Acta Mathematica Sinica, English Series July, 2001, Vol.17, No.3, pp. 527–540

An Infinite Horizon Linear Quadratic Problem with Unbounded Controls in Hilbert Space

Han Zhong WU

Xun Jing LI

Department of Mathematics, Fudan University, Shanghai 200433, P. R. China E-mail: [email protected] [email protected]

Abstract An infinite horizon linear quadratic optimal control problem for analytic semigroup with unbounded control in Hilbert space is considered. The state weight operator is allowed to be indefinite while the control weight operator is coercive. Under the exponential stabilization condition, it is proved that any optimal control and its optimal trajectory are continuous. The positive real lemma as a necessary and sufficient condition for the unique solvability of this problem is established. The closed-loop synthesis of optimal control is given via the solution to the algebraic Riccati equation. Keywords Infinite horizon LQ problem, Unbounded control, Two-point boundary value problem, Algebraic Riccati equation, Frequency characteristic 2000 MR Subject Classification 49N10, 93C25

1

Introduction and Main Results

Linear quadratic optimal control (LQ, for short) problems with bounded control were studied by Lions [1], Curtain-Pritchard [2] and Balakrishnan [3] in the 1970s. Since the 1980s, LasieckaTriggiani ([4] and the papers quoted therein) used the Riccati equation to study LQ problems for parabolic, hyperbolic, and other types of partial differential equations with boundary controls. See Flandoli [5, 6], Da Prato-Ichikawa [7] as well as Bensoussan-Da Prato-Delfour-Mitter [8] also, for a direct study of the Riccati equation. On the other hand, Yakubovich and his colleagues ([9, 10], and the references cited therein) established the frequency theorem, or the so called positive real lemma, as a necessary and sufficient frequency condition for the unique solvability of finite-dimensional LQ problems. Received January 1, 2001, Accepted March 5, 2001 This work is partially supported by the National Key Project of China, the National Nature Science Foundation of China No. 19901030, NSF of the Chinese State Education Ministry and Lab. of Math. for Nonlinear Sciences at Fudan University

H. Z. Wu and X. J. Li

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In LQ differential games, the state weight operators of the cost functional appear to be indefinite in general. Molinari [11] pointed out that the state weight operators could be negative in certain LQ problems related to network analysis. The LQ problems with indefinite cost functional also relate to the analysis of the stability of feedback systems and other problems. For the LQ problems in Hilbert spaces with indefinite cost functional and bounded control, You [12] found a relation between the optimal feedback control and a Fredholm integral equation. Schumitzky [13] and Chen [14] directly discussed an equivalence between the Riccati equation and a Fredholm integral equation. Li-Yong [15] and Wu-Li [16] studied finite horizon LQ problems with unbounded control, which include the Neumann, Dirichlet boundary control and pointwise control of parabolic systems. Lasiecka-Pandolfi-Triggiani [17], Bucci-Pandolfi [18] and others deal with the pathological singular case. On the other hand, the first extension of Yakubovich’s results to distributed systems was given by Louis-Wexler [19]; See Pandolfi [20] for a survey of this approach. Recently, Pandolfi [21, 22] established a frequency condition for the existence of a solution to dissipation inequality, associated with the singular LQ problem for parabolic, hyperbolic equations with boundary controls. This paper will study the infinite horizon LQ problem for analytic semigroup with unbounded control and indefinite cost functional (the control weight operator is coercive) in Hilbert space, including the case for the Neumann, Dirichlet boundary controls and pointwise controls of parabolic systems. Throughout this paper, we assume: (H1 ) Let both X and U be Hilbert spaces, B ∈ L(U, X), −A generate an analytic semigroup e−A· on X with 0 ∈ ρ(A), and the fractional power Aγ be well defined for all γ ∈ R. Then, for any γ > 0, there exists a cγ > 0 such that (see [23])
∀ t > 0; Aγ e−At  ≤ cγ t−γ , (1.1) −At γ γ (I − e )x ≤ cγ t A x, ∀ x ∈ D(Aγ ), t > 0. Denote the adjoint of A by A∗ . Then the above properties of A also hold for A∗ . We assume that (1.1) holds for A∗ and (A∗ )γ with the same constant cγ . Consider the following abstract controlled system:  t y(t) = e−At x + Aα e−A(t−s) Bu(s) ds, (1.2) 0

