regularity of wave-maps in dimension 2 + 1...digital object identiﬁer (doi)...

Digital Object Identifier (DOI) 101007s00220-010-1062-3Commun Math Phys 298 231ndash264 (2010) Communications in

MathematicalPhysics

Regularity of Wave-Maps in Dimension 2 + 1

Jacob Sterbenz1 Daniel Tataru2

1 Department of Mathematics University of California San Diego CA 92093-0112 USAE-mail jsterbenmathucsdedu

2 Department of Mathematics University of California Berkeley CA 94720-3840 USAE-mail tatarumathberkeleyedu

Received 24 July 2009 Accepted 27 December 2009Published online 23 May 2010 ndash copy The Author(s) 2010 This article is published with open accessat Springerlinkcom

Abstract In this article we prove a Sacks-UhlenbeckStruwe type global regularityresult for wave-maps R

2+1 rarr M into general compact target manifolds M

Contents

1 Introduction 23211 The Question of Blowup 23412 The Question of Scattering 236

2 Overview of the Proof 2373 Weighted Energy Estimates for the Wave Equation 2414 Finite Energy Self Similar Wave-Maps 2465 A Simple Compactness Result 2486 Proof of Theorems 1315 249

61 Extension and scaling in the blowup scenario 24962 Extension and scaling in the non-scattering scenario 25163 Elimination of the null concentration scenario 25364 Nontrivial energy in a time-like cone 25565 Propagation of time-like energy concentration 25666 Final rescaling 25767 Concentration scales 25868 The compactness argument 261

References 263

The first author was supported in part by the NSF grant DMS-0701087 The second author was supported in part by the NSF grant DMS-0801261

232 J Sterbenz D Tataru

1 Introduction

In this article we consider large data Wave-Maps from R2+1 into a compact Riemannian

manifold (M m) and we prove that regularity and dispersive bounds persist as longas a soliton-like concentration is absent This is a companion to our concurrent article[27] where the same result is proved under a stronger energy dispersion assumptionsee Theorem 12 below

The set-up we consider is the same as the one in [41] using the extrinsic formulationof the Wave-Maps equation Precisely we consider the target manifold (M m) as anisometrically embedded submanifold of R

N Then we can view the M valued functionsas R

N valued functions whose range is contained in M Such an embedding alwaysexists by Nashrsquos theorem [20] (see also Gromov [8] and Guumlnther [9]) In this context theWave-Maps equation can be expressed in a form which involves the second fundamentalform S of M viewed as a symmetric bilinear form

S T M times T M rarr NM 〈S(X Y ) N 〉 = 〈partX N Y 〉The Cauchy problem for the wave maps equation has the form

φa = minusSabc(φ)partαφbpartαφc (φ1 φN ) = (1a)

(0 x) = 0(x) partt(0 x) = 1(x) (1b)

where the initial data (01) is chosen to obey the constraint

0(x) isin M 1(x) isin T0(x)M x isin R2

In the sequel it will be convenient for us to use the notation [t] = ((t) partt(t))There is a conserved energy for this problem

E[](t) = 1

2

int(|partt|2 + |nablax|2)dx

This is invariant with respect to the scaling that preserves the equation (x t) rarr(λx λt) Because of this we say that the problem is energy critical

The present article contributes to the understanding of finite energy solutions witharbitrarily large initial data a problem which has been the subject of intense investiga-tions for some time now We will not attempt to give a detailed account of the history ofthis subject here Instead we refer the reader to the surveys [40] and [15] and the refer-ences therein The general (ie without symmetry assumptions) small energy problemfor compact targets was initiated in the ground-breaking work of Klainerman-Machedon[1314] and completed by the work of Tao [36] when M is a sphere and Tataru [41]for general isometrically embedded manifolds see also the work of Krieger [17] on thenon-compact hyperbolic plane which was treated from the intrinsic point of view

At a minimum one expects the solutions to belong to the space C(R H1(R2)) capC1(R L2(R2)) However this information does not suffice in order to study the equa-tion and to obtain uniqueness statements Instead a smaller Banach space S sub C(RH1(R2)) cap C1(R L2(R2)) was introduced in [36] modifying an earlier structure in[39] Beside the energy S also contains Strichartz type information in various frequencylocalized contexts The full description of S is not necessary here However we do use

the fact that S cap Linfin is an algebra as well as the Xsb type embedding S sub X1 1

2infin Seeour companion paper [27] for precise definitions The standard small data result is asfollows

Large Data Wave Maps 233

Theorem 11 ([364117]) There is some E0 gt 0 so that for each smooth initial data(01) satisfying E[](0) le E0 there exists a unique global smooth solution =T (01) isin S In addition the above solution operator T extends to a continuousoperator from H1 times L2 to S with

S (01)H1timesL2

We call such solutions ldquostrong finite energy wave-mapsrdquo Furthermore the followingweak Lipschitz stability estimate holds for these solutions for s lt 1 and close to 1

minus C(RH s )capC1(RH sminus1) [0] minus [0]H stimesH sminus1 (2)

Due to the finite speed of propagation the corresponding local result is also valid saywith the initial data in a ball and the solution in the corresponding domain of uniqueness

The aim of the companion [27] to the present work is to provide a conditional Sbound for large data solutions under a weak energy dispersion condition For that wehave introduced the notion of the energy dispersion of a wave map defined on aninterval I

E D[] = supkisinZ

PkLinfintx [I ]

where Pk are spatial Littlewood-Paley projectors at frequency 2k The main result in[27] is as follows

Theorem 12 (Main Theorem in [27]) For each E gt 0 there exist F(E) gt 0 andε(E) gt 0 so that any wave-map in a time interval I with energy E[] le E andenergy dispersion E D[] le ε(E) must satisfy S[I ] le F(E)

In particular this result implies that a wave map with energy E cannot blow-up aslong as its energy dispersion stays below ε(E)

In this article we establish unconditional analogs of the above result in particularsettling the blow-up versus global regularity and scattering question for the large dataproblem We begin with some notations We consider the forward light cone

C = 0 le t lt infin r le tand its subsets

C[t0t1] = t0 le t le t1 r le tThe lateral boundary of C[t0t1] is denoted by partC[t0t1] The time sections of the cone aredenoted by

St0 = t = t0 |x | le tWe also use the translated cones

Cδ = δ le t lt infin r le t minus δas well as the corresponding notations Cδ[t0t1] partCδ[t0t1] and Sδ

t0 for t0 gt δGiven a wave map in C or in a subset C[t0t1] of it we define the energy of on

time sections as

ESt [] = 1

2

intSt

(|partt|2 + |nablax|2)dx


It is convenient to do the computations in terms of the null frame

L = partt + partr L = partt minus partr part = rminus1partθ

We define the flux of between t0 and t1 as

F[t0t1][] =int

partC[t0t1]

(1

4|L|2 +

1

2|part|2

)d A

By standard energy estimates we have the energy conservation relation

ESt1[] = ESt0

[] + F[t0t1][] (3)

This shows that ESt [] is a nondecreasing function of t

11 The Question of Blowup We begin with the blow-up question A standard argu-ment which uses the small data result and the finite speed of propagation shows that ifblow-up occurs then it must occur at the tip of a light-cone where the energy (inside thecone) concentrates After a translation and rescaling it suffices to consider wave maps in the cone C[01] If

limtrarr0

ESt [] le E0

then blow-up cannot occur at the origin due to the local result and in fact it follows that

limtrarr0

ESt [] = 0

Thus the interesting case is when we are in an energy concentration scenario

limtrarr0

ESt [] gt E0 (4)

The main result we prove here is the following

Theorem 13 Let C(01] rarr M be a Cinfin wave map Then exactly one of the follow-ing possibilities must hold

A) There exists a sequence of points (tn xn) isin C[01] and scales rn with

(tn xn) rarr (0 0) lim sup|xn|tn

lt 1 limrn

tn= 0

so that the rescaled sequence of wave-maps

(n)(t x) = (tn + rnt xn + rn x) (5)

converges strongly in H1loc to a Lorentz transform of an entire Harmonic-Map of

nontrivial energy

(infin) R2 rarr M 0 lt (infin) H1(R2) lim

trarr0ESt []


B) For each ε gt 0 there exists 0 lt t0 1 and a wave map extension

R2 times (0 t0] rarr M

with bounded energy

E[] le (1 + ε8) limtrarr0

ESt [] (6)

and energy dispersion

suptisin(0t0]

supkisinZ

(Pk(t)Linfin

x+ 2minuskPkpartt(t)Linfin

x

)le ε (7)

We remark that a nontrivial harmonic map (infin) R2 rarr M cannot have an arbi-

trarily small energy Precisely there are two possibilities Either there are no such har-monic maps (for instance in the case when M is negatively curved see [19]) or thereexists a lowest energy nontrivial harmonic map which we denote by E0(M) gt 0 Fur-thermore a simple computation shows that the energy of any harmonic map will increaseif we apply a Lorentz transformation Hence combining the results of Theorem 13 andTheorem 12 we obtain the following

Corollary 14 (Finite Time Regularity for Wave-Maps) The following statements hold

A) Assume that M is a compact Riemannian manifold so that there are no nontrivialfinite energy harmonic maps (infin) R

2 rarr M Then for any finite energy data[0] R

2 times R2 rarr M times T M for the wave map equation (1) there exists a global

solution isin S(0 T ) for all T gt 0 In addition this global solution retains anyadditional regularity of the initial data

B) Let π M rarr M be a Riemannian covering with M compact and such thatthere are no nontrivial finite energy harmonic maps (infin) R

2 rarr M If [0] R

2 times R2 rarr M times T M is Cinfin then there is a global Cinfin solution to M with this

dataC) Suppose that there exists a lowest energy nontrivial harmonic map into M with

energy E0(M) Then for any data [0] R2 times R

2 rarr M times T M for the wave mapequation (1) with energy below E0(M) there exists a global solution isin S(0 T )

for all T gt 0

We remark that the statement in part B) is a simple consequence of A) and restrictingthe projection π to a sufficiently small section St of a cone where one expects blowupof the original map into M In particular since this projection is regular by part A) itsimage lies in a simply connected set for sufficiently small t Thus this projection canbe inverted to yield regularity of the original map close to the suspected blowup pointBecause of this trivial reduction we work exclusively with compact M in the sequelIt should be remarked however that as a (very) special case of this result one obtainsglobal regularity for smooth Wave-Maps into all hyperbolic spaces H

n which has beena long-standing and important conjecture in geometric wave equations due to its relationwith problems in general relativity (see Chapter 16 of [2])

The statement of Corollary 14 in its full generality was known as the ThresholdConjecture Similar results were previously established for the Wave-Map problem viasymmetry reductions in the works [42630] and [29] General results of this type as


well as fairly strong refinements have been known for the Harmonic-Map heat-flowfor some time (see [28] and [22]) As with the heat flow Theorem 13 does not pre-vent the formation of multiple singularities on top of each other To the contrary suchbubble-trees are to be expected (see [42])

Finally we remark that this result is sharp In the case of M = S2 there exists a lowest

energy nontrivial harmonic map namely the stereographic projection Q The results in[16] assert that blow-up with a rescaled Q profile can occur for initial data with energyarbitrarily close to E[Q] We also refer the reader to [23] for blow-up results near higherenergy harmonic maps

12 The Question of Scattering Next we consider the scattering problem for whichwe start with a finite energy wave map in R

2 times [0infin) and consider its behavior ast rarr infin Here by scattering we simply mean the fact that isin S if that is the case thenthe structure theorem for large energy Wave-Maps in [27] shows that behaves at infinas a linear wave after an appropriate renormalization

We can select a ball B so that outside B the energy is small EBc [] lt 110 E0 Then

outside the influence cone of B the solution behaves like a small data wave mapHence it remains to study it within the influence cone of B After scaling and transla-tion it suffices to work with wave maps in the outgoing cone C[1infin) which have finiteenergy ie

limtrarrinfin ESt [] lt infin (8)

We prove the following result

Theorem 15 Let C[1infin) rarr M be a Cinfin wave map which satisfies (8) Thenexactly one of the following possibilities must hold

A) There exists a sequence of points (tn xn) isin C[1infin) and scales rn with

tn rarr infin lim sup|xn|tn

lt 1 limrn

tn= 0


(n)(t x) = (tn + rnt xn + rn x) (9)


nontrivial energy


trarrinfin ESt []B) For each ε gt 0 there exists t0 gt 1 and a wave map extension

R2 times [t0infin) rarr M

with bounded energy

E[] le (1 + ε8) limtrarrinfin ESt [] (10)


suptisin[t0infin)

supkisinZ

(Pk(t)Linfin


x

)le ε (11)


In case B) Theorem 12 then implies that scattering holds as t rarr infin Thus if scatteringdoes not hold then we must be in case A) As a corollary it follows that scattering canonly fail for wave-maps whose energy satisfies

E[] ge E0(M) (12)

Thus Corollary 14 can be strengthened to

Corollary 16 (Scattering for Large Data Wave-Maps) The following statementshold

A) Assume that there are no nontrivial finite energy harmonic maps (infin) R2 rarr M

Then for any finite energy data [0] for the wave map equation (1) there exists aglobal solution isin S

B) Suppose that there exists a lowest energy nontrivial harmonic map with energyE0(M) Then for any data [0] for the wave map equation (1) with energy belowE0(M) there exists a global solution isin S

Ideally one would also like to have a constructive bound of the form

S le F(E[])This does not seem to follow directly from our results Furthermore our results do notseem to directly imply scattering for non-compact targets in the absence of harmonicmaps (only scattering of the projection) Results similar to Corollary 16 were previouslyestablished in spherically symmetric and equivariant cases see [3] and [5]

Finally we would like to remark that results similar in spirit to the ones of this paperand [27] have been recently announced In the case where M = H

n the hyperbolicspaces globally regularity and scattering follows from the program of Tao [3731ndash3335] and [34] In the case where the target M is a negatively curved Riemann surfaceKrieger and Schlag [18] provide global regularity and scattering via a modification ofthe Kenig-Merle method [12] which uses as a key component suitably defined Bahouri-Gerard [1] type decompositions

2 Overview of the Proof

The proofs of Theorem 13 and Theorem 15 are almost identical The three main build-ing blocks of both proofs are (i) weighted energy estimates (ii) elimination of finiteenergy self-similar solutions and (iii) a compactness result

Our main energy estimates are established in Sect 3 Beside the standard energybounds involving the partt vector field we also use the vector field

X0 = 1

ρ(tpartt + rpartr ) ρ =

radict2 minus r2 (13)

as well as its time translates This leads to a family of weighted energy estimates see (26)below which has appeared in various guises in the literature The first such referencewe are aware of is the work of Grillakis [7] Our approach is closest to the work of Tao[37] and [31] (see also Chap 63 of [38]) These bounds are also essentially identicalto the ldquorigidity estimaterdquo of Kenig-Merle [12] It should be noted that estimates of this


type are probably the only generally useful time-like component concentration boundspossible for non-symmetric wave equations and they will hold for any Lagrangian fieldequation on (2 + 1) Minkowski space

Next we introduce a general argument to rule out the existence of finite energy self-similar solutions to (1) Such results are essentially standard in the literature (eg see thesection on wave-maps in [25]) but we take some care here to develop a version whichapplies to the setup of our work This crucially uses the energy estimates developed inSect 3 as well as a boundary regularity result of J Qing for harmonic maps (see [21])

The compactness result in Proposition 51 proved in Sect 5 allows us to producethe strongly convergent subsequence of wave maps in case A) of Theorems 13 15 Itapplies to local sequences (n) of small energy wave maps with the additional propertythat X(n) rarr 0 in L2 for some time-like vector field X This estimate uses only thestandard small energy theory of [41] and is completely independent of the more involvedregularity result in our companion paper [27]

Given these three building blocks the proof of Theorems 13 and 15 presented inSect 6 proceeds as follows

Step 1 (Extension and scaling) We assume that part B) of Theorem 13 respectivelyTheorem 15 does not hold for a wave map and for some ε gt 0 We construct anextension of as in part B) satisfying (6) respectively (10) Then the energy dispersionrelation (7) respectively (11) must fail Thus we can find sequences tn xn and kn sothat

|Pkn (tn xn)| + 2minuskn |Pkn partt(tn xn)| gt ε

with tn rarr 0 in the case of Theorem 13 respectively tn rarr infin in the case of Theorem 15In addition the flux-energy relation

F[t1t2][] = ESt2[] minus ESt1

[]shows that in the case of Theorem 13 we have

limt1t2rarr0

F[t1t2][] = 0

and in the case of Theorem 15 we have

limt1t2rarrinfin F[t1t2][] = 0

This allows us also to choose εn rarr 0 such that

F[εn tn tn ][] le ε12n E[]

Rescaling to t = 1 we produce the sequence of wave maps

(n)(t x) = (tnt tn x)

in the increasing regions C[εn 1] so that

F[εn 1][(n)] le ε12n E[]

and also points xn isin R2 and frequencies kn isin Z so that

|Pkn (n)(1 xn)| + 2minuskn |Pkn partt

(n)(1 xn)| gt ε (14)


From this point on the proofs of Theorems 1315 are identical

Step 2 (Elimination of null concentration scenario) Using the fixed time portion of theX0 energy bounds we eliminate the case of null concentration

|xn| rarr 1 kn rarr infinin estimate (14) and show that the sequence of maps (n) at time t = 1 must eitherhave low frequency concentration in the range

m(ε E) lt kn lt M(ε E) |xn| lt R(ε E)

or high frequency concentration strictly inside the cone

kn ge M(ε E) |xn| lt γ (ε E) lt 1

Step 3 (Time-like energy concentration) In both remaining cases above we show that anontrivial portion of the energy of (n) at time 1 must be located inside a smaller cone

1

2

intt=1|x |ltγ1

(|partt

(n)|2 + |nablax(n)|2

)dx ge E1

where E1 = E1(ε E) and γ1 = γ1(ε E) lt 1

Step 4 (Uniform propagation of non-trivial time-like energy) Using again the X0 energybounds we propagate the above time-like energy concentration for (n) from time 1 to

smaller times t isin [ε12n ε

14n ]

1

2

int|x |ltγ2(εE)t

(|partt


)dx ge E0(ε E) t isin [ε

12n ε

14n ]

At the same time we obtain bounds for X0(n) outside smaller and smaller neighbor-

hoods of the cone namelyint

Cεn

[ε12

n ε

14

n ]

ρminus1|X0(n)|2dxdt 1

Step 5 (Final rescaling) By a pigeonhole argument and rescaling we end up producinganother sequence of maps denoted still by (n) which are sections of the original wave

map and are defined in increasing regions C[1Tn ] Tn = e| ln εn | 12 and satisfy the

following three properties

ESt [(n)] asymp E t isin [1 Tn] (Bounded Energy)

ES

(1minusγ2)tt

[(n)] ge E2 t isin [1 Tn] (Nontrivial Time-like Energy)int intC

ε

12

n[1Tn ]

1

ρ|X0

(n)|2dxdt | log εn|minus 12 (Decay to Self-similar Mode)


Step 6 (Isolating the concentration scales) The compactness result in Proposition 51only applies to wave maps with energy below the threshold E0 in the small data resultThus we need to understand on which scales such concentration can occur Using sev-eral additional pigeonholing arguments we show that one of the following two scenariosmust occur

i) Energy Concentration On a subsequence there exist (tn xn) rarr (t0 x0) with

(t0 x0) inside C12

[ 12 infin)

and scales rn rarr 0 so that we have

EB(xn rn)[(n)](tn) = 1

10E0

EB(xrn)[(n)](tn) le 1

10E0 x isin B(x0 r)

rminus1n

int tn+rn2

tnminusrn2

intB(x0r)

|X0(n)|2dxdt rarr 0

ii) Non-concentration For each j isin N there exists an r j gt 0 such that for every (t x)

inside C j = C1[1infin) cap 2 j lt t lt 2 j+1 one has

EB(xr j )[(n)](t) le 1

10E0 forall(t x) isin C j

ES

(1minusγ2)tt

[(n)](t) ge E2int intC j

|X0(n)|2dxdt rarr 0

uniformly in nStep 7 (The compactness argument) In case i) above we consider the rescaled wave-maps

(n)(t x) = (n)(tn + rnt xn + rn x)

and show that on a subsequence they converge locally in the energy norm to a finiteenergy nontrivial wave map in R

2 times [minus 12 1

2 ] which satisfies X (t0 x0) = 0 Thus must be a Lorentz transform of a nontrivial harmonic map

In case ii) above we show directly that the sequence (n) converges locally on asubsequence in the energy norm to finite energy nontrivial wave map defined inthe interior of a translated cone C2

[2infin) which satisfies X0 = 0 Consequently inhyperbolic coordinates we may interpret as a nontrivial harmonic map

H2 rarr M

Compactifying this and using conformal invariance we obtain a non-trivial finite energyharmonic map

D2 rarr M

from the unit disk D2 which according to the estimates of Sect 3 obeys the additional

weighted energy bound intD2

|nablax|2 dx

1 minus rlt infin

But such maps do not exist via combination of a theorem of Qing [21] and a theorem ofLemaire [19]


3 Weighted Energy Estimates for the Wave Equation

In this section we prove the main energy decay estimates The technique we use is thestandard one of contracting the energy-momentum tensor

Tαβ [] = mi j ()

[partαφipartβφ j minus 1

2gαβ partγ φipartγ φ j

] (15)

with well chosen vector-fields Here = (φ1 φn) is a set of local coordinates onthe target manifold (M m) and (gαβ) stands for the Minkowski metric The main twoproperties of Tαβ [] are that it is divergence free nablaαTαβ = 0 and also that it obeys thepositive energy condition T (X Y ) 0 whenever both g(X X) 0 and g(Y Y ) 0This implies that contracting Tαβ [] with timelikenull vector-fields will result in goodenergy estimates on characteristic and space-like hypersurfaces

If X is some vector-field we can form its associated momentum density (ie itsNoether current)

(X)Pα = Tαβ []Xβ

This one form obeys the divergence rule

nablaα(X)Pα = 1

2Tαβ [](X)παβ (16)

where (X)παβ is the deformation tensor of X

(X)παβ = nablaα Xβ + nablaβ Xα

A simple computation shows that one can also express

(X)π = LX g

This latter formulation is very convenient when dealing with coordinate derivativesRecall that in general one has

(LX g)αβ = X (gαβ) + partα(Xγ )gγβ + partβ(Xγ )gαγ

Our energy estimates are obtained by integrating the relation (16) over cones Cδ[t1t2]Then from (16) we obtain for δ le t1 le t2

intSδ

t2

(X)P0 dx +1

2

int intCδ[t1t2]

Tαβ [](X)παβ dxdt =int

Sδt1

(X)P0 dx +int

partCδ[t1t2]

(X)PL d A (17)

where d A is an appropriately normalized (Euclidean) surface area element on the lateralboundary of the cone r = t minus δ

The standard energy estimates come from contracting Tαβ [] with Y = partt Then wehave

(Y )π = 0 (Y )P0 = 1

2(|partt|2 + |nablax|2) (Y )PL = 1

4|L|2 +

1

2|part|2


Applying (17) over C[t1t2] we obtain the energy-flux relation (3) used in the IntroductionApplying (17) over Cδ[δ1] yields

intpartCδ[δ1]

1

4|L|2 +

1

2|part|2 d A le E[] (18)

It will also be necessary for us to have a version of the usual energy estimate adaptedto the hyperboloids ρ = radic

t2 minus r2 = const Integrating the divergence of the (Y )Pα

momentum density over regions of the form R = ρ ρ0 t t0 we haveintρ=ρ0captt0

(Y )PαdVα le E[] (19)

where the integrand on the LHS denotes the interior product of (Y )P with the Minkow-ski volume element To express this estimate in a useful way we use the hyperboliccoordinates (CMC foliation)

t = ρ cosh(y) r = ρ sinh(y) θ = (20)

In this system of coordinates the Minkowski metric becomes

minus dt2 + dr2 + r2dθ2 = minusdρ2 + ρ2(

dy2 + sinh2(y)d2) (21)

A quick calculation shows that the contraction on line (19) becomes the one-form

(Y )PαdVα = T (partρ partt )ρ2d AH2 d AH2 = sinh(y)dyd (22)

The area element d AH2 is that of the hyperbolic plane H2 To continue we note that

partt = t

ρpartρ minus r

ρ2 party

so in particular

T (partρ partt ) = cosh(y)

2|partρ|2 minus sinh(y)

ρpartρ middot party

+cosh(y)

2ρ2

(|party|2 +

1

sinh2(y)|part|2

)

Letting t0 rarr infin in (19) we obtain a useful consequence of this namely a weightedhyperbolic space estimate for special solutions to the wave-map equations which willbe used in the sequel to rule out the existence of non-trivial finite energy self-similarsolutions

Lemma 31 Let be a finite energy smooth wave-map in the interior of the cone CAssume also that partρ equiv 0 Then one has

1

2

intH2

|nablaH2|2 cosh(y)d AH2 le E[] (23)

Here

|nablaH2|2 = |party|2 +1

sinh2(y)|part|2

is the covariant energy density for the hyperbolic metric


Our next order of business is to obtain decay estimates for time-like components ofthe energy density For this we use the timelikenull vector-field

Xε = 1

ρε

((t + ε)partt + rpartr ) ρε =radic

(t + ε)2 minus r2 (24)

In order to gain some intuition we first consider the case of X0 This is most readilyexpressed in the system of hyperbolic coordinates (20) introduced above One easilychecks that the coordinate derivatives turn out to be

partρ = X0 party = rpartt + tpartr

In particular X0 is uniformly timelike with g(X0 X0) = minus1 and one should expect it togenerate a good energy estimate on time slices t = const In the system of coordinates(20) one also has that

LX0 g = 2ρ(

dy2 + sinh2(y)d2)

Raising this one then computes

(X0)παβ = 2

ρ3

(party otimes party + sinhminus2(y)part otimes part

)

Therefore we have the contraction identity

1

2Tαβ [](X0)παβ = 1

ρ|X0|2

To compute the components of (X0)P0 and (X0)PL we use the associated optical functions

u = t minus r v = t + r

Notice that ρ2 = uv Also simple calculations show that

X0 = 1ρ

( 12vL + 1

2 uL) partt = 1

2L +

1

2L (25)

Finally we record here the components of Tαβ [] in the null frame

T (L L) = |L|2 T (L L) = |L|2 T (L L) = |part|2By combining the above calculations we see that we may compute

(X0)P0 = T (partt X0) = 1

4

(v

u

) 12 |L|2 +

1

4

[(v

u

) 12

+(u

v

) 12]

|part|2 +1

4

(u

v

) 12 |L|2

(X0)PL = T (L X0) = 1

2

(v

u

) 12 |L|2 +

1

2

(u

v

) 12 |part|2

These are essentially the same as the components of the usual energy currents (partt )P0 and(partt )PL modulo ratios of the optical functions u and v

One would expect to get nice space-time estimates for X0 by integrating (16) overthe interior cone r t 1 The only problem is that the boundary terms degeneraterather severely when ρ rarr 0 To avoid this we simply redo everything with the shifted


version Xε from line (24) The above formulas remain valid with u v replaced by theirtime shifted versions

uε = (t + ε) minus r vε = (t + ε) + r

Furthermore notice that for small t one has in the region r t the bounds

(vε

uε

) 12 asymp 1

(uε

vε

) 12 asymp 1 0 lt t le ε

Therefore one has in r t that

(Xε )P0 asymp (partt )P0 0 lt t le ε

In what follows we work with a wave-map in C[ε1] We denote its total energyand flux by

E = ES1 [] F = F[ε1][]In the limiting case F = 0 ε = 0 one could apply (17) to obtain

intS0

t2

(Xε )P0 dx +int int

C0[t1t2]

1

ρε

|Xε|2 dxdt =int

S0t1

(Xε )P0 dx

By (26) letting t1 rarr 0 followed by ε rarr 0 and taking the supremum over 0 lt t2 1we would get the model estimate

suptisin(01]

intS0

t

(X0)P0 dx +int int

C0[01]

1

ρ|X0|2 dxdt le E

However here we need to deal with a small nonzero flux Observing that

(Xε )PL εminus 12 (partt )PL

from (17) we obtain the weaker boundint

S0t2


C0[t1t2]

1

ρε

|Xε|2 dxdt int

S0t1

(Xε )P0 dx + εminus 12 F

Letting t1 = ε and taking supremum over ε t2 1 we obtain

suptisin(ε1]

intS0

t


C0[01]

1

ρε

|Xε|2 dxdt E + εminus 12 F (26)

A consequence of this is the following which will be used to rule out the case of asymp-totically null pockets of energy

Lemma 32 Let be a smooth wave-map in the cone C(ε1] which satisfies the flux-

energy relation F ε12 E Then

intS0

1

(Xε )P0 dx E (27)


Next we can replace Xε by X0 in (26) if we restrict the integrals on the left tor lt t minus ε In this region we have

(Xε )P0 asymp (X0)P0 ρε asymp ρ

Using the second member above a direct computation shows that in r lt t minus ε

1

ρ|X0|2 1

ρε

|Xε|2 +ε2

ρ3 |partt|2

and also

intCε

(ε1]

ε2

ρ3 |partt|2dxdt leint

Cε(ε1]

ε12

t32

|partt|2dxdt E

Thus we have proved the following estimate which will be used to conclude that rescal-ing of are asymptotically stationary and also used to help trap uniformly time-likepockets of energy


energy relation F ε12 E Then we have

suptisin(ε1]

intSε

t

(X0)P0 dx +int int

Cε[ε1]

1

ρ|X0|2dxdt E (28)

Finally we use the last lemma to propagate pockets of energy forward away fromthe boundary of the cone By (17) for X0 we have

intSδ

1

(X0)P0 dx leint

Sδt0

(X0)P0 dx +int

partCδ[t01]

(X0)PL d A ε le δ lt t0 lt 1

We consider the two components of (X0)PL separately For the angular component by(18) we have the bound

intpartCδ[t01]

(u

v

) 12 |part|2 d A

(δ

t0

) 12int

partCδ[t01]|part|2 d A

(δ

t0

) 12

E

For the L component a direct computation shows that

|L| (u

v

) 12 |X0| +

(u

v

)|L|

Thus we obtain

intSδ

1

(X0)P0 dx int

Sδt0

(X0)P0 dx +

(δ

t0

) 12

E +int

partCδ[t01]

((u

v

) 12 |X0φ|2 +

(u

v

) 32 |L|2

)d A

For the last term we optimize with respect to δ isin [δ0 δ1] to obtain



energy relation F ε12 E Suppose that ε le δ0 δ1 le t0 le 1 Then

intS

δ11

(X0)P0 dx int

Sδ0t0

(X0)P0 dx +

((δ1

t0

) 12

+ (ln(δ1δ0))minus1

)E (29)

To prove this lemma it suffices to choose δ isin [δ0 δ1] so that

intpartCδ[t01]

[(u

v

) 12 |X0φ|2 +

(u

v

) 32 |L|2

]d A | ln(δ1δ0)|minus1 E

This follows by pigeonholing the estimate

intC

δ0[t01]Cδ1[t01]

1

u

[(u

v

) 12 |X0φ|2 +

(u

v

) 32 |L|2

]dxdt E

The first term is estimated directly by (28) For the second we simply use energy boundssince in the domain of integration we have the relation

1

u

(u

v

) 32 le δ

121

t32

4 Finite Energy Self Similar Wave-Maps

The purpose of this section is to prove the following theorem

Theorem 41 (Absence of non-trivial finite energy self similar wave-maps in 2D)Let be a finite energy1 solution to the wave-map equation (1) defined in the forwardcone C Suppose also that partρ equiv 0 Then equiv const

Remark 42 Theorems of this type are standard in the literature The first such referencewe are aware of is in the work of ShatahndashTahvildar-Zadeh [26] on the equivariant caseThis was later extended by Shatah-Struwe [25] to disprove the existence of (initially)smooth self-similar profiles However the authors are not aware of an explicit referencein the literature ruling out the possibility of general (ie non-symmetric) finite energyself-similar solutions to the system (1) although this statement has by now acquired thestatus of a folk-lore theorem It is this latter version in the above form that is necessaryin the context of the present work

Remark 43 It is important to remark that the finite energy assumption cannot be droppedand also that this failure is not due to interior regularity That is there are non-trivial Cinfinself similar solutions to (1) in C but these solutions all have infinite energy Howeverthe energy divergence is marginal ie the energy in Sδ

t only grows as | ln(δ)| as δ rarr 0Further these solutions have finite energy when viewed as harmonic maps in H

2

1 That is to say a weak solution of (1) (in the sense of [10]) such that ess sup0ltt nablatx (t) L2(|x |ltt) lt infin


Proof of Theorem 41 Writing out the system (1) in the coordinates (20) and cancelingthe balanced factor of ρminus2 on both sides we have that obeys

H2a = minusgi jH2Sa

bc()partibpart j

c

where in polar coordinates gH2 = dy2 + sinh2(y)d2 is the standard hyperbolic metricThus a is an entire harmonic map H

2 rarr M By elliptic regularity such a mapis smooth with uniform bounds on compact sets (see [10]) In particular is Cinfin inthe interior of the cone C Therefore from estimate (23) of Lemma 31 enjoys theadditional weighted energy estimate

intH2

|nablaH2|2 cosh(y)d AH2 le 2E() =int

St

(|partt|2 + |nablax|2

)dx (30)

for any fixed t gt 0To proceed further it is convenient to rephrase all of the above in terms of the confor-

mal compactification of H2 Using the pseudo-spherical stereographic projection from

hyperboloids to the unit disk D2 = t = 0 cap x2 + y2 lt 1 in Minkowski space (see

Chap II of [6])

π(t x1 x2) = (minus1 0 0) minus 2ρ(t + ρ x1 x2)lang(t + ρ x1 x2) (t + ρ x1 x2)

rang

as well as the conformal invariance of the 2D harmonic map equation and its associ-ated Dirichlet energy we have from (30) that induces a finite energy harmonic map D

2 rarr M with the additional property that

1

2

intD2

|nablax|2(

1 + r2

1 minus r2

)dx lt infin (31)

To conclude we only need to show that such maps are trivial Notice that the weight(1 minus r)minus1 is critical in this respect for there are many non-trivial2 finite energy har-monic maps from D

2 rarr M regardless of the curvature of M which are also uniformlysmooth up to the boundary partD

2 and may therefore absorb an energy weight of the form(1 minus r)minusα with α lt 1

From the bound (31) we have that there exists a sequence of radii rn 1 such thatint

r=rn

|nabla|2dl = on(1)

From the uniform boundedness of (r) in L2(S1) and the trace theorem this impliesthat

|partD2 L2(S1) lt infin |partD2 H

12 (S1)

= 0

so therefore |partD2 equiv const In particular D2 rarr M is a finite energy harmonic

map with smooth boundary values By a theorem of J Qing (see [21]) it follows that

2 For example if M is a complex surface with conformal metric m = λ2dd then the harmonic mapequation becomes partzpartz = minus2part(ln λ)partzpartz Thus any holomorphic function D

2 rarr M suffices toproduce an infinite energy self similar ldquoblowup profilerdquo for (1) See Chap I of [24]


has uniform regularity up to partD2 Thus is Cinfin on the closure D2 with constant bound-

ary value By Lemairersquos uniqueness theorem (see Theorem 823 of [11] and originally[19]) equiv const throughout D

