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Travelling waves in nonlinear magneto-inductive lattices
M. Agaoglou, M. Fečkan, M. Pospíšil, V.M. Rothos, H. Susanto
To cite this version:
Travelling
waves
in
nonlinear
magneto-inductive
lattices
M. Agaoglou
a,
M. Feˇckan
b,c,
M. Pospíšil
c,
V.M. Rothos
a,∗,
H. Susanto
daLabofNonlinearMathematicsandDepartmentofMechanicalEngineering,FacultyofEngineering,
AristotleUniversityofThessaloniki,Thessaloniki54124,Greece
bDepartmentofMathematicalAnalysisandNumericalMathematics,ComeniusUniversityinBratislava,
Mlynskádolina,84248Bratislava,Slovakia
cMathematicalInstituteofSlovakAcademyofSciences,Štefánikova49,81473Bratislava,Slovakia dDepartmentofMathematicalSciences,UniversityofEssex,WivenhoePark,ColchesterCO43SQ,UnitedKingdom
Weconsider a lattice equation modelling one-dimensional metamaterials formed by a discrete array of nonlinear resonators. We focus on periodic travelling waves due to the presence of a periodic force. The existence and uniqueness results of periodic travelling waves of the system are presented. Our analytical results are found to be in good agreement with direct numerical computations.
Keywords: Travellingwave;Latticewave;Forcedmagneto-inductivelattice
1. Introduction
Inthiswork,weconsider[26] d2 dt2(un− λun−1− λun+1)+ γ d dtun+ un+ d2 dt2(u 2 n− λu2n−1− λu2n+1) + γ d dtu 2 n− h(ωt + pn) = 0 (1) * Correspondingauthor.[email protected],[email protected], [email protected],
whereγ≥ 0,λ,ω > 0,p= 0 areparametersandh∈ C(R,R) issuchthat h(z)= k∈Z hkekız, k∈Z |hk| < ∞.
Theequationmodelsthedynamicsofelectromagneticwavesintheso-calledmagneto-inductive metamaterials.
Metamaterialsareartificialmaterialsthatareengineeredtohavepropertiesthatmaynotbe foundinnature[5].The(structuralratherthanchemical)engineeringisachievedby compos-ing periodicinhomogeneities to createdesirable effective behaviour. The invention igniteda newparadigminelectromagnetism,includingcloakingdevices[20](seealso[17]forarecent comprehensive review of electromagnetic manipulation enabledby metamaterials).Magnetic metamaterialsarenon-magneticmaterialsexhibitingmagneticpropertiesintheTerahertzand op-ticalfrequencies.Theywerepredictedtheoreticallyin[12,28]anddemonstratedexperimentally in,e.g.,[7,11,14,27].Whenthematerialscomposedofmagneticmetamaterials,e.g.,split-ring resonators(SRRs),aremagneticallyweaklycoupledthroughtheirmutualinductances,one ob-tains magneto-inductivemetamaterials[9,21–23].Magneto-inductivemetamaterialsconsisting of periodicarraysofSRRshavebeenmade inone-,two- andthreedimensions[19].Boththe dimensionsofSRRsandtheirinter-spacedistancearesmallrelativetothefreespacewavelength and, thus,the dynamics of electromagneticfields in such settingsare governed by the quasi magnetostaticapproximation[13].
Modelequationsforthepropagationofnonlinearelectromagneticwavesinmetamaterialscan begroupedintotwoclasses.Thefirstapproachassumesthateffectivemetamaterialsare homo-geneousmediawithspecificphysicalpropertiesresultinginpartialdifferentialequations,suchas coupledshort-pulseequation[24]andhigher-ordernonlinearSchrödingerequations[25].Inthe secondclass,metamaterialsaremodelledbyarraysofcoupledoscillators,i.e.latticeequations, suchasanonlinearKlein–Gordonequation[18]andcoupledKlein–Gordonequations[15,16]. The governingequation(1)fallsinto thesecondapproachwithγ representingtheloss coeffi-cient,λ isthecouplingparameter,h isanexternalforcingthatisperiodicintimeandvaryingin thespatialdirection.
Inthelatticeequation(1)thenonlinearityappearsinthecouplingtermsandinthedamping. Thisisduetotheassumptionofthenonlinearityofthecapacitanceofthesplit-ringresonators thatcomposethemagneto-inductivematerials[26].Notethatthenonlinearityisdifferentfrom our previous work[1,6], whereit isinthe onsitepotential. The present workisto studythe effectofsuchnonlinearity.Here,weinvestigatetheexistenceoftravellingperiodicsolutionsof system(1) duetothe periodic forcing,andthe bifurcationof such solutionswithsmall γ , λ andh.WealsostudythemodulationalstabilityoftheperiodicsolutionsbycomputingFloquet multipliersofthelinearized systems.
Notethatinourgoverningequation(1)thenonlinearityinthecouplingtermsbetweenthesites isakintothatintheFermi–Pasta–Ulamlattices[10].However,thereisasignificantdifference infactthatthe couplinginourgoverningequationisalsothederivativeterm.Thebifurcation structures of periodicsolutions ingeneralFPU lattices forced byperiodicdrive werestudied in[8].Itisimperativetostudytheeffectofthepresentnonlinearcouplingstoperiodicsolutions causedbythesameperiodicdrive.UsingLyapunov–Schmidtreduction,wederivetheasymptotic expressionsofthebifurcatingsolutions.
we provethe existence of such waves rigorously. The resonanceconditionfor the parameter valuesisalsoderived.InSection3,westudytheresonanceregion.UsingtheLyapunov–Schmidt reduction,weshowthattheperiodicsolutionspersist.Severalexplicitexamplesofthegeneral resultsobtainedintheprevioussectionsare presentedinSection4.In Section5 wesolvethe governingequationsnumerically.Comparisonswiththeanalyticalresultsarepresentedwhere weobtaingoodagreement.
2. Existence of small periodic solutions
Puttingun(t)= U(ωt + pn),z= ωt + pn in(1),weobtain
ω2(U (z)− λU(z − p) − λU(z + p))+ γ ωU(z)+ U(z)
+ ω2(U2(z)− λU2(z− p) − λU2(z+ p))+ γ ω(U2(z))− h(z) = 0. (2)
WetakeBanachspaces
X:= U∈ C(R, R) | U(z) = k∈Z ckekız, k∈Z |ck| < ∞ , Y:= ⎧ ⎨ ⎩U∈ C1(R, R) | U(z) = k∈Z ckekız, k∈Z\{0} |k||ck| < ∞ ⎫ ⎬ ⎭, Z:= ⎧ ⎨ ⎩U∈ C2(R, R) | U(z) = k∈Z ckekız, k∈Z\{0} k2|ck| < ∞ ⎫ ⎬ ⎭
withthenorms
U := k∈Z |ck|, U1:= |c0| + k∈Z\{0} |k||ck|, U2:= |c0| + k∈Z\{0} k2|ck|,
respectively. It is easy to verify that Z Y X are compact embeddings and U≤
U1∀U ∈ Y ,U1≤ U2∀U ∈ Z.
Thefollowinglemmaisclear.
Lemma 1. IfU1,U2∈ X thenU1U2∈ X andU1U2≤ U1U2.
Bysetting
KU := ω2 U(z)− λU(z− p) − λU(z+ p) + γ ωU(z)+ U(z),
F(U, h) := −ω2 (U2(z))− λ(U2(z− p))− λ(U2(z+ p))− γ ω(U2(z))+ h(z),
equation(2)hastheform
Wehavethenextresult.
Lemma 2. FunctionF : Z × X → X fulfils
F(U, h) ≤ 2ω(2ω + 4λω + γ )U2
2+ h, (3)
F(U1, h)− F(U2, h) ≤ 2ω(2ω + 4λω + γ )U1− U22(U12+ U22), (4)
F(U, h1)− F(U, h2) ≤ h1− h2 (5)
foranyU,U1,U2∈ Z andh,h1,h2∈ X.
