Integrated Fractional Resolvent Operator Function and Fractional Abstract Cauchy Problem

Abstract

We firstly prove that 𝛽-times integrated 𝛼-resolvent operator function ((𝛼,𝛽)-ROF) satisfies a functional equation which extends that of 𝛽-times integrated semigroup and 𝛼-resolvent operator function. Secondly, for the inhomogeneous 𝛼-Cauchy problem ^𝑐𝐷^𝛼_𝑡𝑢(𝑡)=𝐴𝑢(𝑡)+𝑓(𝑡), 𝑡∈(0,𝑇), 𝑢(0)=𝑥₀, 𝑢^′(0)=𝑥₁, if 𝐴 is the generator of an (𝛼,𝛽)-ROF, we give the relation between the function 𝑣(𝑡)=𝑆_𝛼,𝛽(𝑡)𝑥₀+(𝑔₁∗𝑆_𝛼,𝛽)(𝑡)𝑥₁+(𝑔_𝛼−1∗𝑆_𝛼,𝛽∗𝑓)(𝑡) and mild solution and classical solution of it. Finally, for the problem ^𝑐𝐷^𝛼_𝑡𝑣(𝑡)=𝐴𝑣(𝑡)+𝑔_𝛽+1(𝑡)𝑥, 𝑡>0, 𝑣^(𝑘)(0)=0, 𝑘=0,1,…,𝑁−1, where 𝐴 is a linear closed operator. We show that 𝐴 generates an exponentially bounded (𝛼,𝛽)-ROF on a Banach space 𝑋 if and only if the problem has a unique exponentially bounded classical solution 𝑣_𝑥 and 𝐴𝑣_𝑥∈𝐿¹_loc(ℝ⁺,𝑋). Our results extend and generalize some related results in the literature.

1. Introduction

This paper is concerned with the properties of 𝛽-integrated 𝛼-resolvent operator function ((𝛼,𝛽)-ROF) and two inhomogeneous fractional Cauchy problems.

Throughout this paper, ℝ⁺=[0,∞), ℕ denotes the set of natural numbers. ℕ₀=ℕ∪{0}. Let 𝑋,𝑌 be Banach spaces, 𝐵(𝑋,𝑌) denote the space of all bounded linear operators from 𝑋 to 𝑌, 𝐵(𝑋)=𝐵(𝑋,𝑋). If 𝐴 is a closed linear operator, 𝜌(𝐴) denotes the resolvent set of 𝐴 and 𝑅(𝜆,𝐴)=(𝜆𝐼−𝐴)⁻¹ denotes the resolvent operator of 𝐴. 𝐿¹(ℝ⁺,𝑋) denotes the space of 𝑋-valued Bochner integrable functions: 𝑢∶ℝ⁺→𝑋 with the norm ‖𝑢‖_𝐿¹_(ℝ⁺_,𝑋)=∫^∞₀‖𝑢(𝑡)‖𝑑𝑡, it is a Banach space. By ∗ we denote the convolution of functions 𝑓∗𝑔(𝑡)=^𝑡₀𝑓(𝑡−𝜏)𝑔(𝜏)𝑑𝜏,𝑡≥0. (1)𝑔_𝛼 denotes the function 𝑔_𝛼(𝑡)=𝑡^𝛼−1Γ(𝛼),𝑡>0,0,𝑡≤0, (2) and 𝑔₀(𝑡)=𝛿₀(𝑡), the Dirac delta function.

In 1997, Mijatović et al. [1] introduced the concept of 𝛽-times integrated semigroup (𝛽∈ℝ⁺) which extends 𝑘-times integrated semigroup (𝑘∈ℕ₀) [2], they showed an 𝑅(𝜆) to be the pseudoresolvent of a 𝛽-times (𝛽>0) integrated semigroup {𝑆(𝑡)} if and only if {𝑆(𝑡)} satisfies the following functional equation: _𝑡^𝑡+𝑠(𝑠+𝑡−𝑟)^𝛽−1𝑆(𝑟)𝑑𝑟−^𝑠₀(𝑠+𝑡−𝑟)^𝛽−1𝑆(𝑟)𝛽𝑆(𝑡)𝑆(𝑠)𝑑𝑟=Γ,𝑡,𝑠≥0. (3) In the special case of 𝛽=𝑘∈ℕ, the corresponding result is summarized in [2].

For the inhomogeneous Cauchy problem 𝑢^′(𝑡)=𝐴𝑢(𝑡)+𝑓(𝑡),𝑡∈[0,𝑇],𝑢(0)=𝑥, (4) where 𝑇>0, 𝑓∈𝐿¹([0,𝑇],𝑋), 𝑥∈𝑋, and 𝐴 is the generator of a 𝑘-times integrated semigroup {𝑆(𝑡)} on a Banach space 𝑋 for some 𝑘∈ℕ₀. Let 𝑣(𝑡)=𝑆(𝑡)𝑥+∫^𝑡₀𝑆(𝑡−𝑠)𝑓(𝑠)𝑑𝑠,𝑡∈[0,𝑇]. Lemmas 3.2.9 and 3.2.10 of [2] show that if there is a mild(classical) solution 𝑢 of (4), then 𝑣∈𝐶^𝑘([0,𝑇],𝑋)(𝐶^𝑘+1([0,𝑇],𝑋)) and 𝑢=𝑣^(𝑘). On the other hand, if 𝑣∈𝐶^𝑘([0,𝑇],𝑋)(𝐶^𝑘+1([0,𝑇],𝑋)), then 𝑣^(𝑘) is also a mild (classical) solution of it.

Furthermore, if 𝐴 generates an exponential bounded 𝑘-times integrated semigroup on a Banach space 𝑋, then, for any 𝑥∈𝑋, 𝑣(𝑡)=∫^𝑡₀𝑆(𝑠)𝑥𝑑𝑠 is the unique exponential bounded classical solution of the following problem: 𝑢^′(𝑡)=𝐴𝑢(𝑡)+𝑔_𝑘+1(𝑡)𝑥,𝑡≥0,𝑢(0)=0. (5)

In recent years, a considerable interest has been paid to fractional evolution equation due to its applications in different areas such as stochastic, finance, and physics; see [3–8]. One of the most important tools in the theory of fractional evolution equation is the solution operator (fractional resolvent family) [9–15]. The notion of solution operator was developed to study some abstract Volterra integral equations [16] and was first used by Bajlekova [17] to study a class of fractional order abstract Cauchy problem. In [9], Chen and Li introduced 𝛼-resolvent operator functions (𝛼-ROF for short) defined by purely algebraic equation. They showed that a family {𝑆_𝛼(𝑡)}_𝑡≥0⊂𝐵(𝑋) is an 𝛼-ROF if and only if {𝑆_𝛼(𝑡)}_𝑡≥0 is a solution of abstract fractional Cauchy problem ^𝑐𝐷^𝛼_𝑡𝑣(𝑡)=𝐴𝑣(𝑡)+𝑔_𝛽+1(𝑡)𝑣(0)𝑥,𝑡>0,=𝑥,𝑣^(𝑘)(0)=0,𝑘=1,…,𝑁−1. (6) When 0<𝛼<1, Peng and Li [18] proved that the solution operator {𝑆_𝛼(𝑡)}_𝑡≥0 for (6) satisfies the following equality: _𝑡^𝑡+𝑠𝑆_𝛼(𝜏)(𝑡+𝑠−𝜏)^𝛼𝑑𝜏−^𝑠₀𝑆_𝛼(𝜏)(𝑡+𝑠−𝜏)^𝛼𝑑𝜏=𝛼^𝑡₀^𝑠₀𝑆_𝛼𝜏₁𝑆_𝛼𝜏₂𝑡+𝑠−𝜏₁−𝜏₂^1+𝛼𝑑𝜏₁𝑑𝜏₂,𝑡,𝑠≥0. (7) We refer to [5, 15, 16, 19] for further information concerning general resolvent operator functions. In addition, Chen and Li [9] also introduced the concept of integrated fractional resolvent operator function in an algebraic notion as follows.

