The quadratic regulator problem and the Riccati

arXiv:1610.07127v1 [math.OC] 23 Oct 2016

equation for a process governed by a linear Volterra
integrodifferential equations∗
L. Pandolfi†
March 31, 2018

Abstract: In this paper we study the quadratic regulator problem for a process
governed by a Volterra integral equation in IRn . Our main goal is the proof that
it is possible to associate a Riccati differential equation to this quadratic control
problem, which leads to the feedback form of the optimal control. This is in contrast
with previous papers on the subject, which confine themselves to study the Fredholm
integral equation which is solved by the optimal control.
Key words:Quadratic regulator problem, Volterra integrodifferential equations, Riccati equation
AMS classification: 93B22, 45D05, 49N05, 49N35

1

Introduction

The quadratic regulator problem for control processes regulated by linear differential
equations both in finite and infinite dimensional spaces has been at the center
of control theory at least during the last eighty years, after the proof that the
synthesis of dissipative systems amounts to the study of a (singular) quadratic
∗ This

papers fits into the research program of the GNAMPA-INDAM and has been written in

the framework of the “Groupement de Recherche en Contrôle des EDP entre la France et l’Italie
(CONEDP-CNRS)”.
† Dipartimento di Scienze Matematiche “Giuseppe Luigi Lagrange”, Politecnico di Torino, Corso
Duca degli Abruzzi 24, 10129 Torino, Italy (luciano.pandolfi@polito.it)

1

control problem (see [2]). In this period, the theory reached a high level of maturity
and the monographs [1, 8] contain the crucial ideas used in the study of the quadratic
regulator problems for lumped and distributed systems (see [3, 4, 11, 12, 13] for the
singular quadratic regulator problem for distributed systems).
In recent times, the study of controllability of systems described by Voterra
integrodifferential equations (in Hilbert spaces) has been stimulated by several applications (see [14]) while the theory of the quadratic regulator problem for these
systems is still at a basic level. In essence, we can cite only the paper [15] and
some applications of the results in this paper, see for example [7]. In these papers, the authors study a standard regulator problem for a system governed by a
Volterra integral equation (in a Hilbert space and with bounded operators. The
paper [7] and some other applications of the results in [15] studies a stochastic
system) and the synthesis of the optimal control is given by relying on the usual
variational approach and Fredholm integral equation for the optimal control. The
authors of these papers do not develop a Riccati differential equation and this is our
goal here. In order to avoid the technicalities inevitably introduce by the presence
of unbounded operators which are introduced by the action of boundary controls,
we confine ourselves to study Volterra integral equations in IRn .
The control problem we consider is described by
Z t
N (t − s)x(s) ds + Bu(t) ,
x′ =

x(0) = x0

(1)

0

where x ∈ IRn , u ∈ IRm , B is a constant n × m matrix and N (t) is a continuous
n × n matrix (extension to B = B(t) and N = N (t, s) is simple). Our goal is the
study of the minimization of the standard quadratic cost
Z T
 ∗

x (t)Qx(t) + |u(t)|2 dt + x∗ (T )Q0 x(T )

(2)

0

where Q = Q∗ ≥ 0, Q0 = Q∗0 ≥ 0.

Existence of a unique optimal control in L2 (0, T ; IRm ) for every fixed x0 ∈ IRn
is obvious.
The plan of the paper is as follows: in order to derive a Riccati differential
equation, we need a suitable “state space” in which our system evolves. In fact, a
Volterra integral equation is a semigroup system in a suitable infinite dimensional

2

space (see [10, Ch. 6]) and we could relay on this representation of the Volterra
equation to derive a theory of the Riccati equation in a standard way but the shortcoming is that the “state space” is IRn × L2 (0, +∞; IRn ) and the Riccati differential
equation so obtained should be solved in a space with infinite memory, even if the
process is considered on a finite time interval [0, T ]. We wish a “Riccati differential
equation” in a space which has a “short memory”, say of duration at most T , as
required by the optimization problem. So, we need the introduction of a different
“state space approach” to Eq. (1). This is done in Sect. 2 where, using dynamic
programming, we prove that the minimum of the cost is a quadratic form which
satisfy a (suitable version) of the Linear Operator Inequality (LOI).
Differentiability properties of the cost are studied in section 3 (using a variational
approach to the optimal control related to the arguments in [15]). The regularity
properties we obtain finally allows us to write explicitly a system of partial differential equations (with a quadratic nonlinearity) on [0, T ], which is the version of the
Riccati differential equations for our system.
We believe that the introduction of the state space in Sect. 2 is a novelty of this
paper.

