Existence and optimal control of Hilfer fractional evolution equations

Mian Zhou; Yong Zhou; Jia Wei He; Yong Liang

doi:10.1515/dema-2024-0093

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Existence and optimal control of Hilfer fractional evolution equations

Mian Zhou , Yong Zhou , Jia Wei He and Yong Liang

Published/Copyright: January 21, 2025

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From the journal Demonstratio Mathematica Volume 58 Issue 1

Abstract

This article investigates the existence and optimal controls for a class of Hilfer fractional evolution equations of order in ( 0 , 1 ) with type of [ 0 , 1 ]. The first part focuses on analyzing the existence and uniqueness of the mild solution using the fixed point theorem and semigroup theory. In the second part, an optimal control problem that admits an optimal pair is considered. Furthermore, the necessary optimality conditions for an optimal control problem governed by semilinear fractional diffusion equations are presented.

Keywords: existence; uniqueness; optimal control; fractional evolution equations; Hilfer derivative

MSC 2010: 49J20; 49K20; 35R11; 26A33

1 Introduction

Fractional calculus is an area of mathematical analysis that applies the concepts of differentiation and integration to noninteger orders. It has been drawing more attention lately due to its capacity to give more accurate descriptions of systems with complicated dynamics and memory effects that integer-order calculus cannot effectively describe. Fractional differential equations are used in a wide range of fields, including physics, engineering, signal processing, control theory, and economics. The applications and outcomes of fractional calculus have been influenced by the concepts of fractional integrals and derivatives, such as Caputo, Riemann-Liouville, Riesz-Caputo, Grunwald-Letnikov, Caputo-Fabrizio, etc., see the monographs [1–5] and recent references. Hilfer invented the Hilfer fractional derivative, which takes the form D 0 + μ , λ H , where λ ∈ ( 0 , 1 ) denotes the order and μ ∈ [ 0 , 1 ] denotes the type. The type μ allows for interpolation between the Riemann-Liouville fractional derivative and the Caputo fractional derivative when μ = 0 and μ = 1 . The conclusion of the Hilfer fractional derivative is a generalization of the Riemann-Liouville fractional derivative and Caputo fractional derivatives.

There are many results about the existence and optimal control of solutions for fractional systems, see [6–9] and references therein. Initially, Wang and Zhou [6] proved the existence of mild solutions for semilinear fractional evolution equations and optimal controls in the α -norm in the sense of the Caputo fractional derivative. Mophou [7] obtained the existence and the uniqueness of the solution of the fractional diffusion equation in the sense of the Riemann-Liouville fractional derivative. She also showed that that there is a unique solution and an optimality system for optimal control. In a recent study, Xiao and Peng [10] studied solvability and optimal control of semilinear fractional evolution equations with Riemann-Liouville fractional derivatives. Chang et al. [11] considered a fractional stochastic evolution equation of Sobolev type, and they showed the existence of solutions and existence conditions of optimal pairs in optimal control systems. Liu and Wang [12] considered the solvability and the existence result of optimal controls for the Lagrange problem for an optimal control problem with no instantaneous impulses. We remark that the existence of solutions for fractional evolution equations with Hilfer derivative has already been investigated in [13,14]. Gu and Trujillo [13], Zhou and He [14] established the existence of mild solutions on finite and infinite intervals respectively.

Consider the Cauchy problem of fractional evolution equations with Hilfer derivatives

(1) D 0 + μ , λ H y ( t ) = A y ( t ) + g ( t , y ( t ) ) a.e. t > 0 , I 0 + ( 1 − λ ) ( 1 − μ ) y ( 0 + ) = y 0 ,

where D 0 + μ , λ H is the Hilfer fractional derivative of order 0 < λ < 1 and type 0 ≤ μ ≤ 1 , A is the infinitesimal generator of a strongly continuous semigroup of bounded linear operators (i.e., C 0 -semigroup) { T ( t ) } t ≥ 0 in Banach space X , g : [ 0 , T ] × X → X is a nonlinear term which will be specified later. Moreover, I 0 + ( 1 − λ ) ( 1 − μ ) is Riemann-Liouville integral of order ( 1 − λ ) ( 1 − μ ) , y 0 ∈ X and I 0 + ( 1 − λ ) ( 1 − μ ) y ( 0 + ) = lim t → 0 + I 0 + ( 1 − λ ) ( 1 − μ ) y ( t ) , where I 0 + ( 1 − λ ) ( 1 − μ ) stands for the Riemann-Liouville integral of order ( 1 − λ ) ( 1 − μ ) .

In this article, we investigate the existence and optimal controls of this type of problem with the Hilfer fractional derivative. The remaining part of the article is structured as follows: Section 2 provides basic definitions, notations, and significant propositions. Section 3 investigates the solvability and uniform boundedness of the mild solutions to problem (1). Section 4 presents the existence of optimal pairs of the states, and then we derive the necessary optimality condition by the Lagrange multiplier method for the optimal control governed by semilinear fractional diffusion equations.

2 Preliminaries

We first introduce some notations and definitions about fractional calculus, function spaces, and the definition of mild solutions. For more details, we refer to [1,2]. Let J = [ 0 , T ] . Assume that X is a Banach space with the norm ∣ ⋅ ∣ . Let C ( J , X ) be the Banach space of all continuous functions from J into X with the norm

‖ x ‖ C ( J , X ) = sup t ∈ J ∣ x ( t ) ∣ ,

where x ∈ C ( J , X ) . Let A C ( J , X ) be the space of functions that are absolutely continuous on J . Let 1 ≤ p ≤ ∞ . L p ( J , X ) denotes the Banach space of all measurable functions f : J → X . L p ( J , X ) is normed by

‖ f ‖ L p ( J , X ) = ∫ J ∣ f ( t ) ∣ p d t 1 p , 1 ≤ p < ∞ , inf μ ( J ¯ ) = 0 { sup t ∈ J \ J ¯ ∣ f ( t ) ∣ } , p = ∞ .

Definition 2.1

(See [2,4].) The left and right fractional integrals of order λ > 0 for a function u ∈ L 1 ( J , X ) are defined as

I 0 + λ u ( t ) = 1 Γ ( λ ) ∫ 0 t ( t − s ) λ − 1 u ( s ) d s , t > 0 ,

and

I T − λ u ( t ) = 1 Γ ( λ ) ∫ t T ( s − t ) λ − 1 u ( s ) d s , t < T ,

respectively, where Γ ( ⋅ ) is the gamma function.

Definition 2.2

(Hilfer fractional derivative, see [1]). Let 0 < λ < 1 and 0 ≤ μ ≤ 1 . The Hilfer fractional derivative of order λ and type μ for a function u ∈ A C ( J , X ) is defined as

D 0 + μ , λ H u ( t ) = I 0 + μ ( 1 − λ ) d d t I 0 + ( 1 − λ ) ( 1 − μ ) u ( t )

as well as the right Hilfer fractional derivative of order λ and type μ is defined as

D T − μ , λ H u ( t ) = − I T − ( 1 − λ ) ( 1 − μ ) d d t I T − μ ( 1 − λ ) u ( t ) ,

Remark 2.1

In particular, for 0 < λ < 1 , then
D 0 + 0 , λ H u ( t ) = d d t I 0 + 1 − λ u ( t ) ≕ D 0 + λ L u ( t ) , D T − 1 , λ H u ( t ) = − d d t I T − 1 − λ u ( t ) ≕ D T − λ L u ( t ) ,
where D 0 + λ L and D 0 + λ L are the left and right Riemann-Liouville derivatives, respectively.
For 0 < λ < 1 , then
D 0 + 1 , λ H u ( t ) = I 0 + 1 − λ d d t u ( t ) ≕ D 0 + λ C u ( t ) , D T − 0 , λ H u ( t ) = − I T − 1 − λ d d t u ( t ) ≕ D T − 0 , λ C u ( t ) ,
where D 0 + λ C and D T − 0 , λ C u ( t ) are the left and right Caputo derivatives, respectively.

Definition 2.3

The Banach space C λ ( J , X ) is defined by

C λ ( J , X ) = { y : J ′ → X ; t ( 1 − λ ) ( 1 − μ ) y ∈ C ( J , X ) } ,

with the norm

‖ y ‖ C λ ( J , X ) = sup t ∈ J { t ( 1 − λ ) ( 1 − μ ) ∣ y ( t ) ∣ } ,

where J ′ = ( 0 , T ] . Note that t ( 1 − λ ) ( 1 − μ ) y ∈ C ( J , X ) is understood that the limit η = lim t → 0 t ( 1 − λ ) ( 1 − μ ) y exists, and the function t ( 1 − λ ) ( 1 − μ ) y is continuous on J by taking the value at t = 0 with the limit η .

Definition 2.4

[15] Define Wright function W λ ( θ ) by

W λ ( θ ) = ∑ n = 1 ∞ ( − θ ) n − 1 ( n − 1 ) ! Γ ( 1 − λ n ) , 0 < λ < 1 , θ ∈ C ,

with the following property

∫ 0 ∞ θ δ W λ ( θ ) d θ = Γ ( 1 + δ ) Γ ( 1 + λ δ ) for δ ≥ 0 .

We have an identity

∫ 0 ∞ M λ ( θ ) e − z θ d θ = E λ , 1 ( − z ) , ∫ 0 ∞ λ θ M λ ( θ ) e − z θ d θ = E λ , λ ( − z ) , z ∈ C .

