# Data-Completion and Model Correction by Means of Evanescent Regularization
## 2 Correction of the Available Model
### Methods
#### 2.1.1 Dynamical Model  
We consider that a model is available, coming from a system modeling software, for example. For the sake of simplicity, we consider:  
\[\frac{d\mathbf{y}}{dt}=\mathbf{A}\mathbf{y} \tag{1}\]  
where \(\mathbf{A}\) is a no-singular and diagonalizable matrix, with negative eigenvalues to ensure stability. The fact of being diagonalizable and having negative eigenvalues is not a limitation of the proposed methodology, however, they allow us to define sound physical models. The simulated problem is supposed to be less accurate than the exact one, involving the matrix \(\mathbf{B}\) such as:  
\[\mathbf{B}=\mathbf{A}+e\mathbf{A} \tag{2}\]with \(\epsilon\) small enough. Thus, the modeling and its simulation considers:  
\[\frac{d\mathbf{y}}{dt}=\mathbf{B}\mathbf{y} \tag{3}\]  
The discreet counterparts of Equations (1) and (3) lead to \(\mathbf{y}^{\epsilon}\) and \(\mathbf{y}^{s}\), the reference and simulated solutions, respectively. Thus, using a simple Euler explicit integration scheme in time, one can write at the \(i\)-time step:  
\[\left\{ \begin{array}{l}\mathbf{y}_{i+1}^{\epsilon}=\mathbf{y}_{i}^{\epsilon} +\Delta t\mathbf{A}\mathbf{y}_{i}^{\epsilon}\\ \mathbf{y}_{i+1}^{s}=\mathbf{y}_{i}^{s}+\Delta t\mathbf{B}\mathbf{y}_{i}^{s} \end{array} \right. \tag{4}\]  
where \(\Delta t\) is the time step. The model update and correction algorithm takes as input a partial measurements of \(\mathbf{y}\), which is named \(\mathbf{y}^{m}\) and aims to predict both \(\mathbf{y}^{m}\) of dimension \(N_{m}\), as well as \(\mathbf{y}^{nm}\) of dimension \(N_{mm}\), the non-measurable part of \(\mathbf{y}\), such as \(N_{m}+N_{mm}=N\), the dimension of the quantities of interest \(\mathbf{y}\). First, the measurable and non-measurable parts of the model are separated such as:  
\[\left( \begin{array}{c}\mathbf{y}_{i+1}^{s,m}\\ \mathbf{y}_{i+1}^{s,m}\end{array} \right)=\left( \begin{array}{c}\mathbf{y}_{ i}^{s,m}\\ \mathbf{y}_{i}^{s,mm}\end{array} \right)+\Delta t\left( \begin{array}{cc}B^{m,m }&B^{m,nm}\\ B^{mm,m}&B^{mm,mm}\end{array} \right)\left( \begin{array}{c}\mathbf{y}_{i}^{s, m}\\ \mathbf{y}_{i}^{s,mm}\end{array} \right) \tag{5}\]  
Identifying the components of matrix \(\mathbf{B}\) in the previous equation seems a tricky issue, because at each time step \(t_{i}\), the solution involves the hidden variables at the previous time step \(t_{i-1}\). Therefore, one could think that there are too many unknowns to be identified; the components of \(\mathbf{B}\) and the hidden estate at each time step, a number that exceeds the number of available equations, the latter being the number of observable states at each time step. However, by using recurrence, all the solutions at any time step will depend on only the initial one, which consists of some observable and hidden values. Thus, the number of equations (involving the observable state at each time step) should be large enough (when considering a sufficient number of time steps, which it is usually the case) for computing the matrix and the hidden state at only the initial time. Since this work also aims at identifying a correction for both the measurable and non-measurable outputs, a substitution of \(\mathbf{y}^{s,nm}\) through the use of the Schur complement is not the most suitable approach. Instead, a second time iteration will show that measurable and non-measurable quantities are coupled:  
\[\left( \begin{array}{c}\mathbf{y}_{i+2}^{s,m}\\ \mathbf{y}_{i+2}^{s,m}\end{array} \right)=\left( \begin{array}{c}\mathbf{y}_{ i+1}^{s,m}\\ \mathbf{y}_{i+1}^{s,m}\end{array} \right)+\Delta t\left( \begin{array}{cc}B^{m, m}&B^{m,nm}\\ B^{nm,m}&B^{mm,mm}\end{array} \right)\left( \begin{array}{c}\mathbf{y}_{i+1}^{s, m}\\ \mathbf{y}_{i+1}^{s,m}\end{array} \right)= \tag{6}\] \[\left( \begin{array}{c}\mathbf{y}_{i}^{s,m}\\ \mathbf{y}_{i}^{s,mm}\end{array} \right)+\left(2\Delta t\left( \begin{array}{ cc}B^{m,m}&B^{m,nm}\\ \mathbf{y}_{i}^{s,mm}\end{array} \right)+2\Delta t\left( \begin{array}{cc}B^{ m,m}&B^{m,nm}\\ B^{mm,m}&B^{mm,mm}\end{array} \right)\left(