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21

Causal State Feedback Representation

for Linear Quadratic Optimal Control Problems of

Singular Volterra Integral Equations ∗

Shuo Han†, Ping Lin‡ and Jiongmin Yong §

Abstract. This paper is concerned with a linear quadratic optimal control for a class of singular Volterraintegral equations. Under proper convexity conditions, optimal control uniquely exists, and it could becharacterized via Frechet derivative of the quadratic functional in a Hilbert space or via maximum principletype necessary conditions. However, these (equivalent) characterizations have a shortcoming that the currentvalue of the optimal control depends on the future values of the optimal state. Practically, this is not feasible.The main purpose of this paper is to obtain a causal state feedback representation of the optimal control.

AMS 2020 Mathematics Subject Classification. 45D05, 45F15, 49N10, 49N35, 93B52

Keywords. singular Volterra integral equation, quadratic optimal control, causal state feedback

1 Introduction.

Consider the following controlled singular linear Volterra integral equation:

(1.1) X(t) = ϕ(t) +

∫ t

0

A(t, s)X(s) +B(t, s)u(s)

(t− s)1−βds, a.e. t ∈ [0, T ].

In the above, T > 0 is a fixed finite time horizon, ϕ(·) is a given map, called the free term of the stateequation, X(·) is called the state trajectory taking values in the Euclidean space Rn, u(·) is called the controltaking values in the Euclidean space Rm, A(· , ·) and B(· , ·) are called the coefficients, taking values in R

n×n

and Rn×m, respectively, and β > 0.

We denote X = L2(0, T ;Rn), U = L2(0, T ;Rm). Under some mild conditions, for any control u(·) ∈ U ,the state equation (1.1) admits a unique solution X(·) ∈ X . To measure the performance of the control, weintroduce the following quadratic cost functional

(1.2)J(u(·)) =

∫ T

0

(〈Q(t)X(t), X(t)〉+ 2〈S(t)X(t), u(t)〉+ 〈R(t)u(t), u(t)〉

+2〈q(t), X(t)〉+ 2〈ρ(t), u(t)〉)dt+ 〈GX(T ), X(T )〉+ 2〈g,X(T )〉,

where Q(·) ∈ L∞(0, T ; Sn), S(·) ∈ L∞(0, T ;Rm×n), R(·) ∈ L∞(0, T ; Sm), q(·) ∈ X , ρ(·) ∈ U , G ∈ Sn,

g ∈ Rn, with S

k being the set of all (k × k) symmetric (real) matrices. Our optimal control problem can bestated as follows.

Problem (P). Find a control u(·) ∈ U such that

(1.3) J(u(·)) = infu(·)∈U

J(u(·)).

∗This work was partially supported by the National Natural Science Foundation of China under grant 12071067, National

Key R&D Program of China under grant 2020YFA0714102, and NSF grant DMS–1812921.†School of Mathematics and Statistics, Northeast Normal University, Changchun 130024, China. E-mail: hans861@nenu.

edu.cn.‡School of Mathematics and Statistics, Northeast Normal University, Changchun 130024, China. E-mail: linp258@nenu.

edu.cn.§Department of Mathematics, University of Central Florida, Orlando, FL 32816, USA. E-mail: jiongmin.yong@ucf.edu.

1

Any u(·) satisfying (1.3) is called an open-loop optimal control of Problem (P), the corresponding state X(·)is called an open-loop optimal state and (X(·), u(·)) is called an open-loop optimal pair.

Memory exists in many application problems, heat transfer, population growth, disease spread, to mentiona few. Volterra integral equations can be used to describe some dynamics involving memories. Study ofoptimal control problems for Volterra integral equations can be traced back to the works of Vinokurov in thelater 1960s [45], followed by the works of Angell [4], Kamien-Muller [28], Medhin [35], Carlson [17], Burnap–Kazemi [14], and some recent works by de la Vega [20], Belbas [7, 8], and Bonnans–de la Vega–Dupuis [11]. Allof the above-mentioned works are concerned with non-singular Volterra integral equations which exclude thecase of (1.1) with β ∈ (0, 1). On the other hand, in the past several decades, fractional (order) differentialequations have attracted quite a few researchers’ attention due to some very interesting applications inphysics, chemistry, engineering, population dynamics, finance and other sciences; See Oldham–Spanier [38]for some early examples of diffusion processes, Torvik–Bagley [44], Caputo [15], and Caputo–Mainardi [16]for modeling of the mechanical properties of materials, Benson [9] for the advection and the dispersion ofsolutes in natural porous or fractured media, Chern [18], Diethelm–Freed [23] for the modeling behaviorof viscoelastic and viscoplastic materials under external influences, Scalas–Gorenflo–Mainardi [41] for themathematical models in finance, Das–Gupta [19], Demirci–Unal–Ozalp [21], Arafa–Rida–Khalil [5], Diethelm[22] for some population and epidemic models, Metzler et al. [36] for the relaxation in filled polymer networks,and Okyere et al. [37] for a SIR model with constant population. An extensive survey on fractional differentialequations can be found in the book by Kilbas–Srivastava–Trujillo [31]. In the recent years, optimal controlproblems have been studied for fractional differential equations by a number of authors. We mention theworks of Agrawal [1, 2], Agrawal–Defterli–Baleanu [3], Bourdin [12], Frederico–Torres [24], Hasan–Tangpong–Agrawal [27] and Kamocki [29, 30], Gomoyunov [25], Koenig [32].

It turns out that fractional differential equations (of the order no more than 1), in the sense of Riemann–Liouville or in the sense of Caputo, are equivalent to Volterra integral equations with the integrand beingsingular along s = t, and the free term ϕ(·) being possibly discontinuous (blowing up) at t = 0 (See [33] forsome details). More precisely, in the linear case, the corresponding controlled state equation of form (1.1)could have the free term look like the following:

(1.4) ϕ(t) =c

t1−β(or c),

for some constant c ∈ R. In [33], a class of controlled nonlinear singular Volterra integral equations wasconsidered. Well-posedness of the state equation and some regularity of the state trajectory were established,and a Pontryagin type maximum principle for optimal controls was proved.

On the other hand, Pritchard–You [40] considered the quadratic optimal control problems for the followingcontrolled linear Volterra integral equations in a Hilbert space H :

(1.5) y(t) = f(t) +

∫ t

0

F (t, τ)u(τ)dτ, t ∈ [0, T ].

It was assumed in [40] that f(·) ∈ C([0, T ];H) and F : ∆ → L (U ;H) is strongly continuous in the sensethat for each u ∈ U , F (· , ·)u ∈ C(∆;H). Here, U is another Hilbert space and ∆ is the closure of thefollowing set

(1.6) ∆ = {(t, s) ∈ [0, T ]2∣∣ 0 6 s < t 6 T ]}.

Thus, in particular, the following holds

(1.7) ‖F (t, t)‖L (U ;H) <∞, t ∈ [0, T ].

This excludes our state equation (1.1) which has a singular kernel. We will see later that when the variationof constants formula is applied, our state process will have a similar representation as (1.5), but with boththe free term f(·) and the operator F (· , ·) being not necessarily continuous.

Practically, if an optimal control exists, one expects that the optimal could have a state feedback rep-resentation which is non-anticipating. In the case of state equation being an ordinary differential equation

2

(or a partial differential equation, a stochastic differential equation), such kind of representation can be ob-tained, under some proper conditions, via a solution to a Riccati differential equation. Pandolfi [39] derivedan optimal feedback control for a Volterra integro-differential equation by using the corresponding Riccatiequation. That could be done because the state equation in [39] was of a special form which has the semi-group property and thus one could use semigroup representation to derive a theory of Riccati equation ina standard way. But the general Volterra equation does not have a semigroup evolutionary property. Forthe controlled linear Volterra integral equation of form (1.5), a so-called projection causality approach wasintroduced in [40]. The optimal control could be represented as a so-called linear causal feedback (see laterfor a precise definition) of the state trajectory with the feedback operator being determined by a solution toa Fredholm integral equation.

In this paper, we will carry out some careful analysis on the state equation, and pay special attention tocertain continuity of the state trajectory since in the quadratic cost functional, the terminal valueX(T ) of thestate trajectory is involved. Also, we make it clear that the condition β > 1

2 should be assumed in order theLQ problem is well-formulated. By a standard method for minimization of a quadratic functional in Hilbertspace, we obtain a characterization of the (open-loop) optimal control in some abstract form. On the otherhand, by variational method, in the spirit of maximum principle, we may obtain another characterization ofthe open-loop optimal control. We will show that these two characterizations are equivalent. However, fromthose characterizations, the open-loop optimal control is not non-anticipating in the sense that in determiningthe value u(t) of the open-loop optimal control u(·) at time t, some future information {X(s)

∣∣ s ∈ [t, T ]} ofthe optimal state trajectory has to be used. This is not practically realizable. In the classical LQ problemof differential equations, one could get a closed-loop representation of the open-loop optimal control via thesolution to a Riccati equation. However, for general integral equations, such an approach is not working. Infact, the problem we considered in this paper is a nonlocal problem. Thus, we could not obtain a closed-loopoptimal control whose current value only depends on the current state value, by using the standard methodof ODE (or PDE, Volterra integro-differential equation in [39]) in terms of the Riccati equation. Inspiredby [40], we will try to obtain a causal state feedback representation for the open-loop optimal control inthe following sense: The current value u(t) of the open-loop optimal control u(·) is written in terms of thecurrent optimal state value X(t), as well as a causal trajectory Xt(·) and an auxiliary trajectory Xa(t), via afamily of Fredholm integral equations which essentially plays a role of Riccati equation in the classical LQproblems. It is worthy of pointing out that Xt(·) and Xa(·) can be running at the same time as the stateequation, and they are non-anticipating. Note that Xt(·) and Xa(·) are not involved in calculating the costfunctional, but they are used to represent the open-loop optimal control. Although the main idea comesfrom [40], our modified version of the method is more direct which reveals the essence of the problem moreclearly. In the proof of [40], they introduce an abstract operator to establish the interrelations between thestate trajectory and the causal trajectory. In this paper, we do not need to introduce the similar abstractoperator, but give a more direct proof. Furthermore, the trajectory Xt(T ) in [40] may raise the doubt aboutthe causality. In this paper, we can avoid this doubt by introducing the auxiliary trajectory Xa(·).

The rest of the paper is organized as follows. In section 2, we carry out some analysis for the stateequation. Section 3 is devoted to the open-loop optimal control and its characterizations. Causal projectionas well as abstract form of casual state feedback representation of the open-loop optimal control is presented inSection 4. We introduce a family of Fredholm integral equations in Section 5, which makes the representationobtained in Section 4 more practically accessible. In section 6, we briefly present a possible numerical schemewhich is applicable to solve the Fredholm integral equation obtained in Section 5.

