prague sum · awell-posednessresultfortheregularizedsystem theorem(moussa-s.2012) existence and...
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On the motion of rigid bodies immersedin a two dimensional incompressible perfect fluid
Franck Sueur
Laboratoire Jacques-Louis Lions, Université Paris 6 (for 4 more days);Institut de Mathématiques de Bordeaux, Université de Bordeaux (from Sept. 1st)
Summer School and Workshop "Particles in Flows",Prague, August 2014
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Part 1. On the motion of a rigid body immersed in a two dimensionalirrotational and incompressible perfect fluid.
In the first part we revisit the complex-analytic approach of the motion of a rigidbody immersed in a two dimensional irrotational and incompressible perfect fluidinitiated by Blasius, Kutta, Joukowski, Chaplygin and Sedov.
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The equations
Fluid equations :
∂u∂t
+ (u · ∇)u +∇π = 0 for t ∈ (0,∞), x ∈ F(t),
div u = 0 for t ∈ [0,∞), x ∈ F(t),
Solid equations :
m(h)′′(t) =
∫∂S(t)
πn ds for t ∈ (0,∞),
J r ′(t) =
∫∂S(t)
(x − h(t))⊥ · πn ds for t ∈ (0,∞),
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Boundary and initial conditions
Boundary conditions :
u · n =(
(h)′(t) + r(t)(x − h(t))⊥)· n for t ∈ [0,∞), x ∈ ∂S(t),
lim|x|→∞
|u(t, x)| = 0 for t ∈ [0,∞),
Initial data :
u|t=0 = u0 for x ∈ F0,
h(0) = 0, (h)′(0) = `0, θ(0) = 0, r(0) = r0.
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Rotation
There exists a rotation matrix
Rθ(t) :=
(cos θ(t) − sin θ(t)sin θ(t) cos θ(t)
)such that
S(t) := h(t) + Rθ(t)x , x ∈ S0.
Furthermore, the angle satisfies
(θ)′(t) = r(t).
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Change of frame
We apply the following isometric change of variable :v(t, x) = RT
θ(t) u(t,Rθ(t)x + h(t)),
π(t, x) = π(t,Rθ(t)x + h(t)),
`(t) = RTθ(t) (h)′(t).
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Recasting of the equations
The equations become
∂v∂t
+[(v − `− rx⊥) · ∇
]v + rv⊥ +∇π = 0 x ∈ F0,
div v = 0 x ∈ F0,
m`′(t) =
∫∂S0
πn ds −mr`⊥
J r ′(t) =
∫∂S0
x⊥ · πn ds
v · n =(`+ rx⊥
)· n x ∈ ∂S0,
v(0, x) = v0(x) x ∈ F0,
`(0) = `0, r(0) = r0.
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The Kirchoff decomposition
−∆Φi = 0 in F0,∂Φi
∂n= Ki on ∂F0, Φi −→ 0 when x →∞,
where(K1, K2, K3) := (n1, n2, x⊥ · n).
One then assumes that the velocity is of the form
v := `1∇Φ1 + `2∇Φ2 + r∇Φ3.
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Added inertia
We introduceMa :=
[mi,j
]i,j∈1,2,3 .
where, for i , j ∈ 1, 2, 3,
mi,j :=
∫F0∇Φi · ∇Φj dx .
The index a refers to “added”, by opposition to the genuine inertia :
Mg :=
m 0 00 m 00 0 J
,
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Reformulation of the solid equations
We can first reformulate the main solid equations as follows :
(Mg +Ma)
(`r
)′= −(Ai )i=1,2,3 −
(mr`⊥
0
),
where for i = 1, 2, 3,
Ai :=12
∫∂S0|v |2Ki ds −
∫∂S0
(`+ rx⊥) · vKi ds.
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A complex-analytic approach
We identify C and R2 through (x1, x2) = x1 + ix2 = z .
We also use the notation f = f1 − if2 for any f = (f1, f2).
If f is divergence and curl free if and only if f is holomorphic.
Lemma (Blasius)
Let C be a smooth Jordan curve, f := (f1, f2) and g := (g1, g2) two smoothtangent vector fields on C. Then∫
C(f · g)n ds = i
(∫Cf g dz
)∗,∫
C(f · g)(x⊥ · n) ds = Re
(∫Czf g dz
).
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Laurent expansion of the Kirchhoff potentials
Lemma
Let C be a smooth Jordan curve, f := (f1, f2) a smooth vector field on C :∫Cf dz =
∫Cf · τ ds − i
∫Cf · n ds.
We infer from this lemma that ∫∂S0∇Φi dz = 0.
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Laurent expansion of the Kirchhoff potentials
Corollary
Let C be a smooth Jordan curve, f := (f1, f2) a smooth vector field on C :∫Czf dz =
∫C
(x1 + ix2)(f · τ) ds − i∫C
(x1 + ix2)(f · n) ds.
Lemma
We infer from this corollary that∫∂S0
z∇Φi dz = −(mi,2 + |S0|δi,2) + i(mi,1 + |S0|δi,1), for i = 1, 2;∫∂S0
z∇Φ3 dz = −(m3,2 + |S0|xG ,1) + i(m3,1 − |S0|xG ,2).
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Decomposition of the pressure
We start with the following observation :
Ai =12
∫∂S0|v − (`+ rx⊥)|2Ki ds −
12
∫∂S0|`+ rx⊥|2Ki ds,
and replace v by the decomposition
v = v# + r∇Φ3,
to get
Ai =12
∫∂S0|v# − `|2Ki ds +
12
∫∂S0|r(∇Φ3 − x⊥)|2Ki ds
+
∫∂S0
r(v# − `) · (∇Φ3 − x⊥)Ki ds −12
∫∂S0|`+ rx⊥|2Ki ds
=: Ai,a + Ai,b + Ai,c + Ai,d .
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Computation of Ai ,d
Using Stokes formula we easily check that(Ai,d
)i=1,2
= −r`⊥|S0|+ r2xG |S0| and A3,d = −r(` · xG )|S0|,
where |S0| is the Lebesgue measure of S0 and
xG :=1|S0|
∫S0
x dx .
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Computation of Ai ,a, i = 1, 2
We go on with the computation of
Ai,a =12
∫∂S0|v# − `|2Ki ds.
As v# − ` is tangent to the boundary, we can apply the Blasius formula,
v# − `(z) = −`1 + i`2 + `1∇Φ1 + `2∇Φ2,
and Cauchy’s residue theorem, to obtain(Ai,a
)i=1,2
= 0.
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Computation of A3,a
We proceed in the same way for i = 3 :
A3,a =12
∫∂S0|v#,I − `|2K3 ds
=12Re
(∫∂S0
z( v#,I − `)2 dz)
= Re
(∫∂S0
z(−`1 + i`2)(`1∇Φ1 + `2∇Φ2
)dz
)= (−`1)
[− `1m1,2 − `2(m2,2 + |S0|)
]+(−`2)
[`1(m1,1 + |S0|) + `2m2,1
]= `⊥M[`,
whereM[ is the following 2× 2 restriction ofMa :
M[ :=[mi,j
]i,j∈1,2 .
