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
The Jacobian of matrices \(\mathcal{F}\) and ν at the disease-free equilibrium \(E_{0}\) is given by
where \(f_{2}(E_{0})\) is the derivative of f with respect to I at \(E_{0}\). The inverse of V is given by
Thus, the next generation matrix for system (5) is
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 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. □