DIPLOMARBEIT. Long-Time Asymptotics for the KdV Equation. Zur Erlangung des akademischen Grades. Magistra der Naturwissenschaften (Mag. rer. nat.

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1 DIPLOMARBEIT Long-Time Asymptotics for the KdV Equation Zur Erlangung des akademischen Grades Magistra der Naturwissenschaften Mag rer nat Verfasserin: Katrin Grunert Matrikel-Nummer: 4997 Studienrichtung: A 45 Mathematik Betreuer: Ao Univ-Prof Gerald Teschl Wien, am 5 28

2 Contents Introduction ii Scattering theory and the Riemann-Hilbert problem Results from scattering theory 2 The Riemann-Hilbert problem for the KdV equation 3 2 A uniqueness result for symmetric Riemann Hilbert problems 8 3 Conjugation and deformation 3 Conjugation 32 The phase and the partial transmission coefficient 2 33 Deformation 4 34 The long-time asymptotics in the soliton region 7 4 The Riemann-Hilbert problem in the similarity region 2 4 Decoupling 2 42 The long-time asymptotics in the similarity region 2 5 Analytic Approximation 26 A The Riemann-Hilbert problem on a small cross 3 A Approximation 32 A2 Solving the Riemann Hilbert problem on a small cross with constant jumps 35 B Singular integral equations 4 B Properties of the Cauchy-transform 4 B2 Singular integral equations in the context of Riemann Hilbert problems 42 i

3 Chapter Introduction One of the most famous examples of nonlinear wave equations is the Korteweg de Vries KdV equation q t x, t = 6qx, tq x x, t q xxx x, t x, t R R Here the subscripts denote the differentiation with respect to the corresponding variables We will consider real-valued solutions qx, t corresponding to rapidly decaying initial conditions, for example, qx, SR = {f C R sup x α β fx <, α, β N } 2 x The KdV equation occurs in the context of models for small amplitudes, long water waves, collision-free hydromagnetic waves and ion-acoustic waves in plasmas It is well-known that the KdV equation can be solved by the inverse scattering method applied to the associated Schrödinger operator Ht = d2 + qx, t 3 dx2 The scattering data see eg [7], [5] consists of the reflection coefficient Rk, t, a finite number of eigenvalues κ 2 j with < κ < κ 2 < < κ N and norming constants γ j t The aim of my diploma thesis is to compute the long-time asymptotics for the KdV equation in the soliton and the similarity region Our approach is based on the nonlinear steepest descent method for oscillatory Riemann Hilbert problems from Deift and Zhou [9] We closely follow the recent review article [3], where Krüger and Teschl applied this method to compute the long-time asymptotics for the Toda lattice One of the many differences here is the fact that the jump contour of the associated Riemann Hilbert problem is no longer bounded For computing the long-time behavior of the KdV equation there are four cases to distinguish i Soliton Region: For x/t > C for some C >, qx, t 2 N j= κ 2 j cosh 2 κ j x 4κ 3 j t p j, 4 ii

4 Chapter Introduction where the phase shifts p j = 2 log[ γ2 j 2κ j N l=j+ κl κ j κ l +κ j 2] and γj = γ j This result, which will be proven in Theorem 35, has an interesting physical interpretation: In general every term in our asymptotic formula describes a wave Considering only two of them, with the smaller one to the right, after a certain time the waves will overlap ie the bigger one catches up and next the bigger one will separate from the smaller one Then gradually the waves regain their initial speed and shape The only permanent effect of this interaction is that the bigger one is shifted to the right and the smaller one to the left A proof using inverse scattering based on the Gel fand-levitan-marchenko equation can be found in Eckhaus and Schuur [] see also [8] ii Self-Similar Region: x/3t /3 C for some C > The solution in this region is connected with the Painléve II transcendent The investigation of this connection can be found in Segur and Ablowitz [9] iii Collisionless Shock Region: x < and for C < x 3t /3 logt 2/3 < C, for some C > This region only occurs in the generic case ie, when R = and the long-time asymptotic is investigated in Deift, Venakides and Zhou [8] iv Similarity Region: For x/t < C for some C >, /2 4νk k qx, t sin6tk 3 νk log92tk 3 3t + δk, 5 with νk = 2π log Rk 2, δk = π 4 argrk + argγiνk π k N j= k log ζ k d log Rζ 2 arctan κ j k Here k = x 2t denotes the stationary phase point, Rk = Rk, t = the reflection coefficient, and Γ the Gamma function This will be proven in Theorem 43 An analytic discussion can be found in Ablowitz and Segur [] and in Buslaev and Sukhanov [4] Note that if qx, t solves the KdV equation, then so does q x, t Therefore it suffices to investigate the case t The content of this thesis is as follows: In Chapter we derive the Riemann Hilbert problem from scattering theory, where the eigenvalues are added by appropriate pole conditions, which are then turned into jumps iii

5 Chapter Introduction Chapter 2 proves a uniqueness result for symmetric Riemann Hilbert problems In general, there is a well-known non-uniqueness issue for the involved Riemann Hilbert problems see eg [3, Chap 38] Chapter 3 demonstrates how to conjugate our Riemann Hilbert problem and deform our contour such that the jumps are exponentially decreasing away from the stationary phase points Moreover, the asymptotics in the soliton region are computed In Chapter 4 we compute the asymptotics in the similarity region The crucial step is to reduce the given Riemann Hilbert problem to one or more Riemann Hilbert problems localized at the stationary phase points These local Riemann Hilbert problems can be analyzed and controlled individually In Chapter 5 we consider the case where the reflection coefficient has no analytic extension to a neighborhood of the real axis and so we show how to split it in an analytic part plus a small rest Appendix A investigates the solution on a small cross, which occurs in the neighborhood of the stationary phase points In Appendix B we have a close look at the connection between singular integral equations and Riemann Hilbert problems, which is needed for computing the asymptotics Thanks I wish to thank my advisor Gerald Teschl for all the support I had, when writing this thesis, and my friends and colleagues Alice Mikikits-Leitner and Helge Krüger for many discussions and proof reading This work was supported by the Austrian Science Fund FWF under Grant No Y33 and the Faculty of Mathematics of the University of Vienna which provided me with excellent working conditions Errata This version differs a bit from the one, I submitted as my diploma thesis, since I have corrected some small mistakes iv

