ARTICLE IN PRESS
J. Differential Equations 196 (2004) 418–447
Quasilinear evolutionary equations and
continuous interpolation spaces
Philippe Clément,a Stig-Olof Londen,b, and Gieri Simonettc
a
Department of Mathematics and Informatics, TU Delft 2600 GA Delft, The Netherlands
b
Institute of Mathematics, Helsinki University of Technology, 02150 Espoo, Finland
c
Department of Mathematics, Vanderbilt University, Nashville, TN 37240, USA
Received August 26, 2002; revised June 17, 2003
Abstract
In this paper we analyze the abstract parabolic evolutionary equations
Dat ðu xÞ þ AðuÞu ¼ f ðuÞ þ hðtÞ;
uð0Þ ¼ x;
in continuous interpolation spaces allowing a singularity as tk0: Here Dat denotes the timederivative of order aAð0; 2Þ: We first give a treatment of fractional derivatives in the spaces
Lp ðð0; TÞ; X Þ and then consider these derivatives in spaces of continuous functions having (at
most) a prescribed singularity as tk0: The corresponding trace spaces are characterized and
the dependence on a is demonstrated. Via maximal regularity results on the linear equation
Dat ðu xÞ þ Au ¼ f ;
uð0Þ ¼ x;
we arrive at results on existence, uniqueness and continuation on the quasilinear equation.
Finally, an example is presented.
r 2003 Elsevier Inc. All rights reserved.
Keywords: Abstract parabolic equations; Continuous interpolation spaces; Quasilinear evolutionary
equations; Maximal regularity
Corresponding author. Institute of Mathematics, Helsinki University of Technology, P.O. Box 1100,
02015 HUT, Finland. Fax: +35-89-45-13-016.
E-mail address: stig-olof.londen@hut.fi (S.-O. Londen).
0022-0396/$ - see front matter r 2003 Elsevier Inc. All rights reserved.
doi:10.1016/j.jde.2003.07.014
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1. Introduction
In a recent paper, [7], the quasilinear parabolic evolution equation
du
þ AðuÞu ¼ f ðuÞ;
dt
uð0Þ ¼ x;
was considered in continuous interpolation spaces. The analysis was based on
maximal regularity results concerning the linear equation
du
þ Au ¼ f ;
dt
uð0Þ ¼ x:
In particular, the approach allowed for solutions having (at most) a prescribed
singularity as tk0: Thus the smoothing property of parabolic evolution equations
could be incorporated.
In this paper we show that the approach and the principal results of [7] extend, in a
very natural way, to the entire range of abstract parabolic evolutionary equations
Dat ðu xÞ þ AðuÞu ¼ f ðuÞ;
uð0Þ ¼ x:
Here Dat denotes the time-derivative of arbitrary order aAð0; 2Þ:
As in [7], our basic setting is the following. Let E0 ; E1 be Banach spaces, with
E1 CE0 ; and assume that, for each u; AðuÞ is a linear bounded map of E1 into E0
which is positive and satisfies an appropriate spectral angle condition as a map in E0 :
Moreover, AðuÞ and f ðuÞ are to satisfy a specific local continuity assumption with
respect to u:
Problems of fractional order occur in several applications, e.g., in viscoelasticity
[10], and in the theory of heat conduction in materials with memory [17]. For an
entire volume devoted to applications of fractional differential systems, see [16].
Our paper is structured as follows. We first (Section 2) define, and give a brief
treatment of, fractional derivatives in the spaces Lp ðð0; TÞ; X Þ and then (Section 3)
consider these derivatives in spaces of continuous functions having a prescribed
singularity as tk0: In Section 4 we characterize the corresponding trace spaces at
t ¼ 0 and show how these spaces depend on a:
In Section 5 we consider the maximal regularity of the linear equation
Dat ðu xÞ þ Au ¼ f ;
uð0Þ ¼ x;
ð1Þ
where again aAð0; 2Þ and where the setting is the space of continuous functions
having at most a prescribed singularity as tk0: To obtain maximal regularity we
make a further assumption on E0 ; E1 :
In Section 6 we analyze the nonautonomous, A ¼ AðtÞ; version of (1). Here we
assume that for each fixed t the corresponding operator admits maximal regularity
and deduce maximal regularity of the nonautonomous case.
In Sections 7 and 8 we combine our results of the previous sections with a
contraction mapping technique to obtain existence, uniqueness, and continuation
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results on
Dat ðu xÞ þ AðuÞu ¼ f ðuÞ þ hðtÞ;
uð0Þ ¼ x:
Finally, in Section 9, we present an application of our results to the nonlinear
equation
Dat ðu u0 Þ ðsðux ÞÞx ¼ hðtÞ;
xAð0; 1Þ;
tX0;
with u ¼ uðt; xÞ; uð0; xÞ ¼ u0 ðxÞ; aAð0; 2Þ; Dirichlet boundary conditions, s
monotone increasing and sufficiently smooth.
This equation occurs in nonlinear viscoelasticity, and has been studied, e.g., in
[10,12].
Parabolic evolution equations, linear and quasilinear, have been considered by
several authors using different approaches. Of particular interest to our approach are
the references, among others, [1,2,8,15]. The reader may consult [7] for more detailed
comments on the relevant literature.
It should also be observed that we draw upon results of [4], where (1) is considered
in spaces of continuous functions on ½0; T; i.e., without allowance for any
singularity at the origin.
2. Fractional derivatives in Lp
We recall [20, II, pp. 134–136] the following definition and the ensuing properties.
Let X be a Banach space and write
gb ðtÞ ¼
1 b1
t ;
GðbÞ
t40;
b40:
Definition 1. Let uAL1 ðð0; TÞ; X Þ for some T40: We say that u has a fractional
derivative of order a40 provided u ¼ ga f for some f AL1 ðð0; TÞ; X Þ: If this is the
case, we write Dat u ¼ f :
Note that if a ¼ 1; then the above condition is sufficient for u to be absolutely
continuous and differentiable a.e. with u0 ¼ f a.e.
Tradition has that the word fractional is used to characterize derivatives of
noninteger order, although a may of course be any positive real number.
The fractional derivative (whenever existing) is essentially unique. Observe the
consistency; if u ¼ ga f ; and aAð0; 1Þ; then f ¼ Dat u ¼ dtd ðg1a uÞ: Thus, if u has a
fractional derivative of order aAð0; 1Þ; then g1a u is differentiable a.e. and
absolutely continuous. Also note a trivial consequence of the definition; i.e., Dat ðga
uÞ ¼ u:
Suppose aAð0; 1Þ: By the Hausdorff–Young inequality one easily has that if the
fractional derivative f of u satisfies f ALp ðð0; TÞ; X Þ with pA½1; 1aÞ; then
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p
: Furthermore, if f ALp ðð0; TÞ; X Þ; with p ¼ a1 ; then
uALq ðð0; TÞ; X Þ for 1pqo1ap
uALq ðð0; TÞ; X Þ
for
If
qA½1; NÞ:
f ALp ðð0; TÞ; X Þ
with
a1 op;
then
1
ap
uAh0-0
ð½0; T; X Þ
[20, II, p. 138]. In particular note that uð0Þ is now well defined
and that one has uð0Þ ¼ 0: (By hy0-0 we denote the little-Hölder continuous functions
having modulus of continuity y and vanishing at the origin.)
The extension of the last statement to higher order fractional derivatives is
obvious. Thus, if u has a fractional derivative f of order aAð1; 2Þ and f ALp with
1
ða 1Þ1 op; then ut Aha1p
:
0-0
We also note that if uAL1 ðð0; TÞ; X Þ with Dat uALN ðð0; TÞ; X Þ; aAð0; 1Þ; then
a
a
uAC0-0
ð½0; T; X Þ: The converse is not true, for uAC0-0
ð½0; T; X Þ the fractional
derivative of order a of u does not necessarily even exist. To see this, take vAL [20,
vAC a ð½0; T; X Þ: Without loss of
I, p. 43], then [20, II, Theorem 8.14(ii), p. 136] D1a
t
1a
generality, assume Dt v vanishes at t ¼ 0: Assume that there exists f AL1 ðð0; TÞ; X Þ
such that
Dt1a v ¼ t1þa f :
But this implies (convolve by ta ) v ¼ 1 f ; which does not in general hold for vAL
[20, I, p. 433].
