Long-Time Asymptotics of the Toda Lattice for Decaying Initial Data Revisited

LONG-TIME ASYMPTOTICS OF THE TOD A LA TTICE F OR DECA YING INITIAL D A T A REVISITED HELGE KR ¨ UGER AND GERALD TESCH L Abstract. The p urp ose o f this article is to give a streamlined and self- con tained treatmen t of the long-time asymptotics of the T o da lattice f or de- ca ying initial data i n the sol i ton and in the sim ilarity region via the method of nonlinear steepest descen t. 1. Introduction The simplest mo del of a so lid is a chain of par ticles with nearest neighbor inter- action. The Hamiltonian of such a system is g iven b y (1.1) H ( p, q ) = X n ∈ Z  p ( n, t ) 2 2 + V ( q ( n + 1 , t ) − q ( n, t ))  , where q ( n, t ) is the displacement of the n -th particle from its equilibrium p osition, p ( n, t ) is its momentum (mass m = 1), and V ( r ) is the int eractio n po ten tial. Restricting the atten tion to ﬁnitely many par ticles (e.g ., b y impos ing p erio dic bo undary conditions ) and to the harmonic in teraction V ( r ) = r 2 2 , the equa tions o f motion form a linear system of diﬀerent ial equations with co nstant coe ﬃcie nts. The solution is then g iv en by a sup erp o sition of the ass ocia ted normal m o des . Around 1950 it was genera lly believed that a generic nonlinea r pe rturbation would yield to thermalization . That is, for any initial co ndition the energy should even tually be equally distributed ov er all normal mo des. In 1955 Enr ico F ermi, J ohn Pasta, and Stanislaw Ulam carried out a seemingly inno cent computer exp eriment at Los Alamos, [15], to in v estigate the ra te of a pproach to the equipartition of ene r gy . How ev er, m uch to everybo dy’s surpris e , the exper imen t indicated, instead of the exp ected thermalizatio n, a q ua si-p e rio dic mo tion of the system! Many a ttempts were ma de to explain this result but it was no t until ten y ears later that Martin Krusk al and Norman Zabusk y , [43], r ev ealed the connectio ns with solitons (see [2] for further his to rical information and a p edagog ical discussio n). This had a big impa ct on soliton mathematics and led to an explosive g rowth in the last deca des. In par ticular, it le d to the search for a potential V ( r ) for which the a bove system has so liton solutions. By co nsidering addition formulas for elliptic functions , Morik azu T o da came up with the choice V ( r ) = e − r + r − 1. The corres p onding sy s tem is now known a s the T o da e quation, [4 0]. 2000 Mathematics Subje ct Classiﬁc ation. Primary 37K40, 37K45; Secondary 35Q15, 37K10. Key wor ds and phr ases. Riemann–Hilbert problem, T oda lattice, solitons. Researc h supported by the Austri an Science F und (FWF) under Gran t No. Y330. Rev. Math. Phys. 21:1 , 6 1–109 (2009). 1 2 H. KR ¨ UGER AND G. TESCHL The equation of mo tion in this case rea ds explicitly d dt p ( n, t ) = − ∂ H ( p, q ) ∂ q ( n, t ) = e − ( q ( n, t ) − q ( n − 1 ,t )) − e − ( q ( n +1 , t ) − q ( n,t )) , d dt q ( n, t ) = ∂ H ( p, q ) ∂ p ( n, t ) = p ( n, t ) . (1.2) The imp ortant prope rt y of the T o da equatio n is the existence of s o called soliton solutions, that is, pulslike wa ves which spread in time without changing their size or shap e and interact with ea c h other in a particle - lik e wa y . This is a surprising phenomenon, since for a gener ic linear equation one would expect spreading of wa ves (disper sion) and for a generic nonlinear for ce one would ex pect that s olutions only exist for a ﬁnite time (breaking of w av es). Ob viously our particular force is such that b oth phenomena canc e l each o ther giving rise to a stable w av e existing for all time! In fact, in the simplest case o f one soliton, you can easily verify that this solution is g iv en b y (1.3) q 1 ( n, t ) = q + + log 1 + γ 1 − e − 2 κ exp( − 2 κn + 2 σ sinh( κ ) t ) 1 + γ 1 − e − 2 κ exp( − 2 κ ( n + 1) + 2 σ sinh( κ ) t ) ! , with κ, γ > 0 and σ ∈ { ± 1 } . It describ es a single bump trav eling through the - 20 0 20 40 0.0 0.5 1.0 1.5 2.0 Figure 1. One so liton q 1 ( n, 0) with κ = 1, γ = 1, and q 0 = 0 . crystal with sp eed σ sinh( κ ) /κ and width pro por tional to 1 /κ . In other words, the smaller the soliton the faster it propagates . It r esults in a total displacement 2 κ of the crystal. How ev er, this is just the tip of the iceb erg and can b e g eneralized to the N -soliton solution (1.4) q N ( n, t ) = q + + log  det( I + C N ( n, t )) det( I + C N ( n + 1 , t ))  , where (1.5) C N ( n, t ) = p γ i ( n, t ) γ j ( n, t ) 1 − e − ( κ i + κ j ) ! 1 ≤ i,j ≤ N , γ j ( n, t ) = γ j e − 2 κ j n − 2 σ j sinh( κ j ) t , with κ j , γ j > 0 and σ j ∈ {± 1 } . The case N = 1 coincides with the one so lito n solution from ab ov e and asymptotica lly , as t → ∞ , the N - s oliton solution can be written a s a s um of one-s oliton solutions. Historically such solitary waves were ﬁr st obs erved by the nav al ar c hitect Jo hn Scott Russel [35], who follo wed the b ow wav e of a bar ge which moved along a LONG-TIME ASYMPTOTICS OF THE TODA LA TTICE 3 - 200 - 100 0 100 200 - 1 0 1 Figure 2. Numerically computed solution q ( n, 1 50) of the T o da lattice, with initial condition all particles at r est in their equilib- rium p ositions except for the one in the middle which is displace d by 1. channel maintaining its sp eed and size (see the review ar ticle [3 3] for further infor- mation). The impo rtance o f these so litary wav es is that they co nstitute the stable part of the solutions ar ising from arbitrary s hort range initia l conditions and ca n b e used to explain the qua s i-per io dic be haviour found by F ermi, Pasta, and Ulam. In fact, the class ical result discov ered by Zabusky and K rusk al [43] s tates tha t every ”short range” initial condition even tually splits into a num b er of stable solitons and a decaying background radiatio n co mponent. This is illustrated in Figure 2 which shows the numerically computed solutio n q ( n, t ) corresp onding to the initial condition q ( n, 0) = δ 0 ,n , p ( n, 0) = 0 a t some larg e time t = 130. Y ou ca n see the soliton r egion | n t | > 1 w ith t wo single s o liton o n the very left resp ectively right and the similar it y r e g ion | n t | < 1 where there is a contin uous displa cemen t plus some small oscillations whic h decay like t − 1 / 2 and are asymptotically given by (1.6) q ( n, t ) ≍ 2 log( T 0 ( z 0 )) +  2 ν ( z 0 ) − sin( θ 0 ) t  1 / 2 cos  t Φ 0 ( z 0 ) + ν ( z 0 ) log( t ) − δ ( z 0 )  , where z 0 = e i θ 0 is a slow variable dep ending only on n t and the functions T 0 ( z 0 ), ν ( z 0 ), Φ 0 ( z 0 ), and δ ( z 0 ) are explicitly g iv en in terms o f the scattering data asso- ciated with the initial data. Our main g oal will be to mathematica lly justify this formula fo r the solution in the similarit y reg ion | n t | < 1 (Theorem 2.2) and to show that the s olution splits into a num ber of solitons in the soliton reg ion | n t | > 1 (Theorem 2.1). Existence of so liton so lutions is usually connected to complete integrability of the system, and this is also true for the T o da equa tio n. T o see that the T o da equation is indeed integrable w e introduce Flaschk a’s v ariables [16] (1.7) a ( n, t ) = 1 2 e − ( q ( n +1 , t ) − q ( n,t )) / 2 , b ( n, t ) = − 1 2 p ( n, t ) 4 H. KR ¨ UGER AND G. TESCHL and o btain the form mo st convenien t for us d dt a ( t ) = a ( t )  b + ( t ) − b ( t )  , d dt b ( t ) = 2  a ( t ) 2 − a − ( t ) 2  . (1.8) Here w e hav e used the abbreviation (1.9) f ± ( n ) = f ( n ± 1) . Note that if q ( n, t ) → q ± suﬃciently fast as n → ±∞ , the co nverse map is given by (1.10) q ( n, t ) = q + + 2 log   ∞ Y j = n (2 a ( j, t ))   , p ( n, t ) = − 2 b ( n, t ) . Moreov er, q ( n, t ) → q ± , p ( n, t ) → 0 as | n | → ∞ corres ponds to a ( n, t ) → 1 2 , b ( n, t ) → 0. T o show complete integrability it suﬃces to ﬁnd a so-ca lled Lax pair [27], that is, tw o op erators H ( t ), P ( t ) in ℓ 2 ( Z ) such that the Lax equation (1.11) d dt H ( t ) = P ( t ) H ( t ) − H ( t ) P ( t ) is e q uiv alent to (1.8). O ne can easily convince oneself that the right choice is H ( t ) = a ( t ) S + + a − ( t ) S − + b ( t ) , P ( t ) = a ( t ) S + − a − ( t ) S − , (1.12) where ( S ± f )( n ) = f ± ( n ) = f ( n ± 1 ) a re the shift op erator s . No w the Lax equatio n (1.11) implies that the o pera tors H ( t ) for diﬀere nt t ∈ R are unitar ily equiv alent (cf. [38, Thm. 1 2.4]): Theorem 1.1. L et P ( t ) b e a family of b ounde d skew-adj oint op er ators, such that t 7→ P ( t ) is diﬀer entiable. Then ther e exists a family of unitary pr op agators U ( t, s ) for P ( t ) , that is, (1.13) d dt U ( t, s ) = P ( t ) U ( t, s ) , U ( s, s ) = I . Mor e over, t he L ax e quation (1 .11) implies (1.14) H ( t ) = U ( t, s ) H ( s ) U ( t, s ) − 1 . This r esult has several imp ortant consequences. First of all it implies global existence of solutions of the T o da la ttice. In fact, co nsidering the Banach spa ce of all b ounded re a l-v alued c oeﬃcie nts ( a ( n ) , b ( n )) (with the s up norm), lo cal ex istence follows fro m standa rd res ults for diﬀerential equations in Banach space s. Moreov er, Theorem 1.1 implies that the norm k H ( t ) k is constant, which in turn provides a uniform bound on the co eﬃcients of H ( t ), (1.15) k a ( t ) k ∞ + k b ( t ) k ∞ ≤ 2 k H ( t ) k = 2 k H (0) k . Hence solutions o f the T o da lattice cannot blow up a nd are globa l in time (see [38, Sect. 12.2] for deta ils ). LONG-TIME ASYMPTOTICS OF THE TODA LA TTICE 5 Second, it provides a n inﬁnite sequence of conserv ation laws exp ected from a completely in tegrable system. Indeed, if the Lax equatio n (1.11) holds for H ( t ), it automatically also holds for H ( t ) j . T a k ing tra ces shows that (1.16) tr  H ( t ) j − H j 0  , j ∈ N , is an inﬁnite sequence of co nserved quantities, where H 0 is the o pera tor cor r esp o nd- ing to the constant solutio n a 0 ( n, t ) = 1 2 , b 0 ( n, t ) = 0 (it is needed to make the trac e conv erge). Introducing a suitable symplectic str ucture, they can be shown to b e in inv olution as well ([18, Sect. 1.7]). F o r ex a mple, tr  H ( t ) − H 0  = X n ∈ Z b ( n, t ) = − 1 2 X n ∈ Z p ( n, t ) a nd tr  H ( t ) 2 − H 2 0  = X n ∈ Z b ( n, t ) 2 + 2( a ( n, t ) 2 − 1 4 ) = 1 2 H ( p, q ) (1.17) corres p ond to conser v ation of the total momentum and the total ener gy , resp ec- tively . These obser v ations pav e the way for a so lution o f the T o da equa tion via the inv erse scattering transform o riginally inv ented b y Gardner, Green, Krusk al, and Miura [1 7] for the Korteweg–De V r ies eq uation (see [38, Sect. 13 .4] for the ca se of the T o da lattice). In par ticula r, Theor em 1.1 implies that the o p era tors H ( t ), t ∈ R , ar e unita r ily equiv ale n t and that the sp ectrum σ ( H ( t )) is indep enden t of t . Now the g eneral ide a is to ﬁnd suitable spectr al data S ( H ( t )) for H ( t ) which uniquely determine H ( t ). Then equation (1.11) can b e used to derive linear evolu- tion equatio ns for S ( H ( t )) which are eas y to solve. In o ur case these data will be the s o ca lled s c attering data and the forma l pro cedure (which c an b e thoug h t of a s a nonlinear F ourier transfo rm) is summarized b elow: ( a (0) , b (0)) ( a ( t ) , b ( t )) ✻ direct scattering ❄ inv erse scattering S ( H (0 )) S ( H ( t )) ✲ time evolution The in verse sc attering step will b e done b y reformulating the pr oblem as a Riemann– Hilber t factorization problem. This Riemann–Hilb ert pro blem will then be analyzed using the metho d o f nonlinear steep est descent by Deift a nd Zho u [4] (which is the nonlinear analog o f the s tee p est descent for F ourier type integrals). In fact, one of our g oals is to give a complete and exp ository introduction to this metho d. W e are trying to present a strea mlined and simpliﬁed approa c h with complete proo fs. In particular, we have a dded tw o app endices whic h sho w how to so lve the lo calize d Riemann–Hilb ert problem on a s mall cross via para bolic cylinder functions and how to rewrite Rie ma nn–Hilber t problems as sing ular integral equa tions. Only some ba- sic knowledge on Riemann–Hilber t pr oblems, which can be fo und for example in the bea utiful lecture notes by Deift [3], is r e quired. 