SIAM J. MATH. ANAL.
Vol. 43, No. 1, pp. 389–410
c 2011 Society for Industrial and Applied Mathematics
EXISTENCE OF STRONG SOLUTIONS TO A
FLUID-STRUCTURE SYSTEM∗
JULIEN LEQUEURRE†
Abstract. We study a coupled fluid-structure system. The structure corresponds to a part of
the boundary of a domain containing an incompressible viscous fluid. The structure displacement is
modeled by a damped beam equation. We prove the existence of strong solutions to our system for
small data and the existence of local strong solutions for any initial data.
Key words. fluid-structure interaction, Navier–Stokes equations, beam equation, coupled system
AMS subject classifications. 35Q30, 74F10, 76D05, 76D03
DOI. 10.1137/10078983X
1. Introduction. We study a fluid-structure system coupling the Navier–Stokes
equations in a two-dimensional domain with a damped beam equation located on the
boundary of a domain occupied by a fluid flow. For similar systems, the existence of
weak solutions has been established in [4, 8] for two-dimensional domains and in [4, 6]
for three-dimensional domains.
Here we are interested in the existence of local-in-time strong solutions. In [3],
Beirão da Veiga proves the existence of local strong solutions for small data under the
assumption α ≥ 0 (see the beam equation (1.3)). In this paper, we improve this type
of result, with α > 0, by showing the existence of local strong solutions without any
smallness condition (Theorem 3.2), and we also prove the existence of global strong
solutions in a given time interval [0, T ] for small data (Theorem 3.1).
To the best of the author’s knowledge, this problem was introduced in [10] by
Quarteroni, Tuveri, and Veneziani to model cardiovascular systems like blood flow in
large vessels—arteries, for instance.
Let L > 0 and T > 0 be, respectively, a length and a time. Let η be a function
from (0, T ) × (0, L) to (−1, +∞). Let t ∈ (0, T ); we can define a domain Ωη(t)
depending on time by
Ωη(t) = (x, y) ; 0 ≤ x ≤ L and 0 ≤ y ≤ 1 + η(t, x) .
Here η(t) is the displacement of the beam. We note by Γs = (0, L) × {1} the reference
configuration of the beam. The displacement η has to satisfy the assumption
(1.1)
∃δ0 > 0 such that ∀t ≥ 0 ∀x ∈ (0, L) 1 + η(t, x) ≥ δ0 > 0
to ensure that, for every time t, Ωη(t) is a connected domain. Let us set Ω = (0, L) ×
(0, 1) and Γ = ∂Ω, that is,
Γ = {0} × (0, 1) {L} × (0, 1) (0, L) × {0} (0, L) × {1}.
∗ Received
by the editors March 23, 2010; accepted for publication (in revised form) October
15, 2010; published electronically February 1, 2011. This work was supported by the ANR project
CISIFS, 09-BLAN-0213-03.
http://www.siam.org/journals/sima/43-1/78983.html
† Institut de Mathématiques, UMR CNRS 5219, Université Paul Sabatier, 31062 Toulouse Cedex 9,
France ([email protected]).
389
390
JULIEN LEQUEURRE
We also set Γ0 = Γ \ Γs , the fixed boundary part
Γ0 = {0} × (0, 1) {L} × (0, 1) (0, L) × {0}
and
Γη(t) = (x, y) ; 0 ≤ x ≤ L and y = 1 + η(t, x) .
Thus ∂Ωη(t) = Γ0 ∪ Γη(t) . We will use other notation:
Σ0T = (0, T ) × Γ0 ,
QT = (0, T ) × Ω,
ΣsT = (0,
T ) × Γs ,
QT =
{t} × Ωη(t) ,
t∈(0,T )
ΣT = (0, T ) × Γ,
ST =
{t} × Γη(t) .
t∈(0,T )
The velocity u and the pressure p of the fluid in the domain QT are described by the
Navier–Stokes equations
(1.2)
ut + (u · ∇)u − div σ(u, p) = 0
div u = 0
u = ηt e2
u=0
u(0) = u0
(QT ),
(QT ),
(ST ),
(Σ0T ),
(Ωη0 ).
The displacement η satisfies the following beam equation:
(1.3)
ηtt − βηxx − γηtxx + αηxxxx = φ[u, p, η]
η(0) = η 0
ηt (0) = η 1
(ΣsT ),
(Γs ),
(Γs ).
In these equations, σ and φ are defined by
σ(u, p) = −pI + ν(∇u + (∇u)T ),
φ[u, p, η] = −σ(u, p)(−ηx e1 + e2 ) · e2 ,
where e1 = (1, 0) and e2 = (0, 1) and u = u1 e1 + u2 e2 , ν > 0 is the viscosity of the
fluid; α > 0, β ≥ 0, γ > 0 are constants relative to the structure (see [3] for more
details).
2. Functional settings. We have to define the function spaces for the solutions
(u, p, η) of (1.2)–(1.3).
In the fixed domain Ω, we define the classical Hilbert space in two dimensions
L2 (Ω) = L2 (Ω; R2 ) and in the same way the Sobolev spaces Hs (Ω) = H s (Ω; R2 ). We
introduce
Vσ (Ω) = u ∈ Hσ (Ω) ; div u = 0 in Ω ,
Hσ,τ (QT ) = L2 (0, T ; Hσ (Ω)) ∩ H τ (0, T ; L2 (Ω)),
Vσ,τ (QT ) = L2 (0, T ; Vσ (Ω)) ∩ H τ (0, T ; V0 (Ω)).
We need a definition of Sobolev spaces in the time-dependent
domain Ωη(t) .
Definition 2.1. We say that u belongs to H τ ( t∈(0,T ) {t} × Hσ (Ωη(t) )) (resp.,
to H τ ( t∈(0,T ) {t} × Vσ (Ωη(t) ))) if
391
STRONG SOLUTIONS TO A FLUID-STRUCTURE SYSTEM
• for almost every t in (0, T ), u(t) belongs to Hσ (Ωη(t) ) (resp., in Vσ (Ωη(t) )),
• t → u(t)Hσ (Ωη(t) ) (resp., t → u(t)Vσ (Ωη(t) ) ) is in H τ (0, T ; R).
We finally define
⎛
⎛
⎞
⎞
Hσ,τ (QT ) = L2 ⎝
{t} × Hσ (Ωη(t) )⎠ H τ ⎝
{t} × L2 (Ωη(t) )⎠ ,
t∈(0,T )
⎛
⎞
Vσ,τ (QT ) = L2 ⎝
t∈(0,T )
{t} × Vσ (Ωη(t) )⎠
⎛
Hτ ⎝
t∈(0,T )
⎞
{t} × V0 (Ωη(t) )⎠ .
t∈(0,T )
Solutions (u, p, η) of (1.2)–(1.3) satisfy
div u(t) =
u(t) · n(t) =
u(t) · n(t) +
u(t) · n0
0=
Ωη(t)
∂Ωη(t)
Γη(t)
Γ0
L
=
ηt (t) + 0 =
ηt (t, x)dx,
Γs
where n(t) = √
1
2 (t)
1+ηx
0
− ηx (t)e1 + e2 is the unit normal to Γη(t) outward Ωη(t) and
n0 is the unit normal to each part of Γ0 outward Ωη(t) , that is,
n0 = e1 on {L}×(0, 1),
n0 = −e1 on {0}×(0, 1), or n0 = −e2 on (0, L)×{0}.
Thus we must choose η 1 in L20 (Γs ) = η ∈ L2 (Γs ) ; Γs η = 0 . Furthermore, we can
choose η 0 ∈ L20 (Γs ), and then we shall have
η(t) = 0 and
ηt (t) = 0 ∀t ≥ 0.
(2.1)
Γs
Γs
We have to choose boundary conditions for η too. Here, we decide to fix η and ηx on
(0, T ) × {0, L} as follows:
(2.2)
η(t, 0) = η(t, L) = 0
and
ηx (t, 0) = ηx (t, L) = 0
∀t ∈ (0, T ).