6 Chapter The Inverse scattering transform and the Riemann Hilbert problem In this chapter we want to derive the Riemann Hilbert problem from scattering theory The eigenvalues will be added by appropriate pole conditions which are then turned into jumps following Deift, Kamvissis, Kriecherbauer, and Zhou [6] in the case of the Toda lattice see also Krüger and Teschl [4] Results from scattering theory For the necessary results from scattering theory respectively the inverse scattering transform for the KdV equation we refer to [7] and [5] We consider real-valued solutions qx, t of the KdV equation, which decay rapidly, that is max t T R + x qx, t dx <, for all T >, for q together with its first three derivatives q x, q xx, and q xxx with respect to x For existence of solutions with such initial data we refer to Section 42 in [5] Associated with the KdV equation is the following Lax-pair: Lt = 2 x + q, t, 2 P t = 4 3 x + 6q, t x + 3q x, t, 3 and d dt Lt = [P t, Lt] = P tlt LtP t on H5 R, which is equivalent to the KdV equation Moreover, P t is skew adjoint with DP t = H 3 R We are more interested in the self-adjoint Schrödinger operator Lt = Ht = d2 dx 2 + q, t, DH = H2 R L 2 R 4

7 Chapter Scattering theory and the Riemann-Hilbert problem L 2 R denotes the Hilbert space of square integrable complex-valued functions over R By our assumption the spectrum of H consists of an absolutely continuous part [, plus a finite number of eigenvalues κ 2 j, ], j N In addition, there exist two Jost solutions ψ ± k, x, t, which solve the differential equation Htψ ± k, x, t = k 2 ψ ± k, x, t, Imk >, 5 and asymptotically look like the free solutions lim x ± e ikx ψ ± k, x, t = 6 Both ψ ± k, x, t are analytic for Imk > and continuous for Imk Theorem The asymptotics of the two Jost solutions are given by ψ ± k, x, t = e ±ikx + Q ± x, t 2ik + O k 2, 7 as k with Imk, where Q + x, t = x qy, tdy, x Q x, t = qy, tdy 8 Proof The Jost solutions ψ ± k, x, t are the unique solutions of the following integral equation for all k with Imk ± ψ ± k, x, t = e ±ikx x sinkx y qy, tψ ± k, y, tdy 9 k We furthermore know that lim x ± e ikx ψ ± k, x, t = + Ok We will proof the asymptotic only for ψ + k, x, t: e ikx ψ + k, x, t = + = + x x = 2ik = 2ik e 2iky x qy, te iky ψ + k, y, tdy 2ik e 2iky x qy, tdy + O 2ik x x e 2iky x qy, tdy + x 2ik qy, tdy + O k 2 k x as k For the first equality we used sink = eik e ik 2i e 2ikx y qy, tdy 2ik qy, tdy + O k 2 Considering the Wronskian W ψ ±, ψ ± = ψ ± ψ ± ψ ± ψ±, where denotes the derivation with respect to x, we see that it is independent of x and t along the real axis Thus we can compute W ψ ±, ψ ± k = ±2ik, which shows that these two functions are linearly independent, but the Schrödinger equation can have at most two linearly independent solutions, which in our case are represented 2

8 Chapter Scattering theory and the Riemann-Hilbert problem by ψ + and ψ Therefore we can write ψ ± as a linear combination of the given Jost solutions In particular, one has the scattering relations T kψ k, x, t = ψ ± k, x, t + R ± k, tψ ± k, x, t, k R, where T k, R ± k, t are the transmission respectively reflection coefficients The transmission and reflection coefficients have the following well-known properties: Lemma 2 The transmission coefficient T k is meromorphic for Imk > with simple poles at iκ,, iκ N and is continuous up to the real line The residues of T k are given by where and ψ + iκ j, x, t = µ j tψ iκ j, x, t Moreover, Res iκj T k = iµ j tγ +,j t 2 = iµ j γ 2 +,j, γ +,j t = ψ + iκ j,, t 2 2 T kr + k, t + T kr k, t =, T k 2 + R ± k, t 2 = 3 The functions qx, t, x R for fixed t R are uniquely determined by their right scattering data, that is, by the right reflection coefficient R + k, t, k R and the eigenvalues κ j,, j =,, N, together with the corresponding norming constants γ +,j t >, j =,, N So in particular, one reflection coefficient, say Rk, t = R + k, t, and one set of norming constants, say γ j t = γ +,j t, suffices Moreover, the time dependence is given by: Lemma 3 The time evolutions of the quantities R + k, t, T k, t and γ +,j t are given by where Rk = Rk,, T k = T k, and γ j = γ j Rk, t = Rke 8ik3t, 4 γ j t = γ j e 4κ3 j t, 5 T k, t = T k, 6 2 The Riemann-Hilbert problem for the KdV equation We will define a Riemann Hilbert problem as follows: { T kψ k, x, te mk, x, t = ikx ψ + k, x, te ikx, Imk >, ψ+ k, x, te ikx T kψ k, x, te ikx, Imk < 7 We are interested in the jump condition of mk, x, t on the real axis R oriented from negative to positive To formulate our jump condition we use the following convention: When representing functions on R, the lower subscript denotes the non-tangential limit from different sides By m + k we denote the limit from above and by m k the one from below Using the notation above implicitly assumes that these limits exist in the sense that mk extends to a continuous function on the real axis away from 3

