FEEDBACK BOUNDARY STABILIZATION OF THE ... - Semantic Scholar
FEEDBACK BOUNDARY STABILIZATION OF THE TWO DIMENSIONAL NAVIER-STOKES EQUATIONS JEAN-PIERRE RAYMOND∗ Abstract. We study the exponential stabilization of the linearized Navier-Stokes equations around an unstable stationary solution, by means of a feedback boundary control, in dimension 2 or 3. The feedback law is determined by solving a Linear-Quadratic control problem. We do not assume that the normal component of the control is equal to zero. In that case the state equation, satisfied by the velocity field y, is decoupled into an evolution equation satisfied by P y, where P is the so-called Helmholtz projection operator, and a quasi-stationary elliptic equation satisfied by (I − P )y. Using this decomposition we show that the feedback law can be expressed only in function of P y. In the two dimensional case we show that the linear feedback law provides a local exponential stabilization of the Navier-Stokes equations. Key words. equation
Dirichlet control, feedback control, stabilization, Navier-Stokes equations, Oseen equations, Riccati
AMS subject classifications. 93B52, 93C20, 93D15, 35Q30, 76D55, 76D05, 76D07
1. Setting of the problem. Let Ω be a bounded and connected domain in R2 or R3 with a regular boundary Γ, ν > 0, and consider a couple (w, χ) – a velocity field and a pressure – solution to the stationary Navier-Stokes equations in Ω: −ν∆w + (w · ∇)w + ∇χ = f
and div w = 0 in Ω,
w = u∞ s
on Γ.
(1.1)
We assume that w is regular and is an unstable solution of the instationary Navier-Stokes equations. The purpose of this paper is to determine a Dirichlet boundary control u, in feedback form, localized in a part of the boundary Γ, so that the corresponding controlled system: ∂y − ν∆y + (y · ∇)w + (w · ∇)y + (y · ∇)y + ∇p = 0, ∂t y = M u on Σ∞ , y(0) = y0 in Ω,
div y = 0
in Q∞ ,
(1.2)
be stable for initial values y0 small enough in an appropriate space X(Ω). In this setting, Q∞ = Ω×(0, n o ∞), 2 0 Σ∞ = Γ × (0, ∞), X(Ω) is a subspace of Vn (Ω) = y ∈ L (Ω) | div y = 0 in Ω, y · n = 0 on Γ , and the operator M is a restriction operator precisely defined in section 2. If we set (z, q) = (w + y, χ + p) and if u = 0, we see that (z, q) is the solution to the Navier-Stokes equations ∂z − ν∆z + (z · ∇)z + ∇q = f , div z = 0 ∂t z = u∞ on Σ∞ , z(0) = w + y0 in Ω. s
in Q∞ ,
Thus y0 is a perturbation of the stationary solution w. To study the local feedback stabilization of system (1.2), we first study the feedback stabilization of the corresponding linearized system ∂y − ν∆y + (w · ∇)y + (y · ∇)w + ∇p = 0, in Q∞ , ∂t div y = 0 in Q∞ , y = M u on Σ∞ , y(0) = y0 in Ω.
(1.3)
To stabilize this system we can look for a control u belonging either to L2 (0, ∞; V0 (Γ)) or to L2 (0, ∞; Vn0 (Γ)), where n o n o V0 (Γ) = y ∈ L2 (Γ) | hy · n, 1iH −1/2 (Γ),H 1/2 (Γ) = 0 and Vn0 (Γ) = y ∈ L2 (Γ) | y · n = 0 on Γ . ∗ Laboratoire MIP, UMR CNRS 5640, [email protected]
Universit´ e Paul Sabatier, 1
31062 Toulouse Cedex 9,
France,
email:
A linear feedback law stabilizing (1.3) in X(Ω) ⊂ Vn0 (Ω), with a control u ∈ L2 (0, ∞; V0 (Γ)), is determined by a bounded operator K ∈ L(X(Ω), V0 (Γ)) such that the solution of the closed loop system ∂y − ν∆y + (w · ∇)y + (y · ∇)w + ∇p = 0, in Ω × (0, T ), ∂t div y = 0 in Ω × (0, T ), y = M Ky on Γ × (0, T ), y(0) = y0
in Ω,
obeys |y(t)|X(Ω) ≤ C|y0 |X(Ω)
for all t ∈ (0, ∞).
(1.4)
The feedback control law is given by u(t) = Ky(t) for all t ∈ (0, ∞).
(1.5)
In place of (1.4) we can look for an exponential decay |y(t)|X(Ω) ≤ Ce−σt |y0 |X(Ω) for all t ∈ (0, ∞),
with σ > 0.
(1.6)
Let us underline that a feedback law of the form (1.5) is a pointwise (in time) feedback law. A feedback law may be of a different form, for example of the form u = L0 y0 ,
L0 ∈ L(X(Ω), L2 (0, ∞; V0 (Γ))).
In engineering applications pointwise feedback law are needed because they are more robust with respect to perturbations in the models. This paper is only focused on the characterization of pointwise feedback laws for the Oseen and the Navier-Stokes equations. Several important questions must be addressed when we look for a pointwise feedback law able to stabilize system (1.3) or (1.2): (Q1 ) Does there exist a control u ∈ L2 (0, ∞; V0 (Γ)) such that the solution of (1.3) obeys (1.6) or (1.4) ? In other words, is the system (1.3) stabilizable ? (Q2 ) Assume that K is a pointwise feedback law able to stabilize system (1.3) in X(Ω). Does K also stabilize the nonlinear system (1.2) for |y0 |X(Ω) small enough ? (Q3 ) Assume that we have proved the existence of a feedback law stabilizing system (1.3). Can we find an equation characterizing K which can be numerically solved by classical methods ? We stop the list here, but lot of other questions are very important in applications, such as the robustness of feedback laws, the numerical accuracy of approximations. In this paper, we do not address the stabilizability of system (1.3), and we shall use the existing results in the literature. This paper is mainly devoted to (Q2 ) and (Q3 ). But firstly mention some results answering to (Q1 ). When w ∈ L∞ (Ω) and X(Ω) = Vn0 (Ω) ∩ L4 (Ω), the existence of a boundary control u ∈ L2 (0, ∞; 0 V (Γ)) such that the solution of system (1.3) obeys (1.6) may be deduced, by using an extension method, from exact controllability results with internal controls n stated in [9]. o In the three dimensional case, and when X(Ω) = y ∈ H1 (Ω) | div y = 0 in Ω , the existence of a feedback control law exponentially stabilizing (1.2) or (1.3) is proved in [13]. But the feedback operator constructed in [13] is of the form L0 (it is not a pointwise feedback operator, and it requires the knowledge of the eigenfunctions and the eigenvalues of the Oseen operator of equation (1.3)). In the three dimensional cases, when Ω is simply connected, Barbu, Lasiecka and Triggiani [4] have proved the stabilizability of system (1.3) in X(Ω) = y ∈ H1/2+ε (Ω) | div y = 0 in Ω, y · n = 0 on Γ with controls u ∈ L2 (0, ∞; Vn0 (Γ)) acting everywhere on the bondary Γ. Some additional results in the two dimensional case are also stated in [4] (see also section 7.1). Let us now focus on (Q2 ) and (Q3 ). One way to address these questions is to use the optimal control theory, where the pointwise feedback law is obtained by solving an infinite time horizon control problem of the form n o (Q) inf J(y, u) | (y, u) satisfies (1.3), u ∈ L2 (0, T ; U) . 2
More precisely one has to prove that the value function of problem (Q) obeys inf(Q) = Πy0 , y0 , 0 (Ω) Vn
where Π ∈ L(Vn0 (Ω)), Π = Π∗ ≥ 0. Next the feedback law is defined thanks to the operator Π. In this framework, question (Q3 ) can be reformulated as follows: Does the operator Π satisfy an algebraic Riccati equation in the domain of the Oseen operator ? The answer depends on the choice of J and u and is not necessarily obvious in the case of a boundary control problem. According to the stabilizability results mentioned above, we have two possible choices for U: U = Vn0 (Γ)
or U = V0 (Γ).
Both choices are interesting for applications. Even if U = V0 (Γ) leads to a little bit complicated analysis, this case is interesting since the normal velocity is used as a control in many applications [17, 21, 29]. These papers are devoted to the suppression of vortex shedding past a cylinder. It corresponds to the case where Ω = Ωe \ Ωi , Ωi is a bounded domain in R2 with a regular boundary Γi , and Ωe is an other bounded domain in R2 with a regular boundary Γe such that Ωi ⊂ Ωe (Ωi is a disk in the case of a circular cylinder). Given a stationary velocity u∞ s , we can take (w, χ) as the solution to the stationary NavierStokes equation (1.1) with f = 0. In that case the control has to be localized in a part of the boundary Γi , and Γe must be far enough from the cylinder in order that the solution to the stationary or instationary Navier-Stokes equations give a good approximation of the corresponding equation in the exterior domain R2 \ Ωi . Some heuristic feedback laws have been successfully tested in numerical experiments [29, 21], but in general these feedback laws, which are designed for a very specific value of u∞ s , are not robust , contrarily to feedback laws obtained by solving Riccati equations. with respect to perturbations of u∞ s This paper is written in the case when U = V0 (Γ) and the adaptation to the case when U = Vn0 (Γ) is given in section 7.1. We would like to underline that the choice of J is critical. Indeed once X(Ω), U and J are chosen, K is generally uniquely determined. Thus we have to choose J so that the corresponding feedback law also stabilizes system (1.2). The functional J is usually of the form Z Z 1 ∞ 1 ∞ J(y, u) = |Cy(t)|2V0 (Ω) dt + |u(t)|2U dt, 2 0 2 0 where the observation operator C may be a bounded or an unbounded operator in V0 (Ω) = y ∈ L2 (Ω) | div y = 0 when U = V0 (Γ) (or in Vn0 (Ω) when U = Vn0 (Γ)). If D(C) denotes the domain of C in V0 (Ω) (or in Vn0 (Ω) when U = Vn0 (Γ)), the solution y of (1.3) is searched in L2 (0, ∞; D(C)). In [4, 2], C is chosen so that |Cy|V0 (Ω) be a norm in Vn0 (Ω) equivalent to the usual norm of the space H3/2+ε (Ω), for some ε > 0. The idea in [4, 2] is to choose the operator C so that the norm |Cy|V0 (Ω) be strong enough to dominate the nonlinearity of the Navier-Stokes equations. In that way a Lyapunov function for the nonlinear closed loop system can be defined thanks to Π. The price to pay is that the operator Π, corresponding this problem, does not satisfy a Riccati equation in the domain of the Oseen operator. An algebraic Riccati equation is stated only in D((AΠ )2 ) [4, section 4.5], where AΠ is the infinitesimal generator of the associated closed loop system. Thus the domain in which the Riccati equation is stated depends on the unknown Π of the equation. This is a serious drawback. Here we follow a different approach. We choose J as follows J(y, u) =
1 2
Z 0
T
Z
|y|2 dxdt +
Ω
1 2
Z
T
0
Z
|u|2 dxdt.
(1.7)
Γ
One difficulty comes from the fact that, choosing U = V0 (Γ), the system (1.3) cannot be written in the form of a classical evolution equation. We have shown in [30] that (1.3) can be rewritten in the form: P y0 = AP y + BM u,
y(0) = y0 ,
(I − P )y = (I − P )DA γn M u, 3
(1.8)
where A is the corresponding Oseen operator, B is a boundary control operator, P is the so-called Helmholtz or Leray projector, and DA the Dirichlet operator associated with A. For the precise definitions and properties of these operators we refer to section 2. Using the writing of system (1.3) in the form (1.8), we show that the control problem (Q) can be rewritten in the form of another control problem in which the state variable is P y and not y. This transformation is essential in our approach. It leads to a Riccati equation which is the natural one for the new control problem, but which is not the expected one if we only consider problem (Q). Moreover the Riccati equation is satisfied in the domain of A. Now the difficulty is to prove that the linear feedback law determined in this way also stabilizes the nonlinear equation (1.2). This is not obvious because the norm of y involved in J is to weak to dominate the nonlinearity of the Navier-Stokes equations. Thus we cannot follow the Lyapunov function approach as in [2] and [4]. We develop a completely different approach. We study the regularizing properties of the operator Π. Thanks to these results – which, to the best of our knowledge, are new even in the case of other parabolic equations like the heat equation – we are able to establish a local feedback stabilization result for the Navier-Stokes equation for initial data y0 small enough in the space 1/2−ε Vn (Ω) = Vn0 (Ω) ∩ H1/2−ε (Ω), for all 0 < ε < 1/4. We show that the solution to the closed loop nonlinear system obeys an exponential decay in the corresponding space V1/2−ε (Ω) = V0 (Ω)∩H1/2−ε (Ω). Finally mention that in the three dimensional case, the stabilization of the nonlinear problem cannot be treated with the same tools and requires a more delicate analysis [31, 32]. The paper is organized as follows. The Oseen operator and some associated boundary control operators are studied in section 2. Optimality conditions for the finite horizon control problem of the Oseen equations are established in section 3. We study the corresponding infinite time horizon control problem in section 4. We show that the optimal solution of this problem is characterized by an optimality system. This kind of characterization, which is known in the case of bounded control operators [1, 3] is, to the best of our knowledge, new in the case of unbounded operators. Thanks to this optimality system we are able to study the regularity properties of the feedback operator. In order to study the nonlinear problem, we first study in section 5 a nonhomogeneous linear-quadratic control problem. The nonhomogeneous term will play the role of the nonlinearity in the next section. The local stabilization of the Navier-Stokes equations is studied in section 6. Some additional results are stated in section 7. We have collected some regularity results needed throughout the paper in an appendix. 2. Functional framework. 2.1. Notation and assumptions. Let us introduce the following spaces: H s (Ω; RN ) = Hs (Ω), L (Ω; RN ) = L2 (Ω), the same notation conventions will be used for trace spaces and for the spaces H0s (Ω; RN ). We also introduce different spaces of free divergence functions and some corresponding trace spaces: n o Vs (Ω) = y ∈ Hs (Ω) | div y = 0 in Ω, hy · n, 1iH −1/2 (Γ),H 1/2 (Γ) = 0 for s ≥ 0, n o Vns (Ω) = y ∈ Hs (Ω) | div y = 0 in Ω, y · n = 0 on Γ for s ≥ 0, n o V0s (Ω) = y ∈ Hs (Ω) | div y = 0 in Ω, y = 0 on Γ for s > 1/2, n o Vs (Γ) = y ∈ Hs (Γ) | hy · n, 1iH −1/2 (Γ),H 1/2 (Γ) = 0 for s ≥ −1/2 . 2
In the above setting n denotes the unit normal to Γ outward Ω. We shall use the following notation QT = Ω × (0, T ), ΣT = Γ × (0, T ), Qt¯,T = Ω × (t¯, T ) and Σt¯,T = Γ × (t¯, T ) for t¯ > 0, and 0 < T ≤ ∞. For spaces of time dependent functions we set Vs,σ (QT ) = H σ (0, T ; V0 (Ω)) ∩ L2 (0, T ; Vs (Ω)), and Vs,σ (ΣT ) = H σ (0, T ; V0 (Γ)) ∩ L2 (0, T ; Vs (Γ)). We assume that Ω is of class C 4 and w ∈ V3 (Ω). 4
In order to find a control u, supported in an open subset Γc of Γ, we introduce a weight function m ∈ C 2 (Γ) with values in [0, 1], with support in Γc , equal to 1 in Γ0 , where Γ0 is an open subset in Γc . Associated with this function m we introduce the operator M ∈ L(V0 (Γ)) defined by Z m R M u(x) = m(x)u(x) − mu · n n(x), m Γ Γ where |Γ| is the (N − 1)-dimensional Lebesgue measure of Γ. By this way, we can replace the condition supp(u) ⊂ Γc by considering a boundary condition of the form z − w = Mu
on
Σ∞ .
The main interest of this operator M is that if u ∈ L2 (0, ∞; H s (Γc ; RN )) ∩ H s/2(0, ∞; L2 (Γc ; RN )) for ˜ denotes the extension of u by zero to Σ∞ \ Γc × (0, ∞) , then M u ˜ belongs to some 0 < s ≤ 2, and if u 2 s N s/2 2 N ˜. L (0, ∞; H (Γ; R )) ∩ H (0, ∞; L (Γ; R )), which is not true for u 0 For all ψ ∈ H 1/2+ε (Ω), with ε0 > 0, we denote by c(ψ) and c(mψ) the constants defined by Z Z 1 1 ψ and c(mψ) = mψ. (2.1) c(ψ) = |Γ| Γ |Γ| Γ 2.2. Properties of some operators. In the following we consider the linearized Navier-Stokes equation ∂y − ν∆y + (w · ∇)y + (y · ∇)w + ∇p = 0, in QT , ∂t div y = 0 in QT , y = M u on ΣT , y(0) = y0 in Ω ,
(2.2)
and the adjoint equation ∂Φ − ν∆Φ − (w · ∇)Φ + (∇w)T Φ + ∇ψ = y, in QT , ∂t div Φ = 0 in QT , Φ = 0 on ΣT , Φ(T ) = 0 in Ω ,
−
(2.3)
where T is finite or infinite. To study these equations, we introduce the Stokes and the Oseen operators associated with equations (2.2) and (2.3). Let P be the orthogonal projector in L2 (Ω) onto Vn0 (Ω), and denote by (A0 , D(A0 )), (A, D(A)) (A∗ , D(A∗ )) and the unbounded operators in Vn0 (Ω) defined by D(A0 ) = H2 (Ω) ∩ V01 (Ω),
A0 y = P ∆y
D(A) = H2 (Ω) ∩ V01 (Ω),
Ay = νP ∆y − P ((w · ∇)y) − P ((y · ∇)w),
∗
2
D(A ) = H (Ω) ∩
V01 (Ω),
for all y ∈ D(A0 ),
A∗ y = νP ∆y + P ((w · ∇)y) − P ((∇w)T y).
Throughout the following we denote by λ0 > 0 an element in the resolvent set of A satisfying (λ0 I − A)y, y V0 (Ω) ≥ ω0 |y|2V1 (Ω) for all y ∈ D(A), 0 n and (λ0 I − A∗ )y, y V0 (Ω) ≥ ω0 |y|2V1 (Ω) for all y ∈ D(A∗ ),
(2.4)
0
n
for some 0 < ω0 < 1. Theorem 2.1. The unbounded operator (A−λ0 I) (respectively (A∗ −λ0 I)) with domain D(A−λ0 I) = D(A) (respectively D(A∗ −λ0 I)) is the infinitesimal generator of a bounded analytic semigroup on Vn0 (Ω). Moreover, for all 0 ≤ α ≤ 1, we have D((λ0 I − A)α ) = D((λ0 I − A∗ )α ) = D((λ0 I − νA0 )α ) = D((−A0 )α ) . Proof. Under condition (2.4) the analyticity of the semigroup generated by (A − λ0 I) is well known ([6, Chapter 1, Theorem 2.1.2]). The characterization of the domains of fractional powers of (λ0 I − A) and (λ0 I − A∗ ) may be deduced from [28]. 5
Observe that the semigroups (et(A−λ0 I) )t≥0 and (et(A and that ket(A−λ0 I) kL(Vn0 (Ω)) ≤ e−ωt
∗
−λ0 I)
and ket(A
∗
)t≥0 are exponentially stable on Vn0 (Ω)
−λ0 I)
kL(Vn0 (Ω)) ≤ e−ωt ,
for all ω < ω0 (see [6, Chapter 1, Theorem 2.12]). Lemma 2.2. If Φ belongs to V3/2+ε (Ω) ∩ V01 (Ω), for some ε > 0, then ∂Φ ∂n · n = 0 on Γ. Proof. This result is proved in [4, Lemma 3.3.1] for functions Φ belonging to V01 (Ω)∩(C 1 (Ω))N . Since V01 (Ω)∩(C 2 (Ω))N is dense in V3/2+ε (Ω)∩V01 (Ω), the result is still true for functions in V3/2+ε (Ω)∩V01 (Ω), when ε > 0. Let us introduce DA and Dp , two Dirichlet operators associated with A, defined as follows. For u ∈ V0 (Γ), set DA u = y and Dp u = q where (y, q) is the unique solution in V1/2 (Ω) × (H 1/2 (Ω)/R)0 to the equation λ0 y − ν∆y + (w · ∇)y + (y · ∇)w + ∇q = 0 div y = 0 in Ω,
in Ω,
y = u on Γ.
Lemma 2.3. (i) The operator DA is a bounded operator from V0 (Γ) into V0 (Ω), moreover it satisfies kDA ukVs+1/2 (Ω) ≤ C(s)kukVs (Ω)
for all 0 ≤ s ≤ 2 .
∗ ∈ L(V0 (Ω), V0 (Γ)), the adjoint operator of DA ∈ L(V0 (Γ), V0 (Ω)), is defined by (ii) The operator DA ∗ DA g = −ν
∂z + πn − c(π)n, ∂n
(2.5)
where (z, π) is the solution of λ0 z − ν∆z − (w · ∇)z + (∇w)T z + ∇π = g
and div z = 0 in Ω,
z=0
on Γ,
(2.6)
and c(π) is defined by (2.1). Proof. Part (i) is well known when w = 0 (see e.g. [35]). Its adaptation in the case when w ∈ V3 (Ω) is given in [30, Corollary 7.1]. Part (ii) is stated in [30, Lemma 7.4]. Let us define the operators γτ ∈ L(V0 (Γ)) and γn ∈ L(V0 (Γ)) by γτ u = u − u · n n and γn u = u · n n = u − γτ u for all u ∈ V0 (Γ) . R Let us also denote by PΓ the projector in L2 (Γ) onto V0 (Γ) defined by PΓ u = u − R mm Γ u · n n. Γ Observe that M = PΓ m, where m denotes the multiplication operator by the function m. Introducing the spaces n o n o Vτ0 (Γ) = u ∈ V0 (Γ) | γτ u = 0 and Vn0 (Γ) = u ∈ V0 (Γ) | γn u = 0 , we have V0 (Γ) = Vτ0 (Γ) ⊕ Vn0 (Γ). Lemma 2.4. The operator M obeys the following properties M = M ∗,
M γτ = γτ M = mγτ ,
and
M γn = γn M .
The operators γτ and γn satisfy: γτ = γτ∗ ,
γn = γn∗
(I − P )DA = (I − P )DA γn .
and
Proof. From the definition of M it follows that Z Z M u, v 0 = mu · v = u · PΓ mv = u, M v V (Γ)
Γ
Γ
6
V0 (Γ)
,
for all u ∈ V0 (Γ) and all v ∈ V0 (Γ). Thus M = M ∗ . Observe that PΓ γτ = γτ = γτ PΓ , and mγτ = γτ m. Thus we have M γτ = PΓ mγτ = mγτ = γτ m = γτ M. Thus the second identity is established. Since γτ + γn = I, where I is the identity operator in V0 (Γ), we have M (γτ + γn ) = (γτ + γn )M , and the identity M γn = γn M follows from the equality M γτ = γτ M . The first two identities satisfied by γτ and γn are obvious. Moreover, if γn u = 0 then DA u ∈ Vn0 (Ω), and (I − P )DA u = 0. This proves the last identity. In the next lemma we study the properties of an operator RA which plays a crucial role in optimality conditions of control problems that we consider. Lemma 2.5. The operator ∗ R A = M DA (I − P )DA M + I
is an isomorphism from V0 (Γ) into itself. Moreover, for all 0 ≤ s ≤ 3/2, its restriction to Vs (Γ) is an isomorphism from Vs (Γ) into itself. In addition RA satisfies RA γn = γn RA γn
and
RA γτ = γτ RA γτ = γτ .
The restriction of RA to Vτ0 (Γ) is an isomorphism from Vτ0 (Γ) into itself, and we have −1 −1 −1 RA u = (γn RA γn )−1 u = γn RA γn u = γn RA u
for all u ∈ Vτ0 (Γ) .
Proof. Let s belong to [0, 3/2]. The operator DA is continuous from Vs (Γ) into Vs+1/2 (Ω), (I − P ) is ∗ ∗ (I − is bounded from Vs+1/2 (Ω) into Vs+1 (Γ). Thus M DA bounded from Vs+1/2 (Ω) into itself, and DA s s+1 s P )DA M is bounded from V (Γ) into V (Γ), and compact in V (Γ). Let us show that the kernel of RA is reduced to 0. If u ∈ V0 (Γ) obeys RA u = 0, then we have 0 = RA u, u = (I − P )DA M u, (I − P )DA M u + u, u ≥ u, u . V0 (Γ)
V0 (Ω)
V0 (Γ)
V0 (Γ)
Thus u = 0, which shows that RA is injective. From the Fredholm alternative it follows that RA is an isomorphism from Vs (Γ) into itself. With the identities stated in Lemma 2.4, we can write RA γn u, v 0 = (I − P )DA M γn u, (I − P )DA γn M v 0 + γn u, γn v 0 V (Γ) V (Ω) V (Γ) = (I − P )DA M γn u, (I − P )DA M γn v 0 + γn u, γn v 0 = RA γn u, γn v 0 V (Ω)
V (Γ)
for all u ∈ V0 (Γ) and all v ∈ V0 (Γ). Thus RA γn = γn RA γn . Similarly, we have RA γτ u, v = γτ u, v V0 (Γ)
V (Γ)
V0 (Γ)
for all u ∈ V0 (Γ) and all v ∈ V0 (Γ), because (I − P )DA M γτ u = 0. Thus RA γτ = γτ , from which we deduce γτ RA γτ = γτ . From the above identities it follows that RA u = RA γn u = γn RA γn u ∈ Vτ0 (Γ)
for all u ∈ Vτ0 (Γ).
Thus the restriction of RA to Vτ0 (Γ) is an isomorphism from Vτ0 (Γ) into itself. Also observe that the restriction of γn RA γn to Vτ0 (Γ) enjoys the same property. The last identities in the lemma follows from these properties. We introduce the operators Bn = (λ0 I − A)P DA γn ,
Bτ = (λ0 I − A)DA γτ , 7
B = Bn + Bτ .
Let us set Bn,α = (λ0 I − A)α−1 Bn = (λ0 I − A)α P DA γn ,
Bτ,α = (λ0 I − A)α−1 Bτ = (λ0 I − A)α DA γτ ,
and Bα = Bn,α + Bτ,α . Theorem 2.6. For all α ∈]0, 41 [, Bn,α and Bτ,α belong to L(V0 (Γ), Vn0 (Ω)). Proof. Let us prove the result for Bn,α , the other one can be shown in a similar way. Let u ∈ V0 (Γ), 1/2 set y = DA γn (u). Then P y = P DA γn (u) belongs to Vn (Ω). We have D E (λ0 I − A)P y, z (D(A∗ ))0 ,D(A∗ ) D E = (λ0 I − A)α P y, (λ0 I − A∗ )1−α z (2.7) (D((−A∗ )1−α ))0 ,D((−A∗ )1−α ) Z = yP (λ0 z − ν∆z − (w · ∇)z + (∇w)T z), for all z ∈ D(A∗ ) = V2 (Ω) ∩ V01 (Ω). Ω
Every z ∈ D(A∗ ) is the solution to the equation λ0 z − ν∆z − (w · ∇)z + (∇w)T z + ∇q = f ,
div z = 0 in Ω,
z = 0 on Γ,
∂z − q n satisfies the estimate where f = P (λ0 z − ν∆z − (w · ∇)z + (∇w)T z). It is well known that ν ∂n
∂z − q n ≤ C|f |D((−A∗ )−α ) ≤ C|z|D((−A∗ )1−α ) , ν ∂n V0 (Γ) provided that 0 < α < 1/4. Thus we have Z Z yf = yP λ0 z − ν∆z − (w · ∇)z + (∇w)T z Ω Ω Z Z ∂z −ν y λ0 z − ν∆z − (w · ∇)z + (∇w)T z + ∇q = = + q n · γn (u) ∂n Γ Ω ∂z ≤ C|u|V0 (Γ) |z|D((−A∗ )1−α ) . − q n ≤ |u|V0 (Γ) ν ∂n V0 (Γ) Combining this inequality with (2.7), we obtain Z yP λ0 z − ν∆z − (w · ∇)z + (∇w)T z |(λ0 I − A)α P y|Vn0 (Ω) ≤ sup|z|D((−A∗ )1−α ) =1 Ω
≤ C|u|V0 (Γ) . Thus the proof is complete. Proposition 2.7. For all Φ ∈ D(A∗ ), B ∗ Φ belongs to V1/2 (Γ), we have ∗ B ∗ Φ = DA (λ0 I − A∗ )Φ,
∗ Bτ∗ Φ = γτ DA (λ0 I − A∗ )Φ,
∗ Bn∗ Φ = γn DA (λ0 I − A∗ )Φ ,
and B ∗ Φ = −ν
∂Φ ∂Φ + ψn − c(ψ)n, Bτ∗ Φ = γτ B ∗ Φ = −ν ∂n ∂n
and
Bn∗ Φ = γn B ∗ Φ = ψn − c(ψ)n ,
with h i ∇ψ = (I − P ) ν∆Φ + (w · ∇)Φ − (∇w)T Φ , and c(ψ) is defined by (2.1). In particular if Φ ∈ Vs (Ω) ∩ V01 (Ω) with s ≥ 2, the following estimate holds |B ∗ Φ|Vs−3/2 (Γ) ≤ C|Φ|Vs (Ω)∩V01 (Ω) . 8
Proof. The first result is a direct consequence of the definition of B and of the fact that B is a bounded operator from V0 (Γ) into (D(A∗ ))0 . The identities for Bτ∗ Φ and Bn∗ Φ follows from Lemma 2.4. ∗ From the definitions of B ∗ and DA it follows that B ∗ Φ = −ν
ˆ ∂Φ ˆ − c(ψ)n ˆ , + ψn ∂n
ˆ is defined by ˆ ψ) where (Φ, ˆ − ν∆Φ ˆ − (w · ∇)Φ ˆ + (∇w)T Φ ˆ + ∇ψˆ = P (λ0 I − A∗ )Φ = (λ0 I − A∗ )Φ, λ0 Φ ˆ =0 div Φ
in Ω,
in Ω,
ˆ = 0 on Γ , Φ
ˆ is defined by (2.1). From the first equation we deduce (λ0 I − A∗ )Φ = (λ0 I − A∗ )Φ. ˆ Thus and c(ψ) 1 ˆ Next we can see that ψˆ is the unique element in H (Ω)/R defined by Φ = Φ. h i ˆ + ν∆Φ ˆ + (w · ∇)Φ ˆ − (∇w)T Φ ˆ ∇ψˆ = (I − P ) − λ0 Φ h i = (I − P ) ν∆Φ + (w · ∇)Φ − (∇w)T Φ . The estimate of B ∗ Φ directly follows from its definition. The proof is complete. 3. A finite time horizon control problem. To deal with the stabilization problem formulated in section 1, we first study the following optimal control problem n o (QTs,ζ ) inf JT (s, y, u) | (y, u) satisfies (3.1), u ∈ V0,0 (Σs,T ) , where 1 JT (s, y, u) = 2
Z s
T
Z
1 |y| dxdt + 2 Ω 2
Z
T
|u(t)|2V0 (Γ) dt,
s
and ∂y − ν∆y + (w · ∇)y + (y · ∇)w + ∇p = 0, in Qs,T , ∂t div y = 0 in Qs,T , y = M u on Σs,T , y(s) = ζ in Ω.
