A series of spectral gaps for the almost Mathieu operator with a small coupling constant
AA SERIES OF SPECTRAL GAPS FOR THE ALMOST MATHIEUOPERATOR WITH A SMALL COUPLING CONSTANT
ALEXANDER FEDOTOV
Abstract.
For the almost Mathieu operator with a small coupling constant λ , for a series of spectral gaps, we describe the asymptotic locations of the gapsand get lower bounds for their lengths. The number of the gaps we considercan be of the order of ln 1 /λ , and the length of the k -th gap is roughly of theorder of λ k . Introduction
We consider the almost Mathieu operator acting in l ( Z ) by the formula(1.1) ( H θ f ) k = f k +1 + f k − + 2 λ cos(2 π ( θ + hk )) f k , k ∈ Z , where λ >
0, 0 ≤ θ <
1, and 0 < h < λ iscalled a coupling constant . The operator (1.1) arises when studying an electronin a crystal submitted to a constant magnetic field when the field is weak, whenit is strong, in semiclassical regime etc, see, e.g., [15] and references therein. Thisoperator attracts attention of mathematicians as well as physicists thanks to its richand unusual properties. One of the most difficult and interesting problems is theproblem of describing the geometry of the spectrum of H θ . During three decades,efforts of many mathematicians have been aimed at proving that for irrational h the spectrum is a Cantor set. Among them are A. Avila, J. Bellissard, B. Helffer,S.Zhitomirskaya, R. Krikorian, Y. Last, J. Puig, B. Simon, J. Sj¨ostrand and manyothers, see [1], where the proof was completed. We mention also [17, 18, 19] thatare ones of the latest papers on the geometry of the spectrum.Among the papers of physicists explaining the cantorian structure of the spec-trum, we single out the paper [23] containing heuristic analysis clear for mathemati-cians. In the semiclassical approximation, the author has successively describedsequences of shorter and shorter spectral gaps, i.e., obtained a constructive de-scription of the spectrum as a Cantor set. According to [23], the spectrum locatedon certain intervals of the real line, being “put under a microscope”, looks likethe spectrum of the almost Mathieu operator with new parameters. That is whythe approach described by Wilkinson is called a renormalization method. Usingmethods of the pseudodifferential operator theory, B. Helffer and J. Sj¨ostrand havedeveloped a rigorous asymptotic renormalization method, and turned the heuristicresults into mathematical theorems.Let us note that the asymptotic renormalization methods are used when the param-eter h can be represented as a continued fraction with sufficiently large elements.Later, V. Buslaev and A. Fedotov suggested the monodromization method, onemore renormalization approach that arose when trying to use the Bloch-Floquettheory ideas to study the geometry of the spectrum of difference operators in L ( R ). Mathematics Subject Classification.
Key words and phrases.
Almost Mathieu operator, small coupling, monodromy matrix, spec-tral gaps, asymptotics.The work was supported by the Russian Science Foundation under the grant No. 17-11-01069. a r X i v : . [ m a t h . SP ] F e b he method was further developed by A.Fedotov and F.Klopp when studying adi-abatic quasiperiodic operators. More details can be found in review [10]. Themonodromization method can be used for one-dimensional two-frequency differ-ence and differential quasiperiodic operators independently of any assumptions onthe continued fraction. If such an equation contains an asymptotic parameter, onecan effectively describe the asymptotic geometry of the spectrum. In [10] we de-scribed how to apply the monodromization method to get a constructive asymptoticdescription of the spectrum as a Cantor set in the case studied by B. Helffer andJ. Sj¨ostrand. The present paper is the first step to a similar constructive asymp-totic description in the case of a small coupling constant. Here, we make only thefirst renormalization. This allows to describe asymptotically a series of the longestspectral gaps. We get asymptotic formulas for the gap centers and lower boundsfor the gap lengths. Note that, the number of the gaps we describe can be of theorder of ln(1 /λ ), and the length of the k -th gap is roughly of the order of λ k . So,it can be difficult to get such results by means of standard perturbation methods.The results we prove in this paper were partially announced in the short note [13].Below C denotes positive constants independent of any parameters, variablesand indices, H denotes expressions of the form Ce C/h . When writing a = O ( b ), wemean that | a | ≤ C | b | , and when writing a = O H ( b ), we mean that | a | ≤ e C/h | b | .Furthermore, for z ∈ C , we often use the notations x = Re z and y = Im z .For v , v ∈ C , we denote by V = ( v v ) the matrix with the columns v and v .2. Main results
It is well known, see, for example, [16], that for the irrational h , as a set, thespectrum of the almost Mathieu operator is independent of the parameter θ andcoincides with the spectrum of the Harper operator acting in L ( R ) by the formula Hψ ( x ) = ψ ( x + h ) + ψ ( x − h ) + 2 λ cos(2 πx ) ψ ( x ). It is a difference Schr¨odingeroperator with a 1-periodic potential. Below we discuss only it.As for the one-dimensional periodic differential operators, for the operator H onecan define a monodromy matrix. Here, we describe asymptotics of a monodromymatrix and spectral results obtained by means of these asymptotics.2.1. Monodromy matrix. ψ ( x + h ) + ψ ( x − h ) + 2 λ cos(2 πx ) ψ ( x ) = Eψ ( x ) , x ∈ R , where E ∈ C is a spectral parameter. Its solution space is invariant with respect totranslation by one. Let us fix a basis in the solution space. The corresponding mon-odromy matrix represents the restriction of the translation operator to the solutionspace. Equation (2.1) is a second order difference equation on R , and its solutionspace is a two-dimensional modul over the ring of h -periodic functions, see, e.g. [6].Thus, the monodromy matrix is a 2 × h -periodic in x . When defining amonodromy matrix, it is convenient to make a linear change of variables, for it tobe 1-periodic. The formal definitions can be found in sections 3.1.1 and 3.1.3.The following two theorems describe the functional structure of one of the mon-odromy matrices. Theorem 2.1.
In the solution space of (2.1) , there exists a basis such that thecorresponding monodromy matrix is of the form (2.2) M ( x ) = a − λ cos(2 πx ) s + t e − πix − s − t e πix st/λ , a = λ − s − t st , λ = λ h , here the coefficients s and t are independent of x and meromorphic in E . This theorem is a part of Theorem 7.3 from [6]. The basis solutions are minimalentire solutions, i.e., solutions that are entire functions of z = x + iy growing themost slowly as | y | → ∞ . These minimal solutions are meromorphic in E . Remark 2.1.
It follows from the proof of this theorem that if, for given h = h ∈ (0 , λ = λ ∈ (0 , ∞ ), and E = E ∈ C , the coefficients s and t are finite, thenthey are continuous in ( h, λ, E ) in a neighborhood of ( h , λ , E ).In Section 6, we check Theorem 2.2.
If for an E ∈ R the coefficient t is finite, then, for this E , (2.3) t ∈ i R , | s | = λ (cid:115) | t | λ + | t | , and the zeroth Fourier coefficient of the trace of the monodromy matrix equals (2.4) L = 2 it (cid:113) (1 + | t | )( λ + | t | ) cos(arg ( is )) . Relations (2.3) and (2.4) reflect the self-adjointness of the Harper operator.2.1.2. Pick a ∈ (0 , π ). The asymptotics of the coefficients s and t as λ → σ a satisfying the equation(2.5) σ a ( z + a ) = (1 + e − iz ) σ a ( z − a ) , z ∈ C . Let S = { z ∈ C : | Re z | < π + a } . The function σ a is uniquely characterized bythe following properties. In the strip S , it is analytic, does not vanish, tends toone as y → −∞ and has the minimal possible growth as y → + ∞ . This functionand functions related to it arose in different areas of mathematical physics, see,e.g., [6, 2, 3, 8, 12, 20, 22]. We discuss σ a in section 8. Let(2.6) F ( p ) = σ πh (4 πp − π + πh ) σ πh (4 πp − π + πh ) . To describe the asymptotics of t and s , we also use the parameter p related to E by the equation(2.7) E = 2 cos(2 πp ) , and the notation ξ = π ln λ . Theorem 2.3.
Fix β ∈ (0 , / . There exists c > such that if λ < e − c/h , then,for p satisfying the inequalities | Im p | ≤ h and h/ < Re p < / − h/ , one has (2.8) t = ie π (1 / − p ) ξh F ( p )2 sin(2 πp ) (1 + O H ( λ β )) , s = − i e πpξh + πiph sin(2 πp ) F ( p ) (1 + O H ( λ β )) . Note that, in the case of this theorem, one has st/λ = e πiph (1 + O H ( λ β )).The asymptotics of s and t near the points p = 0 and p = 1 / s and t , we obtain asymptotics of minimal entiresolutions to equation (2.1) as λ →
0. For this, first, we construct entire solutions tothe auxiliary equations ψ ( x + h ) + ψ ( x − h ) + λe ∓ πix ψ ( x ) = Eψ ( x ). Next, in thehalf-planes C ± respectively, for sufficiently small λ , we construct analytic solutionsto the Harper equation that are close to the solutions to the auxiliary equations.Finally, with the help of a Riemann-Hilbert problem, we make of these analyticsolutions the minimal entire solutions.Our asymptotic method works if λ < e − c/h . If h is so small that λ >> e − c/h ,then the asymptotics of the solutions to the Harper equation can be obtained bysemiclassical methods, see, e.g., [4]. .2. Spectral gaps.
The first renormalization of the monodromization methodconsists in replacing equation (2.1) with the first monodromy equation (2.9) Ψ ( x + h ) = M ( x )Ψ ( x ) , x ∈ C , h = { /h } , where M is a monodromy matrix, and {·} is the fractional part. Equations (2.1)and (2.9) simultaneously have pairs of linearly independent solutions such thatone solution of a pair decays exponentially as x → + ∞ , and the other decays as x → −∞ , see Corollary 3.1. This allows to find gaps in the spectrum of the Harperequation by studying solutions to the monodromy equation. In section 3.2 we prove Theorem 2.4.
Let I ⊂ R be an open interval. There exists such a constant C independent of h and E that if (2.10) ( L/ ≥ (1 + Cλ )(1 + | t | ) , λ ≤ | t | ≤ , ∀ E ∈ I, then I is in a gap of the Harper operator. It is useful to compare this theorem with a well-known theorem from the theoryof the one-dimensional periodic differential Schr¨odinger operators. The latter saysthat the spectrum of a periodic operator is located on the intervals where theabsolute value of the trace of a monodromy matrix is less than or equal to two.Using Theorem 2.4, formula (2.4) and the asymptotics s and t described inTheorem 2.3, one can describe a sequence of the longest gaps in the spectrum ofthe Harper operator. As its spectrum is symmetric with respect to zero, and asthe spectra for the frequencies h and 1 − h coincide, we consider only the spectrumlocated on R + in the case where 0 < h < /
2. Let [ · ] be the integer part. One has Theorem 2.5.
Let h ∈ (0 , / and β ∈ (0 , / . Let λ ≤ e − c/h with a sufficientlylarge c . There exist points E k > , ≤ k ≤ K , K = [1 / h ] , such that E k = 2 cos( πhk + O H ( λ p )) , and, if ≤ k ≤ K − or if k = K and [1 /h ] is odd, then the point E k is locatedinside a gap g k . The length of g k satisfies the estimate (2.11) | g k | ≥ (cid:18) λ (cid:19) k (1 + O H ( λ β ))sin ( πh ) sin (2 πh ) . . . sin ( πh ( k − where for k = 1 the product of the sines has to be replaced with one.If [1 /h ] be even, and λ h ≤ e − c/h , then the point E K also is located in a gap g K ,and the length of g K satisfies (2.11) with β replaced with p = min { h , β } . In the next paper, we will prove that the expression in the right hand side of (2.11)is the leading term of the asymptotics of the length of the k th gap.Theorem 2.5 agree well with the results of computer calculations described in [15].The K th gap in the case where where [1 /h ] is even is the most difficult to describe.For small hh , it is located near zero that is a very special point. For h (cid:54)∈ Q , thecomplexity of the spectrum near zero is well-known. One can find a series of newresults in [17].The right hand side in (2.11) equals λ k e O (1 /h ) , see Corollary 3.4. Therefore, if λ is small, and h is of the order of | ln λ | , then | g K | , the length of the gap closestto zero, is of the order of λ | ln λ | + O (1) . So, it can be rather difficult to compute | g K | using standard perturbation methods.According to [18], for the almost Mathieu equation with λ < k tends to infinity, the gap lengths are boundedfrom above and below by expressions of the form C ( λ, h, (cid:15) ) λ (1 ± (cid:15) ) k , where (cid:15) is a fixedpositive number. And as we mentioned, in our case, one has | g k | ≥ λ k e O (1 /h ) . I am grateful to Qi Zhou for attracting my attention to paper [18] and discussing its results. t is interesting to compare our results with the results obtained in the case ofsmall h and λ = 1, see [10]. In this case, there is a statement similar to Theorem 2.4.However, the asymptotics of the coefficients t and s turn out to be quite different,and, on the most of the interval [ − ,
4] containing the spectrum, L ( E ) oscillateswith an amplitude that is exponentially large with respect to h , whereas in ourcase, for most E ∈ (0 , L ≈ p/h ). Thus, in the case of small h , thespectrum is located on a series of exponentially small intervals, and in the case ofsmall λ , we observe small gaps in the spectrum.2.3. Other gaps.
Since the matrix M is 1-periodic, for equation (2.9), we canalso define a monodromy matrix M and consider the second monodromy equationthat can be obtained from the first one by replacing M with M and h with h = { /h } . Continuing, we can construct an infinite sequence of difference equations.In the next paper, studying consequently the equations of this sequence, we willdescribe consequently series of shorter and shorter gaps. We also obtain upperbounds for the gaps lengths.Note that, to prove Theorem 2.5, we use only the asymptotics of the coefficients s and t from Theorem 2.3, i.e., their asymptotics for E being bounded away from 2.However, the asymptotics of s and t for E close to 2 are crucial to study the geome-try of the spectrum near its edge. All we need to get these asymptotics is preparedwhen proving Theorem 2.3, and they are obtained almost like the ones describedin this theorem. So, omitting elementary details, we get them in section 7.4.2.4. The plan of the paper.
In section 3, we give the definition of a monodromymatrix, and prove Theorems 2.4 and 2.5. For this we use Theorems 2.1–2.3. Themost of the remaining part of the paper is devoted to the proof of Theorem 2.3. Insection 4, we construct and analyze analytic solutions to the model equation (4.1).In section 5, we show that, in the upper half-plane, there are analytic solutions tothe Harper equation that are close to the solutions to the model equation. Recallthat the monodromy matrix described in Theorem 2.1 corresponds to a basis oftwo minimal entire solutions to the Harper equation. In section 6, we recall thedefinition of minimal entire solutions and prove Theorem 2.2. In section 7, for suf-ficiently small λ , using the analytic solutions to the Harper equation constructedin section 5, we construct and study the minimal entire solutions, and prove The-orem 2.3. Section 7.4 is devoted to the asymptotics of s and t for p close to zero(i.e., for E close to 2). In Section 8, we describe the properties of σ a that are usedin this paper.3. Monodromy matrices, monodromy equation and spectral results
Here we remind the definition of a monodromy matrix, describe relations be-tween solutions to a difference equation with periodic coefficients and solutions toa corresponding monodromy equation, and prove Theorems 2.4 and 2.5.3.1.
Monodromy matrices and monodromy equation.
Definition and elementary properties of a monodromy matrix.
Here, follow-ing [10] we discuss the difference equations of the form(3.1) Ψ( x + h ) = M ( x )Ψ( x ) , where x is a real variable, M : R → SL (2 , C ) is a given 1-periodic function, and h ∈ (0 ,
1) is a fixed number.Obviously, for any solution Ψ to (3.1), we have det Ψ( x + h ) = det Ψ( x ), x ∈ R .We call SL (2 , C )-valued solutions to (3.1) fundamental matrix solutions.Note that, to construct a fundamental solution, it suffices to define it arbitrarily n the interval 0 < x < h , and then, to define its values outside of this intervaldirectly with the help of equation (3.1).It can be shown that ˜Ψ R (cid:55)→ M ( C ) is a matrix-valued solution to (3.1) if and onlyif it can be represented in the form˜Ψ( x ) = Ψ( x ) p ( x ) , x ∈ R , where p : R (cid:55)→ M ( C ) is an h -periodic function, and Ψ is a fundamental solution.Note that this representation implies that the space of matrix-valued solutionsto (3.1) is a module over the ring of h -periodic functions.Let ψ , ψ : R → C be two vector-valued solutions to (3.1). We say that theyare linearly independent if det( ψ , ψ ) does not vanish. In this case, a function ψ : R → C is a vector-valued solution to (3.1) if and only if it is a linear combinationof ψ and ψ with h -periodic coefficients.Let Ψ be a fundamental solution. As M is 1-periodic, the function x (cid:55)→ Ψ( x + 1)is also a solution to (3.1), and we can writeΨ( x + 1) = Ψ( x ) p ( x ) , p ( x + h ) = p ( x ) , x ∈ R . The matrix M ( x ) = p t ( hx ), where t denotes transposition, is called the mon-odromy matrix corresponding to the fundamental solution Ψ.Note that, by construction, a monodromy matrix is 1-periodic and unimodular.In early papers, the h -periodic matrix p was called the monodromy matrix. It ismore convenient to consider 1-periodic monodromy matrices.3.1.2. Monodromy equation.