where 0 ≤ α < 1 is a constant, x ∈ X and u(·) ∈ L2 (0, ∞; U ). For the case α = 0, e−At is allowed to be just a C0 -semigroup, not necessarily analytic. The abstract setting (1.2) has many concrete examples. In fact, it is well known (see [4, 15], etc.) that in the case of a secondorder parabolic equation defined on a bounded domain Ω ⊂ Rn , the relevant values of the constant α are as follows: α = 34 + ε, ∀ε > 0, for Dirichlet boundary control with U = L2 (∂Ω), X = L2 (Ω); α = 14 + ε, ∀ε > 0, for Neumann boundary control with U = L2 (∂Ω), X = L2 (Ω); and n4 < α < 1, for pointwise control with U = H −2α (Ω), X = L2 (Ω). Denote U = L2 (0, ∞; U ) and X = L2 (0, ∞; X). A control u(·) ∈ U is said to be admissible at an initial state x ∈ X if the solution y(·) to (1.2) given by u(·) satisfies y(·) ∈ X . Throughout this paper, we assume:

Infinite Horizon Linear Quadratic Problem

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(H2 ) The set of admissible controls at any initial state x ∈ X is nonempty. Assumption (H2 ) is called the stability condition for the finite-dimensional controlled system in [9]. It was proved in [6] that (H2 ) is also equivalent to the exponential stabilizability of the controlled system (1.2). Therefore, without loss of generality, we can assume: (H3 )

e−At is exponentially stable, i.e., there exist C > 0 and λ > 0 such that e−At  ≤ Ce−ωt ,

∀ t ≥ 0.

(1.3)

With the dynamics (1.2), we associate the following quadratic cost functional:  ∞ J(x; u(·)) = {Qy(t), y(t) + 2ReSy(t), u(t) + Ru(t), u(t)} dt,

(1.4)

0

where y(·)
y(·; x, u(·)) is the mild solution of (1.2), and Q, S and R satisfy the following assumption: (H4 ) Q ∈ L(X) is self-adjoint but is allowed to be indefinite; S ∈ L(X, U ), R ∈ L(U ) and R  0, i.e., R = R∗ and there exists δ > 0 such that Ru, u ≥ δu, u, ∀u ∈ U . Our linear quadratic optimal control problem can be stated as follows: Problem LQ

For given x ∈ X, find u(·) ∈ U so that the cost functional (1.4) is minimized.

If it is true that inf u(·)∈U J(x; u(·)) > −∞, then Problem LQ is said to be well-posed at x ∈ X. If there exists a u ¯(·) ∈ U such that J(x; u ¯(·)) = inf J(u(·)) > −∞,

(1.5)

u(·)∈U

we say that Problem LQ is solvable at x ∈ X. In this case, we call the control u ¯(·) an optimal control of Problem LQ, also the corresponding trajectory y¯(·) and the pair (¯ u(·), y¯(·)) are referred to as an optimal trajectory and an optimal pair, respectively. In the following, we use the notations: define a linear operator L as  t [Lu(·)](t)
Aα e−A(t−s) Bu(s) ds, t ≥ 0, ∀u(·) ∈ U. (1.6) 0

Then L ∈ L(U; X ), and the adjoint operator L∗ of L is given by  ∞ ∗ [L∗ y(·)](t) = B ∗ (A∗ )α e−A (s−t) y(s) ds, ∀ y(·) ∈ X .

(1.7)

t

Here, L∗ ∈ L(X ; U). In what follows, we let  ∗ ∗ ∗   Φ
R + L QL + SL + L S ∈ L(U);   Θ
L∗ Qe−A· + Se−A· ∈ L(X; U);  ∞   ∗  Γ
e−A s Qe−As ds ∈ L(X).

(1.8)

0

Therefore J(x; u(·)) = Φu(·), u(·) + 2ReΘx, u(·) + Γx, x,

∀ u(·) ∈ U.

(1.9)

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Define  Φ(ω)
R + B ∗ (A∗ )α (−iω + A∗ )−1 QAα (iω + A)−1 B + SAα (iω + A)−1 B + B ∗ (A∗ )α (−iω + A∗ )−1 S ∗ ,

∀ ω ∈ R,

(1.10)

which is called the frequency characteristic of Problem LQ. It is known that the state trajectory of (1.2) may be discontinuous in general. This paper proves that any optimal control and its optimal state trajectory of Problem LQ must be continuous. We now state our main results as follows: Main Theorem Assume that (H1 ) (H3 ) (H4 ) hold. Then the following statements are equivalent: (I) Problem LQ is uniquely solvable at any x ∈ X with the optimal pair (¯ u(·), y¯(·)) ∈ U × X ; (II) The algebraic Riccati equation (ARE for short) − Ax, P x − P x, Ax + Qx, x − [S + B ∗ (A∗ )α P ]∗ R−1 [S + B ∗ (A∗ )α P ]x, x = 0,