3 Existence results

We introduce the following hypotheses:

Operator A is the infinitesimal generator of a strongly continuous semigroup of exponential bounded linear operators { T ( t ) } t ≥ 0 with ∣ T ( t ) ∣ ≤ M for some constant M > 0 and all t ≥ 0 .
g ( t , ⋅ ) is Lebesgue measurable with respect to t on [ 0 , ∞ ) with ϕ ( ⋅ ) ≔ ‖ g ( ⋅ , 0 ) ‖ ∈ L p ( J ) , and there exists a constant L > 0 such that
(2) ∣ g ( t , y 1 ) − g ( t , y 2 ) ∣ ≤ L ∣ y 1 − y 2 ∣ , a.e. t ∈ J , ∀ y 1 , y 2 ∈ X .

Lemma 3.1

[13] Assume that y ( ⋅ ) satisfies problem (1). Then

y ( t ) = P μ , λ ( t ) y 0 + ∫ 0 t Q λ ( t − s ) g ( s , y ( s ) ) d s , t ∈ ( 0 , ∞ ) ,

where

P μ , λ ( t ) = I 0 + μ ( 1 − λ ) Q λ ( t ) , Q λ ( t ) = t λ − 1 S λ ( t ) , a n d S λ ( t ) = ∫ 0 ∞ λ θ W λ ( θ ) T ( t λ θ ) d θ .

In view of Lemma 3.1, we have the following definition.

Definition 3.1

If y ∈ C ( J ′ , X ) satisfies

y ( t ) = P μ , λ ( t ) y 0 + ∫ 0 t Q λ ( t − s ) g ( s , y ( s ) ) d s , t ∈ J ′ ,

then y ( ⋅ ) is called a mild solution of the initial value problem (1).

By using the similar method as in [4], we have the following lemmas.

Lemma 3.2

For any fixed t > 0 , S λ ( t ) is linear and bounded operator, i.e., for any x ∈ X

∣ S λ ( t ) x ∣ ≤ M Γ ( λ ) ∣ x ∣ .

Lemma 3.3

{ S α ( t ) } t > 0 is strongly continuous, which means that, for ∀ x ∈ X and t ″ > t ′ > 0 , we have

∣ S λ ( t ″ ) x − S λ ( t ′ ) x ∣ → 0 , a s t ″ → t ′ .

Lemma 3.4

Assume that { T ( t ) } t > 0 is a compact operator. Then, { S λ ( t ) } t > 0 is also a compact operator.

Lemma 3.5

Let 1 p < λ < 1 and (H1) hold. Then, the function

φ ( t ) ≔ ∫ 0 t Q λ ( t − s ) k ( s ) d s , t ∈ J ,

belongs to C ( J , X ) for all k ∈ L p ( J , X ) .

Proof

By assumption (H1), for 0 < t < t + h < T with h > 0 , we have

φ ( t + h ) − φ ( t ) = ∫ 0 t + h Q λ ( t + h − s ) k ( s ) d s − ∫ 0 t Q λ ( t − s ) k ( s ) d s = ∫ − h t Q λ ( t − s ) k ( s + h ) d s − ∫ 0 t Q λ ( t − s ) k ( s ) d s = ∫ 0 t Q λ ( t − s ) ( k ( s + h ) − k ( s ) ) d s + ∫ − h 0 Q λ ( t − s ) k ( s + h ) d s .

This means that

∣ φ ( t + h ) − φ ( t ) ∣ ≤ M Γ ( λ ) ∫ 0 t ( t − s ) λ − 1 ∣ k ( s + h ) − k ( s ) ∣ d s + M Γ ( λ ) ∫ − h 0 ( t − s ) λ − 1 ∣ k ( s + h ) ∣ d s .

The Hölder inequality shows that

∫ 0 t ( t − s ) λ − 1 ∣ k ( s + h ) − k ( s ) ∣ d s ≤ ∫ 0 t ( t − s ) ( λ − 1 ) p p − 1 d s p − 1 p ∫ 0 t ∣ k ( s + h ) − k ( s ) ∣ p d s 1 p .

Note that t ( λ − 1 ) p p − 1 is L 1 -integral for λ > 1 p , and

∫ 0 t ∣ k ( s + h ) − k ( s ) ∣ p d s → 0 , as h → 0 + ,

because k ∈ L p ( J , X ) , which also implies that

∣ φ ( t + h ) − φ ( t ) ∣ → 0 , as h → 0 + ,

since

∫ − h 0 ( t − s ) λ − 1 ∣ k ( s + h ) ∣ d s = ∫ 0 h ( t + h − s ) λ − 1 ∣ k ( s ) ∣ d s ≤ ∫ 0 h ( t + h − s ) ( λ − 1 ) p p − 1 d s p − 1 p ∫ 0 h ∣ k ( s ) ∣ d s 1 p → 0 , as h → 0 + .

Conversely, for 0 < t − h < t < T with h > 0 , we have

φ ( t ) − φ ( t − h ) = ∫ 0 t Q λ ( t − s ) k ( s ) d s − ∫ h t Q λ ( t − s ) k ( s − h ) d s = ∫ h t Q λ ( t − s ) ( k ( s ) − k ( s − h ) ) d s + ∫ 0 h Q λ ( t − s ) k ( s ) d s .

Hence, the limit φ ( t − h ) → φ ( t ) as h → 0 − follows the aforementioned arguments. The proof is completed.□

Theorem 3.1

Assume that (H1) and (H2) hold. Then problem (1) admits a unique mild solution y ∈ C ( J ′ , X ) ∩ L p ( J , X ) with p ∈ ( 1 λ , 1 ( 1 − μ ) ( 1 − λ ) ) . Moreover,

‖ y ‖ L p ( J , X ) ≤ ( η 1 ( 1 + η 2 T e η 2 T ) ) 1 ⁄ p ,

where

η 1 = 3 p M p ( Γ ( μ ( 1 − λ ) + λ ) ) p 1 1 − ( 1 − μ ) ( 1 − λ ) p T 1 − ( 1 − μ ) ( 1 − λ ) p ∣ y 0 ∣ p + 3 p M p κ p T ‖ ϕ ‖ L p ( J ) p , κ = 1 Γ ( λ ) p − 1 p − 1 + ( λ − 1 ) p p − 1 p .

Proof

Let us define an operator ℱ by

( ℱ y ) ( t ) = P μ , λ ( t ) y 0 + ∫ 0 t Q λ ( t − s ) g ( s , y ( s ) ) d s , t > 0 .

We first claim that operator ℱ maps L p ( J , X ) into itself. In fact, note that

‖ ℱ y ‖ L p ( J , X ) ≤ ‖ P μ , λ ( ⋅ ) y 0 ‖ L p ( J , X ) + ∫ 0 ⋅ Q λ ( ⋅ − s ) g ( s , y ( s ) ) d s L p ( J , X ) .

From Lemma 3.2, and t − ( 1 − μ ) ( 1 − λ ) p is L 1 -integral for 1 p > ( 1 − μ ) ( 1 − λ ) , it yields

‖ P μ , λ ( ⋅ ) y 0 ‖ L p ( J , X ) ≤ 1 Γ ( μ ( 1 − λ ) ) ∫ 0 ⋅ ( ⋅ − s ) μ ( 1 − λ ) − 1 Q λ ( s ) y 0 d s L p ( J , X ) ≤ M Γ ( λ ) Γ ( μ ( 1 − λ ) ) ∫ 0 ⋅ ( ⋅ − s ) μ ( 1 − λ ) − 1 s λ − 1 d s L p ( J ) ∣ y 0 ∣ = M Γ ( μ ( 1 − λ ) + λ ) 1 1 − ( 1 − μ ) ( 1 − λ ) p 1 ⁄ p T 1 p − ( 1 − μ ) ( 1 − λ ) ∣ y 0 ∣ .

By assumption (H2), for any y ∈ L p ( J , X ) , Young’s inequality shows that

∫ 0 ⋅ Q λ ( ⋅ − s ) g ( s , y ( s ) ) d s L p ( J , X ) ≤ M Γ ( λ ) ∫ 0 ⋅ ( ⋅ − s ) λ − 1 ( ϕ ( s ) + L ∣ y ( s ) ∣ ) d s L p ( J ) ≤ M Γ ( λ + 1 ) T λ ‖ ϕ ‖ L p ( J ) + M L Γ ( λ + 1 ) T λ ‖ y ‖ L p ( J , X ) .

Therefore, ℱ : L p ( J , X ) → L p ( J , X ) . In particular, ℱ is a contraction in L p ( J , X ) . Form assumption (H2), note that for any x , y ∈ L p ( 0 , t ; X ) , Hölder’s inequality shows that

∣ ( ℱ y ) ( s ) − ( ℱ x ) ( s ) ∣ = ∫ 0 t Q λ ( t − s ) ( g ( s , y ( s ) ) − g ( s , x ( s ) ) ) d s ≤ M L Γ ( λ ) ∫ 0 t ( t − s ) λ − 1 ∣ y ( s ) − x ( s ) ∣ d s ≤ M L κ t λ − 1 p ‖ y − x ‖ L p ( 0 , t ; X ) .

