t_{*}}^{t}F(u)du\leq-\int_{t_{*}}^{t}C_{\ref{f't4}}(u-t_{*})du=-\frac{C_{\ref{f't4}}}{2}(t-t_{*})^{2}.$
(5.25)

Let $\epsilon_{1}$ be some number to be determined and define the events

$\Omega_{\ref{62-1}}=\left\\{\sup_{0\leq t\leq
1\wedge\widehat{\tau}}\left|M_{3}(t)\right|\leq\epsilon_{1}n\right\\},$ (5.26)

and

$\Omega_{\ref{620}}=\\{\widehat{\tau}\leq
t_{*}+\epsilon\\}\cup\\{\widehat{\tau}>t_{*}+\epsilon,\widehat{I}(t)\leq\left(t-t_{*}+\epsilon_{1}\right)n,\,\forall
t_{*}\leq t\leq\widehat{\tau}\\}.$ (5.27)

Similarly to (5.14),

$\lim_{n\to\infty}\mathbb{P}(\Omega_{\ref{62-1}})=\lim_{n\to\infty}\mathbb{P}(\Omega_{\ref{620}})=1.$
(5.28)

Consider the event

$\begin{split}\Omega_{\ref{621}}=&\left\\{\sup_{0\leq t\leq
2\wedge\widehat{\tau}^{(n)}}\left|M_{4}(t)\right|\leq\epsilon_{1}n\right\\}\cap\Omega_{\ref{fn-f}}(\epsilon_{1})\\\
\cap&\left(\\{\widehat{\tau}\leq
t_{*}+\epsilon\\}\cup\\{\widehat{\tau}>t_{*}+\epsilon,\widehat{I}_{E}(t_{*})\leq\epsilon_{1}n,\widehat{I}(t)\leq(t-t_{*}+\epsilon_{1})n,\forall
t_{*}\leq t\leq\widehat{\tau}\\}\right)\end{split}$ (5.29)

By the fact that $\widehat{i}_{E}(t_{*})=0$, (5.28), (5.6) (with the $\delta$
there set to be $\epsilon_{1}$) and (4.39),

$\lim_{n\to\infty}\mathbb{P}(\Omega_{\ref{621}})=1.$

On $\Omega_{\ref{621}}$, using the definition of $E(t)$ in (5.7),

$\begin{split}E(t)\leq&\int_{t_{*}}^{t}(n\epsilon_{1})du+\epsilon_{1}n+\int_{t_{*}}^{t}n\frac{\omega\alpha}{\lambda}(u-t_{*}+\epsilon_{1})du+2\epsilon_{1}n\\\
\leq&\left(\frac{\omega\alpha}{2\lambda}(t-t_{*})^{2}+\left(1+\frac{\omega\alpha}{\lambda}\right)\epsilon_{1}(t-t_{*})+3\epsilon_{1}\right)n.\end{split}$
(5.30)

If both $\epsilon$ and $\epsilon_{1}$ are small enough, then

$\frac{\omega\alpha}{2\lambda}(t-t_{*})^{2}+\left(1+\frac{\omega\alpha}{\lambda}\right)\epsilon_{1}(t-t_{*})\leq\frac{C_{\ref{f't4}}\gamma}{4\omega\alpha}(t-t_{*}),\,\forall
t_{*}\leq t\leq t_{*}+\epsilon.$ (5.31)

It follows from (5.30), (5.31), (5.8) and (5.25) that, on
$\Omega_{\ref{621}}$, for $t_{*}\leq t\leq t_{*}+\epsilon$,

$\widehat{I}_{E}(t)\leq
E(t)\leq\frac{C_{\ref{f't4}}\gamma}{4\omega\alpha}(t-t_{*})n+3\epsilon_{1}n.$
(5.32)

Define

$\Omega_{\ref{623}}:=\Omega_{\ref{62-1}}\cap\Omega_{\ref{621}}.$ (5.33)

Using (5.4) and (5.32), on $\Omega_{\ref{623}}$,

$\begin{split}0\leq\widehat{I}(t)&\leq\int_{t_{*}}^{t}n\left(-\frac{\gamma}{\lambda}\frac{\widehat{I}(u)}{\widehat{I}_{E}(u)}+1\right)du+3\epsilon_{1}n\\\
&\leq
n\left(-\frac{\gamma}{\lambda}\int_{t_{*}}^{t}\frac{\widehat{I}(u)}{C_{\ref{f't4}}\gamma(u-t_{*})n/(4\omega\alpha)+3\epsilon_{1}n}du+t-t_{*}+3\epsilon_{1}\right).\end{split}$

