J Optim Theory Appl DOI 10.1007/s10957-017-1119-y

Optimal Controls for Riemann–Liouville Fractional Evolution Systems without Lipschitz Assumption Shouguo Zhu1,2 · Zhenbin Fan1

· Gang Li1

Received: 1 March 2017 / Accepted: 29 May 2017 © Springer Science+Business Media New York 2017

Abstract In this paper, an evolution system with a Riemann–Liouville fractional derivative is proposed and analyzed. With the help of a resolvent technique, a suitable concept of solutions to this system is formulated and the corresponding existence of solutions is demonstrated. Furthermore, without the Lipschitz continuity of the nonlinear term, the optimal control result is derived by setting up minimizing sequences twice. Our work essentially generalizes previous results on optimal controls of all evolution systems. Finally, a simple example is presented to illustrate our theoretical results. Keywords Optimal controls · Resolvent · Riemann–Liouville derivative Mathematics Subject Classification 49J15 · 47A10 · 34K37

1 Introduction Since differential equations with fractional derivatives have become powerful tools in describing many physical phenomena, they have been the active ongoing research

B

Zhenbin Fan [email protected] Shouguo Zhu [email protected] Gang Li [email protected]

1

School of Mathematical Sciences, Yangzhou University, Yangzhou 225002, People’s Republic of China

2

School of Mathematics, Taizhou College of Nanjing Normal University, Taizhou 225300, People’s Republic of China

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issues during the past two decades, and many fruitful achievements have been obtained [1–6]. There are broad applications of Riemann–Liouville fractional differential equations in viscoelasticity [7]. Therefore, many researchers have investigated them extensively [8–15]. In [9], by employing Laplace transformations and probability densities, Liu and Li proposed an appropriate concept of solutions to a semilinear system when A generates a C0 -semigroup. In [13], based on the theory of α-order resolvents (see Sect. 2, Definition 2.3), Li and Peng analyzed solutions to a fractional homogeneous system, when A generates an α-order resolvent. Their results were later extended to an inhomogeneous linear system by Fan [8]. But solutions of semilinear systems have not been paid attention to when A generates an α-order resolvent. Considering that the resolvent approach is efficient and convenient, we will also employ this method to study solutions of semilinear systems. On the other hand, many researchers have shown an increased interest in investigating optimal controls on fractional evolution equations (see [16–26]). However, most of the existing results were obtained under the condition that the nonlinear term is Lipschitz continuous. Naturally, a question arises: How to analyze optimal controls without the Lipschitz assumption? The objective of the article is to deal with this question. There exist two main difficulties in the study. One is how to formulate a suitable concept of solutions to the semilinear system (1). The other is how to seek an optimal state–control pair when the uniqueness of the solution cannot be guaranteed. Here, under suitable assumptions on the resolvent, we formulate the notion of solutions by utilizing a resolvent approach. Moreover, inspired by Zhu and Huang [27], we employ a new approach of setting up minimizing sequences twice to look for the optimal state–control pair, when the Lipschitz continuity of the nonlinear term is not satisfied. It should be pointed out that this approach can be applied to optimal control problems of all evolution systems, including integer order and fractional order. Moreover, our work essentially generalizes previous related results since we weaken relevant conditions on optimal controls. The work is built up as follows. We propose our problem and summarize some preliminaries in Sect. 2. In Sect. 3, we formulate a suitable concept of solutions and display the existence result. Sect. 4 deals with the optimal control problem. A simple example is proposed in Sect. 5 to illustrate the validity of the proposed theoretical results. Finally, in Sect. 6, we provide some conclusions.

2 Problem Statement and Preliminaries We explore the following semilinear evolution system with a Riemann–Liouville fractional derivative: D α x(t) = Ax(t) + f (t, x(t)) + B(t)u(t), α ∈]0, 1[ , t ∈ J  , lim (α)t 1−α x(t) = x0 , t↓0

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where J  :=]0, b], D α stands for the α-order fractional derivative of Riemann– Liouville type, A : D(A) ⊆ X → X generates an α-order resolvent {Rα (t)}t>0 on a Banach space X and f : J × X → X is a nonlinear function. Moreover, B ∈ L ∞ (J, L(Y, X )) and u ∈ Uad , where Uad is an admissible set, J := [0, b] and Y is a reflexive and separable Banach space. Throughout this work, L(X ) is the space of all bounded linear operators from X to itself, C(J, X ) the space of all strongly continuous functions from J to X that are normed by x := sup x(s) for x ∈ C(J, X ), L p (J, X ) the space of all Bochner s∈J