The following proposition shows that the Lp -fractional derivative is the fractional
power of the realization of the derivative in Lp :
Proposition 2. Let 1ppoN and define
def
DðLÞ ¼ W01;p ðð0; TÞ; X Þ;
and
def
Lu ¼ u0 ;
uADðLÞ:
Then L is m-accretive in Lp ðð0; TÞ; X Þ with spectral angle p2: With aAð0; 1Þ we have
La u ¼ Dat u;
uADðLa Þ;
where in fact DðLa Þ coincides with the set of functions u having a fractional derivative
in Lp ; i.e.,
n
o
DðLa Þ ¼ uALp ðð0; TÞ; X Þ j g1a uAW01;p ðð0; TÞ; X Þ :
Moreover, La has spectral angle
ap
2:
We only briefly indicate the proof of this known result. (Cf. the proof of
Proposition 5 below.)
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The fact that L is m-accretive and has spectral angle p2 is well known. See, e.g., [3,
Theorem 3.1]. The representation formula given in the proof of Proposition 5 and
the arguments following give the equality of La and Dat : The reasoning used to prove
[4, Lemma 11(b)] can be adapted to give that La has spectral angle ap
2:
p
We remark that if X has the UMD-property then (in L ðð0; TÞ; X Þ with 1opoN)
we have
DðLa Þ ¼ DðDat Þ ¼ ½Lp ðð0; TÞ; X Þ; W01;p ðð0; TÞ; X Þa :
See [9, p. 20] or [19, pp. 103–104], and observe that
powers in Lp ðð0; TÞ; X Þ:
d
dt
admits bounded imaginary
3. Fractional derivatives in BUC1l
Let X be a Banach space and T40: We consider functions defined on J0 ¼ ð0; T
having (at most) a singularity of prescribed order at t ¼ 0:
Let J ¼ ½0; T; mAð0; 1Þ; and define
BUC1m ðJ; X Þ
¼ fuACðJ0 ; X Þjt1m uðtÞABUCðJ0 ; X Þ; lim t1m jjuðtÞjjX ¼ 0g;
tk0
with
def
jjujjBUC1m ðJ;X Þ ¼ sup t1m jjuðtÞjjX :
ð2Þ
tAJ0
(In this paper, we restrict ourselves to the case mAð0; 1Þ: The case m ¼ 1 was
considered in [4].) It is not difficult to verify that BUC1m ðJ; X Þ; with the norm given
in (2), is a Banach space. Note the obvious fact that for T1 4T2 we may view
BUC1m ð½0; T1 ; X Þ as a subset of BUC1m ð½0; T2 ; X Þ; and also that if
uABUC1m ð½0; T; X Þ for some T40; then (for this same u) one has
lim jjujjBUC1m ð½0;t;X Þ ¼ 0:
ð3Þ
tk0
Moreover, one easily deduces the inequality
jjujjLp ðJ;X Þ pcjjujjBUC1m ðJ;X Þ ;
mAð0; 1Þ;
1ppoð1 mÞ1 ;
and so, for these ðm; pÞ-values,
BUC1m ðJ; X ÞCLp ðJ; X Þ;
with dense imbedding. To see that this last fact holds, recall that CðJ; X Þ is dense in
Lp ðJ; X Þ and that obviously CðJ; X ÞCBUC1m ðJ; X Þ:
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423
We make the following fundamental assumption:
a þ m41:
ð4Þ
To motivate this assumption, suppose we require (as we will do) that both u and
Dat u lie in BUC1m and that uð0Þ (¼ 0) is well defined. The requirement Dat uABUC1m
1
: On the other hand, if
implies, by the above, Dat uALp ðð0; TÞ; X Þ; for 1ppo1m
a
1
p
Dat uALp with a1 op then uAh0-0
and uð0Þ ð¼ 0Þ is well defined. Thus our
requirements motivate the assumption that the interval ða1 ; ð1 mÞ1 Þ be nonempty. But this is (4).
Therefore, under the assumption (4), the following definition makes sense.
def
a
BUC1m
ðJ; X Þ ¼ fuABUC1m ðJ; X Þj
there exist xAX and f ABUC1m ðJ; X Þ such that u ¼ x þ ga f g:
ð5Þ
a
We keep in mind that if uABUC1m
ðJ; X Þ; then (assuming (4)) uð0Þ ¼ x and u is
Hölder-continuous.
a
We equip BUC1m
ðJ; X Þ with the following norm:
jjujjBUC a
1m
def
ðJ;X Þ
¼ jjujjBUC1m ðJ;X Þ þ jjDat ðu xÞjjBUC1m ðJ;X Þ :
ð6Þ
Lemma 3. Let a40; mAð0; 1Þ; and let (4) hold. Space (5), equipped with norm (6), is a
a
Banach space. In particular, BUC1m
ðJ; X ÞCBUCðJ; X Þ:
a
Proof. Take fwn gN
n¼1 to be a Cauchy-sequence in BUC1m ðJ; X Þ: Then, by (6), and as
BUC1m ðJ; X Þ is a Banach space, there exists wABUC1m ðJ; X Þ such that jjwn
def
wjjBUC1m ðJ;X Þ -0: Moreover, fn ¼ Dat ðwn wn ð0ÞÞ converges in BUC1m ðJ; X Þ to
some function z:
We claim that wð0Þ is well defined and that z ¼ Dat ðw wð0ÞÞ: To this end, note
that
wn ðtÞ wn ð0Þ ¼ ga fn ¼ ga z þ ga ½fn z:
ð7Þ
We have limn-N jjt1m ½fn ðtÞ zðtÞjjX ¼ 0; uniformly on J: Thus, by (4),
limn-N jjga ½fn zjjX ¼ 0; uniformly on J: So, uniformly on J;
lim ½wn ðtÞ wn ð0Þ ¼ ga z:
n-N
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N
For each fixed t40; fwn ðtÞgN
n¼1 converges to wðtÞ in X : Thus fwn ð0Þgn¼1 must
converge in X and by (4) and (7) we must have wn ð0Þ-wð0Þ as n-N: For the proof
of the last statement, use the considerations preceding the theorem. &
Our
next
purpose
is
to
consider
in
more
detail
differentiation
on
def
X̃ ¼ BUC1m ðJ; X Þ and to connect the fractional powers of this operation with
that of taking fractional derivatives. First consider the derivative of integer
order.
Take a ¼ 1 in (5), (6), (thus a þ m41Þ and define
n
o
def
1
DðL̃Þ ¼ uABUC1m
ðJ; X Þ j uð0Þ ¼ 0 ;
and
L̃u ¼ u0 ðtÞ;
uADðL̃Þ:
We have
Lemma 4.
(i) DðL̃Þ is dense in X̃;
(ii) L̃ is a positive operator in X̃; with spectral angle p2:
def
Proof. (i) Clearly, Ỹ ¼ uAC 1 ðJ; X Þ j uð0Þ ¼ 0 CDðL̃Þ: It is therefore sufficient to
prove that Ỹ is dense in X̃: Observe that ỸCC0-0 ðJ; X ÞCX̃: It is well known that Ỹ
is dense in C0-0 ðJ; X Þ with respect to the sup-norm (which is stronger than the norm
in X̃). So it suffices to prove that C0-0 ðJ; X Þ is dense in X̃:
Let uAX̃: There exists vAC0-0 ðJ; X Þ such that uðtÞ ¼ tm1 vðtÞ; tAð0; T: Set, for n
large enough,
(
0;
tA½0; 1n;
vn ðtÞ ¼
vðt 1nÞ; tAð1n; T;
un ðtÞ ¼ tm1 vn ðtÞ;
tAð0; T;
un ð0Þ ¼ 0:
Then un ðtÞAC0-0 ðJ; X Þ; and
sup jjt1m ½uðtÞ un ðtÞjjX ¼ sup jjvðtÞ vn ðtÞjjX
tAð0;T
tAð0;T
1
p sup jjvðtÞjjX þ sup jjvðtÞ v t jjX -0;
n
1
1
0ptpn
notpT
as n-N: It follows that C0-0 ðJ; X Þ is dense in X̃ and (i) holds.
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(ii). First, note that X̃CL1 ðJ; X Þ and that for every lAC and every f AL1 ðJ; X Þ;
the problem
lu þ u0 ¼ f ;
uð0Þ ¼ 0;
has a unique solution uAW01;1 ðð0; TÞ; X ÞCC0-0 ð½0; T; X Þ; given by
uðtÞ ¼
t
Z
exp½lðt sÞ f ðsÞ ds;
tAJ:
0
We use this expression to estimate
sup jljt1m jjuðtÞjjX ;
sup
jarg ljpy tAð0;T
in case f AX̃ and yA½0; p2Þ: Thus
jjlt
1m
1m
uðtÞjjX p t
p
def
t
Z
0
jljexp½Rlðt sÞsm1 ds jjf jjX̃
1 1m
t
cos y
Z
t
ðRlÞ exp½Rlðt sÞsm1 dsjjf jjX̃ :
0
def
We write Z ¼ Rl40; t ¼ Zs; to obtain
1 1m
ðcos yÞ t
Z
t
ðRlÞ exp½Rlðt sÞsm1 ds
0
¼ ðcos yÞ1 ðZtÞ1m
Z
Zt
exp½Zt þ ttm1 dtpcy ;
0
where cy is independent of Z40; t40: To see that the last inequality holds, first
observe that the expression to be estimated only depends on the product Zt (and on
m; y). Then split the integral into two parts, over ð0; Zt2 Þ; and over ðZt2 ; ZtÞ; respectively
(cf. [2, p. 106]).
