In this section, we discuss the single-infection-free equilibrium, \(E_{1}\).

### Theorem 4.1

\(E_{1}\)
*is locally asymptotically stable if*
\(1< R_{0}<1+\frac{b\beta k e^{-m\tau_{1}}}{cdu}\), *provided that*
\(\tau_{1}, \tau_{2}\geq0\), *otherwise*
\(E_{1}\)
*is unstable*.

### Proof

The characteristic equation of the Jacobian matrix corresponding to the linearized system (1) at \(E_{0}\) is given by

$$\begin{aligned}& \det\bigl[\gamma I -J(E_{1})\bigr] \\& \quad =\det \left (\textstyle\begin{array}{@{}c@{\quad}c@{\quad}c@{\quad}c@{}} \gamma+d+\beta v_{1} e^{-\tau_{1}(m+\gamma)} & 0 & \beta x_{1} e^{-\tau _{1}(\gamma+m)}&0 \\ -\beta v_{1} e^{-\tau_{1}(m+\gamma)}& \gamma+a & -\beta x_{1} e^{-\tau _{1}(\gamma+m)} & p \\ 0 & -k & \gamma+u & 0 \\ 0 & 0& 0 & \gamma+b -cy_{1}e^{-\gamma\tau_{2}} \end{array}\displaystyle \right )=0. \end{aligned}$$

After some fundamental calculation, we get the above characteristic equation in the following form:

$$ \bigl(\gamma+b -cy_{1}e^{-\gamma \tau_{2}}\bigr) \bigl[ \gamma^{3}+a_{0}\gamma^{2}+a_{1}\gamma +a_{2}+\bigl(b_{0}\gamma^{2}+b_{1} \gamma+b_{2}\bigr)e^{-\tau_{1}\gamma} \bigr]=0, $$

(11)

where

$$\begin{aligned}& a_{0}=a+u+d,\qquad a_{1}=(a+u)d+au,\qquad a_{2}=aud, \\& b_{0}=\beta v_{1} e^{-m\tau_{1}},\qquad b_{1}=(a+u)\beta v_{1} e^{-m \tau _{1}}-k\beta x_{1} e^{-m\tau_{1}}, \qquad b_{2}=au\beta v_{1}e^{-m\tau_{1}}. \end{aligned}$$

First, we discuss the following factor of equation (11):

$$ \gamma^{3}+a_{0}\gamma^{2}+a_{1} \gamma+a_{2}+\bigl(b_{0}\gamma ^{2}+b_{1} \gamma+b_{2}\bigr)e^{-\tau_{1}\gamma}=0. $$

(12)

Now, we present possible cases of delay term \(\tau_{1}\). When \(\tau_{1}=0\), then equation (12) becomes

$$ \gamma^{3}+c_{0}\gamma^{2}+c_{1} \gamma+c_{2}=0, $$

(13)

where

$$\begin{aligned}& c_{0} = a_{0}+b_{0}=a+u+d+d(R_{0}-1)>0, \\& c_{1} = a_{1}+b_{1}=(a+u)d+(a+u)d(R_{0}-1)>0, \\& c_{2} = a_{2}+b_{2}=aud+aud(R_{0}-1)>0, \\& c_{1}-c_{0}c_{2} = dR_{0} \bigl(a^{2}+(a+u) (u+dR_{0})\bigr)>c_{2}=aud+aud(R_{0}-1). \end{aligned}$$

Thus, by using the Routh-Hurtwitz criterion in [23] one has no positive roots when \(\tau_{1}=0\).

