# Integrated Fractional Resolvent Operator Function and Fractional Abstract Cauchy Problem

Authors: Ya-Ning Li; Hong-Rui Sun
Journal: Abstract and Applied Analysis
 (2014)

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## Abstract

We firstly prove thatβ-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 cDtαu(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)=Sα,β(t)x0+(g1*Sα,β)(t)x1+(gα-1*Sα,β*f)(t) and mild solution and classical solution of it. Finally, for the problem cDtαv(t)=Av(t)+gβ+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 vx and Avx∈Lloc1(ℝ+,X). Our results extend and generalize some related results in the literature.


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## Body

## 1. Introduction

This paper is concerned with the properties ofβ-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
(1)(f*g)(t)=∫0t‍f(t-τ)g(τ)dτ,t≥0.gα denotes the function
(2)gα(t)={tα-1Γ(α),t>0,0,t≤0,
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:
(3)∫tt+s‍(s+t-r)β-1S(r)dr-∫0s‍(s+t-r)β-1S(r)dr=Γ(β)S(t)S(s),t,s≥0.
In the special case of β=k∈ℕ, the corresponding result is summarized in [2].For the inhomogeneous Cauchy problem(4)u′(t)=Au(t)+f(t),t∈[0,T],u(0)=x,
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+∫0t‍S(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∈Ck([0,T],X)(Ck+1([0,T],X)) and u=v(k). On the other hand, if v∈Ck([0,T],X)(Ck+1([0,T],X)), then v(k) is also a mild (classical) solution of it.Furthermore, ifA generates an exponential bounded k-times integrated semigroup on a Banach space X, then, for any x∈X, v(t)=∫0t‍S(s)xds is the unique exponential bounded classical solution of the following problem:
(5)u′(t)=Au(t)+gk+1(t)x,t≥0,u(0)=0.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 {Sα(t)}t≥0⊂B(X) is an α-ROF if and only if {Sα(t)}t≥0 is a solution of abstract fractional Cauchy problem
(6)Dctαv(t)=Av(t)+gβ+1(t)x,t>0,v(0)=x,v(k)(0)=0,k=1,…,N-1.
When 0<α<1, Peng and Li [18] proved that the solution operator {Sα(t)}t≥0 for (6) satisfies the following equality:
(7)∫tt+sSα(τ)(t+s-τ)αdτ‍-∫0sSα(τ)(t+s-τ)αdτ‍=α∫0t∫0sSα(τ1)Sα(τ2)(t+s-τ1-τ2)1+α‍dτ1dτ2,t,s≥0.
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]).
Letα>0, β≥0. A function Sα,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)
Sα,β(·) is strongly continuous on ℝ+ and Sα,β(0)=gβ+1(0)I;(b)
Sα,β(s)Sα,β(t)=Sα,β(t)Sα,β(s) for all s,t≥0;(c)
the functional equation(8)Sα,β(s)JtαSα,β(t)-JsαSα,β(s)Sα,β(t)=gβ+1(s)JtαSα,β(t)-gβ+1(t)JsαSα,β(s)holds fors,t≥0, where Jtα is the Riemann-Liouville fractional integral of order α.The generatorA of Sα,β(t) is defined by
(9)D(A)≔{x∈X:limt→0+Sα,β(t)x-gβ+1(t)xgα+β+1(t)exists},Ax≔limt→0+Sα,β(t)x-gβ+1(t)xgα+β+1(t),x∈D(A).
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
(10)Dctαu(t)=Au(t)+f(t),t∈(0,T),ggggggu(0)=x0,u′(0)=x1,
where 1<α<2, T>0, f∈L1((0,T),X), and A is assumed to be the generator of an (α,β)-ROF Sα,β(t) on X. We give the relation between the function v(t)=Sα,β(t)x0+(g1*Sα,β)(t)x1+(gα-1*Sα,β*f)(t) and solution of (10). We also study the problem
(11)Dctαv(t)=Av(t)+gβ+1(t)x,t>0,hv(k)(0)=0,k=0,1,…,N-1,
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 vx and Avx∈LLoc1(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 Section2, 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 orderα>0 of f is defined by
(12)Jtαf(t)=(gα*f)(t)=∫0tgα(t-s)f(s)ds,
and the Caputo fractional derivative of order α>0 of f can be written as
(13)Dctαf(t)=dmdtm(gm-α*(f(t)-∑k=0m-1f(k)(0)gk+1(t)‍)),
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(14)Eα,β(z)=∑n=0∞znΓ(αn+β)‍,Eα(z)=Eα,1(z),gggggggggggggggggReα>0,β,z∈ℂ.
