aa r X i v : . [ m a t h - ph ] M a r Riemann–Hilbert Problems
Percy Deift
Abstract.
These lectures introduce the method of nonlinear steepest descent forRiemann-Hilbert problems. This method finds use in studying asymptotics asso-ciated to a variety of special functions such as the Painlevé equations and orthog-onal polynomials, in solving the inverse scattering problem for certain integrablesystems, and in proving universality for certain classes of random matrix ensem-bles. These lectures highlight a few such applications.
Contents
Lecture 1 1Lecture 2 10Lecture 3 20Lecture 4 29
Lecture 1
These four lectures are an abridged version of 14 lectures that I gave at theCourant Institute on RHPs in 2015. These 14 lectures are freely available on theAMS website AMS Open Notes.Basic references for RHPs are [8, 12, 28]. Basic references for complex functiontheory are [19, 23, 24]. Many more specific references will be given as the courseproceeds.Special functions are important because they provide explicitly solvable models for a vast array of phenomena in mathematics and physics. By “special functions”I mean Bessel functions, Airy functions, Legendre functions, and so on. If youhave not yet met up with these functions, be assured, sooner or later, you surelywill.It works like this. Consider the
Airy equation (see, e.g. [1, 29])(1.1) y ′′ ( x ) = xy ( x ) , − ∞ < x < ∞ .Seek a solution of (1.1) in the form y ( x ) = Z Σ e xs f ( s ) ds ©0000 (copyright holder) Riemann–Hilbert Problems for some functions f ( x ) and some contours Σ in the complex plane C . We have y ′′ ( x ) = Z Σ s e xs f ( s ) ds and x y ( x ) = Z Σ (cid:18) dds e xs (cid:19) f ( s ) ds = − Z e xs f ′ ( s ) ds provided we can drop the boundary terms. In order to solve (1.1) we need tohave − f ′ ( s ) = s f and so f ( s ) = const. e − s .Thus y ( x ) = const. Z Σ e xs − s ds provides a solution of the Airy equation.The particular choice const. = πi and Σ in Figure 1.3 is known as Airy’s integral Ai ( x ) (1.2) Ai ( x ) = πi Z Σ e xz − z dz . ∞ e i π/ ∞ e − i π/ Σ = F igure Σ for Airy’s integral.Other contours provide other, independent solutions of Airy’s equation, suchas Bi ( x ) (see [1]). Now the basic fact of the matter is that the integral representation(1.2) for Ai ( x ) enables us, using the classical method of stationary phase/steepestdescent , to compute the asymptotics of Ai ( x ) as x → + ∞ and − ∞ with any desired ercy Deift 3 accuracy . We find, in particular [1, p. 448], that for ζ = x / (1.4) Ai ( x ) ∼ √ π x − e − ζ ∞ X k = (− ) k c k ζ − k as x → + ∞ , where c = c k = Γ (cid:16) k + (cid:17) k k ! Γ (cid:16) k + (cid:17) = ( k + )( k + ) . . . ( k − )( ) k k ! , k > Ai (− x ) ∼ √ π x − / (cid:18) sin (cid:16) ζ + π (cid:17) ∞ X (− ) k c k ζ − k − cos (cid:16) ζ + π (cid:17) ∞ X (− ) k c k + ζ − k − (cid:19) ,(1.5)as x → + ∞ .Such results for solutions of general 2 nd order equations are very rare. Formu-lae (1.4) and (1.5) solve the fundamental connection problem or scattering problem for solutions of the Airy equation. Thus, if we know that a solution y ( x ) of theAiry equation behaves like y ( x ) = √ π x − / e − ζ (cid:18) − c ζ + . . . (cid:19) as x → + ∞ , then we know precisely how it behaves as x → − ∞ , and vice versa,by (1.4) (1.5), see Figure 1.6. x +−x √ π (− x ) − / h sin (cid:16) ζ + π (cid:17) + O ( /ζ ) i √ π x − / e − ζ ( + . . . ) F igure Exercise 1.7.
Use the classical steepest-descent method to verify (1.4) and (1.5).There are similar precise results for all the classical special functions. The dili-gent student should regard Abramowitz & Stegun [1] as an exercise book for thesteepest descent method — verify all the asymptotic formulae!Now in recent years it has become clear that a new and extremely broad class ofproblems in mathematics, engineering and physics is described by a new class ofspecial functions, the so-called
Painlevé functions . There are six Painlevé equations
Riemann–Hilbert Problems and we will say more about them later on. Whereas the classical special functions,such as Airy functions, Bessel functions, etc. typically arise in linear (or linearizedproblems) such as acoustics or electromagnetism, the Painlevé equations arise innonlinear problems, and they are now recognized as forming the core of modernspecial function theory. Here are some examples of how Painlevé equations arise:
Example 1.8.
Consider solutions of the modified Korteweg–de Vries equation(MKdV) u t − u u x + u xxx = − ∞ < x < ∞ , t > u ( x , 0 ) = u ( x ) → | x | → ∞ .(1.9)Then [16] as t → ∞ , in the region | x | c t / , c < ∞ ,(1.10) u ( x , t ) = ( t ) / p (cid:18) x ( t ) / (cid:19) + O (cid:18) t / (cid:19) where p ( s ) is a particular solution of the Painlevé II (PII) equation p ′′ ( s ) = s p ( s ) + p ( s ) . Example 1.11.
Let π := ( π π . . . π N ) ∈ S N be a permutation of the numbers1, 2, . . . , N . We say that π i , π i , . . . π i k is an increasing subsequence of π of length k if i < i < · · · < i k and π i < π i < · · · < π i k .Thus if N = π = ( ) , then 125 and 136 are increasing subsequencesof π of length 3. Let ℓ N ( π ) denote the length of a longest increasing subsequenceof π , e.g., for N = π as above, ℓ ( π ) =
3, which is the length of the longestincreasing subsequences 125 and 136.Now equip S N with uniform measure. ThusProb ( ℓ N n ) = { π ∈ S N : ℓ N ( π ) n } N ! . Question.
How does ℓ N behave statistically as N , n → ∞ ? Theorem 1.12 ([2]) . Center and scale ℓ N as follows: ℓ N → X N = ℓ N − √ NN / then lim N → ∞ Prob ( X N x ) = e − R ∞ x ( s − x ) u ( s ) ds where u ( s ) is the (unique) solution of Painlevé II (the so-called Hastings-McLeod solu-tion) normalized such that u ( s ) ∼ Ai ( s ) as s → + ∞ .The distribution on the right in Theorem 1.12 is the famous Tracy-Widom dis-tribution for the largest eigenvalue of a GUE matrix in the edge scaling limit. ercy Deift 5 Theorem 1 is one of a very large number of probabilistic problems in combina-torics and related areas, whose solution is expressed in terms of
Random MatrixTheory (RMT) via Painlevé functions (see, e.g., [3]).The key question is the following: Can we describe the solutions of the Painlevéequations as precisely as we can describe the solutions of the classical specialfunctions such as Airy, Bessel, . . . ? In particular, can we describe the solutionsof the Painlevé equations asymptotically with arbitrary precision and solve theconnection/scattering problem as in (1.4) and (1.5) for the Airy equation (or anyother of the classical special functions):known behavior as x → + ∞ ⇒ known behavior as x → − ∞ and vice versa.As we have indicated, at the technical level, connection formulae such as (1.4)and (1.5) can be obtained because of the existence of an integral representationsuch as (1.2) for the solution. Once we have such a representation the asymptoticbehavior is obtained by applying the (classical) steepest descent method to theintegral. There are, however, no known integral representations for solutions ofthe Painlevé equations and we are led to the following questions: Question 1:
Is there an analog of an integral representation for solutions of thePainlevé equations?
