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Onq-DierenceandFractionalEquationsandTheirSolutionsByZeinabSayedIbrahimMansourSUBMITTED FOR THEREQUIREMENTS FOR THE DEGREE OFDOCTOR OF PHILOSOPHYATCAIRO UNIVERSITYGIZA, EGYPTFEBRUARY 2006SupervisorsM.H.Annaby M.E.H.IsmailCairoUniversity UniversityofCenteralFloridaTableofContentsTableofContents iiiAcknowledgements ivNotations vPreface viI ClassicalResults 11 BasicCosineandSineFunctions 21.1 Aq-calculus . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 21.2 Asymptoticsofzerosofcos(z; q)andsin(z; q) . . . . . . . . . . . . . . 101.3 Asymptoticsofcos(z; q)andsin(z; q) . . . . . . . . . . . . . . . . . . 201.4 BasicanaloguesofatheoremofPolya . . . . . . . . . . . . . . . . . . 302 ZerosofFiniteq-HankelTransforms 352.1 Introductionandpreliminaries. . . . . . . . . . . . . . . . . . . . . . 352.2 Mainresults . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 372.3 Applications . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 45II BasicDierenceEquations 503 Introduction 513.1 q-successiveapproximations . . . . . . . . . . . . . . . . . . . . . . . 513.2 q-Initialvalueproblems . . . . . . . . . . . . . . . . . . . . . . . . . 533.3 Linearq-dierenceequations . . . . . . . . . . . . . . . . . . . . . . . 553.4 Aq-typeWronskian . . . . . . . . . . . . . . . . . . . . . . . . . . . 58ii4 BasicSturm-LiouvilleProblems 604.1 Introduction. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 604.2 Fundamentalsolutions . . . . . . . . . . . . . . . . . . . . . . . . . . 634.3 Theselfadjointproblem. . . . . . . . . . . . . . . . . . . . . . . . . 684.4 BasicGreensfunction . . . . . . . . . . . . . . . . . . . . . . . . . . 744.5 Eigenfunctionexpansions. . . . . . . . . . . . . . . . . . . . . . . . . 794.6 Asymptoticsofeigenvaluesandeigenfunctions . . . . . . . . . . . . . 88III BasicFractionalCalculusandEquations 1075 BasicFractionalCalculus 1085.1 Fractionalcalculusandequations. . . . . . . . . . . . . . . . . . . . 1085.2 Somefractionalintegraloperators . . . . . . . . . . . . . . . . . . . . 1125.3 q-Notationsandresults . . . . . . . . . . . . . . . . . . . . . . . . . . 1185.4 BasicRiemannLiouvillefractionalcalculi . . . . . . . . . . . . . . . 1266 BasicMittag-LeerFunctions 1416.1 BasicMittag-Leerfunction. . . . . . . . . . . . . . . . . . . . . . . 1416.2 BasicVolterraintegralequations . . . . . . . . . . . . . . . . . . . . . 1466.3 ZerosofbasicMittag-Leerfunctions . . . . . . . . . . . . . . . . . . 1527 BasicFractionalDierenceEquations 1577.1 BasicRiemannLiouvillefractionalordersystems . . . . . . . . . . . 1577.2 BasicCaputofractionalordersystems . . . . . . . . . . . . . . . . . . 1657.3 Solutionsviaaq-Laplacetransform. . . . . . . . . . . . . . . . . . . 1707.4 Aq-Cauchyproblem . . . . . . . . . . . . . . . . . . . . . . . . . . . 176Bibliography 185iiiAcknowledgementsIwishtothankprofessorsMahmoudAnnaby,CairoUniversity,andMouradIsmail,UniversityofCentral Florida, fortheirguidancethroughoutthepreparationofthisthesis. Idonotndwordsofthankthatmayexpressmygratitudeandappreciationtomyparentsforbeingkeenonteachingmeall moralsneededtoachievesuccessinmylife. Lastbutnotleast, Iwouldliketodedicatethisthesistomyfamily, myhusband Abdallah and daughter Maryam, for their love, patience, and understandingsothattheyallowedmetospendmostofthetimeonthisthesis.ZeinabS.I.MansourFebruary1,2006ivNotationsN := 0, 1, 2,Z+:= 1, 2, 3,Z := N

NZ:= Z 0R := (, )R+:= (0, )Cisthesetofallcomplexnumbers.f.s. isabbreviationforfundamentalset.RLFisabbreviationforRiemannLiouvillefractional.vPrefaceThis thesis is concerned with the fast growing area of basic special functions and basicdierenceequations. Itconsistsofthreeparts. Therstoneisconcernedwithsomebasicanaloguesof classical results. Thispartcontainsaself containedq-calculus,whichhasconsideredbeforein[9, 90]. Wealsoderivesomenewbasictypeidentitieswhichwillbeneededthroughoutthethesis. Therstpartcontainstwonewresultsconcerningzerosof basicfunctionsandq-integral transforms. Westudytheq-sineandq-cosinefunctions,theirasymptoticsandzeros. Thenwederiveaq-analogueofatheoremofGeorgePolyainChapter1. InChapter2westudythezerosofaniteq-Hankel transform. Asanapplicationof theresultsof thischapter weintroduceanotherq-analogueofthetheoremofGeorgePolyaofChapter1.Part II is devoted to study q-dierence equations. We start this part with Chapter3whichcontainssomeknownresultsinthissubject, seee.g. [8, 9, 90], inparticularexistenceanduniquenesstheoremsaswell as, q-linearequations. Ournewcontri-butionlieinChapter 4. Weestablishaq-SturmLiouvilletheoryandinvestigatetheasymptoticbehaviorof eigenvaluesandeigenfunctionsof basicSturmLiouvilleeigenvalueproblems.Inthelastpartwestudyq-analoguesoffractionalcalculusandfractionaldier-ential equations. In Chapter 5 we give a brief account about the history of fractionalcalculi andsomeresultswhichweareintendedinderivingtheirq-analogues. BasicRiemannLiouville fractional calculus introduced in [10,11,13] is studied in a rigorousanalyticway. Wederiveq-analoguesofothertypesoffractionalderivatives,namelyviviiGr unwaldLetinkov derivative and Caputo fractional derivative. q-type fractional dif-ference equations are introduced in Chapter 7. The solutions of fractional q-dierenceequationsarealsostudiedinthispart. Forthistaskwedeneq-Mittag-Leerfunc-tions,andstudysomeoftheirproperties. Wealsosolveq-fractionaldierenceequa-tionswithconstantcoecientsbyusingq-typeLaplacetransform.PartIClassicalResults1Chapter1BasicCosineandSineFunctionsThischapterincludessome q-notationsandresults. Itstartswithaq-calculusandincluding someq-notations and functions. We also considerq-sine andq-cosine functions,some of their properties and their zeros. Then we end up with a basic analogue of a theoremof George Polya.1.1 Aq-calculusThroughoutthisthesisunlessotherwisestatedqisapositivenumberlessthan1andbythe word basic we mean q-analogue. In this section we introduce some of the q-notationsand results. We start with theq-shifted factorial, see [53], fora C,(a; q)n :=___1, n = 0,n1

i=0(1 aqi), n = 1, 2, . . ..(1.1.1)The limit of (a; q)nasn tends to innity exists and will be denoted by (a; q). Moreover(a; q) has the following series representation, cf., e.g. [53, p. 11],(a; q) =

n=0(1)nqn(n1)2an(q; q)n. (1.1.2)The multipleq-shifted factorial for complex numbersa1, . . . , akis dened by(a1, a2, . . . , ak; q)n :=k

j=1(aj; q)n. (1.1.3)23We also use the following notations for theq-binomial coecients_0_q= 1,_n_q=(1 q)(1 q1) . . . (1 qn+1)(q; q)n, n Z+, R. (1.1.4)For C,aq,= qn,n N, we dene (a; q)to be(a; q) :=(a; q)(aq; q). (1.1.5)The-function is dened forz C 0 , 0 < [q[ < 1 to be(z; q) :=

n=qn2zn. (1.1.6)The following identity is introduced by C.G.J. Jacobi in 1829, and it is called Jacobi tripleproduct identity, see [53]

n=qn2zn=_q2, qz, qz1; q2_, z C 0 , 0 < [q[ < 1. (1.1.7)Therefore (z; q) has only real and simple zeros at the points_q2m+1, m Z_. The followingformula was obtained by Euler (1748) and by Gauss (1876), cf. [47, p. 67]m

n=0(1)n_mn_qqn(n1)/2un= (u; q)m. (1.1.8)Letrs denote theq-Hypergeometric seriesrs___a1, . . . , arb1, . . . , bsq, z___=

n=0(a1, . . . , ar; q)n(q, b1, . . . , bs; q)nzn(q(n1)/2)n(s+1r). (1.1.9)Theq-Gamma function, [53, 71], is dened byq(z) :=(q; q)(qz; q)(1 q)1z, z C, 0 < [q[ < 1. (1.1.10)Here we take the principal values of qzand (1q)1z. Then q(z) is a meromorphic functionwith poles atz = n,n N. Because q(z) has no zeros,1q(z)is an entire function withzeros atz = n,n N. q(z) is the unique function that satises the functional equationq(z + 1) =1 qz1 q q(z), q(1) = 1, (1.1.11)4and logarithmically convex. See [14, 53]. Hence, q(n) = (q; q)n1_(1 q)n1. Theq-Betafunction is dened byBq(a, b) :=_10xa1(qx; q)b1dqx, a, b > 0, (1.1.12)wheretheintegrationisdenedin(1.1.31)below. Usingtheq-binomialtheorem,cf.[14,p. 488], we have

n=0(a; q)n(q; q)nzn=(az; q)(z; q); [z[ < 1, (1.1.13)and therefore, cf. [14, p.494] we can prove thatBq(a, b) =q(a)q(b)q(a +b). (1.1.14)The third type ofq-Bessel functions is dened forz C by, cf. e.g. [63, 67, 68],J(3)(z; q) := z(q+1; q)(q; q)11___0q+1q; qz2___= z(q+1; q)(q; q)

n=0(1)nqn(n+1)/2z2n(q; q)n(q+1; q)n(1.1.15)ThisfunctioniscalledinsomeliteraturetheHahn-Extonq-Besselfunction, see[78, 112].Itisalsocalledthe11q-Bessel function, cf. [77]. Sincetheothertypesoftheq-Besselfunctions, i.e. J(1)(; q), J(2)(; q), seee.g. [63, 69, 104], will notbeusedthroughoutthisthesis we use the notationJ(; q) forJ(3)(; q). The basic trigonometric functions cos(z; q)and sin(z; q) are dened on C bycos(z; q) :=

n=0(1)nqn2(z(1 q))2n(q; q)2n=(q2; q2)(q; q2)(zq1/2(1 q))1/2J1/2_z(1 q)/q ; q2_,(1.1.16)sin(z; q) :=

n=0(1)nqn(n+1)(z(1 q))2n+1(q; q)2n+1=(q2; q2)(q; q2)(z(1 q))1/2J1/2_z(1 q);q2_.(1.1.17)5Theyare q-analogues of thecosineandsinefunctions, [14, 53]. Thebasichyperbolictrigonometric functions cosh(z; q) and sinh(z; q) are dened forz C bycosh(z; q) := cos(iz; q), sinh(z; q) := i sin(iz; q). (1.1.18)Aq-analogue of the exponential function is the functionE(z; q) :=

n=0qn2/4znq(n + 1), z C. (1.1.19)One can easily see thatE(iz; q) = cos(z; q) +iq1/4sin(z; q), (1.1.20)andE([z[; q) = cosh([z[; q) +q1/4sin([z[; q). (1.1.21)F. H. Jackson[70, 1904] introducedthefunctionseq(z)andEq(z)asq-analoguesof theexponential functions, whereEq(z) :=

n=0q(n2)zn(q; q)n= (z; q), z C, (1.1.22)eq(z) :=

n=0zn(q; q)n, [z[ < 1. (1.1.23)Eq(z) is an entire function with simple zeros at the points qn,n N. Sinceeq(z)Eq(z) 1, [z[ < 1, (1.1.24)seee.g.[53, 70], thenthedomainofthefunctioneq(z)canbeextendedto Cbydeningeq(z),z C, to beeq(z) :=1

n=0_1 +qnz_. (1.1.25)Hence, the relation (1.1.24) holds in C, and the function eq(z) has simple poles at the pointsqn,n N. Letthebasictrigonometricfunctionssinq z, cosq z, Sinqz, andCosqzbe6dened bysinq z :=eq(iz) eq(iz)2i, cosq z :=eq(iz) eq(iz)2, [z[ < 1, (1.1.26)Sinqz :=Eq(iz) Eq(iz)2i, Cosqz :=Eq(iz) Eq(iz)2, z C. (1.1.27)Inthenextsectionweprovethatthefunctionscos(z; q)andsin(z; q)haveonlyrealandsimple zeros, which will be denoted throughout this thesis by xmm=1, and 0, ymm=1respectively, wherexm andym,m1, are the positive zeros, cf. [7, 28, 78].If R, a subsetA of R is called a-geometric set ifx A for allx A. If a subsetAof Risa-geometric, thenitcontainsall geometricsequences xnn=0, x A. Letfbe a function,real or complex valued,dened on aq-geometric setA. Theq-dierenceoperator is dened byDqf(x) :=f(x) f(qx)x qx, x A 0 . (1.1.28)If 0 A, theq-derivative at zero is dened byDqf(0) :=limnf(xqn) f(0)xqn, x A 0 , (1.1.29)if the limit exists and does not depend onx. In some literature theq-derivative at zero isdened to bef