9 Chapter Scattering theory and the Riemann-Hilbert problem Theorem 4 Vector Riemann Hilbert problem Let S + H = {Rk, k ; κ j, γ j, j N} the right scattering data of the operator H Then mk = mk, x, t defined in 7 is meromorphic away from the real axis with simple poles at iκ j, iκ j and satisfies: i The jump condition Rk 2 Rke m + k = m kvk, vk = tφk Rke tφk, 8 for k R, ii the pole conditions Res iκj mk = lim mk k iκ j iγj 2, etφiκj iγ 2 Res iκj mk = lim mk j e tφiκj, k iκ j 9 iii the symmetry condition m k = mk, 2 iv and the normalization lim miκ = 2 κ Here the phase is given by Φk = 8ik 3 + 2ik x t 22 Proof i For the proof of the jump condition we need the scattering relations and 3 vk Rke = tφk Rke tφk Rk 2 23 and so we can also show m + kvk = m k T kψ k, x, te ikx Rke tφk ψ + k, x, te ikx = T kψ k, x, te ikx Rk, te ikx ψ + k, x, t = ψ + k, x, te ikx = ψ + k, x, te ikx, hence we have proven the jump condition for the first component of m k For the second component of m k we compute T kψ k, x, te ikx Rke tφk + Rk 2 ψ + k, x, te ikx = T kψ k, x, te ikx Rk, t + T k 2 ψ + k, x, te ikx = ψ + k, x, te ikx Rk, t + Rk, t 2 e ikx ψ + k, x, t + T k 2 e ikx ψ + k, x, t = ψ + k, x, trk, te ikx + ψ + k, x, te ikx = T kψ k, x, te ikx = T kψ k, x, te ikx 4

10 Chapter Scattering theory and the Riemann-Hilbert problem ii First of all note that the Jost solutions ψ ± k, x, t are analytic for Imk > and that the transmission coefficient T k has only simple poles at iκ j Hence Res iκj m 2 k = and Res iκj m k = Moreover Res iκj m k = Res iκj T kψ iκ j, x, te κjx = iµ j tγ j t 2 ψ iκ j, x, te κjx = iγ j t 2 ψ + iκ j, x, te κjx = iγ 2 j e tφiκj ψ + iκ j, x, te κjx Similarly Res iκj m 2 k can be computed Here m l k denotes the l th component of mk iii The symmetry condition is obvious from the construction of our function mk, x, t iv The normalization follows immediately from the next Lemma Remark 5 Observe that the pole condition at iκ j is sufficient since the one at iκ j follows by symmetry as the following calculation shows Res iκj mk = lim k + iκ j mk k iκ j = lim k iκ j mk k iκ j = lim mk k iκ j iγj 2etΦiκj = lim m k k iκ j iγj 2etΦiκj iγ 2 = lim mk j e tφiκj k iκ j : Moreover, we have the following asymptotic behavior as k with Imk Lemma 6 The function mk, x, t defined in 7 satisfies mk, x, t = + Qx, t + O 2ik k 2 24 Here Qx, t = Q + x, t is defined in 8 Proof This follows from 7 and T kψ k, x, tψ + k, x, t = + qx,t 2k +O 2 k 4 For details we refer to [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 [6] Choose ε so small 5

11 Chapter Scattering theory and the Riemann-Hilbert problem that the discs k iκ j < ε lie inside the upper half plane and do not intersect Then redefine mk in a neighborhood of iκ j respectively iκ j according to mk mk = mk mk,, k iκ j < ε, iγ2 j etφiκ j k iκ j iγ 2 j etφiκ j k+iκ j, k + iκ j < ε, else 25 Note that for Imk < we redefined mk with respect to our symmetry 2 Then a straightforward calculation using Res iκ mk = lim k iκ k iκmk shows: Lemma 7 Suppose mk is redefined as in 25 Then mk is holomorphic away from the real axis and the small circles around iκ j and iκ j Furthermore it satisfies 8, 2, 2 and the pole condition is replaced by the jump condition m + k = m k, k iκ j = ε, iγ2 j etφiκ j k iκ j m + k = m k iγ 2 j etφiκ j k+iκ j, k + iκ j = ε, where the small circle around iκ j is oriented counterclockwise 26 Proof 8,2, and 2 still hold, because mk is only redefined with respect to our symmetry condition on the small circles around iκ j, j N A simple calculation inserting the definition of the new mk and using the former pole condition shows that the pole condition is replaced by the jump condition 26 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 k C, such that mk =, then nk = k k mk satisfies the same jump and pole conditions as mk 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 look at the one soliton solution verifies that this case indeed can happen Lemma 8 One soliton solution Suppose there is only one eigenvalue and that the reflection coefficient vanishes, that is, S + Ht = {Rk, t, k R; κ, γt, κ >, γ > } Then the unique solution of the Riemann Hilbert problem 8 2 is given by m k = fk f k 27 fk = + k + iκ + 2κ γ 2 e tφiκ k iκ 2κ γ 2 e tφiκ 6

12 Chapter Scattering theory and the Riemann-Hilbert problem In particular, Q + x, t = 2γ 2 e tφiκ κ γ 2 etφiκ Proof By assumption the reflection coefficient vanishes and so the jump along the real axis disappears Therefore and by the symmetry condition, we know that the solution is of the form m k = fk f k, where fk is meromorphic Furthermore the function fk has only a simple pole at iκ, so that we can make the ansatz fk = C + D k+iκ k iκ Then the constants C and D are uniquely determined by the pole conditions and the normalization In fact, observe fk = f k = if and only if k = and 2κ = γ 2 e tφiκ Furthermore, even in the general case mk = can only occur at k = as the following lemma shows Lemma 9 If mk = for m defined as in 7, then k = Moreover, the zero of at least one component is simple in this case Proof By 7 the condition mk = implies that the Jost solutions ψ k, x and ψ + k, x are linearly dependent or that the transmission coefficient T k = This can only happen, at the band edge, k = or at an eigenvalue k = iκ j We begin with the case k = iκ j In this case the derivative of the Wronskian W k = ψ + k, xψ k, x ψ +k, xψ k, x does not vanish by the wellknown formula d dk W k k=k = 2k R ψ +k, xψ k, xdx Moreover, the diagonal Green s function gz, x = W k ψ + k, xψ k, x is Herglotz as a function of z = k 2 and hence can have at most a simple zero at z = k 2 Since z k 2 is conformal away from z = the same is true as a function of k Hence, if ψ + iκ j, x = ψ iκ j, x =, both can have at most a simple zero at k = iκ j But T k has a simple pole at iκ j and hence T kψ k, x cannot vanish at k = iκ j, a contradiction It remains to show that one zero is simple in the case k = In fact, one can show that d dk W k k=k in this case as follows: First of all note that ψ ± k where the dot denotes the derivative with respect to k again solves H ψ ± k = k 2 ψ ± k if k = Moreover, by W k = we have ψ + k = c ψ k for some constant c independent of x Thus we can compute Ẇ k = W ψ + k, ψ k + W ψ + k, ψ k = c W ψ + k, ψ + k + cw ψ k, ψ k by letting x + for the first and x for the second Wronskian in which case we can replace ψ ± k by e ±ikx, which gives Ẇ k = ic + c Hence the Wronskian has a simple zero But if both functions had more than simple zeros, so would the Wronskian, a contradiction 7