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Force and torque in the moving frame
We thus obtain : (A1A2
)= r2
(−m3,2m3,1
)+ r(M[`
)⊥and
A3 = `⊥M[`− r` ·(−m3,2m3,1
).
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Solid motion
Let
p :=
(`r
)and µ :=
m1,3m2,30
,
Let Γg : R3 × R3 → R3 and Γa : R3 × R3 → R3 be the bilinear symmetricmappings defined as follows :
∀p =
(`r
)∈ R3, 〈Γg , p, p〉 = mr
(`⊥
0
)and 〈Γa, p, p〉 =
(r(M[`)
⊥
`⊥ · M[`
)+ rp × µ,
Thus the solid equations read :[Mg +Ma
](p)′ + 〈Γg , p, p〉+ 〈Γa, p, p〉 = 0.
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Harmonic field
Let H the unique solution vanishing at infinity of
div H = 0 in F0, curlH = 0 in F0, H · n = 0 on ∂S0,
∫∂S0
H · τ ds = 1.
the vector field H admits a harmonic stream function ΨH(x) :
H = ∇⊥ΨH ,
which vanishes on the boundary ∂S0, and behaves like 12π ln |x | as x goes to
infinity.
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Harmonic field
the function H is holomorphic (as a function of z = x1 + ix2), and can bedecomposed in Laurent Series :
H(z) =1
2iπz+O(
1z2 ) as z →∞.
Coming back to the variable x ∈ R2, the previous decomposition implies
H(x) = O(
1|x |
)and ∇H = O
(1|x |2
).
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Harmonic field
The so-called conformal center of S0 is :
ξ1 + iξ2 :=
∫∂S0
zH dz .
In the case of a disk, we have H = HR2 where we denote
HR2(x) :=12π
x⊥
|x |2.
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Decomposition of the velocity field
For ` in R2, r and γ in R given, there exists a unique vector field v verifying :
div v = 0 and curl v = 0 in F0,
v · n =(`+ rx⊥
)· n on ∂S0,
limx→∞
v = 0,∫∂S0
v · τ ds = γ,
and it is given by the law :
v = γH + `1∇Φ1 + `2∇Φ2 + r∇Φ3.
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Decomposition of the velocity field
We will denote by
v := v − γH= v# + r∇Φ3,
where, as previously,v# := `1∇Φ1 + `2∇Φ2.
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Reformulation of the solid equations
The solid equations can be rewritten in the form
(Mg +Ma)
(`r
)′= −(Ai + Bi + Ci )i=1,2,3 −
(mr`⊥
0
),
where for i = 1, 2, 3,
Ai :=12
∫∂S0|v |2Ki ds −
∫∂S0
(`+ rx⊥) · vKi ds,
Bi := γ
∫∂S0
(v − (`+ rx⊥)) · HKi ds,
Ci :=γ2
2
∫∂S0|H|2Ki ds.
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Computation of Bi
Let us start with the term Bi . Let us prove the following.
Lemma
One has (B1B2
)= −γ(`)⊥ + γrξ,
andB3 = −γ ξ · `.
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Computation of Ci
From Blasius lemma 1, the Laurent Series of H and Cauchy’s Residue Theorem,we deduce that
C1 = C2 = C3 = 0.
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Conclusion
• Therefore the solid equations can be recast as follows :[Mg +Ma
](p)′ + 〈Γg , p, p〉+ 〈Γa, p, p〉 = γp × B,
where
B :=
(ξ⊥
−1
).
• Let us define, for any p ∈ R3, the energies
Eg (p) =12p · Mgp and Ea(p) =
12p · Map.
Then the total energyEg (p) + Ea(p)
is conserved along time.
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Part 2. On the motion of a rigid body immersed in a two dimensionalincompressible perfect fluid with vorticity.
In the second part we consider the motion of a rigid body immersed in a twodimensional incompressible perfect fluid with vorticity.
We prove a result of global in time existence and uniqueness similar to thecelebrated result by Yudovich about a fluid alone.
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Equations in the body frame
The equations become
∂v∂t
+[(v − `− rx⊥) · ∇
]v + rv⊥ +∇π = 0 x ∈ F0,
div v = 0 x ∈ F0,
m`′(t) =
∫∂S0
πn ds −mr`⊥
J r ′(t) =
∫∂S0
x⊥ · πn ds
v · n =(`+ rx⊥
)· n x ∈ ∂S0,
v(0, x) = v0(x) x ∈ F0,
`(0) = `0, r(0) = r0.
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Vorticity in the body frame
The vorticity, in the body frame, is given by
ω(t, x) := curl v(t, x).
Taking the curl of the Euler equation we get
∂tω +[(v − `− rx⊥) · ∇
]ω = 0 for x ∈ F0
which yields the following conservation laws, at least for smooth solutions : forany t > 0, for any p in [1,+∞],
‖ω(t, ·)‖Lp(F0) = ‖ω0‖Lp(F0).
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Global weak formulation
We introduce the following space
H :=
Ψ ∈ L2loc(R2)
/div Ψ = 0 in R2 and ∇Ψ + (∇Ψ)T = 0 in S0
.
It is classical that the space H can be recast thanks to the property :
∃(`Ψ, rΨ) ∈ R2 × R, ∀x ∈ S0, Ψ(x) = `Ψ + rΨx⊥. (1)
More precisely,
H =
Ψ ∈ L2loc(R2)
/div Ψ = 0 in R2 and satisfies (1)
,
and the ordered pair (`Ψ, rΨ) above is unique.
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Global weak formulation
Let us also introduce
H :=
Ψ ∈ H/
Ψ|F0 ∈ C 1c (F0)
and HT := C 1([0,T ]; H).
When (u, v) ∈ H × H, we denote
< u, v >= m `u · `v + J ru rv +
∫F0
u · v dx .
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Global weak formulation
Definition (Weak Solution)
Let us be given v0 ∈ H and T > 0. We say that v ∈ C ([0,T ];H− w) is a weaksolution in [0,T ] if for any test function Ψ ∈ HT ,
< Ψ(T , ·), v(T , ·) > − < Ψ(0, ·), v0 >=
∫ T
0<∂Ψ
∂t, v > dt
+
∫ T
0
∫F0
v ·((v − `v − rvx⊥
)· ∇)
Ψ dx dt −∫ T
0
∫F0
rv v⊥ ·Ψ dx dt
−∫ T
0mrv `⊥v · `Ψ dt.
We say that v ∈ C ([0,+∞);H− w) is a weak solution on [0,+∞) if it is a weaksolution in [0,T ] for all T > 0.
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Log-Lipschitz functions
The notation LL(F0) refers to the space of log-Lipschitz functions on F0, that isthe set of functions f ∈ L∞(F0) such that
‖f ‖LL(F0) := ‖f ‖L∞(F0) + supx 6=y
|f (x)− f (y)||(x − y)(1 + ln− |x − y |)|
< +∞.
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Result
Theorem (Glass-S. 2012)
For any u0 ∈ C 0(F0;R2), (`0, r0) ∈ R2 × R, such that :
div u0 = 0 in F0 and u0 · n = (`0 + r0x⊥) · n on ∂S0,
ω0 := curl u0 ∈ L∞c (F0),
lim|x|→+∞
u0(x) = 0,
there exists a unique solution (`, r , u) in
C 1(R+;R2 × R)× L∞(R+,LL(F0)).
