\documentclass[reqno]{amsart}
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\AtBeginDocument{{\noindent\small
\emph{Electronic Journal of Differential Equations},
Vol. 2017 (2017), No. 306, pp. 1--17.\newline
ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu}
\thanks{\copyright 2017 Texas State University.}
\vspace{8mm}}

\begin{document}
\title[\hfilneg EJDE-2017/306\hfil Stability and phase portraits]
{Stability and phase portraits of susceptible-infective-removed
epidemic models with vertical transmissions and linear treatment rates}

\author[M. Hoti, X. Huo, K. Lan \hfil EJDE-2017/306\hfilneg]
{Marvin Hoti, Xi Huo, Kunquan Lan}

\address{Marvin Hoti \newline
Faculty of Science,
University of Ontario, Institute of Technology,
Oshawa, Ontario, Canada L1H 7K4. \newline
Department of Mathematics, Ryerson University,
Toronto, Ontario, Canada M5B 2K3}
\email{marvin.hoti@ryerson.ca}

\address{Xi Huo \newline
Department of Mathematics, 
University of Miami,  
1365 Memorial Drive, 
Coral Gables, FL 33146, USA.\newline
Department of Mathematics, Ryerson University,
Toronto, Ontario, Canada M5B 2K3. \newline
Centre for Disease Modelling,
York University,
Toronto, Ontario, Canada M3J 1P3}
\email{huoxi@yorku.ca}

\address{Kunquan Lan \newline
Department of Mathematics, Ryerson University,
Toronto, Ontario, Canada M5B 2K3}
\email{klan@ryerson.ca}

\dedicatory{Communicated by Ratnasingham Shivaji}

\thanks{Submitted October 19, 2016. Published December 14, 2017.}
\subjclass[2010]{34C23, 92D25, 34D20, 34D23}
\keywords{SIR model; vertical transmission; treatment rate; stability;
\hfill\break\indent  node; focus; saddle-node}

\begin{abstract}
 We study stability and phase portraits of susceptible-infective-removed
 (SIR) epidemic models with horizontal and vertical transmission rates and
 linear treatment rates by studying the reduced dynamical planar systems
 under the assumption that the total population keeps unchanged.
 We find out all the ranges of the parameters involved in the models for
 the infection-free equilibrium and the epidemic equilibrium to be positive.
 The novelty of this paper lies in the demonstration and justification of
 the parameter conditions under which the positive equilibria are stable
 focuses or  nodes. These phase portraits provide more detailed descriptions
 of behaviors  and  extra biological understandings of the epidemic diseases
 than local or global stability of the models.
 Previous results  only discussed the stability of the SIR models with
 horizontal or vertical transmission rates and without treatment rates.
 Our results involving vertical transmission and treatment rates will
 exhibit the effect of the vertical transmissions and the linear treatment
 rates on the epidemic models.
\end{abstract}

\maketitle
\numberwithin{equation}{section}
\newtheorem{theorem}{Theorem}[section]
\newtheorem{lemma}[theorem]{Lemma}
\newtheorem{remark}[theorem]{Remark}
\allowdisplaybreaks

\section{Introduction}

The classic susceptible infectious recovered (SIR) models with vital dynamics 
(birth and death) (see Hethcote \cite{Het})
only deal with the horizontal transmission of epidemic diseases, that is, 
diseases are transmitted through contact
between the infectives and the susceptibles.
There are infectious diseases such as rubella, herpes simplex, hepatitis B,
chagas disease, and AIDS which can be transmitted
by horizontal transmission or by vertical transmission, i.e., 
the diseases are transmitted from infective parents  to unborn or newly
born offsprings\cite{bd96,BC2,LSW2}.
Treatments, including isolation, quarantine and hospitalization,
are important control measures to prevent epidemic diseases.
A variety of treatment rates including linear, constant or
saturated treatment rates
have been incorporated into some  epidemic disease models
\cite{jnk2016, lah2015ab, reavga2016, WRu, zhangliu08jmaa}.

Recently, Luo, Zhu and Lan \cite{lzl} generalized
the SIR models with horizontal and vertical transmissions  in \cite{MCh} by
incorporating constant treatment rates on the infectives.
They provided the conditions on the parameters involved and
justified that under these conditions, the equilibria are stable focuses, 
stable nodes, saddle-nodes or cusps with dimension 2, and studied
Bogdanov-Takens bifurcations containing saddle-node bifurcations, 
Hopf bifurcations and homoclinic bifurcations.


In this article, we consider the SIR models with horizontal and vertical 
transmissions by incorporating linear treatment rates on the infectives,
and study the stability and phase portraits of the models.
Some of models like the classic ratio-dependent predator-prey models 
\cite{hhk01siamjam, kb98jmb} have been incorporated with
linear harvesting rates on predator, see for example, the second equation of
 \cite[(1.1) p. 349]{heglan}, the second equation of $(3)$ in
\cite[p.1866]{rg12amm} and the second equation of 
\cite[(2.2) p. 4046]{Chak} and \cite{lramc2010}.
Similar to such predator-prey models, we shall exhibit the impact of the 
linear treatment rates on the SIR models by
seeking the ranges of the treatment rates which show the changes of
the dynamics of the models.

As in \cite{lzl,MCh}, we assume that the birth rate equals the death rate
and the recovered population never becomes susceptible. 
Under these assumptions, the SIR model we study in this paper is
governed by the following system of three first-order
ordinary differential equations
\begin{equation} \label{1.10}
\begin{gathered}
\dot S=b-\beta SI-bS-qbI, \\
\dot I=\beta SI-bI-rI+qbI-h I,\\
\dot R=rI-bR+hI,
\end{gathered}
\end{equation}
where $S(t)\ge 0$, $I(t)\ge0$ and $R(t)\ge 0$ denote the densities of the 
populations of the susceptible, infective and removed, respectively at time $t\ge 0$;
the constant $b>0$ in the first and last term of the right-side of the first 
equation denotes the birth rate  of susceptible population
and the coefficient $b>0$ in the terms $bS$, $bI$ and $bR$ denotes the death 
rate of the corresponding population, respectively
(note that the birth rate equals the death rate).
The parameter $\beta>0$ denotes the effective per capita transmission rate of 
infective individuals, and the term $\beta SI$ is the incidence rate,
which is bilinear, and
$r>0$ is the recovery rate of the infective individuals.
The parameter $q\in [0,1]$ is the fraction of unborn or newly born 
offsprings of the infective parents, and
$h\ge 0$ is  the proportionality of infective receiving treatments.

We emphasize that the three coefficients $b$, $r$ and $h$ in the terms 
$bI$, $rI$ and $hI$ in the second equation of \eqref{1.10}
can not be simply reduced to one term $(b+r+h)I$  with $b+r+h>0$ as one 
parameter. We must consider the ranges of $(b,r,h)$ in the first quadrant of
$\mathbb R^3$ instead of the ranges of one parameter $b+r+h$ in
$\mathbb R^{1}$. We refer to \cite{Chak, heglan, lramc2010, rg12amm} 
for the study of ratio-dependent predator-prey models with linear harvesting 
rates, where the death rate and harvesting rate of predator cannot be 
combined into one parameter.
We refer to \cite{dzl16jnfa,zhulan2014} and the references therein for the 
study of epidemic diseases with nonlinear  incidence rates.


