Models-Partial Differential
Equations-and Location
Problems
Diaraf SECK
[email protected]
Université Cheikh Anta Diop de Dakar
Ecole Doctorale de Mathématiques et
Informatique
Laboratory of Mathematics and Applications
Faculté des Sciences Economiques et de
Gestion
CIMPA, DAKAR DU
05 au 15 AVRIL 2011
April 15, 2011
Chapter I: Models presentation.
Exemples de problèmes d’ingénieurs
Contraintes thermoelastiques dans un échangeur
Croisement de trains ou TGV dans les tunnels
Résonnance magnétique nucléaire
Zone d’action, un champs magnétique intense
et le plus uniforme possible. On suppose un
champs léger fluctuant.
Contacts thermiques electriques et thermoelastiques
Matériaux composites tissés
Tubes(flambage et les essaies techniques)
flambage: quand il y a un conflit de symétrie.
Design ou profil d’ailes d’avions, de maisons
ou ...
Phénomènes biologiques, écologiques
Par exemple : la pollution,les tumeurs, évolution
des insectes, bactéries etc.
Exemples de problèmes pédagogiques
Ecoulement de fluide dans un cylindre
Traction simple d’un échantillon
Ingrédients necessaires
On va avoir des éléments descriptifs, (d’ordre
cinématique)
- intensifs
et des lois
Eléments descriptifs
a) Eléments cinématiques ou géométriques
a = {a1, a2, a3} les coordonnées d’une particule
à l’instant t = t0.
x = {x1, x2, x3} les coordonnées de la même
particule à l’instant t.
x = ϕ(a, t), (équation de la trajectoire)
u = x − a = ϕ(a, t) − a, déplacement, inconnue
fondamentale pour les solides.
∂ϕ
dϕ
v(x, t) = dx
=
=
dt
∂t
dt , inconnue fondamentale
en mécanique des fluides.
b) Eléments intensifs
−
→
−
→
. T (x, k ), vecteur contraintes en x pour une
−
→
direction k de l’espace.
. On peut être interessé par les températures:

conduction (il y a contact)

rayonnement

 convection (il y a mouvement)
Les champs éléctrique et magnétique
les inductions éléctreique et magnétique
Les lois qui relient ces grandeurs entre elles
a) Les lois générales
. Conservation de la masse
. principe fondamental de la dynamique
. 1er principe de la thermodynamique
. 2eme principe de la thermodynamique: loi
de dissipation
. les équations de Maxwell.
Les lois particulières
Nécessité d’outils mathématiques
Constatations physiques
rhéologie (en mécanique)
(les lois de comportement, les lois d’états, les
lois constitutives.)




 newtonnien ou non 



fluides :
gaz






liquides



élastiques





solides :
plastiques





 viscoélastiques
et les phénomènes thermiques.
milieux isolants ou non, aimentés ou non, polarisés ou non.
Ingrédients mathématiques néccessaires à la
description d’un milieu continu
Coordonnées d’une particule
Référentiel:
un instant de référence,
une référence spatiale, par exemple un repére
orthonormé, connaı̂tre l’évolution d’un milieu.
une relation de type x = F{(X, T, t)}, le lieu x
est la trajectoire de la particule (X, T ).
F
 n’est pas quelconque, c’est une relation:

reflexivité x = F{(x, t, t)}

symétrie X = F{(x, t, T )}alorsx = F{(X, T, t)}

 transitivité x = F{(X, T, t)} et si X = F (ξ, τ, t)]
alors
x = F [F (ξ, τ, t), T, t] = x = F (ξ, τ, t)
Coordonnées d’Euler
de
 la particule sont désignées

