existence of insensitizing controls for a semilinear heat equation with a superlinear nonlinearity

Existence of Insensitizing Controls for a Semilinear HeatEquation with a Superlinear Nonlinearity

O Bodart1 M Gonzalez-Burgos2 and R Perez-Garcıa2

1Laboratoire de Mathematiques Appliquees Universite Blaise Pascal

(Clermont-Ferrand 2) Aubiere Cedex France2Dpto de Ecuaciones Diferenciales y Analisis Numerico Universidad de Sevilla

Sevilla Spain

ABSTRACT

In this paper we consider a semilinear heat equation (in a bounded domain O of

RN ) with a nonlinearity that has a superlinear growth at infinity We prove theexistence of a control with support in an open set o O that insensitizes theL2norm of the observation of the solution in another open subset O O when

o O 6frac14 under suitable assumptions on the nonlinear term fethyTHORN and the righthand side term x of the equation The proof involving global Carleman estimatesand regularizing properties of the heat equation relies on the sharp study of a

similar linearized problem and an appropriate fixed-point argument For certainsuperlinear nonlinearities we also prove an insensitivity result of a negativenature The crucial point in this paper is the technique of construction ofLr-controls (r large enough) starting from insensitizing controls in L2

Key Words Controllability Nonlinear PDE of parabolic type

Mathematics Subject Classification 93B05 35K55 35K05

Correspondence O Bodart Laboratoire de Mathematiques Appliquees Universite Blaise

Pascal (Clermont-Ferrand 2) Aubiere Cedex 63177 France Fax thorn33(0)4-73-40-70-64E-mail olivierbodartmathuniv-bpclermontfr

COMMUNICATIONS IN PARTIAL DIFFERENTIAL EQUATIONS

Vol 29 Nos 7 amp 8 pp 1017ndash1050 2004

1017

DOI 101081PDE-200033749 0360-5302 (Print) 1532-4133 (Online)

Copyright 2004 by Marcel Dekker Inc wwwdekkercom

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1 INTRODUCTION AND MAIN RESULTS

Problem Formulation

Let O RN N 1 be a bounded connected open set with boundary O 2 C2Let f be a C1 function defined on R Let o and O be two open subsets of O (thoughtto be small in practice) For T gt 0 we denote Q frac14 O eth0TTHORN and S frac14 O eth0TTHORN

We consider a semilinear heat equation with partially known initial condition

ty Dythorn fethyTHORN frac14 xthorn v1o in Q

y frac14 0 on S yethx 0THORN frac14 y0ethxTHORN thorn t yy0ethxTHORN in O

(eth1THORN

where x 2 L2ethQTHORN is a given heat source y0 2 L2ethOTHORN is a given initial data (althoughby the reasons which will be seen later we will address in this paper the case y0 frac14 0)yy0 2 L2ethOTHORN is unknown with kyy0kL2ethOTHORN frac14 1 t is an unknown small real number andv 2 L2ethQTHORN is a control function to be determined Here 1o is the characteristicfunction of the control set o

Let us define

fethyTHORN frac14 1

2

ZZOeth0TTHORN

jyethx t t vTHORNj2 dx dt eth2THORN

y frac14 yeth t vTHORN being a solution to (1) associated to t and v A control function v issaid to insensitize f if

fethyeth t vTHORNTHORNt

tfrac140

frac14 0 8yy0 2 L2ethOTHORN with kyy0kL2ethOTHORN frac14 1 eth3THORN

This insensitivity condition means that we seek a control function v acting ono eth0TTHORN such that f is locally insensitive to small perturbations in the initialcondition

In Bodart and Fabre (1995) and Lions (1990) the existence of a control v satisfy-ing (3) is proved to be equivalent to the existence of a control v solving the followingproblem


y frac14 0 on S yethx 0THORN frac14 y0ethxTHORN in O

(eth4THORN

tq Dqthorn f 0ethyTHORNq frac14 y1O in Q

q frac14 0 on S qethxTTHORN frac14 0 in O

(eth5THORN

qethx 0THORN frac14 0 in O eth6THORN

Thus the problem of seeking a control that insensitizes f boils down to a non-classical null controllability problem First it is a null controllability problem ofbackward-forward nature for a cascade system of heat equations the first one of

1018 Bodart Gonzalez-Burgos and Perez-Garcıa

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semilinear type In addition the control acts on the second equation only indirectlythrough the first one while q is the function we want to lead to zero after a timeinterval of length T

Preliminaries and Existing Results

This problem addressed by Lions (1990) has been studied for globallyLipschitz-continuous nonlinearities and o O 6frac14 (this last hypothesis is absolutelyessential and to our knowledge nothing is known when the intersection is empty)First in Bodart and Fabre (1995) the authors relaxed the notion of insensitizingcontrols introducing the so-called e-insensitizing controls Given e gt 0 a control vis said to e-insensitize f if


tfrac140

e 8yy0 2 L2ethOTHORN with kyy0kL2ethOTHORN frac14 1

In the above-mentioned paper the existence of e-insensitizing controls for partiallyknown data both in the initial and boundary conditions was proved This problemis equivalent to an approximate controllability problem for a system of coupled heatequations and it was solved by using the techniques in Fabre et al (1995)

The first results on the existence and non-existence of insensitizing controls wereproved in De Teresa (2000) To be precise the author showed that when f 0 andOnoo 6frac14 there exists y0 2 L2ethOTHORN such that for every v 2 L2ethQTHORN the correspondingsolution ethy qTHORN to (4)ndash(5) satisfies qeth0THORN 6frac14 0 that is to say the functional f cannotbe insensitized (see Theorem 2 in De Teresa 2000) On the other hand wheno O 6frac14 y0 frac14 0 and f is a C1 globally Lipschitz-continuous function such thatfeth0THORN frac14 0 in De Teresa (2000) it is proved If x 2 L2ethQ expethM=2tTHORNTHORN with M gt 0large enough there exists v2L2ethQTHORN such that the solution ethyqTHORN to (4)ndash(5) satisfies (6)(see Theorem 1 in De Teresa 2000) Here L2ethQ expethM=2tTHORNTHORN stands for the weightedHilbert space

L2ethQ expethM=2tTHORNTHORN frac14 fx 2 L2ethQTHORN k expethM=2tTHORNxkL2ethQTHORN lt 1g

The proof combines a similar null controllability result for linear coupled parabolicsystems and an appropriate fixed-point argument More precisely the author firstlinearizes the system and shows its null controllability with controls uniformlybounded in L2ethQTHORN when the potentials of the linearized system lie in a boundedset of L1ethQTHORN Since f is a globally Lipschitz-continuous function this fact sufficesfor proving that the fixed-point mapping maps L2ethQTHORN into a convex compact setof L2ethQTHORN

The main goal of the present paper is to analyze the existence of controls thatinsensitize f that is to say to study the null controllability properties of the coupledsystem (4)ndash(5) when f has a superlinear growth at infinity In accordance withTheorems 1 and 2 in De Teresa (2000) we will assume from now on that y0 frac14 0and o O 6frac14

Insensitizing Controls for a Superlinear 1019

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In the study of the null controllability of semilinear parabolic systems withsuperlinear nonlinearities additional technical difficulties arise Let us recall whathappens in the simpler case of one semilinear heat equation During the last yearsthe controllability properties of

ty Dythorn fethyTHORN frac14 v1o in Q


(eth7THORN

with a superlinear nonlinearity fethyTHORN has been thoroughly studied by several authorsAs in the sublinear case the technique to deal with this problem combines a fixed-point reformulation together with the study of the null controllability of linearproblems of the form

ty Dythorn ay frac14 v1o in Q


(

where a 2 L1ethQTHORN Due to the superlinear growth of the nonlinearity it is necessaryto perform a fixed point argument that the solution of the controlled linear problembelongs to L1ethQTHORN To this aim controls in LrethQTHORN with r gt ethN thorn 2THORN=2 must be builtMoreover in the linear case it is necessary to analyze how the Lr-norm of thecontrols depends on kak1 Let us mention some papers on this issue

(1) Barbu (2000) obtained null LrethNTHORN-controls v such that

kvkLrethNTHORNethQTHORN Cethkak1THORNky0kL2ethOTHORN

with rethNTHORN 2 2 2ethNthorn2THORNN2

if N 3 rethNTHORN 2 frac1221THORN if N frac14 2 and rethNTHORN 2 frac1221 if N frac14 1

The proof of this estimate is based on a global Carleman inequality for the adjointproblem

tj Djthorn aj frac14 0 in Q

j frac14 0 on S jethxTTHORN frac14 j0ethxTHORN in O

(eth8THORN

due to Fursikov and Imanuvilov (1996) In fact the author proved a sharp estimateof the constant appearing on the right-hand side of the Carleman inequality withrespect to kak1 Nevertheless this technique can only be applied to the study ofthe null controllability of the superlinear heat equation (7) when N lt 6 For othercontrollability results proved in a similar way see Anita and Barbu (2000)

(2) A second approach was developed by Fernandez-Cara and Zuazua (2000)They proved the lsquolsquorefinedrsquorsquo observability inequality

kjeth0THORNk2L2ethOTHORN CethT kak1THORNZ Z

oeth0TTHORNjjjdx dt

2

eth9THORN


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for the solutions to (8) which implies the existence of a null control v 2 L1ethQTHORNsatisfying

kvkL1ethQTHORN CethT kak1THORNky0kL2ethOTHORN

In this case the observability inequality (9) was deduced combining a globalCarleman estimate for the adjoint problem and the regularizing effect of the heatequation The authors give the explicit dependence on T and kak1 of the constantCethT kak1THORN in (9) which is essential to prove their nonlinear null controllabilityresult This technique was later used in Doubova et al (2002) for a nonlinear heatequation with a superlinear term fethyHyTHORN

The study of the null controllability properties of the superlinear coupled system(4)ndash(5) is more intricate In this case as in Anita and Barbu (2000) Barbu (2000) andFernandez-Cara and Zuazua (2000) null Lr-controls (with r gt ethN thorn 2THORN=2) for the cor-responding coupled linear systemmust also be built Again Lr-estimates of the controlsare needed The technique introduced by V Barbu could be applied in this case if thecondition N lt 6 is imposed (which does not seem to be a natural restriction on N )On the other hand the approach proposed byE Fernandez-Cara andE Zuazua cannotbe applied since the regularizing effect of the associated linear adjoint problem involvestwo functions j and c (see (21)ndash(22)) while the corresponding lsquolsquorefinedrsquorsquo observabilityinequality should only involve c (recall that the control v only appears in (4))

In Bodart et al (2002) the authors introduced a new technique of constructionof null Lr-controls for coupled linear parabolic systems This strategy made it pos-sible to generalize Theorem 1 in De Teresa (2000) to more general nonlinearitiesThe proof of this insensitivity result as well as the above-mentioned techniquesketched in Bodart et al (2002) are developed in the present paper To our knowl-edge this is the first insensitivity result in the literature for semilinear heat equationswith a superlinear nonlinearity

Main Results

The first relevant result in this paper is the following one

Theorem 11 Assume that o O 6frac14 and y0 frac14 0 Let f be a C1 function defined onR verifying f 00 2 L1

locethRTHORN feth0THORN frac14 0 and

limjsj1

f 0ethsTHORNlogeth1thorn jsjTHORN frac14 0 eth10THORN

Let r 2 N

2thorn 11

be given Then for any x 2 LrethQTHORN such that

ZZQ

exp1

t3

jxj2 dx dt lt 1 eth11THORN

there exists a control function v 2 LrethQTHORN insensitizing the functional f givenby (2)


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Condition (11) means that the given source term x is asked to decay rapidly tozero close to the initial time t frac14 0 As seen before a similar assumption if a weakerone is required in the case when a globally Lipschitz-continuous nonlinearity isconsidered Observe that hypothesis feth0THORN frac14 0 is in accordance with assumptionson x (see De Teresa 2000 for both considerations)

As usual in the study of controllability problems for nonlinear equations acontrollability result for a linearized version of (4)ndash(6) will be first proved We willthen apply a fixed-point argument to deal with the general case The structure ofthe proof is quite general (for other controllability results proved in a similar waysee Fernandez-Cara and Zuazua 2000 Zuazua 1991 )

Remark 1 Hypothesis (10) is fulfilled by certain superlinear nonlinearities f such as

jfethsTHORNj frac14 jp1ethsTHORNj logaeth1thorn jp2ethsTHORNjTHORN for all jsj s0 gt 0

with a 2 frac120 1THORN where p1 and p2 are real affine functionsFor nonlinearities f 2 C1ethRTHORN satisfying hypothesis (10) system (1) admits a

global solution when the data x v y0 and yy0 are regular enough For instance bylinearization and the later application of a fixed-point argument one can prove thatfor given x and v in LrethQTHORN and y0 yy0 2 W22=rrethOTHORN W

1r0 ethOTHORN with r gt N=2thorn 1

system (1) possesses a unique solution in Lreth0T W2rethOTHORNTHORN Let us remark thatthroughout the paper this regularity will be assumed on the data

Our second main result is of a negative nature

Theorem 12 There exist C1 functions f verifying f 00 2 L1locethRTHORN feth0THORN frac14 0 and

jfethsTHORNj jsj logaeth1thorn jsjTHORN as jsj 1 eth12THORN

with a gt 2 and there exist source terms x 2 LrethQTHORN satisfying (11) for which it is notpossible to find control functions insensitizing the functional f given by (2)

For the proof of this result we choose

fethsTHORN frac14Z jsj

0

loga1thorn jsj2ds for all s 2 R

and we prove a localized estimate in Onoo of the AQ1corresponding solution y of (4) thatshows that for certain source terms x the control v cannot compensate the blow upphenomena occurring in Onoo

In view of Theorems 11 and 12 it would be interesting to analyze what happenswhen f satisfies (12) with 1 a 2 (see Subsec 64 for further comments)

The rest of this paper will be organized as follows In the following sectionwe present some technical results which will be proved in an appendix Section 3provides an exhaustive study of the linear case In Sec 4 we analyze the nonlinearcase and prove Theorem 11 The fifth section is devoted to the proof of Theorem 12In Sec 6 we give other insensitivity results and discuss some open problems


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2 SOME TECHNICAL RESULTS

In this Section we state some technical results which will be used later They areknown results on the local regularity for the solutions to the linear heat equationNevertheless we include the proof of these results in an appendix so as to obtainthe explicit dependence of the constants on the potentials which will be essentialin our analysis

First let us present the following notation which is used all along this paperFor r 2 frac1211 and a given Banach space X kkLrethXTHORN will denote the norm inLreth0T XTHORN For simplicity the norm in LrethQTHORN will be represented by kkLr forr 2 frac1211THORN and kk1 will denote the norm in L1ethQTHORN For r 2 frac1211THORN and any openset V RN we will consider the Banach space

Xreth0T VTHORN frac14 u 2 Lreth0T W2rethVTHORNTHORN tu 2 Lreth0T LrethVTHORNTHORN

and its norm defined by

kukXreth0T VTHORN frac14 kukLrethW2rethVTHORNTHORN thorn ktukLrethLrethVTHORNTHORN

In particular we will consider the space Xr frac14 Xreth0T OTHORN and its norm denoted bykkXr The norm in the space L2eth0T H2ethOTHORNTHORN Cethfrac120T H1ethOTHORNTHORN will be denoted bykkL2ethH2THORNCethH1THORN

On the other hand for a 2 eth0 1THORN and u 2 C0ethQTHORN we define the quantity

frac12uaa2 frac14 supQ

juethx tTHORN uethx0 tTHORNjjx x0ja thorn sup

Q

juethx tTHORN uethx t0THORNjjt t0ja2

We will consider the space Caa2ethQTHORN frac14 u 2 C0ethQTHORN frac12uaa2 lt 1 which is a Banach

space with its natural norm jujaa2Q frac14 kuk1 thorn frac12uaa2 We will also consider the Banachspace defined by

C1thorna1thorna2 ethQTHORN frac14

u 2 C0ethQTHORN u

xi2 Caa2ethQTHORN 8i sup

Q

juethx tTHORN uethx t0THORNjjt t0j1thorna2

lt 1

The following holds

Proposition 21 Let a 2 L1ethQTHORN and F 2 L2ethQTHORN be given Let us consider a solutiony 2 L2eth0T H2ethOTHORNTHORN Cethfrac120T H1ethOTHORNTHORN to

ty Dythorn ay frac14 F in Q

y frac14 0 on S yethx 0THORN frac14 0 in O

eth13THORN

(a) Let V O (resp B O) be an open set Let us suppose thatF 2 Lreth0T LrethVTHORNTHORN (resp F 2 Lreth0T LrethOnBTHORNTHORN with r 2 eth21THORN Then for any


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open set V0 V (resp B B0 O) one has

y 2 Xreth0T V0THORN ethresp y 2 Xreth0T OnB0THORNTHORN

Moreover there exist positive constants C frac14 CethOT N rVV0THORN (resp C frac14CethOT N rBB0THORN) and K frac14 KethNTHORN such that

kykXreth0T V0THORN C eth1thorn kak1THORNKkFkLrethLrethVTHORNTHORN thorn kykL2ethH2THORNCethH1THORN eth14THORN

resp kykXreth0T OnB0THORN C eth1thorn kak1THORNKkFk

LrethLrethOnBTHORNTHORN thorn kykL2ethH2THORNCethH1THORN

(b) Assume in addition that F 2 Lreth0T W1rethVTHORNTHORN with r as above andHa 2 LgethQTHORNN with

g frac14 max r N2 thorn 1

if r 6frac14 N2 thorn 1

N2 thorn 1thorn e if r frac14 N

2 thorn 1

(eth15THORN

and e being an arbitrarily small positive number Then for any open set V0 Vone has

y 2 Lreth0T W3rethV0THORNTHORN ty 2 Lreth0T W1rethV0THORNTHORN

and for a new positive constant C frac14 CethOT N rVV0THORN the following estimateholds

kykLrethW3rethV0THORNTHORN thorn ktykLrethW1rethV0THORNTHORN CHkFkLrethW1rethVTHORNTHORN thorn kykL2ethH2THORNCethH1THORN

where

H frac14 HethN kak1 kHakLgTHORN frac14 eth1thorn kak1THORNKthorn1eth1thorn kHakLgTHORN

K frac14 KethNTHORN being as in (14)

We will also use the following result which is immediately obtained rewritingLemma 33 p 80 in Ladyzenskaya et al (1967) with our notation

Lemma 22 Let O RN N 1 be an open set with O 2 C2 The followingcontinuous embeddings hold

(i) If r lt N2 thorn 1 then Xr LpethQTHORN where 1

pfrac14 1

r 2

Nthorn2

(ii) If r frac14 N2 thorn 1 then Xr LqethQTHORN for all q lt 1

(iii) If N2 thorn 1 lt r lt N thorn 2 then Xr Cbb2ethQTHORN with b frac14 2 Nthorn2

r

(iv) If r frac14 N thorn 2 then Xr Cll2ethQTHORN for all l 2 eth0 1THORN(v) If r gt N thorn 2 then Xr C1thorna1thorna

2 ethQTHORN where a frac14 1 Nthorn2r


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3 THE LINEAR CASE

Let us consider the linear systems

ty Dythorn ay frac14 xthorn v1o in Q


(eth16THORN

tq Dqthorn bq frac14 y1O in Q


(eth17THORN

where a b 2 L1ethQTHORN x 2 L2ethQTHORN and the open sets o O are givenRecall that o O 6frac14 The aim of this Section is to build a control

v 2 LrethQTHORN with r as in the statement of Theorem 11 such that the solution to(16)ndash(17) verifies


