1. Principle of elimination#

In \({ℝ}^{n}\), we are looking to solve the following constrained minimization problem (Pb1):

\(\underset{u\in {U}_{G}}{\text{min}}(\frac{1}{2}{u}^{T}Ku-{u}^{T}f)\) with \({U}_{G}=\left\{u\in {ℝ}^{n},{u\mid }_{G}={u}_{0}\right\}\)

where

  • \({u}_{0}\in {ℝ}^{p}\) is known (\(1\le p\le n\))

  • \(G\) is the subset of \(N=\{\mathrm{1,}\mathrm{...},n\}\), of cardinal \(p\): \(G={g}_{1}\mathrm{...}{g}_{p}\)

  • \({u\mid }_{G}\) is the projection of \(u\) onto the subspace generated by \({\{{u}_{i}\}}_{i\in G}\)

  • where \({({u}_{i})}_{j}={\delta }_{\mathrm{ij}},\forall j\in N\)

  • \(K\) is a symmetric matrix \(n\times n\),

  • \(f\in {ℝ}^{n}\) is fixed.

Constraint \({u\mid }_{G}={u}_{0}\) represents Dirichlet boundary conditions that are homogeneous or not.

If we note \(L={C}_{N}G\) the complementary of \(G\) in \(N\), we can, using the \({u}_{i}\) defined above, we can, using the defined previously, decompose \({ℝ}^{n}\) into the direct sum of \({V}_{G}\) = vector space generated by \({\{{u}_{i}\}}_{i\in G}\) and of \({V}_{L}\) = vector space generated by \({\{{u}_{i}\}}_{i\in L}\);

So we have \({ℝ}^{n}={V}_{G}\oplus {V}_{L}\)

and we note \(u={u}_{G}\oplus {u}_{K}\) where \({u}_{G}={u\mid }_{G}\) and \({u}_{L}={u\mid }_{L}\)

Or again in vector notation \(u=\left\{\begin{array}{}{u}_{G}\\ {u}_{L}\end{array}\right\}\)

The problem (Pb1) can therefore be written in the form of the problem (Pb2):

\(\{\begin{array}{}\text{min}(\frac{1}{2}{u}_{G}^{T}{K}_{\text{GG}}{u}_{G}+\frac{1}{2}{u}_{L}^{T}{K}_{\text{LL}}{u}_{L}+{u}_{L}^{T}{K}_{\mathrm{LG}}{u}_{G}-{u}_{L}^{T}{f}_{L}-{u}_{G}^{T}{f}_{G})\\ {u}_{G}\in {V}_{G}\\ {u}_{L}\in {V}_{L}\\ {u}_{G}={u}_{0}\end{array}\)

Which is the same as writing:

\((\text{Pb1})\iff (\text{Pb2})\) \(\{\begin{array}{}\text{min}(\frac{1}{2}{u}_{L}^{T}{K}_{\text{LL}}{u}_{L}+{u}_{L}^{T}{K}_{\mathrm{LG}}{u}_{0}-{u}_{L}^{T}{f}_{L})\\ {u}_{L}\in {V}_{L}\\ u={u}_{0}\oplus {u}_{L}\end{array}\)

We then eliminated \({u}_{G}\) from the minimization problem.

We will now look for the matrix problem associated with (Pb3).

We’re looking for \({u}_{L}\) minimizing

\(\frac{1}{2}{u}_{L}^{T}{K}_{\text{LL}}{u}_{L}+{u}_{L}^{T}{K}_{\mathrm{LG}}{u}_{0}-{u}_{L}^{T}{f}_{L}\)

which amounts to solving the following matrix problem:

\({K}_{\text{LL}}{u}_{L}={F}_{L}-{K}_{\mathrm{LG}}{u}_{0}\)

So we can write:

\((\text{Pb1})\iff (\text{Pb2})\iff (\text{Pb3})\iff \left[\begin{array}{cc}{K}_{\text{LL}}& 0\\ 0& {I}_{G}\end{array}\right]\left[\begin{array}{}{u}_{L}\\ {u}_{G}\end{array}\right]=\left[\begin{array}{}{f}_{L}-{K}_{\mathrm{LG}}{u}_{0}\\ {u}_{0}\end{array}\right]\), which is \(K\text{'}\left[\begin{array}{}{u}_{L}\\ {u}_{G}\end{array}\right]=f\text{'}\)