Solvability for perturbations of a class of real vector fields on the two-torus

Contents lists available atScienceDirect

Journal

of

Mathematical

Analysis

and

Applications

www.elsevier.com/locate/jmaa

Solvability

for

perturbations

of

a class

of

real

vector

fields

on

the

two-torus

Paulo L. Dattori da Silvaa,RafaelB. Gonzalezb,∗, Marcio A. Jorge Silvac a

DepartamentodeMatemática,InstitutodeCiênciasMatemáticasedeComputação,Universidadede SãoPaulo,CaixaPostal668,SãoCarlos,SP13560-970,Brazil

b_{DepartamentodeMatemática,UniversidadeEstadualde Maringá,Maringá,PR87020-900,Brazil} c_{DepartamentodeMatemática,UniversidadeEstadualdeLondrina, CaixaPostal10.011,Londrina,}

PR 86057-970,Brazil

a r t i cl e i n f o a b s t r a c t

Articlehistory:

Received13April2020 Availableonline5August2020 Submittedby E.Saksman Keywords: Closedrange Globalsolvability Perturbations Fourierseries Diophantineconditions Flatfunctions

LetL= ∂t+ a(x)∂x be a realvectorfield definedon thetwo-dimensional torus

T2_{,wherea isareal-valuedandsmoothfunctiononT}1_{.Wedealwiththeglobal} solvabilityofequationsintheformLu+ pu= f ,wherep,f∈ C∞(T2_{).Solvability} to theequationLu= f iswell-understood.Weshowthat a perturbationof zero ordermayaffecttheglobalsolvabilityofL;wemaymaintain,gainorlosesolvability byadding aperturbation.Thisphenomenonislinkedto theorderofvanishingof thecoefficienta ofL.Weobtainedresultsintheclassofsmoothfunctionson T2 and,also,inthespaceofSchwartzdistributionsD(T2_).

©2020ElsevierInc.Allrightsreserved.

1. Introduction

LetL beanonsingularrealvectorfielddefinedonasmoothn-dimensionalmanifoldM .

Weareinterestedinsolvingequationsintheform

Lu + pu = f, (1.1)

where p,f and u arecomplex-valuedsmoothfunctionsdefined onM .The spaceof suchfunctionswillbe denotedbyC∞(M ).

Itfollowsfrom tubularflow theoremthatlocally(1.1) alwayshassmoothsolutions. Thesolvabilityof(1.1) isstillaninterestingproblemifweseekforglobal solutions. Forp∈ C∞(M ),wedefine Lp:C∞(M )→ C∞(M ) byLpu=Lu+ pu.

* Correspondingauthor.

E-mailaddresses:[email protected](P.L. Dattori da Silva),[email protected](R.B. Gonzalez),[email protected] (M.A. Jorge Silva).

https://doi.org/10.1016/j.jmaa.2020.124467 0022-247X/©2020ElsevierInc. Allrightsreserved.

InthecasewhereM isnoncompact,paracompact,Malgrangein[15] andalsoDuistermaatandHörmander in[10] presented necessaryandsufficientconditionstoobtainLp(C∞(M ))=C∞(M ).

MalgrangeshowedessentiallythatLpC∞(M )=C∞(M ) isequivalenttothefollowinggeometriccondition:

(GC) (i) No completeintegralcurveofL isrelatively compact;

(ii) ForeverycompactsubsetK ofM thereexistsacompactsubsetKofM suchthateverycompact intervalonanintegralcurvewithendpointsinK iscontainedinK.

DuistermaatandHörmanderprovedthat(GC) isequivalentto

(iii) There exists amanifold M0, anopen neighborhood M1 of M0× {0} inM0× R which is convex intheR direction,andadiffeomorphismM → M1 whichcarriesL intotheoperator∂/∂t,where pointsinM1 aredenotedby(x,t).

Comparingwithlocal solvability,under(GC)thezeroorderterminLp doesnotplayarelevantrolein

theglobal solvability.

There isnogeneralresultaboutglobalsolvability ofLp inthecasewhereM is compact.

We willaddress globalsolvability issuesfor zeroorder perturbationsof certainreal vectorfields onthe 2-dimensional torus,T2_{ R}2_/2πZ2_.

DenotingthecoordinatesinT2_{by(x,t)∈ T}1_{× T}1_{,weconsider therealvectorfieldgiven by}

L0= ∂t+ a(x)∂x, (1.2)

where a isareal-valued andsmoothfunctiononT1.

In[6],BergamascoandPetronilhocharacterizedtheclosednessoftherange(globalsolvability)ofL0.By Theorem 1.1in[6],L0is globallysolvableifandonlyifoneofthefollowingconditionsholds:

• a vanishesidentically;

• a nevervanishesand(2π)−1₀2π(1/a) iseitherrationalornon-Liouvilleirrational; • ∅= a−1(0)= T1 _{anda vanishesonlyof finiteorder.}

Oneof themotivationsforthis workwasaquestionposedbyAdalbertoBergamascoand JoséRuidival dosSantosFilhotothesecondauthorconcerningsomeresultsinthepaper[6].Asmentionedin[6] without adetailedproof,theoperator∂t+ a(x)∂x+ a(x) isglobally solvableonC∞(T2).

Ourpurpose hereisto studytheinfluenceofaperturbationofL0 byazeroordertermintheexistence of globalsolutions.Weconsider differentialoperators

Lp= ∂t+ a(x)∂x+ p(x), (1.3)

definedonT2_{,wherea isareal-valuedandsmoothfunctiononT}1_{,andp isacomplex-valuedandsmooth} functiononT1_.

As in[6], we saythatLp is globally solvableon C∞(T2) if its rangeLpC∞(T2) is aclosed subspaceof C∞_(T2_{).BystandardargumentsofFunctionalAnalysis,L}

pisgloballysolvable ifandonlyifLpC∞(T2)=

(kert_L

p)◦,wheretLp denotesthetransposeoperatorand(kertLp)◦isthesetoffunctionsφ∈ C∞(T2) such

thatμ(φ)= 0,forallμ∈ kert_L

p ⊂ D(T2).

We will also consider theglobal solvability problem inthe space ofSchwartz distributionsD(T2_{). We} saythatLp isgloballysolvableonD(T2) ifLpD(T2)=◦(ker La−p),inwhichLa−p:C∞(T2)→ C∞(T2)

is given by La−p = ∂t+ a(x)∂x+ a(x)− p(x) and ◦(ker La−p) isthe space of distributions ν ∈ D(T2)

IncontrasttothepreviouslymentionedresultsofMalgrange,DuistermaatandHörmander,inourcontext thezeroorderperturbationp hasadirectbearingontheclosedrangepropertyforLp.

Ourfirstcontributionisto show thatwe maylose, maintain,or gainsolvability whenweadd a pertur-bationofzeroorderintheoperatorL0. ThisisthesubjectofSection2,wheretheglobal solvabilityofLp

ischaracterized inthecasewherethecoefficienta nevervanishes.

InSection3weshowthatLp isgloballysolvable,providedthata−1(0)= ∅ and a vanishesonlyoffinite

order.This isourmainresultinthis article.Whenp vanishesas muchasa at thepointsina−1(0),then we mayusetheapproachin[6]; however,theproof ofcertain resultsneedsacareful treatment(compare, forinstance,theproofofProposition3.1withtheproofofLemma2.1 in[6]).Whenthedifferencebetween theorder ofvanishing ofa and theorderof vanishingof p,atsomepoint ina−1(0), isgreater thanone, thenwe usecertaintechniques whichappearin[4] and[11]. Thebiggestnoveltyappearswhenthere exist pointsina−1(0), wherethedifferencebetweentheorder ofvanishingofa andtheorderofvanishingofp isone.Thisisacriticalcasewhichisrelatedtoothercriticalcasesthatappearin[2],[4], and[7]. Totreat these criticalcases,we needanimprovement oftechniques of [6] andwe alsoneedanewapproach, which ispresentedinCase5,intheproofofTheorem3.4.

WealsonoticeinSection3thattheglobalsolvabilitymaybecomestrongeronC∞(T2),bytheactionofthe perturbation,i.e.,LpC∞(T2) mayhavefinitecodimension,whileL0C∞(T2) hasinfinitecodimension.Onthe otherhand,weshowthatthesurjectivityofL0:D(T2)→ D(T2) maybeaffected,i.e.,LpD(T2) D(T2)

for certain perturbations p. Hence, the solvability on D(T2_{) may become weaker by the action of the} perturbation.

Finally,inSection4wegivefurtherresultsconsideringtheexistenceofpointsatwhichthecoefficienta

isflat.Inthecasewherea vanishesidenticallyweshowthatthezeroorderperturbationp maydestroythe globalsolvability ofL0 (seeTheorem4.2).

Thequestionsaddressedherearewithinthespiritofthose containedinthearticles[1,5,7–9,12–14,16]. 2. Coefficienta nevervanishing

Whena never vanishes,it follows from [6] that L0= ∂t+ a(x)∂x is globally solvable ifand onlyif the

mean (1/2π)₀2π(1/a) is either a rational or a non-Liouville irrational number. Recall that an irrational numberα is said to be aLiouville numberifthere exists asequence (pn,qn)∈ Z× N suchthatqn → ∞

and|pn+ αqn|< qnn, foralln∈ N.

