LONG-TIME ASYMPTOTICS FOR THE TODA LATTICE IN THE SOLITON REGION

Transcription

1 LONG-TIME ASYMPTOTICS FOR THE TODA LATTICE IN THE SOLITON REGION HELGE KRÜGER AND GERALD TESCHL Abstract. We apply the method of nonlinear steepest descent to compute the long-time asymptotics of the Toda lattice for decaying initial data in the soliton region. In addition, we point out how to reduce the problem in the remaining region to the nown case without solitons.. Introduction In this paper we want to compute the long time asymptotics for the doubly infinite Toda lattice which reads in Flascha s variables see e.g. [6], [7], or [8]. ḃn, t = 2an, t 2 an, t 2, ȧn, t = an, tbn +, t bn, t, n, t Z R. Here the dot denotes differentiation with respect to time. We will consider solutions a, b satisfying.2 n l an, t + bn, t < 2 n for every l N for one and hence for all, see [6] t R. It is well-nown that this initial value problem has unique global solutions which can be computed via the inverse scattering transform [6]. The long-time asymptotics for this problem were first given by Novoshenov and Habibullin [2] and were later made rigorous by Kamvissis [7], however, only in the case without solitons. The purpose of the present paper is to finally fill this gap and show how to include solitons. As in [7], our approach is based on the nonlinear steepest descent analysis for oscillatory Riemann Hilbert problems from Deift and Zhou [4]. It turns out that in the case of solitons, two new phenomena enter the scene which require significant adaptions to the original method of Deift and Zhou. Of course our technique also applies to other soliton equations, e.g., the Korteweg de Vries equation. First of all, it is well-nown that there is a subtle nonuniqueness issue for the involved Riemann Hilbert problems see e.g. [2, Chap. 38]. In fact, in certain exceptional sets the corresponding vanishing Riemann Hilbert problem has a nontrivial solution and hence by Fredholm theory the corresponding matrix Riemann Hilbert problem has no solution at all. This problem does not affect the similarity region, 2 Mathematics Subject Classification. Primary 37K4, 37K45; Secondary 35Q5, 37K. Key words and phrases. Riemann Hilbert problem, Toda lattice, solitons. Research supported by the Austrian Science Fund FWF under Grant No. Y33. Math. Z. 262,

2 2 H. KRÜGER AND G. TESCHL since it is easy to see that there it does not happen for sufficiently large times. However, in the soliton region, this occurs precisely in the neighborhoods of the single solitons. To avoid this problem we will wor directly with the vector Riemann Hilbert problem and impose a symmetry condition in order to ensure uniqueness. This also has the advantage that it eliminates the step of going forth and bac between the vector and matrix Riemann Hilbert problem. It should be pointed out here that even the symmetry condition alone does not guarantee uniqueness, rather we will need existence of a certain solution with an additional property. We will also demonstrate that this additional property is in fact necessary. Secondly, in the regions of the single solitons, the zeroth order asymptotics are not equal to zero but given by a one soliton solution. Hence for the usual perturbation argument based on the second resolvent identity to wor, a uniform bound for the inverse of the singular integral equation associated with the one soliton solution is needed. Unfortunately, such a bound cannot be easily obtained. To overcome this problem we will shift the leading asymptotics from the one soliton solution to the inhomogeneous part of the singular integral equation and craft our Cauchy ernel in such a way that it preserves the pole conditions for this single soliton. To state our main result, we begin by recalling that the sequences an, t, bn, t, n Z, for fixed t R, are uniquely determined by their scattering data, that is, by the right reflection coefficient R + z, t, z = and the eigenvalues λ j,,, j =,..., N together with the corresponding right norming constants γ +,j t >, j =,..., N. Rather than in the complex plane, we will wor on the unit disc using the usual Jouowsi transformation.3 λ = z +, z = λ λ 2 z 2, λ C, z. In these new coordinates the eigenvalues λ j,, will be denoted by ζ j,,. The continuous spectrum [, ] is mapped to the unit circle T. Moreover, the phase of the associated Riemann Hilbert problem is given by.4 Φz = z z + 2 n t logz. and the stationary phase points, Φ z =, are denoted by.5 z = n t n t 2, z = n t + n t 2, λ = n t. For n t < we have z,, for n t we have z T and hence z = z, and for n t > we have z,. For n t > we will also need the value ζ,, defined via ReΦζ =, that is,.6 n t = ζ ζ 2 log ζ. We will set ζ = if n t for notational convenience. A simple analysis shows that for n t < we have < ζ < z < and for n t > we have < z < ζ <. Furthermore, recall that the transmission coefficient T z, z, is time independent and can be reconstructed using the Poisson Jensen formula. In particular,