The estimate of kvkLr with respect to the potentials is also given To this endadditional assumptions on a b and x will be required The regularity of v thus ofy and q will enable us to deal with the nonlinear case

We will proceed in two steps First using an appropriate observability inequal-ity we will construct controls in L2ethQTHORN This will be done in Subsec 31 Then inview of the results in the precedent Section we will exhibit more regular controls(see Subsec 32)

31 Insensitizing Controls in L2(Q)

We actually seek a regular control supported in o O 6frac14 In the sequel B0 willbe a fixed open set such that B0 o O and we will omit the dependence of theconstants on B0

Under suitable assumptions on the given heat source x the following result pro-vides a control function vv in L2ethQTHORN with support in B0 frac120T The L2- norm of vv isalso estimated with respect to x

Proposition 31 Let a b 2 L1ethQTHORN be given and let x 2 L2ethQTHORN verify

ZZQ

expCM

t


with C frac14 CethOoOTHORN and M frac14 MethT kak1 kbk1THORN as in Proposition 32 Then thereexists a control function vv 2 L2ethQTHORN such that supp vv B0 frac120T and the corre-sponding solution ethyy qqTHORN to (16)ndash(17) satisfies (18) Moreover vv can be chosen in


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such a way that

kvvkL2 expC

2H

ZZQ

expCM

t

jxj2 dx dt

1=2

eth20THORN

H frac14 HethT kak1 kbk1THORN being as in Proposition 32

The proof of this result can be found in De Teresa (2000) and it will be omittedhere The key point in this proof is to obtain an appropriate observability inequalityfor the corresponding adjoint systems

tj Djthorn bj frac14 0 in Q

j frac14 0 on S jethx 0THORN frac14 j0ethxTHORN in O

(eth21THORN

and

tc Dcthorn ac frac14 j1O in Q

c frac14 0 on S cethxTTHORN frac14 0 in O

(eth22THORN

with j0 2 L2ethOTHORN and a b 2 L1ethQTHORN The following result establishes the above-mentioned observability inequality

Proposition 32 Let us assume that o O 6frac14 Let B0 be an arbitrary open subsetof o O Then there exist positive constants C M and H such that for everyj0 2 L2ethOTHORN the corresponding solution ethjcTHORN to (21)ndash(22) satisfies

ZZQ

exp CM

t

jcj2 dx dt expethCHTHORN

ZZB0eth0TTHORN

jcj2 dx dt

with

M frac14 MethT kak1 kbk1THORN frac14 1thorn T1thorn kak2=31 thorn kbk2=31 thorn ka bk1=21

H frac14 HethT kak1 kbk1THORN frac14 1thorn 1

Tthorn T 1thorn kak1 thorn kbk1

thorn kak2=31

thorn kbk2=31 thorn ka bk1=21

and C frac14 CethOoOB0THORN

This inequality was first proved in De Teresa (2000) as a consequence of a globalCarleman inequality for the heat equation (cf Fursikov and Imanuvilov 1996) InBodart et al (2004) the result is extended to the heat equation involving first orderterms and the explicit dependence of the constants M and H on T and the size of the


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potentials is given This is crucial to prove the existence of insensitizing controls inthe nonlinear case when f is not a globally Lipschitz-continuous function

32 Insensitizing Controls in Lr(Q)

In this Subsection we prove that under additional assumptions on x and thepotentials a and b starting from an insensitizing control vv 2 L2ethQTHORN supported inB0 frac120T we can build a more regular insensitizing control with a slightly largersupport Furthermore we estimate the norm of the new control with respect to kvvkL2

We proceed as follows Let B B1 and B2 be three open sets such that

B0 B1 B2 B o O

B0 being the open set considered in the previous Subsection For a b 2 L1ethQTHORN andx 2 L2ethQTHORN satisfying (19) let vv be a control (associated to B0) provided byProposition 31 and let ethyy qqTHORN be the corresponding solution to (16)ndash(18)

Our goal is to construct a regular control supported in B frac120T To this end letus consider a function y 2 DethBTHORN such that y 1 in B2 We set

q frac14 eth1 yTHORN qq eth23THORN

and

y frac14 eth1 yTHORN yythorn 2Hy Hqqthorn ethDyTHORNqq eth24THORN

We will see that under appropriate regularity assumptions on the data x a and bethy qTHORN solves (16)ndash(18) with the control function supported in B frac120T given by

v frac14 yxthorn 2Hy Hyythorn ethDyTHORNyythorn etht Dthorn aTHORN 2Hy Hqqthorn ethDyTHORNqqfrac12 eth25THORN

For r 2 eth21THORN let us set

Zr frac14Lreth0T W1rethOTHORNTHORN if r 2 2 N2 thorn 1

C0ethQTHORN Lreth0T W1rethOTHORNTHORN if r gt N2 thorn 1

(eth26THORN

In the sequel unless otherwise specified C will stand for a generic positive con-stant depending on O o O and T (the dependence on N and r will be omitted forsimplicity) whose value may change from line to line One has the following result

Proposition 33 Let x 2 LrethQTHORN verify (19) with r 2 eth21THORN and let vv 2 L2ethQTHORN be acontrol provided by Proposition 31 (associated to B0) The following holds

(a) For a b 2 L1ethQTHORN it holds that y q given by (24) and (23) lie in Zr and

kykZrthorn kqkZr

C1ethOoOT kak1 kbk1THORN kxkLr thorn kvvkL2eth THORN eth27THORN


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where

C1ethOoOT kak1 kbk1THORN frac14 exp Ceth1thorn kak1 thorn kbk1THORN eth28THORN

(b) Suppose in addition that Hb 2 LgethQTHORNN with g given by (15) Then v

defined by (25) satisfies v 2 LrethQTHORN supp v B frac120T and

kvkLr C2ethOoOT kak1 kbk1 kHbkLgTHORN kxkLr thorn kvvkL2eth THORN eth29THORN

where C2ethOoOT kak1 kbk1 kHbkLgTHORN is given by

C2 frac14 exp Ceth1thorn kak1 thorn kbk1THORN 1thorn kHbkLgeth THORN eth30THORN

Proof (a) First let us see that for a b 2 L1ethQTHORN y and q lie in Lreth0T W1rethOTHORNTHORNLet us write

y frac14 y1 thorn y2 thorn y3

with

y1 frac14 eth1 yTHORNyy y2 frac14 2Hy Hqq and y3 frac14 ethDyTHORNqq

Recalling that ethyy qqTHORN solves (16)ndash(17) with a control vv in L2ethQTHORN provided by Proposi-tion 31 classical energy estimates give

yy qq 2 L2eth0T H2ethOTHORN H10ethOTHORNTHORN Cethfrac120T H1

0ethOTHORNTHORNkyykL2ethH2H1

0THORN thorn kyykCethH1

0THORN exp Ceth1thorn kak1THORN kxkL2 thorn kvvkL2eth THORN eth31THORN

and

kqqkL2ethH2H10THORN thorn kqqkCethH1

0THORN exp Ceth1thorn kak1 thorn kbk1THORN kxkL2 thorn kvvkL2eth THORN eth32THORN

We can apply Proposition 21 to yy and the open sets B0 and B1 together with (31) andconclude

yy 2 Xreth0T OnB1THORN

and

kyykXreth0T OnB1THORN exp Ceth1thorn kak1THORN kxkLr thorn kvvkL2eth THORN eth33THORN

Then since suppeth1 yTHORN OnB1 one has

y1 2 Xr and ky1kXr exp Ceth1thorn kak1THORN kxkLr thorn kvvkL2eth THORN eth34THORN


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On the other hand as in particular yy1O 2 LrethethOnB1THORN eth0TTHORNTHORN Proposition 21(applied this time to qq and the open sets B1 and B2) together with (33) and (32) yield

qq 2 Xreth0T OnB2THORN

and

kqqkXreth0T OnB2THORN C1ethOoOT kak1 kbk1THORN kxkLr thorn kvvkL2eth THORN

with C1ethOoOT kak1 kbk1THORN given by (28) Then arguing as above one deducesthat

y2 2 Lreth0T W1rethOTHORNTHORN y3 q 2 Xr eth35THORN

and

ky2kLrethW1rethOTHORNTHORN thorn ky3kXr thorn kqkXr C1 kxkLr thorn kvvkL2eth THORN eth36THORN

In particular for r 2 2 N2 thorn 1

one has

y q 2 Zr frac14 Lreth0T W1rethOTHORNTHORN ethindeed q 2 XrTHORN

and inequality (27) holdsLet us now suppose that r gt N=2thorn 1 In this case in view of Lemma 22

the space Xr embeds continuously (and also compactly) in C0ethQTHORN Hence from(34)ndash(36) one has

y1 y3 q 2 Zr frac14 C0ethQTHORN Lreth0T W1rethOTHORNTHORN

and

ky1kZrthorn ky3kZr

thorn kqkZr C1 kxkLr thorn kvvkL2eth THORN

In order to complete the proof of point (a) it suffices to see that y2 2 C0ethQTHORN Tothis end let us observe that for r greater than N=2thorn 1 the function yy1O lies inL1ethethOnB1THORN eth0TTHORNTHORN Then from Proposition 21 one deduces that

qq 2 Xpeth0T BnB2THORN for all p lt 1

with

kqqkXpeth0T BnB2THORN C1 kxkLr thorn kvvkL2eth THORN


ORDER REPRINTS

Thus for fixed p gt N thorn 2 once again in view of Lemma 22 it holds that

qq 2 C1thorna1thorna2 ethBnB2 frac120T THORN with a frac14 1 N thorn 2

p

This gives y2 2 Caa2ethQTHORN This space being continuously embedded in C0ethQTHORN one has

y2 2 C0ethQTHORN and ky2kC0ethQTHORN C1 kxkLr thorn kvvkL2eth THORN

Is is then infered that y 2 Zr frac14 C0ethQTHORN Lreth0T W1rethOTHORNTHORN and (27) holds

(b) Let us suppose in addition that Hb 2 LgethQTHORNN with g given by (15) We firstobserve that v defined by (25) is supported in B frac120T Moreover yy lies inXreth0T OnB1THORN and (33) holds We can apply point (b) of Proposition 21 to qq forthe open set OnB1 and deduce

qq 2 Lreth0T W3rethBnB2THORNTHORN tqqjBnB22 Lreth0T W1rethBnB2THORNTHORN

and

kqqkLrethW3rethBnB2THORNTHORN thorn ktqqkLrethW1rethBnB2THORNTHORN C2 kxkLr thorn kvvkL2eth THORN

with C2 frac14 C2ethOoOT kak1 kbk1 kHbkLgTHORN given by (30)Now in view of (25) and taking into account the previous considerations on yy

and qq and the choice of y one concludes that v lies in LrethQTHORN and satisfies estimate(29) This ends the proof amp

Finally from the regularity of y q and v defined by (23)ndash(25) we deduce thatethy qTHORN solves (16)ndash(18) with control function v One then has the following insensi-tivity result for the linear case

Corollary 34 Let x 2 LrethQTHORN satisfy (19) with r 2 eth21THORN Let us assume thata 2 L1ethQTHORN and b 2 L1ethQTHORN Lgeth0T W1gethOTHORNTHORN with g given by (15) Then y q

defined by (24) and (23) lie in the space Zr introduced in (26) and solve (16)ndash(18)for the control function v 2 LrethQTHORN given by (25) Furthermore there exists a positiveconstant C depending on O o O and T such that the following estimates hold

kykZrthorn kqkZr

C1 kxkLr thorn expC

2H

exp

CM

2t

x

L2

and

kvkLr C2 kxkLr thorn expC

2H

exp

CM

2t

x

L2

with C1 and C2 of the form (28) and (30) respectively and H M being as inProposition 32


ORDER REPRINTS

Let us remark that in the context of the heat equation the previous techniqueprovides a new method of construction of regular controls starting from controlsin L2ethQTHORN This will make it possible to give a new proof of known null controllabilityresults for nonlinear heat equations A local result on the null controllability for theclassical heat equation and a local result on insensitizing controls for a semilinearheat equation both with nonlinear Fourier boundary conditions can also beobtained by using a similar construction (see Bodart et al accepted for publicationDoubova et al 2004)

4 THE NONLINEAR CASE PROOF OF THEOREM 11

In this Section we will apply an appropriate fixed-point argument to treat thenonlinear case In a first step f will be assumed to be a C2 function The general casewill be studied in Subsec 42

41 The Case When f is a C2 Function

Let f 2 C2ethRTHORN be a function verifying feth0THORN frac14 0 and (10) Let x 2 LrethQTHORN satisfyhypothesis (11) with r gt N=2thorn 1

Let us define

gethsTHORN frac14fethsTHORNs

if s 6frac14 0

f 0eth0THORN if s frac14 0

(

Then g f 0 2 C0ethRTHORN and fethsTHORN frac14 gethsTHORN s for all s 2 R Since feth0THORN frac14 0 hypothesis (10)on f 0 implies a similar one on g that is

limjsj1

gethsTHORNlogeth1thorn jsjTHORN frac14 0

Thus for each e gt 0 there exists a positive constant Ce (which only depends on e andon the function f) such that

jgethsTHORNj thorn jf 0ethsTHORNj Ce thorn e logeth1thorn jsjTHORN for all s 2 R eth37THORN

Let us recall that for r gt N=2thorn 1 we defined

Zr frac14 C0ethQTHORN Lreth0T W1rethOTHORNTHORN

For any z 2 Beth0RTHORN Zr R gt 0 to be determined later we consider the linearcontrollability problem

ty Dythorn gethzTHORNy frac14 xthorn v1o in Q


(eth38THORN


ORDER REPRINTS

tq Dqthorn f 0ethzTHORNq frac14 y1O in Q


(eth39THORN


Let us observe that (38)ndash(39) are of the form (16)ndash(17) with potentials

a frac14 az frac14 gethzTHORN 2 L1ethQTHORNb frac14 bz frac14 f 0ethzTHORN 2 L1ethQTHORN Lreth0T W1rethOTHORNTHORN ethg frac14 r in this caseTHORN

(

In view of Corollary 34 for any z 2 Beth0RTHORN Zr there exists a controlvz 2 LrethQTHORN such that the corresponding solution ethyz qzTHORN to (38)ndash(39) lies in Zr Zr

and satisfies (40) Moreover estimates

kyzkZr C1ethOoOT zTHORN kxkLr thorn exp

C

2Hz

exp

CMz

2t

x

L2

eth41THORN

and

kvzkLr C2ethOoOT zTHORN kxkLr thorn expC

2Hz

exp

CMz

2t

x

L2

eth42THORN

hold where

C1ethOoOT zTHORN frac14 exp Ceth1thorn kgethzTHORNk1 thorn kf 0ethzTHORNk1THORN

C2ethOoOT zTHORN frac14 exp Ceth1thorn kgethzTHORNk1 thorn kf 0ethzTHORNk1THORN 1thorn kf 00ethzTHORNHzkLreth THORN

Mz frac14 1thorn kgethzTHORNk2=31 thorn kf 0ethzTHORNk2=31 thorn kgethzTHORN f 0ethzTHORNk1=21

Hz frac14 1thorn kgethzTHORNk1 thorn kf 0ethzTHORNk1 thorn kgethzTHORNk2=31 thorn kf 0ethzTHORNk2=31 thorn kgethzTHORN f 0ethzTHORNk1=21

and C frac14 CethOoOTTHORN gt 0Due to hypothesis (11) on x one hasZZ

Q

expCMz

t

jxj2 dx dt frac14

ZZQ

expCMz

t 1

t3

exp

1

t3

jxj2 dx dt

exp CM3=2z

13 ZZQ

exp1

t3

jxj2 dx dt eth43THORN

C frac14 CethOoOTTHORN being a new positive constantThen from inequalities (41) (42) and (43) and using the convexity of the real

function s 7 s3=2 it can be estimated

jjyzjjZr C1ethOoOT zTHORN kxkLr thorn exp

1

2t3

x

L2

eth44THORN


ORDER REPRINTS

and

jjvzjjLr C2ethOoOT zTHORN kxkLr thorn exp1

2t3

x

L2

eCCethOoOT RTHORN kxkLr thorn exp

1

2t3

x

L2

eth45THORN

where eCCethOoOT RTHORN is a positive constant independent of zLet us define

A z 2 Beth0RTHORN Zr 7AethzTHORN LrethQTHORN

with

AethzTHORN frac14 v 2 LrethQTHORN ethy qTHORN satisfieseth38THORNeth40THORN v verifying eth45THORNf g

and let L be the set-valued mapping defined on Zr as follows

L z 2 Beth0RTHORN Zr 7LethzTHORN Zr

with

LethzTHORN frac14 y 2 Zr ethy qTHORN solves eth38THORNeth39THORN with v 2 AethzTHORN y satisfying eth44THORNf g

Let us prove that L fulfills the assumptions of Kakutanirsquos fixed-point TheoremIn the first place one can check that LethzTHORN is a non-empty closed convex subset of

Zr for fixed z 2 Zr due to the linearity of systems (38) and (39)In fact in view of Theorem 91 in Ladyzenskaya et al (1967) and (45) LethzTHORN is

uniformly bounded in Xr the space introduced in Sec 2 Recall that forr gt N=2thorn 1 Xr is continuously embedded in the Holder space Cbb2ethQTHORN withb frac14 2 ethN thorn 2THORN=r (see Lemma 22) Then there exists a compact set K Zr K onlydepending on R such that

LethzTHORN K 8z 2 Beth0RTHORN eth46THORNLet us now prove that L is an upper hemicontinuous multivalued mapping that

is to say for any bounded linear form m 2 Z0r the real-valued function

z 2 Beth0RTHORN Zr 7 supy2LethzTHORN

hm yi

is upper semicontinuous Correspondingly let us see that

Blm frac14z 2 Beth0RTHORN sup

y2LethzTHORNhm yi l


ORDER REPRINTS

is a closed subset of Zr for any l 2 R m 2 Z0r (see Doubova et al 2002 for a similar

proof) To this end we consider a sequence znf gn1 Blm such that

zn z in Zr

Our aim is to prove that z 2 Blm Since all the LethznTHORN are compact sets by the defini-tion of Blm for any n 1 there exists yn 2 LethznTHORN such that

hm yni frac14 supy2LethznTHORN

hm yi l eth47THORN

Recalling now the definition ofAethznTHORN and LethznTHORN let vn 2 AethznTHORN qn 2 Zr be such thatethyn qnTHORN solves (38)ndash(40) with control vn and potentials gethznTHORN f 0ethznTHORN From (44) and(45) yn and vn satisfy

jjynjjZr C1ethOoOT znTHORN kxkLr thorn exp

1

2t3

x

L2

and

jjvnjjLr C2ethOoOT znTHORN kxkLr thorn exp1

2t3

x

L2

Thus ynf g (resp vnf g) is uniformly bounded in Zr (resp in LrethQTHORN) In particular(46) gives us fyng K Hence there exist subsequences still denoted by ynf g andvnf g such that

yn y strongly in Zr

and

vn v weakly in LrethQTHORN

Since g and f 00 are continuous functions one also has

gethznTHORN gethzTHORN in C0ethQTHORN

f 0ethznTHORN f 0ethzTHORN in C0ethQTHORN

and

f 00ethznTHORNHzn f 00ethzTHORNHz in LrethQTHORNN

Passing to the limit one deduces that yy and the associated function Q solve (38)ndash(40)with control function vv (and potentials gethzTHORN f 0ethzTHORN) Moreover yy and vv satisfy