WewillseethattheperturbationsLp= ∂t+ a(x)∂x+ p(x) may maintain,loseorgainsolvability.

Inorderto presentthemainresultofthis section,wedenotebyθ0 themeanofafunctionθ∈ C∞(T1), thatis,θ0= (1/2π)

2π 0 θ(s)ds.

Theorem2.1. LetLpbetheoperatorgivenby(1.3).Supposethata nevervanishes.Inthiscase,Lpisglobally solvable(on either C∞(T2_{) orD}_(T2_{)) ifandonly if thereexistpositiveconstantsC andγ suchthat}

|j + k(1/a)0+ (p/a)0− i(p/a)0| ≥ C(|j| + |k|)−γ,

forall(j,k)∈ Z2_{\ {(0,0)} suchthatj + k(1/a)}

0+ (p/a)0− i(p/a)0= 0.

Beforewritingtheproof, wewouldliketo discusscertainimplicationsoftheaboveresult. Nextexampleshowsthattheperturbationp maymaintaintheglobalsolvabilityofL0.

Example 2.2.If (1/a)0 is a rationalnumber, then L0 is globally solvable and any perturbation Lp is still

globallysolvable(theaprioriestimatepresentedinTheorem2.1issatisfied).Forinstance,∂t+ (1+ π)(1+

cos2_(x))−1_∂

The samesituation occurs (both L0 and any perturbationLp are globally solvable) if (1/a)0 isa non-Liouville irrationalnumber and (p/a)0 is arational number. Forinstance, ∂t−

√ 2(2− sin(x))−1∂x and ∂t− √ 2(2− sin(x))−1∂x+ i(π √

2)−1sin(x) aregloballysolvable.

In certain cases, we may have anon-globally solvable vector field L0 with a perturbation Lp which is

globally solvable.

Example 2.3.When(p/a)0= 0, thenTheorem 2.1 impliesthatanyperturbation Lp is globallysolvable.

Hence,if(1/a)0 iseitherrationaloranon-Liouvilleirrationalnumber,thentheperturbationLpmaintains

the global solvability of L0. On the other hand, if(1/a)0 is a Liouville irrational number,then L0 is not globally solvable, but Lp is globally solvable. It follows that we mayobtain a globally solvable operator

byconsidering aperturbationofzeroorderofthenon-globally solvablevectorfield L0.Thissituationalso occurs when (p/a)0 = 0. Forinstance,if(p0/a) = 0,(p/a)0=−1/2, and(1/a)0 =−β, inwhichβ is theLiouvillenumberconstructedinExample4.4in[1],thenL0isnotgloballysolvable,whileLpisglobally

solvable.Inparticular,∂t− 2β−1(2− sin(x))−1∂xisnotgloballysolvable,while∂t− 2β−1(2− sin(x))−1∂x+ i(2β)−1 isglobally solvable.

Mayweobtainanon-globally solvableoperatorbyconsideringaperturbationofzeroorderofaglobally solvable vector field L0? The answer is positive. As we have already noticed,to find operators satisfying this property,p mustsatisfy(p/a)0= 0 and (p/a)0 mustbeanirrationalnumber.Inaddition,if(1/a)0 is anirrationalnumber,then theaprioriestimate presentedinTheorem2.1, alsocharacterizes theglobal hypoellipticityoftheoperator∂t+(1/a)0∂x+i(p/a)0,whichisanissuetreatedin[1].ApplyingProposition 3.5in[1],weseethatthesetoffunctionsp suchthatLp isnotgloballysolvablecontainsadenseGδ.Hence,

inthesense of Bairecategory,there aremany perturbations(ofagloballysolvable vectorfield) whichare notgloballysolvable.

Wenow proceedto proveTheorem 2.1:Sincea nevervanishes, theglobal solvabilityofLp is equivalent

totheglobalsolvabilityof(1/a)Lp.Ontheotherhand,inthenewcoordinatesX = x andT = t+ (1/a)0− x

0(1/a), operator(1/a)Lp becomes ∂X+ (1/a)0∂T + (p/a). The global solvability ofthis last operatoris equivalent totheglobalsolvabilityof∂T + α∂X+ α(p/a),inwhichα = (1/a)−10 .Theproofnowfollowsby applying Propositions2.4 and2.5below.

Proposition 2.4.The operatorL= ∂t+ λ∂x+ ρ(x),whereλ∈ R\ {0} andρ∈ C∞(T1),isglobally solvable on C∞(T2_{) if andonly ifthere existpositiveconstantsC andγ suchthat}

|k + jλ + (ρ)0− i(ρ)0| ≥ C(|j| + |k|)−γ, (2.1)

forall(j,k)∈ Z2_{\ {(0,0)} suchthat k + jλ+ (ρ)}

0− i(ρ)0= 0. Proof. For eachk∈ Z,define

ck . = 1 2π 2π  0 ρ(s) + i(k + ρ(s)) λ ds.

Applying Lemma 3.1 of [3], it follows that (2.1) is equivalent to the following condition: there exist positive constantsC andγ such that

|1 − exp{−2πck}| ≥ C(|k| + 1)−γ,

Necessity:

If(2.1) fails tohold, thenthere exists asequence (kn)⊂ Z such that|kn+1|>|kn|> n,ckn ∈ iZ,/ and

0<|1− exp{−2πckn}|< (1+|kn|)−n,foralln∈ N.

Letf (x, kn) bethe2π-periodicextensionof

λ(1 +|kn|)−n/2φ(x) exp ⎧ ⎨ ⎩ π  x ρ(s) + i(kn+ρ(s)) λ ds ⎫ ⎬ ⎭, t∈ [0, 2π],

inwhich φ∈ Cc∞((π/2)− ,(π/2)+ ),0≤ φ(x) ≤ 1,φ ≡ 1 on [(π− )/2,(π + )/2],and  > 0 is small

enoughsothat((π/2)− ,(π/2)+ )⊂ (0,π).

Weclaimthatthefunctionf ,givenby

f (x, t) = ∞  n=1  f (x, kn) exp{iknt}, belongsto(kert_L)◦_{\ LC}∞_(T2_).

Noticethattheterm(1+|kn|)−n/2yieldstherapiddecayingoff (·,kn).Hencef ∈ C∞(T2).Inaddition,

ifμ∈ kert_{L,then μ(x,−k}

n)= 0,sinceckn ∈ iZ./ Therefore,f ∈ (ker t_L)◦_.

Tocompletetheproofof necessity,wewill showthatf cannotbelongto LC∞(T2_).

Supposethatthereexistsu∈ C∞(T2_{) solutionto Lu= f .Byusingpartial Fourierseriesinthevariable}

t we obtain

∂xu(x, kn) +

1

λ(ikn+ p(x))u(x, kn) = f (x, kn)/λ, x∈ T

1_{, n∈ N.}

Sinceckn∈ iZ,/ eachof theaboveequationshasauniquesolutioninC∞(T

1_{),whichmaybewrittenas}

u(x, kn) = Cn 2π  0 exp ⎧ ⎨ ⎩− x  x−y ρ(s) + i(kn+ρ(s)) λ ds ⎫ ⎬ ⎭λ−1f (x − y, kn)dy, (2.2) withCn= (1− exp{−2πckn})−1.

Thedefinitionof f (x, kn) impliesthat,forx∈ [π,2π],wehave

C_n−1exp ⎧ ⎨ ⎩ x  π ρ(s) + i(kn+ρ(s)) λ ds ⎫ ⎬ ⎭ u(x, kn)≥ (1 + |kn|)−n/2. (2.3) Since|Cn|≥ (1+|kn|)n,we obtain |u(π, kn)| ≥ (1 + |kn|)n/2, n∈ N,

whichisacontradiction,sinceu(π,kn) decaysrapidly.

Sufficiency:

Byassuming (2.1) wemust provethatforeach f ∈ (kert_L)◦ _{⊂ C}∞_(T2_{), there exists u∈ C}∞_(T2_{) such} thatLu= f .

PartialFourierseriesinthevariable t leadsustotheequations

∂xu(x, k) +

1

λ(ik + p(x))u(x, kn) = f (x, k)/λ, x∈ T

Weobtainsolutionstotheseequations defining u(x, k) = x  0 exp ⎧ ⎨ ⎩− x  y ρ(s) + i(k + ρ(s)) λ ds ⎫ ⎬ ⎭λ−1f (y, k)dy, (2.4)

ifk is suchthatck ∈ iZ,and, intheothercase,wedefine

u(x, k) = Ck 2π  0 exp ⎧ ⎨ ⎩− x  x−y ρ(s) + i(k + ρ(s)) λ ds ⎫ ⎬ ⎭λ−1f (x − y, k)dy, (2.5) inwhichCk= (1− exp{−2πck})−1.