(0) if it exists, cf. [78, 112], but the above denition is more suitable for ourapproaches. Also theq1-derivative of a functionf(x) at zero is dened byDq1f(0) := limnf(xqn) f(0)xqn= Dqf(0), x A 0 , (1.1.30)provided that the limit exists and does not depend on x. A right inverse to Dq, the Jacksonq-integration, cf. [72], is_x0f(t) dqt := x(1 q)

n=0qnf(xqn), x A, (1.1.31)provided that the series converges, and_baf(t) dqt :=_b0f(t) dqt _a0f(t) dqt, a, b A. (1.1.32)7IfA isq1-geometric, then theq-integration over [x, ),x A, is dened by_xf(t) dqt :=

n=1xqn(1 q)f(xqn). (1.1.33)Lemma1.1.1. Theq-integrals(1.1.31) and (1.1.33) exist iflimnxqnf(xqn) = 0 and limnxqnf(xqn) = 0,respectively.Proof. Based on the facts that, forx Climkxqkf(xqk) = 0 = [0, 1) C> 0, [f(xqk)[C[xqk[,k N, (1.1.34)limkxqkf(xqk) = 0 = (1, ) C> 0, [f(xqk)[C[xqk[,k N. (1.1.35)Kac and Cheung [75, p. 68] have proved that ifxf(x) is bounded on [0, a] for some0 < 1, then_x0f(t) dqt exists for allx [0, a].Denition1.1.1. Letfbeafunctiondenedonaq-geometricsetA. Wesaythatfisq-integrable onA if and only if_x0f(t) dqt exists for allx A.Denition1.1.2. A functionfwhich is dened on aq-geometric setA, 0 A, is said tobeq-regular at zero iflimnf(xqn) = f(0), for every x A.Theorem 1.1.2 below, which we call a fundamental theorem of q-calculus, shows that theq-integral operator and theq-dierence operator are, in certain sense, inverse operations.Theorem1.1.2. Letf : [0, b] C beq-regular at zero. DeneF(x) =_xcf(t) dqt, 0cb. (1.1.36)8ThenFisq-regular at zero. Furthermore,DqF(x) exists for everyx [0, b] andDqF(x) = f(x) for every x [0, b].Conversely if 0ab, then_baDqf(t) dqt = f(b) f(a). (1.1.37)Proof. See [9].It is worth mentioning that in Theorem 1.1.2 iffis notq-regular at zero, then_b0Dqf(t) dqt = f(b) limnf(bqn). (1.1.38)Consequently (1.1.37) would take the form_baDqf(t) dqt =_f(b) limnf(bqn)_f(a) limnf(aqn). (1.1.39)The non-symmetric Leibniz rule isDq(fg)(x) = g(x)Dqf(x) +f(qx)Dqg(x). (1.1.40)Relation(1.1.40)canbesymmetrizedusingtherelationf(qx)=f(x) x(1 q)Dqf(x),giving the additional term x(1 q)Dqf(x)Dqg(x). Also the rule of q-integration by parts is_a0g(x)Dqf(x) dqx = (fg)(a) limn(fg)(aqn) _a0Dqg(x)f(qx) dqx. (1.1.41)Iff,gareq-regular at zero, then limn(fg)(aqn) on the right hand side of (1.1.41) willbe replaced by (fg)(0).In the following we dene Hilbert spaces where a q-analogue of SturmLiouville problemswill be considered. Let L2q(0, a) be the space of all complex valued functions dened on [0, a]such that|f| :=__a0[f(x)[2dqx_1/2< . (1.1.42)9The spaceL2q(0, a) is a separable Hilbert space with the inner productf, g) :=_a0f(x)g(x) dqx, f, g L2q(0, a), (1.1.43)and the orthonormal basisn(x) =___1x(1q), x = aqn,0, otherwise,(1.1.44)n=0, 1, 2, . . ., cf. [15]. ThespaceL2q_(0, a)(0, a)_isthespaceof all complexvaluedfunctionsf(x, t) dened on [0, a] [0, a] such that|f(, )|2 :=__a0_a0[f(x, t)[2dqxdqt_1/2< . (1.1.45)The elements of L2q_(0, a) (0, a)_ are equivalence classes where f,g are in the same equiv-alenceclassif f(aqm, aqn) =g(aqm, aqn), m, n N. Thezeroelementistheequivalenceclass of all functionsf(x, t) which satisfyf(aqm, aqn) = 0, for allm, n N.Lemma1.1.3. L2q_(0, a) (0, a)_ is a separable Hilbert space with the inner productf, g)2 :=_a0_a0f(x, t)g(x, t) dqxdqt. (1.1.46)Proof. Similar to[15, pp. 217-218], L2q_(0, a)(0, a)_is aBanachspace. Toproveseparability, it suces to prove thatij(x, t) := i(x)j(t), i, j = 1, 2, . . . , (1.1.47)is an orthonormal basis of L2q_(0, a)(0, a)_ whenever i()i=1is an orthonormal basisofL2q(0, a). Indeed,jk, mn)2=_a0_a0j(x)k(t)m(x)n(t) dqxdqt=_a0j(x)m(x) dqx_a0k(t)n(t) dqt = jmkn,10proving orthogonality. To prove thatij is a basis, we prove that if there existsf L2q_(0, a) (0, a)_ such that f, ij)2 = 0, thenfis the zero element. Indeed,0 = f, ij) =_a0_a0f(x, t)i(x)j(t) dqxdqt=_a0j(t)__a0f(x, t)i(x) dqx_dqt =_a0h(t)j(t) dqt.Thush(t) :=_a0f(x, t)i(x) dqx (1.1.48)isorthogonal tothejswhichimpliesthath(aqn)=0, forall n N. So, f(x, aqn)isorthogonal to eachi. Consequently,f(aqm, aqn) = 0, for allm, n N.ByL1q(0, a),a> 0, we mean the Banach space of all complex valued functions denedon (0, a] such that|f| :=_a0[f(t)[ dqt < . (1.1.49)Let L1q(0, a) denote the space of all functionsfdened on (0, a] such thatf L1q(0, x) forallx (qa, a].1.2 Asymptoticsofzerosofcos(z; q)andsin(z; q)Therearestudiesonthezerosof q-trigonometricfunctions, seereferencesbelow. Inthissection we establish other formulae for the asymptotic behavior of the zeros of cos(z; q) andsin(z; q), i.e. xmandymrespectively. Our asymptotic formulae t with the study for theeigenvalues of q-SturmLiouville problems as well as for deriving a q-analogue of a theoremofG.Polya. Wealsostateandprovesomeinequalitiesforexpressionsinvolvingxmandym. Werststatetheasymptoticformulaeintheexistingliterature, namely[7, 28]. Westartwiththefollowingpreliminariesconcerningentirefunctions, cf. e.g[27, 61, 84]. Letf,g be entire functions , we say thatf(z) = O_g(z)_, as z , (1.2.1)11iff(z)_g(z) is bounded in a neighborhood of . On the other hand we writef(z) g(z), as z , if limz[f(z)[[g(z)[= 1. (1.2.2)Denition1.2.1. Letf(z) := n=0anznbe an entire function. The maximum modulusis dened byM(r; f) := sup [f(z)[ : [z[ = r , r > 0. (1.2.3)The order off,(f), is dened by, cf. [27, 84](f) := limsuprlog log M(r, f)log r= limsupnnlog nlog a1n. (1.2.4)Theorem1.2.1. [27] If (f)isniteandisnot equal toapositiveinteger, thenf hasinnitely many zeros, or is a polynomial.OnecaneasilyseethatJ(; q2)isanentirefunctionsoforderzero. Therefore, fromTheorem1.2.1, J(; q2)hasinnitelymanyzeros. KoelinkandSwarttouwprovedthatJ(; q2) has only real and simple zeros, cf. [78, 1994]. Let w()m, > 1, denote the positivezeros ofJ(; q2) in an increasing order ofm Z+. In [7] L.D. Abreu et al. proved that ifq2+2< (1 q2)2, thenw()m(q) = qm+2m(), 0 < m() 1. In other words, in this case the NewtonPuiseuxdiagram of (1.2.9) has at least one vertex which is not (0, 0), i.e. K1. Fork = 1, . . . , Ksetk :=djk djk1jkjk1.13Then 1> 2> . . . > K> 0. The ks are the slopes of the segments which together withthe negative part of they-axis form the boundary ofP. It is shown in [26, Theorem 1.2]and also in [105] that iffis an entire transcendental solution of (1.2.9), then there existsk0 1, . . . , K such thatlog M(r; f) k02 log [c[(log r)2as r . (1.2.12)The following example is a special case of (1.2.9) and is important in our study since thesolutions are nothing but basic trigonometric functions dened in (1.1.16) and (1.1.17).Example1.2.1. Consider the functional equationqy(z) +_(1 +q) +q2z2(1 q)2_y(qz) +y(q2z) = 0, z C. (1.2.13)Thefunctionscos(z; q),sin(z; q)aretranscendentalentiresolutionsof(1.2.13). Usingtheprevious notations we havel = 2, a0(z) = q, a1(z) =_(1 +q) +q2z2(1 q)2_, a2(z) = 1, Q(z) = 0. (1.2.14)Thereforep0 = p2 = 0, p1 = 2, d0 = d2 = 0, d1 = 2. (1.2.15)HencePis the minimal convex set containing_(x, y) R2: x0, y0__(x, y) R2: x1, y2__(x, y) R2: x2, y0_. (1.2.16)ThereforePis the set in R2whose boundary consists of the negative part of they-axis, theline segmenty = 2x, 0x1 and the liney = 2,x1. This convex hull is illustrated inFigure 1 below. Clearly Phas only two vertices (0, 0) and (1, 2). Thus there is only one linesegment contained in the boundary of P with a slope,1,1 = 2. So from(1.2.12) it followsthat the functions cos(z; q),sin(z; q) and E(z; q) have the following asymptotic behavior asr log M_r; cos(z; q)_, log M_r; sin(z; q)_, log M(r; E(z; q)) 1log q(log r)2. (1.2.17)14Figure1.1: TheNewton-PuiseuxdiagramPof (1.2.13).Notice that the asymptotic ofE(z; q) is obtained via (1.1.20).The following lemma is taken from [25] and will be needed in the sequel. It is concernedwith the asymptotic behavior of the-function dened in (1.1.6).LemmaA. Supposethat0< [q[< 1andthatz C, [z[1. Form Z+letAmbetheannulus dened byAm :=_z C, q2m+2 [z[ < [q[2m_. (1.2.18)Then we have, uniformly asm ,log [(z; q)[ = (log [z[)24 log [q[+ log [1 +q2m1z[ +O(1), z Am. (1.2.19)Usingthetechniqueof Bergweiler andHayman[25], wewill derivetheasymptoticformulae for the zeros of cos(z; q) and sin(z; q) in the following theorem.Theorem 1.2.2. If xm are the positive zeros of cos(z; q) and ym are the positive zeros15of sin(z; q), then we have for suciently largem,xm= qm+1/2(1 q)1_1 +O(qm)_, (1.2.20)ym= qm(1 q)1_1 +O(qm)_. (1.2.21)Proof. We prove the theorem only for the zeros of cos(z; q) since the proof for the zeros ofsin(z; q) is similar. LetR(z) be the dierenceR(z) := cos(z; q) 1(q; q)(z2(1 q)2; q), z C 0 . (1.2.22)HenceR(z) =

k=2kz2k, 2k :=___(1)kqk2(1 q)2k(q; q)_(q2k+1; q)1_, k0,(1)k+1qk2(1 q)2k(q; q), k < 0.(1.2.23)We shall prove that(q2k+1; q)1 = O(q2k) as k +. (1.2.24)Indeed from (1.1.2), we have fork Z+(q2k+1; q)1 =

j=1(1)jqj(j1)/2q(2k+1)j(q; q)j q2k+1

j=1qj(j1)/2(q; q)j

q2k+1(q; q)

j=1q(j1)/2=q2k+1(1 q)(q; q). (1.2.25)This proves (1.2.24). We conclude from (1.2.23) and (1.2.25) that[2k[ qk2+2k(1 q)2k(1 q)(q; q)2, k Z. (1.2.26)Thus forz C 0,[R(z)[ 1(1 q)(q; q)2

qk2+2k(1 q)2k[z2k[=1(1 q)(q; q)2(q2(1 q)2[z[2; q).(1.2.27)16Taking the logarithm of both sides of (1.2.27) yieldslog [R(z)[log (q2(1 q)2[z[2; q) + log1(1 q)(q; q)2, z C 0 . (1.2.28)We deduce from (1.2.19) that there existsm0 N and a constantC> 0 such that ifq2m+2 [z[ < q2m, mm0,then(log [z[)24 log q+log [1+q2m1z[Clog [(z; q)[(log [z[)24 log q+log [1+q2m1z[+C. (1.2.29)Thus ifq2m+2 q2[z[2(1 q)2< q2m, mm0, (1.2.30)then replacingz in (1.2.29) byq2(1 q)2[z[2, we obtain formm0log [(q2(1 q)2[z[2; q)[(log [z[2(1 q)2q2)24 log q+log1 +q2m+1(1 q)2[z[2+C. (1.2.31)Since log(1 +x) < x, x > 0, then (1.2.30) implieslog1 +q2m+1(1 q)2[z[2 q2m+1(1 q)2[z[2 q1. (1.2.32)Substituting in (1.2.31), we obtain forz C, which satises (1.2.30),log [(q2(1 q)2[z[2; q)[(log [z[(1 q)q)2log q+q1+C. (1.2.33)Consequently, by (1.2.28), forz satises (1.2.30)log [R(z)[_log [z[(1 q)q_2log q+q1+C + log1(1 q)(q; q)2= _log [z[(1 q)_2log qlog [z[2+C1,(1.2.34)whereC1 := C log(1 q)(q; q)22 log(1 q) log q +q1. (1.2.35)17Notice that (1.2.34) holds uniformly for z satises (1.2.30), mm0 and C1 is positive. LetAm,mm0, be the annulus dened byAm :=_z C: q2m [z[2(1 q)2< q2m2_. (1.2.36)The other side of (1.2.29) implies forz Am_log r(1 q)_2log q log_[((1 q)2z2; q)[_log1 q2m+1z2(1 q)2+C. (1.2.37)Therefore by (1.2.34),z Am,log [R(z)[log_(q; q)1[((1q)2z2; q)[_log1q2m+1z2(1 q)2log r2+C2, (1.2.38)whereC2 = C +C1 + log(q; q). Ifz Am,mm0, then[1 q2m+1z2(1 q)2[q2m+1[z[2(1 q)21 = q1(1 q), [z[ =qm11 q, (1.2.39)[1 q2m+1z2(1 q)2[1 q2m+1[z[2(1 q)2= (1 q), [z[ =qm1 q. (1.2.40)That islog [1 q2m+1z2(1 q)2[log(1 q), z Am. (1.2.41)Substituting in (1.2.38) we obtainlog [R(z)[log_(q; q)1((1 q)2z2; q)_log(1 q) log r2+C2, z Am. (1.2.42)We can also choosem0 suciently large such that log(1 q) log r2+C2< 0. Hence[R(z)[ _(q; q)1[((1 q)2z2; q)[_, z Am, mm0. (1.2.43)Hence an application of Rouche theorem shows that cos(z; q) and ((1q)2z2; q) have thesame number of zeros inAm. Since ((1q)2z2; q) has two simple symmetric zeros inAm,we deduce that cos(z; q) has exactly two zeros inAm. Since cos(z; q) is an even function,18then these two zeros are real, symmetric and simple. Assume now thatxm,mm0, is thepositive zero of cos(z; q) inAm. Then from (1.2.22)[R(xm)[ =_(q; q)1[((1 q)2x2m; q)[_and we obtain from (1.2.38) thatlog1 q2m+1x2m(1 q)2 2 log xm +C2. (1.2.44)In other wordslog1 qm+(1/2)xm(1 q)+ log1 +qm+(1/2)xm(1 q) 2 log xm +C2. (1.2.45)Thus, eitherlog1 qm+(1/2)xm(1 q) log xm +C22, (1.2.46)orlog1 +qm+(1/2)xm(1 q) log xm +C22. (1.2.47)We can assume without any loss of generality that (1.2.46) holds. Then for allmm01 qm+(1/2)xm(1 q)= O(x1m ). (1.2.48)This implies thatxm qm1/2(1 q)1for suciently largem. More precisely, we havexm = qm1/2(1 q)1_1 +O(qm)_, as m . (1.2.49)Lemma1.2.3. Asm , we haveq2kx2m+kx2m1= O(qm), k N. (1.2.50)TheO-term is uniform ink N.19Proof. From (1.2.20) we see that form xm =qm+1/21 q_1 +O(qm)_. (1.2.51)Hence there exists a constantC> 0 andm0 N such that[qm1/2(1 q)xm1[Cqm, for all mm0. (1.2.52)Consequently for allk N we have[qm+k1/2(1 q)xm+k1[Cqm+k, for all mm0. (1.2.53)We can choose m0 suciently large such that 1Cqm> 0 for all mm0. So from (1.2.52)and (1.2.53) we obtain(1 Cqm+k)2(1 +Cqm)2 q2kx2m+kx2m

