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 [superscript]c[/superscript] [superscript]Dtα[/superscript] u(t)=Au(t)+f(t) , t∈(0,T) , u(0)=[subscript]x0[/subscript] , u'(0)=[subscript]x1[/subscript] , if A is the generator of an (α,β) -ROF, we give the relation between the function v(t)=[subscript]Sα,β[/subscript] (t)[subscript]x0[/subscript] +([subscript]g1[/subscript] *[subscript]Sα,β[/subscript] )(t)[subscript]x1[/subscript] +([subscript]gα-1[/subscript] *[subscript]Sα,β[/subscript] *f)(t) and mild solution and classical solution of it. Finally, for the problem [superscript]c[/superscript] [superscript]Dtα[/superscript] v(t)=Av(t)+[subscript]gβ+1[/subscript] (t)x , t>0 , [superscript]v(k)[/superscript] (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 [subscript]vx[/subscript] and A[subscript]vx[/subscript] ∈[superscript]L loc 1[/superscript] ([superscript]...+[/superscript] ,X). Our results extend and generalize some related results in the literature.

(ProQuest: ... denotes non-US-ASCII text omitted.)

Academic Editor:Yong Zhou

School of Mathematics and Statistics, Lanzhou University, Lanzhou, Gansu 730000, China

Received 31 October 2013; Accepted 11 December 2013; 8 January 2014

This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.

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, ^...+ =[0,∞) , ... denotes the set of natural numbers. _...0 =...∪{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 (^...+ ,X) denotes the space of X -valued Bochner integrable functions: u:^...+ [arrow right]X with the norm _{||u||^L1 (^...+ ,X)} =^∫0∞ ...||u(t)||dt , it is a Banach space. By * we denote the convolution of functions [figure omitted; refer to PDF] _gα denotes the function [figure omitted; refer to PDF] and _g0 (t)=_δ0 (t) , the Dirac delta function.

In 1997, Mijatovic et al. [1] introduced the concept of β -times integrated semigroup (β∈^...+ ) which extends k -times integrated semigroup (k∈_...0 ) [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: [figure omitted; refer to PDF] In the special case of β=k∈... , the corresponding result is summarized in [2].

For the inhomogeneous Cauchy problem [figure omitted; refer to PDF] 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∈_...0 . Let v(t)=S(t)x+^∫0t ...S(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, if A generates an exponential bounded k -times integrated semigroup on a Banach space X , then, for any x∈X , v(t)=^∫0t ...S(s)x ds is the unique exponential bounded classical solution of the following problem: [figure omitted; refer to PDF]

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...5;0 ⊂B(X) is an α -ROF if and only if {_Sα (t)_}t...5;0 is a solution of abstract fractional Cauchy problem [figure omitted; refer to PDF] When 0<α<1 , Peng and Li [18] proved that the solution operator {_Sα (t)_}t...5;0 for (6) satisfies the following equality: [figure omitted; refer to PDF] 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...5;0 . A function _Sα,β :^...+ [arrow right]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 ^...+ 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...5;0 ;

(c) the functional equation [figure omitted; refer to PDF]

: holds for s,t...5;0 , where ^Jtα is the Riemann-Liouville fractional integral of order α .

The generator A of _Sα,β (t) is defined by [figure omitted; refer to PDF] 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 [figure omitted; refer to PDF] 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 [figure omitted; refer to PDF] 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 A_vx ∈^LLoc1 (^...+ ,X) . If α[arrow right]¹⁺ , β=k∈... , 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 [figure omitted; refer to PDF] and the Caputo fractional derivative of order α>0 of f can be written as [figure omitted; refer to PDF] 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 [figure omitted; refer to PDF] And if 0<α<2 , β>0 , then [figure omitted; refer to PDF] where [figure omitted; refer to PDF] and the O -term is uniform in argz if |arg(-z)|...4;(1-(α/2)-...)π .

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

Lemma 2 (see [9, Proposition 3.10]).