2 By inspection of the coordinates (20) and the fact thatpartρ equiv 0 this easily implies that is trivial in all of C

5 A Simple Compactness Result

The main aim of this section is the following result

Proposition 51 Let Q be the unit cube and let (n) be a family of wave maps in 3Qwhich have small energy E[(n)] le 1

10 E0 and so that

X(n)L2(3Q) rarr 0 (32)

for some smooth time-like vector field X Then there exists a wave map isin H32 minusε(Q)

with E[] le 110 E0 so that on a subsequence we have the strong convergence

(n) rarr in H1(Q)

Proof The argument we present here is inspired by the one in Struwersquos work onHarmonic-Map heat flow (see [28]) The main point there is that compactness is gainedthrough the higher regularity afforded via integration in time For a parabolic equationintegrating L2 in time actually gains a whole derivative by scaling so in the heat flowcase one can control a quantity of the form

int int |nabla2|2dxdt which leads to control ofint |nabla2|2dx for many individual points in timeBy the small data result in [41] we have a similar (uniform) space-time bound (n)

in any compact subset K of 3Q which gains us 12 a derivative over energy namely

χ(n) X

1 12infin

1 supp χ sub 3Q (33)

where X1 1

2infin here denotes the infin Besov version of the critical inhomogeneous Xsb spaceWe obtain a strongly converging H1

tx sequence of wave-maps through a simple fre-quency decomposition argument as follows The vector field X is timelike therefore itssymbol is elliptic in a region of the form τ gt (1 minus 2δ)|ξ | with δ gt 0 Hence givena cutoff function χ supported in 3Q and with symbol equal to 1 in 2Q there exists amicrolocal space-time decomposition

χ = Qminus1(Dxt x t)X + Q0(Dxt x t) + R(Dxt x t)

where the symbols qminus1 isin Sminus1 and q0 isin S0 are supported in 3Q times τ gt (1 minus 2δ)|ξ |respectively 3Q times τ lt (1 minus δ)|ξ | while the remainder R has symbol r isin Sminusinfin withspatial support in 3Q This yields a decomposition for χ(n)

χ(n) = (Q0(D x) + R(D x))(n) + Qminus1(D x)X(n) = (n)bulk + Rn (34)

Due to the support properties of q0 for the main term we have the bound

(n)bulk

H32 minusε

tx

χ(n) X

1 12infin

1


On the other hand the remainder decays in norm

Rn H1tx

X(n)L2(3Q) minusrarrnrarrinfin 0

Hence on a subsequence we have the strong convergence

χ(n) rarr in H1tx (3Q)

In addition must satisfy both

isin H32 minusε X = 0 in 2Q

It remains to show that is a wave-map in Q (in fact a ldquostrongrdquo finite energy wave-mapaccording to the definition of Theorem 11) There exists a time section 2Qt0 close tothe center of 2Q such that both

[t0] (H1timesL2)(2Qt0 ) lt infin (n)[t0] minus [t0] (H stimesH sminus1)(2Qt0 ) rarr 0

for some s lt 1 Letting be the solution to (1) with data [t0] from the weak stabilityresult (2) in Theorem 11 we have (n) rarr in Hs

x (Qt ) at fixed time for s lt 1 Thus = in Hs

tx (Q) which suffices We consider now the two cases we are interested in namely when X = partt or

X = tpartt + xpartx If X = partt (as will be the case for a general time-like X vector af-ter boosting) then is a harmonic map

Q rarr Mand is therefore smooth (see [10])

If X = tpartt + xpartx and 3Q is contained within the cone t gt |x | then can beinterpreted as a portion of a self-similar Wave-Map and therefore it is a harmonic mapfrom a domain

H2 supe rarr M

and is again smooth Note that since the Harmonic-Map equation is conformally invari-ant one could as well interpret this as a special case of the previous one However inthe situation where we have similar convergence on a large number of such domains

that fill up H2 will be globally defined as an H

2 harmonic map to M and we willtherefore be in a position to apply Theorem 41

6 Proof of Theorems 1315

We proceed in a series of steps

61 Extension and scaling in the blowup scenario We begin with Theorem 13 Let

be a wave map in C(01] with terminal energy

E = limtrarr0

ESt []

Suppose that the energy dispersion scenario B) does not apply Let ε gt 0 be so thatB) does not hold We can choose ε arbitrarily small We will take advantage of this to


construct an extension of outside the cone which satisfies (6) therefore violating (7)on any time interval (0 t0]

For this we use energy estimates Setting

Ft0 [] =int

partC(0t0]

(1

4|L|2 +

1

2|part|2

)d A

it follows that as t0 rarr 0 we have

Ft0 [] = ESt0[] minus E rarr 0 (35)

Then by pigeonholing we can choose t0 arbitrarily small so that we have the bounds

Ft0 [] ε8 E

intpartSt0

|part|2ds ε8

t0E (36)

The second bound allows us to extend the initial data for at time t0 from St0 to all ofR

2 in such a way that

E[](t0) minus ESt0[] ε8 E

We remark that by scaling it suffices to consider the case t0 = 1 The second bound in(36) shows by integration that the range of restricted to partSt0 is contained in a smallball of size ε8 in M Thus the extension problem is purely local in M and can be carriedout in a suitable local chart by a variety of methods

We extend the solution outside the cone C between times t0 and 0 by solving thewave-map equation By energy estimates it follows that for t isin (0 t0] we have

E[](t) minus ESt [] = E[](t0) minus ESt0[] + Ft0 [] minus Ft [] le 1

2ε8 E (37)

Hence the energy stays small outside the cone and by the small data result in Theo-rem 11 there is no blow-up that occurs outside the cone up to time 0 The extension wehave constructed is fixed for the rest of the proof

Since our extension satisfies (6) but B) does not hold it follows that we can find asequence (tn xn) with tn rarr 0 and kn isin Z so that

|Pkn (tn xn)| + 2minusk |Pkn partt(tn xn)| ge ε (38)

The relation (35) shows that Ftn rarr 0 Hence we can find a sequence εn rarr 0 such that

Ftn lt ε12n E Thus

F[εntn tn ] lt ε12n E (39)

Rescaling we obtain the sequence of wave maps


in the increasing family of regions R2 times [εn 1] with the following properties

a) Uniform energy size

E[(n)] asymp E (40)


b) Small energy outside the cone

E[(n)] minus ESt [(n)] ε8 E (41)

c) Decaying flux

F[εn 1][(n)] lt ε12n E (42)

d) Pointwise concentration at time 1


(n)(1 xn)| gt ε (43)

for some xn isin R2 kn isin Z

62 Extension and scaling in the non-scattering scenario Next we consider the case ofTheorem 15 Again we suppose that the energy dispersion scenario B) does not applyfor a finite energy wave map in C[1infin) Let ε gt 0 be so that B) does not hold Setting

E = limt0rarrinfin ESt0

[] Ft0 [] =int

partC[t0infin)

(1

4|L|2 +

1

2|part|2

)d A

it follows that

Ft0 [] = ESinfin[] minus ESt0[] rarr 0 (44)

We choose t0 gt 1 so that

Ft0 [] le ε8 E

We obtain our extension of to the interval [t0infin) from the following lemma

Lemma 61 Let be a finite energy wave-map in C[1infin) and E t0 as above Then thereexists a wave-map extension of to R

2 times [t0infin) which has energy E

We remark that as t rarr infin all the energy of the extension moves inside the cone It islikely that an extension with this property is unique We do not pursue this here as it isnot needed

Proof By pigeonholing we can choose a sequence tk rarr infin so that we have the bounds

Ftk [] rarr 0 tk

intpartStk

|part|2d A rarr 0

The second bound allows us to obtain an extension (k)[tk] of the initial data [tk] for at time tk from the circle Stk to all of R


E[(k)](tk) minus EStk[] rarr 0

By rescaling this extension problem is equivalent to the one in the case of Theorem 13


We solve the wave map equation backwards from time tk to time t0 with data (k)[tk]We obtain a wave map (k) in the time interval [t0 tk] which coincides with in C[t0tk ]The above relation shows that

E[(k)] rarr E (45)

By energy estimates it also follows that for large k and t isin [t0 tk] we have

E[(k)](t) minus ESt [(k)] = E[(k)](tk) minus EStk[] + F[ttk ][] ε8 E (46)

Hence the energy stays small outside the cone which by the small data result inTheorem 11 shows that no blow-up can occur outside the cone between times tk and t0

We will obtain the extension of as the strong limit in the energy norm of a subse-quence of the (k)

= limkrarrinfin (k) in C(t0infin H1) cap C1(t0infin L2) (47)

We begin with the existence of a weak limit By uniform boundedness and the Banach-Alaoglu theorem we have weak convergence for each fixed time

(k)[t] [t] weakly in H1 times L2

Within C[t0infin) all the (k)rsquos coincide so the above convergence is only relevantoutside the cone But by (46) outside the cone all (k) have small energy This placesus in the context of the results in [41] Precisely the weak stability bound (2) in Theo-rem 11 and an argument similar to that used in Sect 5 shows that the limit is a regularfinite energy wave-map in [t0infin) times R

2It remains to upgrade the convergence On one hand (45) and weak convergence

shows that E[] le ESinfin[] On the other hand E[] ge ESt [] and the latter convergesto ESinfin[] Thus we obtain

E[] = ESinfin[] = limkrarrinfin E[(k)]

From weak convergence and norm convergence we obtain strong convergence

(k)[t] rarr [t] in H1 times L2

The uniform convergence in (47) follows by applying the energy continuity in the smalldata result of Theorem 11 outside C[t0infin)

By energy estimates for it follows that

E[] minus ESt [] = Ft [] le Ft0 [] le ε8 E t isin [t0infin)

Hence the extended satisfies (10) so it cannot satisfy (11) By (35) we obtain thesequence (tn xn) and kn isin Z with tn rarr infin so that (38) holds On the other hand fromthe energy-flux relation we have


This allows us to select again εn rarr 0 so that (39) holds clearly in this case we must alsohave εntn rarr infin The same rescaling as in the previous subsection leads to a sequence(n) of wave maps which satisfy the conditions a)ndashd) above


63 Elimination of the null concentration scenario Due to (41) the contribution of theexterior of the cone to the pointwise bounds for (n) is negligible Precisely for lowfrequencies we have the pointwise bound

|Pk(n)(1 x)| + 2minusk |Pkpartt

(n)(1 x)| (

2k(1 + 2k |x |)minusN + ε4)

E12 k le 0

where the first term contains the contribution from the interior of the cone and the secondis the outside contribution On the other hand for large frequencies we similarly obtain

|Pk(n)(1 x)| + 2k |Pkpartt

(n)(1 x)| ((1 + 2k(|x | minus 1)+)minusN + ε4

)E

12 k ge 0

(48)

Hence in order for (43) to hold kn must be large enough

2kn gt m(ε E)

while xn cannot be too far outside the cone

|xn| le 1 + 2minuskn g(ε E)

This allows us to distinguish three cases

(i) Wide pockets of energy This is when

2kn lt C lt infin

(ii) Sharp time-like pockets of energy This is when

2kn rarr infin |xn| le γ lt 1

(iii) Sharp pockets of null energy This is when

2kn rarr infin |xn| rarr 1

Our goal in this subsection is to eliminate the last case Precisely we will show that

Lemma 62 There exists M = M(ε E) gt 0 and γ = γ (ε E) lt 1 so that for anywave map (n) as in (40)ndash(43) with a small enough εn we have

2kn gt M rArr |xn| le γ (49)

Proof We apply the energy estimate (27) with ε replaced by εn to (n) in the timeinterval [εn 1] This yields

intS1

(1 minus |x | + εn)minus12

(|L(n)|2 + |part(n)|2

)dx E (50)

The relation (49) would follow from the pointwise concentration bound in (43) if wecan prove that for k gt 0 the bound (50) together with (40) and (41) at time t = 1 implythe pointwise estimate


(n)(1 x)| ((

(1 minus |x |)+ + 2minusk + εn

) 18

+ ε2)

E12 (51)


In view of (48) it suffices to prove this bound in the case where 12 lt |x | lt 2 After a

rotation we can assume that x = (x1 0) with 12 lt x1 lt 2

In general we have

(1 minus |x |)+ le (1 minus x1)+ + x22

and therefore from (50) we obtainintS1

1

(|1 minus x1| + x22 + εn)

12

(|L(n)|2 + |part(n)|2

)dx E (52)

At the spatial location (1 0) we have part(n) = part2 so we obtain the rough bound

|part2(n)| |part(n)| + (|x2| + |1 minus x1|)|nablax

(n)|Similarly

|partt(n) minus part1

(n)| |L(n)| + (|x2| + |1 minus x1|)|nablax(n)|

Hence taking into account (40) and (41) from (52) we obtainintt=1

1

((1 minus x1)+ + x22 + εn)

12 + ε8

(|partt

(n) minus part1(n)|2 + |part2

(n)|2)

dx E (53)

Next given a dyadic frequency 2k ge 1 we consider an angular parameter 2minus k2 le

θ le 1 and we rewrite the multiplier Pk in the form

Pk = P1kθ part1 + P2

kθ part2

where P1kθ and P2

kθ are multipliers with smooth symbols supported in the sets |ξ | asymp2k |ξ2| 2kθ respectively |ξ | asymp 2k |ξ2| 2kθ The size of their symbols is givenby

|p1kθ (ξ)| 2minusk |p2

kθ (ξ)| 1

|ξ2|

Therefore they satisfy the L2 rarr Linfin bounds

P1kθL2rarrLinfin θ

12 P2

kθL2rarrLinfin θminus 12 (54)

In addition the kernels of both P1kθ and P2

kθ decay rapidly on the 2minusk times θminus12minusk scaleThus one can add weights in (54) provided that they are slowly varying on the samescale The weight in (52) is not necessarily slowly varying but we can remedy this byslightly increasing the denominator to obtain the weaker boundint

t=1

1

((1 minus x1)+ + x22 + 2minusk + εn)

12 + ε8

(|partt

(n)minus part1(n)|2 + |part2

(n)|2)

dx E (55)

Using (40) (55) and the weighted version of (54) we obtain

|Pk(n)(1 x)| le|P1

kθ part1(n)(1 x)| + |P2

kθ part2(n)(1 x)|

(θ

12 + θminus 1

2

(((1 minus x1)+ + x2

2 + 2minusk + εn)14 + ε4

))E

12


We set x2 = 0 and optimize with respect to θ isin [2minusk2 1] to obtain the desired bound(51) for (n)

|Pk(n)(1 x1 0)|

(((1 minus x1)+ + 2minusk + εn

) 18

+ ε2)

E12

1

2lt x1 lt 2

A similar argument yields

2minusk |Pkpart1(n)(1 x1 0)|


) 18

+ ε2)

E12

1

2lt x1 lt 2

On the other hand from (55) we directly obtain

2minusk |Pk(partt minus part1)(n)(1 x1 0)|


) 18

+ ε2)

E12

1

2lt x1 lt2

Combined with the previous inequality this yields the bound in (51) for partt(n)

64 Nontrivial energy in a time-like cone According to the previous step the pointsxn and frequencies kn in (43) satisfy one of the following two conditions


c(ε E) lt 2kn lt C(ε E)


2kn gt C(ε E) |xn| le γ (ε E) lt 1

Using only the bounds (40) and (41) we will prove that there existsγ1 = γ1(ε E) lt 1and E1 = E1(ε E) gt 0 so that in both cases there is some amount of uniform time-likeenergy concentration

1

2

intt=1|x |ltγ1

(|partt


)dx ge E1 (56)

For convenience we drop the index n in the following computations Denote

E(γ1) = 1

2

intt=1|x |ltγ1


)dx

where γ1 isin (γ +1

2 1) will be chosen later Here γ is from line (49) of the previous sub-section Thus at time 1 the function has energy E(γ1) in |x | lt γ1 energy le Ein γ1 lt |x | lt 1 and energy le ε8 E outside the unit disc Then we obtain differentpointwise estimates for Pk[1] in two main regimes

(a) 2k(1 minus γ1) lt 1 Then we obtain

Pk(1)Linfinx

+ 2minuskPkpartt(1)Linfinx

E(γ1)12 +

((2k(1 minus γ1))

12 + ε4

)E

12

with a further improvement if both


(a1) 2k(1 minus γ1) lt 1 lt 2k(1 minus γ ) and |x | lt γ namely

|Pk(1 x)| + 2minusk |Pkpartt(1 x)| E(γ1)12 +

((2k(1 minus γ1))

12 (2k(1 minus γ ))minusN + ε4

)E

12

(b) 2k(1 minus γ1) ge 1 Then


((2k(1 minus γ ))minusN + ε4

)E

12 |x | lt γ

We use these estimates to bound E(γ1) from below We first observe that if 1 minus γ1 issmall enough then case (i) above implies we are in regime (a) and from (43) we obtain

ε E(γ1)12 +

((C(ε E)(1 minus γ1))

12 + ε4

)E

12

which gives a bound from below for E(γ1) if 1 minus γ1 and ε are small enoughConsider now the remaining case (ii) above If we are in regime (a) but not (a1) then

the bound in (a) combined with (43) gives

ε E(γ1)12 +

((1 minus γ1)

12

(1 minus γ )12

+ ε4

)E

12

which suffices if 1 minus γ1 is small enough If we are in regime (a1) then we obtain exactlythe same inequality directly Finally if we are in regime (b) then we achieve an evenbetter bound

ε E(γ1)12 +

((1 minus γ1

1 minus γ

)N

+ ε4

)E

12

Thus (56) is proved in all cases for a small enough 1 minus γ1

65 Propagation of time-like energy concentration Here we use the flux relation (39)to propagate the time-like energy concentration in (56) uniformly to smaller times t isin[ε

12n ε

14n ] Precisely we show that there exists γ2 = γ2(ε E) lt 1 and E2 = E2(ε E) gt 0

so that

1

2

int|x |ltγ2t

(|partt


)dx ge E2 t isin [ε

12n ε

14n ] (57)

At the same time we also obtain uniform weighted L2tx bounds for X0

(n) outsidesmaller and smaller neighborhoods of the cone namely

intCεn

[ε12

n ε

14

n ]

ρminus1|X0(n)|2dxdt E (58)


The latter bound (58) is a direct consequence of (28) so we turn our attention to

(57) Given a parameter γ2 = γ2(γ1 E1) and any t0 isin [ε12n ε

14n ] we define δ0 and δ1

according to

(1 minus γ2)t0 = δ0 δ1 le t0

We apply Proposition 34 to (n) with this set of small constantsOptimizing the right-hand side in (29) with respect to the choice of δ1 it follows that

intS

t01

(X0)P0[(n)] dx int

Sδ0t0

(X0)P0[(n)] dx + | ln(t0δ0)|minus1 E

Converting the X0 momentum density into the partt momentum density it follows that

(1 minus γ1)12

intS

1minusγ11

(partt )P0[(n)] dx (1 minus γ2)minus 1

2

intS

δ0t0

(partt )P0[(n)] dx + | ln(1 minus γ2)|minus1 E

Hence by (56) we obtain

(1 minus γ1)12 E1 (1 minus γ2)

minus 12 E

Sδ0t0

[(n)] + | ln(1 minus γ2)|minus1 E

We choose γ2 so that

| ln(1 minus γ2)|minus1 E (1 minus γ1)12 E1

Then the second right-hand side term in the previous inequality can be neglected andfor

0 lt E2 (1 minus γ1)12 (1 minus γ2)

12 E1

we obtain (57)

66 Final rescaling The one bound concerning the rescaled wave maps (n) which isnot yet satisfactory is (57) where we would like to have decay in n instead of uniform

boundedness This can be achieved by further subdividing the time interval [ε12n ε

14n ]

For 2 lt N lt εminus 14 we divide the time interval [ε

12n ε

14n ] into about | ln εn| ln N sub-

intervals of the form [t Nt] By pigeonholing there exists one such subinterval whichwe denote by [tn Ntn] so that

int intCεn[tn Ntn ]

1

ρ|X0

(n)|2dxdt ln N

| ln εn| E (59)

We assign to N = Nn the value

Nn = eradic| ln εn |

Rescaling the wave maps (n) from the time interval [tn Nntn] to the time interval[1 Nn] we obtain a final sequence of rescaled wave maps still denoted by (n) definedon increasing sets C[1Tn ] where Tn rarr infin with the following properties


a) Bounded energy

ESt [(n)](t) asymp E t isin [1 Tn] (60)

b) Uniform amount of nontrivial time-like energy

ES

(1minusγ2)tt

[(n)](t) ge E2 t isin [1 Tn] (61)

c) Decay to self-similar modeint int

Cε

12

n[1Tn ]

1

ρ|X0

(n)|2dxdt | log εn|minus 12 E (62)

67 Concentration scales We partition the set C1[1infin) into dyadic subsets

C j = (t x) isin C1[1infin) 2 j lt t lt 2 j+1 j isin N

We also consider slightly larger sets

C j = (t x) isin C12

[ 12 infin)

2 j lt t lt 2 j+1 j isin N

Then we prove that

Lemma 63 Let (n) be a sequence of wave maps satisfying (60) (61) and (62) Thenfor each j isin N one of the following alternatives must hold on a subsequence

(i) Concentration of non-trivial energy There exist points (tn xn) isin C j a sequenceof scales rn rarr 0 and some r = r j with 0 lt r lt 1

4 so that the following threebounds hold


10E0 (63)


10E0 x isin B(xn r) (64)

rminus1n

int tn+rn2

tnminusrn2

intB(xn r)

|X0(n)|2dxdt rarr 0 (65)

(ii) Nonconcentration of uniform energy There exists some r = r j with 0 lt r lt 14 so

that the following three bounds hold

EB(xr)[(n)](t) le 1

10E0 forall(t x) isin C j (66)

ES

(1minusγ2)tt

[(n)](t) ge E2 when B (0 (1 minus γ2)t) sube C1[1infin) (67)int int

C j



Proof The argument boils down to some straightforward pigeonholing and is essen-tially identical for all j isin N which is now fixed throughout the proof Given any largeparameter N isin N we partition the time interval [2 j 2 j+1] into about N2 j equal intervals

Ik = [2 j + (k minus 1)(10N ) 2 j + k(10N )] k = 1 10N2 j

Then it suffices to show that the conclusion of the lemma holds with C j and C j replacedby

Ckj = C j cap Ik times R

2 Ckj = C j cap Ik times R

2

We begin by constructing a low energy barrier around Ckj To do this we partition Ck

j Ckj

into N sets

Cklj =

(t x) isin Ck

j 1

4+

l minus 1

4Nlt t minus |x | lt

1

4+

l

4N

l = 1 N

By integrating energy estimates we have

Nsuml=1

intIk

ECklj

[(n)](t)dt leint

Ik

ECkj[(n)](t)dt le 1

10NE

Thus by pigeonholing for each fixed n there must exist ln so that

ln+1suml=lnminus1

intIk

ECklj

[(n)](t)dt le 3

10N 2 E

and further there must be some tn isin Ik so that

jn+1sumj= jnminus1

ECklj

[(n)](tn) le 3

NE

For t isin Ik we have |t minus tn| lt 1(10N ) and therefore the t section of Cklnj lies within

the influence cone of the tn section of Cklnminus1j cap Ckln

j cap Ckln+1j Hence it follows that

one has the uniform bound

ECklnj

[(n)(t)] le 3

NE t isin Ik

We choose N large enough so that we beat the perturbation energy

3

NE le 1

20E0

Then the set Cklnj acts as an energy barrier for (n) within Ck

j separating the evolutioninside from the evolution outside with a small data region We denote the inner regionby Ckltln

j and its union with Cklnj by Ckleln

j We fix r0 independent of n so that

(t x) isin Ckltlnj rArr t times B(x 4r0) sub Ckleln

j


To measure the energy concentration in balls we define the functions

fn [0 r0] times Ik rarr R+ fn(r t) = sup

xttimesB(xr)subCklelnj

EB(xr)[(n)](t)

The functions fn are continuous in both variables and nondecreasing with respect to r We also define the functions

rn Ik rarr (0 r0] rn(t) =

infr isin [0 r0] fn(t r) ge E010 if fn(t r0) ge E0

10 r0 otherwise

which measure the lowest spatial scale on which concentration occurs at time t Dueto the finite speed of propagation it follows that the rn are Lipschitz continuous withLipschitz constant 1

|rn(t1) minus rn(t2)| le |t1 minus t2|The nonconcentration estimate (66) in case ii) of the lemma corresponds to the case

when all functions rn admit a common strictly positive lower bound (note that (61) and(62) give the other conclusions)

It remains to consider the case when on a subsequence we have

limnrarrinfin inf

Ikrn = 0

and show that this yields the concentration scenario i) We denote ldquokinetic energyrdquo inC j by

α2n =

intC j

|X0(n)|2dxdt

By (62) we know that αn rarr 0 Using αn as a threshold for the concentration functionsrn after passing to a subsequence we must be in one of the following three cases

Case 1 (rn Dominates) For each n we have rn(t) gt αn in Ik Then we let tn be the mini-mum point for rn in Ik and set r0

n = rn(tn) rarr 0 By definition we have fn(tn rn(tn)) =E010 We choose a point xn where the maximum of fn(tn rn(tn)) is attained Thisdirectly gives (63) For (64) we observe that due to the existence of the energy bar-rier xn must be at least at distance 3r0 from the lateral boundary of Ckleln

j Hence if

x isin B(xn r0) then x is at distance at least 2r0 from partCklelnj and (64) follows For (65)

it suffices to know that rn(tn)minus1α2n rarr 0 which is straightforward from the assumptions

of this case

Case 2 (Equality) For each n there exists tn isin Ik such that αn = rn(tn) This argumentis a repeat of the previous one given that we define r0

n = rn(tn) and set up the estimate(64) around this tn as opposed to the minimum of rn(t)

Case 3 (αn Dominates) For each n we have rn(t) lt αn in Ik For t isin Ik set

g(t) =int

Cklelnj

|X0(n)(t)|2dx


Then by definitionint

Ik

g(t) = α2n

Let Ik be the middle third of Ik and consider the localized averages

I =int

Ik

1

rn(t)

int t+rn(t)2

tminusrn(t)2g(s)dsdt

Since rn(t) is Lipschitz with Lipschitz constant 1 if s isin [t minus rn(t)2 t + rn(t)2] then12rn(s) lt rn(t) lt 2rn(s) and t isin [s minus rn(s) s + rn(s)] Hence changing the order ofintegration in I we obtain

I le 2int

Ik

int t+rn(t)2

tminusrn(t)2

1

rn(s)g(s)dsdt le 4

intIk

g(s)ds = 4α2n

Hence by pigeonholing there exists some tn isin Ik so that

1

rn(tn)

int tn+rn(tn)2

tnminusrn(tn)2g(s)ds le 4α2

n | Ik |minus1

Then let xn be a point where the supremum in the definition of f (tn rn) is attainedThe relations (63)-(66) follow as above

68 The compactness argument To conclude the proof of Theorems 1315 we considerseparately the two cases in Lemma 63

(i) Concentration on small scales Suppose that the alternative (i) in Lemma 63 holdsfor some j isin N On a subsequence we can assume that (tn xn) rarr (t0 x0) isin C j Then we define the rescaled wave maps


in the increasing sets B(0 r0rn) times [minus 12 1

2 ] They have the following propertiesa) Bounded energy

E[(n)](t) le E[] t isin [minus1

2

1

2]

b) Small energy in each unit ball

supx

EB(x1)[(n)](t) le 1

10E0 t isin [minus1

2

1

2]

c) Energy concentration in the unit ball centered at t = 0 x = 0

EB(01)[(n)](0) = 1

10E0


d) Time-like energy decay There exists a constant time-like vector X0(t0 x0)

such that for each x we haveint int

[minus 12 1

2 ]timesB(x1)

|X0(t0 x0)(n)|2dxdt rarr 0

Note in part (d) we used the fact that (tn xn) rarr (t0 x0)By the compactness result in Proposition 51 it follows that on a subsequence wehave strong uniform convergence on compact sets

(n) rarr in H1loc

(B(0 r0(2rn) times [minus1

2

1

2])

where isin H32 minusε

loc is a wave-map Thus we have obtained a wave map definedon all of [minus 1

2 12 ] times R

2 with the additional properties that

1

10E0 le E[] le E[]

and

X0(t0 x0) = 0

Then extends uniquely to a wave map in RtimesR2 with the above properties (eg

by transporting its values along the flow of X0(t0 x0)) After a Lorentz transformthat takes X0(t0 x0) to partt the function is turned into a nontrivial finite energyharmonic map with energy bound E[] le E[]

(ii) Nonconcentration Assume now that the alternative (ii) in Lemma 63 holds forevery j isin N There there is no need to rescale Instead we successively use directlythe compactness result in Proposition 51 in the interior of each set C j cap C2

[2infin)We obtain strong convergence on a subsequence

(n) rarr in H1loc(C

2[2infin))

with isin H32 minusε

loc (C2[2infin)) From (67) and energy bounds (eg (61)) we obtain

0 lt E2 le supt2

EB(0tminus2)[](t) le E[]

From (68) it follows also that

X0 = 0

By rescaling (ie extending via homogeneity) we may replace the interior of thetranslated cone C2

[2infin) with the interior of the full cone t gt r and retain the assumptionson in particular that it is non-trivial with finite energy up to the boundary t = r Butthis contradicts Theorem 41 and therefore shows that scenario (i) above is in fact theonly alternative


Acknowledgements The authors would like to thank Manos Grillakis Sergiu Klainerman JoachimKrieger Matei Machedon Igor Rodnianski and Wilhelm Schlag for many stimulating discussions over theyears regarding the wave-map problem We would also especially like to thank Terry Tao for several keydiscussions on the nature of induction-on-energy type proofs

Open Access This article is distributed under the terms of the Creative Commons Attribution NoncommercialLicense which permits any noncommercial use distribution and reproduction in any medium provided theoriginal author(s) and source are credited

References

1 Bahouri H Geacuterard P High frequency approximation of solutions to critical nonlinear wave equa-tions Amer J Math 121(1) 131ndash175 (1999)

2 Choquet-Bruhat Y General relativity and the Einstein equations Oxford Mathematical MonographsOxford Oxford University Press 2009

3 Christodoulou D Tahvildar-Zadeh AS On the asymptotic behavior of spherically symmetric wavemaps Duke Math J 71(1) 31ndash69 (1993)

4 Christodoulou D Tahvildar-Zadeh AS On the regularity of spherically symmetric wave maps CommPure Appl Math 46(7) 1041ndash1091 (1993)

5 Cote R Kenig CE Merle F Scattering below critical energy for the radial 4d Yang-Mills equationand for the 2d corotational wave map system httparxivorgabs07093222v1[mathAP] 2007

6 Gallot S Hulin D Lafontaine J Riemannian geometry Universitext Berlin Springer-Verlag Secondedition 1990

7 Grillakis MG On the wave map problem In Nonlinear wave equations (Providence RI 1998) Volume263 of Contemp Math Providence RI Amer Math Soc 2000 pp 71ndash84

8 Gromov ML Isometric imbeddings and immersions Dokl Akad Nauk SSSR 192 1206ndash1209 (1970)9 Guumlnther M Isometric embeddings of Riemannian manifolds In Proceedings of the International Con-

gress of Mathematicians Vol I II (Kyoto 1990) Tokyo Math Soc Japan 1991 pp 1137ndash114310 Heacutelein F Harmonic maps conservation laws and moving frames Volume 150 of Cambridge Tracts in

Mathematics Cambridge Cambridge University Press second edition 200211 Jost J Riemannian geometry and geometric analysis Universitext Berlin Springer-Verlag 199512 Kenig CE Merle F Global well-posedness scattering and blow-up for the energy-critical focusing

non-linear wave equation Acta Math 201(2) 147ndash212 (2008)13 Klainerman S Machedon M Space-time estimates for null forms and the local existence theo-

rem Comm Pure Appl Math 46(9) 1221ndash1268 (1993)14 Klainerman S Machedon M Smoothing estimates for null forms and applications Duke Math

J 81(1) 99ndash133 (1996)15 Krieger J Global regularity and singularity development for wave maps In Surveys in differential

geometry Vol XII Geometric flows Volume 12 of Surv Differ Geom Somerville MA Int Press 2008pp 167ndash201

16 Krieger J Schlag W Tataru D Renormalization and blow up for charge one equivariant critical wavemaps Invent Math 171(3) 543ndash615 (2008)

17 Krieger J Global regularity of wave maps from R2+1 to H2 Small energy Commun Math

Phys 250(3) 507ndash580 (2004)18 Krieger J Schlag W Concentration compactness for critical wave-maps httparxivorgabs0908

2474[mathAP] 200919 Lemaire L Applications harmoniques de surfaces riemanniennes J Diff Geom 13(1) 51ndash78 (1978)20 Nash J The imbedding problem for Riemannian manifolds Ann of Math (2) 63 20ndash63 (1956)21 Qing J Boundary regularity of weakly harmonic maps from surfaces J Funct Anal 114(2) 458ndash

466 (1993)22 Qing J Tian G (1993) Bubbling of the heat flows for harmonic maps from surfaces Comm Pure Appl

Math 50(4) 295ndash310 (1997)23 Rodnianski I Sterbenz J On the formation of singularities in the critical O(3) sigma-model

httparxivorgabsmath0605023v3[mathAP] 200824 Schoen R Yau ST Lectures on harmonic maps Conference Proceedings and Lecture Notes in Geom-

etry and Topology II Cambridge MA International Press 199725 Shatah J Struwe M Geometric wave equations Volume 2 of Courant Lecture Notes in Mathematics

New York New York University Courant Institute of Mathematical Sciences 199826 Shatah J Tahvildar-Zadeh AS On the Cauchy problem for equivariant wave maps Comm Pure Appl

Math 47(5) 719ndash754 (1994)


27 Sterbenz J Tataru D Energy dispersed large data wave maps in 2 + 1 dimensions httparxivorgabs09063384v1[mathAP] 2009 doi101007s00220-010-1061-4

28 Struwe M On the evolution of harmonic mappings of Riemannian surfaces Comment MathHelv 60(4) 558ndash581 (1985)

29 Struwe M Equivariant wave maps in two space dimensions Comm Pure Appl Math 56(7) 815ndash823(2003)

30 Struwe M Radially symmetric wave maps from (1 + 2)-dimensional Minkowski space to general tar-gets Calc Var Part Diff Eqs 16(4) 431ndash437 (2003)

31 Tao T Global regularity of wave maps III Large energy from R2+1 to hyperbolic spaces httparxiv

orgabs08054666v3[mathAP] 200932 Tao T Global regularity of wave maps IV Absence of stationary or self-similar solutions in the energy

class httparxivorgabs08063592v2[mathAP] 200933 Tao T Global regularity of wave maps V Large data local well-posedness in the energy class http

arxivorgabs08080368v2[mathAP] 200934 Tao T Global regulrity of wave maps VI Minimal energy blowup solutions httparxivorgabs0906

2833v2[mathAP] 200935 Tao T An inverse theorem for the bilinear L2 Strichartz estimate for the wave equation httparxivorg

abs09042880v1[mathNT] 200936 Tao T Global regularity of wave maps II Small energy in two dimensions Commun Math