Let M be the monodromy matrix corresponding to afundamental solution Ψ to (3.1). Let us consider the first monodromy equation (2.9).It appears that the behavior of solutions to (3.1) at infinity “copies” the behaviorof solutions to (2.9). Let us formulate the precise statement.Let M be a SL (2 , C )-valued function of real variable, and h >
0. Let k ∈ Z and x ∈ R . We put P k ( M, x, h ) = (cid:40) M ( x + h ( k − . . . M ( x + h ) M ( x ) , k ≥ ,M − ( x + hk ) . . . M − ( x − h ) M − ( x − h ) , k < . Clearly, if ψ : R → C satisfies (3.1), then ψ ( x + hk ) = P k ( M, x, h ) ψ ( x )One has Theorem 3.1. [11]
Let Ψ be a fundamental solution to (3.1) , and let M be thecorresponding monodromy matrix. Then, for all N ∈ Z , P N ( M, h, x ) = Ψ( { x + N h } ) σ P N ( M , h , x ) σ Ψ − ( x ) , (3.2) N = − [ θ + N h ] , h = { /h } , x = { x/h } , where σ is the Pauli matrix (cid:18) − ii (cid:19) . In this paper we use
Corollary 3.1.
In the case of Theorem 3.1, we assume that Ψ ∈ L ∞ loc ( R , SL (2 , C )) ,and that ψ (1) ± are two vector-valued solutions to the monodromy equation such that (3.3) (cid:107) ψ (1) ± ( ± x ) (cid:107) C ≤ C e ∓ κ x , x ≥ , with some positive constants C and κ . Then there are two vector-valued solutions ψ (0) ± to equation (3.1) such that (3.4) det (cid:16) ψ (0)+ ( x ) , ψ (0) − ( x ) (cid:17) = det (cid:16) ψ (1)+ ( { x/h } ) , ψ (1) − ( { x/h } ) (cid:17) , ∀ x ∈ R , (cid:107) ψ (0) ± ( ± x ) (cid:107) C ≤ C e ∓ κ h x , ∀ x ≥ , with a positive constant C .Proof. As P N ( M, h, x )Ψ( x ) = Ψ( x + N h ), formula (3.2) implies that(3.6) Ψ( x + N h ) σ = Ψ( { x + N h } ) σ P N ( M , h , x ) . Let us define the solutions ψ (0) ± : R → C to (3.1) by the formulas(3.7) ψ (0) ± ( x ) = Ψ( x ) σ ψ (1) ∓ ( { x/h } ) . As det Ψ ≡
1, relation (3.4) is obvious. Furthermore, as x = { x/h } , and N = − [ x + N h ], formulas (3.7) and (3.6) lead to the relation ψ (0) ± ( x + N h ) = Ψ( { x + N h } ) σ ψ (1) ∓ ( x − [ x + N h ] h ) , N ∈ Z , that can be rewritten in the form ψ (0) ± ( x ) = Ψ( { x } ) σ ψ (1) ∓ ( x − [ x ] h ) , x ∈ R . This formula and estimates (3.3) imply (3.5). The proof is complete. (cid:3)
Monodromy matrices for difference Schr¨odinger equations.
Let h > v : R → C . The difference Schr¨odinger equation(3.8) ψ ( x + h ) + ψ ( x − h ) + v ( x ) ψ ( x ) = Eψ ( x ) , x ∈ R , is equivalent to (3.1) with(3.9) M ( z ) = (cid:18) E − v ( x ) −
11 0 (cid:19) . More precisely, a function Ψ : R (cid:55)→ C satisfies (3.1) with this matrix if andonly if Ψ( x ) = (cid:18) ψ ( x ) ψ ( x − h ) (cid:19) , and ψ is a solution to (3.8). This allows to turn theobservations made for (3.1) into observations for (3.8) .Let ψ and ψ be two solutions to (3.1). The expression { ψ ( x ) , ψ ( x ) } = ψ ( x + h ) ψ ( x ) − ψ ( x ) ψ ( x + h ) , their Wronskian, is h -periodic in x .Assume that the Wronskian is constant and nonzero. Then ψ , form a basis in thespace of solutions, and a function ψ satisfies (3.8) if and only if(3.10) ψ ( x ) = a ( x ) ψ ( x ) + b ( x ) ψ ( x ) , where a and b are h -periodic coefficients. One easily proves that(3.11) a ( x ) = { ψ ( x ) , ψ ( x ) }{ ψ ( x ) , ψ ( x ) } , b ( x ) = { ψ ( x ) , ψ ( x ) }{ ψ ( x ) , ψ ( x ) } . If v is 1-periodic, the functions x → ψ ( x + 1) and x → ψ ( x + 1) are also solutionsto (3.8), and one can write(3.12) Ψ( x + 1) = M ( x/h )Ψ( x ) , Ψ( x ) = (cid:18) ψ ( x ) ψ ( x ) (cid:19) , x ∈ R , where M is a 1-periodic 2 × ψ and ψ . It coincides with a monodromy matrix for (3.1) with the matrix (3.9). .2. Gaps in the spectrum of the Harper equation: proof of Theorem 2.4.
Let us consider equation (2.9) with the matrix M described in Theorem 2.1. The-orem 2.4, a sufficient condition for E to be in gap, follows from Proposition 3.1.
In the case of Theorem 2.4, for all E ∈ I , there exist two vectorsolutions ψ (1) ± to (2.9) such that det( ψ (1)+ ( x ) , ψ (1) − ( x )) is a nonzero constant, and,for x ≥ , (cid:107) ψ (1) ± ( x ) (cid:107) C ≤ Ce ∓ x ln | L | . This proposition implies
Lemma 3.1.
In the case of Theorem 2.4, for any E ∈ I , the only polynomiallybounded solution to the Harper equation (2.1) is zero. First, using these proposition and lemma, we prove Theorem 2.4.Assume that the Harper operator has some spectrum on I . As it is a direct integralof the almost Mathieu operators with respect to θ , then, for some θ , the almostMathieu operator, has a nontrivial spectrum on I , and on I almost everywhere withrespect to the spectral measure, the almost Mathieu equation f k +1 + f k − + 2 λ cos(2 π ( θ + hk )) f k = Ef k , k ∈ Z . has a polynomially bounded solution f (see section 2.4 in [7]). One defines asolution ψ to the Harper equation so that ψ ( x ) = f k ( θ ) if x = θ + kh with k ∈ Z ,and ψ ( x ) = 0 otherwise. The ψ is a non-trivial polynomially bounded solution tothe Harper equation. This contradicts Lemma 3.1.Now, let us prove Lemma 3.1 and Proposition 3.1. Proof of Lemma 3.1 . Let us assume that, for an E ∈ I , there is a nontrivialpolynomially bounded solution ψ to (2.1). For this E we construct the solutionsto the monodromy equation described in Proposition 3.1. Then, in terms of thesesolutions, we construct the solutions ψ (0) ± to equation (3.1) with matrix (3.9) asdescribed in Corollary 3.1 (this is possible as the fundamental solution used todefine the monodromy matrix from Theorem 2.1 is entire in x ).Let ψ be the first entry of ψ (0)+ , and ψ be the one of ψ (0) − . One has { ψ ( x ) , ψ ( x ) } = det( ψ (0)+ ( x ) , ψ (0) − ( x )) = det( ψ (1)+ ( x ) , ψ (1) − ( x )) , where we used Corollary 3.1. Thus, by Proposition 3.1, { ψ ( x ) , ψ ( x ) } is a nonzeroconstant. Therefore, one has (3.10)–(3.11). Now, it suffices to show that the Wron-skians { ψ ( x ) , ψ j ( x ) } , j = 1 ,
2, equal zero. But this is obvious, as these Wronskiansare periodic and, on the other hand, { ψ ( x ) , ψ ± ( x ) } → x → ±∞ since ψ ispolynomially bounded, and ψ ± are exponentially decreasing as x → ±∞ . Theproof of Lemma 3.1 is complete. (cid:3) Now, let us prove Proposition 3.1.
Proof.
Below we assume that E ∈ I , and that conditions (2.10) are satisfied.In view of Theorem 2.1, we can represent the monodromy matrix in the form(3.13) M ( x ) = M + ˜ M ( x ) , ˜ M ( x ) = O ( t ) , M = (cid:18) λ st (1 − s − t ) s − s stλ (cid:19) . The plan of the proof is the following. First, we transform the monodromy equationwith a matrix M of the form (3.13) to the equation(3.14) φ ( x + h ) = p ( D + ∆( x )) φ ( x ) , D = (cid:18) /U U (cid:19) , x ∈ R , where p and U are parameters, and ∆ is a “sufficiently small” matrix. Then, weconstruct two solutions to (3.14) by means of roposition 3.2. Let us consider equation (3.14) with parameters h > , p ≥ and U ∈ R , and a function p ( D + ∆) ∈ L ∞ ( R , SL (2 , C )) . Let (3.15) | U + U − | > , and | U − U − | > m, m = max ≤ i,j ≤ sup x ∈ R | ∆ ij ( x ) | . There exist φ ± ∈ L ∞ loc ( R , C ) , vector-valued solutions to (3.14) , such that (3.16) (cid:107) φ ± ( ± x ) (cid:107) C ≤ Ce ∓ xh ln p | U + U − | ∀ x ≥ , inf x ∈ R | det( φ + ( x ) , φ − ( x )) | > . Mutatis mutandis, the proof of Proposition 3.2 repeats the proof of Proposition 4.1from [9] where we have considered the case of p = 1.The det( φ + , φ − ) being h -periodic, the function x (cid:55)→ φ + ( x ) / det( φ + ( x ) , φ − ( x )) sat-isfies (3.14). We keep for this new function the old notation φ + . It belongs to L ∞ loc ( R , C ), satisfies estimate (3.16), and we have det( φ + ( x ) , φ − ( x )) = 1.To complete the proof of Proposition 3.1, we return from (3.14) to the monodromyequation constructing ψ (1) ± in terms of φ ± .Let us transform the monodromy equation to the form (3.14). Therefore, wecompute the eigenvalues and eigenvectors of M . In view of Theorem 2.2, one has t = iτ, τ ∈ R , s = − iλ (cid:115) τ τ + λ e iα , α ∈ R , Let p = (cid:112) τ , q = 1 τ (cid:113) τ + λ , Q = (cid:112) q cos α − . Then tr M = L = 2 pq cos α, det M = p . The eigenvalues ν ± and the corresponding eigenvectors v ± are given by the formulae ν ± = p ( q cos α ± Q ) , v ± = (cid:32) − ps (cid:16) q cos α − q e iα ∓ Q (cid:17)(cid:33) . Let V = ( v + v − ). We represent a vector-valued solution to (2.9) in the form(3.17) ψ ( x ) = V φ ( x ) . Then φ satisfies equation (3.14) with(3.18) U = q cos α + Q and ∆( x ) = 1 p V − ˜ M ( x ) V. Let us determine the conditions under which U and ∆ from (3.18) satisfy theassumptions of Proposition 3.2. Let(3.19) q cos α > . We can and do assume that
Q > q cos α >
1. Thecomplementary case is analyzed similarly.We have
U >
1, and the first condition in (3.15) is satisfied.Now, let us estimate the entries of ∆. One has | sin α | = (cid:112) − cos α ≤ (cid:114) − q ≤ (cid:12)(cid:12)(cid:12)(cid:12) λ τ (cid:12)(cid:12)(cid:12)(cid:12) , Q = (cid:115)(cid:18) λ τ (cid:19) cos α − ≤ (cid:12)(cid:12)(cid:12)(cid:12) λ τ (cid:12)(cid:12)(cid:12)(cid:12) . Therefore and as λ ≤ | τ | , the second entries of v ± are uniformly bounded : (cid:12)(cid:12)(cid:12)(cid:12) ps (cid:18) q cos α − q e iα ∓ Q (cid:19)(cid:12)(cid:12)(cid:12)(cid:12) = (cid:12)(cid:12)(cid:12)(cid:12) pqs (cid:12)(cid:12)(cid:12)(cid:12) (cid:12)(cid:12)(cid:12)(cid:12) λ τ cos α − i sin α ∓ qQ (cid:12)(cid:12)(cid:12)(cid:12) ≤ C (cid:12)(cid:12)(cid:12)(cid:12) pλ qsτ (cid:12)(cid:12)(cid:12)(cid:12) = C. s det V = − pQ/s , this estimate implies that, for all i, j ∈ { , } ,max x ∈ R | ∆ ij ( x ) | ≤ C (cid:12)(cid:12)(cid:12)(cid:12) sτp Q (cid:12)(cid:12)(cid:12)(cid:12) = C (cid:12)(cid:12)(cid:12)(cid:12) λ pqQ (cid:12)(cid:12)(cid:12)(cid:12) < C λ Q .
Therefore, the second condition from (3.15) is satisfied if(3.20) U − U − Q ≥ C λ Q ⇐⇒ q cos α ≥ Cλ . Clearly, this condition implies (3.19) and, as L = 2 pq cos α , it is equivalent to (2.10).Let assume that (3.20) is satisfied. Then, using Proposition 3.2 and formula (3.17),we construct two vector-valued solutions ψ ± = V φ ± to the monodromy equation.As p ( U − + U ) / pq cos α = L/
2, estimates (3.16) imply the estimates for ψ ± from Proposition 3.1. Furthermore, one hasdet( ψ + ( x ) , ψ − ( x )) = det V det( φ + ( x ) , φ − ( x )) = det V (cid:54) = 0 . The proof is complete. (cid:3)
Gaps in the spectrum of the Harper equation: proof of Theorem 2.5.
Theorem 2.5 describing a sequence of gaps in the spectrum of the Harper operatorfollows from Theorem 2.4, a sufficient condition for E to be in a gap written in termsof the coefficients s and t of a monodromy matrix, and Theorem 2.3 describing theasymptotics of the monodromy matrix as λ → p ∈ I p = [ h/ , / . where p is the parameter related to E by (2.7). In view of (2.4), condition (2.10)can be written in the form(3.21) (1 + X ) cos α ≥ C λ , X = λ /τ = iλ /t, where α = arg is , s and t being the coefficients from Theorem 2.1, λ = λ /h , and C is a certain positive constant.Below, when analyzing (3.21), we assume that the following hypothesis is true. Hypothesis 1.
One has λ ≤ e − c /h , where c > is a sufficiently large constant. This hypothesis is needed in particular, to use Theorem 2.3 on the asymptotics ofthe monodromy matrix coefficients s and t .Also, we fix β ∈ (0 , / Locations of gaps.
Inequality (3.21) can be rewritten as X − Cλ X ≥ sin α ,and assuming that λ /X and X are sufficiently small, we transform it to the form(3.22) | X | (1 + O ( λ /X ) + O ( X )) ≥ | sin α | , and see that there are gaps containing the points E k defined by the relations(3.23) E k = 2 cos(2 πp k ) , α ( p k ) = πk, k ∈ Z To continue, we need the following two lemmas:
Lemma 3.2.
For p ∈ I p , one has (3.24) α = 2 πph + O H (cid:0) λ β (cid:1) . If p > Ch > h/ , the error term is analytic in p and satisfies the estimate (3.25) (cid:16) O H (cid:0) λ β (cid:1)(cid:17) (cid:48) p = O H (cid:0) λ β (cid:1) . roof. Pick c sufficiently small. Let V ch be the ( ch )-neighborhood of I p .Recall that F is given by (2.6). The description of the zeros and poles of thefunction σ a from section 8.1.2 implies that in V ch the function F is real analyticand does not vanish.Formula (3.24) follows from the second formula in (2.8).As the function p (cid:55)→ sin(2 πp ) /F ( p ) from (2.8) is analytic and does not vanishin V ch , and as s is meromorphic in E , the second formula in (2.8) implies that s is bounded and thus analytic in p ∈ V ch . And the analyticity of s and the sameformula imply that the error term in (2.8) is also analytic in V ch .The estimate (3.25) follows from the analyticity of the error term and the Cauchyrepresentation for the derivatives of analytic functions. (cid:3) Lemma 3.3.
For p ∈ I p , one has X = X ( p ) (cid:0) O H (cid:0) λ β (cid:1)(cid:1) , X ( p ) = e p ln λ/h πp ) F ( p ) , (3.26) ln X ( p ) = 2 p ln λh + O (cid:16) h (cid:17) , (cid:18) O (cid:16) h (cid:17)(cid:19) (cid:48) p = O (cid:16) h (cid:17) . (3.27) Proof.