13 Chapter 2 A uniqueness result for symmetric vector Riemann Hilbert problems In this chapter we want to investigate uniqueness for the holomorphic vector Riemann Hilbert problem m + k = m kvk, k Σ, m k = mk, 2 lim miκ = κ Hypothesis H2 Let Σ consist of a finite number of smooth oriented curves in C such that the distance between Σ and {iy y y } is positive for some y > Assume that the contour Σ is invariant under k k Moreover, suppose the jump matrix v can be factorized according to v = b b + = I w I+w +, where w ± are continuous, bounded, square integrable and satisfy w ± k = w k, k Σ 22 Now we are ready to show that the symmetry condition in fact guarantees uniqueness Theorem 22 Suppose there exists a solution mk of the Riemann Hilbert problem 2 for which mk = can happen at most for k = in which k case lim sup k m jk is bounded from any direction for j = or j = 2 Then the Riemann Hilbert problem 2 with norming condition replaced by lim miκ = α α 23 κ for given α C, has a unique solution m α k = α mk Proof Let m α k be a solution of 2 normalized according to 23 Then we can construct a matrix valued solution via M = m, m α and there are two possible cases: Either det Mk is nonzero for some k or it vanishes identically 8

14 Chapter 2 A uniqueness result for symmetric Riemann Hilbert problems We start with the first case By Lemma 7, we can rewrite all poles as jumps with determinant one Hence, the determinant of this modified Riemann Hilbert problem has no jump Next consider a triangle, which intersects our contour Σ We can now apply the same method as in the proof of the Schwarz reflection principle to conclude that the function det Mk is holomorphic on C Since, it is bounded at infinity, we can apply Liouville s theorem and hence the determinant is constant But taking determinants in M k = Mk, gives a contradiction It remains to investigate the case where detm In this case we have m α k = δkmk with a scalar function δ Moreover, δk must be holomorphic for k C\Σ and continuous for k Σ except possibly at the points where mk = Since it has no jump across Σ, δ + km + k = m α,+ k = m α, kvk = δ km kvk = δ km + k, we can conclude by the same method as in the first case that it is even holomorphic in C\{} with at most a simple pole at k = Hence it must be of the form δk = A + B k Since δ has to be symmetric, δk = δ k, we obtain B = Now, by the normalization, we obtain δk = A = α This finishes the proof Furthermore, the requirements cannot be relaxed to allow eg second order zeros instead of simple zeros In fact, if mk is a solution for which both components vanish of second order at, say, k =, then mk = k mk is a 2 nontrivial symmetric solution of the vanishing problem ie for α = By Lemma 9 we have Corollary 23 The function mk, x, t defined in 7 is the only solution of the vector Riemann Hilbert problem 8 2 Proof The function mk, x, t defined in 7 satisfies the assumptions of Theorem 22 and therefore, is the unique solution of our Riemann-Hilbert problem Observe that there is nothing special about k where we normalize, any other point would do as well However, observe that for the one soliton solution 27, fk vanishes at k = iκ 2κ2 γ 2 e tφiκ + 2κ 2 γ 2 e tφiκ 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 are 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 B 9

15 Chapter 3 Conjugation and deformation This chapter demonstrates how to conjugate our Riemann-Hilbert problem and how to deform our jump contour, such that the jumps will be exponentially decreasing away from the stationary phase points Furthermore the asymptotics in the soliton region are computed We will further study how poles can be dealt with in this chapter, because solitons are represented in a Riemann-Hilbert problem by pole conditions 3 Conjugation For easy reference we note the following result: Lemma 3 Conjugation Assume that Σ Σ Let D be a matrix of the form dk Dk =, 3 dk where d : C\ Σ C is a sectionally analytic function Set then the jump matrix transforms according to mk = mkdk, 32 ṽk = D k vkd + k 33 If d satisfies d k = dk and dk = + O k as k Then the transformation mk = mkdk respects our symmetry, that is, mk satisfies 2 if and only if mk does, and our normalization condition In particular, we obtain v v ṽ = 2 d 2 v 2 d 2, k Σ\ Σ, 34 v 22 respectively ṽ = d d + v v 2 d + d v 2 d + d, k Σ Σ 35 d + d v 22

16 Chapter 3 Conjugation and deformation Proof For k Σ we compute m + k = m + kd + k = m kvkd + k = m kd kd k vkd + k = m kṽk and thus ṽk = D k vkd + k The symmetry condition follows by the next calculation m k = m kdk = mk Dk = mkdkdk Dk = mk In order to remove the poles there are two cases to distinguish Some jumps are already exponentially decaying and in this case there is nothing to do Otherwise we use conjugation to turn the jumps into exponentially decaying ones, again following Deift, Kamvissis, Kriecherbauer, and Zhou [6] It turns out that we will have to handle the poles at iκ j and iκ j in one step in order to preserve symmetry and in order to not add additional poles elsewhere Lemma 32 Assume that the Riemann Hilbert problem for m has jump conditions near iκ and iκ given by m + k = m k, k iκ = ε, iγ2 k iκ m + k = m k iγ 2 k+iκ, k + iκ = ε 36 Then this Riemann Hilbert problem is equivalent to a Riemann Hilbert problem for m which has jump conditions near iκ and iκ given by k+iκ2 m + k = m k iγ 2 k iκ, k iκ = ε, m + k = m k k iκ 2, k + iκ = ε, iγ 2 k+iκ and all remaining data conjugated as in Lemma 3 by k iκ Dk = k+iκ k+iκ 37 k iκ Proof To turn γ 2 into γ 2, introduce D by k iκ iγ 2 iγ 2 Dk = k iκ iγ2 k+iκ k+iκ iγ 2 k iκ k+iκ k+iκ k iκ k iκ k+iκ k+iκ k iκ k iκ k+iκ k+iκ k iκ, k iκ < ε,, else,, k + iκ < ε,

