### Existence of equilibria points

System (5) has a disease-free equilibrium

$$ E_{0} = (S_{0}, 0),\quad \mbox{with } S_{0} = \frac{(1-(1-\epsilon )p)b}{\mu +d}. $$

(6)

On the other hand, using the next generation method [36], the basic reproduction number should be as follows.

### Lemma 3.1

*The basic reproduction number is*

$$ {\mathcal{R}}_{0}=\frac{\beta k (\frac{(1-(1-\epsilon )p)b}{( \mu +d)} )}{(\mu +\gamma +c)+T^{\prime }(0)} = \frac{ \beta k(S_{0})}{( \mu +\gamma +c)+T^{\prime }(0)}. $$

*Note that*
\(S_{0} \)*depends on the vaccination of susceptible population and the treatment terms*.

### Proof

Let \(X=(I,S)^{T}\), then it follows from system (5) that

\begin{array}{rl}\frac{dX}{dt}=& \left(\begin{array}{c}\beta {\int}_{0}^{h}g(\tau )f(S(t),I(t-\tau ))\phantom{\rule{0.2em}{0ex}}d\tau \\ 0\end{array}\right)\\ & -\left(\begin{array}{c}(\mu +c+\gamma )I(t)+T(I)\\ \beta {\int}_{0}^{h}g(\tau )f(S(t),I(t-\tau ))\phantom{\rule{0.2em}{0ex}}d\tau -(1-(1-\u03f5)p)b+(\mu +d)S(t)\end{array}\right),\\ =& \mathcal{F}-\nu .\end{array}

The Jacobian of matrices \(\mathcal{F}\) and *ν* at the disease-free equilibrium \(E_{0}\) is given by

F=\left(\begin{array}{cc}\beta {f}_{2}({E}_{0})& 0\\ 0& 0\end{array}\right)\phantom{\rule{1em}{0ex}}\text{and}\phantom{\rule{1em}{0ex}}V=\left(\begin{array}{cc}(\mu +c+\gamma )+{T}^{\prime}(0)& 0\\ \beta {f}_{2}({E}_{0})& \mu +d\end{array}\right),

where \(f_{2}(E_{0})\) is the derivative of *f* with respect to *I* at \(E_{0}\). The inverse of *V* is given by

{V}^{-1}=\left(\begin{array}{cc}\frac{1}{(\mu +c+\gamma )+{T}^{\prime}(0)}& 0\\ \frac{-\beta {f}_{2}({E}_{0})}{((\mu +c+\gamma )+{T}^{\prime}(0))(\mu +d)}& \frac{1}{\mu +d}\end{array}\right).

Thus, the next generation matrix for system (5) is

F{V}^{-1}=\left(\begin{array}{cc}\frac{\beta {f}_{2}({E}_{0})}{(\mu +c+\gamma )+{T}^{\prime}(0)}& 0\\ 0& 0\end{array}\right).

Since \(R_{0}\) is the spectral radius of the matrix \(FV^{-1}\), it follows that the basic reproduction number is

$$ \mathcal{R}_{0}=\frac{\beta f_{2}(E_{0})}{(\mu +\gamma +c)+T^{\prime }(0)} = \frac{ \beta k(S_{0})}{(\mu +\gamma +c)+T^{\prime }(0)}. $$

□

To prove the existence of an endemic equilibrium, we need the following lemma.

### Lemma 3.2

*Assume that assumptions*
\((\mathbf{T}_{1})\)*and*
\(( \mathbf{T}_{2})\)*are satisfied*. *Then the equation*

*for*
\(a>0\)*and*
\(b>0\), *has a unique positive solution*.

### Proof

Let \(\mathcal{K}\) be the function defined on \(\mathbb{R_{+}}\) by

$$ \mathcal{K}(u)=b-a u-T(u). $$

We have

$$ \mathcal{K}(0)=b>0 \quad \mbox{and} \quad \mathcal{K}\biggl(\frac{b}{a} \biggr)=-T \biggl( \frac{b}{a}\biggr)< 0. $$

Since \(\mathcal{K}\) is continuous, the equation \(\mathcal{K}(u)=0\) has a unique positive solution in the interval \((0,\frac{b}{a})\). □

Next result shows the existence of the endemic equilibrium.

