Proceedings of the Royal Society of Edinburgh, 121A, 5-32, 1992
An optimal control problem in exterior hydrodynamics
S. S. Sritharan*
Department of Aerospace Engineering, University of Southern California, Los
Angeles, CA 90089-1191, U.S.A.
(MS received 10 January 1991)
Synopsis
In this paper we consider the problem of accelerating an obstacle in an incompressible viscous fluid
from rest to a given speed in a given time with minimum energy expenditure. An existence theorem
for the speed trajectory which corresponds to the absolute minimum is provided. The results are valid
for arbitrary Reynolds numbers.
1. Introduction
Optimal control theory of distributed parameter systems is a rapidly developing
subject [3,10,9,11]. So far the impact of this growth to continuum mechanics has
been essentially in the branch of solid mechanics. In this paper, we develop some
of these ideas for a problem in fluid mechanics which has profound practical
applications such as aero/hydro maneuvering of vehicles. We consider the task of
finding the optimal way to accelerate an obstacle from rest to a given speed in an
infinite medium of viscous fluid. Here the objective is to find the speed trajectory
which corresponds to a global minimum for the energy expenditure functional.
We note here that the above problem can be transformed into a nonstationary
problem in the exterior domain by a simple change of coordinates. We only deal
with translatory motions of the obstacle in this paper. The first step towards the
theory is a well-posedness theorem which relates the generalised solution of the
nonstationary Navier Stokes problem in the exterior domain to the prescribed
speed of the obstacle. We have shown that for the two-dimensional problem this
correspondence is unique and analytic. The central result of this paper is an
existence theorem for the speed trajectory which corresponds to the absolute
minimum for the energy expenditure. The Reynolds number in this theorem is
arbitrary. The existence theorem for absolute minimum is the fundamental step in
establishing a (Pontryagin type) Maximum principle to derive the necessary
conditions and in the dynamic programming method (using the Hamilton-JacobiBellman equations) for the control synthesis in the feedback form [14, 13, 2].
2. Mathematical formulation
The governing equations for the moving obstacle problem can be transformed
to describe a nonstationary exterior problem in the following way. First the
* This research is supported by the office of Naval Research through the URI-Contract
No: N00014-86-K-0679.
Present address: Program in Applied Mathematics, University of Colorado at Boulder, Campus
Box 526, Boulder, Colorado 80309-0526, U.S.A.
6
5. 5. Sritharan
Navier Stokes equations are written in an inertial coordinate system. We then
transform these equations to the noninertial coordinate system fixed on the
obstacle. Let Q obs be a bounded open set in R2 with uniformly C2 boundary dQ.
Let us denote by Q:=R 2 \{Q o b s U dQ} the domain exterior to Qobs. Here Q
contains the complete neighbourhood of infinity. The nonstationary exterior
problem obtained using the above transformation method is the following: given
the speed of the obstacle U: [0, T]-^-R + with £/(0) = 0 and U(T)=UT, find
(u,p): Q x [ 0 , T ] ^ R 2 X R such that
dU
u, + u . Vu = -Vp + v Au + ex —-
in Q x [0, T]
at
V . u = 0 in Q x [0, T]
u{x, 0 = 0 on 3Q x [0, T]
u^U(t)e1
as |*|—»» and
u(x, 0) = 0, xeQ.
(2.1)
Here u is the velocity field, p is the pressure field, v > 0 is the coefficient of
kinematic viscosity, ex is the unit vector in the direction of the jcx-coordinate and
UT and T are two given positive numbers.
Let us first derive an expression for the energy expenditure assuming sufficient
smoothness and decay properties for the solution. We shall of course prove later
that there are generalised solutions for which this (energy expenditure) functional
is finite.
Let us denote by def u the deformation tensor,
and by o,y the stress tensor
<^ = - p 6 i y +
If F denotes the force on the body, then
F,=
THEOREM
Jaa
ounjdS.
2.1. Let the solution u be smooth and satisfy the decay conditions
lim I
oik(Ui - U dn)nk dS = 0
lim f
ui{ni-U5n)2nidS
and
= Q.
Then the total energy expenditure JJ [/(0Fi(0 dt is given by
\ U(t)Yx{t)dt = \ \ \a{x,T)-UTel\2dx
Jo
2. Ja
+ 2v \ f [def u: def u] dxdt.
Jo Ja
An optimal control problem in exterior hydrodynamics
7
Proof. If z is any solenoidal vectorfield, then the following identity holds:
-— [oikXi] =
dxk
z, + vz, Au, + 2v def u: def z.
dXj
Let us denote Q R = Q n { | x | ^ i ? } . Integrating the identity over QR and
applying the divergence theorem, we obtain
J
oikZink dS —
Jan
\X\=R
oikZjnk dS =
— —— + v Au, z, dx + 2v
'nR\-
"Xj
J
def u: def z dx.
JQK
We now set z = u — Uex and take the limit R —> °° in the integrals
l/(f)F,(f) = lim f [ - | ^ + v A u , l [ u , - ( / d , , ] d j c + 2v f def u: def u<£c.
«-»«> JnRL dx,J
JQ
Let us consider the first integral on the right-hand side. Substituting from the
Navier Stokes equations, we get
=~ (
2 dt Ja
(», - U Sn)2 dx + lim bR(u, u, u - Uex).
/?-•«.
Here,
6«(u, u, u - Uex) = f u ; | ^ (u, - (/ <5(1) dx.
Note that
bR(u, u, u — USx) = bR(u, u — f/e,, u — {/?,)
2
=2UJ SJC,
£[«>,-(/a,,)
] ^.
Q)r
By applying the divergence theorem and noting that u| 3 Q = 0, we get
bR(u, u, u - Uex) = \
-+0
uy(u, - t/ d n ) 2 n ; dS
as /?-*<».
Collecting all these results we get
l/(r)F,(/) = J 4 f (»i- - U 5 a) 2 dx + 2v f def u: def u rfx.
2 dt JQ
Ja
Integrating this equality with respect to t from 0 to T and noting that u(x, 0) = 0
and f/(0) = 0, we get the desired relationship for the energy expenditure. •
Let us now introduce the functional framework that is used in this paper.
8
5. 5. Sritharan
CQ(Q) = the space of infinitely differentiable vectorfields with compact support
in Q.
= {<t>: Q^R2;<t>eL2(Q),
Vtf>eL 2 (Q) and <p\3a = 0},
W0(Q) = completion of CQ(Q) in the norm ||V0|| L 2 (n) <<»,
/(Q) = {<f>: Q-»R 2 ; <p e CQ(Q) and V • <f> = 0},
/(Q) = completion of ;"(Q) in the L2(Q) norm,
/i(Q) = completion of /(Q) in the /^(Q) norm,
/ 0 (Q) = completion of/(Q) in the norm
For the exterior type domains such as Q defined earlier, we have the following
results [7]:
PROPOSITION
2.2. Let QcR 2 i>e an exterior domain with dQ e C2. Then,
J(Q) = {$: Q^R2;
<j> e L 2 (Q); V • <f> = 0 and <p • n\3a = 0}
^(Q) = {</>: Q ^ R 2 ; $ e Hh(Q); andV-<f> = 0}
and
J0(Q) = {(t>: Q^-R 2 ; ^ € W0(Q); and V • $ = 0}.