Definition 1 (see [9, Definition 3.7]). Let 𝛼>0, 𝛽≥0. A function 𝑆_𝛼,𝛽∶ℝ⁺→𝐵(𝑋) is called a 𝛽-times integrated 𝛼-resolvent operator function or an (𝛼,𝛽)-resolvent operator function ((𝛼,𝛽)-ROF for short) if the following conditions hold:(a)𝑆_𝛼,𝛽(⋅) is strongly continuous on ℝ⁺ and 𝑆_𝛼,𝛽(0)=𝑔_𝛽+1(0)𝐼;(b)𝑆_𝛼,𝛽(𝑠)𝑆_𝛼,𝛽(𝑡)=𝑆_𝛼,𝛽(𝑡)𝑆_𝛼,𝛽(𝑠) for all 𝑠,𝑡≥0;(c)the functional equation 𝑆_𝛼,𝛽(𝑠)𝐽^𝛼_𝑡𝑆_𝛼,𝛽(𝑡)−𝐽^𝛼_𝑠𝑆_𝛼,𝛽(𝑠)𝑆_𝛼,𝛽(𝑡)=𝑔_𝛽+1(𝑠)𝐽^𝛼_𝑡𝑆_𝛼,𝛽(𝑡)−𝑔_𝛽+1(𝑡)𝐽^𝛼_𝑠𝑆_𝛼,𝛽(𝑠) (8) holds for 𝑠,𝑡≥0, where 𝐽^𝛼_𝑡 is the Riemann-Liouville fractional integral of order 𝛼.

The generator 𝐴 of 𝑆_𝛼,𝛽(𝑡) is defined by 𝐷(𝐴)∶=𝑥∈𝑋∶lim_𝑡→0+𝑆_𝛼,𝛽(𝑡)𝑥−𝑔_𝛽+1(𝑡)𝑥𝑔_𝛼+𝛽+1(𝑡)exists,𝐴𝑥∶=lim_𝑡→0+𝑆_𝛼,𝛽(𝑡)𝑥−𝑔_𝛽+1(𝑡)𝑥𝑔_𝛼+𝛽+1(𝑡),𝑥∈𝐷(𝐴). (9) Note that an (𝛼,0)-ROF is just an 𝛼-ROF.

In this paper, we firstly show that (𝛼,𝛽)-ROF satisfies an equality which extends (3) and (7) for 𝛽-integrated semigroup and 𝛼-ROF, respectively. Then, we consider the inhomogeneous fractional order abstract Cauchy problem ^𝑐𝐷^𝛼_𝑡𝑢(𝑡)(𝑡)(𝑡)(),𝑢(0)=𝐴𝑢+𝑓,𝑡∈0,𝑇=𝑥₀,𝑢^′(0)=𝑥₁, (10) where 1<𝛼<2, 𝑇>0, 𝑓∈𝐿¹((0,𝑇),𝑋), and 𝐴 is assumed to be the generator of an (𝛼,𝛽)-ROF 𝑆_𝛼,𝛽(𝑡) on 𝑋. We give the relation between the function 𝑣(𝑡)=𝑆_𝛼,𝛽(𝑡)𝑥₀+(𝑔₁∗𝑆_𝛼,𝛽)(𝑡)𝑥₁+(𝑔_𝛼−1∗𝑆_𝛼,𝛽∗𝑓)(𝑡) and solution of (10). We also study the problem ^𝑐𝐷^𝛼_𝑡𝑣(𝑡)=𝐴𝑣(𝑡)+𝑔_𝛽+1(𝑡)𝑣𝑥,𝑡>0,_(𝑘)(0)=0,𝑘=0,1,…,𝑁−1, (11) where 𝛼>0, 𝑥∈𝑋, 𝑁 is the smallest integer greater than or equal to 𝛼. We prove that if 𝐴 generates an exponentially bounded (𝛼,𝛽)-ROF on 𝑋 if and only if the problem (11) has a unique exponentially bounded classical solution 𝑣_𝑥 and 𝐴𝑣_𝑥∈𝐿¹_Loc(ℝ⁺,𝑋). If 𝛼→1⁺, 𝛽=𝑘∈ℕ, our Theorem 13 reduces to Lemma 3.2.10 in [2]. When 𝛼=1,  𝛽=𝑘, it is easy to see that our Theorem 15 extends and generalizes Theorem 3.2.13 in [2].

This paper is organized as follows. In Section 2, we provide some preliminaries of the fractional calculus and (𝛼,𝛽)-ROF. Section 3 is devoted to present an equality characteristic of the (𝛼,𝛽)-ROF. Finally, as an application of (𝛼,𝛽)-ROF, we discuss the solutions of fractional abstract Cauchy problem in Section 4.

2. Preliminary

Recall that the Riemann-Liouville fractional integral of order 𝛼>0 of 𝑓 is defined by 𝐽^𝛼_𝑡𝑓(𝑡)=𝑔_𝛼∗𝑓(𝑡)=^𝑡₀𝑔_𝛼(𝑡−𝑠)𝑓(𝑠)𝑑𝑠, (12) and the Caputo fractional derivative of order 𝛼>0 of 𝑓 can be written as ^𝑐𝐷^𝛼_𝑡𝑓(𝑡)=𝑑^𝑚𝑑𝑡^𝑚𝑔_𝑚−𝛼∗𝑓(𝑡)−^𝑚−1_𝑘=0𝑓^(𝑘)(0)𝑔_𝑘+1(𝑡), (13) where 𝑚 is the smallest integer greater than or equal to 𝛼. For more details in fractional calculus, we refer to [5, 20, 21].

The Mittag-Leffler function is defined by 𝐸_𝛼,𝛽(𝑧)=^∞_𝑛=0𝑧^𝑛Γ𝛼𝑛+𝛽,𝐸_𝛼(𝑧)=𝐸_𝛼,1(𝑧),Re𝛼>0,𝛽,𝑧∈ℂ. (14) And if 0<𝛼<2, 𝛽>0, then 𝐸_𝛼,𝛽(𝑧)=1𝛼𝑧^{(1−𝛽)/𝛼}exp𝑧^1/𝛼+𝜀_𝛼,𝛽(𝑧),≤1arg𝑧2𝛼𝜋,𝐸_𝛼,𝛽(𝑧)=𝜀_𝛼,𝛽(𝑧),(<1arg−𝑧)1−2𝛼𝜋, (15) where 𝜀_𝛼,𝛽(𝑧)=−^𝑁−1_𝑛=1𝑧^−𝑛Γ|𝑧|𝛽−𝛼𝑛+𝑂^−𝑁as𝑧⟶∞, (16) and the 𝑂-term is uniform in arg𝑧 if |arg(−𝑧)|≤(1−(𝛼/2)−𝜖)𝜋.

We now recall some properties of (𝛼,𝛽)-ROF.

Lemma 2 (see [9, Proposition 3.10]). Let 𝑆_𝛼,𝛽∶ℝ⁺→𝐵(𝑋) be an (𝛼,𝛽)-ROF generated by 𝐴. The following assertions hold:(a)𝑆_𝛼,𝛽(𝑡)𝐷(𝐴)⊂𝐷(𝐴) and 𝐴𝑆_𝛼,𝛽(𝑡)𝑥=𝑆_𝛼,𝛽(𝑡)𝐴𝑥 for 𝑥∈𝐷(𝐴) and 𝑡≥0;(b)for all 𝑥∈𝑋, 𝐽^𝛼_𝑡𝑆_𝛼,𝛽(𝑡)𝑥∈𝐷(𝐴) and 𝑆_𝛼,𝛽(𝑡)𝑥=𝑔_𝛽+1(𝑡)𝑥+𝐴𝐽^𝛼_𝑡𝑆_𝛼,𝛽(𝑡)𝑥, 𝑡≥0;(c)𝑥∈𝐷(𝐴) and 𝐴𝑥=𝑦 if and only if 𝑆_𝛼,𝛽(𝑡)𝑥=𝑔_𝛽+1(𝑡)𝑥+𝐽^𝛼_𝑡𝑆_𝛼,𝛽(𝑡)𝑦, 𝑡≥0;(d)𝐴 is closed.

Lemma 3 (see [9, Proposition 3.5, Theorem 3.11]). Let 𝛼>0, 𝛽≥0. 𝐴 generates an (𝛼,𝛽)-ROF 𝑆_𝛼,𝛽 satisfying ‖𝑆_𝛼,𝛽(𝑡)‖≤𝑀𝑒^𝜔𝑡, 𝑡≥0, for some constants 𝑀>0 and 𝜔≥0, if and only if (𝜔^𝛼,∞)⊂𝜌(𝐴) and there exists a strongly continuous function 𝑆∶ℝ⁺→𝐵(𝑋) such that ‖𝑆(𝑡)‖≤𝑀𝑒^𝜔𝑡 for all 𝑡≥0 and ∫^∞₀𝑒^−𝜆𝑡𝑆(𝑡)𝑥𝑑𝑡=𝜆^{𝛼−𝛽−1}𝑅(𝜆^𝛼,𝐴)𝑥, 𝜆>𝜔, for all 𝑥∈𝑋. Furthermore, 𝑆(𝑡) is 𝑆_𝛼,𝛽(𝑡).

