On Integrable Models for the Spread of Disease

Abstract

The integrable versions of SIR epidemic models are introduced. The exact solutions of these models are derived. The advantage of these models is the possibility of full analysis of obtained solutions and the simplicity of explicit formulas for the important metrics of spread of disease. The effectiveness of these formulas is illustrated by applications to the spread of COVID-19.

1. Introduction

In 1760, to show the effectiveness of inoculation, Daniel Bernoulli [1] introduced a mathematical model to estimate the life expectancy of a population affected by smallpox virus.

A general model of spread of a disease was developed in 1927 by Kermack and McKendrick [2] with its special case: SIR model. This model splits a population into three compartments: susceptible (S) to the infection, infectious (I) and removed (R) (recovered, isolated or deceased).

In [3], Norman Bailey studied deterministic and stochastic mathematical models in epidemiology. He introduced an integrable version of the SIR model.

For a presentation of mathematical models in epidemiology before the year 2000, see the survey article of H. W. Hethcote [4].

The SIR model has been solved numerically by various numerical methods: Taylor approximation, Pade approximations, time series, spectral truncation method and so on. Disadvantages of approximation methods are the error estimates requirements, which are not always possible to satisfy.

In 2014, T. Harko, F. S. N. Lobo and M. K. Mak [5] obtained an exact analytical solution to the SIR model (containing integrals that can only be evaluated numerically), and the effectiveness of various control strategies for epidemic models was studied in [6].

Intensive biological, medical and mathematical studies of the COVID-19 pandemic were conducted in 2020–2025 [7,8,9,10,11,12,13,14,15,16,17,18,19].

An accurate analytic expression and approximation of solutions for the SIR model with applications to COVID-19 was obtained in 2021 by Kröger and Schlickeiser [14,15,16].

In [13], G. Gaeta introduced A-SIR model with numerical analysis of the spread of disease by taking into account the presence of asymptomatic (undetected) infectives.

In [11], a new class of kinetic epidemic models was introduced which was capable of incorporating the social heterogeneity of a population. The numerical simulations confirmed the ability of the model to describe the rapid spread of infection. In [8], the discrete-time epidemic model is considered with the goal of investigating the daily behavior of reproduction number. The approximation methods used in this paper are the time series and spectral truncation. In [9], the kinetic model was proposed and studied for disease spread where the disease transmission depends on viral loads of compartments. Analysis of epidemic dynamics was performed based on stability and bifurcation theory. In [10], a SIR-like kinetic model was proposed where contact rates depend on the behavioral patterns adopted across the population. In [19], a comprehensive overview is given about the capabilities of artificial intelligence in disease forecasting with a large bibliography therein.

In this paper, we introduce new integrable versions of Gaeta’s model. The spread of diseases of these models is similar to the dynamics of the SIR model with additional details and practical formulas for important metrics. The model’s efficiency is illustrated by applications to COVID-19.

In Section 2, the modified Gaeta’s epidemic model is studied. The results of Section 2 are formulated in Theorem 1.

In Section 3, another integrable epidemic model is considered, which is a combination of Bailey’s and Gaeta’s model. The results of Section 3 are formulated in Theorem 2.

This research did not receive any specific grant from funding agencies in the public, commercial, or not-for-profit sectors.

2. Modified Gaeta Model

Consider A-SIR (or Gaeta’s) model [13]:

$$S^{\prime }\left(t\right)=-aS\left(t\right)(I\left(t\right)+J\left(t\right)),I^{\prime }\left(t\right)=a\xi S\left(t\right)(I\left(t\right)+J\left(t\right))-bI\left(t\right),$$

$$J^{\prime }\left(t\right)=a(1-\xi )S\left(t\right)(I\left(t\right)+J\left(t\right))-\eta J\left(t\right)R^{\prime }\left(t\right)=bI\left(t\right),U^{\prime }\left(t\right)=\eta J\left(t\right),t\ge t_0,$$

$$S\left(t_0\right)+I\left(t_0\right)+J\left(t_0\right)+R\left(t_0\right)+U\left(t_0\right)=1.$$

Here, $S\left(t\right)=s\left(t\right)/N$ is a proportion of susceptible population at time t, $s\left(t\right)$ is a number of susceptible people out of local population N, and $I\left(t\right)$ and $J\left(t\right)$ are the proportions of symptomatic and asymptomatic infectious people (infectives), correspondingly. $R\left(t\right)$ and $U\left(t\right)$ are the proportions of removed (recovered or isolated) infectives out of symptomatic and asymptomatic compartments correspondingly with constant removal rates $b>0$ and $\eta >0$. $a>0$ is the constant rate of transmission of infection, $\xi$ is the probability to detect infected individuals, and $T=1/b$, $T_1=1/\eta$ are the duration times of a disease (in days). In this paper, we will assume that the rates of removal are the same: $b=\eta ,T_1=T$.

Note that the functions $S\left(t\right),I\left(t\right),J\left(t\right),R\left(t\right),U\left(t\right)$ are less or equal to 1 and non-negative.

A ratio of

$${\alpha }_0:=aS\left(t_0\right)/b$$

is called a basic reproduction number.

The spread of disease starts if $S\left(t_0\right)<1,I^{\prime }\left(t_0\right)>0$, which in the case $J\left(t_0\right)\approx 0$ means

$${\alpha }_0=aS\left(t_0\right)/b>1.$$

Indeed, from

$$I^{\prime }\left(t_0\right)=aS\left(t_0\right)(I\left(t_0\right)+J\left(t_0\right))-bI\left(t_0\right)>0$$

or

$$aS\left(t_0\right)(I\left(t_0\right)+J\left(t_0\right))-bI\left(t_0\right)\approx I\left(t_0\right)(S\left(t_0\right)a-b)>0,$$

we get

$$aS\left(t_0\right)>b,S\left(t_0\right)>b/a.$$

A number

$$k:=b/a$$

is called the epidemic threshold.