Solutions givenby(2.4) are2π-periodic, sincef ∈ (kert_L)◦ _{andthedistributions}

exp ⎧ ⎨ ⎩ x  0 ρ(s) + i(k + ρ(s)) λ ds ⎫ ⎬ ⎭exp{−ikt} belong tokert_L,wheneverc

k∈ iZ.Inaddition, therapiddecayingoff (·,k) implies therapiddecayingof

thesequence ofsolutionsgivenby(2.4).

Finally,condition(2.1) impliesthatthereexistpositiveconstantsC andγ suchthat|1− exp{−2πck}|≥ C(1+|k|)−γ,for allk∈ Z suchthatck ∈ iZ;/ hencethe rapiddecayingof f (·,k) yieldthe rapiddecaying

of thesequence givenby(2.5). 

Proposition 2.5.The operatorL= ∂t+ λ∂x+ ρ(x),whereλ∈ R\ {0} andρ∈ C∞(T1),isglobally solvable on D(T2_{) if andonlyif condition (2.1) holds.}

Proof. Notice thattheoperatorP :C∞(T2)→ C∞(T2),givenbyP = ∂t+ λ∂x− ρ(x),satisfiestP =−L.

SincetheglobalsolvabilityofP onC∞(T2) impliestheglobalsolvabilityofL onD(T2),thesufficiencyof condition(2.1) followsfrom Proposition2.4.

On theother hand,ifcondition(2.1) fails tohold,then wedefine f ∈ C∞(T2_{) by proceedingasinthe} proof of necessityin Proposition2.4. Weclaim thatthis functionf belongs to ◦(ker P )\ LD(T2_{), which} means thatL isnotglobally solvable onD(T2_{). Noticethat,if u∈ C}∞_(T2_{) andP u = 0,then u(·,−k}

n)

vanishes identically, sinceckn ∈ iZ./ Hence f ∈◦(ker P ).Finally, ifμ∈ D(T

2_{) andLμ= f , then μ(·,k}

n)

belongstoC∞(T1_{) anditisgivenby(2.2).Henceestimate(2.3) holdstrue.Bytakingψ∈ C}∞_(T1_{,R) such} thatψ≥ 0,suppψ⊂ [π,2π],and_π2πψ = 1,itfollowsthat

   μ(x, kn), Cn−1exp ⎧ ⎨ ⎩ x  π ρ(s) + i(kn+ρ(s)) λ (s)ds ⎫ ⎬ ⎭ψ(x)   ≥ (1 + |kn|)−n/2, for all n∈ N. (2.6)

Since μ(·,kn) increasesslowly,weobtainconstantsK > 0 andM∈ Z+,whichdonotdependonn,such that    μ(x, kn), Cn−1exp ⎧ ⎨ ⎩ x  π ρ(s) + i(kn+ρ(s)) λ (s)ds ⎫ ⎬ ⎭ψ(x)    ≤ |Cn|−1K(1 +|kn|)M, for all n∈ N. (2.7)

Theestimates(2.6) and(2.7) implythat

foralln∈ N,whichisacontradiction. 

Remark2.6.Whenλ= 0,wehaveoperatorsofthetype∂t+ ρ(x).Theglobalsolvabilityoftheseoperators

ischaracterized inSection4.SeeTheorem4.2and Theorem4.3. 3. Coefficienta withonlyzerosoffiniteorder

Throughout this section,we assumethata−1(0)= ∅ andthata vanishesonly offinite order. Thus, we may write a−1(0) = {x1 < · · · < xN}, and xN +1 = x1+ 2π. In this case, we denote by nj the order of

vanishingofa at xj,j = 1,. . . ,N .

According to [6], under these assumptions the operator L0 is globally solvable on both C∞(T2) and

D_(T2_).Moreover,L

0 issurjectiveonD(T2).HenceL0C∞(T2)= (kertL0)◦ andL0D(T2)=D(T2). Byusing partial Fourierseries inthevariable t,wemaydescribe thefunctionsp= p(x) whichbelongs to L0C∞(T2).Note that,ifadistributionμ=



k∈Zμ(x,k)exp{ikt} belongsto kertL0,then thedivision theoremimpliesthat

μ(x, 0) =C0 a + N  j=1 nj−1 =0 Cjδ()(x− xj),

whereC0 andCjareconstants.Conversely,foranyconstantsthedistribution

⎛ ⎝C0 a + N  j=1 nj−1 =0 Cjδ()(x− xj) ⎞ ⎠ ⊗ 1t

belongsto kertL0. Therefore,p= p(x) belongstoL0C∞(T2)= (kertL0)◦ ifand onlyifp vanishesat each

xj,withorder ofvanishinggreaterthannj− 1,foreachj = 1,. . . ,N ,andmoreover,

2π

0 (p/a)(x)dx= 0. IfL0isgloballysolvableonC∞(T2) (respectivelyD(T2))andp∈ L0C∞(T2),thensimplecomputations showthatLp isstillgloballysolvable onC∞(T2) (respectivelyD(T2)).

Also,accordingto[6],La isglobally solvableonC∞(T2).Notethata∈ L/ 0C∞(T2). Ourmainresultinthissectionwillimplythat,foranyp∈ C∞(T1

x),theoperatorLp isgloballysolvable

onbothC∞(T2) andD(T2).

Wefirstpresentaresultconcerningsolvability modulofunctionswhichareflatata−1(0)× T1_.

Proposition3.1.LetLp begivenby(1.3).Supposethata vanishesonlyoffiniteorderanda−1(0)= ∅.Given f ∈ (kert_L

p)◦,thereexistsu∈ C∞(T2) such thatLpu− f isflat ata−1(0)× T1.

Proof. Byusingcutofffunctionsweseethatitisenoughtoworkonasmallneighborhood(x0−,x0+)×T1, withx0∈ a−1(0).

Givenf,u∈ C∞((x0− ,x0+ )× T1),we useformal Taylorseriestowrite

u(x, t) ∞  j=0 uj(t)(x− x0)j, uj(t) = 1 j!∂ j xu(x0, t), f (x, t) ∞  j=0 fj(t)(x− x0)j, fj(t) = 1 j!∂ j xf (x0, t), a(x) ∞  j=0 aj(x− x0)j, aj= 1 j!a (j)_(x 0),

and p(x) ∞  j=0 pj(x− x0)j, pj = 1 j!p (j)_(x 0).

TheformalTaylor seriesofLpu− f is

Lpu− f  ∞  j=0  u_j+ j  k=0 [(j + 1− k)akuj+1−k+ pkuj−k]− fj  (x− x0)j.

It followsthatLpu− f isflatat {x0}× T1 ifandonlyif

u_j+

j

k=0

[(j + 1− k)akuj+1−k+ pkuj−k] = fj, j = 0, 1, 2, . . . . (3.1)

After finding the sequence of solutions (uj)j∈Z+, the required function u is obtained by using Borel’s

Lemma.Hencetheproofreducesto solvetheequations(3.1).Inordertosolvethem(wewantsolutionsin

C∞_(T1

t)),wedenote byn (1≤ n<∞)theorder ofvanishing ofa at x0 andwesplit theproof into three cases:

Case 1: Assumethatn= 1.Thentheequations(3.1) reduce to

u₀+ p0u0= f0, (3.2) u₁+ (a1+ p0)u1= f1− p1u0, (3.3) u_j+ (ja1+ p0)uj = fj− j  k=1 pkuj−k− j  k=2 (j + 1− k)akuj+1−k, (3.4) forallj > 1.

Alltheseequations maybe solved,withuniquesolutions,ifp0∈ (−a/ 1)Z++ iZ.

If p0 =−im, m ∈ Z, then δ(x− x0)⊗ exp{−imt} belongs to kertLp. Since f ∈ (kertLp)◦, it follows

that f0(m) = (2π)−1δ(x− x0)⊗ exp{−imt},f = 0. Hence equation (3.2) has asolution. All the other equations, givenby(3.3) and(3.4),haveuniquesolutions,sincea1= 0.

If p0 = −a1− im, (m ∈ Z), then equation (3.2) has a unique solution u0 and the next equation is

u₁− imu1= f1− p1u0,whichhassolutionifandonlyif0= f1(m)− p1u0(m)= f1(m)+ p1f0(m)/a1.Since [−a1δ(x− x0)+ p1δ(x− x0)]⊗ exp{−imt}∈ kertLp, itfollows thata1f1(m)+ p1f0(m)= 0 andthen we canfindasolutionu1 to(3.3).Theotherequations,givenby(3.4),haveuniquesolutionsuj,j ≥ 2.

In the casep0 =−2a1− im,(m ∈ Z), the firsttwo equationshave uniquesolutions u0 and u1, which satisfy  u0(m) =− f0(m)/2a1 andu1(m) =− 1 a1   f1(m) + p1f0(m) 2a1  .

Thenextequationis

u₂+ (p0+ 2a1)u2= f2− (p1+ a2)u1− p2u0, whichhassolutionsifandonlyif

0 = f2(m) + (p1+ a2) a1  f1(m) +  (p1+ a2)p1 2a2 1 + p2 2a1   f0(m). (3.5) Setting ν(x) = 1 2δ _{(x− x} 0)− (p1+ a2) a1 δ(x− x0) +  (p1+ a2)p1 2a2 1 + p2 2a1  δ(x− x0),

directcomputations show thatν(x)⊗ exp{−imt} ∈ kertLp. Since f ∈ (kertLp)◦, we obtain0=ν(x)⊗

exp{−imt},f .Thislast identityisequivalentto (3.5).Hencewecanfindasolutionu2. Again,allthenextequations, givenby(3.4),haveuniquesolutionsuj,j≥ 3.