Let _Sα,β :^...+ [arrow right]B(X) be an (α,β) -ROF generated by A . The following assertions hold:

(a) _Sα,β (t)D(A)⊂D(A) and A_Sα,β (t)x=_Sα,β (t)Ax for x∈D(A) and t...5;0 ;

(b) for all x∈X , ^Jtα_Sα,β (t)x∈D(A) and _Sα,β (t)x=_gβ+1 (t)x+A^Jtα_Sα,β (t)x , t...5;0 ;

(c) x∈D(A) and Ax=y if and only if _Sα,β (t)x=_gβ+1 (t)x+^Jtα_Sα,β (t)y , t...5;0 ;

(d) A is closed.

Lemma 3 (see [9, Proposition 3.5, Theorem 3.11]).

Let α>0 , β...5;0 . A generates an (α,β) -ROF _Sα,β satisfying ||_Sα,β (t)||...4;M^eωt , t...5;0 , for some constants M>0 and ω...5;0 , if and only if (^ωα ,∞)⊂ρ(A) and there exists a strongly continuous function S:^...+ [arrow right]B(X) such that ||S(t)||...4;M^eωt for all t...5;0 and ^∫0∞ ...^e-λt S(t)x dt=^λα-β-1 R(^λα ,A)x , λ>ω , for all x∈X . Furthermore, S(t) is _Sα,β (t) .

Lemma 4 (see [2, Proposition B.6]).

Let U⊂... . If function R:U[arrow right]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...5;0} in this theorem.

Theorem 5.

Let α∈^...+ \_...0 , β∈^...+ satisfy β-α>-1 . If {S(t)_}t...5;0 is an (α, β) -ROF, then it satisfies the following equality: [figure omitted; refer to PDF]

Proof.

Denote by L(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...4;t...4;a and _fa (t)=0 otherwise.

We will show that the Laplace transform of _La (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 of _La (t,s) with respect to s as follows [figure omitted; refer to PDF] then taking Laplace transform with respect to t , we have [figure omitted; refer to PDF] where the last equality follows from Lemma 3.

On the other hand, observing that [figure omitted; refer to PDF] Then taking Laplace transform with respect to t and s , respectively, we deduce [figure omitted; refer to PDF] 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...5;0 . The arbitrariness of a implies L(t,s)=R(t,s) for t,s...5;0 .

Remark 6.

(a) If β=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 each x∈X , the map t[arrow right]_Sα,β (t)x is continuously differentiable on [0, ∞) and the limit of (α,β) -ROF _Sα,β (t) exists as α[arrow right]^1- , then multiplying both sides of (17) with 1-α and integrating by parts to the right side of (17) and letting α[arrow right]^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 ^λα-β-1 R(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[arrow right]^1- , β=k∈... . Our proof is different since we could not use the binomial formula as in [2].

Theorem 7.

Let S~:^...+ [arrow right]B(X) be a strongly continuous function satisfying ||S~(t)||...4;M^eωt (t...5;0) for some M, ω...5;0 . Let α∈^...+ \_...0 , β∈^...+ satisfy that β-α>-1 , set [figure omitted; refer to PDF] Then the following assertions are equivalent.

(i) There exists an operator A such that (^ωα ,∞)⊂ρ(A) and R(^λα )=(^λα I-A^)-1 for λ>ω .

(ii) For s,t...5;0 , the equality [figure omitted; refer to PDF]

: holds and S~(t)x=0 for all t...5;0 implies that x=0 .

Proof.