(1 +Cqm+k)2(1 Cqm)2, mm0. (1.2.54)ThusC(1 +qk)(2 +Cqm(1 qk))(1 +Cqm)2

_q2kx2m+kx2m1_qm

C(1 +qk)(2 Cqm(1 qk))(1 Cqm)2. (1.2.55)That is4C(1 +Cqm)(1 +Cqm)2

_q2kx2m+kx2m1_qm

4C(1 +Cqm)(1 Cqm)2, (1.2.56)holds for allk N andmm0. Hence_q2kx2m+kx2m1_qm max_4C(1 +Cqm)(1 +Cqm)2, 4C(1 +Cqm)(1 Cqm)2_=4C(1 +Cqm)(1 Cqm)2, (1.2.57)forall mm0andk N. Thusthesequence__q2k x2m+kx2m1_qm_m=1isuniformlybounded fork N since4C(1 +Cqm)(1 Cqm)24C as m . (1.2.58)20Lemma 1.2.4.Forn N, ifz C andx R are such that [zx[ qn1 q, then forr Z+,we haver

m=1x2m +q2n[z[2x2x2m +[z[2x2 q2nrr

m=1(1 +x2m). (1.2.59)Proof. Assume thatz C,x R such that [zx[ qn1 q,n N. Thenx2m +q2n[z[2x2x2m +[z[2x2q2n=(1 q2n)x2mx2m +[zx[2

x2m[zx[2q2nx2m. (1.2.60)Consequentlyr

m=1x2m +q2n[z[2x2x2m +[z[2x2

r

m=1q2n(1 +x2m) = q2nrr

m=1(1 +x2m), (1.2.61)proving the lemma.Similarly we haveLemma 1.2.5. Forn N, ifz C andy R are such that [zy[ qn1 q, then forr Z+,we haver

m=1y2m +q2m[z[2y2y2m +[z[2y2 q2nrr

m=1(1 +y2m). (1.2.62)1.3 Asymptoticsofcos(z; q)andsin(z; q)Inthis section, westudytheasymptoticbehavior of thebasictrigonometricfunctionscos(z; q), and sin(z; q). The results of this section are contained in [19]. We start our investi-gations with the following lemma that exhibits a major dierence between the trigonometricfunctions and their basic analogues.Lemma1.3.1. Thebasictrigonometricfunctionssin(x; q)andcos(x; q)arenotboundedon R.21Proof. Daalhuis in [38] and Chen et al. in [35] proved thatJ(x; q2) has the asymptoticbehaviorJ(x; q2) [x[(x2q2; q2)(q2; q2), as[x[ . (1.3.1)But (x2; q2) is not bounded on R. Indeed, takern := qn1(1 +q)2, n Z+. (1.3.2)Thenqn< rn< qn1, n Z+. (1.3.3)Therefore

k=n+2(1 r2nq2k)

k=n+2(1 q2k2n2) =

j=1(1 q2j) = (q2; q2), (1.3.4)andn1

k=0(1 r2nq2k) n1

k=0(1 q2n2k) = (q2; q2)n> (q2; q2). (1.3.5)But(r2n; q2)=

k=0[1 r2nq2k[= (r2nq2n1)(1 r2nq2n+2)n1

k=0[1 r2nq2k[

k=n+2(1 r2nq2k)=q216(1 + 2q 3q2)(3 2q q2)r2nnqn(n1)n1

k=0(1 r2nq2k)

k=n+2(1 r2nq2k).(1.3.6)Combining equations (1.3.3)(1.3.6) yields(r2n; q2)

q216 (1 + 2q 3q2)(3 2q q2)(q2; q2)2qn2nr2nn=q216 (1 + 2q 3q2)(3 2q q2)(q2; q2)2qn23n_1 +q2_2n.(1.3.7)But_1 +q2_2n q2n, for all n N. (1.3.8)22Thus[(r2n; q2)[ q216 (1 + 2q 3q2)(3 2q q2)(q2; q2)2qn2n. (1.3.9)Hence(r2n; q2) as n .This proves that (x2; q2) is not bounded on R. By (1.3.1), J(x; q2) is also not bounded onR. Applying this result to the cases = 1/2 respectively and using (1.1.16) and (1.1.17),we deduce that the functions cos(x; q) and sin(x; q) are not bounded on R.The following two lemmas will be needed in the sequel.Lemma1.3.2. For a positive sequence zmm=1, ifzm 1 asm , then infmNzm> 0.Proof. Sincezm 1asm , thenfor0 0.Lemma1.3.3. Ifz C such that 0[z[ < 1, thenlog [1 z[ log11 [z[. (1.3.10)Proof. Letz C such that 0[z[ < 1. Since [1 z[1 [z[, then we havelog [1 z[log(1 [z[) = log11 [z[. (1.3.11)Also [1 z[1 +[z[, thenlog [1 z[(1 [z[)log(1 [z2[)0.Consequentlylog [1 z[log(1 [z[) = log1(1 [z[). (1.3.12)23Combining (1.3.11) and (1.3.12) we obtain (1.3.10).In Lemma A the asymptotic behavior of log [(z; q)[ as z is introduced by studyingthe asymptotic behavior of log [(z; q)[ in the set of annuli Amm=1 dened in (1.2.18) whenm is large enough. Similarly we shall study the asymptotic behavior of log [ cos(z; q)[ andlog [ sin(z; q)[. Webeginwiththeq-cosinefunction. WedividethecomplexplaneexceptforanitediskDwithcenteratzerointoaninnitesetofannuli Acmm=1, wherethediskDwill be determined later. Then we study the asymptotic behavior of log [ cos(z; q)[asz via the study of its behavior in eachAcm, for suciently largem. Letm :=log (xm/xm+1)/ log q, m Z+. (1.3.13)Thenm> 0 for allm Z+. Moreover, from (1.2.20)m = 1 +_log_1 +O(qm)1 +O(qm+1)___log q,for suciently largem. That is limmm = 1. Then from Lemma 1.3.20 < := infmZ+m1. (1.3.14)We dene two sequences, (am)m=1, (bm)m=1, to beam :=___m +2, if m ,= ,2, if m = ,(1.3.15)andb1 :=2, bm+1 :=___m2, if m ,= ,2, if m = ,(1.3.16)wherem1. Then we havexmqam= xm+1qbm+1, m1. (1.3.17)Thus the set of annuli Acmm=1 whereAcm is dened to beAcm :=_z C : xmqbm [z[ < xmqam_, m1 (1.3.18)24Figure1.2: TheannulusAc2divide the region _z C : [z[q/2x1_.In the following we shall study the asymptotic behavior of log [ cos(z; q)[ in each of theannuli described above whenm is suciently large.Theorem 1.3.4.Let be the positive number dened in (1.3.14). Assume that [z[q/2x1and let Acm, m1, be the annulus dened by (1.3.18). Then we have, uniformly as m ,the asymptotic relationlog [ cos(z; q)[ = (log [z[)2log q+ log [1 z2x2m[ +O(1), z Acm. (1.3.19)Proof. First of all, it should be noted thatz Acm,m1, if and only if[z[ = xmqgm(t), gm(t) := (am +bm)t +bm, t [0, 1). (1.3.20)Letm1 andz Acm be xed. Then there existst [0, 1) such thatr := [z[ = xmqgm(t)= xmq(am+bm)t+bm. (1.3.21)25Since cos(z; q) is an entire function of order zero, then, see e.g. [27],cos(z; q) =

n=1_1 z2x2n_. (1.3.22)Thuslogcos(z; q)1 (zxm)2=m1

n=1log1 z2x2n+

n=m+1log1 z2x2n=m1

n=1log [zxn[2+m1

n=1log1 x2nz2+

n=m+1log1 z2x2n.(1.3.23)Setn := qn12(1 q)xn1, n = 1, 2, . . . . (1.3.24)From (1.2.20),n = O(qn) asn . Dene the constantsl,L to bel :=infnN[1 +n[, L := supnN[1 +n[. (1.3.25)Since 1 +n> 0, for alln N and 1 +n 1 asn , thenl > 0. We dene constantsK1 andK2 as followsK1 :=Ll , K2 := qa; a =supnZ+an. (1.3.26)Thenqgm(t) qam K2 for allt [0, 1). From (1.3.24), we have fornm+ 1zxn=xmqgm(t)xn=qnm1 +m1 +nqgm(t) K1qamqnm K1K2qnm.(1.3.27)Consequently

n=m+1log1 z2x2n

n=m+1log_1 +zxn2_

n=m+1zxn2 K21K22

n=m+1q2(nm) K21K22q21 q2. (1.3.28)As forn = 1, 2, . . . , m1,xnz =xnqgm(t)xm= qmn 1 +n1 +mqgm(t) K1qbmqmn. (1.3.29)26From (1.3.15) and (1.3.16) we obtainqbm= qqam1 qqa= K2q. (1.3.30)Combining (1.3.29) and (1.3.30) yieldsxnzK1K2qmn. (1.3.31)Moreoverxnz=xnqgm(t)xm

xm1qgm(t)xm, n = 1, 2, . . . , m1. (1.3.32)But from (1.3.17)xm1xm= qam1+bm, and qgm(t)= qamt(1t)bm qbm.Thenxnzqam1< q< 1, z Acm. (1.3.33)So from (1.3.10) we obtainlog1 x2nz2 log1_1 [x2nz2[_= log__1 +

j=1[xnz [2j__= log_1 +[xnz [21 [xnz [2_

[xnz [21 [xnz [2 [xnz [21 q2

K21K221 q2q2(mn),(1.3.34)Thereforem1

n=1log1 x2nz2

K21K221 q2m1

n=1q2(mn)

K21K22q2(1 q2)(1 q2). (1.3.35)Hence from (1.3.23), (1.3.28), and (1.3.35) we obtainlogcos(z; q)1 (zxm)2=m1

n=1log [zxn[2+O(1), as z . (1.3.36)27Nowm1

n=1logzxn2=m1

n=1log [z[2m1

n=1log [xn[2= 2(m1) log [z[ 2m1

n=1(n + 1/2) log q 2m1

n=1log [1 +n[= 2(m1) log [z[ + (m1)2log q 2m1

n=1log [1 +n[. (1.3.37)Sincen=O(qn)asn ,thenthereexistsaconstantK3> 0andn0 Nsuchthat[n[K3qn, for allnn0. LetK4 := max_q1[1[, . . . , qm0+1[n01[, K3_.Thus [n[K4qnfor alln Z+, andm1

n=1log [1 +n[ m1

n=1log_1 +[n[_

m1

n=1[n[

n=1[n[K4

n=1qn K4q1 q.(1.3.38)Sincer =qm+1/2(1q)(1 +m)qgm(t), thenm1 = log rlog q +dm, dm := 1/2 + log(1 +m)log q+gm(t) log(1 q)log q. (1.3.39)A simple computation yields2(m1) log r + (m1)2log q = (log r)2log q+d2m log q = (log r)2log q+O(1), (1.3.40)as m since dmm=1 is a bounded sequence. Hence the theorem follows by combining(1.3.36), (1.3.37), and (1.3.40).Corollary1.3.5.M(r; cos(z; q)) = O_(exp_(log [z[)2log q__, as z . (1.3.41)28Proof. From (1.3.19) we conclude that there existsm0 N such that for all m m0wehave(log r)2log q+ log [1 z2x2m[ log [ cos(z; q)[(log r)2log q+ log [1 z2x2m[ +, z Acm.(1.3.42)From (1.3.18) we obtainq2bm

[z[2x2m q2am. (1.3.43)Hencelog1 z2x2m log1 + [z[2x2m log q2am, m1, (1.3.44)Since the sequence amm=1 is bounded, then there exists a constantR > 0 such thatlog [ cos(z; q)[(log [z[)2log q+R, z Acm, mm0. (1.3.45)Thus[ cos(z; q)[eRe(log [z[)2log q, z Acm, mm0, (1.3.46)proving the corollary.The same study may be carried out for sin(z; q) and for this purpose we dene suitableannuli. Letm :=log (ym/ym+1)/ log q, m Z+. (1.3.47)Sincem 1 asm , then := infmZ+m> 0. (1.3.48)Let (cm)m=1 and (dm)m=1 be the sequence dened bycm :=___m +2, if m ,= ,2, if m = ,(1.3.49)29andd1 :=2, dm+1 :=___m2, if m ,= ,2, if m = ,(1.3.50)wherem1. Then we haveymqcm= ym+1qdm+1, m1. (1.3.51)Thus the set of annuli Asmm=1 whereAsm is dened to beAsm :=_z C : ymqdm [z[ < ymqcm_, m1 (1.3.52)divide the region _z C : [z[q/2y1_.Theorem 1.3.6. Let be the positive number dened in (1.3.48). Assume that [z[q/2x1.LetAsm,m1, be the annulus dened in (1.3.52). Then we have, uniformly asm log [ sin(z; q)[ = (log [z[)2log q+ log [1 z2y2m[ +O(1), z Asm. (1.3.53)The proof is similar to that of Theorem 1.3.4, and so it will not be given here.Also we have the following corollaryCorollary1.3.7.M(r; sin(z; q)) = O_(exp_(log [z[)2log q__, as z . (1.3.54)Lemma1.3.8. The function tanh([z[; q) :=sinh([z[; q)cosh([z[; q)is bounded on C.Proof. From (1.3.19) and (1.3.53) we conclude that there existsm0 N such that for allmm0 we have(log r)2log q+ log [1 + [z[2x2m[ log [ cosh([z[; q)[(log r)2log q+ log [1 + [z[2x2m[ +, z Acm(1.3.55)30and(log r)2log q+log1 + [z[2y2mlog [ sinh([z[; q)[(log r)2log q+log1 + [z[2y2m+, z Asm. (1.3.56)Butif z Asm, m 1, then [z[2/y2mq2cm, wherethesequence cmm=1denedin(1.3.49) is bounded. Hence there exists a constantL > 0 such that1 + [z[2y2m L, for all m Z+.Letr0 := max_qbm0xm0, qdm0ym0_. Thus by (1.3.55) and (1.3.56) we obtain for allz C,[z[r0sinh([z[; q)cosh([z[; q)e2[1 + [z[2y2m[Le2. (1.3.57)1.4 BasicanaloguesofatheoremofPolyaIn this section we give aq-analogue of Theorem 1.4.1 below of George Polya, cf. [102, 103].Theseq-analogues are the main results of [18].Theorem1.4.1. If the functionf L1(0, 1) is positive and increasing, then1. the zeros of the entire functions of exponential typeU(z) =_10f(t) cos(zt) dt, V (z) =_10f(t) sin(zt) dt (1.4.1)are real, innite and simple.2. U(z) is an even function having no zeros in [0,2), and its positive zeros are situated inthe intervals (k/2, k+/2), 1k < , one in each. The odd function V (z) has onlyonezeroz = 0in[0, ),anditspositivezerosaresituatedintheintervals(k, (k + 1)),1k < , one zero in each interval.31Theorem1.4.2. Iff L1q(0, a) andUf(z) is dened forz C byUf(z) :=_10f(t) cos(tz; q) dqt, 0 < q< 1. (1.4.2)ThenUf(z) has at most a nite number of non real zeros and it has an innite number ofreal zeros mm=1,m> 0, such thatm xmasm . More preciselym = xm(1 +O(qm)) as m . (1.4.3)Proof. From the denition of theq-integral (1.1.31)Uf(z) can be written asUf(z) = H(z) +R(z), (1.4.4)whereH(z) := (1 q)f(1) cos(z; q), and R(z) :=

k=1qk(1 q)f(qk) cos(qkz; q). (1.4.5)From (1.3.41) we deduce that there exists constantsr0,C> 0 such thatmax|z|=r[ cos(z; q)[Ce(log r)2/log q, for all r > r0. (1.4.6)Letz C such that [z[ > q1r0, then[R(z)[Ce(log qr)2log q