### Theorem 3.1

*Assume that assumptions*
\((\mathbf{H}_{1})\), \((\mathbf{H}_{2})\), \((\mathbf{T}_{1})\), *and*
\((\mathbf{T}_{2})\)*hold*. *If*
\(\mathcal{R}_{0} > 1 \), *then system* (5) *admits a unique endemic equilibrium*
\(E^{*} = (S^{*}, I^{*}) \).

### Proof

At the equilibrium point, we have

$$ \bigl(1-(1-\epsilon )p\bigr)b- (\mu +d) S^{*} -(\mu +c+\gamma )I^{*} -T\bigl(I^{*}\bigr)=0, $$

and so

$$ S^{*}=\frac{(1-(1-\epsilon )p)b-(\mu +c+\gamma )I^{*}-T(I^{*})}{ \mu +d}. $$

Let \(\overline{\mathcal{K}} \) be the function defined for \(\mathbb{R}^{+}\setminus \{0\} \) to \(\mathbb{R}\) by

$$ \overline{\mathcal{K}}(I)=\beta \frac{f (S^{*},I )}{I}-(\mu +c+ \gamma )- \frac{T(I)}{I}. $$

By hypotheses \((\mathbf{H}_{2})\) and \((\mathbf{T}_{2})\), \(\overline{ \mathcal{K}}\) is strictly monotone decreasing on \(\mathbb{R}^{+} \setminus \{0\}\) satisfying

$$ \lim_{I \rightarrow 0^{+}} \overline{\mathcal{K}}(I)=\beta k \biggl( \frac{(1-(1-\epsilon )p)b}{\mu +d} \biggr)-(\mu +c+\gamma )-T^{\prime }(0)= \bigl(\mu +c+ \gamma +T^{\prime }(0)\bigr) (\mathcal{R}_{0}-1)>0. $$

Moreover, by Lemma 3.2, there exists a unique solution \(I^{0}\) of the following equation:

$$ \frac{(1-(1-\epsilon )p)b}{\mu +d}-\frac{1}{{\mu +d}}\bigl((\mu +c+ \gamma )I+T(I) \bigr)=0, $$

and then

$$ \overline{\mathcal{K}}\bigl(I^{0}\bigr)=-\biggl((\mu +c+\gamma )+ \frac{T(I^{0})}{I ^{0}}\biggr)< 0. $$

Hence, there exists a unique positive real \(I^{*}\) such that

$$ 0 < I^{*} < I^{0}\quad \mbox{and} \quad \overline{\mathcal{K}} \bigl(I^{*}\bigr)=0, $$

which allows us to conclude that \(E^{*} = (S^{*}, I ^{*})\) is the unique endemic equilibrium of system (5). □

### Local stability analysis

In this section, we discuss the local stability of the disease-free equilibrium of system (5). We have the following result.

### Theorem 3.2

*If*
\(\mathcal{R}_{0} < 1 \), *then the disease*-*free equilibrium*
\(E_{0} = (S_{0}, 0)\)*is locally asymptotically stable*.

### Proof

We consider the following linearization equation of system (5) at \(E_{0}\):

$$ \textstyle\begin{cases} \frac{dS(t)}{dt}=- (\mu +d) S(t)- \beta \int _{0}^{h}g(\tau )f_{2}(E_{0})I(t-\tau )\,d\tau , \\ \frac{dI(t)}{dt}= \beta \int _{0}^{h}g(\tau )f_{2}(E_{0})I(t-\tau )\,d\tau -(\mu +c+\gamma )I(t)-T ^{\prime }(0)I(t). \end{cases} $$

(7)