Phys 224(2) 443ndash544 (2001)37 Tao T Geometric renormalization of large energy wave maps In Journeacutees ldquoEacutequations aux Deacuteriveacutees

Partiellesrdquo pages Exp No XI 32 Eacutecole Polytech Palaiseau 200438 Tao T Nonlinear dispersive equations Volume 106 of CBMS Regional Conference Series in Mathemat-

ics Washington DC Conference Board of the Mathematical Sciences 200639 Tataru D On global existence and scattering for the wave maps equation Amer J Math 123(1) 37ndash

77 (2001)40 Tataru D The wave maps equation Bull Amer Math Soc (NS) 41(2) 185ndash204 (electronic) (2004)41 Tataru D Rough solutions for the wave maps equation Amer J Math 127(2) 293ndash377 (2005)42 Topping P An example of a nontrivial bubble tree in the harmonic map heat flow In Harmonic mor-

phisms harmonic maps and related topics (Brest 1997) Volume 413 of Chapman amp HallCRC ResNotes Math Boca Raton FL Chapman amp HallCRC 2000 pp 185ndash191

Communicated by P Constantin

Regularity of Wave-Maps in Dimension 2+1

Abstract

1 Introduction

11 The Question of Blowup

12 The Question of Scattering






61 Extension and scaling in the blowup scenario

62 Extension and scaling in the non-scattering scenario

63 Elimination of the null concentration scenario

64 Nontrivial energy in a time-like cone

65 Propagation of time-like energy concentration

66 Final rescaling

67 Concentration scales

68 The compactness argument

Acknowledgements

References

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ltFEFF4f7f75288fd94e9b8bbe5b9a521b5efa7684002000410064006f006200650020005000440046002065876863900275284e8e9ad88d2891cf76845370524d53705237300260a853ef4ee54f7f75280020004100630072006f0062006100740020548c002000410064006f00620065002000520065006100640065007200200035002e003000204ee553ca66f49ad87248672c676562535f00521b5efa768400200050004400460020658768633002gt CHT ltFEFF4f7f752890194e9b8a2d7f6e5efa7acb7684002000410064006f006200650020005000440046002065874ef69069752865bc9ad854c18cea76845370524d5370523786557406300260a853ef4ee54f7f75280020004100630072006f0062006100740020548c002000410064006f00620065002000520065006100640065007200200035002e003000204ee553ca66f49ad87248672c4f86958b555f5df25efa7acb76840020005000440046002065874ef63002gt CZE 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 DAN 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 ESP ltFEFF005500740069006c0069006300650020006500730074006100200063006f006e0066006900670075007200610063006900f3006e0020007000610072006100200063007200650061007200200064006f00630075006d0065006e0074006f00730020005000440046002000640065002000410064006f0062006500200061006400650063007500610064006f00730020007000610072006100200069006d0070007200650073006900f3006e0020007000720065002d0065006400690074006f007200690061006c00200064006500200061006c00740061002000630061006c0069006400610064002e002000530065002000700075006500640065006e00200061006200720069007200200064006f00630075006d0065006e0074006f00730020005000440046002000630072006500610064006f007300200063006f006e0020004100630072006f006200610074002c002000410064006f00620065002000520065006100640065007200200035002e003000200079002000760065007200730069006f006e0065007300200070006f00730074006500720069006f007200650073002egt ETI 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 FRA 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 GRE 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 HEB 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 HRV (Za stvaranje Adobe PDF dokumenata najpogodnijih za visokokvalitetni ispis prije tiskanja koristite ove postavke Stvoreni PDF dokumenti mogu se otvoriti Acrobat i Adobe Reader 50 i kasnijim verzijama) HUN ltFEFF004b0069007600e1006c00f30020006d0069006e0151007300e9006701710020006e0079006f006d00640061006900200065006c0151006b00e90073007a00ed007401510020006e0079006f006d00740061007400e100730068006f007a0020006c006500670069006e006b00e1006200620020006d0065006700660065006c0065006c0151002000410064006f00620065002000500044004600200064006f006b0075006d0065006e00740075006d006f006b0061007400200065007a0065006b006b0065006c0020006100200062006500e1006c006c00ed007400e10073006f006b006b0061006c0020006b00e90073007a00ed0074006800650074002e0020002000410020006c00e90074007200650068006f007a006f00740074002000500044004600200064006f006b0075006d0065006e00740075006d006f006b00200061007a0020004100630072006f006200610074002000e9007300200061007a002000410064006f00620065002000520065006100640065007200200035002e0030002c0020007600610067007900200061007a002000610074007400f3006c0020006b00e9007301510062006200690020007600650072007a006900f3006b006b0061006c0020006e00790069007400680061007400f3006b0020006d00650067002egt ITA 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 JPN ltFEFF9ad854c18cea306a30d730ea30d730ec30b951fa529b7528002000410064006f0062006500200050004400460020658766f8306e4f5c6210306b4f7f75283057307e305930023053306e8a2d5b9a30674f5c62103055308c305f0020005000440046002030d530a130a430eb306f3001004100630072006f0062006100740020304a30883073002000410064006f00620065002000520065006100640065007200200035002e003000204ee5964d3067958b304f30533068304c3067304d307e305930023053306e8a2d5b9a306b306f30d530a930f330c8306e57cb30818fbc307f304c5fc59808306730593002gt KOR ltFEFFc7740020c124c815c7440020c0acc6a9d558c5ec0020ace0d488c9c80020c2dcd5d80020c778c1c4c5d00020ac00c7a50020c801d569d55c002000410064006f0062006500200050004400460020bb38c11cb97c0020c791c131d569b2c8b2e4002e0020c774b807ac8c0020c791c131b41c00200050004400460020bb38c11cb2940020004100630072006f0062006100740020bc0f002000410064006f00620065002000520065006100640065007200200035002e00300020c774c0c1c5d0c11c0020c5f40020c2180020c788c2b5b2c8b2e4002egt LTH 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 LVI 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 NLD (Gebruik deze instellingen om Adobe PDF-documenten te maken die zijn geoptimaliseerd voor prepress-afdrukken van hoge kwaliteit De gemaakte PDF-documenten kunnen worden geopend met Acrobat en Adobe Reader 50 en hoger) NOR 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 POL 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 PTB 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 RUM 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 RUS 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 SKY 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 SLV 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 SUO 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 SVE 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1 Introduction

In this article we consider large data Wave-Maps from R2+1 into a compact Riemannian

manifold (M m) and we prove that regularity and dispersive bounds persist as longas a soliton-like concentration is absent This is a companion to our concurrent article[27] where the same result is proved under a stronger energy dispersion assumptionsee Theorem 12 below

The set-up we consider is the same as the one in [41] using the extrinsic formulationof the Wave-Maps equation Precisely we consider the target manifold (M m) as anisometrically embedded submanifold of R

N Then we can view the M valued functionsas R

N valued functions whose range is contained in M Such an embedding alwaysexists by Nashrsquos theorem [20] (see also Gromov [8] and Guumlnther [9]) In this context theWave-Maps equation can be expressed in a form which involves the second fundamentalform S of M viewed as a symmetric bilinear form

S T M times T M rarr NM 〈S(X Y ) N 〉 = 〈partX N Y 〉The Cauchy problem for the wave maps equation has the form

φa = minusSabc(φ)partαφbpartαφc (φ1 φN ) = (1a)

(0 x) = 0(x) partt(0 x) = 1(x) (1b)

where the initial data (01) is chosen to obey the constraint

0(x) isin M 1(x) isin T0(x)M x isin R2

In the sequel it will be convenient for us to use the notation [t] = ((t) partt(t))There is a conserved energy for this problem

E[](t) = 1

2

int(|partt|2 + |nablax|2)dx

This is invariant with respect to the scaling that preserves the equation (x t) rarr(λx λt) Because of this we say that the problem is energy critical

The present article contributes to the understanding of finite energy solutions witharbitrarily large initial data a problem which has been the subject of intense investiga-tions for some time now We will not attempt to give a detailed account of the history ofthis subject here Instead we refer the reader to the surveys [40] and [15] and the refer-ences therein The general (ie without symmetry assumptions) small energy problemfor compact targets was initiated in the ground-breaking work of Klainerman-Machedon[1314] and completed by the work of Tao [36] when M is a sphere and Tataru [41]for general isometrically embedded manifolds see also the work of Krieger [17] on thenon-compact hyperbolic plane which was treated from the intrinsic point of view

At a minimum one expects the solutions to belong to the space C(R H1(R2)) capC1(R L2(R2)) However this information does not suffice in order to study the equa-tion and to obtain uniqueness statements Instead a smaller Banach space S sub C(RH1(R2)) cap C1(R L2(R2)) was introduced in [36] modifying an earlier structure in[39] Beside the energy S also contains Strichartz type information in various frequencylocalized contexts The full description of S is not necessary here However we do use

the fact that S cap Linfin is an algebra as well as the Xsb type embedding S sub X1 1

2infin Seeour companion paper [27] for precise definitions The standard small data result is asfollows



S (01)H1timesL2





E D[] = supkisinZ

PkLinfintx [I ]










time sections as

ESt [] = 1

2

intSt






F[t0t1][] =int

partC[t0t1]

(1

4|L|2 +

1

2|part|2

)d A


ESt1[] = ESt0

[] + F[t0t1][] (3)



limtrarr0

ESt [] le E0


limtrarr0

ESt [] = 0


limtrarr0

ESt [] gt E0 (4)





lt 1 limrn

tn= 0


(n)(t x) = (tn + rnt xn + rn x) (5)


nontrivial energy


trarr0ESt []




with bounded energy


ESt [] (6)


suptisin(0t0]

supkisinZ

(Pk(t)Linfin


x

)le ε (7)









2 rarr M If [0] R





for all T gt 0

















lt 1 limrn

tn= 0


(n)(t x) = (tn + rnt xn + rn x) (9)


nontrivial energy




with bounded energy



suptisin[t0infin)

supkisinZ

(Pk(t)Linfin


x

)le ε (11)



E[] ge E0(M) (12)













X0 = 1














limt1t2rarr0

F[t1t2][] = 0








F[εn 1][(n)] le ε12n E[]



(n)(1 xn)| gt ε (14)









1

2

intt=1|x |ltγ1

(|partt


)dx ge E1




14n ]

1

2

int|x |ltγ2(εE)t

(|partt



12n ε

14n ]



Cεn

[ε12

n ε

14

n ]






ES

(1minusγ2)tt


ε

12

n[1Tn ]

1

ρ|X0





(t0 x0) inside C12

[ 12 infin)



10E0


10E0 x isin B(x0 r)

rminus1n

int tn+rn2

tnminusrn2

intB(x0r)

|X0(n)|2dxdt rarr 0





ES

(1minusγ2)tt


|X0(n)|2dxdt rarr 0




2 times [minus 12 1




H2 rarr M


D2 rarr M



|nablax|2 dx

1 minus rlt infin





Tαβ [] = mi j ()



] (15)





nablaα(X)Pα = 1

2Tαβ [](X)παβ (16)




(X)π = LX g




intSδ

t2

(X)P0 dx +1

2

int intCδ[t1t2]


Sδt1

(X)P0 dx +int

partCδ[t1t2]

(X)PL d A (17)



(Y )π = 0 (Y )P0 = 1


4|L|2 +

1

2|part|2



intpartCδ[δ1]

1

4|L|2 +

1

2|part|2 d A le E[] (18)









dy2 + sinh2(y)d2) (21)




partt = t

ρpartρ minus r

ρ2 party

so in particular




+cosh(y)

2ρ2

(|party|2 +

1

sinh2(y)|part|2

)



1

2

intH2


Here


sinh2(y)|part|2




Xε = 1

ρε


(t + ε)2 minus r2 (24)




LX0 g = 2ρ(

dy2 + sinh2(y)d2)


(X0)παβ = 2

ρ3


)


1

2Tαβ [](X0)παβ = 1

ρ|X0|2




X0 = 1ρ

( 12vL + 1

2 uL) partt = 1

2L +

1

2L (25)



(X0)P0 = T (partt X0) = 1

4

(v

u

) 12 |L|2 +

1

4

[(v

u

) 12

+(u

v

) 12]

|part|2 +1

4

(u

v

) 12 |L|2

(X0)PL = T (L X0) = 1

2

(v

u

) 12 |L|2 +

1

2

(u

v

) 12 |part|2







(vε

uε

) 12 asymp 1

(uε

vε






intS0

t2


C0[t1t2]

1

ρε

|Xε|2 dxdt =int

S0t1

(Xε )P0 dx


suptisin(01]

intS0

t

(X0)P0 dx +int int

C0[01]

1

ρ|X0|2 dxdt le E




S0t2


C0[t1t2]

1

ρε

|Xε|2 dxdt int

S0t1



suptisin(ε1]

intS0

t


C0[01]

1

ρε





intS0

1

(Xε )P0 dx E (27)





1

ρ|X0|2 1

ρε

|Xε|2 +ε2

ρ3 |partt|2

and also

intCε

(ε1]

ε2


Cε(ε1]

ε12

t32

|partt|2dxdt E




suptisin(ε1]

intSε

t

(X0)P0 dx +int int

Cε[ε1]

1

ρ|X0|2dxdt E (28)


intSδ

1

(X0)P0 dx leint

Sδt0

(X0)P0 dx +int

partCδ[t01]



intpartCδ[t01]

(u

v

) 12 |part|2 d A

(δ

t0

) 12int


(δ

t0

) 12

E


|L| (u

v

) 12 |X0| +

(u

v

)|L|

Thus we obtain

intSδ

1

(X0)P0 dx int

Sδt0

(X0)P0 dx +

(δ

t0

) 12

E +int

partCδ[t01]

((u

v

) 12 |X0φ|2 +

(u

v

) 32 |L|2

)d A





intS

δ11

(X0)P0 dx int

Sδ0t0

(X0)P0 dx +

((δ1

t0

) 12


)E (29)


intpartCδ[t01]

[(u

v

) 12 |X0φ|2 +

(u

v

) 32 |L|2



intC

δ0[t01]Cδ1[t01]

1

u

[(u

v

) 12 |X0φ|2 +

(u

v

) 32 |L|2

]dxdt E


1

u

(u

v

) 32 le δ

121

t32







2




H2a = minusgi jH2Sa

bc()partibpart j

c



intH2


St


)dx (30)




Chap II of [6])


rang



1

2

intD2

|nablax|2(

1 + r2

1 minus r2

)dx lt infin (31)





r=rn

|nabla|2dl = on(1)



12 (S1)

= 0












10 E0 and so that

X(n)L2(3Q) rarr 0 (32)



(n) rarr in H1(Q)




χ(n) X

1 12infin


where X1 1







(n)bulk

H32 minusε

tx

χ(n) X

1 12infin

1



Rn H1tx














H2 supe rarr M








E = limtrarr0

ESt []





Ft0 [] =int

partC(0t0]

(1

4|L|2 +

1

2|part|2

)d A




Ft0 [] ε8 E

intpartSt0

|part|2ds ε8

t0E (36)







2ε8 E (37)





Ftn lt ε12n E Thus






E[(n)] asymp E (40)




c) Decaying flux

F[εn 1][(n)] lt ε12n E (42)



(n)(1 xn)| gt ε (43)




[] Ft0 [] =int

partC[t0infin)

(1

4|L|2 +

1

2|part|2

)d A

it follows that



Ft0 [] le ε8 E






Ftk [] rarr 0 tk

intpartStk

|part|2d A rarr 0







E[(k)] rarr E (45)























(n)(1 x)| (

2k(1 + 2k |x |)minusN + ε4)

E12 k le 0




)E

12 k ge 0

(48)


2kn gt m(ε E)





2kn lt C lt infin









intS1


(|L(n)|2 + |part(n)|2

)dx E (50)



(n)(1 x)| ((


) 18

+ ε2)

E12 (51)




In general we have



1

(|1 minus x1| + x22 + εn)

12

(|L(n)|2 + |part(n)|2

)dx E (52)



(n)|Similarly




1

((1 minus x1)+ + x22 + εn)

12 + ε8

(|partt


(n)|2)

dx E (53)




kθ part2

where P1kθ and P2



kθ (ξ)| 1

|ξ2|



12 P2




t=1

1


12 + ε8

(|partt


(n)|2)

dx E (55)


|Pk(n)(1 x)| le|P1

kθ part1(n)(1 x)| + |P2

kθ part2(n)(1 x)|

(θ

12 + θminus 1

2

(((1 minus x1)+ + x2

2 + 2minusk + εn)14 + ε4

))E

12



|Pk(n)(1 x1 0)|


) 18

+ ε2)

E12

1

2lt x1 lt 2




) 18

+ ε2)

E12

1

2lt x1 lt 2




) 18

+ ε2)

E12

1

2lt x1 lt2








1

2

intt=1|x |ltγ1

(|partt


)dx ge E1 (56)


E(γ1) = 1

2

intt=1|x |ltγ1


)dx




Pk(1)Linfinx


E(γ1)12 +

((2k(1 minus γ1))

12 + ε4

)E

12





((2k(1 minus γ1))


)E

12




)E

12 |x | lt γ


ε E(γ1)12 +


12 + ε4

)E

12



ε E(γ1)12 +

((1 minus γ1)

12

(1 minus γ )12

+ ε4

)E

12


ε E(γ1)12 +

((1 minus γ1

1 minus γ

)N

+ ε4

)E

12



12n ε


so that

1

2

int|x |ltγ2t

(|partt



12n ε

14n ] (57)



intCεn

[ε12

n ε

14

n ]






according to



intS

t01

(X0)P0[(n)] dx int

Sδ0t0



(1 minus γ1)12

intS

1minusγ11


2

intS

δ0t0




minus 12 E

Sδ0t0






12 E1

we obtain (57)



14n ]


12n ε




1

ρ|X0

(n)|2dxdt ln N

| ln εn| E (59)





a) Bounded energy



ES

(1minusγ2)tt

[(n)](t) ge E2 t isin [1 Tn] (61)


Cε

12

n[1Tn ]

1

ρ|X0






[ 12 infin)


Then we prove that





10E0 (63)



rminus1n

int tn+rn2

tnminusrn2

intB(xn r)




EB(xr)[(n)](t) le 1


ES

(1minusγ2)tt


C j








2


j Ckj

into N sets

Cklj =

(t x) isin Ck

j 1

4+

l minus 1


1

4+

l

4N

l = 1 N


Nsuml=1

intIk

ECklj

[(n)](t)dt leint

Ik

ECkj[(n)](t)dt le 1

10NE


ln+1suml=lnminus1

intIk

ECklj

[(n)](t)dt le 3

10N 2 E


jn+1sumj= jnminus1

ECklj

[(n)](tn) le 3

NE





ECklnj

[(n)(t)] le 3

NE t isin Ik


3

NE le 1

20E0






j





EB(xr)[(n)](t)




10 r0 otherwise





limnrarrinfin inf

Ikrn = 0


α2n =

intC j

|X0(n)|2dxdt




j Hence if



of this case




g(t) =int

Cklelnj

|X0(n)(t)|2dx



Ik

g(t) = α2n


I =int

Ik

1

rn(t)

int t+rn(t)2



I le 2int

Ik

int t+rn(t)2

tminusrn(t)2

1

rn(s)g(s)dsdt le 4

intIk

g(s)ds = 4α2n


1

rn(tn)

int tn+rn(tn)2


n | Ik |minus1








2

1

2]


supx

EB(x1)[(n)](t) le 1

10E0 t isin [minus1

2

1

2]


EB(01)[(n)](0) = 1

10E0




[minus 12 1

2 ]timesB(x1)



(n) rarr in H1loc


2

1

2])



2 12 ] times R


1

10E0 le E[] le E[]

and

X0(t0 x0) = 0





(n) rarr in H1loc(C

2[2infin))



0 lt E2 le supt2



X0 = 0






References
























Math 47(5) 719ndash754 (1994)



















Abstract

1 Introduction













66 Final rescaling



Acknowledgements

References

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StartPage 1 SubsetFonts false TransferFunctionInfo Apply UCRandBGInfo Preserve UsePrologue false ColorSettingsFile () AlwaysEmbed [ true ] NeverEmbed [ true ] AntiAliasColorImages false CropColorImages true ColorImageMinResolution 149 ColorImageMinResolutionPolicy Warning DownsampleColorImages true ColorImageDownsampleType Bicubic ColorImageResolution 150 ColorImageDepth -1 ColorImageMinDownsampleDepth 1 ColorImageDownsampleThreshold 150000 EncodeColorImages true ColorImageFilter DCTEncode AutoFilterColorImages true ColorImageAutoFilterStrategy JPEG ColorACSImageDict ltlt QFactor 040 HSamples [1 1 1 1] VSamples [1 1 1 1] gtgt ColorImageDict ltlt QFactor 015 HSamples [1 1 1 1] VSamples [1 1 1 1] gtgt JPEG2000ColorACSImageDict ltlt TileWidth 256 TileHeight 256 Quality 30 gtgt JPEG2000ColorImageDict ltlt TileWidth 256 TileHeight 256 Quality 30 gtgt AntiAliasGrayImages false CropGrayImages true GrayImageMinResolution 149 GrayImageMinResolutionPolicy Warning DownsampleGrayImages true GrayImageDownsampleType Bicubic GrayImageResolution 150 GrayImageDepth -1 GrayImageMinDownsampleDepth 2 GrayImageDownsampleThreshold 150000 EncodeGrayImages true GrayImageFilter DCTEncode AutoFilterGrayImages true GrayImageAutoFilterStrategy JPEG GrayACSImageDict ltlt QFactor 040 HSamples [1 1 1 1] VSamples [1 1 1 1] gtgt GrayImageDict ltlt QFactor 015 HSamples [1 1 1 1] VSamples [1 1 1 1] gtgt JPEG2000GrayACSImageDict ltlt TileWidth 256 TileHeight 256 Quality 30 gtgt JPEG2000GrayImageDict ltlt TileWidth 256 TileHeight 256 Quality 30 gtgt AntiAliasMonoImages false CropMonoImages true MonoImageMinResolution 599 MonoImageMinResolutionPolicy Warning DownsampleMonoImages true MonoImageDownsampleType Bicubic MonoImageResolution 600 MonoImageDepth -1 MonoImageDownsampleThreshold 150000 EncodeMonoImages true MonoImageFilter CCITTFaxEncode MonoImageDict ltlt K -1 gtgt AllowPSXObjects false CheckCompliance [ None ] PDFX1aCheck false PDFX3Check false PDFXCompliantPDFOnly false PDFXNoTrimBoxError true PDFXTrimBoxToMediaBoxOffset [ 000000 000000 000000 000000 ] PDFXSetBleedBoxToMediaBox true PDFXBleedBoxToTrimBoxOffset [ 000000 000000 000000 000000 ] PDFXOutputIntentProfile (None) PDFXOutputConditionIdentifier () PDFXOutputCondition () PDFXRegistryName () PDFXTrapped False CreateJDFFile false Description ltlt ARA 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 BGR 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 CHS ltFEFF4f7f75288fd94e9b8bbe5b9a521b5efa7684002000410064006f006200650020005000440046002065876863900275284e8e9ad88d2891cf76845370524d53705237300260a853ef4ee54f7f75280020004100630072006f0062006100740020548c002000410064006f00620065002000520065006100640065007200200035002e003000204ee553ca66f49ad87248672c676562535f00521b5efa768400200050004400460020658768633002gt CHT ltFEFF4f7f752890194e9b8a2d7f6e5efa7acb7684002000410064006f006200650020005000440046002065874ef69069752865bc9ad854c18cea76845370524d5370523786557406300260a853ef4ee54f7f75280020004100630072006f0062006100740020548c002000410064006f00620065002000520065006100640065007200200035002e003000204ee553ca66f49ad87248672c4f86958b555f5df25efa7acb76840020005000440046002065874ef63002gt CZE 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 DAN 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 ESP 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 ETI 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 FRA 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 GRE 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 HEB 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 HRV (Za stvaranje Adobe PDF dokumenata najpogodnijih za visokokvalitetni ispis prije tiskanja koristite ove postavke Stvoreni PDF dokumenti mogu se otvoriti Acrobat i Adobe Reader 50 i kasnijim verzijama) HUN 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 ITA 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 JPN ltFEFF9ad854c18cea306a30d730ea30d730ec30b951fa529b7528002000410064006f0062006500200050004400460020658766f8306e4f5c6210306b4f7f75283057307e305930023053306e8a2d5b9a30674f5c62103055308c305f0020005000440046002030d530a130a430eb306f3001004100630072006f0062006100740020304a30883073002000410064006f00620065002000520065006100640065007200200035002e003000204ee5964d3067958b304f30533068304c3067304d307e305930023053306e8a2d5b9a306b306f30d530a930f330c8306e57cb30818fbc307f304c5fc59808306730593002gt KOR ltFEFFc7740020c124c815c7440020c0acc6a9d558c5ec0020ace0d488c9c80020c2dcd5d80020c778c1c4c5d00020ac00c7a50020c801d569d55c002000410064006f0062006500200050004400460020bb38c11cb97c0020c791c131d569b2c8b2e4002e0020c774b807ac8c0020c791c131b41c00200050004400460020bb38c11cb2940020004100630072006f0062006100740020bc0f002000410064006f00620065002000520065006100640065007200200035002e00300020c774c0c1c5d0c11c0020c5f40020c2180020c788c2b5b2c8b2e4002egt LTH 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 LVI 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 NLD (Gebruik deze instellingen om Adobe PDF-documenten te maken die zijn geoptimaliseerd voor prepress-afdrukken van hoge kwaliteit De gemaakte PDF-documenten kunnen worden geopend met Acrobat en Adobe Reader 50 en hoger) NOR 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 POL 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 PTB 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 RUM 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 RUS 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 SKY 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 SLV 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 SUO 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 SVE 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 TUR ltFEFF005900fc006b00730065006b0020006b0061006c006900740065006c0069002000f6006e002000790061007a006401310072006d00610020006200610073006b013100730131006e006100200065006e0020006900790069002000750079006100620069006c006500630065006b002000410064006f006200650020005000440046002000620065006c00670065006c0065007200690020006f006c0075015f007400750072006d0061006b0020006900e70069006e00200062007500200061007900610072006c0061007201310020006b0075006c006c0061006e0131006e002e00200020004f006c0075015f0074007500720075006c0061006e0020005000440046002000620065006c00670065006c0065007200690020004100630072006f006200610074002000760065002000410064006f00620065002000520065006100640065007200200035002e003000200076006500200073006f006e0072006100730131006e00640061006b00690020007300fc007200fc006d006c00650072006c00650020006100e70131006c006100620069006c00690072002egt UKR 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 ENU (Use these settings to create Adobe PDF documents best suited for high-quality prepress printing Created PDF documents can be opened with Acrobat and Adobe Reader 50 and later) DEU 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 gtgt Namespace [ (Adobe) (Common) (10) ] OtherNamespaces [ ltlt AsReaderSpreads false CropImagesToFrames true ErrorControl WarnAndContinue FlattenerIgnoreSpreadOverrides false IncludeGuidesGrids false IncludeNonPrinting false IncludeSlug false Namespace [ (Adobe) (InDesign) (40) ] OmitPlacedBitmaps false OmitPlacedEPS false OmitPlacedPDF false SimulateOverprint Legacy gtgt ltlt AddBleedMarks false AddColorBars false AddCropMarks false AddPageInfo false AddRegMarks false ConvertColors ConvertToCMYK DestinationProfileName () DestinationProfileSelector DocumentCMYK Downsample16BitImages true FlattenerPreset ltlt PresetSelector MediumResolution gtgt FormElements false GenerateStructure false IncludeBookmarks false IncludeHyperlinks false IncludeInteractive false IncludeLayers false IncludeProfiles false MultimediaHandling UseObjectSettings Namespace [ (Adobe) (CreativeSuite) (20) ] PDFXOutputIntentProfileSelector DocumentCMYK PreserveEditing true UntaggedCMYKHandling LeaveUntagged UntaggedRGBHandling UseDocumentProfile UseDocumentBleed false gtgt ]gtgt setdistillerparamsltlt HWResolution [2400 2400] PageSize [595276 841890]gtgt setpagedevice



S (01)H1timesL2





E D[] = supkisinZ

PkLinfintx [I ]










time sections as

ESt [] = 1

2

intSt






F[t0t1][] =int

partC[t0t1]

(1

4|L|2 +

1

2|part|2

)d A


ESt1[] = ESt0

[] + F[t0t1][] (3)



limtrarr0

ESt [] le E0


limtrarr0

ESt [] = 0


limtrarr0

ESt [] gt E0 (4)





lt 1 limrn

tn= 0


(n)(t x) = (tn + rnt xn + rn x) (5)


nontrivial energy


trarr0ESt []




with bounded energy


ESt [] (6)


suptisin(0t0]

supkisinZ

(Pk(t)Linfin


x

)le ε (7)









2 rarr M If [0] R





for all T gt 0

















lt 1 limrn

tn= 0


(n)(t x) = (tn + rnt xn + rn x) (9)


nontrivial energy




with bounded energy



suptisin[t0infin)

supkisinZ

(Pk(t)Linfin


x

)le ε (11)



E[] ge E0(M) (12)













X0 = 1














limt1t2rarr0

F[t1t2][] = 0








F[εn 1][(n)] le ε12n E[]



(n)(1 xn)| gt ε (14)









1

2

intt=1|x |ltγ1

(|partt


)dx ge E1




14n ]

1

2

int|x |ltγ2(εE)t

(|partt



12n ε

14n ]



Cεn

[ε12

n ε

14

n ]






ES

(1minusγ2)tt


ε

12

n[1Tn ]

1

ρ|X0





(t0 x0) inside C12

[ 12 infin)



10E0


10E0 x isin B(x0 r)

rminus1n

int tn+rn2

tnminusrn2

intB(x0r)

|X0(n)|2dxdt rarr 0





ES

(1minusγ2)tt


|X0(n)|2dxdt rarr 0




2 times [minus 12 1




H2 rarr M


D2 rarr M



|nablax|2 dx

1 minus rlt infin





Tαβ [] = mi j ()



] (15)





nablaα(X)Pα = 1

2Tαβ [](X)παβ (16)




(X)π = LX g




intSδ

t2

(X)P0 dx +1

2

int intCδ[t1t2]


Sδt1

(X)P0 dx +int

partCδ[t1t2]

(X)PL d A (17)



(Y )π = 0 (Y )P0 = 1


4|L|2 +

1

2|part|2



intpartCδ[δ1]

1

4|L|2 +

1

2|part|2 d A le E[] (18)









dy2 + sinh2(y)d2) (21)




partt = t

ρpartρ minus r

ρ2 party

so in particular




+cosh(y)

2ρ2

(|party|2 +

1

sinh2(y)|part|2

)



1

2

intH2


Here


sinh2(y)|part|2




Xε = 1

ρε


(t + ε)2 minus r2 (24)




LX0 g = 2ρ(

dy2 + sinh2(y)d2)


(X0)παβ = 2

ρ3


)


1

2Tαβ [](X0)παβ = 1

ρ|X0|2




X0 = 1ρ

( 12vL + 1

2 uL) partt = 1

2L +

1

2L (25)



(X0)P0 = T (partt X0) = 1

4

(v

u

) 12 |L|2 +

1

4

[(v

u

) 12

+(u

v

) 12]

|part|2 +1

4

(u

v

) 12 |L|2

(X0)PL = T (L X0) = 1

2

(v

u

) 12 |L|2 +

1

2

(u

v

) 12 |part|2







(vε

uε

) 12 asymp 1

(uε

vε






intS0

t2


C0[t1t2]

1

ρε

|Xε|2 dxdt =int

S0t1

(Xε )P0 dx


suptisin(01]

intS0

t

(X0)P0 dx +int int

C0[01]

1

ρ|X0|2 dxdt le E




S0t2


C0[t1t2]

1

ρε

|Xε|2 dxdt int

S0t1



suptisin(ε1]

intS0

t


C0[01]

1

ρε





intS0

1

(Xε )P0 dx E (27)





1

ρ|X0|2 1

ρε

|Xε|2 +ε2

ρ3 |partt|2

and also

intCε

(ε1]

ε2


Cε(ε1]

ε12

t32

|partt|2dxdt E




suptisin(ε1]

intSε

t

(X0)P0 dx +int int

Cε[ε1]

1

ρ|X0|2dxdt E (28)


intSδ

1

(X0)P0 dx leint

Sδt0

(X0)P0 dx +int

partCδ[t01]



intpartCδ[t01]

(u

v

) 12 |part|2 d A

(δ

t0

) 12int


(δ

t0

) 12

E


|L| (u

v

) 12 |X0| +

(u

v

)|L|

Thus we obtain

intSδ

1

(X0)P0 dx int

Sδt0

(X0)P0 dx +

(δ

t0

) 12

E +int

partCδ[t01]

((u

v

) 12 |X0φ|2 +

(u

v

) 32 |L|2

)d A





intS

δ11

(X0)P0 dx int

Sδ0t0

(X0)P0 dx +

((δ1

t0

) 12


)E (29)


intpartCδ[t01]

[(u

v

) 12 |X0φ|2 +

(u

v

) 32 |L|2



intC

δ0[t01]Cδ1[t01]

1

u

[(u

v

) 12 |X0φ|2 +

(u

v

) 32 |L|2

]dxdt E


1

u

(u

v

) 32 le δ

121

t32







2




H2a = minusgi jH2Sa

bc()partibpart j

c



intH2


St


)dx (30)




Chap II of [6])


rang



1

2

intD2

|nablax|2(

1 + r2

1 minus r2

)dx lt infin (31)





r=rn

|nabla|2dl = on(1)



12 (S1)

= 0












10 E0 and so that

X(n)L2(3Q) rarr 0 (32)



(n) rarr in H1(Q)




χ(n) X

1 12infin


where X1 1







(n)bulk

H32 minusε

tx

χ(n) X

1 12infin

1



Rn H1tx














H2 supe rarr M








E = limtrarr0

ESt []





Ft0 [] =int

partC(0t0]

(1

4|L|2 +

1

2|part|2

)d A




Ft0 [] ε8 E

intpartSt0

|part|2ds ε8

t0E (36)







2ε8 E (37)





Ftn lt ε12n E Thus






E[(n)] asymp E (40)




c) Decaying flux

F[εn 1][(n)] lt ε12n E (42)



(n)(1 xn)| gt ε (43)




[] Ft0 [] =int

partC[t0infin)

(1

4|L|2 +

1

2|part|2

)d A

it follows that



Ft0 [] le ε8 E






Ftk [] rarr 0 tk

intpartStk

|part|2d A rarr 0







E[(k)] rarr E (45)























(n)(1 x)| (

2k(1 + 2k |x |)minusN + ε4)

E12 k le 0




)E

12 k ge 0

(48)


2kn gt m(ε E)





2kn lt C lt infin









intS1


(|L(n)|2 + |part(n)|2

)dx E (50)



(n)(1 x)| ((


) 18

+ ε2)

E12 (51)




In general we have



1

(|1 minus x1| + x22 + εn)

12

(|L(n)|2 + |part(n)|2

)dx E (52)



(n)|Similarly




1

((1 minus x1)+ + x22 + εn)