(3.1)
The main objective of this section is to establish optimality conditions for (QTs,ζ ) and to show that ¯ ) obeys a pointwise feedback formulation defined thanks to the solution to a the optimal solution (¯ y, u differential Riccati equation. To look for the solution u to problem (QTs,ζ ) in feedback form, we rewrite equation (3.1) in the form P y0 = AP y + BM u in (s, T ),
y(s) = ζ,
(I − P )y = (I − P )DA γn M u in (s, T ). As we see the situation is more complicated than in the case when we have a single equation of the form ¯ (t) = −B ∗ Π(t)¯ y(t) (Π being the solution of y0 = Ay + Bu and when the feedback law is of the form u some differential Riccati equation). To overcome the difficulty coming from the presence of the second equation (I − P )y = (I − P )DA γn M u, which is not an evolution equation, we also transform the writing of the functional JT in the following way: Z Z Z Z 1 T 1 T 2 JT (s, y, u) = |y| + |u|2 2 s Ω 2 s Γ Z Z Z Z Z Z 1 T 1 T 1 T = |P y|2 + |(I − P )y|2 + |u|2 2 s Ω 2 s Ω 2 s Γ 9
and 1 2
T
Z
Z
s
1 = 2
Z
1 = 2
Z
|(I − P )y|2 +
Ω T Z
s
1 2
T
Z
Z
s
|u|2
Γ
|(I − P )DA M γn u|2 +
Ω T
Z
s
1 2
Z
T
Z
s
(|γτ u|2 + |γn u|2 )
Γ
∗ |(M DA (I − P )DA M )1/2 γn u|2 + |γn u|2 + |γτ u|2 .
Γ
Thus, if y is the solution of (3.1), we have Z Z Z Z Z Z 1 T 1 T 1 T 1/2 2 2 JT (s, y, u) = IT (s, y, u) = |P y| + |R γn u| + |γτ u|2 , 2 s Ω 2 s Γ A 2 s Γ where RA is the operator introduced in Lemma 2.5. The control problem (QTs,ζ ) is equivalent to n o T (Ps,ζ ) inf IT (s, y, u) | (y, u) satisfies (3.2), u ∈ V0,0 (Σs,T ) , where P y0 = AP y + BM u in (s, T ),
y(s) = ζ.
(3.2)
T Even if problems (QTs,ζ ) and (Ps,ζ ) are equivalent, we are going to see that the optimality conditions for T ¯ ), that we cannot obtain with (Ps,ζ ) allows us to prove regularity results for the optimal solution (¯ y, u T the optimality system of problem (Qs,ζ ). T ) admits a unique solution (yζs , usζ ) Theorem 3.1. For all s ∈ [0, T ] and all ζ ∈ Vn0 (Ω), problem (Ps,ζ and −1 usζ = −M Bτ∗ Φsζ − RA M Bn∗ Φsζ
in (s, T ),
(3.3)
Φ(T ) = 0 .
(3.4)
where Φsζ is solution to the equation −Φ0 = A∗ Φ + P yζs
in (s, T ),
Conversely the system −1 M Bn∗ Φ P y0 = AP y − Bτ M 2 Bτ∗ Φ − Bn M RA 0
∗
−Φ = A Φ + P y
in (s, T ),
in (s, T ),
P y(s) = ζ ,
(3.5)
Φ(T ) = 0 ,
admits a unique solution (P yζs , Φsζ ) in L2 (s, T ; Vn0 (Ω)) × (V2,1 (Qs,T ) ∩ L2 (s, T ; V01 (Ω)), and (P yζs − (I − −1 −1 T P )DA M RA M Bn∗ Φsζ , −M Bτ∗ Φsζ − RA M Bn∗ Φsζ ) is the optimal solution to (Ps,ζ ). Remark 3.2. Due to Lemma 2.5, we have −1 −1 M Bτ∗ Φsζ + RA M Bn∗ Φsζ = RA M B ∗ Φsζ ,
but we prefer to keep the writing using the decomposition of B ∗ in the form Bτ∗ + Bn∗ to give the respective expressions of the tangential and the normal components. T Proof. (i) The existence of a unique solution (yζs , usζ ) to problem (Ps,ζ ) is obvious. Let u be in 2 0 2 0 L (s, T ; V (Γ)) and v ∈ L (s, T ; V (Γ)). Denote by (yu , pu ) the solution to equation (3.2) corresponding to u, and set IT (s, yu , u) = IT (u). We have I0T (u)v =
Z
T
Z
Z
T
Z
P yu · P z + s
Ω
Z
T
Z
RA γn u · γn v + s
Γ
10
γτ u · γτ v, s
Γ
where z is the solution to P z0 = AP z + BM v
in (s, T ),
(I − P )z = (I − P )DA M v .
P z(s) = 0,
Let Φ be the solution to the equation −Φ0 = A∗ Φ + P yu ,
Φ(T ) = 0 .
Due to Lemma 8.6 Φ belongs to V2,1 (Qs,T ). Thus B ∗ Φ belongs to L2 (s, T ; V0 (Γ)). The functions z and Φ obey the following identity Z TZ Z TD Z TZ E P yu · P z = BM v(τ ), Φ(τ ) dτ = v · M B∗Φ . s
Ω
(D(A∗ ))0 ,D(A∗ )
s
s
Γ
Thus I0T (u)v
Z
T
Z
T
Z
∗
Z
Z
v · MB Φ +
= s
Γ
T
Z
RA γn u · γn v + s
Γ
γτ u · γτ v. s
(3.6)
Γ
T If (yζs , usζ ) is the solution to problem (Ps,ζ ), we have I0T (usζ ) = 0, which gives −1 −1 γn usζ = −γn RA M B ∗ Φsζ = −RA M Bn∗ Φsζ
γτ usζ = −γτ M B ∗ Φsζ = −M Bτ∗ Φsζ .
and
(ii) From part (i) of the proof it follows that (yζs , Φsζ ) is a solution of system (3.5) (Φsζ is the solution to ¯ with ¯ − R−1 M B ∗ Φ, ¯ is a solution of system (3.5), and if we set u ¯ = −M Bτ∗ Φ equation (3.4)). If (¯ y, Φ) n A s s s 0 ¯ ¯ = uζ . Thus y ¯ = yζ , and Φ = Φζ . The proof is u) = 0, which implies that u (3.6) we can verify that IT (¯ complete. T There is another way to characterize the optimal solution of (Ps,ζ ), it consists in writing the optimality T conditions for (Qs,ζ ). More precisely we can state the following theorem. Theorem 3.3. For all s ∈ [0, T ] and all ζ ∈ Vn0 (Ω), the unique solution (yζs , usζ ) of problem (QTs,ζ ) is characterized by ∂Φs ζ usζ = m ν − ψζs n + c(mψζs )n , ∂n
(3.7)
where (Φsζ , ψζs ) is the solution to equation ∂Φ − ν∆Φ − (w · ∇)Φ + (∇w)T Φ + ∇ψ = yζs , in Qs,T , ∂t div Φ = 0 in Qs,T , Φ = 0 on Σs,T , Φ(T ) = 0 in Ω ,
−
(3.8)
and c(mψζs ) is the constant defined by (2.1) corresponding to mψζs . Remark 3.4. It is clear that the solution Φsζ of equation (3.8) is equal to the solution of equation (3.4). It is the reason why they are denoted in the same way. Proof. Let u be in L2 (s, T ; V0 (Γ)) and v ∈ L2 (s, T ; V0 (Γ)). Denote by yu the solution to equation (3.1), and set JT (s, yu , u) = JT (u). We have J0T (u)v
Z
T
Z
Z
T
Z
yu · z +
= s
Ω
u · v, s
Γ
where (z, q) is the solution to ∂z − ν∆z + (w · ∇)z + (z · ∇)w + ∇q = 0, in Qs,T , ∂t div z = 0 in Qs,T , y = M v on Σs,T , z(s) = 0 in Ω. 11
Let (Φ, ψ) be the solution to the equation ∂Φ − ν∆Φ − (w · ∇)Φ + (∇w)T Φ + ∇ψ = yu , ∂t div Φ = 0 in Qs,T , Φ = 0 on Σs,T , Φ(T ) = 0 −
in Qs,T , in Ω .
We can verify that Z
T
Z
T
Z
Z
yu · z = s
Ω
s
Γ
Z TZ ∂Φ ∂Φ −ν + ψn · M v = m −ν + ψn · v. ∂n ∂n s Γ
Thus usζ is characterized by Z s
T
Z Γ
∂Φsζ −mν + mψζs n + usζ ∂n
·v =0
for all v ∈ V0 (Γ), that is γτ usζ = mν
∂Φsζ ∂n
and γn usζ = −mψζs n + c(mψζs )n.
The proof is complete. T Since (Ps,ζ ) and (QTs,ζ ) admit the same solution we have ∂Φs ζ −1 − ψζs n + c(mψζs )n . −M Bτ∗ Φsζ − RA M Bn∗ Φsζ = m ν ∂n The characterization given in (3.7) seems to be easier to use than the one in (3.3). However we are going to see that we are able to characterize the regularity of the solution (yζs , usζ ) in an optimal way by using (3.3), which is not the case with (3.7). To understand the relationship between (3.3) and (3.7), let us rewrite the second equation in (3.5) in the form ∂Φsζ − ν∆Φsζ − (w · ∇)Φsζ + (∇w)T Φsζ + ∇ψ˜ζs = P yζs , in Qs,T , ∂t div Φsζ = 0 in Qs,T , Φsζ = 0 on Σs,T , Φ(T ) = 0 in Ω.
−
(3.9)
Observe that ∇ψ˜ζs = ∇ψζs − (I − P )yζs , where ψζs is the pressure appearing in (3.8). Lemma 3.5. If (Φsζ , ψ˜ζs ) is the solution to equation (3.9), then B ∗ Φsζ (t) = −ν
∂Φsζ (t) + ψ˜ζs (t)n − c(ψ˜ζs (t))n ∂n
for almost all t ∈ (s, T ),
and the optimal control usζ is defined by usζ = mν
∂Φsζ −1 − RA M ψ˜ζs n. ∂n
Proof. From Proposition 2.7 it follows that B ∗ Φsζ (t) = −ν
∂Φsζ (t) + ψˆζs (t)n − c(ψˆζs (t))n , ∂n
where ∇ψˆζs is defined by h i ∇ψˆζs = (I − P ) ν∆Φsζ + (w · ∇)Φsζ − (∇w)T Φsζ . 12
(3.10)
s ∂Φ Thus ∇ψˆζs = ∇ψ˜ζs because (I − P ) ∂tζ = 0. Therefore usζ is defined by
−1 usζ = −M Bτ∗ Φsζ − RA M Bn∗ Φsζ =
−1 − M γτ − RA M γn
∂Φs ζ − ψ˜ζs n , ν ∂n
and due to Lemma 2.5 the proof is complete. In the following theorem we improve the regularity result of the optimal solution. Theorem 3.6. The solution (yζs , Φsζ ) to system (3.5) belongs to V1,1/2 (Qs,T ) × L2 (s, T ; 3 1 3/2 V (Ω) ∩ V0 (Ω)) ∩ H (s, T ; Vn0 (Ω)). In particular Φsζ belongs to C([s, T ]; V2 (Ω) ∩ V01 (Ω)). Proof. Since B ∗ Φsζ ∈ L2 (s, T ; V0 (Γ)), due to Lemmas 8.1 and 8.3, yζs belongs to 1/2−ε,1/4−ε V (Qs,T ) for all ε > 0. From Lemmas 8.6 and 8.8 it follows that ∂Φsζ ∈ V1−ε,1/2−ε/2 (Σs,T ) ∂n
∇ψ˜ζs ∈ H1/2−ε,1/4−ε/2 (Qs,T ) ,
and
for all ε > 0, where ψ˜ζs is the pressure appearing in equation (3.9). From this regularity result and from Lemma 3.5, we deduce that B ∗ Φsζ ∈ V1/2−ε,1/4−ε/2 (Σs,T ) and usζ ∈ V1/2−ε,1/4−ε/2 (Σs,T ) for all ε > 0, −1 T ). Still applying Lemma M Bn∗ Φsζ is the solution of problems (QTs,ζ ) and (Ps,ζ where usζ = −M Bτ∗ Φsζ − RA s 1−ε,1/2−ε 8.3, we obtain yζ ∈ V (Qs,T ) for all ε > 0, and repeating the above analysis for B ∗ Φsζ , we can s 1−ε,1/2−ε/2 show that uζ ∈ V (Σs,T ) for all ε > 0. Still with Lemmas 8.1 and 8.3, we prove that yζs belongs to V1,1/2 (Qs,T ). Due to Lemma 8.6, Φsζ belongs to V3,3/2 (Qs,T ). In particular Φsζ belongs to C([s, T ]; V2 (Ω) ∩ V01 (Ω)). The proof is complete. Remark 3.7. Observe that the regularity result stated in Theorem 3.6 has been obtained by using the expression of usζ given in (3.10). In (3.10), only the pressure ψ˜ζs appears, and ψ˜ζs only depends on Φsζ . Thus improving the regularity of Φsζ we are able to improve the regularity of ψ˜ζs , and this is what is done in Lemma 8.7. If we use the expression of usζ given in (3.7), the pressure ψζs obeys ∇ψζs = ∇ψ˜ζs + (I − P )yζs . Since the regularity with respect to the time variable of (I − P )yζs cannot be improved, we are not able to improve the regularity of usζ by using (3.7). Corollary 3.8. For all s ∈ [0, T ] and all zeta ∈ Vn0 (Ω), the unique solution (yζs , usζ ) to problem T (Ps,ζ ) and the corresponding solution (yζs , Φsζ ) to system (3.5) obeys IT (s, yζs , usζ ) =
1 2
Z
Φsζ (s) · ζ .
Ω
Proof. By multiplying by Φsζ the equation (3.1) satisfied by (yζs , usζ ), integrating in space and time, making integration by parts, and using equation (3.9) satisfied by Φsζ , we can prove that Z Ω
Φsζ (s)
Z
T
Z
·ζ = s
|P yζs |2
Ω
Z + s
T
Z ∂Φsζ ν − ψ˜ζs n + c(ψ˜ζs )n · M usζ . ∂n Γ
Next using (3.10) we can show that Z s
T
Z Z TZ Z TZ ∂Φsζ 1/2 ν − ψ˜ζs n + c(ψ˜ζs )n · M usζ = |RA γn usζ |2 + |γτ usζ |2 , ∂n Γ s Γ s Γ
and the proof is complete. Let Π(s) be the operator defined by Π(s) : ζ 7−→ Φsζ (s) ,
(3.11)
where (P yζs , Φsζ ) is the unique solution to system (3.5). From Theorem 3.6 it follows that Π(s) ∈ L(Vn0 (Ω), V2 (Ω) ∩ V01 (Ω)). Using Theorem 3.6 we can prove that the family of operators (Π(s))s∈[0,T ] defined by (3.11) belongs to Cs ([0, T ]; L(Vn0 (Ω)) (the space of functions Π from [0, T ] into L(Vn0 (Ω)) such 13
that, for all y ∈ Vn0 (Ω), Π(·)y be continuous from [0, T ] into Vn0 (Ω)). Next, using the optimality system (3.5) we can show that Π is the unique solution in Cs ([0, T ]; L(Vn0 (Ω))) to the Riccati equation Π∗ (t) = Π(t) and Π(t) ≥ 0, for all y ∈ Vn0 (Ω), t ∈ [0, T ], Π(t)y ∈ V2 (Ω) ∩ V01 (Ω) and |Π(t)y|V2 (Ω) ≤ C|y|Vn0 (Ω) , −1 −Π0 (t) = A∗ Π(t) + Π(t)A − Π(t)Bτ M 2 Bτ∗ Π(t) − Π(t)Bn M RA M Bn∗ Π(t) + I, Π(T ) = 0.
(3.12)
From the definition of Π, from Theorem 3.1 and Corollary 3.8, we deduce the following theorem. We also refer to [24, Theorem 1.2.2.1] where the existence of a unique solution to equation (3.12) is established. T ) belongs to C([0, T ]; V0 (Ω)) × C([0, T ]; V0 (Γ)), Theorem 3.9. The solution (y, u) to problem (P0,y 0 it obeys the feedback formula −1 u(t) = − M Bτ∗ + RA M Bn∗ Π(t)P y(t), and the optimal cost is given by J(y, u) =
1 Π(0)y0 , y0 0 . 2 Vn (Ω)
b b is the unique solution in Cs ([0, T ]; L(Vn0 (Ω))) to the Riccati equation If we set Π(t) = Π(T − t), then Π b ∗ (t) = Π(t) b b Π and Π(t) ≥ 0, 0 b b 0 (Ω) , ∈ V2 (Ω) ∩ V01 (Ω) and |Π(t)y| for all y ∈ Vn (Ω), t ∈ [0, T ], Π(t)y V2 (Ω) ≤ C|y|Vn −1 ∗ ∗ 0 ∗ b b b b + Π(t)A b b b (t) = A Π(t) Π(t) + I, Π(t) − Π(t)B R B − Π(t)B B Π n A τ τ n b Π(0) = 0.
(3.13)
b it follows that Π(0) = Π(T b ). From the definition of Π ∞ 4. An infinite time horizon problem. In this section we want to study problem (P0,y ), and we 0 want to study the regularity of its solution in function of the regularity of y0 . For notational simplicity ∞ problem (P0,y ) will now be denoted by (P0,y0 ), and the state variable by y and not P y as in the previous 0 section. With this notation the problem we consider is n o (P0,y0 ) inf I(y, u) | (y, u) satisfies (4.1), u ∈ V0,0 (Σ∞ ) ,
where I(y, u) =
1 2
Z
∞
Z
0
|y|2 dxdt +
Ω
1 2
∞
Z
0
1/2 |γτ u(t)|2V0 (Γ) + |RA γn u(t)|2V0 (Γ) dt,
and y0 = Ay + BM u in (0, ∞),
y(0) = y0 .
(4.1)
k Accordingly, for 0 ≤ s < k < ∞, problem (Ps,ζ ) will now be defined by k (Ps,ζ )
n o inf Ik (s, y, u) | (y, u) satisfies (4.2), u ∈ V0,0 (Σs,k ) ,
where Ik (s, y, u) =
1 2
Z s
k
Z Ω
|y|2 dxdt +
1 2
Z
k
s
1/2 |γτ u(t)|2V0 (Γ) + |RA γn u(t)|2V0 (Γ) dt,
and y0 = Ay + BM u
in (s, k), 14
y(s) = ζ.
(4.2)
Theorem 4.1. For all y0 ∈ Vn0 (Ω), problem (P0,y0 ) admits a unique solution (yy0 , uy0 ). There exists Π ∈ L(Vn0 (Ω)), obeying Π = Π∗ , such that the optimal cost is given by 1 Πy0 , y0 . I(yy0 , uy0 ) = 0 (Ω) 2 Vn Proof. From the null controllability results stated in [9], we can deduce that there exist controls u ∈ L2 (0, ∞; V0 (Γ)) such that I(yu , u) < ∞, where yu is the solution of equation (4.1) corresponding to u. The null controllability results in [9] are stated for a distributed control. Using an extension of the domain, this result also provides a null controllability result for a control localized on the boundary. The existence of a unique solution (yy0 , uy0 ) to (P0,y0 ) follows from classical arguments. From the dynamic programming principle, it follows that the mapping b )y0 , y0 T 7−→ Π(T 0 Vn (Ω)
is nondecreasing, and we have 1b Π(T )y0 , y0 0 ≤ I(yy0 , uy0 ) < ∞ . 2 Vn (Ω) As in [7], or in [24], we can show that there exists an operator Π ∈ L(Vn0 (Ω)) satisfying Π = Π∗ ≥ 0 and b )y0 Πy0 = limT →∞ Π(T for all y0 ∈ Vn0 (Ω). k Let us show that I(yy0 , uy0 ) = 21 Πy0 , y0 . Problem (P0,y ) admits a unique solution (yk , uk ) 0 0 Vn (Ω)
characterized by yk0 = Ayk + BM uk
in (0, k),
yk (0) = y0 ,
−Φ0k
in (0, k),
Φk (k) = 0,
∗
= A Φk + yk
γτ uk = −M Bτ∗ Φk ,
(4.3)
−1 γn uk = −RA M Bn∗ Φk .
˜ k the extension by zero of uk to (k, ∞), and by y ˜ k the extension Convergence of yk and uk . Denote by u by zero of yk to (k, ∞). Since we have Z kZ Z k 1/2 |yk |2 dxdt + |γτ uk (t)|2V0 (Γ) + |RA γn uk (t)|2V0 (Γ) dt 0
Z
Ω ∞
0
Z
≤ 0
|yy0 |2 dxdt +
Ω
Z
∞
0
1/2 |γτ uy0 (t)|2V0 (Γ) + |RA γn uy0 (t)|2V0 (Γ) dt ,
the sequences (˜ yk )k and (˜ uk )k are respectively bounded in L2 (0, ∞; Vn0 (Ω)) and L2 (0, ∞; V0 (Γ)). Thus 2 there exist y∞ ∈ L (0, ∞; Vn0 (Ω)) and u∞ ∈ L2 (0, ∞; V0 (Γ)) such that ˜ k * u∞ u
weakly in L2 (0, ∞; V0 (Γ)),
˜ k * y∞ y
weakly in L2 (0, ∞; Vn0 (Ω)).
By passing to the limit in the above inequality we obtain Z ∞Z Z ∞ 1/2 |y∞ |2 dxdt + |γτ u∞ (t)|2V0 (Γ) + |RA γn u∞ (t)|2V0 (Γ) dt 0
Z
Ω ∞Z
≤
0
2
Z
|yy0 | dxdt + 0
Ω
0
∞
1/2 |γτ uy0 (t)|2V0 (Γ) + |RA γn uy0 (t)|2V0 (Γ) dt .
And by passing to the limit in the equation satisfied by (yk , uk ), we have 0 y∞ = Ay∞ + BM u∞
in (0, ∞), 15
y∞ (0) = y0 .
Thus the pair (y∞ , u∞ ) is admissible and we have (y∞ , u∞ ) = (yy0 , uy0 ), because I(y∞ , u∞ ) ≤ I(yy0 , uy0 ). Therefore we can claim that in L2 (0, ∞; V0 (Γ))
˜ k → yy0 u
in L2 (0, ∞; Vn0 (Ω)).
˜ k → uy0 and y
Since Ik (0, yk , uk ) =
1b Π(k)y0 , y0 0 , 2 Vn (Ω)
by passing to the limit when k tends to infinity, we obtain 1 Πy0 , y0 . I(yy0 , uy0 ) = 0 (Ω) 2 Vn We denote by ϕ(y0 ) the value function of problem (P0,y0 ), that is: ϕ(y0 ) = I(yy0 , uy0 ). Lemma 4.2. For every y0 ∈ Vn0 (Ω), the system −1 y0 = Ay − Bτ M 2 Bτ∗ Φ − Bn M RA M Bn∗ Φ
−Φ0 = A∗ Φ + y
in (0, ∞),
in (0, ∞),
y(0) = y0 , (4.4)
Φ(∞) = 0 ,
for all t ∈ (0, ∞) ,
Φ(t) = Πy(t)
admits a unique solution in L2 (0, ∞; Vn0 (Ω)) × V2,1 (Q∞ ). This solution belongs to Cb (R+ ; Vn0 (Ω)) ∩ V1,1/2 (Q∞ ) × V3,3/2 (Q∞ ) and it satisfies: kykCb (R+ ;Vn0 (Ω)) + kykV1,1/2 (Q∞ ) + kΦkV3,3/2 (Q∞ ) ≤ C|y0 |Vn0 (Ω) . −1 M Bn∗ Φ) is the solution of (P0,y0 ). The pair (y, −M Bτ∗ Φ − RA ˆ ). We denote by Proof. For notational simplicity the solution to (P0,y0 ) will now be denoted by (ˆ y, u k ϕk (0, y0 ) the value function of problem (P0,y ) and by ϕk (t¯, ζ) the value function of problem (Pt¯k,ζ ). 0 k Let (ykt¯ , utk¯ ) be the solution of (Pt¯k,yk (t¯) ), and let (yk , uk ) be the solution of (P0,y ) characterized by 0 t¯ t¯ t¯ (4.3). Denote by Φk the adjoint state corresponding to (yk , uk ), and by Φk the adjoint state corresponding to (yk , uk ). From the dynamic programming principle it follows that (ykt¯ , utk¯ , Φtk¯ )(t) = (yk , uk , Φk )(t) b − t¯)yk (t¯). Thus for all t ∈ (t¯, k). Therefore we have Φtk¯ (t¯) = Φk (t¯) ∈ ∂y ϕk (t¯, yk (t¯)), that is Φk (t¯) = Π(k
b − t¯)k|yk (t¯)|V0 (Ω) ≤ C|yk (t¯)|V0 (Ω) , |Φk (t¯)|Vn0 (Ω) ≤ kΠ(k n n and ˜ k kL2 (0,∞;V0 (Ω)) ≤ Ck˜ kΦ yk kL2 (0,∞;Vn0 (Ω)) , n ˜ k (respectively y ˜ k ) is the extension by zero of Φk (respectively yk ) to (k, ∞). In the proof of where Φ ˆ in Theorem 4.1 we have shown that (˜ yk )k is bounded in L2 (0, ∞; Vn0 (Ω)) and that it converges to y 2 0 ˜ ˜ k is also the L (0, ∞; Vn (Ω)). Therefore (Φk )k is also bounded in L2 (0, ∞; Vn0 (Ω)). Observe that Φ solution of the equation ˜ 0 = (A∗ − λ0 I)Φ ˜k + y ˜ k, ˜ k + λ0 Φ −Φ k
˜ k (∞) = 0. Φ
Thus ˜ k (t) = Φ
Z
∞
e(A
∗
−λ0 I)(τ −t)
˜ k (τ )) dτ (˜ yk (τ ) + λ0 Φ
t
16
for all t ≥ 0.
˜ k )k is also bounded in L∞ (0, ∞; V0 (Ω)). From Young’s inequality for convolutions it follows that (Φ n ∞ 0 2 0 ˆ ∈ L (0, ∞; V (Ω)) ∩ L (0, ∞; V (Ω)) such that, after extraction of a subsequence, There then exists Φ n n we have ˜k * Φ ˆ Φ
weakly in L2 (0, ∞; Vn0 (Ω))
weak-star in L∞ (0, ∞; Vn0 (Ω)),
and
ˆ obeys the equation and Φ ˆ Φ(t) =
Z
∞
e(A
∗
−λ0 I)(τ −t)
ˆ )) dτ (ˆ y(τ ) + λ0 Φ(τ
for all t ≥ 0.
t
ˆ We have Step 3. Regularity of Φ. ˆ 0 = (A∗ − λ0 I)Φ ˆ + λ0 Φ ˆ +y ˆ −Φ
ˆ Φ(∞) = 0.
in (0, ∞),
ˆ belongs to L2 (0, ∞; V0 (Ω)), we deduce that Φ ˆ belongs to L2 (0, ∞; V1 (Ω)) ∩ V2,1 (Q∞ ) (see Since Φ n 0 ˜ k − R−1 M B ∗ Φ ˜ Lemma 8.5). Moreover due to Lemma 8.9, the sequence (˜ uk )k = (−M Bτ∗ Φ n k )k converges A −1 −1 2 0 ∗ˆ ∗ˆ ∗ ∗ ˆ − R M B Φ, ˆ and (ˆ ˆ obey ˆ = −M Bτ Φ weakly in L (0, ∞; V (Γ)) to −M Bτ Φ − RA M Bn Φ. Thus u y, Φ) n A the first two equations in (4.4). Step 4. Let us show that if (y, Φ) ∈ L2 (0, ∞; Vn0 (Ω)) × V2,1 (Q∞ ) is a solution to system (4.4), then y belongs Cb ([0, ∞); Vn0 (Ω) ∩ V1,1/2 (Q∞ ), Φ ∈ V3,3/2 (Q∞ ), and kykCb ([0,∞);Vn0 (Ω)) + kykV1,1/2 (Q∞ ) + kΦkV3,3/2 (Q∞ )
(4.5)
≤ C(|y0 |Vn0 (Ω) + kykL2 (0,∞;Vn0 (Ω)) + kΦkL2 (0,∞;Vn0 (Ω)) ) . To establish this result we rewrite system (4.4) as follows −1 M Bn∗ Φ + λ0 y y0 = (A − λ0 I)y − Bτ M 2 Bτ∗ Φ − Bn M RA
−Φ0 = (A∗ − λ0 I)Φ + y + λ0 Φ in (0, ∞),
in (0, ∞),
y(0) = y0 ,
(4.6)
Φ(∞) = 0.
Due to Lemma 8.9 we know that B ∗ Φ ∈ L2 (0, ∞; V0 (Γ)). Applying Lemmas 8.1 and 8.3, we obtain: kykV1/2−ε0 ,1/4−ε0 /2 (Q∞ ) ≤ C(|y0 |Vn0 (Ω) + kykL2 (0,∞;Vn0 (Ω)) + kΦkL2 (0,∞;Vn0 (Ω)) ) 0
for all ε0 > 0.
0
Still from Lemma 8.9, we deduce that B ∗ Φ ∈ V1/2−ε ,1/4−ε /2 (Σ∞ ) for all ε0 > 0. Applying successively 0 0 Lemmas 8.1, 8.3 and Lemma 8.9 we can prove that y belongs to V1−ε ,1/2−ε /2 (Q∞ ) and B ∗ Φ belongs to 0 0 V1−ε ,1/2−ε /2 (Σ∞ ) for all ε0 > 0. Another iteration gives y ∈ V1,1/2 (Q∞ ) ∩ Cb ([0, ∞); Vn0 (Ω)) (because y0 ∈ Vn0 (Ω)). From Lemma 8.5 we deduce that Φ ∈ V3,3/2 (Q∞ ), and the estimate (4.5) holds true. ˆ obeys the third equation in (4.4). With Lemma 8.5 we can show Step 5. We show that the pair (ˆ y, Φ) that ˜ k (t) * Φ(t) ˆ Φ
weakly in Vn0 (Ω)
for all t ≥ 0.
Since Φk (t) ∈ ∂y ϕk (t, yk (t)),
ˆ Φk (t) * Φ(t) weakly in Vn0 (Ω),
and ϕk (t, yk (t)) → ϕ(ˆ y(t)) as k → ∞ , we deduce that ˆ Φ(t) ∈ ∂ϕ(ˆ y(t)),
ˆ i.e. Φ(t) = Πˆ y(t).
