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arXiv:1909.07662v1 [math.FA] 17 Sep 2019

A Hilbert space approach to

fractional diﬀerential equations

Kai Diethelm1, Konrad Kitzing2, Rainer Picard2, Stefan Siegmund2, Sascha Trostorﬀ3,

and Marcus Waurick4

1Fakult¨at Angewandte Natur- und Geisteswissenschaften, Hochschule f¨ur angewandte

Wissenschaften W¨urzburg-Schweinfurt, Germany

2Institute of Analysis, Faculty of Mathematics, TU Dresden, Germany

3Mathematisches Seminar, Christian-Albrechts-Universit¨at zu Kiel, Germany

4Department of Mathematics and Statistics, University of Strathclyde, Glasgow,

Scotland

Abstract

We study fractional diﬀerential equations of Riemann-Liouville and Caputo type in Hilbert spaces.

Using exponentially weighted spaces of functions deﬁned on R, we deﬁne fractional operators by means

of a functional calculus using the Fourier transform. Main tools are extrapolation- and interpolation

spaces. Main results are the existence and uniqueness of solutions and the causality of solution

operators for non-linear fractional diﬀerential equations.

Keywords: fractional diﬀerential equations; Caputo derivative, Riemann-Liouville derivative; causal-

ity;

MSC 2010: 26A33 Fractional derivatives and integrals; 45D05 Volterra integral equations

1 Introduction

The concept of a fractional derivative ∂α

0,α∈]0,1], which we utilize, will be based on inverting a suitable

continuous extension of the Riemann-Liouville fractional integral of continuous functions f∈Cc(R) with

compact support given by

t7→ 1

√2πˆt

−∞

1

Γ(α)(t−s)α−1f(s) ds

as an apparently natural interpolation suggested by the iterated kernel formula for repeated integration.

The choice of the lower limit as −∞ is determined by our wish to study dynamical processes, for which

causality1should play an important role. It is a pleasant fact that the classical deﬁnition of ∂α

0in the sense

1Other frequent choices such as

t7→ 1

√2πχ]a,∞[(t)ˆt

a

1

Γ(α)(t−s)α−1f(s) ds

for a∈R, would lose time-shift invariance (a suggestive choice is a= 0), which we consider undesirable. For our choice

of the limit case a=−∞ it should be noted that the Riemann-Liouville and the Caputo fractional derivative essentially

coincide.

1

of [1] coincides with the other natural choice of ∂α

0as a function of ∂0in the sense of a spectral function

calculus of a realization of ∂0as a normal operator in a suitable Hilbert space setting. This is speciﬁed

below. The Hilbert space framework is based on observations in [5] and has already been exploited for

linear fractional partial diﬀerential equations in [2] and [6]. Since fractional diﬀerential equations are

routinely discussed as integral equations, we shall establish the connection of typical Riemann-Liouville

or Caputo type fractional diﬀerential equations to the above choice of fractional derivative, by working

our way backward to the fractional diﬀerential equation it represents in our chosen terminology.

2 Fractional derivative in a Hilbert space setting

In the present section, we introduce the necessary operators to be used in the following. We will formulate

all results in the vector-valued, more speciﬁcally, in the Hilbert space-valued situation. On a ﬁrst read,

one may think of scalar-valued functions.

To begin with, we introduce an L2-variant of the exponentially weighted space of continuous functions

that proved useful in the proof of the Picard–Lindel¨of Theorem and is attributed to Morgenstern, [3].

We denote by Lp(R;H) and L1

loc(R;H) the space of p-Bochner integrable functions and the space of

locally Bochner integrable functions on a Hilbert space H, respectively.

Deﬁnition. Let Hbe a Hilbert space, ∈Rand p∈[1,∞]. For f∈L1

loc(R;H) we denote e−m f:=

(R∋t7→ e−tf(t)). We deﬁne the normed spaces

Lp

(R;H):=f∈L1

loc(R;H); e−m f∈Lp(R;H),

with norm

kfkLp

(R;H):=

e−mf

Lp(R;H)=ˆRkf(t)kp

He−pt dt1/p

(p < ∞),

kfkLp

(R;H):=

e−mf

Lp(R;H)= ess sup

e−mf

H(p=∞).

Remark 2.1.The operator e−m :Lp

(R;H)→Lp(R;H), f 7→ e−mfis an isometric isomorphism from

Lp

(R;H) to Lp(R;H). Moreover L2

(R;H) is a Hilbert space with scalar product

(f, g)7→ hf, g iL2

(R;H)=ˆRhf(t), g(t)iHe−2t dt.

Next, we introduce the time derivative.

Deﬁnition. Let Hbe a Hilbert space.

(a) Let f, g ∈L1

loc(R;H). We say that f′=g, if for all φ∈C∞

c(R)

−ˆR

fφ′=ˆR

gφ.

(b) Let ∈R. We deﬁne

∂0, :H1

(R;H)⊆L2

(R;H)→L2

(R;H)

f7→ f′,

where H1

(R;H):={f∈L2

(R;H); f′∈L2

(R;H)}.

2

The index 0 in ∂0, shall indicate that the derivative is with respect to time. We will introduce the

fractional derivatives and fractional integrals by means of a functional calculus for ∂0, . For this, we

introduce the Fourier–Laplace transform.

Deﬁnition. Let Hbe a complex Hilbert space. Let ∈R.

(a) We deﬁne the Fourier transform of f∈L1(R;H) by

Ff(ξ) = 1

√2πˆR

f(t)e−iξt dt, ξ ∈R.

(b) We deﬁne the Fourier–Laplace transform on L1

(R;H) by L:=Fe−m.

(c) We deﬁne the Fourier transform on L2(R;H) denoted F:L2(R;H)→L2(R;H) to be the unitary

extension of the operator F:L1(R;H)∩L2(R;H)→L2(R;H).

(d) We deﬁne the Fourier–Laplace transform on L2

(R;H) as the unitary mapping L:=Fe−m :

L2

(R;H)→L2(R;H)

From now on, Hdenotes a complex Hilbert space. With the latter notion at hand, we provide the spectral

representation of ∂0, as the multiplication-by-argument operator

dom(m):={f∈L2(R;H); (R∋ξ7→ ξf(ξ)) ∈L2(R;H)},

m:L2(R;H)⊇dom(m)→L2(R;H), f 7→ (R∋ξ7→ ξf (ξ)).

Theorem 2.2. Let ∈R.Then

(a) ∂0,0=F∗imF,

(b) (e−m)∗∂0,0e−m =∂0, −,

(c) ∂0, =L∗

im +L.

Proof. For the proof of (a), we observe that the equality holds on the Schwartz space S(R;H) of smooth,

rapidly decaying functions. In fact, this is an easy application of integration by parts. The result thus

follows from using that Fis a bijection on S(R;H) and that S(R;H) is an operator core, for both mand

∂0,0.

For the statement (b) let f∈H1

(R;H) and ϕ∈C∞

c(R). Then e−m ϕ∈C∞

c(R) and

ˆR

(e−mf)ϕ′=ˆR

f(e−mϕ′)

=ˆR

f(e−mϕ)′+e−mϕ

=−ˆR

∂0,fe−m ϕ+ˆR

fe−mϕ

=−ˆR

(e−m∂0, f−e−mf)ϕ.

Hence e−mf∈H1

0(R;H) and ∂0,0e−mf= e−m ∂0,f−e−mf.

