The constraint of inextensibility
imposed on a curve by a Lagrange multiplier
...with an Inf-Sup Puzzle
Jocelyn Étienne∗
∗
Jérôme Lohéac∗∗∗
Pierre Saramito∗∗
Laboratoire de Spectrométrie Physique,
∗∗
Laboratoire Jean-Kuntzman
CNRS – Université Joseph Fourier
Grenoble
The Setting

Z t

 γ(t, s) = γ 0 (s) +
u(τ, γ(τ, s)) dτ,
0


J(u) =
min
v∈V div (t)
J(v)
with V div (t) = v ∈ H01 (Ω)2 , div s v = 0 in Z 0 (Γ(t)) ∩ V 0 + uD
The Setting: the flow around a flag

Z t

 γ(t, s) = γ 0 (s) +
u(τ, γ(τ, s)) dτ,
0


J(u) =
min
v∈V div (t)
J(v)
with V div (t) = v ∈ H01 (Ω)2 , div s v = 0 in Z 0 (Γ(t))
∩ div v = 0 in L2 (Ω) + uD
Z
1
and J(v) =
|D(v)|2 dx, where D(v) =
∇v + (∇v)T
2
Ω
The Setting: which functional spaces?

Z t

 γ(t, s) = γ 0 (s) +
u(τ, γ(τ, s)) dτ,
0


J(u) =
min
v∈V div (t)
J(v)
with V div (t) = v ∈ H01 (Ω)2 , div s v = 0 in Z 0 (Γ(t)) ∩ V 0 + uD
we have Z (t) ⊂ H 1/2 (Γ(t)), there may be boundary conditions
The Variational Setting: saddle-point approach
Introduce ξ ∈ Z (t), and let:
Z
L(v, ξ) = J(v) −
ξ div s v ds,
Γ
the solution is such that:
L(u, ζ) = min max L(v, ξ).
v
ξ
The Variational Setting: saddle-point
1
Assume J(v) = a(v, v), then it is characterised by:
2
Z
1
L(u, ζ) = min max a(v, v) − ξ div s v ds
v
ξ
2
Γ
Z



 a(u, v) − ζ div s v ds = 0 ∀v ∈ V0
ZΓ
⇔



− ξ div s u ds = 0 ∀ξ ∈ Z (t)
Γ
Lagrange multiplier ζ is called the tension of the curve (e.g., flag)
The Variational Setting: tension force
By a generalisation of Green’s formula on curved domains,
Z
Z
ξ div s v ds =
Γ
−
Γ
∂ξ
t · v ds
∂s
.
Z
κξn · v ds
−
Γ
&
+ [ξt · v]B
A
The Variational Setting : space Z (t)
If ξ ∈ Z (t) ⊂ H 1/2 (Γ), then
Z
Z
ξ div s v ds =
Γ
−
Γ
∂ξ
1/2
1/2
∈ (H00 (Γ))0 but not (H0 (Γ))0
∂s
∂ξ
t · v ds
∂s
Z
−
κξn · v ds
Γ
1/2
1/2
+ [ξt · v]B
A
ζ, ξ ∈ Z (t) = H00 (Γ)
u, v ∈ H 3/2 (Γ)
ζ, ξ ∈ Z (t) = H 1/2 (Γ)
u, v ∈ H0
3/2
Note: H00 (0, +∞) is the space such that ξ ∈ H 1/2 (0, +∞) and
ξ
√ ∈ L2 (0, +∞).
x
1/2
⊂ H00 (Γ)
2D Navier-Stokes flow around 1D inextensible membrane
Z
Z
Z
d
d
Let a(u, v) = 2D(u) : D(v) dx+Re
u · v dx + ReΓ
u · v ds,
dt Ω
dt Γ
Ω
we have
Z
Z ∂ζ
a(u, v) −
p div v dx +
t + κζn v ds = 0 ∀v ∈ V0
∂s
Ω
Γ
Z
− q div u dx = 0 ∀q ∈ L20 (Ω)
Ω
Z ∂ξ
t + κξn · u ds = 0 ∀ξ ∈ Z (t)
Γ ∂s
Equivalent strong formulation (under regularity assumptions)
du
− div 2D(u) + ∇p = 0
in Ω \ Γ(t)
Re
dt
div u = 0
in Ω
div s u = 0
on Γ(t)
with a boundary condition on Γ(t)
−[p]n + [2D(u)]n = ReΓ
du ∂ζ
+
t + κζn
dt
∂s
Discretisation : 1. interface tracking
Discretisation : 1. interface tracking
Discretisation : 1. interface tracking
Discretisation : 2. finite elements
ux , uy ∈ P2 (Ω) ∩ C 0 (Ω),
p ∈ P1 (Ω) ∩ C 0 (Ω \ Γ(t)),
ζ ∈ P1 (Γ(t)) ∩ C 0 (Γ(t))
Resolution : augmented Lagrangian method
of thelinear

 :
Structure
 system
T
T
Fh
Uh
A B
C
 B 0
0   Ph  =  0 
0
Zh
C 0
0
I
momentum
incompressibility
inextensibility
I
In search of the saddle point: augmented Lagrangian (uh , { ph , ζh })
I
Uzawa algorithm:
h
i
A + r (B T B + C T C ) Uhn+1 = Fh − B T Phn − C T Zhn
Phn+1 = Phn + rBUhn+1
Zhn+1 = Zhn + rCUhn+1
I
Residuals at convergence of order 10−8 (momentum eq.) and 10−12
(incompressibility, inextensibility eqs.)
Closed membrane: object rendered undeformable
by incompressibility and inextensibility constraints
Closed membrane: object rendered undeformable
by incompressibility and inextensibility constraints
H 1 error
‘Open’ membrane...
‘Open’ membrane... and open problem: inf-sup condition
With ζ ∈ Zh1 (t) = ξ ∈ C 0 (Γ), ξ|e ∈ P1 (e), ∀e ∈ Γh
‘Open’ membrane... and open problem: inf-sup condition
With ζ ∈ Zh1 (t) = ξ ∈ C 0 (Γ), ξ|e ∈ P1 (e), ∀e ∈ Γh
With ζ ∈ Zh2 (t) = ξ ∈ C 0 (Γ), ξ|e ∈ P2 (e), ∀e ∈ Γh
Stationary inertia-less flag aligned in a laminar flow field
I
I
Solution is identical to the flow past a plate
Comparison with the asymptotic solution at lee point