Formulas (3.26) follow from Theorem 2.3.For p ∈ I p , we get(3.28) ln X = 2 p ln λh − ln F ( p ) + ln p + O (1) , dO (1) dp = O (1) , where ln F ( p ) and ln p are real. It suffices to show thatln F ( p ) = O (1 /h ) , (ln F )( p ) = O (1 /h ) , p ∈ I p . Let us fix δ ∈ (0 , π ). By Corollary 8.1 we get(3.29) ln F ( p ) = O (1 /h ) if | πp + πh | ≥ δ, − h/ ≤ Re p ≤ | Im p | ≤ . This estimate and the Cauchy representations for the derivatives of analytic func-tions imply that, in any fixed subinterval of the interval [ δ/ π − h/ , F ) (cid:48) ( p ) = O (1 /h ). However, as δ is chosen quite arbitrarily, we can write(3.30) (ln F ) (cid:48) ( p ) = O (1 /h ) , δ/ π − h/ ≤ p ≤ / . If max { δ/ π − h/ , h/ } ≤ p ≤ /
4, estimates (3.29), (3.30), (3.28) imply (3.27).Let us assume that | πp + πh | ≤ δ and p ∈ I p . For p ∈ I p , one has 2 p/h > / F ( p ) = 4 p ln hh + 2 ln Γ (cid:16) ph + 1 (cid:17) + O (cid:16) h (cid:17) , (cid:18) O (cid:16) h (cid:17)(cid:19) (cid:48) p = O (cid:16) h (cid:17) , p ∈ I p , where the values of the logarithms are real. When deriving this formula, we haveestimated the derivative of the error term from (8.7) arguing as when proving (3.30).As 2 p/h > /
2, we estimate ln Γ(2 p/h +1) using the Stirling formula (ln Γ)( x +1) = x (ln x − O (ln x ), where the error term satisfies the estimate ( O (ln x )) (cid:48) x = O (1 /x ).We getln F ( p ) = 4 p ln ph + O (cid:16) h (cid:17) = O (cid:16) h (cid:17) , (cid:18) O (cid:16) h (cid:17)(cid:19) (cid:48) p = O (cid:16) h (cid:17) , h ≤ p ≤ δ π − h , and thus, in view of (3.28), estimate (3.27) . The proof is complete. (cid:3) Lemma 3.2 immediately implies orollary 3.2. For p k ∈ I p , one has (3.31) p k = hk O H (cid:0) λ β (cid:1) where the error is uniform uniform in k One can easily see that, for sufficiently small λ hk/ ∈ I p if and only if(3.32) 1 ≤ k ≤ K, K = (cid:20) h (cid:21) , where [ x ] denotes the integer part of x ∈ R . We note also that − hK = h · (cid:40) h if [1 /h ] is even , h if [1 /h ] is odd . For the points p k one has Corollary 3.3.
Let us fix a positive constant C so that the error terms in (3.24) and (3.25) be bounded by δ = λ β e C /h . If δ /π ≤ min { / , h / } , and k satis-fies (3.32) , then there is p k ∈ I p .Proof. By Lemma 3.2, the function α is monotonous on the interval [ h/ , / π/ δ , π/ (2 h ) − δ ] ⊂ α (cid:0) [ h/ , / (cid:1) . Thus, for any k satisfying23 + δ π ≤ k ≤ h − δ π , the equation α ( p k ) = πk has a unique solution in [ h/ , / δ /π ≤ /
3, the the minimal possible value of k equals 1. Recall that 1 /h =[1 /h ] + h , h ∈ (0 , h = (cid:40)(cid:2) h (cid:3) + h , if [1 /h ] is even , (cid:2) h (cid:3) + h , if [1 /h ] is odd , and as δ /π ≤ h /
2, the maximal value of k equals (cid:2) h (cid:3) . The proof is complete. (cid:3) Remark 3.1.
As seen from the proof, either if (cid:2) h (cid:3) is odd, or if k < (cid:2) h (cid:3) , then thecondition on δ can be weakened and replaced with | δ /π | ≤ / Lemma 3.4.
For any k = 1 , , . . . K − , the point E k = 2 cos(2 πp k ) is located in agap. The point E K is in a gap if either (cid:2) h (cid:3) is odd, or (cid:2) h (cid:3) is even, and λ h < e − c h where c > is a certain constant.Proof. In view of (3.21) and the definition of p k , see (3.23), it suffices to prove that X ( p k ) ≥ C λ , where λ = λ /h , and C > c in Hypothesis 1 is sufficiently large, then using Lemma 3.3 and (3.31), we get(3.33) X ( p k ) = e pk ln λh + O ( h ) = e k ln λ + O H ( kλ β ln λ )+ O ( h ) = λ k e O ( h ) . So, λ X ( p k ) ≤ λ h − k + Ch , and we get(3.34) λ X ( p k ) ≤ e Ch · λ if 1 ≤ k ≤ K − ,λ h if k = K, and (cid:2) h (cid:3) is odd ,λ h if k = K, and (cid:2) h (cid:3) is even . This implies the needed. (cid:3)
Below, we denote by g k the gap containing E k . We have proved the statement ofTheorem 2.5 on the location of g k , k = 1 , , . . . K . .3.2. The lengths of gaps.
Here we prove (2.11). Let us fix 1 ≤ k ≤ K . To get alower bound for | g k | , the length of g k , we first assume that | p − p k | ≤ ChX ( p k )and prove Lemma 3.5.
Under hypothesis 1, one has (3.35) X ( p ) = X ( p k )(1 + O H ( λ β )) , sin α ( p ) = 2 π ( p − p k ) h (1 + O H ( λ β )) . Proof.
As when proving Lemma 3.4, we see that X ( p k ) = λ k e O (1 /h ) . Therefore,(3.36) | p − p k | ≤ hλ k e C/h . Lemma 3.3 and (3.36) imply that one has X ( p ) = X ( p )(1 + O H ( λ β )) = X ( p k ) e ln λ λ k e C/h (1 + O H ( λ β )) . This implies the first formula in (3.35). Lemma 3.2 implies that α ( p ) = π ( p − p k ) h (1+ O H ( λ β )). Thus, in view of (3.36), we getsin α = 2 π ( p − p k ) h (cid:16) O (cid:16) ( p − p k ) h (cid:17)(cid:17) (1 + O H ( λ β )) = π ( p − p k ) h (1 + O H ( λ β )) . We proved the second formula in (3.35). (cid:3)
Now, we can readily prove
Proposition 3.3.
Let ≤ k ≤ K , let λ h e c /h be sufficiently small if k = K and [1 /h ] is even. Then (3.22) can be transformed to the form (3.37) | p − p k | ≤ h π X ( p k )(1 + O H ( λ p )) where p = β if ≤ k ≤ K − or [1 /h ] is odd, and p = min { h , β } if k = K and [1 /h ] is even.Proof. The statement follows from Lemma 3.5 and formulas (3.33) and (3.34). (cid:3)
Now, to complete the proof of (2.11), we need to compute X ( p k ) and towrite (3.37) in terms of E . We begin with Lemma 3.6.
For ≤ k ≤ K , one has (3.38) X ( p k ) = (cid:18) λ (cid:19) k sin( πhk ) h sin ( πhk ) . . . sin ( πh ) (1 + O H ( λ β )) . Proof.
By the second estimate in (3.27), X ( p k ) = X (cid:0) hk (cid:1) (1 + O H ( λ β ))as β in Lemma 3.3 can be chosen greater then in this lemma. Let us recall that X is defined in (3.26) with F given by (2.6). Using (2.5) and (8.5), we get F (cid:0) hk (cid:1) = (cid:81) kκ =1 | e − i (2 πhκ − π ) | | σ πh ( πh − π ) | = 4 k h (cid:81) kκ =1 | sin( πhκ ) | . This formula and formulas (3.26) and (2.6) imply (3.38). (cid:3)
Finally, we rewrite (3.37) in terms of E . We denote the right hand side in (3.37)by ∆ k . As E = 2 cos(2 πp ), we get | g k | ≥ (cid:0) π ( p k − ∆ k ) (cid:1) − cos (cid:0) π ( p k + ∆ k ) (cid:1) = 4 sin(2 πp k ) sin(2 π ∆ k ) . As X ( p k ) = O H ( λ k ), we get | ∆ k | = O H ( λ k ) , and, by means of (3.31), we obtain | g k | ≥ π ∆ k sin(2 πp k )(1 + O H ( λ k )) = 8 π ∆ k sin( πhk )(1 + O H ( λ β )) . ubstituting into this estimate instead of ∆ k the right hand side from (3.37) andusing formula (3.38), we come to (2.11). This completes the proof Theorem 2.5. (cid:3) We will use the following immediate corollary from the above proof:
Corollary 3.4.
The right hand side in (2.11) satisfies equals to λ k e O (1 /h ) .Proof. As X ( p k ) = O H ( λ k ), it suffices to compare (2.11) with (3.38). (cid:3) Model equation
Entire solutions to the equation(4.1) µ ( z + h ) + µ ( z − h ) + e − πiz µ ( z ) = 2 cos(2 πp ) µ ( z ) , z ∈ C . were constructed in [14]. Here, we briefly recall construction of these solutions, andget for them estimates uniform in h .4.1. Construction of solutions.
Integral representation.
We construct solutions in the form:(4.2) µ ( z ) = 1 √ h (cid:90) γ e πizkh v ( k ) dk, where γ is a curve in the complex plane that we describe later, and v is a functionanalytic in a sufficiently large neighborhood of γ . The function µ satisfies (4.1) if v ( k + h ) = 2(cos(2 πp ) − cos(2 πk )) v ( k ) . We note that2(cos(2 πp ) − cos(2 πk )) = − e πik (1 − e − πi ( k − p ) ) (1 − e − πi ( k + p ) ) . Therefore, one can choose(4.3) v ( k ) = e iπk h − πikh − πik σ πh (2 π ( k − p − h − )) σ πh (2 π ( k + p − h − )) , where σ a is a solution to (2.5). In this paper, we use the meromorphic solutiondescribed in Section 8.4.1.2. Properties of the function v . As σ a the function v is meromorphic. In Sec-tion 8.1.2 we list all the poles of σ a . This description implies that the poles of v are located at the points(4.4) k = ± p − lh − m, l, m = 0 , , , . . . . Let l ± = ± p − ( −∞ , l ± contain all the poles of v .Below, for k ∈ C , we write r = Re k and q = Im k .Let | q | > C | r | . Corollary 8.1 implies the uniform asymptotic representations v ( k ) = v − e iπ (cid:32) k − − h (cid:33) h + o (1) , v − = − ie − iπ h − πih , q → −∞ , (4.5) v ( k ) = v + e − iπ (cid:32) k − − h (cid:33) h + o (1) , v + = − ie − πip h − iπ h − πih , q → + ∞ . (4.6)4.1.3. Integration path.
We choose the integration path in (4.2) so that it does notintersect l ± , comes from infinity from bottom to top along the line e iπ/ R and goesto infinity upward along the line e iπ/ R . This completes the construction of µ .4.1.4. Notations.
Fix
X >
0. We study µ in the strip | x | ≤ X assuming that p ∈ P , P = { p ∈ C : 0 ≤ Re p ≤ / , | Im p | ≤ h } . .2. Estimates in the upper half-plane.
Let ξ ± ( z ) = e ± πiz h + iπzh + πiz . One has
Proposition 4.1.
Let us pick
X, Y > . Assume that p ∈ P . One has (4.7) µ = ν + ξ + + ν − ξ − , ν ± = ∓ e ∓ iπ v ± F ± ; and for | x | ≤ X the functions F ± satisfy the uniform in x estimates: (4.8) F ± = O ( H ) if y ≥ − Y, and F ± = 1 + o (1) as y → + ∞ . Proof.
In view of (4.5)– (4.6), as q → ±∞ the behavior of the integrand in (4.2) isdescribed by the exponentials(4.9) e πikzh e ∓ iπ ( k − − h )2 h = ξ ± ( z ) e ∓ πi ( k − k ± ( z ))2 h , where k ± ( z ) = ± z + h + . Let us consider the straight lines L ± ( z ) = k ± ( z ) + e ∓ πi/ R . The lines L ± ( z ) are lines of steepest descent for the functions k (cid:55)→ | e ∓ πi ( k − k ± ( z ))2 h | .They intersect one another at k ∗ = y + ix + h/ / d >
0. There is an
Y > y > Y , then for all( h, p, x ) ∈ (0 , × P × [ − X, X ], the distance from L ± to l ± is greater than d .First, we assume that y > Y . In this case, we choose the integration path γ in (4.2) so that it go upwards first along L − ( z ) from infinity to k ∗ and then along L + ( z ) from k ∗ to infinity. We denote by γ − ( γ + ), the part of γ below (resp., above) k ∗ ( z ). We define two functions µ ± by same the formula as µ , i.e. by (4.2), but withthe integration path γ replaced with γ ± . It suffices to show that(4.10) µ ± = ∓ e ∓ iπ v ± ξ ± F ± , with F ± satisfying the estimates (4.8). We prove (4.10) only for µ − ; µ + is esti-mated similarly. Substituting (4.3) into (4.2) and using (4.9), we get(4.11) µ − = ξ − ( z ) v − ν, ν = 1 √ h (cid:90) γ − e πi ( k − k − ( z ))2 h F ( k, p, h ) dk, where F ( k, p, h ) = σ πh (2 π ( k − p − h − )) σ πh (2 π ( k + p − h − )).We remind that k = r + iq , r, q ∈ R . Let 0 < κ <
1. By Corollary 8.1, for k ∈ γ − ,(4.12) F ( k, p, h ) = e O ( h − e − πκ | q | (1+ | r | ) ) . As along γ − one has k − k − ( z ) = √ e iπ/ Im ( k − k − ( z )) , Im ( k − k − ( z )) = q + y, | r | ≤ C + | q + y | , formulas (4.11) and (4.12) imply that ν = e iπ (cid:114) h (cid:90) x −∞ e − π ( q + y )2 h + O (cid:18) e − πκ | q | (1+ | q + y | ) h (cid:19) dq = e iπ (cid:114) h (cid:90) x + y −∞ e − πt h + O (cid:18) e − πκ | t − y | (1+ | t | ) h (cid:19) dt. (4.13)Therefore, | ν | ≤ C √ h (cid:90) ∞−∞ e − πt h + C (1+ | t | ) h dt ≤ H. his proves the first estimate in (4.8) for y > Y . To prove the second one, we notethat, for sufficiently large y , one has e − πκ | t − y | (1 + | t | ) ≤ Cy (cid:40) e − πκy if t ≤ y , y ≤ t ≤ y + X. So, representation (4.13) implies that as y → ∞ ν = e iπ (cid:114) h (cid:32)(cid:90) y −∞ e − πt h + O ( e − πκy y/h ) dt + (cid:90) x + y y e − πt h + O ( y/h ) dt (cid:33) = e iπ/ + o (1) . This proves the second estimate in (4.8).To complete the proof, it suffices to check that if | x | ≤ X and | y | ≤ Y , then µ ≤ H . In this case, we pick δ, r > γ in (4.2) thatgoes along L − ( z ) upwards from infinity to the circle c r with radius r and center at k ∗ , then along c r in the anticlockwise direction to the upper point of intersectionof L + ( z ) and c r , and finally along L + ( z ) upwards from this point to infinity. Weassume that r is sufficiently large so that the distance between γ and the rays l ± begreater than δ . In view of Corollary 8.1, on γ ∩ c r the integrand in (4.2) is boundedby H . On γ \ c r it is estimated as when proving the first estimate in (4.8). (cid:3) Estimates in the lower half-plane.
Set(4.14) a ( p ) = e − iπ h + iπ − πih e πip h − ipπh − πip σ πh (4 πp − πh − π )) . We note that a is meromorphic in p , and its poles in ( −∞ ,
0] . One has
Proposition 4.2.
Pick positive X . Let p ∈ P . There is an Y > independent of p and h and such that, for y ≤ − Y and | x | ≤ X , one has the following results • Fix α , β so that < β < α < . Then, (4.15) µ ( z ) = e πipzh (cid:16) a ( p ) + O H (cid:0) e − πβ | y | (cid:1)(cid:17) , if Re p ≥ αh/ , • Fix α and β (cid:48) so that < α < and < β (cid:48) < . Then µ ( z ) = a ( p ) e πipzh + a ( − p ) e − πipzh + O H (cid:16) e πipzh − πβ (cid:48) | y | (cid:17) , if 0 ≤ Re p ≤ αh/ . (4.16) Proof.