Hence we deduce that on $\Omega_{\ref{623}}$, for $\epsilon_{1}$ small enough,

$\begin{split}\int_{t_{*}}^{t}\widehat{I}(u)du&\leq\left(\frac{C_{\ref{f't4}}\gamma}{4\omega\alpha}(t-t_{*})+3\epsilon_{1}n\right)\int_{t_{*}}^{t}\frac{\widehat{I}(u)}{C_{\ref{f't4}}\gamma(u-t_{*})/(4\omega\alpha)+3\epsilon_{1}}du\\\
&\leq\left(\frac{C_{\ref{f't4}}\gamma}{4\omega\alpha}(t-t_{*})+3\epsilon_{1}n\right)\frac{\lambda}{\gamma}(t-t_{*}+3\epsilon_{1})n\\\
&\leq
n\left(\frac{C_{\ref{f't4}}\lambda}{4\omega\alpha}(t-t_{*})^{2}+\epsilon_{1}\right).\end{split}$
(5.34)

Apply (5.34) to (5.7), on $\Omega_{\ref{623}}$,

$\widehat{I}_{E}(t)\leq\int_{t_{*}}^{t}F(u)du+\frac{\omega\alpha}{\lambda}\left(n\frac{C_{\ref{f't4}}\lambda}{4\omega\alpha}(t-t_{*})^{2}+\epsilon_{1}n\right)+\epsilon_{1}n(t-t_{*})+3\epsilon_{1}n.$
(5.35)

Using (5.25),

$\widehat{I}_{E}(t)\leq-\frac{C_{\ref{f't4}}}{4}(t-t_{*})^{2}+\left(\frac{\omega\alpha}{\lambda}+4\right)\epsilon_{1}n.$
(5.36)

Using (5.36), by choosing $\epsilon_{1}$ small enough so that

$\epsilon_{1}<\frac{C_{\ref{f't4}}}{4}\left(\frac{\omega\alpha}{\lambda}+4\right)^{-1}\epsilon^{2},$

we can achieve $\widehat{I}_{E}(t_{*}+\epsilon)\leq 0$ on
$\Omega_{\ref{623}}$, which then implies that $\widehat{\tau}<t_{*}+\epsilon$
on this event. Since $\mathbb{P}(\Omega_{\ref{623}})\to 1$ as $n\to\infty$,
(5.2) follows. Hence we have completed the proof of (5.2) in Case ②
($F(t_{*})=0$) and also the proof of part (a) of Theorem 1.4. As we mentioned
in the beginning of this section, part (b) can be proved in the same way. We
omit the details.

##  6\. Continuity of Phase transitions of the SIR-$\omega$ model

By Theorem 1.4, it suffices to prove that $t_{*}\to 0$ as
$\lambda\to\lambda_{c}=(\gamma+\omega)/(\mu-1)$. Recall the definition of the
function $F(t)$ in (5.5). Using the expression of $F$ we can write

$\begin{split}\frac{d\widehat{i}}{dt}&=-\frac{\gamma}{\lambda}\frac{\widehat{i}}{\widehat{i}_{E}}+1,\\\
\frac{d\widehat{i}_{E}}{dt}&=F-\frac{\widehat{i}_{E}}{\widehat{s}}+\frac{\omega\alpha}{\lambda}\widehat{i}.\end{split}$
(6.1)

Using the equation for $\widehat{i}^{\prime}(t)$ and the fact
$\widehat{i}(0)=0$ (initially only one vertex is infected),

$\widehat{i}(t)=-\int_{0}^{t}\frac{\gamma}{\lambda}\frac{\widehat{i}(u)}{\widehat{i}_{E}(u)}du+t\geq
0.$ (6.2)

Hence we have, for $t\leq t_{*}$,

$\int_{0}^{t}\frac{\widehat{i}(u)}{\widehat{i}_{E}(u)}du\leq\frac{\lambda}{\gamma}t.$
(6.3)

