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Article

Fractional Integral Equations Tell Us How to Impose Initial Values in Fractional Differential Equations

by
Daniel Cao Labora
Department of Applied Mathematics I, School of Forest Engineering, Universidade de Vigo, Campus Universitario da Xunqueira, S/N, 36005 Pontevedra, Spain
Mathematics 2020, 8(7), 1093; https://doi.org/10.3390/math8071093
Submission received: 31 May 2020 / Revised: 30 June 2020 / Accepted: 2 July 2020 / Published: 4 July 2020
(This article belongs to the Special Issue Fractional Integrals and Derivatives: “True” versus “False”)

Abstract

:
One major question in Fractional Calculus is to better understand the role of the initial values in fractional differential equations. In this sense, there is no consensus about what is the reasonable fractional abstraction of the idea of “initial value problem”. This work provides an answer to this question. The techniques that are used involve known results concerning Volterra integral equations, and the spaces of summable fractional differentiability introduced by Samko et al. In a few words, we study the natural consequences in fractional differential equations of the already existing results involving existence and uniqueness for their integral analogues, in terms of the Riemann–Liouville fractional integral. In particular, we show that a fractional differential equation of a certain order with Riemann–Liouville derivatives demands, in principle, less initial values than the ceiling of the order to have a uniquely determined solution, in contrast to a widely extended opinion. We compute explicitly the amount of necessary initial values and the orders of differentiability where these conditions need to be imposed.

1. Introduction

One of the most typical trademarks involving Fractional Calculus is the wide range of opinions about the notions of what is a natural fractional version of some integer order concept and what is not. On the one hand, this plurality leads interesting debates and fosters a critical thinking about whether research is going “in the right direction” or not. On the other hand, it is difficult to handle such an amount of different notions and ideas in the extant literature. In particular, it is common to find lots of generalized fractional versions of a single integer order concept, some of them not very accurate or incoherent between them. These debates are still very alive nowadays [1].
In this frame, the task of this paper is to point out some relevant facts concerning the imposition of initial values for Riemann–Liouville fractional differential equations (FDE). Although the existence and study of FDE has been widely described, for instance in [2,3], in this paper, we provide strong reasons to reconsider the way of imposing initial values.
We have to highlight that our research has been conducted for the Riemann–Liouville fractional derivative, which is the most classical extension for the usual derivative. In addition, the results have been developed for the particular case of linear equations with constant coefficients. However, it seems natural that the ideas described here could be extended to much more general cases.
The main reason to study this issue for the Riemann–Liouville fractional derivative, and not for other fractional versions, is that it is the left inverse of the Riemann–Liouville fractional integral. In this sense, if we restrict the study of Fractional Calculus to functions defined on finite length intervals [ a , b ] , it is a big consensus that Riemann–Liouville fractional integral with base point a is the unique reasonable extension for the integral operator a t . The previous asseveration is not a simple opinion, since the Riemann–Liouville fractional integral can be characterized axiomatically in very reasonable terms.
Theorem 1
(Cartwright-McMullen, [4]). Given a fixed a R , there is only one family of operators I a + α α > 0 on L 1 [ a , b ] satisfying the following conditions:
1. 
The operator of order 1 is the usual integral with base point a. (Interpolation property)
2. 
The Index Law holds. That is, I a + α I a + β = I a + α + β for all α , β > 0 . (Index Law)
3. 
The family is continuous with respect to the parameter. That is, the following map Ind a : R + End B L 1 [ a , b ] given by Ind a ( α ) = I a + α is continuous, where End B L 1 [ a , b ] denotes the Banach space of bounded linear endomorphisms on L 1 [ 0 , b ] . (Continuity)
This family is precisely given by the Riemann–Liouville fractional integrals, whose expression will be recalled during this paper.
Hence, it makes sense to study in detail fractional integral problems for the Riemann–Liouville fractional integral to derive consequences for the corresponding fractional equations afterwards. Finally, to draw the attention of curious readers, we mention again that one of the most interesting results that we have found out is that a FDE of order α > 0 with Riemann–Liouville derivatives can demand, in principle, less initial values than α to have a uniquely determined solution. This result differs from a widely held opinion (see Theorem 1, Section 5.5, in [5]) which states that the necessary amount of initial values is α . The reason for this discrepancy is that the question involving the “fractional smoothness” of the solutions is often neglected, since many results are derived after a not totally rigorous usage of certain mathematical concepts or results. In other words, it is important to build first the space where solutions lie in, to later seek solutions in that space.
A complete range of highlighted results with their implications can be consulted in Section 5, while the previous sections are devoted to the corresponding deductions.

Goal of the Work

The goal of this work is to study how should we impose initial values in fractional problems with a Riemann–Liouville derivative to ensure that they have a smooth and unique solution, where smooth simply means that the solution lies in a certain suitable space of fractional differentiability. To achieve this, we will depart from some known results involving the Riemann–Liouville fractional integral, since it arises as the natural generalization of the usual integral operator; recall Theorem 1.
First, we will recall some results that imply that fractional integral problems have always a unique solution. We also recall the fundamental notions concerning Fractional Calculus, and we pay special attention to the functional spaces where calculations are performed, and especially where fractional derivatives are well defined. Note that this point of “where are functions defined” is crucial to talk about existence or uniqueness of solution and is often neglected in the extant literature concerning Fractional Calculus. Indeed, to avoid this problem, much research has been conducted for Caputo derivatives instead of Riemann–Liouville, see, for instance, [6,7] or general comments in [8]. The ideas of this paragraph are developed in the second section, and most of them are available in the extant literature, except (to the best of the author’s knowledge) Lemma 2, which plays a key role in the rest of the paper.
Second, we see how each FDE of order β is linked with a family of fractional integral problems, whose source term lives in a β dimensional affine subspace of L 1 [ 0 , b ] . This means that each solution to the FDE is a solution to one (and only one!) fractional integral problem of the β dimensional family. Conversely, any solution to a fractional integral problem of the family solves the FDE, provided that the solution is smooth enough. In general, the set of source terms of the family of fractional integral problems that provide a smooth solution will consist in an affine subspace of L 1 [ 0 , b ] of a dimension lower than β . This is done in the third section.
Third, we characterize when a source term of the β dimensional family induces a smooth solution, and thus a solution to the associated FDE. In particular, the affine space of such source terms is shown to have dimension β β , where β is the highest order of differentiability in the FDE such that β β N . This characterization induces a natural correspondence between each source term inducing a smooth solution for the integral problem and a vector of β β initial values fulfilled by the solution. More specifically, if we denote the solution to the FDE by y ( t ) , the initial values ensuring existence and uniqueness of solution are D 0 + β k y ( 0 ) for k { 1 , 2 , , β β } . The content of this paragraph is discussed in the fourth section.
Finally, we establish a section of conclusions to highlight the most relevant obtained results, and to point out to some relevant work that should be performed in the future to continue with this approach.