ˆ obeys the third equation in (4.4). Thus we have shown that (ˆ y, Φ) 17
Step 6. Uniqueness. If a solution (y, Φ) ∈ L2 (0, ∞; Vn0 (Ω)) × V2,1 (Q∞ ) to system (4.4), due to step 4 it obeys (4.5), and we can show that Z k Z k Z Z k Z −1 |M Bτ∗ Φ(t)|2V0 (Γ) dt + |RA |y(t)|2Vn0 (Ω) dt + M Bn∗ Φ(t)|2V0 (Γ) dt = y0 Φ(0) − y(k)Φ(k). 0
0
0
Ω
Passing to the limit when k tends to infinity we obtain Z ∞ Z ∞ Z |y(t)|2Vn0 (Ω) dt + |M Bτ∗ Φ(t)|2V0 (Γ) dt + 0
0
0
∞ −1 |RA M Bn∗ Φ(t)|2V0 (Γ) dt =
Ω
Z y0 Φ(0). Ω
Thus if y0 = 0 we have y = 0. From the relation Φ = Πy we deduce that Φ = 0, and the uniqueness is established. ˆ is the unique solution to system Step 7. Final estimate. From the previous steps it follows that (ˆ y, Φ) (4.4). Since kˆ ykL2 (0,∞;Vn0 (Ω)) ≤ C|y0 |Vn0 (Ω) , we have ˆ L2 (0,∞;V0 (Ω)) ≤ C|y0 |V0 (Ω) , kΦk n n the estimate of the lemma follows from (4.5). 1/2−ε (Ω) for some 0 < ε ≤ 1/2, then the solution (y, Φ) of system (4.4) Corollary 4.3. If y0 ∈ Vn 3/2−ε,3/4−ε/2 belongs to V (Q∞ ) × V7/2−ε,7/4−ε/2 (Q∞ ), and we have: kykV3/2−ε,3/4−ε/2 (Q∞ ) + kΦkV7/2−ε,7/4−ε/2 (Q∞ ) + kB ∗ ΦkV3/2−ε,3/4−ε/2 (Q∞ ) ≤ C|y0 |V1/2−ε (Ω) . n
Proof. With Lemmas 4.2 and 8.9, we first prove that u ∈ V1,1/2 (Σ∞ ). Applying Lemmas 8.1 and 8.3, it follows that y belongs to V3/2−ε,3/4−ε/2 (Q∞ ). The estimate for Φ follows from Lemma 8.5, and we deduce that B ∗ Φ ∈ V3/2−ε,3/4−ε/2 (Q∞ ) with Lemma 8.9. Remark 4.4. Due to Lemma 4.2, if y0 ∈ Vn0 (Ω), then Φ belongs to C([0, T ]; V2 (Ω) ∩ V01 (Ω)). In 1/2−ε (Ω) then particular Φ(0) belongs to V2 (Ω) ∩ V01 (Ω). From Corollary 4.3 we deduce that if y0 ∈ Vn 5/2−ε 1 5/2−ε 1 (Ω) ∩ V0 (Ω). This means that Π belongs Φ ∈ C([0, T ]; V (Ω) ∩ V0 (Ω)) and Φ(0) belongs to V 1/2−ε (Ω), V5/2−ε (Ω) ∩ V01 (Ω)). to L(Vn0 (Ω), V2 (Ω) ∩ V01 (Ω)), and Π also belongs to L(Vn ∗ 2 1 1/2 Since B is continuous from V (Ω) ∩ V0 (Ω) into V (Γ) (see Proposition 2.7), the operator B ∗ Π 1/2−ε (Ω), V1−ε (Γ)) for all ε > 0. belongs to L(Vn0 (Ω), V1/2 (Γ)), and B ∗ Π also belongs to L(Vn 0 From Lemma 4.2 it follows that, for all y0 ∈ Vn (Ω), the evolution equation −1 y0 = Ay − Bτ M 2 Bτ∗ Πy − Bn M RA M Bn∗ Πy
in (0, ∞),
y(0) = y0 ,
(4.7)
admits at least one weak solution belonging to L2 (0, ∞; Vn0 (Ω)) ∩ Cb (R; Vn0 (Ω)). It is easy to show that this solution is unique. Due to Lemma 4.2, it is equal to yy0 , where (yy0 , uy0 ) is the solution of (P0,y0 ). Still from Lemma 4.2, we deduce that the family of operators (S(t))t≥0 defined by S(t) : y0 7−→ yy0 (t) , is a strongly continuous exponentially stable semigroup on Vn0 (Ω). Let (AΠ , D(AΠ )) be its infinitesimal generator, and let us denote by (etAΠ )t≥0 the semigroup (S(t))t≥0 . Since the semigroup (etAΠ )t≥0 is exponentially stable on Vn0 (Ω), we can define D(AΠ ) by Z ∞ n o D(AΠ ) = y ∈ Vn0 (Ω) | y = − eτ AΠ f dτ for some f ∈ Vn0 (Ω) , 0
R∞
τ AΠ
and AΠ y = f if y = − 0 e f dτ ∈ D(AΠ ). We want to find a more explicit characterization of D(AΠ ) and AΠ . From Remark 4.4, it follows that −1 the operator A − BM RA M B ∗ Π may be considered as a bounded operator from Vn0 (Ω) into (D(A0 ))0 , the dual of D(A0 ) with respect to the pivot space Vn0 (Ω). We want to prove that n o −1 D(AΠ ) = y ∈ Vn0 (Ω) | A − BM RA M B ∗ Π ∈ Vn0 (Ω) . (4.8) 18
For that we are going to use the characterization of (A∗Π , D(A∗Π )), the adjoint of (AΠ , D(AΠ )). Let us define the unbounded operator (A]Π , D(A]Π )) in Vn0 (Ω) by D(A]Π ) = D(A0 )
−1 and A]Π Φ = A∗ Φ − (B ∗ Π)∗ M RA M B ∗ Φ for all Φ ∈ D(A]Π ),
where (B ∗ Π)∗ ∈ L(V−1/2 (Γ), Vn0 (Ω)) is the adjoint of B ∗ Π ∈ L(Vn0 (Ω), V1/2 (Γ)). We know that (A∗ − λ0 I, D(A∗ )) is the infinitesimal generator of a bounded analytic semigroup on Vn0 (Ω). Let (L, D(L)) be the unbounded operator in Vn0 (Ω) defined by D(L) = V3/2+ε (Ω) ∩ V01 (Ω) with 0 < ε < 1/2, and −1 LΦ = (B ∗ Π)∗ M RA M B ∗ Φ. We can easily verify that L is (A∗ − λ0 I)-bounded with relative bound zero. From [22, Corollary 2.5, page 500] it follows that (A]Π − λ0 I, D(A0 )) is the infinitesimal generator of a quasi-bounded analytic semigroup on Vn0 (Ω). Thus (A]Π , D(A0 )) is also the infinitesimal generator of an analytic semigroup on Vn0 (Ω). Consider the equations Φ0 = A]Π Φ,
Φ(0) = Φ0 ,
(4.9)
and ˆ 0 = A] Φ, ˆ −Φ Π
ˆ ) = Φ0 . Φ(T
(4.10)
ˆ We clearly have Φ(T − t) = Φ(t) for all t ∈ [0, T ]. If Φ0 ∈ D(A0 ), then the solution Φ to equation (4.9) belongs to C([0, T ]; D(A0 )) ∩ C 1 ([0, T ]; D(A0 )). With equations (4.7) and (4.10), we can show that ]
y0 , eT AΠ Φ0
0 (Ω) Vn
= eT AΠ y0 , Φ0
∗
0 (Ω) Vn
= y0 , eT AΠ Φ0
0 (Ω) Vn ]
,
∗ and eT AΠ t≥0 t≥0 we have (A]Π , D(A0 )) = (A∗Π , D(A∗Π )). For all g ∈ Vn0 (Ω), the equation on Vn0 (Ω) are identical, R ∞ and ∗ τ A∗ AΠ Φ = g admits − 0 e Π g dτ as unique solution in D(A∗Π ). Thus, for all g ∈ Vn0 (Ω), the solution R∞ ∗ − 0 eτ AΠ g dτ to the equation A∗Π Φ = A]Π Φ = g belongs to D(A0 ).
for all y0 ∈ Vn0 (Ω), Φ0 ∈ Vn0 (Ω), and T > 0. Thus the two semigroups eT AΠ
Let us come back to the characterization of D(AΠ ). Let y be in D(AΠ ), set AΠ y = f . Setting Rt Rt z(t) = − 0 eτ AΠ f dτ = − 0 e(t−s)AΠ f ds, using the definition of weak solution for z, by passing to the limit when t tends to infinity, we can show that y satisfies the equation −1 Ay − BM RA M B ∗ Πy = f ,
(4.11)
where f = AΠ y. Thus n o −1 D(AΠ ) ⊂ y ∈ Vn0 (Ω) | A − BM RA M B ∗ Π ∈ Vn0 (Ω) . −1 In equation (4.11), Ay − BM RA M B ∗ Πy is considered as an element in (D(A0 ))0 . This means that y is a solution to equation (4.11) if and only if Z Z Z Z −1 yA∗ Φ − M RA M B∗Π y B∗Φ = yA∗Π Φ = f Φ for all Φ ∈ D(A0 ). (4.12) Ω
Γ
Ω
Ω
To prove (4.8), it is now sufficient to show that the solution to equation (4.11) or (4.12) is unique. Assume that y ∈ Vn0 (Ω) obeys (4.12) with f = 0. If we take the solution Φ of A∗Π Φ = y in (4.12), we obtain R |y|2 = 0. Thus y = 0 and the equality (4.8) is established. Ω Theorem 4.5. The unbounded operator (AΠ , D(AΠ )) defined by: n o −1 D(AΠ ) = y ∈ Vn0 (Ω) | A − BM RA M B ∗ Π ∈ Vn0 (Ω) , −1 AΠ y = Ay − BM RA M B ∗ Πy
for all y ∈ D(AΠ ),
is the infinitesimal generator of an analytic exponentially stable semigroup on Vn0 (Ω). 19
The operator Π is the unique weak solution to the algebraic Riccati equation Π∗ = Π ∈ L(Vn0 (Ω)) for all y ∈
Vn0 (Ω),
Π ≥ 0,
and 2
Πy ∈ V (Ω) ∩ V01 (Ω)
and
|Πy|V2 (Ω) ≤ C|y|Vn0 (Ω) ,
−1 A∗ Π + ΠA − ΠBτ M 2 Bτ∗ Π − ΠBn M RA M Bn∗ Π + I = 0. 1/2−ε
Moreover Π ∈ L(Vn (Ω); V5/2−ε (Ω) ∩ V01 (Ω)). Proof. The first part of the theorem is already proved. We have already shown that Π∗ = Π ≥ 0 and that Π ∈ L(Vn0 (Ω), V2 (Ω) ∩ V01 (Ω)). Since Π ∈ L(Vn0 (Ω), V2 (Ω) ∩ V01 (Ω)), it is relatively easy to prove that Π is the unique solution to the algebraic Riccati equation stated in the theorem. The uniqueness is shown in [24, Theorem 2.2.1] under a weaker condition than Π ∈ L(Vn0 (Ω), V2 (Ω) ∩ V01 (Ω)). 5. Problems with a nonhomogeneous source term. For all t¯ ≥ 0 and all y0 ∈ Vn0 (Ω) we consider the problem n o (Pt¯,y0 ,f ) inf I(t¯, y, u) | (y, u) satisfies (5.1), u ∈ V0,0 (Σt¯,∞ ) , where I(t¯, y, u) =
1 2
∞
Z t¯
Z
|y|2 dxdt +
Ω
1 2
Z
∞
t¯
1/2 |γτ u(t)|2V0 (Γ) + |RA γn u(t)|2V0 (Γ) dt,
and y0 = Ay + BM u + f
in (t¯, ∞),
y(t¯) = y0 .
(5.1)
The function f belongs to L2/(1+2ε) (0, ∞; Vn0 (Ω)) ∩ L2 (0, ∞; (V2ε (Ω))0 ), with 0 < ε < 1/2. In this section we want to study the regularity of solutions to the control problem (P0,y0 ,f ) in function of the regularity of y0 . This result will be used in the next section to study the local stabilization of the two dimensional Navier-Stokes equations. Theorem 5.1. For all t¯ ≥ 0, y0 ∈ Vn0 (Ω), f ∈ L2/(1+2ε) (0, ∞; Vn0 (Ω)), problem (Pt¯,y0 ,f ) admits a unique solution (yt¯,y0 ,f , ut¯,y0 ,f ) and the optimal cost ϕ(t¯, y0 , f ) = J(t¯, yt¯,y0 ,f , ut¯,y0 ,f ) obeys ϕ(t¯, y0 , f ) ≤ C |y0 |2Vn0 (Ω) + kf k2L2/(1+2ε) (t¯,∞;V0 (Ω)) , (5.2) n
where the constant C is independent of t¯. Proof. The semigroup generated by AΠ on Vn0 (Ω) is analytic and exponentially stable. Thus the solution to the equation z0 = AΠ z + f ,
z(t¯) = y0 ,
is defined by ¯
z(t) = e(t−t)AΠ y0 +
Z
t
e(t−τ )AΠ f (τ ) dτ,
t¯
and it obeys ¯
|z(t)|Vn0 (Ω) ≤ Ce−ω(t−t) |y0 |Vn0 (Ω) + C
Z t¯
t
e−ω(t−τ ) |f (τ )|Vn0 (Ω) dτ,
for some C > 0 and some ω > 0. It follows that kzkL2 (t¯,∞;Vn0 (Ω)) ≤ C |y0 |Vn0 (Ω) + kf kL2/(1+2ε) (t¯,∞;Vn0 (Ω)) , −1 where the constant C is independent of t¯. Since M Bτ∗ Π and RA M Bn∗ Π are bounded operators from Vn0 (Ω) into V0 (Γ) (see Remark 4.4 and Lemma 2.5), we also have −1 kM Bτ∗ ΠzkL2 (t¯,∞;V0 (Γ)) + kRA M Bn∗ ΠzkL2 (t¯,∞;V0 (Γ)) ≤ C |y0 |Vn0 (Ω) + kf kL2/(1+2ε) (t¯,∞;Vn0 (Ω)) .
20
−1 The pair (z, −M Bτ∗ Πz − RA M Bn∗ Πz) is admissible for (Pt¯,y0 ,f ) and we have
−1 I(t¯, z, −M Bτ∗ Πz − RA M Bn∗ Πz) ≤ C |y0 |2Vn0 (Ω) + kf k2L2/(1+2ε) (t¯,∞;V0 (Ω)) . n
Therefore by classical arguments we can prove that (Pt¯,y0 ,f ) admits a unique solution (yt¯,y0 ,f , ut¯,y0 ,f ) and that the optimal cost ϕ(t¯, y0 , f ) = J(t¯, yt¯,y0 ,f , ut¯,y0 ,f ) obeys (5.2). Lemma 5.2. For all u ∈ L2 (0, ∞; V0 (Γ)), y ∈ L2 (0, ∞; Vn0 (Ω)), the equation −1 −Φ0 = A∗Π Φ − (M Bτ∗ Π)∗ γτ u − (RA M Bn∗ Π)∗ γn u + y
in (0, ∞),
Φ(∞) = 0,
(5.3)
admits a unique solution in L2 (0, ∞; Vn0 (Ω)) and kΦkL2 (0,∞;Vn0 (Ω)) + kΦkL∞ (0,∞;Vn0 (Ω)) ≤ C(kykL2 (0,∞;Vn0 (Ω)) + kukL2 (0,∞;V0 (Γ)) ). −1 Proof. We already know that (M Bτ∗ Π)∗ ∈ L(V0 (Γ), Vn0 (Ω)) and (RA M Bn∗ Π)∗ ∈ L(V0 (Γ), ∗ Vn0 (Ω)). Thus from the exponential stability of the semigroup (etAΠ )t≥0 it yields Z ∞ e−ω(τ −t) (|y(τ )|Vn0 (Ω) + |u(τ )|V0 (Γ) )dτ. |Φ(t)|Vn0 (Ω) ≤ C t
The estimates of the lemma follows from Young’s inequality for convolutions. The uniqueness of solution is obvious. ˆ ) the solution to y, u Lemma 5.3. Let f be in L2/(1+2ε) (0, ∞; Vn0 (Ω)), y0 be in Vn0 (Ω), denote by (ˆ ˆ be the solution to equation (5.3) corresponding to (ˆ ˆ is also ˆ ). Then (ˆ problem (P0,y0 ,f ), and let Φ y, u y, Φ) 2 0 2 0 solution in L (0, ∞; Vn (Ω)) × L (0, ∞; Vn (Ω)) to the system −1 M Bn∗ Φ + f y0 = Ay − Bτ M 2 Bτ∗ Φ − Bn M RA
−Φ0 = A∗ Φ + y
in (0, ∞),
in (0, ∞),
y(0) = y0 ,
(5.4)
Φ(∞) = 0.
The following estimate holds ˆ V2,1 (Q ) ≤ C |y0 |V0 (Ω) + kf kL2/(1+2ε) (0,∞;V0 (Ω)) . kˆ ykL2 (0,∞;Vn0 (Ω)) + kΦk ∞ n n Proof. Step 1. Due to estimate (5.2), we have kˆ ykL2 (0,∞;Vn0 (Ω)) + kˆ ukL2 (0,∞;V0 (Γ)) ≤ C |y0 |Vn0 (Ω) + kf kL2/(1+2ε) (0,∞;Vn0 (Ω)) . Consider the problem k (P0,y ) 0 ,f
n o inf Ik (0, y, u) | (y, u) satisfies (5.5), u ∈ V0,0 (Σ0,k ) ,
where Ik (0, y, u) =
1 2
Z
k
0
Z
|y|2 dxdt +
Ω
1 2
k
Z
0
1/2 |γτ u(t)|2V0 (Γ) + |RA γn u(t)|2V0 (Γ) dt.
and y0 = Ay + BM u + f
in (0, k),
y(0) = y0 .
(5.5)
k Problem (P0,y ) admits a unique solution (yk , uk ) characterized by 0 ,f
yk0 = Ayk + BM uk + f −Φ0k
∗
= A Φk + yk
γτ uk = −M Bτ∗ Φk ,
in (0, k),
in (0, k),
yk (0) = y0 , Φk (k) = 0,
−1 γn uk = −RA M Bn∗ Φk .
21
(5.6)
Since we have Z
k
Z
k
Z
2
|yk | dxdt + 0
Z
Ω ∞
0
Z
Z
2
≤
1/2 |γτ uk (t)|2V0 (Γ) + |RA γn uk (t)|2V0 (Γ) dt
∞
|ˆ y| dxdt + 0
Ω
0
1/2
ˆ (t)|2V0 (Γ) + |RA γn u ˆ (t)|2V0 (Γ) ) dt, |γτ u
as in the proof of Theorem 4.1, we can show that ˜k → u ˆ u
in L2 (0, ∞; V0 (Γ))
in L2 (0, ∞; V0 (Ω)),
˜k → y ˆ and y
(5.7)
˜ k and y ˜ k denote the extensions by zero of uk and yk to (k, ∞). where u ˜ k be the extension by zero of Φk to (k, ∞). We have Step 2. Passage to the limit for Φk . Let Φ ˜ 0 = A∗ Φ ˜k + y ˜k −Φ k
in (0, ∞),
˜ k (∞) = 0 . Φ
−1 We already know that (M Bτ∗ Π)∗ ∈ L(V0 (Γ), Vn0 (Ω)) and (RA M Bn∗ Π)∗ ∈ L(V0 (Γ), Vn0 (Ω)). We can rewrite the above equation in the form −1 ∗ ∗ ˜ 0 = A∗ Φ ˜ ˜ k − (RA ˜k + y ˜k −Φ M Bn∗ Π)∗ γn u k Π k − (M Bτ Π) γτ u
˜ k (∞) = 0. Φ
in (0, ∞),
(5.8)
Due to (5.7) and to Lemma 5.2, we can claim that ˜ k − Φk ˆ L∞ (0,∞;V0 (Ω)) + kΦ ˜ k − Φk ˆ L2 (0,∞;V0 (Ω)) −→ 0 kΦ n n
when k −→ ∞,
ˆ is the solution of equation (5.3) corresponding to (ˆ ˆ ). Notice that where Φ y, u ˆ L2 (0,∞;V0 (Ω)) ≤ C kˆ kΦk ukL2 (0,∞;V0 (Γ)) + kˆ ykL2 (0,∞;Vn0 (Ω)) n ≤ C |y0 |Vn0 (Ω) + kf kL2/(1+2ε) (0,∞;Vn0 (Ω)) .
(5.9)
By passing to the limit in the equation Z ∞ ∗ ˜ ˜ k (τ )) dτ, Φk (t) = e(A −λ0 I)(τ −t) (˜ yk (τ ) + λ0 Φ t
˜ k )k converges to Φ ˆ in V2,1 (Q∞ ), (B ∗ Φ ˜ k )k converges to with Lemmas 8.6 and 8.9, we can show that (Φ ∗ˆ 2 0 ˆ satisfies B Φ in L (0, ∞; V (Γ)), and Φ ˆ Φ(t) =
Z
∞
e(A
∗
−λ0 I)(τ −t)
ˆ )) dτ. (ˆ y(τ ) + λ0 Φ(τ
t
ˆ satisfies the second equation in (5.4) corresponding to y ˜k ˆ . Since (˜ Thus Φ uk )k = (−M Bτ∗ Φ −1 −1 ∗˜ ∗ ∗ ˆ ˆ ˆ ˆ , we have u ˆ = −M Bτ Φ − RA M Bn Φ, and (ˆ − RA M Bn Φk )k converges to u y, Φ) obeys the system ˆ follows from Lemma 8.5 and from (5.9). (5.4). The estimate for Φ Theorem 5.4. Assume that f ∈ L2/(1+2ε) (0, ∞; Vn0 (Ω)) ∩ L2 (0, ∞; (V2ε (Ω))0 ). If y0 ∈ Vn0 (Ω), then the solution (y, Φ) of system (5.4) belongs to V1,1/2 (Q) × V3,3/2 (Q) and we have: kykCb ([0,∞);Vn0 (Ω)) + kykV1,1/2 (Q∞ ) + kΦkV3,3/2 (Q∞ ) ≤ C(|y0 |Vn0 (Ω) + kf kL2/(1+2ε) (0,∞;Vn0 (Ω)) + kf kL2 (0,∞;(V2ε (Ω))0 ) ) . 1/2−ε
If y0 ∈ Vn (Ω) for some 0 < ε < 1/2, then the solution (y, Φ) of system (4.4) belongs to V3/2−ε,3/4−ε/2 (Q) × V7/2−ε,7/4−ε/2 (Q), we have: kykV3/2−ε,3/4−ε/2 (Q) + kΦkV7/2−ε,7/4−ε/2 (Q) ≤ C(|y0 |V1/2−ε (Ω) + kf kL2/(1+2ε) (0,∞;Vn0 (Ω)) + kf kL2 (0,∞;(V2ε (Ω))0 ) ) , n
22
and kB ∗ ΦkV3/2−ε,3/4−ε/2 (Q) ≤ C(|y0 |V1/2−ε (Ω) + kf kL2/(1+2ε) (0,∞;Vn0 (Ω)) + kf kL2 (0,∞;(V2ε (Ω))0 ) ) . n
Proof. Assume that y0 ∈ Vn0 (Ω). Applying Lemmas 8.1, 8.2, 8.3, we first obtain that y belongs 0 0 0 0 to V1/2−ε ,1/4−ε /2 (Q∞ ) for all ε0 > 0. From Lemma 8.9, we deduce that B ∗ Φ ∈ V1/2−ε ,1/4−ε /2 (Σ∞ ) 0 for all ε > 0. As in the proof of Lemma 4.2, we can use a bootstrap argument to show that y ∈ V1,1/2 (Q∞ ) ∩ Cb ([0, ∞); Vn0 (Ω)) and Φ ∈ V3,3/2 (Q∞ ), with the corresponding estimates for y and Φ. 1/2−ε The same procedure can be used to prove the other estimates in the case when y0 ∈ Vn (Ω). 6. Stabilization of the two dimensional Navier-Stokes equations. Throughout this section, we assume that N = 2. 6.1. First stabilization result. Consider the Navier-Stokes equations with the linear feedback law determined in section 4: P y0 = AΠ P y + P F (y) (I − P )y = −(I −
in (0, ∞),
P y(0) = y0 ,
−1 P )DA M RA M Bn∗ ΠP y
(6.1)
in (0, ∞),
where F (y) = −(y · ∇)y. Theorem 6.1. For all 0 < ε < 1/4, There exists µ0 > 0 and a nondecreasing function η from R+ into itself, such that if µ ∈ (0, µ0 ) and |y0 |V1/2−ε (Ω) ≤ η(µ), then equation (6.1) admits a unique solution n in the set n o Dµ = y ∈ V3/2−ε,3/4−ε/2 (Q∞ ) | kykV3/2−ε,3/4−ε/2 (Q∞ ) ≤ µ . Moreover (I − P )y belongs to H 3/4−ε/2 (0, ∞; V1/2 (Ω)) ∩ L2 (0, ∞; V3/2−ε (Ω)). 1/2−ε (Ω) and all f ∈ L2/(1+2ε) (0, ∞; Vn0 (Ω)) ∩ L2 (0, ∞; (V2ε (Ω))0 ) the Lemma 6.2. For all y0 ∈ Vn solution to the equation y0 = AΠ y + f ,
y(0) = y0 ,
(6.2)
obeys kykV3/2−ε,3/4−ε/2 (Q∞ ) ≤ C1 |y0 |V1/2−ε (Ω) + kf kL2/(1+2ε) (0,∞;Vn0 (Ω)) + kf kL2 (0,∞;(V2ε (Ω))0 ) . n
Proof. Since the semigroup (etAΠ )t≥0 is exponentially stable on Vn0 (Ω), equation (6.2) admits a unique solution y belonging to Cb ([0, ∞); Vn0 (Ω)) ∩ L2 (0, ∞; Vn0 (Ω)) which satisfies kykL∞ (0,∞;Vn0 (Ω)) + kykL2 (0,∞;Vn0 (Ω)) ≤ C |y0 |V1/2−ε (Ω) + kf kL2/(1+2ε) (0,∞;Vn0 (Ω)) . n
ˆ the solution to system (5.4). We set We denote by (ˆ y, Φ) ˆ r(t) = Φ(t) − Πy(t) . ˜ the solution to the equation We denote by y −1 ˜ 0 = A˜ y y + Bτ M 2 Bτ∗ r + Bn M RA M Bn∗ r,
˜ (0) = 0 . y
From the definition of r it follows that ˆ − Bn M R−1 M B ∗ Φ ˆ −Bτ M 2 Bτ∗ Φ n A −1 −1 = −Bτ M 2 Bτ∗ Πy − Bn M RA M Bn∗ Πy − Bτ M 2 Bτ∗ r − Bn M RA M Bn∗ r .
Thanks to this identity, we can rewrite equation (6.2) in the form ˆ − Bn M R−1 M B ∗ Φ ˆ + Bτ M 2 B ∗ r + Bn M R−1 M B ∗ r + f , y0 = Ay − Bτ M 2 Bτ∗ Φ n τ n A A 23
y(0) = y0 .
ˆ+y ˜ . Due to Theorem 5.4, we notice that Thus y = y ˆ V3/2−ε,3/4−ε/2 (Σ ) ≤ C |y0 | 1/2−ε kB ∗ Φk + kf kL2/(1+2ε) (0,∞;Vn0 (Ω)) + kf kL2 (0,∞;(V2ε (Ω))0 ) . ∞ V (Ω) n
Moreover y belongs to L2 (0, ∞; Vn0 (Ω)) and B ∗ Π ∈ L(Vn0 (Ω), V1/2 (Γ)) (see Remark (4.4)). Thus B ∗ r = ˆ − B ∗ Πy belongs to V0,0 (Σ∞ ) and B∗Φ kB ∗ rkV0,0 (Σ∞ ) ≤ C |y0 |V1/2−ε (Ω) + kf kL2/(1+2ε) (0,∞;Vn0 (Ω)) + kf kL2 (0,∞;(V2ε (Ω))0 ) . n
From Lemma 8.3 it follows that k˜ ykV1/2−ε0 ,1/4−ε0 (Q∞ ) ≤ CkB ∗ rkV0,0 (Σ∞ ) 0
for all ε0 > 0.
0
ˆ + y ˜ belongs to V1/2−ε ,1/4−ε /2 (Q∞ ) for all ε0 > 0. Therefore B ∗ r belongs to Thus y = y 0 0 0 1/2−ε0 V1/2−ε ,1/4−ε /2 (Σ∞ ) because B ∗ Π ∈ L(Vn (Ω), V1−ε (Γ)) for all ε0 > 0 (see Remark (4.4)), and we have kB ∗ rkV1/2−ε0 ,1/4−ε0 /2 (Σ∞ ) ≤ C |y0 |V1/2−ε (Ω) + kf kL2/(1+2ε) (0,∞;Vn0 (Ω)) + kf kL2 (0,∞;(V2ε (Ω))0 ) ∀ε0 > 0. n
0
0
With another iteration we can prove an estimate of B ∗ r in V1−ε ,1/2−ε /2 (Σ∞ ). Still with Lemma 8.3 we obtain k˜ ykV3/2−ε00 ,3/4−ε00 (Q∞ ) ≤ CkB ∗ rkV1−ε0 ,1/2−ε0 /2 (Σ) , for all ε00 > ε0 > 0. With Theorem 5.4, we have kˆ ykV3/2−ε,3/4−ε/2 (Q∞ ) ≤ C |y0 |V1/2−ε (Ω) + kf kL2/(1+2ε) (0,∞;Vn0 (Ω)) + kf kL2 (0,∞;(V2ε (Ω))0 , n
and the proof is complete. Lemma 6.3. If z belongs to V3/2−ε,3/4−ε/2 (Q) then (1−2ε)
kzkL1/ε (0,∞;V1/2+ε (Ω)) ≤ CkzkL∞ (0,∞;V1/2−ε (Ω)) kzk2ε L2 (0,∞;V3/2−ε (Ω)) ≤ CkzkV3/2−ε,3/4−ε/2 (Q) . Proof. We first observe that (1−2ε)
|z(t)|V1/2+ε (Ω) ≤ C|z(t)|V1/2−ε (Ω) |z(t)|2ε V3/2−ε (Ω) . Therefore we have Z 0
∞
1/ε
(1−2ε)/ε
|z(t)|V1/2+ε (Ω) dt ≤ CkzkL∞ (0,∞;V1/2−ε (Ω))
Z 0
∞
|z(t)|2V3/2−ε (Ω) dt.
