Annals of Mathematics, 150 (1999), 185–266

Semiclassical asymptotics of orthogonal polynomials, Riemann-Hilbert problem, and universality in the model

By Pavel Bleher and Alexander Its

Abstract

We derive semiclassical asymptotics for the orthogonal polynomials Pn(z) on the line with respect to the exponential weight exp(NV(z)), where V (z) is a double-well quartic polynomial, in the limit when n, N →∞. We assume that ε (n/N) cr ε for some ε>0, where cr is the critical value which separates orthogonal polynomials with two cuts from the ones with one cut. Simultaneously we derive semiclassical asymptotics for the recursive coecients of the orthogonal polynomials, and we show that these coecients form a cycle of period two which drifts slowly with the change of the ratio n/N. The proof of the semiclassical asymptotics is based on the methods of the theory of integrable systems and on the analysis of the appropriate matrix Riemann- Hilbert problem. As an application of the semiclassical asymptotics of the orthogonal polynomials, we prove the universality of the local distribution of eigenvalues in the matrix model with the double-well quartic interaction in the presence of two cuts. Contents

1. Introduction and formulation of the main theorem 2. Universality of the local distribution of eigenvalues in the matrix model 3. The Lax pair for the Freud equation 4. The Stokes phenomenon 5. The Riemann-Hilbert problem 6. Formal asymptotic expansion for Rn 7. Proof of the main theorem: Approximate solution to the Riemann-Hilbert problem Appendix A. Proof of formula (7.8) Appendix B. Proof of Lemma 7.1 Appendix C. Proof of Lemmas 7.2, 7.4, and 7.6 Appendix D. Proof of estimate (7.45) Appendix E. The Riemann-Hilbert approach to orthogonal polynomials References

186 PAVEL BLEHER AND ALEXANDER ITS

1. Introduction and formulation of the main theorem

About thirty years ago Freeman Dyson found an exact solution for the scaling limit of correlations between eigenvalues in the Gaussian unitary en- semble of random matrices. He conjectured that this scaling limit should appear in a much broader class of non-Gaussian unitary ensembles of random matrices. This constitutes the famous conjecture of universality in the theory of random matrices. Dyson found also a remarkable formula which expresses the eigenvalue correlations for nite N (before the scaling limit) in terms of orthogonal polynomials on the line with respect to an exponential weight (see [Dys]; see also [Meh], [BIPZ], and others). This reduces the universality con- jecture to semiclassical asymptotics of orthogonal polynomials. The theory of orthogonal polynomials is a classical branch of mathemat- ical analysis and it nds applications in dierent areas of pure and applied mathematics. Among the most exciting recent applications of orthogonal poly- nomials related to the random matrices are those to quantum gravity, string theory, and integrable PDEs (see, e.g., [ASM], [Dem], [Wit] and references therein). In the present paper we show how these recent developments can be benecial for the theory of orthogonal polynomials itself and, in particular, for the aspects of the theory related to the matrix models. The Dyson formula gives the eigenvalue correlations for nite N. To get the universal correlation functions in the limit when N →∞one needs to obtain semiclassical asymptotics for orthogonal polynomials. For the Gauss- ian matrix model the problem reduces to the semiclassical asymptotics for the Hermite polynomials which were obtained in the classical work of Plancherel and Rotach [PR]. The Hermite polynomials are the eigenfunctions of the quan- tum linear oscillator (up to a Gaussian function) and their asymptotics follow from the semiclassical formulae for the Schrodinger operator with quadratic potential. Using the Plancherel-Rotach asymptotics in the inner region, be- tween the turning points, Dyson evaluated the scaling limit of correlations in the Gaussian matrix model inside the support of the limiting spectral measure. The scaling limit is expressed in terms of the sine-kernel. Dyson’s conjecture of universality states that the same scaling limit in the bulk of the spectrum should appear in the general case of non-Gaussian unitary ensembles of ran- dom matrices. The conjecture of universality was then extended to the scaling limit of correlations at the edges of the support of the limiting spectral mea- sure. For the Gaussian matrix model the scaling limit of correlations at the edge is expressed in terms of the Airy kernel (see [BB], [For], [TW2]), and this follows from the Plancherel-Rotach asymptotics near the turning points. The conjecture of universality states that the same scaling limit is to appear at all regular edges of the support of the limiting spectral measure for a gen-

SEMICLASSICAL ASYMPTOTICS 187 eral model. Here regular means that the spectral density has a square-root singularity at the edge. The conjecture of universality has, in fact, an even more general formulation. One should expect universal scaling limits to exist at regular critical points, tricritical points, etc., each being re- lated to its own class of universality. The critical points do not occur in the Gaussian matrix model but they do in non-Gaussian matrix models, and the critical points correspond usually to bifurcations of the support of the limiting spectral measure. In this paper we prove the universality conjecture inside the spectrum and at the edges for the matrix model with quartic interaction. We will discuss the universality at the critical points in another publication. A very dierent approach using estimates for resolvents, to the universality conjecture for a large class of potentials has been given in the paper [PS] by Pastur and Shcherbina. We derive the scaling limit of correlations from the semiclassical asymp- totics of orthogonal polynomials. For the general matrix model the problem of semiclassical asymptotics is much more dicult than for the Gaussian model. The corresponding Schrodinger operator is very complicated and, what is even more important, it depends on unknown recursive coecients for orthogonal polynomials. So one has to solve the two problems simultaneously: to nd semiclassical asymptotics for the recursive coecients and for the orthogonal polynomials. The recursive coecients for orthogonal polynomials satisfy a nontrivial nonlinear equation which was discovered independently in mathe- matics and in physics. Mathematicians call it the Freud equation (and its generalizations), in the name of Geza Freud who rst derived it for orthogonal polynomials with some exponential weights on the line. For physicists this equation is known as a discrete string equation, and it was derived and studied in the papers by Bessis, Brezin, Itzykson, Parisi, and Zuber, and others (see, e.g., [BPIZ], [BIZ], [IZ]), devoted to the problem of enumeration of Feynman graphs in string theory. The Freud equation does not admit a direct solution, and this is the main reason why despite many interesting and deep achieve- ments in the theory of orthogonal polynomials (see, e.g., [LS], [Mag4], [Nev3] and references therein), the semiclassical asymptotics for general exponential weights remains one of the central unsolved problems in the theory. Our new technique is based on the important observation made in [FIK1] that the Freud equation is integrable in the sense that it admits a Lax pair. This motivates use of the powerful tools of the theory of integrable systems, the method de- veloped in the present paper. The central role in our scheme is played by the Riemann-Hilbert setting of orthogonal polynomials suggested in [FIK4]. To formulate the problem let us consider a polynomial V (z) with real coecients, 2p V (z)=a0 + a1z + + a2pz ,a2p > 0,

188 PAVEL BLEHER AND ALEXANDER ITS and orthogonal polynomials on the real line, n Pn(z)=z + ..., n=0, 1, 2,..., with respect to the exponential weight eNV(z)dz, i.e., Z ∞ NV(z) (1.1) Pn(z)Pm(z)e dz = hnnm,hn > 0. ∞ We normalize the orthogonal polynomials by taking the leading coecient equal to 1. The number N in the weight is a large parameter and we are inter- ested in the asymptotics of the polynomials Pn(z) in the limit when n, N →∞ in such a way that n ε< <ε 1, N for some xed ε>0. In this approach the large parameter N plays the same role as the inverse Planck constant plays in the semiclassical asymptotics of quantum mechanics. The orthogonal polynomials Pn(z) satisfy the basic recursive equation

(1.2) zPn(z)=Pn+1(z)+QnPn(z)+RnPn1(z), where Qn and Rn are some coecients which depend on N. The semiclassical asymptotics of the coecients Qn and Rn are tied to one of the polynomials Pn(z) and we solve these two problems together. Although our approach is quite general and can be applied to a general polynomial V (z), in this paper we will work out all the details for the quartic polynomial tz2 gz4 (1.3) V (z)= + ,g>0, 2 4 and we will consider the most interesting case when t<0 (a double-well potential). Since V (z) is even, (1.2) is simplied to

(1.4) zPn(z)=Pn+1(z)+RnPn1(z), where

hn (1.5) Rn = . hn1 In addition, integration by parts gives 0 (1.6) Pn(z)=NRn[t + g(Rn1 + Rn + Rn+1)]Pn1(z) 0 d + g(NR R R )P (z), ( ) . n 2 n 1 n n 3 dz 0 n1 Since Pn(z)=nz + ..., this implies the Freud equation

(1.7) n = NRn[t + g(Rn1 + Rn + Rn+1)],n 1,

SEMICLASSICAL ASYMPTOTICS 189

(cf. [Fre]). The dierence equation (1.7) is supplemented by the following initial conditions: R ∞ 2 NV(z) 0 z e dz h1 R =0,R= R ∞ . 0 1 NV(z) 0 e dz h0 From (1.5) and (1.7) it follows that p t + t2 +4g n (1.8) 0

1 NV (z)/2 (1.9) n(z)=√ Pn(z)e . hn Then Z ∞ (1.10) n(z)m(z) dz = nm. ∞

Our main goal is to prove semiclassical asymptotics for the functions n(z) and for the coecients Rn in the limit when N,n →∞in such a way that there exists ε>0 such that the ratio = n/N satises the inequalities n (1.11) ε<< ε, = , cr N where t2 (1.12) = . cr 4g In what follows the potential function (gz2 + t)2 U(z)=z2 g , 4 is important. We introduce the turning points z1 and z2 as zeros of U(z), √ t 2 g 1/2 (1.13) z = . 1,2 g √ The condition (1.11) implies that z1 and z2 are real, and z2 >z1 >C ε.We prove the following main theorem.

Theorem 1.1. Assume that N,n →∞in such a way that (1.11) holds. Then there exists C = C(ε) > 0 such that p n 2 t (1) t 4g n (1.14) R CN 1,= . n 2g N

190 PAVEL BLEHER AND ALEXANDER ITS

In addition, for every >0, in the interval z1 +

If z>z2 + or 0 zz + , h i 2 = n  = k if 0 z

SEMICLASSICAL ASYMPTOTICS 191 and p t (1)n t2 4g R0 = . n 2g

The constant factor Cn in (1.15) and (1.17) satises the asymptotic equation 1 g 1/4 (1.21) C = √ (1 + O(N 1)), n 2 and Dn in (1.18) is h i √ n (1.22) D = N 1/6 g (1)0 ,=(2 j) . n 0 2

Finally, the asymptotics of hn in (1.1), (1.3) is p 2 Nt N g (1.23) h =2 R0 exp 1+ln + O(N 1) . n n 4g 2

The asymptotic formulae (1.15), (1.17), and (1.18) are an extension of the classical Plancherel–Rotach asymptotics for the Hermite polynomials (see [PR] and [Sze]), to the orthogonal polynomials with respect to the weight eNV(z) where V (z) is the quartic polynomial (1.3). These formulae are extended into the complex plane in z as well (see Theorems 7.5, 7.7 in Section 7 below). The formula (1.15) can be rewritten as the semiclassical formula p " Z ! # z z g/ 1/2 1 (1.24) n(z)= cos N |UN (v)| dv + + O(N ) , |U(z)|1/4 N 4 z2 where UN (z) is as dened in (1.20). Asymptotics (1.14) of the coecients Rn are Freud-type asymptotics. For the homogeneous function V (z)=|z| and some of its generalizations, the asymptotics of Rn are obtained in the papers of Freud [Fre], Nevai [Nev1], Magnus [Mag1,2], Lew and Quarles [LQ], Mate, Nevai, and Zaslavsky [MNZ], Bauldry, Mate, and Nevai [BMN]. Semiclassical 4 asymptotics of the functions n(z) are proved then for V (z)=z by Nevai [Nev1] and for V (z)=z6 by Sheen [She]. For general homogeneous V (z), somewhat weaker asymptotics are obtained in the works of Lubinsky and Sa [LS], Lubinsky, Mhaskar, and Sa [LMS], Lubinsky [Lub], Levin and Lubinsky [LL], Rahmanov [Rah], and others. Application of these asymptotics to ran- dom matrices is discussed in the work of Pastur [Pas]. The distribution of zeros and related problems for orthogonal polynomials corresponding to general ho- mogeneous V (z) are studied in the recent work [DKM] by Deift, Kriecherbauer, and McLaughlin. Many results and references on the asymptotics of orthogo- nal polynomials are given in the comprehensive review article [Nev3] of Nevai. The problem of nding asymptotics of Rn for a quartic nonconvex polynomial is discussed in [Nev2, 3], and is known as “Nevai’s problem.”

192 PAVEL BLEHER AND ALEXANDER ITS

Equation (1.14) shows that if 0 <<cr then p t t2 4g (1.25) lim R2m = L()= , N→∞;(2m)/N → 2g p t + t2 4g lim R2m+1 = R()= . N→∞;(2m+1)/N → 2g Both L() and R() satisfy the quadratic equation gu2 + tu + =0, so that, when n grows, Rn jumps back and forth from one sheet of the parabola to another (see Fig.1). In other words, the sequence { Rn,n=0, 1, 2,...} forms a period two cycle which is slowly drifting with the change of the ratio n/N.At = cr the two sheets of the parabola merge, i.e., L(cr)=R(cr). For >cr,

(1.26) lim Rn = Q(), N→∞; n/N→ where u = Q() satises the quadratic equation 3gu2 + tu =0

Rn

...... n λ λ = cr N

Figure 1. The qualitative behavior of the recurrence coecients

SEMICLASSICAL ASYMPTOTICS 193

(which follows from the Freud equation (1.7) if we put u = Rn1 = Rn = Rn+1). We consider semiclassical asymptotics for >cr and in the vicinity of cr (double scaling limit) in a separate work. The dierence in the asymptotics between the cases <cr and >cr is that for <cr the function n(z) is concentrated on two intervals, or two cuts, [z2, z1] and [z1,z2], and it is exponentially small outside of these intervals, while for >cr, z1 becomes pure imaginary, and n(z) is concentrated on one cut [z2,z2]. The transition from a two-cut to a one-cut regime is discussed in physical papers by Cicuta, Molinari, and Montaldi [CMM], Crnkovic and Moore [CM], Douglas, Seiberg, Shenker [DSS], Periwal and Shevitz [PeS], and others. Another proof of the limits (1.25) was recently given by Albeverio, Pastur, and Shcherbina (see [APS]) in a completely dierent approach based on some estimates of the Stieltjes transform of the spectral measure. A general ansatz on the structure of the semiclassical asymptotics of the functions n(z) for a “generic” polynomial V (z) was proposed in the work [BZ] of Brezin and Zee. They considered n close to N, n = N +O(1), and suggested that for these n’s, 1 (1.27) n(z)=p cos N(z) (N n)ϕ(z)+(z) , f(z) for some functions f(z),(z),ϕ(z), and (z). This ts in well the asymptotics (1.15), except for the factor (1)n at in (1.15), which is related to the two-cut structure of n(z). Equation (1.7) also appears in the planar Feynman diagram expansions of models, which were introduced and studied in the classical papers [BIPZ], [BIZ], [IZ] by Brezin, Bessis, Itzykson, Parisi, and Zuber and in the well-known recent works by Brezin, Kazakov [BK], Douglas, Shenker [DS], and Gross, Migdal [GM] devoted to the matrix models for 2D quantum gravity (see also [Dem] and [Wit]). In fact, it is the latter context that broad- ened the interest to the Freud equation (1.7) and brought to the area new powerful analytic methods from the theory of integrable systems. It turns out [FIK1,2] that equation (1.7) admits a 2 2 matrix Lax pair representation (see equation (3.15) below), which allows one to identify the Freud equation (1.7) as a discrete Painleve I equation and imbeds it in the framework of the isomonodromy deformation method suggested in 1980 by Flaschka and Newell [FN] and by Jimbo, Miwa, and Ueno [JMU] (about analytical aspects of the method see, e.g., [IN] and [FI]). The relevant Riemann-Hilbert formalism for (1.7) was developed in [FIK1,2] as well. It was used in [FIK1-3] together with the isomonodromy method for the asymptotic analysis of the solution of (1.7), which is related to the double-scaling limit in the 2D quantum gravity studied in [BK], [DS], [GM].

