On a linear-quadratic problem with Caputo derivative - Opuscula ...

Opuscula Math. 36, no. 1 (2016), 49–68 http://dx.doi.org/10.7494/OpMath.2016.36.1.49

Opuscula Mathematica

ON A LINEAR-QUADRATIC PROBLEM WITH CAPUTO DERIVATIVE Dariusz Idczak and Stanislaw Walczak Communicated by Marek Galewski Abstract. In this paper, we study a linear-quadratic optimal control problem with a fractional control system containing a Caputo derivative of unknown function. First, we derive the formulas for the differential and gradient of the cost functional under given constraints. Next, we prove an existence result and derive a maximum principle. Finally, we describe the gradient and projection of the gradient methods for the problem under consideration. Keywords: fractional Caputo derivative, linear-quadratic problem, existence and uniqueness of a solution, maximum principle, gradient method, projection of the gradient method. Mathematics Subject Classification: 26A33, 49J15, 49K15, 49M37.

1. INTRODUCTION In this paper, we consider the following fractional linear control system ( C α Da+ x(t) = Ax(t) + Bu(t), t ∈ [a, b] a.e., x(a) = 0

(1.1)

with a quadratic performance index J(u) = J1 (u) + J2 (u) + J3 (u) 1 1 1 2 2 2 = |xu (b) − c|Rn + kxu (·) − y(·)kL2 + ku(·)kI 1−α (L2 ) , a+ 2 2 2

(1.2)

where α ∈ ( 12 , 1) is a fixed number, A ∈ Rn×n , B ∈ Rn×m , n, m ∈ N, [a, b] ⊂ R, c is a fixed vector in Rn , y : [a, b] → Rn – a fixed function belonging to L2 = L2 ([a, b], Rn ), 1−α α Ia+ (L2 ) – a space of controls defined below. By C Da+ x we denote the left Caputo n derivative of a function x : [a, b] → R and by xu : [a, b] → Rn – a solution of c AGH University of Science and Technology Press, Krakow 2016

49

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Dariusz Idczak and Stanislaw Walczak

problem (1.1), corresponding to a control u : [a, b] → Rm . We shall consider the above optimal control problem in the set AC 2 = AC 2 ([a, b], Rn ) = {x : [a, b] → Rn ; x is absolutely continuous and x0 ∈ L2 } of trajectories. Problems of such a type can be used in the examination of the pointwise and functional controllabilities of systems (1.1) with taking an energy cost into consideration. The paper is organized as follows. First, we calculate the differential and the gradient of the functional J. Next, we prove existence of a unique optimal control and derive a maximum principle for problem (1.1)–(1.2). Finally, we use the formula for the gradient of J to describe the gradient and projection of the gradient methods for the approximative solving of this problem. Results of such a type for systems containing the classical derivative of the first order can be found in [22] (cf. also [11] for the comprehensive study of the classical linear-quadratic problems). The case of the fractional Riemann-Liouville derivative is α,2 studied in [10] for J containing only the pointwise term, in the space ACa+ (defined 2 below) of trajectories and in the space L of controls. The presence of the Caputo derivative means that the problem has to be investigated in quite different (given above) spaces of trajectories and controls. Optimal control problems for systems of fractional order are studied for over ten years. For the first time, fractional optimal control problems of type ( α Da+ x(t) = G(t, x(t), u(t)), x(a) = x0 , α where Da+ x denotes the left Riemann-Liouville derivative, with cost functional of integral type Z1 I(u) = F (t, x(t), u(t))dt 0

and without control constraints, were investigated in [2]. The author formulated the necessary optimality conditions for such a problem and described a numerical scheme for finding an approximative solution to such systems in the case of linear control systems and quadratic cost functionals 1 I(u) = 2

Z1

(q(t)x2 (t) + r(t)u2 (t))dt.

0

Necessary optimality conditions are given in the form of a system of equations containing the Lagrange multipliers and do not contain any minimum condition. Unconstrained problems are considered also in [5] where second order optimality conditions are derived and in [19] where the final time is not fixed. A numerical scheme presented in [2] is adopted from the case of positive integer order problems, given in [1]. Numerical

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On a linear-quadratic problem with Caputo derivative

schemes for fractional optimal control problems, consisting in an approximation of a fractional derivative can be found in [3, 17, 19, 21]. These schemes lead to problems of positive integer order. In paper [15], system (1.1) with a non-zero initial condition and cost functional Z1 I(u) = (hγ, x(t)i + g(t, u(t)))dt, 0 n

where γ ∈ R , is considered. Authors obtained existence of an optimal solution and necessary optimality conditions in the form of a maximum principle. To the best of our knowledge results presented in our paper has not been obtained by other authors.