Since $\left|\widehat{i}^{\prime}_{E}\right|$ and $\left|F^{\prime}(t)\right|$
are uniformly bounded for all $\lambda$ in a neighborhood of $\lambda_{c}$ and
$t\leq t_{*}$, for some constant $C$, $\widehat{i}_{E}(t)\leq Ct$ and
$F(t)\leq F(0)+Ct$. Using this fact together with (6.1) and (6.3), we see that
for some constant $C_{\ref{ie1}}$,

$\begin{split}\widehat{i}_{E}(t)&\leq\int_{0}^{t}F(u)du+\int_{0}^{t}\frac{\omega\alpha}{\lambda}\widehat{i}(u)du,\\\
&\leq\int_{0}^{t}(F(0)+Cu)du+\frac{\omega\alpha}{\lambda}\int_{0}^{t}\frac{\widehat{i}(u)}{\widehat{i}_{E}(u)}Cudu,\\\
&\leq F(0)t+C_{\ref{ie1}}t^{2}.\end{split}$ (6.4)

Using the expression of $F(t)$ and (5.21),

$\begin{split}F(0)&=-\frac{1+\gamma+\omega}{\lambda}+\mu,\\\
F^{\prime}(0)&=-\mu+2\frac{\omega\alpha}{\lambda},\\\
F^{\prime\prime}(t)&=\frac{-2\omega\alpha}{\lambda(1-t)}.\end{split}$ (6.5)

Note that $F(0)\to 0$ as $\lambda\to\lambda_{c}$, while $F^{\prime\prime}(t)$
is bounded from above by $-2\omega\alpha/\lambda$.

We now divide the proof of Theorem 1.1 into two cases.

###  6.1. Proof of Case ①

Suppose either $\omega(2\alpha-1)\leq\gamma$ or

$\omega(2\alpha-1)>\gamma\mbox{ and
}\mu<2\omega\alpha/(\omega(2\alpha-1)-\gamma).$

Then by (5.22),

$-\mu+\frac{2\omega\alpha}{\lambda_{c}}=-\mu+\frac{2\omega\alpha(\mu-1)}{\gamma+\omega}<0.$

By (6.5) and (5.22),

$\lim_{\lambda\to\lambda_{c}}F^{\prime}(0)=-\mu+\frac{2\omega\alpha}{\lambda_{c}}<0.$
(6.6)

Hence there exists $\epsilon_{0},\delta_{0}>0$, such that for
$\lambda-\lambda_{c}<\epsilon_{0}$ and $t<\delta_{0}$,

$F(t)\leq F(0)-C_{\ref{f't6}}t$ (6.7)

for some $C_{\ref{f't6}}>0$. By (6.3), (6.4) and (6.7), for $t\leq\delta_{0}$,

$\begin{split}\widehat{i}_{E}(t)&\leq\int_{0}^{t}F(u)du+\int_{0}^{t}\frac{\omega\alpha}{\lambda}\widehat{i}(u)du\\\
&\leq\int_{0}^{t}(F(0)-C_{\ref{f't6}}u)du+\frac{\omega\alpha}{\lambda}\int_{0}^{t}\frac{\widehat{i}(u)}{\widehat{i}_{E}(u)}(F(0)u+C_{\ref{ie1}}u^{2})du\\\
&\leq
F(0)t-\frac{C_{\ref{f't6}}}{2}t^{2}+\frac{\omega\alpha}{\lambda}(F(0)t+C_{\ref{ie1}}t^{2})\frac{\lambda}{\gamma}t\\\
&=F(0)t-\left(\frac{C_{\ref{f't6}}}{2}-\frac{\omega\alpha}{\gamma}F(0)\right)t^{2}+\frac{\omega\alpha}{\gamma}t^{3},\end{split}$
(6.8)

where we have used
$\widehat{i}(u)=(\widehat{i}(u)/\widehat{i}_{E}(u))\widehat{i}_{E}(u)$ in the
second inequality. Note that

$\lim_{\lambda\to\lambda_{c}}F(0)=0\mbox{ and
}\lim_{\lambda\to\lambda_{c}}\left(\frac{C_{\ref{f't6}}}{2}-\frac{\omega\alpha}{\gamma}F(0)\right)=\frac{C_{\ref{f't6}}}{2}>0.$

Hence there exist $\delta_{1}<\delta_{0}$ and $\epsilon_{1}<\epsilon_{0}$ such
that for $t\leq\delta_{1}$ and $\lambda<\lambda_{c}+\epsilon_{1}$,

