Appendix A: Proof of Lemma 1
Proof
Let \(t\) be a fixed positive real value. Since \(V(t,.)\) is an increasing function for \(a\ge 0\) such that \(V(t,a)\le A,\) then there exist a unique \(\tau (t)\ge 0\) which is the first instant such that \(V(t,\tau (t))=A\). Furthermore, \(V(t,a)\le A\) is equivalent to \(0\le a\le \tau (t)\). Now, differentiating both sides of the equation \(V(t,\tau (t))=A\) with respect to \(t,\) we obtain
$$\begin{aligned} \frac{\partial }{\partial a}V(t,\tau (t))\frac{d}{dt}\left( t-\tau (t)\right) =rV(t,\tau (t))+F(I(t)). \end{aligned}$$
Thus, since \(V(t,a)\) is increasing for \(a\le \tau (t)\) and the right hand side of the above equation is positive, the function \(t-\tau (t)\) is increasing for \(t\ge 0\). Using the characteristic method (See, Webb 1985), we obtain, for \(a\le \tau (t)\) and \(0\le a\le t,\)
$$\begin{aligned} V(t,a)=ce^{ra}+\int \limits _{-a}^{0}e^{-ru}F(I(t+u))du. \end{aligned}$$
Thus, \(V(t,\tau (t))=A\) leads to
$$\begin{aligned} ce^{r\tau (t)}+\int \limits _{-\tau (t)}^{0}e^{-ru}F(I(t+u))du=A. \end{aligned}$$
Furthermore, \(\tau \) is bounded on \([0,\infty [\) and, for \(t\in [0,\infty [,\)
$$\begin{aligned} \tau (t)\le \frac{1}{r}\ln \left( \frac{A}{c}\right) . \end{aligned}$$
\(\square \)
Appendix B: Proof of Lemma 3
Proof
To prove this Lemma, we will use a global implicit function theorem (GIFT) in Ichiraku (1985, Theorem 3) (See Appendix 1). In what follows, we will prove conditions 1–3 of this theorem:
-
1)
Let \(\phi \in C\). We need to solve the equation \(\Psi (s,\phi )=0\) for \(s>0\). This last equation can be written as
$$\begin{aligned} \int \limits _{-s}^{0}e^{ru}F(\phi (u))du+ce^{rs}-A=0. \end{aligned}$$
Remember that \(c<A\). Thus, \(\Delta (0,\phi )=c-A<0\). Moreover, since \(\Psi (s,\phi )>ce^{rs}-A,\)
\(\lim _{s\rightarrow \infty }\Psi (s,\phi )=\infty \). Hence, there exists \(s_{0}>0\) such that \(\Psi (s_{0},\phi )=0\). On the other hand, the derivative \(D_{s}\Psi (s,\phi )\) is given by
$$\begin{aligned} D_{s}\Psi (s,\phi )=rce^{rs}+e^{rs}F\left( \phi (-s)\right) -F\left( \phi (0)\right) . \end{aligned}$$
However, since \(k>\frac{b}{r},\) we have,
$$\begin{aligned} \max _{x\ge 0}(F\left( x\right) )=\frac{bc}{k}<rce^{rs},\quad \hbox { for all }s>0. \end{aligned}$$
Thus, \(D_{s}\Psi (s,\phi )>0\) for all \(s>0\) and \(\Psi (.,\phi )\) is increasing. Consequently, there exists unique \(s_{0}>0\) such that \(\Psi (s_{0},\phi )=0\).
-
2)
The second partial derivative \(D_{\phi }\Delta (s,\phi )\) is given by
$$\begin{aligned} D_{\phi }\Delta (s,\phi )=\int \limits _{-s}^{0}e^{-ru}DF(\phi (u))du \end{aligned}$$
for \(s\ge 0\hbox { and}\phi \in C\). This map is invertible since \(F\) is an increasing function.