1 b integrable functions normed by  f  L p := ( 0  f (s) p ds) p for f ∈ L p (J, X ), where 1 ≤ p < ∞, and L ∞ (J, X ) the space of essentially bounded functions f : J → X . Let   C1−α (J, X ) := x ∈ C(J  , X ) : z(s) := s 1−α x(s), z(0) := lim z(s), z ∈ C(J, X )

s↓0

with the norm xC1−α := sup z(s). Then, C1−α (J, X ) is a Banach space. Furs∈J

thermore, we employ the notation P f c (Y ) to mean a class of nonempty, convex and closed subsets of Y and the symbol ∗ to denote the convolution, i.e., ( f ∗ h)(s) := s f (s − τ )h(τ )dτ . 0 Definition 2.1 [28] For any f ∈ L 1 (J, X ), the α-order fractional integral is Itα

1 f (t) := (α)



t

(t − τ )α−1 f (τ )dτ, α > 0, t > 0.

0

For the convenience of presentation, we write Itα f (t) := (gα ∗ f )(t), where gα (τ ) :=

τ α−1 , τ > 0. (α)

Definition 2.2 [28] Let 0 < α < 1. For any f ∈ L 1 (J, X ), the α-order fractional derivative of Riemann–Liouville type is d 1 D f (t) := (1 − α) dt α



t

(t − τ )−α f (τ )dτ, t > 0.

0

Definition 2.3 [13] Let 0 < α < 1. A family {Rα (s)}s>0 ⊆ L(X ) is said to be a resolvent of α-order, if it fulfills (a) Rα (·)x ∈ C(R+ , X ) and lim (α)s 1−α Rα (s)x = x, x ∈ X ; s↓0

(b) Rα (t)Rα (s) = Rα (s)Rα (t), s, t > 0; (c) Rα (t)Isα Rα (s)− Itα Rα (t)Rα (s) = gα (t)Isα Rα (s)−gα (s)Itα Rα (t), s, t > 0. The generator A : D(A) ⊆ X → X of the resolvent {Rα (s)}s>0 is the operator Ax := (2α) lim s↓0

s 1−α Rα (s)x − sα

x (α)

,

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where  D(A) := x ∈ X : lim s↓0

s 1−α Rα (s)x − sα

x (α)

 exists .

Remark 2.1 It is easy to check the boundedness of s 1−α Rα (s) on J . For notational simplicity, denote M := sup s 1−α Rα (s). s∈J

Lemma 2.1 [8,13] Let {Rα (s)}s>0 be a resolvent with generator A. Then, (a) For any x ∈ X , lim (2α)s 1−2α Isα Rα (s)x = x; s↓0

(b) For any s > 0 and x ∈ X , Rα (s)x = gα (s)x + A(gα ∗ Rα )(s)x. Similar to the proof of Lemmas 3.4 and 3.5 of [29], we can get the following properties of the operator s 1−α Rα (s). Lemma 2.2 Let {s 1−α Rα (s)}s>0 be equicontinuous and compact. Then, for every s ∈ J , (a) lim (s + h)1−α Rα (s + h) − (h 1−α Rα (h))(s 1−α Rα (s)) = 0; h↓0

(b) lim s 1−α Rα (s) − (h 1−α Rα (h))((s − h)1−α Rα (s − h)) = 0. h↓0

Lemma 2.3 [6] Let 0 < σ ≤ 1 and a, b > 0. Then, |a σ − bσ | ≤ |a − b|σ . Lemma 2.4 [30] Let φ ∈ L p (J, X ) with 1 ≤ p < ∞. Then,  lim

s→0 0

b

φ(s + τ ) − φ(τ ) p dτ = 0,

where φ(t) = 0 for t ∈ / J.

3 Existence Results In this section, by utilizing a resolvent method, we formulate a concept of solutions to system (1) and investigate the corresponding existence of solutions. The conditions required are listed as follows: (H A) {s 1−α Rα (s)}s>0 is equicontinuous and compact. (H f ) f : J × X → X satisfies: (i) z → f (s, z) is continuous for a.e. s ∈ J and s → f (s, z) is measurable for each z ∈ X ; (ii) for a.e. s ∈ J and all z ∈ X ,  f (s, z) ≤ ϕ(s) + s 1−α z with ϕ ∈ α . L p (J, R+ ), p > α1 and 0 <  < Mb ∞ (H B) B ∈ L (J, L(Y, X )).