Moregeneral,assumethatp0=−j0a1− im (j0∈ N,j0≥ 2,andm∈ Z).Fork = 1,. . . ,j0,weset

Ck= k  =1  α∈N |α|=k  o=1(pαo+ (j0− α1− · · · − αo)aαo+1) a 1α1(α1+ α2)· · · (α1+· · · + α) ,

whereα= (α1,. . . ,α) isamulti-indexinNand |α|= α1+· · · + α.Note thattheequation

u_j0+ (j0a1+ p0)uj0 = fj0− j0  k=1 pkuj0−k− j0  k=2 (j0+ 1− k)akuj0+1−k

hasasolutionuj0 ifandonlyif

0 = fj0(m)− j0  k=1 pkuj0−k(m)− j0  k=2 (j0+ 1− k)akuj0+1−k(m), (3.6)

whilealltheotherequations,givenby(3.2)–(3.4) (withj= j0),haveuniquesolutions.Byusingtheprevious equations(indicesj = 0,1,. . . ,j0− 1)weseethatcondition(3.6) isequivalent to

0 = j0  k=0 Ckfj0−k(m), (3.7) inwhichC0= 1.. Define μ=.  _j0  k=0 ˜ Ckδ(j0−k)(x− x0)  ⊗ exp{−imt}, withC˜k= (−1)j0−k_(j0C_−k)!k ,k = 0,1,. . . ,j0. Ifμ belongstokert_L

p,then0=μ,f ,sincef ∈ (kertLp)◦.Ontheotherhand,bytheaboveexpression

whichdefinesμ,weseethatthecondition0=μ,f isequivalentto(3.7).Hencetheproofreducestoshow thatμ∈ kert_L

p.

Byusing inductionwe mayverifytheidentity

ka1Ck= k−1

=0

Thedistributionμ belongstokert_L p ifandonlyif d dx  a(x) j0  k=0 ˜ Ckδ(j0−k)(x− x0)  − (im + p) j0  k=0 ˜ Ckδ(j0−k)(x− x0) = 0.

Above identityisequivalentto

0 = j0  =1 _j0−  k=0 (−1)j0−k+−1_C˜ k  j0− k − 1  a(j0−k−+1)_(x 0)  δ()(x− x0) + j0  =0 _j0−  k=0 (−1)j0−k+−1_C˜ k  j0− k   (im + p)(j0−k−)_(x 0)  δ()(x− x0).

Since thederivatives of δ(x− x0) are linearlyindependent, it follows thatthe above identity becomes equivalent to(3.8).

Therefore, μ∈ kert_L p.

Case 2: Assumethat1< n<∞ andpj= 0,forallj≥ 1. Thentheequations(3.1) become

u_j+ p0uj = fj, j = 1, . . . , n− 1, (3.9) u_j+ p0uj = fj− j  k=n (j + 1− k)akuj+1−k, j≥ n. (3.10)

If p0 ∈ iZ,/ theneach equationineither(3.9) or (3.10) hasauniquesolution.We thensolverecursively these equations.

If p0 =−im, m ∈ Z, then δ(j)(x− x0)⊗ exp{−imt} ∈ kertLp, j = 0,. . . ,n− 1, and we maychoose

a solutionuj,which isnot unique,to uj+ p0uj = fj. After this, weproceed to solveequations (3.10).In

order tofindasolutionun to

u_n+ p0un = fn− anu1, we mustadjustthesolutionu1 so that

 u1(m) =  fj(m) an .

Similarly,to findasolutionuj (j > n)to (3.10),wemustadjustuj+1−n setting

 u_j+1−n(m) = 1 (j + 1− n)an   fj(m)− j  k=n+1 (j + 1− k)akuj+1−k(m)  .

Case 3: Supposenowthat1< n<∞ and thereexists j ≥ 1 suchthatpj= 0.Setj0= 1 ifp1= 0.Inthe other case,letj0 be thesmallestoftheindices1< j <∞ suchthatpj= 0 andp= 0,= 1,. . . ,j− 1.

Whenj0≥ n,theequations(3.1) become

u_j+ p0uj=fj, j = 0, . . . , n− 1, (3.11) u_j+ p0uj=fj− j  k=n (j + 1− k)akuj+1−k, n≤ j < j0, (3.12)

u_j+ p0uj =fj− j  k=j0 pkuj−k− j  k=n (j + 1− k)akuj+1−k, j≥ j0. (3.13)

Noticethatequations(3.12) donotappearwhenj0= n.

Applyingargumentssimilar tothoseused above,we seethatwemaysolverecursivelyequations(3.11), (3.12) and(3.13).

When1≤ j0< n,theequations(3.1) reduce to

u_j+ p0uj =fj, j = 0, . . . , j0− 1, (3.14) uj+ p0uj =fj− j  k=j0 pkuj−k, j0≤ j < n, (3.15) u_j+ p0uj =fj− j  k=j0 pkuj−k− j  k=n (j + 1− k)akuj+1−k, j≥ n. (3.16)

Again, if eitherj0 < n− 1 or p0 ∈ iZ,/ then we maysolve recursively these equations without further difficulties.

Whenj0 = n− 1 and p0=−im,m∈ Z, then wedonot findfurtherdifficultiesto solvethe equations providedthatpj0 ∈ −a/ nN.

Thelastcasetobecheckedisj0= n−1,p0=−im,m∈ Z,andpj0=−an,∈ N.Forinstance,if= 1,

then we maysolvethe first j0 equations in(3.14), sinceδ(j)(x− x0)⊗ exp{−imt} belongs to kertLp,for j = 0,. . . ,j0− 1.Sincej0= n− 1,thenextequationisu_n−1+ p0un−1 = fn−1− pj0u0(see(3.15)).Inorder

tosolvethis equation,wemust choosethesolutionu0suchthatu0(m)= fn−1(m)/pj0.Thenextequation

isu_n+ p0un= fn− pnu0(see(3.16)).Thisequationhasasolutionifandonlyifpj0fn(m)− pnfn−1(m)= 0.

Thislast identityfollowsfrom thefactthatthedistribution  pj0 n!δ (n)_{(x− x} 0) + pn (n− 1)!δ (n−1)_{(x− x} 0)  ⊗ exp{−imt}

belongs tokertLp. Hencewe havesolutions to theequation un+ p0un = fn− pnu0. Theother equations to be solved are given by (3.16), with j ≥ n+ 1. Each of these equations has a solution. Indeed, since

pj0+ (j + 1− n)an = 0,forallj≥ n+ 1,wemaychoosetheprevioussolutionuj−j0 so that

 fj(m)− j  k=j0 pkuj−k(m)− j  k=n (j + 1− k)akuj+1−k(m) = 0.

Noticethat,tofindsolutionsun+1,un+2,. . .,wemustadjustu2,u3,. . ..

Suppose now that j0 = n− 1, p0 = −im, m ∈ Z, and pj0 = −an,  ∈ N,  ≥ 2. Again, we solve

the first j0 equations in(3.14), since δ(j)(x− x0)⊗ exp{−imt} belongs to kertLp, for j = 0,. . . ,j0− 1. Thenextequationisu_n−1+ p0un−1= fn−1− pj0u0,whichwemaysolvebychoosingthesolutionu0 such

thatu0(m)= fn−1(m)/pj0. Similarly,we may recursivelyfind solutions un,. . . ,un+−2, to the respective

equationin(3.16),providedthatweadjusttheprevioussolutionsu1,. . . ,u−1 bychoosingthem sothat

 uk(m) = 1 pj0+ kan ⎛ ⎝ fj0+k(m)− k  j=1 [pj0+j+ (k− j)aj0+j+1]uk−j(m) ⎞ ⎠ ,

u_n+−1+ p0un+−1 = fn+−1− (pj0+1+ (− 1)aj0+2)u−1− · · · − (pj0+−1+ aj0+)u1− pj0+u0.

This equationhassolutions ifandonlyif

0 = fj0+(m)−

−1

k=0

(pj0+−k+ kaj0+−k+1)uk(m). (3.17)

Ifthereexistsasolutionu_n+−1,thenalltheotherequationsgivenby(3.16) (withj≥ n+ )maybesolved recursivelybyadjustingprevioussolutions,withaprocedureanalogoustotheonewhichgivesthesolutions

un,. . . ,un+−2.

Hencetheproofreduces toverify(3.17). Define F0= 1 andfork = 1,. . . ,,define

Fk = k  j=1 (−1)j  αj∈Nj |αj|=k j o=1(pj0+αoj+ (− α1j− · · · − αoj)aj0+αoj+1) j o=1(pj0+ (− α1j− · · · − αoj)an) , inwhichαj= (α1j,. . . ,αjj)∈ Nj and|αj|= α1j+· · · + αjj.