Question 2:
Is there an analog of the classical steepest descent method whichwill enable us to extract precise asymptotic information about solutions of thePainlevé equations from this analog representation?The answer to both questions is yes : In place of an integral representation suchas (1.2), we have a
Riemann–Hilbert Problem (RHP) , and in place of the classicalsteepest descent method we have the nonlinear (or non-commutative) steepest descentmethod for RHPs (introduced by P. Deift and X. Zhou [16]).So what is a RHP? Let Σ be an oriented contour in the plane, see Figure 1.13. −+ +− +−+− Σ F igure Σ in the direction of the orientation,the ( ± ) -sides lie on the left (resp. right). Let v : Σ → GL ( k , C ) , the jump matrix , bean invertible k × k matrix function defined on Σ with v , v − ∈ L ∞ ( Σ ) . Riemann–Hilbert Problems
We say that an n × k matrix function m ( z ) is a solution of the RHP ( Σ , v ) if m ( z ) is analytic in C /Σ , m + ( z ) = m − ( z ) v ( z ) , z ∈ Σ ,where m ± ( z ) = lim z ′ → z ± m ( z ′ ) . +− zz ′ → z − z ′ → z + If, in addition, n = k and m ( z ) → I k as z → ∞ ,we say that m ( z ) solves the normalized RHP ( Σ , v ) .RHPs involve a lot of technical issues. In particular • How smooth should Σ be? • What measure theory/function spaces are suitable for RHPs? • What happens at points of self intersection (see Figure 1.14)?F igure • In what sense are the limits m ± ( z ) achieved? • In the case n = k , in what sense is the limit m ( z ) → I k achieved? • Does an n × k solution exist? • In the normalized case, is the solution unique?And most importantly • at the analytical level, what kind of problem is a RHP? As we will see, theproblem reduces to the analysis of singular integral equations on Σ .There is not enough time in these 4 lectures to address all these issues system-atically. Rather we will address specific issues as they arise.As an example of how things work, we now show how PII is related to a RHP(see, e.g. [22]). Let Σ denote the union of six rays Σ k = e i ( k − ) π/ ρ , ρ >
0, 1 k p , q , r be complex numbers satisfying the relation(1.15) p + q + r + pqr = v ( z ) , z ∈ Σ , be constant on each ray as indicated in Figure 1.16 and for fixed x ∈ C set v x ( z ) = e − iθ e iθ ! v ( z ) e iθ e − iθ ! , z ∈ Σ where θ = θ x ( z ) = z + xz . ercy Deift 7 Thus for z ∈ Σ v x ( z ) = r e − iθ ! and so on. Σ Σ Σ Σ Σ Σ Σ p ! p ! q ! r ! r ! q ! F igure fixed x , let m x ( z ) be the 2 × ( Σ , v x ) .Then u ( x ) = i ( m ( x )) is a solution of the PII equation where m x ( z ) = I + m ( x ) z + O (cid:18) z (cid:19) as z → ∞ . (This result is due to Jimbo and Miwa [27], and independently toFlaschka and Newell [20].) The asymptotic behavior of u ( x ) as x → ∞ is thenobtained from the RHP ( Σ , v x ) by the nonlinear steepest descent method.In the classical steepest descent method for integrals such as (1.2) above, thecontour Σ is deformed so that the integral passes through a stationary phase pointwhere the integrand is maximal and the main contribution to the integral thencomes from a neighborhood of this point. The nonlinear (or non-commutative)steepest descent method for RHPs involves the same basic ideas as in the clas-sical scalar case in that one deforms the RHP, Σ → Σ ′ , in such a way that theexponential terms (see e.g. e iθ above) in the RHP have maximal modulus atpoints of the deformed contour Σ ′ . The situation is far more complicated thanthe scalar integral case, however, as the problem involves matrices that do notcommute. In addition, terms of the form e − iθ also appear in the problem andmust be separated algebraically from terms involving e iθ , so that in the endthe terms involving e iθ and e − iθ both have maximal modulus along Σ ′ (see[16–18]). A simple example of the nonlinear steepest descent method is given atthe end of Lecture 4. Riemann–Hilbert Problems
One finds, in particular, ([18], and also [22, 25]) the following:Let − < q < p = − q , r =
0. Then as x → − ∞ ,(1.17) u ( x ) = √ ν (− x ) / cos (cid:18) (− x ) / − ν log (− x ) + φ (cid:19) + O (cid:18) log (− x )(− x ) / (cid:19) where(1.18) ν = ν ( q ) = − π log (cid:16) − q (cid:17) and(1.19) φ = − ν log 2 + arg Γ ( iν ) + π ( q ) − π x → + ∞ (1.20) u ( x ) = q Ai ( x ) + O e − / x / x / ! .These asymptotics should be compared with (1.4), (1.5) for the Airy function.Note from (1.4) that as x → + ∞ Ai ( x ) ∼ x − / e − / x / .Also observe that PII u ′′ ( x ) = x u ( x ) + u ( x ) is a clearly a nonlinearization of the Airy equation u ′′ ( x ) = x u ( x ) and so we expect similar solutions when the nonlinear term 2 u ( x ) is small.Also note that (1.17) and (1.18) solve the connection problem for PII. If weknow the behavior of the solutions u ( x ) of PII as x → + ∞ , then we certainlyknow q from (1.20). But then we know ν = ν ( q ) and φ = φ ( q ) in (1.18) and (1.19)and hence we know the asymptotics of u ( x ) as x → − ∞ from (1.17). Conversely,if we know the asymptotics of u ( x ) as x → − ∞ , we certainly know ν > q from (1.18), q = − e − π ν . But then again from(1.17), we know φ , and hence sgn ( q ) from (1.19). Thus we know q , and hencethe asymptotics of the solution u ( x ) as x → + ∞ from (1.20). Finally note thesimilarity of the multiplier(1.21) e x z − z for the Airy equation with the multiplier(1.22) e iθ = e i ( x z + z ) in the RHP for PII. Setting z → i z in (1.21) e x z − z → e i ( x z + z ) which agrees with (1.22) up to appropriate scalings.Also note from (1.15) that PII is parameterized by parameters lying on a 2-dimvariety: this corresponds to the fact that PII is second order. ercy Deift 9 The fortunate and remarkable fact is that the class of problems in physics,mathematics, and engineering expressible in terms of a RHP is very broad andgrowing. Here is one more, with more to come!The RHP for the MKdV equation (1.9) is as follows (see e.g., [16]): Let Σ = R ,oriented from − ∞ to + ∞ . For fixed x , t ∈ R let(1.23) v x , t ( z ) = − | r ( z ) | − r ( z ) e − iτ r ( z ) e iτ ! , z ∈ R where τ = τ x , t ( z ) = xz + tz and r = r ( z ) is a given function in L ∞ ( R ) ∩ L ( R ) with k r k ∞ < r ( z ) = − r (− z ) , z ∈ R .There is a bijection from the initial data u ( x , t = ) = u ( x ) for MKdV onto suchfunctions r ( z ) — see later. The function r ( z ) is called the reflection coefficient for u , see (4.13).Let m = m x , t ( z ) be the solution of the normalized RHP ( Σ , v x , t ) . Then(1.24) u ( x , t ) = ( m ( x , t )) ,is the solution of MKdV with initial condition u ( x , t = ) = u ( x ) correspondingto r ( z ) . Here m x , t ( z ) = I + m ( x , t ) z + O (cid:18) z (cid:19) as z → ∞ . Σ = R →→ r = ! r =
00 1 ! F igure ( Σ , v x , t ) in the region | x | c t / . In this case PIIemerges as the RHP ( Σ , v x , t ) is “deformed” into the RHP ( Σ , v x ) in Figure 1.16.As we will see, RHPs are useful not only for asymptotics, but also they can beused to determine symmetries and formulae/identities/equations, and also foranalytical purposes. Lecture 2
We now consider some of technical issues that arise for RHPs, which are listedwith bullet points above.A key role in RH theory is played by the
Cauchy operator . We first consider thecase when Σ = R . Here the Cauchy operator C = C R is given by C f ( z ) = Z R f ( s ) s − z ¯ ds , z ∈ C / R , ¯ ds = ds πi for suitable functions f on R (General refs for the case Σ = R , and also when Σ = {| z | = } , are [19] and [23].) Assume first that f ∈ S ( R ) , the Schwartz space offunctions on R . Let z = x + i ǫ , x ∈ R , ǫ >
0. Then
C f ( x + i ǫ ) = Z R f ( s ) s − x − i ǫ ¯ ds = Z R f ( s ) i ǫ ( s − x ) + ǫ ¯ ds + Z R f ( s ) s − x ( s − x ) + ǫ ¯ ds = Z R f ( s ) π ǫ ( s − x ) + ǫ ds + πi Z R f ( s ) s − x ( s − x ) + ǫ ds := I + II .Now I = I ǫ = Z R f ( x + ǫ u ) π ( u + ) du .Then, by dominated convergence,lim ǫ ↓ I ǫ = f ( x ) π Z R duu + = f ( x ) .Write II = II ǫ = II <ǫ + II >ǫ where II <ǫ = πi Z | x − s | <ǫ f ( s ) s − x ( s − x ) + ǫ ds and II >ǫ = πi Z | x − s | >ǫ f ( s ) s − x ( s − x ) + ǫ ds .As s − x ( s − x ) + ǫ is an odd function about s = x , II <ǫ can be written as II <ǫ = πi Z | s − x | <ǫ s − x ( s − x ) + ǫ ( f ( s ) − f ( x )) ds and so | II <ǫ | π k f ′ k L ∞ Z | s − x | <ǫ ( s − x ) ( s − x ) + ǫ ds k f ′ k L ∞ ǫ π ercy Deift 11 which goes to 0 as ǫ ↓
0. Finally II >ǫ = πi Z | s − x | >ǫ (cid:18) s − x ( s − x ) + ǫ − s − x (cid:19) f ( s ) ds + πi Z | s − x | >ǫ s − x f ( s ) ds = III ǫ + IV ǫ .We have | III ǫ | = π (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) Z | s − x | >ǫ ǫ ( s − x ) + ǫ f ( s ) s − x ds (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) = π (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) Z | u | > u + f ( x + ǫ u ) u du (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) and so as ǫ ↓
0, again by dominated convergence, | III ǫ | → π | f ( x ) | (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) Z | u | > duu ( u + ) (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) = Σ = R and f ∈ S ( R ) C + f ( x ) ≡ lim ǫ ↓ Cf ( x + i ǫ ) = f ( x ) + i Hf ( x ) where Hf ( x ) = lim ǫ ↓ π Z | s − x | >ǫ f ( s ) x − s ds . Hf ( x ) is called the Hilbert transform of f . Note that1 π Z | s − x | >ǫ f ( s ) x − s ds = π Z | s − x | > f ( s ) x − s ds + π Z ǫ< | s − x | < f ( s ) − f ( x ) x − s ds which converges to1 π Z | s − x | > f ( s ) x − s ds + π Z | s − x | < f ( s ) − f ( x ) x − s ds as ǫ ↓
0, so that lim ǫ ↓ π R | s − x | >ǫ f ( s ) x − s ds indeed exists pointwise for f ∈ S .Similarly one finds C − f ( x ) ≡ lim ǫ ↓ Cf ( x − iǫ ) = − f ( x ) + i Hf ( x ) , x ∈ R and we obtain the fundamental relations for f ∈ S (2.1) C + f − C − f = f and C + f + C − f = iHf . Exercise 2.2.