-
3)
Let \(\left( s_{i},\phi _{i}\right) _{i\ge 1}\) be a sequence of vectors in \( {R}^{+}\times C\) such that \(\Psi (s_{i},\phi _{i})=0\) and the sequence \(\phi _{i}\) converge to \(\phi \) in \(C\) as \(i\) tends to infinity. Thus, for \(i\ge 1,\)
$$\begin{aligned} \int \limits _{-s_{i}}^{0}e^{-ru}F(\phi (u))du=-ce^{rs}+A>0 \end{aligned}$$
and, consequently,
$$\begin{aligned} ce^{rs_{i}}<A. \end{aligned}$$
This implies that the sequence \(s_{i}\) is bounded and, thus, there xists a subsequence of \(s_{i}\) which converges to a point in \( {R}^{+}\).
Finally, using the GIFT theorem in Ichiraku (1985, Theorem 3), there exists a unique and continuous map \(\sigma :C\rightarrow {R}^{+}\) such that \(\Psi (\sigma (\phi ),\phi )=0\) for any \(\phi \in C\). The function \(\sigma \) is given, for \(\phi \in C,\) by
$$\begin{aligned} \int \limits _{-\sigma (\phi )}^{0}e^{-ru}F(\phi (u))du=-ce^{r\sigma (\phi )}+A. \end{aligned}$$
Thus, since \(F(0)=0, e^{r\sigma (0)}=\frac{A}{c}\). Consequently,
$$\begin{aligned} \sigma (0)=\frac{1}{r}\ln \left( \frac{A}{c}\right) >0. \end{aligned}$$
Using the global function theorem, for Banach spaces, in Dieudonné et al. (1960, chp. 6, pp. 270), we conclude that the map \(\sigma \) s continuously differentiable and satisfy, for all \(\psi \in C,\)
$$\begin{aligned} \sigma '(\phi )\psi =-\frac{D_{\phi }\Psi (\sigma (\phi ),\phi )\psi }{D_{s}\Psi (\sigma (\phi ),\phi )}=-\frac{\int _{-\sigma (\phi )}^{0}e^{-ru}DF(\phi (u))\psi (u)du}{ce^{r\sigma (\phi )}+e^{r\sigma (\phi )}F\left( \phi (-\sigma (\phi ))\right) -F\left( \phi (0)\right) } \end{aligned}$$
(18)
for \(\phi \in C\). Since \(D_{\phi }\Psi (\sigma (\phi ),\phi )\) and \(D_{\phi }\Psi (\sigma (\phi ),\phi )\) are positive for all \(\phi \in C,\) the function \(\sigma (.)\) is decreasing. This achieves the proof of the Lemma. \(\square \)
Appendix C: Proof of Proposition 5
Proof
Define, for \(y\ge 0,\) the function
$$\begin{aligned} \chi (y):=\frac{\beta \pi e^{-\int _{0}^{\sigma (y)}\delta (s)ds}}{\beta y+d}. \end{aligned}$$
(19)
Note that any nontrivial equilibrium \(\mathcal {E}^{*}=\left( x^{*},y^{*}\right) \) atisfy \(\chi (y^{*})=\mu \).We have
$$\begin{aligned} \chi (0)=\frac{\beta \pi e^{-\int _{0}^{\sigma (0)}\delta (s)ds}}{d}\qquad \text {and}\qquad \lim _{y\rightarrow +\infty }\chi (y)=0. \end{aligned}$$
Moreover, condition (9) is equivalent to \(\chi (z)<\chi (0)\) for all \(z\) positive. Hence, the existence and number of positive equilibria depends on the sign of \(\chi (0)-\mu \). This prove all the assertions of the proposition. \(\square \)