k=1qk(1 q)[f(qk)[ = q1Ce(log r)2/ log qr2_q0[f(t)[ dqt. (1.4.7)Thuslog [R(z)[C12 log r _log r_2log q, (1.4.8)whereC1:=log_q1C_q0 [f(t) dqt_. Let Acm, m 1, betheannulsdenedin(1.3.18).Thenlog [ cos(z; q)[ = _log r_2log q+ log [1 z2x2m[ +O(1), z Acm(1.4.9)uniformly asm . Hence there exists a constantC2> 0 andm0 N such that for allz Acm,mm0 we have_log r_2log q+log1 z2x2mC2log [ cos(z; q)[_log r_2log q+log1 z2x2m+C2. (1.4.10)32Consequently by (1.4.8)log [R(z)[C1 +C22 log r + log [ cos(z; q)[ log1 z2x2m= log [H(z)[ 2 log r log1 z2x2m+C3, (1.4.11)whereC3 := C1 +C2log(1 q)[f(1)[. LetDm0be the disk dened byDm0 :=_z C : [z[qbm0xm0_. (1.4.12)ClearlyC := Dm0 mm0Am.Ifz Dm0, i.e. [z[ = qbm0xm0, then by (1.4.11)log [R(z)[log [H(z)[ 2 log r log1 q2bm0[ +C3. (1.4.13)We can choose the m0 suciently large such that 2 log r log1q2bm0 +C3< 0. That islog [R(z)[log [H(z)[, [z[ = qbm0xm0. (1.4.14)SoapplyingRouchetheoremonDm0, weconcludethat H(z)andUf(z)havethesamenumber of zeros insideDm0. As cos(z; q) has exactly 2m02 zeros insideDm0, thenUf(z)has at most 2m02 zeros insideDm0.Ifz Acm,mm0, then1 z2x2m 1 q2bm; [z[ = qbmxm,1 z2x2m q2am1; [z[ = qamxm. (1.4.15)Thuslog [1 z2x2m[infmm0_log(1 q2bm) log(q2am1)_, z Acm. (1.4.16)Sincethesequences amm=1and bmm=1areboundedandpositive, thenthereexitsaconstantC4> 0 such thatlog [1 z2x2m[C4, z Acm, mm0. (1.4.17)33That islog [R(z)[log [H(z)[ 2 log r +C3C4. (1.4.18)Again, we choose them0large enough such that ifr = [z[ Acm,mm0, then 2 log r +C3C4< 0. That is[R(z)[[H(z)[, z Acm, mm0.ApplyingRouchetheoremonAcm, m m0, weconcludethatH(z)andUf(z)havethesamenumberofzerosinsideAcm. Ascos(z; q)hastwosimplesymmetriczerosthere, wededuce that the even functionUf(z) has only two real, symmetric and simple zeros insideAcm,mm0.Now we give the asymptotic behavior of such zeros. Letm, be a positive zero ofUf(z)inAcm, then log [H(m)[ = log [R(m)[. Consequently by (1.4.11)log1 2mx2m C32 log [m[, mm0. (1.4.19)Thus eitherlog1 mxm

C32log [m[, (1.4.20)orlog1 +mxm

C32log [m[. (1.4.21)We can assume without any loss of generality thatm satises (1.4.20), then1 mxm= O([m[1), as m . (1.4.22)Thusm xm for all suciently largem. More preciselym = xm(1 +O(qm)), as m . (1.4.23)SinceUf(z) is even function then mare both zeros ofUf(z) inAcm. This completes theproof.Similarly we have the following theorem when we replace cos(tz; q) by sin(tz; q) in (1.4.2).34Theorem1.4.3. Iff L1q(0, a) andVf(z) be dened forz C byVf(z) :=_10f(t) sin(tz; q) dqt, 0 < q< 1. (1.4.24)ThenVf(z)hasat most anitenumberof nonreal zerosandit hasaninnitenumberof real zeros mm=1, m>0, forall m Z+suchthat m ymasm . Morepreciselym = ym(1 +O(qm)) as m . (1.4.25)Chapter2ZerosofFiniteq-HankelTransformsIn this chapter we prove that the basic nite Hankel transform whose kernel is the third-typeJacksonq-Bessel functionhas innitelymanyreal andsimplezeros under certainconditions onq. We also study the asymptotic behavior of the zeros. The obtained resultsare applied to investigate the zeros ofq-Bessel functions as well as zeros ofq-trigonometricfunctions. A basic analogue of a theorem of G. Polya (1918) on the zeros of sine and cosinetransformations is given as a special case.2.1 IntroductionandpreliminariesThis chapter is devoted to a study the zeros of theq-type nite Hankel transformF(z) =_10f(t)J(tz; q2) dqt, z C, (2.1.1)where J(; q2) is the third Jackson q-Bessel function dened in (1.1.15), cf. [17]. Throughoutthischapterweassumethat > 1. Nextwestateandprovethemainresultsof thischapter. These results prove that, under certain conditions onfandq, the niteq-Hankeltransform(2.1.1)hasonlyrealandsimplezerosandthereareinnitelymanyzeros. Wealsoinvestigatethedistributionofthesezeros. Section2.3containssomeapplicationsof3536the results of Section 2.2. Among these applications we derive another basic analogue of atheorem of Polya. We also derive in a manner dierent from the analysis of [7, 28, 63, 112]the distribution of zeros ofq-Bessel andq-trigonometric functions. It is worth mentioningthatq-type Hankel transforms are studied in [80] where the Hankel transform is dened on(0, )simulatingtheclassicalsituation. Thefollowinglemmaisneededfortheproofofthe main results of this chapter. It is taken from[103, p. 143].LemmaA.Let g1(z),g2(z), . . . ,gn(z), . . . beentirefunctionswhichhavereal zerosonly.Iflimngn(z) = g(z),uniformly in any compact subset of C, theng(z) is an entire function with only real zeros.Weusethefollowingconvenientnotations. Forf L1q(0, 1),wedenotebyAk(f)theq-moments off, i.e.Ak(f) :=_10tkf(t) dqt, k N, (2.1.2)see (1.1.31) for the denition of theq-integration. Let alsock,(f),bk,(f),> 1,k N,denote the numbersck,(f) :=A2k+2(f)A2k(f)(1 q2k+2+2)(1 q2k+2), k N, (2.1.3)andbk,(f) :=A2k+3(f)A2k+1(f)(1 q2k+2+2)(1 q2k+2), k N, (2.1.4)We start our investigations with the following lemma.Lemma2.1.1. Letf L1q(0, 1) be positive on 0, qn, n N. Then the numbersc,f:=infkNck,(f), C,f:= supkNck,(f), (2.1.5)andb,f:=infkNbk,(f), B,f:= supkNbk,(f), (2.1.6)37exist and they are nite positive numbers.Proof. Wewillproveonly(2.1.5). Theproofof(2.1.6)issimilar. Firstofallitshouldbe noted that the sequence Ak(f)k=0is positive and strictly decreasing. Using Cauchy-Schwarz inequality, fork N,k2, we obtain_A2k(f)_2=__10t2kf(t) dqt_2=__

j=0qj(1 q)q2kjf(qj)__2

j=0qj(1 q)q(2k+2)jf(qj)

j=0qj(1 q)q(2k2)jf(qj)= A2k+2(f)A2k2(f).That is,_A2k+2(f)A2k(f)_k=0is an increasing sequence of positive numbers. Since A2k(f)k=0is strictly decreasing, then the above inequality impliesA2(f)A0(f) A2k+2(f)A2k(f)< 1, k = 0, 1, . . . . (2.1.7)Therefore, fork N we haveA2(f)A0(f) A2k+2(f)A2k(f)

A2k+2(f)A2k(f)(1 q2k+2+2)(1 q2k+2)= ck,(f) 1 andf L1q(0, 1) be positive on 0, qn,n N. Ifq1(1 q) c,fC,f> 1, (2.2.9)thenthezerosof U,f(z)arereal,simpleandinniteinnumber. Moreover, U,f(z)isaneven function with no zeros in _0, q1/_C,f_, and its positive zeros lie in the intervals_qr+1/2_C,f, qr1/2_C,f_, r = 1, 2, 3, . . . , (2.2.10)one zero in each interval.Proof. We prove the theorem in three steps.1. We show thatU,f(z) has no zeros inBR(0), whereR := 1/q_C,fandBR(0) := z C : [z[R, z ,= R .40First0 0,[c0 +c1z[2= c20 + 2c0c1r cos +c21r2< c20 + 2c0c1r +c21r2= (c0 +c1[z[)2. (2.2.13)Consequently [c0 +c1z[ < c0 +c1[z[. Thus for non realz BR(0), we obtain

k=1(1)kz2k(ak1R2ak)

z4(a1a2R2) +z6(a2a3R2)+

k=1k=2,3[z[2k(ak1R2ak)< [z4[_(a1a2R2) +[z[2(a2a3R2)_+

k=1k=2,3[z[2k(ak1R2ak) (2.2.14)=

k=1[z[2k(ak1R2ak)

k=1(R2kak1R2k+2ak)= a0R2limmR2m+2am.From (2.2.11)m1

k=0ak+1ak=m1

k=0q2k+2ck,(f) m1

k=0q2k+2C,f= qm2+mCm,f. (2.2.15)Thusamqm2+mCm,fa0, R2m+2amqm2m2a0C,f. (2.2.16)41That is limmR2m+2am = 0, and therefore we have for a non realz BR(0),z ,= iR,(R2+z2)U,f(z)>limmR2m+2am = 0. (2.2.17)Ifz = iR, then by (2.2.3)U,f(iR) =

k=0akR2k> 0. (2.2.18)ThusU,f(z)> 0 for all non real z BR(0). On the other hand if 0< [z[ a0R2

k=1R2k(ak1R2ak),(2.2.19)since if 0 < n< n, and n=0n< , then n=0n 0forallreal z BR(0). ConsequentlyU,f(z)hasnozerosinBR(0).2. WeprovethatU,f(z)hasaninnitenumberofrealzerosonly. LetU2n,,f(z)bethepolynomialU2n,,f(z) :=(q2+2; q2)(q2; q2)2n

k=0(1)kqk2+kA2k(f)(q2+2; q2)k(q2; q2)kz2k,n N,z C. (2.2.20)ObviouslyU2n,,f(z)approaches U,f(z)uniformlyas n onanycompactsubsetofC. Takezr:=qr1/2/_C,f, r N. We prove thatU2n,,f(z) has a zero in the interval42(zr1, zr),r = 1, 2, . . . , 2n. Indeed,U2n,,f(zr) =2n

k=0(1)kk(r), (2.2.21)wherek(r) is the positive sequencek(r) :=(q2+2; q2)(q2; q2)qk22krCk,f(q2; q2)k(q2+2; q2)kA2k(f), k N. (2.2.22)If 0kr 1, then from (2.2.9)k+1(r)k(r)= q2k2r+1ck,(f)C,f

c,fC,fq1> (1 q)1> 1, (2.2.23)andk+1(r)k(r) q< 1, if kr. (2.2.24)From (2.2.23), and (2.2.24), we obtain forr1r+1(r)r(r)+r1(r)r(r)< q + (1 q) = 1. (2.2.25)Therefore, we have the following inequalityr+1(r) < r(r) r1(r), r = 1, 2, 3, . . . . (2.2.26)Now we prove thatsign U2n,,f(zr) = (1)r, r = 0, 1, . . . , 2n. (2.2.27)Ifr = 0, then from (2.2.24)k+1(0)k(0)< 1 for allk0. ConsequentlyU2n,,f(z0) =n1

j=0_2j(0) 2j+1(0)_+2n(0) > 0.Taker = 2m, 1mn. ThenU2n,,f(zr) =m1

k=1(2k(r) 2k1(r)) +n1

k=m+1(2k(r) 2k+1(r))+0(r) +2n(r) +_r(r) r1(r)_r+1(r).(2.2.28)43Hence from (2.2.23), (2.2.24), and (2.2.26),U2n,,f_zr_

_r(r) r1(r)_r+1(r) > 0.Similarly ifr = 2m1, 1mn, we haveU2n,,f_zr_= m2

k=0_2k+1(r) 2k(r)_n

k=m+1_2k1(r) 2k(r)__r(r) r1(r)_+r+1(r) < 0.(2.2.29)Thus, U2n,,f(z)hasatleastonezeroineachoftheintervals(zr1, zr), r=1, 2, . . . , 2n,i.e. ithasatleast2n positivezeros. SinceU2n,,f(z)isanevenpolynomialofdegree4n,thenitalsohas2nnegativezeros. So, the4nzerosof thepolynomial U2n,,f(z)areallreal and simple. Consequently, from Lemma A,U,f(z) has only real zeros. Sincek(r) isindependent ofn, the same argument holds whenn is replaced by , and noting that allseries converge absolutely, shows thatsign U,f(zr) = (1)r, r N. (2.2.30)proving thatU,f(z) has an innite number of real zeros w()r,w()r (zr, zr+1),r N.3. Finally we prove that the zeros ofU,f(z) are simple. Consider the annulus Trdenedforr Z+byTr := z C : zr1[z[zr . (2.2.31)SinceU,f(z)hasnozerosontheboundaryTrof Tr, where [z[=qr(1/2)C,f, thenthereexists> 0 such that [U,f(z)[>onTr. For thiswe can ndN0 N, N0>rsuchthat[U2m,,f(z) U,f(z)[, mN0, z Tr. (2.2.32)Hence,[U2m,,f(z) U,f(z)[ < [U,f(z)[, mN0, z Tr. (2.2.33)44ApplyingRouchesTheorem,weconcludethatU,f(z)andU2m,,f(z), m N0,havethesame numbers of zeros inside Tr. Since for all mN0, U2m,,f(z) has exactly two symmetricsimple real zeros inside Tr, then so is U,f(z). This completes the proof of the theorem.Remark 2.2.1. Step2oftheproofofTheorem2.2.2includesaproofofthefactthatthefunctionU,fhasinnitelymanyzeros. However, accordingtoTheorem1.2.1wedonotneed to give a proof because (U,f) = 0.Theorem2.2.3. Let> 1 andf L1q(0, 1) be positive on 0, qn, n N. Ifq1(1 q) b,fB,f> 1, (2.2.34)then the zeros ofV,f(z) are real, simple and innite in number. The odd functionV,f(z)has only one zeroz = 0 in [0,q1B,f), and its positive zeros are situated in the intervals(qr+1/2_B,f, qr1/2_B,f), r = 1, 2, 3, . . . , (2.2.35)one zero in each interval.Proof. The proof is similar to that of the previous theorem.Corollary 2.2.4.Let > 1 and f L1q(0, 1) be positive and decreasing on 0, qn, n N.Ifq1(1 q)2(1 q2)(1 q2+2)f2(1)f2(0)> 1, (2.2.36)thenthezerosof U,f(z)arereal, simpleandinniteinnumber. Moreover, thepositivezeros lie in the intervals_qr+1/2__C,f, qr1/2__C,f_, r = 1, 2, . . . ,onezeroineachinterval. Wehavethesameresult if f is increasingandpositiveon0, qn, n N andqsatisesq1(1 q)2(1 q2)(1 q2+2)f2(0)f2(1)> 1, (2.2.37)instead of (2.2.36).45Proof. We notice that ifq1(1 q) l,fL,f> 1, (2.2.38)wherel,fandL,fare lower and upper bounds of the sequence ck,(f)k=0respectively,then (2.2.9) is satised and the results of Theorem 2.2.2 hold. This observation suces toprove the corollary because if fis decreasing and positive on 0, qn, n N, then by (2.1.2)f(1)(1 q)(1 q2k+1)A2k(f) f(0)(1 q)(1 q2k+1), k N. (2.2.39)Hencef(1)(1 q2k+1)f(0)(1 q2k+3) A2k+2(f)A2k(f)