Show that the limits C ± f ( x ) = lim ǫ ↓ Cf ( x ± iǫ ) are in fact non-tangential limits i.e. C + f ( x ) = lim z ′ → x Cf ( z ′ ) where z ′ lies in a cone of arbitraryopening angle α < π (see Figure 2.3), and similarly for C − f ( x ) (see refs. [5, 6, 8]). x z ′ α F igure α .A critical property of the singular integral operator H , and hence the operators C ± , is that, as we now show, H is a bounded operator from L p ( R ) → L p ( R ) forall 1 < p < ∞ . To prove the result for L , recall that the Fourier transformˆ f ( z ) = F f ( z ) ≡ lim R → ∞ Z R − R e − izt f ( t ) dt √ π , z ∈ R and the inverse Fourier transformˇ f ( x ) = F − f ( x ) ≡ lim R → ∞ Z R − R e izt f ( t ) dt √ π , x ∈ R are unitary maps k f k L = k ˆ f k L = k ˇ f k L from L onto L . Moreover F and F − are indeed inverse to each other, (cid:0) ˆ f (cid:1) ∨ = f = (cid:0) ˇ f (cid:1) ∧ , f ∈ L ( R ) .For f ∈ S ( R ) , fix ǫ >
0, and set ( C ǫ f ) ( x ) ≡ Cf ( x + iǫ ) .Then ( F C ǫ f ) ( z ) = lim R → ∞ Z R − R (cid:18) Z R f ( s ) s − x − iǫ ¯ ds (cid:19) e − ixz dx √ π = lim R → ∞ Z R f ( s ) √ π Z R − R e − ixz s − x − iǫ ¯ dx ! ds ,(2.4)by Fubini’s Theorem. ercy Deift 13 Now for s fixed and R large, and z > Z R − R e − ixz s − x − iǫ ¯ dx = − Z R − R e − ixz x − ( s − iǫ ) ¯ dx = Z R − R s − iǫ e − ixz x − ( s − iǫ ) ¯ dx − Z R − R e − ixz x − ( s − iǫ ) ¯ dx = e − i ( s − iǫ ) z − Z R − R e − ixz x − ( s − iǫ ) ¯ dx . Exercise 2.5.
Show that, for z >
0, we havelim R → ∞ Z R − R e − ixz x − ( s − iǫ ) ¯ dx = s fixed and z > R → ∞ Z R − R e − ixz s − x − iǫ ¯ dx = e − isz e − ǫz .But we also have Exercise 2.6.
For z > Z R − R e − ixz s − x − iǫ ¯ dx is bounded in s uniformly for R > R → ∞ in (2.4) in the s -integral and so for z > F C ǫ f ( z ) = Z R f ( s ) e − isz e − ǫz ds √ π = e − ǫz F f ( z ) . Exercise 2.7.
Show, by a similar argument, that F C ǫ f ( z ) = z < C ǫ f = F − (cid:0) χ > ( · ) e − ǫ · (cid:1) F f where χ > ( z ) e − ǫz = (cid:14) e − ǫz for z > z < S ( R ) is dense in L , and as F − ( χ > ( · ) e − ǫ · ) F is clearly bounded in L it follows that C ǫ f extends to a bounded operator on L .Moreover ˆ C f ( x ) ≡ F − χ > ( · ) F f is clearly also a bounded operator in L and for f ∈ L k C ǫ f − ˆ C f k L = k F − χ > ( · ) (cid:0) e − ǫ · − (cid:1) F f k L = k χ > ( · ) (cid:0) e − ǫ · − (cid:1) F f k L which converges to 0 as ǫ ↓
0, again by dominated convergence. In other words,for f ∈ L , C f ( x + iǫ ) = Z R f ( s ) s − x − iǫ ¯ ds → ˆ C f ( x ) in L ( dx ) .In particular, it follows by general measure theory, that for some sequence ǫ n ↓ C f ( x + i ǫ n ) → ˆ C f ( x ) (2.8)pointwise a.e. In particular (2.8) holds for f ∈ S ( R ) . But then by our previouscalculations, C f ( x + i ǫ n ) converges pointwise for all x , and we conclude that for f ∈ S and a.e. xC + f ( x ) = f ( x ) + i H f ( x ) = ˆ C f ( x ) a.e. x .Thus C + f and, hence H f , extend to bounded operators on L ( R ) and12 ˆ f + i F H f = F ˆ C f = χ > ˆ f and so F H f ( z ) = i (cid:18) χ > ( z ) − (cid:19) ˆ f ( z )= − i sgn ( z ) ˆ f ( z ) where sgn ( z ) = + z > ( z ) = − z < f ∈ L , C + f = f + i H f = f + (cid:0) ˆ f sgn ( · ) (cid:1) ∨ and similarly C − f = − f + i H f = − f + (cid:0) ˆ f sgn ( · ) (cid:1) ∨ . C R = R F igure C ± , and hence H , arebounded in L p ( R ) , for all 1 < p < ∞ . Consider first the case p =
4. Suppose f ∈ C ∞ ( R ) , the infinitely differentiable functions with compact support. Then as z → ∞ , C f ( z ) = Z R f ( s ) s − z ¯ ds = O (cid:18) z (cid:19) ercy Deift 15 and C f ( z ) is continuous down to the axis. By Cauchy’s theorem Z C R ( C f ( z )) dz = C R is given in Figure 2.9, and as Z R − R ( C f ( z )) dz → R → ∞ ,we conclude that Z R (cid:0) C + f ( x ) (cid:1) dx = C + f = f + i H f we obtain(2.10) 0 = Z R (cid:16) f + f ( Hf ) i + f ( Hf ) i + f ( Hf ) i + ( Hf ) i (cid:17) dx .Now suppose that f is real. Then Hf is real and the real part of (2.10) yields0 = Z R (cid:16) f − f ( Hf ) + ( Hf ) (cid:17) dx ,hence Z R ( Hf ) dx = Z R f ( Hf ) dx − Z R f dx (cid:18) Z c f dx + Z R c ( Hf ) dx (cid:19) − Z R f dx for any c >
0. Take c =
6. Then12 Z R ( Hf ) dx ( − ) Z R f dx or Z R ( Hf ) dx Z R f dx .The case when f is complex valued is handled by taking real and imaginary parts.Thus, by density, H maps L boundedly to L . Exercise 2.11.
Show that H maps L p → L p for all 1 < p < ∞ . Hints:(1) Show that the above argument works for all even integers p .(2) Show that the result follows for all p > < p < zz ′ Σ Σ Σ Σ F igure Exercise 2.13.
Show that H is not bounded from L → L . (However H maps L → weak L .) As indicated in Lecture 1, RHPs take place on contours whichself-intersect (see Figure 2.12).We will need to know, for example, that if f is supported on Σ , say, and weconsider Cf ( z ′ ) = Z Σ f ( z ) z − z ′ ¯ dz for z ′ ∈ Σ , say, then Cf ( z ′ ) ∈ L ( Σ ) if f ∈ L ( Σ ) . Here is a prototype resultwhich one can prove using the Mellin transform, which we recall is the Fouriertransform for the multiplicative group { x > } . We have [5, p. 88] the following:For f ∈ L ( ∞ ) and r >
0, set C θ f ( r ) = Z ∞ f ( s ) s − ˆ z r ¯ ds , ˆ z = e iθ where 0 < θ < π . Then(2.14) k C θ f k L ( dr ) c θ k f k L ( ds ) where c θ = γ γ ( − γ ) − γ , γ = θ π .One can also show that for any 1 < p < ∞ k C θ f k L p ( dr ) C θ , p k f k L p ( ds ) for some c θ , p < ∞ .Results such as (2.14) are useful in many ways. For example, we have thefollowing result. Theorem 2.15.