∀ x ∈ D(A),

admits a unique solution P satisfying (A∗ )α P ∈ L(X) and

−Aα {A1−α + BR−1 [S + B ∗ (A∗ )α P ]}, A  = {x ∈ D(A1−α )|{A1−α + BR−1 [S + B ∗ (A∗ )α P ]}x ∈ D(Aα )}, D(A)

(1.11)

(1.12)

generates an analytic semigroup, which is exponentially stable. In this case, the synthesis for optimal control of Problem LQ is given by u ¯(t) = −R−1 [S + B ∗ (A∗ )α P ]¯ y(t),

t ≥ 0;

(1.13)

(III) There exists some constant σ > 0 such that Φu(·), u(·) ≥ σu(·), u(·),

∀ u(·) ∈ U;

(1.14)

∀ u ∈ U, ω ∈ R.

(1.15)

(IV) There exists some constant σ > 0 such that  Φ(ω)u, u ≥ σu, u,

In Section 6, a simple example is given as an application of our Main Theorem, in which the operators in (1.4) take the form R = I, S = 0 and Q = −ρI with ρ ∈ (0, ∞). If ρ ∈ (0, 3), then Problem LQ is uniquely solvable at any initial state; however, if ρ ∈ (3, ∞), then Problem LQ is not even well-posed.

2

LQ Problems and Two-Point Boundary Value Problems

In general, the state trajectory of (1.2) is not necessarily continuous if α ≥ 12 . Applying the inequalities established in Appendix, we will prove (Theorem 2.4) that any solution to the twopoint boundary value problem is uniformly continuous, namely, any optimal control and its optimal trajectory must be uniformly continuous.

Infinite Horizon Linear Quadratic Problem

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Theorem 2.1 Assume that (H1 ) (H3 ) (H4 ) hold. Then u ¯(·) ∈ U is an optimal control for Problem LQ at x ∈ X if and only if

Proof

Φ ≥ 0, i.e., Φu(·), u(·) ≥ 0, ∀ u(·) ∈ U;

(2.1)

Φ¯ u(·) + Θx = 0.

(2.2)

The proof is similar to [15, pp. 371–372].

Theorem 2.2 Assume that (H1 ) (H3 ) (H4 ) hold. Then Problem LQ is solvable at x ∈ X if and only if (2.1) holds and there exists a pair (y(·), ψ(·)) ∈ X × U satisfying the following two-point boundary value problem:   t  −At  x+ Aα e−A(t−s) {BR−1 ψ(s) − BR−1 Sy(s)} ds, y(t) = e    0   ∞

B ∗ (A∗ )α e−A ψ(t) = −    t    + S ∗ R−1 ψ(σ)} dσ,

∗

(σ−t)

{(Q − S ∗ R−1 S)y(σ)

(2.3)

t ≥ 0.

In this case, u ¯(t) = R−1 {ψ(t) − Sy(t)},

t ≥ 0,

(2.4)

gives an optimal control and y(·) is the corresponding optimal trajectory. Proof

It can follow from Theorem 2.1, similarly to [15, pp. 381–382] or [16].

Theorem 2.3 (2.3). Then

Assume that (H1 ) (H3 ) (H4 ) hold and (y(·), ψ(·)) ∈ X × U is a solution of  y(t) + ψ(t) ≤ k(x + y(·)L2 + ψ(·)L2 );       ∞ e−η|t−τ | (y(τ ) + ψ(τ )) dτ |t − τ |γ  0     ≤ l(γ, η)(x + y(·)L2 + ψ(·)L2 ),

(2.5)

for 0 ≤ t < ∞, where γ ∈ [α, 1), η ∈ (0, ∞), and k, l(γ, η) > 0 are constants independent of t ∈ [0, ∞). Proof Denote S(t)
y(t) + ψ(t) for 0 ≤ t < ∞. By (1.1) and (1.3), it follows from (2.3) that for some absolute constant c > 0,   −λt S(t) ≤ c e x +

0

∞

|t−τ | e−λ 2 S(τ ) dτ , |t − τ |α

for 0 ≤ t < ∞.

(2.6)

By Theorem A.2 in Appendix, we obtain (2.5) from (a.5) and (a.6). Theorem 2.4 Assume that (H1 ) (H3 ) (H4 ) hold and (y(·), ψ(·)) ∈ X × U is a solution of (2.3). Then both y(·) and ψ(·) are uniformly continuous on [0, ∞).