We also have

∣ ( ℱ 2 y ) ( s ) − ( ℱ 2 x ) ( s ) ∣ = ∫ 0 t Q λ ( t − s ) ( g ( s , ( ℱ y ) ( s ) ) − g ( s , ( ℱ x ) ( s ) ) ) d s ≤ M L Γ ( λ ) ∫ 0 t ( t − s ) λ − 1 ∣ ( ℱ y ) ( s ) − ( ℱ x ) ( s ) ∣ d s ≤ ( M L ) 2 κ Γ ( λ − 1 p + 1 ) Γ ( 2 λ − 1 p + 1 ) t 2 λ − 1 p ‖ y − x ‖ L p ( 0 , t ; X ) .

It can be shown, by mathematical induction, that for every n ∈ N

∣ ( ℱ n y ) ( s ) − ( ℱ n x ) ( s ) ∣ ≤ ( M L ) n κ Γ ( λ − 1 p + 1 ) Γ ( n λ − 1 p + 1 ) t n λ − 1 p ‖ y − x ‖ L p ( 0 , t ; X ) .

The above inequality implies that

‖ ℱ n y − ℱ n x ‖ L p ( J , X ) ≤ 1 λ n p 1 p ( M L ) n κ Γ ( λ − 1 p + 1 ) Γ n λ − 1 p + 1 t n λ ‖ y − x ‖ L p ( J , X ) .

Since

lim n → ∞ 1 λ n p 1 p ( M L ) n κ Γ ( λ − 1 p + 1 ) Γ ( n λ − 1 p + 1 ) T n λ = 0 ,

there exists a positive integer n 0 large enough such that

1 λ n p 1 p ( M L ) n κ Γ ( λ − 1 p + 1 ) Γ ( n λ − 1 p + 1 ) t n λ ≤ 1 λ n 0 p 1 p ( M L ) n 0 κ Γ ( λ − 1 p + 1 ) Γ ( n 0 λ − 1 p + 1 ) T n 0 λ = k < 1 .

Then, ‖ ℱ n y − ℱ n x ‖ L p ( J , X ) ≤ k ‖ y − x ‖ L p ( J , X ) , and ℱ n 0 is contractive in L p ( J , X ) . For any y ∈ L p ( J , X ) , we know that ℱ y ∈ L p ( J , X ) , the induction also shows that ℱ n 0 y ∈ L p ( J , X ) . In fact, we just show n = 2 as follows:

‖ ℱ 2 y ‖ L p ( J , X ) ≤ ‖ P μ , λ ( ⋅ ) y 0 ‖ L p ( J , X ) + ∫ 0 ⋅ Q λ ( ⋅ − s ) g ( s , ( ℱ y ) ( s ) ) d s L p ( J , X ) ≤ M Γ ( μ ( 1 − λ ) + λ ) 1 1 − ( 1 − μ ) ( 1 − λ ) p 1 ⁄ p T 1 p − ( 1 − μ ) ( 1 − λ ) ∣ y 0 ∣ + M Γ ( λ + 1 ) T λ ‖ ϕ ‖ L p ( J ) + M L Γ ( λ + 1 ) T λ ‖ ℱ y ‖ L p ( J , X ) < + ∞ .

We thus prove that ℱ n 0 has a unique fixed point y = ℱ n 0 y in L p ( J , X ) by the Banach fixed point theorem. Since ℱ ( ℱ n 0 y ) = ℱ n 0 + 1 y = ℱ n 0 ( ℱ y ) and the uniqueness of fixed point, we obtain ℱ y = y in L p ( J , X ) .

We next claim that y is a mild solution of problem (1), it suffices to check that y ∈ C ( J ′ , X ) . Note that f ( ⋅ , y ( ⋅ ) ) ∈ L p ( J , X ) for all y ∈ L p ( J , X ) by (H2), Lemma 3.5 implies that

∫ 0 ⋅ Q λ ( ⋅ − s ) g ( s , y ( s ) ) d s ∈ C ( J , X ) .

Moreover, when t → 0 + , one can see that

P μ , λ ( t ) y 0 = I μ ( 1 − λ ) Q λ ( t ) y 0 = 1 Γ ( μ ( 1 − λ ) ) ∫ 0 t ( t − s ) μ ( 1 − λ ) − 1 Q λ ( s ) y 0 d s = t − ( 1 − μ ) ( 1 − λ ) Γ ( μ ( 1 − λ ) ) ∫ 0 1 ( 1 − s ) μ ( 1 − λ ) − 1 s λ − 1 S λ ( t s ) y 0 d s → ∞ ,

because S λ ( t s ) → I as t → 0 and the integrand is L 1 -integral. Hence, let us consider 0 < t ′ < t ″ ≤ T , then

∣ P μ , λ ( t ″ ) y 0 − P μ , λ ( t ′ ) y 0 ∣ = ∣ I μ ( 1 − λ ) Q λ ( t ″ ) y 0 − I μ ( 1 − λ ) Q λ ( t ′ ) y 0 ∣

≤ 1 Γ ( μ ( 1 − λ ) ) ∫ t ′ t ″ ( t ″ − s ) μ ( 1 − λ ) − 1 Q λ ( s ) y 0 d s + 1 Γ ( μ ( 1 − λ ) ) ∫ 0 t ′ ( ( t ″ − s ) μ ( 1 − λ ) − 1 − ( t ′ − s ) μ ( 1 − λ ) − 1 ) Q λ ( s ) y 0 d s ≤ M ∣ y 0 ∣ Γ ( λ ) Γ ( μ ( 1 − λ ) ) ∫ t ′ t ″ ( t ″ − s ) μ ( 1 − λ ) − 1 s λ − 1 d s + M ∣ y 0 ∣ Γ ( λ ) Γ ( μ ( 1 − λ ) ) ∫ 0 t ′ ( ( t ′ − s ) μ ( 1 − λ ) − 1 − ( t ″ − s ) μ ( 1 − λ ) − 1 ) s λ − 1 d s .

On the one hand, we have

∫ t ′ t ″ ( t ″ − s ) μ ( 1 − λ ) − 1 s λ − 1 d s = ∫ 0 t ″ − t ′ ( t ″ − t ′ − s ) μ ( 1 − λ ) − 1 ( s + t ′ ) λ − 1 d s ≤ ∫ 0 t ″ − t ′ ( t ″ − t ′ − s ) μ ( 1 − λ ) − 1 s λ − 1 d s = Γ ( μ ( 1 − λ ) ) Γ ( λ ) Γ ( μ ( 1 − λ ) + λ ) ( t ″ − t ′ ) μ ( 1 − λ ) + λ − 1 → 0 ,

as t ″ → t ′ . On the other hand, in view of ( t ′ − s ) μ ( 1 − λ ) − 1 → ( t ″ − s ) μ ( 1 − λ ) − 1 a.e. s ∈ [ 0 , t ′ ] , ( x − s ) μ ( 1 − λ ) − 1 s λ − 1 ∈ L 1 [ 0 , t ′ ] for x ∈ { t ′ , t ″ } . Then, the Lebesgue dominated convergence theorem shows that

∫ 0 t ′ ( ( t ′ − s ) μ ( 1 − λ ) − 1 − ( t ″ − s ) μ ( 1 − λ ) − 1 ) s λ − 1 d s → 0 , as t ″ → t ′ ,

which means that P μ , λ ( ⋅ ) y 0 ∈ C ( J ′ , X ) .

To obtain the prior estimate, we note that

∣ y ( t ) ∣ ≤ ∣ P μ , λ ( t ) y 0 ∣ + ∫ 0 t ∣ Q λ ( t − s ) g ( s , y ( s ) ) ∣ d s ≤ M t − ( 1 − μ ) ( 1 − λ ) Γ ( λ + μ ( 1 − λ ) ) ∫ 0 1 ( 1 − s ) μ ( 1 − λ ) − 1 s λ − 1 d s ∣ y 0 ∣ + M Γ ( λ ) ∫ 0 t ( t − s ) λ − 1 ϕ ( s ) d s + M L Γ ( λ ) ∫ 0 t ( t − s ) λ − 1 ∣ y ( s ) ∣ d s ≤ M t − ( 1 − μ ) ( 1 − λ ) Γ ( λ + μ ( 1 − λ ) ) ∣ y 0 ∣ + M κ ‖ ϕ ‖ L p ( J ) + M L Γ ( λ ) ∫ 0 t ( t − s ) λ − 1 ∣ y ( s ) ∣ d s .

Timing exponent number p both sides on the above inequalities, and integrating over ( 0 , t ) with t ∈ J , we have

∫ 0 t ∣ y ( s ) ∣ p d s ≤ 3 p M p ( Γ ( μ ( 1 − λ ) + λ ) ) p 1 1 − ( 1 − μ ) ( 1 − λ ) p t 1 − ( 1 − μ ) ( 1 − λ ) p ∣ y 0 ∣ p + 3 p M p κ p t ‖ ϕ ‖ L p ( J ) p + ( 3 M L κ ) p t λ p − 1 ∫ 0 t ∫ 0 τ ∣ y ( s ) ∣ p d s d τ ,

where we have used the Hölder inequality and ( a + b + c ) p ≤ 3 p ( a p + b p + c p ) for a , b , c ∈ R + . Now, let

η 1 = 3 p M p ( Γ ( μ ( 1 − λ ) + λ ) ) p 1 1 − ( 1 − μ ) ( 1 − λ ) p T 1 − ( 1 − μ ) ( 1 − λ ) p ∣ y 0 ∣ p + 3 p M p κ p T ‖ ϕ ‖ L p ( J ) p ,

and ξ ( t ) = ∫ 0 t ‖ y ( s ) ‖ p d s , we have

ξ ( t ) ≤ η 1 + η 2 ∫ 0 t ξ ( s ) d s .