6 H. KR ¨ UGER AND G. TESCHL F or further infor mation on the history o f the steep est descent metho d, which was inspired by earlier work of Manakov [28] and I ts [19], and the pr oblem of ﬁnding the long-time a symptotics fo r integrable no nlinear w av e eq uations, we refer to the survey by Deift, Its, and Zhou [7 ]. More information on the T o da lattice can b e found in the monog raphs b y F addeev and T a k h ta jan [14], Gesz tesy , Ho lden, Michor, a nd T eschl [1 8], T eschl [38], or T o da [40]. Here we partly follow ed the re view article [3 9]. A muc h more compr e hensive guide to the litera ture can b e found in Section 1.8 of [18]. First results on the lo ng-time asymptotics of the do ubly inﬁnite T o da lattice were g iv en by Nov okshenov a nd Habibullin [32] and Kamvissis [20]. Long -time asymptotics for the ﬁnite and se mi inﬁnite T o da lattice can b e found in Moser [30] and Deift, Li, and T omei [9], resp ectively . The long- time b ehaviour of T o da sho ck problem was inv estigated by K am vissis [21] and V enakide s , Deift, a nd O ba [41] and of the T o da ra refaction pro ble m by Deift, Kamvissis, K r iecherbauer, and Zhou [11]. F or the case of a p erio dic driving force see Deift, Kriecherbauer, and V enakides [8]. Finally , we also wan t to mention that one c ould replac e the constant background solution by a p erio dic one. Howev er, this c ase exhibits a m uch diﬀerent behaviour, as was p ointed out b y Kamvissis and T eschl in [22] (see also [12], [13], [23], [24], and [2 6] for a rigo rous mathematical treatment). 2. Main resul ts As stated in the introductio n, we wan t to compute the long-time asymptotics for the doubly inﬁnite T o da lattice whic h reads in Flaschk a ’s v ariables (2.1) ˙ b ( n, t ) = 2 ( a ( n, t ) 2 − a ( n − 1 , t ) 2 ) , ˙ a ( n, t ) = a ( n, t )( b ( n + 1 , t ) − b ( n, t )) , ( n, t ) ∈ Z × R . Here the dot denotes diﬀerentiation with resp ect to time. W e will consider solutions ( a, b ) s atisfying (2.2) X n (1 + | n | ) l +1 ( | a ( n, t ) − 1 2 | + | b ( n, t ) | ) < ∞ for s ome l ∈ N for one (and hence for a ll, see [38]) t ∈ R . It is w ell-known that the cor resp onding initial v alue pr oblem has uniq ue globa l solutions which can be computed v ia the inv erse scattering transform [38]. The lo ng -time asymptotics were ﬁrs t derived b y Nov okshenov and Habibullin [3 2] and were la ter ma de rigo rous by Kamvissis [20] under the additional ass umption that no solitons a re prese n t. The ca se of solitons was r ecen tly investigated by us in [25]. As one o f our main simpliﬁcations in contradistinction to [20] we will w ork with the vector Riema nn– Hilbert problem which arises naturally from the inv erse scatter - ing theory , th us avoiding the detour over the a s so ciated matrix Riemann–Hilb ert problem. This also av oids the singular ities appearing in the matrix Riemann– Hilber t pr oblem in ca se the reﬂection co eﬃcient is − 1 at the band edges. T o s tate the main results, we b egin by reca lling that the sequences a ( n, t ), b ( n, t ), n ∈ Z , for ﬁxed t ∈ R , a r e uniquely determined by its scattering data, that is, by its right r eﬂection co eﬃcient R + ( z , t ), | z | = 1, and its eigenv alues λ j ∈ ( − ∞ , − 1) ∪ (1 , ∞ ), j = 1 , . . . , N , together with the cor resp onding r ig h t norming constan ts γ + ,j ( t ) > 0, j = 1 , . . . , N . It is well-kno wn that under the a ssumption (2.2) the LONG-TIME ASYMPTOTICS OF THE TODA LA TTICE 7 reﬂection co eﬃcients are C l +1 ( T ). Rather than in the complex plane, w e will work on the unit disc using the usual Joukowski trans fo rmation (2.3) λ = 1 2  z + 1 z  , z = λ − p λ 2 − 1 , λ ∈ C , | z | ≤ 1 . In these new co or dinates the eigenv alues λ j ∈ ( −∞ , − 1) ∪ (1 , ∞ ) will b e denoted by ζ j ∈ ( − 1 , 0 ) ∪ (0 , 1). The contin uous sp ectrum [ − 1 , 1] is mapp ed to the unit circle T . Moreover, the phase of the a sso ciated Riema nn–Hilber t problem is given by (2.4) Φ( z ) = z − z − 1 + 2 n t log( z ) and the s ta tionary phas e p oints, Φ ′ ( z ) = 0 , ar e denoted by (2.5) z 0 = − n t − r ( n t ) 2 − 1 , z − 1 0 = − n t + r ( n t ) 2 − 1 and c orresp ond to (2.6) λ 0 = − n t . Here the branch of the squa re ro ot is chosen such that Im( √ z ) ≥ 0. F or n t < − 1 we hav e z 0 ∈ (0 , 1), for − 1 ≤ n t ≤ 1 we hav e z 0 ∈ T (and hence z − 1 0 = z 0 ), and for n t > 1 we hav e z 0 ∈ ( − 1 , 0). F or | n t | > 1 we will als o need the v alue ζ 0 ∈ ( − 1 , 0) ∪ (0 , 1) deﬁned via Re(Φ( ζ 0 )) = 0, that is, (2.7) n t = − ζ 0 − ζ − 1 0 2 log( | ζ 0 | ) . W e will set ζ 0 = − 1 if | n t | ≤ 1 for notational co n venience. A simple analysis shows that fo r n t < − 1 we hav e 0 < ζ 0 < z 0 < 1 a nd for n t > 1 we hav e − 1 < z 0 < ζ 0 < 0. F urthermo re, recall that the trans mis s ion co eﬃcient T ( z ), | z | ≤ 1, is time inde- pendent and can b e reco ns tructed us ing the Poisson–Je ns en fo rm ula. In particular , we deﬁne the partial tra nsmission co eﬃcien t with r espect to z 0 by T ( z , z 0 ) =                    Q ζ k ∈ ( ζ 0 , 0) | ζ k | z − ζ − 1 k z − ζ k , z 0 ∈ ( − 1 , 0) , Q ζ k ∈ ( − 1 , 0) | ζ k | z − ζ − 1 k z − ζ k ! exp 1 2 π i z 0 R z 0 log( | T ( s ) | ) s + z s − z ds s ! , | z 0 | = 1 , Q ζ k ∈ ( − 1 , 0) ∪ ( ζ 0 , 1) | ζ k | z − ζ − 1 k z − ζ k ! exp  1 2 π i R T log( | T ( s ) | ) s + z s − z ds s  , z 0 ∈ (0 , 1 ) . (2.8) Here, in the case z 0 ∈ T , the integral is to be ta ken alo ng the ar c Σ( z 0 ) = { z ∈ T | Re ( z ) < Re( z 0 ) } oriented counterclockwise. F or z 0 ∈ ( − 1 , 0 ) w e s et Σ( z 0 ) = ∅ and for z 0 ∈ (0 , 1 ) we set Σ( z 0 ) = T . Then T ( z , z 0 ) is mer omorphic for z ∈ C \ Σ( z 0 ). Observe that T ( z , z 0 ) = T ( z ) once z 0 ∈ (0 , 1) and (0 , ζ 0 ) contains no eigenv alue s . Moreov er, T ( z , z 0 ) ca n b e c o mputed in terms of the s cattering da ta since | T ( z ) | 2 = 1 − | R + ( z , t ) | 2 = 1 − | R + ( z , 0) | 2 . 8 H. KR ¨ UGER AND G. TESCHL Moreov er, we set T 0 ( z 0 ) = T (0 , z 0 ) =                  Q ζ k ∈ ( ζ 0 , 0) | ζ k | − 1 , z 0 ∈ ( − 1 , 0) , Q ζ k ∈ ( − 1 , 0) | ζ k | − 1 ! exp 1 2 π i z 0 R z 0 log( | T ( s ) | ) ds s ! , | z 0 | = 1 , Q ζ k ∈ ( − 1 , 0) ∪ ( ζ 0 , 1) | ζ k | − 1 ! exp  1 2 π i R T log( | T ( s ) | ) ds s  , z 0 ∈ (0 , 1 ) , (2.9) and T 1 ( z 0 ) = ∂ ∂ z log T ( z , z 0 )    z =0 =              P ζ k ∈ ( ζ 0 , 0) ( ζ − 1 k − ζ k ) , z 0 ∈ ( − 1 , 0) , P ζ k ∈ ( − 1 , 0) ( ζ − 1 k − ζ k ) + 1 π i z 0 R z 0 log( | T ( s ) | ) ds s 2 , | z 0 | = 1 , P ζ k ∈ ( − 1 , 0) ∪ ( ζ 0 , 1) ( ζ − 1 k − ζ k ) + 1 π i R T log( | T ( s ) | ) ds s 2 , z 0 ∈ (0 , 1) . (2.10) In o ther words, T ( z , z 0 ) = T 0 ( z 0 )(1 + T 1 ( z 0 ) z + O ( z 2 )). Theorem 2 .1 (Soliton r egion) . Assum e (2.2) for some l ∈ N and abbr evia te by c k = − ζ k − ζ − 1 k 2 log ( | ζ k | ) the velo city of the k ’th soliton determine d by Re(Φ( ζ k )) = 0 . Then the asymptotics in the soliton r e gion, | n/ t | ≥ 1 + C /t log( t ) 2 for any C > 0 , ar e as fol lows. L et ε > 0 suﬃciently smal l such that the intervals [ c k − ε , c k + ε ] , 1 ≤ k ≤ N , ar e disjoint and lie inside ( −∞ , − 1) ∪ (1 , ∞ ) . If | n t − c k | < ε for some k , the solution is asymptotic al ly given by a single soliton ∞ Y j = n (2 a ( j, t )) = T 0 ( z 0 ) s 1 − ζ 2 k + γ k ( n, t ) 1 − ζ 2 k + γ k ( n + 1 , t ) + O ( t − l ) ! , ∞ X j = n +1 b ( j, t ) = 1 2 T 1 ( z 0 ) + γ k ( n, t ) ζ k (1 − ζ 2 k ) 2(( γ k ( n, t ) − 1) ζ 2 k + 1) + O ( t − l ) , (2.11) wher e (2.12) γ k ( n, t ) = γ k T ( ζ k , − c k − q c 2 k − 1) − 2 e t ( ζ k − ζ − 1 k ) ζ 2 n k . If | n t − c k | ≥ ε , for al l k , one has ∞ Y j = n (2 a ( j, t )) = T 0 ( z 0 )  1 + O ( t − l )  , ∞ X j = n +1 b ( j, t ) = 1 2 T 1 ( z 0 ) + O ( t − l ) . (2.13) Note that one can choos e | n t − c k | < ε 1 for the reg io ns where (2.11) is v alid, resp ectively | n t − c k | ≥ ε 2 for the r egions whe r e (2.13) is v a lid, such that the reg ions LONG-TIME ASYMPTOTICS OF THE TODA LA TTICE 9 ov erlap if ε 1 > ε 2 . Due to the expo nen tial decay of the one-s o liton solution, b oth formulas of course pro duce the sa me result on the ov erlap. In particular, we recover the well-known fact that the solution splits into a sum of independent s olitons where the pr esence of the o ther solitons and the radia tion part corres p onding to the co n tinuous sp ectrum manifests itself in phas e shifts given by T ( ζ k , − c k − p c 2 k − 1) − 2 . Indeed, notice that for ζ k ∈ ( − 1 , 0) this ter m just co n tains the pro duct ov er the Blaschk e factor s corresp onding to solitons ζ j with ζ k < ζ j . F or ζ k ∈ (0 , 1) we hav e the pro duct o v er the Blaschke factors cor resp onding to solitons ζ j ∈ ( − 1 , 0), the integral over the full unit circle, plus the pro duct ov er the Blaschk e factor s corre s ponding to solitons ζ j with ζ k > ζ j . F urthermo re, this result shows that in the region n t > 1 the solution is asymp- totically given by a N − -soliton solution, wher e N − is the n umber of ζ j ∈ ( − 1 , 0), formed from the data ζ j , γ j for all ζ k ∈ ( − 1 , 0). Similarly , in the region n t < − 1 the solution is asymptotically given by a N + -soliton solution, where N + is the n umber of ζ j ∈ (0 , 1 ), formed from the data ζ j , ˜ γ j for all ζ j ∈ (0 , 1), where (2.14) ˜ γ j = γ j   Y ζ k ∈ ( − 1 , 0) | ζ k | ζ j − ζ − 1 k ζ j − ζ k   exp   1 2 π i Z T log( | T ( s ) | ) s + ζ j s − ζ j ds s   . In the remaining region, w e will show Theorem 2.2 (Similar it y region) . Assum e (2.2) with l ≥ 5 , then, away fr om the soliton r e gion, | n/t | ≤ 1 − C for any C > 0 , the asymptotics ar e given by ∞ Y j = n (2 a ( j, t )) = T 0 ( z 0 ) 1 +  ν ( z 0 ) − 2 sin( θ 0 ) t  1 / 2 cos  t