We note here that [6] if (f> e W0(Q) then 0 | s n = 0 and 0 e Lfoc(Q).
If we identify /(Q) with its dual J(Q)' = ££(J(Q); R) using the Riesz representation theorem, we get the continuous and dense embedding
/(Q)
= J(Q)'
c
3. Characterisation of the Stokes operator and other relevant linear and
multilinear operators
Let us first define the Stokes operator for the two-dimensional exterior problem
and analyse its properties. Note that unlike the case of bounded domains, the
Stokes operator for the exterior domain has an unbounded inverse. We denote by
a(-, •) the symmetric bilinear form,
a(u, v) =
Vu . Vv dx.
Let us define the Stokes operator in the following way.
Given u e /i(Q), if there exists an element g e /(Q) such that
a(u, v) = (g, v) 7(Q) ,
for all v e /,
then we say u e D(A) and An = g. We have, in fact, the following theorem:
THEOREM 3.1. There exists a maximal accretive self-adjoint operator Ae
£(D(A);J(Q)) such that
a(u, v) = (An, v) 7(n) , for all u e D(A),
and for all v
An optimal control problem in exterior hydrodynamics
with D(A) cii(Q) dense. Moreover, it is possible to extend this operator as
such that
a(u, v) = (^u, v)/,(Q)'X7l(£j), for all u , v e h
We recall here that a densely defined linear accretive operator is maximal if
and only if it is closed. Let us now prove the theorem.
Proof. We shall obtain A as the strong limit of regularly accretive operators A£
as e—»0. Let us denote by ae(-, •) the symmetric bilinear form
aE(u,v)=
Vu-V\dx + e\ u.\dx,
wi
withO<e<l.
Using the Schwartz inequality, we obtain the continuity as
|a,(u, v)| ^ ||u|| /l(n) ||v|U(Q),
for all u, v € /X(Q),
for 0 ^ e < 1.
The coerciveness (.^(Q^-elliptic) property is obtained as:
a,(u, u) § £ ||u||?l(Q),
for all u e MQ),
for 0 < e < 1.
Note that at £ = 0, we have the following situation. The form is continuous:
|a(u, v)| =S ||u|U(Q) ||v|| /l(Q) , for all u, v € J^Q),
but it is not /1(Q)-elliptic. This difficulty is due to the fact that the Poincare lemma
does not hold for exterior domains. At s = 0, the bilinear form satisfies
«(n, u) = || Vu||i2(Q) = ||n||5l(Q) - ||u||5(Q),
for all u e /,(Q),
which is known as the Garding inequality. Note that although a(-, •) is not
positive definite in Ji(Q), it is in fact positive:
a(u, u) = || Vu |||2 (Q) i? 0, for all u e A(Q).
Now, since the form aE(-, •) is symmetric, continuous and positive definite, we
have the following standard results:
PROPOSITION 3.2. There exists a self-adjoint, regularly accretive onto map
Ae e £(D(AE); J(Q)) such that
ae(u, v) = (Aeu, v)y(Q), for all u e D(Ae), and for all v 6 .A
with D(Ae) cJr(Q) dense. Moreover, it is possible to extend this operator as an
isomorphism on to map
such that
ae(u, v) = (Aeu, v)y l(Q) . x y l(n) ,
for all u, v € J
We now set,
Au = Aeu — eu, for all u e D(A) = D(Ae) and
An = Aen — EU, for all u e J\(Q.).
10
S. S. Sritharan
We can then easily show [15] that the properties of A and A defined in the
theorem hold. •
We note finally that since a(-, •): J0(Q) x J0(Q)-+ R satisfies
|a(u, v)| ^ ||u||7o(n) ||v||/o(Q), for all u, v e J0(Q)
and
a(u, u) = ||u||5o(Q),
for all u e J0(Q),
we can define an isomorphism on to map
such that
a(u, v) = (Au, v)MayxMa),
for all u, v e J0(Q).
The operators A, A and A are different characterisations of the Stokes operator
and are denoted simply as
n i?(7 1 (Q);/ 1 (Q)') n <e(UQ);J0(Q)').
A e 2{D(A);J(Q))
Remark 3.3. Since J1(Q)(=J0(Q), we have J0(Q)' cJ^Q)'. However, since
J0(Q)(jiJ(Q), we have J(Q)' t^/ 0 (Q)' and care must be taken in the interpretation of the Stokes operator in the above form.
Let us now note certain properties of the range R(A) and the domain D{A) of
the Stokes operator.
LEMMA
3.4. R(A) cz/(Q) is dense.
Proof. Let g e / ( Q ) be an arbitrary element. Since for any 0 < £ < l ,
J£(D(A);J(Q)) is an onto map, there exists ue € D(A) such that
Aee
Now,
Hence R(A)czJ(Q) is dense. Let us now prove a regularity theorem. This will
give us an explicit representation of D(A).
3.5. Let Q be an exterior domain with boundary 3Q e C2. Then for a
given g e R(A) c J(Q), the Stokes problem of finding (u,p): Q—» R2 x R such that
THEOREM
- A u + V/> = g
V.u = 0
v|sn = 0
and
in Q,
in Q,
v—»0 as
has a unique solution u which satisfies,
|*|—»°°,
An optimal control problem in exterior hydrodynamics
11
Moreover, if we define the graph norm,
M\D(A)
= ||u||i.2(Q) + HAull^n), for all u e D(A),
then there exists C > 0 such that
Proof. Existence and uniqueness of u e D ( j 4 ) c ; , ( Q ) have been established
earlier. Let us analyse the regularity. For geR(A) <=/(Q), the following
fundamental estimate holds [6]:
(3.1)
with
g = An.
Moreover, since u e /i(Q) and solves
a(u,v) = (g,v),
for all v e ^ ( Q ) ,
we have, setting v = u,
a(u, u) = ||Vu||^ (n) = (g, u) = (Au, u)z.2(Q).
Hence, by the Schwartz inequality,
||Vu|| L 2 (Q) ^ ||u||^2(Q) HAull^Q),
for all u e D{A).