ORDER REPRINTS

(44) and (45) that is vv 2 AethzTHORN and yy 2 LethzTHORN Then taking limits in (47) it holds that

supy2LethzTHORN

hm yi hm yi l

whence it is deduced that z 2 Blm and hence L is upper hemicontinuousFinally let us see that there exists R gt 0 such that

L Beth0RTHORN Beth0RTHORN eth48THORN

Let R gt 0 be to be determined For any z 2 Beth0RTHORN Zr from (44) and (37) it isobserved that each y 2 LethzTHORN satisfies

jjyjjZr exp Ceth1thorn Ce thorn e logeth1thorn jjzjj1THORNTHORN kxkLr thorn exp

1

2t3

x

L2

exp Ceth1thorn CeTHORNfrac12 eth1thorn RTHORNCe kxkLr thorn exp

1

2t3

x

L2

with C frac14 CethOoOTTHORN gt 0 Thus choosing e frac14 1=2C we get

jjyjjZr Ceth1thorn RTHORN1=2 kxkLr thorn exp

1

2t3

x

L2

from which we infer the existence of R gt 0 large enough such that (48) is satisfiedThe Kakutani Fixed-point Theorem thus applies which ends the proof when f is

a C2 function

Remark 2 Let us notice that two different nonlinearities f for which the positiveconstant Ce in (37) coincide lead to the same R gt 0 This fact will be used in thefollowing Subsection when studying the general case

42 The General Case

It is now assumed that f is a C1 function satisfying f 00 2 L1locethRTHORN feth0THORN frac14 0 and

(10) and let x be as aboveWe consider a function r 2 DethRTHORN such that

r 0 in R supp r frac121 1 and

ZRrethsTHORNds frac14 1

For any n 1 let us set

rnethsTHORN frac14 nrethnsTHORN for all s 2 R

Fn frac14 rn f fnethTHORN frac14 FnethTHORN Fneth0THORN


ORDER REPRINTS

and

gnethsTHORN frac14fnethsTHORNs

if s 6frac14 0

f 0neth0THORN if s frac14 0

(

Due to properties of rn and of the convolution as well as the hypothesis on f one can prove that fn and gn have the following properties

(i) gn and f 00n are continuous functions and fneth0THORN frac14 0 for all n 1

(ii) fn f in C1ethKTHORN for all compact set K R(iii) gn g uniformly on compact sets of R(iv) For any given M gt 0 there exists a positive constant CethMTHORN such that

supjsjM

jgnethsTHORNj thorn jf 0nethsTHORNj thorn jf 00

n ethsTHORNj CethMTHORN

for all n 1(v) It also holds that

limjsj1

jf 0nethsTHORNj thorn jgnethsTHORNjlogeth1thorn jsjTHORN frac14 0 uniformly in n

that is for any e gt 0 there exists Me gt 0 such that

jgnethsTHORNj thorn jf 0nethsTHORNj e logeth1thorn jsjTHORN for all jsj Me and n 1

In particular the last two properties imply that for any e gt 0 there existsCe gt 0 only depending on e and not on the function fn such that

jgnethsTHORNj thorn jf 0nethsTHORNj Ce thorn e logeth1thorn jsjTHORN for all s 2 R and n 1 eth49THORN

As it was proved in the previous Subsection for any n 1 there exists a controlfunction vn 2 LrethQTHORN with supp vn o frac120T such that the following cascade ofsystems

tyn Dyn thorn fnethynTHORN frac14 xthorn vn1o in Q

yn frac14 0 on S ynethx 0THORN frac14 0 in O

(eth50THORN

tqn Dqn thorn f 0nethynTHORNqn frac14 yn1O in Q

qn frac14 0 on S qnethxTTHORN frac14 0 in O

(eth51THORN

admits a solution ethyn qnTHORN 2 Zr Zr satisfying

qnethx 0THORN frac14 0 in O eth52THORN


ORDER REPRINTS

Moreover estimates

jjynjjZr C1ethOoOT ynTHORN kxkLr thorn exp

1

2t3

x

L2

eth53THORN

and

jjvnjjLr C2ethOoOT ynTHORN kxkLr thorn exp1

2t3

x

L2

eth54THORN

hold where

C1ethOoOT ynTHORN frac14 exp Ceth1thorn kgnethynTHORNk1 thorn kf 0nethynTHORNk1THORN

C2ethOoOT ynTHORN frac14 exp Ceth1thorn kgnethynTHORNk1 thorn kf 0nethynTHORNk1THORN eth1thorn kf 00

n ethynTHORNHynkLr THORN

and C frac14 CethOoOTTHORN gt 0Let us recall that yn is for any n 1 a fixed point of a set-valued mapping Ln

defined from Zr onto itself In view of estimates (53) and (54) and taking intoaccount (49) and Remark 2 one deduces arguing as in the previous subsection thatthere exists R gt 0 large enough and independent of n such that

Ln Beth0RTHORN Beth0RTHORN

In addition ynf g (resp vnf g) is uniformly bounded in Zr (resp in LrethQTHORN) Indeedreasoning as in Subsec 41 ynf g is uniformly bounded in the space Xr and hencer being grater than N=2thorn 1 there exists a compact set K in Zr such that fyng K

Thus up to a subsequence one has

yn y strongly in Zr

and


with v 2 LrethQTHORN and y 2 K Zr Taking now into account properties (ii) and (iii) onealso has

gnethynTHORN gethyTHORN and f 0nethynTHORN f 0ethyTHORN in C0ethQTHORN

Hence passing to the limit in (50)ndash(52) one infers that y and the corresponding qsolve (38)ndash(40) that is we have found a control v in LrethQTHORN insensitizing f This endsthe proof of Theorem 11


ORDER REPRINTS

5 PROOF OF THEOREM 12

This Section is devoted to proving the insensitivity result of a negative naturestated in Theorem 12 To do so we will show that for certain functions f as inthe statement and certain source terms x 2 LrethQTHORN vanishing for t 2 eth0 t0THORN witht0 2 eth0TTHORN whatever the control v is the corresponding solution y to (4) blows upbefore the time t frac14 T and hence the functional f cannot be insensitized We willfollow the proof of Theorem 11 in Fernandez-Cara and Zuazua (2000) where thelack of null controllability of a semilinear heat equation is proved

Let us consider the following function


0

logaeth1thorn sTHORNds for all s 2 R

with a gt 2 It is easy to check that f is a convex function fethsTHORNs lt 0 for all s lt 0 and

jfethsTHORNj jsj logaeth1thorn jsjTHORN as jsj 1

Let r 2 DethOTHORN be such that

r 0 in O r 0 in o and

ZOrethxTHORNdx frac14 1

For a fixed t0 2 eth0TTHORN we set

xethx tTHORN frac140 if t 2 frac120 t0ethM thorn kTHORN if t 2 etht0T

(

where M is a positive constant which will be chosen later and k is given by

k frac14 1

2

ZOrf 2

jDrjr

dx

Here f is the convex conjugate of the convex function f (the function r can betaken in such a way that rfeth2jDrj=rTHORN 2 L1ethOTHORN see Fernandez-Cara and Zuazua2000)

Let y be a solution to (4) associated to a control v and x defined in the maximalinterval frac120TTHORN Multiplying the equation in (4) by r and integrating in O we get

d

dt

ZOry dx frac14

ZOrDy dx

ZOrfethyTHORNdxthorn

ZOrxdx

We also set

zethtTHORN frac14 ZOrethxTHORNyethx tTHORNdx 8t 2 frac12t0TTHORN


ORDER REPRINTS

Using the properties of f (see Fernandez-Cara and Zuazua 2000 for the details) weobtain

z0ethtTHORN M thorn 1

2fethzethtTHORNTHORN t 2 frac12t0TTHORN

zetht0THORN frac14 z0 frac14 ZOrethxTHORNyethx t0THORNdx

8gtgtltgtgtDefining

Gethz0 sTHORN frac14Z s

z0

2

fethsTHORN thorn 2Mds 8s z0

we can prove that

T t0 thorn supt2frac12t0TTHORN

Gethz0 zethtTHORNTHORN t0 thornZ 1

z0

2

fethsTHORN thorn 2Mds

Thus for M gt 0 large enough the solution y blows up in L1ethOTHORN before T Thisends the proof

6 FURTHER COMMENTS RESULTS AND OPEN PROBLEMS

61 On the Construction of Regular Controls for ParabolicNull Controllability Problems

The technique of construction of regular controls (starting from L2-controls)introduced in the present paper can be applied to the study of the null controllabilityof (7) not only when f just depends on the state y but also when f frac14 fethyHyTHORN In factcontrols in Caa2ethQTHORN with a 2 eth0 1 can be obtained for potentials regular enough Letus describe this strategy in the linear case We consider the null controllabilityproblem

ty Dythorn B Hythorn ay frac14 v1o in Q

y frac14 0 on S yethx 0THORN frac14 y0 in O

(eth55THORN

yethxTTHORN frac14 0 in O

where y0 2 L2ethOTHORN a 2 L1ethQTHORN and B 2 L1ethQTHORNN The previous null controllabilityproblem is equivalent to

tq Dqthorn B Hqthorn aq frac14 Z0ethtTHORNY thorn v1o in Q

q frac14 0 on S qethx 0THORN frac14 0 in O

qethxTTHORN frac14 0 in O

8gtltgt eth56THORN


ORDER REPRINTS

where q frac14 y ZethtTHORNY Z 2 C1ethfrac120T THORN satisfies

Z 1 in frac120T=3 Z 0 in frac122T=3T

and Y solves (55) with v frac14 0 Suppose that there exists a control vv 2 L2ethQTHORN solving(56) with supp vv B0 frac120T and let qq be the associated state (here B0 o is anon-empty open set) Then q frac14 eth1 yethxTHORNTHORNqq together with

v frac14 yethxTHORNZ0ethtTHORNY thorn 2Hy Hqqthorn ethDyTHORNqq ethB HyTHORNqq

solve the null controllability problem (56) where y 2 DethoTHORN verifies y 1 in B0Following Subsec 32 one can prove that v 2 L1ethQTHORN and

kvk1 CethT Oo kak1 kBk1THORNkvvkL2

The explicit dependence of the constant C on kak1 and kBk1 can also be obtained

Moreover if B 2 Caa2ethQTHORN the control v is proved to lie in Caa2ethQTHORN and v verifies a

Caa2-estimate similar to the previous one

This strategy has been used in Bodart et al (Accepted for publication) to con-struct holderian controls for a null controllability problem for the heat equation withnonlinear boundary Fourier conditions (see also Doubova et al 2004)

62 Other Insensitivity Results

The proof of Theorem 11 can be adapted to prove other insensitivity results forsystem (1)

1 Theorem 11 is still true under a slightly more general condition on f To be precise there exists l1ethOoOTTHORN gt 0 such that if hypothesis (10) is replacedby

lim supjsj1

jf 0ethsTHORNjlogeth1thorn jsjTHORN l1

a control function insensitizing the functional given by (2) can be found for any givensource term x as in Theorem 11

2 The following local insensitivity result can also be proved under no restric-tions on the increasing of f

Theorem 61 Suppose that o O 6frac14 and y0 frac14 0 Let f 2 C1ethRTHORN be such thatf 00 2 L1

locethRTHORN and feth0THORN frac14 0 Let r 2 N2 thorn 11

be given Then there exist two posi-tive constants M frac14 MethOoOT fTHORN and Z frac14 ZethOoOT fTHORN such that for any


ORDER REPRINTS

x 2 LrethQTHORN verifying

kxkLr thorn expM

2t

x

L2

Z

one can find a control function v 2 LrethQTHORN insensitizing the functional definedin (2)

63 Simultaneous Null and Insensitizing Controls

As an extension of Theorem 11 one can prove the following result on the exis-tence of simultaneous null and insensitizing controls in other words controls suchthat the solution ethy qTHORN to the cascade of systems (4)ndash(5) (with y0 frac14 0) verifies (6) and

yethxTTHORN frac14 0 in O eth57THORN

Theorem 62 Let us assume that o O 6frac14 and y0 frac14 0 Let f be a functionsatisfying hypothesis in Theorem 11 and let r 2 N

2 thorn 11 be given Then for

any x 2 LrethQTHORN verifyingZZ

Q

exp1

t3ethT tTHORN3


there exists a control function v 2 LrethQTHORN insensitizing the functional given by (2)and such that the solution yeth t vTHORNjtfrac140 to (1) (associated to t frac14 0 and v) satisfies

yethxT t vTHORNjtfrac140 frac14 0 in O

Proof The scheme of the proof is similar to that of Theorem 11 Let us observethat in this case the source term x is required to decay rapidly to zero not onlyin the neighbourhood of t frac14 0 but also close to the final time t frac14 T which is a nat-ural assumption We will assume that f is a C2 function verifying feth0THORN frac14 0 and (10)The general case is studied following Subsec 42 Let x 2 LrethQTHORN satisfy (58) withr gt N=2thorn 1

Let us see that there exists a control v regular enough such that the solutionethy qTHORN to the cascade of systems (4)ndash(5) (with y0 frac14 0) simultaneously verifies(6) and (57) As usual in a first step a similar insensitivity result in the linearcase is shown A fixed-point argument is then applied to solve the nonlinearproblem

Let us start with the linear case We consider the systems (16) and (17) with a

and b in L1ethQTHORN and the corresponding adjoint systems



(eth59THORN


ORDER REPRINTS

and


c frac14 0 on S cethxTTHORN frac14 c0ethxTHORN in O

8lt eth60THORN

with j0 and c0 in L2ethOTHORN We first construct controls in L2ethQTHORN Let B0 be a fixednon-empty open subset of o O In this case one has the following observabilityinequality for the solutions to the previous systems whose proof is similar to thatof Proposition 32 (see Bodart et al 2004)

Lemma 63 There exist two positive constants C and M such that

ZZQ

exp CM

tethT tTHORN

jcj2 dx dt C

ZZB0eth0TTHORN

jcj2 dx dt eth61THORN

for all j0c0 2 L2ethOTHORN More precisely C only depends on O o and O and M isgiven by

M frac14 1thorn T thorn T21thorn kak2=31 thorn kbk2=31 thorn ka bk1=21

Observe that in (60) cethTTHORN is allowed to be different from zero so that in this case adifferent observability inequality is obtained

For each e gt 0 we consider the continuous and convex functional eJJe defined onL2ethOTHORN L2ethOTHORN by

eJJeethj0c0THORN frac14 1

2

ZZB0eth0TTHORN

jcj2 dx dtthorn ekj0kL2ethOTHORN thorn ekc0kL2ethOTHORN thornZZ

Q

xcdx dt

where j and c solve (59)ndash(60) The following unique continuation property for thesolutions to (59)ndash(60) holds

(If j0c0 2 L2ethOTHORN ethjcTHORN is the associated solution to (59)ndash(60) andc frac14 0 in o eth0TTHORN then j c 0)

Since B0 o O Lemma 63 gives that c 0 in Q and hence j 0 inO eth0TTHORN Classical unique continuation properties for the heat equation impliesj 0 in Q

As a consequence of the previous continuation property for the solutions of(59)ndash(60) the functional eJJe is indeed strictly convex and satisfies

lim infkj0k

L2ethOTHORNthornkc0kL2ethOTHORNthorn1

eJJeethj0c0THORNkj0kL2ethOTHORN thorn kc0kL2ethOTHORN

e


ORDER REPRINTS

Consequently eJJe admits a unique minimizer ethj0e c

0e THORN 2 L2ethOTHORN L2ethOTHORN Let ethjeceTHORN

be the solution to (59)ndash(60) associated to ethj0e c

0e THORN Then the control

ve frac14 ce1B0

is such that the corresponding solution ethye qeTHORN to (16)ndash(17) satisfies

kqeeth0THORNkL2ethOTHORN e and kyeethTTHORNkL2ethOTHORN e eth62THORN

In view of (61) the controls fveg are proved to be uniformly bounded in L2ethQTHORN(observe that if x verifies (58) then it satisfies

ZZQ

expCM

tethT tTHORN


with C and M as in Lemma 63) Thus reasoning as in the proof of Proposition 31and using (62) one infers the existence of a control vv 2 L2ethQTHORN with supp vv B0 frac120T such that the associated solution ethyy qqTHORN to (16)ndash(17) satisfies (57) and(18) Moreover one can estimate

kvvk2L2 C

ZZQ

expCM

tethT tTHORN


with C and M as aboveThe construction of a regular control starting from vv 2 L2ethQTHORN is the same as in

Subsec 32 More precisely let us define y and q by (24) and (23) respectively As inProposition 33 for a 2 L1ethQTHORN and b 2 L1ethQTHORN Lreth0T W1rethOTHORNTHORN the functions y

and q lie in Zr frac14 C0ethQTHORN Lreth0T W1rethOTHORNTHORN they solve (16)ndash(18) for the controlv 2 LrethQTHORN given by (25) and (57) is satisfied (we use here that yyethTTHORN frac14 qqethTTHORN frac14 0 inL2ethOTHORN and qq 2 Cethfrac120T H1

0 ethOTHORNTHORN) In addition from (27) (29) (63) and (64) oneobtains the following estimates similar to those in Corollary 34

kykZrthorn kqkZr

C1 kxkLr thorn expCM

2tethT tTHORN

x

L2

and

kvkLr C2 kxkLr thorn expCM

2tethT tTHORN

x

L2

with C1 and C2 as in (28) and (30) M provided by Lemma 63 and C being a newpositive constant depending on O o O and T


ORDER REPRINTS

Finally by applying a fixed-point argument completely analogous to theone used in the previous Section one can infer the existence of a controlv 2 LrethQTHORN solving the nonlinear case when f is a C2 function As stated above itis enough to follow Subsec 42 to deal with the general case This ends the proofof Theorem 62

64 Open Problems

1 Theorem 11 has been proved under the essential assumption y0 frac14 0 Itwould be interesting to give sufficient conditions on the initial data y0 in order toinsensitize (2) When f 0 and O frac14 O a positive result on the existence of insensitiz-ing controls is proved for a small class of initial data (see Lemma 2 in De Teresa2000) The proof of this result is strongly based on properties of the semigroup ofthe classical heat equation The general case is an open problem for a linear heatequation with an L1 potential and even in the case of the classical heat equationwhen O 6frac14 O

2 Hypothesis o O 6frac14 is required in order to prove the existence of bothendashinsensitizing and insensitizing controls In the first case the hypothesis is used toprove that certain functional is coercive using a unique continuation property Inthe case of insensitizing controls this hypothesis is used to prove an observabilityinequality At present both problems are far from being solved for disjoint controlset and observation set (see De Teresa 2000)

3 In view of known null controllability results for a semilinear heat equationwith homogeneous Dirichlet boundary conditions (see Theorem 12 in Fernandez-Cara and Zuazua 2000) one could think of extending Theorem 11 to nonlinearitiesf with growth at infinity of order jsj logaeth1thorn jsjTHORN with 1 a lt 3=2 Notice that inthis case in the absence of control blow up phenomena occur under suitable signconditions The idea in Fernandez-Cara and Zuazua (2000) is to control the systemto zero in a short time to prevent the solution from blowing up and to set v 0 forthe rest of the time interval Observe that this argument cannot be applied when asource term x appears on the right-hand side of (7) This strategy also fails in ourproblem since in insensitivity problems both initial and final times are fixed andin addition there is a right-hand side term x in the equation The problem thenremains open for such nonlinearities

