Appl Math Optim https://doi.org/10.1007/s00245-018-9493-x
Shape Sensitivity Analysis for a Viscous Flow with Navier Boundary Condition Chaima Bsaies1 · Raja Dziri1
© Springer Science+Business Media, LLC, part of Springer Nature 2018
Abstract The shape derivability analysis of the flow of a viscous and incompressible fluid surrounding a rigid body B is considered. The novelty being in the choice of the boundary condition on the body B where we impose the so-called Navier boundary condition. Well-posedness of the time-dependent Navier–Stokes equations with mixed boundary conditions, of Navier and Dirichlet type, is established under regularity and smallness assumptions. After proving the shape differentiability of the state system, we compute the first order necessary optimality condition associated to drag shape minimization problem. Keywords Shape optimization · Navier–Stokes equations · Navier boundary conditions Mathematical Subject Classification 76D05 · 49K20
1 Introduction A body B, completely immersed in a viscous fluid in motion, undergoes a friction force, the drag, which tends to slow it down. Generally, the drag is approximated by the viscous energy dissipated in the fluid [1].
B
Raja Dziri
[email protected] Chaima Bsaies
[email protected]
1
Department of Mathematics, Faculty of Mathematical, Physical and Natural Sciences of Tunis, Tunis-El Manar University, 2092-El Manar Tunis, Tunisia
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Appl Math Optim
The obstacle B is three-dimensional and its wall is impermeable. Thus the normal component of the flow velocity U shall satisfy, on the boundary of B, U · n = 0 on ΓB . In 1823, Navier [15] proposed a slipping condition with friction on the wall which makes it possible to take into account the slip of the fluid near the boundary: the tangential component of the stress tensor is proportional to the tangential component of the flow velocity. [σ(U)n + βU]tg = 0 on ΓB . The condition proposed by Navier describes the mean behavior of the fluid near a rough boundary: when the size of the roughness tends to 0, the Dirichlet condition tends to the Navier condition. Cf., for example, [10,13]. Notice that as the friction coefficient β goes to infinity, we recover the classical Dirichlet condition. The shape differentiability of Navier–Stokes system is widely studied. However, the boundary condition considered is, generally, rather of Dirichlet or free boundary type. For the standard Dirichlet boundary condition, we mention, among others, the works of Simon [17] and Bello et al. [1] where shape derivative was proved in both cases: stationary Stokes and Navier–Stokes flows using a variant of the implicit function theorem and for W 2,∞ domains. In [2], the same result is extended to Lipschitz domains. In [3], the authors studied shape differentiability of functionals which are not defined when the pressure and velocity fields are with minimal regularity. Concerning time-dependent Navier–Stokes system, the issue is addressed for example in [9], for the two dimensional case. The authors proved shape differentiability of the state system using the weak form of the implicit function theorem. In our study, for the shape sensitivity analysis, we adopt the velocity method introduced by Sokolowski and Zolésio [18]. The outline of this paper is as follows. In Sect. 3, we study the well-posedness of time-dependant Navier–Stokes equations with Navier and non-homogeneous Dirichlet boundary conditions. This study is based on classical tools: the construction of an approximate solution by the Galerkin method and compactness method for the existence. A specific attention is given to the choice of the lift used to come back to homogeneous boundary conditions. Among many alternatives, we choose it to be the solution of a quasi-stationary Stokes system with non-homogeneous Dirichlet boundary condition. So we benefit of already known regularity results [19]. In Sect. 4, we recall few basic results of the velocity method. This section contains also some technical lemmas such as a weak form of the implicit function theorem [20] and the Piola transformation. Assuming data smooth enough, we are able to give a characterization of the shape derivative of the flow velocity and to compute the shape gradient of the drag functional.
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2 Statement of the Problem A fixed three dimensional rigid body B immersed in a viscous and incompressible Newtonian fluid is considered. We are interested in the motion of the fluid around the solid B. Therefore it is reasonable to reduce the study to a smooth and bounded domain D ⊂ R3 (hold-all), of boundary ΓD , which is sufficiently large to contain the solid B ¯ with boundary Γ = ΓB ∪ ΓD , be (B ⊂⊂ D) and the perturbed flow. Let Ω = D\B, the domain occupied by the studied fluid. The flow velocity U and its pressure π obey the time-dependent Navier–Stokes system with Navier condition on the boundary of obstacle B and non-homogeneous Dirichlet condition on ΓD : ⎧ ⎪ ⎪ ∂t U − μΔU + DU · U + ∇π = f ⎪ ⎪ divU = 0 ⎪ ⎪ ⎨ U=g U ·n=0 ⎪ ⎪ ⎪ ⎪ [2με(U) · n + βU]tg = 0 ⎪ ⎪ ⎩ U(0) = U0
in Ω × (0, T ), in Ω × (0, T ), on ΓD × (0, T ), on ΓB × (0, T ), on ΓB × (0, T ) in Ω
(1)
We denoted by f the external force per unit mass, μ the kinematic viscosity coefficient, β is a friction coefficient and ε(U) = 12 (DU +∗ DU) is the strain tensor. The tangential component of a vector field is denoted [·]tg , and n is the unit outward normal to Ω on Γ. The Dirichlet boundary condition on ΓD is given in terms of g which is the trace, on ΓD , of the flow velocity in absence of the solid B. It should satisfy the compatibility condition ΓD g · n = 0. In this work, we are focused on establishing a shape derivability result of the couple (U, π) solution of (1). As an application the drag functional is considered. We give the expression of the shape gradient by introducing a suitable adjoint state and deduce the first order necessary optimality condition.
3 Existence and Uniqueness Result We introduce some notations that will be used throughout this work. 3.1 Notation and Functional Spaces We denote by (·, ·) the inner product in (L2 (Ω))3 ; Ω ⊂ R3 being an open bounded set. The space [H1 (Ω)]3 is equipped with the scalar product:
Du · ·Dv +
((u, v)) = Ω
u · v, Ω
where Du · ·Dv = 3i=1 3j=1 ∂j ui ∂j vi . We set L2 (Ω) = (L2 (Ω))3 , | · | the L2 ¯ = (D(Ω)) ¯ 3 , D(Ω) ¯ being the space norm, Hm (Ω)=[Hm (Ω)]3 (m 0) and D(Ω) ∞ ¯ of C functions with compact support in Ω.
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Let
H1 (div, Ω) = v ∈ H1 (Ω), divv = 0 in Ω . We denote by Lp (0, T ; X), 1 p < +∞, the space of LP integrable functions f : [0, T ] −→ X where X is a Banach space. It is endowed with the norm fLp (0, T ; X) =
T 0
1/p p f(t)X dt
.
For p = ∞, L∞ (0, T ; X) denotes the space of essentially bounded functions and fL∞ (0, T ; X) = ess sup[0,T ] f(t)X . Also, we introduce
¯ V(Ω) = v ∈ D(Ω)|divv = 0, v|ΓD = 0, v · n|ΓB = 0 . L2 (Ω)
H1 (Ω)
and V(Ω) = V = {v ∈ H1 (Ω)|divv = 0, v|ΓD = H(Ω) = V 0, v · n|ΓB = 0}. We identify H(Ω) and its dual H(Ω) . So we obtain the following continuous injections V(Ω) → H(Ω) → V(Ω) . We set, ∀u, v ∈ H1 (Ω), a(u, v) =
ε(u) · ·ε(v), Ω
and ∀u, v, w ∈ H1 (Ω) b(u, v, w) =
ui Di vj wj dx =
1i, j3 Ω
(Dv · u) · wdx. Ω
The space V(Ω) is endowed with the norm v2V =
ε(v) · ·ε(v) Ω
which is equivalent to the norm of H1 (Ω) in V(Ω) (cf. [7]). The norm in (L2 (Ω))d (d ≥ 1) is denoted by | |.
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3.2 Existence of Solutions We begin our study by proving an existence result of weak solutions for system (1). The first step is to transform (1) into a system with homogeneous Dirichlet boundary condition on ΓD . 3.2.1 Properties of the Lift 1 Assume g ∈ W 1,∞ 0, T ; H 2 (ΓD ) and ΓD g(t) · n dΓ = 0. Lemma 3.1 For any t ∈ [0, T ], there exists a unique couple of functions (G(t), h(t)) ∈ H1 (Ω) × L2 (Ω)/R solution of: ⎧ ⎨ −μΔG(t) + ∇h(t) = 0 (Pt ) div G(t) = 0 ⎩ G(t) = g(t) ˜ where g(t) ˜ =
in Ω, in Ω, on Γ,
g(t) on ΓD , 0 on ΓB .