Lemma 4 (see [2, Proposition B.6]). Let 𝑈⊂ℂ. If function 𝑅∶𝑈→𝐵(𝑋) satisfies 𝑅(𝜆)−𝑅(𝜇)=(𝜇−𝜆)𝑅(𝜆)𝑅(𝜇), then there is an operator 𝐴 on 𝑋 such that 𝑅(𝜆)=(𝜆𝐼−𝐴)⁻¹ for all 𝜆∈𝑈 if and only if ker𝑅(𝜆)={0}.

3. An Novel Equality Characteristic for (𝛼,𝛽)-ROF

The following theorem shows that an (𝛼,𝛽)-ROF satisfies a functional equation and the treatment bases on the technique of Laplace transform. For convenience, we drop the subscript 𝛼,𝛽 from {𝑆_𝛼,𝛽}_𝑡≥0 in this theorem.

Theorem 5. Let 𝛼∈ℝ⁺⧵ℕ₀,𝛽∈ℝ⁺ satisfy 𝛽−𝛼>−1. If {𝑆(𝑡)}_𝑡≥0 is an (𝛼,𝛽)-ROF, then it satisfies the following equality: _𝑡^𝑡+𝑠(𝑠+𝑡−𝑟)^𝛽−𝛼𝑆(𝑟)𝑑𝑟−^𝑠₀(𝑠+𝑡−𝑟)^𝛽−𝛼𝑆(𝑟)=𝑑𝑟𝛼Γ𝛽−𝛼+1Γ(1−𝛼)^𝑠₀^𝑡₀𝑆𝑟₁𝑆𝑟₂𝑡+𝑠−𝑟₁−𝑟₂^1+𝛼𝑑𝑟₁𝑑𝑟₂. (17)

Proof. Denote by 𝐿(𝑡,𝑠) and 𝑅(𝑡,𝑠) the left and right sides of equality (17), respectively, and denote by 𝑓_𝑎(𝑡) the truncation of 𝑓(𝑡) at 𝑎, that is, 𝑓_𝑎(𝑡)=𝑓(𝑡) for 0≤𝑡≤𝑎 and 𝑓_𝑎(𝑡)=0 otherwise.
We will show that the Laplace transform of 𝐿_𝑎(𝑡,𝑠) and 𝑅_𝑎(𝑡,𝑠) with respect to 𝑡 and 𝑠 is equivalent, and by the uniqueness of Laplace transform, we can get that 𝐿_𝑎(𝑡,𝑠)=𝑅_𝑎(𝑡,𝑠).
Taking Laplace transform of 𝐿_𝑎(𝑡,𝑠) with respect to 𝑠 as follows 𝐿_𝑎(𝑡,𝜆)=^∞₀𝑒^−𝜆𝑠_𝑡^𝑡+𝑠(𝑠+𝑡−𝑟)^𝛽−𝛼𝑆_𝑎(𝑟)𝑑𝑟−^𝑠₀(𝑠+𝑡−𝑟)^𝛽−𝛼𝑆_𝑎(𝑟)=𝑑𝑟𝑑𝑠_∞𝑡𝑆_𝑎(𝑟)^∞_𝑟−𝑡𝑒^−𝜆𝑠(𝑠+𝑡−𝑟)^𝛽−𝛼−𝑑𝑠𝑑𝑟_∞0𝑆_𝑎(𝑟)^∞_𝑟𝑒^−𝜆𝑠(𝑠+𝑡−𝑟)^𝛽−𝛼=𝑑𝑠𝑑𝑟_∞𝑡𝑆_𝑎(𝑟)𝑒^{−𝜆(𝑟−𝑡)∞}₀𝑒^−𝜆𝜏𝜏^𝛽−𝛼−𝑑𝜏𝑑𝑟_∞0𝑆_𝑎(𝑟)𝑒^{−𝜆(𝑟−𝑡)∞}_𝑡𝑒^−𝜆𝜏𝜏^𝛽−𝛼=Γ𝑑𝜏𝑑𝑟𝛽−𝛼+1𝜆^{(𝛽−𝛼+1)∞𝑡}𝑆_𝑎(𝑟)𝑒^{−𝜆(𝑟−𝑡)}𝑑𝑟−𝑒^𝜆𝑡𝑆_𝑎(𝜆)^∞_𝑡𝑒^−𝜆𝜏𝜏^𝛽−𝛼𝑑𝜏, (18) then taking Laplace transform with respect to 𝑡, we have 𝐿_𝑎𝜇,𝜆=^∞₀𝑒^−𝜇𝑡Γ𝛽−𝛼+1𝜆^{(𝛽−𝛼+1)∞𝑡}𝑆_𝑎(𝑟)𝑒^{−𝜆(𝑟−𝑡)}𝑑𝑟−𝑒^𝜆𝑡𝑆_𝑎(𝜆)^∞_𝑡𝑒^−𝜆𝜏𝜏^𝛽−𝛼=Γ𝑑𝜏𝑑𝑡𝛽−𝛼+1𝜆^{(𝛽−𝛼+1)∞0}𝑒^{−𝜇𝑡∞}_𝑡𝑆_𝑎(𝑟)𝑒^{−𝜆(𝑟−𝑡)}−𝑆𝑑𝑟𝑑𝑡_𝑎(𝜆)^∞₀𝑒^{(𝜆−𝜇)𝑡∞}_𝑡𝑒^−𝜆𝜏𝜏^𝛽−𝛼=Γ𝑑𝜏𝑑𝑡𝛽−𝛼+1𝜆^{(𝛽−𝛼+1)∞0}𝑒^−𝜆𝑟𝑆_𝑎(𝑟)^𝑟₀𝑒^{(𝜆−𝜇)𝑡}𝑑𝑡𝑑𝑟−^∞₀𝑒^−𝜆𝜏𝜏^{𝛽−𝛼𝜏}₀𝑒^{(𝜆−𝜇)𝑡}𝑑𝑡𝑑𝜏𝑆_𝑎(𝜆)=Γ𝛽−𝛼+1𝜆−𝜇𝜆^{(𝛽−𝛼+1)}×^∞₀𝑒^−𝜇𝑟𝑆_𝑎(𝑟)𝑑𝑟−^∞₀𝑒^−𝜆𝑟𝑆_𝑎(𝑟)−1𝑑𝑟×𝜆−𝜇_∞0𝑒^−𝜇𝜏𝜏^𝛽−𝛼𝑑𝜏−^∞₀𝑒^−𝜆𝜏𝜏^𝛽−𝛼𝑑𝜏𝑆_𝑎(𝜆)=Γ𝛽−𝛼+1𝜆−𝜇𝜆^{(𝛽−𝛼+1)}𝑆_𝑎𝜇−𝑆_𝑎(𝜆−1)Γ𝜆−𝜇𝛽−𝛼+1𝜇^{(𝛽−𝛼+1)}−Γ𝛽−𝛼+1𝜆^{(𝛽−𝛼+1)}𝑆_𝑎(𝜆)=Γ𝛽−𝛼+1𝜆𝜆−𝜇^{(𝛼−𝛽−1)}𝑆_𝑎𝜇−𝜇^{(𝛼−𝛽−1)}𝑆_𝑎(𝜆=Γ)𝛽−𝛼+1𝜆−𝜇𝜆𝜇^{𝛽−𝛼+1}×𝜇^{(𝛽−𝛼+1)}𝑆_𝑎𝜇−𝜆^{(𝛽−𝛼+1)}𝑆_𝑎(𝜆)=Γ𝛽−𝛼+1𝜆−𝜇𝜆𝜇^{𝛽−𝛼+1}𝑅𝜇^𝛼,𝐴−𝑅𝜆^𝛼,𝐴, (19) where the last equality follows from Lemma 3.
On the other hand, observing that 𝑅_𝑎(𝑡,𝑠)=𝛼Γ𝛽−𝛼+1Γ(1−𝛼)^𝑡₀𝑆_𝑎(𝑟)(𝑡+𝑠−𝑟)^1+𝛼𝑑𝑟∗𝑆_𝑎(𝑠), (20) Then taking Laplace transform with respect to 𝑡 and 𝑠, respectively, we deduce 𝑅_𝑎(𝑡,𝜆)=𝛼Γ𝛽−𝛼+1Γ(1−𝛼)×^∞₀𝑒^{−𝜆𝑠𝑡}₀𝑆_𝑎(𝑟)(𝑡+𝑠−𝑟)^1+𝛼𝑑𝑟𝑑𝑠𝑆_𝑎(𝜆)=𝛼Γ𝛽−𝛼+1Γ(1−𝛼)×^∞₀𝑒^−𝜆𝑠(𝑡+𝑠)^−𝛼−1∗𝑆_𝑎(𝑡)𝑑𝑠𝑆_𝑎(𝜆),𝑅_𝑎𝜇,𝜆=𝛼Γ𝛽−𝛼+1Γ()×1−𝛼_∞0𝑒^{−𝜇𝑡∞}₀𝑒^−𝜆𝑠(𝑡+𝑠)^−𝛼−1∗𝑆_𝑎(𝑡)𝑑𝑠𝑑𝑡𝑆_𝑎(𝜆)=𝛼Γ𝛽−𝛼+1Γ(1−𝛼)×^∞₀𝑒^{−𝜆𝑠∞}₀𝑒^−𝜇𝑡(𝑡+𝑠)^−𝛼−1∗𝑆_𝑎(𝑡)𝑑𝑠𝑑𝑡𝑆_𝑎(𝜆)=𝛼Γ𝛽−𝛼+1Γ()×1−𝛼_∞0𝑒^{−𝜆𝑠∞}₀𝑒^−𝜇𝑡(𝑡+𝑠)^−𝛼−1𝑑𝑡𝑑𝑠𝑆_𝑎𝜇𝑆_𝑎(𝜆)=Γ𝛽−𝛼+1Γ()Γ()1−𝛼1−𝛼𝜆𝜆−𝜇^𝛼−𝜇^𝛼𝑆_𝑎𝜇𝑆_𝑎(𝜆)=Γ𝛽−𝛼+1𝜆𝜆−𝜇^𝛼−𝜇^𝛼𝑆_𝑎𝜇𝑆_𝑎(𝜆)=Γ𝛽−𝛼+1𝜆𝜆−𝜇^𝛼−𝜇^𝛼×1𝜆𝜇^{𝛽−𝛼+1}𝑅𝜇^𝛼,𝐴𝑅𝜆^𝛼=Γ,𝐴𝛽−𝛼+1𝜆−𝜇𝜆𝜇^{𝛽−𝛼+1}𝑅𝜇^𝛼,𝐴−𝑅𝜆^𝛼,𝐴, (21) where the last equality follows from the resolvent identity. In view of (19), (21), and the uniqueness of Laplace transform, we obtain 𝐿_𝑎(𝑡,𝑠)=𝑅_𝑎(𝑡,𝑠),𝑡,𝑠≥0. The arbitrariness of 𝑎 implies 𝐿(𝑡,𝑠)=𝑅(𝑡,𝑠) for 𝑡,𝑠≥0.

