### Basic properties

### Theorem 1

(Positivity)

*Suppose that the related solution of the initial data* \((S(0), I(0), R(0))\in \mathbf{R}_{+}^{3}\) *is* \((S(t), I(t), R(t) )\). *Then for model* (1) *the positively invariant set is* \(\mathbf{R}^{3}_{+}\).

### Proof

By setting \(\Lambda =\mu N\) and \(\lambda _{1}=\frac{\beta I}{1+\alpha I}\), the first equation of system (1) implies that

$$\begin{aligned} \frac{dS}{dt}=\Lambda -\lambda _{1}S-\mu S. \end{aligned}$$

Suppose that the solution exists of system (1) for a certain interval \(J\in [0;+\infty [\), then the above equation can be solved, for all \(t\in J\), as

$$\begin{aligned} & \frac{dS}{dt}+(\lambda _{1}+\mu )S= \Lambda, \\ &\frac{d}{dt} \bigl(S(t)e^{\mu t+\int _{0}^{t}\lambda _{1}(s)\,ds} \bigr) \geq \Lambda e^{\mu t+\int _{0}^{t}\lambda _{1}(s)\,ds}, \end{aligned}$$

which implies that

$$\begin{aligned} & S(t)e^{\mu t+\int _{0}^{t}\lambda _{1}(s)\,ds}-S(0)\geq \int _{0}^{t} \Lambda e^{\mu t+\int _{0}^{w}\lambda _{1}(u)\,du}\,dw, \\ &S(t)e^{\mu t+\int _{0}^{t}\lambda _{1}(s)\,ds}\geq S(0)+ \int _{0}^{t} \Lambda e^{\mu w+\int _{0}^{w}\lambda _{1}(u)\,du}\,dw, \\ &S(t)\geq S(0)e^{-\mu t-\int _{0}^{t}\lambda _{1}(s)\,ds}+e^{-\mu t- \int _{0}^{t}\lambda _{1}(s)\,ds}\times \int _{0}^{t}\Lambda e^{\mu t+ \int _{0}^{w}\lambda _{1}(u)\,du}\,dw>0. \end{aligned}$$

Hence, \(\forall t\in J\), S(t) is positive. Now the second equation of the model (1) implicitly shows that

$$\begin{aligned} \frac{dI}{dt} =\frac{\beta SI}{1+\alpha I}-(\gamma +\mu )I\geq -( \gamma +\mu )I, \end{aligned}$$

or

$$\begin{aligned} \frac{dI}{dt}\geq -(\gamma +\mu )I, \end{aligned}$$

which can be written as

$$\begin{aligned} \frac{dI}{I}\geq -(\gamma +\mu )\,dt\quad (I\neq 0). \end{aligned}$$

By integrating, we get

$$\begin{aligned} &\ln {I}\geq -(\gamma +\mu )t+C, \\ &I\geq e^{-(\gamma +\mu )t+C}, \\ &I\geq C_{1}e^{-(\gamma +\mu )t}, \end{aligned}$$

at \(t=0\),

$$\begin{aligned} I\geq I(0)e^{-(\gamma +\mu )0}\geq 0. \end{aligned}$$

Therefor, for all values of *t*, \(I(t)\) is positive. Similarly, the last equation of system (1) implies that

$$\begin{aligned} \frac{dR}{dt} =\gamma I-\mu R\geq -\mu R, \end{aligned}$$

or

$$\begin{aligned} \frac{dR}{dt}\geq -\mu R, \end{aligned}$$

which can be written as

$$\begin{aligned} \frac{dR}{R}\geq -\mu dt \quad(R\neq 0). \end{aligned}$$

By integrating, we get

$$\begin{aligned} &\ln {R}\geq -\mu t+K, \\ &R\geq e^{-\mu t+K}, \\ &R\geq K_{1}e^{-\mu t}, \end{aligned}$$

at \(t=0\),

$$\begin{aligned} R\geq R(0)e^{-\mu 0}\geq 0. \end{aligned}$$

Hence, \(R(t)\) also is positive in the given interval. □

### Existence and uniqueness of the solution

The first-order ODE in general form is

$$\begin{aligned} \acute{z}=g(t,z),\qquad z(t_{0})=z_{0}. \end{aligned}$$

(5)

With the help of the theorem below, we can establish the existence and uniqueness of the solution for the considered model.