194 PAVEL BLEHER AND ALEXANDER ITS

As a matter of fact, the solution of (1.7) which is analysed in [FIK] corre- sponds to the system of orthogonal polynomials on certain rays in the complex plane. Nevertheless, the basic elements of the Riemann-Hilbert isomonodromy scheme suggested in [FIK] can be easily extended to an arbitrary system of orthogonal polynomials corresponding to a rational potential V (z) (cf. [FIK4]; see also Appendix E below). The proof of Theorem 1.1 is based on the ap- proach of [FIK] combined with the nonlinear steepest descent method proposed recently by Deift and Zhou [DZ1] for analyzing the asymptotics of oscillatory matrix Riemann-Hilbert problems. We appeal to the Deift-Zhou method in Section 7 where we construct explicitly and then justify rigorously the as- ymptotic solution of the master Riemann-Hilbert problem associated to the orthogonal polynomials (see conditions (i)–(iii) and (5.16)–(5.18) below). The paper is organized as follows: In the next section we use the results of Theorem 1.1 for proving the universality of the local distribution of eigenvalues in the matrix model with quartic potential. We prove both the sine-kernel universality at internal points and the Airy-kernel universality at the end-points of the support of the limiting spectral measure. It is important to note that for internal points the univer- sality was recently proved by a dierent technique in the paper by Pastur and Shcherbina [PS], for a general class of matrix models. In Sections 3–5 we reproduce in a slightly dierent way the results of [FIK] concerning the Lax pair representation of equation (1.7) and the matrix Riemann-Hilbert reformulation of the orthogonal polynomials. In particular, we show that there is an exact and simple relation between orthogonal polyno- mials Pn(z) and the 2 2 matrix-valued function n(z) which solves the fol- lowing matrix Riemann-Hilbert problem on a line (the problem (5.16)–(5.18) below):

(i) n(z) is analytic in C \ R, and it has a jump at the real line. ! X∞ k NV (z) n ln z+ (ii) (z) e 2 n 3 ,z→∞, where n zk k=0 1 10 = ln h ,= , n 2 n 3 0 1 10 01 0 = 1/2 , 1 = 1/2 . 0 Rn Rn 0 1 2i (iii) (z)= (z)S, Im z =0,S= . n+ n 01 As explained at the end of Section 5, in the setting of the Riemann-Hilbert problem (i)–(iii), the real quantities Rn and n are not the given data. They

SEMICLASSICAL ASYMPTOTICS 195 are evaluated via the solution n(z), which is determined by conditions (i)– (iii) uniquely without any prior specication of Rn and n. Simultaneously, the function n(z) satises the Lax pair (see equation (3.15) below) whose second equation is the linear dierential equation: d (z) (1.28) n = NA (z) (z), dz n n where tz gz3 1/2 2 ( 2 + 2 + gzRn) Rn [t + gz + g(Rn + Rn+1)] A (z)= 3 . n 1/2 2 tz gz Rn [t + gz + g(Rn1 + Rn)] 2 + 2 + gzRn The jump matrix S in (iii) constitutes the only nontrivial Stokes matrix (for more details see Sections 4, 5) corresponding to the system (1.28) with Rn generated by (1.4) and (1.5). This reduces the problem of the asymptotic analysis of the quantities Pn(z) and Rn to the asymptotic solution of the matrix Riemann-Hilbert problem (i)–(iii), i.e. to the asymptotic solution of the corresponding inverse monodromy problem for dierential equation (1.28). The short but important Section 6 provides a formal asymptotic ansatz indicated in (1.14) for the recurrence coecients Rn. The proof of Theorem 1.1 is given in Section 7. The central point of the proof is a construction of an approximate solution 0(z) to the Riemann– Hilbert problem (i)–(iii). To that end we consider an approximate dierential equation (1.28) in which the coecients R are replaced by the numbers p n t (1)n t2 4(n/N)g R0 = , n 2g that are taken from our formal analysis in Section 6. The function 0(z)is constructed then as a semiclassical solution to this approximate dierential equation with the large parameter N. Our semiclassical construction is based on the version of the complex WKB method which was recently suggested in [Kap] for asymptotic solution of the direct monodromy problems for the 2 2 systems with rational coecients. We partition the whole complex plane into several regions which separate dierent turning points of the approximate dierential equation, and we construct WKB- and turning point semiclassical solutions in each of these regions (see details in Section 7). This provides us with an explicit matrix-valued function 0(z) which solves asymptotically, as N →∞, the basic Riemann-Hilbert problem (i)–(iii). Then we prove that the 0 1 1 1 quotient n(z) (z) is equal to I + O(N (1 + |z|) ) and this completes the proof of Theorem 1.1. We emphasize that we only use equation (1.28) to motivate our choice of the function 0(z). The uniform estimate for the 0 1 dierence, n(z) (z) I, is proved by means independent of the WKB theory of dierential equations. The main ideas and technique used in Section 7 are based on the Deift-Zhou nonlinear steepest descent method [DZ].

196 PAVEL BLEHER AND ALEXANDER ITS

The Riemann-Hilbert reformulation of the orthogonal polynomials (1.1), (1.3) suggested in [FIK] plays the central role in our approach and in its ex- tention to the general rational potentials V (z). For that reason we decided to present with more detail the scheme of [FIK] in Appendix E. As mentioned above, the Freud equation (1.7) has a meaning similar to the discrete Painleve I equation. We refer the reader to the papers [FIZ], [NPCQ], [GRP], [Mag3,4], [Meh2] for more on the subject. As rst noticed by Kitaev, equation (1.7) can also be interpreted as the Backlund-Schlezinger transform of the classical Painleve IV equation so that the coecients Rn coincide, in fact, with the special PIV function (see [FIK1,3] for more details). This PIV function, in turn, can be expressed in terms of certain n n involving the parabolic cylinder functions (see [Mag4]). In this work however we do not use these algebraic connections to the modern Painleve theory. Instead, we use its analytical methods. The present paper is a revised and shortened version of our earlier preprint [BI].

2. Universality of the local distribution of eigenvalues in the matrix model

Theorem 1.1 can be applied to proving the universality of the local distri- bution of eigenvalues in the matrix model with quartic potential. The matrix model is dened as follows. Let M =(Mjk)j,k=1,...N be a Hermitian , with the probability distribution 1 NTr V (M) (2.1) N (dM)=ZN e dM, where 2p V (M)=a0 + a1M + + a2pM ,a2p > 0, is a polynomial, Y Y dM = (d Re Mjk d Im Mjk) dMjj, j

SEMICLASSICAL ASYMPTOTICS 197

In the matrix model we are interested in the following problems:

(1) To calculate the limit density p(z) = limN→∞ pN (z). (2) To calculate the limit local distribution (scaling limit) of eigenvalues at regular points, where p(z) is positive, and at end-points, where p(z) vanishes. (3) To calculate the free energy

log ZN (a0,...,a2p) f(a0,...,a2p)= lim N→∞ N 2 and to nd the points of nonanalyticity of f (critical points) in the space of the parameters a0,...,a2p. A further problem is to calculate the critical asymptotics of the recursive coecients Rn and of the local distribution of eigenvalues (double scaling limit). Dyson [Dys] (see also [Meh1] and [TW1]) proved a formula which ex- presses the correlations between the eigenvalues of M in terms of orthogonal polynomials. Namely, the m-point correlation function is written as (2.2) KNm(z1,...,zm) = det QN (zj,zk) j,k=1,...,m where XN (2.3) QN (z,w)= j(z)j(w), j=1 and j(z) is as dened in (1.9). When m = 1 the correlation function reduces to the function NpN (z); hence XN 1 2 pN (z)=N j (z). j=1

By the Christoel-Darboux formula (see, e.g., [Sze]), the kernel QN (z,w) can be written as p R [ (z) (w) (z) (w)] (2.4) Q (z,w)= N+1 N+1 N N N+1 , N z w and p 0 0 RN+1 (z)N (z) (z)N+1(z) (2.5) p (z)= N+1 N . N N The formula (1.24) is valid in a complex neighborhood of the interval [z1 + , z2 ] and this allows us to dierentiate it. We will assume that √ t

198 PAVEL BLEHER AND ALEXANDER ITS

(two-cut case); hence we can use n = N in the asymptotic formulae (1.15)– (1.18). For the sake of brevity we rewrite (1.24), (1.15) as Cz (2.6) n = √ cos(N + ), z where (2.7) Z p z 0 0 C = g/ ; = (z; )= |U (v; )|1/2dv + ; 0 4N z2 0 2 2 ∂(z; ) 0 1/2 0 2 (gz + t) 0 z = = |U0(z; )| ; U0(z; )=z g ; ∂z √ 4 n n 0 2 (1) 0 (1) 2 g tq gz + t = (z; )= arccos r, r = √ ,q= √ , 4 4 2 0gq t 2 0g and we drop terms of the order of N 1. In addition, (1.24) gives that modulo terms of the order of N 1, Cz (2.8) n1 = √ cos(N ), z where ∂(z; 0) 1 gz2 + t (2.9) = = arccos q, q = √ . ∂0 2 2 0g

The functions n satisfy the recursive equation p p zn = Rn+1 n+1 + Rn n1 (see (1.4)), hence from (2.6) and (2.8) we obtain that z cos(N ) cos(2) z sin(N ) sin(2) p p = R cos(N ) cos R sin(N ) sin pn+1 p n+1 + Rn cos(N ) cos + Rn sin(N ) sin . Equating the coecients at cos(N ) and sin(N ), we obtain that p p (2.10) z cos 2 =( R + R ) cos , p n+1 p n z sin 2 =( Rn+1 Rn) sin . These formulae can be checked directly from (1.14), (2.7) and (2.9). Dieren- tiating (2.6) and (2.8) in z, we get that p 0 = Cz sin(N + ) N + O(1), n zp 0 n+1 = Cz sin(N + ) N z + O(1);

SEMICLASSICAL ASYMPTOTICS 199 hence by (2.5), modulo terms of the order of N 1, (2.11) p 2 2 pN = RN+1 C z [ sin(N + ) cos(N + ) p + cos(N + p ) sin(N + )] 2 2 2 2 = RN+1 C z sin(2 )= RN+1 C z (sin 2 cos cos 2 sin ), and by (2.10), p h p p 2 pN = RN+1 C z ( RN+1 RN ) sin cos p p i ( RN+1 + RN ) sin cos p 2 = RN+1RN C z sin 2. Since modulo terms of the order of N 1, 1 g p R R = ; C2 = ; sin 2 = sin( arccos q)= 1 q2, N+1 N g we obtain that √ p g p = z 1 q2 + O(N 1). N Substitution of the value of q gives that

1 pN (z)=p(z)+O(N ), where (2.12) " # 1/2 q 1 |z| gz2 + t 2 g|z| p(z)= |U (z;1)|1/2 = g = (z2 z2)(z2 z2) 0 2 2 1 2 and √ t 2 g 1/2 (2.13) z = . 1,2 g This gives an explicit formula for the limiting density p = p(z) of eigenvalues (integrated density of states). In a completely dierent approach, based on the Coulomb gas representation of the matrix model, this formula is derived in the work [BPS] of Boutet de Monvel, Pastur, and Shcherbina, as an application of the proven (in [BPS]) variational principle for the integrated density of states. The scaling limit of the correlation function KNm(z1,...,zm) at a regular point z, where p(z) > 0, is dened as m u1 um Km(u1,...,um) = lim Np(z) KNm z + ,...,z+ . N→∞ Np(z) Np(z)

200 PAVEL BLEHER AND ALEXANDER ITS

Observe that Km(u1,...,um) is the limiting m-point correlation function of the rescaled eigenvalues j = Np(z)(j z).

The rescaling reduces the mean value of the spacing j+1 j to 1. From Dyson’s formula (2.2), (2.14) Km(u1,...,um) = det Q(uj,uk) j,k=1,...,m, where 1 u v (2.15) Q(u, v) = lim Np(z) QN z + ,z+ . N→∞ Np(z) Np(z) By (2.4), 1 u v (2.16) Np(z) Q z + ,z+ N Np(z) Np(z) p R u v = N+1 T z + ,z+ , u v N Np(z) Np(z) where TN (z,w)=N+1(z)N (w) N (z)N+1(w). By (2.6) and (2.8), modulo terms of the order of N 1, (2.17) u Cz zu N z + = √ cos(N + + ),= , Np(z) z Np(z) u Cz N+1 z + = √ cos(N + + ); Np(z) z hence (2.18) u v T z + ,z+ N Np(z) Np(z) C2z2 = [cos(N + + ) cos(N + + ) z cos(N + + ) cos(N + + )] C2z2 = [cos( + 2) cos( +2)] 2z C2z2 = sin(2 ) sin( ), z where u |U (z)|1/2u (2.19) = z = 0 = u, = v . 1/2 1 p(z) |U0(z)|

SEMICLASSICAL ASYMPTOTICS 201

By (2.11) and (2.12),

p 1 p R C2z2 sin(2 )=p(z)= |U (z;1)| = (z; 1); N+1 0 z hence (2.18) implies that p u v p C2z2 RN+1 TN z + ,z+ = RN+1 sin(2 ) sin( ) Np(z) Np(z) z sin( ) sin (u v) = = , and, by (2.15), (2.16), sin (u v) Q(u, v)= . (u v)

This proves the Dyson sine-kernel for the local distribution of eigenvalues at a regular point z. In a completely dierent approach, the sine-kernel at regular points is proved in [PS].

Remark. It follows from the Dyson sine-kernel, due to the Gaudin formula (see, e.g., [Meh1]), that the spacing distribution of eigenvalues is determined by the Fredholm det(1 Q(x, y))x,y∈J . The asymptotics of this determinant as |J|→∞has been studied intensively since the classical works by des Cloizeaux, Dyson, Gaudin, Mehta, and Widom (see [Meh1] for the history of the subject). The Riemann-Hilbert approach to this asymptotics has been developed in the paper [DIZ].

At the endpoints of the spectrum we use the semiclassical asymptotics (1.18), and it leads to the Airy kernel (cf. the papers of Bowick and Brezin [BB], Forrester [For], Moore [Mo], and Tracy and Widom [TW2], where the Airy kernel is discussed for the Gaussian matrix model and some other related models, and, in addition, some nonrigorous arguments are given for general matrix models). Consider for the sake of deniteness z = z2. By (1.18),

1/6 h i √ DN z 2/3 1 (2.20) n = √ Ai N w + O(N ) ,D= g, w0 where w is dened as in (1.19). From (1.19), Z √ ∂w ∂ z p w 0 = 0 UN (v) dv. ∂ ∂ (N) z2

202 PAVEL BLEHER AND ALEXANDER ITS

This allows us to derive from (2.20) that (2.21) 1/6 h i DNp z 2/3 1/3 1 n = 0 Ai (N ϕ0 + N ω)+O(N ) , ϕ0 1/6 h i DNp z 2/3 1/3 1/3 1 n1 = 0 Ai (N ϕ0 N N ω)+O(N ) , ϕ0 where Z z p 2/3 0 3 0 (2.22) ϕ0 = ϕ0(z; )= U0(v; ) dv , 2 z2 0 0 0 p(z; ) 0 p(z; ) = (z; )= 0 ,ω= ω(z; )= 0 , ϕ0(z; ) ϕ0(z; ) and 1 2 0 cosh q gz√ + t (2.23) (z; )= ,q= 0 ; 2 2 g √ n 0 0 (1) 2 g tq (z; )= cosh 1 r, r = √ . 4 2 0gq t 0 0 0 The formulae (2.22), (2.23) dene the functions ϕ0(z; ),(z; ) and ω(z; ) for z z2. It is easy to check that these functions are analytic in z at z = z2, and they can be continued analytically to the interval z>z1. In addition,

0 ∂U 0 0 (2.24) U (z ; )=0, 0 (z ; )={ =2( )1/2g3/2z3; 0 2 ∂z 2 2 0 ∂ϕ 0 0 ϕ (z ; )=0, 0 (z ; )={1/3 =21/3( )1/6g1/2z ; 0 2 ∂z 2 2 2/3 0 1/3 (z2)=2 ( ) ; n (1) 0 ω(z )= 21/3( ) 1/3z z 1. 2 4 1 2 We will consider

2/3 2/3 (2.26) z = z2 + N , w = z2 + N , where and are xed. Substitution of (2.21) into the recursive equation p p zn = Rn+1n+1 + Rnn1 gives the equations p p (2.27) z = R + R , 2 p n+1 n p z2ω = Rn+1( ω)+ Rn( ω),

SEMICLASSICAL ASYMPTOTICS 203 from whence p p p p (2.28) ( Rn+1 + Rn)2ω =( Rn+1 Rn) . Similarly, p p n z1 =(1) ( Rn+1 Rn); hence (1)nz (2.29) 2ω = 1 , z2 which agrees with (2.24). Substituting the formulae (2.21) into (2.4) and throwing away terms of the lower order, we obtain that (2.30) p R D2N 1/3z2 Q (z,w)= N+1 2 N (z w)ϕ0 h 0 2/3 1/3 1/3 2/3 1/3 Ai N ϕ0(z)+N N ω Ai N ϕ0(w)+N ω i 2/3 1/3 2/3 1/3 1/3 Ai N ϕ0(z)+N ω Ai N ϕ0(w)+N N ω , 0 where ϕ0,and ω are taken at z2. Taking the linear part of Ai we obtain that p 2 1/3 2 RN+1 D N z2 0 0 1/3 QN (z,w)= 0 Ai (u)Ai (v) Ai (u)Ai(v) (2ω )N , (z w)ϕ0 where 0 0 u = ϕ0 , v = ϕ0 . By (2.28) and (2.24), modulo terms of the order of N 1/3, √ p p p ( 2 Rn√) 2/3 1/3 1/2 1 RN+1 (2ω )= RN+1 ( 2 )=2 g z2 ; RN+1 + Rn hence Ai (u)Ai0(v) Ai 0(u)Ai(v) Q (z,w)=N 2/321/3g1/2z + O(N 1/3). N 2 u v Thus, 1 u v Ai (u)Ai0(v) Ai 0(u)Ai(v) lim QN z2 + ,z2 + = , N→∞ cN 2/3 cN 2/3 cN 2/3 u v where 0 1/3 1/2 c = ϕ0(z2;1)=2 g z2.

