Scalar and matrix Riemann-Hilbert approach to the strong ... - CiteSeerX

Scalar and matrix Riemann-Hilbert approach to the strong asymptotics of Pad´e approximants and complex orthogonal polynomials with varying weight∗ A. I. Aptekarev and W. Van Assche Keldysh Institute of Applied Mathematics Katholieke Universiteit Leuven

Abstract We describe methods for the derivation of strong asymptotics for the denominator polynomials and the remainder of Pad´e approximants for a Markov function with a complex and varying weight. Two approaches, both based on a Riemann-Hilbert problem, are presented. The first method uses a scalar Riemann-Hilbert boundary value problem on a two-sheeted Riemann surface, the second approach uses a matrix Riemann-Hilbert problem. The result for a varying weight is not with the most general conditions possible, but the loss of generality is compensated by an easier and transparent proof.

1

Introduction

Recently, there has been considerable progress in proving strong asymptotics for general orthogonal polynomials using methods of complex analysis and a Riemann-Hilbert boundary value problem (BVP) for matrix analytic functions. This new approach has been used extensively by P. Deift and his collaborators and an important reference is [3]. The matrix Riemann-Hilbert problem for orthogonal polynomials was first formulated by Fokas, Its and Kitaev [8], and analyzed by Bleher and Its [2], Deift et al. [4, 5, 6], and Kriecherbauer and McLaughlin [10]. We recommend the exposition of Kuijlaars [11] for an introduction to the use of the RiemannHilbert approach for the asymptotic analysis of orthogonal polynomials. In the present paper we explain special versions of BVP based approaches. The roots of these versions lie in the research of rational approximants. The first presentation of a version based on BVP on a Riemann surface was in the paper [13] of J. Nuttall, although the ideas were in [14] and even in earlier papers of Nuttall (related references can be found in [14], see also the paper of H. Stahl [16]). A substantial development of the approach based on the BVP on a Riemann surface has been done by S.P. Suetin in a recent paper [17], where Nuttall’s version of the asymptotic analysis has been extended to Pad´e approximants for hyperelliptic functions. Also we would like to add to the list of references an older paper [1] where the main ingredient, i.e., a boundary value Riemann problem on a Riemann surface, has been considered in connection with strong asymptotics of orthogonal polynomials. ∗

This research was partly supported by INTAS Research Network NeCCA 03-51-6637 and FWO research grants G.0184.02 and G.0455.04 and was finished while the first author visited K.U.Leuven with senior fellowship F/99/009 and F/01/061. The first author was also partly supported by grant RFBR 02-01-00564 of the Russian Foundation for Fundamental Research and by the program of Modern Problems of Theoretical Mathematics, RAS (DMS).

1

The present paper is intended to describe the Riemann-Hilbert boundary value problem and its relation to strong asymptotics of Pad´e approximants in a rather general setting. In the Appendix we give an introduction to the Riemann boundary value problem. So a reader not familiar with the Riemann boundary value problem is encouraged to start reading the paper from the end. The aim of Appendix is to provide a ‘shortcut’ in complex analysis to give an idea of how to find a solution of a boundary value problem on a Riemann surface, based just on the notion of Cauchy’s integral formula and its generalization, the Cauchy residue theorem. The Riemann surfaces, which will be used in this paper, have genus zero, which allows us to avoid the non-trivial part of the theory of Riemann surfaces and requires from the reader just a ‘naive’ understanding of a Riemann surface as a two-dimensional manifold in four dimensional space. In Section 2 we solve a special Riemann problem, which will be used for the derivation of the strong asymptotics. Then in Section 3 we prove strong asymptotics of Pad´e approximants for a Markov function generated by an analytic, complex valued weight function, i.e., strong asymptotics for polynomials orthogonal with respect to an analytic complex valued weight. This result is just a repetition of the corresponding theorem of Nuttall from [13], using for its proof ideas and details from Suetin’s paper [17]. Nevertheless there are some new methodological insights which make the proof easier and which indicate that the method can be developed for a wider class of applications. Next in Section 4 we prove a generalization of Nuttall’s theorem when the complex valued weight has a varying real valued component. An equilibrium problem in the presence of an external field plays an important role here. Finally (see Section 5) we consider P. Deift’s (with co-authors) version of the Riemann-Hilbert problem approach to the strong asymptotics for orthogonal polynomials (see [3]). The starting point of the approach is a reformulation of the orthogonality relations in terms of a matrix Riemann-Hilbert problem which has been introduced and developed in [7], [2]. We choose for this presentation of the approach a model problem of asymptotics for polynomials orthogonal on [−1, 1] with respect to a complex weight function — the same problem as in Section 3. Here (again as in Section 3) we assume that the analytic weight function has the same behavior at the end points of the interval of orthogonality as the classical Chebyshev weight. This allows us to get a very easy proof (in the framework of Deift’s approach) for the asymptotic formulas. This shortcut avoids most of one delightful ingredient of matrix Riemann-Hilbert problem approach, which is the analysis around the end points. Concluding the Introduction we have to say that the paper has more emphasis on explaining a method than on exposition of new results. We pay more attention to showing the different points of view on the same subject and to transparency of the proofs than to generality of the results proven here.

2

Auxiliary BVP

In this section we start from the preliminary material presented in the Appendix (see below) to study a function which will be used in the formulation of the asymptotic formulas later on. This finction is introduced as a solution of some boundary value problem. The function contains a generalisation of the well known Szeg˝o function for the complex weight.

2.1

Statement of the problem, properties of the solution

Let %(z) be a complex valued function, %(x) 6= 0 for x ∈ [−1, 1], and assume that % has an analytic continuation in the domain δ ⊃ [−1, 1], so that % ∈ H(δ). We consider the following

2

BVP: (   ψ(z) |∞ = O(z −n ),   1) ψ, ϕ ∈ H(C \ [−1, 1]);   ϕ(z) |∞ = O(z n ),   ( Find ϕ, ψ such that %ϕ+ = ψ− ,  2) on [−1, 1] ,   %ϕ− = ψ+ ,     3) ψϕ |∞ = 1 .

.

(1)

The boundary values ϕ± and ψ± of ϕ and ψ are assumed to be uniformly bounded on [−1, 1]. We highlight several properties of the solutions of the problem (1): Property 1 For every z ∈ C we have ψ(z)ϕ(z) = 1.

(2)

Proof: From 2) in (1) it follows that (ϕψ)+ = (ϕψ)− on [−1, 1], so that 1) in (1) implies that ϕψ is analytic in C. The maximum principle (or Liouville’s theorem) then shows that ϕψ is constant, and by 3) of (1) we see that the constant is 1. ¥ From Property 1 we also have %ϕ+ ϕ− = 1 on [−1, 1].

(3)

ψ, ϕ 6= 0 in C.

(4)

Property 2 We have Proof: This is an immediate consequence of 1) in (1) and Property 1. ¥ Let us consider a piecewise analytic function F on the two-sheeted Riemann surface R (see Appendix, Figure 4) consisting of two analytic pieces ϕ and ψ placed on the sheets R(+) and R(−) respectively: ( ϕ(z), z ∈ R(+) , F(z) = (5) ψ(z), z ∈ R(−) . Let ∆ be a contour on the Riemann surface splitting R into two pieces {R(+) \ [−1, 1]} and {R(−) \ [−1, 1]}. For ∆ we can choose a Jordan contour which goes along the upper side of [−1, 1] on {R(+) } and along the lower side of [−1, 1] on {R(+) } in the clockwise direction. Then the BVP (1) for the functions ϕ and ψ is equivalent to the BVP on R for F given by (5):  + (±) ±n  1) F ∈ H(R \ {∆, ∞ }), F (z) |∞ = O(z ), Find F such that 2) %F+ = F− on ∆, (6)   3) F (+) F (−) |∞ = 1. Properties 1 and 2 for (ϕ, ψ) transform into (see (2) and (3)) F(z (+) )F(z (−) ) = 1,

∀z ∈ C ,

and F 6= 0 on R. Finally we mention the uniqueness property of the solution of (1) and (6). 3

Property 3 If the solution of the BVP (6) exists, then it is unique. Proof: Let F and Fe be two solutions of (6). Then 2) of (6) shows that

³ ´ e F F

e

³ ´ +

=

e F F

−

on ∆, and hence F is holomorphic on the whole Riemann surface R, and by the maximum F e e F principle on R we have that F = const, and by 3) of (6) we conclude that F = 1. ¥ F Now we construct a solution of the BVP (6). First we proceed with a special case and after that we consider the general case.

