Description of document versions
====================================


.. csv-table::

    "**Version** **Aster**", "**Author (s)**
    **Organization (s)**", "**Description of changes**"
    "5", "**J. PELLET (EDF -R&D/ AMA)**", ""


**Be problem 1**

:math:`\underset{u\in V}{\text{min}}J(u)`

:math:`J(u)=\frac{1}{2}(\text{Au,}u)-(\text{b,}u)`

:math:`V` under :math:`{ℝ}^{n}\text{=}\left\{u\in {ℝ}^{n}\text{tel que}{C}_{i}u-{d}_{i}=\mathrm{0,}\forall i=\mathrm{1,}p\right\}` refined space

:math:`A` and :math:`b` are set to :math:`{ℝ}^{n}`

:math:`A` positive symmetric matrix of order :math:`n`.

**Problem 2**

Find :math:`u\in V` such as: :math:`((\text{Au,}{v}_{0})-(\text{b,}{v}_{0})=0\text{}\forall {v}_{0}\in {V}_{0})`

:math:`V=\left\{u\in {ℝ}^{n}\text{tel que}{C}_{i}u-{d}_{i}=0\text{,}\forall i=\mathrm{1,}p\right\}`

:math:`{V}_{0}=\left\{{v}_{0}\in {ℝ}^{n}\text{tel que}{C}_{i}u=0\text{,}\forall i=\mathrm{1,}p\right\}`

We will show that the two previous problems are equivalent.

First of all, let's note that problem 2 is equivalent to problem 2'.

**Problem 2'**

Find :math:`u\in V` such as: :math:`((\text{Au,}v-u)-(b,v-u)=0\text{}\forall v\in V)`

:math:`V=\left\{u\in {ℝ}^{n}\text{tel que}{C}_{i}u-{d}_{i}=0\text{,}\forall i=\mathrm{1,}p\right\}`

There is indeed a bijection between the set of :math:`\left\{v-\text{u,}u\in \text{V,}v\in V\right\}` and the set :math:`{V}_{0}`.

**Let's show that problem 2' is equivalent to problem1:**

Let's be :math:`u` solution of 2'

So, :math:`\forall v\in V`


.. math::
: label: EQ-None

    \ begin {array} {} J (v) -J (u) =\ frac {1} {u) =\ frac {1} {u) =\ frac {1} {2} (\ text {Au,} u) =\ frac {1} {2} (\ text {Au,} u) + (\ text {b,} u) + (\ text {b,} v) - (\ text {b,} v) - (\ text {b,} v) -\ frac {1} {2} (\ text {Au,} u) + (\ text {b,} u) + (\ text {b,} u) + (\ text {b,} u)\\ text {b,} v) - (\ frac {1} {2} (\ text {Av,},} v) - (\ text {Au,} v-u) -\ frac {1} {2} (\ text {Au,} u)\\ text {} =\ frac {1} {2} ((\ text {Av,} v) -2 (\ text {Av,} v) -2 (\ text {Au,} v) + (\ text {Au,} u)) =\ frac {1} {2} ((A (u-2)) v, u-v))\ ge 0\ end {array}

First, let's calculate the derivative of :math:`J(u)`:

:math:`\forall u\in V,\forall {v}_{0}\in {V}_{0},`

:math:`\begin{array}{}J\text{'}(u)\cdot {v}_{0}=\underset{\varepsilon \to 0}{\text{lim}}\frac{J(u+\varepsilon {v}_{0})-J(u)}{\text{}\varepsilon }\\ \text{}=\underset{\varepsilon \to 0}{\text{lim}}[(\text{Au},{v}_{0})+\frac{\varepsilon }{2}({\text{Av}}_{0},{v}_{0})-(b,{v}_{0})]=(\text{Au}-b){v}_{0}\end{array}`

Let :math:`u` be the Pb 1 solution

:math:`\forall V\in V`, let's say :math:`{\mathrm{v}}_{0}\mathrm{=}\mathrm{v}\mathrm{-}\mathrm{u}\text{;}{\mathrm{v}}_{0}\mathrm{\in }{\mathrm{V}}_{0}`.

:math:`\forall \varepsilon \text{}\frac{J(u+\varepsilon {v}_{0})-J(u)}{\varepsilon }\ge 0\text{}\Rightarrow \text{}(J\text{'}(u)\cdot {v}_{0}\ge 0\text{,}\forall {v}_{0}\in {V}_{0})`

We can see that :math:`J\text{'}(u)`, which is a linear form over :math:`{V}_{0}`, must always be positive. This is only possible if this form is identically null.