The Gronwall inequality shows that

ξ ( t ) ≤ η 1 ( 1 + η 2 t e η 2 t ) , t ∈ J .

The proof is completed.□

Remark 3.1

An admissible condition of λ , μ in Theorem 3.1 is 1 − μ 2 − μ < λ < 1 . The fractional problem (1) in the sense of Riemann-Liouville’s type with μ = 0 is proved in [10]. We note that the result in the sense of Caputo’s type with μ = 1 may be new, the admissible condition will reduce to p > 1 λ .

Remark 3.2

Note that the existing space of solution in Theorem 3.1 is more precise compared to [13], where they obtained the existence of mild solution on a bounded, closed, and convex set in C λ ( J , X ) , and the proofs were shown under the slightly strong assumptions that T ( t ) is continuous in the uniform operator topology for t > 0 and ∣ f ( t , x ) ∣ ≤ m ( t ) for all x ∈ X and almost all t ∈ J with I 0 + μ m ∈ C ( J ′ , X ) , t ( 1 − μ ) ( 1 − λ ) I 0 + μ m ( t ) → 0 as t → 0 .

The embedding C λ ( J , X ) ⊂ L p ( J , X ) holds for 1 p > ( 1 − μ ) ( 1 − λ ) . In fact, let y ∈ C λ ( J , X ) , it follows that

∫ 0 T ∣ y ( s ) ∣ p d s = ∫ 0 T s − ( 1 − μ ) ( 1 − λ ) p s ( 1 − μ ) ( 1 − λ ) p ∣ y ( s ) ∣ p d s ≤ ‖ y ‖ C λ ( J , X ) p 1 1 − ( 1 − μ ) ( 1 − λ ) p T 1 − ( 1 − μ ) ( 1 − λ ) p ,

which implies that y ∈ L p ( J , X ) . Thus, the desired embedding follows. Since P μ , λ ( t ) y 0 = O ( t − ( 1 − μ ) ( 1 − λ ) ) as t → 0 + , from the proof in Theorem 3.1, by Henry’s inequality in [16, Lemma 7.1.2.], we thus have the following conclusion.

Corollary 3.1

Assume that (H1) and (H2) hold. Then, problem (1) admits a unique mild solution y ∈ C λ ( J , X ) . Moreover,

‖ y ‖ C λ ( J , X ) ≤ η 1 E λ , 1 − ( 1 − μ ) ( 1 − λ ) ( ( η 2 ′ Γ ( λ ) ) 1 ⁄ ν T ) ,

where ν = λ − ( 1 − μ ) ( 1 − λ ) > 0 , and for p λ > 1

η 1 ′ = M Γ ( λ + μ ( 1 − λ ) ) ‖ y 0 ‖ + M Γ ( λ ) p − 1 p − 1 + ( λ − 1 ) p p − 1 p ‖ ϕ ‖ L p ( J ) , η 2 ′ = M L Γ ( λ ) .

By repeating the proof in Theorem 3.1, the following conclusion is not difficult to prove by using the generalized Gronwall inequality.

Corollary 3.2

Let y 0 = 0 . Assume that (H1) and (H2) hold. Then, problem (1) admits a unique mild solution y ∈ C ( J , X ) . Moreover,

‖ y ‖ C ( J , X ) ≤ M Γ ( λ ) p − 1 p − 1 + ( λ − 1 ) p p − 1 p ‖ ϕ ‖ L p ( J ) E λ , 1 ( M L T λ ) ,

for p λ > 1 .

4 Optimal control problem

In this section, we consider an optimal control problem governed by a fractional evolution equation, such as a fractional diffusion equation. We consider p = 2 in this section, we set X is a real Hilbert space, and Y = L 2 ( J , U ) , where U is a reflexive Banach space. Next, consider an optimal control problem:

Minimize the functional

J ( u ) = 1 2 ‖ y − z d ‖ L 2 ( J , X ) 2 + N 2 ‖ u ‖ Y 2 ,

for all y , u ∈ L 2 ( J , X ) × Y a d , subject to

(3) D 0 + μ , λ H y ( t ) = A y ( t ) + B u ( t ) + g ( t , y ( t ) ) , a.e. t > 0 I 0 + ( 1 − λ ) ( 1 − μ ) y ( 0 + ) = y 0 ,

where y is the state, u stands for the control, z d ∈ L 2 ( J , X ) , N ∈ R + , operator A is a self-adjoint operator generated a compact C 0 -semigroup in X , and Y a d is a nonempty, closed, and convex subset of the control space Y . In addition, the operator B : U → X is linear and continuous.

Definition 4.1

If y ∈ C ( J ′ , X ) satisfies

y ( t ) = P μ , λ ( t ) y 0 + ∫ 0 t Q λ ( t − s ) ( B u ( s ) + g ( s , y ( s ) ) ) d s , t ∈ J ′ .

Then, y is called a mild solution of the initial value problem (3).

We set the solution set of problem (3) by D .

Lemma 4.1

[17] Let D ⊂ L p ( J , X ) . Then, D is relatively compact in L p ( J , X ) for 1 ≤ p < ∞ if and only if

the set
∫ t 1 t 2 a ( t ) d t , a ∈ D , ∀ 0 < t 1 < t 2 < T ,
is relatively compact in X, and
the limit
∫ 0 T − h ‖ a ( t + h ) − a ( t ) ‖ p d t → 0 a s h → 0 +
is uniform for a ∈ D .

Lemma 4.2

[18] For each f ∈ L p ( J , X ) with 1 ≤ p < + ∞ , then

lim h → 0 ∫ 0 T ∣ f ( t + h ) − f ( t ) ∣ p d t = 0 ,

where f ( s ) = 0 for s does not belong to J .

Lemma 4.3

If there is a constant c > 0 such that ‖ u ‖ Y ≤ c and operator A generates a compact C 0 -semigroup in X, then D is relatively compact in L 2 ( J , X ) .

Proof

We next divide this proof into two claims. Let

T y = ∫ t 1 t 2 y ( τ ) d τ , 0 < t 1 < t 2 < T , y ∈ D y .

We first claim that { T y , y ∈ D } is relatively compact in X . Since B is a bounded and linear operator, we set v = B u + g ( ⋅ , y ) , from (H2), it yields

‖ g ( ⋅ , y ) ‖ L 2 ( J , X ) ≤ ‖ ϕ ‖ L 2 ( J ) + ‖ y ‖ L 2 ( J , X ) ,

which implies that there exists a constant C > 0 such that ‖ v ‖ L 2 ( J , X ) ≤ C .

For small ε ∈ ( 0 , t 1 ) , and δ > 0 , let

T ε , δ y = 1 Γ ( μ ( 1 − λ ) ) ∫ t 1 t 2 ∫ ε t ∫ δ ∞ ( t − s ) μ ( 1 − λ ) − 1 s λ − 1 λ θ M λ ( θ ) T ( s λ θ ) y 0 d θ d s d t + ∫ t 1 t 2 ∫ 0 t − ε ∫ δ ∞ ( t − s ) λ − 1 λ θ M λ ( θ ) T ( ( t − s ) λ θ ) v ( s ) d θ d s d t ≕ I 1 + I 2 ,

we next show that { T ε , δ y , y ∈ D } is a relatively compact set in X . The proof of case I 1 is similar to that of I 2 for which v ( s ) deduces to s μ ( 1 − λ ) − 1 y 0 ⁄ Γ ( μ ( 1 − λ ) ) by a variable substitution; hence, it is sufficient to show { I 2 , y ∈ D } is a relatively compact set in X . By using the semigroup property of T ( t ) , we have

I 2 = T ( ε λ δ ) ∫ t 1 t 2 ∫ 0 t − ε ∫ δ ∞ ( t − s ) λ − 1 λ θ M λ ( θ ) T ( ( t − s ) λ θ − ε λ δ ) v ( s ) d θ d s d t ≕ T ( ε λ δ ) I 21 .

Since ε λ δ > 0 and T ( ⋅ ) is a compact C 0 -semigroup, it remains to check that I 21 is bounded. In fact, by using the Hölder inequality, we have

∣ I 21 ∣ = ∫ t 1 t 2 ∫ 0 t − ε ∫ δ ∞ ( t − s ) λ − 1 λ θ M λ ( θ ) T ( ( t − s ) λ θ − ε λ δ ) v ( s ) d θ d s d t ≤ M Γ ( λ ) ∫ t 1 t 2 ∫ 0 t − ε ( t − s ) λ − 1 ∣ v ( s ) ∣ d s d t ≤ M ( t 2 − t 1 ) Γ ( λ ) 1 2 λ − 1 ( t 2 2 λ − 1 − ε 2 λ − 1 ) 1 ⁄ 2 ‖ v ‖ L 2 ( J , X ) < + ∞ ,

where we have used the inequality

0 ≤ ∫ δ ∞ λ θ M λ ( θ ) d θ ≤ ∫ 0 ∞ λ θ M λ ( θ ) d θ = 1 Γ ( λ ) .