2.2

Special case. Representation of the mapping function and Szeg˝ o function for a Bernstein weight by rational functions on R

Let the multiplicative jump in the BVP (6) be given by % :=

1 , p

where p is an arbitrary polynomial of degree m with√complex coefficients. Polynomials orthogonal on [−1, 1] with respect to the weight 1/(p(x) 1 − x2 ) have been considered by S.N. Bernstein, see, e.g., [18, §2.6]. Let B(z) be a function meromorphic on R, which is (like a rational function on C) defined by its zeros and poles (divisors) as follows:  (+) (+)  there is a pole of order n, • On R at ∞ (−) B : • On R we have the m zeros of p(z) = cm z m + · · · ,   and at ∞(−) there is a pole of order m − n.

(∞)n (+)

(−)

u

u

u

u

−1

+1

z

e

m

}|

e

e

{

e

∞

Figure 1: Zeros and poles for the function B The normalization of B is chosen as B (+) (z)B (−) (z) = cm . z→∞ zm lim

It is easy to check that the function ( F :=

B (+) , %B (−) , 4

(∞)m−n

on R(+) , on R(−) ,

satisfies the BVP(6) for % = 1/p. Indeed, the condition 1) of (6) is valid by construction of B. Condition 2) becomes %B+ = %B− on ∆ , and it is valid since B (−) is an analytic continuation of B (+) . The condition 3) is valid because of the normalization of B at ∞. The function B can be decomposed into the product of two rational functions on R B = FB Φ n , where the zeros of FB are on the sheet R(−) at the zeros of p and FB has a pole of order m at ∞(−) : FB : (+)

(−)

u

u

u

u

−1

+1

z

e

m

}|

e

e

{

e

(∞)m ∞

Figure 2: Zeros and poles for the function FB The function Φ has just a simple pole at ∞(+) and a simple zero at ∞(−) Φ: (∞) (+)

u

u

(−)

u

u

−1

+1

(0) ∞

Figure 3: Zeros and poles for the function Φ The normalization is Φ(+) Φ(−) |∞ = 1. For the function Φ we know the explicit expression √ Φ(z) = z + z 2 − 1 , Φ(+) |∞ = 2z + · · · .

(7)

This is the inverse Zhukovsky function: Φ(z) maps the following regions to the inside and outside of the unit circle: Φ(−) : {C \ [−1, 1]} → {| w |< 1} =: U, Φ(+) : {C \ [−1, 1]} → {|w| > 1} = C \ U . 5

2.3

General case. Determination of the Szeg˝ o function for a complex valued weight by means of the solution of the Riemann problem on R

First we normalize the solution of the BVP (6) on R: F Φn

F :=

on R .

For the bounded function F on R (taking into account the properties of F and Φ) we have   1) F ∈ H(R \ ∆) ; F 6= 0 on R, 2) %F+ = F− on ∆ ⊂ R,   3) F (+) F (−) = 1 on C. Observe the following equality, which follows from 2) and 3) of the above relation: (+)

(+)

%F+ F−

= 1 on [−1, 1] ⊂ C.

(8)

Since F does not vanish on R and % 6= 0 on ∆, we can choose a single-valued branch of the complex logarithm, connecting the functions ln F (+) , ln F (−) and ln % on ∆. Thus, for the single-valued function ln F on R, we have the standard Riemann problem (see Appendix (A.7)) on R:   1) ln F ∈ H(R \ ∆) , 2) (ln F )+ − (ln F )− = − ln % on ∆ ,   3) ln F (z (+) ) = − ln F (z (−) ) , which can be solved using the Cauchy integral (A.8) Z 1 (+) ln %(ζ) dΩ(ζ; z (+) , z (−) ) + ln F (z (−) ) . ln F (z ) = − 2πi ∆ Finally, substituting the explicit expression for the meromorphic differential (A.9), we have Z Z √ 2 1 1 z −1 1 p ln F (z (+) ) = − ln %(ζ) dΩ(ζ; z (+) , z (−) ) = − ln %(ζ) dζ. 4πi ∆ 4πi ∆ ζ 2 − 1 ζ − z √ Let us denote by w(z) a branch of z 2 − 1 on C \ [−1, 1] such that 1

w(x) = (x2 − 1) 2 > 0 for x ∈ (1, ∞]. Then we have ln F

(+)

1 (z) = 4πi

(9)

Z

w(z) 1 p ln %(ζ) dζ ζ2 − 1 z − ζ ∆ Z 1 Z 1 w(z) ln %(x) 1 w(z) ln %(x) 1 dx + dx = 4πi −1 w(x)+ z − x 4πi −1 w(x)− z − x Z 1 1 w(z) ln %(x) = dx . 2πi −1 w(x)+ z − x

The function

½

F

(+)

1√ 2 z −1 (z) = exp − 2π 6

Z

1 −1

ln %(x) dx √ z − x 1 − x2

¾ (10)

is the reciprocal of the so-called Szeg˝o function, which satisfies the boundary condition (8) and is normalized at infinity as ¾ ½ Z 1 1 ln %(x) dx (+) √ F (∞) = exp − . 2π −1 1 − x2 Thus we obtained ϕ(z) = Φn (z)F (+) (z) ,

ϕ(z) |∞ = F (+) (∞)(2z)n + · · · ,

(11)

where Φ and F are given by (7) and (10).

3 3.1

Strong asymptotics for Pad´ e approximants of a Markov function with complex weight Jump condition for Pad´ e approximants

We consider a Markov function 1 %b(z) := 2π

Z

1 −1

%(x) dx √ , z − x 1 − x2

(12)

and its Pad´e approximants: a polynomial denominator Q, a polynomial numerator P , and the remainder function R such that 1) deg Q, deg P ≤ n,

R |z→∞ = O(

1 z n+1

),

(13)

2) Qb %+P =R . We assume that % satisfies the conditions of Section 2. Also, for the moment, we will assume that the complex-valued function % is such that deg Q = n. Later on we will show that this is always the case for large enough n. This was already proved by Magnus [12], who also proved the convergence of the Pad´e approximants using Toeplitz matrix techniques. We take the normalization of Q as Q(z) = z n + · · · . From the definition of a Markov function (12), we have by the Sokhotsky-Plemelj (or StieltjesPerron) formula % on [−1, 1], %b+ − %b− = w+ which gives us the jump condition for the remainder function (this follows from the definition (13)) % R+ − R− = Q on [−1, 1]. (14) w+

3.2

Riemann problem on R for Pad´ e approximants and its solution

We have from (14) and (9) (wR)+ − Q% = −(wR)−

on [−1, 1].

(15)

The idea of what follows is: using the decomposition of % by means of the auxiliary BVP considered in Section 2, we rewrite the above jump condition as a jump for functions analytic 7

in C \ [−1, 1], for which the Riemann problem on R can be stated and solved, giving as a result an integral equation for the remainder function R. The analysis of this integral equation leads to the asymptotics for Q and R. So, (15) and (1) give us on [−1, 1] %=

ψ− ⇒ (wRϕ)+ − (Qψ)− = −ϕ+ (wR)− , ϕ+

%=

ψ+ ⇒ (wRϕ)− − (Qψ)+ = −ϕ− (wR)+ . ϕ−

and

Thus, for the functions (wRϕ) , (Qψ) ∈ H(C \ [−1, 1]) , we have the following jump condition (wRϕ)± − (Qψ)∓ = (− We define on R

( f (z) :=

ψwR )∓ . % z ∈ R(+) , z ∈ R(−) .

w(z)R(z)ϕ(z), Q(z)ψ(z),

Then for f we have the following Riemann problem (see (A.7)) on R   1) f ∈ H(R \ ∆) , 2) f+ − f− = j− on ∆ := {[−1, 1]+ ∪ [−1, 1]− } ⊂ R(+) ,   3) f (∞(−) ) = cn , where j = −( and

(16)

ψwR ) , on R(−) ∩ π −1 (δ), %

F (−) (∞) 1 cn = = n exp n 2 2

½

1 2π

Z

1 −1

ln %(x) dx √ 1 − x2

The solution of this problem (see (A.8)) is Z 1 j(ζ) dΩ(ζ; z, ∞(−) ) + cn , f (z) = 2πi ∆

¾ .