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Moreover, we introduce an admissible set Uad := {u ∈ L p (J, F) : u(s) ∈ U (s) a.e.}, where p > α1 , F ⊂ Y is a bounded set and U : J → P f c (Y ) is a measurable multivalued mapping satisfying U (·) ⊆ F. It follows from Proposition 2.1.7 of [31] that Uad = ∅. Furthermore, in view of the definition of Uad and condition (H B), it is not difficult to see that Bu ∈ L p (J, X ) for all u ∈ Uad . Definition 3.1 For fixed u ∈ Uad , by a solution to system (1) depending on u, we understand a function x ∈ C1−α (J, X ) satisfying that x(t) := gα (t)x0 + AItα x(t) + Itα ( f (t, x(t)) + B(t)u(t)), t ∈ J  .

(2)

We first give the properties of Rα ∗ g, where g ∈ L p (J, X ) with p > α1 . Lemma 3.1 Let g ∈ L p (J, X ) with p > α1 and condition (H A) be fulfilled. Then, for t t ∈ J , the convolution (Rα ∗g)(t) := 0 Rα (t −s)g(s)ds exists and Rα ∗g ∈ C(J, X ). Proof Following the proof of Proposition 1.3.4 in [32], for t ∈ J  , we can derive the measurability of Rα (t − ·)g(·) on ]0, t[. Additionally, we have  t    1−α α−1  (Rα ∗ g)(t) =  ((t − s) Rα (t − s))(t − s) g(s)ds   0

1− 1 p p−1 α− 1 ≤ Mg L p b p . αp − 1 Hence, (Rα ∗ g)(t) exists. We verify that Rα ∗ g ∈ C(J, X ). For convenience, denote ψ(t2 , t1 , s) := (t2 − s)1−α Rα (t2 − s) − (t1 − s)1−α Rα (t1 − s). Let 0 < ε < t1 < t2 ≤ b. In light of Lemma 2.3, we get (Rα ∗ g)(t2 ) − (Rα ∗ g)(t1 )  t1 −ε    α−1  ≤ ψ(t2 , t1 , s)(t2 − s) g(s)ds   0 t1    α−1  + ψ(t2 , t1 , s)(t2 − s) g(s)ds   t −ε  1t     1  1−α α−1 α−1  + (t g(s)ds − s) R (t − s) (t − s) − (t − s) 1 α 1 2 1   0  t    2  + (t2 − s)1−α Rα (t2 − s) (t2 − s)α−1 g(s)ds    t1

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≤

sup

s∈[0,t1 −ε]

ψ(t2 , t1 , s)g L p b

α− 1p





1− 1

p−1 αp − 1

p

1 p − 1 1− p α− 1p + 2Mg ε αp − 1  t1

1− 1  p p α−1 α−1 p−1 p (t2 − s) + Mg L − (t1 − s) ds Lp

+ Mg

Lp

0

p−1 αp − 1

1− 1

p

(t2 − t1 )

α− 1p

.

Thus, based on condition (H A), Lemma 2.4 and the arbitrariness of ε, we can infer that Rα ∗ g ∈ C(J  , X ). Moreover, the continuity of Rα ∗ g at t = 0 is obvious. Hence,   Rα ∗ g ∈ C(J, X ). We now display an equivalent concept of solutions of (1). Lemma 3.2 Assume that conditions (H A), (H f ) and (H B) are satisfied. Then, x ∈ C1−α (J, X ) is a solution to system (1) depending on u ∈ Uad if and only if x satisfies  x(t) = Rα (t)x0 +

t 0

Rα (t − s)( f (s, x(s)) + B(s)u(s))ds, t ∈ J  .

(3)

Proof For fixed u ∈ Uad , let x be a solution to system (1). Then, by applying Definition 3.1 and Lemma 2.1, we get gα ∗ x = (Rα − Agα ∗ Rα ) ∗ x = Rα ∗ (x − Agα ∗ x) = Rα ∗ (gα x0 + gα ∗ f (·, x(·)) + gα ∗ (Bu)) = gα ∗ (Rα x0 + Rα ∗ f (·, x(·)) + Rα ∗ (Bu)), which yields  x(t) = Rα (t)x0 +

0

t

 Rα (t − s) f (s, x(s))ds +

0

t

Rα (t − s)B(s)u(s)ds.