2 Preliminary Results.

In this section, we will present some preliminary results which will be useful later. Let us recall ∆ definedby (1.6). Note that the “diagonal line” {(t, t) | t ∈ [0, T ]} is not contained in ∆. Thus if (t, s) 7→ f(t, s)is continuous on ∆, f(· , ·) is allowed to be unbounded as |t − s| → 0. Throughout this paper, we denotet1 ∨ t2 = max{t1, t2} and t1 ∧ t2 = min{t1, t2}, for any t1, t2 ∈ R. The characteristic function of any set E isdenoted by 1E(·). For any set E ⊆ R and a function ϕ : E → R, we extend it to be zero in R \ E. We calla strictly increasing continuous function ω(·) : [0,∞) → [0,∞) a modulus of continuity if ω(0) = 0. Also, Kwill be a generic constant which could be different from line to line.

3

Let us recall the Young’s inequality for convolution (Theorem 3.9.4 in [10]).

Lemma 2.1. Let p, q, r ∈ [1,+∞] satisfy 1p+ 1 = 1

q+ 1

r. Then for any f(·) ∈ Lq(Rn), g(·) ∈ Lr(Rn),

(2.1) ‖f(·) ∗ g(·)‖Lp(Rn) 6 ‖f(·)‖Lq(Rn)‖g(·)‖Lr(Rn).

From the above lemma, we have the following corollary which is a refinement of that found in [33].

Corollary 2.2. Let θ : ∆ → Rn and θ0 : [0, T ] → R be measurable such that

(2.2) |θ(t, τ)| 6 θ0(τ), a.e. (t, τ) ∈ ∆.

For any s ∈ [0, T ), define

η(t, s) =

∫ t

s

θ(t, τ)

(t− τ)1−βdτ, t ∈ (s, T ].

(i) Let β ∈ (0, 1), 1 6 r < 11−β

, 1p+ 1 = 1

q+ 1

r, p, q ∈ [1,∞], and θ0(·) ∈ Lq(s, T ). Then

(2.3) ‖η(·, s)‖Lp(s,T ;Rn) 6

( (T − s)1−r(1−β)

1− r(1 − β)

) 1

r ‖θ0(·)‖Lq(s,T ).

In particular, with q = 2,

(2.4) ‖η(·, s)‖Lp(s,T ;Rn) 6(T − s)β−( 1

2− 1

p)

(β−( 1

2− 1

p)

1

2+ 1

p

) 1

2+ 1

p

‖θ0(·)‖L2(s,T ).

Further, with both p = q = 2,

(2.5) ‖η(·, s)‖L2(s,T ;Rn) 6(T − s)β

β‖θ0(·)‖L2(s,T ),

and with p = ∞, q = 2, β > 12 ,

(2.6) ‖η(·, s)‖L∞(s,T ;Rn) 6(T − s)β−

1

2

√2β − 1

‖θ0(·)‖L2(s,T ).

(ii) Let there exist a δ0 ∈ (0, T−s2 ), and a modulus of continuity ω such that ∀t ∈ [T − 2δ0, T ],

(2.7)|θ(t, τ) − θ(T, τ)| 6 ω(T − t)θ1(τ), a.e. τ ∈ [s, t),

|θ(t, τ)| 6 θ0(τ), a.e. τ ∈ [s, t),

for some θ0(·), θ1(·) ∈ Lq(s, T ), q ∈ (1,∞], 1 > β > 1q. Then η(·, s) is continuous at T .

Proof. (i) By Lemma 2.1 with f(τ) = θ0(τ)1[s,T ](τ) and g(τ) = 1τ1−β 1(0,T−s](τ), we can obtain our

conclusion. The rest is clear.

(ii) Pick any δ ∈ (0, δ0). For any t ∈ (T − δ, T ), we look at the following, assuming first that q ∈ (1,∞)and setting κ = (1− β) q

q−1 < 1 (since β > 1q):

|η(T, s)− η(t, s)| =∣∣∣∫ T

s

θ(T, τ)

(T − τ)1−βdτ −

∫ t

s

θ(t, τ)

(t− τ)1−βdτ

∣∣∣

6

∫ t−δ

s

|θ(t, τ)|( 1

(t− τ)1−β− 1

(T − τ)1−β

)dτ +

∫ t−δ

s

|θ(t, τ) − θ(T, τ)|(T − τ)1−β

dτ

+

∫ t

t−δ

θ0(τ)

(t− τ)1−βdτ +

∫ T

t−δ

θ0(τ)

(T − τ)1−βdτ

6 (T − t)1−β

∫ t−δ

s

θ0(τ)

(t− τ)1−β(T − τ)1−βdτ + ω(T − t)

∫ t−δ

s

θ1(τ)

(T − τ)1−βdτ

+‖θ0(·)‖Lq(s,T )

( ∫ t

t−δ

dτ

(t− τ)κ

) q−1

q

+ ‖θ0(·)‖Lq(s,T )

(∫ T

t−δ

dτ

(T − τ)κ

) q−1

q

6(T − t)1−β

δ2(1−β)‖θ0(·)‖L1(s,T ) + ω(T − t)

( (T − s)1−κ

1− κ

) q−1

q ‖θ1(·)‖Lq(s,T )

4

+‖θ0(·)‖Lq(s,T )

[( δ1−κ

1− κ

) q−1

q

+( (T − t+ δ)1−κ

1− κ

) q−1

q].

Hence, for any ε > 0, we first take δ > 0 sufficiently small so that

‖θ0(·)‖Lq(s,T )

[( δ1−κ

1− κ

) q−1

q

+((2δ)1−κ

1− κ

) q−1

q]<ε

2.

Since the modulus of continuity ω(·) is continuous and ω(0) = 0, we can take δ ∈ (0, δ) even smaller so that

δ1−β

δ2(1−β)‖θ0(·)‖L1(s,T ) + ω(δ)

( (T − s)1−κ

1− κ

) q−1

q ‖θ1(·)‖Lq(s,T ) <ε

2.

Combining the above, we see that η(·, s) is continuous at T .In the case q = ∞, we have

|η(T, s)− η(t, s)| =∣∣∣∫ T

s

θ(T, τ)

(T − τ)1−βdτ −

∫ t

s

θ(t, τ)

(t− τ)1−βdτ

∣∣∣

6(T − t)1−β

δ2(1−β)‖θ0(·)‖L1(s,T ) + ω(T − t)

(T − s)β

β‖θ1(·)‖L∞(s,T )

+‖θ0(·)‖L∞(s,T )

[δββ

+(T − t+ δ)β

β

].

Then, similar to the above, we obtain the continuity of η(·, s) at T .We now look at the following linear Volterra integral equation

(2.8) X(t) = ξ(t) +

∫ t

0

A(t, s)X(s)

(t− s)1−βds, a.e. t ∈ [0, T ].

Note that (1.1) is a case of the above with

(2.9) ξ(t) = ϕ(t) +

∫ t

0

B(t, s)u(s)

(t− s)1−βds, a.e. t ∈ [0, T ].

Before going further, we introduce the following assumption for the coefficients of (1.1).

(A1) The coefficients A(· , ·) ∈ L∞(∆;Rn×n) and B(· , ·) ∈ L∞(∆;Rn×m). The free term ϕ(·) ∈ X .

For convenience, throughout the paper, we assume that

|A(t, s)| 6 ‖A‖∞, |B(t, s)| 6 ‖B‖∞, ∀(t, s) ∈ ∆.

Then, under (A1), for any u(·) ∈ U , by Corollary 2.2, (i), we have that if β ∈ (0, 1),

(2.10)( ∫ T

0

∣∣∣∫ t

0

B(t, s)u(s)

(t− s)1−βds∣∣∣2

dt) 1

2

6 ‖B‖∞[ ∫ T

0

(∫ t

0

|u(s)|(t− s)1−β

ds)2

dt] 1

2

6 ‖B‖∞(T β

β

)‖u(·)‖U .

Consequently, for the free term ξ(·) defined by (2.9), one has

(2.11) ‖ξ(·)‖X 6 ‖ϕ(·)‖X +T β‖B‖∞

β‖u(·)‖U .

The following gives the well-posedness of (2.8) as well as its variation of constants formula.

Theorem 2.3. Let (A1) hold. Then for any ξ(·) ∈ X , (2.8) admits a unique solution X(·) ∈ X .Moreover, there exists a measurable function Φ(· , ·) : ∆ → R

n×n satisfying that for any s ∈ [0, T ),

(2.12) Φ(t, s) =A(t, s)

(t− s)1−β+

∫ t

s

A(t, τ)Φ(τ, s)

(t− τ)1−βdτ, t ∈ (s, T ],

5

such that for some constant K > 0,

(2.13) |Φ(t, s)| 6 K

(t− s)1−β, (t, s) ∈ ∆,

and the solution X(·) to (2.8) can be represented by the following variation of constants formula:

(2.14) X(t) = ξ(t) +

∫ t

0

Φ(t, s)ξ(s)ds, a.e. t ∈ [0, T ],

with the following estimate:

(2.15) ‖X(·)‖X 6 K‖ξ(·)‖X .

Moreover, the function Φ(· , ·) also satisfies that for any s ∈ [0, T ),

(2.16) Φ(t, s) =A(t, s)

(t− s)1−β+

∫ t

s

Φ(t, τ)A(τ, s)

(τ − s)1−βdτ, t ∈ (s, T ].

Proof. First of all, by a standard contraction mapping argument, making use of Corollary 2.2, (i), we seethat for any ξ(·) ∈ X , (2.8) admits a unique solution X(·) ∈ X .

Next, by (A1), if Φ(· , ·) is a solution of (2.12), then

|Φ(t, s)| 6 ‖A‖∞(t− s)1−β

+

∫ t

s

‖A‖∞|Φ(τ, s)|(t− τ)1−β

dτ, (t, s) ∈ ∆.

Thus, by Gronwall’s inequality, we have

|Φ(t, s)| 6 ‖A‖∞(t− s)1−β

+K

∫ t

s

‖A‖2∞(t− τ)1−β(τ − s)1−β

dτ

=‖A‖∞

(t− s)1−β+K‖A‖2∞B(β, β)(t− s)1−2β

=‖A‖∞ +K‖A‖2∞B(β, β)(t− s)β

(t− s)1−β6

K

(t− s)1−β, (t, s) ∈ ∆.

This proves (2.13). In the above, B(· , ·) is the Beta function, and recall that K stands for a generic constantwhich could be different from line to line.

Now, we inductively define the following sequence of measurable functions:

(2.17) F1(t, s) =A(t, s)

(t− s)1−β, Fk+1(t, s) =

∫ t

s

F1(t, τ)Fk(τ, s)dτ, k = 1, 2, 3, · · · , (t, s) ∈ ∆.

Then

|F1(t, s)| 6‖A‖∞

(t− s)1−β, (t, s) ∈ ∆,

|F2(t, s)| 6∫ t

s

|F1(t, τ)F1(τ, s)|dτ 6 ‖A‖2∞∫ t

s

dτ

(t− τ)1−β(τ − s)1−β=

‖A‖2∞B(β, β)(t− s)1−2β

, (t, s) ∈ ∆.