Appendix D: Proof of Theorem 7
Proof
Let \(\alpha \in R\) be given by \(\alpha =e^{\int _{\sigma ^{*}}^{\sigma (0)}\delta (s)ds}\) such that \(\sigma ^{*}=\inf _{\phi \in C}\sigma (\phi )\ge 0\). Assume that \(\alpha R_{0}<1,\) then for \(\epsilon \) small enough
$$\begin{aligned} \alpha \left( \frac{\pi }{d}+\epsilon \right) \beta e^{-\int _{0}^{\sigma (0)}\delta (s)ds}<\mu . \end{aligned}$$
(20)
Furthermore we already have
$$\begin{aligned} \limsup _{t\rightarrow \infty }x(t)\le \frac{\pi }{d}. \end{aligned}$$
Then for \(\epsilon >0,\) there is \(T_{1}>0\) such that \(x(t)\le \frac{\pi }{d}+\epsilon \).Thus, for \(t\ge T_{1}+\sigma (0)\)
$$\begin{aligned} \dot{y}(t)&\le -\mu y(t)+\beta e^{-\int _{0}^{\sigma (y_{t})}\delta (s)ds}y_{t}\left( \frac{\pi }{d}+\epsilon \right) y(t-\sigma (y_{t}))\\&\le -\mu y(t)+\beta \left( \frac{\pi }{d}+\epsilon \right) e^{-\int _{0}^{\sigma ^{*}}\delta (s)ds}y(t-\sigma (y_{t}))\nonumber \\&\le -\mu y(t)+\alpha \beta \left( \frac{\pi }{d}+\epsilon \right) e^{-\int _{0}^{\sigma (0)}\delta (s)ds}y(t-\sigma (y_{t}))\nonumber \end{aligned}$$
(21)
Now, consider the linear scalar equation
$$\begin{aligned} \dot{y}(t)=-\mu y(t)+\alpha \beta \left( \frac{\pi }{d}+\epsilon \right) e^{-\int _{0}^{\sigma (0)}\delta (s)ds}y(t-\sigma (0)). \end{aligned}$$
(22)
We claim that the trivial equilibrium \(y=0\) of Eq. (22) is globally asymptotically stable (GAS). Indeed, Eq. (22) is linear and its characteristic equation around the trivial equilibrium \(y=0\) is given by
$$\begin{aligned} \Delta (\lambda )=\lambda +\mu -\alpha \beta \left( \frac{\pi }{d}+\epsilon \right) e^{-\int _{0}^{\sigma (0)}\delta (s)ds}e^{-\lambda \sigma (0)}=0. \end{aligned}$$
Let \(\lambda =a+ib\) a root of \(\Delta (\lambda )\) with \(a\ge 0\). Then \(\mid e^{-\lambda \sigma (0)}\mid \le 1\). Therefore, from the above characteristic equation and (20),
$$\begin{aligned} \begin{array}{lcl} \mid \lambda +\mu \mid &{}=&{} \alpha \beta \left( \frac{\pi }{d}+\epsilon \right) e^{-\int _{0}^{\sigma (0)}\delta (s)ds}\mid e^{-\lambda \sigma (0)}\mid \\ &{}\le &{} \alpha \beta \left( \frac{\pi }{d}+\epsilon \right) e^{-\int _{0}^{\sigma (0)}\delta (s)ds}\\ &{}< &{} \mu . \end{array} \end{aligned}$$
However, since \(a\ge 0, \mid \lambda +\mu \mid \ge \mu \) hold true. This is a contradiction and, therefore, every root of the equation \(\Delta (\lambda )=0\) has negative real part and the trivial equilibrium \(y=0\) of the linear equation (22) is locally asymptotically stable. It follows directly that \(y=0\) of this equation is also GAS.