Let us begin with justifying (4.15). Remind that µ has the integral rep-resentation (4.2). For y <
0, the behavior of µ appears to be determined by therightmost poles of v .The poles of v are at the points listed in (4.4). As p ∈ P , they are inside thestrip | Im k | ≤ h . As Re p ≥ αh/
2, 0 < α <
1, we see that, to the right of the lineRe k = Re p − αh , the function v has only one simple pole; it is situated at k = p .We deform γ , the integration contour in (4.2), to a curve that has the sameasymptotes as γ , but, instead of staying to the right of all the poles of v , it goesbetween the pole at k = p and the other ones (they stay to the left of this curve).We keep for the new integration curve the old notation γ . One has µ = A + B,A = 2 πi √ h res k = p I ( k ) , B = 1 √ h (cid:90) γ I ( k ) dk, I = e πikzh v ( k ) . (4.17)Using the representation (4.3), the information on the poles of the function σ πh ( z )from section 8.1.2, and formula (8.5), we get A = a ( p ) e πipzh with a ( p ) given by (4.14). igure 1. Curves γ and γ Now, to complete the proof of the proposition, we need only to estimate theterm B in (4.17). Let γ + ( γ − ) be the part of γ located above (resp., below) theline Im k = Im p . First, we choose γ ± , and then, we prove that(4.18) (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:90) γ ± I ( k ) dk (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) ≤ H (cid:12)(cid:12)(cid:12) e πipzh (cid:12)(cid:12)(cid:12) e − πβ | y | . We begin with estimating the integral along γ − . We remind that the exponential e πi ( k − k − ( z ))2 h governs the behavior of I ( k ), the integrand in (4.2), as Im k → −∞ ,see the beginning of the proof of Proposition 4.1. We assume that y < − Y with apositive Y . Therefore, Im k − ( z ) > Y .For ξ ∈ C , we denote by H i ( ξ ) the smooth curve described by an equation ofthe form Im ( k − k − ( z )) = c , c ∈ R , and containing ξ . If c = 0 this curve is oneof the straight lines k − ( z ) + R and k − ( z ) + i R , otherwise it is a hyperbola locatedin one of the sectors bounded by these lines. Its asymptotes are two half lines ofthese straight lines. Let H r ( ξ ) be the smooth curve described by an equation ofthe form Re ( k − k − ( z )) = c , c ∈ R , and containing ξ . If c = 0 this curve is one ofthe straight lines k − ( z ) + e ± iπ R , otherwise it is a hyperbola located in one of thesectors bounded by these lines, and they are its asymptotes.Set k = p − βh . As 0 < β < α <
1, the point k is to the right of the lineRe k = p − αh and to the left of p . As p ∈ P , one has | Im k | ≤ h < Y be sufficiently large, and y < − Y . Then the hyperbola H i ( k ) stays inthe half plane Im k < Im k − ( z ) and intersects the line Im k = − k .Wedenote by γ its segment of H i ( k ) between k and k .Furthermore, if Y is sufficiently large, H r ( k ) is a hyperbola located below k − ( z ).We denote by γ its segment between k and ∞ along which Re k → −∞ as k → ∞ .The curve γ − is the union of γ and γ , see Fig. 1.If Y is sufficiently big, then (1) the curve γ − does stays between p and all theother poles of v ; (2) its segment γ is located below the poles of the integrand at adistance greater than 1.Let us estimate (cid:82) γ I ( p ) dp . We note that, by the definition of H i and by (4.9), the xpression Im (cid:0) kz + k − k − kh (cid:1) is constant on H i ( k ). By (4.3) we get(4.19) (cid:12)(cid:12)(cid:12)(cid:12) √ h (cid:90) γ e πizkh v ( k ) dk (cid:12)(cid:12)(cid:12)(cid:12) ≤ C (cid:12)(cid:12)(cid:12) e − C ( Hi ) h (cid:12)(cid:12)(cid:12) √ h ×× (cid:90) γ | σ πh (2 π ( k + p − h − )) σ πh (2 π ( k − p − h − )) dk | where C ( H i ) = π Im (cid:0) kz + k − k − kh (cid:1)(cid:12)(cid:12) H i ( k ) , Computing C ( H i ) at the point k , we get(4.20) (cid:12)(cid:12)(cid:12) e − C ( Hi ) h (cid:12)(cid:12)(cid:12) ≤ C (cid:12)(cid:12)(cid:12) e πipzh (cid:12)(cid:12)(cid:12) e − πβ | y | . Let us estimate the integrand in the right hand side of (4.19). Using (2.5), we get σ πh (2 π ( k − p − h − )) = σ πh (2 π ( k − p + h − ))1 − e − πi ( k − p ) . For k ∈ γ , one hasRe ( k + p − h/ − / ≥ − / − h/ p − βh − C ( Y ) | Im ( k − p ) |≥ − / − h/ α − β ) h − C ( Y ) | Im ( k − p ) | , Re ( k − p + h/ − / ≥ − / − h/ − β ) h − C ( Y ) | Im ( k − p ) | . where C ( Y ) > Y → ∞ . These observations and Corollaries 8.1–8.2 imply that, for sufficiently large Y and k ∈ γ , | σ πh (2 π ( k + p − h − )) σ πh (2 π ( k − p − h − )) | ≤ H. This estimate and (4.20) imply estimate (4.18) with γ instead of γ ± .Consider the integral (cid:82) γ I ( k ) dk . As γ stays below the poles of I , at a distancegreater than 1, by means of Corollary 8.1, one immediately obtains | σ πh (2 π ( k + p − h − )) σ πh (2 π ( k − p − h − )) | ≤ H, k ∈ γ , and (cid:12)(cid:12)(cid:12)(cid:12)(cid:90) γ I dk (cid:12)(cid:12)(cid:12)(cid:12) ≤ H (cid:12)(cid:12)(cid:12)(cid:12) e πih (2 kz + k − k − kh ) | k = k (cid:12)(cid:12)(cid:12)(cid:12) (cid:90) γ (cid:12)(cid:12)(cid:12) e πih (( k − k − ( z )) − ( k − k − ( z )) ) dk (cid:12)(cid:12)(cid:12) . Clearly, (cid:12)(cid:12)(cid:12)(cid:12) e πih (2 kz + k − k − kh ) | k = k (cid:12)(cid:12)(cid:12)(cid:12) = e − C ( Hi ) h ≤ C (cid:12)(cid:12)(cid:12) e πipzh (cid:12)(cid:12)(cid:12) e − πβ | y | . We remind that curve γ ⊂ H r ( k ) goes to infinity approaching the asymptote e iπ/ ( −∞ , (cid:90) γ (cid:12)(cid:12)(cid:12) e πih (( k − k − ( z )) − ( k − k − ( z )) ) dk (cid:12)(cid:12)(cid:12) ≤ Ch/ | k − k − | . These estimates imply that (cid:82) γ I dk satisfies an estimate of the form (4.18). Thisimplies (4.18) with γ − .The estimates of the integral along γ + , the part of γ above the line Im k = Im p are similar. We omit the details and mention only that now the role of e i ( k − k − ( z ))2 h is played by the exponential e − i ( k − k +( z ))2 h . The obtained estimates and the formulafor A imply (4.15). This completes the proof of (4.15). Representation (4.16) isobtained similarly. (cid:3) .4. Rough estimates.
We shall need
Lemma 4.1.
Pick
X > . Let p ∈ P and | x | ≤ X . One has (4.21) | µ ( z ) | , | µ (cid:48) ( z ) | ≤ Hw ( x, y ) , w ( x, y ) = (1 + | y | ) (cid:40) e π ( | x |− / yh − πy , y ≥ ,e π Re p | y | h , y ≤ . When proving this lemma we use
Lemma 4.2.
Pick α ∈ (0 , . For p ∈ P one has (4.22) | a ( p ) | ≤ H if | p | ≥ αh/ , (4.23) | a ( p ) + a ( − p ) | ≤ H, | pa ( p ) | ≤ H if | p | ≤ αh/ . Proof.
Let | p | ≥ αh/ p ∈ P . Corollaries 8.1 and 8.2 imply that | σ πh (4 πp − πh − π )) | ≤ H . This and (4.14) lead to (4.22).Assume that | p | ≤ αh/ p is not necessarily in P ). In view of (4.14), it sufficesto check that | σ πh (4 πp − πh − π ) + σ πh ( − πp − πh − π ) | ≤ Ch , | pσ πh (4 πp − πh − π ) | ≤ C. Both the functions p (cid:55)→ σ πh (4 πp − πh − π )+ σ πh ( − πp − πh − π ) and p (cid:55)→ pσ πh (4 πp − πh − π ) are analytic in p in the αh -neighborhood of zero (see section 8.1.2). ByTheorem 8.2 σ πh (2 π (2 p − h − )) is bounded by C/h at the boundary of thisneighborhood. This and the maximum principle imply the needed estimates. (cid:3)
Now we can check Lemma 4.1.
Proof.
Let p ∈ P . Pick Y sufficiently large. For y ≥ − Y , the estimate for µ followsdirectly from Proposition 4.1. For | y | ≤ Y , the estimate for µ (cid:48) follows from theestimate for µ and the Cauchy estimates for the derivatives of analytic functions(as X and Y were chosen rather arbitrarily). Let y ≥ Y and | x | ≤ X . The firstestimate in (4.21) implies that, in the ( h/y )-neighborhood of z , µ is bounded by Hw ( x, y ) (we again use the fact that X and Y were chosen rather arbitrarily). Thisand the Cauchy estimates for the derivatives of analytic functions lead then to theestimate | µ (cid:48) ( z ) | ≤ yHw ( x, y ). This completes the proof of (4.21) for y ≥ − Y .Let us prove (4.21) for y ≤ − Y . Pick 0 < α <
1. Let | x | ≤ X , y ≤ − Y andRe p ≥ αh/
2. Estimate (4.22) and Proposition 4.2 imply that(4.24) | µ ( z ) e − πipzh | ≤ H. By means of the Cauchy estimates for the derivatives of analytic functions we getthe estimate(4.25) (cid:12)(cid:12)(cid:12)(cid:12) ddz ( µ ( z ) e − πipzh ) (cid:12)(cid:12)(cid:12)(cid:12) ≤ H. Estimates (4.24)–(4.25) lead to (4.21) for | x | ≤ X , y ≤ − Y and Re p ≥ αh/
2. If0 ≤ Re p ≤ αh/
2, the estimates for µ and its derivative are deduced from (4.16)and (4.23) similarly. We omit further details. (cid:3) One more solution to the model equation.
Let(4.26) ˜ µ ( z, p ) = e − iπz/h µ ( z + 1 / , / − p ) , were we indicate the dependence of µ on p explicitly. Together with µ , ˜ µ is asolution to (4.1). It is entire in z and p . We use it to construct entire solutions tothe Harper equation. Here, we compute the Wronskian { µ, ˜ µ } = µ ( z + h )˜ µ ( z ) − µ ( z )˜ µ ( z + h ). emma 4.3. For all p ∈ C , (4.27) { µ, ˜ µ } = ie − πip h − iπ h − πih . Proof.
Pick
X > < α < . Assume that αh/ ≤ p ≤ / − αh/
2. Then,by means of (4.26), (4.15) and (4.22), we check that, uniformly in x ∈ [ − X, X ], as y → −∞ the Wronskian { µ, ˜ µ } tends to(4.28) 2 ie iπ h − iπph a (1 / − p ) a ( p ) sin(2 πp ) . By means of (4.21), we check that, as y → + ∞ , uniformly in x ∈ [ − h, |{ µ ( z ) , ˜ µ ( z ) }| ≤ H | y | (cid:16) e πyh ( x − + | x +1 / | ) + 1 (cid:17) ≤ H | y | . The Wronskian being entire (as µ and ˜ µ ) and h -periodic in z (as the Wronskian ofany two solutions to a one-dimensional difference Schr¨odinger equation, see sec-tion 3.1.3), we conclude that it is independent of z and equals the expressionin (4.28). As the Wronskian is entire in p , this statement is valid for all p . Fi-nally, using the definition of a , equation (2.5) and formula (8.3), we check that(4.29) 2 i a (1 / − p ) a ( p ) sin(2 πp ) = ie − πip h + iπph − iπ h − πih . This leads to the statement of the lemma. (cid:3) Analytic solution to Harper equation
Preliminaries.
For Y ∈ R , we set C + ( Y ) = { y > Y } . Here, we pick Y > λ > C + ( − Y ).Below, we represent the spectral parameter E in the form E = 2 cos(2 πp ) andconsider solutions to (2.1) as functions of the parameter p .As λ is small, then, when constructing solutions to (2.1) in C + ( − Y ), it is naturalto rewrite this equation in the form(5.1) ψ ( z + h ) + ψ ( z − h ) + λ e − πiz ψ ( z ) − πp ) ψ ( z ) = − λ e πiz ψ ( z ) , so that the term in the right hand side could be considered as a perturbation. Let ξ = π ln λ . Then µ ( z + iξ ) is a solution to the unperturbed equation(5.2) ψ ( z + h ) + ψ ( z − h ) + λ e − πiz ψ ( z ) − πp ) ψ ( z ) = 0 . We construct ψ , an analytic solution to equation (5.1) close to µ ( z + iξ ). We prove Theorem 5.1.
Pick positive Y . There exists C such that if λ ≤ e − Ch , then: • There is ψ , a solution to (2.1) analytic in ( z, p ) ∈ C + ( − Y ) × P ; • Pick positive X . As y → + ∞ , uniformly in x ∈ [ − X, X ](5.3) ψ ( z ) = e iπ ξ − ( z + iξ ) v − (1 + κ + o (1)) − e − iπ ξ + ( z + iξ ) v + (1 + o (1)) , where κ is a constant satisfying the estimate | κ | ≤ Hλ ν ( p ) (1 + | ξ | ) , ν ( p ) = min (cid:110) , − ph (cid:111) . • Fix
X > . For | x | ≤ X and y ≤ Y / , one has (5.4) | ψ ( z ) − µ ( z + iξ ) | ≤ Hλ (1 + | ξ | ) e π Re p | ξ | /h . The rest of the section is devoted to the proof of Theorem 5.1.Let us explain the idea of the proof of Theorem 5.1. Let γ = { z ∈ i R : y ≥ − Y } .We study the integral equation(5.5) ψ ( z ) = µ ( z + iξ ) − ( Kψ )( z ) , Kψ ( z ) = (cid:90) γ κ ( z, z (cid:48) ) ψ ( z (cid:48) ) dz (cid:48) , z ∈ γ. he kernel κ is constructed in terms of µ ( · + iξ ) and ˜ µ ( · + iξ ), two linearly inde-pendent solutions to the unperturbed equation (5.2), κ ( z, z (cid:48) ) = λ ih Θ ( z, z (cid:48) ) [ µ ( z + iξ ) ˜ µ ( z (cid:48) + iξ ) − µ ( z (cid:48) + iξ ) ˜ µ ( z + iξ ) ] { µ, ˜ µ } e πiz (cid:48) , Θ ( z, z (cid:48) ) = c ot π ( z (cid:48) − z ) h − i. Similar integral operators have appeared in [5]. The kernel κ can be consideredas a difference analog of the resolvent kernel arising in the theory of differentialequations. First, we construct a solution to the integral equation (5.5), and then,we check that it is analytic in z and satisfies the difference equation (2.1). Finally,we obtain the asymptotics of this solution for y → + ∞ and for y ∼ Integral equation.
Here, we prove the existence of a solution (continuous in z and analytic in p ) to the integral equation. Below, q ( y ) = (1 + | y | ) (cid:40) e − πyh − πy , y ≥ ,e π Re p | y | h , y ≤ , ˜ q ( y ) = (1 + | y | ) (cid:40) e + πyh − πy , y ≥ ,e − π Re p | y | h , y ≤ . Estimates of µ and ˜ µ . To estimate the norm of the integral operator, we use
Corollary 5.1.
On the curve γ , the functions µ and ˜ µ satisfy the estimates (5.6) | µ ( z ) | , | µ (cid:48) ( z ) | ≤ Hq ( y ) , | ˜ µ ( z ) | , | ˜ µ (cid:48) ( z ) | ≤ H ˜ q ( y ) , This Corollary follows directly from Lemma 4.1.5.2.2.
A solution to the integral equation.
One has
Proposition 5.1.
Fix positive α < . There is a positive constant C such that if λ ≤ e − Ch , then the integral equation (5.5) has a solution ψ continuous in z ∈ γ ,analytic in p ∈ P and satisfying the estimate (5.7) | ψ ( z ) − µ ( z + iξ ) | ≤ λ α Hq ( y + ξ ) , z ∈ γ. Proof.
Let C ( γ, q ) be the space of functions defined and continuous on γ and havingthe finite norm (cid:107) f (cid:107) = sup z ∈ γ | q − ( y + ξ ) f ( z ) | . The proof is based on Lemma 5.1.
For z, z (cid:48) ∈ γ , one has (5.8) | q − ( y + ξ ) κ ( z, z (cid:48) ) q ( y (cid:48) + ξ ) | ≤ λ H e − πy (cid:48) (1 + ξ ) . First, we prove the proposition, and then, we check estimate (5.8). This estimateimplies that the norm of K as an operator acting in C ( γ, q ) is bounded by λ H (1 + ξ ). By Corollary 5.1, µ ( . + ξ ) ∈ C ( γ, q ). So, there is a positive constant C suchthat, if λ (1 + ξ ) < e − C/h , then there is ψ , a solution to (5.5) from C ( γ, q ). Theestimate of the norm of the integral operator implies that | ψ ( z ) − µ ( z ) | ≤ λ (1 + ξ ) Hq ( y + ξ ) , z ∈ γ. This implies (5.7) for any fixed positive α <
1. The analyticity of ψ in p followsfrom one of µ and the uniformity of the estimates. The proof is complete. (cid:3) Let us prove Lemma 5.1. Below, z, z (cid:48) ∈ γ . We note that by (4.27), for p ∈ P one has C − ≤ { µ ( z ) , ˜ µ ( z ) } ≤ C .First, we consider the case where | y − y (cid:48) | ≥ h . In view of Corollary 5.1, we get | q − ( y + ξ ) κ ( z, z (cid:48) ) q ( y (cid:48) + ξ ) | ≤ λHe − πy (cid:48) ( E + E ) ,E = | Θ( z, z (cid:48) ) | ˜ q ( y (cid:48) + ξ ) q ( y (cid:48) + ξ ) , E = | Θ( z, z (cid:48) ) | ˜ q ( y + ξ ) q ( y + ξ ) q ( y (cid:48) + ξ ) . o justify (5.8), it suffices to check that E , ≤ C (1 + | ξ | ) . Note that(5.9) | Θ( z, z (cid:48) ) | ≤ C (cid:40) e − π ( y − y (cid:48) ) h if y − y (cid:48) ≥ h, y (cid:48) − y ≥ h. Clearly, E ≤ C ˜ q ( y (cid:48) + ξ ) q ( y (cid:48) + ξ ). For y (cid:48) ≥ − Y , we have˜ q ( y (cid:48) + ξ ) q ( y (cid:48) + ξ ) ≤ (1 + | y (cid:48) + ξ | ) e − π ( y (cid:48) + ξ ) ≤ C, if y (cid:48) + ξ ≥ , ˜ q ( y (cid:48) + ξ ) q ( y (cid:48) + ξ ) ≤ (1 + | y (cid:48) + ξ | ) ≤ C (1 + | ξ | ) otherwise . (5.10)This implies that E ≤ C (1 + | ξ | ) . To estimate E , we have to consider four cases.If y + ξ, y (cid:48) + ξ ≥
0, we have E ≤ | Θ( z, z (cid:48) ) | (1 + | y (cid:48) + ξ | ) e π ( y − y (cid:48) ) h − π ( y (cid:48) + ξ ) ≤ C | Θ( z, z (cid:48) ) | e π ( y − y (cid:48) ) h ≤ C. If y + ξ ≥ ≥ y (cid:48) + ξ , then E ≤ (1 + | y (cid:48) + ξ | ) | Θ( z, z (cid:48) ) | e π ( y + ξ ) h − π Re p ( y (cid:48) + ξ ) h ≤ C (1 + ξ ) e π (1 − p )( y (cid:48) + ξ ) h ≤ C (1 + ξ ) . If y (cid:48) + ξ ≥ ≥ y + ξ , then E ≤ (1 + | y (cid:48) + ξ | ) | Θ( z, z (cid:48) ) | e π Re p ( y + ξ ) h − π ( y (cid:48) + ξ ) h − π ( y (cid:48) + ξ ) ≤ C | Θ( z, z (cid:48) ) | e π Re p ( y + ξ ) h − π ( y (cid:48) + ξ ) h ≤ C | Θ( z, z (cid:48) ) | ≤ C. Finally, if y + ξ, y (cid:48) + ξ ≤
0, we get E ≤ (1 + | y (cid:48) + ξ | ) | Θ( z, z (cid:48) ) | e π Re p ( y − y (cid:48) ) h ≤ C (1 + | ξ | ) . These estimates imply that E ≤ C (1 + | ξ | ) . This completes the proof in the casewhere | y − y (cid:48) | ≥ h .Let us consider the case where | y − y (cid:48) | ≤ h . Let η = Im ζ . Using (4.21) we get | Θ( z, z (cid:48) ) ( µ ( z + iξ ) ˜ µ ( z (cid:48) + iξ ) − µ ( z (cid:48) + iξ ) ˜ µ ( z + iξ ) ) |≤ Ch max ζ ∈ γ, | y − η |≤ h | µ ( z + iξ ) ˜ µ (cid:48) ( ζ + iξ ) − µ (cid:48) ( ζ + iξ ) ˜ µ ( z + iξ ) |≤ H (cid:18) q ( y + ξ ) max | y − η |≤ h ˜ q ( η + ξ ) + ˜ q ( y + ξ ) max | y − η |≤ h q ( η + ξ ) (cid:19) ≤ Hq ( y + ξ )˜ q ( y + ξ ) , and, using (5.10), we again come to (5.8). This completes the proof. (cid:3) Note that Corollary 5.1 and Proposition 5.1 imply
Corollary 5.2.