2

The state of the Volterra integral equation, and
the (LOI)

According to the general definition in [9]), the state at time τ is the information
at time τ needed to uniquely solve the equation for t > τ (assuming the control is
known for t > τ ).
It is clear that if τ = 0 then the sole vector x0 is sufficient to solve equation (1)
in the future, and the state space at τ = 0 is IRn . Things are different if we solve
the equation till time τ and we want to solve it in the future. In this case, Eq. (1)
for t > τ takes the form
Z t
Z
x′ =
N (t − s)x(s) ds + Bu(t) +
τ

τ

N (t − s)x(s) ds .

(3)

0

In order to solve this equation for t > τ we must know the pair1 Xτ = (x(τ ), xτ (·))
1 Remark

on the notation: xτ = xτ (s) is a function on (0, τ ) while Xτ (upper case letter) is the

3

where xτ (s) = x(s), s ∈ (0, τ ).
Note that in order to uniquely solve (3), xτ (·) needs not be a segment of previously computed trajectory. It can be an “arbitrary” function. This observation
suggests the definition of the following state space at time τ :
Mτ2 = IRn × L2 (0, τ ; IRn )
(to be compare with the state space of differential equations with a fixed delay h
which is IRn × L2 (−h, 0; IRn )).
Eq. (3) defines, for every fixed u and τ1 > τ , a solution map from Mτ2 to Mτ21
which is affine linear and continuous. An explicit expression of this map can be
obtained easily. Let us fix an initial time τ ≥ 0. Let t ≥ τ and let Z(t, τ ) be the
n × n matrix solution of
d
Z(t, τ ) =
dt

Z

t

Z(ξ, τ )N (t − ξ) dξ ,

Z(τ, τ ) = I .

(4)

τ

Then,

x(t) = Z(t, τ )x̂ +

Z

τ

Y (t, s; τ )x̃(s) ds +

0

Z

t

Z(t − r + τ, τ )Bu(r) dr

(5)

τ

where
Y (t, s; τ ) =

Z

t

Z(t − ξ + τ, τ )N (ξ − s) dξ .

τ

This way, for every τ1 > τ we define two linear continuous transformations:
E(τ1 ; τ ) from Mτ2 to Mτ21 (when u = 0) and Λ(τ1 ; τ ) from L2 (τ, τ1 ; IRm ) to Mτ21
(when Xτ = 0), as follows:
E(τ1 ; τ )(x̂, x̃(·)) = (x(τ1 ), y)


 x(t) given by (5)
y=
 x̃(t)

if

τ < t < t1

if

t ∈ (0, τ ) .

The operator Λ(τ1 ; τ ) is defined by the same formula as E(τ1 ; τ ), but when Xτ = 0
and u 6= 0.
The evolution of the system is describe by the operator
E(t1 ; τ )Xτ + Λ(t1 ; τ )u .
pair (x(τ ), xτ ).

4

(6)

The evolutionary properties of this operator follow from the unicity of solutions
of the Volterra integral equation. Let us consider Eq. (3) on [τ, T ] with initial
condition (x̂, x̃(·)), whose solution is given by (5). Let τ1 ∈ (τ, T ) and let us consider
Eq. (3) on [τ1 , T ] but with initial condition (x(τ1 ), xτ1 ). Eq. (3) on [τ1 , T ] and this
initial condition takes the form
Z
Z t
N (t − s)x(s) ds + Bu(t) +
x′ (t) =

τ1

N (t − s)xτ1 (s) ds ,

0

τ1

and so, on [τ1 , T ] we have
Z
x′ (t) = Z(t, τ1 )x(τ1− ) +

τ1

Y (t, s; τ1 )xτ1 (s) ds +

Z

x(τ1+ ) = x(τ1− )

t

Z(t − s − τ1 , τ1 )Bu(s) ds .