Remark 6. (a) If 𝛽=0, then (𝛼,0)-ROF 𝑆_𝛼,0(𝑡) is an 𝛼-ROF and the equality (17) degenerates to be equality (7).
(b) If we assume that, for each 𝑥∈𝑋, the map 𝑡→𝑆_𝛼,𝛽(𝑡)𝑥 is continuously differentiable on [0,∞) and the limit of (𝛼,𝛽)-ROF 𝑆_𝛼,𝛽(𝑡) exists as 𝛼→1⁻, then multiplying both sides of (17) with 1−𝛼 and integrating by parts to the right side of (17) and letting 𝛼→1⁻, we can get that (3) is just the limit state of (17).

By Lemma 3, (𝛼,𝛽)-ROF generated by operator 𝐴 is exactly operator valued functions whose Laplace transforms are 𝜆^{𝛼−𝛽−1}𝑅(𝜆,𝐴). In the following theorem, we show that this property corresponds to the functional equation (17) for 𝑆_𝛼,𝛽(𝑡). The proof of this theorem is proved by Ardent [2, proposition 3.2.4] for 𝛼→1⁻,𝛽=𝑘∈ℕ. Our proof is different since we could not use the binomial formula as in [2].

Theorem 7. Let 𝑆∶ℝ⁺→𝐵(𝑋) be a strongly continuous function satisfying ‖𝑆(𝑡)‖≤𝑀𝑒^𝜔𝑡(𝑡≥0) for some 𝑀,𝜔≥0. Let 𝛼∈ℝ⁺⧵ℕ₀,𝛽∈ℝ⁺ satisfy that 𝛽−𝛼>−1, set 𝑅𝜆^𝛼∶=𝜆^{−𝛼+𝛽+1∞}₀𝑒^−𝜆𝑡𝑆(𝑡)𝑑𝑡. (22) Then the following assertions are equivalent.(i)There exists an operator 𝐴 such that (𝜔^𝛼,∞)⊂𝜌(𝐴) and 𝑅(𝜆^𝛼)=(𝜆^𝛼𝐼−𝐴)⁻¹ for 𝜆>𝜔.(ii)For 𝑠,𝑡≥0, the equality _𝑡^𝑡+𝑠(𝑠+𝑡−𝑟)^𝛽−𝛼𝑆(𝑟)𝑑𝑟−^𝑠₀(𝑠+𝑡−𝑟)^𝛽−𝛼𝑆(𝑟)=𝑑𝑟𝛼Γ𝛽−𝛼+1Γ(1−𝛼)^𝑠₀^𝑡₀𝑆𝑟₁𝑆𝑟₂𝑡+𝑠−𝑟₁−𝑟₂^1+𝛼𝑑𝑟₁𝑑𝑟₂ (23) holds and 𝑆(𝑡)𝑥=0 for all 𝑡≥0 implies that 𝑥=0.

Proof. Assume that (i) holds; then (𝜔^𝛼,∞)⊂𝜌(𝐴), (𝜆^𝛼𝐼−𝐴)⁻¹=𝜆^{−𝛼+𝛽+1}∫^∞₀𝑒^−𝜆𝑡𝑆(𝑡)𝑑𝑡 for 𝜆>𝜔; from Lemma 3, we know that 𝑆(𝑡) is the (𝛼,𝛽)-ROF generated by 𝐴; then Theorem 5 shows that equality (23) holds. It follows from (𝜔^𝛼,∞)⊂𝜌(𝐴) and 𝑅(𝜆^𝛼)=(𝜆^𝛼𝐼−𝐴)⁻¹ for 𝜆>𝜔 that 𝑅(𝜆^𝛼) is injective. If 𝑆(𝑡)𝑥=0 for all 𝑡≥0, from 𝑅(𝜆^𝛼)∶=𝜆^{−𝛼+𝛽+1}∫^∞₀𝑒^−𝜆𝑡𝑆(𝑡)𝑑𝑡, we have 𝑅(𝜆^𝛼)𝑥=0; thus 𝑥=0.
If (ii) is satisfied, similar as the calculations of (19) and (21), we can get that the Laplace transform of the left side and the right side of (17) are Γ𝛽−𝛼+1𝜆−𝜇𝜆𝜇^{𝛽−𝛼+1}𝑅𝜇^𝛼−𝑅𝜆^𝛼,Γ𝛽−𝛼+1𝜆𝜆−𝜇^𝛼−𝜇^𝛼𝜆𝜇^{𝛽−𝛼+1}𝑅𝜇^𝛼𝑅𝜆^𝛼, (24) respectively. So, 𝑅𝜇^𝛼−𝑅𝜆^𝛼=𝜆^𝛼−𝜇^𝛼𝑅𝜇^𝛼𝑅𝜆^𝛼. (25)
On the other hand, if 𝑅(𝜆^𝛼)𝑥=0, by 𝑅(𝜆^𝛼)=𝜆^{−𝛼+𝛽+1}∫^∞₀𝑒^−𝜆𝑡𝑆(𝑡)𝑑𝑡 and uniqueness of Laplace transform, we have 𝑆(𝑡)𝑥=0 for all 𝑡≥0, then from (ii) we know 𝑥=0, so, Ker𝑅(𝜆^𝛼)=0, by (25) and Lemma 4, we get the conclusion.

4. Fractional Abstract Cauchy Problems

In this section, we study the following inhomogeneous fractional abstract Cauchy problem: ^𝑐𝐷^𝛼_𝑡𝑢(𝑡)(𝑡)(𝑡)(),𝑢(0)=𝐴𝑢+𝑓,𝑡∈0,𝑇=𝑥₀,𝑢^′(0)=𝑥₁, (26) where 1<𝛼<2,𝑇>0,𝑓∈𝐿¹((0,𝑇),𝑋),𝑥₀,𝑥₁∈𝑋, 𝐴 is a linear closed operator.

First, we give the definitions of solutions to (26).