2. PRELIMINARIES In all the paper we consider linear spaces over R. Let α ∈ (0, 1). By the left Caputo derivative of order α of an absolutely continuous function x : [a, b] → Rn we mean the left Riemann-Liouville derivative of the function x(·) − x(a). Of course, C

α α Da+ x(t) = Da+ x(t) −

x(a) 1 , Γ(1 − α) (t − a)α

t ∈ [a, b] a.e.,

α where Da+ x is the left Riemann-Liouville derivative of order α of x. It is easy to see that 1−α 0 C α Da+ x(t) = Ia+ x (t), t ∈ [a, b] a.e., 1−α where Ia+ is the left Riemann-Liouville integral operator of order 1 − α defined on the space L1 = L1 ([a, b], Rm ) of integrable functions. Indeed, if x is absolutely continuous on [a, b], then C

d 1−α (I (x(·) − x(a))(t) dt a+ d 1−α 1 0 d 1 1−α 0 1−α 0 = (Ia+ (Ia+ x ))(t) = (Ia+ (Ia+ x ))(t) = Ia+ x (t) dt dt

α α Da+ x(t) = Da+ (x(·) − x(a))(t) =

1−α 1 1−α 0 1 for t ∈ [a, b] a.e. (we used here the fact that Ia+ (Ia+ x0 ) = Ia+ (Ia+ x ) everywhere on [a, b] (cf. [20, formula (2.21) and the subsequent comments])). More properties of the fractional integrals and derivatives can be found in [20] and [16]. α,2 α,2 By ACa+ = ACa+ ([a, b], Rn ) we denote the set of all functions x : [a, b] → Rn of the form 1 c α x(t) = + Ia+ ϕ(t), t ∈ [a, b] a.e., Γ(α) (t − a)1−α α,2 with c ∈ Rn , ϕ ∈ L2 . One can show ([4]) that x ∈ ACa+ if and only if x possesses the α left Riemann-Liouville derivative Da+ x ∈ L2 . In such a case 1−α α Ia+ x(a) = c and Da+ x = ϕ.

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Dariusz Idczak and Stanislaw Walczak

α,2 It is easy to show that ACa+ with natural operations and the scalar product

1−α 1−α Ia+ x(a)Ia+ y(a)

hx, yi =

Zb +

α α Da+ x(t)Da+ y(t)dt

a

is the Hilbert space. α,2 α,2 Similarly, by ACb− = ACb− ([a, b], Rn ) we mean the set of all functions n x : [a, b] → R of the form x(t) =

1 d α + Ib− ψ(t), Γ(α) (b − t)1−α

t ∈ [a, b] a.e.,

α with d ∈ Rn , ψ ∈ L2 , where Ib− is the right Riemann-Liouville integral operator of α,2 order 1 − α defined on the space L1 . As in the “left” case x ∈ ACb− if and only if x α possesses the right Riemann-Liouville derivative Db− x ∈ L2 and, in such a case, 1−α α Ib− x(b) = d and Db− x = ψ. α,2 ACb− with the scalar product

hx, yi =

1−α 1−α Ib− x(b)Ib− y(b)

Zb +

α α Db− x(t)Db− y(t)dt

a

is complete. 1,2 1,2 Of course, ACa+ = ACb− = AC 2 . α,2 α 2 α By Ia+ (L ) we denote the set {Ia+ (ϕ); ϕ ∈ L2 } which is contained in ACa+ . It is α 2 clear that Ia+ (L ) with the scalar product Zb hx, yi =

α α Da+ x(t)Da+ y(t)dt

a

is the Hilbert space. In what follows, we shall use the following variant of a fractional theorem on integration by parts. α,2 Theorem 2.1. Let α ∈ (0, 1). If f ∈ AC02 = {x ∈ AC 2 ; x(a) = 0}, g ∈ ACb− ∩ L2 , then Zb Zb 1−α α α f (t)Db− g(t)dt = −f (b)Ib− g(b) + Da+ f (t)g(t)dt. a

Proof. Since f ∈

a

AC02 ,

α,1 therefore (cf. [9, Theorem 6]) f ∈ ACa+ and 1−α 0 α Da+ f (t) = Ia+ (f ).

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On a linear-quadratic problem with Caputo derivative

So, Zb

α Da+ f (t)g(t)dt

Zb =

a

1−α 0 Ia+ (f )(t)g(t)dt

Zb =

a

1−α f 0 (t)Ib− g(t)dt

a

and Zb

α f (t)Db− g(t)dt

Zb =

a

 d 1−α  g (t)dt f (t) − Ib− dt

a

=

1−α −f (b)Ib− g(b)

Zb +

1−α f 0 (t)Ib− g(t)dt

a

and the proof is complete. Using the method applied in [7] one can obtain (cf. [12]) the following result. Theorem 2.2. If α ∈ ( 12 , 1), c ∈ Rn , v ∈ L2 , then the problem ( α x(t) = Ax(t) + v(t), t ∈ [a, b] a.e., Da+ 1−α Ia+ x(a) = c α,2 has a unique solution xv in ACa+ . It is given by

xv (t) =

ΦA α,α (t

Zt − a)c +

ΦA α,α (t − s)v(s)ds,

t ∈ [a, b] a.e.,

a

where ΦA α,β (t) =

P∞

k=0

k (k+1)α−1

A t Γ(kα+β)

.

Remark 2.3. If c = 6 0, then condition α > 21 can not be omitted. Indeed, let us consider problem ( α Da+ x(t) = x(t), t ∈ [a, b] a.e., 1−α Ia+ x(a) = c If x(t) =

1 c α + Ia+ ϕ(t), Γ(α) (t − a)1−α

t ∈ [a, b] a.e.,

α,2 where ϕ ∈ L2 , is a solution to the above system, belonging to ACa+ , then c α α x = Da+ x ∈ L2 , Ia+ ϕ ∈ L2 , and, consequently, (t−a)1−α ∈ L2 . It means that α > 12 .