H. Z. Wu and X. J. Li

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Proof Denote W (t)
(Q − S ∗ R−1 S)y(t) + S ∗ R−1 ψ(t) for 0 ≤ t < ∞. Take 0 ≤ t ≤ t˜ < ∞ and consider the following:   t˜ ∗ ∗ ˜ ψ(t) − ψ(t) ≤ B  (A∗ )α e−A (σ−t) W (σ) dσ t  ∞ ∗ α −A∗ (σ−t˜) −A∗ (t˜−t) + (A ) e (I − e )W (σ) dσ

(2.7)

t˜


B ∗ (I1 + I2 ). Let ε ∈ (0, 1 − α) be fixed. Then it follows from (1.1) and the first inequality of (2.5) that I1 ≤

ck ˜ |t − t|1−α [x + y(·)L2 + ψ(·)L2 ]. 1−α

Substituting γ
α + ε and η
 I2 =

t˜

∞

λ 2

(2.8)

into the second inequality of (2.5) yields

(A∗ )γ e−A

˜ ∗ σ−t 2

e−A

˜ ∗ σ−t 2

[(I − e−A

∗

(t˜−t)

)(A∗ )−ε ]W (σ) dσ

(2.9)

≤ c|t˜ − t|ε [x + y(·)L2 + ψ(·)L2 ]. Using (2.8) and (2.9) in (2.7), we obtain that ψ(·) is uniformly continuous on [0, ∞). Similarly, we can prove the uniform continuity of y(·). Theorems 2.2 and 2.4 imply that Theorem 2.5 Assume that (H1 ) (H3 ) (H4 ) hold and Problem LQ is solvable at x ∈ X with an optimal pair (¯ u(·), y¯(·)) ∈ U × X . Then both the optimal control u ¯(·) and the corresponding optimal trajectory y¯(·) are uniformly continuous on [0, ∞).

3

Necessary Conditions of Solvability of Problem LQ

In the following, we use the notation CU
C([0, ∞); U )



U.

Arguing by Lemma 3.3 in [18], we have Lemma 3.1 Assume that (H1 ) (H3 ) (H4 ) hold and Problem LQ is uniquely solvable at any x ∈ X. Then Φ admits an inverse operator Φ−1 from R(Φ) ⊂ U to U and Φ−1 Θ ∈ L(X; U), where R(Φ) is the range of Φ. If the assumption of Lemma 3.1 holds, then (2.2) in Theorem 2.1 and (1.9) yield that J(x; u ¯(·)) = (Γ − Θ∗ Φ−1 Θ)x, x.

(3.1)

P
Γ − Θ∗ Φ−1 Θ.

(3.2)

We call P the value operator of Problem LQ. Theorem 3.2 If the assumption of Lemma 3.1 holds, then the value operator P of Problem LQ satisfies (A∗ )β P ∈ L(X) for any β ∈ [0, 1).

Infinite Horizon Linear Quadratic Problem

Proof

533

By (1.1) and (1.3), it follows that  ∞     ∗ β ∗ β −A∗ t −At   (A ) Γ =  (A ) e Qe dt ≤ cβ CQ 0

For any u(·) ∈ CU, it is true that  ∞      ∗ β −A∗ t ∗   (A ) e S u(t) dt  = 0

∞

0

∞ 0

∗ β −A∗ 2t −A∗ 2t

(A ) e

e

e−λt dt < ∞. tβ

  S u(t) dt  < ∞; ∗

 t  1−β   β −A(t−s)  ≤ cβ B t [Lu(·)](t) =  max u(s). A e Bu(s) ds   1 − β s∈[0,t] 0

(3.3)

(3.4)

(3.5)

Therefore,    

0

∞

∗ β −A∗ t

(A ) e

    Q[Lu(·)](t) dt ≤ c

0

∞

λt

e− 2 [Lu(·)](t) dt < ∞. tβ

(3.6)

Arguing by Theorem 2.5, we obtain Φ−1 Θx ∈ CU for any x ∈ X. Thus, combining (3.3), (3.4) and (3.6) yields P x ∈ D((A∗ )β ) for any x ∈ X. The Closed Graph Theorem implies that (A∗ )β P is bounded.