This proves the Airy kernel at the endpoint z2. The endpoint z1 is treated similarly.

204 PAVEL BLEHER AND ALEXANDER ITS

3. The Lax pair for the Freud equation

Let

1 NV (z)/2 (3.1) n(z)=√ Pn(z)e . hn Then Z ∞ (3.2) n(z)m(z) dz = nm. ∞

A recursive equation for n(z) follows from (1.4):

1/2 1/2 (3.3) zn(z)=Rn+1n+1(z)+Rn n1(z). In addition, (3.4) 0 g 1/2 1/2 1/2 n(z)= N R +1R +2R +3 n+3(z) 2 n n n t 1/2 g 1/2 N Rn+1 + N Rn+1(Rn + Rn+1 + Rn+2) n+1(z) 2 2 t 1/2 g 1/2 + N Rn + N Rn (Rn1 + Rn + Rn+1) n1(z) 2 2 g 1/2 1/2 1/2 + N R R R (z). 2 n2 n1 n n 3 Let n(z) (3.5) ~ n(z)= . n1 Then combining (3.3) with (3.4), one can obtain (cf. (3.1–7) in [FIK2]) that ( ~ n+1(z)=Un(z)~ n(z), (3.6) ~ 0 ~ n(z)=NAn(z)n(z), where R 1/2z R 1/2R1/2 (3.7) U (z)= n+1 n+1 n , n 10 and (3.8) tz gz3 1/2 2 ( 2 + 2 + gzRn) Rn [t + gz + g(Rn + Rn+1)] A (z)= 3 . n 1/2 2 tz gz Rn [t + gz + g(Rn1 + Rn)] 2 + 2 + gzRn Observe that tr An(z)=0

SEMICLASSICAL ASYMPTOTICS 205 and (3.9) 3 2 tz gz 2 det A (z)= + +gR (t + gR + gR + gR ) z +R , n 2 2 n n 1 n n+1 n n 1 n where

(3.10) n = t + gRn + gRn+1.

Due to (1.7), we can rewrite det An(z)as 3 2 2 0 tz gz gnz (3.9 ) det A (z)= + + + R . n 2 2 N n n 1 n The compatibility condition of equations (3.6) is 0 (3.11) Un(z)=NAn+1(z)Un(z) NUn(z)An(z). 0 Restricting this equation to the matrix element Un,11(z) we obtain that

(3.12) Jn+1 Jn =1, where Jn = NRn[t + g(Rn1 + Rn + Rn+1)].

Hence Jn = n + const. Since J0 = 0, in fact, Jn = n. This means, that the compatibility condition (3.10), together with the initial value J0 = 0 imply (1.7), and thus equations (3.6) give the Lax pair for the nonlinear dierence Freud equation (1.7). In addition, equation (3.12) gives the recursive equation

n (3.13) R = R + . n+1 n n+1 n n 1 n N

Equations (3.6) have two linear independent solutions and ~ n(z) is one of them. We will consider another solution, ϕn(z) ~ n(z)= , ϕn1(z) and the 2 2 matrix n(z) ϕn(z) (3.14) n(z)= n1(z) ϕn1(z) which satises the same equations, ( n+1(z)=Un(z)n(z), (3.15) 0 n(z)=NAn(z)n(z).

To dene ~ n(z) we consider an arbitrary, linearly independent with ~ 1(z), ~ 0 ~ solution of the dierential equation 1(z)=NA1(z)1(z), and then dene

206 PAVEL BLEHER AND ALEXANDER ITS

~ n(z),n 2, with the help of the recursive equation ~ n+1(z)=Un(z)~ n(z). ~ 0 ~ Equation (3.11) then leads to the dierential equation n(z)=NAnn(z) for n 2. Note that the equation ~ n+1(z)=Un(z)~ n(z) means that ϕn(z) satises the recursive equation (3.3); i.e.,

1/2 1/2 (3.16) zϕn(z)=Rn+1ϕn+1(z)+Rn ϕn1(z).

Since tr An(z) = 0, the second equation in (3.15) implies that

(3.17) det n(z)=C =06 is independent of z; i.e.,

(3.18) n(z)ϕn1(z) n1(z)ϕn(z)=C, C = C(n).

This enables us to derive a rst order dierential equation for ϕn(z). Namely, 0 from the rst row in n(z)=NAnn(z) we have that 0 (3.19) ϕn(z)=Na11(z)ϕn(z)+Na12(z)ϕn1(z), 0 n(z)=Na11(z)n(z)+Na12(z)n1(z), where An(z)= aij(z) i,j=1,2; hence 0 0 n(z)ϕn(z) ϕn(z)n(z)=Na12(z)[n(z)ϕn1(z) n1(z)ϕn(z)] = CNa12(z), and 0 ϕn(z) CNa12(z) (3.20) = 2 . n(z) n(z) 0 In a similar way from the second row in (3.15), n(z)=NAn(z)n(z), we obtain that 0 ϕn1(z) CNa21(z) (3.21) = 2 ,n1. n1(z) n1(z)

It is useful to note that (3.18) allows us to express ϕn1(z) in terms of ϕn(z):

n1(z) C (3.22) ϕn1(z)= ϕn(z)+ . n(z) n(z)

Remark. The system of two dierential equations of the rst order, ~ 0 ~ n = NAnn, can be reduced to one equation of the second order (cf. [Sho]). Namely, from the rst equation of the system we can express n1 in terms of n,

1 1 0 a11 (3.23) n1 = N n n, a12 a12

SEMICLASSICAL ASYMPTOTICS 207 and then we can substitute this expression into the second equation of the system, which gives 0 0 00 a12 0 2 a11 (3.24) n n + N (a11a22 a12a21) n Na12 n =0. a12 a12 With the help of the substitution 1/2 (3.25) n = a12 n we reduce (3.24) to the Schrodinger equation 00 2 ˆ (3.26) n + N Un =0, where (3.27) 0 00 0 2 ˆ 1 0 a12 2 a12 3(a12) U = (a11a22 a12a21)+N a11 a11 N 2 . a12 2a12 4a12 By (3.8), tz gz3 (3.28) a = a = + + gzR , 11 22 2 2 n 1/2 2 1/2 2 a12 = Rn (n + gz ),a21 = Rn (n1 + gz ), which gives 2 6 4 2 g z tgz t n 2 (3.29) Uˆ(z)= + + g z Rnn1n 4 2 4 N 2 2 2 1 t 3gz gz (t + gz +2gRn) N + + gRn 2 2 2 gz + n 2 2 g(2gz n) + N 2 2 . (gz + n) It is convenient to write Uˆ(z)as

(3.30) Uˆ(z)=U0(z)+U1(z)+U2(z), where

(3.31) " # 2 2 1 gz + t 0 0 n + U (z)=z2 g ,= 2 , 0 2 N 1 t U1(z)=N + gRn , 2 2 2 1 n(t + gz +2gRn) 2 g(2gz n) U2(z)= Rnn 1n N 2 + N 2 2 . gz + n (gz + n)

208 PAVEL BLEHER AND ALEXANDER ITS

4. The Stokes phenomenon

Consider the sectors (4.1) (j 1) 3 (j 1) = z : + + ε0, on a complex plane where the function

1 N( t z2+ g z4) (4.2) n(z)=√ Pn(z) e 4 8 hn goes to innity as z →∞. Let bj be the bisector of j, i + (j 1) (4.3) bj = {z : z = rωj,r>0} ,ωj = e 4 2 ,j=1, 2, 3, 4.

For a xed n 1, let Cn > 0 be a constant which will be chosen later, and let us consider the following special solution to equation (3.20) with C = Cn: (4.4) Z z a12(u) t 2 g 4 1/2 N( 4 z + 8 z ) ϕnj(z)=CnNn(z) 2 du = CnNhn Pn(z) e ωj ∞ n(u) Z z 1/2 2 Rn [t + gu + g(Rn + Rn+1)] t 2 g 4 N( 2 u + 4 u ) 2 e du. ωj ∞ Pn (u)

Then ϕnj(z) → 0asz →∞in j. Evaluating the integral in (4.4) with the help of the Laplace method, we obtain the asymptotic expansion of ϕnj(z)in j: (4.5) ! X∞ 1/2 1/2 n1 NV (z) 2k ϕ (z) C h R z e 2 1+ ,= (n). nj n n n z2k 2k 2k k=1

Let us dene the function ϕn1,j(z) by the formula (3.22),

n1(z) Cn (4.6) ϕn1,j(z)= ϕnj(z)+ . n(z) n(z) Then the vector-function ϕnj(z) ~ nj(z)= ϕn1,j(z) satises the dierential equation ~ 0 ~ nj(z)=NAn(z)nj(z).

Hence the function ϕn1,j(z) satises the rst-order dierential equation (3.21) with C = Cn. In addition, (4.6) gives that ϕn1,j(z) → 0asz →∞in j.

SEMICLASSICAL ASYMPTOTICS 209

This implies that Z z a21(u) (4.7) ϕn 1,j(z)= CnNn 1(z) 2 du ωj ∞ n1(u) 1/2 t 2 g 4 N( 4 z + 8 z ) = CnNhn1Pn1(z) e Z z 1/2 2 Rn [t + gu + g(Rn1 + Rn)] t 2 g 4 N( 2 u + 4 u ) 2 e du , ωj ∞ Pn1(u) which for n 2 can be rewritten as 1/2 1/2 t 2 g 4 1/2 N( 4 z + 8 z ) ϕn1,j(z)= CnRn Rn1 Nhn1Pn1(z) e Z z 1/2 2 Rn1[t + gu + g(Rn1 + Rn)] t 2 g 4 N( 2 u + 4 u ) 2 e du . ωj ∞ Pn1(u) The last expression is similar to (4.4) if we take

1/2 1/2 Cn1 = CnRn Rn1 ,n 2, or 1/2 1/2 Cn1Rn1 = CnRn ,n 2.

We solve the last equation recursively. The initial constant C1 is a free param- 1/2 1/2 eter. We put C1 = R1 . This gives Cn = Rn , so that from (4.4) and (4.7) we derive

1/2 N t z2+ g z4 ϕ (z)=Nh P (z) e ( 4 8 ) (4.8) nj Zn n z 2 t + gu + g(Rn + Rn+1) t 2 g 4 N( 2 u + 4 u ) 2 e du , n 0, ωj ∞ Pn (u) and ! X∞ 1/2 n1 NV (z) 2k (4.9) ϕ (z) h z e 2 1+ . nj n z2k k=1

Observe that ~ nj(z) satises the recursive equation

~ n+1,j(z)=Un(z)~ nj(z). Indeed, consider the vector-function 0 ~ ~ 0 ϕn+1,j(z) Un(z)nj(z) n+1,j(z)= . ϕnj(z)

Then, since nj(z) satises the dierential equation ~ 0 ~ nj(z)=NAn(z)nj(z),

210 PAVEL BLEHER AND ALEXANDER ITS

~ 0 the equation (3.11) implies that n+1,j(z) satises the dierential equation 0 ~ 0 ~ 0 n+1,j(z) = NAn+1(z)n+1,j(z),

~ ~ 0 → →∞ the same as for n+1,j(z). In addition, n+1,j(z) 0asz in j. Hence ~ 0 ~ n+1,j(z) coincides with n+1,j(z) up to a constant factor, but their second ~ 0 ~ components are the same; hence n+1,j(z) and n+1,j(z) coincide, which was stated. It is also worth noticing that ~ nj(z) as a solution of linear ODE with polynomial coecients is an entire function of z. As a matter of fact, the Laplace method combined with a deformation of the contour of integration gives the asymptotic expansion (4.9) in a bigger domain, (4.10) (j 1) 5 (j 1) = + + ε0. j 8 2 8 2 Thus, the matrix-valued function n(z) ϕnj(z) (4.11) nj(z)= = ~ n(z), ~ nj(z) n1(z) ϕn1,j(z) is an entire function of z, and according to (4.2) and (4.9) it has the asymptotic expansion ! X∞ k NV (z) n ln z+ (4.12) (z) e 2 n 3 ,z→∞,z∈ , nj zk j k=0 where 10 (4.13) = 3 0 1 is the Pauli matrix, 1 (4.14) = ln h , n 2 n and k are some 2 2 matrices which only depend on n. From (4.2) and (4.9) we obtain that 10 01 (4.15) 0 = 1/2 , 1 = 1/2 . 0 Rn Rn 0

The independence on j of the coecients k in the asymptotic series (4.12) constitutes the so-called Stokes phenomenon, and the functions nj(z) form the set of canonical solutions (see, e.g., [Sib]) for the equation 0(z)= NAn(z)(z). The corresponding Stokes matrices (see equation (5.3) below) are evaluated in the next section.

SEMICLASSICAL ASYMPTOTICS 211

5. The Riemann-Hilbert problem

Since both ~ nj(z) for dierent j and ~ n(z) satisfy the same dierential equation (3.6), they are linearly dependent,

~ n,j+1(z)=q~ nj(z)+s~ n(z), where j is dened mod 4. The domains j and j+1 intersect and in the intersection the functions ~ nj(z) and ~ n,j+1(z) grow to innity and have the same asymptotic expansion (see (4.12)). On the other hand ~ n(z) goes to zero in this intersection, hence q = 1, so that

(5.1) ~ n,j+1(z)=~ nj(z)+s~ n(z).

Since both ~ nj(z) and ~ n(z) satisfy the same recursive equation (3.6), the coecient s does not depend on n, but in general it depends on j, s = sj. The equation (5.1) implies that

(5.2) ϕn,j+1(z)=ϕnj(z)+sjn(z). We can rewrite (5.1) in matrix form as

(5.3) n,j+1(z)=nj(z) Sj, where 1 s (5.4) S = j . j 01

To determine sj consider (5.2) at n = 0. From (4.8) it follows that Z z 1/2 t 2 g 4 t 2 g 4 N( 4 z + 8 z ) 2 N( 2 u + 4 u ) (5.5) ϕ0j(z)=Nh0 e (t + gu + gR1)e du. ωj ∞ Putting z = 0 in (5.2) we get

ϕ0,j+1(0) ϕ0j(0) sj = . 0(0) Since 1/2 0(0) = h0 (see (4.2)) and Z 0 1/2 t 2 g 4 2 N( 2 u + 4 u ) ϕ0j(0) = Nh0 (t + gu + gR1)e du ωj ∞ we obtain that Z ωj ∞ 2 N( t u2+ g u4) (5.6) sj = Nh0 (t + gu + gR1)e 2 4 du, j =1, 2, 3, 4. ωj+1∞

212 PAVEL BLEHER AND ALEXANDER ITS

The change of variable u →u gives

(5.7) s3 = s1,s4 = s2.

Another way to compute sj is to use the Cauchy type integral. The function NV (z) (5.8) ynj(z)=e 2 ϕnj(z),j=1, 2, 3, 4, is an entire function of z and in it has the asymptotics j ! X∞ (5.9) y (z) h1/2z n 1 1+ 2k . nj n z2k k=1 In addition, NV (z) (5.10) yn,j+1(z)=ynj(z)+sje 2 n(z). Observe that the domain contains the jth quadrant, j (j 1) j = z : arg z ; j 2 2 hence (5.9) holds in j. This allows us to solve (5.10) with the help of the Cauchy type integral. Namely, (5.11) Z Z ∞ NV (u) i∞ NV (u) s4 e 2 n(u) s1 e 2 n(u) yn(z)= du + du 2i ∞ u z 2i i∞ u z Z Z ∞ 1/2 NV(u) i∞ 1/2 NV(u) s4 hn Pn(u)e s1 hn Pn(u)e = du + du 2i ∞ u z 2i i∞ u z where yn(z) is a piecewise analytic function which coincides with ynj(z)inthe quadrant j. Expanding X∞ 1 1 uk = , u z z zk k=0 we obtain from (5.11) the asymptotic expansion of yn(z)asz →∞: X∞ 1 s a + s b (5.12) y (z) 4 nk 1 nk , n z zk k=0 where Z ∞ 1 1/2 k NV(u) (5.13) ank = hn Pn(u)u e du, 2i ∞ Z i∞ 1 1/2 k NV(u) bnk = hn Pn(u)u e du. 2i i∞

SEMICLASSICAL ASYMPTOTICS 213

n1 Since by (5.9), yn(z)=O(z ),

s4ank + s1bnk =0,k=0, 1,...,n 1, but ank = 0 for k =0, 1,...,n1 in virtue of the orthogonality property (1.3); hence s1bnk = 0 for these k’s. Let us take n = 2 and k = 0. In this case Z Z i∞ ∞ NV (u) NV(iu) i P2(u)e du = P2(iu)e du i∞ ∞ is obviously positive; hence b20 =6 0. This implies

(5.14) s1 = s3 =0.