Suppose f ∈ H ( R ) = { f ∈ L ( R ) : f ′ ∈ L ( R ) } . Then C f ( z ) isuniformly Hölder- in C + and in C − . In particular, Cf is continuous down to the axisin C + and in C − .Proof. For z ∈ C \ R ddz Cf ( z ) = Z R f ( s )( s − z ) ¯ ds = − Z R (cid:18) dds (cid:18) s − z (cid:19)(cid:19) f ( s ) ¯ ds = Z R f ′ ( s ) s − z ¯ ds .Now suppose z ′ , z ′′ ∈ C + , and the straight line L through z ′ , z ′′ intersects theline R at x at an angle θ as in Figure 2.16. x z ′ z ′′ θ L F igure R . ercy Deift 17 Then as R R f ′ ( s ) s − z ¯ ds = R ∞ x f ′ ( s ) s − z ¯ ds + R x − ∞ f ′ ( s ) s − z ¯ ds , and as f ′ ∈ L (− ∞ , x ) ⊕ L ( x , ∞ ) it follows from (2.14) that Z ∞ (cid:12)(cid:12)(cid:12)(cid:12) ddz Cf (cid:16) re iθ (cid:17)(cid:12)(cid:12)(cid:12)(cid:12) dr = Z ∞ (cid:12)(cid:12)(cid:12) C f ′ ( re iθ ) (cid:12)(cid:12)(cid:12) dr c k f ′ k L .But (cid:12)(cid:12) Cf ( z ′′ ) − Cf ( z ′ ) (cid:12)(cid:12) = (cid:12)(cid:12)(cid:12)(cid:12) Z z ′ → z ′′ in L ddz Cf ( z ) dz (cid:12)(cid:12)(cid:12)(cid:12) (cid:12)(cid:12) z ′′ − z ′ (cid:12)(cid:12) k ddz Cf k L ( ∞ ) c | z ′′ − z ′ | k f ′ k L ( R ) . (cid:3) We now consider general contours Σ ⊂ C = C ∪ { ∞ } , which are composedcurves : By definition a composed curve Σ is a finite union of arcs { Σ i } ni = whichcan intersect only at their end points. Each arc Σ i is homeomorphic to an interval [ a i , b i ] ⊂ R : (cid:14) ϕ i : [ a i , b i ] → Σ i ⊂ C , [ a i , b i ] ∋ t → ϕ i ( t ) ∈ Σ i , ϕ i ( a i ) = ϕ i ( b i ) .Here C has the natural topology generated by the sets {| z | < R } , {| z | > R } where R , R >
0. A loop, in particular the unit circle T = {| z | = } , is a composed curveon the understanding that it is a union of (at least) two arcs.Although it is possible, and sometimes useful, to consider other function spaces(e.g. Hölder continuous functions), we will only consider RHPs in the sense of L p ( Σ ) for 1 < p < ∞ .So the first question is “What is L p ( Σ ) ?”. The natural measure theory foreach arc Σ i is generated by arc length measure µ as follows. If z = ϕ ( t ) and z n = ϕ ( t n ) are the end-points of some arc Σ ⊂ C , and z , z , . . . , z n is anypartition of [ z , z n ] = { ϕ ( t ) : t t t n } (we assume z i + succeeds z i in theordering induced on Σ by ϕ , symbolically z i < z i + , etc. ) then L = L [ z , z n ] ≡ sup all partitions { z i } n − X i = | z i + − z i | .If L < ∞ we say that the arc Σ = [ z , z n ] is rectifiable and L [ z , z n ] is its arc length .We will only consider composed curves Σ that are locally rectifiable i.e. for any R > Σ ∩ {| z | < R } is rectifiable (note that the latter set is an at most countableunion of simple arcs and rectifiability of the set means that the sum of the arclengths of these arcs is finite. In particular, the unit circle T as a union of 2rectifiable subarcs, is rectifiable, and R is locally rectifiable.) For any interval [ α , β ) on Σ i ⊂ C (the case where Σ i passes through ∞ , must be treated separately— exercise!) define µ i ([ α , β )) = arc length α → β .Now the sets { [ α , β ) : α < β on Σ i } form a semi-algebra (see [30]) and hence µ i can be extended to a complete measure on a σ -algebra A containing the Borel setson Σ i . The restriction of the measure to the Borel sets is unique. For 1 p < ∞ , we can define L p ( Σ i , dµ i ) to be the set of f measurable with respect to A on Σ forwhich, Z Σ i | f ( z ) | p dµ i ( z ) < ∞ ,and then all the “usual” properties go through. One usually writes dµ = | dz | . For Σ = S ni = Σ i , L p ( Σ , dµ ) is simply the direct sum of L p ( Σ i , dµ i ) ni = . Exercise 2.17. | dz | is also equal to Hausdorff-1 measure on Σ .Note that if Σ = R and Σ = (cid:10) (cid:16) x , x sin x (cid:17) : x ∈ R (cid:11) then Σ = Σ ∪ Σ is not acomposed curve, although Σ and Σ are both locally rectifiable.For Σ as above we define the Cauchy operator for h ∈ L p ( Σ , | dz | ) , 1 p < ∞ ,by(2.18) Cf ( z ) = C Σ f ( z ) = Z Σ f ( ζ ) ζ − z ¯ dζ , z ∈ C \ Σ .Given the homeomorphisms ϕ i : [ a i , b i ] → Σ i , the contour Σ carries a naturalorientation, and the integral here is a line integral following the orientation; if weparametrize the arcs Σ i in Σ by arc length s ,0 s s i , ζ = ζ ( s ) , then (cid:12)(cid:12)(cid:12)(cid:12) dζ ( s ) ds ( s ) (cid:12)(cid:12)(cid:12)(cid:12) = Σ i of integrals of the form Z s i f ( ζ ( s )) ζ ( s ) − z dζ ( s ) ds ¯ ds , z ∈ C \ Σ for each i , the integrand (clearly) lies in L p ( ds : [ s i )) .Now the fact of the matter is that many of the properties that were true for C Σ when Σ = R , go through for C Σ in the general situation. (See, in particular, [24].)In particular for f ∈ L p ( Σ , dµ ) , the non-tangential limits(2.19) C ± Σ f ( z ) = lim z ′ → z ± C Σ f ( z ′ ) exist pointwise a.e. on Σ . Figure 2.20 demonstrates non-tangential limits. +− ˆ z +− z z ′ → zz ′ → ˆ z − z ′ → z F igure Σ i is locally rectifiable, the tangent vector to the arc dζds exists ata.e. point ζ = ζ ( s ) : the normal to dζds bisects the cone. ercy Deift 19 The normal ξ F igure C ± Σ f ( z ) = ± f ( z ) + i Hf ( z ) where the Hilbert transform is now given by(2.22) Hf ( z ) = π lim ǫ ↓ Z | s − z | >ǫs ∈ Σ f ( s ) z − s ds , z ∈ Σ and the points z ∈ Σ for which the non-tangential limits (2.19) exists are preciselythe points for which the limit in (2.22) exists.Again, for f ∈ L p ( Σ , dµ ) with 1 p < ∞ , C + f ( z ) − C − f ( z ) = f ( z ) and C + f ( z ) + C − f ( z ) = i Hh ( z ) .The following issue is crucial for the analysis of RHPs: Question.
For which locally rectifiable contours Σ are the operators C ± and H bounded in L p , 1 < p < ∞ ?Quite remarkably, it turns out that there are necessary and sufficient conditionson a simple rectifiable curve for C ± , H to be bounded in L p ( Σ ) , 1 < p < ∞ .The result is due to many authors, starting with Calderón [7], and then Coifman,Meyer and McIntosh [9], with Guy David [10] (see [6] for details and historicalreferences) making the final decisive contribution.Let Σ be a simple, rectifiable curve in C . For any z ∈ Σ , and any r >
0, let ℓ r ( z ) = arc length of ( Σ ∩ D r ( z )) where D r ( z ) is the ball of the radius r centered at z , see Figure 2.23. zD r ( z ) Σ F igure D r ( z ) of radius r centered at z . Set λ = λ Σ = sup z ∈ Σ , r> ℓ r ( z ) r . Theorem 2.24.
Suppose λ Σ < ∞ . Then for any < p < ∞ , the limit in (2.22) existsfor a.e. z ∈ Σ and defines a bounded operator for any < p < ∞ (2.25) k H f k L p c p k f k L p , f ∈ L p , c p < ∞ . Conversely if the limit in (2.22) exists a.e. and defines a bounded operator H in L p ( Σ ) forsome < p < ∞ , then H gives rise to a bounded operator for all p , < p < ∞ , and λ Σ < ∞ . An excellent reference for the above Theorem, and more, is [6].
Remarks.
Additional remarks:(1) Locally rectifiable curves Σ for which λ = λ Σ < ∞ are called Carlesoncurves,(2) the constant c p in (2.25) has the form c p = φ p ( λ Σ ) for some continuous,increasing function, φ p ( t ) > independent of Σ , such that φ p ( ) = φ p is independent of Σ , is very important for the nonlinear steep-est descent method, where one deforms curves in a similar way to the classicalsteepest descent method for integrals.Carleson curves are sometimes called AD-regular curves: the A and D denoteAhlfors and David. To get some sense of the subtlety of the above result, considerthe following curve Σ with a cusp at the origin (see Figure 2.26): Σ = { x y = } ∪ { ( x , x ) : x } . (
1, 0 ) (
1, 1 ) Σ F igure λ Σ < ∞ so that the Hilbert transform H Σ is bounded in L p , 1 < p < ∞ . Exercise 2.27.
For Σ in Figure 2.26, prove directly that H Σ is bounded in L . Thepresence of the cusp makes the proof surprisingly difficult. Lecture 3
We now make the notion of a RHP precise (see [8,17,28]). Let Σ be a composite,oriented Carleson contour in C and let v : Σ → GL ( n , C ) be a jump matrix on Σ , ercy Deift 21 with v , v − ∈ L ∞ ( Σ ) . Let Ch ( z ) = C Σ h ( z ) , C ± Σ h , H Σ h be the associated Cauchyand Hilbert operators.We say that a pair of L p ( Σ ) function f ± ∈ ∂C ( L p ) if there exists a (unique)function h ∈ L p ( Σ ) such that f ± ( z ) = ( C ± h )( z ) , z ∈ Σ .In turn we call f ( z ) ≡ Ch ( z ) , z ∈ C \ Σ , the extension of f ± = C ± h ∈ ∂C ( L p ) off Σ . Definition 3.1.
Fix < p < ∞ . Given Σ , v and a measurable function f on Σ , we saythat m ± ∈ f + ∂C ( L p ) solves an inhomogeneous RHP of the first kind ( IRHP p ) if m + ( z ) = m − ( z ) v ( z ) , z ∈ Σ . Definition 3.2.
Fix < p < ∞ . Given Σ , v and a function F ∈ L p ( Σ ) , we say that M ± ∈ ∂ C ( L p ) solves an inhomogeneous RHP of the second kind ( IRHP p ) if M + ( z ) = M − ( z ) v ( z ) + F ( z ) , z ∈ Σ .Recall that m solves the normalized RHP ( Σ , v ) if, at least formally, • m ( z ) is a n × n analytic function in C \ Σ , • m + ( z ) = m − ( z ) v ( z ) , z ∈ Σ , • m ( z ) → I as z → ∞ .(3.3)More precisely, we make the following definition. Definition 3.4.