We conclude that the spectral angle of L̃ is not strictly greater that p2:
Finally, assume that the spectral angle is less than p2: Then L̃ would generate an
analytic semigroup. To obtain a contradiction, observe that L̃ is the restriction to X̃
def
1;1
of L̃1 considered on L1 ðð0; TÞ; X Þ; where DðL̃1 Þ ¼ W0-0
ðð0; TÞ; X Þ; L̃1 u ¼ u0 ;
uADðL̃1 Þ: Thus the analytic semigroup TðtÞ generated by L̃ would be the
restriction to X̃ of right translation, i.e.,
ðTðtÞf ÞðsÞ ¼
f ðs tÞ; 0ptps;
0;
sot:
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But X̃ is not invariant under right translation. By this contradiction, (ii) follows and
Lemma 4 is proved. &
Proceeding next to the fractional powers and fractional derivatives we have:
Proposition 5. Let a; mAð0; 1Þ: Then
def
DðL̃a Þ ¼ DðDat Þ ¼ fuAX̃ j u ¼ ga f for some f AX̃g;
and L̃a u ¼ Dat u; for uADðL̃a Þ: Moreover,
Dat is positive; densely defined on X̃; and has spectral angle
ap
:
2
ð8Þ
Proof. We first show that
ðL̃1 Þa f ¼ ga f ;
ð9Þ
for f AX̃:
Observe that 0ArðL̃Þ; and that L̃ is positive. Thus
Z N
1
1 a
a
ðL̃ Þ f ¼ L̃ f ¼
sa ðsI þ L̃Þ1 f ds;
GðaÞGð1 aÞ 0
where the integral converges absolutely. But
ðsI þ L̃Þ1 f ¼
Z
t
exp½sðt sÞ f ðsÞ ds;
0ptpT;
0
and so, after a use of Fubini’s theorem,
Z t Z N
1
sa exp½ssds f ðt sÞ ds:
ðL̃1 Þa f ¼
GðaÞGð1 aÞ
0
0
To obtain (9), note that the inner integral equals ga ðsÞ:
Let uADðDat Þ: Then u ¼ ga f ; with Dat u ¼ f AX̃: So, by (9), u ¼ ðL̃1 Þa f ; which
implies uADðL̃a Þ and L̃a u ¼ f :
Conversely, let uADðL̃a Þ: Then, for some f AX̃; L̃a u ¼ f ; and so u ¼ ðL̃a Þ1 f : By
(9), this gives u ¼ ga f and so uADðDat Þ:
We conclude that DðL̃a Þ ¼ DðDat Þ and that L̃a u ¼ Dat u; uADðL̃a Þ:
To get that Dat is densely defined, use (i) of Lemma 4 and apply, e.g., [18,
Proposition 2.3.1]. The fact that the spectral angle is ap
2 follows, e.g., by the same
arguments as those used to prove [4, Lemma 11(b)]. &
Analogously, higher order fractional derivatives may be connected to fractional
powers. We have, e.g., the following statement.
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Proposition 6. Let a; mAð0; 1Þ: Define
n
o
def
1
a
DðD1þa
Þ
¼
uABUC
ð½0;
T;
X
Þ
j
uð0Þ
¼
0;
u
ADðD
Þ
;
t
t
1m
t
u ¼ Dat ut ; for uADðD1þa
Þ: Then
and D1þa
t
t
L̃1þa u ¼ Dat ut ;
uADðD1þa
Þ:
t
Moreover, L̃1þa is positive, densely defined on X̃ with spectral angle
with (cf. (9)),
ðL̃1þa Þ1 f ¼ g1þa f ; for f AX̃:
ð1þaÞp
2
and
For the proof of Proposition 6, first use Proposition 5 and the definition D1þa
u¼
t
uADðD1þa
Þ:
To
obtain
the
size
of
the
spectral
angle
one
may
argue
as
in
the
t
proof of [5, Lemma 8(a)].
Dat ut ;
4. Trace spaces
Let E1 ; E0 be Banach spaces with E1 CE0 and dense imbedding and let A
be an isomorphism mapping E1 into E0 : Take aAð0; 2Þ; mAð0; 1Þ: Further, let
A as an operator in E0 be nonnegative with spectral angle fA satisfying
fA op 1
a
:
2
Assume (4) holds and write J ¼ ½0; T:
We consider the spaces
def
Ẽ0 ðJÞ ¼ BUC1m ðJ; E0 Þ;
ð10Þ
a
Ẽ1 ðJÞ ¼ BUC1m ðJ; E1 Þ-BUC1m
ðJ; E0 Þ;
ð11Þ
def
and equip Ẽ1 ðJÞ with the norm
h
i
def
jjujjẼ1 ðJÞ ¼ sup t1m jjf ðtÞjjE0 þ jjuðtÞjjE1 ;
tAð0;T
where f is defined through the fact that uAẼ1 ðJÞ implies u ¼ x þ ga f ; for some
f AẼ0 ðJÞ:
Without loss of generality, we take jjyjjE1 ¼ jjAyjjE0 ; for yAE1 ; and note that by
Lemma 3, Ẽ1 ðJÞ is a Banach space. We write
def
def
Ey ¼ ðE0 ; E1 Þy ¼ ðE0 ; E1 Þ0y;N ;
yAð0; 1Þ;
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for the continuous interpolation spaces between E0 and E1 : Recall that if Z is some
number such that 0pZop fA ; then
xAEy
iff
lim
jlj-N;jarg ljpZ
jjly AðlI þ AÞ1 xjjE0 ¼ 0;
ð12Þ
and that we may take
def
jjxjjy ¼
sup
jarg ljpZ;la0
jjly AðlI þ AÞ1 xjjE0
as norm on Ey (see [13, Theorem 3.1, p. 159] and [14, p. 314]).
Our purpose is to investigate the trace space of Ẽ1 ðJÞ:
We define
g : Ẽ1 ðJÞ-E0
by gðuÞ ¼ uð0Þ;
def
and the trace space gðẼ1 ðJÞÞ ¼ ImðgÞ; with
def
jjxjjgðẼ1 ðJÞÞ ¼ inffjjvjjẼ1 ðJÞ j vAẼ1 ðJÞ; gðvÞ ¼ xg:
It is straightforward to show that this norm makes gðẼ1 ðJÞÞ a Banach space.