2. Fundamental Notions

In this section, we will introduce the fundamental notions of Fractional Calculus that we are going to use, together with their more relevant properties and some results of convolution theory that will be useful for our purposes. We assume that the reader is familiar with the basic theory of Banach spaces, Special Functions, and Integration Theory, especially the main facts involving the space of integrable functions over a finite length interval, denoted by L 1 [ a , b ] , and the main properties of the Γ function.

2.1. The Riemann–Liouville Fractional Integral

We briefly introduce the Riemann–Liouville fractional integral, together with its most relevant properties. We make this introduction from the perspective of convolutions, since it will be relevant to notice that the Riemann–Liouville fractional integral is no more than a particular convolution operator, to apply later some adequate results of convolution theory.
Definition 1.
Given f L 1 [ a , b ] , we defined its associated convolution operator as C a ( f ) : L 1 [ a , b ] L 1 [ a , b ] defined as
( f a g ) ( t ) : = ( C a ( f ) g ) ( t ) : = a t f ( t s + a ) · g ( s ) d s
for g L 1 [ a , b ] and t [ a , b ] . Under the previous notation, we say that f is the kernel of the convolution operator C a ( f ) .
Definition 2.
We define the left Riemann–Liouville fractional integral of order α > 0 of a function f L 1 [ a , b ] with base point a as
I a + α g ( t ) = a t ( t s ) α 1 Γ ( α ) · g ( s ) d s ,
for almost every t [ a , b ] . In case that α = 0 , we just define
I a + 0 g ( t ) = Id g ( t ) = g ( t ) .
Without loss of generality, we will assume that a = 0 , since the results for a generic value of a can be achieved after developing them for a = 0 and applying a suitable translation. Moreover, when using the expression “Riemann–Liouville fractional integral”, we will be referring to the left Riemann–Liouville fractional integral with base point a = 0 .
Remark 1.
We observe that, for α > 0 , the Riemann–Liouville fractional integral operator I 0 + α can be written as a convolution operator C 0 ( f ) , with kernel
f ( t ) = t α 1 Γ ( α ) .
It is well known that the Riemann–Liouville fractional integral fulfills the following properties, see [9].
Proposition 1.
For every α , β 0 :
  • I 0 + α is well defined, meaning that I 0 + α L 1 [ 0 , b ] L 1 [ 0 , b ] .
  • I 0 + α is a continuous operator (equivalently, a bounded operator) from the Banach space L 1 [ 0 , b ] to itself.
  • I 0 + α is an injective operator.
  • I 0 + α preserves continuity, meaning that I 0 + α C [ 0 , b ] C [ 0 , b ] .
  • We have the Index Law I 0 + β I 0 + α = I 0 + α + β for α , β 0 . In particular, I 0 + α + β L 1 [ 0 , b ] I 0 + β L 1 [ 0 , b ] .
  • Given f L 1 [ 0 , b ] and α 1 , we have that I 0 + α f is absolutely continuous and, moreover, I 0 + α f ( 0 ) = 0 .
Furthermore, we will also use several times the following well known and straightforward remark, see [9].
Remark 2.
We have that, for β > 1 and α 0 ,
I 0 + α t β = Γ ( β + 1 ) Γ ( α + β + 1 ) t α + β .
Indeed, I 0 + α t β I 0 + γ L 1 [ 0 , b ] if and only if α + β > γ 1 .

2.2. The Riemann–Liouville Fractional Derivative

In this subsection, we will indicate the most relevant points when constructing the Riemann–Liouville fractional derivative. We will begin with a short introduction to absolutely continuous functions of order n, since the spaces where Riemann–Liouville differentiability is well defined can be understood as their natural generalization for the fractional case, see [9].