17 Chapter 3 Conjugation and deformation and note that Dk is analytic away from the two circles Now set mk = mkdk The new jump conditions can be verified with the same method as in the previous Lemma 32 The phase and the partial transmission coefficient The jump along the real axis is of oscillatory type and our aim is to apply a contour deformation such that all jumps will be moved into regions where the oscillatory terms will decay exponentially Since the jump matrix v contains both exptφ and exp tφ we need to separate them in order to be able to move them to different regions of the complex plane We recall that the phase of the associated Riemann Hilbert problem is given by Φk = 8ik 3 + 2ik x 38 t and the stationary phase points, Φ k =, are denoted by k = x 2t, k = x 2t, λ = x 2t 39 For x t > we have k ir, and for x t < we have k R For x t > we will also need the value iκ ir defined via ReΦiκ =, that is, x t = 4κ2 3 We will set κ = if x t < for notational convenience A simple analysis shows that for x t > we have < k /i < κ As mentioned above we will need the following factorization of the jump condition 8 The correct factorization for Rek < k is given by vk = b k b + k, 3 where Rke tφk b k =, b + k = Rke tφk For k < Rek the factorization is given by vk = B k Rk 2 B + k, 32 Rk 2 where B k =, B RketΦk + k = Rk 2 Rke tφk Rk 2 To get rid of the diagonal part in the factorization corresponding to k < Rek and to conjugate the jumps near the eigenvalues we need the partial transmission coefficient 2

18 Chapter 3 Conjugation and deformation We define the partial transmission coefficient with respect to k by T k, k = N κ j κ, j= k+iκ j k+iκ j k iκ j exp k iκ j, k ir +, k 33, k R + 2πi k log T ζ 2 ζ k dζ for k C\Σk, where Σk = [ Rek, Rek ] Note that T k, k can be computed in terms of the scattering data since T k 2 = R + k, t 2 Moreover, we conclude that T k, k = + T k i k + O k 2, as k where T k = κ j κ, κ j κ, 2κ j, k ir +, 2κ j + k 2π k log T ζ 2 dζ, k R + For the next theorem, we need the Plemelj formula Theorem 33 Plemelj Let L be a simple smooth oriented arc If ψt is a function satisfying a Hölder condition on L, then Ψ ± t = ± 2 ψt + ψτ dτ, 34 2πi τ t or, equivalently L Ψ + t Ψ t = ψt, 35 Ψ + t + Ψ t = ψτ dτ 36 πi L τ t Here the connection between ψ and Ψ is given by Ψt = ψτ dτ 37 2πi τ t Proof A proof can be found in Muskhelishvili [6] Theorem 34 The partial transmission coefficient T k, k is meromorphic in C\Σk, where Σk = [ Rek, Rek ], with simple poles at iκ j and simple zeros at iκ j for all j with κ < κ j, and satisfies the jump condition Moreover, L T + k, k = T k, k Rk 2, for k Σk 38 i T k, k = T k, k, k C\Σk and lim k T k, k > ii T k, k = T k, k Proof That iκ j are simple poles and iκ j are simple zeros is obvious from the Blaschke factors and that T k, k has the given jump follows from Plemelj formula i and ii are straightforward to check 3

19 Chapter 3 Conjugation and deformation 33 Deformation Now we are ready to perform our conjugation step Introduce k iκj iγ j 2etΦiκ j D iγ 2 j etφiκ j k, k iκ j < ε, κ < κ j, k iκ j Dk = iγ2 j etφiκ j k+iκ j k+iκ j D k, k + iκ j < ε, κ < κ j, iγj 2etΦiκ j D k, else, where Note that we have D k = D k = T k, k T k, k Dk Now we conjugate our problem using Dk and set mk = mkdk Then using Lemma 3 and Lemma 32 the jump corresponding to κ < κ j if any is given by k iκ j ṽk = iγj 2etΦiκ j T k,k 2, k iκ j = ε, ṽk = k+iκ j iγ 2 j etφiκ j T k,k 2, k + iκ j = ε, 39 and corresponding to κ > κ j if any by ṽk = iγ2 j etφiκ j T k,k 2, k iκ j = ε, k iκ j iγ ṽk = 2 j etφiκ j T k,k 2 k+iκ j, k + iκ j = ε 32 In particular, all jumps corresponding to poles, except for possibly one if κ j = κ, are exponentially decreasing In this case we will keep the pole condition which now reads Res iκj mk = lim mk k iκ j Res iκj mk = lim mk k iκ j Furthermore, the jump along R is given by iγj 2etΦiκj T iκ j, k 2, iγ 2 j e tφiκj T iκ j, k 2 32 { b k b + k, ṽk = B k B+ k, Rek < k, Rek > k, 322 4

20 Chapter 3 Conjugation and deformation + k ir + + k R + k k + Figure 3: Sign of ReΦk for different values of k where b k = R ke tφk T k,k 2, b+ k = Rke tφk, 323 T k,k 2 and B k = T k,k 2 Rke tφk = T k,k, Rk 2 T k,k RketΦk B T+k,k2 R ke + k = tφk T+k,k Rk 2 = T + k,k R ke tφk Here we have used T ± k, k = T k, k, k Σk and the jump condition for the partial transmission coefficient T k, k along Σk in the last step, which shows that the matrix entries are bounded for k R Note also that we have used T k, k = T k, k and R k = Rk for k R to show that there exists an analytic continuation into the neighborhood of the real axis Now we deform the jump along R to move the oscillatory terms into regions where they are decaying There are two cases to distinguish: Case : k ir, k : We set Σ ± = {k C Imk = ±ε} for some small ε such that Σ ± lies in the region with ±Rek < and such that we don t intersect any other contours Then we can split our jump by redefining mk according to mk b + k, < Imk < ε, mk = mk b k, ε < Imk <, 324 mk, else Thus the jump along the real axis disappears and the jump along Σ ± is given by { b+ k, k Σ + vk = b k 325, k Σ 5