Similarly, we have the following theorem. Theorem 2.4. If α ∈ ( 12 , 1), c ∈ Rn , v ∈ L2 , then problem ( α Db− y(t) = Ay(t) + v(t), t ∈ [a, b] a.e., 1−α Ib− y(b) = c

(2.1)

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Dariusz Idczak and Stanislaw Walczak

α,2 has a unique solution yv ∈ ACb− and it is given by

yv (t) =

ΦA α,α (b

Zb − t)c +

ΦA α,α (s − t)v(s)ds,

t ∈ [a, b] a.e.

(2.2)

t

In particular, a unique solution y0 corresponding to control v(·) ≡ 0 is given by y0 (t) = ΦA α,α (b − t)c,

t ∈ [a, b] a.e.

Remark 2.5. If c = 0, then the assumption α ∈ ( 12 , 1) in Theorems 2.2, 2.4 can be replaced by α ∈ (0, 1). Theorem 2.4 implies the following corollary. Corollary 2.6. If α ∈ ( 12 , 1), then the function n×n [a, b] 3 t 7−→ ΦA α,α (b − t) ∈ R

belongs to L2 ([a, b], Rn×n ). 3. REMARKS ON SYSTEM (1.1) Let α ∈ (0, 1). From the results contained in [8, Theorem 4] it follows that if 1−α u ∈ Ia+ (L2 ), then there exists a unique in AC 1 = AC 1 ([a, b], Rn ) = {x : [a, b] → Rn ; x is absolutely continuous and x0 ∈ L1 } solution xu of problem (1.1). Moreover (cf. [8, Theorem 6]), the following result holds. 1−α Lemma 3.1. If uj −→ u0 in Ia+ (L2 ), then xj ⇒ x0 uniformly on [a, b] (here xj is a unique solution to problem (1.1), corresponding to control uj ). 1−α Of course, a solution xu of problem (1.1) , corresponding to u ∈ Ia+ (L2 ), is a solution of system α Da+ x(t) = Ax(t) + Bu(t),

t ∈ [a, b] a.e.

(3.1)

α Since Ax(·) + Bu(·) belongs to L2 , Da+ xu ∈ L2 , too. Using this fact and [8, Proof of Theorem 4, formula (16)] we deduce that xu ∈ AC 2 . Consequently, xu ∈ AC02 . Thus, 1−α if u ∈ Ia+ (L2 ) and xu ∈ AC 2 satisfies (1.1), then xu belongs to AC02 and satisfies 1−α (3.1). Conversely, if u ∈ Ia+ (L2 ) and xu ∈ AC02 satisfies (3.1), then xu belongs to 2 AC and satisfies (1.1). Now, let us consider the operator 1−α F : AC02 → Ia+ (L2 ), α F (x) = Da+ x − Ax.

The above operator is well defined because 1−α 0 1−α α Da+ x = Ia+ x ∈ Ia+ (L2 )

and

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On a linear-quadratic problem with Caputo derivative 1−α α 1−α α 1−α 1 x = Ia+ x0 = Ia+ Ia+ x0 ∈ Ia+ (Ia+ (L2 )) ⊂ Ia+ (L2 )

for any x ∈ AC02 . Of course, it is linear. Moreover, since

1−α 0

α

Da+ x 1−α 2 = Ia+ x I 1−α (L2 ) = kx0 kL2 = kxkAC 2 , I (L ) a+

0

a+

1−α α 0

α 0 kxkI 1−α (L2 ) = Ia+ x L2 ≤ d kx0 kL2 = d kxkAC 2 Ia+ x I 1−α (L2 ) = Ia+ 0 a+ a+

1−α for any x ∈ AC02 and some d > 0, where kxkI 1−α (L2 ) = Da+ x L2 , kxkAC 2 = kx0 kL2 , 0 a+ therefore F is bounded. From its bijectivity and from the Banach inverse mapping theorem it follows that F is a homeomorphism.

4. GRADIENT OF THE POINTWISE TERM Let us assume that α ∈ ( 12 , 1) and consider the pointwise term J1 (u) =

1 2 |xu (b) − c|Rn , 2

1−α u ∈ Ia+ (L2 ),

of functional (1.2). J1 can be written as the following superposition 1−α (L2 ) 7−→ xu ∈ AC02 7−→ xu (b) − c ∈ Rn 7−→ u ∈ Ia+

1 2 |xu (b) − c|Rn ∈ R. 2

(4.1)

The interior mapping 1−α λ : u ∈ Ia+ (L2 ) 7−→ xu ∈ AC02

is linear and continuous. It follows from the fact that λ is the following superposition 1−α 1−α u ∈ Ia+ (L2 ) 7−→ Bu ∈ Ia+ (L2 ) 7−→ F −1 (Bu) ∈ AC02 . 1−α So, the differential λ0 (u) of λ at a point u ∈ Ia+ (L2 ) is the following mapping 1−α λ0 (u) : v ∈ Ia+ (L2 ) 7−→ hv ∈ AC02 ,

where hv ∈ AC02 is such that α Da+ hv (t) = Ahv (t) + Bv(t),

t ∈ [a, b] a.e.