Therefore, { T ε , δ y , y ∈ D y } is a relatively compact set in X for each ε ∈ ( 0 , t 1 ) and δ > 0 . Furthermore, we note that

∣ T ε , δ y − T y ∣ ≤ 1 Γ ( μ ( 1 − λ ) ) ∫ t 1 t 2 ∫ 0 t ∫ 0 δ ( t − s ) μ ( 1 − λ ) − 1 s λ − 1 λ θ M λ ( θ ) T ( s λ θ ) y 0 d θ d s d t + 1 Γ ( μ ( 1 − λ ) ) ∫ t 1 t 2 ∫ 0 ε ∫ δ ∞ ( t − s ) μ ( 1 − λ ) − 1 s λ − 1 λ θ M λ ( θ ) T ( s λ θ ) y 0 d θ d s d t

+ ∫ t 1 t 2 ∫ 0 t − ε ∫ 0 δ ( t − s ) λ − 1 λ θ M λ ( θ ) T ( ( t − s ) λ θ ) v ( s ) d θ d s d t + ∫ t 1 t 2 ∫ t − ε t ∫ δ ∞ ( t − s ) λ − 1 λ θ M λ ( θ ) T ( ( t − s ) λ θ ) v ( s ) d θ d s d t ≕ I 3 + I 4 + I 5 + I 6 .

Obviously, we have

I 3 ≤ M ∣ y 0 ∣ Γ ( μ ( 1 − λ ) ) ∫ t 1 t 2 ∫ 0 t ∫ 0 δ ( t − s ) μ ( 1 − λ ) − 1 s λ − 1 λ θ M λ ( θ ) d θ d s d t = M ∣ y 0 ∣ Γ ( λ ) Γ ( μ ( 1 − λ ) + λ + 1 ) ( t 2 μ ( 1 − λ ) + λ − t 1 μ ( 1 − λ ) + λ ) ∫ 0 δ λ θ M λ ( θ ) d θ → 0 as δ → 0 .

For ε ∈ ( 0 , t 1 ) and t ∈ ( t 1 , t 2 ) , we note that

∫ 0 ε t ( 1 − s ) μ ( 1 − λ ) − 1 s λ − 1 d s ≤ ∫ 0 ε t 1 ( 1 − s ) μ ( 1 − λ ) − 1 s λ − 1 d s ≤ Γ ( μ ( 1 − λ ) ) Γ ( λ ) Γ ( μ ( 1 − λ ) + λ ) .

Hence, as ε → 0 , we obtain

I 4 ≤ M ∣ y 0 ∣ Γ ( μ ( 1 − λ ) ) ∫ t 1 t 2 ∫ 0 ε ∫ 0 ∞ ( t − s ) μ ( 1 − λ ) − 1 s λ − 1 λ θ M λ ( θ ) d θ d s d t = M ∣ y 0 ∣ Γ ( μ ( 1 − λ ) ) Γ ( λ ) ∫ t 1 t 2 ∫ 0 ε ( t − s ) μ ( 1 − λ ) − 1 s λ − 1 d s d t = M ∣ y 0 ∣ Γ ( μ ( 1 − λ ) ) Γ ( λ ) ∫ t 1 t 2 ∫ 0 ε t t μ ( 1 − λ ) + λ − 1 ( 1 − s ) μ ( 1 − λ ) − 1 s λ − 1 d s d t → 0 .

By using the Hölder inequality, we have

I 5 ≤ M ∫ t 1 t 2 ∫ 0 t − ε ( t − s ) λ − 1 ∣ v ( s ) ∣ d s d t ∫ 0 δ λ θ M λ ( θ ) d θ ≤ M ( t 2 − t 1 ) 1 2 λ − 1 ( t 2 2 λ − 1 − ε 2 λ − 1 ) 1 ⁄ 2 ‖ v ‖ L 2 ( J , X ) ∫ 0 δ λ θ M λ ( θ ) d θ → 0 as δ → 0 .

As for I 6 , we have

I 6 ≤ M ∫ t 1 t 2 ∫ t − ε t ∫ 0 ∞ ( t − s ) λ − 1 λ θ M λ ( θ ) ∣ v ( s ) ∣ d θ d s d t = M Γ ( λ ) ∫ t 1 t 2 ∫ t − ε t ( t − s ) λ − 1 ∣ v ( s ) ∣ d s d t ≤ M ( t 2 − t 1 ) Γ ( λ ) ε 2 λ − 1 2 λ − 1 1 ⁄ 2 ∣ v ∣ L 2 ( J , X ) → 0 as ε → 0 .

Therefore, we obtain T ε , δ y → T y , as ε , δ → 0 . This means that the set { T y , y ∈ D y } is closed to an arbitrary compact set in X . As a result, { T y , y ∈ D y } is a relatively compact in X .

We next claim that

∫ 0 T − h ∣ y ( t + h ) − y ( t ) ∣ 2 d t → 0 as h → 0 .

Obviously, we have

∣ y ( t + h ) − y ( t ) ∣ ≤ ∣ P μ , λ ( t + h ) y 0 − P μ , λ ( t ) y 0 ∣ + ∫ 0 t ∣ Q λ ( t − s ) ( g ( s + h , y ( s + h ) ) − g ( s , y ( s ) ) ) ∣ d s + ∫ − h 0 ∣ Q λ ( t − s ) g ( s + h , y ( s + h ) ) d s + ∫ t t + h ∣ Q λ ( t + h − s ) B u ( s ) ∣ d s + ∫ 0 t ∣ ( Q λ ( t + h − s ) − Q λ ( t − s ) ) B u ( s ) ∣ d s ≤ L M Γ ( λ ) ∫ 0 t ( t − s ) λ − 1 ∣ y ( s + h ) − y ( s ) ∣ d s + J 1 + J 2 + J 3 + J 4 ,

where

J 1 = ∣ P μ , λ ( t + h ) y 0 − P μ , λ ( t ) y 0 ∣ J 2 = M Γ ( λ ) ∫ 0 h ( t + h − s ) λ − 1 ( L ϕ ( s ) + ∣ y ( s ) ∣ ) d s J 3 = M Γ ( λ ) ∫ t t + h ( t + h − s ) λ − 1 ∣ B u ( s ) ∣ d s J 4 = ∫ 0 t ∣ ( Q λ ( t + h − s ) − Q λ ( t − s ) ) B u ( s ) ∣ d s .

Therefore, we have

∣ y ( t + h ) − y ( t ) ∣ 2 ≤ 5 L M Γ ( λ ) 2 t 2 λ − 1 2 λ − 1 ∫ 0 t ∣ y ( s + h ) − y ( s ) ∣ 2 d s + 5 ( J 1 2 + J 2 2 + J 3 2 + J 4 2 ) .

The Gronwall inequality shows that

∫ 0 T − h ∣ y ( s + h ) − y ( s ) ∣ 2 d s ≤ 5 e 5 L M Γ ( λ ) 2 ∫ 0 T − h t 2 λ − 1 2 λ − 1 d t ∫ 0 T − h ( J 1 2 + J 2 2 + J 3 2 + J 4 2 ) d t = 5 e 5 L M Γ ( λ ) 2 ( T − h ) 2 λ ( 2 λ ) ( 2 λ − 1 ) ∫ 0 T − h ( J 1 2 + J 2 2 + J 3 2 + J 4 2 ) d t .

From the proof in Theorem 3.1, we have

∫ 0 T − h ∣ P μ , λ ( t + h ) y 0 − P μ , λ ( t ) y 0 ∣ 2 d t ≤ 8 M ∣ y 0 ∣ Γ ( λ ) Γ ( μ ( 1 − λ ) ) 2 ∫ 0 T − h ∫ t t + h ( t + h − s ) μ ( 1 − λ ) − 1 s λ − 1 d s 2 d t + 2 M ∣ y 0 ∣ Γ ( λ ) Γ ( μ ( 1 − λ ) ) 2 ∫ 0 T − h ∫ 0 t ( t − s ) μ ( 1 − λ ) − 1 s λ − 1 d s − ∫ 0 t + h ( t + h − s ) μ ( 1 − λ ) − 1 s λ − 1 d s 2 d t .

Observe that

∫ t t + h ( t + h − s ) μ ( 1 − λ ) − 1 s λ − 1 d s = ∫ 0 h ( h − s ) μ ( 1 − λ ) − 1 ( s + t ) λ − 1 d s ≤ 1 μ ( 1 − λ ) t λ − 1 h μ ( 1 − λ ) ,

it follows that

∫ 0 T − h ∫ t t + h ( t + h − s ) μ ( 1 − λ ) − 1 s λ − 1 d s 2 d t ≤ 1 μ ( 1 − λ ) h μ ( 1 − λ ) 2 1 2 λ − 1 ( T − h ) 2 λ − 1 → 0 , as h → 0 .

Since

∫ 0 t ( t − s ) μ ( 1 − λ ) − 1 s λ − 1 d s ∈ L 2 ( J ) ,

for ( 1 − μ ) ( 1 − λ ) < 1 2 , it follows from Lemma 4.2 that

∫ 0 T − h ∫ 0 t ( t − s ) μ ( 1 − λ ) − 1 s λ − 1 d s − ∫ 0 t + h ( t + h − s ) μ ( 1 − λ ) − 1 s λ − 1 d s 2 d t → 0 , as h → 0 .

Therefore, we have

∫ 0 T − h ∣ P μ , λ ( t + h ) y 0 − P μ , λ ( t ) y 0 ∣ 2 d t → 0 , as h → 0 .