x1


par les variables  x2 
x3
Les coordonnées de Lagrange sont les trois
variables qui sont liées à la position de référence,
une description plus simple, on prend T = 0
x = F (a, 0, t) = ϕ(a, t)
a désigne la coordonnée de Lagrange
x désigne la coordonnée d’Euler
La fonction ϕ est inversible, on peut trouver a
en fonction de x: a = ψ(x, t)
x = ϕ(a, t) =⇒ x = ϕ[ψ(x, T ), t]
Comportement des fluides en évolution isotherme
ou adiabatique
−
→
∂ρ
1. ∂t + divρ V = 0 conservation de la masse
(2)
−
→
∂V
−−→ V 2
−−→
→ −
→
−→−
+ grad
+ rot V ∧ V ] + gradp =
ρ[
∂t
2
−−→ −
→
→
−
→ −
(λ + µ)graddiv V + µ∆ V + f
−
→
−
→
−
→
dV
∂V
∂V
−
→−
→
=
+ grad V . V =
+ (V.∇)V
dt
∂t
∂t
Remark: Modélisation du problème de pollution dans un milieu poreux non saturé
Lois de comportement élastiques en milieu isotherme
et adiabatique
ρ0 = ρdet(1 + ∇u); les hypothèses de petites
→
pertbations entraı̂nent: ρ ' (1 − div −
u )ρ0 ' ρ0
et
∂ 2ui
∂
∂u
ρ0 2 +
[aijkh k ] = fi
∂t
∂xj
∂xh
Ω0
Other Importants models:
Boltzmann equation:
∂f
+ v∇xf = Q(f, f ), t ≥ 0, x ∈ RN , v ∈ RN
∂t
Q depends on the Boltzmann collision kernel,
v is the velocity of particles
and f : a distribution function (density of particles)
for more details see for example: Boltzmann,
....., Cercignani ,P.L. Lions, Di Perna, N. Masmoudi, C.D. Levermore, W. Greenberg, L St
Raymond, F. Golse, C. Bardos, T. Goudon,
E.Carlen, I. Gamba, ......,
C. Villani (A review of mathematical topics in
collisional kinetic theory,:Handbook of Mathematical Fluid Dynamics (Vol. 1), edited by
S. Friedlander and D. Serre, published by Elsevier Science (2002), ......), ........ references
therein
Plasma:
∂f
+ v∇xf + F (x)∇v f = 0,
∂t
F = −∇V ,
R
2
e
V = 4π r ∗x ρ, ρ(t, x) = f (t, x, v)dv
0
equation for one species of particles,
and also there is not not the effect of a magnetic field, which leads to the Vlasov- Maxwell
system
e: charge of the particle
and : permittivity of vacuum
for more details Balescu Delcroix and Bers ,
Decoster, C. Villani, Clément Mouhot, P.A.
Raviart, E. Sonnendrucker, E. Frénod, .......
references therein
Chapter II: Presentation of shape and
topological optimization
II.1. Shape Optimization
General Formulation:
min J(Ω, uΩ)
Ω∈Θ
(1)
Θ est un ensemble de domaines admissibles,
et uΩ est solution d’une certaine équation
(dans beaucoup de cas c’est une équation aux
dérivées partielles) posée dans Ω. et des conditions aux limites.
Beaucoup de travaux dans ce domaine ont fait
l’objet de plusieurs papiers . On souhaiterait
donc avoir toujours des solutions pour ces types
de problémes, mais lorsqu’il n’y a pas des hypothéses restrictives. Ces types de problémes
n’admettent pas toujours de solution:
Soient D =]0, 1[x]0, 1[, f > 0, d ∈ R, d > 0 et
w est solution de
(
−∆w + 2π
d w = = f dans D
w ∈ H01(D)
(2)
et considérons
J(Ω) =
Z
Ω
(uΩ − w)2dx
avec
(
−∆uΩ =
= f dans Ω
uΩ ∈ H01(Ω)
(3)
Alors le probleme
min J(Ω)
Ω∈O
avec
1
O = {ω ⊂ D :
dx ≥ , }
2
ω
ne posséde pas de solution.
Z
Hadamard, D. Henry, M. Schiffer, G. Szego,
J.L. Lions, D. Chenais, J.P. Zolesio, J. Sokolowski,
F. Murat, J. Simon, O. Pironneau, , Ashbaugh,
Benguria, M. Crouzeix, J. Ca, M. Pierre, A.
Henrot, M. Masmoudi, P. Guillaume, G. Allaire, D. Bucur, G. Buttazzo, G. Dalmaso, G.
Bouchitt, A. Acker, H.W.Alt, L. Caffarelli, E.