The proof is complete. Lemma 6.4. Let ε be in (0, 1/4). If z belongs to V3/2−ε,3/4−ε/2 (Q) then k(z · ∇)zkL2/(1+2ε) (0,∞;L2 (Ω)) + kP div(z ⊗ z)kL2 (0,∞;(V2ε (Ω))0 ) ≤ C2 kzk2V3/2−ε,3/4−ε/2 (Q∞ ) . Proof. We have k(z · ∇)zkL2 (Ω) ≤ kzkL4/(1−2ε) (Ω) k∇zkL4/(1+2ε) (Ω) ≤ CkzkV1/2+ε (Ω) kzkV3/2−ε (Ω) . Therefore with Lemma 6.3, it follows that k(z · ∇)zkL2/(1+2ε) (0,∞;L2 (Ω)) ≤ CkzkL1/ε (0,∞;V1/2+ε (Ω)) kzkL2 (0,∞;V3/2−ε (Ω)) 24
≤ Ckzk2V3/2−ε,3/4−ε/2 (Q) . From [20, Theorem B.3] we deduce that kz ⊗ zkH1−2ε,1/2−ε (Q∞ ) ≤ Ckzk2V3/2−ε,3/4−ε/2 (Q∞ ) . The divergence operator is continuous from L2 (Ω) into H−1 (Ω), and from H1 (Ω) to L2 (Ω). Thus it is also continuous from H1−2ε (Ω) into (H2ε (Ω))0 . The projector P is continuous from L2 (Ω) into Vn0 (Ω), and from H−1 (Ω) into V−1 (Ω) (see e.g. [36]). Thus it is also continuous from (H2ε (Ω))0 into (Vn2ε (Ω))0 . Therefore we have kP div(z ⊗ z)kL2 (0,∞;(Vn2ε (Ω))0 ) ≤ Ckz ⊗ zkL2 (0,∞;H1−2ε (Ω)) ≤ Ckzk2V3/2−ε,3/4−ε/2 (Q∞ ) . Observe that the condition 0 < ε < 1/4 is needed to have (H2ε (Ω))0 ,→ H−1 (Ω). Lemma 6.5. Let ε be in (0, 1/4). The mapping P F is locally Lispchitz continuous from V3/2−ε,3/4−ε/2 (Q∞ ) into L2/(1+2ε) (0, ∞; Vn0 (Ω)) ∩ L2 (0, ∞; (V2ε (Ω))0 ). More precisely we have kP F (z1 ) − P F (z2 )kL2/(1+2ε) (0,∞;Vn0 (Ω)) + kP F (z1 ) − P F (z2 )kL2 (0,∞;(V2ε (Ω))0 ) ≤ C2 kz1 kV3/2−ε,3/4−ε/2 (Q∞ ) + kz2 kV3/2−ε,3/4−ε/2 (Q∞ ) kz1 − z2 kV3/2−ε,3/4−ε/2 (Q∞ ) . Proof. From Lemma 6.4 it follows that P F is a mapping from V3/2−ε,3/4−ε/2 (Q∞ ) into L (0, ∞; Vn0 (Ω)) ∩ L2 (0, ∞; (V2ε (Ω))0 ). By calculations similar as those in Lemma 6.4, we obtain: 2/(1+2ε)
kP F (z1 ) − P F (z2 )kL2/(1+2ε) (0,∞;Vn0 (Ω)) + kP F (z1 ) − P F (z2 )kL2 (0,∞;(V2ε (Ω))0 ) ≤ kP ((z1 · ∇)(z1 − z2 ))kL2/(1+2ε) (0,∞;Vn0 (Ω)) + kP (((z1 − z2 ) · ∇)z2 )kL2/(1+2ε) (0,∞;Vn0 (Ω)) +kP div(z1 ⊗ (z1 − z2 ))kL2 (0,∞;(Vn2ε (Ω))0 ) + kP div((z1 − z2 ) ⊗ z2 )kL2 (0,∞;(Vn2ε (Ω))0 ) ≤ C2 kz1 kV3/2−ε,3/4−ε/2 (Q∞ ) + kz2 kV3/2−ε,3/4−ε/2 (Q∞ ) kz1 − z2 kV3/2−ε,3/4−ε/2 (Q∞ ) . The proof is complete. Lemma 6.6. If P y belongs to V3/2−ε,3/4−ε/2 (Q∞ ) for some 0 < ε < 1/2, then −1 k(I − P )DA M RA M Bn∗ ΠP ykV3/2−ε,3/4−ε/2 (Q∞ ) ≤ C3 kP ykV3/2−ε,3/4−ε/2 (Q∞ ) .
1/2−ε
(Ω), V1−ε (Γ)) (see Remark 4.4). Thus Proof. The operator B ∗ Π belongs to L(Vn 1/2−ε 1/2−ε −1 −1 1−ε ∗ ∗ (Ω), (Ω), V (Γ)), and DA M RA M Bn Π belongs to L(Vn M RA M Bn Π belongs to L(Vn V3/2−ε (Ω)). The lemma follows from the continuity properties of the operator (I − P ). Proof. Proof of Theorem 6.1. We set µ0 = V
3/2−ε,3/4−ε/2
1 4C1 C2 (1+C3 )
and η(µ) =
3µ 4C1 (1+C3 ) .
For z ∈
(Ω)), we denote by yz the solution to the equation P y0 = AΠ P y + P F (z),
y(0) = y0 ,
(6.3)
−1 (I − P )y = −(I − P )DA M RA M Bn∗ ΠP y .
We are going to prove that the mapping M : z 7→ yz is a contraction in Dµ . (i) From Lemma 6.2, Lemma 6.4, and Lemma 6.6 it follows that kyz kV3/2−ε,3/4−ε/2 (Q∞ ) ≤ kP yz kV3/2−ε,3/4−ε/2 (Q∞ ) + k(I − P )yz kV3/2−ε,3/4−ε/2 (Q∞ ) ≤ C1 (1 + C3 ) |y0 |V1/2−ε (Ω) + kP F (z)kL2/(1+2ε) (0,∞;Vn0 (Ω)) + kP F (z)kL2 (0,∞;(V2ε (Ω))0 ) 3µ 2 + C2 kzk2V3/2−ε,3/4−ε/2 (Q∞ ) ≤ 3µ ≤ C1 (1 + C3 ) 4C1 (1+C 4 + C1 (1 + C3 )C2 µ ≤ µ, 3) if µ < µ0 . Thus M is a mapping from Dµ into itself. 25
(ii) From Lemma 6.2 and Lemma 6.5 it follows that kP yz1 − P yz2 kV3/2−ε,3/4−ε/2 (Q∞ ) ≤ C1 kP F (z1 ) − P F (z2 )kL2/(1+2ε) (0,∞;Vn0 (Ω)) + kP F (z1 ) − P F (z2 )kL2 (0,∞;(V2ε (Ω))0 ) ≤ C2 C1 kz1 kV3/2−ε,3/4−ε/2 (Q∞ ) + kz2 kV3/2−ε,3/4−ε/2 (Q∞ ) kz1 − z2 kV3/2−ε,3/4−ε/2 (Q∞ )
≤ 2C2 C1 µkz1 − z2 kV3/2−ε,3/4−ε/2 (Q∞ ) . Thus with Lemma 6.6 and the previous estimate we obtain kyz1 − yz2 kV3/2−ε,3/4−ε/2 (Q∞ ) ≤ 2C2 C1 µ(1 + C3 )kz1 − z2 kV3/2−ε,3/4−ε/2 (Q∞ ) . Thus if µ < µ0 , the mapping M is a contraction in Dµ , and the system (6.1) admits a unique solution in Dµ . 6.2. Second stabilization result. To obtain a feedback law providing an exponential stabilization of the Navier-Stokes, we are going to use the linear feedback law determined thanks to an auxiliary problem. For that, we set ˆ = eωt y, y
ˆ = eωt u. u
If P y0 = AP y + P F (y) + BM u,
P y(0) = y0 ,
(I − P )y = (I − P )DA γn M u, ˆ is the solution to the system then y ˆ 0 = AP y ˆ + ωP y ˆ + e−ωt P F (ˆ ˆ, Py y) + BM u
ˆ (0) = y0 , Py
ˆ. (I − P )ˆ y = (I − P )DA γn M u Set Aω = A + ωI, and let Πω ∈ L(Vn0 (Ω)) be the solution to the algebraic Riccati equation: Πω = Π∗ω ≥ 0,
−1 Πω Aω + A∗ω Πω − Πω Bτ M 2 Bτ∗ Πω − Πω Bn M RA M Bn∗ Πω + I = 0.
(6.4)
The existence of a unique solution in L(Vn0 (Ω); V2 (Ω) ∩ V01 (Ω)) to this equation may be proved as in section 4. As in section 4, it can be shown that 1 I(zy0 , vy0 ) = Πω y0 , y0 0 , 2 Vn (Ω) where (zy0 , vy0 ) is the solution of the control problem n o (P0,y0 ) inf I(z, v) | (z, v) satisfies (6.5), v ∈ V0,0 (Σ∞ ) , where I is the functional of section 4, and z0 = Aω z + BM v
in (0, ∞),
z(0) = y0 .
(6.5)
Consider the Navier-Stokes equations with the linear feedback law: ˆ 0 = Aω,Πω P y ˆ + e−ωt P F (ˆ Py y),
ˆ (0) = y0 , Py
−1 ˆ, (I − P )ˆ y = −(I − P )DA M RA M Bn∗ Πω P y
where D(Aω,Πω ) = D(AΠω )
and Aω,Πω y = AΠω y + ωy for all y ∈ D(AΠω ). 26
(6.6)
ˆ is a solution to (6.6), then y = e−ωt y ˆ is the solution of As previously, if y −1 P y0 = AP y − Bτ M 2 Bτ∗ Πω P y − Bn M RA M Bn∗ Πω P y + P F (y),
(I − P )y = −(I −
P y(0) = y0 ,
−1 P )DA RA M Bn∗ Πω P y.
(6.7)
Theorem 6.7. For all 0 < ε < 1/4, there exists µ0 > 0 and a nondecreasing function η0 from R+ into itself, such that if µ ∈ (0, µ0 ) and ky0 kV1/2−ε (Ω) ≤ η0 (µ), the equation (6.7) admits a unique solution n in the set n o Dµ = y ∈ V3/2−ε,3/4−ε/2 (Q∞ ) | keω (·) ykV3/2−ε,3/4−ε/2 (Q∞ ) ≤ µ . Moreover y, which belongs to Cb ([0, ∞); V1/2−ε (Ω)), satisfies |y(t)|V1/2−ε (Ω) ≤ C(w, µ, ω) e−ωt . Proof. Substituting F (y) by e−ωt F (ˆ y) in the proof of Theorem 6.1, we can show that there exists µ0 > 0 and a nondecreasing function η0 from R+ into itself, such that if µ ∈ (0, µ0 ) and ky0 kV1/2−ε (Ω) ≤ η0 (µ) n n b µ = y ∈ V3/2−ε,3/4−ε/2 (Q∞ ) | ˆ in D for some 0 < ε < 1/4, the equation (6.6) admits a unique solution y o ˆ is the solution of equation kykV3/2−ε,3/4−ε/2 (Q∞ ) ≤ µ , and |ˆ y(t)|V1/2−ε (Ω) ≤ C(w, µ, ω). Thus y = e−ωt y (6.7), it belongs to Dµ and it satisfies obeys |y(t)|V1/2−ε (Ω) ≤ C(w, µ, ω) e−ωt . The proof is complete. 7. Additional results. 7.1. Control by a tangential velocity. Assume that Ω and w are such that (1.3) be stabilizable 1/2−ε (Ω) with controls u ∈ L2 (0, ∞; Vn0 (Γ)) satisfying supp(u) ⊂ Γ0 × (0, ∞), where Γ0 is an open in Vn set in Γ in which the function m is equal to 1 (see the assumptions on the function m at the beginning of section 2). Such a stabilizability result is stated in [4, Theorem B.1.1]. With such a result, as in section 6.2 (and section 4), we can show that the algebraic Riccati equation Πω ∈ L(Vn0 (Ω)), Πω = Π∗ω ≥ 0,
Πω Aω + A∗ω Πω − Πω Bτ M 2 Bτ∗ Πω + I = 0,
L(Vn0 (Ω); V2 (Ω)
(7.1)
V01 (Ω)),
admits a unique solution satisfying Πω ∈ ∩ and which enjoys the same regularizing properties as in Theorem 4.5. Since we deal with controls belonging to Vn0 (Γ), the operator RA is now the identity. As in Theorem 6.7 we can prove that, for all 0 < ε < 1/4, there exists µ0 > 0 and a nondecreasing function η0 from R+ into itself, such that if µ ∈ (0, µ0 ) and ky0 kV1/2−ε (Ω) ≤ η0 (µ), the n equation y0 = Ay − Bτ M 2 Bτ∗ Πω y + P F (y),
y(0) = y0 ,
admits a unique solution in the set n o Dµ = y ∈ V3/2−ε,3/4−ε/2 (Q∞ ) ∩ L2 (0, ∞; Vn0 (Ω)) | keω (·) ykV3/2−ε,3/4−ε/2 (Q∞ ) ≤ µ . 1/2−ε
Moreover y, which belongs to Cb ([0, ∞); Vn
(Ω)), satisfies
|y(t)|V1/2−ε (Ω) ≤ C(w, µ, ω) e−ωt . n
Notice that the local feedback stabilization of the two dimensional Navier-Stokes equations is not studied in [4] (only the stabilization of the linearized Navier-Stokes equations is studied in two dimension in [4, Appendix B]). If we transpose the results obtained in [4] in the 3D case to the 2D case, the results are completely different from the ones we obtain. Indeed we here prove a local stabilization result for the Navier-Stokes equations thanks to the solution of a classical Riccati equation (equation (7.1)), while the Riccati equation obtained in [4, Section 4] is only defined in the domain of the square of the feedback operator, which is unknown. 27
7.2. Dependence of solutions with respect to ν for the stabilization of the Stokes equation. The dependence of solutions to our control problems with respect to ν is very complicated because the stationary solution w involved in the Oseen operator also depends on ν in a complicated way. There is one particular situation for which we can clarify this dependence, it is the one corresponding to w = 0. Assume that the semigroup generated by (νP ∆ + ωI, V2 (Ω) ∩ V01 (Ω)) on Vn0 (Ω) be unstable for a given ω > 0. What has been done in section 6.2 is still valid if w ≡ 0. Let Πω be the solution to equation (6.4) ˆ to the closed loop system corresponding to A = νA0 and Aω = νA0 + ωI. Consider the solution y −1 ˆ 0 = AP y ˆ + ωP y ˆ − B M RA ˆ, Py M B ∗ Πω P y
ˆ (0) = y0 , Py
−1 ˆ. (I − P )ˆ y = −(I − P )DA RA M Bn∗ Πω P y
From the previous sections we know that −1 ˆ) = I(ˆ y, −M RA M B ∗ Πω P y
1 Πω y0 , y0 0 , 2 Vn (Ω)
and |ˆ y(t)|V1/2−ε (Ω) ≤ C(ε, ω)e−σt |y0 |V1/2−ε (Ω) for all 0 < ε ≤ 1/2, where C(ε, ω) > 0 depends on ε and ω, and σ is positive, but not precisely known. ˆ , we can check that y is the solution to the closed loop system Setting y = e−ωt y −1 P y0 = AP y − B M RA M B ∗ Πω P y,
(I − P )y = −(I −
P y(0) = y0 ,
−1 P )DA RA M Bn∗ Πω P y,
(7.2)
and y obeys the decay rate |y(t)|V1/2−ε (Ω) ≤ C(ε)e−(ω+σ)t |y0 |V1/2−ε (Ω) ≤ C(ε)e−ωt |y0 |V1/2−ε (Ω) . Thus we have solved the boundary feedback stabilization problem of the Stokes equations with a prescribed exponential decay rate ω. Now we would like to clarify the dependence of the solution y of equation (7.2) with respect to ω. We have to clarify the dependence of Πω with respect to ν. For that we first have to clarify the dependence of RA , B, and B ∗ with respect to ν. We can choose λ0 = 0 in the definition of DA . We denote by DA0 u the solution y to the equation −∆y + ∇q = 0 and div y = 0 in Ω,
y = u on Γ,
∗ ∗ ∗ (I − P )DA M + I = and we have DA = ν1 DA0 u because λ0 = 0. Thus DA = ν1 DA u, and RA = M DA 0 1 ∗ ∗ ν 2 M DA0 (I − P )DA0 M + I. Since B = (−A)P DA = (−A0 )P DA0 , we observe that B and B do not depend on ν. Let ν1 and ν2 be two viscosity coefficients such that ν1 > ν2 . Assume that the semigroup generated by (ν1 P ∆ + ωI, V2 (Ω) ∩ V01 (Ω)) on Vn0 (Ω) is unstable for a given ω > 0. For i = 1, 2, let Πνi ,ω be the solution to equation (6.4) corresponding to Aω = νi A0 + ωI. With this notation we can observe that Πν1 ,ω is also the solution to the equation
Πν1 ,ω Aν2 ,ω + A∗ν2 ,ω Πν1 ,ω − νν12 Πν1 ,ω BM
1 ∗ M DA (I ν12 0
− P )DA0 M + I
−1
M B ∗ Πν1 ,ω +
ν2 ν1 I
+ ωI( νν12 − 1) = 0.
Since ν1 > ν2 , we can verify that ν2 ν1 BM
= BM ≥ BM
−1 1 ∗ M DA (I − P )DA0 M + I M B∗ ν12 0 −1 1 ∗ M B∗ ν1 ν2 M DA0 (I − P )DA0 M + I 1 ∗ −1 ( ν 2 M DA0 (I − P )DA0 M + I) M B ∗ 2
28
(where the inequality ≤ is the inequality between quadratic forms), and ν2 ν2 I + ωI( − 1) ≤ I. ν1 ν1 Thus, as in [7], we can claim that Πν1 ,ω ≤ Πν2 ,ω . If yνi is the solution to the closed loop system (7.2) for A = νi A0 , we have 1 ∗ Π y , y I(e−ωt yν1 , −e−ωt M Rν−1 M B Π P y ) = ν ,ω 0 0 ν ,ω ν 1 1 1 2 1 A0 1 ∗ ≤ I(e−ωt yν2 , −e−ωt M Rν−1 Π y , y , M B Π P y ) = ν ,ω 0 0 ν ,ω ν 2 2 2 2 2 A0 which provides a comparison between the L2 of the solution to the stabilization problems in function of ν. 8. Appendix. In this section we prove some regularity results for the state and the adjoint equations. 1/2−ε (Ω) with 0 < ε ≤ 1/2, the weak solution to the equation Lemma 8.1. If y0 ∈ Vn y0 = (A − λ0 )y
in (0, ∞),
y(0) = y0 ,
obeys kykC
1/2−ε (Ω)) b ([0,∞);Vn
+ kykV3/2−ε,3/4−ε/2 (Q∞ ) ≤ C|y0 |V1/2−ε (Ω) . n
Proof. The exponential stability of the semigroup (et(A−λ0 I) )t≥0 gives a bound for y in L2 (0, ∞; 1/2−ε (Ω)), the other estimates may be obtained by using Using the estimate in L2 (0, ∞; Vn proofs in [5] and interpolation results. Lemma 8.2. If f ∈ L2/(1+2ε) (0, ∞; Vn0 (Ω)), the solution y to the equation 1/2−ε (Ω)). Vn
y0 = (A − λ0 I)y + f
in (0, ∞),
y(0) = 0,
obeys kykL2 (0,∞;V2−2ε (Ω)) ≤ Ckf kL2/(1+2ε) (0,∞;Vn0 (Ω)) . If 0 < ε < 1/4 and f ∈ L2 (0, ∞; (Vn2ε (Ω))0 ), then kykV2−2ε,1−ε (Q∞ ) ≤ Ckf kL2 (0,∞;(Vn2ε (Ω))0 ) . Proof. From the identity (−A + λ0 I)1−ε y(t) =
t
Z
(−A + λ0 I)1−ε e(t−τ )(A−λ0 I) f (τ ) dτ,
0
it follows that Z ky(t)kV2−2ε (Ω) ≤ C 0
t
e−ω(t−τ ) (t − τ )−1+ε |f (τ )|Vn0 (Ω) dτ.
Thus, with the generalized Young inequality for convolutions (see [34, page 32]), we have: kykLr (0,∞;V2−2ε (Ω)) ≤ Ckf kL2/(1+2ε) (0,∞;Vn0 (Ω)) , for all 2/(1 + 2ε) ≤ r ≤ 2. 29
(8.1)
(ii) Now we assume that f belongs to L2 (0, ∞; (Vn2ε (Ω))0 ). To prove the estimate in that case we proceed by interpolation. We know that kykV2,1 (Q∞ ) ≤ Ckf kL2 (0,∞;Vn0 (Ω))
kykV1,1/2 (Q∞ ) ≤ Ckf kL2 (0,∞;V−1 (Ω)) .
and
Since we have: [L2 (0, ∞; Vn0 (Ω)), L2 (0, ∞; V−1 (Ω))]2ε = ([L2 (0, ∞; Vn0 (Ω)), L2 (0, ∞; V01 (Ω))]2ε )0 = L2 (0, ∞; (Vn2ε (Ω))0 ), we obtain the desired result by interpolation. (Notice that [Vn0 (Ω), V01 (Ω)]2ε = Vn2ε (Ω) because 0 < ε < 1/4.) Lemma 8.3. If u belongs to Vs,s/2 (Σ∞ ) with 0 ≤ s < 1, then the weak solution to the equation y0 = (A − λ0 )y + BM u
in (0, ∞),
y(0) = 0,
obeys kykV1/2+s−ε,1/4+s/2−ε/2 (Q∞ ) ≤ CkukVs,s/2 (Σ∞ )
for all ε > 0.
If u belongs to Vs,s/2 (Σ∞ ) with 1 < s ≤ 2, and if u(0) = 0, then kykV1/2+s−ε,1/4+s/2−ε/2 (Q∞ ) ≤ CkukVs,s/2 (Σ∞ )
for all ε > 0.
Proof. Similar estimates are proved in [30, Theorem 2.3] for a finite time interval. It is sufficient to rewrite the proof and to use the exponential stability of the semigroup (et(A−λ0 I) )t≥0 to obtain the desired results. Lemma 8.4. For all y ∈ Vn2 (Ω), the solution Φ ∈ V2 (Ω) ∩ V01 (Ω) to the stationary equation λ0 Φ − A∗ Φ = y obeys |Φ|V4 (Ω) ≤ C|y|Vn2 (Ω) Proof. We rewrite the equation in the form λ0 Φ − νP ∆Φ = y − P ((w · ∇)Φ) + P ((∇w)T Φ). Since w ∈ V3 (Ω) and Φ ∈ V2 (Ω), then P ((w · ∇)Φ) and P ((∇w)T Φ) belong to V1 (Ω), which gives an estimate of Φ in V3 (Ω). Knowing that Φ ∈ V3 (Ω), P ((w · ∇)Φ) and P ((∇w)T Φ) belong to V2 (Ω), and the proof is complete. Lemma 8.5. If the function y belongs to Vs,s/2 (Q∞ ) ∩ L2 (0, ∞; Vn0 (Ω)) with 0 ≤ s ≤ 2, then the solution Φ to the equation −Φ0 = (A∗ − λ0 I)Φ + y
in (0, ∞),
Φ(∞) = 0,
(8.2)
satisfies kΦkVs+2,s/2+1 (Q∞ ) ≤ CkykVs,s/2 (Q∞ ) .
(8.3)
Proof. (i) Let us first prove estimate (8.3) for s = 0. Assume that y ∈ L2 (0, ∞; Vn0 (Ω)). The existence of a unique solution Φ to equation (8.2) in L2 (0, ∞; Vn0 (Ω)) follows from the exponential stability of the ∗ semigroup (et(A −λ0 I) )t≥0 on Vn0 (Ω). Moreover we have kΦkL2 (0,∞;Vn0 (Ω)) ≤ CkykL2 (0,∞;Vn0 (Ω)) .
(8.4)
It is sufficient to prove estimate (8.3) for s = 0 and y ∈ Cc (0, ∞; Vn0 (Ω)). In that case, it can be deduced from (8.4) with the method of proof in [5, Theorem 1.1]. 30
(ii) Now let us assume that y belongs to V2,1 (Q∞ ) and let us prove estimate (8.3) for s = 2. We have Z ∞ ∗ ∗ (−A + λ0 I)Φ(t) = (−A∗ + λ0 I)e(τ −t)(A −λ0 I) y(τ ) dτ. t
Integrating by parts we obtain (−A∗ + λ0 I)Φ(t) = y(t) +
Z
∞
e(τ −t)(A
∗
−λ0 I) 0
y (τ ) dτ.
t
Since y0 belongs to L2 (0, ∞; Vn0 (Ω)), with estimate (8.3) for s = 0 we have
Z ∞
(τ −t)(A∗ −λ0 I) 0
≤ Cky0 kL2 (0,∞;Vn0 (Ω)) . e y (τ ) dτ
t
V2,1 (Q∞ )
Thus k(−A∗ + λ0 I)ΦkL2 (0,∞;V2 (Ω)) ≤ C(kykL2 (0,∞;V2 (Ω)) + ky0 kL2 (0,∞;Vn0 (Ω)) ) ≤ CkykV2,1 (Q∞ ) . From regularity results for the stationary Oseen equation (Lemma 8.4), it follows that kΦkL2 (0,∞;V4 (Ω)) ≤ Ck(−A∗ + λ0 I)ΦkL2 (0,∞;V2 (Ω)) . Thus kΦkL2 (0,∞;V4 (Ω)) ≤ CkykV2,1 (Q∞ ) . From equation (8.2) we deduce kΦ0 kL2 (0,∞;V2 (Ω)) ≤ C(k(−A∗ + λ0 I)ΦkL2 (0,∞;V2 (Ω)) + kykL2 (0,∞;V2 (Ω)) ) ≤ CkykV2,1 (Q∞ ) . The estimate (8.3) is obtained by interpolation between the estimates obtained for s = 0 and s = 2. Lemma 8.6. For all y ∈ L2 (0, T ; Vn0 (Ω)), the solution to the equation −Φ0 = A∗ Φ + y
in (0, T ),
Φ(T ) = 0,
(8.5)
satisfies kΦkV2,1 (QT ) ≤ CkykL2 (0,T ;Vn0 (Ω)) .
(8.6)
If the function y belongs to Vs,s/2 (QT ) with 0 ≤ s < 3/2, then the function Φ belongs to Vs+2,s/2+1 (QT ) and the following estimate holds: kΦkVs+2,s/2+1 (QT ) ≤ CkykVs,s/2 (QT ) .
(8.7)
Proof. The regularity result when y ∈ L2 (0, T ; Vn0 (Ω)) and estimate (8.6) follow from [5, Theorem 1.1]. If y belongs to V2,1 (QT ) and if y(T ) ∈ V01 (Ω), we have kΦ0 kL2 (0,T ;V2 (Ω)) + kΦkL2 (0,T ;V4 (Ω)) ≤ CkykV2,1 (QT ) , see e.g. [33]. The estimate for 0 ≤ s < 3/2 is obtained by interpolation between the estimates obtained for s = 0 and s = 2. For s < 3/2 the regularity condition y(T ) ∈ V01 (Ω) is not needed. Lemma 8.7. Let Φ be the solution to equation (8.2) and let ψ be the pressure associated with Φ, that is the function satisfying ∂Φ − ν∆Φ − (w · ∇)Φ + (∇Φ)T w + λ0 Φ + ∇ψ = y in Q∞ , ∂t div Φ = 0 in Q∞ , Φ = 0 on Σ∞ , Φ(∞) = 0 in Ω. −
31
(8.8)
If in (8.2) y belongs to L2 (0, ∞; Vn0 (Ω)), then the function ψ belongs to L2 (0, ∞; H 1 (Ω)). If in addition y belongs to Vs,s/2 (Q∞ ) with 0 ≤ s ≤ 2, then the function ψ belongs to L2 (0, ∞; H s+1 (Ω)) ∩ H s/2 (0, ∞; H 1 (Ω)). Proof. Assume that y belongs to L2 (0, ∞; Vns (Ω))∩H s/2 (0, ∞; Vn0 (Ω)) with 0 ≤ s ≤ 2. From Lemma 8.5 it follows that ∂Φ + ν∆Φ + (w · ∇)Φ − (∇Φ)T w ∈ Hs,s/2 (Q∞ ). y+ ∂t Thus ∇ψ belongs to Hs,s/2 (Q∞ ), and the proof is complete. Lemma 8.8. If in (8.5) y belongs to Vs,s/2 (QT ) with 0 ≤ s < 3/2, then the function ψ, the pressure associated with Φ, belongs to L2 (0, T ; H s+1 (Ω)) ∩ H s/2 (0, T ; H 1 (Ω)). Proof. The proof is similar to the previous one. Lemma 8.9. Let Φ ∈ V2,1 (Q) be the solution to equation (8.2), and set −1 u = −(M γτ + RA M γn )B ∗ Φ .
If in (8.2) the function y belongs to L2 (0, ∞; Vns (Ω)) ∩ H s/2 (0, ∞; Vn0 (Ω)) with 0 ≤ s ≤ 2, then kB ∗ ΦkL2 (0,∞;Vs+1/2 (Γ))∩H s/2 (0,∞;V1/2 (Γ)) + kukL2 (0,∞;Vs+1/2 (Γ))∩H s/2 (0,∞;V1/2 (Γ)) ≤ CkykVs,s/2 (Q∞ ) .
(8.9)
Proof. As in Lemma 3.5 we can show that ∂Φ −1 u = −(M γτ + RA M γn ) ν − ψ n + c(ψ)n , ∂n where ψ is the pressure associated with Φ and c(ψ) is the constant defined in (2.1). Since y belongs to L2 (0, ∞; Vns (Ω)) ∩ H s/2 (0, ∞; Vn0 (Ω)), from Lemma 8.5 we deduce that Φ belongs to Vs+2,s/2+1 (Q∞ ), and from Lemma 8.7 it follows that ψ belongs to L2 (0, ∞; H s+1 (Ω)) ∩ H s/2 (0, ∞; H 1 (Ω)). Thus u belongs to L2 (0, ∞; H s+1/2 (Γ)) ∩ H s/2 (0, ∞; H 1/2 (Γ)). Acknowledgements. The author thanks the referees for their valuable questions and suggestions that have contributed to improve the first version of the paper. After this paper was submitted, I. Lasiecka informed the author of her joint work with V. Barbu and R. Triggiani [4]. The author thanks I. Lasiecka for providing him a copy of this joint paper and for discussions in Oberwolfach. REFERENCES [1] [2] [3] [4] [5] [6] [7] [8] [9] [10] [11] [12] [13] [14]
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Barbu, Mathematical Methods in Optimization of Differential Systems, Kluwer Academic Publishers, 1994. Barbu, Feedback stabilization of the Navier-Stokes equations, ESAIM COCV, Vol. 9 (2003), 197-206. Barbu, S. S. Sritharan, H ∞ - control theory of fluid dynamics, Proc. R. Soc. Lond. A 454 (1998), 3009-3033. Barbu, I. Lasiecka, R. Triggiani, Boundary stabilization of Navier-Stokes equations, Memoirs of the A.M.S., 2005, to appear. C. Bardos, A regularity theorem for parabolic equations, Journal of Functional Analysis Vol. 7 (1971), 311-322. A. Bensoussan, G. Da Prato, M. C. Delfour, S. K. Mitter, Representation and Control of Infinite Dimensional Systems, Vol. 1, Birkh¨ auser, 1992. A. Bensoussan, G. Da Prato, M. C. Delfour, S. K. Mitter, Representation and Control of Infinite Dimensional Systems, Vol. 2, Birkh¨ auser, 1993. H. O. Fattorini, Infinite dimensional optimization and control theory, Cambridge University Press, 1999. E. Fernandez-Cara, S. Guerrero, O. Yu. Imanuvilov, J.-P. Puel, Local exact controllability of the Navier-Stokes system, J. Math. Pures Appl., 83 (2004), 1501-1542. A. V. Fursikov, M. D. Gunzburger, L. S. Hou, Inhomogeneous boundary value problems for the three-dimensional evolutionary Navier-Stokes equations, J. Math. Fluid Mech., 4 (2002), 45-75. A. V. Fursikov, M. D. Gunzburger, L. S. Hou, Boundary value problems and optimal boundary control for the Navier-Stokes system: the two-dimensional case, SIAM J. Cont. Optim., 36 (1998), 852-894. A. V. Fursikov, Stabilizability of two-dimensional Navier-Stokes equations with help of a boundary feedback control, J. Math. Fluid Mech., 3 (2001), 259-301. A. V. Fursikov, Stabilization for the 3D Navier-Stokes system by feedback boundary control, Discrete and Cont. Dyn. Systems, 10 (2004), 289-314. B. B. King, Y.-R. Ou, Nonlinear dynamic compensator design for flow control in a driven cavity, Proc. 34th CDC, Dec. 1995, 3741-3746. 32
[15] Ou Y.-R., Design of feedback compensators for viscous flow, in Optimal control of viscous flow, S. S. Sritharan Ed., SIAM, 1998, 151-180. [16] A. V. Fursikov, M. Gunzburger, L. S. Hou, S. Manservisi, Optimal control for the Navier-Stokes equations, in Lecture on applied Mathematics, Bungartz et al. Eds, Springer 2000. [17] J.-W. He, R. Glowinski, R. Metcalfe, A. Nordlander, J. Periaux, Active control and drag optimization for flow past a circular cylinder, JCP 163, 83-117 (2000). [18] G. P. Galdi, An introduction to the mathematical theory of the Navier-Stokes equations, Volume 1, Springer-Verlag, 1994. [19] G. Grubb, Nonhomogeneous Dirichlet Navier-Stokes problems in low regularity Lp Sobolev spaces, J. Math. Fluid Mech., Vol. 3 (2001), 57-81. [20] G. Grubb, V. A. Solonnikov, Boundary value problems for the nonstationary Navier-Stokes equations treated by pseudo-differential methods, Math. Scand., Vol. 69 (1991), 217-290. [21] M. D. Gunzburger, H. C. Lee, Feedback control of vortex shedding, Transactions of the ASME, Vol. 63 (1996), 828-835. [22] T. Kato, Perturbation theory for linear operators, Reprint of the 1980 Edition, Springer-Verlag, 1995. [23] I. Lasiecka, R. Triggiani, The regulator problem for parabolic equations with Dirichlet boundary control, part 1, Appl. Math. Optim., Vol. 16 (1987), 147-168. [24] I. Lasiecka, R. Triggiani, Control Theory for Partial Differential Equations, Vol. 1, Cambridge University Press, 2000. [25] I. Lasiecka, R. Triggiani, Control Theory for Partial Differential Equations, Vol. 2, Cambridge University Press, 2000. [26] J.-L. Lions, E. Magenes, Probl` emes aux limites non homog` enes, Vol. 1, Dunod, Paris 1968. [27] J.-L. Lions, E. Magenes, Probl` emes aux limites non homog` enes, Vol. 2, Dunod, Paris 1968. [28] J.-L. Lions, Espaces d’interpolation et domaines de puissances fractionnaires d’op´ erateurs, J. Math. Soc. Japan, 14 (1962), 233-241. [29] D. S. Park, D. M. Ladd, and E. W. Hendricks, Feedback control of von Karman vortex shedding behind a circular cylinder at low Reynolds numbers, Phys. Fluids, Vol. 6 (1994), 2390-2405. [30] J.-P. Raymond, Stokes and Navier-Stokes equations with nonhomogeneous boundary conditions, 2005, submitted. [31] J.-P. Raymond, Boundary feedback stabilization of the three dimensional Navier-Stokes equations, in preparation. [32] J.-P. Raymond, Local boundary feedback stabilization of the Navier-Stokes equations, in Control Systems: Theory, Numerics and Applications, http://pos.sissa.it. [33] V. A. Solonnikov, Estimates of the solutions of a nonstationary linearized system of Navier-Stokes equations, Am. Math. Soc. Transl., Ser. 2, Vol. 75 (1968), 1-116. [34] M. Reed, B. Simon, Fourier Analysis, Self-Adjointness, Academic Press, San Diego, California, 1975. [35] R. Temam, Navier-Stokes equations, North-Holland, 1984. [36] W. von Wahl, The equations of Navier-Stokes and abstract parabolic equations, Vieweg and Sohn, Braunschweig, Wiesbaden, 1985.