The Hölder inequality implies that

∫ 0 T − h ∫ 0 h ( t + h − s ) λ − 1 ( L ϕ ( s ) + ∣ y ( s ) ∣ ) 2 d s d t ≤ ∫ 0 T − h ∫ 0 h ( t + h − s ) 2 λ − 2 d s ∫ 0 h ( L ϕ ( s ) + ∣ y ( s ) ∣ ) 2 d s d t ≤ 1 2 λ − 1 ∫ 0 T − h ( ( t + h ) 2 λ − 1 − t 2 λ − 1 ) d t ( L ‖ ϕ ‖ L 2 [ 0 , h ] + ‖ y ‖ L 2 ( 0 , h ; X ) ) ,

since t 2 λ − 1 ∈ L 1 ( J ) for λ > 1 2 , it follows from Lemma 4.2 that

∫ 0 T − h M Γ ( λ ) ∫ 0 h ( t + h − s ) λ − 1 ( L ϕ ( s ) + ∣ y ( s ) ∣ ) 2 d s d t → 0 , as h → 0 .

Since B : U → X is a bounded linear operator ‖ u ‖ Y ≤ C , it yields ‖ B u ‖ L 2 ( J , X ) < + ∞ . Hence, the Hölder inequality implies that

∫ 0 T − h M Γ ( λ ) ∫ t t + h ( t + h − s ) λ − 1 ∣ B u ( s ) ∣ d s 2 d t ≤ M Γ ( λ ) 2 1 2 λ − 1 h 2 λ − 1 ( T − h ) ‖ B u ‖ L 2 ( J , X ) 2 → 0 , as h → 0 .

We also observe that

∫ 0 t ∣ ( Q λ ( t + h − s ) − Q λ ( t − s ) ) B u ( s ) ∣ d s

≤ ∫ 0 t ∣ ( ( t + h − s ) λ − 1 − ( t − s ) λ − 1 ) Q λ ( t + h − s ) B u ( s ) ∣ d s + ∫ 0 t ∣ ( t − s ) λ − 1 ( S λ ( t + h − s ) − S λ ( t − s ) ) B u ( s ) ∣ d s ≤ M Γ ( λ ) ∫ 0 t ( ( t + h − s ) λ − 1 − ( t − s ) λ − 1 ) 2 d s 1 ⁄ 2 ‖ B u ‖ L 2 ( 0 , t ; X ) + ∫ 0 t ( t − s ) 2 λ − 2 ‖ S λ ( t + h − s ) − S λ ( t − s ) ‖ 2 d s 1 ⁄ 2 × ‖ B u ‖ L 2 ( 0 , t ; X ) .

This implies that

∫ 0 T − h ∫ 0 t ∣ ( Q λ ( t + h − s ) − Q λ ( t − s ) ) B u ( s ) ∣ d s 2 d t ≤ 2 M Γ ( λ ) 2 ∫ 0 T − h ∫ 0 t ( ( t + h − s ) λ − 1 − ( t − s ) λ − 1 ) 2 d s d t ‖ B u ‖ L 2 ( J , X ) 2 + 2 ∫ 0 T − h ∫ 0 t ( t − s ) 2 λ − 2 ‖ S λ ( t + h − s ) − S λ ( t − s ) ‖ 2 d s d t ‖ B u ‖ L 2 ( J , X ) 2 = 2 M Γ ( λ ) 2 ∫ 0 T − h ∫ 0 t ( ( s + h ) λ − 1 − s λ − 1 ) 2 d s d t ‖ B u ‖ L 2 ( J , X ) 2 + 2 ∫ 0 T − h ∫ 0 t s 2 λ − 2 ‖ S λ ( s + h ) − S λ ( s ) ‖ 2 d s d t ‖ B u ‖ L 2 ( J , X ) 2 .

In view of t λ − 1 ∈ L 2 ( J ) , it follows from Lemma 4.2 that

∫ 0 T ( ( s + h ) λ − 1 − s λ − 1 ) 2 d s → 0 , as h → 0 .

In addition, from Lemma 3.3, we know that operator S λ ( t ) is uniform continuous for t > 0 , and then for any ε ∈ ( 0 , δ ) with δ ∈ ( 0 , T − h ) ,

∫ 0 T − h ∫ 0 t s 2 λ − 2 ∣ S λ ( s + h ) − S λ ( s ) ∣ 2 d s d t = ∫ 0 δ ∫ 0 t s 2 λ − 2 ∣ S λ ( s + h ) − S λ ( s ) ∣ 2 d s d t + ∫ δ T − h ∫ ε t s 2 λ − 2 ∣ S λ ( s + h ) − S λ ( s ) ∣ 2 d s d t + ∫ δ T − h ∫ 0 ε s 2 λ − 2 ∣ S λ ( s + h ) − S λ ( s ) ∣ 2 d s d t ≤ M Γ ( λ ) 2 δ 2 λ ( 2 λ ) ( 2 λ − 1 ) + sup t ∈ [ δ , T − h ] sup s ∈ [ ε , t ] ∣ S λ ( s + h ) − S λ ( s ) ∣ 2 ( T − h ) 2 λ ( 2 λ ) ( 2 λ − 1 ) + M Γ ( λ ) 2 1 2 λ − 1 ε 2 λ − 1 ( T − h − δ ) → 0 , as ε , δ , h → 0 .

Hence, we have

∫ 0 T − h ∫ 0 t ∣ ( Q λ ( t + h − s ) − Q λ ( t − s ) ) B u ( s ) ∣ d s 2 d t → 0 , as h → 0 .

Together with the above arguments, we have proved the desired claims. The proof follows Lemma 4.1.□

Theorem 4.1

Given u ∈ Y a d , assume that (H1) and (H2) hold. Then problem (3) admits a unique mild solution y u ∈ C ( J ′ , X ) ∩ L p ( J , X ) with p ∈ ( 1 λ , 1 ( 1 − μ ) ( 1 − λ ) ) .

Theorem 4.2

Assume that (H1) and (H2) hold. Self-adjoint operator A generates a compact C 0 -semigroup in X. Then problem (3) admits an optimal pair ( y ¯ , u ¯ ) .

Proof

According to the nonnegativity of functional J , there exists a minimizing sequence u n ∈ Y a d such that

J ( u n ) = 1 2 ‖ y n − z d ‖ 2 + N 2 ‖ u n ‖ Y 2 → inf u ∈ Y a d J ( u ) ,

as n → ∞ , here y n ∈ D is a unique mild solution of

y n ( t ) = P μ , λ ( t ) y 0 + ∫ 0 t Q λ ( t − s ) ( B u n ( s ) + g ( s , y n ( s ) ) ) d s .

Since Y is a reflexive Banach space, it follows that there exist a N > 0 large enough and a constant c > 0 such that for all n ≥ N ,

‖ y n ‖ L 2 ( J , X ) ≤ c , ‖ u n ‖ Y ≤ c .

Moreover, there are subsequences (still denote { y n } and { u n } ) such that

y n ⇀ y ¯ in L 2 ( J , X ) , u n ⇀ u ¯ in Y .

By the Marzur lemma, it yields u ¯ ∈ Y a d because Y a d is closed and convex. In addition, since { y n } ⊂ D and D is relatively compact in L 2 ( J , X ) , it means that

y n → y ¯ in L 2 ( J , X ) .

We next prove that y ¯ = y ¯ u is the mild solution of problem (3). In fact, by using (H2), we have

g ( ⋅ , y n ) → g ( ⋅ , y ¯ ) in L 2 ( J , X ) ,

it follows that

∫ 0 t Q λ ( t − s ) ( g ( s , y n ( s ) ) − g ( s , y ¯ ( s ) ) ) d s ≤ M L Γ ( λ ) ∫ 0 t ( t − s ) λ − 1 ∣ y n ( s ) − y ¯ ( s ) ∣ d s ≤ M L Γ ( λ ) 1 2 λ − 1 T 2 λ − 1 1 ⁄ 2 ‖ y n − y ¯ ‖ L 2 ( J , X ) .

This implies that

∫ 0 t Q λ ( t − s ) g ( s , y n ( s ) ) d s → ∫ 0 t Q λ ( t − s ) g ( s , y ¯ ( s ) ) d s , a.e. t ∈ J .

By the boundedness of B , we also obtain

B u n ⇀ B u , in L 2 ( J , X ) .

Let φ ∈ L 2 ( J , X ) and set

ψ ( t , x ) = φ ( x ) Q λ * ( t − s ) 1 0 < s < t , for t ∈ J ,

where Q λ * ( ⋅ ) is the dual operator of Q λ ( ⋅ ) . Clearly, ψ ∈ L 2 ( J , X ) , consider the inner product ( ⋅ , ⋅ ) X of real Hilbert space X , we have

φ , ∫ 0 t Q λ ( t − s ) ( B u n ( s ) − B u ( s ) ) d s X = ∫ 0 t ( φ , Q λ ( t − s ) ( B u n ( s ) − B u ( s ) ) ) X d s

= ∫ 0 t ( φ Q λ * ( t − s ) , B u n ( s ) − B u ( s ) ) X d s = ( ψ , B u n ( s ) − B u ( s ) ) X ,

where Q λ * ( ⋅ ) is the dual operator of Q λ ( ⋅ ) associated with T * ( ⋅ ) , which is generated by self-adjoint operator A * . The last identity tends to zero as n → ∞ from B u n ⇀ B u . This implies that

∫ 0 t Q λ ( t − s ) B u n ( s ) d s ⇀ ∫ 0 t Q λ ( t − s ) B u ¯ ( s ) d s , a.e. t ∈ J .

Therefore, we have

y n ( t ) ⇀ P μ , λ ( t ) y 0 + ∫ 0 t Q λ ( t − s ) ( B u ¯ ( s ) + g ( s , y ¯ ( s ) ) ) d s , a.e. t ∈ J .