By induction, we can show that

(2.18) |Fk(t, s)| 6∫ t

s

|F1(t, τ)Fk−1(τ, s)|dτ 6‖A‖k∞

(t− s)1−kβ

k−1∏

j=1

B(β, jβ), k > 1, (t, s) ∈ ∆.

According to [43] (p.102), Gamma function Γ(·) admits the following asymptotic expansion:

Γ(z + 1) =√2πz zze−z

(1 +R(z)

), z >> 1; R(z) =

1

12z+

1

288z2− 139

51840z3+ · · · .

6

Thus, for j large enough, we have

B(β, jβ) = Γ(β)Γ(jβ)

Γ(jβ + β)=

Γ(β)√

2π(jβ − 1)(jβ − 1)jβ−1e−(jβ−1)(1 +R(jβ − 1)

)

√2π(jβ + β − 1)(jβ + β − 1)jβ+β−1e−(jβ+β−1)

(1 +R(jβ + β − 1)

)

= Γ(β)

√jβ − 1

jβ + β − 1

( jβ − 1

jβ + β − 1

)jβ−1 eβ

(jβ + β − 1)β

(1 + R(j)

)

6Γ(β)eβ

ββ

( j

j + 1− 1β

)β(1 + R(j)

) 1

jβ6

Γ(β)eβ

ββ

2β+1

jβ,

for some R(j) → 0 as j → ∞. Consequently, there exists a k0 such that for k > k0,

|Fk(t, s)|6‖A‖k∞

(t−s)1−kβ

( k0−1∏

j=1

B(β, jβ))(Γ(β)eβ2β+1

ββ

)k−k0 [(k0−1)!]β

[(k−1)!]β6

K0Kk

(t−s)1−kβ

1

[(k−1)!]β, (t, s)∈∆,

for some constants K0,K > 0. Then, for any δ > 0, series∞∑

k=1

Fk(t, s) is uniformly and absolutely convergent

for (t, s) ∈ ∆δ with∆δ = {(t, s) ∈ ∆

∣∣ t− s > δ}.

Now we define

Φ(t, s) =

∞∑

k=1

Fk(t, s), (t, s) ∈ ∆,

which is measurable (since each Fk(· , ·) is measurable) and bounded on each ∆δ, δ > 0. We can easily checkthat the above defined Φ(· , ·) is the unique solution of (2.12), and therefore, estimate (2.13) holds. Further,for any ξ(·) ∈ L2(0, T ;Rn), similar to (2.10), we see that

X(t) = ξ(t) +

∫ t

0

Φ(t, s)ξ(s)ds, a.e. t ∈ [0, T ],

is well-defined as an element in L2(0, T ;Rn). In addition, for such defined X(·), one has

∫ t

0

A(t, s)X(s)

(t− s)1−βds =

∫ t

0

A(t, s)

(t− s)1−β

[ξ(s) +

∫ s

0

Φ(s, τ)ξ(τ)dτ]ds

=

∫ t

0

A(t, s)ξ(s)

(t− s)1−βds+

∫ t

0

∫ t

τ

A(t, s)Φ(s, τ)ξ(τ)

(t− s)1−βdsdτ

=

∫ t

0

( A(t, s)

(t− s)1−β+

∫ t

s

A(t, τ)Φ(τ, s)

(t− τ)1−βdτ

)ξ(s)ds =

∫ t

0

Φ(t, s)ξ(s)ds = X(t)− ξ(t), a.e. t ∈ [0, T ].

This proves (2.14). Making use of (2.13), and similar to (2.10), we obtain

‖X(·)‖X 6 ‖ξ(·)‖X +( ∫ T

0

∣∣∣∫ t

0

Φ(t, s)ξ(s)ds∣∣∣2

dt) 1

2

6 ‖ξ(·)‖X +( ∫ T

0

(∫ t

0

|ξ(s)|(t− s)1−β

ds)2

dt) 1

2

6 ‖ξ(·)‖X +K‖ξ(·)‖X .

Finally, we prove (2.16). To this end, we make the following observation:

F3(t, s) =

∫ t

s

F1(t, τ)F2(τ, s)dτ =

∫ t

s

F1(t, τ)

∫ τ

s

F1(τ, r)F1(r, s)drdτ

=

∫ t

s

∫ t

r

F1(t, τ)F1(τ, r)F1(r, s)dτdr =

∫ t

s

F2(t, r)F1(r, s)dr, (t, s) ∈ ∆.

7

Hence, by induction, we see that (comparing with (2.17))

Fk+1(t, s) =

∫ t

s

Fk(t, r)F1(r, s)dr, (t, s) ∈ ∆, k > 1.

Then we see that (2.16) holds. This completes the proof.

According to the above theorem, for state equation (1.1), we have the following representation of thestate process X(·) in terms of the control u(·) and the free term ϕ(·):

(2.19)

X(t) = ϕ(t) +

∫ t

0

B(t, s)u(s)

(t− s)1−βds+

∫ t

0

Φ(t, s)[ϕ(s) +

∫ s

0

B(s, τ)u(τ)

(s− τ)1−βdτ

]ds

= ϕ(t) +

∫ t

0

Φ(t, s)ϕ(s)ds +

∫ t

0

B(t, s)u(s)

(t− s)1−βds+

∫ t

0

∫ t

τ

Φ(t, s)B(s, τ)u(τ)

(s− τ)1−βdsdτ

= ϕ(t) +

∫ t

0

Φ(t, s)ϕ(s)ds +

∫ t

0

( B(t, s)

(t− s)1−β+

∫ t

s

Φ(t, τ)B(τ, s)

(τ − s)1−βdτ

)u(s)ds

≡ ψ(t) +

∫ t

0

Ψ(t, s)u(s)ds,

where

(2.20)ψ(t) = ϕ(t) +

∫ t

0

Φ(t, s)ϕ(s)ds, a.e. t ∈ [0, T ],

Ψ(t, s) =B(t, s)

(t− s)1−β+

∫ t

s

Φ(t, τ)B(τ, s)

(τ − s)1−βdτ, (t, s) ∈ ∆.

Clearly, Ψ : ∆ → Rn×m, and

(2.21)|Ψ(t, s)| 6 ‖B‖∞

(t− s)1−β+ ‖B‖∞

∫ t

s

|Φ(t, τ)|(τ − s)1−β

dτ

6K

(t− s)1−β+K

∫ t

s

dτ

(t− τ)1−β(τ − s)1−β6

K

(t− s)1−β, (t, s) ∈ ∆.

Moreover, noting ξ(·) defined by (2.9) and the estimate (2.11),

(2.22) ‖X(·)‖X 6 K‖ξ(·)‖X 6 K(‖ϕ(·)‖X + ‖u(·)‖U

).

We call (2.19) the variation of constants formula for the state X(·). From the above, we see that under (A1),for any control u(·) ∈ U , the state equation (1.1) is well-posed in X . Thus, the running cost in (1.2) iswell-defined. However, the terminal cost is still not necessarily defined. We need the state process X(·) tobe continuous at t = T . To achieve this, we need a little more assumption which we now introduce.

(A2) Let (A1) hold, and in addition, there exists a modulus of continuity ω(·) and some δ0 ∈ (0, T ],

|A(T, s)−A(t, s)|+ |B(T, s)−B(t, s)|+ |ϕ(T )− ϕ(t)| 6 ω(T − t), t ∈ [T − δ0, T ], (t, s) ∈ ∆.

(A3) β > 12 .

We have the following result.

Theorem 2.4. (i) Let (A2)–(A3) hold. Then for any s ∈ [0, T ), t 7→ Φ(t, s) is continuous at t = T .

(ii) Let (A2)–(A3) hold. Then for any control u(·) ∈ U , the corresponding state processX(·) is continuousat t = T .

8

Proof. (i) By Corollary 2.2 (ii), we can get (i).

(ii) Now, we let (A2)–(A3) hold. Then for any u(·) ∈ U ,

|X(t)| 6 |ψ(t)|+∫ t

0

|Ψ(t, s)u(s)|ds 6 |ϕ(t)|+∫ t

0

|Φ(t, s)ϕ(s)|ds +( ∫ t

0

|Ψ(t, s)|2ds) 1

2 ‖u(·)‖U

6 |ϕ(t)|+(∫ t

0

|Φ(t, s)|2ds) 1

2 ‖ϕ(·)‖X +K(∫ t

0

ds

(t− s)2(1−β)

) 1

2 ‖u(·)‖U

6 |ϕ(t)|+K(∫ t

0

ds

(t− s)2(1−β)

) 1

2 ‖ϕ(·)‖X +Ktβ−

1

2

√2β − 1

‖u(·)‖U

6 |ϕ(t)|+ Ktβ−1

2

√2β − 1

‖ϕ(·)‖X +Ktβ−

1

2

√2β − 1

‖u(·)‖U 6 |ϕ(t)| +K(‖ϕ(·)‖X + ‖u(·)‖U

).

Thus, X(t) is defined at all points where ϕ(t) is defined. To obtain the continuity of X(·) at t = T , sinceϕ(·) is continuous at T , it suffices to obtain the continuity of the following expression at t = T :

∫ t

0

A(t, τ)X(τ) +B(t, τ)u(τ)

(t− τ)1−βdτ.

Since

|A(t, τ)X(τ) +B(t, τ)u(τ)| 6 K(|X(τ)|+ |u(τ)|

), (t, τ) ∈ ∆,

|A(T, τ)X(τ) +B(T, τ)u(τ) −A(t, τ)X(τ) −B(t, τ)u(τ)| 6 ω(T − t)(|X(τ)| + |u(τ)|

),

t ∈ [T − δ0, T ], (t, τ) ∈ ∆,

with |X(·)|+ |u(·)| ∈ L2(0, T ), by Corollary 2.2 (ii), we obtain the continuity.

Note that if β 612 , then in general, we do not expect to have a continuity of the state process at t = T

for some control u(·) ∈ U . The following example illustrates this.

Example 2.5. Let T = 1, A(·, ·) = 0, B(·, ·) = 1, and ϕ(·) = 0. Then we have

X(t) =

∫ t

0

u(s)

(t− s)1−βds, t ∈ [0, 1].

Let

u(s) =1[ 1

2,1)(s)

(1− s)1

2 log(1− s), s ∈ [0, 1).

Then, ∫ 1

0

|u(s)|2ds =∫ 1

1

2

ds

(1− s)[log(1− s)]2=

1

log 2.

Thus, u(·) ∈ U . However,

X(1) =

∫ 1

0

u(s)

(1 − s)1−βds =

∫ 1

1

2

ds

(1− s)3

2−β log(1− s)

= ∞, ∀β 61

2.

Having the above result, we see that under (A2)–(A3), for any u(·) ∈ U , the cost functional is well-defined, and therefore, Problem (P) is well-formulated.