The existence and uniqueness of (G(t), h(t)) is a classical result. The proof can be found in [19, Proposition 2.2, p. 33 and Proposition 2.3, p. 35]. Also it is proved that there exists a constant K(μ, Ω) such that G(t)H1 (Ω) Kg(t)
1
H 2 (ΓD )
(2)
.
Finally we mention that to establish uniqueness and shape derivability results for system (1), we shall need more regularity assumptions on the lift G. That’s why we recall the following regularity result and remark. 3 Corollary 3.2 If Ω is of class Cm+2 (m 0) and g ∈ W k,∞ 0, T ; Hm+ 2 (ΓD ) (k ≥ 1), then (G, h) ∈ W k,∞ 0, T ; Hm+2 (Ω) × W k,∞ 0, T ; Hm+1 (Ω)/R . Remark 3.3 Notice that (∂t G, ∂t h) is solution of:
⎧ ⎨ −μΔ(∂t G(t)) + ∇(∂t h(t)) = 0 div (∂t G(t)) = 0 Pt ⎩ ˜ ∂t G(t) = ∂t g(t)
in Ω, in Ω, on Γ.
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3.2.2 Weak Formulation In system (1), we replace U by u + G and π by p + h. Then the problem expressed in terms of (u, p) is the following: ⎧ ∂t u − μΔu + Du · u + Du · G + DG · u + ∇p = FG ⎪ ⎪ ⎪ ⎪ div u = 0 ⎪ ⎪ ⎨u = 0 u·n=0 ⎪ ⎪ ⎪ ⎪ (2με(u) · n + βu)tg = (−2με(G) · n)tg ⎪ ⎪ ⎩ def u(0) = u0 = U0 − G(0)
in Ω × (0, T ), in Ω × (0, T ), on ΓD × (0, T ), on ΓB × (0, T ), (3) on ΓB × (0, T ), in Ω,
where FG = f − ∂t G − DG · G. The corresponding weak formulation can be stated as follows: Assume 1 f ∈ L2 (0, T ; V(Ω) ), g ∈ W 1,∞ 0, T ; H 2 (ΓD ) , U0 ∈ H(Ω) such that U0 · n = g(0) · n in H−1/2 (ΓD ) ,
(4)
g(t) · n = 0. ΓD
(5) Find u satisfying u ∈ L2 (0, T ; V(Ω)), ∂t u ∈ L1 (0, T ; V(Ω) ), d (u, v) + 2μa(u, v) + β(u, v)ΓB + b(u, u, v) + b(G, u, v) dt +b(u, G, v) = FG , v Ω − 2μa(G, v) ∀v ∈ V(Ω), u(0) = U0 − G(0),
(6)
(7) (8)
where ·, · Ω = ·, · V(Ω) ×V(Ω) and (·, ·)ΓB denotes the inner product in L2 (ΓB ). 3.2.3 Existence of Weak Solutions To prove existence of weak solutions [solutions of system (6)–(8)], we proceed, as it is classically done, by Galerkin method and using the following compactness result which involves fractional derivatives. Cf. [14], for example. Let, for B0 and B1 two Hilbert spaces with B0 ⊂ B1 ,
γ Hγ (R; B0 , B1 ) = v ∈ L2 (R; B0 ) | Dt v ∈ L2 (R; B1 ) , γ
where Dt v is the fractional derivative in t of order γ > 0 of v which is defined as the γ inverse Fourier transform of (2iπτ)γ ^v: D^t v(τ) = (2iπτ)γ ^v and γ
H[0,T ] (R; B0 , B1 ) = {u ∈ Hγ (R; B0 , B1 ) , support u ⊂ [0, T ]} .
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Lemma 3.4 Let B0 , B, B1 , be Hilbert spaces. Assume that • the injections B0 ⊂ B ⊂B1 , are continuous, • the injection B0 → B is compact. γ Then the injection H[0,T ] (R; B0 , B1 ) → L2 (0, T ; B) is continuous and compact. Theorem 3.5 Assume Ω of class C1,1 . For f, U0 and g satisfying (4) and (5), there exists a vector function u, solution of the weak formulation (6)–(8). Proof (1) We apply the Faedo–Galerkin method. Since V is separable and V is dense in V, then there exists {w1 , w2 , . . . , wm , . . .} a family of elements of V which is free and dense in V. Set Vm = vect w1 , w2 , . . . , wm ∀m ∈ N∗ , we look for um (t) =
m
gim (t)wi
i=1
such that: ∀t ∈ [0, T ], ∀j = 1, . . . , m, um (t), wj + 2μa (um (t), wj ) + b (um (t), um (t), wj ) +b (G(t), um (t), wj ) + b (um (t), G(t), wj ) + β (um (t), wj )ΓB = FG (t), wj − 2μa (G(t), wj ) , (9) um (0) = u0m ,
(10)
where u0m is the orthogonal projection, in H(Ω), of u0 on Vm . This leads to a nonlinear ordinary differential equation on gm = (g1m , . . . , gmm ): m
(wi , wj ) gim (t) + 2μ
i=1
m
a (wi , wj ) gim (t)
i=1
+ +
m
b (wi , wl , wj ) gim (t)glm (t)
i,l=1 m
b (G(t), wi , wj ) gim (t) +
i=1 m
+β
m
b (wi , G(t), wj ) gim (t)
i=1
gim (t) (wi , wj )ΓB
i=1
= FG (t), wj − 2μa (G(t), wj ) ∀t ∈ [0, T ], ∀j = 1, . . . , m
(11)
with initial condition gim (0) = the ith component of u0m ; i = 1, . . . , m,
(12)
which, by standard existence results for ODE, admits a unique solution defined in the whole interval [0, T ] (as it will be shown by a priori estimates below).
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Appl Math Optim
(2) A priori estimates We multiply Eq. (9) by gj m (t) and add these equations for j=1,…,m. Since b(um , um , um ) = b(G, um , um ) = 0, we get: 1 d |um (t)|2 + 2μa (um (t), um (t)) + b (um (t), G(t), um (t)) 2 dt +β (um , um )ΓB = FG (t), um (t) − 2μa (G(t), um (t)) . We have
2 |b (um (t), G(t), um (t))| |DG(t)| um (t) L4 (Ω) . By interpolation inequality
um (t)
L4 (Ω)
3 1
|um (t)| 4 um (t) L4 6 (Ω) .
Then by Young inequality, for some α > 0, |b (um (t), G(t), um (t))|
2 1 3 4
|DG(t)|4 |um (t)|2 + α 3 u m (t) L6 (Ω) . 4 4 4α
Therefore 1 d |um (t)|2 + 2μa (um (t), um (t)) + b (um (t), G(t), um (t)) 2 dt
2
1 d |um (t)|2 + 2μa (um (t), um (t)) +β um (t) L2 (Γ ) B 2 dt
2 1 3 4
− 4 |DG(t)|4 |um (t)|2 − α 3 um (t) L6 (Ω) . 4 4α Thus, by estimation (2), 1 d |um (t)|2 + 2μa (um (t), um (t)) + b (um (t), G(t), um (t)) 2 dt
2 1 d +β |um (t)|2 + 2μ um (t) V(Ω) |um (t)|2 2 dt Γ B 4
2 3 4
K + − 4 d4 |um (t)|2 − c2 α 3 um (t) V(Ω) , 4 4α where c2 is the norm of the embedding V(Ω) → L6 (Ω) and d = . g 1 L∞ 0,T ;H 2 (ΓD )
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Appl Math Optim
By choosing α =
3
2 3c2 μ
4
, we obtain
2 1
d K4 4 μ um (t) V(Ω) + |um (t)|2 − d |um (t)|2 2 dt 2α4
2 2
FG (t) V (Ω) + 2c 2 μG(t)2H1 (Ω) . μ
(13)
The latter inequality can be rewritten as follows 4 4
2 d μ − K4 d4 t
−K d t e 2α4 |um (t)|2 + e 2α4 um (t) V(Ω) dt 2 K4 d4
− t 2
2 2 2 4
2α FG (t) V (Ω) + 2c μG(t)H1 (Ω) . e μ
(14)
Integrating from 0 to s (s Tmax ) we deduce 2
|um (s)| e
K4 d4 T 2α4
2 |u0 | + μ 2
T
FG (t) 2
V(Ω)
0
2
+ 2c μ
T 0
G(t)2H1 (Ω)
.