By the uniqueness of the limit, it follows that

y ¯ ( t ) = P μ , λ ( t ) y 0 + ∫ 0 t Q λ ( t − s ) ( B u ¯ ( s ) + g ( s , y ¯ ( s ) ) ) d s , a.e. t ∈ J .

Therefore, y ¯ is a mild solution of problem (3) associated with the control u ¯ ∈ Y a d .

Due to the weak lower semi-continuity of the functional J ( u ) , i.e., liminf n → ∞ J ( u n ) ≥ J ( u ¯ ) . Additionally, from J ( u n ) → inf u ∈ Y a d J ( u ) , we have

J ( u ¯ ) ≤ inf u ∈ Y a d J ( u ) ,

this also implies that

J ( u ¯ ) = inf u ∈ Y a d J ( u ) .

Thus, we deduce that ( y ¯ , u ¯ ) is an optimal pair. The proof is completed.□

An example is given to illustrate our theory. Consider the following problem:

(4) D 0 + μ , λ H y ( t , x ) − Δ y ( t , x ) = 2 u ( t , x ) + y ( t , x ) a.e. t ∈ [ 0 , 1 ] , x ∈ Ω , y ( t , x ) = 0 , t > 0 , x ∈ ∂ Ω , I 0 + ( 1 − λ ) ( 1 − μ ) y ( 0 , x ) = 0 , x ∈ Ω ,

where Ω is a bounded subset of R 3 and ∂ Ω ∈ C 3 .

Define X = U = L 2 ( Ω ) , D ( A ) = H 2 ( Ω ) ∩ H 0 1 ( Ω ) , and A = − ( ∂ x 1 x 1 + ∂ x 2 x 2 + ∂ x 3 x 3 ) for x ∈ D ( A ) . The admissible control Y a d = { u ∈ U , ‖ u ‖ L 2 ( 0 , 1 ; U ) ≤ 1 } . Find the control u that minimizes the performance index

J ( u ) = 1 2 ‖ y ‖ L 2 ( J , X ) 2 + 1 2 ‖ u ‖ Y 2 .

Define y ( t ) ( x ) = y ( t , x ) , B u ( t ) ( x ) = 2 u ( t , x ) , g ( t , y ( t ) ) ( x ) = y ( t ) ( x ) . Thus, problem (4) can be rewritten as problem (1) for J = [ 0 , 1 ] , with the cost function

J ( u ) = 1 2 ∫ 0 1 ( ‖ y ( t ) ‖ X 2 d t + ‖ u ( t ) ‖ U 2 ) d t ,

Then, it satisfies all the assumptions given in Theorem 4.2. Therefore, problem (4) has an optimal pair.

Lemma 4.4

Assume that ( 1 − μ ) ( 1 − λ ) < 1 2 < λ < 1 , (H1) and r ∈ L 2 ( J , X ) hold. Then, the following problem

D T − μ , λ H p ( t ) = A p ( t ) + r ( t ) , t ∈ [ 0 , T ] I T − μ ( 1 − λ ) p ( T ) = 0

has a unique solution p ∈ L 2 ( J , X ) with p ( t ) = q ( T − t ) given by

(5) q ( t ) = ∫ 0 t Q λ ( t − s ) r ( T − s ) d s .

Proof

We set a = μ ( 1 − λ ) and b = ( 1 − μ ) ( 1 − λ ) , let σ T p ( t ) = p ( T − t ) for t ∈ [ 0 , T ] , then by changing the variable t → T − t in I T − a p ( t ) ,

I T − a p ( T − t ) = 1 Γ ( a ) ∫ T − t T ( τ − ( T − t ) ) a − 1 p ( τ ) d τ = 1 Γ ( a ) ∫ 0 t ( t − τ ) a − 1 p ( T − τ ) d τ = 1 Γ ( a ) ∫ 0 t ( t − τ ) a − 1 σ T p ( τ ) d τ = I 0 + a σ T p ( t ) ,

which means that

D T − μ , λ H p ( T − t ) = − 1 Γ ( b ) ∫ T − t T ( s − ( T − t ) ) b − 1 d d s 1 Γ ( a ) ∫ s T ( τ − s ) a − 1 p ( τ ) d τ d s = − 1 Γ ( b ) ∫ T − t T ( t − ( T − s ) ) b − 1 d d s 1 Γ ( a ) ∫ s T ( τ − s ) a − 1 p ( τ ) d τ d s = s → T − u 1 Γ ( b ) ∫ 0 t ( t − u ) b − 1 d d u 1 Γ ( a ) ∫ T − u T ( τ − ( T − u ) ) a − 1 p ( τ ) d τ d u = I 0 + b d d t I 0 + a σ T p ( t ) .

This means that

σ T D T − μ , λ H p ( t ) = D 0 + 1 − μ , λ H σ T p ( t ) .

By changing the variable t → T − t in the current problem, we obtain

D 0 + 1 − μ , λ H σ T p ( t ) = A σ T p ( t ) + σ T r ( t ) , T − t ∈ [ 0 , T ] , I 0 + μ ( 1 − λ ) σ T p ( 0 ) = 0 .

That is

D 0 + 1 − μ , λ H p ˜ ( t ) = A p ˜ ( t ) + r ˜ ( t ) , t ∈ [ 0 , T ] , I 0 + μ ( 1 − λ ) p ˜ ( 0 ) = 0 ,

where p ˜ ( t ) = σ T p ( t ) and r ˜ ( t ) = σ T r ( t ) . Theorem 3.1 shows that there is a unique mild solution to the above problem in L 2 ( J , X ) given by (5).□

We next consider the necessary optimality conditions of a special optimal control problem (3), subjected to A = − Δ on D ( A ) = H 2 ( Ω ) ∩ H 0 1 ( Ω ) and Ω is a bounded subset of R d for d ≥ 1 with a zero Dirichlet boundary condition, ∂ Ω ∈ C 2 , i.e.,

(6) D 0 + μ , λ H y ( t , x ) − Δ y ( t , x ) = B u ( t , x ) + g ( t , x , y ( t , x ) ) , a.e. t > 0 , x ∈ Ω , y ( t , x ) = 0 , t ∈ J , x ∈ ∂ Ω , I 0 + ( 1 − λ ) ( 1 − μ ) y ( 0 , x ) = 0 , x ∈ Ω .

We also assume that U is a Hilbert space, and then Y is a Hilbert space too. And, we further introduce the following hypothesis:

g ( t , ⋅ ) is Fréchet differentiable with respect to y .

By a variable substitution, the following lemma is not difficult to check.

Lemma 4.5

Let f , g ∈ L 1 [ 0 , T ] . For γ ∈ ( 0 , 1 ) , then

∫ 0 T ( I 0 + γ f ) ( t ) g ( t ) d t = ∫ 0 T f ( t ) ( I T − γ g ) ( t ) d t .

Lemma 4.6

For any φ ∈ C ∞ ( [ 0 , T ] × Ω ¯ ) , we have

∫ 0 T ∫ Ω ( D 0 + μ , λ H y ( t , x ) − Δ y ( t , x ) ) φ ( t , x ) d x d t = ∫ Ω I 0 + ( 1 − μ ) ( 1 − λ ) y ( T , x ) I T − μ ( 1 − λ ) φ ( T , x ) d x − ∫ Ω I 0 + ( 1 − μ ) ( 1 − λ ) y ( 0 , x ) I T − μ ( 1 − λ ) φ ( 0 , x ) d x − ∫ 0 T ∫ Ω y ( t , x ) ( D T − μ , λ H φ ( t , x ) + Δ φ ( t , x ) ) d x d t − ∫ 0 T ∫ ∂ Ω ∂ y ( t , x ) ∂ ν φ ( t , x ) d σ d t + ∫ 0 T ∫ ∂ Ω y ( t , x ) ∂ φ ( t , x ) ∂ ν d σ d t .

Proof

Let φ ∈ C ∞ ( [ 0 , T ] × Ω ¯ ) . We have

∫ 0 T ∫ Ω ( D 0 + μ , λ H y ( t , x ) − Δ y ( t , x ) ) φ ( t , x ) d x d t = ∫ 0 T ∫ Ω D 0 + μ , λ H y ( t , x ) φ ( t , x ) d x d t − ∫ 0 T ∫ Ω Δ y ( t , x ) φ ( t , x ) d x d t .

By using the Green formula, we have

(7) − ∫ 0 T ∫ Ω Δ y ( t , x ) φ ( t , x ) d x d t = − ∫ 0 T ∫ ∂ Ω ∂ y ∂ ν φ d σ d t + ∫ 0 T ∫ ∂ Ω y ∂ φ ∂ ν d σ d t − ∫ 0 T ∫ Ω y ( t , x ) Δ φ ( t , x ) d x d t .

We note from Lemma 4.5 that

∫ 0 T ∫ Ω D 0 + μ , λ H y ( t , x ) φ ( t , x ) d x d t = ∫ Ω ∫ 0 T I 0 + μ ( 1 − λ ) d d t ( I 0 + ( 1 − μ ) ( 1 − λ ) y ( t , x ) ) φ ( t , x ) d t d x = ∫ Ω ∫ 0 T d d t ( I 0 + ( 1 − μ ) ( 1 − λ ) y ( t , x ) ) I T − μ ( 1 − λ ) φ ( t , x ) d t d x .