2.2.1. A Short Reminder Involving Absolutely Continuous Functions and the Fundamental Theorem of Calculus

It is widely known that absolutely continuous functions play a key role in several theories of Mathematical Analysis. These functions can be characterized via a “ ε , δ ” definition, but we only recall that absolutely continuous functions are, essentially, antiderivatives of functions in L 1 [ 0 , b ] up to addition with a constant. We will see later how this idea is highly relevant to construct the maximal spaces where Riemann–Liouville fractional derivatives are well defined.
Theorem 2
(Fundamental Theorem of Calculus). Consider a real function f defined on an interval [ 0 , b ] R . Then, f A C [ 0 , b ] if and only if there exists φ L 1 [ 0 , b ] such that
f ( t ) = f ( 0 ) + 0 t φ ( s ) d s .
This last result allows us to define the derivative of an absolutely continuous function on [ 0 , b ] as a certain function in L 1 [ 0 , b ] . If f A C [ 0 , b ] , we define its derivative D 1 f as the unique function φ L 1 [ 0 , b ] that makes (1) hold.
Remark 3.
It is relevant to have in mind that the previous definition makes sense because the antiderivative operator I 0 + 1 is injective, recall Proposition 1. In particular,
A C [ 0 , b ] = { 1 } I 0 + 1 L 1 [ 0 , b ] ,
where “1” denotes the constant function with value 1.
Of course, it is possible to talk about absolutely functions of order n, for n Z + . In this case, for any n Z + , we say that f A C n [ 0 , b ] if and only if f C n 1 [ 0 , b ] and D n 1 f A C [ 0 , b ] .
Thus, A C n [ 0 , b ] consists of functions that can be differentiated n times, but the last derivative might be computable only in the weak sense of Fundamental Theorem of Calculus. Analogously to the previous remark, we have the following result, see page 3 of [9].
Remark 4.
We have that
A C n [ 0 , b ] = { 1 , t , , t n 1 } I 0 + n L 1 [ 0 , b ] ,
after applying n times the Fundamental Theorem of Calculus 2. Moreover, the sum is direct since the property f I 0 + n L 1 [ 0 , b ] implies that f ( 0 ) = f ( 0 ) = = f n 1 ) ( 0 ) = 0 , and the only polynomial of degree at most n 1 satisfying those conditions is the zero one.
The relevant observation is that the vector space of functions that can be differentiated n times, in the sense of Fundamental Theorem of Calculus 2, has two distinct parts that only share the zero function.
  • The left part { 1 , t , , t n 1 } consists of polynomials of degree strictly lower than n. These functions describe, indeed, the kernel of the operator D n and, thus,
    ker D n = { 1 , t , , t n 1 } .
  • The right part I 0 + n L 1 [ 0 , b ] consists of functions that are obtained after integrating n times an element of L 1 [ 0 , b ] , and hence it contains functions of trivial initial values until the derivative of order n 1 .
At this point, we recall the following result, which is widely known and can be proved immediately with the previous definition.
Proposition 2.
If n > m > 0 , we have that
A C n [ 0 , b ] A C m [ 0 , b ] .
Although Proposition 2 seems irrelevant, it hides the key for a successful treatment of the fractional case. In the next part of the paper, we will reproduce a natural construction for the fractional analogue of the spaces A C n [ 0 , b ] .
We will arrive to the same definition that was already presented in [9]. However, we will emphasize that the space of fractional differentiability of order α will never be contained in the space of order β , except if α β N . This fact will cause FDEs to have less solutions than expected.

2.2.2. The Fractional Abstraction of the Spaces A C n [ 0 , b ]

It is reasonable to define the Riemann–Liouville fractional derivative in such a way that it is the left inverse operator for the Riemann–Liouville fractional integral of the same order. After doing this, an easy analytical expression for its computation follows. Moreover, this explicit description can be extended to a bigger space, and it coincides with the definition available in the classical literature [9].
Property 1.
Consider α 0 . The Riemann–Liouville fractional derivative of order α (and base point 0) fulfils that it is the left inverse of the Riemann–Liouville fractional integral of order α, meaning
D 0 + α I 0 + α f = f ,
for every f L 1 [ 0 , b ] .
We should note that, due to the injectivity of the fractional integral, Property 1 defines the Riemann–Liouville fractional derivative on the space I 0 + α L 1 [ 0 , b ] . Moreover, it will be a surjective operator from I 0 + α L 1 [ 0 , b ] to L 1 [ 0 , b ] .
However, it is clear that we are ignoring something if we pretend that D 0 + α matches perfectly the usual derivative when α is an integer. In particular, we observe that, for an integer value of α , Property 1 only describes the behavior of D α over the space I 0 + α L 1 [ 0 , b ] . Nevertheless, we are missing its definition over the supplementary part of I 0 + α L 1 [ 0 , b ] in A C α [ 0 , b ] , which is ker D α .
This problem is easily solved, since it is possible to describe Property 1 more explicitly. We just observe that the left inverse for I 0 + α is given by the expression D 0 + α = D 0 + α I 0 + α α , due to the Fundamental Theorem of Calculus 2 and the Index Law in Proposition 1. Thus, one could define D 0 + α in a more general space than I 0 + α L 1 [ 0 , b ] , since the only necessary condition to define D 0 + α I 0 + α α f is to ensure that I 0 + α α f A C α [ 0 , b ] . Hence, the following definition is natural.
Definition 3.
For each α > 0 , we construct the following space
X α = I 0 + α α 1 A C α [ 0 , b ] ,
which will be called the space of functions with summable fractional derivative of order α. If α = 0 , we define X α = L 1 [ 0 , b ] .
Remark 5.
Therefore, functions of X α are defined as the ones producing a function in A C α [ 0 , b ] after being integrated α α times. This new function can be differentiated α times in the weak sense of Fundamental Theorem of Calculus 2.
It is relevant to point out that the previous definition, although sometimes related, is different from other notions of “fractional spaces” available in the literature like, for instance, the Fractional Sobolev Spaces in Gagliardo’s sense [10]. In our case, Definition 3 coincides with the one already presented in [9], and we can make it totally explicit.
Lemma 1.
For any α > 0 , we have that
X α = t α α , , t α 2 , t α 1 I 0 + α L 1 [ 0 , b ] .
Proof. 
First, we check t α α , , t α 2 , t α 1 I 0 + α L 1 [ 0 , b ] = { 0 } . If there is a function f in both summands, then I 0 + α α f will be simultaneously a polynomial of degree at most α 1 , and a function in I 0 + α L 1 [ 0 , b ] . Therefore, I 0 + α α f has to be the zero function after repeating the argument in Remark 4 and, since fractional integrals are injective (Proposition 1), f 0 .
It is clear that, applying I 0 + α α to the right-hand side, we will produce a function A C α [ 0 , b ] . Moreover, it is trivial that any function in A C α [ 0 , b ] can be obtained in this way by virtue of Remark 2. Since the operator I 0 + α α is injective, the result follows. □
From the previous lemma, we get this immediate corollary.
Corollary 1.
Given f L 1 [ 0 , b ] , we have that f I 0 + α L 1 [ 0 , b ] if, and only if, f X α and also D s I 0 + α α f ( 0 ) = 0 , for each s { 0 , , α 1 } .
Hence, we can use Property 1 and Corollary 1 to define the Riemann–Liouville fractional derivative, coinciding with Definition 2.4 in [9].
Definition 4.
Consider α 0 and f X α . We define the Riemann–Liouville fractional derivative of order α (and base point 0) as
D 0 + α f : = D 0 + α I 0 + α α f ,
where the last derivative might be only computed in the weak sense of the Fundamental Theorem of Calculus.