f(0)(1 q2k+1)f(1)(1 q2k+3), k N. (2.2.40)Now the sequence_(1 q2k+1)((1 q2k+2)(1 q2k+3)(1 q2k+2+2))_k=0is bounded above by 1/_(1 q2)(1 q2+2)_ and it is bounded below by (1 q). Letl,fandL,fbe the numbersl,f=f(1)(1 q)f(0), L,f=f(0)f(1)(1 q2)(1 q2+2). (2.2.41)Thenfrom(2.1.3), l,fandL,farelower andupper bounds of ck,,frespectively. Adirect substitution withl,fandL,fof (2.2.41) in (2.2.38) yields (2.2.36). Similarly iffisincreasing and positive on 0, qn, n N, thenl,fandL,fcan be taken to bel,f=f(0)(1 q)f(1), L,f=f(1)f(0)(1 q2)(1 q2+2).A similar result holds forV,fwhenfis positive and monotonic on 0, qn, n N.2.3 ApplicationsIn this section we introduce some applications of the results of Section 2.2. First we inves-tigate the zeros ofJ(z; q2) and theq-trigonometric functions.46Lemma2.3.1. Ifq (0, 1) satises the inequalityq1(1 q)(1 q2)(1 q2+2) > 1, > 1, (2.3.1)then the zeros of the third Jacksonq-Bessel functionJ(; q2) are real, simple, and innitein number. MoreoverJ(; q2) has a zero atz = 0 if and only if> 0. The positive zerosofJ(; q2) belong to the intervals_qr+1/2_(1 q2+2)(1 q2), qr1/2_(1 q2+2)(1 q2)_, r = 1, 2, . . . , (2.3.2)where each interval contains only one zero.Proof. Applying Theorem 2.2.2 withf(t) :=___(1 q)1, t = 1,0, otherwise,thenU,f(z) = zJ(z; q2). Fork N, we haveAk(f) 1, k N, C,f=1(1 q2)(1 q2+2), c,f= 1.Therefore condition (2.2.9) is nothing but (2.3.1) and the lemma is proved.Remark2.3.1. In [66] asymptotics of theq-Bessel functionsJ(1)(; q) andJ(2)(; q) is given.Also, astudyforthezerosof thesecondJacksonq-Bessel functionJ(2)(x; q)isgivenin[60].Corollary2.3.2. i. If 0 < q< 0, where0 0.429052 is the root of(1 q)(1 q2)(1 q3) q, q (0, 1),thenthezerosofsin(z; q)arereal,inniteandsimpleandthepositivezerosbelongtotheintervals_qr+1/2_(1 q2)(1 q3), qr1/2_(1 q2)(1 q3)_, r = 1, 2, . . . , (2.3.3)47and each interval contains only one zero.ii. If 0 < q< 0, where0 0.38197 is the zero of (1 q)3(1 + q) qin 0 < q< 1, thenthe zeros of cos(z; q) are real and simple and the positive zeros belong to the intervals_qr(1 q)_1 +q, qr1(1 q)_1 +q_, r = 0, 1, 2, . . . , (2.3.4)and each interval contains only one zero.Proof. To prove i, substitute with = 1/2 in (2.3.1). Then (2.3.1) becomesq1(1 q)(1 q2)(1 q)3> 1whichholdsif0 1, then the zeros urr=0 of theq-functionU,f:= z_10tf(t)J(tz; q2) dqt = z_J(z; q2) +qJ(qz; q2)_, (2.3.8)are real, innite, simple andur _qr+1/2/_C,f, qr1/2/_C,f_,r = 0, 1, 2, . . ..In the following we introduce anotherq-analogue of Theorem 1.4.1.Theorem2.3.3. Letf L1q(0, 1) be positive on 0, qn, n N. Ifq1(1 q) c1/2,fC1/2,f> 1, (2.3.9)then the zeros of the entire function of order zeroUf(z) :=_10f(t) cos(tz; q) dqt, z C, (2.3.10)arereal, simpleandinnite. Moreover Uf(z) is anevenfunctionwithnozeros intheinterval _0, q1/2/__C1/2,f(1 q)__, and its positive zeros lie in the intervals_qr/_(1 q)_C1/2,f), qr1/_(1 q)_C1/2,f)_, r = 0, 1, . . . , (2.3.11)one zero in each interval.Proof. From (1.1.16) we conclude thatUf(z) =(q2; q2)(q; q2)U1/2,f(z(1 q)q1/2). (2.3.12)49ApplyingTheorems 2.2.2toU1/2,f(z(1 q)q1/2) where q-satises (2.3.9), proves thetheorem.Similarly, we have the following theorem.Theorem2.3.4. Letf L1q(0, 1) be positive on 0, qn, n N. Ifq1(1 q) b1/2,fB1/2,f> 1, (2.3.13)then the zeros of the entire function of order zeroVf(z) :=_10f(t) sin(tz; q) dqt, z C, (2.3.14)are real, simple and innite. Moreover Vf(z) is an odd function with exactly one zero, z = 0,in the interval [0,q1B1/2,f(1q)), and its positive zeros are located in the intervals_qr+1/2(1 q)_B1/2,f,qr1/2(1 q)_B1/2,f_, r = 0, 1, . . . , (2.3.15)one in each interval.A remarkable dierence between Polyas theorem and its basic analogue here is that inTheorems2.3.3and2.3.4wedonothavethemonotonicitycondition. Howeverthepricefor this are the restrictions (2.3.9) and (2.3.13) onq. This allowed us to prove thatUf(z)andVf(z) have no non-real zeros.PartIIBasicDierenceEquations50Chapter3IntroductionIn this introductory chapter we briey introduce the main concepts and theorems of linearq-dierence equations as stated in[8,9,90]. We start with the existence and uniqueness the-orem of q-Cauchy problems based on q-analogue of the celebrated successive approximationmethod. Then we state some results concerningq-linear dierence equations.3.1 q-successiveapproximationsIn this section we dene the q-successive approximations associated with q-Cauchy problems.This technique is the main method we use to prove the theorems of existence as in [9, 90].Proofs will not be given here. The interested reader could nd them in [9, 90].Denition3.1.1. Letr, s, andni,i = 0, 1, . . . , r, be positive integers and letN= (n0 + 1) +. . . + (nr + 1) 1.LetFj(x, y0, y1, . . . , yN) ,j = 0, 1, . . . , s, be real or complexvalued functions, wherex is areal variable lying in some intervalIand eachyi is a complex variable lying in some regionDiofthecomplexplane. Ifthereareasubinterval Jof Iandfunctionsi, 0 i r,dened inJsuch that51521. i hasniq-derivatives inJfor 0 i r ;2. Dmqiexists and lies in the regionDifor all x inJ, 0 m ni, and 0 i r, forwhich the lefthand side in equation (3.1.1) below is dened;3. for allx inJand 0 j s, the following equations holdFj_x, 0(x), Dq0(x), . . . , Dn0q0(x), . . . , r(x), . . . , Dnrqr(x)_= 0; (3.1.1)then we say that iri=0 is a solution of the system of theq-dierence equationsFj_x, y0(x), Dqy0(x), . . . , Dn0qy0(x), . . . , yr(x), . . . , Dnrqyr(x)_= 0, (3.1.2)0 j s, valid inJ, or that the set iri=0satises (3.1.2) inJ. If there are no suchJand functionsi, we say that the system (3.1.2) has no solutions.System (3.1.2) is said to be of ordern, wheren := max0ipni. If the functionsFjaresuch that the equations (3.1.2) can be solved for theDniqyi in the formDniqyi(x) = fi (x, y0(x), Dqy0(x), . . . , y1(x) . . .) , 0ip, (3.1.3)or in the formDniqyi(x) = fi (qx, y0(qx), Dq,qxy0(qx), . . . , y1(qx) . . .) , 0ip, (3.1.4)then the systems (3.1.3) and (3.1.4) will be called normal systems. The following systemsare examples of normal rst order systems,Dqyi(x) = fi_qx, y0(qx), y1(qx), . . . , yp(qx)_, i = 0, 1, . . . , p, (3.1.5)andDqyi(x) = fi_x, y0(x), y1(x), . . . , yp(x)_, i = 0, 1, . . . , p. (3.1.6)In a more restrictive manner, rst order normal linear systems have been dened in [34].Theexistenceof asolutionof system(3.1.5) or (3.1.6) inaneighborhoodof azerois53establishedin[9, 90] byusingaq-analogueof thePicard-Lindelof methodof successiveapproximations , see e.g. [37, 41]. The following is a description of thisq-analogue.Dene sequences of functions i,mm=1,i = 0, 1, . . . , p, by the equationsi,m(x) =___bi, m = 1bi +_x0fi (t, 0,m(t),. . . , p,m(t))dqt, m2,(3.1.7)wherebiareconstantswhichlieinDi. Itisworthmentioningherethattheconstantsbipi=0 should be replaced byq-periodic functions if we are looking for solutions which arenot necessarily continuous at zero.AsisstatedinTheorem3.2.2belowundersuitableconditionsonthefis, therearefunctions i and an interval J contained in I and containing zero such that i,m(x) i(x)on J as m , where by we mean the uniform convergence. Then, it will be possibleto apply Lemma 3.2.1 below to obtaini(x) = bi +_x0fi (t, 0(t), . . . , p(t))dqt. (3.1.8)Hencei(0) = bi, and ipi=0 is a solution of (3.1.6). Similarly for (3.1.5).3.2 q-InitialvalueproblemsHerewestateanexistenceanduniquenesstheoremof q-Cauchyproblemsastreatedin[9, 90].Denition3.2.1. Let biDibearbitraryvalues. Byaq-initial valueprobleminaneighborhood of zero we mean the problem of nding functions yipi=0, continuous at zeroand satisfying the system (3.1.5) or (3.1.6) and the initial conditionsyi(0) = bi, bi Di, 0ip. (3.2.1)The proof of the following lemma is straightforward and will be omitted.54Lemma3.2.1. Let I andJbeintervalscontainingzero, suchthat J I. Let fn, fbefunctions dened inI,n N, such thatlimnfn(t) = f(t), when t I, and fn f on J. (3.2.2)Then,limn_x0fn(t) dqt =_x0f(t) dqt, for all x I. (3.2.3)Theorem3.2.2. LetIbe an interval containing zero and letErbe disks of the formEr := y C : [y br[ < , > 0, br C,andr = 0, 1, . . . , p. Letfi(x, y0, y1, . . . , yp),i = 0, 1, . . . , p, be functions dened onI E0

. . . Epfor which:(i) foryr Er, 0rp,fi(x, y0, y1, . . . , yp) is continuous atx = 0, 0ip,(ii)thereisapositiveconstant Asuchthat,forx Iandyr, yr Er,0 r, i p,thefollowing Lipschiz condition is fullledfi(x, y0, . . . , yp) fi(x, y0, . . . , yp) A_[ y0y0[ +. . . +[ ypyp[_. (3.2.4)Then, if 0 isnot anendpoint of I, thereexistsapositive h suchthat the q-Cauchyproblem(3.1.6), (3.2.1) has a unique solution valid for [x[ h. Moreover if 0 is the left orright end point ofI, the result holds, except that the interval [h, h] is replaced by [0, h] or[h, 0], respectivelyTheorem3.2.3(Rangeof validity). Assumethatall conditionsofTheorem3.2.2aresatisedwithEr= Cforall r; r = 0, 1, . . . , p. Thenproblem(3.1.6),(3.2.1)hasauniquesolution valid at least inI _1A(p+1)(1q),1A(p+1)(1q)_.Remark3.2.1. Theorem3.2.2holdsfortheq-Cauchyproblem(3.1.5),(3.2.1),buttheso-lution will be valid throughout the whole intervalIiffis satisfy the conditions (i), (ii) ofTheorem 3.2.2 withEr = C, 0r, ip.55Example 3.2.1. It is known, [53], that both of the functions eq(x) and Eq(x) are q-analoguesof the exponential function. While Eq(x) is entire, eq(x) is analytic only when [x[ < 1. Theyare dened and related byeq(x) =1Eq(x)=:1

