Model framework
In this section, we describe an epidemic transmission SEIR model with demographic changes. The model is used in epidemiology to compute the amount of susceptible, exposed, visibly infected, recovered people in a population (N). Since the asymptomatic and presymptomatic people can transmit the virus but their symptoms are not visible, they are grouped into the E compartment and infection from E is referred to as latent infection. This model is used under the following assumptions:

The population is constant but large.

The only way a person can leave the susceptible state (S) is to become infected either from the exposed (E) or from visibly infected (I) state or die of natural causes.

The only way a person can leave the E state is to show signs and symptoms of the illness or die of natural death.

The only way a person can leave the I state is to recover from the disease or die from natural death or die as a result of the disease.

A person who recovered (R) from the illness received permanent immunity.

Age, sex, social status, and race do not affect the probability of being infected.

The member of the population has the same contacts with one another equally.

All births are into the susceptible state, and it is assumed that the birth and natural death rates are equal.
The transmission is measured at \(S\beta \left(I+\kappa E\right)/N\), where \(\beta\) is the direct transmission rate, and \(\kappa\) is the proportional rate constant when an uninfected individual comes into contact with an individual from state E. We assume natural birth and death rate to be measured at an equal rate \(\mu\) and induced death rate measured at \(\delta\). The rate for an individual to move from state E to state I is measured at rate \(\sigma ,\), and the rate of recovery is measured at \(\gamma\). Figure 1 represents the latent infection SEIR model, which is described using the system of nonlinear ordinary differential equations
$$\begin{aligned}&\frac{\mathrm{d}S\left(t\right)}{\mathrm{d}t}=\mu N\beta \frac{S\left(I+\kappa E\right)}{N}\mu S,\\ &\frac{\mathrm{d}E\left(t\right)}{\mathrm{d}t}=\beta \frac{S\left(I+\kappa E\right)}{N}\left(\mu +\sigma \right)E, \\ &\frac{\mathrm{d}I\left(t\right)}{\mathrm{d}t}=\sigma E\left(\mu +\gamma +\delta \right)I,\\ &\frac{\mathrm{d}R(t)}{\mathrm{d}t}=\gamma I\mu R,\end{aligned}$$
(1)
where \(S\left(t\right)=S,E\left(t\right)=E,I\left(t\right)=I\) and \(R\left(t\right)=R\) denote the number of susceptible, exposed, infectious, and remove individuals at time \(t\), respectively, and \(N=S+E+I+R\). System (1) is subjected to the initial condition
$$S\left(0\right)\ge 0, E\left(0\right)\ge 0, I\left(0\right)\ge 0,\mathrm{and} R\left(0\right)\ge 0$$
(2)
For simplicity system (1) is reduced to a proportional framework given as
$$\begin{aligned}&\frac{\mathrm{d}s\left(t\right)}{\mathrm{d}t}=\mu \beta s\left(i+\kappa e\right)\mu s,\\ &\frac{\mathrm{d}e\left(t\right)}{\mathrm{d}t}=\beta s\left(i+\kappa e\right)\left(\mu +\sigma \right)e,\\ &\frac{\mathrm{d}i\left(t\right)}{\mathrm{d}t}=\sigma e\left(\mu +\gamma +\delta \right)i,\\ &\frac{\mathrm{d}r(t)}{\mathrm{d}t}=\gamma i\mu r, \end{aligned}$$
(3)
where \(s=S/N,e=E/N,i=I/N\), and \(r=r/N\). By considering the total population
$$s+e+i+r=1 \Rightarrow r=1sei,$$
therefore, system (3) can be reduced to
$$\begin{aligned}&\frac{\mathrm{d}s(t)}{\mathrm{d}t}=\mu \beta s\left(i+\kappa e\right)\mu s,\\ &\frac{\mathrm{d}e(t)}{\mathrm{d}t}=\beta s\left(i+\kappa e\right)\left(\mu +\sigma \right)e,\\ &\frac{\mathrm{d}i(t)}{\mathrm{d}t}=\sigma e\left(\mu +\gamma +\delta \right)i. \end{aligned}$$
(4)
Positivity of the solution
Assume that system (4) has a global solution corresponding to nonnegative initial conditions. Then, the following Lemma confirms that the solution is nonnegative at all times.