And if 0<α<2, β>0, then
(15)Eα,β(z)=1αz(1-β)/αexp(z1/α)+εα,β(z),|argz|≤12απ,Eα,β(z)=εα,β(z),|arg(-z)|<(1-12α)π,
where
(16)εα,β(z)=-∑n=1N-1z-nΓ(β-αn)‍+O(|z|-N)asz⟶∞,
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]).
LetSα,β:ℝ+→B(X) be an (α,β)-ROF generated by A. The following assertions hold:(a)
Sα,β(t)D(A)⊂D(A) and ASα,β(t)x=Sα,β(t)Ax for x∈D(A) and t≥0;(b)
for allx∈X, JtαSα,β(t)x∈D(A) and Sα,β(t)x=gβ+1(t)x+AJtαSα,β(t)x, t≥0;(c)
x∈D(A) and Ax=y if and only if Sα,β(t)x=gβ+1(t)x+JtαSα,β(t)y, t≥0;(d)
A is closed.Lemma 3 (see [9, Proposition 3.5, Theorem 3.11]).
Letα>0, β≥0. A generates an (α,β)-ROF Sα,β satisfying ∥Sα,β(t)∥≤Meωt, t≥0, for some constants M>0 and ω≥0, if and only if (ωα,∞)⊂ρ(A) and there exists a strongly continuous function S:ℝ+→B(X) such that ∥S(t)∥≤Meωt for all t≥0 and ∫0∞‍e-λtS(t)xdt=λα-β-1R(λα,A)x, λ>ω, for all x∈X. Furthermore, S(t) is Sα,β(t).Lemma 4 (see [2, Proposition B.6]).
LetU⊂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 {Sα,β}t≥0 in this theorem.Theorem 5.
Letα∈ℝ+∖ℕ0,β∈ℝ+ satisfy β-α>-1. If {S(t)}t≥0 is an (α,β)-ROF, then it satisfies the following equality:
(17)∫tt+s(s+t-r)β-αS(r)dr-∫0s(s+t-r)β-αS(r)dr‍=αΓ(β-α+1)Γ(1-α)∫0s∫0tS(r1)S(r2)(t+s-r1-r2)1+α‍dr1dr2.Proof.
Denote byL(t,s) and R(t,s) the left and right sides of equality (17), respectively, and denote by fa(t) the truncation of f(t) at a, that is, fa(t)=f(t) for 0≤t≤a and fa(t)=0 otherwise.
We will show that the Laplace transform ofLa(t,s) and Ra(t,s) with respect to t and s is equivalent, and by the uniqueness of Laplace transform, we can get that La(t,s)=Ra(t,s).