(17)

where the explicit expression for the differential dΩ is given in (A.12). If we consider (17) for z ∈ R(+) , we get an integral equation for R. Now an important point! Taking into account that the jump function j(ζ) has an analytic continuation from ∆ to the R(−) sheet, we can deform the contour ∆ to the contour ∆0 = {z :| Φ(z) |2 = 1 − ε , ε > 0} ⊂ R(−) ∩ π −1 (δ), and for z outside a ring A, bounded by ∆ and ∆0 , we have (see Remark 1 in Appendix, Section A.1) Z 1 j(ζ) dΩ(ζ; z, ∞(−) ) + cn , z ∈ R \ A. (18) f (z) = 2πi ∆0

8

3.3

Outer asymptotics for Q

First we estimate kRkC(∆) . We denote Mn := kwRϕkC(∆) . Consider the equation (18) for z ∈ R(+) \ ∆ ¶ Z µ 1 wψR (wRϕ)(z) = − (ζ) dΩ(ζ; z, ∞(−) ) + cn . (19) 2πi ∆0 % Note that the integral here is not singular anymore, therefore wRϕ has a continuous limit on ∆ and (19) is valid also for z ∈ ∆. Thus, (19) implies ¯ Z ¯ µ ¶ ¯1 ¯ ψ (−) (wRϕ)(ζ) Mn = |(wRϕ)(z0 )| = ¯¯ (ζ) dΩ(ζ; z0 , ∞ ) + cn ¯¯ , 2π ∆0 %φ for some z0 ∈ ∆. Taking into account the expression for dΩ has no singularities in R(−) \ (∆ ∪ {∞(−) }), we find Mn ≤ Mn C(1 − ε)n + cn ⇒ Mn ≤

ψ ϕ

(see (7) and (11)) and the fact that

cn . 1 − C(1 − ε)n

(20)

Now we can get the asymptotics of Q on compact sets K ⊂ C \ [−1, 1]. Fix K and choose ε such that K ⊂ C \ A. Considering (18) for z ∈ R(−) \ A, we have µ ¶ Z 1 ψ (Qψ)(z) − cn = − (ζ) dΩ(ζ; z, ∞(−) ) . (hϕR)(ζ) 2πi ∆0 %ϕ Therefore e n (1 − ε)n ≤ cn kQψ − cn kC(K) ≤ CM

e C (1 − ε)n . 1 − C(1 − ε)n

Dividing by cn , we obtain the desired asymptotics: (Qψ)(z) n = 1 + O(qK ), cn

qK < 1,

or n Q(z) = cn ϕ(z)(1 + O(qK )) .

3.4

Asymptotics on [−1, 1]: statement of the theorem

As we already mentioned, the integral equation (19) remains valid for z ∈ ∆. Thus µ ¶ Z ψ 1 (wRϕ)(ζ) f+ (z)|z∈∆ = − (ζ) dΩ(ζ; z, ∞(−) ) + cn . 2πi ∆0 %ϕ Therefore, for z ∈ ∆ ¯ Z ¯ µ ¶ ¯1 ¯ ψ (−) ¯ ¯ (wRϕ)(ζ) |f+ (z) − cn | = ¯ dΩ(ζ; z, ∞ )¯ 2π 0 %ϕ ∆

≤ Mn C(1 − ε)n ≤ cn

C(1 − ε)n . 1 − C(1 − ε)n

So, we have obtained uniform asymptotics for the remainder R(z) on the interval [−1, 1]: (wRϕ)± (x) f+ (x) |x∈∆ = |x∈[−1,1] = 1 + O(q n ) , cn cn 9

where 0 < q < 1 and q depends on the size of the domain of analyticity of %(z), (i.e., on δ). Now, for the polynomials Q we have from the boundary value Pad´e problem (15) µ ¶ µ ¶ wR wR Q= + on [−1, 1], % + % − and because of (wR)± cn cn = (1 + O(q n )) = (1 + O(q n )) = cn ϕ∓ (1 + O(q n )) , % ϕ± % ψ∓ we finally obtain (taking into account the boundedness of ϕ on [−1, 1]) Q = ϕ+ + ϕ− + O(q n ), cn

on [−1, 1] .

Now we can make the following remark about the normality of Pad´e approximants. We can omit our assumption for % that deg Q = n, for n large enough. Indeed, if deg Q < n, then everything above remains valid apart from the condition 3) in the Riemann problem (16), which becomes 3)0 F (∞(−) ) = cn = 0 , but this leads to a contradiction with (20) for large n. Thus we have proved the following theorem Theorem 1 Let % be a non vanishing complex valued function on [−1, 1], which is analytic in some δ neighborhood of [−1, 1]. Then for the Markov function Z 1 %(ζ) 1 dζ p %b(z) := 2π −1 z − ζ 1 − ζ 2 and sufficiently large n, there exists a unique polynomial Q, with Q(z) = z n + · · · , such that µ ¶ 1 Qb % + P = R, , deg P = n − 1 , R(z)|z→∞ = O . z n+1 The polynomial Q and the remainder function R have the following asymptotic formulas 1)

2)

Q(z) n = 1 + O(qK,% ), z ∈ K ⊂ C \ [−1, 1], cn ϕ(z) Q(x) x ∈ [−1, 1], = ϕ+ (x) + ϕ− (x) + O(q%n ) , cn R± (x) 1 = √ x ∈ [−1, 1], + O(q%n ) cn ( x2 − 1ϕ(x))±

where ϕ(z) = (z +

√

½

z2

1√ 2 − 1) exp − z −1 2π n

and

Z

1 −1

ln %(x) dx √ z − x 1 − x2

¾ ,

½ Z 1 µ ¶n ¾ 1 ln %(x) 1 √ exp cn = dx . 2 2π −1 1 − x2 The constants qK,% , qs are such that 0 < qK,% , qs < 1 and depend on the compact set K and on the size of the domain of analyticity of % (i.e., on δ). 10

4 4.1

Strong asymptotics for polynomials with respect to a varying weight Conditions on the varying weight. Equilibrium in external field. Statement of the problem

Here we assume the weight generating a Markov function in (12) has a dependence on n of the form ρn := e−2nqn ρ, ρ ∈ H(δ), ρ 6= 0 on [−1, 1] ⊂ δ and for qn we assume 1 kqn − qkC(δ) = O( ) , qn ∈ H(δ), n

=q = 0 on [−1, 1].

These settings for ρn can be rewritten as ρn = e−2nq ρ˜n ,

ρn ∈ H(δ),

ρn 6= 0 in δ

and k˜ ρn − ρkC(δ) = O(1). We note that for further consideration it is enough to assume that {˜ ρn } is a compact family in H(δ). Next we have to put extra conditions on q. For this we consider the equilibrium problem in the presence of the external field q for the logarithmic potential Z 1 Vµ (z) = − ln |x − z| dµ(x) −1

of positive probability measures µ ∈ M+ ([−∞, ∞]), supported on [−1, 1]. It is well-known (see [9], [15]) that there exists a unique measure λ (called equilibrium measure in external field) such that ½ γ on supp λ, Vλ + q = (21) ≥ γ on [−1, 1]. In what follows we assume q is such that

suppλ = [−1, 1].

(22)

Remark. A sufficient condition for (22) is convexity of q on [−1, 1]. It is not so difficult to see (see, for example [9]), that (22) and q ∈ H(δ) imply absolute continuity of λ and λ0 ∈ A(δ \ ({−1} ∪ {+1})), i.e., λ0 has analytic continuation in the punctured (at ±1) domain δ. In what follows we assume more, namely that q is such that

m(z) , m ∈ H(δ) , m(±1) 6= 0. λ0 (z) = √ z2 − 1

(23)

Remark. If the equilibrium measure for the problem on M+ (I), for some I : [−1, 1] ⊂ I ⊂ δ has its support equal to I, then condition (23) is fullfiled. A proof of these remarks can be 11

found in [9]. We also mention that if q is a convex analytic function on δ ∩ R, then both our extra conditions (22) and (23) hold true (this follows from the previous two remarks). Now we state a problem about polynomials orthogonal with respect to the varying weight. These polynomials have a variety of very important applications. Suppose the family of Markov functions Z 1 1 ρn (x)dx √ ρˆn (z) = 2π −1 (z − x) 1 − x2 is generated by the family of weight functions of the form ρn = e−2nq ρ˜n ,

ρn ∈ H(δ),

ρn 6= 0 in δ,

=q = 0 on [−1, 1],

where q is such that equilibrium measure λ of the problem (21) satisfies conditions (22), (23) and {˜ ρn } is a compact family in H(δ). For each n we consider Pad´e approximants of index n to the function ρˆn µ ¶ 1 n Qn (z) = z + · · · , Qn ρˆn + Pn = Rn = O . z n+1 The polynomial denominators Qn satisfy the following orthogonality relations Z 1 Qn (x)ρn (x)xν dx = 0, ν = 0, . . . , n − 1. −1

The investigation of the strong asymptotics for these polynomials Qn and for the remainder function Rn goes along the same lines as we did in the previous chapter for non-varying weight. Only two new ingredients will be introduced. In the next section, using the equilibrium problem (21) we present another expression for the solution of auxiliary BVP (1) for a varying weight. Then in Subsection 4.3 we do a more delicate analytic continuation of the jump function for the remainder function’s Riemann problem (16) on the second sheet R(−) of the Riemann surface.