3 LONG-TIME ASYMPTOTICS FOR THE TODA LATTICE 3 we define the partial transmission coefficient with respect to z by T z, z =.7 ζ ζ, ζ, ζ z ζ z ζ, z,, ζ z ζ z ζ ζ, ζ, ζ z ζ z ζ exp 2πi exp z log T s s+z s z z 2πi T ds s log T s s+z s z, z =, ds s, z,. Here, in the case z T, the integral is to be taen along the arc Σz = {z T Rez < Rez } oriented counterclocwise. For z, we set Σz = and for z, we set Σz = T. Then T z, z is meromorphic for z C\Σz. Observe T z, z = T z once z, and, ζ contains no eigenvalues. Moreover, T z, z can be computed in terms of the scattering data since T z 2 = R + z, t 2. Moreover, we set T z = T, z ζ, z,, ζ ζ, = ζ z.8 exp 2πi log T s ds s, z =, ζ, z ζ exp 2πi log T s ds s, z,, ζ, ζ, and T z = z log T z, z z= ζ ζ, z,, ζ ζ, = ζ ζ + z.9 πi log T s ds s, z 2 =, ζ, z ζ ζ + πi log T s ds s, z 2,. ζ, ζ, Theorem.. Assume.2 and abbreviate by c = ζ ζ 2 log ζ the velocity of the th soliton determined by ReΦζ =. Then the asymptotics in the soliton region, n/t + C/t logt 2 for some C >, are as follows. Let ε > sufficiently small such that the intervals [c ε, c + ε], N, are disjoint and lie inside,,. If n t c < ε for some, one has. 2aj, t = T z j=n j=n+ T T ζ 2 + γ n, t ζ 2 + γ n, tζ 2 + Ot l bj, t = 2 T z γ n, tζ ζ 2 2γ n, t ζ Ot l,,

4 4 H. KRÜGER AND G. TESCHL for any l, where. γ n, t = γ T ζ, c c 2 2 e tζ ζ If n t c ε, for all, one has 2aj, t = T z + Ot l,.2 for any l. j=n j=n+ bj, t = 2 T z + Ot l, ζ 2n. In particular, we recover the well-nown fact that the solution splits into a sum of independent solitons where the presence of the other solitons and the radiation part corresponding to the continuous spectrum manifests itself in phase shifts given by T ζ, c c 2 2. Indeed, notice that for ζ, this term just contains product over the Blasche factors corresponding to solitons ζ j with ζ < ζ j. For ζ, we have the product over the Blasche factors corresponding to solitons ζ j,, the integral over the full unit circle, plus the product over the Blasche factors corresponding to solitons ζ j with ζ > ζ j. The proof will be given at the end of Section 4. Furthermore, in the remaining regions the analysis in Section 4 also shows that the Riemann Hilbert problem reduces to one without solitons. In fact, away from the soliton region, the asymptotics are given by 2aj, t = T 2ãj, t + Ot l,.3 j=n j=n+ j=n bj, t = 2 T + j=n+ bj, t + Ot l, where ãn, t, bn, t are the solutions corresponding to the case without solitons and with R + z, replaced by.4 R+ z, = T z, 2 R + z,. Note that the Blasche product T z, = ζ, ζ z ζ z ζ satisfies T z, = for z T. Hence everything is reduced to the case studied in [7]. Finally we remar that the same method can be used to handle solitons on a periodic bacground [] cf. also [6], [8], [9]. 2. The Inverse scattering transform and the Riemann Hilbert problem In this section we want to derive the Riemann Hilbert problem from scattering theory. The special case without eigenvalues was first given in Kamvissis [7]. The