Next, we address (c). By part (a) and (b), we compute

∂0, = (e−m)∗∂0,0e−m +

3

= (e−m)∗F∗imFe−m +

= (e−m)∗F∗imFe−m + (e−m )∗F∗Fe−m

=L∗

im+L.

Theorem 2.2 tells us that ∂0, is unitarily equivalent to a multiplication operator with spectrum equal to

iR+={z∈C; Re z=}. In particular, we are now in the position to deﬁne functions of ∂0, .

Deﬁnition. Let ∈Rand F: dom(F)⊆ {it+;t∈R} → Cbe measurable such that {t∈R; it+ /∈

dom(F)}has Lebesgue measure zero. We deﬁne

F(∂0,):=L∗

Fim+L,

where

Fim+f:=R∋ξ7→ Fiξ+f(ξ)

in case f∈L2(R;H) is such that ξ7→ Fiξ+f(ξ)∈L2(R;H).

We record an elementary fact on multiplication operators.

Proposition 2.3. Let Fbe as in the previous deﬁnition. We denote kFk,∞:= ess supξ∈R|F(iξ+)| ∈

[0,∞]. The operator F(∂0,)is bounded, if and only if kFk,∞<∞. If F(∂0,)is bounded, then

kF(∂0,)k=kFk,∞.

Proof. Since Lis unitary we may prove that F(im+) is bounded on L2(R;H) if and only if kFk,∞<∞.

Suppose kFk,∞<∞. Then for f∈L2(R;H) we have ´RkF(iξ+)f(ξ)k2

Hdξ≤ kFk2

,∞kfk2

L2. Hence

F(im+) is bounded with kF(im+)k ≤ kFk,∞. Let F(im+) be bounded. For f∈L2(R;H) with

kfkL2= 1 we have

∞>kF(im+)k2≥ˆR|F(iξ+)|2kf(ξ)k2

Hdξ.

There is a sequence (fn)n∈N∈L2(R;H)Nwith kfnkL2= 1 (n∈N) and ´R|F(iξ+)|2kfn(ξ)k2

Hdξ→

ess sup |F(i ·+)|2for n→ ∞. This shows ∞>kF(im+)k ≥ kFk,∞. To construct (fn)n∈Nwlog. we

may assume that ess sup |F(i ·+)|>0. Let x∈Hwith kxkH= 1 and let (cn)n∈N∈RNbe a positive

sequence with cn↑ess sup |F(i ·+)|. Then for n∈Nset An:= [−n, n]∩ {|F(i ·+)|> cn}. By the

deﬁnition of the essential supremum, we may assume that λ(An)>0 and for n∈Nwe set

fn:= 1An

x

pλ(An).

One important class of operators that can be rooted to be of the form just introduced are fractional

derivatives and fractional integrals:

Example 2.4. Let α∈R>0and ∈R. Then the fractional derivative of order αis given by

∂α

0, =L∗

im+αL

and the fractional integral of order αis given by

∂−α

0, =L∗

1

im+αL.

Note that both expressions are well-deﬁned in the sense of functions of ∂0, deﬁned above and that ∂−α

0,

is bounded iﬀ 6= 0. Moreover, ∂α

0,−1=∂−α

0, . We set ∂0

0, as the identity operator on L2

(R;H).

4

In order to provide the connections to the more commonly known integral representation formulas for

the fractional integrals, we recall the multiplication theorem, that is,

√2πFf· Fg=F(f∗g),

for f∈L1(R) and g∈L2(R;H).

We recall the cut-oﬀ function

χR>0(t):=(1, t > 0,

0, t ≤0.

Lemma 2.5. For all , α > 0, and ξ∈R, we have

√2πLt7→ 1

Γ(α)tα−1χR>0(t)(ξ) = 1

iξ+α.(2.1)

Proof. We start by deﬁning the function

f(ξ):=

∞

ˆ

0

e−(iξ+)ssα−1ds

for ξ∈R. Then we have

f′(ξ) =

∞

ˆ

0−ie−(iξ+)ssαds

=−iα

iξ+f(ξ),

where we have used integration by parts. By separation of variables, it follows that

f(ξ) = f(0) α

(iξ+)α

for ξ∈R. Now, since

f(0) =

∞

ˆ

0

e−ssα−1ds=1

αΓ(α),

we infer

f(ξ) = Γ(α)1

(iξ+)α.

Since the left hand side of (2.1) equals 1

Γ(α)f(ξ), the assertion follows.

Next, we draw the connection from our fractional integral to the one used in the literature.

Theorem 2.6. For all , α > 0,f∈L2

(R;H), we have

∂−α

0, f(t) = ˆt

−∞

1

Γ(α)(t−s)α−1f(s) ds.

5

Proof. For the proof we set g:=t∈R7→ 1

Γ(α)tα−1χR>0(t). Then g∈L1

(R). For f∈L2

(R;H) we

have by Youngs convolution inequality

(e−mg)∗(e−mf) = e−m(g∗f)∈L2(R;H).

Using the convolution property of the Fourier transform we obtain

√2πLg· Lf=L(g∗f).

Using Lemma 2.5 we compute

∂−α

0, f=L∗

1

im +α

Lf

=L∗

√2πLg· Lf

=L∗

L(g∗f)

=ˆ(·)

−∞

1

Γ(α)(·)−sα−1f(s) ds.

Corollary 2.7. Let , α > 0. Then for all t∈R, we have for h∈H

(∂−α

0, χR>0h)(t) = (1

Γ(α+1) tαh, t > 0,

0, t ≤0.

Proof. We use Theorem 2.6 and obtain for t∈R

(∂−α

0, χR>0h)(t) = ˆt

−∞

1

Γ(α)(t−s)α−1χR>0(s)hds

=ˆR

χR>0(t−s)χR>0(s)1

Γ(α)(t−s)α−1ds h.

Thus, if t > 0, we obtain

(∂−α

0, χR>0h)(t) = ˆt

0

1

Γ(α)(t−s)α−1ds h

=ˆt

0

1

Γ(α)sα−1ds h =1

Γ(α)

1

αtαh.

For t≤0, we infer χR>0(t−s)χR>0(s) = 0 for s∈R, i.e. (∂−α

0, χR>0h)(t) = 0.

Remark 2.8.It seems to be hard to determine analog formulas for the case < 0, although the operator

∂−α

0, for < 0, α > 0 is bounded. The reason for this is that the corresponding multiplier (im+)−αis

not deﬁned in 0 and has a jump there. In particular, it cannot be extended to an analytic function on

some right half plane of C. This, however, corresponds to the causality or anticausality of the operator

∂−α

0, by a Paley-Wiener result ([4] or [7, 19.2 Theorem]) and hence, we cannot expect to get a convolution

formula as in the case > 0.

3 A reformulation of classical Riemann–Liouville and Caputo

diﬀerential equations

As it has been slightly touched in the introduction, there are two main concepts of fractional diﬀerentiation

(or integration). In this section we shall start to identify both these notions as being part of the same

6

solution theory. More precisely, equipped with the results from the previous section, we will consider the

initial value problems for the Riemann–Liouville and for the Caputo derivative. In order to avoid subtelties

as much as possible, we will consider the associated integral equations for both the Riemann–Liouville

diﬀerential equation and the Caputo diﬀerential equation and reformulate these equivalently with the

description of the time-derivative from the previous section. Well-posedness results for these equations

are postponed to Section 6.

