On the existence and uniqueness of solutions of optimal ... - Turpion

c 2011 RAS(DoM) and LMS

Izvestiya: Mathematics 75:2 395–412 Izvestiya RAN : Ser. Mat. 75:2 177–194

DOI 10.1070/IM2011v075n02ABEH002538

On the existence and uniqueness of solutions of optimal control problems of linear distributed systems which are not solved with respect to the time derivative M. V. Plekhanova and V. E. Fedorov Abstract. We investigate optimal control problems for linear distributed systems which are not solved with respect to the time derivative and whose homogeneous part admits a degenerate strongly continuous solution semigroup. To this end, we first obtain theorems on the existence of a unique strong solution of the Cauchy problem. This enables us to formulate sufficient conditions for the solubility of the optimal control problems under consideration. In contrast to earlier papers on a similar topic, we substantially weaken the conditions on the quality functional with respect to the state function. The abstract results thus obtained are illustrated by an example of an optimal control problem for the linearized system of Navier–Stokes equations. Keywords: optimal control problem, distributed system, equation of Sobolev type, degenerate operator semigroup, unique solubility.

Introduction Let X , Y be Hilbert spaces, L ∈ L(X ; Y) a continuous linear operator from X to Y with ker L 6= {0}, M ∈ Cl(X ; Y) a densely defined closed linear operator from X to Y, and N ∈ L(X ) a continuous linear operator from X to X . We shall consider optimal control problems for distributed systems that are not solved with respect to the time derivative: J0 (x) + ku − u0 k2H p+1 (0,T ;U ) → inf,

(0.1)

Lx(t) ˙ = M x(t) + y(t) + Bu(t),

(0.2)

N (x(0) − x0 ) = 0,

(0.3)

u ∈ U∂ .

(0.4)

The initial condition (0.3) will either be the Cauchy condition (when N = I) or the generalized Showalter condition [1] (when N = P is the projection onto the phase space of the homogeneous equation (0.2)). We note that, when considering systems of the form (0.2), the latter condition is in a sense the more natural of the two. Indeed, in contrast to the Cauchy problem, the solubility of the generalized Supported by the Russian Foundation for Basic Research (grant no. 07-01-96030-r ural a) and the Programme of the President of the Russian Federation ‘Support of Young Doctors of Sciences’ (grant no. MD-4312.2006.1). AMS 2010 Mathematics Subject Classification. Primary 49J20. Secondary 34H05, 35Q93, 49J15, 49K20, 93C20.

396

M. V. Plekhanova and V. E. Fedorov

Showalter problem needs neither additional conditions nor the agreement of the initial value with the function on the right-hand side of the equation. We shall use the apparatus of the theory of degenerate operator semigroups [2]–[4] to obtain sufficient conditions for the existence of a strong solution of (0.2), (0.3) in the case when the pair (L, M ) generates a degenerate strongly continuous operator semigroup. We first obtain necessary and sufficient conditions on the right-hand side for the equation to be soluble when the operator at the derivative is nilpotent. The necessity of these conditions forces us to avoid non-smooth control functions and to choose the space H p+1 (0, T ; U) as the space of control functions. When considering similar problems in the case of finite-dimensional spaces X and Y, many papers have pointed out the necessity of choosing sufficiently smooth control functions (see, for example, [5] and [6]). For example, a partition was introduced in [5] of the controllable descriptor systems (that is, systems of the form (0.2)) in the finite-dimensional case into proper systems, in which the state does not depend on the derivatives of the control functions (p = 0), and improper systems, in which a state is defined, in particular, by derivatives of the control function (p ∈ N). Various approaches have been used in the study of optimal control problems for linear operator equations that are not solved with respect to the derivative. In [7]–[9] and in the papers [10], [11] of the present authors, additional conditions on the initial value x0 and on the control function at the initial moment of time were used when proving the unique solubility of such control problems. The unique solubility of the optimal control problem was proved in [12] in the sense of a weak solution of the equation (0.2), and in [13] for a system with generalized Showalter condition, while the paper [14] deals only with problems of start-up control. As in [14], when considering the optimal control problem for equations of the form (0.2), we shall use the general scheme of investigation suggested in [15]. This makes it possible to get rid of the aforementioned additional conditions on the initial value and the control function. Moreover, as compared with the earlier papers listed above, we have weakened the conditions on the quality functional with respect to the state function. As to the functional J0 , we shall only assume that it is convex, non-negative, and lower semicontinuous with respect to weak convergence. As special cases, we consider the functional containing the squared norm of the function in the Lebesgue space (in contrast to the functional with the squared norm in the Sobolev space H 1 (0, T ; X ) used in all previous studies) and the terminal functional, that is, the functional containing the squared norm of the state function at a fixed moment of time. We obtain sufficient conditions for the unique solubility of the problems (0.1)–(0.4) provided that a strongly continuous solution semigroup for the homogeneous equation (0.2) exists. Our results are illustrated by the example of an optimal control problem for the linearized system of Navier–Stokes equations. § 1. Relatively p-radial operators In this section we list the results needed below on the existence and properties of strongly continuous solution semigroup for a first-order linear operator differential equation with degenerate operator at the derivative. The proofs of these results can be found in [2]–[4].

On the existence and uniqueness of solutions

397

Let X and Y be Banach spaces, and let L ∈ L(X ; Y), ker L 6= {0}, and M ∈ Cl(X ; Y) be operators. We also introduce the following notation: N0 = {0} ∪ N, R+ = {0} ∪ R+ , ρL (M ) = {µ ∈ C : (µL − M )−1 ∈ L(Y; X )}, σ L (M ) = C \ ρL (M ), and RµL (M ) = (µL − M )−1 L, L R(µ,p) (M )

=

p Y

RµLk (M ),

−1 LL , µ (M ) = L(µL − M )

LL (µ,p) (M )

=

p Y

LL µk (M ).

k=0

k=0

L 1 1 Let X 0 (Y 0 ) denote the kernel ker R(µ,p) (M ) (ker LL (µ,p) (M )), X (Y ) the closure L of the image im R(µ,p) (M ) (im LL (µ,p) (M )) with respect to the norm of the space X (Y), and Mk (Lk ) the restriction of M (L) to dom Mk = X k ∩dom M (X k ), k = 0, 1.