33
Dirichlet control, feedback control, stabilization, Navier-Stokes equations, Oseen equations, Riccati
AMS subject classifications. 93B52, 93C20, 93D15, 35Q30, 76D55, 76D05, 76D07
1. Setting of the problem. Let Ω be a bounded and connected domain in R2 or R3 with a regular boundary Γ, ν > 0, and consider a couple (w, χ) – a velocity field and a pressure – solution to the stationary Navier-Stokes equations in Ω: −ν∆w + (w · ∇)w + ∇χ = f
and div w = 0 in Ω,
w = u∞ s
on Γ.
(1.1)
We assume that w is regular and is an unstable solution of the instationary Navier-Stokes equations. The purpose of this paper is to determine a Dirichlet boundary control u, in feedback form, localized in a part of the boundary Γ, so that the corresponding controlled system: ∂y − ν∆y + (y · ∇)w + (w · ∇)y + (y · ∇)y + ∇p = 0, ∂t y = M u on Σ∞ , y(0) = y0 in Ω,
div y = 0
in Q∞ ,
(1.2)
be stable for initial values y0 small enough in an appropriate space X(Ω). In this setting, Q∞ = Ω×(0, n o ∞), 2 0 Σ∞ = Γ × (0, ∞), X(Ω) is a subspace of Vn (Ω) = y ∈ L (Ω) | div y = 0 in Ω, y · n = 0 on Γ , and the operator M is a restriction operator precisely defined in section 2. If we set (z, q) = (w + y, χ + p) and if u = 0, we see that (z, q) is the solution to the Navier-Stokes equations ∂z − ν∆z + (z · ∇)z + ∇q = f , div z = 0 ∂t z = u∞ on Σ∞ , z(0) = w + y0 in Ω. s
in Q∞ ,
Thus y0 is a perturbation of the stationary solution w. To study the local feedback stabilization of system (1.2), we first study the feedback stabilization of the corresponding linearized system ∂y − ν∆y + (w · ∇)y + (y · ∇)w + ∇p = 0, in Q∞ , ∂t div y = 0 in Q∞ , y = M u on Σ∞ , y(0) = y0 in Ω.
(1.3)
To stabilize this system we can look for a control u belonging either to L2 (0, ∞; V0 (Γ)) or to L2 (0, ∞; Vn0 (Γ)), where n o n o V0 (Γ) = y ∈ L2 (Γ) | hy · n, 1iH −1/2 (Γ),H 1/2 (Γ) = 0 and Vn0 (Γ) = y ∈ L2 (Γ) | y · n = 0 on Γ . ∗ Laboratoire MIP, UMR CNRS 5640, [email protected]
Universit´ e Paul Sabatier, 1
31062 Toulouse Cedex 9,
France,
email:
A linear feedback law stabilizing (1.3) in X(Ω) ⊂ Vn0 (Ω), with a control u ∈ L2 (0, ∞; V0 (Γ)), is determined by a bounded operator K ∈ L(X(Ω), V0 (Γ)) such that the solution of the closed loop system ∂y − ν∆y + (w · ∇)y + (y · ∇)w + ∇p = 0, in Ω × (0, T ), ∂t div y = 0 in Ω × (0, T ), y = M Ky on Γ × (0, T ), y(0) = y0
in Ω,
obeys |y(t)|X(Ω) ≤ C|y0 |X(Ω)
for all t ∈ (0, ∞).
(1.4)
The feedback control law is given by u(t) = Ky(t) for all t ∈ (0, ∞).
(1.5)
In place of (1.4) we can look for an exponential decay |y(t)|X(Ω) ≤ Ce−σt |y0 |X(Ω) for all t ∈ (0, ∞),
with σ > 0.
(1.6)
Let us underline that a feedback law of the form (1.5) is a pointwise (in time) feedback law. A feedback law may be of a different form, for example of the form u = L0 y0 ,
L0 ∈ L(X(Ω), L2 (0, ∞; V0 (Γ))).
In engineering applications pointwise feedback law are needed because they are more robust with respect to perturbations in the models. This paper is only focused on the characterization of pointwise feedback laws for the Oseen and the Navier-Stokes equations. Several important questions must be addressed when we look for a pointwise feedback law able to stabilize system (1.3) or (1.2): (Q1 ) Does there exist a control u ∈ L2 (0, ∞; V0 (Γ)) such that the solution of (1.3) obeys (1.6) or (1.4) ? In other words, is the system (1.3) stabilizable ? (Q2 ) Assume that K is a pointwise feedback law able to stabilize system (1.3) in X(Ω). Does K also stabilize the nonlinear system (1.2) for |y0 |X(Ω) small enough ? (Q3 ) Assume that we have proved the existence of a feedback law stabilizing system (1.3). Can we find an equation characterizing K which can be numerically solved by classical methods ? We stop the list here, but lot of other questions are very important in applications, such as the robustness of feedback laws, the numerical accuracy of approximations. In this paper, we do not address the stabilizability of system (1.3), and we shall use the existing results in the literature. This paper is mainly devoted to (Q2 ) and (Q3 ). But firstly mention some results answering to (Q1 ). When w ∈ L∞ (Ω) and X(Ω) = Vn0 (Ω) ∩ L4 (Ω), the existence of a boundary control u ∈ L2 (0, ∞; 0 V (Γ)) such that the solution of system (1.3) obeys (1.6) may be deduced, by using an extension method, from exact controllability results with internal controls n stated in [9]. o In the three dimensional case, and when X(Ω) = y ∈ H1 (Ω) | div y = 0 in Ω , the existence of a feedback control law exponentially stabilizing (1.2) or (1.3) is proved in [13]. But the feedback operator constructed in [13] is of the form L0 (it is not a pointwise feedback operator, and it requires the knowledge of the eigenfunctions and the eigenvalues of the Oseen operator of equation (1.3)). In the three dimensional cases, when Ω is simply connected, Barbu, Lasiecka and Triggiani [4] have proved the stabilizability of system (1.3) in X(Ω) = y ∈ H1/2+ε (Ω) | div y = 0 in Ω, y · n = 0 on Γ with controls u ∈ L2 (0, ∞; Vn0 (Γ)) acting everywhere on the bondary Γ. Some additional results in the two dimensional case are also stated in [4] (see also section 7.1). Let us now focus on (Q2 ) and (Q3 ). One way to address these questions is to use the optimal control theory, where the pointwise feedback law is obtained by solving an infinite time horizon control problem of the form n o (Q) inf J(y, u) | (y, u) satisfies (1.3), u ∈ L2 (0, T ; U) . 2
More precisely one has to prove that the value function of problem (Q) obeys inf(Q) = Πy0 , y0 , 0 (Ω) Vn
where Π ∈ L(Vn0 (Ω)), Π = Π∗ ≥ 0. Next the feedback law is defined thanks to the operator Π. In this framework, question (Q3 ) can be reformulated as follows: Does the operator Π satisfy an algebraic Riccati equation in the domain of the Oseen operator ? The answer depends on the choice of J and u and is not necessarily obvious in the case of a boundary control problem. According to the stabilizability results mentioned above, we have two possible choices for U: U = Vn0 (Γ)
or U = V0 (Γ).
Both choices are interesting for applications. Even if U = V0 (Γ) leads to a little bit complicated analysis, this case is interesting since the normal velocity is used as a control in many applications [17, 21, 29]. These papers are devoted to the suppression of vortex shedding past a cylinder. It corresponds to the case where Ω = Ωe \ Ωi , Ωi is a bounded domain in R2 with a regular boundary Γi , and Ωe is an other bounded domain in R2 with a regular boundary Γe such that Ωi ⊂ Ωe (Ωi is a disk in the case of a circular cylinder). Given a stationary velocity u∞ s , we can take (w, χ) as the solution to the stationary NavierStokes equation (1.1) with f = 0. In that case the control has to be localized in a part of the boundary Γi , and Γe must be far enough from the cylinder in order that the solution to the stationary or instationary Navier-Stokes equations give a good approximation of the corresponding equation in the exterior domain R2 \ Ωi . Some heuristic feedback laws have been successfully tested in numerical experiments [29, 21], but in general these feedback laws, which are designed for a very specific value of u∞ s , are not robust , contrarily to feedback laws obtained by solving Riccati equations. with respect to perturbations of u∞ s This paper is written in the case when U = V0 (Γ) and the adaptation to the case when U = Vn0 (Γ) is given in section 7.1. We would like to underline that the choice of J is critical. Indeed once X(Ω), U and J are chosen, K is generally uniquely determined. Thus we have to choose J so that the corresponding feedback law also stabilizes system (1.2). The functional J is usually of the form Z Z 1 ∞ 1 ∞ J(y, u) = |Cy(t)|2V0 (Ω) dt + |u(t)|2U dt, 2 0 2 0 where the observation operator C may be a bounded or an unbounded operator in V0 (Ω) = y ∈ L2 (Ω) | div y = 0 when U = V0 (Γ) (or in Vn0 (Ω) when U = Vn0 (Γ)). If D(C) denotes the domain of C in V0 (Ω) (or in Vn0 (Ω) when U = Vn0 (Γ)), the solution y of (1.3) is searched in L2 (0, ∞; D(C)). In [4, 2], C is chosen so that |Cy|V0 (Ω) be a norm in Vn0 (Ω) equivalent to the usual norm of the space H3/2+ε (Ω), for some ε > 0. The idea in [4, 2] is to choose the operator C so that the norm |Cy|V0 (Ω) be strong enough to dominate the nonlinearity of the Navier-Stokes equations. In that way a Lyapunov function for the nonlinear closed loop system can be defined thanks to Π. The price to pay is that the operator Π, corresponding this problem, does not satisfy a Riccati equation in the domain of the Oseen operator. An algebraic Riccati equation is stated only in D((AΠ )2 ) [4, section 4.5], where AΠ is the infinitesimal generator of the associated closed loop system. Thus the domain in which the Riccati equation is stated depends on the unknown Π of the equation. This is a serious drawback. Here we follow a different approach. We choose J as follows J(y, u) =
1 2
Z 0
T
Z
|y|2 dxdt +
Ω
1 2
Z
T
0
Z
|u|2 dxdt.
(1.7)
Γ
One difficulty comes from the fact that, choosing U = V0 (Γ), the system (1.3) cannot be written in the form of a classical evolution equation. We have shown in [30] that (1.3) can be rewritten in the form: P y0 = AP y + BM u,
y(0) = y0 ,
(I − P )y = (I − P )DA γn M u, 3
(1.8)
where A is the corresponding Oseen operator, B is a boundary control operator, P is the so-called Helmholtz or Leray projector, and DA the Dirichlet operator associated with A. For the precise definitions and properties of these operators we refer to section 2. Using the writing of system (1.3) in the form (1.8), we show that the control problem (Q) can be rewritten in the form of another control problem in which the state variable is P y and not y. This transformation is essential in our approach. It leads to a Riccati equation which is the natural one for the new control problem, but which is not the expected one if we only consider problem (Q). Moreover the Riccati equation is satisfied in the domain of A. Now the difficulty is to prove that the linear feedback law determined in this way also stabilizes the nonlinear equation (1.2). This is not obvious because the norm of y involved in J is to weak to dominate the nonlinearity of the Navier-Stokes equations. Thus we cannot follow the Lyapunov function approach as in [2] and [4]. We develop a completely different approach. We study the regularizing properties of the operator Π. Thanks to these results – which, to the best of our knowledge, are new even in the case of other parabolic equations like the heat equation – we are able to establish a local feedback stabilization result for the Navier-Stokes equation for initial data y0 small enough in the space 1/2−ε Vn (Ω) = Vn0 (Ω) ∩ H1/2−ε (Ω), for all 0 < ε < 1/4. We show that the solution to the closed loop nonlinear system obeys an exponential decay in the corresponding space V1/2−ε (Ω) = V0 (Ω)∩H1/2−ε (Ω). Finally mention that in the three dimensional case, the stabilization of the nonlinear problem cannot be treated with the same tools and requires a more delicate analysis [31, 32]. The paper is organized as follows. The Oseen operator and some associated boundary control operators are studied in section 2. Optimality conditions for the finite horizon control problem of the Oseen equations are established in section 3. We study the corresponding infinite time horizon control problem in section 4. We show that the optimal solution of this problem is characterized by an optimality system. This kind of characterization, which is known in the case of bounded control operators [1, 3] is, to the best of our knowledge, new in the case of unbounded operators. Thanks to this optimality system we are able to study the regularity properties of the feedback operator. In order to study the nonlinear problem, we first study in section 5 a nonhomogeneous linear-quadratic control problem. The nonhomogeneous term will play the role of the nonlinearity in the next section. The local stabilization of the Navier-Stokes equations is studied in section 6. Some additional results are stated in section 7. We have collected some regularity results needed throughout the paper in an appendix. 2. Functional framework. 2.1. Notation and assumptions. Let us introduce the following spaces: H s (Ω; RN ) = Hs (Ω), L (Ω; RN ) = L2 (Ω), the same notation conventions will be used for trace spaces and for the spaces H0s (Ω; RN ). We also introduce different spaces of free divergence functions and some corresponding trace spaces: n o Vs (Ω) = y ∈ Hs (Ω) | div y = 0 in Ω, hy · n, 1iH −1/2 (Γ),H 1/2 (Γ) = 0 for s ≥ 0, n o Vns (Ω) = y ∈ Hs (Ω) | div y = 0 in Ω, y · n = 0 on Γ for s ≥ 0, n o V0s (Ω) = y ∈ Hs (Ω) | div y = 0 in Ω, y = 0 on Γ for s > 1/2, n o Vs (Γ) = y ∈ Hs (Γ) | hy · n, 1iH −1/2 (Γ),H 1/2 (Γ) = 0 for s ≥ −1/2 . 2
In the above setting n denotes the unit normal to Γ outward Ω. We shall use the following notation QT = Ω × (0, T ), ΣT = Γ × (0, T ), Qt¯,T = Ω × (t¯, T ) and Σt¯,T = Γ × (t¯, T ) for t¯ > 0, and 0 < T ≤ ∞. For spaces of time dependent functions we set Vs,σ (QT ) = H σ (0, T ; V0 (Ω)) ∩ L2 (0, T ; Vs (Ω)), and Vs,σ (ΣT ) = H σ (0, T ; V0 (Γ)) ∩ L2 (0, T ; Vs (Γ)). We assume that Ω is of class C 4 and w ∈ V3 (Ω). 4
In order to find a control u, supported in an open subset Γc of Γ, we introduce a weight function m ∈ C 2 (Γ) with values in [0, 1], with support in Γc , equal to 1 in Γ0 , where Γ0 is an open subset in Γc . Associated with this function m we introduce the operator M ∈ L(V0 (Γ)) defined by Z m R M u(x) = m(x)u(x) − mu · n n(x), m Γ Γ where |Γ| is the (N − 1)-dimensional Lebesgue measure of Γ. By this way, we can replace the condition supp(u) ⊂ Γc by considering a boundary condition of the form z − w = Mu
on
Σ∞ .
The main interest of this operator M is that if u ∈ L2 (0, ∞; H s (Γc ; RN )) ∩ H s/2(0, ∞; L2 (Γc ; RN )) for ˜ denotes the extension of u by zero to Σ∞ \ Γc × (0, ∞) , then M u ˜ belongs to some 0 < s ≤ 2, and if u 2 s N s/2 2 N ˜. L (0, ∞; H (Γ; R )) ∩ H (0, ∞; L (Γ; R )), which is not true for u 0 For all ψ ∈ H 1/2+ε (Ω), with ε0 > 0, we denote by c(ψ) and c(mψ) the constants defined by Z Z 1 1 ψ and c(mψ) = mψ. (2.1) c(ψ) = |Γ| Γ |Γ| Γ 2.2. Properties of some operators. In the following we consider the linearized Navier-Stokes equation ∂y − ν∆y + (w · ∇)y + (y · ∇)w + ∇p = 0, in QT , ∂t div y = 0 in QT , y = M u on ΣT , y(0) = y0 in Ω ,
(2.2)
and the adjoint equation ∂Φ − ν∆Φ − (w · ∇)Φ + (∇w)T Φ + ∇ψ = y, in QT , ∂t div Φ = 0 in QT , Φ = 0 on ΣT , Φ(T ) = 0 in Ω ,
−
(2.3)
where T is finite or infinite. To study these equations, we introduce the Stokes and the Oseen operators associated with equations (2.2) and (2.3). Let P be the orthogonal projector in L2 (Ω) onto Vn0 (Ω), and denote by (A0 , D(A0 )), (A, D(A)) (A∗ , D(A∗ )) and the unbounded operators in Vn0 (Ω) defined by D(A0 ) = H2 (Ω) ∩ V01 (Ω),
A0 y = P ∆y
D(A) = H2 (Ω) ∩ V01 (Ω),
Ay = νP ∆y − P ((w · ∇)y) − P ((y · ∇)w),
∗
2
D(A ) = H (Ω) ∩
V01 (Ω),
for all y ∈ D(A0 ),
A∗ y = νP ∆y + P ((w · ∇)y) − P ((∇w)T y).
Throughout the following we denote by λ0 > 0 an element in the resolvent set of A satisfying (λ0 I − A)y, y V0 (Ω) ≥ ω0 |y|2V1 (Ω) for all y ∈ D(A), 0 n and (λ0 I − A∗ )y, y V0 (Ω) ≥ ω0 |y|2V1 (Ω) for all y ∈ D(A∗ ),
(2.4)
0
n
for some 0 < ω0 < 1. Theorem 2.1. The unbounded operator (A−λ0 I) (respectively (A∗ −λ0 I)) with domain D(A−λ0 I) = D(A) (respectively D(A∗ −λ0 I)) is the infinitesimal generator of a bounded analytic semigroup on Vn0 (Ω). Moreover, for all 0 ≤ α ≤ 1, we have D((λ0 I − A)α ) = D((λ0 I − A∗ )α ) = D((λ0 I − νA0 )α ) = D((−A0 )α ) . Proof. Under condition (2.4) the analyticity of the semigroup generated by (A − λ0 I) is well known ([6, Chapter 1, Theorem 2.1.2]). The characterization of the domains of fractional powers of (λ0 I − A) and (λ0 I − A∗ ) may be deduced from [28]. 5
Observe that the semigroups (et(A−λ0 I) )t≥0 and (et(A and that ket(A−λ0 I) kL(Vn0 (Ω)) ≤ e−ωt
∗
−λ0 I)
and ket(A
∗
)t≥0 are exponentially stable on Vn0 (Ω)
−λ0 I)
kL(Vn0 (Ω)) ≤ e−ωt ,
for all ω < ω0 (see [6, Chapter 1, Theorem 2.12]). Lemma 2.2. If Φ belongs to V3/2+ε (Ω) ∩ V01 (Ω), for some ε > 0, then ∂Φ ∂n · n = 0 on Γ. Proof. This result is proved in [4, Lemma 3.3.1] for functions Φ belonging to V01 (Ω)∩(C 1 (Ω))N . Since V01 (Ω)∩(C 2 (Ω))N is dense in V3/2+ε (Ω)∩V01 (Ω), the result is still true for functions in V3/2+ε (Ω)∩V01 (Ω), when ε > 0. Let us introduce DA and Dp , two Dirichlet operators associated with A, defined as follows. For u ∈ V0 (Γ), set DA u = y and Dp u = q where (y, q) is the unique solution in V1/2 (Ω) × (H 1/2 (Ω)/R)0 to the equation λ0 y − ν∆y + (w · ∇)y + (y · ∇)w + ∇q = 0 div y = 0 in Ω,
in Ω,
y = u on Γ.
Lemma 2.3. (i) The operator DA is a bounded operator from V0 (Γ) into V0 (Ω), moreover it satisfies kDA ukVs+1/2 (Ω) ≤ C(s)kukVs (Ω)
for all 0 ≤ s ≤ 2 .
∗ ∈ L(V0 (Ω), V0 (Γ)), the adjoint operator of DA ∈ L(V0 (Γ), V0 (Ω)), is defined by (ii) The operator DA ∗ DA g = −ν
∂z + πn − c(π)n, ∂n
(2.5)
where (z, π) is the solution of λ0 z − ν∆z − (w · ∇)z + (∇w)T z + ∇π = g
and div z = 0 in Ω,
z=0
on Γ,
(2.6)
and c(π) is defined by (2.1). Proof. Part (i) is well known when w = 0 (see e.g. [35]). Its adaptation in the case when w ∈ V3 (Ω) is given in [30, Corollary 7.1]. Part (ii) is stated in [30, Lemma 7.4]. Let us define the operators γτ ∈ L(V0 (Γ)) and γn ∈ L(V0 (Γ)) by γτ u = u − u · n n and γn u = u · n n = u − γτ u for all u ∈ V0 (Γ) . R Let us also denote by PΓ the projector in L2 (Γ) onto V0 (Γ) defined by PΓ u = u − R mm Γ u · n n. Γ Observe that M = PΓ m, where m denotes the multiplication operator by the function m. Introducing the spaces n o n o Vτ0 (Γ) = u ∈ V0 (Γ) | γτ u = 0 and Vn0 (Γ) = u ∈ V0 (Γ) | γn u = 0 , we have V0 (Γ) = Vτ0 (Γ) ⊕ Vn0 (Γ). Lemma 2.4. The operator M obeys the following properties M = M ∗,
M γτ = γτ M = mγτ ,
and
M γn = γn M .
The operators γτ and γn satisfy: γτ = γτ∗ ,
γn = γn∗
(I − P )DA = (I − P )DA γn .
and
Proof. From the definition of M it follows that Z Z M u, v 0 = mu · v = u · PΓ mv = u, M v V (Γ)
Γ
Γ
6
V0 (Γ)
,
for all u ∈ V0 (Γ) and all v ∈ V0 (Γ). Thus M = M ∗ . Observe that PΓ γτ = γτ = γτ PΓ , and mγτ = γτ m. Thus we have M γτ = PΓ mγτ = mγτ = γτ m = γτ M. Thus the second identity is established. Since γτ + γn = I, where I is the identity operator in V0 (Γ), we have M (γτ + γn ) = (γτ + γn )M , and the identity M γn = γn M follows from the equality M γτ = γτ M . The first two identities satisfied by γτ and γn are obvious. Moreover, if γn u = 0 then DA u ∈ Vn0 (Ω), and (I − P )DA u = 0. This proves the last identity. In the next lemma we study the properties of an operator RA which plays a crucial role in optimality conditions of control problems that we consider. Lemma 2.5. The operator ∗ R A = M DA (I − P )DA M + I
is an isomorphism from V0 (Γ) into itself. Moreover, for all 0 ≤ s ≤ 3/2, its restriction to Vs (Γ) is an isomorphism from Vs (Γ) into itself. In addition RA satisfies RA γn = γn RA γn
and
RA γτ = γτ RA γτ = γτ .
The restriction of RA to Vτ0 (Γ) is an isomorphism from Vτ0 (Γ) into itself, and we have −1 −1 −1 RA u = (γn RA γn )−1 u = γn RA γn u = γn RA u
for all u ∈ Vτ0 (Γ) .
Proof. Let s belong to [0, 3/2]. The operator DA is continuous from Vs (Γ) into Vs+1/2 (Ω), (I − P ) is ∗ ∗ (I − is bounded from Vs+1/2 (Ω) into Vs+1 (Γ). Thus M DA bounded from Vs+1/2 (Ω) into itself, and DA s s+1 s P )DA M is bounded from V (Γ) into V (Γ), and compact in V (Γ). Let us show that the kernel of RA is reduced to 0. If u ∈ V0 (Γ) obeys RA u = 0, then we have 0 = RA u, u = (I − P )DA M u, (I − P )DA M u + u, u ≥ u, u . V0 (Γ)
V0 (Ω)
V0 (Γ)
V0 (Γ)
Thus u = 0, which shows that RA is injective. From the Fredholm alternative it follows that RA is an isomorphism from Vs (Γ) into itself. With the identities stated in Lemma 2.4, we can write RA γn u, v 0 = (I − P )DA M γn u, (I − P )DA γn M v 0 + γn u, γn v 0 V (Γ) V (Ω) V (Γ) = (I − P )DA M γn u, (I − P )DA M γn v 0 + γn u, γn v 0 = RA γn u, γn v 0 V (Ω)
V (Γ)
for all u ∈ V0 (Γ) and all v ∈ V0 (Γ). Thus RA γn = γn RA γn . Similarly, we have RA γτ u, v = γτ u, v V0 (Γ)
V (Γ)
V0 (Γ)
for all u ∈ V0 (Γ) and all v ∈ V0 (Γ), because (I − P )DA M γτ u = 0. Thus RA γτ = γτ , from which we deduce γτ RA γτ = γτ . From the above identities it follows that RA u = RA γn u = γn RA γn u ∈ Vτ0 (Γ)
for all u ∈ Vτ0 (Γ).
Thus the restriction of RA to Vτ0 (Γ) is an isomorphism from Vτ0 (Γ) into itself. Also observe that the restriction of γn RA γn to Vτ0 (Γ) enjoys the same property. The last identities in the lemma follows from these properties. We introduce the operators Bn = (λ0 I − A)P DA γn ,
Bτ = (λ0 I − A)DA γτ , 7
B = Bn + Bτ .
Let us set Bn,α = (λ0 I − A)α−1 Bn = (λ0 I − A)α P DA γn ,
Bτ,α = (λ0 I − A)α−1 Bτ = (λ0 I − A)α DA γτ ,
and Bα = Bn,α + Bτ,α . Theorem 2.6. For all α ∈]0, 41 [, Bn,α and Bτ,α belong to L(V0 (Γ), Vn0 (Ω)). Proof. Let us prove the result for Bn,α , the other one can be shown in a similar way. Let u ∈ V0 (Γ), 1/2 set y = DA γn (u). Then P y = P DA γn (u) belongs to Vn (Ω). We have D E (λ0 I − A)P y, z (D(A∗ ))0 ,D(A∗ ) D E = (λ0 I − A)α P y, (λ0 I − A∗ )1−α z (2.7) (D((−A∗ )1−α ))0 ,D((−A∗ )1−α ) Z = yP (λ0 z − ν∆z − (w · ∇)z + (∇w)T z), for all z ∈ D(A∗ ) = V2 (Ω) ∩ V01 (Ω). Ω
Every z ∈ D(A∗ ) is the solution to the equation λ0 z − ν∆z − (w · ∇)z + (∇w)T z + ∇q = f ,
div z = 0 in Ω,
z = 0 on Γ,
∂z − q n satisfies the estimate where f = P (λ0 z − ν∆z − (w · ∇)z + (∇w)T z). It is well known that ν ∂n
∂z − q n ≤ C|f |D((−A∗ )−α ) ≤ C|z|D((−A∗ )1−α ) , ν ∂n V0 (Γ) provided that 0 < α < 1/4. Thus we have Z Z yf = yP λ0 z − ν∆z − (w · ∇)z + (∇w)T z Ω Ω Z Z ∂z −ν y λ0 z − ν∆z − (w · ∇)z + (∇w)T z + ∇q = = + q n · γn (u) ∂n Γ Ω ∂z ≤ C|u|V0 (Γ) |z|D((−A∗ )1−α ) . − q n ≤ |u|V0 (Γ) ν ∂n V0 (Γ) Combining this inequality with (2.7), we obtain Z yP λ0 z − ν∆z − (w · ∇)z + (∇w)T z |(λ0 I − A)α P y|Vn0 (Ω) ≤ sup|z|D((−A∗ )1−α ) =1 Ω
≤ C|u|V0 (Γ) . Thus the proof is complete. Proposition 2.7. For all Φ ∈ D(A∗ ), B ∗ Φ belongs to V1/2 (Γ), we have ∗ B ∗ Φ = DA (λ0 I − A∗ )Φ,
∗ Bτ∗ Φ = γτ DA (λ0 I − A∗ )Φ,
∗ Bn∗ Φ = γn DA (λ0 I − A∗ )Φ ,
and B ∗ Φ = −ν
∂Φ ∂Φ + ψn − c(ψ)n, Bτ∗ Φ = γτ B ∗ Φ = −ν ∂n ∂n
and
Bn∗ Φ = γn B ∗ Φ = ψn − c(ψ)n ,
with h i ∇ψ = (I − P ) ν∆Φ + (w · ∇)Φ − (∇w)T Φ , and c(ψ) is defined by (2.1). In particular if Φ ∈ Vs (Ω) ∩ V01 (Ω) with s ≥ 2, the following estimate holds |B ∗ Φ|Vs−3/2 (Γ) ≤ C|Φ|Vs (Ω)∩V01 (Ω) . 8
Proof. The first result is a direct consequence of the definition of B and of the fact that B is a bounded operator from V0 (Γ) into (D(A∗ ))0 . The identities for Bτ∗ Φ and Bn∗ Φ follows from Lemma 2.4. ∗ From the definitions of B ∗ and DA it follows that B ∗ Φ = −ν
ˆ ∂Φ ˆ − c(ψ)n ˆ , + ψn ∂n
ˆ is defined by ˆ ψ) where (Φ, ˆ − ν∆Φ ˆ − (w · ∇)Φ ˆ + (∇w)T Φ ˆ + ∇ψˆ = P (λ0 I − A∗ )Φ = (λ0 I − A∗ )Φ, λ0 Φ ˆ =0 div Φ
in Ω,
in Ω,
ˆ = 0 on Γ , Φ
ˆ is defined by (2.1). From the first equation we deduce (λ0 I − A∗ )Φ = (λ0 I − A∗ )Φ. ˆ Thus and c(ψ) 1 ˆ Next we can see that ψˆ is the unique element in H (Ω)/R defined by Φ = Φ. h i ˆ + ν∆Φ ˆ + (w · ∇)Φ ˆ − (∇w)T Φ ˆ ∇ψˆ = (I − P ) − λ0 Φ h i = (I − P ) ν∆Φ + (w · ∇)Φ − (∇w)T Φ . The estimate of B ∗ Φ directly follows from its definition. The proof is complete. 3. A finite time horizon control problem. To deal with the stabilization problem formulated in section 1, we first study the following optimal control problem n o (QTs,ζ ) inf JT (s, y, u) | (y, u) satisfies (3.1), u ∈ V0,0 (Σs,T ) , where 1 JT (s, y, u) = 2
Z s
T
Z
1 |y| dxdt + 2 Ω 2
Z
T
|u(t)|2V0 (Γ) dt,
s
and ∂y − ν∆y + (w · ∇)y + (y · ∇)w + ∇p = 0, in Qs,T , ∂t div y = 0 in Qs,T , y = M u on Σs,T , y(s) = ζ in Ω.