Define
1m
m# ¼ 1
a
for mAð0; 1Þ; aAð0; 2Þ with a þ m41: Observe that this very last condition is
equivalent to m40
#
and that ao1 implies mom;
#
whereas aAð1; 2Þ gives mom:
# Thus
0omomo1;
#
aAð0; 1Þ;
0omomo1;
#
aAð1; 2Þ:
Obviously, if a ¼ 1; then m# ¼ m:
We claim
Theorem 7. For mAð0; 1Þ; aAð0; 2Þ; a þ m41; one has
gðẼ1 ðJÞÞ ¼ Em# :
Proof. The case a ¼ 1 is treated in [7]. Thus let aa1 and first consider the case
aAð0; 1Þ:
Let xAEm# : We define u as the solution of
u x þ ga Au ¼ 0;
tAJ;
ð13Þ
tAJ:
ð14Þ
or, equivalently, as the solution of
Dat ðu xÞ þ Au ¼ 0;
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By Clément et al. [4, Lemma 7], u is well defined and given by
Z
1
exp½ltðla I þ AÞ1 la1 x dl; t40;
uðtÞ ¼
2pi G1;c
429
ð15Þ
A
Here cAðp2; minðp; pf
a ÞÞ and
def
Gr;c ¼ freit j jtjpcg,freic j roroNg,freic j roroNg:
Note that limtk0 jjuðtÞ xjjE0 ¼ 0: We assert that limtk0 jjt1m Dat ðu xÞjjE0 ¼ 0; i.e.,
that
Z
lim t1m
exp½ltAðla I þ AÞ1 la1 x dl ¼ 0
ð16Þ
t-0
G1;c
in E0 : To show this assertion, we take t40 arbitrary and rewrite the expression in
def
(16) ð ¼ IÞ as follows:
I ¼ t1m
¼
Z
exp½ltAðla I þ AÞ1 la1 x dl
G1
t ;c
s
exp½s
t
G1;c
Z
am#
A
n s a
o1
x sm ds:
I þA
t
ð17Þ
The first equality followed by analyticity; to obtain the second we made the variable
def
transform s ¼ lt and used the definition of m:
#
Now recall that xAEm# and use (12) in (17) to get (16). Observe also that by the
above one has
sup jjt1m Dat ðu xÞjjE0 pcjjxjjEm# ;
ð18Þ
tAJ0
where c ¼ cðm; cÞ but where c does not depend on T:
By (14), (16), (18),
sup jjt1m AuðtÞjjE0 pcjjxjjEm# ;
tAJ0
lim jjt1m AuðtÞjjE0 ¼ 0:
tk0
ð19Þ
Continuity of AuðtÞ and Dat ðu xÞ in E0 for tAð0; T follows from (15). One
concludes that
Em# CgðẼ1 ðJÞÞ:
ð20Þ
Observe that we also have:
If
xAEm# ; and u solves ð13Þ; then uAẼ1 ðJÞ:
ð21Þ
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Conversely, take xAgðẼ1 ðJÞÞ and take vAẼ1 ðJÞ such that vð0Þ ¼ x: Then
def
H0 ðtÞ ¼ t1m Dat ðv xÞABUC0-0 ðJ; E0 Þ;
def
H1 ðtÞ ¼ t1m AvðtÞABUC0-0 ðJ; E0 Þ:
def
It follows that, with H ¼ H0 þ H1 ;
Dat ðv xÞ þ AvðtÞ ¼ tm1 HðtÞ:
ð22Þ
We take the Laplace transform ðl40Þ of tm1 HðtÞ (take HðtÞ ¼ 0; t4T), to obtain,
in E0 ;
Z
T
exp½lttm1 HðtÞ dt ¼ lm
0
Z
lT
exp½ssm1 H
0
s
ds ¼ oðlm Þ
l
ð23Þ
for l-N: For the last equality, use HAC0-0 ðJ; E0 Þ:
Obviously, (23) holds with H replaced by H0 : Hence, by the way H0 was defined
and after some straightforward calculations,
ṽ l1 x ¼ la oðlm Þ for l-N:
ð24Þ
Take transforms in (22), use (23), (24) to obtain
Aðla I þ AÞ1 x ¼ l1a oðlm Þ;
and so, in E0 ;
lam# Aðla I þ AÞ1 x-0;
l-N:
Hence xAEm# :
The case aAð1; 2Þ follows in the same way. Again, define u by (13) (or (14))
but now use [5, Lemma 3] instead of [4, Lemma 7]. Note that one in fact
takes ut ð0Þ ¼ 0: Relations (15)–(19) remain valid and (20) follows. The proof
of the converse part also carries over from the case where aAð0; 1Þ: &
We next show that uAẼ1 ðJÞ implies that the values of u remain in Em# : In
particular, we have:
Theorem 8. Let mAð0; 1Þ; aAð0; 2Þ and let (4) hold. Then
Ẽ1 ðJÞCBUCðJ; Em# Þ:
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Proof. Take uAẼ1 ðJÞ: By Theorem 7, uð0ÞAEm# : We split u into two parts, writing
u ¼ v þ w where v; w satisfy
Dat ðv uð0ÞÞ þ AvðtÞ ¼ 0;
vð0Þ ¼ uð0ÞAEm# ;
Dat w þ AwðtÞ ¼ tm1 hðtÞ;
wð0Þ ¼ 0:
ð25Þ
ð26Þ
The function hABUC0-0 ðJ; E0 Þ is defined through Eqs. (25), (26).
We consider the equations separately, beginning with the former. The claim is then
that vAẼ1 ðJÞ-BUCðJ; Em# Þ:
Take transforms in (25), use analyticity and invert to get, for t40;
Z
1
vðtÞ uð0Þ ¼
exp½ltl1 Aðla I þ AÞ1 uð0Þ dl;
2pi G1
t ;c
and so
Zm# AðZI þ AÞ1 ðvðtÞ uð0ÞÞ
Z
1
¼
exp½ltl1 Aðla I þ AÞ1 Zm# AðZI þ AÞ1 uð0Þ dl:
2pi G1
t ;c
Thus, using uð0ÞAEm# ;
1
m#
jjZ AðZI þ AÞ ðvðtÞ uð0ÞÞjjE0 pe
¼ e
Z
Z
jexp½ltl1 j djlj
G1
t ;c
jexp½tjjtj1 djtjpce;
G1;c
where e40 arbitrary, and ZXZðeÞ sufficiently large.
The conclusion is that ½vðtÞ uð0ÞAEm# ; for all t40: Moreover, jjvðtÞ
uð0ÞjjEm# pcjjuð0ÞjjEm# ; and so
jjvðtÞjjEm# pjjvðtÞ uð0ÞjjEm# þ jjuð0ÞjjEm# p½c þ 1jjuð0ÞjjEm# :
Continuity in Em# follows as in the proof of [4, Lemma 12f]. We infer that
vABUCðJ; Em# Þ:
The fact that vAẼ1 ðJÞ is stated in (21).
We proceed to (26).
By assumption, uAẼ1 ðJÞ: Hence, w ¼ u vAẼ1 ðJÞ: We claim that
wABUCðJ; Em# Þ: To show this, first note that wAẼ1 ðJÞ; wð0Þ ¼ 0; implies that
Dat w ¼ tm1 hðtÞ;
where hABUC0-0 ðJ; E0 Þ;
ð27Þ
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and where suptAJ jjhðtÞjjE0 pjjwjjẼ1 ðJÞ : So, after convolving (27) by t1þa and
estimating in E0 ;
Z t
1
jjwðtÞjjE0 pðGðaÞÞ jjwjjẼ1 ðJÞ
ðt sÞ1þa sm1 dspGð1 aÞtaþm1 jjwjjẼ1 ðJÞ : ð28Þ
0
Moreover,
jjwðtÞjjE1 ¼ jjAwðtÞjjE0 ptm1 jjwjjẼ1 ðJÞ :
ð29Þ
We interpolate between the two estimates (28),(29). To this end, recall that
def
Kðt; wðtÞ; E0 ; E1 Þ ¼
inf
wðtÞ¼aþb
jjajjE0 þ tjjbjjE1 ;
a
t wðtÞ
t
fix t; and choose a ¼ tþt
a wðtÞ; b ¼ tþta : Then, by (28), (29),
2Gð1 aÞttaþm1
Kðt; wðtÞ; E0 ; E1 Þp
jjwjjẼ1 ðJÞ :
t þ ta
So, without loss of generality,
jjwðtÞjjEm# ¼ sup tm# Kðt; wðtÞ; E0 ; E1 Þ
tAð0;1
p sup
tAð0;1
2Gð1 aÞt1m# taþm1
jjwjjẼ1 ðJÞ :
t þ ta
It is not hard to show that from this follows:
jjwðtÞjjEm# p2Gð1 aÞjjwjjẼ1 ðJÞ ;
tAJ:
ð30Þ
Finally observe that the same estimate holds with J ¼ ½0; T replaced by
J1 ¼ ½0; T1 for any 0oT1 oT; and recall (3). Thus wðtÞ is continuous in Em#
at t ¼ 0:
To have continuity for t40 it suffices to observe that since wAẼ1 ðJÞ; then
wABUC1m ðJ; DðAÞÞ; and so, (with DðAÞ ¼ E1 ) a fortiori, wACðð0; T; Em# Þ: Thus
wABUCð½0; T; Em# Þ:
Adding up, we have u ¼ v þ wABUCðJ; Em# Þ: Theorem 8 is proved. &
Corollary 9. For uAẼ1 ðJÞ with gðuÞ ¼ 0 one has
jjujjBUCðJ;Em# Þ p2Gð1 aÞjjujjẼ1 ðJÞ :
ð31Þ
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Proof. It suffices to note that if uAẼ1 ðJÞ; with gðuÞ ¼ 0; then v in (25) vanishes
identically and u ¼ w; (w as in (26)) and to recall (30). &
Next, we consider Hölder continuity.
Theorem 10. Let mAð0; 1Þ; aAð0; 2Þ; a þ m41: Then
Ẽ1 ðJÞCBUC a½1s½1m ðJ; Es Þ;
0pspm:
#
Note that if a þ m42; then the Hölder exponent exceeds 1; provided s40 is
sufficiently small.