Definition 1.1. The operator M is said to be strongly (L, p)-radial, p ∈ N0 , if (i) ∃ a ∈ R such that (a, +∞) ⊂ ρL (M ), (ii) ∃ K ∈ R+ such that K n k=0 (µk − a)

L n max k(R(µ,p) (M ))n kL(X ) , k(LL (µ,p) (M )) kL(Y) 6 Qp for any µk ∈ (a, +∞), k = 0, . . . , p, and any n ∈ N, ˚ dense in Y such that (iii) there is a vector subspace Y

M (λL − M )−1 LL (M )y 6 (µ,p) Y

const(y) Qp (λ − a) k=0 (µk − a)

for any λ, µ0 , µ1 , . . . , µp ∈ (a, +∞), (iv)

L

−1

R 6 (µ,p) (M )(λL − M ) L(Y;X )

(λ − a)

K Qp

k=0 (µk

˚ ∀y ∈ Y

− a)

for any λ, µ0 , µ1 , . . . , µp ∈ (a, +∞). Theorem 1.1. Let the operator M be strongly (L, p)-radial. Then (i) X = X 0 ⊕ X 1 , Y = Y 0 ⊕ Y 1 , (ii) Lk ∈ L(X k ; Y k ), Mk ∈ Cl(X k ; Y k ), k = 0, 1, 1 1 (iii) there are operators M0−1 ∈ L(Y 0 ; X 0 ) and L−1 1 ∈ L(Y ; X ), (iv) the operator G = M0−1 L0 ∈ L(X 0 ) is nilpotent of degree not exceeding p ∈ N0 , (v) there is a strongly continuous solution semigroup {X t ∈ L(X ): t ∈ R+ } × of the homogeneous equation Lx(t) ˙ = M x(t), and X t = s- limk→∞ k(p+1) t k(p+1) L R k(p+1) (M ) , t

(vi) the infinitesimal generator of the C0 -continuous semigroup {X1t ∈ L(X 1 ): 1 t t ∈ R+ } is the operator S = L−1 1 M1 ∈ Cl(X ), where X1 stands for the restriction t 1 of X to X .

398


The identity element of the semigroup X 0 = P = s- limµ→+∞ (µRµL (M ))p+1 is p+1 the projection along the subspace X 0 onto X 1 , and Q = s- limµ→+∞ (µLL µ (M )) 0 1 is the projection along Y onto Y . Remark 1.1. The semigroup in part (v) of Theorem 1.1 satisfies the bound kX t kL(X ) 6 Keat for t > 0, where the constants K and a are as in Definition 1.1. Remark 1.2. By the Hille–Yosida theorem, it follows from part (vi) of Theorem 1.1 that the operator S satisfies the conditions ∃ a ∈ R : (a, +∞) ⊂ ρ(S), ∃ K ∈ R+ :

∀ µ ∈ (a, +∞),

∀n ∈ N

k(µI − S)−n kL(X 1 ) 6

K . (µ − a)n

Let ker L 6= {0} and ϕ0 ∈ ker L \ {0}. An ordered set of vectors {ϕ0 , ϕ1 , . . . } is referred to as a chain of M -adjoint vectors if Lϕk+1 = M ϕk ,

k ∈ N0 ,

ϕl ∈ / ker L,

l ∈ N.

The position of a vector in such a chain (beginning with zero) is referred to as the height of the vector. Remark 1.3. Under the condition that the operator M is strongly (L, p)-radial, the subspace X 0 is the linear span of the set of M -adjoint vectors of L of height not exceeding p and, in this case, there are no M -adjoint vectors of greater height. § 2. A solubility criterion for a singular equation The investigation of an equation of Sobolev type, Lx(t) ˙ = M x(t) + y(t), is carried out by reducing it to a system of two equations, namely, a regular equation on the subspace X 1 and a singular equation on the subspace X 0 with a nilpotent operator at the derivative. In this section we obtain necessary and sufficient conditions on the function y for the solubility of the singular equation, and these lead to the choice of the space H p+1 (0, T ; U) (rather than, say, L2 (0, T ; U)) as the space of control functions in optimal control problems of the form (0.1)–(0.4). Consider the following equation on the space X 0 : GDx(t) = x(t) + f (t),

(2.1)

d where G ∈ L(X 0 ) stands for a nilpotent operator of degree p, D = dt for the formal operator of differentiation, and f : (0, T ) → X 0 for a given function. We introduce the following notation for Sobolev spaces:

Wqk (X ) = Wqk (0, T ; X ),

H k (X ) = W2k (X )

for k ∈ N0 , 1 6 q < ∞. Here Lq (X ) = Wq0 (X ), L2 (X ) = H 0 (X ). A function x ∈ Wq1 (X 0 ) is said to be a strong solution of the equation (2.1) if it satisfies the equation almost everywhere on (0, T ).


399

Theorem 2.1. Let f ∈ Lq (X 0 ). Then a strong solution x ∈ Wq1 (X 0 ) of the equation (2.1) exists if and only if the following conditions hold: Gp f ∈ Wq1 (X 0 ),

(Gp−1 + DGp )f ∈ Wq1 (X 0 ), . . . ..., Gp−k+1 + D(Gp−k+2 + · · · + D(Gp−1 + DGp ) . . . ) f ∈ Wq1 (X 0 ), . . . ..., I + D(G + D(G2 + D(G3 + · · · + D(Gp−1 + DGp ) . . . ))) f ∈ Wq1 (X 0 ). (2.2) This strong solution is unique and has the form x(t) = − I + D(G + D(G2 + D(G3 + · · · + D(Gp−1 + DGp ) . . . ))) f (t). (2.3) Proof. Suppose there exists a solution x ∈ Wq1 (X 0 ) of the equation (2.1). Multiply the equation (2.1) by Gp . Then, since G is nilpotent, we obtain −Gp x = Gp f ∈ Wq1 (X 0 ). Differentiating the equation, we obtain DGp (x + f ) = 0. Since x ∈ Wq1 (X 0 ), we can write this equation in the form Gp Dx = −DGp f . Applying the operator Gp−1 to (2.1), we obtain from the last equation that −DGp f = Gp−1 (x + f ). Then Gp−1 x = −Gp−1 f − DGp f, and both the sides of this equation are elements of Wq1 (X 0 ). Arguing in the same way, we obtain the formula (2.2) at the kth step, conclude at the (p + 1)st step that all conditions of the theorem are necessary for the existence of a strong solution x ∈ Wq1 (X 0 ), and also obtain the form (2.3) of the solution. This also implies the uniqueness. Indeed, if f (t) = 0 almost everywhere on the interval (0, T ), then it also follows from (2.3) that x(t) = 0 almost everywhere on (0, T ). The sufficiency of the conditions of the theorem is verified by the direct substitution of the function (2.3) into (2.1). This proves the theorem. Remark 2.1. If f ∈ Wqp+1 (X 0 ), then the solution of (2.1) becomes x(t) = Pp − k=0 Gk Dk f (t). Corollary 2.1. Let the operator M be strongly (L, p)-radial, G = M0−1 L0 . Then there are functions f ∈ Lq (X 0 ) for which (2.1) has no solution in the space Wq1 (X 0 ). Proof. Let f (t) = g(t)ϕp , where ϕp stands for the M -adjoint vector of height p for the operator L and g ∈ Lq (0, T ; C) \ Wq1 (0, T ; C). It follows from the definition of M -adjoint vectors and Remark 1.3 that Gϕp = ϕp−1 , where ϕp−1 stands for the previous vector in some chain of M -adjoint vectors of L. Hence, applying G to the last equation p − 1 times, we obtain Gp ϕp = ϕ0 for the chain, where ϕ0 stands for an eigenvector of L. Thus, Gp f (t) = g(t)ϕ0 ∈ / Wq1 (X 0 ) by the choice of g(t). The first of the necessary conditions in Theorem 2.1 for the existence of a solution x ∈ Wq1 (X 0 ) fails to hold. Remark 2.2. If f (t) = g(t)ϕk , where ϕk stands for an M -adjoint vector of height k, 0 6 k 6 p, and g ∈ Lq (0, T ; C) \ Wq1 (0, T ; C), then the kth necessary condition in (2.2) for the existence of a solution x ∈ Wq1 (X 0 ) fails to hold, although the previous conditions are satisfied.

