In this section, we investigate the existence of equilibria of system (1.4).

In order to find the equilibria of system (1.4), we put

\{\begin{array}{l}s-{\mu}_{T}T(t)+rT(t)(1-\frac{T(t)+I(t)}{{T}_{max}})-kT(t)V(t)+\rho I(t)=0,\\ {k}^{\prime}T(t-\tau )V(t-\tau )+rI(t)(1-\frac{T(t)+I(t)}{{T}_{max}})-({\mu}_{I}+\rho )I(t)=0,\\ N{\mu}_{b}I(t)-{\mu}_{v}V(t)=0.\end{array}

(3.1)

Following the analysis in [12], we find that system (3.1) has always the uninfected equilibrium {E}_{0}=({T}_{0},0,0), where

{T}_{0}=\frac{{T}_{max}}{2r}(r-{\mu}_{T}+\sqrt{{(r-{\mu}_{T})}^{2}+\frac{4rs}{{T}_{max}}}).

We define the parameter {N}_{\mathrm{crit}} as

{N}_{\mathrm{crit}}=\frac{{\mu}_{V}}{{k}^{\prime}{\mu}_{I}{T}_{0}}(\frac{s}{{T}_{0}}+{\mu}_{I}-{\mu}_{T}+\rho )

and we also find that, if N>{N}_{\mathrm{crit}}, system (3.1) has a unique positive equilibrium, {E}^{\ast}({T}^{\ast},{I}^{\ast},{V}^{\ast}). If N=\frac{r{\mu}_{V}}{{k}^{\prime}{\mu}_{I}{T}_{max}}, system (3.1) has a unique positive equilibrium {E}^{\ast}({T}^{\ast},{I}^{\ast},{V}^{\ast}), where

\begin{array}{c}{T}^{\ast}=\frac{{T}_{max}}{2r}[2({\mu}_{I}+\rho )-r-{\mu}_{T}+\sqrt{{[r+{\mu}_{T}-2({\mu}_{I}+\rho )]}^{2}+4\rho (r-{\mu}_{I}-\rho )+\frac{4rs}{{T}_{max}}}],\hfill \\ {I}^{\ast}=\frac{{\mu}_{V}{V}^{\ast}}{N{\mu}_{I}},\phantom{\rule{2em}{0ex}}{V}^{\ast}=\frac{r-{\mu}_{I}-\rho}{k}.\hfill \end{array}

Next, we shall discuss the stability for the local asymptotic stability of the viral free equilibrium {E}_{0} and the infected equilibrium {E}^{\ast}.

To discuss the stability of system (1.4), let us consider the following coordinate transformation:

x(t)=T(t)-\overline{T},\phantom{\rule{2em}{0ex}}y(t)=I(t)-\overline{I},\phantom{\rule{2em}{0ex}}z(t)=V(t)-\overline{V},

where (\overline{T},\overline{I},\overline{V}) denotes any equilibrium of (1.4). So we see that the corresponding linearized system of (1.4) is of the form

\{\begin{array}{l}{D}^{\alpha}x(t)=x(t)(-{\mu}_{T}+r-\frac{2r\overline{T}+r\overline{I}}{{T}_{max}}-k\overline{V})+y(t)(\rho -\frac{r\overline{T}}{{T}_{max}})-k\overline{T}z(t),\\ {D}^{\alpha}y(t)={k}^{\prime}\overline{V}x(t-\tau )+{k}^{\prime}\overline{T}z(t-\tau )\\ \phantom{{D}^{\alpha}y(t)=}+y(t)(r-\frac{r\overline{T}+2r\overline{I}}{{T}_{max}}-({\mu}_{I}+\rho ))-\frac{r\overline{I}}{{T}_{max}}x(t),\\ {D}^{\alpha}z(t)=N{\mu}_{b}y(t)-{\mu}_{V}z(t).\end{array}

(3.2)

The characteristic equation of system (3.2) at (\overline{T},\overline{I},\overline{V}) is given by