(15) Therefore um is defined in [0, T ] and remains in a bounded set of L∞ (0, T ; H(Ω)). Now integrating (14), over [0, T ], gives T
um (t) 2 dt V 0 T
2 2 K4 d44 T 2 T
2 2 2
FG (t) V(Ω) + 2c μ e 2α G(t)H1 (Ω) . |u0 | + μ μ 0 0 (16)
Hence {um } remains in a bounded set of L2 (0, T ; V(Ω)). (3) Let um denotes the function from R into V(Ω), which is equal to um on (0, T ) ^ m . In order to apply the and 0 outside. The Fourier transform of um is denoted by u compactness result of Lemma 3.4, we shall prove that um belongs to a bounded γ set of H[0,T ] (R; V(Ω), H(Ω)) for some γ > 0. That is to say we have to prove the existence of γ > 0 such that R
γ D u ^ m (τ) = t
R
|τ|2γ |^ um (τ)|2 Const.
For that we need to introduce the continuous operators B : H1 (Ω) × H1 (Ω) −→ (H1 (Ω)) defined by B(u, v), w (H1 (Ω)) ×H1 (Ω) = b(u, v, w) ∀u, v, w ∈ H1 (Ω),
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Appl Math Optim
E : H1 (Ω) −→ (H1 (Ω)) such that E(u), v (H1 (Ω)) ×H1 (Ω) = 2μ
ε(u) · ·ε(v)dx + β
u · vdΓ,
Ω
ΓB
and A : H1 (Ω) −→ (H1 (Ω)) A(u), v (H1 (Ω)) ×H1 (Ω) =
ε(u) · ·ε(v)dx. Ω
Then the differential system (9)–(10) becomes: ∀j = 1, . . . , m d ( u m , wj ) = R m , wj + (u0m , wj ) δ0 − (um (T ), wj ) δT , dt where Rm = FG − E (um ) − B (um , um ) − B (G, um ) − B (um , G) − 2μA(G), and Rm is equal to Rm on (0, T) and to 0 outside this interval, δ0 and δT are Dirac distributions at 0 and T. By the Fourier transform, we get ^ m , wj + (u0m , wj ) − (um (T ), wj ) e(−2iπTτ) . 2iπτ (^ um , wj ) = R After multiplying by g ^jm (τ) and add the resulting equations for j = 1, . . . , m, we get ^ m (τ), u ^ m (τ) + (u0m , u ^ m (τ)) i2πτ |^ um (τ)|2 = R ^ m (τ)) e(−2iπτT ) . − (um (T ), u
(17)
We have T T T T Rm V (Ω) FG V (Ω) + c um V(Ω) + c1 um 2V(Ω) 0
0
+ 2c1
T 0
0
GH1 (Ω) um V(Ω) + 2μc
Therefore, thanks to a priori estimates (16) T 0
Rm V (Ω) Cst.
Thus
Rm (τ) V(Ω) c1 , ∀m. sup ^
τ∈R
123
0
T 0
GH1 (Ω) .
Appl Math Optim
Moreover, as by (15), there exists c0 > 0 such that |um (0)| c0 and |um (T )| c0 . We deduce, from (17), that
^ m (τ) V(Ω) + 2c0 |^ |τ| |^ um (τ)|2 c1 u um (τ)| . Then
^ m (τ) V(Ω) . |τ| |^ um (τ)|2 k3 u
For 0 < γ < 14 , we have |τ|2γ k4 (γ) So
+∞ −∞
1 + |τ| ∀τ ∈ R. 1 + |τ|1−2γ +∞
1 + |τ| |^ um (τ)|2 dτ 1−2γ −∞ 1 + |τ| +∞
2
u ^ m (τ) V(Ω) dτ k5
|τ|2γ |^ um (τ)|2 dτ k4 (γ)
−∞ +∞
+ k6
−∞
^ um (τ)V(Ω) 1 + |τ|1−2γ
.
By Parseval equality ^ um L2 (R; V(Ω)) = um L2 (R,V(Ω)) = um L2 (0,T ;V(Ω)) . By the Schwarz inequality we obtain +∞
^ um (τ)V(Ω)
−∞
1 + |τ|1−2γ
+∞ −∞
1 12 T 2
2
1
um (t)
dτ dt . V(Ω) (1 + |τ|1−2γ )2 0
The right-hand side is finite and bounded since γ ∈ 0,
1 4
.
As a consequence, there a exists a constant k7 > 0 such that R
|τ|2γ |^ um (τ)|2 k7 .
γ
So um is bounded in H[0,T ] (R; V(Ω), H(Ω)).
Since um is bounded in L2 (0, T ; V(Ω)) ∩ L∞ (0, T ; H(Ω)), by compactness theorem 3.4 there exists a subsequence {um } of {um } such that:
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Appl Math Optim
um u in L2 (0, T ; V) weakly, um u in L∞ (0, T ; H) weak-star, and um −→ u in L2 (0, T, H) strongly. The rest of the proof is straight forward. See for example (cf. [14]). Remark 3.6 Once existence of u solution of (1) is established, we deduce, by a result due to De Rahm [19], existence of a corresponding pressure p ∈ L2 (Q) (defined up to a constant). Notice that since u ∈ L∞ (0, T ; H(Ω)) and H(Ω) ⊂ V (Ω) with continuous injection, u : [0, T ] −→ V (Ω) is weakly continuous. 3.3 Uniqueness As for Dirichlet boundary conditions, uniqueness of solution of system (6)–(8) is also closely related to regularity. Indeed, below it is shown that under regularity assumptions on the data and a constraint on the viscosity size, solvability took place in a set where we have uniqueness. Lemma 3.7 The variational problem (6)–(8) has at most one solution u ∈ L8 (0, T ; L4 (Ω)). The proof is similar to the one done for Dirichlet boundary conditions in [19, p. 297]. 3 Theorem 3.8 Assume Ω of class C2 , g ∈ W 2,∞ 0, T ; H 2 (ΓD ) , f ∈ L∞ (0, T ; H(Ω)), ∂t f ∈ L1 (0, T ; H(Ω)) and U0 ∈ H2 (Ω), with divU0 = 0, U0 = g(0) ˜ on Γ. The variational problem (6)–(8) possesses, for a viscosity coefficient μ sufficiently large, a unique solution u which satisfies ∂t u ∈ L2 (0, T ; V(Ω)) ∩ L∞ (0, T ; H(Ω)) . Proof Thanks to Lemma 3.7, it is sufficient to prove that u ∈ L8 (0, T ; L4 (Ω)). 3 (1) Since g ∈ W 2,∞ 0, T ; H 2 (ΓD ) , the lift G ∈ W 2,∞ (0, T ; H2 (Ω)). Therefore FG ∈ L∞ (0, T ; H(Ω)) and ∂t FG ∈ L1 (0, T ; H(Ω)). Set d = g
3 L∞ 0, T ; H 2 (ΓD )
and d1 = FG L∞ (0, T ; V(Ω) ) and u0m is the
orthogonal projection of u0 in H2 (Ω) ∩ V on Vm =vect{w1 , . . . , wm }. We have u0m −→ u0 in H2 (Ω), as m → ∞, and u0m H2 (Ω) u0 H2 (Ω) .
123
Appl Math Optim (0) remains in a bounded set of H(Ω). First we prove that um We have
um (t), wj + 2μa (um (t), wj ) + b (um (t), um (t), wj ) + b (G(t), um (t), wj ) + b (um (t), G(t), wj ) + β (um (t), wj )ΓB = (FG (t), wj ) − 2μa (G(t), wj ) j = 1, . . . , m, ∀t ∈ [0, T ]. Multiply by gjm and add the resulting equations for j = 1, . . . , m: ∀t ∈ [0, T ]
u (t)2 + 2μa um (t), u (t) + b um (t), um (t), u (t) m m m + b G(t), um (t), um (t) + b um (t), G(t), um (t) + β um (t), um (t) Γ B = FG (t), um (t) − 2μa G(t), um (t) . Then at t = 0 u (0)2 − μ Δum (0) · um (0) + b u0m , u0m , um (0) m Ω + b G(0), u0m , um (0) + b u0m , G(0), um (0) = FG (0), um (0) . But |Δu0m |L2 (Ω) c1 u0m H2 (Ω) c1 u0 H2 (Ω) and (0))| c u 2 |b (u0m , u0m , um 3 0 H2 (Ω) |um (0) |L2 (Ω) .