Conversely, suppose that x satisfies the expression (3). Then, according to the imposed conditions and employing Lemma 3.1, the expression (3) is well defined. By virtue of (c) of Definition 2.3, we have

1 1−α s I α x(t) Rα (s) − (α) t

 α  1 1−α It Rα (t)x0 + (gα ∗ Rα ∗ ( f (·, x(·))+ Bu))(t) Rα (s)− = s (α)  1−α α Is Rα (s) Rα (t)x0 − gα (t)x0 + (Rα ∗ f (·, x(·)))(t) =s

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 −(gα ∗ f (·, x(·)))(t) + (Rα ∗ (Bu))(t) − (gα ∗ (Bu))(t)  = s 1−α Isα Rα (s) x(t) − gα (t)x0 − Itα f (t, x(t)) − Itα (B(t)u(t))), which together with Lemma 2.1 gives

lim (2α)

 s 1−α Rα (s) −

1 (α)



Itα x(t)

sα

s↓0

= lim (2α)s 1−2α Isα Rα (s)(x(t)− gα (t)x0 − Itα f (t, x(t)) − Itα (B(t)u(t))) s↓0

= x(t) − gα (t)x0 − Itα f (t, x(t)) − Itα (B(t)u(t)). Hence, we obtain x(t) = gα (t)x0 + AItα x(t) + Itα f (t, x(t)) + Itα (B(t)u(t)),  

that is, x is a solution to system (1).

Set Br := {x ∈ C1−α (J, X ) : xC1−α ≤ r }, where r is a positive number. With the aid of Lemma 3.2, we are in a position to exhibit the existence result. Theorem 3.1 Suppose that hypotheses (H A), (H f ) and (H B) hold. Then, for every u ∈ Uad , problem (1) possesses at least one solution. Proof For fixed u ∈ Uad , define a map Q : C1−α (J, X ) → C1−α (J, X ) as  (Qx)(t) := Rα (t)x0 +

t 0

Rα (t − s)( f (s, x(s)) + B(s)u(s))ds.

Based on Lemma 3.1, the operator Q is well defined. Obviously, our problem becomes to show that for every u ∈ Uad , the map Q admits a fixed point in C1−α (J, X ). To make our later analysis more transparent, we discuss the proof in three steps. Step 1. We claim that Q(Br ) ⊆ Br , where  

1 α p − 1 1− p r≥ Mx0  + M b (ϕ L p + Bu L p ) . α − Mb αp − 1 For each x ∈ Br , according to hypotheses (H A), (H f ) and (H B), t 1−α (Qx)(t)

 t    1−α α−1  ((t − s) R (t − s))(t − s) f (s, x(s))ds ≤ t 1−α Rα (t)x0  + t 1−α  α   0  t    1−α + t 1−α  Rα (t − s))(t − s)α−1 B(s)u(s)ds   ((t − s)  0

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t

≤ Mx0  + Mb1−α

(t − s)α−1 (ϕ(s) + r + B(s)u(s))ds

0

1 p − 1 1− p Mbr ≤ Mx0  + M b (ϕ L p + Bu L p ) + αp − 1 α ≤ r, which shows that Q(Br ) ⊆ Br . Step 2. We check the continuity of Q on Br . For a sequence {xm }m≥1 ⊆ Br satisfying lim xm = x in Br , in view of (H f ), for t ∈ J , we obtain m→∞

(t − s)α−1 ( f (s, xm (s)) − f (s, x(s))) → 0 a.e. s ∈ [0, t] and (t − s)α−1  f (s, xm (s)) − f (s, x(s)) ≤ 2(t − s)α−1 (ϕ(s) + r ), s ∈ [0, t]. Thus, by employing the dominated convergence theorem, one has t 1−α (Qxm )(t) − (Qx)(t)  t 1−α (t − s)1−α Rα (t − s)(t − s)α−1  f (s, xm (s)) − f (s, x(s))ds ≤t 0  t 1−α ≤ Mb (t − s)α−1  f (s, xm (s)) − f (s, x(s))ds 0

→ 0, m → ∞, which leads to the continuity of Q on Br . Step 3. We demonstrate the compactness of the mapping Q : Br → Br . Put W := {z ∈ C(J, X ) : z(t) := t 1−α (Qx)(t), x ∈ Br , t ∈ J } and   W := z ∈ C(J, X ) : z(t) := t 1−α (Q 1 x)(t), x ∈ Br , t ∈ J , where 

t

(Q 1 x)(t) := 0

Rα (t − s)( f (s, x(s)) + B(s)u(s))ds.

