Principle of elimination
=========================

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

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

where


* :math:`{u}_{0}\in {ℝ}^{p}` is known (:math:`1\le p\le n`)


* :math:`G` is the subset of :math:`N=\{\mathrm{1,}\mathrm{...},n\}`, of cardinal :math:`p`: :math:`G={g}_{1}\mathrm{...}{g}_{p}`


* :math:`{u\mid }_{G}` is the projection of :math:`u` onto the subspace generated by :math:`{\{{u}_{i}\}}_{i\in G}`


* where :math:`{({u}_{i})}_{j}={\delta }_{\mathrm{ij}},\forall j\in N`


* :math:`K` is a symmetric matrix :math:`n\times n`,


* :math:`f\in {ℝ}^{n}` is fixed.

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

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

So we have :math:`{ℝ}^{n}={V}_{G}\oplus {V}_{L}`

and we note :math:`u={u}_{G}\oplus {u}_{K}` where :math:`{u}_{G}={u\mid }_{G}` and :math:`{u}_{L}={u\mid }_{L}`

Or again in vector notation :math:`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):

:math:`\{\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:

:math:`(\text{Pb1})\iff (\text{Pb2})` :math:`\{\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 :math:`{u}_{G}` from the minimization problem.

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

We're looking for :math:`{u}_{L}` minimizing

:math:`\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:

:math:`{K}_{\text{LL}}{u}_{L}={F}_{L}-{K}_{\mathrm{LG}}{u}_{0}`

So we can write:

:math:`(\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 :math:`K\text{'}\left[\begin{array}{}{u}_{L}\\ {u}_{G}\end{array}\right]=f\text{'}`