Proof. Property (5) is obvious. Next, we use U= U1 ≤ U2, U = U2 with
Lemma 1toderive
(U2(z+ c)) = 2(U(z+ c))2+ U(z + c)U(z+ c) ≤ 4U(z + c)2 2 forc= −p,0,p.Moreover, U(z ± p) = k∈Z |ck|| ekıp| = k∈Z |ck| = U
forU (z)=k∈Zckekız.Similarlywederive
(U2(z)) = 2U(z)U(z) ≤ 2U2 2.
Hence,property(3)isobtained.Analogically,weget
(U2 1(z+ c) − U 2 2(z+ c)) ≤ (U1(z+ c) − U2(z+ c))(U1(z+ c) + U2(z+ c)) + 2(U1(z+ c) − U2(z+ c))(U1(z+ c) + U2(z+ c)) + (U1(z+ c) − U2(z+ c))(U1(z+ c) + U2(z+ c)) ≤ 4U1− U22U1+ U22 forc= −p,0,p,and (U2 1(z)− U22(z)) ≤ (U1(z)− U2(z))(U1(z)+ U2(z))
+ (U1(z)− U2(z))(U1(z)+ U2(z)) ≤ 2U1− U22U1+ U22
sincea2− b2= (a − b)(a + b) foranya,b∈ C.Now,(4)followseasily. 2
Next,ifU∈ Z withU (z)=k∈Zckekızthen
KL(Z,X)≤ 1 + ω2(1+ 2λ) + γ ω. If := inf k∈Z\{0} ⎧ ⎨ ⎩1, 1 k2+ ω 2(2λ cos kp− 1) 2 +γ2ω2 k2 ⎫ ⎬ ⎭> 0 (7)
isaconstantdependingonγ ,λ,ω andp,thenwealsohaveK−1∈ L(X,Z)⊂ L(X,Y )⊂ L(X). SoK−1: X → Z iscontinuoussuchthat
K−1L(X,Z)≤
1
. (8)
Nowwe canprovethefollowingexistence result on(2)when all parametersexcept h are fixed.
Theorem 1. Assume(7)alongwith
0 <h <
2
8ω(2ω+ 4λω + γ ). (9)
Thenequation(2)hasauniquesolutionU (h)∈ B(ρh) inaclosedball
B(ρh):= {U ∈ Z | U2≤ ρh},
where
ρh=
−2− 8ω(2ω + 4λω + γ )h
4ω(2ω+ 4λω + γ ) . (10)
Moreover,U (h) canbeapproximatedbyaniterationprocess.Finally,itholds
U(h1)− U(h2)2≤ h 1− h2 2− 8ω(2ω + 4λω + γ ) max{h 1, h2} (11)
foranyh1,h2∈ X satisfying(9).
Proof. We rewrite(2)asaparameterizedfixedpointproblem
U= R(U, h) := K−1F(U, h)
inZ.WealreadyknowthatR: Z × X → Z iscontinuousandby(3),(8)suchthat
A(ρh):= ρh− 2ω(2ω + 4λω + γ )ρh2= h, (12)
thenR(·,h) mapsB(ρh) intoitself.Soit remainstostudy (12).In ordertofindH > 0 –the
largest right-hand side of (12) when this equation has asolution ρH > 0, we need to solve
A(r)= H togetherwithDA(r)= 0.Thisimplies
ρH=
4ω(2ω+ 4λω + γ )
and(9).Soassuming(7),(9),weknowthat(12)hasapositivesolutionρh< ρH.Wetakethe
smallestonewhichclearlyhastheform(10).SoR(·,h) mapsB(ρh) intoitselfand,moreover,
by(4),(8)
R(U1, h)− R(U2, h)2≤
4ω(2ω+ 4λω + γ )U1− U22ρh
foranyU1,U2∈ B(ρh).Hence,R(·,h) isacontractiononB(ρh) withacontractionconstant
4ω(2ω+ 4λω + γ )ρh
<
4ω(2ω+ 4λω + γ )ρH
= 1.
The proof ofthe existenceanduniqueness isfinished bytheBanach fixedpointtheorem[2]. Next, let h1,h2 ∈ X satisfy (9). Then U (hi)∈ B(ρhi)⊂ B(ρH) for i = 1,2 and H :=
max{h1, h2}.NoteH satisfies (9).By(4),(5)and(8),wederive
U(h1)− U(h2)2= R(U(h1), h1)− R(U(h2), h2)2
≤ R(U(h1), h1)− R(U(h2), h1)2+ R(U(h2), h1)− R(U(h2), h2)2
≤4ω(2ω+ 4λω + γ )U(h1)− U(h2)2ρH+ h1− h2
whichimplies(11). 2
Remark 1. Inthefollowingspecialcases,(7)holdsandwecanreplace withthecorresponding i intheaboveconsiderations:
1. Ifp∈ 2πZ,λ>12,then
≥ 1:= min{1, ω2(2λ− 1)} > 0.
2. Letγ= 0.Ifp∈ πQ,thenwecanwritep=p1
p2π forsomep1∈ Z,p2∈ N,wherep1,p2
arerelativelyprime(theironlycommondivisoris1).Moreover,
M1:= {cos kp | k ∈ Z\{0}} ⊂ coskπ p2 | k ∈ {0, 1, . . . , 2p2− 1} =: M2. (13)
Indeed,foreachk∈ Z thereexisti,j∈ Z suchthat0≤ i ≤ 2p2− 1 andkp1= 2jp2+ i.Hence,
cos kp= coskp1π p2 = cos
iπ p2∈ M2
.
TofindabetterrelationshipbetweenM1andM2,weneedtosolvecosk1pp1
2 π= cos k2π
p2 for
k1∈ Z\{0} andk2∈ {0,1,. . . ,2p2− 1}.Thisisequivalentto k1pp1
2 π= k2π
p2 + 2πl forl∈ Z and
k1= ±k1,i.e.,k1p1= k2+ 2lp2.Ifp1isodd,thenp1and2p2arerelativelyprime.Thus,we
know[3]thatthereexistˆk,ˆl ∈ Z suchthat ˆkp1= 1+ 2ˆlp2.Consequently, ˆkk2p1= k2+ 2ˆlk2p2
andM1= M2.Ifp1iseven,thenp1= 2 ¯p1 andk12¯p1= k2+ 2lp2,sok2= 2¯k2 andwehave
k1¯p1= ¯k2+ lp2.Clearly ¯p1 andp2 are relativelyprime.Thus,thereexist ¯k,¯l ∈ Z such that
¯k ¯p1= 1+ ¯lp2.Then¯k ¯k2¯p1= ¯k2+ ¯l¯k2p2.HenceM1= cos2kπp 2 | k ∈ {0, 1, . . . , p2− 1} M2. Nevertheless,assuming ω2∈/ 1 k2(1− 2λ cos kp) k∈Z\{0} (14) and 0 /∈ 2λM2− 1 (15)
forM2definedin(13),wegettheexistenceofδ > 0 suchthat
1 k2+ ω 2(2λ cos kp− 1) 2 +γ2ω2 k2 ≥ ω 2 1 ω2k2+ 2λ cos kp − 1 ≥ δ > 0
foreachk∈ Z\{0}.Thus,ifp∈ πQ and(14),(15)arevalid,then
≥ min 1, inf k∈Z\{0}ω 2 1 ω2k2+ 2λ cos kp − 1 =: 2> 0. Notethat 1
k2(1−2λ cos kp)→ 0 if|k|→ ∞.Soifp∈ πQ and(15)holds,thereisatmostafinite
numberofresonantmodesk0∈ Z\{0} determinedbyequation
ω2= 1
k02(1− 2λ cos k0p)
. (16)
Moreprecisely,forgivenλ,ω thereisnotmorethan2(p2+ 1) resonantmodesk0.Indeed,let
{k0j}Jj=1beanincreasingsequenceofpositiveresonantmodescorrespondingtothefixedλ,ω.