4

Algebraic Riccati Equation

If Problem LQ is uniquely solvable at any x ∈ X with the optimal pair (¯ u(·), y¯(·)), then it follows from Lemma 3.1 and Theorem 3.2 that J(x; u ¯(·)) = P x, x
V (x), which is called the value function of Problem LQ. By Theorem 2.5, we see that the optimal control and optimal trajectory are always continuous. Therefore, the following result follows from Theorem 3.2: Lemma 4.1 Assume that (H1 ) (H3 ) (H4 ) hold and Problem LQ is uniquely solvable at any x ∈ X. Then the value function V (x) of Problem LQ satisfies V (x) =

inf

 V (y(t; x, u(·))) +

t

[Qy(s; x, u(·)), y(s; x, u(·))

+ 2ReSy(s; x, u(·)), u(s) + Ru(s), u(s)] ds .

u(·)∈CU

0

(4.1)

Theorem 4.2 Assume that (H1 ) (H3 ) (H4 ) hold and Problem LQ is uniquely solvable at any x ∈ X with the optimal pair (¯ u(·), y¯(·)) ∈ U × X . Then the value operator P of Problem LQ satisfies the ARE (1.11) and (A∗ )α P ∈ L(X). Proof

First, let us fix a u ∈ U and an x ∈ D(A). By (4.1) and (A∗ )α P ∈ L(X),  0 ≤ V (y(t; x, u)) − V (x) +

0

t

[Qy(s; x, u), y(s; x, u)

+ 2ReSy(s; x, u), u + Ru, u] ds.

(4.2)

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Hence, dividing by t in (4.2) and letting t ↓ 0 yields that 0 ≤ −Ax, P x − P x, Ax + 2Re(A∗ )α P x, Bu + [Qx, x + 2ReSx, u + Ru, u].

(4.3)

Thus, it follows that 0 ≤ −Ax, P x − P x, Ax + Qx, x − [S + B ∗ (A∗ )α P ]∗ R−1 [S + B ∗ (A∗ )α P ]x, x
H(x, x). On the other hand, let x ∈ D(A) be fixed.  t [Q¯ y (s), y¯(s) + 2ReS y¯(s), u ¯(s) + R¯ u(s), u ¯(s)] ds. 0 = V (¯ y (t)) − V (x) +

(4.4)

(4.5)

0

Dividing by t and letting t ↓ 0 yields that 0 ≥ −Ax, P x − P x, Ax + Qx, x + inf {2Re[S + B ∗ (A∗ )α P ]x, u + Ru, u} = H(x, x).

(4.6)

u∈U

Combining (4.4) and (4.6), the desired result, (1.11), follows. Now we adapt Lemma 1 of [12] to the unbounded control case as follows: Lemma 4.3

Assume that N ∈ L(X) is a self-adjoint operator. Then N e−A(T −t) y(t), e−A(T −t) y(t) = N y(T ), y(T )  T −2 ReN e−A(T −s) y(s), Aα e−A(T −s) Bu(s) ds,

(4.7)

t

where 0 ≤ t ≤ T < ∞ and y(·) is the solution to (1.2) given by x ∈ X and u(·) ∈ U. Lemma 4.4

Assume that (H1 ) holds and K ∈ L(X). Then the feedback system  t y(t) = e−At x + Aα e−A(t−s) BKy(s) ds,

(4.8)

0

determines an analytic semigroup S(t) by S(t)x
y(t; x),

∀ x ∈ X, t ∈ [0, ∞).

 of S(t) is given as follows: The infinitesimal generator A
 = Aα [−A1−α + BK], A  = {x ∈ D(A1−α )|[−A1−α + BK]x ∈ D(Aα )}. D(A)

(4.9)

(4.10)

Relation (4.10) is given in Theorem 2.2 of [24] while the analyticity is proved in [25, Part 2, Section 4]. The proof of the relation (I) =⇒ (II) It follows from Theorem 4.2 that P , given by (3.2), is a solution to the ARE (1.11) with (A∗ )α P ∈ L(X). By (1.3), the ARE (1.11) yields that  ∞ ∗ P = e−A t {Q − [S + B ∗ (A∗ )α P ]∗ R−1 [S + B ∗ (A∗ )α P ]}e−At dt. (4.11) 0

Infinite Horizon Linear Quadratic Problem

535

Similarly to [12] or [15], it follows from (4.11), Lemma 4.3 and Fubini’s Theorem that, for any x ∈ X and u(·) ∈ U,  J(x; u(·)) − P x, x =

∞

0

1

R 2 {u(t) + R−1 [S + B ∗ (A∗ )α P ]y(t)}2 dt.