12 + ε8

(|partt


(n)|2)

dx E (53)




kθ part2

where P1kθ and P2



kθ (ξ)| 1

|ξ2|



12 P2




t=1

1


12 + ε8

(|partt


(n)|2)

dx E (55)


|Pk(n)(1 x)| le|P1

kθ part1(n)(1 x)| + |P2

kθ part2(n)(1 x)|

(θ

12 + θminus 1

2

(((1 minus x1)+ + x2

2 + 2minusk + εn)14 + ε4

))E

12



|Pk(n)(1 x1 0)|


) 18

+ ε2)

E12

1

2lt x1 lt 2




) 18

+ ε2)

E12

1

2lt x1 lt 2




) 18

+ ε2)

E12

1

2lt x1 lt2








1

2

intt=1|x |ltγ1

(|partt


)dx ge E1 (56)


E(γ1) = 1

2

intt=1|x |ltγ1


)dx




Pk(1)Linfinx


E(γ1)12 +

((2k(1 minus γ1))

12 + ε4

)E

12





((2k(1 minus γ1))


)E

12




)E

12 |x | lt γ


ε E(γ1)12 +


12 + ε4

)E

12



ε E(γ1)12 +

((1 minus γ1)

12

(1 minus γ )12

+ ε4

)E

12


ε E(γ1)12 +

((1 minus γ1

1 minus γ

)N

+ ε4

)E

12



12n ε


so that

1

2

int|x |ltγ2t

(|partt



12n ε

14n ] (57)



intCεn

[ε12

n ε

14

n ]






according to



intS

t01

(X0)P0[(n)] dx int

Sδ0t0



(1 minus γ1)12

intS

1minusγ11


2

intS

δ0t0




minus 12 E

Sδ0t0






12 E1

we obtain (57)



14n ]


12n ε




1

ρ|X0

(n)|2dxdt ln N

| ln εn| E (59)





a) Bounded energy



ES

(1minusγ2)tt

[(n)](t) ge E2 t isin [1 Tn] (61)


Cε

12

n[1Tn ]

1

ρ|X0






[ 12 infin)


Then we prove that





10E0 (63)



rminus1n

int tn+rn2

tnminusrn2

intB(xn r)




EB(xr)[(n)](t) le 1


ES

(1minusγ2)tt


C j








2


j Ckj

into N sets

Cklj =

(t x) isin Ck

j 1

4+

l minus 1


1

4+

l

4N

l = 1 N


Nsuml=1

intIk

ECklj

[(n)](t)dt leint

Ik

ECkj[(n)](t)dt le 1

10NE


ln+1suml=lnminus1

intIk

ECklj

[(n)](t)dt le 3

10N 2 E


jn+1sumj= jnminus1

ECklj

[(n)](tn) le 3

NE





ECklnj

[(n)(t)] le 3

NE t isin Ik


3

NE le 1

20E0






j





EB(xr)[(n)](t)




10 r0 otherwise





limnrarrinfin inf

Ikrn = 0


α2n =

intC j

|X0(n)|2dxdt




j Hence if



of this case




g(t) =int

Cklelnj

|X0(n)(t)|2dx



Ik

g(t) = α2n


I =int

Ik

1

rn(t)

int t+rn(t)2



I le 2int

Ik

int t+rn(t)2

tminusrn(t)2

1

rn(s)g(s)dsdt le 4

intIk

g(s)ds = 4α2n


1

rn(tn)

int tn+rn(tn)2


n | Ik |minus1








2

1

2]


supx

EB(x1)[(n)](t) le 1

10E0 t isin [minus1

2

1

2]


EB(01)[(n)](0) = 1

10E0




[minus 12 1

2 ]timesB(x1)



(n) rarr in H1loc


2

1

2])



2 12 ] times R


1

10E0 le E[] le E[]

and

X0(t0 x0) = 0





(n) rarr in H1loc(C

2[2infin))



0 lt E2 le supt2



X0 = 0






References
























Math 47(5) 719ndash754 (1994)



















Abstract

1 Introduction













66 Final rescaling



Acknowledgements

References






F[t0t1][] =int

partC[t0t1]

(1

4|L|2 +

1

2|part|2

)d A


ESt1[] = ESt0

[] + F[t0t1][] (3)



limtrarr0

ESt [] le E0


limtrarr0

ESt [] = 0


limtrarr0

ESt [] gt E0 (4)





lt 1 limrn

tn= 0


(n)(t x) = (tn + rnt xn + rn x) (5)


nontrivial energy


trarr0ESt []




with bounded energy


ESt [] (6)


suptisin(0t0]

supkisinZ

(Pk(t)Linfin


x

)le ε (7)









2 rarr M If [0] R





for all T gt 0

















lt 1 limrn

tn= 0


(n)(t x) = (tn + rnt xn + rn x) (9)


nontrivial energy




with bounded energy



suptisin[t0infin)

supkisinZ

(Pk(t)Linfin


x

)le ε (11)



E[] ge E0(M) (12)













X0 = 1














limt1t2rarr0

F[t1t2][] = 0








F[εn 1][(n)] le ε12n E[]



(n)(1 xn)| gt ε (14)









1

2

intt=1|x |ltγ1

(|partt


)dx ge E1




14n ]

1

2

int|x |ltγ2(εE)t

(|partt



12n ε

14n ]



Cεn

[ε12

n ε

14

n ]






ES

(1minusγ2)tt


ε

12

n[1Tn ]

1

ρ|X0





(t0 x0) inside C12

[ 12 infin)



10E0


10E0 x isin B(x0 r)

rminus1n

int tn+rn2

tnminusrn2

intB(x0r)

|X0(n)|2dxdt rarr 0





ES

(1minusγ2)tt


|X0(n)|2dxdt rarr 0




2 times [minus 12 1




H2 rarr M


D2 rarr M



|nablax|2 dx

1 minus rlt infin





Tαβ [] = mi j ()



] (15)





nablaα(X)Pα = 1

2Tαβ [](X)παβ (16)




(X)π = LX g




intSδ

t2

(X)P0 dx +1

2

int intCδ[t1t2]


Sδt1

(X)P0 dx +int

partCδ[t1t2]

(X)PL d A (17)



(Y )π = 0 (Y )P0 = 1


4|L|2 +

1

2|part|2



intpartCδ[δ1]

1

4|L|2 +

1

2|part|2 d A le E[] (18)









dy2 + sinh2(y)d2) (21)




partt = t

ρpartρ minus r

ρ2 party

so in particular




+cosh(y)

2ρ2

(|party|2 +

1

sinh2(y)|part|2

)



1

2

intH2


Here


sinh2(y)|part|2




Xε = 1

ρε


(t + ε)2 minus r2 (24)




LX0 g = 2ρ(

dy2 + sinh2(y)d2)


(X0)παβ = 2

ρ3


)


1

2Tαβ [](X0)παβ = 1

ρ|X0|2




X0 = 1ρ

( 12vL + 1

2 uL) partt = 1

2L +

1

2L (25)



(X0)P0 = T (partt X0) = 1

4

(v

u

) 12 |L|2 +

1

4

[(v

u

) 12

+(u

v

) 12]

|part|2 +1

4

(u

v

) 12 |L|2

(X0)PL = T (L X0) = 1

2

(v

u

) 12 |L|2 +

1

2

(u

v

) 12 |part|2







(vε

uε

) 12 asymp 1

(uε

vε






intS0

t2


C0[t1t2]

1

ρε

|Xε|2 dxdt =int

S0t1

(Xε )P0 dx


suptisin(01]

intS0

t

(X0)P0 dx +int int

C0[01]

1

ρ|X0|2 dxdt le E




S0t2


C0[t1t2]

1

ρε

|Xε|2 dxdt int

S0t1



suptisin(ε1]

intS0

t


C0[01]

1

ρε





intS0

1

(Xε )P0 dx E (27)





1

ρ|X0|2 1

ρε

|Xε|2 +ε2

ρ3 |partt|2

and also

intCε

(ε1]

ε2


Cε(ε1]

ε12

t32

|partt|2dxdt E




suptisin(ε1]

intSε

t

(X0)P0 dx +int int

Cε[ε1]

1

ρ|X0|2dxdt E (28)


intSδ

1

(X0)P0 dx leint

Sδt0

(X0)P0 dx +int

partCδ[t01]



intpartCδ[t01]

(u

v

) 12 |part|2 d A

(δ

t0

) 12int


(δ

t0

) 12

E


|L| (u

v

) 12 |X0| +

(u

v

)|L|

Thus we obtain

intSδ

1

(X0)P0 dx int

Sδt0

(X0)P0 dx +

(δ

t0

) 12

E +int

partCδ[t01]

((u

v

) 12 |X0φ|2 +

(u

v

) 32 |L|2

)d A





intS

δ11

(X0)P0 dx int

Sδ0t0

(X0)P0 dx +

((δ1

t0

) 12


)E (29)


intpartCδ[t01]

[(u

v

) 12 |X0φ|2 +

(u

v

) 32 |L|2



intC

δ0[t01]Cδ1[t01]

1

u

[(u

v

) 12 |X0φ|2 +

(u

v

) 32 |L|2

]dxdt E


1

u

(u

v

) 32 le δ

121

t32







2




H2a = minusgi jH2Sa

bc()partibpart j

c



intH2


St


)dx (30)




Chap II of [6])


rang



1

2

intD2

|nablax|2(

1 + r2

1 minus r2

)dx lt infin (31)





r=rn

|nabla|2dl = on(1)



12 (S1)

= 0












10 E0 and so that

X(n)L2(3Q) rarr 0 (32)



(n) rarr in H1(Q)




χ(n) X

1 12infin


where X1 1







(n)bulk

H32 minusε

tx

χ(n) X

1 12infin

1



Rn H1tx














H2 supe rarr M








E = limtrarr0

ESt []





Ft0 [] =int

partC(0t0]

(1

4|L|2 +

1

2|part|2

)d A




Ft0 [] ε8 E

intpartSt0

|part|2ds ε8

t0E (36)







2ε8 E (37)





Ftn lt ε12n E Thus






E[(n)] asymp E (40)




c) Decaying flux

F[εn 1][(n)] lt ε12n E (42)



(n)(1 xn)| gt ε (43)




[] Ft0 [] =int

partC[t0infin)

(1

4|L|2 +

1

2|part|2

)d A

it follows that



Ft0 [] le ε8 E






Ftk [] rarr 0 tk

intpartStk

|part|2d A rarr 0







E[(k)] rarr E (45)























(n)(1 x)| (

2k(1 + 2k |x |)minusN + ε4)

E12 k le 0




)E

12 k ge 0

(48)


2kn gt m(ε E)





2kn lt C lt infin









intS1


(|L(n)|2 + |part(n)|2

)dx E (50)



(n)(1 x)| ((


) 18

+ ε2)

E12 (51)




In general we have



1

(|1 minus x1| + x22 + εn)

12

(|L(n)|2 + |part(n)|2

)dx E (52)



(n)|Similarly




1

((1 minus x1)+ + x22 + εn)

12 + ε8

(|partt


(n)|2)

dx E (53)




kθ part2

where P1kθ and P2



kθ (ξ)| 1

|ξ2|



12 P2




t=1

1


12 + ε8

(|partt


(n)|2)

dx E (55)


|Pk(n)(1 x)| le|P1

kθ part1(n)(1 x)| + |P2

kθ part2(n)(1 x)|

(θ

12 + θminus 1

2

(((1 minus x1)+ + x2

2 + 2minusk + εn)14 + ε4

))E

12



|Pk(n)(1 x1 0)|


) 18

+ ε2)

E12

1

2lt x1 lt 2




) 18

+ ε2)

E12

1

2lt x1 lt 2




) 18

+ ε2)

E12

1

2lt x1 lt2








1

2

intt=1|x |ltγ1

(|partt


)dx ge E1 (56)


E(γ1) = 1

2

intt=1|x |ltγ1


)dx




Pk(1)Linfinx


E(γ1)12 +

((2k(1 minus γ1))

12 + ε4

)E

12





((2k(1 minus γ1))


)E

12




)E

12 |x | lt γ


ε E(γ1)12 +


12 + ε4

)E

12



ε E(γ1)12 +

((1 minus γ1)

12

(1 minus γ )12

+ ε4

)E

12


ε E(γ1)12 +

((1 minus γ1

1 minus γ

)N

+ ε4

)E

12



12n ε


so that

1

2

int|x |ltγ2t

(|partt



12n ε

14n ] (57)



intCεn

[ε12

n ε

14

n ]






according to



intS

t01

(X0)P0[(n)] dx int

Sδ0t0



(1 minus γ1)12

intS

1minusγ11


2

intS

δ0t0




minus 12 E

Sδ0t0






12 E1

we obtain (57)



14n ]


12n ε




1

ρ|X0

(n)|2dxdt ln N

| ln εn| E (59)





a) Bounded energy



ES

(1minusγ2)tt

[(n)](t) ge E2 t isin [1 Tn] (61)


Cε

12

n[1Tn ]

1

ρ|X0






[ 12 infin)


Then we prove that





10E0 (63)



rminus1n

int tn+rn2

tnminusrn2

intB(xn r)




EB(xr)[(n)](t) le 1


ES

(1minusγ2)tt


C j








2


j Ckj

into N sets

Cklj =

(t x) isin Ck

j 1

4+

l minus 1


1

4+

l

4N

l = 1 N


Nsuml=1

intIk

ECklj

[(n)](t)dt leint

Ik

ECkj[(n)](t)dt le 1

10NE


ln+1suml=lnminus1

intIk

ECklj

[(n)](t)dt le 3

10N 2 E


jn+1sumj= jnminus1

ECklj

[(n)](tn) le 3

NE





ECklnj

[(n)(t)] le 3

NE t isin Ik


3

NE le 1

20E0






j





EB(xr)[(n)](t)




10 r0 otherwise





limnrarrinfin inf

Ikrn = 0


α2n =

intC j

|X0(n)|2dxdt




j Hence if



of this case




g(t) =int

Cklelnj

|X0(n)(t)|2dx



Ik

g(t) = α2n


I =int

Ik

1

rn(t)

int t+rn(t)2



I le 2int

Ik

int t+rn(t)2

tminusrn(t)2

1

rn(s)g(s)dsdt le 4

intIk

g(s)ds = 4α2n


1

rn(tn)

int tn+rn(tn)2


n | Ik |minus1








2

1

2]


supx

EB(x1)[(n)](t) le 1

10E0 t isin [minus1

2

1

2]


EB(01)[(n)](0) = 1

10E0




[minus 12 1

2 ]timesB(x1)



(n) rarr in H1loc


2

1

2])



2 12 ] times R


1

10E0 le E[] le E[]

and

X0(t0 x0) = 0





(n) rarr in H1loc(C

2[2infin))



0 lt E2 le supt2



X0 = 0






References
























Math 47(5) 719ndash754 (1994)



















Abstract

1 Introduction













66 Final rescaling



Acknowledgements

References





with bounded energy


ESt [] (6)


suptisin(0t0]

supkisinZ

(Pk(t)Linfin


x

)le ε (7)









2 rarr M If [0] R





for all T gt 0

















lt 1 limrn

tn= 0


(n)(t x) = (tn + rnt xn + rn x) (9)


nontrivial energy




with bounded energy



suptisin[t0infin)

supkisinZ

(Pk(t)Linfin


x

)le ε (11)



E[] ge E0(M) (12)













X0 = 1














limt1t2rarr0

F[t1t2][] = 0








F[εn 1][(n)] le ε12n E[]



(n)(1 xn)| gt ε (14)









1

2

intt=1|x |ltγ1

(|partt


)dx ge E1




14n ]

1

2

int|x |ltγ2(εE)t

(|partt



12n ε

14n ]



Cεn

[ε12

n ε

14

n ]






ES

(1minusγ2)tt


ε

12

n[1Tn ]

1

ρ|X0





(t0 x0) inside C12

[ 12 infin)



10E0


10E0 x isin B(x0 r)

rminus1n

int tn+rn2

tnminusrn2

intB(x0r)

|X0(n)|2dxdt rarr 0





ES

(1minusγ2)tt


|X0(n)|2dxdt rarr 0




2 times [minus 12 1




H2 rarr M


D2 rarr M



|nablax|2 dx

1 minus rlt infin





Tαβ [] = mi j ()



] (15)





nablaα(X)Pα = 1

2Tαβ [](X)παβ (16)




(X)π = LX g




intSδ

t2

(X)P0 dx +1

2

int intCδ[t1t2]


Sδt1

(X)P0 dx +int

partCδ[t1t2]

(X)PL d A (17)



(Y )π = 0 (Y )P0 = 1


4|L|2 +

1

2|part|2



intpartCδ[δ1]

1

4|L|2 +

1

2|part|2 d A le E[] (18)









dy2 + sinh2(y)d2) (21)




partt = t

ρpartρ minus r

ρ2 party

so in particular




+cosh(y)

2ρ2

(|party|2 +

1

sinh2(y)|part|2

)



1

2

intH2


Here


sinh2(y)|part|2




Xε = 1

ρε


(t + ε)2 minus r2 (24)




LX0 g = 2ρ(

dy2 + sinh2(y)d2)


(X0)παβ = 2

ρ3


)


1

2Tαβ [](X0)παβ = 1

ρ|X0|2




X0 = 1ρ

( 12vL + 1

2 uL) partt = 1

2L +

1

2L (25)



(X0)P0 = T (partt X0) = 1

4

(v

u

) 12 |L|2 +

1

4

[(v

u

) 12

+(u

v

) 12]

|part|2 +1

4

(u

v

) 12 |L|2

(X0)PL = T (L X0) = 1

2

(v

u

) 12 |L|2 +

1

2

(u

v

) 12 |part|2







(vε

uε

) 12 asymp 1

(uε

vε






intS0

t2


C0[t1t2]

1

ρε

|Xε|2 dxdt =int

S0t1

(Xε )P0 dx


suptisin(01]

intS0

t

(X0)P0 dx +int int

C0[01]

1

ρ|X0|2 dxdt le E




S0t2


C0[t1t2]

1

ρε

|Xε|2 dxdt int

S0t1



suptisin(ε1]

intS0

t


C0[01]

1

ρε





intS0

1

(Xε )P0 dx E (27)





1

ρ|X0|2 1

ρε

|Xε|2 +ε2

ρ3 |partt|2

and also

intCε

(ε1]

ε2


Cε(ε1]

ε12

t32

|partt|2dxdt E




suptisin(ε1]

intSε

t

(X0)P0 dx +int int

Cε[ε1]

1

ρ|X0|2dxdt E (28)


intSδ

1

(X0)P0 dx leint

Sδt0

(X0)P0 dx +int

partCδ[t01]



intpartCδ[t01]

(u

v

) 12 |part|2 d A

(δ

t0

) 12int


(δ

t0

) 12

E


|L| (u

v

) 12 |X0| +

(u

v

)|L|

Thus we obtain

intSδ

1

(X0)P0 dx int

Sδt0

(X0)P0 dx +

(δ

t0

) 12

E +int

partCδ[t01]

((u

v

) 12 |X0φ|2 +

(u

v

) 32 |L|2

)d A





intS

δ11

(X0)P0 dx int

Sδ0t0

(X0)P0 dx +

((δ1

t0

) 12


)E (29)


intpartCδ[t01]

[(u

v

) 12 |X0φ|2 +

(u

v

) 32 |L|2



intC

δ0[t01]Cδ1[t01]

1

u

[(u

v

) 12 |X0φ|2 +

(u

v

) 32 |L|2

]dxdt E


1

u

(u

v

) 32 le δ

121

t32







2




H2a = minusgi jH2Sa

bc()partibpart j

c



intH2


St


)dx (30)




Chap II of [6])


rang



1

2

intD2

|nablax|2(

1 + r2

1 minus r2

)dx lt infin (31)





r=rn

|nabla|2dl = on(1)



12 (S1)

= 0












10 E0 and so that

X(n)L2(3Q) rarr 0 (32)



(n) rarr in H1(Q)




χ(n) X

1 12infin


where X1 1







(n)bulk

H32 minusε

tx

χ(n) X

1 12infin

1



Rn H1tx














H2 supe rarr M








E = limtrarr0

ESt []





Ft0 [] =int

partC(0t0]

(1

4|L|2 +

1

2|part|2

)d A




Ft0 [] ε8 E

intpartSt0

|part|2ds ε8

t0E (36)







2ε8 E (37)





Ftn lt ε12n E Thus






E[(n)] asymp E (40)




c) Decaying flux

F[εn 1][(n)] lt ε12n E (42)



(n)(1 xn)| gt ε (43)




[] Ft0 [] =int

partC[t0infin)

(1

4|L|2 +

1

2|part|2

)d A

it follows that



Ft0 [] le ε8 E






Ftk [] rarr 0 tk

intpartStk

|part|2d A rarr 0







E[(k)] rarr E (45)























(n)(1 x)| (

2k(1 + 2k |x |)minusN + ε4)

E12 k le 0




)E

12 k ge 0

(48)


2kn gt m(ε E)





2kn lt C lt infin









intS1


(|L(n)|2 + |part(n)|2

)dx E (50)



(n)(1 x)| ((


) 18

+ ε2)

E12 (51)




In general we have



1

(|1 minus x1| + x22 + εn)

12

(|L(n)|2 + |part(n)|2

)dx E (52)



(n)|Similarly




1

((1 minus x1)+ + x22 + εn)

12 + ε8

(|partt


(n)|2)

dx E (53)




kθ part2

where P1kθ and P2



kθ (ξ)| 1

|ξ2|



12 P2




t=1

1


12 + ε8

(|partt


(n)|2)

dx E (55)


|Pk(n)(1 x)| le|P1

kθ part1(n)(1 x)| + |P2

kθ part2(n)(1 x)|

(θ

12 + θminus 1

2

(((1 minus x1)+ + x2

2 + 2minusk + εn)14 + ε4

))E

12



|Pk(n)(1 x1 0)|


) 18

+ ε2)

E12

1

2lt x1 lt 2




) 18

+ ε2)

E12

1

2lt x1 lt 2




) 18

+ ε2)

E12

1

2lt x1 lt2








1

2

intt=1|x |ltγ1

(|partt


)dx ge E1 (56)


E(γ1) = 1

2

intt=1|x |ltγ1


)dx




Pk(1)Linfinx


E(γ1)12 +

((2k(1 minus γ1))

12 + ε4

)E

12





((2k(1 minus γ1))


)E

12




)E

12 |x | lt γ


ε E(γ1)12 +


12 + ε4

)E

12



ε E(γ1)12 +

((1 minus γ1)

12

(1 minus γ )12

+ ε4

)E

12


ε E(γ1)12 +

((1 minus γ1

1 minus γ

)N

+ ε4

)E

12



12n ε


so that

1

2

int|x |ltγ2t

(|partt



12n ε

14n ] (57)



intCεn

[ε12

n ε

14

n ]






according to



intS

t01

(X0)P0[(n)] dx int

Sδ0t0



(1 minus γ1)12

intS

1minusγ11


2

intS

δ0t0




minus 12 E

Sδ0t0






12 E1

we obtain (57)



14n ]


12n ε




1

ρ|X0

(n)|2dxdt ln N

| ln εn| E (59)





a) Bounded energy



ES

(1minusγ2)tt

[(n)](t) ge E2 t isin [1 Tn] (61)


Cε

12

n[1Tn ]

1

ρ|X0






[ 12 infin)


Then we prove that





10E0 (63)



rminus1n

int tn+rn2

tnminusrn2

intB(xn r)




EB(xr)[(n)](t) le 1


ES

(1minusγ2)tt


C j








2


j Ckj

into N sets

Cklj =

(t x) isin Ck

j 1

4+

l minus 1


1

4+

l

4N

l = 1 N


Nsuml=1

intIk

ECklj

[(n)](t)dt leint

Ik

ECkj[(n)](t)dt le 1

10NE


ln+1suml=lnminus1

intIk

ECklj

[(n)](t)dt le 3

10N 2 E


jn+1sumj= jnminus1

ECklj

[(n)](tn) le 3

NE





ECklnj

[(n)(t)] le 3

NE t isin Ik


3

NE le 1

20E0






j





EB(xr)[(n)](t)




10 r0 otherwise





limnrarrinfin inf

Ikrn = 0


α2n =

intC j

|X0(n)|2dxdt




j Hence if



of this case




g(t) =int

Cklelnj

|X0(n)(t)|2dx



Ik

g(t) = α2n


I =int

Ik

1

rn(t)

int t+rn(t)2



I le 2int

Ik

int t+rn(t)2

tminusrn(t)2

1

rn(s)g(s)dsdt le 4

intIk

g(s)ds = 4α2n


1

rn(tn)

int tn+rn(tn)2


n | Ik |minus1








2

1

2]


supx

EB(x1)[(n)](t) le 1

10E0 t isin [minus1

2

1

2]


EB(01)[(n)](0) = 1

10E0




[minus 12 1

2 ]timesB(x1)



(n) rarr in H1loc


2

1

2])



2 12 ] times R


1

10E0 le E[] le E[]

and

X0(t0 x0) = 0





(n) rarr in H1loc(C

2[2infin))



0 lt E2 le supt2



X0 = 0






References
























Math 47(5) 719ndash754 (1994)



















Abstract

1 Introduction













66 Final rescaling



Acknowledgements

References















lt 1 limrn

tn= 0


(n)(t x) = (tn + rnt xn + rn x) (9)


nontrivial energy




with bounded energy



suptisin[t0infin)

supkisinZ

(Pk(t)Linfin


x

)le ε (11)



E[] ge E0(M) (12)













X0 = 1














limt1t2rarr0

F[t1t2][] = 0








F[εn 1][(n)] le ε12n E[]



(n)(1 xn)| gt ε (14)









1

2

intt=1|x |ltγ1

(|partt


)dx ge E1




14n ]

1

2

int|x |ltγ2(εE)t

(|partt



12n ε

14n ]



Cεn

[ε12

n ε

14

n ]






ES

(1minusγ2)tt


ε

12

n[1Tn ]

1

ρ|X0





(t0 x0) inside C12

[ 12 infin)



10E0


10E0 x isin B(x0 r)

rminus1n

int tn+rn2

tnminusrn2

intB(x0r)

|X0(n)|2dxdt rarr 0





ES

(1minusγ2)tt


|X0(n)|2dxdt rarr 0




2 times [minus 12 1




H2 rarr M


D2 rarr M



|nablax|2 dx

1 minus rlt infin





Tαβ [] = mi j ()



] (15)





nablaα(X)Pα = 1

2Tαβ [](X)παβ (16)




(X)π = LX g




intSδ

t2

(X)P0 dx +1

2

int intCδ[t1t2]


Sδt1

(X)P0 dx +int

partCδ[t1t2]

(X)PL d A (17)



(Y )π = 0 (Y )P0 = 1


4|L|2 +

1

2|part|2



intpartCδ[δ1]

1

4|L|2 +

1

2|part|2 d A le E[] (18)









dy2 + sinh2(y)d2) (21)




partt = t

ρpartρ minus r

ρ2 party

so in particular




+cosh(y)

2ρ2

(|party|2 +

1

sinh2(y)|part|2

)



1

2

intH2


Here


sinh2(y)|part|2




Xε = 1

ρε


(t + ε)2 minus r2 (24)




LX0 g = 2ρ(

dy2 + sinh2(y)d2)


(X0)παβ = 2

ρ3


)


1

2Tαβ [](X0)παβ = 1

ρ|X0|2




X0 = 1ρ

( 12vL + 1

2 uL) partt = 1

2L +

1

2L (25)



(X0)P0 = T (partt X0) = 1

4

(v

u

) 12 |L|2 +

1

4

[(v

u

) 12

+(u

v

) 12]

|part|2 +1

4

(u

v

) 12 |L|2

(X0)PL = T (L X0) = 1

2

(v

u

) 12 |L|2 +

1

2

(u

v

) 12 |part|2







(vε

uε

) 12 asymp 1

(uε

vε






intS0

t2


C0[t1t2]

1

ρε

|Xε|2 dxdt =int

S0t1

(Xε )P0 dx


suptisin(01]

intS0

t

(X0)P0 dx +int int

C0[01]

1

ρ|X0|2 dxdt le E




S0t2


C0[t1t2]

1

ρε

|Xε|2 dxdt int

S0t1



suptisin(ε1]

intS0

t


C0[01]

1

ρε





intS0

1

(Xε )P0 dx E (27)





1

ρ|X0|2 1

ρε

|Xε|2 +ε2

ρ3 |partt|2

and also

intCε

(ε1]

ε2


Cε(ε1]

ε12

t32

|partt|2dxdt E




suptisin(ε1]

intSε

t

(X0)P0 dx +int int

Cε[ε1]

1

ρ|X0|2dxdt E (28)


intSδ

1

(X0)P0 dx leint

Sδt0

(X0)P0 dx +int

partCδ[t01]



intpartCδ[t01]

(u

v

) 12 |part|2 d A

(δ

t0

) 12int


(δ

t0

) 12

E


|L| (u

v

) 12 |X0| +

(u

v

)|L|

Thus we obtain

intSδ

1

(X0)P0 dx int

Sδt0

(X0)P0 dx +

(δ

t0

) 12

E +int

partCδ[t01]

((u

v

) 12 |X0φ|2 +

(u

v

) 32 |L|2

)d A





intS

δ11

(X0)P0 dx int

Sδ0t0

(X0)P0 dx +

((δ1

t0

) 12


)E (29)


intpartCδ[t01]

[(u

v

) 12 |X0φ|2 +

(u

v

) 32 |L|2



intC

δ0[t01]Cδ1[t01]

1

u

[(u

v

) 12 |X0φ|2 +

(u

v

) 32 |L|2

]dxdt E


1

u

(u

v

) 32 le δ

121

t32







2




H2a = minusgi jH2Sa

bc()partibpart j

c



intH2


St


)dx (30)




Chap II of [6])


rang



1

2

intD2

|nablax|2(

1 + r2

1 minus r2

)dx lt infin (31)





r=rn

|nabla|2dl = on(1)



12 (S1)

= 0












10 E0 and so that

X(n)L2(3Q) rarr 0 (32)



(n) rarr in H1(Q)




χ(n) X

1 12infin


where X1 1







(n)bulk

H32 minusε

tx

χ(n) X

1 12infin

1



Rn H1tx














H2 supe rarr M








E = limtrarr0

ESt []





Ft0 [] =int

partC(0t0]

(1

4|L|2 +

1

2|part|2

)d A




Ft0 [] ε8 E

intpartSt0

|part|2ds ε8

t0E (36)







2ε8 E (37)





Ftn lt ε12n E Thus






E[(n)] asymp E (40)




c) Decaying flux

F[εn 1][(n)] lt ε12n E (42)



(n)(1 xn)| gt ε (43)




[] Ft0 [] =int

partC[t0infin)

(1

4|L|2 +

1

2|part|2

)d A

it follows that



Ft0 [] le ε8 E






Ftk [] rarr 0 tk

intpartStk

|part|2d A rarr 0







E[(k)] rarr E (45)























(n)(1 x)| (

2k(1 + 2k |x |)minusN + ε4)

E12 k le 0




)E

12 k ge 0

(48)


2kn gt m(ε E)





2kn lt C lt infin









intS1


(|L(n)|2 + |part(n)|2

)dx E (50)



(n)(1 x)| ((


) 18

+ ε2)

E12 (51)




In general we have



1

(|1 minus x1| + x22 + εn)

12

(|L(n)|2 + |part(n)|2

)dx E (52)



(n)|Similarly




1

((1 minus x1)+ + x22 + εn)

12 + ε8

(|partt


(n)|2)

dx E (53)




kθ part2

where P1kθ and P2



kθ (ξ)| 1

|ξ2|



12 P2




t=1

1


12 + ε8

(|partt


(n)|2)

dx E (55)


|Pk(n)(1 x)| le|P1

kθ part1(n)(1 x)| + |P2

kθ part2(n)(1 x)|

(θ

12 + θminus 1

2

(((1 minus x1)+ + x2

2 + 2minusk + εn)14 + ε4

))E

12



|Pk(n)(1 x1 0)|


) 18

+ ε2)

E12

1

2lt x1 lt 2




) 18

+ ε2)

E12

1

2lt x1 lt 2




) 18

+ ε2)

E12

1

2lt x1 lt2








1

2

intt=1|x |ltγ1

(|partt


)dx ge E1 (56)


E(γ1) = 1

2

intt=1|x |ltγ1


)dx




Pk(1)Linfinx


E(γ1)12 +

((2k(1 minus γ1))

12 + ε4

)E

12





((2k(1 minus γ1))


)E

12




)E

12 |x | lt γ


ε E(γ1)12 +


12 + ε4

)E

12



ε E(γ1)12 +

((1 minus γ1)

12

(1 minus γ )12

+ ε4

)E

12


ε E(γ1)12 +

((1 minus γ1

1 minus γ

)N

+ ε4

)E

12



12n ε


so that

1

2

int|x |ltγ2t

(|partt



12n ε

14n ] (57)



intCεn

[ε12

n ε

14

n ]






according to



intS

t01

(X0)P0[(n)] dx int

Sδ0t0



(1 minus γ1)12

intS

1minusγ11


2

intS

δ0t0




minus 12 E

Sδ0t0






12 E1

we obtain (57)



14n ]


12n ε




1

ρ|X0

(n)|2dxdt ln N

| ln εn| E (59)





a) Bounded energy



ES

(1minusγ2)tt

[(n)](t) ge E2 t isin [1 Tn] (61)


Cε

12

n[1Tn ]

1

ρ|X0






[ 12 infin)


Then we prove that





10E0 (63)



rminus1n

int tn+rn2

tnminusrn2

intB(xn r)




EB(xr)[(n)](t) le 1


ES

(1minusγ2)tt


C j








2


j Ckj

into N sets

Cklj =

(t x) isin Ck

j 1

4+

l minus 1


1

4+

l

4N

l = 1 N


Nsuml=1

intIk

ECklj

[(n)](t)dt leint

Ik

ECkj[(n)](t)dt le 1

10NE


ln+1suml=lnminus1

intIk

ECklj

[(n)](t)dt le 3

10N 2 E


jn+1sumj= jnminus1

ECklj

[(n)](tn) le 3

NE





ECklnj

[(n)(t)] le 3

NE t isin Ik


3

NE le 1

20E0






j





EB(xr)[(n)](t)




10 r0 otherwise





limnrarrinfin inf

Ikrn = 0


α2n =

intC j

|X0(n)|2dxdt




j Hence if



of this case




g(t) =int

Cklelnj

|X0(n)(t)|2dx



Ik

g(t) = α2n


I =int

Ik

1

rn(t)

int t+rn(t)2



I le 2int

Ik

int t+rn(t)2

tminusrn(t)2

1

rn(s)g(s)dsdt le 4

intIk

g(s)ds = 4α2n


1

rn(tn)

int tn+rn(tn)2


n | Ik |minus1








2

1

2]


supx

EB(x1)[(n)](t) le 1

10E0 t isin [minus1

2

1

2]


EB(01)[(n)](0) = 1

10E0




[minus 12 1

2 ]timesB(x1)



(n) rarr in H1loc


2

1

2])



2 12 ] times R


1

10E0 le E[] le E[]

and

X0(t0 x0) = 0





(n) rarr in H1loc(C

2[2infin))



0 lt E2 le supt2



X0 = 0






References
























Math 47(5) 719ndash754 (1994)



















Abstract

1 Introduction













66 Final rescaling



Acknowledgements

References




E[] ge E0(M) (12)













