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Article

Fractional Calculus and Time-Fractional Differential Equations: Revisit and Construction of a Theory †

by
Masahiro Yamamoto
1,2,3,4
1
Graduate School of Mathematical Sciences, The University of Tokyo, Komaba, Meguro, Tokyo 153-8914, Japan
2
Academy of Romanian Scientists, Ilfov, nr. 3, 062217 Bucuresti, Romania
3
Accademia Peloritana dei Pericolanti, Palazzo Università, Piazza S. Pugliatti 1, 98122 Messina, Italy
4
Peoples’ Friendship University of Russia (RUDN University), 6 Miklukho-Maklaya St., 117198 Moscow, Russia
Dedicated to the memory of Professor Dr. Rudolf Gorenflo who invited the author to studies of fractional calculus.
Mathematics 2022, 10(5), 698; https://doi.org/10.3390/math10050698
Submission received: 21 January 2022 / Revised: 9 February 2022 / Accepted: 13 February 2022 / Published: 23 February 2022

Abstract

:
For fractional derivatives and time-fractional differential equations, we construct a framework on the basis of operator theory in fractional Sobolev spaces. Our framework provides a feasible extension of the classical Caputo and the Riemann–Liouville derivatives within Sobolev spaces of fractional orders, including negative ones. Our approach enables a unified treatment for fractional calculus and time-fractional differential equations. We formulate initial value problems for fractional ordinary differential equations and initial boundary value problems for fractional partial differential equations to prove well-posedness and other properties.

1. Motivations

Let Ω R d be a bounded domain with smooth boundary Ω , and let ν = ν ( x ) be the unit outward normal vector to Ω at x Ω . We set
A v ( x ) = i , j = 1 d i ( a i j ( x ) j v ( x ) ) + j = 1 d b j ( x ) j v ( x ) + c ( x ) v ( x ) , x Ω , 0 < t < T ,
where a i j = a j i , b j , c C 1 ( Ω ¯ ) , 1 i , j d and we assume that there exists a constant κ > 0 such that
i , j = 1 d a i j ( x ) ξ i ξ j κ j = 1 d | ξ j | 2 , x Ω , ξ 1 , . . . , ξ d R .
Henceforth, Γ ( γ ) denotes the gamma function for γ > 0 : Γ ( γ ) : = 0 e t t γ 1 d t .
Our eventual purpose is to construct a theoretical framework for treating initial boundary value problems for time-fractional diffusion equations with source term F ( x , t ) , which can be described for the case of 0 < α < 1 :
d t α u ( x , t ) = A u ( x , t ) + F ( x , t ) , x Ω , 0 < t < T , u ( x , 0 ) = 0 , x Ω , u ( x , t ) = 0 , x Ω , 0 < t < T .
Here, for 0 < α < 1 , we can formally define the pointwise Caputo derivative
d t α v ( t ) = 1 Γ ( 1 α ) 0 t ( t s ) α d v d s ( s ) d s ,
as long as the right-hand side exists.
For classical treatments on fractional calculus and equations, we can refer to monographs from Gorenflo, Kilbas, Mainardi and Rogosin [1], Kilbas, Srivastava and Trujillo [2], Podlubny [3] and Samko, Kilbas and Marichev [4] for examples.
Henceforth, let L p ( 0 , T ) : = { v ; 0 T | v ( t ) | p d t < } with p 1 and W 1 , 1 ( 0 , T ) : = { v L 1 ( 0 , T ) ; d v d t L 1 ( 0 , T ) } . We define the norm by
v L p ( 0 , T ) : = 0 T | v ( t ) | p d t 1 p , v W 1 , 1 ( 0 , T ) : = v L 1 ( 0 , T ) + d v d t L 1 ( 0 , T ) .
In view of the Young inequality on the convolution, we can directly verify that the classical Caputo derivative (3) can be well-defined for v W 1 , 1 ( 0 , T ) and d t α v L 1 ( 0 , T ) . However, everything is not clear for v W 1 , 1 ( 0 , T ) , and it is not a feasible assumption that any solutions to (2) have the W 1 , 1 ( 0 , T ) -regularity in t. The Young inequality is well known and we can refer to Lemma A.1 in Appendix in Kubica, Ryszewska and Yamamoto [5] for an example.
In (3), the pointwise Caputo derivative d t α requires the first-order derivative d v d s in any sense. Therefore, in order to discuss d t α v for a function v which keeps apparently reasonable regularity such as “ α -times” differentiability, we should formulate d t α v in a suitable distribution space. Moreover, such a formulation is not automatically unique. There have been several works: for example, Kubica, Ryszewska and Yamamoto [5], Kubica and Yamamoto [6] and Zacher [7]. Here, we restrict ourselves to an extremely limited number of references. In this article, we extend the approach in Kubica, Ryszewska and Yamamoto [5] to discuss fractional derivatives in fractional Sobolev spaces of arbitrary real number orders and construct a convenient theory for initial value problems and initial boundary value problems for time-fractional differential equations. In [5], the orders α are restricted to 0 < α < 1 , but here we study fractional orders α R .
The main purpose of this article is to define such fractional derivatives and establish the framework which enables us to uniformly consider weaker and stronger solutions with exact specification of classes of solutions in terms of fractional Sobolev spaces. Thus, we intend to construct a comprehensive theory for time-fractional differential equations within Sobolev spaces.
In (3), we take the first-order derivative and then ( 1 α ) -times integral operator 1 Γ ( 1 α ) 0 t ( t s ) α d s to finally reach the α -times derivative. As is described in Section 2, our strategy for the definition of the fractional derivative in t is to consider the α -times derivative of v as the inverse to the α -times integral, not via d v d t .
Moreover, for initial value problems for time-fractional differential equations, we meet other complexities for how to pose initial conditions, as the following examples show.
Example 1. 
We consider an initial value problem for a simple time-fractional ordinary differential equation:
d t α v ( t ) = f ( t ) , 0 < t < T , v ( 0 ) = a .
We remark that (4) is not necessarily well-posed for all α ( 0 , 1 ) , f L 2 ( 0 , T ) and a R because of the initial condition. The order α 0 , 1 2 especially causes difficulty: Choosing γ > 1 and
f ( t ) = t γ L 1 ( 0 , T ) ,
in (4), we consider
d t α v ( t ) = t γ , v ( 0 ) = a .
A solution formula is known:
v ( t ) = a + 1 Γ ( α ) 0 t ( t s ) α 1 s γ d s
(e.g., (3.1.34), Kilbas, Srivastava and Trujillo [2], p. 141). For 0 < α < 1 and γ > 1 , the right-hand side of (6) makes sense and we can obtain
v ( t ) = a + Γ ( γ + 1 ) Γ ( α + γ + 1 ) t α + γ , t > 0 .
However, the initial condition v ( 0 ) = a is delicate:
(i) 
Let α + γ > 0 . Then we can readily see that v given by (7) satisfies (5).
(ii) 
Let α + γ = 0 . Then (7) provides that v ( t ) = a + Γ ( 1 α ) . However, this v does not satisfy d t α v = t α .
(iii) 
Let α + γ < 0 . Then, for v ( t ) defined by (7), we see that lim t 0 v ( t ) = . Therefore, (7) cannot give a solution to (5), and, moreover, we are not sure whether there exists a solution to (5).
We have another classical fractional derivative called the Riemann–Liouville derivative:
D t α v ( t ) = 1 Γ ( 1 α ) d d t 0 t ( t s ) α v ( s ) d s , 0 < t < T
for 0 < α < 1 , provided that the right-hand side exists.
Let f L 2 ( 0 , T ) be arbitrarily given. In terms of D t α , we can formulate an initial value problem:
D t α u ( t ) = f ( t ) , 0 < t < T , ( J 1 α u ) ( 0 ) = a .
The solution formula is
u ( t ) = a Γ ( α ) t α 1 + J α f ( t ) , 0 < t < T
(e.g., Kilbas, Srivastava and Trujillo [2], p. 138). Here, we set
J α f ( t ) : = 1 Γ ( α ) 0 t ( t s ) α 1 f ( s ) d s , 0 < t < T , f L 1 ( 0 , T ) .
Indeed, we can directly verify that u given by (10) satisfies (9):
D t α J α f ( t ) = 1 Γ ( α ) Γ ( 1 α ) d d t 0 t ( t s ) α 0 s ( s ξ ) α 1 f ( ξ ) d ξ d s = 1 Γ ( α ) Γ ( 1 α ) d d t 0 t f ( ξ ) ξ t ( t s ) α ( s ξ ) α 1 d s d ξ = d d t 0 t f ( ξ ) d ξ = f ( t ) f o r f L 1 ( 0 , T )
and
D t α a Γ ( α ) t α 1 = 1 Γ ( 1 α ) d d t 0 t ( t s ) α s α 1 d s = 0 , 0 < t < T ,
which means that D t α u ( t ) = f ( t ) for 0 < t < T . We can similarly verify
J 1 α a Γ ( α ) t α 1 + J α f ( t ) ( 0 ) = a .
By 0 < α < 1 2 , the formula (10) makes sense in L 1 ( 0 , T ) , but u L 2 ( 0 , T ) for each f L 2 ( 0 , T ) if a 0 . This suggests that the initial value problem (9) may not be well-posed in L 2 ( 0 , T ) .
Furthermore, we can consider other formulation:
D t α u ( t ) = f ( t ) , 0 < t < T , u ( 0 ) = a .
We see that u ( t ) given by (10) satisfies D t α u = f in ( 0 , T ) , but it follows from α 1 < 0 that lim t 0 | u ( t ) | = by lim t 0 t α 1 = if a 0 , while if a = 0 , then (10) can satisfy u ( 0 ) = 0 , that is, (11) by f in some class such as f L ( 0 , T ) .
The delicacies in the above examples are more or less known and motivate us to construct a uniform framework for time-fractional derivatives. Moreover, we are concerned with the range space of the corresponding solutions for a prescribed function space of f, e.g., L 2 ( 0 , T ) , not only calculations of the derivatives of an individually given function u. Naturally, for the well-posedness of an initial value problem, we are required to characterize the function space of solutions corresponding to a space of f.
In other words, one of our main interests is to define a fractional derivative, denoted by t α , and characterize the space of u satisfying t α u L 2 ( 0 , T ) . Moreover, such t α should be an extension of d t α and D t α in a minimum sense in order that important properties of these classical fractional derivatives should be inherited to t α .
Throughout this article, we treat fractional integrations and fractional derivatives as operators from specified function spaces to others, that is, we always attach them with their domains and ranges.
Thus we will define a time-fractional derivative, denoted by t α , as a suitable extension of d t α satisfying the requirements:
  • Such an extended derivative t α admits usual rules of differentiation as much as possible. For example, t α t β = t α + β for all α , β 0 .
  • It admits a relevant formulation of initial conditions, even for α R .
  • In L 2 -based Sobolev spaces, there exists a unique solution to an initial value problem for a time-fractional ordinary differential equation and an initial boundary value problem for a time-fractional partial differential equation for α > 0 and even α R .
In this article, we intend to outline foundations for a comprehensive theory for time-fractional differential equations. Some arguments are based on Gorenflo, Luchko and Yamamoto [8] and Kubica, Ryszewska and Yamamoto [5].
This article is composed of eight sections and Appendix A:
  • Section 2: Definition of the extended derivative t α :
    We extend d t α as operator so that it is well-defined as an isomorphism in relevant Sobolev spaces. We emphasize that our fractional derivative coincides with the classical Riemann–Liouville derivative and the Caputo derivative in suitable spaces, and we never aim at creating novel notions of fractional derivatives but we are concerned with a formulation of a fractional derivative allowing us convenient applications to time-fractional differential equations, as Section 5 and Section 6 discuss.
  • Section 3: Basic properties in fractional calculus.
  • Section 4: Fractional derivatives of the Mittag-Leffler functions.
  • Section 5: Initial value problem for fractional ordinary differential equations.
  • Section 6: Initial boundary value problem for fractional partial differential equations: selected topics.
  • Section 7: Application to an inverse source problem:
    For illustrating the feasibility of our approach, we consider one inverse source problem of determining a time-varying function.
  • Section 8: Concluding remarks.

2. Definition of the Extended Derivative t α

2.1. Introduction of Function Spaces and Operators

We set
( J α v ) ( t ) : = 1 Γ ( α ) 0 t ( t s ) α 1 v ( s ) d s , 0 < t < T , D ( J α ) = L 2 ( 0 , T ) , ( J α v ) ( t ) : = 1 Γ ( α ) t T ( ξ t ) α 1 v ( ξ ) d ξ , 0 < t < T , D ( J α ) = L 2 ( 0 , T ) .
We can consider J α , J α for v L 1 ( 0 , T ) , but for the moment, we consider them for v L 2 ( 0 , T ) . There are many works on the Riemann–Liouville time-fractional integral operator J α , and here we refer only to two monographs: Gorenflo and Vessella [9] and Samko, Kilbas and Marichev [4]. See also Gorenflo and Yamamoto [10] for an operator theoretical approach.
We can directly prove
Lemma 1.
Let α , β 0 . Then
J α J β v = J α + β v , J α J β v = J α + β v f o r   v L 1 ( 0 , T ) .
For 0 < α < 1 , we define an operator τ : L 2 ( 0 , T ) L 2 ( 0 , T ) by
( τ v ) ( t ) : = v ( T t ) , v L 2 ( 0 , T ) .
Then it is readily seen that τ is an isomorphism between L 2 ( 0 , T ) and itself.
Throughout this article, we call K an isomorphism between two Banach spaces X and Y if the mapping K is injective, K X = Y and there exists a constant C > 0 such that C 1 K v Y v X C K v Y for all v X . Here, · X and · Y denote the norms in X and Y, respectively.
We set
0 C 1 [ 0 , T ] : = { v C 1 [ 0 , T ] ; v ( 0 ) = 0 } , 0 C 1 [ 0 , T ] : = { v C 1 [ 0 , T ] ; v ( T ) = 0 } = τ ( 0 C 1 [ 0 , T ] ) .
By H α ( 0 , T ) with 0 < α < 1 , we denote the Sobolev–Slobodecki space with the norm · H α ( 0 , T ) defined by
v H α ( 0 , T ) : = v L 2 ( 0 , T ) 2 + 0 T 0 T | v ( t ) v ( s ) | 2 | t s | 1 + 2 α d t d s 1 2
(e.g., Adams [11]).
By X ¯ Y , we denote the closure of X Y in a normed space Y. We set
H α ( 0 , T ) : = 0 C 1 [ 0 , T ] ¯ H α ( 0 , T ) , α H ( 0 , T ) : = 0 C 1 [ 0 , T ] ¯ H α ( 0 , T ) .
Henceforth, we set H 0 ( 0 , T ) : = L 2 ( 0 , T ) and 0 H ( 0 , T ) : = L 2 ( 0 , T ) . Then we have the following proposition.
Proposition 1.
Let 0 < α < 1 .
(i) 
H α ( 0 , T ) : = H α ( 0 , T ) , 0 < α < 1 2 , v H 1 2 ( 0 , T ) ; 0 T | v ( t ) | 2 t d t < , α = 1 2 , { v H α ( 0 , T ) ; v ( 0 ) = 0 } , 1 2 < α 1 .
Moreover, the norm in H α ( 0 , T ) is equivalent to
v H α ( 0 , T ) : = v H α ( 0 , T ) , α 1 2 , v H 1 2 ( 0 , T ) 2 + 0 T | v ( t ) | 2 t d t 1 2 , α = 1 2 .
(ii) 
Similarly, we have
α H ( 0 , T ) : = H α ( 0 , T ) , 0 < α < 1 2 , v H 1 2 ( 0 , T ) ; 0 T | v ( t ) | 2 T t d t < , α = 1 2 , { v H α ( 0 , T ) ; v ( T ) = 0 } , 1 2 < α 1 .
Moreover, the norm in α H ( 0 , T ) is equivalent to
v α H ( 0 , T ) : = v H α ( 0 , T ) , α 1 2 , v H 1 2 ( 0 , T ) 2 + 0 T | v ( t ) | 2 T t d t 1 2 , α = 1 2 .
We can refer to [5] for the proof of Proposition 1 (i). Since τ : H α ( 0 , T ) α H ( 0 , T ) is an isomorphism, we can verify part (ii) of the proposition.