4 In Bodart et al (2004) the authors prove a result on the existence ofinsensitizing controls for a semilinear heat equation with a nonlinear termfethyHyTHORN with f RRN 7R a globally Lipschitz-continuous function It wouldbe interesting to generalize this result to functions f with a superlinear growth atinfinity To do this controls in Lr with r gt N thorn 2 should be built for the nullcontrollability problem

ty Dythorn B Hythorn ay frac14 xthorn v1o in Q


8lt eth65THORN


ORDER REPRINTS

tq Dq H ethDqTHORN thorn cq frac14 y1O in Q


(eth66THORN

qethx 0THORN frac14 0 in O

where a c 2 L1ethQTHORN and BD 2 L1ethQTHORNN Unfortunately the technique introduced inthe present paper cannot be applied in this case because of the lack of regularityintroduced by the term H ethDqTHORN To be precise assume that a null controlvv 2 L2ethQTHORN for (65)ndash(66) with supp vv B0 frac120T has already been obtained (B0 asin Subsec 61) Starting from vv and the associated ethyy qqTHORN the expression of a newcontrol obtained by means of this strategy would be

v frac14 yxthorn 2Hy Hyythorn ethDyTHORNyy Hy ethByyTHORNthorn etht Dthorn athorn B HTHORN 2Hy Hqqthorn ethDyTHORNqqthorn Hy ethDqqTHORNfrac12

Here y 2 DethoTHORN verifies y 1 in B0 Observe that if D 2 L1ethQTHORNN some terms in thisformula are not regular enough to make the state y lie in a suitable space to apply afixed-point argument Thus the problem remains open for such nonlinearities

APPENDIX

A Proof of Proposition 21

Let a 2 L1ethQTHORN F 2 L2ethQTHORN and y be as in Proposition 21 Let V and V0 bearbitrary open sets such that V0 V O Throughout the proof C will be apositive constant whose value may change from one line to another The dependenceof the constant C on O T N r V and V0 which is not used in any essential wayin our analysis will not be specified for the sake of simplicity

We will restrict our attention to the case when N gt 2 the discussion beingsimilar but more direct when N frac14 1 or N frac14 2

(a) Suppose that F 2 Lreth0T LrethVTHORNTHORN with r 2 eth21THORN We will proceed inseveral steps Let us consider a family of open sets fVig0iI such that

V0 frac14 VI VI1 V1 V0 V

where the integer I will be determined laterIn the first place let z0 2 DethVTHORN be a function such that z0 1 in V0 Setting

w0 frac14 z0y it is easy to check that w0 solves the following problem

tw Dw frac14 G in Q

w frac14 0 on S wethx 0THORN frac14 0 in O

(eth67THORN


ORDER REPRINTS

with G frac14 G0 given by

G0 frac14 z0F frac12z0aythorn 2Hz0 Hythorn ethDz0THORNyFrom the regularity assumptions on F and y one has z0F 2 LrethQTHORN and

z0aythorn 2Hz0 Hythorn ethDz0THORNy 2 L2eth0T L2YethOTHORNTHORN L1eth0T L2ethOTHORNTHORN

with 12Z frac14 1

2 1N Using classical interpolation inequalities one obtains

G0 2 Lreth0T Lp0ethOTHORNTHORN with p0 frac14 min r2Nr

Nr 4

and

kG0kLrethLp0 ethOTHORNTHORN Ceth1thorn kak1THORNkFkLrethLrethVTHORNTHORN thorn kykL2ethH2THORNCethH1THORN eth68THORN

Due to the regularizing effect and the properties of the semigroup generated bythe heat equation with Dirichlet boundary conditions (see for instance Giga andSohr 1991 Hieber and Pruss 1997) one deduces that

w0 2 Lreth0T W2p0ethOTHORNTHORN and kw0kLrethW2p0 ethOTHORNTHORN CkG0kLrethLp0 ethOTHORNTHORN eth69THORN

Then recalling that w0jV0frac14 yjV0

one infers from (69) and (68) that

y 2 Lreth0T W2p0ethV0THORNTHORN

and the following estimate is satisfied

kykLrethW2p0 ethV0THORNTHORN Ceth1thorn kak1THORNkFkLrethLrethVTHORNTHORN thorn kykL2ethH2THORNCethH1THORN

If r 2thorn 4=N that is if p0 frac14 r the first point is already proved Let us now supposethat r gt 2thorn 4=N (ie 2 lt p0 frac14 2Nr

Nr4 lt r) We will now use the following Lemmawhich will be proved at the end of this Appendix

Lemma A1 Let a 2 L1ethQTHORN and F 2 L2ethQTHORN Lreth0T LrethVTHORNTHORN with V O an arbi-trary open set and r 2 eth21THORN Let y 2 L2eth0T H2ethOTHORNTHORN Cethfrac120T H1ethOTHORNTHORN satisfy(13) Let o0 and o1 be two open subsets of O such that o1 o0 V Let usassume that y 2 Lreth0T W2r0etho0THORNTHORN with r0 2 2 rfrac12 THORN Then

y 2 Lreth0T W2r1etho1THORNTHORN and ty 2 Lreth0T Lr1etho1THORNTHORN

with

r1 frac14 min r Nr0Nr0

13 if r0 lt N

r if r0 N

(eth70THORN


ORDER REPRINTS

Furthermore the following estimate holds

kykLrethW2r1 etho1THORNTHORN thorn ktykLrethLr1 etho1THORNTHORN Ceth1thorn kak1THORNkFkLrethLrethVTHORNTHORN thorn kykLrethW2r0 etho0THORNTHORN

where C is a positive constant depending on O T N r o0 and o1

We apply this Lemma for i frac14 1 I replacing o0 o1 r0 and r1 respectivelyby Vi1 Vi pi1 and pi with

1

pi

frac14 1

p0 i

Nfor i frac14 1 I 1 and pI frac14 r

This yields y 2 Lreth0T W2piethViTHORNTHORN ty 2 Lreth0T LpiethViTHORNTHORN 1 i I and the corre-sponding estimates In order to determine I observe that we may go on applying

Lemma A1 while pi1 lt N and pi lt r that is to say while i lt Nethr2THORN42r Thus in I

steps with

I frac14 Nethr 2THORN 4

2r

thorn 1

(frac12s being the integer part of the real number s) we have y 2 Xreth0T V0THORN togetherwith estimate (14) with K frac14 N=2thorn 2 which is a uniform bound of I thorn 1

(b) Suppose now that F 2 Lreth0T W1rethVTHORNTHORN r as above and Ha 2 LgethQTHORNN with g given by (15) Let us consider a new open set fVV such that

V0 fVV V

In view of point (a) y 2 Xreth0T fVVTHORN and

kykXreth0T ~VVTHORN C eth1thorn kak1THORNKkFkLrethLrethVTHORNTHORN thorn kykL2ethH2THORNCethH1THORN eth71THORN

To end the proof let us see that in addition

iy 2 Xreth0T V0THORN 1 i N

and that the following estimate is satisfied

kiykXreth0T V0THORN C eth1thorn kak1THORNKthorn1eth1thorn kiakLgTHORNkFkLrethW1rethVTHORNTHORN thorn kykL2ethH2THORNCethH1THORN

where iy denotes the derivative of y with respect to xi 1 i N To do so let us set

wi frac14 z1iy for a fixed i 2 f1 Ng


ORDER REPRINTS

with z1 2 DethfVVTHORN a function such that z1 1 in V0 Then wi solves (67) with G frac14 Gi

given by

Gi frac14 z1iF z1aiy z1yia 2Hz1 H iyeth THORN ethDz1THORNiy eth72THORN

Let us see that Gi 2 LrethQTHORN We will study in detail the term z1yia Underassumptions on y a and F it is direct to see that the other terms in (72) lie in LrethQTHORN

Let us observe that z1y 2 Xr In view of Lemma 22 since Ha 2 LgethQTHORNN with ggiven by (15) and recalling that the Holder space Cll2ethQTHORN is continuously embeddedin C0ethQTHORN for all r 2 eth21THORN one may infer that the term into consideration z1yialies in LrethQTHORN and one has

kz1yiakLr Ckz1ykXrkiakLg

Hence coming back to (72) Gi 2 LrethQTHORN and one can estimate

kGikLr CiFk kLrethLrethVTHORNTHORNthorn kak1 thorn kiakLg thorn 1

kykXreth0T eVTHORN

eth73THORN

The regularizing effect of the heat equation (see Giga and Sohr 1991 Hieber andPruss 1997) yields wi 2 Xr with

kwikXr CkGikLr eth74THORN

Then from (74) (73) and (71) one deduces

kwikXr Ceth1thorn kak1THORNKthorn1 1thorn iak kLg

kFkLrethW1rethVTHORNTHORN thorn kykL2ethH2THORNCethH1THORN

with C frac14 CethOT N rVV0THORN and K frac14 KethNTHORN the same as aboveFinally just taking into account that wi iy in V0 point (b) is proved

Remark 3 Following the previous proof one easily observes that the same resultcan be obtained when replacing hypothesis y 2 L2eth0T H2ethOTHORNTHORN Cethfrac120T H1ethOTHORNTHORN by

y 2 L2eth0T H1locethOTHORNTHORN L1eth0T L2

locethOTHORNTHORN

This fact just affects the number I of steps required to prove point (a)

We end this Appendix by giving the

Proof of Lemma A1 Let us consider a function z 2 Detho0THORN such that z 1 in o1

and let us set u frac14 zy Then u solves (67) with

G frac14 zF frac12zaythorn 2Hz Hythorn ethDzTHORNy


ORDER REPRINTS

The regularity of y and usual Sobolev embeddings give G 2 Lreth0T Lr1ethOTHORNTHORNwhere r1 is given by (70) and the estimate

kGkLrethLr1 ethOTHORNTHORN Ceth1thorn kakTHORN1kFkLrethLrethVTHORNTHORN thorn kykLrethW2r0 etho0THORNTHORN

eth75THORN

(here C depends on the open sets o0 and o1)Then again due to the regularizing properties of the heat equation one deduces

that

u 2 Lreth0T W2r1ethOTHORNTHORN tu 2 Lreth0T Lr1ethOTHORNTHORN

and

kukLrethW2r1 ethOTHORNTHORN thorn ktukLrethLr1 ethOTHORNTHORN CkGkLrethLr1 ethOTHORNTHORN

Finally taking into account that ujo1frac14 yjo1

and inequality (75) the resultfollows

ACKNOWLEDGMENTS

The authors thank the referees for their interesting comments and suggestionsThis work has been partially financed by DGES Spain Grant PB98ndash1134

REFERENCES

Anita S Barbu V (2000) Null controllability of nonlinear convective heatequations ESAIMCOCV 5157ndash173

Barbu V (2000) Exact controllability of the superlinear heat equation Appl MathOptim 42(1)73ndash89

Bodart O Fabre C (1995) Controls insensitizing the norm of the solution of asemilinear heat equation J Math Anal Appl 195(3)658ndash683

Bodart O Gonzalez-Burgos M Perez-Garcıa R (2002) Insensitizing controls fora semilinear heat equation with a superlinear nonlinearity CR Acad SciParis 335(8)677ndash682

Bodart O Gonzalez-Burgos M Perez-Garcıa R (2004) Insensitizing controlsfor a heat equation with a nonlinear term involving the state and the gradientNonlinear Anal 57(5-6)687ndash711

Bodart O Gonzalez-Burgos M Perez-Garcıa R A local result on insensitizingcontrols for a semilinear heat equation with nonlinear boundary Fourierconditions Accepted for publication in SIAM J Control Optim

De Teresa L (2000) Insensitizing controls for a semilinear heat equation CommPartial Differential Equations 25(1ndash2)39ndash72

Doubova A Fernandez-Cara E Gonzalez-Burgos M Zuazua E (2002) On thecontrollability of parabolic systems with a nonlinear term involving the stateand the gradient SIAM J Control Optim 41(3)798ndash819


ORDER REPRINTS

Doubova A Fernandez-Cara E Gonzalez-Burgos M (2004) On the con-trollability of the heat equation with nonlinear boundary Fourier conditionsJ Differential Equations 196(2)385ndash417

Fabre C Puel J-P Zuazua E (1995) Approximate controllability of thesemilinear heat equation Proc R Soc Edinburgh Sect A 125(1)31ndash61

Fernandez-Cara E Zuazua E (2000) Null and approximate controllability forweakly blowing up semilinear heat equations Ann Inst H Poincare AnalNon Lineaire 17(5)583ndash616

Fursikov A Imanuvilov O Yu (1996) Controllability of Evolution EquationsLecture Notes Series 34 Seoul Seoul National University

Giga Y Sohr H (1991) Abstract Lp estimates for the Cauchy problem withapplications to the NavierndashStokes equations in exterior domains J FunctAnal 102(1)72ndash94

Hieber M Pruss J (1997) Heat kernels and maximal Lp-Lq estimates for parabolicevolution equations Comm Partial Differential Equations 22(9ndash10)1647ndash1669

Ladyzenskaya O A Solonnikov V A Uraltzeva N N (1967) Linear and Quasi-linear Equations of Parabolic Type Trans Mathematical Monograph Vol 23Moscow American Mathematical Society

Lions J L (1990) Quelques notions dans lrsquoanalyse et le controle de systemes adonnees incompletes In Proceedings of the XIth Congress on DifferentialEquations and Applications=First Congress on Applied Mathematics MalagaUniversity of Malaga pp 43ndash54

Zuazua E (1991) Exact boundary controllability for the semilinear wave equationIn Nonlinear Partial Differential Equations and Their Applications Vol XPitman Research Notes in Mathematical Series 220 Harlow LongmanScience and Technology pp 357ndash391 (Paris 1987ndash1988)

Received January 2003Revised April 2004


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ORDER REPRINTS

1 INTRODUCTION AND MAIN RESULTS

Problem Formulation

Let O RN N 1 be a bounded connected open set with boundary O 2 C2Let f be a C1 function defined on R Let o and O be two open subsets of O (thoughtto be small in practice) For T gt 0 we denote Q frac14 O eth0TTHORN and S frac14 O eth0TTHORN

We consider a semilinear heat equation with partially known initial condition


y frac14 0 on S yethx 0THORN frac14 y0ethxTHORN thorn t yy0ethxTHORN in O

(eth1THORN

where x 2 L2ethQTHORN is a given heat source y0 2 L2ethOTHORN is a given initial data (althoughby the reasons which will be seen later we will address in this paper the case y0 frac14 0)yy0 2 L2ethOTHORN is unknown with kyy0kL2ethOTHORN frac14 1 t is an unknown small real number andv 2 L2ethQTHORN is a control function to be determined Here 1o is the characteristicfunction of the control set o

Let us define

fethyTHORN frac14 1

2

ZZOeth0TTHORN

jyethx t t vTHORNj2 dx dt eth2THORN

y frac14 yeth t vTHORN being a solution to (1) associated to t and v A control function v issaid to insensitize f if


tfrac140

frac14 0 8yy0 2 L2ethOTHORN with kyy0kL2ethOTHORN frac14 1 eth3THORN

This insensitivity condition means that we seek a control function v acting ono eth0TTHORN such that f is locally insensitive to small perturbations in the initialcondition

In Bodart and Fabre (1995) and Lions (1990) the existence of a control v satisfy-ing (3) is proved to be equivalent to the existence of a control v solving the followingproblem



(eth4THORN

tq Dqthorn f 0ethyTHORNq frac14 y1O in Q


(eth5THORN


Thus the problem of seeking a control that insensitizes f boils down to a non-classical null controllability problem First it is a null controllability problem ofbackward-forward nature for a cascade system of heat equations the first one of