τ1

0

Unicity of the solutions of the Volterra integral equation shows that, for t ∈ (τ1 , T ]
the following equality holds
E(t, τ ) (x̂, x̃) + Λ(t; τ )u = E(t, τ1 ) [E(τ1 , τ ) (x̂, x̃) + Λ(τ1 ; τ )u] + Λ(t; t1 )u .
Remark 1 The solution Z(t, τ ) of Eq. (4) solves the following Volterra integral
equation on [τ, T ]:
Z(t) = 1 +

Z

t

Z(ξ)M (t − ξ) dξ ,

τ

M (t) =

Z

t

N (s) ds .

0

The usual Picard iteration gives
Z t
Z tZ ξ
Z(t, τ ) = 1 +
M (t − ξ) dξ +
M (ξ − ξ1 ) dξ1 M (t − ξ) dξ + · · · =
τ
τ
τ
Z t
Z t Z t−s
=1+
M (t − ξ) dξ +
M (t − s − r)M (r) dr ds + · · ·
τ

τ

0

The properties of these integrals is that, once exchanged, we have
Z t
Z(t, τ ) = 1 +
H(t − s) ds
τ

where H(t) does not depend on τ and it is differentiable. It follows that the function
(τ, t) 7→ Z(t, τ ) is continuously differentiable on 0 < τ < t < T and the derivative
has continuous extension to 0 ≤ τ ≤ t ≤ T .
Now we begin our study of the quadratic regulator problem and of the Riccati
equation.

5

One of the possible ways to derive an expression of the optimal control and
possibly a Riccati differential equation for the quadratic regulator problem is via
dynamic programming. We follow this way. For every fixed τ < T we introduce
Z T

 ∗
x (t)Qx(t) + |u(t)|2 dt + x∗ (T )Q0 x(T )
Jτ (Xτ , u) =
τ

where x(t) is the solution of (3) (given by (5)) and we define
W (τ ; Xτ ) =

min

u∈L2 (τ,T ;IRm )

Jτ (Xτ , u) .

(7)

Existence of the minimum is obvious and we denote u+ (t) = u+ (t; τ, Xτ ) the optimal
control. The corresponding solution is denoted x+ (t) = x+ (t; τ, Xτ ) while we put

Xt+ = x+ (t), x+
t (·) .

Let us fix any τ1 ∈ (τ, T ) and let u(t) = u1 (t) if t ∈ (τ, τ1 ), u(t) = u2 (t) if

t ∈ (τ1 , T ), while
Xt1 = E(t, τ )Xτ +Λ(t, τ )u1

t ∈ [τ, τ1 ] ,

Xt2 = E(t, τ1 )Xτ11 +Λ(t, τ1 )u2

t ∈ [τ1 , T ] .

We noted that X(t; τ, Xτ ) given by (6) on [τ, T ] is equal to Xt1 on [τ, τ1 ] and to Xt2
on [τ1 , T ].
Let xi be the IRn component of X i . Then, for every u we have (we use the
crochet to denote the inner product instead of the more cumberstome notation
∗
x1 (t) Qx1 (t))
Z τ1



hQx1 (t), x1 (t)i + |u1 (t)|2 dt + Jτ1 Xτ11 , u2 .
(8)
W (τ, Xτ ) ≤
τ

This inequality holds for every u1 and u2 and equality holds when u1 and u2 are
restrictions of the optimal control u+ .
We keep u1 fixed and we compute the minumum of the right hand side respect
to u2 . We get the Linear Operator Inequality (LOI):
Z τ1



hQx1 (t), x1 (t)i + |u1 (t)|2 dt + W τ1 , Xτ11 .
W (τ, Xτ ) ≤

(9)

τ

This inequality holds for every control u ∈ L2 (τ, τ1 ; IRn ). Let in particular u1 be the
restriction to (τ, τ1 ) of u+ (·) = u+ (·; τ, Xτ ). Inequality (8) shows that the minimum


of Jτ1 Xτ11 , u2 cannot be strictly less then Jτ1 Xτ11 , u+ , i.e. the optimal control

of the cost Jτ1 Xτ11 , u2 is the restriction to (τ1 , T ) of u+ (t), the optimal control of

Jτ (Xτ , u).