(3.1)
The main objective of this section is to establish optimality conditions for (QTs,ζ ) and to show that ¯ ) obeys a pointwise feedback formulation defined thanks to the solution to a the optimal solution (¯ y, u differential Riccati equation. To look for the solution u to problem (QTs,ζ ) in feedback form, we rewrite equation (3.1) in the form P y0 = AP y + BM u in (s, T ),
y(s) = ζ,
(I − P )y = (I − P )DA γn M u in (s, T ). As we see the situation is more complicated than in the case when we have a single equation of the form ¯ (t) = −B ∗ Π(t)¯ y(t) (Π being the solution of y0 = Ay + Bu and when the feedback law is of the form u some differential Riccati equation). To overcome the difficulty coming from the presence of the second equation (I − P )y = (I − P )DA γn M u, which is not an evolution equation, we also transform the writing of the functional JT in the following way: Z Z Z Z 1 T 1 T 2 JT (s, y, u) = |y| + |u|2 2 s Ω 2 s Γ Z Z Z Z Z Z 1 T 1 T 1 T = |P y|2 + |(I − P )y|2 + |u|2 2 s Ω 2 s Ω 2 s Γ 9
and 1 2
T
Z
Z
s
1 = 2
Z
1 = 2
Z
|(I − P )y|2 +
Ω T Z
s
1 2
T
Z
Z
s
|u|2
Γ
|(I − P )DA M γn u|2 +
Ω T
Z
s
1 2
Z
T
Z
s
(|γτ u|2 + |γn u|2 )
Γ
∗ |(M DA (I − P )DA M )1/2 γn u|2 + |γn u|2 + |γτ u|2 .
Γ
Thus, if y is the solution of (3.1), we have Z Z Z Z Z Z 1 T 1 T 1 T 1/2 2 2 JT (s, y, u) = IT (s, y, u) = |P y| + |R γn u| + |γτ u|2 , 2 s Ω 2 s Γ A 2 s Γ where RA is the operator introduced in Lemma 2.5. The control problem (QTs,ζ ) is equivalent to n o T (Ps,ζ ) inf IT (s, y, u) | (y, u) satisfies (3.2), u ∈ V0,0 (Σs,T ) , where P y0 = AP y + BM u in (s, T ),
y(s) = ζ.
(3.2)
T Even if problems (QTs,ζ ) and (Ps,ζ ) are equivalent, we are going to see that the optimality conditions for T ¯ ), that we cannot obtain with (Ps,ζ ) allows us to prove regularity results for the optimal solution (¯ y, u T the optimality system of problem (Qs,ζ ). T ) admits a unique solution (yζs , usζ ) Theorem 3.1. For all s ∈ [0, T ] and all ζ ∈ Vn0 (Ω), problem (Ps,ζ and −1 usζ = −M Bτ∗ Φsζ − RA M Bn∗ Φsζ
in (s, T ),
(3.3)
Φ(T ) = 0 .
(3.4)
where Φsζ is solution to the equation −Φ0 = A∗ Φ + P yζs
in (s, T ),
Conversely the system −1 M Bn∗ Φ P y0 = AP y − Bτ M 2 Bτ∗ Φ − Bn M RA 0
∗
−Φ = A Φ + P y
in (s, T ),
in (s, T ),
P y(s) = ζ ,
(3.5)
Φ(T ) = 0 ,
admits a unique solution (P yζs , Φsζ ) in L2 (s, T ; Vn0 (Ω)) × (V2,1 (Qs,T ) ∩ L2 (s, T ; V01 (Ω)), and (P yζs − (I − −1 −1 T P )DA M RA M Bn∗ Φsζ , −M Bτ∗ Φsζ − RA M Bn∗ Φsζ ) is the optimal solution to (Ps,ζ ). Remark 3.2. Due to Lemma 2.5, we have −1 −1 M Bτ∗ Φsζ + RA M Bn∗ Φsζ = RA M B ∗ Φsζ ,
but we prefer to keep the writing using the decomposition of B ∗ in the form Bτ∗ + Bn∗ to give the respective expressions of the tangential and the normal components. T Proof. (i) The existence of a unique solution (yζs , usζ ) to problem (Ps,ζ ) is obvious. Let u be in 2 0 2 0 L (s, T ; V (Γ)) and v ∈ L (s, T ; V (Γ)). Denote by (yu , pu ) the solution to equation (3.2) corresponding to u, and set IT (s, yu , u) = IT (u). We have I0T (u)v =
Z
T
Z
Z
T
Z
P yu · P z + s
Ω
Z
T
Z
RA γn u · γn v + s
Γ
10
γτ u · γτ v, s
Γ
where z is the solution to P z0 = AP z + BM v
in (s, T ),
(I − P )z = (I − P )DA M v .
P z(s) = 0,
Let Φ be the solution to the equation −Φ0 = A∗ Φ + P yu ,
Φ(T ) = 0 .
Due to Lemma 8.6 Φ belongs to V2,1 (Qs,T ). Thus B ∗ Φ belongs to L2 (s, T ; V0 (Γ)). The functions z and Φ obey the following identity Z TZ Z TD Z TZ E P yu · P z = BM v(τ ), Φ(τ ) dτ = v · M B∗Φ . s
Ω
(D(A∗ ))0 ,D(A∗ )
s
s
Γ
Thus I0T (u)v
Z
T
Z
T
Z
∗
Z
Z
v · MB Φ +
= s
Γ
T
Z
RA γn u · γn v + s
Γ
γτ u · γτ v. s
(3.6)
Γ
T If (yζs , usζ ) is the solution to problem (Ps,ζ ), we have I0T (usζ ) = 0, which gives −1 −1 γn usζ = −γn RA M B ∗ Φsζ = −RA M Bn∗ Φsζ
γτ usζ = −γτ M B ∗ Φsζ = −M Bτ∗ Φsζ .
and
(ii) From part (i) of the proof it follows that (yζs , Φsζ ) is a solution of system (3.5) (Φsζ is the solution to ¯ with ¯ − R−1 M B ∗ Φ, ¯ is a solution of system (3.5), and if we set u ¯ = −M Bτ∗ Φ equation (3.4)). If (¯ y, Φ) n A s s s 0 ¯ ¯ = uζ . Thus y ¯ = yζ , and Φ = Φζ . The proof is u) = 0, which implies that u (3.6) we can verify that IT (¯ complete. T There is another way to characterize the optimal solution of (Ps,ζ ), it consists in writing the optimality T conditions for (Qs,ζ ). More precisely we can state the following theorem. Theorem 3.3. For all s ∈ [0, T ] and all ζ ∈ Vn0 (Ω), the unique solution (yζs , usζ ) of problem (QTs,ζ ) is characterized by ∂Φs ζ usζ = m ν − ψζs n + c(mψζs )n , ∂n
(3.7)
where (Φsζ , ψζs ) is the solution to equation ∂Φ − ν∆Φ − (w · ∇)Φ + (∇w)T Φ + ∇ψ = yζs , in Qs,T , ∂t div Φ = 0 in Qs,T , Φ = 0 on Σs,T , Φ(T ) = 0 in Ω ,
−
(3.8)
and c(mψζs ) is the constant defined by (2.1) corresponding to mψζs . Remark 3.4. It is clear that the solution Φsζ of equation (3.8) is equal to the solution of equation (3.4). It is the reason why they are denoted in the same way. Proof. Let u be in L2 (s, T ; V0 (Γ)) and v ∈ L2 (s, T ; V0 (Γ)). Denote by yu the solution to equation (3.1), and set JT (s, yu , u) = JT (u). We have J0T (u)v
Z
T
Z
Z
T
Z
yu · z +
= s
Ω
u · v, s
Γ
where (z, q) is the solution to ∂z − ν∆z + (w · ∇)z + (z · ∇)w + ∇q = 0, in Qs,T , ∂t div z = 0 in Qs,T , y = M v on Σs,T , z(s) = 0 in Ω. 11
Let (Φ, ψ) be the solution to the equation ∂Φ − ν∆Φ − (w · ∇)Φ + (∇w)T Φ + ∇ψ = yu , ∂t div Φ = 0 in Qs,T , Φ = 0 on Σs,T , Φ(T ) = 0 −
in Qs,T , in Ω .
We can verify that Z
T
Z
T
Z
Z
yu · z = s
Ω
s
Γ
Z TZ ∂Φ ∂Φ −ν + ψn · M v = m −ν + ψn · v. ∂n ∂n s Γ
Thus usζ is characterized by Z s
T
Z Γ
∂Φsζ −mν + mψζs n + usζ ∂n
·v =0
for all v ∈ V0 (Γ), that is γτ usζ = mν
∂Φsζ ∂n
and γn usζ = −mψζs n + c(mψζs )n.
The proof is complete. T Since (Ps,ζ ) and (QTs,ζ ) admit the same solution we have ∂Φs ζ −1 − ψζs n + c(mψζs )n . −M Bτ∗ Φsζ − RA M Bn∗ Φsζ = m ν ∂n The characterization given in (3.7) seems to be easier to use than the one in (3.3). However we are going to see that we are able to characterize the regularity of the solution (yζs , usζ ) in an optimal way by using (3.3), which is not the case with (3.7). To understand the relationship between (3.3) and (3.7), let us rewrite the second equation in (3.5) in the form ∂Φsζ − ν∆Φsζ − (w · ∇)Φsζ + (∇w)T Φsζ + ∇ψ˜ζs = P yζs , in Qs,T , ∂t div Φsζ = 0 in Qs,T , Φsζ = 0 on Σs,T , Φ(T ) = 0 in Ω.
−
(3.9)
Observe that ∇ψ˜ζs = ∇ψζs − (I − P )yζs , where ψζs is the pressure appearing in (3.8). Lemma 3.5. If (Φsζ , ψ˜ζs ) is the solution to equation (3.9), then B ∗ Φsζ (t) = −ν
∂Φsζ (t) + ψ˜ζs (t)n − c(ψ˜ζs (t))n ∂n
for almost all t ∈ (s, T ),
and the optimal control usζ is defined by usζ = mν
∂Φsζ −1 − RA M ψ˜ζs n. ∂n
Proof. From Proposition 2.7 it follows that B ∗ Φsζ (t) = −ν
∂Φsζ (t) + ψˆζs (t)n − c(ψˆζs (t))n , ∂n
where ∇ψˆζs is defined by h i ∇ψˆζs = (I − P ) ν∆Φsζ + (w · ∇)Φsζ − (∇w)T Φsζ . 12
(3.10)
s ∂Φ Thus ∇ψˆζs = ∇ψ˜ζs because (I − P ) ∂tζ = 0. Therefore usζ is defined by
−1 usζ = −M Bτ∗ Φsζ − RA M Bn∗ Φsζ =
−1 − M γτ − RA M γn
∂Φs ζ − ψ˜ζs n , ν ∂n
and due to Lemma 2.5 the proof is complete. In the following theorem we improve the regularity result of the optimal solution. Theorem 3.6. The solution (yζs , Φsζ ) to system (3.5) belongs to V1,1/2 (Qs,T ) × L2 (s, T ; 3 1 3/2 V (Ω) ∩ V0 (Ω)) ∩ H (s, T ; Vn0 (Ω)). In particular Φsζ belongs to C([s, T ]; V2 (Ω) ∩ V01 (Ω)). Proof. Since B ∗ Φsζ ∈ L2 (s, T ; V0 (Γ)), due to Lemmas 8.1 and 8.3, yζs belongs to 1/2−ε,1/4−ε V (Qs,T ) for all ε > 0. From Lemmas 8.6 and 8.8 it follows that ∂Φsζ ∈ V1−ε,1/2−ε/2 (Σs,T ) ∂n
∇ψ˜ζs ∈ H1/2−ε,1/4−ε/2 (Qs,T ) ,
and
for all ε > 0, where ψ˜ζs is the pressure appearing in equation (3.9). From this regularity result and from Lemma 3.5, we deduce that B ∗ Φsζ ∈ V1/2−ε,1/4−ε/2 (Σs,T ) and usζ ∈ V1/2−ε,1/4−ε/2 (Σs,T ) for all ε > 0, −1 T ). Still applying Lemma M Bn∗ Φsζ is the solution of problems (QTs,ζ ) and (Ps,ζ where usζ = −M Bτ∗ Φsζ − RA s 1−ε,1/2−ε 8.3, we obtain yζ ∈ V (Qs,T ) for all ε > 0, and repeating the above analysis for B ∗ Φsζ , we can s 1−ε,1/2−ε/2 show that uζ ∈ V (Σs,T ) for all ε > 0. Still with Lemmas 8.1 and 8.3, we prove that yζs belongs to V1,1/2 (Qs,T ). Due to Lemma 8.6, Φsζ belongs to V3,3/2 (Qs,T ). In particular Φsζ belongs to C([s, T ]; V2 (Ω) ∩ V01 (Ω)). The proof is complete. Remark 3.7. Observe that the regularity result stated in Theorem 3.6 has been obtained by using the expression of usζ given in (3.10). In (3.10), only the pressure ψ˜ζs appears, and ψ˜ζs only depends on Φsζ . Thus improving the regularity of Φsζ we are able to improve the regularity of ψ˜ζs , and this is what is done in Lemma 8.7. If we use the expression of usζ given in (3.7), the pressure ψζs obeys ∇ψζs = ∇ψ˜ζs + (I − P )yζs . Since the regularity with respect to the time variable of (I − P )yζs cannot be improved, we are not able to improve the regularity of usζ by using (3.7). Corollary 3.8. For all s ∈ [0, T ] and all zeta ∈ Vn0 (Ω), the unique solution (yζs , usζ ) to problem T (Ps,ζ ) and the corresponding solution (yζs , Φsζ ) to system (3.5) obeys IT (s, yζs , usζ ) =
1 2
Z
Φsζ (s) · ζ .
Ω
Proof. By multiplying by Φsζ the equation (3.1) satisfied by (yζs , usζ ), integrating in space and time, making integration by parts, and using equation (3.9) satisfied by Φsζ , we can prove that Z Ω
Φsζ (s)
Z
T
Z
·ζ = s
|P yζs |2
Ω
Z + s
T
Z ∂Φsζ ν − ψ˜ζs n + c(ψ˜ζs )n · M usζ . ∂n Γ
Next using (3.10) we can show that Z s
T
Z Z TZ Z TZ ∂Φsζ 1/2 ν − ψ˜ζs n + c(ψ˜ζs )n · M usζ = |RA γn usζ |2 + |γτ usζ |2 , ∂n Γ s Γ s Γ
and the proof is complete. Let Π(s) be the operator defined by Π(s) : ζ 7−→ Φsζ (s) ,
(3.11)
where (P yζs , Φsζ ) is the unique solution to system (3.5). From Theorem 3.6 it follows that Π(s) ∈ L(Vn0 (Ω), V2 (Ω) ∩ V01 (Ω)). Using Theorem 3.6 we can prove that the family of operators (Π(s))s∈[0,T ] defined by (3.11) belongs to Cs ([0, T ]; L(Vn0 (Ω)) (the space of functions Π from [0, T ] into L(Vn0 (Ω)) such 13
that, for all y ∈ Vn0 (Ω), Π(·)y be continuous from [0, T ] into Vn0 (Ω)). Next, using the optimality system (3.5) we can show that Π is the unique solution in Cs ([0, T ]; L(Vn0 (Ω))) to the Riccati equation Π∗ (t) = Π(t) and Π(t) ≥ 0, for all y ∈ Vn0 (Ω), t ∈ [0, T ], Π(t)y ∈ V2 (Ω) ∩ V01 (Ω) and |Π(t)y|V2 (Ω) ≤ C|y|Vn0 (Ω) , −1 −Π0 (t) = A∗ Π(t) + Π(t)A − Π(t)Bτ M 2 Bτ∗ Π(t) − Π(t)Bn M RA M Bn∗ Π(t) + I, Π(T ) = 0.
(3.12)
From the definition of Π, from Theorem 3.1 and Corollary 3.8, we deduce the following theorem. We also refer to [24, Theorem 1.2.2.1] where the existence of a unique solution to equation (3.12) is established. T ) belongs to C([0, T ]; V0 (Ω)) × C([0, T ]; V0 (Γ)), Theorem 3.9. The solution (y, u) to problem (P0,y 0 it obeys the feedback formula −1 u(t) = − M Bτ∗ + RA M Bn∗ Π(t)P y(t), and the optimal cost is given by J(y, u) =
1 Π(0)y0 , y0 0 . 2 Vn (Ω)
b b is the unique solution in Cs ([0, T ]; L(Vn0 (Ω))) to the Riccati equation If we set Π(t) = Π(T − t), then Π b ∗ (t) = Π(t) b b Π and Π(t) ≥ 0, 0 b b 0 (Ω) , ∈ V2 (Ω) ∩ V01 (Ω) and |Π(t)y| for all y ∈ Vn (Ω), t ∈ [0, T ], Π(t)y V2 (Ω) ≤ C|y|Vn −1 ∗ ∗ 0 ∗ b b b b + Π(t)A b b b (t) = A Π(t) Π(t) + I, Π(t) − Π(t)B R B − Π(t)B B Π n A τ τ n b Π(0) = 0.
(3.13)
b it follows that Π(0) = Π(T b ). From the definition of Π ∞ 4. An infinite time horizon problem. In this section we want to study problem (P0,y ), and we 0 want to study the regularity of its solution in function of the regularity of y0 . For notational simplicity ∞ problem (P0,y ) will now be denoted by (P0,y0 ), and the state variable by y and not P y as in the previous 0 section. With this notation the problem we consider is n o (P0,y0 ) inf I(y, u) | (y, u) satisfies (4.1), u ∈ V0,0 (Σ∞ ) ,
where I(y, u) =
1 2
Z
∞
Z
0
|y|2 dxdt +
Ω
1 2
∞
Z
0
1/2 |γτ u(t)|2V0 (Γ) + |RA γn u(t)|2V0 (Γ) dt,
and y0 = Ay + BM u in (0, ∞),
y(0) = y0 .
(4.1)
k Accordingly, for 0 ≤ s < k < ∞, problem (Ps,ζ ) will now be defined by k (Ps,ζ )
n o inf Ik (s, y, u) | (y, u) satisfies (4.2), u ∈ V0,0 (Σs,k ) ,
where Ik (s, y, u) =
1 2
Z s
k
Z Ω
|y|2 dxdt +
1 2
Z
k
s
1/2 |γτ u(t)|2V0 (Γ) + |RA γn u(t)|2V0 (Γ) dt,
and y0 = Ay + BM u
in (s, k), 14
y(s) = ζ.
(4.2)
Theorem 4.1. For all y0 ∈ Vn0 (Ω), problem (P0,y0 ) admits a unique solution (yy0 , uy0 ). There exists Π ∈ L(Vn0 (Ω)), obeying Π = Π∗ , such that the optimal cost is given by 1 Πy0 , y0 . I(yy0 , uy0 ) = 0 (Ω) 2 Vn Proof. From the null controllability results stated in [9], we can deduce that there exist controls u ∈ L2 (0, ∞; V0 (Γ)) such that I(yu , u) < ∞, where yu is the solution of equation (4.1) corresponding to u. The null controllability results in [9] are stated for a distributed control. Using an extension of the domain, this result also provides a null controllability result for a control localized on the boundary. The existence of a unique solution (yy0 , uy0 ) to (P0,y0 ) follows from classical arguments. From the dynamic programming principle, it follows that the mapping b )y0 , y0 T 7−→ Π(T 0 Vn (Ω)
is nondecreasing, and we have 1b Π(T )y0 , y0 0 ≤ I(yy0 , uy0 ) < ∞ . 2 Vn (Ω) As in [7], or in [24], we can show that there exists an operator Π ∈ L(Vn0 (Ω)) satisfying Π = Π∗ ≥ 0 and b )y0 Πy0 = limT →∞ Π(T for all y0 ∈ Vn0 (Ω). k Let us show that I(yy0 , uy0 ) = 21 Πy0 , y0 . Problem (P0,y ) admits a unique solution (yk , uk ) 0 0 Vn (Ω)
characterized by yk0 = Ayk + BM uk
in (0, k),
yk (0) = y0 ,
−Φ0k
in (0, k),
Φk (k) = 0,
∗
= A Φk + yk
γτ uk = −M Bτ∗ Φk ,
(4.3)
−1 γn uk = −RA M Bn∗ Φk .
˜ k the extension by zero of uk to (k, ∞), and by y ˜ k the extension Convergence of yk and uk . Denote by u by zero of yk to (k, ∞). Since we have Z kZ Z k 1/2 |yk |2 dxdt + |γτ uk (t)|2V0 (Γ) + |RA γn uk (t)|2V0 (Γ) dt 0
Z
Ω ∞
0
Z
≤ 0
|yy0 |2 dxdt +
Ω
Z
∞
0
1/2 |γτ uy0 (t)|2V0 (Γ) + |RA γn uy0 (t)|2V0 (Γ) dt ,
the sequences (˜ yk )k and (˜ uk )k are respectively bounded in L2 (0, ∞; Vn0 (Ω)) and L2 (0, ∞; V0 (Γ)). Thus 2 there exist y∞ ∈ L (0, ∞; Vn0 (Ω)) and u∞ ∈ L2 (0, ∞; V0 (Γ)) such that ˜ k * u∞ u
weakly in L2 (0, ∞; V0 (Γ)),
˜ k * y∞ y
weakly in L2 (0, ∞; Vn0 (Ω)).
By passing to the limit in the above inequality we obtain Z ∞Z Z ∞ 1/2 |y∞ |2 dxdt + |γτ u∞ (t)|2V0 (Γ) + |RA γn u∞ (t)|2V0 (Γ) dt 0
Z
Ω ∞Z
≤
0
2
Z
|yy0 | dxdt + 0
Ω
0
∞
1/2 |γτ uy0 (t)|2V0 (Γ) + |RA γn uy0 (t)|2V0 (Γ) dt .
And by passing to the limit in the equation satisfied by (yk , uk ), we have 0 y∞ = Ay∞ + BM u∞
in (0, ∞), 15
y∞ (0) = y0 .
Thus the pair (y∞ , u∞ ) is admissible and we have (y∞ , u∞ ) = (yy0 , uy0 ), because I(y∞ , u∞ ) ≤ I(yy0 , uy0 ). Therefore we can claim that in L2 (0, ∞; V0 (Γ))
˜ k → yy0 u
in L2 (0, ∞; Vn0 (Ω)).
˜ k → uy0 and y
Since Ik (0, yk , uk ) =
1b Π(k)y0 , y0 0 , 2 Vn (Ω)
by passing to the limit when k tends to infinity, we obtain 1 Πy0 , y0 . I(yy0 , uy0 ) = 0 (Ω) 2 Vn We denote by ϕ(y0 ) the value function of problem (P0,y0 ), that is: ϕ(y0 ) = I(yy0 , uy0 ). Lemma 4.2. For every y0 ∈ Vn0 (Ω), the system −1 y0 = Ay − Bτ M 2 Bτ∗ Φ − Bn M RA M Bn∗ Φ
−Φ0 = A∗ Φ + y
in (0, ∞),
in (0, ∞),
y(0) = y0 , (4.4)
Φ(∞) = 0 ,
for all t ∈ (0, ∞) ,
Φ(t) = Πy(t)
admits a unique solution in L2 (0, ∞; Vn0 (Ω)) × V2,1 (Q∞ ). This solution belongs to Cb (R+ ; Vn0 (Ω)) ∩ V1,1/2 (Q∞ ) × V3,3/2 (Q∞ ) and it satisfies: kykCb (R+ ;Vn0 (Ω)) + kykV1,1/2 (Q∞ ) + kΦkV3,3/2 (Q∞ ) ≤ C|y0 |Vn0 (Ω) . −1 M Bn∗ Φ) is the solution of (P0,y0 ). The pair (y, −M Bτ∗ Φ − RA ˆ ). We denote by Proof. For notational simplicity the solution to (P0,y0 ) will now be denoted by (ˆ y, u k ϕk (0, y0 ) the value function of problem (P0,y ) and by ϕk (t¯, ζ) the value function of problem (Pt¯k,ζ ). 0 k Let (ykt¯ , utk¯ ) be the solution of (Pt¯k,yk (t¯) ), and let (yk , uk ) be the solution of (P0,y ) characterized by 0 t¯ t¯ t¯ (4.3). Denote by Φk the adjoint state corresponding to (yk , uk ), and by Φk the adjoint state corresponding to (yk , uk ). From the dynamic programming principle it follows that (ykt¯ , utk¯ , Φtk¯ )(t) = (yk , uk , Φk )(t) b − t¯)yk (t¯). Thus for all t ∈ (t¯, k). Therefore we have Φtk¯ (t¯) = Φk (t¯) ∈ ∂y ϕk (t¯, yk (t¯)), that is Φk (t¯) = Π(k
b − t¯)k|yk (t¯)|V0 (Ω) ≤ C|yk (t¯)|V0 (Ω) , |Φk (t¯)|Vn0 (Ω) ≤ kΠ(k n n and ˜ k kL2 (0,∞;V0 (Ω)) ≤ Ck˜ kΦ yk kL2 (0,∞;Vn0 (Ω)) , n ˜ k (respectively y ˜ k ) is the extension by zero of Φk (respectively yk ) to (k, ∞). In the proof of where Φ ˆ in Theorem 4.1 we have shown that (˜ yk )k is bounded in L2 (0, ∞; Vn0 (Ω)) and that it converges to y 2 0 ˜ ˜ k is also the L (0, ∞; Vn (Ω)). Therefore (Φk )k is also bounded in L2 (0, ∞; Vn0 (Ω)). Observe that Φ solution of the equation ˜ 0 = (A∗ − λ0 I)Φ ˜k + y ˜ k, ˜ k + λ0 Φ −Φ k
˜ k (∞) = 0. Φ
Thus ˜ k (t) = Φ
Z
∞
e(A
∗
−λ0 I)(τ −t)
˜ k (τ )) dτ (˜ yk (τ ) + λ0 Φ
t
16
for all t ≥ 0.
˜ k )k is also bounded in L∞ (0, ∞; V0 (Ω)). From Young’s inequality for convolutions it follows that (Φ n ∞ 0 2 0 ˆ ∈ L (0, ∞; V (Ω)) ∩ L (0, ∞; V (Ω)) such that, after extraction of a subsequence, There then exists Φ n n we have ˜k * Φ ˆ Φ
weakly in L2 (0, ∞; Vn0 (Ω))
weak-star in L∞ (0, ∞; Vn0 (Ω)),
and
ˆ obeys the equation and Φ ˆ Φ(t) =
Z
∞
e(A
∗
−λ0 I)(τ −t)
ˆ )) dτ (ˆ y(τ ) + λ0 Φ(τ
for all t ≥ 0.
t
ˆ We have Step 3. Regularity of Φ. ˆ 0 = (A∗ − λ0 I)Φ ˆ + λ0 Φ ˆ +y ˆ −Φ
ˆ Φ(∞) = 0.
in (0, ∞),
ˆ belongs to L2 (0, ∞; V0 (Ω)), we deduce that Φ ˆ belongs to L2 (0, ∞; V1 (Ω)) ∩ V2,1 (Q∞ ) (see Since Φ n 0 ˜ k − R−1 M B ∗ Φ ˜ Lemma 8.5). Moreover due to Lemma 8.9, the sequence (˜ uk )k = (−M Bτ∗ Φ n k )k converges A −1 −1 2 0 ∗ˆ ∗ˆ ∗ ∗ ˆ − R M B Φ, ˆ and (ˆ ˆ obey ˆ = −M Bτ Φ weakly in L (0, ∞; V (Γ)) to −M Bτ Φ − RA M Bn Φ. Thus u y, Φ) n A the first two equations in (4.4). Step 4. Let us show that if (y, Φ) ∈ L2 (0, ∞; Vn0 (Ω)) × V2,1 (Q∞ ) is a solution to system (4.4), then y belongs Cb ([0, ∞); Vn0 (Ω) ∩ V1,1/2 (Q∞ ), Φ ∈ V3,3/2 (Q∞ ), and kykCb ([0,∞);Vn0 (Ω)) + kykV1,1/2 (Q∞ ) + kΦkV3,3/2 (Q∞ )
(4.5)
≤ C(|y0 |Vn0 (Ω) + kykL2 (0,∞;Vn0 (Ω)) + kΦkL2 (0,∞;Vn0 (Ω)) ) . To establish this result we rewrite system (4.4) as follows −1 M Bn∗ Φ + λ0 y y0 = (A − λ0 I)y − Bτ M 2 Bτ∗ Φ − Bn M RA
−Φ0 = (A∗ − λ0 I)Φ + y + λ0 Φ in (0, ∞),
in (0, ∞),
y(0) = y0 ,
(4.6)
Φ(∞) = 0.
Due to Lemma 8.9 we know that B ∗ Φ ∈ L2 (0, ∞; V0 (Γ)). Applying Lemmas 8.1 and 8.3, we obtain: kykV1/2−ε0 ,1/4−ε0 /2 (Q∞ ) ≤ C(|y0 |Vn0 (Ω) + kykL2 (0,∞;Vn0 (Ω)) + kΦkL2 (0,∞;Vn0 (Ω)) ) 0
for all ε0 > 0.