4.2

Another representation for the solution of the auxiliary BVP for varying weight ρn

Here we consider the solution of BVP problem (1) for the varying weight ρn . As we know (see (11), (10)) the solution is Z 1 Z 1 n 1 w(z) 2q(x) w(z) ln ρ˜n (x) n ϕ(z) = Φ (z) exp{− dx + dx}, z∈C 2πi −1 w+ (x) z − x 2πi −1 w+ (x) z − x 1 ψ(z) = . ϕ(z) We denote by G = ln Φ the complex Green function, H is the first integral in the exponential above and F (˜ρn ) stands for the last multiplier of ϕ: ϕ =: exp{n(G − H)}F (˜ρn )

(24)

We will also use the notation F (ρq ) := e−H , ϕ(q) := exp{n(G − H)}

12

and

ψ (q) =

1 . ϕ(q)

From (3) and (8) it follows that on [−1, 1] (q)

(q)

ϕ+ ϕ− e−2nq = 1, (ρ )

(ρ )

F+ q F− q e−2q = 1.

(25)

The last relation gives us H+ + H− + 2q = 0 ,

on [−1, 1],

which, taking into account the symmetry with respect to R leads to 0 we analytically continue the above formula from some point z ∈ [1, ∞] to some point x ∈ [−1, 1] along some path belonging to the upper half plane and along some path from the lower half plane. As a result we will have Z 1 V± (x) = V (x) ∓ iπ λ0 (t)dt , x ∈ [−1, 1]. x

Thus

Z

1

`± (x) = ∓

λ0 (t)dt.

x

2. Then we see that the function

Z

1

`(z) = −

λ0 (ξ)dξ

(32)

z

gives a holomorphic continuation of `± (x) in the upper and lower neighborhood of [−1, 1]. We denote these lense-shaped simply connected domains as δ˜+ and δ˜− . There the function `(z) is a primitive of the holomorphic branch of the analytic function λ0 (z) Z 1 `± (z) := − λ0 (ξ)dξ , z ∈ δ˜± . z

Using a local representation of `(z) in the neighborhoods O+1 and O−1 of the end points of [−1, 1] (which follows from (32) and (23)) ½ √ m+1 (z)√1 − z, m+1 ∈ H(O+1 ), = m+1 = 0 on R ∩ O+1 , z ∈ O+1 , `(z) = (33) m−1 (z) z − 1, m−1 ∈ H(O−1 ), = m−1 = 0 on R ∩ O−1 , z ∈ O−1 . we see that analytic continuation `+ (z) from δ˜+ to some point x ∈ R such that x ∈ ((Q+1 ∪ O−1 ) \ [−1, 1]) ∩ R) coincides with the analytic continuation of `− (z) from δ˜− to the same point x. Thus the function (32) is holomorphic in a neighborhood of [−1, 1] ` ∈ H(δ˜ \ [−1, 1]) , δ˜ := δ˜+ ∪ δ˜− ∪ O+1 ∪ O−1 , and satisfies the boundary condition (30) on [−1, 1]. 3. It remains to check (31). In the domains δ˜+ and δ˜− inequality (31) is true because of the Cauchy-Riemann equations ∂ 0. ∂y ∂x To check (31) in the domain (O+1 ∪ O−1 ) \ [−1, 1], we use the local representation (33) of `(z) there. Take for example O+1 , we can choose O+1 small enough such that the argument of m+1 (which is zero on R ∩ O+1 ) does not make a substantial contribution to the argument of `(z), so arg `− (x) = 0 < arg `(z) < π = arg `+ (x) ,

z ∈ O+ \ [−1, 1] , x ∈ [−1, +1].

Thus =` > 0 in (O+ ∪ O−1 ) \ ∆, and we have (31) for ` given by (32). 15

Summarizing our transformations of the jump function in (29), we have j=

wψRn wRn (F (˜ρn ) )−2 2ni` = e on ∆ = [−1, 1]± ⊂ R. J , J := ρn ψ ρ˜n

So now we can lift the domain δ˜ on the sheet R(−) , we denote it by ˜ ˜ (−) ∈ R(−) , π(D ˜ (−) ) = δ, D ˜ (−) it is possible to deform contour ∆ in (29) to some and using analytic continuation of j to D 0 (−) ˜ contour ∆ ⊂ D : Z wRn 1 ( fn (z) = − (ξ)J(ξ)dΩ(ξ; z, ∞(−) ) + Cn , z ∈ R \ A, (34) 2πi ∆0 ψ where A ⊂ R(−) is a ring domain bounded by ∆ ∪ ∆0 . ˜ and (31) we have Finally, taking into account compactness of ρ˜n in H(δ) kJkC(∆0 ) ≤ Cκn ,

κ < 1,

(35)

where the constants C and κ depend on analytic properties of the varying weight ρn .

4.4

Statement of the theorem (for varying weight)

Thus using the estimation (35) in the integral equation (34) and repeating all arguments we used to prove the corresponding theorem for the non-varying weight, we arrive at the following theorem Theorem 2 Let {ρn } be a family of holomorphic functions in the domain δ ⊃ [−1, 1] ρn = e−2nq ρ˜n , ρn ∈ H(δ) , ρn 6= 0 in δ ⊃ [−1, 1], where {˜ ρn } is a compact family in H(δ), and q is real valued on [−1, 1], =q = 0

on [−1, 1].

Furthermore ρn is such that the equilibrium measure λ in the external field q ½ γ on supp λ Vλ + q = ≥ γ on [−1, 1] is absolutely continuous on supp λ = [−1, 1] and its derivative has the form m(x) λ0 (x) = √ , 1 − x2

m ∈ H(δ) , m(±1) 6= 0.

Then for sufficiently large n, there exists a unique polynomial Qn (z) = z n + · · · which is orthogonal with respect to the weight function ρn Z 1 ρn (x)dx Qn (x)xν √ = 0 , ν = 0, . . . , n − 1, 1 − x2 −1 16

and for the polynomials Qn and the functions of the second kind Z 1 Qn (x) ρn (x) dx √ R(z) = 1 − x2 −1 z − x the following asymptotics formulas hold Qn (z) = 1 + O(κnK ) , cn ϕ(z)

z ∈ K ⊂ C \ [−1, 1],

Qn (x) = ϕ+ (x) + ϕ− (x) + O(κn1 ) , cn

x ∈ [−1, 1]

R± (x) 1 = √ + O(κn2 ) , 2 cn ( x − 1ϕ(x))±

x ∈ [−1, 1],

where the constants κ ∈ (0, 1), ϕ, and cn are ϕ(z) = e

n(γ−Vλ (z))

1√ 2 exp{− z −1 2π −nγ

cn = e

Z exp{

1 −1

and

5

1 γ = ln 2 + π

Z

1 −1

ln ρ˜n (x) dx √ } (z − x) 1 − x2

ln ρ˜ (x) √ n dx}, 1 − x2

1

√ −1

Z

q(x) dx. 1 − x2

A matrix Riemann-Hilbert problem approach

We start here with the matrix Riemann-Hilbert problem formulation of the orthogonality relations. Then we recall the auxiliary boundary value problem, which we studied in Section 2 and which will be used here for the normalization of the matrix Riemann-Hilbert problem. Next, according to Deift [3], we proceed with the transformation of the original matrix RiemannHilbert problem to a problem with a jump matrix function which tends to the identity matrix as n (the degree of the polynomials) tends to infinity. Finally, to make the presentation self contained, we give a proof of the lemma stating that, if the jump matrix for the matrix valued homogeneous Riemann-Hilbert problem tends to the identity matrix, then the solution also tends to the identity matrix. At this point, we again take a shortcut by assuming that the jump matrix is analytic, it is possible for us to give a trivial proof, just based on the Cauchy theorem, without applying the harmonic analysis which was used in its original version in [3].

5.1

Orthogonal polynomials and a matrix Riemann-Hilbert problem

Let Qn (x) := xn + · · · be a monic orthogonal polynomial Z 1 Qn (x)xk h(x) dx = 0, −1

17

k = 0, . . . , n − 1,

(36)

with respect to a weight function h, which we assume to have the form h(x) := √

%(x) , 1 − x2

x ∈ [−1, 1],

(37)

where % is a complex valued function, non-vanishing on [-1,1], which is holomorphic in some domain δ containing the interval [−1, 1] % 6= 0 on [−1, 1],

% ∈ H(δ),

[−1, 1] ⊂ δ.

Let Rn be the function of the second kind associated with Qn Z 1 1 Qn (x)h(x) dx Rn (z) = , Rn ∈ H(C \ [−1, 1]). 2πi −1 x−z

(38)

(39)

It is easy to verify that the orthogonality relations for Qn are equivalent to the fact that Rn (z) = O(

1 z n+1

),

z → ∞.

(40)

As before (see Section 3), our starting point is the Sokhotsky-Plemelj formula for the boundary values of a Cauchy type integral, which we apply to the function of the second kind (39) Rn+ − Rn− = hQn

on [−1, 1].