Taking Laplace transform ofLa(t,s) with respect to s as follows
(18)L^a(t,λ)=∫0∞e-λs‍[∫tt+s(s+t-r)β-αSa(r)dr‍=∫0∞e-λs=.-∫0s(s+t-r)β-αSa(r)dr‍]ds=∫t∞Sa(r)∫r-t∞e-λs(s+t-r)β-αdsdr‍-∫0∞Sa(r)∫r∞e-λs(s+t-r)β-αdsdr‍=∫t∞Sa(r)e-λ(r-t)∫0∞e-λττβ-αdτdr‍-∫0∞Sa(r)e-λ(r-t)∫t∞e-λττβ-αdτdr‍=Γ(β-α+1)λ(β-α+1)∫t∞Sa(r)e-λ(r-t)dr‍-eλtS^a(λ)∫t∞e-λττβ-αdτ,
then taking Laplace transform with respect to t, we have
(19)L^a(μ,λ)=∫0∞e-μt[Γ(β-α+1)λ(β-α+1)∫t∞Sa(r)e-λ(r-t)dr‍‍hhhhhhhh.hΓ(β-α+1)λ(β-α+1)-eλtS^a(λ)∫t∞e-λττβ-αdτ]dt=Γ(β-α+1)λ(β-α+1)∫0∞e-μt∫t∞Sa(r)e-λ(r-t)drdt‍-S^a(λ)∫0∞e(λ-μ)t∫t∞e-λττβ-αdτdt‍=Γ(β-α+1)λ(β-α+1)∫0∞e-λrSa(r)‍∫0re(λ-μ)tdtdr‍-∫0∞e-λττβ-α‍∫0τe(λ-μ)tdtdτS^a(λ)‍=Γ(β-α+1)(λ-μ)λ(β-α+1)×(∫0∞e-μrSa(r)dr‍-∫0∞e-λrSa(r)dr)-1λ-μ×(∫0∞e-μττβ-αdτ-∫0∞e-λττβ-αdτ)S^a(λ)=Γ(β-α+1)(λ-μ)λ(β-α+1)(S^a(μ)-S^a(λ))-1λ-μ(Γ(β-α+1)μ(β-α+1)-Γ(β-α+1)λ(β-α+1))S^a(λ)=Γ(β-α+1)λ-μ(λ(α-β-1)S^a(μ)-μ(α-β-1)S^a(λ))=Γ(β-α+1)(λ-μ)(λμ)β-α+1×(μ(β-α+1)S^a(μ)-λ(β-α+1)S^a(λ))=Γ(β-α+1)(λ-μ)(λμ)β-α+1(R(μα,A)-R(λα,A)),
where the last equality follows from Lemma 3.
On the other hand, observing that(20)Ra(t,s)=αΓ(β-α+1)Γ(1-α)∫0tSa(r)(t+s-r)1+α‍dr*Sa(s),
Then taking Laplace transform with respect to t and s, respectively, we deduce
(21)R^a(t,λ)=αΓ(β-α+1)Γ(1-α)×∫0∞e-λs‍∫0tSa(r)(t+s-r)1+αdrdsS^a(λ)‍=αΓ(β-α+1)Γ(1-α)×∫0∞e-λs(t+s)-α-1*Sa(t)dsS^a(λ),R^a(μ,λ)=αΓ(β-α+1)Γ(1-α)×∫0∞e-μt‍∫0∞e-λs(t+s)-α-1*Sa(t)dsdtS^a(λ)=αΓ(β-α+1)Γ(1-α)×∫0∞e-λs∫0∞‍e-μt(t+s)-α-1*Sa(t)dsdtS^a(λ)=αΓ(β-α+1)Γ(1-α)×∫0∞e-λs∫0∞e-μt(t+s)-α-1‍dtdsS^a(μ)S^a(λ)=Γ(β-α+1)Γ(1-α)Γ(1-α)λ-μ(λα-μα)S^a(μ)S^a(λ)=Γ(β-α+1)λ-μ(λα-μα)S^a(μ)S^a(λ)=Γ(β-α+1)(λ-μ)(λα-μα)×1(λμ)β-α+1R(μα,A)R(λα,A)=Γ(β-α+1)(λ-μ)(λμ)β-α+1(R(μα,A)-R(λα,A)),
where the last equality follows from the resolvent identity. In view of (19), (21), and the uniqueness of Laplace transform, we obtain La(t,s)=Ra(t,s),t,s≥0. The arbitrariness of a implies L(t,s)=R(t,s) for t,s≥0.Remark 6.
(a) Ifβ=0, then (α,0)-ROF Sα,0(t) is an α-ROF and the equality (17) degenerates to be equality (7).
(b) If we assume that, for eachx∈X, the map t→Sα,β(t)x is continuously differentiable on [0,∞) and the limit of (α,β)-ROF Sα,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 Lemma3, (α,β)-ROF generated by operator A is exactly operator valued functions whose Laplace transforms are λα-β-1R(λ,A). In the following theorem, we show that this property corresponds to the functional equation (17) for Sα,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.
LetS~:ℝ+→B(X) be a strongly continuous function satisfying ∥S~(t)∥≤Meωt(t≥0) for some M,ω≥0. Let α∈ℝ+∖ℕ0,β∈ℝ+ satisfy that β-α>-1, set
(22)R(λα)≔λ-α+β+1∫0∞e-λtS~(t)dt‍.
