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Abstract

We study fractional differential equations of Riemann-Liouville and Caputo type in Hilbert spaces. Using exponentially weighted spaces of functions defined on $\mathbb{R}$, 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.
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 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 fCc(R) with
compact support given by
t7→ 1
2πˆt
−∞
1
Γ(α)(ts)α1f(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)(ts)α1f(s) ds
for aR, 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 fL1
loc(R;H) we denote em f:=
(Rt7→ etf(t)). We define the normed spaces
Lp
(R;H):=fL1
loc(R;H); em fLp(R;H),
with norm
kfkLp
(R;H):=
emf
Lp(R;H)=ˆRkf(t)kp
Hept 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)iHe2t 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 φC
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):={fL2
(R;H); fL2
(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 fL1(R;H) by
Ff(ξ) = 1
2πˆR
f(t)eiξt 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):={fL2(R;H); (Rξ7→ ξf(ξ)) L2(R;H)},
m:L2(R;H)dom(m)L2(R;H), f 7→ (Rξ7→ ξf (ξ)).
Theorem 2.2. Let R.Then
(a) 0,0=FimF,
(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 fH1
(R;H) and ϕC
c(R). Then em ϕC
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m0, femf)ϕ.
Hence emfH1
0(R;H) and 0,0emf= em 0,femf.
Next, we address (c). By part (a) and (b), we compute
0, = (em)0,0em +
3
= (em)FimFem +
= (em)FimFem + (em )FFem
=L
im+L.
Theorem 2.2 tells us that 0, is unitarily equivalent to a multiplication operator with spectrum equal to
iR+={zC; Re z=}. In particular, we are now in the position to define functions of 0, .
Definition. Let Rand F: dom(F) {it+;tR} Cbe measurable such that {tR; it+ /
dom(F)}has Lebesgue measure zero. We define
F(0,):=L
Fim+L,
where
Fim+f:=Rξ7→ Fiξ+f(ξ)
in case fL2(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 supξR|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 fL2(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 fL2(R;H) with
kfkL2= 1 we have
>kF(im+)k2ˆR|F(iξ+)|2kf(ξ)k2
Hdξ.
There is a sequence (fn)nNL2(R;H)Nwith kfnkL2= 1 (nN) and ´R|F(iξ+)|2kfn(ξ)k2
Hdξ
ess sup |F(i ·+)|2for n . This shows >kF(im+)k kFk,. To construct (fn)nNwlog. we
may assume that ess sup |F(i ·+)|>0. Let xHwith kxkH= 1 and let (cn)nNRNbe a positive
sequence with cness sup |F(i ·+)|. Then for nNset An:= [n, n] {|F(i ·+)|> cn}. By the
definition of the essential supremum, we may assume that λ(An)>0 and for nNwe 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 αR>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(fg),
for fL1(R) and gL2(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 ξR, we have
2πLt7→ 1
Γ(α)tα1χR>0(t)(ξ) = 1
+α.(2.1)
Proof. We start by defining the function
f(ξ):=
ˆ
0
e(iξ+)ssα1ds
for ξR. Then we have
f(ξ) =
ˆ
0ie(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 ξR. Now, since
f(0) =
ˆ
0
essα1ds=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,fL2
(R;H), we have
α
0, f(t) = ˆt
−∞
1
Γ(α)(ts)α1f(s) ds.
5
Proof. For the proof we set g:=tR7→ 1
Γ(α)tα1χR>0(t). Then gL1
(R). For fL2
(R;H) we
have by Youngs convolution inequality
(emg)(emf) = em(gf)L2(R;H).
Using the convolution property of the Fourier transform we obtain
2πLg· Lf=L(gf).
Using Lemma 2.5 we compute
α
0, f=L
1
im +α
Lf
=L
2πLg· Lf
=L
L(gf)
=ˆ(·)
−∞
1
Γ(α)(·)sα1f(s) ds.