0
Still from Lemma 8.9, we deduce that B ∗ Φ ∈ V1/2−ε ,1/4−ε /2 (Σ∞ ) for all ε0 > 0. Applying successively 0 0 Lemmas 8.1, 8.3 and Lemma 8.9 we can prove that y belongs to V1−ε ,1/2−ε /2 (Q∞ ) and B ∗ Φ belongs to 0 0 V1−ε ,1/2−ε /2 (Σ∞ ) for all ε0 > 0. Another iteration gives y ∈ V1,1/2 (Q∞ ) ∩ Cb ([0, ∞); Vn0 (Ω)) (because y0 ∈ Vn0 (Ω)). From Lemma 8.5 we deduce that Φ ∈ V3,3/2 (Q∞ ), and the estimate (4.5) holds true. ˆ obeys the third equation in (4.4). With Lemma 8.5 we can show Step 5. We show that the pair (ˆ y, Φ) that ˜ k (t) * Φ(t) ˆ Φ
weakly in Vn0 (Ω)
for all t ≥ 0.
Since Φk (t) ∈ ∂y ϕk (t, yk (t)),
ˆ Φk (t) * Φ(t) weakly in Vn0 (Ω),
and ϕk (t, yk (t)) → ϕ(ˆ y(t)) as k → ∞ , we deduce that ˆ Φ(t) ∈ ∂ϕ(ˆ y(t)),
ˆ i.e. Φ(t) = Πˆ y(t).
ˆ obeys the third equation in (4.4). Thus we have shown that (ˆ y, Φ) 17
Step 6. Uniqueness. If a solution (y, Φ) ∈ L2 (0, ∞; Vn0 (Ω)) × V2,1 (Q∞ ) to system (4.4), due to step 4 it obeys (4.5), and we can show that Z k Z k Z Z k Z −1 |M Bτ∗ Φ(t)|2V0 (Γ) dt + |RA |y(t)|2Vn0 (Ω) dt + M Bn∗ Φ(t)|2V0 (Γ) dt = y0 Φ(0) − y(k)Φ(k). 0
0
0
Ω
Passing to the limit when k tends to infinity we obtain Z ∞ Z ∞ Z |y(t)|2Vn0 (Ω) dt + |M Bτ∗ Φ(t)|2V0 (Γ) dt + 0
0
0
∞ −1 |RA M Bn∗ Φ(t)|2V0 (Γ) dt =
Ω
Z y0 Φ(0). Ω
Thus if y0 = 0 we have y = 0. From the relation Φ = Πy we deduce that Φ = 0, and the uniqueness is established. ˆ is the unique solution to system Step 7. Final estimate. From the previous steps it follows that (ˆ y, Φ) (4.4). Since kˆ ykL2 (0,∞;Vn0 (Ω)) ≤ C|y0 |Vn0 (Ω) , we have ˆ L2 (0,∞;V0 (Ω)) ≤ C|y0 |V0 (Ω) , kΦk n n the estimate of the lemma follows from (4.5). 1/2−ε (Ω) for some 0 < ε ≤ 1/2, then the solution (y, Φ) of system (4.4) Corollary 4.3. If y0 ∈ Vn 3/2−ε,3/4−ε/2 belongs to V (Q∞ ) × V7/2−ε,7/4−ε/2 (Q∞ ), and we have: kykV3/2−ε,3/4−ε/2 (Q∞ ) + kΦkV7/2−ε,7/4−ε/2 (Q∞ ) + kB ∗ ΦkV3/2−ε,3/4−ε/2 (Q∞ ) ≤ C|y0 |V1/2−ε (Ω) . n
Proof. With Lemmas 4.2 and 8.9, we first prove that u ∈ V1,1/2 (Σ∞ ). Applying Lemmas 8.1 and 8.3, it follows that y belongs to V3/2−ε,3/4−ε/2 (Q∞ ). The estimate for Φ follows from Lemma 8.5, and we deduce that B ∗ Φ ∈ V3/2−ε,3/4−ε/2 (Q∞ ) with Lemma 8.9. Remark 4.4. Due to Lemma 4.2, if y0 ∈ Vn0 (Ω), then Φ belongs to C([0, T ]; V2 (Ω) ∩ V01 (Ω)). In 1/2−ε (Ω) then particular Φ(0) belongs to V2 (Ω) ∩ V01 (Ω). From Corollary 4.3 we deduce that if y0 ∈ Vn 5/2−ε 1 5/2−ε 1 (Ω) ∩ V0 (Ω). This means that Π belongs Φ ∈ C([0, T ]; V (Ω) ∩ V0 (Ω)) and Φ(0) belongs to V 1/2−ε (Ω), V5/2−ε (Ω) ∩ V01 (Ω)). to L(Vn0 (Ω), V2 (Ω) ∩ V01 (Ω)), and Π also belongs to L(Vn ∗ 2 1 1/2 Since B is continuous from V (Ω) ∩ V0 (Ω) into V (Γ) (see Proposition 2.7), the operator B ∗ Π 1/2−ε (Ω), V1−ε (Γ)) for all ε > 0. belongs to L(Vn0 (Ω), V1/2 (Γ)), and B ∗ Π also belongs to L(Vn 0 From Lemma 4.2 it follows that, for all y0 ∈ Vn (Ω), the evolution equation −1 y0 = Ay − Bτ M 2 Bτ∗ Πy − Bn M RA M Bn∗ Πy
in (0, ∞),
y(0) = y0 ,
(4.7)
admits at least one weak solution belonging to L2 (0, ∞; Vn0 (Ω)) ∩ Cb (R; Vn0 (Ω)). It is easy to show that this solution is unique. Due to Lemma 4.2, it is equal to yy0 , where (yy0 , uy0 ) is the solution of (P0,y0 ). Still from Lemma 4.2, we deduce that the family of operators (S(t))t≥0 defined by S(t) : y0 7−→ yy0 (t) , is a strongly continuous exponentially stable semigroup on Vn0 (Ω). Let (AΠ , D(AΠ )) be its infinitesimal generator, and let us denote by (etAΠ )t≥0 the semigroup (S(t))t≥0 . Since the semigroup (etAΠ )t≥0 is exponentially stable on Vn0 (Ω), we can define D(AΠ ) by Z ∞ n o D(AΠ ) = y ∈ Vn0 (Ω) | y = − eτ AΠ f dτ for some f ∈ Vn0 (Ω) , 0
R∞
τ AΠ
and AΠ y = f if y = − 0 e f dτ ∈ D(AΠ ). We want to find a more explicit characterization of D(AΠ ) and AΠ . From Remark 4.4, it follows that −1 the operator A − BM RA M B ∗ Π may be considered as a bounded operator from Vn0 (Ω) into (D(A0 ))0 , the dual of D(A0 ) with respect to the pivot space Vn0 (Ω). We want to prove that n o −1 D(AΠ ) = y ∈ Vn0 (Ω) | A − BM RA M B ∗ Π ∈ Vn0 (Ω) . (4.8) 18
For that we are going to use the characterization of (A∗Π , D(A∗Π )), the adjoint of (AΠ , D(AΠ )). Let us define the unbounded operator (A]Π , D(A]Π )) in Vn0 (Ω) by D(A]Π ) = D(A0 )
−1 and A]Π Φ = A∗ Φ − (B ∗ Π)∗ M RA M B ∗ Φ for all Φ ∈ D(A]Π ),
where (B ∗ Π)∗ ∈ L(V−1/2 (Γ), Vn0 (Ω)) is the adjoint of B ∗ Π ∈ L(Vn0 (Ω), V1/2 (Γ)). We know that (A∗ − λ0 I, D(A∗ )) is the infinitesimal generator of a bounded analytic semigroup on Vn0 (Ω). Let (L, D(L)) be the unbounded operator in Vn0 (Ω) defined by D(L) = V3/2+ε (Ω) ∩ V01 (Ω) with 0 < ε < 1/2, and −1 LΦ = (B ∗ Π)∗ M RA M B ∗ Φ. We can easily verify that L is (A∗ − λ0 I)-bounded with relative bound zero. From [22, Corollary 2.5, page 500] it follows that (A]Π − λ0 I, D(A0 )) is the infinitesimal generator of a quasi-bounded analytic semigroup on Vn0 (Ω). Thus (A]Π , D(A0 )) is also the infinitesimal generator of an analytic semigroup on Vn0 (Ω). Consider the equations Φ0 = A]Π Φ,
Φ(0) = Φ0 ,
(4.9)
and ˆ 0 = A] Φ, ˆ −Φ Π
ˆ ) = Φ0 . Φ(T
(4.10)
ˆ We clearly have Φ(T − t) = Φ(t) for all t ∈ [0, T ]. If Φ0 ∈ D(A0 ), then the solution Φ to equation (4.9) belongs to C([0, T ]; D(A0 )) ∩ C 1 ([0, T ]; D(A0 )). With equations (4.7) and (4.10), we can show that ]
y0 , eT AΠ Φ0
0 (Ω) Vn
= eT AΠ y0 , Φ0
∗
0 (Ω) Vn
= y0 , eT AΠ Φ0
0 (Ω) Vn ]
,
∗ and eT AΠ t≥0 t≥0 we have (A]Π , D(A0 )) = (A∗Π , D(A∗Π )). For all g ∈ Vn0 (Ω), the equation on Vn0 (Ω) are identical, R ∞ and ∗ τ A∗ AΠ Φ = g admits − 0 e Π g dτ as unique solution in D(A∗Π ). Thus, for all g ∈ Vn0 (Ω), the solution R∞ ∗ − 0 eτ AΠ g dτ to the equation A∗Π Φ = A]Π Φ = g belongs to D(A0 ).
for all y0 ∈ Vn0 (Ω), Φ0 ∈ Vn0 (Ω), and T > 0. Thus the two semigroups eT AΠ
Let us come back to the characterization of D(AΠ ). Let y be in D(AΠ ), set AΠ y = f . Setting Rt Rt z(t) = − 0 eτ AΠ f dτ = − 0 e(t−s)AΠ f ds, using the definition of weak solution for z, by passing to the limit when t tends to infinity, we can show that y satisfies the equation −1 Ay − BM RA M B ∗ Πy = f ,
(4.11)
where f = AΠ y. Thus n o −1 D(AΠ ) ⊂ y ∈ Vn0 (Ω) | A − BM RA M B ∗ Π ∈ Vn0 (Ω) . −1 In equation (4.11), Ay − BM RA M B ∗ Πy is considered as an element in (D(A0 ))0 . This means that y is a solution to equation (4.11) if and only if Z Z Z Z −1 yA∗ Φ − M RA M B∗Π y B∗Φ = yA∗Π Φ = f Φ for all Φ ∈ D(A0 ). (4.12) Ω
Γ
Ω
Ω
To prove (4.8), it is now sufficient to show that the solution to equation (4.11) or (4.12) is unique. Assume that y ∈ Vn0 (Ω) obeys (4.12) with f = 0. If we take the solution Φ of A∗Π Φ = y in (4.12), we obtain R |y|2 = 0. Thus y = 0 and the equality (4.8) is established. Ω Theorem 4.5. The unbounded operator (AΠ , D(AΠ )) defined by: n o −1 D(AΠ ) = y ∈ Vn0 (Ω) | A − BM RA M B ∗ Π ∈ Vn0 (Ω) , −1 AΠ y = Ay − BM RA M B ∗ Πy
for all y ∈ D(AΠ ),
is the infinitesimal generator of an analytic exponentially stable semigroup on Vn0 (Ω). 19
The operator Π is the unique weak solution to the algebraic Riccati equation Π∗ = Π ∈ L(Vn0 (Ω)) for all y ∈
Vn0 (Ω),
Π ≥ 0,
and 2
Πy ∈ V (Ω) ∩ V01 (Ω)
and
|Πy|V2 (Ω) ≤ C|y|Vn0 (Ω) ,
−1 A∗ Π + ΠA − ΠBτ M 2 Bτ∗ Π − ΠBn M RA M Bn∗ Π + I = 0. 1/2−ε
Moreover Π ∈ L(Vn (Ω); V5/2−ε (Ω) ∩ V01 (Ω)). Proof. The first part of the theorem is already proved. We have already shown that Π∗ = Π ≥ 0 and that Π ∈ L(Vn0 (Ω), V2 (Ω) ∩ V01 (Ω)). Since Π ∈ L(Vn0 (Ω), V2 (Ω) ∩ V01 (Ω)), it is relatively easy to prove that Π is the unique solution to the algebraic Riccati equation stated in the theorem. The uniqueness is shown in [24, Theorem 2.2.1] under a weaker condition than Π ∈ L(Vn0 (Ω), V2 (Ω) ∩ V01 (Ω)). 5. Problems with a nonhomogeneous source term. For all t¯ ≥ 0 and all y0 ∈ Vn0 (Ω) we consider the problem n o (Pt¯,y0 ,f ) inf I(t¯, y, u) | (y, u) satisfies (5.1), u ∈ V0,0 (Σt¯,∞ ) , where I(t¯, y, u) =
1 2
∞
Z t¯
Z
|y|2 dxdt +
Ω
1 2
Z
∞
t¯
1/2 |γτ u(t)|2V0 (Γ) + |RA γn u(t)|2V0 (Γ) dt,
and y0 = Ay + BM u + f
in (t¯, ∞),
y(t¯) = y0 .
(5.1)
The function f belongs to L2/(1+2ε) (0, ∞; Vn0 (Ω)) ∩ L2 (0, ∞; (V2ε (Ω))0 ), with 0 < ε < 1/2. In this section we want to study the regularity of solutions to the control problem (P0,y0 ,f ) in function of the regularity of y0 . This result will be used in the next section to study the local stabilization of the two dimensional Navier-Stokes equations. Theorem 5.1. For all t¯ ≥ 0, y0 ∈ Vn0 (Ω), f ∈ L2/(1+2ε) (0, ∞; Vn0 (Ω)), problem (Pt¯,y0 ,f ) admits a unique solution (yt¯,y0 ,f , ut¯,y0 ,f ) and the optimal cost ϕ(t¯, y0 , f ) = J(t¯, yt¯,y0 ,f , ut¯,y0 ,f ) obeys ϕ(t¯, y0 , f ) ≤ C |y0 |2Vn0 (Ω) + kf k2L2/(1+2ε) (t¯,∞;V0 (Ω)) , (5.2) n
where the constant C is independent of t¯. Proof. The semigroup generated by AΠ on Vn0 (Ω) is analytic and exponentially stable. Thus the solution to the equation z0 = AΠ z + f ,
z(t¯) = y0 ,
is defined by ¯
z(t) = e(t−t)AΠ y0 +
Z
t
e(t−τ )AΠ f (τ ) dτ,
t¯
and it obeys ¯
|z(t)|Vn0 (Ω) ≤ Ce−ω(t−t) |y0 |Vn0 (Ω) + C
Z t¯
t
e−ω(t−τ ) |f (τ )|Vn0 (Ω) dτ,
for some C > 0 and some ω > 0. It follows that kzkL2 (t¯,∞;Vn0 (Ω)) ≤ C |y0 |Vn0 (Ω) + kf kL2/(1+2ε) (t¯,∞;Vn0 (Ω)) , −1 where the constant C is independent of t¯. Since M Bτ∗ Π and RA M Bn∗ Π are bounded operators from Vn0 (Ω) into V0 (Γ) (see Remark 4.4 and Lemma 2.5), we also have −1 kM Bτ∗ ΠzkL2 (t¯,∞;V0 (Γ)) + kRA M Bn∗ ΠzkL2 (t¯,∞;V0 (Γ)) ≤ C |y0 |Vn0 (Ω) + kf kL2/(1+2ε) (t¯,∞;Vn0 (Ω)) .
20
−1 The pair (z, −M Bτ∗ Πz − RA M Bn∗ Πz) is admissible for (Pt¯,y0 ,f ) and we have
−1 I(t¯, z, −M Bτ∗ Πz − RA M Bn∗ Πz) ≤ C |y0 |2Vn0 (Ω) + kf k2L2/(1+2ε) (t¯,∞;V0 (Ω)) . n
Therefore by classical arguments we can prove that (Pt¯,y0 ,f ) admits a unique solution (yt¯,y0 ,f , ut¯,y0 ,f ) and that the optimal cost ϕ(t¯, y0 , f ) = J(t¯, yt¯,y0 ,f , ut¯,y0 ,f ) obeys (5.2). Lemma 5.2. For all u ∈ L2 (0, ∞; V0 (Γ)), y ∈ L2 (0, ∞; Vn0 (Ω)), the equation −1 −Φ0 = A∗Π Φ − (M Bτ∗ Π)∗ γτ u − (RA M Bn∗ Π)∗ γn u + y
in (0, ∞),
Φ(∞) = 0,
(5.3)
admits a unique solution in L2 (0, ∞; Vn0 (Ω)) and kΦkL2 (0,∞;Vn0 (Ω)) + kΦkL∞ (0,∞;Vn0 (Ω)) ≤ C(kykL2 (0,∞;Vn0 (Ω)) + kukL2 (0,∞;V0 (Γ)) ). −1 Proof. We already know that (M Bτ∗ Π)∗ ∈ L(V0 (Γ), Vn0 (Ω)) and (RA M Bn∗ Π)∗ ∈ L(V0 (Γ), ∗ Vn0 (Ω)). Thus from the exponential stability of the semigroup (etAΠ )t≥0 it yields Z ∞ e−ω(τ −t) (|y(τ )|Vn0 (Ω) + |u(τ )|V0 (Γ) )dτ. |Φ(t)|Vn0 (Ω) ≤ C t
The estimates of the lemma follows from Young’s inequality for convolutions. The uniqueness of solution is obvious. ˆ ) the solution to y, u Lemma 5.3. Let f be in L2/(1+2ε) (0, ∞; Vn0 (Ω)), y0 be in Vn0 (Ω), denote by (ˆ ˆ be the solution to equation (5.3) corresponding to (ˆ ˆ is also ˆ ). Then (ˆ problem (P0,y0 ,f ), and let Φ y, u y, Φ) 2 0 2 0 solution in L (0, ∞; Vn (Ω)) × L (0, ∞; Vn (Ω)) to the system −1 M Bn∗ Φ + f y0 = Ay − Bτ M 2 Bτ∗ Φ − Bn M RA
−Φ0 = A∗ Φ + y
in (0, ∞),
in (0, ∞),
y(0) = y0 ,
(5.4)
Φ(∞) = 0.
The following estimate holds ˆ V2,1 (Q ) ≤ C |y0 |V0 (Ω) + kf kL2/(1+2ε) (0,∞;V0 (Ω)) . kˆ ykL2 (0,∞;Vn0 (Ω)) + kΦk ∞ n n Proof. Step 1. Due to estimate (5.2), we have kˆ ykL2 (0,∞;Vn0 (Ω)) + kˆ ukL2 (0,∞;V0 (Γ)) ≤ C |y0 |Vn0 (Ω) + kf kL2/(1+2ε) (0,∞;Vn0 (Ω)) . Consider the problem k (P0,y ) 0 ,f
n o inf Ik (0, y, u) | (y, u) satisfies (5.5), u ∈ V0,0 (Σ0,k ) ,
where Ik (0, y, u) =
1 2
Z
k
0
Z
|y|2 dxdt +
Ω
1 2
k
Z
0
1/2 |γτ u(t)|2V0 (Γ) + |RA γn u(t)|2V0 (Γ) dt.
and y0 = Ay + BM u + f
in (0, k),
y(0) = y0 .
(5.5)
k Problem (P0,y ) admits a unique solution (yk , uk ) characterized by 0 ,f
yk0 = Ayk + BM uk + f −Φ0k
∗
= A Φk + yk
γτ uk = −M Bτ∗ Φk ,
in (0, k),
in (0, k),
yk (0) = y0 , Φk (k) = 0,
−1 γn uk = −RA M Bn∗ Φk .
21
(5.6)
Since we have Z
k
Z
k
Z
2
|yk | dxdt + 0
Z
Ω ∞
0
Z
Z
2
≤
1/2 |γτ uk (t)|2V0 (Γ) + |RA γn uk (t)|2V0 (Γ) dt
∞
|ˆ y| dxdt + 0
Ω
0
1/2
ˆ (t)|2V0 (Γ) + |RA γn u ˆ (t)|2V0 (Γ) ) dt, |γτ u
as in the proof of Theorem 4.1, we can show that ˜k → u ˆ u
in L2 (0, ∞; V0 (Γ))
in L2 (0, ∞; V0 (Ω)),
˜k → y ˆ and y
(5.7)
˜ k and y ˜ k denote the extensions by zero of uk and yk to (k, ∞). where u ˜ k be the extension by zero of Φk to (k, ∞). We have Step 2. Passage to the limit for Φk . Let Φ ˜ 0 = A∗ Φ ˜k + y ˜k −Φ k
in (0, ∞),
˜ k (∞) = 0 . Φ
−1 We already know that (M Bτ∗ Π)∗ ∈ L(V0 (Γ), Vn0 (Ω)) and (RA M Bn∗ Π)∗ ∈ L(V0 (Γ), Vn0 (Ω)). We can rewrite the above equation in the form −1 ∗ ∗ ˜ 0 = A∗ Φ ˜ ˜ k − (RA ˜k + y ˜k −Φ M Bn∗ Π)∗ γn u k Π k − (M Bτ Π) γτ u
˜ k (∞) = 0. Φ
in (0, ∞),
(5.8)
Due to (5.7) and to Lemma 5.2, we can claim that ˜ k − Φk ˆ L∞ (0,∞;V0 (Ω)) + kΦ ˜ k − Φk ˆ L2 (0,∞;V0 (Ω)) −→ 0 kΦ n n
when k −→ ∞,
ˆ is the solution of equation (5.3) corresponding to (ˆ ˆ ). Notice that where Φ y, u ˆ L2 (0,∞;V0 (Ω)) ≤ C kˆ kΦk ukL2 (0,∞;V0 (Γ)) + kˆ ykL2 (0,∞;Vn0 (Ω)) n ≤ C |y0 |Vn0 (Ω) + kf kL2/(1+2ε) (0,∞;Vn0 (Ω)) .
(5.9)
By passing to the limit in the equation Z ∞ ∗ ˜ ˜ k (τ )) dτ, Φk (t) = e(A −λ0 I)(τ −t) (˜ yk (τ ) + λ0 Φ t
˜ k )k converges to Φ ˆ in V2,1 (Q∞ ), (B ∗ Φ ˜ k )k converges to with Lemmas 8.6 and 8.9, we can show that (Φ ∗ˆ 2 0 ˆ satisfies B Φ in L (0, ∞; V (Γ)), and Φ ˆ Φ(t) =
Z
∞
e(A
∗
−λ0 I)(τ −t)
ˆ )) dτ. (ˆ y(τ ) + λ0 Φ(τ
t
ˆ satisfies the second equation in (5.4) corresponding to y ˜k ˆ . Since (˜ Thus Φ uk )k = (−M Bτ∗ Φ −1 −1 ∗˜ ∗ ∗ ˆ ˆ ˆ ˆ , we have u ˆ = −M Bτ Φ − RA M Bn Φ, and (ˆ − RA M Bn Φk )k converges to u y, Φ) obeys the system ˆ follows from Lemma 8.5 and from (5.9). (5.4). The estimate for Φ Theorem 5.4. Assume that f ∈ L2/(1+2ε) (0, ∞; Vn0 (Ω)) ∩ L2 (0, ∞; (V2ε (Ω))0 ). If y0 ∈ Vn0 (Ω), then the solution (y, Φ) of system (5.4) belongs to V1,1/2 (Q) × V3,3/2 (Q) and we have: kykCb ([0,∞);Vn0 (Ω)) + kykV1,1/2 (Q∞ ) + kΦkV3,3/2 (Q∞ ) ≤ C(|y0 |Vn0 (Ω) + kf kL2/(1+2ε) (0,∞;Vn0 (Ω)) + kf kL2 (0,∞;(V2ε (Ω))0 ) ) . 1/2−ε
If y0 ∈ Vn (Ω) for some 0 < ε < 1/2, then the solution (y, Φ) of system (4.4) belongs to V3/2−ε,3/4−ε/2 (Q) × V7/2−ε,7/4−ε/2 (Q), we have: kykV3/2−ε,3/4−ε/2 (Q) + kΦkV7/2−ε,7/4−ε/2 (Q) ≤ C(|y0 |V1/2−ε (Ω) + kf kL2/(1+2ε) (0,∞;Vn0 (Ω)) + kf kL2 (0,∞;(V2ε (Ω))0 ) ) , n
22
and kB ∗ ΦkV3/2−ε,3/4−ε/2 (Q) ≤ C(|y0 |V1/2−ε (Ω) + kf kL2/(1+2ε) (0,∞;Vn0 (Ω)) + kf kL2 (0,∞;(V2ε (Ω))0 ) ) . n
Proof. Assume that y0 ∈ Vn0 (Ω). Applying Lemmas 8.1, 8.2, 8.3, we first obtain that y belongs 0 0 0 0 to V1/2−ε ,1/4−ε /2 (Q∞ ) for all ε0 > 0. From Lemma 8.9, we deduce that B ∗ Φ ∈ V1/2−ε ,1/4−ε /2 (Σ∞ ) 0 for all ε > 0. As in the proof of Lemma 4.2, we can use a bootstrap argument to show that y ∈ V1,1/2 (Q∞ ) ∩ Cb ([0, ∞); Vn0 (Ω)) and Φ ∈ V3,3/2 (Q∞ ), with the corresponding estimates for y and Φ. 1/2−ε The same procedure can be used to prove the other estimates in the case when y0 ∈ Vn (Ω). 6. Stabilization of the two dimensional Navier-Stokes equations. Throughout this section, we assume that N = 2. 6.1. First stabilization result. Consider the Navier-Stokes equations with the linear feedback law determined in section 4: P y0 = AΠ P y + P F (y) (I − P )y = −(I −
in (0, ∞),
P y(0) = y0 ,
−1 P )DA M RA M Bn∗ ΠP y
(6.1)
in (0, ∞),
where F (y) = −(y · ∇)y. Theorem 6.1. For all 0 < ε < 1/4, There exists µ0 > 0 and a nondecreasing function η from R+ into itself, such that if µ ∈ (0, µ0 ) and |y0 |V1/2−ε (Ω) ≤ η(µ), then equation (6.1) admits a unique solution n in the set n o Dµ = y ∈ V3/2−ε,3/4−ε/2 (Q∞ ) | kykV3/2−ε,3/4−ε/2 (Q∞ ) ≤ µ . Moreover (I − P )y belongs to H 3/4−ε/2 (0, ∞; V1/2 (Ω)) ∩ L2 (0, ∞; V3/2−ε (Ω)). 1/2−ε (Ω) and all f ∈ L2/(1+2ε) (0, ∞; Vn0 (Ω)) ∩ L2 (0, ∞; (V2ε (Ω))0 ) the Lemma 6.2. For all y0 ∈ Vn solution to the equation y0 = AΠ y + f ,
y(0) = y0 ,
(6.2)
obeys kykV3/2−ε,3/4−ε/2 (Q∞ ) ≤ C1 |y0 |V1/2−ε (Ω) + kf kL2/(1+2ε) (0,∞;Vn0 (Ω)) + kf kL2 (0,∞;(V2ε (Ω))0 ) . n
Proof. Since the semigroup (etAΠ )t≥0 is exponentially stable on Vn0 (Ω), equation (6.2) admits a unique solution y belonging to Cb ([0, ∞); Vn0 (Ω)) ∩ L2 (0, ∞; Vn0 (Ω)) which satisfies kykL∞ (0,∞;Vn0 (Ω)) + kykL2 (0,∞;Vn0 (Ω)) ≤ C |y0 |V1/2−ε (Ω) + kf kL2/(1+2ε) (0,∞;Vn0 (Ω)) . n
ˆ the solution to system (5.4). We set We denote by (ˆ y, Φ) ˆ r(t) = Φ(t) − Πy(t) . ˜ the solution to the equation We denote by y −1 ˜ 0 = A˜ y y + Bτ M 2 Bτ∗ r + Bn M RA M Bn∗ r,
˜ (0) = 0 . y
From the definition of r it follows that ˆ − Bn M R−1 M B ∗ Φ ˆ −Bτ M 2 Bτ∗ Φ n A −1 −1 = −Bτ M 2 Bτ∗ Πy − Bn M RA M Bn∗ Πy − Bτ M 2 Bτ∗ r − Bn M RA M Bn∗ r .
Thanks to this identity, we can rewrite equation (6.2) in the form ˆ − Bn M R−1 M B ∗ Φ ˆ + Bτ M 2 B ∗ r + Bn M R−1 M B ∗ r + f , y0 = Ay − Bτ M 2 Bτ∗ Φ n τ n A A 23
y(0) = y0 .
ˆ+y ˜ . Due to Theorem 5.4, we notice that Thus y = y ˆ V3/2−ε,3/4−ε/2 (Σ ) ≤ C |y0 | 1/2−ε kB ∗ Φk + kf kL2/(1+2ε) (0,∞;Vn0 (Ω)) + kf kL2 (0,∞;(V2ε (Ω))0 ) . ∞ V (Ω) n
Moreover y belongs to L2 (0, ∞; Vn0 (Ω)) and B ∗ Π ∈ L(Vn0 (Ω), V1/2 (Γ)) (see Remark (4.4)). Thus B ∗ r = ˆ − B ∗ Πy belongs to V0,0 (Σ∞ ) and B∗Φ kB ∗ rkV0,0 (Σ∞ ) ≤ C |y0 |V1/2−ε (Ω) + kf kL2/(1+2ε) (0,∞;Vn0 (Ω)) + kf kL2 (0,∞;(V2ε (Ω))0 ) . n
From Lemma 8.3 it follows that k˜ ykV1/2−ε0 ,1/4−ε0 (Q∞ ) ≤ CkB ∗ rkV0,0 (Σ∞ ) 0
for all ε0 > 0.
0
ˆ + y ˜ belongs to V1/2−ε ,1/4−ε /2 (Q∞ ) for all ε0 > 0. Therefore B ∗ r belongs to Thus y = y 0 0 0 1/2−ε0 V1/2−ε ,1/4−ε /2 (Σ∞ ) because B ∗ Π ∈ L(Vn (Ω), V1−ε (Γ)) for all ε0 > 0 (see Remark (4.4)), and we have kB ∗ rkV1/2−ε0 ,1/4−ε0 /2 (Σ∞ ) ≤ C |y0 |V1/2−ε (Ω) + kf kL2/(1+2ε) (0,∞;Vn0 (Ω)) + kf kL2 (0,∞;(V2ε (Ω))0 ) ∀ε0 > 0. n
0
0
With another iteration we can prove an estimate of B ∗ r in V1−ε ,1/2−ε /2 (Σ∞ ). Still with Lemma 8.3 we obtain k˜ ykV3/2−ε00 ,3/4−ε00 (Q∞ ) ≤ CkB ∗ rkV1−ε0 ,1/2−ε0 /2 (Σ) , for all ε00 > ε0 > 0. With Theorem 5.4, we have kˆ ykV3/2−ε,3/4−ε/2 (Q∞ ) ≤ C |y0 |V1/2−ε (Ω) + kf kL2/(1+2ε) (0,∞;Vn0 (Ω)) + kf kL2 (0,∞;(V2ε (Ω))0 , n
and the proof is complete. Lemma 6.3. If z belongs to V3/2−ε,3/4−ε/2 (Q) then (1−2ε)
kzkL1/ε (0,∞;V1/2+ε (Ω)) ≤ CkzkL∞ (0,∞;V1/2−ε (Ω)) kzk2ε L2 (0,∞;V3/2−ε (Ω)) ≤ CkzkV3/2−ε,3/4−ε/2 (Q) . Proof. We first observe that (1−2ε)
|z(t)|V1/2+ε (Ω) ≤ C|z(t)|V1/2−ε (Ω) |z(t)|2ε V3/2−ε (Ω) . Therefore we have Z 0
∞
1/ε
(1−2ε)/ε
|z(t)|V1/2+ε (Ω) dt ≤ CkzkL∞ (0,∞;V1/2−ε (Ω))
Z 0
∞
|z(t)|2V3/2−ε (Ω) dt.