X0 = 1














limt1t2rarr0

F[t1t2][] = 0








F[εn 1][(n)] le ε12n E[]



(n)(1 xn)| gt ε (14)









1

2

intt=1|x |ltγ1

(|partt


)dx ge E1




14n ]

1

2

int|x |ltγ2(εE)t

(|partt



12n ε

14n ]



Cεn

[ε12

n ε

14

n ]






ES

(1minusγ2)tt


ε

12

n[1Tn ]

1

ρ|X0





(t0 x0) inside C12

[ 12 infin)



10E0


10E0 x isin B(x0 r)

rminus1n

int tn+rn2

tnminusrn2

intB(x0r)

|X0(n)|2dxdt rarr 0





ES

(1minusγ2)tt


|X0(n)|2dxdt rarr 0




2 times [minus 12 1




H2 rarr M


D2 rarr M



|nablax|2 dx

1 minus rlt infin





Tαβ [] = mi j ()



] (15)





nablaα(X)Pα = 1

2Tαβ [](X)παβ (16)




(X)π = LX g




intSδ

t2

(X)P0 dx +1

2

int intCδ[t1t2]


Sδt1

(X)P0 dx +int

partCδ[t1t2]

(X)PL d A (17)



(Y )π = 0 (Y )P0 = 1


4|L|2 +

1

2|part|2



intpartCδ[δ1]

1

4|L|2 +

1

2|part|2 d A le E[] (18)









dy2 + sinh2(y)d2) (21)




partt = t

ρpartρ minus r

ρ2 party

so in particular




+cosh(y)

2ρ2

(|party|2 +

1

sinh2(y)|part|2

)



1

2

intH2


Here


sinh2(y)|part|2




Xε = 1

ρε


(t + ε)2 minus r2 (24)




LX0 g = 2ρ(

dy2 + sinh2(y)d2)


(X0)παβ = 2

ρ3


)


1

2Tαβ [](X0)παβ = 1

ρ|X0|2




X0 = 1ρ

( 12vL + 1

2 uL) partt = 1

2L +

1

2L (25)



(X0)P0 = T (partt X0) = 1

4

(v

u

) 12 |L|2 +

1

4

[(v

u

) 12

+(u

v

) 12]

|part|2 +1

4

(u

v

) 12 |L|2

(X0)PL = T (L X0) = 1

2

(v

u

) 12 |L|2 +

1

2

(u

v

) 12 |part|2







(vε

uε

) 12 asymp 1

(uε

vε






intS0

t2


C0[t1t2]

1

ρε

|Xε|2 dxdt =int

S0t1

(Xε )P0 dx


suptisin(01]

intS0

t

(X0)P0 dx +int int

C0[01]

1

ρ|X0|2 dxdt le E




S0t2


C0[t1t2]

1

ρε

|Xε|2 dxdt int

S0t1



suptisin(ε1]

intS0

t


C0[01]

1

ρε





intS0

1

(Xε )P0 dx E (27)





1

ρ|X0|2 1

ρε

|Xε|2 +ε2

ρ3 |partt|2

and also

intCε

(ε1]

ε2


Cε(ε1]

ε12

t32

|partt|2dxdt E




suptisin(ε1]

intSε

t

(X0)P0 dx +int int

Cε[ε1]

1

ρ|X0|2dxdt E (28)


intSδ

1

(X0)P0 dx leint

Sδt0

(X0)P0 dx +int

partCδ[t01]



intpartCδ[t01]

(u

v

) 12 |part|2 d A

(δ

t0

) 12int


(δ

t0

) 12

E


|L| (u

v

) 12 |X0| +

(u

v

)|L|

Thus we obtain

intSδ

1

(X0)P0 dx int

Sδt0

(X0)P0 dx +

(δ

t0

) 12

E +int

partCδ[t01]

((u

v

) 12 |X0φ|2 +

(u

v

) 32 |L|2

)d A





intS

δ11

(X0)P0 dx int

Sδ0t0

(X0)P0 dx +

((δ1

t0

) 12


)E (29)


intpartCδ[t01]

[(u

v

) 12 |X0φ|2 +

(u

v

) 32 |L|2



intC

δ0[t01]Cδ1[t01]

1

u

[(u

v

) 12 |X0φ|2 +

(u

v

) 32 |L|2

]dxdt E


1

u

(u

v

) 32 le δ

121

t32







2




H2a = minusgi jH2Sa

bc()partibpart j

c



intH2


St


)dx (30)




Chap II of [6])


rang



1

2

intD2

|nablax|2(

1 + r2

1 minus r2

)dx lt infin (31)





r=rn

|nabla|2dl = on(1)



12 (S1)

= 0












10 E0 and so that

X(n)L2(3Q) rarr 0 (32)



(n) rarr in H1(Q)




χ(n) X

1 12infin


where X1 1







(n)bulk

H32 minusε

tx

χ(n) X

1 12infin

1



Rn H1tx














H2 supe rarr M








E = limtrarr0

ESt []





Ft0 [] =int

partC(0t0]

(1

4|L|2 +

1

2|part|2

)d A




Ft0 [] ε8 E

intpartSt0

|part|2ds ε8

t0E (36)







2ε8 E (37)





Ftn lt ε12n E Thus






E[(n)] asymp E (40)




c) Decaying flux

F[εn 1][(n)] lt ε12n E (42)



(n)(1 xn)| gt ε (43)




[] Ft0 [] =int

partC[t0infin)

(1

4|L|2 +

1

2|part|2

)d A

it follows that



Ft0 [] le ε8 E






Ftk [] rarr 0 tk

intpartStk

|part|2d A rarr 0







E[(k)] rarr E (45)























(n)(1 x)| (

2k(1 + 2k |x |)minusN + ε4)

E12 k le 0




)E

12 k ge 0

(48)


2kn gt m(ε E)





2kn lt C lt infin









intS1


(|L(n)|2 + |part(n)|2

)dx E (50)



(n)(1 x)| ((


) 18

+ ε2)

E12 (51)




In general we have



1

(|1 minus x1| + x22 + εn)

12

(|L(n)|2 + |part(n)|2

)dx E (52)



(n)|Similarly




1

((1 minus x1)+ + x22 + εn)

12 + ε8

(|partt


(n)|2)

dx E (53)




kθ part2

where P1kθ and P2



kθ (ξ)| 1

|ξ2|



12 P2




t=1

1


12 + ε8

(|partt


(n)|2)

dx E (55)


|Pk(n)(1 x)| le|P1

kθ part1(n)(1 x)| + |P2

kθ part2(n)(1 x)|

(θ

12 + θminus 1

2

(((1 minus x1)+ + x2

2 + 2minusk + εn)14 + ε4

))E

12



|Pk(n)(1 x1 0)|


) 18

+ ε2)

E12

1

2lt x1 lt 2




) 18

+ ε2)

E12

1

2lt x1 lt 2




) 18

+ ε2)

E12

1

2lt x1 lt2








1

2

intt=1|x |ltγ1

(|partt


)dx ge E1 (56)


E(γ1) = 1

2

intt=1|x |ltγ1


)dx




Pk(1)Linfinx


E(γ1)12 +

((2k(1 minus γ1))

12 + ε4

)E

12





((2k(1 minus γ1))


)E

12




)E

12 |x | lt γ


ε E(γ1)12 +


12 + ε4

)E

12



ε E(γ1)12 +

((1 minus γ1)

12

(1 minus γ )12

+ ε4

)E

12


ε E(γ1)12 +

((1 minus γ1

1 minus γ

)N

+ ε4

)E

12



12n ε


so that

1

2

int|x |ltγ2t

(|partt



12n ε

14n ] (57)



intCεn

[ε12

n ε

14

n ]






according to



intS

t01

(X0)P0[(n)] dx int

Sδ0t0



(1 minus γ1)12

intS

1minusγ11


2

intS

δ0t0




minus 12 E

Sδ0t0






12 E1

we obtain (57)



14n ]


12n ε




1

ρ|X0

(n)|2dxdt ln N

| ln εn| E (59)





a) Bounded energy



ES

(1minusγ2)tt

[(n)](t) ge E2 t isin [1 Tn] (61)


Cε

12

n[1Tn ]

1

ρ|X0






[ 12 infin)


Then we prove that





10E0 (63)



rminus1n

int tn+rn2

tnminusrn2

intB(xn r)




EB(xr)[(n)](t) le 1


ES

(1minusγ2)tt


C j








2


j Ckj

into N sets

Cklj =

(t x) isin Ck

j 1

4+

l minus 1


1

4+

l

4N

l = 1 N


Nsuml=1

intIk

ECklj

[(n)](t)dt leint

Ik

ECkj[(n)](t)dt le 1

10NE


ln+1suml=lnminus1

intIk

ECklj

[(n)](t)dt le 3

10N 2 E


jn+1sumj= jnminus1

ECklj

[(n)](tn) le 3

NE





ECklnj

[(n)(t)] le 3

NE t isin Ik


3

NE le 1

20E0






j





EB(xr)[(n)](t)




10 r0 otherwise





limnrarrinfin inf

Ikrn = 0


α2n =

intC j

|X0(n)|2dxdt




j Hence if



of this case




g(t) =int

Cklelnj

|X0(n)(t)|2dx



Ik

g(t) = α2n


I =int

Ik

1

rn(t)

int t+rn(t)2



I le 2int

Ik

int t+rn(t)2

tminusrn(t)2

1

rn(s)g(s)dsdt le 4

intIk

g(s)ds = 4α2n


1

rn(tn)

int tn+rn(tn)2


n | Ik |minus1








2

1

2]


supx

EB(x1)[(n)](t) le 1

10E0 t isin [minus1

2

1

2]


EB(01)[(n)](0) = 1

10E0




[minus 12 1

2 ]timesB(x1)



(n) rarr in H1loc


2

1

2])



2 12 ] times R


1

10E0 le E[] le E[]

and

X0(t0 x0) = 0





(n) rarr in H1loc(C

2[2infin))



0 lt E2 le supt2



X0 = 0






References
























Math 47(5) 719ndash754 (1994)



















Abstract

1 Introduction













66 Final rescaling



Acknowledgements

References












limt1t2rarr0

F[t1t2][] = 0








F[εn 1][(n)] le ε12n E[]



(n)(1 xn)| gt ε (14)









1

2

intt=1|x |ltγ1

(|partt


)dx ge E1




14n ]

1

2

int|x |ltγ2(εE)t

(|partt



12n ε

14n ]



Cεn

[ε12

n ε

14

n ]






ES

(1minusγ2)tt


ε

12

n[1Tn ]

1

ρ|X0





(t0 x0) inside C12

[ 12 infin)



10E0


10E0 x isin B(x0 r)

rminus1n

int tn+rn2

tnminusrn2

intB(x0r)

|X0(n)|2dxdt rarr 0





ES

(1minusγ2)tt


|X0(n)|2dxdt rarr 0




2 times [minus 12 1




H2 rarr M


D2 rarr M



|nablax|2 dx

1 minus rlt infin





Tαβ [] = mi j ()



] (15)





nablaα(X)Pα = 1

2Tαβ [](X)παβ (16)




(X)π = LX g




intSδ

t2

(X)P0 dx +1

2

int intCδ[t1t2]


Sδt1

(X)P0 dx +int

partCδ[t1t2]

(X)PL d A (17)



(Y )π = 0 (Y )P0 = 1


4|L|2 +

1

2|part|2



intpartCδ[δ1]

1

4|L|2 +

1

2|part|2 d A le E[] (18)









dy2 + sinh2(y)d2) (21)




partt = t

ρpartρ minus r

ρ2 party

so in particular




+cosh(y)

2ρ2

(|party|2 +

1

sinh2(y)|part|2

)



1

2

intH2


Here


sinh2(y)|part|2




Xε = 1

ρε


(t + ε)2 minus r2 (24)




LX0 g = 2ρ(

dy2 + sinh2(y)d2)


(X0)παβ = 2

ρ3


)


1

2Tαβ [](X0)παβ = 1

ρ|X0|2




X0 = 1ρ

( 12vL + 1

2 uL) partt = 1

2L +

1

2L (25)



(X0)P0 = T (partt X0) = 1

4

(v

u

) 12 |L|2 +

1

4

[(v

u

) 12

+(u

v

) 12]

|part|2 +1

4

(u

v

) 12 |L|2

(X0)PL = T (L X0) = 1

2

(v

u

) 12 |L|2 +

1

2

(u

v

) 12 |part|2







(vε

uε

) 12 asymp 1

(uε

vε






intS0

t2


C0[t1t2]

1

ρε

|Xε|2 dxdt =int

S0t1

(Xε )P0 dx


suptisin(01]

intS0

t

(X0)P0 dx +int int

C0[01]

1

ρ|X0|2 dxdt le E




S0t2


C0[t1t2]

1

ρε

|Xε|2 dxdt int

S0t1



suptisin(ε1]

intS0

t


C0[01]

1

ρε





intS0

1

(Xε )P0 dx E (27)





1

ρ|X0|2 1

ρε

|Xε|2 +ε2

ρ3 |partt|2

and also

intCε

(ε1]

ε2


Cε(ε1]

ε12

t32

|partt|2dxdt E




suptisin(ε1]

intSε

t

(X0)P0 dx +int int

Cε[ε1]

1

ρ|X0|2dxdt E (28)


intSδ

1

(X0)P0 dx leint

Sδt0

(X0)P0 dx +int

partCδ[t01]



intpartCδ[t01]

(u

v

) 12 |part|2 d A

(δ

t0

) 12int


(δ

t0

) 12

E


|L| (u

v

) 12 |X0| +

(u

v

)|L|

Thus we obtain

intSδ

1

(X0)P0 dx int

Sδt0

(X0)P0 dx +

(δ

t0

) 12

E +int

partCδ[t01]

((u

v

) 12 |X0φ|2 +

(u

v

) 32 |L|2

)d A





intS

δ11

(X0)P0 dx int

Sδ0t0

(X0)P0 dx +

((δ1

t0

) 12


)E (29)


intpartCδ[t01]

[(u

v

) 12 |X0φ|2 +

(u

v

) 32 |L|2



intC

δ0[t01]Cδ1[t01]

1

u

[(u

v

) 12 |X0φ|2 +

(u

v

) 32 |L|2

]dxdt E


1

u

(u

v

) 32 le δ

121

t32







2




H2a = minusgi jH2Sa

bc()partibpart j

c



intH2


St


)dx (30)




Chap II of [6])


rang



1

2

intD2

|nablax|2(

1 + r2

1 minus r2

)dx lt infin (31)





r=rn

|nabla|2dl = on(1)



12 (S1)

= 0












10 E0 and so that

X(n)L2(3Q) rarr 0 (32)



(n) rarr in H1(Q)




χ(n) X

1 12infin


where X1 1







(n)bulk

H32 minusε

tx

χ(n) X

1 12infin

1



Rn H1tx














H2 supe rarr M








E = limtrarr0

ESt []





Ft0 [] =int

partC(0t0]

(1

4|L|2 +

1

2|part|2

)d A




Ft0 [] ε8 E

intpartSt0

|part|2ds ε8

t0E (36)







2ε8 E (37)





Ftn lt ε12n E Thus






E[(n)] asymp E (40)




c) Decaying flux

F[εn 1][(n)] lt ε12n E (42)



(n)(1 xn)| gt ε (43)




[] Ft0 [] =int

partC[t0infin)

(1

4|L|2 +

1

2|part|2

)d A

it follows that



Ft0 [] le ε8 E






Ftk [] rarr 0 tk

intpartStk

|part|2d A rarr 0







E[(k)] rarr E (45)























(n)(1 x)| (

2k(1 + 2k |x |)minusN + ε4)

E12 k le 0




)E

12 k ge 0

(48)


2kn gt m(ε E)





2kn lt C lt infin









intS1


(|L(n)|2 + |part(n)|2

)dx E (50)



(n)(1 x)| ((


) 18

+ ε2)

E12 (51)




In general we have



1

(|1 minus x1| + x22 + εn)

12

(|L(n)|2 + |part(n)|2

)dx E (52)



(n)|Similarly




1

((1 minus x1)+ + x22 + εn)

12 + ε8

(|partt


(n)|2)

dx E (53)




kθ part2

where P1kθ and P2



kθ (ξ)| 1

|ξ2|



12 P2




t=1

1


12 + ε8

(|partt


(n)|2)

dx E (55)


|Pk(n)(1 x)| le|P1

kθ part1(n)(1 x)| + |P2

kθ part2(n)(1 x)|

(θ

12 + θminus 1

2

(((1 minus x1)+ + x2

2 + 2minusk + εn)14 + ε4

))E

12



|Pk(n)(1 x1 0)|


) 18

+ ε2)

E12

1

2lt x1 lt 2




) 18

+ ε2)

E12

1

2lt x1 lt 2




) 18

+ ε2)

E12

1

2lt x1 lt2








1

2

intt=1|x |ltγ1

(|partt


)dx ge E1 (56)


E(γ1) = 1

2

intt=1|x |ltγ1


)dx




Pk(1)Linfinx


E(γ1)12 +

((2k(1 minus γ1))

12 + ε4

)E

12





((2k(1 minus γ1))


)E

12




)E

12 |x | lt γ


ε E(γ1)12 +


12 + ε4

)E

12



ε E(γ1)12 +

((1 minus γ1)

12

(1 minus γ )12

+ ε4

)E

12


ε E(γ1)12 +

((1 minus γ1

1 minus γ

)N

+ ε4

)E

12



12n ε


so that

1

2

int|x |ltγ2t

(|partt



12n ε

14n ] (57)



intCεn

[ε12

n ε

14

n ]






according to



intS

t01

(X0)P0[(n)] dx int

Sδ0t0



(1 minus γ1)12

intS

1minusγ11


2

intS

δ0t0




minus 12 E

Sδ0t0






12 E1

we obtain (57)



14n ]


12n ε




1

ρ|X0

(n)|2dxdt ln N

| ln εn| E (59)





a) Bounded energy



ES

(1minusγ2)tt

[(n)](t) ge E2 t isin [1 Tn] (61)


Cε

12

n[1Tn ]

1

ρ|X0






[ 12 infin)


Then we prove that





10E0 (63)



rminus1n

int tn+rn2

tnminusrn2

intB(xn r)




EB(xr)[(n)](t) le 1


ES

(1minusγ2)tt


C j








2


j Ckj

into N sets

Cklj =

(t x) isin Ck

j 1

4+

l minus 1


1

4+

l

4N

l = 1 N


Nsuml=1

intIk

ECklj

[(n)](t)dt leint

Ik

ECkj[(n)](t)dt le 1

10NE


ln+1suml=lnminus1

intIk

ECklj

[(n)](t)dt le 3

10N 2 E


jn+1sumj= jnminus1

ECklj

[(n)](tn) le 3

NE





ECklnj

[(n)(t)] le 3

NE t isin Ik


3

NE le 1

20E0






j





EB(xr)[(n)](t)




10 r0 otherwise





limnrarrinfin inf

Ikrn = 0


α2n =

intC j

|X0(n)|2dxdt




j Hence if



of this case




g(t) =int

Cklelnj

|X0(n)(t)|2dx



Ik

g(t) = α2n


I =int

Ik

1

rn(t)

int t+rn(t)2



I le 2int

Ik

int t+rn(t)2

tminusrn(t)2

1

rn(s)g(s)dsdt le 4

intIk

g(s)ds = 4α2n


1

rn(tn)

int tn+rn(tn)2


n | Ik |minus1








2

1

2]


supx

EB(x1)[(n)](t) le 1

10E0 t isin [minus1

2

1

2]


EB(01)[(n)](0) = 1

10E0




[minus 12 1

2 ]timesB(x1)



(n) rarr in H1loc


2

1

2])



2 12 ] times R


1

10E0 le E[] le E[]

and

X0(t0 x0) = 0





(n) rarr in H1loc(C

2[2infin))



0 lt E2 le supt2



X0 = 0






References
























Math 47(5) 719ndash754 (1994)



















Abstract

1 Introduction













66 Final rescaling



Acknowledgements

References










1

2

intt=1|x |ltγ1

(|partt


)dx ge E1




14n ]

1

2

int|x |ltγ2(εE)t

(|partt



12n ε

14n ]



Cεn

[ε12

n ε

14

n ]






ES

(1minusγ2)tt


ε

12

n[1Tn ]

1

ρ|X0





(t0 x0) inside C12

[ 12 infin)



10E0


10E0 x isin B(x0 r)

rminus1n

int tn+rn2

tnminusrn2

intB(x0r)

|X0(n)|2dxdt rarr 0





ES

(1minusγ2)tt


|X0(n)|2dxdt rarr 0




2 times [minus 12 1




H2 rarr M


D2 rarr M



|nablax|2 dx

1 minus rlt infin





Tαβ [] = mi j ()



] (15)





nablaα(X)Pα = 1

2Tαβ [](X)παβ (16)




(X)π = LX g




intSδ

t2

(X)P0 dx +1

2

int intCδ[t1t2]


Sδt1

(X)P0 dx +int

partCδ[t1t2]

(X)PL d A (17)



(Y )π = 0 (Y )P0 = 1


4|L|2 +

1

2|part|2



intpartCδ[δ1]

1

4|L|2 +

1

2|part|2 d A le E[] (18)









dy2 + sinh2(y)d2) (21)




partt = t

ρpartρ minus r

ρ2 party

so in particular




+cosh(y)

2ρ2

(|party|2 +

1

sinh2(y)|part|2

)



1

2

intH2


Here


sinh2(y)|part|2




Xε = 1

ρε


(t + ε)2 minus r2 (24)




LX0 g = 2ρ(

dy2 + sinh2(y)d2)


(X0)παβ = 2

ρ3


)


1

2Tαβ [](X0)παβ = 1

ρ|X0|2




X0 = 1ρ

( 12vL + 1

2 uL) partt = 1

2L +

1

2L (25)



(X0)P0 = T (partt X0) = 1

4

(v

u

) 12 |L|2 +

1

4

[(v

u

) 12

+(u

v

) 12]

|part|2 +1

4

(u

v

) 12 |L|2

(X0)PL = T (L X0) = 1

2

(v

u

) 12 |L|2 +

1

2

(u

v

) 12 |part|2







(vε

uε

) 12 asymp 1

(uε

vε






intS0

t2


C0[t1t2]

1

ρε

|Xε|2 dxdt =int

S0t1

(Xε )P0 dx


suptisin(01]

intS0

t

(X0)P0 dx +int int

C0[01]

1

ρ|X0|2 dxdt le E




S0t2


C0[t1t2]

1

ρε

|Xε|2 dxdt int

S0t1



suptisin(ε1]

intS0

t


C0[01]

1

ρε





intS0

1

(Xε )P0 dx E (27)





1

ρ|X0|2 1

ρε

|Xε|2 +ε2

ρ3 |partt|2

and also

intCε

(ε1]

ε2


Cε(ε1]

ε12

t32

|partt|2dxdt E




suptisin(ε1]

intSε

t

(X0)P0 dx +int int

Cε[ε1]

1

ρ|X0|2dxdt E (28)


intSδ

1

(X0)P0 dx leint

Sδt0

(X0)P0 dx +int

partCδ[t01]



intpartCδ[t01]

(u

v

) 12 |part|2 d A

(δ

t0

) 12int


(δ

t0

) 12

E


|L| (u

v

) 12 |X0| +

(u

v

)|L|

Thus we obtain

intSδ

1

(X0)P0 dx int

Sδt0

(X0)P0 dx +

(δ

t0

) 12

E +int

partCδ[t01]

((u

v

) 12 |X0φ|2 +

(u

v

) 32 |L|2

)d A





intS

δ11

(X0)P0 dx int

Sδ0t0

(X0)P0 dx +

((δ1

t0

) 12


)E (29)


intpartCδ[t01]

[(u

v

) 12 |X0φ|2 +

(u

v

) 32 |L|2



intC

δ0[t01]Cδ1[t01]

1

u

[(u

v

) 12 |X0φ|2 +

(u

v

) 32 |L|2

]dxdt E


1

u

(u

v

) 32 le δ

121

t32







2




H2a = minusgi jH2Sa

bc()partibpart j

c



intH2


St


)dx (30)




Chap II of [6])


rang



1

2

intD2

|nablax|2(

1 + r2

1 minus r2

)dx lt infin (31)





r=rn

|nabla|2dl = on(1)



12 (S1)

= 0












10 E0 and so that

X(n)L2(3Q) rarr 0 (32)



(n) rarr in H1(Q)




χ(n) X

1 12infin


where X1 1







(n)bulk

H32 minusε

tx

χ(n) X

1 12infin

1



Rn H1tx














H2 supe rarr M








E = limtrarr0

ESt []





Ft0 [] =int

partC(0t0]

(1

4|L|2 +

1

2|part|2

)d A




Ft0 [] ε8 E

intpartSt0

|part|2ds ε8

t0E (36)







2ε8 E (37)





Ftn lt ε12n E Thus






E[(n)] asymp E (40)




c) Decaying flux

F[εn 1][(n)] lt ε12n E (42)



(n)(1 xn)| gt ε (43)




[] Ft0 [] =int

partC[t0infin)

(1

4|L|2 +

1

2|part|2

)d A

it follows that



Ft0 [] le ε8 E






Ftk [] rarr 0 tk

intpartStk

|part|2d A rarr 0







E[(k)] rarr E (45)























(n)(1 x)| (

2k(1 + 2k |x |)minusN + ε4)

E12 k le 0




)E

12 k ge 0

(48)


2kn gt m(ε E)





2kn lt C lt infin









intS1


(|L(n)|2 + |part(n)|2

)dx E (50)



(n)(1 x)| ((


) 18

+ ε2)

E12 (51)




In general we have



1

(|1 minus x1| + x22 + εn)

12

(|L(n)|2 + |part(n)|2

)dx E (52)



(n)|Similarly




1

((1 minus x1)+ + x22 + εn)

12 + ε8

(|partt


(n)|2)

dx E (53)




kθ part2

where P1kθ and P2



kθ (ξ)| 1

|ξ2|



12 P2




t=1

1


12 + ε8

(|partt


(n)|2)

dx E (55)


|Pk(n)(1 x)| le|P1

kθ part1(n)(1 x)| + |P2

kθ part2(n)(1 x)|

(θ

12 + θminus 1

2

(((1 minus x1)+ + x2

2 + 2minusk + εn)14 + ε4

))E

12



|Pk(n)(1 x1 0)|


) 18

+ ε2)

E12

1

2lt x1 lt 2




) 18

+ ε2)

E12

1

2lt x1 lt 2




) 18

+ ε2)

E12

1

2lt x1 lt2








1

2

intt=1|x |ltγ1

(|partt


)dx ge E1 (56)


E(γ1) = 1

2

intt=1|x |ltγ1


)dx




Pk(1)Linfinx


E(γ1)12 +

((2k(1 minus γ1))

12 + ε4

)E

12





((2k(1 minus γ1))


)E

12




)E

12 |x | lt γ


ε E(γ1)12 +


12 + ε4

)E

12



ε E(γ1)12 +

((1 minus γ1)

12

(1 minus γ )12

+ ε4

)E

12


ε E(γ1)12 +

((1 minus γ1

1 minus γ

)N

+ ε4

)E

12



12n ε


so that

1

2

int|x |ltγ2t

(|partt



12n ε

14n ] (57)



intCεn

[ε12

n ε

14

n ]






according to



intS

t01

(X0)P0[(n)] dx int

Sδ0t0



(1 minus γ1)12

intS

1minusγ11


2

intS

δ0t0




minus 12 E

Sδ0t0






12 E1

we obtain (57)



14n ]


12n ε




1

ρ|X0

(n)|2dxdt ln N

| ln εn| E (59)





a) Bounded energy



ES

(1minusγ2)tt

[(n)](t) ge E2 t isin [1 Tn] (61)


Cε

12

n[1Tn ]

1

ρ|X0






[ 12 infin)


Then we prove that





10E0 (63)



rminus1n

int tn+rn2

tnminusrn2

intB(xn r)




EB(xr)[(n)](t) le 1


ES

(1minusγ2)tt


C j








2


j Ckj

into N sets

Cklj =

(t x) isin Ck

j 1

4+

l minus 1


1

4+

l

4N

l = 1 N


Nsuml=1

intIk

ECklj

[(n)](t)dt leint

Ik

ECkj[(n)](t)dt le 1

10NE


ln+1suml=lnminus1

intIk

ECklj

[(n)](t)dt le 3

10N 2 E


jn+1sumj= jnminus1

ECklj

[(n)](tn) le 3

NE





ECklnj

[(n)(t)] le 3

NE t isin Ik


3

NE le 1

20E0






j





EB(xr)[(n)](t)




10 r0 otherwise





limnrarrinfin inf

Ikrn = 0


α2n =

intC j

|X0(n)|2dxdt




j Hence if



of this case




g(t) =int

Cklelnj

|X0(n)(t)|2dx



Ik

g(t) = α2n


I =int

Ik

1

rn(t)

int t+rn(t)2



I le 2int

Ik

int t+rn(t)2

tminusrn(t)2

1

rn(s)g(s)dsdt le 4

intIk

g(s)ds = 4α2n


1

rn(tn)

int tn+rn(tn)2


n | Ik |minus1








2

1

2]


supx

EB(x1)[(n)](t) le 1

10E0 t isin [minus1

2

1

2]


EB(01)[(n)](0) = 1

10E0




[minus 12 1

2 ]timesB(x1)



(n) rarr in H1loc


2

1

2])



2 12 ] times R


1

10E0 le E[] le E[]

and

X0(t0 x0) = 0





(n) rarr in H1loc(C

2[2infin))



0 lt E2 le supt2



X0 = 0






References
























Math 47(5) 719ndash754 (1994)



















Abstract

1 Introduction













66 Final rescaling



Acknowledgements

References





(t0 x0) inside C12

[ 12 infin)



10E0


10E0 x isin B(x0 r)

rminus1n

int tn+rn2

tnminusrn2

intB(x0r)

|X0(n)|2dxdt rarr 0





ES

(1minusγ2)tt


|X0(n)|2dxdt rarr 0




2 times [minus 12 1




H2 rarr M


D2 rarr M



|nablax|2 dx

1 minus rlt infin





Tαβ [] = mi j ()



] (15)





nablaα(X)Pα = 1

2Tαβ [](X)παβ (16)




(X)π = LX g




intSδ

t2

(X)P0 dx +1

2

int intCδ[t1t2]


Sδt1

(X)P0 dx +int

partCδ[t1t2]

(X)PL d A (17)



(Y )π = 0 (Y )P0 = 1


4|L|2 +

1

2|part|2



intpartCδ[δ1]

1

4|L|2 +

1

2|part|2 d A le E[] (18)









dy2 + sinh2(y)d2) (21)




partt = t

ρpartρ minus r

ρ2 party

so in particular




+cosh(y)

2ρ2

(|party|2 +

1

sinh2(y)|part|2

)



1

2

intH2


Here


sinh2(y)|part|2




Xε = 1

ρε


(t + ε)2 minus r2 (24)




LX0 g = 2ρ(

dy2 + sinh2(y)d2)


(X0)παβ = 2

ρ3


)


1

2Tαβ [](X0)παβ = 1

ρ|X0|2




X0 = 1ρ

( 12vL + 1

2 uL) partt = 1

2L +

1

2L (25)



(X0)P0 = T (partt X0) = 1

4

(v

u

) 12 |L|2 +

1

4

[(v

u

) 12

+(u

v

) 12]

|part|2 +1

4

(u

v

) 12 |L|2

(X0)PL = T (L X0) = 1

2

(v

u

) 12 |L|2 +

1

2

(u

v

) 12 |part|2







(vε

uε

) 12 asymp 1

(uε

vε






intS0

t2


C0[t1t2]

1

ρε

|Xε|2 dxdt =int

S0t1

(Xε )P0 dx


suptisin(01]

intS0

t

(X0)P0 dx +int int

C0[01]

1

ρ|X0|2 dxdt le E




S0t2


C0[t1t2]

1

ρε

|Xε|2 dxdt int

S0t1



suptisin(ε1]

intS0

t


C0[01]

1

ρε





intS0

1

(Xε )P0 dx E (27)





1

ρ|X0|2 1

ρε

|Xε|2 +ε2

ρ3 |partt|2

and also

intCε

(ε1]

ε2


Cε(ε1]

ε12

t32

|partt|2dxdt E




suptisin(ε1]

intSε

t

(X0)P0 dx +int int

Cε[ε1]

1

ρ|X0|2dxdt E (28)


intSδ

1

(X0)P0 dx leint

Sδt0

(X0)P0 dx +int

partCδ[t01]



intpartCδ[t01]

(u

v

) 12 |part|2 d A

(δ

t0

) 12int


(δ

t0

) 12

E


|L| (u

v

) 12 |X0| +

(u

v

)|L|

Thus we obtain

intSδ

1

(X0)P0 dx int

Sδt0

(X0)P0 dx +

(δ

t0

) 12

E +int

partCδ[t01]

((u

v

) 12 |X0φ|2 +

(u

v

) 32 |L|2

)d A





intS

δ11

(X0)P0 dx int

Sδ0t0

(X0)P0 dx +

((δ1

t0

) 12


)E (29)


intpartCδ[t01]

[(u

v

) 12 |X0φ|2 +

(u

v

) 32 |L|2



intC

δ0[t01]Cδ1[t01]

1

u

[(u

v

) 12 |X0φ|2 +

(u

v

) 32 |L|2

]dxdt E


1

u

(u

v

) 32 le δ

121

t32







2




H2a = minusgi jH2Sa

bc()partibpart j

c



intH2


St


)dx (30)




Chap II of [6])


rang



1

2

intD2

|nablax|2(

1 + r2

1 minus r2

)dx lt infin (31)





r=rn

|nabla|2dl = on(1)



12 (S1)

= 0












10 E0 and so that

X(n)L2(3Q) rarr 0 (32)



(n) rarr in H1(Q)




χ(n) X

1 12infin


where X1 1







(n)bulk

H32 minusε

tx

χ(n) X

1 12infin

1



Rn H1tx














H2 supe rarr M








E = limtrarr0

ESt []





Ft0 [] =int

partC(0t0]

(1

4|L|2 +

1

2|part|2

)d A




Ft0 [] ε8 E

intpartSt0

|part|2ds ε8

t0E (36)







2ε8 E (37)





Ftn lt ε12n E Thus






E[(n)] asymp E (40)




c) Decaying flux

F[εn 1][(n)] lt ε12n E (42)



(n)(1 xn)| gt ε (43)




[] Ft0 [] =int

partC[t0infin)

(1

4|L|2 +

1

2|part|2

)d A

it follows that



Ft0 [] le ε8 E






Ftk [] rarr 0 tk

intpartStk

|part|2d A rarr 0







E[(k)] rarr E (45)























(n)(1 x)| (

2k(1 + 2k |x |)minusN + ε4)

E12 k le 0




)E

12 k ge 0

(48)


2kn gt m(ε E)





2kn lt C lt infin









intS1


(|L(n)|2 + |part(n)|2

)dx E (50)



(n)(1 x)| ((


) 18

+ ε2)

E12 (51)




In general we have



1

(|1 minus x1| + x22 + εn)

12

(|L(n)|2 + |part(n)|2

)dx E (52)



(n)|Similarly




1

((1 minus x1)+ + x22 + εn)