By (5.2) this means that

ϕn2(z)=ϕn1(z),ϕn4(z)=ϕn3(z).

Let us nd s4. From (5.9) we know that

1/2 1 y0(z)=h0 (z + ...);

1/2 hence by (5.12), s4a00 = h0 . In addition, by (5.13), 1 a = h1/2. 00 2i 0 This gives

s4 = 2i.

It is interesting to notice that the values of s1 = s3 = 0 and s2 = s4 = 2i can be obtained directly from formula (5.6) as well, although the calculations are more involved. Namely, the recurrent coecient R1 is expressed in terms of the parabolic cylinder function: h i 1/2 2 d z2/4 R1 = ln e D1/2(z) q . Ng dz N z=t 2g

Then, using two dierent integral representations of the function D1/2(z) (see, e.g., [BE]), Z ∞ √1 z2/4 ztt2/2 1/2 D1/2(z)= e e t dt 0 r Z ∞ 2 2 2 = ez /4 e t /2 cos zt + t 1/2dt, 0 4

214 PAVEL BLEHER AND ALEXANDER ITS one derives from (5.6) that s1 and s4 are expressed in terms of the , √ D1/2(z) D1/2(z) s1 = 2 0 0 =0, D1/2(z) D1/2(z) √ D1/2(z) D1/2( z) s = 2 i = 2i, 4 0 0 D1/2(z) D1/2( z) as stated above. Now we can formulate the Riemann-Hilbert problem. Dene (5.15) n(z) ϕn1(z) n(z) ϕn3(z) n+(z)= , n(z)= . n1(z) ϕn1,1(z) n1(z) ϕn1,3(z) Let ( n+(z), if Im z 0, n(z)= n(z), if Im z 0. Then (z) has the asymptotic expansion n ! X∞ k NV (z) n ln z+ (5.16) (z) e 2 n 3 ,z→∞, n zk k=0 1 where n = ln hn and 2 10 01 (5.17) 0 = 1/2 , 1 = 1/2 . 0 Rn Rn 0 On the real line

(5.18) n+(z)=n(z)S, Im z =0, where 1 2i (5.19) S = . 01 In addition,

1/2 6 (5.20) det n(z)=Rn =0.

It must be emphasized that in the setting of the Riemann-Hilbert problem (5.16)–(5.18) the real quantities Rn and n are not the given data. They are evaluated via the solution n(z). Indeed, suppose that ˜ n(z) is another function satisfying (5.16)–(5.18) with perhaps another R˜n and another ˜n. Consider the matrix ratio: ˜ n3 1 n3 ˜ 1 ˜ 1 X (z)=[e 0 n(z)][e 0 n(z)] .

SEMICLASSICAL ASYMPTOTICS 215

Since the jump matrix S for both n(z) and ˜ n(z) is the same, function X (z) has no jump across the real line. Therefore, it is an entire function equal to I at z = ∞. Hence X (z) I, or ˜ n3 1 n3 ˜ 1 ˜ 1 [e 0 n(z)][e 0 n(z)] I.

Substituting into this identity asymptotic expansion (5.16) and equating the terms of order z1 we come up with the matrix equation, ˜ ˜ n3 1 n3 n3 ˜ 1 ˜ n3 e 0 1e = e 0 1e , or ˜ 0 e2n 0 e2n 2n = 2˜n ; Rne 0 R˜ne 0 hence n = ˜n and Rn = R˜n. The Riemann-Hilbert problem (5.16)–(5.18) for the orthogonal polyno- mials (1.1), (1.3) rst appeared in [FIK2,3] as a particular case of the more general Riemann-HilbertS problem related to the system of orthogonal polyno- mials on the cross R iR with the same weight function . Also, in [FIK2,3] an alternative approach to the derivation of the basic equations (5.15), which can be easily extended on an arbitrary system of orthogonal polynomials (cf. [FIK4]), was suggested. We present the method of [FIK] in Appendix E.

6. Formal asymptotic expansion for Rn

To guess a behavior of Rn let us evaluate Rn for small n. We have R0 =0 and R ∞ 2 gN (z2M 2)2 h z e 4 dz R = 1 = R0 ,M= |t|1/2g 1/2. 1 ∞ gN 2 2 2 h0 4 (z M ) 0 e dz Using the change of variable gN 1/2 u = (z2 M 2) 2 we obtain that R 2 ∞ 1/2 u (1 + u) e 2 du 1/2 2 1/ (2g) R1 = M R ,= . ∞ u2 | | 1/2 1/2 2 t N 1/(1 + u) e du

216 PAVEL BLEHER AND ALEXANDER ITS

Expanding (1 + u)1/2 into Taylor’s series and evaluating the Gaussian inte- grals we derive that 2 4 3 |t| 1 3g R = M 2 1 + O(N 3)= + O(N 3). 1 2 4 g |t|N |t|3N 2 Now from the Freud equation (1.7) we obtain subsequently that

2 4g R = + + O(N 3), 2 |t|N |t|3N 2 |t| 3 R = + O(N 2), 3 g |t|N 4 R = + O(N 2). 4 |t|N These formulae suggest that for a xed n as N →∞,   |t| n  + O(N 2)ifn =2k +1, g |t|N Rn =  n  + O(N 2)ifn =2k. |t|N n More generally we expect that if is bounded and N →∞then N   n  R + O(N 2)ifn =2k +1, (6.1) R = N n  n  L + O(N 2)ifn =2k, N where R() and L() are smooth functions. Substituting this ansatz into the Freud equation (1.7) and neglecting terms of the order of O(N 1), we obtain the equations (6.2) = R[t + g(2L + R)], = L[t + g(2R + L)]. Equating the expressions on the right we obtain that (R L)[t + g(R + L)]=0. We assume that R =6 L; hence (6.3) t + g(R + L)=0. Combining this with (6.2) we obtain (6.4) = gRL, so that R, L are solutions of the quadratic equation t (6.5) u2 + u + =0, g g

SEMICLASSICAL ASYMPTOTICS 217 which are p t t2 4g (6.6) R, L = . 2g

Actually, we can nd a formal asymptotic expansion of Rn in powers of N 2. Put ( R(n/N)ifn =2k +1, Rn = L(n/N)ifn =2k. Then (1.7) is equivalent to (6.7) = R(t + gR +2gL + N 2gL), = L(t + gL +2gR + N 2gR), where f( 1 ) 2f()+f( + 1 ) f()= N N . (1/N )2 Let us substitute the expansions 2 2 L()=L0()+N L1()+...; R()=R0()+N R1()+... into (6.7) and equate coecients at powers of N 2. Then the equations on L0(),R0() coincide with (6.2), hence L0(),R0() are given by (6.6). Equat- ing coecients at N 2 we obtain the system of equations ( (R0 + L0)R1 +2R0L1 = R0L0, (6.8) 2L0R1 +(R0 + L0)L1 = L0R0.

Solving this system we nd R1,L1 and so on.

7. Proof of the main theorem: The asymptotic Riemann-Hilbert problem

In this section we construct explicitly, using semiclassical formulae, a piecewise analytic function 0(z) which is an approximate solution to the basic Riemann-Hilbert problem (5.16)–(5.18). Then we prove that 0(z) is close to n(z) (the Deift–Zhou method, see [DZ]). Dene the numbers (cf. (6.6)) t (1)n(t2 4g)1/2 n (7.1) R0 = ,= , n 2g N n 2 1/2 0 t +( 1) (t 4g) R = , n 1 2g and consider an auxiliary matrix dierential equation, 0 (7.2) (z)=NA0(z)(z),

218 PAVEL BLEHER AND ALEXANDER ITS where the matrix 0 0 0 0 a11 a12 A = A (z)= 0 0 a21 a22 is dened as

(7.3) tz gz3 gz3 (1)n(t2 4g)1/2 a0 = + + gzR0 = z ,= , 11 2 2 n n 2 n 2 0 0 1/2 2 a12 =(Rn) gz , 0 0 0 0 a21 = a12,a22 = a11.

0 0 The matrix A (z) is obtained from the matrix An(z) by the substitution of Rj 0 for Rj,j= n1,n,n+1. The function (z) is constructed as a semiclassical approximation to the dierential equation (7.2). A direct calculation gives that

z2 det A0(z)= (t + gz2)2 4g . 4 Dene the -function as (7.4) q p z p zg (z)= det A0(z)= (t + gz2)2 4g = (z2 z2)(z2 z2), 2 2 1 2 where we take the branch of the square root which is positive on the positive half-axis. The zeros of the -function are the turning points for the equation (7.2). There are ve turning points: a double turning point at z = 0 and four simple turning points at z1 and z2 where √ t 2 g 1/2 z = . 1,2 g We assume that t2 ε<< ε 4g

(cf. (1.11)), which ensures that both z1 and z2 are real and separated from zero. Let be a neighborhood of the interval [z1,z2]. For the sake of deniteness we dene as an ellipse with foci at z1 and z2, (x z )2 y2 = z = x + iy : 0 + 1 , a2 b2 z + z z = 1 2 ,a2 =(z z )2 + b2 . 0 2 0 1

SEMICLASSICAL ASYMPTOTICS 219

We will assume that b>0 is independent of N and it is suciently small so that does not contain the origin. Let

1 =∩{Re z z0}, 2 =∩{Re z z0}, u =∩{Im z 0}, d =∩{Im z 0},

u and assume that the boundaries ∂,∂1, and ∂ are positively oriented (see Fig.2). Denote

C1 = ∂ ∩{Re z z0},C2 = ∂ ∩{Re z z0}.

Let v1,2 = z0 ib be two vertices of the ellipse and

u d l0 =[v1,v2],l0 =[v1,z0],l0 =[z0,v2]. We shall also take the following convention: given any set C we denote by u and d the closed upper and lower parts of set with respect to the real axis, so that u ∩{ } d ∩{ } 1,2 =1,2 Im z 0 , 1,2 =1,2 Im z 0 , u ∩{ } d ∩{ } C1,2 = C1,2 Im z 0 ,C1,2 = C1,2 Im z 0 , etc. (see Fig. 2). Our plan is the following: (1) Outside of the domain ∪(), where ={z : z ∈ }, we dene 0(z) by a WKB formula.

Im z

u u C v C 1 1 2 Ω u u Ω u . 1 l0 2 . z z z 1 Ω d 0 d Ω d 2 Re z 1 l0 2 d C d v C 1 2 2

Figure 2. The set Ω

220 PAVEL BLEHER AND ALEXANDER ITS

u,d 0 (2) In the domains 1,2 we dene (z) by turning point formulae, with 0 n 0 the help of the Airy function. In we dene (z)=(1) 3 (z)3. (3) We check that the multiplicative jump matrix of 0(z) between u and d coincides with the matrix S in (5.19), and the multiplicative jump matrix on ∂ is close to the unit matrix, as well as the multiplicative jump matrix between 1 and 2. This will give us an approximate explicit solution 0(z) to the Riemann- Hilbert problem (5.16)–(5.18), and then we will establish the closeness of 0(z) to n(z). As the rst step of our plan, we dene 0(z) outside of by a WKB formula. To get this formula we diagonalize iteratively the equation (7.2). The diagonalization is described in detail by Kapaev [Kap]. We will use the WKB formula with the error term O(N 1). In this approximation we get that

0 N(z)3(z)+C13 (7.5) (z)=WKB(z) C0T (z)e , where  0  a12 Z 1 z a0 (7.6) T (z)= 11  ,(z)= (u) du, (a0 ) 21 1 z2 a0 Z 11 z 0 (z)= diag T 1(z)T (z) dz, z2 and h i p 1/2 0 0 2 0 0 (7.7) (z)= det A (z)= a11(z) + a12(z)a21(z) . In (7.6) diag M stands for a which coincides with M on the diagonal. The formula (7.5) gives an approximate solution of the dierential equation (7.2), up to the terms of the order of N 1, and this is a motivation for us to dene 0(z) outside of by (7.5). However, it is important to underscore here that our ultimate goal is a construction of an approximate solution to the Riemann-Hilbert problem (5.16)–(5.18) and we are not going to prove any results concerning the dierential equation (7.2). It serves only as an auxiliary tool in our construction of 0(z). Observe that p p t + gz2 (z)=z g x2 1,x= √ , 2 g and the values z = z1 and z = z2 correspond to x = 1 and x = 1, respectively. We x the branches of the multivalued functions in (7.6) by the inequalities p 2 < arg z< , |z| >z2,

SEMICLASSICAL ASYMPTOTICS 221

This denes (z) as an analytic function on C, with two cuts [z2, z1] and [z1,z2]. The function (z), in addition, saties the symmetry equations, (z)=(z),(z)=(z). The formula for (z) leads to the following expression for the function (z): (7.70) h p p i t + gz2 n (z)= x x2 1 ln(x + x2 1) ,x= √ ,= . 2 2 g N

The function (z) is analytic and it has one cut (∞,z2]. The matrix T (z) diagonalizes the matrix A0(z),

1 0 T (z)A (z)T (z)=(z)3, so that (z) are eigenvalues of the matrix A0(z) and T (z) is the matrix of corresponding eigenvectors. The matrix diag T 1(z)T 0(z) turns out to be proportional to the unit matrix and a straightforward calculation gives that up to a nonessential constant factor, 0 1/2 a 10 (7.8) e (z) = 11 I, I = 2 01 (see Appendix A below). Thus (7.5) reduces to

0 N(z)3+C13 (7.9) (z)=WKB(z) C0T0(z)e , where C0 =06 ,C1 are constants and  a0  1/2 12 0 1 0 a11  a11  (7.10) T0(z)= 0 . 2 a12 0 1 a11 We shall x the branch of the square root by the condition, a0 1/2 11 > 0,z>z. 2 2

It is easy to check that T0(z) has no singularities except for those at z = z1,2, and that det T0(z)=1. The last equation in turn implies that 2 6 det WKB(z)=C0 =0. Consider the region C = C \ (∪ ()).

222 PAVEL BLEHER AND ALEXANDER ITS

Lemma 7.1. The function WKB(z) dened in (7.9), is a single-valued analytic function on C, which, in addition, satises the symmetry equation

n (7.11) WKB(z)=(1) 3WKB(z)3.

Proof of Lemma 7.1 is given in Appendix B below. As z →∞, 0 1/2 10 1 0(Rn) | |2 →∞ (7.12) T0(z)= +z 0 1/2 +O z ,z . 01 (Rn) 0 In addition, from (7.6), (7.70), V (z) n ln z (7.13) (z)= + + O |z| 2 ,z→∞, 2 N n where t2 g (7.14) = ln . n 8g 4 4 Thus, ! ∞ X 0 NV (z) 0 n ln z+NnC1 3 (7.15) (z) k C e 2 , zk 0 k=0 with 0 1/2 0 10 0 0(Rn) (7.16) 0 = , 1 = 0 1/2 . 01 (Rn) 0 0 Turning Point Functions. To construct (z) in the sets 1,2 we use turning point functions which are expressed in terms of the Airy function. Let us consider rst the set 2. On this set we are looking for a solution to (7.2) in the form (7.17) (z)=W (z)(w(z)). Following [Kap] (see also the earlier work [Ble]), we want to nd an analytical matrix-valued function W (z) (gauge matrix) and an analytical function w(z)of the change of variable such that (7.2) reduces to the following model equation for (w): 0 01 (7.18) (w)=N (w). w 0 This model equation is nothing else than the Airy equation 00(w)=N 2w(w) written in the matrix form. The functions W (z) and w(z) are determined iteratively in powers of N 1 and we consider approximation of the order of N 1. In this approximation the functions w(z) and W (z) are dened as follows.

SEMICLASSICAL ASYMPTOTICS 223

1. Function of change of variable. Put ! Z 2/3 3 z (7.19) w(z; z2)= (z) dz , 2 N z2 where 2 1/2 t gz (7.20) (z)= 2(z)+N 1 + gR0 , 2 n 2

N and z2 isarootof(z) close to z2,

N N 1 (z2 )=0,z2 z2 = O(N ).

The argument z2 in w(z; z2) reminds us that this function is constructed in a 2 neighborhood of the turning point z2. Observe that the function (z) can be written in terms of the elements of the matrix A0(z)as 0 0 0 2 0 1 0 0 0 (a12) (z) (7.20 ) (z)= det A (z)+N (a11) (z) a11(z) 0 a12(z) which coincides (up to terms of the order of N 2) with the potential of the Schrodinger equation equivalent to (7.2) (cf. (3.26,7)). In (7.19) and (7.20) we take the branches of the power functions that are positive on the positive N 1 half-axis. Since z2 z2 = O(N ), for large N the function w(z; z2) is analytic for z ∈ 2 and 0 w (z; z2) =06 ,z∈ 2.