The stability of the model \eqref{1.10} with
$q\in (0,1)$ and $h=0$ was  studied in \cite{MCh}.
It was shown that when the basic reproductive rate $R_0>1$
the model has a unique positive infection-free unstable equilibrium,
and one positive interior (epidemic) locally stable equilibrium; and
when $R_0<1$,  the infection-free equilibrium
is locally stable and  the interior equilibrium is unstable. But
neither the case $R_0=1$ nor the phase portraits near the
positive equilibria of \eqref{1.10} with $h=0$ was studied in \cite{MCh}. 
Our results will fill in the gap. It is well known that the phase portraits near the
positive equilibria provide
detailed descriptions of behaviors  and  extra biological understandings of 
the epidemic diseases
 We prove that when the basic reproduction number $\mathcal{R}_0\le 1$,
the model \eqref{1.10} with
$q\in (0,1)$ and $h>0$ has a unique
disease-free equilibrium $(1,0)$, and when $\mathcal{R}_0>1$,
the model has both a disease-free equilibrium $(1,0)$ and an interior 
(epidemic) equilibrium
$(\bar{x},\bar{y})$. (The symbols used in the Introduction will be given later).
We show that  when
$\mathcal{R}_0>1$, $(1,0)$ is a saddle, when
$\mathcal{R}_0<1$,  $(1,0)$ is a stable node  and when
$\mathcal{R}_0=1$, it is a saddle-node and is stable in the triangle region
$\{(u,v)\in \mathbb R_{+}: u+v\le 1\}$.
For the equilibrium $(\bar{x},\bar{y})$, we provide sufficient conditions on 
the parameters involved and prove that under these conditions,  the
equilibria are stable focuses or stable nodes. The latter results are new and
their proofs  follow from several lemmas.
Some simulations on our results will be provided
to  understand the phase portraits of the infection-free equilibrium and the positive
equilibrium.


\section{Positive equilibria}

Since the birth rate equals the death rate,
the total population keeps unchanged and can be normalized to $1$.
Hence, we have $S(t)+I(t)+R(t)=1$ for $t\ge 0$ and
\eqref{1.10} is equivalent to the following system
\begin{equation} \label{1.10-1}
\begin{gathered}
\dot S=b-\beta SI-bS-qbI, \\
\dot I=\beta SI-bI-rI+qbI-h I.
\end{gathered}
\end{equation}
As mentioned in the Introduction, the three parameters $b,r,h$ in the 
second equation of \eqref{1.10-1} can not be combined into one parameter 
since they have their own biological meanings.
For simplification of symbols,  we let $x(t)=S(t)$ and $y(t)=I(t)$ for 
$t\ge 0$. Using $x(t)$ and $y(t)$, we rewrite \eqref{1.10-1} as follows
\begin{equation}\label{1.1}
\begin{gathered}
\dot x = b-\beta xy-bx-qby:= f(x,y),\\
\dot y = \beta xy-by-ry+qby-hy:= g(x,y),
\end{gathered}
\end{equation}
where $x(t)$ and $y(t)$ denote the densities of the populations of the 
susceptible and infective, respectively at time $t\ge 0$.

Recall that $(x, y)\in \mathbb R^2$ is an equilibrium of \eqref{1.1}
if $f(x,y)=0$ and $g(x,y)=0$.
An equilibrium  $(x,y)$ of \eqref{1.1} is said to be positive if $x \ge 0$ 
and $y\ge 0$, and a
positive interior (endemic) equilibrium if $x>0$ and $y>0$.

\subsection*{Notation}  Let
$$
\eta:=\eta(r,b,q)=r+(1-q)b\quad\text{and $q_{1}=(b+r-\beta)/b$.}
$$
We denote by $\mathcal{R}_0$ the basic reproduction number,
representing the average number of cases
that are induced by one infective individual, of the model \eqref{1.1} as
\begin{equation}
\label{brnexpression}
\mathcal{R}_0=\beta/(\eta+h),
\end{equation}
where $1/(\eta+h)$ is the average time for an infected individual staying 
in the infectious class.

We shall use the conditions: $\mathcal{R}_0>1, =1$ or $< 1$, so the following
two lemmas provide the ranges of the parameters $b, r, \beta, q$ under which 
the above conditions hold and give better understanding on $\mathcal{R}_0$.
The proofs are straightforward and
we omit them.

\begin{lemma} \label{betaeta}
\begin{itemize}
\item[(1)] $\beta<\eta$ if and only if either $r\le \beta<b+r$ and 
$0\le q<q_{1}$ or $\beta<r$ and $0\le q\le 1$.

\item[(2)] $\eta<\beta$ if and only if $b+r<\beta$ and $0\le q\le 1$ or 
$r<\beta\le b+r$ and $q_{1}<q\le 1$.

\item[(3)] $\beta\le \eta$ if and only if either $b+r<\beta$ and 
$0\le q\le 1$ or $r\le \beta\le b+r$ and $q_{1}\le q\le 1$.

\item[(4)] $\beta\le \eta<\eta+b$ if and only if either 
$b+r<\beta<2b+r-qb$ and $0\le q\le 1$ or $r\le \beta\le b+r$ and 
$q_{1}\le q\le 1$.

\item[(5)] $\beta=\eta$ if and only if $r\le \beta\le b+r$ and $q=q_{1}$.
\end{itemize}
\end{lemma}

\begin{lemma} \label{brn}
\begin{itemize}
\item[(1)] $\mathcal{R}_0>1$ if and only if $\eta<\beta$ and $0\le h<\beta-\eta$.

\item[(2)] $\mathcal{R}_0<1$ if and only if either  $0<\beta<\eta$ and $h\ge 0$ or
 $\eta\le \beta$
and $h>\beta-\eta$.

\item[(3)] $\mathcal{R}_0=1$ if and only if  $\eta\le \beta$ and $h=\beta-\eta$.
\end{itemize}
\end{lemma}

We prove the following main result on the number of equilibria of \eqref{1.1}.

\begin{theorem} \label{main1}
\begin{itemize}
\item[(1)]
If $\mathcal{R}_0\le 1$, then $(1,0)$ is the unique positive equilibrium of \eqref{1.1}.

\item[(2)]
If $\mathcal{R}_0>1$, then \eqref{1.1} has only two positive  equilibria: $(1,0)$ and $(\bar{x}, \bar{y})$, where
\begin{equation}
\label{ps}
\bar{x}=\frac{\eta+h}{\beta} \quad\text{and}\quad
\bar{y}=\frac{b(\beta-\eta-h)}{\beta(b+r+h)}.
\end{equation}
\end{itemize}
\end{theorem}

\begin{proof}
It is clear that $(x,y)$ is an equilibrium  of \eqref{1.1} if and only if $(x,y)$ 
satisfies  the system
\begin{equation} \label{yley0}
\begin{gathered}
b-\beta xy-bx-qby=0,\\
\beta xy-by-ry+qby- hy=0.
\end{gathered}
\end{equation}
For $b, r, \beta>0$, $q\in [0,1]$ and $h\ge 0$, 
it is clear that $(1,0)$ is a solution of \eqref{yley0} and is  a  positive
equilibrium of \eqref{1.1}.
It is easy to see that \eqref{yley0} with $y\ne 0$ is equivalent to the 
 system
\begin{equation}\label{simp1}
\begin{gathered}
b-\beta xy-bx-qby=0,\\
\beta x-r-b-h+qb=0.
\end{gathered}
\end{equation}
Solving the second equation of \eqref{simp1} we obtain
\begin{equation} \label{x}
x=\frac{r+(1-q)b+h}{\beta}=\frac{\eta+h}{\beta}.
\end{equation}
 This, together with the first equation of \eqref{simp1}, implies
\begin{equation} \label{y}
y=\frac{b[\beta+qb-(r+b+h)]}{\beta(r+b+h)}
=\frac{b(\beta-\eta-h)}{\beta(r+b+h)}.
\end{equation}
If $\mathcal{R}_0>1$, then by Lemma \ref{brn} (1) we have
 $\eta<\beta$ and $0\le h<\beta-\eta$. This, together with
\eqref{y}, implies $y>0$. Hence,  $(\bar{x}, \bar{y})$ given in \eqref{ps} is
a positive interior equilibrium of \eqref{1.1}.
If $\mathcal{R}_0<1$, then by Lemma \ref{brn} (2) we have
 either  $0<\beta<\eta$ and $h\ge 0$ or
 $\eta\le \beta$
and $h>\beta-\eta$. This implies that $\beta-\eta-h<0$ and $y<0$.
Hence, \eqref{1.1} has no positive interior equilibria.
If $\mathcal{R}_0=1$, then by Lemma \ref{brn} (3) we have
 $\eta\le \beta$ and $h=\beta-\eta$. This, together with
\eqref{x} and \eqref{y}, implies $y=0$ and $x=1$.
The results follow.
\end{proof}

Theorem \ref{main1} improves \cite[Theorem 6.1]{Het} and the result on 
the number of positive equilibria obtained in \cite[section 2]{MCh}.