We could have chosen periodic boundary conditions as in [3]. The result obtained in
the following may be directly translated to this situation.
With (2.1) and (2.2), we get
ηtt = 0 ,
ηxx = 0, and
ηtxx = 0 ∀ t ≥ 0.
Γs
Γs
Γs
We use Ms the orthogonal projection from L2 (Γs ) onto L20 (Γs ) to rewrite (1.3). We
will use a special trace function γs defined by
1
1
γs p = Ms (p|Γs ) = p|Γs −
p|Γs ∀p ∈ H σ (Ω) with σ > .
|Γs | Γs
2
Equation (1.3) becomes
ηtt − βηxx − γηtxx + αMs ηxxxx = γs p + φ[u, η],
with φ[u, η] = −νγs ∇u + (∇u)T (−ηx e1 + e2 ) · e2 .
(2.3)
392
JULIEN LEQUEURRE
Let us introduce the spaces
(2.4)
⎧ σ
2
⎪
μ
∈
H
(Γ
)
∩
L
(Γ
)
s.t.
μ
=
μ
=
0
at
x
=
0,
L
⎪
s
s
x
0
⎨ σ
σ
2
H(0)
(Γs ) =
μ
∈
H
(Γ
)
∩
L
s
0 (Γs ) s.t. μ = 0 at x = 0, L
⎪
⎪
⎩ σ
H (Γs ) ∩ L20 (Γs )
for
3
2
< σ,
1
2
for < σ ≤ 32 ,
for 0 ≤ σ ≤ 12 .
Due to (2.1) and (2.2), we look for η in the spaces
σ,τ
σ
H(0)
(ΣsT ) = L2 (0, T ; H(0)
(Γs )) ∩ H τ (0, T ; L20(Γs )).
The pressure term p is defined in the Navier–Stokes equations up to an additive
constant. Then we define the space Hσ (Ω) by
Hσ (Ω) = p ∈ H σ (Ω) s.t.
p=0 .
Ω
We will look for p in L ( t∈(0,T ) {t} × H1 (Ωη(t) )) (see Definition 2.1).
2
3. Main results. We can now state the two main theorems of this paper. First,
we consider global strong solutions of system (1.2)–(1.3) with a condition on the size
of the initial data only. Second, we prove the existence of a local strong solution for
the same system.
3
1
(Γs )×H(0)
(Γs ). There exists R >
Theorem 3.1. Let (u0 , η 0 , η 1 ) ∈ V1 (Ωη0 )×H(0)
0 2
0 2
0 such that for any initial data satisfying u V1 (Ω 0 ) + η H 3 (Γs ) + η 1 2H 1 (Γs ) ≤
R2 and the compatibility condition
(3.1)
u0 = 0
u0 = η 1 e 2
η
(0)
(0)
(Γ0 ),
(Γη0 ),
system (1.2)–(1.3) has a unique global strong solution (u, p, η) in
⎛
⎞
4,2
V2,1 (QT ) × L2 ⎝
{t} × H1 (Ωη(t) )⎠ × H(0)
(ΣsT ).
t∈(0,T )
3
1
Theorem 3.2. Let (u0 , η 0 , η 1 ) ∈ V1 (Ωη0 ) × H(0)
(Γs ) × H(0)
(Γs ), satisfying the
compatibility condition (3.1). There exists a time T0 > 0 such that system (1.2)–(1.3)
has a unique strong solution (u, p, η) ∈ V2,1 (QT0 ) × L2 ( t∈(0,T0 ) {t} × H1 (Ωη(t) )) ×
4,2
H(0)
(ΣsT0 ).
The core of the paper consists in the proof of these theorems. First of all, thanks to
a suitable change of variables, we introduce an equivalent problem (4.5) in a cylindrical
domain QT . Due to the change of variables, new nonlinear terms appear in the
equations. The proof of existence of solutions for system (4.5) is split into different
steps:
(i) We study the nonhomogeneous linearized system (5.1), where the nonlinearities in (4.5) are now considered as right-hand sides. The proof of existence for this
system uses a fixed point method for another equivalent system (5.10) introduced in
section 5.2 thanks to the splitting method due to Raymond [11]. Indeed, we see in
section 5.1 that we cannot apply a fixed point method directly to system (5.1).
STRONG SOLUTIONS TO A FLUID-STRUCTURE SYSTEM
393
(ii) From the linearized system, we prove the existence of strong solutions for
system (4.5) thanks to another fixed point method in section 6.
In section 7, we complete the proof by checking that the change of variables
defined in section 4 is suitable in the sense of Definition 4.1.
4. An equivalent problem in the fixed domain Ω. We want to use a change
of variables to rewrite system (1.2)–(1.3) in the domain QT = (0, T ) × Ω. This change
of variables introduces nonlinear terms in the variables (u, p, η) that we will treat
as right-hand sides in section 5. As in [3], for a fixed t ∈ (0, T ), we introduce the
following change of variables:
Ωη(t)
(4.1)
−→ Ω,
(x, y) −→ (x, z) =
x,
y
1 + η(t, x)
.
Setting fˆ(x, z) = f (x, y), we have the formulas
⎧
⎨fˆ(x, z) = f (x,
(1 + η(t, x))z),
y
⎩f (x, y) = fˆ x,
.
1 + η(t, x)
Then we can calculate the derivatives of f (x, y) using the derivatives of fˆ(x, z):
⎧
ηt ˆ
⎪
ft = fˆt − z
fz ,
⎪
⎪
1
+η
⎪
⎪
⎪
⎪
ηx ˆ
⎪
⎪
fx = fˆx − z
fz ,
⎪
⎪
⎪
1
+η
⎪
⎪
⎨
1 ˆ
fz ,
fy =
1
+
η
⎪
2
⎪
⎪
2
⎪
ηx ˆ
ηx
⎪
ˆ
ˆzz − z (1 + η)ηxx − ηx fˆz ,
⎪
f
=
f
−
2z
+
z
f
f
⎪
xx
xx
xz
⎪
1+η
1+η
(1 + η)2
⎪
⎪
⎪
⎪
1
⎪
⎪
fˆzz .
⎩ fyy =
(1 + η)2
Now, we state the system satisfied by û(x, z) = u(x, y) and p̂(x, z) = p(x, y):
(4.2)
ût − div σ(û, p̂) = F̂ [û, p̂, η]
div û = div ŵ [û, η]
û(0) = û0
û = ηt (t, x)e2
û = 0
(QT ),
(QT ),
(Ω),
(ΣsT ),
(Σ0T ),
with
(4.3)
F̂(t, x, z) = F̂ [û, p̂, η]
2
ηx
− ηxx ûz
= −η ût + zηt + νz
1+η
z 2 ηx2 − η
ûzz
+ ν −2zηx ûxz + η ûxx +
1+η
+ z(ηx p̂z − η p̂x )e1 − (1 + η)û1 ûx + (zηx û1 − û2 )ûz ,
ŵ(t, x) = ŵ [û, η]
= −ηû1 e1 + zηx û1 e2 .
394
JULIEN LEQUEURRE
For instance, to calculate the divergence term, we write u1,x + u2,z in terms of û, and
taking 1 + η as a multiplier, we get
0 = (1 + η)û1,x − zηx û1,z + û2,z .
Then we see that
û1,x + û2,z = div û = −ηû1,x + zηx û1,z = div ŵ.
The beam equation (2.3) becomes
ηtt − βηxx − γηtxx + αMs ηxxxx = γs p̂ − 2νγs û2,z + γs Ĥ[û, η],
with
(4.4)
Ĥ[û, η] = ν
ηx
ηx2 − 2η
û1,z + ηx û2,x −
û2,z .
1+η
1+η
To simplify the notation, we drop out the symbol ˆ· and obtain the system
(4.5)
ut − div σ(u, p) = F[u, p, η]
div u = div w[u, η]
u=0
u = ηt e2
u(0) = u0
ηtt − γηtxx − βηxx + αMs ηxxxx = γs p − 2νγs u2,z + γs H[u, η]
η(0) = η 0
ηt (0) = η 1
(QT ),
(QT ),
(Σ0T ),
(ΣsT ),
(Ω),
(ΣsT ),
(Γs ),
(Γs ).