In this section, we will discuss an integrable version of Gaeta’s model:

$$S^{\prime }\left(t\right)=-a(I\left(t\right)+J\left(t\right))\sqrt{S\left(t\right)},I^{\prime }\left(t\right)=a\xi (I\left(t\right)+J\left(t\right))\sqrt{S\left(t\right)}-bI\left(t\right),R^{\prime }\left(t\right)=bI\left(t\right),$$

$$J^{\prime }\left(t\right)=a(1-\xi )(I\left(t\right)+J\left(t\right))\sqrt{S\left(t\right)}-\eta J\left(t\right),U^{\prime }\left(t\right)=\eta J\left(t\right),t>t_0.$$

(1)

$$S\left(t_0\right)+I\left(t_0\right)+J\left(t_0\right)+R\left(t_0\right)+U\left(t_0\right)=1,a,b,\eta \mathrm{are}\mathrm{positive}\mathrm{constants},0\le \xi \le 1.$$

(2)

In this model, it is assumed that the rate of increase of infectives $I^{\prime }\left(t\right)$ is proportional to $(I\left(t\right)+J\left(t\right))\sqrt{S\left(t\right)}.$ The advantage of this model is that it is integrable with simple explicit formulas for the behavior of the spread of infection.

From (1), we have

$$I_S:={\displaystyle \frac{dI}{dS}}={\displaystyle \frac{I^{\prime }\left(t\right)}{S^{\prime }\left(t\right)}}=-\xi +{\displaystyle \frac{I\left(t\right)b}{a(I\left(t\right)+J\left(t\right))\sqrt{S\left(t\right)}}},$$

$$J_S:={\displaystyle \frac{dJ}{dS}}={\displaystyle \frac{J^{\prime }\left(t\right)}{S^{\prime }\left(t\right)}}=\xi -1+{\displaystyle \frac{J\left(t\right)\eta }{a(I\left(t\right)+J\left(t\right))\sqrt{S\left(t\right)}}},$$

$$\eta I_S+b{J}_S=-\eta \xi +b(\xi -1)+{\displaystyle \frac{b\eta }{a\sqrt{S\left(t\right)}}}.$$

By integration

$$\eta I\left(t\right)+bJ\left(t\right)=(-\eta \xi +b\xi -b)S\left(t\right)+2\sqrt{S\left(t\right)}\eta b/a+b{C}_1.$$

If

$$b=\eta$$

we get

$$b(I\left(t\right)+J\left(t\right))=-bS\left(t\right)+2b^2\sqrt{S\left(t\right)}/a+b{C}_1$$

$$I\left(t\right)+J\left(t\right)=C_1-S\left(t\right)+2k\sqrt{S\left(t\right)},b=\eta .$$

(3)

Applying the initial condition at $t=t_0$ to (3) and assuming $R\left(t_0\right)=U\left(t_0\right)=0$, we have

$$C_1=J_0+I_0+S_0-2k\sqrt{S_0}=1-2k\sqrt{S_0},$$

or

$$C_1=1-2k\sqrt{S_0}=D^2-k^2,D:=\sqrt{1-2k\sqrt{S_0}+k^2}.$$

(4)

From the first Equation (1),

$${\left(2\sqrt{S\left(t\right)}\right)}_t:={\displaystyle \frac{d}{dt}}2\sqrt{S\left(t\right)}=-a(I\left(t\right)+J\left(t\right)).$$

Further,

$$-b{\left(2\sqrt{S\left(t\right)}\right)}_t=ab(I\left(t\right)+J\left(t\right))=ab{C}_1-abS\left(t\right)+2abk\sqrt{S\left(t\right)},$$

and by substitution $v=\sqrt{S\left(t\right)}$, we get

$$-2v_t=a{C}_1-a{v}^2+2akv.$$

By separation of variables

$${\displaystyle \frac{dv}{v^2-2kv-C_1}}={\displaystyle \frac{adt}{2}}.$$

After integration, we get

$${\displaystyle \frac{1}{2\sqrt{k^2+C_1}}}ln\left|{\displaystyle \frac{v-k-\sqrt{k^2+C_1}}{v-k+\sqrt{k^2+C_1}}}\right|=C_2+a(t-t_0)/2,v=\sqrt{S\left(t\right)}.$$

or

$${\displaystyle \frac{1}{2D}}ln\left|{\displaystyle \frac{v-k-D}{v-k+D}}\right|=C_2+a(t-t_0)/2,D=\sqrt{k^2+C_1}=\sqrt{k^2-2k\sqrt{S_0}+1}.$$

(5)

From (5) at initial time $t=t_0$ and notation $S_0:=S\left(t_0\right)<1$, we get

$$C_2={\displaystyle \frac{1}{2D}}ln\left|{\displaystyle \frac{\sqrt{S_0}-k-D}{\sqrt{S_0}-k+D}}\right|={\displaystyle \frac{1}{2D}}ln\left|{\displaystyle \frac{D+k-\sqrt{S_0}}{D-k+\sqrt{S_0}}}\right|.$$

(6)

Note that if

$${\alpha }_0:=S_0/k>1,\mathrm{then}{k}^2<k<S_0<1.$$

From (1), it follows that $S\left(t\right)$ is a decreasing function and

$$k<\sqrt{S\left(t\right)}<\sqrt{S\left(t_0\right)}<1.$$

(7)

Denoting $v=\sqrt{S\left(t\right)}$, we get

$$k-v+D=k-\sqrt{S}+\sqrt{1-S_0+{(\sqrt{S_0}-k)}^2}>k-\sqrt{S}+(\sqrt{S_0}-k)>0$$

and

$$k-v-D=k-\sqrt{S}-\sqrt{1-S_0+{(\sqrt{S_0}-k)}^2}<k-\sqrt{S}-(\sqrt{S_0}-k)<0,$$

so, in the case of (7), we have

$$\left|{\displaystyle \frac{v-k-D}{v-k+D}}\right|={\displaystyle \frac{k-v+D}{v-k+D}}={\displaystyle \frac{2D}{v-k+D}}-1>0.$$

Solving

$${\displaystyle \frac{1}{2D}}ln\left({\displaystyle \frac{k-v+D}{v-k+D}}\right)=C_2+{\displaystyle \frac{a(t-t_0)}{2}}$$

for v, we get

$$\sqrt{S\left(t\right)}=v=k-D+{\displaystyle \frac{2D}{1+exp\left\{D(2C_2+a(t-t_0))\right\}}},$$

(8)

or

$$\sqrt{S\left(t\right)}=k-Dtanh\left[D\left(C_2+a(t-t_0)/2\right)\right],D=\sqrt{k^2-2k\sqrt{S_0}+1}.$$

(9)

Note that a sigmoid function (8) appeared in the original Bernoulli paper [1] and later in the logistic equation of Verhulst.