2.2. Extension of d t α to H α ( 0 , T ) : Intermediate Step

We can prove the following (e.g., [5,8]).
Proposition 2.
Let 0 < α < 1 . Then
J α : L 2 ( 0 , T ) H α ( 0 , T )
is an isomorphism. In particular, H α ( 0 , T ) = J α L 2 ( 0 , T ) .
The monographs [4,9] show comprehensive results on J α , in particular on characterizations of J α L p ( 0 , T ) with p 1 . In the case of 1 < p < 1 α , for the characterizations, we can refer to Section 6 of Chapter 2 and Section 13 of Chapter 3 in [4] and we highlight Theorem 6.2 in [4], p. 127. We remark that the condition 1 < p < 1 α implies 0 < α < 1 2 in the case where p = 2 , which does not cover the range 0 < α < 1 of the orders.
By the proposition, we can easily verify that some functions belong to H α ( 0 , T ) , although the verification by (14) is complicated.
Example 2.
In view of Proposition 2, we will verify that
t β H α ( 0 , T ) i f   β > α 1 2 .
Indeed, setting γ : = β α , we see γ > 1 2 , and so t γ L 2 ( 0 , T ) .
Moreover,
J α t γ = 1 Γ ( α ) 0 t ( t s ) α 1 s γ d s = Γ ( γ + 1 ) Γ ( α + γ + 1 ) t α + γ ,
that is,
t β = t α + γ = J α Γ ( α + γ + 1 ) Γ ( γ + 1 ) t γ J α L 2 ( 0 , T ) .
Thus we see that t β H α ( 0 , T ) .
In Proposition 2.4 in [5], the direct proof for t β H α ( 0 , T ) is given for more restricted β > 0 and 0 < α < 1 2 , which is more complicated than the proof here.
We set
t α : = ( J α ) 1 = J α w i t h D ( t α ) = H α ( 0 , T ) = J α L 2 ( 0 , T ) .
The essence of this extension of t α is that we define t α as the inverse to the isomorphism J α on L 2 ( 0 , T ) onto H α ( 0 , T ) . For estimating or treating t α v = g later, we will often consider through J α g , as is already calculated in Example 2.
From Proposition 2, we can directly derive
Proposition 3.
There exists a constant C > 0 such that
C 1 v H α ( 0 , T ) t α v L 2 ( 0 , T ) C v H α ( 0 , T )
for all v H α ( 0 , T ) .
Example 3.
We return to Example 2. Let β > α 1 2 . Then t β H α ( 0 , T ) and
J α t β α = Γ ( β α + 1 ) Γ ( β + 1 ) t β .
Hence, by the definition of t α , we obtain
t β α = Γ ( β α + 1 ) Γ ( β + 1 ) t α t β ,
that is,
t α t β = Γ ( β + 1 ) Γ ( β α + 1 ) t β α i f   β > α 1 2   a n d   0 < α < 1 .
If 0 < α < 1 2 , then β < 0 is possible and, for β < 0 , we have t β W 1 , 1 ( 0 , T ) . Therefore, d t α t β cannot be defined directly.
Next, we will define t α for general α > 0 . Let α : = m + σ with m N { 0 } and 0 < σ 1 . Then we define
H m ( 0 , T ) : = v H m ( 0 , T ) ; v ( 0 ) = = d m 1 v d t m 1 ( 0 ) = 0 , m 1 , L 2 ( 0 , T ) , m = 0 , H m + σ ( 0 , T ) : = v H m ( 0 , T ) ; d m v d t m H σ ( 0 , T )
and
v H m + σ ( 0 , T ) : = d m d t m v L 2 ( 0 , T ) 2 + d m d t m v H σ ( 0 , T ) 2 1 2 .
We can easily verify that H m + σ ( 0 , T ) is a Banach space.
We similarly define α H ( 0 , T ) for each α > 0 . Then, in view of Proposition 2, we can prove the following.
Proposition 4.
Let m N { 0 } and 0 < σ 1 . Then J m + σ : L 2 ( 0 , T ) H m + σ ( 0 , T ) is an isomorphism.
Proof. 
We can assume that m 1 . By definition, v H m + σ ( 0 , T ) if and only if
d m d t m v H σ ( 0 , T ) , v ( 0 ) = = d m 1 d t m 1 v ( 0 ) = 0 ,
which implies that d m d t m v = J σ w for some w L 2 ( 0 , T ) . By v ( 0 ) = = d m 1 d t m 1 v ( 0 ) = 0 , we see
v ( t ) = 1 ( m 1 ) ! 0 t ( t s ) m 1 d m d t m v ( s ) d s , 0 < t < T .
Therefore, exchanging the order of the integral, we have
v ( t ) = 1 ( m 1 ) ! 0 t ( t s ) m 1 1 Γ ( σ ) 0 s ( s ξ ) σ 1 w ( ξ ) d ξ d s = 1 ( m 1 ) ! Γ ( σ ) 0 t ξ t ( t s ) m 1 ( s ξ ) σ 1 d s w ( ξ ) d ξ = 1 Γ ( m + σ ) 0 t ( t ξ ) m + σ 1 w ( ξ ) d ξ = J m + σ w ( t ) f o r 0 < t < T .
Hence, v J m + σ L 2 ( 0 , T ) , which means that J m + σ L 2 ( 0 , T ) H m + σ ( 0 , T ) . The converse inclusion is direct, and we see that J m + σ L 2 ( 0 , T ) = H m + σ ( 0 , T ) . The norm equivalence between v L 2 ( 0 , T ) and J m + σ v H m + σ ( 0 , T ) readily follows from the definition and Proposition 2. □
For α = m + σ with m N and 0 < σ 1 , we now define t α as an extension of the Caputo derivative
d t α v ( t ) = 1 Γ ( m + 1 α ) 0 t ( t s ) m α d m + 1 d s m + 1 v ( s ) d s , 0 < t < T .
We note that d t α requires ( m + 1 ) -times differentiability of v.
By Proposition 4, the inverse to J α exists for each α 0 , and by J α we denote the inverse:
J α : = ( J α ) 1 .
As the extension of such d t α to H m + σ ( 0 , T ) , we define
t m + σ = J m σ w i t h D ( t m + σ ) = H m + σ ( 0 , T ) .
Thus we have the following proposition.
Proposition 5.
Let α , β 0 .
(i)
t α : H α + β ( 0 , T ) H β ( 0 , T )
and
J α : H β ( 0 , T ) H α + β ( 0 , T )
are isomorphisms.
(ii)
It holds that
J α J β = J α + β o n   D ( J α + β ) .
Here, we set
D ( J α + β ) = L 2 ( 0 , T ) i f α + β 0 , H α β ( 0 , T ) i f α + β < 0 .
Proof. 
Part (i) is seen by (18) and Proposition 4. Part (ii) can be proved as follows: Let α + β 0 . If α = β , then J α J α = I : the identity operator on L 2 ( 0 , T ) by (18), and the conclusion is trivial. Let β > α . Set γ : = β α > 0 . Let v L 2 ( 0 , T ) be arbitrary. Then, by Lemma 1, we have
J α J β v = J α ( J α + γ v ) = J α ( J α J γ ) v = ( J α J α ) J γ v = J γ v .
Since J γ v = J α + β v , we obtain J α J β v = J α + β v for each v L 2 ( 0 , T ) .
Next, let α + β < 0 . Given v H α β ( 0 , T ) arbitrarily, by Proposition 4 we can find w L 2 ( 0 , T ) such that v = J α β w . Then
J α J β v = J α J β ( J α β w ) = J α ( J β + ( α β ) w ) = J α J α w = w
by β , α β > 0 and Lemma 1. Therefore, J α J β v = w . Since v = J α β w implies w = ( J α β ) 1 v , we have
w = ( J α β ) 1 v = J ( α β ) v = J α + β v ,
that is, J α J β v = J α + β v for each v H α β ( 0 , T ) . Thus the proof of Proposition 5 is complete. □
Next, for 0 < α < 1 , we characterize t α as an extension over H α ( 0 , T ) of the operator d t α defined on 0 C 1 [ 0 , T ] . Let X and Y be Banach spaces with the norms · X and · Y , respectively, and let K : X Y be a densely defined linear operator. We call K a closed operator if the following is satisfied: if v n D ( K ) , lim n v n = v and K v n converges to some w in Y, then v D ( K ) and K v = w . By K ¯ , we denote the closure of an operator K from X to Y, that is, K ¯ is the minimum closed extension of K in the sense that if K ˜ is a closed operator such that D ( K ˜ ) D ( K ) , then D ( K ˜ ) D ( K ¯ ) . It is trivial that K ¯ = K for a closed operator K. Moreover, we say that K is closable if there exists K ¯ .
Here, we always consider the operator d t α with the domain 0 C 1 [ 0 , T ] : d t α : 0 C 1 [ 0 , T ] H α ( 0 , T ) L 2 ( 0 , T ) . Then we can prove the following (e.g., [5]).
Proposition 6.
The operator d t α with the domain 0 C 1 [ 0 , T ] is closable and d t α ¯ = t α .
Proposition 6 is not used later but means that our derivative t α is reasonable as the minimum extension of the classical Caputo derivative d t α for v 0 C 1 [ 0 , T ] .
Interpretation of D ( t α ) = H α ( 0 , T ) . Let α > 1 2 and let v H α ( 0 , T ) . By the definition of H α ( 0 , T ) in (14), we can choose an approximating sequence v n 0 C 1 [ 0 , T ] , n N such that lim n v n v H α ( 0 , T ) = 0 . By the Sobolev embedding (e.g., Adams [11]), we see that H α ( 0 , T ) H α ( 0 , T ) C [ 0 , T ] if α > 1 2 so that v C [ 0 , T ] and lim n v n v C [ 0 , T ] = 0 . Therefore,
lim n | v ( 0 ) v n ( 0 ) | lim n v n v C [ 0 , T ] = 0 ,
that is, v ( 0 ) = lim n v n ( 0 ) . Therefore, v H α ( 0 , T ) with α > 1 2 yields v ( 0 ) = 0 . In other words, if α > 1 2 , then v D ( t α ) means that v = v ( t ) satisfies the zero initial condition, which is useful for the formulation of the initial value problem in Section 5 and Section 6. We can similarly consider and see that v D ( t α ) with α > 3 2 yields v ( 0 ) = t v ( 0 ) = 0 .
As is directly proved by the Young inequality on the convolution, we see
D t α v L 1 ( 0 , T ) i f   v W 1 , 1 ( 0 , T ) .
However, we do not necessarily have D t α v L 2 ( 0 , T ) . Indeed,
D t α t β = Γ ( 1 + β ) Γ ( 1 α + β ) t β α f o r   β > 0 .
If β α < 1 2 and β > 0 , then D t α t β L 2 ( 0 , T ) , in spite of t β W 1 , 1 ( 0 , T ) . This means that if we want to keep the range of D t α within L 2 ( 0 , T ) , then even W 1 , 1 ( 0 , T ) of a space of differentiable functions is not sufficient, although D t α are concerned with α -times differentiability.
In general, we can readily prove the following.
Proposition 7.
We have
t α v = D t α v = d t α v f o r   v 0 C 1 [ 0 , T ]
and
t α v ( t ) = 1 Γ ( 1 α ) d d t 0 t ( t s ) α v ( s ) d s = D t α v ( t ) f o r   v H α ( 0 , T ) .
Proof. 
Equation (19) is easily seen. We will prove (20) as follows: For an arbitrary v H α ( 0 , T ) , Proposition 2 yields that v = J α w with some w L 2 ( 0 , T ) and w = t α v . On the other hand, Lemma 1 implies
D t α v = d d t J 1 α v = d d t J 1 α ( J α w ) = d d t J w = w .
Therefore, D t α v = t α v , and then the proof of Proposition 7 is complete. □
Equation (20) means that t α coincides with the Riemann–Liouville fractional derivative, provided that we consider H α ( 0 , T ) as the domain of t α and D t α . This extension t α of d t α | 0 C 1 [ 0 , T ] is not yet complete, and in the next subsection, we will continue to extend.

2.3. Definition of t α : Completion of the Extension of d t α

By the current extension of t α , we understand that t α 1 = 0 for 0 < α < 1 2 , but t α 1 cannot be defined for 1 2 α < 1 :
1 H α ( 0 , T ) = D ( t α ) 0 < α < 1 2 , 1 D ( t α ) 1 2 α < 1 .
Moreover, we note that d t α 1 = 0 and D t α 1 = t α Γ ( 1 α ) for all α ( 0 , 1 ) . Therefore, t α 1 is consistent with neither the classical fractional derivative d t α nor D t α , which suggests that our current extension from d t α to t α is not sufficient as a fractional derivative. Moreover, as is seen in Section 7, the current t α is not convenient for treating less regular source terms in fractional differential equations. In order to define t α v in more general spaces such as L 2 ( 0 , T ) , we should continue to extend t α .
We recall (12) and (14) for α > 0 . We can readily verify that τ : H α ( 0 , T ) α H ( 0 , T ) is an isomorphism with τ defined by (13). Then we have the following proposition.
Proposition 8.
Let α > 0 and β 0 . Then J α : β H ( 0 , T ) α + β H ( 0 , T ) is an isomorphism.
We recall that J α : H β ( 0 , T ) H α + β ( 0 , T ) is an isomorphism by Proposition 5 (i). Proposition 8 can be proved via the mapping τ : L 2 ( 0 , T ) L 2 ( 0 , T ) defined by (13).
When J α remains an operator defined over L 2 ( 0 , T ) , the operator J α is not defined for f L 1 ( 0 , T ) , but the integral 1 Γ ( α ) 0 t ( t s ) α 1 f ( s ) d s itself exists as a function in L 1 ( 0 , T ) for any f L 1 ( 0 , T ) . This is a substantial inconvenience, and we have to make a suitable extension of J α | L 2 ( 0 , T ) . Our formulation is based on L 2 -space and we cannot directly treat L 1 ( 0 , T ) -space. Thus we introduce the dual space of H α ( 0 , T ) , which contains L 1 ( 0 , T ) .
The key ideas for the further extension of t α are the family { H α ( 0 , T ) } α > 0 and the isomorphism t α : H α + β ( 0 , T ) H β ( 0 , T ) with α , β > 0 , which is called a Hilbert scale. We can refer to Chapter V in Amann [12] for a general reference.
Let X be a Hilbert space over R , V X be a dense subspace of X and the embedding V X be continuous. By the dual space X of X, we call the space of all the bounded linear functionals defined on X. Then, identifying the dual space X with itself, we can conclude that X is a dense subspace of the dual space V of V:
V X V .
By V < f , v > V , we denote the value of f V at v V . We note that
V < f , v > V = ( f , v ) X i f   f X ,
where ( f , v ) X is the scalar product in X.
We note that H α ( 0 , T ) and α H ( 0 , T ) are both dense in L 2 ( 0 , T ) . Henceforth, identifying the dual space L 2 ( 0 , T ) with itself, by the above manner, we can define ( H α ( 0 , T ) ) and ( α H ( 0 , T ) ) . Then
H α ( 0 , T ) L 2 ( 0 , T ) ( H α ( 0 , T ) ) = : H α ( 0 , T ) , α H ( 0 , T ) L 2 ( 0 , T ) ( α H ( 0 , T ) ) = : α H ( 0 , T ) , α > 0 ,
where the above inclusions mean dense subsets. We refer to Brezis [13] and Yosida [14] for examples of general treatments for dual spaces.
Let X , Y be Hilbert spaces and let K : X Y be a bounded linear operator with D ( K ) = X . Then we recall that the dual operator K is the maximum operator among operators K ^ : Y X with D ( K ^ ) Y such that X < K ^ y , x > X = Y < y , K x > Y for each x X and y D ( K ^ ) Y (e.g., [13]).
Henceforth, we consider the dual operator ( J α ) of J α : L 2 ( 0 , T ) H α ( 0 , T ) and the dual ( J α ) of J α : L 2 ( 0 , T ) α H ( 0 , T ) by setting X = L 2 ( 0 , T ) and Y = H α ( 0 , T ) or Y = α H ( 0 , T ) .
Then we can show the following.
Proposition 9.
Let α > 0 and β 0 . Then
(i) 
( J α ) : H α β ( 0 , T ) H β ( 0 , T ) is an isomorphism.
In particular, ( J α ) : H α ( 0 , T ) L 2 ( 0 , T ) is an isomorphism.
(ii) 
( J α ) : α β H ( 0 , T ) β H ( 0 , T ) is an isomorphism.
In particular, ( J α ) : α H ( 0 , T ) L 2 ( 0 , T ) is an isomorphism.
(iii) 
It holds that
J α v = ( J α ) v f o r   v L 2 ( 0 , T ) .
Henceforth, we write J α : = ( J α ) . We see
J α ( J α ) 1 w = w f o r   w L 2 ( 0 , T )
and
( J α ) 1 J α w = w f o r   w α H ( 0 , T ) .
Proof. 
From Proposition 8, in terms of the closed range theorem (e.g., Section 7 of Chapter 2 in [13]), we can see (i) and (ii).
We now prove (iii). We can directly verify ( J α v , w ) L 2 ( 0 , T ) = ( v , J α w ) L 2 ( 0 , T ) for each v , w L 2 ( 0 , T ) . Hence, by the maximality property of the dual operator J α of J α , we see that J α J α . Thus the proof of Proposition 9 is complete. □
Due to Propositions 5, 8 and 9, for β 0 , we can regard the operators J α with D ( J α ) = β H ( 0 , T ) and J α with D ( J α ) = H β ( 0 , T ) and, accordingly, also the operators J α with D ( J α ) = α β H ( 0 , T ) and ( J α ) with D ( ( J α ) ) = H α β ( 0 , T ) . We do not specify the domains if we need not emphasize them.
We can define J α u for u L 1 ( 0 , T ) as follows: We note that J α is the dual operator of J α : γ H ( 0 , T ) α + γ H ( 0 , T ) , where we choose γ > 0 such that D ( J α ) L 1 ( 0 , T ) .
To this end, choosing γ > 0 such that α + γ > 1 2 , we regard J α as an operator J α : α γ H ( 0 , T ) γ H ( 0 , T ) . Then we can define J α u for u L 1 ( 0 , T ) . Indeed, the Sobolev embedding yields that
α + γ H ( 0 , T ) H α + γ ( 0 , T ) C [ 0 , T ]
by α + γ > 1 2 . Therefore, any u L 1 ( 0 , T ) can be considered an element in ( α + γ H ( 0 , T ) ) by
α γ H ( 0 , T ) < u , φ > α + γ H ( 0 , T ) = 0 T u ( t ) φ ( t ) d t f o r   φ α + γ H ( 0 , T ) ,
which means that L 1 ( 0 , T ) D ( J α ) = α γ H ( 0 , T ) and J α u is well-defined for u L 1 ( 0 , T ) .
Henceforth, we always make the above definition of J α u for u L 1 ( 0 , T ) , if not specified.
Then we can improve Proposition 9 (iii) with the following.
Proposition 10.
We have
J α u = 1 Γ ( α ) 0 t ( t s ) α 1 u ( s ) d s , 0 < t < T f o r   u L 1 ( 0 , T ) .
We can write (22) as J α u = J α u for u L 1 ( 0 , T ) , when we do not specify the domain D ( J α ) .
Proof of Proposition 10.
Let γ > 1 2 . We consider J α as an operator J α : α γ H ( 0 , T ) γ H ( 0 , T ) .
Let u L 1 ( 0 , T ) be arbitrary. Since L 1 ( 0 , T ) γ H ( 0 , T ) and L 2 ( 0 , T ) is dense in L 1 ( 0 , T ) , we can choose u n L 2 ( 0 , T ) , n N such that u n u in L 1 ( 0 , T ) as n . By Proposition 9 (iii), we have J α u n = J α u n for n N . By Proposition 9 (ii), we obtain
J α u n J α u i n   γ H ( 0 , T ) .
On the other hand, by u L 1 ( 0 , T ) , the Young inequality on the convolution implies
J α u n J α u L 1 ( 0 , T ) 1 Γ ( α ) s α 1 L 1 ( 0 , T ) u n u L 1 ( 0 , T ) ,
which yields J α u n = J α u n J α u in L 1 ( 0 , T ) , that is, J α u n J α u in γ H ( 0 , T ) . Therefore, with (23), we obtain J α u = J α u . Thus the proof of Proposition 10 is complete. □
Now we complete the extension of d t α with the domain 0 C 1 [ 0 , T ] .
Definition 1.
For β 0 , we define
t α : = ( J α ) 1 w i t h D ( t α ) = H β ( 0 , T ) o r D ( t α ) = β H ( 0 , T ) .
For definitions of fractional derivatives for less regular functions, including generalized functions, we can refer to Section 8 of Chapter 2 in [4], but our approach is different and admits transparent applications to fractional differential equations in Section 5, Section 6.4 and Section 7.
Thus we have the following theorem.
Theorem 1.
Let α > 0 and β 0 . Then t α : β H ( 0 , T ) α β H ( 0 , T ) and
t α : H α + β ( 0 , T ) H β ( 0 , T ) are both isomorphisms.
The extension t α : β H ( 0 , T ) α β H ( 0 , T ) is not only a theoretical interest but also useful for studies of fractional differential equations even if we consider all the functions within L 2 ( 0 , T ) , as we will do in Section 6 and Section 7.
We consider a special case β = 0 . Then ( J α ) 1 : D ( ( J α ) 1 ) = L 2 ( 0 , T ) α H ( 0 , T ) . Then, since Proposition 9 (iii) yields
( J α ) 1 ( J α ) 1 ,
we see that this ( J α ) 1 = t α defined on L 2 ( 0 , T ) is an extension defined previously in H α ( 0 , T ) . In particular, this extended t α still satisfies
( J α ) 1 H α ( 0 , T ) = L 2 ( 0 , T ) .
Our completely extended derivative t α of d t α with D ( d t α ) = 0 C 1 [ 0 , T ] operates similarly to the Riemann–Liouville fractional derivative D t α and is equivalent to d d t J 1 α associated with the domain β H ( 0 , T ) and the range α β H ( 0 , T ) with α > 0 and β 0 .
Proposition 10 enables us to calculate t α u = f provided that u , f L 1 ( 0 , T ) , and we show the following example.
Example 4.
Let 0 < α < 1 . Choosing α + γ > 1 2 , we consider t α with D ( t α ) = γ α H ( 0 , T ) L 1 ( 0 , T ) . Then 1 D ( t α ) , and we have
t α 1 = 1 Γ ( 1 α ) t α .
Indeed, since 1 Γ ( 1 α ) t α L 1 ( 0 , T ) , Proposition 10 yields
J α 1 Γ ( 1 α ) t α = 1 Γ ( 1 α ) 1 Γ ( α ) 0 t ( t s ) α 1 s α d s = 1 .
Therefore, the definition justifies t α 1 = 1 Γ ( 1 α ) t α .
Before proceeding to the next section, we will provide two propositions.
Proposition 11.
Let 0 < α < 1 . Then
t α u ( t ) = d d t 1 Γ ( 1 α ) 0 t ( t s ) α u ( s ) d s i n   ( C 0 ( 0 , T ) )
for u L 1 ( 0 , T ) , where d d t is taken in ( C 0 ( 0 , T ) ) .
In the proposition, we note 1 Γ ( 1 α ) 0 t ( t s ) α u ( s ) d s L 1 ( 0 , T ) ( C 0 ( 0 , T ) ) . We remark that we cannot take the pointwise differentiation of 0 t ( t s ) α u ( s ) d s in general for u L 1 ( 0 , T ) .
Proposition 11 enables us to calculate t α u = f for u L 1 ( 0 , T ) even if t α u cannot be defined within L 1 ( 0 , T ) , as the following Example 5 (a) shows.
Example 5.
(a) Let 1 2 < α < 1 and let
h t 0 ( t ) = 0 , t t 0 , 1 , t > t 0
with arbitrary t 0 ( 0 , T ) . We can verify that h t 0 H 1 α ( 0 , T ) if α > 1 2 . Indeed,
0 T 0 T | h t 0 ( t ) h t 0 ( s ) | 2 | t s | 3 2 α d s d t = 2 0 t 0 t 0 T 1 | t s | 3 2 α d t d s = 1 1 α 0 t 0 ( ( t 0 s ) 2 + 2 α ( T s ) 2 + 2 α ) d s <
by 2 + 2 α > 1 .
Therefore, by Proposition 2, we can choose w L 2 ( 0 , T ) such that J 1 α w = h t 0 . By Proposition 11, we can calculate t α w because w L 2 ( 0 , T ) L 1 ( 0 , T ) :
t α w = d d t 1 Γ ( 1 α ) 0 t ( t s ) α w ( s ) d s = d d t J 1 α w = d d t h t 0 = δ t 0 ,
which is a Dirac delta function satisfying
( C 0 ( 0 , T ) ) < δ t 0 , ψ > C 0 ( 0 , T ) = ψ ( t 0 ) f o r   a l l   ψ C 0 ( 0 , T ) .
(b)  We have
t α t β = Γ ( 1 + β ) Γ ( 1 α + β ) t β α 0 < α < 1 , β > 1
in α γ H ( 0 , T ) with some α + γ > 1 2 . See Example 4 as (28) with β = 0 . We remark that β α < 1 is possible, and so t β α L 1 ( 0 , T ) may occur.
Indeed, by β > 1 , we can see
1 Γ ( 1 α ) 0 t ( t s ) α s β d s = Γ ( 1 + β ) Γ ( 2 α + β ) t 1 α + β .
Taking the derivative in the sense of ( C 0 ( 0 , T ) ) , we see
d d t Γ ( 1 + β ) Γ ( 2 α + β ) t 1 α + β = Γ ( 1 + β ) Γ ( 2 α + β ) ( 1 α + β ) t β α = Γ ( 1 + β ) Γ ( 1 α + β ) t β α .
Thus (28) is verified.
We cannot define d t α t β in general for 1 < β 0 , but we can calculate D t α t β . We remark that t α t β in the operator sense, coincides with the result calculated by D t α t β . We here emphasize that our interest is not only computations of fractional derivatives but also formulating t α as an operator defined on α γ H ( 0 , T ) with the isomorphism.
(c) For any constant t 0 ( 0 , T ) , we consider a function of the Heaviside type defined by (27). By Proposition 10 or 11, we can see
t α h t 0 ( t ) = 0 , t t 0 , ( t t 0 ) α Γ ( 1 α ) , t > t 0 .
We compare with the case α = 1 :
d d t h t 0 ( t ) = δ t 0 ( t ) : D i r a c   d e l t a   f u n c t i o n   a t   t 0
in ( C 0 ( 0 , T ) ) . On the other hand, t α h t 0 for α < 1 does not generate the singularity of a Dirac delta function.
Moreover,
t α h t 0 δ t 0 a s   α 1   i n   ( C 0 ( 0 , T ) ) ,
which is taken in the distribution sense. More precisely,
lim α 1 ( C 0 ( 0 , T ) ) < t α h t 0 , ψ > C 0 ( 0 , T ) = ( C 0 ( 0 , T ) ) < δ t 0 , ψ > C 0 ( 0 , T ) = ψ ( t 0 )
for each ψ C 0 ( 0 , T ) .
Indeed, by integration by parts, we obtain
( C 0 ( 0 , T ) ) < t α h t 0 , ψ > C 0 ( 0 , T ) = t 0 T ( t t 0 ) α Γ ( 1 α ) ψ ( t ) d t = ( t t 0 ) 1 α ( 1 α ) Γ ( 1 α ) ψ ( t ) t = t 0 t = T 1 ( 1 α ) Γ ( 1 α ) t 0 T ( t t 0 ) 1 α ψ ( t ) d t = 1 Γ ( 2 α ) T t 0 ( t t 0 ) 1 α ψ ( t ) d t .
Letting α 1 and applying the Lebesgue convergence theorem, we know that the right-hand side tends to
1 Γ ( 1 ) T t 0 ψ ( t ) d t = ψ ( t 0 ) .
Proof of Proposition 11.
We prove by approximating u L 1 ( 0 , T ) by u n 0 C 1 [ 0 , T ] , n N and using Proposition 7. As before, we choose γ > 1 2 α so that L 1 ( 0 , T ) α γ H ( 0 , T ) by the Sobolev embedding: α + γ H ( 0 , T ) C [ 0 , T ] .
Since we can choose u n 0 C 1 [ 0 , T ] , n N such that u n u in L 1 ( 0 , T ) as n , we see that u n u in α γ H ( 0 , T ) . Hence, Theorem 1 yields that t α u n t α u in 2 α γ H ( 0 , T ) . Since C 0 ( 0 , T ) 2 α + γ H ( 0 , T ) yields 2 α γ H ( 0 , T ) ( C 0 ( 0 , T ) ) , we see that t α u n t α u in 2 α γ H ( 0 , T ) implies t α u n t α u in ( C 0 ( 0 , T ) ) , that is,
lim n ( C 0 ( 0 , T ) ) < t α u n , ψ > C 0 ( 0 , T ) = ( C 0 ( 0 , T ) ) < t α u , ψ > C 0 ( 0 , T )
for all ψ C 0 ( 0 , T ) .
On the other hand, by Proposition 7, we have t α u n = D t α u n , n N . Therefore,
lim n ( C 0 ( 0 , T ) ) < D t α u n , ψ > C 0 ( 0 , T ) = ( C 0 ( 0 , T ) ) < t α u , ψ > C 0 ( 0 , T )
for all ψ C 0 ( 0 , T ) .
Then, for any ψ C 0 ( 0 , T ) , we obtain
( C 0 ( 0 , T ) ) < D t α u n , ψ > C 0 ( 0 , T ) = ( D t α u n ψ ) L 2 ( 0 , T ) = 1 Γ ( 1 α ) d d t 0 t ( t s ) α u n ( s ) d s , ψ L 2 ( 0 , T ) = 1 Γ ( 1 α ) 0 t ( t s ) α u n ( s ) d s , d ψ d t L 2 ( 0 , T ) .
Since the Young inequality on the convolution yields
0 t ( t s ) α u n ( s ) d s 0 t ( t s ) α u ( s ) d s L 1 ( 0 , T ) s α L 1 ( 0 , T ) u n u L 1 ( 0 , T ) 0
as n , we have
lim n ( D t α u n , ψ ) L 2 ( 0 , T ) = 1 Γ ( 1 α ) 0 t ( t s ) α u ( s ) d s , d ψ d t L 2 ( 0 , T ) .
Hence, with (29), we obtain
( C 0 ( 0 , T ) ) < t α u , ψ > C 0 ( 0 , T ) = lim n ( C 0 ( 0 , T ) ) < D t α u n , ψ > C 0 ( 0 , T )
= 1 Γ ( 1 α ) 0 t ( t s ) α u ( s ) d s , d ψ d t L 2 ( 0 , T )
for all ψ C 0 ( 0 , T ) . Consequently, (30) means that
t α u ( t ) = d d t 1 Γ ( 1 α ) 0 t ( t s ) α u ( s ) d s
in the sense of the derivative of a distribution. Thus the proof of Proposition 11 is complete. □