In view of the relation between (Q Br ;  · C1−α ) and (W ;  · ), it suffices to examine the precompactness of the set W in C(J, X ). Firstly, by virtue of Step 1, we get z ≤ r .

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Then, we show the precompactness of W (t) := {z(t), z ∈ W } in X . For convenience, denote gt (ε, s) := (t − s)1−α Rα (t − s) − ((t − s − ε)1−α Rα (t − s − ε))(ε1−α Rα (ε)) and h t (ε, s) := ((t − s − ε)1−α Rα (t − s − ε))(ε1−α Rα (ε)). x0 } and t 1−α Rα (t)x0 , t ∈ J  are precompact, we only need to Since W (0) = { (α) prove that W (t) is precompact on J  . For any ε ∈ ]0, t[, one can deduce from the ε compactness of ε1−α Rα (ε) that W (t) := {z ε (t), x ∈ Br , t ∈ J  } is precompact, where

z ε (t) := (ε1−α Rα (ε))t 1−α



t−ε 0

Rα (t − s − ε)( f (s, x(s)) + B(s)u(s))ds.

Furthermore, for t ∈ J  and δ ∈ ]ε, t[, we get z(t) − z ε (t)  t−δ 1−α gt (ε, s)(t − s)α−1 (ϕ(s) + r + B(s)u(s))ds ≤b 0  t−ε + gt (ε, s)(t − s)α−1 (ϕ(s) + r + B(s)u(s))ds t−δ t−ε

 +

0



t

+

t−ε

h t (ε, s)((t − s − ε)α−1 − (t − s)α−1 )(ϕ(s) + r + B(s)u(s))ds

1−α α−1 (t − s) Rα (t − s)(t − s) (ϕ(s) + r + B(s)u(s))ds



t−δ

≤ b1−α 0



+ (M + M ) 2

 +M

2

gt (ε, s)(t − s)α−1 (ϕ(s) + r + B(s)u(s))ds

t−ε

p−1 αp − 1

1− 1

((t − s − ε)

p

α−1

(ϕ L p α−1

− (t − s)

0

+

δ

α− 1p

r Mεα 2r M 2 εα + + M α α



p−1 αp − 1

)

r δ α + Bu L p ) + α p p−1

1− 1

p

ε

p−1 p ds α− 1p



(ϕ L p + Bu L p )

 (ϕ L p + Bu L p ) .

Thanks to Lemma 2.2, for s ∈ [0, t − δ], we have lim gt (ε, s) = 0. ε↓0

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Thus, in view of the dominated convergence theorem, the arbitrariness of δ and Lemma 2.4, we obtain lim z(t) − z ε (t) = 0, t ∈ J  . ε↓0

Hence, we derive the precompactness of the set W (t) in X , t ∈ J  . Finally, we show the equicontinuity of W on J . For every z ∈ W , let 0 ≤ t1 < t2 ≤ b. If t1 = 0, in view of (a) of Definition 2.3 and the imposed conditions, we get z(t2 ) − z(0)    t2  1−α x0  1−α   Rα (t2 − s)( f (s, x(s)) + B(s)u(s))ds − ≤ t2 Rα (t2 )x0 + t2 (α)  0    

1  1−α p − 1 1− p x0  t2 r   ≤ t2 Rα (t2 )x0 − +M t2 (ϕ L p + Bu L p ) + (α)  αp − 1 α → 0, t2 → 0. If t1 > 0, z(t2 ) − z(t1 )       ≤ t21−α Rα (t2 )x0 − t11−α Rα (t1 )x0  + t21−α (Rα ∗ ( f (·, x(·)) + Bu))(t2 )   − t11−α (Rα ∗ ( f (·, x(·)) + Bu))(t1 )     ≤ t21−α Rα (t2 )x0 − t11−α Rα (t1 )x0      +  t21−α − t11−α (Rα ∗ ( f (·, x(·)) + Bu))(t2 ) + t11−α (Rα ∗ ( f (·, x(·)) + Bu))(t2 ) − (Rα ∗ ( f (·, x(·)) + Bu))(t1 )     ≤ t21−α Rα (t2 )x0 − t11−α Rα (t1 )x0   

1  p − 1 1− p Mr bα α− 1p 1−α 1−α + t2 − t1 Mb (ϕ L p + Bu L p ) + αp − 1 α + b1−α (Rα ∗ ( f (·, x(·)) + Bu)) (t2 ) − (Rα ∗ ( f (·, x(·)) + Bu)) (t1 )) . Thus, due to the imposed conditions as well as Lemma 3.1, we infer that lim z(t2 ) − z(t1 ) = 0,

t2 →t1

which leads to the equicontinuity of W on J . Thus, by the Arzela–Ascoli theorem, the precompactness of the set W is established in C(J, X ). Therefore, we derive the compactness of the map Q : Br → Br . Hence, by applying Schauder’s fixed point theorem, Q possesses a fixed point, which is a solution to system (1).  