6

Equality holds in (9) if u1 = u+ .
In conclusion, we divide with τ1 − τ (which is positive) and we find the following
inequality, which holds with equality if u = u+ :


1
1 
W τ1 ; Xτ11 − W (τ ; Xτ ) ≥ −
τ1 − τ
τ1 − τ

Z

τ

τ1




hQx1 (t), x1 (t)i + |u(t)|2 dt .

So, the following inequality holds when τ is a Lebesgue point of u(t) (every τ if u
is continuous):
lim inf
τ1

→τ +





1 
W τ1 ; Xτ11 − W (τ ; Xτ ) ≥ − hQx(τ ), x(τ )i + |u(τ )|2 . (10)
τ1 − τ

Equality holds if u = u+ and τ is a Lebesgue point of u+ and in this case we can

even replace lim inf with lim, i.e. W τ1 ; Xτ+1 is differentiable if τ is a Lebesgue
points of u+ .

The previous argument can be repeated for every τ so that the previous inequalities/equalities holds a.e. on [0, T ] and we might even replace τ with the generic
notation t.
Remark 2 If it happens that ker N (t) = S, a subspace of IRn , we might also
consider as the second component of the “state” Xτ the projection of x̃ on (any
fixed) complement of S, similar to the theory developed in [5, 6]. We dont’t pursue
this approach here.

3

The regularity properties of the value function,
the synthesis of the optimal control and the Riccati equation

We prove that W is a continuous quadratic form with smooth coefficients and
we prove that u+ (t) is continuous (so that every time t is a Lebesgue point of
u+ (t)). We arrive at this result via the variational characterization of the optimal
pair (u+ , x+ ) (x+ is the IRn -component of X + ) in the style of [15]. The standard
perturbation approach gives a representation of the optimal control (and a definition

7

of the adjoint state p(t)):
"Z
T

u+ (t) = −B ∗

#

Z ∗ (s − t + τ, τ )Qx+ (r) dr + Z ∗ (T − t + τ, τ )Q0 x+ (T ) = −B ∗ p(t)

t

(11)

where p, the function in the bracket, solves the adjoint equation
Z T
N ∗ (s − t)p(s) ds ,
p(T ) = Q0 x+ (T ) .
p′ (t) = −Qx+ (t) −

(12)

t

Note that p depends on τ and that Eq. (12) has to be solved (backward) on the
interval [τ, T ].
The simplest way to derive the differential equation (12) is to note that the
function q(t) = p(T − t) is given by
Z T
q(t) =
Z ∗ (s − T + τ + t, t)Qx+ (s) ds + Z ∗ (t + τ, τ )Q0 x+ (T ) =
T −t
t

=

Z

Z ∗ (t − r + τ, τ )Qx+ (T − r) dr + Z ∗ (t + τ, τ )Q0 x+ (T ) .

0

Comparison with (5) shows that q(t) solves
Z t
q ′ (t) =
N ∗ (t − s)q(s) ds + Qx+ (T − t) ,

q(0) = Q0 x+ (T )

0

from which the equation of p(t) is easily obtained.
We recapitulate: the equations which characterize (x+ , u+ ) when the initial time
is τ and Xτ = (x̂, x̃(·)) is the following system of equations on the interval [τ, T ]:
x′ =

Rt
τ

N (t − s)x(s) ds − BB ∗ p(t) +

p′ (t) = −Qx(t) −

RT
t

Rτ
0

N (t − s)x̃(s) ds , x(τ ) = x̂

N ∗ (s − t)p(s) ds ,

p(T ) = Q0 x(T )

u+ (t) = −B ∗ p(t) .

(13)
We replace u+ (t) = u+ (t; τ, Xτ ) in (5). The solution is x+ (t). Then we replace
the resulting expression in (11). We get the Fredholm integral equation for u+ (t):
Z T
u+ (t) + B ∗ Z ∗ (T − t + τ, τ )Q0
Z(T − r + τ, τ )Bu+ (r) dr+
τ

+ B∗

Z

T

Z ∗ (s − t + τ, τ )Q

t

= −B

∗

"

Z

s

Z(s − r + τ, τ )Bu+ (r) dr ds =

τ

∗

Z (T − t + τ, τ )Q0 F (T, τ ) +

Z

t

8

T
∗

Z (s − t + τ, τ )QF (s, τ ) ds

#

where
F (t, τ ) = Z(t, τ )x̂ +

Z

τ

Y (t, s; τ )x̃(s) ds .