k=0_1 xqk_, x C _qn, n N_.Theorem3.2.2andRemark 3.2.1gaveusanexplanationsince eq(x) andEq(x) arethesolutions of theq-Cauchy problemsDqy(x) =11 qy(x),y(0) = 1; Dqy(x) =11 qy(qx), y(0) = 1, (3.2.5)respectively. From Theorem 3.2.3 it follows that the solution of the rst problem is valid atmost in [x[ < 1 and from Remark 3.2.1 the solution of the second one is valid in R.3.3 Linearq-dierenceequationsThissectioncontainsabrief studyof linear q-dierenceequationsof order n, n Z+.Complete treatments could be found in [8, 9, 90].For n Z+, bi C, Theorem 3.2.2 can be applied to discuss the existence and uniquenessof then-th orderq-initial value problem___Dnqy(x) = f_x, y(x), Dqy(x), . . . , Dn1qy(x)_Di1qy(0) = bi, bi C; 1in .(3.3.1)Corollary3.3.1. Letp,I,Er,brbe as in the previous theorem. Letf(x, y0, y1, . . . , yp) bea function dened onI E0. . . Epsuch that the following conditions are satised(i) For any xed values of theyrinEr, f(x, y0, y1, . . . , yp) is continuous at zero.(ii)fsatises Lipschitz condition with respect toy0, . . . , yp.56If 0 isaninteriorpointof I, thenthereexistsapositive hsuchthat q-Cauchyproblem(3.3.1) has a unique solution valid for[x[ h. If 0 is a left or right endpoint of I theresult holds, except that the interval[x[ his replaced by [0, h] or [h, 0] respectively.Proof. Assume that 0 is an interior point of I. The q-Cauchy problem (3.3.1) is equivalentto the rst orderq-initial value problemDqyi(x) = fi(x, y0, . . . , yn1) yi(0) = bi, ; 1in,(3.3.2)inthesensethat in1i=0isasolutionof(3.3.2)ifandonlyif 0isasolutionof(3.3.1).Herefi are the functionsfi(x, y0, . . . , yn1) =___yi+1, 0in 2f(x, y0, . . . , yn1), i = n.(3.3.3)By Theorem (3.2.2) there existsh > 0 such that system (3.3.2) has a unique solution validin [x[h.Corollary3.3.2. Consider theq-Cauchy problem___a0(x)Dnqy(x) +a1(x)Dn1qy(x) +. . . +an(x)y(x) = b(x)Diq(0) = bi, 0in 1 .(3.3.4)Lettheaj(x), 0 j n, andb(x)bedenedonaninterval Icontainingzerosuchthata0(x) ,= 0forall xinIandthefunctionsaj(x), b(x)arecontinuousatzeroandboundedonI. Thenforanycomplexnumbersbr,thereexistsasubinterval Jof Icontainingzerosuch that (3.3.4) has a unique solution valid inJ.Proof. Dividing bya0(x), we get the equationDnqy(x) = An(x)Dn1qy(x) +. . . +A1(x)y(x) +B(x), (3.3.5)57where Aj(x) = aj(x)/a0(x), 1jn, and B(x) = b(x)/a0(x). Equation (3.3.5) is of theform (3.3.2) withf(x, y0, . . . , yn1) = An(x)y0 +. . . +A1(x)yn1 +B(x).Since Aj(x) and B(x) are continuous at zero and bounded on I, then f(x, y0, . . . , yp) satisestheconditionsofTheorem3.3.1. Hence,thereexistsasubinterval Jof Icontainingzerosuch that equation (3.3.5) has a unique solution valid inJ.Consider the non-homogeneous linearq-dierence equation of orderna0(x)Dnqy(x) +a1(x)Dn1qy(x) +. . . +an(x)y(x) = b(x), x I, (3.3.6)for whichai, 0in, andb are continuous at zero functions dened onIanda0(x) ,= 0for allx I. From Corollary 3.3.2 equation (3.3.6) together with the initial conditionsDi1qy(0) = bi, bi C, i = 1, . . . , n, (3.3.7)has a unique solution which is continuous at zero in a subinterval J of I, J = [h, h], h > 0.The correspondingn-th order homogeneous linear equation of (3.3.6) isa0(x)Dnqy(x) +a1(x)Dn1qy(x) +. . . +an(x)y(x) = 0, x I. (3.3.8)LetMdenotethesetofsolutionsof(3.3.8)validinasubsetJ I,0 J. Then,cf.[8],Mis a linear space over C of dimensionn. Also from the existence and uniqueness of thesolutions, if MandDiq(0) =0, 0in 1, then(x) 0onJ. Moreover,_Diq_n1i=0are continuous at zero for any M. A set ofn solutions of (3.3.8) is said tobe a fundamental set (f.s.)for (3.3.8) valid inJor a f.s. ofMif it is linearly independentinJ. Moreover, as in dierential equations, if bij, 1 i, j n, are numbers, and, for eachj,jis the unique solution of (3.3.8) which satises the initial conditionsDi1qj(0) = bij, 1 i n,then jnj=1 is a f.s. of (3.3.8) if and only if det(bij) ,= 0.583.4 Aq-typeWronskianIn the following we give a q-analogue of the Wronskian for n-th order qdierence equations.A study for a second order Wronskian is established by R.F. Swarttouw and H.G. Meijerin [113]. The results of the present section are also included in [8, 90].Denition3.4.1. Letyi, 1 i n, befunctionsdenedonaq-geometricsetA. Theq-Wronskian of the functions yi which will be denoted by Wq(y1, . . . , yn)(x) is dened to beWq(y1, . . . , yn)(x) :=y1(x) . . . yn(x)Dqy1(x) . . . Dqyn(x).........Dn1qy1(x) . . . Dn1qyn(x), (3.4.1)providedthat the derivatives exist inI. For convenience we write Wq(x) insteadofWq(y1, . . . , yn)(x).Theorem 3.4.1.Ify1,y2, . . . ,yn are solutions of (3.3.8) inJ I, then theirq-Wronskiansatises the rst orderq-dierence equationDqWq(x) = R(x)Wq(x), x J 0 ; R(x) =n1

k=0(x qx)kak+1(x)a0(x). (3.4.2)The following theorems gives aq-type Liouvilles formula for theq-Wronskian.Theorem3.4.2. Suppose thatx(1 q)R(x) ,= 1 whenx J. Then theq-Wronskian ofany set of solutions ini=1of equation (3.3.8) is given byWq(x) = Wq(1, . . . , n)(x) =1

k=0(1 +x(1 q)qkR(xqk))Wq(0), x J. (3.4.3)Corollary3.4.3. Let ini=1beasetof solutionsof (3.3.8)insomesubinterval Jof Iwhichcontainszero. Then Wq(x) iseitherneverzerooridenticallyzeroinI. Therstcase occurs when ini=1is a fundamental set of (3.3.8) and the second when it is not.59Example3.4.1. In this example we calculate theq-Wronskian of1qDq1Dqy(x) +y(x) = 0, x R. (3.4.4)The solutions of (3.4.4) subject to the initial conditionsy(0) = 0,Dqy(0) = 1 and y(0) = 1,Dqy(0) = 0,are sin(x; q), cos(x; q), x R, respectively. Since (3.4.4) can be written asD2qy(x) +qx(1 q)Dqy(x) qy(x) = 0, (3.4.5)thena0(x) 1,a1(x) = qx(1 q) anda2(x) = q. ThusR(x) 0 on R and Wq(x) Wq(0).ButWq(0) = Wq_cos(; q), sin(; q)_(0)= _cos(x; q) cos(qx; q) +q sin(x; q) sin(qx; q)_x=0 = 1.Then, Wq(x) 1for all x R.Example3.4.2. We calculate theq-Wronskian of solutions of theq-dierence equationD2qy(x) +y(x) = 0, x R. (3.4.6)The functions sinq x(1 q), cosq x(1 q), [x[(1 q) < 1, are solutions of (3.4.6) subjectto the initial conditionsy(0) = 0,Dqy(0) = 1 and y(0) = 1,Dqy(0) = 0,respectively. HereR(x) = x(1 q). So,x(1 q)R(x) ,= 1 for all x in R. Hence,Wq(x) =Wq(0)

n=0(1 +q2nx(1 q)2), [x[(1 q) < 1.ButWq(0) = Wq_cosq, sinq_(0) =_cosq2x + sinq2x_x=0 = 1. (3.4.7)Therefore,Wq(x) 1/

n=0(1 +q2nx(1 q)2), [x[(1 q) < 1.Chapter4BasicSturm-LiouvilleProblemsThis chapter is devoted to a study a q-analogue of the Sturm-Liouville eigenvalue prob-lem. We formulate a formally self adjointq-dierence operator in a Hilbert space. Some oftheproperties oftheeigenvaluesandtheeigenfunctions are discussed. TheGreensfunc-tion is constructed and the problem in question is inverted into aq-type Fredholm integraloperatorwithasymmetrickernel. Thesetof eigenfunctionsisshowntobeacompleteorthogonal set in the Hilbert space. Examples involving basic trigonometric functions willbe given. A study for the asymptotics of the eigenvalues and eigenfunctions will be givenat the end of this Chapter. The results of this Chapter also exists in [16, 19].4.1 IntroductionLet[a, b] beaniteclosedinterval inRandlet beacontinuousreal-valuedfunctiondened on [a, b]. By a Sturm-Liouville problem we mean the problem of nding a functiony and a number in C satisfying the dierential equationLy := y

+(x)y(x) = y(x), axb, (4.1.1)together with the boundary conditionsU1(y) := a1y(a) +a2y

(a) = 0, (4.1.2)6061U2(y) := b1y(b) +b2y

(b) = 0, (4.1.3)whereai andbi,i = 1, 2 are real numbers for which[a1[ +[a2[ , = 0 ,= [b1[ +[b2[. (4.1.4)This problem has been extensively studied. It is known that the dierential equation (4.1.1)andtheboundaryconditions(4.1.2)(4.1.3)determineaselfadjointoperatorinL2(a, b).Thereisasequenceof real numbers nn=0withlimnn=+, suchthatcorre-sponding to each n there is one and only one linearly independent solution of the problem(4.1.1)(4.1.3). Thesequence nn=0isthesequenceofeigenvaluesandthesequenceofcorrespondingsolutions nn=0issaidtobeasequenceof eigenfunctions. Oneof themost important properties of these eigenfunctions is that, nn=0is an orthogonal basisof L2(a, b). Forexample,let(x) 0on[a, b]. Ifwetakea = 0, b =, a1= 1, a2= 0,b1 = 1, andb2 = 0, we getn = n2, n(x) = sin nx, n = 1, 2, . . . , (4.1.5)leadingtothewellknownfactthat sin nxn=1isacompleteorthogonal setof L2(0, ),while takinga = 0,b = ,a1 = 0,a2 = 1,b1 = 0, andb2 = 1, we getn = n2, n(x) = cos nx, n = 0, 1, 2, . . . , (4.1.6)whichleadstothecompletenessof cos nxn=0inL2(0, ). WementionherethattheFourierorthogonal basis _einx_n=0of L2(, )doesnotariseinthissettingbutarisesfrom a simpler situation, namely the rst order problemiy

= y, y() = y(). (4.1.7)Amongseveral referencesfortheabovementionedfactswementionthemonographsofCoddington and Levinson [37], Estham [41], Levitan and Sargsjan [85, 86], Marchenko [91]and Titchmarsh [114].62The discrete analogue of the theory outlined above, i.e. when the dierential operatord/dx is replaced by the forward dierence operator y(n) = y(n+1)y(n) or the backwardoperator y(n) = y(n) y(n 1) wheren is a positive integer belonging to a nite set ofintegers of the form m, m+ 1, m+ 2, . . . , m+N,m1, is treated in Atkinsons [23], seealso [73].The aim of this chapter is to study a basic analogue of SturmLiouville systems whenthedierentialoperatorisreplacedbytheq-dierenceoperatorDq,see(1.1.28)(1.1.29).In [48, Chapter 5] and [49], a basic Sturm-Liouville system is dened. It is the systemDq(r(x)Dqy) +_l(x) +w(x)_y(qx) = 0, axb, (4.1.8)h1y(a) +h2Dqy(a) = 0, (4.1.9)k1y(b) +k2Dqy(b) = 0. (4.1.10)wherer, landware real-valued functions which posses appropriateq-derivatives, h1, h2,k1, k2 are constants. It is proved[48, pp.164-170] that all eigenvalues of this system are realand the eigenfunctions satisfy an orthogonality relation, [48, equation(5.1.5)]. Exton, [48],considered only formal computational aspects to prove an orthogonality relation of some q-special functions. No attention is paid to several points, which may lead to several mistakes.First the existence of eigenvalues is not proved and it is not indicated how to determine theeigenvalues and the eigenfunctions. A basic point here is that ifa ,= 0 ,=b,then it is notguaranteedthatinitialconditionsateitheraorbdetermineauniquesolutionof(4.1.8),see [9, 90]. The geometric and algebraic simplicity of the eigenvalues, which plays a majorruleinprovingtherealityoftheeigenvaluesandtheorthogonalityoftheeigenfunctionsarenotprovedorevenassumed. Moreoverthespacewheretheproblemisdenedisnotspecied. Ifaninnerproductisdenedintheviewof [48, equation(5.1.5)], therewillbe no orthogonality ifh1,h2,k1andk2are not real. For more information concerning themonograph [48], see the Review by Mourad Ismail in [SIAM Rev. 16(1985), pp. 279-281].63SeealsotheReviewof [49]byWolfgangHahnin[MathRev. 84i:39002(1982)]. Inthepresentpaperweformulateaself adjointbasicSturmLiouvilleeigenvalueprobleminaHilbert space. We prove the existence of a sequence of real eigenvalues with no nite limitpoints. The associated Greens function is constructed and the equivalence between the basicSturm-Liouvilleproblemandaq-typeFredholmintegral operatorisproved. Illustrativeexamples are given at the end of the paper.Thereareseveral physical modelsinvolvingq-derivatives, q-integrals, q-functionsandtheir related problems,see e.g. [36, 51, 52, 55, 108, 109]. Also the problem of expansions offunctions in terms of q-orthogonal functions, which seems to be rst discussed by Carmichaelin [33, 34] has attracted the work of several authors, see e.g. [6, 28, 29, 64, 109, 111]. However,asfarweknow,thegeneralprobleminthepresentsettinghasnotbeenstudied. Atthispoint, itisworthmentioningthatourworkbasedontheq-dierenceoperatorwhichisattributed to Jackson, see [69], and a similar study of the Sturm-Liouville systems generatedby the Askey-Wilson divided dierence operator, cf. [21] is very much needed. The resultsof this chapter except Section 4.6 are contained in [16].4.2 FundamentalsolutionsIn this section we investigate the fundamental solutions of the basic Sturm-Liouville equa-tion1qDq1Dqy(x) +(x)y(x) = y(x), 0xa < , C, (4.2.1)where is dened on [0, a] and continuous at zero. Let C2q(0, a) be the space of all functionsydened on [0, a] such thaty,Dqyare continuous at zero. Clearly,C2q(0, a) is a subspaceof the Hilbert spaceL2q(0, a).Theorem4.2.1. Forc1, c2 C,equation(4.2.1)hasauniquesolutioninC2q(0, a)which64satises(0, ) = c1, Dq1(0, ) = c2, C. (4.2.2)Moreover(x, ) is entire in for all x [0, a].Proof. The functions1(x, ) = cos(sx; q), and 2(x, ) =___sin(sx; q)s, ,= 0x, = 0, (4.2.3)wheres := is dened with respect to the principal branch, is a fundamental set of1qDq1Dqy(x) +y(x) = 0, (4.2.4)withWq_1(, ), 2(, )_ 1on[0, a], C. Forall x [0, a], C, wedeneasequence ym(, )m=1 of successive approximations byy1(x, ) = c11(x, ) +c22(x, ), (4.2.5)ym+1(x, ) = c11(x, ) +c22(x, )+q_x02(x, )1(qt, ) 1(x, )2(qt, ) (qt) ym(qt, ) dqt. (4.2.6)Weshall provethatforeachxed Ctheuniformlimitof ym(, )m=1asm exists and denes a solution of (4.2.1) and (4.2.2). Let C be xed. There exist positivenumbersK() andA such that[(x)[A, [i(x, )[ _K()2; i = 1, 2; x [0, a]. (4.2.7)Let K():= _[c1[ + [c2[__K()2. Then, from(4.2.7), [y1(x, )[ K(), forall x [0, a].Using mathematical induction, we have[ym+1(x, ) ym(x, )[ K()qm(m+1)2(AK()x(1 q))m(q; q)m, m = 1, 2, . . . . (4.2.8)Consequently by Weierstrass test the seriesy1(x, ) +