ORDER REPRINTS





tfrac140








ORDER REPRINTS




(eth7THORN




(









(eth8THORN




oeth0TTHORNjjjdx dt

2

eth9THORN


ORDER REPRINTS






Main Results




limjsj1


Let r 2 N

2thorn 11


ZZQ

exp1

t3




ORDER REPRINTS















0






ORDER REPRINTS











Q





u 2 C0ethQTHORN u


Q


lt 1

The following holds




eth13THORN



ORDER REPRINTS











2 thorn 1

(eth15THORN





where






pfrac14 1

r 2

Nthorn2



r




ORDER REPRINTS

3 THE LINEAR CASE




(eth16THORN



(eth17THORN










ZZQ

expCM

t




ORDER REPRINTS

such a way that

kvvkL2 expC

2H

ZZQ

expCM

t

jxj2 dx dt

1=2

eth20THORN





(eth21THORN

and



(eth22THORN



ZZQ

exp CM

t


ZZB0eth0TTHORN

jcj2 dx dt

with




thorn kak2=31





ORDER REPRINTS





B0 B1 B2 B o O




and







(eth26THORN




kykZrthorn kqkZr



ORDER REPRINTS

where









with







and





and





ORDER REPRINTS



and




and



one has





and

ky1kZrthorn ky3kZr




with



ORDER REPRINTS



p






and







kykZrthorn kqkZr

C1 kxkLr thorn expC

2H

exp

CM

2t

x

L2

and


2H

exp

CM

2t

x

L2



ORDER REPRINTS






Let us define


if s 6frac14 0


(


limjsj1









(eth38THORN


ORDER REPRINTS



(eth39THORN




(




C

2Hz

exp

CMz

2t

x

L2

eth41THORN

and


2Hz

exp

CMz

2t

x

L2

eth42THORN

hold where






Q

expCMz

t

jxj2 dx dt frac14

ZZQ

expCMz

t 1

t3

exp

1

t3

jxj2 dx dt

exp CM3=2z

13 ZZQ

exp1

t3





1

2t3

x

L2

eth44THORN


ORDER REPRINTS

and


2t3

x

L2


1

2t3

x

L2

eth45THORN



with




with








hm yi



y2LethzTHORNhm yi l


ORDER REPRINTS



zn z in Zr



hm yi l eth47THORN



1

2t3

x

L2

and


2t3

x

L2


yn y strongly in Zr

and





and




ORDER REPRINTS


supy2LethzTHORN

hm yi hm yi l





1

2t3

x

L2


1

2t3

x

L2



1

2t3

x

L2


a C2 function


42 The General Case









ORDER REPRINTS

and


if s 6frac14 0


(




supjsjM




limjsj1









(eth50THORN



(eth51THORN




ORDER REPRINTS

Moreover estimates


1

2t3

x

L2

eth53THORN

and


2t3

x

L2

eth54THORN

hold where









yn y strongly in Zr

and






ORDER REPRINTS





0









(


k frac14 1

2

ZOrf 2

jDrjr

dx



d

dt

ZOry dx frac14

ZOrDy dx


ZOrxdx

We also set



ORDER REPRINTS





8gtgtltgtgtDefining


z0

2


we can prove that



z0

2








(eth55THORN






8gtltgt eth56THORN


ORDER REPRINTS














lim supjsj1








ORDER REPRINTS


kxkLr thorn expM

2t

x

L2

Z








Q

exp1

t3ethT tTHORN3










(eth59THORN


ORDER REPRINTS

and



8lt eth60THORN



ZZQ

exp CM

tethT tTHORN

jcj2 dx dt C

ZZB0eth0TTHORN







2

ZZB0eth0TTHORN


Q

xcdx dt





lim infkj0k



e


ORDER REPRINTS





ve frac14 ce1B0




ZZQ

expCM

tethT tTHORN



kvvk2L2 C

ZZQ

expCM

tethT tTHORN






kykZrthorn kqkZr


2tethT tTHORN

x

L2

and


2tethT tTHORN

x

L2



ORDER REPRINTS


64 Open Problems







8lt eth65THORN


ORDER REPRINTS



(eth66THORN





APPENDIX








tw Dw frac14 G in Q


(eth67THORN


ORDER REPRINTS




with 12Z frac14 1



Nr 4

and













with


13 if r0 lt N

r if r0 N

(eth70THORN


ORDER REPRINTS





1

pi

frac14 1

p0 i




steps with


2r

thorn 1



V0 fVV V










ORDER REPRINTS


given by







kykXreth0T eVTHORN

eth73THORN









locethOTHORNTHORN







ORDER REPRINTS



eth75THORN


that


and




ACKNOWLEDGMENTS


REFERENCES










ORDER REPRINTS




















ORDER REPRINTS





tfrac140








ORDER REPRINTS




(eth7THORN




(









(eth8THORN




oeth0TTHORNjjjdx dt

2

eth9THORN


ORDER REPRINTS






Main Results




limjsj1


Let r 2 N

2thorn 11


ZZQ

exp1

t3




ORDER REPRINTS















0






ORDER REPRINTS











Q





u 2 C0ethQTHORN u


Q


lt 1

The following holds




eth13THORN



ORDER REPRINTS











2 thorn 1

(eth15THORN





where






pfrac14 1

r 2

Nthorn2



r




ORDER REPRINTS

3 THE LINEAR CASE




(eth16THORN



(eth17THORN










ZZQ

expCM

t




ORDER REPRINTS

such a way that

kvvkL2 expC

2H

ZZQ

expCM

t

jxj2 dx dt

1=2

eth20THORN





(eth21THORN

and



(eth22THORN



ZZQ

exp CM

t


ZZB0eth0TTHORN

jcj2 dx dt

with




thorn kak2=31





ORDER REPRINTS





B0 B1 B2 B o O




and







(eth26THORN




kykZrthorn kqkZr



ORDER REPRINTS

where









with







and





and





ORDER REPRINTS



and




and



one has





and

ky1kZrthorn ky3kZr




with



ORDER REPRINTS



p






and







kykZrthorn kqkZr

C1 kxkLr thorn expC

2H

exp

CM

2t

x

L2

and


2H

exp

CM

2t

x

L2



ORDER REPRINTS






Let us define


if s 6frac14 0


(


limjsj1









(eth38THORN


ORDER REPRINTS



(eth39THORN




(




C

2Hz

exp

CMz

2t

x

L2

eth41THORN

and


2Hz

exp

CMz

2t

x

L2

eth42THORN

hold where






Q

expCMz

t

jxj2 dx dt frac14

ZZQ

expCMz

t 1

t3

exp

1

t3

jxj2 dx dt

exp CM3=2z

13 ZZQ

exp1

t3





1

2t3

x

L2

eth44THORN


ORDER REPRINTS

and


2t3

x

L2


1

2t3

x

L2

eth45THORN



with




with








hm yi



y2LethzTHORNhm yi l


ORDER REPRINTS



zn z in Zr



hm yi l eth47THORN



1

2t3

x

L2

and


2t3

x

L2


yn y strongly in Zr

and





and




ORDER REPRINTS


supy2LethzTHORN

hm yi hm yi l





1

2t3

x

L2


1

2t3

x

L2



1

2t3

x

L2


a C2 function


42 The General Case









ORDER REPRINTS

and


if s 6frac14 0


(




supjsjM




limjsj1









(eth50THORN



(eth51THORN




ORDER REPRINTS

Moreover estimates


1

2t3

x

L2

eth53THORN

and


2t3

x

L2

eth54THORN

hold where









yn y strongly in Zr

and






ORDER REPRINTS





0









(


k frac14 1

2

ZOrf 2

jDrjr

dx



d

dt

ZOry dx frac14

ZOrDy dx


ZOrxdx

We also set



ORDER REPRINTS





8gtgtltgtgtDefining


z0

2


we can prove that



z0

2








(eth55THORN






8gtltgt eth56THORN


ORDER REPRINTS














lim supjsj1








ORDER REPRINTS


kxkLr thorn expM

2t

x

L2

Z








Q

exp1

t3ethT tTHORN3










(eth59THORN


ORDER REPRINTS

and



8lt eth60THORN



ZZQ

exp CM

tethT tTHORN

jcj2 dx dt C

ZZB0eth0TTHORN







2

ZZB0eth0TTHORN


Q

xcdx dt





lim infkj0k



e


ORDER REPRINTS





ve frac14 ce1B0




ZZQ

expCM

tethT tTHORN



kvvk2L2 C

ZZQ

expCM

tethT tTHORN






kykZrthorn kqkZr


2tethT tTHORN

x

L2

and


2tethT tTHORN

x

L2



ORDER REPRINTS


64 Open Problems







8lt eth65THORN


ORDER REPRINTS



(eth66THORN





APPENDIX








tw Dw frac14 G in Q


(eth67THORN


ORDER REPRINTS




with 12Z frac14 1



Nr 4

and













with


13 if r0 lt N

r if r0 N

(eth70THORN


ORDER REPRINTS





1

pi

frac14 1

p0 i




steps with


2r

thorn 1



V0 fVV V










ORDER REPRINTS


given by







kykXreth0T eVTHORN

eth73THORN









locethOTHORNTHORN







ORDER REPRINTS



eth75THORN


that


and




ACKNOWLEDGMENTS


REFERENCES










ORDER REPRINTS




















ORDER REPRINTS




(eth7THORN




(









(eth8THORN




oeth0TTHORNjjjdx dt

2

eth9THORN


ORDER REPRINTS






Main Results




limjsj1


Let r 2 N

2thorn 11


ZZQ

exp1

t3




ORDER REPRINTS















0






ORDER REPRINTS











Q





u 2 C0ethQTHORN u


Q


lt 1

The following holds




eth13THORN



ORDER REPRINTS











2 thorn 1

(eth15THORN





where






pfrac14 1

r 2

Nthorn2



r




ORDER REPRINTS

3 THE LINEAR CASE




(eth16THORN



(eth17THORN










ZZQ

expCM

t




ORDER REPRINTS

such a way that

kvvkL2 expC

2H

ZZQ

expCM

t

jxj2 dx dt

1=2

eth20THORN





(eth21THORN

and



(eth22THORN



ZZQ

exp CM

t


ZZB0eth0TTHORN

jcj2 dx dt

with




thorn kak2=31





ORDER REPRINTS





B0 B1 B2 B o O




and







(eth26THORN




kykZrthorn kqkZr



ORDER REPRINTS

where









with







and





and





ORDER REPRINTS



and




and



one has





and

ky1kZrthorn ky3kZr




with



ORDER REPRINTS



p






and







kykZrthorn kqkZr

C1 kxkLr thorn expC

2H

exp

CM

2t

x

L2

and


2H

exp

CM

2t

x

L2



ORDER REPRINTS






Let us define


if s 6frac14 0


(


limjsj1









(eth38THORN


ORDER REPRINTS



(eth39THORN




(




C

2Hz

exp

CMz

2t

x

L2

eth41THORN

and


2Hz

exp

CMz

2t

x

L2

eth42THORN

hold where






Q

expCMz

t

jxj2 dx dt frac14

ZZQ

expCMz

t 1

t3

exp

1

t3

jxj2 dx dt

exp CM3=2z

13 ZZQ

exp1

t3





1

2t3

x

L2

eth44THORN


ORDER REPRINTS

and


2t3

x

L2


1

2t3

x

L2

eth45THORN



with




with








hm yi



y2LethzTHORNhm yi l


ORDER REPRINTS



zn z in Zr



hm yi l eth47THORN



1

2t3

x

L2

and


2t3

x

L2


yn y strongly in Zr

and





and




ORDER REPRINTS


supy2LethzTHORN

hm yi hm yi l





1

2t3

x

L2


1

2t3

x

L2



1

2t3

x

L2


a C2 function


42 The General Case









ORDER REPRINTS

and


if s 6frac14 0


(




supjsjM




limjsj1









(eth50THORN



(eth51THORN




ORDER REPRINTS

Moreover estimates


1

2t3

x

L2

eth53THORN

and


2t3

x

L2

eth54THORN

hold where









yn y strongly in Zr

and






ORDER REPRINTS





0









(


k frac14 1

2

ZOrf 2

jDrjr

dx



d

dt

ZOry dx frac14

ZOrDy dx


ZOrxdx

We also set



ORDER REPRINTS





8gtgtltgtgtDefining


z0

2


we can prove that



z0

2








(eth55THORN






8gtltgt eth56THORN


ORDER REPRINTS














lim supjsj1








ORDER REPRINTS


kxkLr thorn expM

2t

x

L2

Z








Q

exp1

t3ethT tTHORN3










(eth59THORN


ORDER REPRINTS

and



8lt eth60THORN



ZZQ

exp CM

tethT tTHORN

jcj2 dx dt C

ZZB0eth0TTHORN







2

ZZB0eth0TTHORN


Q

xcdx dt





lim infkj0k



e


ORDER REPRINTS





ve frac14 ce1B0




ZZQ

expCM

tethT tTHORN



kvvk2L2 C

ZZQ

expCM

tethT tTHORN






kykZrthorn kqkZr


2tethT tTHORN

x

L2

and


2tethT tTHORN

x

L2



ORDER REPRINTS


64 Open Problems







8lt eth65THORN


ORDER REPRINTS



(eth66THORN





APPENDIX








tw Dw frac14 G in Q


(eth67THORN


ORDER REPRINTS




with 12Z frac14 1



Nr 4

and













with


13 if r0 lt N

r if r0 N

(eth70THORN


ORDER REPRINTS





1

pi

frac14 1

p0 i




steps with


2r

thorn 1



V0 fVV V










ORDER REPRINTS


given by







kykXreth0T eVTHORN

eth73THORN









locethOTHORNTHORN







ORDER REPRINTS



eth75THORN


that


and




ACKNOWLEDGMENTS


REFERENCES










ORDER REPRINTS




















ORDER REPRINTS






Main Results




limjsj1


Let r 2 N

2thorn 11


ZZQ

exp1

t3




ORDER REPRINTS















0






ORDER REPRINTS











Q





u 2 C0ethQTHORN u


Q


lt 1

The following holds




eth13THORN



ORDER REPRINTS











2 thorn 1

(eth15THORN





where






pfrac14 1

r 2

Nthorn2



r




ORDER REPRINTS

3 THE LINEAR CASE




(eth16THORN



(eth17THORN










ZZQ

expCM

t




ORDER REPRINTS

such a way that

kvvkL2 expC

2H

ZZQ

expCM

t

jxj2 dx dt

1=2

eth20THORN





(eth21THORN

and



(eth22THORN



ZZQ

exp CM

t


ZZB0eth0TTHORN

jcj2 dx dt

with




thorn kak2=31





ORDER REPRINTS





B0 B1 B2 B o O




and







(eth26THORN




kykZrthorn kqkZr



ORDER REPRINTS

where









with







and





and





ORDER REPRINTS



and




and



one has





and

ky1kZrthorn ky3kZr




with



ORDER REPRINTS



p






and







kykZrthorn kqkZr

C1 kxkLr thorn expC

2H

exp

CM

2t

x

L2

and


2H

exp

CM

2t

x

L2



ORDER REPRINTS






Let us define


if s 6frac14 0


(


limjsj1









(eth38THORN


ORDER REPRINTS



(eth39THORN




(




C

2Hz

exp

CMz

2t

x

L2

eth41THORN

and


2Hz

exp

CMz

2t

x

L2

eth42THORN

hold where






Q

expCMz

t

jxj2 dx dt frac14

ZZQ

expCMz

t 1

t3

exp

1

t3

jxj2 dx dt

exp CM3=2z

13 ZZQ

exp1

t3





1

2t3

x

L2

eth44THORN


ORDER REPRINTS

and


2t3

x

L2


1

2t3

x

L2

eth45THORN



with




with








hm yi



y2LethzTHORNhm yi l


ORDER REPRINTS



zn z in Zr



hm yi l eth47THORN



1

2t3

x

L2

and


2t3

x

L2


yn y strongly in Zr

and





and




ORDER REPRINTS


supy2LethzTHORN

hm yi hm yi l





1

2t3

x

L2


1

2t3

x

L2



1

2t3

x

L2


a C2 function


42 The General Case









ORDER REPRINTS

and


if s 6frac14 0


(




supjsjM




limjsj1









(eth50THORN



(eth51THORN




ORDER REPRINTS

Moreover estimates


1

2t3

x

L2

eth53THORN

and


2t3

x

L2

eth54THORN

hold where









yn y strongly in Zr

and






ORDER REPRINTS





0









(


k frac14 1

2

ZOrf 2

jDrjr

dx



d

dt

ZOry dx frac14

ZOrDy dx


ZOrxdx

We also set



ORDER REPRINTS





8gtgtltgtgtDefining


z0

2


we can prove that



z0

2








(eth55THORN






8gtltgt eth56THORN


ORDER REPRINTS














lim supjsj1








ORDER REPRINTS


kxkLr thorn expM

2t

x

L2

Z








Q

exp1

t3ethT tTHORN3










(eth59THORN


ORDER REPRINTS

and



8lt eth60THORN



ZZQ

exp CM

tethT tTHORN

jcj2 dx dt C

ZZB0eth0TTHORN







2

ZZB0eth0TTHORN


Q

xcdx dt





lim infkj0k



e


ORDER REPRINTS





ve frac14 ce1B0




ZZQ

expCM

tethT tTHORN



kvvk2L2 C

ZZQ

expCM

tethT tTHORN






kykZrthorn kqkZr


2tethT tTHORN

x

L2

and


2tethT tTHORN

x

L2



ORDER REPRINTS


64 Open Problems







8lt eth65THORN


ORDER REPRINTS



(eth66THORN





APPENDIX








tw Dw frac14 G in Q


(eth67THORN


ORDER REPRINTS




with 12Z frac14 1



Nr 4

and













with


13 if r0 lt N

r if r0 N

(eth70THORN


ORDER REPRINTS





1

pi

frac14 1

p0 i




steps with


2r

thorn 1



V0 fVV V










ORDER REPRINTS


given by







kykXreth0T eVTHORN

eth73THORN









locethOTHORNTHORN







ORDER REPRINTS



eth75THORN


that


and




ACKNOWLEDGMENTS


REFERENCES










ORDER REPRINTS




















ORDER REPRINTS















0






ORDER REPRINTS











Q





u 2 C0ethQTHORN u


Q


lt 1

The following holds




eth13THORN



ORDER REPRINTS











2 thorn 1

(eth15THORN





where






pfrac14 1

r 2

Nthorn2



r




ORDER REPRINTS

3 THE LINEAR CASE




(eth16THORN



(eth17THORN










ZZQ

expCM

t




ORDER REPRINTS

such a way that

kvvkL2 expC

2H

ZZQ

expCM

t

jxj2 dx dt

1=2

eth20THORN





(eth21THORN

and



(eth22THORN



ZZQ

exp CM

t


ZZB0eth0TTHORN

jcj2 dx dt

with




thorn kak2=31





ORDER REPRINTS





B0 B1 B2 B o O




and







(eth26THORN




kykZrthorn kqkZr



ORDER REPRINTS

where









with







and





and





ORDER REPRINTS



and




and



one has





and

ky1kZrthorn ky3kZr




with



ORDER REPRINTS



p






and







kykZrthorn kqkZr

C1 kxkLr thorn expC

2H

exp

CM

2t

x

L2

and


2H

exp

CM

2t

x

L2



ORDER REPRINTS






Let us define


if s 6frac14 0


(


limjsj1









(eth38THORN


ORDER REPRINTS



(eth39THORN




(




C

2Hz

exp

CMz

2t

x

L2

eth41THORN

and


2Hz

exp

CMz

2t

x

L2

eth42THORN

hold where






Q

expCMz

t

jxj2 dx dt frac14

ZZQ

expCMz

t 1

t3

exp

1

t3

jxj2 dx dt

exp CM3=2z

13 ZZQ

exp1

t3





1

2t3

x

L2

eth44THORN


ORDER REPRINTS

and


2t3

x

L2


1

2t3

x

L2

eth45THORN



with




with








hm yi



y2LethzTHORNhm yi l


ORDER REPRINTS



zn z in Zr



hm yi l eth47THORN



1

2t3

x

L2

and


2t3

x

L2


yn y strongly in Zr

and





and




ORDER REPRINTS


supy2LethzTHORN

hm yi hm yi l





1

2t3

x

L2


1

2t3

x

L2



1

2t3

x

L2