Lemma 1
If s(0) ≥ 0, e(0) ≥ 0 and i(0) ≥ 0 then the solution s(t), e(t) and i(t) are all positive for all \(t\ge 0\).
Proof.
We use the contradiction: we assuming there exists positive real \({t}_{1},{t}_{2}\) and \({t}_{3}\) for which one of the conditions hold:

1.
\(s\left({t}_{1}\right)=0, \mathrm{d}s\left({t}_{1}\right)/\mathrm{d}t<0\), and for all \(0\le t\le {t}_{1}\) one has that \(e\left(t\right)\ge 0\) and \(i\left(t\right)\ge 0\);

2.
\(e\left({t}_{2}\right)=0, \mathrm{d}e\left({t}_{2}\right)/\mathrm{d}t<0\), and for all \(0\le t\le {t}_{2}\) one has that \(s\left(t\right)\ge 0\) and \(i\left(t\right)\ge 0\);

3.
\(i\left({t}_{3}\right)=0, \mathrm{d}i\left({t}_{3}\right)/\mathrm{d}t<0\), and for all \(0\le t\le {t}_{3}\) one has that \(s\left(t\right)\ge 0\) and \(e\left(t\right)\ge 0\).
Condition (I) contradicts if \(s\left(t\right)\ge 0\), \(\mathrm{d}s\left({t}_{1}\right)/\mathrm{d}t=\mu >0\). Also, condition (II) contradicts because \(e\left(t\right)\ge 0\), \(\mathrm{d}e\left({t}_{2}\right)/\mathrm{d}t=\beta si\ge 0\). Finally, condition (III) contradicts since for \(i\left(t\right)\ge 0\), \(\mathrm{d}i\left({t}_{3}\right)/\mathrm{d}t=\sigma E\ge 0.\). Thus, the solutions of \(s(t), e(t) and i(t)\) remain positive for all \(t>0\).
Hence, the positively invariant for the system (4) is
$$\varOmega =\left\{s\left(t\right), e\left(t\right), i\left(t\right)\epsilon {R}_{+}^{3},s\left(t\right)+e\left(t\right)+ i\left(t\right)\le 1\right\}.$$
(5)
\(\square\)
The equilibrium points and reproduction number calculations of the model
There are two equilibrium points for the system (4), i.e. the diseasefree equilibrium (DFE), the state when the disease is absent, and the endemic equilibrium (EE), which is the state when the disease continues to persist in the population.