\section{Stability and phase portraits of the model}

In this section, we study the stability and phase portraits of each positive 
equilibrium of \eqref{1.1}.
We  recall some results  on stability and phase portraits of planar systems
 near equilibria in the qualitative theory \cite{AI, Per, Halbook}.
We consider the following  planar system
\begin{equation} \label{geneq}
\begin{gathered}
\dot x(t)=f(x(t),y(t)),\\
\dot y(t)=g(x(t),y(t))
\end{gathered}
\end{equation}
subject to the initial value condition:
\begin{equation} \label{bc}
(x(0),y(0))=(x_0,y_0)
 \end{equation}
where $f,g\in C^{1}(\mathbb R^2)$.

Recall that $(x,y)$ is said to be a solution of \eqref{geneq}-\eqref{bc} 
if $x,y\in C^{1}(\mathbb R_{+})$ and satisfy
\eqref{geneq}-\eqref{bc}. A solution $(x,y)$ is said to be positive if $x,y\in P$,
where
$$
P=\{x\in C^{1}(R_{+}): x(t)\ge 0\quad\text{for } t\in R_{+}\}.
$$
Since $f,g\in C^{1}(\mathbb R^2)$, it is well known that for each initial value
$(x_0,y_0)\in \mathbb R^2$, \eqref{geneq}-\eqref{bc}
has a unique solution. Moreover, if $f$ and $g$ satisfy
$$
f(0,y)\ge 0 \quad\text{and $g(x,0)\ge 0$ for $x,y\in \mathbb R_{+}$},
$$
then for each initial value $(x_0,y_0)\in \mathbb R_{+}^2$, the unique solution
of \eqref{geneq}-\eqref{bc} is positive (see \cite[Proposition B.7]{Halbook}).

We denote by $A(x,y)$ the Jacobian matrix of $f$ and $g$ at $(x,y)$, that is,
\begin{equation} \label{matrix}
A(x,y)
= \begin{pmatrix}
\frac{\partial f}{\partial x}&\frac{\partial f}{\partial y}
\\
\frac{\partial g}{\partial x}&\frac{\partial g}{\partial y}
\end{pmatrix}
\end{equation}
  and by  $|A(x,y)|$ and  $\operatorname{tr}(A(x,y))$ the determinant and trace of $A(x,y)$,
respectively.

The following results can be found in \cite{Per} and have been used in 
\cite{heglan,lanzhu2,llz, lzl,zhulan09}.

\begin{lemma} \label{gen}
If $(x^{*},y^{*})$ is an equilibrium of \eqref{geneq}, then the following 
assertions hold.
\begin{itemize}
\item[(i)] If $|A(x^{*},y^{*})|<0$, then $(x^{*},y^{*})$ is  a 
saddle of \eqref{geneq}.

\item[(ii)] If $|A(x^{*},y^{*})|>0$, $\operatorname{tr}(A(x^{*},y^{*}))<0$ and
$(\operatorname{tr}(A(x^{*},y^{*})))^2-4|A(x^{*},y^{*})|\ge 0$ then
$(x^{*},y^{*})$ is a stable node of \eqref{geneq}.

\item[(iii)] If $|A(x^{*},y^{*})|>0$, $\operatorname{tr}(A(x^{*},y^{*}))<0$ and
$(\operatorname{tr}(A(x^{*},y^{*})))^2-4|A(x^{*},y^{*})| < 0$ then $(x^{*},y^{*})$
is a stable focus of \eqref{geneq}.

\item[(iv)] If $|A(x^{*},y^{*})|>0$ and $\operatorname{tr}(A(x^{*},y^{*}))<0$,
then $(x^{*},y^{*})$ is  locally asymptotically stable.
\end{itemize}
\end{lemma}

Recall that an equilibrium $(x^{*},y^{*})$ is said to be globally asymptotically 
stable if it is locally asymptotically stable and each solution $(x,y)$ 
of \eqref{geneq}-\eqref{bc} with $(x_0,y_0)\in \mathbb R_{+}^2$
converges to $(x^{*},y^{*})$ in $\mathbb R^2$;
 that is, $\lim_{t\to \infty}(x(t),y(t))=(x^{*},y^{*})$.

The following result is a special case of the well-known Poincar\'{e}-Bendixson 
Trichotomy theorem, see \cite[Theorem 8.8 and Lemma 8.9]{Teschl}.

\begin{lemma} \label{PB}
Assume that each positive solution of \eqref{geneq}-\eqref{bc} with 
$(x_0,y_0)\in \mathbb R_{+}^2$ is contained in a bounded closed subset $B$
of $\mathbb R^2$.
Assume that $B$ contains only one equilibrium $(x^{*},y^{*})$ of 
\eqref{geneq} and $(x^{*},y^{*})$ belongs to the boundary of $B$.
Then each positive solution of \eqref{geneq}-\eqref{bc} converges to 
$(x^{*},y^{*})$.
\end{lemma}

Recall that a map $T:\mathbb R^2 \to \mathbb R^2$ defined by
$T(x,y)=(f(x,y), g(x,y))$ is said to be regular if $T$ is  one to
one and onto, $T$ and $T^{-1}$ are continuous and  $|A(x,y)|\ne 0$ 
on $\mathbb R^2$.
If $T$ is  regular, then the following transformation
 \begin{equation} \label{trf1}
\begin{gathered}
 u=f(x,y), \\
 v=g(x,y)
\end{gathered}
\end{equation}
is said to be a regular transformation. If \eqref{geneq} is changed into
another system under suitable regular transformations, then the two systems 
are said to be equivalent.
It is known that under regular transformations, the topological structures 
of solutions of a planar system  near  equilibria including a variety of 
dynamics like  saddles, topological saddles, nodes, saddle-nodes, foci,  
centers, or cusps remain  unchanged.

\begin{lemma} \label{implicit1} \cite{lanzhu2}
Let $(x^{*},y^{*})$ be an equilibrium of \eqref{geneq}.
Assume that  $|A(x^{*},y^{*})|=0$, $\operatorname{tr}(A(x^{*},y^{*}))\ne 0$ and
\eqref{geneq} is equivalent to the following system
\begin{equation} \label{sadd-node111}
\begin{gathered}
\dot u=p(u,v), \\
\dot v=\varrho v+q(u,v)
\end{gathered}
\end{equation}
with an isolated equilibrium point $(0,0)$, where $\varrho\ne 0$,
$p(u,v)=\sum_{i+j=2,i,j\ge 0}^{\infty}a_{ij}u^{i}v^{j}$ and
$q(u,v)=\sum_{i+j=2,i,j\ge 0}^{\infty}b_{ij}u^{i}v^{j}$ are convergent
power series.
If $a_{20}\ne 0$, then  $(x^{*},y^{*})$ is a saddle-node of \eqref{geneq}.
\end{lemma}

Now, we begin to study the stability and phase portraits near each of the 
positive equilibrium  of \eqref{1.1}.

Let $A(x,y)$ be the Jacobian matrix of $f$ and $g$ defined in \eqref{1.1}. 
By \eqref{1.1}  and \eqref{matrix}, we have
\begin{equation*} % \label{jm1}
A(x,y)
= \begin{pmatrix}
-\beta y-b & -\beta x-qb \\
\beta y &  \beta x-\eta-h
\end{pmatrix}.
\end{equation*}
Note that $\eta=r+(1-q)b$, we have
\begin{equation} \label{det}
|A(x,y)|=\beta y(b+r+h)-b\beta x+b(\eta+h)
\end{equation}
and
\begin{equation}
\label{tr}
\operatorname{tr}(A(x,y))=-\beta(y-x)-(b+\eta+h).
\end{equation}
We first prove the following result on the global stability and phase 
portraits near the infection-free equilibrium $(1,0)$  of \eqref{1.1}.

\begin{theorem} \label{st1}
\begin{itemize}
\item[(1)] If $\mathcal{R}_0>1$, then $(1,0)$ is a saddle of \eqref{1.1}.