12 + ε8

(|partt


(n)|2)

dx E (53)




kθ part2

where P1kθ and P2



kθ (ξ)| 1

|ξ2|



12 P2




t=1

1


12 + ε8

(|partt


(n)|2)

dx E (55)


|Pk(n)(1 x)| le|P1

kθ part1(n)(1 x)| + |P2

kθ part2(n)(1 x)|

(θ

12 + θminus 1

2

(((1 minus x1)+ + x2

2 + 2minusk + εn)14 + ε4

))E

12



|Pk(n)(1 x1 0)|


) 18

+ ε2)

E12

1

2lt x1 lt 2




) 18

+ ε2)

E12

1

2lt x1 lt 2




) 18

+ ε2)

E12

1

2lt x1 lt2








1

2

intt=1|x |ltγ1

(|partt


)dx ge E1 (56)


E(γ1) = 1

2

intt=1|x |ltγ1


)dx




Pk(1)Linfinx


E(γ1)12 +

((2k(1 minus γ1))

12 + ε4

)E

12





((2k(1 minus γ1))


)E

12




)E

12 |x | lt γ


ε E(γ1)12 +


12 + ε4

)E

12



ε E(γ1)12 +

((1 minus γ1)

12

(1 minus γ )12

+ ε4

)E

12


ε E(γ1)12 +

((1 minus γ1

1 minus γ

)N

+ ε4

)E

12



12n ε


so that

1

2

int|x |ltγ2t

(|partt



12n ε

14n ] (57)



intCεn

[ε12

n ε

14

n ]






according to



intS

t01

(X0)P0[(n)] dx int

Sδ0t0



(1 minus γ1)12

intS

1minusγ11


2

intS

δ0t0




minus 12 E

Sδ0t0






12 E1

we obtain (57)



14n ]


12n ε




1

ρ|X0

(n)|2dxdt ln N

| ln εn| E (59)





a) Bounded energy



ES

(1minusγ2)tt

[(n)](t) ge E2 t isin [1 Tn] (61)


Cε

12

n[1Tn ]

1

ρ|X0






[ 12 infin)


Then we prove that





10E0 (63)



rminus1n

int tn+rn2

tnminusrn2

intB(xn r)




EB(xr)[(n)](t) le 1


ES

(1minusγ2)tt


C j








2


j Ckj

into N sets

Cklj =

(t x) isin Ck

j 1

4+

l minus 1


1

4+

l

4N

l = 1 N


Nsuml=1

intIk

ECklj

[(n)](t)dt leint

Ik

ECkj[(n)](t)dt le 1

10NE


ln+1suml=lnminus1

intIk

ECklj

[(n)](t)dt le 3

10N 2 E


jn+1sumj= jnminus1

ECklj

[(n)](tn) le 3

NE





ECklnj

[(n)(t)] le 3

NE t isin Ik


3

NE le 1

20E0






j





EB(xr)[(n)](t)




10 r0 otherwise





limnrarrinfin inf

Ikrn = 0


α2n =

intC j

|X0(n)|2dxdt




j Hence if



of this case




g(t) =int

Cklelnj

|X0(n)(t)|2dx



Ik

g(t) = α2n


I =int

Ik

1

rn(t)

int t+rn(t)2



I le 2int

Ik

int t+rn(t)2

tminusrn(t)2

1

rn(s)g(s)dsdt le 4

intIk

g(s)ds = 4α2n


1

rn(tn)

int tn+rn(tn)2


n | Ik |minus1








2

1

2]


supx

EB(x1)[(n)](t) le 1

10E0 t isin [minus1

2

1

2]


EB(01)[(n)](0) = 1

10E0




[minus 12 1

2 ]timesB(x1)



(n) rarr in H1loc


2

1

2])



2 12 ] times R


1

10E0 le E[] le E[]

and

X0(t0 x0) = 0





(n) rarr in H1loc(C

2[2infin))



0 lt E2 le supt2



X0 = 0






References
























Math 47(5) 719ndash754 (1994)



















Abstract

1 Introduction













66 Final rescaling



Acknowledgements

References





Tαβ [] = mi j ()



] (15)





nablaα(X)Pα = 1

2Tαβ [](X)παβ (16)




(X)π = LX g




intSδ

t2

(X)P0 dx +1

2

int intCδ[t1t2]


Sδt1

(X)P0 dx +int

partCδ[t1t2]

(X)PL d A (17)



(Y )π = 0 (Y )P0 = 1


4|L|2 +

1

2|part|2



intpartCδ[δ1]

1

4|L|2 +

1

2|part|2 d A le E[] (18)









dy2 + sinh2(y)d2) (21)




partt = t

ρpartρ minus r

ρ2 party

so in particular




+cosh(y)

2ρ2

(|party|2 +

1

sinh2(y)|part|2

)



1

2

intH2


Here


sinh2(y)|part|2




Xε = 1

ρε


(t + ε)2 minus r2 (24)




LX0 g = 2ρ(

dy2 + sinh2(y)d2)


(X0)παβ = 2

ρ3


)


1

2Tαβ [](X0)παβ = 1

ρ|X0|2




X0 = 1ρ

( 12vL + 1

2 uL) partt = 1

2L +

1

2L (25)



(X0)P0 = T (partt X0) = 1

4

(v

u

) 12 |L|2 +

1

4

[(v

u

) 12

+(u

v

) 12]

|part|2 +1

4

(u

v

) 12 |L|2

(X0)PL = T (L X0) = 1

2

(v

u

) 12 |L|2 +

1

2

(u

v

) 12 |part|2







(vε

uε

) 12 asymp 1

(uε

vε






intS0

t2


C0[t1t2]

1

ρε

|Xε|2 dxdt =int

S0t1

(Xε )P0 dx


suptisin(01]

intS0

t

(X0)P0 dx +int int

C0[01]

1

ρ|X0|2 dxdt le E




S0t2


C0[t1t2]

1

ρε

|Xε|2 dxdt int

S0t1



suptisin(ε1]

intS0

t


C0[01]

1

ρε





intS0

1

(Xε )P0 dx E (27)





1

ρ|X0|2 1

ρε

|Xε|2 +ε2

ρ3 |partt|2

and also

intCε

(ε1]

ε2


Cε(ε1]

ε12

t32

|partt|2dxdt E




suptisin(ε1]

intSε

t

(X0)P0 dx +int int

Cε[ε1]

1

ρ|X0|2dxdt E (28)


intSδ

1

(X0)P0 dx leint

Sδt0

(X0)P0 dx +int

partCδ[t01]



intpartCδ[t01]

(u

v

) 12 |part|2 d A

(δ

t0

) 12int


(δ

t0

) 12

E


|L| (u

v

) 12 |X0| +

(u

v

)|L|

Thus we obtain

intSδ

1

(X0)P0 dx int

Sδt0

(X0)P0 dx +

(δ

t0

) 12

E +int

partCδ[t01]

((u

v

) 12 |X0φ|2 +

(u

v

) 32 |L|2

)d A





intS

δ11

(X0)P0 dx int

Sδ0t0

(X0)P0 dx +

((δ1

t0

) 12


)E (29)


intpartCδ[t01]

[(u

v

) 12 |X0φ|2 +

(u

v

) 32 |L|2



intC

δ0[t01]Cδ1[t01]

1

u

[(u

v

) 12 |X0φ|2 +

(u

v

) 32 |L|2

]dxdt E


1

u

(u

v

) 32 le δ

121

t32







2




H2a = minusgi jH2Sa

bc()partibpart j

c



intH2


St


)dx (30)




Chap II of [6])


rang



1

2

intD2

|nablax|2(

1 + r2

1 minus r2

)dx lt infin (31)





r=rn

|nabla|2dl = on(1)



12 (S1)

= 0












10 E0 and so that

X(n)L2(3Q) rarr 0 (32)



(n) rarr in H1(Q)




χ(n) X

1 12infin


where X1 1







(n)bulk

H32 minusε

tx

χ(n) X

1 12infin

1



Rn H1tx














H2 supe rarr M








E = limtrarr0

ESt []





Ft0 [] =int

partC(0t0]

(1

4|L|2 +

1

2|part|2

)d A




Ft0 [] ε8 E

intpartSt0

|part|2ds ε8

t0E (36)







2ε8 E (37)





Ftn lt ε12n E Thus






E[(n)] asymp E (40)




c) Decaying flux

F[εn 1][(n)] lt ε12n E (42)



(n)(1 xn)| gt ε (43)




[] Ft0 [] =int

partC[t0infin)

(1

4|L|2 +

1

2|part|2

)d A

it follows that



Ft0 [] le ε8 E






Ftk [] rarr 0 tk

intpartStk

|part|2d A rarr 0







E[(k)] rarr E (45)























(n)(1 x)| (

2k(1 + 2k |x |)minusN + ε4)

E12 k le 0




)E

12 k ge 0

(48)


2kn gt m(ε E)





2kn lt C lt infin









intS1


(|L(n)|2 + |part(n)|2

)dx E (50)



(n)(1 x)| ((


) 18

+ ε2)

E12 (51)




In general we have



1

(|1 minus x1| + x22 + εn)

12

(|L(n)|2 + |part(n)|2

)dx E (52)



(n)|Similarly




1

((1 minus x1)+ + x22 + εn)

12 + ε8

(|partt


(n)|2)

dx E (53)




kθ part2

where P1kθ and P2



kθ (ξ)| 1

|ξ2|



12 P2




t=1

1


12 + ε8

(|partt


(n)|2)

dx E (55)


|Pk(n)(1 x)| le|P1

kθ part1(n)(1 x)| + |P2

kθ part2(n)(1 x)|

(θ

12 + θminus 1

2

(((1 minus x1)+ + x2

2 + 2minusk + εn)14 + ε4

))E

12



|Pk(n)(1 x1 0)|


) 18

+ ε2)

E12

1

2lt x1 lt 2




) 18

+ ε2)

E12

1

2lt x1 lt 2




) 18

+ ε2)

E12

1

2lt x1 lt2








1

2

intt=1|x |ltγ1

(|partt


)dx ge E1 (56)


E(γ1) = 1

2

intt=1|x |ltγ1


)dx




Pk(1)Linfinx


E(γ1)12 +

((2k(1 minus γ1))

12 + ε4

)E

12





((2k(1 minus γ1))


)E

12




)E

12 |x | lt γ


ε E(γ1)12 +


12 + ε4

)E

12



ε E(γ1)12 +

((1 minus γ1)

12

(1 minus γ )12

+ ε4

)E

12


ε E(γ1)12 +

((1 minus γ1

1 minus γ

)N

+ ε4

)E

12



12n ε


so that

1

2

int|x |ltγ2t

(|partt



12n ε

14n ] (57)



intCεn

[ε12

n ε

14

n ]






according to



intS

t01

(X0)P0[(n)] dx int

Sδ0t0



(1 minus γ1)12

intS

1minusγ11


2

intS

δ0t0




minus 12 E

Sδ0t0






12 E1

we obtain (57)



14n ]


12n ε




1

ρ|X0

(n)|2dxdt ln N

| ln εn| E (59)





a) Bounded energy



ES

(1minusγ2)tt

[(n)](t) ge E2 t isin [1 Tn] (61)


Cε

12

n[1Tn ]

1

ρ|X0






[ 12 infin)


Then we prove that





10E0 (63)



rminus1n

int tn+rn2

tnminusrn2

intB(xn r)




EB(xr)[(n)](t) le 1


ES

(1minusγ2)tt


C j








2


j Ckj

into N sets

Cklj =

(t x) isin Ck

j 1

4+

l minus 1


1

4+

l

4N

l = 1 N


Nsuml=1

intIk

ECklj

[(n)](t)dt leint

Ik

ECkj[(n)](t)dt le 1

10NE


ln+1suml=lnminus1

intIk

ECklj

[(n)](t)dt le 3

10N 2 E


jn+1sumj= jnminus1

ECklj

[(n)](tn) le 3

NE





ECklnj

[(n)(t)] le 3

NE t isin Ik


3

NE le 1

20E0






j





EB(xr)[(n)](t)




10 r0 otherwise





limnrarrinfin inf

Ikrn = 0


α2n =

intC j

|X0(n)|2dxdt




j Hence if



of this case




g(t) =int

Cklelnj

|X0(n)(t)|2dx



Ik

g(t) = α2n


I =int

Ik

1

rn(t)

int t+rn(t)2



I le 2int

Ik

int t+rn(t)2

tminusrn(t)2

1

rn(s)g(s)dsdt le 4

intIk

g(s)ds = 4α2n


1

rn(tn)

int tn+rn(tn)2


n | Ik |minus1








2

1

2]


supx

EB(x1)[(n)](t) le 1

10E0 t isin [minus1

2

1

2]


EB(01)[(n)](0) = 1

10E0




[minus 12 1

2 ]timesB(x1)



(n) rarr in H1loc


2

1

2])



2 12 ] times R


1

10E0 le E[] le E[]

and

X0(t0 x0) = 0





(n) rarr in H1loc(C

2[2infin))



0 lt E2 le supt2



X0 = 0






References
























Math 47(5) 719ndash754 (1994)



















Abstract

1 Introduction













66 Final rescaling



Acknowledgements

References




intpartCδ[δ1]

1

4|L|2 +

1

2|part|2 d A le E[] (18)









dy2 + sinh2(y)d2) (21)




partt = t

ρpartρ minus r

ρ2 party

so in particular




+cosh(y)

2ρ2

(|party|2 +

1

sinh2(y)|part|2

)



1

2

intH2


Here


sinh2(y)|part|2




Xε = 1

ρε


(t + ε)2 minus r2 (24)




LX0 g = 2ρ(

dy2 + sinh2(y)d2)


(X0)παβ = 2

ρ3


)


1

2Tαβ [](X0)παβ = 1

ρ|X0|2




X0 = 1ρ

( 12vL + 1

2 uL) partt = 1

2L +

1

2L (25)



(X0)P0 = T (partt X0) = 1

4

(v

u

) 12 |L|2 +

1

4

[(v

u

) 12

+(u

v

) 12]

|part|2 +1

4

(u

v

) 12 |L|2

(X0)PL = T (L X0) = 1

2

(v

u

) 12 |L|2 +

1

2

(u

v

) 12 |part|2







(vε

uε

) 12 asymp 1

(uε

vε






intS0

t2


C0[t1t2]

1

ρε

|Xε|2 dxdt =int

S0t1

(Xε )P0 dx


suptisin(01]

intS0

t

(X0)P0 dx +int int

C0[01]

1

ρ|X0|2 dxdt le E




S0t2


C0[t1t2]

1

ρε

|Xε|2 dxdt int

S0t1



suptisin(ε1]

intS0

t


C0[01]

1

ρε





intS0

1

(Xε )P0 dx E (27)





1

ρ|X0|2 1

ρε

|Xε|2 +ε2

ρ3 |partt|2

and also

intCε

(ε1]

ε2


Cε(ε1]

ε12

t32

|partt|2dxdt E




suptisin(ε1]

intSε

t

(X0)P0 dx +int int

Cε[ε1]

1

ρ|X0|2dxdt E (28)


intSδ

1

(X0)P0 dx leint

Sδt0

(X0)P0 dx +int

partCδ[t01]



intpartCδ[t01]

(u

v

) 12 |part|2 d A

(δ

t0

) 12int


(δ

t0

) 12

E


|L| (u

v

) 12 |X0| +

(u

v

)|L|

Thus we obtain

intSδ

1

(X0)P0 dx int

Sδt0

(X0)P0 dx +

(δ

t0

) 12

E +int

partCδ[t01]

((u

v

) 12 |X0φ|2 +

(u

v

) 32 |L|2

)d A





intS

δ11

(X0)P0 dx int

Sδ0t0

(X0)P0 dx +

((δ1

t0

) 12


)E (29)


intpartCδ[t01]

[(u

v

) 12 |X0φ|2 +

(u

v

) 32 |L|2



intC

δ0[t01]Cδ1[t01]

1

u

[(u

v

) 12 |X0φ|2 +

(u

v

) 32 |L|2

]dxdt E


1

u

(u

v

) 32 le δ

121

t32







2




H2a = minusgi jH2Sa

bc()partibpart j

c



intH2


St


)dx (30)




Chap II of [6])


rang



1

2

intD2

|nablax|2(

1 + r2

1 minus r2

)dx lt infin (31)





r=rn

|nabla|2dl = on(1)



12 (S1)

= 0












10 E0 and so that

X(n)L2(3Q) rarr 0 (32)



(n) rarr in H1(Q)




χ(n) X

1 12infin


where X1 1







(n)bulk

H32 minusε

tx

χ(n) X

1 12infin

1



Rn H1tx














H2 supe rarr M








E = limtrarr0

ESt []





Ft0 [] =int

partC(0t0]

(1

4|L|2 +

1

2|part|2

)d A




Ft0 [] ε8 E

intpartSt0

|part|2ds ε8

t0E (36)







2ε8 E (37)





Ftn lt ε12n E Thus






E[(n)] asymp E (40)




c) Decaying flux

F[εn 1][(n)] lt ε12n E (42)



(n)(1 xn)| gt ε (43)




[] Ft0 [] =int

partC[t0infin)

(1

4|L|2 +

1

2|part|2

)d A

it follows that



Ft0 [] le ε8 E






Ftk [] rarr 0 tk

intpartStk

|part|2d A rarr 0







E[(k)] rarr E (45)























(n)(1 x)| (

2k(1 + 2k |x |)minusN + ε4)

E12 k le 0




)E

12 k ge 0

(48)


2kn gt m(ε E)





2kn lt C lt infin









intS1


(|L(n)|2 + |part(n)|2

)dx E (50)



(n)(1 x)| ((


) 18

+ ε2)

E12 (51)




In general we have



1

(|1 minus x1| + x22 + εn)

12

(|L(n)|2 + |part(n)|2

)dx E (52)



(n)|Similarly




1

((1 minus x1)+ + x22 + εn)

12 + ε8

(|partt


(n)|2)

dx E (53)




kθ part2

where P1kθ and P2



kθ (ξ)| 1

|ξ2|



12 P2




t=1

1


12 + ε8

(|partt


(n)|2)

dx E (55)


|Pk(n)(1 x)| le|P1

kθ part1(n)(1 x)| + |P2

kθ part2(n)(1 x)|

(θ

12 + θminus 1

2

(((1 minus x1)+ + x2

2 + 2minusk + εn)14 + ε4

))E

12



|Pk(n)(1 x1 0)|


) 18

+ ε2)

E12

1

2lt x1 lt 2




) 18

+ ε2)

E12

1

2lt x1 lt 2




) 18

+ ε2)

E12

1

2lt x1 lt2








1

2

intt=1|x |ltγ1

(|partt


)dx ge E1 (56)


E(γ1) = 1

2

intt=1|x |ltγ1


)dx




Pk(1)Linfinx


E(γ1)12 +

((2k(1 minus γ1))

12 + ε4

)E

12





((2k(1 minus γ1))


)E

12




)E

12 |x | lt γ


ε E(γ1)12 +


12 + ε4

)E

12



ε E(γ1)12 +

((1 minus γ1)

12

(1 minus γ )12

+ ε4

)E

12


ε E(γ1)12 +

((1 minus γ1

1 minus γ

)N

+ ε4

)E

12



12n ε


so that

1

2

int|x |ltγ2t

(|partt



12n ε

14n ] (57)



intCεn

[ε12

n ε

14

n ]






according to



intS

t01

(X0)P0[(n)] dx int

Sδ0t0



(1 minus γ1)12

intS

1minusγ11


2

intS

δ0t0




minus 12 E

Sδ0t0






12 E1

we obtain (57)



14n ]


12n ε




1

ρ|X0

(n)|2dxdt ln N

| ln εn| E (59)





a) Bounded energy



ES

(1minusγ2)tt

[(n)](t) ge E2 t isin [1 Tn] (61)


Cε

12

n[1Tn ]

1

ρ|X0






[ 12 infin)


Then we prove that





10E0 (63)



rminus1n

int tn+rn2

tnminusrn2

intB(xn r)




EB(xr)[(n)](t) le 1


ES

(1minusγ2)tt


C j








2


j Ckj

into N sets

Cklj =

(t x) isin Ck

j 1

4+

l minus 1


1

4+

l

4N

l = 1 N


Nsuml=1

intIk

ECklj

[(n)](t)dt leint

Ik

ECkj[(n)](t)dt le 1

10NE


ln+1suml=lnminus1

intIk

ECklj

[(n)](t)dt le 3

10N 2 E


jn+1sumj= jnminus1

ECklj

[(n)](tn) le 3

NE





ECklnj

[(n)(t)] le 3

NE t isin Ik


3

NE le 1

20E0






j





EB(xr)[(n)](t)




10 r0 otherwise





limnrarrinfin inf

Ikrn = 0


α2n =

intC j

|X0(n)|2dxdt




j Hence if



of this case




g(t) =int

Cklelnj

|X0(n)(t)|2dx



Ik

g(t) = α2n


I =int

Ik

1

rn(t)

int t+rn(t)2



I le 2int

Ik

int t+rn(t)2

tminusrn(t)2

1

rn(s)g(s)dsdt le 4

intIk

g(s)ds = 4α2n


1

rn(tn)

int tn+rn(tn)2


n | Ik |minus1








2

1

2]


supx

EB(x1)[(n)](t) le 1

10E0 t isin [minus1

2

1

2]


EB(01)[(n)](0) = 1

10E0




[minus 12 1

2 ]timesB(x1)



(n) rarr in H1loc


2

1

2])



2 12 ] times R


1

10E0 le E[] le E[]

and

X0(t0 x0) = 0





(n) rarr in H1loc(C

2[2infin))



0 lt E2 le supt2



X0 = 0






References
























Math 47(5) 719ndash754 (1994)



















Abstract

1 Introduction













66 Final rescaling



Acknowledgements

References




Xε = 1

ρε


(t + ε)2 minus r2 (24)




LX0 g = 2ρ(

dy2 + sinh2(y)d2)


(X0)παβ = 2

ρ3


)


1

2Tαβ [](X0)παβ = 1

ρ|X0|2




X0 = 1ρ

( 12vL + 1

2 uL) partt = 1

2L +

1

2L (25)



(X0)P0 = T (partt X0) = 1

4

(v

u

) 12 |L|2 +

1

4

[(v

u

) 12

+(u

v

) 12]

|part|2 +1

4

(u

v

) 12 |L|2

(X0)PL = T (L X0) = 1

2

(v

u

) 12 |L|2 +

1

2

(u

v

) 12 |part|2







(vε

uε

) 12 asymp 1

(uε

vε






intS0

t2


C0[t1t2]

1

ρε

|Xε|2 dxdt =int

S0t1

(Xε )P0 dx


suptisin(01]

intS0

t

(X0)P0 dx +int int

C0[01]

1

ρ|X0|2 dxdt le E




S0t2


C0[t1t2]

1

ρε

|Xε|2 dxdt int

S0t1



suptisin(ε1]

intS0

t


C0[01]

1

ρε





intS0

1

(Xε )P0 dx E (27)





1

ρ|X0|2 1

ρε

|Xε|2 +ε2

ρ3 |partt|2

and also

intCε

(ε1]

ε2


Cε(ε1]

ε12

t32

|partt|2dxdt E




suptisin(ε1]

intSε

t

(X0)P0 dx +int int

Cε[ε1]

1

ρ|X0|2dxdt E (28)


intSδ

1

(X0)P0 dx leint

Sδt0

(X0)P0 dx +int

partCδ[t01]



intpartCδ[t01]

(u

v

) 12 |part|2 d A

(δ

t0

) 12int


(δ

t0

) 12

E


|L| (u

v

) 12 |X0| +

(u

v

)|L|

Thus we obtain

intSδ

1

(X0)P0 dx int

Sδt0

(X0)P0 dx +

(δ

t0

) 12

E +int

partCδ[t01]

((u

v

) 12 |X0φ|2 +

(u

v

) 32 |L|2

)d A





intS

δ11

(X0)P0 dx int

Sδ0t0

(X0)P0 dx +

((δ1

t0

) 12


)E (29)


intpartCδ[t01]

[(u

v

) 12 |X0φ|2 +

(u

v

) 32 |L|2



intC

δ0[t01]Cδ1[t01]

1

u

[(u

v

) 12 |X0φ|2 +

(u

v

) 32 |L|2

]dxdt E


1

u

(u

v

) 32 le δ

121

t32







2




H2a = minusgi jH2Sa

bc()partibpart j

c



intH2


St


)dx (30)




Chap II of [6])


rang



1

2

intD2

|nablax|2(

1 + r2

1 minus r2

)dx lt infin (31)





r=rn

|nabla|2dl = on(1)



12 (S1)

= 0












10 E0 and so that

X(n)L2(3Q) rarr 0 (32)



(n) rarr in H1(Q)




χ(n) X

1 12infin


where X1 1







(n)bulk

H32 minusε

tx

χ(n) X

1 12infin

1



Rn H1tx














H2 supe rarr M








E = limtrarr0

ESt []





Ft0 [] =int

partC(0t0]

(1

4|L|2 +

1

2|part|2

)d A




Ft0 [] ε8 E

intpartSt0

|part|2ds ε8

t0E (36)







2ε8 E (37)





Ftn lt ε12n E Thus






E[(n)] asymp E (40)




c) Decaying flux

F[εn 1][(n)] lt ε12n E (42)



(n)(1 xn)| gt ε (43)




[] Ft0 [] =int

partC[t0infin)

(1

4|L|2 +

1

2|part|2

)d A

it follows that



Ft0 [] le ε8 E






Ftk [] rarr 0 tk

intpartStk

|part|2d A rarr 0







E[(k)] rarr E (45)























(n)(1 x)| (

2k(1 + 2k |x |)minusN + ε4)

E12 k le 0




)E

12 k ge 0

(48)


2kn gt m(ε E)





2kn lt C lt infin









intS1


(|L(n)|2 + |part(n)|2

)dx E (50)



(n)(1 x)| ((


) 18

+ ε2)

E12 (51)




In general we have



1

(|1 minus x1| + x22 + εn)

12

(|L(n)|2 + |part(n)|2

)dx E (52)



(n)|Similarly




1

((1 minus x1)+ + x22 + εn)

12 + ε8

(|partt


(n)|2)

dx E (53)




kθ part2

where P1kθ and P2



kθ (ξ)| 1

|ξ2|



12 P2




t=1

1


12 + ε8

(|partt


(n)|2)

dx E (55)


|Pk(n)(1 x)| le|P1

kθ part1(n)(1 x)| + |P2

kθ part2(n)(1 x)|

(θ

12 + θminus 1

2

(((1 minus x1)+ + x2

2 + 2minusk + εn)14 + ε4

))E

12



|Pk(n)(1 x1 0)|


) 18

+ ε2)

E12

1

2lt x1 lt 2




) 18

+ ε2)

E12

1

2lt x1 lt 2




) 18

+ ε2)

E12

1

2lt x1 lt2








1

2

intt=1|x |ltγ1

(|partt


)dx ge E1 (56)


E(γ1) = 1

2

intt=1|x |ltγ1


)dx




Pk(1)Linfinx


E(γ1)12 +

((2k(1 minus γ1))

12 + ε4

)E

12





((2k(1 minus γ1))


)E

12




)E

12 |x | lt γ


ε E(γ1)12 +


12 + ε4

)E

12



ε E(γ1)12 +

((1 minus γ1)

12

(1 minus γ )12

+ ε4

)E

12


ε E(γ1)12 +

((1 minus γ1

1 minus γ

)N

+ ε4

)E

12



12n ε


so that

1

2

int|x |ltγ2t

(|partt



12n ε

14n ] (57)



intCεn

[ε12

n ε

14

n ]






according to



intS

t01

(X0)P0[(n)] dx int

Sδ0t0



(1 minus γ1)12

intS

1minusγ11


2

intS

δ0t0




minus 12 E

Sδ0t0






12 E1

we obtain (57)



14n ]


12n ε




1

ρ|X0

(n)|2dxdt ln N

| ln εn| E (59)





a) Bounded energy



ES

(1minusγ2)tt

[(n)](t) ge E2 t isin [1 Tn] (61)


Cε

12

n[1Tn ]

1

ρ|X0






[ 12 infin)


Then we prove that





10E0 (63)



rminus1n

int tn+rn2

tnminusrn2

intB(xn r)




EB(xr)[(n)](t) le 1


ES

(1minusγ2)tt


C j








2


j Ckj

into N sets

Cklj =

(t x) isin Ck

j 1

4+

l minus 1


1

4+

l

4N

l = 1 N


Nsuml=1

intIk

ECklj

[(n)](t)dt leint

Ik

ECkj[(n)](t)dt le 1

10NE


ln+1suml=lnminus1

intIk

ECklj

[(n)](t)dt le 3

10N 2 E


jn+1sumj= jnminus1

ECklj

[(n)](tn) le 3

NE





ECklnj

[(n)(t)] le 3

NE t isin Ik


3

NE le 1

20E0






j





EB(xr)[(n)](t)




10 r0 otherwise





limnrarrinfin inf

Ikrn = 0


α2n =

intC j

|X0(n)|2dxdt




j Hence if



of this case




g(t) =int

Cklelnj

|X0(n)(t)|2dx



Ik

g(t) = α2n


I =int

Ik

1

rn(t)

int t+rn(t)2



I le 2int

Ik

int t+rn(t)2

tminusrn(t)2

1

rn(s)g(s)dsdt le 4

intIk

g(s)ds = 4α2n


1

rn(tn)

int tn+rn(tn)2


n | Ik |minus1








2

1

2]


supx

EB(x1)[(n)](t) le 1

10E0 t isin [minus1

2

1

2]


EB(01)[(n)](0) = 1

10E0




[minus 12 1

2 ]timesB(x1)



(n) rarr in H1loc


2

1

2])



2 12 ] times R


1

10E0 le E[] le E[]

and

X0(t0 x0) = 0





(n) rarr in H1loc(C

2[2infin))



0 lt E2 le supt2



X0 = 0






References
























Math 47(5) 719ndash754 (1994)



















Abstract

1 Introduction













66 Final rescaling



Acknowledgements

References






(vε

uε

) 12 asymp 1

(uε

vε






intS0

t2


C0[t1t2]

1

ρε

|Xε|2 dxdt =int

S0t1

(Xε )P0 dx


suptisin(01]

intS0

t

(X0)P0 dx +int int

C0[01]

1

ρ|X0|2 dxdt le E




S0t2


C0[t1t2]

1

ρε

|Xε|2 dxdt int

S0t1



suptisin(ε1]

intS0

t


C0[01]

1

ρε





intS0

1

(Xε )P0 dx E (27)





1

ρ|X0|2 1

ρε

|Xε|2 +ε2

ρ3 |partt|2

and also

intCε

(ε1]

ε2


Cε(ε1]

ε12

t32

|partt|2dxdt E




suptisin(ε1]

intSε

t

(X0)P0 dx +int int

Cε[ε1]

1

ρ|X0|2dxdt E (28)


intSδ

1

(X0)P0 dx leint

Sδt0

(X0)P0 dx +int

partCδ[t01]



intpartCδ[t01]

(u

v

) 12 |part|2 d A

(δ

t0

) 12int


(δ

t0

) 12

E


|L| (u

v

) 12 |X0| +

(u

v

)|L|

Thus we obtain

intSδ

1

(X0)P0 dx int

Sδt0

(X0)P0 dx +

(δ

t0

) 12

E +int

partCδ[t01]

((u

v

) 12 |X0φ|2 +

(u

v

) 32 |L|2

)d A





intS

δ11

(X0)P0 dx int

Sδ0t0

(X0)P0 dx +

((δ1

t0

) 12


)E (29)


intpartCδ[t01]

[(u

v

) 12 |X0φ|2 +

(u

v

) 32 |L|2



intC

δ0[t01]Cδ1[t01]

1

u

[(u

v

) 12 |X0φ|2 +

(u

v

) 32 |L|2

]dxdt E


1

u

(u

v

) 32 le δ

121

t32







2




H2a = minusgi jH2Sa

bc()partibpart j

c



intH2


St


)dx (30)




Chap II of [6])


rang



1

2

intD2

|nablax|2(

1 + r2

1 minus r2

)dx lt infin (31)





r=rn

|nabla|2dl = on(1)



12 (S1)

= 0












10 E0 and so that

X(n)L2(3Q) rarr 0 (32)



(n) rarr in H1(Q)




χ(n) X

1 12infin


where X1 1







(n)bulk

H32 minusε

tx

χ(n) X

1 12infin

1



Rn H1tx














H2 supe rarr M








E = limtrarr0

ESt []





Ft0 [] =int

partC(0t0]

(1

4|L|2 +

1

2|part|2

)d A




Ft0 [] ε8 E

intpartSt0

|part|2ds ε8

t0E (36)







2ε8 E (37)





Ftn lt ε12n E Thus






E[(n)] asymp E (40)




c) Decaying flux

F[εn 1][(n)] lt ε12n E (42)



(n)(1 xn)| gt ε (43)




[] Ft0 [] =int

partC[t0infin)

(1

4|L|2 +

1

2|part|2

)d A

it follows that



Ft0 [] le ε8 E






Ftk [] rarr 0 tk

intpartStk

|part|2d A rarr 0







E[(k)] rarr E (45)























(n)(1 x)| (

2k(1 + 2k |x |)minusN + ε4)

E12 k le 0




)E

12 k ge 0

(48)


2kn gt m(ε E)





2kn lt C lt infin









intS1


(|L(n)|2 + |part(n)|2

)dx E (50)



(n)(1 x)| ((


) 18

+ ε2)

E12 (51)




In general we have



1

(|1 minus x1| + x22 + εn)

12

(|L(n)|2 + |part(n)|2

)dx E (52)



(n)|Similarly




1

((1 minus x1)+ + x22 + εn)

12 + ε8

(|partt


(n)|2)

dx E (53)




kθ part2

where P1kθ and P2



kθ (ξ)| 1

|ξ2|



12 P2




t=1

1


12 + ε8

(|partt


(n)|2)

dx E (55)


|Pk(n)(1 x)| le|P1

kθ part1(n)(1 x)| + |P2

kθ part2(n)(1 x)|

(θ

12 + θminus 1

2

(((1 minus x1)+ + x2

2 + 2minusk + εn)14 + ε4

))E

12



|Pk(n)(1 x1 0)|


) 18

+ ε2)

E12

1

2lt x1 lt 2




) 18

+ ε2)

E12

1

2lt x1 lt 2




) 18

+ ε2)

E12

1

2lt x1 lt2








1

2

intt=1|x |ltγ1

(|partt


)dx ge E1 (56)


E(γ1) = 1

2

intt=1|x |ltγ1


)dx




Pk(1)Linfinx


E(γ1)12 +

((2k(1 minus γ1))

12 + ε4

)E

12





((2k(1 minus γ1))


)E

12




)E

12 |x | lt γ


ε E(γ1)12 +


12 + ε4

)E

12



ε E(γ1)12 +

((1 minus γ1)

12

(1 minus γ )12

+ ε4

)E

12


ε E(γ1)12 +

((1 minus γ1

1 minus γ

)N

+ ε4

)E

12



12n ε


so that

1

2

int|x |ltγ2t

(|partt



12n ε

14n ] (57)



intCεn

[ε12

n ε

14

n ]






according to



intS

t01

(X0)P0[(n)] dx int

Sδ0t0



(1 minus γ1)12

intS

1minusγ11


2

intS

δ0t0




minus 12 E

Sδ0t0






12 E1

we obtain (57)