### Theorem 2

(Uniqueness of solution)

*Let us use D to denote the domain*

$$\begin{aligned} \vert t-t_{0} \vert \leq a,\qquad \vert z-z_{0} \vert \leq b,\quad z=(z_{1},z_{2}, \ldots,z_{n}), z_{0}=(z_{10},z_{20}, \ldots,z_{n0}), \end{aligned}$$

(6)

*and assume that the Lipschitz condition is satisfied by* \(h(t,z)\):

$$\begin{aligned} \bigl\Vert h (t, z_{1}) - h(t, z_{2}) \bigr\Vert \leq c \Vert z_{1}- z_{2} \Vert , \end{aligned}$$

(7)

*and the two pairs* \((t, z_{1})\) *and* \((t, z_{2})\) *are in* *D*, *where* *c* *is a positive constant*. *Then there exists a constant* \(\delta > 0\) *such that for the interval* \(|t-t_{0}|\leq \delta \) *there exists a unique continuous vector solution* \(z(t)\) *of the system* (5). *It should be noticed that condition* (6) *is satisfied with*

$$\begin{aligned} \frac{\partial h_{i}}{\partial z_{j}},\quad i,j=1,2,\ldots,n, \end{aligned}$$

(8)

*in the domain D*, *being continuous and bounded*.

### Lemma 1

*If the continuous partial derivative of* \(h (t, z)\) (*i*.*e*., \(\frac{\partial h_{i}}{\partial z_{j}}\)) *exists for a bounded closed convex domain* ℜ, *then*, *for* ℜ, *it satisfies a Lipschitz condition*. *We are interested in the domain*

$$\begin{aligned} 1\leq \epsilon \leq \Re. \end{aligned}$$

(9)

*Hence*, *a solution in the form of condition* (10) *is searched*:

$$\begin{aligned} 0< \Re < \infty. \end{aligned}$$

(10)

Now the existence theorem can be proved as follows.

### Theorem 3

*Assume D represents the domain of* (6) *in such a manner that* (7) *and* (8) *hold*. *Then the bounded solution in domain D of* (1) *exists*.

### Proof

Let

$$\begin{aligned} & h_{1}=\mu N-\frac{\beta SI}{1+\alpha I}-\mu S, \end{aligned}$$

(11)

$$\begin{aligned} & h_{2}=\frac{\beta SI}{1+\alpha I}-(\gamma +\mu )I, \end{aligned}$$

(12)

$$\begin{aligned} &h_{3}=\gamma I-\mu R. \end{aligned}$$

(13)

To show that \(\frac{\partial h_{i}}{\partial z_{j}}\) \(j = 1, 2, 3\) are continuous and bounded, the following partial derivatives for the proposed model are performed. By taking the partial derivative of Eq. (11) we have

$$\begin{aligned} & \frac{\partial h_{1}}{\partial S}=- \frac{\beta I}{1+\alpha I}-\mu, \\ &\biggl\vert \frac{\partial h_{1}}{\partial S} \biggr\vert = \biggl\vert - \frac{\beta I}{1+\alpha I}-\mu \biggr\vert < \infty, \\ & \frac{\partial h_{1}}{\partial I}=- \frac{\beta S}{(1+\alpha I)^{2}} \\ &\biggl\vert \frac{\partial h_{1}}{\partial I} \biggr\vert = \biggl\vert - \frac{\beta S}{(1+\alpha I)^{2}} \biggr\vert < \infty, \end{aligned}$$

and

$$\begin{aligned} &\frac{\partial h_{1}}{\partial R}=0, \\ &\biggl\vert \frac{\partial h_{1}}{\partial R} \biggr\vert = \vert 0 \vert < \infty. \end{aligned}$$