Definition 8. A function 𝑢∈𝐶([0,𝑇);𝑋) is called a mild solution of (26), if (𝑔_𝛼∗𝑢)(𝑡)∈𝐷(𝐴) and 𝑢(𝑡)=𝑥₀+𝑡𝑥₁+𝐴(𝑔_𝛼∗𝑢)(𝑡)+(𝑔_𝛼∗𝑓)(𝑡),𝑡∈[0,𝑇).

Definition 9. A function 𝑢∈𝐶([0,𝑇);𝑋) is called a classical solution of (26) if 𝑢 satisfies the following.(a)𝑢∈𝐶([0,𝑇);𝐷(𝐴))∩𝐶¹([0,𝑇);𝑋).(b)𝑔_2−𝛼∗(𝑢−𝑥₀−𝑡𝑥₁)∈𝐶²([0,𝑇);𝑋).(c)𝑢 satisfies (26).

From the above definitions, it is clear that a classical solution of (26) is a mild solution of it. The following assertion shows that a mild solution of the problem (26) with suitable regularity is also a classical solution.

Theorem 10. Let 𝑢 be a mild solution of (26) and 𝑓∈𝐶([0,𝑇);𝑋), if 𝑔_2−𝛼∗(𝑢−𝑥₀−𝑡𝑥₁)∈𝐶²([0,𝑇);𝑋), and for any 𝑡∈(0,𝑇), 𝑔_𝛼∗𝑢∈𝐿¹((0,𝑡),𝐷(𝐴)); then 𝑢 is also a classical solution of (26).

Proof. Since 𝑢 is a mild solution of (26), we have 𝑔_𝛼(𝑡)(𝐴),𝑢(𝑡)∗𝑢∈𝐷=𝑥₀+𝑡𝑥₁+𝐴𝑔_𝛼∗𝑢(𝑡)+𝑔_𝛼𝑡),[).∗𝑓(𝑡∈0,𝑇 (27) If we denote 𝑤(𝑡)∶=𝑢(𝑡)−𝑥₀−𝑡𝑥₁, then it follows from (27) that 𝑔_2−𝛼∗𝑤(𝑡)=𝑔_2−𝛼∗𝐴𝑔_𝛼∗𝑢(𝑡)+𝑔_𝛼(𝑡)𝑔∗𝑓=𝐴₂∗𝑢(𝑡)+𝑔₂∗𝑓(𝑡). (28)
Since 𝑔_2−𝛼∗𝑤∈𝐶²([0,𝑇);𝑋), then ^𝑐𝐷^𝛼_𝑡𝑢(𝑡)=(𝑑²/𝑑𝑡²)(𝑔_2−𝛼∗𝑤)(𝑡) is well defined, and by (28), we have ^𝑐𝐷^𝛼_𝑡𝑢(𝑡)=𝑑²𝑑𝑡²𝑔_2−𝛼∗𝑤(𝑡)=lim_ℎ→01ℎ²𝑔_2−𝛼∗𝑤(𝑡)−2𝑔_2−𝛼()+𝑔∗𝑤𝑡−ℎ_2−𝛼∗𝑤(𝑡−2ℎ)=lim_ℎ→01ℎ²𝐴𝑔₂∗𝑢(𝑡)−2𝐴𝑔₂()𝑔∗𝑢𝑡−ℎ+𝐴₂∗𝑢(𝑡−2ℎ)+lim_ℎ→01ℎ²𝑔₂∗𝑓(𝑡)−2𝑔₂()+𝑔∗𝑓𝑡−ℎ₂∗𝑓(𝑡−2ℎ)=lim_ℎ→01ℎ²𝐴𝑔₂∗𝑢(𝑡)−2𝐴𝑔₂()𝑔∗𝑢𝑡−ℎ+𝐴₂∗𝑢(𝑡−2ℎ)+𝑓(𝑡). (29)Thus, lim_ℎ→01ℎ²𝐴𝑔₂∗𝑢(𝑡)−2𝐴𝑔₂()𝑔∗𝑢𝑡−ℎ+𝐴₂∗𝑢(𝑡−2ℎ)=^𝑐𝐷^𝛼_𝑡𝑢(𝑡)−𝑓(𝑡). (30) On the other hand, from the closeness of 𝐴 and 𝑔_𝛼∗𝑢∈𝐿¹((0,𝑡),𝐷(𝐴)) for 𝑡∈[0,𝑇), by Proposition 1.1.7 in [2], we have 𝑔₂∗𝑢(𝑡)=𝑔_2−𝛼∗𝑔_𝛼∗𝑢(𝑡)∈𝐷(𝐴), (31) Then from (30) and the closeness of 𝐴, we obtain 𝑢(𝑡)=lim_ℎ→01ℎ²𝑔₂∗𝑢(𝑡)−2𝑔₂()+𝑔∗𝑢𝑡−ℎ₂∗𝑢(𝑡−2ℎ)∈𝐷_𝑐(𝐴),𝐷^𝛼_𝑡𝑢(𝑡)=𝐴𝑢(𝑡)+𝑓(𝑡),𝑡∈[0,𝑇). (32) It is clear that 𝑢(0)=𝑥₀,𝑢^′(0)=𝑥₁. Thus, 𝑢 is a classical solution of (26).

Lemma 11. Let 1<𝛼<2, 𝑓∈𝐿¹((0,𝑇),𝑋). Suppose 𝐴 is the generator of an (𝛼,𝛽)-ROF 𝑆_𝛼,𝛽(𝑡) on 𝑋 for some 𝛽∈ℝ⁺. Then, for every 𝑡∈[0,𝑇), (𝑔_𝛼−1∗𝑆_𝛼,𝛽∗𝑓)(𝑡) exists, and (𝑔_𝛼−1∗𝑆_𝛼,𝛽∗𝑓)∈𝐶([0,𝑇),𝑋).

Proof. For every 𝑡∈[0,𝑇), since 𝑔_𝛼−1∈𝐿¹((0,𝑡),ℝ⁺),𝑓∈𝐿¹((0,𝑡),𝑋), we get 𝑔_𝛼−1∗𝑓∈𝐿¹((0,𝑡),𝑋), hence, from 𝑔_𝛼−1∗𝑆_𝛼,𝛽∗𝑓(𝑡)=𝑆_𝛼,𝛽∗𝑔_𝛼−1(𝑡)=∗𝑓_𝑡0𝑆_𝛼,𝛽(𝑡−𝑠)𝑔_𝛼−1∗𝑓(𝑠)𝑑𝑠, (33) we obtain that (𝑔_𝛼−1∗𝑆_𝛼,𝛽∗𝑓)(𝑡) exists.
For ℎ∈ℝ,|ℎ|≪1 and 𝑡+ℎ∈[0,𝑇), we have 𝑔_𝛼−1∗𝑆_𝛼,𝛽∗𝑓(𝑡+ℎ)−𝑔_𝛼−1∗𝑆_𝛼,𝛽(𝑡)=∗𝑓_0𝑡+ℎ𝑆_𝛼,𝛽(𝑡+ℎ−𝑠)𝑔_𝛼−1(𝑠)−∗𝑓𝑑𝑠_𝑡0𝑆_𝛼,𝛽(𝑡−𝑠)𝑔_𝛼−1(𝑠)=∗𝑓𝑑𝑠_0𝑡+ℎ𝑆_𝛼,𝛽(𝑡+ℎ−𝑠)−𝑆_𝛼,𝛽(𝑡−𝑠)𝑔_𝛼−1(𝑠)−∗𝑓𝑑𝑠_𝑡𝑡+ℎ𝑆_𝛼,𝛽(𝑡−𝑠)𝑔_𝛼−1∗𝑓(𝑠)𝑑𝑠. (34) From the dominated convergence theorem and absolute continuity of integral, we deduce lim_ℎ→0𝑔_𝛼−1∗𝑆_𝛼,𝛽∗𝑓(𝑡+ℎ)−𝑔_𝛼−1∗𝑆_𝛼,𝛽∗𝑓(𝑡)=0. (35) So, (𝑔_𝛼−1∗𝑆_𝛼,𝛽∗𝑓)∈𝐶([0,𝑇),𝑋).

Let 𝑣(𝑡)=𝑆_𝛼,𝛽(𝑡)𝑥₀+𝑔₁∗𝑆_𝛼,𝛽(𝑡)𝑥₁+𝑔_𝛼−1∗𝑆_𝛼,𝛽∗𝑓(𝑡). (36) From Lemma 11, we know that 𝑣 is well defined, and 𝑣∈𝐶([0,𝑇),𝑋).

The following theorem is proved by Arendt [2, Lemma 3.2.9] for 𝛼=1,𝛽=𝑙∈ℕ. Our proof is different because we could not use the formula of integration by parts as [2, Lemma 3.2.9].