In the case of Proposition 5.1, there is C such that, for λ ≤ e − Ch , (5.11) | ψ ( z ) | ≤ Hq ( y + ξ ) , z ∈ γ. Analytic continuation of the solution to the integral equation.
Here,we prove the first point of Theorem 5.1. One has
Lemma 5.2.
The solution ψ can be analytically continued in C + ( − Y ) . Similar statements were checked in [5] and [6], we outline the proof only for theconvenience of the reader.
Proof.
For z (cid:48) ∈ γ , the kernel κ ( z, z (cid:48) ) is analytic in z ∈ S h = {| Re z | < h, y > − Y } ,and the function Kψ can be analytically continued in S h ; ψ , being a solution to(5.5), can be also analytically continued in S h .Having proved that ψ is analytic in S h , one can deform the integration contourin the formula for Kψ inside S h , and check that, in fact, ψ can be analytically ontinued in S h = {| Re z | < h, y > − Y } . Continuing in this way, one comes tothe statement of the Lemma. (cid:3) Below, we denote by ψ also the analytic continuation of the old ψ .5.3.1. Function ψ and the difference equation. To check that ψ satisfies (5.1), weagain borrow an argument from [5] and [6].We call a curve in C vertical if along it x is a smooth function of y . For z ∈ C + ( − Y ) we denote by γ ( z ) a vertical curve that begins at − iY , goes upward to z ,then comes back to the imaginary axis and goes along it to + i ∞ . One can compute Kψ ( z ) by the formula in (5.5) with the integration path γ replaced with γ ( z ).Set ( H f ) ( z ) = f ( z + h ) + f ( z − h ) + λ e − πiz f ( z ) − πp ) f ( z ). Then, H ψ = −H Kψ . Using (5.5) with γ replaced with γ ( z ), z ∈ C + ( − Y ), we get( H K ψ ) ( z ) = 2 πi res z (cid:48) = z κ ( z + h, z (cid:48) ) ψ ( z (cid:48) ) = λ e πiz ψ ( z ) . Thus, we come to
Lemma 5.3.
The solution ψ satisfies equation (2.1) in C + ( − Y ) . The last two lemmas imply the first point of the Theorem 5.1.5.4.
Asymptotics in the upper half-plane.
We get asymptotics (5.3) using theintegral equation for ψ . First, we pick δ ∈ (0 ,
1) and sufficiently large
Y > | x | ≤ δh and y > Y , and represent ( Kψ )( z ) in the form(5.12) ( Kψ )( z ) = λ ih { µ, ˜ µ } ( µ ( z + iξ )( I (˜ µ ) + J (˜ µ )) − ˜ µ ( z + iξ )( I ( µ ) + J ( µ ))) , where(5.13) I ( f ) = (cid:90) γ, | y − y (cid:48) |≥ h Θ ( z, z (cid:48) ) f ( z (cid:48) + iξ ) e πiz (cid:48) ψ ( z (cid:48) ) dz (cid:48) , (5.14) J ( f ) = (cid:90) γ, | y − y (cid:48) |≤ h Θ ( z, z (cid:48) ) ( f ( z (cid:48) + iξ ) − f ( z + iξ )) e πiz (cid:48) ψ ( z (cid:48) ) dz (cid:48) . Let us estimate I ( µ ), I (˜ µ ), J ( µ ) and J (˜ µ ). The first two are defined ∀ x . One has Lemma 5.4.
Let p ∈ P . We pick X > . As y → + ∞ , uniformly in x ∈ [ − X, X ](5.15) I (˜ µ )( z ) = o (1) , I ( µ )( z ) = e πi ( z + iξ ) h ( a + o (1)) , where (5.16) a = 2 i (cid:90) γ e − πi ( z (cid:48) + iξ ) h µ ( z (cid:48) + iξ ) e πiz (cid:48) ψ ( z (cid:48) ) dz (cid:48) . One has (5.17) a = O (cid:16) (1 + | ξ | ) λ ν ( p ) H (cid:17) , ν ( p ) = min (cid:110) , − ph (cid:111) . Note that the integral in (5.16) converges in view of (5.6) and (5.11).
Proof.
Let us estimate I (˜ µ ). We assume that y + ξ >
0. Using (5.9), (5.6) and (5.11),we get | I (˜ µ )( z ) | ≤ H (cid:18)(cid:90) ∞ y (1 + | y (cid:48) + ξ | ) e − π ( y (cid:48) + ξ ) e − πy (cid:48) dy (cid:48) + (cid:90) y − ξ e − π ( y − y (cid:48) ) h (1 + | y (cid:48) + ξ | ) e − π ( y (cid:48) + ξ ) e − πy (cid:48) dy (cid:48) + (cid:90) − ξ − Y e − π ( y − y (cid:48) ) h (1 + | y (cid:48) + ξ | ) e − πy (cid:48) dy (cid:48) (cid:33) . he expression in the brackets is bounded by (cid:90) ∞ y e − πy (cid:48) dy (cid:48) + (cid:90) y − ξ e − π ( y − y (cid:48) ) h e − πy (cid:48) dy (cid:48) + (1 + | Y − ξ | ) (cid:90) − ξ − Y e − π ( y − y (cid:48) ) h e − πy (cid:48) dy (cid:48) . As h <
1, this implies the first estimate from (5.15).Let us turn to I ( µ ). Arguing as when estimating I (˜ µ ), we check that(5.18) (cid:12)(cid:12)(cid:12)(cid:12)(cid:90) γ, y (cid:48) >y + h Θ ( z, z (cid:48) ) µ ( z (cid:48) + iξ ) e πiz (cid:48) ψ ( z (cid:48) ) dz (cid:48) (cid:12)(cid:12)(cid:12)(cid:12) ≤≤ H (cid:90) ∞ y + h (1 + | y (cid:48) + ξ | ) e − π ( y (cid:48) + ξ ) h − π ( y (cid:48) + ξ ) e − πy (cid:48) dy (cid:48) = o ( e − πyh )uniformly in x as y → + ∞ . Then, using the estimate (cid:12)(cid:12)(cid:12) Θ( z, z (cid:48) ) − ie − πi ( z (cid:48)− z ) h (cid:12)(cid:12)(cid:12) ≤ Ce − π ( y − y (cid:48) ) h , y − y (cid:48) ≥ h, and again arguing as before, we prove that as y → ∞ , uniformly in x (5.19) (cid:90) γ, y>y (cid:48) + h Θ ( z, z (cid:48) ) µ ( z (cid:48) + iξ ) e πiz (cid:48) ψ ( z (cid:48) ) dz (cid:48) = e πi ( z + iξ ) h a + o (cid:16) e − πyh (cid:17) with a from (5.16). Estimates (5.18)–(5.19) imply the second estimate in (5.15).Let us prove (5.17). Using (5.16) and estimates (5.6) and (5.11), we get | a | ≤ H (1 + | ξ | ) − ξ (cid:90) − Y e π (1 − p )( y (cid:48) + ξ ) h − πy (cid:48) dy (cid:48) + H + ∞ (cid:90) − ξ e − π (2 y (cid:48) + ξ ) (1 + | y (cid:48) + ξ | ) dy (cid:48) . One has max ≤ y (cid:48) ≤− ξ e π (1 − p )( y (cid:48) + ξ ) h − πy (cid:48) = e πν ( p ) ξ = λ ν ( p ) with ν ( p ) defined bythe second formula in (5.17). Furthermore, the second integral in the last esti-mate for a equals λ (cid:82) + ∞ e − πt (1 + t ) dt . These observations lead to (5.17). Thiscompletes the proof of the lemma. (cid:3) Let us turn to the terms J ( µ ) and J (˜ µ ). One has Lemma 5.5.
Fix δ ∈ (0 , . Let p ∈ P . As y → + ∞ , one has (5.20) J (˜ µ )( z ) = o (1) , J ( µ )( z ) = o ( e − πyh ) . These estimates are uniform in x ∈ [ − δh, δh ] .Proof. Let y + ξ ≥ h . Using estimates (4.21) for dµ/dz and estimate (5.11), we get | J ( µ ) | ≤ H (1+ | y + ξ | ) e πδ ( y + ξ ) − π ( y + ξ ) h − π ( y + ξ ) − πy (1+ | y + ξ | ) ≤ He − π ( y + ξ ) h − πy , where we used the inequalities | x | ≤ δh and 0 < δ <
1. This implies the firstestimate in (5.20). Similarly one proves the second one. (cid:3)
Now we are ready to prove (5.3). We do it in three steps. Below we assume that p ∈ P . All the o (1) are uniform in x . Fix δ ∈ (0 , | x | ≤ δh . Estimates (5.15) and (5.20) imply that as y → ∞ (5.21) ψ = µ ( z + iξ ) − ( Kψ )( z ) = µ ( z + iξ )(1+ o (1))+ e πi ( z + iξ ) h ˜ µ ( z + iξ )(˜ a + o (1))where ˜ a = λa/ (2 hi { µ, ˜ µ } ). Representation (5.21), and formulas (4.26) and (4.7)imply (5.3) with κ = ie iπ h ˜ a . This completes the proof of (5.3) for x ∈ [ − δh, δh ].We note that (5.3) implies that, for sufficiently large y ,(5.22) | ψ ( z ) | ≤ C (1 + | κ | ) e π | x | ( y + ξ ) h − π ( y + ξ ) h − π ( y + ξ ) . Now, we pick δ ∈ (1 / ,
1) and justify (5.3) assuming that hδ ≤ x ≤ h (1 + δ ).Let (cid:15) ∈ (0 , − δ ). We denote by γ ( z ) the curve that goes first along a straight ine from i ( y − h ) to the point z − h + (cid:15)h and next along a straight line from thispoint to i ( y + h ). One obtains representation (5.12) with J ( f ) defined by the newformula:(5.23) J ( f )( z ) = (cid:90) γ ( z ) Θ ( z, z (cid:48) ) f ( z (cid:48) + iξ ) e πiz (cid:48) ψ ( z (cid:48) ) dz (cid:48) . We note that on γ ( z ), one has | Θ ( z, z (cid:48) ) | ≤ C . As − δh ≤ ( δ − h < x − h + (cid:15)h ≤ ( δ + (cid:15) ) h, and as δ + (cid:15) <
1, we can and do assume that on γ ( z ) for sufficiently large y solution ψ satisfies (5.22). Using (5.22) and (4.21), we check that as y → + ∞| J ( µ )( z ) | ≤ H (1 + | κ | ) e π ( δ + (cid:15) − y + ξ ) − π ( y + ξ ) h +2 πξ (1 + | y + ξ | ) = o ( e − π ( y + ξ ) h ) . Reasoning analogously, we also prove that J (˜ µ )( z ) = o (1) as y → + ∞ . Thesetwo estimates and (5.15) lead again to (5.3). This completes the proof of (5.3) for0 ≤ x ≤ (1 + δ ) h . The case of (1 + δ ) h ≤ x ≤ To justify (5.3) for larger | x | , one uses equation (5.1). We discuss only thecase of x ≥ δ ∈ (0 ,
1) and
X >
0. In thecase of Theorem 5.3, we can assume that | κ | ≤ /
2. Then, 1 + κ (cid:54) = 0 and for x ∈ [ δh, X ] (5.3) actually means that ψ ( z ) = Aξ − ( z + iξ )(1 + o (1)) , A = e iπ (1 + κ ) v − . By (5.1), one can write ψ ( z ) = − λe − πi ( z − h ) ψ ( z − h ) (cid:16) o (1) + e πi ( z + iξ − h ) ψ ( z − h ) /ψ ( z − h ) (cid:17) . Let (1 + δ ) h ≤ x ≤ (2 + δ ) h . Then ψ ( z − h ) = Aξ − ( z + iξ − h )(1 + o (1)). Thisand (5.22) (that is valid on any given compact subinterval of ( − h, h )) imply that e πiz ψ ( z − h ) /ψ ( z − h ) = o (1). Therefore, ψ ( z ) = − λ e − πi ( z − h ) ψ ( z − h ) (1 + o (1)) = Aξ − ( z + iξ )(1 + o (1)) . So, we proved (5.3) for x ∈ [0 , (2 + δ ) h ]. Continuing in this way, one proves (5.3)for all x such that 0 ≤ x ≤ X . This completes the proof of (5.3). (cid:3) Asymptotics in the lower half-plane.
Here, we prove the third statementof Theorem 5.1, i.e., estimate (5.4). We pick
X >
Y > | x | ≤ X and | y | ≤ Y /
2. We also assume that λ is so small (or λ ≤ e C/h with C so large) that Y < − ξ .The proof follows the same plan and uses the same estimates for µ , ˜ µ and ψ asfor studying ψ as y → ∞ . So, we omit elementary details. First, we represent Kψ in the form (5.12) with I and J given by (5.13)– (5.14).Then, using (5.6) and (5.11) and the rough estimate | Θ( z, z (cid:48) ) | ≤ C for | y − y (cid:48) | ≥ h ,we get(5.24) | I (˜ µ )( z ) | ≤ H (1 + | ξ | ) , | I ( µ )( z ) | ≤ H (1 + | ξ | ) e π Re p | ξ | /h . Let us turn to the terms J ( µ ) and J (˜ µ ). We first fix a positive δ <
1, andconsider the case where | x | ≤ δh . By means of Lemma 4.1 and (5.11) , we get(5.25) | J (˜ µ )( z ) | ≤ H (1 + | ξ | ) , | J ( µ )( z ) | ≤ H (1 + | ξ | ) e π Re p | ξ | /h . We recall that ψ ( z ) − µ ( z + iξ ) = Kψ ( z ). Estimating the right hand side bymeans of (5.24)–(5.25), the estimate from Lemma 4.1 for µ and (4.26), the definitionof ˜ µ , we get(5.26) | ψ ( z ) − µ ( z + iξ ) | ≤ λH (1 + | ξ | ) e π Re p | ξ | /h , i.e., representation (5.4) for | x | ≤ δh Now, we pick δ ∈ (1 / ,
1) and justify (5.26) for | x | ≤ (1 + δ ) h . s | y | < Y /
2, we can assume that the point z − h is above the lower end of γ . Thisallows to choose the curve γ ( z ) as in section 5.4, and redefine J by (5.23).We can and do assume that estimate (5.26) is proved on γ ( z ). This estimateand (4.21) imply that on γ ( z ) one has | ψ ( z ) | ≤ H (1 + | ξ | ) e π Re p | ξ | /h . Using thisand (4.21) we obtain for the new J ( µ ) and J (˜ µ ) the old estimates (5.25), and,therefore, we come to (5.26) for δh ≤ x ≤ (1 + δ ) h . The case of negative x istreated similarly. Let us prove (5.26) for all | x | ≤ X . Therefore, we use a difference analog of theGr¨onwall’s inequality. We discuss only the case where x >
0. The case of x < δ ( z ) = ψ ( z ) − µ ( z + iξ ). Equations (2.1) and (4.1) for ψ and µ imply that δ ( z + h ) + δ ( z − h ) + 2( λ cos(2 πz ) − cos(2 πp )) δ ( z ) = − λe πiz µ ( z + iξ ) . Therefore, ∆( z + h ) = M ( z )∆( z ) − λe πiz µ ( z + iξ ) e , where∆( z ) = (cid:18) δ ( z ) δ ( z − h ) (cid:19) , M ( z ) = (cid:18) πp ) − λ cos(2 πz )) −
11 0 (cid:19) , e = (cid:18) (cid:19) . Let n ( z ) = (cid:107) ∆( z ) (cid:107) C and A = max | y |≤ Y (cid:107) M ( z ) (cid:107) , where (cid:107) · (cid:107) is the Hilbert-Schmidtnorm. In view of (4.21) we obtain(5.27) n ( z ) ≤ B + A n ( z − h ) , B = λ (1 + | ξ | ) He π Re p | ξ | /h . Therefore, for N ∈ N , we have(5.28) n ( z ) ≤ B (1 + A + ...A N − ) + A N n ( z − N h ) =
B A N − A − A N n ( z − N h ) . Assume that h ≤ x ≤ X and choose N so that h ≤ x − N h ≤ h . Then, • by (5.26), n ( z − N h ) ≤ λH (1 + | ξ | ) e π Re p | ξ | /h ; • ≤ N ≤ C/h , and A N ≤ H .Using these observations and the estimate for B from (5.27), we deduce from (5.28)estimate (5.26) for 0 ≤ x ≤ X . This completes the proof of Theorem 5.1.6. A monodromy matrix for the Harper equation
Here we give a definition of the minimal entire solutions to the Harper equa-tion and describe the monodromy matrix corresponding to a basis of two minimalentire solutions. This is the matrix described in Theorem 2.1. Then, we proveTheorem 2.2.6.1.