It is concluded that:


.. math::
: label: EQ-None

    J\ text {'} (u)\ cdot (v-u) = (\ text {Au} -b) (v-u) =0\ text {,}\ forall v\ in V

**Definitions, notations**

:math:`A` is the unconstrained stiffness matrix :math:`(n´n)` (symmetric and positive)

:math:`C` is the blocking matrix: :math:`\text{Cu}-d=0` (:math:`C` matrix :math:`n´p(p<n)`)

:math:`u` is the vector for physical degrees of freedom :math:`\mathrm{\in }{R}^{n}`

:math:`{\lambda }^{1}` is the vector of the first Lagrange degrees of freedom :math:`\mathrm{\in }{\mathrm{ℝ}}^{p}`

:math:`{\lambda }^{2}` is the Lagrange second degrees of freedom vector :math:`\in {ℝ}^{p}`

:math:`x=(u,{\lambda }^{1},{\lambda }^{2})\in {ℝ}^{n}\times {ℝ}^{p}\times {ℝ}^{p}`

We note:


* :math:`U` the set of physical degrees of freedom,


* :math:`{\Lambda }^{1}` the set of the first Lagrange degrees of freedom,


* :math:`{\Lambda }^{2}` the set of Lagrange's second degrees of freedom.

:math:`\alpha \in {ℝ}^{+}`

:math:`K` symmetric :math:`2p+n` order matrix

:math:`K=\left[\begin{array}{ccc}A& {C}^{T}& {C}^{T}\\ C& -\alpha l& \alpha l\\ C& \alpha l& -\alpha l\end{array}\right]`

The matrix :math:`K` written above corresponds to a certain numbering of the unknowns:

:math:`\mathrm{x}\mathrm{=}(\mathrm{u},{\lambda }^{1},{\lambda }^{2})`

The real matrix :math:`\mathrm{K}`, which we are trying to show can be factorized by :math:`{\mathit{LDL}}^{T}` without a permutation, is not written with this numbering. The only numbering rule taken into account is the following:

**Rule** :math:`\mathit{R0}` **:**

The two Lagrange degrees of freedom associated with a bond equation :math:`{C}_{i}U-{d}_{i}=0` frame the physical ddls constrained by this equation.

Then, to simplify the writing, we will use :math:`\alpha \mathrm{=}1`.

We want to show that any submatrix :math:`{K}_{i}` of :math:`K` is invertible.

Let's say a given :math:`{K}_{i}` submatrix. It corresponds to a sharing of degrees of freedom: those of rank :math:`\le i`, those of rank :math:`\ge i`.

We will note:

:math:`\tilde{U}` the subset of :math:`\mathrm{U}` corresponding to the degrees of freedom of rank :math:`\mathrm{\le }i`.

:math:`\tilde{\tilde{U}}` the subset of :math:`\mathrm{U}` corresponding to the degrees of freedom of rank :math:`>i`.

:math:`{\mathrm{L}}_{1}` is the set of :math:`({\lambda }_{1}^{1}\text{,}{\lambda }_{1}^{2})` couples such as :math:`\text{rang}({\lambda }_{1}^{1})<\text{rang}({\lambda }_{1}^{2})\mathrm{\le }i`

:math:`{L}_{1}^{1}=\left\{{\lambda }_{1}^{1}\right\}\text{;}{L}_{1}^{2}=\left\{{\lambda }_{1}^{2}\right\}`

:math:`{\mathrm{L}}_{3}` is the set of :math:`({\lambda }_{3}^{1}\text{,}{\lambda }_{3}^{2})` couples such as :math:`i<\text{rang}({\lambda }_{3}^{1})<\text{rang}({\lambda }_{3}^{2})`

:math:`{L}_{3}^{1}=\left\{{\lambda }_{3}^{1}\right\}\text{;}{L}_{3}^{2}=\left\{{\lambda }_{3}^{2}\right\}`

:math:`{\mathrm{L}}_{2}` is the set of :math:`({\lambda }_{2}^{1}\text{,}{\lambda }_{2}^{2})` couples such as :math:`\text{rang}({\lambda }_{2}^{1})\mathrm{\le }i<\text{rang}({\lambda }_{2}^{2})`

:math:`{L}_{2}^{1}=\left\{{\lambda }_{2}^{1}\right\}\text{;}{L}_{2}^{2}=\left\{{\lambda }_{2}^{2}\right\}`

We have :math:`L=\underset{\begin{array}{}i=1\text{,}3\\ j=1\text{,}2\end{array}}{\cup }{L}_{i}^{j}`.

The matrix :math:`C` can be divided into 3 parts corresponding to the division :math:`({L}_{1}\text{,}{L}_{2}\text{,}{L}_{3})`


.. csv-table::

    ":math:`{C}_{1}`"
    ":math:`{C}_{2}`"
    ":math:`{C}_{3}`"


Each :math:`{C}_{i}` matrix can be divided into :math:`2` parts corresponding to division :math:`(\tilde{U},\tilde{\tilde{U}})`


.. math::
: label: EQ-None

    {\ tilde {C}} _ {i}

Matrix :math:`\mathrm{A}` can be divided into 4 parts corresponding to division :math:`(\tilde{U},\tilde{\tilde{U}})`

:math:`A=\left[\begin{array}{cc}\tilde{A}& \stackrel{ˉ}{A}\\ \stackrel{ˉ}{{A}^{T}}& \tilde{\tilde{A}}\end{array}\right]`

Using these notations, the problem to be solved is to show that the matrix Ki is invertible.

:math:`{K}_{i}=\left[\begin{array}{cccc}-I& I& 0& \tilde{{C}_{1}}\\ I& -I& 0& \tilde{{C}_{1}}\\ 0& 0& -I& \tilde{{C}_{2}}\\ \tilde{{C}_{1}^{T}}& \tilde{{C}_{1}^{T}}& \tilde{{C}_{2}^{T}}& \tilde{A}\end{array}\right]`

This matrix corresponds to vector :math:`{\mathrm{X}}_{\mathrm{i}}\mathrm{=}\left[\begin{array}{c}{\lambda }_{1}^{1}\\ {\lambda }_{1}^{2}\\ {\lambda }_{2}^{1}\\ \tilde{\mathrm{U}}\end{array}\right]`

It must be shown that: :math:`{\mathrm{K}}_{\mathrm{i}}\text{.}{\mathrm{X}}_{\mathrm{i}}\mathrm{=}0\mathrm{\Rightarrow }{\mathrm{X}}_{\mathrm{i}}\mathrm{=}0`

The problem is equivalent to:

Problem 1:

:math:`(S)=\{\begin{array}{}-{\lambda }_{1}^{1}+{\lambda }_{1}^{2}+\tilde{{C}_{1}}\tilde{u}=0\\ {\lambda }_{1}^{1}-{\lambda }_{1}^{2}+{\tilde{C}}_{1}\tilde{u}=0\\ -{\lambda }_{2}^{1}+\tilde{{C}_{2}}\tilde{u}=0\\ \tilde{{C}_{1}^{T}}({\lambda }_{1}^{1}+{\lambda }_{1}^{2})+\tilde{{C}_{2}^{T}}{\lambda }_{2}^{1}+\tilde{A}\text{.}\tilde{u}=0\end{array}\Rightarrow \{\begin{array}{}\tilde{u}=0\\ {\lambda }_{1}^{1}={\lambda }_{1}^{2}=0\\ {\lambda }_{2}^{1}=0\end{array}`

**General case:**

Assume that :math:`\tilde{\mathrm{U}}\mathrm{\ne }\mathrm{\varnothing }`; :math:`{L}_{1}\ne \varnothing`; :math:`{L}_{2}\ne \varnothing`

:math:`(S)\mathrm{\Rightarrow }`


.. _RefEquation An2-1:

:math:`{\lambda }_{1}^{1}={\lambda }_{1}^{2}` eq Year2-1


.. _RefEquation An2-2:

:math:`\tilde{{C}_{1}}\tilde{u}=0` eq An2-2


.. _RefEquation An2-3:

:math:`{\lambda }_{2}^{1}=\tilde{{C}_{2}}\tilde{u}` eq An2-3


.. _RefEquation An2-4:

:math:`2\tilde{{C}_{1}^{T}}{\lambda }_{1}^{1}+({\tilde{C}}_{2}^{T}{\tilde{C}}_{2}+\tilde{A})\tilde{u}=0` eq Year2-4

From [:ref:`éq An2-4 <éq An2-4>`], we deduce:

:math:`2{\tilde{u}}^{T}{\tilde{C}}_{1}^{T}{l}_{1}^{1}+{\tilde{u}}^{T}({\tilde{C}}_{2}^{T}{\tilde{C}}_{2}+\tilde{A})\tilde{u}=0`

From [:ref:`éq An2-2 <éq An2-2>`], we get:

:math:`\tilde{{u}^{T}}\tilde{{C}_{1}^{T}}=0\text{}\Rightarrow \text{}\tilde{{u}^{T}}({\tilde{C}}_{2}^{T}{\tilde{C}}_{2}+\tilde{A})\tilde{u}=0`

:math:`\Rightarrow \tilde{{u}^{T}}\tilde{{C}_{2}^{T}}\tilde{{C}_{2}}\tilde{u}+\tilde{{u}^{T}}\tilde{A}\tilde{u}=0`

Now :math:`\tilde{A}` is symmetric positive (sub-matrix of a positive matrix) and :math:`{\tilde{C}}_{2}^{T}{\tilde{C}}_{2}` is also a symmetric positive matrix, so this sum can only be zero if both terms are zero.

:math:`\Rightarrow \{\begin{array}{}{\tilde{u}}^{T}{\tilde{C}}_{2}^{T}\text{.}{\tilde{C}}_{2}\tilde{u}=0\\ {\tilde{u}}^{T}\tilde{A}\tilde{u}=0\end{array}`


.. _RefEquation An2-5:

:math:`\Rightarrow {\tilde{C}}_{2}\tilde{u}=0\Rightarrow {\lambda }_{2}^{1}=0` eq An2-5

:math:`\tilde{A}` is a positive matrix, we want to show that:


.. _RefEquation An2-6:

:math:`{\tilde{u}}^{T}\tilde{A}\tilde{u}=0\Rightarrow \tilde{u}=0` eq An2-6

All we have to do is demonstrate that: :math:`\tilde{u}=0\text{et}{\lambda }_{1}^{1}=0`


* :math:`\tilde{u}=0`

Let's extend :math:`\tilde{u}` on :math:`{ℝ}^{n}` by :math:`\tilde{\tilde{u}}=0\text{}u=(\tilde{u},\tilde{\tilde{u}})`:

             :math:`{u}^{T}\text{Au}={\tilde{u}}^{T}\tilde{A}\tilde{u}=0\Rightarrow u\in \text{ker}A`

:math:`\text{Cu}=\left[\begin{array}{}{C}_{1}u\\ {C}_{2}u\\ {C}_{3}u\end{array}\right]=\left[\begin{array}{}{\tilde{C}}_{1}\tilde{u}\\ {\tilde{C}}_{2}\tilde{u}\\ {\tilde{C}}_{3}\tilde{u}\end{array}\right]=\left[\begin{array}{}0\\ 0\\ 0\end{array}\right]`

Indeed :math:`{\tilde{C}}_{3}=0` because otherwise, there would be :math:`\tilde{u}` ddls constrained by equations not yet taken into account (of rank :math:`>i`) which is contrary to :math:`{R}_{0}`.

The :math:`u` extension of :math:`\tilde{u}` is therefore in the cores of :math:`A` and :math:`C`. We're going to show that he sucks then. Let's go back to the problem with the "simple Lagranges".

:math:`({S}_{2})=\{\begin{array}{}\text{Au}+{C}^{T}\lambda =b\\ \text{Cu}=d\end{array}`

If :math:`{u}_{0}\ne 0` is such as :math:`{\text{Au}}_{0}=0` and :math:`{\text{Cu}}_{0}=0`.

If :math:`{u}_{1}` is a solution to S2, we can see that :math:`{u}_{1}+\mu {u}_{0}` is also a solution. This is impossible because we assume our problem is physically well posed.

We conclude that :math:`u=0\text{}\Rightarrow \tilde{u}=0`.


* :math:`{\lambda }_{1}^{1}=0`

[:ref:`éq An2-4 <éq An2-4>`] gives:


.. _RefEquation An2-7:

:math:`{\tilde{C}}_{1}^{T}{\lambda }_{1}^{1}=0` eq An2-7

In the same way as the :math:`{R}_{0}` rule imposing :math:`{\tilde{C}}_{3}=0`, we see that :math:`{\tilde{\tilde{C}}}_{1}=0`.

[:ref:`éq An2-6 <éq An2-6>`] gives:

:math:`{C}_{1}^{T}{\lambda }_{1}^{1}=\left[\begin{array}{c}{\tilde{C}}_{1}^{T}{\lambda }_{1}^{1}\\ {\tilde{\tilde{C}}}_{1}^{T}{l}_{1}^{1}\end{array}\right]=\left[\begin{array}{c}0\\ 0\end{array}\right]=0` eq An2-8

Let's reason with the absurd: if :math:`{\lambda }_{1}^{1}\ne 0` is such as :math:`{C}_{1}^{T}{\lambda }_{1}^{1}=0`, it's because there is a linear combination of the lines of :math:`{C}_{1}` that is zero, which is contradictory to the fact that the lines of :math:`{C}_{1}` are independent of each other (a well-posed physical problem).

So :math:`{\lambda }_{1}^{1}=0`.

**Special case:**

When one (or more) of the sets :math:`\tilde{u}`, :math:`{L}_{1}`, :math:`{L}_{2}` is empty, the :math:`(S)` system simplifies. We can verify that the reasoning we made in the general case makes it possible to demonstrate similar results.