Corollary 2.7. Let , α > 0. Then for all tR, we have for hH
(α
0, χR>0h)(t) = (1
Γ(α+1) tαh, t > 0,
0, t 0.
Proof. We use Theorem 2.6 and obtain for tR
(α
0, χR>0h)(t) = ˆt
−∞
1
Γ(α)(ts)α1χR>0(s)hds
=ˆR
χR>0(ts)χR>0(s)1
Γ(α)(ts)α1ds h.
Thus, if t > 0, we obtain
(α
0, χR>0h)(t) = ˆt
0
1
Γ(α)(ts)α1ds h
=ˆt
0
1
Γ(α)sα1ds h =1
Γ(α)
1
αtαh.
For t0, we infer χR>0(ts)χR>0(s) = 0 for sR, 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 y0Cnis a given initial value; f:R>0×CnCnis continuous, satisfying
|f(t, y1)f(t, y2)| c|y1y2|(3.1)
for some c0 and all y1, y2Cn, t > 0. For definiteness, we shall also assume that
(t7→ f(t, 0)) L2
0(R>0;Cn) (3.2)
for some 0R. 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
(ts)α1f(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×CnCn,(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
−∞
(ts)α1e
f(s, y(s)) ds(t > 0),(3.4)
which in turn can be (trivially) stated for all tR. Next, we present the desired rewriting of equation
(3.4).
Theorem 3.1. Let > max{0, 0}. Assume that yL2
(R). Then the following statements are equiva-
lent:
(i) y(t) = y0χR>0(t) + 1
Γ(α)´t
−∞(ts)α1e
f(s, y(s)) dsfor almost every tR,
(ii) y=α
0, e
f(·, y(·)) + y0χR>0,
(iii) α
0,(yy0χR>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×RnRnwe may consider the Caputo differential
equation with f:R>0×CnCn,(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)),
dα1
dxα1y(x)x=0+
=y0,
where as before fsatisfies (3.1) and (3.2) and y0R,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
tα1
Γ(α)+1
Γ(α)ˆt
0
(ts)α1f(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 gα1L2
(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=gα1y0+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
gα1y0=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,
αR. 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 αRwe define the space
Hα(im+):=
fL1
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 Xx7→ xY, if XY(under a canonical identification,
which will always be obvious from the context).
Lemma 4.1. For 6= 0 and αRthe space Hα(im+)is a Hilbert space. Moreover, for β > α we
have
jβα:Hβ(im+)֒Hα(im+)
where the embedding is dense and continuous with kjβαk ||αβ.
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 fHβ(im+). Then
ˆ
R
k(it+)αf(t)k2
Hdt=ˆ
R
(t2+2)αβk(it+)βf(t)k2
Hdt2αβkfk2
Hβ(im+),
which proves the continuity of the embedding jβαand the asserted norm estimate. The density follows,
since C
c(R;H) lies dense in Hγ(im+) for each γR.
Definition. Let 6= 0 and αR.We consider the space
Wα
(R;H):=uL2
(R;H) ; LuHα(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 ,·i,α.
Lemma 4.2. Let 6= 0.
(a) For α0we 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
ιβα:Hβ
(R;H)֒Hα
(R;H)
is continuous and dense with kιβαk ||αβ.
(d) For each β > 0and αRthe operator
β
0, :Hβ+|α|
(R;H)Hα
(R;H)Hαβ
(R;H)
has a unique unitary extension, which will again be denoted by β
0,.
Proof.
9
(a) Let α0. For uHα(im+), i.e. uL1
loc(R;H) and (im+)αuL2(R;H), we infer that
uL2(R;H). It follows that udom((im+)α). Hence Hα(im+) = dom((im+)α). Moreover,
uWα
(R;H)uL2
(R;H) LuHα(im+)
uL2
(R;H) Ludom ((im+)α)
udom(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 ϕC
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 ιβα=L
jβαL,the assertion follows from Lemma 4.1.