\left|\begin{array}{ccc}\lambda -(-{\mu}_{T}+r-\frac{2r\overline{T}+r\overline{I}}{{T}_{max}}-k\overline{V})& -\rho +\frac{r\overline{T}}{{T}_{max}}& k\overline{T}\\ -{k}^{\prime}\overline{V}{e}^{-\lambda \tau}+\frac{r\overline{I}}{{T}_{max}}& \lambda -(r-({\mu}_{I}+\rho )-\frac{r\overline{T}+2r\overline{I}}{{T}_{max}})& -{k}^{\prime}\overline{T}{e}^{-\lambda \tau}\\ 0& -N{\mu}_{b}& \lambda +{\mu}_{V}\end{array}\right|=0.

For the local asymptotic stability of the viral free equilibrium {E}_{0}, we have the following result.

**Theorem 3.1** *If* N<{N}_{\mathrm{crit}}, *the uninfected state* {E}_{0}=({T}_{0},0,0) *is locally asymptotically stable for* \tau \ge 0.

*Proof* The associated transcendental characteristic equation at {E}_{0}=({T}_{0},0,0)=(\overline{T},\overline{I},\overline{V}) is given by

(\lambda +{\mu}_{T}-r+\frac{2r{T}_{0}}{{T}_{max}})((\lambda +{\mu}_{V})(\lambda -r+{\mu}_{I}+\rho +\frac{r{T}_{0}}{{T}_{max}})-N{\mu}_{b}{k}^{\prime}{T}_{0}{e}^{-\lambda \tau})=0.

Obviously, the above equation has the characteristic root

{\lambda}_{1}=r-{\mu}_{T}-\frac{2r{T}_{0}}{{T}_{max}}<0,

where {T}_{0}=\frac{{T}_{max}}{2r}(r-{\mu}_{T}+\sqrt{{(r-{\mu}_{T})}^{2}+\frac{4rs}{{T}_{max}}}).

Next, we consider the transcendental polynomial

{\lambda}^{2}+\lambda ({\mu}_{V}-r+{\mu}_{I}+\rho +\frac{r{T}_{0}}{{T}_{max}})+{\mu}_{I}{\mu}_{V}-r{\mu}_{V}+{\mu}_{V}\rho +\frac{r{T}_{0}{\mu}_{V}}{{T}_{max}}-N{\mu}_{b}{k}^{\prime}{T}_{0}{e}^{-\lambda \tau}=0.

For \tau =0, we get

{\lambda}^{2}+\lambda ({\mu}_{V}-r+{\mu}_{I}+\rho +\frac{r{T}_{0}}{{T}_{max}})+{\mu}_{I}{\mu}_{V}-r{\mu}_{V}+{\mu}_{V}\rho +\frac{r{T}_{0}{\mu}_{V}}{{T}_{max}}-N{\mu}_{b}{k}^{\prime}{T}_{0}=0.

Then we note that

{\mu}_{I}-r+\rho +\frac{r{T}_{0}}{{T}_{max}}=\frac{{\mu}_{b}}{{\mu}_{I}}(\frac{s}{{T}_{0}}+({\mu}_{I}-{\mu}_{T})+\rho )>0,

we easily see that

{\mu}_{V}({\mu}_{I}-r+\rho +\frac{r{T}_{0}}{{T}_{max}})={N}_{\mathrm{crit}}{\mu}_{b}{k}^{\prime}{T}_{0}.

We have

{\lambda}_{1,2}=\frac{-({\mu}_{V}-r+{\mu}_{I}+\rho +\frac{r{T}_{0}}{{T}_{max}})\pm \sqrt{{({\mu}_{V}-r+{\mu}_{I}+\rho +\frac{r{T}_{0}}{{T}_{max}})}^{2}-4({N}_{\mathrm{crit}}-N){\mu}_{b}{k}^{\prime}{T}_{0}}}{2},

if N<{N}_{\mathrm{crit}}, the characteristic roots have negative real parts for \tau =0.

