Null Controllability of the Incompressible Stokes Equations in a 2-D Channel Using Normal Boundary Control

In this paper, we consider the Stokes equations in a two-dimen- sional channel with periodic conditions in the direction of the channel. We establish null controllability of this system using a boundary control which acts on the normal component of the velocity only. We show null controllability of the system, subject to a constraint of zero average, by proving an observability inequality with the help of a Muntz-Szasz Theorem.

Definition 1.1. The system (1.2) is null controllable in a Hilbert space Z at time T > 0, if for any initial condition U 0 ∈ Z, there exists a control ψ such that the solution U of (1.2) with control ψ hits 0 at time T , i.e. U (T ) = 0.
Our goal in this paper is to study the null controllability of the linearized system (1.2) by using a control ψ acting only in the normal direction on the upper part of boundary. Due to the incompressibility condition div U = 0, we necessarily have L 0 ψ(x, t) dx = 0, ∀t ∈ (0, T ). (1. 3) The main obstacle to null controllability using only one control acting in the normal direction of the boundary is as follows. In (1.2), we denote U (x, y, 0) = U 0 (x, y) = (u 0 , v 0 ) and U = (u, v). Taking a dot product between (1.2) 1 and (sin(nπy), 0) and then using an integration by parts on (0, L) × (0, 1) and the condition (1.3), we get d dt Thus U (x, y, T ) = (0, 0) implies that the initial condition U 0 (x, y) has to satisfy L 0 1 0 u 0 (x, y) sin(nπy) dy dx = 0, ∀ n ∈ N. (1.4) So there are infinitely many directions, namely (sin(lπy), 0), l ∈ N, which are not null controllable by the control acting in the normal direction of the boundary. Note NORMAL BOUNDARY CONTROL OF STOKES EQUATIONS 449 that the subspace spanned by these directions is infinite-dimensional. The question then arises whether the system is null controllable in the orthogonal complement of this subspace, and this paper will answer this question affirmatively. If we do not restrict the initial condition, we cannot control the flow to zero, but we can control to a purely horizontal flow. This might be of interest, for instance, if the objective of the control is to preserve stratification. In view of this discussion and (1.3), we study the null controllability of the linearized system by using the control acting in the normal direction of the boundary with appropriate constraints on the control and initial condition, i.e.
Using spectral methods we prove null controllability for U 0 belonging to V 0 ,n (Ω) with L 0 u 0 (x, y) dx = 0, for all y ∈ (0, 1), where V 0 ,n (Ω) denotes space of Lperiodic divergence free L 2 vector functions which have normal trace zero. (For details of the function spaces, see Section 2.1). This is the main result (see Theorem 3.1 in Section 3) of this paper.
As far as we know there are no prior null controllability results using one normal boundary control for the Stokes system (1.2).
The proof of the null controllability result relies on an observability inequality (see Section 3) for the solutions of the adjoint system and the spectral analysis of the linearized operator. The spectrum of the Stokes operator lies on the left side of the complex plane. It consists of a family of real eigenvalues, which diverges to −∞. Moreover, explicit expressions of eigenvalues and eigenfunctions in terms of a Fourier basis are obtained. This helps to split the observability inequality into observability inequalities corresponding to each Fourier mode of the adjoint system. The observability inequality for each mode will be established using a parabolic type of Ingham inequality. The proof that the observability inequality (Lemma 3.3) for the kth Fourier mode holds with a positive constant C T , independent of k, is the key result in this work.
The stabilization of the incompressible Navier-Stokes system in a 2-D channel (with periodic conditions along the x axis) linearized around a steady-state parabolic laminar flow profile (L(y), 0) has been studied in Munteanu [12] and Barbu [4]. In particular, Munteanu in [12] proved that the linearized system of (1.1) around (L(y) = C(y 2 − y), 0) is exponentially stabilizable with some decay rate ω, 0 < ω ≤ νπ 2 by a normal boundary, finite-dimensional feedback controller acting on the 450 SHIRSHENDU CHOWDHURY, DEBANJANA MITRA AND MICHAEL RENARDY upper wall Γ 1 (y = 1) only. A similar stability result for this linearized system when the normal velocity is controlled on the both walls Γ 0 (y = 0) and Γ 1 (y= 1) of the channel is proved by Barbu in [4]. In [3], Barbu established that the exponential stability of the linearized system around (L(y), 0) can be achieved using a finite number of Fourier modes and a boundary feedback stochastic controller which acts on the normal component of velocity only. In the present paper (see Section 2.2), we also notice that the incompressible Navier-Stokes system (1.1) linearized around the origin (0, 0) is stabilizable using only a normal boundary control, with any decay rate ω such that 0 < ω ≤ νπ 2 . Thus we know that the exponential decay rate can be at most νπ 2 for the linearized system. (See also equation (96) in Section 9 in [12] or Section 2 in [6]). A similar situation occurs also in Triggiani [16] for a linearized system where homogeneous boundary conditions on the tangential component u of the velocity are of Neumann type.