(d) Since,
(im+)β:Hα(im+)Hαβ(im+)
f7→ t7→ (it+)βf(t)
is obviously unitary, we infer that for uHβ+|α|
(R;H)
kβ
0,uk,αβ=kLβ
0,ukHαβ(im+)
=k(im+)βLukHαβ(im+)
=kLukHα(im+)
=kuk,α,
which shows that β
0, is an isometry. Moreover, for ϕC
c(R;H), we have that (im+)γϕ
C
c(R;H) for all γRand thus, in particular L
(im+)βϕTγRHγ
(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 Hαβ(im+), we infer
that L
[C
c(R;H)] is dense in Hαβ
(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
αR(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 α0.
10
Lemma 4.3. Let 6= 0 and αR. Then
σ1:Wα
(R;H)Hα
(R;H)Hα
(R;H)
f7→ (t7→ f(t))
extends to a unitary operator. Moreover, for fHα
(R;H)we have
Lσ1f=σ1Lfand L
σ1f=σ1L
f.
Proof. For fWα
(R;H) we have that
Lσ1f=σ1Lf
and hence,
ˆ
R
k(it)α(Lσ1f) (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 σ1.Moreover, σ1has dense range, since σ1[Wα
(R;H)] = Wα
(R;H).
Hence, σ1extends to a unitary operator. The equality Lσ1f=σ1Lfholds for fHα
(R;H),
since Wα
(R;H) is dense in its completion Hα
(R;H).
Proposition 4.4. Let 6= 0, α Rand fHα
(R;H).Then
hf, ·i :C
c(R;H)C
given by
hf, ϕi:=ˆ
R
hLf(t),Lϕ(t)iHdt
defines a distribution. Moreover, for fHα
(R;H)and ϕC
c(R;H)we have
hf, ϕi=hf , α
0, e2mσ1α
0, σ1ϕi,α.
In particular, for α= 0
hf, ϕi=ˆ
R
hf(t), ϕ(t)iHdt.
Note that the operator α
0, e2mσ1α
0, σ1maps Hα
(R;H)to Hα
(R;H)unitarily.
Proof. Let fHα
(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 fHk
(R;H) for some kN.Indeed, if fHk
(R;H),then we
know that t7→ (it+)k(Lf) (t)L2(R;H)
and hence, for ϕC
c(R;H) we obtain using older’s inequality and the fact that Lϕ(k)= (im+
)kLϕ
|hf, ϕi| ˆ
R
|h(it+)k(Lf)(t),(it+)k(Lϕ) (t)iH|dt
11
kLfkHk(im+)kLϕ(k)kL2(R;H)
kLfkHk(im+)
ˆ
spt ϕ
e2t 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
σ1Lϕ=Le2mσ1ϕ
for ϕL2
(R;H). Let fHα
(R;H) and compute
hf, ϕi=hLf , LϕiL2(R;H)
=hLf, σ1Lσ1ϕiL2(R;H)
=h(im+)αLf, (im+)ασ1Lσ1ϕiL2(R;H)
=h(im+)αLf, σ1(im+)αLσ1ϕiL2(R;H)
=h(im+)αLf, σ1Lα
0, σ1ϕiL2(R;H)
=h(im+)αLf, Le2mσ1α
0, σ1ϕiL2(R;H)
=h(im+)αLf, (im+)αLα
0, e2mσ1α
0, σ1ϕiL2(R;H)
=hf, α
0, e2mσ1α
0, σ1ϕi,α
for each ϕC
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=0RHα
(R;H) D(R;H).In particular, the sup-
port of an element in Hα
(R;H) is then well-defined by
\{R\U;URopen,ϕC
c(U;H) : hf, ϕi= 0},
and the second part of the latter proposition shows, that it coincides with the usual L2-support if α0.