Let the DFE points of the model are denoted as \({E}^{0}=\left({s}^{0},{e}^{0},{i}^{0}\right)\) and represent a system (4) at \({E}^{0}\) as
$$\begin{aligned} &\mu \beta {s}^{0}\left({i}^{0}+\kappa {e}^{0}\right)\mu {s}^{0}=0,\\ & \beta {s}^{0}\left({i}^{0}+\kappa {e}^{0}\right)\left(\mu +\sigma \right){e}^{0}=0,\\ &\sigma {e}^{0}\left(\mu +\gamma +\delta \right){i}^{0}=0. \end{aligned}$$
(6)
In terms of \({i}^{0}\), from the last equation of (6), we get
$${e}^{0}=\frac{\left(\mu +\gamma +\delta \right){i}^{0}}{\sigma }.$$
Adding the first two equations of (6) and substituting for \({e}^{0}\), we get
$${s}^{0}=1\frac{\left(\mu +\sigma \right)\left(\mu +\gamma +\delta \right){i}^{0}}{\mu \sigma }.$$
Because at the diseasefree state no one has the infection then, \({i}^{0}={e}^{0}=0\). We can see that \({E}^{0}=\left({s}^{0},{e}^{0},{i}^{0}\right)=\left(\mathrm{1,0},0\right).\)
Also, the EE points are denoted as\({E}^{*}=\left({s}^{*},{e}^{*},{i}^{*}\right)\), where \({s}^{*},{e}^{*}\) and \({i}^{*}\) are calculated by letting \({s}^{0}={s}^{*} , {e}^{0}={e}^{*} , {i}^{0}={i}^{*}\), and then, the second equation of (6) becomes
$$\begin{aligned} &\beta \left(1+\frac{\kappa \left(\mu +\gamma +\delta \right)}{\sigma }\right)\beta \left(\frac{\left(\mu +\sigma \right)\left(\mu +\gamma +\delta \right)}{\mu \sigma }\right)\left(1+\frac{\kappa \left(\mu +\gamma +\delta \right)}{\sigma }\right){i}^{*}\\ & \quad \,\frac{\left(\mu +\sigma \right)\left(\mu +\gamma +\delta \right)}{\sigma }=0. \end{aligned}$$
(7)
Multiplying both sides of (7) by \(\sigma /\left(\mu +\sigma \right)\left(\mu +\gamma +\delta \right)\), we get
$$\frac{\sigma \beta }{\left(\mu +\sigma \right)\left(\mu +\gamma +\delta \right)}\left(1+\frac{\kappa \left(\mu +\gamma +\delta \right)}{\sigma }\right)\frac{\beta }{\mu }\left(1+\frac{\kappa \left(\mu +\gamma +\delta \right)}{\sigma }\right){i}^{*}1=0,$$
which implies that
$$\frac{\beta }{\mu }\left(1+\frac{\kappa \left(\mu +\gamma +\delta \right)}{\sigma }\right){i}^{*}=\frac{\sigma \beta }{\left(\mu +\sigma \right)\left(\mu +\gamma +\delta \right)}\left(1+\frac{\kappa \left(\mu +\gamma +\delta \right)}{\sigma }\right)1,$$
Hence
$${i}^{*}=\frac{\mu }{\beta }\left(\frac{\sigma }{\sigma +\kappa \left(\mu +\gamma +\delta \right)}\right)\left(\frac{\sigma \beta }{\left(\mu +\sigma \right)\left(\mu +\gamma +\delta \right)}\left(1+\frac{\kappa \left(\mu +\gamma +\delta \right)}{\sigma }\right)1\right),$$
also
$$\begin{aligned} &{e}^{*}=\left(\frac{\left(\mu +\gamma +\delta \right)}{\sigma }\right){i}^{*},\\ &{s}^{*}=1\frac{\left(\mu +\sigma \right)\left(\mu +\gamma +\delta \right){i}^{*}}{\mu \sigma }. \end{aligned}$$
To understand the stability of the model, we need an expression to estimate the basic reproduction number (\({R}_{0})\).