m=1ym+1(x, ) ym(x, ) (4.2.9)65converges uniformlyin[0, a]. Sincethe m-thpartial sums of theseries is nothingbutym+1(, ), thenym+1(, )approachesafunction(, )uniformlyin[0, a] as m ,where(x, ) is the sum of the series. We can also prove by induction onm thatym(x, )andDqym(x, ) are continuous at zero, whereDqym+1(x, ) = c1Dq1(x, ) +c2Dq2(x, )+q_x0Dq2(x, )1(qt, ) Dq1(x, )2(qt, ) (qt)ym(qt, ) dqt,(4.2.10)m Z+. Hence, both(, ) andDq(, ) are continuous at zero. i.e. (, ) C2q(0, a).Because of the uniform convergence, lettingm in (4.2.6) we obtain(x, ) = c11(x, ) +c22(x, )+q_x02(x, )1(qt, ) 1(x, )2(qt, ) (qt) (qt, ) dqt.(4.2.11)We prove that(, ) satises (4.2.1) and (4.2.2). Clearly(0, ) = c1 andDq1(0, ) =limn(xqn, ) (0, )xqn= c1Dq11(0, ) +c2Dq12(0, ) = c2, (4.2.12)i.e. (, )satises(4.2.2). Toprovethat(, )satises(4.2.1),Wedistinguishbetweentwo cases,x ,= 0 andx = 0. Ifx ,= 0, then from (1.1.28),Dq(x, ) = c1Dq1(x, ) +c2Dq2(x, )+q_x0Dq2(x, )1(qt, ) Dq1(x, )2(qt, ) (qt)(qt, ) dqt(4.2.13)and hence1qDq1Dq(x, ) = 1qDq1Dq1(x, )_c1_x02(qt, )(qt)(qt, ) dqt_ 1qDq1Dq2(x, )_c2 +_x01(qt, )(qt)(qt, ) dqt_(x)(x, ).(4.2.14)Substituting from (4.2.3), (4.2.4) and (4.2.11) in (4.2.14), we conclude that(x, ) satises(4.2.1) forx ,= 0. Ifx = 0, then (4.2.1) is nothing butD2qy(0) q(0)y(0) = qy(0). (4.2.15)66From (1.1.29), one can see thatD2q(0, ) = c1D2q1(0, ) +c2D2q2(0, ) +q(0)(0, )= c1q1(0, ) c2q2(0, ) +q(0)(0, )= q(0, ) +q(0)(0, ). (4.2.16)Consequently (4.2.1) is satised atx = 0.To prove that problem (4.2.1), (4.2.2) has a unique solution, assume the contrary, thatisi(, ),i = 1, 2, are two solutions of (4.2.1), (4.2.2). Let(x, ) = 1(x, ) 2(x, ), x [0, a].Then(, ) is a solution of (4.2.1) subject to the initial conditions(0, ) = Dq1(0, ) = 0.Applying theq-integration process on (4.2.1) twice yields(x, ) =_x0(x t)_ (t)_(t, ) dqt. (4.2.17)Since(x, ), (x)arecontinuousatzero, thenthereexistpositivenumbersNx,, Mx,such thatNx, = supnN[(xqn, )[, Mx, = supnN (xqn). (4.2.18)Again we can prove by mathematical induction onk that[(x, )[Nx,Mkx,qk2(1 q)2kx2k(q; q)2k, k N, x [0, a]. (4.2.19)Indeed, if (4.2.19) holds atk N, then from (4.2.17)[(x, )[Nx,Mk+1x,qk2 (1 q)2k(q; q)2k_x0(x t)t2kdqt= Nx,Mk+1x,qk2q2k+1(1 q)2k+2(q; q)2k+2x2k+2= Nx,Mk+1x,q(k+1)2 (1 q)2k+2(q; q)2k+2x2k+2. (4.2.20)67Hence (4.2.19) holds at k+1 and therefore (4.2.19) holds for all k N because from (4.2.18)it holds at k = 0. Since limkMkx,qk2(1q)2kx2k(q; q)2k= 0, then (x, ) = 0, for all x [0, a].This proves the uniqueness. Now, letM> 0 be arbitrary but xed. To prove that(x, ),x [0, a], isentirein, itissucienttoprovethat(x, )isanalyticineachdiskM;M:= C : [[M. We prove by induction onm thatfor all x [0, a] ym(x, ) is analytic on M, (4.2.21)for all Mym(x, ) is continuous at (0, ). (4.2.22)Clearly, 1(x, ),2(x, )areentirefunctionsof forx [0, a]. Moreoveri(x, )iscontinuous at (0, ) for each C. Then (4.2.21) and (4.2.22 ) holds atm = 1. Assumethat (4.2.21) and (4.2.22) hold atm1. Then forx0 [0, a],0 M, we obtainym+1(x0, )=0= q2(x0, )=0_x001(qt, )ym(qt, ) dqt+y1(x0, )=0q1(x0, )=0_x002(qt, )ym(qt, ) dqt+q2(x0, )__x001(qt, )ym(qt, ) dqt_=0q1(x0, )__x002(qt, )ym(qt, ) dqt_=0(4.2.23)From (4.2.22) we conclude that /_i(qt, )ym(qt, )_, i = 1, 2, are continuous at (0, 0).Therefore there exist constantsC, > 0 such that_i(x0qn, )ym(x0qn, )_ C, n N, [ 0[. (4.2.24)Hencex0(1 q)qn_i(x0qn+1, )ym(x0qn+1)_ x0A(1 q)qn, n N,for all in the disk [ 0[, i.e. the series corresponding to theq-integrals_x00_i(qt, )ym(qt, )_dqt, i = 1, 2 (4.2.25)68areuniformlyconvergent inaneighborhoodof =0. Thus, wecaninterchangethedierentiation and integration processes in (4.2.23). Sincex0, 0 are arbitrary, thenym+1(x, ) =y1(x, ) q_x0_2(x, )1(qt, )ym(qt, )_(qt) dqt+q_x0_1(x, )2(qt, )ym(qt, )_(qt) dqt,(4.2.26)for all x [0, a], M. From (4.2.22) the integrals in (4.2.26) are continuous at (0, ).Consequentlyym+1(x, ) is continuous at (0, ). Letx0 [0, a] be arbitrary. Then thereexistsB(x0), B(x0) > 0 such that[i(x0, )[ _B(x0)2, i = 1, 2, [y1(x, )[ B(x0), M. (4.2.27)Finally the use of the mathematical induction yields[ym+1(x0, ) ym(x0, )[ B(x0)qm(m+1)2(AB(x0)(1 q))m(q; q)m. (4.2.28)Consequently the series (4.2.9), with x = x0, converges uniformly in Mto (x0, ). Hence(x0, ) is analytic in M, i.e. it is entire.4.3 TheselfadjointproblemIn this section we dene a basic Sturm-Liouville problem and prove that it is formally selfadjoint in L2q(0, a). The following lemma which is needed in the sequel indicates that unliketheclassical dierential operator d/dx, Dqis neither self adjoint nor skewself adjoint.Equation (4.3.2) below indicates that the adjoint ofDqis 1qDq1.Lemma4.3.1. Letf,ginL2q(0, a) be dened on [0, q1a]. Then, forx (0, a], we have(Dqg)(xq1) = Dq,xq1g(xq1) = Dq1g(x), (4.3.1)Dqf, g) = f(a)g(aq1) limnf(aqn)g(aqn1) +f,1qDq1g_, (4.3.2) 1qDq1f, g_= limnf(aqn1)g(aqn) f(aq1)g(a) +f, Dqg) . (4.3.3)69Proof. Relation (4.3.1) follows fromDq1g(x) =g(x) g(q1x)x(1 q1)=g(xq1) g(x)xq1(1 q)= (Dqg)(xq1) = Dq,xq1g(xq1). (4.3.4)Using the formula (1.1.41) ofq-integration by parts we obtainDqf, g) =_a0Dqf(x)g(x) dqx= f(a)g(a) limnf(aqn)g(aqn) _a0f(qt)Dqg(t) dqt= f(a)g(a) limnf(aqn)g(aqn) _qa0f(t)1qDq1g(t) dqt= f(a)g(a) limnf(aqn)g(aqn) +aq1(1 q)f(a)Dq1g(a)+_a0f(t)1qDq1g(t) dqt= f(a)g(aq1) limnf(aqn)g(aqn1) +f,1qDqg_, (4.3.5)proving (4.3.2). Equation (4.3.3) can be proved by a use of (4.3.2).Now consider the basic Sturm-Liouville problem(y) := 1qDq1Dqy(x) +(x)y(x) = y(x), 0xa < , C, (4.3.6)U1(y) := a11y(0) +a12Dq1y(0) = 0, (4.3.7)U2(y) := a21y(a) +a22Dq1y(a) = 0, (4.3.8)where is a continuous at zero real valued function and aij , i, j 1, 2 are arbitrary realnumbers such that the rank of the matrix (aij)1i,j2 is 2. Problem (4.3.6)(4.3.8) is said tobe formally self adjoint if for any functions y and z of C2q(0, a) which satisfy (4.3.7)(4.3.8),y, z) = y, z) . (4.3.9)Theorem 4.3.2. The basic Sturm- Liouville eigenvalue problem (4.3.6)(4.3.8) is formallyself adjoint.70Proof. We rst prove that fory,z inL2q(0, a), we have the followingq-Lagranges identity_a0_y(x)z(x) y(x)z(x)_dqx = [y, z](a) limn[y, z](aqn), (4.3.10)where[y, z](x) := y(x)Dq1z(x) Dq1y(x)z(x). (4.3.11)Applying (4.3.3) withf(x) = Dqy(x) andg(x) = z(x), we obtain 1qDq1Dqy(x), z(x)_= (Dqy)(aq1)z(a) +limn(Dqy)(aqn1)z(aqn) +Dqy, Dqz_= Dq1y(a)z(a) +limnDq1y(aqn)z(aqn) +Dqy, Dqz_.(4.3.12)Applying (4.3.2) withf(x) = y(x),g(x) = Dqz(x),Dqy, Dqz_= y(a)Dqz(aq1) limny(aqn)Dqz(aqn1) +y, 1qDq1Dqz_= y(a)Dq1z(a) limny(aqn)Dq1z(aqn) +y, 1qDq1Dqz_. (4.3.13)Therefore, 1qDq1Dqy(x), z(x)_= [y, z](a) limn[y, z](aqn) +y, 1qDq1Dqz_. (4.3.14)Lagrangesidentity(4.3.10)resultsfrom(4.3.14)andtherealityof (x). Lettingy, zinC2q(0, a) and assuming that they satisfy (4.3.7)(4.3.8), we obtaina11y(0) +a12Dq1y(0) = 0, a11z(0) +a12Dq1z(0) = 0. (4.3.15)The continuity ofy,z at zero implies that limn[y, z](aqn) = [y, z](0). Then (4.3.14) willbe 1qDq1Dqy(x), z(x)_= [y, z](a) [y, z](0) +y, 1qDq1Dqz_. (4.3.16)Sincea11 anda12 are not both zero, it follows from (4.3.15) that[y, z](0) = y(0)Dq1z(0) Dq1y(0)z(0) = 0.71Similarly,[y, z](a) = y(a)Dq1z(a) Dq1y(a)z(a) = 0.Since(x) is real valued, then(y), z) = 1qDq1Dqy(x) +(x)y(x), z(x)_= 1qDq1Dqy(x), z(x)_+(x)y, z(x))= y, 1qDq1Dqz(x)_+y, (x)z(x)_= y, (z)) ,i.e. (4.3.6)(4.3.8) is a formally self adjoint operator.A complex number is said to be an eigenvalue of the problem (4.3.6)(4.3.8) if thereis a non trivial solutionwhich satises the problem at this. In this case we say that is an eigenfunction of the basic Sturm-Liouville problem corresponding to the eigenvalue. Thegeometricmultiplicityof aneigenvalueisdenedtobethenumberof linearlyindependent solutions corresponding to it. In particular an eigenvalue is simple if and onlyif it has only one linearly independent solution.Lemma 4.3.3. The eigenvalues andthe eigenfunctions of the boundaryvalue problem(4.3.6)(4.3.8) have the following properties:i. The eigenvalues are real.ii. Eigenfunctions that belong to dierent eigenvalues are orthogonal.iii. All eigenvalues are simple from the geometric point of view.Proof. i. Let0 be an eigenvalue with an eigenfunctiony0(). From (4.3.9), we have(y0), y0) = y0, (y0)) . (4.3.17)But(y0) = 0y0, implying_00__a0[y(x)[2dqx = 0. (4.3.18)72Sincey0() is non-trivial then0 = 0, which proves i.ii. Let, be two (real) distinct eigenvalues with corresponding eigenfunctionsy(),z(),respectively. Again, we make use of (4.3.9) to get( )_a0y(x)z(x) dqx = 0. (4.3.19)Since ,= , theny() andz() are orthogonal, proving ii.iii. Let0beaneigenvaluewithtwoeigenfunctionsy1()andy2(). Weprovethattheyarelinearlydependentbyprovingthattheirq-Wronskianvanishesatx = 0,see[8]foracomplete treatment of linearq-dierence equations . Indeed,Wq(y1, y2)(0) = y1(0)Dqy2(0) y2(0)Dqy1(0) = 0, (4.3.20)since bothy1 andy2 satisfy (4.3.7).In the following we indicate how to obtain the eigenvalues and the corresponding eigen-functions. Let1(, ),2(, ) be the linearly independent solutions of (4.3.6) determinedby the initial conditionsDj1qi(0, ) = ij, i, j = 1, 2, C. (4.3.21)Thus1(, ) is determined by (4.2.5) by takingc1= 1, c2= 0 and2(, ) is determinedby takingc1 = 0,c2 = 1. Then, every solution of (4.3.6) is of the formy(x, ) = A11(x, ) +A22(x, ), (4.3.22)where A1 and A2 does not depend on x. A solution y(, ) of (4.3.6) will be an eigenfunctionif it satises the boundary conditions (4.3.7)(4.3.8), i.e. if we can nd a non trivial solutionof the linear systemA1U1(1) +A2U1(2) = 0,A1U2(1) +A2U2(2) = 0,(4.3.23)73Hence, in R is an eigenvalue if and only if() =U1(1) U1(2)U2(1) U2(2)= 0. (4.3.24)Thefunction()denedin(4.3.24)iscalledthecharacteristicdeterminantassociatedwith the basic Sturm- Liouville problem (4.3.6)(4.3.8). The zeros of () are exactly theeigenvalues of the problem. Since1(x, ) and2(x, ) are entire in for each xedx in[0, a], then()isalsoentire. ThustheeigenvaluesofthebasicSturm-Liouvillesystem(4.3.6)(4.3.8)areatmostcountablewithnonitelimitpoints. FromLemma4.3.3weknow that all eigenvalues are simple from the geometric point of view. We can prove thatthe eigenvalues are also simple algebraically, i.e. they are simple zeros of (). Indeed, let1(, ) and2(, ) be dened by the relations1(x, ) := U1(2)1(x, ) U1(1)2(x, ),2(x, ) := U2(2)1(x, ) U2(1)2(x, ).(4.3.25)Hence,1(, ),2(, ) are solutions of (4.3.6) such that1(0, ) = a12, Dq11(0, ) = a11; 2(a, ) = a22, Dq12(a, ) = a21. (4.3.26)One can verify thatWq_1(, ), 2(, )_(x, ) = ()Wq_1(, ), 2(, )_= (). (4.3.27)Let0 be an eigenvalue of (4.3.6)(4.3.8). Then0 is a real number and thereforei(x, 0)can be taken to be real valued, i = 1, 2. From (4.3.27), we conclude that 1(x, 0), 2(x, 0)are linearly dependent eigenfunctions. So, there exists a non zero constantk0 such that1(x, 0) = k02(x, 0). (4.3.28)From (4.3.25) and (4.3.26)1(a, 0) = k0a22 = k01(a, ), Dq11(a, 0) = k0a21 = k0Dq11(a, ). (4.3.29)74In theq-Lagrange identity (4.3.10), takingy(x) = 1(x, ), andz(x) = 1(x, 0) implies_ 0__a01(x, )1(x, 0) dqx = 1(a, )Dq11(a, 0) Dq11(a, )1(a, 0)= k0_1(a, )Dq12(a, ) 2(a, )Dq11(a, )_= k0Wq(1(, ), 2(, ))(q1a) = k0().Since () is entire in,