The previous system is equivalent to system (1.2)–(1.3). More precisely, we state
the following definition.
4,2
(ΣsT )
Definition 4.1. (u, p, η) in H2,1 (QT ) × L2 ( t∈(0,T ) {t} × H1 (Ωη(t) )) × H(0)
is a solution of (1.2)–(1.3) when the following conditions are satisfied:
(i) (û, p̂, η) obtained for the change of variables û(x, z) = u(x, y), p̂(x, z) =
y
is a solution of (4.5);
p(x, y) with z = 1+η(t,x)
(ii) for any time t in (0, T ), the previous change of variables is a C 1 -diffeomorphism
from Ωη(t) into Ω;
(iii) η satisfies condition (1.1).
If we set u = v + w[u, η], we notice that div v = 0 and the system satisfied by
(v, p, η) is
(4.6)
vt + div σ(v, p) = f [u, p, η]
div v = 0
v = −w[u, η]
v = ηt e2 − w[u, η]
v(0) = u0 − w[u, η](0)
ηtt − γηtxx − βηxx + αMs ηxxxx = γs p − 2νγs v2,z − 2νγs w2,z [u, η] + γs H[u, η]
η(0) = η 0
ηt (0) = η 1
with
(4.7)
f [u, p, η] = F[u, p, η] + νΔw[u, η] − ∂t w[u, η],
(QT ),
(QT ),
(Σ0T ),
(ΣsT ),
(Ω),
(ΣsT ),
(Γs ),
(Γs ),
STRONG SOLUTIONS TO A FLUID-STRUCTURE SYSTEM
v = v1 e1 + v2 e2
and
395
w[u, η] = w1 [u, η]e1 + w2 [u, η]e2 .
The explicit expression of w[u, η] = −ηu1 e1 + zηx u1 e2 depends only on u1 and η.
Thus, the boundary conditions on Σ0T and ΣsT are
v=0
v = ηt e2
(Σ0T ),
(ΣsT ).
Moreover, the term −2νγs v2,z in (4.6)6 vanishes. Indeed, v1,x + v2,z = 0 in QT and
v1 = 0 on ΣsT . Furthermore, if v is in H2,1 (QT ), then v1,x|ΣsT = 0 and v2,z|ΣsT = 0.
That is why we are considering the following system:
(4.8)
vt + div σ(v, p) = f [u, p, η]
div v = 0
v=0
v = ηt e2
v(0) = v0
ηtt − γηtxx − βηxx + αMs ηxxxx = γs p + h[u, η]
η(0) = η 0
ηt (0) = η 1
(QT ),
(QT ),
(Σ0T ),
(ΣsT ),
(Ω),
(ΣsT ),
(Γs ),
(Γs ),
where
(4.9)
h[u, η] = −2νγs w2,z [u, η] + γs H[u, η]
and
(4.10)
v0 = u0 − w[u, η](0) = u0 + η 0 u01 e1 − zηx0 u01 e2 .
On the other hand, to have continuity on [0,T), the previous conditions on v must
be checked at time t = 0. Thus, we have to add a compatibility condition at time
t = 0:
(4.11)
div v0 = 0
v0 = 0
v 0 = η 1 e2
(Ω),
(Γ0 ),
(Γs ),
which is written in terms of (u0 , η 0 , η 1 ) as follows:
(4.12)
div (u0 + η 0 u01 e1 − zηx0 u01 e2 ) = 0
u0 = 0
u0 = η 1 e 2
(Ω),
(Γ0 ),
(Γs ).
5. Study of an auxiliary linear system. In this section, we prove existence
and uniqueness of solutions to the system
(5.1)
vt + div σ(v, p) = f
div v = 0
v=0
v = ηt e2
v(0) = v0
ηtt − γηtxx − βηxx + αMs ηxxxx = γs p + h
η(0) = η 0
ηt (0) = η 1
(QT ),
(QT ),
(Σ0T ),
(ΣsT ),
(Ω),
(ΣsT ),
(Γs ),
(Γs )
396
JULIEN LEQUEURRE
for a right-hand side
(5.2)
1/2
(f , h) ∈ ZT = L2 (QT ) × L2 (0, T ; H(0) (Γs )),
0
, where
and initial data (v0 , η 0 , η 1 ) in Xcc
3
1
(Γs ) × H(0)
(Γs )
X 0 = H1 (Ω) × H(0)
and
0
Xcc
= (z0 , μ0 , μ1 ) ∈ X 0 s.t. (z0 , μ0 , μ1 ) satisfies (4.11) .
The space X 0 will be equipped with the norm
1/2
.
(z0 , k 0 , k 1 )X 0 = z0 2H1 (Ω) + η 0 2H 3 (Γs ) + η 1 2H 1 (Γs )
The main result of this section is the following theorem.
0
and (f , h) be in ZT . Then system (5.1)
Theorem 5.1. Let (v0 , η 0 , η 1 ) be in Xcc
admits one and only one solution (v, p, η) in
4,2
XT = (z, q, μ) ∈ H2,1 (QT ) × L2 (0, T ; H1(Ω)) × H(0)
(ΣsT )
(5.3)
s.t. z = 0 on Σ0T and z = μt e2 on ΣsT .
Moreover, we get the estimate
(5.4)
(v, p, η)XT ≤ C((v0 , η 0 , η 1 )X 0 + (f , h)ZT ).
5.1. Why a fixed point method on the pressure term p does not work.
A way to find solutions of the coupled system (5.1) is to use a fixed point method.
For a given p in L2 (0, T ; H1 (Ω)), we consider the following system:
(5.5)
vt + div σ(v, p) = f
div v = 0
v=0
v = ηt e2
v(0) = v0
ηtt − γηtxx − βηxx + αMs ηxxxx = γs p + h
η(0) = η 0
ηt (0) = η 1
(QT ),
(QT ),
(Σ0T ),
(ΣsT ),
(Ω),
(ΣsT ),
(Γs ),
(Γs ).
Given fixed (η 0 , η 1 ) and h, we can solve the beam equation. Next, knowing η, we
can find solutions to the Stokes system with right-hand side f , initial data v0 , and a
boundary condition depending on ηt .
This idea cannot be applied directly with isomorphism theorems for the two
equations separately. Indeed, we first get the following proposition.
3
1
(Γs ) × H(0)
(Γs ). For γs p, h in
Proposition 5.2. Let (η 0 , η 1 ) be in Hs = H(0)
2
2
L (0, T ; L0(Γs )), equation
ηtt − βηxx − γηtxx + αMs ηxxxx = γs p + h
η(0) = η 0
ηt (0) = η 1
(ΣsT ),
(Γs ),
(Γs )
STRONG SOLUTIONS TO A FLUID-STRUCTURE SYSTEM
397
4,2
admits a solution η in H(0)
(ΣsT ) satisfying the estimate
ηH 4,2 (Σs ) ≤ C (η 0 , η 1 )H s + hL2 (ΣsT ) + γs pL2 (ΣsT ) .
(0)
T
Then the result for the Stokes equations is the following.
Proposition 5.3. Let v0 be in V1 (Ω). For f and g, respectively, in L2 (QT ) and
2,1
H(0) (ΣsT ) with the compatibility condition v0 = g(0)e2 on Γs and v0 = 0 on Γ0 , the
system
(5.6)
vt − div σ(v, p) = f
div v = 0
v = ge2
v=0
v(0) = v0
(QT ),
(QT ),
(ΣsT ),
(Σ0T ),
(Ω)
admits a unique solution (v, p) in V2,1 (QT ) × L2 (0, T ; H1 (Ω)) and furthermore
(v, p)V2,1 (QT )×L2 (0,T ;H1 (Ω)) ≤ C v0 V1 (Ω) + gH 2,1 (Σs ) + f L2 (QT ) .
(0)
T
The first proposition comes from regularity results for the beam equation proved
in Proposition 5.9. The second proposition is a result from [11] in the case when g
2,1
belongs to H(0)
(ΣsT ).