Fix < p < ∞ . We say that m ± solves the normalized RHP ( Σ , v ) p if m ± solves the IRHP p with f ≡ I . In the above definition, if m ± − I ∈ C ± h , then clearly the extension m ( z ) = I + Ch ( z ) , z ∈ C \ Σ ,off Σ solves the normalized RHP in the formal sense of (3.3). ≡ IRHP2 p IRHP1 p ↑ ↑ Invertibilityof 1 − C ω Useful for de-formationsof RHP in C F igure IRHP p and IRHP p .Let v = (cid:0) v − (cid:1) − v + = (cid:0) I − ω − (cid:1) − (cid:0) I + ω + (cid:1) ω + ≡ v + − I , ω − ≡ I − v − ,be a pointwise a.e. factorization of v , i.e., v ( x ) = ( v − ( x )) − v + ( x ) for a.e. x , with v ± , ( v ± ) − ∈ L ∞ , and let ω = ( ω − , ω + ) . Let C ω denote the basic associated operator C ω h ≡ C + ( h ω − ) + C − ( h ω + ) acting on L p ( Σ ) − n × n matrix valued functions h . As ω ± ∈ L ∞ , C ω ∈ L ( L p ) ,the bounded operators on L p , for all 1 < p < ∞ . The utility of IRHP p and IRHP p will soon become clear, see Figure 3.5. Theorem 3.6. If f and v are such that f ( v − I ) ∈ L p ( Σ ) for some < p < ∞ , then m ± = M ± + f solves IRHP p if M ± solves IRPH p with F = f ( v − I ) . Conversely if F ∈ L p ( Σ ) , then M + = m + + F , M − = m − solves IRHP p if m ± solves IRHP p with f = C − F . The first part of this result is straightforward: Suppose M ± ∈ ∂C ( L p ) solves M + = M − v + F on Σ with F = f ( v − I ) ∈ L p . Then M + = M − v + f ( v − I ) = ( M − + f ) v − f or m + = m − v with m ± = f + M ± ∈ f + ∂C ( L p ) . The converse is more subtle andis left as an exercise: Exercise 3.7.
Show
IRHP p ⇒ IRHP p .We now show that the RHPs IRHP p and IRHP p , and, in particular, the nor-malized RHP ( Σ , v ) p are intimately connected with the singular integral operator1 − C ω .Let f ∈ L p ( Σ ) and let m ± = f + C ± h for some h ∈ L p ( Σ ) . Also suppose m + = m − v = m − ( v − ) − v + . Set µ = m − (cid:0) v − (cid:1) − = m + (cid:0) v + (cid:1) − ∈ L p ( Σ ) and define H ( z ) = (cid:0) C (cid:0) µ (cid:0) ω + + ω − (cid:1)(cid:1)(cid:1) ( z ) , z ∈ C \ Σ .Then we have on Σ , using C + − C − = H + = C + (cid:0) µ (cid:0) ω + + ω − (cid:1)(cid:1) = C + µ ω + + C + µ ω − = C + µ ω − + C − µ ω + + µ ω + = C ω µ + µ ω + = ( C ω − ) µ + µ (cid:0) I + ω + (cid:1) = ( C ω − ) µ + µ v + = ( C ω − ) µ + m + .i.e. H + = ( C ω − I ) µ + m + .Similarly H − = ( C ω − ) µ + m − .Thus(3.8) m ± − f − H ± = ( − C ω ) µ − f . ercy Deift 23 But m ± − f − H ± ∈ ∂C ( L p ) ; i.e. m ± − f − H ± = C ± h for some h ∈ L p . However,from (3.8) C + h = C − h ⇒ h = C + h − C − h = ( − C ω ) µ = f , µ ∈ L p .Conversely, if µ ∈ L p ( Σ ) solves ( − C ω ) µ = f , then the above calculationsshow that H ≡ C ( µ ( ω + + ω − )) satisfies H ± = − f + µ v ± .Thus setting m ± = µ v ± , we see that m + = m − v and m ± − f ∈ ∂ C ( L p ) . In par-ticular µ ∈ L p solves ( − C ω ) µ = m ± = µ v ± solves the homogeneous RHP.(3.9) m + = m − v , m ± ∈ ∂C ( L p ) .We summarize the above calculations as follows: Proposition 3.10.
Let < p < ∞ . Then ( − C ω ) is a bijection in L p ( Σ ) ⇐⇒ IRHP p has a unique solution for all f ∈ L p ( Σ ) ⇐⇒ IRHP p has a unique solution for all F ∈ L p ( Σ ) . Moreover, if one, and hence all three of the above conditions, is satisfied, then for all f ∈ L p ( Σ ) (3.11) ( − C ω ) − f = m + ( v + ) − = m − ( v − ) − = ( M + + f )( v + ) − = ( M − + f )( v − ) − where m ± solves IRHP p with the given f and M ± solves IRHP p with F = f ( v − I ) ( ∈ L p !), and if M ± solves IRHP p with F ∈ L p ( Σ ) , then M + = (cid:16) ( − C ω ) − ( C − F ) (cid:17) v + + F and M − = (cid:16) ( − C ω ) − ( C − F ) (cid:17) v − .Finally, if f ∈ L ∞ ( Σ ) and v ± − I ∈ L p ( Σ ) , then (3.11) remains valid providedwe interpret(3.12) ( − C ω ) − f ≡ f + ( − C ω ) − C ω f .This is true, in particular, for the normalized RHP ( Σ , v ) p where f ≡ I . Remark.
If 1 − C ω is invertible, for one choice of v ± , then ( exercise ) it is invertiblefor all choices of v ± such that v = ( v − ) − v + , v ± , ( v ± ) − ∈ L ∞ ( Σ ) .Note that if we take v + = v , v − = I , in particular, then C ω h = C − ( h ( v − I )) . The above Proposition implies, in particular, that if µ ∈ I + L p solves(3.13) ( − C ω ) µ = I in the sense of (3.12) i.e. µ = I + ν , ν ∈ L p (3.14) ( − C ω ) ν = C ω I = C + ω − + C − ω + ∈ L p then m ± = µ v ± solves the normalized RHP ( Σ , v ) p . It is in this precise sense thatthe solution of the normalized RHP is equivalent to the solution of a singular integralequation (3.13) , (3.14) on Σ .One very important consequence of the proof of Proposition 3.10 is given bythe following Corollary 3.15.
Let f ∈ L p ( Σ ) .Let m ± solve IRHP p with the given f and let M ± solve IRHP p with F = f ( v − I ) .Then k ( − C ω ) − f k p c k m ± k p (3.16) and k ( − C ω ) − f k p c ′ ( k M ± k p + k f k p ) (3.17) for some constants c = c p , c ′ = c ′ p . In particular if we know, or can show, that k m ± k p const k f k p , or k M ± k p const k f k p , then we can conclude from (3.16) or (3.17) that ( − C ω ) − is bounded in L p with a corresponding bound. Conversely if weknow that ( − C ω ) − exists, then the above calculations show that k m ± k p ˜ c k f k p and k M ± k p ˜˜ c k f k p for corresponding constants ˜ c , ˜˜ c . Finally we consider uniqueness for the solution of the normalized RHP ( Σ , v ) p as given in Definition 3.4. Observe first that if F ( z ) = ( Cf )( z ) for f ∈ L p ( Σ ) and G ( z ) = ( Cg )( z ) for g ∈ L q ( Σ ) , r = p + q
1, 1 < p , q < ∞ , then a simplecomputation shows that(3.18) FG ( z ) = Ch ( z ) where(3.19) h ( s ) = − i ( g ( s )( Hf )( s ) + f ( s )( Hg )( s )) where again H f ( s ) = Hilbert transform = lim ǫ ↓ R | s ′ − s | >ǫ f ( s ′ ) s − s ′ ds ′ π , and simi-larly for Hg ( s ) . As h clearly lies in L r ( Σ ) , r >
1, it follows that F + G + ( z ) − F − G − ( z ) = h ( z ) for a.e. z ∈ Σ .( Note: C + h ( z ) − C − h ( z ) = h ( z ) even if h is in L , even though C ± is not boundedin L .) Theorem 3.20.
Fix < p < ∞ . Suppose m ± solves the normalized RHP ( Σ , v ) p .Suppose that m − ± exists a.e. on Σ and m − ± ∈ I + ∂ C ( L q ) , < q < ∞ , r = p + q .Then the solution of the normalized RHP ( Σ , v ) p is unique. ercy Deift 25 Proof.
Suppose ˆ m ± = I + C ± ˆ h , ˆ h ∈ L p ( Σ ) is a 2 nd solution of the normalizedRHP. We have, by assumption, m − ± = I + C ± k for some k ∈ L q ( Σ ) . (It is an Exercise to show that I + ( Ck )( z ) , the extension of m − ± to C \ Σ , is in fact m ( z ) − .).Then arguing as aboveˆ m ± m − ± − I = ( ˆ m ± − I ) (cid:16) m − ± − I (cid:17) + ( ˆ m ± − I ) + (cid:16) m − ± − I (cid:17) = C ± h for some h ∈ L r ( Σ ) + L p ( Σ ) + L q ( Σ ) .Hence ˆ m + m − + − ˆ m − m − − = h .But ˆ m + m − + = ( ˆ m − v ) ( m − v ) − = ˆ m − m − − and so h =
0. Thus ˆ m ± m − ± − I = m ± = m ± . (cid:3) Theorem 3.21. If n = p = and det v ( z ) = a.e. on Σ , then the solution of thenormalized RHP ( Σ , v ) is unique.Proof. Because n = p =
2, (3.18), (3.19) = ⇒ ( det m ( z )) ± = + C ± h , where h ∈ L ( Σ ) + L ( Σ ) and so ( det m ) + − ( det m ) − = h ( z ) a.e. But det m + = det m − as det v =
1, and so h ≡
0. But then det m ( z ) ± =
1. Hence, if m ± = m ± m ± m ± m ± ! we have m − ± = m ± − m ± − m ± m ± ! and so clearly m − ± ∈ I + ∂ C ( L ) . The result now follows from Theorem 3.20. (cid:3) These results immediately imply that the normalized RHP ( Σ = R , v x , t ) forMKdV with v x , t given by (1.23) has a unique solution in L ( R ) . Indeed, factorize v x , t ( z ) = ( v − x , t ) − v + x , t = (cid:0) I − w − x , t (cid:1) − (cid:0) I + w + x , t (cid:1) = − ¯ r e − iτ ! re iτ ! so that w x , t = (cid:0) w − x , t , w + x , t (cid:1) = − ¯ re − i τ ! , re i τ !! But for Σ = R , we have Exercise 3.22.
