sums. We focus on proving that $\mathcal S:=0$. Indeed, recall the expression of $c_{v,1}$ in terms of $c_1$, given in (\[c0cvc1\]): $$c_{v,1}\frac{aS_v^\ast}{N_h}=c_1\frac{\nu_hE_h^*}{\sum_{l=1}^n  \beta_l I_l^\ast }.$$ By replacing $c_{v,1}$ by its value in $\mathcal S$, we obtain,

$$\begin{aligned}
\label{S}
\mathcal S &= c_{v,1}aS_v^\ast\sum_{i=1}^n \frac{\beta_i I_i^\ast}{N_h} \ln\frac{ I_1}{I_1^\ast}\frac{ I_i^\ast }{I_i}   
  +\sum_{i=1}^nc_i \sum_{j=1}^{n}m_{ij}I_j^\ast \ln\frac{I_iI_j^\ast}{I_i^\ast I_j}\nonumber\\
   & =c_1\frac{\nu_hE_h^*}{\sum_{l=1}^n  \beta_l I_l^\ast }\sum_{i=1}^n  \beta_i I_i^\ast \ln\frac{ I_1}{I_1^\ast}\frac{ I_i^\ast }{I_i}   
  +\sum_{i=1}^nc_i \sum_{j=1}^{n}m_{ij}I_j^\ast \ln\frac{I_iI_j^\ast}{I_i^\ast I_j}\nonumber\\
  & =c_1\sum_{j=1}^{n}\ln\frac{I_1I_j^\ast}{I_1^\ast I_j}\left[\frac{\nu_hE_h^*}{\sum_{l=1}^n  \beta_l I_l^\ast } \beta_j I_j^\ast    
 + m_{1j}I_j^\ast \right]
    +\sum_{i=2}^nc_i \sum_{j=1}^{n}m_{ij}I_j^\ast \ln\frac{I_iI_j^\ast}{I_i^\ast I_j} 
 \end{aligned}$$

However, since $c_i$ are the components of the solution of $Bc=0$ where $B$ is given in (\[B\]), it follows that, for any $j\geq2,$

$$c_1\left( \nu_hE_h^*\frac{\beta_j I_j^\ast}{\sum_{l=1}^n \beta_l I_l^\ast} +m_{1j}I_{j}^\ast\right)=
   c_j\left(\sum_{k=1}^ n m_{jk}I_k^\ast \right)
   -\sum_{i=2}^{n}c_im_{ij}I_j^\ast,$$ Plugging this expression into , and using again the properties of natural logarithms, we obtain:

$$\begin{aligned}
\label{S2}
\mathcal S
& =\sum_{j=1}^{n}\ln\frac{I_1I_j^\ast}{I_1^\ast I_j}\left[c_j\left(\sum_{k=1}^ n m_{jk}I_k^\ast \right)
   -\sum_{i=2}^{n}c_im_{ij}I_j^\ast \right]
  +\sum_{i=2}^nc_i \sum_{j=1}^{n}m_{ij}I_j^\ast \ln\frac{I_iI_j^\ast}{I_i^\ast I_j}\nonumber\\
   &= \sum_{i=1}^{n}c_i\ln\frac{I_1I_i^\ast}{I_1^\ast I_i}\left(\sum_{k=1}^ n m_{ik}I_k^\ast \right)+\sum_{i=2}^nc_i \sum_{j=1}^{n}m_{ij}I_j^\ast \left[ -\ln\frac{ I_1 I_j^\ast}{I_1^\ast I_j}+\ln\frac{I_i I_j^\ast} {I_i^\ast I_j}\right]\nonumber\\
      &= \sum_{i=2}^{n}c_i\ln\frac{I_1I_i^\ast}{I_1^\ast I_i}\left(\sum_{j=1}^ n m_{ij}I_j^\ast \right)+\sum_{i=2}^nc_i \sum_{j=1}^{n}m_{ij}I_j^\ast \left[ \ln\frac{I_1^\ast I_i}{ I_1 I_i^\ast}\right]\nonumber\\
  &:=0, \end{aligned}$$

since for $i=1$, the coefficient of the sum is $\ln1=0$. Finally using and , the derivative of $\mathcal V$ along the trajectories of 