2. Benchmark solution

2.1. Calculation method used for the reference solution

2.1.1. Reference model

For the reference model, the following displacement vector is assumed:

\(x=\left(\begin{array}{c}{x}_{1}\\ {x}_{2}\\ {x}_{3}\end{array}\right)\)

Thus, the structural matrices of mass (M) and stiffness (K) are:

\(M=\left(\begin{array}{ccc}1& 0& 0\\ 0& 1& 0\\ 0& 0& 1\end{array}\right)\) \(K=\left(\begin{array}{ccc}2& -1& 0\\ -1& 2& -1\\ 0& -1& 2\end{array}\right)\)

The natural pulsations of the mass-spring system are equal to:

\({\omega }_{1}^{2}=(2-\sqrt{2})\) \({\omega }_{2}^{2}=2\) \({\omega }_{3}^{2}=(2+\sqrt{2})\)

of respective modal deformations (normalized to 1 according to the greatest amplitude):

\({\varphi }_{1}=\frac{1}{2}\left(\begin{array}{c}\sqrt{2}\\ 2\\ \sqrt{2}\end{array}\right)\) \({\varphi }_{2}=\left(\begin{array}{c}-1\\ 0\\ 1\end{array}\right)\) \({\varphi }_{3}=\frac{1}{2}\left(\begin{array}{c}-\sqrt{2}\\ 2\\ -\sqrt{2}\end{array}\right)\)

The matrix that makes it possible to observe x1 and x2 perfectly is:

\(H=\left(\begin{array}{ccc}1& 0& 0\\ 0& 1& 0\end{array}\right)\)

Moreover, the standard matrix (Gr) is selected by combining the reduced mass and stiffness Guyan matrices. For their construction, the static modes associated with x1 and x2 are:

\({\psi }_{1}=\left(\begin{array}{c}1\\ 0\\ 0\end{array}\right)\) \({\psi }_{2}=\left(\begin{array}{c}0\\ 1\\ 0.5\end{array}\right)\)

forming the basis for static \(\psi =\left[{\psi }_{1}{\psi }_{2}\right]\) modes. This leads to:

\({G}_{r}={\psi }^{T}[K+M]\psi\);

\(\mathit{Gr}=\left(\begin{array}{cc}3& -1\\ -1& 2.75\end{array}\right)\)

The weighting parameters for \({e}_{\omega }^{2}(u,v,w)\) are chosen \(\alpha =0.5\) and \(\gamma =0.5\).

Construction of the linear system associated with the problem of the error functional in relation to behavior. ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ ~~~~~~~~~~~~~~~~

The problem making it possible to find the fields associated with the error-type functional in a behavioral relationship leads to the resolution of the following linear matrix system:

\(Al=b\)

with, for each natural pulsation \({\omega }_{i}\):

\({A}_{i}=\left(\begin{array}{cc}\gamma (K+\gamma /(1-\gamma ){\omega }_{i}^{2}M)& -\gamma (K-{\omega }_{i}^{2}M)\\ -\gamma (K-{\omega }_{i}^{2}M)& (-2\alpha /(1-\alpha )){H}^{T}{G}_{r}H\end{array}\right)\) and \({b}_{i}=\left(\begin{array}{c}{0}_{3}\\ (-2\alpha /(1-\alpha )){H}^{T}{G}_{r}\widehat{{u}_{i}}\end{array}\right)\)

where \(\widehat{{u}_{i}}\) represents the observation of the \({\varphi }_{i}\) mode associated with the estimated natural pulsation \({\omega }_{i}\). For their part, \(\alpha\) and \(\gamma\) are the weighting parameters associated with the error functional.

Finally, the solution vector \(l\) is the concatenation of two fields associated with the error functional so that:

\(l=\left(\begin{array}{c}u-v\\ u\end{array}\right)\)

2.2. Benchmark results

Two reference cases are tested:

1. First clean mode, perfect observations:

\(\widehat{{u}_{1}}=\left(\begin{array}{c}\frac{\sqrt{2}}{2}\\ 1\end{array}\right)\); \(\widehat{{\omega }_{1}}=(2-\sqrt{2})\)

In this case, the result is trivial because it should lead to a perfectly zero error functional result associated with:

\((u-v)=\left(\begin{array}{c}0\\ 0\\ 0\end{array}\right)\); \(u=\frac{1}{2}\left(\begin{array}{c}\sqrt{2}\\ 2\\ \sqrt{2}\end{array}\right)\)

1. Second clean mode, perfect movements, and error-prone natural pulsation:

\(\widehat{{u}_{2}}=\left(\begin{array}{c}-1\\ 0\end{array}\right)\); \(\widehat{{\omega }_{2}}=1.25\ast 2\)

In this case, the construction of the matrix problem \(Al=b\) leads to:

\(A=\left(\begin{array}{cccccc}1+{1.25}^{2}& -0.5& 0& -1+{1.25}^{2}& 0.5& 0\\ -0.5& 1+{1.25}^{2}& -0.5& 0.5& -1+{1.25}^{2}& 0.5\\ 0& -0.5& 1+{1.25}^{2}& 0& 0.5& -1+{1.25}^{2}\\ -1+{1.25}^{2}& 0.5& 0& -6& 2& 0\\ 0.5& -1+{1.25}^{2}& 0.5& 2& -5.5& 0\\ 0& 0.5& -1+{1.25}^{2}& 0& 0& 0\end{array}\right)\)

\(b=\left(\begin{array}{c}0\\ 0\\ 0\\ 6\\ -2\\ 0\end{array}\right)\)

This system of equations was solved using the Matlab software, leading to the solution:

\(l=\left(\begin{array}{c}0.223608826207038\\ 0.107013222975753\\ -0.095122864867336\\ -0.957415448053491\\ 0.038110367724860\\ 0.494584477951991\end{array}\right)\)

In this case, the value of the functional is equal to:

\({e}_{{\omega }_{2}}^{2}(u,v,w)=0.089643288114668\)

whose part associated with the error fields is:

\(\frac{\gamma }{2}{(u-v)}^{T}[K](u-v)+\frac{1-\gamma }{2}{(u-w)}^{T}{\omega }^{2}[M](u-w)=0.083454681437031\)

2.3. Uncertainty about the solution

First case: analytical solution.

Second case: semi-analytical solution.