Both C + and − C − are orthogonal projections in L ( R ) and so k C ± k L = Using the Hilbert-Schmidt matrix norm k M k = (cid:16) P i , j | M ij | (cid:17) , we have k C ω x , t h k L = (cid:13)(cid:13)(cid:13)(cid:13)(cid:13) C + h h h h ! − ¯ r e − iτ ! + C − h h h h ! re iτ !(cid:13)(cid:13)(cid:13)(cid:13)(cid:13) L = (cid:13)(cid:13)(cid:13)(cid:13)(cid:13) C + h (− ¯ r ) e iτ C + h (− ¯ r ) e − iτ ! + C − h re iτ C − h re iτ !(cid:13)(cid:13)(cid:13)(cid:13)(cid:13) = k C − h re iτ k L + k C − h re iτ k L + k C + h (− ¯ r ) e − iτ k L + k C + h (− ¯ r ) e − iτ k L k r k ∞ (cid:16) k h k L + k h k L + k h k L + k h k L (cid:17) = k r ∞ k k h k L and so, as k r k ∞ < k C ω x , t k < x , t ∈ R , (cid:0) − C ω x , t (cid:1) − exists in L ( R ) and (cid:13)(cid:13)(cid:13)(cid:0) − C ω x , t (cid:1) − (cid:13)(cid:13)(cid:13) L − k r k ∞ < ∞ and the proof of the existence and uniqueness for ( Σ , v x , t ) follows from Proposi-tion 3.10. On the other hand, just uniqueness alone follows from Theorem 3.21 asdet v ( z ) ≡ R .Now it turns out that a key role in the theory of RHPs is played by Fredholmoperators . Recall that a bounded linear operator T from a Banach space X to aBanach space Y is Fredholm if dim ker
T < ∞ and dim coker T < ∞ i.e. Y /ran T is a finite dimensional space.If T is Fredholm, we define index T ≡ dim ker T − dim coker T . Exercise 3.23. If T : X → Y is Fredholm, then ran T is closed in Y . Exercise 3.24. T : X → Y is Fredholm iff it has a pseudo-inverse S ∈ L ( Y , X ) suchthat ST = X + K and T S = Y + L where K is a compact operator in L ( X ) and L is a compact operator in L ( Y ) . ercy Deift 27 We know that a normalized RHP ( Σ , v ) p , say, has a (unique) solution if ( − C ω ) − exists. The situation where we know, for example, that k C ω k L <
1, asin the example ( Σ = R , v x , t ) above so that ( − C ω ) − exists, is very rare. Forexample, for the KdV equation on R u t + uu x + u xxx = u ( x , 0 ) = u ( x ) → | x | → ∞ ,the associated RHP is exactly the same as ( R , v x , t ) for MKdV, except that now,generically,(3.25) | r ( z ) | < | z | > | r ( ) | = k r k ∞ = • Prove 1 − C ω is Fredholm. • Prove ind ( − C ω ) = • Prove dim ker ( − C ω ) = − C ω is a bijection, and hence the normalized RHP ( Σ , v ) has a unique solution.Let’s see how this goes for KdV with normalized RHP ( Σ = R , v x , t ) , but now r satisfies (3.25), (3.26). By our previous comments (see Remark above), it isenough to consider the special case v + = v , v − = I so that ω + = v − I and ω − =
0. Thus C ω h = C − h ( v − I ) .We assume r ( z ) is continuous and r ( z ) → | z | → ∞ . Let S be the operator Sh = C − h (cid:16) v − − I (cid:17) .Then C ω Sh = C − ( Sh ( v − I ))= C − h(cid:16) C − h (cid:16) v − − I (cid:17)(cid:17) ( v − I ) i = C − h(cid:16) C + h (cid:16) v − − I (cid:17)(cid:17) ( v − I ) i − C − h h (cid:16) v − − I (cid:17) ( v − I ) i as C + − C − =
1. But h (cid:0) v − − I (cid:1) ( v − I ) = h (cid:0) I − v − v − (cid:1) = h ( I − v ) + h (cid:0) I − v − (cid:1) .Thus C ω Sh = C − h(cid:16) C + h (cid:16) v − − I (cid:17)(cid:17) ( v − I ) i + C − (cid:16) h (cid:16) v − − I (cid:17)(cid:17) + C − ( h ( v − I )) = C − h(cid:16) C + h (cid:16) v − − I (cid:17)(cid:17) ( v − I ) i + C ω h + Sh and we see that ( − C ω )( − S ) h = h + C − h(cid:16) C + h (cid:16) v − − I (cid:17)(cid:17) ( v − I ) i .But Exercise 3.27.
K h = C − (cid:2)(cid:0) C + h (cid:0) v − − I (cid:1)(cid:1) ( v − I ) (cid:3) is compact in L ( R ) .Hint: v − I is a continuous function which → | z | → ∞ and hence can beapproximated in L ∞ ( R ) by finite linear combinations of functions of the form a/ ( z − z ′ ) for suitable constants a and points z ′ ∈ C \ R . Then use the followingfact: Exercise 3.28. If T n , n > L ( X , Y ) and k T n − T k → n → ∞ for some operator T ∈ L ( X , Y ) , then T is compact.Similarly ( − S )( − C ω ) = + L , L compact.Thus ( − C ω ) is Fredholm.Now we use the following fact: Exercise 3.29.
Suppose that for γ ∈ [
0, 1 ] , T ( γ ) is a norm-continuous family ofFredholm operators. Then for γ ∈ [
0, 1 ] ,ind T ( γ ) = const = ind T ( ) = ind T ( ) .Apply this fact to C ω ( γ ) , where we replace r by γr in v x , t , v x , t , γ = − γ | r | − γ ¯ re − iτ γ re iτ ! .The proof above shows that C ω ( γ ) is a norm continuous family of Fredholmoperators and so ind ( − C ω ) = ind (cid:16) − C ω ( γ = ) (cid:17) = ind (cid:16) − C ω ( γ = ) (cid:17) = C ω ( γ = ) = ( − C ω ) µ = m + = µv and m − = µ solve m + = m − v , m ± ∈ ∂C ( L ) .Consider P ( z ) = m ( z ) ( m ( ¯ z )) ∗ for z ∈ C + where m ( z ) is the extension of m ± off R i.e. if m ± = C ± h , h ∈ L , then m ( z ) = ( Ch )( z ) . Then for a contour Γ R , ǫ ,pictured in Figure 3.30, R Γ R , ǫ P ( z ) dz = P ( z ) is analytic.0 R ǫ ↓↑ F igure ǫ above the real axis. ercy Deift 29 Letting ǫ ↓ R → ∞ , we obtain ( exercise ) R ∞ − ∞ P + ( z ) dz =
0; i.e.0 = Z R m + ( z ) m − ( z ) ∗ dz = Z R m − ( z ) v ( z ) m − ( z ) ∗ dz .Taking adjoints and adding, we find0 = Z R m − ( z ) ( v + v ∗ ) ( z ) m − ( z ) ∗ dz .But a direct calculation shows that v + v ∗ is diagonal and ( v + v ∗ ) ( z ) = − | r ( z ) |
00 1 ! .Now since | r ( z ) | < z = m − ( z ) =
0. But µ = m − and so we see that ker ( − C ω ) = { } .The result of the above chain of arguments is that the solution of the normal-ized RHP ( Σ , v x , t ) for KdV exists and is unique. Such Fredholm arguments havewide applicability in Riemann–Hilbert Theory [22].One last general remark. The scalar case n = m + = m − v , then itfollows that ( log m ) + = ( log m − ) + log v and hence log m ( z ) is given by Plemelj’sformula, which provides the general solution of additive RHPs, vialog m = C ( log v ) ( z ) = Z Σ log v ( s ) s − z ¯ ds and so(3.31) m ( z ) = exp (cid:18) Z Σ log v ( s ) s − z ¯ ds (cid:19) a formula which is easily checked directly. However, there is a hidden subtlety inthe business: On R , say, although v ( s ) may go rapidly to 0 as s → ± ∞ , v ( s ) maywind around 0 and so log v ( s ) may not be integrable at both ± ∞ . Thus there isa topological obstacle to the existence of a solution of the RHP. If n >
1, there aremany more such “hidden” obstacles.
Lecture 4
RHP’s arise in many difference ways. For example, consider orthogonal poly-nomials: we are given a measure µ on R with finite moments, Z R | x | m dµ ( x ) < ∞ for m =
0, 1, 2, . . .Performing Gram-Schmidt on 1, x , x , . . . with respect to dµ ( x ) , we obtain (monic)orthogonal polynomials π n ( x ) = x n + . . . , , n > Z R π n ( x ) π m ( x ) dµu ( x ) = n = m , n , m > (Here we assume that dµ has infinite support: otherwise there are only a finitenumber of such polynomials.)Associated with the π n ’s are the orthonormal polynomials(4.1) P n ( x ) = γ n π n ( x ) , γ n > n > Z R P n ( x ) P m ( x ) dµ ( x ) = δ n , m , n , m > Z R H n ( x ) H m ( x ) e − x dx = n = m , n , m > H n ( x ) = n ! Z C ω − n − e xω − ω dω where C is a (small) circle enclosing the origin, (Note: the H n ’s are not monic,but are proportional to the π n ’s, H n ( x ) = c n π n ( x ) where the c n ’s are explicit)and the asymptotic behavior of the H n ’s follow from the classical steepest de-scent method. For general weights, however, no such integral representations areknown.The Hermite polynomials play a key role in random matrix theory in the so-called Gaussian Unitary, Orthogonal and Symplectic Ensembles. However it waslong surmised that local properties of random matrix ensembles were universal ,i.e., independent of the underlying weights. In other words if one considers gen-eral weights such as e − x dx , e −( x + x ) dx , etc., instead of the weight e − x dx for the Hermite polynomials, the local properties of the random matrices, at thetechnical level, boil down to analyzing the asymptotics of the polynomials orthog-onal with respect to the weights e − x dx , e −( x + x ) dx , etc., for which no integralrepresentations are known. What to do?It turns out however, that orthogonal polynomials with respect to an arbitrary weight can be expressed in terms of a RHP. Suppose dµ ( x ) = ω ( x ) dx , for some ω ( x ) > Z R | x | m ω ( x ) dx < ∞ , m =
0, 1, 2, . . . .and suppose for simplicity that(4.2) ω ∈ H ( R ) = { f ∈ L : f ′ ∈ L } . ercy Deift 31 Fix n > Y ( n ) = { Y ( n ) ij ( z ) } i , j solve the RHP (cid:0) Σ = R , v = (cid:0) ω (cid:1)(cid:1) normalized so that Y ( n ) ( z ) z − n z n ! → I as z → ∞ . Exercise 4.3.