3 Open-loop Optimal Control.

In this section, we will present the unique existence of open-loop optimal control and its characterizations.Let us first introduce the the following operators:

(3.1)

(Θu)(t) =

∫ t

0

Ψ(t, s)u(s)ds, u ∈ U , t ∈ [0, T ],

ΘTu =

∫ T

0

Ψ(T, s)u(s)ds = (Θu)(T ), u ∈ U .

9

Then

(3.2) X(t) = ψ(t) + (Θu)(t), t ∈ [0, T ]; X(T ) = ψ(T ) + ΘTu.

From (2.21), we see that (only need β ∈ (0, 1)) for any u(·) ∈ U ,

(3.3) ‖Θu‖X =( ∫ T

0

∣∣∣∫ t

0

Ψ(t, s)u(s)ds∣∣∣2

dt) 1

2

6 K[ ∫ T

0

(∫ t

0

|u(s)|(t− s)1−β

ds)2

dt] 1

2

6 K‖u(·)‖U ,

and (noting β ∈ (12 , 1))

(3.4) |ΘTu| =∣∣∣∫ T

0

Ψ(T, s)u(s)ds∣∣∣ 6 K

∫ T

0

|u(s)|(T − s)1−β

ds 6 K‖u(·)‖U .

Thus, Θ ∈ L (U ;X ) and ΘT ∈ L (U ;Rn). Consequently, their adjoint operators Θ∗ ∈ L (X ;U ) andΘ∗

T ∈ L (Rn;U ) are well-defined. Let us identify them as follows. For any X(·) ∈ X ,

〈X(·), (Θu)(·)〉X =

∫ T

0

〈X(t), (Θu)(t)〉dt =∫ T

0

〈X(t),

∫ t

0

Ψ(t, s)u(s)ds〉dt

=

∫ T

0

∫ t

0

〈X(t),Ψ(t, s)u(s)〉dsdt =∫ T

0

∫ T

s

〈X(t),Ψ(t, s)u(s)〉dtds

=

∫ T

0

〈∫ T

s

Ψ(t, s)⊤X(t)dt, u(s)〉ds = 〈∫ T

s

Ψ(t, s)⊤X(t)dt, u(s)〉U = 〈(Θ∗X)(·), u(·)〉U .

This gives

(3.5) (Θ∗X)(s) =

∫ T

s

Ψ(t, s)⊤X(t)dt, X(·) ∈ X .

From Corollary 2.2, (i), we see that (only need β ∈ (0, 1))

( ∫ T

0

∣∣∣∫ T

s

Ψ(t, s)⊤X(t)dt∣∣∣2

ds) 1

2

6 K[ ∫ T

0

(∫ T

s

|X(t)|(t− s)1−β

dt)2

ds] 1

2

6 K‖X(·)‖X .

Likewise (noting β ∈ (12 , 1)),

(3.6) (Θ∗Tx)(s) = Ψ(T, s)⊤x, ∀x ∈ R

n,

with (∫ T

0

|(Θ∗Tx)(s)|2ds

) 1

2

6 K|x|(∫ T

0

ds

(T − s)2(1−β)

) 1

2

= K( T 2β−1

2β − 1

) 1

2 |x|.

Remark 3.1. By (2.21), noting β > 12 , we see that for X(·) ∈ X ,

|(Θ∗X)(s)| 6∫ T

s

|Ψ(t, s)||X(t)|dt 6 K

∫ T

s

|X(t)|(t− s)1−β

dt 6 K(T − s)β−

1

2

(2β − 1)1

2

‖X(·)‖X , s ∈ [0, T ],

which is bounded. But,

|(Θ∗Tx)(s)| 6

K|x|(T − s)1−β

, s ∈ [0, T ),

which may be unbounded. This is different from the case that (1.7) holds as in [40]. Consequently, we needto modify the technique used there so that it works for us.

Now, we would like to represent the cost functional. We will use Q to denote the bounded operator fromX to itself induced by Q(·), and so on. Then

(3.7)

J(u(·)) = 〈QX,X〉+ 2〈SX,u〉+ 〈Ru, u〉+ 2〈q, X〉+ 2〈ρ, u〉+ 〈GX(T ), X(T )〉+ 2〈g, X(T )〉= 〈Q(ψ +Θu), ψ +Θu〉+ 2〈S(ψ +Θu), u〉+ 〈Ru, u〉+ 2〈q, ψ +Θu〉+ 2〈ρ, u〉+〈G[ψ(T ) + ΘTu], ψ(T ) + ΘTu〉+ 2〈g, ψ(T ) + ΘTu〉

= 〈Λu, u〉+ 2〈λ1, u〉+ λ0,

10

where

Λ = Θ∗QΘ+ SΘ+Θ∗S∗ +R+Θ∗TGΘT ,

λ1 = Θ∗Qψ + Sψ +Θ∗q+ ρ+Θ∗TGψ(T ) + Θ∗

Tg,

λ0 = 〈Qψ, ψ〉+ 2〈q, ψ〉+ 〈Gψ(T ), ψ(T )〉+ 2〈g, ψ(T )〉.The above shows that our Problem (P) can be regarded as an optimization of the functional J(u(·)) on theHilbert space U . In order the functional u(·) 7→ J(u(·)) to be bounded from below, it is necessary thatΛ > 0. In what follows, we want J(u(·)) to admit a unique minimum. To guarantee this, we may assumethe following stronger condition:

(3.8) Λ > δ,

for some δ > 0. Since it is not the main theme of this paper to discuss the conditions under which (3.8) holds,we are satisfied to assume proper conditions to guarantee (3.8). For simplicity, we introduce the followingstandard assumption:

(A4) Let{Q(·) ∈ L∞(0, T ; Sn), S(·) ∈ L∞(0, T ;Rm×n), R(·) ∈ L∞(0, T ; Sm),

q(·) ∈ L2(0, T ;Rn), ρ(·) ∈ L2(0, T ;Rm), G ∈ Sn, g ∈ R

n,

and the following holds:

(3.9) R(t) > δ, Q(t)− S(t)⊤R(t)−1S(t) > 0, G > 0, t ∈ [0, T ],

for some δ > 0.

Now, we have the following result for Problem (P).

Theorem 3.2. Suppose that (A2)–(A4) hold. Then Problem (P) admits a unique open-loop optimalpair (X(·), u(·)) ∈ X × U . Moreover, the following relation is satisfied:

(3.10)

u(t) = −R(t)−1{Ψ(T, t)⊤

(GX(T ) + g

)+ S(t)X(t) + ρ(t)

+

∫ T

t

Ψ(s, t)⊤(Q(s)X(s) + S(s)⊤u(s) + q(s)

)ds}, a.e. t ∈ [0, T ].

Proof. Under (A4), one has (3.8). Thus, from the representation (3.7) of the cost functional, we see thatu(·) 7→ J(u(·)) admits a unique minimum u(·) which is given by the solution to the following:

0 = Λu+ λ1 =(Θ∗QΘ+ SΘ+Θ∗S∗ +R+Θ∗

TGΘT

)u+Θ∗Qψ + Sψ +Θ∗q+ ρ+Θ∗

TGψ(T ) + Θ∗Tg

= Θ∗Q(ψ +Θu) + S(ψ +Θu) + Θ∗S∗u+Ru+Θ∗TG(ψ(T ) + ΘT u) + Θ∗q+ ρ+Θ∗

Tg

= Θ∗T [GX(T ) + g] + Θ∗(QX + S∗u+ q) +Ru+ SX + ρ.

Thus,

(3.11) u = −R−1(Θ∗

T [GX(T ) + g] + SX + ρ+Θ∗(QX + S∗u+ q)),

which is the same as (3.10).

Let us take a closer look at the above result. Note that

Ψ(s, t)⊤ =B(s, t)⊤

(s− t)1−β+

∫ s

t

B(τ, t)⊤Φ(s, τ)⊤

(τ − t)1−βdτ, (s, t) ∈ ∆.

Thus,

Ψ(T, t)⊤(GX(T ) + g

)+

∫ T

t

Ψ(s, t)⊤[Q(s)X(s) + S(s)⊤u(s) + q(s)

]ds

=B(T, t)⊤[GX(T ) + g]

(T − t)1−β+

∫ T

t

B(s, t)⊤Φ(T, s)⊤[GX(T ) + g]

(s− t)1−βds

+

∫ T

t

B(s, t)⊤

(s− t)1−β

{Q(s)X(s) + S(s)⊤u(s) + q(s) +

∫ T

s

Φ(τ, s)⊤[Q(τ)X(τ) + S(τ)⊤u(τ) + q(τ)

]dτ

}ds

11

=B(T, t)⊤[GX(T ) + g]

(T − t)1−β+

∫ T

t

B(s, t)⊤

(s− t)1−β

{Φ(T, s)⊤[GX(T ) + g] +Q(s)X(s) + S(s)⊤u(s) + q(s)

+

∫ T

s

Φ(τ, s)⊤[Q(τ)X(τ) + S(τ)⊤u(τ) + q(τ)

]dτ

}ds, t ∈ [0, T ].

Consequently, we have the following relation for the optimal control u(·):

(3.12)

u(t) = −R(t)−1[B(T, t)⊤[GX(T ) + g]

(T − t)1−β+ S(t)X(t) + ρ(t)

+

∫ T

t

B(s, t)⊤

(s− t)1−β

(Φ(T, s)⊤[GX(T ) + g] +Q(s)X(s) + S(s)⊤u(s) + q(s)

+

∫ T

s

Φ(τ, s)⊤[Q(τ)X(τ) + S(τ)⊤u(τ) + q(τ)

]dτ

)ds], a.e. t ∈ [0, T ],

with X(·) being the optimal state trajectory.

On the other hand, we know that optimal control can also be characterized by the variational method.The following is the result for Problem (P) from a different angle.

Theorem 3.3. Let (A2)–(A4) hold. Then Problem (P) admits a unique open-loop optimal pair (X(·), u(·))such that

(3.13) u(t) = −R(t)−1[ ∫ T

t

B(s, t)⊤Y (s)

(s− t)1−βds+ S(t)X(t) + ρ(t) +

B(T, t)⊤[GX(T ) + g]

(T − t)1−β

], a.e. t ∈ [0, T ],

where Y (·) is the solution to the following adjoint equation:

(3.14) Y (t) = Q(t)X(t)+S(t)⊤u(t)+q(t)+A(T, t)⊤[GX(T ) + g]

(T − t)1−β+

∫ T

t

A(s, t)⊤Y (s)

(s− t)1−βds, a.e. t ∈ [0, T ].

Proof. Let (X(·), u(·)) be the optimal pair. Then for any u(·) ∈ U , we have

0 = limε→0

1

2ε

[J(u(·) + εu(·))− J(u(·))

]

=

∫ T

0

(〈Q(t)X(t) + S(t)⊤u(t) + q(t), X(t)〉+ 〈S(t)X(t) +R(t)u(t) + ρ(t), u(t)〉

)dt+ 〈GX(T ) + g,X(T )〉,

where

X(t) =

∫ t

0

A(t, s)X(s) +B(t, s)u(s)

(t− s)1−βds, t ∈ [0, T ].