To start oﬀ, we recall the Caputo diﬀerential equation. In [1], the author treated the following initial

value problem of Caputo type for 1 ≥α > 0:

Dα

∗y(t) = f(t, y(t)) (t > 0)

y(0) = y0,

where y0∈Cnis a given initial value; f:R>0×Cn→Cnis continuous, satisfying

|f(t, y1)−f(t, y2)| ≤ c|y1−y2|(3.1)

for some c≥0 and all y1, y2∈Cn, t > 0. For deﬁniteness, we shall also assume that

(t7→ f(t, 0)) ∈L2

0(R>0;Cn) (3.2)

for some 0∈R. In order to circumvent discussions of how to interpret the initial condition, we shall

rather put [1, Equation (6)] into the perspective of the present exposition. In fact, this equation reads

y(t) = y0+1

Γ(α)ˆt

0

(t−s)α−1f(s, y(s)) ds(t > 0).(3.3)

First of all, we remark that in contrast to the setting in the previous section, the diﬀerential equation

just discussed ‘lives’ on R>0, only. To this end we put

e

f:R×Cn→Cn,(t, y)7→ χR>0(t)f(t, y),

with the apparent meaning that e

fvanishes for negative times t. We note that by (3.1) and (3.2) it follows

that

L2

(R)∋y7→ (t7→ e

f(t, y(t))) ∈L2

(R)

is a well-deﬁned Lipschitz continuous mapping for all ≥0. Obviously, (3.3) is equivalent to

y(t) = y0χR>0(t) + 1

Γ(α)ˆt

−∞

(t−s)α−1e

f(s, y(s)) ds(t > 0),(3.4)

which in turn can be (trivially) stated for all t∈R. Next, we present the desired rewriting of equation

(3.4).

Theorem 3.1. Let > max{0, 0}. Assume that y∈L2

(R). Then the following statements are equiva-

lent:

(i) y(t) = y0χR>0(t) + 1

Γ(α)´t

−∞(t−s)α−1e

f(s, y(s)) dsfor almost every t∈R,

(ii) y=∂−α

0, e

f(·, y(·)) + y0χR>0,

(iii) ∂α

0,(y−y0χR>0) = e

f(·, y(·)).

Proof. The assertion follows trivially from Theorem 2.6.

7

Remark 3.2.For a real valued-function g:R>0×Rn→Rnwe may consider the Caputo diﬀerential

equation with f:R>0×Cn→Cn,(t, z)7→ g(t, Re(z)).

Next we introduce Riemann–Liouville diﬀerential equations. Using the exposition in [8], we want to

discuss the Riemann–Liouville fractional diﬀerential equation given by

dα

dxαy(x) = f(x, y(x)),

dα−1

dxα−1y(x)x=0+

=y0,

where as before fsatisﬁes (3.1) and (3.2) and y0∈R,and α∈(0,1]. Again, not hinging on too much of

an interpretation of this equation, we shall rather reformulate the equivalent integral equation related to

this initial value problem. According to [8, Chapter 42] this initial value problem can be formulated as

y(t) = y0

tα−1

Γ(α)+1

Γ(α)ˆt

0

(t−s)α−1f(s, y(s)) ds(t > 0).

We abbreviate gβ(t):=1

Γ(β+1) tβχR>0(t) for t, β ∈R. For α > 1/2 we have gα−1∈L2

(R;H). Let us

assume that α > 1/2. Invoking the cut-oﬀ function χR>0and deﬁning e

fas before, we may provide a

reformulation of the Riemann–Liouville equation on the space L2

(R;H) by

y=gα−1y0+∂−α

0, f(·, y(·)), y ∈L2

(R;H).

By a formal calculation and when applying Corollary 2.7, i.e. ∂−α

0, χR>0y0=gαy0, we would obtain

gα−1y0=∂0,gαy0=∂0, ∂−α

0, χR>0y0=∂−α

0, ∂0,χR>0y0=∂−α

0, y0δ0,

where ∂0,χR>0y0is, when understood distributionally, the delta function y0δ0and we could reformulate

the Riemann–Liouville equation by

∂α

0,y=y0δ0+e

f(·, y(·)).(3.5)

However, the calculation indicates that we have to extend the L2

(R;H) calculus to understand Rie-

mann–Liouville diﬀerential equations. This will be done in the coming sections.

4 Extra- and interpolation spaces

We begin to deﬁne extra- and interpolation spaces associated with the fractional derivative ∂α

0, for 6= 0,

α∈R. Since by deﬁnition

∂α

0, =L∗

(im+)αL,

we will deﬁne the extra- and interpolation spaces in terms of the multiplication operators (im+)αon

L2(R;H).

Deﬁnition. Let 6= 0. For each α∈Rwe deﬁne the space

Hα(im+):=

f∈L1

loc(R;H) ; ˆ

R

k(it+)αf(t)k2

Hdt < ∞

and equip it with the natural inner product

hf, giHα(im+):=ˆ

R

h(it+)αf(t),(it+)αg(t)iHdt

for each f, g ∈Hα(im+).

8

We shall use X ֒→Yto denote the mapping X∋x7→ x∈Y, if X⊆Y(under a canonical identiﬁcation,

which will always be obvious from the context).

Lemma 4.1. For 6= 0 and α∈Rthe space Hα(im+)is a Hilbert space. Moreover, for β > α we

have

jβ→α:Hβ(im+)֒→Hα(im+)

where the embedding is dense and continuous with kjβ→αk ≤ ||α−β.

Proof. Note that Hα(im+) = L2(µ;H), where µis the Lebesgue measure on Rweighted with the

function t7→ |it+|2α. Thus, Hα(im+) is a Hilbert space by the Fischer–Riesz theorem. Let now

β > α and f∈Hβ(im+). Then

ˆ

R

k(it+)αf(t)k2

Hdt=ˆ

R

(t2+2)α−βk(it+)βf(t)k2

Hdt≤2α−βkfk2

Hβ(im+),

which proves the continuity of the embedding jβ→αand the asserted norm estimate. The density follows,

since C∞

c(R;H) lies dense in Hγ(im+) for each γ∈R.

Deﬁnition. Let 6= 0 and α∈R.We consider the space

Wα

(R;H):=u∈L2

(R;H) ; Lu∈Hα(im+)

equipped with the inner product

hu, vi,α :=hLu, LviHα(im+)

and set Hα

(R;H) as its completion with respect to the norm induced by h·,·i,α.

Lemma 4.2. Let 6= 0.

(a) For α≥0we have that Hα

(R;H) = Wα

(R;H) = dom(∂α

0,).

(b) The operator

L:Wα

(R;H)⊆Hα

(R;H)→Hα(im+)

has a unique unitary extension, which will again be denoted by L.

(c) For α, β ∈Rwith β > α we have that

ιβ→α:Hβ

(R;H)֒→Hα

(R;H)

is continuous and dense with kιβ→αk ≤ ||α−β.

(d) For each β > 0and α∈Rthe operator

∂β

0, :Hβ+|α|

(R;H)⊆Hα

(R;H)→Hα−β

(R;H)

has a unique unitary extension, which will again be denoted by ∂β

0,.

Proof.

9

(a) Let α≥0. For u∈Hα(im+), i.e. u∈L1

loc(R;H) and (im+)αu∈L2(R;H), we infer that

u∈L2(R;H). It follows that u∈dom((im+)α). Hence Hα(im+) = dom((im+)α). Moreover,

u∈Wα

(R;H)⇔u∈L2

(R;H)∧ Lu∈Hα(im+)

⇔u∈L2

(R;H)∧ Lu∈dom ((im+)α)

⇔u∈dom(∂α

0,),

by Example 2.4. Moreover, since

L:Wα

(R;H)→Hα(im+)

is unitary, we infer that Wα

(R;H) is complete with respect to k · k,α =kL· kHα(im+),and thus

Hα

(R;H) = Wα

(R;H).

