Proof
The simple integration of the equation (8) gives the solution\(S\left(t\right)=e^{\text{tM}}S\left(0\right)\), which is
bounded, since the eigenvalues of the matrix M (given by (9) are
negative)\(\ \)
Let \(W\left(t\right)=T+T_{i}\), then,
\(\dot{W}(t)=\lambda-d^{\prime}T-kTV+2\left(1-a_{n-1}\right)p_{n-1}S_{n-1}+kTV-\rho T_{i}\)
\begin{equation}
=\lambda+2\left(1-a_{n-1}\right)p_{n-1}S_{n-1}-d^{\prime}T-\rho T_{i}\leq\lambda^{\prime}-\min\left(d^{\prime},\rho\right)W\left(t\right)\nonumber \\
\end{equation}and,
\begin{equation}
\dot{W}(t)\leq a-\text{bW}\left(t\right)\nonumber \\
\end{equation}we deduce, by corollary 3.2,
\(W\left(t\right)\leq\frac{a}{b}+(W\left(0\right)-\frac{a}{b})e^{-bt}\leq\frac{2a}{b}\)+\(\text{\ W}\left(0\right)=c_{1}\)
Finally,
\begin{equation}
\dot{V}(t)=\sigma T_{i}-\text{cV}\leq{\sigma c}_{1}-cV\nonumber \\
\end{equation}and we conclude by corollary 3.2, in a similar way.