21 Chapter 3 Conjugation and deformation ReΦ < Σ + R ReΦ > ReΦ > ReΦ < Σ Figure 32: Deformed contour for k ir + ReΦ< Σ ReΦ> ReΦ> Σ 2 + Σ + Σ + k k Σ 2 ReΦ< ReΦ< Σ R ReΦ> Figure 33: Deformed contour for k R + All other jumps are unchanged Note that the resulting Riemann-Hilbert problem still satisfies our symmetry condition 2, since we have b± k = b k 326 By construction the jump along Σ ± is exponentially decreasing as t Case 2: k R, k : We set Σ ± = Σ ± Σ 2 ± Again note that Σ ± respectively Σ 2 ± lie in the region with ±ReΦk < Then we can split our jump by redefining mk according to mk b + k, k between R and Σ +, mk b k, k between R and Σ, mk = mk B + k, k between R and Σ 2 +, 327 mk B k, k between R and Σ 2, mk, else One checks that the jump along R disappears and the jump along Σ ± is 6

22 Chapter 3 Conjugation and deformation given by b+ k, k Σ +, b k, k Σ vk =, B + k, k Σ 2 +, B k, k Σ All other jumps are unchanged Again the resulting Riemann-Hilbert problem still satisfies our symmetry condition 2 and the jump along Σ ± \{k, k } is exponentially decreasing as t 34 The long-time asymptotics in the soliton region Now we are ready to state and proof one of our main results: Theorem 35 Assume and abbreviate by c j = 4κ 2 j the velocity of the j th soliton determined by ReΦiκ j = Then the asymptotics in the soliton region, x/t C for some C >, are as follows: Let ε > sufficiently small such that the intervals [c j ε, c j + ε], j N, are disjoint and lie inside R + If x t c j < ε for some j, one has respectively x for any l, where qy, tdy = qx, t = If x t c j ε, for all j, one has respectively for any l 2γj 2 x, t + 2κ j γj 2x, t 2T k + Ot l, 4κ j γj 2 x, t + 2κ j γj 2x, + t2 Ot l 329 γ 2 j x, t = γ 2 j e tφiκj T iκ j, i κ j Proof m has the following asymptotic mk = m + m k + O k 2 = T k, k T k, k x qy, tdy = 2T k + Ot l, qx, t = Ot l 33 + T k, k Q + x, t2ik T k, k Q + x, t2ik + O = + it k + Q + x, t + O k 2ik k 2 k 2 7

23 Chapter 3 Conjugation and deformation Thus we have m = it k + Q +x, t 332 2i By construction the jump along Σ ± is exponentially decreasing as t Hence we can apply Theorem B6 as follows: If x t c j > ε for all j we can choose γ = and w t by removing all jumps corresponding to poles from w t The error between the solutions of w t and w t is exponentially small in the sense of Theorem B6 In particular w t w t Ot l as t for all l and w t w t 2 Ot l as t for all l and so the associated Riemann-Hilbert problems only differ by Ot l From Lemma 8, we have the one soliton solution m = fk f k with fk for Imk big enough and so Q + x, t = +2T k + Ot l If x t c j < ε for some j, we choose γ t = γ k x, t and w t As before we conclude that the error between the solutions of w t and w t is exponentially small in the sense of Theorem B6 and so the associated solutions of the Riemann-Hilbert problems only differ by Ot l From Lemma 8, we have the one soliton solution m = fk f k with fk = +2κ j γj 2x,t + k+iκ j k iκ j 2κ j γj 2 x, t for Imk big enough where γ 2 j x, t = γ 2 j e tφiκj T iκ j, i κ j 3 2 and hence, plugging in the power series expansion, Q + x, t = +2T k + For the second part recall from Lemma 6 that 2γj 2 x, t + 2κ j γj 2x, t + Ot l 333 qx, t T kψ + k, x, tψ k, x, t = + 2k 2 + O 334 k4 In the first case x t c j > ε for all j we know that the solution mk is of the form mk = + Ot l and we can therefore conclude by multiplying the two components that qx, t = Ot l 335 In the second case x t c j < ε for some j we know from the first part that the solution mk is of the form mk = fk f k + Ot l, where fk is defined as in the first part Multiplying the first and the second component and plugging in the power series expansion we get 4κ j γj 2 x, t qx, t = + 2κ j γj 2x, t + 2γj 4 x, t + 2κ j γj 2x, + t2 Ot l = 4κ j γj 2 x, t + 2κ j γj 2x, + t2 Ot l 8

24 Chapter 3 Conjugation and deformation Corollary 36 Assume, then the asymptotic in the soliton region, x/t C for some C >, is as follows qx, t = N j= where γ 2 j x, t = γ2 j etφiκj T iκ j, i κj 3 2 4κ j γj 2 x, t + 2κ j γj 2x, + t2 Ot l, 336 Proof The claim follows immediately after some easy calculations: i If x t c j < ε for some j, we can use the last theorem to obtain the corresponding term ii If the eigenvalue κ j lies in the region where ReΦk <, we see that γj 2 x, t is exponentially decreasing as t and therefore the whole, corresponding term is exponentially decreasing as t iii If the eigenvalue κ j lies in the region where ReΦk >, we look at the corresponding term: 4κ j γj 2 x, t + 2κ j γj 2x, = 6κ 3 j γ 2 j x, t t2 + 2κ j γ 2 j x, t Thus these term is also exponentially decaying as t, because γ 2 j x, t is exponentially decreasing This finishes the proof Remark 37 This is exactly the same result as mentioned in the introduction as the following calculation shows: qx, t = N j= = 2 = 2 4κ j γj 2 x, t + 2κ j γj 2 x, t2 N j= N j= 4κ 2 j e tφiκj 2pj + e tφiκj+2pj + 2 κ 2 j cosh 2 κ j x 4κ 3 j p j, 338 where we used that T iκ j, i κ j 3 = κ l κ j, κ j + κ l κ j κ l 339 9