(b) Obviously,

L:Wα

(R;H)⊆Hα

(R;H)→Hα(im+)

is isometric by the deﬁnition of the norm on Hα

(R;H).Moreover, its range is dense, since L∗

ϕ∈

Wα

(R;H) for each ϕ∈C∞

c(R;H) and thus, C∞

c(R;H)⊆ LWα

(R;H).Hence, the continuous

extension of Lto Hα

(R;H) is onto and, thus, unitary.

(c) Since ιβ→α=L∗

jβ→αL,the assertion follows from Lemma 4.1.

(d) Since,

(im+)β:Hα(im+)→Hα−β(im+)

f7→ t7→ (it+)βf(t)

is obviously unitary, we infer that for u∈Hβ+|α|

(R;H)

k∂β

0,uk,α−β=kL∂β

0,ukHα−β(im+)

=k(im+)βLukHα−β(im+)

=kLukHα(im+)

=kuk,α,

which shows that ∂β

0, is an isometry. Moreover, for ϕ∈C∞

c(R;H), we have that (im+)γϕ∈

C∞

c(R;H) for all γ∈Rand thus, in particular L∗

(im+)−βϕ∈Tγ∈RHγ

(R;H)⊆Hβ+|α|

(R;H).

Next,

∂β

0,L∗

(im+)−βϕ=L∗

(im+)βLL∗

(im+)−βϕ=L∗

ϕ

and thus, L∗

[C∞

c(R;H)] ⊆∂β

0,[Hβ+|α|

(R;H)]. Since C∞

c(R;H) is dense in Hα−β(im+), we infer

that L∗

[C∞

c(R;H)] is dense in Hα−β

(R;H) and thus, ∂β

0, has dense range. This completes the

proof.

We conclude this section by providing an alternative perspective to elements lying in Hα

(R;H) for some

α∈R(with a particular focus on α < 0). In particular, we aim for a deﬁnition of a support for those

elements which coincides with the usual support of L2functions in the case α≥0.

10

Lemma 4.3. Let 6= 0 and α∈R. Then

σ−1:Wα

(R;H)⊆Hα

(R;H)→Hα

−(R;H)

f7→ (t7→ f(−t))

extends to a unitary operator. Moreover, for f∈Hα

(R;H)we have

L−σ−1f=σ−1Lfand L∗

−σ−1f=σ−1L∗

f.

Proof. For f∈Wα

(R;H) we have that

L−σ−1f=σ−1Lf

and hence,

ˆ

R

k(it−)α(L−σ−1f) (t)k2

Hdt=ˆ

Rt2+2αk(Lf) (−t)k2

Hdt

=ˆ

Rt2+2αk(Lf) (t)k2

Hdt=kfk2

Hα

(R;H),

which proves the isometry of σ−1.Moreover, σ−1has dense range, since σ−1[Wα

(R;H)] = Wα

−(R;H).

Hence, σ−1extends to a unitary operator. The equality L−σ−1f=σ−1Lfholds for f∈Hα

(R;H),

since Wα

(R;H) is dense in its completion Hα

(R;H).

Proposition 4.4. Let 6= 0, α ∈Rand f∈Hα

(R;H).Then

hf, ·i :C∞

c(R;H)→C

given by

hf, ϕi:=ˆ

R

hLf(t),L−ϕ(t)iHdt

deﬁnes a distribution. Moreover, for f∈Hα

(R;H)and ϕ∈C∞

c(R;H)we have

hf, ϕi=hf , ∂−α

0, e2mσ−1∂−α

0, σ−1ϕi,α.

In particular, for α= 0

hf, ϕi=ˆ

R

hf(t), ϕ(t)iHdt.

Note that the operator ∂−α

0, e2mσ−1∂−α

0, σ−1maps H−α

−(R;H)to Hα

(R;H)unitarily.

Proof. Let f∈Hα

(R;H).We ﬁrst prove that the expression hf , ·i is indeed a distribution. Due to Lemma

4.2(c) it suﬃces to prove this for f∈H−k

(R;H) for some k∈N.Indeed, if f∈H−k

(R;H),then we

know that t7→ (it+)−k(Lf) (t)∈L2(R;H)

and hence, for ϕ∈C∞

c(R;H) we obtain using H¨older’s inequality and the fact that L−ϕ(k)= (im+

)kL−ϕ

|hf, ϕi| ≤ ˆ

R

|h(it+)−k(Lf)(t),(−it+)k(L−ϕ) (t)iH|dt

11

≤ kLfkH−k(im+)kL−ϕ(k)kL2(R;H)

≤ kLfkH−k(im+)

ˆ

spt ϕ

e2t dt

1

2

kϕ(k)k∞,

which proves that hf, ·i is indeed a distribution. Next, we prove the asserted formula. For this, we note

the following elementary equality

σ−1Lϕ=Le2mσ−1ϕ

for ϕ∈L2

(R;H). Let f∈Hα

(R;H) and compute

hf, ϕi=hLf , L−ϕiL2(R;H)

=hLf, σ−1Lσ−1ϕiL2(R;H)

=h(im+)αLf, (−im+)−ασ−1Lσ−1ϕiL2(R;H)

=h(im+)αLf, σ−1(im+)−αLσ−1ϕiL2(R;H)

=h(im+)αLf, σ−1L∂−α

0, σ−1ϕiL2(R;H)

=h(im+)αLf, Le2mσ−1∂−α

0, σ−1ϕiL2(R;H)

=h(im+)αLf, (im+)αL∂−α

0, e2mσ−1∂−α

0, σ−1ϕiL2(R;H)

=hf, ∂ −α

0, e2mσ−1∂−α

0, σ−1ϕi,α

for each ϕ∈C∞

c(R;H).In particular, in the case α= 0 we obtain

hf, ϕi=hf , e2m ϕi,0=ˆ

R

hf(t), ϕ(t)iHdt.

Remark 4.5.The latter proposition shows that S6=0,α∈RHα

(R;H)⊆ D(R;H)′.In particular, the sup-

port of an element in Hα

(R;H) is then well-deﬁned by

\{R\U;U⊆Ropen,∀ϕ∈C∞

c(U;H) : hf, ϕi= 0},

and the second part of the latter proposition shows, that it coincides with the usual L2-support if α≥0.

Moreover, we can now compare elements in Hα

(R;H) and Hβ

µ(R;H) by saying that those elements are

equal if they are equal as distributions. We shall further elaborate on this matter in Proposition 4.9. In

particular, we shall show that f7→ hf , ·i is injective. We shall also mention that the notation hf, ϕiis

justiﬁed, as it does not depend on nor α.

Example 4.6. Let f∈L2

(R;H). Then, by deﬁnition, ∂0,f∈H−1

(R;H). We shall compute the action

of ∂0,fas a distribution. For this let ϕ∈C∞

c(R;H) and we compute with the formula outlined in

Proposition 4.4 for α=−1:

h∂0,f , ϕi=h∂0,f, ∂0, e2mσ−1∂0, σ−1ϕi,−1

=h(im+)−1L∂0,f, (im+)−1L∂0,e2m σ−1∂0,σ−1ϕiL2(R;H)

=hLf, Le2m σ−1∂0,σ−1ϕiL2(R;H)

=−hLf, Le2m ∂0,ϕiL2(R;H)

=−hf, e2m ϕ′iL2

(R;H)

=−ˆRhf(t), ϕ′(t)iHdt.