(41)

Here, as usually, (+) denotes the boundary values of the function from the upper side of [−1, 1], and (−) from the lower side. Choosing m as a normalization constant so that mRn−1 =

1 + ··· , zn

i.e.,

z→∞,

2πi

m = −R 1

Q2 (x)h(x) dx −1 n−1

,

and applying (41) to Rn−1 , then for the matrix valued analytic function µ ¶ Qn Rn Y = mQn−1 mRn−1 we obtain the following Riemann-Hilbert problem:   Y ∈ H(C \ [−1, 1]),    Y (z) = (I + O( 1 ))diag(z n , z −n ) , as z 7→ ∞, z ! Ã Y :=  1 h   on [−1, 1].   Y+ = Y− 0 1

(42)

(43)

Our goal is to show that the solution (42) of (43), which depends on n, tends as n → ∞ to a solution of some boundary value problem which does not depend on n. First we mention that our restriction for the weight function to be of the form (37) and (38) allows us to make an analytic continuation of h(x) in the domain δ \ [−1, 1] h(z) :=

%(z) 1

i(z 2 − 1) 2 18

∈ H(δ \ [−1, 1]),

(44)

1

where the branch for (z 2 − 1) 2 is chosen such that 1

(z 2 − 1) 2 > 0 for z > 1. It gives for the limiting values of h(z) h± (x) = ∓ √

%(x) 1 − x2

and if we denote

µ

then we have

µ W− =

on [−1, 1] ,

W (z) =

1 h(z) 0 1

¶

µ

1 h 0 1

,

W+ =

¶ ,

1 −h 0 1

¶ = W−−1 .

(45)

The latter remarkable relation allows us to present the jump condition in (43) as Y+ = Y− W −

5.2

or Y− = Y+ W+

on [−1, 1].

Auxiliary boundary value problem recall

Before stating an asymptotic result for the solution of the matrix Riemann-Hilbert problem, we recall the auxiliary BVP which we studied in Section 2. In terms of the solution of this BVP we will write an answer (i.e., final asymptotic formulas) and the properties of the solution will be used in the proof. The problem consist on finding ϕ such that   ϕ ∈ H(C \ [−1, 1]), ∃ϕ+ , ϕ− ∈ C([−1, 1]),   ϕ(z) ϕ : ϕ(z) = O(z n ), z → ∞, (46) |∞ > 0,  zn   ϕ+ ϕ− % = 1 on [−1, 1]. The unique solution of problem (46) is (see (10)-(11) ) ½ ¾ Z 1 √ √ ln %(x) 1 dx n √ ϕ(z) = (z + z 2 − 1) exp − z2 − 1 , 2π 1 − x2 −1 z − x

(47)

n

and the normalization constant for ϕ|∞ = zc + · · · is ½ Z 1 ¾ 1 1 ln %(x) √ c = n exp dx . 2 2π −1 z 2 − 1

(48)

We also note that the scalar problem (46) admits several equivalent reformulations, i.e., the vector version   (ϕ, ψ) ∈ H(C \ [−1, 1]),    (ϕ, ψ)(z) = ( zn + · · · , c + · · · ), for some c > 0, n → ∞, c à zn !  0 %   on [−1, 1],  (ϕ, ψ)+ = (ϕ, ψ)− %−1 0 and the matrix version

 Ã !  0 ϕ    Ψ = ψ 0 , ϕ, ψ Ã !  % 0  T   Ψ+ = Ψ− 0 %−1 .

as above,

We are not going to use these versions here and mention them just for completeness. 19

5.3

Statement of the asymptotic result

Here we prove the following Theorem 3 Suppose that the weight function (37) satisfies conditions (38). Then for the matrix Y (see (42)) of the orthogonal polynomials (36) and for the functions of the second kind (39), the following asymptotic formula holds µ ¶ O(q n ) C Y S(z) = I + X, n → ∞, 0 < q < 1, (49) z uniformly outside any Jordan contour Γ ⊂ δ around [−1, 1], and C Y S(z) = (I + O(q n ))XD−1 ,

(50)

uniformly inside the domain bounded by Γ, where C = diag(c−1 , c) ,

S = diag(ϕ−1 , ϕ)

(51)

are given by (46), (48), and µ D=  X=

1 0 (hϕ2 )−1 1

¶ , 

i

1 1

(z−(z 2 −1) 2 ) 2i

1 (z 2 −1) 2

1

z+(z 2 −1) 2

.

1

2(z 2 −1) 2

The matrix asymptotic formulas (49) and (50) lead to the following asymptotic formulas for the polynomials Qn and for the functions of the second kind Rn 1)

2)

5.4

Qn (z) = 1 + O(q n ), z ∈ K ⊂ C \ [−1, 1], cϕ(z) Qn (x) = ϕ+ (x) + ϕ− (x) + O(q n ), x ∈ [−1, 1], c Rn± (x) 1 = √ + O(q n ), x ∈ [−1, 1]. c ( x2 − 1ϕ(x))±

Proof of the theorem

1. First we normalize the Riemann-Hilbert problem (43), forcing the solution to be holomorphic at ∞. We define (see (42), (51))   Qn Rn ϕ   cϕ c . (52) Z := C Y S =    cmQn−1 cmϕRn−1 ϕ Now the matrix-valued function Z is holomorphic in C\[−1, 1] and it satisfies the following Riemann-Hilbert problem   Z ∈ H(C \ [−1, 1]), (53) Z+ = Z− J on [−1, 1],   Z(z) = I + O( z1 ), as z → ∞, 20

where the jump matrix J, because of (C −1 ZS −1 )+ = (C −1 ZS −1 )− W− is given by

 ϕ  − ϕ+ ϕ− h   J = S−−1 W− S+ =  ϕ+ ϕ+  0 ϕ−

on [−1, 1] ,

on [−1, 1].

Our choice of the solution of the auxiliary BVP (46) for the normalization of the problem (43) gives ! Ã J=

ϕ− ϕ+

0

√ 1 1−x2 ϕ+ ϕ−

on [−1, 1].

This ‘lucky’ expression for the jump matrix of the problem (53) allows us to decompose it as Ã Ã ! ! 1 1 0 1 0 √ 0 √ ϕ+ √  J= 1 − x2  ϕ− 1 − x2 1 √ 1 − x2 1 ϕ− ϕ+ − 1 − x2 0     1 1 0 1 0 √ 0   . 1 = (54) 1 − x2   −1 √ 1 1 2 − 1−x 0 (hϕ2 )− (hϕ2 )+ We see that the first matrix here admits an holomorphic continuation in the lower half of the complex plane and the last matrix does the same in the upper half plane. After the analytic continuation, and because of the exponential decrease of ϕ−1 outside [−1, 1] as n → ∞, we see that J will be close to a very ‘friendly’ central matrix. This is the essence of the method! 2. Now, in order to develop this idea about the analytic continuation of the parts of the jump, we transform the Riemann-Hilbert problem (53) into the following problem. We denote µ ¶ 1 0 D := ∈ H(δ \ [−1, 1]) , (55) (hϕ2 )−1 1 and let Γ be a contour in δ such that ∆ := [−1, 1] is in the domain Int(Γ) which is bounded by Γ and ∞ ∈ / Int(Γ): Γ ⊂ δ, We define

( Ze =

For Ze we have on Γ

Z, ZD,

∆ ∈ Int(Γ).

in Out(Γ) := C \ Int(Γ), in Int(Γ). Ze+ = Ze− D,

and on [−1, 1] this Ze satisfies −1 −1 f+ = Z+ D+ = Z− JD+ = Z− D− D− JD+ ) . JD+ = Ze− (D− Z

21

(56)

If we substitute here the decomposition of J from (54), we see that   µ ¶µ 1 √ 0 1 0 −1 −1   2 D− JD+ = D− D− 1−x √ − (hϕ12 )+ 1 − 1 − x2 0   1 √ 0 = √ 1 − x2  . 2 − 1−x 0 Denoting the last matrix as

 j := 

√

0

− 1 − x2

√

1 1 (hϕ2 )+

0 1

¶

 1 1 − x2  , 0

we conclude that the matrix function Ze from (56) is a solution of the following RiemannHilbert problem  Ze ∈ H(C \ {[−1, 1] ∪ Γ}), Γ ⊂ δ, [−1, 1] ⊂ Int(Γ),    Ze = Z f− D on Γ, + (57) e+ = Ze− j on [−1, 1],  Z   e Z(z) = I + O( z1 ), as z → ∞. 3. The next step is to consider the limiting problem for (57), i.e., without the jump on Γ, because, as we already mentioned, this jump function tends to the identity matrix as n → ∞. We consider the following Riemann-Hilbert problem for the determination of the function X:   X ∈ H(C \ [−1, 1]), (58) X+ = X− j on [−1, 1],   1 X(z) = I + O( z ), as z → ∞. Writing the entries for the matrix jump condition  √ x11−  2 µ ¶ −x12− 1 − x √ x11+ x12+  1 − x2  = √  , x 21− x21+ x22+ −x22− 1 − x2 √ 1 − x2 we see that matrix-valued Riemann-Hilbert problem (56) reduces to two scalar problems   x11 , x12 ∈ H(C \ [−1, 1]),     x11 (z) = 1 + O( z1 ) as z → ∞,  (59) x12 = O( z1 ) as z → ∞,  √  2  x11+ = x12− (− 1 − x ) on [−1, 1],   √  x 2 11− = x12+ ( 1 − x ) on [−1, 1], and

  x21 , x22 ∈ H(C \ [−1, 1]),    1   x21 (z) = O( z ) as z → ∞, x22 = 1 + O( z1 ) as z → ∞,  √   1 − x2 ) on [−1, 1], x = x (−  21+ 22−  √  x 2 21− = x22+ ( 1 − x ) on [−1, 1]. 22