The boundary stabilization of Navier-Stokes equations, with tangential controllers or normal controllers was studied in two dimensions, for example by Barbu [5,4], Munteanu [13], Coron [17], Krstic [18], [1], [2], Raymond [15]. In most of these papers, either there are sufficiently many boundary controls so there are no missing directions, or stabilizability is proved but with no specific decay rate (except in [15]). In contrast, in this work we are using only one boundary control (acting on the normal component of velocity) on the upper part of the boundary and we are looking for null controllability instead of stabilizability.
In [6], Chowdhury and Ervedoza proved a local stabilization result for the viscous incompressible Navier-Stokes equations (1.1) at any exponential decay rate by a normal boundary control acting at the upper boundary. The linearized system around zero is exponentially stable with decay rate νπ 2 but not stabilizable at a higher decay rate. To overcome this difficulty, the idea is to use the nonlinear term to stabilize the system in the directions which are not stabilizable for the linearized equations. The argument is based on the power series expansion method introduced by J.-M. Coron and E. Crépeau in [8] and described in [7,Chapter 8].
Coron and Guerrero in [9] consider the two-dimensional Navier-Stokes system in a torus. They establish the local null controllability with internal controls having one vanishing component. Note that in their case also the linearized control system around the origin is not null controllable. In fact for the linearized system infinitely many missing directions are there corresponding to like our case here for sin(nπy). But in [9] the nonlinear term helps to get this null controllability using the return method. Coron and Lissy proved in [10] local null controllability for the three-dimensional Navier-Stokes equations on a smooth bounded domain of R 3 using localized interior control with two vanishing components. In this case also, the linearized system is not necessarily null controllable even if the control is distributed on the entire domain. They show local null controllability using the return method together with a new algebraic method inspired by the works of M. Gromov. For our system also the study of null controllability for the nonlinear system (1.1) is an open question. Moreover null controllability of the linearized and nonlinear system when control is localized is an interesting issue. These are interesting challenges for future research. This paper is organized as follows. In Section 2, we introduce function spaces required for our analysis. Then we study the behavior of the spectrum of the linearized operator and well-posedness of the linearized system (1.2) and the corresponding adjoint system. In Section 3, we split the observability inequality into observability inequalities corresponding to each Fourier mode of the adjoint system. Thereafter we give the completion of the proof of the observability inequality using a Müntz-Szász Theorem for each mode and showing uniformity of the constant arising in the observability inequality. In this fashion, null controllability (Theorem 3.1) is proved.
2. Spectral analysis of the Stokes system. In this section we study the Stokes system using its modal description. In particular, we identify the modes which are null controllable for the linearized system.

Functional framework.
Let We also define the space Let us denote the vector valued functional spaces: We now introduce the following spaces of divergence free vector fields: (Ω); div U = 0, U.n = 0 on Γ , (here the subscript n indicates the vanishing of the normal component) and We also denote the dual space of V 1 (Ω) by V −1 (Ω).
We also define the space of L 2 functions in (0, L) having mean-value zero bẏ Let us denote by P the Helmholtz orthogonal projection operator from L 2 (Ω) × L 2 (Ω) on to V 0 ,n (Ω), defined as where q is the weak solution of Further, taking ψ = 0 in (1.5), the linear operator associated to (1.5) is the ,n (Ω). The next lemma recalls well known properties of the Stokes operator.
Linearized system and its modal description. Here we study the linearized system with a normal boundary control ψ and some details of the spectrum of the corresponding linearized operator.
The adjoint problem corresponding to (1.5) is Let us consider the eigenvalue problem λΦ = AΦ = νP ∆Φ (to limit the number of different symbols, we use the same letters for the eigenfunctions as for the solution of the adjoint), i.e.
We expand (Φ, q) = (φ, ξ, q) into Fourier series: The eigenvalue problem for (φ k , ξ k , q k ) is The cases k = 0 and k = 0 need to be considered separately.