Thus, ∂0,fcoincides with the distibutional derivative of L2

(R;H) functions.

12

Lemma 4.7. Let α∈R. Let H∞

(R;H):=Tk∈NHk

(R;H)for 6= 0.

(a) Let ϕ∈C∞

c(R;H). For all > 0we have ∂α

ϕ∈C∞(R;H)∩H∞

(R;H)and inf spt ∂α

ϕ≥inf spt ϕ.

For , µ > 0,α∈Rwe have ∂α

0,ϕ=∂α

0,µϕ.

(b) Let α∈Rand µ, 6= 0. Let ψ∈C∞(R;H)∩H∞

(R;H)∩H∞

µ(R;H). Then there is (ϕn)n∈N∈

C∞

c(R;H)Ns.t. ϕn→ψfor n→ ∞ in Hα

(R;H)and Hα

µ(R;H)and spt(ϕn)⊆spt(ψ)for n∈N.

Proof. (a): Let α∈R. Let µ, > 0. For α > 0 it holds that ∂α

=∂α−⌈α⌉

∂⌈α⌉

and ∂⌈α⌉

ϕ=ϕ(⌈α⌉)=

∂⌈α⌉

µϕ∈C∞

c(R;H) and α− ⌈α⌉<0. Thus we may assume that α < 0. By Theorem 2.6 we have

∂α

0,ϕ=∂α

0,µϕand inf spt ∂α

0,ϕ > −∞. From ϕ∈H∞

(R;H) we deduce ∂α

0,ϕ∈H∞

(R;H).

(b): Let k∈Nwith k≥α. We choose a sequence (χn)n∈Nin C∞

c(R) such that spt χn⊆[−n−1, n + 1],

χn= 1 on [−n, n] and

sup nkχ(j)

nk∞;j∈ {0,...,k}, n ∈No<∞.

Set ϕn:=χnψ∈C∞

c(R;H). Then spt(ϕn)⊆spt(ψ) for n∈N. Since Hk

ν(R;H) is dense and continuously

embedded into Hα

ν(R;H) (ν6= 0), it suﬃces to show that ϕn→ψ(n→ ∞) in Hk

(R;H) and Hk

µ(R;H).

Indeed, by the product rule, the choice of χnand dominated convergence we obtain

ϕ(k)

n=

k

X

j=0

χ(k−j)

nψ(j)=χnψ(k)+

k−1

X

j=0

χ(k−j)

nψ(j)→ψ(k)

for n→ ∞ in L2

(R;H) and in L2

µ(R;H).

Lemma 4.8. Let 6= 0 and α∈R.Then C∞

c(R;H)is dense in Hα

(R;H).

Proof. We ﬁrst note that it suﬃces to prove the assertion for > 0,since the operator σ−1from Lemma

4.3 leaves C∞

c(R;H) invariant. It is well known that C∞

c(R;H) is dense in L2

(R;H). We have ∂α

0,f∈

L2

(R;H). Let (ψn)n∈N∈C∞

c(R;H)Nwith ψn→∂α

0,f(n→ ∞) in L2

(R;H). By Lemma 4.7(a)

we have ∂−α

0, ψn∈C∞(R;H)∩H∞

(R;H) and by Lemma 4.7(b) we ﬁnd (ϕn)n∈N∈C∞

c(R;H)Nwith

∂−α

0, ψn−ϕn

,α →0 (n→ ∞). Then

kf−ϕnk,α ≤

f−∂−α

0, ψn

,α +

∂−α

0, ψn−ϕn

,α →0 (n→ ∞).

With this result at hand, we can characterize those distributions, which belong to Hα

(R;H) for some

α∈R, 6= 0, in the following way.

Proposition 4.9. Let ψ∈ D(R;H)′and α∈R, 6= 0.Then, there exists f∈Hα

(R;H)such that

ψ(ϕ) = hf, ϕi(ϕ∈C∞

c(R;H))

in the sense of Propositions 4.4 if and only if there is C≥0such that

|ψ(ϕ)| ≤ Ckϕk−,−α

for each ϕ∈C∞

c(R;H).

Proof. Assume ﬁrst that there is f∈Hα

(R;H) representing ψ. Then we estimate

|ψ(ϕ)|=|hf, ϕi|

13

=ˆ

R

hLf(t),L−ϕ(t)iHdt

=ˆ

R

h(it+)αLf(t),(−it+)−αL−ϕ(t)iHdt

≤ kLfkHα(im+)kL−ϕkH−α(im−)

=kfk,αkϕk−,−α

for each ϕ∈C∞

c(R;H).Let C≥0 such that ψsatisﬁes for ϕ∈C∞

c(R;H)

|ψ(ϕ)| ≤ Ckϕk−,−α.

The operator A:=∂−α

0, e2mσ−1∂−α

0, σ−1:H−α

−(R;H)→Hα

(R;H) (cf. Proposition 4.4) is unitary. Thus

for ϕ∈C∞

c(R;H)ψ(A−1ϕ)≤C

A−1ϕ

−,−α=Ckϕk,α .

Moreover, C∞

c(R;H)⊆Hα

(R;H) is dense. Thus ψ(A−1·) can be extended continuously to Hα

(R;H).

By the Riesz representation theorem, there is a f∈Hα

(R;H) such that for ϕ∈Hα

(R;H)

ψ(A−1ϕ) = hf, ϕi,α .

By Theorem 4.4 we have for ϕ∈C∞

c(R;H)

ψ(ϕ) = ψ(A−1Aϕ) = hf, Aϕi,α =hf , ϕi.

In the next proposition, we shall also obtain the announced uniqueness statement, that is, the injectivity

of the mapping f7→ hf , ·i.

Proposition 4.10. Let α∈Rand µ, > 0.Moreover, let f∈Hα

(R;H)and g∈Hα

µ(R;H).Then the

following statements are equivalent:

(i) f=gin the sense of distributions, i.e., for each ϕ∈C∞

c(R;H)we have that

ˆ

R

hLf(t),L−ϕ(t)iHdt=ˆ

R

hLµg(t),L−µϕ(t)iHdt.

(ii) ∂α

0,f=∂α

0,µgas functions in L1

loc(R;H).

(iii) There is a sequence (ϕn)n∈Nin C∞

c(R;H)with ϕn→fin Hα

(R;H)and ϕn→gin Hα

µ(R;H)as

n→ ∞.

Proof. (i)⇒(ii): Let ψ∈C∞

c(R;H) and e

ψ:=σ−1∂α

σ−1ψ=σ−1∂α

µσ−1ψ. Then by Lemma 4.7(a)

e

ψ∈C∞(R;H)∩H∞

−(R;H)∩H∞

−µ(R;H). By Lemma 4.7(b) there’s (ϕn)n∈N∈C∞

c(R;H)Nwith ϕn→e

ψ

(n→ ∞) in H−α

−(R;H) and in H−α

−µ(R;H). Thus

ˆRh∂α

0,f(t), ψ(t)iHdt=h∂α

0,f , e2mψi,0

=hf, ∂ −α

0, e2mσ−1∂−α

0, σ−1(σ−1∂α

0,σ−1ψ)i,α

14

= lim

n→∞hf, ∂ −α

0, e2mσ−1∂−α

0, σ−1ϕni,α

= lim

n→∞hf, ϕni

= lim

n→∞hg, ϕni

= lim

n→∞hf, ∂ −α

0,µ e2µmσ−1∂−α

0,µ σ−1ϕniµ,α

=ˆRh∂α

0,µf(t), ψ(t)iHdt.

