## 4 Parameter Estimation with One Output (New Hospitalizations)
### Adaptive Observer
Even if the estimation error given by (64)-(68) depends on unknown variables such as \(q(t)\), \(\phi_{w}(t)\), \(\theta(t)\), and \(w(t)\), considering that their values are bounded (see Remark 8), their upper values exist, so they might be used to calculate the maximum estimation error, whose value is important in the sense that it is certain that the real value will lie within the range defined by the estimated value plus and minus the maximum estimation error. Before giving the expression for the maximum estimation error, which is provided in Proposition 4, let us clarify that we will refer to the norm \(\|\cdot\|_{2}\) as \(\|\cdot\|\) for simplicity reasons. Additionally, consider the inequality for the matrix exponential norm provided in Lemma 1  
**Lemma 1** (Matrix Exponential Norm [39]). : _Given a matrix \(A\in\mathbb{C}^{n\times n}\) with spectrum \(\sigma(A)=\{\lambda\in\mathbb{C}\ :\ \det(A-\lambda\mathbb{I}_{n\times n})=0\}\), there exists a unitary matrix \(U\in\mathbb{C}^{n\times n}\) such that \(A\) can be expressed as_  
\[U^{*}AU=D+T_{r}, \tag{69}\]  
_where \(D=\operatorname{\mathbf{diag}}(\lambda_{i})\) and \(\lambda_{i}\in\sigma(A)\) (including multiplicities) for all \(i=1,\ldots,n\), and \([T_{r}]_{i,j}=0\) if \(i\leq j\) for all \(i,j=1,\ldots,n\). Then, the norm of the matrix exponential \(e^{At}\) satisfies the inequality_  
\[\|e^{At}\|\leq e^{v(A)t}\sum_{k=0}^{n-1}\frac{\left(\|T_{r}\|t\right)^{k}}{k!} \tag{70}\]  
_where \(v(A)=\max\{\operatorname{Re}(\lambda)\ :\ \lambda\in\sigma(A)\}\)._  
Note that in the particular case where \(A\) is diagonalizable, as a result of the expression (70) in Lemma 1, the inequality  
\[\|e^{At}\|\leq e^{v(A)t} \tag{71}\]  
holds. **Lemma 2**. : _Assuming that the initial values \(\hat{\theta}(t_{0})\) and \(\hat{x}_{i}(t_{0})\) for \(i=1,2,3\) are chosen such that (i) \(\hat{\theta}(t_{0})\in[\beta_{m},\beta_{M}]\); (ii) \(0<\hat{x}_{1}(t_{0})\leq 1\), \(0\leq\hat{x}_{3}(t_{0})\leq 1\), and \(\hat{x}_{2}(t_{0})=y_{1}(t_{0})/\sigma\); and (iii) \(\hat{x}_{1}(t_{0})+\hat{x}_{2}(t_{0})+\hat{x}_{3}(t_{0})\leq 1\), then_  
\[\dot{\widetilde{z}}(t)=(A_{c}-KC_{c})\widetilde{z}(t)+\phi(t)\widetilde{ \theta}(t)+\mathrm{Y}(t)\dot{\widetilde{\theta}}(t)+\mathrm{Y}(t)\dot{\theta }(t)-\phi_{w}(t)\big{[}\dot{\theta}(t)+\theta(t)\big{]}-w(t) \tag{63}\]0  
_where_  
\[\dot{\widetilde{z}}(t)=(A_{c}-KC_{c})\widetilde{z}(t)+\phi(t)\widetilde{ \theta}(t)+\mathrm{Y}(t)\dot{\widetilde{\theta}}(t)+\mathrm{Y}(t)\dot{\theta }(t)-\phi_{w}(t)\big{[}\dot{\theta}(t)+\theta(t)\big{]}-w(t) \tag{63}\]1  
_and_  
\[\dot{\widetilde{z}}(t)=(A_{c}-KC_{c})\widetilde{z}(t)+\phi(t)\widetilde{ \theta}(t)+\mathrm{Y}(t)\dot{\widetilde{\theta}}(t)+\mathrm{Y}(t)\dot{\theta }(t)-\phi_{w}(t)\big{[}\dot{\theta}(t)+\theta(t)\big{]}-w(t) \tag{63}\]2  
Proof. : The first result (see Equation (72)) follows directly from the properties of absolute values and the fact that \(\hat{\theta}(t_{0})\) and \(\theta(t_{0})\) are constrained to the same interval. Regarding the second result (see Equation (74)), first consider the expression (64) whose norm leads to  
\[\dot{\widetilde{z}}(t)=(A_{c}-KC_{c})\widetilde{z}(t)+\phi(t)\widetilde{ \theta}(t)+\mathrm{Y}(t)\dot{\widetilde{\theta}}(t)+\mathrm{Y}(t)\dot{\theta }(t)-\phi_{w}(t)\big{[}\dot{\theta}(t)+\theta(t)\big{]}-w(t) \tag{63}\]3  
Since \(\tilde{z}(t)=\hat{z}(t)-z(t)\), where \(\hat{z}=T(\hat{x}(t),\hat{\theta}(t))\) and \(z=T(x(t),\theta(t))\), and considering \(\hat{x}_{2}(t)=y_{1}(t)/\sigma=x_{2}(t)\), it follows that  
\[\|\widetilde{z}(t)\|=\left\|\left[ \begin{array}{c}-C\big{[}\hat{\theta}(t)( \hat{x}_{1}(t)+\hat{x}_{2}(t))-\theta(t)(x_{1}(t)+x_{2}(t))\big{]}-\rho(\hat{ \theta}(t)\hat{x}_{3}-\theta x_{3})\\ \hat{\theta}(t)(\hat{x}_{1}(t)+\hat{x}_{2}(t))-\theta(t)(x_{1}(t)+x_{2}(t)) \\ 0\end{array} \right]\right\|.