For k = 0 we have The eigenvalue problem for (φ k , ξ k , q k ) when k = 0 is We have Since (D sin(nπy), 0, 0) is a solution of the eigenvalue problem (2.5) for eigenvalue λ = −νπ 2 n 2 , n ∈ N, the solution of (1.2) with control zero and initial condition (D sin(nπy), 0), for any n ∈ N, is Thus, the solution is exponentially decaying at the rate −νπ 2 . But we cannot get any arbitrary decay for the system (1.5) using only the normal control ψ and the mode for k = 0 is not null controllable. To control it, we would require some additional tangential control.
The following lemma summarizes some elementary facts about the eigenvalues and eigenfunctions for nonzero k.
Lemma 2.2. We have the following results regarding the spectrum of the linear operator associated to (2.3) and its eigenfunctions.
1. The spectrum of the linear operator associated to (2.3) is real. The resolvent of the linear operator associated to (2.3) is compact and hence its spectrum consists of a set of isolated eigenvalues. 2. If λ ≥ −νk 2 , for all k ∈ 2π L Z, k = 0, then ξ k = 0 for all k = 0. Thus, the spectrum of the linear operator associated to (2.3) is a subset of (−∞, − 4π 2 L 2 ν) and in particular, eigenvalues for the kth mode satisfy λ k < −νk 2 for all k ∈ 2π L Z − {0}. In the following, let l be a natural number which counts the eigenvalues for fixed k in order of increasing magnitude. Since we have λ k,l < −νk 2 , we may

6)
and, for all k ∈ 2π L Z − {0} and for all l ∈ N, λ k,l satisfies where µ k,l = i µ k,l , and λ k,l < −νk 2 is necessary for a nontrivial solution.
In the following lemma we give some important properties of the positive, real roots of (2.7). (In addition, (2.7) has negative real roots which lead to same eigenvalues λ, and a root at zero, which does not lead to an eigenvalue but is due to the fact that the functions in (2.6) fail to be linearly independent in that case). Lemma 2.3. The solution of (2.7) for all k ∈ 2π L Z − {0} satisfies: 1. For any small δ > 0, there exists a (sufficiently large) k 0 ∈ N, such that for all k ∈ 2π L Z − {0} with |k| ≥ k 0 , we have the following: a) If any root µ k,j of (2.7), with |k| ≥ k 0 and some j ∈ N, satisfies µ 2 k,j k 2 − 1 < δ, then it is unique between two consecutive zeros of sin(·).
b) If the root µ k,j of (2.7), with |k| ≥ k 0 and some j ∈ N satisfies µ 2 k,j k 2 −1 ≥ δ, then it is unique between two consecutive zeros of cos(·). c) Moreover, k 0 can be chosen large enough such that the gap between two consecutive roots of (2.7) for k ∈ 2π L Z−{0} with |k| ≥ k 0 is always greater than π − 0 , for some positive small constant 0 > 0. d) There exists a unique root µ k,l in (lπ, (l + 1)π), for all l ∈ N. 2. For all k ∈ 2π L Z − {0}, k = 0 and |k| < k 0 , there exist l k ∈ N and N k ∈ N, where N k is the number of roots of (2.7) in (0, (l k +1)π − π 4 ), and the following hold: a) for all l ≥ l k + 1, there exists a unique root µ k,j of (2.7) in the ball B(lπ, π/4), where j = l + N k − l k and the root in fact lies in the interval (lπ − π/4, lπ + π/4). b) there is no root µ of (2.7) between µ k,j and µ k,j+1 , where j = l + N k − l k , for l ≥ l k + 1. c) µ k,l+N k −l k − lπ → 0 as l → ∞. 3. For each fixed k ∈ 2π L Z−{0} and for all l ∈ N, µ k,l , the root of (2.7) is simple. 4. For any k ∈ 2π L Z − {0}, there exists a l 0 ∈ N, independent of k, such that µ k,j > jπ/4, for all j > l 0 . 5. The solution µ k,l of (2.7) corresponds to λ k,l ν = − µ 2 k,l −k 2 for all k ∈ 2π L Z−{0} and for all l ∈ N. For each fixed k ∈ 2π L Z − {0}, and for all l ∈ N, λ k,l is a solution of the eigenvalue problem (2.3) with multiplicity one and there exist positive constants C 1 and C independent of k, l, such that inf k,l {λ k,l − λ k,l+1 } > C > 0, and where l 0 is introduced in the previous property.
