DOI: 10.1007/s002459910014
Appl Math Optim 41:255–308 (2000)
© 2000 Springer-Verlag New York Inc.
Deterministic and Stochastic Control of Navier–Stokes Equationwith Linear, Monotone, and Hyperviscosities∗
S. S. Sritharan
Code D73H, Space and Naval Warfare Systems Center,San Diego, CA 92152, [email protected]
Abstract. This paper deals with the optimal control of space–time statistical be-havior of turbulent fields. We provide a unified treatment of optimal control prob-lems for the deterministic and stochastic Navier–Stokes equation with linear andnonlinear constitutive relations. Tonelli type ordinary controls as well as Young typechattering controls are analyzed. For the deterministic case with monotone viscositywe use the Minty–Browder technique to prove the existence of optimal controls.For the stochastic case with monotone viscosity, we combine the Minty–Browdertechnique with the martingale problem formulation of Stroock and Varadhan to es-tablish existence of optimal controls. The deterministic models given in this paperalso cover some simple eddy viscosity type turbulence closure models.
Key Words. Stochastic control, Control of fluids, Turbulence control, Navier–Stokes equation, Minty–Browder theory.
AMS Classification. 76, 49, 60.
1. Introduction
Optimal control theory of viscous flow has been a rapidly developing subject during thepast several years [64]. Most of the work is concerned with the case of deterministic con-trol. Stochastic dynamic programming is developed in [64] by the author and nonlinearstochastic filtering is developed in [63]. In this paper we obtain the existence of optimalcontrols for stochastically forced fluid flow with Newtonian and non-Newtonian consti-
∗ This research was supported by the Office of Naval Research.
256 S. S. Sritharan
tutive relations. The deterministic Navier–Stokes equation with nonlinear viscosity wasintroduced in [39], [40], [42], and [43] and hyperviscosity was introduced in [43] (seealso [52] and [53]) to demonstrate global unique solvability in three dimensions. We con-sider the stochastic versions of these equations in the martingale problem frame work ofStroock–Varadhan [65] as formulated for infinite-dimensional problems in Lusin spacesin the works of Viot [69] and Metivier [47]. The existence of admissible martingalesolutions (space–time statistical solutions) are proven by Galerkin approximations andusing weak limits and the Minty–Browder technique [48], [8]. Uniqueness is proven fortwo- and three-dimensional stochastic Navier–Stokes equation with linear and nonlinearviscosities under appropriate additional conditions implicitly dictated by the diffusionterm. The method here is first to prove pathwise uniqueness and then use the Yamada–Watanabe [71] technique to get uniqueness in law. We then establish the main resultsof this paper which concerns the existence of optimal controls. An existence result foroptimal control for stochastic semilinear equations with bounded nonlinearities (hencecannot cover the Navier–Stokes equation) was obtained in [28]. For the deterministicNavier–Stokes equation without control action, statistical extremal problems have beenintroduced and studied to establish uniqueness theorems for the Hopf equation [21], [22],uniqueness of space–time statistical solutions for the three-dimensional problem withinitial probability distribution concentrated on dense data [70], and to prove momentclosure type theorems [27], [17].
We use chattering controls as in the deterministic case [19], [20]. The abstractstochastic system formulated in this paper also covers the Boussinesq system of coupledNavier–Stokes and temperature equations [12], magnetohydrodynamic equations [12],[16], constant density combustion models [45], and multidimensional Burger’s equation[23]. Moreover, fluid or diffusion limits of the dynamics and control of interactingsystems [23] and stochastic networks [37] also lead to equations and control problemsof the type studied in this paper.
2. Formulation of Stochastic Control Problems in Viscous Flow
Let G ⊂ Rn, n = 2,3, be an arbitrary bounded open set withC∞ boundary∂G. Weformulate a stochastic control problem where the control is actuated as a “body force”in the Navier–Stokes equation. One way to achieve this in practice is by electromagneticforcing for slightly conducting fluids such as salt water [30], [31], [11], [12], [33], [58]. Inthis case theLorentz forceacts as a body force (distributed control). Body-force controlis also important in combustion control by laser heating and Tokamak MHD control byradio and microwave heating [9], [49]. Let(Θ, 6,m) be a complete probability space.The velocity and pressure fields are denoted as(u, p): G× (0, T)×Θ→ Rn× R. Weconsider the stochastic Navier–Stokes problem:
∂tu+u · ∇u = ∇ · σ(u)+N (U )+ g(u)Υ, (x, t,ω) ∈ G× (0, T)×Θ, (1)
∇ · u = 0, (x, t,ω) ∈ G× (0, T)×Θ, (2)
u(x, t,ω) = 0, (x, t,ω) ∈ ∂G× (0, T)×Θ, (3)
u(x,0,ω) = u0(x,ω), (x,ω) ∈ G×Θ. (4)
Control of Navier–Stokes Equation 257
Hereσ(u) is the (possibly nonlinear) stress tensor,U is the control which takes itsvalue in some metrizable Lusin spaceU [14], g is some vectorfield which depends onthe velocity distribution, and the noise termΥ is taken as the generalized derivativeof an infinite-dimensionalQ-Wiener process [13]W. The form of the noise here ischosen to model exogeneous forces such as structural vibrations, magnetic fields, andother body forces.N (·) here can be a linear or nonlinear operator representing possiblenonlinearities in the actuator system. The task is to find the optimal controlU whichminimizes, for example, a cost functional of the form
12 E
[∫ T
0
∫G
(|u(x, t)|2α1 + |∇ × u(x, t)|2α2 + |U (x, t)|2α3)
dx dt
]→ inf, (5)
with αi ∈ R+, i = 1,2,3. We actually consider much more general cost functionals ofthe form
E
[∫ T
0
∫GF(u(x, t),U (x, t))dx dt
]→ inf, (6)
for a suitable functionF . The deterministic case will be obtained as a special case whereg= 0 and cost functional∫ T
0
∫GF(u(x, t),U (x, t))dx dt→ inf. (7)
It is in fact possible to extend the results of this paper to other cost functionals involv-ing physical quantities such as entropy [68], where the dependence of the cost functionalon the joint probability distribution of velocity and control is lower semicontinuous.
2.1. Nonlinear Constitutive Relations, Monotone Operators,and Other Nonlinearities
We consider the following cases for the stress tensorσ(u). The theory developed inthis paper, however, covers more general constitutive relationships of monotone typegiven in [40] (described in Supplement 1 of this book entitled “New equations for thedescription of the motion of viscous incompressible flow”). Moreover, there is also hopethat future generalizations to more realistic non-Newtonian fluids [57], [44] would bepossible. The nonlinear viscosity models in cases II and III below can also be regardedas simple moment closure models (eddy viscosities) for the turbulent Reynolds stress[50], [51].
Case I: Newtonian(Linear) Constitutive Relationship.
σI(u) := −pI + ν0(∇u+ (∇u)T
), ν0 > 0. (8)
This will correspond to the conventional Navier–Stokes equation with∇ · σI(u) =−∇ p+ ν01u.
Case II: Nonlinear Constitutive Relationship[42], [39], [40], [43].
σII (u) = −pI + ν0(∇u+ (∇u)T
)+ ν1|∇u|q−2∇u, (9)
whereν0, ν1 > 0 andq ≥ 2. In this case,∇·σII (u) = −∇ p+ν01u+ν1∇·(|∇u|q−2∇u).
258 S. S. Sritharan
Case III: Nonlinear Nonlocal Viscosity[40].
σIII (u) = −pI + (ν0+ ν1‖∇u‖2) (∇u+ (∇u)T), (10)
so that∇ · σIII (u) = −∇ p + ν(‖∇u‖)1u, where the nonlinear viscosity is given byν(‖∇u‖) := ν0+ ν1‖∇u‖2, with ν0, ν1 > 0 and‖∇u‖2 := ∫G |∇u|2 dx.
Case IV: Hyperviscosity[43], [52]–[54]. This type of regularization has been used inatmospheric dynamics models and also in the study of vortex reconnections [46], [41], [6].
σIV (u) = −pI + ν0(∇u+ (∇u)T
)− ν1(−1)m∇(1m−1u), (11)
with m ≥ 2, ν0, ν1 > 0. In this case,∇ · σIV (u) = −∇ p + ν01u − ν1(−1)m1mu.In this case we also prescribe additional boundary conditions,(∂u/∂n)|
∂G = · · · =(∂m−1u/∂nm−1)|
∂G = 0.
We now define the forms
aI(u, v) := ν0
∫G∇u · ∇vdx, (12)
aII (u, v) := ν0
∫G∇u · ∇vdx+ ν1
∫G|∇u|q−2∇u · ∇vdx, (13)
aIII (u, v) := ν0
∫G∇u · ∇vdx+ ν1‖∇u‖2
∫G∇u · ∇vdx, (14)
and
aIV (u, v) := ν0
∫G∇u · ∇vdx+ ν1
∑α∈N,[α]=m
∫G
Dαu · Dαvdx. (15)
We have the following estimates,∀u, v ∈ D(G):|aI(u, v)| ≤ ν0‖∇u‖‖∇v‖, (16)
aI(u,u) = ν0‖∇u‖2. (17)
|aII (u, v)| ≤ ν0‖∇u‖‖∇v‖ + ν1‖∇u‖q−1Lq(G)‖∇v‖Lq(G), (18)
aII (u,u) ≥ ν0‖∇u‖2+ ν1‖∇u‖qLq(G). (19)
|aIII (u, v)| ≤ ν0‖∇u‖‖∇v‖ + ν1‖∇u‖3‖∇v‖, (20)
aIII (u,u) ≥ ν0‖∇u‖2+ ν1‖∇u‖4. (21)
|aIV (u, v)| ≤ ν0‖∇u‖‖∇v‖ + ν1
∑α∈N,[α]=m
‖Dαu‖‖Dαv‖, (22)
aIV (u,u) ≥ ν0‖∇u‖2+ ν1C‖u‖2Hm(G). (23)
Control of Navier–Stokes Equation 259
We define the function spaces [40], [66], [61]
H := u ∈ L2(G); ∇ · u = 0,u · n|∂G = 0
,
Vr,p := u ∈ Wr,p0 (G); ∇ · u = 0
. (24)
Estimates (16)–(23) ensure (by Riesz representation theorem) the existence of anelementA(u) := ν0Au+ ν1A(u) ∈ V′r,q (linear for cases I and IV) such that
aI(u, v) = ν0〈Au, v〉, ∀u, v ∈ V1,2, (25)
aII (u, v) = ν0〈Au, v〉 + ν1〈AII (u), v〉, ∀u, v ∈ V1,q, (26)
aIII (u, v) = ν0〈Au, v〉 + ν1〈AIII (u), v〉, ∀u, v ∈ V1,2, (27)
aIV (u, v) = ν0〈Au, v〉 + ν1〈AIV (u), v〉, ∀u, v ∈ Vm,2. (28)
Here A is the well-known Stokes operator which is self-adjoint and positive definite.AIV is also self-adjoint and positive definite. We now show that the operatorsAII andAIII are demicontinuous, maximal monotone, and surjective.
Definition 1. A mappingA: V → 2V ′ is said to be monotone if,∀x, y ∈ D(A) andu ∈ A(x), v ∈ A(y), we have〈u− v, x− y〉 ≥ 0. HereA is called strictly monotone if,equality impliesx = y and maximal monotone if, for(x,u) ∈ V × V′, the inequalities〈u − v, x − y〉 ≥ 0, ∀(y, v) ∈ G(A), imply (x,u) ∈ G(A), whereG(A) is the graphof A.
Theorem 1. The nonlinear viscous operatorsAII and AIII are strictly monotone:∀u, v ∈ D(G),
〈AII (u)−AII (v),u− v〉 ≥ γ (n,q)‖∇u−∇v‖qLq(G) (29)
and
〈AIII (u)−AIII (v),u− v〉= 1
2
(‖∇u‖2+ ‖∇v‖2) ‖∇u−∇v‖2+ 12
(‖∇u‖2− ‖∇v‖2)2 , (30)
where〈·, ·〉 is the integral(duality pairing) in G andD(G) is the class of test func-tions.
Proof. Note that, foru, v ∈ D(G),
〈AII (u)−AII (v),u−v〉 = ν1
∫G
(|∇u|q−2∇u−|∇v|q−2∇v) · (∇u−∇v)dx. (31)
The integrand is estimated from below by the following algebraic inequality.
260 S. S. Sritharan
Lemma 1 [15, Section 1.4]. Let q≥ 2. Then, ∀x, y ∈ Rn, we have(|x|q−2x− |y|q−2y) · (x− y) ≥ γ (q,n)|x− y|q. (32)
Using this lemma we get〈AII (u)−AII (v),u− v〉 ≥ γ (q,n) ∫G |∇u−∇v|q dx.We now consider
〈AIII (u)−AIII (v),u− v〉 =∫
G
(‖∇u‖2∇u− ‖∇v‖2∇v) · (∇u−∇v)dx. (33)
We write the integral as
ν1
2
(‖∇u‖2+ ‖∇u‖2) ∫G(∇u−∇v)2 dx
+ ν1
2
(‖∇u‖2− ‖∇u‖2) ∫G(∇u−∇v) · (∇u+∇v)dx, (34)
from which (30) follows.
Theorem 2. The operatorAII (·) is demicontinuous: let un→ u inV1,q, thenAII (un)→AII (u) in the weak-star(weak) topology ofV′1,q.
Proof. Let un→ u strongly inV1,q. Then noting that (18)
‖AII (un)‖V ′rq≤ ‖∇un‖q−1
Lq(G), (35)
taking a subsequence (again denoting)un ∈ V1,q we getAII (un)→ κ weakly inV′1,q.We will now show thatκ = AII (u). In fact un → u strongly in W1,q(G) implies∇un → ∇u strongly inLq(G) and hence∇un(x) → ∇u(x), x a.e. inG. Hence also|∇un(x)|q−2∇un(x) → |∇u(x)|q−2∇u(x), x a.e. inG. We now use Lemma 1.3 inSection 1.1 of [43] to conclude that,∀v ∈ V1,q,
〈AII (un), v〉 = aII (un, v)
=∫
G|∇un|q−2∇un ·∇vdx→
∫G|∇u|q−2∇u·∇vdx. (36)
Now noting thatAIII (·): V1,2→ V′1,2 is continuous and combining the above resultswe have the following properties.
Corollary 1. The operatorsAII (·) andAIII (·) are maximal monotone.
This is a consequence of demicontinuity and monotonicity [5]. The following result isproved in Theorem 2.1 of Section 2.1 in [43].
Corollary 2. The operatorsAII (·): V1,q → V′1,q andAIII (·): V1,2 → V′1,2 are sur-jective.
Corollary 3. AII (·) andAIII (·) are strongly hemicontinuous. That is, ∀VN ⊂ V, VN
finite-dimensional, the mapsu→ AII (u) andu→ AIII (u) are continuous fromVN →V′.
Control of Navier–Stokes Equation 261
This result can be verified directly and also follows from Lemma 1.1 in Section 1 ofChapter II of [5].
Hypothesis 1. We suppose that q and r are chosen in the following way for cases I–IV:
Case I: G⊂ R2, with r = 1 and q= 2.Case II: G⊂ Rn,n = 2,3, with r = 1 and q≥ 3.
Case III: G⊂ Rn,n = 2,3, with r = 1 and q= 4.Case IV: G⊂ Rn,n = 2,3, with r = m and q= 2.
We note a property of the nonlinearityB(u).
Lemma 2. For cases II and III, if u ∈ Lq(0, T;Vrq), then B(u) ∈ Lq′(0, T;V′rq), and
for cases I and IV, if u ∈ Lq(0, T;Vrq) ∩ L∞(0, T;H), then B(u) ∈ Lq′(0, T;V′rq).