Substituting \((S(t),I(t))=\exp (\lambda t) (S_{0},I_{0})\) into (7), we have

$$ \textstyle\begin{cases} \lambda S_{0} \exp (\lambda t) =- (\mu +d) S_{0} \exp (\lambda t)- \beta \int _{0}^{h}g(\tau )f_{2}(E_{0})I_{0} \exp \lambda ( t-\tau )\,d\tau , \\ \lambda I_{0} \exp (\lambda t)= \beta \int _{0}^{h}g(\tau )f_{2}(E_{0})I_{0} \exp \lambda ( t-\tau )\,d\tau -( \mu +c+\gamma +T^{\prime }(0))I_{0} \exp (\lambda t), \end{cases} $$

hence

$$ \textstyle\begin{cases} - (\mu +d +\lambda ) S_{0} - \beta \int _{0}^{h}g(\tau )f_{2}(E_{0})I_{0} \exp ( -\lambda \tau ) \,d\tau =0, \\ \beta \int _{0}^{h}g(\tau )f_{2}(E_{0})I_{0} \exp ( -\lambda \tau ) \,d\tau -( \mu +c+\gamma +T^{\prime }(0)+ \lambda )I_{0}=0. \end{cases} $$

(8)

We can write (8) in the following abstract form:

where X=\left(\begin{array}{c}{S}_{0}\\ {I}_{0}\end{array}\right) and

$$ B= \begin{pmatrix} -(\mu +d+\lambda ) & - \beta f_{2}(E_{0}) \int _{0}^{h}g(\tau ) \exp ( -\lambda \tau ) \,d\tau \\ 0 & \beta f_{2}(E_{0}) \int _{0}^{h}g(\tau )\exp ( -\lambda \tau ) \,d\tau - (\mu +c+\gamma +T ^{\prime }(0) + \lambda ) \end{pmatrix} . $$

Then the characteristic equation of system (8) at \(E_{0} \) is of the form

$$ (\mu +d +\lambda ) \biggl(- \beta f_{2}(E_{0}) \int _{0}^{h}g(\tau )\exp ( - \lambda \tau ) \,d \tau + \bigl(\lambda +\mu +c+\gamma +T^{\prime }(0)\bigr)\biggr)=0. $$

(9)

It is clear that \(\lambda = -(\mu +d)\) is a root of (9). All other roots *λ* of (9) are determined by the following equation:

$$ - \beta f_{2}(E_{0}) \int _{0}^{h}g(\tau )\exp ( -\lambda \tau ) \,d \tau + \bigl(\lambda +\mu +c+T^{\prime }(0)+\gamma \bigr)=0. $$

Then by separating real \((\Re )\) and imaginary \((\Im )\) parts, we derive

$$ \textstyle\begin{cases} - \beta f_{2}(E_{0}) \int _{0}^{h}g(\tau ) \exp ( -\Re (\lambda ) \tau )\cos (\Im (\lambda ) \tau ) \,d\tau + (\Re (\lambda ) +\mu +c+\gamma +T^{\prime }(0))=0, \\ - \beta f_{2}(E_{0}) \int _{0}^{h}g(\tau )\exp ( -\Re (\lambda ) \tau )\sin (\Im (\lambda ) \tau ) \,d\tau + (\Im (\lambda ) +\mu +c+\gamma +T^{\prime }(0))=0. \end{cases} $$

Using the first equation of the above system, we obtain

$$ \Re (\lambda ) = \beta f_{2}(E_{0}) \int _{0}^{h}g(\tau ) \exp \bigl( - \Re (\lambda ) \tau \bigr)\cos \bigl(\Im (\lambda ) \tau \bigr) \,d\tau - \bigl(\mu +c+ \gamma +T^{\prime }(0)\bigr). $$

(10)

We suppose, by contradiction, that there exists \(\lambda \in \mathbb{C}\) such that \(\Re (\lambda )\geq 0\), and it satisfies equality (10). Then

$$ \beta f_{2}(E_{0}) \int _{0}^{h}g(\tau ) \exp \bigl( -\Re (\lambda ) \tau \bigr)\cos \bigl(\Im (\lambda ) \tau \bigr) \,d\tau \geq \mu +c+\gamma +T^{\prime }(0). $$

(11)

Since the function *T* is concave down, it follows that \(T^{\prime }(0) \geq 0\).