2.2.3. Properties of the Space X α

We want to fully understand how D 0 + α works over X α , and the most natural way is to split the problem into two parts, as suggested by Lemma 1. We already know that D 0 + α is the left inverse for I 0 + α , so we should study how D 0 + α behaves when applied to t α α , , t α 2 , t α 1 . It is a well known and straightforward computation that
D 0 + α t α α , , t α 2 , t α 1 = { 0 } .
Hence, the kernel of D 0 + α has dimension α and is given by
ker D 0 + α = t α α , , t α 2 , t α 1 .
Moreover, we should note that, if f ( t ) = a 0 t α α + + a α 1 t α 1 with a j R for each j 0 , 1 , , α 1 , it is immediately necessary to do the following calculations from Remark 2, where j { 1 , , α 1 } ,
I 0 + α α f ( 0 ) = a 0 Γ ( α α + 1 ) , D 0 + α α + j f ( 0 ) = a j Γ ( α α + j + 1 ) .
The previous formula generalizes the obtention of the Taylor coefficients for a fractional case and it can be used to codify functions in X α modulo I 0 + α L 1 [ 0 , b ] , since
I 0 + α α g ( 0 ) = 0 , D 0 + α α + j g ( 0 ) = 0 ,
for g I 0 + α L 1 [ 0 , b ] , due to Proposition 1.

2.2.4. Intersection of Fractional Summable Spaces

In general, fractional differentiation presents some extra problems that do not exist when dealing with fractional integrals. One of the most famous ones is that there is no Index Law for fractional differentiation. One underlying reason for all these complications is the following one.
Remark 6.
The condition α > β does not ensure X α X β , although the condition α β N trivially does. This makes Riemann–Liouville fractional derivatives somehow tricky, since the differentiability for a higher order does not imply, necessarily, the differentiability for a lower order with a different decimal part. In particular, this fact has critical implications when considering fractional differential equations, as we shall see in the paper, since the unknown function has to be differentiable for each order involved in the equation. These problems give an idea of why it can be a reasonable approach to work with fractional integrals instead, and try to inherit the obtained results for the case of fractional derivatives, instead of proving them for fractional derivatives directly.
Consequently, it is interesting to compute the exact structure of a finite intersection of such spaces of fractional differentiability of different orders. To the best of our knowledge, this result is not available in the extant literature.
Lemma 2.
Considering β n > > β 1 0 , we have that
j = 1 n X β j = t β n β n β , , t β n 1 I 0 + β n L 1 [ 0 , b ] ,
where β is the maximum β j such that β n β j N . If such a β j does not exist, the result still holds after defining β = 0 .
In particular, I 0 + β n L 1 [ 0 , b ] j = 1 n X β j and it has codimension β n β .
Proof. 
It is obvious that j = 1 n X β j X β n . Hence,
j = 1 n X β j X β n = t β n β n , , t β n 1 I 0 + β n L 1 [ 0 , b ] .
It is clear that I 0 + β n L 1 [ 0 , b ] lies in j = 1 n X β j , for any j { 1 , , n } . This is simply because, due to the Index Law, I 0 + β n L 1 [ 0 , b ] I 0 + β j L 1 [ 0 , b ] , since β n β j .
Thus, the remaining question is to see when a linear combination of the t β n k , where k { 1 , , β n } , lies in j = 1 n X β j . The key remark is to realize that, for any finite set F ( 1 , + ) ,
γ F c γ t γ X β j , where   c γ 0 for   each   γ F ,
if and only if γ β j > 1 or γ β j Z , for every γ { 1 , , r } with c γ 0 . Consequently, it is enough study when t β n k lies in X β j , and there are two options:
  • If β n β j N , we know that t β n k X β j always. This happens because either β n k β j β j , , β j 1 or β n k > β j 1 .
  • In other case, β n and β j do not share decimal parts and we need to have β n k > β j 1 that can be rewritten as k < β n β j + 1 . If we want this to happen for every j such that β n β j N , the condition is equivalent to k < β n β + 1 , where β is the greatest β j such that β n β j N . Indeed, it can be rewritten as 1 k β n β .
Therefore, the coefficients which are not necessarily null are the ones associated with t β n k , where k { 1 , , β n β } . □
Remark 7.
Due to Lemma 2, any affine subspace of j = 1 n X β j with dimension strictly higher than β n β contains two distinct functions whose difference lies in I 0 + β n L 1 [ 0 , b ] . Thus, in any vector subspace of j = 1 n X β j with dimension strictly higher than β n β , there are infinitely many functions that lie in I 0 + β n L 1 [ 0 , b ] .

2.3. Fractional Integral Equations

Consider the fractional integral equation
c n I 0 + γ n + + c 1 I 0 + γ 1 + I 0 + γ 0 x ( t ) = f ˜ ( t ) ,
where f L 1 [ 0 , b ] , γ n > > γ 0 0 and assume that it has a solution x L 1 [ 0 , b ] . Since I 0 + γ n L 1 [ 0 , b ] I 0 + γ 0 L 1 [ 0 , b ] , and the left-hand side lies in I 0 + γ 0 L 1 [ 0 , b ] , the condition f ˜ I 0 + γ 0 L 1 [ 0 , b ] is mandatory to ensure the existence of solution. In that case, we can apply the operator D 0 + γ 0 to the previous equation and we obtain
c n I 0 + α n + + c 1 I 0 + α 1 + Id x ( t ) = f ( t ) ,
where D 0 + γ 0 f ˜ = f and α j = γ j γ 0 for j { 1 , , n } . If we use the notation
Υ = c n I 0 + α n + + c 1 I 0 + α 1 ,
Equation (5) can be rewritten as
Υ + Id x ( t ) = f ( t ) .
Therefore, it is relevant to study the properties of the operator Υ + I d from L 1 [ 0 , b ] to itself, in order to understand Equation (6).