a C2 function


42 The General Case









ORDER REPRINTS

and


if s 6frac14 0


(




supjsjM




limjsj1









(eth50THORN



(eth51THORN




ORDER REPRINTS

Moreover estimates


1

2t3

x

L2

eth53THORN

and


2t3

x

L2

eth54THORN

hold where









yn y strongly in Zr

and






ORDER REPRINTS





0









(


k frac14 1

2

ZOrf 2

jDrjr

dx



d

dt

ZOry dx frac14

ZOrDy dx


ZOrxdx

We also set



ORDER REPRINTS





8gtgtltgtgtDefining


z0

2


we can prove that



z0

2








(eth55THORN






8gtltgt eth56THORN


ORDER REPRINTS














lim supjsj1








ORDER REPRINTS


kxkLr thorn expM

2t

x

L2

Z








Q

exp1

t3ethT tTHORN3










(eth59THORN


ORDER REPRINTS

and



8lt eth60THORN



ZZQ

exp CM

tethT tTHORN

jcj2 dx dt C

ZZB0eth0TTHORN







2

ZZB0eth0TTHORN


Q

xcdx dt





lim infkj0k



e


ORDER REPRINTS





ve frac14 ce1B0




ZZQ

expCM

tethT tTHORN



kvvk2L2 C

ZZQ

expCM

tethT tTHORN






kykZrthorn kqkZr


2tethT tTHORN

x

L2

and


2tethT tTHORN

x

L2



ORDER REPRINTS


64 Open Problems







8lt eth65THORN


ORDER REPRINTS



(eth66THORN





APPENDIX








tw Dw frac14 G in Q


(eth67THORN


ORDER REPRINTS




with 12Z frac14 1



Nr 4

and













with


13 if r0 lt N

r if r0 N

(eth70THORN


ORDER REPRINTS





1

pi

frac14 1

p0 i




steps with


2r

thorn 1



V0 fVV V










ORDER REPRINTS


given by







kykXreth0T eVTHORN

eth73THORN









locethOTHORNTHORN







ORDER REPRINTS



eth75THORN


that


and




ACKNOWLEDGMENTS


REFERENCES










ORDER REPRINTS




















ORDER REPRINTS











Q





u 2 C0ethQTHORN u


Q


lt 1

The following holds




eth13THORN



ORDER REPRINTS











2 thorn 1

(eth15THORN





where






pfrac14 1

r 2

Nthorn2



r




ORDER REPRINTS

3 THE LINEAR CASE




(eth16THORN



(eth17THORN










ZZQ

expCM

t




ORDER REPRINTS

such a way that

kvvkL2 expC

2H

ZZQ

expCM

t

jxj2 dx dt

1=2

eth20THORN





(eth21THORN

and



(eth22THORN



ZZQ

exp CM

t


ZZB0eth0TTHORN

jcj2 dx dt

with




thorn kak2=31





ORDER REPRINTS





B0 B1 B2 B o O




and







(eth26THORN




kykZrthorn kqkZr



ORDER REPRINTS

where









with







and





and





ORDER REPRINTS



and




and



one has





and

ky1kZrthorn ky3kZr




with



ORDER REPRINTS



p






and







kykZrthorn kqkZr

C1 kxkLr thorn expC

2H

exp

CM

2t

x

L2

and


2H

exp

CM

2t

x

L2



ORDER REPRINTS






Let us define


if s 6frac14 0


(


limjsj1









(eth38THORN


ORDER REPRINTS



(eth39THORN




(




C

2Hz

exp

CMz

2t

x

L2

eth41THORN

and


2Hz

exp

CMz

2t

x

L2

eth42THORN

hold where






Q

expCMz

t

jxj2 dx dt frac14

ZZQ

expCMz

t 1

t3

exp

1

t3

jxj2 dx dt

exp CM3=2z

13 ZZQ

exp1

t3





1

2t3

x

L2

eth44THORN


ORDER REPRINTS

and


2t3

x

L2


1

2t3

x

L2

eth45THORN



with




with








hm yi



y2LethzTHORNhm yi l


ORDER REPRINTS



zn z in Zr



hm yi l eth47THORN



1

2t3

x

L2

and


2t3

x

L2


yn y strongly in Zr

and





and




ORDER REPRINTS


supy2LethzTHORN

hm yi hm yi l





1

2t3

x

L2


1

2t3

x

L2



1

2t3

x

L2


a C2 function


42 The General Case









ORDER REPRINTS

and


if s 6frac14 0


(




supjsjM




limjsj1









(eth50THORN



(eth51THORN




ORDER REPRINTS

Moreover estimates


1

2t3

x

L2

eth53THORN

and


2t3

x

L2

eth54THORN

hold where









yn y strongly in Zr

and






ORDER REPRINTS





0









(


k frac14 1

2

ZOrf 2

jDrjr

dx



d

dt

ZOry dx frac14

ZOrDy dx


ZOrxdx

We also set



ORDER REPRINTS





8gtgtltgtgtDefining


z0

2


we can prove that



z0

2








(eth55THORN






8gtltgt eth56THORN


ORDER REPRINTS














lim supjsj1








ORDER REPRINTS


kxkLr thorn expM

2t

x

L2

Z








Q

exp1

t3ethT tTHORN3










(eth59THORN


ORDER REPRINTS

and



8lt eth60THORN



ZZQ

exp CM

tethT tTHORN

jcj2 dx dt C

ZZB0eth0TTHORN







2

ZZB0eth0TTHORN


Q

xcdx dt





lim infkj0k



e


ORDER REPRINTS





ve frac14 ce1B0




ZZQ

expCM

tethT tTHORN



kvvk2L2 C

ZZQ

expCM

tethT tTHORN






kykZrthorn kqkZr


2tethT tTHORN

x

L2

and


2tethT tTHORN

x

L2



ORDER REPRINTS


64 Open Problems







8lt eth65THORN


ORDER REPRINTS



(eth66THORN





APPENDIX








tw Dw frac14 G in Q


(eth67THORN


ORDER REPRINTS




with 12Z frac14 1



Nr 4

and













with


13 if r0 lt N

r if r0 N

(eth70THORN


ORDER REPRINTS





1

pi

frac14 1

p0 i




steps with


2r

thorn 1



V0 fVV V










ORDER REPRINTS


given by







kykXreth0T eVTHORN

eth73THORN









locethOTHORNTHORN







ORDER REPRINTS



eth75THORN


that


and




ACKNOWLEDGMENTS


REFERENCES










ORDER REPRINTS




















ORDER REPRINTS











2 thorn 1

(eth15THORN





where






pfrac14 1

r 2

Nthorn2



r




ORDER REPRINTS

3 THE LINEAR CASE




(eth16THORN



(eth17THORN










ZZQ

expCM

t




ORDER REPRINTS

such a way that

kvvkL2 expC

2H

ZZQ

expCM

t

jxj2 dx dt

1=2

eth20THORN





(eth21THORN

and



(eth22THORN



ZZQ

exp CM

t


ZZB0eth0TTHORN

jcj2 dx dt

with




thorn kak2=31





ORDER REPRINTS





B0 B1 B2 B o O




and







(eth26THORN




kykZrthorn kqkZr



ORDER REPRINTS

where









with







and





and





ORDER REPRINTS



and




and



one has





and

ky1kZrthorn ky3kZr




with



ORDER REPRINTS



p






and







kykZrthorn kqkZr

C1 kxkLr thorn expC

2H

exp

CM

2t

x

L2

and


2H

exp

CM

2t

x

L2



ORDER REPRINTS






Let us define


if s 6frac14 0


(


limjsj1









(eth38THORN


ORDER REPRINTS



(eth39THORN




(




C

2Hz

exp

CMz

2t

x

L2

eth41THORN

and


2Hz

exp

CMz

2t

x

L2

eth42THORN

hold where






Q

expCMz

t

jxj2 dx dt frac14

ZZQ

expCMz

t 1

t3

exp

1

t3

jxj2 dx dt

exp CM3=2z

13 ZZQ

exp1

t3





1

2t3

x

L2

eth44THORN


ORDER REPRINTS

and


2t3

x

L2


1

2t3

x

L2

eth45THORN



with




with








hm yi



y2LethzTHORNhm yi l


ORDER REPRINTS



zn z in Zr



hm yi l eth47THORN



1

2t3

x

L2

and


2t3

x

L2


yn y strongly in Zr

and





and




ORDER REPRINTS


supy2LethzTHORN

hm yi hm yi l





1

2t3

x

L2


1

2t3

x

L2



1

2t3

x

L2


a C2 function


42 The General Case









ORDER REPRINTS

and


if s 6frac14 0


(




supjsjM




limjsj1









(eth50THORN



(eth51THORN




ORDER REPRINTS

Moreover estimates


1

2t3

x

L2

eth53THORN

and


2t3

x

L2

eth54THORN

hold where









yn y strongly in Zr

and






ORDER REPRINTS





0









(


k frac14 1

2

ZOrf 2

jDrjr

dx



d

dt

ZOry dx frac14

ZOrDy dx


ZOrxdx

We also set



ORDER REPRINTS





8gtgtltgtgtDefining


z0

2


we can prove that



z0

2








(eth55THORN






8gtltgt eth56THORN


ORDER REPRINTS














lim supjsj1








ORDER REPRINTS


kxkLr thorn expM

2t

x

L2

Z








Q

exp1

t3ethT tTHORN3










(eth59THORN


ORDER REPRINTS

and



8lt eth60THORN



ZZQ

exp CM

tethT tTHORN

jcj2 dx dt C

ZZB0eth0TTHORN







2

ZZB0eth0TTHORN


Q

xcdx dt





lim infkj0k



e


ORDER REPRINTS





ve frac14 ce1B0




ZZQ

expCM

tethT tTHORN



kvvk2L2 C

ZZQ

expCM

tethT tTHORN






kykZrthorn kqkZr


2tethT tTHORN

x

L2

and


2tethT tTHORN

x

L2



ORDER REPRINTS


64 Open Problems







8lt eth65THORN


ORDER REPRINTS



(eth66THORN





APPENDIX








tw Dw frac14 G in Q


(eth67THORN


ORDER REPRINTS




with 12Z frac14 1



Nr 4

and













with


13 if r0 lt N

r if r0 N

(eth70THORN


ORDER REPRINTS





1

pi

frac14 1

p0 i




steps with


2r

thorn 1



V0 fVV V










ORDER REPRINTS


given by







kykXreth0T eVTHORN

eth73THORN









locethOTHORNTHORN







ORDER REPRINTS



eth75THORN


that


and




ACKNOWLEDGMENTS


REFERENCES










ORDER REPRINTS




















ORDER REPRINTS

3 THE LINEAR CASE




(eth16THORN



(eth17THORN










ZZQ

expCM

t




ORDER REPRINTS

such a way that

kvvkL2 expC

2H

ZZQ

expCM

t

jxj2 dx dt

1=2

eth20THORN





(eth21THORN

and



(eth22THORN



ZZQ

exp CM

t


ZZB0eth0TTHORN

jcj2 dx dt

with




thorn kak2=31





ORDER REPRINTS





B0 B1 B2 B o O




and







(eth26THORN




kykZrthorn kqkZr



ORDER REPRINTS

where









with







and





and





ORDER REPRINTS



and




and



one has





and

ky1kZrthorn ky3kZr




with



ORDER REPRINTS



p






and







kykZrthorn kqkZr

C1 kxkLr thorn expC

2H

exp

CM

2t

x

L2

and


2H

exp

CM

2t

x

L2



ORDER REPRINTS






Let us define


if s 6frac14 0


(


limjsj1









(eth38THORN


ORDER REPRINTS



(eth39THORN




(




C

2Hz

exp

CMz

2t

x

L2

eth41THORN

and


2Hz

exp

CMz

2t

x

L2

eth42THORN

hold where






Q

expCMz

t

jxj2 dx dt frac14

ZZQ

expCMz

t 1

t3

exp

1

t3

jxj2 dx dt

exp CM3=2z

13 ZZQ

exp1

t3





1

2t3

x

L2

eth44THORN


ORDER REPRINTS

and


2t3

x

L2


1

2t3

x

L2

eth45THORN



with




with








hm yi



y2LethzTHORNhm yi l


ORDER REPRINTS



zn z in Zr



hm yi l eth47THORN



1

2t3

x

L2

and


2t3

x

L2


yn y strongly in Zr

and





and




ORDER REPRINTS


supy2LethzTHORN

hm yi hm yi l





1

2t3

x

L2


1

2t3

x

L2



1

2t3

x

L2


a C2 function


42 The General Case









ORDER REPRINTS

and


if s 6frac14 0


(




supjsjM




limjsj1









(eth50THORN



(eth51THORN




ORDER REPRINTS

Moreover estimates


1

2t3

x

L2

eth53THORN

and


2t3

x

L2

eth54THORN

hold where









yn y strongly in Zr

and






ORDER REPRINTS





0









(


k frac14 1

2

ZOrf 2

jDrjr

dx



d

dt

ZOry dx frac14

ZOrDy dx


ZOrxdx

We also set



ORDER REPRINTS





8gtgtltgtgtDefining


z0

2


we can prove that



z0

2








(eth55THORN






8gtltgt eth56THORN


ORDER REPRINTS














lim supjsj1








ORDER REPRINTS


kxkLr thorn expM

2t

x

L2

Z








Q

exp1

t3ethT tTHORN3










(eth59THORN


ORDER REPRINTS

and



8lt eth60THORN



ZZQ

exp CM

tethT tTHORN

jcj2 dx dt C

ZZB0eth0TTHORN







2

ZZB0eth0TTHORN


Q

xcdx dt





lim infkj0k



e


ORDER REPRINTS





ve frac14 ce1B0




ZZQ

expCM

tethT tTHORN



kvvk2L2 C

ZZQ

expCM

tethT tTHORN






kykZrthorn kqkZr


2tethT tTHORN

x

L2

and


2tethT tTHORN

x

L2



ORDER REPRINTS


64 Open Problems







8lt eth65THORN


ORDER REPRINTS



(eth66THORN





APPENDIX








tw Dw frac14 G in Q


(eth67THORN


ORDER REPRINTS




with 12Z frac14 1



Nr 4

and













with


13 if r0 lt N

r if r0 N

(eth70THORN


ORDER REPRINTS





1

pi

frac14 1

p0 i




steps with


2r

thorn 1



V0 fVV V










ORDER REPRINTS


given by







kykXreth0T eVTHORN

eth73THORN









locethOTHORNTHORN







ORDER REPRINTS



eth75THORN


that


and




ACKNOWLEDGMENTS


REFERENCES










ORDER REPRINTS




















ORDER REPRINTS

such a way that

kvvkL2 expC

2H

ZZQ

expCM

t

jxj2 dx dt

1=2

eth20THORN





(eth21THORN

and



(eth22THORN



ZZQ

exp CM

t


ZZB0eth0TTHORN

jcj2 dx dt

with




thorn kak2=31





ORDER REPRINTS





B0 B1 B2 B o O




and







(eth26THORN




kykZrthorn kqkZr



ORDER REPRINTS

where









with







and





and





ORDER REPRINTS



and




and



one has





and

ky1kZrthorn ky3kZr




with



ORDER REPRINTS



p






and







kykZrthorn kqkZr

C1 kxkLr thorn expC

2H

exp

CM

2t

x

L2

and


2H

exp

CM

2t

x

L2



ORDER REPRINTS






Let us define


if s 6frac14 0


(


limjsj1









(eth38THORN


ORDER REPRINTS



(eth39THORN




(




C

2Hz

exp

CMz

2t

x

L2

eth41THORN

and


2Hz

exp

CMz

2t

x

L2

eth42THORN

hold where






Q

expCMz

t

jxj2 dx dt frac14

ZZQ

expCMz

t 1

t3

exp

1

t3

jxj2 dx dt

exp CM3=2z

13 ZZQ

exp1

t3





1

2t3

x

L2

eth44THORN


ORDER REPRINTS

and


2t3

x

L2


1

2t3

x

L2

eth45THORN



with




with








hm yi



y2LethzTHORNhm yi l


ORDER REPRINTS



zn z in Zr



hm yi l eth47THORN



1

2t3

x

L2

and


2t3

x

L2


yn y strongly in Zr

and





and




ORDER REPRINTS


supy2LethzTHORN

hm yi hm yi l





1

2t3

x

L2


1

2t3

x

L2



1

2t3

x

L2


a C2 function


42 The General Case









ORDER REPRINTS

and


if s 6frac14 0


(




supjsjM




limjsj1









(eth50THORN



(eth51THORN




ORDER REPRINTS

Moreover estimates


1

2t3

x

L2

eth53THORN

and


2t3

x

L2

eth54THORN

hold where









yn y strongly in Zr

and






ORDER REPRINTS





0









(


k frac14 1

2

ZOrf 2

jDrjr

dx



d

dt

ZOry dx frac14

ZOrDy dx


ZOrxdx

We also set



ORDER REPRINTS





8gtgtltgtgtDefining


z0

2


we can prove that



z0

2








(eth55THORN






8gtltgt eth56THORN


ORDER REPRINTS














lim supjsj1








ORDER REPRINTS


kxkLr thorn expM

2t

x

L2

Z








Q

exp1

t3ethT tTHORN3










(eth59THORN


ORDER REPRINTS

and



8lt eth60THORN



ZZQ

exp CM

tethT tTHORN

jcj2 dx dt C

ZZB0eth0TTHORN







2

ZZB0eth0TTHORN


Q

xcdx dt





lim infkj0k



e


ORDER REPRINTS





ve frac14 ce1B0




ZZQ

expCM

tethT tTHORN



kvvk2L2 C

ZZQ

expCM

tethT tTHORN






kykZrthorn kqkZr


2tethT tTHORN

x

L2

and


2tethT tTHORN

x

L2



ORDER REPRINTS


64 Open Problems







8lt eth65THORN


ORDER REPRINTS



(eth66THORN





APPENDIX








tw Dw frac14 G in Q


(eth67THORN


ORDER REPRINTS




with 12Z frac14 1



Nr 4

and













with


13 if r0 lt N

r if r0 N

(eth70THORN


ORDER REPRINTS





1

pi

frac14 1

p0 i




steps with


2r

thorn 1



V0 fVV V










ORDER REPRINTS


given by







kykXreth0T eVTHORN

eth73THORN









locethOTHORNTHORN







ORDER REPRINTS



eth75THORN


that


and




ACKNOWLEDGMENTS


REFERENCES










ORDER REPRINTS




















ORDER REPRINTS





B0 B1 B2 B o O




and







(eth26THORN




kykZrthorn kqkZr



ORDER REPRINTS

where









with







and





and





ORDER REPRINTS



and




and



one has





and

ky1kZrthorn ky3kZr




with



ORDER REPRINTS



p






and







kykZrthorn kqkZr

C1 kxkLr thorn expC

2H

exp

CM

2t

x

L2

and


2H

exp

CM

2t

x

L2



ORDER REPRINTS






Let us define


if s 6frac14 0


(


limjsj1









(eth38THORN


ORDER REPRINTS



(eth39THORN




(




C

2Hz

exp

CMz

2t

x

L2

eth41THORN

and


2Hz

exp

CMz

2t

x

L2

eth42THORN

hold where






Q

expCMz

t

jxj2 dx dt frac14

ZZQ

expCMz

t 1

t3

exp

1

t3

jxj2 dx dt

exp CM3=2z

13 ZZQ

exp1

t3





1

2t3

x

L2

eth44THORN


ORDER REPRINTS

and


2t3

x

L2


1

2t3

x

L2

eth45THORN



with




with








hm yi



y2LethzTHORNhm yi l


ORDER REPRINTS



zn z in Zr



hm yi l eth47THORN



1

2t3

x

L2

and


2t3

x

L2


yn y strongly in Zr

and





and




ORDER REPRINTS


supy2LethzTHORN

hm yi hm yi l





1

2t3

x

L2


1

2t3

x

L2



1

2t3

x

L2


a C2 function


42 The General Case









ORDER REPRINTS

and


if s 6frac14 0


(




supjsjM




limjsj1









(eth50THORN



(eth51THORN




ORDER REPRINTS

Moreover estimates


1

2t3

x

L2

eth53THORN

and


2t3

x

L2

eth54THORN

hold where









yn y strongly in Zr

and






ORDER REPRINTS





0









(


k frac14 1

2

ZOrf 2

jDrjr

dx



d

dt

ZOry dx frac14