For class *I*, from Eq. (12)

$$\begin{aligned} & \frac{\partial h_{2}}{\partial S}=\frac{\beta I}{1+\alpha I} , \\ &\biggl\vert \frac{\partial h_{2}}{\partial S} \biggr\vert =\biggl\vert \frac{\beta I}{1+\alpha I} \biggr\vert < \infty, \\ & \frac{\partial h_{2}}{\partial I}=\frac{\beta S}{(1+\alpha I)^{2}}-( \gamma +\mu ), \\ &\biggl\vert \frac{\partial h_{2}}{\partial I} \biggr\vert = \biggl\vert \frac{\beta S}{(1+\alpha I)^{2}}-(\gamma +\mu ) \biggr\vert < \infty, \end{aligned}$$

and

$$\begin{aligned} & \frac{\partial h_{2}}{\partial R}=0, \\ &\biggl\vert \frac{\partial h_{2}}{\partial R} \biggr\vert = \vert 0 \vert < \infty. \end{aligned}$$

Similarly, for class *R*, from Eq. (13)

$$\begin{aligned} & \frac{\partial h_{3}}{\partial S}=0, \\ &\biggl\vert \frac{\partial h_{3}}{\partial S} \biggr\vert = \vert 0 \vert < \infty, \\ & \frac{\partial h_{3}}{\partial I}=\gamma, \\ &\biggl\vert \frac{\partial h_{3}}{\partial I} \biggr\vert = \vert \gamma \vert < \infty, \end{aligned}$$

and

$$\begin{aligned} & \frac{\partial h_{3}}{\partial R}=-\mu, \\ &\biggl\vert \frac{\partial h_{3}}{\partial R} \biggr\vert =\vert -\mu \vert < \infty, \end{aligned}$$

therefore, it can be concluded that all partial derivatives are bounded in the considered domain and are continuous. Hence, from Theorem 2, it is proved that there exists a unique solution of system (1) in D. □

### Equilibria and the reproduction number

For finding the equilibrium points of model (1), we take the algebraic system

$$\begin{aligned} \begin{aligned} & \mu N-\frac{\beta SI}{1+\alpha I}-\mu S = 0, \\ &\frac{\beta SI}{1+\alpha I}-(\gamma +\mu )I = 0, \\ &\gamma I-\mu R = 0.\end{aligned} \end{aligned}$$

(14)

Two solutions of the system (1) are obtained by some algebraic manipulations, one is \(D_{0} = (N, 0, 0)\), a disease free equilibrium (DFE) point and the second will be discussed after computing the reproduction number of the model (1). The reproductive number is computed with the help of the next generation matrix approach presented by van den Driessche and Watmough [26].

Let \(x = (I, S)\) and rewrite the model (1) for the susceptible and infected classes in the general form

$$\begin{aligned} \frac{dx}{dt} = \mathcal{F}(x) - \mathcal{V}(x), \end{aligned}$$

(15)

where

$$\begin{aligned} \mathcal{F}(x) = \begin{pmatrix} \frac{\beta SI}{1+\alpha I} \\ 0 \end{pmatrix}, \qquad\mathcal{V}(x) = \begin{pmatrix} (\gamma +\mu )I \\ -\mu N+\frac{\beta SI}{1+\alpha I}+\mu S \end{pmatrix}. \end{aligned}$$

(16)

Now the Jacobian of \(\mathcal{F}(x)\) and \(\mathcal{V}(x)\) of the disease free equilibrium point is

$$\begin{aligned} F = \begin{pmatrix} \beta N &0 \\ 0 &0 \end{pmatrix},\qquad V = \begin{pmatrix} \gamma +\mu & 0 \\ \beta N&\mu \end{pmatrix}, \end{aligned}$$

(17)

and further by using the idea of van den Driessche and Watmough [26], the reproduction number of the model (1) is follows:

$$\begin{aligned} \mathcal{R}_{0} = \rho \bigl(FV^{-1}\bigr) = \frac{\beta N}{\mu +\gamma }. \end{aligned}$$

(18)

### Theorem 4

*For system* (1), *there exists a unique positive endemic equilibrium point* \(D_{*}\), *if* \(\mathcal{R}_{0}>1\).