Moreover for all t > 0,ω0 := curl u(t) ∈ L∞c (F0).
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Green’s function and Biot-Savart operator
LetG (x , y) be the Green’s function of F0 with Dirichlet boundary conditions,K (x , y) = ∇⊥G (x , y),K [ω] be the so-called Biot-Savart operator acting on ω ∈ L∞c (F0) throughthe formula
K [ω](x) =
∫F0
K (x , y)ω(y) dy .
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Green’s function and Biot-Savart operator
Proposition
Let ω ∈ L∞c (F0). Then K [ω] is in LL(F0), divergence-free, tangent to theboundary and such that curlK [ω] = ω.Moreover, it satisfies
K [ω](x) = O(
1|x |2
)as x →∞,
and its circulation around ∂S0 is given by∫∂S0
K [ω] · τ ds = −∫F0ω dx .
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Velocity decomposition
Proposition
Let be given ω in L∞c (F0), ` in R2, r and γ in R. Then there is a unique solutionv in LL(F0)
div v = 0, for x ∈ F0,curl v = ω for x ∈ F0,v · n =
(`+ rx⊥
)· n for x ∈ ∂S0,
v −→ 0 as x →∞,∫∂S0 v · τ ds = γ.
Moreover v is given by
v = K [ω] + (γ + α)H + `1∇Φ1 + `2∇Φ2 + r∇Φ3,
withα :=
∫F0ω dx .
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Hydrodynamic Biot-Savart operator and Green function
Let us also introduce the so-called hydrodynamic Biot-Savart operator
KH := K + αH,
which satisfies
div KH [ω] = 0, and curlKH [ω] = ω, for x ∈ F0,
KH [ω] · n = 0, for x ∈ ∂S0,∫∂S0
KH [ω] · τ ds = 0.
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Hydrodynamic Biot-Savart operator and Green function
Then v can be decomposed as
v = KH [ω] + γH + `1∇Φ1 + `2∇Φ2 + r∇Φ3.
We also introduce the hydrodynamic Green function GH as
GH(x , y) := G (x , y) + ΨH(x) + ΨH(y).
Consequently one has
KH [ω](x) =
∫F0∇⊥GH(x , y)ω(y) dy .
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A priori estimates. Vorticity transport and circulationconservation
In addition to the conservation of Lp norms of vorticity, Kelvin’s theorem alsoholds true, at least for smooth solutions, and the total vorticity is conserved aswell :
γ :=
∫∂S0
v(t, ·) · τ ds =
∫∂S0
v0 · τ ds.,
α =
∫F0ω(t, x) dx =
∫F0ω0(x) dx .
We introducev := v − βH,
whereβ = α + γ.
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A priori estimates. Energy-like estimate
Proposition
There exists a constant C > 0 (depending only on S0, m and J ) such that forany solution (`, r , v) of the problem on the time interval [0,T ], with
(`, r) ∈ C 1([0,T ];R3) and v ∈ C 1([0,T ]; (L2 ⊗ RH) ∩ C∞(F0)),
and with compactly supported vorticity,
E (t) :=12
(m|`(t)|2 + J r(t)2 +
∫F0
v(t, ·)2dx),
satisfies the inequalityE (t) 6 E (0)eC |β|t .
In the case where β = 0, that is, when the solution is of finite energy, the energyis conserved.
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A priori estimates. Bound of the body acceleration
Proposition
There exists a constant C > 0 depending only on S0, m, J , β and E (0) such thatany classical solution satisfies the estimate
‖(`′, r ′)‖L∞(0,T ) ≤ C .
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Reformulation
The system can be recast as follows :
∂tω +[(v − `− rx⊥) · ∇
]ω = 0,
(Mg +Ma)
[`r
]′=[∫F0
(v ·[((v − `− rx⊥) · ∇)∇Φi
]− rv⊥ · ∇Φi
)dx]i∈1,2,3
+
[−mr`⊥
0
],
wherev = KH [ω] + γH + `1∇Φ1 + `2∇Φ2 + r∇Φ3.
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Existence
Theorem (Schauder)
Let E denotes a Banach space and let C be a nonempty closed convex set in E .Let F : C 7→ C be a continuous map such that F (C ) ⊂ K, where K is a compactsubset of C . Then F has a fixed point in K.
F : (ω, `, r) 7→ (ω, ˜, r) with
∂t ω +[(v − `− rx⊥) · ∇
]ω = 0,
(Mg +Ma)
[˜
r
]′=[∫F0
(v ·[((v − `− rx⊥) · ∇)∇Φi
]− rv⊥ · ∇Φi
)dx]i∈1,2,3
+
[−mr`⊥
0
].
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Energy conservation
Let us define
E(ω) := −12
∫F0×F0
GH(x , y)ω(x)ω(y) dx dy − γ∫F0ω(x)ΨH(x) dx ,
andH := Eg (p) + Ea(p) + E(ω).
Proposition
The quantity H is conserved along the motion.
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Short-range and longe-range interactions
Lemma
Let f in L1(R2) ∩ L∞(R2). We denote by
ρf := inf d > 1 / Supp(f ) ⊂ B(0, d).
Then there exists C > 0 such that∫R2
∣∣∣ln |x − y |f (x)∣∣∣ dx ≤ C‖f ‖L∞ + ln(2ρf )‖f ‖L1 ,
for any y ∈ B(0, ρf ).
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Short/long-range interactions
Proposition
For some constant C, depending only on m,J , ‖ω0‖L1∩L∞ , |`0|, |r0|, |γ|, ρω0 andthe geometry,
|`(t)|+ |r(t)| ≤ C [1 + ln(ρ(t))],
whereρ(t) := ρω(t,·) = infd > 1 / Supp(ω(t, ·)) ⊂ B(0, d).
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Part 3. A vanishingly small body in an rotational 2d flow.
In the third part we still consider the motion of a rigid body immersed in a twodimensional incompressible perfect fluid with vorticity but we are now interested inthe limit where the body shrinks to a pointwise particle.
We will consider two cases depending on whether the body shrinks to a massive ora massless particle.
In both cases the circulation is assumed to be fixed.
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A vanishingly small body
Sε0 := εS0,
The body moves rigidly so that at times t it occupies a domain Sε(t) which isisometric to Sε0 . We denote
Fε(t) := R2 \ Sε(t)
the domain occupied by the fluid at time t starting from the initial domain
Fε0 := R2 \ Sε0 .
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The equations
∂uε
∂t+ (uε · ∇)uε +∇πε = 0 for t ∈ (0,∞), x ∈ Fε(t),
div uε = 0 for t ∈ [0,∞), x ∈ Fε(t),
mε(hε)′′(t) =
∫∂Sε(t)
πεn ds for t ∈ (0,∞),
J ε(rε)′(t) =
∫∂Sε(t)
(x − hε(t))⊥ · πεn ds for t ∈ (0,∞),
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Initial and boundary conditions
uε · n =(
(hε)′(t) + rε(t)(x − hε(t))⊥)· n for t ∈ [0,∞), x ∈ ∂Sε(t),
lim|x|→∞
|uε(t, x)| = 0 for t ∈ [0,∞),
uε|t=0 = uε0 for x ∈ Fε0 ,hε(0) = 0, (hε)′(0) = `ε0, θε(0) = 0, rε(0) = rε0 .