0

This Fredholm integral equation has to be solved on [τ, T ].
By solving the Fredholm integral equation we find an expression for u+ (t), of
the following form:
+

+

u (t) = u (t; τ, Xτ ) = Φ1 (t, τ )x̂ +

Z

τ

Φ2 (t, s; τ )x̃(s) ds ,

t≥τ

(14)

Z2 (t, r; τ )x̃(r) dr ,

t≥τ.

(15)

0

and so also
+

+

x (t) = x (t; τ, Xτ ) = Z1 (t, τ )x̂ +

Z

τ
0

The explicit form of the matrices Φ1 (t, τ ), Φ2 (t, s; τ ), Z1 (t, τ ), Z2 (t, r; τ ) (easily derived using the resolvent operator of the Fredholm integral equation) is not needed.
The important fact is that these matrices have continuous partial derivative respect
to their arguments t, s and τ . In particular, u+ (t) = u+ (t; τ, Xτ ) is a continuous
function of t for t ≥ τ . The derivative has continuous extensions to s = τ and to
t = τ . Differentiability respect to τ follows from Remark 1.
We replace (14) and (15) in (7) and we get
W (τ ; Xτ ) =
+

Z

T

τ
Z T

Q1/2 Z1 (s, τ )x̂ + Q1/2

Z

2

τ

Z2 (s, r; τ )x̃(r) dr

ds+

0

Φ1 (s; τ )x0 +

τ

Z

τ

2

Φ2 (s, r; τ )x̃(r) dr

ds .

(16)

0

This equality shows that Xτ 7→ W (τ, Xτ ) is a continuous quadratic form of Xτ ∈
Mτ .
We use dynamic programming again, in particular the fact that u+ (·; τ1 , Xτ+1 ) is
the restriction to [τ1 , T ] of u+ (·; τ, Xτ ). Hence, for every τ1 ≥ τ we have
W (τ1 ; Xτ+1 ) =
+

Z

T

τ1
Z T

Q1/2 Z1 (s, τ1 )x+ (τ1 ) + Q1/2

Z

τ1

2

Z2 (s, r; τ1 )x+ (r) dr

ds+

0

+

Φ1 (s; τ1 )x (τ1 ) +

τ1

Z

τ1

2

Φ2 (s, r; tτ1 )x+ (r) dr

ds .

(17)

0

We simplify the notations: from now on we drop the

+

and we replace τ1 with

t but we must recall that we are computing for t ≥ τ and, when we use equality
in (9), on the optimal evolution.
9

By expanding the squares we see that W (τ1 ; Xτ1 ) has the following general form:
Z t
W (t; Xt ) = x∗ (t)P0 (t)x(t) + x∗ (t)
P1 (t, s)x(s) ds+
0

+

Z

t

P1 (t, s)x(s) ds

0

∗

x(t) +

Z tZ
0

t

x∗ (r)K(t, ξ, r)x(ξ) dξ dr .

(18)

0

For example,
P0 (t) =

Z

t

T

[Z1∗ (s, t)QZ1 (s, t) + Φ∗1 (s, t)Φ1 (s, t)] ds .

Note that P0 (t) is a selfadjoint differentiable matrix.
Now we consider the matrix K(t, ξ, r). We consider the contribution of the first
line in (18) (the contribution of the second line is similar). Exchanging the order
of integration and the names of the variables of integration, we see that
"Z
#
Z
Z Z
t

t

t

x∗ (r)K(t, ξ, r)x(ξ) dξ dr =

0

=

0

Z tZ
0

t

0

∗

x (ξ)

"Z

T

x∗ (r)

0

t

T
t

Z2∗ (s, ξ, t)QZ2 (s, r, t)

Z2∗ (s, r, t)QZ2 (s, ξ, t) ds x(ξ) dr dξ =

#

ds x(r) dξ dr =

Z tZ
0

t

x∗ (ξ)K ∗ (t, r, ξ)x(r) dξ dr

0

so that we have
K(t, ξ, r) = K ∗ (t, r, ξ)
and this matrix function is differentiable respect to its arguments t, r and ξ.
Analogously we see differentiability of P1 (t, s).
We whish a differential equations for the matrix functions P0 (t), P1 (t, s),
K(t, s, r). In order to achieve this goal, we compute the right derivative of W (t; Xt )
(and any continuous control) for t > τ and we use inequality (10). We use explicitly that equality holds in (10) when the derivative is computed along an optimal
evolution.