To conclude, the solution (v, p, η) of system (5.5) obeys
(v, p, η)XT ≤ C (v0 , η 0 , η 1 )X 0 + (f , h)ZT + pL2 (0,T ;H1 (Ω)) .
Thus this method gives directly the solution of system (5.5) in the expected spaces
(thanks to the isomorphism theorems), but we cannot act on the constant C to get a
contraction. That is why we have to consider a new equivalent system.
5.2. New equivalent system. Let us define the so-called Leray projection P
from L2 (Ω) in Vn0 (Ω), where
Vn0 (Ω) = u ∈ L2 (Ω) s.t. u · n = 0 on Γ and div u = 0 in Ω .
We want to split system (5.1) into two parts in order to construct a contraction
mapping acting on a part of the pressure term only. More precisely, following the idea
of [11, 12], the Stokes system can be expressed in terms of ve = P v, vs = (I − P )v
and their associated pressures pe , ps ; then we will construct a contraction mapping
acting on pe to obtain the expected result.
To express simply the Stokes system in the variables (ve , vs , pe , ps ), we have to
introduce some operators. Let us denote by N the operator defined from H σ (Γ) to
H σ+3/2 (Ω) (for σ ≥ −1/2) by q = N (g) (for g in H σ (Γ)) if
Δq(t) = 0 (Ω),
(5.7)
∂q(t)
= g (Γ).
∂n
Then, for z in L2 (Ω), the solution π of
Δπ = div z
∂π
= z·n
∂n
(Ω),
(Γ)
398
JULIEN LEQUEURRE
is a sum of two terms π1 and π2 in H 1 (Ω) satisfying
π1 ∈ H01 (Ω),
and π2 = N (z − ∇π1 ) · n .
Δπ1 = div z (Ω)
Setting π1 = −(−ΔD )−1 (div z), we get π2 = N ((z + ∇(−ΔD )−1 (div z)) · n). Thus,
we can define the operator π from L2 (Ω) into H 1 (Ω) by
(5.8)
π(z) = −(−ΔD )−1 (div z)+N ((z+∇(−ΔD )−1 (div z))·n)
for z ∈ L2 (Ω).
Finally we denote by Ns the restriction on H σ (Γs ) of N , that is, Ns (g) = N (gχs ) for
any g in H σ (Γs ) (σ ≥ −1/2).
With these notations, system (5.1)1−5 is equivalent to
(5.9)
ve,t − νΔve + ∇pe = P f
ve = −γτ vs
ve (0) = P v0
vs = ∇Ns (ηt )
ps = π(f ) − Ns (ηtt )
p = pe + ps
(QT ),
(ΣT ),
(Ω),
(QT ),
(QT ),
(QT ).
The explications to obtain this system are detailed in [11].
The pressure term on the right-hand side of the beam equation is
γs p = γs pe + γs π(f ) − γs Ns (ηtt ).
System (5.1) is equivalent to the following system in terms of (ve , pe , vs , ps , η):
(5.10)
ve,t − νΔve + ∇pe = P f
ve = −γτ vs
ve (0) = P v0
vs = ∇Ns (ηt )
(I + γs Ns )ηtt − βηxx − γηtxx + αMs ηxxxx = γs pe + h̃
η(0) = η 0
ηt (0) = η 1
p = pe + ps
v = ve + vs
ps = π(f ) − Ns (ηtt )
(QT ),
(ΣT ),
(Ω),
(QT ),
(ΣsT ),
(Γs ),
(Γs ),
(QT ),
(QT ),
(QT ),
with
(5.11)
h̃ = h + γs π(f ).
We want to find solutions to system (5.10). With (f , h) and (v0 , η 0 , η 1 ) fixed, our
method is to set the pressure term pe ∈ L2 (0, T ; H1(Ω)) only on the right-hand side
of the beam equation. Then, considering pe only in L2−ε (0, T, H1 (Ω)) (for a small
parameter ε > 0), we find a solution η of the modified beam equation in a space
ETε . The next step is, with η in ETε , to get ve , vs , and pe , respectively, in V2,1 (QT ),
L2 (0, T ; H2(Ω)) ∩ H 3/4 (0, T ; H1/2 (Ω)), and L2 (0, T ; H1 (Ω)). All of these results will
allow us to define a contraction mapping from a ball of the space of pressure term
L2 (0, T ; H1 (Ω)) into itself for a small time T0 in (0, T ). Then, because of the linearity
of system (5.10), we will have the existence and uniqueness of a strong solution in
0
and right-hand members
(0, T ) corresponding with fixed initial data (v0 , η 0 , η 1 ) in Xcc
(f , h) in ZT .
399
STRONG SOLUTIONS TO A FLUID-STRUCTURE SYSTEM
5.3. Existence of solutions for each part of (5.1) and estimates. We
begin this loop by fixing a pressure term pe in the beam equation. We will suppose that pe is in L2 (0, T ; H1 (Ω)). By a classic embedding theorem, we get γs pe ∈
1/2
L2−ε (0, T ; H(0) (Γs )) for any 0 < ε < 1. Then we have the estimate
(5.12)
γs pe L2−ε (0,T ;H 1/2 (Γs )) ≤ CT θ γs pe L2 (0,T ;H 1/2 (Γs ))
(0)
(0)
for θ =
1
1
− .
2−ε 2
Thus, we can prove the following proposition.
Proposition 5.4. Let 0 < ε < 1, let (η 0 , η 1 ) be in Hs , and let (f , h) be in
ZT , as defined in Proposition 5.2 and in (5.2). Then first h̃ defined by (5.11) is in
1/2
L2 (0, T ; H(0) (Γs )) and, second, with pe in L2−ε (0, T ; H1 (Ω)), the equation
(5.13)
(I + γs Ns )ηtt − βηxx − γηtxx + αMs ηxxxx = γs pe + h̃
η(0) = η 0
ηt (0) = η 1
(ΣsT ),
(Γs ),
(Γs )
admits a unique solution η in
(5.14)
4+ε/2
ETε = L2−ε (0, T ; H(0)
ε/2
(Γs )) ∩ W 2,2−ε (0, T ; H(0) (Γs )),
satisfying
(5.15)
ηETε ≤ C (η 0 , η 1 )Hs + pe L2−ε (0,T ;H1 (Ω)) + h̃L2−ε (0,T ;H 1/2 (Γs )) .
(0)
3/2,3/4
Furthermore, ηt belongs to H(0)
(ΣsT ).
1/2
Proof. First, h̃ is in L2 (0, T ; H(0) (Γs )) thanks to the regularity of f and h via
formula (5.11).
We want to rewrite (5.13) as a first order system. For that, we set
0 η(t)
η
Y (t) =
, Y0 =
,
ηt (t)
η1
and
(5.16)
A=
0
(I + γs Ns )−1 (−αMs Δ2 + βΔ)
I
γ(I + γs Ns )−1 Δ
.
4
2
2
Then D(−A) = H(0)
(Γs ) × H(0)
(Γs ). We denote H = H(0)
(Γs ) × L20 (Γs ). Y is the
solution of the equation
0
Y (t) = AY (t) +
(ΣsT ),
(5.17)
(I + γs Ns )−1 (γs pe + h̃)
Y (0) = Y0
(Γs ).
We use the well-known Duhamel’s formula, with B =
Y (t) = etA Y0 +
0
t
0
(I+γs Ns )−1 (γs pe +h̃) ,
e(t−τ )A B(τ )dτ .
400
JULIEN LEQUEURRE
For κ > 0, we have formally
(−A)κ Y (t) = (−A)κ etA Y0 +
t
0
(−A)κ e(t−τ )A B(τ )dτ ,
and because Y0 is in [D(−A), H]1/2 and B(τ ) is in [D(−A), H]3/4 , we get
(−A)κ Y (t) = (−A)κ−1/2 etA (−A)1/2 Y0 +
0
t
(−A)κ−1/4 e(t−τ )A (−A)1/4 B(τ )dτ .