14n ]


12n ε




1

ρ|X0

(n)|2dxdt ln N

| ln εn| E (59)





a) Bounded energy



ES

(1minusγ2)tt

[(n)](t) ge E2 t isin [1 Tn] (61)


Cε

12

n[1Tn ]

1

ρ|X0






[ 12 infin)


Then we prove that





10E0 (63)



rminus1n

int tn+rn2

tnminusrn2

intB(xn r)




EB(xr)[(n)](t) le 1


ES

(1minusγ2)tt


C j








2


j Ckj

into N sets

Cklj =

(t x) isin Ck

j 1

4+

l minus 1


1

4+

l

4N

l = 1 N


Nsuml=1

intIk

ECklj

[(n)](t)dt leint

Ik

ECkj[(n)](t)dt le 1

10NE


ln+1suml=lnminus1

intIk

ECklj

[(n)](t)dt le 3

10N 2 E


jn+1sumj= jnminus1

ECklj

[(n)](tn) le 3

NE





ECklnj

[(n)(t)] le 3

NE t isin Ik


3

NE le 1

20E0






j





EB(xr)[(n)](t)




10 r0 otherwise





limnrarrinfin inf

Ikrn = 0


α2n =

intC j

|X0(n)|2dxdt




j Hence if



of this case




g(t) =int

Cklelnj

|X0(n)(t)|2dx



Ik

g(t) = α2n


I =int

Ik

1

rn(t)

int t+rn(t)2



I le 2int

Ik

int t+rn(t)2

tminusrn(t)2

1

rn(s)g(s)dsdt le 4

intIk

g(s)ds = 4α2n


1

rn(tn)

int tn+rn(tn)2


n | Ik |minus1








2

1

2]


supx

EB(x1)[(n)](t) le 1

10E0 t isin [minus1

2

1

2]


EB(01)[(n)](0) = 1

10E0




[minus 12 1

2 ]timesB(x1)



(n) rarr in H1loc


2

1

2])



2 12 ] times R


1

10E0 le E[] le E[]

and

X0(t0 x0) = 0





(n) rarr in H1loc(C

2[2infin))



0 lt E2 le supt2



X0 = 0






References
























Math 47(5) 719ndash754 (1994)



















Abstract

1 Introduction













66 Final rescaling



Acknowledgements

References






1

ρ|X0|2 1

ρε

|Xε|2 +ε2

ρ3 |partt|2

and also

intCε

(ε1]

ε2


Cε(ε1]

ε12

t32

|partt|2dxdt E




suptisin(ε1]

intSε

t

(X0)P0 dx +int int

Cε[ε1]

1

ρ|X0|2dxdt E (28)


intSδ

1

(X0)P0 dx leint

Sδt0

(X0)P0 dx +int

partCδ[t01]



intpartCδ[t01]

(u

v

) 12 |part|2 d A

(δ

t0

) 12int


(δ

t0

) 12

E


|L| (u

v

) 12 |X0| +

(u

v

)|L|

Thus we obtain

intSδ

1

(X0)P0 dx int

Sδt0

(X0)P0 dx +

(δ

t0

) 12

E +int

partCδ[t01]

((u

v

) 12 |X0φ|2 +

(u

v

) 32 |L|2

)d A





intS

δ11

(X0)P0 dx int

Sδ0t0

(X0)P0 dx +

((δ1

t0

) 12


)E (29)


intpartCδ[t01]

[(u

v

) 12 |X0φ|2 +

(u

v

) 32 |L|2



intC

δ0[t01]Cδ1[t01]

1

u

[(u

v

) 12 |X0φ|2 +

(u

v

) 32 |L|2

]dxdt E


1

u

(u

v

) 32 le δ

121

t32







2




H2a = minusgi jH2Sa

bc()partibpart j

c



intH2


St


)dx (30)




Chap II of [6])


rang



1

2

intD2

|nablax|2(

1 + r2

1 minus r2

)dx lt infin (31)





r=rn

|nabla|2dl = on(1)



12 (S1)

= 0












10 E0 and so that

X(n)L2(3Q) rarr 0 (32)



(n) rarr in H1(Q)




χ(n) X

1 12infin


where X1 1







(n)bulk

H32 minusε

tx

χ(n) X

1 12infin

1



Rn H1tx














H2 supe rarr M








E = limtrarr0

ESt []





Ft0 [] =int

partC(0t0]

(1

4|L|2 +

1

2|part|2

)d A




Ft0 [] ε8 E

intpartSt0

|part|2ds ε8

t0E (36)







2ε8 E (37)





Ftn lt ε12n E Thus






E[(n)] asymp E (40)




c) Decaying flux

F[εn 1][(n)] lt ε12n E (42)



(n)(1 xn)| gt ε (43)




[] Ft0 [] =int

partC[t0infin)

(1

4|L|2 +

1

2|part|2

)d A

it follows that



Ft0 [] le ε8 E






Ftk [] rarr 0 tk

intpartStk

|part|2d A rarr 0







E[(k)] rarr E (45)























(n)(1 x)| (

2k(1 + 2k |x |)minusN + ε4)

E12 k le 0




)E

12 k ge 0

(48)


2kn gt m(ε E)





2kn lt C lt infin









intS1


(|L(n)|2 + |part(n)|2

)dx E (50)



(n)(1 x)| ((


) 18

+ ε2)

E12 (51)




In general we have



1

(|1 minus x1| + x22 + εn)

12

(|L(n)|2 + |part(n)|2

)dx E (52)



(n)|Similarly




1

((1 minus x1)+ + x22 + εn)

12 + ε8

(|partt


(n)|2)

dx E (53)




kθ part2

where P1kθ and P2



kθ (ξ)| 1

|ξ2|



12 P2




t=1

1


12 + ε8

(|partt


(n)|2)

dx E (55)


|Pk(n)(1 x)| le|P1

kθ part1(n)(1 x)| + |P2

kθ part2(n)(1 x)|

(θ

12 + θminus 1

2

(((1 minus x1)+ + x2

2 + 2minusk + εn)14 + ε4

))E

12



|Pk(n)(1 x1 0)|


) 18

+ ε2)

E12

1

2lt x1 lt 2




) 18

+ ε2)

E12

1

2lt x1 lt 2




) 18

+ ε2)

E12

1

2lt x1 lt2








1

2

intt=1|x |ltγ1

(|partt


)dx ge E1 (56)


E(γ1) = 1

2

intt=1|x |ltγ1


)dx




Pk(1)Linfinx


E(γ1)12 +

((2k(1 minus γ1))

12 + ε4

)E

12





((2k(1 minus γ1))


)E

12




)E

12 |x | lt γ


ε E(γ1)12 +


12 + ε4

)E

12



ε E(γ1)12 +

((1 minus γ1)

12

(1 minus γ )12

+ ε4

)E

12


ε E(γ1)12 +

((1 minus γ1

1 minus γ

)N

+ ε4

)E

12



12n ε


so that

1

2

int|x |ltγ2t

(|partt



12n ε

14n ] (57)



intCεn

[ε12

n ε

14

n ]






according to



intS

t01

(X0)P0[(n)] dx int

Sδ0t0



(1 minus γ1)12

intS

1minusγ11


2

intS

δ0t0




minus 12 E

Sδ0t0






12 E1

we obtain (57)



14n ]


12n ε




1

ρ|X0

(n)|2dxdt ln N

| ln εn| E (59)





a) Bounded energy



ES

(1minusγ2)tt

[(n)](t) ge E2 t isin [1 Tn] (61)


Cε

12

n[1Tn ]

1

ρ|X0






[ 12 infin)


Then we prove that





10E0 (63)



rminus1n

int tn+rn2

tnminusrn2

intB(xn r)




EB(xr)[(n)](t) le 1


ES

(1minusγ2)tt


C j








2


j Ckj

into N sets

Cklj =

(t x) isin Ck

j 1

4+

l minus 1


1

4+

l

4N

l = 1 N


Nsuml=1

intIk

ECklj

[(n)](t)dt leint

Ik

ECkj[(n)](t)dt le 1

10NE


ln+1suml=lnminus1

intIk

ECklj

[(n)](t)dt le 3

10N 2 E


jn+1sumj= jnminus1

ECklj

[(n)](tn) le 3

NE





ECklnj

[(n)(t)] le 3

NE t isin Ik


3

NE le 1

20E0






j





EB(xr)[(n)](t)




10 r0 otherwise





limnrarrinfin inf

Ikrn = 0


α2n =

intC j

|X0(n)|2dxdt




j Hence if



of this case




g(t) =int

Cklelnj

|X0(n)(t)|2dx



Ik

g(t) = α2n


I =int

Ik

1

rn(t)

int t+rn(t)2



I le 2int

Ik

int t+rn(t)2

tminusrn(t)2

1

rn(s)g(s)dsdt le 4

intIk

g(s)ds = 4α2n


1

rn(tn)

int tn+rn(tn)2


n | Ik |minus1








2

1

2]


supx

EB(x1)[(n)](t) le 1

10E0 t isin [minus1

2

1

2]


EB(01)[(n)](0) = 1

10E0




[minus 12 1

2 ]timesB(x1)



(n) rarr in H1loc


2

1

2])



2 12 ] times R


1

10E0 le E[] le E[]

and

X0(t0 x0) = 0





(n) rarr in H1loc(C

2[2infin))



0 lt E2 le supt2



X0 = 0






References
























Math 47(5) 719ndash754 (1994)



















Abstract

1 Introduction













66 Final rescaling



Acknowledgements

References





intS

δ11

(X0)P0 dx int

Sδ0t0

(X0)P0 dx +

((δ1

t0

) 12


)E (29)


intpartCδ[t01]

[(u

v

) 12 |X0φ|2 +

(u

v

) 32 |L|2



intC

δ0[t01]Cδ1[t01]

1

u

[(u

v

) 12 |X0φ|2 +

(u

v

) 32 |L|2

]dxdt E


1

u

(u

v

) 32 le δ

121

t32







2




H2a = minusgi jH2Sa

bc()partibpart j

c



intH2


St


)dx (30)




Chap II of [6])


rang



1

2

intD2

|nablax|2(

1 + r2

1 minus r2

)dx lt infin (31)





r=rn

|nabla|2dl = on(1)



12 (S1)

= 0












10 E0 and so that

X(n)L2(3Q) rarr 0 (32)



(n) rarr in H1(Q)




χ(n) X

1 12infin


where X1 1







(n)bulk

H32 minusε

tx

χ(n) X

1 12infin

1



Rn H1tx














H2 supe rarr M








E = limtrarr0

ESt []





Ft0 [] =int

partC(0t0]

(1

4|L|2 +

1

2|part|2

)d A




Ft0 [] ε8 E

intpartSt0

|part|2ds ε8

t0E (36)







2ε8 E (37)





Ftn lt ε12n E Thus






E[(n)] asymp E (40)




c) Decaying flux

F[εn 1][(n)] lt ε12n E (42)



(n)(1 xn)| gt ε (43)




[] Ft0 [] =int

partC[t0infin)

(1

4|L|2 +

1

2|part|2

)d A

it follows that



Ft0 [] le ε8 E






Ftk [] rarr 0 tk

intpartStk

|part|2d A rarr 0







E[(k)] rarr E (45)























(n)(1 x)| (

2k(1 + 2k |x |)minusN + ε4)

E12 k le 0




)E

12 k ge 0

(48)


2kn gt m(ε E)





2kn lt C lt infin









intS1


(|L(n)|2 + |part(n)|2

)dx E (50)



(n)(1 x)| ((


) 18

+ ε2)

E12 (51)




In general we have



1

(|1 minus x1| + x22 + εn)

12

(|L(n)|2 + |part(n)|2

)dx E (52)



(n)|Similarly




1

((1 minus x1)+ + x22 + εn)

12 + ε8

(|partt


(n)|2)

dx E (53)




kθ part2

where P1kθ and P2



kθ (ξ)| 1

|ξ2|



12 P2




t=1

1


12 + ε8

(|partt


(n)|2)

dx E (55)


|Pk(n)(1 x)| le|P1

kθ part1(n)(1 x)| + |P2

kθ part2(n)(1 x)|

(θ

12 + θminus 1

2

(((1 minus x1)+ + x2

2 + 2minusk + εn)14 + ε4

))E

12



|Pk(n)(1 x1 0)|


) 18

+ ε2)

E12

1

2lt x1 lt 2




) 18

+ ε2)

E12

1

2lt x1 lt 2




) 18

+ ε2)

E12

1

2lt x1 lt2








1

2

intt=1|x |ltγ1

(|partt


)dx ge E1 (56)


E(γ1) = 1

2

intt=1|x |ltγ1


)dx




Pk(1)Linfinx


E(γ1)12 +

((2k(1 minus γ1))

12 + ε4

)E

12





((2k(1 minus γ1))


)E

12




)E

12 |x | lt γ


ε E(γ1)12 +


12 + ε4

)E

12



ε E(γ1)12 +

((1 minus γ1)

12

(1 minus γ )12

+ ε4

)E

12


ε E(γ1)12 +

((1 minus γ1

1 minus γ

)N

+ ε4

)E

12



12n ε


so that

1

2

int|x |ltγ2t

(|partt



12n ε

14n ] (57)



intCεn

[ε12

n ε

14

n ]






according to



intS

t01

(X0)P0[(n)] dx int

Sδ0t0



(1 minus γ1)12

intS

1minusγ11


2

intS

δ0t0




minus 12 E

Sδ0t0






12 E1

we obtain (57)



14n ]


12n ε




1

ρ|X0

(n)|2dxdt ln N

| ln εn| E (59)





a) Bounded energy



ES

(1minusγ2)tt

[(n)](t) ge E2 t isin [1 Tn] (61)


Cε

12

n[1Tn ]

1

ρ|X0






[ 12 infin)


Then we prove that





10E0 (63)



rminus1n

int tn+rn2

tnminusrn2

intB(xn r)




EB(xr)[(n)](t) le 1


ES

(1minusγ2)tt


C j








2


j Ckj

into N sets

Cklj =

(t x) isin Ck

j 1

4+

l minus 1


1

4+

l

4N

l = 1 N


Nsuml=1

intIk

ECklj

[(n)](t)dt leint

Ik

ECkj[(n)](t)dt le 1

10NE


ln+1suml=lnminus1

intIk

ECklj

[(n)](t)dt le 3

10N 2 E


jn+1sumj= jnminus1

ECklj

[(n)](tn) le 3

NE





ECklnj

[(n)(t)] le 3

NE t isin Ik


3

NE le 1

20E0






j





EB(xr)[(n)](t)




10 r0 otherwise





limnrarrinfin inf

Ikrn = 0


α2n =

intC j

|X0(n)|2dxdt




j Hence if



of this case




g(t) =int

Cklelnj

|X0(n)(t)|2dx



Ik

g(t) = α2n


I =int

Ik

1

rn(t)

int t+rn(t)2



I le 2int

Ik

int t+rn(t)2

tminusrn(t)2

1

rn(s)g(s)dsdt le 4

intIk

g(s)ds = 4α2n


1

rn(tn)

int tn+rn(tn)2


n | Ik |minus1








2

1

2]


supx

EB(x1)[(n)](t) le 1

10E0 t isin [minus1

2

1

2]


EB(01)[(n)](0) = 1

10E0




[minus 12 1

2 ]timesB(x1)



(n) rarr in H1loc


2

1

2])



2 12 ] times R


1

10E0 le E[] le E[]

and

X0(t0 x0) = 0





(n) rarr in H1loc(C

2[2infin))



0 lt E2 le supt2



X0 = 0






References
























Math 47(5) 719ndash754 (1994)



















Abstract

1 Introduction













66 Final rescaling



Acknowledgements

References




H2a = minusgi jH2Sa

bc()partibpart j

c



intH2


St


)dx (30)




Chap II of [6])


rang



1

2

intD2

|nablax|2(

1 + r2

1 minus r2

)dx lt infin (31)





r=rn

|nabla|2dl = on(1)



12 (S1)

= 0












10 E0 and so that

X(n)L2(3Q) rarr 0 (32)



(n) rarr in H1(Q)




χ(n) X

1 12infin


where X1 1







(n)bulk

H32 minusε

tx

χ(n) X

1 12infin

1



Rn H1tx














H2 supe rarr M








E = limtrarr0

ESt []





Ft0 [] =int

partC(0t0]

(1

4|L|2 +

1

2|part|2

)d A




Ft0 [] ε8 E

intpartSt0

|part|2ds ε8

t0E (36)







2ε8 E (37)





Ftn lt ε12n E Thus






E[(n)] asymp E (40)




c) Decaying flux

F[εn 1][(n)] lt ε12n E (42)



(n)(1 xn)| gt ε (43)




[] Ft0 [] =int

partC[t0infin)

(1

4|L|2 +

1

2|part|2

)d A

it follows that



Ft0 [] le ε8 E






Ftk [] rarr 0 tk

intpartStk

|part|2d A rarr 0







E[(k)] rarr E (45)























(n)(1 x)| (

2k(1 + 2k |x |)minusN + ε4)

E12 k le 0




)E

12 k ge 0

(48)


2kn gt m(ε E)





2kn lt C lt infin









intS1


(|L(n)|2 + |part(n)|2

)dx E (50)



(n)(1 x)| ((


) 18

+ ε2)

E12 (51)




In general we have



1

(|1 minus x1| + x22 + εn)

12

(|L(n)|2 + |part(n)|2

)dx E (52)



(n)|Similarly




1

((1 minus x1)+ + x22 + εn)

12 + ε8

(|partt


(n)|2)

dx E (53)




kθ part2

where P1kθ and P2



kθ (ξ)| 1

|ξ2|



12 P2




t=1

1


12 + ε8

(|partt


(n)|2)

dx E (55)


|Pk(n)(1 x)| le|P1

kθ part1(n)(1 x)| + |P2

kθ part2(n)(1 x)|

(θ

12 + θminus 1

2

(((1 minus x1)+ + x2

2 + 2minusk + εn)14 + ε4

))E

12



|Pk(n)(1 x1 0)|


) 18

+ ε2)

E12

1

2lt x1 lt 2




) 18

+ ε2)

E12

1

2lt x1 lt 2




) 18

+ ε2)

E12

1

2lt x1 lt2








1

2

intt=1|x |ltγ1

(|partt


)dx ge E1 (56)


E(γ1) = 1

2

intt=1|x |ltγ1


)dx




Pk(1)Linfinx


E(γ1)12 +

((2k(1 minus γ1))

12 + ε4

)E

12





((2k(1 minus γ1))


)E

12




)E

12 |x | lt γ


ε E(γ1)12 +


12 + ε4

)E

12



ε E(γ1)12 +

((1 minus γ1)

12

(1 minus γ )12

+ ε4

)E

12


ε E(γ1)12 +

((1 minus γ1

1 minus γ

)N

+ ε4

)E

12



12n ε


so that

1

2

int|x |ltγ2t

(|partt



12n ε

14n ] (57)



intCεn

[ε12

n ε

14

n ]






according to



intS

t01

(X0)P0[(n)] dx int

Sδ0t0



(1 minus γ1)12

intS

1minusγ11


2

intS

δ0t0




minus 12 E

Sδ0t0






12 E1

we obtain (57)



14n ]


12n ε




1

ρ|X0

(n)|2dxdt ln N

| ln εn| E (59)





a) Bounded energy



ES

(1minusγ2)tt

[(n)](t) ge E2 t isin [1 Tn] (61)


Cε

12

n[1Tn ]

1

ρ|X0






[ 12 infin)


Then we prove that





10E0 (63)



rminus1n

int tn+rn2

tnminusrn2

intB(xn r)




EB(xr)[(n)](t) le 1


ES

(1minusγ2)tt


C j








2


j Ckj

into N sets

Cklj =

(t x) isin Ck

j 1

4+

l minus 1


1

4+

l

4N

l = 1 N


Nsuml=1

intIk

ECklj

[(n)](t)dt leint

Ik

ECkj[(n)](t)dt le 1

10NE


ln+1suml=lnminus1

intIk

ECklj

[(n)](t)dt le 3

10N 2 E


jn+1sumj= jnminus1

ECklj

[(n)](tn) le 3

NE





ECklnj

[(n)(t)] le 3

NE t isin Ik


3

NE le 1

20E0






j





EB(xr)[(n)](t)




10 r0 otherwise





limnrarrinfin inf

Ikrn = 0


α2n =

intC j

|X0(n)|2dxdt




j Hence if



of this case




g(t) =int

Cklelnj

|X0(n)(t)|2dx



Ik

g(t) = α2n


I =int

Ik

1

rn(t)

int t+rn(t)2



I le 2int

Ik

int t+rn(t)2

tminusrn(t)2

1

rn(s)g(s)dsdt le 4

intIk

g(s)ds = 4α2n


1

rn(tn)

int tn+rn(tn)2


n | Ik |minus1








2

1

2]


supx

EB(x1)[(n)](t) le 1

10E0 t isin [minus1

2

1

2]


EB(01)[(n)](0) = 1

10E0




[minus 12 1

2 ]timesB(x1)



(n) rarr in H1loc


2

1

2])



2 12 ] times R


1

10E0 le E[] le E[]

and

X0(t0 x0) = 0





(n) rarr in H1loc(C

2[2infin))



0 lt E2 le supt2



X0 = 0






References
























Math 47(5) 719ndash754 (1994)



















Abstract

1 Introduction













66 Final rescaling



Acknowledgements

References









10 E0 and so that

X(n)L2(3Q) rarr 0 (32)



(n) rarr in H1(Q)




χ(n) X

1 12infin


where X1 1







(n)bulk

H32 minusε

tx

χ(n) X

1 12infin

1



Rn H1tx














H2 supe rarr M








E = limtrarr0

ESt []





Ft0 [] =int

partC(0t0]

(1

4|L|2 +

1

2|part|2

)d A




Ft0 [] ε8 E

intpartSt0

|part|2ds ε8

t0E (36)







2ε8 E (37)





Ftn lt ε12n E Thus






E[(n)] asymp E (40)




c) Decaying flux

F[εn 1][(n)] lt ε12n E (42)



(n)(1 xn)| gt ε (43)




[] Ft0 [] =int

partC[t0infin)

(1

4|L|2 +

1

2|part|2

)d A

it follows that



Ft0 [] le ε8 E






Ftk [] rarr 0 tk

intpartStk

|part|2d A rarr 0







E[(k)] rarr E (45)























(n)(1 x)| (

2k(1 + 2k |x |)minusN + ε4)

E12 k le 0




)E

12 k ge 0

(48)


2kn gt m(ε E)





2kn lt C lt infin









intS1


(|L(n)|2 + |part(n)|2

)dx E (50)



(n)(1 x)| ((


) 18

+ ε2)

E12 (51)




In general we have



1

(|1 minus x1| + x22 + εn)

12

(|L(n)|2 + |part(n)|2

)dx E (52)



(n)|Similarly




1

((1 minus x1)+ + x22 + εn)

12 + ε8

(|partt


(n)|2)

dx E (53)




kθ part2

where P1kθ and P2



kθ (ξ)| 1

|ξ2|



12 P2




t=1

1


12 + ε8

(|partt


(n)|2)

dx E (55)


|Pk(n)(1 x)| le|P1

kθ part1(n)(1 x)| + |P2

kθ part2(n)(1 x)|

(θ

12 + θminus 1

2

(((1 minus x1)+ + x2

2 + 2minusk + εn)14 + ε4

))E

12



|Pk(n)(1 x1 0)|


) 18

+ ε2)

E12

1

2lt x1 lt 2




) 18

+ ε2)

E12

1

2lt x1 lt 2




) 18

+ ε2)

E12

1

2lt x1 lt2








1

2

intt=1|x |ltγ1

(|partt


)dx ge E1 (56)


E(γ1) = 1

2

intt=1|x |ltγ1


)dx




Pk(1)Linfinx


E(γ1)12 +

((2k(1 minus γ1))

12 + ε4

)E

12





((2k(1 minus γ1))


)E

12




)E

12 |x | lt γ


ε E(γ1)12 +


12 + ε4

)E

12



ε E(γ1)12 +

((1 minus γ1)

12

(1 minus γ )12

+ ε4

)E

12


ε E(γ1)12 +

((1 minus γ1

1 minus γ

)N

+ ε4

)E

12



12n ε


so that

1

2

int|x |ltγ2t

(|partt



12n ε

14n ] (57)



intCεn

[ε12

n ε

14

n ]






according to



intS

t01

(X0)P0[(n)] dx int

Sδ0t0



(1 minus γ1)12

intS

1minusγ11


2

intS

δ0t0




minus 12 E

Sδ0t0






12 E1

we obtain (57)



14n ]


12n ε




1

ρ|X0

(n)|2dxdt ln N

| ln εn| E (59)





a) Bounded energy



ES

(1minusγ2)tt

[(n)](t) ge E2 t isin [1 Tn] (61)


Cε

12

n[1Tn ]

1

ρ|X0






[ 12 infin)


Then we prove that





10E0 (63)



rminus1n

int tn+rn2

tnminusrn2

intB(xn r)




EB(xr)[(n)](t) le 1


ES

(1minusγ2)tt


C j








2


j Ckj

into N sets

Cklj =

(t x) isin Ck

j 1

4+

l minus 1


1

4+

l

4N

l = 1 N


Nsuml=1

intIk

ECklj

[(n)](t)dt leint

Ik

ECkj[(n)](t)dt le 1

10NE


ln+1suml=lnminus1

intIk

ECklj

[(n)](t)dt le 3

10N 2 E


jn+1sumj= jnminus1

ECklj

[(n)](tn) le 3

NE





ECklnj

[(n)(t)] le 3

NE t isin Ik


3

NE le 1

20E0






j





EB(xr)[(n)](t)




10 r0 otherwise





limnrarrinfin inf

Ikrn = 0


α2n =

intC j

|X0(n)|2dxdt




j Hence if



of this case




g(t) =int

Cklelnj

|X0(n)(t)|2dx



Ik

g(t) = α2n


I =int

Ik

1

rn(t)

int t+rn(t)2



I le 2int

Ik

int t+rn(t)2

tminusrn(t)2

1

rn(s)g(s)dsdt le 4

intIk

g(s)ds = 4α2n


1

rn(tn)

int tn+rn(tn)2


n | Ik |minus1








2

1

2]


supx

EB(x1)[(n)](t) le 1

10E0 t isin [minus1

2

1

2]


EB(01)[(n)](0) = 1

10E0




[minus 12 1

2 ]timesB(x1)



(n) rarr in H1loc


2

1

2])



2 12 ] times R


1

10E0 le E[] le E[]

and

X0(t0 x0) = 0





(n) rarr in H1loc(C

2[2infin))



0 lt E2 le supt2



X0 = 0






References
























Math 47(5) 719ndash754 (1994)



















Abstract

1 Introduction













66 Final rescaling



Acknowledgements

References




Rn H1tx














H2 supe rarr M








E = limtrarr0

ESt []





Ft0 [] =int

partC(0t0]

(1

4|L|2 +

1

2|part|2

)d A




Ft0 [] ε8 E

intpartSt0

|part|2ds ε8

t0E (36)







2ε8 E (37)





Ftn lt ε12n E Thus






E[(n)] asymp E (40)




c) Decaying flux

F[εn 1][(n)] lt ε12n E (42)



(n)(1 xn)| gt ε (43)




[] Ft0 [] =int

partC[t0infin)

(1

4|L|2 +

1

2|part|2

)d A

it follows that



Ft0 [] le ε8 E






Ftk [] rarr 0 tk

intpartStk

|part|2d A rarr 0







E[(k)] rarr E (45)























(n)(1 x)| (

2k(1 + 2k |x |)minusN + ε4)

E12 k le 0




)E

12 k ge 0

(48)


2kn gt m(ε E)





2kn lt C lt infin









intS1


(|L(n)|2 + |part(n)|2

)dx E (50)



(n)(1 x)| ((


) 18

+ ε2)

E12 (51)




In general we have



1

(|1 minus x1| + x22 + εn)

12

(|L(n)|2 + |part(n)|2

)dx E (52)



(n)|Similarly




1

((1 minus x1)+ + x22 + εn)

12 + ε8

(|partt


(n)|2)

dx E (53)




kθ part2

where P1kθ and P2



kθ (ξ)| 1

|ξ2|



12 P2




t=1

1


12 + ε8

(|partt


(n)|2)

dx E (55)


|Pk(n)(1 x)| le|P1

kθ part1(n)(1 x)| + |P2

kθ part2(n)(1 x)|

(θ

12 + θminus 1

2

(((1 minus x1)+ + x2

2 + 2minusk + εn)14 + ε4

))E

12



|Pk(n)(1 x1 0)|


) 18

+ ε2)

E12

1

2lt x1 lt 2




) 18

+ ε2)

E12

1

2lt x1 lt 2




) 18

+ ε2)

E12

1

2lt x1 lt2








1

2

intt=1|x |ltγ1

(|partt


)dx ge E1 (56)


E(γ1) = 1

2

intt=1|x |ltγ1


)dx




Pk(1)Linfinx


E(γ1)12 +

((2k(1 minus γ1))

12 + ε4

)E

12





((2k(1 minus γ1))


)E

12




)E

12 |x | lt γ


ε E(γ1)12 +


12 + ε4

)E

12



ε E(γ1)12 +

((1 minus γ1)

12

(1 minus γ )12

+ ε4

)E

12


ε E(γ1)12 +

((1 minus γ1

1 minus γ

)N

+ ε4

)E

12



12n ε


so that

1

2

int|x |ltγ2t

(|partt



12n ε

14n ] (57)



intCεn

[ε12

n ε

14

n ]






according to



intS

t01

(X0)P0[(n)] dx int

Sδ0t0



(1 minus γ1)12

intS

1minusγ11


2

intS

δ0t0




minus 12 E

Sδ0t0






12 E1

we obtain (57)



14n ]


12n ε




1

ρ|X0

(n)|2dxdt ln N

| ln εn| E (59)





a) Bounded energy



ES

(1minusγ2)tt

[(n)](t) ge E2 t isin [1 Tn] (61)


Cε

12

n[1Tn ]

1

ρ|X0






[ 12 infin)


Then we prove that





10E0 (63)



rminus1n

int tn+rn2

tnminusrn2

intB(xn r)




EB(xr)[(n)](t) le 1


ES

(1minusγ2)tt


C j








2


j Ckj

into N sets

Cklj =

(t x) isin Ck

j 1

4+

l minus 1


1

4+

l

4N

l = 1 N


Nsuml=1

intIk

ECklj

[(n)](t)dt leint

Ik

ECkj[(n)](t)dt le 1

10NE


ln+1suml=lnminus1

intIk

ECklj

[(n)](t)dt le 3

10N 2 E


jn+1sumj= jnminus1

ECklj

[(n)](tn) le 3

NE





ECklnj

[(n)(t)] le 3

NE t isin Ik


3

NE le 1

20E0






j





EB(xr)[(n)](t)




10 r0 otherwise





limnrarrinfin inf

Ikrn = 0


α2n =

intC j

|X0(n)|2dxdt




j Hence if



of this case




g(t) =int

Cklelnj

|X0(n)(t)|2dx



Ik

g(t) = α2n


I =int

Ik

1

rn(t)

int t+rn(t)2



I le 2int

Ik

int t+rn(t)2

tminusrn(t)2

1

rn(s)g(s)dsdt le 4

intIk

g(s)ds = 4α2n


1

rn(tn)

int tn+rn(tn)2


n | Ik |minus1








2

1

2]


supx

EB(x1)[(n)](t) le 1

10E0 t isin [minus1

2

1

2]


EB(01)[(n)](0) = 1

10E0




[minus 12 1

2 ]timesB(x1)



(n) rarr in H1loc


2

1

2])



2 12 ] times R


1

10E0 le E[] le E[]

and

X0(t0 x0) = 0





(n) rarr in H1loc(C

2[2infin))



0 lt E2 le supt2



X0 = 0






References
























Math 47(5) 719ndash754 (1994)



















Abstract

1 Introduction













66 Final rescaling



Acknowledgements

References





Ft0 [] =int

partC(0t0]

(1

4|L|2 +

1

2|part|2

)d A




Ft0 [] ε8 E

intpartSt0

|part|2ds ε8

t0E (36)







2ε8 E (37)





Ftn lt ε12n E Thus






E[(n)] asymp E (40)




c) Decaying flux

F[εn 1][(n)] lt ε12n E (42)



(n)(1 xn)| gt ε (43)




[] Ft0 [] =int

partC[t0infin)

(1

4|L|2 +

1

2|part|2

)d A

it follows that



Ft0 [] le ε8 E






Ftk [] rarr 0 tk

intpartStk

|part|2d A rarr 0







E[(k)] rarr E (45)























(n)(1 x)| (

2k(1 + 2k |x |)minusN + ε4)

E12 k le 0




)E

12 k ge 0

(48)


2kn gt m(ε E)





2kn lt C lt infin









intS1


(|L(n)|2 + |part(n)|2

)dx E (50)



(n)(1 x)| ((


) 18

+ ε2)

E12 (51)




In general we have



1

(|1 minus x1| + x22 + εn)

12

(|L(n)|2 + |part(n)|2

)dx E (52)



(n)|Similarly




1

((1 minus x1)+ + x22 + εn)

12 + ε8

(|partt


(n)|2)

dx E (53)




kθ part2

where P1kθ and P2



kθ (ξ)| 1

|ξ2|



12 P2




t=1

1


12 + ε8

(|partt


(n)|2)

dx E (55)


|Pk(n)(1 x)| le|P1

kθ part1(n)(1 x)| + |P2

kθ part2(n)(1 x)|

(θ

12 + θminus 1

2

(((1 minus x1)+ + x2

2 + 2minusk + εn)14 + ε4

))E

12



|Pk(n)(1 x1 0)|


) 18

+ ε2)

E12

1

2lt x1 lt 2




) 18

+ ε2)

E12

1

2lt x1 lt 2




) 18

+ ε2)

E12

1

2lt x1 lt2








1

2

intt=1|x |ltγ1

(|partt


)dx ge E1 (56)


E(γ1) = 1

2

intt=1|x |ltγ1


)dx




Pk(1)Linfinx


E(γ1)12 +

((2k(1 minus γ1))

12 + ε4

)E

12





((2k(1 minus γ1))


)E

12




)E

12 |x | lt γ


ε E(γ1)12 +


12 + ε4

)E

12



ε E(γ1)12 +

((1 minus γ1)

12

(1 minus γ )12

+ ε4

)E

12


ε E(γ1)12 +

((1 minus γ1

1 minus γ

)N

+ ε4

)E

12



12n ε


so that

1

2

int|x |ltγ2t

(|partt



12n ε

14n ] (57)



intCεn

[ε12

n ε

14

n ]






according to



intS

t01

(X0)P0[(n)] dx int

Sδ0t0



(1 minus γ1)12

intS

1minusγ11


2

intS

δ0t0




minus 12 E

Sδ0t0






12 E1

we obtain (57)



14n ]


12n ε




1

ρ|X0

(n)|2dxdt ln N

| ln εn| E (59)





a) Bounded energy



ES

(1minusγ2)tt

[(n)](t) ge E2 t isin [1 Tn] (61)


Cε

12

n[1Tn ]

1

ρ|X0






[ 12 infin)


