# Trajectory of individual immunity and vaccination required for SARS-CoV-2 community immunity: a conceptual investigation

###### tags: `articles`

[ToC]

## Introduction
In this paper, we develop a simple analytical model to examine the importance of vaccination rates and the duration of immunity for a trajectory towards community immunity.

Let's analyze stage-by-stage:
- **Stage 1:** consider an epidemic without vaccination but under strong non-pharmaceutical mitigation strategies
- **Stage 2:** consider the deployment of a vaccine
- **Stage 3:** consider cases where individuals may have preexisting immunity

## General framework (Stage 1)
:::info
**Notation**
- $I(t)$: the number of infectious individuals at time $t$
- $\gamma$: the average number of infections caused by each infectious individual in a time unit
- ${\cal M}$: the number of individuals become immune per time
- $\delta$: the waning rate of immunity
- $P(t)$: the number of people protected at time $t$
- ${\cal H}$: the herd immunity threshold, which is given before the model is applied
:::

Assume that the system has reached a **flat epidemic curve** due to control measures _(such as social distancing)_, that is, let $I(t) = {\Bbb M}$, where ${\Bbb M}$ is a constant.
- ${\cal M} = \gamma{\Bbb M}$, which is also the number of infections per time
- It follows that
$$
\frac{{\rm d}P}{{\rm d}t} = {\cal M} - \delta P
$$
Integrating this equation and assuming that $P(0) = 0$ gives
$$
P(t) = \frac{\cal M}{\delta}(1 - e^{-\delta t})
$$
as $t\to \infty, P(t)\to\frac{\cal M}{\delta}$

- In order to reach herd immunity eventually, it is necessary that
$$
\lim_{t\to \infty}(P(t)+I(t)) \geq {\cal H} \\
\Rightarrow \frac{\cal M}{\delta} + {\Bbb M} \geq {\cal H}
$$
which gives
$$
\frac{1}{\delta} \geq \frac{{\cal H}-{\Bbb M}}{\cal M} = \frac{\cal H}{\cal M} - \frac{1}{\gamma}
$$
This relation implies that
$$
\text{Length of immunity}\geq \frac{\text{Herd immunity threshold}}{\text{Infections per time}} - \text{Duration of infection}
$$

- Another key parameter is the time until the population reaches herd immunity $t_\text{protected}$. Let's set $P(t)+I(t)={\cal H}$ and solve for $t$:
$$
\begin{aligned}
&\ P(t)+{\Bbb M}={\cal H} \\
\Rightarrow &\ \frac{\cal M}{\delta}(1 - e^{-\delta t}) + \frac{\cal M}{\gamma} = {\cal H} \\
\Rightarrow &\ t_\text{protected} = -\frac{1}{\delta}\log{(1-\delta(\frac{\cal H}{\cal M} - \frac{1}{\gamma}))}
\end{aligned}
$$

## Introducing vaccination (Stage 2)
Let's make an assumption that natural immunity alone cannot lead to herd immunity, i.e.
$$
\frac{1}{\delta} < \frac{\cal H}{\cal M} - \frac{1}{\gamma}\ \text{ or }\ \frac{\cal M}{\delta} + {\Bbb M} < {\cal H}
$$

- Suppose that immunity from vaccination of individuals without natural immunity occurs at a constant rate ${\cal V}$ and starts at time $t_v$. We also assume that vaccine-derived immunity wanes at rate $\rho$, then the number of individuals with vaccine-derived immunity $Y(t)$ at time $t$ follows:
$$
\frac{{\rm d}Y}{{\rm d}t} = {\cal V} - \rho Y
$$
Since vaccination is started at time $t_v$, it follows by integration that
$$
Y(t) = \begin{cases}0 & \text{, if $t<t_v$}\\ \frac{\cal V}{\rho}(1 - e^{-\rho (t-t_v)}) & \text{, if $t\geq t_v$}\end{cases}
$$
Thus, the total number of individuals with immunity is
$$
\begin{aligned}
T(t) & = P(t) + Y(t) \\
& = \begin{cases}\frac{\cal M}{\delta}(1 - e^{-\delta t}) & \text{, if $t<t_v$}\\ \frac{\cal M}{\delta}(1 - e^{-\delta t}) + \frac{\cal V}{\rho}(1 - e^{-\rho (t-t_v)}) & \text{, if $t\geq t_v$}\end{cases}
\end{aligned}
$$