ZOrDy dx


ZOrxdx

We also set



ORDER REPRINTS





8gtgtltgtgtDefining


z0

2


we can prove that



z0

2








(eth55THORN






8gtltgt eth56THORN


ORDER REPRINTS














lim supjsj1








ORDER REPRINTS


kxkLr thorn expM

2t

x

L2

Z








Q

exp1

t3ethT tTHORN3










(eth59THORN


ORDER REPRINTS

and



8lt eth60THORN



ZZQ

exp CM

tethT tTHORN

jcj2 dx dt C

ZZB0eth0TTHORN







2

ZZB0eth0TTHORN


Q

xcdx dt





lim infkj0k



e


ORDER REPRINTS





ve frac14 ce1B0




ZZQ

expCM

tethT tTHORN



kvvk2L2 C

ZZQ

expCM

tethT tTHORN






kykZrthorn kqkZr


2tethT tTHORN

x

L2

and


2tethT tTHORN

x

L2



ORDER REPRINTS


64 Open Problems







8lt eth65THORN


ORDER REPRINTS



(eth66THORN





APPENDIX








tw Dw frac14 G in Q


(eth67THORN


ORDER REPRINTS




with 12Z frac14 1



Nr 4

and













with


13 if r0 lt N

r if r0 N

(eth70THORN


ORDER REPRINTS





1

pi

frac14 1

p0 i




steps with


2r

thorn 1



V0 fVV V










ORDER REPRINTS


given by







kykXreth0T eVTHORN

eth73THORN









locethOTHORNTHORN







ORDER REPRINTS



eth75THORN


that


and




ACKNOWLEDGMENTS


REFERENCES










ORDER REPRINTS




















ORDER REPRINTS

where









with







and





and





ORDER REPRINTS



and




and



one has





and

ky1kZrthorn ky3kZr




with



ORDER REPRINTS



p






and







kykZrthorn kqkZr

C1 kxkLr thorn expC

2H

exp

CM

2t

x

L2

and


2H

exp

CM

2t

x

L2



ORDER REPRINTS






Let us define


if s 6frac14 0


(


limjsj1









(eth38THORN


ORDER REPRINTS



(eth39THORN




(




C

2Hz

exp

CMz

2t

x

L2

eth41THORN

and


2Hz

exp

CMz

2t

x

L2

eth42THORN

hold where






Q

expCMz

t

jxj2 dx dt frac14

ZZQ

expCMz

t 1

t3

exp

1

t3

jxj2 dx dt

exp CM3=2z

13 ZZQ

exp1

t3





1

2t3

x

L2

eth44THORN


ORDER REPRINTS

and


2t3

x

L2


1

2t3

x

L2

eth45THORN



with




with








hm yi



y2LethzTHORNhm yi l


ORDER REPRINTS



zn z in Zr



hm yi l eth47THORN



1

2t3

x

L2

and


2t3

x

L2


yn y strongly in Zr

and





and




ORDER REPRINTS


supy2LethzTHORN

hm yi hm yi l





1

2t3

x

L2


1

2t3

x

L2



1

2t3

x

L2


a C2 function


42 The General Case









ORDER REPRINTS

and


if s 6frac14 0


(




supjsjM




limjsj1









(eth50THORN



(eth51THORN




ORDER REPRINTS

Moreover estimates


1

2t3

x

L2

eth53THORN

and


2t3

x

L2

eth54THORN

hold where









yn y strongly in Zr

and






ORDER REPRINTS





0









(


k frac14 1

2

ZOrf 2

jDrjr

dx



d

dt

ZOry dx frac14

ZOrDy dx


ZOrxdx

We also set



ORDER REPRINTS





8gtgtltgtgtDefining


z0

2


we can prove that



z0

2








(eth55THORN






8gtltgt eth56THORN


ORDER REPRINTS














lim supjsj1








ORDER REPRINTS


kxkLr thorn expM

2t

x

L2

Z








Q

exp1

t3ethT tTHORN3










(eth59THORN


ORDER REPRINTS

and



8lt eth60THORN



ZZQ

exp CM

tethT tTHORN

jcj2 dx dt C

ZZB0eth0TTHORN







2

ZZB0eth0TTHORN


Q

xcdx dt





lim infkj0k



e


ORDER REPRINTS





ve frac14 ce1B0




ZZQ

expCM

tethT tTHORN



kvvk2L2 C

ZZQ

expCM

tethT tTHORN






kykZrthorn kqkZr


2tethT tTHORN

x

L2

and


2tethT tTHORN

x

L2



ORDER REPRINTS


64 Open Problems







8lt eth65THORN


ORDER REPRINTS



(eth66THORN





APPENDIX








tw Dw frac14 G in Q


(eth67THORN


ORDER REPRINTS




with 12Z frac14 1



Nr 4

and













with


13 if r0 lt N

r if r0 N

(eth70THORN


ORDER REPRINTS





1

pi

frac14 1

p0 i




steps with


2r

thorn 1



V0 fVV V










ORDER REPRINTS


given by







kykXreth0T eVTHORN

eth73THORN









locethOTHORNTHORN







ORDER REPRINTS



eth75THORN


that


and




ACKNOWLEDGMENTS


REFERENCES










ORDER REPRINTS




















ORDER REPRINTS



and




and



one has





and

ky1kZrthorn ky3kZr




with



ORDER REPRINTS



p






and







kykZrthorn kqkZr

C1 kxkLr thorn expC

2H

exp

CM

2t

x

L2

and


2H

exp

CM

2t

x

L2



ORDER REPRINTS






Let us define


if s 6frac14 0


(


limjsj1









(eth38THORN


ORDER REPRINTS



(eth39THORN




(




C

2Hz

exp

CMz

2t

x

L2

eth41THORN

and


2Hz

exp

CMz

2t

x

L2

eth42THORN

hold where






Q

expCMz

t

jxj2 dx dt frac14

ZZQ

expCMz

t 1

t3

exp

1

t3

jxj2 dx dt

exp CM3=2z

13 ZZQ

exp1

t3





1

2t3

x

L2

eth44THORN


ORDER REPRINTS

and


2t3

x

L2


1

2t3

x

L2

eth45THORN



with




with








hm yi



y2LethzTHORNhm yi l


ORDER REPRINTS



zn z in Zr



hm yi l eth47THORN



1

2t3

x

L2

and


2t3

x

L2


yn y strongly in Zr

and





and




ORDER REPRINTS


supy2LethzTHORN

hm yi hm yi l





1

2t3

x

L2


1

2t3

x

L2



1

2t3

x

L2


a C2 function


42 The General Case









ORDER REPRINTS

and


if s 6frac14 0


(




supjsjM




limjsj1









(eth50THORN



(eth51THORN




ORDER REPRINTS

Moreover estimates


1

2t3

x

L2

eth53THORN

and


2t3

x

L2

eth54THORN

hold where









yn y strongly in Zr

and






ORDER REPRINTS





0









(


k frac14 1

2

ZOrf 2

jDrjr

dx



d

dt

ZOry dx frac14

ZOrDy dx


ZOrxdx

We also set



ORDER REPRINTS





8gtgtltgtgtDefining


z0

2


we can prove that



z0

2








(eth55THORN






8gtltgt eth56THORN


ORDER REPRINTS














lim supjsj1








ORDER REPRINTS


kxkLr thorn expM

2t

x

L2

Z








Q

exp1

t3ethT tTHORN3










(eth59THORN


ORDER REPRINTS

and



8lt eth60THORN



ZZQ

exp CM

tethT tTHORN

jcj2 dx dt C

ZZB0eth0TTHORN







2

ZZB0eth0TTHORN


Q

xcdx dt





lim infkj0k



e


ORDER REPRINTS





ve frac14 ce1B0




ZZQ

expCM

tethT tTHORN



kvvk2L2 C

ZZQ

expCM

tethT tTHORN






kykZrthorn kqkZr


2tethT tTHORN

x

L2

and


2tethT tTHORN

x

L2



ORDER REPRINTS


64 Open Problems







8lt eth65THORN


ORDER REPRINTS



(eth66THORN





APPENDIX








tw Dw frac14 G in Q


(eth67THORN


ORDER REPRINTS




with 12Z frac14 1



Nr 4

and













with


13 if r0 lt N

r if r0 N

(eth70THORN


ORDER REPRINTS





1

pi

frac14 1

p0 i




steps with


2r

thorn 1



V0 fVV V










ORDER REPRINTS


given by







kykXreth0T eVTHORN

eth73THORN









locethOTHORNTHORN







ORDER REPRINTS



eth75THORN


that


and




ACKNOWLEDGMENTS


REFERENCES










ORDER REPRINTS




















ORDER REPRINTS



p






and







kykZrthorn kqkZr

C1 kxkLr thorn expC

2H

exp

CM

2t

x

L2

and


2H

exp

CM

2t

x

L2



ORDER REPRINTS






Let us define


if s 6frac14 0


(


limjsj1









(eth38THORN


ORDER REPRINTS



(eth39THORN




(




C

2Hz

exp

CMz

2t

x

L2

eth41THORN

and


2Hz

exp

CMz

2t

x

L2

eth42THORN

hold where






Q

expCMz

t

jxj2 dx dt frac14

ZZQ

expCMz

t 1

t3

exp

1

t3

jxj2 dx dt

exp CM3=2z

13 ZZQ

exp1

t3





1

2t3

x

L2

eth44THORN


ORDER REPRINTS

and


2t3

x

L2


1

2t3

x

L2

eth45THORN



with




with








hm yi



y2LethzTHORNhm yi l


ORDER REPRINTS



zn z in Zr



hm yi l eth47THORN



1

2t3

x

L2

and


2t3

x

L2


yn y strongly in Zr

and





and




ORDER REPRINTS


supy2LethzTHORN

hm yi hm yi l





1

2t3

x

L2


1

2t3

x

L2



1

2t3

x

L2


a C2 function


42 The General Case









ORDER REPRINTS

and


if s 6frac14 0


(




supjsjM




limjsj1









(eth50THORN



(eth51THORN




ORDER REPRINTS

Moreover estimates


1

2t3

x

L2

eth53THORN

and


2t3

x

L2

eth54THORN

hold where









yn y strongly in Zr

and






ORDER REPRINTS





0









(


k frac14 1

2

ZOrf 2

jDrjr

dx



d

dt

ZOry dx frac14

ZOrDy dx


ZOrxdx

We also set



ORDER REPRINTS





8gtgtltgtgtDefining


z0

2


we can prove that



z0

2








(eth55THORN






8gtltgt eth56THORN


ORDER REPRINTS














lim supjsj1








ORDER REPRINTS


kxkLr thorn expM

2t

x

L2

Z








Q

exp1

t3ethT tTHORN3










(eth59THORN


ORDER REPRINTS

and



8lt eth60THORN



ZZQ

exp CM

tethT tTHORN

jcj2 dx dt C

ZZB0eth0TTHORN







2

ZZB0eth0TTHORN


Q

xcdx dt





lim infkj0k



e


ORDER REPRINTS





ve frac14 ce1B0




ZZQ

expCM

tethT tTHORN



kvvk2L2 C

ZZQ

expCM

tethT tTHORN






kykZrthorn kqkZr


2tethT tTHORN

x

L2

and


2tethT tTHORN

x

L2



ORDER REPRINTS


64 Open Problems







8lt eth65THORN


ORDER REPRINTS



(eth66THORN





APPENDIX








tw Dw frac14 G in Q


(eth67THORN


ORDER REPRINTS




with 12Z frac14 1



Nr 4

and













with


13 if r0 lt N

r if r0 N

(eth70THORN


ORDER REPRINTS





1

pi

frac14 1

p0 i




steps with


2r

thorn 1



V0 fVV V










ORDER REPRINTS


given by







kykXreth0T eVTHORN

eth73THORN









locethOTHORNTHORN







ORDER REPRINTS



eth75THORN


that


and




ACKNOWLEDGMENTS


REFERENCES










ORDER REPRINTS




















ORDER REPRINTS






Let us define


if s 6frac14 0


(


limjsj1









(eth38THORN


ORDER REPRINTS



(eth39THORN




(




C

2Hz

exp

CMz

2t

x

L2

eth41THORN

and


2Hz

exp

CMz

2t

x

L2

eth42THORN

hold where






Q

expCMz

t

jxj2 dx dt frac14

ZZQ

expCMz

t 1

t3

exp

1

t3

jxj2 dx dt

exp CM3=2z

13 ZZQ

exp1

t3





1

2t3

x

L2

eth44THORN


ORDER REPRINTS

and


2t3

x

L2


1

2t3

x

L2

eth45THORN



with




with








hm yi



y2LethzTHORNhm yi l


ORDER REPRINTS



zn z in Zr



hm yi l eth47THORN



1

2t3

x

L2

and


2t3

x

L2


yn y strongly in Zr

and





and




ORDER REPRINTS


supy2LethzTHORN

hm yi hm yi l





1

2t3

x

L2


1

2t3

x

L2



1

2t3

x

L2


a C2 function


42 The General Case









ORDER REPRINTS

and


if s 6frac14 0


(




supjsjM




limjsj1









(eth50THORN



(eth51THORN




ORDER REPRINTS

Moreover estimates


1

2t3

x

L2

eth53THORN

and


2t3

x

L2

eth54THORN

hold where









yn y strongly in Zr

and






ORDER REPRINTS





0









(


k frac14 1

2

ZOrf 2

jDrjr

dx



d

dt

ZOry dx frac14

ZOrDy dx


ZOrxdx

We also set



ORDER REPRINTS





8gtgtltgtgtDefining


z0

2


we can prove that



z0

2








(eth55THORN






8gtltgt eth56THORN


ORDER REPRINTS














lim supjsj1








ORDER REPRINTS


kxkLr thorn expM

2t

x

L2

Z








Q

exp1

t3ethT tTHORN3










(eth59THORN


ORDER REPRINTS

and



8lt eth60THORN



ZZQ

exp CM

tethT tTHORN

jcj2 dx dt C

ZZB0eth0TTHORN







2

ZZB0eth0TTHORN


Q

xcdx dt





lim infkj0k



e


ORDER REPRINTS





ve frac14 ce1B0




ZZQ

expCM

tethT tTHORN



kvvk2L2 C

ZZQ

expCM

tethT tTHORN






kykZrthorn kqkZr


2tethT tTHORN

x

L2

and


2tethT tTHORN

x

L2



ORDER REPRINTS


64 Open Problems







8lt eth65THORN


ORDER REPRINTS



(eth66THORN





APPENDIX








tw Dw frac14 G in Q


(eth67THORN


ORDER REPRINTS




with 12Z frac14 1



Nr 4

and













with


13 if r0 lt N

r if r0 N

(eth70THORN


ORDER REPRINTS





1

pi

frac14 1

p0 i




steps with


2r

thorn 1



V0 fVV V










ORDER REPRINTS


given by







kykXreth0T eVTHORN

eth73THORN









locethOTHORNTHORN







ORDER REPRINTS



eth75THORN


that


and




ACKNOWLEDGMENTS


REFERENCES










ORDER REPRINTS




















ORDER REPRINTS



(eth39THORN




(




C

2Hz

exp

CMz

2t

x

L2

eth41THORN

and


2Hz

exp

CMz

2t

x

L2

eth42THORN

hold where






Q

expCMz

t

jxj2 dx dt frac14

ZZQ

expCMz

t 1

t3

exp

1

t3

jxj2 dx dt

exp CM3=2z

13 ZZQ

exp1

t3





1

2t3

x

L2

eth44THORN


ORDER REPRINTS

and


2t3

x

L2


1

2t3

x

L2

eth45THORN



with




with








hm yi



y2LethzTHORNhm yi l


ORDER REPRINTS



zn z in Zr



hm yi l eth47THORN



1

2t3

x

L2

and


2t3

x

L2


yn y strongly in Zr

and





and




ORDER REPRINTS


supy2LethzTHORN

hm yi hm yi l





1

2t3

x

L2


1

2t3

x

L2



1

2t3

x

L2


a C2 function


42 The General Case









ORDER REPRINTS

and


if s 6frac14 0


(




supjsjM




limjsj1









(eth50THORN



(eth51THORN




ORDER REPRINTS

Moreover estimates


1

2t3

x

L2

eth53THORN

and


2t3

x

L2

eth54THORN

hold where









yn y strongly in Zr

and






ORDER REPRINTS





0









(


k frac14 1

2

ZOrf 2

jDrjr

dx



d

dt

ZOry dx frac14

ZOrDy dx


ZOrxdx

We also set



ORDER REPRINTS





8gtgtltgtgtDefining


z0

2


we can prove that



z0

2








(eth55THORN






8gtltgt eth56THORN


ORDER REPRINTS














lim supjsj1








ORDER REPRINTS


kxkLr thorn expM

2t

x

L2

Z








Q

exp1

t3ethT tTHORN3










(eth59THORN


ORDER REPRINTS

and



8lt eth60THORN



ZZQ

exp CM

tethT tTHORN

jcj2 dx dt C

ZZB0eth0TTHORN







2

ZZB0eth0TTHORN


Q

xcdx dt





lim infkj0k



e


ORDER REPRINTS





ve frac14 ce1B0




ZZQ

expCM

tethT tTHORN



kvvk2L2 C

ZZQ

expCM

tethT tTHORN






kykZrthorn kqkZr


2tethT tTHORN

x

L2

and


2tethT tTHORN

x

L2



ORDER REPRINTS


64 Open Problems







8lt eth65THORN


ORDER REPRINTS



(eth66THORN





APPENDIX








tw Dw frac14 G in Q


(eth67THORN


ORDER REPRINTS




with 12Z frac14 1



Nr 4

and













with


13 if r0 lt N

r if r0 N

(eth70THORN


ORDER REPRINTS





1

pi

frac14 1

p0 i




steps with


2r

thorn 1



V0 fVV V










ORDER REPRINTS


given by







kykXreth0T eVTHORN

eth73THORN









locethOTHORNTHORN







ORDER REPRINTS



eth75THORN


that


and




ACKNOWLEDGMENTS


REFERENCES










ORDER REPRINTS




















ORDER REPRINTS

and


2t3

x

L2


1

2t3

x

L2

eth45THORN



with




with








hm yi



y2LethzTHORNhm yi l


ORDER REPRINTS



zn z in Zr



hm yi l eth47THORN



1

2t3

x

L2

and


2t3

x

L2


yn y strongly in Zr

and





and




ORDER REPRINTS


supy2LethzTHORN

hm yi hm yi l





1

2t3

x

L2


1

2t3

x

L2



1

2t3

x

L2


a C2 function


42 The General Case









ORDER REPRINTS

and


if s 6frac14 0


(




supjsjM




limjsj1









(eth50THORN



(eth51THORN




ORDER REPRINTS

Moreover estimates


1

2t3

x

L2

eth53THORN

and


2t3

x

L2

eth54THORN

hold where









yn y strongly in Zr

and






ORDER REPRINTS





0









(


k frac14 1

2

ZOrf 2

jDrjr

dx



d

dt

ZOry dx frac14

ZOrDy dx


ZOrxdx

We also set



ORDER REPRINTS





8gtgtltgtgtDefining


z0

2


we can prove that



z0

2








(eth55THORN






8gtltgt eth56THORN


ORDER REPRINTS














lim supjsj1








ORDER REPRINTS


kxkLr thorn expM

2t

x

L2

Z








Q

exp1

t3ethT tTHORN3










(eth59THORN


ORDER REPRINTS

and



8lt eth60THORN



ZZQ

exp CM

tethT tTHORN

jcj2 dx dt C

ZZB0eth0TTHORN







2

ZZB0eth0TTHORN


Q

xcdx dt





lim infkj0k



e


ORDER REPRINTS





ve frac14 ce1B0




ZZQ

expCM

tethT tTHORN



kvvk2L2 C

ZZQ

expCM

tethT tTHORN






kykZrthorn kqkZr


2tethT tTHORN

x

L2

and


2tethT tTHORN

x

L2



ORDER REPRINTS


64 Open Problems







8lt eth65THORN


ORDER REPRINTS



(eth66THORN





APPENDIX








tw Dw frac14 G in Q


(eth67THORN


ORDER REPRINTS




with 12Z frac14 1



Nr 4

and













with


13 if r0 lt N

r if r0 N

(eth70THORN


ORDER REPRINTS





1

pi

frac14 1

p0 i




steps with


2r

thorn 1



V0 fVV V










ORDER REPRINTS


given by







kykXreth0T eVTHORN

eth73THORN









locethOTHORNTHORN







ORDER REPRINTS