Proof. Case I is classical and it follows from the estimate
|b(u,u, v)| = |b(u, v,u)| ≤ C‖u‖‖u‖1/2‖v‖1/2, (37)
which gives∣∣∣∣∫ T
0〈B(u), v〉dt
∣∣∣∣ ≤ C‖u‖L∞(0,T;H)‖u‖L2(0,T;V1,2)‖v‖L2(0,T;V1,2), (38)
and case IV is similar. For cases II and III, we start with the observation that [1]
W1,q(G) ⊂ Lr (G), for1
r= 1
q− 1
n≥ 0. (39)
Now applying the Holder inequality,|b(u, v,u)| ≤ ‖u‖2Lr (G)‖∇v‖Lq(G), with 2/r +1/q = 1. Thus,∣∣∣∣∫ T
0〈B(u), v〉dt
∣∣∣∣ ≤ ∫ T
0‖u(s)‖2Lr (G)‖∇v(s)‖Lq(G) ds
≤ C
(∫ T
0‖u‖2q′
Lr (G) ds
)1/q′ (∫ T
0‖∇v‖qLq(G) ds
)1/q
,
with1
q′+ 1
q= 1. (40)
Thus, using the embedding (39) noted earlier,
‖B(u)‖Lq′ (0,T;V ′1,q) ≤ C
(∫ T
0‖u‖2q′
Lr (G) ds
)1/q′
≤ C
(∫ T
0‖u‖2q′
V1,qds
)1/q′
. (41)
Now, taking 2q′ ≤ q, we get‖B(u)‖Lq′ (0,T;V ′1,q) ≤ C(∫ T
0 ‖u‖qV1,qds)2/q, and the two
conditionsq ≥ 2q′ and 1/q′ + 1/q = 1 are combined to giveq ≥ 3.
262 S. S. Sritharan
2.2. Unified Control Theoretic Formulation
We can thus describe our main goal abstractly as follows. We have an abstract nonlinearstochastic partial differential equation
du+ (ν0Au+ ν1A(u)+ B(u))dt = N (U )dt + g(u)dW, u(0) = u0, (42)
with the cost functionalE[J ]: = E[∫ T
0 F(u,U )dt] → inf.Our analysis will also include the special case where the control input operator is
linear,
N (U ) = LNU, LN ∈ L(H;H), (43)
with U = H, and the integrand of the cost functional is convex inU , that is,F(u, ·): H →R is convex. In the general case of nonlinearN and nonconvexF we will construct(following the well-known measure-valued convexification method of L. C. Young andJ. Warga) a relaxed formulation with suitable probability meaure-valued controlsµt inthe following way. We disintegrate [60] a Borel probability measureµ on the BorelalgebraB(U × [0, T ]) for the projectionU × [0, T ] → U as
µ(dU,dt) = µt (dU)dt, µ ∈M(U × [0, T ]), (44)
whereM(U× [0, T ]) is the space of Borel probability measures onB(U× [0, T ]). Then
du+ (ν0Au+ ν1A(u)+ B(u))dt = Nµt dt + g(u)dW, (45)
whereNµt := ∫UN (U )µt (dU), u(0) = u0, andE[∫ T
0
∫U F(u,U )µt (dU)dt] → inf.
The deterministic cases in the above will be obtained by simply setting the randomforcing to be zero (i.e.,g= 0).
We use the Lusin space (see the definition at the end of this section)Ω = Ω× ˜Ω:
Ω := Lq(0, T;H) ∩ D([0, T ];V′r,q) ∩ Lq(0, T;Vr,q)σ ∩ L∞(0, T;H)w∗ ,˜Ω :=M(U × [0, T ]).
Here D([0, T ];V′r,q) is the V′r,q-valued Skorohod space with J-topology [47], andLq(0, T;Vr,q)σ is the spaceLq(0, T;Vr,q) endowed with weak topology. The canonicalfiltration is
Ft := σ
u(s),∫ s
0
∫Cµr (dU)dr, C ∈ B(U), s ≤ t
. (46)
Solvability of (45) will be resolved through the martingale problem of finding a RadonmeasureP on the Borel algebraB(Ω) such that
M t (u,µ) := u(t)+∫ t
0
(ν0Au+ ν1A(u)+ B(u)− Nµs
)ds (47)
is a V′r,q-valued (Ω,B(Ω),Ft ,P)-martingale (i.e., anFt adapted process such that
Control of Navier–Stokes Equation 263
E[M t |Fs] = Ms) with quadratic variation,
〈〈M t 〉〉 :=∫ t
0g(u(s))Qg∗(u(s))ds. (48)
The optimal control problem will then be to find the optimal probability lawPwhich solves the above martingale problem and minimizes the cost functionalEP[
∫ T0
∫U F(u,U )µt (dU)dt] → inf.
We now recall for convenience that a topological space which is a one to onecontinuous image of a Polish space is called a Lusin space [47], [59]. Moreover, atopological space is Lusin if and only if it is homeomorphic to a topological spacewhich is a Borel subset of a Polish space [3]. A metrizable topological space is calleda metrizable Lusin space if it is homeomorphic to a Borel subset of a compact metricspace (Chapter III, Definition 16(a) of [14]).
A Radon measure on a topological space is a measure on the Borel algebra whichis locally finite and inner regular (Part I, Chapter I, Definition R3 of [59]). Every Borelmeasure on a Lusin space is Radon (Part I, Chapter II, Theorem 9 of [59]).
3. The Deterministic Control Problem
In this section we extend the earlier existence theories for ordinary [24]–[26], [62], [18],[64] and chattering optimal controls [20] for various flow control problems to the casesof nonlinear viscosities utilizing the method of Minty–Browder.
3.1. Relaxed(Young Measure Valued) Controls and Stable Topology
As in [20] we takeN in the following class:
Hypothesis 2. N is an admissible nonlinear input(control) operator ifN : U → His continuous and there exists aninf-compact function[67] κ(·): U → R+ such that‖N (U )‖ ≤ κ(U )+ C, U ∈ U.
The constantC does not play any significant role in the subsequent estimates concerningN and hence will be ignored. Note that in the special case ofN (U ) := LNU withU := H,
‖N (U )‖ ≤ C1‖U‖ + C2,
and we takeκ(U ) := C1‖U‖. Note that with the weak topologyHσ is Lusin [47] andalso the sets,Uα = U ∈ H; κ(U ) ≤ α , ∀α ∈ R+, are compact inHσ and henceκ(·)is inf-compact [67].
We will take relaxed controls in the following class:
Hypothesis 3. We sayMad(U × [0, T ]) is the class of admissible relaxed controls ifµ ∈Mad(U×[0, T ]) takes values inM(U×[0, T ])and
∫ T0
∫U κ(U )
2µ(dU,dt) < +∞.
264 S. S. Sritharan
We note that with such measuresNµt satisfy∫ T
0 ‖Nµt‖2 dt < +∞. In fact,∫ T
0‖Nµt‖2 dt ≤
∫ T
0
(∫U‖N (U )‖µt (dU)
)2
dt ≤∫ T
0
(∫Uκ(U )µt (dU)
)2
dt
≤∫ T
0
(∫Uκ(U )2µt (U )
)(∫Uµt (dU)
)dt
=∫ T
0
∫Uκ(U )2µt (dU)dt < +∞. (49)
Definition 2. Let U be a topological space and letM(U × [0, T ]) be the class ofbounded nonnegative Borel measures onB(U× [0, T ]). Stable topologyis the weakesttopology for which the map
µ→∫ T
0
∫Uϕ(U, t)µ(dU,dt) (50)
is continuous for all bounded, real-valued measurableϕ which is continuous inU .
It has been proven in [4] that
Lemma 3 [4]. If U is a metrizable Lusin space, thenM(U × (0, T)) with stabletopology is a metrizable Lusin space.
Lemma 4 [34]. Let µn → µ in the stable topology ofM(U × (0, T)). Then thefollowing convergence results hold:
(i) Let ϕ(·, ·): U × (0, T) → R+ be a bounded measurable function such that,∀t ∈ (0, T), ϕ(·, t): U→ R+ is lower semicontinuous. Then
lim infn→∞
∫ T
0
∫Uϕ(U, t)µn(dU,dt) ≥
∫ T
0
∫Uϕ(U, t)µ(dU,dt).
(ii) Let A ⊂ U × [0, T ] be measurable and each t-section ofA is closed. Thenlim supn→∞ µ
n(A) ≤ µ(A).(iii) Letψn(·, ·): U× (0, T)→ R be a sequence of uniformly bounded measurable
functions such that
lim supn→∞
µn (U, t) ∈ U × (0, T); |ψn(U, t)| > γ = 0, ∀γ > 0, (51)
thenlimn→∞∫ T
0
∫U ψn(U, t)µn(dU,dt) = 0.
Lemma 5. The sequenceµn ∈M(U×[0, T ])which satisfies∫ T
0
∫U κ(U )
2µn(dU,dt)≤ C, is tight in the stable topology. That is there exists a subsequenceµn′ such thatµn′ → µ in the stable topology and∫ T
0
∫Uκ(U )2µ(dU,dt) ≤ C. (52)
Proof. The existence of convergence subsequence is proven in Theorem 5.1(V) of [4]and the estimate (52) can then be obtained by truncation ofκ(U )2 as in Lemma 10below.
Control of Navier–Stokes Equation 265
Lemma 6. Letµn→ µ in the stable topology with∫ T
0
∫U κ(U )
2µn(dU,dt) ≤ C, thentakingN as the admissible input operator we have the following convergence: ∀z∈ H,∫ T
0
∫U(N (U ), z)µn(dU,dt)→
∫ T
0
∫U(N (U ), z)µ(dU,dt). (53)
Proof. LetϕR ∈ Cb(U) such that
ϕR(U ) =
1 for κ(U ) ≤ R,0 for κ(U ) ≥ 2R,
∣∣∣∣∫ T
0
∫U(N (U ), z)µn(dU,dt)−
∫ T
0
∫U(N (U ), z)µ(dU,dt)
∣∣∣∣≤∫ T
0
∫U|(1− ϕR(U ))(N (U ), z)|µn(dU,dt)
+∫ T
0
∫U|(1− ϕR(U ))(N (U ), z)|µ(dU,dt)
+∣∣∣∣∫ T
0
∫UϕR(U )(N (U ), z)µn(dU,dt)
−∫ T
0
∫UϕR(U )(N (U ), z)µ(dU,dt)
∣∣∣∣ := I1+ I2+ I3.
I3→ 0 due to stable convergence ofµn. Note that, forUR := U ∈ U; κ(U ) ≥ R,
I1 =∫ T
0
∫UR
|(1− ϕR(U ))(N (U ), z)|µn(dU,dt)
≤∫ T
0
∫UR
‖z‖κ(U )−1κ(U )2µn(dU,dt) ≤ 1
R
∫ T
0
∫UR
‖z‖κ(U )2µn(dU,dt)
≤ C
R‖z‖ → 0. (54)
Similarly we can show thatI2→ 0.
Lemma 7. Letµn→ µ ∈M(U × [0, T ]) in the stable topology with∫ T
0
∫Uκ(U )2µn(dU,dt) ≤ C
andyn→ ystrongly in L2(0, T;H). Then∫ T
0
∫U(y
n(t),N (U ))µn(dU,dt)makes sense
and converges to∫ T
0
∫U(y(t),N (U ))µ(dU,dt).
Proof. We have
In :=∫ T
0
∫U(y(t),N (U ))µ(dU,dt)−
∫ T
0
∫U(yn(t),N (U ))µn(dU,dt)
= In,1+ In,2,
266 S. S. Sritharan
where
In,1 :=∫ T
0
∫U(y(t)− yn(t),N (U ))µn(dU,dt)
and
In,2 :=∫ T
0
∫U(y(t),N (U ))(µ(dU,dt)− µn(dU,dt)).
ConsiderIn,1,
|In,1| ≤∫ T
0
∫U‖y(t)− yn‖‖N (U )‖µn(dU,dt)
=∫ T
0‖y(t)− yn(t)‖
(∫Uκ(U )µn
t (dU)
)dt
≤(∫ T
0‖y(t)− yn(t)‖2 dt
)1/2(∫ T
0
∫Uκ(U )2µn
t (dU)dt
)1/2
→ 0
as n→∞, (55)
sinceyn→ y strongly inL2(0, T;H) and for admissible classMad(0, T;U) of controlmeasures, the second integral is finite. We now come toIn,2. Sincey(·) ∈ L2(0, T;H)is strongly measurable from [0, T ] → H, there exists a sequence ofH-valued simplefunctions,yk(t) :=∑m
l=1 ak,l I1k,l (t), ak,l ∈ H, such thatyk(·)→ y(·) in L2(0, T;H),and also fort a.s in [0, T ], yk(t)→ y(t) in H. Thus,
In,2,k :=∫ T
0
∫U(yk(t),N (U ))µ(dU,dt)−
∫ T
0
∫U(yk(t),N (U ))µn(dU,dt), (56)
with In,2,k → In,2 ask→∞, due to the previous arguements used forIn,1. Now,
In,2,k =m∑
l=1
∫1k,l
∫U(ak,l ,N (U ))µ(dU,dt)
−∫1k,l
∫U(ak,l ,N (U ))µn(dU,dt)
. (57)
Thus,In,2,k → 0 asn→∞ due to Lemma 6. Since we can takek andn independentlyto infinity, we have proved thatIn,2→ 0 asn→∞.
Definition 3. A functionalF(·, ·, ·): [0, T ]×Vr,q×U→ R is anadmissible integrandfor the cost functional if it satisfies the following conditions:
(1) F(·, ·, ·): [0, T ] × Vr,q × U→ R is measurable. (58)(2) ∀t ∈ [0, T ], F(t, ·, ·): Vr,q,w × U→ R (59)
is lower semicontinuous whereVr,q,w is the spaceVr,q endowed with the weaktopology.
(3) F(t,u,U ) ≥ Cκ(U )2,∀(t,u,U ) ∈ [0, T ] × Vrq × U. (60)
Definition 4. We denote byτ -topology the supremum of the topologiesτ 1, . . . , τ 4,τ = τ 1 ∨ τ 2 ∨ τ 3 ∨ τ 4, whereτ 1 := L∞(0, T;H)-weak star,τ 2 := Lq(0, T;Vr,q)-weak,τ 3 := C([0, T ];V′rq)-strong, andτ 4 := Lq(0, T;H)-strong.
Control of Navier–Stokes Equation 267
We define the functional (which later in the stochastic case becomes a martingale)
Mθ
t (u,χ,µ) := 〈u(t),θ〉 − 〈u(0),θ〉+∫ t
0〈ν0Au(r )+ ν1χ(r )+ B(u(r ))− Nµ(r ),θ〉dr,
∀θ ∈ Vrq , t ∈ [0, T ]. (61)
Lemma 8 Continuity of the Martingale, I. Suppose that q and r are chosen as in Hy-pothesis1.If un→ u in theτ -topology,µn→ µ in the stable topology ofM(U×[0, T ]),
andχn→ χ in the weak topology of Lq′(0, T;V′r,q), thenM
θ
t (un,χn,µn) = 0, ∀θ ∈
Vrq , t ∈ [0, T ],∀n, impliesMθ
t (u,χ,µ) = 0, ∀θ ∈ Vrq , t ∈ [0, T ].
Proof. The proof is essentially the same as that for Lemma 16 in Section 5.3.
Lemma 9. Suppose that q and r are chosen as in Hypothesis1. Then on the classof functionsu ∈ Lq(0, T;Vr,q) ∩ C([0, T ];H), u0 ∈ H with ∂tu ∈ Lq(0, T;V′rq),χ ∈ Lq′(0, T;V′rq), andµ(·) ∈M(U × [0, T ]), the equation of evolution,
∂tu+ ν0Au+ ν1χ+ B(u) = Nµ(t), t > 0, u(0) = u0 ∈ H, (62)
is equivalent to the identityMθ
t (u,χ,µ) = 0, ∀θ ∈ Vrq , t ∈ [0, T ].