Then{1− 2λcos k0jp}Jj=1hastobedecreasing,since
k0j2 (1− 2λ cos k0jp)=
1
ω2= const. ∀j = 1, . . . , J,
i.e.,{cos k0jp}Jj=1 isincreasing.From propertiesof cos x and(13),itfollows thatthe longest
sequenceoftheformcos k0jp hasthevalues{coskπp
2 | k = p2,. . . ,2p2} whichhasp2+ 1
3. Ifγ > 0,condition(7)issatisfiedevenwithoutthenon-resonancecondition(14),sincein eachresonantmodek0wehave
1 k20+ ω 2(2λ cos k 0p− 1) 2 +γ2ω2 k20 = γ ω |k0| > 0
andthereisonlythefinitenumberofresonantmodes.Foreachnon-resonantmodek itholds
1 k2+ ω2(2λ cos kp− 1) 2 +γ2ω2 k2 ≥ k12+ ω 2(2λ cos kp− 1) > 0.
Summarizing,ifγ > 0,p∈ πQ and(15)holds,condition(7)isfulfilledand
≥ min 1, min k∈M γ ω |k|,k∈Z\({0}∪M)inf ω 2 1 ω2k2+ 2λ cos kp − 1 =: 3> 0
whereM isthesetofresonantmodesk0. 3. Simple resonances
Inthissection,weinvestigatethebifurcationofasmallsolutionof(2)undertheassumption ofasimpleresonancedescribedinRemark 1(2) So,wesuppose
(R) p∈ πQ,condition(15)isvalid,forsomek0∈ Z\{0} equation(16)holdsandhastheonly
solutions±k0,
andweconsidertheequation
A(U, ε):= KU − F(U) − εG(U) = 0 (17) with
KU := ω2(U(z)− λU(z− p) − λU(z+ p)) + U(z)
F(U) := −ω2((U2(z))− λ(U2(z− p))− λ(U2(z+ p)))
G(U)(z) := −γ ω(U(z)+ (U2(z)))+ h(z) (18)
forU andε small.
We shall solve equation (17) via Lyapunov–Schmidt reduction method [4]. Clearly, A(0,0)= 0 andDUA(0,0)= K.Forsimplicitywedenoteκ:= |k0| andR1κ:= N K,R2κ:= RK
thenullspaceandtherangeoftheoperatorK ∈ L(Z,X),respectively.Equation(6)withγ= 0 togetherwith(R) yieldsthatifU∈ Z withU (z)=k∈Zckekız,then
KU(z) =
k∈Z\{±k0}
Consequently, R1κ= c eκız+¯c e−κız| c ∈ C , R2κ= ⎧ ⎨ ⎩U∈ X | U(z) = k∈Z\{±k0} ckekız ⎫ ⎬ ⎭.
LetQ: X → Rκ2betheprojectionontoRκ2⊂ X givenby
QU (z)=
k∈Z\{±k0}
ckekız.
Now,wetakethedecompositionU= U1+ U2,U1∈ R1κ,U2∈ Rκ2∩ Z forU∈ Z and
decou-ple(17)toequations
QA(U1+ U2, ε)= 0, (19)
(I − Q)A(U1+ U2, ε)= 0. (20)
Applyingtheimplicitfunctiontheoremtothefirstequationimpliestheexistenceof neighbour-hoodsV1× V0⊂ R1κ× R ofthepoint(0,0) inR1κ× R, V2⊂ R2κ∩ Z of 0 inRκ2∩ Z anda
uniqueC∞-functionU2: V1× V0→ V2 suchthatQA(U1+ U2,ε)= 0 for(U1,ε)∈ V1× V0
andU2∈ V2ifandonlyifU2= U2(U1,ε).Moreover,U2(0,0)= 0.
Thus,equation(20)hastheform
(I − Q)A(U1+ U2(U1, ε), ε)= 0 (21)
forU1∈ V1andε∈ V0.Differentiatingequation(19)weobtain
DU1U2(0, 0)= 0, DεU2(0, 0)= −K−1QDεA(0, 0)= K−1QG(0). (22)
Hence
U2(U1, ε)= K−1QG(0)ε + O((U1, ε)2). (23)
ExpandingF andG of(18)intoTaylorseriesusingthelastexpansionofU2(U1,ε) yields
F(U1+ U2(U1, ε))=
1 2D
2F(0)(U
1+ DεU2(0, 0)ε)2+ O((U1, ε)3),
G(U1+ U2(U1, ε))= G(0) + DG(0)(U1+ DεU2(0, 0)ε)+ O((U1, ε)2).
Note that D2F(0)V2= 2F(V ) and εO((U1,ε)2)= O((U1,ε)3). Thusequation (21) is
equivalentto
(I − Q)(F(U1+ DεU2(0, 0)ε)+ εG(0) + εDG(0)(U1+ DεU2(0, 0)ε)+ O((U1, ε)3))= 0
(I − Q)(ω2(((U1(z)+ K−1QG(0)(z)ε)2)− λ((U1(z− p) + K−1QG(0)(z − p)ε)2) − λ((U1(z+ p) + K−1QG(0)(z + p)ε)2))− εG(0)(z) − εDG(0)(z)(U1+ K−1QG(0)ε) + O((U1, ε)3))= 0. (24) Notethat ((U1(z)+ K−1QG(0)(z)ε)2)= 2(U1(z)+ (K−1QG(0)(z))ε) 2 + 2(U1(z)+ K−1QG(0)(z)ε)(U1(z)+ (K−1QG(0)(z))ε)
foranyz∈ R.LetU1(z)= c ek0ız+¯c e−k0ız∈ Rκ1andh(z)=
k∈Zhkekız∈ X.Then U1(z)= ˜c ek0ız+˜c e−k0ız∈ Rκ 1, K−1QG(0)(z) = k∈Z\{±k0} dkekız∈ Rκ2, (K−1QG(0)(z))= k∈Z\{0,±k0} ˜dkekız∈ Rκ2 (25) where ˜c = ık0c, dk= hk 1− ω2k2+ 2λω2k2cos kp, ˜dk= ıkdk.
Clearly,d−k= dk, ˜d−k= ˜dkforeachk∈ Z\{0,±k0}.Itcanbeeasilyshownthat
(I − Q)(U1(z))2= 0 = (I − Q)(U1(z)U1(z)),
(I − Q)(U1(z)(K−1QG(0)(z)))= ˜c ˜d2k0e−k0 ız+˜c ˜d 2k0e k0ız. (26) Next,setting ˜d±k0= ˜d0= 0, k∈Z\{0,±k0} ˜dkekız j∈Z\{0,±k0} ˜djej ız= l∈Z k+j=l ˜dk˜djelız.