3. Basic Properties in Fractional Calculus

In this section, we present fundamental properties of t α in the case of D ( t α ) = H α ( 0 , T ) . We can consider for t α with the domain L 2 ( 0 , T ) , but we here omit.
We set t 0 = I : the identity operator on L 2 ( 0 , T ) .
Theorem 2.
Let α , β 0 . Then
t α ( t β v ) = t α + β v f o r   a l l   v H α + β ( 0 , T ) .
This kind of sequential derivative is more complicated for d t α and D t α when we do not specify the domains. For t α , the domain is already installed in a convenient way.
Proof of Theorem 2.
By (18), we have t α = ( J α ) 1 in D ( t α ) = H α ( 0 , T ) with α 0 . Hence, it suffices to prove
J α ( J β v ) = J ( α + β ) v , v H α + β ( 0 , T ) .
Setting w = J β v , we have w H α ( 0 , T ) by Proposition 5 (i). Then J α J β v = J α w and
J ( α + β ) v = J ( α + β ) J β w = J ( α + β ) + β w = J α w .
For the second equality to the last, we applied Proposition 5 (ii) in terms of w D ( J α ) = H α ( 0 , T ) . Hence, J α J β v = J ( α + β ) v for v H α + β ( 0 , T ) . Thus the proof of Theorem 2 is complete. □
We define the Laplace transform by
( L v ) ( p ) = v ^ ( p ) : = lim T 0 T v ( t ) e p t d t
provided that the limit exists.
Theorem 3
(Laplace transform of t α v ). Let u H α ( 0 , T ) with arbitrary T > 0 . If ( | t α u | ^ ) ( p ) exists for p > p 0 , which is some positive constant, then u ^ ( p ) exists for p > p 0 and
t α u ^ ( p ) = p α u ^ ( p ) , p > p 0 .
For initial value problems for fractional ordinary differential equations and initial boundary value problems for fractional partial differential equations, it is known that the Laplace transform is useful if one can verify the existence of the Laplace transform of solutions to these problems. The existence of the Laplace transform is concerned with the asymptotic behavior of unknown solution u as t , which may not be easy to verify for solutions to be determined. The method of Laplace transform is definitely helpful in finding solutions heuristically.
Corollary 1.
Let u H α ( 0 , T ) with arbitrary T > 0 . If t α u L r ( 0 , ) with some r 1 , then
t α u ^ ( p ) = p α u ^ ( p ) , p > p 0 .
Proof of Theorem 3.
We divide the proof into two steps.
First Step. We first prove the following lemma.
Lemma 2.
Let w L 2 ( 0 , T ) with arbitrary T > 0 . If | w | ^ ( p ) exists for p > p 0 . Then
( J α w ^ ) ( p ) = p α w ^ ( p ) , p > p 0 .
The same conclusion as in Lemma 2 is shown, for example, as in Lemma 2.14 in [2], p. 84, and in Theorem 7.2 in [4], p. 141, with the different assumption that there exists some constant p 0 0 such that | w ( t ) | = O ( e p 0 t ) for large t > 0 . In Theorem 3, if we replace the current assumption by | t α u ( t ) | = O ( e p 0 t ) for large t > 0 , then we can directly apply these lemmata in [2,4] to complete the proof. However, we keep the current assumption only on the existence of ( | t α u | ^ ) ( p ) in Theorem 3, and accordingly, we should formulate Lemma 2 with the corresponding assumption, which requires only the absolute convergence of the Laplace transform.
Proof of Lemma 2.
We recall that
( J α w ) ( t ) = 1 Γ ( α ) 0 t ( t s ) α 1 w ( s ) d s , t > 0 f o r   w L 2 ( 0 , T ) .
By w L 2 ( 0 , T ) , we see that J α w L 2 ( 0 , T ) for arbitrarily fixed T > 0 . By the assumption on the existence of w ^ , we obtain
lim T 0 T e p t w ( t ) d t = w ^ ( p ) , p > p 0 .
Choose T > 0 arbitrarily. Then w L 2 ( 0 , T ) , and
0 T e p t 1 Γ ( α ) 0 t ( t s ) α 1 w ( s ) d s d t = 1 Γ ( α ) 0 T w ( s ) s T e p t ( t s ) α 1 d t d s .
Changing the variables, t ξ by ξ : = ( t s ) p , we have
s T e p t ( t s ) α 1 d t = p α e p s 0 ( T s ) p ξ α 1 e ξ d ξ ,
and so
0 T e p t 1 Γ ( α ) 0 t ( t s ) α 1 w ( s ) d s d t = 1 Γ ( α ) p α 0 T e p s w ( s ) 0 ( T s ) p ξ α 1 e ξ d ξ d s = 1 Γ ( α ) p α T 2 T e p s w ( s ) 0 ( T s ) p ξ α 1 e ξ d ξ d s + 1 Γ ( α ) p α 0 T 2 e p s w ( s ) 0 ( T s ) p ξ α 1 e ξ d ξ d s = : I 1 + I 2 .
We remark that
0 ( T s ) p ξ α 1 e ξ d ξ 0 ξ α 1 e ξ d ξ = Γ ( α ) .
Then, since 0 e p t | w ( t ) | d t exists, we see
| I 1 | = 1 Γ ( α ) p α T 2 T e p s w ( s ) 0 ( T s ) p ξ α 1 e ξ d ξ d s C p α T 2 T e p s | w ( s ) | d s 0 a s T .
Since 0 < s < T 2 implies
T 2 p < ( T s ) p < T p ,
we obtain
0 T p 2 ξ α 1 e ξ d ξ 0 ( T s ) p ξ α 1 e ξ d ξ 0 T p ξ α 1 e ξ d ξ 0 ξ α 1 e ξ d ξ = Γ ( α )
and so
lim T 0 ( T s ) p ξ α 1 e ξ d ξ = 0 ξ α 1 e ξ d ξ = Γ ( α ) i f   0 < s < T 2
for any p > p 0 > 0 . Since
lim T 0 T w ( s ) e p s d s
exists for p > p 0 in view of (32) and (33), the Lebesgue convergence theorem yields
lim T 0 T 2 e p s w ( s ) 0 ( T s ) p ξ α 1 e ξ d ξ d s = Γ ( α ) 0 e p s w ( s ) d s ,
and we reach
lim T I 2 = p α Γ ( α ) 0 w ( s ) e p s Γ ( α ) d s = p α w ^ ( p ) , p > p 0 .
Thus,
lim T 0 T e p t ( J α w ) ( t ) d t = lim T ( I 1 + I 2 ) = p α w ^ ( p ) , p > p 0 .
The proof of Lemma 2 is complete. □
Second Step. We will complete the proof of the theorem. Since u H α ( 0 , T ) for any T > 0 , we can find w T L 2 ( 0 , T ) such that J α w T = u in ( 0 , T ) . For any t ( 0 , T ) , we can define w satisfying w ( t ) = w T ( t ) for 0 < t < T . Therefore, for all t > 0 , we can define w ( t ) and J α w ( t ) = u ( t ) for any t ( 0 , T ) , that is, for all t > 0 . Therefore, t α u ( t ) = w ( t ) for t > 0 and w L 2 ( 0 , T ) with arbitrary T > 0 .
Since t α u ( t ) = w ( t ) for t > 0 and | t α u | ^ ( p ) exists for p > p 0 , we know that | w | ^ ( p ) exists for p > p 0 . Therefore, Lemma 2 yields
( J α w ^ ) ( p ) = p α w ^ ( p ) , p > p 0 ,
that is,
u ^ ( p ) = p α t α u ^ ( p ) , p > p 0 .
Thus the proof of Theorem 3 is complete. □
Moreover, t α with D ( t α ) = H α ( 0 , T ) is consistent also with the convolution of two functions. We set
( u g ) ( t ) = 0 t u ( t s ) g ( s ) d s , 0 < t < T
for u L 2 ( 0 , T ) and g L 1 ( 0 , T ) . Then the Young inequality on the convolution yields
u g L 2 ( 0 , T ) u L 2 ( 0 , T ) g L 1 ( 0 , T ) .
We now prove the following theorem.
Theorem 4.
Let α 0 . Then
J α ( u g ) = ( J α u ) g f o r   u L 1 ( 0 , T )   a n d   g L 1 ( 0 , T )
u g H α ( 0 , T ) C u H α ( 0 , T ) g L 1 ( 0 , T ) f o r   u H α ( 0 , T )   a n d   g L 1 ( 0 , T ) .
t α ( u g ) = ( t α u ) g f o r   u H α ( 0 , T )   a n d   g L 1 ( 0 , T ) .
Proof of Theorem 4.
We prove (36)–(38) respectively.
Proof of (36). By exchange of the order of the integrals and change of variables s η : = s ξ , we can derive that
J α ( u g ) ( t ) = 1 Γ ( α ) 0 t ( t s ) α 1 0 s u ( s ξ ) g ( ξ ) d ξ d s = 1 Γ ( α ) 0 t g ( ξ ) ξ t ( t s ) α 1 u ( s ξ ) d s d ξ = 1 Γ ( α ) 0 t g ( ξ ) 0 t ξ ( t ξ η ) α 1 u ( η ) d η d ξ = 0 t g ( ξ ) ( J α u ) ( t ξ ) d ξ = ( g J α u ) ( t ) , 0 < t < T ,
which completes the proof of (36).
Proof of (37). Let u H α ( 0 , T ) and g L 1 ( 0 , T ) . Then, by Proposition 5, there exists w L 2 ( 0 , T ) such that u = J α w and w L 2 ( 0 , T ) C u H α ( 0 , T ) . By (36), we obtain u g = J α w g = J α ( w g ) . Since w L 2 ( 0 , T ) and g L 1 ( 0 , T ) imply w g L 2 ( 0 , T ) , by u g J α L 2 ( 0 , T ) , we see that u g H α ( 0 , T ) . Moreover, by Proposition 5 and u g = J α ( w g ) , we have
u g H α ( 0 , T ) C w g L 2 ( 0 , T ) C w L 2 ( 0 , T ) g L 1 ( 0 , T ) C u H α ( 0 , T ) g L 1 ( 0 , T ) .
which completes the proof of (37).
Proof of (38). For arbitrary u H α ( 0 , T ) , Proposition 5 yields that there exists w L 2 ( 0 , T ) such that u = J α w . Applying (36), we have J α ( w g ) = J α w g . Therefore,
t α ( J α w g ) = t α J α ( w g ) = w g .
Since u = J α w is equivalent to w = t α u , we reach t α ( u g ) = ( t α u ) g . Thus the proof of Theorem 4 is complete. □
We will show a useful variant of Theorem 4.
Theorem 5.
Let γ > 1 2 and β + γ > 1 2 . Then for t β : γ H ( 0 , T ) β γ H ( 0 , T ) , we have
t β v u = t β ( v u )
for all u L 1 ( 0 , T ) and v L 1 ( 0 , T ) satisfying t β v L 1 ( 0 , T ) .
We note that L 1 ( 0 , T ) γ H ( 0 , T ) by the Sobolev embedding and γ > 1 2 , and so if v L 1 ( 0 , T ) D ( t β ) = γ H ( 0 , T ) , then t β v β γ H ( 0 , T ) is well-defined. The assumption of the theorem further requires t β v L 1 ( 0 , T ) . On the other hand, for v , u L 1 ( 0 , T ) , the Young inequality implies v u L 1 ( 0 , T ) , and so v u γ H ( 0 , t ) . Hence, in the theorem, t β ( v u ) β γ H ( 0 , t ) is well-defined.
Proof of Theorem 5.
The Young inequality yields t β v u L 1 ( 0 , T ) . Consequently, by Proposition 10, we see
J β ( t β v u ) = J β ( t β v u ) .
By (36) in Theorem 4, we have
J β ( t β v u ) = ( J β t β v ) u .
Since t β v L 1 ( 0 , T ) , by using the definition given in (24), again, Proposition 10 implies
J β t β v = J β t β v = J β ( J β ) 1 v = v .
Hence, J β ( t β v u ) = v u . Operating ( J β ) 1 to both sides and noting v u L 1 ( 0 , T ) γ H ( 0 , T ) , by (24), we have
( J β ) 1 ( J β ( t β v u ) ) = ( J β ) 1 ( v u ) = t β ( v u ) ,
that is, t β v u = t β ( v u ) . Thus the proof of Theorem 5 is complete. □
Theorem 6
(Coercivity). Let 0 < α < 1 . Then
0 T v ( t ) t α v ( t ) d t 1 2 Γ ( 1 α ) T α v L 2 ( 0 , T ) 2 f o r   v H α ( 0 , T )
and
1 Γ ( α ) 0 t ( t s ) α 1 v ( s ) s α v ( s ) d s 1 2 | v ( t ) | 2 f o r   v H α ( 0 , T ) .
The proof of Theorem 6 can be found in [5]. Theorem 6 is not used in this article and is useful for proving the well-posedness for initial boundary value problems for fractional partial differential equations (e.g., [5,6,7]).
We can generalize Theorem 6 to an arbitrary v L 2 ( 0 , T ) , but we omit the details for conciseness.