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Remark 3.1 In Theorem 3.1, the uniqueness of solutions to system (1) cannot be acquired. For the convenience of later analysis, by S(u) we denote all solutions to system (1) depending on u ∈ Uad .

4 Optimal Controls In this section, without the Lipschitz continuity on the nonlinear term f , we utilize the technique of setting up minimizing sequences twice to seek an optimal state–control pair of the following limited Lagrange optimal control problem: Seek a state–control pair (x, ¯ u) ¯ ∈ C1−α (J, X ) × Uad satisfying J (x, ¯ u) ¯ ≤ J (x, u), for all u ∈ Uad , where x ∈ S(u) and



b

J (x, u) :=

(4)

L(τ, x(τ ), u(τ ))dτ.

0

On the cost integrand L : J × X × Y → R ∪ {+∞}, we suppose that (H L) (1) (τ, x, u) → L(τ, x, u) is measurable; (2) for a.e. τ ∈ J , (x, u) → L(τ, x, u) is sequentially lower semicontinuous; (3) for a.e. τ ∈ J and each x ∈ X , u → L(τ, x, u) is convex; p (4) L(τ, x, u) ≥ φ(τ ) + cx + duY with φ ∈ L 1 (J ; R), c ≥ 0 and d > 0. Lemma 4.1 Assume that conditions (H A), (H f ) and (H B) hold. Then, for fixed u ∈ Uad , there is a number λ > 0 such that xC1−α ≤ λ, where x ∈ S(u). Proof Since x ∈ S(u),  x(t) = Rα (t)x0 +

t 0

Rα (t − s)( f (s, x(s)) + B(s)u(s))ds, t ∈ J  .

By virtue of the imposed conditions, we obtain t 1−α x(t)

 t    1−α α−1  ((t − s) R (t − s))(t − s) f (s, x(s))ds ≤ t 1−α Rα (t)x0  + t 1−α  α   0  t    1−α + t 1−α  Rα (t − s))(t − s)α−1 B(s)u(s)ds   ((t − s)  0  t  ≤ Mx0  + Mb1−α (t − s)α−1 s 1−α x(s) + ϕ(s) + B(s)u(s) ds 0



1 p − 1 1− p ≤ Mx0  + Mb (ϕ L p + Bu L p ) αp − 1  t 1−α + Mb  (t − s)α−1 s 1−α x(s)ds. 1− 1p

(5)

0

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Setting ω(t) := t 1−α x(t) and k := Mx0  + Mb

1− 1p



p−1 αp − 1

1− 1

p

(ϕ L p + Bu L p )

together with (5) implies that 

t

ω(t) ≤ k + Mb1−α 

(t − s)α−1 ω(s)ds.

0

Applying Corollary 2 of [34] gives rise to ω(t) ≤ k E α (Mb(α)) := λ. This yields that xC1−α ≤ λ.

 

Lemma 4.2 Let conditions (H A) and (H B) be satisfied. Then, the operator  : L p (J, Y ) → C1−α (J, X ), p > α1 , defined by  (u)(·) := 0

·

Rα (· − s)B(s)u(s)ds,

is compact. Proof Let {u k }k≥1 be a bounded sequence in L p (J, Y ), p > α1 . Then, condition (H B) leads to the boundedness of {Bu k }k≥1 ⊆ L p (J, X ). Therefore, by a similar approach as employed in Step 3 of Theorem 3.1, one can derive the compactness of the operator .   Theorem 4.1 Suppose that conditions (H A), (H f ), (H B) and (H L) hold. Then the limited Lagrange optimal control problem possesses at least one optimal state–control pair. Proof For clarity, we proceed in the following steps to seek an optimal state–control pair. Step 1 Let u ∈ Uad be fixed and J (u) := inf J (x, u), and we will look for x∈S(u)

 x ∈ S(u) satisfying J ( x , u) = J (u). If S(u) contains only finitely many elements, the proof is obvious. If S(u) contains infinitely many elements, we can suppose that J (u) < +∞, since it is trivial for the case of J (u) = +∞. By means of condition (H L), we obtain J (u) > −∞. Thus, we can choose a sequence {xk }k≥1 ⊆ S(u) satisfying lim J (xk , u) = J (u). k→∞

Thanks to {xk }k≥1 ⊆ S(u), it holds that  xk (t) = Rα (t)x0 +

123

t 0

Rα (t − s)( f (s, xk (s)) + B(s)u(s))ds, t ∈ J  .