The proof is complete. Lemma 6.4. Let ε be in (0, 1/4). If z belongs to V3/2−ε,3/4−ε/2 (Q) then k(z · ∇)zkL2/(1+2ε) (0,∞;L2 (Ω)) + kP div(z ⊗ z)kL2 (0,∞;(V2ε (Ω))0 ) ≤ C2 kzk2V3/2−ε,3/4−ε/2 (Q∞ ) . Proof. We have k(z · ∇)zkL2 (Ω) ≤ kzkL4/(1−2ε) (Ω) k∇zkL4/(1+2ε) (Ω) ≤ CkzkV1/2+ε (Ω) kzkV3/2−ε (Ω) . Therefore with Lemma 6.3, it follows that k(z · ∇)zkL2/(1+2ε) (0,∞;L2 (Ω)) ≤ CkzkL1/ε (0,∞;V1/2+ε (Ω)) kzkL2 (0,∞;V3/2−ε (Ω)) 24
≤ Ckzk2V3/2−ε,3/4−ε/2 (Q) . From [20, Theorem B.3] we deduce that kz ⊗ zkH1−2ε,1/2−ε (Q∞ ) ≤ Ckzk2V3/2−ε,3/4−ε/2 (Q∞ ) . The divergence operator is continuous from L2 (Ω) into H−1 (Ω), and from H1 (Ω) to L2 (Ω). Thus it is also continuous from H1−2ε (Ω) into (H2ε (Ω))0 . The projector P is continuous from L2 (Ω) into Vn0 (Ω), and from H−1 (Ω) into V−1 (Ω) (see e.g. [36]). Thus it is also continuous from (H2ε (Ω))0 into (Vn2ε (Ω))0 . Therefore we have kP div(z ⊗ z)kL2 (0,∞;(Vn2ε (Ω))0 ) ≤ Ckz ⊗ zkL2 (0,∞;H1−2ε (Ω)) ≤ Ckzk2V3/2−ε,3/4−ε/2 (Q∞ ) . Observe that the condition 0 < ε < 1/4 is needed to have (H2ε (Ω))0 ,→ H−1 (Ω). Lemma 6.5. Let ε be in (0, 1/4). The mapping P F is locally Lispchitz continuous from V3/2−ε,3/4−ε/2 (Q∞ ) into L2/(1+2ε) (0, ∞; Vn0 (Ω)) ∩ L2 (0, ∞; (V2ε (Ω))0 ). More precisely we have kP F (z1 ) − P F (z2 )kL2/(1+2ε) (0,∞;Vn0 (Ω)) + kP F (z1 ) − P F (z2 )kL2 (0,∞;(V2ε (Ω))0 ) ≤ C2 kz1 kV3/2−ε,3/4−ε/2 (Q∞ ) + kz2 kV3/2−ε,3/4−ε/2 (Q∞ ) kz1 − z2 kV3/2−ε,3/4−ε/2 (Q∞ ) . Proof. From Lemma 6.4 it follows that P F is a mapping from V3/2−ε,3/4−ε/2 (Q∞ ) into L (0, ∞; Vn0 (Ω)) ∩ L2 (0, ∞; (V2ε (Ω))0 ). By calculations similar as those in Lemma 6.4, we obtain: 2/(1+2ε)
kP F (z1 ) − P F (z2 )kL2/(1+2ε) (0,∞;Vn0 (Ω)) + kP F (z1 ) − P F (z2 )kL2 (0,∞;(V2ε (Ω))0 ) ≤ kP ((z1 · ∇)(z1 − z2 ))kL2/(1+2ε) (0,∞;Vn0 (Ω)) + kP (((z1 − z2 ) · ∇)z2 )kL2/(1+2ε) (0,∞;Vn0 (Ω)) +kP div(z1 ⊗ (z1 − z2 ))kL2 (0,∞;(Vn2ε (Ω))0 ) + kP div((z1 − z2 ) ⊗ z2 )kL2 (0,∞;(Vn2ε (Ω))0 ) ≤ C2 kz1 kV3/2−ε,3/4−ε/2 (Q∞ ) + kz2 kV3/2−ε,3/4−ε/2 (Q∞ ) kz1 − z2 kV3/2−ε,3/4−ε/2 (Q∞ ) . The proof is complete. Lemma 6.6. If P y belongs to V3/2−ε,3/4−ε/2 (Q∞ ) for some 0 < ε < 1/2, then −1 k(I − P )DA M RA M Bn∗ ΠP ykV3/2−ε,3/4−ε/2 (Q∞ ) ≤ C3 kP ykV3/2−ε,3/4−ε/2 (Q∞ ) .
1/2−ε
(Ω), V1−ε (Γ)) (see Remark 4.4). Thus Proof. The operator B ∗ Π belongs to L(Vn 1/2−ε 1/2−ε −1 −1 1−ε ∗ ∗ (Ω), (Ω), V (Γ)), and DA M RA M Bn Π belongs to L(Vn M RA M Bn Π belongs to L(Vn V3/2−ε (Ω)). The lemma follows from the continuity properties of the operator (I − P ). Proof. Proof of Theorem 6.1. We set µ0 = V
3/2−ε,3/4−ε/2
1 4C1 C2 (1+C3 )
and η(µ) =
3µ 4C1 (1+C3 ) .
For z ∈
(Ω)), we denote by yz the solution to the equation P y0 = AΠ P y + P F (z),
y(0) = y0 ,
(6.3)
−1 (I − P )y = −(I − P )DA M RA M Bn∗ ΠP y .
We are going to prove that the mapping M : z 7→ yz is a contraction in Dµ . (i) From Lemma 6.2, Lemma 6.4, and Lemma 6.6 it follows that kyz kV3/2−ε,3/4−ε/2 (Q∞ ) ≤ kP yz kV3/2−ε,3/4−ε/2 (Q∞ ) + k(I − P )yz kV3/2−ε,3/4−ε/2 (Q∞ ) ≤ C1 (1 + C3 ) |y0 |V1/2−ε (Ω) + kP F (z)kL2/(1+2ε) (0,∞;Vn0 (Ω)) + kP F (z)kL2 (0,∞;(V2ε (Ω))0 ) 3µ 2 + C2 kzk2V3/2−ε,3/4−ε/2 (Q∞ ) ≤ 3µ ≤ C1 (1 + C3 ) 4C1 (1+C 4 + C1 (1 + C3 )C2 µ ≤ µ, 3) if µ < µ0 . Thus M is a mapping from Dµ into itself. 25
(ii) From Lemma 6.2 and Lemma 6.5 it follows that kP yz1 − P yz2 kV3/2−ε,3/4−ε/2 (Q∞ ) ≤ C1 kP F (z1 ) − P F (z2 )kL2/(1+2ε) (0,∞;Vn0 (Ω)) + kP F (z1 ) − P F (z2 )kL2 (0,∞;(V2ε (Ω))0 ) ≤ C2 C1 kz1 kV3/2−ε,3/4−ε/2 (Q∞ ) + kz2 kV3/2−ε,3/4−ε/2 (Q∞ ) kz1 − z2 kV3/2−ε,3/4−ε/2 (Q∞ )
≤ 2C2 C1 µkz1 − z2 kV3/2−ε,3/4−ε/2 (Q∞ ) . Thus with Lemma 6.6 and the previous estimate we obtain kyz1 − yz2 kV3/2−ε,3/4−ε/2 (Q∞ ) ≤ 2C2 C1 µ(1 + C3 )kz1 − z2 kV3/2−ε,3/4−ε/2 (Q∞ ) . Thus if µ < µ0 , the mapping M is a contraction in Dµ , and the system (6.1) admits a unique solution in Dµ . 6.2. Second stabilization result. To obtain a feedback law providing an exponential stabilization of the Navier-Stokes, we are going to use the linear feedback law determined thanks to an auxiliary problem. For that, we set ˆ = eωt y, y
ˆ = eωt u. u
If P y0 = AP y + P F (y) + BM u,
P y(0) = y0 ,
(I − P )y = (I − P )DA γn M u, ˆ is the solution to the system then y ˆ 0 = AP y ˆ + ωP y ˆ + e−ωt P F (ˆ ˆ, Py y) + BM u
ˆ (0) = y0 , Py
ˆ. (I − P )ˆ y = (I − P )DA γn M u Set Aω = A + ωI, and let Πω ∈ L(Vn0 (Ω)) be the solution to the algebraic Riccati equation: Πω = Π∗ω ≥ 0,
−1 Πω Aω + A∗ω Πω − Πω Bτ M 2 Bτ∗ Πω − Πω Bn M RA M Bn∗ Πω + I = 0.
(6.4)
The existence of a unique solution in L(Vn0 (Ω); V2 (Ω) ∩ V01 (Ω)) to this equation may be proved as in section 4. As in section 4, it can be shown that 1 I(zy0 , vy0 ) = Πω y0 , y0 0 , 2 Vn (Ω) where (zy0 , vy0 ) is the solution of the control problem n o (P0,y0 ) inf I(z, v) | (z, v) satisfies (6.5), v ∈ V0,0 (Σ∞ ) , where I is the functional of section 4, and z0 = Aω z + BM v
in (0, ∞),
z(0) = y0 .
(6.5)
Consider the Navier-Stokes equations with the linear feedback law: ˆ 0 = Aω,Πω P y ˆ + e−ωt P F (ˆ Py y),
ˆ (0) = y0 , Py
−1 ˆ, (I − P )ˆ y = −(I − P )DA M RA M Bn∗ Πω P y
where D(Aω,Πω ) = D(AΠω )
and Aω,Πω y = AΠω y + ωy for all y ∈ D(AΠω ). 26
(6.6)
ˆ is a solution to (6.6), then y = e−ωt y ˆ is the solution of As previously, if y −1 P y0 = AP y − Bτ M 2 Bτ∗ Πω P y − Bn M RA M Bn∗ Πω P y + P F (y),
(I − P )y = −(I −
P y(0) = y0 ,
−1 P )DA RA M Bn∗ Πω P y.
(6.7)
Theorem 6.7. For all 0 < ε < 1/4, there exists µ0 > 0 and a nondecreasing function η0 from R+ into itself, such that if µ ∈ (0, µ0 ) and ky0 kV1/2−ε (Ω) ≤ η0 (µ), the equation (6.7) admits a unique solution n in the set n o Dµ = y ∈ V3/2−ε,3/4−ε/2 (Q∞ ) | keω (·) ykV3/2−ε,3/4−ε/2 (Q∞ ) ≤ µ . Moreover y, which belongs to Cb ([0, ∞); V1/2−ε (Ω)), satisfies |y(t)|V1/2−ε (Ω) ≤ C(w, µ, ω) e−ωt . Proof. Substituting F (y) by e−ωt F (ˆ y) in the proof of Theorem 6.1, we can show that there exists µ0 > 0 and a nondecreasing function η0 from R+ into itself, such that if µ ∈ (0, µ0 ) and ky0 kV1/2−ε (Ω) ≤ η0 (µ) n n b µ = y ∈ V3/2−ε,3/4−ε/2 (Q∞ ) | ˆ in D for some 0 < ε < 1/4, the equation (6.6) admits a unique solution y o ˆ is the solution of equation kykV3/2−ε,3/4−ε/2 (Q∞ ) ≤ µ , and |ˆ y(t)|V1/2−ε (Ω) ≤ C(w, µ, ω). Thus y = e−ωt y (6.7), it belongs to Dµ and it satisfies obeys |y(t)|V1/2−ε (Ω) ≤ C(w, µ, ω) e−ωt . The proof is complete. 7. Additional results. 7.1. Control by a tangential velocity. Assume that Ω and w are such that (1.3) be stabilizable 1/2−ε (Ω) with controls u ∈ L2 (0, ∞; Vn0 (Γ)) satisfying supp(u) ⊂ Γ0 × (0, ∞), where Γ0 is an open in Vn set in Γ in which the function m is equal to 1 (see the assumptions on the function m at the beginning of section 2). Such a stabilizability result is stated in [4, Theorem B.1.1]. With such a result, as in section 6.2 (and section 4), we can show that the algebraic Riccati equation Πω ∈ L(Vn0 (Ω)), Πω = Π∗ω ≥ 0,
Πω Aω + A∗ω Πω − Πω Bτ M 2 Bτ∗ Πω + I = 0,
L(Vn0 (Ω); V2 (Ω)
(7.1)
V01 (Ω)),
admits a unique solution satisfying Πω ∈ ∩ and which enjoys the same regularizing properties as in Theorem 4.5. Since we deal with controls belonging to Vn0 (Γ), the operator RA is now the identity. As in Theorem 6.7 we can prove that, for all 0 < ε < 1/4, there exists µ0 > 0 and a nondecreasing function η0 from R+ into itself, such that if µ ∈ (0, µ0 ) and ky0 kV1/2−ε (Ω) ≤ η0 (µ), the n equation y0 = Ay − Bτ M 2 Bτ∗ Πω y + P F (y),
y(0) = y0 ,
admits a unique solution in the set n o Dµ = y ∈ V3/2−ε,3/4−ε/2 (Q∞ ) ∩ L2 (0, ∞; Vn0 (Ω)) | keω (·) ykV3/2−ε,3/4−ε/2 (Q∞ ) ≤ µ . 1/2−ε
Moreover y, which belongs to Cb ([0, ∞); Vn
(Ω)), satisfies
|y(t)|V1/2−ε (Ω) ≤ C(w, µ, ω) e−ωt . n
Notice that the local feedback stabilization of the two dimensional Navier-Stokes equations is not studied in [4] (only the stabilization of the linearized Navier-Stokes equations is studied in two dimension in [4, Appendix B]). If we transpose the results obtained in [4] in the 3D case to the 2D case, the results are completely different from the ones we obtain. Indeed we here prove a local stabilization result for the Navier-Stokes equations thanks to the solution of a classical Riccati equation (equation (7.1)), while the Riccati equation obtained in [4, Section 4] is only defined in the domain of the square of the feedback operator, which is unknown. 27
7.2. Dependence of solutions with respect to ν for the stabilization of the Stokes equation. The dependence of solutions to our control problems with respect to ν is very complicated because the stationary solution w involved in the Oseen operator also depends on ν in a complicated way. There is one particular situation for which we can clarify this dependence, it is the one corresponding to w = 0. Assume that the semigroup generated by (νP ∆ + ωI, V2 (Ω) ∩ V01 (Ω)) on Vn0 (Ω) be unstable for a given ω > 0. What has been done in section 6.2 is still valid if w ≡ 0. Let Πω be the solution to equation (6.4) ˆ to the closed loop system corresponding to A = νA0 and Aω = νA0 + ωI. Consider the solution y −1 ˆ 0 = AP y ˆ + ωP y ˆ − B M RA ˆ, Py M B ∗ Πω P y
ˆ (0) = y0 , Py
−1 ˆ. (I − P )ˆ y = −(I − P )DA RA M Bn∗ Πω P y
From the previous sections we know that −1 ˆ) = I(ˆ y, −M RA M B ∗ Πω P y
1 Πω y0 , y0 0 , 2 Vn (Ω)
and |ˆ y(t)|V1/2−ε (Ω) ≤ C(ε, ω)e−σt |y0 |V1/2−ε (Ω) for all 0 < ε ≤ 1/2, where C(ε, ω) > 0 depends on ε and ω, and σ is positive, but not precisely known. ˆ , we can check that y is the solution to the closed loop system Setting y = e−ωt y −1 P y0 = AP y − B M RA M B ∗ Πω P y,
(I − P )y = −(I −
P y(0) = y0 ,
−1 P )DA RA M Bn∗ Πω P y,
(7.2)
and y obeys the decay rate |y(t)|V1/2−ε (Ω) ≤ C(ε)e−(ω+σ)t |y0 |V1/2−ε (Ω) ≤ C(ε)e−ωt |y0 |V1/2−ε (Ω) . Thus we have solved the boundary feedback stabilization problem of the Stokes equations with a prescribed exponential decay rate ω. Now we would like to clarify the dependence of the solution y of equation (7.2) with respect to ω. We have to clarify the dependence of Πω with respect to ν. For that we first have to clarify the dependence of RA , B, and B ∗ with respect to ν. We can choose λ0 = 0 in the definition of DA . We denote by DA0 u the solution y to the equation −∆y + ∇q = 0 and div y = 0 in Ω,
y = u on Γ,
∗ ∗ ∗ (I − P )DA M + I = and we have DA = ν1 DA0 u because λ0 = 0. Thus DA = ν1 DA u, and RA = M DA 0 1 ∗ ∗ ν 2 M DA0 (I − P )DA0 M + I. Since B = (−A)P DA = (−A0 )P DA0 , we observe that B and B do not depend on ν. Let ν1 and ν2 be two viscosity coefficients such that ν1 > ν2 . Assume that the semigroup generated by (ν1 P ∆ + ωI, V2 (Ω) ∩ V01 (Ω)) on Vn0 (Ω) is unstable for a given ω > 0. For i = 1, 2, let Πνi ,ω be the solution to equation (6.4) corresponding to Aω = νi A0 + ωI. With this notation we can observe that Πν1 ,ω is also the solution to the equation
Πν1 ,ω Aν2 ,ω + A∗ν2 ,ω Πν1 ,ω − νν12 Πν1 ,ω BM
1 ∗ M DA (I ν12 0
− P )DA0 M + I
−1
M B ∗ Πν1 ,ω +
ν2 ν1 I
+ ωI( νν12 − 1) = 0.
Since ν1 > ν2 , we can verify that ν2 ν1 BM
= BM ≥ BM
−1 1 ∗ M DA (I − P )DA0 M + I M B∗ ν12 0 −1 1 ∗ M B∗ ν1 ν2 M DA0 (I − P )DA0 M + I 1 ∗ −1 ( ν 2 M DA0 (I − P )DA0 M + I) M B ∗ 2
28
(where the inequality ≤ is the inequality between quadratic forms), and ν2 ν2 I + ωI( − 1) ≤ I. ν1 ν1 Thus, as in [7], we can claim that Πν1 ,ω ≤ Πν2 ,ω . If yνi is the solution to the closed loop system (7.2) for A = νi A0 , we have 1 ∗ Π y , y I(e−ωt yν1 , −e−ωt M Rν−1 M B Π P y ) = ν ,ω 0 0 ν ,ω ν 1 1 1 2 1 A0 1 ∗ ≤ I(e−ωt yν2 , −e−ωt M Rν−1 Π y , y , M B Π P y ) = ν ,ω 0 0 ν ,ω ν 2 2 2 2 2 A0 which provides a comparison between the L2 of the solution to the stabilization problems in function of ν. 8. Appendix. In this section we prove some regularity results for the state and the adjoint equations. 1/2−ε (Ω) with 0 < ε ≤ 1/2, the weak solution to the equation Lemma 8.1. If y0 ∈ Vn y0 = (A − λ0 )y
in (0, ∞),
y(0) = y0 ,
obeys kykC
1/2−ε (Ω)) b ([0,∞);Vn
+ kykV3/2−ε,3/4−ε/2 (Q∞ ) ≤ C|y0 |V1/2−ε (Ω) . n
Proof. The exponential stability of the semigroup (et(A−λ0 I) )t≥0 gives a bound for y in L2 (0, ∞; 1/2−ε (Ω)), the other estimates may be obtained by using Using the estimate in L2 (0, ∞; Vn proofs in [5] and interpolation results. Lemma 8.2. If f ∈ L2/(1+2ε) (0, ∞; Vn0 (Ω)), the solution y to the equation 1/2−ε (Ω)). Vn
y0 = (A − λ0 I)y + f
in (0, ∞),
y(0) = 0,
obeys kykL2 (0,∞;V2−2ε (Ω)) ≤ Ckf kL2/(1+2ε) (0,∞;Vn0 (Ω)) . If 0 < ε < 1/4 and f ∈ L2 (0, ∞; (Vn2ε (Ω))0 ), then kykV2−2ε,1−ε (Q∞ ) ≤ Ckf kL2 (0,∞;(Vn2ε (Ω))0 ) . Proof. From the identity (−A + λ0 I)1−ε y(t) =
t
Z
(−A + λ0 I)1−ε e(t−τ )(A−λ0 I) f (τ ) dτ,
0
it follows that Z ky(t)kV2−2ε (Ω) ≤ C 0
t
e−ω(t−τ ) (t − τ )−1+ε |f (τ )|Vn0 (Ω) dτ.
Thus, with the generalized Young inequality for convolutions (see [34, page 32]), we have: kykLr (0,∞;V2−2ε (Ω)) ≤ Ckf kL2/(1+2ε) (0,∞;Vn0 (Ω)) , for all 2/(1 + 2ε) ≤ r ≤ 2. 29
(8.1)
(ii) Now we assume that f belongs to L2 (0, ∞; (Vn2ε (Ω))0 ). To prove the estimate in that case we proceed by interpolation. We know that kykV2,1 (Q∞ ) ≤ Ckf kL2 (0,∞;Vn0 (Ω))
kykV1,1/2 (Q∞ ) ≤ Ckf kL2 (0,∞;V−1 (Ω)) .
and
Since we have: [L2 (0, ∞; Vn0 (Ω)), L2 (0, ∞; V−1 (Ω))]2ε = ([L2 (0, ∞; Vn0 (Ω)), L2 (0, ∞; V01 (Ω))]2ε )0 = L2 (0, ∞; (Vn2ε (Ω))0 ), we obtain the desired result by interpolation. (Notice that [Vn0 (Ω), V01 (Ω)]2ε = Vn2ε (Ω) because 0 < ε < 1/4.) Lemma 8.3. If u belongs to Vs,s/2 (Σ∞ ) with 0 ≤ s < 1, then the weak solution to the equation y0 = (A − λ0 )y + BM u
in (0, ∞),
y(0) = 0,
obeys kykV1/2+s−ε,1/4+s/2−ε/2 (Q∞ ) ≤ CkukVs,s/2 (Σ∞ )
for all ε > 0.
If u belongs to Vs,s/2 (Σ∞ ) with 1 < s ≤ 2, and if u(0) = 0, then kykV1/2+s−ε,1/4+s/2−ε/2 (Q∞ ) ≤ CkukVs,s/2 (Σ∞ )
for all ε > 0.
Proof. Similar estimates are proved in [30, Theorem 2.3] for a finite time interval. It is sufficient to rewrite the proof and to use the exponential stability of the semigroup (et(A−λ0 I) )t≥0 to obtain the desired results. Lemma 8.4. For all y ∈ Vn2 (Ω), the solution Φ ∈ V2 (Ω) ∩ V01 (Ω) to the stationary equation λ0 Φ − A∗ Φ = y obeys |Φ|V4 (Ω) ≤ C|y|Vn2 (Ω) Proof. We rewrite the equation in the form λ0 Φ − νP ∆Φ = y − P ((w · ∇)Φ) + P ((∇w)T Φ). Since w ∈ V3 (Ω) and Φ ∈ V2 (Ω), then P ((w · ∇)Φ) and P ((∇w)T Φ) belong to V1 (Ω), which gives an estimate of Φ in V3 (Ω). Knowing that Φ ∈ V3 (Ω), P ((w · ∇)Φ) and P ((∇w)T Φ) belong to V2 (Ω), and the proof is complete. Lemma 8.5. If the function y belongs to Vs,s/2 (Q∞ ) ∩ L2 (0, ∞; Vn0 (Ω)) with 0 ≤ s ≤ 2, then the solution Φ to the equation −Φ0 = (A∗ − λ0 I)Φ + y
in (0, ∞),
Φ(∞) = 0,
(8.2)
satisfies kΦkVs+2,s/2+1 (Q∞ ) ≤ CkykVs,s/2 (Q∞ ) .
(8.3)
Proof. (i) Let us first prove estimate (8.3) for s = 0. Assume that y ∈ L2 (0, ∞; Vn0 (Ω)). The existence of a unique solution Φ to equation (8.2) in L2 (0, ∞; Vn0 (Ω)) follows from the exponential stability of the ∗ semigroup (et(A −λ0 I) )t≥0 on Vn0 (Ω). Moreover we have kΦkL2 (0,∞;Vn0 (Ω)) ≤ CkykL2 (0,∞;Vn0 (Ω)) .
(8.4)
It is sufficient to prove estimate (8.3) for s = 0 and y ∈ Cc (0, ∞; Vn0 (Ω)). In that case, it can be deduced from (8.4) with the method of proof in [5, Theorem 1.1]. 30
(ii) Now let us assume that y belongs to V2,1 (Q∞ ) and let us prove estimate (8.3) for s = 2. We have Z ∞ ∗ ∗ (−A + λ0 I)Φ(t) = (−A∗ + λ0 I)e(τ −t)(A −λ0 I) y(τ ) dτ. t
Integrating by parts we obtain (−A∗ + λ0 I)Φ(t) = y(t) +
Z
∞
e(τ −t)(A
∗
−λ0 I) 0
y (τ ) dτ.
t
Since y0 belongs to L2 (0, ∞; Vn0 (Ω)), with estimate (8.3) for s = 0 we have
Z ∞
(τ −t)(A∗ −λ0 I) 0
≤ Cky0 kL2 (0,∞;Vn0 (Ω)) . e y (τ ) dτ
t
V2,1 (Q∞ )
Thus k(−A∗ + λ0 I)ΦkL2 (0,∞;V2 (Ω)) ≤ C(kykL2 (0,∞;V2 (Ω)) + ky0 kL2 (0,∞;Vn0 (Ω)) ) ≤ CkykV2,1 (Q∞ ) . From regularity results for the stationary Oseen equation (Lemma 8.4), it follows that kΦkL2 (0,∞;V4 (Ω)) ≤ Ck(−A∗ + λ0 I)ΦkL2 (0,∞;V2 (Ω)) . Thus kΦkL2 (0,∞;V4 (Ω)) ≤ CkykV2,1 (Q∞ ) . From equation (8.2) we deduce kΦ0 kL2 (0,∞;V2 (Ω)) ≤ C(k(−A∗ + λ0 I)ΦkL2 (0,∞;V2 (Ω)) + kykL2 (0,∞;V2 (Ω)) ) ≤ CkykV2,1 (Q∞ ) . The estimate (8.3) is obtained by interpolation between the estimates obtained for s = 0 and s = 2. Lemma 8.6. For all y ∈ L2 (0, T ; Vn0 (Ω)), the solution to the equation −Φ0 = A∗ Φ + y
in (0, T ),
Φ(T ) = 0,
(8.5)
satisfies kΦkV2,1 (QT ) ≤ CkykL2 (0,T ;Vn0 (Ω)) .
(8.6)
If the function y belongs to Vs,s/2 (QT ) with 0 ≤ s < 3/2, then the function Φ belongs to Vs+2,s/2+1 (QT ) and the following estimate holds: kΦkVs+2,s/2+1 (QT ) ≤ CkykVs,s/2 (QT ) .
(8.7)
Proof. The regularity result when y ∈ L2 (0, T ; Vn0 (Ω)) and estimate (8.6) follow from [5, Theorem 1.1]. If y belongs to V2,1 (QT ) and if y(T ) ∈ V01 (Ω), we have kΦ0 kL2 (0,T ;V2 (Ω)) + kΦkL2 (0,T ;V4 (Ω)) ≤ CkykV2,1 (QT ) , see e.g. [33]. The estimate for 0 ≤ s < 3/2 is obtained by interpolation between the estimates obtained for s = 0 and s = 2. For s < 3/2 the regularity condition y(T ) ∈ V01 (Ω) is not needed. Lemma 8.7. Let Φ be the solution to equation (8.2) and let ψ be the pressure associated with Φ, that is the function satisfying ∂Φ − ν∆Φ − (w · ∇)Φ + (∇Φ)T w + λ0 Φ + ∇ψ = y in Q∞ , ∂t div Φ = 0 in Q∞ , Φ = 0 on Σ∞ , Φ(∞) = 0 in Ω. −
31
(8.8)
If in (8.2) y belongs to L2 (0, ∞; Vn0 (Ω)), then the function ψ belongs to L2 (0, ∞; H 1 (Ω)). If in addition y belongs to Vs,s/2 (Q∞ ) with 0 ≤ s ≤ 2, then the function ψ belongs to L2 (0, ∞; H s+1 (Ω)) ∩ H s/2 (0, ∞; H 1 (Ω)). Proof. Assume that y belongs to L2 (0, ∞; Vns (Ω))∩H s/2 (0, ∞; Vn0 (Ω)) with 0 ≤ s ≤ 2. From Lemma 8.5 it follows that ∂Φ + ν∆Φ + (w · ∇)Φ − (∇Φ)T w ∈ Hs,s/2 (Q∞ ). y+ ∂t Thus ∇ψ belongs to Hs,s/2 (Q∞ ), and the proof is complete. Lemma 8.8. If in (8.5) y belongs to Vs,s/2 (QT ) with 0 ≤ s < 3/2, then the function ψ, the pressure associated with Φ, belongs to L2 (0, T ; H s+1 (Ω)) ∩ H s/2 (0, T ; H 1 (Ω)). Proof. The proof is similar to the previous one. Lemma 8.9. Let Φ ∈ V2,1 (Q) be the solution to equation (8.2), and set −1 u = −(M γτ + RA M γn )B ∗ Φ .
If in (8.2) the function y belongs to L2 (0, ∞; Vns (Ω)) ∩ H s/2 (0, ∞; Vn0 (Ω)) with 0 ≤ s ≤ 2, then kB ∗ ΦkL2 (0,∞;Vs+1/2 (Γ))∩H s/2 (0,∞;V1/2 (Γ)) + kukL2 (0,∞;Vs+1/2 (Γ))∩H s/2 (0,∞;V1/2 (Γ)) ≤ CkykVs,s/2 (Q∞ ) .
(8.9)
Proof. As in Lemma 3.5 we can show that ∂Φ −1 u = −(M γτ + RA M γn ) ν − ψ n + c(ψ)n , ∂n where ψ is the pressure associated with Φ and c(ψ) is the constant defined in (2.1). Since y belongs to L2 (0, ∞; Vns (Ω)) ∩ H s/2 (0, ∞; Vn0 (Ω)), from Lemma 8.5 we deduce that Φ belongs to Vs+2,s/2+1 (Q∞ ), and from Lemma 8.7 it follows that ψ belongs to L2 (0, ∞; H s+1 (Ω)) ∩ H s/2 (0, ∞; H 1 (Ω)). Thus u belongs to L2 (0, ∞; H s+1/2 (Γ)) ∩ H s/2 (0, ∞; H 1/2 (Γ)). Acknowledgements. The author thanks the referees for their valuable questions and suggestions that have contributed to improve the first version of the paper. After this paper was submitted, I. Lasiecka informed the author of her joint work with V. Barbu and R. Triggiani [4]. The author thanks I. Lasiecka for providing him a copy of this joint paper and for discussions in Oberwolfach. REFERENCES [1] [2] [3] [4] [5] [6] [7] [8] [9] [10] [11] [12] [13] [14]
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