Let us assume that the root of (2.7), denoted by µ k,j , for some j ∈ N, where δ > 0 is small enough, and k ∈ 2π L Z − {0} satisfying |k| ≥ k 0 , where k 0 is large enough chosen later. We claim that, for large enough k 0 and small enough δ, this root is unique. For any > 0, choosing k 0 large enough and δ small enough, we get that µ, any zero of f (·) satisfying µ 2 k 2 − 1 < δ, obeys | cos(µ)| < , and hence this µ must belong to a small neighbourhood N r (with radius 0 < r < π/4) of the zero of cos(·) between two consecutive zeros of sin(·). Now, if there are multiple zeros of f (·) between two consecutive zeros of sin(·) satisfying µ 2 k 2 − 1 < δ, the zeros of f (·) will be in the neighbourhood N r and between of two zeros of f (·), there must be a zero of f (·) in N r . Further, we check and the zeros of f (·) are in a small neighbourhood of the zeros of sin(·) due to the choice of k 0 and δ and we can make that the neighborhoods around the zeros of sin(·) and N r disjoint by a suitable refinement of k 0 and δ, if necessary. Thus, f (·) cannot have any zeros in N r and hence there exists unique zero of f (·) between two consecutive zeros of sin(·). 1.b) Now let the root µ k,j for some j ∈ N and for all |k| ≥ k 0 , where k 0 is large enough, satisfy | µ 2 k,j k 2 − 1| ≥ δ, where δ is as mentioned above. At the two consecutive zeros of cos(·), from (2.8), it follows that for j = m − 1, m, where m ∈ N, Choosing k 0 large enough in the above relation such that for all |k| ≥ k 0 , 1 cosh(k) and π/k are small enough, we get that f changes sign between two consecutive zeros of cos(·). To prove that µ k,j satisfying | µ 2 k,j k 2 − 1| ≥ δ is the unique root of (2.7) between to consecutive zeros of cos(·), we notice that any nonzero roots of (2.7) satisfy sin(µ)(sinh(k)/ cosh(k)) (1/ cosh(k)) − cos(µ) and for any > 0, there exists k 0 , large enough, such that for all |k| ≥ k 0 , between to consecutive zeros of cos(·), the zeros of (2.9) satisfy tan(µ) + 2 cosh(k) sinh(k) µ/k (µ 2 /k 2 ) − 1 = tan(µ) + sin(µ) (1/ cosh(k)) − cos(µ) < . (2.10)

SHIRSHENDU CHOWDHURY, DEBANJANA MITRA AND MICHAEL RENARDY
Now, if (2.9) has multiple roots between two consecutive zeros of cos(·), as argued in the first part, f (·) should have zeros between any two consecutive zeros of f (·). From the representation of f (·), we see that for µ, any zero of f (·), tan(µ) satisfies for any arbitrary > 0, choosing k 0 large enough. In any interval between two consecutive roots of cos(·), there is one subinterval where tan µ+ cosh(k) cannot also be small. Hence, (2.7) has a unique zero between two consecutive zeros of cos(·). 1.c) Let us assume that |k| ≥ k 0 and k 0 is large enough. If µ k,j , the root of (2.7) is such that | µ k,j /k| is close to 1, then the roots are close to the zeros of cos(·) and hence they are approximately π apart. If |µ/k| is not close to 1, the roots are close to those of tan µ = −2 cosh(k) . For large k, cosh k/ sinh k is close to 1, and µ/k changes slowly with µ. Hence tan µ changes little between successive roots. Therefore, the roots in this case are also spaced approximately π apart. 1.d) Since, for all |k| ≥ k 0 , where k 0 is large enough, f , defined in (2.8), changes sign between two consecutive zeros of sin(·), there can be an odd number of roots of (2.7) between two consecutive zeros of sin(·). Now, by 1.c), we have that the gap between two consecutive roots of (2.8) is greater than π − 0 . Choosing 0 < 0 < π/2, we can derive that each interval (lπ, (l + 1)π), for all l ∈ N, contains a unique root of (2.7) and we denote the root by µ k,l . This argument does not apply to the interval (0, π), since atμ = 0, there is a zero in the denominator of the last term in (2.8). For large k, the dominant terms in f are − cos(μ) + k sin(μ) 2μ .