Next, we consider the root distribution of equation (12) when \(\tau_{1}\neq0\). If *iκ* (\(\kappa>0\)) is a solution of equation (12), then, separating real and imaginary parts, we get the following equations:

$$\begin{aligned} \begin{aligned} &a_{1}\kappa-\kappa^{3} = \bigl(b_{2}-b_{0} \kappa^{2}\bigr)\sin\kappa\tau_{1} -b_{1}\kappa \cos\kappa\tau_{1}, \\ &a_{0}\kappa^{2}-a_{2} = \bigl(b_{2}-b_{0} \kappa^{2}\bigr)\cos\kappa\tau_{1} +b_{1}\kappa \sin\kappa\tau. \end{aligned} \end{aligned}$$

By squaring and adding the above equations, we get

$$ \kappa^{6}+m_{1}\kappa^{4}+m_{2} \kappa^{2}+m_{3}=0, $$

(14)

where

$$\begin{aligned}& m_{1} = a_{0}^{2}-2a_{1}-b_{0}^{2}, \\& m_{2} = a_{1}^{2}-2a_{0}a_{2}+2b_{0}b_{2}-b_{1}^{2}, \\& m_{3} = a_{2}^{2}-b_{2}^{2}. \end{aligned}$$

Let us suppose that \(\sigma=\kappa^{2}>0\), then equation (14) becomes

$$ \sigma^{3}+m_{1}\sigma^{2}+m_{2} \sigma+m_{3}=0, $$

(15)

where

$$\begin{aligned}& m_{1} = a_{0}^{2}-2a_{1}-b_{0}^{2}=a^{2}+u^{2}+d^{2}-(R_{0}-1)^{2}d^{2}>a^{2}+u^{2}, \\& m_{2} = a_{1}^{2}-2a_{0}a_{2}+2b_{0}b_{2}-b_{1}^{2} \\& \hphantom{m_{2}}= (ad)^{2}+(ud)^{2}+2au\bigl(d(R_{0}-1) \bigr)^{2}+(a+u)d(R_{0}-1) \bigl(2au-(a+u)d(R_{0}-1) \bigr) \\& \hphantom{m_{2}}> (ad)^{2}+(ud)^{2} +2au(a+u)d(R_{0}-1), \\& m_{3} = (aud)^{2}-\bigl(au d(R_{0}-1)^{2} \bigr)>(aud)^{2}(R_{0}-1). \end{aligned}$$

Hence, if \(R_{0} > 1\), then equation (15) has no positive roots. It is to be noted that the equilibrium \(E_{1}\) is locally asymptotically stable by the general theory on characteristic equations of delay differential equations [23].

To find the other root, we consider the second factor of equation (11) to be given by

$$ \gamma+b -cy_{1}e^{-\gamma\tau_{2}}=0. $$

(16)

If \(\tau_{2}=0\), then for \(1< R_{0}<1+\frac{b\beta k e^{-m\tau _{1}}}{cdu}\), we have

$$ \gamma=cy_{1}-b=\frac{c}{k\beta e^{-m\tau_{1}}} \biggl(R_{0} -\biggl(1+ \frac {b\beta k e^{-m\tau_{1}}}{cdu}\biggr) \biggr)< 0. $$

This shows that the root of equation (16) is negative for \(\tau_{2}=0\). To discuss the roots in the case \(\tau_{2}>0\), we assume \(\gamma=\kappa i\) (\(\kappa>0\)) to be a pure imaginary root of equation (16), to get

$$\begin{aligned}& \kappa=c\biggl(\frac{\beta N k e^{-m\tau_{1}}-aud}{a\beta k e^{-m\tau _{1}}}\biggr)\sin\kappa\tau_{2}, \\& b=c\biggl(\frac{\beta N k e^{-m\tau_{1}}-aud}{a\beta k e^{-m\tau_{1}}}\biggr)\cos \kappa \tau_{2}, \end{aligned}$$

which implies that

$$ \kappa^{2}= \biggl[c\biggl(\frac{\beta N k e^{-m\tau_{1}}-aud}{a\beta k e^{-m\tau_{1}}}\biggr) \biggr]^{2}-b^{2}. $$

However, for \(1< R_{0}<1+\frac{b\beta k e^{-m\tau_{1}}}{cdu}\) this implies that \(\kappa^{2}<0\), which is a contradiction.