Hencethecoefficientbyek0ızin((K−1QG(0)(z)))2is
wehave U1(z)= ˜˜c ek0ız+˜˜c e−k0ız∈ Rκ 1, (K−1QG(0)(z))= k∈Z\{0,±k0} ˜˜dkekız∈ Rκ2,
andthenextidentitiescanbeproved
(I − Q)(U1(z)(K−1QG(0)(z)))= c ˜˜d2k0e k0ız+c ˜˜d 2k0e−k 0ız, (I − Q)(U1(z)K−1QG(0)(z)) = ˜˜cd2k0+ ˜˜cd0 ek0ız+ ˜˜cd2k0+ ˜˜cd0 e−k0ız, (I − Q)(K−1QG(0)(z)(K−1QG(0)(z)))= ˜b ek0ız+ ˜b e−k0ız. (28) Moreover,bydefinitionofG in(18), (I − Q)G(0)(z) = hk0e k0ız+h k0e−k0 ız, (I − Q)DG(0)(U1+ K−1QG(0)(z)ε) = −γ ω(˜c ek0ız+˜c e−k0ız). (29)
Now,weput(26),(27),(28),(29)in(24)andcollectcoefficientsbyek0ızande−k0ıztoobtain
H ek0ız+H e−k0ız+O((U 1, ε)3)= 0 (30) where H= 2ω2 2˜c ˜d2k0+ c ˜˜d2k0+ ˜˜cd2k0+ ˜˜cd0 (1− λ e−k0ıp−λ ek0ıp)ε − (hk0− γ ω ˜c)ε + 2ω 2(b+ ˜b)(1 − λ e−k0ıp−λ ek0ıp)ε2. Equation(30)isequivalentto H+ O((c, c, ε)3)= 0. (31) Usingresonancecondition(16),weget
(1− λ e−k0ıp−λ ek0ıp)= 1 − 2λ cos k 0p= 1 ω2k2 0 .
Droppingtildesyields
2˜c ˜d2k0+ c ˜˜d2k0+ ˜˜cd2k0+ ˜˜cd0= −k 2 0cd2k0− k 2 0cd0, b+ ˜b = k∈Z\{0,±k0,2k0} k0(k− k0)dkdk−k0.
− 2(cd2k0+ cd0)+ hk0− γ ω ˜c ε+2ε 2 k20 k∈Z\{0,±k0,2k0} k0(k− k0)dkdk−k0 + O((c, c, ε)3)= 0. Sincec,hk∈ C,wehave c= x + ıy, hk= μk+ ıνk, dk= μk+ ıνk δk
wherex,y,μk,νk∈ R andδk= 1− ω2k2+ 2λω2k2cos kp∈ R.Obviously,δ0= 1,h0= d0=
μ0∈ R andμk= μ−k,νk= −ν−k∀k ∈ N.So,weget
− 2(x− ıy)(μ2k0+ ıν2k0) δ2k0 + 2(x + ıy)μ0+ μk0+ ıνk0− γ ωık0(x+ ıy) ε +2ε2 k02 k∈Z\{0,±k0,2k0} k0(k− k0)(μk+ ıνk)(μk−k0− ıνk−k0) δkδk−k0 + O((x, y, ε)3)= 0
or,separatingtherealandimaginaryparts,
− 2(xμ2k0+ yν2k0) δ2k0 + 2xμ0+ μk0+ γ ωk0y ε +2ε2 k02 k∈Z\{0,±k0,2k0} k0(k− k0)(μkμk−k0+ νkνk−k0) δkδk−k0 + O((x, y, ε)3)= 0, − 2(xν2k0− yμ2k0) δ2k0 + 2yμ0+ νk0− γ ωk0x ε +2ε2 k2 0 k∈Z\{0,±k0,2k0} k0(k− k0)(μk−k0νk− μkνk−k0) δkδk−k0 + O((x, y, ε)3)= 0.
Multiplyingbothequationsby (−1) andcollectingtermsbyx andy weobtain
2 μ2k0 δ2k0 + μ0 εx+ 2ν2k0 δ2k0 + γ ωk0 εy+ μk0ε −2ε2 k02 k∈Z\{0,±k0,2k0} k0(k− k0)(μkμk−k0+ νkνk−k0) δkδk−k0 + O((x, y, ε)3)= 0, 2ν2k0 δ2k0 − γ ωk0 εx+ 2 −μ2k0 δ2k0 + μ0 εy+ νk0ε −2ε2 k20 k∈Z\{0,±k0,2k0} k0(k− k0)(μk−k0νk− μkνk−k0) δkδk−k0 + O((x, y, ε)3)= 0. (32)
Now we intend to derive the cubic O((U1,0)3) of (30) as follows. We denote by B :=
−1 2D
2F(0) andu
Q(Ku2(U1)+ B(U1+ u2(U1))2)= 0, (33)
(I − Q)B(U1+ u2(U1))2= 0. (34)
Firstnotethatu2(U1)= O(U12) by(23),(I− Q)BU12= 0 andthen(34)hastheform
0= (I − Q)B(U1+ u2(U1))2= (I − Q)BU12+ 2(I − Q)B(U1, u2(U1))+ (I − Q)B(u2(U1))2
= (I − Q)B(U1, D2U1u2(0)U
2
1)+ O(U14),
henceO((U1,0)3)= (I − Q)B(U1,D2U
1u2(0)U
2
1).ToderiveD2U1u2(0),wetwice
differenti-ate(33)toobtain KDU1u2(U1)V1+ 2QB(U1+ u2(U1), V1+ DU1u2(U1)V1)= 0 andthen KD2 U1u2(0)(V1, V2)+ 2QB(V1, V2)= 0, hence D2U 1u2(0)U 2 1= −2K−1QBU 2 1= 2K−1QF(U1), consequently,wearriveat
O((U1, 0)3)= (I − Q)B(U1, D2U1u2(0)U
U1(z)[K−1QF(U1)(z)]= −4k02U1(z)[K−1QF(U1)(z)] = −4k2 0 μc3e3k0ız+μ¯c3e−3k0ız+μc2¯c ek0ız+μc ¯c2e−k0ız , U1(z)[K−1QF(U1)(z)] = −k02U1(z)[K−1QF(U1)(z)] = −k2 0 μc3e3k0ız+μ¯c3e−3k0ız+μc2¯c ek0ız+μc ¯c2e−k0ız , U1(z)[K−1QF(U1)(z)]= k0ıc ek0ız−k0ı¯c e−k0ız 2k0ıμc2e2k0ız−2k0ıμ¯c2e−2k0ız = −2k2 0μc 3e3k0ız−2k2 0μ¯c 3e−3k0ız+2k2 0μc 2¯c ek0ız+2k2 0μc¯c 2e−k0ız. So (I− Q) U1(z)[K−1QF(U1)(z)]+ 2U1(z)[K−1QF(U1)(z)] = 0. Hence −(I − Q)D2F(0)(U 1,K−1QF(U1))= 2ω2 μc2¯c ek0ız(−k2 0)+ μc ¯c 2e−k0ız(−k2 0) − λμc2¯c ek0ı(z−p)(−k2 0)− λμc ¯c 2e−k0ı(z−p)(−k2 0)− λμc 2¯c ek0ı(z+p)(−k2 0) − λμc ¯c2e−k0ı(z+p)(−k2 0) = −2k2 0ω 2μ|c|2 U 1(z)− λU1(z− p) − λU1(z+ p) = −2k2 0ω 2μ|c|2 (1− λ ek0ıp−λ e−k0ıp)U 1(z) = −2μ|c|2U 1(z)= − 2(1− δ2k0) δ2k0 |c|2 c ek0ız+¯c e−k0ız .
Thus,by(35)andthelastidentity,in(31)wehave
O((c, c, 0)3)= −2c|c|
2(1− δ 2k0)
δ2k0
.
Summarizing,thebifurcationequation(32)hasaform
Now,wescale(36)
x←→ εx, y←→ εy, ε←→ ε3, γ←→ γ (37) anddividebothequationsbyε3.Thisleadsto
μk0+ 2(1− δ2k0)x(x 2+ y2) δ2k0 + O(ε) = 0 νk0+ 2(1− δ2k0)y(x 2+ y2) δ2k0 + O(ε) = 0 (38)
andthefollowingresultholds.
Theorem 2. Assume(R) and(μk0,νk0)= (0,0).Thenequation(17)hasasolutionU∈ Z close
to0 foranyε= 0 sufficientlysmall. Proof. Let usdenote
H1(x, y, ε):= μk0+ 2(1− δ2k0)x(x 2+ y2) δ2k0 + O(ε), νk0+ 2(1− δ2k0)y(x 2+ y2) δ2k0 + O(ε) .