Φ 0 ( z 0 ) + ν ( z 0 ) log( t ) − δ ( z 0 )  + O ( t − α ) ! , (2.15) ∞ X j = n +1 b ( j, t ) = 1 2 T 1 ( z 0 ) +  ν ( z 0 ) − 2 sin( θ 0 ) t  1 / 2 cos  t Φ 0 ( z 0 ) + ν ( z 0 ) log( t ) − δ ( z 0 ) + θ 0  + O ( t − α ) , z 0 = e i θ 0 , (2.16) for any α < 1 . Her e ν ( z 0 ) = − 1 π log( | T ( z 0 ) | ) , Φ 0 ( z 0 ) = 2(sin( θ 0 ) − θ 0 cos( θ 0 )) , δ ( z 0 ) = π / 4 − 3 ν ( z 0 ) log | 2 sin( θ 0 ) | + 2 arg( ˜ T ( z 0 )) − arg( R + ( z 0 , 0)) (2.17) + arg(Γ(i ν ( z 0 ))) , ˜ T ( z 0 ) = Y ζ k ∈ ( − 1 , 0) | ζ k | z − ζ − 1 k z − ζ k · exp   1 2 π i z 0 Z z 0 log  | T ( s ) | | T ( z 0 ) |  s + z 0 s − z 0 ds s   , and Γ( z ) is the gamma fun ction. F or a ( n, t ) resp ectively b ( n, t ) we obtain a s a simple conse q uence: 10 H. KR ¨ UGER AND G. TESCHL - 200 - 100 0 100 200 0.5 1.0 1.5 2.0 Figure 3. Numerically computed solution a ( n, 150) of the T o da lattice in Flaschk a’s v ariables. Corollary 2.3. Under the same assumptions as in The or em 2.2 we have a ( n, t ) = 1 2 +  − sin( θ 0 ) ν ( z 0 ) 2 t  1 / 2 cos  t Φ 0 ( z 0 ) + ν ( z 0 ) log( t ) − δ ( z 0 ) − θ 0  + O ( t − α ) , (2.18) b ( n, t ) =  − 2 sin( θ 0 ) ν ( z 0 ) t  1 / 2 sin  t Φ 0 ( z 0 ) + ν ( z 0 ) log( t ) − δ ( z 0 ) + 2 θ 0  + O ( t − α ) . (2.19) Pr o of. T o get the ﬁrst formula for we use a ( n, t ) = 1 2 Q ∞ j = n (2 a ( j, t )) / Q ∞ j = n +1 (2 a ( j, t )). Now set x = n t and o bserve θ 0 ( n +1 t ) = θ 0 ( x + 1 t ) = θ 0 ( x ) ± θ ′ ( x ) 1 t + O ( t − 2 ) uniformly in | x | ≤ 1 − C . Similar ly for for the other terms and hence on ch ecks that the only diﬀerence up to O ( t − α ) er rors in the ab ov e formulas for n and n ± 1 is a ∓ 2 θ 0 in the a rgument of the cosine (stemming fro m the t Φ 0 ( z 0 ) ter m). The second formula follows in the sa me manner from b ( n, t ) = P ∞ j = n b ( j, t ) − P ∞ j = n +1 b ( j, t ).  This is illus trated in Figure 3, whic h shows the sa me so lution a s in Figure 2 but in Flaschk a ’s v ar iables. I t is also interesting to lo ok at the relation b etw een the energy λ of the underlying Lax op erator H and the propag ation sp eed at which the corres p onding parts of the T o da la ttice trav el, tha t is, the a nalog of the classica l disp e rsion relation. By the ab ov e theorems, the nonlinear disp ersion relation is given by (see Fig ure 4 ) (2.20) v ( λ ) = n t , where (2.21) v ( λ ) = ( − λ, λ ∈ [ − 1 , 1] , √ λ 2 − 1 log( | λ − √ λ 2 − 1 | ) , λ ∈ ( −∞ , − 1] ∪ [1 , ∞ ) . LONG-TIME ASYMPTOTICS OF THE TODA LA TTICE 11 - 4 - 2 2 4 Λ - 2 - 1 1 2 v H Λ L Figure 4. Nonlinear disp ersion relatio n for the T o da lattice. W e will not a ddress the asymptotics in the missing regio n ar ound | n | ≈ t . In the case | R + ( z , 0) | < 1 the solution can be given in terms of Painlev´ e I I transcendents. If | R + ( z , 0) | = 1 (which is the generic case), a n additional region, the collis ionless sho ck reg ion, will appea r where the solution can be describ ed in terms of elliptic functions. F or the Painlev ´ e region we refer to [4 ], [20]. F or the collisionles s sho c k region an outline using the g -function metho d w as given in [10] (for the case of the Korteweg–de Vires equation). The case of the T o da lattice will b e dealt with in [29]. W e als o r emark that the pres en t methods can also b e used to obtain further terms in the a symptotic ex pansion [5]. Finally , no te that one can obtain the asymptotics for n ≥ 0 from the ones for n ≤ 0 by virtue o f a simple reﬂection. Similarly for t ≥ 0 versus t ≤ 0. Lemma 2.4. Supp ose a ( n, t ) , b ( n, t ) satisfy the T o da e quation (2 .1) , then so do ˜ a ( n, t ) = a ( − n − 1 , t ) , ˜ b ( n, t ) = − b ( − n, t ) r esp e ctively ˜ a ( n, t ) = a ( n, − t ) , ˜ b ( n, t ) = − b ( n, − t ) . 3. The I nverse sca ttering transform and the Riemann–Hilber t pr oblem In this section we want to derive the Riemann–Hilb ert pro ble m fro m scattering theory . The sp e cial case without eig en v a lues was ﬁrst given in Kamvissis [20]. How eigenv alues ca n b e added w as ﬁrst shown in Deift, Kamvissis, Kr iec herbauer , and Zhou [11]. W e essentially follow [25] in this section. F or the necessary results from s cattering theory resp ectively the inv erse sca tter- ing transform for the T o da lattice w e refer to [36], [37], [38]. Asso ciated with a ( t ) , b ( t ) is a self-adjoint Jaco bi op erato r (3.1) H ( t ) = a ( t ) S + + a − ( t ) S − + b ( t ) in ℓ 2 ( Z ), where S ± f ( n ) = f ± ( n ) = f ( n ± 1 ) are the usual shift op erato rs and ℓ 2 ( Z ) denotes the Hilb e rt space of squar e summable (complex -v alued) sequence s ov er Z . By our as sumption (2.2) the sp ectrum of H co nsists of an abso lutely contin uous part 12 H. KR ¨ UGER AND G. TESCHL [ − 1 , 1] plus a ﬁnite num ber o f eig en v a lues λ k ∈ R \ [ − 1 , 1], 1 ≤ k ≤ N . In addition, there exist tw o Jost functions ψ ± ( z , n, t ) which s olve the recur rence equatio n (3.2) H ( t ) ψ ± ( z , n, t ) = z + z − 1 2 ψ ± ( z , n, t ) , | z | ≤ 1 , and a symptotically lo ok like the fr e e solutio ns (3.3) lim n →±∞ z ∓ n ψ ± ( z , n, t ) = 1 . Both ψ ± ( z , n, t ) are analytic for 0 < | z | < 1 with smo oth boundar y v alues for | z | = 1. The asy mptotics of the tw o Jos t function are (3.4) ψ ± ( z , n, t ) = z ± n A ± ( n, t )  1 + 2 B ± ( n, t ) z + O ( z 2 )  , as z → 0, where (3.5) A + ( n, t ) = ∞ Y j = n 2 a ( j, t ) , B + ( n, t ) = − ∞ X j = n +1 b ( j, t ) , A − ( n, t ) = n − 1 Y j = −∞ 2 a ( j, t ) , B − ( n, t ) = − n − 1 X j = −∞ b ( j, t ) . One has the scattering relations (3.6) T ( z ) ψ ∓ ( z , n, t ) = ψ ± ( z , n, t ) + R ± ( z , t ) ψ ± ( z , n, t ) , | z | = 1 , where T ( z ), R ± ( z , t ) are the tr ansmission resp ectively reﬂectio n co eﬃcients. The transmission and reﬂection coeﬃcients hav e the following well-known pr oper ties ([38, Sect. 10.2]): Lemma 3. 1. The tr ansmission c o eﬃcient T ( z ) has a mer omorphic ext ension to the interior of the un it cir cle with simple p oles at the images of the eigenvalues ζ j . The r esidues of T ( z ) ar e given by (3.7) Res ζ k T ( z ) = − ζ k γ + ,k ( t ) µ k ( t ) = − ζ k γ − ,k ( t ) µ k ( t ) , wher e (3.8) γ ± ,k ( t ) − 1 = X n ∈ Z | ψ ± ( ζ k , n, t ) | 2 and ψ − ( ζ k , n, t ) = µ k ( t ) ψ + ( ζ k , n, t ) . Mor e over, (3.9) T ( z ) R + ( z , t ) + T ( z ) R − ( z , t ) = 0 , | T ( z ) | 2 + | R ± ( z , t ) | 2 = 1 . In par ticular one reﬂection coeﬃcient, say R ( z , t ) = R + ( z , t ), and one set of norming co nstant s, say γ k ( t ) = γ + ,k ( t ), suﬃces. Mor eov er, the time dependence is given by ([38, Thm. 13.4]): Lemma 3.2. The time evolutions of the qu antities R + ( z , t ) , γ + ,k ( t ) ar e given by R ( z , t ) = R ( z )e t ( z − z − 1 ) (3.10) γ k ( t ) = γ k e t ( ζ k − ζ − 1 k ) , (3.11) wher e R ( z ) = R ( z , 0) and γ k = γ k (0) . LONG-TIME ASYMPTOTICS OF THE TODA LA TTICE 13 Now we deﬁne the sectionally meromorphic vector (3.12) m ( z , n, t ) =   T ( z ) ψ − ( z , n, t ) z n ψ + ( z , n, t ) z − n  , | z | < 1 ,  ψ + ( z − 1 , n, t ) z n T ( z − 1 ) ψ − ( z − 1 , n, t ) z − n  , | z | > 1 . W e are in terested in the jump condition o f m ( z , n, t ) on the unit circle T (ori- ent ed counterclo ckwise). T o for m ulate our jump condition we use the following conv en tion: W hen representing functions on T , the low er subscript deno tes the non-tangential limit from diﬀeren t sides, (3.13) m ± ( z ) = lim ζ → z , | ζ | ± 1 < 1 m ( ζ ) , | z | = 1 . In general, for a n orie nted contour Σ, m + ( z ) (resp. m − ( z )) will denote the limit of m ( ζ ) a s ζ → z from the pos itiv e (re sp. neg ative) s ide o f Σ. Here the pos itiv e (resp. negative) side is the one which lies to the left (res p. right) as one trav erses the contour in the direction of the or ien tation. Using the notation ab ov e implicitly assumes that these limits exist in the sense that m ( z ) extends to a con tinuous function on the b oundar y . Theorem 3.3 (V ector Riemann–Hilb ert problem) . L et S + ( H (0)) = { R ( z ) , | z | = 1; ( ζ k , γ k ) , 1 ≤ k ≤ N } the left sc attering data of t he op er ator H (0) . Then m ( z ) = m ( z , n, t ) deﬁne d in (3.12) is a solution of the fol lowing ve ctor Riema nn–Hilb ert pr oblem. Find a function m ( z ) which is mer omorphic away fr om the unit cir cle with simple p oles at ζ k , ζ − 1 k and satisﬁes: (i) The jump c ondition (3.14) m + ( z ) = m − ( z ) v ( z ) , v ( z ) =  1 − | R ( z ) | 2 − R ( z )e − t Φ( z ) R ( z )e t Φ( z ) 1  , for z ∈ T , (ii) the p ole c onditions (3.15) Res ζ k m ( z ) = lim z → ζ k m ( z )  0 0 − ζ k γ k e t Φ( ζ k ) 0  , Res ζ − 1 k m ( z ) = lim z → ζ − 1 k m ( z )  0 ζ − 1 k γ k e t Φ( ζ k ) 0 0  , (iii) the symmetry c ondition (3.16) m ( z − 1 ) = m ( z )  0 1 1 0  (iv) and the n ormaliza tion (3.17) m (0) = ( m 1 m 2 ) , m 1 · m 2 = 1 m 1 > 0 . Her e the phase is given by (3.18) Φ( z ) = z − z − 1 + 2 n t log z . Pr o of. The jump condition (3.1 4) is a simple calcula tion us ing the scattering re- lations (3.6) plus (3.9). The pole c o nditions follow since T ( z ) is mero morphic in | z | < 1 with s imple p oles at ζ k and residues given by (3.7). The symmetry co nditio n holds by construction and the nor malization (3 .17) is immediate from the following lemma.  14 H. KR ¨ UGER AND G. TESCHL Observe that the pole condition at ζ k is suﬃcient since the one at ζ − 1 k follows by symmetry . Mor eov er, it can b e shown that the so lutio n of the ab ov e Riemann– Hilber t problem is unique [25]. How ever, we will not need this fact here and it will follow a s a bypro duct o f our analysis at least for suﬃciently larg e t . Moreov er, we hav e the following a s ymptotic b ehaviour nea r z = 0 : Lemma 3.4. The function m ( z , n, t ) deﬁne d in (3.12) satisﬁ es (3.19) m ( z , n, t ) =  A ( n, t )(1 − 2 B ( n − 1 , t ) z ) 1 A ( n,t ) (1 + 2 B ( n, t ) z )  + O ( z 2 ) . Her e A ( n, t ) = A + ( n, t ) and B ( n, t ) = B + ( n, t ) ar e deﬁne d in (3.5) . Pr o of. This follows from (3 .4) and T ( z ) = A + A − (1 − 2( B + − b + B − ) z + O ( z 2 )).  