1−α Consequently, the differential J10 (u) of J1 at a point u ∈ Ia+ (L2 ) is the mapping 1−α J10 (u) : v ∈ Ia+ (L2 ) 7−→ (xu (b) − c)hv (b) ∈ R. α,2 Now (cf. Theorem 2.4), let Ψ1u ∈ ACb− be a unique solution of problem

(

α Db− Ψ(t) = AT Ψ(t), 1−α Ib− Ψ(b) = xu (b) − c.

t ∈ [a, b] a.e.,

(4.2)

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Dariusz Idczak and Stanislaw Walczak

We have (cf. Theorem 2.1 and Corollary 2.6) 1−α 1 (xu (b) − c)hv (b) = Ib− Ψu (b)hv (b)

Zb =

α Da+ hv (t)Ψ1u (t)dt

a

−

α hv (t)Db− Ψ1u (t)dt

a

Zb =

(Ahv (t) +

Bv(t)) Ψ1u (t)dt

a

Zb −

hv (t)AT Ψ1u (t)dt

(4.3)

a

Zb =

Zb

Ψ1u (t)Bv(t)dt

a

and Zb

Ψ1u (t)Bv(t)dt

Zb =

a

1−α 1−α Ia+ Da+ (Bv)(t)Ψ1u (t)dt

a

Zb =

1−α 1−α Da+ v(t)Ib− (B T Ψ1u )(t)dt

a

Zb =

1−α 1−α 1−α 1−α Da+ v(t)Da+ Ia+ Ib− (B T Ψ1u )(t)dt

a

 1−α 1−α = v, Ia+ Ib− (B T Ψ1u ) I 1−α (L2 ) .

a+

So, J10 (u)v

Zb =

1−α T 1−α 1  Ψ1u (t)Bv(t)dt = v, Ia+ (B Ib− Ψu ) I 1−α (L2 )

(4.4)

a+

a 1−α for v ∈ Ia+ (L2 ) and, consequently, 1−α 1−α 1 ∇J1 (u) = Ia+ (B T Ib− Ψu ),

where

T

Ψ1u (t) = ΦA α,α (b − t)(xu (b) − c)

(4.5) (4.6)

for t ∈ [a, b] a.e. Now, we shall show that the gradient ∇J1 of J1 is Lipschitzian. In the proof of this fact we shall use the integrability of the function Zt 2 T [a, b] 3 t 7−→ ΦA (t − s) ds ∈ R+ α,α 0, a

proved in [10].

(4.7)

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On a linear-quadratic problem with Caputo derivative

We have the following result. Theorem 4.1. Gradient ∇J1 satisfies the Lipschitz condition, i.e. k∇J1 (u) − ∇J1 (w)kI 1−α (L2 ) ≤ a+

(b − a)1−α L ku − wkI 1−α (L2 ) a+ Γ(2 − α)

1−α for u, w ∈ Ia+ (L2 ), where

2

L = (2 |A|

 Zb  Zt

 1 2 A Φα,α (t − s) 2 ds 2 dt

a

a

1 2

2

(4.8)

2

1−α A Φα,α (b − ·) 2 . + 2 |B| (b − a)) |B| Ib− L 1−α Proof. Let us fix u, w ∈ Ia+ (L2 ). We have

1−α 1−α Ψu − B T Ib− Ψw L2 . k∇J1 (u) − ∇J1 (w)kI 1−α (L2 ) = B T Ib− a+

In [10] it has been shown that

T 1−α

T 1−α

B I

b− Ψu − B Ib− Ψw L2 ≤ L ku − wkL2 with L given by (4.8). But (cf. [20, formula (2.72)])

1−α 1−α 1−α 1−α Da+ u − Ia+ Da+ w L2 ku − wkL2 = Ia+

(b − a)1−α (b − a)1−α 1−α 1−α

Da+ ≤ u − Da+ w L2 = ku − wkI 1−α (L2 ) a+ Γ(2 − α) Γ(2 − α) and the proof is complete.

5. GRADIENT OF THE FUNCTIONAL TERM Let α ∈ ( 12 , 1) and consider the functional term J2 (u) =

1 2 1−α kxu (·) − y(·)kL2 , u ∈ Ia+ (L2 ), 2

of the functional (1.2). J2 can be written as the following superposition 1−α u ∈ Ia+ (L2 ) 7−→ xu ∈ AC02 7−→

1 2 kxu (·) − y(·)kL2 ∈ R. 2

It is easy to see that J20 (u)v

Zb (xu (t) − y(t))hv (t)dt

= a

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Dariusz Idczak and Stanislaw Walczak

1−α for u, v ∈ Ia+ (L2 ), where hv ∈ AC02 is such that α Da+ hv (t) = Ahv (t) + Bv(t),

t ∈ [a, b] a.e.