3.1

The Riccati equation

In order to derive a set of differential equations for the matrices P0 (t), P1 (t, s),
K(t, ξ, r) we proceed as follows: we fix (any) τ ∈ [0, T ] and the initial condition
Xτ = (x̂, x̃(·)). We consider (18) with any continuous control u(t) on [τ, T ] (the corresponding solution of the Volterra equation is x(t)). We consider the quadratic form
W with the control u(t) and the corresponding solution Xt given in in (18). In this
10

form we separate the contribution of the functions on (0, τ ) and the contribution on
Rt
[τ, t]. For example x∗ (t)P0 (t)x(t) remains unchanged while x∗ (t) 0 P1 (t, ξ)x(ξ) dξ
is written as
Z t
Z
∗
∗
x (t)
P1 (t, ξ)x(ξ) dξ = x (t)
0

τ
∗

P1 (t, s)x̃(s) ds + x (t)
0

Z

t

P1 (t, s)x(s) ds .
τ

The other addenda are treated analogously.
We obtain a function of t which is continuously differentiable. Its derivative at
t = τ is the left hand side of (10) and so it satisfies the inequality (10), with equality
if it happens that we compute with u = u+ . So, the function of u ∈ IRm


d
W (τ ; Xτ ) + u∗ (τ )u(τ )
u 7→
dt
reaches a minimum at u = u+
τ . Note that τ ∈ [0, T ] is arbitrary and so by computing
this minimum we get an expression for u+ (τ ), for every τ ∈ [0, T ].
It turns out that d W (τ ; Xτ ) + u∗ (τ )u(τ ) is the sum of several terms. Some
dt
of them do not depend on u and the minimization concerns solely the terms which
depends on u. We get (we recall that P0 (τ ) is selfadjoint)

Z τ
+
∗ ∗
∗ ∗
u (τ ) = arg min u B P0 (τ )x̂ + u B
P1 (τ, s)x̃(s) ds+
0
Z τ


∗
∗
∗
∗
+x̂ P0 (τ )Bu +
x̃ (s)P1 (τ, s) ds Bu + u u .

(19)

0

The minimization gives

Z
u+ (τ ) = −B ∗ P0 (τ )x̂ +

τ


P1 (τ, s)x̃(s) ds .

0

(20)

If the system is solved up to time t along an optimal evolution (so that x+ (t) is
equal to x̃(t) when t < τ and it is the solution which corresponds to the optimal
control for larger times) we have

Z
u+ (t) = −B ∗ P0 (τ )x+ (t) +

0

τ

P1 (t, s)x+ (s) ds



and this is the feedback form of the optimal control (compare [15]).
We repalce (20) in the brace in (19) and we see that the minimum is
Z τ
− x̂∗ P0 (τ )BB ∗ P0 (τ )x̂ − x̂∗ P0 (τ )BB ∗
P1 (τ, ξ)x̃(ξ) dξ−
0
Z τ

Z τZ τ
−
x̃∗ (r)P1∗ (τ, r) dr BB ∗ P0 (τ )x̂ −
x̃(r)P1 (τ, r)BB ∗ P1 (τ, ξ)x̃(ξ) dξ dr .
0

0

0

(21)