Proof. Noting that the vector-valued forcing for the evolution equation (62) isNµ ∈Lq′(0, T;V′rq), the proof is essentially same as that for Proposition 1.1 in Chapter IV of[70].
We begin with the following solvability theorem:
Theorem 3. Suppose that q and r are chosen as in Hypothesis1.Letu0 ∈ H andµ(·) ∈M(U × [0, T ]). Then there exists a unique solutionu ∈ Lq(0, T;Vr,q) ∩C([0, T ];H)to the equation of evolution,
∂tu+ ν0Au+ ν1A(u)+ B(u) = Nµ(t), t > 0, (63)
u(0) = u0 ∈ H, (64)
which satisfies the energy equality,
12‖u(t)‖2+ ν0
∫ t
0‖A1/2u(s)‖2 ds+ ν1
∫ t
0〈A(u(s)),u(s)〉ds
= 12‖u0‖2+
∫ t
0〈Nµ(s),u(s)〉ds, t ≥ 0, and ∂tu ∈ Lq′(0, T;V′r,q). (65)
In this theorem we have
Nµt =∫
UN (U )µt (dU) ∈ L2(0, T;H) ⊂ Lq′(0, T;V′r,q), (66)
268 S. S. Sritharan
and hence the results follow from Remark 6.11 of Chapter 1 and Section 5 of Chapter 2in [43]. Moreover, the last integral in the energy equality (65) is represented as∫ t
0〈Nµ(s),u(s)〉ds=
∫ t
0
∫U〈N (U ),u(s)〉µs(dU)ds (67)
and ∣∣∣∣∫ t
0〈Nµ(s),u(s)〉ds
∣∣∣∣ ≤ (∫ t
0
∫Uκ(U )2µs(dU)ds
)1/2(∫ t
0‖u(s)‖2 ds
)1/2
. (68)
3.2. Existence of Optimal Controls: Combining Minty–Browder withTonelli–Young
We first treat a special case of our general control problem, namely, the case of the linearcontrol operator and convex cost, to understand the steps involved. We then indicate thegeneralization necessary to deal with the general case. We consider the control problemof finding (u,U ) such that
∂tu+ ν0Au+ ν1A(u)+ B(u) = LNU, t > 0, (69)
u(0) = u0 ∈ H, U ∈ L2(0, T;H), (70)
andJ(u,U ) := ∫ T0 (‖u(s)‖2Vr,q
+ ‖U (s)‖2)ds→ inf.
Theorem 4. There exists a pair
(u, U ) ∈ (Lq(0, T;Vr,q) ∩ L∞(0, T;H))× L2(0, T;H), (71)
which solves(69)–(70),and
J(u, U ) = minJ(u,U ); (u,U ) ∈ (Lq(0, T;Vr,q) ∩ L∞(0, T;H))× L2(0, T;H), (72)
where the pair(u,U ) solves(69)–(70). (73)
Proof. We restrict ourselves to admissible pairs(u,U ) ∈ (Lq(0, T;Vr,q) ∩L∞(0, T;H)) × L2(0, T;H) which solve (69)–(70) and such thatJ(u,U ) < +∞.Then noting thatJ ≥ 0 and
J(u,U )→+∞, ‖u‖ + ‖U‖ → ∞, (74)
there exists a minimizing sequence of admissible pairs(un,Un) such thatJ(un,Un) ≤R, for some largeR> 0 and lim supn→∞ J(un,Un) ≤ inf J. Here infJ is the infimumof the cost among the admissible class. We will show that the limit pair(u, U ) solves(69)–(70) and satisfiesJ(u, U ) ≤ inf J. We have
∫ T0 ‖Un(s)‖2 ds≤ R and, hence, we
also have, due to the solvability theorem and the energy equality,
‖un‖L∞(0,T;H) + ‖un‖Lq(0,T;Vr,q) ≤ C(R) (75)
Control of Navier–Stokes Equation 269
and‖∂tun‖Lq′ (0,T;V ′r,q) ≤ C(R). If we define the spaceY asY := u ∈ Lq(0, T;Vr,q);∂tu ∈ Lq′(0, T;V′r,q), then the embeddingY → Lq(0, T;H) is compact [43, Chap-ter 1, Section 5.2, Theorem 5.1]. Henceun belongs to a compact subset inLq(0, T;H).Moreover, the embeddingY → C([0, T ];V′rq) is continuous [66, Chapter III, Sec-tion 2.2]. Hence we can extract a subsequence of admissible pairs (again denoted)(un,Un) which converge to some pair(u, U ) in the following sense:un → u in theτ -topology andUn → U in the weak-topology ofL2(0, T;H). Note that due to theweak lower semicontinuity ofJ we haveJ(u, U ) ≤ lim inf J(un,Un) ≤ inf J, andhence to complete the proof we need to show that(u, U ) solves (69)–(70). We can de-
duce from Lemma 8 thatMθ
t (un,A(un),Un) = 0, ∀n, impliesM
θ
t (u,χ, U ) = 0,whereχ is the weak limit ofA(un) in Lq′(0, T;V′r,q). We need to show thatχ = A(u). Dueto Lemma 9 we also have
∂t u+ ν0Au+ ν1χ+ B(u) = LNU , t > 0. (76)
We now use the following result to deduce an energy equality for the above limit equation.
Proposition 1 [43]. Let y ∈ Lq(0, T;Vr,q), z ∈ Lq′(0, T;V′r,q), q > 1, and y0 ∈ Hbe such that
y(t) = y0+∫ t
0z(s)ds, t ∈ [0, T ]. (77)
Theny ∈ C([0, T ];H) and‖y(t)‖2 = ‖y0‖2+ 2∫ t
0 〈z(s), y(s)〉ds.
Applying this result to (69)–(70) gives the energy equality for the limit problem as
12‖u(t)‖2+ ν0
∫ t
0‖A1/2u(s)‖2 ds+ ν1
∫ t
0〈χ(r ), u(r )〉dr
=∫ t
0〈LNU (r ), u(r )〉dr + 1
2‖u0‖2. (78)
We define, forv(·) ∈ Lq(0, T;Vr,q),
Xn = ν1
∫ t
0〈A(un(s))−A(v(s)),un(s)− v(s)〉ds
+ 12‖un(t)‖2+ ν0
∫ t
0‖A1/2un(s)‖2 ds≥ 0. (79)
We have, from the energy equality for each pair in the minimizing sequence, thatun(t)and
∫ t0 ‖un(r )‖2 dr are uniformly bounded. Hence (taking subsequence)un(t) converges
weakly to u(t) in H andun(·) converges weakly tou(·) in L2(0, T; D(A1/2)) (this isactually implied by the convergence in theτ -topology). Hence, using the monotonicityproperty ofA(·), we get
lim infn→∞ Xn ≥ 1
2‖u(t)‖2+ ν0
∫ t
0‖A1/2u(s)‖2 ds. (80)
270 S. S. Sritharan
We now use the fact that each pair(un,Un) satisfies the energy equality, we rewriteXn
as
Xn = 12‖u0‖2+
∫ t
0(LNUn(r ),un(r ))dr
− ν1
∫ t
0〈A(un(s)), v(s)〉ds− ν1
∫ t
0〈A(v(s)),un(s)− v(s)〉ds. (81)
Noting thatLNUn→ LNU weakly inL2(0, T;H) andun→ u strongly inL2(0, T;H)we can take the limit in the first integral in the right-hand side. For the other two integrals,the limit can be taken due to convergence in theτ -topology and we getXn→ X with
X = 12‖u0‖2+
∫ t
0(LNU (r ), u(r ))dr
− ν1
∫ t
0〈χ(s), v(s)〉ds− ν1
∫ t
0〈A(v(s)), u(s)− v(s)〉ds≥ 0. (82)
Now using the energy equality (78) for the limit problem we finally get
ν1
∫ t
0〈χ(s)−A(v(s)), u(s)− v(s)〉ds≥ 0. (83)
From this we getχ = A(u) sinceA(·) is maximal monotone. Actually we couldsubstitutev= u+ λw with w ∈ Lq(0, T;Vr,q) to get
λ
∫ t
0〈χ(s)−A(u(s)+ λw(s)),w(s)〉ds≤ 0, ∀λ ∈ R, (84)
and takeλ → 0 and use the hemicontinuity to get∫ t
0 〈χ(s) − A(u(s)),w(s)〉ds ≤ 0,which impliesχ = A(u) sincew is arbitrary.
We now state the main existence theorem of this section.
Theorem 5. Let the control input operatorN and the integrand F(·, ·, ·) of the costfunctional satisfy the admissibility conditions(Hypothesis2 and Definition3), andlet q, r be chosen as in Hypothesis1. Then there exists an optimal pair(u, µ) ∈Y ×Mad((0, T) × U) which satisfies(63), (66)or, rather, M
θ
t (u,A(u), µ) = 0, andminimizes the relaxed cost functional:∫ T
0
∫U
F(t, u,U )µ(dt,dU) = min
∫ T
0
∫U
F(t,u,U )µ(dt,dU)
, (85)
where the minimium is taken over all admissible pairs(u,µ) ∈ Y ×Mad((0, T)× U)which satisfy
Mθ
t (u,A(u),µ) = 0. (86)
Control of Navier–Stokes Equation 271
Proof. We note that the relaxed cost functional is
J(u,µ) :=∫ T
0
∫U
F(t,u,U )µ(dU,dt)
≥∫ T
0
∫Uκ(U )2µ(dU,dt) := ‖µ‖2κ . (87)
Thus, as before, noting that 0≤ J(u,µ) and thatJ(u,µ)→ +∞ as‖µ‖κ →∞, weconsider∫ T
0
∫Uκ(U )2µ(dU,dt) ≤ R, (88)
which will give us (due to Lemma 5) a minimizing sequence of tight measuresµn in thestable topology and (sinceNµn
(·) ∈ L2(0, T;H)) a corresponding sequence of solutionsun ∈ Y such that
Mθ
t (un,A(un),µn) = 0. (89)
We haveun→ u in theτ -topology andµn→ µ in the stable topology. Hence, using the
Lemmas 7–9, we can follow the same steps as Theorem 4 to getMθ
t (u,A(u), µ) = 0.To complete the proof we need to show
Lemma 10. The relaxed cost functional is lower semicontinuous. That is, for un→ uin theτ -topology andµn→ µ in the stable topology,
lim infn→∞
∫ T
0
∫U
F(t,un,U )µn(dU,dt) ≥∫ T
0
∫U
F(t,u,U )µ(dU,dt). (90)
Proof. We show this using certain ideas from [32] and [34]. LetFN(t,u,U ) :=F(t,u,U ) ∧ N, and note that
lim infn→∞
∫ T
0
∫U
F(t,un,U )µn(dU,dt)
≥ lim infn→∞
∫ T
0
∫U
FN(t,un,U )µn(dU,dt)
≥ − lim supn→∞
∫ T
0
∫U
(FN(t,un,U )− FN(t,u,U )
)−µn(dU,dt)
+ lim infn→∞
∫ T
0
∫U
FN(t,u,U )µn(dU,dt), (91)
where, for anyϕ,ϕ(x)− := [ϕ(x)∧0].The first integral on the right-hand side is zero dueto Lemma 11 below. We consider the second integral and note that, due to Lemma 4(i),we have
lim infn→∞
∫ T
0
∫U
FN(t,u,U )µn(dU,dt) ≥∫ T
0
∫U
FN(t,u,U )µ(dU,dt), (92)
272 S. S. Sritharan
hence
lim infn→∞
∫ T
0
∫U
F(t,un,U )µn(dU,dt) ≥∫ T
0
∫U
FN(t,u,U )µ(dU,dt). (93)
Now using the Beppo–Levi theorem on the right-hand side we get the required lowersemicontinuity (90).
We now show that the first integral in (91) is zero.
Lemma 11. Letµn → µ in the stable topology andun → u in theτ -topology. Letψ(·, ·, ·): (0, T) × Vr,q,w × U → R+ be a bounded measurable function such that,∀t ∈ (0, T), ψ(t, ·, ·): Vr,q,w × U→ R+ is lower semicontinuous. Then
limn→∞
∫ T
0
∫U
(ψ(t,un,U )− ψ(t,u,U ))− µn(dU,dt) = 0. (94)
Proof. We set2(t, z,U ) := ψ(t, z,U )− ψ(t,u,U ), and, forγ > 0 andy ∈ V′rq ,
ϒm :=(t,U ) ∈ [0, T ] × U; inf
|〈y,u(t)〉−〈y,z〉|≤1/m2(t, z,U ) ≤ −γ
. (95)
Then the lower semicontinuity ofψ implies that eacht-section ofϒm is closed. Moreover,ϒm+1 ⊆ ϒm. Furthermore,ψ(t, ·,U ): Vrqw → R+ lower semicontinuous implies thataszn→ u in Vrqw we have
lim infn
2(t, zn,U ) ≥ 0. (96)
This implies that⋂m
ϒm = ∅. (97)
We have, by Lemma 4(ii),
limm
lim supn
µn(ϒm) ≤ lim sup
mµ(ϒm) ≤ µ
(⋂r>0
⋃m>r
ϒm
)
= µ(⋂
m
ϒm
)= µ(∅) = 0. (98)
Note that asun(·)→ u(·) in the weak topology ofLq(0, T;Vrq) we haveun(t)→ u(t)in the weak topology ofVrq for t a.s. Hence, if we define
ϒn := (t,U );2(t,un(t),U )− > γ = (t,U );2(t,un(t),U ) < −γ , (99)
then forn sufficiently large we have
µ(ϒn) ≤ µ(ϒm). (100)
Control of Navier–Stokes Equation 273
Thus,
lim supn
µ(ϒn) = 0. (101)
Now Lemma 4(iii) gives us
limn
∫ T
0
∫U2(t,un(t),U )µn(dU,dt) = 0, (102)
from which (94) follows.
4. The Stochastic Control Problem
4.1. A Priori Estimates, Energy Equality, and Higher-Order Moments
We begin by noting the disintegration ([60] or Corollary 3.1.2 of [7]) of the measureP for the projectionΩ → Ω, P(dµ,du) = P(µ;du)Π(dµ), whereP(µ;du) is theregular conditional distribution ofu givenµ. The martingale problem formulated in (47)reduces to that of finding the martingale solutionP(µ;du) for a fixedµt . Hence in thissection and the subsequent sections we resolve the following martingale problem: findthe Radon measureP(µ;du) onB(Ω) depending measurably onµ· such that
M t (u) := u(t)+∫ t
0
(ν0Au+ ν1A(u)+ B(u)− Nµs
)ds (103)
is aV′r,q-valued(Ω,B(Ω), Ft ,P)-martingale (whereFt = σ u(s), s ≤ t) with quadratic
variation,
〈〈M t 〉〉 :=∫ t
0g(u(s))Qg∗(u(s))ds. (104)
Definition 5. A mapu→ g(u) from H → L(H;H) is an admissible diffusion oper-ator if it satisfies the following conditions. There exist constantsC1,C2,C3 such that
(i) ‖g(u)‖ ≤ C1‖u‖+C2, ∀u ∈ H, (105)
(ii) ‖g(u1)− g(u2)‖ ≤ C3‖u1− u2‖, ∀u1,u2 ∈ H. (106)
A priori estimates for cases II–IV and also for case I in the two-dimensional domainscan be derived from a stochastic energy equality. We begin with the following result.