On the other hand, from Lemma (3), and (18), \(\sigma \) is continuously differentiable and its derivative \(\sigma '\) is bounded on \(C\). Thus, using Theorem 2.4 of Gyori and Hartung (2007) paper, we deduce that the trivial equilibrium of equation
$$\begin{aligned} \dot{y}(t)=-\mu y(t)+\alpha \beta \left( \frac{\pi }{d}+\epsilon \right) e^{-\int _{0}^{\sigma (0)}\delta (s)ds}y(t-\sigma (y_{t})) \end{aligned}$$
is GAS. Let \((x(t),y(t))\) be a solution of Eq. (7) with positive initial condition. It follows, by comparison, that \(y(t)\) converges to \(0\) as \(t\) tends to \(\infty \). Furthermore, integrating the first equation in (7) and taking the limit, we obtain \(\lim _{t\rightarrow \infty }x(t)=\frac{\pi }{d}\). The proof of Theorem 7 is complete. \(\square \)
Appendix E: Proof of Theorem 8
Proof
The characteristic equation (13), when \(y^{e}=0\) and \(x^{e}=\overline{x}\), is given by
$$\begin{aligned} \Delta (\lambda )=\lambda +\mu -\beta \overline{x}e^{-\int _{0}^{\sigma (0)}\delta (s)ds}e^{-\lambda \sigma (0)}=0. \end{aligned}$$
(23)
Thus,
$$\begin{aligned} \Delta (0)&= -\beta \overline{x}e^{-\int _{0}^{\sigma (0)}\delta (s)ds}+\alpha =\mu \left( 1-\frac{\beta \overline{x}}{\mu }e^{-\int _{0}^{\sigma (0)}\delta (s)ds}\right) ,\\&= \mu \left( 1-R_{0}\right) \end{aligned}$$
and \(\lim _{\lambda \rightarrow \infty }\Delta (\lambda )=+\infty \). When \(R_{0}>1\) holds, then \(\Delta (0)<0\), so there exists \(\lambda _{0}>0\) such that \(\Delta (\lambda _{0})=0\). This proves the instability of the trivial steady state.
Conversely, assume that \(R_{0}<1\). Let \(\lambda =a+ib\) a root of \(\Delta (\lambda )\) with \(a\ge 0\). Then \(\mid e^{-\lambda \sigma (0)}\mid \le 1\). Therefore, from (23),
$$\begin{aligned} \begin{array}{rcl} \mid \lambda +\mu \mid &{}=&{} \beta \overline{x}e^{-\int _{0}^{\sigma (0)}\delta (s)ds}\mid e^{-\lambda \sigma (0)}\mid \\ &{}\le &{} \beta \overline{x}e^{-\int _{0}^{\sigma (0)}\delta (s)ds}\\ &{}=&{} R_{0}\mu <\mu . \end{array} \end{aligned}$$
However, since \(a\ge 0, \mid \lambda +\mu \mid \ge \mu \) hold true. This is a contradiction and, therefore, every root of the equation \(\Delta (\lambda )=0\) has negative real part and the DFE \(\mathcal {E}_{f}\) is locally asymptotically stable. \(\square \)
Appendix E: Proof of Theorem 9
Proof
The stability and uniqueness of the DFE follows from Theorems 7 and 8. We investigate local asymptotic stability of \(\mathcal {E}^{*}(\nu )\) in a neighborhood of \(\nu _{0}\). We have