F or our further analy s is it will b e conv enient to rewrite the p ole condition a s a jump c ondition a nd hence turn our meromorphic Riemann– Hilbert pr o blem into a holomorphic Riema nn– Hilbert pr oblem following [11]. Cho os e ε so s ma ll that the discs | z − ζ k | < ε a re inside the s e t { z | 0 < | z | < 1 } and do not intersect. Then redeﬁne m in a neighbor hoo d of ζ k resp ectively ζ − 1 k according to (3.20) m ( z ) =                  m ( z ) 1 0 ζ k γ k e t Φ( ζ k ) z − ζ k 1 ! , | z − ζ k | < ε, m ( z ) 1 − z γ k e t Φ( ζ k ) z − ζ − 1 k 0 1 ! , | z − 1 − ζ k | < ε, m ( z ) , else . Then a straightforw ard calculation using Res ζ m = lim z → ζ ( z − ζ ) m ( z ) shows Lemma 3.5. S upp ose m ( z ) is r e deﬁne d as in (3.20) . Then m ( z ) is holomorph ic away fr om the un it cir cle and satisﬁes (3.14) , (3.16) , (3.17) and the p ole c onditions ar e r eplac e d by the jump c onditions (3.21) m + ( z ) = m − ( z ) 1 0 ζ k γ k e t Φ( ζ k ) z − ζ k 1 ! , | z − ζ k | = ε, m + ( z ) = m − ( z ) 1 z γ k e t Φ( ζ k ) z − ζ − 1 k 0 1 ! , | z − 1 − ζ k | = ε , wher e the sm al l cir cle ar ound ζ k is oriente d c ounter clo ckwise and t he one ar ound ζ − 1 k is oriente d clo ckwise. Finally , we note that the case of just o ne eigenv alue and zer o re ﬂection co eﬃcient can be so lv ed explicitly . Lemma 3.6 (One soliton solution) . Supp ose ther e is only one eigenvalue and a vanishing r eﬂe ction c o eﬃcient, that is, S + ( H ( t )) = { R ( z ) ≡ 0 , | z | = 1; ( ζ , γ ) } with ζ ∈ ( − 1 , 0) ∪ (0 , 1) and γ ≥ 0 . Then the Ri emann–Hilb ert pr oblem (3.14) – (3.17) has a unique solution is given by m 0 ( z ) =  f ( z ) f (1 /z )  (3.22) f ( z ) = 1 p 1 − ζ 2 + γ ( n, t ) p 1 − ζ 2 + ζ 2 γ ( n, t )  γ ( n, t ) ζ 2 z − ζ − 1 z − ζ + 1 − ζ 2  , LONG-TIME ASYMPTOTICS OF THE TODA LA TTICE 15 wher e γ ( n, t ) = γ e t Φ( ζ ) . In p articular, (3.23) A + ( n, t ) = s 1 − ζ 2 + γ ( n, t ) 1 − ζ 2 + γ ( n, t ) ζ 2 , B + ( n, t ) = γ ( n, t ) ζ ( ζ 2 − 1) 2(1 − ζ 2 + γ ( n, t ) ζ 2 ) . F u rthermor e, the zer o solution is the only solution of t he c orr esp onding vanishi ng pr oblem wher e the normalization is r eplac e d by m (0) = (0 m 2 ) with m 2 arbitr ary. Pr o of. By symmetry , the solution m ust b e of the form m 0 ( z ) =  f ( z ) f (1 /z )  , where f ( z ) is mer omorphic in C ∪ {∞} with the only p oss ible p ole at ζ . Hence f ( z ) = 1 A  1 + 2 B z − ζ  , where the unknown constants A a nd B are uniquely determined by the p ole con- dition Res ζ f ( z ) = − ζ γ ( n, t ) f ( ζ − 1 ) and the normalization f (0) f ( ∞ ) = 1, f (0) > 0.  4. Conjuga tion and def orma tion This section demo nstrates how to conjuga te o ur Riema nn–Hilber t problem and deform the jump cont ours, such that the jumps will b e exp onentially clo se to the ident ity aw ay fro m the stationary pha s e p oin ts. In order to do this we will assume that R ( z ) has an analytic extension to a strip a round the unit circle throug hout this and the fo llo wing section. This is fo r example the cas e if the decay in (2.2) is exp onent ially . W e will even tually show how to remove this assumption in Section 6. F or easy reference we note the following result which ca n be chec k ed by a straight- forward ca lculation. Lemma 4.1 (Co njugation) . Assume t hat e Σ ⊆ Σ . L et D b e a matrix of the form (4.1) D ( z ) =  d ( z ) − 1 0 0 d ( z )  , wher e d : C \ e Σ → C is a se ct ional ly analytic funct ion. Set (4.2) ˜ m ( z ) = m ( z ) D ( z ) , then the jump m atrix tra nsforms ac c or ding to (4.3) ˜ v ( z ) = D − ( z ) − 1 v ( z ) D + ( z ) . If d satisﬁes d ( z − 1 ) = d ( z ) − 1 and d (0) > 0 . Then the tr ansformation ˜ m ( z ) = m ( z ) D ( z ) r esp e cts ou r symmetry, that is, ˜ m ( z ) s atisﬁ es (3.16) if and only if m ( z ) do es. In particular, w e obtain (4.4) ˜ v =  v 11 v 12 d 2 v 21 d − 2 v 22  , z ∈ Σ \ e Σ , resp ectively (4.5) ˜ v = d − d + v 11 v 12 d + d − v 21 d − 1 + d − 1 − d + d − v 22 ! , z ∈ e Σ . In o rder to remov e the p oles ther e are tw o cases to distinguish. If Re (Φ( ζ k )) < 0 the corresp onding jumps (3.21) are exp onent ially close to the identit y as t → ∞ and there is nothing to do. Otherwise, if Re(Φ( ζ k )) < 0, we use conjugation to 16 H. KR ¨ UGER AND G. TESCHL turn the jumps into exp onentially decaying ones, again following Deift, Kamvissis, Kriecherbauer, a nd Zhou [11] (see also [2 5]). F or this purp ose we will use the next lemma which shows how γ k e t Φ( ζ k ) can b e replaced b y its inv erse. It turns out that we will hav e to handle the p oles at ζ k and ζ − 1 k in o ne step in or de r to pres e rve symmetry and in o rder to not add additional p oles elsewhere. Lemma 4.2. A ssume that the Riemann–Hilb ert pr oblem for m has ju m p c onditions ne ar ζ and ζ − 1 given by (4.6) m + ( z ) = m − ( z )  1 0 γ ζ z − ζ 1  , | z − ζ | = ε, m + ( z ) = m − ( z )  1 γ z z − ζ − 1 0 1  , | z − 1 − ζ | = ε. Then this Riemann–Hilb ert pr oblem is e quivalent to a Riemann–Hilb ert pr oblem for ˜ m which has jump c onditions ne ar ζ and ζ − 1 given by ˜ m + ( z ) = ˜ m − ( z ) 1 ( ζ z − 1) 2 ζ ( z − ζ ) γ 0 1 ! , | z − ζ | = ε, ˜ m + ( z ) = ˜ m − ( z ) 1 0 ( z − ζ ) 2 ζ z ( ζ z − 1) γ 1 ! , | z − 1 − ζ | = ε, and al l r emaining data c onjugate d (as in L emma 4.1) by (4.7) D ( z ) = z − ζ ζ z − 1 0 0 ζ z − 1 z − ζ ! . Pr o of. T o turn γ into γ − 1 , in tro duce D b y D ( z ) =                      1 1 γ z − ζ ζ − γ ζ z − ζ 0 ! z − ζ ζ z − 1 0 0 ζ z − 1 z − ζ ! , | z − ζ | < ε , 0 γ z ζ z ζ − 1 − 1 γ z ζ − 1 z ζ 1 ! z − ζ ζ z − 1 0 0 ζ z − 1 z − ζ ! , | z − 1 − ζ | < ε, z − ζ ζ z − 1 0 0 ζ z − 1 z − ζ ! , else , and note that D ( z ) is a nalytic aw ay from the tw o circles. Now set ˜ m ( z ) = m ( z ) D ( z ), whic h is a gain symmetric by D ( z − 1 ) =  0 1 1 0  D ( z )  0 1 1 0  . The jumps along | z − ζ | = ε and | z − 1 − ζ | = ε follow b y a s traightforw ard calculation a nd the remaining jumps follow from Lemma 4 .1.  The jumps alo ng T a r e of o scillatory type and our a im is to apply a contour deformation which will move them in to regions where the oscillatory ter ms will decay exp onentially . Since the jump matrix v contains b oth exp( t Φ) and exp( − t Φ) we need to separa te them in orde r to be able to mov e them to diﬀere nt regio ns of the complex plane. F or this we will need the following factor izations o f the jump condition (3.14). First of all (4.8) v ( z ) = b − ( z ) − 1 b + ( z ) , LONG-TIME ASYMPTOTICS OF THE TODA LA TTICE 17 where b − ( z ) =  1 R ( z )e − t Φ( z ) 0 1  , b + ( z ) =  1 0 R ( z )e t Φ( z ) 1  . This will b e the prop er factoriza tion for z > z 0 . Here z > z 0 has to be under sto o d as λ ( z ) > λ 0 . Similarly , w e hav e (4.9) v ( z ) = B − ( z ) − 1  1 − | R ( z ) | 2 0 0 1 1 −| R ( z ) | 2  B + ( z ) , where B − ( z ) = 1 0 − R ( z )e t Φ( z ) 1 −| R ( z ) | 2 1 ! , B + ( z ) = 1 − R ( z )e − t Φ( z ) 1 −| R ( z ) | 2 0 1 ! . This will b e the prop er factoriza tion for z < z 0 . T o get rid of the diagonal part w e need to solve the c o rresp onding scalar Riemann– Hilber t problem. Moreover, for z 0 ∈ ( − 1 , 0) we hav e Re(Φ( z )) > 0 for z ∈ ( ζ 0 , 0) and Re(Φ( z )) < 0 for z ∈ ( − 1 , ζ 0 ) ∪ (0 , 1), for z 0 ∈ T we hav e Re(Φ( z )) > 0 for z ∈ ( − 1 , 0) and Re(Φ( z )) < 0 for z ∈ (0 , 1), a nd for z 0 ∈ (0 , 1 ) w e have Re(Φ( z )) > 0 for z ∈ ( − 1 , 0) ∪ ( ζ 0 , 1) and Re(Φ( z )) < 0 for z ∈ (0 , ζ 0 ) (compare Figure 5 a nd note that by Re(Φ( z − 1 )) = − Re(Φ( z )) the curves Re(Φ( z )) = 0 are symmetric with resp ect to z 7→ z − 1 ). T og ether with the Blaschk e factors needed to conjugate the jumps near the eigen- v alues, this is just the partial transmiss ion co eﬃcien t T ( z , z 0 ) intro duced in (2.8). In fact, it sa tisﬁes the following scalar mero mo rphic Riema nn– Hilbert problem: Lemma 4.3. Set Σ( z 0 ) = ∅ for z 0 ∈ ( − 1 , 0) , Σ( z 0 ) = { z ∈ T | Re( z ) < Re( z 0 ) } for z 0 ∈ T , and Σ( z 0 ) = T for z 0 ∈ (0 , 1) . Then the p artial tr ansmission c o eﬃcient T ( z , z 0 ) is mer omorphic for z ∈ C \ Σ( z 0 ) , with simple p oles at ζ j and simple zer os at ζ − 1 j for al l j with 1 2 ( ζ j + ζ − 1 j ) < λ 0 , and satisﬁes the jump c ondition T + ( z , z 0 ) = T − ( z , z 0 )(1 − | R ( z ) | 2 ) , z ∈ Σ( z 0 ) . Mor e over, (i) T ( z − 1 , z 0 ) = T ( z , z 0 ) − 1 , z ∈ C \ Σ( z 0 ) , and T (0 , z 0 ) > 0 , (ii) T ( z , z 0 ) = T ( z , z 0 ) , z ∈ C and, in p articular, T ( z , z 0 ) is re al-value d for z ∈ R , (iii) T ( z , z 0 ) = T ( z )( C + o (1)) with C 6 = 0 for | z | ≤ 1 ne ar ± 1 if ± 1 ∈ Σ( z 0 ) and c ontinu ous otherwise. Pr o of. That ζ j are simple p oles and ζ − 1 j are simple zero s is o b vious fr o m the Blaschk e factors and that T ( z , z 0 ) has the given jump follo ws from Plemelj’s for- m ulas. (i)–(iii) are straig htforward to chec k.  Observe that for ζ 0 < ζ N if ζ N ∈ (0 , 1) resp ectively ζ 0 < 1 else w e have T ( z ) = T ( z , z 0 ). Moreov er, note that (i) and (ii) imply (4.10) | T ( z , z 0 ) | 2 = T ( z , z 0 ) T ( z , z 0 ) = T ( z − 1 , z 0 ) T ( z , z 0 ) = 1 , z ∈ T \ Σ( z 0 ) . 18 H. KR ¨ UGER AND G. TESCHL Now we are ready to per form o ur conjugation step. Introduce D ( z ) =                  1 z − ζ k ζ k γ k e t Φ( ζ k ) − ζ k γ k e t Φ( ζ k ) z − ζ k 0 ! D 0 ( z ) , | z − ζ k | < ε, λ k < 1 2 ( ζ 0 + ζ − 1 0 ) , 0 z ζ k γ k e t Φ( ζ k ) z ζ k − 1 − z ζ k − 1 z ζ k γ k e t Φ( ζ k ) 1 ! D 0 ( z ) , | z − 1 − ζ k | < ε , λ k < 1 2 ( ζ 0 + ζ − 1 0 ) , D 0 ( z ) , else , where D 0 ( z ) =  T ( z , z 0 ) − 1 0 0 T ( z , z 0 )  . Note that we have D ( z − 1 ) =  0 1 1 0  D ( z )  0 1 1 0  . Now we conjugate o ur vector m ( z ) deﬁned in (3.12) r espe c tiv ely (3.20) using D ( z ), (4.11) ˜ m ( z ) = m ( z ) D ( z ) . Since, by Lemma 4.3 (iii), T ( z , z 0 ) is either nonzero and co n tin uous near z = ± 1 (if ± 1 / ∈ Σ( z 0 )) or it has the same behaviour as T ( z ) near z = ± 1 (if ± 1 ∈ Σ( z 0 )), the new vector ˜ m ( z ) is ag ain contin uous nea r z = ± 1 (even if T ( z ) v anishes ther e). Then using Lemma 4.1 a nd Lemma 4.2 the jumps corres ponding to eige nv alues λ k < 1 2 ( ζ 0 + ζ − 1 0 ) (if a n y) are given by (4.12) ˜ v ( z ) = 1 z − ζ k ζ k γ k T ( z ,z 0 ) − 2 e t Φ( ζ k ) 0 1 ! , | z − ζ k | = ε, ˜ v ( z ) = 1 0 ζ k z − 1 ζ k z γ k T ( z ,z 0 ) 2 e t Φ( ζ k ) 1 ! , | z − 1 − ζ k | = ε, and c orresp onding to eigenv alues λ k > 1 2 ( ζ 0 + ζ − 1 0 ) (if any) by (4.13) ˜ v ( z ) = 1 0 ζ k γ k T ( z ,z 0 ) − 2 e t Φ( ζ k ) z − ζ k 1 ! , | z − ζ k | = ε, ˜ v ( z ) = 1 z γ k T ( z ,z 0 ) 2 e t Φ( ζ k ) z − ζ − 1 k 0 1 ! , | z − 1 − ζ k | = ε. In pa rticular, an inv estigation of the sign of Re(Φ( z )) (see Figur e 5 b elow) s ho ws that a ll o ﬀ-diagonal entries of thes e jump ma trices, except for p ossibly one if ζ k 0 = ζ 0 for some k 0 , are exp onentially decreasing. In the latter case we will keep the po le conditio n fo r ζ k 0 = ζ 0 which now reads (4.14) Res ζ k 0 ˜ m ( z ) = lim z → ζ k 0 ˜ m ( z )  0 0 − ζ k 0 γ k 0 T ( ζ k 0 , z 0 ) − 2 e t Φ( ζ k 0 ) 0  , Res ζ − 1 k 0 ˜ m ( z ) = lim z → ζ − 1 k 0 ˜ m ( z )  0 ζ − 1 k 0 γ k 0 T ( ζ k 0 , z 0 ) − 2 e t Φ( ζ k 0 ) 0 0  . F urthermo re, the jump along T is given by (4.15) ˜ v ( z ) = ( ˜ b − ( z ) − 1 ˜ b + ( z ) , λ ( z ) > λ 0 , ˜ B − ( z ) − 1 ˜ B + ( z ) , λ ( z ) < λ 0 , LONG-TIME ASYMPTOTICS OF THE TODA LA TTICE 19 z 0 ∈ ( − 1 , 0) + − − + ✲ ✻ q ζ 0 q ζ − 1 0 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . z 0 ∈ T − + − + ✲ ✻ q z − 1 0 q z 0 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . z 0 ∈ (0 , 1) − + − − + ✲ ✻ q ζ 0 q ζ − 1 0 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Figure 5. Sign of Re(Φ( z )) for diﬀerent v alues of z 0 where (4.16) ˜ b − ( z ) = 1 R ( z − 1 )e − t Φ( z ) T ( z − 1 ,z 0 ) 2 0 1 ! , ˜ b + ( z ) = 1 0 R ( z )e t Φ( z ) T ( z ,z 0 ) 2 1 ! , and ˜ B − ( z ) = 1 0 − T − ( z ,z 0 ) − 2 1 − R ( z ) R ( z − 1 ) R ( z )e t Φ( z ) 1 ! , ˜ B + ( z ) = 1 − T + ( z ,z 0 ) 2 1 − R ( z ) R ( z − 1 ) R ( z − 1 )e − t Φ( z ) 0 1 ! . (4.17) Here we hav e used T ± ( z − 1 , z 0 ) = T ± ( z , z 0 ) = T ± ( z , z 0 ) and R ( z − 1 ) = R ( z ) = R ( z ) for z ∈ T to show that ther e exis ts a n a na lytic contin uation into a neighborho o d of the unit circle. Moreov er, using T ± ( z , z 0 ) = T ∓ ( z − 1 , z 0 ) − 1 , z ∈ Σ( z 0 ) , we can write (4.18) T − ( z , z 0 ) − 2 1 − R ( z ) R ( z − 1 ) = T − ( z , z 0 ) T − ( z , z 0 ) , T + ( z , z 0 ) 2 1 − R ( z ) R ( z − 1 ) = T + ( z , z 0 ) T + ( z , z 0 ) for z ∈ T , which shows that the matrix entries are in fact bo unded. Now we deform the jump a long T to move the oscillato ry terms in to regions where they ar e decaying. There are thr ee cases to dis ting uish (see Figur e 5): Case 1 : z 0 ∈ ( − 1 , 0) . In this ca se we will se t Σ ± = { z | | z | = (1 − ε ) ± 1 } for so me small ε ∈ (0 , 1 ) such that Σ ± lies in the r egion with ± Re(Φ( z )) < 0 and such that we do no t intersect the origina l co n tours (i.e., w e stay aw a y fro m ζ ± 1 j ). Then we can split our jump by rede ﬁning ˜ m ( z ) according to (4.19) ˆ m ( z ) =      ˜ m ( z ) ˜ b + ( z ) − 1 , (1 − ε ) < | z | < 1 , ˜ m ( z ) ˜ b − ( z ) − 1 , 1 < | z | < (1 − ε ) − 1 , ˜ m ( z ) , else . It is stra ight forward to chec k that the jump alo ng T disapp ears and the jump along Σ ± is given by (4.20) ˆ v ( z ) = ( ˜ b + ( z ) , z ∈ Σ + , ˜ b − ( z ) − 1 , z ∈ Σ − . The other jumps (4.1 2), (4.13) as well as the p ole co ndition (4.14) (if prese n t) are unch anged. 