4. Fractional Derivatives of the Mittag-Leffler Functions

Related to time-fractional differential equations, we introduce the Mittag-Leffler functions:
E α , β ( z ) = k = 0 z k Γ ( α k + β ) , α , β > 0 , z C .
It is known (e.g., [3]) that E α , β ( z ) is an entire function in z C .
Henceforth, let λ R be a constant and α > 0 and α N . We fix an arbitrary constant Λ 0 > 0 and we assume that λ > Λ 0 .
First, we prove the following proposition.
Proposition 12.
Let 0 < α < 2 . We fix constants T > 0 and λ > Λ 0 arbitrarily. Then
(i) 
We have E α , 1 ( λ t α ) 1 H α ( 0 , T ) and
t α ( E α , 1 ( λ t α ) 1 ) = λ E α , 1 ( λ t α ) , t > 0
and there exist constants C 1 = C 1 ( α , Λ 0 , T ) > 0 and C 2 = C 2 ( α ) > 0 such that
| E α , 1 ( λ t α ) | C 1 f o r 0 t T i f Λ 0 λ < 0 , C 2 1 + λ t α f o r t 0 i f λ 0 .
(ii) 
We have t E α , 2 ( λ t α ) t H α ( 0 , T ) ,
t α ( t E α , 2 ( λ t α ) t ) = λ t E α , 2 ( λ t α ) , t > 0 .
Furthermore, we can find constants C 1 = C 1 ( α , Λ 0 , T ) > 0 and C 2 = C 2 ( α ) > 0 such that
| E α , 2 ( λ t α ) | C 1 f o r 0 t T i f Λ 0 λ < 0 , C 2 1 + λ t α f o r t 0 i f λ 0 .
A direct proof for the inclusions in H α ( 0 , T ) is complicated and through the operator J α , we will provide simpler proofs.
Proof of Proposition 12.
(i) Since
E α , 1 ( λ t α ) = k = 0 ( λ ) k t α k Γ ( α k + 1 ) ,
where the series is uniformly convergent for 0 t T , termwise integration yields
J α ( E α , 1 ( λ t α ) ) = k = 0 ( λ ) k Γ ( α ) 0 t ( t s ) α 1 s α k Γ ( α k + 1 ) d s = k = 0 ( λ ) k Γ ( α ) Γ ( α ) Γ ( α k + α + 1 ) t α k + α = 1 λ j = 1 ( λ t α ) j Γ ( α j + 1 ) .
Here, we set j = k + 1 to change the indices of the summation. Therefore,
J α ( E α , 1 ( λ t α ) ) = 1 λ ( E α , 1 ( λ t α ) 1 ) .
Since t α v = ( J α ) 1 v for v H α ( 0 , T ) and E α , 1 ( λ t α ) L 2 ( 0 , T ) , we obtain
1 λ ( E α , 1 ( λ t α ) 1 ) H α ( 0 , T )
and
( J α ) 1 J α E α , 1 ( λ t α ) = 1 λ t α ( E α , 1 ( λ t α ) 1 ) ,
that is, t α ( E α , 1 ( λ t α ) 1 ) = λ E α , 1 ( λ t α ) .
On the other hand, for 0 < α < 2 and β > 0 , by Theorems 1.5 and 1.6 in [3], p. 35, we can find a constant C 0 = C 0 ( α , β ) > 0 such that
| E α , β ( λ t α ) | C 0 ( 1 + | λ | t α ) 1 β α exp ( | λ | 1 α t ) i f   Λ 0 λ < 0 , C 0 1 + λ t α i f   λ 0
for all t 0 . Hence, we can choose constants C 3 = C 3 ( α , β , Λ 0 , T ) > 0 and C 4 = C 4 ( α , β ) > 0 such that
| E α , β ( λ t α ) | C 3 f o r   0 t T   i f   Λ 0 λ < 0 , C 4 1 + λ t α f o r   t 0   i f   λ 0 .
In terms of (43), we can finish the proof of (40). Thus the proof of (i) is complete.
(ii) Since t E α , 2 ( λ t α ) t L 2 ( 0 , T ) , noting Γ ( 2 ) = 1 × Γ ( 1 ) = 1 , we have
J α ( t E α , 2 ( λ t α ) ) = 1 Γ ( α ) 0 t ( t s ) α 1 k = 0 s ( λ ) k s α k Γ ( α k + 2 ) d s = t k = 0 ( λ ) k Γ ( α k + α + 2 ) t α k + α = t λ j = 1 ( λ ) j Γ ( α j + 2 ) t α j = t λ j = 0 ( λ ) j t α j Γ ( α j + 2 ) 1 Γ ( 2 ) ,
that is,
J α ( t E α , 2 ( λ t α ) ) = 1 λ ( t E α , 2 ( λ t α ) t ) , 0 < t < T .
Therefore, t E α , 2 ( λ t α ) t H α ( 0 , T ) and
λ t E α , 2 ( λ t α ) = t α ( t E α , 2 ( λ t α ) t ) .
In terms of (43), the proof of the estimate is similar to part (i). Thus we can complete the proof of Proposition 12. □
We set
( B λ f ) ( t ) : = 0 t ( t s ) α 1 E α , α ( λ ( t s ) α ) f ( s ) d s .
Next, we show the following.
Proposition 13.
Let f L 2 ( 0 , T ) . Then
B λ f H α ( 0 , T ) a n d ( t α B λ f ) ( t ) = λ ( B λ f ) ( t ) + f ( t ) , 0 < t < T .
Moreover, there exist constants C 5 = C 5 ( α , Λ 0 , T ) > 0 and C 6 = C 6 ( α , T ) > 0 such that
B λ f H α ( 0 , T ) C 5 ( α , Λ 0 , T ) f L 2 ( 0 , T ) f o r   a l l   f L 2 ( 0 , T )   a n d   λ > Λ 0 , B λ f H α ( 0 , T ) C 6 ( α , T ) f L 2 ( 0 , T ) f o r   a l l   f L 2 ( 0 , T )   a n d   λ 0 .
The uniformity of estimate (46) on λ 0 plays an important role in Lemma 5 (ii) in Section 6. Such uniformity can be derived from the complete monotonicity of E α , 1 ( λ t α ) , which is characteristic only for 0 < α < 1 , see (47) below.
Here, we prove by using t α = ( J α ) 1 in H α ( 0 , T ) , although another proof by Theorem 5 is possible.
Proof. 
Since, in view of (43), we can choose constants C 7 = C 7 ( α , Λ 0 , T ) > 0 and C 8 = C 8 ( α , T ) > 0 such that
| E α , α ( λ ( t s ) α ) | C 7 ( α , Λ 0 , T ) f o r   0 t T   a n d   λ > Λ 0 , C 8 ( α , T ) f o r   0 t T   a n d   λ 0 ,
we estimate
0 t ( t s ) α 1 E α , α ( λ ( t s ) α ) f ( s ) d s C 0 t ( t s ) α 1 | f ( s ) | d s , 0 < t < T .
Here and henceforth, C > 0 denotes a generic constant which depends on α , Λ 0 and T when we consider λ > Λ 0 , and does not depend on Λ 0 if λ 0 .
Hence, the Young inequality yields that 0 t ( t s ) α 1 | f ( s ) | d s L 2 ( 0 , T ) , and so, B λ f L 2 ( 0 , T ) .
Now,
J α ( B λ f ) ( t ) = 1 Γ ( α ) 0 t ( t s ) α 1 ( B λ f ) ( s ) d s = 1 Γ ( α ) 0 t ( t s ) α 1 0 s ( s ξ ) α 1 E α , α ( λ ( s ξ ) α ) f ( ξ ) d ξ d s = 0 t f ( ξ ) 1 Γ ( α ) ξ t ( t s ) α 1 ( s ξ ) α 1 E α , α ( λ ( s ξ ) α ) d s d ξ .
Here,
1 Γ ( α ) ξ t ( t s ) α 1 ( s ξ ) α 1 E α , α ( λ ( s ξ ) α ) d s = ξ t 1 Γ ( α ) ( t s ) α 1 ( s ξ ) α 1 k = 0 ( λ ) k ( s ξ ) α k Γ ( α k + α ) d s = 1 Γ ( α ) k = 0 ( λ ) k Γ ( α k + α ) ξ t ( t s ) α 1 ( s ξ ) α k + α 1 d s = 1 Γ ( α ) k = 0 ( λ ) k Γ ( α k + α ) Γ ( α ) Γ ( α k + α ) Γ ( α k + 2 α ) ( t ξ ) α k + 2 α 1 = 1 λ ( t ξ ) α 1 j = 1 ( λ ( t ξ ) α ) j Γ ( α j + α ) = 1 λ ( t ξ ) α 1 E α , α ( λ ( t ξ ) α ) 1 Γ ( α ) .
Therefore, we have
J α ( B λ f ) ( t ) = 1 λ ( B λ f ) ( t ) + 1 λ 0 t ( t ξ ) α 1 1 Γ ( α ) f ( ξ ) d ξ = 1 λ ( B λ f ) ( t ) + 1 λ ( J α f ) ( t ) , 0 < t < T ,
that is,
( B λ f ) ( t ) = λ J α ( B λ f ) ( t ) + ( J α f ) ( t ) = J α ( λ B λ f + f ) ( t ) , 0 < t < T .
Hence, B λ f J α L 2 ( 0 , T ) = H α ( 0 , T ) by λ B λ f + f L 2 ( 0 , T ) . Therefore, we apply t α = ( J α ) 1 to reach
t α ( B λ f ) = λ B λ f + f i n   ( 0 , T ) .
On the other hand, termwise differentiation of the power series of E α , 1 ( z ) yields
d d t E α , 1 ( λ t α ) = λ t α 1 E α , α ( λ t α ) .
Furthermore, we note the complete monotonicity:
E α , 1 ( λ t α ) > 0 , d d t E α , 1 ( λ t α ) 0 , t 0
(e.g., Gorenflo, Kilbas, Mainardi and Rogosin [1]). Therefore, for λ 0 , we have
0 T | λ t α 1 E α , α ( λ t α ) | d t = 0 T λ t α 1 E α , α ( λ t α ) d t
= 0 T d d t E α , 1 ( λ t α ) d t = 1 E α , 1 ( λ T α ) 1 .
For Λ 0 < λ < 0 , by (43), we have
0 T | λ t α 1 E α , α ( λ t α ) | d t | Λ 0 | C 0 T t α 1 d t ,
where the constant C > 0 depends on α , Λ 0 and T. Consequently,
λ s α 1 E α , α ( λ s α ) L 1 ( 0 , T ) C f o r   λ > Λ 0 .
Applying the Young inequality in (44), we obtain
λ B λ f L 2 ( 0 , T ) λ s α 1 E α , α ( λ s α ) L 1 ( 0 , T ) f L 2 ( 0 , T ) C f L 2 ( 0 , T ) .
Therefore,
B λ f H α ( 0 , T ) C t α ( B λ f ) L 2 ( 0 , T ) = C λ B λ f + f L 2 ( 0 , T ) C ( λ B λ f L 2 ( 0 , T ) + f L 2 ( 0 , T ) ) C f L 2 ( 0 , T ) f o r   a l l   f L 2 ( 0 , T ) .
Thus the proof of Proposition 13 is complete. □
We close this section with a lemma which is used in Section 6.
Lemma 3.
Let 0 < β < α and λ > 0 . Fixing γ > 1 2 , we consider t β : γ H ( 0 , T ) β γ H ( 0 , T ) . Then
(i) 
t α 1 E α , α ( λ t α ) L 1 ( 0 , T ) γ H ( 0 , T ) .
(ii) 
t β ( t α 1 E α , α ( λ t α ) ) = t α β 1 E α , α β ( λ t α ) L 1 ( 0 , T ) .
(iii) 
t β ( t α 1 E α , α ( λ t α ) v ) = ( t α β 1 E α , α β ( λ t α ) v ) f o r e a c h v L 2 ( 0 , T ) .
Proof. 
(i) By (42), we see that | t α 1 E α , α ( λ t α ) | C t α 1 for t > 0 and t α 1 E α , α ( λ t α ) L 1 ( 0 , T ) . The embedding L 1 ( 0 , T ) γ H ( 0 , T ) is derived by the Sobolev embedding.
(ii) By α β > 0 , we can similarly verify that t α β 1 E α , α β ( λ t α ) L 1 ( 0 , T ) . Therefore, Proposition 10 yields
J β ( t α β 1 E α , α β ( λ t α ) ) ( t ) = 1 Γ ( β ) 0 t ( t s ) β 1 s α β 1 E α , α β ( λ s α ) d s .
By β > 0 and α β > 0 , we apply the formula (1.100) in [3], p. 25, which can be also directly verified by the expansion of the power series of E α , α β ( λ s α ) so that
1 Γ ( β ) 0 t ( t s ) β 1 s α β 1 E α , α β ( λ s α ) d s = t α 1 E α , α ( λ t α ) .
Therefore, J β ( t α β 1 E α , α β ( λ t α ) ) ( t ) = t α 1 E α , α ( λ t α ) . By t β = ( J β ) 1 in L 1 ( 0 , T ) , we see part (ii).
Since t α 1 E α , α ( λ t α ) , t β ( t α 1 E α , α ( λ t α ) are both in L 1 ( 0 , T ) , part (iii) follows from Theorem 5. Thus the proof of Lemma 3 is complete. □