\item[(2)] If $\mathcal{R}_0<1$, then $(1,0)$ is a stable node of \eqref{1.1}. 
Moreover, the infection-free equilibrium $(1,0)$  of \eqref{1.1} is
 globally asymptotically stable.

\item[(3)] If $\mathcal{R}_0=1$,
then $(1,0)$ is a saddle-node of \eqref{1.1}.
\end{itemize}
\end{theorem}

\begin{proof}
By \eqref{det} and \eqref{tr}  with $(x,y)=(1,0)$, we have
\begin{gather} \label{det(1,0)}
|A(1,0)|=b(h-\beta+\eta), \\
\label{tr(1,0)}
\operatorname{tr}(A(1,0))=\beta-\eta-h-b.
\end{gather}

(1) Since $\mathcal{R}_0>1$,
by \eqref{det(1,0)} and Lemma \ref{brn} (1), we have $|A(1,0)|<0$. 
The result follows from Lemma \ref{gen} (i).

(2) Since $\mathcal{R}_0<1$, by Lemma \ref{brn} (2), \eqref{det} and \eqref{tr}  
we obtain $|A(1,0)|>0$ and $\operatorname{tr}(A(1,0))<0$.
Moreover, we have
\begin{align*}
\operatorname{tr}(A(1,0))^2-4|A(1,0)|
&=(\beta-\eta-h-b)^2-4b(h-\beta+\eta)\\
&=(\beta-\eta-h)^2-2b(\beta-\eta-h)+b^2+4b(\beta-h-\eta)\\
&=(\beta-\eta-h)^2+2b(\beta-\eta-h)+b^2\\
&=(\beta-\eta-h+b)^2\ge 0.
\end{align*}
The first result follows from Lemma \ref{gen} (ii).

Let $B=\{(u,v)\in \mathbb R_{+}^2: u+v\le 1\}$, which is a positive
invariant set of \eqref{1.1}. $B$ is a bounded closed
subset of $\mathbb R^2$ and contains only the equilibrium $(1,0)$ of \eqref{1.1}.
Since $(1,0)$ is on the boundary of $B$, it follows from
Lemma \ref{PB} that every positive solution of \eqref{1.1} converges to
$(1,0)$ as $t\to \infty$. Hence, $(1,0)$  is globally asymptotically stable.

(3) Since $\mathcal{R}_0=1$, by Lemma \ref{brn} (3), \eqref{det} 
and \eqref{tr}, we have
$|A(1,0)|=0$ and $\operatorname{tr}(A(1,0))<0$.
We change the equilibrium $(1,0)$ to the origin $(0,0)$ by the change of variables 
$u_{1}=x-1$ and $v_{1}=y$. Note that $h=\beta-\eta$.
 Then system \eqref{1.1} becomes
\begin{gather*} %\label{equi00}
\dot u_{1}=\dot x=b-\beta (u_{1}+1)v_{1}-b(u_{1}+1)-qbv_{1}
=-\beta u_{1}v_{1}-(\beta+qb)v_{1}-bu_{1}, \\
\dot v_{1}=\dot y=\beta (u_{1}+1)v_{1}-(\eta+h)v_{1}
=\beta u_{1}v_{1}-(\eta+h-\beta)v_{1}=\beta u_{1}v_{1}.
\end{gather*}
Let $\xi=(\beta+qb)b^{-1}$, $u_2=u_{1}+\xi v$ and $v_2=v_{1}$.
Then the last system becomes
\begin{align*}
\dot u_2
&=\dot u_{1}+\xi \dot v_{1}=-\beta u_{1}v_{1}-(\beta+qb)v_{1}-bu_{1}
+\xi\beta u_{1}v_{1}\\
&=(\xi-1)\beta u_{1}v_{1}-(\beta+qb)v_{1}-bu_{1}\\
&=(\xi-1)\beta\bigl[u_{2}v_{2}-\xi v_{2}^2\bigr]
 -(\beta+qb)v_{2}-b(u_{2}-\xi v_{2})\\
&=(\xi-1)\beta u_{2}v_{2}-\xi(\xi-1)\beta v_{2}^2-bu_{2}
\end{align*}
and
$\dot v_{2}=\beta v_{2}[u_{2}-\xi v_{2}]=-\xi b v_{2}^2+\beta u_{2} v_{2}$.

Let $u=v_{2}$ and $v=u_{2}$. Then the above last two equations become
\begin{gather*}
\dot u=-\xi b u^2+\beta u v,\\
\dot v=-bv+(\xi-1)\beta uv-\xi(\xi-1)\beta u^2.
\end{gather*}
Since $\varrho:=-b\ne 0$ and $a_{20}:=-\xi b\ne 0$, it follows from Lemma 
\ref{implicit1} that $(1,0)$ is a saddle-node of \eqref{1.1}.
\end{proof}

\begin{remark} \label{rmk3.1} \rm
When $\mathcal{R}_0<1$, Theorem \ref{st1} (2) shows that the infection-
free equilibrium $(1,0)$   is a stable node and is  globally asymptotically stable.
By Lemma \ref{brn} (2), we see that the biological interpretation of
 Theorem \ref{st1} (2) is that if $0<\beta<\eta$ with any
treatment rate $h\ge 0$ or $\eta\le \beta$ and
the treatment rate $h>\beta-\eta$, then the epidemic disease will be eradicated 
and the epidemic can not maintain itself
(see Figure \ref{fig1} (a) as an example).

Theorem \ref{st1} (3) is new and  from its proof, we see that $(1,0)$ is 
a saddle-node in its neighborhood, so it is unstable.
But since we only consider the biologically meaningful solutions in the 
triangle $B=\{(u,v)\in \mathbb R_{+}^2: u+v\le 1\}$, from
the Figure 1 $(b)$ below,
we see that all the positive solutions in $B$ converge to
$(1,0)$ and $(1,0)$ is stable in $B$. This conclusion is consistent with the 
result in \cite[Theorem 6.1]{Het} which falls into
the special case of $q=h=0$.
Results in Theorem \ref{st1} (1) and (2) were obtained in \cite[section 2]{MCh} 
which corresponds to the special condition of $h=0$ in our model.
\end{remark}

\begin{figure}
\begin{center}
 \includegraphics[height=5cm,width=6cm]{fig1a} % thm31a.eps
\quad
 \includegraphics[height=5cm,width=6cm ]{fig1b} % thm31b.eps
\end{center}
\caption{ (a) shows that $(1,0)$ is globally asymptotically stable, where  
$b=0.5$, $r=2$, $\beta=2$, $q=0.5$, $h=1$ and
$\mathcal{R}_0=\beta/(b+r-qr+h)=2/3.25<1$.  (b) shows that all the positive 
solutions in $B$ converge to $(1,0)$, where $b,r,q,h$ are same as in 
(a) and  $\beta=3.25$, so $\mathcal{R}_0=1$.}
\label{fig1} % \label{diseasefree1}
\end{figure}

Now, we turn our attention to the positive endemic equilibrium $(\bar x,\bar y)$ 
given in \eqref{ps} of \eqref{1.1}.
We first prove that $(\bar x,\bar y)$  is locally asymptotically stable under 
suitable conditions.