5 LONG-TIME ASYMPTOTICS FOR THE TODA LATTICE 5 eigenvalues will be added by appropriate pole conditions which are then turned into jumps following Deift, Kamvissis, Kriecherbauer, and Zhou [5]. For the necessary results from scattering theory respectively the inverse scattering transform for the Toda lattice we refer to [4], [5], [6]. Associated with at, bt is a self-adjoint Jacobi operator 2. Ht = ats + + a ts + bt in l 2 Z, where S ± fn = f ± n = fn ± are the usual shift operators and l 2 Z denotes the Hilbert space of square summable complex-valued sequences over Z. By our assumption.2 the spectrum of H consists of an absolutely continuous part [, ] plus a finite number of eigenvalues λ R\[, ], N. In addition, there exist two Jost functions ψ ± z, n, t which solve the recurrence equation 2.2 Htψ ± z, n, t = z + z ψ ± z, n, t, z, 2 and asymptotically loo lie the free solutions 2.3 lim n ± z n ψ ± z, n, t =. Both ψ ± z, n, t are analytic for < z < with smooth boundary values for z =. The asymptotics of the two projections of the Jost function are 2.4 ψ ± z, n, t = z±n + 2B ± n, tz + Oz 2, A ± n, t as z, where 2.5 A + n, t = A n, t = 2aj, t, j=n n j= One has the scattering relations 2aj, t, B + n, t = B n, t = j=n+ n j= bj, t, bj, t. 2.6 T zψ z, n, t = ψ ± z, n, t + R ± z, tψ ± z, n, t, z =, where T z, R ± z, t are the transmission respectively reflection coefficients. The transmission and reflection coefficients have the following well-nown properties: Lemma 2.. The transmission coefficient T z has a meromorphic extension to the interior of the unit circle with simple poles at the images of the eigenvalues ζ. The residues of T z are given by 2.7 Res ζ T z = ζ γ +, t µ t where 2.8 γ ±, t = n Z and ψ ζ, n, t = µ tψ + ζ, n, t. Moreover, = ζ γ, tµ t, ψ ± ζ, n, t T zr + z, t + T zr z, t =, T z 2 + R ± z, t 2 =.

6 6 H. KRÜGER AND G. TESCHL In particular one reflection coefficient, say Rz, t = R + z, t, and one set of norming constants, say γ t = γ +, t, suffices. Moreover, the time dependence is given by: Lemma 2.2. The time evolutions of the quantities R + z, t, γ +, t are given by where Rz = Rz, and γ = γ. Rz, t = Rze tz z γ t = γ e tζ ζ, Now we define the sectionally meromorphic vector { T zψ z, n, tz 2.2 mz, n, t = n ψ + z, n, tz n, z <, ψ+ z, n, tz n T z ψ z, n, tz n, z >. We are interested in the jump condition of mz, n, t on the unit circle T oriented counterclocwise. To formulate our jump condition we use the following convention: When representing functions on T, the lower subscript denotes the non-tangential limit from different sides, 2.3 m ± z = lim mζ, z =. ζ z, ζ ± < In general, for an oriented contour Σ, m + z resp. m z will denote the limit of mζ as ζ z from the positive resp. negative side of Σ. Using the notation above implicitly assumes that these limits exist in the sense that mz extends to a continuous function on the boundary. Theorem 2.3 Vector Riemann Hilbert problem. Let S + H = {Rz, z = ; ζ, γ, N} the right scattering data of the operator H. Then mz = mz, n, t defined in 2.2 is meromorphic away from the unit circle with simple poles at ζ, ζ and satisfies: i The jump condition Rz 2 Rze 2.4 m + z = m zvz, vz = tφz Rze tφz 2.5 for z T, ii the pole conditions Res ζ mz = lim z ζ mz Res ζ mz = iii the symmetry condition lim z ζ mz 2.6 mz = mz iv and the normalization ζ γ e tφζ, ζ γ e tφζ, 2.7 m = m m 2, m m 2 = m >. Here the phase is given by 2.8 Φz = z z + 2 n log z. t,

7 LONG-TIME ASYMPTOTICS FOR THE TODA LATTICE 7 Proof. The jump condition 2.4 is a simple calculation using the scattering relations 2.6 plus 2.9. The pole conditions follow since T z is meromorphic in z < with simple poles at ζ and residues given by 2.7. The symmetry condition holds by construction and the normalization 2.7 is immediate from the following lemma. Observe that the pole condition at ζ is sufficient since the one at ζ follows by symmetry. Moreover, we have the following asymptotic behaviour near z = : Lemma 2.4. The function mz, n, t defined in 2.2 satisfies 2.9 mz, n, t = An, t 2Bn, tz An,t + 2Bn, tz + Oz 2. Here An, t = A + n, t and Bn, t = B + n, t are defined in 2.5. Proof. This follows from 2.4 and T z = A + A 2B + b+b z +Oz 2. For our further analysis it will be convenient to rewrite the pole condition as a jump condition and hence turn our meromorphic Riemann Hilbert problem into a holomorphic Riemann Hilbert problem following [5]. Choose ε so small that the discs z ζ < ε are inside the unit circle and do not intersect. Then redefine m in a neighborhood of ζ respectively ζ according to mz, mz ζ γ e tφζ, z ζ < ε, z ζ 2.2 mz = zγ e tφζ mz, z ζ < ε, z ζ else. Then a straightforward calculation using Res ζ m = lim z ζ z ζmz shows Lemma 2.5. Suppose mz is redefined as in 2.2. Then mz is holomorphic away from the unit circle and satisfies 2.4, 2.6, 2.7 and the pole conditions are replaced by the jump conditions 2.2 m + z = m z m + z = m z ζ γ e tφζ z ζ zγ e tφζ z ζ, z ζ = ε,, z ζ = ε, where the small circle around ζ is oriented counterclocwise and the one around is oriented clocwise. ζ Next we turn to uniqueness of the solution of this vector Riemann Hilbert problem. This will also explain the reason for our symmetry condition. We begin by observing that if there is a point z C, such that mz =, then nz = z z mz satisfies the same jump and pole conditions as mz. However, it will clearly violate the symmetry condition! Hence, without the symmetry condition, the solution of our vector Riemann Hilbert problem will not be unique in such a situation. Moreover, a loo at the one soliton solution verifies that this case indeed can happen.