(6)

J Optim Theory Appl

According to Lemma 4.1, we have xk C1−α ≤ λ. Similar to the proof of Step 3 in Theorem 3.1, we can infer that {xk }k≥1 is a precompact set in C1−α (J, X ). Hence, there is  x ∈ C1−α (J, X ) and a subsequence extracted from {xk }k≥1 , still relabeled as x in C1−α (J, X ) and  x C1−α ≤ λ. Note that condition (H f ) itself, such that xk →  results in x (s))) Rα (t − s)( f (s, xk (s)) − f (s, 

= ((t − s)1−α Rα (t − s))(t − s)α−1 ( f (s, xk (s)) − f (s,  x (s))) α−1 ≤ 2M(t − s) (ϕ(s) + λ)

and Rα (t − s)( f (s, xk (s)) − f (s,  x (s))) 1−α = ((t − s) Rα (t − s))(t − s)α−1 ( f (s, xk (s)) − f (s,  x (s))) ≤ M(t − s)α−1  f (s, xk (s)) − f (s,  x (s)) → 0, a.e. s ∈ [0, t].

Hence, by taking the limit k → ∞ to both sides of (6) and exploiting the dominated convergence theorem, for t ∈ J  , one has   x (t) = Rα (t)x0 +

t

Rα (t − s)( f (s,  x (s)) + B(s)u(s))ds,

0

that is,  x ∈ S(u). Thus, due to (H L) and Theorem 2.1 of [33], we get  J (u) = lim J (xk , u) = lim k→∞

k→∞ 0



b

≥

b

L(τ, xk (τ ), u(τ ))dτ

L(τ,  x (τ ), u(τ ))dτ = J ( x , u)

0

≥ J (u), which yields J ( x , u) = J (u) = inf J (x, u). x∈S(u)

Step 2 We shall seek u 0 ∈ Uad which satisfies J (u 0 ) := inf J (u). u∈Uad

The case

inf J (u) = +∞ is easy to deal with. Now, we assume that

u∈Uad

inf J (u) < +∞. Using (H L) again, one yields inf J (u) > −∞. Hence, one

u∈Uad

u∈Uad

can take a sequence {u k }k≥1 ⊆ Uad satisfying lim J (u k ) = inf J (u). k→∞

u∈Uad

It follows from {u k }k≥1 ⊆ Uad that {u k }k≥1 is bounded in L p (J, Y ), p > α1 . Thus, w

we can extract a subsequence from {u k }k≥1 , still denoted by it, such that u k → u 0 , for some u 0 ∈ L p (J, Y ), p > α1 . By utilizing the closedness and convexity of Uad together with Mazur’s theorem, we obtain u 0 ∈ Uad .

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On the other hand, in view of Step 1, for any k ≥ 1, we can find  xk ∈ S(u k ) xk ∈ S(u k ), satisfying J ( xk , u k ) = J (u k ). Since    xk (t) = Rα (t)x0 +

t 0

Rα (t − s)( f (s,  xk (s)) + B(s)u k (s))ds.

(7)

Moreover, a similar argument as in Step 1 gives rise to the precompactness of { xk }k≥1 ⊆ xk → x 0 in C1−α (J, X ). Letting k → ∞ C1−α (J, X ). Hence, we may suppose that  w in (7), by virtue of the fact that u k → u 0 in L p (J, Y ) together with Lemma 4.2, we can infer that  x 0 (t) = Rα (t)x0 +

t

0

Rα (t − s)( f (s, x 0 (s)) + B(s)u 0 (s))ds, t ∈ J  ,

that is, x 0 ∈ S(u 0 ). Thus, exploiting Theorem 2.1 of [33] yields xk , u k ) inf J (u) = lim J (u k ) = lim J (

u∈Uad

k→∞



= lim ≥

k→∞ 0  b

k→∞

b

L(τ,  xk (τ ), u k (τ ))dτ

L(τ, x 0 (τ ), u 0 (τ ))dτ = J (x 0 , u 0 )

0

≥ J (u 0 ) ≥ inf J (u), u∈Uad

which shows that J (x 0 , u 0 ) = inf J (u) = inf u∈Uad

inf J (x, u).

u∈Uad x∈S(u)

Therefore, (x 0 , u 0 ) is an optimal state–control pair.