Proof. The case a ¼ 1 was in fact covered in [7]. The case s ¼ m# was already
considered above. In case s ¼ 0; the claim is
Ẽ1 ðJÞCBUC aþm1 ðJ; E0 Þ:
To see that this claim is true, note that if uAẼ1 ðJÞ; then Dat ðu uð0ÞÞ ¼ tm1 hðtÞ;
where hABUC0-0 ðJ; E0 Þ and suptAJ jjhðtÞjjE0 pjjuðtÞjjẼ1 ðJÞ : Then
jjuðtÞ uð0ÞjjE0 pGð1 aÞtaþm1 jjujjẼ1 ðð0;tÞÞ :
ð32Þ
So we have the desired Hölder continuity at t ¼ 0 for s ¼ 0: The case t40 is
straightforward and left to the reader.
There remains the case sAð0; mÞ:
# By the Reiteration theorem, Es ¼ ðE0 ; Em# Þs ; and
m#
by the interpolation inequality,
s
1 #
s
#
jjuðtÞ uðsÞjjEs pcjjuðtÞ uðsÞjjE0 m jjuðtÞ uðsÞjjmEm# ;
Hence, for s ¼ 0; using (32) and the fact that jjuðtÞjjEm# is bounded,
jjuðtÞ uð0ÞjjEs pct
s
½aþm1½1m#
We leave the case 0osot to the reader.
¼ cta½1s½1m :
&
5. Maximal regularity
Let E1 ; E0 ; A be as in Section 4. Let mAð0; 1Þ; aAð0; 2Þ; a þ m41: We have shown
that given uAẼ1 ðJÞ we have uð0ÞAEm# : Also, by definition, if uAẼ1 ðJÞ; then
def
f ¼ Dat ðu uð0ÞÞ þ AuAẼ0 ðJÞ:
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We now consider the converse question, i.e., the maximal regularity. We ask
whether there exists c40 such that
h
i
jjujjẼ1 ðJÞ pc jjf jjẼ0 ðJÞ þ jjxjjEm# ;
where u solves Dat ðu xÞ þ Au ¼ f :
By (21) and linearity we may obviously take x ¼ 0: Thus we let u solve
Dat u þ Au ¼ f ;
uð0Þ ¼ 0;
ð33Þ
with f AẼ0 ðJÞ; and claim that uAẼ1 ðJÞ: This will follow only under a particular
additional assumption on E0 ; E1 :
We first need to formulate some definitions. We write, for oX0;
n
def
def
Ha ðE1 ; E0 ; oÞ ¼ AALðE1 ; E0 Þ j Ao ¼ oI þ A
is a nonnegative closed operator in E0 with spectral angle opð1 a2Þ and
def
Ha ðE1 ; E0 Þ ¼
[
Ha ðE1 ; E0 ; oÞ:
oX0
Note that as Ha ðE1 ; E0 ; o1 ÞCHa ðE1 ; E0 ; o2 Þ; for o1 oo2 ; we may as well take the
union over, e.g., o40: Also note that Ha ðE1 ; E0 Þ is open in LðE1 ; E0 Þ:
Furthermore, we let
def
Mam ðE1 ; E0 Þ ¼ fAAHa ðE1 ; E0 Þj Dat u þ Au ¼ f ;
uð0Þ ¼ 0; has maximal regularity in Ẽ0 ðJÞg:
Observe that using the assumption a þ m41 one can show that if Dat u þ Au ¼ f has
maximal regularity in Ẽ0 ðJÞ; then Dat u þ ðoI þ AÞu ¼ f has maximal regularity in
Ẽ0 ðJÞ for any oAR:
We equip Mam ðE1 ; E0 Þ with the topology of LðE1 ; E0 Þ and make the following
assumptions on E0 ; E1 :
Let F1 ; F0 be Banach spaces such that
E1 CF1 CE0 CF0 ;
ð34Þ
and assume that there is an isomorphism à : F1 -F0 such that à (as an operator in
F0 ) is nonnegative with spectral angle fà satisfying
fà op 1
a
;
2
ð35Þ
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and such that for some yAð0; 1Þ;
def
E0 ¼ Fy ¼ ðF0 ; F1 Þy0;N
ð36Þ
Ax ¼ Ãx
ð37Þ
and such that
for xAE1 :
Our claim is that if f AẼ0 ðJÞ ¼ BUC1m ðJ; Fy Þ; then Aw lies in the same space and we
have a norm estimate. Specifically:
Theorem 11. Let mAð0; 1Þ; aAð0; 2Þ; a þ m41: Assume (34), let à be as in (35) and
suppose (36), (37) hold. Then AAMam ðE1 ; E0 Þ:
Proof. We define
F̃0 ¼ BUC1m ðJ; F0 Þ;
F̃1 ¼ BUC1m ðJ; F1 Þ:
Then
ðF̃0 ; F̃1 Þy ¼ BUC1m ðJ; ðF0 ; F1 Þy Þ ¼ BUC1m ðJ; E0 Þ ¼ Ẽ0 ðJÞ:
To get the first equality above one recalls the characterization of F0 ; F1 ; and that by
Clément et al. [4, Lemma 9(c)] the statement holds for m ¼ 1: The cases mAð0; 1Þ
follow by an easy adaptation of the proof of [4, Lemma 9(c)]. The second equality
above is (36), the third is the definition of Ẽ0 ðJÞ:
Write, for aAð0; 2Þ;
def
*
ðAuÞðtÞ
¼ ÃuðtÞ;
def
*
ðBuÞðtÞ
¼ Dat uðtÞ;
* def
uADðAÞ
¼ F̃1 ;
n
o
a
* def
uADðBÞ
¼ u j uABUC1m
ð½0; T; F0 Þ; uð0Þ ¼ 0 :
One then has, using (8), (35), and Proposition 6,
* is positive; densely defined in F̃0 ; with spectral angle op 1 a ;
A
2
* is positive densely defined in F̃0 with spectral angle ¼ pa:
B
2
* B
* are resolvent commuting and 0ArðAÞ-rð
*
*
Moreover, the operators A;
BÞ:
Consider the equation
* þ Au
* ¼ f;
Bu
ð38Þ
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where f AẼ0 ðJÞ: By the Da Prato–Grisvard Method of Sums (in particular see [6,
*
* such that (38) holds, and such
Theorem 4]) there exists a unique uADðAÞ-Dð
BÞ
*
*
that Au; BuAẼ0 with
*
jjAujj
Ẽ0 pcjjf jjẼ0 ;
where c is independent of f : Thus, recall (37), the function u satisfies (33), uAẼ1 ðJÞ;
and there exists c such that
jjujjẼ1 ðJÞ pcjjf jjẼ0 ðJÞ :
Observe that c ¼ cðTÞ but can be taken the same for all intervals ½0; T1 ; with
T1 pT: &
6. Linear nonautonomous equations
As earlier, we take mAð0; 1Þ; aAð0; 2Þ; a þ m41; and define m# ¼ 1 1m
a : Consider
the equation
u þ ga BðtÞu ¼ u0 þ ga h:
ð39Þ
We prove
Theorem 12. Let E0 ; E1 be as in Section 4, let TAð0; NÞ; J ¼ ½0; T and assume that
BACðJ; Mam ðE1 ; E0 Þ-Ha ðE1 ; E0 ; 0ÞÞ;
u0 AEm# ;
hAẼ0 ðJÞ:
ð40Þ
Then there exists a unique uAẼ1 ðJÞ solving (39) such that BðtÞuðtÞAẼ0 ðJÞ and there
exists c40 such that
jjujjBUC1m ðJ;E1 Þ þ jjDat ðu u0 ÞjjẼ0 ðJÞ pc jju0 jjEm# þ jjhjjẼ0 ðJÞ :
Proof. From (40) it follows that the norms
def
jjxjjEm# ¼ sup jjlm# BðsÞðlI þ BðsÞÞ1 xjjE0
l40
are all uniformly equivalent for sA½0; T:
ð41Þ
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Fix sA½0; T; T 0 Að0; T; and write J 0 ¼ ½0; T 0 : Let uðsÞ ¼ uðsÞ ðtÞ be the solution of
Dat ðuðsÞ u0 Þ þ BðsÞuðsÞ ¼ h;
on J 0 :
We claim that there exists c1 40; independent of s; T 0 ; such that
jjDat ðuðsÞ u0 ÞjjẼ0 ðJ 0 Þ þ jjBðsÞuðsÞ ðtÞjjẼ0 ðJ 0 Þ pc1 jju0 jjEm# þ jjhjjẼ0 ðJ 0 Þ :
ðsÞ
ð42Þ
ðsÞ
To prove (42), write uðsÞ ¼ u1 þ u2 ; where
ðsÞ
ðsÞ
Dat ðu1 u0 Þ þ BðsÞu1 ¼ 0;
ðsÞ
ðsÞ
Dat u2 þ BðsÞu2 ¼ h;
ðsÞ
u1 ð0Þ ¼ u0 ;
ðsÞ
u2 ð0Þ ¼ 0:
By (18),
ðsÞ
jjDat ðu1 u0 ÞjjẼ0 ðJ 0 Þ pcjju0 jjEm# ;
where c ¼ cðm; cðsÞÞ: By (40), cðsÞ; hence c; can be taken independent of s:
By the fact that B takes values in Mam ðE1 ; E0 Þ one has
ðsÞ
ðsÞ
jjDat u2 jjẼ0 ðJ 0 Þ þ jjBðsÞu2 jjẼ0 ðJ 0 Þ pc̃jjhjjẼ0 ðJ 0 Þ ;
and from the fact that BACðJ; LðE1 ; E0 ÞÞ one concludes that c̃ can be taken
independent of s: Hence claim (42) holds.