(60)

1

Note that, in accordance with the chosen branch of (z 2 − 1) 2 , we have 1 √ 1 2 (z − 1)∓2 = ∓ 1 − x2 i

therefore

1 1 x11± = ( x12 (z 2 − 1) 2 )∓ i

on [−1, 1],

1 1 and x21± = ( x22 (z 2 − 1) 2 )∓ . i

1

Thus the function ( 1i x12 (z 2 − 1) 2 ) is the analytic continuation of the function x11 on the second sheet of the Riemann surface, obtained from the two extended complex planes cut along the interval [−1, 1] and pasted together ‘cross by cross’. Analogously, the function 1 ( 1i x22 (z 2 − 1) 2 ) is the analytic continuation of x21 on the second sheet of the Riemann surface. If we choose, in accordance with the normalization at infinity, x11 ≡ 1 in C \ [−1, 1], then its continuation on the second sheet will again be equal to the constant function 1, which gives i x12 = 1 . 2 (z − 1) 2 Checking the normalization of x12 at ∞, we see that x11 and x12 as above satisfy the problem in (59). As for the problem in (60), we have to find the function x21 having a zero at infinity, and the analytic continuation of x21 on the second sheet of the Riemann surface which needs 1 to have a pole at infinity (because x22 is regular at ∞ and (z 2 − 1) 2 has a pole over there). The choice of x21 as 1 (z − (z 2 − 1) 2 ) x21 = 2i satisfies these conditions. Then 1

x22 =

(z + (z 2 − 1) 2 ) 1

2(z 2 − 1) 2

,

and the verification of the normalization of x22 at infinity x22 = 1 + O( z1 ) shows that we have found the solution for the problem in (60). Hence   i 1 z 2 −1 X =  (z−(z2 −1)) 12 (z+(z2 −1)) 12  (61) 2i

1

2(z 2 −1) 2

is a solution for the problem (58). 4. Finally, we have to show that the solution of the Riemann problem with a jump on Γ close to the identity (see (56)), which presents orthogonal polynomials and functions of the second kind (see (55), (56) and (52)), tends as n → ∞ to the solution of the problem (58) without the jump on Γ, which is the matrix function (61). To do this we define the function e −1 . = := ZX (62) The matrix = has a jump on Γ =+ = Ze− DX −1 = Ze− X −1 (XDX −1 ) , 23

so that

e =+ = =− D,

e = XDX −1 , D

(63)

and on [−1, 1] we have =+ = Ze− j (X− j)−1 = Ze− j j −1 X−−1 = =− . Thus the function = satisfies the following Riemann-Hilbert problem (note that there is no jump on [−1, 1]):   = ∈ H(C \ Γ), e on Γ, (64) =+ = =− D   1 =(z) = I + O( z ), as z → ∞, e on Γ we have and for the jump matrix D e = I + O(q n ) on Γ , D since the entries of X and X −1 in (63) do not depend on n and for D the asymptotics on Γ follows from the representation of ϕ in (47) substituted in (55). To finish the proof we apply to the function (62) the following well-known lemma e n for the Riemann problem (64) is analytic Lemma 1 Suppose that the jump matrix D in the domain A containing the contour Γ fn ∈ H(A), D and satisfies

Γ ⊂ A,

e n = I + εn , D

(65) (66)

where εn → 0 uniformly on compact subsets of A as n → ∞. Then = = I + O(εn ).

(67)

A proof of the lemma, with a weaker condition on the jump than (65), can be found in [3]. We note that for the case of scalar =, the proof of the lemma is rather trivial by considering a problem for log(=) for which we can write a solution by means of the e However, the function log of a matrix is not properly defined Cauchy integral of log(D). so that another approach is required for proving the lemma in the matrix case. This will be done in the next section. Finally, substituting in (67) the expressions (62), (58), (56), (55) and (52) we get the desired asymptotic formulas. The theorem is proved. ¥

5.5

Proof of the Lemma

Assuming the more restrictive condition (65) on the jump matrix in (64) rather than the one stated in [3], we have an opportunity to give a more elementary proof of Lemma 1 than the proof in [3].

24

Substituting (66) in the jump condition of (64), we have the following Riemann problem for the matrix function   = ∈ H(C \ Γ), =+ = =− + εn =− on Γ,   =(∞) = I. Applying Cauchy’s integral formula (see (A.1)-(A.2)) for each entry of =, we have Z 1 dξ =(z) = (εn =− )(ξ) + I, z ∈ C \ Γ. 2πi Γ ξ−z

(68)

Then, because of (65), we can deform Γ to Γ0 which is lying inside the domain A ∩ Out(Γ) Z 1 dξ e ∂ Ae = Γ ∪ Γ0 , =(z) = (εn =− )(ξ) + I, z ∈ C \ A, (69) 2πi Γ0 ξ−z and now the integral is not singular anymore for z ∈ Γ, and both sides of (69) have a limit when z → Γ+ . Hence Z dξ e =− D = =+ = (εn =− )(ξ) + I, z ∈ Γ. (70) ξ−z Γ0 Let z0 ∈ Γ be such that k=− (z0 )k = max k=− (z)k =: Mn , Γ

then from (70) we have Mn ≤ Mn εn constΓ,Γ0 + 1 , and therefore Mn ≤

1 , 1 − O(εn )

and substituting this estimate in (68), we arrive at (67). ¥

APPENDIX: The Cauchy residue theorem on a Riemann surface and the solution of the Riemann problem A.1

Representation of piecewise analytic functions in C by Cauchy’s integral formula

Let us consider the piecewise analytic function f ( f2 (z), z ∈ D, f (z) := f1 (z), z ∈ C \ D, where f1 , f2 are holomorphic functions in the closed domains of their definition f1 ∈ H(C \ D),

f2 ∈ H(D) .

We assume that D is bounded, with piecewise smooth boundary ∂D which coincides (as a set) with ∂(C \ D). The Cauchy integral formula allows us to write f as Z dξ 1 (f2 (ξ) − f1 (ξ)) + f1 (∞). f (z) = 2πi ∂D ξ−z 25

Indeed, the integral on the right hand side is Z Z 1 1 dξ dξ + f1 (ξ) + f1 (∞) f2 (ξ) 2πi ∂D ξ − z 2πi ∂(C\D) ξ−z ( f2 (z) − f1 (∞) + f1 (∞), = 0 + f1 (z) − f1 (∞) + f1 (∞),

z ∈ D, z ∈ C \ D.