25 Chapter 4 The Riemann-Hilbert problem in the similarity region In the previous section we have seen that for k R we can reduce everything to a Riemann-Hilbert problem for mk such that the jumps are exponentially decaying except in small neighborhoods of the stationary phase points k and k Hence we need to continue our investigation of this case in this chapter 4 Decoupling Denote by Σ c ±k the parts of Σ + Σ inside a small neighborhood of ±k We will now show that solving the two problems on the small crosses Σ c k respectively Σ c k will lead us to the solution of our original problem Theorem 4 Decoupling Consider the Riemann Hilbert problem m + k = m kvk, k Σ, m =, 4 and let < α < β 2α, ρt be given Suppose that for every sufficiently small ε > both the L 2 and the L norms of v are Ot β away from some ε neighborhoods of some points k j, j n Moreover, suppose that the solution of the matrix problem with jump vk restricted to the ε 2 neighborhood has a solution given by M j k = I + M j ρt α + Oρt β, k k j > ε k k j 2 42 Then the solution mk is given by mk = + ρt α n j= M j k k j + Oρt β, 43 where the error term depends on the distance of k to Σ 2

26 Chapter 4 The Riemann-Hilbert problem in the similarity region Proof In this proof we will use the theory developed in Appendix B with m k = I and the usual Cauchy kernel Ω s, k = I ds s k Assume that mk exists Introduce mk by { mkm j k, k k j ε, mk = 44 mk, else The Riemann Hilbert problem for mk has jumps given by M j k, k k j = ε, M j kvkm j k, k Σ, ε ṽk = 2 < k k j < ε, I, k Σ, k k j < ε 2, vk, else 45 By assumption the jumps are I + Oρt α on the circles k k j = ε and even I+Oρt β on the rest both in L 2 and L norms In particular as in Lemma A4 we infer µ 2 = Oρt α 46 Thus we can conclude mk = + µs ws ds 2πi Σ s k = + n µsm j s I ds 2πi j= k k j =ε s k + Oρt β = ρt α n ds M j 2πi s k j s k + Oρt β = + ρt α n and hence the claim is proven j= j= k k j =ε M j k k j + Oρt β, The long-time asymptotics in the similarity region Now let us turn to the solution of the problem on Σ c k = Σ + Σ {k k k < ε} for some small ε > Since, we no longer impose the symmetry condition, we can also deform our contour slightly such that Σ c k consists of two straight lines Next, Φk = 6ik 3, Φ k = 48ik As a first step we make a change of coordinates ζ = 48k k k, k = k + ζ 48k 48 such that the phase reads Φk = Φk + i 2 ζ2 + Oζ 3 The corresponding Riemann-Hilbert problem will be solved in Appendix A To apply this result, 2

27 Chapter 4 The Riemann-Hilbert problem in the similarity region we will need the behavior of our jump matrix near k, that is, the behavior of T k, k near k Lemma 42 Let k R, then iν k k T k, k = T k, k, 49 k + k where ν = π log T k and the branch cut of the logarithm is chosen along the negative real axis Here N k + iκ j T k, k = exp k T ζ 2 log k iκ j 2πi T k 2 ζ k dζ 4 j= k is Hölder continuous of any exponent less then at the stationary phase point k = k and satisfies T k, k T Proof First of all note that k iν exp log T k 2 k 2πi k ζ k dζ k = 4 k + k Moreover we know from Theorem 34 T k, k = T k, k T k, k = T k, k T k, k = T k, k T k, k 42 for k C\Σk Furthermore the Blaschke products N k+iκ j j= k iκ j are continuous for k iκ j, j =,, N and k T ζ 2 log 2πi k T k 2 ζ k dζ + k T ζ 2 log 2πi k T k 2 ζ + k dζ = k k k T ζ 2 log 2πi k T k 2 ζ k 2 dζ, which tends to as k k with k C\Σk Hence T k, k as k k and so T k, k T For the proof of the Hölder continuity of any exponent less than at k = k, we refer to Muskhelishvili [6] If kζ is defined as in 48 and < α <, then there is an L > such that T kζ, k ζ iν T k, k e iν log2k 48k L ζ α, 43 where the branch cut of ζ iν is the negative real axis Here we used the following observations: i iν iν k k = ζ iν = ζ iν F ζ, k 44 k + k 2k 48k + ζ This function F ζ, k is Hölder continuous of any exponent less than at ζ = 22

28 Chapter 4 The Riemann-Hilbert problem in the similarity region ii With the same idea as for differentiable functions at ζ =, we can show that the product of two Hölder continuous functions is again Hölder continuous We also have Rkζ Rk L ζ α 45 and thus the assumptions of Theorem A are satisfied with r = Rk T k, k 2 e 2iν log2k 48k 46 Therefore we can conclude that the solution on Σ c k is given by M c k = I + ζ i t /2 = I + 48k k k β + Ot α β i t /2 β + Ot α, β 47 where β is given by β = νe iπ/4 argr+argγiν e tφk t iν = νe iπ/4 argrk+argγiν T k, k 2 92k 3 iν e tφk t iν 48 and /2 < α < We also need the solution M2k c on Σ c k We make the following ansatz, which is inspired by the symmetry condition for the vector Riemann Hilbert problem outside the two small crosses: From this we conclude M c 2k = M2k c = I 48k k + k Applying Theorem 4 leads to mk = + 48k x i t /2 M c k i t /2 49 β + Ot α 42 β k k β β k + k β β We are now ready to state and proof our second main result: + Ot α 42 Theorem 43 Assume with l = 5, then the asymptotics in the similarity region, x/t C for some C >, are given by ν qy, tdy = 2T k 3k t /2 cos6tk3 ν log92tk+δk 3 +Ot α respectively 422 4νk qx, t = sin6tk 3 ν log92tk 3 3t + δk + Ot α