### Proof

By some algebraic manipulations, the second solution of the system (14) yields

$$\begin{aligned} &S_{*}=\frac{\mu \alpha N+\mu +\gamma }{\beta +\alpha \mu }, \\ &I_{*}=\frac{\mu (\mathcal{R}_{0} -1)}{\beta +\alpha \mu }, \\ &R_{*}=\frac{\gamma (\mathcal{R}_{0} -1)}{\beta +\alpha \mu }. \end{aligned}$$

It is clear from the values of \(I_{*}\) and \(R_{*}\) that is there exists a unique positive endemic equilibrium point \(D_{*}\), if \(\mathcal{R}_{0}>1\). □

### Stability analysis of the model

### Theorem 5

*The system* (1) *is locally stable related to the virus free equilibrium point* \(E_{0}\), *if* \(\mathcal{R}_{0}<1\) *and unstable if* \(\mathcal{R}_{0}>1\).

### Proof

For local stability the Jacobian of system (1) is

$$\begin{aligned} J= \begin{pmatrix} -\mu - \frac{\beta I}{1+\alpha I}& -\frac{\beta S}{(1+\alpha I)^{2}} & 0 \\ \frac{\beta I}{1+\alpha I} & \frac{\beta S}{(1+\alpha I)^{2}}-(\mu + \gamma )& 0 \\ 0 & \gamma & -\mu \end{pmatrix}. \end{aligned}$$

(19)

At \(E_{0}\), the Jacobian becomes

$$\begin{aligned} J(E_{0})= \begin{pmatrix} -\mu & -\beta N & 0 \\ 0 & \beta N -(\mu +\gamma )& 0 \\ 0 & \gamma & -\mu \end{pmatrix}, \end{aligned}$$

(20)

from which it follows that the eigenvalues are \(\lambda _{1}= -\mu <0, \lambda _{3}= -\mu <0\) and \(\lambda _{2}=\beta N -(\mu +\gamma )\), implying that \(\lambda _{2}<0\), if \(\mathcal{R}_{0}<1\). So the system (1) is locally stable for \(\mathcal{R}_{0}<1\) and unstable for \(\mathcal{R}_{0}>1\). The proof is complete. □

### Theorem 6

*If* \(\mathcal{R}_{0}<1\), *then the DFE point of the system* (1) *is globally stable*.

### Proof

For the proof of this theorem, first we construct the Lyapunov function *L*:

$$\begin{aligned} L(I)=\ln {\frac{I}{I_{0}}}. \end{aligned}$$

(21)

Differentiating Eq. (21) with respect to time, we have

$$\begin{aligned} \begin{aligned} \frac{d}{dt}\bigl(L(I)\bigr)={}&\frac{1}{I}\frac{dI}{dt}, \\ \frac{d}{dt}\bigl(L(I)\bigr)={}&\frac{1}{I} \biggl( \frac{\beta SI}{1+\alpha I}-(\gamma +\mu )I \biggr) \\ ={}&\frac{\beta S}{1+\alpha I}-(\gamma +\mu ) \\ ={}&(\gamma +\mu ) \biggl(\frac{\beta S}{(\gamma +\mu )(1+\alpha I)}-1 \biggr) \\ \leq {}&(\gamma +\mu ) \biggl(\frac{\beta S}{\gamma +\mu }-1 \biggr) \\ ={}&(\gamma +\mu ) (\mathcal{R}_{0}-1 ) \\ \leq{} &0 \quad\text{for } \mathcal{R}_{0}< 1. \end{aligned} \end{aligned}$$

(22)

Therefore, if \(\mathcal{R}_{0}<1\), then \(\frac{d}{dt}(L(I))<0\), which implies that, for \(\mathcal{R}_{0}<1\), the DFE point of the system (1) is globally stable. □

### Theorem 7

*For* \(\mathcal{R}_{0}>1\), *the system* (1) *at the positive endemic equilibrium point* \(E_{*}\) *is locally stable*.

### Proof

The Jacobian matrix of system (1) is

$$\begin{aligned} J= \begin{pmatrix} -\mu - \frac{\beta I}{1+\alpha I}& -\frac{\beta S}{(1+\alpha I)^{2}} & 0 \\ \frac{\beta I}{1+\alpha I} & \frac{\beta S}{(1+\alpha I)^{2}}-(\mu + \gamma )& 0 \\ 0 & \gamma & -\mu \end{pmatrix}. \end{aligned}$$

(23)