Theorem 12. Suppose that 𝐴 is the generator of an (𝛼,𝛽)-ROF 𝑆_𝛼,𝛽(𝑡) on 𝑋 for some 𝛽∈ℝ⁺. Let 𝑣 be defined by (36). Then one has the following results.(a)If (26) has a mild solution 𝑢, then 𝑔_𝑚−𝛽∗(𝑣−∑^𝑚−1_𝑘=0𝑣^(𝑘)(0)𝑔_𝑘+1(𝑡))∈𝐶^𝑚([0,𝑇);𝑋) and 𝑢(𝑡)=^𝑐𝐷^𝛽_𝑡𝑣(𝑡).(b)If there is a classical solution 𝑢 of (26), then 𝑔_2−𝛼∗(^𝑐𝐷^𝛽_𝑡𝑣(𝑡)−𝑥₀−𝑡𝑥₁)∈𝐶²([0,𝑇);𝑋) and 𝑢(𝑡)=^𝑐𝐷^𝛽_𝑡𝑣(𝑡).

Proof. If 𝑢 is a mild solution of (26), then (𝑔_𝛼∗𝑢)(𝑡)∈𝐷(𝐴) and 𝑢(𝑡)=𝑥₀+𝑡𝑥₁+𝐴𝑔_𝛼∗𝑢(𝑡)+𝑔_𝛼∗𝑓(𝑡),𝑡∈[0,𝑇). (37) Using Lemma 2(b) and the closeness of 𝐴, we have 𝑔_𝛽+1∗𝑢(𝑡)=𝑆_𝛼,𝛽−𝐴𝑔_𝛼∗𝑆_𝛼,𝛽(𝑡)=𝑆∗𝑢_𝛼,𝛽∗𝑢(𝑡)−𝐴𝑔_𝛼∗𝑆_𝛼,𝛽(𝑡)=𝑆∗𝑢_𝛼,𝛽∗𝑢(𝑡)−𝑆_𝛼,𝛽∗𝐴𝑔_𝛼𝑡)=𝑆∗𝑢(_𝛼,𝛽∗𝑢(𝑡)−𝑆_𝛼,𝛽∗𝑢−𝑥₀−𝑡𝑥₁−𝑔_𝛼∗𝑓(𝑡)=𝑆_𝛼,𝛽∗𝑥₀+𝑆_𝛼,𝛽∗𝑡𝑥₁+𝑆_𝛼,𝛽∗𝑔_𝛼∗𝑓(𝑡); (38) that is, (𝑔_𝛽+1∗𝑢)(𝑡)=(1∗𝑆_𝛼,𝛽)(𝑡)𝑥₀+(𝑔₂∗𝑆_𝛼,𝛽)(𝑡)𝑥₁+(𝑔_𝛼∗𝑆∗𝑓)(𝑡). So 𝐽^𝛽_𝑡𝑢(𝑡)=𝑔_𝛽(𝑡)=𝑑∗𝑢𝑔𝑑𝑡^𝛽+1∗𝑢(𝑡)=𝑆_𝛼,𝛽(𝑡)𝑥₀+𝑔₁∗𝑆_𝛼,𝛽(𝑡)𝑥₁+𝑔_𝛼−1∗𝑆_𝛼,𝛽∗𝑓(𝑡)=𝑣(𝑡). (39) Thus, it follows from 𝑢∈𝐶([0,𝑇),𝑋) that 𝑔_𝑚−𝛽∗(𝑣−∑^𝑚−1_𝑘=0𝑣^(𝑘)(0)𝑔_𝑘+1(𝑡))∈𝐶^𝑚([0,𝑇);𝑋) and 𝑢(𝑡)=^𝑐𝐷^𝛽_𝑡𝑣(𝑡). Hence (a) holds. If 𝑢 is a classical solution of (26), then 𝑢 is a mild solution of (26). So, assertion (b) follows immediately from (a).

Theorem 13. Let 𝑣 be defined by (36). Assume that 𝑣∈𝐶^𝑚−1([0,𝑇);𝑋), 𝑣^(𝑘)(0)=0 for 𝑘=0,1,…,𝑚−1, and 𝑔_𝑚−𝛽∗𝑣∈𝐶^𝑚([0,𝑇);𝑋); then ^𝑐𝐷^𝛽_𝑡𝑣(𝑡) is a mild solution of the problem (26). Moreover, if 𝑔_2−𝛼∗(^𝑐𝐷^𝛽_𝑡𝑣(𝑡)−𝑥₀−𝑡𝑥₁)∈𝐶²([0,𝑇);𝑋), and for any 𝑡∈(0,𝑇), 𝑔_𝛼∗^𝑐𝐷^𝛽_𝑡𝑣(𝑡)∈𝐿¹((0,𝑡),𝐷(𝐴)), then ^𝑐𝐷^𝛽_𝑡𝑣(𝑡) is also a classical solution of (26).