However, from (6) let\({s}^{0}=s , {e}^{0}=e , {i}^{0}=i\), in terms of \(i\) the second equation of (6) becomes
$$\beta i\left(1+\frac{\kappa \left(\mu +\gamma +\delta \right)}{\sigma }\right)\left(1\frac{\left(\mu +\sigma \right)\left(\mu +\gamma +\delta \right)i}{\mu \sigma }\right)\frac{\left(\mu +\sigma \right)\left(\mu +\gamma +\delta \right)i}{\sigma }=0.$$
(8)
By the factorising method, we get
$$i\left(\beta \left(1+\frac{\kappa \left(\mu +\gamma +\delta \right)}{\sigma }\right)\left(1\frac{\left(\mu +\sigma \right)\left(\mu +\gamma +\delta \right)i}{\mu \sigma }\right)\frac{\left(\mu +\sigma \right)\left(\mu +\gamma +\delta \right)}{\sigma }\right)=0,$$
(9)
either \(i=0\) or
$$\begin{aligned}&\beta \left(1+\frac{\kappa \left(\mu +\gamma +\delta \right)}{\sigma }\right)\beta \left(\frac{\left(\mu +\sigma \right)\left(\mu +\gamma +\delta \right)}{\mu \sigma }\right)\left(1+\frac{\kappa \left(\mu +\gamma +\delta \right)}{\sigma }\right)i\frac{\left(\mu +\sigma \right)\left(\mu +\gamma +\delta \right)}{\sigma }=0 , \\ &i=\left(\frac{{\mu \sigma }^{2}}{\left(\mu +\sigma \right)\left(\mu +\gamma +\delta \right)\left(\sigma +\kappa \left(\mu +\gamma +\delta \right)\right)}\right)\left[\left(1+\frac{\kappa \left(\mu +\gamma +\delta \right)}{\sigma }\right)\frac{\left(\mu +\sigma \right)\left(\mu +\gamma +\delta \right)}{\beta \sigma }\right], \end{aligned}$$
Equating the highest \(i\) value from (9) to zero, we get
$$\frac{\sigma \beta }{\left(\mu +\sigma \right)\left(\mu +\gamma +\delta \right)}\left(1+\frac{\kappa \left(\mu +\gamma +\delta \right)}{\sigma }\right)=1.$$
(10)
Since the threshold for \({R}_{0}\) is unity, we then assume
$${R}_{0}=\frac{\sigma \beta }{\left(\mu +\sigma \right)\left(\mu +\gamma +\delta \right)}\left(1+\frac{\kappa \left(\mu +\gamma +\delta \right)}{\sigma }\right).$$
(11)
Equation (11) is justified using the nextgeneration matrix method defined in [16] as \(K=\rho (F{\mathcal{V}}^{1})\), where \(\rho (F{\mathcal{V}}^{1})\) is the spectral radius of the matrix \(F{\mathcal{V}}^{1}\) and the largest eigenvalue of \(K\) is the \({R}_{0}\). \(F\) and \(\mathcal{V}\) are the matrices associated with the DFE points defined as
$$\begin{aligned}&F=\left(\begin{array}{cc}\kappa \beta & \beta \\ 0& 0\end{array}\right) \mathrm{and}\mathcal{V}=\left(\begin{array}{cc}\mu +\sigma & 0\\ \sigma & \mu +\delta +\gamma \end{array}\right),\\ & K=\left(\begin{array}{cc}\kappa \beta & \beta \\ 0& 0\end{array}\right)\left(\begin{array}{cc}\frac{1}{\mu +\sigma }& 0\\ \frac{\sigma }{\left(\mu +\sigma \right)\left(\mu +\gamma +\delta \right)}& \frac{1}{\left(\mu +\gamma +\delta \right)}\end{array}\right)\\ &K=\left(\begin{array}{cc}\frac{\beta }{\mu +\sigma }\left(\kappa +\frac{\sigma }{\left(\mu +\gamma +\delta \right)}\right)& \frac{\beta }{\left(\mu +\gamma +\delta \right)}\\ 0& 0\end{array}\right) \end{aligned}$$
The largest eigenvalues of \({\rm K}\) is as (11) given as
$${R}_{0}=\frac{\sigma \beta }{\left(\mu +\sigma \right)\left(\mu +\gamma +\delta \right)}\left(1+\frac{\kappa \left(\mu +\gamma +\delta \right)}{\sigma }\right).$$
Hence
$${s}^{*}=\frac{1}{{R}_{0}}, {e}^{*}=\frac{\mu \left({R}_{0}1\right)}{{R}_{0}\left(\mu +\sigma \right)}, {i}^{*}=\frac{\mu \sigma \left({R}_{0}1\right)}{{R}_{0}\left(\mu +\sigma \right)\left(\mu +\gamma +\delta \right)}.$$
Stability analysis of the diseasefree equilibrium points
Theorem 1
If \({R}_{0}<1\) and \(\kappa \beta <\left(\mu +\sigma \right)+\left(2\mu +\gamma +\delta +\sigma \right)\) , then the DFE is locally asymptotically stable in \(\varOmega\).
