Proof
We use the Lyapunov method. Let
\(L=\frac{\sigma}{\rho}\int_{\frac{\lambda}{d}}^{T}\frac{\xi-\frac{\lambda}{d}}{\xi}\)d\(\ \xi+\frac{\sigma}{\rho}T_{i}+V\)
on \(\mathbb{R}_{+}^{3}\)
Evidently, \(L\left({P_{u}}^{*}\right)=0\) and \(L>0\), in\(\mathbb{R}_{+}^{3}\backslash\left\{{P_{u}}^{*}\right\}\)
The derivative of L along the solutions of (1) is
\begin{equation}
\dot{L}=\frac{\sigma}{\rho}\left(\lambda-dT-kTV\right)\frac{T-\frac{\lambda}{d}}{T}+\frac{\sigma}{\rho}\left(kTV-\rho T_{i}\right)+\sigma T_{i}-\text{cV}\nonumber \\
\end{equation}\begin{equation}
=\frac{\sigma}{\rho}\left(\lambda-dT-kTV\right)\frac{T-\frac{\lambda}{d}}{T}+\frac{\sigma}{\rho}\left(\text{kTV}\right)-\text{cV}\nonumber \\
\end{equation}\begin{equation}
=\frac{-\sigma}{\text{dTρ}}(\lambda-dT)^{2}-\frac{\text{σk}}{\text{ρd}}TV(d+\frac{-\lambda}{T})+\frac{\sigma k}{\rho}TV-\text{cV}\nonumber \\
\end{equation}\begin{equation}
=\frac{-\sigma}{\text{dTρ}}(\lambda-dT)^{2}+\left(\frac{\text{σkλ}}{\text{ρd}}+c\right)V=\frac{-\sigma}{\text{dTρ}}(\lambda-dT)^{2}+\left(\frac{\text{σkλ}}{\text{ρdc}}-1\right)\text{cV}\nonumber \\
\end{equation}\begin{equation}
\dot{L}=\frac{-\sigma}{d^{{}^{\prime}}\text{Tρ}}(\lambda-dT)^{2}+\left(R_{0}-1\right)\text{cV}\nonumber \\
\end{equation}And \(\dot{L}<0\) in\(\mathbb{R}_{+}^{3}\backslash\left\{{P_{u}}^{*}\right\}\) if and only
if \(R_{0}\leq 1\)
Now, for the endemic point, we use LaSalle’s principle. let\(T^{*}=\frac{\lambda}{d}\frac{1}{R_{0}}\),\({T_{i}}^{*}=\frac{dc}{\text{kσ}}\left(R_{0}-1\right),\ V^{*}=\frac{d}{k}\left(R_{0}-1\right)\),
and
\(L=\int_{T^{*}}^{T}\frac{\xi-T^{*}}{\xi}\)d\(\ \xi+\int_{{T_{i}}^{*}}^{T_{i}}\frac{\xi-{T_{i}}^{*}}{\xi}\)d\(\ \xi+\frac{\rho}{\sigma}\int_{V^{*}}^{V}\frac{\xi-V^{*}}{\xi}\)d ξ
The derivative of L along the solutions of (1) is
\(\dot{L}=\left(\lambda-dT-kTV\right)\frac{T-T^{*}}{T}+\frac{T_{i}-{T_{i}}^{*}}{T_{i}}\left(kTV-\rho T_{i}\right)+\frac{\rho}{\sigma}\frac{V-V^{*}}{V}\)(\(\sigma T_{i}-\text{cV}\))
\(=\left(\lambda-dT\right)-\left(\lambda-dT-kTV\right)\frac{T^{*}}{T}-\frac{{T_{i}}^{*}}{T_{i}}\left(kTV-\rho T_{i}\right)+\frac{\rho}{\sigma}\left(-\text{cV}\right)-\frac{\rho}{\sigma}\frac{V^{*}}{V}\)(\(\sigma T_{i}-\text{cV}\))
=\(\lambda-dT\ -\lambda\frac{T^{*}}{T}+dT^{*}-\text{kTV}\frac{{T_{i}}^{*}}{T_{i}}-\rho T_{i}\frac{V^{*}}{V}+{\frac{\text{cρ}}{\sigma}V}^{*}+\rho{T_{i}}^{*}\)
=\(\text{\ d}T^{*}\left[2-\frac{T^{*}}{T}-\frac{T}{T^{*}}\right]+\rho{T_{i}}^{*}\left[3-\frac{k}{\rho}\frac{\text{TV}}{T_{i}}-\frac{\rho}{k}\frac{{T_{i}}^{*}}{TV^{*}}-\frac{V^{*}T_{i}}{V{T_{i}}^{*}}\right]\)
Since
\(x+\frac{1}{x}-2\) and \(x+y+\frac{1}{\text{xy}}-3\)
are non negative for all x, y positive, we deduce \(\dot{L}\leq 0\)
\begin{equation}
\dot{L}=0\Leftrightarrow\left[2-\frac{T^{*}}{T}-\frac{T}{T^{*}}\right]=\left[3-\frac{k}{\rho}\frac{\text{TV}}{T_{i}}-\frac{\rho}{k}\frac{{T_{i}}^{*}}{TV^{*}}-\frac{V^{*}T_{i}}{V{T_{i}}^{*}}\right]=0\nonumber \\
\end{equation}\begin{equation}
\Leftrightarrow\left(T-T^{*}\right)^{2}=\left[3-\frac{k}{\rho}\frac{\text{TV}}{T_{i}}-\frac{\rho}{k}\frac{{T_{i}}^{*}}{TV^{*}}-\frac{V^{*}T_{i}}{V{T_{i}}^{*}}\right]=0\nonumber \\
\end{equation}\begin{equation}
\Leftrightarrow T=T^{*}\text{and}\left[2-\frac{c}{\sigma}\frac{V}{T_{i}}-\frac{\sigma T_{i}}{\text{cV}}\right]=0\nonumber \\
\end{equation}\begin{equation}
\Leftrightarrow T=T^{*}\text{\ and\ }\left(cV-\sigma T_{i}\right)^{2}=0\nonumber \\
\end{equation}\(\Leftrightarrow T=T^{*}\) and\(V=\frac{\sigma}{c}T_{i}\)
The largest invariant subset of \(\left\{\dot{L}=0\right\}\) by (1)
in the positive octant, satisfies \(T=T^{*}\), then \(\dot{T}=0\)and \(\lambda-d^{{}^{\prime}}T^{*}-kT^{*}V=0\Leftrightarrow V=V^{*}\)
So, the largest invariant subset of \(\left\{\dot{L}=0\right\}\) by
(1) is reduced to \(\left\{{P_{e}}^{*}\right\}\)
then from LaSalle’s invariance principal, we conclude the global
asymptotic stability.