\begin{theorem} \label{thm3.4}
If $\mathcal{R}_0>1$,  then $(\bar x,\bar y)$ is locally asymptotically stable.
\end{theorem}

\begin{proof}
Since  $\mathcal{R}_0>1$, it follows from
Theorem \ref{main1} (2) that $(\bar x,\bar y)$ given in \eqref{ps} is 
well defined. By  \eqref{det} and \eqref{tr} with $(x,y)=(\bar x, \bar y)$, we have
\begin{gather} \label{det(x,y)}
|A(\bar x,\bar y)|=\beta \bar y(b+r+h)+b(\eta+h)-\beta b\bar x
=\beta \bar{y}(b+r+h),\\
\label{tr(x,y)}
\operatorname{tr}(A(\bar x,\bar y))=-\beta (\bar y-\bar x)-(b+h+\eta)
=-(\beta \bar y +b)\,.
\end{gather}
Since $\bar y>0$ and $b,r,\beta>0$, by \eqref{det(x,y)} and \ref{tr(x,y)}, 
we have $|A(\bar x,\bar y)|>0$ and
$\operatorname{tr}(A(\bar x,\bar y))<0$. The result follows from 
Lemma \ref{gen} (iv).
\end{proof}

\begin{remark} \label{rmk3.2} \rm
By Lemma \ref{brn} (1), we see that $\mathcal{R}_0>1$ if and only if  
$\eta<\beta$ and $0\le h<\beta-\eta$. Hence,
Theorem \ref{thm3.4} generalizes the result in \cite[section 2]{MCh} from  $h=0$ to
$h\in [0,\beta-\eta)$. The biological interpretation of Theorem \ref{thm3.4} 
is that if $\eta<\beta$, then the epidemic can not be eradicated if the treatment
 rate $h$ is smaller than  $\beta-\eta$.
\end{remark}

It is not easy to determine if $(\bar x,\bar y)$ is a stable node or stable focus. 
Our main goal in the rest of this paper is to
 find sufficient conditions on the parameters $b,r,\beta, q, h$
under which $(\bar x,\bar y)$ is a stable node or stable focus.
All the results obtained below are new even when $q=0$ or $h=0$.

Our first result shows that  $(\bar x,\bar y)$ could be a stable node or 
stable focus for sufficiently small $h$.
To do that, we first prove the following lemmas.

Let
$$
\beta_1:=\frac{2(b+r)^{3/2}}{\sqrt{b+r}+\sqrt{r}}\quad\text{and}\quad
\beta_2:=\frac{2(b+r)^{3/2}}{\sqrt{b+r}-\sqrt{r}}.
$$
The following simple result will be useful in the proof of Lemma \ref{ratio}.
Its proof is straightforward and we omit it.

\begin{lemma} \label{function}
Let $h(x)=16x(1+x)^3-1$ for $x\in \mathbb R_{+}$.
Then the following assertions hold.
\begin{itemize}
\item[(i)]  The equation $h(x)=0$ has a unique solution $\gamma_{1}\in (0.05, 0.055)$.

\item[(ii)]  $h(x)<0$ for  $x\in [0,\gamma_{1})$ and $h(x)>0$ for   
$x\in (\gamma_{1}, \infty)$.
\end{itemize}
\end{lemma}

\begin{lemma} \label{ratio}
\textrm{(1)} If $b>0$, $r>0$, then
\begin{equation} \label{ineqin0}
r<\beta_{1}-b<b+r<\beta_2-b
\end{equation}
and
\begin{itemize}
\item[(i)] If $r<\beta<\beta_{1}-b$, then 
$0<q_1<1<\frac{\beta_{1}-\beta}{b}<\frac{\beta_{2}-\beta}{b}$;

\item[(ii)] If $\beta_{1}-b<\beta\le b+r$, then 
$0<q_1<\frac{\beta_{1}-\beta}{b}<1$ and $\frac{\beta_{2}-\beta}{b}>1$.
\end{itemize}

\noindent\textrm{(2)} If $b>0$ and $r>\gamma_1 b$, then
\begin{equation} \label{ineqin}
r<\beta_{1}-b<b+r<\beta_1<\beta_2-b<\beta_{2}
\end{equation}
and the following assertions hold:
\begin{itemize}
\item[(i)] If $b+r<\beta\le \beta_1$, then $0\le \frac{\beta_{1}-\beta}{b}<1$ 
and $\frac{\beta_{2}-\beta}{b}>1$;

\item[(ii)] If $\beta_{1}<\beta<\beta_2-b$, then $\frac{\beta_{1}-\beta}{b}<0$ 
and $\frac{\beta_{2}-\beta}{b}>1$;

\item[(iii)] If $\beta_2-b<\beta<\beta_2$, then 
$\frac{\beta_{1}-\beta}{b}<0$ and $0<\frac{\beta_{2}-\beta}{b}<1$.
\end{itemize}

\noindent\textrm{(3)} If $0<r<\gamma_1 b$, then
\begin{equation} \label{ineqin2}
r<\beta_1-b<b+r<\beta_2-b<\beta_1<\beta_{2}
\end{equation}
and the following assertions hold:
\begin{itemize}
\item[(i)] If $b+r<\beta<\beta_2-b$, then $0<\frac{\beta_{1}-\beta}{b}<1$ 
and $\frac{\beta_{2}-\beta}{b}>1$;

\item[(ii)] If $\beta_2-b<\beta\le \beta_1$, then 
$0\le \frac{\beta_{1}-\beta}{b}<1$ and $0<\frac{\beta_{2}-\beta}{b}<1$;

\item[(iii)] If $\beta_1<\beta<\beta_2$, then $\frac{\beta_{1}-\beta}{b}<0$ 
and $0<\frac{\beta_{2}-\beta}{b}<1$.
\end{itemize}

\noindent\textrm{(4)} If $r=\gamma_1 b$, then
\begin{equation} \label{ineqin3}
r<\beta_1-b<b+r<\beta_2-b=\beta_1<\beta_{2}
\end{equation}
and  the following assertions hold:
\begin{itemize}
\item[(i)] If $b+r<\beta<\beta_2-b$, then $0<\frac{\beta_{1}-\beta}{b}<1$ and 
$\frac{\beta_{2}-\beta}{b}>1$;

\item[(ii)] If $\beta_1<\beta<\beta_2$, then $\frac{\beta_{1}-\beta}{b}<0$ and 
$0<\frac{\beta_{2}-\beta}{b}<1$.
\end{itemize}
\end{lemma}

\begin{proof}
(1) Let $b>0$ and $r>0$. Since
\[ %\label{b1br}
\beta_1
=\frac{2(b+r)^{3/2}}{\sqrt{b+r}+\sqrt{r}}
=\frac{2(b+r)}{1+\sqrt{\frac{r}{b+r}}}>\frac{2(b+r)}{1+1}
=b+r,
\]
it follows that  $r<\beta_1-b$.
Since
\begin{align*}
\beta_1-(2b+r)&=\frac{2(b+r)}{1+\sqrt{\frac{r}{b+r}}}-(2b+r)=\frac{2(b+r)-(2b+r)\bigl[1+\sqrt{\frac{r}{b+r}}\bigr]}{1+\sqrt{\frac{r}{b+r}}}\\
&=\frac{r-(2b+r)\sqrt{\frac{r}{b+r}}}{1+\sqrt{\frac{r}{b+r}}}=\frac{r^2-(2b+r)^2\frac{r}{b+r}}{\bigl[1+\sqrt{\frac{r}{b+r}}\bigr]
\bigl[r+(2b+r)\sqrt{\frac{r}{b+r}}\bigr]}\\
&=\frac{r^2(b+r)-r(2b+r)^2}{(b+r)\bigl[1+\sqrt{\frac{r}{b+r}}\bigr]\bigl[r+(2b+r)\sqrt{\frac{r}{b+r}}\bigr]}\\
&=-\frac{rb(4b+3r)}{(b+r)\bigl[1+\sqrt{\frac{r}{b+r}}\bigr]\bigl[r+(2b+r)\sqrt{\frac{r}{b+r}}\bigr]}<0,
\end{align*}
we have $\beta_1-b<b+r$.
Since
\begin{align*}
\beta_2-(2b+r)
&=\frac{2(b+r)}{1-\sqrt{\frac{r}{b+r}}}-(2b+r)=\frac{2(b+r)-(2b+r)
 \bigl[1-\sqrt{\frac{r}{b+r}}\bigr]}{1-\sqrt{\frac{r}{b+r}}}\\
&=\frac{r+(2b+r)\sqrt{\frac{r}{b+r}}}{1-\sqrt{\frac{r}{b+r}}}>0,
\end{align*}
we have $b+r<\beta_{2}-b$.  Hence, \eqref{ineqin0} holds. 
Note that $q_{1}=(b+r-\beta)b^{-1}$.
By \eqref{ineqin0}, it is steadily verified that the results (i)-(ii) hold.