11

Now we compute the derivative of the function τ 7→ W (τ ; Xτ ) along an optimal
evolution and we consider its limit for t → τ +. We insert this quantity in (10), which
is an equality since we are computing the limit along an optimal evolution. We take
into account that the terms which contains u sum up to the expression (21) and we
get the following equality. In this equality, a superimposed dot denotes derivative
with respect to the variable τ :
Ṗ0 (τ ) =

d
P0 (τ ) ,
dτ

Ṗ1 (τ, ξ) =

∂
P1 (τ, ξ) ,
∂τ

K̇(τ, ξ, r) =

∂
K(τ, ξ, r) .
∂τ

The equality is:
∗

∗

∗

∗

Z

τ

− x̂ P0 (τ )BB P0 (τ )x̂ − x̂ P0 (τ )BB
P1 (τ, ξ)x̃(ξ) dξ−
0
Z τ

Z τZ τ
−
x̃∗ (r)P1∗ (τ, r) dr BB ∗ P0 (τ )x̂ −
x̃(r)P1 (τ, r)BB ∗ P1 (τ, ξ)x̃(ξ) dξ dr+
0
0
Z0 τ

Z τ
∗
∗
∗
+
x̃ (r)N (τ − r) ds P0 (τ )x̂ + x̂ Ṗ0 (τ )x̂ + x̂∗
N (τ − ξ)x̃(ξ) dξ + x̂∗ P1 (τ, τ )x̂+
0
0
Z τ
Z τ
Z τ
∗ ∗
∗
∗
+ x̂ P1 (τ, τ )x̂ +
x̃ (r)N (τ − r) dr
P1 (τ, s)x̃(s) ds + x̂∗
Ṗ1 (τ, ξ)x̃(ξ) dξ+
0
0
0
Z τ

Z τ
 Z τ

+
x̃∗ (r)Ṗ1∗ (τ, r) dr x̂ +
x̃∗ (r)P1 (τ, r) dr
N (τ − ξ)x̃(ξ) dξ +
0
0
Z0 τ

Z τ
∗
∗
+
x̃ (r)K(τ, τ, r) dr x̂ + x̂
K(τ, ξ, τ )x̃(ξ) dξ+
0
0
Z τ
Z τ
x̃∗ (r)
+
K̇(τ, ξ, r)x̃(ξ) dξ dr + x̂∗ Qx̂ = 0
0

0

The vector x̂ and the function x̃(·) are arbitrary. So, we first impose x̃(·) = 0
and x̂ arbitrary, then the converse and finally both nonzero arbitrary. We find that
the three matrix functions P0 (τ ), P1 (τ, r), K(τ, ξ, r) solve the following system of
differential equations in the arbitrary variable τ . The variables r and ξ belong to
[0, τ ] for every τ ∈ [0, T ].

P0′ (τ ) − P0 (τ )B ∗ BP0 (τ ) + Q(τ ) + P1 (τ, τ ) + P1∗ (τ, τ ) = 0
∂
P1 (τ, ξ) − P0 (τ )BB ∗ P1 (τ, ξ) + P0 (τ )N (τ − ξ) + K(τ, ξ, τ ) = 0
∂τ
∂
K(τ, ξ, r) − P1∗ (τ, r)BB ∗ P1 (τ, ξ)+
∂τ
+ P1∗ (τ, r)N (τ − ξ) + N ∗ (τ − r)P1 (τ, ξ) = 0
P0 (T ) = Q0 ,

P1 (T, ξ) = 0 ,

K(T, ξ, r) = 0
12

(22)

The final conditions are obtained by noting that when τ = T i.e. with XT =
(x̂, x̃T (·)) arbitrary in MT2 = IRn × L2 (0, T ; IRn ), the expression W (T, XT ) in (18)
is equal to JT (XT ; u) = x̂∗ Q0 x̂ for every XT .
This is the Riccati differential equation of our optimization problem.
Remark 3 We note the following facts:
• We take into account the fact that P0 is selfadjoint and K ∗ (τ, ξ, τ ) =
K(τ, τ, ξ). We compute the adjoint of the second line in (22) and we find:
∂ ∗
P (τ, r) − P1∗ (τ, r)BB ∗ P0 (τ ) + N ∗ (τ − r)P0 (τ ) + K(τ, τ, r) = 0 .
∂τ 1
• The form of the Riccati differential equations we derived for the Volterra
integral equation (1) has to be compared with the Riccati differential equation
“ in decoupled form” which was once fashionable in the study of the quadratic
regulator problem for systems with finite delays, see [16].

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