400


§ 3. A strong solution of the Cauchy problem Let X and Y be Banach spaces and let L ∈ L(X ; Y) and M ∈ Cl(X ; Y), where M is strongly (L, p)-radial. We introduce the operators A0 : y → −

p X

G

k

M0−1 (I

− Q)y

(k)

t

Z

X t−s L−1 1 Qy(s) ds.

A1 : y →

(t),

0

k=0

Lemma 3.1. Let M be strongly (L, p)-radial. Then (i) A0 ∈ L(Wqp+1 (Y); Wq1 (X )), (ii) A1 ∈ L(Wq1 (Y); Wq1 (X )), ˙ 0 the function X t x0 belongs to C 1 ([0, T ]; X ). (iii) for any x0 ∈ dom M1 +X Proof. (i) We have kA0 ykWq1 (X ) 6

p X

kGk M0−1 (I − Q)kL(Y,X ) ky (k) kLq (Y) + ky (k+1) kLq (Y)

k=0

6

p+1 X

Ck ky (k) kLq (Y) 6

k=0

max

k=0,...,p+1

Ck · kykWqp+1 (Y) .

(ii) The following inequalities hold:

Z t

t−s −1

X L1 Qy(s) ds

0

T Z t

Z

kX

6 0

Lq (X )

Z

L−1 1 Qy(s)kX

q ds

1/q dt

0

T

t

6

t−s

q−1

0

Z

1/q

t

kX 0

t−s

q L−1 1 Qy(s)kX

ds dt

6 T q −1/q Ke|a|T kL−1 1 QkL(Y;X ) kykLq (Y) = C1 kykLq (Y) . We now claim that Z t 0 Z t t−s −1 0 t −1 X L1 Qy(s) ds = X t−s L−1 1 Qy (s) ds + X L1 Qy(0) 0

0

for any y ∈ Wq1 (Y). For ∆ > 0 we have 1 ∆

Z

t+∆

X

t+∆−s

L−1 1 Qy(s) ds

t

Z −

0

X

t−s

L−1 1 Qy(s) ds

0

Z =

t

X 0

t−s

y(s L−1 1 Q

+ ∆) − y(s) 1 ds + ∆ ∆

Z

∆

X t+∆−s L−1 1 Qy(s) ds.

0

Since y ∈ Wq1 (Y), it follows that y is differentiable in the classical sense almost everywhere on (0, T ). Therefore,

y(s + ∆) − y(s)

< ky 0 (s)kY + ε

∆ Y


401

for almost all s ∈ (0, T ) and for any ε > 0 provided that |∆| is sufficiently small. Since y 0 ∈ Lq (Y) ⊂ L1 (Y), the limit Z t Z t t−s −1 y(s + ∆) − y(s) 0 lim X L1 Q ds = X t−s L−1 1 Qy (s) ds ∆→0+ 0 ∆ 0 exists by Lebesgue’s theorem and, since y is continuous because of the embedding Wq1 (Y) ,→ C([0, T ); Y), it follows that Z 1 ∆ t+∆−s −1 t −1 X L1 Qy(s) ds = lim X t+∆−θ L−1 lim 1 Qy(θ) = X L1 Qy(0). ∆→0+ ∆→0+ ∆ 0 Here θ ∈ (0, ∆) by the mean value theorem. We define y(s) for s ∈ (−ε, 0), where ε > 0, in such a way that y ∈ Wq1 (−ε, T ; Y). Then for ∆ ∈ (−ε, 0) we have Z t+∆ Z t 1 t−s −1 X t+∆−s L−1 Qy(s) ds − X L Qy(s) ds 1 1 ∆ 0 0 Z t Z 1 −∆ t−s −1 t−s −1 y(s + ∆) − y(s) = X L1 Q ds − X L1 Qy(s + ∆) ds. ∆ ∆ 0 0 Passing to the limit in the last expression as ∆ → 0−, we obtain the desired result: Z t 0 t −1 X t−s L−1 1 Qy (s) ds + X L1 Qy(0). 0

Further,

Z t

t−s −1 0 t −1

X L1 Qy (s) ds + X L1 Qy(0)

0

Lq (X )

T

Z 6 0

1/q Z q kX t L−1 Qy(0)k dt + 1 X

|a|T

6 Ke

Z T

t

0

T

1/q

kL−1 1 Qy(0)kX

Z +

T

t

q−1

X

t−s

0

Z

0

q

−1 0 L1 Qy (s) ds

X

1/q dt 1/q

T

kX 0

t−s

q 0 L−1 1 Qy (s)kX

ds dt

−1/q 0 6 Ke|a|T T 1/q kL−1 Ke|a|T kL−1 1 QkL(Y;X ) ky(0)kY + T q 1 QkL(Y;X ) ky kLq (Y) .

The above considerations and the embedding Wq1 (X ) ,→ C([0, T ); X ) yield kA1 ykWq1 (X ) 6 C1 kykLq (Y) + C2 ky 0 kLq (Y) + C3 kykC([0,T ];Y) 6 CkykWq1 (Y) . (iii) If M is strongly (L, p)-radial for any x0 in the set dom M1 = dom M ∩ X 1 , then X1t x0 ∈ C 1 ([0, T ]; X ) by Theorem 1.1, (vi). Since X t x0 = X1t P x0 for any x0 ∈ X , this completes the proof. Consider the Cauchy problem x(0) = x0 ,

(3.1)

Lx(t) ˙ = M x(t) + y(t).