$$\begin{aligned} \Delta (0,\nu )=\mu -\beta x^{*}(\nu )e^{-\int _{0}^{\nu \tilde{\sigma }(y^{*}(\nu ))}\delta (s)ds}+Q(\nu ), \end{aligned}$$
where
$$\begin{aligned} \begin{array}{rcl} Q(\nu ) &{}=&{} \beta e^{-\int _{0}^{\nu \tilde{\sigma }(y(\nu ))}\delta (s)ds}\frac{\beta x^{*}(\nu )y^{*}(\nu )}{d+\beta y^{*}(\nu )}\\ &{}&{}+\,\nu \beta y^{*}(\nu )x^{*}(\nu )e^{-\int _{0}^{\nu \tilde{\sigma }(y^{*}(\nu ))}\delta (s)ds}\delta (\nu \tilde{\sigma }(y^{*}(\nu )))\tilde{\sigma }'(y^{*}(\nu ))\\ &{}=&{} \beta e^{-\int _{0}^{\nu \tilde{\sigma }(y^{*}(\nu ))}\delta (s)ds}x^{*}(\nu )y^{*}(\nu )\left( \frac{\beta }{d+\beta y^{*}(\nu )}+\nu \delta (\nu \tilde{\sigma }(y^{*}(\nu )))\tilde{\sigma }'(y^{*}(\nu ))\right) . \end{array} \end{aligned}$$
However, \(\frac{\beta }{d+\beta y^{*}(\nu )}+\nu \delta (\nu \tilde{\sigma }(y^{*}(\nu )))\tilde{\sigma }'(y^{*}(\nu ))>0\). Therefore, \(Q(\nu )>0\). On the other hand, from (15),
$$\begin{aligned} \mu -\beta x^{*}(\nu )e^{-\int _{0}^{\nu \tilde{\sigma }(y^{*}(\nu ))}\delta (s)ds}=\mu -\frac{\beta e^{-\int _{0}^{\nu \tilde{\sigma }(y^{*}(\nu ))}\delta (s)ds}}{d+\beta y^{*}(\nu )}=0. \end{aligned}$$
Thus, \(\Delta (0,\nu )>0,\) which prove that \(\lambda =0\) is not a root of \(\Delta (\lambda ,\nu )=0\) for \(\nu \) close to \(\overline{\nu .}\)
Therefore, \(\Delta (\lambda ,\nu )=0\) is equivalent to
$$\begin{aligned} \lambda +\mu +\beta x_{M}(\nu )e^{-\int _{0}^{\nu \tilde{\sigma }(y_{M}(\nu ))}\delta (s)ds}e^{-\lambda \nu \tilde{\sigma }(y^{*}(\nu ))}=Q_{1}+Q_{2}, \end{aligned}$$
(24)
where
$$\begin{aligned} Q_{1}=\frac{\beta y^{*}(\nu )}{\lambda +d+\beta y^{*}(\nu )}\beta x^{*}(\nu )e^{-\int _{0}^{\nu \tilde{\sigma }(y^{*}(\nu ))}\delta (s)ds}e^{-\lambda \nu \tilde{\sigma }(y^{*}(\nu ))} \end{aligned}$$
and
$$\begin{aligned} Q_{2}=\nu \beta x^{*}(\nu )\delta (\nu \tilde{\sigma }(y^{*}(\nu )))y^{*}(\nu )\tilde{\sigma }'(y^{*}(\nu ))e^{-\int _{0}^{\nu \tilde{\sigma }(y^{*}(\nu ))}\delta (s)ds}. \end{aligned}$$
Now, let \(\lambda =a+ib\) be a root of \(\Delta (\lambda ,\nu )\) with \(a\ge 0\). Then \(\lambda \ne 0\) and \(\mid e^{-\lambda \nu \tilde{\sigma }(y^{*}(\nu ))}\mid \le 1\). Thus
$$\begin{aligned}&\mid Q_{1}+Q_{2}\mid = \beta x^{*}(\nu )e^{-\int _{0}^{\nu \tilde{\sigma }(y^{*}(\nu ))}\delta (s)ds}\mid \frac{\beta y^{*}(\nu )}{\lambda +d+\beta y^{*}(\nu )}e^{-\lambda \nu \tilde{\sigma }(y^{*}(\nu ))}\\&\qquad \qquad \qquad \qquad +\nu y^{*}(\nu )\delta (\nu \tilde{\sigma }(y^{*}(\nu )))\tilde{\sigma }'(y^{*}(\nu ))\mid \\&\qquad \le \beta x^{*}(\nu )y^{*}(\nu )e^{-\int _{0}^{\nu \tilde{\sigma }(y^{*}(\nu ))}\delta (s)ds}\mid \frac{\beta }{d+\beta y^{*}(\nu )}+\nu \delta (\nu \tilde{\sigma }(y^{*}(\nu )))\tilde{\sigma }'(y^{*}(\nu ))\mid . \end{aligned}$$