From the above estimate and (3.1) we can also get:
\\D*u\\L\a> ^ C2(\\Au\\LHO) + Hnl
Hence,
The reverse inequality is immediate since Au = — P, Au, for all ue D(A), where
P y : L2(£2)—*J(Q) is the orthogonal projection operator. D
Let us define a trilinear form as
b{u,v, (/>)= 2
ij=i Ja
Ui-^fydx.
ox
LEMMA 3.6. The form
b(-, •, •):J1(Q)xJ0(Q)xJl(Q)-*R
Moreover, for all u, v, <j> e Ji(Q), we have
continuously.
b(u, v, (j>) = — b(u, <j>, v) and
Proof. We have by Holder's inequality
b(u, v, 0) g Hnll^Q) ||Vv||^ (
( )
We now use the following well-known inequality [6] which is valid for exterior
two-dimensional domains:
^ S C \\u\\iHa) ||Vu||[ 2(Q) for all u e W0(Q).
12
5. 5. Sritharan
This gives
|6(u, v, 0 ) | =g C2 ||u||l 2 ( Q ) ||Vu||[ 2 ( n ) ||Vv|| i2(Q)
From this, we get
\b(n, v, <t>)\ ^ C2 ||u|| 7 l ( n ) H
for all, u, ^ e /j(Q)
and for all v e •/<,
We now note that by simple integration of b(-, •, •) by parts,
b(u, \,<p) = -b(u, (f>, v) and
b(u, v, v) = 0, for all u, v, $ e/(Q).
Hence these results also hold for all <f>, v, u e / ^ Q ) due to the continuity of the
trilinear form and the density of j(Q) in Ji(Q). •
LEMMA
3.7. 77iere exwf continuous bilinear operators,
and
6(U, V, </)) = <B(U, V),
= <B*(v, </>), u)yl(a)'xyl(Q), /or a// u, v, 0 e/,(Q).
Moreouer,
\\B*(y, <t>)\\MQy^C ||Vv||L2(Q)
B*(v, 0) = -B*(0, v), /or a// </>, ve/,(Q)
Proof. The continuity estimate for the trilinear form and the results of the
previous lemma gives us the required results using the Riesz Representation
Theorem. •
Let us now establish an important estimate on the trilinear form.
LEMMA
3.8. For all 6 >0, there exist w6 € J0(Q) n C\Q) and Ro>0 such that
|6(B,W«, V)| ^
6 \\u\\MaK) ||v||yo(nR), for all u,v6/ 0 (Q f i ),
and for all R ^ Ro.
Proof. The key to the proof is the construction of a certain cutoff function
which has its origins in the works of Leray [8] and Hopf [4]. We write w6 in the
form
w a (*) = curl [e3x2(l - 6E(x))],
where 6e is a positive scalar function satisfying the following properties. If the
distance function is denoted as p(x) = dist (dQ, x), x e Q, then for a given e > 0,
An optimal control problem in exterior hydrodynamics
13
2
there exists a C (Q) function 8e(x) such that 8e(x) = 1 for x in some neighbourhood of dQ which depends on £, 6E = 0 if p(x) §? 2 exp (1/f) and in the strip in
which 8C varies, its derivatives satisfy
1
p=s2exp(--),
-p(x)'
dxk
A: = 1,2.
The properties of the Leray-Hopf cutoff function 8E imply the following
properties for the solenoidal vectorfield ws: w6 = 0 in a neighbourhood of dQ
which depends on e; ws = ex outside a neighbourhood of dQ which again depends
on £. In the strip around dQ where w6 is variable, the following estimate holds:
Cl£
lwa(x)|gQ 1
Let us now indicate briefly how to complete the proof of this lemma. The details
are similar to those found in [6] for a similar lemma. Note that if we denote by Qd
the subset of Q in which the vectorfield ex — wd is nonzero, then
o(u, w6, v) = o(u, Wg - ex, v)
= y
f
("/ == 1 *^£2A
u
J_
( w
-
8
)vdx = -b(u v w
-e)
wA
by integration by parts. Thus,
2
|o(u, w 6 , v)| ^ ||Vv||L2(Qs) 2
ll«/(way - ^
We can now use the estimate for wa in Q6 along with the Hardy inequality,
u
P
to get
|o(u,w 6 , v)|^<
with <5(e)-*0 as e—»0. Note that in this construction the parameter e > 0 can be
taken arbitrarily small and its size is chosen by the given size of 6 in the
theorem. •
Let us now obtain certain properties of the forms b(-, w6, •) and b(w6, •, •).
LEMMA
3.9. b(w6, •, •): J^Q) x/i(Q)—»R continuously and satisfies,
(i) b(ws, u, v) = -b(wd, v, u), for all u, v e
(ii) fc(wfi, u, u) = 0, for all u e /
Proof. Note that since w6 € Cl(Q),
|6(wa, u, v p c . || Vu|| L , (n) ||v|| i2(fi) g C . ||n|| /l(U) ||v|| /l(Q) , for all u, v e U
Now, note that (i) and (ii) hold for all u, v €;(Q) and hence they are valid for all
u, v e /i(Q) by the continuity of 6(w6, •, •)• ^
14
S. S. Sritharan
LEMMA
3.10. The form b(-, ws, •): / 0 (Q) x / 0 (Q)-» R satisfies
(i) 6(-,w a , (ii) o(u, w a , v) = - 6 ( 0 , v, v/6 - e^), for all u , v e /i(Q).
PTOO/.
Since w6 is C\Q) with support Q6, we easily have
\b(u, yvs, v)| ^ d ||u||L2(£2a) ||v|| L2(na) ^ C, llull^Q, IMI^o,,
for all u, v € /(Q).
Now, to see (ii) note that,
\b(u, w 6 , v) + b(u, v, w.5 - eOl ^ C3 ||u|| 7l(n) ||v||yl(n)
and
b(u, w a , v) + 6(u,v,w 6 - ^ = 0,
for a l l u , v e / ( Q ) .
D
LEMMA 3.11. 7/iere exists a continuous linear operator Bx e 3?(J0(Q);J(Q)) such
that
(i) b(\vs, u, v) = (#i(u), v)i.2(Q>, /or a// u e / 0 (Q), and /or a// v e J(Q).
Moreover,
(ii) (B^u), v)L2(a) = - ( u , fii(v))i.2(Q) for all u, v e ^(Q).
Proof. Since the form o(w 6 , u, v) satisfies the estimate
\b(wa, u, v)| ^ C ||u||7o(Q) ||v|| y(Q) ,
for all u eJ0(Q),
we have by the Riesz representation
i?(/ 0 (fi);/(Q)) such that
6(w a , u, v) = (5i(u), v) y(n) ,
theorem
for all u e/0(&)>
and for all v e/(Q),
that
there exists fij €
and for all v €
Property (ii) then follows from the property
o(w a ,u, v) = -b(w6, v,u),
for allu,v£/ 1 (Q).
•
LEMMA 3.12. 77iere e^«f continuous linear operators B2, B* e
(i) (B2(u), v)L2(n) = (u, B 2 '(v)) t!(a) = 6(n, w a , v), for all u, v e /(Q).