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 fL2
(R;H). Then, by definition, 0,fH1
(R;H). We shall compute the action
of 0,fas a distribution. For this let ϕC
c(R;H) and we compute with the formula outlined in
Proposition 4.4 for α=1:
h0,f , ϕi=h0,f, 0, e2mσ10, σ1ϕi,1
=h(im+)1L0,f, (im+)1L0,e2m σ10,σ1ϕiL2(R;H)
=hLf, Le2m σ10,σ1ϕiL2(R;H)
=−hLf, Le2m 0,ϕiL2(R;H)
=−hf, e2m ϕiL2
(R;H)
=ˆRhf(t), ϕ(t)iHdt.
Thus, 0,fcoincides with the distibutional derivative of L2
(R;H) functions.
12
Lemma 4.7. Let αR. Let H
(R;H):=TkNHk
(R;H)for 6= 0.
(a) Let ϕC
c(R;H). For all > 0we have α
ϕC(R;H)H
(R;H)and inf spt α
ϕinf spt ϕ.
For , µ > 0,αRwe have α
0,ϕ=α
0ϕ.
(b) Let αRand µ, 6= 0. Let ψC(R;H)H
(R;H)H
µ(R;H). Then there is (ϕn)nN
C
c(R;H)Ns.t. ϕnψfor n in Hα
(R;H)and Hα
µ(R;H)and spt(ϕn)spt(ψ)for nN.
Proof. (a): Let αR. Let µ, > 0. For α > 0 it holds that α
=α−⌈α
α
and α
ϕ=ϕ(α)=
α
µϕC
c(R;H) and α α<0. Thus we may assume that α < 0. By Theorem 2.6 we have
α
0,ϕ=α
0ϕand inf spt α
0,ϕ > −∞. From ϕH
(R;H) we deduce α
0,ϕH
(R;H).
(b): Let kNwith kα. We choose a sequence (χn)nNin C
c(R) such that spt χn[n1, n + 1],
χn= 1 on [n, n] and
sup nkχ(j)
nk;j {0,...,k}, n No<.
Set ϕn:=χnψC
c(R;H). Then spt(ϕn)spt(ψ) for nN. Since Hk
ν(R;H) is dense and continuously
embedded into Hα
ν(R;H) (ν6= 0), it suffices to show that ϕnψ(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(kj)
nψ(j)=χnψ(k)+
k1
X
j=0
χ(kj)
nψ(j)ψ(k)
for n in L2
(R;H) and in L2
µ(R;H).
Lemma 4.8. Let 6= 0 and αR.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 σ1from 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)nNC
c(R;H)Nwith ψnα
0,f(n ) in L2
(R;H). By Lemma 4.7(a)
we have α
0, ψnC(R;H)H
(R;H) and by Lemma 4.7(b) we find (ϕn)nNC
c(R;H)Nwith
α
0, ψnϕn
,α 0 (n ). Then
kfϕnk,α
fα
0, ψn
,α +
α
0, ψnϕn
,α 0 (n ).
With this result at hand, we can characterize those distributions, which belong to Hα
(R;H) for some
αR, 6= 0, in the following way.
Proposition 4.9. Let ψ D(R;H)and αR, 6= 0.Then, there exists fHα
(R;H)such that
ψ(ϕ) = hf, ϕi(ϕC
c(R;H))
in the sense of Propositions 4.4 if and only if there is C0such that
|ψ(ϕ)| Ckϕk,α
for each ϕC
c(R;H).
Proof. Assume first that there is fHα
(R;H) representing ψ. Then we estimate
|ψ(ϕ)|=|hf, ϕi|
13
=ˆ
R
hLf(t),Lϕ(t)iHdt
=ˆ
R
h(it+)αLf(t),(it+)αLϕ(t)iHdt
kLfkHα(im+)kLϕkHα(im)
=kfk,αkϕk,α
for each ϕC
c(R;H).Let C0 such that ψsatisfies for ϕC
c(R;H)
|ψ(ϕ)| Ckϕk,α.