8 8 H. KRÜGER AND G. TESCHL Lemma 2.6 One soliton solution. Suppose there is only one eigenvalue and a vanishing reflection coefficient, that is, S + Ht = {Rz, z = ; ζ, γ} with ζ,, and γ. Then the Riemann Hilbert problem has a unique solution is given by 2.22 m z = fz f/z fz = ζ2 + γn, t ζ 2 + ζ 2 γn, t γn, tζ 2 z ζ z ζ + ζ 2, where γn, t = γe tφζ. In particular, ζ 2.23 A + n, t = 2 + γn, t ζ 2 + γn, tζ 2, B +n, t = γn, tζζ2 2 ζ 2 + γn, tζ 2. Proof. By symmetry, the solution must be of the form m z = fz f/z, where fz is meromorphic in C { } with the only possible pole at ζ. Hence fz = + 2 B, A z ζ where the unnown constants A and B are uniquely determined by the pole condition Res ζ fz = ζγn, tfζ and the normalization ff =, f >. In fact, observe fz = fz = if and only if z = ± and γ = ±ζ ζ. Furthermore, even in the general case mz = can only occur at z = ± as the following lemma shows. Lemma 2.7. If mz = for m defined as in 2.2, then z = ±. Moreover, the zero of at least one component is simple in this case. Proof. By 2.2 the condition mz = implies that either the Jost solutions ψ z, n and ψ + z, n are linearly dependent or T z =. This can only happen, at a band edge, z = ±, or at an eigenvalue z = ζ j. We begin with the case z = ζ j. In this case the derivative of the Wronsian W z = anψ + z, nψ z, n + ψ + z, n + ψ z, n does not vanish d dz W z z=z [6, Chap. ]. Moreover, the diagonal Green s function gλ, n = W z ψ + z, nψ z, n is Herglotz and hence can have at most a simple zero at z = z. Hence, if ψ + ζ j, n = ψ ζ j, n =, both can have at most a simple zero at z = ζ j. But T z has a simple pole at ζ j and hence T zψ z, n cannot vanish at z = ζ j, a contradiction. It remains to show that one zero is simple in the case z = ±. In fact, one can show that d dz W z z=z in this case as follows: First of all note that ψ ±z where denotes the derivative with respect to z again solves Hψ ±z = λ ψ ±z if z = ±. Moreover, by W z = we have ψ + z = cψ z for some constant c independent of n. Thus we can compute W z = W ψ +z, ψ z + W ψ + z, ψ z = c W ψ +z, ψ + z + cw ψ z, ψ z

9 LONG-TIME ASYMPTOTICS FOR THE TODA LATTICE 9 by letting n + for the first and n for the second Wronsian in which case we can replace ψ ± z by z ±n, which gives W z = c + c. 2 Hence the Wronsian has a simple zero. But if both functions had more than simple zeros, so would the Wronsian, a contradiction. Finally, it is interesting to note that the assumptions ζ,, and γ are crucial for uniqueness. Indeed, if we choose γ = ζ 2 <, then every solution is a multiple of fz = zζ z ζ which cannot be normalized at. 3. A uniqueness result for symmetric vector Riemann Hilbert problems In this section we want to investigate uniqueness for the holomorphic vector Riemann Hilbert problem 3. m + z = m zvz, z Σ, mz = mz, m = m 2. where Σ is a nice oriented contour see Hypothesis A., symmetric with respect to z z, and v is continuous satisfying detv = and 3.2 vz = vz, z Σ. The normalization used here will be more convenient than 2.7. In fact, 2.7 will be satisfied by m /2 2 mz. Now we are ready to show that the symmetry condition in fact guarantees uniqueness. Theorem 3.. Suppose there exists a solution mz of the Riemann Hilbert problem 3. for which mz = can happen at most for z = ± in which case z lim sup z ± m jz is bounded from any direction for j = or j = 2. Then the Riemann Hilbert problem 3. with norming condition replaced by 3.3 m = α m 2 for given α C, has a unique solution m α z = α mz. Proof. Let m α z be a solution of 3. normalized according to 3.3. Then we can construct a matrix valued solution via M = m, m α and there are two possible cases: Either det Mz is nonzero for some z or it vanishes identically. We start with the first case. Since the determinant of our Riemann Hilbert problem has no jump and is bounded at infinity, it is constant. But taing determinants in Mz = Mz. gives a contradiction. It remains to investigate the case where detm. In this case we have m α z = δzmz with a scalar function δ. Moreover, δz must be holomorphic