Moreover,
Proof The estimate
|6(v, w«, u)| ^ C ||v|| y(n) ||0|| y(Q) ,
for all u, v e /(Q)
immediately gives the characterisations of B2 and 6* via the Riesz representation
theorem. Note that B2 is the adjoint of B2- The result (ii) can be deduced from
|6(v, w a , 0)| ^ 5 ||v||7o(n) ||u||yo(Q), u, v 6/ 0 (Q),
which is proved in Lemma 3.8.
•
4. A priori estimates
A key result to be used in obtaining the a priori estimates is Lemma 3.8 where
the number 6 can be taken as small as we wish. Let us introduce the following
An optimal control problem in exterior hydrodynamics
15
change of variables: u(x, t) — v(x, t) + U(t)w6(x). The problem then is to find
(v,p): Q x [0, T] : ^ R 2 X R such that
v, + v . Vv + U(t)v. Vw6 + U(t)v/S . Vv
= -Vp + vAv + flv in Q x (0, T)
V . v = 0 in Q x (0, T)
v|aa = 0, v—»0 as |JC|—»°° and
\(x, 0) = 0,
xeQ.
Here,
fw(x, t) = vU(t) Aw6 - i/(f)2(w6 . Vwa) + — (e, - w a ).
Note that the support {f^,} c c Q . A priori estimates for the solution can be
obtained by assuming v e Cl([0, T]; D(A)).
2
THEOREM 4.1. Let Q. be an exterior domain with 3Q.eC .
Then the smooth
solution v of the above problem satisfies the following a priori estimates:
(i) sup \\y(;t)\\hw+
v f | | V v ( - , 0 l l i W ^ C i ( r , v, ||f/|U,(O,r)).
(4.1)
Moreover
(ii) sup ||Vv(-,0lli:(Q, + v [ \\Av(-,t)\\2LHa)dt^C2(T,
«e[0, 7")
v, ||f/||Hl((),r))-
(4.2)
Jo
Proof. We begin with the result
(n)
= 0,
which holds for smooth solutions with sufficient decay. Taking the inner product
by v in the governing equations, we get
~ IMfoa) + b(y, v, v) + v ||Vv||i2(Q) + U(t)b(y, wa> v) + U(t)b(yt6, v, v)
We have
b(y, v, v) = 6(wa, v, v) = 0 and
\U(t)b(\, Y/6, v)| ^ UT \b(\, w6, v)| ^ 6UT ||Vv|||2(Q), for all \ eJ0(Q),
using Lemma 3.8. Since 6 > 0 here is arbitrary, we take d < v/(4UT) so that,
\U(t)b(\, w6, v)| ^ ^ ||Vv||i2(Q), for all veJ0(Q),
and for all t e [0, T].
Let us now estimate
\(L, V)z.2(£2)| ^ V(/(O |(AW6, v)z.*(Q)l + t/(0 2 I(W6 • V w 6 . V)L2(Q)|
16
S. S. Sritharan
Noticing that the support of Aw6, y/6 . Vwa and ea — w6 is Q6 c c Q, we get
Since v vanishes on the inner boundary of Q6, we have, using the Poincare
Lemma,
\U(t)\2+
,
for all V6/,(Q).
Using Young's inequality, we get
),
•+
for all
Combining these results, we get
IU(t)\4
Integrating and noting that v(0) = 0 and U e H^O, T) a C(0, T), we get
This gives the first a priori estimate, which is known as the energy estimate. We
note here that the constant CX{T, v, \\U\\H^Oi T)) grows with T.
We will see later that the above a priori estimate alone is sufficient to establish
the existence and uniqueness of the weak solution. Let us now obtain additional a
priori estimates. Notice that the governing equations can be written in the
evolution form as
vAv + B(y, v)
U(t)B2(y) = U(t% + U(t)% + ^f 3 ,
at
where tu f2 and f3 are independent of time and belong to J(Q6). This form can be
obtained by simply projecting the governing equation in to /(Q) using the
orthogonal projector P 7 : L2(Q)—>J(Q).
We now take inner product with A\ to obtain
(•„ Av)Lna) + v ||Av|||2 (n) + b(\, v, Av) + U(t)b(y, v>6, Av) + U(t)b(yrd, v, ^v)
= U(t)(tuAv)LHa)
+ U(t)2(t2,Av)L2(a)+^(t3,Ay)LHn).
Note first that since f1? f2 and f3 are square integrable in Q, we have
\U(t)(tuAv)LHa)\ ^ 11/(01 llftll^Q) \\Av\\LHo),
\U(t)\t2,Av)LHQ)\ ^ \U(t)\2 \\t2\\LHa) \\Av\\LHa), and
dU
(f3,
dt
\dU
\~dt WUWL \a)
(4.3)
An optimal control problem in exterior hydrodynamics
17
We now use Young's inequality to get
t/(O2(f2, Av) L2(n) + ^ (f3, Av)L>(i
• 2
^C1(Q6,v)\\U(t)\2
+
Let us now estimate the other terms:
\b(y, w6, Ay)\ ^ C2(Q6) ||V||Z.2(Q6) HAVH^Q^
^ C 2 (Q 6 , v) ||v||| 2(na) + ^ ||/lv||| 2(Q6) .
Similarly,
=iC3(Qa, v) |N
We now estimate the term
|&(v, v,i4v)|| ^ Hvll^Q) IIVvllz.^) \\Av\\LHmWe have the following well-known inequality [6]:
IMIz.«(Q)^C Hvll^Q) HVvll^Q), for all v e / ^ Q ) ,
valid for any arbitrary open domain Q with constant C independent of the size of
Q. However, for the term ||Vv||L<(n) we cannot use such an estimate since Vv
does not vanish on dQ. In this case we use the following estimate [6]:
*, for all v e H 2 (Q).
^ C \\Vy\\{Ha)
Hence, using the regularity theorem for the Stokes Operator, we get
||Vv||L4(n)=§ C HVvllk^fllWH^Q) + \\Av\\LHa)}'i,
for all v € H\Q) n/ x (Q).
We thus have
l \\Av\\L
\b(y, v, Av)\ ^ C \\y\\[2(a) ||Vv||L2(fi)
^ C \\y\\h(a) II
Now, using Young's inequality, we get
\b(y, v, ,4v)| ^ d ( v ) \\y\\LHf» \\Vy\\iHa) + C2(v) ||v||i 2(n) ||Vv||la(Q)
Combining all these estimates in (4.3), we get
^
|
i
v
) \\v\\LHQ)
\\Vy\\lHa>
+ 2C2(v)||v|||2(n)||Vv||l,(n)
a)
18
5. 5. Sritharan
Integrating with respect to t from 0 to t and noting that Vv(x, 0) = 0, we get using
the a priori estimate (4.1)
^C4(T,v,
||U\\HHfS.T)) + C 5 ( v ) J \ ( r ) IIVv|||2(Q)dT,
(4.4)
where
h(t) = \\y\\LHa) ||Vv|| L2(Q)
Note that h(-) e L\0, T) due to the first a priori estimate (i). Dropping the
second term in the left-hand side of (4.4) and using Gronwall's inequality, we get
jjup^ ||Vv(-, t)\\L2(a)^C6(T, v, \\U\\HHo, T)).