Moreover, we know that \(f_{2}(E_{0})>0\), which implies

$$ 0\leq \int _{0}^{h}g(\tau ) \exp \bigl( -\Re (\lambda ) \tau \bigr)\cos \bigl(\Im (\lambda ) \tau \bigr) \,d\tau \leq 1. $$

If \(\mathcal{R}_{0} <1 \), then \(\beta f_{2}(E_{0}) < \mu +c+\gamma +T ^{\prime }(0)\) and

$$ \beta f_{2}(E_{0}) \int _{0}^{h}g(\tau ) \exp \bigl( -\Re (\lambda ) \tau \bigr) \cos \bigl(\Im (\lambda ) \tau \bigr) \,d\tau < \mu +c+\gamma +T^{\prime }(0), $$

which gives a contradiction with inequality (11). Then the real parts of all the eigenvalues of (9) are negative. Therefore, if \(\mathcal{R}_{0} < 1\), the disease-free equilibrium \(E_{0} \) of system (5) is locally asymptotically stable. Now, let

$$ P(\lambda )= - \beta f_{2}(E_{0}) \int _{0}^{h}g(\tau )\exp ( -\lambda \tau ) \,d \tau + \bigl( \lambda +\mu +c+\gamma +T^{\prime }(0)\bigr). $$

From the fact that \(P(0)= (\mu +c+\gamma +T^{\prime }(0))(1-\mathcal{R} _{0})<0 \) if \(\mathcal{R}_{0} > 1 \) and \(\lim_{\lambda \longrightarrow + \infty } P(\lambda )= + \infty \), we conclude that there is at least one positive root of (9). Hence, if \(\mathcal{R}_{0}> 1\), \(E_{0}\) is unstable. □

### Global stability of the disease-free equilibrium

The next result gives the condition of the global asymptotic stability of the disease-free equilibrium \(E_{0}\) of system (5).

### Theorem 3.3

*If hypotheses*
\((\mathbf{H}_{1})\), \((\mathbf{H}_{2})\), \((\mathbf{T}_{1})\), *and*
\((\mathbf{T}_{2})\)*hold and*
\(\mathcal{R}_{0} \leq 1 \), *then the disease*-*free equilibrium*
\(E_{0} \)*of system* (5) *is globally asymptotically stable*.

### Proof

To prove this result, we consider the following Lyapunov function:

$$ V(t)=V_{1}(t)+I(t)+V_{2}(t)+V_{3}(t), $$

where

$$\begin{aligned}& V_{1}(t)= \int _{\frac{(1-(1-\epsilon )p)b}{ \mu +d }}^{S(t)} \biggl(1-\frac{k( \frac{(1-(1-\epsilon )p)b}{ \mu +d})}{k( \sigma )} \biggr)\,d\sigma , \\& V_{2}(t)= \sigma \int _{0}^{h} g( \tau ) \int _{t-\tau }^{t} I(u)\,du\,d\tau , \end{aligned}$$

where \(\sigma =\mu +c+\gamma \), and

$$ V_{3}(t)= \int _{0}^{h} g( \tau ) \int _{t-\tau }^{t} T\bigl(I(u)\bigr)\,du\,d\tau . $$