2.3.1. Υ + Id Is Bounded

This claim is a very well-known result, since each summand in Υ is a bounded operator, and Id too, see [2]. It is also possible to prove this, just recalling that Υ is a convolution operator with kernel in L 1 [ 0 , b ] and, thus, a bounded operator.

2.3.2. Υ + Id Is Injective

To prove that Υ + Id is injective, we will need a result concerning the annulation of a convolution. In a few words, we need to know what are the possibilities for the factors of a convolution, provided that the obtained result is the zero function. Roughly speaking, the classical result in this direction, known as the Titchmarsh Theorem, states that the integrand of the convolution from 0 to t is always zero, independently of t.
Theorem 3
(Titchmarsh, [11]). Suppose that f , g L 1 [ 0 , b ] are such that f 0 g 0 . Then, there exist λ , μ R + such that the following three conditions hold:
  • f 0 in the interval [ 0 , λ ] ,
  • g 0 in the interval [ 0 , μ ] ,
  • λ + μ b .
Remark 8.
In particular, the Titchmarsh Theorem states that the operator C 0 ( f ) : L 1 [ 0 , b ] L 1 [ 0 , b ] is injective, provided that f L 1 [ 0 , b ] and that f is not null at any interval [ 0 , λ ] for λ > 0 .
In particular, we need the following result.
Corollary 2.
The operator Υ + I d described in (6) is injective.
Proof. 
Note that we can not apply Theorem 3 directly to Υ + I d , since it is not a convolution operator due to the “Id” term. However, I 0 + 1 T + I d is a convolution operator and we conclude, following Remark 8, that I 0 + 1 Υ + I d is injective. If the previous composition is injective, the right factor Υ + I d has to be injective. □

2.3.3. Υ + Id Is Surjective

In this case, we will use the following result, concerning Volterra integral equations of the second kind. This result essentially states that some family of integral equations do always have a continuous solution, provided that the source term is continuous.
Theorem 4
(Rust, [12]). Given k L 1 [ 0 , b ] , the Volterra integral equation
C 0 ( k ) v ( t ) + v ( t ) : = 0 t k ( t s ) · v ( s ) d s + v ( t ) = w ( t )
has exactly one continuous solution v C [ 0 , b ] , provided that w C [ 0 , b ] and that the following two conditions hold:
  • If h C [ 0 , b ] , then C a ( k ) h C [ 0 , b ] .
  • If n Z + is big enough, then C 0 ( k ) n = C 0 ( k ˜ ) for some k ˜ C [ 0 , b ] ,
Remark 9.
We know that the image of Υ + I d will contain C [ 0 , b ] , since fractional integrals map continuous functions into continuous functions (Proposition 1). Moreover, Υ n will be defined by a continuous kernel when n α 1 1 ; recall that α 1 > 0 was the least integral order in Υ.
We need to conclude that, indeed, the image of Υ + Id is L 1 [ 0 , b ] .
Corollary 3.
The operator Υ + I d described in (6) is surjective.
Proof. 
Consider f L 1 [ 0 , b ] and the equation
Υ + I d x ( t ) = f ( t ) .
We need to show that there is x L 1 [ 0 , b ] solving this problem. Observe that x solves the previous equation if and only if it solves
Υ + I d ( x ( t ) f ( t ) ) = Υ f ( t ) ,
but now the source term is in I 0 + α 1 L 1 [ 0 , b ] . If we repeat this idea inductively, we see that x solves the original equation if and only if
Υ + I d ( x ( t ) ( I d Υ + + ( 1 ) n Υ ) f ( t ) ) = ( 1 ) n + 1 Υ n + 1 f ( t ) .
The right-hand side will be continuous for n α 1 1 and, by Remark 9, it will have a solution. □

2.3.4. Υ + Id Is a Bounded Automorphism in L 1 [ 0 , b ]

We have already seen that Υ + Id is bounded and bijective, and hence the inverse is also bounded due to the Bounded Inverse Theorem for Banach spaces. Therefore, we have the following result.
Theorem 5.
The operator Υ + I d , described in (6) is an invertible bounded linear map from the Banach space L 1 [ 0 , b ] to itself, whose inverse is also bounded.
In particular, we get the following corollary
Theorem 6.
Given f L 1 [ 0 , b ] , the equation
Υ + I d x ( t ) = f ( t )
has exactly one solution x L 1 [ 0 , b ] .
Although it is not the scope of this paper, we highlight that such an equation can be solved using classical techniques for integral equations or specifical tools for the particular case of fractional integral equations, like the one exposed in [13].