Proof
The Jacobian matrix of system (3) associated with DFE is given as
$${J}_{\left(\mathrm{1,0},0\right)}=\left(\begin{array}{ccc}\mu & \kappa \beta & \beta \\ 0& \kappa \beta \left(\mu +\sigma \right)& \beta \\ 0& \sigma & \left(\mu +\gamma +\delta \right)\end{array}\right),$$
(12)
with characteristic polynomial
$$P\left(\leftthreetimes \right)=\left(\mu +\leftthreetimes \right)\left[{\leftthreetimes }^{2}+\left(\left(\mu +\sigma \right)\kappa \beta +\left(\mu +\gamma +\delta \right)\right)\leftthreetimes +\left(\left(\mu +\sigma \right)\kappa \beta \right)\left(\mu +\gamma +\delta \right)\sigma \beta \right],$$
where \(\leftthreetimes\) is an eigenvalue. It is easy to see that for Theorem 1 to satisfy
$$\left(\mu +\sigma \right)\kappa \beta +\left(\mu +\gamma +\delta \right)>0\Rightarrow \kappa \beta <\left(\mu +\sigma \right)+\left(2\mu +\gamma +\delta +\sigma \right),$$
and
$$\left(\left(\mu +\sigma \right)\kappa \beta \right)\left(\mu +\gamma +\delta \right)\sigma \beta >0\Rightarrow {R}_{0}=\frac{\beta }{\left(\mu +\sigma \right)}\left(\frac{\sigma }{\left(\mu +\gamma +\delta \right)}+\kappa \right)<1.$$
The proof of Theorem 1 is complete.
Theorem 2
If \({R}_{0}\le 1\), the DFE is globally asymptotically stable in \(\varOmega\).
Proof
To prove the global asymptotic stability (GAS) of the DFE, we construct the following Lyapunov function \(V{:}\,\varOmega \to R\), where \(V\left(s,e,i\right)=i\left(t\right)\). Then, the time derivative of \(V\) is given as
$$\frac{\mathrm{d}V}{\mathrm{d}t}=\frac{\mathrm{d}{i}^{0}}{\mathrm{d}t}=\frac{\mathrm{d}{e}^{0}}{\mathrm{d}t},$$
since at the equilibrium points \(\mathrm{d}{i}^{0}/\mathrm{d}t=\mathrm{d}{e}^{0}/\mathrm{d}t=0\). Therefore
$$\frac{\mathrm{d}V}{\mathrm{d}t}=\beta {s}^{0}\left({i}^{0}+\kappa {e}^{0}\right)\left(\mu +\sigma \right){e}^{0}.$$
(13)
Substituting \({s}^{0}\) and \({e}^{0}\) into (13), we get
$$\begin{aligned} & \frac{\mathrm{d}V}{\mathrm{d}t}=\beta \left(1+\frac{\kappa \left(\mu +\gamma +\delta \right)}{\sigma }\right){i}^{0}\beta \left(\frac{\left(\mu +\sigma \right)\left(\mu +\gamma +\delta \right)}{\mu \sigma }\right)\left(1+\frac{\kappa \left(\mu +\gamma +\delta \right)}{\sigma }\right){\left({i}^{0}\right)}^{2}\\ &\quad \,\frac{\left(\mu +\sigma \right)\left(\mu +\gamma +\delta \right){i}^{0}}{\sigma } , \end{aligned}$$
(14)
By the factorisation method