Here we use a version of the Cauchy theorem for a domain Ω (with ∞ ∈ Ω) Z 1 c−1 F ∈ H(Ω) ⇒ F (ξ)dξ = res F = −c−1 , F (z) = F (∞) + + ··· . ∞ 2πi ∂Ω z Thus, if we introduce a jump function j(ξ) := f2 (ξ) − f1 (ξ) =: f+ (ξ) − f− (ξ), then the integral

1 f (z) = 2πi

Z j(ξ) Γ

ξ ∈ ∂D =: Γ ,

dξ +C ξ−z

(A.1)

gives a solution to the following boundary value problem (BVP) for the piecewise analytic function f (with continuous boundary values f+ and f− ):   a) f ∈ H(C \ Γ), (A.2) Find f such that b) (f+ − f− )|Γ = j ∈ H(Γ),   c) f (∞) = C. This is a BVP which is usually called the non homogeneous Riemann problem or ‘jump’ problem. e Remark 1 If j ∈ H(δ), where δ is such that Γ ⊂ δ and if Γ0 ⊂ δ, then for every z ∈ C \ δ, S with δe ⊂ δ and ∂ δe = Γ Γ0 , the solution for (A.2) can be represented as Z 1 dξ e f (z) = j(ξ) + C, z ∈ C \ δ. (A.3) 2πi Γ0 ξ−z e the right-hand side of (A.3) does not represent the solution of (A.2). Note that for z ∈ δ, For further insight it would be useful also to understand the Cauchy integral formula as a corollary of the more general Cauchy residue theorem. Indeed, for f ∈ H(D), with D ⊂ C, the formula Z f (ξ) f (ξ) 1 dξ = res[ ] = f (z) z ξ −z 2πi ∂D ξ − z is a corollary of

1 2πi

Z f (ξ)w(ξ) dξ = ∂D

X

f (zj ) res w =

j

zj

X j

res f w, zj

where w ∈ M is a meromorphic function in D, i.e., w ∈ H(D \ {zj }), with zj ∈ D. The expression h(ξ) dξ is usually called a meromorphic differential in D ⊂ C. Thus the Cauchy differential ¶ µ 1 dξ ξ−z is an example of a meromorphic differential in C. 26

A.2

Riemann surfaces, meromorphic differentials and Cauchy’s theorem on a Riemann surface

A mysterious fact: why is an integral around infinity, for a function holomorphic at infinity, not equal to zero and why do we have to count the contribution of the function at ∞ despite of its holomorphicity there? This fact has an explanation in terms of the Riemann surfaces and meromorphic differentials on them. We start with an explanation of what a Riemann surface means. First of all it is a surface, i.e., a two-dimensional topological manifold, which means that the surface R is locally a result of a continuous deformation of the extended complex plane (or a piece of the plane). Moreover, there is a correspondence between the points ξ ∈ R on the surface and the points τ ∈ C of the plane which is established by an open continuous mapping τ : R → C. The mapping τ (ξ) is locally one to one everywhere except for a discrete set of points, which are called branch points of R. All the other points of R are called regular points. This mapping is called a projection of R. Next, to become a Riemann surface, the surface R needs to have a complex structure. To explain what this means, we recall the notion of a holomorphic function on R. In a neighborhood of a regular point ξ0 ∈ R the projection τ (ξ) is invertible, therefore the mapping ξ(τ ) is defined in a neighborhood of the point τ (ξ0 ) ∈ C. A function f is called holomorphic in a neighborhood of a regular point ξ0 on R if the function f (ξ(τ )) is holomorphic in the neighborhood of τ (ξ0 ) on C. A function f is called holomorphic in a neighborhood U of a branch point ξ0 ∈ R if f is continuous in U and f is holomorphic in U \ ξ0 . A pair K = (U, τ (ζ)) is called a holomorphic chart of the surface R if • U is a domain in R, • τ (ζ) is holomorphic in U and maps U to the plane disk u = {|τ | < r} ∈ C, establishing a one to one correspondence. In practice, Riemann surfaces are presented as a union of holomorphic charts together with formulas of transformation from one local coordinate to another, and it is convenient to consider a function f (ζ) on R locally as a function f (ζ(τ )) on C using the local variables τ ∈ u. Example A. The extended complex plane R = C is considered as a Riemann surface with an atlas of the following holomorphic charts : K1 = ({|ζ| < R}, τ (ζ) = ζ), K2 = ({|ζ| > r}, τ (ζ) = ζ1 ), where R ≥ r. Example B. A two-sheeted Riemann surface (see Figure 4 on the right) is a union of the following holomorphic charts Ã

(

1

(ζ − a) 2 = τ Ka = Ua , ζ = τ2 + a à ( 1 (ζ − b) 2 = τ Kb = Ub , ζ = τ2 + b

! ,

(A.4)

! ,

Kj = (Uj : {a, b, ∞(+) , ∞(−) } ∈ / Uj , τ (ζ) = ζ), . . . K∞(+) = ({|ζ (+) | > r}, ζ(τ ) =

1 ), τ

27

K∞(−) = ({|ζ (−) | > r}, ζ(τ ) =

1 ). τ

¡ ¡

ζ

¡

¨ s §a

(+)

¡

(+) ¶¶ /

as

¡

¡ ¡

sb

¡

¡

¡

¡

¡ ¡

¡ º· P P := Ua −→ ³ ³

τ ¹¸ ¡

¡

¨ s §a

(−)

(−) ¡¡

¡ ¡

¡

¡

as

­ À­

¡

ζ

¡ ¡

sb

¡

¡ ¡

Figure 4: A two-sheeted Riemann surface Now a function f (ζ) on R can be understood locally in the domain of the chart (U, τ (ζ)) as a function of the local variable τ and its holomorphicity (meromorphicity) in U means that the function f (ζ(τ )) is holomorphic (meromorphic) in u = {|τ | < r}. In the same way, using a local variable, we define meromorphic differentials on R as dΩ(ζ) := ω(ζ)dζ, where ω(ζ) is a meromorphic function on R. For correctness of this definition, i.e., to be independent from the choice of local coordinates, it is necessary (and sufficient) to satisfy a condition of correspondence : dΩi (ζi ) = dΩj (ζj ) in the intersection of Ui and Uj . A singularity of the differential dΩ = ωdζ in the chart (U, τ (ζ)) is a point ζ0 ∈ U , ζ0 = ζ(τ0 ), where τ0 is a singularity of the function (in the variable τ ) w(ζ(τ ))ζ 0 (τ ). The residue of the differential dΩ = ωdζ at the point ζ0 = ζ(τ0 ) is res dΩ = c−1 , ζ0

P where c−1 is the coefficient of (τ − τ0 )−1 in the Laurent expansion w(ζ(τ ))ζ 0 (τ ) = ∞ j=−∞ cj (τ − j τ0 ) . Note that, because of the holomorphic correspondence of the charts, the position of the singularity ζ0 of dΩ(ζ) and its residue resζ0 dΩ do not depend on the choice of the chart covering ζ0 . Now we are able to state Cauchy’s theorem on a Riemann surface. Let f ∈ H(U ), U ⊂ R (R is a compact Riemann surface) and dΩ be a meromorphic differential in U . Then Z X 1 f (ζ) dΩ(ζ) = res[f dΩ], (A.5) ζj 2πi ∂U j where {ζj } are the singularities of dΩ in U .

A.3

The Riemann surface R = C. Examples of meromorphic differentials on R = C, and normalization of the Riemann problem at an arbitrary point

Here we consider differentials on R = C (see Example A from the previous section). Example A1. dΩ(ζ) = dζ. In the chart K1 = ({|ζ| < R|} , ζ(τ ) = τ ) we have ζ 0 (τ ) = 1, hence dζ has no singularities for |ζ| < R. 28

In the chart K2 = ({|ζ| > r}, ζ(τ ) = τ1 ) we have ζ 0 (τ ) = − τ12 , hence dζ has a pole of second order at ζ = ∞. Thus, in accordance with (A.5), the integral for a function f , which is holomorphic at ∞, f (z) = c0 +

c−1 + ··· , z

is given by Z f (ζ) dζ = res [f dζ] = res [(− ∂(|ζ|>r)

τ =0

ζ=∞

1 )(c0 + c−1 τ + . . .)] = −c−1 . τ2

This explains the usual convention concerning integration around infinity (see Section 2.1). Example A2. Consider the meromorphic differential dΩ(ζ) = 1. In K1 we have w(ζ(τ ))ζ 0 (τ ) = resζ=z dΩ(ζ) = 1.

1 , τ −z

dζ . ζ−z

hence ζ = z is a pole with

1 2. In K2 we have w(ζ(τ ))ζ 0 (τ ) = − τ (1−τ , hence z)

(

ζ = z is a pole, ζ = ∞ is a pole,

resζ=z dΩ(ζ) = 1, resζ=∞ dΩ(ζ) = −1.