Thus,

〈GX(T ) + g,X(T )〉 =∫ T

0

(〈A(T, t)

⊤[GX(T ) + g]

(T − t)1−β, X(t)〉+ 〈B(T, t)⊤[GX(T ) + g]

(T − t)1−β, u(t)〉

)dt.

This yields

0 =

∫ T

0

(〈Q(t)X(t) + S(t)⊤u(t) + q(t) +

A(T, t)⊤[GX(T ) + g]

(T − t)1−β, X(t)〉

+〈S(t)X(t) +R(t)u(t) + ρ(t) +B(T, t)⊤[GX(T ) + g]

(T − t)1−β, u(t)〉

)dt.

Now, we let Y (·) be the solution to the following:

Y (t) = γ(t) +

∫ T

t

A(s, t)⊤Y (s)

(s− t)1−βds, a.e. t ∈ [0, T ],

12

with

γ(t) = Q(t)X(t) + S(t)⊤u(t) + q(t) +A(T, t)⊤[GX(T ) + g]

(T − t)1−β, a.e. t ∈ [0, T ].

Then for any u ∈ U ,

0 =

∫ T

0

(〈Y (t)−

∫ T

t

A(s, t)⊤Y (s)

(s− t)1−βds,X(t)〉

+〈S(t)X(t) +R(t)u(t) + ρ(t) +B(T, t)⊤[GX(T ) + g]

(T − t)1−β, u(t)〉

)dt

=

∫ T

0

〈Y (t), X(t)〉dt−∫ T

0

〈Y (s),

∫ s

0

A(s, t)X(t)

(s− t)1−βdt〉ds

+

∫ T

0

〈S(t)X(t) +R(t)u(t) + ρ(t) +B(T, t)⊤[GX(T ) + g]

(T − t)1−β, u(t)〉dt

=

∫ T

0

(〈Y (t),

∫ t

0

B(t, s)u(s)

(t− s)1−βds〉+ 〈S(t)X(t) +R(t)u(t) + ρ(t) +

B(T, t)⊤[GX(T ) + g]

(T − t)1−β, u(t)〉

)dt

=

∫ T

0

〈∫ T

t

B(s, t)⊤Y (s)

(s− t)1−βds+ S(t)X(t) +R(t)u(t) + ρ(t) +

B(T, t)⊤[GX(T ) + g]

(T − t)1−β, u(t)〉dt.

Hence,∫ T

t

B(s, t)⊤Y (s)

(s− t)1−βds+ S(t)X(t) +R(t)u(t) + ρ(t) +

B(T, t)⊤[GX(T ) + g]

(T − t)1−β= 0, a.e. t ∈ [0, T ].

Then (3.13) follows.

The above is actually the Pontryagin type maximum principle. Comparing (3.12) and (3.13), we see thatthey coincide if the following is true:

(3.15)Y (s) = Φ(T, s)⊤[GX(T ) + g] +Q(s)X(s) + S(s)⊤u(s) + q(s)

+

∫ T

s

Φ(τ, s)⊤[Q(τ)X(τ) + S(τ)⊤u(τ) + q(τ)

]dτ, a.e. s ∈ [0, T ].

This can be shown as follows. By (2.16), we have

(3.16) Φ(t, s)⊤ =A(t, s)⊤

(t− s)1−β+

∫ t

s

A(τ, s)⊤Φ(t, τ)⊤

(τ − s)1−βdτ, a.e. t ∈ [s, T ].

Denotez(t) = Q(t)X(t) + S(t)⊤u(t) + q(t), a.e. t ∈ [0, T ]; ζ = GX(T ) + g.

Then, we have

Y (t) = z(t) +A(T, t)⊤ζ

(T − t)1−β+

∫ T

t

A(s, t)⊤Y (s)

(s− t)1−βds, a.e. t ∈ [0, T ],

and we need to check:

Y (t) = z(t) + Φ(T, t)⊤ζ +

∫ T

t

Φ(s, t)⊤z(s)ds, a.e. t ∈ [0, T ].

This can be checked as follows:

∫ T

t

A(s, t)⊤Y (s)

(s− t)1−βds =

∫ T

t

A(s, t)⊤[z(s) + Φ(T, s)⊤ζ +

∫ T

sΦ(τ, s)⊤z(τ)dτ

]

(s− t)1−βds

=

∫ T

t

A(s, t)⊤z(s)

(s− t)1−βds+

∫ T

t

A(s, t)⊤Φ(T, s)⊤ζ

(s− t)1−βds+

∫ T

t

(∫ s

t

A(τ, t)⊤Φ(s, τ)⊤

(τ − t)1−βdτ

)z(s)ds

=

∫ T

t

Φ(s, t)⊤z(s)ds+

∫ T

t

A(s, t)⊤Φ(T, s)⊤ζ

(s− t)1−βds

= Y (t)− z(t)−(Φ(T, t)⊤ −

∫ T

t

A(s, t)⊤Φ(T, s)⊤

(s− t)1−βds)ζ = Y (t)− z(t)− A(T, t)⊤ζ

(T − t)1−β, a.e. t ∈ [0, T ].

13

Hence, (3.12) and (3.13) are equivalent.

4 Causal State Feedback Representation.

In the relation (3.12) (or (3.13)) for the open-loop optimal control, the current-time value u(t) of the optimalcontrol u(·) is given in terms of the future-time values {X(s)

∣∣ s ∈ [t, T ]} of the corresponding optimal statetrajectory X(·). Practically, this is not realizable. Thus, our next goal is to seek a causal state feedback

representation of optimal control, by which we mean that the value u(t) of u(·) at time t can be written interms of {X(s)

∣∣ s ∈ [0, t]} and Xa(t) for some non-anticipating auxiliary process Xa(·). Recall that forstandard LQ problems of differential equations, optimal control could admit a closed-loop representation bymeans of differential Riccati equations. Here, we borrow the idea from [40], but with more straightforwardapproach which more naturally reveals the essence of the problem.

Since our cost functional contains the cross term of the state and control, as well as a linear term in thecontrol, we would like to make a reduction first. Let

(4.1) u(t) = v(t)−R(t)−1[S(t)X(t) + ρ(t)

], a.e. t ∈ [0, T ].

Then, the state equation becomes

(4.2) X(t) = ϕ(t) +

∫ t

0

A(t, s)X(s) +B(t, s)v(s)

(t− s)1−βds, a.e. t ∈ [0, T ],

where

A(t, s) = A(t, s)−B(t, s)R(s)−1S(s), ϕ(t) = ϕ(t) −∫ t

0

B(t, s)R(s)−1ρ(s)

(t− s)1−βds, a.e. (t, s) ∈ ∆.

Also, the running cost rate becomes

〈QX,X〉+ 2〈SX, u〉+ 〈Ru, u〉+ 2〈q,X〉+ 2〈ρ, u〉= 〈(Q − S⊤R−1S)X,X〉+ 〈Rv, v〉+ 2〈q − S⊤R−1ρ,X〉 − 〈R−1ρ, ρ〉≡ 〈QX,X〉+ 〈Rv, v〉+ 2〈q, X〉 − 〈R−1ρ, ρ〉.

Thus, if we define

J(v(·)) =∫ T

0

(〈Q(t)X(t), X(t)〉+ 〈R(t)v(t), v(t)〉 + 2〈q(t), X(t)〉

)dt+ 〈GX(T ), X(T )〉+ 2〈g,X(T )〉,

then the corresponding optimal control problem is equivalent to the original one. Namely, (X(·), u(·)) is theopen-loop optimal pair of the original problem if and only if (X(·), v(·)) is the optimal pair of the reducedproblem with u(·) and v(·) being related by the following:

(4.3) u(t) = v(t)−R(t)−1[S(t)X(t) + ρ(t)

], a.e. t ∈ [0, T ].

Therefore, if the optimal control v(·) of the reduced problem has a causal state feedback representation, thenso is u(·). Hence, for simplicity, we introduce the following hypothesis.

(A5) Let (A2)–(A4) hold with

(4.4) S(t) = 0, ρ(t) = 0, t ∈ [0, T ].

Under (A5), we have

(4.5)

Λ = Θ∗QΘ+Θ∗TGΘT +R,

λ1 = Θ∗Qψ +Θ∗q+Θ∗TGψ(T ) + Θ∗

Tg,

λ0 = 〈Qψ, ψ〉+ 2〈q, ψ〉+ 〈Gψ(T ), ψ(T )〉+ 2〈g, ψ(T )〉,

14

and the open-loop optimal control is given by

(4.6) u = −R−1(Θ∗(QX + q) + Θ∗

T [GX(T ) + g])= −R−1

[Θ∗QX +Θ∗

TGX(T )]−R−1

[Θ∗q+Θ∗

Tg],

or more precisely,

(4.7)u(t) = −R(t)−1

[ ∫ T

t

Ψ(τ, t)⊤Q(τ)X(τ)dτ +Ψ(T, t)⊤GX(T )]

−R(t)−1[ ∫ T

t

Ψ(τ, t)⊤q(τ)dτ +Ψ(T, t)⊤g], a.e. t ∈ [0, T ].

In the above, Ψ(· , ·), which is defined by (2.20), characterizes the control system and Q(·), q(·), G and g areall known a priori. The only unrealistic terms are {X(τ)

∣∣ τ ∈ [t, T ]} and X(T ) on the right-hand side ofthe above, since at the time t of determining the value u(t) of u(·), these are not available.

The idea of getting a feasible representation of the optimal control is to introduce the following simpledecomposition for the state trajectory: For any σ ∈ [0, T ),

X(t) = ψ(t) +

∫ t

0

Ψ(t, s)u(s)1[0,σ)(s)ds+

∫ t

0

Ψ(t, s)u(s)1[σ,T ](s)ds

= ψ(t) +

∫ t∧σ

0

Ψ(t, s)u(s)ds+

∫ t

t∧σ

Ψ(t, s)u(s)ds ≡ Xσ(t) +

∫ t

t∧σ

Ψ(t, s)u(s)ds, a.e. t ∈ [0, T ].

Also,

X(T ) = ψ(T ) +

∫ σ

0

Ψ(T, s)u(s)ds+

∫ T

σ

Ψ(T, s)u(s)ds ≡ Xa(σ) +

∫ T

σ

Ψ(T, s)u(s)ds.

Here, Xσ(t) and Xa(σ) do not use the information of u(·) beyond σ. With such a decomposition, we canrewrite (4.7) as follows:

(4.8)

u(t) = −R(t)−1[ ∫ T

t

Ψ(τ, t)⊤Q(τ)(Xσ(τ) +

∫ τ

τ∧σ

Ψ(τ, s)u(s)ds)dτ

+Ψ(T, t)⊤G(Xa(σ) +

∫ T

σ

Ψ(T, s)u(s)ds)]

−R(t)−1[ ∫ T

t

Ψ(τ, t)⊤q(τ)dτ +Ψ(T, t)⊤g], a.e. t ∈ [0, T ].