(3.2)

By a strong solution of the problem (3.1), (3.2) we mean a function x ∈ Wq1 (X ) satisfying the equation (3.2) almost everywhere on (0, T ) and the condition (3.1).

402


Lemma 3.2. Let A ∈ L(V), where V is a Banach space. Then for any f ∈ Lq (V) and v0 ∈ V there is a unique strong solution of the problem v(t) ˙ = Av(t) + f (t),

v(0) = v0 .

(3.3)

Rt Proof. It can readily be seen that the operator B : v(t) → 0 Av(s) ds is contracting on the space Lq (0, T1 ; V) for T1 < kAk−1 L(V) . If there are two solutions v1 (t), v2 (t) of the problem (3.3), then their difference v(t) = v1 (t) − v2 (t) is a solution of the problem v(t) ˙ = Av(t), v(0) = 0. Hence, v(t) is the unique fixed point of B, and therefore v(t) = 0 almost everywhere on the interval (0, T1 ). Considering the operator B on the space Lq (T1 , 2T1 ; V), we see that v(t) = 0 almost everywhere on (T1 , 2T1 ). We exhaust the interval (0, T ) in finitely many steps. The unique solution of the problem (3.3) is obviously of the form At

At

Z

v(t) = e v0 + e

t

e−As f (s) ds ∈ Wq1 (V),

0

P∞ k k where eAt = k=0 Ak!t stands for the analytic group generated by A. This completes the proof of the lemma. Theorem 3.1. Let M be strongly (L, p)-radial. Then for any y ∈ Wqp+1 (Y) and p X −1 k (k) x0 ∈ My = x ∈ dom M: (I − P )x = − G M0 (I − Q)y (0) , k=0

there is a unique strong solution of the problem (3.1), (3.2), and this solution is of the form x(t) = (A0 + A1 )y(t) + X t x0 and satisfies the condition kxkWq1 (X ) 6 C kx0 kX + kSP x0 kX + kykWqp+1 (Y) . (3.4) Proof. Applying first the projection I − Q and then the operator M0−1 ∈ L(Y 0 ; X 0 ) to (3.2), we obtain the equation Gx(t) ˙ = x(t) + M0−1 (I − Q)y(t)

(3.5)

on the space X 0 with the initial condition x(0) = (I − P )x0 . It follows from Theorem 2.1 and Remark 2.1 using the continuity of the operator M0−1 that the Pp unique solution of this equation is x(t) = − k=0 Gk M0−1 (I − Q)y (k) (t) = A0 y(t). It obviously satisfies the initial condition x(0) = (I − P )x0 if and only if x0 ∈ My . Now applying the operator L−1 1 Q to (3.2), we obtain the equation x(t) ˙ = Sx(t) + L−1 1 Qy(t)

(3.6)

on the space X 1 with the initial condition x(0) = P x0 , where S = L−1 1 M1 . By the hypothesis of the theorem we have y ∈ Wqp+1 (Y) ⊂ L1 (Y), and hence X t−s L−1 1 Qy(s) ∈ L1 (X ). Further, in view of the claims proved in Lemma 3.1,


403

we obtain from Theorem 1.1, (vi) that Z

t

X

S

t−s

L−1 1 Qy(s) ds

0

X1∆ − I = lim ∆→0+ ∆

t

Z

X t−s L−1 1 Qy(s) ds

0

Z t Z t 1 t−s −1 X t+∆−s L−1 Qy(s) ds − X L Qy(s) ds 1 1 ∆→0+ ∆ 0 0 Z Z d t t−s −1 1 t+∆ t+∆−s −1 = X L1 Qy(s) ds − lim X L1 Qy(s) ds ∆→0+ ∆ t dt 0 Z d t t−s −1 = X L1 Qy(s) ds − L−1 1 Qy(t). dt 0 = lim

This implies that the function A1 y(t) + X1t P x0 = A1 y(t) + X t x0 is a solution of (3.6). We claim the uniqueness of a strong solution of (3.6) satisfying the initial condition (3.1). The operator Jλ = (λ − a)(λI − S)−1 exists for λ > a by Remark 1.2. A strong solution xλ (t) of the Cauchy problem xλ (0) = x0 for the equation x˙ λ (t) = (SJλ )xλ (t) + L−1 1 Qy(t) is uniquely determined by the initial value x0 in view of Lemma 3.2, because the operator SJλ = λ(λ − a)(λI − S)−1 − (λ − a)I is bounded. Let us write x(t) − xλ (t) = vλ (t), where x(t) is a solution of the problem (3.1), (3.6). Then the identity v˙ λ (t) = (SJλ )vλ (t) + (S − SJλ )x(t) holds. This yields Z vλ (t) = 0

t

Xλt−s (S − SJλ )x(s) ds =

because vλ (0) = 0. Here Xλt−s = e(t−s)SJλ = kXλt k = ke(λ(λ−a)(λI−S)

−1

−(λ−a)I)t

Z

t

0

P∞

Xλt−s (I − Jλ )Sx(s) ds

k=0

(t−s)k (SJλ )k . k!

k 6 et(a−λ)

For t > 0 we have

∞ X (tλ)k kJ k k λ

k=0

k!

= Keat

by Remark 1.2. By Remark 1.2 again, limλ→+∞ λ(λI − S)−1 x = x for any x ∈ X . Hence, k(I − Jλ )xk 6 k(I − λ(λI − S)−1 )xk + |a| k(λI − S)−1 xk → 0 for any x ∈ X as λ → +∞. Therefore, if s is fixed and λ → +∞, then kXλt−s (I − Jλ )Sx(s)k 6 Kea(t−s) k(I − Jλ )Sx(s)k → 0. Moreover, there is a C > 0 such that kI − Jλ k 6 C for any λ ∈ (a, +∞). Thus, kXλt−s (I − Jλ )Sx(s)k 6 C1 kSx(s)k for the pair (λ, s) ∈ (a, +∞) × (0, T ). Since Sx(s) = x(s) ˙ − L−1 1 Qy(s) ∈ Lq (X ) ⊂ L1 (X ), it follows from Lebesgue’s theorem that vλ (t) → 0, that is, the function x(t) = limλ→+∞ xλ (t) is uniquely defined.