Thus, we conclude that all the roots of equation (16) have a negative real part when \(\tau_{2}\geq0\). Therefore, the equilibrium \(E_{1}\) is locally asymptotically stable, when \(1< R_{0}<1+\frac{b\beta k e^{-m\tau_{1}}}{cdu}\) and \(\tau_{1},\tau_{2}\geq0\). □

### Theorem 4.2

*The single*-*infection*-*free equilibrium*, \(E_{1}\), *is globally asymptotically stable*, *if*
\(1< R_{0}<1+\frac{b\beta k e^{-m\tau_{1}}}{cdu}\), *while for*
\(R_{0}>1+\frac{b\beta k e^{-m\tau_{1}}}{cdu}\), \(E_{1}\)
*is unstable*.

### Proof

Denote \(f(\rho)=\rho-1-\ln\rho\), \(\rho\in R^{+}\). Let us construct the Lyapunov functional

$$\begin{aligned} L_{1}(t) =&x_{1}f\biggl(\frac{x}{x_{1}} \biggr)+y_{1}f\biggl(\frac{x}{y_{1}}\biggr)+\frac {a}{k}v_{1}f \biggl(\frac{v}{v_{1}}\biggr)+\frac{p}{c}z_{1}f\biggl( \frac{z}{z_{1}}\biggr) \\ &{}+\beta x_{1}v_{1}e^{-m\tau_{1}} \int_{t-\tau_{1}}^{t}f\biggl(\frac{x(\mu)v(\mu )}{x(\tau_{1}+\mu)v_{1}}\biggr)\,d \mu+p \int_{t-\tau_{2}}^{t}y(\mu)z(\mu)\,d\mu. \end{aligned}$$

(17)

By taking the derivative of the equation (17), we obtain

$$\begin{aligned} \dot{L}_{1}(t) =&\biggl(1-\frac{x_{1}}{x}\biggr) \bigl(\lambda-d x(t)-\beta e^{-m\tau_{1}} x(t-\tau_{1}) v(t- \tau_{1}) \bigr) \\ &{}+\biggl(1-\frac{y_{1}}{y}\biggr) \bigl(\beta e^{-m\tau_{1}} x(t-\tau_{1}) v(t-\tau_{1})-a y(t)-p y(t)z(t) \bigr) \\ &{} +\frac{p}{c}\biggl(1-\frac{z}{z_{1}} \biggr) \bigl(c y(t-\tau_{2})z(t-\tau_{2})-b z(t) \bigr)+ \frac{a}{k}\biggl(1-\frac{v_{1}}{v}\biggr) \bigl(ky(t)-u v(t) \bigr) \\ &{} +\beta x_{1}e^{-m\tau_{1}}\frac{x(t)v(t)}{x(\tau_{1}+t)} -\beta x_{1}e^{-m\tau_{1}} v_{1}\ln\biggl(\frac{x(t)v(t)}{x(\tau_{1}+t)v_{1}} \biggr) \\ &{}-\beta x_{1}e^{-m\tau_{1}}\frac{x(t-\tau_{1})v(t-\tau_{1})}{x(\tau_{1}+t)} + \beta x_{1}e^{-m\tau_{1}} v_{1}\ln\biggl(\frac{x(t-\tau_{1})v(t-\tau_{1})}{x(\tau_{1}+t)v_{1}} \biggr). \end{aligned}$$

(18)

Using \(E_{1}\) in the system (1) yields the following identities:

$$\begin{aligned}& \lambda = d x_{1}+\beta x_{1}v_{1}e^{-m\tau_{1}}, \\& ay_{1} = \beta e^{-m\tau_{1}} x_{1}v_{1}, \\& ky_{1} = uv_{1}. \end{aligned}$$