20 H. KR ¨ UGER AND G. TESCHL Re(Φ) < 0 Re(Φ) > 0 Re(Φ) > 0 Re(Φ) < 0 Σ 2 − Σ 1 + T Σ 2 + Σ 1 − z − 1 0 z 0 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 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. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Figure 6. Deformed contour Note that the resulting Riemann–Hilb ert problem still satisﬁes our symmetry condition (3.16) since we have ˜ b ± ( z − 1 ) =  0 1 1 0  ˜ b ∓ ( z )  0 1 1 0  . By cons truction all jumps (4.12), (4.1 3), a nd (4.19) are exp onentially close to the ident ity as t → ∞ . The only no n- decaying part b eing the p ole conditio n (4 .14) (if present). Case 2: z 0 ∈ T \ { ± 1 } . In this cas e we will set Σ ± = Σ 1 ± ∪ Σ 2 ± as indicated in Figure 6. Again note that Σ 1 ± resp ectively Σ 2 ∓ lies in the re gion with ± Re(Φ( z )) < 0 and must b e c hosen such that we do not int ersect any other parts of the contour. Then we can split our jump by redeﬁning ˜ m ( z ) according to (4.21) ˆ m ( z ) =                ˜ m ( z ) ˜ b + ( z ) − 1 , z betw een T and Σ 1 + , ˜ m ( z ) ˜ b − ( z ) − 1 , z b etw een T a nd Σ 1 − , ˜ m ( z ) ˜ B + ( z ) − 1 , z b et ween T and Σ 2 + , ˜ m ( z ) ˜ B − ( z ) − 1 , z b et ween T and Σ 2 − , ˜ m ( z ) , else . One c hecks that the jump along T disa ppea r s and the jump along Σ ± is given by (4.22) ˆ v ( z ) =          ˜ b + ( z ) , z ∈ Σ 1 + , ˜ b − ( z ) − 1 , z ∈ Σ 1 − , ˜ B + ( z ) , z ∈ Σ 2 + , ˜ B − ( z ) − 1 , z ∈ Σ 2 − . All o ther jumps (4.12) and (4.13) are unchanged. Aga in the resulting Riemann– Hilber t pro blem s till satisﬁes our symmetry condition (3.16) and the jump a long Σ ± aw a y fr om the stationar y phas e p oints z 0 , z − 1 0 is exp onentially close to the ident ity as t → ∞ . Case 3: z 0 ∈ (0 , 1) . In this case we will s et Σ ± = { z | | z | = (1 − ε ) ± 1 } for some small ε ∈ (0 , 1 ) such that Σ ± lies in the r egion with ∓ Re(Φ( z )) < 0 and such that LONG-TIME ASYMPTOTICS OF THE TODA LA TTICE 21 we do not in tersect the or ig inal contours. Then w e can split our jump by r edeﬁning ˜ m ( z ) acco rding to (4.23) ˆ m ( z ) =      ˜ m ( z ) ˜ B + ( z ) − 1 , (1 − ε ) < | z | < 1 , ˜ m ( z ) ˜ B − ( z ) − 1 , 1 < | z | < (1 − ε ) − 1 , ˜ m ( z ) , else . One c hecks that the jump along T disa ppea r s and the jump along Σ ± is given by (4.24) ˆ v ( z ) = ( ˜ B + ( z ) , z ∈ Σ + , ˜ B − ( z ) − 1 , z ∈ Σ − . The other jumps (4.12), (4.1 3) as well as the pole condition (4.14) (if present) are unchanged. Again the resulting Riemann–Hilber t problem still satisﬁes our symmetry conditio n (3.16) and all jumps (4.12), (4.1 3), and (4.23) are e x ponentially close to the identit y as t → ∞ . The o nly non-decaying part b eing the p ole co ndition (4.14) (if present). In Case 1 and 3 we can immediately apply Theo rem B.6 to ˆ m as follows: If | n t − c k | > ε for all k we can choose γ 0 = 0. Since the error b etw een ˆ w t and ˆ w t 0 is expo nen tially small, this prov es the second part of Theorem 2.1 in the a nalytic case up on compar ing (4.25) m ( z ) = ˆ m ( z )  T ( z , z 0 ) 0 0 T ( z , z 0 ) − 1  with (3.19). The changes necessar y for the general case will b e given in Section 6. Otherwise, if | n t − c k | < ε for some k , we choo se γ t 0 = γ k ( n, t ). Again we conclude that the e rror betw een ˆ w t and ˆ w t 0 is expo nen tially small, proving the ﬁrst par t of Theorem 2.1. The c hanges necessary for the general case will also be given in Section 6. In Case 2 the jump will not decay on the tw o s ma ll cr osses co n taining the sta- tionary phase p o in ts z 0 and z − 1 0 . Hence w e will need to contin ue the inv estigation of this pr oblem in the next sectio n. 5. Reduction to a Riemann–Hilber t problem on a smal l cross In the previous sec tion we hav e shown that for z 0 ∈ T \{± 1 } we can reduce everything to a Riemann– Hilbert problem for ˆ m ( z ) such that the jumps are of order O ( t − 1 ) except in a small neighborho o ds of the stationary phase p oints z 0 and z − 1 0 . Denote b y Σ C ( z ± 1 0 ) the parts o f Σ + ∪ Σ − inside a small neig h b orho o d of z ± 1 0 . In this section we will show that everything can reduced to so lving the tw o problems in the tw o small c rosses Σ C ( z 0 ) resp ectively Σ C ( z − 1 0 ). It will be s lig h tly more conv enien t to use the alternate normaliza tio n (5.1) ˇ m ( z ) = 1 ˜ A ˆ m ( z ) , A = T 0 ˜ A, such that (5.2) ˇ m (0) =  1 1 ˜ A 2  . Without loss of ge ne r ality we can also assume that ˆ Σ consists o f tw o s traight lines in a suﬃcien tly small neig h b orho o d of z 0 . 22 H. KR ¨ UGER AND G. TESCHL W e will need the solution of the corresp onding 2 × 2 ma trix (5.3) M C + ( z ) = M C − ( z ) ˜ v ( z ) , z ∈ Σ C , M C ( ∞ ) = I , where the jump ˜ v is the s a me as for ˜ m ( z ) but restric ted to a neighbor hoo d of one of the tw o crosses Σ C = (Σ + ∪ Σ − ) ∩ { z | | z − z 0 | < ε / 2 } for some small ε > 0. As a ﬁrst step we make a change of co or dinates (5.4) ζ = p − 2 sin( θ 0 ) z 0 i ( z − z 0 ) , z = z 0 + z 0 i p − 2 sin( θ 0 ) ζ such that the phase reads Φ( z ) = iΦ 0 + i 2 ζ 2 + O ( ζ 3 ). Here we hav e set z 0 = e i θ 0 , θ 0 ∈ ( − π , 0 ) , resp ectively co s( θ 0 ) = − n/t , which implies Φ 0 = 2(sin( θ 0 ) − θ 0 cos( θ 0 )) , Φ ′′ ( z 0 ) = 2ie − 2i θ 0 sin( θ 0 ) . The corre sponding Riemann–Hilb ert problem will b e solved in Sec tio n A. T o ap- ply this res ult we need the b ehaviour of our jump matrices near z 0 , tha t is, the behaviour of T ( z , z 0 ) near z → z 0 . Lemma 5.1. L et z 0 ∈ T , t hen (5.5) T ( z , z 0 ) =  − z 0 z − z 0 z − z 0  i ν ˜ T ( z , z 0 ) wher e ν = − 1 π log( | T ( z 0 ) | ) and the br anch cut of the lo garithm use d t o deﬁne z i ν = e i ν log( z ) is chosen along the ne gative r e al axis. Her e ˜ T ( z , z 0 ) = Y ζ k ∈ ( − 1 , 0) | ζ k | z − ζ − 1 k z − ζ k · exp   1 2 π i z 0 Z z 0 log  | T ( s ) | | T ( z 0 ) |  s + z s − z ds s   , is H¨ older c ontinuous of any exp onent less than 1 at z = z 0 and satisﬁes ˜ T ( z 0 , z 0 ) ∈ T . Pr o of. This follows since exp   1 2 π i z 0 Z z 0 log  | T ( z 0 ) |  s + z s − z ds s   =  − z 0 z − z 0 z − z 0  i ν . The pr oper t y ˜ T ( z 0 , z 0 ) ∈ T follows after letting z → z 0 in (4.10).  Now if z ( ζ ) is deﬁned a s in (5.4) and 0 < α < 1, then there is an L > 0 suc h that | T ( z ( ζ ) , z 0 ) − ζ i ν ˜ T ( z 0 , z 0 )e − 3 2 i ν log( − 2 sin( θ 0 )) | ≤ L | ζ | α , where the br anch c ut of ζ i ν is tangent to the nega tiv e r eal ax is. Clea rly we also hav e | R ( z ( ζ )) − R ( z 0 ) | ≤ L | ζ | α and thus the assumptions of Theo rem A.1 are satisﬁed with r = R ( z 0 ) ˜ T ( z 0 , z 0 ) − 2 e 3i ν log( − 2 sin( θ 0 )) LONG-TIME ASYMPTOTICS OF THE TODA LA TTICE 23 and the s o lution of (5.3) is given by M C ( z ) = I − z 0 ( − 2 sin( θ 0 ) t ) 1 / 2 M 0 z − z 0 + O  1 t α  , (5.6) M 0 =  0 − β β 0  , β = √ ν e i( π / 4 − arg( R ( z 0 ))+arg(Γ(i ν ))) ( − 2 sin( θ 0 )) − 3i ν ˜ T ( z 0 , z 0 ) 2 e − i t Φ 0 t − i ν , (5.7) where 1 / 2 < α < 1, and cos( θ 0 ) = − λ 0 . Note | r | = | R ( z 0 ) | and hence ν = − 1 2 π log(1 − | R ( z 0 ) | 2 ). Now we are ready to show Theorem 5.2. The solution ˇ m ( z ) is given by (5.8) ˇ m ( z ) =  1 1  − 1 ( − 2 sin( θ 0 ) t ) 1 / 2 ( m 0 ( z ) + ¯ m 0 ( z )) + O  1 t α  , wher e (5.9) m 0 ( z ) =  β z z − z 0 − β z 0 z − z 0  , ¯ m 0 ( z ) = m 0 ( z ) = m 0 ( z − 1 )  0 1 1 0  . Pr o of. Intro duce m ( z ) by m ( z ) =      ˇ m ( z ) M C ( z ) − 1 , | z − z 0 | ≤ ε , ˇ m ( z ) ˜ M C ( z ) − 1 , | z − 1 − z 0 | ≤ ε, ˇ m ( z ) , else , where ˜ M C ( z ) =  0 1 1 0  M C ( z − 1 )  0 1 1 0  = I − z ( − 2 sin( θ 0 ) t ) 1 / 2 M 0 z − z 0 + O  1 t α  . The Riema nn–Hilbert pr oblem for m has jumps given by v ( z ) =                        M C ( z ) − 1 , | z − z 0 | = ε, M C ( z ) ˆ v ( z ) M C ( z ) − 1 , z ∈ ˆ Σ , ε 2 < | z − z 0 | < ε, I , z ∈ Σ , | z − z 0 | < ε 2 , ˜ M C ( z ) − 1 , | z − 1 − z 0 | = ε, ˜ M C ( z ) ˆ v ( z ) ˜ M C ( z ) − 1 , z ∈ ˆ Σ , ε 2 < | z − 1 − z 0 | < ε, I , z ∈ Σ , | z − 1 − z 0 | < ε 2 , ˆ v ( z ) , else . The jumps are I + O ( t − 1 / 2 ) o n the lo o ps | z − z 0 | = ε , | z − 1 − z 0 | = ε and e ven I + O ( t − α ) on the rest (in the L ∞ norm, hence a lso in the L 2 one). In pa rticular, as in Lemma A.3 we infer k µ −  1 1  k 2 = O ( t − 1 / 2 ) . 24 H. KR ¨ UGER AND G. TESCHL Thu s we hav e with Ω ∞ as in (B.8) m ( z ) =  1 1  + 1 2 π i Z Σ µ ( s ) w ( s )Ω ∞ ( s, z ) =  1 1  + 1 2 π i Z | s − z 0 | = ε µ ( s )( M C ( s ) − 1 − I )Ω ∞ ( s, z ) + 1 2 π i Z | s − 1 − z 0 | = ε µ ( s )( ˜ M C ( s ) − 1 − I )Ω ∞ ( s, z ) + O ( t − α ) =  1 1  + 1 ( − 2 sin( θ 0 ) t ) 1 / 2  1 1  M 0 1 2 π i Z | s − z 0 | = ε z 0 s − z 0 Ω ∞ ( s, z ) + 1 ( − 2 sin( θ 0 ) t ) 1 / 2  1 1  M 0 1 2 π i Z | s − 1 − z 0 | = ε s s − z 0 Ω ∞ ( s, z ) + O ( t − α ) =  1 1  − 1 ( − 2 sin( θ 0 ) t ) 1 / 2 ( m 0 ( z ) + ¯ m 0 ( z )) + O  1 t α  ﬁnishing the pr o o f.  