Now, using triangular inequality in H, we have for r > 1
r r
(−A)κ Y (t)rH ≤ c (−A)κ−1/2 etA (−A)1/2 Y0 L(H)
H
t
r κ−1/4 (t−τ )A
1/4
+ (−A)
e
(−A) B(τ )dτ .
0
H
Because (−A) is a generator of an analytic semigroup (see the proof in [12], which
relies on a result in [5]), we get the estimates (see [9])
c
≤ κ for κ > 0.
(−A)κ etA t
L(H)
With Young’s formula, we have
T
r
(−A)κ Y (t) dt
H
0
T
r
r
dt(−A)1/2 Y0 ≤
(−A)κ−1/2 etA L(H)
H
0
r/p q r/q
p
T T κ−1/4 (·)A
1/4
+
e
(−A)
(−A) B(·)
L(H)
H
0
0
r
T
r
1 dt r
1/2
1/4
t →
≤c
Y
+
c
B(·)
,
(−A)
(−A)
q
0
1
1
(κ−
)r
(κ−
)p
H
L (0,T ;H)
2
4
t
0 t
Lp (0,T )
with 1 +
1
r
=
1
p
+ 1q , and κ satisfying
(κ − 12 )r < 1,
(κ − 14 )p < 1.
Then the triplet (p, q, r) = (1, 2 − ε, 2 − ε) is suitable. For this choice, κ has only to
ε
, and thus κ = 1 + ε/4 is convenient. This gives us first
obey κ < 1 + 4−2ε
4+ε/2
Y ∈ L2−ε (0, T ; [D((−A)2 ), D(−A)]1−ε/4 ) = L2−ε (0, T ; H(0)
2+ε/2
(Γs ) × H(0)
(Γs ))
and second
2+ε/2
Y ∈ L2−ε (0, T ; [D(−A), H]1−ε/4 ) = L2−ε (0, T ; H(0)
4+ε/2
ε/2
(Γs ) × H(0) (Γs )).
ε/2
Thus, the solution η of (5.13) belongs to L2−ε (0, T ; H(0) (Γs ))∩W 2,2−ε (0, T ; H(0) (Γs )).
The estimate comes from Duhamel’s formula and the different calculations above.
STRONG SOLUTIONS TO A FLUID-STRUCTURE SYSTEM
401
3/2,3/4
The last part is to prove that ηt is in H(0)
(ΣsT ). We use different interpo2+ε/2
lation formulas: η belongs to ETε , and thus ηt is in L2−ε (0, T ; H(0) (Γs )) ∩ W 1,2−ε
ε/2
2+ε/2
(0, T ; H(0) (Γs )), which can be embedded continuously in W λ,2−ε (0, T ; [H(0) (Γs ),
ε/2
H(0) (Γs )]λ ) for 0 < λ < 1. A quick calculation gives us
2+ε/2
[H(0)
ε/2
2+ε/2−2λ
(Γs ), H(0) (Γs )]λ = H(0)
(Γs ).
An injection formula in Sobolev spaces of fractional order (see [1]) gives
W λ,2−ε (0, T ) → W 0,2 (0, T )
2+ε/2
So W λ,2−ε (0, T ; [H(0)
when λ =
ε/2
1
1
− .
2−ε 2
3/2
(Γs ), H(0) (Γs )]λ ) → L2 (0, T ; H(0) (Γs )). In the same way we
2+ε/2
ε/2
can prove that W λ,2−ε (0, T ; [H(0) (Γs ), H(0) (Γs )]λ ) → H 3/4 (0, T ; L20 (Γs )).
We use a new definition of solutions for the Stokes system (5.6). Indeed, we look
for a solution (ve , vs , pe ) of the equivalent system (see section 5.2)
(5.18)
ve,t − νΔve + ∇pe = P f
ve = −γτ vs
ve (0) = P v0
vs = ∇Ns (g)
v = ve + vs
p = ps + pe
ps = π(f ) − Ns (gt )
(QT ),
(ΣT ),
(Ω),
(QT ),
(QT ),
(QT ),
(QT ),
where π(f ) is given in (5.8). We now can state the following result on solutions of the
Stokes equivalent system (5.18).
3/2,3/4
(ΣsT ), f in L2 (QT ), and v0 in V1 (Ω)
Proposition 5.5. Let g be in H(0)
0
with the compatibility condition v = 0 on Γ0 and v0 = g(0)e2 on Γs . Then
(5.18) admits a unique solution (ve , vs , pe ) in XTe,s = V2,1 (QT ) × L2 (0, T ; H2 (Ω)) ∩
H 3/4 (0, T ; H1/2 (Ω)) × L2 (0, T ; H1 (Ω)). We have the estimate
(ve , vs , pe )XTe,s ≤ c v0 V1 (Ω) + gH 3/2,3/4 (Σs ) + f L2 (QT ) .
(0)
T
Proof. In [11], system (5.18) is solved with f = 0. Thus, we have to look for
solution (ve , vs , pe ) of (5.18) as a sum of two terms, one (ve1 , vs1 , p1e ) solution of system
(5.19)
1
− νΔve1 + ∇p1e = 0
ve,t
ve1 = −γτ vs1
1
ve (0) = P v0
vs1 = ∇Ns (g)
v1 = ve1 + vs1
p1 = p1s + p1e
p1s = −Ns (gt )
(QT ),
(ΣT ),
(Ω),
(QT ),
(QT ),
(QT ),
(QT )
and the other (ve2 , vs2 , p2e ) solution of system
(5.20)
2
− νΔve2 + ∇p2e = P f
ve,t
ve2 = 0
ve2 (0) = 0
vs2 = 0
v2 = ve2 + vs2
p2 = p2s + p2e
p2s = π(f )
(QT ),
(ΣT ),
(Ω),
(QT ),
(QT ),
(QT ),
(QT ).
402
JULIEN LEQUEURRE
Thanks to the compatibility condition, g and v0 satisfy the hypothesis of Theorem 2.7
in [11], and then there exists a solution (ve1 , vs1 , p1e ) in XTe,s of (5.19) satisfying the
estimate
(ve1 , vs1 , p1e )XTe,s ≤ C P v0 V1 (Ω) + gH 3/2,3/4 (Σs ) .
(0)
T
Thanks to classical results on the Stokes system, there exists a solution (ve2 , vs2 , p2e )
of (5.20) with (ve2 , p2e ) in V2,1 (QT ) × L2 (0, T ; H1 (Ω)) for f in L2 (QT ) and vs2 = 0.
Furthermore (ve2 , vs2 , p2e ) obeys
(ve2 , vs2 , p2 )XTe,s ≤ CP f L2 (QT ) ≤ Cf L2 (QT ) .
Finally, (ve , vs , p), defined by ve = ve1 + ve2 , vs = vs1 + vs2 , and pe = p1e + p2e , is a
solution of (5.18), belongs to XTe,s , and satisfies the estimate
(ve , vs , pe )XTe,s ≤ C P v0 V1 (Ω) + gH 3/2,3/4 (ΣT ) + f L2 (QT ) .
(0)
5.4. Construction of a solution of system (5.1). In order to prove the
existence of solutions for system (5.1), we have to construct a contraction mapping
0
and right-hand sides
for the equivalent system (5.10). Initial data (v0 , η 0 , η 1 ) in Xcc
(f , h) in ZT are fixed in this section. For pe , we consider the mapping G defined by
G : L2 (0, T ; H1 (Ω)) −→
XTε = (ve , vs , pe , η) ∈ XTe,s × ETε ,
pe
(5.21)
−→
(ve , vs , pe , η) the solution of system (5.21),
ve,t − νΔve + ∇pe = P f
ve = −γτ vs
ve (0) = P v0
vs = ∇Ns (ηt )
(I + γs Ns )ηtt − βηxx − γηtxx + αMs ηxxxx = γs pe + h̃
η(0) = η 0
ηt (0) = η 1
p = pe + ps
v = ve + vs
ps = π(f ) − Ns (ηtt )
(QT ),
(ΣT ),
(Ω),
(QT ),
(ΣsT ),
(Γs ),
(Γs ),
(QT ),
(QT ),
(QT ),
where h̃ is defined from f and h in (5.11), and ETε and XTe,s are defined, respectively,
in (5.14) and in Proposition 5.5.