10 H. KRÜGER AND G. TESCHL for z C\Σ and continuous for z Σ except possibly at the points where mz =. Since it has no jump across Σ, δ + zm + z = m α,+ z = m α, zvz = δ zm zvz = δ zm + z, it is even holomorphic in C\{±} with at most simple poles at z = ±. Hence it must be of the form δz = A + B z + C z +. Since δ has to be symmetric, δz = δz, we obtain B = C =. Now, by the normalization we obtain δz = A = α. This finishes the proof. Furthermore, note that the requirements cannot be relaxed to allow e.g. second order zeros in stead of simple zeros. In fact, if mz is a solution for which both components vanish of second order at, say, z = +, then mz = z z mz is a 2 nontrivial symmetric solution of the vanishing problem i.e. for α =. By Lemma 2.7 we have Corollary 3.2. The function mz, n, t defined in 2.2 is the only solution of the vector Riemann Hilbert problem Observe that there is nothing special about the point z = where we normalize, any other point would do as well. However, observe that for the one soliton solution 2.22, fz vanishes at z = ζ + γ ζ2 γ ζ 2 + and hence the Riemann Hilbert problem normalized at this point has a nontrivial solution for α = and hence, by our uniqueness result, no solution for α =. This shows that uniqueness and existence connected, a fact which is not surprising since our Riemann Hilbert problem is equivalent to a singular integral equation which is Fredholm of index zero see Appendix A. 4. Solitons and the soliton region This section demonstrates the basic method of passing from a Riemann Hilbert problem involving solitons to one without. Furthermore, the asymptotics inside the soliton region are computed. Solitons are represented in a Riemann Hilbert problem by pole conditions, for this reason we will further study how poles can be dealt with in this section. For easy reference we note the following result which can be checed by a straightforward calculation. Lemma 4. Conjugation. Assume that Σ Σ. Let D be a matrix of the form dz 4. Dz =, dz where d : C\ Σ C is a sectionally analytic function. Set 4.2 mz = mzdz, then the jump matrix transforms according to 4.3 ṽz = D z vzd + z.

11 LONG-TIME ASYMPTOTICS FOR THE TODA LATTICE If d satisfies dz = dz and d >. Then the transformation mz = mzdz respects our symmetry, that is, mz satisfies 2.6 if and only if mz does. In particular, we obtain 4.4 ṽ = respectively 4.5 ṽ = v v 2 d 2 v 2 d 2, z Σ\ Σ, v 22 d d + v v 2 d + d v 2 d + d, z Σ Σ. d + d v 22 In order to remove the poles there are two cases to distinguish. If λ > 2 ζ +ζ the jump is exponentially decaying and there is nothing to do. Otherwise we use conjugation to turn the jumps into exponentially decaying ones, again following Deift, Kamvissis, Kriecherbauer, and Zhou [5]. It turns out that we will have to handle the poles at ζ and ζ in one step in order to preserve symmetry and in order to not add additional poles elsewhere. Lemma 4.2. Assume that the Riemann Hilbert problem for m has jump conditions near ζ and ζ given by m + z = m z, z ζ = ε, 4.6 m + z = m z γζ z ζ γz z ζ, z ζ = ε. Then this Riemann Hilbert problem is equivalent to a Riemann Hilbert problem for m which has jump conditions near ζ and ζ given by ζz 2 m + z = m z ζz ζγ, z ζ = ε, m + z = m z z ζ 2, z ζ = ε, ζzζz γ and all remaining data conjugated as in Lemma 4. by 4.7 Dz = z ζ ζz Proof. To turn γ into γ, introduce D by z ζ z ζ γ ζ γ ζ z ζ γ zζ z ζ zζ Dz = zζ γ zζ z ζ ζz ζz z ζ ζz z ζ ζz ζz z ζ ζz. ζz z ζ, z ζ < ε,, else,, z ζ < ε, and note that Dz is analytic away from the two circles. Now set mz = mzdz, which is again symmetric by Dz = Dz. The jumps