Hence, using (3.19) and (5.1), (5.10) ( ˇ m ( z )) 2 = 1 ˜ A 2  1 + ( T 1 + 2 B ) z + O ( z 2 )  and c omparing with (5.11) ( ˇ m ( z )) 2 =  1 − 2 Re( β ) ( − 2 sin( θ 0 ) t ) 1 / 2  −  2 Re( z 0 β ) ( − 2 sin( θ 0 ) t ) 1 / 2  z + O ( z 2 ) + O  1 t α  , we obtain (5.12) ˜ A 2 = 1 + 2 Re( β ) ( − 2 sin( θ 0 ) t ) 1 / 2 + O  1 t α  and (5.13) T 1 + 2 B = − 2 Re( z 0 β ) ( − 2 sin( θ 0 ) t ) 1 / 2 + O  1 t α  . In summar y we have A = T 0  1 + Re( β ) ( − 2 sin( θ 0 ) t ) 1 / 2 + O  1 t α  , (5.14) B = − 1 2 T 1 − Re( z 0 β ) ( − 2 sin( θ 0 ) t ) 1 / 2 + O  1 t α  , (5.15) which prov es Theore m 2.2 in the analytic case. Remark 5.3. Note that, in c ontr adistinction to The or em B.6, The or em 5.2 do es not re quir e uniform b ounde dness of the asso ciate d inte gr al op er ators, but only s ome know le dge of the solution of the Riemann-Hilb ert pr oblem. However, it r e quir es that the solut ion is of the form I + o (1) and henc e c annot b e use d in the soliton r e gion. LONG-TIME ASYMPTOTICS OF THE TODA LA TTICE 25 6. Anal ytic A ppr oxima tion In this section we wan t to present the necessary changes in the case where the reﬂection co e ﬃcie n t do es not hav e an a nalytic extensio n. The idea is to use an analytic approximation and split the r eﬂection in an analytic part plus a small rest. The analytic part will b e mov ed to the complex plane while the rest remains on the unit circle. This nee ds to be done in such a wa y that the r est is of O ( t − l ) and the growth of the ana lytic par t can b e controlled by the decay of the phase. In the soliton region a stra ight forward splitting based on the F ourier ser ies (6.1) R ( z ) = ∞ X k = −∞ ˆ R ( k ) z k will be suﬃcient. It is well-kno wn that our assumption (2.2) implies k l ˆ R ( − k ) ∈ ℓ 1 ( N ) (this follows from the estimate [38, eq. (1 0 .83)]) a nd R ∈ C l ( T ). Lemma 6.1. Supp ose ˆ R ( k ) ∈ ℓ 1 ( Z ) , k l ˆ R ( − k ) ∈ ℓ 1 ( N ) and let 0 < ε < 1 , β > 0 b e given. Then we c an split the r eﬂe ct ion c o eﬃcient ac c or ding to R ( z ) = R a,t ( z ) + R r,t ( z ) such that R a,t ( z ) is analytic in 0 < | z | < 1 and (6.2) | R a,t ( z )e − β t | = O ( t − l ) , 1 − ε ≤ | z | ≤ 1 , | R r,t ( z ) | = O ( t − l ) , | z | = 1 . Pr o of. W e choose R a,t ( z ) = P ∞ k = − K ( t ) ˆ R ( k ) z k with K ( t ) = ⌊ β 0 − log(1 − ε ) t ⌋ for some po sitiv e β 0 < β . Then, for 1 − ε ≤ | z | , | R a,t ( z )e − β t | ≤ ∞ X k = − K ( t ) | ˆ R ( k ) | e − β t (1 − ε ) k ≤ k ˆ R k 1 e − β t (1 − ε ) − K ( t ) ≤ k ˆ R k 1 e − ( β − β 0 ) t Similarly , for | z | = 1, | R r,t ( z ) | ≤ − K ( t ) − 1 X k = −∞ | ˆ R ( k ) | ≤ const ∞ X k = K ( t )+1 k l K ( t ) l | ˆ R ( − k ) | ≤ const K ( t ) l ≤ const t l .  T o apply this lemma in the soliton region z 0 ∈ ( − 1 , 0) we c ho ose (6.3) β = min | z | =1 − ε − Re(Φ( z )) > 0 and s plit R ( z ) = R a,t ( z ) + R r,t ( z ) according to Lemma 6.1 to o btain ˜ b ± ( z ) = ˜ b a,t, ± ( z ) ˜ b r,t, ± ( z ) = ˜ b r,t, ± ( z ) ˜ b a,t, ± ( z ) . Here ˜ b a,t, ± ( z ), ˜ b r,t, ± ( z ) denote the matrice s obtained from ˜ b ± ( z ) as deﬁned in (4 .16) by replacing R ( z ) with R a,t ( z ), R r,t ( z ), res pectively . Now we can mov e the a nalytic parts into the complex plane as in Se c tio n 4 while leaving the rest o n T . Hence, rather then (4 .2 0), the jump now r eads (6.4) ˆ v ( z ) =      ˜ b a,t, + ( z ) , z ∈ Σ + , ˜ b a,t, − ( z ) − 1 , z ∈ Σ − , ˜ b r,t, − ( z ) − 1 ˜ b r,t, + ( z ) , z ∈ T . By constructio n we hav e ˆ v ( z ) = I + O ( t − l ) on the whole co n tour and the r est follows as in Section 4. 26 H. KR ¨ UGER AND G. TESCHL In the other soliton regio n z 0 ∈ (0 , 1) we pr o c eed similarly , with the only dif- ference that the jump matrices ˜ B ± ( z ) have a t ﬁrst sight more complicated oﬀ diagonal entries. T o r e medy this w e will rewr ite them in terms o f left rather then right sc attering data. F o r this purpose let us use the notation R r ( z ) ≡ R + ( z ) for the r ig h t and R l ( z ) ≡ R − ( z ) for the left reﬂection co eﬃcient. Moreov er, let T r ( z , z 0 ) ≡ T ( z , z 0 ) b e the right and T l ( z , z 0 ) ≡ T ( z ) /T ( z , z 0 ) be the left partial transmission co eﬃcien t. With this notation we hav e (6.5) ˜ v ( z ) = ( ˜ b − ( z ) − 1 ˜ b + ( z ) , λ ( z ) > λ 0 , ˜ B − ( z ) − 1 ˜ B + ( z ) , λ ( z ) < λ 0 , where ˜ b − ( z ) = 1 R r ( z − 1 )e − t Φ( z ) T r ( z − 1 ,z 0 ) 2 0 1 ! , ˜ b + ( z ) = 1 0 R r ( z )e t Φ( z ) T r ( z ,z 0 ) 2 1 ! , and ˜ B − ( z ) = 1 0 − T r, − ( z ,z 0 ) − 2 | T ( z ) | 2 R r ( z )e t Φ( z ) 1 ! , ˜ B + ( z ) = 1 − T r, + ( z ,z 0 ) 2 | T ( z ) | 2 R r ( z − 1 )e − t Φ( z ) 0 1 ! . Using (3.9) together with (4.18) we can further write ˜ B − ( z ) = 1 0 R l ( z − 1 )e − t Φ( z ) T l ( z − 1 ,z 0 ) 2 1 ! , ˜ B + ( z ) = 1 R l ( z )e t Φ( z ) T l ( z ,z 0 ) 2 0 1 ! . Now we can pro ceed as be fore with ˜ B ± ( z ) as with ˜ b ± ( z ) b y splitting R l ( z ) rather than R r ( z ). In the similarity regio n we need to take the small vicinities of the stationary phase p oin ts into account. Since the phas e is quadratic near these points, w e cannot use it to dominate the exp onential g rowth of the analy tic part aw a y fr o m the unit cir cle. Hence we will take the phase as a ne w v ar iable and use the F o urier transform with resp ect to this new v ariable. Since this change of co or dinates is singular near the stationary phase p oints, there is a price we hav e to pay , na mely , requiring additional smo othness for R ( z ). W e b egin with Lemma 6.2. Supp ose R ( z ) ∈ C 5 ( T ) . Then we c an split R ( z ) ac c or ding to (6.6) R ( z ) = R 0 ( z ) + ( z − z 0 )( z − z 0 ) H ( z ) , z ∈ Σ( z 0 ) , wher e R 0 ( z ) is a r e al p olynomia l in z such that H ( z ) vanishes at z 0 , z 0 of or der thr e e and has a F ourier series (6.7) H ( z ) = ∞ X k = −∞ ˆ H k e kω 0 Φ( z ) , ω 0 = π π co s( θ 0 ) + Φ 0 , with k ˆ H k summable. H er e Φ 0 = Φ( z 0 ) / i . LONG-TIME ASYMPTOTICS OF THE TODA LA TTICE 27 ❅ ❅ ❅ ❅ ❅ ❅ ❅ ❅ ❅ ❅ ❅ ❅ ❅ ❅ ❅ ❅ ❅ ❘ ❅ ❅ ❘                  ✒   ✒ Σ 1 Σ 2 Σ 3 Σ 4  1 − R 1 ( z ) · · · 0 1   1 0 R 2 ( z ) · · · 1   1 − R 3 ( z ) · · · 0 1   1 0 R 4 ( z ) · · · 1  Figure 7. Contours of a cros s Pr o of. By cho osing a p olynomial R 0 we can match the v alues of R and its ﬁr st four deriv a tiv es at z 0 , z 0 . Hence H ( z ) ∈ C 4 ( T ) and v anishes tog ether with its ﬁr st three deriv a tiv es at z 0 , z 0 . When r estricted to Σ( z 0 ) the phas e Φ( z ) / i g ives a one to one co ordina te tr ans- form Σ ( z 0 ) → [iΦ 0 , iΦ 0 + i ω 0 ] and we can hence expres s H ( z ) in this new co or dina te. The co o r dinate tra nsform loca lly lo oks like a squar e ro ot near z 0 and z 0 , how ev er, due to our ass umption that H v anis hes there, H is still C 2 in this new co o rdinate and the F ourier tra nsform with resp ect to this new co ordinates exists a nd ha s the required prop erties.  Moreov er, as in Lemma 6.1 we obtain: Lemma 6. 3. L et H ( z ) b e as in the pr evious lemma. Then we c an split H ( z ) ac c or ding to H ( z ) = H a,t ( z ) + H r,t ( z ) such that H a,t ( z ) is analytic in the re gion Re(Φ( z )) < 0 and (6.8) | H a,t ( z )e Φ( z ) t/ 2 | = O (1) , Re(Φ( z )) < 0 , | z | ≤ 1 , | H r,t ( z ) | = O ( t − 1 ) , | z | = 1 . Pr o of. W e c ho ose H a,t ( z ) = P ∞ k = − K ( t ) ˆ H k e kω Φ( z ) with K ( t ) = ⌊ t / (2 ω ) ⌋ . The re s t follows a s in Lemma 6.1.  By co nstruction R a,t ( z ) = R 0 ( z ) + ( z − z 0 )( z − z 0 ) H a,t ( z ) will sa tisfy the requir ed Lipschitz estimate in a vicinity of the statio na ry phase po in ts (uniformly in t ) and all jumps will b e I + O ( t − 1 ). Hence we can pro ceed a s in Section 5. Appendix A. The sol ution on a small cr oss Int ro duce the cross Σ = Σ 1 ∪ · · · ∪ Σ 4 (see Fig ure 7 ) by Σ 1 = { u e − i π / 4 , u ∈ [0 , ∞ ) } Σ 2 = { u e i π / 4 , u ∈ [0 , ∞ ) } Σ 3 = { u e 3i π / 4 , u ∈ [0 , ∞ ) } Σ 4 = { u e − 3i π / 4 , u ∈ [0 , ∞ ) } . (A.1) Orient Σ such that the real part of z increases in the positive direction. Denote by D = { z , | z | < 1 } the op en unit disc. Throughout this s ection z i ν will denote the function e i ν log( z ) , whe r e the branch c ut o f the logarithm is chosen a long the negative real axis ( −∞ , 0). 28 H. KR ¨ UGER AND G. TESCHL Int ro duce the following jump matrices ( v j for z ∈ Σ j ) v 1 =  1 − R 1 ( z ) z 2i ν e − t Φ( z ) 0 1  , v 2 =  1 0 R 2 ( z ) z − 2i ν e t Φ( z ) 1  , v 3 =  1 − R 3 ( z ) z 2i ν e − t Φ( z ) 0 1  , v 4 =  1 0 R 4 ( z ) z − 2i ν e t Φ( z ) 1  . (A.2) Now consider the RHP given b y m + ( z ) = m − ( z ) v j ( z ) , z ∈ Σ j , j = 1 , 2 , 3 , 4 , (A.3) m ( z ) → I , z → ∞ . W e hav e the next theor e m, in which we follow the co mputations o f Sectio ns 3 a nd 4 in [4]. The method can b e found in earlier liter ature, see for example [19]. One can also ﬁnd ar gumen ts lik e this in Section 5 in [20] or (3.65 ) to (3.76) in [6]. W e will allow so me v ariation, in a ll par a meters as indicated in the next result. Theorem A. 1. Ther e is some ρ 0 > 0 such that v j ( z ) = I for | z | > ρ 0 . Mor e over, supp ose that within | z | ≤ ρ 0 the fol lowing estimates hold: (i) The phase s atisﬁ es Φ(0) = iΦ 0 ∈ i R , Φ ′ (0) = 0 , Φ ′′ (0) = i and ± Re  Φ( z )  ≥ 1 4 | z | 2 , ( + for z ∈ Σ 1 ∪ Σ 3 , − else , (A.4) | Φ( z ) − Φ(0) − i z 2 2 | ≤ C | z | 3 . (A.5) (ii) Ther e is some r ∈ D and c onstants ( α, L ) ∈ (0 , 1] × (0 , ∞ ) such that R j , j = 1 , . . . , 4 , satisfy H¨ older c onditions of the form | R 1 ( z ) − r | ≤ L | z | α , | R 2 ( z ) − r | ≤ L | z | α , | R 3 ( z ) − r 1 − | r | 2 | ≤ L | z | α , | R 4 ( z ) − r 1 − | r | 2 | ≤ L | z | α . (A.6) Then the solution of t he RHP (A.3) satisﬁes (A.7) m ( z ) = I + 1 z i t 1 / 2  0 − β β 0  + O ( t − 1+ α 2 ) , for | z | > ρ 0 , wher e (A.8) β = √ ν e i( π / 4 − arg( r )+ arg (Γ(i ν ))) e − i t Φ 0 t − i ν , ν = − 1 2 π log(1 − | r | 2 ) . F u rthermor e, if R j ( z ) and Φ( z ) dep end on some p ar ameter, the erro r term is u ni- form with r esp e ct to this p ar ameter as long as r r emains within a c omp act subset of D and the c onstants in the ab ove estimates c an b e chosen indep endent of the p ar ameters. W e remar k that the so lution of the RHP (A.3) is unique. This follows fro m the usual Liouville argument [3, Lem. 7.18] s ince det ( v j ) = 1. Note that the actual v alue of ρ 0 is of no impor tance. In fact, if w e choose 0 < ρ 1 < ρ 0 , then the solution ˜ m of the problem with jump ˜ v , where ˜ v is equal to v for | z | < ρ 1 and I otherwise, diﬀers from m only by an e x ponentially small error. This a lready indicates, that we should b e able to replace R j ( z ) by their resp ective v alues at z = 0 . T o see this we sta rt by rewriting our RHP as a s ingular in tegra l equation. W e will use the theor y develop ed in App endix B for the cas e of 2 × 2 LONG-TIME ASYMPTOTICS OF THE TODA LA TTICE 29 matrix v alued functions with m 0 ( z ) = I and the usual Cauch y kernel (since we won’t require symmetry in this s ection) Ω( s, z ) = I ds s − z . Moreov er, since our contour is un b ounded, we will assume w ∈ L 1 (Σ) ∩ L 2 (Σ). All results fr om App endix B s till hold in this case with so me straightforward mo diﬁca- tions if one observes that µ − I ∈ L 2 (Σ). Indeed, a s in Theorem B.3, in the sp ecial case b + ( z ) = v j ( z ) and b − ( z ) = I for z ∈ Σ j , we obtain (A.9) m ( z ) = I + 1 2 π i Z Σ µ ( s ) w ( s ) ds s − z , where µ − I is the s o lution of the singular integral equatio n (A.10) ( I − C w )( µ − I ) = C w I , that is, (A.11) µ = I + ( I − C w ) − 1 C w I , C w f = C − ( wf ) . Here C denotes the us ual Cauch y op erator