However
$$\begin{aligned} \frac{\beta }{d+\beta y^{*}(\nu )}+\delta (\nu \tilde{\sigma }(y^{*}(\nu )))\nu \tilde{\sigma }'(y^{*}(\nu ))>0 \end{aligned}$$
and \(\tilde{\sigma }'(y^{*}(\nu ))<0\). Thus,
$$\begin{aligned} \mid Q_{1}\!+\!Q_{2}\mid&\le \beta x^{*}(\nu )e^{-\int _{0}^{\nu \tilde{\sigma }(y^{*}(\nu ))}\delta (s)ds}\left( \frac{\beta }{d\!+\!\beta y^{*}(\nu )}\!+\!\delta (\nu \tilde{\sigma }(y^{*}(\nu )))\nu \tilde{\sigma }'(y^{*}(\nu ))\right) y^{*}(\nu )\\&\le \beta x^{*}(\nu )e^{-\int _{0}^{\mu \tilde{\sigma }(y^{*}(\nu ))}\delta (s)ds}. \end{aligned}$$
Moreover, \(\mu =\beta x^{*}(\nu )e^{-\int _{0}^{\nu \tilde{\sigma }(y^{*}(\nu ))}\delta (s)ds}\) and \(\mid \lambda +\mu \mid >\mu \). Thus,
$$\begin{aligned} \begin{array}{llll} &{}\beta x^{*}(\nu )e^{-\int _{0}^{\nu \tilde{\sigma }(y^{*}(\nu ))}\delta (s)ds}\mid 1+e^{-\lambda \nu \tilde{\sigma }(y^{*}(\nu ))}\mid \\ &{}\quad \le \mid \mu +\beta x^{*}(\nu )e^{-\int _{0}^{\nu \tilde{\sigma }(y*(\nu ))}\delta (s)ds}e^{-\lambda \nu \tilde{\sigma }(y^{*}(\nu ))}\mid \\ &{}\quad \le \mid \lambda +\mu +\beta p\gamma x^{*}(\nu )e^{-\int _{0}^{\nu \tilde{\sigma }(y^{*}(\nu ))}\delta (s)ds}e^{-\lambda \nu \tilde{\sigma }(y^{*}(\nu ))}\mid \\ &{}\quad = \mid Q_{1}+Q_{2}\mid \\ &{}\quad \le \beta x^{*}(\nu )e^{-\int _{0}^{\nu \tilde{\sigma }(y^{*}(\nu ))}\delta (s)ds}. \end{array} \end{aligned}$$
Consequently,
$$\begin{aligned} \mid 1+e^{-\lambda \nu \tilde{\sigma }(y^{*}(\mu ))}\mid \le 1, \end{aligned}$$
which is false since \(\lambda =0\) is not a root of \(\Delta (\lambda ,\nu )\). Therefore, every root has negative real part and the local asymptotic stability of the positive steady state immediately follows \(\square \)
Appendix F: Proof of Theorem 10
Proof
We will prove that the equilibrium \(\mathcal {E}_{M}(\nu )=\left( x_{M}(\nu ),y_{M}(\nu )\right) \) is locally asymptotically stable and \(\mathcal {E}_{m}(\nu )=\left( x_{m}(\nu ),y_{m}(\nu )\right) \) is unstable when \(R_{0}<1\) or equivalently \(\nu >\nu _{0}\). First of all, we prove that \(\lambda =0\) is not a root of \(\Delta (\lambda ,\nu )=0\) for \(\nu \) close to \(\overline{\nu .}\) Using the above definition of \(\overline{\alpha }(\nu )\) and \(\overline{\beta }(\nu ,0)\),