(0) :=lim0() 0=1k0_a021(x, 0) dqx ,= 0. (4.3.30)Therefore0 is a simple zero of ().4.4 BasicGreensfunctionThe q-type Greens function arises when we seek a solution of the nonhomogeneous equation1qDq1Dqy(x) + +(x) y(x) = f(x), x [0, a], C, (4.4.1)which satises the boundary conditions (4.3.7)(4.3.8), wheref L2q(0, a) is given. First,we note that if is not an eigenvalue of the Sturm-Liouville problem (4.3.6)(4.3.8), thenthesolutionof (4.4.1), if itexists, wouldbeunique. Toseethis, assumethat 1(x, ),2(x, ) are two solutions of (4.4.1). Then1(x, ) 2(x, ) is a solution of the problem(4.3.6)(4.3.8). So, itisidenticallyzeroif isnotaneigenvalue. Anotherproofofthisassertion is included in the proof of the next theorem.Theorem4.4.1. Supposethat isnot aneigenvalueof (4.3.6)(4.3.8). Let (, )sat-isfythe q-dierenceequation(4.4.1) andtheboundaryconditions (4.3.7)(4.3.8), wheref L2q(0, a). Then(x, ) =_a0G(x, t, )f(t) dqt, x 0, aqm,m N , (4.4.2)75whereG(x, t, ) is the Greens function of problem (4.3.6)(4.3.8) and it given byG(x, t, ) =1()___2(x, )1(t, ) , 0tx,1(x, )2(t, ) , x < ta.(4.4.3)Conversely the function (, ) dened by (4.4.2) satises (4.4.1) and (4.3.7)(4.3.8). GreensfunctionG(x, t, ) is unique in the sense that if there exists another function G(x, t, ) suchthat (4.4.2) is satised, thenG(x, t, ) = G(x, t, ), in L2q_(0, a) (0, a)_. (4.4.4)Iffisq-regular at zero, then(4.4.2) holds for all x [0, a].Proof. Using aq-analogue of the methods of variation of constants, a particular solutionof the non-homogenous equation (4.4.1) may be given by(x, ) = c1(x)1(x, ) +c2(x)2(x, ), (4.4.5)wherec1(x),c2(x) are solutions of the rst orderq-dierence equationsDq,xc1(x) =q()2(qx, )f(qx), Dq,xc2(x) = q()1(qx, )f(qx). (4.4.6)Dene theq-geometric setAfbyAf:=_x [0, a] :_x0i(qt, )f(qt) dqt exists, i = 1, 2_. (4.4.7)Afisaq-geometricsetcontaining 0,aqm,m NbecausefL2q(0, a). FromHolderinequality we obtain_x0[i(qt, )f(qt)[ dqt __x0[i(qt, )[2dqt_1/2__x0[f(t)[2dqt_1/2, x Af.Sincef,i L2q(0, a), i=1, 2and() ,=0, thenDqci(), i=1, 2, areqintegrableon[0, x] for allx Af. Hence direct computations lead to the following appropriate solutionsof (4.4.6)c1(x) = c1(0) +q()_x02(qt, )f(qt) dqt, x Af(4.4.8)76c2(x) = c2(a) +q()_ax1(qt, )f(qt) dqt, x Af. (4.4.9)That is the general solution of (4.4.1) is given by(x, ) = c11(x, ) +c22(x, ) +q()1(x, )_x02(qt, )f(qt) dqt+q()2(x, )_ax1(qt, )f(qt) dqt, (4.4.10)wherex Af, andc1,c2arearbitraryconstants. Now, wedeterminec1, c2forwhich(x, ) satises (4.3.7)(4.3.8). It easy to see that(0, ) = c11(0, ) +_c2 +q()_a01(qt, )f(qt) dqt_2(0, ),Dq1(0, ) = limnxAf(xqn, ) (0, )xqn= c1Dq11(0, ) +_c2 +q()_a01(qt, )f(qt) dqt_Dq12(0, ).The boundary conditiona11(0, ) +a12Dq1(0, ) = 0 implies that_c2 +q()_a01(qt, )f(qt) dqt_Wq(1, 2)(0) = 0. (4.4.11)Therefore,c2 =q()_a01(qt, )f(qt) dqt. (4.4.12)Hence,(x, ) = c11(x, ) +q()_x0_1(x, )2(qt, ) 2(x, )1(qt, )_f(qt) dqt. (4.4.13)Now we compute(a, ) andDq1(a, ). Indeed, from the denition of theq-integration(1.1.31) and relation (4.4.10)(a, ) = c11(a, ) +q()_a0_1(a, )2(qt, ) 2(a, )1(qt, )_f(qt) dqt= c11(a, ) +q()_q1a0_1(a, )2(qt, ) 2(a, )1(qt, )_f(qt) dqt77andDq1(a, ) = Dq11(a, )_c1 +q()_q1a02(qt, )f(qt) dqt_q()Dq12(a, )_q1a01(qt, )f(qt) dqt.The boundary conditiona212(a, ) +a22Dq12(a, ) = 0 implies_c1 +q()_q1a02(qt, )f(qt) dqt_Wq(1, 2)(a) = 0. (4.4.14)Hencec1 =q()_q1a02(qt, )f(qt) dq. (4.4.15)So forx Af(x, ) =q()2(x, )_x01(qt, )f(qt) dqt q()1(x, )_q1ax2(qt, )f(qt) dqt=1()2(x, )_qx01(t, )f(t) dqt 1()1(x, )_aqx2(t, )f(t) dqt=1()2(x, )_x01(t, )f(t) dqt 1()1(x, )_ax2(t, )f(t) dqt.proving(4.4.2)(4.4.3). Conversely, bydirectcomputations, if (x, )isgivenby(4.4.2),then it is a solution of (4.4.1) and satises the boundary conditions (4.3.7)(4.3.8). To provethe uniqueness, suppose that there exists another function, G(x, t, ), such that(x, ) =_a0G(x, t, ) f(t) dqt, (4.4.16)is a solution of (4.4.1) which satises (4.3.7)(4.3.8). For convenience, letG(x, t, ) =___G1(x, t, ), 0tx,G2(x, t, ), xta.,G(x, t, ) =___G1(x, t, ), 0tx,G2(x, t, ), xta.By subtraction, we see that_a0_G(x, t, ) G(x, t, )_f(t) dqt = 0, x 0,aqm,m N , (4.4.17)78holds for all functionsf(t) L2q(0, a). Let us takef(t) := G(x, t, ) G(x, t, ), x aqm,m N .Then_a0G(aqm, t, ) G(aqm, t, )2dqt=_aqm0G1(aqm, t, ) G1(aqm, t, )2dqt +_aaqmG2(aqm, t, ) G2(aqm, t, )2dqt= a(1 q)

n=0qnG(aqm, aqn, ) G(aqm, aqn, )2= 0 (4.4.18)Therefore, from (4.4.18) we conclude thatG(aqm, aqn, ) = G(aqm, aqn, ), m,n N.Iffisq-regular at zero, thenAf [0, a] and (4.4.2) will be dened for allx [0, a].Theorem4.4.2. Basic Greens function has the following properties(i)G(x, t, ) is continuous at the point (0, 0).(ii)G(x, t, ) = G(t, x, ).(iii) For each xed t (0, qa], as a function of x, G(x, t, ) satises the q-dierence equation(4.3.6) intheintervals[0, t), (t, a] andit alsosatisestheboundaryconditions(4.3.7)(4.3.8).(iv) Let0be a zero of (). Then0can be a simple pole of the functionG(x, t, ), andin this caseG(x, t, ) = 0(x)0(t) 0+ G(x, t, ), (4.4.19)where G(x, t, )isanalyticfunctionof inaneighborhoodof 0and0isanormalizedeigenfunction corresponding to0.79Proof. (i) Follows from the continuity of1(, ),2(, ) at zero for each xed C and(ii) is easily checked. Now, we prove (iii). Lett (0, qa] be xed. Ifx [0, t], thenG(x, t, ) =1()1(x, )2(t, ).So,G(x, t, ) =1()2(t, )1(x, ) =()2(t, )1(x, ) = G(x, t, ).Similarly ifx [t, a]. From (4.3.26) and (4.4.3), we havea11G(0, t, ) +a12Dq1G(0, t, ) =1()2(t, )_a111(0, ) +a12Dq11(0, )_= 0,a21G(a, t, ) +a22Dq1G(a, t, ) =1()1(t, )_a211(a, ) +a22Dq11(a, )_= 0.(iv) Let0 be a pole ofG(x, t, ), andR(x, t) be the residue ofG(x, t, ) at = 0. From(4.3.28) and (4.3.30), we obtainR(x, t) = lim0( 0)G(x, t, ) = k101(x, 0)1(t, 0) lim0 0()= 1(x, 0)1(t, 0)_a0 [1(u, )[2dqu= 0(x, 0)1(t, 0).4.5 EigenfunctionexpansionsIn this section, the existence of a countable sequence of eigenvalues of with no nite limitpointswill beprovedbyusingthespectral theoremofcompactselfadjointoperatorsinHilbert spaces, see e.g. [24]. Moreover it will be proved that the corresponding eigenfunctionsformanorthonormal basisof L2q(0, a). Wedenetheoperator L: TL L2q(0, a)tobeLy=yfor all y TL,where TLis the subspace ofL2q(0, a) consisting of those complexvalued functionsy that satises (4.3.7)(4.3.8) such thatDqy isq-regular at zero andD2qylies inL2q(0, a). Thus L is the dierence operator generated by the dierence expressionand the boundary conditions (4.3.7)(4.3.8). By L(y) =y, we mean that(y) =yand80ysatises(4.3.7)(4.3.8). Theoperator Lhasthesameeigenvaluesof thebasicSturm-Liouville problem (4.3.6)(4.3.8). We assume without any loss of generality that = 0 isnot an eigenvalue. Thus ker L = 0. From the previous section the solution of the problem(Ly)(x) = f(x), f L2q(0, a), (4.5.1)is given uniquely inL2q(0, a) byy(x) =_a0G(x, t)f(t) dqt, (4.5.2)whereG(x, t) = G(x, t, 0) =___c1(t)2(x), 0txc1(x)2(t), xta,, c := 1Wq(1, 2). (4.5.3)Replacingfbyy in (4.5.1). Then the eigenvalue problem(Ly)(x) = y(x), (4.5.4)is equivalent to the following basic Fredholm integral equation of the second kindy(x) = _a0G(x, t)y(t) dqt, in L2q(0, a). (4.5.5)Let (be the integral operator( : L2q(0, a) L2q(0, a), ((f)(x) =_a0G(x, t)f(t) dqt. (4.5.6)We prove that(L()f= f, f L2q(0, a). (4.5.7)We show rst thaty = (f TL. From (4.5.5) and (4.5.6)y(x) = ((f)(x) = 2(x)y1(x) +1(x)y2(x), (4.5.8)81wherey1(x) = c_x01(t)f(t) dqt, y2(x) = c_ax2(t)f(t) dqt.Thus, for allx Af, cf. (4.4.7),Dqy(x) = Dq2(x)y1(qx) +Dq1(x)y2(qx), (4.5.9)D2qy(x) = q(qx)(y)(qx) qf(qx) L2q(0, a). (4.5.10)SinceDqi(x, ),yi(x),i = 1, 2, areq-regular at zero, then so isDqy andDq1y(0) = Dqy(0) =limnxAfy(xqn) y(0)xqn= Dq12(0)y2(0),y1(0) = 0, andy2(a) = 0, thena11y(0) +a12Dq1y(0) =_a111(0) +a12Dq11(0)_y2(0) = 0,anda21y(a) +a22Dq1y(a) =_a212(a) +a22Dq12(a)_y1(a) = 0.Thusy TL. It follows from (4.5.10) that Ly = (L()(f) = f. Hence we have established(4.5.7). Also, we can see that((L)(y) = y, y TL. (4.5.11)Indeed, replacing fin (4.5.7) by Ly, we get Ly = L(Ly. Thus y = (Ly since L is assumedto be injective. It follows from (4.5.7) and (4.5.11) that ker ( = 0 and is an eigenfunctionof (with eigenvalue if and only if is an eigenfunction of L with eigenvalue 1/.Theorem4.5.1. The operator (is compact and self adjoint.Proof. Letf,h, L2q(0, a). SinceG(x, t) is a real valued function dened on [0, a] [0, a]andG(x, t) = G(t, x), then forf, h L2q(0, a),((f), h) =_a0((f)(x)h(x) dqx =_a0_a0G(x, t)f(t)h(x) dqt dqx=_a0f(t)__a0G(t, x)h(x) dqx_dqt = f, ((h)) ,82i.e. (is self adjoint. Letij(x, t) = i(x)j(t)i,j=1be an orthonormal basis of L2q_(0, a)(0, a)_. ConsequentlyG = i,j=1G, ij_ij. Forn Z+setGn =

ni,j=1G, ij_ij, and let (n be the nite rank integral operator denedonL2q(0, a) by(n(f)(x) :=_a0Gn(x, t)f(t) dqt, in L2q(0, a). (4.5.12)Obviously (n is compact for alln N. From the Cauchy-Schwarz inequality|(( (n)(f)| =__a0[(( (n)(f)(x)[2dqt_1/2=__a0_a0(GGn)(x, t)(f)(t) dqt2dqx_1/2

__a0_a0[(GGn)(x, t)[2dqt dqx_1/2__a0[f(x)[2dqx_1/2= |GGn|2|f| ,then|( (n||GGn|2 0 as n .This completes the proof.Corollary4.5.2. The eigenvalues of the operator L form an innite sequence kk=1ofreal numbers which can be ordered so that[1[[2[. . .[n[. . . as n .The set of all normalized eigenfunctions of L form an orthonormal basis forL2q(0, a).Proof. Since (is acompact self adjoint operator onL2q(0, a), then (has aninnitesequenceof nonzeroreal eigenvalues nn=1, R, n 0as n . Let nn=1denote an orthonormal set of eigenfunctions corresponding to nn=1. From the spectral83theorem of compact self adjoint operators, we have,((f) =

n=0nf, n_n. (4.5.13)Since the eigenvalues nn=1 of the operator L are the reciprocal of those of (, then[n[ =1[n[ as n . (4.5.14)Lety TL. Then,y = G(f), for somef L2q(0, a). Consequently,y =

n=0nf, n) n =

n=0nly, n) n=

n=0ny, n) n =

n=0y, n) n.If zero is an eigenvalue of L. Then, we can chooser R such thatr is not an eigenvalue ofL. Now, applying the above result on L rIin place of L yields the corollary.Example4.5.1. Consider theq-Sturm-Liouville boundary value problem1qDq1Dqy(x) = y(x), (4.5.15)with theq-Dirichlet conditionsU1(y) = y(0) = 0, U2(y) = y(1) = 0. (4.5.16)A fundamental set of solutions of (4.5.15) is1(x, ) = cos(x; q), 2(x, ) =sin(x; q). (4.5.17)Now, the eigenvalues of problem (4.5.15) are the zeros of the determinant() =U1(1) U2(1)U1(2) U2(2)= 2(1, ) =sin(; q). (4.5.18)84Hence, the eigenvalues mm=1are the zeros of sin(; q). From(1.2.21),m =q2m(1 q)2_1 +O(qm)_, m1, (4.5.19)for suciently largem and the corresponding set of eigenfunctions_sin(mx;q)m_m=1is anorthogonal basis ofL2q(0, 1). In the previous notations1(x, ) =sin(x; q), (4.5.20)and2(x, ) =sin(; q)cos(x; q) + cos(; q)sin(x; q). (4.5.21)So, if is not an eigenvalue, Greens function is given byG(x, t, ) =sin(t; q)sin(; q)_cos(x; q)sin(; q)cos(; q)sin(x; q)_, (4.5.22)for 0tx, andG(x, t, ) =sin(x; q)sin(; q)_cos(t; q)sin(; q)cos(; q)sin(t; q)_, (4.5.23)forx t1. Since = 0isnotaneigenvalue,thenGreensfunctionG(x, t)isnothingbutG(x, t) = G(x, t, 0) =___t(1 x), 0tx,x(1 t), xt1.Hence the boundary value problem (4.5.15)(4.5.16) is equivalent to the basic Fredholm in-