and we se t w ( z ) = w + ( z ) (since w − ( z ) = 0). As our ﬁr st step we will get rid of some constants and rescale the entire problem by setting (A.12) ˆ m ( z ) = D ( t ) − 1 m ( z t − 1 / 2 ) D ( t ) , where (A.13) D ( t ) =  d ( t ) − 1 0 0 d ( t )  , d ( t ) = e i t Φ 0 / 2 t i ν / 2 , d ( t ) − 1 = d ( t ) . Then o ne easily chec ks that ˆ m ( z ) solves the RHP ˆ m + ( z ) = ˆ m − ( z ) ˆ v j ( z ) , z ∈ Σ j , j = 1 , 2 , 3 , 4 , (A.14) ˆ m ( z ) → I , z → ∞ , z / ∈ Σ , where ˆ v j ( z ) = D ( t ) − 1 v j ( z t − 1 / 2 ) D ( t ), j = 1 , . . . , 4, explicitly ˆ v 1 ( z ) =  1 − R 1 ( z t − 1 / 2 ) z 2i ν e − t (Φ( z t − 1 / 2 ) − Φ(0)) 0 1  , ˆ v 2 ( z ) =  1 0 R 2 ( z t − 1 / 2 ) z − 2i ν e t (Φ( zt − 1 / 2 ) − Φ(0)) 1  , ˆ v 3 ( z ) =  1 − R 3 ( z t − 1 / 2 ) z 2i ν e − t (Φ( z t − 1 / 2 ) − Φ(0)) 0 1  , ˆ v 4 ( z ) =  1 0 R 2 ( z t − 1 / 2 ) z − 2i ν e t (Φ( zt − 1 / 2 ) − Φ(0)) 1  . (A.15) Our next aim is to show that the solution ˆ m ( z ) of the resc a led pro blem is clos e to the solution ˆ m c ( z ) of the RHP ˆ m c + ( z ) = ˆ m c − ( z ) ˆ v c j ( z ) , z ∈ Σ j , j = 1 , 2 , 3 , 4 , (A.16) ˆ m c ( z ) → I , z → ∞ , z / ∈ Σ , 30 H. KR ¨ UGER AND G. TESCHL asso ciated with the following jump matr ices ˆ v c 1 ( z ) =  1 − rz 2i ν e − i z 2 / 2 0 1  , ˆ v c 2 ( z ) =  1 0 rz − 2i ν e i z 2 / 2 1  , ˆ v c 3 ( z ) = 1 − r 1 −| r | 2 z 2i ν e − i z 2 / 2 0 1 ! , ˆ v c 4 ( z ) = 1 0 r 1 −| r | 2 z − 2i ν e i z 2 / 2 1 ! . (A.17) The diﬀere nc e betw een these jump matr ices ca n b e estimated as follows. Lemma A.2. The matric es ˆ w c and ˆ w ar e close in the sense that (A.18) ˆ w j ( z ) = ˆ w c j ( z ) + O ( t − α/ 2 e −| z | 2 / 8 ) , z ∈ Σ j , j = 1 , . . . 4 . F u rthermor e, the err or term is uniform with r esp e ct to p ar ameters as s t ate d in The or em A.1. Pr o of. W e only give the pro of z ∈ Σ 1 , the other cases b eing similar. T he r e is o nly one nonzero ma trix entry in ˆ w j ( z ) − ˆ w c j ( z ) given by W = ( − R 1 ( z t − 1 / 2 ) z 2i ν e − t (Φ( z t − 1 / 2 ) − Φ(0)) + r z 2i ν e − i z 2 / 2 , | z | ≤ ρ 0 t 1 / 2 , r z 2i ν e − i z 2 / 2 | z | > ρ 0 t 1 / 2 . A straightforw ard estimate for | z | ≤ ρ 0 t 1 / 2 shows | W | = e ν π / 4 | R 1 ( z t − 1 / 2 )e − t ˆ Φ( zt − 1 / 2 ) − r | e −| z | 2 / 2 ≤ e ν π / 4 | R 1 ( z t − 1 / 2 ) − r | e Re( − t ˆ Φ( zt − 1 / 2 )) −| z | 2 / 2 + e ν π / 4 | e − t ˆ Φ( z t − 1 / 2 ) − 1 | e −| z | 2 / 2 ≤ e ν π / 4 | R 1 ( z t − 1 / 2 ) − r | e −| z | 2 / 4 + e ν π / 4 t | ˆ Φ( z t − 1 / 2 ) | e −| z | 2 / 4 , where ˆ Φ( z ) = Φ( z ) − Φ(0) − i 2 z 2 = Φ ′′′ (0) 6 z 3 + . . . . Her e w e hav e used i 2 z 2 = 1 2 | z | 2 for z ∈ Σ 1 and Re( − t ˆ Φ( z t − 1 / 2 )) ≤ | z | 2 / 4 by (A.4). F urthermore, by (A.5) and (A.6), | W | ≤ e ν π / 4 Lt − α/ 2 | z | α e −| z | 2 / 4 + e ν π / 4 C t − 1 / 2 | z | 3 e −| z | 2 / 4 , for | z | ≤ ρ 0 t 1 / 2 . F or | z | > ρ 0 t 1 / 2 we hav e | W | ≤ e ν π / 4 e −| z | 2 / 2 ≤ e ν π / 4 e − ρ 2 0 t/ 4 e −| z | 2 / 4 which ﬁnishes the pro of.  The next lemma allows us, to replace ˆ m ( z ) by ˆ m c ( z ). Lemma A.3. Consider the R HP m + ( z ) = m − ( z ) v ( z ) , z ∈ Σ , (A.19) m ( z ) → I , z → ∞ , z / ∈ Σ . Assume t hat w ∈ L 2 (Σ) ∩ L ∞ (Σ) . Then (A.20) k µ − I k 2 ≤ c k w k 2 1 − c k w k ∞ pr ovide d c k w k ∞ < 1 , wher e c is the norm of the Cauchy op er ator on L 2 (Σ) . Pr o of. This follows since ˜ µ = µ − I ∈ L 2 (Σ) satisﬁes ( I − C w ) ˜ µ = C w I .  LONG-TIME ASYMPTOTICS OF THE TODA LA TTICE 31 Lemma A.4. The solution ˆ m ( z ) has a c onver gent asymptotic exp ansion (A.21) ˆ m ( z ) = I + 1 z ˆ M ( t ) + O ( 1 z 2 ) for | z | > ρ 0 t 1 / 2 with the err or t erm uniformly in t . Mor e over, (A.22) ˆ M ( t ) = ˆ M c + O ( t − α/ 2 ) . Pr o of. Cons ider ˆ m d ( z ) = ˆ m ( z ) ˆ m c ( z ) − 1 , whose jump ma trix is g iv en by ˆ v d ( z ) = ˆ m c − ( z ) ˆ v ( z ) ˆ v c ( z ) − 1 ˆ m c − ( z ) − 1 = I + ˆ m c − ( z )  ˆ w ( z ) − ˆ w c ( z )  ˆ m c − ( z ) − 1 . By Lemma A.2, w e hav e tha t ˆ w − ˆ w c is decaying of order t − α/ 2 in the norms of L 1 and L ∞ and thus the same is true for ˆ w d = ˆ v d − I . Hence by the previous lemma k ˆ µ d − I k 2 = O ( t − α/ 2 ) . F urthermo re, by ˆ µ d = ˆ m d − = ˆ m − ( ˆ m c − ) − 1 = ˆ µ ( ˆ µ c ) − 1 we infer k ˆ µ − ˆ µ c k 2 = O ( t − α/ 2 ) since ˆ µ c is b ounded. Now ˆ m ( z ) = I − 1 2 π i 1 z Z Σ ˆ µ ( s ) ˆ w ( s ) ds + 1 2 π i 1 z Z Σ s ˆ µ ( s ) ˆ w ( s ) ds s − z shows (recall that ˆ w is supp orted inside | z | ≤ ρ 0 t 1 / 2 ) ˆ m ( z ) = I + 1 z ˆ M ( t ) + O ( k ˆ µ ( s ) k 2 k s ˆ w ( s ) k 2 z 2 ) , where ˆ M ( t ) = − 1 2 π i Z Σ ˆ µ ( s ) ˆ w ( s ) ds. Now the rest follows fro m ˆ M ( t ) = ˆ M c − 1 2 π i Z Σ ( ˆ µ ( s ) ˆ w ( s ) − ˆ µ c ( s ) ˆ w c ( s )) ds using k ˆ µ ˆ w − ˆ µ c ˆ w c k 1 ≤ k ˆ w − ˆ w c k 1 + k ˆ µ − I k 2 k ˆ w − ˆ w c k 2 + k ˆ µ − ˆ µ c k 2 k ˆ w c k 2 .  Finally , it re ma ins to solve (A.16) and to show: Theorem A.5. The solution of the RHP (A.16) is of t he form (A.23) ˆ m c ( z ) = I + 1 z ˆ M c + O ( 1 z 2 ) , wher e ˆ M c = i  0 − β β 0  , β = √ ν e i( π / 4 − arg( r )+ arg (Γ(i ν ))) . (A.24) The err or t erm is uniform with r esp e ct t o r in c omp act subsets of D . Mor e over, the solution is b ounde d (again uniformly with r esp e ct to r ). Given this result, Theorem A.1 follows from Lemma A.4 m ( z ) = D ( t ) ˆ m ( z t 1 / 2 ) D ( t ) − 1 = I + 1 t 1 / 2 z D ( t ) ˆ M ( t ) D ( t ) − 1 + O ( z − 2 t − 1 ) = I + 1 t 1 / 2 z D ( t ) ˆ M c D ( t ) − 1 + O ( t − (1+ α ) / 2 ) (A.25) for | z | > ρ 0 , since D ( t ) is b ounded. 32 H. KR ¨ UGER AND G. TESCHL ❅ ❅ ❅ ❅ ❅ ❅ ❅ ❅ ❅ ❅ ❅ ❅ ❅ ❅ ❅ ❅ ❅ ❘ ❅ ❅ ❘                  ✒   ✒ ✲ ✲ Σ 1 Σ 2 Σ 3 Σ 4 R Ω 1 Ω 2 Ω 3 Ω 4 Figure 8. Deforming back the cro s s The pro of of this r esult will be given in the remainder o f this section. In or de r to so lve (A.16) we b egin with a deformation which moves the jump to R as follows. Denote the r egion enclosed b y R and Σ j as Ω j (cf. Figure 8) and deﬁne (A.26) ˜ m c ( z ) = ˆ m c ( z ) ( D 0 ( z ) D j , z ∈ Ω j , j = 1 , . . . , 4 , D 0 ( z ) , else , where D 0 ( z ) = z i ν e − i z 2 / 4 0 0 z − i ν e i z 2 / 4 ! , and D 1 =  1 r 0 1  D 2 =  1 0 r 1  D 3 =  1 − r 1 −| r | 2 0 1  D 4 =  1 0 − r 1 −| r | 2 1  . Lemma A.6. The function ˜ m c ( z ) deﬁne d in (A.26) satisﬁes the RHP ˜ m c + ( z ) = ˜ m c − ( z )  1 − | r | 2 − r r 1  , z ∈ R (A.27) ˜ m c ( z ) = ( I + 1 z ˆ M c + . . . ) D 0 ( z ) , z → ∞ , π 4 < a rg( z ) < 3 π 4 . Pr o of. Firs t, o ne c hecks that ˜ m c + ( z ) = ˜ m c − ( z ) D 0 ( z ) − 1 ˆ v c 1 ( z ) D 0 ( z ) D 1 = ˜ m c − ( z ), z ∈ Σ 1 and similarly for z ∈ Σ 2 , Σ 3 , Σ 4 . T o compute the jump alo ng R observe that, by our choice of br anc h cut for z i ν , D 0 ( z ) has a jump a long the nega tiv e real axis given by D 0 , ± ( z ) = e (log | z |± i π )i ν e − i z 2 / 4 0 0 e − (log | z |± i π )i ν e i z 2 / 4 ! , z < 0 . Hence the jump a long R is given by D − 1 1 D 2 , z > 0 and D − 1 4 D − 1 0 , − ( z ) D 0 , + ( z ) D 3 , z < 0 , and (A.27 ) follows after reca lling e − 2 π ν = 1 − | r | 2 .  LONG-TIME ASYMPTOTICS OF THE TODA LA TTICE 33 Now, we can follow (4.17) to (4 .5 1) in [4] to construct an approximate solution. The idea is as follows, since the jump matrix for (A.27), the der iv ative d dz ˜ m c ( z ) has the same jump and hence is given by n ( z ) ˜ m c ( z ), where the entire matrix n ( z ) can b e determined fro m the b ehaviour z → ∞ . Since this will just serv e as a motiv atio n for our ansatz, we will not w orr y ab out justifying any steps. F or z in the sector π 4 < arg( z ) < 3 π 4 (enclosed by Σ 2 and Σ 3 ) we hav e ˜ m c ( z ) = ˆ m c ( z ) D 0 ( z ) and hence  d dz ˜ m c ( z ) + i z 2 σ 3 ˜ m c ( z )  ˜ m c ( z ) − 1 =  i( ν z − z 2 ) ˆ m c ( z ) σ 3 + d dz ˆ m c ( z ) + i z 2 σ 3 ˆ m c ( z )  ˆ m c ( z ) − 1 = i 2 [ σ 3 , ˆ M c ] + O ( 1 z ) , σ 3 =  1 0 0 − 1  . Since the left hand side ha s no jump, it is entire and hence by Liouv ille’s theorem a constant given by the right hand side. In other words, (A.28) d dz ˜ m c ( z ) + i z 2 σ 3 ˜ m c ( z ) = β ˜ m c ( z ) , β =  0 β 12 β 21 0  = i 2 [ σ 3 , ˆ M c ] . This diﬀeren tial equation c an be solved in terms of parab olic cylinder function which then gives the solution of (A.27). Lemma A.7. The RHP (A.27) has a unique solution, and the term ˆ M c is given by ˆ M c = i  0 − β 12 β 21 0  , β 12 = β 21 = √ ν e i( π / 4 − arg( r )+ arg (Γ(i ν ))) . (A.29) Pr o of. Uniqueness follows by the sta ndard Liouville argument since the deter mina n t of the jump ma trix is equal to 1. T o ﬁnd the so lution w e use the a ns atz ˜ m c ( z ) =  ψ 11 ( z ) ψ 12 ( z ) ψ 21 ( z ) ψ 22 ( z )  , where the functions ψ j k ( z ) satisfy ψ ′′ 11 ( z ) = −  i 2 + 1 4 z 2 − β 12 β 21  ψ 11 ( z ) , ψ 12 ( z ) = 1 β 21  d dz − i z 2  ψ 22 ( z ) , ψ 21 ( z ) = 1 β 12  d dz + i z 2  ψ 11 ( z ) , ψ ′′ 22 ( z ) =  i 2 − 1 4 z 2 + β 12 β 21  ψ 22 ( z ) . That is , ψ 11 (e 3 π i / 4 ζ ) sa tisﬁes the par abo lic cylinder equation D ′′ ( ζ ) +  a + 1 2 − 1 4 ζ 2  D ( ζ ) = 0 with a = i β 12 β 21 and ψ 22 (e i π / 4 ζ ) satisﬁes the parab olic cylinder equation with a = − i β 12 β 21 . 34 H. KR ¨ UGER AND G. TESCHL Let D a be the entire pa rab olic cylinder function of § 16.5 in [42] and set ψ 11 ( z ) = ( e − 3 π ν / 4 D i ν ( − e i π / 4 z ) , Im( z ) > 0 , e π ν / 4 D i ν (e i π / 4 z ) , Im( z ) < 0 , ψ 22 ( z ) = ( e π ν / 4 D − i ν ( − ie i π / 4 z ) , Im( z ) > 0 , e − 3 π ν / 4 D − i ν (ie i π / 4 z ) , Im( z ) < 0 . Using the a symptotic b ehavior D a ( z ) = z a e − z 2 / 4  1 − a ( a − 1) 2 z 2 + O ( z − 4 )  , z → ∞ , | ar g( z ) | ≤ 3 π / 4 , shows that the c hoice β 12 β 21 = ν ensures the correct asymptotics ψ 11 ( z ) = z i ν e − i z 2 / 4 (1 + O ( z − 2 )) , ψ 12 ( z ) = − i β 12 z − i ν e i z 2 / 4 ( z − 1 + O ( z − 3 )) , ψ 21 ( z ) = i β 21 z i ν e − i z 2 / 4 ( z − 1 + O ( z − 3 )) , ψ 22 ( z ) = z − i ν e i z 2 / 4 (1 + O ( z − 2 )) , as z → ∞ inside the half plane Im( z ) ≥ 0. In particula r , ˜ m c ( z ) =  I + 1 z ˆ M c + O ( z − 2 )  D 0 ( z ) with ˆ M c = i  0 − β 12 β 21 0  . It r emains to chec k that we hav e the co rrect jump. Since b y construction bo th limits ˜ m c + ( z ) and ˜ m c − ( z ) satisfy the same diﬀer e n tial equation (A.28), there is a constant matrix v such that ˜ m c + ( z ) = ˜ m c − ( z ) v . Moreover, since the co eﬃcien t matrix of the linear diﬀerential equatio n (A.28) has trace 0, the determinant of ˜ m c ± ( z ) is constant and hence det( ˜ m c ± ( z )) = 1 by o ur asy mptotics. Moreov er, a straightforward calculation shows v = ˜ m c − (0) − 1 ˜ m c + (0) = e − 2 π ν − √ 2 π e − i π/ 4 e − πν / 2 √ ν Γ(i ν ) γ − 1 √ 2 π e i π/ 4 e − πν / 2 √ ν Γ( − i ν ) γ 1 ! where γ = √ ν β 12 = β 21 √ ν . Here we hav e used D a (0) = 2 a/ 2 √ π Γ((1 − a ) / 2) , D ′ a (0) = − 2 (1+ a ) / 2 √ π Γ( − a/ 2) plus the duplica tion formula Γ( z )Γ( z + 1 2 ) = 2 1 − 2 z √ π Γ(2 z ) for the Gamma function. Hence, if w e c ho ose γ = √ ν Γ( − i ν ) √ 2 π e i π / 4 e − π ν / 2 r , we hav e v =  1 − | r | 2 − r r 1  since | γ | 2 = 1. T o see this use | Γ( − i ν ) | 2 = Γ(1 − i ν )Γ(i ν ) − i ν = π ν sinh( π ν ) , which follows from Euler’s r eﬂection for m ula Γ(1 − z )Γ( z ) = π sin( π z ) for the Gamma function. In particular, β 12 = β 21 = √ ν e i( π / 4 − arg( r )+arg(Γ(i ν ))) , which ﬁnishes the pro of.  