Then

$$ \begin{aligned} \frac{d}{dt}V(t)={}& \biggl(1- \frac{k ( \frac{(1-(1-\epsilon )p)b}{ \mu +d } )}{k(S(t))} \biggr) \biggl(\bigl(1-(1- \epsilon )p\bigr)b- (\mu +d) S(t) \\ &{}- \beta \int _{0}^{h}g(\tau )f\bigl(S(t),I(t-\tau ) \bigr)\,d\tau \biggr) \\ &{} + \beta \int _{0}^{h} g(\tau )f\bigl(S(t),I(t-\tau ) \bigr)\,d\tau -(\mu +c+\gamma )I(t)-T(I) \\ &{}+ \sigma \int _{0}^{h} g( \tau ) \bigl( i(t) - i(t-\tau ) \bigr) \,d\tau + \int _{0}^{h} g( \tau ) \bigl( T\bigl(i(t)\bigr) - T\bigl(i(t-\tau )\bigr)\bigr) \,d\tau \\ = {}& -\mu \biggl(1- \frac{k ( \frac{(1-(1-\epsilon )p)b}{ \mu +d } )}{k(S(t))} \biggr) \biggl(S(t)- \frac{(1-(1-\epsilon )p)b}{ \mu +d } \biggr) \\ &{} + \int _{0}^{h} g(\tau ) \biggl( \beta \frac{k( \frac{(1-(1-\epsilon )p)b}{ \mu +d })}{k(S(t))} \frac{f(S(t),I(t-\tau ))}{I(t -\tau )} -\sigma - \frac{T(I(t-\tau ))}{I(t-\tau )} \biggr)I(t- \tau )\,d\tau . \end{aligned} $$

From hypothesis \((\mathbf{T}_{2})\), it follows that

$$ T^{\prime }(0) \leq \frac{T(I(t-\tau ))}{I(t-\tau )}. $$

Then

$$ \begin{aligned} \frac{d}{dt}V(t) \leq{} & -\mu \biggl(1- \frac{k ( \frac{(1-(1-\epsilon )p)b}{ \mu +d } )}{k(S(t))} \biggr) \biggl(S(t)- \frac{(1-(1-\epsilon )p)b}{ \mu +d } \biggr) \\ &{}+ \int _{0}^{h} g(\tau ) \biggl( \frac{\phi (S(t),I(t - \tau ))}{\sigma +T^{\prime }(0)} \frac{k ( \frac{(1-(1-\epsilon )p)b}{ \mu +d } )}{k(S(t))} -1 \biggr) \bigl(\sigma +T^{\prime }(0) \bigr)I(t - \tau )\,d\tau . \end{aligned} $$

Hypothesis \((\mathbf{H}_{1})\) implies that

$$ -\mu \biggl(1-\frac{k ( \frac{(1-(1-\epsilon )p)b}{ \mu +d } )}{k(S(t))} \biggr) \biggl(S(t)- \frac{(1-(1-\epsilon )p)b}{ \mu +d } \biggr) \leq 0, $$

and hypothesis \((\mathbf{H}_{2})\) gives that

$$ \beta \frac{\phi (S(t),I(t - \tau ))}{\sigma +T^{\prime }(0)} \frac{k ( \frac{(1-(1-\epsilon )p)b}{ \mu +d } )}{k(S(t))} \leq \beta \frac{k(S(t))}{ \sigma +T^{\prime }(0)} \frac{k ( \frac{(1-(1-\epsilon )p)b}{ \mu +d } )}{k(S(t))} = \mathcal{R} _{0}. $$

Hence,

$$ \begin{aligned} \frac{d}{dt}V(t) \leq{} & -\mu \biggl(1- \frac{k ( \frac{(1-(1-\epsilon )p)b}{ \mu +d } )}{k(S(t))} \biggr) \biggl(S(t)- \frac{(1-(1-\epsilon )p)b}{ \mu +d } \biggr) \\ &{}+ ( \mathcal{R}_{0} -1 ) \bigl(\sigma +T^{\prime }(0)\bigr) \int _{0}^{h} g(\tau )I(t-\tau )\,d\tau . \end{aligned} $$

Then the condition \(\mathcal{R}_{0} \leq 1\) implies that

$$ \frac{d}{dt}V(t)\leq 0 \quad \mbox{for all } t\geq 0. $$

Moreover, we have

$$ \frac{d}{dt}V(t)=0 \quad \mbox{holds if}\quad (S, I) = (S_{0}, 0). $$

Hence, it follows from system (5) that the set \(\{E_{0}\} \) is the largest invariant set in \(\{(S, I): \frac{d}{dt}V(t)=0 \}\). By the Lyapunov–LaSalle principle, we conclude that the disease-free equilibrium \(E_{0} \) of system (5) is globally asymptotically stable. □

### Global stability of the endemic equilibrium

In this section, we aim to show the global asymptotic stability of the endemic equilibrium \(E^{*} \) of system (5) via a Lyapunov stability approach.