Then the following assertions are equivalent.(i)
There exists an operatorA such that (ωα,∞)⊂ρ(A) and R(λα)=(λαI-A)-1 for λ>ω.(ii)
Fors,t≥0, the equality
(23)∫tt+s(s+t-r)β-αS~(r)dr-∫0s(s+t-r)β-αS~(r)dr‍=αΓ(β-α+1)Γ(1-α)∫0s∫0tS~(r1)S~(r2)(t+s-r1-r2)1+αdr1dr2‍‍holds andS~(t)x=0 for all t≥0 implies that x=0.Proof.
Assume that (i) holds; then(ωα,∞)⊂ρ(A), (λαI-A)-1=λ-α+β+1∫0∞‍e-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 (ωα,∞)⊂ρ(A) and R(λα)=(λαI-A)-1 for λ>ω that R(λα) is injective. If S~(t)x=0 for all t≥0, from R(λα):=λ-α+β+1∫0∞‍e-λtS~(t)dt, we have R(λα)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
(24)Γ(β-α+1)(λ-μ)(λμ)β-α+1(R(μα)-R(λα)),Γ(β-α+1)λ-μλα-μα(λμ)β-α+1R(μα)R(λα),
respectively. So,
(25)R(μα)-R(λα)=(λα-μα)R(μα)R(λα).
On the other hand, ifR(λα)x=0, by R(λα)=λ-α+β+1∫0∞‍e-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(λα)=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:(26)Dctαu(t)=Au(t)+f(t),t∈(0,T),hhhhu(0)=x0,u′(0)=x1,
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 functionu∈C([0,T);X) is called a mild solution of (26), if (gα*u)(t)∈D(A) and u(t)=x0+tx1+A(gα*u)(t)+(gα*f)(t),t∈[0,T).Definition 9.
A functionu∈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)
g2-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.
Letu be a mild solution of (26) and f∈C([0,T);X), if g2-α*(u-x0-tx1)∈C2([0,T);X), and for any t∈(0,T), gα*u∈L1((0,t),D(A)); then u is also a classical solution of (26).Proof.
Sinceu is a mild solution of (26), we have
(27)ggggggg(gα*u)(t)∈D(A),u(t)=x0+tx1+A(gα*u)(t)+(gα*f)(t),hhhhhhhhhhhhhhhhhhhhhhhhhhhh.t∈[0,T).
If we denote w(t):=u(t)-x0-tx1, then it follows from (27) that
(28)(g2-α*w)(t)=g2-α*(A(gα*u)(t)+(gα*f)(t))=A(g2*u)(t)+(g2*f)(t).
Sinceg2-α*w∈C2([0,T);X), then cDtαu(t)=(d2/dt2)(g2-α*w)(t) is well defined, and by (28), we have
(29)Dctαu(t)=d2dt2(g2-α*w)(t)=limh→01h2[(g2-α*w)(t)-2(g2-α*w)(t-h)hhhhhhhh+(g2-α*w)(t-2h)]=limh→01h2[A(g2*u)(t)-2A(g2*u)(t-h)hhhhhhhh+A(g2*u)(t-2h)]+limh→01h2[(g2*f)(t)-2(g2*f)(t-h)hhhhhhhl+(g2*f)(t-2h)]=limh→01h2[A(g2*u)(t)-2A(g2*u)(t-h)hhhhhhhh+A(g2*u)(t-2h)]+f(t).Thus,
(30)limh→01h2[A(g2*u)(t)-2A(g2*u)(t-h)hhhhhh+A(g2*u)(t-2h)]=cDtαu(t)-f(t).
On the other hand, from the closeness of A and gα*u∈L1((0,t),D(A)) for t∈[0,T), by Proposition 1.1.7 in [2], we have
(31)(g2*u)(t)=(g2-α*(gα*u))(t)∈D(A),
Then from (30) and the closeness of A, we obtain
(32)u(t)=limh→01h2[(g2*u)(t)-2(g2*u)(t-h)hhhhhhhh+(g2*u)(t-2h)]∈D(A),Dctαu(t)=Au(t)+f(t),t∈[0,T).
It is clear that u(0)=x0,u′(0)=x1. Thus, u is a classical solution of (26).Lemma 11.