- Since we assume that herd immunity is not attained at $t_v$, it follows that herd immunity is eventually reached if and only if
$$
\lim_{t\to \infty}(T(t)+I(t)) \geq {\cal H} \\
\Rightarrow \frac{\cal M}{\delta} + \frac{\cal V}{\rho} + {\Bbb M} \geq {\cal H}
$$
which gives
$$
\frac{1}{\delta} + \frac{1}{\gamma} \geq \frac{\cal H}{\cal M} - \frac{1}{\rho} \frac{\cal V}{\cal M}
$$
This relation implies that
$$
\begin{aligned}
& \ \text{Length of natural immunity} + \text{Duration of infection} \\
\geq & \ \frac{\text{Herd immunity threshold}}{\text{Infections per time}} - \text{Length of vaccine-derived immunity}\cdot \frac{\text{Vaccination rate}}{\text{Infections per time}}
\end{aligned}
$$

- Assume vaccine-derived and natural immunity last for similar time periods, i.e. $\rho \approx \delta$. Now we solve $T(t)+I(t)={\cal H}$ for $t_\text{protected}$

$$
\begin{aligned}
&\ T(t)+{\Bbb M}={\cal H} \\
\Rightarrow &\ \frac{\cal M}{\delta}(1 - e^{-\delta t}) + \frac{\cal V}{\delta}(1 - e^{-\delta (t-t_v)}) + \frac{\cal M}{\gamma} = {\cal H} \\
\Rightarrow &\ t_\text{protected} = -\frac{1}{\delta}\log{(1-\delta(\frac{\cal H}{\cal M} - \frac{1}{\gamma}) + \frac{\cal V}{\cal M})} + \frac{1}{\delta}\log{(1+\frac{\cal V}{\cal M}e^{\delta t_v})}
\end{aligned}
$$

## Preexisting immunity (Stage 3)
We generalize the model to include $P(0) = P_0$ individuals with immunity at $t = 0$, where $P_0 < \frac{\cal M}{\delta}$
- With $\frac{{\rm d}P}{{\rm d}t} = {\cal M} - \delta P$, and $P(0) = P_0$, we have
$$
P(t) = \frac{\cal M}{\delta}(1 - e^{-\delta t}) + P_0e^{-\delta t}
$$

- Since $\lim_{t\to \infty}P_0e^{-\delta t} = 0$, $\lim_{t\to \infty}P(t) = \frac{\cal M}{\delta}$, the duration of immunity required to reach herd immunity does not change:
$$
\frac{1}{\delta} \geq \frac{\cal H}{\cal M} - \frac{1}{\gamma}
$$

- The time until protection is now
$$
t_\text{protected} = \frac{1}{\delta}\log{(1-\frac{\delta P_0}{\cal M})} - \frac{1}{\delta}\log{(1-\delta(\frac{\cal H}{\cal M} - \frac{1}{\gamma}))}
$$

- With vaccination,
$$
T(t) = \begin{cases}\frac{\cal M}{\delta}(1 - e^{-\delta t})+P_0e^{-\delta t} & \text{, if $t<t_v$}\\ \frac{\cal M}{\delta}(1 - e^{-\delta t}) + \frac{\cal V}{\rho}(1 - e^{-\rho (t-t_v)})+P_0e^{-\delta t} & \text{, if $t\geq t_v$}\end{cases}
$$
and the duration of immunity required to reach herd immunity:
$$
\frac{1}{\delta} + \frac{1}{\gamma} \geq \frac{\cal H}{\cal M} - \frac{1}{\rho} \frac{\cal V}{\cal M}
$$
, since $\lim_{t\to \infty}P_0e^{-\delta t} = 0$

