WELL-POSEDNESS AND OPTIMAL CONTROL OF A HEMIVARIATIONAL INEQUALITY FOR NONSTATIONARY STOKES FLUID FLOW

A time-dependent Stokes fluid flow problem is studied with nonlinear boundary conditions described by the Clarke subdifferential. We present equivalent weak formulations of the problem, one of them in the form of a hemivariational inequality. The existence of a solution is shown through a limiting procedure based on temporally semi-discrete approximations. Uniqueness of the solution and its continuous dependence on data are also established. Finally, we present a result on the existence of a solution to an optimal control problem for the hemivariational inequality.


Introduction.
Let Ω be a bounded simply connected domain in R d (d = 2 or 3) with a C 2 boundary Γ.Let T 0 > 0 and define Q = Ω × (0, T 0 ).In this paper, we consider hemivariational inequalities for the nonstationary Stokes system where u is the flow velocity field, ν > 0 the kinematic viscosity, h = p + |u| 2 /2 the dynamic pressure (p the pressure), f the density of external forces.The system (1)-( 2) is to be supplemented by initial and boundary conditions.For simplicity in writing, we use u(t) to stand for the function Ω x → u(x, t).Let u 0 denote the initial velocity.Then the initial condition is 5. Section 6 is devoted to the optimal control problem for which we establish the existence of an optimal solution.
2. Preliminaries.For a normed space X, we denote by • X its norm, by X * its topological dual, and by •, • X * ×X the duality pairing between X * and X.
The symbol X w is used for the space X endowed with the weak topology.Weak convergence will be indicated by the symbol .We denote the Euclidean norm in R n by |•|.The symbol 2 X * represents the set of all subsets of X * .We always assume X is a Banach space, unless stated otherwise.We first recall some definitions.Let f : X → R be a locally Lipschitz function.Following [8], we define the generalized directional derivative of f at x ∈ X in the direction v ∈ X by We then define the generalized gradient or subdifferential of f at x by We say f is regular (in the sense of Clarke) at x ∈ X if for all v ∈ X, the one-sided directional derivative f (x; v) exists and f 0 (x; v) = f (x; v).The concept of pseudomonotonicity plays an important role in this paper.We say a single-valued operator F : X → X * is pseudomonotone, if (i) F is bounded (i.e., it maps bounded subsets of X into bounded subsets of X * ); (ii) u n u in X and lim sup n→∞ F u n , u n − u X * ×X ≤ 0 imply It can be proved (see [25], for example) that an operator F : X → X * is pseudomonotone iff it is bounded and u n u in X together with lim sup n→∞ F u n , u n − u X * ×X ≤ 0 imply F u n F u in X * and lim n→∞ F u n , u n − u X * ×X = 0. Now let X be a reflexive Banach space.We say a multi-valued operator F : X → 2 The following proposition is usually used to check the pseudomonotonicity of a operator.
Proposition 1. ( [11]) Let X be a real reflexive Banach space, and assume that F : X → 2 X * satisfies the following conditions: (i) for each v ∈ X, F (v) is a nonempty, closed and convex subset of Then the operator F is pseudomonotone.
We need the notion of coercivity.We say an operator F : The following surjectivity result for pseudomonotone and coercive operators will be applied later in the paper.
For a Banach space X and a finite time interval I = (0, T 0 ), we will use the spaces L p (I; X), 1 ≤ p ≤ ∞.Denote by BV (I; X) the space of functions of bounded total variation on I defined as follows.Let π denote a finite partition of I: 0 = a 0 < a 1 < • • • < a n = T 0 , and let F be the collection of all such partitions.Then we define the total variation as For 1 ≤ q < ∞, we similarly define Now for Banach spaces X, Z such that X ⊂ Z we introduce a vector space M p,q (I; X, Z) = L p (I; X) ∩ BV q (I; Z).
It is a Banach space for 1 ≤ p, q < ∞ with the norm given by • L p (I;X) + • BV q (I;Z) .The following result is crucial in proving the convergence of the Rothe method (cf.Theorem 4.5).
Theorem 2.2.( [18]) Let 1 ≤ p, q < ∞.Let X 1 ⊂ X 2 ⊂ X 3 be real Banach spaces such that X 1 is reflexive, the embedding X 1 ⊂ X 2 is compact and the embedding X 2 ⊂ X 3 is continuous.Then a bounded subset of M p,q (I; X 1 , X 3 ) is relatively compact in L p (I; X 2 ).