### Theorem 3.4

*If hypotheses*
\((\mathbf{H}_{1})\), \((\mathbf{H}_{2})\), \((\mathbf{T}_{1})\), *and*
\((\mathbf{T}_{2})\)*hold and*
\(\mathcal{R}_{0} > 1 \), *then the endemic equilibrium of system* (5) *is the only equilibrium and is globally asymptotically stable*.

### Proof

Let *G* be the function defined from \(\mathbb{R}^{+} \) to \(\mathbb{R}\) by

It is clear that \(G(x)\geq 0 \) if \(x > 0\) and \(G(x)=0 \) if \(x=1 \). Let us consider the following Lyapunov function:

$$ U(t)=U_{1}(t)+U_{2}(t), $$

where

$$ U_{1}(t) = S(t)-S^{*} - \int _{S^{*}} ^{S(t)} \frac{f(S^{*},I^{*})}{f( \sigma ,I^{*})}\,d\sigma + I(t)-I^{*} -I^{*} \ln \biggl(\frac{I(t)}{I^{*}}\biggr) $$

and

$$ U_{2}(t) = \beta f\bigl(S^{*},I^{*}\bigr) \int _{0}^{h} g(\tau ) \int _{t-\tau } ^{t} G\biggl( \frac{ I(U)}{I^{*}} \biggr)\,du\,d\tau . $$

Then

$$ \begin{aligned} \frac{d}{dt}U_{1}(t) ={} & \biggl(1- \frac{f(S^{*},I^{*})}{f(S(t),I^{*})} \biggr) \biggl(\bigl(1-(1-\epsilon )p\bigr)b- (\mu +d) S(t) \\ &{}- \beta \int _{0}^{h} g(\tau )f\bigl(S(t),I(t- \tau ) \bigr)\,d\tau \biggr) \\ &{} + \biggl(1-\frac{I^{*}}{I(t)} \biggr) \biggl(\beta \int _{0}^{h}g(\tau )f\bigl(S(t),I(t-\tau ) \bigr)\,d\tau -(\mu +c+ \gamma )I(t)-T(I) \biggr). \end{aligned} $$

Moreover, we have

$$ \frac{d}{dt}U_{2}(t)= \beta f\bigl(S^{*},I^{*} \bigr) \int _{0}^{h}g(\tau ) \biggl(G\biggl( \frac{I(t)}{I^{*}}\biggr)-G\biggl( \frac{I(t-\tau )}{I^{*}}\biggr) \biggr) \,d\tau $$

and

$$ G\biggl( \frac{I(t)}{I^{*}}\biggr)-G\biggl( \frac{I(t-\tau )}{I^{*}}\biggr)= \frac{I}{I ^{*}}- \frac{I(t- \tau )}{I^{*}} + \ln \biggl( \frac{I(t- \tau )}{I^{*}} \biggr). $$

Since

$$ \textstyle\begin{cases} (1-(1-\epsilon )p)b= (\mu +d) S^{*} + \beta f(S^{*},I^{*}), \\ \beta f(S^{*},I^{*})= ( \mu +c+ \gamma )I^{*}+T(I^{*}), \end{cases} $$

