6. Note on system conditioning

When we look at the expression for the matrix that we will finally factor \(\text{K'}\) (Cf. [§3]), we see that its various sub-matrices \(A\), \(C\), \(\alpha I\) can be of very different orders of magnitude. We know that in general this situation is not numerically favorable (limited precision of computers).

Note that the link equations \(\text{Cu}-d=0\) can be multiplied by an arbitrary constant \((\beta )\) without changing the problem. Additionally, we saw that the \(\alpha \mathrm{I}\) matrices were also arbitrary \((\alpha >0)\). We therefore have two parameters that allow us to « adjust » the conditioning of the matrix.

We will not do a general demonstration but we are content to examine the most trivial case there is: a spring, a ddl, a connection.

The \(\text{K'}\) matrix is written if \(k\) is the spring stiffness:

\(\text{K'}=\left[\begin{array}{ccc}k& \beta & \beta \\ \beta & -\alpha & +\alpha \\ \beta & \alpha & -\alpha \end{array}\right]\)

The conditioning of this matrix is linked to the dispersion of its eigenvalues \({m}_{i}\):

Let’s calculate the characteristic polynomial of \(\text{K'}\):

\(P(\mu )=(\mu +\mathrm{2\alpha })(-{\mu }^{2}+k\mu +{\mathrm{2\beta }}^{2})\)

\(\iff \begin{array}{}{\mu }_{1}=-\mathrm{2\alpha }<0\\ {\mu }_{2}=\frac{+k+\sqrt{{k}^{2}+{\mathrm{8\beta }}^{2}}}{2}>0\\ {\mu }_{3}=\frac{+k-\sqrt{{k}^{2}+{\mathrm{8\beta }}^{2}}}{2}<0\end{array}\)

\(k\) is the eigenvalue of the unconstrained system. This eigenvalue is the order of magnitude sought for \({\mu }_{1},{\mu }_{2},{\mu }_{3}\).

We note that \({\mu }_{1}<\mathrm{0,}{\mu }_{3}<0\) and \({\mu }_{2},>0\), that is to say that the 2 eigenvalues added by the Lagrange coefficients are \(\text{<}0\) (it is also because of this that \({\mathit{LDL}}^{T}\mathrm{-}\mathit{SP}\) is not guaranteed without precautions).

We want to obtain eigenvalues of the same order of magnitude:

\(\mid {\mu }_{1}\mid \simeq \mid {\mu }_{2}\mid \simeq \mid {\mu }_{3}\mid\)

\(\mid {\mu }_{2}{\mu }_{3}\mid \simeq {\mid {\mu }_{1}\mid }^{2}\Rightarrow {\mathrm{2\beta }}^{2}\simeq {\mathrm{4\alpha }}^{2}\) eq 6-1

If \(\beta \text{<<}k\) then \({\mu }_{3}\mathrm{\simeq }0{\mu }_{2}\mathrm{\simeq }k\): this is not the result you were looking for.

If \(\beta \text{>>}k\), then \(\mathrm{\mid }{\mu }_{2}\mathrm{\mid }\mathrm{\simeq }\mathrm{\mid }{\mu }_{3}\mathrm{\mid }\mathrm{\simeq }\sqrt{2}\beta \mathrm{\simeq }\mathrm{\mid }{\mu }_{1}\mathrm{\mid }\)

The three eigenvalues are then in absolute values of the order of \(\beta\) which is a very large arbitrary constant in front of \(k\). This solution is not the one that will be retained because the value \(k\) is in the general case (with a large number of degrees of freedom) of an order of magnitude comparable to the other eigenvalues of the system.

Instead, we will choose:

\(\begin{array}{}\beta =\alpha \simeq k\Rightarrow {\mu }_{1}\simeq -\mathrm{2k}\\ \text{}{\mu }_{2}\simeq \mathrm{2k}\\ \text{}{\mu }_{3}\simeq -k\end{array}\)

Practically in*Code_Aster*, you choose a value of \(\alpha\) that is unique for the whole system. This value is the average of the extreme values of the diagonal terms associated with the physical degrees of freedom: \((\text{min}({a}_{\text{ii}})+\text{max}({a}_{\text{ii}}))/2\). In addition, we take \(\beta =\alpha\).