Similary, b u0m , G(0), u (0) , b G(0), u0m , u (0) m m c3 u0m H2 (Ω) G(0)H2 (Ω) |um (0)|L2 (Ω) . So 2 u (0) μc1 + 2c3 Kd u0 2 H (Ω) + c3 u0 H2 (Ω) + |FG (0)| . m (0)} remains in a bounded set of H(Ω). Then {um Set
d2 = μc1 + 2c3 Kd u0 H2 (Ω) + c3 u0 2H2 (Ω) + d1 .
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Appl Math Optim
(2) By differentiating (9) in the t variable, we get um , wj + 2μa um , wj + b um , um , wj + b um , um , wj + b G , um , wj + b G, um , wj + b um , G, wj + b um , G , wj + β um , wj Γ B = FG , wj − 2μa G , wj . Multiply by gjm and add the resulting equations for j = 1, . . . , m:
2
2 2 1 d um (t) + 2μ um (t) V + β um (t)L2 (Γ ) B 2 dt + b um (t), um (t), um (t) + b G , um (t), um (t) + b um (t), G, um (t) + b um (t), G , um (t) = FG , um (t) − 2μa G , um (t) . Thus 2
2
2
1 d um (t) + 2μ um (t) V − c4 um (t) V um (t) V 2 dt
2
− c5 G(t)H2 (Ω) um (t) V
F G (t) + c6 + c6 G (t)H2 (Ω) um (t) V(Ω) um (t)
+ 2μc2 G (t)H2 (Ω) um (t) V(Ω) . Finally 2
2
1 d um (t) + (μ − c4 um (t) V − c5 GH2 (Ω) ) um (t) V 2 dt μc2 2 G 2L∞ (0, T ; H2 (Ω)) 1
2 2
c6 + c6 G (t)H2 (Ω) + um (t) V(Ω) um + FG (t) + (t) . 2 (18) (3) Now we need to prove that (μ − c4 um (t)V − c5 G(t)H2 (Ω) ) is positive in some interval [0, Tm ], Tm T. Thanks to (13), we have:
μ
d K4 d4 2
um (t) 2 |u + (t)| − |um (t)|2 m V(Ω) 2 dt 2α4
2 2
FG (t) V(Ω) + 2c 2 μG(t)2H1 (Ω) . μ
123
Appl Math Optim
Then
μ
2
K4 d4 2 2
um (t) 2
FG (t) 2 + 2c μG(t) + |um (t)|2 1 V(Ω) V(Ω) H (Ω) 4 2 μ 2α +2 |um (t)| um (t) . But 2
|um (t)| e Then we obtain
2 μ um (t)
V(Ω)
K4 d4 T 2α4
2T 2 2 2 2 2 |u0 | + d + 2c μTd K . μ 1
4 2 d1 + 4c 2 μK2 d2 μ K4 d4 K4 d44 T 2T 2 2 2 2 2 2α |u d + e | + + 2c μTd K 0 μ 1 α4 1 2 K4 d4 2T 2 T 2 2 2 2 4 u (t) . 4α |u0 | + d1 + 2c μTd K + 4e m μ (19)
At t=0, we get
2 4 μ um (0) V(Ω) d1 2 + 4c 2 μK2 d2 μ K4 d4 K4 d44 T 2T 2 2 2 2 2 2α |u d + e | + + 2c μTd K 0 μ 1 α4 1 2 K4 d4 2T 2 T 2 2 2 2 4 |u0 | + d1 + 2c μTd K + 4e 4α d2 . μ Set, (α4 = co μ3 )
K4 d4 4 2 K4 d4 2c 2T 2 2 2 2 2 2 2 2 3T μ o d3 = d1 + 4c μK d + |u0 | + d + 2c Td K μ e μ μ 1 co μ3 1 K4 d4 2 2T 2 T |u0 |2 + d1 + 2c 2 μTd2 K2 d2 +4e 4co μ3 μ and d4 =
K4 d4 4 2 K4 d4 2c 2T 2 T 2 2 2 2 o μ3 |u d1 + 4c 2 μK2 d2 + d e | + + 2c μTd K 0 μ μ 1 co μ3 1 K4 d4 2 2T 2 T +2e 4co μ3 |u0 |2 + d1 + 2c 2 μTd2 K2 μ T ×eK0 + 0 a(s)ds 1 + d22 + 2Tμc2 2 b ,
123
Appl Math Optim
where K0 =
(c6 + c6 ) K4 d44 T e 2α μ T
2 2 T
2
FG (t) × |u0 | + + 2c 2 μ G(t)2H1 (Ω) . V (Ω) μ 0 0
(s)| + 1 (c + c )G (s)2 a(s) = |FG , and b = G 2L∞ (0,T ;H2 (Ω)) . 6 2 6 H2 (Ω) Assume μ large enough so that
Kc5 d < μ and d4 <
2 μ μ − Kc5 d . 2 c4
(20)
Since d3 d4 , thus 2
2 μ μ um (0) V d3 d4 < 2 μ − c5 G(0)H2 (Ω) . c4 Therefore
μ − c5 G(0)H2 (Ω) − c4 um (0) V > 0. We deduce that, on some interval with origin 0,
μ − c5 G(t)H2 (Ω) − c4 um (t) V > 0. One of two alternatives should happen: there exists Tm ( T ) the first time such that
μ − c5 G (Tm ) H2 (Ω) − c4 um (Tm ) V = 0, or it doesn’t and at this time we set Tm = T. In the sequel we will show that it is the second alternative that will take place. (4) We have
μ − c5 G(t)H2 (Ω) − c4 um (t) V 0, for 0 t Tm . Let us prove that
μ − c5 G(t)H2 (Ω) − c4 um (t) V > 0, ∀t ∈ [0, T ].
123
(21)
Appl Math Optim
According to (21) and (18), we deduce that 2 1 d um (t) + 1 2 dt (t) + μc2 2 G 2L∞ (0,T ;H2 (Ω)) , ∀t ∈ [0, Tm ] , c(t) um (t)| + 1 (c + c )(G (t)2 + um (t)2V(Ω) ). where c(t) = |FG 6 2 6 H2 (Ω) Then
2 2 d um (t) + 1 c(t) 1 + um (t) + 2μc22 b. dt Therefore t 2 2 (t) e 0 c(s)ds 1 + um (0) 1 + um t t + 2μc22 b e τ c(s)ds dτ, ∀t ∈ [0, Tm ] . 0
Hence T 2 1 + um (t) eK0 e 0
a(s)ds
1 + d22 + 2Tμc22 b , ∀t ∈ [0, Tm ] .
As a consequence, inequality (19) gives
2 μ um (t) V d4 , 0 t Tm . Using (20), we get
μ − c5 G(t)H2 (Ω) − c4 um (t) V > 0, ∀ t ∈ [0, Tm ] . Then we can conclude that Tm = T and {um } is bounded in L∞ (0, T ; V(Ω)). This implies that u ∈ L8 (0, T ; L4 (Ω)) and the uniqueness is obtained by Lemma 3.7. Finally and as it was done for A priori estimates, we come back to inequality } remains in a bounded set of L2 (0, T ; V(Ω)) ∩ (18) and deduce easily that {um ∞ L (0, T ; H(Ω)). Remark 3.9 Under hypotheses of Theorem 3.8 and if Ω is of class C2,1 , there exists a unique p ∈ L2 (0, T ; L2 (Ω)/R) such that (u, p) ∈ L∞ 0, T ; H2 (Ω) × L∞ 0, T ; H1 (Ω) . This is a consequence of regularity results for Stokes system with Navier boundary condition, cf. [16].