404


By Theorem 1.1, the equation X = X 0 ⊕ X 1 holds. Therefore, the sum x(t) = x0 (t) + x1 (t), where x0 (t) = (I − P )x(t) and x1 (t) = P x(t), is a solution (3.1), (3.2). We showed above that the function x0 (t) = Ppof thek problem −1 − k=0 G M0 (I − Q)y (k) (t) = A0 y(t) is a solution of the equation on X 0 , and so is x1 (t) = A1 y(t) + X t x0 on the subspace X 1 . Thus, x(t) = (A0 + A1 )y(t) + X t x0 . Let us prove the above bound for the solution x(t). By Lemma 3.1, we have kxkWq1 (X ) 6 kA0 ykWq1 (X ) + kA1 ykWq1 (X ) + kX t x0 kWq1 (X ) Z T 1/q Z T 1/q q q t t 6 C2 kykWqp+1 (X ) + kX x0 kX dt + kX1 SP x0 kX dt 0

Z 6 C2 kykWqp+1 (X ) +

0

T

1/q K q eaqt dt kx0 kX + kSP x0 kX ,

0

which implies the inequality (3.4). Remark 3.1. In Theorem 3.1, some standard tools for proving the existence and uniqueness of a classical solution of an equation solved with respect to the derivative (see, for example, [16]) are used in the case of a strong solution. Remark 3.2. A result similar to Theorem 3.1 was obtained in [10] under other conditions on the function y. Arguing as in the proof of Theorem 3.1, one can readily prove the existence of a unique strong solution of (3.2) satisfying the generalized Showalter condition [1] P x(0) = P x0 .

(3.7)

Theorem 3.2. Let the operator M be strongly (L, p)-radial. Then for any y ∈ ˙ X 0 there is a unique strong solution of the problem Wqp+1 (Y) and x0 ∈ dom M1 + (3.2), (3.7), and this solution is of the form x(t) = (A0 + A1 )y(t) + X t x0 and satisfies the condition kxkWq1 (X ) 6 C kP x0 kX + kSP x0 kX + kykWqp+1 (Y) . § 4. Unique solubility of optimal control problems Let X , Y and U be Hilbert spaces, B ∈ L(U; Y), L ∈ L(X ; Y), and let M ∈ Cl(X ; Y) be strongly (L, p)-radial. For x0 ∈ dom M and y ∈ H p+1 (Y), let H∂ (x0 , y) denote the set of all functions u ∈ H p+1 (U) such that p X

Gk M0−1 (I − Q)Bu(k) (0) = −(I − P )x0 −

k=0

p X

Gk M0−1 (I − Q)y (k) (0).

k=0

Consider the optimal control problem N ku − u0 k2H p+1 (U ) → inf, 2 Lx(t) ˙ = M x(t) + y(t) + Bu(t),

J1 (x, u) = J0 (x) +

(4.1) (4.2)

x(0) = x0 ,

(4.3)

u ∈ U∂ .

(4.4)


405

Here the set of admissible controls U∂ is a non-empty closed convex subset of the control space H p+1 (U), u0 ∈ H p+1 (U) and y ∈ H p+1 (Y) are given functions, x0 ∈ dom M is a given vector, and N > 0. We also assume that the functional J0 is convex, non-negative on some linear normed space Y , and lower semicontinuous with respect to weak convergence in Y , and there is a continuous embedding Zp+1 ⊂ Y of the space Zp+1 = {z ∈ H 1 (X ) : Lz˙ − M z ∈ H p+1 (Y)} in which it is natural to seek strong solutions of the equation (4.2) due to the form of this equation and the smoothness of the functions y and u. We now define an inner product in this space. Lemma 4.1. The space Zp+1 is a Hilbert space with respect to the inner product hx, ziZp+1 = hx, ziH 1 (X ) + hLx˙ − M x, Lz˙ − M ziH p+1 (Y) . Proof. Choose a Cauchy sequence {xn } in Zp+1 . Then the expression hxn − xm , xn − xm iZp+1 = kxn − xm k2H 1 (X ) + kL(x˙ n − x˙ m ) − M (xn − xm )k2H p+1 (Y) tends to zero as n, m → ∞. Therefore, the limits limn→∞ xn = x ∈ H 1 (X ) and limn→∞ (Lx˙ n − M xn ) = z exist in the space H p+1 (Y). Thus, Lx˙ n → Lx˙ as n → ∞ in L2 (Y), and hence the convergence M xn → Lx˙ − z holds in L2 (Y) as n → ∞. Therefore, the set of values t ∈ (0, T ) for which xn (t) does not converge to x(t) in X or M xn (t) does not converge to Lx(t) ˙ − z(t) in Y has measure zero. Since M is closed, we see that x(t) ∈ dom M for almost all t ∈ (0, T ) and M x = L˙x − z in L2 (Y). Hence, Lx−M ˙ x = z ∈ H p+1 (Y). This completes the proof of the lemma. Consider the operator γ0 : H 1 (X ) → X , γ0 x = x(0), which is obviously continuous. Remark 4.1. It is clear that the space Zp+1 is continuously embedded in H 1 (X ), and therefore the map γ0 : Zp+1 → X is also continuous. We refer to the set W of all pairs (x, u) ∈ Zp+1 × H p+1 (U) satisfying (4.2)–(4.4) as the set of admissible pairs of the problem (4.1)–(4.4). Since the set U∂ is convex, it follows from the form of the conditions (4.2), (4.3) that the set of admissible pairs W is also convex. To solve the problem (4.1)–(4.4) is to find the pairs (ˆ x, u ˆ) ∈ W minimizing the cost functional J1 : J1 (ˆ x, u ˆ) =

inf (x,u)∈W

J1 (x, u).

Theorem 4.1. Let U∂ ∩ H∂ (x0 , y) 6= ∅. Then there is a unique solution (ˆ x, u ˆ) ∈ Zp+1 × H p+1 (U) of the problem (4.1)–(4.4). Proof. We shall show that the conditions of Theorem 1.2.3 in [15] are satisfied. To this end, we set U = H p+1 (U), V = H l (Y) × X , l ∈ {0, . . . , p + 1}, and Y1 = Zp+1 . Then the linear operator L : Y1 × U → V acting by the rule L(x, u) = (Lx˙ − M x − Bu, γ0 x) is continuous. Indeed, in view of Remark 4.1 we have kLx˙ − M x − Buk2H l (Y) + kγ0 xk2X 6 2kLx˙ − M xk2H p+1 (Y) + 2kBuk2H p+1 (Y) + C1 kxk2Zp+1 6 C(kxk2Zp+1 + kuk2H p+1 (U ) ) = Ck(x, u)k2Zp+1 ×H p+1 (U ) .