Assume that (i) holds; then (^ωα ,∞)⊂ρ(A) , (^λα I-A^)-1 =^{λ-α+β+1∫0∞} ...^e-λt S~(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 )=(^λα I-A^)-1 for λ>ω that R(^λα ) is injective. If S~(t)x=0 for all t...5;0 , from R(^λα ):=^{λ-α+β+1∫0∞} ...^e-λt S~(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 [figure omitted; refer to PDF] respectively. So, [figure omitted; refer to PDF]

On the other hand, if R(^λα )x=0 , by R(^λα )=^{λ-α+β+1∫0∞} ...^e-λt S~(t) dt and uniqueness of Laplace transform, we have S~(t)x=0 for all t...5;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: [figure omitted; refer to PDF] 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 (_gα *u)(t)∈D(A) and u(t)=_x0 +t_x1 +A(_gα *u)(t)+(_gα *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) _g2-α *(u-_x0 -t_x1 )∈^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 _g2-α *(u-_x0 -t_x1 )∈^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.

Since u is a mild solution of (26), we have [figure omitted; refer to PDF] If we denote w(t):=u(t)-_x0 -t_x1 , then it follows from (27) that [figure omitted; refer to PDF]

Since _g2-α *w∈^C2 ([0,T);X) , then ^cDtα u(t)=(^d2 /d^t2 )(_g2-α *w)(t) is well defined, and by (28), we have [figure omitted; refer to PDF] Thus, [figure omitted; refer to PDF] 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 [figure omitted; refer to PDF] Then from (30) and the closeness of A , we obtain [figure omitted; refer to PDF] It is clear that u(0)=_x0 , ^{u[variant prime]} (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 _Sα,β (t) on X for some β∈^...+ . 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 every t∈[0,T) , since _gα-1 ∈^L1 ((0,t),^...+ ), f∈^L1 ((0,t),X) , we get _gα-1 *f∈^L1 ((0,t),X) , hence, from [figure omitted; refer to PDF] we obtain that (_gα-1 *_Sα,β *f)(t) exists.

For h∈..., |h|...a;1 and t+h∈[0,T) , we have [figure omitted; refer to PDF] From the dominated convergence theorem and absolute continuity of integral, we deduce [figure omitted; refer to PDF] So, (_gα-1 *_Sα,β *f)∈C([0,T),X) .

Let [figure omitted; refer to PDF] 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∈... . 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 _Sα,β (t) on X for some β∈^...+ . 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-1 ...^v(k) (0)_gk+1 (t))∈^Cm ([0,T);X) and u(t)= ^cDtβ v(t) .

(b) If there is a classical solution u of (26), then _g2-α *^(cDtβ v(t)-_x0 -t_x1 )∈^C2 ([0,T);X) and u(t)= ^cDtβ v(t) .

Proof.

If u is a mild solution of (26), then (_gα *u)(t)∈D(A) and [figure omitted; refer to PDF] Using Lemma 2(b) and the closeness of A , we have [figure omitted; refer to PDF] that is, (_gβ+1 *u)(t)=(1*_Sα,β )(t)_x0 +(_g2 *_Sα,β )(t)_x1 +(_gα *S*f)(t) . So [figure omitted; refer to PDF] Thus, it follows from u∈C([0,T),X) that _gm-β *(v-^∑k=0m-1 ...^v(k) (0)_gk+1 (t))∈^Cm ([0,T);X) and u(t)=...^Dctβ v(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∈^Cm-1 ([0,T);X) , ^v(k) (0)=0 for k=0,1,...,m-1 , and _gm-β *v∈^Cm ([0,T);X) ; then ...^Dctβ v(t) is a mild solution of the problem (26). Moreover, if _g2-α *(...^Dctβ v(t)-_x0 -t_x1 )∈^C2 ([0,T);X) , and for any t∈(0,T) , _gα *...^Dctβ v(t)∈^L1 ((0,t),D(A)) , then ...^Dctβ v(t) is also a classical solution of (26).

Proof.

Consider the following steps.

S tep 1. We first claim that ^Jtα v(t)∈D(A) and [figure omitted; refer to PDF]

In view of definition of v(t) , we have [figure omitted; refer to PDF] for t∈[0,T) .

From Lemma 2(b), for 0<τ<t , we have [figure omitted; refer to PDF] combining with the closeness of A , one has [figure omitted; refer to PDF] Thus ^Jtα v(t)∈D(A) , and [figure omitted; refer to PDF] So [figure omitted; refer to PDF]

Step 2. We prove ...^DctβJtα v(t)∈D(A) and A^DctβJtα v(t)=^Dctβ A^Jtα v(t) .