12 2 H. KRÜGER AND G. TESCHL along z ζ = ε and z ζ = ε follow by a straightforward calculation and the remaining jumps follow from Lemma 4.. Now we are ready to prove our main result: Proof of Theorem.. We begin by observing that the partial transmission coefficient T z, z introduced in.7 satisfies the following scalar meromorphic Riemann Hilbert problem: i T z, z is meromorphic in C\Σz, where Σz is the arc given by Σz = {z T Rez < Rez }, with simple poles at ζ zeros at ζ for all with λ < 2 ζ + ζ, ii T + z, z = T z, z Rz 2 for z Σz, iii T z, z = T z, z, z C\Σz, and T, z >. and simple Note also T z, z = T z, z and in particular T z, z is real-valued for z R. Next introduce z ζ ζ γ e tφζ D ζ γ e tφζ z, z ζ < ε, λ < 2 z ζ ζ + ζ, Dz = zζ γ e tφζ zζ zζ D z, z ζ < ε, λ < 2 ζ + ζ, zζ γ e tφζ D z, else, where T z, z D z =. T z, z Note that we have Dz = Dz. Now we conjugate our vector mz defined in 2.2 respectively 2.2 using Dz, mz = mzdz. Since T z, z is either nonzero and continuous near z = ± if ± / Σz or it has the same behaviour as T z near z = ± if ± Σz, the new vector mz is again continuous near z = ± even if T z vanishes there. Then using Lemma 4. and Lemma 4.2 the jump corresponding λ < 2 ζ +ζ if any is given by 4.8 z ζ ζ γ T z,z 2 e tφζ ṽz = ṽz = ζ z ζ zγ T z,z 2 e tφζ, z ζ = ε,, z ζ = ε, and corresponding λ 2 ζ + ζ if any by ṽz = ζ γ T z,z 2 e tφζ, z ζ = ε, 4.9 ṽz = z ζ z ζ zγ T z,z 2 e tφζ, z ζ = ε. In particular, an investigation of the sign of ReΦz shows that all off-diagonal entries of these jump matrices, except for possibly one if ζ = ζ for some, are

13 LONG-TIME ASYMPTOTICS FOR THE TODA LATTICE 3 exponentially decreasing. In this case we will eep the pole condition which now reads 4. Res ζ mz = lim z ζ mz Res ζ mz = lim mz z ζ ζ γ T ζ, z 2 e tφζ, ζ γ T ζ, z 2 e tφζ. Furthermore, the jump along T is given by 4. ṽz = { b z b + z, λz > λ, B z B+ z, λz < λ, where b z = Rz e tφz T z,z 2, b+ z = Rze tφz, T z,z 2 and B z = T z,z 2 Rz Rze tφz, B+ z = 2 T+z,z2 Rz Rze tφz 2. Now the jump along T can also be made arbitrarily small following the nonlinear steepest descent method developed by Deift and Zhou [4]: Split the Fourier transform of the reflection coefficient into a part which has an analytic extension to a neighborhood of the unit circle plus a small error. Move the analytic part away from the unit circle using the factorizations from above. Since the Fourier coefficients decay faster than any polynomial, the errors from both parts can be made Ot l for any l N. We refer to [4] respectively [7] for details. Hence we can apply Theorem A.6 as follows: If n t c > ε for all we can choose γ = and w t. Since the error between w t and w t is exponentially small, this proves the second part of Theorem. upon comparing T z, z 4.2 mz = ˆmz T z, z with 2.9. Otherwise, if n t c < ε for some, we choose γ t = γ n, t and w t. Again we conclude that the error between w t and w t is exponentially small, proving the first part of Theorem..

14 4 H. KRÜGER AND G. TESCHL Appendix A. Singular integral equations In this section we show how to transform a meromorphic vector Riemann Hilbert problem with simple poles at ζ, ζ, m + z = m zvz, z Σ, A. Res ζ mz = lim mz z ζ ζγ mz = mz, m = m 2,, Res ζ mz = lim z ζ mz ζ γ, where ζ,, and γ, into a singular integral equation. Since we require the symmetry condition 2.6 for our Riemann Hilbert problems we need to adapt the usual Cauchy ernel to preserve this symmetry. Moreover, we eep the single soliton as an inhomogeneous term which will play the role of the leading asymptotics in our applications. Hypothesis H. A.. Let Σ consist of a finite number of smooth oriented finite curves in C which intersect at most finitely many times with all intersections being transversal. Assume that the contour Σ does not contain, ζ and is invariant under z z. It is oriented such that under the mapping z z sequences converging from the positive sided to Σ are mapped to sequences converging to the negative side. Moreover, suppose the jump matrix v can be factorized according to v = b b + = I w I + w +, where w ± = ±b ± I are continuous and satisfy A.2 w ± z = w z, z Σ. The classical Cauchy-transform of a function f : Σ C which is square integrable is the analytic function Cf : C\Σ C given by A.3 Cfz = fs 2πi s z ds, z C\Σ. Σ Denote the non-tangential boundary values from both sides taen possibly in the L 2 -sense see e.g. [3, eq. 7.2] by C + f respectively C f. Then it is well-nown that C + and C are bounded operators L 2 Σ L 2 Σ, which satisfy C + C = I and C + C = see e.g. []. Moreover, one has the Plemelj Sohotsy formula [] C ± = ih ± I, 2 where A.4 Hft = π fs ds, t Σ, Σ t s is the Hilbert transform and denotes the principal value integral. In order to respect the symmetry condition we will restrict our attention to the set L 2 sσ of square integrable functions f : Σ C 2 such that A.5 fz = fz.