3. Implications of Fractional Integral Equations in Fractional Differential Equations

It would be desirable to have a similar result to the previous one for the case of FDE, ensuring existence and uniqueness of solution. Of course, in order to ensure uniqueness of solution, it is necessary to add some “extra conditions” to the differential version of the equation. Namely, one possibility is to impose initial values. In particular, given an ODE of order n Z + with unknown function u, we know that, after fixing the values u ( 0 ) , u ( 0 ) , , u n 1 ) ( 0 ) , we can ensure uniqueness of solution under general hypotheses. The question is: what is the reasonable “fractional analogue” of the previous idea?
Before answering the previous question, we study the solutions of this general linear problem with constant coefficients
c 1 D 0 + β 1 + + c n 1 D 0 + β n 1 + D 0 + β n u ( t ) = w ( t ) ,
where β n > > β 1 0 and w L 1 [ 0 , b ] . Of course, the first point to answer is where should we look for the solution to (7). It is very relevant to clarify completely this issue, since there are classical references; for instance, see [5] (Theorem 1, Section 5.5), which state the following theorem or equivalent versions, which are not totally accurate as we shall comment on in the follow-up.
Theorem 7.
Consider a linear homogeneous fractional differential equation (for Riemann–Liouville derivatives) with constant coefficients and rational orders. If the highest order of differentiation is α, then the equation has α linearly independent solutions.
It is important to note that several references are not clear enough about the notion of solution to a fractional differential equation. With the previous sentence, we mean that it is desirable to introduce a suitable space of differentiable functions first, to later discuss about the solvability of the fractional differential equation. We devote the rest of the paper to show that the previous theorem is only true in some weak sense. Indeed, after defining formally the notion of “strong solution”, we will see that, in general, there are less than α linearly independent solutions. Indeed, only for those “strong” solutions it will be coherent to talk about initial values.
The inaccuracy involving Theorem 7, and similar results, relies in the fact that the notion of solution is not completely specified. Moreover, it is common to find “proofs” that use Laplace Transform techniques without enough mathematical rigor (the final step of inverting the transform is neglected), the order of infinite sums and linear operators is interchanged (without regarding if there are sufficient hypotheses that make it legit),...
If we go back to (7), we can make the following vital remark.
Remark 10.
We recall that, in the usual case of integer orders, we look for the solutions in X β n . Although it is quite common to forget it, the underlying reason to do this is that j = 1 n X β j = X β n when every β j is a non-negative integer. However, in general, this does not necessarily happen when the involved orders are non-integers. Thus, we may have j = 1 n X β j X β n and, of course, a solution to Equation (7) has to lie in j = 1 n X β j .
Consequently, it is convenient to know the structure of the set j = 1 n X β j , which has already been described in Lemma 2, to study existence and uniqueness of solution. Of course, to expect uniqueness of solution, some initial conditions have to be added to Equation (7), but this will be discussed in the next section. The fundamental remark is that solutions to Equation (7) fulfill
D 0 + β n c 1 I 0 + β n β 1 + + c n 1 I 0 + β n β n 1 + I d u ( t ) = w ( t ) .
Consequently, it is quite natural to make the following reflection. If u ( t ) solves (7), it is because
c 1 I 0 + β n β 1 + + c n 1 I 0 + β n β n 1 + I d u ( t ) I 0 + β n w ( t ) + ker D 0 + β n .
We will refer to the set of solutions to Equation (9), as the set of weak solutions. The previous terminology obeys the following reason: although a solution to (7) solves (9), the converse does not hold in general. Namely, a solution to (9) may not lie in j = 1 n X β j . Of course, if the weak solution lies in j = 1 n X β j , then it solves (7). The set of solutions to (7) will be called the set of strong solutions.
At this moment, we know two vital things:
  • We have already described ker D 0 + β n = t β n 1 , , t β n β n that is a vector space of dimension β n . Therefore, the set of weak solutions has dimension β n too, since it is the image of the affine space I 0 + β n w ( t ) + ker D 0 + β n via the automorphism T 1 A u t B ( L 1 [ 0 , b ] ) .
  • The dimension of the set of strong solutions is bounded from above by β n β . If the dimension were higher, due to Remark 7, we could find two different solutions to (7) whose difference would lie in I 0 + β n L 1 [ 0 , b ] . After writing their difference as I 0 + β n g , where g 0 , it would trivially fulfill
    c 1 I 0 + β n β 1 + + c n 1 I 0 + β n β n 1 + I d g ( t ) = 0 ,
    which is not possible since the linear operator on the left-hand side is injective.
From these remarks, there are some remaining points that need to be studied in detail. First, we prove that the bound β n β is sharp by inspecting which of elements in ker D 0 + β n guarantee that the weak solution associated with those elements is, indeed, a strong one.
Remark 11.
We have j = 1 n X β j I 0 + β n β n β + 1 ε L 1 [ 0 , b ] for every increment ε > 0 , but not for ε = 0 . Moreover, if f ker D 0 + β n is chosen as the right summand on the right-hand side in (9), we have that, for γ β n , f I 0 + γ L 1 [ 0 , b ] if and only if u I 0 + γ L 1 [ 0 , b ] .
Therefore, after putting together the two previous ideas, we arrive to the following conclusion. In order to have a strong solution in (9), it is mandatory to select a source term f t β n 1 , , t β n β n β .
We see that the converse also holds, since f t β n 1 , , t β n β n β can be shown to be sufficient, in order to have a strong solution.
Lemma 3.
If u L 1 [ 0 , b ] solves
c 1 I 0 + β n β 1 + + c n 1 I 0 + β n β n 1 + Id u ( t ) = I 0 + β n w ( t ) + f ( t )
for f t β n 1 , , t β n β n β ker D 0 + β n , then u j = 1 n X β j .
Proof. 
If we use the notation Υ : = c 1 I 0 + β n β 1 + + c n 1 I 0 + β n β n 1 , we deduce from Equation (10) that
Υ + Id u ( t ) f ( t ) = I 0 + β n w ( t ) Υ f ( t ) .
Observe now that the summand Υ f ( t ) can be decomposed into two parts, since two different situations can happen:
  • If β n β j N , we see that I 0 + β n β j t β n k will be always in the space I 0 + β n + ( β n β + 1 ) β n β ε L 1 [ 0 , b ] for every ε > 0 . This simply occurs because the worst choice is β j = β and k = β n β . Indeed, for ε small enough, the previous space is contained in I 0 + β n L 1 [ 0 , b ] , since β n β + 1 β n β is strictly positive.
  • If β n β j N , there are two options:
    -
    If β n β j > β n β , we have that I 0 + β n β j t β n k lies again in I 0 + β n L 1 [ 0 , b ] , since the maximum value admitted for k is β n β .
    -
    If β n β j < β n β , we have that I 0 + β n β j t β n k t β n k for some k < k .
Thus, we can write I 0 + β n w ( t ) Υ f ( t ) = I 0 + β n w 1 ( t ) + f 1 ( t ) , and arrive to the equation
c 1 I 0 + β n β 1 + + c n 1 I 0 + β n β n 1 + Id u ( t ) f ( t ) = I 0 + β n w 1 ( t ) + f 1 ( t ) .
Note that f lived in a β n β dimensional vector space, but f 1 lives in a (at most) β n β 1 dimensional vector space.
If we repeat the process, we obtain
c 1 I 0 + β n β 1 + + c n 1 I 0 + β n β n 1 + Id u ( t ) f ( t ) f 1 ( t ) = I 0 + β n w 2 ( t ) + f 2 ( t ) ,
with f 2 lying in a (at most) β n β 2 dimensional vector space. After enough iterations, the vector space has to be zero dimensional and we will have the situation
Υ + Id u ( t ) f ( t ) f r 1 ( t ) = I 0 + β n w r ( t ) I 0 + β n L 1 [ 0 , b ] .
Therefore, u ( t ) f ( t ) f r 1 ( t ) I 0 + β n L 1 [ 0 , b ] . Finally, if we use that
f ( t ) + f 1 ( t ) + + f r 1 ( t ) t β n 1 , , t β n β n β ,
it follows u t β n 1 , , t β n β n β I 0 + β n L 1 [ 0 , b ] = j = 1 n X β j . □