Show that we then have (see e.g. [12]) Y ( n ) ( z ) = π n ( z ) C ( π n ω )− πi γ n − π n − ( z ) C (cid:0) − π iγ n − π n − ω (cid:1)! where C = C R is the Cauchy operator on R , π n , π n − are the monic orthogonalpolynomials with respect to ω ( x ) dx and γ n − is the normalization coefficientfor π n − as in (4.1). (Note that by (4.2) and Theorem 2.15, Y ( n ) ( z ) is continuousdown to the axis for all z .) This discovery is due to Fokas, Its and Kitaev [21].Moreover this is just exactly the kind of problem to which the nonlinear steepestdescent method can be applied to obtain ([14, 15]) the asymptotics of the π n ’swith comparable precision to the classical cases, Hermite, Legendre, . . . , and soprove universality for unitary ensembles (and later, Deift and Gioev, Shcherbina,for Orthogonal & Symplectic Ensembles of random matrices, see [13] and thereferences therein).As mentioned earlier, RHPs are useful not only for asymptotic analysis, butalso to analyze analytical and algebraic issues. Here we show how RHPs giverise to difference equations, or differential equations, in other situations.Consider the solution Y ( n ) for the orthogonal polynomial RHP (cid:0) R , v = (cid:0) ω (cid:1)(cid:1) .The key fact is that the jump matrix (cid:0) ω (cid:1) is independent of n : the dependence on n is only in the boundary condition Y ( n ) z − n z + n ! → I .So we have Y ( n + )+ = Y ( n + )− v and Y ( n )+ = Y ( n )− v .Let R ( z ) = Y ( n + ) ( z ) (cid:16) Y ( n ) ( z ) (cid:17) − , z ∈ C \ R . Then R + ( z ) = Y ( n + )+ ( z ) (cid:16) Y ( n )+ ( z ) (cid:17) − = (cid:16) Y ( n + )− ( z ) v ( z ) (cid:17) (cid:16) Y ( n )− ( z ) v ( z ) (cid:17) − = Y ( n + )− ( z ) (cid:16) v ( z ) v ( z ) − (cid:17) (cid:16) Y ( n )− ( z ) (cid:17) − = R − ( z ) .Hence R ( z ) has no jump across R and so, by an application of Morera’s Theorem, R ( z ) is in fact entire. But as z → ∞ R ( z ) = " Y ( n + ) ( z ) z − n − z n + ! z z − ! " Y ( n ) ( z ) z − n z n ! − = (cid:18) I + O (cid:18) z (cid:19)(cid:19) z z − ! (cid:18) I + O (cid:18) z (cid:19)(cid:19) = O ( z ) .Thus R ( z ) must be a polynomial of order 1, Y ( n + ) ( z ) (cid:16) Y ( n ) ( z ) (cid:17) − = R ( z ) = Az + B for suitable A and B , or,(4.4) Y ( n + ) ( z ) = ( Az + B ) Y ( n ) ( z ) which is a difference equation for orthogonal polynomials with respect to a fixedweight. Exercise 4.5.
Make the argument leading to (4.4) rigorous (why does (cid:16) Y ( n ) (cid:17) − exist, etc.) Exercise 4.6.
Show that (4.4) implies the familiar three term recurrence relationfor orthogonal polynomials p n ( z ) b n p n + ( z ) + ( a n − z ) p n ( z ) + b n − p n − = n > a n ∈ R , b n > b − ≡ " ∂ x − izσ + q ( x ) ¯ q ( x ) !! ψ = − ∞ < x < ∞ (see e.g. [17]). Here z ∈ C , σ = (cid:0) − (cid:1) and q ( x ) → | x | → ∞ . Equation (4.7) is intimately connected with the defocusingNonlinear Schrödinger Equation (NLS) by virtue of the fact that the operator(4.8) L = ( iσ ) − ∂ x − q ¯ q !! undergoes an isospectral deformation if q = q ( t ) = q ( x , t ) solves NLS(4.9) iq t + q xx − | q | q = q ( x , t = ) = q ( x ) .In other words, if q = q ( t ) solves NLS then the spectrum of L ( t ) = ( iσ ) − (cid:16) ∂ x − (cid:16) q ( x , t ) q ( x , t ) (cid:17)(cid:17) is constant: Thus the spectrum of L ( t ) provides constants of the motion for (4.9),and so NLS is “integrable”. The key fact is that there is a RHP naturally associ-ated with L which expresses the integrability of NLS in a form that is useful for ercy Deift 33 analysis. Here we follow Beals and Coifman, see [4]. Let q ( x ) in (4.8) be givenwith q ( x ) → | x | → ∞ sufficiently rapidly. Then for any z ∈ C \ R , Exercise 4.10.
The equation ( L − z ) ψ = unique solution ψ such that ψ ( x , z ) e − ixzσ → I as x → − ∞ and is bounded x → ∞ . Such ψ ( x , z ) arecalled Beals-Coifman solutions.
Remark 4.11.
These solutions have the following properties:(1) For fixed x , ψ ( x , z ) is analytic in C \ R , and is continuous down to the axis.That is ψ ± ( x , z ) = lim ǫ ↓ ψ ( x , z ± i ǫ ) exist for all x , z ∈ R .(2) For fixed x , ψ ( x , z ) e − ixzσ → I as z → ∞ ,(4.12) ψ ( x , z ) e − ixzσ = I + m ( x ) z + O (cid:18) z (cid:19) , as z → ∞ for some matrix residue term m ( x ) .Now clearly ψ ± ( x , z ) , z ∈ R , are two fundamental solutions of ( L − z ) ψ = z ∈ R , ψ + ( x , z ) = ψ − ( x , z ) v ( z ) for all x ∈ R , where v ( z ) is independent of x . In other words, by (1) of Remark 4.11, ψ ( x , · ) solves a RHP ( Σ = R , v ) , normalized as in (4.12). In this way differentialequations give rise to RHPs in a systematic way.One can calculate ( exercise ) the precise form of v ( z ) and one finds v ( z ) = − | r ( z ) | r ( z )− r ( z ) ! , z ∈ R where, again (cf. (1.23) for MKdV) we have for r , the reflection coefficient , k r k ∞ < q r = R ( q ) is a bijection between suitable spaces: r = R ( q ) , the direct map, is constructedfrom q via the solutions ψ ( x , z ) as above. The inverse map r R − ( r ) = q isconstructed by solving the RHP ( Σ , v ) normalized by (4.12) for any fixed x . Oneobtains ψ ( x , z ) e − izxσ = I + m ( x ; r ) z + O (cid:18) z (cid:19) as z → ∞ and q ( x ) = − i ( m ( x , r )) (cf (1.24) for MKdV).Now if q = q ( t ) = q ( x , t ) solves NLS then r ( t ) = R ( q ( t )) evolves simply, r ( t ) = r ( t , z ) = r ( t = z ) e − itz , z ∈ R i.e. t → q ( t ) → r ( t ) → log r ( t ) = log r ( t = ) − itz linearizes NLS. This leads tothe following formula for the solution of NLS with initial data q (4.14) q ( t ) = R − (cid:16) e − it ( · ) R ( q )( · ) (cid:17) .The effectiveness of this representation, which one should view as the RHP analogof NLS of the integral representation (1.2) for the Airy equation, depends on theeffectiveness of the nonlinear steepest descent method for RHPs. Question.