Further, by substitution of (4) and (8) into (3), we get

$$I\left(t\right)+J\left(t\right)={\displaystyle \frac{4D^{2}exp\left\{D(2C_2+a(t-t_0))\right\}}{{(1+exp\left\{D(2C_2+a(t-t_0))\right\})}^2}},exp\left(2D{C}_2\right)=\left|{\displaystyle \frac{(D+k-\sqrt{S_0})}{D-k+\sqrt{S_0}}}\right|,$$

(10)

or

$$I\left(t\right)+J\left(t\right)=D^{2}sec{h}^2\left(D\left(C_2+a(t-t_0)/2\right)\right).$$

(11)

Note that the function $sec{h}^2(·)$ appeared also in paper [2] and later in the theory of solitons.

Since the maximum value $D^2$ of $D^{2}sec{h}^2(D(C_2+a(t-t_0)/2)$ is achieving at time when $C_2+a(t_{max}-t_0)/2=0$, we get

$$I_{max}+J_{max}=D^2,$$

or a (new) formula for the maximum value of infectives in terms of $k,S_0,N$:

$$i_{max}+j_{max}=N{D}^2=N(k^2-2k\sqrt{S_0}+1),$$

(12)

and an explicit (new) formula for the peak time:

$$t_{max}-t_0=-2C_2/a=-{\displaystyle \frac{1}{aD}}ln\left|{\displaystyle \frac{D+k-\sqrt{S_0}}{D-k+\sqrt{S_0}}}\right|.$$

(13)

Note that from the SIR model ($\xi =1,J\left(t\right)\equiv 0$), another formula is well known for the maximum value of infectives:

$$i_{maxSIR}=N\left(1-k+kln\left({\displaystyle \frac{k}{1-I_0}}\right)\right),$$

(14)

which is close to (12) if $S_0\approx 1,I_0\approx 0,k\approx 1,ln\left(k\right)\approx (k-1)$.

Introducing a rate of new cases of infectives

$$L\left(t\right)=I^{\prime }\left(t\right)+J^{\prime }\left(t\right)={\displaystyle \frac{4a{D}^{3}exp\left\{D(2C_2+a(t-t_0))\right\}(1-exp\left\{D(2C_2+a(t-t_0))\right\})}{{(1+exp\left\{D(2C_2+a(t-t_0))\right\})}^3}},$$

we have in view of (6)

$$L={\displaystyle \frac{4a{D}^3{e}^{aD(t-t_0)}(D+k-\sqrt{S_0})(D-k+\sqrt{S_0})(D-k-e^{aD(t-t_0)}(D+k-\sqrt{S_0})+\sqrt{S_0})}{{(D-k+e^{aD(t-t_0)}(D+k-\sqrt{S_0})+\sqrt{S_0})}^3}}$$

or

$$L\left(t\right)={\displaystyle \frac{4a{D}^{3}exp\left(aD(t-t_0)\right)(1-S_0)(D-k-exp\left(aD(t-t_0)\right)(D+k-\sqrt{S_0})+\sqrt{S_0})}{{(D(exp\left(aD(t-t_0)\right)+1)-(exp\left(aD(t-t_0)\right)-1)(\sqrt{S_0}-k))}^3}}.$$

(15)

Further, from the equation of critical numbers of $L\left(t\right)$

$$L^{\prime }\left(t\right)={(I+J)}^{\prime \prime }={\displaystyle \frac{4a^2{D}^4{e}^{D(2C_2+a(t-t_0))}(e^{4C_{2}D+2atD}+e^{2aD{t}_0}-4e^{D(2C_2+a(t-t_0))})}{{(e^{2C_{2}D+atD)}+e^{aD{t}_0})}^4}}=0$$

we get an explicit formula for a peak time $t={\tau }_{max}$ of new cases

$${\tau }_{max}={\displaystyle \frac{1}{aD}}ln\left|(2\pm \sqrt{3})e^{aD{\tau }_0-2C_{2}D}\right|,{\tau }_{max}-{\tau }_0={\displaystyle \frac{1}{aD}}ln\left|(2\pm \sqrt{3})e^{-2C_{2}D}\right|$$

or

$${\tau }_{max}-{\tau }_0={\displaystyle \frac{1}{aD}}ln\left|{\displaystyle \frac{(2-\sqrt{3})(D-k+\sqrt{S_0})}{D+k-\sqrt{S_0}}}\right|.$$

(16)

The peak value of the rate of new cases is given by formula

$$L_{max}=L\left({\tau }_m\right)=2a{D}^3{3}^{-3/2}=2a{(1-2k\sqrt{S_0}+k^2)}^{3/2}{3}^{-3/2}.$$

(17)

Furthermore, from

$$I^{\prime }\left(t\right)=a\xi (I\left(t\right)+J\left(t\right))\sqrt{S\left(t\right)}-bI\left(t\right),$$

using (8) and (10), we get

$$I^{\prime }\left(t\right)={\displaystyle \frac{4a\xi D^{2}exp\left[D(2C_2+a(t-t_0))\right]}{{(1+exp\left[D(2C_2+a(t-t_0))\right])}^2}}\left(k-D+{\displaystyle \frac{2D}{1+exp\left[D(2C_2+a(t-t_0))\right]}}\right)-bI\left(t\right),$$

and by the substitutions

$$y:=D(2C_2+a(t-t_0)),z:=e^y=\left|{\displaystyle \frac{D+k-\sqrt{S_0}}{D-k+\sqrt{S_0}}}\right|exp\left[aD(t-t_0)\right],$$

(18)

we have

$$I^{\prime }\left(t\right)+bI\left(t\right)={\displaystyle \frac{4a\xi D^{2}z}{{(z+1)}^2}}\left(k-D+{\displaystyle \frac{2D}{z+1}}\right).$$

Multiplying by integrating factor $exp\left[b(t-t_0)\right]$, we get

$${(I\left(t\right)exp\left[b(t-t_0)\right])}_t={\displaystyle \frac{4a\xi D^{2}zexp\left[b(t-t_0)\right]}{{(z+1)}^2}}\left(k-D+{\displaystyle \frac{2D}{z+1}}\right).$$