For \tau \ne 0, we get

{\lambda}^{2}+\lambda ({\mu}_{V}-r+{\mu}_{I}+\rho +\frac{r{T}_{0}}{{T}_{max}})+{\mu}_{I}{\mu}_{V}-r{\mu}_{V}+{\mu}_{V}\rho +\frac{r{T}_{0}{\mu}_{V}}{{T}_{max}}-N{\mu}_{b}{k}^{\prime}{T}_{0}{e}^{-\lambda \tau}=0.

Assume that the above equation has roots \lambda =\omega (cos\frac{\beta \pi}{2}\pm isin\frac{\beta \pi}{2}), for \omega >0 and \tau >0; we get

\begin{array}{r}{\omega}^{2}{(cos\frac{\beta \pi}{2}\pm isin\frac{\beta \pi}{2})}^{2}+\omega (cos\frac{\beta \pi}{2}\pm isin\frac{\beta \pi}{2})({\mu}_{V}-r+{\mu}_{I}+\rho +\frac{r{T}_{0}}{{T}_{max}})\\ \phantom{\rule{1em}{0ex}}+{\mu}_{I}{\mu}_{V}-r{\mu}_{V}+{\mu}_{V}\rho +\frac{r{T}_{0}{\mu}_{V}}{{T}_{max}}-N{\mu}_{b}{k}^{\prime}{T}_{0}{e}^{-\tau \omega (cos\frac{\beta \pi}{2}\pm isin\frac{\beta \pi}{2})}=0.\end{array}

Separating the real and imaginary parts gives

\{\begin{array}{l}{\omega}^{2}({cos}^{2}\frac{\beta \pi}{2}-{sin}^{2}\frac{\beta \pi}{2})+\omega cos\frac{\beta \pi}{2}({\mu}_{V}-r+{\mu}_{I}+\rho +\frac{r{T}_{0}}{{T}_{max}})\\ \phantom{\rule{1em}{0ex}}+{\mu}_{I}{\mu}_{V}-r{\mu}_{V}+{\mu}_{V}\rho +\frac{r{T}_{0}{\mu}_{V}}{{T}_{max}}-N{\mu}_{b}{k}^{\prime}{T}_{0}{e}^{-\tau \omega cos\frac{\beta \pi}{2}}cos(\mp \tau \omega sin\frac{\beta \pi}{2})=0,\\ \pm 2{\omega}^{2}sin\frac{\beta \pi}{2}cos\frac{\beta \pi}{2}\pm \omega sin\frac{\beta \pi}{2}({\mu}_{V}-r+{\mu}_{I}+\rho +\frac{r{T}_{0}}{{T}_{max}})\\ \phantom{\rule{1em}{0ex}}-sin(\mp \tau \omega sin\frac{\beta \pi}{2})N{\mu}_{b}{k}^{\prime}{T}_{0}{e}^{-\tau \omega cos\frac{\beta \pi}{2}}=0.\end{array}

(3.3)

From the second equation of (3.3), we have

sin\frac{\beta \pi}{2}=0,

that is \frac{\beta \pi}{2}=k\pi, k=0,1,2,\dots .

For \frac{\beta \pi}{2}=k\pi, k=0,2,4,\dots , substituting into the first equation of (3.3), we have

\begin{array}{r}{\omega}^{2}+\omega ({\mu}_{V}-r+{\mu}_{I}+\rho +\frac{r{T}_{0}}{{T}_{max}})+{\mu}_{I}{\mu}_{V}-r{\mu}_{V}+{\mu}_{V}\rho +\frac{r{T}_{0}{\mu}_{V}}{{T}_{max}}\\ \phantom{\rule{1em}{0ex}}=N{\mu}_{b}{k}^{\prime}{T}_{0}{e}^{-\tau \omega}.\end{array}

(3.4)

For the parameter values given in Table 1, we take any N<{N}_{\mathrm{crit}}, the infected equilibrium {E}_{0}=(1,000,0,0), and we find that the above equation is unequal for \omega >0. Therefore, \beta \ge 2>\alpha.