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Indeed, since Sε(t) is isometric to Sε0 there exists a rotation matrix
Rθε(t) :=
(cos θε(t) − sin θε(t)sin θε(t) cos θε(t)
)such that
Sε(t) := hε(t) + Rθε(t)x , x ∈ Sε0.
Furthermore, the angle satisfies
(θε)′(t) = rε(t).
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Two pointwise limits
Case (i) : the solid occupying the domain Sε(q) is assumed to have a massand a moment of inertia of the form
mε = m and J ε = ε2J ,
where m > 0 and J > 0 are fixed, so that the solid tends to a massivepointwise particle.Case (ii) : the solid occupying the domain Sε(q) is assumed to have a massand a moment of inertia of the form
mε = εαm and J ε = εα+2 J ,
where α > 0 and m > 0 and J > 0 are fixed, so that the solid tends to amassless pointwise particle.
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Vorticity and circulation
We will consider an initial fluid vorticity wε0 = w0 ∈ L∞c (R2 \ 0) independent of
ε and an initial circulationγ :=
∫∂Sε
0
uε0 · τ ds
around the solid independent of ε as well.
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Initial solid velocity
We will consider an initial solid velocity (`ε0, rε0 ) independent of ε :
(`ε0, rε0 ) = (`0, r0) ∈ R2 × R.
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Initial fluid velocity
The initial fluid velocity uε0 is then defined as the unique log-Lipschitz solution ofthe div-curl type system :
div uε0 = 0, curl uε0 = wε0 in Fε0 ,
uε0 · n = (`0 + r0x⊥) · n on ∂Sε0 ,lim|x|→∞ |uε0(x)| = 0,
∫∂Sε
0uε0 · τ ds = γ,
where wε0 := w0|Fε
0, hence, for ε small enough (depending on dist(supp w0; h0)
and the size of S0), we havewε
0 := w0.
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Biot-Savart operator in R2
We will use the following notation for the Biot-Savart operator in R2 :
KR2 [w ] :=
∫R2
HR2(x − y)w(t, y) dy ,
which is the convolution operator with the vector field HR2 (already) defined by
HR2(x) :=12π
x⊥
|x |2.
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Case (i). The massive limit
Theorem (Glass-Lacave-S. 2012)
Consider T > 0. Then, up to a subsequence, one has the following :hε converges to h weakly-∗ in W 2,∞(0,T ;R2),εθε converges to 0 weakly-∗ in W 2,∞(0,T ;R),wε converges to w in C 0([0,T ]; L∞(R2)− w∗),(h,w) satisfy
∂w∂t
+ div([
u +γ
2π(x − h(t))⊥
|x − h(t)|2
]w)
= 0,
mh′′(t) = γ(h′(t)− u(t, h(t))
)⊥,
w |t=0 = w0, h(0) = 0, h′(0) = `0.
with u(t, x) := KR2 [w(t, ·)](x).
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Case (ii). The massless limit
Theorem (Glass-Lacave-S. Soon)
For any T > 0, as ε→ 0+,hε converges to h weakly-∗ in W 1,∞(0,T ;R2),wε converges to w in C 0([0,T ]; L∞(R2)− w∗),(h,w) satisfy
∂w∂t
+ div([
u +γ
2π(x − h(t))⊥
|x − h(t)|2
]w)
= 0,
h′(t) = u(t, h(t)) in [0,T ],
w |t=0 = w0, h(0) = 0.
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Change of frame
We apply again the following isometric change of variable :vε(t, x) = RT
θε(t) u(t,Rθε(t)x + hε(t)),
ωε(t, x) = wε(t,Rθε(t)x + hε(t)) = curl vε(t, x),
`ε(t) = RTθε(t) (hε)′(t).
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A few basic remarks on scaling laws
Hε(x) =1εH1(xε
),
(∇Φεi )(x) = (∇Φ1
i )(xε
) for i = 1, 2,
(∇Φε3(x) = ε (∇Φ1
3)(xε
),
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A few basic remarks on scaling laws
LetMε
a :=[mε
i,j]i,j∈1,2,3 ,
withmε
i,j :=
∫Fε0
∇Φεi · ∇Φε
j = ε2+δi≥3+δj≥3
∫F0∇Φ1
i · ∇Φ1j .
Therefore
Mεa = ε2 IεMaIε, with Iε :=
1 0 00 1 00 0 ε
.
This has to compared with
Mεg :=
[mε Id2 0
0 J ε]
= εα IεMg Iε.
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Energy estimates
We have seen that the following quantity is conserved along the motion :
Hε =12
(Iεpε)T(εαMg + ε2Ma
)(Iεpε)
−12
∫Fε0×Fε
0
G εH(x , y)ωε(x)ωε(y) dx dy − γ∫Fε0
ωε(x)ΨHε(x) dx ,
where
pε :=
(`ε
rε
)and Iεpε :=
(`ε
εrε
).
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Energy estimates
Using the short/long-range decomposition of the interactions we deduce that
|ε`ε(t)|+ |ε2rε(t)| ≤ C [1 + ln(ρε(t))],
whereρε(t) := ρωε(t,·) = infd > 1 / Supp(ωε(t, ·)) ⊂ B(0, d),
for some constant C = C (m,J0, ‖w0‖L1∩L∞ , |`0|, |r0|, |γ|, ρw0), depending only onthese values and the geometry for ε = 1.
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Decomposition of the pressure
In the presence of vorticity the solid equations can be rewritten in the form
Mε
(`ε
rε
)′= −(Aεi + Bεi + C εi )i=1,2,3 − (Dε
i )i=1,2,3 −(mrε(`ε)⊥
0
),
where for i = 1, 2, 3,
Aεi :=12
∫∂Sε
0
|vε|2Ki ds −∫∂Sε
0
(`ε + rεx⊥) · vεKi ds,
Bεi := γ
∫∂Sε
0
(vε − (`ε + rεx⊥)) · HεKi ds,
C εi :=γ2
2
∫∂Sε
0
|Hε|2Ki ds,
andDε
i :=
∫Fε0
ωε[vε − `ε − rεx⊥]⊥ · ∇Φεi (x) dx .
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Obstacle to the complex-analytical strategy
In the previous slide,vε := vε − γHε
is not curl free.
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Two extra Kirchhoff potentials
−∆Φεi = 0 in Fε0 , Φε
i −→ 0 when x →∞, ∂Φεi
∂n= Ki on ∂Fε0 ,
where
(K4, K5) :=
((−x1x2
)· n,(x2x1
)· n).
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Approximation of the velocity
We have
(∇KR2 [ωε]|x=0)x = aε(−x1x2
)+ bε
(x2x1
).
We introduce
vε# := KR2 [ωε]|x=0 + (∇KR2 [ωε]|x=0)x
−2∑
i=1
(KR2 [ωε]|x=0)i∇Φεi − aε∇Φε
4 − bε∇Φε5.
which is a good approximation of
K εH [ωε].