Since

$$b(t-t_0)=k\left(y/D-2C_2\right),exp\left[b(t-t_0)\right]=exp(ky/D)exp(-2k{C}_2),$$

$$exp\left[b(t-t_0)\right]=z^{k/D}exp(-2k{C}_2)=z^{k/D}{\left|{\displaystyle \frac{D+k-\sqrt{S_0}}{D-k+\sqrt{S_0}}}\right|}^{-k/D},dz=y_{t}zdt=azDdt,$$

we have

$$d\left(I\left(t\right)e^{b(t-t_0)}\right)={\displaystyle \frac{4a\xi D^{2}z{e}^{-2k{C}_2}{z}^{\frac{k}{D}}}{{(z+1)}^2}}\left(k-D+{\displaystyle \frac{2D}{z+1}}\right)dt$$

$$={\displaystyle \frac{4a\xi D^{2}z{e}^{-2k{C}_2}{z}^{\frac{k}{D}}}{{(z+1)}^{2}azD}}\left(k-D+{\displaystyle \frac{2D}{z+1}}\right)dz,$$

and by integration

$$I\left(t\right)z^{k/D}{e}^{-2k{C}_2}=4\xi D{e}^{-2k{C}_2}{\int }_{z_0}^z{\displaystyle \frac{z_1^{\frac{k}{D}}}{{(z_1+1)}^2}}\left(k-D+{\displaystyle \frac{2D}{z_1+1}}\right)d{z}_1+I_0$$

or

$$I\left(t\right)z^{k/D}=4\xi D{\int }_{z_0}^z{\displaystyle \frac{z_1^{\frac{k}{D}}}{{(z_1+1)}^2}}\left(k-D+{\displaystyle \frac{2D}{z_1+1}}\right)d{z}_1+I_0{e}^{2k{C}_2}.$$

By substitution

$$z=exp\left[(2C_2+a(t-t_0))D\right]=z_0{e}^{aD(t-t_0)},z_0=z\left(t_0\right)=\left|{\displaystyle \frac{D+k-\sqrt{S_0}}{D-k+\sqrt{S_0}}}\right|$$

we have

$$I\left(t\right){\left(z_{0}exp\left[aD(t-t_0)\right]\right)}^{k/D}=4\xi D{\int }_{z_0}^z{\displaystyle \frac{z_1^{k/D}}{{(z_1+1)}^2}}\left(k-D+{\displaystyle \frac{2D}{z_1+1}}\right)d{z}_1+I_0{z}_0^{k/D},$$

or

$$I\left(t\right)=exp\left[ak(t_0-t)\right]\left(4\xi D{\int }_{z_0}^z{\displaystyle \frac{{(z_1/z_0)}^Q}{{(z_1+1)}^2}}\left(k-D+{\displaystyle \frac{2D}{z_1+1}}\right)d{z}_1+I_0\right),Q=k/D.$$

(19)

The integral above in general is a combination of hypergeometric functions.

Theorem 1. 

Let $b=\eta$ and (7) is satisfied. Then, the solutions of (1) are given by formulas

$$I\left(t\right)=exp\left[ak(t_0-t)\right]\left(4\xi D{\int }_{z_0}^z{\displaystyle \frac{{(z_1/z_0)}^{Q}exp\left\{ak(t-t_0)\right\}}{{(z_1+1)}^2}}\left(k-D+{\displaystyle \frac{2D}{z_1+1}}\right)d{z}_1+I_0\right),$$

$$S\left(t\right)={\left(k-D+{\displaystyle \frac{2D}{1+exp\left\{D(2C_2+a(t-t_0))\right\}}}\right)}^2,$$

$$J\left(t\right)={\displaystyle \frac{4D^{2}exp\left\{D(2C_2+a(t-t_0))\right\}}{{(1+exp\left\{D(2C_2+a(t-t_0))\right\})}^2}}-I\left(t\right),$$

$$U\left(t\right)=b{\int }_{t_0}^{t}J\left(t\right)1)d{t}_1,R\left(t\right)=1-I\left(t\right)-J\left(t\right)-S\left(t\right)-U\left(t\right),R\left(t_0\right)=U\left(t_0\right)=0,$$

$$C_2={\displaystyle \frac{1}{2D}}ln\left(z_0\right),z_0=\left|{\displaystyle \frac{D+k-\sqrt{S_0}}{D-k+\sqrt{S_0}}}\right|,D=\sqrt{k^2-2k\sqrt{S_0}+1},Q=k/D.$$

The maximum number of infectives

$$i_{max}+j_{max}=N(k^2-2k\sqrt{S_0}+1)$$

is attained at the peak time:

$$t_{max}=t_0+{\displaystyle \frac{1}{aD}}ln\left|{\displaystyle \frac{D-k+\sqrt{S_0}}{D+k-\sqrt{S_0}}}\right|.$$

The rate of new cases of infectives is given by

$$L\left(t\right)=I^{\prime }\left(t\right)+J^{\prime }\left(t\right)$$

$$={\displaystyle \frac{4a{D}^{3}exp\left(aD(t-t_0)\right)(1-S_0)(D-k-exp\left(aD(t-t_0)\right)(D+k-\sqrt{S_0})+\sqrt{S_0})}{{(D(exp\left(aD(t-t_0)\right)+1)-(exp\left(aD(t-t_0)\right)-1)(\sqrt{S_0}-k))}^3}}.$$

The maximum value of the rate of new cases

$$L_{max}=2a{(1-2k\sqrt{S_0}+k^2)}^{3/2}{3}^{-3/2}$$

is attained at the peak time:

$$t={\tau }_{max}={\tau }_0+{\displaystyle \frac{1}{aD}}ln\left|{\displaystyle \frac{(2-\sqrt{3})(D-k+\sqrt{S_0})}{D+k-\sqrt{S_0}}}\right|.$$

Remark 1. 

For the modified SIR model, we have $\xi =1$ so, $J\left(t\right)=U\left(t\right)\equiv 0$ and the solutions are simplified:

$$I\left(t\right)={\displaystyle \frac{4D^{2}exp\left\{D(2C_2+a(t-t_0))\right\}}{{(1+exp\left\{D(2C_2+a(t-t_0))\right\})}^2}},$$

$$S\left(t\right)={\left(k-D+{\displaystyle \frac{2D}{1+exp\left\{D(2C_2+a(t-t_0))\right\}}}\right)}^2,R\left(t\right)=1-I\left(t\right)-S\left(t\right).$$

Remark 2. 

If $Q=k/D$ is given, then one can find k from the quadratic

$$Q=k{(k^2-2k\sqrt{S_0}+1)}^{-1/2}.$$

Further, if $C_1=1-2k\sqrt{S_0}>0,$ or ${\alpha }_0=S_0/k>2S_0\sqrt{S_0}>2$ then

$$D=\sqrt{k^2+1-2k\sqrt{S_0}}>k,Q=k/D<1.$$

Remark 3. 