(4.12)

Since P is the value operator, (4.12) implies the optimal feedback (1.13). By the result of [26], we have, from (¯ u(·), y¯(·)) ∈ U × X , that the analytic semigroup S(t), given by (4.8) and (4.9) −1 with K
−R [S + B ∗ (A∗ )α P ], is exponentially stable. Now, assume that P is another solution to the ARE (1.11) such that (A∗ )α P ∈ L(X) and the analytic semigroup S(t), given by (4.8) and (4.9) with K
−R−1 [S + B ∗ (A∗ )α P], is exponentially stable. It follows, similarly, that the feedback u(·) = Kx(·) is also an optimal control. Therefore P is also the value operator of Problem LQ, which yields that P = P . The proof is completed. The proof of the relation (II) =⇒ (I) Since the feedback system (4.8) with K determines an exponentially stable C0 semigroup S(t), we have S(·)x ∈ X . Therefore −R−1 [S +B ∗ (A∗ )α P ]S(·)x is an admissible control at x. The conclusion easily follows from (4.12).

5

Frequency Domain Characteristics

For any u(·) ∈ U, define u(t) = 0 for t < 0. Then we have the natural embedding U ⊂ L2 (−∞, ∞; U ). Fourier transformation of u(·) is defined as follows: 1  u(·)(ω)
√ 2π



∞

−∞

−iωt

e

1 u(t) dt = √ 2π

 0

∞

e−iωt u(t) dt,

∀ u(·) ∈ U.

(5.1)

By Fubini’s Theorem, it follows from (1.3) that   [Lu(·)](ω) = Aα (iω + A)−1 B u(·)(ω),

∀ u(·) ∈ U.

(5.2)

Therefore, similarly to [15, pp. 409], (1.8) and the Parseval equality yields that 

∞ 0

 [Φu(·)](t), u(t) dt =

∞

−∞

 Φ(ω) u(ω), u (ω) dω,

∀ u(·) ∈ U,

(5.3)

 where Φ(ω) is given by (1.10). Lemma 5.1 Assume that (H1 ) (H3 ) (H4 ) hold and the ARE (1.11) admits a solution P satisfying (A∗ )α P ∈ L(X). Then it is true that  Φ(ω) = (I + E ∗ (ω))R(I + E(ω)) ≥ 0,

∀ ω ∈ R,

(5.4)

with E(ω)
R−1 [S + B ∗ (A∗ )α P ]Aα (iω + A)−1 B,

∀ ω ∈ R.

(5.5)

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536

Proof

For any y ∈ D(Aα ), let x
(iω + A)−1 y ∈ D(A1+α ). Thus, we have, from (1.11), that (A∗ )α (−iω + A∗ )−1 QAα (iω + A)−1 y, y = QAα x, Aα x = {(A∗ )α (−iω + A∗ )−1 [(A∗ )α P ]∗ + (A∗ )α P Aα (iω + A)−1 }y, y + R−1 [S + B ∗ (A∗ )α P ]Aα (iω + A)−1 y, [S + B ∗ (A∗ )α P ]Aα (iω + A)−1 y.

(5.6)

By a simple limit argument, it follows from the density of D(Aα ) that (A∗ )α (−iω + A∗ )−1 QAα (iω + A)−1 = (A∗ )α (−iω + A∗ )−1 [S + B ∗ (A∗ )α P ]∗ R−1 [S + B ∗ (A∗ )α P ]Aα (iω + A)−1 + (A∗ )α (−iω + A∗ )−1 [(A∗ )α P ]∗ + (A∗ )α P Aα (iω + A)−1 .

(5.7)

Hence, pulting (5.7) into (1.10) yields (5.4). Lemma 5.2 If the assumption of Theorem 4.2 holds, then the ARE (1.11) admits a solution P , given by (3.2), which satisfies (A∗ )α P ∈ L(X) and  ∀ x ∈ X, ω ∈ R, {(iω + A)A−α + BR−1 [S + B ∗ (A∗ )α P ]}−1 x ≤ Cx,

(5.8)

 > 0 is an absolute constant. Moreover, it is true that, for ω ∈ R where C I + E(ω) = (I − G(ω)B)−1 ,

(5.9)

with G(ω)
R−1 [S + B ∗ (A∗ )α P ]{(iω + A)A−α + BR−1 [S + B ∗ (A∗ )α P ]}−1 .

(5.10)

 defined by (1.12), also generates an exponentially stable and analytic Proof By Lemma 4.4, A, semigroup. It is easily shown from Theorem 5.2(c) in [23, pp. 61–62] that there exists an absolute constant C1 > 0 such that  −1 Ax  ≤ C1 x, (iω − A)

∀ ω ∈ R,

 x ∈ D(A).