Let $b>0$ and $r>0$. By definition of $\beta_1$ and $\beta_2$ we have
\begin{equation} \label{ineq1}
\begin{aligned}
\beta_2-b-\beta_1
&=\frac{2(b+r)^{3/2}}{\sqrt{b+r}-\sqrt{r}}
-\frac{2(b+r)^{3/2}}{\sqrt{b+r}+\sqrt{r}}-b
=\frac{4\sqrt{r}(b+r)^{3/2}}{b}-b \\
&=\frac{4\sqrt{r}(b+r)^{3/2}-b^2}{b}
=\frac{16r(b+r)^3-b^{4}}{b[4\sqrt{r}(b+r)^{3/2}+b^2]}
=\frac{b^3\bigl[16\frac{r}{b}(1+\frac{r}{b})^3-1\bigr]}{4\sqrt{r}(b+r)^{3/2}+b^2}
 \\
&=\frac{b^3h(\frac{r}{b})}{4\sqrt{r}(b+r)^{3/2}+b^2}
\end{aligned}
\end{equation}

(2) If $b>0$ and $r>\gamma_1 b$, then $\frac{r}{b}>\gamma_{1}$ and by 
Lemma \ref{function}, $h(\frac{r}{b})>0$. It follows from
\eqref{ineq1} that $\beta_1<\beta_2-b$.
Since $\beta_1>b+r$, we have $r<\beta_{1}-b$.
Hence, \eqref{ineqin} holds. By \eqref{ineqin}, it is steadily verified 
that the results (i)-(iii) hold.

(3) If $0<r<\gamma_1 b$,  then $\frac{r}{b}<\gamma_{1}$ and by Lemma \ref{function}, 
$h(\frac{r}{b})<0$. It follows from
\eqref{ineq1} that $\beta_2-b<\beta_1$. It is obvious that $\beta_{1}<\beta_2$. Since
$$
\beta_2=\frac{2(b+r)^{3/2}}{\sqrt{b+r}-\sqrt{r}}
=\frac{2(b+r)}{1-\sqrt{\frac{r}{b+r}}}>\frac{2(b+r)}{1-0}>2b+r,
$$
we obtain  $b+r<\beta_2-b$. It has been proved in $(1)$ that $\beta_1-b<b+r$. 
Hence, \eqref{ineqin2} holds.
By \eqref{ineqin2}, it is steadily verified that the results (i)-(iii) hold.

(4) If $r=\gamma_1 b$, then $\beta_1=\beta_2-b$. Hence,
\eqref{ineqin3} holds and $(i)$ and $(ii)$ hold.
\end{proof}

Let 
\[
(\bar x,\bar y)=
\Bigl(\frac{\eta+h}{\beta}, \frac{b(\beta-\eta-h)}{\beta(b+r+h)}\Bigr)
\]
 be same as in \eqref{ps} and let
\begin{equation} \label{delta(h)}
\Delta(q,h)=\operatorname{tr}(A(\bar x,\bar y))^2-4|A(\bar x,\bar y)|.
\end{equation}

\begin{lemma} \label{deltahlem}
If $\mathcal{R}_0>1$, then
\begin{equation}
\label{deltah}
\Delta (q,h)
=\frac{b}{(b+r+h)^2}\Bigl[b(\beta+qb)^2-4(b+r+h)^2(\beta+qb)+4(b+r+h)^3\Bigr].
\end{equation}
\end{lemma}

\begin{proof}
Noting that $\eta=b+r-qb$, we have
\begin{equation*}%\label{betaybarb}
\beta\bar{y}+b=\frac{b(\beta+qb)}{b+r+h}\quad\text{and}\quad
4\beta\bar{y}(b+r+h)=4b[(\beta+qb)-(b+r+h)].
\end{equation*}
This, \eqref{det(x,y)} and \eqref{tr(x,y)} imply
\begin{align*}
\Delta(q,h)
&:=\operatorname{tr}(A(\bar x,\bar y))^2-4|A(\bar x,\bar y)|
 =(\beta\bar{y}+b)^2-4\beta\bar{y}(b+r+h)\\
& =\frac{b^2(\beta+qb)^2}{(b+r+h)^2}-4b[(\beta+qb)-(b+r+h)]\\
&=\frac{b}{(b+r+h)^2}\Bigl[b(\beta+qb)^2-4(b+r+h)^2(\beta+qb)+4(b+r+h)^3\Bigr].
\end{align*}
The result follows.
\end{proof}

For the next theorem we use the following conditions
\begin{itemize}
\item[(H1)] $b>0$, $r>0$, $\beta_{1}-b<\beta\le b+r$ and 
$\frac{\beta_{1}-\beta}{b}< q\le 1$.

\item[(H2)] $b>0$, $r>\gamma_1 b$ and one of the following conditions holds:
\begin{itemize}
\item[(i)] $b+r<\beta\le \beta_{1}$ and $\frac{\beta_{1}-\beta}{b}< q\le 1$;

\item[(ii)]  $\beta_{1}<\beta<\beta_2-b$ and $0\le q\le 1$;

\item[(iii)]  $\beta=\beta_2-b$ and $0\le q<1$;

\item[(iv)] $\beta_{2}-b<\beta<\beta_{2}$ and $0\le q<\frac{\beta_{2}-\beta}{b}$.
\end{itemize}

\item[(H3)]  $b>0$, $0<r<\gamma_1 b$ and one of the following conditions holds:
\begin{itemize}
\item[(i)] $b+r<\beta<\beta_2-b$ and $\frac{\beta_{1}-\beta}{b}< q\le 1$;

\item[(ii)] $\beta=\beta_2-b$ and $\frac{\beta_{1}-\beta}{b}< q<1$;

\item[(iii)] $\beta_2-b<\beta\le \beta_1$ and 
 $\frac{\beta_{1}-\beta}{b}< q<\frac{\beta_{2}-\beta}{b}$;

\item[(iv)] $\beta_1<\beta<\beta_2$ and $0\le q<\frac{\beta_{2}-\beta}{b}$.
\end{itemize}

\item[(H4)]  $b>0$, $r=\gamma_1 b$ and one of the following conditions holds:
\begin{itemize}
\item[(i)] $b+r<\beta<\beta_2-b$ and $\frac{\beta_{1}-\beta}{b}< q\le 1$;

\item[(ii)] $\beta=\beta_2-b$ and $\frac{\beta_{1}-\beta}{b}< q<1$.

\item[(iii)] $\beta_{2}-b<\beta<\beta_{2}$ and $0\le q<\frac{\beta_{2}-\beta}{b}$.
\end{itemize}
\end{itemize}

\begin{theorem} \label{main22}
\textrm{(1)} Assume that one of the  conditions {\rm (H1)--(H4)} holds.
Then there exists $h_0\in (0,\beta-\eta)$ such that
$(\bar{x},\bar{y})$ is a stable focus of \eqref{1.1} for $h\in [0,h_0)$.

\textrm{(2)} Assume that $b>0$, $r>0$ and one of the following conditions holds:
\begin{itemize}
\item[(i)] $r<\beta<\beta_1-b$ and $q_{1}<q\le 1$.

\item[(ii)] $\beta=\beta_1-b$ and $q_{1}<q<1$.

\item[(iii)]  $\beta_2<\beta<\infty$ and $0\le q\le 1$.

\item[(iv)]  $\beta_2=\beta$ and $0<q\le 1$.
\end{itemize}
Then there exists $h_{1}\in (0,\beta-\eta)$ such that
$(\bar{x},\bar{y})$ is a stable node of \eqref{1.1} for $h\in [0,h_{1})$.
\end{theorem}

\begin{proof}
Under  condition (H1), by (ii) of Lemma \ref{ratio} (1), 
$r<\beta\le b+r$ and $q_{1}<q\le 1$. This, together with
 Lemma \ref{betaeta} (2), implies $\eta<\beta$. Similarly, by 
Lemma \ref{ratio} (2), (3), (4),
 each of the hypotheses in (H2)--(H4) implies
$b+r<\beta$ and $0\le q\le 1$.
 By Lemma \ref{betaeta} (2), we obtain $\eta<\beta$.