eth75THORN


that


and




ACKNOWLEDGMENTS


REFERENCES










ORDER REPRINTS




















ORDER REPRINTS



zn z in Zr



hm yi l eth47THORN



1

2t3

x

L2

and


2t3

x

L2


yn y strongly in Zr

and





and




ORDER REPRINTS


supy2LethzTHORN

hm yi hm yi l





1

2t3

x

L2


1

2t3

x

L2



1

2t3

x

L2


a C2 function


42 The General Case









ORDER REPRINTS

and


if s 6frac14 0


(




supjsjM




limjsj1









(eth50THORN



(eth51THORN




ORDER REPRINTS

Moreover estimates


1

2t3

x

L2

eth53THORN

and


2t3

x

L2

eth54THORN

hold where









yn y strongly in Zr

and






ORDER REPRINTS





0









(


k frac14 1

2

ZOrf 2

jDrjr

dx



d

dt

ZOry dx frac14

ZOrDy dx


ZOrxdx

We also set



ORDER REPRINTS





8gtgtltgtgtDefining


z0

2


we can prove that



z0

2








(eth55THORN






8gtltgt eth56THORN


ORDER REPRINTS














lim supjsj1








ORDER REPRINTS


kxkLr thorn expM

2t

x

L2

Z








Q

exp1

t3ethT tTHORN3










(eth59THORN


ORDER REPRINTS

and



8lt eth60THORN



ZZQ

exp CM

tethT tTHORN

jcj2 dx dt C

ZZB0eth0TTHORN







2

ZZB0eth0TTHORN


Q

xcdx dt





lim infkj0k



e


ORDER REPRINTS





ve frac14 ce1B0




ZZQ

expCM

tethT tTHORN



kvvk2L2 C

ZZQ

expCM

tethT tTHORN






kykZrthorn kqkZr


2tethT tTHORN

x

L2

and


2tethT tTHORN

x

L2



ORDER REPRINTS


64 Open Problems







8lt eth65THORN


ORDER REPRINTS



(eth66THORN





APPENDIX








tw Dw frac14 G in Q


(eth67THORN


ORDER REPRINTS




with 12Z frac14 1



Nr 4

and













with


13 if r0 lt N

r if r0 N

(eth70THORN


ORDER REPRINTS





1

pi

frac14 1

p0 i




steps with


2r

thorn 1



V0 fVV V










ORDER REPRINTS


given by







kykXreth0T eVTHORN

eth73THORN









locethOTHORNTHORN







ORDER REPRINTS



eth75THORN


that


and




ACKNOWLEDGMENTS


REFERENCES










ORDER REPRINTS




















ORDER REPRINTS


supy2LethzTHORN

hm yi hm yi l





1

2t3

x

L2


1

2t3

x

L2



1

2t3

x

L2


a C2 function


42 The General Case









ORDER REPRINTS

and


if s 6frac14 0


(




supjsjM




limjsj1









(eth50THORN



(eth51THORN




ORDER REPRINTS

Moreover estimates


1

2t3

x

L2

eth53THORN

and


2t3

x

L2

eth54THORN

hold where









yn y strongly in Zr

and






ORDER REPRINTS





0









(


k frac14 1

2

ZOrf 2

jDrjr

dx



d

dt

ZOry dx frac14

ZOrDy dx


ZOrxdx

We also set



ORDER REPRINTS





8gtgtltgtgtDefining


z0

2


we can prove that



z0

2








(eth55THORN






8gtltgt eth56THORN


ORDER REPRINTS














lim supjsj1








ORDER REPRINTS


kxkLr thorn expM

2t

x

L2

Z








Q

exp1

t3ethT tTHORN3










(eth59THORN


ORDER REPRINTS

and



8lt eth60THORN



ZZQ

exp CM

tethT tTHORN

jcj2 dx dt C

ZZB0eth0TTHORN







2

ZZB0eth0TTHORN


Q

xcdx dt





lim infkj0k



e


ORDER REPRINTS





ve frac14 ce1B0




ZZQ

expCM

tethT tTHORN



kvvk2L2 C

ZZQ

expCM

tethT tTHORN






kykZrthorn kqkZr


2tethT tTHORN

x

L2

and


2tethT tTHORN

x

L2



ORDER REPRINTS


64 Open Problems







8lt eth65THORN


ORDER REPRINTS



(eth66THORN





APPENDIX








tw Dw frac14 G in Q


(eth67THORN


ORDER REPRINTS




with 12Z frac14 1



Nr 4

and













with


13 if r0 lt N

r if r0 N

(eth70THORN


ORDER REPRINTS





1

pi

frac14 1

p0 i




steps with


2r

thorn 1



V0 fVV V










ORDER REPRINTS


given by







kykXreth0T eVTHORN

eth73THORN









locethOTHORNTHORN







ORDER REPRINTS



eth75THORN


that


and




ACKNOWLEDGMENTS


REFERENCES










ORDER REPRINTS




















ORDER REPRINTS

and


if s 6frac14 0


(




supjsjM




limjsj1









(eth50THORN



(eth51THORN




ORDER REPRINTS

Moreover estimates


1

2t3

x

L2

eth53THORN

and


2t3

x

L2

eth54THORN

hold where









yn y strongly in Zr

and






ORDER REPRINTS





0









(


k frac14 1

2

ZOrf 2

jDrjr

dx



d

dt

ZOry dx frac14

ZOrDy dx


ZOrxdx

We also set



ORDER REPRINTS





8gtgtltgtgtDefining


z0

2


we can prove that



z0

2








(eth55THORN






8gtltgt eth56THORN


ORDER REPRINTS














lim supjsj1








ORDER REPRINTS


kxkLr thorn expM

2t

x

L2

Z








Q

exp1

t3ethT tTHORN3










(eth59THORN


ORDER REPRINTS

and



8lt eth60THORN



ZZQ

exp CM

tethT tTHORN

jcj2 dx dt C

ZZB0eth0TTHORN







2

ZZB0eth0TTHORN


Q

xcdx dt





lim infkj0k



e


ORDER REPRINTS





ve frac14 ce1B0




ZZQ

expCM

tethT tTHORN



kvvk2L2 C

ZZQ

expCM

tethT tTHORN






kykZrthorn kqkZr


2tethT tTHORN

x

L2

and


2tethT tTHORN

x

L2



ORDER REPRINTS


64 Open Problems







8lt eth65THORN


ORDER REPRINTS



(eth66THORN





APPENDIX








tw Dw frac14 G in Q


(eth67THORN


ORDER REPRINTS




with 12Z frac14 1



Nr 4

and













with


13 if r0 lt N

r if r0 N

(eth70THORN


ORDER REPRINTS





1

pi

frac14 1

p0 i




steps with


2r

thorn 1



V0 fVV V










ORDER REPRINTS


given by







kykXreth0T eVTHORN

eth73THORN









locethOTHORNTHORN







ORDER REPRINTS



eth75THORN


that


and




ACKNOWLEDGMENTS


REFERENCES










ORDER REPRINTS




















ORDER REPRINTS

Moreover estimates


1

2t3

x

L2

eth53THORN

and


2t3

x

L2

eth54THORN

hold where









yn y strongly in Zr

and






ORDER REPRINTS





0









(


k frac14 1

2

ZOrf 2

jDrjr

dx



d

dt

ZOry dx frac14

ZOrDy dx


ZOrxdx

We also set



ORDER REPRINTS





8gtgtltgtgtDefining


z0

2


we can prove that



z0

2








(eth55THORN






8gtltgt eth56THORN


ORDER REPRINTS














lim supjsj1








ORDER REPRINTS


kxkLr thorn expM

2t

x

L2

Z








Q

exp1

t3ethT tTHORN3










(eth59THORN


ORDER REPRINTS

and



8lt eth60THORN



ZZQ

exp CM

tethT tTHORN

jcj2 dx dt C

ZZB0eth0TTHORN







2

ZZB0eth0TTHORN


Q

xcdx dt





lim infkj0k



e


ORDER REPRINTS





ve frac14 ce1B0




ZZQ

expCM

tethT tTHORN



kvvk2L2 C

ZZQ

expCM

tethT tTHORN






kykZrthorn kqkZr


2tethT tTHORN

x

L2

and


2tethT tTHORN

x

L2



ORDER REPRINTS


64 Open Problems







8lt eth65THORN


ORDER REPRINTS



(eth66THORN





APPENDIX








tw Dw frac14 G in Q


(eth67THORN


ORDER REPRINTS




with 12Z frac14 1



Nr 4

and













with


13 if r0 lt N

r if r0 N

(eth70THORN


ORDER REPRINTS





1

pi

frac14 1

p0 i




steps with


2r

thorn 1



V0 fVV V










ORDER REPRINTS


given by







kykXreth0T eVTHORN

eth73THORN









locethOTHORNTHORN







ORDER REPRINTS



eth75THORN


that


and




ACKNOWLEDGMENTS


REFERENCES










ORDER REPRINTS




















ORDER REPRINTS





0









(


k frac14 1

2

ZOrf 2

jDrjr

dx



d

dt

ZOry dx frac14

ZOrDy dx


ZOrxdx

We also set



ORDER REPRINTS





8gtgtltgtgtDefining


z0

2


we can prove that



z0

2








(eth55THORN






8gtltgt eth56THORN


ORDER REPRINTS














lim supjsj1








ORDER REPRINTS


kxkLr thorn expM

2t

x

L2

Z








Q

exp1

t3ethT tTHORN3










(eth59THORN


ORDER REPRINTS

and



8lt eth60THORN



ZZQ

exp CM

tethT tTHORN

jcj2 dx dt C

ZZB0eth0TTHORN







2

ZZB0eth0TTHORN


Q

xcdx dt





lim infkj0k



e


ORDER REPRINTS





ve frac14 ce1B0




ZZQ

expCM

tethT tTHORN



kvvk2L2 C

ZZQ

expCM

tethT tTHORN






kykZrthorn kqkZr


2tethT tTHORN

x

L2

and


2tethT tTHORN

x

L2



ORDER REPRINTS


64 Open Problems







8lt eth65THORN


ORDER REPRINTS



(eth66THORN





APPENDIX








tw Dw frac14 G in Q


(eth67THORN


ORDER REPRINTS




with 12Z frac14 1



Nr 4

and













with


13 if r0 lt N

r if r0 N

(eth70THORN


ORDER REPRINTS





1

pi

frac14 1

p0 i




steps with


2r

thorn 1



V0 fVV V










ORDER REPRINTS


given by







kykXreth0T eVTHORN

eth73THORN









locethOTHORNTHORN







ORDER REPRINTS



eth75THORN


that


and




ACKNOWLEDGMENTS


REFERENCES










ORDER REPRINTS




















ORDER REPRINTS





8gtgtltgtgtDefining


z0

2


we can prove that



z0

2








(eth55THORN






8gtltgt eth56THORN


ORDER REPRINTS














lim supjsj1








ORDER REPRINTS


kxkLr thorn expM

2t

x

L2

Z








Q

exp1

t3ethT tTHORN3










(eth59THORN


ORDER REPRINTS

and



8lt eth60THORN



ZZQ

exp CM

tethT tTHORN

jcj2 dx dt C

ZZB0eth0TTHORN







2

ZZB0eth0TTHORN


Q

xcdx dt





lim infkj0k



e


ORDER REPRINTS





ve frac14 ce1B0




ZZQ

expCM

tethT tTHORN



kvvk2L2 C

ZZQ

expCM

tethT tTHORN






kykZrthorn kqkZr


2tethT tTHORN

x

L2

and


2tethT tTHORN

x

L2



ORDER REPRINTS


64 Open Problems







8lt eth65THORN


ORDER REPRINTS



(eth66THORN





APPENDIX








tw Dw frac14 G in Q


(eth67THORN


ORDER REPRINTS




with 12Z frac14 1



Nr 4

and













with


13 if r0 lt N

r if r0 N

(eth70THORN


ORDER REPRINTS





1

pi

frac14 1

p0 i




steps with


2r

thorn 1



V0 fVV V










ORDER REPRINTS


given by







kykXreth0T eVTHORN

eth73THORN









locethOTHORNTHORN







ORDER REPRINTS



eth75THORN


that


and




ACKNOWLEDGMENTS


REFERENCES










ORDER REPRINTS




















ORDER REPRINTS














lim supjsj1








ORDER REPRINTS


kxkLr thorn expM

2t

x

L2

Z








Q

exp1

t3ethT tTHORN3










(eth59THORN


ORDER REPRINTS

and



8lt eth60THORN



ZZQ

exp CM

tethT tTHORN

jcj2 dx dt C

ZZB0eth0TTHORN







2

ZZB0eth0TTHORN


Q

xcdx dt





lim infkj0k



e


ORDER REPRINTS





ve frac14 ce1B0




ZZQ

expCM

tethT tTHORN



kvvk2L2 C

ZZQ

expCM

tethT tTHORN






kykZrthorn kqkZr


2tethT tTHORN

x

L2

and


2tethT tTHORN

x

L2



ORDER REPRINTS


64 Open Problems







8lt eth65THORN


ORDER REPRINTS



(eth66THORN





APPENDIX








tw Dw frac14 G in Q


(eth67THORN


ORDER REPRINTS




with 12Z frac14 1



Nr 4

and













with


13 if r0 lt N

r if r0 N

(eth70THORN


ORDER REPRINTS





1

pi

frac14 1

p0 i




steps with


2r

thorn 1



V0 fVV V










ORDER REPRINTS


given by







kykXreth0T eVTHORN

eth73THORN









locethOTHORNTHORN







ORDER REPRINTS



eth75THORN


that


and




ACKNOWLEDGMENTS


REFERENCES










ORDER REPRINTS




















ORDER REPRINTS


kxkLr thorn expM

2t

x

L2

Z








Q

exp1

t3ethT tTHORN3










(eth59THORN


ORDER REPRINTS

and



8lt eth60THORN



ZZQ

exp CM

tethT tTHORN

jcj2 dx dt C

ZZB0eth0TTHORN







2

ZZB0eth0TTHORN


Q

xcdx dt





lim infkj0k



e


ORDER REPRINTS





ve frac14 ce1B0




ZZQ

expCM

tethT tTHORN



kvvk2L2 C

ZZQ

expCM

tethT tTHORN






kykZrthorn kqkZr


2tethT tTHORN

x

L2

and


2tethT tTHORN

x

L2



ORDER REPRINTS


64 Open Problems







8lt eth65THORN


ORDER REPRINTS



(eth66THORN





APPENDIX








tw Dw frac14 G in Q


(eth67THORN


ORDER REPRINTS




with 12Z frac14 1



Nr 4

and













with


13 if r0 lt N

r if r0 N

(eth70THORN


ORDER REPRINTS





1

pi

frac14 1

p0 i




steps with


2r

thorn 1



V0 fVV V










ORDER REPRINTS


given by







kykXreth0T eVTHORN

eth73THORN









locethOTHORNTHORN







ORDER REPRINTS



eth75THORN


that


and




ACKNOWLEDGMENTS


REFERENCES










ORDER REPRINTS




















ORDER REPRINTS

and



8lt eth60THORN



ZZQ

exp CM

tethT tTHORN

jcj2 dx dt C

ZZB0eth0TTHORN







2

ZZB0eth0TTHORN


Q

xcdx dt





lim infkj0k



e


ORDER REPRINTS





ve frac14 ce1B0




ZZQ

expCM

tethT tTHORN



kvvk2L2 C

ZZQ

expCM

tethT tTHORN






kykZrthorn kqkZr


2tethT tTHORN

x

L2

and


2tethT tTHORN

x

L2



ORDER REPRINTS


64 Open Problems







8lt eth65THORN


ORDER REPRINTS



(eth66THORN





APPENDIX








tw Dw frac14 G in Q


(eth67THORN


ORDER REPRINTS




with 12Z frac14 1



Nr 4

and













with


13 if r0 lt N

r if r0 N

(eth70THORN


ORDER REPRINTS





1

pi

frac14 1

p0 i




steps with


2r

thorn 1



V0 fVV V










ORDER REPRINTS


given by







kykXreth0T eVTHORN

eth73THORN









locethOTHORNTHORN







ORDER REPRINTS



eth75THORN


that


and




ACKNOWLEDGMENTS


REFERENCES










ORDER REPRINTS




















ORDER REPRINTS





ve frac14 ce1B0




ZZQ

expCM

tethT tTHORN



kvvk2L2 C

ZZQ

expCM

tethT tTHORN






kykZrthorn kqkZr


2tethT tTHORN

x

L2

and


2tethT tTHORN

x

L2



ORDER REPRINTS


64 Open Problems







8lt eth65THORN


ORDER REPRINTS



(eth66THORN





APPENDIX








tw Dw frac14 G in Q


(eth67THORN


ORDER REPRINTS




with 12Z frac14 1



Nr 4

and













with


13 if r0 lt N

r if r0 N

(eth70THORN


ORDER REPRINTS





1

pi

frac14 1

p0 i




steps with


2r

thorn 1



V0 fVV V










ORDER REPRINTS


given by







kykXreth0T eVTHORN

eth73THORN









locethOTHORNTHORN







ORDER REPRINTS



eth75THORN


that


and




ACKNOWLEDGMENTS


REFERENCES










ORDER REPRINTS




















ORDER REPRINTS


64 Open Problems







8lt eth65THORN


ORDER REPRINTS



(eth66THORN





APPENDIX








tw Dw frac14 G in Q


(eth67THORN


ORDER REPRINTS




with 12Z frac14 1



Nr 4

and













with


13 if r0 lt N

r if r0 N

(eth70THORN


ORDER REPRINTS





1

pi

frac14 1

p0 i




steps with


2r

thorn 1



V0 fVV V










ORDER REPRINTS


given by







kykXreth0T eVTHORN

eth73THORN









locethOTHORNTHORN







ORDER REPRINTS



eth75THORN


that


and




ACKNOWLEDGMENTS


REFERENCES










ORDER REPRINTS




















ORDER REPRINTS



(eth66THORN





APPENDIX








tw Dw frac14 G in Q


(eth67THORN


ORDER REPRINTS




with 12Z frac14 1



Nr 4

and













with


13 if r0 lt N

r if r0 N

(eth70THORN


ORDER REPRINTS





1

pi

frac14 1

p0 i




steps with


2r

thorn 1



V0 fVV V










ORDER REPRINTS


given by







kykXreth0T eVTHORN

eth73THORN









locethOTHORNTHORN







ORDER REPRINTS



eth75THORN


that


and




ACKNOWLEDGMENTS


REFERENCES










ORDER REPRINTS




















ORDER REPRINTS




with 12Z frac14 1



Nr 4

and













with


13 if r0 lt N

r if r0 N

(eth70THORN


ORDER REPRINTS





1

pi

frac14 1

p0 i




steps with


2r

thorn 1



V0 fVV V










ORDER REPRINTS


given by







kykXreth0T eVTHORN

eth73THORN









locethOTHORNTHORN







ORDER REPRINTS



eth75THORN


that


and




ACKNOWLEDGMENTS


REFERENCES










ORDER REPRINTS




















ORDER REPRINTS





1

pi

frac14 1

p0 i




steps with


2r

thorn 1



V0 fVV V










ORDER REPRINTS


given by







kykXreth0T eVTHORN

eth73THORN









locethOTHORNTHORN







ORDER REPRINTS



eth75THORN


that


and




ACKNOWLEDGMENTS


REFERENCES










ORDER REPRINTS




















ORDER REPRINTS


given by







kykXreth0T eVTHORN

eth73THORN









locethOTHORNTHORN







ORDER REPRINTS



eth75THORN


that


and




ACKNOWLEDGMENTS


REFERENCES










ORDER REPRINTS




















ORDER REPRINTS



eth75THORN


that


and




ACKNOWLEDGMENTS


REFERENCES










ORDER REPRINTS




















ORDER REPRINTS




















existence of insensitizing controls for a semilinear heat equation with a superlinear nonlinearity

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