(5.11)

A1
−A∗ − [S + B ∗ (A∗ )α P ]∗ R−1 B ∗ (A∗ )α also generates (see [23]) an exponentially stable and analytic semigroup. It follows from (1.12) that  y = x, A1 y,  y ∈ D(A1 ). Ax, ∀ x ∈ D(A), (5.12)  ∗ and A−1 = [(A)  ∗ ]−1 . By D(A1 ) = D(A∗ ), it is true that Hence A1 ⊂ (A) 1  ∗ ]−1 = (A∗ )α A−1 ∈ L(X). (A∗ )α [(A) 1

(5.13)

Combining (5.11) and (5.13) yields that, for some absolute constant C2 > 0,  −1 x {(iω + A)A−α + BR−1 [S + B ∗ (A∗ )α P ]}−1 x = [A−α (iω − A)]  −1 Aα x ≤ C2 x, = (iω − A)

∀ x ∈ D(Aα ), ω ∈ R.

(5.14)

Infinite Horizon Linear Quadratic Problem

537

By the density of D(Aα ) in X, (5.8) follows. Let the operator I + E(ω) be left-multiplied by G(ω)B. Noting the following: G(ω)BR−1 [S + B ∗ (A∗ )α P ]Aα (iω + A)−1 B = R−1 [S + B ∗ (A∗ )α P ]Aα (iω + A)−1 B − G(ω)B,

(5.15)

yields (5.9). The proof of the equivalences among (I), (III) and (IV) in Lemma 5.2 that

If (I) holds, then it follows from (5.8)

 −1 [S + B ∗ (A∗ )α P ]B
K < ∞, I − G(ω)B ≤ 1 + CR

∀ ω ∈ R,

(5.16)

where G(ω) is defined by (5.10). Thus, (5.16) and (5.9) yields u = (I − G(ω)B){I + R−1 [S + B ∗ (A∗ )α P ]Aα (iω + A)−1 B}u ≤ KI + R−1 [S + B ∗ (A∗ )α P ]Aα (iω + A)−1 Bu,

∀ u ∈ U, ω ∈ R.

Combining (5.4) in Lemma 5.1 and (5.17) yields (IV). If (IV) holds, then it follows from (5.3) that  ∞  ∞  Φu(·), u(·) = [Φu(·)](t), u(t) dt = Φ(ω) u(ω), u (ω) dω 0 −∞  ∞ ≥σ  u(ω), u (ω) dω = σu(·)2U ,

(5.17)

(5.18)

−∞

which yields that Φ−1 ∈ L(U). Therefore, (I) follows from Lemma 3.3 in [18]. Thus (I), (III) and (IV) are equivalent.

6

An Example of Application of the Main Theorem

In this section, we give a simple explanatory example as an application of our results. Example

Consider the one-dimensional heat equation with Dirichlet control action:  ∂ 2y ∂y   (z, t) = (z, t), 0 ≤ z ≤ 1, t ≥ 0,   ∂t ∂z 2 (6.1) y(0, t) = u(t), y(1, t) = 0,     y(z, 0) = y0 (z), 0 ≤ z ≤ 1,

where y0 (·) ∈ L2 (0, 1), and define the following cost functional:  ∞  1 |u(t)|2 − ρ |y(z, t)|2 dz dt, J(y0 (·); u(·)) = 0

0

2

ρ > 0.

(6.2)

Define X = L2 (0, 1), U = C and −A = ddz 2 with D(A) = H 2 ∩ H01 . Both X and U are complex Hilbert spaces with the usual inner products. −A generates an exponentially stable

H. Z. Wu and X. J. Li

538

and analytic semigroup e−At on X. The fractional power Aγ (γ ∈ R) is defined by:  1 ∞  γ 2γ [A ϕ(·)](z) = 2 (nπ) ϕ(z) sin(nπz)dz sin(nπz), a.e. in [0, 1], if

∞

n=1 (nπ)

4γ

(6.3)

0

n=1

1 [ 0 ϕ(z) sin(nπz) dz]2 < ∞. Consider the following equation:  2  ∂ y (z) = 0, 0 ≤ z ≤ 1; ∂z 2  y(0) = u, y(1) = 0.

(6.4)

The Dirichlet map G of (6.4) is as follows: [Gu](z) = (1 − z)u = 2 For a fixed ε ∈ (0, 14 ), denote α =

3 4

a.e. in [0, 1].

(6.5)

1

+ 6 and B = A 4 −ε G, which satisfies

+∞ 

[Bu](z) = 2

+∞  u sin(nπz), nπ n=1

u 1

n=1

(nπ) 2 +2ε

sin(nπz),

a.e. in

[0, 1].