Aninductionprocessshowsthat

−(pj0+ (− k)an)Fk =

k−1 j=0

(pj0+k−j+ (− k)aj0+k−j+1)Fj, (3.18)

fork = 1,. . . ,.

Withnotationabove,(3.17) isequivalentto

0 =



k=0

Fkfj0+−k(m).

Since f ∈ (kertLp)◦, theproof willbecompletedbyshowingthatthedistribution

 _  k=0 ˜ Fkδ(j0+−k)(x− x0)  ⊗ exp{−imt} belongsto kert_L p,inwhichF˜k= (−1)j0+−k_(j0+Fk_−k)!,k = 0,. . . ,. Bytheexpressionoft_L

p,itfollowsthatthis distributionbelongsto kertLp ifand onlyif

d dx  a   k=0 ˜ Fkδ(j0+−k)(x− x0)  − (im + p)   k=0 ˜ Fkδ(j0+−k)(x− x0) = 0. (3.19)

Byusingthelineardependenceofthederivativesofδ(x−x0),wemayseethatidentity(3.19) isequivalent to thefollowingidentities

0 =   k=0 ˜ Fk  j0+  0  (−1)j0+−k_{(im + p)}(j0+−k)_(x 0),

0 = −ω k=0 ˜ Fk(−1)j0+−k+ω−1  j0+ − k ω− 1  a(j0+−k−ω+1)_(x 0) + −ω k=0 ˜ Fk(−1)j0+−k+ω−1  j0+ − k ω  (im + p)(j0+−k−ω)_(x 0), ω = 1,. . . ,.

Finally,weuse(3.18) in ordertoverifythevalidityoftheaboveidentities. Thiscompletestheproof. 

Remark 3.2.Given f ∈ (kert_L

p)◦, by Proposition3.1 it follows that there exists u ∈ C∞(T2) such that Lpu− f is flatat a−1(0)× T1. Notice thatLpu− f ∈ (kertLp)◦.Hence, ifthere exists v ∈ C∞(T2) such

that Lpv = Lpu− f, then Lp(u− v) = f . Inother words, Proposition 3.1 allows us to assume that the

right-handsideoftheequationLpu= f isflatat a−1(0)× T1.

BeforesolvingtheequationLpu= f forf flat, weneedatechnicalresult,whichwillbeusedtoproduce

certaindistributionsinthekernelofthetranspose operator.

Lemma3.3.Leta∈ C∞(T1,R) andassumethat a vanishesof finiteordern≥ 1 atx1.Letk∈ Z,m0∈ Z+

andp∈ C∞(T1_{). Denoteby q theorderof vanishingof p+ ik atx}

1 (q = 0 ifp(x1)+ ik= 0).Supposethat 0≤ q ≤ n− 1 andwheneverq = n− 1 wehave(p+ k)(q)_(x

1)= 0 orp(q)(x1)+ (/(q + 1))a(n)(x1)= 0,

forall= 0,. . . ,m0.

Under theaboveassumptions, given

ν = m0  =0 cδ()(x− x1), c∈ C, thereexists μ = m0+q =q dδ()(x− x1), d∈ C,

suchthat −∂x(aμ)+ (p+ ik)μ= ν.

Proof. Given μ = m0+q =q dδ()(x− x1), wemaywrite −∂x(aμ) + (p + ik)μ = m0  =0 Dδ()(x− x1),

inwhichtheconstants D aregivenas follows:denotingCj,= (−1)j

 j  ,wehave D0= m0+q j=q Cj,jdj(p + ik)(j)(x1),

and (whenm0> 0) D= m0+q j=+q dj[Cj−,j(p + ik)(j−)(x1)− Cj−+1,ja(j−+1)(x1)], for= 1,. . . ,m0.

Inorderto have−∂x(aμ)+ (p+ ik)μ= ν,wemustchoosedso thatD= c,for= 0,. . . ,m0. Ifm0= 0,thenitisenoughtochoose

dq=

c0 (−1)q_{(p + ik)}(q)_(x

1)

.

Ifm0= 1,thenitisenoughtochoose

dq+1= c1 Cq,q+1(p + ik)q(x1)− Cq+1,q+1a(q+1)(x1) and dq= c0− Cq+1,q+1dq+1(p + ik)(q+1)(x1) Cq,q(p + ik)(q)(x1) .

Finally,ifm0≥ 2, thenwerecursivelychoose

dm0+q= cm0 Cq,m0+q(p + ik)(q)(x1)− Cq+1,m0+qa(q+1)(x1) , d+q= c Cq,+q(p + ik)(q)(x1)− Cq+1,+qa(q+1)(x1) + m0+q j=+q+1 dj[Cj−+1,ja(j−+1)(x1)− Cj−,j(p + ik)(j−)(x1)] Cq,+q(p + ik)(q)(x1)− Cq+1,+qa(q+1)(x1) , for= m0− 1,m0− 2,. . . ,1,and dq= c0−m_j=q+10+q Cj,jdj(p + ik)(j)(x1) (−1)q_{(p + ik)}(q)_(x 1) . 

Wearenow inpositionto stateand proveourmainresult.

Theorem3.4.Ifa−1(0)= ∅ anda vanishesonlyoffiniteorder,thentheoperatorLpgivenby(1.3) isglobally solvable onC∞(T2_).

Proof. Write a−1(0)={x1 <· · · < xN},xN +1= x1+ 2π,anddenote bynj theorder ofvanishingof a at xj, j = 1,. . . ,N .

Given f ∈ (kert_L

p)◦, wemustshow thatthere existsu such thatLpu= f .ByProposition3.1,wemay

assumethatf isflatata−1(0)× T1_{(seeRemark3.2).}

Byusingpartial Fourierseriesinthevariablet,weareledto theequations

a(x)∂xu(x, k) + (ik + p(x))u(x, k) = f (x, k), x∈ T1, k∈ Z. (3.20)

Wewill firstsolvethese equationsandafter wewillshowthatthesequence ofsolutionsdecaysrapidly. Supposethatik + p vanishesata−1(0) asmuchasa.

Ifthemean((ik + p)/a)0= (2π)−1 2π

0 ((ik + p)/a)(x)dx belongstoiZ,thenasolutionto(3.20) isgiven by u(x, k) = exp ⎧ ⎨ ⎩− x  0 ik + p a (s)ds ⎫ ⎬ ⎭ x  0  f (y, k) a(y) exp ⎧ ⎨ ⎩ y  0 ik + p a (s)ds ⎫ ⎬ ⎭dy. Inordertosee thatu(x,k)∈ C∞(T1_{),itis enoughto showthat}

x  0  f (y, k) a(y) exp ⎧ ⎨ ⎩ y  0 ik + p a (s)ds ⎫ ⎬ ⎭dy = 0. Thislast identityfollowsfrom thefollowingtwofacts:f ∈ (kert_L

p)◦ and ⎡ ⎣exp ⎧ ⎨ ⎩ x  0 ((ik + p)/a)(s)ds ⎫ ⎬ ⎭/a(x) ⎤ ⎦ ⊗ exp{−ikt} ∈ kert_L p⊂ D(T1).

Ifthemean((ik + p)/a)0 doesnotbelongtoiZ, thentheuniquesolutionto (3.20) isgivenby

u(x, k) = C exp ⎧ ⎨ ⎩− x  0 ik + p a (s)ds ⎫ ⎬ ⎭+ x  0  f (y, k) a(y) exp ⎧ ⎨ ⎩− x  y ik + p a (s)ds ⎫ ⎬ ⎭dy, where C = 1 exp!₀2π ik+p_a (s)ds"− 1# 2π  0  f (y, k) a(y) exp ⎧ ⎨ ⎩ y  0 ik + p a (s)ds ⎫ ⎬ ⎭dy.

Note that ik + p does not vanish, for |k| large enough. Hence, there is no need to discuss the rapid decayingofthesolutionsgiven above.

Supposenowthatik + p vanisheslessthana at somepoint ina−1(0)={x1< . . . < xN}.

Inthiscase,inordertosolvetheequations(3.20),itisenoughtoworkonintervalsoftheform(xj,xj+σ),

in which ik + p(x) vanishes less than a on both xj and xj+σ. In addition, either σ = 1 or σ > 1 and ik + p(x) vanishes as much as a at the intermediate zeros xj+1,. . . ,xj+σ−1. Here σ ≤ N and we set xN +1= x1+ 2π,xN +2= x2+ 2π,. . . ,xN +j = xj+ 2π.

Onaninterval(xj,xj+σ) asdescribedabove,wehave

∂xu(x, k) + ik + p(x) a(x) u(x, k) =  f (x, k) a(x) , (3.21)

in which the right-hand side is smooth on [xj,xj+σ] and flat at the points xj,. . . ,xj+σ. We will seek a

solution u(x,k) to(3.21) which is smooth on (xj,xj+σ) and flat at {xj,xj+σ}, so that we may jointhis

solutionwithsolutionsonotherintervals withthesametype,inordertoobtainasolutionto(3.20) onT1. Inaddition,ifthesequence{u(x,k)}k∈Zdecaysrapidlyoneach[xj,xj+σ],thenitdecaysrapidlyonT1

and

u(x, t) = k∈Z

will beasolutionto Lpu= f .