Using the above identities in equation (18), we get

$$\begin{aligned} \dot{L}_{1}(t) =&e^{-m\tau_{1}}\biggl(2- \frac{x_{1}}{x}-\frac{x}{x_{1}}\biggr)+ \beta x_{1}v_{1}e^{-m\tau_{1}} \biggl(3-\frac{x_{1}}{x}-\frac{yv_{1}}{y_{1}v} -\frac{y_{1}x(t-\tau_{1})v(t-\tau_{1})}{ x_{1}v_{1}y} \\ &{} - \ln\biggl(\frac{x(t)v(t)}{x(t-\tau_{1})v(t-\tau_{1})}\biggr) \biggr)+ \frac{bp}{c} \biggl(R_{0}-\biggl(1+\frac{b\beta k e^{-m\tau_{1}}}{cdu}\biggr) \biggr)z \\ &{} + \biggl( \frac{au}{k}\frac{x(t)}{x(t+\tau _{1})}-\frac{au}{k} \biggr)v. \end{aligned}$$

(19)

When \(\tau_{1}\) is very large, then \(x(t)=x(t+\tau_{1})\). Thus, the above equation (19) can be written as

$$\begin{aligned} \begin{aligned}[b] \dot{L}_{1}(t)={}&e^{-m\tau_{1}}\biggl(2- \frac{x_{1}}{x}-\frac{x}{x_{1}}\biggr)+ \beta x_{1}v_{1}e^{-m\tau_{1}} \biggl(3-\frac{x_{1}}{x}-\frac{yv_{1}}{y_{1}v} -\frac{y_{1}x(t-\tau_{1})v(t-\tau_{1})}{ x_{1}v_{1}y} \\ &{} - \ln\biggl(\frac{x(t)v(t)}{x(t-\tau_{1})v(t-\tau_{1}) }\biggr) \biggr)+\frac{bp}{c} \biggl(R_{0}-\biggl(1+\frac{b\beta k e^{-m\tau_{1}}}{cdu}\biggr) \biggr)z. \end{aligned} \end{aligned}$$

(20)

To show that \(\dot{L}_{1}(t)<0 \), we need to prove that the following inequalities hold:

$$ \begin{aligned} &e^{-m\tau_{1}}\biggl(2-\frac{x_{1}}{x}- \frac{x}{x_{1}}\biggr)\leq0, \\ & \biggl(3-\frac{x_{1}}{x}-\frac{yv_{1}}{y_{1}v} -\frac{y_{1}x(t-\tau_{1})v(t-\tau_{1})}{x_{1}v_{1}y}- \ln\biggl( \frac{x{t}(t)v(t)}{x(t-\tau_{1})v(t-\tau_{1})}\biggr) \biggr)\leq0. \end{aligned} $$

(21)

Thus, the above results are satisfied only if \(R_{0}<1+\frac{b\beta k e^{-m\tau_{1}}}{cdu}\), then equation (21) implies that \(\frac{dL_{1}}{dt}\leq0\). Moreover, the equality holds when \(x=x_{1}\) and \(y=y_{1}\), \(v=v_{1}\), and \(z=z_{1}\). Thus, by LaSalle’s invariance principle [22], we conclude that \(E_{1}\) is globally asymptotically stable. □

### Theorem 4.3

*If*
\(\tau_{1}\neq0\)
*and*
\(\tau_{2}\neq0\)
*and*
\(R_{0}> 1+\frac{b\beta k e^{-m\tau_{1}}}{cdu}\), *then the double*-*infection equilibrium*, \(E_{2}\), *is globally asymptotically stable*.