Choose nX1 such that with q ¼ n1 T one has
c1
1
jjBðtÞ Bððj 1ÞqÞjjLðE1 ;E0 Þ p ;
2
j¼1;y;n;ðj1Þqptpjq
max
ð43Þ
where c1 as in (42). Fix jAf1; 2; y; ng; and assume we have a unique solution u% j1 of
(39) on ½0; ðj 1Þq (for j ¼ 1; take u% 0 ¼ u0 ). Then define (recall (11))
Z̃j ¼ uAẼ1 ð½0; jqÞ; uð0Þ ¼ u0 j uðtÞ ¼ u% j1 ðtÞ; 0ptpðj 1Þq :
Given an arbitrary vAZ̃j ; we let uj be the unique solution of
u þ ga Bððj 1ÞqÞu ¼ u0 þ ga h þ ga ½Bððj 1ÞqÞ BðtÞv
on ½0; jq: Clearly, ½Bððj 1ÞqÞ BðtÞvABUC1m ð½0; jq; E0 Þ: By uniqueness, uj AZ̃j :
Denote the map vAZ̃j -uj AZ̃j by Fj : By (42),(43), and observing that v1 ¼ v2 on
½0; ðj 1Þq;
1
jjFj ðv1 Þ Fj ðv2 ÞjjẼ1 ð½0;jqÞ p jjv1 v2 jjẼ1 ð½0;jqÞ :
2
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Observe that Z̃j is closed in Ẽ1 ð½0; jqÞ; hence it is a complete metric space with
respect to the induced metric. Consequently we may apply the Contraction mapping
Theorem and conclude that there exists a unique fixed point of Fj in Z̃j : Denote this
fixed point by u% j : Clearly u% j solves (39) on ½0; jq:
Proceeding by induction we have the existence of a solution uAẼ1 ðJÞ of (39). The
induction procedure also gives c40 such that (41) holds. &
7. Local nonlinear theory
We consider the quasilinear equation
Dat ðu u0 Þ þ AðuÞu ¼ f ðuÞ þ hðtÞ;
t40;
ð44Þ
under the following assumptions. Let
mAð0; 1Þ
aAð0; 2Þ;
a þ m41;
ð45Þ
and define m# as earlier by m# ¼ a1 ða þ m 1Þ: For X ; Y Banach spaces, and g a
mapping of X into Y ; write gAC 1 ðX ; Y Þ if every point xAX has a neighbourhood
U such that g restricted to U is globally Lipschitz continuous.
Let E0 ; E1 be Banach spaces such that E1 CE0 with dense imbedding and suppose
ðA; f ÞAC 1 ðEm# ; Mam ðE1 ; E0 Þ
u0 AEm# ;
hABUC1m ð½0; T; E0 Þ;
E0 Þ;
for any T40:
ð46Þ
ð47Þ
Observe that by (46), for ũAEm# there exists oðũÞX0 such that
def
Ao ðũÞ ¼ AðũÞ þ oðũÞIAHa ðE1 ; E0 ; 0Þ-Mam ðE1 ; E0 Þ:
We define a solution u of (44) on an interval JCRþ containing 0 as a function u
satisfying uACðJ; E0 Þ-Cðð0; T; E1 Þ; uð0Þ ¼ u0 ; and such that the fractional
derivative of u u0 of order a satisfies Dat ðu u0 ÞACðð0; T; E0 Þ and such that
(44) holds on 0otpT:
Our result is:
Theorem 13. Let (45)–(47) hold, where Em# ¼ ðE0 ; E1 Þm0;N
is a continuous interpolation
#
space. Then there exists a unique maximal solution u defined on the maximal interval of
existence ½0; tðu0 ÞÞ; where tðu0 ÞAð0; N; and such that for every Totðu0 Þ one has
a
(i) uABUC1m ð½0; T; E1 Þ-BUCð½0; T; Em# Þ-BUC1m
ð½0; T; E0 Þ;
(ii) u þ ga AðuÞu ¼ u0 þ ga ðf ðuÞ þ hÞ; 0ptpT;
(iii) If tðu0 ÞoN; then ueUCð½0; tðu0 ÞÞ; Em# Þ;
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(iv) If tðu0 ÞoN and E1 CCE0 ; then
lim sup jjuðtÞjjEd ¼ N;
for any dAðm;
# 1Þ:
tmtðu0 Þ
We recall that u defined on an interval J is called a maximal solution if there
does not exist a solution v on an interval J 0 strictly containing J such that v restricted
to J equals u: If u is a maximal solution, then J is called the maximal interval of
existence.
In this section, we prove existence and uniqueness of u satisfying (i), (ii) for some
T40: The continuation is dealt with in Section 8.
Proof of Theorem 13 (i), (ii). Choose o such that Ao ðu0 ÞAHa ðE1 ; E0 ; 0Þ: Then
Ao ðu0 ÞAMa ðE1 ; E0 Þ and there exists a constant cu0 ; independent of F ; such that if
F AẼ0 ðJÞ and u ¼ uðF Þ solves
Dat u þ Ao ðu0 Þu ¼ F ðtÞ;
0otpT;
with uð0Þ ¼ 0; then
jjujjẼ1 ð½0;TÞ pcu0 ðGð1 aÞÞ1 jjF jjẼ0 ðJÞ :
ð48Þ
Define
BðuÞ ¼ Aðu0 Þ AðuÞ;
uAEm# :
Then BAC 1 ðEm# ; LðE1 ; E0 ÞÞ; and so, by (46) there exists r0 40; LX1 such that
jjðB; f Þðz1 Þ ðB; f Þðz2 ÞjjLðE1 ;E0 Þ
E0 pLjjz1
z2 jjEm# ;
ð49Þ
for z1 ; z2 AB% Em# ðu0 ; r0 Þ; and such that
1
;
12cu0
zAB% Em# ðu0 ; r0 Þ:
ð50Þ
jjf ðzÞ þ oðu0 ÞzjjE0 pb;
zAB% Em# ðu0 ; r0 Þ;
ð51Þ
jjBðzÞjjLðE1 ;E0 Þ p
Define b by
and
1
e0 ¼ min r0 ;
:
12cu0 L
ð52Þ
Let ũ solve
Dat ðũ u0 Þ þ Ao ðu0 Þũ ¼ 0;
on ½0; T:
ð53Þ
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Take t40 small enough so that (ũ as in (53))
e0
jjũ u0 jjEm# p ;
2
tA½0; t;
e0
jjũjjẼ1 ðJt Þ p ;
2
Gð1 aÞt1m pmin
e0
1
;
;
12cu0 b 12cu0 ðL þ oðu0 ÞÞ
jjhjjẼ0 ðJt Þ p
e0
;
12cu0
where Jt ¼ ½0; t: Define
n
o
Wu0 ðJt Þ ¼ vAẼ1 ðJt Þ j vð0Þ ¼ u0 ; jjv u0 jjCðJt ;Em# Þ pe0 -B% Ẽ1 ðJt Þ ð0; e0 Þ
ð54Þ
ð55Þ
ð56Þ
ð57Þ
ð58Þ
and give this set the topology of Ẽ1 ðJt Þ: Then Wu0 ðJt Þ is a closed subset of Ẽ1 ðJt Þ;
and therefore a complete metric space. Moreover, Wu0 ðJt Þ is nonempty, because
ũAWu0 ðJt Þ:
Consider now the map
Gu0 : Wu0 ðJt Þ-Ẽ1 ðJt Þ
defined by u ¼ Gu0 ðvÞ; vAWu0 ðJt Þ; where u solves
Dat ðu u0 Þ þ Ao ðu0 Þu ¼ BðvÞv þ f ðvÞ þ oðu0 Þv þ hðtÞ:
ð59Þ
Our first claim is that this map is well defined. To see this, note that as
BAC 1 ðEm# ; LðE1 ; E0 ÞÞ and v is continuous in Em# ; and by the assumption on f ; h it
follows that the right-hand side of (59) is in Cðð0; t; E0 Þ: Also, by (50), (51),(53),
(56)–(58),
sup t1m jjBðvðtÞÞvðtÞ þ f ðvðtÞÞ þ oðu0 ÞvðtÞ þ hðtÞjjE0
0otpt
p sup ðt1m jjBðvðtÞÞjjLðE1 ;E0 Þ jjvðtÞjjE1 Þ þ t1m b þ jjhjjẼ0 ðJt Þ
0otpt
1
e0
e0
e0
p
jjvjjẼ1 ðJt Þ þ
þ
p
:
12cu0
12cu0 12cu0 4cu0
ð60Þ
So the right-hand side of (59) is in Ẽ0 ðJt Þ; and hence, by (21),(48), (53), the map is
well defined.