LONG-TIME ASYMPTOTICS OF THE TODA LA TTICE 35 Remark A.8. A n insp e ction of the pr o of shows that ˆ m c is given by the solution of a diﬀer ential e quation dep ending analytic al ly on ν . Henc e, ˆ m c dep ends analytic al ly on ν = − 1 2 π log(1 − | r | 2 ) . This implies lo c al Lipschitz dep endenc e on r as long as r ∈ D . Appendix B. Singular integral equa tions In this se ction we show ho w to transfor m a mer omorphic vector Riemann–Hilb ert problem with simple po le s at ζ , ζ − 1 , m + ( z ) = m − ( z ) v ( z ) , z ∈ Σ , Res ζ m ( z ) = lim z → ζ m ( z )  0 0 − ζ γ 0  , Res ζ − 1 m ( z ) = lim z → ζ − 1 m ( z )  0 ζ − 1 γ 0 0  , (B.1) m ( z − 1 ) = m ( z )  0 1 1 0  , m (0) =  1 m 2  , where ζ ∈ ( − 1 , 0) ∪ (0 , 1) and γ ≥ 0, in to a singula r integral equation. Since we r equire the symmetry condition for our Riemann–Hilb ert problems we need to adapt the usua l Cauchy kernel to pres erve this symmetr y . Mor eov er, w e keep the single soliton as a n inhomogeneous term which will pla y the role of the leading asymptotics in o ur a pplications. Hyp othesis B. 1. S u pp ose the jump data (Σ , v ) satisfy t he fol lowing assumptions: (i) Σ c onsist of a ﬁnite nu mb er of smo oth oriente d ﬁ nite curves in C which interse ct at most ﬁnitely many times with al l interse ct ions b eing tr ansver- sal. (ii) Σ do es not c ont ain 0 , ζ ± 1 . (iii) Σ is invariant under z 7→ z − 1 and is oriente d such that under t he mapping z 7→ z − 1 se qu enc es c onver ging fr om the p ositive side d to Σ ar e mapp e d t o se qu enc es c onver ging to the ne gative side. (iv) The jump matrix v is invertib le and c an b e factorize d ac c or ding to v = b − 1 − b + = ( I − w − ) − 1 ( I + w + ) , wher e w ± = ± ( b ± − I ) ar e c ont inu ous and satisfy (B.2) w ± ( z − 1 ) =  0 1 1 0  w ∓ ( z )  0 1 1 0  , z ∈ Σ . The classical Cauc hy-transform of a function f : Σ → C which is squar e inte- grable is the ana lytic function C f : C \ Σ → C given by (B.3) ( C f )( z ) = 1 2 π i Z Σ f ( s ) s − z ds, z ∈ C \ Σ . Denote the non-tang en tial bounda ry v alues from b oth sides (tak en p ossibly in the L 2 -sense — see e .g. [3, eq . (7.2)]) by C + f re s pectively C − f . The n it is well-kno wn that C + and C − are b ounded op erator s L 2 (Σ) → L 2 (Σ), which satisfy C + − C − = I and C + C − = 0 (see e.g . [1]). Mo reov er, one has the Plemelj–Sokhots k y formula ([31]) C ± = 1 2 (i H ± I ) , 36 H. KR ¨ UGER AND G. TESCHL where (B.4) ( H f )( t ) = 1 π − Z Σ f ( s ) t − s ds, t ∈ Σ , is the Hilb ert trans fo rm and − R denotes the principal v alue integral. In order to re spect the symmetry condition we will restr ic t our attent ion to the set L 2 s (Σ) of square integrable functions f : Σ → C 2 such that (B.5) f ( z − 1 ) = f ( z )  0 1 1 0  . Clearly this will only be p ossible if w e require our jump data to be sy mmetric as well (i.e., Hypo thesis B.1 holds). Next we in tro duce the Cauch y op erator (B.6) ( C f )( z ) = 1 2 π i Z Σ f ( s )Ω ζ ( s, z ) acting on vector-v alued functions f : Σ → C 2 . Here the Cauch y kernel is given by (B.7) Ω ζ ( s, z ) = z − ζ − 1 s − ζ − 1 1 s − z 0 0 z − ζ s − ζ 1 s − z ! ds =  1 s − z − 1 s − ζ − 1 0 0 1 s − z − 1 s − ζ  ds, for some ﬁxed ζ / ∈ Σ. In the case ζ = ∞ we set (B.8) Ω ∞ ( s, z ) =  1 s − z − 1 s 0 0 1 s − z  ds. and o ne ea s ily chec ks the sy mmetry prop erty: (B.9) Ω ζ (1 /s, 1 /z ) =  0 1 1 0  Ω ζ ( s, z )  0 1 1 0  . The pr oper ties o f C are summarized in the next lemma. Lemma B.2. Assume Hyp othesis B.1. The Cauchy op er ator C has the pr op erties, that the b oun dary values C ± ar e b ounde d op er ators L 2 s (Σ) → L 2 s (Σ) which satisfy (B.10) C + − C − = I and (B.11) ( C f )( ζ − 1 ) = (0 ∗ ) , ( C f )( ζ ) = ( ∗ 0) . Her e ∗ is a plac ehold er for an u nsp e ciﬁe d value. F urthermor e, C r estricts to L 2 s (Σ) , that is (B.12) ( C f )( z − 1 ) = ( C f )( z )  0 1 1 0  , z ∈ C \ Σ for f ∈ L 2 s (Σ) and if w ± satisfy (B.2) we also have (B.13) C ± ( f w ∓ )(1 /z ) = C ∓ ( f w ± )( z )  0 1 1 0  , z ∈ Σ . Pr o of. Everything follows from (B.9) a nd the fact that C inherits all pro per ties from the c lassical Ca uc hy op erator.  LONG-TIME ASYMPTOTICS OF THE TODA LA TTICE 37 W e hav e th us obtained a Cauch y transform with the required pro p erties. F ol- lowing Section 7 and 8 of [1] resp ectively [25], w e can solve our Riemann– Hilbert problem using this Cauch y op erator. Int ro duce the op e r ator C w : L 2 s (Σ) → L 2 s (Σ) b y (B.14) C w f = C + ( f w − ) + C − ( f w + ) , f ∈ L 2 s (Σ) and recall fro m Lemma 3.6 that the unique so lution co rresp onding to v ≡ I is given by m 0 ( z ) =  f ( z ) f ( 1 z )  , f ( z ) = 1 1 − ζ 2 + γ  γ ζ 2 z − ζ − 1 z − ζ + 1 − ζ 2  Observe that for γ = 0 we hav e f ( z ) = 1 and for γ = ∞ we hav e f ( z ) = ζ 2 z − ζ − 1 z − ζ . In par ticular, m 0 ( z ) is uniformly bo unded awa y from ζ for all γ ∈ [0 , ∞ ]. Then we hav e the next result. Theorem B.3. Assume Hyp othesis B.1. Supp ose m solves the Riemann–Hilb ert pr oblem (B.1) . Then (B.15) m ( z ) = (1 − c 0 ) m 0 ( z ) + 1 2 π i Z Σ µ ( s )( w + ( s ) + w − ( s ))Ω ζ ( s, z ) , wher e µ = m + b − 1 + = m − b − 1 − and c 0 =  1 2 π i Z Σ µ ( s )( w + ( s ) + w − ( s ))Ω ζ ( s, 0)  1 . Her e ( m ) j denotes the j ’th c omp onent of a ve ctor. F u rthermor e, µ solves (B.16) ( I − C w ) µ = (1 − c 0 ) m 0 . Conversely, supp ose ˜ µ solves (B.17) ( I − C w ) ˜ µ = m 0 , and ˜ c 0 =  1 2 π i Z Σ ˜ µ ( s )( w + ( s ) + w − ( s ))Ω ζ ( s, 0)  1 6 = − 1 , then m deﬁne d via (B.1 5) , with (1 − c 0 ) = (1 + ˜ c 0 ) − 1 and µ = (1 + ˜ c 0 ) − 1 ˜ µ , solves the Riemann–Hilb ert pr oblem (B.1) and µ = m ± b − 1 ± . Pr o of. If m solves (B.1) and we set µ = m ± b − 1 ± , then m sa tisﬁes a n additive jump given by m + − m − = µ ( w + + w − ) . Hence, if we denote the left hand side of (B.15) by ˜ m , both functions sa tisfy the same a dditiv e jump. F urthermore, Hyp othesis B.1 implies tha t µ is symmetric and hence so is ˜ m . Using (B.1 1) we also see that ˜ m satisﬁes the sa me p ole conditions as m 0 . In summary , m − ˜ m has no jump and solves (B.1) with v ≡ I except for the normalizatio n which is given by m (0) − ˜ m (0) =  0 ∗  . Hence Lemma 3 .6 implies m − ˜ m = 0. Moreov er, if m is given by (B.15), then (B.1 0) implies m ± = (1 − c 0 ) m 0 + C ± ( µw − ) + C ± ( µw + ) (B.18) = (1 − c 0 ) m 0 + C w ( µ ) ± µw ± = (1 − c 0 ) m 0 − ( I − C w ) µ + µb ± . 38 H. KR ¨ UGER AND G. TESCHL F rom this we conclude that µ = m ± b − 1 ± solves (B.16). Conv ersely , if ˜ µ so lv es (B.1 7), then set ˜ m ( z ) = m 0 ( z ) + 1 2 π i Z Σ ˜ µ ( s )( w + ( s ) + w − ( s ))Ω ζ ( s, z ) , and the sa me calculation as in (B.18) implies ˜ m ± = ˜ µb ± , whic h shows that m = (1 + ˜ c 0 ) − 1 ˜ m solves the Riemann–Hilb ert problem (B.1).  Note that in the sp ecial ca se γ = 0 w e hav e m 0 ( z ) =  1 1  and we can cho ose ζ as w e please, say ζ = ∞ such that c 0 = ˜ c 0 = 0 in the ab ove theo rem. Hence we hav e a fo r m ula for the solution of o ur Riemann–Hilb ert pr oblem m ( z ) in terms of ( I − C w ) − 1 m 0 and this clearly r aises the question of bo unded invertibilit y of I − C w . This follows from F redholm theor y (cf. e.g. [44]): Lemma B.4. Assume Hyp othesis B.1. The op er ator I − C w is F r e dholm of index zer o, (B.19) ind( I − C w ) = 0 . Pr o of. Since one can easily chec k (B.20) ( I − C w )( I − C − w ) = ( I − C − w )( I − C w ) = I − T w , where T w = T ++ + T + − + T − + + T −− , T σ 1 σ 2 ( f ) = C σ 1 [ C σ 2 ( f w − σ 2 ) w − σ 1 ] , it suﬃces to chec k that the op erators T σ 1 σ 2 are compac t ([34, Thm. 1.4 .3 ]). By Mergelyan’s theo rem we can approximate w ± by r ational functions and, since the norm limit of compact op era tors is compact, we can ass ume w itho ut loss that w ± hav e an analytic extension to a neigh b orho o d of Σ. Indeed, supp ose f n ∈ L 2 (Σ) c o n verges weakly to zer o. Without loss we ca n assume f n to b e contin uous. W e will show that k T w f n k L 2 → 0. Using the analyticity of w in a neig h b orho o d of Σ and the deﬁnitio n of C ± , we can slightly deform the contour Σ to so me cont our Σ ± close to Σ, on the left, a nd hav e, by Cauch y’s theo rem, T ++ f n ( z ) = 1 2 π i Z Σ + ( C ( f n w − )( s ) w − ( s ))Ω ζ ( s, z ) . Now ( C ( f n w − ) w − )( z ) → 0 as n → ∞ . Also | ( C ( f n w − ) w − )( z ) | < const k f n k L 2 k w − k L ∞ < const and thus, b y the dominated conv ergence theorem, k T ++ f n k L 2 → 0 as desired. Moreov er, considering I − εC w = I − C εw for 0 ≤ ε ≤ 1 we obtain ind( I − C w ) = ind( I ) = 0 from ho motopy inv aria nc e of the index.  By the F redholm alternative, it follows that to show the b ounded in vertibilit y of I − C w we only need to show that ker( I − C w ) = 0. The latter b eing equiv alent to unique s o lv ability of the corres ponding v anishing Riemann–Hilb ert problem in the case γ = 0 (where we ca n c ho ose ζ = ∞ such that c 0 = ˜ c 0 = 0). Corollary B. 5. Assu me Hyp othesis B.1. A u nique solution of t he Riemann–Hilb ert pr oblem (B.1) with γ = 0 exists if and only if t he c orr esp onding vanishing Rie mann– Hilb ert pr oblem, wher e the normalization c ondition is r epla c e d by m (0) =  0 m 2  , with m 2 arbitr ary, has at most one solution. LONG-TIME ASYMPTOTICS OF THE TODA LA TTICE 39 W e are interested in co mparing a Riemann–Hilb ert problem for w hich k w k ∞ is small with the o ne-soliton pr o blem, where (B.21) k w k ∞ = k w + k L ∞ (Σ) + k w − k L ∞ (Σ) . F or such a situation we hav e the following r esult: Theorem B .6. Fix a c ontour Σ and cho ose ζ , γ = γ t , v t dep ending on some p ar ameter t ∈ R such that Hyp othesis B.1 holds. Assume t hat w t satisﬁes (B.22) k w t k ∞ ≤ ρ ( t ) for some function ρ ( t ) → 0 as t → ∞ . Then ( I − C w t ) − 1 : L 2 s (Σ) → L 2 s (Σ) ex ists for suﬃciently lar ge t and the solut ion m ( z ) of the Riemann–Hilb ert pr oblems (B.1) diﬀers fr om the one-soliton solution m t 0 ( z ) only by O ( ρ ( t )) , wher e the err or term dep ends on t he distanc e of z to Σ ∪ { ζ ± 1 } . Pr o of. By b oundedness of the Cauch y trans form, one has k C w t k ≤ const k w t k ∞ . Thu s, by the Neumann series, w e infer that ( I − C w t ) − 1 exists for suﬃciently lar ge t a nd k ( I − C w t ) − 1 − I k = O ( ρ ( t )) . This implies k ˜ µ t − m t 0 k L 2 s = O ( ρ ( t )) and ˜ c t 0 = O ( ρ ( t )) (note ˜ µ t 0 = µ t 0 = m t 0 ). Consequently c t 0 = O ( ρ ( t )) a nd k µ t − m t 0 k L 2 s = O ( ρ ( t )) a nd thus m t ( z ) − m t 0 ( z ) = O ( ρ ( t )) uniformly in z as long as it stays a po sitive distance aw ay from Σ ∪ { ζ ± 1 } .  Ac kno wledgment s. 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