Using this again in (4.4), we get the second a priori estimate (4.2).
•
LEMMA 4.2. The energy estimate (4.1) implies the following a priori estimate for
the time derivative:
\
Jo
Proof. We begin with the evolution form:
v, = - V A Y - B(v, v) - U(t)B1(w) - U{t)B2{y) +1, U(t) + f2U{tf +
lM.
at
We have noted earlier that ^ e i?(/ 1 (Q);/ 1 (Q)') ) Bx e i?(/ 1 (Q);/(Q)) and B2e
). Thus we have
-Ay - t/(0Bi(v) - U(t)B2(y) e L2(0, r;/,(
if veL°°(0, T;J(Q))nL2(0, TJ^Q)) and UeH\0,T). Moreover, using the
properties of B(-, •),
f
Jo
due to the energy estimate. Combining these results and noting that f,e/(Q),
i = 1, 2, 3, we get the desired a priori estimate. •
5. Well-posedness in the sense of Hadamard
In this section we study the well-posedness of the problem. We prove in
particular that for each given U e H\0, T), there exists a unique solution v which
depends continuously on U. Moreover, this dependence is analytic. Existence and
An optimal control problem in exterior hydrodynamics
19
uniqueness can be proved using the methods in [6] and we only sketch the details.
The analytic dependence result is new and will be proved in complete detail.
Let us define the function space 3? as
3T={veL 2 (0,
T
;
2
We then have the following embedding theorems:
PROPOSITION 5.1. Let Q be an admissible exterior domain. Then
C([0, T]);/(Q)) is continuous. Moreover, if\e2£, then
(i) - (v, *)LHm
= <v,, a>yl(O).xj1(Q)
in 9>(0, r ) ' ,
for all z e A
and
(ii) ^
I
w/iere Si(0, 7*)' is tfie space of distributions on (0, T).
PROPOSITION
5.2. Let ®cciQ
be an open subset with compact closure. Let 2E&
be defined as
^ = {ve L2(0, T; H\e));
v, € L2(0, T;
(H\&))')}.
Then the embedding
(i) %&<-+ C([0, T]; L 2 (0)) is continuous, and the embedding
(ii) SQ <-» L2(0, T; L 2 (0)) is compact.
These theorems can be deduced from well-known and more general results on
vector-valued distributions [12].
Let us define a nonlinear map G(-, •) and the linear trace y0 as
G(v, U) = v, + vA\ + f/(0fi,(v) + U(t)B2(y) + B(\, v) - PA,
7o (v(-,0) = v(-,0).
and
DEFINITION 5.3. A vector-valued distribution v is called a generalised solution
corresponding to the data U if (v, U)s!Zx //'(O, T) and satisfy
G(v, U) = 0 in 3)(]0, r [ ;
and
Here S(]0, T[;Ji(Q)')' is the class of/,(Q)'-valued distributions on the interval
(0, T).
2
THEOREM 5.4. Uniqueness theorem. Let Q c R
be an admissible exterior
domain. Then the generalised solution is unique.
Proof. We suppose that v t , \\ e 2E correspond to data U e Hl(0, T). Then
G(v,, U) = G(v2, U) = 0, and
20
S. 5. Sritharan
Subtracting the two equations and denoting V] — v2 := v, we get
v, + vAv + I/(r)B,(v) + U(t)B2(y) + B(yu v,) - B(v2, v2) = 0,
y o (v(-,0) = 0.
Taking duality pairing with v, we get
<v() v)MaYxMn)
+ va{\, v) + U(t)b(\, wa> v) + U(t)b(yr6, v, v)
+ b(vu V], v) - b(\2, v2, v) = 0.
This simplifies to (using the properties of b{-, -, •))
v ||Vv|||2(Q) + U(t)b(y, yy6, v) + b(v, v,, v) = 0.
Proposition 5.1 stated above can now be used to take the time derivative out of
the first term. This gives
IMli
+
l|V|||
U(t)b(y, wfl, v) = -b(y, • „ v).
Using the estimates of the trilinear form established earlier, we get
~T IMli2(n) + 2v ||Vv|||2(£2)-2<5 ||Vv||l2 (Q) ^C ||v||L2(n) ||Vv||z,2(£2) HVviHz^n),
with 6 > 0 arbitrary. We now use Young's inequality to the right-hand side to get
y, IMIl*(Q) + 2(v - 6) ||Vv|||2(fi) § 2(v - 6) ||Vv|||2(Q) + C(v, 8) ||v||| 2(n) | | V v ^ l ^ .
at
Noting that v(-,0) = 0, we integrate the inequality (after cancelling the second
term on the left with the first term on the right) from 0 to t to obtain
-,0lli«Q)^C(v, 6) f||v|||2(Q) IIVv,!^) dt.
Now using the fact that Vj e L2(0, T;J0(Q)), we deduce by Gronwall's inequality
that Vj = v 2 .
D
THEOREM 5.5 Existence theorem. For each UeH\0,
U(T) = UT, there exists a unique v e
T) with t/(0) = 0 and
G(v,f/) = 0
and
Proo/. The proof is very similar to the constructions in [6], and here we only
sketch the steps involved. Let ©i <= 0 2 c . . be bounded open subsets of Q, each
having dQ as one of the boundaries and
An optimal control problem in exterior hydrodynamics
21
We can take @k as (for example) Q n {\x\ ^ rk}. In each of the subdomains, we
shall solve the following problem: (v*, pk): &k x [0, T] : - ^ R 2 x R such that
yk + v* . Vv* + U(t)vk . Vw6 + U(t)yfB . Vv*
= - Vpk + v Av* + (w in 0* x (0, T)
V . v = 0 in 0* x (0, T)
v*|SQt = 0, and
v*(*, 0) = 0,
xe@k.
Here d@k = dQ U {|#| = rk}. In each of these subdomains &k the solutions v*
satisfy the same a priori estimates with constants independent of the size of @k.
We can prove the existence of a unique generalised solution using Hopf s
Galerkin-type construction. This gives us a sequence {v*}£=1 € 2£, if we extend
each v* as zero outside &k. The fact that each of these weak solutions satisfies the
same a priori estimate with constants independent of the size of @k allows us to
take the limit to find the generalised solution in Q using the following theorem:
5.6. Let {\k} converge to v weakly in L2(0, T;Jr(Q)), weak star in
U°(0, T;J(Q)) and strongly in L2(0, T; L 2 (e,)); then the equation
THEOREM
G ( v \ [ / ) = 0 in 2(0, r;/,(Q)')',
*>(•*) = 0 in J(Q)
converges to
G(y,U)=0
in 0(0, r;/,(O)')',
yo(v) = 0 inJ(Q).