In the case $Q=1$ from

$$Q=k/D=k{(k^2-2k\sqrt{S_0}+1)}^{-1/2}=1$$

we get

$$k=1/\left(2\sqrt{S_0}\right),\mathit{that}\mathit{is},C_1=1-2k\sqrt{S_0}=0$$

and integral in (19) is simplified:

$$I\left(t\right)exp\left[ak(t-t_0)\right]={\displaystyle \frac{4k^2\xi }{z_0}}{\int }_{z_0}^z{\displaystyle \frac{2z}{{(z+1)}^3}}dz+I_0=I_0-{\displaystyle \frac{4k^2\xi }{z_0}}\left({\displaystyle \frac{1+2z}{{(z+1)}^2}}-{\displaystyle \frac{1+2z_0}{{(z_0+1)}^2}}\right),$$

where

$$z_0=\left|{\displaystyle \frac{D+k-\sqrt{S_0}}{D-k+\sqrt{S_0}}}\right|={\displaystyle \frac{2k-\sqrt{S_0}}{\sqrt{S_0}}}={\displaystyle \frac{1-S_0}{S_0}},$$

or

$$I\left(t\right)=exp\left[ak(t_0-t)\right]\left(I_0+{\displaystyle \frac{4\xi k^2{S}_0^2(1-S_0)(e^{ak(t-t_0)}-1)\left(S_0+(2-S_0)e^{ak(t-t_0)}\right)}{{(S_0+(1-S_0)e^{ak(t-t_0)})}^2}}\right).$$

(20)

This formula may be simplified in case $J_0\approx 0,S_0=1-I_0-J_0\approx 1-I_0$:

$$I\left(t\right)\approx I_{0}exp\left[ak(t_0-t)\right]$$

$$\times \left(1+{\displaystyle \frac{4k^2\xi {(1-I_0)}^2(exp\left[ak(t-t_0)\right]-1)(1-I_0+(1+I_0)exp\left[ak(t-t_0)\right])}{{(1-I_0+I_{0}exp\left[ak(t-t_0)\right])}^2}}\right).$$

Furthermore, if $S_0\approx 1-I_0\approx 1,$ we get

$$I\left(t\right)\approx I_{0}exp\left[ak(t-t_0)\right]\left(1+{\displaystyle \frac{4k^2\xi (exp\left[2ak(t-t_0)\right]-1)}{{(1+I_{0}exp\left[ak(t-t_0)\right])}^2}}\right).$$

(21)

Example 1. 

To illustrate our formulas for the spread of COVID-19, consider statistics of infectious people [20] for Niue Island with population $N=1622$:

$$(11/20/2022,14),(12/18/2022,54),(12/27/2022,174),$$

$$(1/5/2023,185),(1/8/2023,149),(1/24/2023,21),$$

which means that on 20 November 2022, there were 14 infectious people and so on.

Also, the new cases maximum increase $l_{max}=43$ happened on 27 December 2022.

By taking 20 November 2022 as the initial day t = 0, we get adjusted data shifted in time:

$$(0,14),(28,54),(37,174),(46,185),(49,149),(65,21)$$

From the exponential model

$$i\left(t\right)\approx i\left(0\right)k^{-t/T}$$

one can obtain a simple estimate for the epidemic threshold

$$k\approx {\left(i_1/i_2\right)}^{T/\Delta t},\Delta t=t_2-t_1,$$

where $i_1,i_2$ are the numbers of infectives at days $t_1,t_2$ correspondingly and T is a duration period of infection.

By taking data (see statistics above): $i_1=14,i_2=54,\Delta t=28$ days and choosing $T=14$ days, we get the estimate

$$k={\left(14/54\right)}^{14/28}\approx 0.5.$$

From (10), by taking the parameters $k=a=0.5,S_0=1-14/N,N=1622,{\tau }_0\approx t_0=27$, we get the time behavior for the number of infectious (symptomatic and asymptomatic) people:

$$i\left(t\right)+j\left(t\right)=N(I\left(t\right)+J\left(t\right))\approx {\displaystyle \frac{1650exp\left[0.5\right(-9.42+0.5(t-27)\left)\right]}{{(1+exp\left[0.5(-9.42+0.5(t-27))\right])}^2}}.$$

(22)

Further from (20), by taking the parameters mentioned above and $\xi =0.45$, we get the time behavior for the number of symptomatic infectious people:

$$i\left(t\right)=NI\left(t\right)\approx {\displaystyle \frac{8341.98(7.6\ast {10}^8+\mathrm{15,497.05}{e}^{0.25t}+854.06e^{0.5t})}{e^{0.25t}{(\mathrm{98,094.75}+e^{0.25t})}^2}}$$

$$i\left(t\right)=NI\left(t\right)\approx {\displaystyle \frac{8341.98(7.6\ast {10}^8+\mathrm{15,497.05}exp\left[0.25t\right]+854.06exp\left[0.5t\right])}{{(\mathrm{98,094.75}+exp\left[0.25t\right])}^{2}exp\left[0.25t\right]}}$$

(23)

with the maximum value

$$i_{max}\left(46\right)\approx 183.$$

From (12) and (13)

$$i_{max}+j_{max}=N(k^2-2k\sqrt{S_0}+1)\approx 413,t_{max}\approx 45.85.$$

From (16)

$${\tau }_{max}={\tau }_0+{\displaystyle \frac{1}{aD}}ln\left|{\displaystyle \frac{(2-\sqrt{3})(D-k+\sqrt{S_0})}{D+k-\sqrt{S_0}}}\right|\approx 27+13.6\approx 40.$$

From (17)

$$l_{max}=2aN{(1-2k\sqrt{S_0}+k^2)}^{3/2}{3}^{-3/2}\approx 41.$$

Our estimates for $i_{max},t_{max},l_{max}$ are close to the statistical values:

$$i_{statmax}\approx 185,t_{statmax}\approx 46,,l_{statmax}\approx 43,{\tau }_{statmax}\approx 37.$$

From our estimates, one can see that the symptomatic infectives (185) are about $45\%$ of infectives (413).

Note that from statistics in Italy, only $10\%$ of infectives are symptomatic [13].

This example is just an illustration of simplicity of application of this model. The actual number of asymptomatic infectives for Niue Island is unknown.

For illustration of Formulas (22) and (23), see the Figure 1 below.