Minimal entire solutions and monodromy matrices.
In [6] the authorsstudied entire solutions to equation (3.1) with an SL (2 , C )-valued 2 π -periodic entirefunction M . Using the equivalence described in Section 3.1.3, we turn their resultsinto results for the Harper equation.Let Y ∈ R . Below, C + ( Y ) = { z ∈ C : y > Y } and C − ( Y ) = { z ∈ C : y < Y } .6.1.1. Solutions with the simplest behavior as y → ±∞ . To characterize the behav-ior of an entire solution as y → ±∞ , we express it in terms of solutions having thesimplest asymptotic behavior as y → ±∞ . Let us describe these solutions. Thenext theorem follows from Theorem 1.1a from [6]. Theorem 6.1. If Y > is sufficiently large, there exist two solutions u ± to (2.1) that are analytic in the half-plane C + ( Y ) and admit there the representations (6.1) u ± ( z ) = e ± iπh ( z − + iξ ) + iπz + o (1) , y → + ∞ . ne has (6.2) { u + ( z ) , u − ( z ) } = λ. Moreover, u ± are Bloch solutions in the sense of [6] , i.e., the ratios u ± ( z +1) /u ± ( z ) are h -periodic in z . Remark 6.1.
The expressions u ± ( z ) = e ± iπh ( z − + iξ ) + iπz , the leadingterms in (6.1), satisfy the equations u ± ( z ∓ h ) + λe − πiz u ± ( z ) = 0 (compare itwith the Harper equation!).We construct two solutions with the simplest asymptotic behavior as y → −∞ bythe formulas(6.3) d ± ( z ) = − u ± (1 − z ) , z ∈ C − ( − Y ) . We use
Lemma 6.1.
One has (6.4) d ∗± ( z ) = α ± ( z ) u ∓ ( z ) , z ∈ C + ( Y ) , where α ± are analytic and h -periodic functions, and, as y → ∞ , α ( z ) → uni-formly in x .Proof. We check (6.4) for d + . Mutatis mutandis, for d − , the analysis is the same.As { u + ( z ) , u − ( z ) } , the Wronskian of u + and u − , does not vanish in C ( Y ), one has(see section 3.1.3) d ∗ + ( z ) = α ( z ) u − ( z )+ β ( z ) u + ( z ) , α ( z ) = { u + ( z ) , d ∗ + ( z ) }{ u + ( z ) , u − ( z ) } , β ( z ) = { d ∗ + ( z ) , u − ( z ) }{ u + ( z ) , u − ( z ) } . The coefficients α and β are h -periodic and analytic in C + ( Y ).We recall that u − is a Bloch solution. It means that the ratio r ( z ) = u − ( z +1) /u − ( z )is h -periodic. Using (6.1), we get the asymptotic representation r ( z ) = − λ /h e − πiz/h + o (1) , y → ∞ , where the error estimate is uniform in x . This implies that, for sufficiently large y ,the solution u − tends to zero as x → −∞ . Similarly one proves that d ∗ + does thesame. Therefore, being periodic, the Wronskian { d ∗ + ( z ) , u − ( z ) } equals zero. Thisimplies that β = 0.Using (6.1), we check that as y → ∞ one has α ( z ) = 1 + o (1) uniformly in x . (cid:3) The minimal solutions.
Let ψ be an entire solution to (2.1), and let Y be asin Theorem 6.1. Then, ψ admits the representations: ψ ( z ) = A ( z ) u + ( z ) + B ( z ) u − ( z ) , z ∈ C + ( Y ) , (6.5) ψ ( z ) = C ( z ) d + ( z ) + D ( z ) d − ( z ) , z ∈ C − ( − Y ) , (6.6)where A , B , C and D are analytic and h -periodic in z . The solution ψ is called minimal if A , B , C and D are bounded and one of them tends to zero as y tendsto either −∞ or + ∞ .Let ψ be a minimal solution such that lim y →−∞ D ( z ) = 0 and C ( − i ∞ ) =lim y →−∞ C ( z ) (cid:54) = 0. We set ψ D ( z ) = ψ ( z ) /C ( − i ∞ ). In section 7, for sufficientlysmall λ , we construct ψ D in terms of the solution ψ from section 5.Let A , B , C and D be defined for ψ = ψ D by (6.5) and (6.6). The limits A D = lim y →∞ A ( z ) , B D = lim y →∞ B ( z ) , C D = lim y →−∞ C ( z ) , D D = lim y →−∞ e πiz/h D ( z )are called the asymptotic coefficients of ψ D . By definition of ψ D one has C D = 1. .1.3. The monodromy matrix.
In terms of ψ D , we define one more solution to theHarper equation (2.1) by the formula(6.7) ψ B ( z ) = ψ D (1 − z ) . Clearly, ψ B is one more minimal entire solution.Theorem 7.2 from [6] can be formulated as follows: Theorem 6.2.
The minimal entire solutions ψ D and ψ B exist. They, their asymp-totic coefficients and their Wronskian are nontrivial meromorphic functions of E .The Wronskian is independent of z . The monodromy matrix corresponding to ψ D and ψ B is of the form (2.2) , and (6.8) s = − λ D D B D , t = − λ A D , λ = λ h = e πξh . In the case of λ = 1, the analysis of the poles of s and t was done in [5]. Mutatismutandis, it can be done similarly in the general case. One can see that s and t areanalytic on the interval I = [ − − λ, λ ]. In section 7, as λ →
0, we computethe asymptotics of s and t for E ∈ I and so check independently their analyticity.6.2. Symmetries and the monodromy matrix.
For a function f of z and E ,we let f ∗ ( z, E ) = f (¯ z, ¯ E ). It is clear that f and f ∗ satisfy (2.1) simultaneously.Here, using this symmetry, we prove Theorem 2.2.6.2.1. A relation for the monodromy matrix.
Let us consider E such that ψ D and ψ B form a basis in the space of solutions to Harper equation. The monodromymatrix corresponding to this basis is defined by (3.12) with ψ = ψ D and ψ = ψ B .Below we denote it by M (instead of M ).As ψ ∗ D and ψ ∗ B also are solutions to Harper equation, one can write(6.9) Ψ ∗ ( z ) = S ( z/h )Ψ( z ) , Ψ ∗ ( z ) = (cid:18) ψ ∗ D ( z ) ψ ∗ B ( z ) (cid:19) , where S is a 2 × z and meromorphicin E like the basis solutions. One has Lemma 6.2.
The matrices M and S satisfy the relation (6.10) S ( z + h ) M ( z ) = M ∗ ( z ) S ( z ) , h = { /h } , where M ∗ is obtained from M by applying the operation ∗ to each of its entries.Proof. The definition of the monodromy matrix and (6.9) imply that S (( z + 1) /h ) M ( z/h ) Ψ( z ) = M ∗ ( z/h ) S ( z/h ) Ψ( z ) . Let (Ψ( z + h ) , Ψ( z )) be the matrix made of the column vectors Ψ( z + h ) and Ψ( z ).As the functions S and M are 1-periodic, we get S (( z + 1) /h ) M ( z/h ) (Ψ( z + h ) , Ψ( z )) = M ∗ ( z/h ) S ( z/h ) (Ψ( z + h ) , Ψ( z )) . As det(Ψ( z + h ) , Ψ( z )) = { ψ D ( z ) , ψ B ( z ) } , the determinant of (Ψ( z + h ) , Ψ( z )) is nontrivial, and we come to the relation S ( z/h + 1 /h ) M ( z/h ) = M ∗ ( z/h ) S ( z/h ) . As S is 1-periodic, it implies (6.10). (cid:3) .3. Properties of the matrix S . We start with the following elementary obser-vation.
Lemma 6.3.
One has (6.11) σ S ( h − z ) σ = S ( z ) , σ = (cid:18) (cid:19) . Proof.
In view of (6.7), one has Ψ(1 − z ) = σ Ψ( z ). This and (6.9) imply thatΨ ∗ ( z ) = σ S ( h − z/h ) σ Ψ( z ). Using (6.9) once more, we obtain the relation S ( z/h )Ψ( z ) = σ S ( h − z/h ) σ Ψ( z ). Using the argument from the end of the proofof Lemma 6.2, we deduce (6.11) from this relation. The proof is complete. (cid:3) Now, we check
Proposition 6.1.
The entries of S are independent of z . One has S = S = 1 B D , S = S = A D B D , (6.12) A ∗ D = A D , B D B ∗ D − A D A ∗ D = 1 . (6.13) Proof.
We prove formulas (6.12) for S and S . These formulas and relation (6.11)imply the formulas for the other entries of M .According to (3.10)– (3.11), relation (6.9) implies that(6.14) S ( z/h ) = { ψ ∗ D ( z ) , ψ B ( z ) }{ ψ D ( z ) , ψ B ( z ) } , S ( z/h ) = { ψ D ( z ) , ψ ∗ D ( z ) }{ ψ D ( z ) , ψ B ( z ) } . Below, we compute the Wronskians in (6.14) in terms of the asymptotic coefficientsof the solution Ψ D .Let us begin with { ψ D ( z ) , ψ B ( z ) } . We recall that, for sufficiently large Y ,solution ψ = ψ D admits representations (6.5) and (6.6) with h -periodic coefficients A , B , C and D , and D = e − πiz/h D ( z ), where D is bounded in C − ( − Y ). Bymeans of (6.7) and (6.3) we get for z ∈ C + ( Y ) { ψ D ( z ) , ψ B ( z ) } == (cid:110) A ( z ) u + ( z ) + B ( z ) u − ( z ) , − C (1 − z ) u + ( z ) − e − πi (1 − z ) h D (1 − z ) u − ( z ) (cid:111) = (cid:16) − e − πi (1 − z ) h A ( z ) D (1 − z ) + B ( z ) C (1 − z ) (cid:17) { u + ( z ) , u − ( z ) } . Using this representation, (6.2) and the definitions of the asymptotic coefficients of ψ D , see (6.1.2), we check that { ψ D ( z ) , ψ B ( z ) } → λB D C D = λB D as y → ∞ . Similarly, we prove that { ψ D ( z ) , ψ B ( z ) } == (cid:110) C ( z ) d + ( z ) + e − πiz/h D ( z ) d − ( z ) , − A (1 − z ) d + ( z ) − B (1 − z ) d − ( z ) (cid:111) −→ λB D as y → −∞ . As { ψ D ( z ) , ψ B ( z ) } is an h -periodic entire function, these observations imply that(6.15) { ψ D ( z ) , ψ B ( z ) } = λB D . By means of Lemma 6.1, one similarly computes the Wronskians { ψ ∗ D ( z ) , ψ B ( z ) } and { ψ D ( z ) , ψ ∗ D ( z ) } , and obtains the formulas { ψ ∗ D ( z ) , ψ B ( z ) } = λ + o (1) as y → + ∞ , { ψ ∗ D ( z ) , ψ B ( z ) } = λ ( B D B ∗ D − A D A ∗ D ) + o (1) as y → −∞ , nd { ψ D ( z ) , ψ ∗ D ( z ) } = λA D + o (1) as y → + ∞ , { ψ D ( z ) , ψ ∗ D ( z ) } = − λA ∗ D + o (1) as y → −∞ . The last four formulas imply (6.13) and formulas { ψ ∗ D ( z ) , ψ B ( z ) } = λ, { ψ D ( z ) , ψ ∗ D ( z ) } = λA D . This, (6.15) and (6.14) imply (6.12). (cid:3)
Proof of Theorem 2.2.
Formulas (6.10) and (6.12) imply the relation˜
S M ( z ) = M ∗ ( z ) ˜ S, ∀ z ∈ C , ˜ S = (cid:18) A D A D (cid:19) . This relation implies that M ( z ) + A D M ( z ) = M ∗ ( z ) A D + M ∗ ( z ) , ∀ z ∈ C . Substituting into this formula the expressions for the monodromy matrix entriesfrom (2.2), we get s + te − πiz + A D stλ = A D (cid:0) a ∗ − λ cos(2 πz ) (cid:1) + s ∗ + t ∗ e πiz , ∀ z ∈ C . This equality of two trigonometric polynomials leads to the relations t = − λ A D , s + A D stλ = a ∗ A D + s ∗ . The first of these two relations and the first formula in (6.13) imply that t = − t ∗ ,and substituting in the second one the formula A D = − t/λ and the formula for a from (2.2), one easily checks that ss ∗ = λ − t λ − t . These two observations imply (2.3).Let E ∈ R . One has t = iτ and s = − iλ (cid:113) − t λ − t e iα with real τ and α . Using theserepresentations, we get the formula (2.4): L = λ st (cid:0) − s − t (cid:1) + stλ = 2 τ (cid:112) (1 + τ )( λ + τ ) cos α. The proof of Theorem 2.2 is complete. (cid:3) Asymptotics of the monodromy matrix
Formulation of the Riemann-Hilbert problem.
Fix
Y >
0. Below C + = C + ( − Y ) and C − = C − ( Y ). To construct the solution ψ D , we paste it of so-lutions analytic in C + and solutions analytic in C − by means of a Riemann-Hilbertproblem. Here, we formulate this problem.7.1.1. Relations between entire solutions and solutions analytic in C ± . Let σ beeither the sign “+” or the sign “ − ”. Let S σ be the set of solutions to Harperequation that are analytic in C σ , and let K σ be the set of the complex valuedfunctions that are analytic and h -periodic in in C σ .Assume that ψ σ and φ σ belong S σ . Let w σ ( z ) = { ψ σ ( z ) , φ σ ( z ) } . One has w σ ∈ K σ .We assume that w σ does not vanish in C σ . Then, the pair ψ σ , φ σ is a basis in S σ .Any entire solution ψ to (3.1) admits the representations(7.1) ψ ( z ) = a + ( z ) ψ + ( z ) + b + ( z ) φ + ( z ) , z ∈ C + , a + , b + ∈ K + , (7.2) ψ ( z ) = a − ( z ) ψ − ( z ) + b − ( z ) φ − ( z ) , z ∈ C − , a − , b − ∈ K − , ith a ± ( z ) = 1 w ± ( z ) { ψ ( z ) , φ ± ( z ) } , b ± ( z ) = 1 w ± ( z ) { ψ ± ( z ) , ψ ( z ) } . It follows from (7.1) and (7.2) that(7.3) a + ( z ) ψ + ( z ) + b + ( z ) φ + ( z ) = a − ( z ) ψ − ( z ) + b − ( z ) φ − ( z ) , z ∈ R . This implies that(7.4) V + = G V − , z ∈ R , V + = (cid:18) a + b + (cid:19) , V − = (cid:18) a − b − (cid:19) , where G is the matrix with the entries(7.5) G ( z ) = w + ( z ) { ψ − ( z ) , φ + ( z ) } , G ( z ) = w + ( z ) { φ − ( z ) , φ + ( z ) } ,G ( z ) = w + ( z ) { ψ + ( z ) , ψ − ( z ) } , G ( z ) = w + ( z ) { ψ + ( z ) , φ − ( z ) } . Remark 7.1.
One has(7.6) det G ( z ) = w − ( z ) / w + ( z ) , z ∈ R . Indeed, (7.3) implies also the relation V − = 1 w − ( z ) (cid:18) { ψ + ( z ) , φ − ( z ) } { φ + ( z ) , φ − ( z ) }{ ψ − ( z ) , ψ + ( z ) } { ψ − ( z ) , φ + ( z ) } (cid:19) V + . One also can express V − via V + by inverting the matrix G in (7.4). Comparing theresults, one comes to (7.6).We have checked Lemma 7.1.