Bythecommentsabove,itisenoughtoworkonlyonintervalsofthetype(xj,xj+σ).

Wenow splittheconstructionofthesolutionsu(x,k) on(xj,xj+σ) intofivecases:

Case1:p vanishesas muchasa atbothxj andxj+σ,and,consequently,p+ k eitherdoesnotvanish or

itvanisheslessthana atboth xj andxj+σ.

Inthiscase,forη∈ (xj,xj+σ),thefunctions

Ek(x) = exp ⎧ ⎨ ⎩− x  η p(s) + i(k + p(s)) a(s) ds ⎫ ⎬ ⎭ and 1/Ek(x) = exp ⎧ ⎨ ⎩ x  η p(s) + i(k + p(s)) a(s) ds ⎫ ⎬ ⎭ are boundedon(xj,xj+σ).Hence,theexpression

u(x, k) = Ek(x) x  xj  f (y, k) a(y)Ek(y) dy (3.22)

defines asmoothfunctionon(xj,xj+σ) whichisflatatxj and solve(3.21) on[xj,xj+σ).

Thenextstepistoshowthatu(x,k) isalsoflatat xj+σ.This isobtainedbyshowingthat

0 = xj+σ xj  f (y, k) a(y)Ek(y) dy = xj+σ xj  f (y, k) a(y) exp ⎧ ⎨ ⎩ y  η ik + p(s) a(s) ds ⎫ ⎬ ⎭dy.

Asin[6],wewillseethattheaboveidentityfollowsfromtheexistenceofcertaindistributionsinkert_L p. Notice that ψk(x) = $ 1/Ek(x), x∈ (xj, xj+σ) 0, x∈ T1_{\ (x} j, xj+σ)

belongsto D(T1_),since1/E

k(x) is bounded.Bythedivisiontheorem,there existsψk/a∈ D(T1), whose

order isatmostm, themaximumof theorderofvanishingof a on[xj,xj+σ]. Henceωk=−∂x(ψk)+ (p+ ik)(ψk/a) isadistributionoforder atmostm+ 1.Inaddition,supp ωk⊂ {xj,xj+σ}.It followsthat

ωk = m+1 =0 c1δ()(x− xj) + m+1 =0 c2δ()(x− xj+σ),

where c1 andc2 areconstants.ByLemma3.3, thereis

νk= m+1+q j =0 d1δ()(x− xj) + m+1+qj+σ =0 d2δ()(x− xj+σ)

(inwhichq istheorderof vanishingofp+ k atx,= j,j + σ),suchthatωk=−∂x(aνk)+ (p+ ik)νk.

Hence[(ψk/a)− νk]⊗ exp{−ikt} belongstokertLp.Consequently,

0 =(ψk/a)− νk, f (·, k) = xj+σ xj  f (y, k) a(y) exp ⎧ ⎨ ⎩ y  η ik + p(s) a(s) ds ⎫ ⎬ ⎭dy, sincef ∈ (kertLp)◦ andf isflatat {xj,xj+σ}.

Therefore, u(x,k), givenby(3.22),isasolutionto (3.21),whichissmooth on[xj,xj+σ] andflatat the

extremes.

Case 2: Supposethattheorder of vanishing ofa is greaterthan theorder of vanishingof p plusone,at bothxj and xj+σ.Inparticular,p doesnotvanishor itvanishesoffiniteorder.

If(p/a)(x)> 0 onsmallintervals(xj,xj+ ) and(xj+σ− ,xj+σ),weset

u(x, k) =Ek(x) x  xj  f (y, k) a(y)Ek(y) dy = x  xj  f (y, k) a(y) exp ⎧ ⎨ ⎩− x  y p + ik a (s) ⎫ ⎬ ⎭dy, x∈ (xj, xj+σ).

Noticethatu(·,k) iswelldefinedon(xj,xj+σ),since(p/a)(x)> 0 on(xj,xj+ ).Inaddition,u(·,k)∈ C∞_([x

j,xj+σ)) anditisflatatxj.Wenowproceedtoshow thatu(·,k) isalsoflatatxj+σ.It isenoughto

showthatu(xj+σ+ h,k)= O(|h|n),foralln∈ Z+.Forh< 0 sufficientlysmall, wehave

u(xj+σ+ h, k) = xj+σ+2h xj  f (y, k) a(y) exp ⎧ ⎪ ⎨ ⎪ ⎩− xj+σ +h y p + ik a (s) ⎫ ⎪ ⎬ ⎪ ⎭dy + xj+σ +h xj+σ+2h  f (y, k) a(y) exp ⎧ ⎪ ⎨ ⎪ ⎩− xj+σ +h y p + ik a (s) ⎫ ⎪ ⎬ ⎪ ⎭dy . =I1+ I2. Wehave |I1| ≤ Ch && &&f_a&&&&

∞ xj+σ+2h xj exp ⎧ ⎪ ⎨ ⎪ ⎩− xj+σ +h xj+σ+2h p a (s) ⎫ ⎪ ⎬ ⎪ ⎭dy, inwhich Ch= sup xj≤y≤xj+σ+2h ⎛ ⎜ ⎝exp ⎧ ⎪ ⎨ ⎪ ⎩− xj+σ+2h y p a (s) ⎫ ⎪ ⎬ ⎪ ⎭ ⎞ ⎟ ⎠ ; noticethatCh isapositiveconstantthatdoesnotdependonk.

Since (p/a)(x)> 0 on(xj+σ− ,xj+σ),itfollows thattheconstants Ch arebounded. Inaddition, on

(xj+σ− ,xj+σ) wemaywrite (p/a)(s)= (xj+σ− s)−ρβ(s),inwhichρ≥ 2 and0< r ≤ β(s)≤ M.We

then obtain     xj+σ+2h xj  f (y, k) a(y) exp ⎧ ⎪ ⎨ ⎪ ⎩− xj+σ +h y p + ik a (s) ⎫ ⎪ ⎬ ⎪ ⎭dy    ≤ C && &&f_a&&&&

∞|xj+σ− xj| exp ) −r ρ− 1  2ρ−1_{− 1} 2ρ|h|ρ−1 * = O(|h|n), foralln∈ Z+.

Wecanestimate|I2| similarly, sincef isflatat xj+σ.

Inthesequelwetreattheother possibilitiesto thesignofp/a.

If(p/a)(x)< 0 onsmall intervals(xj,xj+ ) and (xj+σ− ,xj+σ),wepickη∈ (xj,xj+σ) andweset

u(x, k) = − xj+σ x  f (y, k) a(y) exp ⎧ ⎨ ⎩− x  y p + ik a (s) ⎫ ⎬ ⎭dy, x∈ (xj, xj+σ). (3.23)

As before,thissolutionu(·,k) issmoothon[xj,xj+σ] and itisflatat theextremes.

If(p/a)(x)< 0 onasmallinterval(xj,xj+ ) and(p/a)(x)> 0 on(xj+σ− ,xj+σ),thenweset

u(x, k) = x  η  f (y, k) a(y) exp ⎧ ⎨ ⎩− x  y p + ik a (s) ⎫ ⎬ ⎭dy, x∈ (xj, xj+σ).

Proceeding as in the first situation treated in this Case 2, we may verify that this solution u(·,k) is

smoothon[xj,xj+σ] anditisflatattheextremes.

If(p/a)(x)> 0 onasmallinterval(xj,xj+ ) and(p/a)(x)< 0 on(xj+σ− ,xj+σ),thenweset

u(x, k) =Ek(x) x  xj  f (y, k) a(y)Ek(y) dy = x  xj  f (y, k) a(y) exp ⎧ ⎨ ⎩− x  y p + ik a (s) ⎫ ⎬ ⎭dy, x∈ (xj, xj+σ).

As before, thesecond identity abovemaybe used inorder to showthatthis solution u(·,k) is smoothon [xj,xj+σ) anditisflatatxj.Moreover,proceedingasintheCase1,viaLemma3.3weseethatthissolution

is alsoflatat xj+σ.

Case 3: Theorderofvanishing ofp isequaltotheorderofvanishing ofa,atboth xj andxj+σ.

In this case, we pick φ(x) = [1− cos(x− xj)]M[1− cos(x− xj+σ)]M, M ∈ N, which vanishes only at {xj,xj+σ},with finite orderof vanishing.PickingM large enough,it followsthatφ(x)/Ek(x) is bounded

on(xj,xj+σ).Hence, ψk(x) = $ φ(x)/Ek(x), x∈ (xj, xj+σ) 0, x∈ T1_{\ (x} j, xj+σ)

belongstoD(T1_{).Bythedivisiontheorem,thereexistsψ}

k/(aφ)∈ D(T1),whoseorderisatmost2M + m,

ωk=−∂x(aψk/(aφ)) + (p + ik)(ψk/(aφ))

isadistributionof orderatmost2M + m+ 1,whichisequalto

−∂x(1/Ek) + (p + ik)(1/(Eka)) = 0 on (xj, xj+σ).