By letting σ = t in the above, we obtain

(4.9)

u(t) = −R(t)−1[ ∫ T

t

Ψ(τ, t)⊤Q(τ)Xt(τ)dτ +Ψ(T, t)⊤GXa(t)]

−R(t)−1[ ∫ T

t

Ψ(τ, t)⊤Q(τ)

∫ τ

τ∧t

Ψ(τ, s)u(s)dsdτ +Ψ(T, t)⊤G

∫ T

t

Ψ(T, s)u(s)ds]

−R(t)−1[ ∫ T

t

Ψ(τ, t)⊤q(τ)dτ +Ψ(T, t)⊤g], a.e. t ∈ [0, T ].

According to the definition of (Xt(·), Xa(t)), no information of u(·) beyond t is used, or they are non-anticipating. Therefore, our goal is to rewrite the last two terms on the right-hand side to be non-anticipating.

To achieve our goal, let us introduce a family of projection operators Πσ : U → U by the following:

(4.10) [Πσu(·)](t) = 1[0,σ)(t)u(t), t ∈ [0, T ],

with σ ∈ [0, T ) being the parameter. Clearly,

(4.11) [(I −Πσ)u(·)](t) = 1[σ,T ](t)u(t), t ∈ [0, T ].

15

Both Πσ and I −Πσ are idempotents on U . Moreover, for any M(·) ∈ L∞(0, T ;Rk×m),

M(t)[Πσu(·)](t) =M(t)1[0,σ)(t)u(t) = 1[0,σ)(t)M(t)u(t) = [ΠσM(·)u(·)](t), t ∈ [0, T ].

This means that Πσ commutes with multiplication operators. We shall call Πσ a causal projection. Now, let

Uσ = R(I −Πσ

)= (I −Πσ)U ,

which is a closed subspace of U , and define a parameterized operator Λσ ∈ L (Uσ) by

(4.12) Λσ = (I −Πσ)Λ∣∣Uσ.

Clearly, (I − Πσ)Λ(I − Πσ) is a natural extension of Λσ, with the value being 0 on U ⊥σ . We have the

following simple lemma.

Lemma 4.1. Suppose that (A5) holds. Then for any given σ ∈ [0, T ), the operator Λσ ∈ L (Uσ) isself-adjoint and positive definite on Uσ. Moreover

(4.13) Λ−1σ = (I −Πσ)Λ

−1∣∣Uσ

; ‖(Λσ)−1‖L (Uσ) 6

1

δ, ∀σ ∈ [0, T ).

Proof. For any u, v ∈ Uσ, we have

〈Λσu, v〉Uσ= 〈Λσu, v〉U = 〈(I −Πσ)Λu, v〉U = 〈Λu, (I −Πσ)v〉U = 〈Λu, v〉U= 〈u,Λv〉U = 〈(I −Πσ)u,Λv〉U = 〈u,Λσv〉U = 〈u,Λσv〉Uσ

.

Thus, Λσ is self-adjoint on Uσ. Moreover, there exists a constant δ > 0 such that for any u ∈ Uσ,

〈Λσu, u〉Uσ= 〈Λ(I −Πσ)u, (I −Πσ)u〉U > δ‖(I −Πσ)u‖2U = δ‖u‖2Uσ

.

Consequently, Λσ is boundedly invertible. Applying the above to (Λσ)−1u, one has

δ‖(Λσ)−1u‖2Uσ

6 〈Λσ(Λσ)−1u, (Λσ)

−1u〉Uσ= 〈u, (Λσ)

−1u〉Uσ6 ‖u‖Uσ

‖(Λσ)−1u‖Uσ

.

Then the second estimate in (4.13) follows.

From the above, we see that

(4.14)Λσ(I −Πσ) = (I −Πσ)Λ(I −Πσ), Λ−1

σ (I −Πσ) = (I −Πσ)Λ−1(I −Πσ),

(I −Πσ)Λ(I −Πσ)Λ−1(I −Πσ) = I −Πσ, (I −Πσ)Λ

−1(I −Πσ)Λ(I −Πσ) = I −Πσ.

Now, we introduce the following auxiliary trajectory:

(4.15) Xa(t) = ψ(T ) +

∫ t

0

Ψ(T, s)u(s)ds, t ∈ [0, T ],

which catches the anticipating information of the free term ψ(T ) and the dynamic system represented byΨ(T, ·) which are assumed to be a priori known1. Note that no anticipating information of the control isinvolved. At the same time, for each σ ∈ [0, T ), we introduce the following Causal trajectory:

(4.16) Xσ(t) = ψ(t) +

∫ t

0

Ψ(t, s)(Πσu)(s)ds = ψ(t) +

∫ t∧σ

0

Ψ(t, s)u(s)ds, t ∈ [0, T ].

This trajectory truncates the control up to time moment σ which is allowed to be smaller than t. It is clearthat

(4.17) Xσ(t) =

X(t), t ∈ [0, σ],

ψ(t) +

∫ σ

0

Ψ(t, s)u(s)ds, t ∈ [σ, T ];

1In time-varying LQ problems, the differential Riccati equation is solved on the whole time interval and all information of

the system and the cost functional through the coefficients and the weights are allowed to be used. The situation here is similar.

16

and

(4.18) Xt(T ) = Xa(t), t ∈ [0, T ].

We point out that both Xσ(·) and Xa(·) can be running at the same time as the system is running. Hence,t 7→ (Xt(·), Xa(t)) is non-anticipating. We now prove the main result of this section.

Theorem 4.2. Let (A5) hold. Then the unique open-loop optimal control u(·) admits the followingrepresentation:

(4.19)u(t) = −R(t)−1

[I − (Λ−R)(I −Πt)Λ

−1(I −Πt)]([

Θ∗QXt(·)](t) +

[Θ∗

TGXa(t)

](t)

)

−R(t)−1[I − (Λ−R)(I −Πt)Λ

−1(I −Πt)]([

Θ∗q(·)](t) +

[Θ∗

Tg](t)

), a.e. t ∈ [0, T ],

where Xt(·) and Xa(·) are the causal and auxiliary trajectories corresponding to the optimal pair (X(·), u(·)).Proof. By the definition of Xσ(·) and Xa(·), we have the following: for any σ ∈ [0, T ),

(4.20)X(t) = ψ(t) +

∫ t

0

Ψ(t, s){(Πσu)(s) + [(I −Πσ)u](s)

}ds = Xσ(t) + [Θ(I −Πσ)u](t), t ∈ [0, T ],

X(T ) = ψ(T ) +

∫ T

0

Ψ(T, s){(Πσu)(s) + [(I −Πσ)u](s)

}ds = Xa(σ) + ΘT (I −Πσ)u.

Thus, the open-loop optimal control can be written as follows (see (4.6)):

(4.21)

u = −R−1[Θ∗QX +Θ∗

TGX(T )]−R−1(Θ∗q+Θ∗

Tg)

= −R−1(Θ∗Q,Θ∗TG)

(Xσ +Θ(I −Πσ)u

Xa(σ) + ΘT (I −Πσ)u

)−R−1(Θ∗q+Θ∗

Tg)

= −R−1(Θ∗Q,Θ∗TG)

(Xσ

Xa(σ)

)−R−1(Λ−R)(I −Πσ)u −R−1(Θ∗q+Θ∗

Tg).

This leads to

[I +R−1(Λ−R)(I −Πσ)

]u = −R−1(Θ∗Q,Θ∗

TG)

(Xσ

Xa(σ)

)−R−1(Θ∗q+Θ∗

Tg),

which is equivalent to

[Λ(I −Πσ) +RΠσ

]u = −(Θ∗Q,Θ∗

TG)

(Xσ

Xa(σ)

)− (Θ∗q+Θ∗

Tg).

Applying (I −Πσ) to the above gives

Λσ(I −Πσ)u = (I −Πσ)Λ(I −Πσ)u = −(I − Πσ)(Θ∗Q,Θ∗

TG)

(Xσ

Xa(σ)

)− (I −Πσ)(Θ

∗q+Θ∗Tg).

Thus,

(I −Πσ)u = −Λ−1σ (I −Πσ)(Θ

∗Q,Θ∗TG)

(Xσ

Xa(σ)

)−Λ−1

σ (I −Πσ)(Θ∗q+Θ∗

Tg).

Substituting the above into (4.21), one obtains

(4.22)

u = −R−1(Θ∗Q,Θ∗TG)

(Xσ

Xa(σ)

)−R−1(Θ∗q+Θ∗

Tg)

−R−1(Λ−R)[−Λ−1

σ (I −Πσ)(Θ∗Q,Θ∗

TG)

(Xσ

Xa(σ)

)−Λ−1

σ (I −Πσ)(Θ∗q+Θ∗

Tg)]

= −R−1[I − (Λ−R)Λ−1

σ (I −Πσ)][(Θ∗Q,Θ∗

TG)

(Xσ

Xa(σ)

)+ (Θ∗q+Θ∗

Tg)].

Setting σ = t, we obtain our conclusion, making use of (4.14).

The above gives a causal state feedback representation for the open-loop optimal control in an abstractfrom. The appearance of Λ−1 makes the result hard to use since Λ is a complicated nonlocal operator.Hence, our next goal is to make the representation more explicitly accessible. We will achieve this goal innext section.

17

5 Representation via Fredholm Integral Equations.

Based on the result given in Theorem 4.2, we will now focus on the further manipulation of the abstractoperator Λ−1. We want to convert it into another feedback gain operator which can be accessed in acomputational manner. To this aim, for each σ ∈ [0, T ), we define Mσ : [0, T ] × [0, T ] → R

m×m by thefollowing:

(5.1)Mσ(t, s) = −R−1(t)

([I − (Λ−R)(I −Πσ)Λ

−1(I − Πσ)](Θ∗Q,Θ∗

TG)

(Ψ(· , s)Ψ(T, s)

))(t),

(t, s) ∈ [0, T ]× [0, T ].

Note that Ψ(· , ·) is extended to be zero in ([0, T ]× [0, T ])\∆. We would like to find an equation forMσ(· , ·),which is easier to be used.

Lemma 5.1. For any σ ∈ [0, T ), the following Fredholm integral equation

(5.2)R(t)Mσ(t, s) = −

∫ T

σ

(∫ T

t∨ξ

Ψ(τ, t)⊤Q(τ)Ψ(τ, ξ)dτ +Ψ(T, t)⊤GΨ(T, ξ))Mσ(ξ, s)dξ

−∫ T

t

Ψ(τ, t)⊤Q(τ)Ψ(τ, s)dτ −Ψ(T, t)⊤GΨ(T, s), (t, s) ∈ [0, T ]× [0, T ],

admits a unique solution Mσ(·, ·), which is given by the expression in (5.1).