3. Bailey–Gaeta Model

In [1], Norman Bailey introduced a homogeneous and integrable version of the SIR model:

$$S^{\prime }\left(t\right)=-{\displaystyle \frac{aI\left(t\right)}{I\left(t\right)+S\left(t\right)}}S\left(t\right),I^{\prime }\left(t\right)={\displaystyle \frac{aI\left(t\right)}{I\left(t\right)+S\left(t\right)}}S\left(t\right)-bI\left(t\right),R^{\prime }\left(t\right)=bI\left(t\right).$$

Combining this model with Gaeta’s model, consider the Bailey–Gaeta model:

$$S^{\prime }\left(t\right)=-{\displaystyle \frac{a\left(I\right(t)+J(t\left)\right)S\left(t\right)}{I\left(t\right)+J\left(t\right)+S\left(t\right)}},I^{\prime }\left(t\right)={\displaystyle \frac{a\xi \left(I\right(t)+J(t\left)\right)S\left(t\right)}{I\left(t\right)+J\left(t\right)+S\left(t\right)}}-bI\left(t\right),$$

$$J^{\prime }\left(t\right)={\displaystyle \frac{a(1-\xi )\left(I\right(t)+J(t\left)\right)S\left(t\right)}{I\left(t\right)+J\left(t\right)+S\left(t\right)}}-\eta J\left(t\right),R^{\prime }\left(t\right)=bI\left(t\right),U^{\prime }\left(t\right)=\eta J\left(t\right)$$

(24)

$$I\left(t\right)+J\left(t\right)+S\left(t\right)+R\left(t\right)+U\left(t\right)=1,\eta =b,t_0\le t<\infty .$$

(25)

By calculations

$$I_S=I^{\prime }\left(t\right)/S^{\prime }\left(t\right)=-\xi +{\displaystyle \frac{bI\left(t\right)\left(I\right(t)+J(t)+S(t\left)\right)}{a\left(I\right(t)+J(t\left)\right)S\left(t\right)}}$$

$$J_S=J^{\prime }\left(t\right)S^{\prime }\left(t\right)=\xi -1+{\displaystyle \frac{\eta J\left(t\right)\left(I\right(t)+J(t)+S(t\left)\right)}{a\left(I\right(t)+J(t\left)\right)S\left(t\right)}}$$

we get

$$b{J}_S+\eta I_S=b\xi -b-\eta \xi +{\displaystyle \frac{b\eta \left(I\right(t)+J(t)+S(t\left)\right)}{aS\left(t\right)}}.$$

Assuming $\eta =b$, we get

$$b(J_S+I_S)=-b+{\displaystyle \frac{b^2(I\left(t\right)+J\left(t\right)+S\left(t\right))}{aS\left(t\right)}},$$

or

$${(J+I+S)}_S=k(I\left(t\right)+J\left(t\right)+S\left(t\right))/S\left(t\right),k=b/a$$

and by separation of variables

$${\displaystyle \frac{d(J+I+S)}{(I+J+S)}}={\displaystyle \frac{kdS}{S\left(t\right)}}.$$

After integration

$$I\left(t\right)+J\left(t\right)+S\left(t\right)=C_1{S}^k\left(t\right),C_1=const.$$

From the initial conditions

$$R_0=R\left(t_0\right)=0,U_0=U\left(t_0\right)=0,I_0=I\left(t_0\right),J_0=J\left(t_0\right),S_0=S\left(t_0\right),$$

(26)

in view of (25), we get $I_0+J_0+S_0=1$.

So, from the initial condition at $t=t_0$, we have

$$I_0+J_0+S_0=C_1{S}_0^k,C_1={\displaystyle \frac{I_0+J_0+S_0}{S_0^k}}={\displaystyle \frac{1}{S_0^k}}.$$

$$I\left(t\right)+J\left(t\right)+S\left(t\right)=S^k\left(t\right)S_0^{-k},I\left(t\right)+J\left(t\right)=S^k\left(t\right)S_0^{-k}-S\left(t\right).$$

(27)

Further from (24), we have

$$-{\displaystyle \frac{S^{\prime }\left(t\right)}{aS\left(t\right)}}={\displaystyle \frac{I+J}{I+J+S}}=1-{\displaystyle \frac{S}{I+J+S}}=1-S_0^k{S}^{1-k},$$

or by separation of variables

$${\displaystyle \frac{dS}{S(1-S_0^k{S}^{1-k})}}=-adt.$$

By integration

$${\int }_{S_0}^S{\displaystyle \frac{dS}{S(1-S_0^k{S}^{1-k})}}=-a(t-t_0)$$

$$ln\left|{\displaystyle \frac{S^{k-1}-S_0^k}{S_0^{k-1}(1-S_0)}}\right|=a(1-k)(t-t_0)$$

$$S^{k-1}=S_0^{k-1}\left(S_0+(1-S_0)e^{a(1-k)(t-t_0)}\right).$$

Denoting

$$A\left(t\right):=S_0+(1-S_0)exp\left[a(1-k)(t-t_0)\right],$$

(28)

we get

$$S\left(t\right)=S_0{A}^{{\displaystyle \frac{1}{k-1}}}\left(t\right),exp\left[a(1-k)(t-t_0)\right]={\displaystyle \frac{A\left(t\right)-S_0}{1-S_0}}$$

(29)

and from (27)

$$I\left(t\right)+J\left(t\right)=(A\left(t\right)-S_0)A^{{\displaystyle \frac{1}{k-1}}}\left(t\right)$$

or

$$I\left(t\right)+J\left(t\right)=(1-S_0)e^{a(1-k)(t-t_0)}{\left(S_0+(1-S_0)exp\left[a(1-k)(t-t_0)\right]\right)}^{{\displaystyle \frac{1}{k-1}}}.$$

(30)

To find the critical points $S=S_c$ of $I\left(t\right)+J\left(t\right)$, we solve

$$I^{\prime }\left(t\right)+J^{\prime }\left(t\right)=\left(k{S}^{k-1}{S}_0^{-k}-1\right)S^{\prime }\left(t\right)=0,S_c={\left({\displaystyle \frac{S_0^k}{k}}\right)}^{{\displaystyle \frac{1}{k-1}}}.$$

So, the maximum value of $I\left(t\right)+J\left(t\right)$ is

$${(I\left(t\right)+J\left(t\right))}_{max}=I\left(t\right)+J\left(t\right){|}_{S=S_c}={\displaystyle \frac{(1-k)k^{\frac{k}{1-k}}}{S_0^{\frac{k}{1-k}}}}=(1-k){\left(k/S_0\right)}^{{\displaystyle \frac{k}{1-k}}},$$

or

$${(i_B\left(t\right)+j_B\left(t\right))}_{max}=N{(I\left(t\right)+J\left(t\right))}_{max}=N(1-k){\left(k/S_0\right)}^{k/(1-k)}.$$

(31)

where in $i_B\left(t\right)=NI\left(t\right)$ a subindex B means that the formula is obtained by using the Bailey–Gaeta model.