5. Initial Value Problems for Fractional Ordinary Differential Equations

Relevant formulations of initial value problems for time-fractional ordinary differential equations is our main issue in this section, in order to treat not-so-smooth data. To keep the compact descriptions, we are restricted to a simple linear fractional ordinary differential equation. The treatments are similar to those of Chapter 3 in [5]. We formulate an initial value problem as follows:
t α ( u a ) ( t ) = λ u ( t ) + f ( t ) , 0 < t < T , u a H α ( 0 , T ) .
As is mentioned in Section 2, if α > 1 2 , then u a H α ( 0 , T ) 0 C 1 [ 0 , T ] and we know that u ( t ) a is continuous and u ( 0 ) = a . Thus, in the case of α > 1 2 , if u satisfies (49), then a usual initial condition u ( 0 ) = a is satisfied.
Our formulation (49) coincides with a conventional formulation D t α ( u ( t ) a ) = f ( t ) of an initial value problem, provided that we can suitably specify the regularity of u. We emphasize that we always attach fractional derivative operators with the domains such as H α ( 0 , T ) or α H ( 0 , T ) with α 0 , which means that our approach is a typical operator theoretic formulation, for example, similarly to that one which prohibits us from considering the Laplacian Δ in Ω R d not associated with the domain. In other words, the operator Δ with the domain { u H 2 ( Ω ) ; u H 0 1 ( Ω ) } is different from Δ with the domain { u H 0 2 ( Ω ) ; u · ν = 0 o n   Ω } . Here, ν = ν ( x ) is the unit outward normal vector to Ω .
In particular, if we consider t α with the domain H α ( 0 , T ) and both sides of (49) in L 2 ( 0 , T ) , we remark that the equality t α ( u a ) = t α u t α a does not make any sense for α > 1 2 because a constant function a is not in H α ( 0 , T ) . On the other hand, if we consider t α with the domain L 2 ( 0 , T ) , then we can justify
t α ( u a ) = t α u t α a = t α u 1 Γ ( 1 α ) t α
for any u L 2 ( 0 , T ) .
Now we can prove the following theorem.
Theorem 7.
Let f L 2 ( 0 , T ) . Then there exists a unique solution u = u ( t ) to initial value problem (49). Moreover,
u ( t ) = a E α , 1 ( λ t α ) + 0 t ( t s ) α 1 E α , α ( λ ( t s ) α ) f ( s ) d s , 0 < t < T .
Formula (50) itself is well known (e.g., (3.1.34), [2], p. 141) for f in some classes. We can refer also to Gorenflo and Mainardi [15], Gorenflo, Mainardi and Srivastva [16], Gorenflo and Rutman [17] and Luchko and Gorenflo [18].
On the other hand, we should understand that (50) holds in the sense that both sides are in H α ( 0 , T ) for each f L 2 ( 0 , T ) .
Sketch of Proof. 
We see that (49) is equivalent to
J α ( u a ) = λ u + f i n   L 2 ( 0 , T ) a n d u a H α ( 0 , T )
and also to
u = a λ J α u + J α f i n   H α ( 0 , T ) ,
because J α J α ( u a ) = J α ( λ u + f ) .
We conclude that J α : L 2 ( 0 , T ) L 2 ( 0 , T ) is a compact operator because the embedding H α ( 0 , T ) H α ( 0 , T ) L 2 ( 0 , T ) is compact (e.g., [11]). On the other hand, we apply the generalized Gronwall inequality (e.g., Lemma A.2 in [5]) to u = λ J α u in ( 0 , T ) , that is,
u ( t ) = λ Γ ( α ) 0 t ( t s ) α 1 u ( s ) d s , 0 < t < T .
Then we obtain u ( t ) = 0 for 0 < t < T . Therefore, the Fredholm alternative yields the unique existence of u satisfying (51).
Finally, we define u ˜ ( t ) by
u ˜ ( t ) a : = a ( E α , 1 ( λ t α ) 1 ) + 0 t ( t s ) α 1 E α , α ( λ ( t s ) α ) f ( s ) d s = a ( E α , 1 ( λ t α ) 1 ) + ( B λ f ) ( t ) , 0 < t < T .
Here, B λ is defined by (44).
By Propositions 12 and 13, it follows that u ˜ ( t ) satisfies (49). Thus the proof of Theorem 7 is complete. □
We can discuss an initial value problem for a multi-term time-fractional ordinary differential equation in the same way:
t α ( u a ) + k = 1 N c k t α k ( u a ) = λ u + f ( t ) , u a H α ( 0 , T ) ,
where c 1 , . . . , c N 0 , 0 < α 1 < < α N < α < 1 . Theorem 1 implies that J α t α k = J α J α k = J α α k . By Proposition 2, we can obtain the equivalent equation
u a = k = 1 N c k J α α k ( u a ) λ J α u + J α f
and we apply the Fredholm alternative to prove the unique existence of the solution, but we omit the details.
We further consider an initial value problem for f α H ( 0 , T ) :
t α ( u a ) ( t ) = λ u ( t ) + f ( t ) , 0 < t < T , u a L 2 ( 0 , T ) .
Similarly to Theorem 7, we prove the well-posedness of (53) for f α H ( 0 , T ) .
Theorem 8.
Let 0 < α < 1 and α N . For f α H ( 0 , T ) , there exists a unique solution u a L 2 ( 0 , T ) to (53). Moreover, we can choose a constant C > 0 such that
u a L 2 ( 0 , T ) C ( | a | + f α H ( 0 , T ) )
for all a R and f α H ( 0 , T ) .
Example 6.
We consider
t α ( u a ) ( t ) = f ( t ) , 0 < t < T
with f ( t ) = Γ ( 1 + β ) Γ ( 1 α + β ) t β α , where β > 1 2 . Then, u ( t ) = a + t β satisfies (53) with λ = 0 by (28). However, if 1 2 < β < 0 and 1 2 < β α , then u , f L 2 ( 0 , T ) , but u ( t ) does not satisfy lim t 0 u ( t ) = a .
Proof of Theorem 8.
Setting v : = u a L 2 ( 0 , T ) , we rewrite (53) as
( J α ) 1 v = λ v λ a + f , v L 2 ( 0 , T ) .
By the definition given in (24) of t α , Equation (53) is equivalent to
v = λ J α v + J α ( λ a + f ) i n   L 2 ( 0 , T ) .
We set g : = J α ( f λ a ) L 2 ( 0 , T ) and P v : = λ J α v . Then we see that the solution v = u a is a fixed point of P, that is, v = P v + g .
First, the operator P : L 2 ( 0 , T ) L 2 ( 0 , T ) is a compact operator. Indeed, Proposition 9 (iii) yields that J α v = J α v for v L 2 ( 0 , T ) , and J α : L 2 ( 0 , T ) H α ( 0 , T ) is an isomorphism by Proposition 2. Thus, J α is a bounded operator from L 2 ( 0 , T ) to H α ( 0 , T ) . Since the embedding H α ( 0 , T ) L 2 ( 0 , T ) is compact, we see that J α : L 2 ( 0 , T ) L 2 ( 0 , T ) is a compact operator (e.g., [11]).
Next, we have to prove that v = 0 in L 2 ( 0 , T ) from the assumption that v = P v in ( 0 , T ) . Then, since
J α v ( t ) = J α v ( t ) = 1 Γ ( α ) 0 t ( t s ) α 1 v ( s ) d s
by v L 2 ( 0 , T ) , we obtain
v ( t ) = λ Γ ( α ) 0 t ( t s ) α v ( s ) d s , 0 < t < T .
Hence,
| v ( t ) | C 0 t ( t s ) α 1 | v ( s ) | d s f o r   a l m o s t   a l l   t ( 0 , T ) .
The generalized Gronwall inequality (e.g., Lemma A.2 in [5]) implies that v ( t ) = 0 for almost all t ( 0 , T ) . Therefore, the Fredholm alternative yields the unique existence of a fixed point of v = P v + g in L 2 ( 0 , T ) . The estimate of v follows from the application of the generalized Gronwall inequality to (55) and Proposition 9 (ii): J α f L 2 ( 0 , T ) C f α H ( 0 , T ) . Thus the proof of Theorem 8 is complete. □
Remark 1.
Now we compare formulation (49) with (53) for f L 2 ( 0 , T ) .
(a) For f L 2 ( 0 , T ) , formulations (49) and (53) are equivalent.
Indeed, we immediately see that (49) implies (53). Conversely, let u satisfy (53). Then the second condition in (53) yields u L 2 ( 0 , T ) , and the first equation in (53) concludes that t α ( u a ) L 2 ( 0 , T ) . By Proposition 5, we see that u a H α ( 0 , T ) , which means that u satisfies (49).
(b) In formulation (49), as we remarked, we should not decompose t α ( u a ) = t α u t α a , which is wrong for α 1 2 because a H α ( 0 , T ) . We further consider this issue for 0 < α < 1 2 . For clearness, only here by t 0 α we denote t α as operator with the domain D ( t 0 α ) = L 2 ( 0 , T ) , and we use the same notation t α with the domain H α ( 0 , T ) .
By means of (28) with β = 0 , for 0 < α < 1 2 , we see that 1 H α ( 0 , T ) and t 0 α 1 = t α 1 = 1 Γ ( 1 α ) t α and also that u a H α ( 0 , T ) if and only if u a H α ( 0 , T ) . Therefore, we see:
If 0 < α < 1 2 , then (49) is equivalent to
t α u = λ u + a Γ ( 1 α ) t α + f ( t ) i n   L 2 ( 0 , T ) , u H α ( 0 , T ) . ( 49 ) ´
Noting 1 L 2 ( 0 , T ) and using (28) with β = 0 , for all α ( 0 , 1 ) , we can verify that (53) is equivalent to
t 0 α u = λ u + a Γ ( 1 α ) t α + f ( t ) i n   α H ( 0 , T ) , u L 2 ( 0 , T ) . ( 53 ) ´
Assuming that f L 2 ( 0 , T ) and 0 < α < 1 2 , we can conclude that (49)’ is equivalent to (53)’. Indeed, a t α Γ ( 1 α ) L 2 ( 0 , T ) , and so a t α Γ ( 1 α ) + f L 2 ( 0 , T ) . Therefore, by u L 2 ( 0 , T ) , the first equation in (53)’ yields t 0 α u L 2 ( 0 , T ) , which means that u H α ( 0 , T ) and t 0 α u = t α u in ( 0 , T ) . Thus, (49)’ and (53)’ are equivalent, provided that 0 < α < 1 2 .
We emphasize that for 1 2 α < 1 , we cannot make any reformulations of (49) similar to (49)’ by decomposing u a into u and a . We can discuss similar reformulations also for initial boundary value problems for fractional partial differential equations.
Now we take the widest domain of t α according to classes in time of functions under consideration, and we do not distinguish, for example, t 0 α from t α , because there is no fear of confusion.
Example 7.
Let α > 1 2 and let f ( t ) : = δ t 0 ( t ) : the Dirac delta function at t 0 ( 0 , T ) . In particular, we can prove that | φ ( t 0 ) | C φ C [ 0 , T ] , and so δ t 0 ( C [ 0 , T ] ) :
( C [ 0 , T ] ) < δ t 0 , ψ > C [ 0 , T ] = ψ ( t 0 )
for any ψ C [ 0 , T ] . Moreover, by the Sobolev embedding, we can see that α H ( 0 , T ) H α ( 0 , T ) C [ 0 , T ] by α > 1 2 . Therefore, δ t 0 α H ( 0 , T ) , and
α H ( 0 , T ) < δ t 0 ψ > α H ( 0 , T ) = ψ ( t 0 )
defines a bounded linear functional on α H ( 0 , T ) .
This δ t 0 describes an impulsive source term in fractional diffusion. We will search for the representation of the solution to (53) with f = δ t 0 and a = 0 :
t α u = λ u + δ t 0 i n   α H ( 0 , T ) , u L 2 ( 0 , T ) .
Simulating a solution formula for
d t α u = λ u + f ( t ) , u ( 0 ) = 0
(e.g., [2], p. 141), we can give a candidate for the solution, which is formally written by
u ( t ) = 0 t ( t s ) α 1 E α , α ( λ ( t s ) α ) δ t 0 ( s ) d s , 0 < t < T .
Our formal calculation suggests that
u ( t ) = 0 , 0 < t t 0 , ( t t 0 ) α 1 E α , α ( λ ( t t 0 ) α ) , t 0 < t T .
Now we will verify that u ( t ) given by (59) is the solution to (56). First, it is clear that u L 1 ( 0 , T ) . Then we will verify u = λ J α u + J α δ t 0 in L 2 ( 0 , T ) .
Since u L 2 ( 0 , T ) , we apply Proposition 10 to have
J α u ( t ) = J α u ( t ) = 1 Γ ( α ) 0 t ( t s ) α 1 u ( s ) d s .
For 0 < t < t 0 , by (59), we see J α u ( t ) = 0 . Next, for t 0 t T , we have
λ J α u ( t ) = λ Γ ( α ) 0 t ( t s ) α 1 ( s t 0 ) α 1 E α , α ( λ ( s t 0 ) α ) d s = 1 Γ ( α ) t 0 t k = 0 ( t s ) α 1 ( s t 0 ) α 1 ( λ ) k + 1 ( s t 0 ) α k Γ ( α k + α ) d s = 1 Γ ( α ) k = 0 t 0 t ( t s ) α 1 ( s t 0 ) α k + α 1 d s ( λ ) k + 1 Γ ( α k + α ) = 1 Γ ( α ) k = 0 0 t t 0 ( t t 0 η ) α 1 η α k + α 1 d η ( λ ) k + 1 Γ ( α k + α ) = 1 Γ ( α ) k = 0 Γ ( α ) Γ ( α k + α ) Γ ( α + α k + α ) ( t t 0 ) α k + 2 α 1 ( λ ) k + 1 Γ ( α k + α )
= ( t t 0 ) α 1 k = 0 ( λ ( t t 0 ) α ) k + 1 Γ ( α ( k + 1 ) + α ) = ( t t 0 ) α 1 j = 1 ( λ ( t t 0 ) α ) j Γ ( α j + α ) .
For calculations of J α u ( t ) , we can apply the formula (1.100) ([3], p. 25), but we take a direct way here.
On the other hand, by the definition of the dual operator J α , setting v 0 : = J α δ t 0 , we have
( v 0 , ψ ) L 2 ( 0 , T ) = α H ( 0 , T ) < δ t 0 , J α ψ > α H ( 0 , T ) f o r   a l l   ψ L 2 ( 0 , T ) .
Since J α ψ α H ( 0 , T ) C [ 0 , T ] by α > 1 2 and Proposition 8, it follows that v 0 satisfies
α H ( 0 , T ) < δ t 0 , J α ψ > α H ( 0 , T ) = ( J α ψ ) ( t 0 ) = 1 Γ ( α ) t 0 T ( s t 0 ) α 1 ψ ( s ) d s .
Therefore,
1 Γ ( α ) t 0 T ( s t 0 ) α 1 ψ ( s ) d s = ( v 0 , ψ ) L 2 ( 0 , T ) = 0 t 0 v 0 ( s ) ψ ( s ) d s + t 0 T v 0 ( s ) ψ ( s ) d s
for all ψ L 2 ( 0 , T ) . Choosing ψ L 2 ( 0 , T ) satisfying ψ = 0 in ( t 0 , T ) , we obtain
0 t 0 v 0 ( s ) ψ ( s ) d s = 0 ,
and so v 0 ( s ) = 0 for 0 < s t 0 . Hence,
J α δ t 0 ( s ) = v 0 ( s ) = 0 , 0 < s t 0 , ( s t 0 ) α 1 Γ ( α ) , t 0 < s < T .
In other words,
t α v 0 = δ t 0 ,
where v 0 L 2 ( 0 , T ) is defined by (61).
Consequently, since
( t t 0 ) α 1 Γ ( α ) + ( t t 0 ) α 1 k = 1 ( λ ( t t 0 ) α ) k Γ ( α k + α ) = ( t t 0 ) α 1 E α , α ( λ ( t t 0 ) α ) ,
in terms of (60) and (61), we reach
λ J α u ( t ) + J α δ t 0 ( t ) = 0 , 0 < t t 0 , ( t t 0 ) α 1 E α , α ( λ ( t t 0 ) α ) , t 0 < t < T .
By (59), we verify
u ( t ) = λ J α u ( t ) + J α δ t 0 ( t ) .
Thus we verify that u ( t ) given by (59) is the unique solution to (56).
We recall (44):
( B λ f ) ( t ) : = 0 t ( t s ) α 1 E α , α ( λ ( t s ) α ) f ( s ) d s , 0 < t < T , f o r   f L 2 ( 0 , T ) .
By Proposition 13, we know that B λ : L 2 ( 0 , T ) H α ( 0 , T ) is a bounded operator.
We close this section with the following proposition.
Proposition 14. (Representation of solution to (53) with f α H ( 0 , T ) ).
Let λ > Λ 0 be fixed. The operator B λ can be extended to S λ : α H ( 0 , T ) L 2 ( 0 , T ) as follows: For f α H ( 0 , T ) , there exists a sequence f n L 2 ( 0 , T ) , n N such that f n f in α H ( 0 , T ) . Then
lim m , n B λ f n B λ f m L 2 ( 0 , T ) = 0
and l i m n B λ f n is unique in L 2 ( 0 , T ) independently of choices of sequences f n , n N such that f n f in α H ( 0 , T ) . Hence, setting
S λ f : = lim n B λ f n i n   L 2 ( 0 , T ) ,
we have
t α ( S λ f ) = λ S λ f + f i n   α H ( 0 , T ) .
Proof. 
First, since L 2 ( 0 , T ) is dense in α H ( 0 , T ) , we can choose a sequence f n L 2 ( 0 , T ) , n N such that lim n f n = f in L 2 ( 0 , T ) .
Verification of (63). By Proposition 13, we see
B λ f = λ J α B λ f + J α f
and
B λ f L 2 ( 0 , T ) C J α f L 2 ( 0 , T ) f o r   f L 2 ( 0 , T ) .
In view of (66), we obtain
B λ f n B λ f m L 2 ( 0 , T ) C J α f n J α f m L 2 ( 0 , T ) .
Proposition 9 (ii) yields
B λ f n B λ f m L 2 ( 0 , T ) C f n f m α H ( 0 , T ) .
Since lim m , n f n f m α H ( 0 , T ) = 0 , we see that B λ f n , n N converge in L 2 ( 0 , T ) .
Similarly we can prove that lim n B λ f n is determined independently of choices of sequences f n , n N such that lim n f n = f in α H ( 0 , T ) . Therefore, S λ f is well-defined for f L 2 ( 0 , T ) .
Verification of (64). For an approximating sequence f n L 2 ( 0 , T ) , n N such that f n f in α H ( 0 , T ) , we have
B λ f n = λ J α B λ f n + J α f n i n   L 2 ( 0 , T ) f o r   n N .
Since lim n B λ f n = S λ f in L 2 ( 0 , T ) , letting n , we obtain
S λ f = λ lim n J α B λ f n + lim n J α f n i n   L 2 ( 0 , T ) .
We apply Proposition 9 (ii) to see lim n J α f n = J α f in L 2 ( 0 , T ) by lim n f n = f in α H ( 0 , T ) . Finally, Proposition 9 (iii) implies that J α | L 2 ( 0 , T ) : L 2 ( 0 , T ) L 2 ( 0 , T ) is bounded, and we have
lim n J α B λ f n = lim n J α B λ f n = J α S λ f i n   L 2 ( 0 , T ) .
Therefore, we reach
S λ f = λ J α S λ f + J α f i n   L 2 ( 0 , T ) .
By the definition given in (24) of t α , this means
t α ( S λ f ) = λ S λ f + f i n   α H ( 0 , T ) .
Thus the verification of (64) is complete, and the proof of Proposition 14 is finished. □
We can discuss more about the representation formula of solution to (53) with f α H ( 0 , T ) in terms of convolution operators, but we will postpone this to a future work.

6. Initial Boundary Value Problems for Fractional Partial Differential Equations: Selected Topics

On the basis of t α defined in Section 2, we construct a feasible framework also for initial boundary value problems. We recall that an elliptic operator A is defined by (1), and we assume all the conditions as described in Section 1 on the coefficients a i j , b j , c C 1 ( Ω ¯ ) .
Here, we mostly consider the case where α < 1 , but cases where α > 1 can be formulated and studied similarly.
By ν = ( ν 1 , . . . . , ν d ) , we denote the outward unit normal vector to Ω at x and set
ν A v = i , j = 1 d a i j ( j v ) ν i o n   Ω .
We define an operator A in L 2 ( Ω ) by
A v ( x ) = A v ( x ) , x Ω , D ( A ) = { v H 2 ( Ω ) ; v | Ω = 0 } = H 2 ( Ω ) H 0 1 ( Ω ) .
Here, v | Ω = 0 is understood as the sense of the trace (e.g., [11]).
We can similarly discuss other boundary conditions, for example, D ( A ) = { v H 2 ( Ω ) ; ν A v + σ ( x ) v = 0 o n   Ω } with fixed function σ ( x ) , but we concentrate on the homogeneous Dirichlet boundary condition u | Ω = 0 .
We formulate the initial boundary value problem by
t α ( u ( x , t ) a ( x ) ) + A u ( x , t ) = F ( x , t ) i n L 2 ( 0 , T ; L 2 ( Ω ) )
and
u a H α ( 0 , T ; L 2 ( Ω ) ) .
We emphasize that we do not adopt formulation (2).
The formulation in (68) and (69) corresponds to (49) for an initial value problem for a time-fractional ordinary differential equation. The term A u in Equation (68) means that u ( · , t ) D ( A ) = H 2 ( Ω ) H 0 1 ( Ω ) for almost all t ( 0 , T ) , that is,
u ( · , t ) | Ω = 0 f o r   a l m o s t   a l l   t ( 0 , T ) .
In other words, the domain of A describes the boundary condition, which is a conventional way in treating the classical partial differential equations. Like fractional ordinary differential equations in Section 5, we understand that (69) means the initial condition.
We first present a basic well-posedness result for (68) and (69):
Theorem 9.
Let 0 < α < 1 . Let a H 0 1 ( Ω ) and F L 2 ( 0 , T ; L 2 ( Ω ) ) . Then there exists a unique solution u = u ( x , t ) to (68) - (69) such that u a H α ( 0 , T ; L 2 ( Ω ) ) and u L 2 ( 0 , T ; H 2 ( Ω ) H 0 1 ( Ω ) ) . Moreover, there exists a constant C > 0 such that
u a H α ( 0 , T ; L 2 ( Ω ) ) + u L 2 ( 0 , T ; H 2 ( Ω ) H 0 1 ( Ω ) ) C ( a H 0 1 ( Ω ) + F L 2 ( 0 , T ; L 2 ( Ω ) ) )
for all a H 0 1 ( Ω ) and F L 2 ( 0 , T ; L 2 ( Ω ) ) .
Here, we remark that
v H α ( 0 , T ; L 2 ( Ω ) ) : = t α v L 2 ( 0 , T ; L 2 ( Ω ) ) f o r   v H α ( 0 , T ; L 2 ( Ω ) ) .
Unique existence results of solutions are known according to several formulations of initial boundary value problems. In Sakamoto and Yamamoto [19], in the case of a symmetric A where b j = 0 for 1 j d in (1), the unique existence is proved by means of the Fourier method, but the class of solutions is not the same as here. In the case where b j for j = 1 , . . . , d are not necessarily zero, assuming that the initial value a is zero, Theorem 9 is proved in Gorenflo, Luchko and Yamamoto [8]. In both [8,19], it is assumed that all the coefficients are independent of t and and c = c ( x ) 0 for x Ω . The work of Kubica, Ryszewska and Yamamoto [5] proves Theorem 9 in a general case where a i j , b j , c depends both on x and t without the extra assumption c 0 . In other words, Theorem 9 is a special case of Theorem 4.2 in [5]. Furthermore, there have been other works on well-posedness for initial boundary value problems and we are restricted to some of them: Bajlekova [20], Kubica and Yamamoto [6], Luchko [21,22], Luchko and Yamamoto [23] and Zacher [7]. See also Prüss [24] for a monograph on related integral equations and a recent book Jin [25] which mainly studies the symmetric A. For further references up to 2019, the handbooks [26] edited by A.Kochubei and Y. Luchko are helpful. Most of the above works discuss the case of F L 2 ( 0 , T ; L 2 ( Ω ) ) .
Next, we consider less regular F and a in x. To this end, we introduce Sobolev spaces of negative orders in x. Similarly to the triple α H ( 0 , T ) L 2 ( Ω ) α H ( 0 , T ) as is explained in Section 2, we introduce the dual space ( H 0 1 ( Ω ) ) of H 0 1 ( Ω ) by identifying the dual space ( L 2 ( Ω ) ) with L 2 ( Ω ) :
H 0 1 ( Ω ) L 2 ( Ω ) ( H 0 1 ( Ω ) ) = : H 1 ( Ω )
(e.g., [13]). For less regular F and a in the x-variable, we know the following.
Theorem 10.
Let 0 < α < 1 . For the coefficients of A, we assume the same conditions as in Theorem 9. Let a L 2 ( Ω ) and F L 2 ( 0 , T ; H 1 ( Ω ) ) . Then there exists a unique solution u = u ( x , t ) to (68) and (69) such that u a H α ( 0 , T ; H 1 ( Ω ) ) and u L 2 ( 0 , T ; H 0 1 ( Ω ) ) . Moreover, there exists a constant C > 0 such that
u a H α ( 0 , T ; H 1 ( Ω ) ) + u L 2 ( 0 , T ; H 0 1 ( Ω ) ) C ( a L 2 ( Ω ) + F L 2 ( 0 , T ; H 1 ( Ω ) ) )
for all a L 2 ( Ω ) and F L 2 ( 0 , T ; H 1 ( Ω ) ) .
In Theorem 10, we consider both sides of (68) in L 2 ( 0 , T ; H 1 ( Ω ) ) . The proof of Theorem 10 is found in [5] and see also [6].
Remark 2.
As for the general α > 0 , by means of the space defined by (17), we can formulate the initial boundary value problem as follows: For α > 1 , we set α = m + σ with m N and 0 < σ 1 . Then, for α > 1 , we formulate an initial boundary value problem by
t α u k = 0 m a k t k k ! = A u + F ( x , t ) , x Ω , 0 < t < T , u k = 0 m a k t k k ! ( x , · ) H α ( 0 , T ) f o r   a l m o s t   a l l   x Ω .
For σ > 1 2 , we can interpret the second condition as the usual initial conditions. More precisely,
u k = 0 m a k t k k ! ( x , · ) H α ( 0 , T )
if and only if
k u t k ( · , 0 ) = a k , k = 0 , 1 , . . . , m 1 , m u t m ( x , · ) a m H σ ( 0 , T ) .
In this section, we pick up five topics and apply the results in Section 2, Section 3 and Section 4. We postpone general and complete descriptions to a future work. Moreover, we are limited to the following A:
A v ( x ) = j = 1 d j ( a i j ( x ) j v ( x ) ) + j = 1 d b j ( x ) j v + c ( x ) v , x Ω
for v D ( A ) : = H 2 ( Ω ) H 0 1 ( Ω ) with
a i j = a j i C 1 ( Ω ¯ ) , b j C 1 ( Ω ¯ ) f o r   1 i , j d , c C 1 ( Ω ¯ ) , 0 o n   Ω .