Zuazua, Rappaz, J Descloux, C. Bandle, O.
Besson, M. Delfour, Fasano, Sverak, A. Chambolle etc.........
Henrot A. and Pierre M.: Variations et Optimisation de formes, une analyse géométrique,
Mathématiques et Applications, 48, Springer
2005.
A. Henrot: Extremum Problems for eigenvalues of Elliptic Operators, Birkhauser, 2006
D. Bucur and G. Buttazzo: Variational methods in shape optimization problems, Birkhauser
2005
Michel Delfour and J. Paul Zolsio: Shapes
and Geometries: Analysis, Differential Calculus and Optimization, Siam, 2000
J. Sokolowski and J. P. Zolsio:Introduction to
shape optimization: shape sensitivity analysis,
Springer, 1992
II.2. Topological Optimization
For a given x0 ∈ Ω, consider the perforated
open set Ωε = Ω\ωε, ωε = x0 + εω, ω ∈ Rd is
a fixed reference domain. The physical meaning of the small ωε is different from problem
to another: for example, in the case of solid
elastic problem, it means to remove some materials from the initial shape, it can be seen as
an obstacle in the viscous fluid.
There are two popular methods: J. Sokolowski
et al. and M. Masmoudi et al.
We recall here to the general adjoint method
and domain truncation M. Masmoudi et al. in
order to get topological derivative.
Let uΩε be the solution of the equation in the
perturbed domain:
The aim of the topological optimization is to
compute the difference J(Ωε)−J(Ω). For many
cases, the asymptotic expansion of the function J can be obtained in the following form:
J(Ωε) = J(Ω) + ρ(ε)G(x0) + o(ρ(ε))
lim ρ(ε) = 0,
ε→0
(4)
ρ(ε) > 0.
The function ρ(ε) depends of the space dimension and the boundaries conditions on ∂ω.
The function G(x0) is called topological derivative (or topological sensitivity) and provides an
information for creating a small hole located at
x0. Hence the function G can be used like a
descent direction in optimization process.
Chapter III: Application to some
mathematical models
I. Faye, M. NGom, A Sy, D. Seck
III.1. Application to crystals problems
a) Phononics Problems
Definition: Phononic crystals are synthetic
materials that are formed by periodic variation
of the acoustic properties of the material.
In this paragraph, we propose a modeling of
the acoustic wave problem. The modeling is
based on the out-flow of a fluid, barotrope,
occupying a domain Ω of the space and immersing a unsettled domain with a weak amplitude (sound source) occupying himself a domain Ω1(thus Ω1 ⊂ Ω). We study the propagation of these small perturbations in the fluid,
and therefore the propagation of the sound.
→
One assumes that the strengths of weight f
are negligible in front of the strengths of inertia.
→
f=0
(5)
Thus, we can suppose that the out-flow irrotational.
→→
rot u = 0
(6)
It follows that there exists a function ϕ such
that
−
→
→
u = ∇ϕ
(7)
From equations of mass conservation, we have
∂ρ
+ ρ,iϕ,i + ρ∆ϕ = 0
∂t
∂ ∂ϕ
u2
1 ∂p
[
+
]+
= 0 , i = 1, 2, 3.
∂xi ∂t
2
ρ ∂xi
(8)
(9)
with
→
2
u = | ∇ϕ |2.
(10)
We make classical hypotheses of small perturbations, such that :
X The velocity ui is enough small as well as
i
their variations ui,k , ∂u
∂t ,
X The pressure p and the density ρ have small
variations around the constant initial data
p0 and ρ0. The linearization of the equations (8) and (9) gives:
∂ρ
+ ρ∆ϕ = 0
∂t
(11)
∂ ∂ϕ
1 dp ∂ρ
( )+
=0
∂xi ∂t
ρ dρ ∂xi
(12)
∂ρ
as ∂ϕ
and
∂t
∂xi remain small, we can replace ρ
dp
by the constants ρ0 and c2
and dρ
0 (see [?])
then :
∂ρ
+ ρ0∆ϕ = 0
(13)
∂t
and
∂ ∂ϕ
ρ
+ c2
(
(14)
0 )=0
∂xi ∂t
ρ0
ρ
∂ϕ
+ c2
is a function of t
0
∂t
ρ0
Then
∂ϕ
ρ
2
+ c0
= k(t)
∂t
ρ0
(15)
ϕ can be modify by a function of t without
→
changing the value of u ; indeed if we set
Φ=ϕ−
Z t
0
k(s)ds
we get
→
→
u =∇Φ and ∆ϕ = ∆Φ.
(16)
A combination of the relations (14), (15) and
(16) show that Φ is obtained by the wave equation
1 ∂ 2Φ
− ∆Φ = 0
2
c2
0 ∂t
(17)
In order to get a well posed problem, we add
to (17) the following initial conditions
∂Φ
Φ(x, 0) = 0,
(x, 0) = 0
∂t
(18)
Setting Φ = Φ0eiωt, it follows ∂Φ
∂t = iωΦ and
∂ 2 Φ = −ω 2 Φ
∂t2
the equation (17) becomes:
ω2
− 2 Φ − ∆Φ = 0
c0
(19)
and setting k = cω , we can write the final model
0
under the form :
(
−∆Φ = k2Φ in Ω
Φ
=
0
on ∂Ω
(20)
Using one or another approach for computing
topological derivative, we get the following results.
For the phononic problem, the aims is to get
the asymptotic expansion of the functional:
J(u) = J(λΩ) = (λΩ − a)2
(21)
Theorem: Let j(ε) = Jε(uε) the cost function, where J is defined by (21) and u is solution of (20). Then j has the following asymptotic expansion
j(ε) = j(0) + f (ε)g(x0) + o(f (ε))
where the topological derivative at x0 ∈ Ω is
g(x0) = −2πu(x0)v0(x0)
v0 is solution of the adjoint problem
(
−∆v0 = −4(λΩ − a)∆u on Ω
v0
=
0
in ∂Ω
(22)
Step 0: Given an initial domain Ω0
Step 1: Solve the direct problem
Step 2: Solve the adjoint problem
Step 3: Compute the topological sensitivity
NO
Step 4: Create a small hole where the topogical
derivative is the most negative
Step 5: The optimal shape is reatched
Yes
END
Figure 1: The proposed algorithm
Initial mesh
0.8
0.6
0.4
0.2
0
−0.2
−0.4
−0.6
−0.8
−1
−1.5
−1
−0.5
0
0.5
1
1.5
Figure 2: step 0: Domain without hole
λ1(Ω0) = 5, 74
Figure 9: step 7: Domain with seven holes
λ1(Ω7) = 23, 87
Remark: There is any possibility to put another hole, because, in the last domain, the
topological derivative is equal to zeros, almost
everywhere. The optimal shape design is reached
after putting seven holes in the initial domain.
Then all admissible frequencies near the limit
audible one can pass.
b) The photonic problem
Definition: Photonic crystals are composed of
periodic dielectric or metallo-dielectric nanostructures that affect the propagation of electromagnetic waves (EM) in the same way as
the periodic potential in a semiconductor crystal affects the electron motion by defining allowed and forbidden electronic energy bands.
A crystal photonic problem is a periodic dielectric structure that has the feature that there
probability frequencies for the propagation of
the electromagnetic waves insides.
The photonic problems based on the electromagnetic equations.
Let E be the electric field, B the magnetic one,
i, j the current density vector, µ the medium
magnetic permeability, , the medium permittivity and ρ the volume density of the electric
charge.
The Maxwell, magnetic and electric equations
write
I
I
→
→
−
→−
−
→−
E .ds = Q,
B .ds = 0
|
{z
electric
}
|
{z
}
magnetic
dB
→
−
→−
E .ds = −
dt}
|
{z
I
, ∇E = −
∂B
∂t
F arady 0 s Law of electromagnetic induction

−
→
∂E −
→
→
−
→−

B . dl = µ .ds + i,
s ∂t

I
Z
∂D
−
→−
→
∇ ·H = J +
∂t
|
{z
}
Ampere0 s law extended by M axwell
where D = E and B = µH.
It follows that the Maxwell equations write in
differential form:











∇.E
∇H
∇×E
∇×H
=
ρ/
=
0
= −µ ∂H
∂t
= j + ∂E
∂t
(23)
In the special case where ρ = 0, j = 0, µ = µ0,
the Maxwell system writes











∇.E
∇H
∇×E
∇×H
=
0
=
0
= −µ ∂H
∂t
∂E
= ∂t
(24)
Thus
∂H
∂ 2E
∇ × (∇ × E) = ∇ × (−µ0
) = −µ0 2
∂t
∂t
It is well know that
2E
∇ × (∇ × E) = |∇(∇.E)
−∇
{z }
=0
It follows that
∂ 2E
2
−∇ E = −∆E = −µ0 2
∂t
Without lost of generality, we can set µ0 = 1,
we obtain
∂ 2E
− ∆E = 0
2
∂t
(25)
When the solution wave is monochromatic (and
that depends on boundary and initial conditions), E is of the form
E(x, t) = Re(u(x)eikt)
where u is a solution of the Helmholtz equation.
Adding initial conditions, we obtain the final
model of the crystal photonic problem.
(
∆u + k2u = 0 on Ω
u
= 0 in ∂Ω
(26)
Using one or another approach for computing
topological derivative, we get the following results.
For the photonic problem, the aims is to get
the asymptotic expansion of the functional:
min JΩ(uΩ) =
Z
Ω
|∇uΩ|2dxdx
(27)
Theorem: Let j(ε) = Jε(uεΩ) the objective
functional then j have the following asymptotic
expansion :
j(ε) − j(0) = f (ε)g(u(x), v0(x)) + o(f (ε))
where v0 is a solution of the adjoint problem,
which strong formulation is
(
∆v0 + k2v0 = −DJ(uΩ)
v0 = 0
in Ω
in ∂Ω
(28)
In the case where ωε = B(0b1), the topological
derivative is given by:
G(x0) = −2π(∇uΩ(x0).∇vΩ(x0)+uΩ(x0)vΩ(x0))
Initial Mesh
0.8
0.6
0.4
0.2
0
−0.2
−0.4
−0.6
−0.8
−1
−1
−0.5
0
0.5
1
1.5
Figure 10: Simulation for the photonic problem
Remark: In this case, the topological derivative is equal to zeros, almost everywhere in Ω
except in some small parts of the boundary of
the domain. Thus, we can not create a hole
centered to x0 ∈ Ω.
III.2. Application to a thermoelasticity
problem.
C. Diallo, I. Faye, D. Seck
In this work we consider the system as the following form:

−−→
−−→
−
→

−µ∆~
uΩ − (λ + µ)grad div ~
uΩ − 3kαgrad θΩ = f




−∆θΩ = g in Ω
∂θΩ


= h on ∂Ω

∂n


B~
uΩ = v on ∂Ω.
(29)
B is a Dirichlet or Neumann operator defined
on ∂Ω, f~ ∈ L2(Ω, R3), g ∈ L2(Ω), v ∈ L2(∂Ω, R3) and
L2(∂Ω). This system models phenomena of
2
3
thermoelasticity where the vector ~
uΩ = (u1
Ω , uΩ , uΩ )
is the deformations vector and θΩ is the temperature in the domain Ω.
Presentation of the thermo elasticity problem
Let Ω, ω two regular open and bounded sets
of RN , N = 2, 3 and let x0 ∈ ω ⊂ Ω. is
a small positive real. Let us defined the hall
ω = x0 + ω and the perturbed domain Ω() =
Ω \ ω. For the theorical study let us suppose
x0 = 0. Consider

−−→

~ in Ω

−µ∆~
u
−
3kα
grad
θ
=
f
Ω
Ω




div(uΩ) = 0
in Ω


−∆θΩ = g in Ω
(30)

∂θΩ




∂n = h on ∂Ω


∂~
uΩ

∂n = v on ∂Ω.
In Ω() = Ω \ ω we have

−−→

−µ∆~
uΩ() − 3kαgrad θΩ() = f~ in Ω()






div(uΩ()) = 0
in Ω()





−∆θΩ() = g in Ω()




















∂θΩ()
∂n = h on ∂Ω
∂~
uΩ()
= v on ∂Ω
∂n
∂~
uΩ()
~
uΩ() = 0 or ∂n
= 0 on ∂ω
∂θΩ()
θΩ() = 0 or ∂n = 0 on ∂ω.
(31)
In the following we set µ = 1 without loosing generality. Let us consider the functional
J(Ω, ~
uΩ, θΩ) defined by
J(Ω, ~
uΩ, θΩ) =
3
X
αi
Z
Ω
i=1
3
X
i
βi
Z
Ω
|∇ui|2dx + r
δ
Z
Ω
Z
Ω
|ui − ui0|2dx+
|θΩ − θ0|2dx+
|∇θΩ|2dx
where αi, βi, i = 1, 2, 3, r and δ are constants.
For any positive real > 0 we consider j() =
J(Ω(), ~
uΩ(), θΩ()) where (~
uΩ(), θΩ()) is solution to (31) and (~
uΩ, ,θΩ) solution to (30).
The aim of this section is to detimine the
asymptotic analysis of functional j() as → 0.
Results
Theorem
Let us suppose that in the functional (??)
βi = 0, i = 1, 2, 3, δ = 0 ie the functional is not
depending of ∇uΩ, ∇θΩ and the Dirichlet conditions are prescribed in ∂ω. Then functional
J(Ω, uΩ , θΩ ) defined by (??) where (uΩ , θΩ )
is solution to (31) admits the following asymptotic expansion
Z
0
ω
J(Ω)−J(Ω)−
≤ o(2)
F
(x,
v(x))η(x)m
v(x)dx
u
Ω
(32)
where η(x) = (η 1(x), . . . , η 4(x)) is the Green
matrix and v = (uΩ, θΩ) is solution to problem
P
i
i
(30) and F0u = 2 4
i=1 (uω − u0 )
Theorem
Let u be the solution of problem (31) and B ω
is a Neumann condition on ∂ω. Then functional J(u) defined by (??) with r = δ = 0
and αi = βi = 1, i = 1, 2, 3 admits the following asymptotic expansion
J(u) = J0(u) + 32π (u − u0)2 + |∇u|2−
Z
R3 \ω
2∇u.∇z(ξ)dξ − 3mω V · u + o(3+δ )
where u is the solution to problem (30) and V
solution to the adjoint problem
0 (x, u(x), ∇u(x))
−∆V = Fu0 (x, u(x), ∇u(x)) − ∇xF∇u
∂V = 0 on ∂Ω
∂n
(33)
mω is the polarization matrix, z = (z 1, . . . , z 3)
solution to problem (??), Fu0 (x, u(x), ∇u(x)) =
0 (x, u(x), ∇u(x)) = −∆u.
2(u − u0), and F∇u
proof
The proof is essentially based on the asymptotic expansion of the functional.
(
Michell trusses problem
Let us begin this section by a presentation of
the problem. We ask the reader to see for example the interesting and meaning full work
due to Bouchitte et al [?] and their references.
Our aim is to link this problem to shape and
topological optimization and we will end our
work by numerical simulations.
Presentation
For the presentation of the Michell trusses problem, we are going to give some elements which
can be found in Bouchitte et al; and for more
details see this reference and the others therein.
A truss is a finite union of bars (Ai, Aj ), i 6=
j, i, j ∈ {1, . . . , n} subjected to a force F =
Pn
i=1 Fi δM i and result of a tension σ :
σ=
n
X
i,j=1
λij σ [Ai,Aj ],
(34)
where Ai ∈ R3, i = 1, . . . , n and σ [Ai,Aj ] is given
by
σ [Ai,Aj ] =
Ai − Aj
Ai − A j 1
⊗
H[A −A ], (35)
i
j
|Ai − Aj | |Ai − Aj |
δMi is the Dirac mass at a point Mi ∈ R3. The
truss is in equilibrium when
divσ + F = 0.
(36)
The problem of Michell trusses is to find all
n
points A = (Ai)n
i=1 ⊂ R and all reals λ =
(λij )n
i,j=1 ⊂ R, which minimize
C(A, Λ) =
n
X
|λi,j ||Aj − Ai|
(37)
i,j=1
such that
(
σ=
Pn
[Ai ,Aj ]
λ
σ
ij
i,j=1
−divσ = F
(38)
Using the second equation of (37) the problem
is equivalent to a decomposition of F as
n
X
Ai − Aj
F =
λi,j (δAi − δAj )
|Ai − Aj |
i,j=1
(39)
with C(A, Λ) minimal.
Let us introduce
T
X
(s̃) = {σ ∈ M (s̃, S n ∗ S n) such that
F
σ=
n
X
λij σ [Ai,Aj ] and − divσ = F }.
i,j=1
Our aim is to see the Michell trusses as a topological optimization problem. For a deformation (displacement) u for all pair (A, Γ) solution
to the problem of Michell trusses we have
Z
Ω
≤
< F, u > dx =
n
X
λi,j < u(Ai−u(Aj );
i,j
n
X
i,j
λi,j |Ai − Aj | ≤
n
X
Ai − Aj
>
|Ai − Aj |
|λi,j ||Ai − Aj | = C(A, Λ).
i,j
(40)
Let us mention that in Bouchite et al;, they
showed that
Z
min{
Ω
Z
|σ|, σ ∈ ΣF (s)} = max{
Ω
< F, u >, u ∈ U1(s)}
(41)
where
U1(Ω) = {u : Ω̄ → Rn, u ∈ C(Ω̄) and kukΩ ≤ 1}
and
kukΩ = sup{
| < u(x) − u(y); x − y > |
,
2
|x − y|
∀x 6= y (x, y) ∈ Ω2}.
To go back to the objective of this section
which is to study a type of Michell trusses
problem as a topological optimization problem.
Let us introduce the classical model in elasticity in the stationary case: this means that
−divσ = F where σ(u) = λT r(ε(u)) + 2µε(u)
with σij (u) = λ(divu)δij + 2µεij (u) and εij =
1 (u
2 i,j + uj,i) and div(u) = 0. Finally we have
−∆ui = Fi where F = (F1, F2, F3) and u =
(u1, u2, u3). Multipying by ui and integrating
we get
∂ui
∇ui∇uidx−
uidσ =
Ω
∂Ω ∂n
Z
Z
Z
Fiuidx, i = 1, 2, 3.
i
Taking ∂u
∂n = 0 on ∂Ω we have
Z
3
X
Ω i=1
Fiuidx =
3 Z
X
|∇ui|2dx =
i=1
Z
Ω
< F, u > dx.
(42)
Let
J(u, Ω) =
3 Z
X
i=1 Ω
|∇ui|2
(43)
under the constraints divσ = F.
Remark
In the compressible case the topological derivative for a point x ∈ Ω of the compliance is given
by
π(λ + 2µ)
g(x0) =
4µAe(u)·
2µ(λ + µ)
e(u) + (λ − µ)tr(Ae(u))tr(e(u)) (x0) in R2
(44)
and
π(λ + 2µ)
g(x0) =
20µAe(u) · e(u)+
µ(9λ + 14µ)
(3λ − 2µ)tr(Ae(u))tr(e(u)) (x0) in R3. (45)
where ω = B(0, 1).
In the incompressible case ie when div u =
0 in Ω, solving the Michell trusses is therefore
to maximize the functional (43) along a set of
fields, so using the topology optimization. For
mathematical convenience we minimize −J(u, Ω)
where J is defined by (43). This reduces to reconsider cases of topological optimization problem related to the thermoelasticity because the
functional (43) is a particular case of general
functional (??) with αi = 0, βi = 1, r = δ = 0
and we consider only the elasticity problem
with F = (Fi), i = 1, 2, 3.
Theorem
Let J(u) the functional defined by (??) where
u = (u1, . . . , uN ) and ui is solution to (??).
Then we have the following assymptotic expansion
J(u) = J(u)+2π3 |∇u|3−∇u.∇V −u.V +o(3+δ ),
(46)
where δ is a positive integer and u = (u1, . . . , uN ); ui
is solution to (??) and V is solution to the adjoint state
(
0 (x, u(x), ∇u(x))
−∆V = −∇xF∇u
∂V = 0
∂Ω.
∂n
Ω
(47)
Numerical Simulations
Case of the thermoelasticity problem
In this section we consider
(
−∆θ = h in Ω
∂θ = 0 on ∂Ω,
∂n
(48)
where h = 10−2 is a given function and the
deformation vector u = (u1, u2, u3) is solution
to

∂θ = f in Ω

−∆u
−
3kα

i
i

∂xi



(49)
div(u) = 0 in Ω
∂ui
∂n = hi on ∂Ω
and ui0 = 2x + y − 2z is given.
We consider also the topological derivative of
functional (??) defined in theorem by
−2π (u−u0)2+|∇u|2+
Z
R3 \ω
2∇u.∇z(ξ)dξ −4πV ·u,
where V is the adjoint state associated to functional (??).
Let us take Ω = [−1, 1] × [−1, 1] × [−1, 1], α =
k = 1, fi = 10−2, hi = 0, i = 1, 2, 3. Using
finite elements methods and Getfeem++ we
obtain the following numerical simulations for
the topological derivative and the temperature
θ in Ω.
Representation of the temperature θ at the top
of this page and representation of the topological derivative at the bottom.
Case of Michell trusses problem’s
Let us consider a square in R2 and three points
A1, A2 and A3 in the square. Let us consider
also a decomposition of F under the form
F = α1 δ A 1 + α2 δ A 2 + α3 δ A 3 .
(50)
We first give the solution u = (u1, . . . , un) where
ui is solution to


 −∆ui = fi in Ω
div(u) = 0 in Ω
(51)

 ∂ui
∂n = 0 on ∂Ω.
Given a load F with finite support, we minimize numerically the functional J where u is
solution to (51). Let us consider the square
Ω = [−1, 1] × [−1, 1] and a decomposition of F
in the form (50). Let us consider two cases:
A1 = (0, 0), A2 = (1/2, 1/2), A3 = (1, 0)
(52)
and
B1 = (−1, −1), B2 = (−1/2, 1/2), B3 = (1/2, −1/2).
(53)
The results obtained for the topological derivative in the case where points defined by (52)
and (53) are given in figure??. The numerical
simulations show that C(λ) is minimum in the
considered points.
Representation of the topological derivative with
P
P3
in the top F = 3
α
δ
and
F
=
i=1 i Ai
i=1 αi δBi
at the bottom, λij = ji , Ai given by (52) and
bi given by (53).
These figures show that the topological derivative is smaller at the considered points, ie the
points where we have the Dirac distributions.
P
We consider also a decomposition of F = 5
i=1 αi Ai
where the points Ai are given by
A1 = ((−1, −1), A2 = (0, 1/2),
A3 = (1, 1)
A4 = (0, −1/2) and A5 = (1, −1)
and let λij = 1, 1 ≤ i, j ≤ 5 and λij = ji , i 6= j.
Then we obtain for the topological sensitivity
figure ??.
Representation of the topological derivative with
P
j
in the top F = 5
α
δ
,
λ
=
i
ij
A
i=1
i and F =
i
P5
i=1 αi δAi , λij = 1 at the bottom, Ai given by
(??).
In the case where Ω = [−1, 1]3 ⊂ R3 and F =
P5
3 , i = 1, . . . , 5
α
δ
where
the
points
A
∈
R
i
i
A
i=1
i
are given by
A1 = (0, 0, 0), A2 = (0, 1, 0), A3 = (0, 0, 1), A4 = (0, 0
(54)
the topological derivative is given in figure ??.
Representation of the topological derivative with
P
λij = ji and F = 5
i=1 αi δAi , Ai given by (54).
Chapter IV: Evolution de dunes sous marines
I. Faye, E. Frénod, D. Seck
Motivations
1. Objectif du programme de recherche : Simuler
la dynamique des bancs de sable à proximité des côtes dans les zones soumises à
la marée
2. Problématique : A chaque marée une masse
importante de sédiment va et vient avec un
effet résultant faible
3. Objectif de cette étude : Modéle duquel
il sera facile de supprimer la présence explicite de l’oscillation de marée
4. Dit autrement : Modéle traitable avec des
méthodes d’analyse asymptotique ou
d’homogénéisation
Méthodologie
• Modéle du phénomène valable à toutes les
échelles
• Tailles caractéristiques des phénomènes ciblés
• Rapports de ces tailles (→ )
• Modéle adimensionné paramétré par cf les deux papiers sur la question.