Proposition 5.6. The mapping G is well-defined from L2 (0, T ; H1 (Ω)) into XTε .
We have, moreover, the following estimate for θ > 0 defined in (5.12):
(5.22)
G(pe ) ≤ C (v0 , η 0 , η 1 )X 0 + (f , h)ZT + T θ pe L2 (0,T ;H1 (Ω)) .
Furthermore, for two pressures pe,1 and pe,2 in L2 (0, T ; H1 (Ω)), we have the solution
G(pe,1 ) − G(pe,2 ) = (ve,1 − ve,2 , vs,1 − vs,2 , pe,1 − pe,2 , η1 − η2 ) corresponding with
G(pe,1 − pe,2 ) in (5.21), with zero for initial data and right-hand sides. Moreover,
G(pe,1 ) − G(pe,2 ) satisfies the estimate
G(pe,1 ) − G(pe,2 )XTε ≤ cT θ pe,1 − pe,2 L2 (0,T ;H1 (Ω)) .
STRONG SOLUTIONS TO A FLUID-STRUCTURE SYSTEM
403
Thanks to Proposition 5.4 in section 5.3, we get η in ETε and ηt in
together with Proposition 5.5 (for g = ηt ), it follows that (ve , vs , pe , η)
and satisfies estimate (5.22).
The proof of the second part of this proposition relies on the linearity of the
system and the same propositions.
We now are able to construct a contraction mapping from a ball of L2 (0, T ; H1 (Ω))
into itself. Let us consider the linear operator F from L2 (0, T ; H1 (Ω)) into itself
defined by
Proof.
3/2,3/4
H(0)
(ΣsT );
belongs to XTε
F = P ◦ G,
where P is the projection from XTε into L2 (0, T ; H1 (Ω)) defined obviously by
P(ve , vs , pe , η) = pe .
We detail some properties on F in the following proposition.
Proposition 5.7. F is well-defined from L2 (0, T ; H1 (Ω)) into itself, and, for
any R > 0, there exists a time T0 > 0 such that F is a contraction in
BL2 (0,T0 ;H1 (Ω)) (R) = qe ∈ L2 (0, T0 ; H1 (Ω)) s.t. qe L2 (0,T0 ;H1 (Ω)) ≤ R .
Proof.
Step 1. The well-posedness of F comes from Proposition 5.6.
Step 2. Furthermore, from estimates
(ve , vs , pe )XTe,s ≤ c v0 V1 (Ω) + ηETε + f L2 (QT )
and
ηETε ≤ C (η 0 , η 1 )H s + γs pe L2−ε (0,T ;H 1/2 (Γs ))
(0)
we get
(ve , pe , η)XTε ≤ C (v0 , η 0 , η 1 )X 0 + (f , h)ZT + T θ pe L2 (0,T ;H1 (Ω)) ,
and thanks to
pe L2 (0,T ;H1 (Ω)) ≤ C(ve , vs , pe , η)XTε ,
we have finally
(5.23)
p
e L2 (0,T ;H1 (Ω))
≤ C (v0 , η 0 , η 1 )X 0 + (f , h)ZT + T θ pe L2 (0,T ;H1 (Ω)) .
Thus, we now introduce R > 0 such that C((v0 , η 0 , η 1 )X 0 + (f , h)ZT ) ≤ R/2. If
we take pe in BL2 (0,T ;H1 (Ω)) (R), then, for any time T0 , we get
pe L2 (0,T0 ;H1 (Ω)) ≤ R/2 + CT0θ R,
which gives
pe L2 (0,T0 ;H1 (Ω)) < R
for T0 such that CT0θ < 1/2, for instance.
404
JULIEN LEQUEURRE
Step 3. The contraction is obtained for two pressure terms pe,1 , pe,2 thanks to
Proposition 5.6. Indeed, we have for two pressure terms pe,1 and pe,2 in L2 (0, T ; H1 (Ω))
the estimate
F (pe,1 ) − F (pe,2 )L2 (0,T ;H1 (Ω)) ≤ cT θ pe,1 − pe,2 L2 (0,T ;H1 (Ω)) .
Thus, for T0 such that cT0θ < 1/2, we get the contraction.
We have now all the arguments to prove Theorem 5.1.
Proof of Theorem 5.1. By the Banach fixed point theorem, Proposition 5.7 is
equivalent to the existence of a unique solution (ve , vs , pe , η) in XTε0 of system (5.21)
on (0, T0 ). To get the existence of solutions on (0, T ), we use the same idea as that in
Proposition 5.7 but initializing with pe on (0, 2T0 ) defined by pe = pe on (0, T0 ) (with
pe coming from the solution (ve , vs , pe , η) obtained above) and pe = 0 on (T0 , 2T0 ).
By linearity of the system, the same estimates occur, and we have the existence and
ε
uniqueness on (0, 2T0 ) in X2T
. Step by step, we get the existence of a solution
0
ε
(ve , vs , pe , η) of (5.10) in XT .
To conclude the proof of Theorem 5.1, we need to prove the regularity of the
solution (v, p, η) of system (5.1) with v = ve + vs and p = pe + ps . We already
have (ve , vs , pe , η) ∈ XTε = V2,1 (QT ) × L2 (0, T ; H2(Ω)) ∩ H 3/4 (0, T ; H1/2 (Ω)) ×
L2 (0, T ; H1 (Ω)) × ETε . Now, we use the following theorem (see [2]).
Theorem 5.8. Assume that A is the generator of a analytic semigroup, B ∈
L2 (0, T ; H), and Y 0 ∈ [D(−A), H]1/2 . Then the problem
Y (t) = AY (t) + B(t),
Y (0) = Y 0
has a unique solution in H 1 (0, T ; H) ∩ L2 (0, T ; D(−A)).
4
2
In our case, remember that D(−A) = H(0)
(Γs ) × H(0)
(Γs ), where A is defined in
2
(5.16), H = H(0)
(Γs ) × L20 (Γs ), and B = (0, (I + γs Ns )−1 (γs pe + h̃))T . Then we have
[D(−A), H]1/2 = Hs and the following proposition.
Proposition 5.9. Let (η 0 , η 1 ) be in Hs . For pe in L2 (0, T ; H1 (Ω)) and h̃ in
2
L (0, T ; L20(Γs )), the equation
(I + γs Ns )ηtt − βηxx − γηtxx + αMs ηxxxx = γs pe + h̃
η(0) = η 0
ηt (0) = η 1
(ΣsT ),
(Γs ),
(Γs )
4,2
admits a solution η in H(0)
(ΣsT ) satisfying the estimate
ηH 4,2 (Σs ) ≤ C (η 0 , η 1 )H s + h̃L2 (0,T ;L20 (Γs )) + pe L2 (0,T ;H1 (Ω)) .
(0)
T
The regularity of η gives ηt in H 1 (0, T ; L20 (Γs )) and then vs in H 1 (0, T ; H1/2 (Ω)).
Consequently, we have (vs , ps ) in H2,1 (QT ) × L2 (0, T ; H1 (Ω)).
Thus, the solution (v, p, η) of (5.1) belongs to XT . The estimate of (v, p, η) in
XT comes from all of the previous ones.
6. Proof of Theorems 3.1 and 3.2 in the fixed domain QT . In this section,
we want to prove Theorems 3.1 and 3.2 in the fixed domain in the sense of Definition 4.1. That is, we will find a solution (u, p, η) of system (4.5). We will use a fixed
point method from a space of solutions of system (5.1) into itself. We begin the proof
405
STRONG SOLUTIONS TO A FLUID-STRUCTURE SYSTEM
by an estimate on (F, w, h), where (F, w) are defined in (4.3) and h = γs H with H
defined in (4.4).