A similar result (Lemma 6.9) will be proved later in the context of optimal
solutions and the proof of this theorem is very similar. We note here that the a
priori estimates
sup \\yk\\L2(&k)+\\yk\\L2(o,T;M&kn
+ \\yk\\L\o,T;M&ky)^C(T,
v,
\\U\\HHo,T))
[0r]
provide us with the forms of convergence (for the subsequences) required in the
above theorem. The initial data are realised continuously since 2t<z
THEOREM 5.7. Analytic dependence theorem. The operator &(•): /f^O, T)—*2£
which associates with the generalised solution v e 2£, the corresponding datum
U e //'(O, T) is analytic in a neighbourhood of each U e /^(O, T) such that
i/(0) = 0 and U(T) = UT. In other words the series
9{U + V) = 9(U) + 2 X(U;
V,...,V)
converges in the norm of 2£ for \\ V||wi(0, T) < 5 with 6 < 0 some number. Here the
n-linear operators
are continuous.
22
S. S. Sritharan
Proof. Let us define a map M{-, •) as
Note that (v, U)e2£x H\0, T) is a generalised solution if
M(\, U) = (0, 0).
The analytic dependence theorem will be proved by applying the (analytic version
of the) implicit function theorem to the map Jt(-, •). As a first step, we prove
LEMMA 5.8.
The map M(-,-):%x
H\0, T) -+ L2(0, T; /j(Q)') X /(Q) is analytic.
Proof. We show that the map M{-, •) and its Frechet derivatives ^ ( - , •).
-, •), Mu(-, •), MVI]{;, •) and A)t/0» ) are continuous. The higher order
derivatives will be shown to be zero. First note that
Now,
llro(v)||y(Q) = ||v(-, 0)|| y ( o ) S \\v\\Cqo,T];j(n))Since the embedding £5? c C([0, T];J(Q)) is continuous, we have
Let us now estimate the norm of G(-, •). We have
+ U(t)B2(y) + B(v,
j + U(t)% + U
From the properties of the operators A, Bi(-). ^2(0 and B(-, •) established
earlier, we easily show that
We thus get
Similarly, we can show that
\\M(yu UJ - M(y2, U2)\\L,(0iT.May)xHQ)^
C3[||v, -
X [1 + l|£/ill«.(o,7-)
These two estimates establish the continuity of M{-, •). Let us now analyse the
derivatives of M(-, •). We have, for all h e f ,
with
CH,(V,
U; h) = h, + v^h + UB^h) + UB2(h) + B(y, h) + B(h, v).
An optimal control problem in exterior hydrodynamics
23
Using the properties of these operators established earlier, we get, for all h e 2£,
||A(v, U;h)\\LHo.Tiuay)xna)^C\\h\\x
[1 + \\U\\H,{0.T) + ||v||*],
2
hence ^(y, U; •) e <£(%.; L (0, 7V,(Q)') x/(Q)). We now consider, for all
7]eHl(0,T),
Mu(v, U; IJ) = (rj(B,(v) + B2(v)) - ijf, - 2l/ijf2 - r,,f3,0).
We can easily estimate this as, for all r, e //'((), T),
\\Mv(\, U; »j)|L2(o,r;/1(Q)')x7(Q) = C ||T/||H.(0,7-) [1 + ||U\\H<i{),T) + ||v||ar].
Thus, ^ ( v , f/; •) e ^(/^(O, T); L2(0, T; /,(Q)') x J(Q)). Now, for all h e SE,
Hence for all h e 2t,
Therefore ^ ( v , U, •, •): 3T® 2 -*L 2 (0,r;/ t (Q)') x / ( Q ) is continuous and bilinear. We now consider, for all rj e Hl(0, T) and for all h e %
Mvuiy, U; h, IJ) = (j/[B,(h) + B2(h)], 0).
From this we get:
WMvuiy, U;h, r))\\LHOiTMay)xj(a) = C \\h\y \\rf\\H\o.Ty
Thus A(/(v, U; •, •): £ X /^(O, r ) ^ L 2 ( 0 , T;/,(Q)') x/(Q)
Finally we have, for all r] eHl(0, T),
is
continuous.
Mvu(v,U;ri,ri) = (-27,%, 0).
This gives
\\Muu(v, U; IJ, ij)||^(o.T;y1(Q)')xj(Q)^ C ||IJ||?,.(O.D for all JJ e //'(O, T).
The higher-order Frechet derivatives of M(-, •) are zero. Hence we conclude that
M{-, •) is analytic for all (v, U)eS£x H\0, T) with (7(0) = 0 and U(T) = UT. •
5.9. For all (v, U)e2£x Hl(0, T) with (7(0) = 0 and U(T) = UT, the
Frechet derivative A ( v , U;-) is a bijective map from % onto L2(0, T;JX(Q)') X
LEMMA
/(Q).
Proo/. We have already established the continuity of this map. We only need
to show that, for all (v, U) e % x H\0, T) and for all (g t , g2) e L2(0, T; /,(Q)') x
/(Q), the equation
has a unique solution h 6 2£. This is the problem of evolution,
h, + vAh + (/(0#i(l») + U(t)B2(h) + B(h, v) + B(\, h) = gj for t > 0,
and h(0) = g 2 e/(£2).
Let us first obtain an a priori estimate for the solution by taking duality pairing
with h. This gives
~
WHhw + v ||Vh|||2(Q) + U(t)b(h, w6, h) + b(h, v, h) = < gl , *)
24
5. 5. Sritharan
We use the following estimates:
\U(t)b(h, w a , h)| ^ \U(t)\ 6 ||h|| 2 (n) ,
for all h 6/,(Q).
Since 6 is arbitrary this estimate becomes
\U(t)b(h, w a , h)| ^
\\Vh\\2LHa) for all he/,(O).
Moreover,
\b(h, v, h)| ^ C ||h|L 2(Q) \\Vv\\LHa)
^ll|Vh|||2(Q)
\\Vh\\LHo)
+ C(v) ||h||| 2(Q) ||Vv||| 2(n)
and
Kgi,
Combining these estimates gives us
| l|h||i 2(n) + v ||Vh||22(fi) =i C(v)
For h(0) = g2 e/(Q) and gt e L2(0,
lHa>
this inequality gives us
sup
re[0,7-]
= C(T, v,
We can easily prove the existence of h e 3£ using this estimate by the method of
Hopf [4]. The uniqueness is proved in the following way. If the data (gl5 g2) e
L2(0, T;J1(Q)')xJ(Q)
correspond to two solutions h^hje^f, then h = h i - h 2
must satisfy
h, + vAh + f/(0Bi(h) + U(t)B2(h) + B(v, h) + B(h, v) = 0, t>0,
and h(0) = 0.
We can then show by a somewhat similar method to that above that
=g C(v)
1].
From this and the fact that v € % we get h(t) = h^f) - h2(t) = 0.