2. Gauge matrix. Dene s ! 0 10 a12(z) (7.21) W (z; z2)= a0 (z) w0(z;z ) , w0(z; z ) 11 2 2 0 0 a12(z) a12(z) where s 0 a12(z) N 0 > 0,z>z2 . w (z; z2)

The function W (z; z2) is analytic in z ∈ 2 for large N. u d To get a desired multiplicative jump matrix S between 2 and 2 we use u d dierent solutions to the model equation (7.18) in the domains 2 and 2. They are dened as follows.

224 PAVEL BLEHER AND ALEXANDER ITS

3. Model canonical solutions. Dene (7.22) 1/6 2/3 2/3 1/2 N 0 y0(N z) y1(N z) (2) 0 u(z; z2)= 1/6 0 2/3 0 2/3 1/2 , 0 N y0(N z) y1(N z) 0(2)

1/6 2/3 2/3 1/2 N 0 y0(N z) y2(N z) (2) 0 d(z; z2)= 1/6 0 2/3 0 2/3 1/2 , 0 N y0(N z) y2(N z) 0(2) where

(7.23) y0(z) = Ai (z), i/6 2i/3 y1(z)=e Ai e z , i/6 2i/3 y2(z)=e Ai e z .

Here Ai (z) is the Airy function, i.e. the solution of the model equation (7.24) ! 3/2 00 1 2z Ai (z)=z Ai (z); Ai (z) √ exp ,z→∞. 2 z1/4 3

It satises the relation Ai (z)+e2i/3 Ai e2i/3z + e 2i/3 Ai e 2i/3z =0; hence y1(z) y2(z)=iy0(z), and this implies the equation 1 2i (7.25) (z)= (z) . u d 01 Now we dene the turning point functions

u,d (7.26) TP (z; z2)=CW(z; z2)u,d(w(z; z2); z2), where C is a constant which will be chosen later. In 1 we dene u,d (1) (7.27) TP (z; z1)=C W (z; z1)u,d(w(z; z1); z1), (1) where C is a constant to be chosen later, and w(z; z1), W (z; z1), and u,d(z; z1) are dened as follows. The function of change of variable ! Z 2/3 z 0 3 (7.27 ) w(z; z1)= (z) dz , 2 N z1

SEMICLASSICAL ASYMPTOTICS 225

N N 1 N where (z1 )=0,z1 z1 = O(N ). Observe that (z) < 0 when 0 0, 0 0. The model solutions u,d(z; z1) are dened as (7.2700) 1/6 2/3 2/3 1/2 N 0 y0(N z) y2(N z) (2) 0 u(z; z1)= 1/6 0 2/3 0 2/3 1/2 , 0 N y0(N z) y2(N z) 0(2)

1/6 2/3 2/3 1/2 N 0 y0(N z) y1(N z) (2) 0 d(z; z1)= 1/6 0 2/3 0 2/3 1/2 . 0 N y0(N z) y1(N z) 0(2)

Comparing this with the model solutions (7.22) in 2, one can notice that we change y1 for y2 in the formula for u and y2 for y1 in the formula for d. Note that equation (7.25) still holds. The gauge matrix W (z; z1) is dened as s ! 0 10 000 a (z) 12 0 0 (7.27 ) W (z; z1)= i 0 a11(z) w (z;z1) ; w (z; z1) a0 (z) a0 (z) 12 s 12 0 a12(z) N i 0 > 0, 0

Now we want to evaluate the jump matrices on ∂1 and ∂2. Observe that the WKB function (7.9) has two free constants C0 and C1, and each of the turning

226 PAVEL BLEHER AND ALEXANDER ITS point solutions (7.26), (7.27) has one free constant, C and C(1), respectively. We show that we can adjust these constants to make the jump matrices close to I.

Lemma 7.2. If h i C ln 2 (1) k n (7.31) C0 = √ ,C1 = ; C =(1) C, k = , 2 2 2 then 0 1 0 (7.32) +(z)= I + O(N ) (z), uniformly with respect to

z ∈ ∂1 ∪ ∂2 ∪ (∂1) ∪ (∂2) , where

∂j = { z : z ∈ ∂j } ,j=1, 2.

The proof of this lemma is given in Appendix C below. Consider now the asymptotics of 0(z) at innity. According to (7.15) and (7.31), ! X∞ 0 NV (z) ln 2 0 C n ln z+Nn+ 3 (7.33) (z) k √ e 2 2 , zk k=0 2 with 0 1/2 0 10 0 0(Rn) 0 = , 1 = 0 1/2 . 01 (Rn) 0 We want to adjust the asymptotics to the one (5.16), (5.17) in the Riemann- Hilbert problem. Dene

0 ln 2 ln R (7.34) 0 = N + + n ,C=(R0 ) 1/4(2)1/2. n n 2 4 n Then (7.33) is rewritten as ! X∞ 0 NV (z) 0 0 n ln z+ 3 (7.35) (z) k e 2 n , zk k=0

0 with some coecients k such that 0 10 0 01 (7.36) 0 = 0 1/2 , 1 = 0 1/2 . 0(Rn) (Rn) 0

SEMICLASSICAL ASYMPTOTICS 227

Proof of Theorem 1.1. We want to prove that 0(z), dened in (7.9) and (7.28) with C, C0, and C1 given by (7.31), (7.34), is close to n(z). Observe that by (5.16) and (7.35), the limit

0 0 1 ( n)3 0 1 (7.37) = lim n(z) (z) =0e n ( ) z→∞ 0 exists and det =6 0. Denote 1 0 1 (7.38) X(z)= n(z) (z) . Then (7.39) lim X(z)=I. z→∞ 0 The multiplicative jump matrices of n(z) and (z)on∩{Im z =0} coincide (both are equal to S); hence X(z) has no jump on ∩{Im z =0}. On ∂j,j=1, 2, the function n(z) has no jump; hence 1 0 1 1 1 0 1 X+(z)= n(z) (z) = n(z) I + O(N ) (z) + 1 0 1 1 1 1 1 = n(z) (z) I + O(N ) = X(z) I + O(N ) , so that 1 (7.40) X+(z)=X(z) I + O(N ) , if z ∈ ∂1 ∪ ∂2. Let (7.41) L = { z :Imz =0,z6∈ ∪ () } . On L, the function 0(z) has no jump; hence

(7.42) 1 0 1 1 0 1 X+(z)= n+(z) (z) = n(z)S (z) 1 0 1 0 0 1 = n(z) (z) (z)S (z) = X(z)Q(z), where 1 Q(z)=0(z)S 0(z) ,z∈ L. 0 When z ∈ L,(z)=WKB(z); hence by (7.9),

[N(z)+C1]3 [N(z)C1]3 1 Q(z)=T0(z)e Se T0 (z) 2N(z)+2C 1 2i e 1 = T (z) T 1(z) 0 01 0 2N(z) 0 ie = I + T (z) T 1(z). 0 00 0 Since det T (z) = 1, we obtain from (7.10) that |T (z)|, |T 1(z)|C; hence 0 0 0 Q(z)=I + O e 2NRe (z)

228 PAVEL BLEHER AND ALEXANDER ITS and 2NRe (z) (7.43) X+(z)=X(z) I + O e ,z∈ L. Thus, we have the following theorem. 1 0 1 Theorem 7.3. The function X(z)= n(z) (z) satises lim X(z)=I z→∞ and (7.44) 1 X (z)=X(z) I + O(N ) , if z ∈ ∂ ∪ ∂ ∪ (∂ ) ∪ (∂ ), + 1 2 1 2 2NRe (z) X+(z)=X(z) I + O e , if z ∈ L; i.e., there exists a constant C>0 independent of N and z such that 1 1 [X(z)] X+(z) I CN , if z ∈ ∂1 ∪ ∂2 ∪ (∂1) ∪ (∂2), 1 2NRe (z) [X(z)] X+(z) I Ce , if z ∈ L.

Remark. Observe that there exists c0 > 0 such that Re (z) 4 0 c0(1 + |z|) > 0 for all z ∈ L (cf. (7.7 )); therefore the error term on L is exponentially small as N →∞and also as z →∞. By standard methods (see e.g. [DZ2], [DIZ] and also Appendix D below), Theorem 7.3 yields the asymptotic equation, 1 (7.45) X(z)=I + O ,z∈ C, N(1 + |z|) so that 1 (7.46) (z)= I + O 0(z). n N(1 + |z|) This implies that ! X∞ X∞ 0 k 1 0 = I + O k e(n n)3 . zk N(1 + |z|) zk k=0 k=0 Equating the coecients at z0 and at z1, we obtain

0 3 0 (7.47) 0 =0e ,= n n, 0 1 0 3 1 = 1 + O(N )0 e . Excluding we obtain

3 1 3 0 1 0 1 e (0) 1e =(0) 1 + O(N ).

SEMICLASSICAL ASYMPTOTICS 229

We have 1 01 0 1 0 01 (0) 1 = , (0) 1 = 0 ; Rn 0 Rn 0 hence 2 3 1 3 0 e e (0) 1e = 2 . Rne 0 Thus, 2 0 e 01 1 2 = 0 + O(N ), Rne 0 Rn 0 which implies that 0 0 1 0 1 (7.47 ) Rn = Rn + O(N ),n = n + O(N ). 0 In view of the denition of Rn (see (7.1)) the rst equation yields (1.14). By (7.47), 3 0 1 e 0 1 =0e (0) = 1/2 0 1/2 = I + O N ; 0 e (Rn) (Rn) hence by (7.46), 1 (7.48) (z)= I + O 0(z), =I + O N 1 . n N(1 + |z|) All the statements of our main Theorem 1.1 follow directly from equation (7.48). In what follows we are presenting the corresponding detailed deriva- tions. To derive the semiclassical asymptotics (1.15) of the function n(z)be- tween the turning points z1 and z2 we use the following lemma. Let >0be an arbitrary xed number. Dene

0 =∩{z : z1 + Re z z2 } and u ∩{ } d ∩{ } 0 =0 z :Imz 0 , 0 =0 z :Imz 0 . 0 Lemma 7.4. The function (z) has the following asymptotics in 0: ! 1 0 1/2 10 0 I + O(N ) a12(z) (1) (7.49) (z)= a0 (z) (z) Q 0 1/4 2(z) 11 u,d (Rn) 0 0 a12(z) a12(z)

(N(z)+d(z))3 (2) ∈ u,d e Qu,dP, if z 0 , where (z) is dened as in (7.6),(7.7 0), 0 0 1 (z) a11(z) (7.49 ) d(z)= ln 0 , 2 a12(z)

230 PAVEL BLEHER AND ALEXANDER ITS and 1 i i 1 Q(1) = Q(1) = ,Q(2) = Q(2) = , u d 1 i u d i 0 (2)1/2 0 P = . 0(2)1/2

Proof of this lemma is given in Appendix C below. Due to (7.48), we can 0 replace (z) with n(z) in (7.49), which gives the semiclassical asymptotics of n(z)in0. Thus, we have the following theorem.

Theorem 7.5. In 0, ! 1 0 1/2 10 I + O(N ) a12(z) (1) (7.50) (z)= a0 (z) (z) Q n 0 1/4 11 u,d (Rn) 2(z) 0 0 a12(z) a12(z)

(N(z)+d(z))3 (2) ∈ u,d e Qu,dP, if z 0 .

Remark. Let 0(z), 0(z), d0(z) denote the analytic continuation of the u functions (z), (z), d(z) from 0 to 0 across the interval [z1 + , z2 ]. ∈ d (Note that 0(z)= (z), 0(z)= (z), and d0(z)= d(z) for z 0.) Taking into account that by (3.14) the function n(z) is the element 11 of the matrix-valued function n(z), we derive from (7.50) that 0 1/2 1 a (z) (7.500) (z)= I + O(N 1) 12 n 1/2 0 1/4 (2) (R ) 20(z) h n i N0(z)+d0(z) (N0(z)+d0(z)) ie + e , if z ∈ 0.

The function N0(z)+d0(z) is pure imaginary for real z in the interval 0 0 z1 + z z2 (cf. (7.7 ) and (7.49 )); hence, when we restrict the 0 asymptotics (7.50 ) to the interval z1 + z z2 , the equation, 0 1/2 1 a12(z) n(z)= (2)1/2(R0 )1/4 2(z) h n i ieN0(z)+d0(z) + e (N0(z)+d0(z)) + O(N 1) , follows. Since p p i 0 0 1/2 2 2 a12 =(Rn) gz ,(z)=e 2 z g 1 x , we obtain that for z1 + z z2 , √ h i 1/4 1 g z 1 (7.51) n(z)=√ cos (z)+ + O(N ) , (1 x2)1/4 4

SEMICLASSICAL ASYMPTOTICS 231

1 where (z)=i [N0(z)+d0(z)]. To nish the proof of (1.15) we use the following lemma.

Lemma 7.6. If z1 + z z2 then 1 n n + sin 2 (1) (7.52) i 1[N (z)+d (z)] = 2 , 0 0 2 2 4 where = arccos q and = arccos r and √ 2 0 1 gz + t 2 g tq 0 n + q = √ ,r= √ ,= 2 . 2 0g 2 0gq t N

Proof of this lemma is given in Appendix C below. Since 1 0 n + n 1 = 2 = + = + O(N 1), N N 2N we obtain p 1 x2 = sin 1+O(N 1) ; hence for z1 + z z2 , ( " √ 1/4 1 1 g z n + 2 sin 2 n(z)=√ √ cos sin 2 2 n (1) + + O(N 1) . 4 4 Formula (1.15) is proved. To prove formulae (1.17), (1.18) we use the following theorem.

Theorem 7.7. The function n(z) has the following asymptotics: in C \ ( ∪ ()),  0  a12 0 0 1/2 1  a0  ln 2 CnCn(z) a11 11 N(z)3 3 (7.53) n(z)=   e 2 , 0 1/4 a0 (Rn) 2 12 0 1 a11 where the matrix Cn does not depend on z, and 1 1 C = I + O ,C0(z)=I + O . n N n N(1 + |z|)

In j,j=1, 2, √ I + O(N 1) 2(1)0 (7.54) (z)= W (z; z ) (w(z; z ); z ), n 0 1/4 j u,d j j (Rn) ∈ u,d z j ,j=1, 2,

232 PAVEL BLEHER AND ALEXANDER ITS n where 0 =(2 j) 2 , ! Z 2/3 3 z (7.55) w(z; zj)= (z) dz , 2 N zj 2 1/2 t gz (z)= 2(z)+N 1 + gR0 , 2 n 2 and the gauge matrix W (z; zj) and the model canonical solutions u,d(z; zj) are as dened in (7.21), (7.27000) and (7.22), (7.2700) respectively.

The proof of the theorem follows from the denition of 0(z) and (7.48).

Theorems 7.5 and 7.7 provide semiclassical asymptotics of n(z)inthe whole complex plane. The formulae (1.17) are derived from (7.53) in the same way we derived (1.15) from Theorem 7.5. Let us show that (1.18) follows from (7.54). For the sake of deniteness consider j = 2. Restricting (7.54) to the element 11, we obtain that if z ∈ 2 then 1/2 0 1/2 h (2) a (z) (z)= 12 (1 + ε )N 1/6 Ai (w(z; z ))(2) 1/2 n 0 1/4 0 1 2 (Rn) w (z; z2) i 1/6 0 + ε2N Ai (w(z; z2)) ,

1 where ε1,2 = O(N ). This gives √ 1/6 N gz 0 (z)= (1 + r )Ai(w(z; z )) + r Ai (w(z; z )) , n 0 1/2 1 2 2 2 (w (z; z2)) 1 4/3 where r1 = O(N ) and r2 = O(N ). Formula (1.18) is proved. Finally, by (4.14), (7.470), (7.34), and (7.14), 0 1 hn = exp (2n) = exp 2n + O(N ) 0 ln R = exp 2N +ln2 + n + O(N 1) n 2 p 2 Nt N g =2 R0 exp 1+ln + O(N 1) , n 4g 2 which proves (1.23). Theorem 1.1 is proved.