Proposition 2 [55], [29]. Consider the probability space(Ω,F,Ft , P)andFt -adaptedprocessesy, z,M t such thatM t be is anH-valued square integrable, right-continuousmartingale withM0 = 0, y ∈ Lq(0, T;Vr,q) a.s, z∈ Lq′(0, T;V′r,q) a.s, and, for P a.s,
y(t) = y0+∫ t
0z(s)ds+ M t , t ∈ [0, T ], (107)
274 S. S. Sritharan
with y0 ∈ H. Then the paths ofy are a.s in D([0, T ];H) (H-valued Skorohod space)and the Ito formula applies for‖y‖2: forP a.s,
‖y(t)‖2 = ‖y0‖2+ 2∫ t
0〈z(s), y(s)〉ds+ 2
∫ t
0(y(s),dMs)
+∫ t
0tr(〈〈Ms〉〉)ds. (108)
We now derive the energy estimates satisfied by every martingale solution.
Theorem 6. Let P be any probability measure on(Ω,Ft ) such thatP is carried bythe subset of pathsω ∈ Ω with (u(·,ω),µ(·,ω)) ∈ (Lq(0, T;Vrq)× L∞(0, T;H))×Mad(0, T;U). We assume also that
M t = u(t)− u0+∫ t
0
ν0Au(s)+ ν1A(u(s))+ B(u(s))− Nµs
ds (109)
is aV′rq -valued(Ω,Ft ,P)-martingale with quadratic variation,
〈〈M〉〉t =∫ t
0g(u(s))Qg∗(u(s))ds. (110)
Then
EP[‖u(t)‖2+ ν0
∫ t
0‖A1/2u(s)‖2 ds+ ν1
∫ t
0‖u(s)‖qVrq
ds
]≤ E‖u0‖2+ EΠ‖µ‖2κ + C
and
EP
[sup
s∈[0,T ]‖u(s)‖2
]≤ C1E‖u0‖2+ C2EΠ‖µ‖2κ + C3. (111)
Let the initial data satisfy E‖u0‖2l < +∞, for 1< l < +∞, and either let the controlset be bounded in the sense that∃R> 0such that0< κ(U ) ≤ R,∀U ∈ U, or let the2l-order moments be finite for the chattering controls E[
∫ T0
∫U κ(U )
2lµt (dU)dt] < +∞.Then
EP
[sup
s∈[0,T ]‖u(t)‖2l + ν0
∫ t
0‖u(s)‖2l−2‖A1/2u(s)‖2 ds
+ν1
∫ t
0‖u(s)‖2l−2‖u(s)‖qVrq
ds
]≤ C1E
[‖u0‖2l +
∫ T
0
∫Uκ(U )2lµt (dU)dt + C2
], 1≤ l <∞. (112)
Proof. We introduce a stopping time,τN = inf t ≤ T; ‖u(t,ω)‖ ≥ N;ω ∈ Ω .Notethat if u ∈ Lq(0, T;Vrq), then Au,A(u), and by Lemma 2 alsoB(u), all belong to
Control of Navier–Stokes Equation 275
Lq′(0, T;V′rq) andNµ· ∈ L2(0, T;H) ⊂ Lq′(0, T;V′rq). Hence, using Proposition 2,we can write the energy equality as
12‖u(t ∧ τN)‖2+ ν0
∫ t∧τN
0‖A1/2u(s)‖2 ds+ ν1
∫ t∧τN
0〈A(u(s)),u(s)〉ds
= 12‖u0‖2+
∫ t∧τN
0(Nµ(s),u(s))ds+
∫ t∧τN
0(u(s), g(u(s))dW)
+ 12
∫ t∧τN
0tr(g(u(s))Qg(u(s))∗
)ds. (113)
Applying Young’s inequality to the first integral on the right,
12‖u(t ∧ τN)‖2+
∫ t∧τN
0
ν0‖A1/2u(s)‖2+ ν1〈A(u(s)),u(s)〉
ds
≤ 12‖u0‖2+ Cε
∫ t∧τN
0‖Nµ(s)‖2 ds+
∫ t∧τN
0(u(s), g(u(s))dW)
+ ε∫ t∧τN
0‖u(s)‖2 ds+ 1
2
∫ t∧τN
0tr(g(u(s))Qg(u(s))∗
)ds. (114)
Using admissibility condition (105), using estimate (35), and redefiningCε as needed,
12‖u(t ∧ τN)‖2+
∫ t∧τN
0
ν0‖A1/2u(s)‖2+ ν1‖u(s)‖qVrq
ds
≤ 12‖u0‖2+ Cε
∫ t∧τN
0‖Nµ(s)‖2 ds+
∫ t∧τN
0(u(s), g(u(s))dW)
+ Cε
∫ t∧τN
0‖u(s)‖2 ds. (115)
Taking expectation,
EP[
12‖u(t ∧ τN)‖2+
∫ t∧τN
0
ν0‖A1/2u(s)‖2+ ν1‖u(s)‖qVrq
ds
]≤ 1
2 E[‖u0‖2
]+ CεEΠ[‖µ‖2κ]+ CεE
[∫ t∧τN
0‖u(s)‖2 ds
]. (116)
Dropping the second and third terms on the left-hand side and using Gronwall’s inequal-ity, we get
EP [ 12‖u(t ∧ τN)‖2
] ≤ 12 E
[‖u0‖2]+ CεE
Π[‖µ‖2κ]+ C. (117)
Now using this estimate on the right-hand side of (116) we also get
EP[ν0
∫ t∧τN
0‖A1/2u(s)‖2 ds+ ν1
∫ t∧τN
0‖u(s)‖qVrq
ds
]≤ E
[‖u0‖2]+ CεE
Π[‖µ‖2κ]+ C. (118)
276 S. S. Sritharan
Now note that by the monotone convergence theorem we get the estimate
EP[ν0
∫ T
0‖A1/2u(s)‖2 ds+ ν1
∫ T
0‖u(s)‖qVrq
ds
]≤ 1
2 E[‖u0‖2
]+ CεEΠ[‖µ‖2κ]+ C. (119)
Moreover, setting
uN(t) =
u(t) for ‖u(t)‖ ≤ N,0 for ‖u(t)‖ > N,
we have‖uN+1(t)‖ ≥ ‖uN(t)‖ and
EP [ 12‖uN(t)‖2
] = EP [ 12‖u(t ∧ τN)‖2
]≤ 1
2 E[‖u0‖2
]+ CεEΠ[‖µ‖2κ]+ C. (120)
Hence by the monotone convergence theorem we get
EP [ 12‖u(t)‖2
] ≤ 12 E
[‖u0‖2]+ CεE
Π[‖µ‖2κ]+ C. (121)
Taket < τN and note that from (115) we also have
EP[
sup0≤s≤t
12‖u(t)‖2+ ν0
∫ t
0‖A1/2u(s)‖2 ds+ ν1
∫ t
0‖u(s)‖qVrq
ds
]≤ 1
2 E‖u0‖2+ CεEΠ
[∫ t
0‖Nµ(s)‖2 ds
]+ E
[sup
0≤s≤t
∫ s
0(u(r ), g(u(r ))dW(r ))
]+ CεE
[∫ t
0‖u(s)‖2 ds
]. (122)
We now use the Burkholder–Gundy inequality [56] for the last term above,
E
[sup
0≤s≤t
∫ s
0(u(r ), g(u(r ))dW(r ))
]≤ C E
[tr
⟨⟨∫ t
0(u(r ), g(u(r ))dW(r ))
⟩⟩]1/2
≤ C E
(∫ t
0‖u(s)‖2‖g(u(s))‖2 tr Qds
)1/2
≤ C E
(sup
0≤s≤t‖u(s)‖2
)1/2(∫ t
0‖g(u(s))‖2 tr Qds
)1/2
≤ C(tr Q)1/2(
E
[sup
0≤s≤t‖u(s)‖2
])1/2(E
[∫ t
0‖u(s)‖2 ds
])1/2
≤ Cε · E[∫ t
0‖u(s)‖2 ds
]+ ε · E
[sup
0≤s≤t‖u(s)‖2
]. (123)
Control of Navier–Stokes Equation 277
Hence we get, combining (121)–(123),
EP
[sup
0≤t≤τN
‖u(t)‖2]≤ C1E‖u0‖2+ C2EΠ‖µ‖2κ + C3. (124)
As before, applying the Monotone convergence theorem we get
EP[
sup0≤t≤T
‖u(t)‖2]≤ C1E‖u0‖2+ C2EΠ‖µ‖2κ + C3. (125)
This implies that if we define
ΩN :=ω ∈ Ω; sup
0≤t≤T‖u(t,ω)‖ < N
, (126)
we have∫ΩN
sup0≤t≤T
‖u(t)‖2P(du)+∫Ω\ΩN
sup0≤t≤T
‖u(t)‖2P(du) ≤ C. (127)
Hence dropping the first integral and noting that, inΩ\ΩN , sup0≤t≤T ‖u(t,ω)‖ ≥ N,we get
N2P Ω\ΩN ≤ C. (128)
Note also thatP ω ∈ Ω; τ N < T ≤ P Ω\ΩN ≤ C/N2.Hence, lim supN→∞ P ω ∈Ω; τ N < T = 0. Thus,τ N → T as N → ∞. To get the higher-order estimates webegin with the scalar-valued semimartingale
ζt = ζ0+∫ t
0ϕ(s)ds+ Nt , (129)
whereζt = ‖u(t)‖2, ζ0 = ‖u0‖2, Nt = 2∫ t
0 (u, g(u)dW(s)), and
ϕ := −2ν0‖A1/2u‖2− 2ν1〈A(u),u〉 + tr(g(u)Qg(u)∗)+ 2〈Nµt ,u〉. (130)
We now recall the Ito formula for scalar-valued semimartingales [35] to get
ζ lt = ζ l
0 + l∫ t
0ζ l−1
s ϕ(s)ds+ l∫ t
0ζ l−1
s d Ns + 12l (l − 1)
∫ t
0ζ l−2
s d〈N〉s. (131)
Hence,
‖u(t)‖2l + 2l∫ t
0‖u(s)‖2l−2
[ν0‖A1/2u(s)‖2+ ν1〈A(u(s)),u(s)〉
]ds
≤ ‖u0‖2l + l∫ t
0‖u(s)‖2l−2 tr
(g(u(s))Qg∗(u(s))
)ds
+ 2l∫ t
0‖u(s)‖2l−2〈Nµ(s),u(s)〉ds
+ l∫ t
0‖u(s)‖2l−2(u(s), g(u(s)))dW(s)
+ 2l (l − 1)∫ t
0‖u(s)‖2l−2‖g(u(s))‖2 ds.
278 S. S. Sritharan
Taking expectation we get
EP‖u(t ∧ τN)‖2l+2l
∫ t∧τN
0‖u(s)‖2l−2
[ν0‖A1/2u(s)‖2+ν1〈A(u(s)),u(s)〉
]ds
≤ E
[‖u0‖2l]+ C(l , tr Q)EP
∫ t∧τN
0‖u(s)‖2l ds
+ 2l EP
∫ t∧τN
0‖u(s)‖2l−1‖Nµ(s)‖ds
. (132)
That is,
EP‖u(t ∧ τN)‖2l+2l
∫ t∧τN
0‖u(s)‖2l−2
[ν0‖A1/2u(s)‖2+ν1〈A(u(s)),u(s)〉
]ds
≤ E
[‖u0‖2l]+ C(l , tr Q)EP
∫ t∧τN
0‖u(s)‖2l ds
+ 2l
∫ t∧τN
0(EP‖u(s)‖2l )1−1/2l (E‖Nµ(s)‖2l )1/2l ds. (133)
Thus denotingm(t) = EP‖u(t)‖2l we get
m(t ∧ τN) ≤ m0+ C∫ t∧τN
0
[m(s)+ g(s)m(s)1−1/2l
]ds, (134)
whereg(s) := (EP‖Nµ(s)‖2l )1/2l . We now use Lemma 12 below to conclude that
EP‖u(t ∧ τN)‖2l ≤ C · E‖u0‖2l +
∫ T
0‖Nµ(s)‖2l ds
. (135)
That is,
EP‖u(t ∧ τN)‖2l ≤ C · E‖u0‖2l +
∫ T
0
∫Uκ(U )2lµt (dU)ds
. (136)
We can now use the monotone convergence theorem as before to get
EP‖u(t)‖2l ≤ C · E‖u0‖2l+
∫ T
0
∫Uκ(U )2lµt (dU)ds
, t ∈ (0, T). (137)
We now note that
EP∫ T
0‖u(t)‖2l dt ≤ C · T · E
‖u0‖2l +
∫ T
0
∫Uκ(U )2lµt (dU)ds
,
t ∈ (0, T). (138)
Control of Navier–Stokes Equation 279
We setG := E‖u0‖2l + ∫ T0
∫U κ(U )
2lµt (dU)ds. We consider the term∫ t
0(EP‖u(s)‖2l )1−1/2l (E‖Nµ(s)‖2l )1/2l ds≤
∫ T
0G1−1/2l · G1/2l ds
= T G. (139)
Hence we deduce that
EP∫ T
0‖u(s)‖2l−2
[ν0‖A1/2u(s)‖2+ ν1〈A(u(s)),u(s)〉
]ds
≤ C(T)G. (140)
We finally note that
EP
[sup
t∈[0,T ]‖u(t)‖2l
]≤ C · G+C E
[sup
t∈[0,T ]
∫ t
0‖u(s)‖2l−2(u, g(u)dW)
]. (141)
Now using the Buckholder inequality and the fact that‖g(u)‖ ≤ C‖u‖,
EP
[sup
t∈[0,T ]‖u(t)‖2l
]≤ C · G+ C E
[∫ T
0‖u(s)‖4l ds
]1/2
≤ C · G+ εE
[sup
t∈[0,T ]‖u(t)‖2l
]
+ CεE
[∫ T
0‖u(s)‖2l ds
]. (142)
Hence using (138), (142) we get
EP
[sup
t∈[0,T ]‖u(t)‖2l
]
≤ C1(ε)
(E
‖u0‖2l +
∫ T
0
∫Uκ(U )2lµt (dU)ds
). (143)
We now state the lemma used above which is due to Krylov [38, Section 2.5].
Lemma 12. Let m(·) ∈ C[0, T ] satisfy m(t) ≤ m0+C∫ t
0 [m(s)+ g(s)m(s)1−1/2l ] ds,then
m(t) ≤[m1/2l
0 + C∫ t
0expC(t − s)g(s)ds
]2l
, ∀t ∈ [0, T ]. (144)
4.2. Continuity and Uniform Integrability of the Martingale
Corollary 4 (Uniform Integrability of the Martingale). Let g(u) be admissible(Defi-nition 5) and let the initial data and chattering control satisfy the higher-order momentconditions
E[‖u0‖2l
]<∞, E
[∫ T
0
∫Uκ(U )2lµt (dU)dt
]<∞. (145)
280 S. S. Sritharan
Then by suitably choosing l andε we can obtain uniform integrability of the martingaleMθ
t :
EP [|Mθt |1+ε
] ≤ C · E[‖u0‖2l+
∫ T
0
∫Uκ(U )2lµt (dU)dt
], ∀θ ∈ Vrq . (146)
Proof. We analyze the cases to see when we need higher-order moment estimates forthe uniform integrability of the martingale. We have
Mθt = 〈u(t),θ〉 +
∫ t
0〈ν0Au(s)+ ν1A(u(s))+ B(u(s))− Nµ(s),θ〉ds,
θ ∈ Vrq . (147)
Thus, for 0< α ≤ 1,
|Mθt | ≤ C
‖u(t)‖ +
∫ t
0
(ν0‖A1/2u(s)‖ + ν1‖u(s)‖q−1
Vrq
+ ‖u(s)‖α‖u(s)‖2−α1/2 + ‖Nµ(s)‖)
ds
. (148)
We consider the equation term by term. Note that since we have, from the energy in-equality (Theorem 6),
E[‖u(t)‖2] ≤ C E
‖u0‖2+
∫ T
0
∫Uκ(U )2µt (dU)dt + 1
, (149)
in the first term in (148) we can takeε = 1. We now consider
E
∣∣∣∣[∫ T
0‖A1/2u(t)‖dt
]∣∣∣∣1+ε≤C E
‖u0‖2+
∫ T
0
∫Uκ(U )2µt (dU)dt+1
, (150)
again due to the energy inequality. Thus in the second term in (148) we can also takeε = 1. Similarly takingε = 1/(q − 1), E| ∫ T
0 ‖u‖q−1Vrq
dt|1+ε, we get
E
∣∣∣∣∫ T
0‖u‖q−1
Vrqdt
∣∣∣∣q/(q−1)
≤ C E∫ T
0‖u‖qVrq
dt
≤ C E
‖u0‖2+
∫ T
0
∫Uκ(U )2µt (dU)dt + 1
. (151)
Similarly, we consider the termE|[∫ T0 ‖Nµ(t)‖dt]|1+ε, taking ε = 1,
E|[∫ T0 ‖Nµ(t)‖dt]|2 ≤ T E[
∫ T0 ‖Nµ(t)‖2 dt]. Thus up to this point we do not need
to use the higher-order moment condition. We now come to the integral term containingthe inertia term which is estimated in the following way. First consider the caseν1 = 0,
Control of Navier–Stokes Equation 281
and note the corresponding energy estimates (Theorem 6). Now we have
E
∣∣∣∣[∫ T
0‖u(t)‖α‖A1/2u‖2−α dt
]∣∣∣∣1+ε , (152)
with α = 12 for n = 3 andα = 1 for n = 2. Forn = 2 we have
E
∣∣∣∣[∫ T
0‖u(t)‖‖u(t)‖1/2 dt
]∣∣∣∣1+ε≤ E
[sup
t‖u(t)‖(1+ε)(4−q)/2
(∫ T
0‖u(t)‖(q−2)/2‖u(t)‖1/2 dt
)1+ε]
≤ C
E
[sup
t‖u(t)‖r (1+ε)(4−q)/2
]1/r
×
E
[∫ T
0‖u(t)‖(q−2)‖u(t)‖21/2 dt
](1+ε)/2, (153)
where 1/r + (1+ ε)/2 = 1, r = 2/(1− ε). Thus we need higher-order estimates ofthe type (112) with power, 2l = max((1+ ε)/(1− ε))(3− q),q. For n = 3, we canestimate as follows:
E
∣∣∣∣[∫ T
0‖u(t)‖1/2‖u(t)‖3/21/2 dt
]∣∣∣∣1+ε≤ E
[sup
t‖u(t)‖(1+ε)(7−3q)/2
(∫ T
0‖u(t)‖3(q−2)/2‖u(t)‖3/21/2 dt
)1+ε]
≤ C
E
[sup
t‖u(t)‖r (1+ε)(7−3q)/2
]1/r
×
E
[∫ T
0‖u(t)‖(q−2)‖u(t)‖21/2 dt
]3(1+ε)/4, (154)
where 1/r + 3(1+ ε)/4= 1, r = 4/(1− 3ε). Thus we need higher-order estimates ofthe type (112) with power, 2l = max(2(1+ ε)/(1− 3ε))(7− 3q),q.
We now consider the case ofν1 > 0, and look at the viscosity as well as the inertiaterms. Whenν1 > 0 we have an additional bound on the momentE[
∫ T0 ‖u(t)‖qVrq
dt] <+∞. This allows us to get the uniform integrability without the higher-order momentsas we see below. First note that
E
[∣∣∣∣∫ T
0‖u(t)‖q−1
Vrqdt
∣∣∣∣1+ε]≤ C
(E
[∫ T
0‖u(t)‖qVrq
dt
])(1+ε)(q−1)/q
. (155)
Thus we need(1+ ε)(q − 1)/q ≤ 1, that is,q ≤ 1+ 1/ε. Hence for this term we do
282 S. S. Sritharan
not need any higher-order moments. For the 2− D case we start with
E
∣∣∣∣[∫ T
0‖u(t)‖‖u(t)‖1/2 dt
]∣∣∣∣1+ε≤ C E
[sup
t‖u(t)‖(1+ε)
(∫ T
0‖u(t)‖q1/2 dt
)(1+ε)/q]
≤ C
E
[sup
t‖u(t)‖2
](1+ε)/2E
[∫ T
0‖u(t)‖q1/2 dt
](1+ε)/q. (156)
Thus we can take 0< ε < 1 andq ≥ 2 to get the uniform integrability and hence donot need high-order moments. We now look at the 3− D case:
E
∣∣∣∣[∫ T
0‖u(t)‖1/2‖u(t)‖3/21/2 dt
]∣∣∣∣1+ε≤ C E
[sup
t‖u(t)‖(1+ε)/2
(∫ T
0‖u(t)‖3/21/2 dt
)1+ε]
≤ C
E
[sup
t‖u(t)‖2
](1+ε)/4E
[∫ T
0‖u(t)‖q1/2 dt
]3r (1+ε)/2q1/r
, (157)
where 1/r + (1+ ε)/4= 1, r = 4/(3− ε). We need to takeq ≥ 6(1+ ε)/(3− ε) andthus do not need any higher-order moments.
We thus note that only for the caseν1 = 0 do we need high-order moment estimatesto get uniform integrability.
4.3. Tightness of Measures
Definition 6. We denote byτ S-topology the supremum of the topologiesτ S1, . . . , τ
S4,
τ S = τ S1 ∨ τ S
2 ∨ τ S3 ∨ τ S
4, whereτ S1 := L∞(0, T;H)-weak-star,τ S
2 := Lq(0, T;Vr,q)-weak,τ S
3 := D([0, T ];V′rq)-Skorohod J-topology, andτ S4 := Lq(0, T;H)-strong.
We now recall the following results concerning Lusin topology.
Proposition 3 [47]. Let E1, . . . ,En be Lusin topological spaces, with topologies de-noted byτ 1, . . . , τ n. We assume thatEi are subsets of a topological spaceX such thatthe canonical embeddingsEi → X are continuous. LetΩ := E1 ∩ · · · ∩ En and callτthe supremum of the topologies induced onΩ byτ 1, . . . , τ n. Then:
(i) Ω endowed with the topologyτ is a Lusin space.(ii) Let (µk)k∈N be a sequence of Borel probability measures on(Ω,B(Ω)), whereB(Ω) is the Borel algebra, such that the images ofµk on (Ei ,B(Ei )) denotedby (µi
k)k∈N are tight forτ i , for all i = 1, . . . ,n. Then(µk)k∈N is tight forτ .
Control of Navier–Stokes Equation 283
Theorem 7. SupposeΩ∗ := L∞(0, T;H)∩Lq(0, T;Vrq)∩D(0, T;V′rq)andΩ isΩ∗
endowed with the topologyτ S. ThenΩ is a completely regular Lusin space. Moreover,
let ˜Ω :=M(U× [0, T ]) endowed with the stable topology. ThenΩ := Ω× ˜Ω endowedwith the product topology is a completely regular Lusin space.
We note here that a topological space is called completely regular if it is Hausdorffseparated and its topology can be defined by a set of pseudodistances.
Proof. We note first the following continuous embeddings,L∞(0, T;H)w∗ → Lq(0, T;V′rq)σ , Lq(0, T;Vr,q)σ → Lq(0, T;V′rq)σ , D([0, T ];V′rq) → Lq(0, T;V′rq)σ , and
Lq(0, T;H) → Lq(0, T;V′rq)σ . Then(Ω, τ ) is Lusin due to Proposition 3 above. Re-calling Lemma 3 in Section 3.1 we deduce thatΩ is also Lusin due to Lemma 4 inChapter II of Part I of [59] on products of Lusin spaces.
Theorem 8. The class measuresPdefined in Theorem6on the Lusin space(Ω,B(Ω))are tight if
EP [‖µ‖2κ] = EΠ[‖µ‖2κ] ≤ C. (158)
Proof. From the a priori energy estimate (Theorem 6), we deduce thatP satisfy
EP[
sup0≤t≤T
‖u(t)‖2+ ν0
∫ T
0‖A1/2u(s)‖2 ds
+ ν1
∫ T
0‖u(s)‖qVrq
ds+ ‖µ‖2κ]≤ C. (159)
Thus we deduce that theu-marginal ofP is tight in
L∞(0, T;H)w∗ ∩ L2(0, T; D(A1/2))σ ∩ Lq(0, T;Vrq)σ (160)
and, by Lemma 5, theµmarginal ofP is tight inM(U× [0, T ])which is endowed withthe stable topology. We now deduce the tightness inD([0, T ];V′rq) with J-topology.We recall the following results on tightness of the laws of semimartingales of Metivier[47, Chapter 4, Theorem 3]. We need to verify the following two facts:
(i) ∀t ∈ [0, T ], the marginal distributions ofu(t) are tight inV′rq ,(ii) ∀θ ∈ Vrq , ∀N ∈ N, for all stopping timesτN , we should have,∀ε > 0,∃δ > 0,
EP[∫ τN+ε
τN
‖ν0Au(s)+ ν1A(u(s))+ B(u(s))− Nµs‖V ′rqds
]≤ δ, (161)
andEP[∫ τN+ετN
tr〈〈Ms〉〉ds] ≤ δ.
We start by noticing the fact that the estimateEP[‖u(t)‖2] ≤ C,∀t ∈ [0, T ], ensuresthat,∀R > 0, by Tchebycheff’s inequality,P u; ‖u‖ ≥ R ≤ C/R2, and hence the
284 S. S. Sritharan
marginal distributions ofu(t) are tight inHσ . Moreover, sinceH → V′rq is a compactembedding, the marginal distributions ofu(t) are tight inV′rq . We now verify (ii):
EP[∫ τN+ε
τN
‖ν0Au(s)+ ν1A(u(s))+ B(u(s))− Nµs‖V ′rqds
]≤ δ. (162)
Note that
EP[∫ τN+ε
τN
‖Au(s)‖V ′rqds
]≤ C EP
[∫ τN+ε
τN
‖Au(s)‖V ′1,2 ds
](163)
and that
EP[∫ τN+ε
τN
‖Au(s)‖V ′1,2 ds
]≤ ε1/2
(EP[∫ τN+ε
τN
‖A1/2u(s)‖2 ds
])1/2
≤ ε1/2C, (164)
by the energy estimate (Theorem 6). Similarly,
EP[∫ τN+ε
τN
‖A(u(s))‖V ′rqds
]≤ C · EP
[∫ τN+ε
τN
‖u(s)‖q−1Vrq
ds
]≤ ε1/qC ·
(EP[∫ τN+ε
τN
‖u(s)‖qVrqds
])(q−1)/q
≤ ε1/qC, (165)
again by the energy estimate (Theorem 6). Now consider
EP[∫ τN+ε
τN
‖Nµs‖V ′rqds
]≤ C · EP
[∫ τN+ε
τN
‖Nµs‖ds
]≤ C · ε1/2
(EP[∫ τN+ε
τN
‖Nµs‖2 ds
])1/2
≤ C · ε1/2(EP [‖µ‖2κ])1/2 ≤ ε1/2 · C, (166)
due to (158). We next look at the nonlinear term and use Lemma 2,
EP[∫ τN+ε
τN
‖B(u(s))‖V ′rqds
]≤ ε1/q
(EP[∫ τN+ε
τN
‖B(u(s))‖q′V ′rqds
])1/q′
≤ C · ε1/q, (167)
again due to the energy estimate (Theorem 6). For cases I and IV the argument isas follows. We have the estimate [40, Lemmas 1 and 2, Chapter 1]‖B(u)‖V ′1,2 ≤C‖u‖2α‖A1/2u‖2−2α, with α = 1
2 for G ⊂ R2 and α = 14 for G ⊂ R3. We only
Control of Navier–Stokes Equation 285
need the fact that 0< α < 1 in this particular part of the proof. Thus,
EP[∫ τN+ε
τN
‖B(u(s))‖V ′1,2 ds
]≤ C · EP
[sup
0≤t≤T‖u(t)‖2α
∫ τN+ε
τN
‖A1/2u(s)‖2−2α ds
]≤ εαC ·
(EP[
sup0≤t≤T
‖u(t)‖2α])α (
EP[∫ τN+ε
τN
‖A1/2u(s)‖2 ds
])1−α
≤ Cεα, (168)
due to the energy estimate (Theorem 6).We finally estimate
EP[∫ τN+ε
τN
tr〈〈g(u(s))Qg∗(u(s))〉〉ds
]≤ EP
[∫ τN+ε
τN
‖u(s)‖2 ds
]≤ εEP
[sup
0≤t≤T‖u(t)‖2
]≤ ε · C, (169)
due to the energy estimate (Theorem 6). We have thus verified conditions (i) and (ii) andhence the marignal distributions ofu are tight in theJ-topology ofD([0, T ];V′rq).
We now establish the tightness ofP in Lq(0, T;H). Note first that, sinceP istight in D([0, T ];V′rq), ∀ε > 0, there exists a compact setKε ⊂⊂ D([0, T ];V′rq) suchthat,∀P ∈ P,
P(Kε) ≥ 1− ε. (170)
We start with the result:
Lemma 13 [43]. Let N ⊂ Lq(0, T;H) which is included in a compact set ofLq(0, T;V′rq) and such that
supu∈N
∫ T
0‖u(t)‖qVrq
dt <∞, (171)
thenN is relatively compact in Lq(0, T;H).
The way we are going to generate such a subset is by the intersectionLq(0, T;H) ∩D(0, T;V′rq). In fact, note that,∀ε > 0, ∃Lε > 0 such that,∀P ∈ P,
P
u;u ∈ Lq(0, T;H);∫ T
0‖u(t)‖qVrq
dt ≤ Lε
≥ 1− ε, (172)
which follows due to the energy estimate (Theorem 6)EP[∫ T
0 ‖u(t)‖qVrqdt] ≤ C. Now,
using the setKε defined earlier we set
Nε = Kε ∩
u;u ∈ Lq(0, T;H);∫ T
0‖u(t)‖qVrq
dt ≤ Lε
. (173)
286 S. S. Sritharan
We get the result that,∀ε > 0, P(Nε) ≥ 1 − 2ε, ∀P ∈ P. Moreover, due toLemma 13 above,Nε ⊂ Lq(0, T;H) is relatively compact since we started with thesetNε ⊂ Lq(0, T;H) which is included in the compact set ofLq(0, T;V′rq) due to theinclusion
Lq(0, T;H) ∩ D(0, T;V′rq) → Lq(0, T;V′rq). (174)
5. The Martingale Problem of the Stochastic Navier–Stokes Equation
5.1. Equivalent Forms of the Martingale Problem
We now formulate certain equivalent forms of the martingale problem.