Any entire solution to (3.1) can be represented by (7.1) – (7.2) with a + , b + ∈ K + and a − , b − ∈ K − , and these coefficients satisfy the relation (7.4) withthe matrix G given by (7.5) . One can easily prove also
Lemma 7.2. If a + and b + belong to K + , a − and b − belong to K − , and if these fourfunctions satisfy relation (7.4) with the matrix G given by (7.5) , then formulae (7.1) – (7.2) describe an entire solution to (3.1) . Basis solutions for constructing ψ D . Let ψ ( z, p ) be the solution to (2.1)described in Proposition 5.1, and let(7.7) ψ + ( z, p ) = ψ ( z, p ) , φ + ( z, p ) = e − πi ( z + iξ ) h ψ ( z − / , / − p ) , and(7.8) ψ − ( z, p ) = ψ ∗ + ( z, p ) , φ − ( z, p ) = φ ∗ + ( z, p ) . Clearly, ψ + , φ + ∈ S + , and ψ − , φ − ∈ S − .To work with ψ ± and φ ± , we need to describe their behavior for large y and for y ∼
0. Theorem 5.1 and formulas (6.1) and (6.3) imply
Corollary 7.1.
In the case of Theorem 5.1, for sufficiently large Y , and for all z ∈ C + ( Y ) , one has ψ + ( z, p ) = A ψ ( z ) e πi ( z + iξ ) h u + ( z ) + B ψ ( z ) u − ( z ) ,φ + ( z, p ) = A φ ( z ) u + ( z ) + B φ ( z ) u − ( z ) . (7.9) For z ∈ C − ( − Y ) , one has ψ − ( z, p ) = B ∗ ψ ( z ) α ∗ + ( z ) d + ( z ) + A ∗ ψ ( z ) α ∗− ( z ) e − πi ( z − iξ ) h d − ( z ) ,φ − ( z, p ) = B ∗ φ ( z ) α ∗ + ( z ) d + ( z ) + A ∗ φ ( z ) α ∗− ( z ) d − ( z ) , (7.10) here α ± , A ψ , B ψ , A φ and B φ are h -periodic and analytic in z ; α ± are describedin Lemma 6.1, and one has A ψ ( z ) = A ψ, (1 + o (1)) , B ψ ( z ) = B ψ, (1 + o (1)) , y → + ∞ , (7.11) A ψ, = e iπ − πip h − iπ h − iπh e − πξ , B ψ, = e − iπ − iπh − πξ (1 + κ ) ,A φ ( z ) = A φ, (1 + o (1)) , B φ ( z ) = B φ, (1 + o (1)) , y → + ∞ , (7.12) A φ, ( p ) = − iA ψ, (1 / − p ) e − iπ h , B φ, ( p ) = − iB ψ, (1 / − p ) e − iπ h . These asymptotics are uniform in Re z . Furthermore, the third point of Theorem 5.1 and Proposition 4.2 imply
Corollary 7.2.
Let Y , λ and h be as in Theorem 5.1. Fix α ∈ (0 , and X > .For p ∈ P , | y | ≤ Y and | x | ≤ X the following holds. • Pick β ∈ (0 , α ) . If Re p ≥ αh/ , then (7.13) ψ + ( z ) = e πip ( z + iξ ) h (cid:0) a ( p ) + O H ( λ β ) (cid:1) , and if Re p ≤ − αh/ , then (7.14) φ + ( z ) = e − πip ( z + iξ ) h (cid:16) e − iπ (1 / − p ) h a (1 / − p ) + O H ( λ β ) (cid:17) . • Pick β ∈ (0 , . If Re p ≤ αh/ , then ψ + ( z ) = a ( p ) e πip ( z + iξ ) h + a ( − p ) e − πip ( z + iξ ) h + O H (cid:16) λ β e − πpξh (cid:17) , (7.15) and if p ≥ π − αh/ , then φ + ( z ) = e − πip ( z + iξ ) h e − iπ (1 / − p ) h a (1 / − p )+ e − πi (1 / − p )( z + iξ ) h e iπ (1 / − p ) h a ( p − /
2) + O H (cid:16) λ β e πpξh (cid:17) . (7.16)We complete this section by computing the Wronskians w ± = w ± ( z ). As(7.17) w − ( z ) = w ∗ + ( z ) , we need to compute only w + . One has Lemma 7.3.
Let Y , λ and h be as in Theorem 5.1. Fix < β < . For p ∈ P and z ∈ C + ( − Y / , one has (7.18) w + = w + O ( Hλ β ) , w = ie − πip h + ipπh − iπ h − iπh . Remark 7.2.
Lemma 7.3 implies that ψ + and φ + are linearly independent if λ < e − C/h , and C is sufficiently large. Proof.
First, we compute w + as y → + ∞ . Using (7.9) we get { ψ + , φ + } = (cid:110) A ψ e πi ( z + iξ ) h u + + B ψ u − , A φ u + + B φ u − (cid:111) = (cid:16) A ψ B φ e πi ( z + iξ ) h − B ψ A φ (cid:17) { u + , u − } = − λA φ, B ψ, + o (1) . So, the Wronskian is bounded as y → + ∞ .Let us pick β ∈ (0 ,
1) and check that(7.19) w + = 2 i sin(2 πp ) a ( p ) a (1 / − p ) e − iπ (1 / − p ) h + O H ( λ β ) , | y | ≤ Y / . Note that in view of (4.29) this already implies representation (7.18) for | y | ≤ Y / o prove (7.19), we pick α ∈ ( β,
1) and consequently consider four cases. In thecase where αh ≤ Re p ≤ / − αh formula (7.19) follows from representations (7.13),(7.14) and estimate (4.22). In the case where Re p ≤ min { αh , − αh } , we userepresentations (7.14)–(7.15) and estimates (4.22)–(4.23), and get w + = 2 i sin(2 πp ) a ( p ) a (1 / − p ) e − iπ (1 / − p ) h + O H ( ξλ β ) . As we can assume that β in the last formula is larger than in (7.19), we againobtain (7.19). The case where Re p ≥ max { αh , − αh } is treated similarly (bymeans of (7.13) and (7.16)). Finally, let − αh ≤ Re p ≤ αh . Note that in thiscase one has h > α > /
2. Now we get w + = 2 i sin(2 πp ) a ( p ) a (1 / − p ) e − iπ (1 / − p ) h − i sin(2 πp ) a ( − p ) a ( p − / e iπ (1 / − p ) h e − πi ( z + iξ ) h + O ( ξλ β ) . (7.20)Let us estimate the second term in the right hand side of (7.20). We denote it by T . According to (4.14), for p ∈ P , we have | T | ≤ Cλ /h (cid:12)(cid:12) sin(2 πp ) σ πh ( − πp − πh − π ) σ πh (4 π ( p − / − πh − π ) (cid:12)(cid:12) . By means of (2.5) and (8.2), we get | T | ≤ Cλ /h (cid:12)(cid:12)(cid:12)(cid:12) σ πh (4 π ( h/ − p ) − πh − π ) σ πh (4 πp − πh − π )1 − e − πiph (cid:12)(cid:12)(cid:12)(cid:12) . As − αh ≤ Re p ≤ αh and < h <
1, we have4 π ( h/ − Re p ) ≥ πh (1 − α ) > π (1 − α ) , π Re p ≥ π (1 − αh ) ≥ π (1 − α ) ,π (1 − α ) ≤ Re 2 πp/h ≤ απ. The first two inequalities and Corollary 8.1 imply that | σ πh (4 π ( h/ − p ) − πh − π ) σ πh (4 πp − πh − π ) | ≤ C . The third inequality implies that | − e − πip/h | ≥ C .These observations prove that | T | ≤ Cλ /h . As 1 / < h <
1, the term T can beincluded in the error term. This completes the proof of (7.18) for | y | ≤ Y / w + − w is h -periodic and analytic in z . Since it is bounded as y → + ∞ , representation (7.18) justified for y = − Y / w + − w = O ( Hλ β ) for all z ∈ C + ( − Y /
2) uniformly in p ∈ P . Thiscompletes the proof. (cid:3) Riemann-Hilbert problem for constructing ψ D . Let ψ ± and φ ± be the baseschosen in section 7.1.2. The minimal solution ψ = ψ D admits the representa-tions (7.1)- (7.2). The coefficients a ± , b ± ∈ K ± satisfy the equation (7.4) with thematrix G defined by (7.5). To formulate the Riemann-Hilbert problem for thesecoefficients, we need to study their behavior at ± i ∞ .The coefficients a ± and b ± being h -periodic, we shall regard them as functions ofthe variable ζ = e πiz/h . Let T = { ζ ∈ C : | ζ | = 1 } , B o = { ζ ∈ C : | ζ | ≤ } , B ∞ = { ζ ∈ C : | ζ | ≥ } ∪ {∞} . The function V + is analytic in B o \ { } , and V − is analytic in B ∞ \ {∞} .Substituting (7.9) into (7.1), we see that if a + and b + are bounded as ζ → y → + ∞ ), then ψ D admits representation (6.5) with A and B staying bounded as y → + ∞ , and one has(7.21) A D = A φ, b + (0) , B D = B ψ, a + (0) + B φ, b + (0) . Substituting (7.10) into (7.2) and taking into account Lemma 6.1, we see thatif, as ζ → ∞ , the coefficient a − is bounded and b − ( ζ ) →
0, then ψ D admits epresentation (6.6) with C staying bounded and D vanishing as ζ → ∞ . One has(7.22) 1 = C D = B ∗ ψ, a − ( ∞ ) , D D = e − πξh A ∗ ψ, a − ( ∞ ) + A ∗ φ, b − , ,b − , = lim ζ →∞ ( ζb − ( ζ )) . Let us collect the obtained information on the coefficients a ± and b ± . One has V + ( ζ ) = G ( ζ ) V − ( ζ ) , ζ ∈ T , (7.23) V + is analytic in B o , V − is analytic in B ∞ , V − ( ∞ ) = B ∗ ψ, (cid:18) (cid:19) , (7.24)Equation (7.23) and conditions (7.24) form a Riemann-Hilbert problem. We shallsee that, for sufficiently small λ , this problem has a unique solution. Having solvedthis problem, we shall reconstruct the coefficients of the minimal solution ψ D bythe formulae (7.21) and (7.22).7.2. Matrix G.
In this section, we study the matrix G .7.2.1. Functional relations.
The properties of the matrix G we discuss here imme-diately follow from (7.5), (7.7) and (7.8). When describing these properties, we usethe variable z , assume that λ < e C/h where C is sufficiently large and that p ∈ P .As w ± are bounded away from zero in the domain ( z, p ) ∈ {| y | ≤ Y / } × P , thematrix G is analytic there. Let g ij ( z, p ) = w + ( z, p ) G ij ( z, p ) , i, j ∈ { , } . As ψ − = ψ ∗ + and φ − = φ ∗ + , (7.5) implies that g ( z, p ) = g ∗ ( z, p ) , g ( z, p ) = − g ∗ ( z, p ) , g ( z, p ) = − g ∗ ( z, p ) . (7.25)Furthermore, relation (7.17) and the formula (7.6) imply thatdet G ∗ = det G − , g g − g g = w ∗ + w + . Finally, as φ + ( z, p ) = e − iπ ( z + iξ ) h ψ + ( z − / , / − p ), we get(7.26) g ( z, p ) = e πξh g ( z − / , / − p ) . The asymptotics of G for < p < / . Here we prove
Proposition 7.1.
Let < β < / . There is a constant C such that if λ < e − C/h ,then for p ∈ P such that h/ ≤ Re p ≤ / − h/ , and for | y | ≤ Y , one has G = w (cid:32) δ e πξph (2 iF (1 / − p ) sin(2 πp ) + δ ) e − πξph (2 iF ( p ) sin(2 πp ) + δ ) δ (cid:33) , where w is defined in (7.18) , δ denotes O ( λ β H ) , and F is the meromorphic func-tion such that (7.27) F ( p ) = | σ πh (4 πp − π − πh ) | , p ∈ R . Moreover, one has | F ( p ) | ≤ H for p ∈ P such that h/ ≤ Re p . So in the case of this proposition, for sufficiently small λ the matrix G appears tobe close to a constant one. Proof.
Below, we assume that all the hypotheses of the proposition are satisfied.First, we estimate the Wronskian g = { ψ − , φ + } . Using (7.13)– (7.14) and thedefinition ψ − from (7.8), we get g = (cid:110) e − πip ( z − iξ ) h ( a ∗ ( p ) + δ ) , e − πip ( z + iξ ) h ( e − iπ (1 / − p ) h a (1 / − p ) + δ ) (cid:111) . bviously, the leading term equals zero, and using estimate (4.22) we prove that g = O ( λ β H ). By means of the first relation from (7.25) we also see that g = O ( λ β H ). As G jj = ( w + ) − g jj , j = 1 ,
2, and in view of Lemma 7.3, we get theannounced estimate for the diagonal elements of the matrix G .Now consider g = { φ − ( z ) , φ + ( z ) } . Using (7.14), (7.8) and (4.22), we get g = { e πip ( z − iξ ) h ( a ∗ (1 / − p ) + δ ) , e − πip ( z + iξ ) h ( a (1 / − p ) + δ ) } == e πpξh (cid:16) i sin(2 πp ) | a (1 / − p ) | + O ( λ β H ) (cid:17) . Let us note that, to get this formula, instead of h/ ≤ p ≤ / − h/
4, we have onlyto assume that Re p ≤ / − h/
4. The definition of a , formula (4.14), implies that(7.28) F ( p ) = | a ( p ) | . Therefore,(7.29) g = e πpξh (cid:16) i F (1 / − p ) sin(2 πp ) + δ (cid:17) , and also, in view of (4.22) | F ( p ) | ≤ H for p ∈ P such that Re p ≥ h/
4. Thisestimate and representations (7.29) and (7.18) imply the formula for G announcedin the proposition. We note that it is valid for all p ∈ P such that Re p ≤ π − h/ G announced in theproposition. It is valid for all p ∈ P such that h/ ≤ Re p .We have checked all the statements of the proposition. (cid:3) To use Proposition 7.1, we need
Lemma 7.4.
One has (7.30) F ( p + 1 /
2) = 4 sin πph F ( p ) , (7.31) 4 sin (2 πp ) F (1 / − p ) F ( p ) = 1 . Proof.
Formula (7.30) follows from (8.2), and (7.31) follows from (8.3) and (2.5). (cid:3)
Proof of Theorem 2.3.
Here, we compute the coefficients s and t of the mon-odromy matrix in the case where p ∈ P is bounded away from 0 and 1 /
2. Therefore,first, we solve the Riemann-Hilbert problem (7.23)–(7.24) to find the asymptoticsof a + (0), b + (0), a − ( ∞ ) and lim ζ →∞ ζb − ( ζ ). Then, by means of formulae (7.21)and (7.22), we compute the coefficients A D , B D and D D of the minimal entiresolution ψ D . Finally, using formulae (6.8), we compute s and t .7.3.1. Solving the Riemann-Hilbert problem.
The leading term of the asymptoticsof G being independent of z , the analysis of the Riemann-Hilbert problem is ele-mentary. Assume that λ , z and p satisfy assumptions of Proposition 7.1. Let G = 2 i sin(2 πp ) w (cid:18) F (1 / − p ) F ( p ) 0 (cid:19) , T = (cid:18) τ τ − (cid:19) , τ = e πξph . Relation (7.31) implies that det G = 1 /w . In view of Proposition 7.1, we have(7.32) G = T G ( I + ∆) T − , ∆ = O ( λ β H ) . The term ∆ is analytic in ( z, p ) ∈ {| y | ≤ Y } × { p ∈ P : h/ ≤ Re p ≤ / − h/ } .Now, we pass to the variable ζ = e πiz/h . Let (cid:107) . (cid:107) be a matrix norm. Pick α ∈ (0 , T = {| ζ | = 1 } denote by (cid:107) . (cid:107) α the standardH¨older norm defined in terms of (cid:107) . (cid:107) . One has emma 7.5. Let ∆ be a matrix-valued function on T . If (cid:107) ∆ (cid:107) α is sufficiently small,then there exist unique matrix functions W ± such that W + is analytic in B o ,W − is analytic in B ∞ , W − ( ∞ ) = I,W + ( ζ ) = ( I + ∆( ζ )) W − ( ζ ) , ζ ∈ T . (7.33) These functions satisfy the estimates: (7.34) (cid:107) W + ( ζ ) − I (cid:107) ≤ C (cid:107) ∆ (cid:107) α , | ζ | ≤ (cid:107) W − ( ζ ) − I (cid:107) ≤ C (cid:107) ∆ (cid:107) α | ζ | , | ζ | ≥ . Proof.