Implicitfunctiontheorem appliedtoequation(38),i.e. H1(x,y,ε)= 0 givesthe existenceof
neighbourhoodsW0⊂ V0⊂ R ofthepoint0 inR,W⊂ R2of−3
!
δ2k0
2(1−δ2k0)(μ2k0+νk02)(μk0,νk0) in
R2andauniquepairofcontinuousfunctionsx(ε),y(ε) definedinW
0suchthatH1(˜x,˜y,ε)= 0
for ε∈ W0, (˜x,˜y)∈ W if and only if ˜x = x(ε) and ˜y = y(ε). Moreover, (x(0),y(0))=
−3
!
δ2k0
2(1−δ2k0)(μ2k0+ν2k0)(μk0,νk0). Afterscalingbackwardsin(37),we havethedesired solution
ofequation(17). 2
Now,assumeεh insteadofh in(17).Thenwescale(36)
x←→ εx, y←→ εy, hk←→ εhk, γ←→ γ (39)
anddividebothequationsbyε2.Thisleadsto
γ ωk0y+ μk0+ O(ε) = 0
−γ ωk0x+ νk0+ O(ε) = 0 (40)
andthefollowingresult.
Theorem 3. Assume(R),γ= 0 andεh insteadofh in(17)with(μk0,νk0)= (0,0).Then
Proof. Let usdenote H2(x, y, ε):= γ ωk0y+ μk0+ O(ε), −γ ωk0x+ νk0+ O(ε) .
Implicit functiontheoremappliedtoequation(40),i.e.H2(x,y,ε)= 0 gives theexistence of
neighbourhoodsW0⊂ V0⊂ R ofthepoint0 inR,W⊂ R2of( νk0 γ ωk0,−
μk0 γ ωk0) inR
2andaunique
pair of continuousfunctionsx(ε), y(ε) defined inW0 such that H2(˜x,˜y,ε)= 0 forε∈ W0,
(˜x,˜y)∈ W ifandonlyif ˜x = x(ε) and ˜y = y(ε).Moreover,x(0)=γ ωkνk0
0,y(0)= − μk0 γ ωk0.After
scalingbackwardsin(39),wehavethedesiredsolutionofequation(17). 2
Finally, we deal witha gapbetween Theorems 2 and 3, namely that h is not scaled and μk0= νk0= 0.Now,wescale(36)
x←→ εx, y←→ εy, ε←→ ±ε2, γ←→ γ (41) anddividebothequationsbyε3.Thisleadsto
±2 μ2k0 δ2k0 + μ0 x± 2ν2k0 δ2k0 + γ ωk0 y+2(1− δ2k0)x(x 2+ y2) δ2k0 + O(ε) = 0, ± 2ν2k0 δ2k0 − γ ωk0 x± 2 −μ2k0 δ2k0 + μ0 y+2(1− δ2k0)y(x 2+ y2) δ2k0 + O(ε) = 0 (42)
andthefollowingresult.
Theorem 4. Assume(R) and(μk0,νk0)= (0,0).Thenequation(17)hasasolutionU∈ Z close
to0 foranyε= 0 sufficientlysmallprovided
δ2k2 0(4μ 2 0+ γ2ω2k02)− 4(μ22k0+ ν 2 2k0)= 0. (43)
TheorderofthissolutionisO(ε).If,inaddition,itholds
δ22k 0(4μ 2 0+ γ 2ω2k2 0)− 4(μ 2 2k0+ ν 2 2k0) < 0, (44)
then(17)hasanothersolution.ThisoneisoforderO(√|ε|).
Proof. Let usdenote
H3(x, y, ε):= ± 2 μ2k0 δ2k0 + μ0 x± 2ν2k0 δ2k0 + γ ωk0 y+2(1− δ2k0)x(x 2+ y2) δ2k0 + O(ε), ± 2ν2k0 δ2k0 − γ ωk0 x± 2 −μ2k0 δ2k0 + μ0 y+2(1− δ2k0)y(x 2+ y2) δ2k0 + O(ε) .
First we note that H3(x,y,ε)= A(x,y)+ B(x,y)+ O(ε) where A is linear and B is
i.e.H3(x,y,ε)= 0 gives theexistence of neighbourhoodsW0⊂ V0⊂ R of thepoint0 in R,
W⊂ R2of(0,0) inR2andauniquepairofcontinuousfunctionsx(ε),y(ε) definedinW0such
that H3(˜x,˜y,ε)= 0 forε∈ W0,(˜x,˜y)∈ W if andonlyif ˜x = x(ε) and ˜y = y(ε).Moreover,
(x(0),y(0))= (0,0), so x(ε)= O(ε),y(ε)= O(ε), whichby scaling (41)gives O(ε) solu-tionfor (17).Ontheotherhand,if (44)holdsthenthe Brouwerdegreedeg( H3,B1,(0,0))=
sgn det A= −1 whiledeg( H3,B2,(0,0))= 1 forsmallandlargeballsB1,B2inR2centred at
(0,0).Thisgivesanothersolutionof(42)oforderO(1) placedinthesetB2\ B1.Afterscaling
backwardsin(41),wehavethedesiredO(√|ε|) solutionof(17). 2
Remark 2. TheaboveO(ε) solutionfor(17)canbederivedbyputtingTaylorexpansionsx(ε)= x(0)ε+ O(ε2),y(ε)= y(0)ε+ O(ε2) into(36)andcomparingε2-terms.Sowederive
2 μ2k0 δ2k0 + μ0 x(0)+ 2ν2k0 δ2k0 + γ ωk0 y(0) = 2 k20 k∈Z\{0,±k0,2k0} k0(k− k0)(μkμk−k0+ νkνk−k0) δkδk−k0 , 2ν2k0 δ2k0 − γ ωk0 x(0)+ 2 −μ2k0 δ2k0 + μ0 y(0) = 2 k20 k∈Z\{0,±k0,2k0} k0(k− k0)(μk−k0νk− μkνk−k0) δkδk−k0 , (45)
whichdeterminesthefirstorderapproximationofthissolution.Henceby(23)and(25),weget
U (z)= U1(z)+ U2(U1, ε)= ε ⎛ ⎝c(0) ek0ız+c(0) e−k0ız+ k∈Z\{±k0} dkekız ⎞ ⎠ + O(ε2) (46) forc(0)= x(0)+ y(0)ı.
Remark 3. 1.NowwespecifytheO(√|ε|) solution(solutions)ofTheorem 4.Nontrivialroots ofequation
H3(x, y, 0)= 0 (47)
canbeexplicitlycalculatedasfollows.Denoting
a:= ±2 μ2k0 δ2k0 + μ0 , b:= ± 2ν2k0 δ2k0 + γ ωk0 , c:= ± 2ν2k0 δ2k0 − γ ωk0 , d:= ±2 −μ2k0 δ2k0 + μ0 , υ:=2(1− δ2k0) δ2k0 (48)
ax+ by+ x(x2 + y2)= 0
cx+ dy+ y(x2 + y2)= 0 (49)
fora= a/υ,b= b/υ,c= c/υ,d= d/υ.Noteυ= 0,i.e.,δ2k0 = 1.Itiseasytoseethaty= 0
wheneverx= 0 andviceversa.Henceweassumex= 0= y.Dividingthefirstequationbyx, thesecondequationbyy,andcomparingtheresultingequations,onederives
a+ by x =c x y + d. Therefromx= q1,2y for q1,2= (a− d)±(a− d)2+ 4bc 2c
(hereq1correspondsto“+”-signandq2to“−”-sign).Asaconsequence,puttingx inthesecond
equationof(49)oneobtains
y1,2= ± −cq1,2+ d 1+ q1,22
(againy1correspondsto“+”,y2to“−”),i.e.,onehasfourpoints(qiyj(qi),yj(qi)),i,j = 1,2.