Let ψ ( t , x ) = I T − μ ( 1 − λ ) φ ( t , x ) , we have

∫ 0 T d d t ( I 0 + ( 1 − μ ) ( 1 − λ ) y ( t , x ) ) ψ ( t , x ) d t = I 0 + ( 1 − μ ) ( 1 − λ ) y ( T , x ) ψ ( T , x ) − I 0 + ( 1 − μ ) ( 1 − λ ) y ( 0 , x ) ψ ( 0 , x ) − ∫ 0 T I 0 + ( 1 − μ ) ( 1 − λ ) y ( t , x ) ψ ′ ( t , x ) d t ,

it follows from Lemma 4.5 again that

∫ 0 T I 0 + ( 1 − μ ) ( 1 − λ ) y ( t , x ) ψ ′ ( t , x ) d t = ∫ 0 T y ( t , x ) I T − ( 1 − μ ) ( 1 − λ ) ψ ′ ( t , x ) d t = − ∫ 0 T y ( t , x ) H D T − μ , λ φ ( t , x ) d t .

Therefore, we have

(8) ∫ 0 T ∫ Ω D 0 + μ , λ H y ( t , x ) φ ( t , x ) d x d t = ∫ Ω I 0 + ( 1 − μ ) ( 1 − λ ) y ( T , x ) ψ ( T , x ) d x − ∫ Ω I 0 + ( 1 − μ ) ( 1 − λ ) y ( 0 , x ) ψ ( 0 , x ) d x − ∫ Ω ∫ 0 T y ( t , x ) H D T − μ , λ φ ( t , x ) d t d x .

Together (7) and (8), we derive the desired conclusion.□

Lemma 4.7

Assume that ( 1 − μ ) ( 1 − λ ) < 1 2 < λ < 1 , and r ∈ L 2 ( J , L 2 ( Ω ) ) hold. Then, the following problem

(9) D T − μ , λ H p ( t ) − Δ p ( t ) − g ′ ( t , y ) p ( t ) = r ( t ) , ( t , x ) ∈ ( 0 , T ) × Ω p = 0 , ( t , x ) ∈ ( 0 , T ) × ∂ Ω , I T − μ ( 1 − λ ) p ( T , x ) = 0 , x ∈ Ω ,

has a unique mild solution p ∈ C ( J , X ) with p ( t ) = q ( T − t ) given by

(10) q ( t ) = ∫ 0 t Q λ ( t − s ) ( g ′ ( T − s , y ) v ( T − s ) + r ( T − s ) ) d s .

Proof

Let σ T p ( t ) = p ( T − t ) , for t ∈ J . Replacing g ( t , y ) in (H2) with g ′ ( σ T t , σ T y ) y + σ T r , the inequality in (H2) holds with L = ∥ g ′ ( σ T t , σ T y ) ∥ since y ¯ , r are fixed, and g ′ ( σ T t , σ T y ) is a linear and continuous mapping. Let D ( A ) = H 2 ( Ω ) ∩ H 0 1 ( Ω ) , and let A be the Laplace operator with the homogeneous Dirichlet boundary condition, the current problem can be abstracted as an evolution system

D T − μ , λ H p ( t ) = A p ( t ) + g ′ ( t , y ) p ( t ) + r ( t ) , t ∈ ( 0 , T ) , I T − μ ( 1 − λ ) p ( T ) = 0 .

Lemma 4.4 deduces that (9) has a unique mild solution σ T p ∈ L 2 ( J , X ) given by (10). The proof is completed.□

Let Q = ( 0 , T ) × Ω , we set

D ( Q ) = { φ , φ ∈ C ∞ ( Q ) such that φ ∣ ∂ Ω = 0 , φ ( 0 , x ) = φ ( T , x ) = 0 in Ω } .

We first introduce a Lagrangian function ℒ associated with problem (6) as follows:

ℒ ( y , u , w ) ≔ J ( u ) − ∫ 0 T ∫ Ω ( D 0 + μ , λ H y − Δ y − g ( t , y ) − B u ) w d x d t ,

for w ∈ L 2 ( Q ) ≔ L 2 ( 0 , T ; L 2 ( Ω ) ) such that I T − μ ( 1 − λ ) w ( T , x ) = 0 in Ω . We next establish the necessary optimality conditions for optimal control problem (6).

Theorem 4.3

If ( y ¯ , u ¯ ) is an optimal pair of the problem (6), then, there exists a triple ( y ¯ , u ¯ , w ¯ ) for w ¯ ∈ L 2 ( J , X ) satisfying

D T − μ , λ H w ¯ − Δ w ¯ − g ′ ( t , y ¯ ) w ¯ = y ¯ − z d , ( t , x ) ∈ ( 0 , T ) × Ω , w ¯ = 0 , ( t , x ) ∈ ( 0 , T ) × ∂ Ω I T − μ ( 1 − λ ) w ¯ ( T ) = 0 , x ∈ Ω

and

(11) ( B ( v − u ¯ ) , w ) L 2 ( Q ) + N ( u ¯ , v − u ¯ ) Y ≥ 0 , ∀ v ∈ Y a d .

Proof

By using the formal Lagrange multiplier method, the following first-order necessary optimality conditions hold in view of the optimal pair of ( y ¯ , u ¯ ) ,

(12) D y ℒ ( y ¯ , u ¯ , w ) φ = 0 , ∀ φ ∈ D ( Q )

and

(13) D u ℒ ( y ¯ , u ¯ , w ) ( v − u ¯ ) ≥ 0 , ∀ v ∈ Y a d ,

Clearly, I 0 + ( 1 − λ ) ( 1 − μ ) φ ( 0 ) = 0 for all φ ∈ D ( Q ) , and from (12), we have

D y ℒ ( y ¯ , u ¯ , w ) φ = ∫ 0 T ∫ Ω ( y ¯ − z d ) φ d x d t − ∫ 0 T ∫ Ω ( D 0 + μ , λ H φ − Δ φ − g ′ ( t , y ¯ ) φ ) w d x d t .

By using Lemma 4.6 and I T − μ ( 1 − λ ) w ( T , x ) = 0 in Ω , it follows that

0 = ∫ 0 T ∫ Ω ( y ¯ − z d + D T − μ , λ H w + Δ w + g ′ ( t , y ¯ ) w ) φ d x d t + ∫ 0 T ∫ ∂ Ω ∂ φ ∂ v w d σ d t + ∫ Ω I 0 + ( 1 − μ ) ( 1 − λ ) φ ( 0 , x ) I T − μ ( 1 − λ ) w ( 0 , x ) d x + ∫ 0 T ∫ ∂ Ω ∂ w ∂ v φ d σ d t − ∫ Ω I T − μ ( 1 − λ ) w ( T , x ) I 0 + ( 1 − μ ) ( 1 − λ ) φ ( T , x ) d x .

Since for all φ ∈ C 0 ∞ ( Q ¯ ) , the expressions φ ∣ ∂ Ω = 0 and ∂ φ ∂ ν ∣ ∂ Ω = 0 , we have

∫ 0 T ∫ Ω ( y ¯ − z d + D T − μ , λ H w + Δ w + g ′ ( t , y ¯ ) w ) φ d x d t = 0 .

Recall that C 0 ∞ ( Q ¯ ) is dense in L 2 ( ( 0 , T ) × Ω ) ; hence, we have

(14) D T − μ , λ H w − Δ w − g ′ ( t , y ¯ ) w = y ¯ − z d in Q .

For all φ ∈ C ∞ ( Q ¯ ) with φ ∣ ∂ Ω = 0 , Lemma 4.6 also shows that

∫ 0 T ∫ ∂ Ω ∂ φ ∂ ν w d σ d t = 0 ,

which implies that

(15) w ∣ ∂ Ω = 0 .

Thus, we obtain that problems (14) and (15) and I T − μ ( 1 − λ ) w ( T , x ) = 0 in Ω , regards as the adjoint problem to (3). From Lemma 4.7, we know that this adjoint problem has a unique mild solution w ¯ ∈ C ( J , X ) , i.e., ( y ¯ , u ¯ , w ¯ ) . Moreover, from condition (13) with w = w ¯ , we have

D u ℒ ( y ¯ , u ¯ , w ¯ ) ( v − u ¯ ) = ( B ( v − u ¯ ) , w ¯ ) L 2 ( Q ) + N ( u ¯ , v − u ¯ ) Y ≥ 0 , ∀ v ∈ Y a d .

Thus, inequality (11) follows. The proof is completed.□

Acknowledgements

We express our sincere thanks to the anonymous reviewers for their valuable comments. This work was supported by the the Fundo para o Desenvolvimento das Ciências e da Tecnologia of Macau (No. 0092/2022/A) and the Guangxi Science and Technology Program (AD23026249).

Funding information: This study was funded by the Fundo para o Desenvolvimento das Ciências e da Tecnologia of Macau (No. 0092/2022/A) and the Guangxi Science and Technology Program (AD23026249).
Author contributions: All authors read and approved the final manuscript.
Conflict of interest: The authors declare that they have no competing interest.
Ethical approval: The conducted research is not related to either human or animal use.
Data availability statement: Data sharing is not applicable to this article as no datasets were generated or analyzed during this study.

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Received: 2024-07-22

Accepted: 2024-10-29

Published Online: 2025-01-21

This work is licensed under the Creative Commons Attribution 4.0 International License.

Articles in the same Issue

https://doi.org/10.1515/dema-2024-0093

Keywords for this article

existence; uniqueness; optimal control; fractional evolution equations; Hilfer derivative

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