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Next, we assert that uAWu0 ðJt Þ: We show first
sup jjGu0 ðvÞðtÞ u0 jjEm# pe0 :
ð61Þ
Gu0 ðvÞ ¼ ũ þ G̃u0 ðvÞ;
ð62Þ
tA½0;t
Split Gu0 ðvÞ:
where G̃u0 ðvÞ solves (zero initial value)
Dat ðG̃u0 ðvÞÞ þ Ao ðu0 ÞG̃u0 ðvÞ ¼ BðvÞv þ f ðvÞ þ oðu0 Þv þ hðtÞ:
By (31), (48), (60),
sup jjG̃u0 ðvÞðtÞjjEm# p 2Gð1 aÞjjG̃u0 ðvÞjjẼ1 ðJt Þ
tA½0;t
p 2cu0 jjBðvÞv þ f ðvÞ þ oðu0 Þv þ hjjẼ0 ðJt Þ p2cu0
e0
e0
¼ :
4cu0
2
ð63Þ
Combining (54) and (63) we have (61).
Next, we assert that
jjGu0 ðvÞjjẼ1 ðJt Þ pe0 :
To show this, split as in (62) and recall (55),(63). So Gu0 ðvÞAWu0 ðJt Þ:
Finally, we claim that Gu0 is a contraction. We have, by linearity and (31), (48),
(49), (50),
jjGu0 ðv1 Þ Gu0 ðv2 ÞjjẼ1 ðJt Þ
pcu0 jjBðv1 Þv1 Bðv2 Þv2 jjẼ0 ðJt Þ þ cu0 jjf ðv1 Þ f ðv2 ÞjjẼ0 ðJt Þ
þ cu0 oðu0 Þjjv1 v2 jjẼ0 ðJt Þ
pcu0 jj½Bðv1 Þ Bðv2 Þv1 jjẼ0 ðJt Þ þ cu0 jjBðv2 Þ½v1 v2 jjẼ0 ðJt Þ
þ cu0 t1m ½L þ oðu0 Þ sup jjv1 ðtÞ v2 ðtÞjjEm#
t
1
jjv1 v2 jjẼ1 ðJt Þ
12
1
þ 2Gð1 aÞcu0 t1m ½L þ oðu0 Þjjv1 v2 jjẼ1 ðJt Þ p jjv1 v2 jjẼ1 ðJt Þ ;
2
pcu0 Ljjv1 v2 jjẼ1 ðJt Þ 2Gð1 aÞjjv1 jjẼ1 ðJt Þ þ
where the last step follows by (52) and(56). Thus the map v-Gðu0 Þv is a contraction
and has a unique fixed point.
We conclude that there exists u satisfying (i), (ii), for some T40:
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We proceed to the proof of uniqueness. Assume there exist two functions u1 ; u2 ;
both satisfying (i), (ii) on ½0; T for some T40 and u1 ðtÞ not identically equal to u2 ðtÞ
on ½0; T:
Define
t1 ¼ sup tA½0; T j ð44Þ has a unique solution in Ẽ1 ð½0; tÞ :
def
Then 0pt1 oT: Also, for any tAðt1 ; T there exists a solution u of (44) on Jt ¼ ½0; t;
such that uðtÞ ¼ u1 ðtÞ on ½0; t1 but u does not equal u1 everywhere on t1 otpt: Let,
for tAðt1 ; T; Jt ¼ ½0; t;
n
Wu1 ðJt Þ ¼ vAẼ1 ðJt Þ j vðtÞ ¼ u1 ðtÞ; 0ptpt1 ;
o
jjv u1 jjCðJ ;E Þ pe0 -B% Ẽ ðJ Þ ðu1 ðtÞ; e0 Þ:
t
m#
1
t
Give this set the topology of Ẽ1 ðJt Þ: Then Wu1 ðJt Þ is a complete metric space which is
nonempty because u1 AWu1 ðJt Þ:
Consider the map Gu1 : Wu1 ðJt Þ-Ẽ1 ðJt Þ defined bu u ¼ Gu1 ðvÞ for vAWu1 ðJt Þ;
where u solves
Dat ðu u0 Þ þ Ao ðu1 ðt1 ÞÞuðtÞ ¼ BðvðtÞÞvðtÞ þ f ðvðtÞÞ þ oðu1 ðt1 ÞÞvðtÞ þ hðtÞ;
def
with BðvðtÞÞ ¼ Aðu1 ðt1 ÞÞ AðvðtÞÞ and where we have chosen oðuðt1 ÞÞ such that
Ao ðu1 ðt1 ÞÞAHa ðE1 ; E0 ; 0Þ: By (46), Ao ðu1 ðt1 ÞÞAMam ðE1 ; E0 Þ: Proceed as in the
existence part to show that the map Gu1 is welldefined, and that for t sufficiently
close to t1 one has that Gu1 maps Wu1 ðJt Þ into itself. Finally show that the map is a
contraction if t t1 is sufficiently small and so the map has a unique fixed point. On
the other hand, any solution of (44) is a fixed point of the map, provided t (depends
on the particular solution) is taken sufficiently close to t1 : A contradiction results
and uniqueness follows.
Thus we have shown that (i), (ii), and uniqueness hold for some T40:
8. Continuation of solutions
We proceed to the final part of the proof of Theorem 13.