Let1<α<2, f∈L1((0,T),X). Suppose A is the generator of an (α,β)-ROF Sα,β(t) on X for some β∈ℝ+. Then, for every t∈[0,T), (gα-1*Sα,β*f)(t) exists, and (gα-1*Sα,β*f)∈C([0,T),X).Proof.
For everyt∈[0,T), since gα-1∈L1((0,t),ℝ+),f∈L1((0,t),X), we get gα-1*f∈L1((0,t),X), hence, from
(33)(gα-1*Sα,β*f)(t)=(Sα,β*gα-1*f)(t)=∫0tSα,β(t-s)(gα-1*f)(s)ds‍,
we obtain that (gα-1*Sα,β*f)(t) exists.
Forh∈ℝ,|h|≪1 and t+h∈[0,T), we have
(34)(gα-1*Sα,β*f)(t+h)-(gα-1*Sα,β*f)(t)=∫0t+hSα,β(t+h-s)(gα-1*f)(s)ds‍-∫0tSα,β(t-s)(gα-1*f)(s)ds‍=∫0t+h(Sα,β(t+h-s)-Sα,β(t-s))(gα-1*f)(s)ds‍-∫tt+hSα,β(t-s)(gα-1*f)(s)ds.‍
From the dominated convergence theorem and absolute continuity of integral, we deduce
(35)limh→0((gα-1*Sα,β*f)(t+h)-(gα-1*Sα,β*f)(t))=0.
So, (gα-1*Sα,β*f)∈C([0,T),X).Let(36)v(t)=Sα,β(t)x0+(g1*Sα,β)(t)x1+(gα-1*Sα,β*f)(t).
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 thatA is the generator of an (α,β)-ROF Sα,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 gm-β*(v-∑k=0m-1‍v(k)(0)gk+1(t))∈Cm([0,T);X) and u(t)=cDtβv(t).(b)
If there is a classical solutionu of (26), then g2-α*(cDtβv(t)-x0-tx1)∈C2([0,T);X) and u(t)=cDtβv(t).Proof.
Ifu is a mild solution of (26), then (gα*u)(t)∈D(A) and
(37)u(t)=x0+tx1+A(gα*u)(t)+(gα*f)(t),t∈[0,T).
Using Lemma 2(b) and the closeness of A, we have
(38)(gβ+1*u)(t)=(Sα,β-A(gα*Sα,β))*u(t)=(Sα,β*u)(t)-(A(gα*Sα,β)*u)(t)=(Sα,β*u)(t)-Sα,β*A(gα*u)(t)=(Sα,β*u)(t)-Sα,β*(u-x0-tx1-(gα*f)(t))=Sα,β*x0+Sα,β*tx1+(Sα,β*gα*f)(t);
that is, (gβ+1*u)(t)=(1*Sα,β)(t)x0+(g2*Sα,β)(t)x1+(gα*S*f)(t). So
(39)Jtβu(t)=(gβ*u)(t)=ddt(gβ+1*u)(t)=Sα,β(t)x0+(g1*Sα,β)(t)x1+(gα-1*Sα,β*f)(t)=v(t).
Thus, it follows from u∈C([0,T),X) that gm-β*(v-∑k=0m-1‍v(k)(0)gk+1(t))∈Cm([0,T);X) and u(t)=Dctβv(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.
Letv be defined by (36). Assume that v∈Cm-1([0,T);X), v(k)(0)=0 for k=0,1,…,m-1, and gm-β*v∈Cm([0,T);X); then Dctβv(t) is a mild solution of the problem (26). Moreover, if g2-α*(Dctβv(t)-x0-tx1)∈C2([0,T);X), and for any t∈(0,T), gα*Dctβv(t)∈L1((0,t),D(A)), then Dctβv(t) is also a classical solution of (26).Proof.
Consider the following steps.
S
tep
1. We first claim that Jtαv(t)∈D(A) and
(40)DctβAJtαv(t)=Dctβv(t)-x0-tx1-gα*f.