406


By Theorem 3.1, the condition U∂ ∩ H∂ (x0 , y) 6= ∅ ensures that the set W is non-empty. The lower semicontinuity of the functional J1 with respect to weak convergence in the space Y × U follows from the semicontinuity of J0 in Y . We now use the bound (3.4) to prove that this functional is coercive on the admissible pairs in Y1 × U . Indeed, since J0 is non-negative, we have kxk2Zp+1 + kuk2H p+1 (U ) = kxk2H 1 (X ) + kBu + yk2H p+1 (Y) + kuk2H p+1 (U ) 6 C2 kx0 k2X + kSP x0 k2X + kyk2H p+1 (Y) + kuk2H p+1 (U ) 6 C3 J1 (x, u) + C4 . Suppose that the problem (4.1)–(4.4) has two solutions, (ˆ x1 , u ˆ1 ) and (ˆ x2 , u ˆ2 ). This means that u ˆ1 6= u ˆ2 by the uniqueness of solution of the problem (4.2), (4.3). x2 u u2 Since W is convex, it follows that the pair ( xˆ1 +ˆ , ˆ1 +ˆ ) is admissible. Then, in 2 2 2 view of the fact that the functional ku − u0 kH p+1 (U ) on H p+1 (U) is strictly convex and J0 is convex, we obtain ˆ1 + u ˆ2 1 x ˆ1 + x ˆ2 u , x1 , u ˆ1 ) + J1 (ˆ x2 , u ˆ2 ) . J1 < J1 (ˆ 2 2 2 This contradicts the assumption that J1 attains its minimum at the pairs (ˆ x1 , u ˆ1 ) and (ˆ x2 , u ˆ2 ), and thus completes the proof of the theorem. Consider the following special cases of the functional J1 : 1 N kx − wk2L2 (X ) + ku − u0 k2H p+1 (U ) → inf, 2 2 1 N 2 J3 (x, u) = kx(T ) − wkX + ku − u0 k2H p+1 (U ) → inf, 2 2 J2 (x, u) =

(4.5) (4.6)

where w ∈ L2 (X ) for (4.5) and w ∈ X for (4.6). Corollary 4.1. Let U∂ ∩ H∂ (x0 , y) 6= ∅. Then there is a unique solution (ˆ x, u ˆ) ∈ Zp+1 × H p+1 (U) of the problem (4.2)–(4.5). Proof. We write U = H p+1 (U), V = H l (Y) × X , l ∈ {0, 1, . . . , p + 1}, Y = L2 (X ), and Y1 = Zp+1 . It is clear that the embedding Y1 ⊂ Y is continuous. The strict convexity and the continuity of the functional J0 = 12 kx − wk2L2 (X ) on the space Y are obvious, and therefore J0 is lower semicontinuous with respect to weak convergence. Reference to Theorem 4.1 completes the proof. Corollary 4.2. Let U∂ ∩ H∂ (x0 , y) 6= ∅. Then there is a unique solution (ˆ x, u ˆ) ∈ Zp+1 × H p+1 (U) of the problem (4.2)–(4.4), (4.6). Proof. We shall prove the continuity of the functional J0 (x) = the space Y = H 1 (X ). If {xn } converges to x in H 1 (X ), then

1 2 kx(T )

− wk2X on

|J0 (xn ) − J0 (x)| 1 = kxn (T ) − wkX − kx(T ) − wkX (kxn (T ) − wkX + kx(T ) − wkX ) 2 6 C1 kxn (T ) − x(T )kX 6 Ckxn − xkH 1 (X ) → 0, n → ∞.


407

It can readily be proved that the functional J0 (x) = kx(T ) − wk2X with a function w ∈ X is convex on the space H 1 (X ). This implies that it is lower semicontinuous in the sense of weak convergence in H 1 (X ). Therefore by Theorem 4.1, there is a solution, and this solution is unique. Applying the line of reasoning used in the proof of Theorem 4.1, we obtain the following assertion. Theorem 4.2. There is a unique solution (ˆ x, u ˆ) ∈ Zp+1 × H p+1 (U) of the problem (3.7), (4.1), (4.2), (4.4). Remark 4.2. In contrast to Theorem 4.1, the fact that the set U∂ ⊂ U = H p+1 (U) is non-empty already ensures the validity of the non-triviality condition because of Theorem 3.2. Remark 4.3. Theorem 4.2 implies the existence of a unique solution (ˆ x, u ˆ) ∈ Zp+1 × H p+1 (U) of the problem (3.7), (4.2), (4.4), (4.5) and also of the problem (3.7), (4.2), (4.4), (4.6). § 5. Reduction of the linearized Navier–Stokes system Consider the initial-boundary-value problem for the linearized Navier–Stokes system w(x, 0) = w0 (x), w(x, t) = 0,

(x, t) ∈ ∂Ω × (0, T ),

wt (x, t) = ν∆w(x, t) − q(x, t) + y(x, t), ∇ · w = 0,

x ∈ Ω,

q(x, 0) = q0 (x),

(5.1) (5.2)

(x, t) ∈ Ω × (0, T ),

(x, t) ∈ Ω × (0, T ).

(5.3) (5.4)

Here Ω ⊂ Rn is a bounded domain with boundary ∂Ω of class C ∞ , ν > 0, and q = ∇p is the gradient of the pressure. The desired vector functions are w(x, t) and q(x, t). ˚ 1 = (H ˚1 (Ω))n , L2 = (L2 (Ω))n and L = We write H2 = (H 2 (Ω))n , H ∞ n {w ∈ (C0 (Ω)) : ∇·w = 0}, and let Hσ denote the closure of the vector subspace L with respect to the norm of L2 . We have L2 = Hσ ⊕Hπ , where Hπ is the orthogonal complement to Hσ . Let Π : L2 → Hπ denote the orthogonal projection associated ˚1 ∩ H2 ⊂ L2 is with this decomposition. The restriction of Π to the subspace H 1 2 1 2 1 ˚ ˚ ˚ a continuous operator Π1: H ∩ H → H ∩ H . Therefore, H ∩ H2 = H2σ ⊕ H2π , where H2σ = ker Π1 and H2π = im Π1 . The formula A = diag{∆, . . . , ∆} obviously defines a continuous linear operator A : H2σ ⊕ H2π → L2 with discrete spectrum σ(A) (of finite multiplicity) concentrated at −∞ only. We write Σ = I − Π, Aσ = A H2 and Aπ = A H2 . σ

Lemma 5.1. The two operators are Aσ :

H2σ

π

→ Hσ and Aπ : H2π → Hπ .

Proof. We claim that ΠAΣ ≡ O. Let v ∈ L. Then ∇ · Av = ∆(∇ · v) = 0. This equation can be extended by continuity to the space H2σ . Hence, Aσ : H2σ → Hσ .

408


We now claim that ΣAΠ ≡ O. Take a v ∈ H2σ and a u ∈ H2π , and let h · , · i be the inner product on L2 . We have hv, Aui = hAv, ui = 0 because A is self-adjoint and Av ∈ Hσ (as proved above), and the spaces Hσ and Hπ are orthogonal to each other. Since H2σ is dense in Hσ , we have Au ⊥ Hσ , and therefore Aπ : H2π → Hπ . This completes the proof of the lemma. ˚1 ∩ H2 satisfies the condition (5.4) if and only if z ∈ H2 or, A function z ∈ H σ equivalently, Π1 z = 0. Let us replace the equation (5.4) by the corresponding inhomogeneous equation (x, t) ∈ Ω × (0, T ).