6.1. Mild Solution and Strong Solution

We define an operator L as a symmetric part of A by
L v ( x ) = j = 1 d j ( a i j ( x ) j v ( x ) ) + c ( x ) v , x Ω , D ( L ) = H 2 ( Ω ) H 0 1 ( Ω ) .
Then there exist eigenvalues of L and, according to the multiplicities, we can arrange all the eigenvalues as
0 < λ 1 λ 2 λ 3 .
Here, by c 0 in Ω , we can prove that λ 1 > 0 . Moreover, we can choose eigenfunctions φ n for λ n , n N such that { φ n } n N is an orthonormal basis in L 2 ( Ω ) . Henceforth, ( · , · ) and · L 2 ( Ω ) denote the scalar product and the norm in L 2 ( Ω ) , respectively, and we write ( · , · ) L 2 ( Ω ) to specify the space, if necessary. Thus, L φ n = λ n φ n , φ n L 2 ( Ω ) = 1 and ( φ n φ m ) = 0 if n m .
For γ R , we can define a fractional power L γ of L by
( L γ v ) ( x ) = n = 1 λ n γ ( v , φ n ) φ n ( x ) , D ( L γ ) : = L 2 ( Ω ) i f   γ 0 , v L 2 ( Ω ) ; n = 1 λ n 2 γ | ( v , φ n ) | 2 < i f   γ > 0 .
We set
v D ( L γ ) : = n = 1 λ n 2 γ | ( v , φ n ) | 2 1 2 i f   γ > 0 .
Then it is known that
D ( L 1 2 ) = H 0 1 ( Ω ) , C 1 v H 0 1 ( Ω ) L 1 2 v L 2 ( Ω ) C v H 0 1 ( Ω ) , v H 0 1 ( Ω ) .
Here, C > 0 is independent of choices of v H 0 1 ( Ω ) .
We further define operator S ( t ) and K ( t ) from L 2 ( Ω ) to L 2 ( Ω ) by
S ( t ) a : = n = 1 E α , 1 ( λ n t α ) ( a , φ n ) φ n
and
K ( t ) a : = n = 1 t α 1 E α , α ( λ n t α ) ( a , φ n ) φ n
for all a L 2 ( Ω ) . Then, for 0 γ 1 , we can find constants C 1 > 0 and C 2 = C 2 ( γ ) > 0 such that
S ( t ) a L 2 ( Ω ) C 1 a L ( Ω ) , L γ K ( t ) a L 2 ( Ω ) C 2 t α ( 1 γ ) 1 a L 2 ( Ω ) f o r   t > 0   a n d   a L 2 ( Ω ) .
The proof of (75) is direct by the definition of S ( t ) and K ( t ) and can be found in [8].
Here and henceforth, we write u ( t ) : = u ( · , t ) as a mapping from ( 0 , T ) to L 2 ( Ω ) . The proof of (75) can be found in [8].
We can show the following proposition.
Proposition 15.
Let a H 0 1 ( Ω ) and F L 2 ( 0 , T ; L 2 ( Ω ) ) , and let (71) hold. The following are equivalent:
(i) 
u L 2 ( 0 , T ; H 2 ( Ω ) H 0 1 ( Ω ) ) satisfies (68) and (69).
(ii) 
u L 2 ( 0 , T ; L 2 ( Ω ) ) satisfies
u ( t ) = S ( t ) a + 0 t K ( t s ) j = 1 d b j j u ( · , s ) d s + 0 t K ( t s ) F ( s ) d s , 0 < t < T .
Equation (76) corresponds to formula (50) for an initial value problem for fractional ordinary differential equations.
According to the parabolic equation (e.g., Pazy [27]), the solution guaranteed by Theorem 9 is called a strong solution of the fractional partial differential equation, while the solution to an integrated Equation (76) is called a mild solution. Proposition 15 asserts the equivalence between these two kinds of solutions under assumption that a H 0 1 ( Ω ) and F L 2 ( 0 , T ; L 2 ( Ω ) ) .
For the proof of Proposition 15, we show two lemmata. Henceforth, for v L 2 ( Ω ) , we define j v H 1 ( Ω ) : = ( H 0 1 ( Ω ) ) , 1 j d , by
H 1 ( Ω ) < j v , φ > H 0 1 ( Ω ) : = ( v , j φ ) L 2 ( Ω ) f o r   a l l   φ H 0 1 ( Ω ) .
Then, by the definition of the norm of the linear functional, we have
j v H 1 ( Ω ) = sup φ H 0 1 ( Ω ) = 1 | H 1 ( Ω ) < j v , φ > H 0 1 ( Ω ) |
sup φ H 0 1 ( Ω ) = 1 | ( v , j φ ) L 2 ( Ω ) | v L 2 ( Ω ) .
We recall that L 1 2 is defined as an operator from L 2 ( Ω ) to L 2 ( Ω ) by (73), while also L 1 2 : H 0 1 ( Ω ) L 2 ( Ω ) is defined by (73).
Then we can state the first lemma, which involves the extension of L 1 2 : L 2 ( Ω ) L 2 ( Ω ) .
Lemma 4.
(i) The operator L 1 2 can be extended as a bounded operator from H 1 ( Ω ) to L 2 ( Ω ) . Henceforth, by the same notation, we denote the extension. There exists a constant C > 0 such that
C 1 v H 1 ( Ω ) L 1 2 v L 2 ( Ω ) C v H 1 ( Ω ) f o r   a l l   v H 1 ( Ω ) .
(ii)  K ( t ) v = L 1 2 K ( t ) L 1 2 v for t > 0 and v H 1 ( Ω ) .
(iii) 
L 1 2 ( b j j v ) L 2 ( Ω ) C v L 2 ( Ω ) f o r   a l l   v L 2 ( Ω ) .
We now state the second lemma.
Lemma 5.
(i) For a L 2 ( Ω ) , we have S ( t ) a D ( L ) for t > 0 and
t α ( S ( t ) a a ) + L S ( t ) a = 0 , 0 < t < T .
Moreover, there exists a constant C 0 > 0 such that
S ( t ) a a H α ( 0 , T ; L 2 ( Ω ) ) C 0 a H 0 1 ( Ω )
for all a H 0 1 ( Ω ) . Here, C 0 > 0 is independent of a H 0 1 ( Ω ) and T > 0 .
(ii) We have
t α 0 t K ( t s ) F ( s ) d s + L 0 t K ( t s ) F ( s ) d s = F ( t ) , 0 < t < T
for F L 2 ( 0 , T ; L 2 ( Ω ) ) . Moreover, there exists a constant C 1 = C 1 ( T ) > 0 such that
0 t K ( t s ) F ( s ) d s H α ( 0 , T ; L 2 ( Ω ) ) C 1 F L 2 ( 0 , T ; L 2 ( Ω ) )
for each F L 2 ( 0 , T ; L 2 ( Ω ) ) .
We note that Lemma 5 corresponds to Propositions 12 (i) and 13.
Let Lemmata 4 and 5 be proved. Then the proof of (i) ⟶ (ii) of Proposition 15 can be conducted similarly to [19]. The proof of (ii) ⟶ (i) of the proposition can be derived from Lemmata 4 and 5, and we omit the details. The proofs of the lemmata are provided in Appendix A.

6.2. Continuity at t = 0

As is discussed in Section 1, Section 2 and Section 5, the continuity of solutions to fractional differential equations at t = 0 is delicate and requires careful treatments for initial conditions. However, if we assume F = 0 in (68), then we can prove the following sufficient continuity.
Theorem 11.
Under (70) and (71), for a L 2 ( Ω ) , the solution u to (68) and (69) satisfies
u C ( [ 0 , T ] ; L 2 ( Ω ) ) .
The theorem means that lim t 0 u ( · , t ) a L 2 ( Ω ) = 0 , but the continuity at t = 0 breaks if a non-homogeneous term F 0 is attached, which is already shown in Example 6 in Section 5 concerning a fractional ordinary differential equation.
In the case of b j = 0 , 1 j d , the same result is proved in [19] and we can refer also to [25]. However, it seems no proof of a non-symmetric elliptic operator A, although one naturally expects the same continuity. The proof is typical, as arguments by operator theory apply to the classical partial differential equations, and we carry out similar arguments for fractional differential equations within our framework.
Proof. 
We divide the proof into four steps.
First Step. From Lemma 4, it follows that the solution u to (68) and (69) with F = 0 is given by
u ( · , t ) = S ( t ) a + 0 t K ( t s ) j = 1 d b j j u ( · , s ) d s = : S ( t ) a + M ( u ) ( t ) , 0 < t < T .
The proof of Theorem 11 is based on an approximating sequence for the solution u ( t ) constructed by
u 1 ( t ) : = S ( t ) a , u k + 1 ( t ) : = S ( t ) a + ( M u k ) ( t ) , k N , 0 < t < T .
First, we can prove
S ( · ) a C ( [ 0 , T ] ; L 2 ( Ω ) ) .
Verification of (79). For m , using (42) with λ n > 0 , for 0 t T we estimate
n = m ( a , φ n ) E α , 1 ( λ n t α ) φ n L 2 ( Ω ) 2 n = m | ( a , φ n ) | 2 | E α , 1 ( λ n t α ) | 2 C n = m | ( a , φ n ) | 2 .
Therefore,
lim , m n = m ( a , φ n ) E α , 1 ( λ n t α ) φ n C ( [ 0 , T ] ; L 2 ( Ω ) ) = 0 ,
which means that
n = 1 N ( a , φ n ) E α , 1 ( λ n t α ) φ n C ( [ 0 , T ] ; L 2 ( Ω ) )
converges to S ( t ) a in C ( [ 0 , T ] ; L 2 ( Ω ) ) . Thus the verification of (79) is complete.
Second Step. We prove
0 t L 1 2 K ( s ) w ( t s ) d s C ( [ 0 , T ] ; L 2 ( Ω ) ) f o r   w C ( [ 0 , T ] ; L 2 ( Ω ) ) .
Verification of (80). Let 0 < t < T be arbitrarily fixed. For small h > 0 , we have
0 t + h L 1 2 K ( s ) w ( t + h s ) d s 0 t L 1 2 K ( s ) w ( t s ) d s = t t + h L 1 2 K ( s ) w ( t + h s ) d s + 0 t L 1 2 K ( s ) ( w ( t + h s ) w ( t s ) ) d s = : I 1 ( t , h ) + I 2 ( t , h ) .
Then, by (75), we can estimate
I 1 ( t , h ) C t t + h s 1 2 α 1 d s w C ( [ 0 , T ] ; L 2 ( Ω ) ) = 2 C α ( ( t + h ) 1 2 α t 1 2 α ) w C ( [ 0 , T ] ; L 2 ( Ω ) ) 0 a s h 0 .
Next,
I 2 ( t , h ) C 0 t s 1 2 α 1 w ( t + h s ) w ( t s ) d s .
We have max 0 s T w ( t + h s ) w ( t s ) 0 as h 0 with fixed t by w C ( [ 0 , T ] ; L 2 ( Ω ) ) . Hence, since s 1 2 α 1 L 1 ( 0 , T ) , the Lebesgue convergence theorem yields that lim h 0 I 2 ( t , h ) = 0 . We can argue similarly also for h < 0 , t = 0 and t = T , and so the verification of (80) is complete.
Next, we proceed to the proofs of (81) and (82):
u k C ( [ 0 , T ] ; L 2 ( Ω ) ) , k N
and
u k c o n v e r g e s i n C ( [ 0 , T ] ; L 2 ( Ω ) ) a s k .
Third Step: Proof of (81). We will prove by induction. By (79), we see that u 1 ( t ) = S ( t ) a C ( [ 0 , T ] ; L 2 ( Ω ) ) . We assume that u k C ( [ 0 , T ] ; L 2 ( Ω ) ) . Then, by Lemma 4 (ii), we have
M ( u k ) ( t ) = 0 t K ( t s ) j = 1 d b j j u k ( s ) d s = 0 t L 1 2 K ( t s ) L 1 2 j = 1 d b j j u k ( s ) d s .
Lemma 4 (iii) yields
L 1 2 ( b j j u k ( t ) ) L 1 2 ( b j j u k ( t ) ) L 2 ( Ω ) C u k ( t ) u k ( t ) L 2 ( Ω ) , t , t [ 0 , T ]
for 1 j d , so that
L 1 2 b j j u k C ( [ 0 , T ] ; L 2 ( Ω ) ) , 1 j d .
Therefore, the application of (80) to M ( u k ) implies
M ( u k ) C ( [ 0 , T ] ; L 2 ( Ω ) ) .
Consequently,
u k + 1 = S ( · ) a + M ( u k ) C ( [ 0 , T ] ; L 2 ( Ω ) ) .
Thus the induction completes the proof of (81).
Fourth Step: Proof of (82). We have
( u k + 2 u k + 1 ) ( t ) = M ( u k + 1 u k ) ( t ) = 0 t L 1 2 K ( t s ) j = 1 d L 1 2 ( b j j ( u k + 1 u k ) ( s ) ) d s , k N , 0 < t < T .
We set v k : = u k + 1 u k . Then Lemma 4 (iii) and (75) yield
v k + 1 ( t ) 0 t L 1 2 K ( t s ) j = 1 d L 1 2 ( b j j v k ( s ) ) d s
C 0 t ( t s ) 1 2 α 1 v k ( s ) d s , k N .
Setting M 0 : = v 1 C ( [ 0 , T ] ; L 2 ( Ω ) ) , we apply (83) with k = 1 to obtain
v 2 ( t ) C 0 t ( t s ) 1 2 α 1 d s M 0 = C M 0 Γ 1 2 α t 1 2 α , 0 < t < T .
Therefore, substituting this into (83) with k = 2 , we obtain
v 3 ( t ) C 2 M 0 Γ 1 2 α 0 t ( t s ) 1 2 α 1 s 1 2 α d s = C 2 M 0 Γ 1 2 α + 1 Γ ( α + 1 ) t α , 0 < t < T .
Continuing this estimate, we can prove
v k + 1 ( t ) 1 2 α M 0 C k Γ 1 2 α k 1 Γ 1 2 k α + 1 t 1 2 k α , 0 < t < T
for all k N . Consequently,
v k + 1 C ( [ 0 , T ] ; L 2 ( Ω ) ) C 1 Γ 1 2 k α + 1 C Γ 1 2 α T 1 2 α k , k N .
By the asymptotic behavior of the gamma function, we can verify
lim k C Γ 1 2 α T 1 2 α k Γ 1 2 k α + 1 = 0 ,
so that k = 0 v k C ( [ 0 , T ] ; L 2 ( Ω ) ) converges. Therefore, lim k u k exists in C ( [ 0 , T ] ; L 2 ( Ω ) ) . Thus the proof of (82) is complete.
From the uniqueness of solution to (82), we can derive that its limit is u, the solution to (68) and (69) with F = 0 . Since u k C ( [ 0 , T ] ; L 2 ( Ω ) ) for k N , the limit is also in C ( [ 0 , T ] ; L 2 ( Ω ) ) . Thus the proof of Theorem 11 is complete. □

6.3. Stronger Regularity in Time of Solution

Again, we consider (68) and (69) with F 0 . Theorem 9 provides a basic result on the unique existence of u for F L 2 ( 0 , T ; L 2 ( Ω ) ) and a H 0 1 ( Ω ) , while Theorem 10 is the well-posedness for a and F, which are less regular in x.
Here, we consider a stronger time-regular F to improve the regularity of solution u. Due to the framework of t α defined by (15), the argument for improving the regularity of solution is automatic.
Theorem 12.
Let 0 < α < 1 and β > 0 . We assume a H 2 ( Ω ) H 0 1 ( Ω ) and
F A a H β ( 0 , T ; L 2 ( Ω ) ) .
Then there exists a unique solution u to (68) and (69) such that
u a H β ( 0 , T ; H 2 ( Ω ) H 0 1 ( Ω ) ) H α + β ( 0 , T ; L 2 ( Ω ) ) ,
and we can find a constant C > 0 such that
u a H β ( 0 , T ; H 2 ( Ω ) ) + u a H α + β ( 0 , T ; L 2 ( Ω ) ) C ( F A a H β ( 0 , T ; L 2 ( Ω ) ) + a H 2 ( Ω ) ) .
If 0 < β < 1 2 , then (84) is equivalent to F H β ( 0 , T ; L 2 ( Ω ) ) under the condition a H 2 ( Ω ) H 0 1 ( Ω ) . If 1 2 < β 1 , then (84) is equivalent to
F H β ( 0 , T ; L 2 ( Ω ) ) , A a = F ( · , 0 ) i n   Ω
under the condition that a H 2 ( Ω ) H 0 1 ( Ω ) . Condition (84) is a compatibility condition which is necessary for lifting up the regularity of the solution. As for the parabolic equation, see Theorem 6 in Chapter 7, Section 1 in Evans [28] for instance.
From Theorem 12, we can easily derive the following corollary.
Corollary 2.
Let a = 0 and 0 < α < 1 . We assume
F H α ( 0 , T ; L 2 ( Ω ) ) L 2 ( 0 , T ; H 2 ( Ω ) H 0 1 ( Ω ) ) .
Then
A 2 u , t 2 α u L 2 ( 0 , T ; L 2 ( Ω ) ) .
Proof of Theorem 12.
We will gain the regularity of the solution u by means of an equation which can be expected to be satisfied by t α u , although such an equation is not justified for the moment.
We consider
t α v + A v = β ( F A a ) , v H α ( 0 , T ; L 2 ( Ω ) ) .
Then, by t β ( F A a ) L 2 ( 0 , T ; L 2 ( Ω ) ) , Theorem 9 yields that v exists and
v H α ( 0 , T ; L 2 ( Ω ) ) L 2 ( 0 , T ; H 2 ( Ω ) H 0 1 ( Ω ) ) .
We set u ˜ : = J β v + a . Then Proposition 5 (i) implies
u ˜ a H α + β ( 0 , T ; L 2 ( Ω ) ) H β ( 0 , T ; H 2 ( Ω ) H 0 1 ( Ω ) ) .
In view of Proposition 5 (ii) and v H α ( 0 , T ; L 2 ( Ω ) ) , we have
J α ( u ˜ a ) = J α J β v = J β J α v ,
and so J α ( u ˜ a ) = J β ( J α v ) . Moreover, by v L 2 ( 0 , T ; H 2 ( Ω ) H 0 1 ( Ω ) ) , we see
A u ˜ = A ( J β v + a ) = J β A v + A a .
Therefore,
J α ( u ˜ a ) + A u ˜ = J β ( J α v + A v ) + A a .
In terms of (87), we obtain
J α ( u ˜ a ) + A u ˜ = J β ( t β ( F A a ) ) + A a = F .
Therefore, combining (88), we can verify that u ˜ a satisfies (68) and (69), and the uniqueness of solution yields u = u ˜ a . Thus the proof of Theorem 12 is complete. □
Proof of Corollary 2.
Theorem 12 yields
u H α ( 0 , T ; H 2 ( Ω ) H 0 1 ( Ω ) ) H 2 α ( 0 , T ; L 2 ( Ω ) ) .
Consequently, applying also Theorem 2, we deduce that t 2 α u L 2 ( 0 , T ; L 2 ( Ω ) ) . Moreover, since t α t α = t 2 α by Theorem 2, operating t α to (68), we obtain
t α A u = t 2 α u + t α F L 2 ( 0 , T ; L 2 ( Ω ) ) .
On the other hand, A ( t α u + A u F ) = 0 in L 2 ( 0 , T ; ( C 0 ( Ω ) ) ) . Consequently, in terms of (89), we reach
A 2 u = A t α + A F L 2 ( 0 , T ; L 2 ( Ω ) ) .
Thus the proof of Corollary 2 is complete. □

6.4. Weaker Regularity in Time of Solution

First, we introduce a function space α H ( 0 , T ; X ) for a Banach space X. Indeed, due to Proposition 9 or Theorem 1, the operator J α : α H ( 0 , T ) L 2 ( 0 , T ) is surjective and an isomorphism for α > 0 , so we define
α H ( 0 , T ; X ) : = { v L 2 ( 0 , T ; X ) ; J α v L 2 ( 0 , T ; X ) } , w i t h t h e n o r m v α H ( 0 , T ; X ) : = J α v L 2 ( 0 , T ; X ) .
By t α : = ( J α ) 1 : L 2 ( 0 , T ) α H ( 0 , T ) , we see also that
lim n v n = v   i n   L 2 ( 0 , T ; L 2 ( Ω ) ) i m p l i e s lim n t α v n = t α v   i n   α H ( 0 , T ) .
In this subsection, we consider
t α ( u a ) + A u = F i n   α H ( 0 , T ; L 2 ( Ω ) )
and
u a L 2 ( 0 , T ; L 2 ( Ω ) ) , u α H ( 0 , T ; H 2 ( Ω ) H 0 1 ( Ω ) ) .
As is argued in Example 7, for α > 1 2 , a source term
F ( x , t ) = δ t 0 ( t ) f ( x ) α H ( 0 , T ; L 2 ( Ω ) )
with f L 2 ( Ω ) describes an impulsive source at t = t 0 , and it is not only mathematically but also physically meaningful to treat a singular source F α H ( 0 , T ; L 2 ( Ω ) ) with α > 0 . We here state one result on the well-posedness for (92) and (93).
Theorem 13.
Let 0 < α < 1 . We assume
F α H ( 0 , T ; L 2 ( Ω ) ) , a L 2 ( Ω ) .
Then there exists a unique solution u L 2 ( 0 , T ; L 2 ( Ω ) ) α H ( 0 , T ; H 2 ( Ω ) H 0 1 ( Ω ) ) to (92) and (93), and we can find a constant C > 0 such that
u L 2 ( 0 , T ; L 2 ( Ω ) ) + u α H ( 0 , T ; H 2 ( Ω ) H 0 1 ( Ω ) ) C ( a L 2 ( Ω ) + F α H ( 0 , T ; L 2 ( Ω ) ) )
for each a L 2 ( Ω ) and F α H ( 0 , T ; L 2 ( Ω ) ) .
Proof. 
We will prove this by creating an equation which should be satisfied by J α u , and such an equation can be given by
t α J α ( u a ) + A J α u = J α F ,
where we have to make justification. Thus we consider the solution v to
t α v + A v = J α F + a , v H α ( 0 , T ; L 2 ( Ω ) ) .
Since J α F + a L 2 ( 0 , T ; L 2 ( Ω ) ) , Theorem 9 implies the unique existence of solution v H α ( 0 , T ; L 2 ( Ω ) ) L 2 ( 0 , T ; H 2 ( Ω ) H 0 1 ( Ω ) ) . Therefore, Proposition 10 yields ( J α ) 1 v = J α v = t α v by v H α ( 0 , T ; L 2 ( Ω ) ) . Setting
u ˜ : = ( J α ) 1 v = J α v ,
we readily verify
u ˜ L 2 ( 0 , T ; L 2 ( Ω ) ) α H ( 0 , T ; H 2 ( Ω ) H 0 1 ( Ω ) )
by Propositions 2 and 9 (ii). Furthermore, since (96) implies that ( J α ) 1 v = t α v and v = J α u ˜ , we obtain
( J α ) 1 ( u ˜ a ) + A u ˜ F = ( J α ) 1 ( u ˜ a + A J α u ˜ J α F ) = ( J α ) 1 ( ( J α ) 1 v a + A v J α F ) = ( J α ) 1 ( t α v + A v J α F a ) = 0
in α H ( 0 , T ; L 2 ( Ω ) ) .
Since J α is injective, the application of (95) implies that t α ( u ˜ a ) + A u ˜ = F and u ˜ a L 2 ( 0 , T ; L 2 ( Ω ) ) . Hence, u ˜ is a solution to (92) and (93), and (94) holds.
We finally prove the uniqueness of solution. Let w L 2 ( 0 , T ; L 2 ( Ω ) ) α H ( 0 , T ; H 2 ( Ω ) H 0 1 ( Ω ) ) and t α w + A w = 0 hold in α H ( 0 , T ; L 2 ( Ω ) ) . Then
J α t α w + J α A w = J α t α w + A J α w = 0 i n   L 2 ( 0 , T ; L 2 ( Ω ) ) .
By w L 2 ( 0 , T ; L 2 ( Ω ) ) and (24), we see
J α t α w = J α ( J α ) 1 w = w .
On the other hand,
w = ( J α ) 1 J α w = t α J α w
by (15) and J α w H α ( 0 , T ) . Hence, J α t α w = t α J α w . Therefore, in terms of (97), we have J α w H α ( 0 , T ; L 2 ( Ω ) ) L 2 ( 0 , T ; H 2 ( Ω ) H 0 1 ( Ω ) ) and t α ( J α w ) + A ( J α w ) = 0 in L 2 ( 0 , T ; L 2 ( Ω ) ) . Hence, the uniqueness asserted by Theorem 9 implies J α w = 0 in ( 0 , T ) . Since J α : L 2 ( 0 , T ) L 2 ( 0 , T ) is injective, we obtain w = 0 in ( 0 , T ) . Thus the proof of Theorem 13 is complete. □
Finally, in this subsection, in order to describe the flexibility of our approach, we show the following.
Proposition 16.
Let 0 < α < 1 . In (92) and (93), we assume
a = 0 , F α H ( 0 , T ; L 2 ( Ω ) ) 2 α H ( 0 , T ; H 2 ( Ω ) H 0 1 ( Ω ) ) .
Then the solution u satisfies
u 2 α H ( 0 , T ; D ( A 2 ) ) 2 α H ( 0 , T ; H 4 ( Ω ) ) .
The proposition means that under (98), the solution u can hold the H 4 ( Ω ) -regularity in x at the expense of the weaker regularity 2 α H ( 0 , T ) in t.
Proof of Proposition 16.
By Theorem 13, the solution u to (92) and (93) with a = 0 exists uniquely. More precisely, we have ( J α ) 1 u + A u = F in α H ( 0 , T ; L 2 ( Ω ) ) and
u α H ( 0 , T ; H 2 ( Ω ) H 0 1 ( Ω ) ) L 2 ( 0 , T ; L 2 ( Ω ) ) .
Therefore, J α ( J α ) 1 = ( J α ) 1 J α in L 2 ( 0 , T ; L 2 ( Ω ) ) and J α A u = A J α u , and so
( J α ) 1 J α u + A J α u = J α F L 2 ( 0 , T ; L 2 ( Ω ) ) .
Hence, operating J α again, we obtain
( J α ) ( J α ) 1 J α u + A J α J α u = J α J α F .
Consequently, we obtain J α ( J α ) 1 J α u = ( J α ) 1 ( J α J α u ) . Setting v : = J α J α u , we have
t α v + A v = J α J α F , v H α ( 0 , T ; L 2 ( Ω ) ) .
We note that Proposition 9 (iii) yields J α ( J α u ) = J α ( J α u ) by J α u L 2 ( 0 , T ; L 2 ( Ω ) ) and that J α F L 2 ( 0 , T ; L 2 ( Ω ) ) by F α H ( 0 , T ; L 2 ( Ω ) ) and Proposition 9 (ii). Moreover, J α ( J α F ) H α ( 0 , T ; L 2 ( Ω ) ) by Propositions 9 (iii) and 2.
Applying Proposition 9 (ii) twice to F 2 α H ( 0 , T ; H 2 ( Ω ) H 0 1 ( Ω ) ) , we obtain J α F α H ( 0 , T ; H 2 ( Ω ) H 0 1 ( Ω ) ) , so that J α ( J α F ) L 2 ( 0 , T ; H 2 ( Ω ) H 0 1 ( Ω ) ) . Therefore, the application of Corollary 2 to (99) yields
A 2 v = A 2 J α J α u L 2 ( 0 , T ; L 2 ( Ω ) ) .
By u L 2 ( 0 , T ; L 2 ( Ω ) ) , we see that J α J α u L 2 ( 0 , T ; D ( A 2 ) ) L 2 ( 0 , T ; H 4 ( Ω ) ) . In terms of Proposition 9 (ii), it follows that u 2 α H ( 0 , T ; D ( A 2 ) ) . Thus the proof of Proposition 16 is complete. □