Thus, the meromorphic Cauchy differential on the Riemann surface C has two poles ζ = z and ζ = ∞ with ( +1 at z , res dΩ = −1 at ∞ . To indicate the singularities and their residues we will use the notation dΩ(ζ) = dΩ(ζ; z, ∞) = dΩ(ζ; z(+1) , ∞(−1) ). By means of the Cauchy differential we can construct the differential dΩ(ζ; z(1) , w(−1) ) = (

1 1 z−w − )dζ = dζ . ζ −z ζ −w (ζ − w)(ζ − z)

Similar to the Cauchy differential in (A.1), this differential can be used for the solution of the Riemann problem with normalization at any point w ∈ C :   a) f ∈ H(C \ Γ), Find f (z) such that b) (f+ − f− ) |Γ = j ∈ H(Γ),   c) f (w) = Cw , ⇒

29

1 f (z) = 2πi

Z j(ζ) dΩ(ζ; z, w) + Cw . (A.6) Γ

A.4

The Riemann problem on R. Examples of meromorphic differentials on the two-sheeted Riemann surface

The last example of the Riemann problem on C (see (A.6)) can be extended to an arbitrary compact Riemann surface R and oriented, piecewise smooth curve Γ on R :   a) f ∈ H(R \ Γ), Find f (z) such that b) (f+ − f− ) |Γ = j ∈ H(Γ), (A.7)   c) f (w) = Cw , w ∈ R \ Γ. Here f+ and f− are the uniformly bounded limit values of f on Γ from the “positive” and “negative” side with respect to the orientation of Γ. By analogy with (A.6), from Cauchy’s theorem on R it follows : if j(ζ)dΩ(ζ; z, w) has no singularities on Γ, then Z 1 f (z) = j(ζ) dΩ(ζ; z, w) + Cw . (A.8) 2πi Γ Note the uniqueness of the solution of (A.7): If R is compact, then the solution of the Riemann problem (A.7) is unique. Proof: Let f and fe be the solutions of (A.7), then for (f − fe) we have j = 0, therefore (f − fe) ∈ H(R) and if R is compact, then, by the maximum principle, (f − fe) is constant. Moreover, by c) we see that this constant is zero. ¥ Now we present some important meromorphic differentials on the two-sheeted Riemann surface (see Example B , Figure 4, we put a = −1, b = 1). Example B1. We will use the notation z (+) and z (−) for the points belonging to the “opposite” sheets, but having the same projection z ∈ C. We have

√ z2 − 1 1 (+) (−) dΩ(ζ) := p dζ = dΩ(ζ; z(1) , z(−1) ). 2 ζ − z ζ −1

(A.9)

is a meromorphic differential on R with singularities described by the right-hand side of (A.9). Proof: It is evident that ζ = z (+) and ζ = z (−) are the poles of dΩ with residue +1 and −1 respectively. Next, (√ √ ζ −1=τ 2τ z 2 − 1 0 K+1 : ⇒ w(ζ(τ )ζ (τ ) = √ ζ = τ2 + 1 τ τ 2 + 2(τ 2 + 1 − z) ⇒

no singularity at ζ = 1.

The same at ζ = −1 . There is no singularity at ζ = ∞(±) : because of √ τ 2 z2 − 1 1 1 0 (− 2 ) . K∞(±) : ζ = , w(ζ(τ ))ζ (τ ) = √ τ 1 − τ 2 (1 − τ z) τ Example B2. 1 dΩ(ζ) := 2

p

√ ζ 2 − 1 + z2 − 1 (+) (−) p dζ = dΩ(ζ; z(1) , ∞(− 1 ) , ∞(− 1 ) ). 2 2 2 (ζ − z) ζ − 1 30

(A.10)

Proof: There is a singularity at z, with resz dΩ = 1, but there is no singularity at ze. There are no singularities at ζ = ±1. Indeed (take +1 for example) √ √ τ τ 2 + 2 + z2 − 1 0 √ K+1 ⇒ w(ζ(τ ))ζ (τ ) = 2τ . 2(τ 2 + 1 − z)τ τ 2 + 2 At ∞(±) we have 1 w(ζ(τ ))ζ 0 (τ ) = 2

√

√ 1 − τ 2 + τ z2 − 1 1 √ τ (− 2 ) , 2 τ (1 − τ z) 1 − τ

therefore res dΩ = −

∞(±)

1 . 2

From (A.10) we obtain the expression for (+)

(−)

(+)

(−)

2

2

2

2

dΩ(ζ; z(1) , w(−1) ) = dΩ(ζ; z(1) , ∞(− 1 ) , ∞(− 1 ) ) − dΩ(ζ; w(1) , ∞(− 1 ) , ∞(− 1 ) ) . Example B3. (Green’s differential) dG(ζ) := p Proof:

1 τ

⇒

ζ2

(+)

−1

(−)

= dΩ(ζ; ∞(−1) , ∞(+1) ).

(A.11)

2τ w(ζ(τ ))ζ 0 (τ ) = √ . τ τ2 + 2 1 τ (− 2 ) ⇒ res w(ζ(τ ))ζ 0 (τ ) = √ dG = −1. ∞+ 1 − τ2 τ

K+1 : ζ = τ 2 + 1 K ∞+ : ζ =

dζ

⇒

From (A.10) and Green’s differential (A.11) we obtain the expression for the differential 1 (−) (+) (−) dΩ(ζ; z(1) , ∞(−1) ) = dΩ(ζ; z(1) , ∞(− 1 ) , ∞(− 1 ) ) − dG . 2 2 2

(A.12)

References [1] A. Aptekarev, “Asymptotical properties of polynomials orthogonal on a system of contours and periodical motions of the Toda lattices,” Mat. Sb. 125 (167) (1984), 231–258, (in Russian); English translation in Math USSR Sb. 53 (1986). [2] P. Bleher and A. R. Its, “Semiclassical asymptotics of orthogonal polynomials, Riemann-Hilbert problems, and universality in the matrix model”, Ann. of Math. 150 (1999), 185–266. [3] P. Deift, “Orthogonal Polynomials and Random Matrices: A Riemann-Hilbert Approach,” Courant Lecture Notes 3, New York University, 1999. [4] P. Deift, T. Kriecherbauer, K. T-R McLaughlin, S. Venakides, and X. Zhou, “Uniform asymptotics for polynomials orthogonal with respect to varying exponential weights and applications to universality questions in random matrix theory,”, Comm. Pure Appl. Math. 52 (1999), 1335–1425. 31

[5] P. Deift, T. Kriecherbauer, K. T-R McLaughlin, S. Venakides, and X. Zhou, “Strong asymptotics of orthogonal polynomials with respect to exponential weights,”, Comm. Pure Appl. Math. 52 (1999), 1491–1552. [6] P. Deift, T. Kriecherbauer, K. T-R McLaughlin, S. Venakides, and X. Zhou, “A Riemann-Hilbert approach to asymptotic questions for orthogonal polynomials,”, J. Comput. Appl. Math. 133 (2001), 47–63. [7] A. R. Its, A. V. Kitaev, and A. S. Fokas, “An isomonodromy approach to the theory of two-dimensional quantum gravity,” Uspekhi Mat. Nauk 45 (1990) no. 6(276), pp. 135–136; translation in Russian Math. Surveys 45 (1990), no. 6, 155–157. [8] A. S. Fokas, A. R. Its, and A. V. Kitaev, “The isomonodromy approach to matrix models in 2D quantum gravity,” Commun. Math. Phys. 147 (1992), 395–430. [9] A. Gonchar and E. Rakhmanov, “The equilibrium measure and the distribution of zeros of extremal polynomials,” Mat. Sb. 125 (167) (1984), 177–127; English transl. in Math. USSR Sb. 53 (1986). [10] T. Kriecherbauer and K. T-R McLaughlin, “Strong asymptotics of polynomials orthogonal with respect to Freud weights,” Internat. Math. Res. Notices 1999, no. 6, (1999), 299-324. [11] A. B. J. Kuijlaars, “Riemann-Hilbert analysis for orthogonal polynomials,” Orthogonal Polynomials and Special Functions, Leuven 2002 (E. Koelink and W. Van Assche, eds.), Lecture Notes in Mathematics 1817, Springer, Berlin, 2003, pp. 167–210. [12] A. P. Magnus, “Toeplitz matrix techniques and convergence of complex weight Pad´e approximants,” J. Comput. Appl. Math. 19 (1987), 23–38. [13] J. Nuttall, “Pad´e polynomial asymptotics from a singular integral equation,” Constructive Approx. 6 (1990), 157–166 [14] J. Nuttall, “Asymptotics of diagonal Hermite-Pad´e polynomials,” J. Approx. Theory 42 (1984), 299–386 [15] E. Saff and V. Totik, “Logarithmic Potentials with External Fields,” Springer Verlag, 1997. [16] H. Stahl, “Three approaches to a proof of convergence of Pad´e approximants,” Rational Approximation and its Applications in Mathematics and Physics (J. Gilewicz et al., eds.), Lecture Notes in Mathematics 1237, Springer, Berlin, 1986, pp. 79–124. [17] S. Suetin, “On the uniform convergence of diagonal Pad´e approximants for hyperelliptic functions”, Mat. Sb. 191, No. 9 (2000), 81–114 (in Russian), English translation in Sbornik: Mathematics 191, No. 9 (2000), 1339–1373. ˝ , “Orthogonal Polynomials”, Amer. Math. Soc. Colloq. Publ. vol. 23, Providence, [18] G. Szego RI, 1975 (4th edition). [19] V. Totik, “Weighted Approximation With Varying Weights,” Lecture Notes in Mathematics 1569, Springer, New York, 1994.

32

Keldysh Institute of Applied Mathematics Russian Academy of Sciences Miusskaya Sq. 4 Moscow RUSSIA [email protected] Department of Mathematics Katholieke Universiteit Leuven Celestijnenlaan 200 B B-3001 Heverlee BELGIUM [email protected]

33