 

Remark 4.1 Since the Lipschitz assumption on f is not required, our result essentially generalizes some recent works about optimal controls of all evolution systems.

5 An Application In this section, a simple example is presented to illustrate the validity of the proposed theoretical results. Consider optimal controls for the following partial differential system with a Riemann–Liouville fractional derivative: ∂2 z(t, x) + f (t, z(t, x)) + u(t, x), x ∈ ]0, 1[ , t ∈ ]0, 1], ∂x2 z(t, 0) = z(t, 1) = 0, D α z(t, x) =

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lim (α)t 1−α z(t, x) = h(x) = t↓0

∞ 

ck sin kπ x,

(8)

k=1

with a cost function 

1 1

J (z, u) := 0

0

where 0 < α < 1. Let X := L 2 (0, 1) and A :=

∂2 ∂x2

(|z(t, x)|2 + |u(t, x)|2 )dxdt,

with domain

D(A) := {v ∈ X : v  , v  ∈ X, v(0) = v(1) = 0}. Then, A generates an analytic compact semigroup T (t), t > 0, which is given by T (t)v :=

∞ 

e−k

2π 2t

v, ek ek , v ∈ X,

(9)

k=1

√ where ek (x) := 2 sin(kπ x) and {ek , k = 1, 2, · · · } is an orthonormal basis of X . Obviously, expression (9) implies that T (t) ≤ 1 and T (t)h(x) =

∞ 

e−k

2π 2t

ck sin kπ x.

k=1

On the other hand, in view of [13], A also generates an α-order resolvent {Rα (t)}t>0 : Rα (t)h(x) :=

∞ 

t α−1 E α,α (−k 2 π 2 t α )ck sin kπ x.

k=1

By employing the technique of Laplace transformations and probability densities [9], for any h ∈ X , we can conclude that  t 1−α Rα (t)h(x) = α

∞ 0

sξα (s)T (t α s)h(x)ds,

which implies that  t

1−α

Rα (t) = α

∞ 0

sξα (s)T (t α s)ds,

where ξα (s) :=

1 1 −1− 1 α α (s − α ), s α

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α (s) :=

∞ (kα + 1) 1 sin(kπ α), s ∈ ]0, ∞[. (−1)k−1 s −kα−1 π k! k=1

Similar to the proof in [35], we can infer that {t 1−α Rα (t)}t>0 is equicontinuous 1 . Therefore, (H A) is and compact. Furthermore, it is clear that t 1−α Rα (t) ≤ (α) satisfied. Now, for every t, x ∈ [0, 1], let z(t)(x) := z(t, x), B(t)u(t)(x) := u(t, x). Moreover, we assume that (H f ) f : [0, 1] × X → X , given by f (t, y)(x) := f (t, y(x)), is continuous and  f (t, y) ≤ a(t) + ct 1−α y with a ∈ L p ([0, 1], R+ ), p > α1 and 0 < c < α(α). Then, problem (8) can be modeled as the abstract system (1) with the cost function  J (z, u) :=

1

(z(t)2 + u(t)2 )dt,

0

and all the conditions of Theorem 4.1 hold. Therefore, problem (8) possesses optimal state–control pairs.

6 Conclusions In this paper, by utilizing an approach of resolvent, we have introduced an appropriate concept of solutions to system (1) and investigated the existence of the system. Moreover, by setting up minimizing sequences twice, we have displayed the optimal control result. It is worth pointing out that this technique can be applied to all previous works on optimal controls, including integer-order and fractional-order systems. Furthermore, our work essentially generalizes previous results on optimal controls of all evolution systems, since we have removed the Lipschitz assumption on the nonlinear function without any additional conditions. Acknowledgements The authors are grateful to the editor and the referees for their constructive comments and suggestions for the improvement in the paper. Furthermore, the work was supported by the NSF of China (11571300, 11271316), the Qing Lan Project of Jiangsu Province of China, the Graduate Research, Innovation Projects in Jiangsu Province (KYLX16-1382) and the High-Level Personnel Support Program of Yangzhou University.

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