The operator A:=α
0, e2mσ1α
0, σ1:Hα
(R;H)Hα
(R;H) (cf. Proposition 4.4) is unitary. Thus
for ϕC
c(R;H)ψ(A1ϕ)C
A1ϕ
,α=Ckϕk,α .
Moreover, C
c(R;H)Hα
(R;H) is dense. Thus ψ(A1·) can be extended continuously to Hα
(R;H).
By the Riesz representation theorem, there is a fHα
(R;H) such that for ϕHα
(R;H)
ψ(A1ϕ) = hf, ϕi,α .
By Theorem 4.4 we have for ϕC
c(R;H)
ψ(ϕ) = ψ(A1) = hf, i,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 αRand µ, > 0.Moreover, let fHα
(R;H)and gHα
µ(R;H).Then the
following statements are equivalent:
(i) f=gin the sense of distributions, i.e., for each ϕC
c(R;H)we have that
ˆ
R
hLf(t),Lϕ(t)iHdt=ˆ
R
hLµg(t),Lµϕ(t)iHdt.
(ii) α
0,f=α
0gas functions in L1
loc(R;H).
(iii) There is a sequence (ϕn)nNin C
c(R;H)with ϕnfin Hα
(R;H)and ϕngin Hα
µ(R;H)as
n .
Proof. (i)(ii): Let ψC
c(R;H) and e
ψ:=σ1α
σ1ψ=σ1α
µσ1ψ. Then by Lemma 4.7(a)
e
ψC(R;H)H
(R;H)H
µ(R;H). By Lemma 4.7(b) there’s (ϕn)nNC
c(R;H)Nwith ϕne
ψ
(n ) in Hα
(R;H) and in Hα
µ(R;H). Thus
ˆRhα
0,f(t), ψ(t)iHdt=hα
0,f , e2mψi,0
=hf, α
0, e2mσ1α
0, σ1(σ1α
0,σ1ψ)i,α
14
= lim
n→∞hf, α
0, e2mσ1α
0, σ1ϕni,α
= lim
n→∞hf, ϕni
= lim
n→∞hg, ϕni
= lim
n→∞hf, α
0 e2µmσ1α
0 σ1ϕniµ,α
=ˆRhα
0f(t), ψ(t)iHdt.
(ii) (iii): Define e
fn:=χ[n,n]·α
0,f=χ[n,n]·α
0gfor nN. Without loss of generality let < µ.
Take a function ψnC
c(R;H) with spt ψn[n, n] such that
ke
fnψnk,01
ne(µ)n.
Then, we estimate
ke
fnψnk2
µ,0=
n
ˆ
nke
fn(t)ψn(t)k2
He2µt dt=
n
ˆ
nke
fn(t)ψn(t)k2
He2te2(µ)tdt
ke
fnψnk2
,0e2(µ)n1
n2.
Hence ψnα
0,f=α
0gin L2
(R;H) and in L2
µ(R;H) by the triangle inequality and dominated
convergence. We set eϕn:=α
0, ψn=α
0 ψnC(R;H)H
(R;H)H
µ(R;H). Then eϕnf
and eϕngin Hα
(R;H) and in Hα
µ(R;H) respectively. We use Lemma 4.7(b) and choose a sequence
(ϕn)nNC
c(R;H)Nwith keϕnϕnk,α 0. Then
kfϕnk,α kfeϕnk,α +keϕnϕnk,α 0 (n ),
kgϕnkµ,α kgeϕnkµ,α +keϕnϕnkµ,α 0 (n ).