Theorem 9. Consider the filtered probability space(Ω,F,Ft ) whereFt := σ (u(s),µ(s)),0 ≤ s ≤ t. Then the following martingale formulations are equivalent. Find aRadon measureP(·): B(Ω)→ [0,1] such that:
(I) M t = u(t)+∫ t
0[ν0Au(s)+ ν1A(u(s))+ B(u(s))− Nµ(s)] ds (175)
is a V′rq -valued right-continuous(Ω,F,Ft ,P)-martingale with quadratic
variation,
〈M〉t :=∫ t
0g(u(s))Qg(u(s))∗ ds. (176)
(II) ∀θ ∈ Vrq , Mθt := 〈M t ,θ〉 (177)
is a right-continuous(Ω,F,Ft ,P)-martingale with quadratic variation,
〈Mθ 〉t :=∫ t
0(g(u(s))θ,Qg(u(s))∗θ)ds. (178)
(III) If f (·) is a cylindrical(tame) function defined as
f (u) := ϕ(〈u,θ1〉, . . . , 〈u,θm〉), (179)
with θi ∈ Vrq andϕ(·) ∈ C∞0 (Rm), thenM f
t := f (u(t))− ∫ t0 L f (u(s))ds is
an (Ω,F,Ft ,P)-martingale where
L f (u) := 12 tr
(g(u)Qg∗(u)
∂2 f (u)∂u2
)−(ν0Au+ ν1A(u)+ B(u)− Nµ,
∂ f
∂u
). (180)
Letψ ∈ Cb(Ω) beFs-measurable, thenMθt , being an(Ω,F,Ft ,P)-martingale, is
the same as
EP [ψ(·)(Mθt − Mθ
s)] = 0. (181)
In the case ofg(u) = I , ν1 = 0, it is possible to express this martingale problem in a
Control of Navier–Stokes Equation 287
more convenient way. Namely,
EP [ψ(·)exp(i 〈M t − Ms,θ〉)] = exp(− 1
2(θ,Qθ))EP[ψ(·)]. (182)
The proof of this result is same as in the finite-dimensional case [65], [69]
Remark 1. The equivalent form in terms of the complex exponential function givenin (182) can be used in the special case whereg= I andν1 = 0 to take the limit of theapproximate martingale solutions and also in proving the existence of optimal controlswithout additional uniform integrability conditions which would require higher-ordermoments bounds.
5.2. Equivalence Between a Martingale Solution and a Weak Solution
Definition 7. A weak solution of the stochastic Navier–Stokes equation with initialdataµ0, law of the controlΠ, and a trace class operatorQ onH, consists of a stochasticbasis(Θ, 6,6t ,m) and on this basis an adaptedV′rq -valued processy, an H-valuedWiener processW with covarianceQ, anH-valued random variabley0 with lawµ0, andan adapted measure-valued stochastic processµ with law Π such that
(i) Em
sup
s∈[0,T ]‖y(s)‖2+ ‖y(t)‖2+
∫ t
0
(ν0‖A1/2y(s)‖2+ ν1〈A(y(s)), y(s)〉)ds
≤ C
Em‖y0‖2+ Em‖µ‖2κ + 1
. (183)
(ii) y(t) = y0−∫ t
0
ν0Ay(s)+ ν1A(y(s))+ B(y(s))− Nµs
ds
+∫ t
0g(y(s))dW(s), ma.s. (184)
Theorem 10. P is a martingale solution toMc(Vrq ,H,V′rq , A,A, B,Q,µ0,Π
)if
and only if there exists a weak solution(Θ, 6,6t ,m, y, y0,µ,W) such thatP⊗ µ0 isthe image of the measurem due to the mapω→ ((y(·,ω),µ(·,ω)),u0(ω)).
Proof. Let (Θ, 6,6t ,m, y,W,µ,u0) be a weak solution. We defineP as the imageof the measurem under the mapω → (y(ω),µ(ω)). Then for each6s measurablefunctionϕ ∈ Cb(Θ), we have
EP [ϕ(·) · (M t (·)− Ms(·))]= Em [ϕ((y,µ)) · (M t ((y,µ))− Ms((y,µ)))] (185)
= Em[ϕ((y,µ)) ·
(∫ t
sg(y(r ))dW(r )
)]= 0, (186)
sinceNt =∫ t
0 g(y(r ))dW(r ) is a square integrableH-valued martingale due to theadmissibility condition (105) ong and the energy estimate (183) on the weak solution.
288 S. S. Sritharan
We also check that
EP [ϕ(·) · (‖M t (·)‖2− ‖Ms(·)‖2)]
= Em [ϕ((y,µ)) · (‖M t ((y,µ))‖2− ‖Ms((y,µ))‖2)]
= Em[ϕ((y,µ)) ·
(∫ t
sg(y(r ))Qg(y(r ))∗ dr
)]= EP
[ϕ(·) ·
(∫ t
sg(·)Qg(·)∗ dr
)]. (187)
ThusP is a martingale solution to (184).To prove the converse letPbe a martingale solution toMc(Vrq ,H,V
′rq , A,A, B,Q,
µ0,Π). Then we have, by Theorem 6, each martingale solution satisfies the energyinequality (183). Moreover,
M t = y(t)− y0+∫ t
0
(ν0Ay(s)+ ν1A(y(s))+ B(y(s))− Nµs
)ds (188)
is aV′rq -valued martingale with quadratic variation,〈〈M〉〉t =∫ t
0 g(y(s))Qg(y(s))∗ ds,with
EP[∫ T
0‖g(y(s))‖2 ds
]≤ C · EP
[∫ T
0‖y(s)‖2 ds
]< +∞. (189)
Then from Theorem 3.35 of [36] and Theorem 1 and Lemma 2 in Chapter V of [47],there exists anH-valued Wiener processβt on a standard extension of the stochasticbasis(Ω,F,Ft ,P) in the form(Θ, 6,6t ,m) = (Ω,F,Ft ,P)× (Ω′,F ′,F ′t ,P′), with〈〈β〉〉t = Qt , such thatM t =
∫ t0 g(y(s))dβs. Thus each martingale solution is a weak
solution.
5.3. Existence of Martingale Solutions
The method for establishing the existence of a martingale solution is as follows. Weconstruct approximate martingale solutionsPN(du,dµ) which solve according to thetheory of Stroock and Varadhan [65] the Galerkin approximated martingale problems of(47). We then show the tightness of these measures and take the limit to get the solutionP of the martingale problem. We begin with
Lemma 14. The mappingy(·)→ A(y(·)) fromΩ→ Lq′(0, T;V′rq) is Borel measur-able.
Proof. Note that∀v(·) ∈ Lq(0, T;Vrq) the mappingy(·) → ∫ T0 〈A(y(t)), v(t)〉dt is
continuous fromLq(0, T;Vrq)→ R. In fact, consider a sequenceyn(·) ∈ Lq(0, T;Vrq)
converging strongly toy(·) ∈ Lq(0, T;Vrq). Then we have, fort almost everywhere in[0, T ], yn(t)→ y(t) in Vrq . Now, by the demicontinuity ofA proved in Theorem 2, wehave, fort almost everywhere in [0, T ],A(yn(t))→ A(y(t)), in the weak-star topologyof V′rq . In particular,∀v(t) ∈ Vrq ,
〈A(yn(t)), v(t)〉 → 〈A(y(t)), v(t)〉, t a.e in [0, T ]. (190)
Control of Navier–Stokes Equation 289
Note also that|〈A(yn(t)), v(t)〉| ≤ gn(t), for gn(t) := C‖yn(t)‖q−1Vrq‖v(t)‖Vrq and
gn(·) → g(·) in L1(0, T), due to the strong convergence ofyn(·) in Lq(0, T;Vrq).
Hence,∫ T
0 〈A(yn(t)), v(t)〉dt → ∫ T0 〈A(y(t)), v(t)〉dt, due to the dominated conver-
gence theorem and this proves thaty(·) → ∫ T0 〈A(y(t)), v(t)〉dt is continuous from
Lq(0, T;Vrq)→ R. Hence,A(·): Lq(0, T;Vrq)→ Lq′(0, T;V′rq)σ is continuous and
we conclude thatA(·) is Borel measurable fromΩ→ Lq′(0, T;V′rq)σ .We use this result to define the image ofPN under the map(y,µ)→ (y,µ,A(y))
as QN
:= PN (I , I ,A(·))−1 or QN(A) := PN ω ∈ Ω; (ω,A(ω)) ∈ A, for A ∈
B(Ω× Lq′(0, T;V′rq)σ ).By Theorem 8,PN are tight onΩ with Lusin topology and the energy estimate
(Theorem 6) gives
EPN
[∫ T
0‖Ay(t)‖q′V ′rq
dt
]≤ C, (191)
and, thus,∃ρε > 0 such that
supN
QN(ω, v) ∈ Ω× Lq′(0, T;V′rq);
∫ T
0‖v(t)‖q′V ′rq
dt ≥ ρε≤ ε. (192)
Sincev ∈ Lq′(0, T;V′rq);∫ T
0 ‖v(t)‖q′
V ′rqdt ≤ ρε is a compact set inLq′(0, T;V′rq)σ ,
the measuresQN
form a tight family onΩ× Lq′(0, T;V′rq)σ .
We now take the limits. LetΩ := Ω× Lq′(0, T;V′rq)σ . We denoteω := (ω1,ω2)
andy(t,ω, v) := ω1(t),µ(t,ω, v) := ω2(t),χ(t,ω, v) := v(t), ∀(ω, v) ∈ Ω. We alsodenote byGt the canonical right continuous filtration generated onΩ by ((y,µ),χ). We
haveQN
satisfying
(i) QN(ω, v) ∈ Ω;A(ω1) = v
= 1, (193)
(ii) Mθ
t (·, ·, ·) is uniformly integrable (Corollary 4), continuous in the Lusin topol-
ogy of Ω, and is a(QN,Gt , Ω)-martingale with
〈〈Mθ
t 〉〉 =∫ t
0(g(y(t))θ,Qg∗(y(t))θ)ds. (194)
Note that, because of the above continuity and uniform integrability properties of
Mθt , ∀ϕ ∈ Cb(Ω) which isGs-measurable,EQ
N
[ϕ · (Mθ
t − Mθ
s)] = 0 will produce, in
the limit, EQ[ϕ · (Mθ
t − Mθ
s)] = 0. We need to show that
Q(ω, v) ∈ Ω;A(ω1) = v
= 1. (195)
We will show an equivalent statement, namely that, for every measurableζ(·, ·, t): Ω→Lq(0, T;Vrq), such that
∫Ω[∫ T
0 ‖ζ(ω, v, t)‖qVrqdt]Q(dω,dv) < +∞, one has∫
Ω
[∫ T
0〈χ(ω, v, s)−A(ζ(ω, v, s)),
y(ω, s)− ζ(ω, v, s)〉ds
]Q(dω,dv) ≥ 0. (196)
290 S. S. Sritharan
In fact, if such an equality holds, then settingζ(ω, v, t) = y(ω, t)− λw(ω, v, t), whereλ > 0 andw(·, ·, t): Ω→ Lq(0, T;Vrq) is any bounded measurable mapping,∫
Ω
[∫ T
0〈χ(ω, v, s)−A(y(ω, s)
− λw(ω, v, s)),w(ω, v, s)〉ds
]Q(dω,dv) ≥ 0. (197)
Now using the hemicontinuity ofA, asλ→ 0,
〈χ(ω, v, s)−A(y(ω, s)− λw(ω, v, s)),w(ω, v, s)〉→ 〈χ(ω, v, s)−A(y(ω, s)),w(ω, v, s)〉, (198)
with
|〈χ−A(y− λw),w〉| ≤ C‖w‖Vrq
(‖y‖Vrq + ‖w‖Vrq
)q−1+ ‖χ‖V ′rq
. (199)
Thus we get, by the dominated convergence theorem,∫Ω
[∫ T
0〈χ(ω, v, s)−A(y(ω, s)),w(ω, v, s)〉dt
]Q(dω,dv) ≥ 0. (200)
Since the above inequality (200) holds for each bounded measurablew we conclude(195). To prove the identity (196) we consider functions of the form
ζ(ω, v, t) =k∑
i=1
ϕi (ω, v, t)ei , ei ∈ Vrq , (201)
which form a dense set inLq(Ω, Q; Lq(0, T;Vrq)). Hereϕi (·, ·, t) are continuous inΩ with paths inLq(0, T). We restrict ourselves to the special case ofζ(ω, v, t) =ϕ(ω, v, t)e0, e0 ∈ Vrq . We now set
9 =∫ T
0
〈χ(ω, v, s)−A(ζ(ω, v, s)), y(ω, s)− ζ(ω, v, s)〉+ ν0‖A1/2y(ω, s)‖2 ds. (202)
Hence,
EQN
[9(·, ·)] = EQN[∫ T
0
〈χ(·, ·, s)−A(ζ(·, ·, s)), y(·, s)− ζ(·, ·, s)〉
+ ν0‖A1/2y(·, s)‖2ds
]. (203)
Now write9 as91+92 where
91(ω, v)=∫ T
0
〈χ(ω, v, s), y(ω, s)〉− 1
2 tr(g(y(ω, s))Qg∗(y(ω, s))
)+ ν0‖A1/2y(ω, s)‖2ds (204)
Control of Navier–Stokes Equation 291
and
92(ω, v) = −∫ T
0〈A(ϕ(ω, v, s)e0), y(ω, s)〉ds
−∫ T
0
〈χ(ω, v, s)−A(ϕ(ω, v, s)e0), ϕ(ω, v, s)e0〉+ 1
2 tr(g(y(ω, s))Qg∗(y(ω, s))
)ds. (205)
Note that92(·, ·) is continuous in the Lusin topology ofΩ. Here the strong hemiconti-nuity (see Corollary 3 for the definition) ofA(·) is used. We have also used the fact thatϕ is continuous. Thus, noting the energy estimate (Theorem 6) and also using Lemma 15below, we get
limN
EQN
[92(·, ·)] = EQ [92(·, ·)] . (206)
For91 which is not continuous in the Lusin topology ofΩ, we use the energy equality(108) to get
EQ [91(·, ·)] = EQ[91(·, ·)
], (207)
where
91(ω, v) := 12
(‖y0‖2− ‖y(T)‖2)+ ∫ T
0〈Nµ(s), y(ω, s)〉ds. (208)
Note that91 is upper semicontinuous on the Lusin topology ofΩ (the integral term in91
is actually continuous due to Lemma 7) and hence using the fact that if we integrate anupper semicontinous function with a Radon measure we will get an upper semicontinuousfunctional with respect to the measure (Theorem 55, Chapter III of [14]) we get
lim supN
EQN [91(·, ·)
]≤ EQ
[91(·, ·)
]. (209)
We note now that, by the definition ofQN
,
EQN
[9] = EQN[∫ T
0
〈χ(ω, v, s)−A(ζ(ω, v, s)), y(ω, s)− ζ(ω, v, s)〉
+ ν0‖A1/2y(ω, s)‖2 ds
]= EQ
N[∫ T
0
〈A(y(ω, v, s))−A(ζ(ω, v, s)), y(ω, s)− ζ(ω, v, s)〉
+ ν0‖A1/2y(ω, s)‖2ds
]≥ 0, (210)
due to the monotonicity property ofA(·). Hence,
0 ≤ lim supN
EQN
[9] = lim supN
EQN [91+92
]≤ EQ
[91+92
]= EQ [91+92] , (211)
292 S. S. Sritharan
where in the last step we have used the upper semicontinuity of91 and thus the uppersemicontinuity of the functional with respect to the measure noted earlier. Hence we getthe desired result (196),
EQ [9] ≥ 0. (212)
Remark 2. We start with (200),∫Ω
[∫ T
0〈χ(ω, v, s)−A(y(ω, s)),w(ω, v, s)〉dt
]Q(dω,dv) ≥ 0.
Take the special case ofw(ω, v, t) = ϕ(ω)z(ω, v, t), whereϕ(·) ∈ Cb(Ä). Moreover,by the disintegration theorem for Radon measures on Lusin spaces [60],Q(dω,dv) =Qω(dv)P(dω). Thus we end up with∫
Ω
ϕ(ω)
∫Lq′ (0,T;V ′rq )σ
(∫ T
0〈χ(ω, v, s)−A(y(ω, s)),
z(ω, v, s)〉ds
)Qω(dv)
]P(dω) ≥ 0,
which holds for arbitraryϕ ∈ Cb(Ω) and, hence,∫Lq′ (0,T;V ′rq )σ
(∫ T
0〈χ(ω, v, s), z(ω, v, s)〉ds
)Qω(dv)
=∫ T
0〈A(y(ω, s)), z(ω, v, s)〉ds, P a.s.