$$\begin{aligned} \Delta (0,\nu )=\mu -\beta x_{M}(\nu )e^{-\int _{0}^{\nu \tilde{\sigma }(y_{M}(\nu ))}\delta (s)ds}+Q(\nu ), \end{aligned}$$
where
$$\begin{aligned} \begin{array}{lll} Q(\nu ) = \beta e^{-\int _{0}^{\nu \tilde{\sigma }(y{}_{M}(\nu ))}\delta (s)ds}\frac{\beta x_{M}(\nu )y_{M}(\nu )}{d+\beta y_{M}(\nu )}\\ \qquad \qquad +\nu \beta y_{M}(\nu )x_{M}(\nu )e^{-\int _{0}^{\nu \tilde{\sigma }(y_{M}(\nu ))}\delta (s)ds}\delta (\nu \tilde{\sigma }(y_{M}(\nu )))\tilde{\sigma }'(y_{M}(\nu ))\\ \quad \;\; \;\;\;= \beta e^{-\int _{0}^{\nu \tilde{\sigma }(y_{M}(\nu ))}\delta (s)ds}x_{M}(\nu )y_{M}(\nu )\left( \frac{\beta }{d+\beta y_{M}(\nu )}+\nu \delta (\nu \tilde{\sigma }(y_{M}(\nu )))\tilde{\sigma }'(y_{M}(\nu ))\right) . \end{array} \end{aligned}$$
However, from (17), \(\frac{\beta }{d+\beta y_{M}(\nu )}+\nu \delta (\nu \tilde{\sigma }(y_{M}(\nu )))\tilde{\sigma }'(y_{M}(\nu ))>0\). Therefore, \(Q(\nu )>0\). On the other hand, from (15),
$$\begin{aligned} \mu -\beta x_{M}(\nu )e^{-\int _{0}^{\nu \tilde{\sigma }(y_{M}(\nu ))}\delta (s)ds}=\mu -\frac{\pi \beta e^{-\int _{0}^{\nu \tilde{\sigma }(y_{M}(\nu ))}\delta (s)ds}}{d+\beta y_{M}(\nu )}=0. \end{aligned}$$
Thus, \(\Delta (0,\nu )>0,\) which prove that \(\lambda =0\) is not a root of \(\Delta (\lambda ,\nu )=0\) for \(\nu \) close to \(\nu _{0},\)
Therefore, \(\Delta (\lambda ,\nu )=0\) is equivalent to
$$\begin{aligned} \lambda +\mu +\beta x_{M}(\nu )e^{-\int _{0}^{\nu \tilde{\sigma }(y_{M}(\nu ))}\delta (s)ds}e^{-\lambda \nu \tilde{\sigma }(y_{M}(\nu ))}=Q_{1}+Q_{2}, \end{aligned}$$
(25)
where
$$\begin{aligned} Q_{1}=\frac{\beta y_{M}(\nu )}{\lambda +d+\beta y_{M}(\nu )}\beta x_{M}(\nu )e^{-\int _{0}^{\nu \tilde{\sigma }(y_{M}(\nu ))}\delta (s)ds}e^{-\lambda \nu \tilde{\sigma }(y_{M}(\nu ))} \end{aligned}$$
and
$$\begin{aligned} Q_{2}=\nu \beta x_{M}(\nu )\delta (\nu \tilde{\sigma }(y_{M}(\nu )))y_{M}(\nu )\tilde{\sigma }'(y_{M}(\nu ))e^{-\int _{0}^{\nu \tilde{\sigma }(y_{M}(\nu ))}\delta (s)ds}. \end{aligned}$$
Now, let \(\lambda =a+ib\) be a root of \(\Delta (\lambda ,\nu )\) with \(a\ge 0\). Then \(\lambda \ne 0\) and \(\mid e^{-\lambda \nu \tilde{\sigma }(z_{M}(\nu ))}\mid \le 1\). Thus,
$$\begin{aligned} \mid Q_{1}+Q_{2}\mid&= \beta x_{M}(\nu )e^{-\int _{0}^{\nu \tilde{\sigma }(y_{M}(\nu ))}\delta (s)ds}\mid \frac{\beta y_{M}(\nu )}{\lambda +d+\beta y_{M}(\nu )}e^{-\lambda \nu \tilde{\sigma }(y_{M}(\nu ))}\\&+\nu y_{M}(\nu )\delta (\nu \tilde{\sigma }(y_{M}(\nu )))\mid \\&\le \beta x_{M}(\nu )y_{M}(\nu )e^{-\int _{0}^{\nu \tilde{\sigma }(y_{M}(\nu ))}\delta (s)ds}\mid \frac{\beta }{d+\beta y_{M}(\nu )}\\&+\delta (\nu \tilde{\sigma }(y_{M}(\nu )))\nu \tilde{\sigma }'(y_{M}(\nu ))\mid . \end{aligned}$$
