## 3 Parameter Estimation with Two Outputs  
In this section, the outputs \(y_{1}\) and \(y_{2}\) are considered. Then, an exponential observer for the estimation of the state variable \(x_{1}(t)\) is proposed in Theorem 1 and its proof is provided. Then, the exponential convergence of the estimated \(\hat{\beta}(t)\) is guaranteed in Proposition 3. Considering that the outputs \(y_{1}(t)\) and \(y_{2}(t)\) are available, an exponential observer for the dynamical system \(\dot{x}_{1}(t)\) in (11) and (12) can be calculated from the solution of an auxiliary ODE, as stated in the subsequent Theorem 1 and its proof. **Theorem 1**. : _Under Assumptions 1-4, and considering the outputs \(y_{1}(t)=\sigma x_{2}(t)\) and \(y_{2}(t)=\alpha x_{3}(t)\), if \(A+y_{2}(t)<-L\) for all \(t\in[t_{0},\infty)\) and some \(0<L<\lambda+\rho\), given the solution of the dynamical equation_  
\[\dot{z}(t)=[A+y_{2}(t)]\dot{z}(t)-\frac{(\gamma_{1}+\sigma)}{\sigma}y_{1}(t)- \frac{\rho}{\alpha}y_{2}(t)+u_{1}(t)+u_{2},\ \forall\dot{z}(t_{0})\in\left(\frac{y_{1}(t_{0})}{\sigma},1\right],\ \forall t\in[t_{0},\infty), \tag{16}\]  
_the function_  
\[\dot{x}_{1}(t)=\mathrm{sat}_{\mathrm{c}}^{1}\bigg{(}\dot{z}(t)-\frac{y_{1}(t )}{\sigma}\bigg{)},\ \forall t\in[t_{0},\infty), \tag{17}\]  
_where \(\epsilon\) refers to a positive and small value, and is an exponential observer for the dynamical equation_  
\[\dot{x}_{1}(t)=Ax_{1}(t)-\theta(t)x_{1}(t)x_{2}(t)+\alpha x_{1}(t)x_{3}(t)- \rho(x_{2}(t)+x_{3}(t)),\ \forall x_{1}(t_{0})\in(0,1],\ \forall t\in[t_{0},\infty). \tag{18}\]  
**Proof of Theorem 1. ** Let us define \(z=x_{1}+x_{2}\), whose time derivative yields  
\[\dot{z}=Az+zy_{2}-\frac{(\gamma_{1}+\sigma)}{\sigma}y_{1}-\frac{\rho}{\alpha} y_{2}+u_{1}+u_{2}. \tag{19}\]  
Defining the error as \(\widetilde{z}=\dot{z}-z\), its time derivative gives  
\[\dot{\widetilde{z}}=(A+y_{2})\widetilde{z}, \tag{20}\]  
whose solution verifies  
\[\widetilde{z}(t)=\widetilde{z}(t_{0})e^{\int_{0}^{t}(A+y_{2}(\tau))d\tau}. \tag{21}\]  
Then, considering the condition \(A+y_{2}(t)<-L\), \(\forall t\in[t_{0},\infty)\), where \(L\) is positive, it follows that  
\[|\widetilde{z}(t)|<\widetilde{z}(t_{0})e^{-L(t-t_{0})}. \tag{22}\]  
From the definition of \(z\) and considering that \(x_{2}=y_{1}/\sigma\), the expression \(x_{1}=z-y_{1}/\sigma\) (equivalently \(\dot{x}_{1}=\dot{z}-y_{1}/\sigma\)) is obtained, so  
\[|\dot{x}_{1}-x_{1}|=|\widetilde{z}|. \tag{23}\]  
Thus, the condition for an exponential observer is verified. Note that the function \(\mathrm{sat}_{\epsilon}^{1}()\) has been applied to the expression \(\dot{z}(t)-\frac{y_{1}(t)}{\sigma}\) since \(x_{1}(t)\in[\epsilon,1],\ \forall\ t\in[t_{0},\infty),\ \forall\ x_{1}(t_{0})\in(0,1]\); see Proposition 1, where \(\epsilon\) stands for a positive and small value. **Remark 3**. : _The upper and lower bounds established for \(L\) are essential to ensure the exponential observer and fulfill the positivity of subpopulations, that is, by considering the expression given in (21), the conditions \(A+y_{2}(t)<-L\) and \(0<L\) ensure the exponential convergence of \(\widetilde{z}(t)\); see Equation (22). From the inequality \(A+y_{2}(t)<-L\), the condition_  
\[H(t)<N(t)\bigg{(}\frac{\lambda+\rho-L}{\alpha}\bigg{)}, \tag{24}\]is obtained, and the non-negativity of \(H(t)\) (see Proposition 1) leads to \(L<\lambda+\rho\). Hence, the condition \(0<L<\lambda+\rho\) established in Theorem 2 is satisfied. Additionally, if \(\alpha\leq\lambda+\rho\), the condition (24) is satisfied for any \(L\in(0,\lambda+\rho)\) and all \(t\in[t_{0},\infty)\)._  
At this point, note that all the state variables are known so an explicit expression for \(\theta(t)\) in terms of those variables can be obtained from the equations in (11) and (12). By doing this, the following result is obtained. **Proposition 3**.