This allows us to connect the two martingale problems (that is the one in terms ofthe measureP and martingaleMθ (in Section 5.1) and the one with measureQ with
martingaleMθ) explicitly as
Mθt ((y,µ)) =
∫Lq′ (0,T;V ′rq )σ
Mθ
t ((y,µ),χ)Q(y,µ)(dχ).
This relationship is central for the developments of this paper.
We will now prove the continuity result used in the above arguments and alsoelsewhere.
Lemma 15. LetΩ be a Lusin space and letPnbe a tight sequence of Radon measuresonB(Ω) converging weakly toP onB(Ω). Let f(·) ∈ C(Ω) be a possibly unboundedfunction such that the following uniform integrability holds: For someε > 0,
supn
EPn [| f |1+ε] ≤ C. (213)
Then EPn[ f ] → EP [ f ] , as n→∞.
Control of Navier–Stokes Equation 293
Proof. We write
EP[ f ]−EPn
[ f ] = EP[ f − fm]+EP[ fm]−EPn
[ fm]+EPn
[ fm− f ], (214)
where
fm :=
f ∧m,− f ∨−m
and thusfm ∈ Cb(Ω). Note first that, due to the weak convergence ofPn, we have
EPn
[ fm] → EP [ fm] . (215)
We need to take care of the termsEP[ f − fm] and EPn[ f − fm]. By the Beppo–Levi
lemma we can deduce thatEP[| f |1+ε] ≤ C. This can be seen as follows. We define thetruncationϕk := | f |1+ε ∧ k ∈ Cb(Ω), which, by (213), gives usEPn
[ϕk] ≤ C. Weakconvergence givesEP[ϕk] ≤ C. This and the monotonicity,ϕk ≤ ϕk+1 ≤ · · · ≤ | f |1+ε,implies, by the Beppo–Levi lemma, limk EP[ϕk] = EP[| f |1+ε] ≤ C. Thus,
|EP [ f − fm] | ≤ EP [| f − fm|]= EP
[IΩC
m| f − fm|
]≤(
EP[IΩC
m
]ε/(1+ε))1+1/ε (EP [| f − fm|1+ε
])1/(1+ε), (216)
whereΩCm := y ∈ Ω, | f (y)| ≥ m . However,PΩC
m ≤ C1/m1+ε, as seen below.Hence,|EP [ f − fm] | ≤ C2/mε → 0, asm→ ∞ if ε > 0. Note here that we have∫Ω | f (y)|1+εP(dy) ≤ C implies
∫ΩC
m| f (y)|1+εP(dy) ≤ C, and, hence,m1+εP(ΩC
m) ≤C, which was the estimate used above.
We now come to the termEPn[ f − fm]. The convergence argument will be similar,
in fact,
|EPn
[ f − fm] | ≤ EPn
[| f − fm|]= EPn
[IΩC
m| f − fm|
]≤(
EPn[IΩC
m
]ε/(1+ε))1+1/ε (EPn [| f − fm|1+ε
])1/(1+ε)≤ C1
mε· C→ 0, as m→∞. (217)
Lemma 16 (Continuity of the Martingale, II). Suppose that q and r are chosen as inHypothesis1. If un → u in the Lusin topology ofΩ, if µn → µ in the stable topologyofM(U × [0, T ]), and ifχn→ χ in the weak topology of Lq
′(0, T;V′r,q), then
Mθ
t (un,χn,µn) = 0, ∀θ ∈ Vrq , t ∈ [0, T ], ∀n, (218)
implies
Mθ
t (u,χ,µ) = 0, ∀θ ∈ Vrq , t ∈ [0, T ]. (219)
294 S. S. Sritharan
Proof. We consider sequencesun → u in the Lusin topology ofΩ, χn → χ in theweak topologyLq′(0, T;V′r,q)σ , andµn → µ in the stable topologyM(U × [0, T ]).We have
Mθ
t (un,χn,µn)
= 〈un(t),θ〉 +∫ t
0〈ν0Aun(s)+ ν1χ
n(s)+ B(un(s))+ Nµn(s),θ〉ds, (220)
and for the above form of convergence, the only term that is not obvious is∫ t
0〈B(un(s)),θ〉ds=
∫ t
0b(un(s),un(s),θ)ds. (221)
We will now show this convergence. We have
〈B(un),θ〉 − 〈B(u),θ〉 = b(un,un,θ)− b(u,u,θ)
= b(un − u,un,θ)+ b(u,un − u,θ)
= b(un − u,un,θ)− b(u,θ,un − u). (222)
We will now show that asun → u in the Lusin topology ofΩ, the integral limits∫ t0 b(un − u,un,θ)dr → 0 and
∫ t0 b(u,θ,un − u)dr → 0 hold. We consider the
estimate for the three-dimensional case,∫ T
0|b(un − u,un,θ)|dr
≤ C‖θ‖Vrq
∫ T
0‖un − u‖1/4‖un − u‖3/4Vrq
‖un‖Vrq dr
≤ C‖θ‖Vrq
(∫ T
0‖un − u‖2 dr
)1/8(∫ T
0‖un − u‖2Vrq
dr
)3/8
×(∫ T
0‖un‖2dr
)1/2
. (223)
Note that the first integral tends to zero due to the strong convergence inL2(0, T;H)and the other two integrals are finite and hence the right-hand side tends to zero. We nowlook at the easier term∣∣∣∣∫ T
0b(u,θ,un − u)dr
∣∣∣∣≤ C‖θ‖Vrq
∫ T
0‖un − u‖1/4‖un − u‖3/4Vrq
‖u‖Vrq dr
≤ C‖θ‖Vrq
(∫ T
0‖un − u‖2 dr
)1/8(∫ T
0‖un − u‖2Vrq
dr
)3/8
×(∫ T
0‖u‖2dr
)1/2
, (224)
which tends to zero as before.
Control of Navier–Stokes Equation 295
5.4. Pathwise Uniqueness and the Yamada–Watanabe Technique
We begin with the pathwise uniqueness theorem.
Theorem 11. We consider two pathsu1,u2 ∈ L∞(0, T;H)∩Lq(0, T;Vrq)∩D([0, T ];V′rq), both defined on the same probability space(Θ, 6,6t ,m) with sameQ-WienerprocessW such that
du1+ (ν0Au1+ ν1A(u1)+ B(u1))dt = Nµt dt + g(u1)dW (225)
and
du2+ (ν0Au2+ ν1A(u2)+ B(u2))dt = Nµt dt + g(u2)dW, (226)
with the same initial data, u1(0) = u2(0) = u0, where E[‖u0‖2] < +∞. Then we havepathwise uniquenessu1 = u2 m-a.s. for the following cases:
(i) g is a constant operator(independent ofy) and(1) ν1 > 0, ν0 ≥ 0, n = 2,3;
(2a) ν1 = 0, ν0 > 0, n = 2;(2b) ν1 = 0, ν0 > 0, n = 3 and the solution satisfies the following additional
finite moment condition:
E
[∫ T
0‖A1/2u(t)‖4 dt
]< +∞. (227)
(ii) g= g(u) satisfying the admissibility condition(105),(1) ν1 > 0, ν0 ≥ 0, n = 2,3;(2) for ν1 = 0, ν0 > 0, n = 2,3, and the solution satisfies the following
additional finite moment condition:
E
[sup
0≤t≤T‖A1/2u(t)‖
]< +∞. (228)
Proof. Settingz := u1− u2 we have
dz+(ν0Az+ν1(A(u1)−A(u2))+B(u1)−B(u2))dt=(g(u1)−g(u2))dW. (229)
Thus, applying Proposition 2, we get
12‖z(t)‖2+ ν0
∫ t
0‖A1/2z(s)‖2 ds+ ν1
∫ t
0〈A(u1(s))−A(u2(s)), z(s)〉ds
= 12‖z0‖2+
∫ t
0(b(u2,u2, z)− b(u1,u1, z)) ds
+∫ t
0〈z, (g(u1(s))− g(u2(s)))dW(s)〉
+ 12
∫ t
0tr(g(u1(s))− g(u2(s))Q(g(u1(s))− g(u2(s)))
∗)ds. (230)
296 S. S. Sritharan
Case(i)(2a): g = const., ν1 = 0,n = 2. In this case (230) becomes exactly as in thedeterministic case,
‖z(t)‖2+ 2ν0
∫ t
0‖A1/2z(s)‖2 ds= ‖z0‖2−
∫ t
0b(z,u1, z)ds, (231)
and the pathwise uniqueness follows immediately due to the estimate
|b(z,u1, z)| ≤ C‖z‖‖z‖1/2‖u1‖1/2 ≤ ε‖z‖21/2+ Cε‖z‖2‖u1‖21/2. (232)
We now consider
Case(i)(2b): g= const., ν1 = 0,n = 3. In this case we have the estimate
|b(z,u1, z)| ≤ C‖z‖1/2‖z‖3/21/2‖u1‖1/2 ≤ ε‖z‖21/2+ Cε‖z‖2‖u1‖41/2. (233)
We thus end up with
‖z(t)‖2 ≤ Cε
∫ t
0‖z(s)‖2‖u1‖41/2 ds+ ‖z0‖2, (234)
‖z(t)‖2 ≤ ‖z(0)‖2 exp
∫ t
0‖u1(s)‖41/2 ds
. (235)
Assume that we have the estimateE[∫ T
0 ‖u1(s)‖41/2 ds]< +∞. Then we note that we
have∫ T
0 ‖u1(s)‖41/2 ds< +∞ a.s and we conclude that ifz(0) = 0, thenz(t) = 0, a.s.
We now come toCase(i)(1): ν1 > 0, ν0 ≥ 0, n = 2,3; and g is constant. We first considern = 2. Wewill then have
‖z(t)‖2+ 2ν0
∫ t
0‖A1/2z(s)‖2 ds+ 2ν1
∫ t
0〈A(u1(s))−A(u2(s)), z(s)〉ds
≤ 2‖z0‖2+ C∫ t
0‖z‖‖z‖1/2‖u1‖1/2 ds. (236)
We have, by monotonicity,〈A(u1(s))−A(u2(s)), z(s)〉 ≥ C‖z‖qVrq, and, hence,
‖z(t)‖2+ 2ν0
∫ t
0‖A1/2z(s)‖2 ds+ 2ν1C
∫ t
0‖z‖qVrq
ds
≤ 2‖z0‖2+ Cν1
∫ t
0‖z‖21/2 ds+ Cν1
∫ t
0‖z‖2‖u1‖21/2 ds. (237)
This gives‖z(t)‖2 ≤ 2‖z0‖2 + Cν1
∫ t0 ‖z‖2‖u1‖21/2 ds, and pathwise uniqueness holds
due to Gronwall’s inequality and the fact that the energy estimate (Theorem 6) in thiscase gives
E
[∫ T
0‖u1(s)‖qVrq
ds
]< +∞, with q ≥ 2, r = 1. (238)
Control of Navier–Stokes Equation 297
We now consider the case whenν0 ≥ 0, ν1 > 0, andn = 3. In this case we have
‖z(t)‖2+ 2ν0
∫ t
0‖A1/2z(s)‖2 ds+ 2ν1C
∫ t
0‖z‖qVrq
ds
≤ 2‖z0‖2+ C∫ t
0‖z‖1/2‖z(s)‖3/21/2‖u1(s)‖1/2 ds, (239)
with r = 1 andq ≥ 3. We write
‖z‖1/2‖z‖3/21/2‖u1‖1/2 ≤ ε1‖z‖31/2+ Cε1‖z‖‖u1‖21/2≤ε1‖z‖31/2+ε2‖u1(s)‖41/2+C(ε1, ε2)‖z‖2, ε1, ε2>0. (240)
Thus the pathwise uniqueness follows forn = 3.We now consider the case of nonconstantg.
Case(ii): g= g(u) with
‖g(u1)− g(u2)‖ ≤ C‖u1− u2‖. (241)
First consider the case whenν1 = 0 andn = 2. For this particular case it has been provenin the literature on nonstandard methods in the stochastic Navier–Stokes equation [10,Theorem 6.2] that if
E supt‖u(t)‖21/2 < +∞, or α > 0, (242)
then we have pathwise uniqueness. We will show a slightly improved result, namely, forn = 2,3 andν1 = 0, if
E supt‖u(t)‖α1/2 < +∞, for α > 0, (243)
then pathwise uniqueness holds. Let
τN := inft ≤ T; ‖u(t)‖1/2 ≥ N
. (244)
Note that ifE supt‖u(t)‖α1/2 < +∞, for anyα > 0, thenτN → T asN →∞.We define
B(ω, t,u) :=
B(u) for t ≤ τN,
0 otherwise,
A(ω, t,u) :=A(u) for t ≤ τN,
0 otherwise,
A(ω, t,u) :=
Au for t ≤ τN,
0 otherwise,
g(ω, t,u) :=
g(u) for t ≤ τN,
0 otherwise.