According to Lemma 2.1, the uninfected equilibrium {E}^{\ast} is locally asymptotically stable. The proof is completed. □

**Remark 3.1** ([2])

The stability region of a system with fractional order \alpha \in (0,1) is always larger than that of a corresponding ordinary differential system. This means that a unstable equilibrium of an ordinary differential system may be stable in a fractional differential system.

Next, for the sake of convenience, at {E}^{\ast}=({T}^{\ast},{I}^{\ast},{V}^{\ast}), we define the following symbols:

\begin{array}{c}{M}_{1}={\mu}_{T}-r+\frac{2r{T}^{\ast}+r{I}^{\ast}}{{T}_{max}}+k{V}^{\ast},\phantom{\rule{2em}{0ex}}{M}_{2}=-r+{\mu}_{I}+\rho +\frac{r{T}^{\ast}+2r{I}^{\ast}}{{T}_{max}},\hfill \\ A={\mu}_{V}+{M}_{1}+{M}_{2},\phantom{\rule{2em}{0ex}}B={M}_{1}{\mu}_{V}+{M}_{2}{\mu}_{V}+{M}_{1}{M}_{2}-\frac{{r}^{2}{I}^{\ast}{T}^{\ast}}{{T}_{max}^{2}}+\frac{\rho r{I}^{\ast}}{{T}_{max}},\hfill \\ C=-N{\mu}_{b}{k}^{\prime}{T}^{\ast}+\frac{r{T}^{\ast}{k}^{\prime}{V}^{\ast}}{{T}_{max}}-\rho {k}^{\prime}{V}^{\ast},\hfill \\ D={M}_{1}{M}_{2}{\mu}_{V}-\frac{N{\mu}_{b}k{T}^{\ast}r{I}^{\ast}}{{T}_{max}}-\frac{{\mu}_{V}r{T}^{\ast}r{I}^{\ast}}{{T}_{max}^{2}}+\frac{{\mu}_{V}\rho r{I}^{\ast}}{{T}_{max}},\hfill \\ E={k}^{\prime}{V}^{\ast}N{\mu}_{b}k{T}^{\ast}-N{M}_{1}{\mu}_{b}{k}^{\prime}{T}^{\ast}-{\mu}_{V}\rho {k}^{\prime}{V}^{\ast}+\frac{{k}^{\prime}{V}^{\ast}{\mu}_{V}r{T}^{\ast}}{{T}_{max}}.\hfill \end{array}

Then the characteristic equation of the linear system is

{\lambda}^{3}+A{\lambda}^{2}+(B+C{e}^{-\lambda \tau})\lambda +D+E{e}^{-\lambda \tau}=0.

(3.5)

Using the results in [31], we get

D(\lambda )={\lambda}^{3}+A{\lambda}^{2}+(B+C)\lambda +D+E

and

{D}^{\prime}(\lambda )=3{\lambda}^{2}+2A\lambda +(B+C).

Denote

\begin{array}{rl}D(\lambda )& =-\left|\begin{array}{ccccc}1& A& B+C& D+E& 0\\ 0& 1& A& B+C& D+E\\ 3& 2A& B+C& 0& 0\\ 0& 3& 2A& B+C& 0\\ 0& 0& 3& 2A& B+C\end{array}\right|\\ =18A(B+C)(D+E)-4{A}^{3}(D+E)-27{(D+E)}^{2}-4{(B+C)}^{3}+{A}^{2}{(B+C)}^{2}.\end{array}

**Theorem 3.2** *Let* 1\pm \frac{C{\tau}^{2}}{2}>0, (1\pm \frac{C{\tau}^{2}}{2})(B\pm C-E\tau )>0, D+E\ge 0 *and* N>{N}_{\mathrm{crit}}, *then the infected equilibrium* {E}^{\ast} *is asymptotically stable for any time delay* \tau \ge 0 *if either*

(\mathrm{i})\phantom{\rule{1em}{0ex}}D(\lambda )>0,\phantom{\rule{2em}{0ex}}A>0,\phantom{\rule{2em}{0ex}}D+E>0,\phantom{\rule{2em}{0ex}}A(B+C)>D+E,

*or*

(\mathrm{ii})\phantom{\rule{1em}{0ex}}D(\lambda )<0,\phantom{\rule{2em}{0ex}}A\ge 0,\phantom{\rule{2em}{0ex}}B+C\ge 0,\phantom{\rule{2em}{0ex}}0.5<\alpha <2/3.