α,2 Denoting by Ψ2u a unique in ACb− solution of problem ( α Db− Ψ(t) = AT Ψ(t) + (xu (t) − y(t)), t ∈ [a, b] a.e., 1−α Ib− Ψ(b) = 0

(5.1)

we obtain (cf. Theorem 2.1 and Corollary 2.6) Zb

Zb (xu (t) − y(t))hv (t)dt =

a

α Db− Ψ2u (t)hv (t)dt

Zb −

a

Zb =

=

α Ψ2u (t)Da+ hv (t)dt

Zb −

AT Ψ2u (t)hv (t)dt

a

Ψ2u (t)(Ahv (t)

Zb + Bv(t))dt −

a

Zb =

(5.2)

a

a

Zb

AT Ψ2u (t)hv (t)dt

AT Ψ2u (t))hv (t)dt

a

Ψ2u (t)Bv(t)dt.

a

In the same way as in the previous section Zb

1−α 1−α T 2  Ψ2u (t)Bv(t)dt = Ia+ Ib− (B Ψu ), v I 1−α (L2 ) a+

a 1−α and, consequently, for u, v ∈ Ia+ (L2 ),

J20 (u)v

Zb =

1−α 1−α T 2  Ψ2u (t)Bv(t)dt = Ia+ Ib− (B Ψu ), v I 1−α (L2 ) ,

(5.3)

a+

a 1−α 1−α ∇J2 (u) = Ia+ Ib− (B T Ψ2u ),

(5.4)

α,2 where Ψ2u ∈ ACb− is a unique solution of problem (5.1), given by

Ψ2u (t)

Zb =

T

ΦA α,α (s − t)(xu (s) − y(s))ds,

t ∈ [a, b] a.e.

(5.5)

t

We also have the following result (to calculate D3 we use integrability of function (4.7)).

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On a linear-quadratic problem with Caputo derivative

Theorem 5.1. Gradient ∇J2 satisfies the Lipschitz condition: k∇J2 (u) − ∇J2 (w)kI 1−α (L2 ) ≤ D ku − wkI 1−α (L2 ) a+

a+

1−α for u, w ∈ Ia+ (L2 ) and some D > 0.

Proof. We have

k∇J2 (u) −

2 ∇J2 (w)kI 1−α (L2 ) a+

Zb =

1−α T 2 2 I (B (Ψu − Ψ2w ))(t) dt b−

a

Zb

T 2 B (Ψu − Ψ2w )(t) 2 dt ≤ D2

≤ D1 a

Zb

2  b Z T   ΦA dt α,α (s − t)(xu (s) − xw (s)) ds t

a

 b  Z Zb 2 T 2  ΦA  |(xu (s) − xw (s))| ds α,α (s − t) ds dt

Zb ≤ D2 a

t

a

Zb

2

|(xu (s) − xw (s))| ds

= D3 a

Zb ≤ D3 a

 s  Z Zs A 2  Φα,α (s − t1 ) dt1 |B(u(t1 ) − w(t1 ))|2 dt1  ds a

a

Zb ≤ D4

Zb

2

|B(u(t1 ) − w(t1 ))| dt1 ≤ D5 a

2

2

|(u(t1 ) − w(t1 ))| dt1 = D5 ku − wkL2 a



1−α

1−α 1−α 1−α 2 1−α 1−α 2 u − Da+ Da+ w L2 ≤ D6 Da+ Da+ u − Ia+ = D5 Ia+ w L2 2

= D2 ku − wkI 1−α (L2 ) a+

1−α (L2 ), where for u, w ∈ Ia+

 D1 =

(b − a)1−α Γ(2 − α)

2 ,

2

D2 = D1 |B| ,

Zb D3 = D2

 b  Z 2 T  ΦA  α,α (s − t) ds dt,

a

ZbZs D4 = D3

A Φα,α (s − t1 ) 2 dt1 ds,

t 2

D5 = D4 |B| ,

a a

 D6 = D5 The proof is complete.

(b − a)1−α Γ(2 − α)

2 ,

D=

p

D6 .

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Dariusz Idczak and Stanislaw Walczak

6. GRADIENT OF THE CONTROL TERM Let α ∈ (0, 1) and consider the control term J3 (u) =

1 2 ku(·)kI 1−α (L2 ) , a+ 2

1−α u ∈ Ia+ (L2 ),

of the functional (1.2). It is clear that J30 (u)v

Zb = hu, viI 1−α (L2 ) = a+

1−α 1−α Da+ u(t)Da+ v(t)dt

(6.1)

a 1−α for u, v ∈ Ia+ (L2 ). Consequently,

∇J3 (u) = u.

(6.2)

Of course, k∇J3 (u) − ∇J3 (w)kI 1−α (L2 ) = ku − wkI 1−α (L2 ) a+

for u, w ∈

1−α Ia+ (L2 ).

a+

So, ∇J3 satisfies the Lipschitz condition with the constant 1.

7. EXISTENCE OF A SOLUTION TO (1.1)–(1.2) Let us recall that a function J : U → R where U is a convex subset of a Hilbert space H, is called strongly convex, if 2

J(γu + (1 − γ)v) ≤ γJ(u) + (1 − γ)J(v) − γ(1 − γ)κ ku − vk

for some κ > 0 and all γ ∈ [0, 1], u, v ∈ U . One can show (cf. [23, Lemma 4.3.1] for the method of the proof) that J is strongly convex on U with a constant κ if and 2 only if function g(u) = J(u) − κ kuk is convex on U . So, for example, functional 2 H 3 u 7−→ kuk ∈ R is strongly convex with constant κ = 1. In [22, Theorem 1.3.8] the following theorem is proved. Theorem 7.1. If U is a convex closed subset of a Hilbert space H and a functional J : U → R is strongly convex with a constant κ and lower semicontinuous on U , then J∗ := inf u∈U J(u) > −∞ and there exists a unique point u∗ such that J(u∗ ) = J∗ . Moreover, any minimizing sequence (uk ) (i.e. such that J(uk ) → J∗ ) converges to u∗ and 1 kuk − u∗ k ≤ (J(uk ) − J∗ ), k = 1, 2, . . . κ Now, let α ∈ (0, 1) and M ⊂ Rm be a fixed set. Consider problem (1.1)–(1.2) in 1−α the set Ia+ (L2M ), where L2M = {u ∈ L2 ; u(t) ∈ M for t ∈ [a, b] a.e.}.