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An exercise
Letn be an integerA be a real n × n skew-symmetric matrix,B be a real n × n matrix,y0, y1 ∈ Rn,ε ∈ (0, 1) and yε(t) the solution to
εy ′′ε = A(y ′ε − Byε),yε(0) = y0, y ′ε(0) = y1.
Prove that there exists T > 0 such that (yε)ε∈(0,1) is bounded in C 1([0,T ];Rn).
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Modulation
Let
˜ε(t) := `ε(t)− KR2 [ωε(t, ·)](0)− ε∇KR2 [ωε(t, ·)](0) · ξ,
where
`ε(t) := RTθε(t) (hε)′(t).
Let
pε := (˜ε, εrε).
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Normal Form
Proposition
Let us fix ρ > 0. If for a given T > 0 and an ε ∈ (0, 1) one has for all t ∈ [0,T ] :
d(hε(t), Supp (ωε(t))) ≥ 1/ρ and Supp (ωε(t)) ⊂ B(hε(t), ρ),
then on [0,T ] : [εαMg + ε2Ma
](pε)′ + 〈εα−1Γg + εΓa, pε, pε〉
= γ pε × B+εγG (ε, t) + εmin(α,2)F (ε, t),
where G is weakly gyroscopic and F is weakly nonlinear.
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The function G = G (ε, t) : (0, 1)× [0,T ]→ R3 satisfies∣∣∣∣∫ t
0pε(s) · G (ε, s) ds
∣∣∣∣ ≤ εC (1 + t +
∫ t
0|pε(s)|2 ds
),
and the function F = F (ε, t) : (0, 1)× [0,T ]→ R3 satisfies
|F (ε, t)| ≤ C(1 + |pε(t)|+ ε|pε(t)|2
).
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Part 1.Bis. A new look at the irrotational case
We revisit the irrotational case of the first part following this time a real-analyticapproach after Lamb.
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We have seen that the solid equations can be rewritten in the form
(Mg +Ma)(p)′ + 〈Γg , p, p〉 = −(12
∫∂S0|v |2Ki ds −
∫∂S0
(`+ rx⊥) · vKi ds)i ,
where i runs over the integers 1, 2, 3.
Let us recall that(K1, K2, K3) := (n1, n2, x⊥ · n).
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Lamb’s lemma
Letζ1(x) := e1 , ζ2(x) := e2 and ζ3(x) := x⊥
denote the elementary rigid velocities.
Lemma
For any pair of vector fields (u, v) in C∞(R2 \ S0;R2) satisfyingdiv u = div v = curl u = curl v = 0,u(x) = O(1/|x |) and v(x) = O(1/|x |) as |x | → +∞,
one has, for any i = 1, 2, 3,∫∂S0
(u · v)Kids =
∫∂S0
ζi ·(
(u · n)v + (v · n)u)ds.
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As a consequence, using Lamb’s lemma and the boundary conditions, we obtain :
12
∫∂S0|v |2Ki ds =
∫∂S0
(v · n)(v · ζi ) ds
=
∫∂S0
((`+ rx⊥) · n
)(v · ζi
)ds
=
∫∂S0
((`+ rx⊥) · n
)(v · n
)Ki ds
+
∫∂S0
((`+ rx⊥) · n
)(v · τ
)(ζi · τ
)ds,
so that
12
∫∂S0|v |2Ki ds −
∫∂S0
(`+ rx⊥) · vKi ds
= −∫∂S0
((`+ rx⊥) · τ
)(v · τ
)Ki ds +
∫∂S0
((`+ rx⊥) · n
)(v · τ
)(ζi · τ
)ds
=∑k
pk
∫∂S0
(v · τ
)[(ζi · τ
)Kk −
(ζk · τ
)Ki ] ds
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Computation of the brackets
For instance,
(ζ1 · τ
)K2 −
(ζ2 · τ
)K1 =
(ζ1 · τ
)(ζ2 · n
)−(ζ2 · τ
)(ζ1 · n
)=
(ζ2 · n
)2+(ζ2 · τ
)2= 1
and (ζ1 · τ
)(ζ3 · n
)−(ζ3 · τ
)(ζ1 · n
)=
(ζ2 · n
)(ζ3 · n
)+(ζ3 · τ
)(ζ2 · τ
)= ζ2 · ζ3.
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Conclusion
One obtains exactly the same kind of formulation than previously, that is :[Mg +Ma
](p)′ + 〈Γg , p, p〉+ 〈Γa, p, p〉 = γp × B,
with
B :=
(ξ⊥
−1
),
whereξ :=
∫∂S0
x(H · τ
)ds,
(instead of
ξ1 + iξ2 :=
∫∂S0
zH dz
in the first part).
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Link with the complex-analytic approach
Lemma
Denote ξ := (ξ1, ξ2). Then ξ = ξ1 + i ξ2.
The proof relies on the formula :∫∂S0
(f1 − if2) dz =
∫∂S0
(f · τ − if · n
)ds. (2)
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Back to the original frame
Going back to the original frame the equations read
q′ = p, (Mg +Ma, ϑ) p′ + 〈Γϑ, p, p〉 = Fϑ(p),
where
Ma,ϑ := R(ϑ)MaR(ϑ)t ,
〈Γϑ, p, p〉 := −(Pa, ϑ0
)× p − rMa, ϑ
(0`⊥
),
Fϑ(p) := γ
(`⊥ − rζϑζϑ · `
)= γp × Bϑ,
with
R(ϑ) :=
(R(ϑ) 00 1
)∈ SO(3),
Pa, ϑ for the two first coordinates ofMa, ϑ p,
Bϑ :=
(−1ζ⊥ϑ
).
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A geodesic interpretation
Let
〈Γϑ, p, p〉 =:
∑1≤i,j≤3
(Γϑ)ki,jpipj
1≤k≤3
∈ R3.
Then for every i , j , k ∈ 1, 2, 3,
(Γϑ)ki,j(q) :=
12
((Ma,ϑ)i
k,j + (Ma,ϑ)jk,i − (Ma,ϑ)k
i,j
)(q),
where(Ma,ϑ)k
i,j :=∂(Ma,ϑ)i,j
∂qk.
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Part 4. A rigid body immersed in a bounded domain
We will make use of Lamb’s approach to tackle the case where the fluid-solidsystem occupies a bounded domain, still in the irrotational case.
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An immersed body
Let now S(t) be a rigid body moving in Ω.
S(t)F(t) Ω
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Setting
Ω : bounded open regular connected and simply connected domain of R2
occupied by the system fluid-solid,S0 ⊂ Ω : domain initially occupied by the solid,F0 := Ω \ S0 : domain initially occupied by the fluid,at time t, the solid occupies S(t) := R(ϑ(t))S0 + h(t),
at time t, he fluid occupies F(t) := Ω \ S(t).