(iii) (i): Let (ϕn)nNbe a sequence in C
c(R;H) such that ϕnfand ϕngin Hα
(R;H) and
Hα
µ(R;H), respectively. Let ϕC
c(R;H).Then we have according to Proposition 4.4
ˆ
R
hLf(t),Lϕ(t)iHdt=hf, ϕi
=hf, α
0, e2mσ1α
0, σ1ϕi,α
= lim
n→∞hϕn, α
0, e2mσ1α
0, σ1ϕi,α
= lim
n→∞hϕn, ϕi
= lim
n→∞hϕn, α
0 e2µmσ1α
0 σ1ϕiµ,α
=hg, ϕi,
=ˆ
R
hLµf(t),Lµϕ(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)T0Hβ
(R;H)T0Hγ
(R;H)
eventually (β , γ)-Lipschitz continuous, if dom(F)C
c(R;H) and there exists ν0such that for each
νthe function Fhas a Lipschitz continuous extension
F:Hβ
(R;H)Hγ
(R;H)
satisfying supν|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,gHβ
(R;H), f6=g
kF(f)F(g)k,γ
kfgk,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 fHβ
(R;H) and gHβ
µ(R;H) generate the same distribution, we have that
F(f) = Fµ(g).
Indeed, by Proposition 4.10 there exists a sequence (ϕn)nNin C
c(R;H) with ϕnfand ϕngin
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
0F(ϕn)γ
0Fµ(g)
with convergence in L2
(R;H) and hence almost everywhere for a suitable subsequence of (ϕn)nN. The
assertion follows from Proposition 4.10.
(b) We shall need the following elementary observation later on. Let Fbe evenutally (β, γ)-Lipschitz
continuous, αR. 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
0fα
0g
µ,β =|Fµ|Lip kfgkµ,α+β,
for µν,f, g C
c(R;H).
16
Theorem 5.2. Let α > 0, β R, 0>0and F: dom(F)T0Hβ
(R;H)T0Hβα
(R;H)be
eventually (β , β α)-contracting. Then there exists ν0such that for each νthere is a unique
uHβ
(R;H)satisfying
α
0,u=F(u).(5.1)
Proof. This is a simple consequence of the contraction mapping theorem. Indeed, choosing ν0large
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, :Hβα
(R;H)Hβ
(R;H) is unitary by Lemma 4.2. Hence, the
mapping α
0, Fadmits a unique fixed point uHβ
(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)T0Hβ
(R;H)T0Hβγ
(R;H)for
some γ[0, α[be eventually (β, β γ)-Lipschitz continuous. Then there exists ν0such that for each
νthere is a unique uHβ
(R;H)satisfying
α
0,u=F(u).
Proof. It suffices to prove that ιβγβαFis eventually (β, β α)-contracting by Theorem 5.2. Let
ν, s.t. for ν,Fexists. Then for ν
|ιβγβαF|Lip kιβγβαk|F|Lip γα|F|Lip
by Lemma 4.2. Since |F|Lip is bounded in on [ν, [ by assumption, we infer
lim sup
→∞ |ιβγβαF|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,αRand aR. Let fHα
(R;H)with spt fRa. Then there is a
sequence (ϕn)nNC
c(R;H)Nwith spt ϕnRafor nNand ϕnfin Hα
(R;H)as n .
Proof. Let ( e
ψn)nNC
c(R;H)Nbe such that e
ψnα
0,fin H0
(R;H) as n . We may assume
that spt e
ψnR>a. We set ψn:=α
0, e
ψnfor nN. Then ψnfas 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)nNC
c(R;H)Nwith
spt(ϕn)spt(ψn) for nNand ϕnψn0 in Hα
(R;H) when n . Then
kϕnfk,α kϕnψnk,α +kψnfk,α 0 (n ).
Proposition 5.5. Let fHα
(R;H)for some αR, > 0.Assume that spt fRafor some aR.
Then
spt β
0,fRa
for all βR.
17
Proof. Let ϕC
c(R;H) with spt ϕR<a. By Lemma 5.4 we pick a sequence (ϕn)nNC
c(R;H)N,
s.t. spt ϕnRa(nN) and ϕnfin Hα
(R;H). Then spt β
0,ϕnRaby Lemma 4.7(a). By
Proposition 4.4 we have. Dβ
0,ϕn, ϕE=ˆRDβ
0,ϕn, ϕEH(t) dt= 0.