The following Aubin-Cellina convergence theorem will be used.
3. Weak formulations.We introduce the weak formulations of the problem (1)-(5) in this section.Let We denote by V and H the closure of M in the norms of H 1 (Ω; R d ) and L 2 (Ω; R d ), respectively, and identity H with its dual H * .We define the space Z to be the closure of M in the norm of H δ (Ω; R d ) with some δ ∈ ( 1 2 , 1).Note the relations with all embeddings being dense and compact.We denote by •, • the duality of V and V * , by (•, •) the scalar product in H.The norms in V and H we denote by • V and • H . Denoting by i : V → Z the embedding injection and by γ : Z → L 2 (Γ; R d ) and γ 0 : the trace operators, for all v ∈ V we have γ 0 v = γ(iv).For simplicity we omit the notation of the embedding i and write γ 0 v = γv.Denoting by ι : V → H the embedding injection.For T 0 > 0, we define the spaces Concerning the data, we assume and We proceed to derive weak formulations of the problem (1)-( 5).Recall the identity (see [14]) where the symbol curl denotes the curl operator (see [14] for its definition).From ( 1)-( 2) we derive that It is known from [33] that in the case of simply connected domain Ω, the bilinear form Multiplying the equation of motion ( 7) by v ∈ V and applying the Green formula, we obtain From the relation (5), by using the definition of the Clarke subdifferential, we have where j 0 (t, ξ; η) ≡ j 0 (x, t, ξ; η) denotes the directional derivative of j(x, t, •) at the point ξ ∈ R in the direction η ∈ R. The last two relations yield the following weak formulation.
We will refer to the following equivalent formulation of Problem 3.3.
Remark 1.If the functional J is of the form ( 9) and H(j) holds, it is clear that every solution to Problem 3.3 (or Problem 3.4) is also a solution to Problem 3.1.If either j or −j is regular, then the converse is also true.Indeed, from [27, Theorem 3.47(vii)] we have, for all v ∈ V and a.e.t ∈ (0, T 0 ), By Proposition 3.37(ii) in [27], we obtain which implies (10).
Note that from our problem setting, we have the following properties: 4. Solution existence.In this section we show the existence of a solution to Problem 3.4.This is achieved through the consideration of a temporally semidiscrete approximation of Problem 3.4 based on the backward Euler difference for the time derivative; such an approximation is also known as the Rothe method.For a fixed N ∈ N, define the time step-size τ = T 0 /N .Introduce the piecewise constant interpolant of f by We approximate the initial condition by elements of V .Namely, let {u 0 τ } ⊂ V be such that u 0 τ → u 0 in H as τ → 0, and [31, Theorem 8.9]).The semi-discrete approximation of Problem 3.4 is the following.
Denote by λ the trace constant of V → L 2 (Γ; R d ): First we show an existence result for Problem 4.1.
Proof.It is sufficient to prove that for a given u k−1 τ ∈ V , there exist u k τ ∈ V and η k τ ∈ L 2 (Γ; R d ) satisfying (12).Note that ( 12) is equivalent to where the multivalued operator L : V → 2 V * is defined by Note that it is enough to prove the surjectivity of L. In view of Theorem 2.1, we will show that L is pseudomonotone and coercive.First, we prove the coercivity of L. Let v ∈ V and v * ∈ Lv.Then where η ∈ ∂J(γv).Using H(A), we have From H(J)(iii), we have It follows from ( 13) and ( 14) that Therefore, the operator L is coercive.Next we prove that L is pseudomonotone.Since the operator ι * ι τ is bounded, continuous and monotone, from Theorem 3.69(i) in [27] we deduce that the operator ι * ι τ is pseudomonotone.Since the trace operator γ : V → L 2 (Γ; R d ) is compact, from Lemma 2 in [18] we obtain that γ * ∂J(γ•) is pseudomonotone.Since the sum of two pseudomonotone operators remains pseudomonotone (cf.[11, Proposition 1.3.68]),L is pseudomonotone.
Let us establish a boundedness result for the semi-discrete solutions.Lemma 4.3.Under the assumptions of Theorem 4.2, there is a constant M 1 > 0, independent of τ , such that max k=1,...,N Proof.
Next, using H(J)(iii) we have V , and hence from the bound on u τ V we get the bound on η τ U .