From

$$S=S_0{A}^{1/(k-1)}=S_c={\left(S_0^k/k\right)}^{1/(k-1)},A=S_0/k,$$

or

$$S_0+(1-S_0)exp\left[a(1-k)(t-t_0)\right]=S_0/k$$

one can find the time $t=t_{Bmax}$ when the maximum is achieved:

$$t_{Bmax}=t_0+{\displaystyle \frac{1}{a(1-k)}}ln\left|{\displaystyle \frac{(1-k)S_0}{k(1-S_0)}}\right|.$$

(32)

Introducing a rate of new cases of infectives

$$L\left(t\right):=I^{\prime }\left(t\right)+J^{\prime }\left(t\right)=\left(k{S}^{k-1}{S}_0^{-k}-1\right)S^{\prime }\left(t\right)=\left(kA{S}_0^{-1}-1\right)S^{\prime }\left(t\right),$$

in view of

$$S^{\prime }\left(t\right)=S_0{A}^{(2-k)/(k-1)}{A}^{\prime }\left(t\right)/(k-1),$$

$$A^{\prime }\left(t\right)=(1-S_0)a(1-k)e^{a(1-k)(t-t_0)}=a(1-k))(A-S_0)$$

we get

$$L:=I^{\prime }\left(t\right)+J^{\prime }\left(t\right)=a(S_0-kA)(A-S_0)A^{(2-k)/(k-1)},$$

(33)

and

$$L^{\prime }\left(t\right)={\displaystyle \frac{a(A^2{k}^2-A(1+k)S_0+(2-k)S_0^2)}{1-k}}{A}^{(2k-3)/(1-k)}{A}^{\prime }\left(t\right).$$

To find the peak time, we solve $L^{\prime }\left[A\right]=0$ for $A=A_c$ or

$$a(A^2{k}^2-A(1+k)S_0+(2-k)S_0^2)=0,A={\displaystyle \frac{1+k\pm (1-k)m}{2k^2}}{S}_0,m=\sqrt{1+4k}.$$

So, excluding $A_c$ from

$$A_c=S_0+(1-S_0)exp\left[a(1-k)(t_c-t_0)\right],A_c={\displaystyle \frac{1+k\pm (1-k)m}{2k^2}}{S}_0$$

we get $t_c={\tau }_{Bmax}$:

$${\tau }_{Bmax}={\tau }_0+{\displaystyle \frac{1}{a(1-k)}}ln\left|{\displaystyle \frac{(1-k)(1+2k-m)S_0}{2k^2(1-S_0)}}\right|,m=\sqrt{1+4k},$$

(34)

and a maximum value (for other values, we have nonsense: $L<0$)

$$L_{Bmax}\left(t\right)={\displaystyle \frac{a{S}_0^2{(1-k)}^2(m-1)(1+2k-m)}{2^{k/(k-1)}{k}^3}}{\left({\displaystyle \frac{1+k-(1-k)m}{k^2}}{S}_0\right)}^{{\displaystyle \frac{2-k}{k-1}}}.$$

(35)

Furthermore, from (24)

$${\displaystyle \frac{I^{\prime }\left(t\right)+bI\left(t\right)}{a\xi S\left(t\right)}}={\displaystyle \frac{I+J}{I+J+S}}=1-{\displaystyle \frac{S}{I+J+S}}=1-S_0^k{S}^{1-k}\left(t\right)$$

$$I^{\prime }\left(t\right)+bI\left(t\right)=a\xi S(1-S_0^k{S}^{1-k})$$

$${\left(I{e}^{b(t-t_0)}\right)}^{\prime }=a\xi S(1-S_0^k{S}^{1-k})e^{b(t-t_0)},$$

or in view of (29)

$${(Iexp\left[b(t-t_0)\right])}^{\prime }=a\xi S_0{A}^{1/(k-1)}(1-S_0{A}^{-1})exp\left[b(t-t_0)\right].$$

By integration, we get

$${\displaystyle \frac{I{e}^{b(t-t_0)}-I_0}{a\xi S_0}}={\int }_{t_0}^t{\displaystyle \frac{e^{b(t-t_0)}dt}{{\left(S_0+(1-S_0)e^{(a-b)(t-t_0)}\right)}^{\frac{1}{1-k}}}}-S_0{\int }_{t_0}^t{\displaystyle \frac{e^{b(t-t_0)}dt}{{\left(S_0+(1-S_0)e^{(a-b)(t-t_0)}\right)}^{\frac{2-k}{1-k}}}}.$$

Here, the integrals on the right side are in general the hypergeometric functions.

Theorem 2. 

Let $b=\eta ,0<k<1,0<\xi \le 1$. Then, the solutions of (24) are given by formulas

$$I_B\left(t\right)=\left(I_0+a\xi S_0{\int }_{t_0}^t{\displaystyle \frac{((S_0+(1-S_0)e^{(a-b)(t_1-t_0)})-S_0)e^{b(t_1-t_0)}d{t}_1}{{\left(S_0+(1-S_0)e^{(a-b)(t_1-t_0)}\right)}^{\frac{2-k}{1-k}}}}\right)e^{-b(t-t_0)},$$

$$S_B\left(t\right)=S_0{\left[S_0+(1-S_0)e^{a(1-k)(t-t_0)}\right]}^{{\displaystyle \frac{1}{k-1}}},$$

$$J_B\left(t\right)=(1-S_0)e^{a(1-k)(t-t_0)}{\left[S_0+(1-S_0)e^{a(1-k)(t-t_0)}\right]}^{{\displaystyle \frac{1}{k-1}}}-I_B\left(t\right),$$

$$U_B\left(t\right)=b{\int }_{t_0}^t{J}_B\left(t\right)d{t}_1,R_B\left(t\right)=1-I_B\left(t\right)-J_B\left(t\right)-S_B\left(t\right)-U_B\left(t\right),$$

where a subindex B means that the formula is obtained by using the Bailey–Gaeta model.