Let $0\le h<\beta-\eta$.  By \eqref{deltah}, we have
\begin{equation} \label{delta2}
\Delta(q,h) =\frac{b^2\Gamma(h)}{(b+r+h)^2},
\end{equation}
where
\begin{equation} \label{gammah}
\Gamma(q,h)=(\beta+qb)^2-\frac{4(b+r+h)^2(\beta+qb)}{b}+\frac{4(b+r+h)^3}{b}.
\end{equation}
We prove that
\begin{equation}
\label{gamma0}
\Gamma(q,0)=b^2\Bigl(q-\frac{\beta_1-\beta}{b}\Bigr)
\Bigl(q-\frac{\beta_2-\beta}{b}\Bigr).
\end{equation}
Indeed, by \eqref{gammah} we have
\begin{align*}
&\Gamma(q,0) \\
&=(\beta+qb)^2-\frac{4(b+r)^2(\beta+qb)}{b}+\frac{4(b+r)^3}{b}\\
&=\Bigl[\beta+qb-\frac{2(b+r)^2}{b}\Bigr]^2-\frac{4r(b+r)^3}{b^2}\\
&=\Bigl[\beta+qb-\frac{2(b+r)^2}{b}+\frac{2\sqrt{r}(b+r)^{3/2}}{b}\Bigr]
\Bigl[\beta+qb-\frac{2(b+r)^2}{b}-\frac{2\sqrt{r}(b+r)^{3/2}}{b}\Bigr]\\
&=\Bigl[\beta+qb-\frac{2(\sqrt{b+r}-\sqrt{r})(b+r)^{3/2}}{b}\Bigr]
\Bigl[\beta+qb-\frac{2(\sqrt{b+r}+\sqrt{r})(b+r)^{3/2}}{b}\Bigr]\\
&=\Bigl[\beta+qb-\frac{2b(b+r)^{3/2}}{b(\sqrt{b+r}+\sqrt{r})}\Bigr]
\Bigl[\beta+qb-\frac{2b(b+r)^{3/2}}{b(\sqrt{b+r}-\sqrt{r})}\Bigr]\\
&=(\beta+qb-\beta_1)(\beta+qb-\beta_2)
\end{align*}
and \eqref{gamma0} holds.

We prove that under each of the conditions in (H1)--(H4),
\begin{equation}
\label{abcineq}
\frac{\beta_1-\beta}{b}<q<\frac{\beta_2-\beta}{b}.
\end{equation}

(H1) If $b>0$, $r>0$, $\beta_{1}-b<\beta<b+r$ and 
$\frac{\beta_{1}-\beta}{b}< q\le 1$.
then by (ii) of Lemma \ref{ratio} (1),
we have
$$
\frac{\beta_{1}-\beta}{b}<q\le 1<\frac{\beta_{2}-\beta}{b}.
$$

(H2) (i)  If $b+r<\beta<\beta_{1}$ and $\frac{\beta_{1}-\beta}{b}\le q\le 1$, 
then by (i) of Lemma \ref{ratio} (2),
$$
\frac{\beta_{1}-\beta}{b}<q\le 1<\frac{\beta_{2}-\beta}{b}.
$$

(ii)  If  $\beta_{1}<\beta<\beta_2-b$ and $0\le q\le 1$, then  by (ii) of 
Lemma \ref{ratio} (2),
$$
\frac{\beta_{1}-\beta}{b}<0\le q\le 1<\frac{\beta_{2}-\beta}{b}.
$$

(iii)  If  $\beta_{2}-b<\beta<\beta_{2}$ and $0\le q<\frac{\beta_{2}-\beta}{b}$, 
then  by (iii) of Lemma \ref{ratio} (2),
$$
\frac{\beta_{1}-\beta}{b}<0\le q<\frac{\beta_{2}-\beta}{b}.
$$
Hence, under each of the conditions (i), (ii) and (iii) in (H2), 
\eqref{abcineq} holds.
Similarly, \eqref{abcineq} holds under each of the conditions in (H3) or (H4).
By \eqref{gamma0} and \eqref{abcineq}, we see that $\Gamma(q,0)<0$. 
It follows from the continuity of $\Gamma$ that
there exists $h_0\in (0,\beta-\eta)$ such that $\Gamma(h)<0$ for $h\in [0,h_0)$, 
and by \eqref{delta2},   $\Delta(q,h)<0$ for $h\in [0,h_0)$. 
By Theorem \ref{thm3.4} (1),
\eqref{det(x,y)} and \eqref{tr(x,y)} we see that for $h\in [0,h_0)$,
$|A(\bar x,\bar y)|>0$ and $tr(\bar x,\bar y)<0$. The result follows 
from Lemma \ref{gen} (iii).

(2) (i) If $r<\beta<\beta_1-b$ and $q_{1}<q\le 1$, then
by (i) of Lemma \ref{ratio} (1), we have
\begin{equation} \label{1}
1<\frac{\beta_{1}-\beta}{b}<\frac{\beta_{2}-\beta}{b}.
\end{equation}
(ii) If $\beta=\beta_1-b$ and $q_{1}<q<1$, then
\begin{equation} \label{2}
q_{1}<q<1=\frac{\beta_{1}-\beta}{b}<\frac{\beta_{2}-\beta}{b}.
\end{equation}
(iii) If $\beta_2<\beta<\infty$ and $0\le q\le 1$, then 
\begin{equation} \label{3}
\frac{\beta_{1}-\beta}{b}<\frac{\beta_{2}-\beta}{b}<0.
\end{equation}
By \eqref{gamma0} and each of \eqref{1}, \eqref{2} and \eqref{3}, we have 
$\Gamma(0)>0$.
It follows from the continuity of $\Gamma$ that
there exists $h_{1}\in (0,\beta-\eta)$ such that $\Gamma(h)>0$ for $h\in [0,h_{1})$. 
It follows from \eqref{delta2}
that  $\Delta(h)>0$ for $h\in [0,h_{1})$. The result follows from Lemma 
\ref{gen} (ii).
\end{proof}

Simulation results for Theorem \ref{main22} (H2)(ii) and (2)(iii) are 
given in figure \ref{fig2}.

\begin{figure}[thb]
\begin{center}
 \includegraphics[height=5cm,width=6cm ]{fig2a} % thm33a.eps
\quad
 \includegraphics[height=5cm,width=6cm ]{fig2b} % thm33b.eps
\end{center}
\caption{In (a) and (b), we use $b=20$, $r=\frac{20}{3}>\gamma_1 b$ 
$(> \frac{1}{20} 20 = 1)$,  $q=0.25$ and $h=0$.
Since $\beta_1=\frac{320}{9}$, $\beta_2-b=\frac{260}{3}$ and 
$\beta_2=\frac{320}{3}$,
conditions $(H_2)(ii)$ with $\beta = 60$  and $(2)(iii)$ with $\beta = 110$ hold.}
\label{fig2} % \label{nonzero} 
\end{figure}

By Theorem \ref{main22} with $h=q=0$, we see that if $\beta_1<\beta<\beta_2$, 
then $(\bar{x},\bar{y})$ is a stable focus of \eqref{1.1} with $h=q=0$
and if $\beta>\beta_2$, then $(\bar{x},\bar{y})$ is a stable node  of 
\eqref{1.1} with $h=q=0$.
Theorem \ref{main22} with $h=q=0$ is inconclusive if $b+r<\beta \le \beta_1$ 
or $\beta=\beta_2$.

Using formulas \eqref{delta2}  and \eqref{gamma0},
we can provide a direct proof to the following new result on the classic
 model  \eqref{1.1} with $h=q=0$.

\begin{theorem}
\begin{itemize}
\item[(1)]
If $\beta_1<\beta<\beta_2$, then $(\bar{x},\bar{y})$ is a stable focus 
of \eqref{1.1} with $h=q=0$.