6.5. Initial Boundary Value Problems for Multi-Term Time-Fractional Partial Differential Equations

Let N N , 0 < α 1 < < α N < α < 1 , q k C ( Ω ¯ ) for 1 k N . We recall that A is defined by (1) and (67). We discuss an initial boundary value problem for a multi-term time-fractional partial differential equation
t α ( u a ) + k = 1 N q k ( x ) t α k ( u a ) + A u = F
and
u a H α ( 0 , T ; L 2 ( Ω ) ) .
Then we will prove the following theorem.
Theorem 14.
For each a H 0 1 ( Ω ) and F L 2 ( 0 , T ; L 2 ( Ω ) ) , there exists a unique solution u L 2 ( 0 , T ; H 2 ( Ω ) H 0 1 ( Ω ) ) to (100) and (101). Moreover, there exists a constant C > 0 such that
u L 2 ( 0 , T ; H 2 ( Ω ) H 0 1 ( Ω ) ) + u a H α ( 0 , T ; L 2 ( Ω ) ) C ( a H 0 1 ( Ω ) + F L 2 ( 0 , T ; L 2 ( Ω ) ) )
for each a H 0 1 ( Ω ) and F L 2 ( 0 , T ; L 2 ( Ω ) ) .
Here, the constant C > 0 depends on Ω , T, A, α k and q k for 1 k N .
In the same way as in Subsection 6.3 and Subsection 6.4, we can establish stronger and weaker regular solutions, but we omit the details.
The arguments are direct within our framework based on t α : β H ( 0 , T ) α β H ( 0 , T ) and t α : H α + γ ( 0 , T ) H γ ( 0 , T ) with γ 0 and β 0 . For fractional ordinary differential equations, we mentioned the corresponding initial value problem as (52) in Section 5.
The multi-term time-fractional partial differential equations are considered, for example, in Kian [29], Li, Imanuvilov and Yamamoto [30] and Li, Liu and Yamamoto [31] for a symmetric A, which makes arguments simple. A non-symmetric A contains an advection term j = 1 d b j j u and so it is physically a more feasible model. However, we do not know the existing results on the corresponding well-posedness for a non-symmetric A. In fact, our method for initial boundary value problems is not restricted to a symmetric A. Moreover, we can treat weaker and stronger solutions in the same way as in Subsection 6.3 and Subsection 6.4, and we omit the details.
We can easily extend our method to the case where the coefficients q k , 1 k N , b j , 1 j d , and c depend on time.
Proof. 
We divide the proof into two steps.
First Step. We prove the theorem under the assumption that b j = 0 for 1 j d and c 0 on Ω . By Proposition 15 and Lemma 5, it suffices to prove that there exists a unique u a H α ( 0 , T ; L 2 ( Ω ) ) satisfying
u ( t ) a = S ( t ) a a + 0 t K ( t s ) F ( s ) d s + k = 1 N 0 t K ( t s ) q k t α k ( u ( s ) a ) d s , 0 < t < T .
In this step, we prove
Lemma 6.
Let k = 1 , . . . , N .
(i) 
t α k K ( t ) a L 2 ( Ω ) C t α α k 1 a L 2 ( Ω ) , 0 < t < T , a L 2 ( Ω ) .
(ii) 
For G L 2 ( 0 , T ; L 2 ( Ω ) ) , we have 0 t K ( t s ) G ( s ) d s H α k ( 0 , T ; L 2 ( Ω ) ) and
t α k 0 t K ( t s ) G ( s ) d s = 0 t t α k K ( t s ) G ( s ) d s , 0 < t < T .
Proof of Lemma 6.
(i) We set
K m ( t ) a : = n = 1 m t α 1 E α , α ( λ n t α ) ( a , φ n ) φ n f o r m N a n d a L 2 ( Ω ) .
Then, by Lemma 3 (ii), we see that
t α k K m ( t ) a = n = 1 m t α k ( t α 1 E α , α ( λ n t α ) ) ( a , φ n ) φ n = n = 1 m t α α k 1 E α , α α k ( λ n t α ) ( a , φ n ) φ n .
Therefore, applying (42), we obtain
t α k K m ( t ) a L 2 ( Ω ) 2 = n = 1 m t 2 ( α α k 1 ) | E α , α α k ( λ n t α ) | 2 | ( a , φ n ) | 2
C t 2 ( α α k 1 ) n = 1 m | ( a , φ n ) | 2 .
Letting m , we complete the proof of part (i).
(ii) By Proposition 13, we can readily prove 0 t K ( t s ) G ( s ) d s H α ( 0 , T ; L 2 ( Ω ) ) H α k ( 0 , T ; L 2 ( Ω ) ) . Noting that G ( x , · ) L 2 ( 0 , T ) for almost all x Ω , in terms of Lemma 3 (iii), we have
t α k 0 t K m ( x , t s ) G ( x , s ) d s = 0 t t α k K m ( x , t s ) G ( x , s ) d s , x Ω , 0 < t < T
for m N .
Similarly to (104), we estimate
0 t ( t α k K m t α k K ) ( t s ) G ( · , s ) d s L 2 ( Ω ) 0 t ( t α k K m t α k K ) ( t s ) G ( · , s ) L 2 ( Ω ) d s C 0 t ( t s ) α α k 1 n = m + 1 | ( G ( s ) , φ n ) L 2 ( Ω ) | 2 1 2 d s .
The Young inequality on the convolution yields
0 t ( t α k K m t α k K ) ( t s ) G ( · , s ) d s L 1 ( 0 , T ; L 2 ( Ω ) ) C s α α k 1 L 1 ( 0 , T ) n = m + 1 | ( G ( s ) , φ n ) L 2 ( Ω ) | 2 1 2 L 2 ( 0 , T ) C T α α k 0 T n = m + 1 | ( G ( s ) , φ n ) L 2 ( Ω ) | 2 d s 1 2 0
as m by G L 2 ( 0 , T ; L 2 ( Ω ) ) . Therefore,
lim m 0 t t α k K m ( t s ) G ( s ) d s = 0 t t α k K ( t s ) G ( s ) d s i n   L 1 ( 0 , T ; L 2 ( Ω ) ) .
Similarly, using (42), we can verify
lim m 0 t K m ( t s ) G ( s ) d s = 0 t K ( t s ) G ( s ) d s i n   L 1 ( 0 , T ; L 2 ( Ω ) ) .
Choosing γ > 1 2 , we see that L 1 ( 0 , T ) γ H ( 0 , T ) and the series in (106) converges also in γ H ( 0 , T ; L 2 ( Ω ) ) . Since t α k : γ H ( 0 , T ) γ α k H ( 0 , T ) is an isomorphism by Theorem 1, we obtain
lim m t α k 0 t K m ( t s ) G ( s ) d s = t α k 0 t K ( t s ) G ( s ) d s i n   γ α k H ( 0 , T ; L 2 ( Ω ) ) .
Combining this with (105), we finish the proof of Lemma 6. □
Second Step. We will prove the unique existence of u to (103) by the contraction mapping theorem. We set
Q u ( t ) : = S ( t ) a + 0 t K ( t s ) F ( s ) d s + k = 1 N 0 t K ( t s ) q k s α k ( u ( s ) a ) d s , 0 < t < T .
By Lemma 5, we see
S ( t ) a a , 0 t K ( t s ) F ( s ) d s H α ( 0 , T ; L 2 ( Ω ) )
f o r   a H 0 1 ( Ω )   a n d   F L 2 ( 0 , T ; L 2 ( Ω ) ) .
We estimate k = 1 N Q u Q v H α k ( 0 , T : L 2 ( Ω ) ) for u a , v a H α k ( 0 , T ; L 2 ( Ω ) ) . To this end, we show
t α j 0 t K ( t s ) w ( s ) d s L 2 ( Ω )
C 0 t ( t s ) α α j 1 w ( s ) L 2 ( Ω ) d s , 1 j N , 0 < t < T .
Verification of (108). The application of Lemma 6 (ii) and (i) yields
t α j 0 t K ( t s ) w ( s ) d s L 2 ( Ω ) = 0 t t α j K ( t s ) w ( s ) d s L 2 ( Ω ) 0 t t α j K ( t s ) w ( s ) L 2 ( Ω ) d s C 0 t ( t s ) α α j 1 w ( s ) L 2 ( Ω ) d s .
Thus (108) follows directly.
We note that if u a , v a H α N ( 0 , T ; L 2 ( Ω ) ) H α k ( 0 , T ; L 2 ( Ω ) ) with 1 k N , then ( u a ) ( v a ) = u v H α k ( 0 , T ; L 2 ( Ω ) ) for 1 k N . Therefore, using q k L ( Ω ) for 1 k N , we have
t α j ( Q u ( t ) Q v ( t ) ) L 2 ( Ω ) = k = 1 N t α j 0 t K ( t s ) q k s α k ( u ( s ) v ( s ) ) d s L 2 ( Ω ) k = 1 N t α j 0 t K ( t s ) q k s α k ( u ( s ) v ( s ) ) d s L 2 ( Ω ) C k = 1 N 0 t ( t s ) α α j 1 s α k ( u ( s ) v ( s ) ) L 2 ( Ω ) d s .
Summing up over j = 1 , . . . , N and applying ( t s ) α α j 1 C ( t s ) α α N 1 for 1 j N , we reach
j = 1 N t α j ( Q u ( t ) Q v ( t ) ) L 2 ( Ω )
C 0 t ( t s ) α α N 1 k = 1 N s α k ( u ( s ) v ( s ) ) L 2 ( Ω ) d s .
Hence, applying (108) for estimating j = 1 N t α j ( Q 2 u ( t ) Q 2 v ( t ) ) L 2 ( Ω ) , we obtain
j = 1 N t α j ( Q 2 u ( t ) Q 2 v ( t ) ) L 2 ( Ω ) C 0 t ( t s ) α α N 1 k = 1 N s α k ( Q u ( s ) Q v ( s ) ) L 2 ( Ω ) d s .
Substituting (109) and exchanging the order of the integrals, we obtain
k = 1 N t α k ( Q 2 u ( t ) Q 2 v ( t ) ) L 2 ( Ω ) C 2 0 t ( t s ) α α N 1 0 s ( t ξ ) α α N 1 k = 1 N ξ α k ( u v ) ( ξ ) L 2 ( Ω ) d ξ d s = C 2 0 t ξ t ( t s ) α α N 1 ( t ξ ) α α N 1 d s k = 1 N ξ α k ( u v ) ( ξ ) L 2 ( Ω ) d ξ = C 2 Γ ( α α N ) 2 Γ ( 2 ( α α N ) ) 0 t ( t ξ ) 2 α 2 α N 1 k = 1 N ξ α k ( u v ) ( ξ ) L 2 ( Ω ) d ξ .
Continuing this estimation, we can reach
k = 1 N t α k ( Q m u ( t ) Q m v ( t ) ) L 2 ( Ω ) ( C Γ ( α α N ) ) m Γ ( m ( α α N ) ) 0 t ( t s ) m ( α α N ) 1 k = 1 N s α k ( u v ) ( s ) L 2 ( Ω ) d s , 0 < t < T , m N .
Consequently, the Young inequality yields
0 T k = 1 N t α k ( Q m u ( t ) Q m v ( t ) ) L 2 ( Ω ) 2 d s 1 2 ( C Γ ( α α N ) ) m Γ ( m ( α α N ) ) s m ( α α N ) 1 k = 1 N s α k ( u v ) ( s ) L 2 ( Ω ) L 2 ( 0 , T ) ( C Γ ( α α N ) ) m Γ ( m ( α α N ) ) T m ( α α N ) m ( α α N ) 0 T k = 1 N s α k ( u v ) ( s ) L 2 ( Ω ) 2 d s 1 2 .
For estimating the right-hand side of the above inequality, we find a constant C N > 0 such that
C N 1 k = 1 N | ξ k | 2 k = 1 N | ξ k | 2 C N k = 1 N | ξ k | 2
for all ξ 1 , . . . , ξ N R , and we obtain
k = 1 N t α k ( Q m u Q m v ) L 2 ( 0 , T ; L 2 ( Ω ) ) ( C Γ ( α α N ) T α α N ) m Γ ( m ( α α N ) + 1 ) k = 1 N t α k ( u v ) L 2 ( 0 , T ; L 2 ( Ω ) ) ,
that is,
Q m u Q m v H α N ( 0 , T ; L 2 ( Ω ) ) ( C Γ ( α α N ) T α α N ) m Γ ( m ( α α N ) + 1 ) u v H α N ( 0 , T ; L 2 ( Ω ) )
for all u a , v a H α N ( 0 , T ; L 2 ( Ω ) ) and m N .
By the asymptotic behavior of the gamma function, we know
lim m ( C Γ ( α α N ) T α α N ) m Γ ( m ( α α N ) + 1 ) = 0 ,
and choosing m N sufficiently large, we see that Q m : H α N ( 0 , T ; L 2 ( Ω ) ) H α N ( 0 , T ; L 2 ( Ω ) ) is a contraction. Thus, we have proved the unique existence of a solution u to (103) such that u a H α N ( 0 , T ; L 2 ( Ω ) ) , which completes the proof of Theorem 14, provided that b j = 0 for 1 j d and c 0 in Ω .
For general b j and c, setting
L v ( x ) : = i , j = 1 d i ( a i j ( x ) j v ) ( x )
in place of (72), we replace (103) by
u ( t ) a = S ( t ) a a + 0 t K ( t s ) j = 1 d b j j u ( s ) d s + 0 t K ( t s ) c u ( s ) d s
+ 0 t K ( t s ) F ( s ) d s + k = 1 N 0 t K ( t s ) s α k ( u ( s ) a ) d s , 0 < t < T
and argue similarly. For the estimation of the second term on the right-hand side of (110), in view of Theorem 5, we see
t α k 0 t K ( t s ) b j j u ( s ) = t α k 0 t K ( s ) b j j u ( t s ) d s = 0 t K ( s ) b j t α k j u ( t s ) d s = 0 t L 1 2 K ( s ) L 1 2 ( b j j t α k u ( t s ) ) d s ,
and so we can apply Lemma 4 (iii) to argue similarly. Here, we omit the details. Thus the proof of Theorem 14 is complete.