Proceeding as intheCase1,we wantto applyLemma3.3 inorderto yield theexistenceof νk ∈ D(T1),

whichisalinearcombinationofderivativesofδ(x−xj) andδ(x−xj+σ),suchthatωk=−∂x(aνk)+(p+ik)νk.

Thisispossible assumingthattwo additionalpropertiesoccur:

• ifp+ k vanishesatxj oforderatleastnj,thenp(nj−1)(xj)+ (/nj)a(nj)(xj)= 0,for= 1,...,2M + m+ 1.

• ifp+ k vanishes atxj+σ oforderat leastnj+σ,thenp(nj+σ−1)(xj+σ)+ (/nj+σ)a(nj+σ)(xj+σ)= 0,

for= 1,...,2M + m+ 1.

Beforeproceeding,wementionthatthesituationinwhichtheextremesdonotsatisfytheaboveproperties willbe treatedinCase5.

Undertheassumptions madeinthisCase3,wehave

0 =(ψk/(aφ))− νk, f (·, k) (3.24) =ψk, [ f (·, k)/(aφ)] = xj+σ xj  f (y, k) a(y) exp ⎧ ⎨ ⎩ y  η ik + p(s) a(s) ds ⎫ ⎬ ⎭dy. Therefore,theexpression

u(x, k) = Ek(x) x  xj  f (y, k) a(y)Ek(y) dy, x∈ (xj, xj+σ)

defines asolutionto (3.21),whichis flatat{xj,xj+σ}.Indeed, sinceφ(x)/Ek(x) isbounded on(xj,xj+σ)

andf (x, k) isflatatxj,itfollows that

lim x→x+ j  f (x, k) a(x)Ek(x) = lim x→x+ j  f (x, k) a(x)φ(x) φ(x) Ek(x) = 0.

A similar procedureshows thatall theright-hand side derivativesof the functionf (x, k)/[a(x)Ek(x)] are

zeroatxj.Hence, x→ x  xj  f (y, k) a(y)Ek(y) dy

iswelldefined,smoothon[xj,xj+σ) andflatatxj.Similarly,itfollowsthatu(x,k) iswelldefined,smooth

on[xj,xj+σ) andflatatxj.Finally,byusingthatf (x, k) isflatatxj+σandusing(3.24) weseethatu(x,k)

isalsoflatatxj+σ.

Case4: Supposexj asinCase1 (or3)andxj+σasinCase2.Wethendefine u(x,k) on(xj,xj+σ) byusing

the end of Case 3),to see that this solutionis flatat xj, and we proceed as inthefirst part ofthe Case

2 in order to showthatthis solutionis flatat xj+σ.On theother hand,if(p/a)< 0 on asmall interval

(xj+σ− ,xj+σ),thenwe argueasinCase1 (or3),viaLemma3.3,inordertoverifythatwe mayrewrite

u(x,k) asin(3.23),fromwhichfollowsthatthissolutionisflatat{xj,xj+σ}.

Asimilarprocedureconstructsasolutionu(x,k) whenxj isasinCase2 andxj+σisasinCase1 (or3).

Indeed,itisenoughto consider u(x,k) givenby(3.23).

Whenxj isasinCase1 andxj+σ isasinCase3 orxj isasinCase3 andxj+σ isasinCase1,thenwe

proceed asinCase3 inorderto presentasmoothsolutionu(x,k) on(xj,xj+σ) whichisflatat{xj,xj+σ}.

Case 5: The order of vanishing of p is equal to the order of vanishing of a at some xj, to say xj1.

Moreover, p+ k vanishesat xj1 oforder atleast nj1,and thereexists 1∈ N suchthatp

(n_j1−1)_(x j1)+

(1/nj1)a

(nj1)_(xj

1) = 0. In this case, we look to a previous interval of the type (xj0,xj0+σ0), such that xj0 < xj1 andxj0+σ0 = xj1.Thechoiceofasolutionto (3.20) on (xj1,xj1+σ1) mustbeadjusteddepending

onthefixedsolutionto (3.20) on(xj0,xj0+σ0).This approachleadsusto thefollowing twosituations:

(i) solving(3.20) onachainofintervals

(xj0, xj0+σ0]∪ [xj1, xj1+σ1)∪ · · · ∪ [xjn, xjn+σn),

in which 1 ≤ n ≤ N, and xjm+σm = xjm+1, for m = 0,. . . ,n; in addition, the extremes xj0 and xjn+σn satisfythe assumptionsinLemma3.3,thatis, theyareofthetypestreatedinCases1–4.For

m = 1,. . . ,n,the order of vanishing of p is equalto the order of vanishing of a at xjm. Moreover,

p+ k vanishes at xjm of order at least njm, and there exists m ∈ N such that p

(njm−1)_(x jm)+

(m/njm)a

(njm)_(x

jm)= 0.

(ii) solving(3.20) onachainofintervals

(xj0, xj0+σ0]∪ [xj1, xj1+σ1)∪ · · · ∪ [xjn, xjn+σn),

as inthepreviousitem. However,nowxj0+ 2π = xjn+σn andforallm= 0,. . . ,n,n+ 1,theorder of

vanishing of p is equal to theorder ofvanishing of a at xjm. Moreover, p+ k vanishes at xjm of

orderat leastnjm,and thereexistsm∈ N suchthatp

(njm−1)_(x

jm)+ (m/njm)a

(njm)_(x

jm)= 0.

Foreachm= 0,. . . ,n,wefix ηm∈ (xjm,xjm+σm),andweset

Em,k(x) = exp ⎧ ⎨ ⎩− x  ηm p + ik a (s)ds ⎫ ⎬ ⎭ x∈ (xjm, xjm+σm).

We now split the construction of a solution to (3.20) on (xj0,xjn+σn) into five subcases, which take

into account different situationsconcerning the extremes xj0 and xjn+σn. The Subcases 5.1–5.4 treat the

situation(i),whileSubcase5.5 treatsthesituation(ii).

Subcase 5.1: The extremesxj0 andxjn+σn satisfythe assumptionsinCase1,i.e., p vanishesas much as

a and p+ k vanisheslessthana.

Weconsider u(x, k) = Em,k(x) ⎛ ⎜ ⎝Cm+ x  xjm  f (y, k) a(y)Em,k(y) dy ⎞ ⎟ ⎠ , (3.25)

forx∈ (xjm,xjm+σm),andm= 0,. . . ,n.

WesetC0= 0 andtheotherconstantsC1,. . . ,Cn mustbeadjustedinordertoobtainasmoothsolution

u(·,k) to (3.20) on(xj0,xjn+ σn).Thisis obtained(seetheAppendixA)bysetting

C1= ∂1 −E0,k(xj1) ∂1 +E1,k(xj1) x_j1  xj0  f (y, k) a(y)E0,k(y) dy, C2= ∂2 −E1,k(xj2) ∂2 +E2,k(xj2) ⎛ ⎜ ⎝C1+ xj2  x_j1  f (y, k) a(y)E1,k(y) dy ⎞ ⎟ ⎠ , and,recursively,forallm= 2,. . . ,n,

Cm= ∂m − Em−1,k(xjm) ∂m + Em,k(xjm) ⎛ ⎜ ⎝Cm−1+ xjm  x_jm−1  f (y, k) a(y)E_m−1,k(y)dy ⎞ ⎟ ⎠ . (3.26)

SinceC0= 0,u(x,k) isflatatxj0 (asinCase1).Wethenproceedtoverifythatthissolutionisalsoflat

atxjn+σn.

Itisenoughtoshow that

Cn+ xjn+σn xjn  f (y, k) a(y)En,k(y) dy = 0. (3.27) Setting Dm= ∂1 −E0,k(xj1)· · · ∂ m − Em−1,k(xjm) ∂1 +E1,k(xj1)· · · ∂ m + Em,k(xjm) , m = 1, . . . , n,

and denoting thecharacteristic functionof the interval (xjm,xjm+1) byχm, m = 0,. . . ,n, it follows that

thefunction Gk(x) = χ0(x)E0,k(x) + n  m=1 Dmχm(x)Em,k(x) (3.28)

is smooth on (xj0,xjn+σn) (we mayverify this byusing identity (A.1) in the Appendix A) and vanishes

onlyatthepointsxjm,wheretheorderofvanishingism,m= 1,. . . ,n.Inaddition,a(x)Gk(x)=−[p(x)+ ik]Gk(x),forallx∈ (xj0,xjn+σn).

Considerμk adistributioninD(T1) whichisgivenasfollows:on(xj0,xj1),μk = 1/Ek,0;on(xjn,xjn+

σn),μk= 1/(DnEn,k);on(xj0,xjn+σn),μk= 1/Gk;finally μk= 0 on T

1_{\ [x}

j0,xjn+σn].

