Benchmark solution ===================== Calculation method used for the reference solution -------------------------------------------------------- Reference model ~~~~~~~~~~~~~~~~~~~~~ For the reference model, the following displacement vector is assumed: :math:`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: :math:`M=\left(\begin{array}{ccc}1& 0& 0\\ 0& 1& 0\\ 0& 0& 1\end{array}\right)` :math:`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: :math:`{\omega }_{1}^{2}=(2-\sqrt{2})` :math:`{\omega }_{2}^{2}=2` :math:`{\omega }_{3}^{2}=(2+\sqrt{2})` of respective modal deformations (normalized to 1 according to the greatest amplitude): :math:`{\varphi }_{1}=\frac{1}{2}\left(\begin{array}{c}\sqrt{2}\\ 2\\ \sqrt{2}\end{array}\right)` :math:`{\varphi }_{2}=\left(\begin{array}{c}-1\\ 0\\ 1\end{array}\right)` :math:`{\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: :math:`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: :math:`{\psi }_{1}=\left(\begin{array}{c}1\\ 0\\ 0\end{array}\right)` :math:`{\psi }_{2}=\left(\begin{array}{c}0\\ 1\\ 0.5\end{array}\right)` forming the basis for static :math:`\psi =\left[{\psi }_{1}{\psi }_{2}\right]` modes. This leads to: :math:`{G}_{r}={\psi }^{T}[K+M]\psi`; :math:`\mathit{Gr}=\left(\begin{array}{cc}3& -1\\ -1& 2.75\end{array}\right)` The weighting parameters for :math:`{e}_{\omega }^{2}(u,v,w)` are chosen :math:`\alpha =0.5` and :math:`\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: :math:`Al=b` with, for each natural pulsation :math:`{\omega }_{i}`: :math:`{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 :math:`{b}_{i}=\left(\begin{array}{c}{0}_{3}\\ (-2\alpha /(1-\alpha )){H}^{T}{G}_{r}\widehat{{u}_{i}}\end{array}\right)` where :math:`\widehat{{u}_{i}}` represents the observation of the :math:`{\varphi }_{i}` mode associated with the estimated natural pulsation :math:`{\omega }_{i}`. For their part, :math:`\alpha` and :math:`\gamma` are the weighting parameters associated with the error functional. Finally, the solution vector :math:`l` is the concatenation of two fields associated with the error functional so that: :math:`l=\left(\begin{array}{c}u-v\\ u\end{array}\right)` Benchmark results ---------------------- Two reference cases are tested: 1. First clean mode, perfect observations: 1. * :math:`\widehat{{u}_{1}}=\left(\begin{array}{c}\frac{\sqrt{2}}{2}\\ 1\end{array}\right)`; :math:`\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: :math:`(u-v)=\left(\begin{array}{c}0\\ 0\\ 0\end{array}\right)`; :math:`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: :math:`\widehat{{u}_{2}}=\left(\begin{array}{c}-1\\ 0\end{array}\right)`; :math:`\widehat{{\omega }_{2}}=1.25\ast 2` In this case, the construction of the matrix problem :math:`Al=b` leads to: :math:`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)` :math:`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: :math:`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: :math:`{e}_{{\omega }_{2}}^{2}(u,v,w)=0.089643288114668` whose part associated with the error fields is: :math:`\frac{\gamma }{2}{(u-v)}^{T}[K](u-v)+\frac{1-\gamma }{2}{(u-w)}^{T}{\omega }^{2}[M](u-w)=0.083454681437031` Uncertainty about the solution --------------------------- First case: analytical solution. Second case: semi-analytical solution.