Since β
0, is unitary, we have β
0,ϕnβ
0,fin Hαβ
(R;H). We compute
Dβ
0,f , ϕE=hβ
0,f , (αβ)
0, e2mσ1(αβ)
0, σ1ϕi,αβ
= lim
n→∞hβ
0,ϕn, (αβ)
0, e2mσ1(αβ)
0, σ1ϕi,αβ
= lim
n→∞ Dβ
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(uv)Rafor some aR,it holds that
spt α
0, F(u)α
0, F(v)Ra.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(uv)
Ra.Take ϕC
c(R;H) with spt ϕR<a .Let µ. Then F(u) = Fµ(u) and
α
0, (F(u)F(v)), ϕ=α
0 (Fµ(u)Fµ(v)), ϕ
=Dα
0 (Fµ(u)Fµ(v)), β
0 e2µmσ1β
0 σ1ϕEµ,β
=DFµ(u)Fµ(v), (βα)
0 e2µmσ1β
0 σ1ϕEµ,βα
kFµ(u)Fµ(v)kµ,βα
(βα)
0 e2µmσ1β
0 σ1ϕ
µ,βα
|Fµ|lip kuvkµ,β
β
0 σ1ϕ
µ,0
where we have used that (βα)
0 e2µmσ1:H0
µ(R;H)Hβα
µ(R;H) is unitary and ϕHβ
µ(R;H).
According to Proposition 5.5 we have that spt β
0 σ1ϕR>aand hence, we compute
kβ
0 σ1ϕk2
µ,0=
ˆ
a
β
0 σ1ϕ(t)
2
He2µt dt=
ˆ
0
β
0 σ1ϕ(ta)
2
He2µt dte2µa.
On the other hand
kuvk2
µ,β =kβ
0(uv)k2
µ,0
18
=
ˆ
akβ
0(uv)(t)k2
He2µt dt
=
ˆ
0
β
0(uv)(t+a)
2
He2µt dte2µa
and consequently,
|Fµ|Lipkuvkµ,β kβ
0 σ1ϕkµ,0
=|Fµ|Lip
ˆ
0
β
0(uv)(t+a)
2
He2µt dt
1
2
ˆ
0
β
0 σ1ϕ(ta)
2
He2µt dt0 (µ ),
by dominated convergence. Summarizing, we have shown that spt(α
0, F(u)α
0, F(v)) Rafor
u, v C
c(R;H) satisfying spt(uv)Ra.
Before we conclude the proof, we show that if (wn)nNis a convergent sequence in Hβ
(R;H) with
spt wnRafor each nN, then its limit walso satisfies spt wRa.For doing so, let ϕC
c(R;H)
with spt ϕR<a.Then
hw, ϕi=Dw, β
0, e2mσ1β
0, σ1ϕE,β = lim
n→∞ Dwn, β
0, e2mσ1β
0, σ1ϕE,β = lim
n→∞ hwn, ϕi= 0.
Finally, let u, v Hβ
(R;H) with spt(uv)RaAccording to Lemma 5.4 there is a sequence (ϕn)nN
C
c(R;H)Nwith spt ϕnRaand ϕnuvin Hα
(R;H) as n . Let (vn)nNC
c(R;H) with
vnvin Hα
(R;H) as n . We set un:=ϕn+vn. Then unuin Hα
(R;H) and spt(unvn)Ra.
By the already proved result for C
c(R;H), we infer that spt α
0, F(un)α
0, F(vn)Rafor all
nN. Thus, letting n , we obtain spt α
0, F(u)α
0, F(v)Ra, 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 ueµHβ
eµ(R;H), uµHβ
µ(R;H)satisfying
α
0,eµueµ=Feµ(ueµ)and α
0uµ=Fµ(uµ).
Then ueµ=uµas distributions in the sense of Proposition 4.4.