The Lemma follows from standard results of the theory of singular integraloperators, see, e.g., [21]. So, we describe the proof omitting standard details.First, in H α ( T ), the space of matrix-valued H¨older functions on T , one constructsa solution to the equation(7.35) W − = I + S ( ∆ W − ) , where, for f ∈ H α ( T ), S ( f )( ζ ) = − f ( ζ ) + 12 πi v.p. (cid:90) T f ( ζ (cid:48) ) dζ (cid:48) ζ (cid:48) − ζ , ζ ∈ T , and the orientation of T is positive. As S is a bounded operator in H α ( T ), and asfor f, g ∈ H α ( T ) one has (cid:107) f g (cid:107) α ≤ (cid:107) f (cid:107) α (cid:107) g (cid:107) α , equation (7.35) has a unique solutionprovided (cid:107) ∆ (cid:107) α is sufficiently small.One defines W + ( ζ ) for ζ ∈ B o and W − ( ζ ) for ζ ∈ B ∞ by the formulas W ± ( ζ ) = I + 12 πi (cid:90) T ∆( ζ (cid:48) ) W − ( ζ (cid:48) ) ζ (cid:48) − ζ dζ (cid:48) , and checks that these two function have all the properties described in Lemma 7.5.We omit further details. (cid:3) In our case, in the ring e − πY/h ≤ | ζ | ≤ e πY/h , ∆ is analytic and satisfies theestimate (cid:107) ∆( ζ ) (cid:107) ≤ Hλ β . Therefore, for any fixed α ∈ (0 , (cid:107) ∆ (cid:107) α ≤ C ( α ) Hλ β . So, there is a C > λ ≤ e − C/h , then ∆ satisfies theassumptions of Lemma 7.5. For this ∆, we construct W ± by Lemma 7.5. Thevector-valued functions defined by the formulas(7.36) V + ( ζ ) = T G W + ( ζ ) T − e, V − ( ζ ) = T W − ( ζ ) T − e, e = 1 B ∗ ψ, (cid:18) (cid:19) , are a solution of the Riemann-Hilbert problem (7.24)–(7.23). Indeed, V + is analyticin B , V − is analytic in B ∞ , and V − ( ∞ ) = T W − ( ∞ ) T − e = e . Moreover, by (7.32)and (7.33), for | ζ | = 1, V + ( ζ ) = T G W + ( ζ ) T − e = T G ( I + ∆( ζ )) W − ( ζ ) T − e == T G ( I + ∆( ζ )) T − V − ( ζ ) = G ( ζ ) V − ( ζ ) . We compute the coefficients s and t of the monodromy matrix using (7.21)–(7.22),where a ± and b ± are the first and the second components of the vectors V ± (0). For-mulas (7.36), formula for B ψ, from (7.11) and the estimate for W − − I from (7.34)imply that(7.37) a − ( ∞ ) = 1 B ∗ ψ, , b − , = ζb − ( ζ ) (cid:12)(cid:12)(cid:12) ζ = ∞ = O ( λ β e − πξp/h + πξ H ) . sing also formula (7.18) for w and the estimate κ = O ( Hλ ) (following from thethird point of Proposition 5.1), we get(7.38) a + (0) = O (cid:0) e πξ λ β H (cid:1) , b + (0) = e − πξph w B ∗ ψ, (cid:0) i sin(2 πp ) F ( p ) + O ( λ β H ) (cid:1) . Asymptotics of the coefficients s and t . Using (6.8), (7.21), (7.22), we get(7.39) t = − λ A φ, b + (0) , s = − A ∗ ψ, a − ( ∞ ) + λ A ∗ φ, b − , B ψ, a + (0) + B φ, b + (0) . Using (7.38)–(7.37), estimates for a ± , b + and b − , , the estimate κ = O ( Hλ ),and (7.11) and (7.12), formulas for A ψ, , A φ, , B ψ, and B φ, , we obtain t = 2 ie π (1 / − p ) ξh F ( p ) sin(2 πp ) + O ( λ β H )1 + O ( λH ) , s = 12 i e πpξh e πiph + O ( λ λ β H )) F ( p ) sin(2 πp ) + O ( λ β H ) . Let us simplify these formulae. For h/ ≤ Re p ≤ / − h/
4, one has | F ( p ) | ≤ H (see Proposition 7.1). By this estimate and (7.31), one also has | F ( p ) sin(2 πp ) | − = | πp ) F (1 / − p ) | ≤ H. Using these observations, we get(7.40) t = 2 ie π (1 / − p ) ξh F ( p ) sin(2 πp )(1+ O ( λ β H )) , s = e πpξh e πiph (1 + O ( λ β H ))2 i F ( p ) sin(2 πp ) . Finally, by means of (2.5), we check that, for p ∈ R ,2 i sin(2 πp ) F ( p ) = 2 i sin(2 πp ) | σ πh (4 πp − π − πh ) | = − | σ (4 πp − π + πh ) | i sin(2 πp ) . This relation and (7.40) imply the statement of Theorem 2.3. (cid:3)
Asymptotics of s and t for p close to zero. Here we prove
Theorem 7.1.
Pick β ∈ (0 , . There is a positive constant C such that if λ Asymptotics of the matrix G . We have Proposition 7.2. Pick < β < . There is a positive constant C such that if λ < e − C/h , then, for p ∈ P , ≤ Re p ≤ h/ , and for | y | ≤ Y / , one has G = 1 w (cid:32) e πpξh F d ( p ) + δ e πpξh (2 iF (1 / − p ) sin(2 πp ) + δ ) e − πpξh ( F a ( p ) + δ ) e πpξh F ∗ d ( p ) + δ (cid:33) , where δ denotes O ( λ β H ) , F d ( p ) = − ie πiph − iπ h sin 2 πph sin(2 πp ) F ( − p ) , (7.42) F a ( p ) = 2 i sin(2 πp )( F ( p ) − e πpξh F ( − p )) , (7.43) nd F , F d and F a satisfy the estimates (7.44) p F ( p ) = O ( H ) , F d ( p ) = O ( H ) , F a ( p ) = O ( ξH ) . Proof. First, we note that G was already computed when proving Proposition 7.1:when computing it we assumed that 0 ≤ Re p ≤ / − h/ 4, and now, as 0 < h < ≤ Re p ≤ h/ ≤ / − h/ 4. The formula for G = g /w + follows fromone for G = g /w + , (7.25) and (7.18).So we only have to compute G = g /w + and G = g /w + . Using formu-las (7.15), (7.14) and (7.8), and estimates (4.22) and (4.23), we get g = e πpξh i sin(2 πp ) e − iπ (1 / − p ) h a ∗ ( − p ) a (1 / − p ) + O ((1 + | ξ | ) λ β H ) , (7.45) g = e − πpξh (cid:16) i sin(2 πp ) ( | a ( p ) | − e πpξh | a ( − p ) | ) + O ( λ β (1 + | ξ | ) H ) (cid:17) . By means of (4.14) and (8.2), we transform (7.45) to the form(7.46) g = e πpξh F d ( p ) + O ( Hλ β (1 + | ξ | )) . Furthermore, the definition of F a and relation (7.28) allows to get the formula(7.47) g = e − πpξh (cid:0) F a ( p ) + O ( λ β (1 + | ξ | ) H ) (cid:1) . Estimates (7.44) follow from (4.22) and (4.23).As we can assume that β in formulas (7.46) and (7.47) is larger than in Proposi-tion 7.2, these formulas, (7.18) and estimates (7.44) imply the representations for G and G from Proposition 7.2. This completes its proof. (cid:3) Solving the Riemann-Hilbert problem. Let(7.48) G = 1 w (cid:32) e πpξh F d ( p ) 2 iF (1 / − p ) sin(2 πp ) F a ( p ) e πpξh F ∗ d ( p ) . (cid:33) First, we prove that again det( G ) = 1 /w . This follows from (7.48), the definitionsof F a and F d , see (7.43) and (7.42), and Lemma 7.4.Then, we prove that G admits again representation (7.32) with the new G . Forthis, using the estimates for F d and F a from (7.44) and the estimate for F fromProposition 7.2, we check that G − T − GT = I + O ( λ β H ).Having obtained (7.32), we proceed as in the case where h/ ≤ Re p and obtain a + (0) = e πpξh F d ( p ) + O ( λ β H ) w B ∗ ψ, , b + (0) = e − πpξh ( F a ( p ) + O ( λ β H )) w B ∗ ψ, , (7.49) a − ( ∞ ) = 1 B ∗ ψ, , b − , = O ( e − πpξh + πξ λ β H ) . (7.50)7.4.3. Asymptotics of s and t . Computing the coefficients s and t by means offormulas (7.39), (7.49), and (7.50), and estimates (7.44), we get t = e π (1 / − p ) ξh (cid:0) F a ( p ) + O ( λ β H ) (cid:1) , s = e πiph + πpξh F a ( p ) + ie πi h + πpξh F d ( p ) + O ( λ β H ) . Let us prove (7.41) for t . In view of (7.27), (2.5), (2.6) and (7.43), one has4 sin (2 πp ) F ( p ) = F ( p ) , F ( − p ) F ( p ) = F ( − p ) F ( p ) = ρ ( p ) , F a ( p ) = iF ( p ) 1 − ρ ( p ) e πpξh πp ) . As 1 / − Re p ≥ / − h/ ≥ h/ 4, formula (7.31) and the estimate for F fromProposition 7.1 imply that 1 /F ( p ) = O ( H ). This and the last formulas for t and F a imply the formula for t from (7.41). he last formula for s and the relation F a ( p ) + ie πi h + πpξh F d ( p ) = iF ( p ) 1 − e πph ( ξ + i ) ρ ( p )2 sin(2 πp ) . imply the formula for s from (7.41). The proof of Theorem 7.1 is complete. (cid:3) A trigonometric analog of the Euler Gamma-function Here, following mostly [6, 12], we discuss equation (2.5) with a ∈ (0 , π ).8.1. Definition and elementary properties. S = {| x | < π + a } , equation (2.5) has a unique solution σ a that isanalytic, nonvanishing, and having the representations σ a ( z ) = 1 + o ( e − α | y | ) , y → −∞ , (8.1) σ a ( z ) = e − iz a + iπ a + ia + o ( e − α | y | ) , y → + ∞ , for any fixed α ∈ (0 , x . If a is boundedaway from zero, they are also uniform in a . The function σ a is continuous in a .8.1.2. Using equation (2.5), one can analytically continue σ a to a meromorphicfunction. Its poles are located at the points − π − a − πl − ak, l, k = 0 , , , . . . , and its zeros are at the points π + a + 2 πl + 2 ak, l, k = 0 , , , . . . . Its zero at π + a and its pole at − π − a are simple.8.1.3. The function σ a satisfies the following relations: σ a ( z + π ) = (1 + e − iπa z ) σ a ( z − π ) , (8.2) σ a ( − z ) = e − i a z + iπ a + ia σ a ( z ) , (8.3) σ a ( z ) = e i a z − iπ a − ia σ a ( z ) . (8.4)8.1.4. One also has(8.5) Res z = − π − a σ a = − iσ a ( − π + a ) = (cid:114) aπ e − iπ a − iπ − ia . Quasiclassical asymptotics. Here, we discuss σ a for small a .Thanks to (8.3), it suffices to study this function for y ≤ z (cid:55)→ ln (1 + e − iz ) analytic in C − , the lowerhalfplane, and satisfying the conditionln (1 + e − iz ) → , y → −∞ . We set L ( z ) = (cid:90) z − i ∞ ln (1 + e − iz (cid:48) ) dz (cid:48) , where we integrate in C − , say, along the line Re z (cid:48) = Re z . Theorem 8.1. [12] Pick < δ < π . In C − ∪ R outside the δ -neighborhood of thehalf-lines z ≥ π and z ≤ − π , for sufficiently small a , σ a admits the representation (8.6) σ a ( z ) = exp (cid:18) a L ( z ) + O (cid:16) a (1 + | x | ) e −| y | (cid:17)(cid:19) . et us discuss the behavior of σ a in a neighborhood of the point − π . Theorem 8.2. [12] Let < δ < π . For t in the δ -neighborhood of zero, one has (8.7) σ a ( − π + t ) = e ln(2 a )2 a t √ π Γ (cid:18) t + a a (cid:19) e a (cid:82) t ˜ l ( ζ ) dζ − iπ a + O ( a ) , ˜ l ( ζ ) = ln 1 − e − iζ ζ , where ˜ l is analytic, and ˜ l (0) = iπ/ . The error term in (8.7) is analytic in t . Uniform estimates. Fix δ ∈ (0 , π ) and κ ∈ (0 , Corollary 8.1. Outside the δ -neighborhood of the ray z < − π , one has σ a ( z ) = e O ( a − e − κ | y | (1+ | x | ) ) , y ≤ , (8.8) σ a ( z ) = e − iz a + iπ a + ia + O ( a − e − κ | y | (1+ | x | ) ) , y ≥ . (8.9) Proof. Estimate (8.9) follows from (8.8) and (8.4). Let us prove (8.8). Assumethat y ≤ 0. First, we note that (8.8) is valid for | x | ≤ a . Indeed, let a > < a < a . For these a , formula (8.8) follows directlyfrom (8.6). For a ≤ a ≤ π , it follows from (8.1) that is valid and uniform in a and in x if | x | ≤ a .Now, we assume that z is outside the δ -neighborhood of the ray z < − π . We choose n ∈ Z so that | x − an | ≤ a . By (2.5) σ a ( z ) = | n | (cid:89) j =1 (cid:16) e − i ( z ∓ (2 j − a ) (cid:17) ± σ a ( z − na ) if ± n ≥ . As | n | ≤ | x | a + , we get(8.10) | n | (cid:89) j =1 (cid:16) e − i ( z ∓ (2 j − a ) (cid:17) ± = e O ( | n | e −| y | ) = e O ( xe −| y | /a ) . Formula (8.8) valid for | x | ≤ a and (8.10) imply (8.8) for all z we consider. (cid:3) Fix positive c , c and δ < π . Corollary 8.2. Let | z + π | ≤ δ and x ≥ − π − a + c a − c | y | . Then, | σ a ( z ) | ≤ e C/a .Proof. 1) Let u = z + π + a a . Then, under the hypothesis of the corollary,Re u ≥ c − c | Im u | . Let D be the domain defined by this inequality in the complex plane of u . For u ∈ D , we set Y ( u ) = (cid:114) u π e − u (ln u − Γ( u )where the branches of ln . and √ . are analytic in C \ { z ≤ } and such that ln 1 = 0and √ Y is bounded in D . By (8.7) and the previous steps, under the hypothesis of the Corollary, | σ a ( z ) | ≤ C (cid:12)(cid:12)(cid:12)(cid:12) exp (cid:18) u (ln u − − 12 ln u + ln(2 a )2 a ( z + π ) + 12 a (cid:90) z + π ˜ l ( ζ ) dζ (cid:19)(cid:12)(cid:12)(cid:12)(cid:12) ≤ e C/a (cid:12)(cid:12)(cid:12)(cid:12) exp (cid:18) u ln u + ln(2 a )2 a ( z + π ) (cid:19)(cid:12)(cid:12)(cid:12)(cid:12) ≤ e C/a (cid:12)(cid:12)(cid:12) e u ln(2 au ) (cid:12)(cid:12)(cid:12) , (8.11)as ˜ l is analytic in the δ -neighborhood of zero and as | u | ≥ C in D . As | au | = | z + π + a | ≤ C , then one also has | au ln(2 au ) | ≤ C , and (8.11)implies that | σ a ( z ) | ≤ e C/a . (cid:3) eferences [1] A. Avila, S. Jitomirskaya. The ten martini problem. Ann. Math., 170 (2009), 303-342.[2] V. Babich, M. Lyalinov, V. Grikurov. Diffraction theory: the Sommerfeld-Malyuzhinets tech-nique. Alpha Science, Oxford, 2008.[3] M. Bobrovnikov and V. Firsanov. Wave diffraction in angular domains. Tomsk State Uni-versity, Tomsk, 1988.[4] V. S. Buslaev and A. A. Fedotov. The complex WKB method for the Harper equation. St.Petersburg Math. J., StPetersburg Math. Journal , 8 (1996), 65-97.[6] Buslaev V. and Fedotov A. On the difference equations with periodic coefficients. Adv. Theor.Math. Phys. Schr¨odinger Operators . Springer Verlag,Berlin, 1987.[8] L. Faddeev, R. Kashaev, A. Volkov. Strongly coupled quantum discrete Liouville theory. I:Algebraic approach and duality. Comm. Math. Phys., 219 (2001), 199-219.[9] A. Fedotov and F. Klopp. Strong resonant tunneling, level repulsion and spectral type for one-dimensional adiabatic quasi-periodic Schr¨odinger operators. Annales Scientifiques de l’EcoleNormale Sup´erieure, 4e s´erie , 38(2005), 889-950.[10] A. Fedotov. Monodromization method in the theory of almost-periodic equations. St. Peters-burg Math. J., 25 (2014), 303-325.[11] Alexander Fedotov, Fedor Sandomirskiy. An exact renormalization formula for the Marylandmodel. Communications in Mathematical Physics, 334 (2015), 1083-1099.[12] A. Fedotov. Quasiclassical asymptotics of Malyuzhinets functions. J. of Math. Sciences (NewYork), 226 (2017), 810–816.[13] A. Fedotov. A monodromy matrix for the almost Mathieu equation with a small couplingconstant. To appear in Functional analysis and its applications, v ( z ) = e − πiz . J. of Math. Sciences (New York), to appear in2018.[15] J. P. Guillement, B. Helffer and P. Treton. Walk inside Hofstadter’s butterfly. J. Phys. France, 50 (1989), 2019-2058.[16] B. Helffer, J. Sj¨ostrand. Analyse semi-classique pour l’´equation de Harper (avec application `al’´equation de Schr¨odinger avec champ magn´etique). M´em. Soc. Math. France (nouv. s´erie), 34 (1988), 1-113.[17] I. Krasovsky. Central Spectral Gaps of the Almost Mathieu Operator. Comm. Math. Phys., 351 (2017), 419-439.[18] M. Leguil, J. You, Z. Zhao, Q. Zhou. Asymptotics of spectral gaps of quasi-periodicSchr¨odinger operators. arXiv:1712.04700 [math.DS], 1-54.[19] W. Liu, Y. Shi. Upper bounds on the spectral gaps of quasi-periodic Schr¨odinger operatorswith Liouville frequencies. Journal of spectral theory, Sov. Phys. Dokl. Singular Integral Equations. Noordhoff, Groningen, 1953.[22] S. Ruijsenaars. On Barnes multiple zeta and gamma functions. Adv. in Math. 156 (2000),107-132.[23] M. Wilkinson. Tunneling between tori in phase space. Phys. D, (1986), 341–354.(Alexander Fedotov) St. Petersburg State University, 7/9 Universitetskaya nab.,St.Petersburg, 199034, Russia Email address : [email protected]@spbu.ru