Where in the representation (4.14) is the information encoded that q ( t ) solves NLS?The answer is as follows. Let ψ ( x , z , t ) be the solution of the RHP with jumpmatrix v t ( z ) = − | r | re − itz − ¯ re itz ! normalized as in (4.12). Set H ( x , z , t ) = ψ ( x , z , t ) e − itz σ and observe that(4.15) H + = H − − | r | r − ¯ r ! = H − v for which the jump matrix is independent of x and t . This means that we can differen-tiate (4.15) with respect to x and t , H x + = H x − v , H t + = H t − v and conclude, asin the case of orthogonal polynomials, that H x H − and H t H − are entire, andevaluating these combinations as z → ∞ , we obtain two equations H x = D H , H t = E H for suitable polynomials matrix functions D and E . These functions constitute thefamous Lax pair ( D , E ) for NLS. Compatibility of these two equations requires ∂ t ∂ x H = ∂ x ∂ t H = ⇒ ∂ t ( D H ) = ∂ x ( E H ) = ⇒ D t H + D E H = E x H + E D H = ⇒ D t + [ D , E ] = E x which reduces directly to NLS. In this way RHP’s lead to difference and differen-tial equations.Another systematic way that RHP’s arise is through the distinguished class ofso-called integrable operators . Let Σ be an oriented contour in C and let f , . . . , f n and g , . . . , g n be bounded measurable functions on Σ . We say that an operator K acting on L p ( Σ ) , 1 < p < ∞ , is integrable if it has a kernel of the form K ( z , z ′ ) = Σ ni = f i ( z ) g i ( z ′ ) z − z ′ , z , z ′ ∈ Σ ; z = z ′ for such L ∞ functions f i , g j , ( K h )( z ) = Z Σ K ( z , z ′ ) h ( z ′ ) dz ′ . ercy Deift 35 Integrable operators were first singled out as a distinguished class of operatorsby Sakhnovich [31] in the late 1960’s, and their theory was developed fully by Its,Izergin, Korepin and Slavnov [26] in the early 1990’s (see [11] for a full discussion).The famous sine kernel of random matrix theory K x ( z , z ′ ) = sin x ( z − z ′ ) π ( z − z ′ ) = e ixz e − ixz ′ + (cid:16) − e ixz ′ (cid:17) · e ixz i π ( z − z ′ ) is a prime example of such an operator, as is likewise the well-known Airy kerneloperator.Integrable operators form an algebra, but their most remarkable property is thattheir inverses can be expressed in terms of the solution of a naturally associatedRHP. Indeed, let m ( z ) be the solution of the normalized RHP ( Σ , v ) where(4.16) v ( z ) = I − πif g T , f = ( f , . . . , f n ) T , g = ( g , . . . , g n ) T .(Here we assume for simplicity that Σ ni = f i ( z ) g i ( z ) =
0, for all z ∈ Σ as in thesine-kernel: otherwise (4.16) must be slightly modified).Then ( − K ) − has the form 1 + L where L is an integrable operator L ( z , z ′ ) = Σ ni = F i ( z ) G i ( z ′ ) z − z ′ , z , z ′ ∈ Σ , z = z ′ and(4.17) (cid:14) F = ( F , . . . , F n ) T = m ± fG = ( G , . . . , G n ) T = ( m − ± ) T g .This means that if, for example, K depends on parameters, as in the case of thesine kernel, asymptotic problems involving K as the parameters become large, areconverted into asymptotic problems for a RHP, to which the nonlinear steepestdescent method can be applied.As an example, we show how to use RHP methods to give a proof of Szeg˝o’scelebrated Strong Limit Theorem. Let T be the unit circle. Theorem 4.18 (Szeg˝o Strong Limit Theorem) . Let ϕ ( z ) = e L ( z ) ∈ L ( T ) , ϕ ( z ) > ,where P ∞ k = k | L k | < ∞ and { L k } are Fourier coefficients of L ( z ) . Let D n be the Toeplitzdeterminant generated by ϕ , D n ( ϕ ) = det X ( ϕ ) where X ( ϕ ) is the ( n + ) × ( n + ) matrix with entries { ϕ i − j } i , j n , and { ϕ k } are the Fourier coefficients of ϕ . Then as n → ∞ , D n = e ( n + ) L + Σ ∞ k = k | L k | ( + o ( )) . Sketch of proof.
Let e k , 0 k n , be the standard basis in C n + . Then the map U n : e k → z k , 0 k n , z ∈ T takes C n + onto the trigonometric polynomials P n = (cid:10)P nj = a j z j (cid:11) of degree n and induces a map τ n : P n → P n which is conjugate to X ( ϕ ) . We then calculate τ n z k = U n X U − n z k = U n X e k = U n (cid:16) n X j = ϕ j − k e j (cid:17) = n X j = ϕ j − k z j , 0 k n .(4.19)Now for any p = P nk = a k z k ∈ P n ( τ n p ) ( z ) = n X k = a k n X j = ϕ j − k z j = n X k = a k n X j = (cid:18) Z Γ ( z ′ ) k − j − ϕ ( z ′ ) ¯ dz ′ (cid:19) z j = n X k = a k Z Γ ( z ′ ) k − ϕ ( z ′ ) ( z/z ′ ) n + − ( z/z ′ ) − dz ′ = Z Γ ϕ ( z ′ ) p ( z ′ ) ( z/z ′ ) n + − ( z − z ′ ) ¯ dz ′ .After some simple calculations ( Exercise ) one finds that(4.20) τ n p = ( − K n ) p , p ∈ P n where K n is the integrable operator on T with kernel of the form(4.21) K n (cid:0) z , z ′ (cid:1) = f ( z ) g ( z ′ ) + f ( z ) g ( z ′ ) z − z ′ , z , z ′ ∈ Γ where f = ( f , f ) T = (cid:16) z n + , 1 (cid:17) T g = ( g , g ) T = (cid:18) z − n − − ϕ ( z ) πi , − ( − ϕ ( z )) πi (cid:19) T .(4.22)We have, in particular, from (4.19) and (4.20), for 0 k n , ( − K n ) z k = n X j = ϕ j − k z j and for k < k > n one easily shows that ( − K n ) z k = z k + n X j = ϕ j − k z i .Thus K n is finite rank, and hence trace class, and ( − K n ) has block form withrespect to the orthonormal basis { z k } ∞ − ∞ for L ( Γ ) as given in Figure 4.23. And so D n = det τ n = det X ( ϕ ) = det ( − K n ) ercy Deift 37 I · · · τ n · · · I F igure − K n in the basis { z k } ∞ − ∞ .Associated with the integrable operator K n we have the normalized RHP ( Σ = Γ , v ) where, by (4.16), (4.22)(4.24) v = I − πi fg T = ϕ −( ϕ − ) z n + z − n − ( ϕ − ) − ϕ ! on T . Now log D n = log det ( − K n )= tr log ( − K n )= Z ddt tr log ( − t K n ) dt = − Z tr (cid:18) − t K n K n (cid:19) dt .(4.25)For 0 t
1, set ϕ t ( z ) = ( − t ) + t ϕ ( z ) , z ∈ T .Clearly ϕ t ( z ) > ϕ ( z ) = ϕ ( z ) = ϕ ( z ) . Now ϕ t − = t ( ϕ − ) and so we have from (4.21) t K n = K t , n = h(cid:16) ( z/z ′ ) n + − (cid:17) / ( z − z ′ ) i (cid:2)(cid:0) − ϕ t ( z ′ ) (cid:1) / πi (cid:3) and it follows that in (4.25) 11 − t K n t K n = − K t , n K t , n = − K t , n − = R t , n where R t , n (cid:0) z , z ′ (cid:1) = P j = F t , j ( z ) G t , j ( z ′ ) z − z ′ where by (4.17)(4.26) F t = (cid:0) F t ,1 , F t ,2 (cid:1) T = m t ± f t , G t = (cid:0) G t ,1 , G t ,2 (cid:1) T = (cid:16) m − t ± (cid:17) T g t .Here m t ± refers to the solution of the RHP ( T , v t ) where v t involves ϕ t ratherthan ϕ in (4.24), and similarly for f t , g t . Hence (
Exercise )(4.27) log D n = − Z Z T X j = F ′ t , j ( z ) G t , j ( z ) dz dtt .So we see that in order to evaluate D n as n → ∞ we must evaluate the asymp-totics of the solution m t of the normalized RHP ( T , v t ) as n → ∞ , for each0 t
1, and substitute this information into (4.27) using (4.26). This is preciselywhat can be accomplished [11] using the nonlinear steepest descent method.Here we present the nonlinear steepest descent analysis in the case when ϕ ( z ) is analytic in an annulus A ǫ = { z : − ǫ < | z | < + ǫ } , ǫ > T . The idea of the proof, which is a common feature of all applicationsof the nonlinear steepest descent method, is to move the z n + term (or its analogin the general situation) in v t into | z | < z − n − term into | z | >
1: then as n → ∞ , these terms are exponentially small, and can be neglected.But first we must separate the z n + and z − n − terms of v t algebraically. Thisis done using the lower-upper pointwise factorization of v t (4.28) v t = z − n − (cid:16) − ϕ − t (cid:17) ϕ t ϕ − t ! − (cid:16) − ϕ − t (cid:17) z n + which is easily verified.Extend T = Σ → ˜ Σ = (cid:8) | z | = ρ } ∪ Σ ∪ {| z | = ρ − (cid:9) = Σ ρ ∪ Σ ∪ Σ ρ − where wechoose 1 − ǫ < ρ < < ρ − < + ǫ . Now define a piecewise analytic function ˜ m by the definitions in Figure 4.29. ˜ m = m Σ p Σ Σ p − ˜ m = m ˜ m = m −( − φ t ) z n + ! − ˜ m = m z − n − ( − φ − t ) ! F igure m .This definition is motivated by the fact that m + = m − v t = m − ( · )( · )( · ) ercy Deift 39 as in (4.28). It follows that ˜ m ( z ) solves the normalized RHP (cid:0) ˜ Σ , ˜ v (cid:1) where˜ v ( z ) = z − n − (cid:16) − ϕ − t (cid:17) on Σ ρ − ,˜ v ( z ) = ϕ t ( z ) ϕ t ( z ) − ! on Σ ,˜ v ( z ) = − (cid:16) − ϕ − t (cid:17) z n + on Σ ρ .Now as n → ∞ , ˜ v ( z ) → I on Σ ρ and on Σ ρ − . This means that ˜ m → m ∞ where m ∞ solves the normalized RHP ( Σ , v ∞ ) where v ∞ = v (cid:12)(cid:12) Σ = ϕ t ϕ − t ! .But this RHP is a direct sum of scalar RHP’s and hence can be solved explicitly,as noted earlier (cf. (3.31)). In this way we obtain the asymptotics of m as n → ∞ and hence the asymptotics of the Toeplitz determinant D n . (cid:3) Here is what, alas, I have not done and what I had hoped to do in these lectures(see AMS open notes): • Show that in addition to the usefulness of RHP’s for algebraic and asymp-totic purposes, RHP’s are also useful for analytic purposes. In particular,RHP’s can be used to show that the Painlevé equations indeed have thePainlevé property. • Show that in addition to RHP’s arising “out of the blue” as in the case oforthogonal polynomials and systematically in the case of ODE’s and alsointegrable operators, RHP’s also arise in a systematic fashion in Wiener–Hopf Theory. • Describe what happens to an RHP when the operator 1 − C ω is Fredholm,but not bijective, and • Finally, I have not succeeded in showing you how the nonlinear steepestdescent method works in general. All I have shown is one simple case.
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