then

$$ \begin{aligned} \frac{d}{dt}U(t)={} & \biggl(1- \frac{f(S^{*},I^{*})}{f(S(t),I^{*})} \biggr) \biggl((\mu +d) S^{*} + \beta f \bigl(S^{*},I^{*}\bigr)- (\mu +d) S(t) \\ &{}- \beta \int _{0}^{h} g(\tau )f\bigl(S(t),I(t- \tau ) \bigr)\,d\tau \biggr) \\ & {}+ \biggl(1-\frac{I^{*}}{I(t)}\biggr) \biggl(\beta \int _{0}^{h}g(\tau )f\bigl(S(t),I(t-\tau ) \bigr)\,d\tau \\ &{}-\beta \frac{ f(S^{*},I ^{*})}{I^{*}}I(t)-T(I)+\frac{I(t)T(I^{*})}{I^{*}} \biggr) \\ &{} + \beta f\bigl(S^{*},I^{*}\bigr) \int _{0}^{h}g(\tau ) \biggl( \frac{I}{I^{*}}- \frac{I(t- \tau )}{I^{*}} + \ln \biggl( \frac{I(t- \tau )}{I^{*}} \biggr) \biggr) \,d\tau \\ ={} & (\mu +d) \biggl(1-\frac{f(S^{*},I^{*})}{f(S(t),I^{*})} \biggr) \bigl(S^{*} - S(t) \bigr) \\ &{} + \beta f\bigl(S^{*},I^{*}\bigr) \int _{0}^{h}g(\tau ) \biggl(1- \frac{f(S^{*},I^{*})}{f(S(t),I^{*})} \biggr) \biggl(1-\frac{f(S(t),I(t- \tau ))}{f(S^{*},I^{*})} \biggr) \,d\tau \\ &{} + \beta f\bigl(S^{*},I^{*}\bigr) \int _{0}^{h}g(\tau ) \biggl(1- \frac{I^{*}}{I(t)}\biggr) \biggl(\frac{f(S(t),I(t- \tau ))}{f(S^{*},I^{*})} - \frac{I(t)}{I^{*}} \biggr)\,d\tau \\ &{} + \beta f\bigl(S^{*},I^{*}\bigr) \int _{0}^{h}g(\tau ) \biggl( \frac{I}{I^{*}}- \frac{I(t- \tau )}{I^{*}} + \ln \biggl( \frac{I(t- \tau )}{I^{*}} \biggr) \biggr) \,d\tau \\ &{}+ \biggl(1-\frac{I^{*}}{I(t)} \biggr) \biggl( \frac{I(t)T(I^{*})}{I^{*}}-T(I) \biggr). \end{aligned} $$

It follows that

$$\begin{aligned} \frac{d}{dt}U(t) ={}& (\mu +d) \biggl(1- \frac{f(S^{*},I^{*})}{f(S(t),I ^{*})} \biggr) \bigl(S^{*} - S(t) \bigr) \\ &{} + \beta f\bigl(S^{*},I^{*}\bigr) \int _{0}^{h}g(\tau ) \biggl(2- \frac{f(S^{*},I^{*})}{f(S(t),I^{*})} + \frac{f(S(t),I(t- \tau ))}{f(S(t),I^{*})} \\ &{}- \frac{I^{*}}{I(t)} \frac{f(S(t),I(t- \tau )}{f(S^{*},I^{*})} \biggr) \,d\tau \\ &{} + \beta f\bigl(S^{*},I^{*}\bigr) \int _{0}^{h}g(\tau ) \biggl( - \frac{I(t- \tau )}{I^{*}} + \ln \biggl( \frac{I(t- \tau )}{I^{*}}\biggr) \biggr)\,d\tau \\ &{}+ \biggl(1-\frac{I^{*}}{I(t)} \biggr) \biggl(\frac{I(t)T(I^{*})}{I^{*}}-T(I) \biggr). \end{aligned}$$

Using

$$\begin{aligned} \ln \biggl( \frac{I(t- \tau )}{I^{*}}\biggr) =& \ln \frac{f(S^{*},I^{*})}{f(S(t),I ^{*})} + \ln \biggl(\frac{I^{*}}{I(t)} \frac{f(S(t),I(t- \tau ))}{f(S ^{*}, I^{*})} \biggr) \\ &{}+\ln \biggl( \frac{I(t- \tau )}{I^{*}} \frac{f(S(t),I ^{*})}{f(S(t),I(t- \tau ))} \biggr), \end{aligned}$$