298 S. S. Sritharan
We now consider the stochastic partial differential equation
du =[−ν0 A(t,u)− ν1A(t,u)− B(t,u)+ Nµt
]dt + g(t,u)dW. (245)
We setz= u1− u2 andg12 := g(u1)− g(u2), and following the steps as before we endup with
‖z(t ∧ τN)‖2+ 2ν0
∫ t∧τN
0‖z(s)‖21/2 ds+ 2Cν1
∫ t∧τN
0‖z(s)‖qVrq
ds
≤ ‖z0‖2+ 2∫ t∧τN
0b(z,u1, z)ds+ 2
∫ t∧τN
0(z, g12 dW)
+ tr∫ t∧τN
0
(g12Qg∗12
)ds
≤ ‖z0‖2+ C∫ t∧τN
0‖z‖‖u1‖1/2‖z‖1/2 ds+ 2
∫ t∧τN
0(z, g12 dW)
+ C∫ t∧τN
0‖z‖2 ds. (246)
Taking expectation,
E
[‖z(t ∧ τN)‖2+ 2ν0
∫ t∧τN
0‖z(s)‖21/2 ds
]≤ E
[‖z0‖2]+ N E
[∫ t∧τN
0‖z‖‖z‖1/2 ds
]+ C E
[∫ t∧τN
0‖z‖2 ds
]. (247)
Here in order to conclude that the stochastic integral is zero we need
E
[∫ t∧τN
0‖z‖2‖g12‖2 ds
]< +∞. (248)
That is,
E
[∫ t∧τN
0‖z‖4 ds
]< +∞. (249)
This is indeed satisfied since‖z(t ∧ τN)‖ ≤ N. Applying Young’s inequality to the firstintegral on the right-hand side of (247) we get
E
[‖z(t ∧ τN)‖2+ 2ν0
∫ t∧τN
0‖z(s)‖21/2] ds
]≤ E
[‖z0‖2]+ εE
[∫ t∧τN
0‖z‖21/2 ds
]+ C(N)E
[∫ t∧τN
0‖z‖2 ds
]. (250)
Hence by Gronwall’s inequality,E[‖z(t ∧ τN)‖2
] = 0. Hencez(t ∧τN) = u1(t ∧τN)−u2(t ∧ τN) = 0, on a setΘ′ of measure one. Sinceu1(t) andu2(t)may not be equal fort > τN we have fort, 0≤ t ≤ T ,
u1(t) = u2(t), on the setΘN, (251)
Control of Navier–Stokes Equation 299
where
ΘN := ω ∈ Θ; τN = T ∩Θ′. (252)
Since we have
m(ΘN)→ 1 as N →∞, (253)
we get
u1(t) = u2(t), for ω ∈⋃N
ΘN, (254)
with m(⋃
N ΘN) = 1.In the above we need the fact that
E supt‖u(t)‖α1/2 < +∞, with α > 0. (255)
We now look at the case whenν1 = 0 andn = 3. In this case the estimate for the integralon the trilinear form would be
CE
[∫ t∧τN
0‖z‖1/2‖z‖3/21/2‖u1‖1/2 ds
]≤ CNE
[∫ t∧τN
0‖z‖1/2‖z‖3/21/2 ds
]≤ εE
[∫ t∧τN
0‖z‖21/2 ds
]+ C(ε, N)E
[∫ t∧τN
0‖z‖2 ds
]. (256)
Thus we again get
u1(t) = u2(t) for ω ∈ Θ with m(Θ) = 1. (257)
We now come to the case ofν1 > 0, ν0 ≥ 0. Here we get
‖z(t ∧ τN)‖2+ 2ν0
∫ t∧τN
0‖z(s)‖21/2 ds+ 2Cν1
∫ t∧τN
0‖z(s)‖qVrq
ds
≤ C∫ t∧τN
0‖z‖β‖u1‖1/2‖z‖2−β1/2 ds+ 2
∫ t∧τN
0(z, g12 dW)
+ C∫ t∧τN
0‖z‖2 ds. (258)
We consider the integral containing the trilinear term withβ = 1 andn = 2. We get
C∫ t∧τN
0‖z‖‖u1‖1/2‖z‖1/2 ds
≤ ε1
∫ t∧τN
0‖z‖41/2 ds+ ε2
∫ t∧τN
0‖u1‖41/2 ds+ C(ε1, ε2)
∫ t∧τN
0‖z‖2 ds. (259)
300 S. S. Sritharan
This leads to
E[‖z(t ∧ τN)‖2
] ≤ E[‖z0‖2
]+ C(ε1 · ε2)E
[∫ t∧τN
0‖z‖2 ds
]+ ε2E
[∫ t∧τN
0‖u1‖4 ds
]. (260)
Thus using Gronwall’s inequality we get (noting thatz(0) = 0)
E[‖z(t ∧ τN)‖2
] ≤ ε2
(E
[∫ T
0‖u1‖4 ds
])exp(C(ε1, ε2) · (t ∧ τN)),
∀ε2 > 0. (261)
Thus, we conclude that
E[‖z(t ∧ τN)‖2
] = 0, (262)
and pathwise uniqueness follows as before.We now consider the case whenν1 > 0,n = 3, and the integral contains the trilinear
term withβ = 12. We get
C∫ t∧τN
0‖z‖1/2‖u1‖1/2‖z‖3/21/2 ds
≤ ε1
∫ t∧τN
0‖z‖31/2 ds+ ε2
∫ t∧τN
0‖u1‖41/2 ds+ C(ε1, ε2)
∫ t∧τN
0‖z‖2 ds,
(263)
and pathwise uniqueness follows as before takingq = 4 in the energy estimate(Theorem 6).
Thus we see that the conditions forg= const.,ν1 = 0, n = 3, we need are
E∫ T
0‖u‖41/2 ds< +∞, (264)
and forν1 > 0 no additional condition is needed. Forg= g(u) we need, forν1 = 0, anadditional estimate of the form
E supt‖u‖α1/2 < +∞ (265)
for n = 2,3, and forν1 > 0 no additonal condition is needed.
We now establish the uniqueness in law following the classical Yamada–Watanabemethod which we only outline.
Theorem 12. For the cases in Theorem11which correspond to pathwise uniqueness,uniqueness in law also holds. That is, given the probability distributionµ0 of the initialdata, given the law of the controlΠ satisfying the admissibility criterion, and giventhe covarianceQ of the Wiener process, there is only one martingale solutionP for themartingale problem(175)–(176).
Control of Navier–Stokes Equation 301
Proof. Let (Θ′, 6′, 6′t ,m′, y′, y′0,µ
′,W′) and (Θ′′, 6′′, 6′′t ,m′′, y′′, y′′0,µ
′′,W′′) betwo weak solutions of the stochastic Navier–Stokes equation (184) with initial datumy′0 and y′′0 having the same distributionµ0, control forcesµ′ andµ′′ having the samedistributionΠ, and Wiener processesW′ andW′′ having the same covarianceQ (andhence the same distributionΛ). Let λ
′andλ′′ be the probability measure induced
respectively by((y′,µ′),W′,u′0) and((y′′,µ′′),W′′,u′′0) onΩ× C([0, T ];H)× H:
λ′ := m′ ((y′,µ′),W′,u′0)−1 and λ′′ := m′′ ((y′′,µ′′),W′′,u′′0)−1.
We define a mapping
π : Ω× C([0, T ];H)× H →M(U × [0, T ])× C([0, T ];H)× H (266)
by
π((u,µ),W,u0) = (µ,W,u0). (267)
Then
λ′ π−1 = λ′′ π−1 = Π⊗Λ⊗ µ0. (268)
Now, we have by the disintegration theorem for probability measures on Lusin spaces[60], that there exist regular conditional probabilitiesλ′µ,W,u0
(·) andλ′′µ,W,u0(·),
λ′(du,dµ,dW,du0) = λ′µ,W,u0(du)Π(dµ)Λ(dW)µ0(du0), (269)
λ′′(du,dµ,dW,du0) = λ′′µ,W,u0(du)Π(dµ)Λ(dW)µ0(du0), (270)
such that:
(i) ∀W ∈ C([0, T ];H), ∀µ ∈M(U× [0, T ]),∀u0 ∈ H,λ′µ,W,u0(·) andλ′′µ,W,u0
(·)are Radon probability measures onB(Ω).
(ii) The maps(µ,W,µ0)→ λ′µ,W,u0(·)and(µ,W,µ0)→ λ′′µ,W,u0
(·) fromM(U×[0, T ])× C([0, T ];H)× H →M(Ω) are Borel measurable.
On the Lusin space,
Ω∗ := Ω× Ω×M(U × [0, T ])× C([0, T ];H)× H. (271)
We define a Borel probability measureλ∗: B(Ω∗)→ [0,1] by
λ∗(C) =∫ ∫ ∫ ∫ ∫
1C(ω1,ω2,W,u0,µ)λ′µ,W,u0
(dω1)
λ′′µ,W,u0
(dω2)Π(dµ)Λ(dW)µ0(du0), (272)
for C ∈ B(Ω∗). Recalling that the distribution of(yi ,µi ,Wi , yi0) (superscripti = 1,2
denoting′ and′′, respectively) onΩ×M(U× [0, T ])×C([0, T ];H)×H isλi , we willnow show that this distribution would be same as that of(ωi ,µ,W,u0) on the space
302 S. S. Sritharan
(Ω∗,B(Ω∗),λ∗). That is,λ∗ (ωi ,µ,W,u0)−1 will be shown to be equal toλi . We
consider the map
Φi : (ω1,ω2,µ,W u0)→ (ωi ,µ,W u0) . (273)
Then we need to show thatλ∗ Φ−1i = λi . In fact let
C ∈ B(Ω×M(U × [0, T ])× C([0, T ];H)× H), (274)
then
Φ−1i (C) ∈ B(Ω× Ω×M(U × [0, T ])× C([0, T ];H)× H) (275)
and (λ∗ Φ−1
i
)(C) =
∫ ∫ ∫ ∫C(λ∗ Φ−1
i )(dωi ,dµ,dW,du0)
=∫ ∫ ∫ ∫ ∫
Ω∗1Φ−1
i (C)(ω1,ω2,µ,W,u0)
λ∗(dω1,dω2,dµ,dW,du0). (276)
Now, since
Φ−1i (C) = Ω× C, (277)
and also since
λ′µ,W,u0
(Ω) = λ′′µ,W,u0(Ω) = 1, (278)
we get(λ∗ Φ−1
i
)(C) =
∫ ∫ ∫ ∫Cλiµ,W,u0
(dωi )Π(µ)Λ(dW)µ0(du0) = λi (C). (279)
The following two lemmas (Lemmas 17 and 18) can be proven exactly as in Lem-mas 6.33 and 6.35 of [36] which deal with nuclear space-valued stochastic processesand measures on Polish spaces.
Lemma 17. For any C ∈ Bt (Ω), whereBt (Ω) := σ u(s), s ≤ t,u ∈ Ω, we de-fine two functionsf ′(µ,W,u0) := λ′µ,W,u0
(C) and f ′′(µ,W,u0) := λ′′µ,W,u0(C). Then
f ′ and f ′′ are measurable with respect to the completion of sigma fieldsBt (M(U ×[0, T ])) × Bt (C([0, T ];H)) × Bt (H) under the measureΠ × Λ × µ0. We define aprocess onΩ∗ = Ω× Ω×M(U × [0, T ])× C([0, T ];H)× H as
β(t,ω1,ω2,µ,W,u0) :=W(t). (280)
Also letF ′t be the completion of
Bt (Ω)× Bt (Ω)× Bt (M(U × [0, T ]))× Bt (C([0, T ];H))× Bt (H) (281)
with respect toλ. Thenβ is an H-valued Wiener process on the stochastic basis(Ω∗,F ′,F ′t ,λ) with quadratic variationQ.
Control of Navier–Stokes Equation 303
Lemma 18. Let P′ andP′′ be two probability measures on a Lusin spaceX. If
(P′ × P′′)(ω′,ω′′) ∈ X × X;ω′ = ω′′ = 1,
then there exists a uniqueω ∈ X such thatP′ = P′′ = δω.
We now prove the main theorem. We define
u1(t,ω1,ω2,µ,u0) := ω1(t) (282)
and
u2(t,ω1,ω2,µ,u0) := ω2(t). (283)
Now the weak solutionsy′, y′′ respectively on the stochastic bases(Θ′, 6′, 6′t ,m′) and
(Θ′′, 6′′, 6′′t ,m′′) satisfy
yi (t) = yi0−
∫ t
0
[ν0Ayi (s)+ ν1A(yi (s))+ B(yi (s))− Nµ(s)
]ds
+∫ t
0g(yi (s))dWi (s), mi a.s., i = 1,2. (284)
(Here superscripts 1,2 represent′ and′′, respectively). We then deduce that
ui (t) = u0−∫ t
0[ν0Aui (s)+ ν1A(ui (s))+ B(ui (s))− Nµ(s)] ds
+∫ t
0g(ui (s))dβ(s), λ a.s., i = 1,2. (285)
Now we have the two equations in (285) defined on the same probability space(Ω∗,B(Ω∗))and hence, by pathwise uniqueness,ui = u2, λ a.s. That is, setting
∆ :=(ω1,ω2) ∈ Ω× Ω;ω1 = ω2
(286)
and
C=∆×M(U × [0, T ])× C([0, T ];H)× H, (287)
we have
λ (C) = 1. (288)
That is,
1 =∫ ∫ ∫ ∫ ∫
C1C(ω1,ω2,µ,W,u0)λ
′µ,u0,W(dω1)
λ′′µ,u0,W(dω2)Π(dµ)Λ(dW)µ0(du0)
=∫ ∫ ∫
M(U×[0,T ])×C([0,T ];H)×H
(λ′µ,W,u0
⊗ λ′′µ,W,u0
)(∆)
Π(dµ)Λ(dW)µ0(du0). (289)
304 S. S. Sritharan
Hence, forΠ⊗Λ⊗ µ0 a.s. in(µ,W,u0), we have(λ′µ,W,u0
⊗ λ′′µ,W,u0
)(∆) = 1. (290)
Thus from Lemma 18, there exists a map
F: M(U × [0, T ])× C([0, T ];H)× H → Ω, (291)
such that
λ′µ,W,u0
(·) = λ′′µ,W,u0(·) = δF(µ,W,u0)(·). (292)
This implies that, forB ∈ B(Ω),mi(yi ∈ B
)=∫ ∫ ∫
M(U×[0,T ])×C([0,T ];H)×Hλi
u0,µ,W(B)Π(dµ)Λ(dW)µ0(du0) (293)
= (Π⊗Λ⊗ µ0
)(F−1(B)). (294)
We note thatF is indeed measurable for this to make sense. For anyC ∈ Bt (Ω), wehave, by (292) and Lemma 17,
1F−1(C)(µ,W,u0) = λiµ,W,u0
(C). (295)
HenceF−1(C) is in the completion ofB(M(U × [0, T ]))× Bt (C([0, T ];H))× B(H)with respect toΠ×Λ× µ0 and henceF(·, ·, ·) is adapted. HenceF(µ,W,u0) is alsoa strong solution to (285).
6. Existence of Stochastic Optimal Controls
Definition 8. The admissible martingale solution is a probability measureP on(Ω,B(Ω),Ft ) such that
M t (u,µ) := u(t)+∫ t
0(ν0Au(r )+ ν1A(u(r ))+ B(u(r ))− Nµ(r ))dr (296)
is aV′rq -valued(Ω,B(Ω),Ft ,P) martingale with quadratic variation,
〈〈M t (u,µ)〉〉 =∫ t
0g(u(s))Qg∗(u(s))ds, (297)
and
EP [J ] < +∞. (298)
Theorem 13. The class of admissible martingale solutionsK(Ω) form a compact setin the class of Radon probability measuresM(Ω).
Control of Navier–Stokes Equation 305
Proof. We begin by noting that the class of martingale solutionsP which satisfy thebound
C · EP [‖µ‖2κ] ≤ EP [J ] < +∞ (299)
are tight due to Theorem 8. We will now establish that they are closed. LetPn be asequence of admissible martingale solution which converge weakly toP. We will showthatP is also an admissible martingale solution. We have the lawsPn of (yn,µn) tightonΩ with Lusin topology. We set
u(ω, v) := ω1(t), (ω, v) ∈ Ω× Lq′(0, T;V′rq)σ , (300)
µ(ω, v) := ω2(t), (ω, v) ∈ Ω× Lq′(0, T;V′rq)σ , (301)
χ(ω, v) := v(t), (ω, v) ∈ Ω× Lq′(0, T;V′rq)σ . (302)
Let Gt be the canonical filtration generated onΩ× Lq′(0, T;V′rq)σ by ((u,µ),χ).
Thus, as in Section 5.3, we also have a tight sequence of probability measuresQn
which are the laws of((u,µ),χ) on the Lusin spaceΩ× Lq′(0, T;V′rq)σ . Moreover,
EQn [ϕ(·, ·)
(Mθ
t (·, ·, ·)− Mθ
s(·, ·, ·))]= 0,
∀n,∀ϕ ∈ Cb
(Ω× Lq′(0, T;V′rq)σ
), (303)
such thatϕ is Gs-measurable. We now note the continuity ofϕ, the continuity ofMθ
t
(given by the Lemma 16), and the uniform integrability (Corollary 4) ofMθ
t to concludethat
EQ[ϕ(·, ·, ·)
(Mθ
t (·, ·, ·)− Mθ
s(·, ·, ·))]= 0,
∀n,∀ϕ ∈ Cb
(Ω× Lq′(0, T;V′rq)σ
). (304)
We can thus, using the stochastic Minty–Browder method as before, conclude that
Q(ω, v) ∈ Ω× Lq′(0, T;V′rq)σ ; A(ω1) = v
= 1. (305)
Lemma 19. The mapP→ EP [J ] is lower semicontinuous.
Proof. By Lemma 10, the functionalJ (·, ·) is lower semicontinuous inΩ, with Lusintopology forΩ. Thus, by Theorem 55 in Chapter III of [14], the functionalEP [J ] islower semicontinuous inP:
lim infn
EPn
[J ] ≥ EP [J ] . (306)
We now complete the existence theorem. Simply noting that any lower semicontinu-ous functional which is bounded below on a topological space [2] achieves its minimumin a compact set, the existence of optimal control is proven.
306 S. S. Sritharan
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Accepted7 June1999