Proof. Consider the following steps.
Step 1. We first claim that 𝐽^𝛼_𝑡𝑣(𝑡)∈𝐷(𝐴) and ^𝑐𝐷^𝛽_𝑡𝐴𝐽^𝛼_𝑡𝑣(𝑡)=^𝑐𝐷^𝛽_𝑡𝑣(𝑡)−𝑥₀−𝑡𝑥₁−𝑔_𝛼∗𝑓. (40)
In view of definition of 𝑣(𝑡), we have 𝐽^𝛼_𝑡𝑣(𝑡)=𝑔_𝛼∗𝑆_𝛼,𝛽(𝑡)𝑥₀+𝑔_𝛼∗𝑔₁∗𝑆_𝛼,𝛽(𝑡)𝑥₁+𝑔_𝛼∗𝑔_𝛼−1∗𝑆_𝛼,𝛽(𝑡)=𝑔∗𝑓_𝛼∗𝑆_𝛼,𝛽(𝑡)𝑥₀+^𝑡₀𝑔_𝛼∗𝑆_𝛼,𝛽(𝜏)𝑥₁+𝑑𝜏_𝑡0𝑔_𝛼∗𝑆_𝛼,𝛽(𝑡−𝜏)𝑔_𝛼−1∗𝑓(𝜏)𝑑𝜏, (41) for 𝑡∈[0,𝑇).
From Lemma 2(b), for 0<𝜏<𝑡, we have 𝑔_𝛼∗𝑆_𝛼,𝛽(𝑡)𝑥₀∈𝐷(𝐴),𝑔_𝛼∗𝑆_𝛼,𝛽(𝜏)𝑥₁(𝐴),𝑔∈𝐷_𝛼∗𝑆_𝛼,𝛽(𝑡−𝜏)𝑔_𝛼−1(𝜏)(𝐴),𝐴𝑔∗𝑓∈𝐷_𝛼∗𝑆_𝛼,𝛽(𝜏)𝑥₁=𝑆_𝛼,𝛽(𝜏)𝑥₁−𝑔_𝛽+1(𝜏)𝑥₁∈𝐿¹(),𝐴𝑔0,𝑡_𝛼∗𝑆_𝛼,𝛽(𝑡−𝜏)𝑔_𝛼−1∗𝑓(𝜏)=𝑆_𝛼,𝛽(𝑡−𝜏)𝑔_𝛼−1∗𝑓(𝜏)−𝑔_𝛽+1(𝑡−𝜏)𝑔_𝛼−1∗𝑓(𝜏)∈𝐿¹(0,𝑡), (42) combining with the closeness of 𝐴, one has ^𝑡₀𝑔_𝛼∗𝑆_𝛼,𝛽(𝜏)𝑥₁𝑑𝜏∈𝐷_𝑡0(𝐴),𝑔_𝛼∗𝑆_𝛼,𝛽(𝑡−𝜏)𝑔_𝛼−1∗𝑓(𝜏)𝑑𝜏∈𝐷(𝐴). (43) Thus 𝐽^𝛼_𝑡𝑣(𝑡)∈𝐷(𝐴), and 𝐴𝐽^𝛼_𝑡𝑣(𝑡)=𝐴𝑔_𝛼∗𝑆_𝛼,𝛽(𝑡)𝑥₀+𝑔₁∗𝐴𝑔_𝛼∗𝑆_𝛼,𝛽(𝑡)𝑥₁+𝑔_𝛼−1∗𝐴𝑔_𝛼∗𝑆_𝛼,𝛽∗𝑓(𝑡)=𝑆_𝛼,𝛽(𝑡)𝑥₀−𝑔_𝛽+1(𝑡)𝑥₀+𝑔₁∗𝑆_𝛼,𝛽(𝑡)𝑥₁−𝑔₁∗𝑔_𝛽+1(𝑡)𝑥₁+𝑔_𝛼−1∗𝑆_𝛼,𝛽∗𝑓−𝑔_𝛽+1∗𝑓(𝑡)=𝑆_𝛼,𝛽(𝑡)𝑥₀+𝑔₁∗𝑆_𝛼,𝛽(𝑡)𝑥₁+𝑔_𝛼−1∗𝑆_𝛼,𝛽∗𝑓(𝑡)−𝑔_𝛽+1(𝑡)𝑥₀−𝑔₁∗𝑔_𝛽+1(𝑡)𝑥₁−𝑔_𝛼+𝛽(𝑡)(𝑡)∗𝑓=𝑣−𝑔_𝛽+1(𝑡)𝑥₀−𝑔₁∗𝑔_𝛽+1(𝑡)𝑥₁−𝑔_𝛼+𝛽∗𝑓(𝑡). (44) So 𝐴𝐽^𝛼_𝑡𝑣(𝑡)=𝑣(𝑡)−𝑔_𝛽+1(𝑡)𝑥₀−𝑔₁∗𝑔_𝛽+1(𝑡)𝑥₁−𝑔_𝛼+𝛽∗𝑓_𝑐(𝑡),𝐷^𝛽_𝑡𝐴𝐽^𝛼_𝑡𝑣(𝑡)=^𝑐𝐷^𝛽_𝑡𝑣(𝑡)−𝑥₀−𝑡𝑥₁−𝑔_𝛼∗𝑓. (45)
Step 2. We prove ^𝑐𝐷^𝛽_𝑡𝐽^𝛼_𝑡𝑣(𝑡)∈𝐷(𝐴) and 𝐴^𝑐𝐷^𝛽_𝑡𝐽^𝛼_𝑡𝑣(𝑡)=^𝑐𝐷^𝛽_𝑡𝐴𝐽^𝛼_𝑡𝑣(𝑡).
Since 𝑣∈𝐶^𝑘([0,𝑇);𝑋),𝑣^(𝑘)(0)=0 for  𝑘=0,1,…,𝑚−1, we have 𝑑^𝑘𝑑𝑡^𝑘𝑔_𝛼∗𝑣(𝑡)|_𝑡=0=𝑔_𝛼∗𝑣^(𝑘)(𝑡)|_𝑡=0=0. (46) So ^𝑐𝐷^𝛽_𝑡𝐽^𝛼_𝑡𝑣(𝑡)=𝑑^𝑚𝑑𝑡^𝑚𝑔_𝑚−𝛽∗𝑔_𝛼∗𝑣(𝑡)=lim_ℎ→01ℎ^𝑚𝑚_𝑟=0𝐶^𝑟_𝑚𝑔_𝑚−𝛽∗𝑔_𝛼∗𝑣(𝑡−𝑟ℎ), (47) where 𝐶^𝑟_𝑚=(𝑚(𝑚−1)⋯(𝑚−𝑟+1))/𝑟!. From (45), we know that 𝐴𝐽^𝛼_𝑡𝑣∈𝐿¹((0,𝑡),𝑋), and the closeness of 𝐴 implies that (𝑔_𝑚−𝛽∗𝑔_𝛼∗𝑣)(𝑡)∈𝐷(𝐴), and 𝐴(𝑔_𝑚−𝛽∗𝑔_𝛼∗𝑣)(𝑡)=𝑔_𝑚−𝛽∗𝐴(𝑔_𝛼∗𝑣)(𝑡), by Step 1, ^𝑐𝐷^𝛽_𝑡𝐴𝐽^𝛼_𝑡𝑣(𝑡) exists, then ^𝑐𝐷^𝛽_𝑡𝐽^𝛼_𝑡𝑣(𝑡)∈𝐷(𝐴), and 𝐴^𝑐𝐷^𝛽_𝑡𝐽^𝛼_𝑡𝑣(𝑡)=^𝑐𝐷^𝛽_𝑡𝐴𝐽^𝛼_𝑡𝑣(𝑡). (48)
Step 3. We show that 𝑣^(𝑘)(0)=0 for 𝑘=0,1,…,𝑚−1 implies ^𝑐𝐷^𝛽_𝑡𝐽^𝛼_𝑡𝑣(𝑡)=𝐽^𝛼_𝑡^𝑐𝐷^𝛽_𝑡𝑣(𝑡). (49)
In fact, if 𝛼≥𝛽, we have ^𝑐𝐷^𝛽_𝑡𝐽^𝛼_𝑡𝑣(𝑡)=𝐽_𝑡^𝛼−𝛽𝑣(𝑡), and 𝐽^𝛼_𝑡^𝑐𝐷^𝛽_𝑡𝑣(𝑡)=𝐽_𝑡^𝛼−𝛽𝐽^𝛽_𝑡^𝑐𝐷^𝛽_𝑡𝑣(𝑡)=𝐽_𝑡^𝛼−𝛽𝑣(𝑡)−^𝑚−1_𝑘=0𝑣^(𝑘)(0)𝑔_𝑘+1(𝑡). (50) If 𝛼<𝛽, we have ^𝑐𝐷^𝛽_𝑡𝐽^𝛼_𝑡𝑣(𝑡)=^𝑐𝐷_𝑡^𝛽−𝛼𝑣(𝑡), and 𝐽^𝛼_𝑡^𝑐𝐷^𝛽_𝑡𝑣(𝑡)=^𝑐𝐷_𝑡^𝛽−𝛼𝐽^𝛽_𝑡^𝑐𝐷^𝛽_𝑡𝑣(𝑡)=^𝑐𝐷_𝑡^𝛽−𝛼𝑣(𝑡)−^𝑚−1_𝑘=0𝑣^(𝑘)(0)𝑔_𝑘+1(𝑡). (51) From the above discussion and 𝑣^(𝑘)(0)=0 for 𝑘=0,1,…,𝑚−1, we conclude that (49) holds.
Finally, in view of (40), (48), and (49), we have 𝐴𝐽^𝛼_𝑡^𝑐𝐷^𝛽_𝑡𝑣(𝑡)=𝐴^𝑐𝐷^𝛽_𝑡𝐽^𝛼_𝑡𝑣(𝑡)=^𝑐𝐷^𝛽_𝑡𝐴𝐽^𝛼_𝑡𝑣(𝑡)=^𝑐𝐷^𝛽_𝑡𝑣(𝑡)−𝑥₀−𝑡𝑥₁−𝑔_𝛼∗𝑓(𝑡). (52) Therefore, ^𝑐𝐷^𝛽_𝑡𝑣(𝑡) is a mild solution of (26).
Moreover, if 𝑔_2−𝛼∗(^𝑐𝐷^𝛽_𝑡𝑣(𝑡)−𝑥₀−𝑡𝑥₁)∈𝐶²([0,𝑇);𝑋), and for any 𝑡∈(0,𝑇), 𝑔_𝛼∗^𝑐𝐷^𝛽_𝑡𝑣(𝑡)∈𝐿¹((0,𝑡),𝐷(𝐴)), applying Theorem 10, we have that ^𝑐𝐷^𝛽_𝑡𝑣(𝑡) is a classical solution of (26).

Remark 14. If 𝛼→1⁺,𝛽=𝑘, then (26) becomes (4). Theorem 13 degenerated to Lemma 3.2.10 in [2]. Note that the condition 𝑣^(𝑗)(0)=0 for  𝑗=0,1,…,𝑘−1 is not necessary in Lemma 3.2.10 of [2], since from its proof, it is easy to see that 𝑣(0)=0 implies that 𝑣^(𝑗)(0)=0 for 𝑗=1,…,𝑘−1.

Now, we turn our attention to the problem ^𝑐𝐷^𝛼_𝑡𝑣(𝑡)=𝐴𝑣(𝑡)+𝑔_𝛽+1𝑡(𝑥)𝑣,𝑡>0,_(𝑘)(0)=0,𝑘=0,1,…,𝑁−1, (53) where 𝛼>0,𝑥∈𝑋, 𝐴 is a linear closed operator on 𝑋 and 𝑁 is the smallest integer greater than or equal to 𝛼.

Theorem 15. Let 𝐴 be a closed operator on 𝑋 and 𝛽>0; the following two assertions are equivalent.(i)𝐴 generates an exponentially bounded (𝛼,𝛽)-ROF 𝑆_𝛼,𝛽 on 𝑋.(ii)For every 𝑥∈𝑋, there exists a unique classical solution 𝑣_𝑥 of (53) which is exponentially bounded and 𝐴𝑣_𝑥∈𝐿¹_loc(ℝ⁺;𝑋).