it follows that

$$ \begin{aligned} \frac{d}{dt}U(t)={} & (\mu +d) \biggl(1- \frac{f(S^{*},I^{*})}{f(S(t),I ^{*})} \biggr) \bigl(S^{*} - S(t) \bigr) \\ &{} + \beta f\bigl(S^{*},I^{*}\bigr) \int _{0}^{h}g(\tau ) \biggl(1- \frac{f(S^{*},I^{*})}{f(S(t),I^{*})} + \ln \frac{f(S^{*},I^{*})}{f(S(t),I^{*})} \biggr) \,d\tau \\ &{} + \beta f\bigl(S^{*},I^{*}\bigr) \int _{0}^{h}g(\tau ) \biggl(1- \frac{I^{*}}{I(t)} \frac{f(S(t),I(t- \tau ))}{f(S^{*},I^{*})} \\ &{}+ \ln \biggl( \frac{I^{*}}{I(t)} \frac{f(S(t),I(t- \tau ))}{f(S^{*},I^{*})} \biggr) \biggr)\,d\tau \\ &{} + \beta f\bigl(S^{*},I^{*}\bigr) \int _{0}^{h}g(\tau ) \biggl(1- \frac{i(t- \tau )}{I^{*}} \frac{f(S(t),I ^{*})}{f(S(t),I(t- \tau ))} \\ &{}+ \ln \biggl( \frac{I(t- \tau )}{I^{*}} \frac{f(S(t),I ^{*})}{f(S(t),I(t- \tau ))} \biggr) \biggr)\,d\tau \\ &{} + \beta f\bigl(S^{*},I^{*}\bigr) \int _{0}^{h}g(\tau ) \biggl( \frac{I(t- \tau )}{I^{*}} \frac{f(S(t),I ^{*})}{f(S(t),I(t- \tau ))} - 1 \\ &{}- \frac{I(t- \tau )}{I^{*}} + \frac{f(S(t),I(t- \tau ))}{f(S(t),I^{*})} \biggr) \,d\tau \\ &{} + \bigl(I(t)-I^{*} \bigr) \biggl(\frac{T(I^{*})}{I^{*}}- \frac{T(I)}{I(t)} \biggr). \end{aligned} $$

By hypotheses \((\mathbf{H}_{1})\) and \((\mathbf{H}_{2})\) we have

$$ \begin{aligned} &\frac{I(t- \tau )}{I^{*}} \frac{f(S(t),I^{*})}{f(S(t),I(t- \tau ))} - 1- \frac{I(t- \tau )}{I^{*}} + \frac{f(S(t),I(t- \tau ))}{f(S(t),I ^{*})} \\ &\quad = \biggl( \frac{I(t- \tau )}{I^{*}} - \frac{f(S(t),I(t- \tau ))}{f(S(t),I ^{*})} \biggr) \biggl( \frac{f(S(t),I^{*})}{f(S(t),I(t- \tau ))}-1 \biggr) \\ &\quad = \frac{I(t- \tau )}{I^{*} \phi (S(t),I^{*}) f(S(t),I(t-\tau ))} \bigl(\phi \bigl(S(t),I^{*}\bigr)-\phi \bigl(S(t),I(t-\tau )\bigr) \bigr) \\ &\qquad {}\times\bigl(f\bigl(S(t),I ^{*}\bigr)-f \bigl(S(t),I(t-\tau )\bigr) \bigr), \end{aligned} $$

and then

$$ \frac{I(t- \tau )}{I^{*}} \frac{f(S(t),I^{*})}{f(S(t),I(t- \tau ))} - 1- \frac{I(t- \tau )}{I^{*}} + \frac{f(S(t),I(t- \tau ))}{f(S(t),I ^{*})}\leq 0. $$

Moreover, hypothesis \((\mathbf{H}_{1})\) implies that

$$ (\mu +d) \biggl(1-\frac{f(S^{*},I^{*})}{f(S(t),I^{*})} \biggr) \bigl(S^{*} - S(t) \bigr) \leq 0, $$

and hypothesis \((\mathbf{T}_{2})\) gives

$$ \bigl(I(t)-I^{*} \bigr) \biggl(\frac{T(I^{*})}{I^{*}}- \frac{T(I)}{I(t)} \biggr)\leq 0. $$

Hence, \(\frac{d}{dt}U(t) \leq 0\). We conclude that the endemic equilibrium of system (5) is globally asymptotically stable. □