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Motion of the rigid body
The solid can translate : its center of gravity h(t) evolves in Ω,The solid can rotate of an angle ϑ(t),
The evolutions of h and ϑ are given by Newton’s laws :
mh′′ =
∫∂S(t)
πn ds,
where n denotes the normal,There’s a similar equation for ϑ :
J ϑ′′ =
∫∂S(t)
π(x − h(t))⊥ · n ds,
The walls (of the cavity and of the body) are impermeable so that the normalcomponent of the velocity is continuous at the fluid-solid interface.
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Helmholtz and Kelvin
For any t,curl u(t, ·) = 0 in F(t),
and ∫∂S(t)
u(t) · τds = γ :=
∫∂S0
u0 · τds.
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Leth := (h1, h2), q := (h1, h2, ϑ) ∈ R3,
and their time derivatives
` := (`1, `2), p := (`1, `2, r) ∈ R3.
S(t), F(t) depend on q only, we shall rather denote them S(q) and F(q),
Q := q ∈ R3 : d(S(q), ∂Ω) > 0,
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Kirchhoff potentials
Let, for any q in Q, ζj(q, ·), Kj(q, ·) and Φj(q, ·)/
ζj(q, x) := ej−1, for j = 1, 2 and ζ3(q, x) := (x − h)⊥,
Kj(q, ·) := n · ζj(q, ·) on ∂Ω ∪ ∂S(q).
∆Φj = 0 in F(q),
∂Φj
∂n(q, ·) = Kj(q, ·) on ∂S(q),
∂Φj
∂n(q, ·) = 0 on ∂Ω.
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Kirchhoff potentials
We also denote
K (q, ·) := (K1(q, ·),K2(q, ·),K3(q, ·))t ,
and
Φ(q, ·) := (Φ1(q, ·),Φ2(q, ·),Φ3(q, ·))t .
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Inertia
Let
Mg :=
J 0 00 m 00 0 m
,
Ma(q) :=
∫∂S(q)
Φ(q, ·)⊗ ∂Φ
∂n(q, ·)ds =
(∫F(q)
∇Φi · ∇Φjdx)
16i,j63,
and
M(q) :=Mg +Ma(q).
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Christoffel symbols
Let
〈Γ(q), p, p〉 :=
∑1≤i,j≤3
Γki,j(q)pipj
1≤k≤3
∈ R3,
where, for every i , j , k ∈ 1, 2, 3, we set
Γki,j(q) :=
12
((Ma)i
k,j + (Ma)jk,i − (Ma)k
i,j
)(q),
where(Ma)k
i,j :=∂(Ma)i,j
∂qk.
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Stream function for the circulation term
For every q ∈ Q, there exists a unique C (q) ∈ R such that the unique solutionψ(q, ·) of the Dirichlet problem :
∆ψ(q, ·) = 0 in F(q),
ψ(q, ·) = C (q) on ∂S(q),
ψ(q, ·) = 0 on ∂Ω,
satisfies ∫∂S(q)
∂ψ
∂n(q, ·)ds = −1.
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Force term
Eventually, we also define :
B(q) :=
∫∂S(q)
(∂ψ
∂n
(∂Φ
∂n× ∂Φ
∂τ
))(q, ·) ds,
E (q) := −12
∫∂S(q)
(∣∣∣∣∂ψ∂n∣∣∣∣2 ∂Φ∂n
)(q, ·) ds,
and the force termF (q, p) := γ2E (q) + γ p × B(q).
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Reformulation as an ODE
Theorem (Glass-Munnier-S. 2014)
Up to the first collision, the fluid-body system is equivalent to the second orderODE
q′ = p,M(q)p′ + 〈Γ(q), p, p〉 = F (q, p),
with Cauchy data
q(0) = 0 ∈ Q, p(0) = (r0, `0) ∈ R× R2.
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Scheme of proof
We will use the weak formulation :
m`′ · `∗ + J r ′r∗ +
∫F(q)
(∂u∂t
+12∇|u|2
)· u∗dx = 0,
for all p∗ = (`∗, r∗) ∈ R3, with
u∗ := ∇(Φ(q, ·) · p∗),
which is defined on F(q).
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Scheme of proof
We decomposeu(q, ·) = u1(q, ·) + u2(q, ·),
with
u1(q, ·) := ∇(Φ(q, ·) · p) = ∇
3∑j=1
Φj(q, ·)pj
and
u2(q, ·) := γ∇⊥ψ(q, ·).
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Scheme of proof
This leads to
m`′ · `∗ + J r ′r∗ +
∫F(q)
(∂u1
∂t+
12∇|u1|2
)· u∗dx = −
∫F(q)
(12∇|u2|2
)· u∗dx
−∫F(q)
(∂u2
∂t+
12∇(u1 · u2)
)· u∗dx .
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Scheme of proof
The proof then reduces to proving that
m`′ · `∗ + J r ′r∗ = Mg (q)p′ · p∗,∫F(q)
(∂u1
∂t+
12∇|u1|2
)· u∗dx = Ma(q)p′ · p∗ + 〈Γ(q), p, p〉 · p∗,
−∫F(q)
(12∇|u2|2
)· u∗dx = γ2E (q) · p∗,
−∫F(q)
(∂u2
∂t+∇(u1 · u2)
)· u∗dx = γ
(p × B(q)
)· p∗.
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Scheme of proof of the second identity
First we prove that∫F(q)
(∂u1
∂t+
12∇|u1|2
)· u∗dx =
(ddt∂E1∂p− ∂E1
∂q
)· p∗,
withE1(q, p) :=
12
∫F(q)
|u1|2dx .
Then we observeE1(q, p) =
12Ma(q)p · p.
and deduce that(ddt∂E1∂p− ∂E1
∂q
)· p∗ =Ma(q)p′ · p∗ + 〈Γ(q), p, p〉 · p∗.
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Energy conservation
Proposition
For any solution (q, p) ∈ C∞([0,T ];Q× R3),
ddtE(q, p) = 0, where E(q, p) :=
12M(q)p · p + U(q),
where the potential energy U(q) is given by
U(q) := −12γ2C (q).
Moreover∀q ∈ Q, E (q) =
12DC (q).
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No collision implies that the velocity is bounded
Let, for δ > 0,
Qδ := q ∈ R3 : d(S(q), ∂Ω) > δ.
Then there exists K > 0 (depending only on S0,Ω, p0, γ,m,J , δ) such that|p|R3 ≤ K on [0,T ].
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First proof of the energy conservation : with the PDEformulation
First we prove the conservation of
E :=12
∫F(q)
u2 dx +12m`2 +
12J r2
thanks to the weak formulation of the PDE formulation.
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First proof of the energy conservation : with the PDEformulation
Then we prove that E coincides with E in the following way. We first decomposeagain u into
u(q, ·) = u1(q, ·) + u2(q, ·),
and use again that ∫F(q)
u21 dx =Ma(q)p,
to obtain that
E =12
∫F(q)
u22 dx +
∫F(q)
u1 · u2 dx +12M(q)p · p,
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First proof of the energy conservation : with the PDEformulation
Moreover, by some integration by parts, we have that
12
∫F(q)
u22 dx =
12γ2∫F(q)
∇⊥ψ · ∇⊥ψ dx
=12γ2∫S(q)
∂ψ
∂n(q, ·)ψ(q, ·)ds
= −12γ2C (q),
since ψ(q, ·) is constant equal to C (q) on ∂S(q) and satisfies∫∂S(q)
∂ψ
∂n(q, ·)ds = −1.