$$\begin{aligned}&\frac{\mathrm{d}V}{\mathrm{d}t}=\frac{\left(\mu +\sigma \right)\left(\mu +\gamma +\delta \right){i}^{0}}{\sigma }\left[\frac{\sigma \beta }{\left(\mu +\sigma \right)\left(\mu +\gamma +\delta \right)}\left(1+\frac{\kappa \left(\mu +\gamma +\delta \right)}{\sigma }\right)\right.\\& \quad \left. \frac{\left(\mu +\sigma \right)\left(\mu +\gamma +\delta \right)}{\mu \sigma }\left(\frac{\sigma \beta }{\left(\mu +\sigma \right)\left(\mu +\gamma +\delta \right)}\left(1+\frac{\kappa \left(\mu +\gamma +\delta \right)}{\sigma }\right)\right){i}^{0}1\right].\end{aligned}$$
Substituting \({R}_{0}\), we get
$$\frac{\mathrm{d}V}{\mathrm{d}t}=\frac{\left(\mu +\sigma \right)\left(\mu +\gamma +\delta \right){i}^{0}}{\sigma }\left[{R}_{0}\frac{\left(\mu +\sigma \right)\left(\mu +\gamma +\delta \right){R}_{0}}{\mu \sigma }{i}^{0}1\right].$$
(15)
Thus, \(\mathrm{d}V/\mathrm{d}t\le 0\) for \({R}_{0}\le 1\). Furthermore, if \({R}_{0}<1\) then \(\mathrm{d}V/\mathrm{d}t=0\iff {i}^{0}=0\) and if \({R}_{0}=1\) then \(\mathrm{d}V/\mathrm{d}t=0\). Hence, by Lasalle invariance principle [17], the DFE point is GAS.
Stability analysis of the endemic equilibrium
Theorem 3
If \({R}_{0}>1\) , the endemic equilibrium is locally asymptotically stable.
Proof
To prove the LAS of the endemic equilibrium, we consider the Jacobian matrix associated with \({E}^{*}\), that is
$${J}_{{E}^{*}}=\left(\begin{array}{ccc}\mu \beta \left({i}^{*}+\kappa {e}^{*}\right)& \kappa \beta {s}^{*}& \beta {s}^{*}\\ \beta \left({i}^{*}+\kappa {e}^{*}\right)& \kappa \beta {s}^{*}\left(\mu +\sigma \right)& \beta {s}^{*}\\ 0& \sigma & \left(\mu +\gamma +\delta \right)\end{array}\right),$$
(16)
Substituting for \({s}^{*},{e}^{*}\) and \({i}^{*}\), we get
$${J}_{{E}^{*}}=\left(\begin{array}{ccc}\mu {R}_{0}& \frac{\kappa \beta }{{R}_{0}}& \frac{\beta }{{R}_{0}}\\ \mu \left({R}_{0}1\right)& \frac{\kappa \beta }{{R}_{0}}\left(\mu +\sigma \right)& \frac{\beta }{{R}_{0}}\\ 0& \sigma & \left(\mu +\gamma +\delta \right)\end{array}\right),$$
(17)
if \(\lambda\) is an eigenvalue, then
$${J}_{{E}^{*}}\lambda {\varvec{I}}=\left(\begin{array}{ccc}\mu {R}_{0}\lambda & \frac{\kappa \beta }{{R}_{0}}& \frac{\beta }{{R}_{0}}\\ \mu \left({R}_{0}1\right)& \frac{\kappa \beta }{{R}_{0}}\left(\mu +\sigma \right)\lambda & \frac{\beta }{{R}_{0}}\\ 0& \sigma & \left(\mu +\gamma +\delta \right)\lambda \end{array}\right),$$
where \({\varvec{I}}\) is a threedimensional identity matrix which by matrix simplification method we then get
$$P\left(\lambda \right)={\lambda }^{3}+a{\lambda }^{2}+b\lambda +c,$$
where
$$a=\mu {R}_{0}+\left(\mu +\gamma +\delta \right)+\left(\mu +\sigma \right)\left(1\frac{\kappa \beta }{{\left(\mu +\sigma \right)R}_{0}}\right).$$