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On a linear-quadratic problem with Caputo derivative

Of course, 1−α 1−α 1−α Ia+ (L2M ) = {u ∈ Ia+ (L2 ); Da+ u(t) ∈ M

for t ∈ [a, b] a.e.}.

We have the following theorem. Theorem 7.2. If M ⊂ Rm is a convex closed set, then there exists a unique point 1−α u∗ ∈ Ia+ (L2M ) such that J(u∗ ) = J∗ := inf u∈I 1−α (L2 ) J(u) and any minimizing a+ M sequence (uk ) converges to u∗ and kuk − u∗ kI 1−α (L2 ) ≤ 2(J(uk ) − J∗ ),

k = 1, 2, . . .

a+

1−α 1−α Proof. The set Ia+ (L2M ) is convex and closed in Ia+ (L2 ). Convexity is obvious. 1−α To prove its closedness let us consider a sequence (uk ) ⊂ Ia+ (L2M ) converging in 1−α 1−α 1−α 2 Ia+ (L ) to some u0 . Thus, the sequence (Da+ uk ) converges in L2 to Da+ u0 . So, 1−α there exists a subsequence of (Da+ uk ) that is converging pointwise a.e. on [a, b] to 1−α Da+ u0 . Since elements of this subsequence take their values a.e. on [a, b] in the closed 1−α 1−α set M , therefore Da+ u0 has the same property. It means that u0 ∈ Ia+ (L2M ). 1−α 1−α Lemma 3.1 implies continuity of J on Ia+ (L2 ) (and, in consequence, on Ia+ (L2M )). 1−α 1−α 2 2 So, since J is strongly convex on Ia+ (L ) (and, in consequence, on Ia+ (LM )) with the constant κ = 12 , therefore from Theorem 7.1 the assertion follows.

8. MAXIMUM PRINCIPLE Let us start with the following classical result ([22, Theorem I.2.5]). Lemma 8.1. Let U be a convex subset of a Banach space X, J – a functional of class C 1 on U . If u∗ is a global minimum point of J on U , then J 0 (u∗ )u ≥ J 0 (u∗ )u∗

(8.1)

for any u ∈ U . If, additionally, J is convex on U , then condition (8.1) is sufficient for u∗ to be the global minimum point of J on U . 1−α Now, let us consider problem (1.1)–(1.2) in the set Ia+ (L2M ) with a convex set 1 m M ⊂ R , in the case of α ∈ ( 2 , 1). We have the following theorem. 1−α (L2M ) Theorem 8.2. Control u∗ is a solution to problem (1.1)–(1.2) in the set Ia+ if and only if Zb T 1−α AT min (Ib− ((Φα,α (b − ·)(xu∗ (b) − c) + ΦA α,α (s − ·)(xu∗ (s) − y(s))ds)B)(t) v∈M

· 1−α + Da+ u∗ (t))v Zb T T 1−α = (Ib− ((ΦA ΦA α,α (b − ·)(xu∗ (b) − c) + α,α (s − ·)(xu∗ (s) − y(s))ds)B)(t) · 1−α 1−α + Da+ u∗ (t))(Da+ u∗ )(t)

for t ∈ [a, b] a.e.

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Proof. Let u∗ be a solution of problem (1.1)–(1.2) in the set L2M . From Lemma 8.1, formulas (4.4), (5.3), (6.1) and convexity of functional (1.2) it follows that optimality 1−α of u∗ ∈ Ia+ (L2M ) is equivalent to the functional condition of the form Zb min

1−α u(·)∈Ia+ (L2M )

Zb =

(Ψ1u∗ (t)

+

1−α 1−α Ψ2u∗ (t))BIa+ (Da+ u)(t)dt

a

Zb +

1−α 1−α Da+ u∗ (t)Da+ u(t)dt

a

1−α 1−α (Ψ1u∗ (t) + Ψ2u∗ (t))BIa+ (Da+ u∗ )(t)dt +

a

Zb

1−α 1−α Da+ u∗ (t)Da+ u∗ (t)dt,

a

α,2 α,2 where Ψ1u∗ ∈ ACb− is a unique solution of problem (4.2) and Ψ2u∗ ∈ ACb− is a unique solution to problem (5.1). Both solutions are given by (2.2). Using the classical fractional theorem on the integration by parts, in the integral form (cf. [20]), we can write the above equality in the following form

Zb min 1−α

u(·)∈Ia+

Zb

(L2M )



1−α 1−α 1−α Ib− (Ψ1u∗ (·) + Ψ2u∗ (·))B (t) + Da+ u∗ (t) (Da+ u)(t)dt

a



1−α 1−α 1−α Ib− (Ψ1u∗ (·) + Ψ2u∗ (·))B (t) + Da+ u∗ (t) (Da+ u∗ )(t)dt.