(ii) ⇒(iii): Deﬁne e

fn:=χ[−n,n]·∂α

0,f=χ[−n,n]·∂α

0,µgfor n∈N. Without loss of generality let < µ.

Take a function ψn∈C∞

c(R;H) with spt ψn⊆[−n, n] such that

ke

fn−ψnk,0≤1

ne(−µ)n.

Then, we estimate

ke

fn−ψnk2

µ,0=

n

ˆ

−nke

fn(t)−ψn(t)k2

He−2µt dt=

n

ˆ

−nke

fn(t)−ψn(t)k2

He−2te2(−µ)tdt

≤ ke

fn−ψnk2

,0e2(µ−)n≤1

n2.

Hence ψn→∂α

0,f=∂α

0,µgin L2

(R;H) and in L2

µ(R;H) by the triangle inequality and dominated

convergence. We set eϕn:=∂−α

0, ψn=∂−α

0,µ ψn∈C∞(R;H)∩H∞

(R;H)∩H∞

µ(R;H). Then eϕn→f

and eϕn→gin Hα

(R;H) and in Hα

µ(R;H) respectively. We use Lemma 4.7(b) and choose a sequence

(ϕn)n∈N∈C∞

c(R;H)Nwith keϕn−ϕnk,α →0. Then

kf−ϕnk,α ≤ kf−eϕnk,α +keϕn−ϕnk,α →0 (n→ ∞),

kg−ϕnkµ,α ≤ kg−eϕnkµ,α +keϕn−ϕnkµ,α →0 (n→ ∞).

(iii) ⇒(i): Let (ϕn)n∈Nbe a sequence in C∞

c(R;H) such that ϕn→fand ϕn→gin Hα

(R;H) and

Hα

µ(R;H), respectively. Let ϕ∈C∞

c(R;H).Then we have according to Proposition 4.4

ˆ

R

hLf(t),L−ϕ(t)iHdt=hf, ϕi

=hf, ∂ −α

0, e2mσ−1∂−α

0, σ−1ϕi,α

= lim

n→∞hϕn, ∂−α

0, e2mσ−1∂−α

0, σ−1ϕi,α

= lim

n→∞hϕn, ϕi

= lim

n→∞hϕn, ∂−α

0,µ e2µmσ−1∂−α

0,µ σ−1ϕiµ,α

=hg, ϕi,

=ˆ

R

hLµf(t),L−µϕ(t)idt.

which completes the proof.

15

5 A uniﬁed solution theory – well-posedness and causality of

fractional diﬀerential equations

We are now able to study abstract fractional diﬀerential equations of the form

∂α

0,u=F(u).

In order to obtain well-posedness of the latter problem, we need to restrict the class of admissible right-

hand sides Fin the latter equation.

Deﬁnition. Let 0>0 and β, γ ∈R.We call a function F: dom(F)⊆T≥0Hβ

(R;H)→T≥0Hγ

(R;H)

eventually (β , γ)-Lipschitz continuous, if dom(F)⊇C∞

c(R;H) and there exists ν≥0such that for each

≥νthe function Fhas a Lipschitz continuous extension

F:Hβ

(R;H)→Hγ

(R;H)

satisfying sup≥ν|F|Lip <∞. Moreover, we call Feventually (β, γ )-contracting, if Fis eventually

(β, γ )-Lipschitz continuous and lim sup→∞ |F|Lip <1.Here, we denote by | · |Lip the smallest Lipschitz

constant of a Lipschitz continuous function:

|F|lip := sup

f,g∈Hβ

(R;H), f6=g

kF(f)−F(g)k,γ

kf−gk,β

.

Note that by Lemma 4.8, any eventually Lipschitz continuous function is densely deﬁned. Thus, the

Lipschitz continuous extension Fis unique.

Remark 5.1.(a) If f∈Hβ

(R;H) and g∈Hβ

µ(R;H) generate the same distribution, we have that

F(f) = Fµ(g).

Indeed, by Proposition 4.10 there exists a sequence (ϕn)n∈Nin C∞

c(R;H) with ϕn→fand ϕn→gin

Hβ

(R;H) and Hβ

µ(R;H), respectively. We infer that

F(f) = lim

n→∞ F(ϕn) and Fµ(g) = lim

n→∞ F(ϕn)

with convergence in Hγ

(R;H) and Hγ

µ(R;H) respectively. Consequently

∂γ

0,F(f)←∂γ

0,F(ϕn) = ∂γ

0,µF(ϕn)→∂γ

0,µFµ(g)

with convergence in L2

(R;H) and hence almost everywhere for a suitable subsequence of (ϕn)n∈N. The

assertion follows from Proposition 4.10.

(b) We shall need the following elementary observation later on. Let Fbe evenutally (β, γ)-Lipschitz

continuous, α∈R. Let ≥0. Then

e

F:C∞

c(R;H)∋ϕ7→ F(∂α

0,ϕ)

is eventually (β+α, γ)-Lipschitz continuous. Indeed, the assertion follows from part (a) and

e

F(f)−e

F(g)

µ,γ ≤ |Fµ|Lip

∂α

0,µf−∂α

0,µg

µ,β =|Fµ|Lip kf−gkµ,α+β,

for µ≥ν,f, g ∈C∞

c(R;H).

16

Theorem 5.2. Let α > 0, β ∈R, 0>0and F: dom(F)⊆T≥0Hβ

(R;H)→T≥0Hβ−α

(R;H)be

eventually (β , β −α)-contracting. Then there exists ν≥0such that for each ≥νthere is a unique

u∈Hβ

(R;H)satisfying

∂α

0,u=F(u).(5.1)

Proof. This is a simple consequence of the contraction mapping theorem. Indeed, choosing ν≥0large

enough, such that |F|Lip <1 for each ≥ν, we obtain that

∂−α

0, F:Hβ

(R;H)→Hβ

(R;H)

is a strict contraction, since ∂−α

0, :Hβ−α

(R;H)→Hβ

(R;H) is unitary by Lemma 4.2. Hence, the

mapping ∂−α

0, Fadmits a unique ﬁxed point u∈Hβ

(R;H),which is equivalent to ubeing a solution

of (5.1).

Corollary 5.3. Let α > 0, β ∈R, 0>0and F: dom(F)⊆T≥0Hβ

(R;H)→T≥0Hβ−γ

(R;H)for

some γ∈[0, α[be eventually (β, β −γ)-Lipschitz continuous. Then there exists ν≥0such that for each

≥νthere is a unique u∈Hβ

(R;H)satisfying

∂α

0,u=F(u).

Proof. It suﬃces to prove that ιβ−γ→β−α◦Fis eventually (β, β −α)-contracting by Theorem 5.2. Let

ν≥, s.t. for ≥ν,Fexists. Then for ≥ν

|ιβ−γ→β−α◦F|Lip ≤ kιβ−γ→β−αk|F|Lip ≤γ−α|F|Lip

by Lemma 4.2. Since |F|Lip is bounded in on [ν, ∞[ by assumption, we infer

lim sup

→∞ |ιβ−γ→β−α◦F|Lip = 0 <1.

Next, we want to show that the solution uof (5.1) is actually independent of the particular choice of .

For doing so, we need the concept of causality, which will be addressed in the next propositions.