123
Appl Math Optim
4 Shape Differentiability Existence and uniqueness being established for Navier–Stokes system (3), we can address the issue of existence and characterization of the shape derivative for its solution (uΩ , pΩ ). First, we shall establish the existence of Lagrangian semi-derivative by a weak form of the implicit function theorem introduced by Zolésio [20]. Then, under additional regularity assumptions, we are able to obtain the shape differentiability. Throughout this section, we assume that we work under the hypotheses of Theorem 3.8, Remark 3.9 and that f is defined in the hold-all cylinder (0, T ) × D. 4.1 Preliminaries ¯ R3 )) (α0 > The hold-all D being smooth, for any vector field V ∈ C([0, α0 ); C 3 (D, 0) satisfying divV(·, x) = 0 in D and V(·, x) · nD (x) = 0 on ΓD , there exists 0 < α α0 and a family of one-to-one transformations {Ts (V), s ∈ [0, α)} ¯ −→ D, ¯ X −→ Ts (V)(X) = x(s, X) where x(·, X) is the solution such that Ts (V) : D of the Cauchy problem ⎧ dx ⎪ ⎨ (s, X) = V(s, x(s, X)), ds ⎪ ⎩ x(0, X) = X. Remark 4.1 Notice that the choice of free divergence vector fields is imposed by the incompressibility of the fluid. Then the associated transformations satisfy detDTs = 1, ∀s ∈ [0, α). By setting Ωs = Ts (V)(Ω), Bs = Ts (V)(B) and Γs = ΓBs ∪ ΓD , we denote by (us , ps ) ∈ L2 (0, T ; H1 (Ωs )) ∩ L2 (0, T ; L2 (Ωs )/R) be the unique solution of the perturbed Navier–Stokes system ⎧ ⎪ ⎪ ∂t us − μΔus + Dus · us + Dus · Gs + DGs · us + ∇ps = FGs ⎪ ⎪ divus = 0 ⎪ ⎪ ⎨u = 0 s
us · ns = 0 ⎪ ⎪ ⎪ ⎪ (2με(u ⎪ s ) · ns + βus )tg = (−2με(Gs ) · ns )tg ⎪ ⎩ us (0) = (DTs · u0 ) ◦ Ts −1
in Ωs × (0, T ), in Ωs × (0, T ), on ΓD × (0, T ), on ΓBs × (0, T ), on ΓBs × (0, T ), in Ωs ,
(22)
123
Appl Math Optim
where FGs = f|Ωs − ∂t Gs − DGs · Gs ; Gs being the lift defined as on Sect. 3.2 and verifying ⎧ ⎨ −μΔGs (t) + ∇hs (t) = 0 in Ωs , in Ωs , div Gs (t) = 0 (23) ⎩ on Γs , Gs (t) = g˜ s (t) g(t) on ΓD , where g˜ s (t)= 0 on ΓBs . 4.2 Existence of the Piola Material Derivative The application of the classical implicit function theorem is not appropriate because the forcing term FGs , in L2 (0, T ; L2 (Ωs )) doesn’t admit a strong derivative (with respect to s) in L2 (0, T ; H−1 (Ω)). Indeed, in [18, p. 72], it is proved that strong differentiability of transported L2 functions in H−1 occurs only when weak topology is considered. Therefore, we will only be able to prove existence of the weak material derivative for the state uΩ , by making use of a weak form of the implicit function theorem (cf. [20]). To preserve the free divergence, we make use of the Piola transform, see for example [4]. Lemma 4.2 The Piola transform defined by Ps : H1 (div, Ω) −→ H1 (div, Ωs ) , ϕ −→ (DTs · ϕ) ◦ Ts −1 is an isomorphism. 4.2.1 Strong Material Derivative of the Lift We start our study by the following result Lemma 4.3 The mapping [0, α) −→ W 1,∞ 0, T ; H2 (div, Ω) × W 1,∞ 0, T ; H1 (Ω) , s −→ Gs = Ps−1 (Gs ) , hs = hs ◦ Ts is continuously differentiable. Proof We consider (Gs , hs ) solution of (23). We have, ∀vs ∈ H10 (Ωs ),
−
− → div (D (DTs Gs ) A(s)) · vs ◦ Ts ,
ΔGs · vs = − Ωs
Ω
− → where A(s) = (DTs )−1 (∗ DTs )−1 and div(M) is a vector, its ith component is the divergence of the ith line of the matrix M.
123
Appl Math Optim
It comes that Gs and hs satisfy:
− → −μdiv(D(DTs Gs )A(s)) + ∗ DTs −1 ∇hs = 0 DTs Gs − g˜ s ◦ Ts = 0
in Ω, on Γ.
We consider the mapping φ = (φ1 , φ2 ) : [0, α) × X → Y, (s; (w, z)) → (φ1 (s; (w, z)), φ2 (s; (w, z))) , where X = W 1,∞ (0, T ; H2(div, Ω)) × W 1,∞ (0, T ; H1 (Ω)/R) and Y = 3 W 1,∞ (0, T ; L2 (Ω)) × W 1,∞ 0, T ; H 2 (Γ ) , − → φ1 (s; (w, z)) = −μdiv(D(DTs w)A(s)) + (∗ DTs )−1 ∇z and φ2 (s, (w, z)) = DTs w|Γ − g˜ s ◦ Ts . ¯ R3 )), then • Since Ts ∈ C1 ([0, α); C3 (D, 1,∞ 2 (0, T ; H (div, Ω)) × W 1,∞ (0, T ; H1 (Ω)/R) ∀(w, z) ∈ W s → φ(s; (w, z)) is continuously differentiable. Also (w, z) → φ(s; (w, z)) is of class C1 (by linearity and continuity). • Obviously ∂2 φ(0, (G, h)) ∈ ISOM(X, Y). Thanks to the classical implicit function theorem, the mapping s → (Gs , hs ) is continuously differentiable in a neighbourhood of 0 and the material derivative ˙ satisfies ˙ p , h) (G p ˙ , h˙ = −∂1 φ(0, G, h). ∂2 φ(0, G, H) G 4.2.2 Weak Material Derivative of the State Let H(Ω) = {v ∈ L2 (0, T ; V(Ω)), ∂t v ∈ L2 (0, T ; V(Ω) )}. The solution us , of problem (22), satisfies T 0
∂t us · Ps (v) + 2μ Ωs
+
T
T 0
b (us , us , Ps (v)) +
0
=
−
T
123
∂t Gs · Ps (v) − Ωs
T 0
0
b (Gs , us , Ps (v)) +
(us , Ps (v))ΓB
s
T 0
f · Ps (v) Ωs
0
T
T
0
T 0
aΩs (us , Ps (v)) + β
b (Gs , Gs , Ps (v))
b (us , Gs , Ps (v))
Appl Math Optim
−2μ
T
a (Gs , Ps (v)) , ∀v ∈ L2 (0, T ; V(Ω)),
(24)
0
(us (0) − Ps (u0 )) Ps (w) = 0, ∀w ∈ H(Ω).
(25)
Ωs