Proof. If (𝑖) is satisfied, for every 𝑥∈𝑋, define 𝑣_𝑥∶ℝ⁺→𝑋 by 𝑣_𝑥(𝑡)=(𝑔_𝛼∗𝑆_𝛼,𝛽)(𝑡)𝑥, then 𝑣_𝑥^(𝑘)(0)=0 for 𝑘=0,1,…,𝑁−1. By Lemma 2(b), we have 𝑣_𝑥(𝑡)=(𝑔_𝛼∗𝑆_𝛼,𝛽)(𝑡)𝑥∈𝐷(𝐴), and ^𝑐𝐷^𝛼_𝑡𝑣_𝑥(𝑡)=𝑆_𝛼,𝛽(𝑡)𝑥=𝐴𝑔_𝛼∗𝑆_𝛼,𝛽(𝑡)𝑥+𝑔_𝛽+1(𝑡)𝑥=𝐴𝑣_𝑥(𝑡)+𝑔_𝛽+1(𝑡)𝑥,𝑡>0. (54) Thus, 𝑣_𝑥 is a classical solution of (53); it is unique by Theorem 12. Since 𝑆_𝛼,𝛽 is exponentially bounded, we have that 𝑣_𝑥 is exponentially bounded. From 𝐴𝑣_𝑥(𝑡)=𝐴𝑔_𝛼∗𝑆_𝛼,𝛽(𝑡)𝑥=𝑆_𝛼,𝛽(𝑡)𝑥−𝑔_𝛽+1(𝑡)𝑥, (55) we know that 𝐴𝑣_𝑥(𝑡)∈𝐿¹_loc(ℝ⁺;𝑋). So (ii) is true.
Assume that (ii) holds. From linearity of (53) and the uniqueness of its solution, we get that 𝑣_𝑥 is linear in 𝑥. So, for each 𝑡≥0, there exists a linear mapping 𝑉(𝑡)∶𝑋→𝐷(𝐴) such that 𝑉(𝑡)𝑥=𝑣_𝑥(𝑡) for any 𝑥∈𝑋.
Next, we show that, for each 𝑡≥0, 𝑉(𝑡)∈𝐵(𝑋,𝐷(𝐴)).
We consider the mapping Φ∶𝑋→𝐶(ℝ⁺,𝐷(𝐴)) by Φ(𝑥)=𝑣_𝑥(⋅)=𝑉(⋅)𝑥. Then, Φ is a linear operator defined on 𝑋. Now we show that Φ is closed, if 𝑥_𝑛→𝑥 in 𝑋 and Φ(𝑥_𝑛)→𝑢 in 𝐶(ℝ⁺,𝐷(𝐴)). For 𝑡>0, by the dominated convergence theorem, we have that 𝐽^𝛼_𝑡𝑣_𝑥𝑛(𝑡) converges to 𝐽^𝛼_𝑡𝑢(𝑡), since 𝑣_𝑥𝑛(⋅)=𝑔_𝛼+𝛽+1(𝑡)𝑥_𝑛+𝐽^𝛼_𝑡𝐴𝑣_𝑥𝑛(𝑡), from the closeness of 𝐴, it follows that as 𝑛→∞, 𝑢(𝑡)=𝑔_𝛼+𝛽+1(𝑡)𝑥+𝐽^𝛼_𝑡𝐴𝑢(𝑡), which implies that 𝑢=Φ(𝑥) and Φ is closed. Therefore, by the closed graph theorem, Φ is bounded. So, for each 𝑡≥0, 𝑉(𝑡)∈𝐵(𝑋,𝐷(𝐴)). Then, the exponentially boundedness of 𝑉(𝑡)𝑥 and Lemma 3.2.14 in [2], imply that ‖𝑉(𝑡)‖≤𝑀𝑒^𝜔𝑡(𝑡≥0) for some constants 𝑀,𝜔≥0. So 𝑄(𝜆)𝑥=𝜆^𝛽+1∫^∞₀𝑒^−𝜆𝑡𝑉(𝑡)𝑥𝑑𝑡 is well defined for 𝜆>𝜔,(𝜔^𝛼,∞)⊂𝜌(𝐴).
Since 𝐴𝑉(𝑡)𝑥∈𝐿¹(ℝ⁺,𝑋), then the Laplace transform of 𝐴𝑉(𝑡)𝑥 is well defined, and from the closeness of 𝐴, for 𝜆>𝜔, we have 𝜆^𝛼−𝐴𝑄(𝜆)𝑥=𝜆^{𝛼+𝛽+1∞}₀𝑒^−𝜆𝑡𝑉(𝑡)𝑥𝑑𝑡−𝜆^𝛼+𝛽+1×^∞₀𝑒^−𝜆𝑡𝐴𝑉(𝑡)𝑥𝑑𝑡=𝜆^{𝛼+𝛽+1∞}₀𝑒^−𝜆𝑡𝑉(𝑡)𝑥𝑑𝑡−𝜆^𝛽+1×^∞₀𝑒^{𝑐−𝜆𝑡}𝐷^𝛼_𝑡𝑉(𝑡)𝑥𝑑𝑡+𝜆^𝛽+1×^∞₀𝑒^−𝜆𝑡𝑔_𝛽+1(𝑡)𝑥𝑑𝑡=𝜆^𝛼+𝛽+1𝑉(𝜆)𝑥−𝜆^𝛽+1𝜆^𝛼𝑉(𝜆)𝑥+𝜆^𝛽+1𝜆^−(𝛽+1)𝑥=𝑥. (56)
Now, we show that (𝜆^𝛼−𝐴) is injective for 𝜆>𝜔. Assume that (𝜆^𝛼−𝐴)𝑥=0 for some 𝑥∈𝐷(𝐴) and 𝜆>𝜔. Then, by the method of Laplace transform, we have that the solution of (53) is 𝑡^𝛼+𝛽𝐸_{𝛼,𝛼+𝛽+1}(𝜆^𝛼𝑡^𝛼). Since ‖𝑣_𝑥(𝑡)‖≤𝑀𝑒^𝜔𝑡, for all 𝑡≥0, combine with (15), it follows that 𝑥=0. Hence (𝜆^𝛼−𝐴)⁻¹=𝑄(𝜆) for 𝜆>𝜔 and 𝑉(𝑡) is an (𝛼,𝛼+𝛽)-ROF. Let 𝑆_𝛼,𝛽(𝑡)𝑥∶=^𝑐𝐷^𝛼_𝑡𝑉(𝑡)𝑥=𝐴𝑉(𝑡)𝑥+𝑔_𝛽+1(𝑡)𝑥; (57) then 𝑆_𝛼,𝛽(𝑡)𝑥 exists and 𝑉(𝑡)𝑥=𝐽^𝛼_𝑡𝑆_𝛼,𝛽(𝑡)𝑥 for all 𝑡≥0 and all 𝑥∈𝑋. So 𝑆_𝛼,𝛽(𝑡)𝑥=𝐴𝐽^𝛼_𝑡𝑆_𝛼,𝛽(𝑡)𝑥+𝑔_𝛽+1(𝑡)𝑥, (58) and taking the Laplace transform, we have 𝑆_𝛼,𝛽(𝜆)𝑥=𝐴𝜆^−𝛼𝑆_𝛼,𝛽(𝜆)𝑥+𝜆^−𝛽−1𝑥,𝜆>𝜔; (59) that is, 𝑆_𝛼,𝛽(𝜆)𝑥=𝜆^{𝛼−𝛽−1}𝜆^𝛼−𝐴⁻¹𝑥,𝜆>𝜔. (60) From Lemma 3, we know that 𝑆_𝛼,𝛽 is the (𝛼,𝛽)-ROF generated by 𝐴.

Remark 16. Theorem 15 extends and generalizes Theorem 3.2.13 in [2]. In fact, when 𝛼=1 and 𝛽=𝑘, (53) becomes (5), 𝑆_𝛼,𝛽(𝑡) is a 𝑘-times integrated semigroup. For problem (5), the condition 𝐴𝑣_𝑥∈𝐿¹_loc(ℝ⁺;𝑋) in (ii) is not necessary. Since from the proof of Theorem 3.2.13 in [2], it is easy to see that the assumption that exponentially boundedness of the unique classical solution to the problem (5) imply that 𝐴𝑣_𝑥∈𝐿¹_loc(ℝ⁺;𝑋).

Conflict of Interests

The authors declare that there is no conflict of interests regarding the publication of this paper.

Acknowledgments

This work is supported by the Program for New Century Excellent Talents in University (NECT-12-0246) and FRFCU (lzujbky-2013-k02).