Lemma 5.4. Let > 0,α∈Rand a∈R. Let f∈Hα

(R;H)with spt f⊆R≥a. Then there is a

sequence (ϕn)n∈N∈C∞

c(R;H)Nwith spt ϕn⊆R≥afor n∈Nand ϕn→fin Hα

(R;H)as n→ ∞.

Proof. Let ( e

ψn)n∈N∈C∞

c(R;H)Nbe such that e

ψn→∂α

0,fin H0

(R;H) as n→ ∞. We may assume

that spt e

ψn⊆R>a. We set ψn:=∂−α

0, e

ψnfor n∈N. Then ψn→fas n→ ∞ in Hα

(R;H) and

inf spt ψn> a by Lemma 4.7(a). We use Lemma 4.7(b) and pick a sequence (ϕn)n∈N∈C∞

c(R;H)Nwith

spt(ϕn)⊆spt(ψn) for n∈Nand ϕn−ψn→0 in Hα

(R;H) when n→ ∞. Then

kϕn−fk,α ≤ kϕn−ψnk,α +kψn−fk,α →0 (n→ ∞).

Proposition 5.5. Let f∈Hα

(R;H)for some α∈R, > 0.Assume that spt f⊆R≥afor some a∈R.

Then

spt ∂β

0,f⊆R≥a

for all β∈R.

17

Proof. Let ϕ∈C∞

c(R;H) with spt ϕ⊆R<a. By Lemma 5.4 we pick a sequence (ϕn)n∈N∈C∞

c(R;H)N,

s.t. spt ϕn⊆R≥a(n∈N) and ϕn→fin Hα

(R;H). Then spt ∂β

0,ϕn⊆R≥aby Lemma 4.7(a). By

Proposition 4.4 we have. D∂β

0,ϕn, ϕE=ˆRD∂β

0,ϕn, ϕEH(t) dt= 0.

Since ∂β

0, is unitary, we have ∂β

0,ϕn→∂β

0,fin Hα−β

(R;H). We compute

D∂β

0,f , ϕE=h∂β

0,f , ∂−(α−β)

0, e2mσ−1∂−(α−β)

0, σ−1ϕi,α−β

= lim

n→∞h∂β

0,ϕn, ∂ −(α−β)

0, e2mσ−1∂−(α−β)

0, σ−1ϕi,α−β

= lim

n→∞ D∂β

0,ϕn, ϕE

= 0.

The proof of the following theorem outlining causality of ∂−α

0, F, is in spirit similar to the approach in

[2, Theorem 4.5]. However, one has to adopt the distributional setting and the (diﬀerent) deﬁnition of

eventually Lipschitz continuity here accordingly.

Theorem 5.6. Let the assumptions of Theorem 5.2 be satisﬁed. Then, for each ≥ν, where νis chosen

according to Theorem 5.2, the mapping

∂−α

0, F:Hβ

(R;H)→Hβ

(R;H)

is causal, that is, for each u, v ∈Hβ

(R;H)satisfying spt(u−v)⊆R≥afor some a∈R,it holds that

spt ∂−α

0, F(u)−∂−α

0, F(v)⊆R≥a.Here, the support is meant in the sense of distributions.

Proof. First of all, we shall show the result for u, v ∈C∞

c(R;H). So, let u, v ∈C∞

c(R;H) with spt(u−v)⊆

R≥a.Take ϕ∈C∞

c(R;H) with spt ϕ⊆R<a .Let µ≥. Then F(u) = Fµ(u) and

∂−α

0, (F(u)−F(v)), ϕ=∂−α

0,µ (Fµ(u)−Fµ(v)), ϕ

=D∂−α

0,µ (Fµ(u)−Fµ(v)), ∂−β

0,µ e2µmσ−1∂−β

0,µ σ−1ϕEµ,β

=DFµ(u)−Fµ(v), ∂−(β−α)

0,µ e2µmσ−1∂−β

0,µ σ−1ϕEµ,β−α

≤ kFµ(u)−Fµ(v)kµ,β−α

∂−(β−α)

0,µ e2µmσ−1∂−β

0,µ σ−1ϕ

µ,β−α

≤ |Fµ|lip ku−vkµ,β

∂−β

0,µ σ−1ϕ

µ,0

where we have used that ∂−(β−α)

0,µ e2µmσ−1:H0

µ(R;H)→Hβ−α

µ(R;H) is unitary and ϕ∈H−β

−µ(R;H).

According to Proposition 5.5 we have that spt ∂−β

0,µ σ−1ϕ⊆R>−aand hence, we compute

k∂−β

0,µ σ−1ϕk2

µ,0=

∞

ˆ

−a

∂−β

0,µ σ−1ϕ(t)

2

He−2µt dt=

∞

ˆ

0

∂−β

0,µ σ−1ϕ(t−a)

2

He−2µt dte2µa.

On the other hand

ku−vk2

µ,β =k∂β

0,µ(u−v)k2

µ,0

18

=

∞

ˆ

ak∂β

0,µ(u−v)(t)k2

He−2µt dt

=

∞

ˆ

0

∂β

0,µ(u−v)(t+a)

2

He−2µt dte−2µa

and consequently,

|Fµ|Lipku−vkµ,β k∂−β

0,µ σ−1ϕk−µ,0

=|Fµ|Lip

∞

ˆ

0

∂β

0,µ(u−v)(t+a)

2

He−2µt dt

1

2∞

ˆ

0

∂−β

0,µ σ−1ϕ(t−a)

2

He−2µt dt→0 (µ→ ∞),

by dominated convergence. Summarizing, we have shown that spt(∂−α

0, F(u)−∂−α

0, F(v)) ⊆R≥afor

u, v ∈C∞

c(R;H) satisfying spt(u−v)⊆R≥a.

Before we conclude the proof, we show that if (wn)n∈Nis a convergent sequence in Hβ

(R;H) with

spt wn⊆R≥afor each n∈N, then its limit walso satisﬁes spt w⊆R≥a.For doing so, let ϕ∈C∞

c(R;H)

with spt ϕ⊆R<a.Then

hw, ϕi=Dw, ∂−β

0, e2mσ−1∂−β

0, σ−1ϕE,β = lim

n→∞ Dwn, ∂−β

0, e2mσ−1∂−β

0, σ−1ϕE,β = lim

n→∞ hwn, ϕi= 0.

Finally, let u, v ∈Hβ

(R;H) with spt(u−v)⊆R≥aAccording to Lemma 5.4 there is a sequence (ϕn)n∈N∈

C∞

c(R;H)Nwith spt ϕn⊆R≥aand ϕn→u−vin Hα

(R;H) as n→ ∞. Let (vn)n∈N∈C∞

c(R;H) with

vn→vin Hα

(R;H) as n→ ∞. We set un:=ϕn+vn. Then un→uin Hα

(R;H) and spt(un−vn)⊆R≥a.

By the already proved result for C∞

c(R;H), we infer that spt ∂−α

0, F(un)−∂−α

0, F(vn)⊆R≥afor all

n∈N. Thus, letting n→ ∞, we obtain spt ∂−α

0, F(u)−∂−α

0, F(v)⊆R≥a, which shows the claim.

Finally, we prove that our solution is independent of the particular choice of the parameter > ν in

Theorem 5.2. The precise statement is as follows.

Proposition 5.7. Let the assumptions of Theorem 5.2 be satisﬁed and νbe chosen according to Theorem

5.2. Let eµ, µ > ν and ueµ∈Hβ

eµ(R;H), uµ∈Hβ

µ(R;H)satisfying

∂α

0,eµueµ=Feµ(ueµ)and ∂α

0,µuµ=Fµ(uµ).