Using H(A), from (20) we have Thus, using the bounds on u τ V and η τ U we get the bound on u τ V * .Suppose the BV 2 (0, T 0 ; V * ) seminorm of piecewise constant function u τ is obtained by some division 0 = a 0 < a 1 < . . .< a n = T 0 , and each a i is in different interval ((m i − 1)τ, m i τ ], such that u τ (a i ) = u mi τ with m 0 = 0, m n = N and m i+1 > m i for i = 1, . . ., N − 1.Thus, from the bound on u τ V * we have Thus, from the bounds on u τ V and u τ V * , we deduce that u τ is bounded in M 2,2 (0, T 0 ; V, V * ).Hence, the bound on u τ M 2,2 (0,T0;V,V * ) is proved.This completes the proof.
Theorem 4.5.Keep the assumptions made in Theorem 4.2.Then there exists a pair (u, η) ∈ W × U such that for a subsequence, Proof.From (21), we know that there exist u ∈ V ∩ L ∞ (0, T 0 ; H), u ∈ V ∩ L ∞ (0, T 0 ; H), u 1 ∈ V * and η ∈ U such that, passing to a subsequence if necessary, the following convergence holds First we show that u = u.Note that Thus, u τ − u τ → 0 in V * as τ → 0. On the other hand, from ( 22) and ( 23) we have From H(A), it is clear that A is linear and continuous operator from V to V * and thus also weakly continuous.Since u τ u in V, we get From ( 25) we get Since f τ → f in V * , we have Using ( 26)-( 29), we can pass to the limit in (20) and obtain Since u τ u in V, from H(γ) we have γ u τ → γu in U. Thus, for a subsequence, γ u τ (t) → γu(t) in L 2 (Γ; R d ) for a.e.t ∈ (0, T 0 ).Since ∂J : Finally, we pass to the limit with the initial conditions on the function u τ .Since u τ u in V and u τ u in V * and the embedding W ⊂ C(0, T 0 ; H) is continuous, we have u τ (t) u(t) in H for all t ∈ [0, T 0 ] (cf.[26, Lemma 4(b)]).Therefore, u 0 τ = u τ (0) u(0) in H. Since u 0 τ → u 0 in H, we have u(0) = u 0 .This completes the proof.
If ν − m N λ 2 > 0, where m N > 0 is the constant from H(J)(v), then the solution to Problem 3.4 is unique.
with η k (t) ∈ ∂J(γu k (t)) a.e.t ∈ (0, T 0 ) and u k (0) = u 0 .From H( U) it follows that the sequence {f k } belongs to a bounded subset of U. Therefore, by the reflexivity of U, we may assume, by passing to a subsequence if necessary, that for some f ∈ U, f k f in U.The weak closedness of U 0 implies that the limit f ∈ U 0 .Similarly to the proof of Theorem 5.1 (cf.( 34)), we conclude that for some constant c 1 > 0. From (40) we have Since η k (t) ∈ ∂J(γu k (t)) for a.e.t ∈ (0, T 0 ), from H(J)(iii) we have where c 2 > 0 is an embedding constant of V ⊂ Z.Using (41) and (43), from (42) we deduce that for some constant c 4 > 0, From ( 41) and (44) we conclude that {u k } is bounded in W. Thus, by passing to a subsequence if necessary, we have that u k u in W for some u ∈ W. Since the embedding of V into H is compact, so is the embedding of W into U.Therefore, u k → u in U. Using [3, Theorem 2.1], we obtain Since the embedding of W into C(0, T 0 ; H) is continuous, u k (t) u(t) in H for all t ∈ [0, T 0 ].Hence, we have that u k (0) u(0) = u 0 in H. Similarly as in the proof of Lemma 4.4, using H(J)(iii) we get where M 3 > 0. Thus, we may assume that η k η in U. Hence, γ * η k γ * η in V. Since A is linear and continuous, it is weakly continuous.Hence, Au k Au in V * .
Since {u k } is bounded in W, from (46) we may assume that η k η in U. Thus, Since A is linear and continuous, it is weakly continuous.Therefore, Similarly as in the proof of Theorem 4.

5 .Theorem 5 . 1 .
Uniqueness and continuous dependence on data.In this section we study the uniqueness of a solution to Problem 3.4 and continuous dependence of the solution on f and u 0 .Keep the assumptions of Theorem 4.2.Then, there exists a constant C > 0 such that for any solution u ∈ V to Problem 3.4.