Set us = Ps−1 (us ) and Gs = Ps−1 (Gs ). Using the fact that D Ts−1 ◦ Ts = (DTs )−1 , D ϕ ◦ Ts−1 = Dϕ · (DTs )−1 ◦ Ts−1 the weak form (24)–(25) becomes in the reference cylinder (0, T ) × Ω : ∀v ∈ L2 (0, T ; V(Ω)), def ψ1 (s, us ) , v = +β
T 0
+ −
T
0
∂t (DTs u ) , DTs v V ,V + 2μ
(ωs DTs us , DTs v)ΓB +
T
s
T 0
D (DTs us ) · Gs · DTs v +
0 Ω T
f ◦ Ts · DTs v +
0 Ω T
0
0
ε(s) (us ) · ·ε(s)(v) Ω
D (DTs us ) · us · DTs v
Ω T 0
T
T
D (DTs Gs ) · us · DTs v Ω
DTs ∂t Gs · DTs v Ω
T D (DTs Gs ) · Gs · DTs v + 2μ ε(s) (Gs ) · ·ε(s)(v) = 0, 0 Ω 0 Ω def s s ψ2 (s, u ) , w = (u (0) − u0 ) · w = 0, ∀w ∈ H(Ω), +
(26) (27)
Ω
where ε(s)(v) =
1 −1 ∗ ∗ D (DTs v) · DT−1 D (DTs v) , s + ( DTs ) 2
and ωs = ∗ (DTs )−1 · nR3 . In the sequel and as announced, we shall use the following weak form of implicit function theorem to establish existence of the weak material derivative of uΩ . Theorem 4.4 Weak implicit function theorem [20]. Let X and Y be two Banach spaces and e : I × X −→ Y , (s, x) −→ e(s, x),
123
Appl Math Optim
where Y is the dual of Y. If the following hypotheses hold (1) s −→ e(s, x), y is continuously differentiable for any y ∈ Y, (s, x) −→ ∂s e(s, x), y is continuous. (2) there exists x(·) ∈ C0,1 (I, X) such that e(s, x(s)) = 0 ∀t ∈ I. (3) x −→ e(s, x) is differentiable and (s, x) −→ ∂x e(s, x) is continuous. (4) there exists s0 ∈ I such that ∂x e(s0 , x(s0 )) is an isomorphism from X to Y , then the mapping u : I −→ X, s −→ x(s) is differentiable at s = s0 for the weak topology in X and its weak derivative x˙ (s) is the solution of ∂x e (s0 , x (s0 )) · x˙ (s0 ) , y + ∂s e (s0 , x (s0 )) , y = 0 ∀ y ∈ Y. As a consequence of the above result: Theorem 4.5 The weak Piola material derivative us − u P−1 (us ) − u = lim s s→0 s→0 s s
u˙ p = lim
exists in H(Ω) and is characterized by the following weak formulation: T 0
∂t u˙ p , v V ,V + 2μ
+
T
T 0
Du · u˙ · v + p
−β
0 T 0
− −
−
(ω (0)u, v)ΓB −
Ω
T
ε(V(0))∂t G · v −
0 Ω T
˙ P · v − 2μ DG · G
−2μ
Ω T 0
T
Ω
(ε(V(0))u, v)ΓB
˙p ·v D(u) · G Ω
T
˙p ·v− ∂t G
0 Ω T
[∗ DV(0)f + DfV(0)] · v
Ω T 0
DG · G · DV(0)v Ω
ε (0)(u) · ·ε(v)
Ω
p ˙ ε (0)(G) + ε G · ·ε(v), ∀v ∈ L2 (0, T ; V(Ω)),
and u˙ p (0) = 0,
123
T 0
0
T
0
Ω
DG · u˙ · v Ω
Ω
˙ p · (u + G) · v + D DV(0)G + G
Ω
0
Du˙ p · u · v
D(DV(0)u) · (u + G) · v Ω
0
T
p
ε(u) · ·ε (0)(v) − 2β
0
T
0
0
(u˙ p , v)ΓB +
T 0
T
D(u + G) · u · DV(0)v −
0 Ω T 0
Du˙ · G · v + p
ε(V(0))∂t u · v − 2μ Ω
T
−2
0
Ω T
T 0
Ω
T
0
= −2
ε (u˙ p ) · ·ε(v) + β
Appl Math Optim
where 1 ε (0)(v) = ε(DV(0) · v) − [Dv · DV(0) + ∗ (Dv · DV(0))] 2 and ω (0) = − DV(0)n, n . Proof Apply Theorem 4.4 with (s, w) → ψ(s, w) : [0, α) × H(Ω) → Y(Ω) , where ψ = (ψ1 , ψ2 ) and Y(Ω) = L2 (0, T ; V(Ω)) × H(Ω). ¯ R3 )) and, according to Lemma 4.3, Gs is C1 , then (1) Since Ts ∈ C1 ([0, α); C3 (D, ∀z ∈ H(Ω), ∀(v, w) ∈ Y(Ω), s → ψ(s, z), (v, w) : [0, α) → R is C1 . We denote by ∂s ψ(s, z) its weak derivative. Then: ∀v ∈ L2 (0, T ; V(Ω)), ∂s ψ1 (s, z), v =
T 0
∂t [D (V(s) ◦ Ts ) z] , DTs v V ,V T
+ ∂t [DTs z] , D (V(s) ◦ Ts ) v V ,V + 2μ +2μ +β
T
0 T
+ +
T 0 T 0 T
0
Ω
Ω
+
Ω
+
0
+
+2μ
D (DTs z) · Gs · D (V(s) ◦ Ts ) v Ω
(D (D(V(s) ◦ Ts ) Gs )
D (DTs Gs ) · z · (DV(s) ◦ Ts ) v
[∗ DV(s)f + Df V(s)] ◦ Ts · DTs v + Ω
T 0
T 0
Ω
(DV(s) ◦ Ts ) ∂t Gs · DTs v Ω
DTs (∂t Gs ) · DTs v
Ω
D (DV(s) ◦ Ts ) Gs + DTs (Gs ) · Gs · DTs v Ω
T 0
D (DTs z) · z · (DV(s) ◦ Ts ) v
Ω
Ω
DTs ∂t Gs · (DV(s) ◦ Ts ) v +
B
(ω(s)DTs z, (DV(s) ◦ Ts ) v)ΓB
Ω T 0
T 0
T 0
T
0 T
0
T
D (D (V(s) ◦ Ts ) z) · Gs · DT s v +
+DTs (Gs ) · z · DTs v +
0 T
T
0
D (DTs z) · (Gs ) · DTs v +
ε (s)(z) · ·ε(s)(v) Ω
ω (s)DTs z, DTs v Γ
D ((DV(s) ◦ Ts ) z) · z · DTs v + Ω
0
−
T
(ω(s) (DV(s) ◦ Ts ) z, DTs v)ΓB + β
0
+
ε(s)(z) · ·ε (s)(v) + β
0
D (DTs Gs ) · (Gs ) · DTs v +
Ω T 0
T 0
D (DTs Gs ) · Gs · (DV(s) ◦ Ts ) v Ω
ε (s) (Gs ) + ε(s) (Gs ) · ·ε(s)(v)
Ω
123
Appl Math Optim
T +2μ ε(s) (Gs ) · ·ε (s)(v), 0 Ω ∂s ψ2 (s, z) , w = 0, ∀w ∈ H(Ω),
where 2ε (s)(v) = D [(DV(s)) ◦ Ts · DTs v] · (DTs )−1 − D (DTs v) · (DTs )−1 · (DV(s)) ◦ Ts −∗ (DV(s)) ◦ Ts · ∗ (DTs )−1 · ∗ D (DTs v) +∗ (DTs )−1 ∗ D [(DV(s)) ◦ Ts · DTs v], ω (s) =
− (DTs )−1 (∗ DV(s) ◦ Ts )∗ DT−1 s · n, n ∗ DT−1 s · nR3
.
The weak continuity of (s, z) → ∂s ψ(s, z) : [0, α0 ) × H(Ω) → Y(Ω) is obviously satisfied. (2) To verify that s −→ us is Lipschitz-continuous, let us1 and us2 satisfying ψ (si , usi ) , (v, w) = 0 ∀(v, w) ∈ Y(Ω), i = 1, 2.
(28)
Computing the difference between the two equations, we keep at the left-hand def
term of the obtained equality (δu = us1 − us2 ) T Ω
0
+β
∂t (DTs1 (δu)) · DTs1 v + 2μ
T ΓB
0
+ +
T
0
Ω
0
ε (s1 ) (δu) · ·ε (s1 ) (v) Ω
DTs1 (δu) · DTs1 vω (s1 ) +
D (DTs1 u
s2
0 Ω T
T
T
) · (δu) · DTs1 v +
0
Ω
D (DTs1 (δu)) · us1 · DTs1 v
T 0
Ω
D (DTs1 (δu)) · Gs1 · DTs1 v
D (DTs1 Gs1 ) · (δu) · DTs1 v.