### Proof

Let us construct the Lyapunov functional given by

$$\begin{aligned} L_{2}(t) =&x_{2}f\biggl(\frac{x}{x_{2}} \biggr)+y_{1}f\biggl(\frac{x}{y_{2}}\biggr)+\frac {a}{k}v_{2}f \biggl(\frac{v}{v_{2}}\biggr)+\frac{p}{c}z_{2}f\biggl( \frac{z}{z_{2}}\biggr) \\ &{}+\beta x_{2}v_{2}e^{-m\tau_{1}} \int_{t-\tau_{1}}^{t}f\biggl(\frac{x(\mu)v(\mu )}{x(\tau_{1}+\mu)v_{2}}\biggr)\,d \mu+p \int_{t-\tau_{2}}^{t}y(\mu)z(\mu)\,d\mu. \end{aligned}$$

(22)

Using \(E_{2}\) in the system (1), we get the following identities:

$$\begin{aligned}& N-d x_{2} = \beta x_{2}v_{2}e^{-m\tau_{1}}, \\& \beta x_{2}v_{2}e^{-m\tau_{1}} = ay_{2}+py_{2}z_{2}, \\& ky_{2} = uv_{2}, \\& by_{2} = c. \end{aligned}$$

By taking the derivative of equation (22) and using the above identities, we get

$$\begin{aligned} \dot{L}_{2}(t) =&e^{-m\tau_{1}}\biggl(2- \frac{x_{2}}{x}-\frac{x}{x_{2}}\biggr)+ \beta x_{2}v_{1}e^{-m\tau_{1}} \biggl(3-\frac{x_{2}}{x}-\frac{yv_{2}}{y_{2}v} -\frac{y_{2}x(t-\tau_{1})v(t-\tau_{1})}{ x_{2}v_{2}y} \\ &{} + \ln\biggl(\frac{x(t-\tau_{1})v(t-\tau_{1})}{x(t)v(t)}\biggr) \biggr)+\biggl(\beta x_{2}e^{-m\tau_{1}}\frac{x(t)}{x(t+\tau_{1})}-\frac{au}{k}\biggr)v \\ &{}-py_{2}z_{2}-pyz_{2}+pz_{2}v_{2} \frac{y}{v}+py_{2}z_{2}\frac{v}{v_{2}}. \end{aligned}$$

(23)

When \(\tau_{1}\) is very large and \(x(t)=x(t+\tau_{1})\) then equation (23) becomes

$$\begin{aligned} \dot{L}_{2}(t) =&e^{-m\tau_{1}}\biggl(2- \frac{x_{2}}{x}-\frac{x}{x_{2}}\biggr)+ \beta x_{2}v_{2}e^{-m\tau_{1}} \biggl(3-\frac{x_{2}}{x}-\frac{yv_{2}}{y_{2}v} -\frac{y_{2}x(t-\tau_{1})v(t-\tau_{1})}{ x_{2}v_{2}y} \\ &{} - \ln \biggl(\frac{x(t)v(t)}{x(t-\tau_{1})v(t-\tau_{1})} \biggr) \biggr)-\frac{p}{acdu} \biggl(R_{0}-\biggl(1+\frac{b\beta k e^{-m\tau_{1}}}{cdu} \biggr) \biggr) \biggl(\biggl( \frac{v_{2}}{v}-1\biggr)y \\ &{}- \biggl(\frac{v}{v_{2}}-1\biggr)y_{2} \biggr). \end{aligned}$$

(24)

The following inequalities hold:

$$\begin{aligned}& e^{-m\tau_{1}}\biggl(2-\frac{x_{1}}{x}-\frac{x}{x_{1}}\biggr)\leq0, \\& \biggl(3-\frac{x_{1}}{x}-\frac{yv_{1}}{y_{1}v} -\frac{y_{1}x(t-\tau_{1})v(t-\tau_{1})}{x_{1}v_{1}y}- \ln\biggl( \frac{x{t}(t)v(t)}{x(t-\tau_{1})v(t-\tau_{1})}\biggr) \biggr)\leq0. \end{aligned}$$

Therefore, equation (24) implies that \(\frac{dL_{2}}{dt}\leq0\), when \(R_{0}<1+\frac{b\beta k e^{-m\tau_{1}}}{auc}\). Moreover, the equality holds when \(x=x_{2}\) and \(y=y_{2}\), \(v=v_{2}\), and \(z=z_{2}\). Thus by LaSalle’s invariance principle [22], we conclude that \(E_{2}\) is globally asymptotically stable. □