7. Application to an Inverse Source Problem: Illustrating Example

We construct our framework for fractional derivatives. In this article, within the framework, we study initial value problems for fractional ordinary differential equations and initial boundary value problems for fractional partial differential equations, which are classified into so-called “direct problems”. On the other hand, “inverse problems”, where we are required to determine some of initial values, boundary values, order α , and coefficients in the fractional differential equations by observation data of solution u, are important.
The inverse problem is indispensable for accurate modeling for analyzing phenomena such as anomalous diffusion. After relevant studies of inverse problems, we can identify coefficients, etc. to determine equations themselves and can proceed to initial value problems and initial boundary value problems. Thus, the inverse problem is the premise for the study of the forward problem.
Actually, the main purpose of the current article is to not only establish a theory for direct problems concerning time-fractional differential equations but also apply the theory to inverse problems. The inverse problems are very various and here we discuss only one inverse problem in order to illustrate how our framework for fractional derivatives works.
Let 0 < α < 1 and A be the same elliptic operator as the previous sections, that is, defined by (67) and (70). We consider
t α u + A u = μ ( t ) f ( x ) i n α H ( 0 , T ; L 2 ( Ω ) ) , u L 2 ( 0 , T ; L 2 ( Ω ) ) ,
where μ α H ( 0 , T ) , ¬ 0 and f L 2 ( Ω ) .
Inverse source problem. Let θ L 2 ( Ω ) , ¬ 0 be arbitrarily chosen, and let f L 2 ( Ω ) be given. Then determine μ = μ ( t ) α H ( 0 , T ) by data
Ω u ( x , t ) θ ( x ) d x , 0 < t < T .
By Theorem 13, we know that the solution u exists uniquely and u L 2 ( 0 , T ; L 2 ( Ω ) ) α H ( 0 , T ; H 2 ( Ω ) H 0 1 ( Ω ) ) . Therefore, the data in (112) are well-defined in L 2 ( 0 , T ) . As another choice of data in the case where the spatial dimensions d 3 , we can choose u ( x 0 , t ) for 0 < t < T with fixed point x 0 Ω . By 1 d 3 , the Sobolev embedding implies u α H ( 0 , T ; H 2 ( Ω ) ) α H ( 0 , T ; C ( Ω ¯ ) ) , and so u ( x 0 , · ) α H ( 0 , T ) , which implies that the data are well-defined in α H ( 0 , T ) . However we are restricted to the data in (112).
The data in (112) are spatial average values of u ( x , t ) with the weight function θ ( x ) . We can choose θ ( x ) in (112) such that supp θ is concentrating near one fixed point x 0 Ω , which means that the data are mean values of u ( · , t ) in a neighborhood of x 0 .
If α > 1 2 , then α H ( 0 , T ) C [ 0 , T ] by the Sobolev embedding, and so the space α H ( 0 , T ) can contain a linear combination of Dirac delta functions:
k = 1 N q k δ t k ( t ) ,
where q k R , 0 , t j ( 0 , T ) for 1 k N and H 1 ( Ω ) < δ t k , φ > H 0 1 ( Ω ) = φ ( t k ) for all φ C 0 ( 0 , T ) . Then our inverse source problem is concerned with determination of N, q k and t k for 1 k N .
Now we are ready to state the following.
Theorem 15.
We assume that f satisfies
Ω θ ( x ) f ( x ) d x 0 .
If
Ω θ ( x ) u ( x , t ) d x = 0 , 0 < t < T ,
then μ = 0 in α H ( 0 , T ) .
Proof. 
The following uniqueness is known: Let v H α ( 0 , T ; L 2 ( Ω ) ) L 2 ( 0 , T ; H 2 ( Ω ) H 0 1 ( Ω ) ) satisfy
t α v + A v = g ( t ) f ( x ) , 0 < t < T ,
where g L 2 ( 0 , T ) and f L 2 ( Ω ) . Under assumption (113), condition
Ω θ ( x ) v ( x , t ) d x = 0 , 0 < t < T ,
yields g = 0 in L 2 ( 0 , T ) .
This uniqueness for g L 2 ( 0 , T ) can be proved by combining Duhamel’s principle and the uniqueness result for initial boundary value problems for fractional partial differential equations with f = 0 . We can refer also to pp. 411–464 of a handbook by Li, Liu and Yamamoto [26] and Sakamoto and Yamamoto [19]. In particular, Jiang, Li, Liu and Yamamoto [32] establish the uniqueness for the case of f = 0 with a non-symmetric A. Thus, the uniqueness in the inverse source problem of determining g is classical within the category of L 2 ( 0 , T ) , and so we omit the proof.
Now we can readily reduce the proof of Theorem 15 to the case of g L 2 ( 0 , T ) . We set v : = J α u . Then, by Propositions 5 and 9 (iii), we see that v H α ( 0 , T ; L 2 ( Ω ) ) L 2 ( 0 , T ; H 2 ( Ω ) H 0 1 ( Ω ) ) satisfy
t α v + A v = ( J α μ ) ( t ) f ( x ) i n   L 2 ( 0 , T ; L 2 ( Ω ) ) ,
where J α μ L 2 ( 0 , T ) .
Noting by Proposition 9 (ii) that Ω θ ( x ) u ( x , · ) d x L 2 ( 0 , T ) , by (112), we obtain
0 = J α Ω θ ( x ) u ( x , · ) d x = J α Ω θ ( x ) u ( x , · ) d x = 1 Γ ( α ) 0 t ( t s ) α 1 Ω θ ( x ) u ( x , s ) d x d s = Ω θ ( x ) ( J α u ) ( x , t ) d x = 0 , 0 < t < T .
That is, we reach (115), which yields J α μ = 0 in ( 0 , T ) by the uniqueness in the inverse problem in the case of J α μ L 2 ( 0 , T ) . Since J α : α H ( 0 , T ) L 2 ( 0 , T ) is injective by Proposition 9 (ii), we obtain that μ = 0 in α H ( 0 , T ) . Thus the proof of Theorem 15 is complete. □

8. Concluding Remarks

8.1. Main Messages

(A) 
We establish fractional derivatives as operators in subspaces of the Sobolev–Slobodecki spaces, whose orders are not necessarily integer nor positive, and consider fractional derivatives in spaces of distribution. Accordingly, we extend classes of functions, of which we take fractional derivatives.
In such distribution spaces, we can justify the fractional calculus and we can argue as if all the functions under consideration would be in L 1 ( 0 , T ) . Such a typical example is a fractional derivative of a Heaviside function in Example 5.
(B) 
More importantly, with our framework of fractional calculus, we can construct a fundamental theory for linear fractional differential equations which is easily adjusted to weak solutions and strong solutions and also classical solutions.
We intend for this article to be an introductory account aiming at operator theoretical treatments of time-fractional partial differential equations. Comprehensive descriptions should require more work, and so this article can provide the essence of the foundations. Next, we give a summary and some prospects.

8.2. What We Have Accomplished

(i) 
In Section 2, we introduce Sobolev spaces H α ( 0 , T ) as subspaces of Sobolev–Slobodecki spaces H α ( 0 , T ) , and we define a fractional derivative t α as an isomorphism from the subspace H α ( 0 , T ) to L 2 ( 0 , T ) and finally as an isomorphism from H α + β ( 0 , T ) H β ( 0 , T ) and β H ( 0 , T ) α β H ( 0 , T ) for α > 0 and β 0 , where β H ( 0 , T ) and α β H ( 0 , T ) are defined by the duality and are subspaces of the distribution space. The key for the definition of t α is an operator theory, and we always attach the fractional derivatives with their domains. The extension procedure is governed by the Riemann–Liouville fractional integral operator J α , and t α is defined by the inverse to J α with suitable domain.
(ii) 
In Section 3, we establish several basic properties of t α , which are naturally expected to hold as formulae of derivatives. We apply some of them to fractional differential equations.
(iii) 
In Section 5 and Section 6, we formulate initial value problems for linear fractional ordinary differential equations and initial boundary value problems for linear fractional partial differential equations to prove the well-posedness and the regularity of solutions within two categories: weak solution and strong solution. Our defined t α enables us to treat fractional differential equations in a feasible manner.
(iv) 
Our research is strongly motivated by inverse problems for fractional differential equations. The studies of inverse problems can be well based on the formulation proposed in this article. Taking into consideration the great variety of inverse problems, in Section 7, we are obliged to be limited to one simple inverse problem in order to illustrate the effectiveness of our framework.

8.3. What We Will Accomplish

Naturally, we must skip many important and interesting issues. However, much research is taking place, and here we give comprehensive prospects for future research topics. We write some of them, related to our framework.
(I) 
As fractional partial differential equations, we exclusively consider evolution equations with elliptic operators of the second order for the spatial part, which is classified into a fractional diffusion-wave equation. The mathematical research should not be limited to such evolution equations. For example, the fractional transport equation is also significant:
t α u ( x , t ) + H ( x , t ) · u ( x , t ) = F ( x , t ) .
One can refer to Luchko, Suzuki and Yamamoto [33] for a maximum principle and to Namba [34] for a viscosity solution.
(II) 
In this article, we do not apply our framework to non-linear fractional differential equations, although necessary steps were prepared. We can refer to Luchko and Yamamoto [35] as one recent work.
(III) 
For fractional partial differential equations, we do not consider non-homogeneous boundary values, in spite of the necessity and the importance. We need more delicate treatments; for an example, see Yamamoto [36]. Moreover, we should develop the corresponding arguments also for the cases of the Neumann and the Robin boundary conditions. Luchko and Yamamoto [35] discuss these cases partly.
(IV) 
We should study variants of fractional derivatives according to physical backgrounds. We mention a distributed derivative just as one example:
0 1 t α u ( x , t ) d α .
The work of Yamamoto [37] studies similar treatments for a generalized fractional derivative
1 Γ ( 1 α ) 0 t ( t s ) α g ( t s ) d v d s ( s ) d s
with 0 < α < 1 and suitable function g and discusses corresponding initial value problems for time-fractional ordinary differential equations. This is another future issue about how well our framework works for such derivatives.
(V) 
Our approach is based essentially on the L 2 -space, and our theory is consistent within L 2 -based Sobolev spaces of any order α R . Therefore, for example, for treating L 1 -functions in time as source terms F ( x , t ) , we have to embed such functions to an L 2 -based Sobolev space of negative order. This is not the best possible way for gaining L p -regularity in time with p 2 , and we should construct the corresponding L p -theory for t α . We can refer to Yamamoto [38] for similar work discussing some fundamental properties in the L p -case with 1 p < .
(VI) 
With the aid of our framework for the direct problem, results on inverse problems can be expected to be sharpened, in view of required regularity, for instance. These should be the main future topics after this article.

Funding

This research was funded by Grants-in-Aid for Scientific Research (A) of Japan Society for the Promotion of Science, grant number 20H00117, and by the National Natural Science Foundation of China, grant numbers 11771270 and 91730303. This research has also been supported by the RUDN University Strategic Academic Leadership Program.

Acknowledgments

The author thanks the anonymous referees for useful comments and Bangti Jin (University College London) for the careful reading and comments.

Conflicts of Interest

The author declares no conflict of interest.

Appendix A. Sketches of Proofs of Lemmata 4 and 5

Proof of Lemma 4.
(i) By (73) and (74), noting that
L 1 2 v = n = 1 λ n 1 2 ( v , φ n ) L 2 ( Ω ) φ n
for v L 2 ( Ω ) , we obtain
L 1 2 L 1 2 v = v f o r   v L 2 ( Ω )
and
( L 1 2 y , z ) L 2 ( Ω ) = ( y , L 1 2 z ) L 2 ( Ω ) f o r   y , z D ( L 1 2 ) = H 0 1 ( Ω ) .
Next, we will prove that there exists a constant C > 0 such that
C 1 v H 1 ( Ω ) L 1 2 v L 2 ( Ω ) C v H 1 ( Ω ) f o r   v L 2 ( Ω ) .
Verification of (A3).
Let v L 2 ( Ω ) . Then we have L 1 2 v H 0 1 ( Ω ) by (73) and L 1 2 L 1 2 v = v . Hence, by applying (73) again, equalities (A1) and (A2) yield
H 1 ( Ω ) < v , ψ > H 0 1 ( Ω ) = H 1 ( Ω ) < L 1 2 L 1 2 v , ψ > H 0 1 ( Ω ) = ( L 1 2 L 1 2 v , ψ ) L 2 ( Ω ) = ( L 1 2 v , L 1 2 ψ ) L 2 ( Ω ) f o r   v L 2 ( Ω ) a n d ψ H 0 1 ( Ω ) .
Consequently, applying (74), we obtain
v H 1 ( Ω ) = sup ψ H 0 1 ( Ω ) = 1 | H 1 ( Ω ) < v , ψ > H 0 1 ( Ω ) | = sup ψ H 0 1 ( Ω ) = 1 | ( L 1 2 v , L 1 2 ψ ) L 2 ( Ω ) | L 1 2 v L 2 ( Ω ) sup ψ H 0 1 ( Ω ) = 1 L 1 2 ψ L 2 ( Ω ) C L 1 2 v L 2 ( Ω ) .
Therefore, the first inequality of (A3) is proved.
Next, we have
L 1 2 v L 2 ( Ω ) = sup ψ H 0 1 ( Ω ) = 1 | ( L 1 2 v , ψ ) L 2 ( Ω ) | = sup ψ H 0 1 ( Ω ) = 1 | ( v , L 1 2 ψ ) L 2 ( Ω ) | .
Since
L 1 2 ψ H 0 1 ( Ω ) C L 1 2 ( L 1 2 ψ ) L 2 ( Ω ) = C ψ L 2 ( Ω ) ,
by (A1), we obtain
sup ψ L 2 ( Ω ) = 1 | ( v , L 1 2 ψ ) L 2 ( Ω ) | sup ψ ˜ H 0 1 ( Ω ) C | ( v , ψ ˜ ) L 2 ( Ω ) | C sup μ H 0 1 ( Ω ) = 1 | H 1 ( Ω ) < v , μ > H 0 1 ( Ω ) | C v H 1 ( Ω ) .
Hence, the second inequality of (A3) is proved, and the verification of (A3) is complete.
Since L 2 ( Ω ) is dense in H 1 ( Ω ) , in view of (A3), we can extend L 1 2 : L 2 ( Ω ) L 2 ( Ω ) to L 1 2 : H 1 ( Ω ) L 2 ( Ω ) . More precisely, for any v H 1 ( Ω ) , we choose v n L 2 ( Ω ) , n N such that lim n v n = v in H 1 ( Ω ) . Then (A3) implies lim m , n L 1 2 ( v m v n ) L 2 ( Ω ) = 0 , that is, a sequence { L 1 2 v n } n N L 2 ( Ω ) is a Cauchy sequence and we can define
L 1 2 v : = lim n L 1 2 v n i n   L 2 ( Ω ) .
Then, by (A3), the limit is independent of choices of { v n } n N , and we see that (A3) holds for v H 1 ( Ω ) . Thus the proof of part (i) of Lemma 4 is complete.
(ii) By (73), we can directly verify
K ( t ) v = L 1 2 K ( t ) L 1 2 v f o r   t > 0   a n d   v L 2 ( Ω ) .
Let v H 1 ( Ω ) be arbitrarily given. Since L 2 ( Ω ) is dense in H 1 ( Ω ) , there exists a sequence v n L 2 ( Ω ) , n N such that v n v in H 1 ( Ω ) as n . By part (i) and the extended L 1 2 , estimate (A3) holds for v H 1 ( Ω ) . Hence, L 1 2 v n L 1 2 v in L 2 ( Ω ) .
In terms of (75), we see
L 1 2 K ( t ) L 1 2 v n L 1 2 K ( t ) L 1 2 v i n   L 2 ( Ω )
as n for t > 0 . Since L 1 2 K ( t ) L 1 2 v n = v n by v n L 2 ( Ω ) , we obtain v n L 1 2 K ( t ) L 1 2 v in L 2 ( Ω ) as n for t > 0 . Using that v n v in H 1 ( Ω ) , we reach L 1 2 K ( t ) L 1 2 v = v for t > 0 . Thus the proof of part (ii) is complete.
(iii) Let b j C 1 ( Ω ¯ ) , 1 j d . Then we will verify
b j j u H 1 ( Ω ) C u L 2 ( Ω ) , 1 j d f o r   a l l   u L 2 ( Ω ) .
Here and henceforth, constants C > 0 depend on b j .
If (A4) is verified, then we can complete the proof of part (iii) of the lemma as follows: By combining (A4) with part (i) of the lemma, we obtain
L 1 2 ( b j j u ) L 2 ( Ω ) C b j j u H 1 ( Ω ) C u L 2 ( Ω ) .
Verification of (A4). By the definition of b j j u as an element in H 1 ( Ω ) , in terms of b j C 1 ( Ω ¯ ) , we have
b j j u H 1 ( Ω ) = sup ψ H 0 1 ( Ω ) = 1 H 1 ( Ω ) < b j j u , ψ > H 0 1 ( Ω ) = sup ψ H 0 1 ( Ω ) = 1 | ( u , j ( b j ψ ) ) L 2 ( Ω ) | sup ψ H 0 1 ( Ω ) = 1 u L 2 ( Ω ) j ( b j ψ ) L 2 ( Ω ) C u L 2 ( Ω ) .
Thus the verification of (A4) and, accordingly, the proof of Lemma 4 are complete. □
Proof of Lemma 5.
(i) We set
S m ( t ) a : = n = 1 m E α , 1 ( λ n t α ) ( a , φ n ) φ n f o r   m N .
Proposition 12 (i) yields
t α ( S m ( t ) a a ) = n = 1 m λ n E α , 1 ( λ n t α ) ( a , φ n ) φ n = L n = 1 m E α , 1 ( λ n t α ) ( a , φ n ) φ n ,
that is,
t α ( S m ( t ) a a ) + L S m ( t ) a = 0 , 0 < t < T , m N .
Since
L S m ( t ) a = t α n = 1 m λ n t α E α , 1 ( λ n t α ) ( a , φ n ) , 0 < t < T ,
by (42), we estimate
L S m ( t ) a L 2 ( Ω ) 2 = t 2 α n = 1 m | λ n t α E α , 1 ( λ n t α ) | 2 | ( a , φ n ) | 2 C t 2 α n = 1 m | ( a , φ n ) | 2 , 0 < t < T .
Therefore, lim m L S m ( t ) a = L S ( t ) a in L 2 ( Ω ) for fixed t > 0 . Hence, letting m in (A5), we obtain t α ( S ( t ) a a ) + L S ( t ) a = 0 for 0 < t < T .
Next, let a H 0 1 ( Ω ) . Since
λ n 1 2 ( a , φ n ) = ( a , L 1 2 φ n ) = ( L 1 2 a , φ n ) f o r   a H 0 1 ( Ω ) ,
we have
L S ( t ) a = n = 1 E α , 1 ( λ n t α ) λ n ( a , φ n ) φ n = n = 1 λ n 1 2 E α , 1 ( λ n t α ) λ n 1 2 ( a , φ n ) φ n = t α 2 n = 1 ( λ n t α ) 1 2 E α , 1 ( λ n t α ) ( L 1 2 a , φ n ) φ n ,
so that
L S ( t ) a L 2 ( Ω ) C t α 2 n = 1 | λ n t α | | E α , 1 ( λ n t α ) | 2 | ( L 1 2 a , φ n ) | 2 1 2 C t α 2 L 1 2 a L 2 ( Ω ) C t α 2 a H 0 1 ( Ω )
by (42). Therefore, this means that L S ( t ) a L 2 ( 0 , T ; L 2 ( Ω ) ) for a H 0 1 ( Ω ) and
L S ( t ) a L 2 ( 0 , T ; L 2 ( Ω ) ) C a H 0 1 ( Ω ) .
Since t α ( S ( t ) a a ) + L S ( t ) a = 0 for 0 < t < T , by (A6), the proof of part (i) is complete.
(ii) We set
R ( t ) : = 0 t K ( t s ) F ( s ) d s , R m ( t ) : = 0 t K m ( t s ) F ( s ) d s = n = 1 m 0 t ( t s ) α 1 E α , α ( λ n ( t s ) α ) ( F ( s ) , φ n ) d s φ n ( x ) , F m : = n = 1 m ( F , φ n ) φ n
for 0 < t < T and m N . Then, by Proposition 13, we can readily verify
t α R m + L R m = F m i n   L 2 ( 0 , T ; L 2 ( Ω ) ) f o r   a l l   m N .
Moreover,
L R m ( t ) L 2 ( Ω ) 2 = n = 1 m 0 t λ n ( t s ) α 1 E α , α ( λ n ( t s ) α ) ( F ( s ) , φ n ) d s 2 .
Hence,
L R m L 2 ( 0 , T ; L 2 ( Ω ) ) 2 = n = 1 m 0 T 0 t λ n ( t s ) α 1 E α , α ( λ n ( t s ) α ) ( F ( s ) , φ n ) d s 2 d t = n = 1 m ( λ n s α 1 E α , α ( λ n s α ) ( F ( s ) , φ n ) L 2 ( 0 , T ) 2 n = 1 m 0 T | λ n s α 1 E α , α ( λ n s α ) | d s 2 0 T | ( F ( s ) , φ n ) | 2 d s 0 T n = 1 m | ( F ( s ) , φ n ) | 2 d s = F m L 2 ( 0 , T ; L 2 ( Ω ) ) 2 .
For the last inequality and the second-to-last inequality, we apply the Young inequality on the convolution and (48), respectively.
Consequently,
L R m L 2 ( 0 , T ; L 2 ( Ω ) ) 2 C F m L 2 ( 0 , T ; L 2 ( Ω ) ) 2 , m N .
Since F L 2 ( 0 , T ; L 2 ( Ω ) ) , we see that L R m L R in L 2 ( 0 , T ; L 2 ( Ω ) ) as m . By (A7) we see
lim m t α R m = lim m ( F m L R m ) = F L R i n   L 2 ( 0 , T ; L 2 ( Ω ) ) .
Since lim m R m = R in L 2 ( 0 , T ; L 2 ( Ω ) ) , it follows from (91) that lim m t α R m = t α R in α H ( 0 , T ; L 2 ( Ω ) ) . Hence (A9) yields t α R = F L R in α H ( 0 , T ; L 2 ( Ω ) ) .
In view of (A7) and (A8), we reach
t α R m L 2 ( 0 , T ; L 2 ( Ω ) ) = L R m + F m L 2 ( 0 , T ; L 2 ( Ω ) ) L R m L 2 ( 0 , T ; L 2 ( Ω ) ) + F m L 2 ( 0 , T ; L 2 ( Ω ) ) C F m L 2 ( 0 , T ; L 2 ( Ω ) ) f o r   e a c h   m N .
Letting m , we reach
t α R L 2 ( 0 , T ; L 2 ( Ω ) ) C F L 2 ( 0 , T ; L 2 ( Ω ) ) .
Thus the proof of Lemma 5 is complete. □

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Yamamoto, M. Fractional Calculus and Time-Fractional Differential Equations: Revisit and Construction of a Theory. Mathematics 2022, 10, 698. https://doi.org/10.3390/math10050698

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Yamamoto M. Fractional Calculus and Time-Fractional Differential Equations: Revisit and Construction of a Theory. Mathematics. 2022; 10(5):698. https://doi.org/10.3390/math10050698

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Yamamoto, Masahiro. 2022. "Fractional Calculus and Time-Fractional Differential Equations: Revisit and Construction of a Theory" Mathematics 10, no. 5: 698. https://doi.org/10.3390/math10050698

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