The maximum value of $i_B\left(t\right)+j_B\left(t\right)$

$${(i_B\left(t\right)+j_B\left(t\right))}_{max}=N(1-k){\left(k/S_0\right)}^{k/(1-k)}.$$

is achieved at time

$$t_{Bmax}=t_0+{\displaystyle \frac{1}{a(1-k)}}ln\left|{\displaystyle \frac{(1-k)S_0}{k(1-S_0)}}\right|.$$

The rate of new cases of infectives

$$L_B\left(t\right)=a(S_0-kA)(A-S_0)A^{(2-k)/(k-1)},A:=S_0+(1-S_0)exp\left\{a(1-k)(t-t_0)\right\}$$

has the maximum

$$L_{Bmax}\left(t\right)={\displaystyle \frac{a{S}_0^2{(1-k)}^2(m-1)(1+2k-m)}{2^{k/(k-1)}{k}^3}}{\left({\displaystyle \frac{1+k-(1-k)m}{k^2}}{S}_0\right)}^{{\displaystyle \frac{2-k}{k-1}}},m=\sqrt{1+4k},$$

that is achieved at time:

$${\tau }_{Bmax}={\tau }_0+{\displaystyle \frac{1}{a(1-k)}}ln\left|{\displaystyle \frac{(1-k)(1+2k-\sqrt{1+4k})S_0}{2k^2(1-S_0)}}\right|.$$

Remark 4. 

For the Bailey model, we have $\xi =1$, so $J\left(t\right)=U\left(t\right)\equiv 0$, and the solutions are simplified:

$$I\left(t\right)=(1-S_0)e^{a(1-k)(t-t_0)}{\left[S_0+(1-S_0)exp\left[a(1-k)(t-t_0)\right]\right]}^{{\displaystyle \frac{1}{k-1}}},$$

$$S\left(t\right)=S_0{\left[S_0+(1-S_0)exp\left\{a(1-k)(t-t_0)\right\}\right]}^{{\displaystyle \frac{1}{k-1}}},R\left(t\right)=1-I\left(t\right)-J\left(t\right)-S\left(t\right)-U\left(t\right).$$

Remark 5. 

In a simple case $k=b/a=0.5$ from Theorem 2, we have

$${\displaystyle \frac{I\left(t\right)e^{a(t-t_0)/2}-I_0}{a\xi S_0}}={\int }_{t_0}^t{\displaystyle \frac{e^{a(t-t_0)/2}dt}{{\left(S_0+(1-S_0)e^{a(t-t_0)/2}\right)}^2}}-S_0{\int }_{t_0}^t{\displaystyle \frac{e^{a(t-t_0)/2}dt}{{\left(S_0+(1-S_0)e^{a(t-t_0)/2}\right)}^3}}$$

$$={\displaystyle \frac{(1-S_0)(e^{a(t-t_0)/2}(2-S_0)-2e^{a(t-t_0)/2}(1-S_0)-S_0)}{a{((1-S_0)e^{a(t-t_0)/2}+S_0)}^2}}$$

$$I\left(t\right)=\left(I_0+{\displaystyle \frac{\xi S_0(1-S_0)(e^{a(t-t_0)/2}(2-S_0)-2e^{a(t-t_0)/2}(1-S_0)-S_0)}{{((1-S_0)e^{a(t-t_0)/2}+S_0)}^2}}\right)e^{a(t_0-t)/2}$$

or from $i_B=NI\left(t\right)$, we get

$$i_B\left(t\right)=Nexp[a(t_0-t)/2]$$

$$\times \left(I_0+{\displaystyle \frac{\xi S_0(1-S_0)(exp[a(t-t_0)/2](2-S_0)-2exp[a(t_0-t)/2](1-S_0)-S_0)}{{((1-S_0)exp[a(t_0-t)/2]+S_0)}^2}}\right).$$

(36)

Note that from (36) one can obtain the analytical formula for peak value and peak time of symptomatic infectives $i_B$, but it contains roots of cubic and is complicated.

Example 2. 

For Niue Island, by taking $k=a=0.5,N=1622,S_0=1-14/1622,{\tau }_0\approx t_0=27$, we get from (30)

$$i_B\left(t\right)+j_B\left(t\right)\approx {\displaystyle \frac{14e^{0.25(t-27)}}{(0.99+0.086e^{0.25(t-27)})}}.$$

(37)

From (36), by taking $\xi =0.45$, we get

$$i_B\left(t\right)\approx {\displaystyle \frac{1.29\ast {10}^9{e}^{-0.25t}(\mathrm{49,047.37}+e^{0.25t}+0.056e^{0.5t})}{{(\mathrm{98,094.75}+e^{0.25t})}^2}}.$$

(38)

Further, from (31) and (32)

$$i_{Bmax}+j_{Bmax}\approx 409,t_{Bmax}-t_0={\displaystyle \frac{1}{a(1-k)}}ln\left|{\displaystyle \frac{(1-k)S_0}{k(1-S_0)}}\right|\approx 46.$$

From (34)

$${\tau }_{Bmax}={\tau }_0+{\displaystyle \frac{1}{a(1-k)}}ln\left|{\displaystyle \frac{(1-k)(1+2k-m)S_0}{2k^2(1-S_0)}}\right|\approx 41.$$

Furthermore, from (35) and (38)

$$i_{Bmax}\approx i\left(t\right){|}_{t=46}\approx 184,l_{Bmax}\approx 39.36.$$

Statistical values are

$$i_{statmax}=185l_{statmax}=43,t_{statmax}=46{\tau }_{statmax}=37.$$

For illustration of Formulas (37) and (38), see the Figure 2 below.

4. Conclusions and Discussion

In this paper, the integrable versions of the Gaeta (SIR-A) model and Bailey–Gaeta model are introduced. They describe the dynamics of the spread of infectious disease.

The explicit solutions of these models are derived. The simplicity of these formulas allows us also to derive and analyze the function that describes the rate of new cases of disease transmission. The formulas for critical metrics (maximum values of infectives and their peak times) for the spread of disease are obtained by using the explicit solutions. These formulas could not be derived by using numerical solutions. The analysis is performed by using classical tools of analysis: calculations of derivatives, finding the critical numbers, extrema and so on. The real dynamics of spreading disease depend on numerous factors. The models considered here cannot capture the full picture of the spread of disease since they are deterministic and they contain a small number of parameters. But by using important average metrics that could be found from statistical data (a basic reproduction number, local population, duration time of a disease), one can obtain theoretical behavior of disease spread and make some predictions. An example of Niue Island COVID-19 dynamics shows that the theoretical behavior of disease spread is surprisingly close to the actual COVID-19 statistics.