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First proof of the energy conservation : with the PDEformulation
Finally by another integral by parts,∫F(q)
u1 · u2 dx = 0.
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Second proof : with the ODE formulation
First we observe that(E(q, p)
)′=M(q)p′ · p +
12
(DM(q) · p)p · p − 12γ2DC (q) · p.
Using the ODE we have
M(q)p′ · p = −〈Γ(q), p, p〉 · p + F (q, p) · p,
andF (q, p) · p = γ2E (q) · p,
so that (E(q, p)
)′= −〈Γ(q), p, p〉 · p +
12
(DM(q) · p)p · p
+γ2(E (q) · p − 1
2DC (q) · p
).
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Second proof : with the ODE formulation
The proof then follows from the two following identities :
−〈Γ(q), p, p〉 · p +12
(DM(q) · p)p · p = 0,
E (q) · p − 12DC (q) · p = 0.
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Second proof : with the ODE formulation
The second identity follows from the fact that the matrix
S(q, p) :=
∑1≤i≤3
Γki,j(q)pi
1≤k,j≤3
,
is such that〈Γ(q), p, p〉 = S(q, p)p.
and such that12DM(q) · p − S(q, p)
is skew-symmetric.
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Hamiltonian structure and third proof of the energyconservation
We introduce the impulses :(PgΠg
):=Mgp,
(PaΠa
):=Ma(q)p,
(PΠ
)=
(Pg + PaΠg + Πa
).
LetI := (P,Π) andM(q) :=Mg +Ma(q),
such that (PΠ
):=M(q)p.
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Poisson manifold
Definition
We say that the manifold P of the smooth functions (q, I ) has a Poisson structureif there exists a a bracket ·, · acting on C∞ functionals f : P → R, bilinear andskew-symmetric, satisfying the Jacobi and the Leibniz identities.
Jacobi : f 1, f 2, f 3+ f 2, f 3, f 1+ f 3, f 1, f 2 = 0,
Leibniz : f 1f 2, f 3 = f 1f 2, f 3+ f 2f 1, f 3.
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Non-canonical Hamiltonian structure
We set
f 1, f 2 :=∂f 2
∂q· ∂f
1
∂I− ∂f 1
∂q· ∂f
2
∂I− γB(q) ·
(∂f 1
∂I× ∂f 2
∂I
),
and consider
H = −E = −12M(q)p · p − U(q) = −1
2I · M(q)−1I − U(q),
as a functional on P.
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Hamiltonian structure
Proposition
If (q, p) satisfies the previous ODE then, for any smooth functional f on P,
ddt
f = f ,H.
(Above f and H stand respectively for f (q, I ) and H(q, I )).
As a trivial consequence of the skew-symmetry of the bracket ·, · we get that Hand therefore E is conserved by the solutions.
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Part 5. A vanishingly small body in a bounded domain
Let us now turn our attention to the limit of the dynamics in the setting of Part 4when the size of the solid goes to 0.
As above, we will distinguish the two cases :Case (i) : when the mass of the solid is fixed, and then the solid tends to amassive pointwise particle, andCase (ii) : when the mass tends to 0 along with the size, and then the solidtends to a massless pointwise particle.
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A vortex point in a bounded domain
If initiallyω|t=0 = γδh0
then at time t > 0,ω = γδh(t)
with
(h)′(t) = γuΩ(h),
where uΩ is a smooth vector field on Ω, depending only on Ω.Smooth solutions to Euler initially close to
ω|t=0 = γδh0
stay close toω = γδh(t),
cf. Turkington, Marchioro-Pulvirenti.
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Definition of uΩ
Let ψ(h, ·) be the solution to :
∆ψ(h, ·) = 0 in Ω, ψ(h, ·) = G (· − h) on ∂Ω,
whereG (r) := − 1
2πln |r |.
The Kirchhoff-Routh stream function ψΩ is then defined by :
ψΩ(x) :=12ψ(x , x),
and the Kirchhoff-Routh velocity uΩ by
uΩ := ∇⊥ψΩ.
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Case (i) : Dynamics of a solid shrinking to a pointwisemassive particle
The solid occupying the domain
Sε(q) := R(ϑ)Sε0 with Sε0 := εS0,
is assumed to have a mass and a moment of inertia of the form
mε = m and J ε = ε2J 1,
where m > 0 and J 1 > 0 are fixed.
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Case (i)
We get at the limit :
m(h(i))′′ = γ
((h(i))
′ − γuΩ(h(i)))⊥.
We have : lim inf T ε ≥ T(i), and the convergence holds for any T ∈ (0,T(i)).
The convergence holds in W 2,∞([0,T ];R2) weak-*, and we think that thisconvergence is optimal.
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Case (ii)
We then get the equation of a point vortex of intensity γ :
(h(ii))′ = γuΩ(h(ii)).
The solution is global (no collision with the boundary), and T ε −→ +∞.
The convergence holds in W 1,∞([0,T ];R2) weak-*, and, once again, wethink that this convergence is optimal.
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The normal form
Withp =
(εϑ′, h′ − γ(uΩ(h) + εuc(θ, h))
),
the ODE can be recast as :
εmin(2,α)M∂Ω(ϑ, ε)(p)′
+ ε〈Γ∂Ω(ϑ), p, p〉 =
γp × B∂Ω(ϑ) + εγ2G (q) + O(εmin(2,α)),
where G (q) is weakly gyroscopic so that an extra ε comes out in the energyestimate.The two last lines of p × B∂Ω(ϑ) = 0 reads
h′ = γuΩ(h).
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Multi-scale expansions
S(t)F(t) Ω
Fluid state in Ω \ Sε(t) = Fluid state as if ∂Ω
+ Corrector as if Sε(t)
+ Corrector(Corrector) as if ∂Ω
+ ...
Method : Potential Theory / Fredholm for ...
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We expand uε1 and uε2 in
γ2E ε(qε), γ p × Bε(qε) and the main part of 〈Γε(qε), pε, pε〉,
which are all three expressions of the form∫∂Sε
0
((uε1 or uε2) · (uε1 or uε2))(ξj · n) ds,
and finally repeatedly use Lamb’s lemma.
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Part 6. A mean-field limit.
In this part we introduce an Euler-Vlasov system which seems a plausible relevantcandidate to describe a cloud of small massive particles in a two dimensionalincompressible perfect fluid. We will provide a piece of beginning of justification ina regularized setting.
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A mean-field limit
Let N solid bodies in a two dimensional incompressible perfect fluid, shrunk tomassive pointwise particles, with mass mi > 0, circulation γi and position hi (t),
∂tω + div x(ωu) = 0,
u(t, x) = KR2 [ω](t, x) +N∑
j=1
γjHR2(x − hj(t)),
mih′′i (t) = γi
(h′i (t)− vi (t, hi (t))
)⊥,
vi (t, x) = KR2 [ω](t, x) +∑j 6=i
γjHR2(x − hj(t)),
ω|t=0 = ω0, hi (0) = hi,0, h′i (0) = hi,1.