Condition(44)meansad− bc < 0,i.e.,ad− bc < 0.Soif(44)holds,then
& (a− d)2+ 4bc= & (a+ d)2− 4(ad− bc) >|a+ d|. Accordingly, cq1,2+ d= 1 2 a+ d± & (a− d)2+ 4bc (50)
whichisnegativeonlyfortheminussign(withoutanyrespecttothesignofunscaledε).Thus thenontrivialsolutionsof(47)are
(q2y1,2(q2), y1,2(q2)). (51)
Ontheotherhand,assuming
0 < δ2k2 0(4μ 2 0+ γ2ω2k02)− 4(μ22k0+ ν 2 2k0) < 4μ 2 0, (52)
equation(47)stillhasnontrivialsolutions.Indeed,inthiscase
0 <
&
(a− d)2+ 4bc=
&
soq1,2arebothreal,again.However,(50)isnegativeforbothsignsonsupposethata+ d < 0,
i.e.,±μ0
υ < 0 where the signdependson the signof unscaled ε, e.g.iforiginally ε > 0 and μ0
υ < 0,equation(47)hasfourdistinctnontrivialsolutions
(qiyj(qi), yj(qi)), i, j= 1, 2, (53)
whilefor μ0
υ > 0,(47)hasnonontrivialsolutions.
Inconclusion,ifcondition(44)holds,theremightexisttwosolutionsof(17)oforderO(√|ε|) forε= 0 small,andif(52)isvalid,theremightexistfoursolutionsof(17)oforderO(√|ε|) for ε sgnμ1−δ0δ2k0
2k0 < 0 andnosolutionsoforderO(
√
|ε|) forε sgnμ1−δ0δ2k0
2k0 > 0.WerecallbyTheorem 4,
thatthereexistsasmallsolutioninallcasesoflowerorderthanO(√|ε|),namelyoforderO(ε). Sowemighthave3,5and1smallsolutions,respectively.Toapplytheimplicitfunctiontheorem andsotoconfirmtheexistenceofthesesolutions,onehastoshowthatthederivativeofH3at
points(51)or(53)issurjective,whichisratherawkwardforgenerala,b,c,d,andsowedonot gointodetailsinthispaper.Butsurelythesesolutionsgenericallyexist.Indeed,thedeterminant oftheJacobianofthelefthandsideof(49)atany(qy,y) is
−bc+ad+ 3a+ d+ q(−2(b+c)+ (a+ 3d)q) y2+ 3
1+ q2
2
y4. (54)
Now we fix all parameters only μ0 is variable. Then q = q1,2,b,c are constant
(indepen-dentof μ0)andy1,22 ,a,d linearlydependonμ0:a= ±υ2(r1+ μ0),d= ±2υ(−r1+ μ0) and
y2= y2 1,2=−
cυq±2(r1−μ0)
υ(1+q2) for constantsr1,r2independentof μ0.Insertingtheseexpressions
into(54),wehave 1 υ2(1+ q2) a±(r1, q,b,c, υ)+ b±(r1, q,b,c, υ)μ0 (55) forpolynomials a+(r1, q,b,c, υ)=c −b+ (b+ 5c)q2+ 3cq4 υ2− 4q b+c 5+ 2q2 υr1+ 16r12, b+(r1, q,b,c, υ)= 4 q b+c 2+ q2 υ− 4r1 , a−(r1, q,b,c, υ)= c −b+ (b+ 5c)q2+ 3cq4 υ2+ 4r1 q b+ c 5+ 2q2 υ+ 4r1 , b−(r1, q,b,c, υ)= −4 q b+c 2+ q2 υ+ 4r1 .
Nowitisclearthata±,b±aregenericallynonzeroandthen(55)isnonzeroforgenericμ0.This
Fig. 1. The graph of functions 122537 and 1
x2(1−2λ cos xp)with values at nonzero integers depicted with asterisks.
andcondition(15)isfulfilled.Ifω2=2537,thenk0= ±1 satisfiestheresonancecondition(16).
Notethat 1 k2(1− 2λ cos kp)≤ 1 k2(1− 2λ)= 37 k2(37− 12√2)< 37 13k2< 37 25
whenever|k|≥ 2.Thusk0= ±1 istheonlyresonantcouplecorrespondingtoω2=3725.Thisis
thecaseofasimpleresonancediscussedinthissection.
Ontheotherside,ifω2=122537 ,thenthereareatleasttwocouplesk0= ±5,±7 suchthat(16)
isfulfilled.Since 37 k2(37− 12√2)< 37 20k2< 37 1225 whenever|k|≥ 8,itissufficienttoevaluate 1
k2(1−2λ cos kp)atk= {1,. . . ,7} todeterminetheexact
numberofresonantmodes.FromFig. 1,onecanseethatforω2=122537 therearetworesonant couplesk0= ±5,±7.Thisisadoubleresonancewhichcanbeinvestigatedanalogicallytothe
presentsection.
4. Examples
Inthissectionwepresentseveralexamplestoillustratetheaboveresults.
Example 1. Firstwestudythenon-resonantcasefromSection2.Weconsidertheequation
forz∈ R andparametersγ ,λ,ω > 0, 12= λ=
√
2
2 ,andapplyTheorem 1.Notethatequation(57)
is inthe form of (2) with p= π4, h(z)= ε cos z, and M2 of (13) is equivalent to(56), i.e.,
condition(15)issatisfied.HenceusingRemark 1(3) andTheorem 1,weobtaintheexistenceof auniquesolutionU (ε,z)= O(ε) ofequation(57)foranyε sufficientlycloseto0.SinceU (ε,z) isC∞-smooth,wecanexpandU (·,z) intoTaylorseriesU (ε,z)= εU1(z)+ O(ε2),andlookfor
thefunctionU1.ItiseasytoseethatU1isuniquelydeterminedbytheequation
ω2 U1(z)− λU1 z−π 4 − λU1 z+π 4 + γ ωU1(z)+ U1(z)− cos z = 0.
WelookforitssolutionintheformofU1(z)= Acos z+ B sin z,and,bycomparingcoefficients
atcos z andsin z,weobtain
−γ ωA + (1 + ω2(√2λ− 1))B = 0, (1 + ω2(√2λ− 1))A + γ ωB = 1. Therefrom, A= 1+ ω 2(√2λ− 1) γ2ω2+ (1 + ω2(√2λ− 1))2, B= γ ω γ2ω2+ (1 + ω2(√2λ− 1))2. Inparticular,if λ= 12 37√2, ω= √ 37 5 , (58) thenA= 0,B=5 √ 37 37γ ,i.e., U (ε, z)=5 √ 37 37γ ε sin z+ O(ε 2). (59)
Example 2. Now, moving ontoresonantcasesrepresented byequation(17), weconsiderthe equation ω2 U (z)− λU z−π 4 − λU z+π 4 + εγ ωU(z)+ U(z) + ω2 U2(z)− λU2 z−π 4 − λU2 z+π 4 + εγ ω(U2(z))− ε cos z = 0, (60)
i.e.whenγ in(57)isreplacedbyεγ .Proceedingwithpoint2ofRemark 3forparameters(58), thenweobtainthatk0= ±1,δ2k0= −
123 25 andυ= 2(1−δ2k0) δ2k0 = − 296 123.
Sinceh(z)= cos z = 12eız+21e−ızin(18),thenμk0 =
1
2 andνk0 = 0.SoTheorem 2canbe
Using (22),(25)and(37),we derive that thesolution from Theorem 2of (60)with parame-ters(58)isgivenby 3 √ εc(√3 ε) eız+√3ε¯c(√3 ε) e−ız+U2 √3εc(√3 ε) eız+√3ε¯c(√3 ε) e−ız, ε =√3ε c(0) eız+¯c(0) e−ız+O √3ε =√3ε 2x(0) cos z+ O √3ε = 3 ! 123ε 74 cos z+ O 3 ε2=. √3 ε1.18456 cos z+ O 3 ε2 (61)
forasmoothfunctionc(·) withc(0)= x(0)+ y(0)ı andε= 0 small.