Π1 w(x, t) = yq (x, t),

(5.5)

We write X = Hσ × Hπ × Hq , Y = Hσ × Hπ × H2q , Hq = Hπ , H2q = H2π , and 

1 L = 0 0

0 1 0

 0 0 , 0

 νAσ M = 0 0

0 νAπ −1

 0 −1 . 0

Then L : X → Y is a continuous linear operator, and the closed linear operator M : dom M → Y has domain dom M = H2σ × H2π × Hq , which is dense in X . Remark 5.1. A similar reduction of the problem (5.1)–(5.4) to the problem (3.2), (3.7) was used in [17]. Theorem 5.1. The operator M is strongly (L, 1)-radial. Proof. We have  µ − νAσ 0 µL − M =  0

0 µ − νAπ 1

 0 1 . 0

It is clear that the spectrum of σ(Aσ ) is discrete and of finite multiplicity, and it is concentrated at −∞ only. Let {ϕk } be a family of orthonormal (with respect to Hσ ) eigenfunctions of the operator Aσ indexed in non-increasing of the eigenvalues Porder ∞ {λk } counted according to multiplicity. Then µ − νAσ = k=1 (µ − νλk )h · , ϕk iϕk , where h · , · i is the inner product in Hσ . Let µ 6= νλk . Then there is a continuous operator ∞ X h · , ϕk iϕk −1 Aσµ ≡ (µ − νAσ )−1 = : Hσ → Hσ . µ − νλk k=1

If µ 6= νλk , then the L-resolvent of M can be constructed:

(µL − M )−1

 −1 Aσµ = 0 0

 0 0 . 0 1 1 −µ + νAπ


409

This gives  −1 −1  Aσµ0 Aσµ1 0 0 L (M ) =  R(µ,1) 0 0 0 , 0 0 0  −1 −1  Aσµ0 Aσµ1 0 0  LL 0 0 0 , (µ,1) (M ) = 0 0 0   −1 −1 −1 Aσµ0 Aσµ1 Aσλ 0 0 L R(µ,1) (M )(λL − M )−1 =  0 0 0 , 0 0 0   −1 −1 −1 νAσµ0 Aσµ1 Aσλ Afσ 0 0  M (λL − M )−1 LL 0 0 0 , (µ,1) (M )f = 0 0 0

 −1 Aσµ RµL (M ) =  0 0  −1 Aσµ  0 LL µ (M ) = 0

 0 0 0 0 , 1 0  0 0 0 1 , 0 0

˚ = H2σ × Hπ × H2q . Thus, for µ0 > νλ1 and µ1 > νλ1 we have where f ∈ F

∞

2

X

vk ϕk

(µ0 − νλk )(µ1 − νλk )

=

L2 (Ω)

k=1

∞ X k=1

6

|vk |2 (µ0 − νλk )2 (µ1 − νλk )2 kvk2L2 (Ω)

(µ0 − νλ1 )2 (µ1 − νλ1 )2

,

and therefore 1 , (µ0 − νλ1 )(µ1 − νλ1 ) 1 , 6 (µ0 − νλ1 )(µ1 − νλ1 )(λ − νλ1 )

L max kR(µ,1) (M )kL(X ) , kLL (µ,1) (M )kL(Y) 6

L

−1

R (µ,1) (M )(λL − M ) L(Y;X )

M (λL − M )−1 LL (M )f 6 (µ,1) Y

νkAfσ kL2 . (µ0 − νλ1 )(µ1 − νλ1 )(λ − νλ1 )

This completes the proof of the theorem. The operators of the solution semigroup {X t ∈ L(X ) : t ∈ R+ } of the homogeneous system (5.2), (5.3), (5.5) are of the form P∞ Xt =   1 P = X 0 = 0 0

0 0 0

 0 0 , 0

k=1

eλk t h · , ϕk iϕk 0 0

 0 0 0 0 , 0 0

 1 2 0 Q = s- lim (µLL µ (M )) = µ→+∞ 0

0 0 0

 0 0 . 0

410


For an arbitrary fixed t ∈ (0, T ) we write Σw(x, t) = wσ (x, t), Πw(x, t) = wπ (x, t), Σy(x, t) = yσ (x, t), and Πy(x, t) = yπ (x, t). The set of admissible initial values for this system is of the form My = (w0 , q0 ) ∈ H2 × Hπ : Πw0 (x) = yq (x, 0), ∂yq (x, 0) . q0 (x) = yπ (x, 0) + νAyq (x, 0) − ∂t Theorem 5.2. For any (w0 , q0 ) ∈ My , yσ ∈ L2 (Hσ ), yπ ∈ H 1 (Hπ ) and yq ∈ H 2 (H2q ), there is a unique strong solution of the problem (5.1)–(5.3), (5.5), which is of the form Z tX ∞ ∞ X wσ (x, t) = eλk t hw0 , ϕk iϕk (x) + eλk (t−s) hyσ (ξ, s), ϕk (ξ)iϕk (x) ds, 0 k=1

k=1

wπ (x, t) = yq (x, t), q(x, t) = yπ (x, t) + νAyq (x, t) −

(5.6) ∂yq (x, t) . ∂t

(5.7)

Proof. For v ∈ dom M we have 

   0 0 M (I − P )v = νAπ vπ − vq  = (I − Q)f = fπ  . −vπ fq Hence, vπ = −fq and vq = −fπ − νAπ fq , and therefore M0−1 =

0 −1 , −1 −νAπ

G=

0 −1

0 , 0

GM0−1 =

0 0

0 . 1

Thus, in view of Theorem 3.1, we obtain the equations (5.6) and (5.7). Because of the form of the solution semigroup and of the projection Q, we have Z

t

wσ (x, t) = X w0 +

t

X t−s L−1 1 Qy(s) ds

0

=

∞ X k=1

λk t

e

hw0 , ϕk iϕk (x) +

Z tX ∞

eλk (t−s) hyσ (ξ, s), ϕk (ξ)iϕk (x) ds.

0 k=1

Remark 5.2. In comparison with the abstract theorem (Theorem 3.1), the conditions imposed on y in Theorem 5.2 are weaker. This is because of specific features of the operators of the problem in question. We note that, as in the monograph [15], no additional smoothness is needed for the right-hand side yσ of the Navier–Stokes system in the case when yπ ≡ yq ≡ 0.