Thedistributionμk/a belongsto D(T1);itissupportedon[xj0,xjn+σn] and satisfies

0 = ∂x(aGk(μk/a)) = Gk[∂x(a(μk/a))− (p + ik)(μk/a)]

on(xj0,xjn+σn).Hence

isalinearcombinationofderivativesofthedistributionsδ(x− xjm),m= 0,. . . ,n+ 1,andtheorderofthe

derivatives at xjm is lesser thanm, for allm = 1,. . . ,n. We then apply Lemma3.3 inorder to obtaina

linearcombination ofderivativesof delta,νk,suchthat

−∂x(a(μk/a− νk)) + (p + ik)(μk/a− νk) = 0. Hence 0 =μk/a, f (·, k) = μk, f (·, k)/a = xjn+σn x_j0  f (y, k) a(y)Gk(y) dy = 0. (3.29)

Inorderto seethat(3.29) isequivalentto(3.27),set

Fm= ∂m − Em−1,k(xjm) ∂m + Em,k(xjm) , m = 1, 2, . . . , n. Identity(3.27) is 0 = Dn xj1  x_j0  f (y, k) a(y)E0,k(y) dy + F2F3· · · Fn xj2  x_j1  f (y, k) a(y)E1,k(y) dy+ · · · + Fn xjn  x_jn−1  f (y, k) a(y)En−1,k(y)dy + xjn+1 xjn  f (y, k) a(y)En,k(y) dy. Multiplyingby1/Dn, weobtain 0 = xj1  xj0  f (y, k) a(y)E0,k(y) dy + (1/D1) xj2  xj1  f (y, k) a(y)E1,k(y) dy+ · · · + (1/Dn−1) xjn  x_jn−1  f (y, k) a(y)E_n−1,k(y)dy + (1/Dn) x_jn+1 xjn  f (y, k) a(y)En,k(y) dy, whichis(3.29).

Therefore, we have constructed, via formula (3.25), a smooth solution to (3.20) on the interval (xj0,xjn+σn),whichisflatattheextremes.

Subcase 5.2: Supposethattheextremesxj0 andxjn+σn satisfytheassumptionsinCase2,i.e.,theorderof

vanishing ofa isgreaterthantheorderofvanishingof p plusone,at bothxj0 andxjn+σn.

Ifforsome> 0p/a> 0 on(xj0,xj0+ ) andp/a< 0 on (xjn+1− ,xjn+1),thenwemayrepeatthe

approachinSubcase5.1.

Assumethatp/a> 0 on(xj0,xj0+) andp/a> 0 on(xjn+1−,xjn+1).SetC0= 0 andC1,. . . ,Cngiven

as inSubcase5.1.Withthischoice,(3.25) definesasmoothsolutionto(3.20) on theinterval(xj0,xjn+σn),

which is flatat xj0. As inCase 2, adirectcomputation(without exhibiting distributionsin thekernel of

Assumenowthatp/a< 0 on(xj0,xj0+ ) andp/a< 0 on(xjn+1− ,xjn+1).Weconsider u(x, k) = Em,k(x) ⎛ ⎝Cm− xjm+1 x  f (y, k) a(y)Em,k(y) dy ⎞ ⎠ ,

for x∈ (xjm,xjm+σm), and m= 0,. . . ,n. SettingCn = 0, itfollows that u(·,k) issmooth on (xjn,xjn+1]

andflatat xjn+1. Similartothe previoussubcase,inorderto obtainasmooth solutionon(xj0,xjn+1],we

choose C_n−1=− ∂ n +En,k(xjn) ∂n −En−1,k(xjn) x_jn+1 xjn  f (y, k) a(y)En,k(y) dy, C_n−2= ∂ n−1 + En−1,k(xjn−1) ∂₋n−1En−2,k(xjn−1) ⎛ ⎜ ⎝Cn−1− xjn  x_jn−1  f (y, k) a(y)E_n−1,k(y)dy ⎞ ⎟ ⎠ , and,recursively,forallm= n− 2,. . . ,0,

Cm= ∂m+1 + Em+1,k(xjm+1) ∂m+1 − Em,k(xjm+1) ⎛ ⎜ ⎝Cm+1− x_jm+2 x_jm+1  f (y, k) a(y)Em+1,k(y) dy ⎞ ⎟ ⎠ .

Again,asinCase2,adirectcomputation(withoutexhibitingdistributionsinthekernelofthetranspose operator)shows thatthissolutionisalsoflatatxj0.

We complete this Subcase 5.2 by considering the situation in which p/a < 0 on (xj0,xj0 + ) and p/a> 0 on(xjn+1−,xjn+1).Inthiscase,wemayconstructasolutionu(·,k) withthefollowingprocedure:

SetC0= C1= 0 anddefine

u(x, k) = −E0,k(x) x_j1  x  f (y, k) a(y)E0,k(y) dy, forx∈ (xj0,xj1);define u(x, k) = E1,k(x) x  x_j1  f (y, k) a(y)E1,k(y) dy,

forx∈ (xj1,xj2).Form= 2,. . . ,n andforx∈ (xjm,xjm+1),weset

u(x, k) = Em,k(x) ⎛ ⎜ ⎝Cm+ x  xjm  f (y, k) a(y)Em,k(y) dy ⎞ ⎟ ⎠ ,

in which C2,. . . ,Cn are given as in Subcase 5.1. As before, this procedure gives us asolution u(·,k) to

(3.20),whichis smoothon[xj0,xjn+1] andflatat{xj0,xjn+1}.

Subcase5.3: Wenowtreatthecaseinwhichtheextremesxj0andxjn+σnsatisfytheassumptionsinCase3,

form= 0,n+ 1,eitherp+ k vanisheslessthana atxjm orp (njm−1)_(x jm)+ (/njm)a (njm)_(x jm)= 0,for all∈ N.

The construction ofa solution u(·,k) to (3.20) on (xj0,xjn+σn) isanalogous to the oneinsubcase 5.1.

That is, on eachinterval (xjm,xjm+1), m = 0,. . . ,n, it is given by the formula (3.25), with C0 = 0 and

Cm satisfies(3.26),form= 1,. . . ,n.Thisgivesasmoothsolutionon[xj0,xjn+σn),whichisflatatxj0 (as

explained intheendofCase3).Wenowproceedto showthatthis solutionisflatatxjn+σn.

Pick φ(x)= [1− cos(x− xj0)]

M_{[1− cos(x− x} jn+σn)]

M_{, M ∈ N, which vanishesonly at {x}

j0,xjn+σn},

with finite orderof vanishing.Moreover, ifM is largeenough,thenφ/E0,k is boundedonasmall interval (xj0,xj0+ ) andφ/En,k isboundedonasmallinterval (xjn+σn− ,xjn+σn).

Take μk a distribution in D(T1) which satisfies: on the open interval (xj0,xj1), μk = φ/Ek,0; on

(xjn,xjn+ σn),μk = φ/(DnEn,k);on(xj0,xjn+σn),μk = φ/Gk,withGk given by(3.28);finallyμk = 0 on

T1_{\ [x}

j0,xjn+σn].

Byusingthedivisiontheorem,thereexists ωk = μk/(aφ)∈ D(T1),whichissupportedon[xj0,xjn+σn]

and aGkωk = 1 on(xj0,xjn+σn).Asbefore,on(xj0,xjn+σn) wehave

0 = ∂x(aGkωk) = Gk(∂x(aωk)− (p + ik)ωk).

Itfollowsthat∂x(aωk)− (p+ ik)ωk isalinearcombinationofderivativesofδ(x− xjm),m= 0,. . . ,n+ 1.

ByLemma3.3,weobtainνk suchthat−∂x(a(ωk− νk))+ (p+ ik)(ωk− νk)= 0.Hence

0 =ωk, f (·, k) = μk, f (·, k)/(aφ) = xjn+σn xj0  f (y, k) a(y)Gk(y) dy.

As insubcase5.1,thisimpliesthatu(·,k) isalsoflatatxjn+σn.

Subcase 5.4: Iftheextremexj0 belongsto oneof thesituationstreatedinSubcases5.1–5.3, andtheother

extreme xjn+σn belongs to another situation treated in Subcases 5.1–5.3, we may combine the previous

techniques inorder to construct asolution u(·,k) to (3.20), which is smooth on(xj0,xjn+σn) and flat at

{xj0,xjn+σn}.Thearguments aresimilartotheonesinCase4.

Subcase 5.5: To complete theCase 5,we now turn ourattentionto the othersituation, whenxj0+ 2π = xjn+σn and forall m = 0,. . . ,n,n+ 1, the order of vanishing of p is equal to theorder of vanishing of

a at xjm. Moreover, p(xjm)+ k = 0 andp+ k vanishesat xjm of order at leastnjm, and there exists

m∈ N such thatp(njm−1)(xjm)+ (m/njm)a

(njm)_(x

jm)= 0. Inother words, nowthe extremesxj0 and xjn+σn areofthesametypeasthemiddlezerosxj1,. . . ,xjn.

If∂n+1

− En,k(xjn+1)Dn= ∂ 0

+E0,k(xj0),thenwemustchooseC0 satisfying

+ ∂0 +E0,k(xj0) ∂n+1 − En,k(xjn+1) − Dn , C0= Dn x_j1  x_j0  f (y, k) a(y)E0,k(y) dy+ F2F3· · · Fn x_j2  x_j1  f (y, k) a(y)E1,k(y) dy +· · · + Fn xjn  x_jn−1  f (y, k) a(y)E_n−1,k(y)dy + x_jn+1 xjn  f (y, k) a(y)En,k(y) dy.