*Proof* According to (3.5).

For \tau =0, we have

{\lambda}^{3}+A{\lambda}^{2}+(B+C)\lambda +D+E=0.

Using the result in [31], the infected steady state {E}^{\ast} is asymptotically stable if the Routh-Hurwitz condition is satisfied, *i.e.*

(\mathrm{i})\phantom{\rule{1em}{0ex}}D(\lambda )>0,\phantom{\rule{2em}{0ex}}A>0,\phantom{\rule{2em}{0ex}}D+E>0,\phantom{\rule{2em}{0ex}}A(B+C)>D+E,

or

(\mathrm{ii})\phantom{\rule{1em}{0ex}}D(\lambda )<0,\phantom{\rule{2em}{0ex}}A\ge 0,\phantom{\rule{2em}{0ex}}B+C\ge 0,\phantom{\rule{2em}{0ex}}0.5<\alpha <2/3.

For \tau \ne 0, we get

{\lambda}^{3}+A{\lambda}^{2}+(B+C{e}^{-\lambda \tau})\lambda +D+E{e}^{-\lambda \tau}=0.

Assume that the above equation has roots \lambda =\omega (cos\frac{\beta \pi}{2}\pm isin\frac{\beta \pi}{2}), for \omega >0 and \tau >0; we get

\begin{array}{r}{\omega}^{3}{(cos\frac{\beta \pi}{2}\pm isin\frac{\beta \pi}{2})}^{3}+A{\omega}^{2}{(cos\frac{\beta \pi}{2}\pm isin\frac{\beta \pi}{2})}^{2}\\ \phantom{\rule{1em}{0ex}}+\omega (B+C{e}^{-\tau \omega (cos\frac{\beta \pi}{2}\pm isin\frac{\beta \pi}{2})})(cos\frac{\beta \pi}{2}\pm isin\frac{\beta \pi}{2})\\ \phantom{\rule{1em}{0ex}}+D+E{e}^{-\tau \omega (cos\frac{\beta \pi}{2}\pm isin\frac{\beta \pi}{2})}=0.\end{array}

Separating the real and imaginary parts yields

\{\begin{array}{l}{\omega}^{3}{cos}^{3}\frac{\beta \pi}{2}-3{\omega}^{3}{sin}^{2}\frac{\beta \pi}{2}cos\frac{\beta \pi}{2}+A{\omega}^{2}{cos}^{2}\frac{\beta \pi}{2}-A{\omega}^{2}{sin}^{2}\frac{\beta \pi}{2}\\ \phantom{\rule{2em}{0ex}}+\omega Bcos\frac{\beta \pi}{2}\pm \omega Ccos\frac{\beta \pi}{2}{e}^{-\tau \omega cos\frac{\beta \pi}{2}}cos(\mp \tau \omega sin\frac{\beta \pi}{2})\\ \phantom{\rule{2em}{0ex}}\mp sin\frac{\beta \pi}{2}C{e}^{-\tau \omega cos\frac{\beta \pi}{2}}sin(\mp \tau \omega sin\frac{\beta \pi}{2})+D\\ \phantom{\rule{2em}{0ex}}+E{e}^{-\tau \omega cos\frac{\beta \pi}{2}}cos(\mp \tau \omega sin\frac{\beta \pi}{2})=0,\\ \pm 3{\omega}^{3}{cos}^{2}\frac{\beta \pi}{2}sin\frac{\beta \pi}{2}\mp {\omega}^{3}{sin}^{3}\frac{\beta \pi}{2}\pm 2A{\omega}^{2}sin\frac{\beta \pi}{2}cos\frac{\beta \pi}{2}\\ \phantom{\rule{2em}{0ex}}\pm \omega Bsin\frac{\beta \pi}{2}\pm \omega sin\frac{\beta \pi}{2}C{e}^{-\tau \omega cos\frac{\beta \pi}{2}}cos(\mp \tau \omega sin\frac{\beta \pi}{2})\\ \phantom{\rule{2em}{0ex}}+\omega C{e}^{-\tau \omega cos\frac{\beta \pi}{2}}sin(\mp \tau \omega sin\frac{\beta \pi}{2})cos\frac{\beta \pi}{2}\\ \phantom{\rule{2em}{0ex}}+E{e}^{-\tau \omega cos\frac{\beta \pi}{2}}sin(\mp \tau \omega sin\frac{\beta \pi}{2})=0.\end{array}