•
Let us now complete the proof of the analytic dependence theorem. We have
shown that the map M:2£xH\Q, T)^> L2(0, T;Jl{Q,)')xJ{Q) is analytic and
for all (v, U) e 2T x H\0, T), the Frechet derivative A,(v, U; •) is a bijection
from & on to L2(0, Tj/^Q)') xJ(Q). Hence by the analytic version of the
implicit function theorem [1], we have the correspondence &: £/—»v is analytic in
a neighbourhood of each (v, U) e 2£ x H\0, T) such that M{\, U) = (0, 0). •
An optimal control problem in exterior hydrodynamics
25
6. Optimal control formulation
6.1. A subspace %tAcHi{^,
control functions if
DEFINITION
%d = {U: [0, T]^R+;
T) is called the admissible class of
(7(0) = 0, U(T) = UT,
U,(t) > 0, almost everywhere in [0, T] and {/(•) e H\0,
T)}.
LEMMA 6.2. %lad is a closed convex subset of Hl{Q, T). Moreover, it is closed in
the weak topology of H1^, T).
Proof. Convexity of %d is easy to check. To see that %d is closed in H1^, T),
we only need to use the continuity of the embedding //*((), T) a C[0, T]. Finally,
it follows from a corollary to the Hahn-Banach theorem [5] that a closed convex
subset of a normed space is closed in the weak topology.
DEFINITION 6.3. Let v be the generalised solution corresponding to the datum
U. A map $: 3? x H\0, T)-» R + is called the cost functional if
*(?, V) = \ ||v(-, T) + £/r(wa(-) - eOlfoo) + 2v [ || def (v + wa£/)||i«o) dt
z
Jo
\ut-u?\2dt,
where A > 0 and Ud(-) e H\0, T) are given.
Here the first two terms of £(•, •) represent the total energy expenditure and the
last term is added to reduce oscillations in the optimal control U.
Let us now define the class of solutions to the Navier Stokes equations in which
the optimum is sought.
DEFINITION 6.4. A vector valued distribution v and a scalar valued distribution
U are called an admissible pair if
(i) (v, U)eSx %d,
(ii) My, U) = (0, 0) in L2(0, TJ^Q)') X/(Q), and
(iii) ${y, U) < oo.
Let us denote by %d the set of all admissible pairs.
THEOREM
6.5. %dc2£x
H\0, T) is nonempty.
Proof. For each U e H\0, T) with U(0) = 0 and U(T) = UT, we have by the
well-posedness theorem, that there exists a unique v e 2£ such that
M{\, U) = (0, 0)
in
L2(0, T; J^Q)') x 7(Q).
We thus only need to check the condition (iii). Note first that
||v(-, T) + [/r(w6(-) - eOll^n) ^ ||v(-, T)\\LHa)
Since 2£ c C([0, T]; J(Q)) we get
||v(-, T) + UT(wd(-)
+ UT ||wa(-) -
eJU^
26
5. 5. Sritharan
Now
Hence
/ ( v , £/) ^ d ( v ) ||v||| + C2(v,
Thus, for all (v, (/) e 3? x H\0, T) and £/<* e H\0, T), we have
0 ^ / ( v , £/)<<*>. D
DEFINITION 6.6. An admissible pair (v, f/) e %d is called an optimal solution if
the corresponding value of the cost functional achieves the absolute minimum:
^(v,f>)=
inf
/(v,f/).
(v,f/)e««, d
6.7. Existence theorem. There exists an optimal solution (v, U) e 2£ x
such that
THEOREM
M(\, U) = (0, 0) in L2(0, T; MQ)') x /(Q)
and
Jf(v,C/)=
inf /(v,[/).
(v,l/)e<3/,d
Proof. Let us first establish certain properties of the cost functional $(•, •) and
the map M{-, •)•
LEMMA
6.8. $(-, •): % x H\0, T)-*R+
is weakly sequentially lower
semicontinuous: if\"—*\ weakly in 2E and if Un-*U weakly in H\0, T), then
j?(v,£/)^liminf^(v", U").
n—*o°
Moreover, $(-, •) is coercive. That is, for all (v, U) e %d, with \\y\\x +
\O,T)^<X>, wehave $(y,U)^><x>.
Proof. Let us first establish the coerciveness of the cost functional. Using the
Cauchy inequality, we can easily show that
7U) i ^ | ||v(-, T)\\L2(a) - u
{J ||defv|||2(Q)A} We have, due to the continuity of U, ||£/|L2(o,7-)= UTVf. Moreover, we have
Korn's inequality [16],
||Vv|U 2(Q) ^C ||def Tll^o),
for all v e
An optimal control problem in exterior hydrodynamics
27
Combining these, we can easily show that for large ||v|| a ,
/(v, U) ^ C,[||v(-, T)\\lHa) + 2v j * ||W|||2(Q) dt + k £ U2 dt\
This shows the role of A in the coerciveness condition. Now, the a priori estimate
(4.1) and Lemma 4.2 imply that
MI* = [£ ||v,||3l(o). dt + ££ ||v||??
i(Q) dt\\
*C3(T,
UT, v, \\U\\H<(0,T)).
(6.2)
Here C3 depends linearly and quadratically on ||£/||wi(o,7-). From (6.1) and (6.2),
we can see that as ||v|| z + ||£/||//i(o,7')-*00> w e should have ${y, £/)—»«>. We now
establish the lower semicontinuity property of ${•, •). Let us begin with the
following forms of convergence in the weak topology:
v" -»v
£/"-*!/
in % and
inH\0,T).
Then, since 3. «= C([0, T];J(Q)), we can deduce that v"(-, T ) ^ v ( - , T) weakly in
7(Q). Hence,
||v(-, T) + UT(yvs ~ iMmo)
^ Hm inf ||v(-, T)n + C/T(w6 -
C,)IUQ)-
rt—•a0
Similarly, weak convergence of U" in //'(O, T) implies that,
rT
fT
\Ut- Uf\2 dt ^ lim inf
|U? -
Uf\2dt.
n->v JQ
Jo
In order to analyse the dissipation term in the cost functional, we introduce the
following Hilbert space:
= completion of y(Q) in the norm defined by ||defv|| i 2 (Q )<oo.
Note that the embedding J0(Q)cJf(Q) is continuous. Therefore, v"—»v weakly
in 2t implies that v"—»v weakly in L2(0, T;Jf(Q)). This, in combination with the
fact that Un-*U weakly in Hl(0, T), implies
v" + w6 U" -> v + v/6 U weakly in Jf(Q).
Hence
lim inf f ||def (v» + t/"wa ||| 2(Q) dt ^ f ||def (v + (/w4 ||i 2(Q) dt.
«-*» Jo
Jo
Thus, combining these results, we get
«?(v, t/) Slim inf ^(v",*/"),
which proves the weak-sequential lower semicontinuity of the cost functional.