At the right-hand term, we leave all terms independent of δu − −
T 0 T 0
−2μ
Ω
∂t (DTs2 us2 ) · (DTs1 − DTs2 ) v
Ω T 0
123
∂t ((DTs1 − DTs2 ) us2 ) · DTs1 v
(ε (s1 ) − ε (s2 )) (us2 ) · ·ε (s1 ) (v) Ω
Appl Math Optim
−2μ
T
−β
ΓB
0
−β
T
ΓB
0
− − − − − −
T 0 T 0 T 0 T 0 T 0 T
ε (s2 ) (us2 ) · · (ε (s1 ) − ε (s2 )) (v)
0 Ω T
Ω
Ω
Ω
Ω
Ω
Ω
0
(DTs1 − DTs2 ) (us2 ) · DTs1 vω (s1 )
DTs2 us2 · (DTs1 ω (s1 ) − DTs2 ω (s2 )) v
D ((DTs1 − DTs2 ) us2 ) · us2 · DTs1 v D (DTs2 us2 ) · us2 · (DTs1 − DTs2 ) v D ((DTs1 − DTs2 ) us2 ) · Gs2 · DTs1 v D (DTs1 us2 ) (Gs1 − Gs2 ) · DTs1 v D (DTs2 us2 ) · Gs2 (DTs1 − DTs2 ) v D (DTs1 Gs1 − DTs2 Gs2 ) · us2 · DTs1 v
−D (DTs2 Gs2 ) · us2 · (DTs1 − DTs2 ) v + +
T 0
− − − − −
Ω
T 0 T 0 T 0 T 0 T 0
−2μ −2μ −2μ
Ω
Ω
Ω
Ω
f ◦ Ts2 · (DTs1 − DTs2 ) v −
T
T 0
(DTs1 − DTs2 ) ∂t Gs2 · DTs1 v −
Ω
0
Ω
f ◦ (Ts1 − Ts2 ) · DTs1 v
DTs1 (∂t Gs1 − ∂t Gs2 ) · DTs1 v
T 0
Ω
DTs2 ∂t Gs2 · (DTs1 − DTs2 ) v
D (DTs1 (Gs1 − Gs2 )) · Gs1 · DTs1 v D (DTs2 Gs2 ) · (Gs1 − Gs2 ) · DTs1 v D ((DTs1 − DTs2 ) Gs2 ) · Gs1 · DTs1 v D (DTs2 Gs2 ) (Gs2 ) · (DTs1 − DTs2 ) v
Ω T 0 T 0 T 0
ε (s1 ) (Gs1 − Gs2 ) · ·ε (s1 ) (v)
Ω
Ω
(ε (s1 ) − ε (s2 )) (Gs2 ) · ·ε (s1 ) (v) ε (s2 ) (Gs2 ) · · (ε (s1 ) − ε (s2 )) (v). Ω
It is easy to check that the left-hand term is Lipschitz-continuous in terms of s (using the regularity of transformations Ts and of Gs ). Moreover, δu = us1 −us2
123
Appl Math Optim
appears as the unique solution of a linearization, of problem (7). Thus taking δu is Lipschitz with as a test function, we deduce that s −→ us (∈ L2 (0, T ; V(Ω))) 1 ∂ (DT respect to s. After that, it is possible to show that s1 −s t s1 δu)DTs1 v 2 Q 2 is bounded for all v ∈ L (0, T ; V(Ω)). As a consequence, there exists a constant c > 0 such that ∂t (DTs1 δu)L2 (0, T ; V(Ω) ) ≤ c|s1 − s2 |. (3) It’s obvious that the mapping w −→ ψ(s, w) is differentiable and (s, w) −→ ∂w ψ(s, w) is continuous. Indeed
∂w ψ1 (s, w)(x), v = +2μ
0
+ + +
∂t (DTs x) , DTs v V ,V
0
T
T
T
ε(s)(x) · ·ε(s)(v) + β
T 0
Ω
(ω(s)DTs x, DTs v)ΓB
D (DTs x) · w · DTs v
0 Ω T
D (DTs w) · x · DTs v +
T
0 Ω T
0
D (DTs x) · Gs · DTs v Ω
D (DTs Gs ) · x · DTs v, ∀v ∈ H(Ω), ∂w ψ2 (s, w)(x), z = x(0) · z, ∀z ∈ H(Ω). 0
Ω
Ω
(4) We verify easily that ∂w ψ(0, u) belongs to ISOM(X(Ω), Y(Ω) ) since it is the linearized of (3) around its unique solution u. Finally the hypothesis of Theorem 4.4 are satisfied, and the existence of the weak Piola material derivative of u, in X(Ω), is established. 4.3 Shape Derivative of Navier–Stokes System In the beginning of this section, we recall standard formulae for differential tangential operators using arbitrary extension. Definition 4.1 Let Ω be a smooth open and bounded subset of R3 with boundary Γ of class C2 . Let h ∈ C1 (Γ ), W ∈ C1 (Γ )3 and u ∈ W 2,1 (Γ ), we define on Γ : ∇Γ h = ∇ h˜ |Γ − (∇ h˜ · n)n, ˜ · n, ˜ |Γ − DWn divΓ W = divΓ W ˜ |Γ − Wn ˜ · n, WΓ = W ΔΓ u = divΓ (∇Γ u) ,
123
Appl Math Optim
˜ ∈ C1 (R3 )3 are two extensions of h and W, respectively. where h˜ ∈ C1 (R3 ) and W We denote by H = divΓ n the mean curvature of the boundary Γ. Lemma 4.6 (cf. [8]) Let E be smooth vector field on Γ. Then ∇Γ (V · E) =∗ DΓ V · E +∗ DΓ E · V. Lemma 4.7 (cf. [8]) Let w, v ∈ H2 (Ω). Then
ε(w) · ·ε(v)V · n = − Γ
divΓ ((V(0) · n)ε(w))vdΓ
Γ
+
< ε(w)n, Dv · n + Hv > V · ndΓ. Γ
Lemma 4.8 (cf. [12]) For all f ∈ H1 (Γ ) and W ∈ H1 (Γ ; R3 ) we have:
W · ∇Γ f = − Γ
fdivΓ W + Γ
HfW · n. Γ
Theorem 4.9 Assume hypotheses of Theorem 3.8 and of Remark 3.9 are satisfied and f ∈ L2 (0, T ; H1 (Ω)). Then the shape derivative (U , p ) of (U, p) exists in L2 (0, T ; H1 (Ω)) × L2 (0, T ; L2 (Ω)/R) and is the unique solution of: ⎧ ∂t U − μΔU + DU · U + DU · U + ∇p = 0 ⎪ ⎪ ⎪ =0 ⎪ divU ⎪ ⎪ ⎨ U =0 ⎪ U · n = divΓ ((V(0) · n)U) ⎪ ⎪ ⎪ (2μ(ε(U ) · n + βU )tg = Q(U) ⎪ ⎪ ⎩ U (0) = 0
in Ω × (0, T ), in Ω × (0, T ), on ΓD × (0, T ), on ΓB × (0, T ), on ΓB × (0, T ), in Ω,
where Q(U) is defined, in a distribution sense, by (34) and if the flow velocity U is sufficiently smooth: Q(U) = −[μΔU + βDU · n]tg V(0) · n + 2μdivΓ (V(0) · nε(U)) +2μdivΓ U∇Γ (V(0) · n).
(29)
Proof Under the announced assumptions, it is proved that system (1) written in Ωs , ∀s ∈ [0, α), admits a unique solution (Us , ps ) ∈ L2 (0, T ; H2 (Ωs )) × L2 (0, T ; H1 (Ωs )/R). Therefore the shape derivative (U , p ) of (U, p): ˙ p + (DV · U − DU · V), and p = p˙ − ∇p · V U = U exists in L2 (0, T ; H1 (Ω)) × L2 (0, T ; L2 (Ω)).
123
Appl Math Optim
To characterize U , recall that Us satisfies the following weak formulation T 0
∂t Us , vs V ,V + 2μ T
+
0
ε (Us ) · ·ε (vs ) + β
T 0
Ωs
(Us , vs )ΓB
s
(DUs · Us ) · vs Ωs
0
T
=
T
f · vs ,
∀vs ∈ L2 (0, T, V (Ωs )) .
(30)
Ωs
0
def
Notice that vs = Ps (v), v ∈ L2 (0, T ; V(Ω)) then its shape derivative v is such that div v = 0, v · n = divΓ ((V · n)v) on ΓB (for details see Lemma 4.10) and v = 0 on ΓD . Therefore by computing the Eulerian derivative of (30), we obtain T
∂t U , v V ,V + 2μ
0
T
+ +
0
Ω
0
ΓB
T
DU · U + DU · U , v
Ω
0
a(U, v ) +
T
0
T
0
(U v + Uv ) + β ΓB
0
0
0
T
DU · U, v Ω
[∂t Uv + 2με(U) · ·ε(v) + DU · U, v ] V(0) · n
T
=
a(U , v) +
∂t U · v + 2μ
T
+β
T
fv +
0
T 0
Ω
T ΓB
∂ (Uv) + HUv V(0) · n ∂n
fvV(0) · n.