Suppose we have a unique solution u of (44) on Jt ¼ ½0; t; for some t40;
such that
uACðJt ; Em# Þ-Ẽ1 ðJt Þ:
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Take T4t and let
n
def
Z ¼ wACð½0; T; Em# Þ j wðtÞ ¼ uðtÞ; tA½0; t;
ðt tÞ1m Dat ðw u0 ÞABUCððt; T; E0 Þ; jj½t t1m Dat ðw u0 ÞjjE0 -0; tkt;
o
½t t1m wABUCððt; T; E1 Þ; jj½t t1m wjjE1 -0; tkt :
ð64Þ
Choose e0 sufficiently small. Define
def
Zu ¼ fwAZ j jjw uðtÞjjCð½t;T;Em# Þ pe0 ; jjwjjẼ1 ð½t;TÞ pe0 g:
ð65Þ
Choose oðuðtÞÞ so that Ao ðuðtÞÞAHa ðE1 ; E0 ; 0Þ: For vAZu ; consider ð0ptpTÞ;
Dat ðu u0 Þ þ Ao ðuðtÞÞuðtÞ
¼ AðuðtÞÞvðtÞ AðvðtÞÞvðtÞ þ f ðvðtÞÞ þ oðuðtÞÞvðtÞ þ hðtÞ:
Let uv be the corresponding solution. If uv ¼ v; then we have a solution of (44) on
½0; T; identically equal to u on ½0; t: This solution may however have a singularity
for tkt:
We may repeat the existence proof above to obtain a unique fixed point (of the
map v-uv ) ûðtÞ; 0ptpT; in Zu if T is sufficiently close to t: Clearly, û ¼ u on ½0; t:
Moreover, ûACð½0; T; Em# Þ and so, by (46), AðûðtÞÞ; tA½0; T; is a compact subset
of Ha ðE1 ; E0 Þ: Now use the arguments of [1, Corollary 1.3.2 and proof of Theorem
#
such that
2.6.1; 9, p. 10] to deduce that there exists a fixed oX0
def
#
Ao# ðûðtÞÞ ¼ AðûðtÞÞ þ oIAH
am ðE1 ; E0 ; 0Þ
for every tA½0; T:
Also,
def
Ao# ðtÞ ¼ Ao# ðûðtÞÞACð½0; T; LðE1 ; E0 ÞÞ
and so Ao# ðtÞ satisfies (40) (recall that a þ m41 is assumed.) In addition,
ˆ def
fðtÞ
¼ f ðûðtÞÞABUCð½0; T; E0 ÞCẼ0 ð½0; TÞ;
# ûðtÞACð½0; T; Em# ÞCẼ0 ð½0; TÞ:
o
Then note that û solves
ˆ þo
# ûðtÞ þ hðtÞ;
Dat ðu u0 Þ þ Âo# ðtÞuðtÞ ¼ fðtÞ
tA½0; T;
ð66Þ
and that the earlier result on nonautonomous linear equations can be applied. But by
this result there is a unique function û1 ðtÞ in BUC1m ð½0; T; E1 Þ solving (66) on ½0; T:
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Moreover, there certainly exists T1 4t such that û1 considered on ½0; T1 is contained
in Zu (in the definition of Zu ; take T ¼ T1 ). Thus we must have û1 ¼ û on ½t; T1 and
so û does not have a singularity as tkt: The solution u may therefore be continued to
½0; T1 ; for some T1 4t; so that (i), (ii) are satisfied on ½0; T1 :
(iii) Suppose 0otðu0 ÞoN; and assume uAUCð½0; tðu0 ÞÞ; Em# Þ: Then limtmtðu0 Þ exists
in Em# : Define
ũðtÞ ¼ uðtÞ; tA½0; tðu0 ÞÞ;
ũðtÞ ¼ lim uðtÞ; t ¼ tðu0 Þ:
tmtðu0 Þ
# sufficiently large,
Then ũACð½0; tðu0 Þ; Em# Þ: Define, for o
BðtÞ ¼ Ao# ðũðtÞÞ;
˜ ¼ f ðũðtÞÞ þ o
# ũðtÞ;
fðtÞ
0ptptðu0 Þ:
By (46) and the compactness arguments above we have that BðtÞ satisfies the
assumptions required in our nonautonomous result. Consider then
˜ þ hðtÞ;
Dat ðv u0 Þ þ BðtÞv ¼ fðtÞ
0ptptðu0 Þ:
By the earlier result on linear nonautonomous equations, there exists a unique
vAẼ1 ð½0; tðu0 ÞÞ which solves this equation on ½0; tðu0 Þ: By uniqueness, vðtÞ ¼ uðtÞ;
0ptotðu0 Þ: But vAUCð½0; tðu0 Þ; Em# Þ and so vðtðu0 ÞÞ ¼ ũðtðu0 ÞÞ; hence vðtÞ ¼ ũðtÞ;
0ptptðu0 Þ: Thus
Dat ðv u0 Þ þ AðvðtÞÞvðtÞ ¼ f ðvðtÞÞ þ hðtÞ;
0ptptðu0 Þ:
By earlier results we may now continue the solution past tðu0 Þ and so a contradiction
follows.
(iv) Suppose tðu0 ÞoN and assume lim suptmtðu0 Þ jjuðtÞjjEd oN for some d4m:
#
Consider the set uð½0; tðu0 ÞÞÞ: This set is bounded in Ed ; hence its closure is compact
in Em# :
Take any t%Að0; tðu0 ÞÞ: Consider
Dat ðu u0 Þ þ Ao ðuðt%ÞÞ
¼ ½Aðuðt%ÞÞ AðvðtÞÞvðtÞ þ f ðvðtÞÞ þ oðuðt%ÞÞvðtÞ þ hðtÞ;
and the solution u (which we have on ½0; tðu0 ÞÞ) on ½0; t%: Now let t% play the role of t
in (64), and define the set from which v is picked as in (65). Then, as in the
considerations following (64), (65), we obtain a continuation of uðtÞ to ½t%; t% þ d;
where d ¼ dðuðt%ÞÞ40: (By uniqueness, on ½t%; tðu0 ÞÞ this is of course the solution we
already
have.) On the other hand, d depends continuously on uðt%Þ: But the closure of
S
0pt%otðu0 Þ uðt%Þ is compact in Em# ; and so dðuðt%ÞÞ is bounded away from zero for
0pt%otðu0 Þ: Hence the solution may be continued past tðu0 Þ (take t% sufficiently close
to tðu0 Þ) and a contradiction follows.
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9. An example
In this last section we indicate briefly how our results may be applied to the
quasilinear equation
u ¼ u0 þ ga ðsðux Þx þ hÞ;
tX0;
xAð0; 1Þ;
ð67Þ
with u ¼ uðt; xÞ; and
uðt; 0Þ ¼ uðt; 1Þ ¼ 0;
tX0;
uð0; xÞ ¼ u0 ðxÞ:
As was indicated in the Introduction, this problem occurs in viscoelasticity theory,
see [10].
We require
sAC 3 ðRÞ;
with sð0Þ ¼ 0;
ð68Þ
and impose the growth condition
0os0 ps0 ðyÞps1 ;
yAR;
ð69Þ
for some positive constants s0 s1 :
Take
F0 ¼ fuAC½0; 1 j uð0Þ ¼ uð1Þ ¼ 0g;
and
F1 ¼ fuAC 2 ½0; 1 j uðiÞ ð0Þ ¼ uðiÞ ð1Þ ¼ 0; i ¼ 0; 2g:
We fix m# ¼ 12; then m ¼ 1 a2; and a þ m41 holds. With yAð0; 12Þ; let
E0 ¼ ðF0 ; F1 Þy0;N ¼ fu j uAh2y ½0; 1; uð0Þ ¼ uð1Þ ¼ 0g;
ð70Þ
E1 ¼ fuAF1 j u00 AE0 g:
ð71Þ
and
Then
Em# ¼ E12 ¼ fu j uAh1þ2y ½0; 1; uð0Þ ¼ uð1Þ ¼ 0g:
We take, for uAE12; vAE1 ;
AðuÞv ¼ s0 ðux Þvxx :
Then one has AðuÞvAE0 ; and, more generally, that the well defined map v-AðuÞv
lies in LðE1 ; E0 Þ for every uAE12 :
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We claim that this map satisfies AðuÞAMam ðE1 ; E0 Þ-Ha ðE1 ; E0 ; 0Þ: To this end
one takes (for fixed uAE12 )
def
Ãv ¼ s0 ðux Þv00 ;
vAF1 ;
and observes that this map is an isomorphism F1 -F0 and that Ã; as an operator in
F0 ; is closed, positive, with spectral angle 0: Thus Theorem 11 can be applied and our
claim follows.
The only remaining condition to be verified is that u-AðuÞAC 1 ðE12 ; LðE1 ; E0 ÞÞ:
But this follows after some estimates which make use of the smoothness assumption
(68) imposed on s:
We thus have, applying Theorem 13:
Theorem 14. Let aAð0; 2Þ: Take yAð0; 12Þ and E0 ; E1 as in, (70), (71). Let (68), (69)
hold. Assume hABUCa2 ð½0; T; h2y ½0; 1Þ; with hð0Þ ¼ hð1Þ ¼ 0: Assume u0 Ah1þ2y ½0; 1
with u0 ð0Þ ¼ u0 ð1Þ ¼ 0:
Then (67) has a unique maximal solution u defined on the maximal interval of
existence ½0; tðu0 ÞÞ where tðu0 ÞAð0; N and such that for any Totðu0 Þ one has
uABUCa ð½0; T; h2þ2y ½0; 1Þ-BUCð½0; T; h1þ2y ½0; 1Þ-BUCaa ð½0; T; h2y ½0; 1Þ:
2
2
If tðu0 ÞoN; then lim suptmtðu0 Þ jjuðtÞjjC 1þ2yþd ¼ N for every d40: In particular, since
yAð0; 12Þ is arbitrary, we conclude that if
lim sup jjuðtÞjjC 1þd oN;
ð72Þ
tmtðu0 Þ
for some d40; then tðu0 Þ ¼ N:
Global existence and uniqueness of smooth solutions of (67) under assumptions
(68), (69), is thus seen to follow from (72). However, the verification of (72) is in
general a very difficult task. For ao43 this task is essentially solved (see [10]).
By different methods, the existence, but not the uniqueness, of a solution u
satisfying
1;N
uAWloc
ðRþ ; L2 ð0; 1ÞÞ-L2loc ðRþ ; W02;2 ð0; 1ÞÞ
was proved in [12], for the range aA½43; 32: For 32oao2; only existence of global weak
solutions has been proved [11]. We do however conjecture that unique smooth,
global solutions do exist for the entire range aAð0; 2Þ:
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Acknowledgments
The first author acknowledges the support of the Magnus Ehrnrooth foundation
(Finland). The second author acknowledges the support of the Nederlandse
organisatie voor wetenschappelijk onderzoek (NWO).
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