The second term in this expression is positive and it dominates except nearμ = π, where − cosμ is also positive. Hence the expression is strictly positive on the entire interval [0, π], and it is easy to show that for large k the perturbing terms do not change that. Hence there is no root between 0 and π. 2. For all k ∈ 2π L Z − {0}, k = 0 and |k| < k 0 , we get that there exists a positive natural number l k such that This inequality holds for large enough l since |z 2 | >> |z|, and | sin z/ cos z| is bounded below on the periphery of the circle. Comparing the solutions of (2.7) with the solutions of [sinh(k) sin(z)]z 2 , by Rouché's theorem, we obtain that for all k ∈ 2π L Z − {0}, k = 0 and |k| < k 0 and for all l > l k , there exists a unique solution of (2.7) in the ball B(lπ, π/4), which in fact lies in the interval (lπ − π/4, lπ + π/4) as Property 1. of Lemma 2.2 gives that all roots of (2.7) are real. The other results also can be derived by comparing the solution of (2.7) for all k ∈ 2π L Z − {0}, k = 0 and |k| < k 0 and for all l > l k with the zeros of [sinh(k) sin(z)]z 2 .

SHIRSHENDU CHOWDHURY, DEBANJANA MITRA AND MICHAEL RENARDY
We have the following existence theorem for the solution of (1.5).
1. We note that L 0 v 0 (x, y) dx is automatically zero due to the boundary condition at the bottom wall and the incompressibility condition. Hence the condition on the average of U 0 = (u 0 , v 0 ) is really just a condition on u 0 . 2. We do not claim that the regularity of the solution as stated in the preceding theorem is optimal (for a more careful discussion of regularity for solutions of inhomogeneous Stokes problems, see [14]). Moreover, the choice of function spaces is not essential. If we leave our system uncontrolled for a short period of time, the solution will become infinitely smooth, so we could have assumed in the first place that our initial data are smooth and then chosen a control which is also smooth. 3. If we only know U ∈ C([0, T ], V −1 (Ω)), we can interpret the vanishing of the integral over x as the vanishing of the k = 0 Fourier component. However, for t > 0 and y < 1, U is actually of class C ∞ , and the integral is defined in the classical sense.
3. Null controllability. In this section, we study the null controllability of system (1.5). In particular, we have the following result: all y ∈ (0, 1), there exists a control ψ ∈ L 2 (0, T ;L 2 (0, L)), such that the solution of (1.5) reaches zero at t = T .
To prove the above theorem, it is enough to show the following result holds.
1. We note that only the pressure q appears in the normal component of stress. The viscous stress vanishes as a result of the divergence condition. 2. We have introduced T 0 < T only to avoid any technical issues related to lack of regularity of the solution of the adjoint equation. By choosing T 0 < T , we ensure that the solutions of the adjoint problem are in fact C ∞ smooth for t ∈ [0, T 0 ]. Therefore, at any time, either U or Φ is C ∞ smooth in the preceding proposition. Choosing T > T 0 also guarantees that the integral of q 2 in the following proposition is finite. We note that a posteriori at is clear that U in Proposition 1 actually vanishes at T 0 , since backward uniqueness holds for the Stokes equations.
Next we prove the existence of the bounded operator B, defined in (3.1), by showing that the observability inequality associated to the null control problem of (1.5) holds. Proposition 2. Let us assume that B is as defined by (3.1). For any T > T 0 > 0, B ∈ L(L 2 (0, T 0 ;L 2 (0, L)), L 2 (Ω)), i.e, there exists a positive constant C(T 0 ) > 0, such that
Using this system, to prove the observability inequality (3.4), it is sufficient to show that the observability inequality for each k ∈ 2π L Z − {0} holds with a positive constant C(T 0 ), independent of k. The following lemma gives an estimate for the eigenfunctions which turns out to be crucial in establishing this. Proof. From Lemma 2.4, since we have ξ k,l (1) = −i4k λ k,l ν µ k,l 2k sinh(k) (1 − cosh(k) cos( µ k,l )) + k sinh(k) + k 2 sin( µ k,l ) , then we get: 1. There exists a large k ∈ N, such that for all |k| > k and for all l ∈ N, we have |ξ k,l (1)| ≥ M * k 2 e |k| |λ k,l ||µ k,l |, for some M * > 0. 2. For |k| ≤ k, k = 0, using the fact that µ k,l − (l − N k + l k )π → 0 as l → ∞ from 2.c) in Lemma 2.3, we find positive constants l k and M k , independent of l, such that |ξ k,l (1)| ≥ M k k 2 e |k| |λ k,l ||µ k,l |, ∀ l > l k .
The result now follows from Lemma 3.3.