At \(E_{*}\), the Jacobian becomes

$$\begin{aligned} J(E_{*})= \begin{pmatrix} -\mu - \frac{\beta I_{*}}{1+\alpha I_{*}}& - \frac{\beta S_{*}}{(1+\alpha I_{*})^{2}} & 0 \\ \frac{\beta I_{*}}{1+\alpha I_{*}} & \frac{\beta S_{*}}{(1+\alpha I_{*})^{2}}-(\mu +\gamma )& 0 \\ 0 & \gamma & -\mu \end{pmatrix}, \end{aligned}$$

(24)

which yields one eigenvalue \(\lambda =-\mu \) and the characteristic equation

$$\begin{aligned} &\lambda ^{2}+ \biggl(\mu +\frac{\beta I_{*}}{1+\alpha I_{*}}- \frac{\beta S_{*}}{(1+\alpha I_{*})^{2}}+(\mu +\gamma ) \biggr)\lambda \\ &\quad {}+ \biggl(\mu +\frac{\beta I_{*}}{1+\alpha I_{*}} \biggr) \biggl( \frac{\beta S_{*}}{(1+\alpha I_{*})^{2}}+(\mu +\gamma ) \biggr)+ \biggl( \frac{\beta S_{*}}{(1+\alpha I_{*})^{2}} \biggr) \biggl( \frac{\beta I_{*}}{1+\alpha I_{*}} \biggr)=0. \end{aligned}$$

(25)

It is clear, for \(\mathcal{R}_{0}>1\), that

$$\begin{aligned} &\biggl(\mu +\frac{\beta I_{*}}{1+\alpha I_{*}}- \frac{\beta S_{*}}{(1+\alpha I_{*})^{2}}+(\mu +\gamma ) \biggr) \\ &\quad= \biggl(\mu +\frac{\beta I_{*}}{1+\alpha I_{*}}- \frac{\gamma +\mu }{(1+\alpha I_{*})}+(\mu +\gamma ) \biggr)>0 \end{aligned}$$

and

$$\begin{aligned} \biggl(\mu +\frac{\beta I_{*}}{1+\alpha I_{*}} \biggr) \biggl( \frac{\beta S_{*}}{(1+\alpha I_{*})^{2}}+(\mu +\gamma ) \biggr)+ \biggl( \frac{\beta S_{*}}{(1+\alpha I_{*})^{2}} \biggr) \biggl( \frac{\beta I_{*}}{1+\alpha I_{*}} \biggr)>0. \end{aligned}$$

Hence, the system (1) is locally stable at \(E_{*}\) for \(\mathcal{R}_{0}>1\). The proof is complete. □

### Remark 1

Although the stability analysis of \(E_{*}\) is an interesting and separate mathematical problem, while for prevention of the disease one is to find an effective strategy, the main focus of this work is on the specific condition \(\mathcal{R}_{0} < 1\).

###
\(\mathcal{R}_{0}\) sensitivity analysis

From Theorem 6 it follows that we can control the parameters such that \(\mathcal{R}_{0} < 1\). This leads to the best strategy to prevent and restrain the disease. In detail, when \(\mathcal{R}_{0} < 1\), then

$$\begin{aligned} \lim_{t \to \infty } S(t) = N, \qquad \lim_{t \to \infty } I(t) = \lim _{t \to \infty } R(t) = 0, \end{aligned}$$

which shows that the spreading speed of the coronavirus can be reduced and prevented in the future. Hence, a sensitivity analysis of \(\mathcal{R}_{0}\) is carried out to select the influential parameters to control the rapidly spreading current pandemic.

It is easy to verify that

$$\begin{aligned} \begin{aligned} &\frac{\partial \mathcal{R}_{0}}{\partial \beta } = \frac{N}{\mu +\gamma } > 0, \\ &\frac{\partial \mathcal{R}_{0}}{\partial \mu } = - \frac{\beta N}{(\mu +\gamma )^{2}}< 0, \\ &\frac{\partial \mathcal{R}_{0}}{\partial \gamma } = - \frac{\beta N}{(\mu +\gamma )^{2}} < 0. \end{aligned} \end{aligned}$$

(26)

Equation (26) can be used to obtain different parameters in such a way that \(\mathcal{R}_{0}\) remains less than one. Hence, necessary actions can be taken on the basis of Eq. (26) to reduce the speed of the coronavirus.