Example 3. Similarly,Theorem 3canbeappliedintheequation
ω2 U (z)− λU z−π 4 − λU z+π 4 + εγ ωU(z)+ U(z) + ω2 U2(z)− λU2 z−π 4 − λU2 z+π 4 + εγ ω(U2(z))− ε2cos z= 0, (62) with λ= 12 37√2, ω= √ 37 5 , γ= 0.1. (63) Notethatcos z-termin(60)hasbeenreplacedbyε cos z inourcasehere,i.e.,h(z) in(18)is re-placedbyεh(z).ThenusingtheproofofTheorem 3,weobtainthat(x(0),y(0))=
0,−√25
37k0
, sincek0= ±1. Using(22),(25)and(39),wederivethatthesolutionfromTheorem 3of(62)
withparameters(63)isgivenby
εc(ε) eık0z+ε ¯c(ε) e−ık0z+U
2(εc(ε) eık0z+ε ¯c(ε) e−ık0z, ε)
= ε(c(0) eık0z+¯c(0) e−ık0z+O(ε)) = ε(−2y(0) sin k
0z+ O(ε))
=√50ε
37k0sin k0z+ O(ε
2)= 8.21995ε sin z + O(ε. 2). (64)
Example 4. Next,weconsider
ω2 U (z)− λU z−π 4 − λU z+π 4 + εγ ωU(z)+ U(z) + ω2 U2(z)− λU2 z−π 4 − λU2 z+π 4 + εγ ω(U2(z)) − ε(μ0+ 2(cos 2z − sin 2z)) = 0 (65)
with parameters (63), i.e., equation (17) with the particular driving function h(z) = μ0 +
2(cos 2z− sin 2z).So,μk0 = νk0 = 0 and μ2k0 = ν2= −ν−2= 1.Hence, Theorems 2 and3
−8 +15 129 625 37 2500+ 4μ 2 0 = 0, i.e.|μ0|= √ 11 940 227 12 300 . = 0.280932.Moreover,(44)isequivalentto |μ0| < √ 11 940 227 12 300 . (66)
First,let k0= 1 and considerε > 0,i.e. use“+”-sign in(48).Thenwe canderivethat q2=.
−0.294153 andy1,2= ±
√
0.214884+ 0.764897μ0.Notethat 0< 0.214884+ 0.764897μ0≤
0.429768.Usingformula(54),wecanestimatetheJacobianofH3(q2y1,2(q2),y1,2(q2),0) as
0 < (0.218047+ 0.776154μ0)v2≤ 2.52553, (67)
since the formula (54) is equal to 1
v2det D H3(qy,y,0). Therefore, we have 3 different
small solutions for k0= 1, ε > 0 small. Using k0= −1, we obtain q2= 0.294153,. y1,2=
±√0.214884+ 0.764897μ0andby(54),estimate(67)follows.Hencefork0= ±1,ε > 0 small
andμ0satisfying(66),by(51)weobtainnontrivialsolutionsof(47).Scalingbackwardsin(41),
weobtainthefollowingsmallsolutionsof(65)withparameters(63)
√
εc(√ε) eık0z+√ε¯c(√ε) e−ık0z+U
2(
√
εc(√ε) eık0z+√ε¯c(√ε) e−ık0z, ε)
=√ε(2x(0) cos k0z− 2y(0) sin k0z)+ O(ε) = 2
√
εy(0)(q2cos k0z− sin k0z)+ O(ε)
= ±2ε(0.214884+ 0.764897μ0)(0.294153 cos z+ sin z) + O(ε). (68)
Moreover,thereisanother(smallest)solutionof(65)withparameters(63)oforderU (z)= O(ε). Using(45),weobtainx(0)= y(0)= 0 andby(46),thesolutionis
U (ε)= ε μ0− 50 123cos 2z+ 50 123sin 2z + O(ε2). (69)
Considering k0= 1 and ε < 0, i.e. using “−”-sign in (48),we derive q2= 1.83348 and.
y1,2= ±
√
0.0535297− 0.190543μ0.Notethat0< 0.0535297− 0.190543μ0≤ 0.107059.
For-mula(54)yields
0 < 0.218047− 0.776154μ0≤ 0.436093, (70)
whichimpliesthatwealsohave3differentsmallsolutionsfork0= 1,ε < 0 smallandμ0
satisfy-ing (66). Note that if k0 = −1, ε < 0, we obtain q2
.
= −1.83348, y1,2 =
±√0.0535297− 0.190543μ0and,by(54),estimate(70).The3solutionsof(65)with
parame-ters(63)havetheforms(69)and
√ −εc(√−ε) eık0z+√−ε ¯c(√−ε) e−ık0z+U 2 √ −εc(√−ε) eık0z+√−ε ¯c(√−ε) e−ık0z, ε =√−ε(2x(0) cos k0z− 2y(0) sin k0z)+ O(ε) = 2
√
−εy(0)(q2cos k0z− sin k0z)+ O(ε)
Next,(52)implies 0.280932=. √ 11 940 227 12 300 <|μ0| < & 11 940 227 74 1400 . = 0.286921. (71)
Fromtheabovecomputations,wecanverifythatformula(54)hasoneoftheforms0.218047± 0.776154μ0= 0.Sincesgn υ= −1,wehavethat:ifε issmall,ε sgn μ0> 0 andμ0fulfils(71),
thenthereare5differentsmallsolutions(4× O(√|ε|) and1× O(ε)),ifε sgn μ0< 0 thereis
only1smallsolutionoforderO(√|ε|),actually(69).Forinstance,ifε > 0 andμ0satisfies(71),
weobtain k0 qi yi 1 q1= 1.83348 y1,2(q1)= ± √ −0.0535297 + 0.190543μ0 q2= −0.294153 y1,2(q2)= ± √ 0.214884+ 0.764897μ0 −1 q1= −1.83348 y1,2(q1)= ± √ −0.0535297 + 0.190543μ0 q2= 0.294153 y1,2(q2)= ±√0.214884+ 0.764897μ0
Hence,equation(65)withparameters(63)hasthesolutions(69)and
±2ε(−0.0535297 + 0.190543μ0)(1.83348 cos z− sin z) + O(ε),
±2ε(0.214884+ 0.764897μ0)(0.294153 cos z+ sin z) + O(ε)
forμ0> 0,andtheonlysolution(69)oforderO(
√
ε) forμ0< 0. 5. Numerical results
To illustrate the theoretical results obtained in the previous sections, we have solved the governing equations (1) and(2) numerically. The advance-delayequation(2) is solvedusing apseudo-spectralmethod,i.e.weexpressthesolutionU intheFourierseries
U (z)= J j=1 ' Bjcos (j− 1)˜kz + Cjsin j ˜kz ( , (72)
where ˜k = 2π/L and −L/2< z < L/2.Substituting the series in (2) andprojecting it onto theFouriermodesinareminiscenceoftheGalerkintruncationmethod,weobtainasystemof algebraicequationsfor theFouriercoefficientsBj andCj,whichare thensolvedusing,e.g.,
a Newton–Raphsonmethod.Typically,weuseL= 2π andlargeJ thatwillbeindicatedineach calculationbelow.
Fig. 2.(a)Theoscillationamplitudeofperiodicsolutionsof(2)asafunctionofthedrivingamplitudeh0forω= 1.23 withγ= 0 (blackthicklines)andγ= 0.001 (bluethinlines).(b)Solutionprofilesatthedrivingamplitudesindicatedin thelegendforvanishingγ .(Forinterpretationofthereferencestocolourinthisfigure,thereaderisreferredtotheweb versionofthisarticle.)