(31)
ΓB
But ∂t U − μΔU + DU · U = f − ∇p in Ω. Then T
0
∂t U , v V ,V + 2μ
− +
T
=
T 0
123
U v+β ΓB
fvV(0) · n. ΓB
T 0
T
DU · U + DU · U , v Ω
ε(U) · n, n v · n ΓB
[∂t Uv + 2με(U) · ·ε(v) + DU · U, v ] V(0) · n
T 0
a(U , v) +
0
ΓB
0
0
pv · n + 2μ
0 ΓB T
+β
T
T 0
ΓB
∂ (Uv) + HUv V(0) · n ∂n (32)
Appl Math Optim
Notice that the integral in the right-hand term exists since f ∈ L2 (0, T, H1 (Ω)). As a consequence, we obtain, in a distribution sense, ∂t U − μΔU + DU · U + DU · U + ∇p = 0,
(33)
where p is the shape derivative of the pressure p. Moreover it’s easy to check that div U = 0 on Ω, and U = 0 on ΓD . Also it will be shown in Lemma 4.10 that U · n = divΓ (V(0) · n U) on ΓB . ¯ such that divv = 0 in Ω, v · n = We multiply (33) by v ∈ D((0, T ), D(Ω)) 0 on ΓB and v = 0 on ΓD , we get T
ε(U )n, v =
2μ 0
T 0
ΓB
+
∂t U , v V ,V + 2μ
T 0
T
a(U , v)
0
DU · U + DU · U , v . Ω
By comparison with (32) we obtain: T
U v + 2μ
β ΓB
0
0
T
def
=
0
−β
ε(U )n, v = Q(U), v ΓB
[p − 2μ ε(U)n, n ]divΓ (V(0) · nv) ΓB
T 0
−
T
ΓB
T
∂ (Uv) + HUv V(0) · n ∂n
[∂t U · v + 2με(U) · ·ε(v) + DU · U, v − fv] V(0) · n. (34) ΓB
0
In order to obtain an explicit expression of Q(U), we shall assume (U, p) smooth enough. Thus, Q(U), v = −β
0
T 0
−
T
T 0
−2μ
[ DU · n, v ]V(0) · n ΓB
(∂t U + DU · U − f) vV(0) · n + 2μ
ΓB T 0
[p − 2μ ε(U)n, n ]divΓ (V(0) · nv) ΓB
T 0
divΓ (V(0) · nε(U))v ΓB
divΓ UdivΓ vV(0) · n ΓB
123
Appl Math Optim
=−
T
μΔU vV(0) · n − 2μ ε(U)n, n ∇Γ (V(0) · n) · v
ΓB T
0
+2μ
0
divΓ ((V(0) · n)ε(U))v − β
T 0
ΓB
DU · n, v V(0) · n. ΓB
Hence, (2με(U ) · n + βU )tg = −[μΔU + β(DU · n)]tg V(0) · n + 2μdivΓ (V(0) · nε(U)) −2μ ε(U)n, n ∇Γ (V(0) · n) on ΓB . Lemma 4.10 The shape derivative U satisfies U · n = divΓ ((V · n)U) on (0, T ) × ΓB . Proof On ΓBs , we have: T 0
T
ΓBs
Us · ns ϕdΓs dt = 0 ∀ϕ ∈ L2 0, T ; C1 (ΓBs ) ,
ω(s)Us ◦ Ts · ns ◦ Ts ψdΓdt = 0, ∀ψ ∈ L2 0, T ; C1 (ΓB ) , 0 ΓB
ω(s) = ∗ (DTs )−1 · n 3 . R
˙ · n + U · n˙ = 0 on ΓB . Since ω(0) = 1 and U · n = 0 on ΓB , we obtain U ˙ − DU · V and n = n˙ − DΓ n · VΓ = The shape derivative of U, U = U Γ −∗ DΓ V · n − DΓ n · VΓ the boundary shape derivative of n (cf [18]). Then U · n = − DU · V, n − U nΓ + DΓ n · VΓ = − DU · V, n + U, ∇Γ (V · n) − U, DΓ n · VΓ . But ∗ DU · n, V = (∗ DΓ U + n∗ (DU · n)) n, V = ∗ DΓ U · n, V + DU · n, n V · n, U · n = U, ∇Γ (V · n) − DΓ n · U, VΓ − ∗ DΓ U · n, V − DU · n, n V · n. Using the fact that ∇Γ V · E =∗ DΓ V · E +∗ DΓ E · V and ∗ DΓ n = DΓ n (cf [5]), we obtain:
123
Appl Math Optim
U · n = U, ∇Γ (V · n) − ∇Γ (U · n), VΓ − DU · n, n V · n = U, ∇Γ (V · n) − DU · n, n V · n = U, ∇Γ (V · n) + divΓ U(V · n) = divΓ ((V · n)U).
5 Application: Shape Gradient of the Drag In this section, we would like to compute the necessary optimality condition associated to the minimization of the drag functional J(Ω) = μ
T 0
|ε(U)|2 dxdt. Ω
By Hadamard formula [6], the Eulerian derivative of J(·) in the direction V is given by: dJ(Ω; V) = 2μ
T 0
ε(U ) · ·ε(U) + μ
T
Ω
0
|ε(U)|2 V(0) · n. Γ
In order to obtain the expression of the shape gradient, we need to introduce the corresponding adjoint system. 5.1 Adjoint State System and the Shape Gradient The adjoint problem associated to the functional J is obtained by considering the adjoint operator of the system satisfied by the shape derivative (U , p ) with a forcing term corresponding to the distributed part of the Eulerian derivative dJ(Ω; V). Precisely we have the following result. Lemma 5.1 There exists a unique couple (w, q) ∈ L2 (0, T ; V(Ω)) × L2 (0, T ; L2 (Ω)/R) solution of the adjoint system: ⎧ −∂t w − μΔw − Dw · U + ∗ DU · w + ∇q = 2μA(U) ⎪ ⎪ ⎪ ⎪ divw = 0 ⎪ ⎪ ⎨ w=0 w ·n=0 ⎪ ⎪ ⎪ ⎪ (2με(w) · n + βw)tg = 0 ⎪ ⎪ ⎩ w(T ) = 0
in Ω × (0, T ), in Ω × (0, T ), on ΓD × (0, T ), on ΓB × (0, T ), on ΓB × (0, T ), in Ω, (35)
where the operator A is defined by: A : L2 (0, T ; H1 (Ω)) −→ L2 (0, T ; (H1 (Ω)) ): A(z), v =
T 0
ε(z) · ·ε(v)dxdt. Ω
123
Appl Math Optim
As we will see below the derivative of J at Ω in the direction V depends only on V(0). This suggest to define
¯ R3 | divV = 0 in D, V · n = 0 on ∂D E(D) = V ∈ C3 D, as the set of admissible directions. Thanks to the contribution of the adjoint system, we are able to give the expression of the shape gradient of J at Ω. Theorem 5.2 The shape gradient of J at Ω can be expressed, in E(D) , as follows ∇J(Ω) = γ∗Γ (j (ΓB ) n)
(36)
where γ∗Γ denotes the transpose of the trace operator defined on E(D) and j(ΓB ) = ∇Γ (σ(w)n · n)U + μ|ε(U)|2 + 2μ∇Γ ε(U)n, n w − 2με(U) · ·ε(w) − β[DU · n + HU]tg w − μ(ΔU)tg w + 2μdivΓ UdivΓ w − β Dw · n, U .
(37)
Proof Thanks to problem (35), we have: dJ(Ω; V) =
T 0
+ =
−2με(w) · n + qn, U dΓdt + μ
T 0
ΓB
−2με(w) · n, n divΓ (V · nU)dΓdt + T 0
T
V · n|ε(U)|2 dΓdt ΓB
2με(U ) · n, w dΓdt ΓB
0
0
0
ΓB
T
+μ =
T
V · n|ε(U)|2 dΓdt + ΓB
T 0
T
qdivΓ (V · nU) ΓB
wtg Q(U)dΓdt 0
ΓB
Q(U)w + ∇Γ (σ(w)n · n)U + μ|ε(U)|2 (V · n)dΓdt.
ΓB
Thus the mapping V −→ dJ(Ω; V) is obviously linear and continuous on E(D). By Hadamard structure theorem [18], we can write dJ(Ω; V) =
T 0
∇J(Ω), V E (D) ,E (D) dt
and identify the shape density gradient j(ΓB ) by replacing Q(U) by its expression (29). Finally we can derive the necessary optimality condition associated to the shape minimization problem under consideration.
123
Appl Math Optim
Corollary 5.3 Let Ω be a sufficiently smooth domain solution of the drag minimization. Then there exists a constant Λ0 such that T
∇Γ (σ(w)n · n)U + μ|ε(U)|2 + 2μdivΓ ( ε(U)n, n w) − 2με(U) · ·ε(w)
0
−β[DU · n + HU]tg w − μ(ΔU)tg w + 2μdivΓ UdivΓ w − β Dw · n, U dt ≡ Λ0 .
(38)
Proof At a critical shape we have dJ(Ω; V) = 0 ∀ V ∈ E(D). By Fubini, it can be written, ∀V ∈ E(D) T
ΓB
∇Γ (σ(w)n · n)U + μ|ε(U)|2 + 2μdivΓ ( ε(U)n, n w) − 2με(U) · ·ε(w)
0
−β[DU · n + HU]tg w − μ(ΔU)tg w + 2μdivΓ UdivΓ w − β Dw · n, U dt) V · n dΓ = 0.
Since any V ∈ E(D) satisfies the constraint
V · n dΓ = 0, thus the necessary ΓB
optimality condition (38) is given in terms of a constant.
The computation of the gradient will be helpful for the development of gradientbased algorithms as it is done, for example, in [11] for heat equation or in [9] for Navier–Stokes flow with Dirichlet boundary condition.
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