From \({R}_{0}\), we get
$$\frac{\sigma \beta }{{\left(\mu +\gamma +\delta \right)\left(\mu +\sigma \right)R}_{0}}=1\frac{\kappa \beta }{\left(\mu +\sigma \right){R}_{0}},$$
(18)
hence
$$\begin{aligned}&a=\mu {R}_{0}+\left(\mu +\gamma +\delta \right)+\frac{\sigma \beta }{{\left(\mu +\gamma +\delta \right)R}_{0}}>0,\\ &b=\mu {R}_{0}\left(\mu +\gamma +\delta \right)+\mu {R}_{0}\left(\mu +\sigma \right)+\left(\mu +\gamma +\delta \right)\left(\mu +\sigma \right)\mu \kappa \beta \frac{\kappa \beta \left(\mu +\gamma +\delta \right)}{{R}_{0}}\frac{\kappa \beta }{{R}_{0}},\\ &b=\mu {R}_{0}\left(\mu +\gamma +\delta \right)+\mu {R}_{0}\left(\mu +\sigma \right)\left(1\frac{\kappa \beta }{\left(\mu +\sigma \right){R}_{0}}\right),\end{aligned}$$
using (18) we get
$$\begin{aligned}&b=\mu {R}_{0}\left(\mu +\gamma +\delta \right)+\frac{\mu \beta }{\left(\mu +\gamma +\delta \right)}>0.\\ &c=\frac{\mu \kappa \beta \left({R}_{0}1\right)}{{R}_{0}}\mu \kappa \beta \left(\mu +\gamma +\delta \right)+\mu \left(\mu +\sigma \right)\left(\mu +\gamma +\delta \right){R}_{0}\mu \kappa \beta ,\\ &c=\frac{\mu \kappa \beta \left({R}_{0}1\right)}{{R}_{0}}>0,\end{aligned}$$
and
$$abc=\frac{\mu \kappa \beta }{{R}_{0}}\left(\frac{\beta }{\left(\mu +\gamma +\delta \right)}+1\right)+\mu \beta \left(\frac{\mu {R}_{0}}{\left(\mu +\gamma +\delta \right)}+1\right)+\mu {R}_{0}\left(\mu +\gamma +\delta \right)\left(\mu {R}_{0}+\mu +\gamma +\delta \right).$$
Since \(a>0,b>0,c>0,\) and \(abc>0\), according to the Routh–Hurwitz criterion, the endemic equilibrium of system (4) is LAS.
Theorem 4
The endemic equilibrium point is globally asymptotically stable on \(\varOmega\).
Proof
We construct the following Lyapunov function \({V}_{1}: {\varOmega }_{+}\to R\), where \({\varOmega }_{+}=\left\{s\left(t\right),e\left(t\right),i\left(t\right)\in\varOmega /s\left(t\right)>0,e\left(t\right)>0,i\left(t\right)>0\right\}\) given by
$${V}_{1}\left(X,t\right)=\frac{1}{2}\left({X}_{1}^{2}+{X}_{2}^{2}+{X}_{3}^{2}\right),$$
(19)
\({X}_{1}=s{s}^{*} , {X}_{2}=e{e}^{*}\), and\({X}_{3}=i{i}^{*}\),\({L}_{1}\). We can see that \({V}_{1}\left(X,t\right)>0\) and \({V}_{1}\left(\mathrm{0,0},0\right)=\left(\mathrm{0,0},0\right)\) for all \(\left({X}_{1}, {X}_{2},{X}_{3}\right)\) in the region, that makes \({V}_{1}\) positive definite. We need to verify that \(\mathrm{d}{V}_{1}/\mathrm{d}t\le 0\) (negative definite). The time derivative of \({\mathrm{V}}_{1}\) is