•
28
5. 5. Sritharan
Let us now prove the optimality theorem. We consider the set
Note that the coerciveness estimate will then imply that for all (v, U) e % ,
\\ns£C1(R,T,UT,v,k)
(6.3)
and
\\U\\H,(0,T)^C2(R,T,
UT, v,A).
(6.4)
This result, in combination with the fact that £{•, •) is bounded below by zero,
allows us to deduce the existence of a minimising sequence (v, U) e % , so that
lim ^(v", U") =
n^oo
inf ${y, U): = y0,
(v,f/)e<3/ad
with y0 = 0 and
M(\", Un) = (0, 0), for all n.
Uniform bounds (6.3) (6.4) on the minimising sequence allow us to extract a
subsequence (v"', Un')e% such that v"'-*\ and U"'—*U, respectively, in the
weak topologies of 2C and //*((), T). Let us now analyse the limit pair
(v, U) € 2E X H\0, T). As noted in the Lemma 6.2, the class of controls %d is
closed in the weak topology of Hl{0, T). Hence we have U e % ad .
Now using the convergence Lemma 6.9 below we can deduce that
M(y, U) = (0, 0).
Moreover, using the sequential lower semicontinuity of ${•, •), we get
$(y, U) ^ lim inf ${y"', U"') < «.
(6.5)
We thus conclude that (v, U) € %d. We now argue that (v, U) is the optimal pair
we are looking for. Firstly, since (v, U) is an admissible pair and y0 is by
definition the absolute minimum of the cost, we should have
yo^(v,f/).
(6.6)
However, (v" , £/"') is a minimising sequence. Therefore
lim inf jP(v"', U"') = lim ${f', Un') = y0.
Thus from the lower semicontinuity inequality (6.5) we get
/(».fi)^o.
(6.7)
Inequalities (6.6) and (6.7) imply that
This says that the pair (v, U) corresponds to the absolute minimum and hence
represents the optimal solution.
An optimal control problem in exterior hydrodynamics
29
Now, in order to complete the proof, we state and prove the technical lemma
used in the above proof.
6.9. Let the sequences {v"} and {Un} converge respectively to v and U
in the weak topologies of 2£ and H1^, T). Then the equation
LEMMA
M(\", Un) = (0, 0)
converges to
M(\, U) = (0, 0) in 2(0, T; JX(Q)')' x J(&)Proof. We shall establish the convergence of each term in the map M(-, •) one
by one, Note first that the continuity of the embedding 2£czC([0, T];J(Q))
implies that
v"(-, f)-» v(-, 0 in the weak topology of J(Q),
for all t e [0, T].
Hence in particular
Yo(y"(-> t)) = v"0> °) = ° converges to
Now note that v" —* v weakly in 2E implies the weak convergence of
<^v,
in
^(OJj
Now, the weak convergence of v" in 3? as well as the continuity and the closure
properties of the Stokes operator A e i?(/ 1 (Q);/i(Q)') imply that Ayn converges
weakly to A\ in L2(0, Tj/^Q)'). Similarly, we can conclude easily that U" and
U" converge weakly in L2(0, T) to U and Ut, respectively. Moreover, since
Un(t)^UT, for all (e[0, T], we can deduce that (Un)2 converges weakly to
U2 in L2(0, T). Thus observing that tu f2 and f3 e/(Q), we conclude that
dU"
C/"f, + (U")% + —- f3 converges weakly to
dt
Ut, + (U)% + ^ f 3
dt
in L2(0, T;7,(0)').
We now consider the terms WB^V),
Un(B2(y") and B(v", v"). In order to
establish their convergence, we note that the restriction of v" to 6 c c Q gives us
a uniformly bounded (independent of the size of 0 and n) sequence in %g,. We
then use the compactness of the embedding 2£BczL2(0, T;L2(&)) to conclude
that
v" -> v strongly in L2(0, T; L 2 (0))
for all 0 c c Q.
30
5. S. Sritharan
Let us consider a test function <$> e 3>(0, T;/(©)). Then, for i = 1 or 2,
f ((/"fi,(v") - UB,(Y), 4>)LHa) dt = \ £/"(B,(V - v), <I>)&) dt
Jo
h
+
f(U"-U)(Bi(v),(t>)LH&)dt
•>o
+ f {WJo
We proved earlier that
and B2*(-),S2(-)
We now use these properties of /?,(•) along with the fact that U"(t) ^ UT, for all
f e [0, T], to deduce that
due to the strong convergence of v". Now, since v e 2T contained in a bounded set,
we have easily that (v, B*(<^))z,2(0) e L2(0, T) and hence that
fT
(U"-U)(v,B*(<t>))L2i&)dt-^0
as M-*OO; for all 0 c c Q .
Jo
We have thus shown that
UBXv) in 3(0, r ; / ( 0 ) ' ) ' , for all 0 c c Q.
To complete the proof, we need to establish the convergence of the term
B(V, v"). Again consider a test function in 0 e ®(0, T;/(©)). We have
f (fi(v, V) - B(v", v"),
Jo
\
=
(B(v - V, v),
Jo
<j))M0yxM
-f
Jo
dt
<fi*(^v-v"),v),,(eVxyl(e)A.
Here we have used the properties of B*(-, •) established earlier. We now use the
estimates for B{-, •) and B*(-, •) to get
I f (B(y, v) - B(y, V), 0>yl(Q).xyl(Q) *I
I Jo
^ d f ||v - v j ^ e ) IIVv - V v J I ^ ||v||[ 2(0) ||Vv||i 2(e) dt
Jo
II* - Vn\\iHB) HVv - Vv n |||. 2 ( e )
An optimal control problem in exterior hydrodynamics
31
We now use the Holder inequality and the fact that v and v" are uniformly
bounded in 2T, to get
>,v)-B(yn,vn),<t>)MayxMa)dt
due to the strong convergence of v" in L2(0, T; L 2 (0)). We have thus shown that
B(yn, yn) converges to B(v, v) in 2(0, T;/(0)')', for all 0 <= c Q. This completes
the proof of the convergence lemma as well as the optimality theorem. •
7. Concluding remarks
(I) We note here that although the existence of optimal control is established
here, it is not known whether it is unique. In fact, except for low Reynold's
numbers (large v), it is unlikely that the optimal control is unique.
(II) The proof provided here is not valid for the three-dimensional exterior
domains due to the lack of a global unique solvability theorem.
(HI) In [14,13,2] we address the question of computing this optimal control
using the Pontryagin maximum principle (which provides the necessary conditions) and dynamic programming method (using the infinite dimensional
Hamilton-Jacobi-Bellman equations) to obtain the feedback map.
(IV) The existence theorem gives the solution in the class 2£. However, from
the a priori estimate (4.2) we can prove the regularity v e 2f, = {</> e
L2(0, T; D(A)); <t>, e L2(0, T;/(Q))} c C([0, T];
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(Issued 10 June 1992)