In view of definition ofv(t), we have
(41)Jtαv(t)=(gα*Sα,β)(t)x0+(gα*g1*Sα,β)(t)x1+(gα*gα-1*Sα,β*f)(t)=(gα*Sα,β)(t)x0+∫0t(gα*Sα,β)(τ)x1dτ‍+∫0t(gα*Sα,β)(t-τ)(gα-1*f)(τ)dτ‍,
for t∈[0,T).
From Lemma2(b), for 0<τ<t, we have
(42)(gα*Sα,β)(t)x0∈D(A),(gα*Sα,β)(τ)x1∈D(A),hhhh(gα*Sα,β)(t-τ)(gα-1*f)(τ)∈D(A),A(gα*Sα,β)(τ)x1=Sα,β(τ)x1-gβ+1(τ)x1∈L1(0,t),A(gα*Sα,β)(t-τ)(gα-1*f)(τ)=Sα,β(t-τ)(gα-1*f)(τ)-gβ+1(t-τ)(gα-1*f)(τ)∈L1(0,t),
combining with the closeness of A, one has
(43)∫0t(gα*Sα,β)(τ)x1dτ∈D(A)‍,∫0t(gα*Sα,β)(t-τ)(gα-1*f)(τ)dτ∈D(A)‍.
Thus Jtαv(t)∈D(A), and
(44)AJtαv(t)=A(gα*Sα,β)(t)x0+g1*A(gα*Sα,β)(t)x1+gα-1*A(gα*Sα,β*f)(t)=Sα,β(t)x0-gβ+1(t)x0+(g1*Sα,β)(t)x1-(g1*gβ+1)(t)x1+gα-1*(Sα,β*f-gβ+1*f)(t)=Sα,β(t)x0+(g1*Sα,β)(t)x1+(gα-1*Sα,β*f)(t)-gβ+1(t)x0-(g1*gβ+1)(t)x1-(gα+β*f)(t)=v(t)-gβ+1(t)x0-(g1*gβ+1)(t)x1-(gα+β*f)(t).
So
(45)sssAJtαv(t)=v(t)-gβ+1(t)x0-(g1*gβ+1)(t)x1ssssAJtαv(t)=-(gα+β*f)(t),DctβAJtαv(t)=Dctβv(t)-x0-tx1-gα*f.
Step
2. We prove DctβJtαv(t)∈D(A) and ADctβJtαv(t)=DctβAJtαv(t).
Sincev∈Ck([0,T);X),v(k)(0)=0 for  k=0,1,…,m-1, we have
(46)dkdtk(gα*v)(t)|t=0=(gα*v(k))(t)|t=0=0.
So
(47)DctβJtαv(t)=dmdtm(gm-β*gα*v)(t)=limh→01hm∑r=0mCmr(gm-β*gα*v)(t-rh)‍,
where Cmr=(m(m-1)⋯(m-r+1))/r!. From (45), we know that AJtαv∈L1((0,t),X), and the closeness of A implies that (gm-β*gα*v)(t)∈D(A), and A(gm-β*gα*v)(t)=gm-β*A(gα*v)(t), by Step 1, DctβAJtαv(t) exists, then DctβJtαv(t)∈D(A), and
(48)ADctβJtαv(t)=DctβAJtαv(t).
Step
3. We show that v(k)(0)=0 for k=0,1,…,m-1 implies
(49)DctβJtαv(t)=JtαDctβv(t).
In fact, ifα≥β, we have DctβJtαv(t)=Jtα-βv(t), and
(50)JtαDctβv(t)=Jtα-βJtβDctβv(t)=Jtα-β(v(t)-∑k=0m-1v(k)(0)gk+1(t)).
If α<β, we have DctβJtαv(t)=Dctβ-αv(t), and
(51)JtαDctβv(t)=Dctβ-αJtβDctβv(t)=Dctβ-α(v(t)-∑k=0m-1v(k)(0)gk+1(t)‍).
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
(52)AJtαDctβv(t)=ADctβJtαv(t)=DctβAJtαv(t)=Dctβv(t)-x0-tx1-(gα*f)(t).
Therefore, Dctβv(t) is a mild solution of (26).
Moreover, ifg2-α*(Dctβv(t)-x0-tx1)∈C2([0,T);X), and for any t∈(0,T), gα*Dctβv(t)∈L1((0,t),D(A)), applying Theorem 10, we have that Dctβv(t) is a classical solution of (26).Remark 14.