$$\frac{\mathrm{d}{V}_{1}}{\mathrm{d}t}={X}_{1}\frac{\mathrm{d}{X}_{1}}{\mathrm{d}t}+{X}_{2}\frac{\mathrm{d}{X}_{2}}{\mathrm{d}t}+{X}_{3}\frac{\mathrm{d}{X}_{3}}{\mathrm{d}t},$$
where \(\mathrm{d}{X}_{1}/\mathrm{d}t=\mathrm{d}s/\mathrm{d}t, \mathrm{d}{X}_{2}/\mathrm{d}t=\mathrm{d}(e)/\mathrm{d}t\) and \(\mathrm{d}{X}_{3}/\mathrm{d}t=\mathrm{d}(i)/\mathrm{d}t.\) Hence,
$$\frac{\mathrm{d}{V}_{1}}{\mathrm{d}t}={X}_{1}\left(\mu \beta s\left(i+\kappa e\right)\mu s\right)+{X}_{2}\left(\beta s\left(i+\kappa e\right)\left(\mu +\sigma \right)e\right)+{X}_{3}\left(\sigma e\left(\mu +\gamma +\delta \right)i\right).$$
From
$$\begin{aligned}&\mu \beta {s}^{*}\left({i}^{*}+\kappa {e}^{*}\right)\mu {s}^{*}=0,\\ &\beta {s}^{*}\left({i}^{*}+\kappa {e}^{*}\right)\left(\mu +\sigma \right){e}^{*}=0,\\ & \sigma {e}^{*}\left(\mu +\gamma +\delta \right){i}^{*}=0, \\ &{s}^{*}=\frac{\mu }{\left\{\mu +\beta \left({i}^{*}+\kappa {e}^{*}\right)\right\}}, {e}^{*}=\frac{\beta {s}^{*}\left({i}^{*}+\kappa {e}^{*}\right)}{\left(\mu +\sigma \right)}, {i}^{*}=\frac{\sigma {e}^{*}}{\left(\mu +\gamma +\delta \right)},\end{aligned}$$
Therefore,
$${X}_{1}=s\frac{\mu }{\left\{\mu +\beta \left({i}^{*}+\kappa {e}^{*}\right)\right\}} , {X}_{2}=e\frac{\beta {s}^{*}\left({i}^{*}+\kappa {e}^{*}\right)}{\left(\mu +\sigma \right)}, \mathrm{and}\, {X}_{3}=i\frac{\sigma {e}^{*}}{\left(\mu +\gamma +\delta \right)},$$
Hence
$$\begin{aligned}&\frac{\mathrm{d}{V}_{1}}{\mathrm{d}t}=\left(s\frac{\mu }{\left\{\mu +\beta \left({i}^{*}+\kappa {e}^{*}\right)\right\}}\right)\left[\mu \beta s\left(i+\kappa e\right)\mu s\right]+\left(i\frac{\sigma {e}^{*}}{\left(\mu +\gamma +\delta \right)}\right)\left[\sigma e\left(\mu +\gamma +\delta \right)i\right] \\ &\quad +\,\left(e\frac{\beta {s}^{*}\left({i}^{*}+\kappa {e}^{*}\right)}{\left(\mu +\sigma \right)}\right)\left[\beta s\left(i+\kappa e\right)\left(\mu +\sigma \right)e\right], \end{aligned}$$
If we assume\(s={s}^{*} , e={e}^{*}\), and\(i={i}^{*}\), we get
$$\begin{aligned} &\frac{\mathrm{d}{V}_{1}}{\mathrm{d}t}=\left\{\mu +\beta \left({i}^{*}+\kappa {e}^{*}\right)\right\}{\left({s}^{*}\frac{\mu }{\left\{\mu +\beta \left({i}^{*}+\kappa {e}^{*}\right)\right\}}\right)}^{2}\left(\mu +\sigma \right){\left({e}^{*}\frac{\beta {s}^{*}\left({i}^{*}+\kappa {e}^{*}\right)}{\left(\mu +\sigma \right)}\right)}^{2} \\ &\quad \,\left(\mu +\gamma +\delta \right){\left({i}^{*}\frac{\sigma {e}^{*}}{\left(\mu +\gamma +\delta \right)}\right)}^{2}\le 0 , \end{aligned}$$
which conclude the proof of Theorem 4.