Proposition 6.1. For (u, p, η) in XT , (F[u, p, η], w[u, η], h[u, η]) belongs to
1/2
WT = (G, z, K) ∈ L2 (QT ) × H2,1 (QT ) × L2 (0, T ; H(0) (Γs ))
(6.1)
s.t. z = 0 on Γ ,
and there exists δ > 0 such that
(6.2)
(F[u, p, η], w[u, η], h[u, η])WT ≤ CT δ (1 + (u, p, η)XT )(u, p, η)2XT .
Let (u1 , p1 , η1 ) and (u2 , p2 , η2 ) be two triplets in XT such that for i = 1, 2
(ui , pi , ηi )XT ≤ R
for some R > 0. Thus we get
(6.3)
(F1 , w1 , h1 ) − (F2 , w2 , h2 )WT ≤ C(1 + R)RT δ (u1 , p1 , η1 ) − (u2 , p2 , η2 )XT
with the notation (Fi , wi , hi ) = (F[ui , pi , ηi ], w[ui , ηi ], h[ui , ηi ]).
To prove Proposition 6.1, we use two lemmas.
Lemma 6.2. For 0 < ε < ε, we get H 1/2+ε (0, T ) → H 1/2+ε (0, T ), and if a
belongs to H 1/2+ε (0, T ), then
aH 1/2+ε (0,T ) ≤ cT (1−θ)/2aH 1/2+ε (0,T ) ,
where θ =
1/2 + ε
.
1/2 + ε
Lemma 6.3. Let b and a be, respectively, in H2,1 (QT ) and H 1,1/2 (QT ); then ab
belongs to L2 (QT ) and there exists δ > 0 such that
abL2 (QT ) ≤ CT δ aH 1,1/2 (QT ) bH2,1 (QT ) .
Proof of Lemma 6.2. By interpolation,
H 1/2+ε (0, T ) = [H 1/2+ε (0, T ), L2 (0, T )]1−θ ,
where θ =
1/2 + ε
1/2 + ε
(0 < θ < 1).
Then if a is in H 1/2+ε (0, T ), then a is in the interpolated space H 1/2+ε (0, T ) with
the estimate
aH 1/2+ε (0,T ) ≤ CaθH 1/2+ε (0,T ) a1−θ
L2 (0,T ) .
On the other hand, the embedding L∞ (0, T ) → L2 (0, T ) and a Hölder inequality in (0, T ) of finite mass gives aL2 (0,T ) ≤ CT 1/2 aL∞ (0,T ) . The embedding
H 1/2+ε (0, T ) → L∞ (0, T ) concludes the proof.
Proof of Lemma 6.3. By Theorem B.3 in [7], for b ∈ H2,1 (QT ) and a ∈ H 1,1/2 (QT ),
ab belongs to H1−2κ,1/2−κ (QT ) for 0 ≤ κ < 1/2. We now use the following two classical embeddings:
• H 1/2−κ (0, T ; R) → L1/κ (0, T ; R) (see [1]);
• L1/κ (0, T ; R) → L2 (0, T ; R) (because 2 < 1/κ ≤ +∞) with the estimate
cL2 (0,T ;R) ≤ T 1/2−κ cL1/κ(0,T ;R)
for c ∈ L1/κ (0, T ; R).
406
JULIEN LEQUEURRE
Together, these two estimates give that abL2 (Ω) , which is in H 1/2−κ (0, T ; R), belongs to L2 (0, T ; R) with the estimate (for 1/2 − κ > 0)
abL2 (QT ) ≤ CT 1/2−κ abH 1/2−κ (0,T ;L2 (Ω))
≤ C T 1/2−κ abH1−2κ,1/2−κ (QT )
≤ C T 1/2−κ aH 1,1/2 (QT ) bH2,1 (QT ) .
We can now prove Proposition 6.1.
Proof of Proposition 6.1. Thanks to Lemmas 6.2 and 6.3, we can estimate the
norms of the right-hand sides. We use the strong regularity of η and u. Indeed, η in
4,2
(ΣsT ) gives
H(0)
4(1−κ)
η ∈ H 2κ (0, T ; H(0)
(Γs ))
0 < κ < 1.
for
This gives us directly that
1/2+ε
η ∈ H 7/4−ε/2 (0, T ; H(0)
η∈
(6.4)
η∈
η∈
(Γs )),
3/2+ε
H
(0, T ; H(0) (Γs )),
5/2+ε
3/4−ε/2
H
(0, T ; H(0) (Γs )),
7/2+ε
H 1/4−ε/2 (0, T ; H(0) (Γs )).
5/4−ε/2
The first three equations of (6.4) gives, respectively, η, ηx , and ηxx in L∞ (ΣsT ) with
the following estimates:
ηL∞ (ΣsT ) + ηx L∞ (ΣsT ) + ηxx L∞ (ΣsT ) ≤ cT χ ηH 4,2 (Σs )
(0)
T
for
χ > 0.
From the last equation in (6.4), we get only ηxxx ∈ L2 (0, T ; L∞(Γs )).
Let us check some terms of F[u, p, η], w[u, η], or h[u, η].
• For F[u, p, η], we only need to check that all the terms are in L2 (QT ). The first
term in F[u, p, η] is −ηut :
−ηut L2 (QT ) ≤ ηL∞ (ΣsT ) ut L2 (QT ) .
1
1
Then, via the embeddings H 2 +ε (0, T ) → H 2 +ε (0, T ) → C(0, T ) and the smoothness
of η, we get
1
ηL∞ (ΣsT ) ≤ cT 2 (1−θ) ηH 1/2+ε (0,T ;H 3−2ε (Γs )) for ε < ε and ε s.t. 0 ≤ ε < 1.
1
Thus ηL∞ (ΣsT ) ≤ cT 2 (1−θ) ηH 4,2 (Σs ) and finally
(0)
T
1
−ηut L2 (Qt ) ≤ cT 2 (1−θ) ηH 4,2 (Σs ) uH2,1 (Qt ) .
(0)
T
η2
x
Another term is 1+η
uz , which satisfies
2
ηx
1 ≤
ηx 2L∞ (Σs ) uz L2 (QT )
1 + η uz 2
T
1 + η L∞ (Σs )
L (QT )
T
and becomes, thanks to Lemma 6.2,
2
ηx
u
≤ cT 1−θ η2H 4,2 (Σs ) uH2,1 (QT ) .
1 + η z 2
T
(0)
L (QT )
STRONG SOLUTIONS TO A FLUID-STRUCTURE SYSTEM
407
Terms with a product of u and a derivative of u like (1 + η)u1 ux , ηx u1 uz or u2 uz
must be carefully studied. Thanks to Lemma 6.3, because u belongs to H2,1 (QT )
and then ux and uz are in H1,1/2 (QT ), we get that u1 ux , u2 uz , and u2 uz belong to
L2 (QT ) with, for 0 ≤ κ < 1/2,
−(1 + η)u1 ux + (zηx u1 − u2 )uz L2 (QT )
≤ CT 1/2−κ 1 + ηL∞ (ΣsT ) + ηx L∞ (ΣsT ) uH2,1 (QT ) ∇uH1,1/2 (QT )
≤ CT 1/2−κ 1 + ηH 4,2 (Σs ) u2H2,1 (QT ) .
(0)
T
• For w[u, η], we have to prove that all the terms belong to H2,1 (QT ) with the
expected estimate. First of all, the calculations of the derivatives of w[u, η] are
(6.5)
wx
wz
wxx
wzz
wt
= −ηx u1 e1 − ηu1,x e1 + zηxx u1 e2 + zηx u1,x e2 ,
= −ηu1,z e1 + ηx u1 e2 + zηx u1,z e2 ,
= −ηxx u1 e1 − 2ηx u1,x e1 − ηu1,xx e1 + zηxxx u1 e2 + 2zηxxu1,x e2 + zηx u1,xx e2 ,
= −ηu1,zz e1 + 2ηx u1,z e2 + ηx u1,zz e2 ,
= −ηt u1 e1 − ηu1,t e1 + zηx,t u1 e2 + zηx u1,t e2 .