Appendix A. Proof of formula (7.8)

By (7.6), 2 0 0 0 2 = a12a21 + a11 ,

SEMICLASSICAL ASYMPTOTICS 233 and  0  a0 a0 21 12 0  a0 2  1 0  11  diag T T =  0  . 0 0 a12 a21 0 0 a11 2 0 0 In our case a21 = a12; hence 2 0 2 0 2 (A.1) = a11 a12 and 0 0 0 0 a a 10 diag T 1T = 12 12 . 0 a11 2 01 Now, h i 0 0 0 0 0 0 0 0 0 a0 a0 a12 a11 a12 (a11) a12 12 12 = . a0 0 2 11 a11 By (A.1), 0 2 0 2 2 (a12) =(a11) , 0 0 0 0 0 0 0 a12(a12) = a11(a11) ; hence 0 0 0 0 0 a0 a0 a0 (a0 ) a0 (a0 )2 2 (a0 ) 12 12 = 11 11 11 11 11 a0 0 2 11 a11 a0 (a0 )0 + 0a0 (a0 )02 (a0 )20 = 11 11 11 11 11 0 2 a11 0 0 (a0 )0 a0 0 a0 = 11 11 = 11 = ln . 0 0 0 a11 a11 a11 Thus, 0 0 0 1 a diag T 1T = ln 11 I 2 and Z z 0 z 0 1 (u) a (u) (z)= diag T 1(u)T (u) du = ln 11 I; 2 (u) z2 u=z2 hence 0 (z) a (z) e 2(z) = C 11 I. (z) Formula (7.8) is proved.

234 PAVEL BLEHER AND ALEXANDER ITS

Appendix B. Proof of Lemma 7.1

Note that (B.1) √ 0 2 1/2 p a11 x + x 1+c g 0 n 2 = √ ,c R = x0 (1) x0 1, 2 2 x2 1 n t x0 = √ > 1. 2 g Let denote the region, = {z ∈ C :Rez>0}\[z1,z2], and dene the analytic functions on by the equations, p p 2 2 2 t + gz (B.2) 1(z)=2 x 1 ,2(z)=x + x 1; x = √ , 2 g and p 2 (B.3) 3(z)=x + x 1+c. These functions generate the conformal mappings, z 7→ , of onto the regions of the complex plane depicted in Figure 3 below. Let us also denote by 1/2 the branch of the square root of dened on the complex plane cut along the ray (∞, 0] and xed by the inequality,

SEMICLASSICAL ASYMPTOTICS 235

ζζ

0 . 0 .

ζζ2 2 3 =x+ x-1 + c 3 =x+ x-1 + c

( n = even ) ( n = odd ) z

1 g x+= - x - - z 2 t λ 0 2 λ x =- - - 0 0 2λ g z z 1 2 n c=xx -(-1) 2 -1 0 0

ζ =2 x2 -1 ζ =x+ x2 -1 1 2

ζ ζ

. 0 0 .

Figure 3. The conformal mappings related to Ψ (z) WKB

1/2 1/2 (iii) The product, 3 (z)1 (z), being analytically extended to C \ [z2,z2] across the imaginary line becomes an even function. n 2 C \ (iv) The function, 2 (z), being analytically extended to [ z2,z2] across the imaginary line satises the symmetry equation,

n n 2 n 2 (B.7) 2 ( z)=( 1) 2 (z).

From the properties (i)–(iv), Lemma 7.1 follows.

236 PAVEL BLEHER AND ALEXANDER ITS

Appendix C. Proofs of Lemmas 7.2, 7.4, and 7.6

Proof of Lemma 7.2. It is easy to see that the function 0(z) dened by (7.9) and (7.28) satises the equation,

(C.1) 0(z)=0(z).

Taking also into account the z →z symmetry (cf. (7.29)), we conclude that it is enough to prove the upper-right part of relation (7.32). We shall start u with discussing the jump over the arc C2 (see Fig.2). For every ε>0, in the sector + ε

(C.2)   X∞ 2 5 2 (z) z3/2 1 z 3 + ... = z3/2 1+ (1)j z 3j , 3 32 3 j j=1 2 0 (z)= z3/2 1+O |z| 3 ,(z)=z1/2 1+O |z| 3 ,z→∞, 3 (cf. [Ble]). Hence

(C.3) ! 3/2 1 2Nz N 1/6 Ai N 2/3z = √ exp 1+O N 1 , 2 z1/4 3 ! 1/4 3/2 0 z 2Nz N 1/6 Ai N 2/3z = √ exp 1+O N 1 , 2 3 |z| >ε, + ε

By (7.23) this implies that

(C.4) ! 3/2 1/6 2/3 1 2Nz 1 N y0 N z = √ exp 1+O N , 2 z1/4 3 ! 1/4 3/2 0 z 2Nz N 1/6y N 2/3z = √ exp 1+O N 1 , 0 2 3 |z| >ε, + ε

SEMICLASSICAL ASYMPTOTICS 237 and

(C.5) ! 3/2 1/6 2/3 1 2Nz 1 N y1 N z = √ exp 1+O N , 2 z1/4 3 ! 1/4 3/2 0 z 2Nz N 1/6y N 2/3z = √ exp 1+O N 1 , 1 2 3 2 |z| >ε, + ε

(C.6) ! 1/4 1/4 3/2 1 z z 2Nz (z; z )= √ I + O N 1 exp 3 P, u 2 2 z1/4 z1/4 3 |z| >ε, + ε

Thus,

(C.7) 1/4 1/4 √1 w (z; z2) w (z; z2) u(w(z; z2); z2)= 1/4 1/4 2 w (z; z2) w (z; z2) ! 3/2 2Nw (z; z ) I + O N 1 exp 2 3 P, 3 under the assumption that (C.70) |w(z; z )| >ε, + ε0 (the vertical half-axis of the ellipse ) is suciently small. Indeed, from the denition of the function w(z; z2) given in Section 7 N (see (7.19), (7.20)) it follows that w(z; z2) > 0 when (z z2 ) > 0. In 0 0 addition, w (z; z2) =6 0 for z ∈ 2 and w (z; z2) > 0 for real z ∈ 2. This insures that there exist b>0 and ε>0, such that the assumption (C.70)on ∈ u w(z; z2) holds when z C2 , for all suciently large N.

238 PAVEL BLEHER AND ALEXANDER ITS

By (7.19), (C.8) Z z q 2 3/2 2 1 (1) w (z; z2)= (u)+N U (u) du 3 N z2 ( Z Z ) q 1 = + 2(u)+N 1U (1)(u) du, 2 0 (z) where 0 0 (1) 0 0 0 (a12(z)) U (z)=(a11(z)) a11(z) 0 , a12(z)

z1+z2 and 0 is a contour which starts at the point z0 = 2 on the lower side of N the cut [z1,z2], goes around the point z2 and ends at the point z0 on the upper side of the cut [z1,z2]. The contour (z) is a simple curve which goes from the point z0 to the point z and does not intersect any of the cuts. On 0 ∪ (z), q (1) U (u) (C.9) 2(u)+N 1U (1)(u)=(u)+ + O N 2 ,u∈ ∪ (z); 2N(u) 0 hence (C.10) ( Z Z ) 2 3/2 1 w (z; z2)= + (u) du 3 2 0 (z) ( Z Z ) (1) 1 1 U (u) du + + + O N 2 N 2 (z) 2(u) Z 0 Z z z (1) U (u) du = (u) du + N 1 + O N 2 2(u) z2 z2 = (z)+N 1d(z)+O N 2 where Z h p p i z n t + gz2 (z)= (u) du = x x2 1 ln(x + x2 1) ,x= √ , z2 2N 2 g (cf. (7.6), (7.70)) and

(C.11) Z Z z U (1)(u) du 1 z (a0 (u))0a0 (u) a0 (u)(a0 (u))0 d(z)= p = 11 p12 11 12 du 0 0 0 z 2 U(u) 2 z a (u) (a (u))2 (a (u))2 2  "2 12 11#  12 1/2 z  0 0 2  1 a11(u) a11(u) 1 = ln  0 + 0 1  = ln a(z) , 2 a12(u) a12(u) 2 z2

SEMICLASSICAL ASYMPTOTICS 239 where p 0 0 2 0 2 0 a11(z)+ (a11(z)) (a12(z)) a11(z)+(z) a(z)= 0 = 0 . a12(z) a12(z) Note also that √ 1 1/4 x + x2 1+c (C.12) a(z)= √ , z g c where q 1/2 g 0 n 2 t c R = x0 (1) x 1,x0 = √ . n 0 2 g Substituting (C.10) into (C.7), we obtain the following asymptotic relation: (C.13) 1 w1/4(z; z ) w1/4(z; z ) (w(z; z ); z )= √ 2 2 u 2 2 2 w1/4(z; z ) w1/4(z; z ) 2 2 I + O N 1 e (N(z)+d(z))3 P, 0 u (assuming again (C.7 )). Let us consider the function TP (z; z2) for z in a small u neighborhood of the arc C2 . In this case, as we discussed above, the condition (C.70) holds (for all suciently large N); hence we can use (C.13). We want ∈ u u to check that for z C2 the turning point solution TP (z; z2) coincides, up to terms of the order of N 1, with the WKB solution (7.9), with some constants C0,C1. Dierentiation of (C.10) gives that 0 1/2 0 1 1 (C.14) w (z; z2)w (z; z2)= (z)+O(N )=(z)+O(N ). Hence we obtain from (7.21) that

(C.15) ! 0 1/2 10 a12(z) W (z; z )= a0 (z) (z) 2 (z) 11 a0 (z) a0 (z) 12 12 1/4 w (z; z2)0 1 1/4 (I + O(N )), 0 w (z; z2) and by (C.13), ! 0 1/2 10 1 a12(z) 11 W (z; z ) (w(z; z ); z )= √ a0 (z) (z) 2 u 2 2 11 11 2 (z) 0 0 a12(z) a12(z) (I + O(N 1))e (N(z)+d(z))3 P 0 1/2 1 a12(z) 11 = √ 2 (z) a 1(z) a(z) (I + O(N 1)) e (N(z)+d(z))3 P.

240 PAVEL BLEHER AND ALEXANDER ITS

To adjust this formula to the WKB solution (7.9), we rewrite it as 0 1/2 1/2 1/2 1 (z) a11(z) a (z) a (z) W (z; z2)u(w(z; z2); z2)=√ 2 2(z) a 3/2(z) a1/2(z) (I + O(N 1))e (N(z)+d(z))3 P. In virtue of (C.11), this gives that (C.16) W (z; z2)u(w(z; z2); z2) 0 1/2 1 1 (z) a11(z) 1 a (z) 1 = √ (I + O(N )) 2 2(z) a 1(z)1

N(z)3 1 1 N(z)3 e P = √ T0(z)(I + O(N )) e P 2 (cf. (7.9)). Thus, by (7.26),

u C 1 N(z)3(1/2) (ln 2) 3 (C.17) (z; z2)=√ T0(z)(I + O(N )) e , TP 2 u uniformly for z in a small neighborhood of the arc C2 . Therefore, if we take in (7.9) C (C.18) C0 = √ ,C1 = (1/2) (ln 2) , 2 then u 1 1 ∈ u (C.19) TP (z; z2)[WKB(z)] = I + O(N ) ,zC2 . Because of symmetry relations (C.1), we obtain simultaneously that d 1 1 ∈ d (C.20) TP (z; z2)[WKB(z)] = I + O(N ) ,zC2 . u Consider now the jump across the arc C1 . By (7.23),

(C.21) y2(z)=y1(z); hence by (C.5), (C.22) ! 3/2 1/6 2/3 1 2Nz 1 N y2(N z)= √ exp (1 + O(N )), 2 z1/4 3 ! 1/4 3/2 0 z 2Nz N 1/6y (N 2/3z)= √ exp (1 + O(N 1)), 2 2 3 2 |z| >ε, + ε

SEMICLASSICAL ASYMPTOTICS 241

Combining this formula with (C.5) and recalling the denition of the function u(z; z1) in (7.27) we have

(C.23) ! 1/4 1/4 3/2 1 z z 2Nz (z; z )= √ I + O N 1 exp 3 P, u 1 2 z1/4 z1/4 3 |z| >ε, + εε, + ε 0 when (z z1 ) < 0. In addition, w (z; z1) =0 0 for z ∈ 1 and w (z; z1) < 0 for real z ∈ 1. This secures that there exist 0 ∈ u b>0 and ε>0 such that the assumption (C.24 ) holds when z C1 , for all suciently large N. By (7.270), Z z q 2 3/2 2 1 (1) (C.25) w (z; z1)= (u)+N U (u) du 3 zN Z 1 z q = 2(u)+N 1U (1)(u) du zN 2 I q 1 + 2(u)+N 1U (1)(u) du, 2 1 N N where 1 is a loop which goes clockwise around the interval [z1 ,z2 ]. On the loop 1, the integrand of the last integral can be estimated as in (C.9). Hence

(C.26) I q Z Z z z (1) 1 2 2 U (u) du 2(u)+N 1U (1)(u) du = (u) du + N 1 2 z z 2(u) 1 1 1 + O N 2 1 2 = (z1) N d(z1)+O N ,

242 PAVEL BLEHER AND ALEXANDER ITS where the functions (z) and d(z) are the same as in (C.10), (C.12) and their values at z1 are taken from the upper half-plane. Using the notation introduced in Appendix B, we can rewrite the functions (z) and d(z) in the form h p i n 2 1 3(z) 1 z1 (z)= x x 1 ln 2(z) and d(z)= ln ln . 2N 2 |3(z1)| 2 z This yields (cf. Fig. 3) that in i N(z )= , and d(z )= n , 1 2 1 2 where n =1ifn is even, and n =0ifn is odd. The last equations together with (C.25) and (C.26) lead to the following representation of the function u w(z; z1) in the neighborhood of the arc C1 (cf. (C.10)), h i 2 i i n (C.27) w3/2(z; z )=(z)+N 1d(z)+ + + O(N 2). 3 1 2N N 2 Substituting (C.27) into (C.24) and assuming (C.240), we arrive at the following asymptotic relation (cf. (C.13)): (C.28) k 1/4 1/4 ( √1) w (z; z1) w (z; z1) 1 u(w(z; z1); z1)= 1/4 1/4 I + O N 2 w (z; z1) w (z; z1) h i i (N(z)+d(z)) n e 2 3 e 3 P, k = . 2 u As discussed above, for z in a small neighborhood of the arc C1 and suf- ciently large N, condition (C.240) holds; hence we can use the asymptotic u u formula (C.28) in equation (7.27) for TP (z; z1). To compare TP (z; z1) with u WKB(z), we also need (as in the previous case of the arc C2 ) the asymptotics of the gauge factor W (z; z1) in (7.27). Similar to the previous case (cf. (C.14)), dierentiation of (C.27) gives that 0 1/2 0 1 1 w (z; z1)w (z; z1)= (z)+O(N )=(z)+O(N ), which in turn implies the relation, 0 1/2 1/4 1/2 1 (C.29) (w (z; z1)) w (z; z1)= (z)+O(N ). (The appearance of the sign “” on the left hand is due to the condition 0 1/2 N 000 000 i(w (z; z1)) > 0, 0

(C.30) ! 0 1/2 10 a12(z) W (z; z )=i a0 (z) (z) 1 (z) 11 a0 (z) a0 (z) 12 12 1/4 w (z; z1)0 1 1/4 (I + O(N )), 0 w (z; z1)

SEMICLASSICAL ASYMPTOTICS 243 and by (C.28),

W (z; z1)u(w(z; z1); z1) ! 0 1/2 10 i k a12(z) 1 1 = √ (1) a0 (z) (z) 11 1 1 2 (z) 0 0 a12(z) a12(z) 1 i (N(z)+d(z)) (I + O(N ))e 2 3 e 3 P.

i 3 Since e 2 = i3, we obtain ! k 0 1/2 10 ( 1) a12(z) 1 1 W (z; z ) (w(z; z ); z )= √ a0 (z) (z) 1 u 1 1 2 (z) 11 1 1 a0 (z) a0 (z) 12 12 10 (I + O(N 1))e (N(z)+d(z))3 P 0 1 ! k 0 1/2 10 ( 1) a12(z) 11 = √ a0 (z) (z) 11 11 2 (z) 0 0 a12(z) a12(z) (I + O(N 1))e (N(z)+d(z))3 P k 0 1/2 ( 1) a12(z) 11 = √ 2 (z) a 1(z) a(z) (I + O(N 1)) e (N(z)+d(z))3 P.

u The rest of the derivation is exactly the same as in the case of the arc C2 , and it yields the following formula (cf. (C.16)): k 0 1/2 1 ( 1) (z) a11(z) 1 a (z) W (z; z1)u(w(z; z1); z1)= √ 2 2(z) a 1(z)1 (I + O(N 1))e N(z)3 P k ( 1) 1 N(z)3 = √ T0(z)(I + O(N )) e P. 2 Thus, by (7.27), (1) k u C ( 1) 1 N(z)3(1/2) (ln 2) 3 (C.31) (z; z1)= √ T0(z)(I + O(N )) e , TP 2 u uniformly for z in a small neighborhood of the arc C1 (cf. (C.17)). u u To ensure the same WKB-asymptotics both for TP (z; z2) and TP (z; z1) we take C(1) =(1)kC. Then u,d 1 1 ∈ u,d (C.32) TP (z; z1)[WKB(z)] = I + O(N ),zC1,2 .