˙un= vn, (73)
[˙vn(1+ 2Un)] = −γ vn(1+ 2Un)− un(1+ 2γ ˙Un)− 2 [ ¨Unun+ 2 ˙Unvn], (74)
where [•n]= •n− λ•n−1−λ•n+1andUn(t)= U(ωt + np) (cf.(72))isaperiodicsolution
of(1).ThelinearsystemisintegratedusingaRunge–Kuttamethodoforderfourwithperiodic boundary conditionsover the periodT = 2π/ω and withthesite indexn= 1,2,. . . ,N .The ithcolumnofthemonodromymatrixM isvector[u1(T ),. . . ,uN(T ),v1(T ),. . . ,vN(T )]t,that
correspondstothe initialcondition[u1(0),. . . ,uN(0),v1(0),. . . ,vN(0)]t equal tothe ith
col-umnvectoroftheidentitymatrixI2N.AperiodicsolutionisasymptoticallystableifallFloquet
multipliersexceptthe trivial Floquetmultiplierare strictlysmaller thanoneinmodulus. The (in)stabilityresultobtainedfromcalculatingthemonodromymatrixisalsoconfirmedby inte-grating(1).Notethatthephysicallyrelevantintervalforthecouplingparameteris|λ|< 1/2[6]. Itisbecausetherewillbeabandofunstablemultiplierswhen|λ|> 1/2,implyingthatall solu-tionsofthesystemwillbemodulationallyunstable.FollowingSection4,wesetλ= 12/
37√2
andp= π/4.Computationsfordifferentparametervalueshavebeenperformed,wherewedid notseeanysignificantdifference.
First,weconsiderthecaseof
h(z)= h0cos z. (75)
From (16) for γ = 0 the first resonance of the system is k= ±1, which corresponds to ω=√37/5.In Fig. 2,we depictthe bifurcationdiagramsof periodicsolutionsforavalueof ω= 1.23>√37/5.OntheverticalaxisistheoscillationamplitudeA oftheperiodicsolutions definedas
A=1
Fig. 3.ThesameasFig. 2(a)forω=√37/5 with(a)γ= 0 and(b)γ= 0.1.Inbothpanels,thesolidblackandthe dashedbluelinesarethenumericsandtheanalyticalresults(61)(a)and(59),(64)(b),respectively.
Tochecktheconvergenceanddependenceofthemethodonthenumberofmodes,we com-putedthebifurcationdiagramsfor severalvaluesof J .Interestinglywe observedthat thereis always a criticalvalue of h0 beyondwhichthe bifurcation diagramsdepend significantly on
theparameter.Thediagramsforthreevaluesof J areshowninFig. 2(a).Usingthefigure,we concludethat todeterminethe validityregion ofour numericalmethod,we needtocompute bifurcationdiagramsusingatleasttwodifferentvaluesofJ .The‘breakingpoint’ofthemethod iswhenthecurvesstartbranchingout.Studyingthesolutionprofilesnearthebreakingpointas showninFig. 2(b),thebreakingpointseeminglycorrespondstoanextremepointofU (z) having discontinuousfirstderivative.Wethereforeinferthatourequation(2)mayhavepeaked-periodic solutions.
AsshownasblackthicklinesinFig. 2(a),whenγ= 0 thereisonesaddle-nodebifurcation that occursaswevaryh0.Notethatthebifurcationdiagramhasamirrorsymmetric(with
re-specttotheverticalaxish0= 0)counterpart,i.e.thesystemissymmetricwiththetransformation
h0→ −h0,U→ −U.Whenγ isturnedontoanon-zerovalue,thebifurcationcurveinteracts
withitssymmetryandattheintersectionpointh0= 0 fornon-vanishingA anadditional
turning-pointisformed,i.e.thereisacurve-splitting-and-merging.Inthatcase,as shownas bluethin linesinFig. 2(b),therearenowtwosaddle-nodebifurcations.
When ω is decreased toward the resonancefrequency, the criticaldrive of h0 for the
oc-currence of the first saddle-nodebifurcation, i.e. the bifurcationthat alsoexistswhen γ= 0, decreasestowardszero. Inthelimit ω=√37/5,wepresent inFig. 3(a)thebifurcationcurve ofperiodicsolutionsattheresonance.Thenumerics(shownassolidblackcurves)are ingood agreementwiththeanalyticalresult(dashedline)inExample 2,i.e.(61),thatwhenγ = 0 the slopeoftheexistencecurveissingularath0= 0.
Inpanel(b)ofthesamefigure,wedepictthebifurcationcurveoftheperiodicsolutionsfor thesameparametervalues,butwithγ= 0.InagreementwiththeanalyticalresultinExamples 1 and3,i.e.(59)and(64),addingdissipationwillregularizetheslopeattheorigin.
WehavealsoconsideredthemultipleresonancecasediscussedinSection4.Forthenumerics, wetakethedrive
Fig. 4.ThesameasFig. 2(a),butfortheperiodicdrive(76).Thedashedlinesaretheanalyticalamplitudesfrom(68)
and(69).Here,γ= 0.
withμ0= 1/4,whichsatisfiesthecondition(66).FromExample 4,weknowthatinthis
reso-nancecase,therearethreesmallsolutionsgivenby(68)and(69).ShowninFig. 4arenumerically computedbifurcationcurvesoftwoofthethreeperiodicsolutions.
Inpanel (a),we showonlyonebifurcationdiagramof thesolutions(68).Itisbecause the amplitudesofthetwosolutionsarealmostthesame.Inpanel(b),wedepictthesmallersolution withamplitudesthatscaleasO(h0).Onecannotethatinbothplotsouranalyticalresultsfrom
Lyapunov–Schmidtreductionagreewiththenumerics.
Wehavealsodeterminedthestabilityofthesolutionsobtainednumericallyandfoundthatin thenon-resonantcondition,theperiodicsolutionsaregenerallystableforsmallenoughdriving amplitudeh0.
Shown in Fig. 5 are the Floquet multipliers of someperiodic solutions, calculated using N = 200.Panels (a)–(b) depict themultipliers of theperiodic solutionscorrespondingto the bifurcationcurveinFig. 3(b)forh0= 0.01 and0.04,respectively.Asallthemultipliersinthe
firstpanelareinsidetheunitcircle,wecandeducethatthesolutionisstable.However,ash0
in-creasesfurther,somemultipliersleavethecircleasshowninpanel(b)andinducemodulational instability.Panel (c)of thesamefigurecorrespondstoasolutionalong theexistencecurvein Fig. 3(a),i.e.withh0= 0.01 andγ= 0.Itisclearthatthesolutionisunstable.Forthisresonant
case,wefoundthatallsolutionsforanydrivingamplitudearemodulationallyunstable.
Wehavealsosimulatedthedynamicsoftheunstablesolutionsbysolvingthegoverning equa-tion (1).Weobserved that the typicaldynamics of the instabilityare in the form of solution blow-up,similarlytothatin[1].
Asaconclusion,wefoundthatthequalitativebehaviourofthesystem(1)issimilartothatof forcedFPUlatticesdiscussedin[8],despitethesignificantdifferenceinthecouplingtermsas explainedinSection1.
Acknowledgments
Fig. 5.Floquetmultipliersofperiodicsolutionsforsomeparametervalues(seethetext).Thedashedcurveistheunit circle,showingasguidetotheeye.
fundsthroughtheOperationalProgramEducationandLifelongLearningoftheNational Strate-gicReferenceFramework(NSRF)ResearchFundingProgramD.534MIS:379337:THALES Investinginknowledge society throughthe European Social Fund. VRandHS acknowledge partialsupportfromtheLondonMathematicalSocietythroughavisitorgrant.
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