Then ueµ=uµas distributions in the sense of Proposition 4.4.

Proof. We note that it suﬃces to show vµ:=∂β

0,µuµ=∂β

0,eµueµ=:veµas L1

loc(R;H) functions by Proposition

4.10. We consider the function

e

F: dom( e

F)⊆\

≥0

H0

(R;H)→\

≥0

Hβ−α

(R;H)

given by e

F(v):=F(∂−β

0, v) (v∈dom( e

F)) (5.2)

with maximal domain dom( e

F) = {w∈T≥0H0

(R;H) ; ∀≥0:∂−β

0, w∈dom(F)}.Note that the

expression on the right hand side of (5.2) does not depend on the particular choice of ≥0by Proposition

4.10. Clearly, e

Fis eventually (0, β −α)-contracting (see also Remark 5.1(b)) and

e

F=F(∂−β

0, (·)) (≥0).

19

In particular,

∂α−β

0,µ vµ=∂α

0,µuµ=Fµ(uµ) = e

Fµ(vµ)

and analogously

∂α−β

0,eµveµ=e

Feµ(veµ).

Let now a∈Rand assume without loss of generality that µ < eµ. We note that spt(veµ−χR≤aveµ)⊆χR≥a.

We obtain, applying Theorem 5.6, that

χR≤aveµ=χR≤a∂β−α

0,eµe

Feµ(veµ) = χR≤a∂β−α

0,eµe

Feµ(χR≤aveµ).

Now, since χR≤aveµ∈L2

µ(R;H)∩L2

eµ(R;H), we infer that

χR≤aveµ=χR≤a∂β−α

0,eµe

Feµ(χR≤aveµ) = χR≤a∂β−α

0,µ e

Fµ(χR≤aveµ),

i.e. χR≤aveµis a ﬁxed point of χR≤a∂β−α

0,µ e

Fµ. However, since χR≤avµis also a ﬁxed point of this mapping,

which is strictly contractive, we derive

χR≤aveµ=χR≤avµ

and since a∈Rwas arbitrary, the assertion follows.

6 Riemann–Liouville and Caputo diﬀerential equations revis-

ited

In this section, we shall consider the diﬀerential equations introduced in Section 3 and prove their well-

posedness and causality. First of all, we gather some results ensuring the Lipschitz continuity property

needed to apply either of the well-posedness theorems presented in the previous section. As in Section 3

we ﬁx α∈(0,1].

Proposition 6.1. Let 0>0,n∈N,y0∈Rn,f:R>0×Cn→Cncontinuous. Assume there exists

c≥0such that for all y1, y2∈Rn, t > 0we have

|f(t, y1)−f(t, y2)| ≤ c|y1−y2|.

Moreover, we assume that

(t7→ f(t, 0)) ∈L2

0(R>0;Cn).

Deﬁne e

f:R×Cn→Cnby

e

f(t, y):=(f(t, y)if t > 0,

0else.

Then the mapping F:C∞

c(R;Cn)→C(R;Cn)given by

F(ϕ)(t):=e

f(t, ϕ(t) + y0) (ϕ∈C∞

c(R;Cn), t ∈R)

is eventually (0,0)-Lipschitz continuous.

Proof. Let ≥0. In order to prove that Fattains values in L2

(R;Cn), we shall show F(0) ∈L2

(R;Cn)

ﬁrst. For this we compute

ˆR|F(0)(t)|2e−2t dt=ˆR>0|f(t, y0)|2e−2t dt

20

≤2ˆR>0|f(t, y0)−f(t, 0)|2e−2t dt+ˆR>0|f(t, 0)|2e−2t dt

≤2c2|y0|21

2+|f(·,0)|2

L2

0(R>0;Cn)<∞.

Here we used that L2

(R>0;H)֒→L2

0(R>0;H) as contraction. Next, let ϕ, ψ ∈C∞

c(R;Rn). Then we

obtain

ˆR|F(ϕ)(t)−F(ψ)(t)|2e−2t dt=ˆR|e

f(t, ϕ(t) + y0)−e

f(t, ψ(t) + y0)|2e−2t dt

=ˆR>0|f(t, ϕ(t) + y0)−f(t, ψ(t) + y0)|2e−2t dt

≤ˆR>0

c2(|ϕ(t)−ψ(t)|)2e−2t dt≤c2kϕ−ψk2

L2

.

Since F(0) ∈L2

(R;Cn), the shown estimate yields F(ϕ)∈L2

(R;Cn) for each ϕ∈C∞

c(R;Cn) as well as

the eventual (0,0)-Lipschitz continuity of F.

The next result is concerned with the well-posedness for Caputo fractional diﬀerential equations. We

shall use the characterization of the Caputo diﬀerential equation outlined in Theorem 3.1.

Theorem 6.2. Let y0∈Cn. Then there is 1>0such that for all ≥1there exists a unique

y∈L2

(R;Cn)with y−y0χR>0∈Hα

(R;Cn)satisfying

∂α

0,(y−y0χR>0) = e

f(·, y(·)).

Moreover, spt y⊆R≥0.

Proof. With Fas deﬁned in Proposition 6.1, we may apply Corollary 5.3 with β=γ= 0 to obtain unique

existence of z∈Hα

(R;Cn) such that

∂α

0,z=F(z).

Setting y:=z+y0χR>0, we obtain in turn unique existence of a solution of the desired equation. Since

spt F(z)⊆R≥0, we obtain with Proposition 5.5 that spt z= spt ∂−α

0, F(z)⊆R≥0. Thus, spt y⊆

R≥0.

We remark here that the condition spt y⊆R≥0together with y−y0χR>0∈Hα

(R;Cn) describes, how

the initial value y0is attained. Indeed, if αis large enough (e.g. α > 1/2) so that Hα

(R;Cn) is a subset

of functions for which the limit at 0 exists, then the mentioned conditions imply

0 = (y−y0χR>0)(0−) = (y−y0χR>0)(0+) = y(0+) −y0,

that is, the initial value is attained.

We conclude this section by having a look at the case of the Riemann–Liouville fractional diﬀerential

equation (3.5). To this end, we note that χR>0y0∈H0

(R;H) for > 0 and by Example 4.6 we have

∂0,χR>0y0=δ0y0∈H−1

(R;H).

We also recall the notation gβ(t):=1

Γ(β+1) tβχR>0for β, t ∈R.

21

Proposition 6.3. Let y0∈Cn. Assume that C∞

c(R;H)∋ϕ7→ e

f(·, ϕ(·)) is eventually (α−1, α −1)-

Lipschitz continuous and denote with Hα−1

(R;H)∋y7→ e

f(·, y(·)) its Lipschitz-continuous extension

for some > 0. There is 1>0such that for ≥1we have a unqiue solution y∈Hα−1

(R;H)of the

equation

∂α

0,y=y0δ0+e

f(·, y(·)), y ∈Hα−1

(R;H),

with ∂α−1

0, y−y0χR>0∈H0

(R;H)and spt(y)⊆R≥0.

Proof. The mapping Gdeﬁned by

G(ϕ)(t):=e

f(t, ∂1−α

0, ϕ(t) + gα−1(t)y0), ϕ ∈C∞

c(R;H), t ∈R

is eventually (0, α −1)-Lipschitz continuous. Indeed, this fact follows from gα−1y0∈Hα−1

(R;H) and

the unitarity of ∂α