4. Smooth Solutions for Fractional Differential Equations

Until this point, we have checked that, in general, there are more weak solutions (a β n dimensional space) than strong solutions (a β n β dimensional space). We have also seen how weak solutions are codified depending on the source term, more specifically depending on the element chosen in ker D 0 + β n . Moreover, we know that, if the choice is made in a certain subspace of ker D 0 + β n , then the obtained solution is a strong one. However, one could think about codifying strong solutions directly in the fractional differential equation via initial conditions, instead of using fractional integral problems and selecting a source term linked to a strong solution. Therefore, the last task should consist of relating the choices for ker D 0 + β n that give a strong solution with the corresponding initial conditions for the strong problem.
First, to simplify the notation, we reconsider Equation (7) with the additional hypotheses that each positive integer less then or equal to β n β 1 can be written as β n β j for some j.
D 0 + β n c 1 I 0 + β n β 1 + + c n 1 I 0 + β n β n 1 + I d u ( t ) = w ( t )
This does not imply a loss of generality, since we can assume some c j = 0 , if needed. The only purpose of this assumption is to ease the notation in this proof, in the way that is described in the following paragraph.
If β = β n m , then β n β n m is the least possible non-integer difference β n β j . Thus, we can use the previous notational assumption to check that β n β n j = j for j < m and β n β n m ( m 1 , m ) . Thus, β n β = β n β n m = m , and c n m + 1 , , c n 1 are m 1 constants multiplying integrals of integer order in (10).
Now, we provide the main result of this section.
Lemma 4.
Under the previous notation, Equation (7) with given initial values D 0 + β n m u ( 0 ) , , D 0 + β n 1 u ( 0 ) has a unique solution in j = 1 n X β j . This solution coincides with the unique solution to (10), where the source term is the unique function f t β n 1 , , t β n m fulfilling
D 0 + β n m u ( 0 ) = D 0 + β n m f ( 0 ) , D 0 + β n m + 1 u ( 0 ) + c n 1 D 0 + β n m u ( 0 ) = D 0 + β n m + 1 f ( 0 ) , D 0 + β n 1 u ( 0 ) + c n 1 D 0 + β n 2 u ( 0 ) + + c n m + 1 D 0 + β n m u ( 0 ) = D 0 + β n 1 f ( 0 ) .
Proof. 
Consider again Equation (10)
c 1 I 0 + β n β 1 + + c n 1 I 0 + β n β n 1 + Id u ( t ) = I 0 + β n w ( t ) + f ( t ) ,
Recall that we look for strong solutions to (10) that lie in the functional space t β n 1 , , t β n m I 0 + β n L 1 [ 0 , b ] , so we write
u ( t ) = d 1 t β n m + + d m t β n 1 + I 0 + β n u ˜ ( t ) .
Moreover, take into account that a strong choice for f ker D 0 + β n allows us to write
f ( t ) = b 1 t β n m + + b m t β n 1 .
Now, we will just derive the initial conditions after applying D 0 + β n k , for every k { 1 , , m } , and substituting t = 0 in (10).
On the right-hand side, this is easy, since D 0 + β n k I 0 + β n w ( t ) I 0 + 1 L 1 [ 0 , b ] and, thus, the substitution at t = 0 gives zero. The function D 0 + β n k f ( t ) can be computed trivially, due to the expression of f, obtaining
D 0 + β n k f ( 0 ) = Γ ( β n k + 1 ) b k .
On the left-hand side, on the one hand, we have again a similar situation to the previous one, since D 0 + β n k I 0 + β n β j I 0 + β n u ˜ ( t ) I 0 + k + ( β n β j ) L 1 [ 0 , b ] I 0 + 1 L 1 [ 0 , b ] for any subindex j { 1 , , n } and, thus, the substitution at t = 0 gives zero. On the other hand, D 0 + β n k I 0 + β n β j t β n l has three possibilities:
  • If β n β j > l k , then D 0 + β n k I 0 + β n β j t β n l is a scalar multiple of a power of t with positive exponent. Thus, when we make the substitution at t = 0 , we get 0.
  • If β n β j = l k , then D 0 + β n k I 0 + β n β j t β n l = Γ ( β n l + 1 ) is constant, and it is obviously defined for t = 0 .
  • If β n β j < l k m 1 , then β n β j is an integer and the computation D 0 + β n k I 0 + β n β j t β n l gives the zero function.
The interest of the previous trichotomy is that we never obtain some t γ with γ > 0 . In other cases, we would have huge trouble, since we could not evaluate the expression for t = 0 . Fortunately, we can always apply D 0 + β n k to Equation (10), for every value k { 1 , , m } , and substitute at t = 0 . We arrive to the following linear system of equations:
D 0 + β n m u ( 0 ) = D 0 + β n m f ( 0 ) , D 0 + β n m + 1 u ( 0 ) + c n 1 D 0 + β n m u ( 0 ) = D 0 + β n m + 1 f ( 0 ) , D 0 + β n 1 u ( 0 ) + c n 1 D 0 + β n 2 u ( 0 ) + + c n m + 1 D 0 + β n m u ( 0 ) = D 0 + β n 1 f ( 0 ) .
Note that all the involved derivatives in the initial conditions have the same decimal part, since only coefficients c n 1 , , c n m + 1 appear in the system. We also highlight that the system always has a unique solution, since it is triangular and it has no zero element in the diagonal. Therefore, a choice for f linked to a strong solution determines a vector of initial values ( D 0 + β n m u ( 0 ) , , D 0 + β n 1 u ( 0 ) ) and vice versa in a bijective way. □
We shall give two examples summarizing how to apply all the previous results.
Example 1.
Consider the following fractional differential equation (strong problem):
D 0 + 7 3 + 3 D 0 + 4 3 + 4 D 0 + 1 3 u ( t ) = t 3
and define β 1 = 1 3 , β 2 = 4 3 , β 3 = 7 3 . In this case, note that β = 0 , since all the differences β 3 β j are integers. The strong solutions for the example will lie in j = 1 3 X β j . The dimension of the affine space of strong solutions will be β 3 = 3 , and the initial conditions that ensure existence and uniqueness of solution will be D 0 + 4 3 u ( 0 ) = a 3 , D 0 + 1 3 u ( 0 ) = a 2 and I 0 + 2 3 u ( 0 ) = a 1 .
Moreover, after left-factoring D 0 + 7 3 , we find that the associated family of weak problems is
4 I 0 + 2 + 3 I 0 + 1 + Id u ( t ) = I 0 + 7 3 t 3 + f ( t )
where f ( t ) t 4 3 , t 1 3 , t 2 3 , which lives in a three-dimensional space. The a priori weak, obtained solution is always strong since we have that β 3 β = β 3 .
Finally, the relation between a choice for f ( t ) = b 3 t 4 3 + b 2 t 1 3 + b 1 t 2 3 providing a strong solution and the initial conditions a 1 , a 2 and a 3 is
a 1 = I 0 + 2 3 f ( 0 ) = b 1 · Γ 1 2 3 , a 2 + 3 a 1 = D 0 + 1 3 f ( 0 ) = b 2 · Γ 1 + 1 3 , a 3 + 3 a 2 + 4 a 1 = D 0 + 4 3 f ( 0 ) = b 3 · Γ 1 + 4 3 .
Example 2.
Consider the following fractional differential equation (strong problem):
D 0 + 13 4 + 3 D 0 + 9 4 + D 0 + 2 + D 0 + 5 4 + D 0 + 1 u ( t ) = t
and define β 1 = 1 , β 2 = 5 4 , β 3 = 2 , β 4 = 9 4 , β 5 = 13 4 . In this case, note that β = β 3 , since it fulfills the property that β 5 β is the least possible non-integer difference β 5 β j . The strong solutions for the example will lie in j = 1 5 X β j . The dimension of the affine space of strong solutions will be β 5 β = 2 and the initial conditions that ensure existence and uniqueness of solution will be D 0 + 9 4 u ( 0 ) = a 2 and D 0 + 5 4 u ( 0 ) = a 1 .
Moreover, after left-factoring D 0 + 13 4 , we find that the associated family of weak problems is
I 0 + 9 4 + I 0 + 2 + I 0 + 5 4 + 3 I 0 + 1 + Id u ( t ) = I 0 + 13 4 t + f ( t )
where f ( t ) t 9 4 , t 5 4 , t 1 4 , t 3 4 , which lives in a four-dimensional space. The a priori weak, obtained solution will be strong if f ( t ) t 9 4 , t 5 4 .
Finally, the relation between a choice for f ( t ) = b 2 t 9 4 + b 1 t 5 4 providing a strong solution and the initial conditions a 1 and a 2 is
a 1 = D 0 + 5 4 u ( 0 ) = D 0 + 5 4 f ( 0 ) = b 1 · Γ 1 + 5 4 , a 2 + 3 a 1 = D 0 + 9 4 u ( 0 ) + 3 D 0 + 5 4 u ( 0 ) = D 0 + 9 4 f ( 0 ) = b 2 · Γ 1 + 9 4 .