= a

This condition is equivalent to Zb min2

v(·)∈LM

Zb =



1−α 1−α Ib− (Ψ1u∗ (·) + Ψ2u∗ (·))B (t) + Da+ u∗ (t) v(t)dt

a



1−α 1−α 1−α Ib− (Ψ1u∗ (·) + Ψ2u∗ (·))B (t) + Da+ u∗ (t) (Da+ u∗ )(t)dt.

a

From [6, Lemma 6] it follows that the above condition is equivalent to the pointwise minimum condition given the theorem.

9. GRADIENT METHOD We have the following result concerning the convergence of the gradient method ([22, Theorem I.4.1]). Lemma 9.1. Let a functional J : H → R be of class C 1 , bounded below and with the gradient satisfying the Lipschitz condition. If (uk ) ⊂ H is a sequence described by the formula uk+1 = uk − βk ∇J(uk ), k = 0, 1, . . . , (9.1)

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On a linear-quadratic problem with Caputo derivative

with any fixed u0 ∈ H, where parameter βk is a minimum point of the function fk : [0, ∞) 3 β 7−→ J(uk − β∇J(uk )) ∈ R,

k = 0, 1, . . . ,

then the sequence (J(uk )) is nonincreasing and lim k∇J(uk )k = 0.

(9.2)

k→∞

If, additionally, J is strongly convex with a constant κ > 0, then the sequence (uk ) converges to u∗ – a unique minimum point of J and 0 ≤ J(uk ) − J∗ ≤ (J(u0 ) − J∗ )q k , 2

kuk − u∗ k ≤

1 (J(u0 ) − J∗ )q k , κ

for k = 0, 1, . . ., where J∗ = inf J(u), q = 1 − u∈H

(9.3)

2κ L1

(9.4)

∈ [0, 1) (L1 is a Lipschitz constant

for ∇J). Remark 9.2. One can show that 2κ ≤ L1 (cf. [10]). 1−α Now, let us consider problem (1.1)–(1.2) in the set U = Ia+ (L2 ) with α ∈ ( 12 , 1). Using the above lemma we obtain

Theorem 9.3. A sequence (uk ) given by (9.1) with βk described in Lemma 9.1 1−α converges to u∗ – a unique minimum point of J on Ia+ (L2 ), the sequence (J(uk )) is nonincreasing and conditions (9.2), (9.3), (9.4) are satified with L1 =

(b − a)1−α L + D + 1, Γ(2 − α)

where L is given by (4.8) and D is described in the proof of Theorem 5.1. Remark 9.4. Let us observe that 2

2 1 1 fk (β) = J(uk − β∇J(uk )) = xuk −β∇J(uk ) (b) − c Rn + xuk −β∇J(uk ) (·) − y(·) L2 2 2 2 1 1 2 xuk (b) − c − βx∇J(uk ) (b) Rn + kuk (·) − β∇J(uk )(·)kI 1−α (L2 ) = a+ 2 2

2 1 1 2

+ xuk (·) − y(·) − βx∇J(uk ) (·) L2 + kuk (·) − β∇J(uk )(·)kI 1−α (L2 ) a+ 2 2 2 1 2 = J1 (uk ) − β(xuk (b) − c)x∇J(uk ) (b) + β x∇J(uk ) (b) Rn 2

2

 1 + J2 (uk ) − β xuk (·) − y(·), x∇J(uk ) (·) L2 + β 2 x∇J(uk ) (·) L2 2 1 2 + J3 (uk ) − β huk (·), ∇J(uk )(·)iI 1−α (L2 ) + β 2 k∇J(uk )(·)kI 1−α (L2 ) . a+ a+ 2 So, if

x∇J(u ) (b) n + x∇J(u ) (·) 2 + k∇J(uk )(·)k 1−α 2 = 0 k k I (L ) R L a+

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Dariusz Idczak and Stanislaw Walczak

for some k, then one can put βk = 0 and, consequently, ∇J(uk ) = 0. This means that uk is a unique minimum point of J. If

x∇J(u ) (b) n + x∇J(u ) (·) 2 + k∇J(uk )(·)k 1−α 2 6= 0, k k I (L ) R L a+

then

 (xuk (b) − c)x∇J(uk ) (b) + xuk (·) − y(·), x∇J(uk ) (·) L2 + huk (·), ∇J(uk )(·)iI 1−α (L2 ) a+ βk =

x∇J(u ) (b) 2 n + x∇J(u ) (·) 2 2 + k∇J(uk )(·)k21−α 2 k k I (L ) R L a+

is the unique minimum point of fk . It is worth observing (cf. (4.3), (4.5), (5.4), (6.2)) that Zb (xuk (b) − c)x∇J(uk ) (b) =

T 1−α 1 2 B I b− Ψuk (t) dt

a

Zb +

1−α 1 1−α 2 (B T Ib− Ψuk (t))(B T Ib− Ψuk (t))dt

Zb +

a

1−α 1 1−α (B T Ib− Ψuk (t))(Da+ uk (t))dt,

a

where Ψ1uk is given by (4.6) and Ψ2uk – by (5.5) (with u replaced by uk ). Similarly (cf. (5.2), (4.5), (5.4), (6.2)),

 xuk (·) − y(·), x∇J(uk ) (·) L2 =

Zb

1−α T 2 2 I b− (B Ψuk )(t) dt

a

Zb +

1−α 2 1−α 1 (B T Ib− Ψuk (t))(B T Ib− Ψuk (t))dt

a

Zb +

1−α 2 1−α (B T Ib− Ψuk (t))(Da+ uk (t))dt

a

and Zb huk (·), ∇J(uk )(·)iI 1−α (L2 ) = a+

1−α 1 1−α uk (t))dt (B T Ib− Ψuk (t))(Da+

a

Zb + a

1−α 2 1−α (B T Ib− Ψuk (t))(Da+ uk (t))dt

Zb + a

1−α 1−α (Da+ uk (t))(Da+ uk (t))dt.