Proof. We note that it suffices to show vµ:=β
0uµ=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
Hβα
(R;H)
given by e
F(v):=F(β
0, v) (vdom( e
F)) (5.2)
with maximal domain dom( e
F) = {wT0H0
(R;H) ; 0:β
0, wdom(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,
αβ
0 vµ=α
0uµ=Fµ(uµ) = e
Fµ(vµ)
and analogously
αβ
0,eµveµ=e
Feµ(veµ).
Let now aRand assume without loss of generality that µ < eµ. We note that spt(veµχRaveµ)χRa.
We obtain, applying Theorem 5.6, that
χRaveµ=χRaβα
0,eµe
Feµ(veµ) = χRaβα
0,eµe
Feµ(χRaveµ).
Now, since χRaveµL2
µ(R;H)L2
eµ(R;H), we infer that
χRaveµ=χRaβα
0,eµe
Feµ(χRaveµ) = χRaβα
0 e
Fµ(χRaveµ),
i.e. χRaveµis a fixed point of χRaβα
0 e
Fµ. However, since χRavµis also a fixed point of this mapping,
which is strictly contractive, we derive
χRaveµ=χRavµ
and since aRwas 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,nN,y0Rn,f:R>0×CnCncontinuous. Assume there exists
c0such that for all y1, y2Rn, t > 0we have
|f(t, y1)f(t, y2)| c|y1y2|.
Moreover, we assume that
(t7→ f(t, 0)) L2
0(R>0;Cn).
Define e
f:R×CnCnby
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) (ϕC
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)|2e2t dt=ˆR>0|f(t, y0)|2e2t dt
20
2ˆR>0|f(t, y0)f(t, 0)|2e2t dt+ˆR>0|f(t, 0)|2e2t 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)|2e2t dt=ˆR|e
f(t, ϕ(t) + y0)e
f(t, ψ(t) + y0)|2e2t dt
=ˆR>0|f(t, ϕ(t) + y0)f(t, ψ(t) + y0)|2e2t dt
ˆR>0
c2(|ϕ(t)ψ(t)|)2e2t dtc2kϕψk2
L2
.
Since F(0) L2
(R;Cn), the shown estimate yields F(ϕ)L2
(R;Cn) for each ϕC
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 y0Cn. Then there is 1>0such that for all 1there exists a unique
yL2
(R;Cn)with yy0χR>0Hα
(R;Cn)satisfying
α
0,(yy0χR>0) = e
f(·, y(·)).
Moreover, spt yR0.
Proof. With Fas defined in Proposition 6.1, we may apply Corollary 5.3 with β=γ= 0 to obtain unique
existence of zHα
(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)R0, we obtain with Proposition 5.5 that spt z= spt α
0, F(z)R0. Thus, spt y
R0.
We remark here that the condition spt yR0together with yy0χR>0Hα
(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 = (yy0χR>0)(0) = (yy0χR>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>0y0H0
(R;H) for > 0 and by Example 4.6 we have
0,χR>0y0=δ0y0H1
(R;H).
We also recall the notation gβ(t):=1
Γ(β+1) tβχR>0for β, t R.
21
Proposition 6.3. Let y0Cn. Assume that C
c(R;H)ϕ7→ e
f(·, ϕ(·)) is eventually (α1, α 1)-
Lipschitz continuous and denote with Hα1
(R;H)y7→ e
f(·, y(·)) its Lipschitz-continuous extension
for some > 0. There is 1>0such that for 1we have a unqiue solution yHα1
(R;H)of the
equation
α
0,y=y0δ0+e
f(·, y(·)), y Hα1
(R;H),
with α1
0, yy0χR>0H0
(R;H)and spt(y)R0.
Proof. The mapping Gdefined by
G(ϕ)(t):=e
f(t, 1α
0, ϕ(t) + gα1(t)y0), ϕ C
c(R;H), t R
is eventually (0, α 1)-Lipschitz continuous. Indeed, this fact follows from gα1y0Hα1
(R;H) and
the unitarity of α