- Under the assumption that $\rho \approx \delta$, the time to protection becomes
$$
t_\text{protected} = -\frac{1}{\delta}\log{(1-\delta(\frac{\cal H}{\cal M} - \frac{1}{\gamma}) + \frac{\cal V}{\cal M})} + \frac{1}{\delta}\log{(1+\frac{\cal V}{\cal M}e^{\delta t_v}-\frac{\delta P_0}{\cal M})}
$$

## Discussion and conclusion
- What we have done:
- Consider models with different situations:
1. without vaccination + without preexisting immunity
2. with vaccination in a fixed rate + without preexisting immunity
3. without vaccination + with preexisting immunity of $P_0$
4. with vaccination in a fixed rate + with preexisting immunity of $P_0$
- Analyze the change of number of individuals with immunity over time ($P(t)$ or $T(t)$)
- Analyze the duration of immunity required to reach herd immunity ($\frac{1}{\delta}$ with ${\cal H}$ known)
- Analyze the time elapsing until the population reaches herd immunity ($t_\text{protected}$)
- The models are still not realistic enough since:
- We have assumed a constant influx of infections, and ignored more complex transmission dynamics
- We have omitted specific biological details in our analyses, such as _mortality_
- We have assumed that the herd immunity threshold is independent of the relative level of vaccination to infection incidence

## Appendix: solving first-order differential equation in the standard form
The following is called the **standard form** of a linear differential equation of $P(t)$:
$$
\frac{{\rm d}P}{{\rm d}t} + A(t)\cdot P = B(t)
$$
To solve this kind of equation, we usually use the **integrating factor** $e^{\int A(t)dt}$:
$$
e^{\int A(t)dt}\cdot \frac{{\rm d}P}{{\rm d}t} + A(t)\cdot e^{\int A(t)dt}\cdot P = B(t)\cdot e^{\int A(t)dt}
$$
Then, apply the **product rule** to the left-hand side, we get:
$$
\frac{{\rm d}}{{\rm d}t}(e^{\int A(t)dt}\cdot P) = B(t)\cdot e^{\int A(t)dt}
$$
After integrating both sides, we will finally solve $P(t)$ in general form:
$$
P = e^{-\int A(t)dt}\cdot \int{B(t)\cdot e^{\int A(t)dt}}
$$
If the **initial value** is given, we can go further to solve the unique solution of $P(t)$.

---
Let's look at an example:
$$
\begin{aligned}
&\ \frac{{\rm d}P}{{\rm d}t} = {\cal M} - \delta P \\
\Rightarrow &\ \frac{{\rm d}P}{{\rm d}t} + \delta P = {\cal M} \\
\Rightarrow &\ e^{\delta t}\cdot \frac{{\rm d}P}{{\rm d}t} + \delta e^{\delta t}\cdot P = {\cal M}\cdot e^{\delta t} \\
\Rightarrow &\ \frac{{\rm d}}{{\rm d}t}(e^{\delta t}\cdot P) = {\cal M}\cdot e^{\delta t} \\
\Rightarrow &\ e^{\delta t}\cdot P = \frac{\cal M}{\delta}\cdot e^{\delta t} + C \\
\Rightarrow &\ P = \frac{\cal M}{\delta} + Ce^{-\delta t}
\end{aligned}
$$
With initial value $P(0) = 0$, we can calculate the value of $C$ and get the unique solution of $P(t)$:
$$
\begin{aligned}
&\ P(0) = \frac{\cal M}{\delta} + C\cdot 1 = 0 \\
\Rightarrow &\ C = -\frac{\cal M}{\delta}
\end{aligned}
$$
Thus,
$$
\begin{aligned}
P(t) & = \frac{\cal M}{\delta} -\frac{\cal M}{\delta}e^{-\delta t} \\
& = \frac{\cal M}{\delta}(1-e^{-\delta t})
\end{aligned}
$$