However \(\frac{\beta }{d+\beta y_{M}(\nu )}+\delta (\nu \tilde{\sigma }(y_{M}(\nu )))\nu \tilde{\sigma }'(y_{M}(\nu ))>0\) and \(\tilde{\sigma }'(y_{M}(\nu ))<0\). Thus,
$$\begin{aligned}&\mid Q_{1}+Q_{2}\mid \le \beta x_{M}(\nu )e^{-\int _{0}^{\nu \tilde{\sigma }(y_{M}(\nu ))}\delta (s)ds}\left( \frac{\beta }{d+\beta y_{M}(\nu )}+\delta (\nu \tilde{\sigma }(y_{M}(\nu )))\nu \tilde{\sigma }'(y_{M}(\nu ))\right) y_{M}(\nu )\\&\quad \le \beta x_{M}(\nu )e^{-\int _{0}^{\mu \tilde{\sigma }(y_{M}(\nu ))}\delta (s)ds}. \end{aligned}$$
Moreover, \(\mu =\beta x_{M}(\nu )e^{-\int _{0}^{\nu \tilde{\sigma }(y_{M}(\nu ))}\delta (s)ds}\) and \(\mid \lambda +\mu \mid >\mu \). Thus,
$$\begin{aligned} \begin{array}{llll} &{}\beta x_{M}(\nu )e^{-\int _{0}^{\nu \tilde{\sigma }(y_{M}(\nu ))}\delta (s)ds}\mid 1+e^{-\lambda \nu \tilde{\sigma }(y_{M}(\nu ))}\mid \\ &{}\quad \le \mid \mu +\beta x_{M}(\nu )e^{-\int _{0}^{\nu \tilde{\sigma }(y_{M}(\nu ))}\delta (s)ds}e^{-\lambda \nu \tilde{\sigma }(y_{M}(\nu ))}\mid \\ &{}\quad \le \mid \lambda +\mu +\beta x_{M}(\nu )e^{-\int _{0}^{\nu \tilde{\sigma }(y_{M}(\nu ))}\delta (s)ds}e^{-\lambda \nu \tilde{\sigma }(y_{M}(\nu ))}\mid \\ &{}\quad = \mid Q_{1}+Q_{2}\mid \\ &{}\quad \le \beta x_{M}(\nu )e^{-\int _{0}^{\nu \tilde{\sigma }(y_{M}(\nu ))}\delta (s)ds}. \end{array} \end{aligned}$$
Consequently,
$$\begin{aligned} \mid 1+e^{-\lambda \nu \tilde{\sigma }(y_{M}(\mu ))}\mid \le 1, \end{aligned}$$
which is false since \(\lambda =0\) is not a root of \(\Delta (\lambda ,\nu )\). Therefore, every root has negative real part and the endemic equilibrium \(\mathcal {E}_{M}\) is locally asymptotically stable.
The characterstic equation associated to \(\mathcal {E}_{m}(\nu )=\left( x_{m}(\nu ),y_{m}(\nu )\right) \) satisfies
$$\begin{aligned} \Delta (0,\nu )=\beta e^{-\int _{0}^{\nu \tilde{\sigma }(y_{m}(\nu ))}\delta (s)ds}x_{m}(\nu )y_{m}(\nu )\left( \frac{\beta }{d+\beta y_{m}(\nu )}+\nu \delta (\nu \tilde{\sigma }(y_{m}(\nu )))\tilde{\sigma }'(y_{m}(\nu ))\right) . \end{aligned}$$
However, from (17), \(\frac{\beta }{d+\beta y_{m}(\nu )}+\nu \delta (\nu \tilde{\sigma }y_{m}(\nu )))\tilde{\sigma }'(y_{m}(\nu ))<0\). Therefore, \(\Delta (0,\nu )<0\). Thus, there exists \(\lambda ^{*}>0\) such that \(\Delta (\lambda ^{*},\nu )=0\). This concludes the proof of the theorem. \(\square \)