Then almost all of the estimates of the derivatives in L2 (QT ) are obtained as for
F[u, p, η]. Others terms like ηxxx u1 e1 are estimated as follows:
(6.6)
ηxxxu1 L2 (QT ) ≤ CT θ ηxxx L2 (0,T ;L∞ (Γs )) uL∞ (0,T ;L2 (Ω)) .
• For h[u, η]. We can remark that h defined in Proposition 6.1 is the trace of
function H on Γs , and we can prove that the lifting H of h belongs to L2 (0, T ; H 1 (Ω)).
We have to calculate the different terms of H and their derivatives:
ηx
ηx2 − 2η
u1,z + ηx u2,x −
u2,z ,
H=ν
1+η
1+η
ηxx (1 + η) − ηx2
ηx
u1,xz + ηxx u2,x + ηx u2,xx
Hx = ν
u1,z +
(1 + η)2
1+η
(6.7)
ηxx − 1
ηx2 − 2η
ηx2 − 2η
− ηx
u2,xz ,
− 2ηx
u2,z −
1+η
(1 + η)2
1+η
ηx
η 2 − 2η
u1,zz + ηx u2,xz − x
u2,zz .
Hz = ν
1+η
1+η
Because of the regularity of η, we always get the expected estimates.
The second point comes from the at least quadratic nonlinearity of the right-hand
sides with respect to (u, p, η). Some calculations give estimates (6.3).
Proposition 6.4. For a given triplet (u, p, η) in XT , system (4.5) with righthand sides (F, w, H) = (F[u, p, η], w[u, η], H[u, η]) and initial data (u0 , η 0 , η 1 ) in X 0
¯
satisfying (4.12) admits a unique solution (u, p, η) in XT with the estimate
(6.8)
(u, p, η)XT ≤ c1 ((u0 , η 0 , η 1 )X 0 + c2 T δ (1 + (u, p, η)XT )(u, p, η)2XT ),
where δ > 0 is defined in Proposition 6.1. In other terms, we can construct a mapping
(6.9)
X :
XT
−→
XT
X (u, p, η) = (u, p, η) is a solution of the system (4.5)
(u, p, η) −→
with (F[u, p, η], w[u, η], H[u, η]) for right-hand sides,
408
JULIEN LEQUEURRE
which satisfies
(6.10)
X(u, p, η)XT
≤ c1 (u0 , η 0 , η 1 )X 0 + c2 T δ (1 + (u, p, η)XT )(u, p, η)2XT .
Proof. Let us notice that (u, p, η) is solution of (4.5) with right-hand sides
(F[u, p, η], w[u, η], H[u, η]) if and only if (v, p, η) = (u − w[u, η], p, η) is a solution
of (4.8) with (f [u, p, η], h[u, η]) as right-hand sides and (v0 , η 0 , η 1 ) for initial data (see
(4.7), (4.9), and (4.10) for the definitions of f , h, and v0 ). Then this proposition relies
first on the result of existence of solutions for system (5.1) in Theorem 5.1 and second
on Proposition 6.1 for the estimate.
We can conclude this section showing existence of solutions in the fixed domain.
Proposition 6.5. Let (u0 , η 0 , η 1 ) be in X 0 satisfying (4.12).
(i) There exists a time T0 > 0 such that system (4.5) admits a unique local strong
solution (u, p, η) in XT0 .
(ii) There exists r small enough such that, under condition (u0 , η 0 , η 1 )X 0 ≤ r,
system (4.5) admits a unique global strong solution (u, p, η) in XT .
Proof. Let (u0 , η 0 , η 1 ) be in X 0 satisfying (4.12). We denote r = (u0 , η 0 , η 1 )X 0
and set R = 2c1 r (where c1 is the constant in (6.10)).
δ
1
and
(i) Let us define T0 = 2c1 c2 R(R+1)
BXT0 (R) = (u, p, η) ∈ XT0 with (u, p, η)XT0 ≤ R .
Then X is a contraction mapping in BXT0 (R). Indeed, let (u1 , p1 , η1 ) and
(u2 , p2 , η2 ) be two triplets in BXT0 (R). With the previous notation, we get
solutions X (ui , pi , ηi ) (i = 1, 2) of system (4.5) corresponding with right-hand
sides (F[ui , pi , ηi ], w[ui , ηi ], H[ui , ηi ]) (i = 1, 2) and initial data (u0 , η 0 , η 1 ).
First, each solution obeys the estimate (6.10) thanks to Proposition 6.4, which
gives, for R and T0 as above,
X (ui , pi , ηi )XT0 ≤
R R
+ = R.
2
2
Second, the difference satisfies
(6.11)
X (u1 , p1 , η1 ) − X (u2 , p2 , η2 )XT0
≤ c1 c2 T0δ (1 + R)R(u1 , p1 , η1 ) − (u2 , p2 , η2 )XT0
thanks to (6.3), that is,
X (u1 , p1 , η1 ) − X (u2 , p2 , η2 )XT ≤
1
(u1 , p1 , η1 ) − (u2 , p2 , η2 )XT .
2
(ii) We choose r such that c2 T δ r(1 + 2c1 r) = 1, that is,
r=
c21 c2 T δ (1
1
!
+ 1+
2
)
c1 c2 T δ
.
Then X is a contraction mapping in BXT (R) (see (i) for details).
STRONG SOLUTIONS TO A FLUID-STRUCTURE SYSTEM
409
7. Back to the moving domain. Thanks to Definition 4.1, the proof of Theorems 3.1 and 3.2 in the moving domain consists in proving that the change of variables
φt :
Ω
−→ Ωη(t)
(x, z) −→ (x, y)
is well-defined as a C 1 -diffeomorphism from Ω into Ωη(t) for every t ∈ [0, T ) and that
condition (1.1) is checked for the solution (u, p, η) of (4.5). Then wewill have the
solution (ũ, p̃, η) = (φt (u), φt (p), η) of (1.2)–(1.3) in V2,1 (QT ) × L2 ( t∈(0,T ) {t} ×
4,2
(ΣsT ). Furthermore, by the change of variables, we will be able to
H1 (Ωη(t) )) × H(0)
check which compatibility condition corresponds in QT to (4.12).
We have to show that condition (1.1) is checked. In the case of the existence of
solutions for small data, because we then have
(u, p, η)XT ≤ r,
we easily get from ηL∞ (ΣsT ) ≤ ηH 4,2 (Σs ) (obtained thanks to the continuous em4,2
bedding H(0)
(ΣsT ) → L∞ (ΣsT )) that
(0)
T
ηL∞ (ΣsT ) ≤ r ≤ 1 − δ0
for r small enough.
Condition (1.1) is checked for local solutions too thanks to the continuity of the
embeddings, for 0 < ε < 1 (see the proof of Proposition 6.1),
4,2
H(0)
(ΣsT0 ) →
3−2ε
H 1/2+ε (0, T0 ; H(0)
(Γs )) → L∞ (ΣsT0 ),
which gives ηL∞ (ΣsT ) ≤ cT0θ ηH 4,2 (Σs ) (for θ > 0) and then ηL∞ (ΣsT ) ≤ cT0θ R ≤
T0
0
0
(0)
1 − δ0 for T0 small enough.
4,2
The embedding H(0)
(ΣsT ) → C([0, T ); C 1 (Γs )) together with the condition 1 + η ≥
δ0 > 0 show that φt is a C 1 -diffeomorphism from Ω into Ωη(t) .
All the derivatives of the solutions written in the variable (x, y) are combinations
of those in the variable (x, z) multiplied at most by η or one of its derivatives which
are smooth enough to get (ũ, p̃) in H4,2 (QT ) × L2 ( t∈(0,T ) {t} × H1 (Ωη(t) )) (the
calculations are exactly the ones proving that F[u, p, η] belongs to L2 (QT ) for (u, p, η)
in XT ).
The compatibility conditions became, after the change of variables,
div u0 = 0
u0 = η 1 e 2
u0 = 0
(Ωη0 ),
(Γη0 ),
(Γ0 ).
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