Since v∈^Ck ([0,T);X), ^v(k) (0)=0 for k=0,1,...,m-1 , we have [figure omitted; refer to PDF] So [figure omitted; refer to PDF] where ^Cmr =(m(m-1)...(m-r+1))/r! . From (45), we know that A^Jtα v∈^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, ^Dctβ A^Jtα v(t) exists, then ^DctβJtα v(t)∈D(A) , and [figure omitted; refer to PDF]

Step 3 . We show that ^v(k) (0)=0 for k=0,1,...,m-1 implies [figure omitted; refer to PDF]

In fact, if α...5;β , we have ...^DctβJtα v(t)=^Jtα-β v(t) , and [figure omitted; refer to PDF] If α<β , we have ...^DctβJtα v(t)=...^Dctβ-α v(t) , and [figure omitted; refer to PDF] 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 [figure omitted; refer to PDF] Therefore, ^Dctβ v(t) is a mild solution of (26).

Moreover, if _g2-α *(...^Dctβ v(t)-_x0 -t_x1 )∈^C2 ([0,T);X) , and for any t∈(0,T) , _gα *^Dctβ v(t)∈^L1 ((0,t),D(A)) , applying Theorem 10, we have that ...^Dctβ v(t) is a classical solution of (26).

Remark 14.

If α[arrow right]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 [figure omitted; refer to PDF] 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 _Sα,β on X .

(ii) For every x∈X , there exists a unique classical solution _vx of (53) which is exponentially bounded and A_vx ∈^Lloc1 (^...+ ;X) .

Proof.

If (i) is satisfied, for every x∈X , define _vx :^...+ [arrow right]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 [figure omitted; refer to PDF] 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 [figure omitted; refer to PDF] we know that A_vx (t)∈^Lloc1 (^...+ ;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...5;0 , there exists a linear mapping V(t):X[arrow right]D(A) such that V(t)x=_vx (t) for any x∈X .

Next, we show that, for each t...5;0 , V(t)∈B(X,D(A)) .

We consider the mapping Φ:X[arrow right]C(^...+ ,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 [arrow right]x in X and Φ(_xn )[arrow right]u in C(^...+ ,D(A)) . For t>0 , by the dominated convergence theorem, we have that ^Jtα_vxn (t) converges to ^Jtα u(t) , since _vxn (·)=_gα+β+1 (t)_xn +^Jtα A_vxn (t) , from the closeness of A , it follows that as n[arrow right]∞ , u(t)=_gα+β+1 (t)x+^Jtα Au(t) , which implies that u=Φ(x) and Φ is closed. Therefore, by the closed graph theorem, Φ is bounded. So, for each t...5;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)||...4;M^eωt (t...5;0) for some constants M, ω...5;0 . So Q(λ)x=^{λβ+1∫0∞} ...^e-λt V(t)x dt is well defined for λ>ω, (^ωα ,∞)⊂ρ(A) .

Since AV(t)x∈^L1 (^...+ ,X) , then the Laplace transform of AV(t)x is well defined, and from the closeness of A , for λ>ω , we have [figure omitted; refer to PDF]

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+β_Eα,α+β+1 (^λαtα ) . Since ||_vx (t)||...4;M^eωt , for all t...5;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 [figure omitted; refer to PDF] then _Sα,β (t)x exists and V(t)x=^Jtα_Sα,β (t)x for all t...5;0 and all x∈X . So [figure omitted; refer to PDF] and taking the Laplace transform, we have [figure omitted; refer to PDF] that is, [figure omitted; refer to PDF] From Lemma 3, we know that _Sα,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), _Sα,β (t) is a k -times integrated semigroup. For problem (5), the condition A_vx ∈^Lloc1 (^...+ ;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 A_vx ∈^Lloc1 (^...+ ;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).

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