411

§ 6. An optimal control problem for the linearized Navier–Stokes system Consider the problem 1 1 kw − v0 k2L2 (L2 ) + kq − r0 k2L2 (L2 ) 2 2 N N 2 + ku − u0 kH 2 (L2 ) + kuq − u1 k2H 2 (H2 ) → inf, 2 2 wt (x, t) = ν∆w(x, t) − q(x, t) + u(x, t), (x, t) ∈ Ω × (0, T ),

J2 (w, q, u, uq ) =

Π1 w = uq (x, t), w(x, 0) = w0 (x),

(x, t) ∈ Ω × (0, T ), q(x, 0) = q0 (x),

w(x, t) = 0,

x ∈ Ω,

(x, t) ∈ ∂Ω × (0, T ), (u, uq ) ∈ U∂ ,

(6.1) (6.2) (6.3) (6.4) (6.5) (6.6)

where v0 ∈ L2 (L2 ), r0 ∈ L2 (Hq ), u0 ∈ H 2 (L2 ), u1 ∈ H 2 (H2q ), the constants N and ν are positive, w0 ∈ L2 , q0 ∈ Hπ , and U∂ is a non-empty closed convex subset of the space H 2 (L2 ) × H 2 (H2q ). To reduce the problem (6.1)–(6.6) to the problem (4.2)–(4.5), we set X = Hσ × Hπ × Hq , Y = U = Hσ × Hπ × H2q , y ≡ 0, and B = I ∈ L(Y). We also define the space Z2 = (z, k) ∈ H 1 (L2 ) × H 1 (Hq ) : zt − ν∆z + k ∈ H 2 (L2 ), Π1 z ∈ H 2 (H2q ) . The set H∂ (x0 , y) = H∂ ((w0 , q0 ), 0) is defined as the set of functions (u, uq ) ∈ H 2 (L2 ) × H 2 (H2q ) such that ∂uq (x, 0) Πw0 (x) = uq (x, 0), q0 (x) = uπ (x, 0) + νAuq (x, 0) − . ∂t By Corollary 4.1, we have the following theorem. Theorem 6.1. Let U∂ ∩ H∂ ((w0 , q0 ), 0) 6= ∅. Then there is a unique solution (w, b qˆ, u ˆ, u ˆq ) ∈ Z2 × H 2 (L2 ) × H 2 (H2q ) of the problem (6.1)–(6.6). Remark 6.1. It can be seen from Theorem 5.2 and Remark 5.2 that the specific nature of the Navier–Stokes system enables one to study the problem (6.1)–(6.6) in a space of controls of lower smoothness, namely, L2 (Hσ ) × H 1 (Hπ ) × H 2 (H2q ). Bibliography [1] R. E. Showalter, “Nonlinear degenerate evolution equations and partial differential equations of mixed type”, SIAM J. Math. Anal. 6:1 (1975), 25–42. [2] G. A. Sviridyuk and V. E. Fedorov, “On the identities of analytic semigroups of operators with kernels”, Sibirsk. Mat. Zh. 39:3 (1998), 604–616; English transl., Siberian Math. J. 39:3 (1998), 522–533. [3] V. E. Fedorov, “Degenerate strongly continuous semigroups of operators”, Algebra i Analiz 12:3 (2000), 173–200; English transl., St. Petersburg Math. J. 12:3 (2001), 471–489.

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[4] G. A. Sviridyuk and V. E. Fedorov, Linear Sobolev type equations and degenerate semigroups of operators, Inverse Ill-posed Probl. Ser., VSP, Utrecht 2003. [5] P. C. M¨ uller, “Linear control design of linear descriptor systems”, 14th Triennial World Congress (Beijing), 1999, pp. 31–36. [6] V. F. Chistyakov and A. A. Shcheglova, Selected chapters of the theory of algebro-differential systems, Nauka, Novosibirsk 2003 (Russian). [7] G. A. Sviridyuk and A. A. Efremov, “Optimal control of Sobolev-type linear equations with relatively p-sectorial operators”, Differ. Uravn. 31:11 (1995), 1912–1919; English transl., Differential Equations 31:11 (1995), 1882–1890. [8] G. A. Sviridyuk and A. A. Efremov, “Optimal control problem for a class of linear equations of Sobolev type”, Izv. Vyssh. Uchebn. Zaved. Mat., 1996, no. 12, 75–83; English transl., Russian Math. (Iz. VUZ ) 40:12 (1996), 60–71. [9] G. A. Sviridyuk and A. A. Efremov, “Optimal control of a class of linear degenerate equations”, Dokl. Akad. Nauk 364:3 (1999), 323–325; English transl., Dokl. Math. 59:1 (1999), 157–159. [10] V. E. Fedorov and M. V. Plekhanova, “Optimal control of Sobolev type linear equations”, Differ. Uravn. 40:11 (2004), 1548–1556; English transl., Differ. Equ. 40:11 (2004), 1627–1637. [11] V. E. Fedorov and M. V. Plekhanova, “Investigation of an optimal control problem”, Vestnik Chelyabinsk. Gos. Ped. Univ. Ser. 4, Estestv. Nauki, 2005, no. 6, 23–31 (Russian). [12] V. E. Fedorov and M. V. Plekhanova, “Weak solutions and the quadratic regulator problem for a degenerate differential equation in a Hilbert space”, Vychislitel’nye Tekhnologii 9:2 (2004), 92–102 (Russian). [13] M. V. Plekhanova and V. E. Fedorov, “An optimal control problem for a class of degenerate equations”, Izv. Akad. Nauk Teor. Sist. Upr., 2004, no. 5, 40–44; English transl., J. Comput. Systems Sci. Internat. 43:5, (2004), 698–702. [14] M. V. Plekhanova and V. E. Fedorov, “Start-up control problems for linear equations of Sobolev type”, Vestnik YuUrGU, Ser. Mat., Fiz., Khim., 2005, no. 6, 43–49 (Russian). [15] A. V. Fursikov, Optimal control of distributed systems. Theory and applications, Nauchnaya Kniga, Novosibirsk 1999; English transl., Transl. Math. Monogr., vol. 187, Amer. Math. Soc., Providence, RI 2000. [16] S. Mizohata, The theory of partial differential equations, Cambridge Univ. Press, London 1973; Russian transl., Mir, Moscow 1977. [17] G. A. Sviridyuk and G. A. Kuznetsov, “Relatively strongly p-sectorial linear operators”, Dokl. Akad. Nauk 365:6 (1999), 736–738; English transl., Dokl. Math. 59:2 (1999), 298–300. M. V. Plekhanova Department of Mathematics, Chelyabinsk State Pedagogical University V. E. Fedorov Department of Mathematics, Chelyabinsk State University E-mail: [email protected]

Received 10/MAY/07 20/MAY/08 Translated by A. I. SHTERN