244 PAVEL BLEHER AND ALEXANDER ITS

u,d Our last step is to prove that the turning point solutions TP (z; z1) and u,d 1 ∩ TP (z; z2) coincide, up to the terms of the order of N ,onl0 = ∂1 ∂2 (see u u Fig. 2). By (C.19), (C.20) and (C.32), TP (z; z1) and TP (z; z2) coincide, up 1 to the terms of the order of N , at the end point of l0, z = z0 +ib. Since b>0 can be chosen as small as we want, this is true for all z = z0 + ic,00 be an arbitrary xed number such that <(z2 z1)/2 and let

(C.33) 0 =∩{z1 + Re z z2 }.

Then l0 0. We will prove that

u u 1 1 ∈ (C.34) TP (z; z1)[TP (z; z2)] = I + O(N ),z0, and h i 1 d d 1 ∈ (C.35) TP (z; z1) TP (z; z2) = I + O(N ),z0.

Observe that w(z; z2) is analytic in z ∈ (∩{Re z z1 +})ifN is suciently 0 6 u,d large, and w (z; z2) = 0 in this region. Hence the TP (z; z2) are analytic u,d ∩{ there as well. Similarly, the TP (z; z1) are analytic in the region Re z } u,d z2 . Thus, all the functions TP (z; z1,2) are analytic in 0. Because of the symmetry equation (C.1), it is enough to prove (C.34) only. The idea of the u proof of (C.34) is to nd a WKB function WKB(z) such that for k =1, 2,

u u 1 1 ∈ (C.36) TP (z; zk)[WKB(z)] = I + O(N ),z0.

u To construct the function WKB(z) let us consider an auxiliary turning point solution

0 (C.37) TP (z; z2)=CW(z; z2)0(w(z; z2)), where (C.38) 1/6 2/3 2/3 1/2 N 0 y1(N z) y2(N z) (2) 0 0(z)= 1/6 0 2/3 0 2/3 1/2 . 0 N y1(N z) y2(N z) 0(2)

The key point here is that in contrast to u,d(z; z2) in (7.22), the matrix elements of 0(z) are dened in terms of y1(z),y2(z) only (and not y0(z); cf. 0 →∞ (7.23)), and this allows us to nd the asymptotics of TP (z; z2)asN for all z ∈ 0. The function y2(z) is an entire function. Changing the branch

SEMICLASSICAL ASYMPTOTICS 245 of the root functions on the right-hand side of (C.22) we obtain

(C.39) ! 3/2 1/6 2/3 i 2Nz 1 N y2(N z)= √ exp (1 + O(N )), 2 z1/4 3 ! 1/4 3/2 0 z 2Nz N 1/6y (N 2/3z)= √ exp (1 + O(N 1)), 2 2 i 3 4 |z| >ε, + ε

Combining this formula with (C.5) we obtain ! 1/4 1/4 3/2 1 z iz 2Nz (C.40) (z)= √ (1 + O(N 1)) exp 3 P, 0 2 z1/4 iz1/4 3 5 |z| >ε, + ε

Let 0(z), 0(z), and d0(z) be the analytic continuation of the func- u tions (z), (z), and d(z), respectively, from 0 to 0, across the interval [z1 + , z2 ] (see equations (7.4), (7.6), and (C.11), (C.12)). Then equations (C.10), (C.14), and (C.15), with (z),(z), and d(z) replaced by 0(z),0(z), and d0(z), hold for all z ∈ 0. In addition, there exist b>0 (the vertical half-axis of ) and ε>0 such that for all z in 0 and large N the inequalities

5 (C.41) |w(z; z )| >ε, + ε

(C.42) ! 0 1/2 10 1 a12(z) 1 i W (z; z ) (w(z; z )) = √ a0 (z) (z) 2 0 2 11 0 1 i 2 0(z) 0 0 a12(z) a12(z)

1 (N0(z)+d0(z))3 (I + O(N )) e P, z ∈ 0,

(cf. (C.16)). Since iy1(z) iy2(z)=y0(z), we obtain that 1/2 i 1 (2) 0 (C.43) (z; z )= (z)P 1 P, P = ; u 2 0 i 0 0(2)1/2

246 PAVEL BLEHER AND ALEXANDER ITS hence

(C.44) u TP (z; z2)=CW(z; z2)u(w(z; z2); z2) ! 0 1/2 10 C a12(z) 1 i = √ a0 (z) (z) 11 0 2 0(z) 1 i a0 (z) a0 (z) 12 12 i 1 (I + O(N 1)) e(N0(z)+d0(z))3 P, z ∈ . i 0 0

Thus, if we dene

(C.45) ! C a0 (z) 1/2 10 u √ 12 0 WKB(z)= a11(z) 0(z) 2 0(z) a0 (z) a0 (z) 12 12 1 i i 1 e(N0(z)+d0(z))3 P, 1 i i 0 then

u 1 u (C.46) TP (z; z2)= I + O(N ) WKB(z).

Consider now in 0 the turning point solution TP (z; z1). For z in 0 the same equations (C.27), (C.29), and (C.30), with the functions (z), (z), d(z) replaced by the functions 0(z), 0(z), d0(z), hold. Also, there exist b>0 and ε>0 such that for all z ∈ 0 and suciently large N, w(z; z1) lies in the sector

5 (C.47) + εε. This implies that

! 0 1/2 10 i k a12(z) i 1 W (z; z ) (w(z; z )) = √ (1) a0 (z) (z) 1 0 1 11 0 i 1 2 0(z) 0 0 a12(z) a12(z) 1 i (N (z)+d (z)) (I + O(N ))e 2 3 e 0 0 3 P.

SEMICLASSICAL ASYMPTOTICS 247

i 3 Since e 2 = i3, we obtain that (C.48) W (z; z1)0(w(z; z1)) ! (1)k a0 (z) 1/2 10i 1 √ 12 0 = a11(z) 0(z) 2 0(z) i 1 a0 (z) a0 (z) 12 12 10 (I + O(N 1))e (N0(z)+d0(z))3 P 0 1 ! k 0 1/2 10 ( 1) a12(z) i 1 = √ a0 (z) (z) 11 0 i 1 2 0(z) 0 0 a12(z) a12(z) (I + O(N 1))e (N0(z)+d0(z))3 P ! k 0 1/2 10 ( 1) a12(z) 1 i = √ a0 (z) (z) 11 0 2 0(z) 1 i a0 (z) a0 (z) 12 12 0 1 (I + O(N 1))e(N0(z)+d0(z))3 P, z ∈ . 10 0 Taking into account that (cf. (C.43)) i 0 (z; z )= (z)P 1 P, u 1 0 i 1 we derive from (C.48) the asymptotic equation u k TP (z; z1)=( 1) CW(z; z1)0(w(z; z1)) ! 0 1/2 10 C a12(z) 1 i = √ a0 (z) (z) 11 0 2 0(z) 1 i a0 (z) a0 (z) 12 12 0 1 i 0 (I + O(N 1))e(N0(z)+d0(z))3 P 10 i 1 ! 0 1/2 10 C a12(z) 1 i = √ a0 (z) (z) 11 0 2 0(z) 1 i a0 (z) a0 (z) 12 12 i 1 (I + O(N 1))e(N0(z)+d0(z))3 P, z ∈ . i 0 0 In other words, u 1 u TP (z; z1)= I + O(N ) WKB(z); hence u u 1 1 ∈ TP (z; z1)[TP (z; z2)] = I + O(N ),z0. Lemma 7.2 is proved.

248 PAVEL BLEHER AND ALEXANDER ITS

Proof of Lemma 7.4. This lemma follows directly from the formulae (C.45), (C.46) (and symmetry (C.1)) if we substitute into these formulae the value of C given in (7.34). Proof of Lemma 7.6. By (C.8), (C.10), (C.49) ( Z Z ) q 1 1 2 1 (1) 2 0(z)+N d0(z)= + (u)+N U (u) du + O N , 2 0 (z) where z2(gz2 + t)2 ngz2 2(z)= , 4 N 0 0 2 (1) 0 0 0 (a12(z)) gz t 0 U (z)=(a11(z)) a11(z) 0 = + + gRn, a12(z) 2 2 so that (C.50) " # 2 2 1 (gz + t) n + g t 2(z)+N 1U (1)(z)=z2 2 + N 1 + gR0 4 N 2 n p 2 2 n 2 (gz + t) 0 (1) t 4g = z2 g N 1 4 2 2 1 = 1(z) N n, where p 2 n 2 2 0 2 2 gz + t ( 1) t 4g (z)= gz (q 1),q= √ ,n = . 1 2 0g 2 Hence, ( Z Z ) q 1 1 2 1 2 0(z)+N d0(z)= + 1(u) N n du + O N 2 0 (z) ( Z Z ) 1 1 n 2 = + 1(u) N du + O N 2 0 (z) 21(u) Z z 1 n 2 = 1(u) N du + O N , N 21(u) z20

N where z20 is a close-to-z2-zero of 1(z). Now, Z Z Z z p z p p z p 0 2 0 2 1(u) du = g q 1 udu = i g 1 q udu. N N N z20 z20 z20

SEMICLASSICAL ASYMPTOTICS 249

2 The substitution v = gu√ +t reduces this integral to 2 0g Z Z z q p 0 p 0 2 i 2 1(u) du = i 1 v dv = q 1 q arccos q zN 1 2 20 i0 sin 2 = ,= arccos q. 2 2 Similar computation gives Z Z z z du p 2du = i 0 N 1(u) N 2 2 z20 z20 u 4 g (gu + t) pi arccos r n i = 0 =( 1) , t2 4 g 2n √ 2 0g tq r = √ ,= arccos r. 2 0gq t Thus, 0 n N sin 2 (1) i 1 [N (z)+d (z)] = 0 0 2 2 4 and Lemma 7.6 is proved.

Appendix D. Proof of estimate (7.45)

Denote =L ∪ ∂1 ∪ ∂2 ∪ (∂1) ∪ (∂2). Theorem 7.3 implies that the function X(z) solves the Riemann-Hilbert prob- lem on the contour , (D.1) X(∞) lim X(z)=I, z→∞ (D.2) X+(z)=X(z)G(z),z∈ , with the jump matrix G(z) given by the equations, (D.3) 0 0 1 G(z)=(z) (z) if z ∈ ∂1 ∪ ∂2 ∪ (∂1) ∪ (∂2), + 1 G(z)=0(z)S 1 0(z) if z ∈ L, and satisfying the uniform estimates, (D.4) 1 1 |I G (z)|CN , if z ∈ ∂1 ∪ ∂2 ∪ (∂1) ∪ (∂2), |I G 1(z)|Ce 2NRe (z), if z ∈ L.

250 PAVEL BLEHER AND ALEXANDER ITS

The Riemann-Hilbert problem is depicted in Figure 4.

1 I + O( - ) N

ξ I + O( e -2NRe (z) )

Figure 4. The Riemann-Hilbert problem for the function X(z) and the contour

Similar to the asymptotic problems considered in [DZ2] and [DIZ], the crucial fact, which follows from (D.4) and which is indicated in Figure 4, is that (D.5) |I G 1(z)| = O(N 1), uniformly for z ∈ (compare to the similar analysis in [DZ2] and [DIZ]). Simultaneously, inequalities (D.4) yield the corresponding L2-norm estimate:

|| 1|| 1 (D.6) I G L2() = O(N ).

Remark D.1. The function 0(z) is dened by explicit formulae (7.9) and 0 (7.28) which, in particular, imply that the functions (z) and, in virtue of (D.3), the jump matrix G(z) can be analytically continued in a small neigh- borhood of ∂1 ∪ ∂2 ∪ (∂1) ∪ (∂2). Moreover, in this neighborhood the rst inequality in (D.4) holds. Indeed, the inequality follows from Lemma 7.2, whose proof is based on the estimates (C.17), (C.31), and (C.34), (C.35) taking place in the small neighborhoods of the corresponding pieces of the contour ∂1 ∪ ∂2. When z ∈ L, 2N(z) 0 ie G(z)=I + T (z) T 1(z), 0 00 0

(cf. (7.42), (7.43)) and hence G(z) is analytic and satises the second inequal- ity in (D.4) in a small neighborhood of L, which may include the narrow

SEMICLASSICAL ASYMPTOTICS 251 sectors (the angle less than /8) about the intervals (∞, z2] and [z2, ∞). The indicated properties of the matrix G(z) mean that in the setting of the Riemann-Hilbert problem (D.1), (D.2) the choice of contour is not rigid; one can always slightly deform the contour and slightly rotate its innite parts. By standard techniques in Riemann-Hilbert theory (see e.g. [CG]; see also [BDT] and [DZ]) the solution X(z) of the Riemann-Hilbert problem (D.1), (D.2) is given by the integral representation, Z 1 1 d 6∈ (D.7) X(z)=I + () I G () ,z, 2i z where (z) X+(z) solves the equation

1 (D.8) (z)=I + C+[ I G ](z),z∈ , and C+ is the corresponding Cauchy operator: Z h() d (C+h)(z) = lim . 0 0 0 z →z, z ∈(+)side z 2i

We note that by (7.38), X(z) is built up from the functions which satisfy asymptotic conditions (5.16) and (7.35). Therefore, I (z)=O z1 as z →∞and hence a priori it belongs to the space L2(). Put

0(z)=(z) I.

Then equation (D.8) can be rewritten as an equation in L2(), 0 (D.8 )(1 B)0 = f, where the function f(z) and the operator B are dened by the equations, f = C (I G 1), + 1 B : 7→ C+[ I G ],∈ L2().

The L2-boundness of the operator C+ (see e.g. [LS]; see also [BDT] and [Zh]) and the estimates (D.5) and (D.6) imply that || || || || || 1|| 1 B L2()→L2() C+ L2()→L2() I G L∞() = O(N ), and || || || || || 1|| 1 f L2() C+ L2()→L2() I G L2() = O(N ).

252 PAVEL BLEHER AND ALEXANDER ITS

0 Therefore, equation (D.8 ) is uniquely solvable in L2() for suciently large N, and its solution satises the estimate,

|| || 1 0 L2() = O(N ), or, in terms of (z), || || 1 I L2() = O(N ).

This equation together with (D.5) and (D.7) show that 1 (D.9) X(z)=I + O ,z∈ K, N(1 + |z|) for all closed subsets K C \ satisfying the following property:

dist{z;} c (K) > 0, for all z ∈ K. 1+|z| 0

To complete the proof of estimate (7.45) we only need to notice that the domain of validity of the asymptotic equation (D.9) can be made the whole z-plane because of the indicated exibility in Remark D.1 in the choice of the contour .

Remark D.2. The kernel in integral equation (D.8) is small (estimate (D.5) again), and it is given explicitly in terms of elementary and Airy functions. Therefore, equations (D.7)–(D.9) make it possible (cf. [DZ3] and [DIZ]) to obtain the full asymptotic expansions for all the three main objects, n, Rn, and Pn(z).

Appendix E. The Riemann-Hilbert approach to orthogonal polynomials

In this appendix , for the readers convenience, we reproduce with some more details the Riemann-Hilbert approach to the orthogonal polynomials sug- gested in [FIK]. Let s1,s3 be complex numbers, |s1| + |s3|6= 0, and let L denote the cross, [ L = R iR .

We assume that the cross L is oriented in a natural way, i.e. from ∞ to +∞, and from i∞ to +i∞ (see Fig. 5).

SEMICLASSICAL ASYMPTOTICS 253

i R

1 s 3 01

1 s 1 1 s 1 01 01 R

1 s 3 0 1

σ NV(z) - 3 Figure 5. The Riemann-Hilbert problem for the function Ye(z) 2

Consider the following Riemann-Hilbert problem posed on the cross L for 2 2 matrix function Y (z) Yn(z),n∈ Z+:

(i) The function Y (z) is analytic on C \ L. It has the limits, Y(z), from the sides of L which satisfy the jump equation, (E.1) NV (z) 1 se NV (z) NV (z) Y (z)=Y(z) Y(z)e 2 3 Se 2 3 ,z∈ L, + 01 where s = s1 if z ∈ R ,s= s3 if z ∈ iR , and 1 s S = . 01

(ii) The function Y (z) is normalized by the asymptotic condition, (E.2) lim Y (z)z n3 = I. z→∞ The same arguments as in the case of the Riemann-Hilbert problem (5.16)– (5.18) (see Section 5) show that the solution Y (z) of problem (E.1), (E.2) is unique (if it exists). In fact, we rst notice that since det S = 1, the scalar function det Y (z) is a bounded analytic function on C and such that its limit as z →∞is 1. Hence, (E.3) det Y (z) 1. Suppose now that Y˜ (z) is another function satisfying (i), (ii). Since the jump matrix S is the same for both Y (z) and Y˜ (z), the matrix ratio, Y˜ (z)Y 1(z),

254 PAVEL BLEHER AND ALEXANDER ITS is a bounded analytic matrix function on C whose limit at z = ∞ equals I . Therefore, Y˜ (z)Y 1(z)=I or Y˜ (z) Y (z).