(6.6)

A straight forward calculation shows that [Aα (iω + A)−1 Bu](z) = 2

+∞ 

nπu sin(nπz), iω + n2 π 2 n=1

hence  Φ(ω) = 1 − 2ρ

∞  n=1

n2 π 2 , + ω2

n4 π 4

a.e. in

[0, 1];

∀ ω ∈ R.

(6.7)

(6.8)

  If ρ ∈ (0, 3), then Φ(ω) ≥ Φ(0) = 1 − ρ3 > 0, for ω ∈ R. Thus, our Main Theorem implies that Problem LQ of (6.2), subject to (6.1), is uniquely solvable at any y0 . If ρ ∈ (3, ∞), then  Φ(0) < 0. Therefore, it easily follows that Problem LQ is not well-posed.

Appendix:

Singular Inequalities

Lemma A.1 Let 0 ≤ λ, µ < 1; C1 , C2 , ω1 , ω2 > 0 and −∞ < a < +∞. Let K(·, ·) and L(·, ·) be Lebesgue measurable on [a, ∞) × [a, ∞) satisfying C1 e−ω1 |ξ−ζ| , |ξ − ζ|λ

0 ≤ K(ξ, ζ) ≤ 

and M (ξ, ζ)


0 ≤ L(ξ, ζ) ≤

∞

K(ξ, η)L(η, ζ) dη,

C2 e−ω2 |ξ−ζ| , |ξ − ζ|µ

a ≤ ξ, ζ < ∞.

(a.1)

(a.2)

a

Then there exist C3 , θ > 0, such that for almost all (ξ, ζ) ∈ [a, ∞) × [a, ∞)  −θ|ξ−ζ|  ,  C3 e if θ + µ < 1;    C3 (| ln |ξ − ζ|| + 1)e−θ|ξ−ζ| , M (ξ, ζ) ≤ if θ + µ = 1;    C3 e−θ|ξ−ζ|  if θ + µ > 1. ,  |ξ − ζ|θ+µ−1

(a.3)

Infinite Horizon Linear Quadratic Problem

Proof

539

A straight forward calculation implies this lemma.

Theorem A.2 Let 0 ≤ θ < 1, δ > 0 and C1 , C2 > 0. For any given T ∈ [0, +∞), suppose that f (·) ∈ L2 (T, ∞) satisfies 

∞

0 ≤ f (t) ≤ C1 + C2

T

e−δ|t−s| f (s) ds, |t − s|θ

T ≤ t < ∞.

(a.4)

Then it is true that f (t) ≤ K(C1 + C2 f (·)L2 ), 

∞

T

T ≤ t < ∞,

e−σ|t−s| f (s) ds ≤ L(σ, γ)(C1 + C2 f (·)L2 ), |t − s|γ

(a.5)

T ≤ t < ∞,

(a.6)

where γ ∈ [0, 1), σ ∈ (0, ∞), and K, L(σ, γ) > 0 are constants independent of T ∈ [0, ∞). Remark (a.4) is of Fredholm’s type rather than Volterra’s type. Therefore this inequality is different from Gronwall’s inequality. Proof G(t, s)


e−δ|t−s| , |t − s|θ

0 ≤ t, s < ∞,

and for any nonnegative measurable v(·), define  ∞ [Sv(·)](t)
G(t, s)v(s) ds,

T ≤ t < ∞.

(a.7)

(a.8)

T 1 Denoting n = [ 1−θ ] + 1, it follows from (a.4) that, for some constant C3 > 0,

0 ≤ f (t) ≤ C1 + C2 [Sf (·)](t) ≤ · · · ≤ C3 {C1 + [S n f (·)](t)}.

(a.9)

By Fubini’s Theorem and Lemma A.1, it follows that there exists some C4 > 0 such that [S n f (·)](t) ≤ C4 C2 f (·)L2 .

(a.10)

Combining (a.9) and (a.10) yields (a.5) and  ∞ G(t, s)f (s) ds = [Sf (·)](t) ≤ C5 (C1 + C2 f (·)L2 ).

(a.11)

T

If 0 ≤ γ ≤ θ, then (a.6) holds with the same constants as in (a.11). If θ < γ < 1, then it follows from (a.4) that for some absolute constant C2 > 0 0 ≤ f (t) ≤ C1 +

C2



∞

T

δ

e− 2 |t−s| f (s) ds, |t − s|γ

T ≤ t < ∞.

(a.12)

Taking (a.12) as a new form of (a.4), we obtain (a.6) by applying the proved result (a.11) to the inequality (a.12).

540

H. Z. Wu and X. J. Li

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