Ifα→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(53)Dctαv(t)=Av(t)+gβ+1t(x),t>0,v(k)(0)=0,k=0,1,…,N-1,
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.
LetA be a closed operator on X and β>0; the following two assertions are equivalent.(i)
A generates an exponentially bounded (α,β)-ROF Sα,β on X.(ii)
For everyx∈X, there exists a unique classical solution vx of (53) which is exponentially bounded and Avx∈Lloc1(ℝ+;X).Proof.
If(i) is satisfied, for every x∈X, define vx:ℝ+→X by vx(t)=(gα*Sα,β)(t)x, then vx(k)(0)=0 for k=0,1,…,N-1. By Lemma 2(b), we have vx(t)=(gα*Sα,β)(t)x∈D(A), and
(54)Dctαvx(t)=Sα,β(t)x=A(gα*Sα,β)(t)x+gβ+1(t)x=Avx(t)+gβ+1(t)x,t>0.
Thus, vx is a classical solution of (53); it is unique by Theorem 12. Since Sα,β is exponentially bounded, we have that vx is exponentially bounded. From
(55)Avx(t)=A(gα*Sα,β)(t)x=Sα,β(t)x-gβ+1(t)x,
we know that Avx(t)∈Lloc1(ℝ+;X). So (ii) is true.
Assume that (ii) holds. From linearity of (53) and the uniqueness of its solution, we get that vx is linear in x. So, for each t≥0, there exists a linear mapping V(t):X→D(A) such that V(t)x=vx(t) for any x∈X.
Next, we show that, for eacht≥0, V(t)∈B(X,D(A)).
We consider the mappingΦ:X→C(ℝ+,D(A)) by Φ(x)=vx(·)=V(·)x. Then, Φ is a linear operator defined on X. Now we show that Φ is closed, if xn→x in X and Φ(xn)→u in C(ℝ+,D(A)). For t>0, by the dominated convergence theorem, we have that Jtαvxn(t) converges to Jtαu(t), since vxn(·)=gα+β+1(t)xn+JtαAvxn(t), from the closeness of A, it follows that as n→∞, u(t)=gα+β+1(t)x+JtαAu(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)∥≤Meωt(t≥0) for some constants M,ω≥0. So Q(λ)x=λβ+1∫0∞‍e-λtV(t)xdt is well defined for λ>ω,(ωα,∞)⊂ρ(A).
SinceAV(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
(56)(λα-A)Q(λ)x=λα+β+1∫0∞e-λtV(t)xdt-λα+β+1‍×∫0∞e-λtAV(t)xdt‍=λα+β+1∫0∞e-λtV(t)xdt-λβ+1‍×∫0∞e-λtcDtαV(t)xdt+λβ+1‍×∫0∞e-λtgβ+1(t)xdt‍=λα+β+1V^(λ)x-λβ+1λαV^(λ)x+λβ+1λ-(β+1)x=x.
Now, we show that(λα-A) is injective for λ>ω. Assume that (λα-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 tα+βEα,α+β+1(λαtα). Since ∥vx(t)∥≤Meωt, for all t≥0, combine with (15), it follows that x=0. Hence (λα-A)-1=Q(λ) for λ>ω and V(t) is an (α,α+β)-ROF. Let
(57)Sα,β(t)x≔DctαV(t)x=AV(t)x+gβ+1(t)x;
then Sα,β(t)x exists and V(t)x=JtαSα,β(t)x for all t≥0 and all x∈X. So
(58)Sα,β(t)x=AJtαSα,β(t)x+gβ+1(t)x,
and taking the Laplace transform, we have
(59)S^α,β(λ)x=Aλ-αS^α,β(λ)x+λ-β-1x,λ>ω;
that is,
(60)S^α,β(λ)x=λα-β-1(λα-A)-1x,λ>ω.
From Lemma 3, we know that Sα,β is the (α,β)-ROF generated by A.Remark 16.
Theorem15 extends and generalizes Theorem 3.2.13 in [2]. In fact, when α=1 and β=k, (53) becomes (5), Sα,β(t) is a k-times integrated semigroup. For problem (5), the condition Avx∈Lloc1(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 Avx∈Lloc1(ℝ+;X).



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