5. Conclusions

We summarize the conclusions obtained in this paper.
  • We have recalled the main results involving existence and uniqueness of solution for linear fractional integral equations with constant coefficients.
  • We have seen that, from each linear FDE with constant coefficients of order β n , it is possible to derive a β n dimensional family of associated fractional integral equations, in a natural way. Moreover, each solution to the fractional differential equation fulfills exactly one of these fractional integral equations.
  • We have shown that there exists a β n β dimensional subfamily (of the β n dimensional family of associated fractional integral equations) such that each solution to a problem of the subfamily gives a solution to the original linear fractional differential equation of order β n . This value β is obtained as the greatest differentiation order in the FDE such that β n β is not an integer. If such a value does not exist, the same result holds after defining β = 0 .
  • We have seen how initial values at t = 0 for the derivatives of orders β n β n β , , β n 1 guarantee existence and uniqueness of solution to a linear fractional differential equation with constant coefficients of order β n . We have described the correspondence between such initial values for the FDE and the selection of a source term in the β n β dimensional subfamily of integral equations, in such a way that both problems have the same unique solution. If β n β n β ( 1 , 0 ) , this first initial value is imposed, indeed, for the fractional integral of order β n β β n .
  • We expect that this idea can be extended to different types of fractional differential problems. It would be nice to amplify the scope of this work to a more general case than the one of constant coefficients. Furthermore, the same philosophy could be applied to other type of problems such as, for instance, periodic ones that have relevant applications [14].

Funding

This research received no external funding.

Conflicts of Interest

The author declares no conflict of interest.

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Cao Labora, D. Fractional Integral Equations Tell Us How to Impose Initial Values in Fractional Differential Equations. Mathematics 2020, 8, 1093. https://doi.org/10.3390/math8071093

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Cao Labora D. Fractional Integral Equations Tell Us How to Impose Initial Values in Fractional Differential Equations. Mathematics. 2020; 8(7):1093. https://doi.org/10.3390/math8071093

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Cao Labora, Daniel. 2020. "Fractional Integral Equations Tell Us How to Impose Initial Values in Fractional Differential Equations" Mathematics 8, no. 7: 1093. https://doi.org/10.3390/math8071093

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