Consider also a system {Pn(z)} of orthogonal polynomials on the cross L with the weight i seNV (z), Z 2 NV(z) n (E.4) Pn(z)Pm(z)e dz = hnnm,Pn(z)=z + ..., L where Z Z Z i +∞ i +i∞ s1 + s3 . L 2 ∞ 2 i∞ We shall assume that the nondegeneracy condition, Z k j NV (z) 6 (E.5) det z z e dz =0, for all n =0, 1, 2,..., L j,k=0,...,n is satised. Using the symmetries, z →z and z → z, one can easily show that (E.5) is true if, for example,

(E.6) s1 ∈ iR \{0} and s3 ∈ iR.

Condition (E.5) guarantees that the orthogonal set {Pn(z)} exists. Each of the polynomials Pn(z) is uniquely determined from the system of linear equations, Z j NV(z) (E.7) z Pn(z)e dz =0,j=0, 1,...n 1. L Moreover, it also follows from (E.5) that (E.8) Z Z 2 NV (z) n NV (z) 6 ∀ hn Pn (z)e dz z Pn(z)e dz =0, n =0, 1, 2,... . L L

Indeed, if for some n condition (E.8) is violated, the coecients of Pn(z) would provide a nontrivial element of the kernel of the matrix, Z k j NV(z) z z e dz . L j,k=0,...,n Remark E.1. A truncated condition, (E.9) Z k j NV(z) 6 ∀ ∈ N det z z e dz =0, n =0, 1, 2,...,m 1,m , L j,k=0,...,n { }m yields the existence of the truncated orthogonal system Pn(z) n=0, and the inequalities, hn =06 ,n=0,...,m 1.

SEMICLASSICAL ASYMPTOTICS 255

The orthogonal polynomials considered in the main text correspond to the choice,

(E.10) s3 =0,s1 = 2i. Similar to special case (E.10), orthogonal polynomials (E.7) satisfy the linear equations (cf. (1.4)–(1.6)),

(E.11) zPn(z)=Pn+1(z)+RnPn1(z), and 0 (E.12) Pn(z)=NRn[t + g(Rn1 + Rn + Rn+1)]Pn1(z) 0 d + g(NR R R )P (z), ( ) , n 2 n 1 n n 3 dz where

hn (E.13) Rn = ,n=1, 2,... . hn1

The coecients Rn satisfy the same Freud equation (1.7); i.e.,

(E.14) n = NRn[t + g(Rn1 + Rn + Rn+1)],n=1, 2,... with the initial conditions, R h z2e NV (z)dz R =0,R= 1 = RL . 0 1 NV (z) h0 L e dz The relation between the Riemann-Hilbert problem (E.1), (E.2) and the orthogonal system {Pn(z)} can be formulated as the following proposition: Proposition E.1 (cf. [FIK2,4]). The Riemann-Hilbert problem (E.1), (E.2) is (uniquely) solvable for all n ∈ N if and only if condition (E.5) is sat- ised, and hence the orthogonal system {Pn(z)} exists. The (unique) solution Y (z) of problem (E.1), (E.2) is given by the exact formula, ! R NV () Pn()e Pn(z) L z d (E.15) Y (z)= R NV () , 1 1 Pn1()e Pn1(z) L d R hn1 hn1 z where the symbol L is as introduced in (E.4). Proof. Suppose that (E.5) takes place, and dene the function Y (z)by the right side of equation (E.15). Then Y (z) solves (E.1), (E.2). Indeed, the jump equation (E.1) follows directly from the Plemelj formula for the boundary values of Cauchy integrals, while the orthogonality conditions, Z j NV (z) z Pn(z)e dz = hnnj,j=0, 1,...,n, L

256 PAVEL BLEHER AND ALEXANDER ITS imply that zn + O(zn1) O(zn1) Y (z) ,z→∞, O(zn1) zn + O(zn1) and hence asymptotic equation (E.2) is valid as well. Conversely, assume that the Riemann-Hilbert problem has a solution Y (z) for all n ∈ N. The triangularity of the jump matrix S allows us to represent this solution in closed form (cf. [FIK4, §3.4]). In fact, asymptotic condition (E.2) implies that zn + o(zn) o(zn) (E.16) Y (z) ,z→∞. o(zn) zn + o(zn)

At the same time, the 11 and 21 components of (E.1) yield (Y+(z))11 = (Y(z))11 and (Y+(z))21 =(Y(z))21 respectively. Using these equations and (E.16) we nd

(E.17) (Y (z))11 = Pn(z), (Y (z))21 = Qn1(z), where Pn(z) and Qn1 are the polynomials of the indicated degrees with the only restriction that n Pn(z)=z + ... .

The 12 and 22 entries of equation (E.1) can be written down as the jump conditions,

NV(z) (Y+(z))12 (Y(z))12 = se (Y(z))11, NV(z) (Y+(z))22 (Y(z))22 = se (Y(z))21, which together with (E.17) provide us with the following representation for the function Y (z): ! R NV () Pn()e Pn(z) L z d (E.18) Y (z)= R NV () . Qn1()e Qn1(z) L z d Referring again to (E.16), we conclude that the following equations, Z j NV(z) (E.19) Pn()e d =0,j=0, 1,...n 1, L and Z j NV(z) (E.20) Qn1()e d = l,n1,j=0, 1,...n 1, L hold for all n ∈ N.

SEMICLASSICAL ASYMPTOTICS 257

Our next step is to obtain the nondegeneracy conditions (E.5). Suppose that (E.21) Z k j NV (z) ∈{ } det z z e dz =0, for some m 0, 1, 2,..., . L j,k=0,...,m

Then there is a nonzero tuple, {c0,...,cm} such that Xm Z k j NV(z) ck z z e dz =0, for all j =0,...,m. k=0 This in turn implies that the polynomials, Xm k P˜m+1(z) Pm+1(z)+ ckz , k=0 and Xm k Q˜m(z) Qm(z)+ ckz , k=0 satisfy (E.19), and (E.20) respectively. The pairs (Pm+1(z),Qm(z)), (P˜m+1(z), Q˜m(z)), (Pm+1(z), Q˜m(z)), and (P˜m+1(z),Qm(z)) generate, via (E.18), the dierent functions which solve the same Riemann-Hilbert problem (E.1), (E.2) for n = m + 1. This contradiction shows that assumption (E.21) is false. Condition (E.5), which has just been established, means that equations (E.19) and (E.20) determine polynomials Pn(z) and Qn1(z) uniquely for all n ∈ N. Simultaneously, the equation, 1 Qn1(z)= Pn1(z), hn1 holds. This equation and equations (E.18), (E.19) complete the proof of the theorem.

Remark E.2. Let m ∈ N. Then the solvability of the Riemann-Hilbert problem (E.1), (E.2) and representation (E.15) for n ∈{1,...,m} only need truncated solvability condition (E.9).

Remark E.3. If n = 0, the Riemann-Hilbert problem (E.1), (E.2) is always solvable, and the solution is given by the equation, R NV () 1 e d (E.22) Y (z)= L z . 01 From exact representation (E.15) it follows that the solution Y (z)ofthe Riemann-Hilbert problem (E.1), (E.2) has, in fact, a complete asymptotic series

258 PAVEL BLEHER AND ALEXANDER ITS at z = ∞, ! X∞ (E.23) Y (z) I + k zn3 ,z→∞. zk k=1 Observe also that if Y (z) is a solution of problem (E.1), (E.2), then so is the function, n (1) 3Y (z)3. Because of the uniqueness of the solution of the Riemann-Hilbert problem, the symmetry equation n (E.24) Y (z)=(1) 3Y (z)3 holds. Equation (E.24) in turn implies that in asymptotic series (E.23) all the even coecients, 2l, are diagonal while all the odd coecients, 2l+1, are o- diagonal. Moreover, again from (E.15), we conclude that

1 (E.25) (1)12 = hn, (1)21 = , hn1 and hence

(E.26) (1)12(1)21 = Rn, where the hn and Rn are dened as in (E.8) and (E.13) respectively. Dene NV (z) n3 3 (E.27) n(z)=e 0Y (z)e 2 , where

10 2n 0 = 1/2 , and e = hn. 0 Rn

The function n(z) is analytic on C \ L. It has the limits n(z) from the sides of L which satisfy the jump equation (cf. (E.1)), 1 s (E.28) (z)= (z)S, S = . n+ n 01

In virtue of (E.23), (E.25), and (E.26), the function n(z) has the asymptotic expansion, ! X∞ k NV (z) n ln z+ (E.29) (z) e 2 n 3 ,z→∞, n zk k=0 where 10 01 (E.30) 0 = 1/2 , 1 = 1/2 . 0 Rn Rn 0

SEMICLASSICAL ASYMPTOTICS 259

Also, equation (E.15) yields the following representation for the function (z), n n(z) ϕn(z) (E.31) n(z)= , n1(z) ϕn1(z) where 1 NV (z) n(z)=√ Pn(z)e 2 , hn and Z NV () e 2 n() ϕn(z)= d. L z Equation (E.31) provides an orthogonal polynomial representation for the solution of the Riemann-Hilbert problem (E.28)–(E.30). In the particular case (E.10), this problem is exactly our basic Riemann-Hilbert problem (5.16)– (5.18), and formula (E.31) coincides with formula (5.15). Equation (5.15) for the solution of the Riemann-Hilbert problem (5.16)–(5.18) was derived in Section 5 via the analysis of the monodromy properties of the solutions of the Lax pair (3.6). Matrix equations (3.6) are equivalent to the pair of the scalar equations (E.11), (E.12). In the approach presented in this appendix we do not use these equations at all. It is the Riemann-Hilbert problem (E.1), (E.2), or equivalently (E.28)–(E.30), that we make now a starting point of the analysis of the orthogonal polynomial system (E.7). The Lax pair, and hence the Freud equation, for orthogonal system (E.7) can be obtained directly from the Riemann-Hilbert problem (E.28)–(E.30) by use of the general inverse monodromy problem technique of [JMU] (see also [FIK4]). To this end, put n3 1 W (z)3 n3 (E.32) Z(z)=e 0 n(z)e Y (z)z , where we denote for the sake of brevity, NV(z) W (z)= n ln z. 2 Then

n3 W (z)3 (E.33) n(z)=e 0Z(z)e , and by (E.23), X∞ (E.34) Z(z) I + k ,z→∞. zk k=1 The next lemma shows that the Riemann-Hilbert problem implies a pair of dierential and dierence equations on n(z). We shall use the following usual notation: If Xm k B(z) bkz ,z→∞, k=∞

260 PAVEL BLEHER AND ALEXANDER ITS we denote by Xm k B(z) = bkz , + k=0 the polynomial part of B(z) at innity.

Lemma E.1. Assume that n(z) is a solution of the Riemann-Hilbert problem (E.28)–(E.30). Then n(z) satises the polynomial 22 matrix dier- ential-dierence Lax pair (cf. (3.6)), ( n+1(z)=Un(z)n(z), 0 d (E.35) 0 ( ) , n(z)=NAn(z)n(z), dz with 0 n3 1 1 n3 (E.36) An(z)= (1/2)e 0 Z(z)V (z)3Z (z) 0 e , + and (E.37)

n+13 10 3 1 10 n3 Un(z)=e 1/2 Zn+1(z)z Zn (z) 1/2 e , 0 Rn+1 + 0 Rn where Z(z) Zn(z) is dened as in (E.32).

Proof. Observe that det n(z) is an entire function, since

det n+(z) = det n(z) det S = det n(z). In addition, 1/2 1/2 1/2 lim det n(z)=R lim det Z(z)=R det I = R . z→∞ n z→∞ n n Hence 1/2 6 det (z) Rn =0. Let us check that (z) satises a matrix dierential equation. Dene 0 1 Q(z)=n(z)n (z). Then by (E.28) and the fact that the jump matrix S does not depend on z, 0 1 0 1 1 Q+(z)=n+(z)n+(z)=n(z)SS n(z)=Q(z), so that Q(z) is an entire matrix-valued function. By (E.33), h i n3 0 W (z)3 0 W (z)3 Q(z)=e 0 Z (z)e Z(z)W (z)3e

W (z)3 1 1 n3 e Z (z)0 e 0 0 n3 1 1 n3 = e 0 Z (z) Z(z)W (z)3 Z (z)0 e ;

SEMICLASSICAL ASYMPTOTICS 261 hence Q(z) grows polynomially at innity, and hence Q(z) is a polynomial, 0 0 n3 1 1 n3 Q(z)=e 0 Z (z) Z(z)W (z)3 Z (z) 0 e + 0 n3 1 1 n3 = (N/2)e 0 Z(z)V (z)3Z (z) 0 e . +

Thus we get a polynomial dierential equation on n(z), 0 n(z)=Q(z)n(z), with 0 n3 1 1 n3 Q(z)= (N/2)e 0 Y (z)V (z)3Y (z) 0 e . + This proves equation (E.36). Equation (E.37) is proved similarly when we consider the “discrete” logarithmic derivative, ˜ 1 Q(z)=n+1(z)n (z), and take this time into account the fact that the jump matrix S does not depend on n. The proof is complete.

A straightforward calculation based on asymptotic series (E.34) and sym- metry equation (cf. (E.24)),

Z(z)=3Z(z)3, leads to representations (3.7) and (3.8) for the matrices Un(z) and An(z) re- spectively. Also, substituting (E.33), (E.34) into the second equation in (E.35) 1 and equating the terms of order z , one obtains, for Rn, Freud equation (E.14). The latter simultaneously is a compatibility condition of the Lax pair (E.35) (see §5). One can repeat the arguments used in Section 5 for the particular case (E.10) and show that the Riemann-Hilbert problem (E.29)–(E.31) represents the Stokes phenomenon for the matrix dierential equation 0 (E.38) n(z)=NAn(z)n(z), tz gz3 1/2 2 ( 2 + 2 + gzRn) Rn [t + gz + g(Rn + Rn+1)] A (z)= 3 , n 1/2 2 tz gz Rn [t + gz + g(Rn1 + Rn)] 2 + 2 + gzRn with Rn dened as the solution of initial problem (E.14) for the Freud equation. For the analysis of the general monodromy problem for equation (E.38) corresponding to an arbitrary solution of the Freud equation and for its relation to the fourth Painleve equation we refer the reader to Section 3 of [FIK2]. The monodromy problem corresponding to the polynomial V (z) of an arbitrary even degree is considered in [FIK4].

262 PAVEL BLEHER AND ALEXANDER ITS

Remark E.4. The technique used in this appendix was rst suggested in [FMA] for analyzing the explicit solutions of the Painleve equations. In [FIK4] it was applied to the case of the arbitrary even polynomial V (z). In fact, using the same idea one can easily reduce the analysis of an arbitrary system of the orthogonal polynomials {Pn(z)} on some contour L with some weight ω(z) to the analysis of relevant 2 2 matrix Riemann-Hilbert problem. The RH problem is formulated for a 2 2 matrix function Y (z) which is analytic outside the contour L, normalized by the asymptotic condition 10 Y (z)z n3 → I, z →∞,= , 3 0 1 and whose boundary values Y(z) satisfy equation: 1 2iω(z) Y (z)=Y(z) ,z∈ L. + 01 This Riemann-Hilbert problem can also be used to explain the appearance (see, e.g., [ASM]) of the KP-type hierarchies in the matrix models. Indeed, let us assume that the weight function ω() is of the form N t zk ω(z)=e k=1 k and put 1 (N tkzk) (z)=Y (z)e 2 k=1 3 . Then, the same arguments as the ones we used for proving Lemma E.1 yield (cf. (1.12)–(1.16) in [FIK4]) the system of linear dierential and dierence equations for function (z) (z; n, t1,t2,t3, ...), (z; n +1)=U(z)(z; n)

∂z(z)=A(z)(z)

∂tk (z)=Vk(z)(z),k=1, 2, 3,... , where U(z), A(z), and Vk(z) are polynomial on z (for their exact expressions in terms of the corresponding Rn see [FIK4]). The rst two equations constitute the Lax pair for the relevant Freud equation. The compatibility conditions of the third equation with the dierent k generate the KP-type hierarchy of the integrable PDEs; the compatibility condition of the second and the third equations produces the Virasoro-type constraints; the compatibility condition of the rst and the third equations is related to the Toda-type hierarchy and vertex operators. Acknowledgement. This work was supported in part by the National Sci- ence Foundation, Grant No. DMS-9623214 (P.B.) and Grant No. DMS- 9501559 (A.I.), and this support is gratefully acknowledged. The authors thank the referees for valuable remarks.

SEMICLASSICAL ASYMPTOTICS 263

Indiana University-Purdue University Indianapolis, Indianapolis, IN E-mail addresses: [email protected] [email protected]

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(Received February 11, 1997)

(Revised March 25, 1998)