Then we prove that





10E0 (63)



rminus1n

int tn+rn2

tnminusrn2

intB(xn r)




EB(xr)[(n)](t) le 1


ES

(1minusγ2)tt


C j








2


j Ckj

into N sets

Cklj =

(t x) isin Ck

j 1

4+

l minus 1


1

4+

l

4N

l = 1 N


Nsuml=1

intIk

ECklj

[(n)](t)dt leint

Ik

ECkj[(n)](t)dt le 1

10NE


ln+1suml=lnminus1

intIk

ECklj

[(n)](t)dt le 3

10N 2 E


jn+1sumj= jnminus1

ECklj

[(n)](tn) le 3

NE





ECklnj

[(n)(t)] le 3

NE t isin Ik


3

NE le 1

20E0






j





EB(xr)[(n)](t)




10 r0 otherwise





limnrarrinfin inf

Ikrn = 0


α2n =

intC j

|X0(n)|2dxdt




j Hence if



of this case




g(t) =int

Cklelnj

|X0(n)(t)|2dx



Ik

g(t) = α2n


I =int

Ik

1

rn(t)

int t+rn(t)2



I le 2int

Ik

int t+rn(t)2

tminusrn(t)2

1

rn(s)g(s)dsdt le 4

intIk

g(s)ds = 4α2n


1

rn(tn)

int tn+rn(tn)2


n | Ik |minus1








2

1

2]


supx

EB(x1)[(n)](t) le 1

10E0 t isin [minus1

2

1

2]


EB(01)[(n)](0) = 1

10E0




[minus 12 1

2 ]timesB(x1)



(n) rarr in H1loc


2

1

2])



2 12 ] times R


1

10E0 le E[] le E[]

and

X0(t0 x0) = 0





(n) rarr in H1loc(C

2[2infin))



0 lt E2 le supt2



X0 = 0






References
























Math 47(5) 719ndash754 (1994)



















Abstract

1 Introduction













66 Final rescaling



Acknowledgements

References





c) Decaying flux

F[εn 1][(n)] lt ε12n E (42)



(n)(1 xn)| gt ε (43)




[] Ft0 [] =int

partC[t0infin)

(1

4|L|2 +

1

2|part|2

)d A

it follows that



Ft0 [] le ε8 E






Ftk [] rarr 0 tk

intpartStk

|part|2d A rarr 0







E[(k)] rarr E (45)























(n)(1 x)| (

2k(1 + 2k |x |)minusN + ε4)

E12 k le 0




)E

12 k ge 0

(48)


2kn gt m(ε E)





2kn lt C lt infin









intS1


(|L(n)|2 + |part(n)|2

)dx E (50)



(n)(1 x)| ((


) 18

+ ε2)

E12 (51)




In general we have



1

(|1 minus x1| + x22 + εn)

12

(|L(n)|2 + |part(n)|2

)dx E (52)



(n)|Similarly




1

((1 minus x1)+ + x22 + εn)

12 + ε8

(|partt


(n)|2)

dx E (53)




kθ part2

where P1kθ and P2



kθ (ξ)| 1

|ξ2|



12 P2




t=1

1


12 + ε8

(|partt


(n)|2)

dx E (55)


|Pk(n)(1 x)| le|P1

kθ part1(n)(1 x)| + |P2

kθ part2(n)(1 x)|

(θ

12 + θminus 1

2

(((1 minus x1)+ + x2

2 + 2minusk + εn)14 + ε4

))E

12



|Pk(n)(1 x1 0)|


) 18

+ ε2)

E12

1

2lt x1 lt 2




) 18

+ ε2)

E12

1

2lt x1 lt 2




) 18

+ ε2)

E12

1

2lt x1 lt2








1

2

intt=1|x |ltγ1

(|partt


)dx ge E1 (56)


E(γ1) = 1

2

intt=1|x |ltγ1


)dx




Pk(1)Linfinx


E(γ1)12 +

((2k(1 minus γ1))

12 + ε4

)E

12





((2k(1 minus γ1))


)E

12




)E

12 |x | lt γ


ε E(γ1)12 +


12 + ε4

)E

12



ε E(γ1)12 +

((1 minus γ1)

12

(1 minus γ )12

+ ε4

)E

12


ε E(γ1)12 +

((1 minus γ1

1 minus γ

)N

+ ε4

)E

12



12n ε


so that

1

2

int|x |ltγ2t

(|partt



12n ε

14n ] (57)



intCεn

[ε12

n ε

14

n ]






according to



intS

t01

(X0)P0[(n)] dx int

Sδ0t0



(1 minus γ1)12

intS

1minusγ11


2

intS

δ0t0




minus 12 E

Sδ0t0






12 E1

we obtain (57)



14n ]


12n ε




1

ρ|X0

(n)|2dxdt ln N

| ln εn| E (59)





a) Bounded energy



ES

(1minusγ2)tt

[(n)](t) ge E2 t isin [1 Tn] (61)


Cε

12

n[1Tn ]

1

ρ|X0






[ 12 infin)


Then we prove that





10E0 (63)



rminus1n

int tn+rn2

tnminusrn2

intB(xn r)




EB(xr)[(n)](t) le 1


ES

(1minusγ2)tt


C j








2


j Ckj

into N sets

Cklj =

(t x) isin Ck

j 1

4+

l minus 1


1

4+

l

4N

l = 1 N


Nsuml=1

intIk

ECklj

[(n)](t)dt leint

Ik

ECkj[(n)](t)dt le 1

10NE


ln+1suml=lnminus1

intIk

ECklj

[(n)](t)dt le 3

10N 2 E


jn+1sumj= jnminus1

ECklj

[(n)](tn) le 3

NE





ECklnj

[(n)(t)] le 3

NE t isin Ik


3

NE le 1

20E0






j





EB(xr)[(n)](t)




10 r0 otherwise





limnrarrinfin inf

Ikrn = 0


α2n =

intC j

|X0(n)|2dxdt




j Hence if



of this case




g(t) =int

Cklelnj

|X0(n)(t)|2dx



Ik

g(t) = α2n


I =int

Ik

1

rn(t)

int t+rn(t)2



I le 2int

Ik

int t+rn(t)2

tminusrn(t)2

1

rn(s)g(s)dsdt le 4

intIk

g(s)ds = 4α2n


1

rn(tn)

int tn+rn(tn)2


n | Ik |minus1








2

1

2]


supx

EB(x1)[(n)](t) le 1

10E0 t isin [minus1

2

1

2]


EB(01)[(n)](0) = 1

10E0




[minus 12 1

2 ]timesB(x1)



(n) rarr in H1loc


2

1

2])



2 12 ] times R


1

10E0 le E[] le E[]

and

X0(t0 x0) = 0





(n) rarr in H1loc(C

2[2infin))



0 lt E2 le supt2



X0 = 0






References
























Math 47(5) 719ndash754 (1994)



















Abstract

1 Introduction













66 Final rescaling



Acknowledgements

References




E[(k)] rarr E (45)























(n)(1 x)| (

2k(1 + 2k |x |)minusN + ε4)

E12 k le 0




)E

12 k ge 0

(48)


2kn gt m(ε E)





2kn lt C lt infin









intS1


(|L(n)|2 + |part(n)|2

)dx E (50)



(n)(1 x)| ((


) 18

+ ε2)

E12 (51)




In general we have



1

(|1 minus x1| + x22 + εn)

12

(|L(n)|2 + |part(n)|2

)dx E (52)



(n)|Similarly




1

((1 minus x1)+ + x22 + εn)

12 + ε8

(|partt


(n)|2)

dx E (53)




kθ part2

where P1kθ and P2



kθ (ξ)| 1

|ξ2|



12 P2




t=1

1


12 + ε8

(|partt


(n)|2)

dx E (55)


|Pk(n)(1 x)| le|P1

kθ part1(n)(1 x)| + |P2

kθ part2(n)(1 x)|

(θ

12 + θminus 1

2

(((1 minus x1)+ + x2

2 + 2minusk + εn)14 + ε4

))E

12



|Pk(n)(1 x1 0)|


) 18

+ ε2)

E12

1

2lt x1 lt 2




) 18

+ ε2)

E12

1

2lt x1 lt 2




) 18

+ ε2)

E12

1

2lt x1 lt2








1

2

intt=1|x |ltγ1

(|partt


)dx ge E1 (56)


E(γ1) = 1

2

intt=1|x |ltγ1


)dx




Pk(1)Linfinx


E(γ1)12 +

((2k(1 minus γ1))

12 + ε4

)E

12





((2k(1 minus γ1))


)E

12




)E

12 |x | lt γ


ε E(γ1)12 +


12 + ε4

)E

12



ε E(γ1)12 +

((1 minus γ1)

12

(1 minus γ )12

+ ε4

)E

12


ε E(γ1)12 +

((1 minus γ1

1 minus γ

)N

+ ε4

)E

12



12n ε


so that

1

2

int|x |ltγ2t

(|partt



12n ε

14n ] (57)



intCεn

[ε12

n ε

14

n ]






according to



intS

t01

(X0)P0[(n)] dx int

Sδ0t0



(1 minus γ1)12

intS

1minusγ11


2

intS

δ0t0




minus 12 E

Sδ0t0






12 E1

we obtain (57)



14n ]


12n ε




1

ρ|X0

(n)|2dxdt ln N

| ln εn| E (59)





a) Bounded energy



ES

(1minusγ2)tt

[(n)](t) ge E2 t isin [1 Tn] (61)


Cε

12

n[1Tn ]

1

ρ|X0






[ 12 infin)


Then we prove that





10E0 (63)



rminus1n

int tn+rn2

tnminusrn2

intB(xn r)




EB(xr)[(n)](t) le 1


ES

(1minusγ2)tt


C j








2


j Ckj

into N sets

Cklj =

(t x) isin Ck

j 1

4+

l minus 1


1

4+

l

4N

l = 1 N


Nsuml=1

intIk

ECklj

[(n)](t)dt leint

Ik

ECkj[(n)](t)dt le 1

10NE


ln+1suml=lnminus1

intIk

ECklj

[(n)](t)dt le 3

10N 2 E


jn+1sumj= jnminus1

ECklj

[(n)](tn) le 3

NE





ECklnj

[(n)(t)] le 3

NE t isin Ik


3

NE le 1

20E0






j





EB(xr)[(n)](t)




10 r0 otherwise





limnrarrinfin inf

Ikrn = 0


α2n =

intC j

|X0(n)|2dxdt




j Hence if



of this case




g(t) =int

Cklelnj

|X0(n)(t)|2dx



Ik

g(t) = α2n


I =int

Ik

1

rn(t)

int t+rn(t)2



I le 2int

Ik

int t+rn(t)2

tminusrn(t)2

1

rn(s)g(s)dsdt le 4

intIk

g(s)ds = 4α2n


1

rn(tn)

int tn+rn(tn)2


n | Ik |minus1








2

1

2]


supx

EB(x1)[(n)](t) le 1

10E0 t isin [minus1

2

1

2]


EB(01)[(n)](0) = 1

10E0




[minus 12 1

2 ]timesB(x1)



(n) rarr in H1loc


2

1

2])



2 12 ] times R


1

10E0 le E[] le E[]

and

X0(t0 x0) = 0





(n) rarr in H1loc(C

2[2infin))



0 lt E2 le supt2



X0 = 0






References
























Math 47(5) 719ndash754 (1994)



















Abstract

1 Introduction













66 Final rescaling



Acknowledgements

References





(n)(1 x)| (

2k(1 + 2k |x |)minusN + ε4)

E12 k le 0




)E

12 k ge 0

(48)


2kn gt m(ε E)





2kn lt C lt infin









intS1


(|L(n)|2 + |part(n)|2

)dx E (50)



(n)(1 x)| ((


) 18

+ ε2)

E12 (51)




In general we have



1

(|1 minus x1| + x22 + εn)

12

(|L(n)|2 + |part(n)|2

)dx E (52)



(n)|Similarly




1

((1 minus x1)+ + x22 + εn)

12 + ε8

(|partt


(n)|2)

dx E (53)




kθ part2

where P1kθ and P2



kθ (ξ)| 1

|ξ2|



12 P2




t=1

1


12 + ε8

(|partt


(n)|2)

dx E (55)


|Pk(n)(1 x)| le|P1

kθ part1(n)(1 x)| + |P2

kθ part2(n)(1 x)|

(θ

12 + θminus 1

2

(((1 minus x1)+ + x2

2 + 2minusk + εn)14 + ε4

))E

12



|Pk(n)(1 x1 0)|


) 18

+ ε2)

E12

1

2lt x1 lt 2




) 18

+ ε2)

E12

1

2lt x1 lt 2




) 18

+ ε2)

E12

1

2lt x1 lt2








1

2

intt=1|x |ltγ1

(|partt


)dx ge E1 (56)


E(γ1) = 1

2

intt=1|x |ltγ1


)dx




Pk(1)Linfinx


E(γ1)12 +

((2k(1 minus γ1))

12 + ε4

)E

12





((2k(1 minus γ1))


)E

12




)E

12 |x | lt γ


ε E(γ1)12 +


12 + ε4

)E

12



ε E(γ1)12 +

((1 minus γ1)

12

(1 minus γ )12

+ ε4

)E

12


ε E(γ1)12 +

((1 minus γ1

1 minus γ

)N

+ ε4

)E

12



12n ε


so that

1

2

int|x |ltγ2t

(|partt



12n ε

14n ] (57)



intCεn

[ε12

n ε

14

n ]






according to



intS

t01

(X0)P0[(n)] dx int

Sδ0t0



(1 minus γ1)12

intS

1minusγ11


2

intS

δ0t0




minus 12 E

Sδ0t0






12 E1

we obtain (57)



14n ]


12n ε




1

ρ|X0

(n)|2dxdt ln N

| ln εn| E (59)





a) Bounded energy



ES

(1minusγ2)tt

[(n)](t) ge E2 t isin [1 Tn] (61)


Cε

12

n[1Tn ]

1

ρ|X0






[ 12 infin)


Then we prove that





10E0 (63)



rminus1n

int tn+rn2

tnminusrn2

intB(xn r)




EB(xr)[(n)](t) le 1


ES

(1minusγ2)tt


C j








2


j Ckj

into N sets

Cklj =

(t x) isin Ck

j 1

4+

l minus 1


1

4+

l

4N

l = 1 N


Nsuml=1

intIk

ECklj

[(n)](t)dt leint

Ik

ECkj[(n)](t)dt le 1

10NE


ln+1suml=lnminus1

intIk

ECklj

[(n)](t)dt le 3

10N 2 E


jn+1sumj= jnminus1

ECklj

[(n)](tn) le 3

NE





ECklnj

[(n)(t)] le 3

NE t isin Ik


3

NE le 1

20E0






j





EB(xr)[(n)](t)




10 r0 otherwise





limnrarrinfin inf

Ikrn = 0


α2n =

intC j

|X0(n)|2dxdt




j Hence if



of this case




g(t) =int

Cklelnj

|X0(n)(t)|2dx



Ik

g(t) = α2n


I =int

Ik

1

rn(t)

int t+rn(t)2



I le 2int

Ik

int t+rn(t)2

tminusrn(t)2

1

rn(s)g(s)dsdt le 4

intIk

g(s)ds = 4α2n


1

rn(tn)

int tn+rn(tn)2


n | Ik |minus1








2

1

2]


supx

EB(x1)[(n)](t) le 1

10E0 t isin [minus1

2

1

2]


EB(01)[(n)](0) = 1

10E0




[minus 12 1

2 ]timesB(x1)



(n) rarr in H1loc


2

1

2])



2 12 ] times R


1

10E0 le E[] le E[]

and

X0(t0 x0) = 0





(n) rarr in H1loc(C

2[2infin))



0 lt E2 le supt2



X0 = 0






References
























Math 47(5) 719ndash754 (1994)



















Abstract

1 Introduction













66 Final rescaling



Acknowledgements

References





In general we have



1

(|1 minus x1| + x22 + εn)

12

(|L(n)|2 + |part(n)|2

)dx E (52)



(n)|Similarly




1

((1 minus x1)+ + x22 + εn)

12 + ε8

(|partt


(n)|2)

dx E (53)




kθ part2

where P1kθ and P2



kθ (ξ)| 1

|ξ2|



12 P2




t=1

1


12 + ε8

(|partt


(n)|2)

dx E (55)


|Pk(n)(1 x)| le|P1

kθ part1(n)(1 x)| + |P2

kθ part2(n)(1 x)|

(θ

12 + θminus 1

2

(((1 minus x1)+ + x2

2 + 2minusk + εn)14 + ε4

))E

12



|Pk(n)(1 x1 0)|


) 18

+ ε2)

E12

1

2lt x1 lt 2




) 18

+ ε2)

E12

1

2lt x1 lt 2




) 18

+ ε2)

E12

1

2lt x1 lt2








1

2

intt=1|x |ltγ1

(|partt


)dx ge E1 (56)


E(γ1) = 1

2

intt=1|x |ltγ1


)dx




Pk(1)Linfinx


E(γ1)12 +

((2k(1 minus γ1))

12 + ε4

)E

12





((2k(1 minus γ1))


)E

12




)E

12 |x | lt γ


ε E(γ1)12 +


12 + ε4

)E

12



ε E(γ1)12 +

((1 minus γ1)

12

(1 minus γ )12

+ ε4

)E

12


ε E(γ1)12 +

((1 minus γ1

1 minus γ

)N

+ ε4

)E

12



12n ε


so that

1

2

int|x |ltγ2t

(|partt



12n ε

14n ] (57)



intCεn

[ε12

n ε

14

n ]






according to



intS

t01

(X0)P0[(n)] dx int

Sδ0t0



(1 minus γ1)12

intS

1minusγ11


2

intS

δ0t0




minus 12 E

Sδ0t0






12 E1

we obtain (57)



14n ]


12n ε




1

ρ|X0

(n)|2dxdt ln N

| ln εn| E (59)





a) Bounded energy



ES

(1minusγ2)tt

[(n)](t) ge E2 t isin [1 Tn] (61)


Cε

12

n[1Tn ]

1

ρ|X0






[ 12 infin)


Then we prove that





10E0 (63)



rminus1n

int tn+rn2

tnminusrn2

intB(xn r)




EB(xr)[(n)](t) le 1


ES

(1minusγ2)tt


C j








2


j Ckj

into N sets

Cklj =

(t x) isin Ck

j 1

4+

l minus 1


1

4+

l

4N

l = 1 N


Nsuml=1

intIk

ECklj

[(n)](t)dt leint

Ik

ECkj[(n)](t)dt le 1

10NE


ln+1suml=lnminus1

intIk

ECklj

[(n)](t)dt le 3

10N 2 E


jn+1sumj= jnminus1

ECklj

[(n)](tn) le 3

NE





ECklnj

[(n)(t)] le 3

NE t isin Ik


3

NE le 1

20E0






j





EB(xr)[(n)](t)




10 r0 otherwise





limnrarrinfin inf

Ikrn = 0


α2n =

intC j

|X0(n)|2dxdt




j Hence if



of this case




g(t) =int

Cklelnj

|X0(n)(t)|2dx



Ik

g(t) = α2n


I =int

Ik

1

rn(t)

int t+rn(t)2



I le 2int

Ik

int t+rn(t)2

tminusrn(t)2

1

rn(s)g(s)dsdt le 4

intIk

g(s)ds = 4α2n


1

rn(tn)

int tn+rn(tn)2


n | Ik |minus1








2

1

2]


supx

EB(x1)[(n)](t) le 1

10E0 t isin [minus1

2

1

2]


EB(01)[(n)](0) = 1

10E0




[minus 12 1

2 ]timesB(x1)



(n) rarr in H1loc


2

1

2])



2 12 ] times R


1

10E0 le E[] le E[]

and

X0(t0 x0) = 0





(n) rarr in H1loc(C

2[2infin))



0 lt E2 le supt2



X0 = 0






References
























Math 47(5) 719ndash754 (1994)



















Abstract

1 Introduction













66 Final rescaling



Acknowledgements

References




|Pk(n)(1 x1 0)|


) 18

+ ε2)

E12

1

2lt x1 lt 2




) 18

+ ε2)

E12

1

2lt x1 lt 2




) 18

+ ε2)

E12

1

2lt x1 lt2








1

2

intt=1|x |ltγ1

(|partt


)dx ge E1 (56)


E(γ1) = 1

2

intt=1|x |ltγ1


)dx




Pk(1)Linfinx


E(γ1)12 +

((2k(1 minus γ1))

12 + ε4

)E

12





((2k(1 minus γ1))


)E

12




)E

12 |x | lt γ


ε E(γ1)12 +


12 + ε4

)E

12



ε E(γ1)12 +

((1 minus γ1)

12

(1 minus γ )12

+ ε4

)E

12


ε E(γ1)12 +

((1 minus γ1

1 minus γ

)N

+ ε4

)E

12



12n ε


so that

1

2

int|x |ltγ2t

(|partt



12n ε

14n ] (57)



intCεn

[ε12

n ε

14

n ]






according to



intS

t01

(X0)P0[(n)] dx int

Sδ0t0



(1 minus γ1)12

intS

1minusγ11


2

intS

δ0t0




minus 12 E

Sδ0t0






12 E1

we obtain (57)



14n ]


12n ε




1

ρ|X0

(n)|2dxdt ln N

| ln εn| E (59)





a) Bounded energy



ES

(1minusγ2)tt

[(n)](t) ge E2 t isin [1 Tn] (61)


Cε

12

n[1Tn ]

1

ρ|X0






[ 12 infin)


Then we prove that





10E0 (63)



rminus1n

int tn+rn2

tnminusrn2

intB(xn r)




EB(xr)[(n)](t) le 1


ES

(1minusγ2)tt


C j








2


j Ckj

into N sets

Cklj =

(t x) isin Ck

j 1

4+

l minus 1


1

4+

l

4N

l = 1 N


Nsuml=1

intIk

ECklj

[(n)](t)dt leint

Ik

ECkj[(n)](t)dt le 1

10NE


ln+1suml=lnminus1

intIk

ECklj

[(n)](t)dt le 3

10N 2 E


jn+1sumj= jnminus1

ECklj

[(n)](tn) le 3

NE





ECklnj

[(n)(t)] le 3

NE t isin Ik


3

NE le 1

20E0






j





EB(xr)[(n)](t)




10 r0 otherwise





limnrarrinfin inf

Ikrn = 0


α2n =

intC j

|X0(n)|2dxdt




j Hence if



of this case




g(t) =int

Cklelnj

|X0(n)(t)|2dx



Ik

g(t) = α2n


I =int

Ik

1

rn(t)

int t+rn(t)2



I le 2int

Ik

int t+rn(t)2

tminusrn(t)2

1

rn(s)g(s)dsdt le 4

intIk

g(s)ds = 4α2n


1

rn(tn)

int tn+rn(tn)2


n | Ik |minus1








2

1

2]


supx

EB(x1)[(n)](t) le 1

10E0 t isin [minus1

2

1

2]


EB(01)[(n)](0) = 1

10E0




[minus 12 1

2 ]timesB(x1)



(n) rarr in H1loc


2

1

2])



2 12 ] times R


1

10E0 le E[] le E[]

and

X0(t0 x0) = 0





(n) rarr in H1loc(C

2[2infin))



0 lt E2 le supt2



X0 = 0






References
























Math 47(5) 719ndash754 (1994)



















Abstract

1 Introduction













66 Final rescaling



Acknowledgements

References





((2k(1 minus γ1))


)E

12




)E

12 |x | lt γ


ε E(γ1)12 +


12 + ε4

)E

12



ε E(γ1)12 +

((1 minus γ1)

12

(1 minus γ )12

+ ε4

)E

12


ε E(γ1)12 +

((1 minus γ1

1 minus γ

)N

+ ε4

)E

12



12n ε


so that

1

2

int|x |ltγ2t

(|partt



12n ε

14n ] (57)



intCεn

[ε12

n ε

14

n ]






according to



intS

t01

(X0)P0[(n)] dx int

Sδ0t0



(1 minus γ1)12

intS

1minusγ11


2

intS

δ0t0




minus 12 E

Sδ0t0






12 E1

we obtain (57)



14n ]


12n ε




1

ρ|X0

(n)|2dxdt ln N

| ln εn| E (59)





a) Bounded energy



ES

(1minusγ2)tt

[(n)](t) ge E2 t isin [1 Tn] (61)


Cε

12

n[1Tn ]

1

ρ|X0






[ 12 infin)


Then we prove that





10E0 (63)



rminus1n

int tn+rn2

tnminusrn2

intB(xn r)




EB(xr)[(n)](t) le 1


ES

(1minusγ2)tt


C j








2


j Ckj

into N sets

Cklj =

(t x) isin Ck

j 1

4+

l minus 1


1

4+

l

4N

l = 1 N


Nsuml=1

intIk

ECklj

[(n)](t)dt leint

Ik

ECkj[(n)](t)dt le 1

10NE


ln+1suml=lnminus1

intIk

ECklj

[(n)](t)dt le 3

10N 2 E


jn+1sumj= jnminus1

ECklj

[(n)](tn) le 3

NE





ECklnj

[(n)(t)] le 3

NE t isin Ik


3

NE le 1

20E0






j





EB(xr)[(n)](t)




10 r0 otherwise





limnrarrinfin inf

Ikrn = 0


α2n =

intC j

|X0(n)|2dxdt




j Hence if



of this case




g(t) =int

Cklelnj

|X0(n)(t)|2dx



Ik

g(t) = α2n


I =int

Ik

1

rn(t)

int t+rn(t)2



I le 2int

Ik

int t+rn(t)2

tminusrn(t)2

1

rn(s)g(s)dsdt le 4

intIk

g(s)ds = 4α2n


1

rn(tn)

int tn+rn(tn)2


n | Ik |minus1








2

1

2]


supx

EB(x1)[(n)](t) le 1

10E0 t isin [minus1

2

1

2]


EB(01)[(n)](0) = 1

10E0




[minus 12 1

2 ]timesB(x1)



(n) rarr in H1loc


2

1

2])



2 12 ] times R


1

10E0 le E[] le E[]

and

X0(t0 x0) = 0





(n) rarr in H1loc(C

2[2infin))



0 lt E2 le supt2



X0 = 0






References
























Math 47(5) 719ndash754 (1994)



















Abstract

1 Introduction













66 Final rescaling



Acknowledgements

References






according to



intS

t01

(X0)P0[(n)] dx int

Sδ0t0



(1 minus γ1)12

intS

1minusγ11


2

intS

δ0t0




minus 12 E

Sδ0t0






12 E1

we obtain (57)



14n ]


12n ε




1

ρ|X0

(n)|2dxdt ln N

| ln εn| E (59)





a) Bounded energy



ES

(1minusγ2)tt

[(n)](t) ge E2 t isin [1 Tn] (61)


Cε

12

n[1Tn ]

1

ρ|X0






[ 12 infin)


Then we prove that





10E0 (63)



rminus1n

int tn+rn2

tnminusrn2

intB(xn r)




EB(xr)[(n)](t) le 1


ES

(1minusγ2)tt


C j








2


j Ckj

into N sets

Cklj =

(t x) isin Ck

j 1

4+

l minus 1


1

4+

l

4N

l = 1 N


Nsuml=1

intIk

ECklj

[(n)](t)dt leint

Ik

ECkj[(n)](t)dt le 1

10NE


ln+1suml=lnminus1

intIk

ECklj

[(n)](t)dt le 3

10N 2 E


jn+1sumj= jnminus1

ECklj

[(n)](tn) le 3

NE





ECklnj

[(n)(t)] le 3

NE t isin Ik


3

NE le 1

20E0






j





EB(xr)[(n)](t)




10 r0 otherwise





limnrarrinfin inf

Ikrn = 0


α2n =

intC j

|X0(n)|2dxdt




j Hence if



of this case




g(t) =int

Cklelnj

|X0(n)(t)|2dx



Ik

g(t) = α2n


I =int

Ik

1

rn(t)

int t+rn(t)2



I le 2int

Ik

int t+rn(t)2

tminusrn(t)2

1

rn(s)g(s)dsdt le 4

intIk

g(s)ds = 4α2n


1

rn(tn)

int tn+rn(tn)2


n | Ik |minus1








2

1

2]


supx

EB(x1)[(n)](t) le 1

10E0 t isin [minus1

2

1

2]


EB(01)[(n)](0) = 1

10E0




[minus 12 1

2 ]timesB(x1)



(n) rarr in H1loc


2

1

2])



2 12 ] times R


1

10E0 le E[] le E[]

and

X0(t0 x0) = 0





(n) rarr in H1loc(C

2[2infin))



0 lt E2 le supt2



X0 = 0






References
























Math 47(5) 719ndash754 (1994)



















Abstract

1 Introduction













66 Final rescaling



Acknowledgements

References



a) Bounded energy



ES

(1minusγ2)tt

[(n)](t) ge E2 t isin [1 Tn] (61)


Cε

12

n[1Tn ]

1

ρ|X0






[ 12 infin)


Then we prove that





10E0 (63)



rminus1n

int tn+rn2

tnminusrn2

intB(xn r)




EB(xr)[(n)](t) le 1


ES

(1minusγ2)tt


C j








2


j Ckj

into N sets

Cklj =

(t x) isin Ck

j 1

4+

l minus 1


1

4+

l

4N

l = 1 N


Nsuml=1

intIk

ECklj

[(n)](t)dt leint

Ik

ECkj[(n)](t)dt le 1

10NE


ln+1suml=lnminus1

intIk

ECklj

[(n)](t)dt le 3

10N 2 E


jn+1sumj= jnminus1

ECklj

[(n)](tn) le 3

NE





ECklnj

[(n)(t)] le 3

NE t isin Ik


3

NE le 1

20E0






j





EB(xr)[(n)](t)




10 r0 otherwise





limnrarrinfin inf

Ikrn = 0


α2n =

intC j

|X0(n)|2dxdt




j Hence if



of this case




g(t) =int

Cklelnj

|X0(n)(t)|2dx



Ik

g(t) = α2n


I =int

Ik

1

rn(t)

int t+rn(t)2



I le 2int

Ik

int t+rn(t)2

tminusrn(t)2

1

rn(s)g(s)dsdt le 4

intIk

g(s)ds = 4α2n


1

rn(tn)

int tn+rn(tn)2


n | Ik |minus1








2

1

2]


supx

EB(x1)[(n)](t) le 1

10E0 t isin [minus1

2

1

2]


EB(01)[(n)](0) = 1

10E0




[minus 12 1

2 ]timesB(x1)



(n) rarr in H1loc


2

1

2])



2 12 ] times R


1

10E0 le E[] le E[]

and

X0(t0 x0) = 0





(n) rarr in H1loc(C

2[2infin))



0 lt E2 le supt2



X0 = 0






References
























Math 47(5) 719ndash754 (1994)



















Abstract

1 Introduction













66 Final rescaling



Acknowledgements

References








2


j Ckj

into N sets

Cklj =

(t x) isin Ck

j 1

4+

l minus 1


1

4+

l

4N

l = 1 N


Nsuml=1

intIk

ECklj

[(n)](t)dt leint

Ik

ECkj[(n)](t)dt le 1

10NE


ln+1suml=lnminus1

intIk

ECklj

[(n)](t)dt le 3

10N 2 E


jn+1sumj= jnminus1

ECklj

[(n)](tn) le 3

NE





ECklnj

[(n)(t)] le 3

NE t isin Ik


3

NE le 1

20E0






j





EB(xr)[(n)](t)




10 r0 otherwise





limnrarrinfin inf

Ikrn = 0


α2n =

intC j

|X0(n)|2dxdt




j Hence if



of this case




g(t) =int

Cklelnj

|X0(n)(t)|2dx



Ik

g(t) = α2n


I =int

Ik

1

rn(t)

int t+rn(t)2



I le 2int

Ik

int t+rn(t)2

tminusrn(t)2

1

rn(s)g(s)dsdt le 4

intIk

g(s)ds = 4α2n


1

rn(tn)

int tn+rn(tn)2


n | Ik |minus1








2

1

2]


supx

EB(x1)[(n)](t) le 1

10E0 t isin [minus1

2

1

2]


EB(01)[(n)](0) = 1

10E0




[minus 12 1

2 ]timesB(x1)



(n) rarr in H1loc


2

1

2])



2 12 ] times R


1

10E0 le E[] le E[]

and

X0(t0 x0) = 0





(n) rarr in H1loc(C

2[2infin))



0 lt E2 le supt2



X0 = 0






References
























Math 47(5) 719ndash754 (1994)



















Abstract

1 Introduction













66 Final rescaling



Acknowledgements

References






EB(xr)[(n)](t)




10 r0 otherwise





limnrarrinfin inf

Ikrn = 0


α2n =

intC j

|X0(n)|2dxdt




j Hence if



of this case




g(t) =int

Cklelnj

|X0(n)(t)|2dx



Ik

g(t) = α2n


I =int

Ik

1

rn(t)

int t+rn(t)2



I le 2int

Ik

int t+rn(t)2

tminusrn(t)2

1

rn(s)g(s)dsdt le 4

intIk

g(s)ds = 4α2n


1

rn(tn)

int tn+rn(tn)2


n | Ik |minus1








2

1

2]


supx

EB(x1)[(n)](t) le 1

10E0 t isin [minus1

2

1

2]


EB(01)[(n)](0) = 1

10E0




[minus 12 1

2 ]timesB(x1)



(n) rarr in H1loc


2

1

2])



2 12 ] times R


1

10E0 le E[] le E[]

and

X0(t0 x0) = 0





(n) rarr in H1loc(C

2[2infin))



0 lt E2 le supt2



X0 = 0






References
























Math 47(5) 719ndash754 (1994)



















Abstract

1 Introduction













66 Final rescaling



Acknowledgements

References




Ik

g(t) = α2n


I =int

Ik

1

rn(t)

int t+rn(t)2



I le 2int

Ik

int t+rn(t)2

tminusrn(t)2

1

rn(s)g(s)dsdt le 4

intIk

g(s)ds = 4α2n


1

rn(tn)

int tn+rn(tn)2


n | Ik |minus1








2

1

2]


supx

EB(x1)[(n)](t) le 1

10E0 t isin [minus1

2

1

2]


EB(01)[(n)](0) = 1

10E0




[minus 12 1

2 ]timesB(x1)



(n) rarr in H1loc


2

1

2])



2 12 ] times R


1

10E0 le E[] le E[]

and

X0(t0 x0) = 0





(n) rarr in H1loc(C

2[2infin))



0 lt E2 le supt2



X0 = 0






References
























Math 47(5) 719ndash754 (1994)



















Abstract

1 Introduction













66 Final rescaling



Acknowledgements

References





[minus 12 1

2 ]timesB(x1)



(n) rarr in H1loc


2

1

2])



2 12 ] times R


1

10E0 le E[] le E[]

and

X0(t0 x0) = 0





(n) rarr in H1loc(C

2[2infin))



0 lt E2 le supt2



X0 = 0






References
























Math 47(5) 719ndash754 (1994)



















Abstract

1 Introduction













66 Final rescaling



Acknowledgements

References





References
























Math 47(5) 719ndash754 (1994)



















Abstract

1 Introduction













66 Final rescaling



Acknowledgements

References




















Abstract

1 Introduction













66 Final rescaling



Acknowledgements

References


regularity of wave-maps in dimension 2 + 1...digital object identiﬁer (doi)...

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