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On a linear-quadratic problem with Caputo derivative

So, Zb βk =

T 1−α 1 2 1−α T 1−α 2 (B I b− Ψuk (t)) + (B Ib− Ψuk (t)) + (Da+ uk (t)) dt

a



x∇J(u ) (b) 2 n + x∇J(u ) (·) 2 2 + k∇J(uk )(·)k21−α 2 k k I (L ) R L a+

= x∇J(u ) (b) 2 n + k R

2 k∇J(uk )(·)kI 1−α (L2 ) a+ .

x∇J(u ) (·) 2 2 + k∇J(uk )(·)k21−α 2 k I (L ) L a+

10. PROJECTION OF THE GRADIENT METHOD By PU (u) we denote the projection of a point u ∈ H on a convex closed subset U of a Hilbert space H. We shall use the following result on the convergence of the projection of the gradient method ([22, Theorem I.4.4]). Lemma 10.1. Let U be a convex closed subset of a Hilbert space H and J : U → R – a functional of class C 1 , bounded below and with the gradient ∇J satisfying the Lipschitz condition with a constant L. If (uk ) ⊂ H is a sequence described by the formula uk+1 = PU (uk − βk ∇J(uk )), k = 0, 1, . . . , (10.1) with any fixed u0 ∈ H, where βk , k = 0, 1, . . . , is such that ε0 ≤ β k ≤

2 L + 2ε

(here ε0 , ε are fixed positive parameters such that ε0 ≤ (J(uk )) is nonincreasing and lim kuk − uk+1 k = 0.

k→∞

(10.2) 2 L+2ε ),

then the sequence (10.3)

If, additionally, J is strongly convex on U , then the sequence (uk ) converges to a unique minimum point u∗ of J on U and there exists a constant c ≥ 0 such that 2

kuk − u∗ k ≤

c k

(10.4)

for k = 1, 2, . . . Remark 10.2. The constant c can be calulated (cf. [22, Theorem I.4.4] and [23, Theorem V.2.2] for the method of calculation). 1−α Now, let us consider problem (1.1)–(1.2) with α ∈ ( 12 , 1) in the set Ia+ (L2M ) ⊂ where M is a convex closed subset of Rm . In the proof of Theorem 7.2 it 1−α 1−α has been shown that Ia+ (L2M ) is a convex closed set in Ia+ (L2 ). So, from the above lemma and Theorem 7.2 we obtain the following result. 1−α Ia+ (L2 ),

66

Dariusz Idczak and Stanislaw Walczak

2 Theorem 10.3. If ε0 , ε > 0 are such that ε0 ≤ L+2ε , where L > 0 is a Lipschitz constant for ∇J, then the sequence (uk ) given by (10.1)–(10.2) converges to u∗ – 1−α a unique minimum point of J on Ia+ (L2M ) and condition (10.4) is satisfied. Moreover, the sequence (J(uk )) is nonincreasing.

Remark 10.4. Let M = [n1 , N1 ] × . . . × [nm , Nm ] ⊂ Rm . In [22, Part I.4] one can find the form of the projection PL2M : L2 → L2M of the space L2 on the set L2M . This 1−α 1−α allows us to describe the projection PI 1−α (L2 ) : Ia+ (L2 ) → Ia+ (L2M ) of the space a+

M

1−α 1−α 1−α Ia+ (L2 ) on the set Ia+ (L2M ). Indeed, for any u ∈ Ia+ (L2 ), we have

min

1−α v∈Ia+ (L2M )

ku − vkI 1−α (L2 ) = a+

min

1−α v∈Ia+ (L2M )

1−α

1−α

Da+ u − Da+ v L2



1−α

1−α 1−α = min2 Da+ u − f L2 = Da+ u − PL2M (Da+ u) 2 L f ∈L

M

1−α

1−α 1−α 1−α = Da+ u − Da+ Ia+ (PL2M (Da+ u)) 2 L



1−α 1−α = u − Ia+ (PL2M (Da+ u)) 1−α 2 . Ia+ (L )

So, 1−α 1−α PI 1−α (L2 ) (u) = Ia+ (PL2M (Da+ u)) a+

for u ∈

M

1−α Ia+ (L2 ).

Acknowledgments The project was financed with funds of National Science Centre, granted on the basis of decision DEC-2011/01/B/ST7/03426.

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68 Dariusz Idczak [email protected] University of Lodz Faculty of Mathematics and Computer Science Banacha 22, 90-238 Lodz, Poland

Stanislaw Walczak [email protected] University of Lodz Faculty of Mathematics and Computer Science Banacha 22, 90-238 Lodz, Poland

Received: November 25, 2014. Revised: March 9, 2015. Accepted: April 10, 2015.

Dariusz Idczak and Stanislaw Walczak