(3.6)

From the second equation of (3.6), we have

sin\frac{\beta \pi}{2}=0,

that is, \frac{\beta \pi}{2}=k\pi, k=0,1,2,\dots .

For \frac{\beta \pi}{2}=k\pi, k=0,2,4,\dots , substituting into the first equation of (3.6), we have

{\omega}^{3}+A{\omega}^{2}+\omega (B\pm C{e}^{-\tau \omega})+D+E{e}^{-\tau \omega}=0.

For the parameter values given in Table 1, we take any N>{N}_{\mathrm{crit}}; then we get the specific value on the infected equilibrium {E}^{\ast}=({T}^{\ast},{I}^{\ast},{V}^{\ast}) and we can see that the above equation is unequal for \omega >0.

For \frac{\beta \pi}{2}=k\pi, k=1,3,5,\dots , substituting into the first equation of (3.6), we have

-{\omega}^{3}+A{\omega}^{2}-\omega (B\pm C{e}^{\tau \omega})+D+E{e}^{\tau \omega}=0.

(3.7)

According to the development of Taylor type, we have

{e}^{\tau \omega}\approx 1+\tau \omega +\frac{{(\tau \omega )}^{2}}{2!}.

We take \omega =-\theta, and (3.7) becomes

{\theta}^{3}(1\pm \frac{C{\tau}^{2}}{2})+{\theta}^{2}(A\mp \tau C+\frac{E{\tau}^{2}}{2})+\theta (B\pm C-E\tau )+D+E=0.

(3.8)

Let

\alpha =1\pm \frac{C{\tau}^{2}}{2},\phantom{\rule{2em}{0ex}}\beta =A\mp \tau C+\frac{E{\tau}^{2}}{2},\phantom{\rule{2em}{0ex}}\gamma =B\pm C-E\tau ,\phantom{\rule{2em}{0ex}}\rho =D+E,

then (3.8) becomes

h(\theta )=\alpha {\theta}^{3}+\beta {\theta}^{2}+\gamma \theta +\rho .

(3.9)

Notice that

{h}^{\prime}(\theta )=3\alpha {\theta}^{2}+2\beta \theta +\gamma .

Set

3\alpha {\theta}^{2}+2\beta \theta +\gamma =0.

(3.10)

Then the roots of (3.10) can be expressed as

{\theta}_{1,2}=\frac{-\beta \pm \sqrt{{\beta}^{2}-3\alpha \gamma}}{3\alpha}.

Due to \alpha \gamma >0, we have \sqrt{{\beta}^{2}-3\alpha \gamma}<\beta. Hence, neither {\theta}_{1} nor {\theta}_{2} is positive. Thus, (3.10) does not have positive roots. Since \alpha >0, h(0)=\rho \ge 0, it follows that (3.9) has no positive roots.

Because of \omega =-\theta, the roots of (3.7) are positive, that is, {\omega}_{1,2,3}>0.

The proof is completed. □