15 LONG-TIME ASYMPTOTICS FOR THE TODA LATTICE 5 Clearly this will only be possible if we require our jump data to be symmetric as well i.e., Hypothesis A. holds. Next we introduce the Cauchy operator A.6 Cfz = fsω ζ s, z 2πi acting on vector-valued functions f : Σ C 2. Here the Cauchy ernel is given by z ζ A.7 Ω ζ s, z = s ζ s z ds = s ζ s z ds, s ζ s z z ζ s ζ s z for some fixed ζ / Σ. In the case ζ = we set A.8 Ω s, z = s z s ds. s z and one easily checs the symmetry property: A.9 Ω ζ /s, /z = Ω ζ s, z. The properties of C are summarized in the next lemma. Lemma A.2. Assume Hypothesis A.. The Cauchy operator C has the properties, that the boundary values C ± are bounded operators L 2 sσ L 2 sσ which satisfy A. and Σ C + C = I A. Cfζ =, Cfζ =. Furthermore, C restricts to L 2 sσ, that is A.2 Cfz = Cfz, z C\Σ for f L 2 sσ and if w ± satisfy H.A. we also have A.3 C ± fw /z = C fw ± z, z Σ. Proof. Everything follows from A.9 and the fact that C inherits all properties from the classical Cauchy operator. We have thus obtained a Cauchy transform with the required properties. Following Section 7 and 8 of [], we can solve our Riemann Hilbert problem using this Cauchy operator. Introduce the operator C w : L 2 sσ L 2 sσ by A.4 C w f = C + fw + C fw +, f L 2 sσ and recall from Lemma 2.6 that the unique solution corresponding to v I is given by m z = fz f z, fz = γζ 2 z ζ ζ 2 + ζ 2 + γ z ζ Observe that for γ = we have fz = and for γ = we have fz = ζ 2 z ζ z ζ. In particular, m z is uniformly bounded away from ζ for all γ [, ]. Then we have the next result.

16 6 H. KRÜGER AND G. TESCHL Theorem A.3. Assume Hypothesis A.. Suppose m solves the Riemann Hilbert problem A.. Then A.5 mz = c m z + µsw + s + w sω ζ s, z, 2πi Σ where µ = m + b + = m b and c = µsw + s + w sω ζ s,. 2πi Σ Here m j denotes the j th component of a vector. Furthermore, µ solves A.6 Conversely, suppose µ solves A.7 I C w µ = c m z. I C w µ = m z, and c = µsw + s + w sω ζ s,, 2πi Σ then m defined via A.5, with c = + c and µ = + c µ, solves the Riemann Hilbert problem A. and µ = m ± b ±. Proof. This can be shown as in the non-symmetric case cf. e.g. [3]. Note that in the special case γ = we have m z = and we can choose ζ as we please, say ζ = such that c = c = in the above theorem. Hence we have a formula for the solution of our Riemann Hilbert problem mz in terms of I C w m and this clearly raises the question of bounded invertibility of I C w. This follows from Fredholm theory cf. e.g. [9]: Lemma A.4. Assume Hypothesis A.. The operator I C w is Fredholm of index zero, A.8 indi C w =. By the Fredholm alternative, it follows that to show the bounded invertibility of I C w we only need to show that eri C w =. The latter being equivalent to unique solvability of the corresponding vanishing Riemann Hilbert problem in the case γ = where we can choose ζ = such that c = c =. Corollary A.5. Assume Hypothesis A.. A unique solution of the Riemann Hilbert problem A. with γ = exists if and only if the corresponding vanishing Riemann Hilbert problem, where the normalization condition is replaced by m = m 2, with m2 arbitrary, has at most one solution. We are interested in comparing two Riemann Hilbert problems associated with respective jumps w and w with w w small, where A.9 w = w + L Σ + w L Σ. For such a situation we have the following result: Theorem A.6. Assume that for some data ζ, γ t, w t the operator A.2 I C w t : L 2 sσ L 2 sσ has a bounded inverse, where the bound is independent of t, and suppose that the corresponding c t, is away from again uniformly in t.