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Mean-field
We want to study the limit system obtained by the empirical measure
fN(t) :=1N
N∑i=1
δ(hi (t),h′i (t))
when N goes to infinity, with an appropriate scaling of the amplitudes.
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We therefore consider now the solutions of
∂tω + div x(ωu) = 0,
u(t, x) = KR2 [ω](t, x) +1N
N∑j=1
HR2(x − hj(t)),
h′′i (t) =(h′i (t)− vi (t, hi (t))
)⊥,
vi (t, x) = KR2 [ω](t, x) +1N
∑j 6=i
HR2(x − hj(t)),
ω|t=0 = ω0, hi (0) = hi,0, h′i (0) = hi,1.
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A plausible limit system
Our guess is that one obtains in the limit the following PDE system :
∂tω + div x(ωu) = 0,∂t f + div x(f ξ) + div ξ(f (ξ − u)⊥) = 0,
where
u := KR2 [ω + ρ] and ρ :=
∫R2
fdξ.
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An alternative formulation based on the fluid velocity
An alternative way to describe the system is to use the following velocityformulation :
∂tu + div x(u ⊗ u) +∇p = (ρu − j)⊥,div x u = 0
∂t f + div x(f ξ) + div ξ(f (ξ − u)⊥) = 0,
where
ρ :=
∫R2
fdξ and j :=
∫R2
f ξdξ.
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A regularized version of the system
u := K [ω + ρ] and ρ :=
∫R2
fdξ,
where K is defined by
K [g ](x) :=
∫R2
H(x − y)g(y)dy ,
with H is in W 1,∞(R2) and satisfies H(0) = 0. With this regularization, a properproof of the mean field limit is possible thanks to optimal transportation.
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Measures
We will denote byM(Rd) the set of signed measures,M+(Rd) the set of finite measures on Rd ,P(Rd) the set of the probability measures that is the subset ofM+(Rd)verifying ν(Rd) = 1,M1(Rd) the subspace of signed measures having a finite first moment
M1(Rd) :=
ν ∈M(Rd) :
∫Rd|x |1d |ν| <∞
,
and similarlyM+1 (Rd) and P1(Rd).
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Pushforward
Givena measure ν inM(Rd1)
and a map τ : Rd1 → Rd2
we define the pushforward measure τ#ν ∈M(Rd2) of ν by τ by
τ#ν[B] := ν[τ−1(B)],
for any Borel subset B of Rd2 .
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Measures
Given a measure γ over the product space
Rd × Rd = R2d
and the projections π1 and π2 on it factors, we define the first and secondmarginals of γ as the measures π1
#γ and π2#γ.
This means that for any Borel set B of Rd ,
π1#γ[B] = γ[B × Rd ] and π2
#γ[B] = γ[Rd × B].
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Wasserstein distances for measures
Definition
Let ν1 and ν2 be inM+1 (Rd) two finite measures on Rd such as ν1(Rd) = ν2(Rd).
We define the Wasserstein distance W1(ν1, ν2) of order p between this twomeasures by
W1(ν1, ν2) := infγ
∫Rd×Rd
γ(x , y)|x − y |dxdy
where the infimum is taken over all γ ∈M+(Rd × Rd) with marginals ν1 and ν2.
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Wasserstein distances for signed measures
Definition
We will say that two signed measures ν1 and ν2 on Rd are compatible wheneverν±1 (Rd) = ν±2 (Rd).
Definition
Given two compatible elements ν1, ν2 ofM1(Rd), we define the Wassersteindistance
W1(ν1, ν2) := W1(ν+1 + ν−2 , ν
−1 + ν+
2 ).
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Kantorovitch duality
Proposition
Let ν1 and ν2 be two compatible elements ofM1(Rd). Then
W1(ν1, ν2) = supφ
∫Rdφ d(ν1 − ν2),
where the supremum is taken over the unit ball of Lip(Rd).
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Translation invariance
Lemma
If ν1, ν2 and ν3, ν4 are two pairs of compatible elements ofM1(Rd), then the twomeasures ν1 + ν3 and ν2 + ν4 are compatible and we have
W1(ν1 + ν3, ν2 + ν4) 6 W1(ν1, ν2) + W1(ν3, ν4),
with equality in the particular case where ν3 = ν4.
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Lemma
Consider two compatible elements ν1, ν2 ofM1(Rd1). Let τ : Rd1 → Rd2 be aLipschitz map. Then
W1(τ#ν1, τ#ν2) 6 ‖τ‖LipW1(ν1, ν2).
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Lemma
Consider two compatible elements ν1, ν2 ofM1(Rd × Rd). We have, for i = 1, 2,
W1(πi#ν1, π
i#ν2) 6 W1(ν1, ν2).
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Wasserstein distances for vector measures
Definition
Let d1 and d2 be two positive integers. Consider two pairs ν1, ν2 and σ1, σ2 ofcompatible measures (respectively inM1(Rd1) andM1(Rd2)).We introduce the couples
µi := (νi , σi ) ∈M1(Rd1)×M1(Rd2),
i = 1, 2 and define the associated Wasserstein distance W1(µ1, µ2) between µ1and µ2 by
W1(µ1, µ2) := W1(ν1, ν2) + W1(σ1, σ2).
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A well-posedness result for the regularized system
Theorem (Moussa-S. 2012)
Existence and uniqueness. Assume that (ω0, f0) is inM1(R2)× P1(R2 × R2). Then there exists only one corresponding solution
(ω, f ) ∈ C 0([0,∞);M1(R2)× P1(R2 × R2)).
Stability. Consider two solutions
µ1 := (ω1, f1) and µ2 := (ω2, f2)
associated with two initial data
µ10 := (ω1
0 , f10 ) and µ2
0 := (ω20 , f
20 )
inM1(R2)× P1(R2 × R2).Then, for any t ≥ 0,
W1(µ1(t), µ2(t)) ≤ e2Ct W1(µ10, µ
20),
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Mean-field limit for regularized kernels
Assume that(ω0, f0) ∈M1(R2)× P1(R2 × R2)
and(h0
i , h1i )i∈N∗ ∈ (R2 × R2)N
∗
are such that
f N0 :=
1N
N∑i=1
δ(h0i ,h1i ) ∈ P1(R2 × R2),
satisfiesW1(f N
0 , f0)→ 0 when N → +∞.
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Mean-field limit for regularized kernels
Let us denote byµN := (ωN , f N)N∈N∗ and µ := (ω, f )
the solutions respectively associated with the initial data
(ω0, f N0 )N∈N∗ and (ω0, f0)
given by the previous theorem.
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Mean-field limit for regularized kernels
Then, for any t > 0, for any N ≥ 1,
f N(t) =1N
N∑i=1
δ(hi,N(t),h′i,N(t)),
where
h′′i,N(t) =(h′i,N(t)− uN(t, hi,N(t))
)⊥,
uN = K [ωN ] +1N
N∑j=1
H(· − hj,N(t)),
(hi,N(0), h′i,N(0)) = (h0i , h
1i ).
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Mean-field limit for regularized kernels
Moreover for any T > 0,W1(µN , µ)→ 0
when N → +∞.
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