\item[(2)]
If either $b+r<\beta \le \beta_1$ or $\beta \ge \beta_2$, 
then $(\bar{x},\bar{y})$ is a stable node  of \eqref{1.1} with $h=q=0$.
\end{itemize}
\end{theorem}

\begin{proof}
By \eqref{delta2}  and \eqref{gamma0}, we have
\begin{equation*}
\Delta(0,0)=\frac{b^2}{(b+r)^2}(\beta-\beta_1)(\beta-\beta_2).
\end{equation*}

(1) If $\beta_1<\beta<\beta_2$, then $\Delta(0,0)<0$. 
By Lemma \ref{gen} (iii), the result (1) holds.
 
(2) If either $b+r<\beta \le \beta_1$ or $\beta \ge \beta_2$,
 then $\Delta(0,0)\ge 0$ and the result (2)
 follows from Lemma \ref{gen} (ii).
\end{proof}

Theorem \ref{main22} shows that the interior equilibrium $(\bar{x},\bar{y})$ 
of \eqref{1.1} can be a stable focus or a stable node
for sufficiently small $h$. However, Theorem \ref{main22} does not provide 
any upper bounds for $h$.
 Hence, the question is that under which range of $h$, can the interior 
equilibrium  $(\bar{x},\bar{y})$  be a stable focus or a stable node?

In the following, we enhance the result (iii) of Theorem \ref{main22} (2)
 and partially answer the above question. We provide a
range of $h$ under which
 $(\bar x,\bar y)$ is a stable node of \eqref{1.1}. 
To do this, we first prove the following lemma.
Let
$$
\beta_0=\frac{4(b+r)^2}{b}\quad\text{and}\quad
h_{1}=\frac{\sqrt{b(\beta+qb)}}{2}-b-r.
$$

\begin{lemma} \label{h1}
\textrm{(1)} $\beta_2<\beta_0-b$.

\textrm(2) If $\beta_0-b \leq \beta\le \infty$ and 
$\max\{0,\frac{\beta_0-\beta}{b}\}\le q\le 1$, then
$0\le h_{1}<\beta-\eta$.
\end{lemma}

\begin{proof}
(1) Since
$$
\frac{2(b+r)}{b}=\frac{1}{1+\sqrt{\frac{r}{b+r}}}+\frac{1}{1-\sqrt{\frac{r}{b+r}}},
$$
we have 
\begin{align*}
\beta_0-\beta_{2}-b
&=\frac{4(b+r)^2}{b}-\frac{2(b+r)}{1-\sqrt{\frac{r}{b+r}}}-b
=\frac{2(b+r)}{1+\sqrt{\frac{r}{b+r}}}-b\\
&=\frac{2(b+r)-b\bigl(1+\sqrt{\frac{r}{b+r}}\bigr)}{1+\sqrt{\frac{r}{b+r}}}
=\frac{b\bigl(1-\sqrt{\frac{r}{b+r}}\bigr)+2r}{1+\sqrt{\frac{r}{b+r}}}>0.
\end{align*}

(2) We first prove that under the given hypotheses,
\begin{equation} \label{bqb}
b-4\beta-4qb<0.
\end{equation}
In fact, if $\frac{\beta_0-\beta}{b}\ge 0$, then
 $q\ge \frac{\beta_0-\beta}{b}$ and
$$
b-4\beta-4qb\le b-4\beta-4(\beta_0-\beta)=b-4\beta_0
=\frac{b^2-4(b+r)^2}{b}<0.
$$
If $\frac{\beta_0-\beta}{b}<0$, then $q\ge 0$ and 
$$
b-4\beta-4qb\le b-4\beta\le b-4(\beta_0-b)=
\frac{5b^2-16(b+r)^2}{b}<-11b<0.
$$
Next, we prove that $h_{1}<\beta-\eta$. Indeed, since
\begin{align*} %\label{h1h2}
h_{1}-(\beta-\eta)
&=\frac{\sqrt{b(\beta+qb)}-2(b+r)}{2}-(\beta-b-r+qb) \\
&=\frac{\sqrt{b(\beta+qb)}}{2}-(\beta+qb) \\
&=\frac{(\beta+qb)[b-4\beta-4qb)]}{2\bigl[\sqrt{b(\beta+qb)}+4(\beta+qb)\bigr]}.
\end{align*}
This, together with \eqref{bqb}, implies $h_{1}<\beta-\eta$.
Finally, we prove that $h_{1}\ge 0$. In fact, since
\begin{align*}
h_{1}
&=\frac{\sqrt{b(\beta+qb)}-2(b+r)}{2}=\frac{b(\beta+qb)
 -4(b+r)^2}{2[\sqrt{b(\beta+qb)}+2(b+r)]}\\
&=\frac{b(\beta+qb)-4(b+r)^2}{2[\sqrt{b(\beta+qb)}+2(b+r)]}
 =\frac{b[(\beta+qb)-\beta_0]}{2[\sqrt{b(\beta+qb)}+2(b+r)]}\\
&=\frac{b^2\Bigl(q-\frac{\beta_0-\beta}{b}\Bigr)}{2[\sqrt{b(\beta+qb)}+2(b+r)]},
\end{align*}
it follows from $q\ge \max\{0,\frac{\beta_0-\beta}{b}\}$ that 
$q-\frac{\beta_0-\beta}{b}\ge 0$ and $h_{1}\ge 0$.
\end{proof}

\begin{theorem} \label{thm3.4i}
If $\beta_0-b \leq \beta\le \infty$, $\max\{0,\frac{\beta_0-\beta}{b}\}\le q\le 1$ 
and $0\le h\le h_{1}$, then $(\bar x,\bar y)$ is a stable node of \eqref{1.1}.
\end{theorem}

\begin{proof}
By Lemma \ref{ratio}, we see that $b+r<\beta_{2}$ and by Lemma \ref{h1}, 
$b+r<\beta_2<\beta_0-b\le \beta$.
Hence, by Lemma \ref{betaeta} (ii), Lemma \ref{brn} and 
Theorem \ref{main1} (2) we see that
for $\max\{0,\frac{\beta_0-\beta}{b}\}\le q\le 1$, $(\bar x,\bar y)$ given 
in \eqref{ps} is well defined.
It is easy to verify that
if $h\le h_{1}$, then $2(b+r+h)\le \sqrt{b(\beta+qb)}$ and
\begin{equation*} %\label{hheq}
b(\beta+qb)-4(b+r+h)^2\ge 0.
\end{equation*}
This, together with \eqref{deltah}, implies
$$
\Delta (q,h)=\frac{b}{(b+r+h)^2}\Bigl\{(\beta+qb)\big[b(\beta+qb)
-4(b+r+h)^2\bigr]+4(b+r+h)^3\Bigr\}>0.
$$
The result follows from Lemma \ref{gen} (ii).
\end{proof}

Theorem \ref{thm3.4i} provides a range for $h$ under which $(\bar x,\bar y)$ 
is a stable node of \eqref{1.1}, see Figure 3 below for a simulation result.
By Lemma \ref{h1} (1), we see that under the hypothesis of 
Theorem \ref{thm3.4i}: $\beta_0-b \le \beta\le \infty$, we have
 $\beta_2<\beta_0-b\le \beta$. Hence, Theorem \ref{thm3.4i} strengthens 
the result $(iii)$ of Theorem \ref{main22} (2)
 which holds for sufficiently small $h$.

\begin{figure} \label{thm35}
\begin{center}
 \includegraphics[height=5cm,width=6cm ]{fig3} % thm35.eps
\end{center}
\caption{$b=r=1$, $\beta=24.5$, $q=0.5$ and $h=0.25$. 
Since $\beta_0=16$, $\beta_0-b=15$  and $h_1=\frac{1}{2}$,
 the conditions of Theorem \ref{thm3.4i} are satisfied.}
\label{rig3}
\end{figure}


\subsection*{Acknowledgments}
This research was supported in part by the Natural
Sciences and Engineering Research Council (NSERC) of Canada.

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\end{document}