17 LONG-TIME ASYMPTOTICS FOR THE TODA LATTICE 7 Furthermore, let ζ = ζ, γ t = γ t and assume w t satisfies A.2 w t w t αt for some function αt as t. Then I C w t : L 2 sσ L 2 sσ also exists for sufficiently large t and the associated solutions of the Riemann Hilbert problems A. only differ by Oαt uniformly in z away from Σ {ζ, ζ }. Proof. By boundedness of the Cauchy transform, one has C w t C w t const w. Thus, by the second resolvent identity, we infer that I C w t exists for large t and I C w t I C w t = Oαt. From which the claim follows since this implies c t c t, = Oαt respectively µ t µ t L 2 = Oαt and thus m t z m t z = Oαt uniformly in z away from Σ {ζ, ζ }. Acnowledgments. We than S. Kamvissis for several helpful discussions and I. Egorova, K. Grunert and A. Miiits-Leitner for pointing out errors in a previous version of this article. References [] R. Beals and R. Coifman, Scattering and inverse scattering for first order systems, Comm. in Pure and Applied Math. 37, [2] R. Beals, P. Deift, and C. Tomei, Direct and Inverse Scattering on the Real Line, Math. Surv. and Mon. 28, Amer. Math. Soc., Rhode Island, 988. [3] P. Deift, Orthogonal Polynomials and Random Matrices: A Riemann Hilbert Approach, Courant Lecture Notes 3, Amer. Math. Soc., Rhode Island, 998. [4] P. Deift and X. Zhou, A steepest descent method for oscillatory Riemann Hilbert problems, Ann. of Math. 2 37, [5] P. Deift, S. Kamvissis, T. Kriecherbauer, and X. Zhou, The Toda rarefaction problem, Comm. Pure Appl. Math. 49, no., [6] I. Egorova, J. Michor, and G. Teschl, Soliton solutions of the Toda hierarchy on quasiperiodic bacground revisited, Math. Nach. 282:4, [7] S. Kamvissis, On the long time behavior of the doubly infinite Toda lattice under initial data decaying at infinity, Comm. Math. Phys., 53-3, [8] S. Kamvissis and G. Teschl, Stability of periodic soliton equations under short range perturbations, Phys. Lett. A, 364-6, [9] S. Kamvissis and G. Teschl, Stability of the periodic Toda lattice under short range perturbations, arxiv: [] H. Krüger and G. Teschl, Stability of the periodic Toda lattice in the soliton region, Int. Math. Res. Not. to appear. [] N.I. Mushelishvili, Singular Integral Equations, P. Noordhoff Ltd., Groningen, 953. [2] V.Yu. Novoshenov and I.T. Habibullin, Nonlinear differential-difference schemes integrable by the method of the inverse scattering problem. Asymptotics of the solution for t, Sov. Math. Dolady 23/2, [3] S. Prössdorf, Some Classes of Singular Equations, North-Holland, Amsterdam, 978. [4] G. Teschl, Inverse scattering transform for the Toda hierarchy, Math. Nach. 22, [5] G. Teschl, On the initial value problem of the Toda and Kac-van Moerbee hierarchies, in Differential Equations and Mathematical Physics, R. Weiard and G. Weinstein eds., , AMS/IP Studies in Advanced Mathematics 6, Amer. Math. Soc., Providence, 2. [6] G. Teschl, Jacobi Operators and Completely Integrable Nonlinear Lattices, Math. Surv. and Mon. 72, Amer. Math. Soc., Rhode Island, 2.

18 8 H. KRÜGER AND G. TESCHL [7] G. Teschl, Almost everything you always wanted to now about the Toda equation, Jahresber. Deutsch. Math.-Verein. 3, no. 4, [8] M. Toda, Theory of Nonlinear Lattices, 2nd enl. ed., Springer, Berlin, 989. [9] X. Zhou, The Riemann Hilbert problem and inverse scattering, SIAM J. Math. Anal. 2-4, Department of Mathematics, Rice University, Houston, TX 775, USA address: URL: Faculty of Mathematics, Nordbergstrasse 5, 9 Wien, Austria, and International Erwin Schrödinger Institute for Mathematical Physics, Boltzmanngasse 9, 9 Wien, Austria address: Gerald.Teschl@univie.ac.at URL: