8. Description of document versions

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

Be problem 1

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

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

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

\(A\) and \(b\) are set to \({ℝ}^{n}\)

\(A\) positive symmetric matrix of order \(n\).

Problem 2

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

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

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

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

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

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

Let’s show that problem 2” is equivalent to problem1:

Let’s be \(u\) solution of 2”

So, \(\forall v\in V\)

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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 0end {array}

First, let’s calculate the derivative of \(J(u)\):

\(\forall u\in V,\forall {v}_{0}\in {V}_{0},\)

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

Let \(u\) be the Pb 1 solution

\(\forall V\in V\), let’s say \({\mathrm{v}}_{0}\mathrm{=}\mathrm{v}\mathrm{-}\mathrm{u}\text{;}{\mathrm{v}}_{0}\mathrm{\in }{\mathrm{V}}_{0}\).

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

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

It is concluded that:

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Jtext {”} (u)cdot (v-u) = (text {Au} -b) (v-u) =0text {,}forall vin V

Definitions, notations

\(A\) is the unconstrained stiffness matrix \((n´n)\) (symmetric and positive)

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

\(u\) is the vector for physical degrees of freedom \(\mathrm{\in }{R}^{n}\)

\({\lambda }^{1}\) is the vector of the first Lagrange degrees of freedom \(\mathrm{\in }{\mathrm{ℝ}}^{p}\)

\({\lambda }^{2}\) is the Lagrange second degrees of freedom vector \(\in {ℝ}^{p}\)

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

We note:

• \(U\) the set of physical degrees of freedom,

• \({\Lambda }^{1}\) the set of the first Lagrange degrees of freedom,

• \({\Lambda }^{2}\) the set of Lagrange’s second degrees of freedom.

\(\alpha \in {ℝ}^{+}\)

\(K\) symmetric \(2p+n\) order matrix

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

The matrix \(K\) written above corresponds to a certain numbering of the unknowns:

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

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

Rule \(\mathit{R0}\) :

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

Then, to simplify the writing, we will use \(\alpha \mathrm{=}1\).

We want to show that any submatrix \({K}_{i}\) of \(K\) is invertible.

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

We will note:

\(\tilde{U}\) the subset of \(\mathrm{U}\) corresponding to the degrees of freedom of rank \(\mathrm{\le }i\).

\(\tilde{\tilde{U}}\) the subset of \(\mathrm{U}\) corresponding to the degrees of freedom of rank \(>i\).

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

\({L}_{1}^{1}=\left\{{\lambda }_{1}^{1}\right\}\text{;}{L}_{1}^{2}=\left\{{\lambda }_{1}^{2}\right\}\)

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

\({L}_{3}^{1}=\left\{{\lambda }_{3}^{1}\right\}\text{;}{L}_{3}^{2}=\left\{{\lambda }_{3}^{2}\right\}\)

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

\({L}_{2}^{1}=\left\{{\lambda }_{2}^{1}\right\}\text{;}{L}_{2}^{2}=\left\{{\lambda }_{2}^{2}\right\}\)

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

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

\({C}_{1}\)

\({C}_{2}\)

\({C}_{3}\)

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

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{tilde {C}} _ {i}

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

\(A=\left[\begin{array}{cc}\tilde{A}& \stackrel{ˉ}{A}\\ \stackrel{ˉ}{{A}^{T}}& \tilde{\tilde{A}}\end{array}\right]\)

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

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

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

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

The problem is equivalent to:

Problem 1:

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

General case:

Assume that \(\tilde{\mathrm{U}}\mathrm{\ne }\mathrm{\varnothing }\); \({L}_{1}\ne \varnothing\); \({L}_{2}\ne \varnothing\)

\((S)\mathrm{\Rightarrow }\)

\({\lambda }_{1}^{1}={\lambda }_{1}^{2}\) eq Year2-1

\(\tilde{{C}_{1}}\tilde{u}=0\) eq An2-2

\({\lambda }_{2}^{1}=\tilde{{C}_{2}}\tilde{u}\) eq An2-3

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

From [éq An2-4], we deduce:

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

From [éq An2-2], we get:

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

\(\Rightarrow \tilde{{u}^{T}}\tilde{{C}_{2}^{T}}\tilde{{C}_{2}}\tilde{u}+\tilde{{u}^{T}}\tilde{A}\tilde{u}=0\)

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

\(\Rightarrow \{\begin{array}{}{\tilde{u}}^{T}{\tilde{C}}_{2}^{T}\text{.}{\tilde{C}}_{2}\tilde{u}=0\\ {\tilde{u}}^{T}\tilde{A}\tilde{u}=0\end{array}\)

\(\Rightarrow {\tilde{C}}_{2}\tilde{u}=0\Rightarrow {\lambda }_{2}^{1}=0\) eq An2-5

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

\({\tilde{u}}^{T}\tilde{A}\tilde{u}=0\Rightarrow \tilde{u}=0\) eq An2-6

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

• \(\tilde{u}=0\)

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

\({u}^{T}\text{Au}={\tilde{u}}^{T}\tilde{A}\tilde{u}=0\Rightarrow u\in \text{ker}A\)

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

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

The \(u\) extension of \(\tilde{u}\) is therefore in the cores of \(A\) and \(C\). We’re going to show that he sucks then. Let’s go back to the problem with the « simple Lagranges ».

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

If \({u}_{0}\ne 0\) is such as \({\text{Au}}_{0}=0\) and \({\text{Cu}}_{0}=0\).

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

We conclude that \(u=0\text{}\Rightarrow \tilde{u}=0\).

• \({\lambda }_{1}^{1}=0\)

[éq An2-4] gives:

\({\tilde{C}}_{1}^{T}{\lambda }_{1}^{1}=0\) eq An2-7

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

[éq An2-6] gives:

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

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

So \({\lambda }_{1}^{1}=0\).

Special case:

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