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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