$\widehat{i}_{E}(t)\leq
F(0)t-\left(\frac{C_{\ref{f't6}}}{2}-\frac{\omega\alpha}{\gamma}F(0)\right)t^{2}+\frac{\omega\alpha}{\gamma}t^{3}\leq
F(0)t-\frac{C_{\ref{f't6}}}{4}t^{2}.$ (6.9)

From this we obtain that

$t_{*}=\inf\\{0<t\leq\widehat{s}(0):\widehat{i}_{E}(t)=0\\}\leq\frac{4F(0)}{C_{\ref{f't6}}},$

which converges to 0 as $\lambda\to\lambda_{c}$ since $F(0)\to 0$.

###  6.2. Proof of Case ②

Suppose $\omega(2\alpha-1)>\gamma$ and
$\mu=2\omega\alpha/(\omega(2\alpha-1)-\gamma)$. By (6.5) and (5.22), we have

$F^{\prime}(0)<0\mbox{ and
}\lim_{\lambda\to\lambda_{c}}F^{\prime}(0)=-\mu+\frac{2\omega\alpha}{\lambda_{c}}=0.$
(6.10)

Hence by the third equation in (6.5),

$F(t)\leq F(0)-\frac{\omega\alpha}{\lambda}t^{2}.$

Using (6.4), for some constant $C_{\ref{iebd6}}$ (independent of
$\lambda-\lambda_{c}$),

$\begin{split}\widehat{i}_{E}(t)&\leq\int_{0}^{t}F(u)du+\int_{0}^{t}\frac{\omega\alpha}{\lambda}\widehat{i}(u)du\\\
&\leq\int_{0}^{t}\left(F(0)-\frac{\omega\alpha}{\lambda}u^{2}\right)du+\frac{\omega\alpha}{\lambda}\int_{0}^{t}\frac{\widehat{i}(u)}{\widehat{i}_{E}(u)}(F(0)u+C_{\ref{ie1}}u^{2})du\\\
&\leq F(0)t+C_{\ref{iebd6}}(F(0)t^{2}+t^{3}),\end{split}$ (6.11)

where we have again used
$\widehat{i}(u)=(\widehat{i}(u)/\widehat{i}_{E}(u))\widehat{i}_{E}(u)$ in the
second inequality. Substituting this bound into (6.11) again, we see that

$\begin{split}\widehat{i}_{E}(t)&\leq\int_{0}^{t}\left(F(0)-\frac{\omega\alpha}{\lambda}u^{2}\right)du+\frac{\omega\alpha}{\gamma}t\left(F(0)t+C_{\ref{iebd6}}(F(0)t^{2}+t^{3})\right)\\\
&=F(0)t+\frac{\omega\alpha}{\gamma}F(0)t^{2}+\left(-\frac{\omega\alpha}{3\lambda}+\frac{\omega\alpha}{\gamma}C_{\ref{iebd6}}F(0)\right)t^{3}+\frac{\omega\alpha}{\gamma}C_{\ref{iebd6}}t^{4}.\end{split}$
(6.12)

Since

$\lim_{\lambda\to\lambda_{c}}F(0)=0\mbox{ and
}\lim_{\lambda\to\lambda_{c}}\left(-\frac{\omega\alpha}{3\lambda}+\frac{\omega\alpha}{\lambda}C_{\ref{iebd6}}F(0)\right)=-\frac{\omega\alpha}{3\lambda_{c}}<0,$

we see that there exist $\epsilon_{2}>0$ and $\delta_{2}>0$ such that for
$\lambda<\lambda_{c}+\epsilon_{2}$ and $t\leq\delta_{2}$,

$\widehat{i}_{E}(t)\leq 2F(0)t-\frac{\omega\alpha}{6\lambda_{c}}t^{3}.$ (6.13)

It follows from (6.13) that

$t_{*}=\inf\\{0<t\leq\widehat{s}(0):\widehat{i}_{E}(t)=0\\}\leq\sqrt{\frac{12F(0)\lambda_{c}}{\omega\alpha}},$

which converges to 0 as $\lambda\to\lambda_{c}$ since $F(0)\to 0$.

## Acknowledgements

We thank Rick Durrett for many helpful discussions. Wenze Chen and Yuewen Hou
thank their advisor Yingchao Xie for encouraging them to join this project.

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