Proof. Note that (by (4.14)) and the fact that RΠσ = ΠσR,

(5.3)

(I −Πσ)R−1

[I − (Λ−R)(I −Πσ)Λ

−1(I −Πσ)]

= (I −Πσ)R−1

[(I −Πσ)− (I −Πσ)(Λ−R)(I −Πσ)Λ

−1(I −Πσ)]

= (I −Πσ)R−1

[(I −Πσ)R(I −Πσ)Λ

−1(I −Πσ)]= (I −Πσ)Λ

−1(I −Πσ).

Thus,

Mσ(t, s) = −R−1(t){[I − (Λ−R)(I −Πσ)Λ

−1(I −Πσ)](Θ∗Q,Θ∗

TG)

(Ψ(· , s)Ψ(T, s)

)}(t)

= −R−1(t)([I − (Λ−R)(I −Πσ)R

−1(I − (Λ−R)(I −Πσ)Λ

−1(I −Πσ))]

(Θ∗Q,Θ∗TG)

(Ψ(· , s)Ψ(T, s)

))(t)

= −R−1(t)((Θ∗Q,Θ∗

TG)

(Ψ(· , s)Ψ(T, s)

)+ (Λ−R)(I −Πσ)Mσ(·, s)

)(t)

= −R−1(t)([

Θ∗QΨ(· , s)](t) +

[Θ∗

TGΨ(T, s)](t)

)

−R−1(t)((Θ∗QΘ+Θ∗

TGΘT )(I −Πσ)Mσ(· , s)), (t, s) ∈ [0, T ]× [0, T ].

This is equivalent to the following:

Mσ(t, s) = −R−1(t)( ∫ T

t

Ψ(τ, t)⊤Q(τ)Ψ(τ, s)dτ +Ψ(T, t)⊤GΨ(T, s))

−R−1(t)

∫ T

t

Ψ(τ, t)⊤Q(τ)

∫ τ

0

Ψ(τ, ξ)1[σ,T ](ξ)Mσ(ξ, s)dξdτ

−R−1(t)Ψ(T, t)⊤G

∫ T

0

Ψ(T, ξ)1[σ,T ](ξ)Mσ(ξ, s)dξ

= −R−1(t)

∫ T

σ

(∫ T

t∨ξ

Ψ(τ, t)⊤Q(τ)Ψ(τ, ξ)dτ +Ψ(T, t)⊤GΨ(T, ξ))Mσ(ξ, s)dξ

−R−1(t)( ∫ T

t

Ψ(τ, t)⊤Q(τ)Ψ(τ, s)dτ +Ψ(T, t)⊤GΨ(T, s)), (t, s) ∈ [0, T ]× [0, T ].

18

This means that Mσ(·, ·) is a solution to the Fredholm equation (5.2). Now, for the uniqueness, it suffices toshow that if

R(t)Mσ(t, s) +

∫ T

σ

(∫ T

t∨ξ

Ψ(τ, t)⊤Q(τ)Ψ(τ, ξ)dτ +Ψ(T, t)⊤GΨ(T, ξ))Mσ(ξ, s)dξ = 0,

then Mσ(t, s) = 0, (t, s) ∈ [0, T ]× [0, T ]. Note that

0 = R(t)Mσ(t, s) +((Λ−R)(I −Πσ)Mσ(·, s)

)(t)

=([

RΠσ +Λ(I −Πσ)]Mσ(·, s)

)(t), (t, s) ∈ [0, T ]× [0, T ].

Applying (I −Πσ) to the above, one has

0 =[(I −Πσ)Λ(I −Πσ)Mσ(·, s)

](t) =

[Λσ(I −Πσ)Mσ(·, s)

](t), (t, s) ∈ [0, T ]× [0, T ].

By the invertibility of Λσ, one has((I − Πσ)Mσ(·, s)

)(t) = 0, (t, s) ∈ [0, T ] × [0, T ]. Consequently,

R(t)Mσ(t, s) = 0, (t, s) ∈ [0, T ] × [0, T ]. Hence, it follows from the invertibility of R that Mσ(t, s) = 0,(t, s) ∈ [0, T ]× [0, T ] completing the proof.

We now prove the following theorem.

Theorem 5.2. Let (A5) hold. Let (X(·), u(·)) be the open-loop optimal pair and (Xσ(·), Xa(·)) be thecorresponding truncation and auxiliary trajectories. Then the open-loop optimal control u(·) admits thefollowing representation:

(5.4)

u(t) = −R(t)−1([

Θ∗QXt(·)](t) +

[Θ∗

TGXa(t)

](t)

)−R(t)−1

([Θ∗q(·)

](t) +

[Θ∗

T g](t)

)

−∫ T

t

Mt(t, s)R(s)−1

([Θ∗QXt(·)

](s) +

[Θ∗

TGXa(t)

](s)

)ds

−∫ T

t

Mt(t, s)R(s)−1

([Θ∗q(·)

](s) +

[Θ∗

T g](s)

)ds, a.e. t ∈ [0, T ],

where Mσ(·, ·) is the unique solution of Fredholm equation (5.2).

Proof. First, similar to (5.3), we have

(5.5)

[(I −Πt)R

−1 − (I −Πt)Λ−1(I −Πt)(Λ−R)R−1

](I −Πt)

=[(I −Πt)− (I −Πt)Λ

−1(I −Πt)(Λ(I −Πt)−R

)]R−1(I −Πt)

=[(I −Πt)− (I −Πt) + (I −Πt)Λ

−1(I −Πt)R]R−1(I −Πt) = (I −Πt)Λ

−1(I −Πt).

Let us denote

Γ(t) = (Θ∗Q,Θ∗TG)

(Xt

Xa(t)

)(t) + (Θ∗,Θ∗

T )

(q(·)g

)(t), a.e. t ∈ [0, T ].

Then,

u(t) = −R(t)−1[I − (Λ−R)(I −Πt)Λ

−1(I −Πt)]Γ(t)

= −R−1Γ(t) +R−1(Λ−R)[(I −Πt)R

−1 − (I −Πt)Λ−1(I −Πt)(Λ−R)R−1

](I −Πt)Γ(t)

= −R−1Γ(t) +R−1(Θ∗Q,Θ∗TG)

(ΘΘT

)(I −Πt)R

−1Γ(t)

−R−1(Λ−R)(I −Πt)Λ−1(I −Πt)(Θ

∗Q,Θ∗TG)

(ΘΘT

)(I −Πt)R

−1Γ(t)

= −R−1Γ(t) +R−1[I − (Λ−R)(I −Πt)Λ

−1(I −Πt)](Θ∗Q,Θ∗

TG)

(ΘΘT

)(I −Πt)R

−1Γ(t)

19

= −R−1[(Θ∗Q,Θ∗

TG)

(Xt

Xa(t)

)(t) + (Θ∗,Θ∗

T )

(q(·)g

)(t)

]

+R−1[I − (Λ−R)(I −Πt)Λ

−1(I −Πt)](Θ∗Q,Θ∗

TG)

·∫ T

t

(Ψ(·, s)Ψ(T, s)

)R(s)−1

[([Θ∗QXt(·) + Θ∗

TGXa(t)

)(s) +

[Θ∗q(·)

](s) +

[Θ∗

Tg](s)

]ds

= −R(t)−1([

Θ∗QXt(·)](t) +

[Θ∗

TGXa(t)

](t)

)−∫ T

t

Mt(t, s)R(s)−1

([Θ∗QXt(·)

](s) +

[Θ∗

TGXa(t)

](s)

)ds

−R(t)−1([

Θ∗q(·)](t) +

[Θ∗

T g](t)

)−∫ T

t

Mt(t, s)R(s)−1

([Θ∗q(·)

](s) +

[Θ∗

T g](s)

)ds, a.e. t ∈ [0, T ].

This completes the proof.

Now, we return to the general case, i.e., S(·) and ρ(·) are not necessarily zero. In this case, we summarizethe result as follows:

(5.6)

A(t, s) = A(t, s)−B(t, s)R(s)−1S(s), ϕ(t) = ϕ(t)−∫ t

0

B(t, s)R(s)−1ρ(s)

(t− s)1−βds,

Q(s) = Q(s)− S(s)⊤R(s)−1S(s), q(s) = q(s)− S(s)⊤R(s)−1ρ(s),

Φ(t, s) =A(t, s)

(t− s)1−β+

∫ t

s

A(t, τ)Φ(τ, s)

(t− τ)1−βdτ,

Ψ(t, s) =B(t, s)

(t− s)1−β+

∫ t

s

Φ(t, τ)B(τ, s)

(τ − s)1−βdτ, ψ(t) = ϕ(t) +

∫ t

0

Φ(t, s)ϕ(s)ds,

(Θv)(t) =

∫ t

0

Ψ(t, s)v(s)ds, v ∈ U , t ∈ [0, T ], ΘT v =

∫ T

0

Ψ(T, s)v(s)ds = (Θv)(T ), v ∈ U .

The truncation and auxiliary trajectories are defined by

(5.7)Xσ(t) = ψ(t) +

∫ t∧σ

0

Ψ(t, s)(u(s) +R(s)−1

[S(s)X(s) + ρ(s)

])ds,

Xa(t) = ψ(T ) +

∫ t

0

Ψ(T, s)(u(s) +R(s)−1

[S(s)X(s) + ρ(s)

])ds,

t ∈ [0, T ].

The corresponding Fredholm equation reads

(5.8)R(t)Mσ(t, s) = −

∫ T

σ

(∫ T

t∨ξ

Ψ(τ, t)⊤Q(τ)Ψ(τ, ξ)dτ + Ψ(T, t)⊤GΨ(T, ξ))Mσ(ξ, s)dξ

−∫ T

t

Ψ(τ, t)⊤Q(τ)Ψ(τ, s)dτ + Ψ(T, t)⊤GΨ(T, s), (t, s) ∈ [0, T ]× [0, T ].

Then we can state the following result whose proof is clear.

Theorem 5.3. Let (A2)–(A4) hold. Let (X(·), u(·)) be the open-loop optimal pair of Problem (P). Thenthe open-loop optimal control u(·) admits the following causal state feedback representation:

(5.9)

u(t) = −R(t)−1[S(t)X(t) + ρ(t)

]−R(t)−1

([Θ∗QXt(·)

](t) +

[Θ∗

TGXa(t)

](t)

)

−∫ T

t

Mt(t, s)R(s)−1

([Θ∗QXt(·)

](s) +

[Θ∗

TGXa(t)

](s)

)ds

−∫ T

t

Mt(t, s)R(s)−1

([Θ∗q(·)

](s) +

[Θ∗

T g](s)

)ds,

−R(t)−1([

Θ∗q(·)](t) +

[Θ∗

T g](t)

), a.e. t ∈ [0, T ],

where Θ, ΘT , Mσ(·, ·), Q(·), and q(·) are given by (5.6) and (5.8), Xσ(·) and Xa(·) are defined by (5.7) withu(·) being replaced by u(·).

20

Since the general Volterra integral equation does not have a semigroup evolutionary property, the directfeedback implementation of the optimal control in terms of the actual trajectory X(τ), a.e. τ ∈ [0, T ], isnot possible, because the future information of the function ψ is not counted. In view of this, the causalstated feedback representation is the best that can be hoped for. Because the auxiliary trajectory dependson ψ(T ), one might also call the above semi-causal state feedback representation as in [40].

6 An Iteration Scheme for the Fredholm Integral Equation.

In this section, we will briefly present a possible numerical scheme which is applicable to solve Fredholm inte-gral equation (5.2). This will, in principle, make the approach presented in the previous sections practicallyfeasible.

During the period 1960–1990, there has been much work on developing and analyzing numerical methodsfor solving linear Fredholm integral equations of the second kind. The Galerkin and collocation methodsare the well-established numerical methods (see [6]). Also, it is known that both the iterated Galerkinand the iterated collocation methods exhibit a higher order of convergence than the Galerkin method andcollocation methods, respectively (see, for example [42]). Long–Nelakanti [34] proposed an efficient iterationalgorithm having much higher order of convergence, while they need less additional computational effortsfor the implementation. Making use of the similar idea of [34], we aim to obtain an efficient iteration schemefor the Fredholm integral equation (5.2). Let us make this more precise now. To this end, we denote

K(t, ξ) = −R(t)−1(∫ T

t∨ξ

Ψ(τ, t)⊤Q(τ)Ψ(τ, ξ)dτ +Ψ(T, t)⊤GΨ(T, ξ)), (t, ξ) ∈ [0, T ]× [0, T ],

and

f(t, s) = −R(t)−1(∫ T

t

Ψ(τ, t)⊤Q(τ)Ψ(τ, s)dτ +Ψ(T, t)⊤GΨ(T, s)), (t, s) ∈ [0, T ]× [0, T ].

Next, we denote M = L2(0, T ;Rm×m), and for any σ ∈ [0, T ), define the integral operator Kσ : M → M

by the following:

Kση(t) =

∫ T

σ

K(t, ξ)η(ξ)dξ, t ∈ [0, T ], ∀η ∈ M .

Then, for each σ ∈ [0, T ), (5.2) can be reconsidered as the following: for each s ∈ [0, T ],

(I − Kσ)Mσ(t, s) = f(t, s), t ∈ [0, T ].

We introduce a partition π : 0 = s0 < s1 < s2 < · · · < sN = T of [0, T ]. For each si, 0 6 i 6 N , we considerthe following equation:

(6.1) (I − Kσ)Mσ(t, si) = f(t, si), t ∈ [0, T ].

By assumption (A3), we can easily obtain that the integral operator Kσ is a compact linear operator on M .Indeed,

∫ T

0

∫ T

0

∣∣∣K(t, ξ)∣∣∣2

dξdt

6 K

∫ T

0

∫ T

0

∣∣∣∫ T

t∨ξ

Ψ(τ, t)⊤Q(τ)Ψ(τ, ξ)dτ∣∣∣2

dξdt+K

∫ T

0

∫ T

0

∣∣∣Ψ(T, t)⊤GΨ(T, ξ)∣∣∣2

dξdt

6 K

∫ T

0

∫ T

0

[ ∫ T

t

|Ψ(τ, t)⊤|2dτ ·∫ T

ξ

|Ψ(τ, ξ)|2dτ]dξdt+K

∫ T

0

∫ T

0

1

(T − t)2(1−β)

1

(T − ξ)2(1−β)dξdt

6 K

∫ T

0

∫ T

0

[ ∫ T

t

1

(τ − t)2(1−β)dτ ·

∫ T

ξ

1

(τ − ξ)2(1−β)dτ

]dξdt+K

∫ T

0

1

(T − t)2(1−β)dt <∞,

which implies that K(· , ·) ∈ L2((0, T ) × (0, T );Rm×m). Then, it is easy to see that Kσ is compact on M

(see, for example Theorem 6.12 in [13]). For each si, 0 6 i 6 N, the Fredholm alternative theorem then

21

guarantees the existence of a unique solution of (6.1) in M (see, for example Theorem 1.3.1 in [6]). Let{Mn : n > 1} be a sequence of increasing finite dimensional subspaces of M . Let Pn : M → Mn be theorthogonal projection operator (see, for example Section 3.1.2 in [6]). Then, for each si, 0 6 i 6 N, theGalerkin approximation is the solution of

(6.2) Mσn(t, si) = Pnf(t, si) + PnKσMσn(t, si), t ∈ [0, T ].

We can show that Pny → y, as n→ ∞, for all y ∈ M (see, for example Section 3.3.1 in [6]). Then, it followsfrom the compactness of Kσ that

(6.3) limn→∞

‖Kσ − PnKσ‖L (M ) = 0,

(see, for example Lemma 3.1.2 in [6] or [42]). Then, by (6.3), one has that (I−PnKσ)−1 exist and uniformly

bounded for sufficiently large n, and the approximation scheme is uniquely solvable (see, for example Theorem3.1.1 in [6]).

For each si, 0 6 i 6 N, the iterated Galerkin approximation (see, [42]) may be defined by

(6.4) Mσn(t, si) = f(t, si) + KσMσn(t, si), t ∈ [0, T ].

Applying Pn to both side of (6.4), we have PnMσn =Mσn, and hence for each si, 0 6 i 6 N, it holds that

Mσn(t, si) = f(t, si) + KσPnMσn(t, si), t ∈ [0, T ].

One can show that the iterated Galerkin scheme (6.4) can converge more rapidly than the rate achieved bythe approximation (6.2) (see, [26] and [42]). Further, Long–Nelakanti [34] proposed a more efficient iterationalgorithm which even has much higher order of convergence than the iterated Galerkin scheme. The iteration

algorithm is as follows: for each si, 0 6 i 6 N, set M(0)σn (t, si) = Mσn(t, si), t ∈ [0, T ], then for k = 0, 1, . . . ,

step 1 : M (k)σn (t, si) = f(t, si) + KσM

(k)σn (t, si), t ∈ [0, T ],

step 2 :˜M

(k)

σn (t, si) = f(t, si) + KσM(k)σn (t, si), t ∈ [0, T ],

step 3 : g(k)n (t, si) =˜M

(k)

σn (t, si)− M (k)σn (t, si), t ∈ [0, T ],

step 4 : for each si, 0 6 i 6 N, seeking a unknown function e(k)n (t, si), t ∈ [0, T ] by solving the equation

(I − PnKσ)e(k)n (t, si) = Png

(k)n (t, si), t ∈ [0, T ],

step 5 : M (k+1)σn (t, si) = Kσe

(k)n (t, si) +

˜M

(k)

σn (t, si), t ∈ [0, T ].

By the superconvergence rates for every step of iteration, we obtain that for each k = 0, 1, . . . , for each si,0 6 i < N,

M (k+1)σn (·, si) →Mσ(·, si) in M , as n→ ∞.

For more details, see [34]. Thus, on a subsequence, still denoted in the same way, for each n = 0, 1, . . . ,

(6.5) ‖M (k+1)σn (·, si)−Mσ(·, si)‖M 6

1

n.

Now, for k = 0, 1, . . . , let

MNσn(t, s) =

N∑

i=1

[ si − s

si − si−1M (k+1)

σn (t, si−1) +s− si−1

si − si−1M (k+1)

σn (t, si)]1[si−1,si)(s), (t, s) ∈ [0, T ]× [0, T ].

Then, for k = 0, 1, . . . ,

|MNσn(t, s)−Mσ(t, s)|

6

N∑

i=1

[ si − s

si − si−1

(|M (k+1)

σn (t, si−1)−Mσ(t, si−1)|+ |Mσ(t, si−1)−Mσ(t, s)|)

+s− si−1

si − si−1

(|M (k+1)

σn (t, si)−Mσ(t, si)|+ |Mσ(t, si)−Mσ(t, s)|)]

1[si−1,si)(s)

22

6

N∑

i=1

[ si − s

si − si−1|Mσ(t, si−1)−Mσ(t, s)|+

s− si−1

si − si−1|Mσ(t, si)−Mσ(t, s)|

]1[si−1,si)(s)

+K

N∑

i=1

[|M (k+1)

σn (t, si−1)−Mσ(t, si−1)|+ |M (k+1)σn (t, si)−Mσ(t, si)|

]1[si−1,si)(s), (t, s) ∈ [0, T ]× [0, T ].

Hence, for k = 0, 1, . . . , for each s ∈ [0, T ),

(6.6)

‖MNσn(·, s)−Mσ(·, s)‖M

6 K

N∑

i=1

[ si − s

si − si−1‖Mσ(·, si−1)−Mσ(·, s)‖M +

s− si−1

si − si−1‖Mσ(·, si)−Mσ(·, s)‖M

]1[si−1,si)(s)

+KN∑

i=1

[‖M (k+1)

σn (·, si−1)−Mσ(·, si−1)‖M + ‖M (k+1)σn (·, si)−Mσ(·, si)‖M

]1[si−1,si)(s).

We need a little more assumption which we now introduce.

(A6) There exists a modulus of continuity ω(·) such that

|B(t, s′)−B(t, s)| 6 ω(s′ − s), (t, s′), (t, s) ∈ ∆.

Then, pick any s0 ∈ [0, T ), making use of the similar argument of the proof for Corollary 2.2, (ii), we

obtain that Ψ(·, s) is continuous in L2(0, T ;Rn×m) at s0, and Ψ(T, s) is continuous at s0. Since −R−1[I −

(Λ − R)Λ−1σ (I − Πσ)

]Θ∗Q is a bounded linear operator from L2(0, T ;Rn×m) to M and −R−1

[I − (Λ −

R)Λ−1σ (I − Πσ)

]Θ∗

TG is a bounded linear operator from Rn×m to M , it holds that Mσ(· , s) is continuous

in M at s0 ∈ [0, T ).

Consequently, for k = 0, 1, . . ., for each s ∈ [0, T ), we let N → ∞ and n→ ∞. Then, by (6.5) and (6.6),we obtain that MN

σn(· , s) →Mσ(· , s) in M . Thus, we obtain a feasible numerical scheme which can be usedto solve Fredholm integral equation (5.2).

7 Concluding Remarks.

In this paper, we have studied an optimal control problem, with the state equation being a linear Volterraintegral equation having a singular kernel. The cost functional is a form of quadratic plus linear terms ofthe state and the control. Under proper conditions, the open-loop optimal control uniquely exists. However,normally, the open-loop optimal control is not of non-anticipating form. Our main goal is to obtain a causalstate feedback representation of the open-loop optimal control. In doing that, we have introduced a Fredholmintegral equation which plays a role of Riccati equation in the standard LQ problems for ODE systems. Tomake our result practically feasible (in principle), we have briefly presented a possible numerical scheme forcomputing the solution to the Fredholm integral equation.

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