29 Chapter 4 The Riemann-Hilbert problem in the similarity region for any /2 < α < Here ν = π log T k, 424 δk = π 4 argrk + argγiν + 2 arg T k, k, 425 N k + iκ j T k, k = exp k T ζ 2 log k iκ j 2πi T k 2 dζ 426 ζ k j= k Proof As in the proof of the asymptotics in the soliton region we set mk = m + m k + Ok 2 as k and conclude Therefore we compute mk = + i 48k t /2 = + 48k + Ot α, which leads to with m = it k + Q +x, t 2i i t /2 k l= 427 β β k + k β β k k l k β β k l= + Ot α k l β β k Q + x, t = 2T k k t /2 Reβ + Ot α, 428 β = νe iπ/4 argrk+argγiν T k, k 2 92k 3 iν e tφk t iν 429 Using the fact that β/ ν = proves the first claim For the second part recall from Lemma 6 that T kψ + k, x, tψ k, x, t = + qx, t 2k 2 + O k 4 = m m 2 43 Plugging in the power series expansions of the components of m yields 4k qx, t = Imβ 43 3t Remark 44 The result mentioned in the introduction is the same as the one stated above, because arg T N k, k = 2 arctan κ j k T ζ 2 log k 2π k T k 2 dζ ζ k j= 24

30 Chapter 4 The Riemann-Hilbert problem in the similarity region and π k k log ζ k d log Rζ 2 = lim k k π = lim k k π k log ζ k d log Rζ 2 k log ζ k log Rζ 2 k k k = lim log Rk 2 k k π = k Rζ 2 log π k Rk 2 k k dζ ζ k ζ k dζ = π log Rζ 2 dζ k ζ k k log Rζ 2 dζ ζ k k T ζ 2 log k T k 2 k ζ k dζ Remark 45 Formally the equation 423 for q can be obtained by differentiating the equation 422 for Q with respect to x That this step is admissible could be shown as in Deift and Zhou [], however our approach avoids this step Remark 46 Note that Theorem 4 does not require uniform boundedness of the associated integral operator in contradistinction to Theorem B6 We only need the knowledge of the solution in some small neighborhoods However it cannot be used in the soliton region, because our solution is not of the form I + o 25

31 Chapter 5 Analytic Approximation In this chapter we want to present the necessary changes in the case where the reflection coefficient does not have an analytic extension The idea is to use an analytic approximation and to split the reflection in an analytic part plus a small rest The analytic part will be moved to the complex plane while the rest remains on the real axis This needs to be done in such a way that the rest is of Ot and the growth of the analytic part can be controlled by the decay of the phase In the soliton region a straightforward splitting based on the Fourier transform Rk = e ikx ˆRxdx 5 R will be sufficient If our solution qx, t is decaying rapid enough, we can conclude that ˆR L R and furthermore x l ˆR x L, For details we refer to [5] Lemma 5 Suppose ˆR L R, x l ˆR x L, and let ε, β > be given Then we can split the reflection coefficient according to Rk = R a,t k + R r,t k such that R a,t k is analytic in < Imk < ε and R a,t ke βt = Ot l, < Imk < ε, R r,t k = Ot l, k R 52 Proof We choose R a,t k = Kt eikx ˆRxdx with Kt = β ε t for some positive β < β Then, for < Imk < ε, Ra,t ke βt = e βt Kt e ikx ˆRxdx e βt Kt ˆRx e Imkx dx e βt e KtImk ˆR e βt e Ktε ˆR ˆR e β βt Moreover, we have ˆR = ˆRx dx = + x + x ˆRx dx R R + x 2 + x ˆRx 2 <, 53 26

32 Chapter 5 Analytic Approximation which proves the first claim Similarly, for Imk =, R r,t k Kt ˆRx dx = x l L ˆR x, Kt l Kt const t l x l+ ˆR x x l+ dx 54 To apply this lemma in the soliton region k ir + we choose β = min ReΦk > 55 Imk= ε and split Rk = R a,t k + R r,t k according to Lemma 5 to obtain b± k = b a,t,± k b r,t,± k = b r,t,± k b a,t,± k Here b a,t,± k, b r,t,± k denote the matrices obtained from b ± k as defined in 323 by replacing Rk with R a,t k, R r,t k, respectively Now we can move the analytic parts into the complex plane as in Chapter 3 while leaving the rest on the real axis Hence, rather then 325, the jump now reads ba,t,+ k, k Σ +, ˆvk = ba,t, k, k Σ, 56 br,t, k b r,t,+ k, k R By construction we have ˆvk = I + Ot l on the whole contour and the rest follows as in Section 3 In the similarity region not only b ± occur as jump matrices but also B ± These matrices B ± have at first sight more complicated off diagonal entries, but a closer look shows that they have indeed the same form As we will rewrite them in terms of the left rather then the right scattering data, we will use the following notations: R r k R + k for the right and R l k R k for the left reflection coefficient Moreover, let T r k, k T k, k respectively T l k, k T k/t k, k be the right respectively left partial transmission coefficient With this notation we have { b k b + k, Rek < k, ṽk = B k 57 B+ k, Rek > k, where and b k = R r ke tφk T r k,k 2, b+ k = R rke tφk, T rk,k 2 B k = Tr, k,k 2 T k R 2 r ke tφk, B Tr,+k,k2 + k = T k R 2 r ke tφk 27

33 Chapter 5 Analytic Approximation Using 3 we can further write B k = R l ke tφk, B+ k = T l k,k 2 R l ke tφk T l k,k 2 58 In the similarity region we need to take the small vicinities of the stationary phase points into account Since the phase is cubic near these points, we cannot use it to dominate the exponential growth of the analytic part away from the unit circle Hence we will take the phase as a new variable and use the Fourier transform with respect to this new variable Since this change of coordinates is singular near the stationary phase points, there is a price we have to pay, namely, requiring additional smoothness for Rk If our solution qx, t is decaying rapidly enough then we can conclude Rk C l R Therefore we begin with Lemma 52 Suppose Rk C 5 R Then we can split Rk according to Rk = R k + k k k + k Hk, k Σk, 59 where R k is a real rational function in k such that Hk vanishes at k, k of order three and has a Fourier transform Hk = Ĥxe xφk dx, 5 with xĥx integrable R Proof We can construct a rational function, which satisfies f n k = f n k for k R, by making the ansatz f n k = k2n+4 + n k k 2n+4 + j= j!2k α j j + iβ j k k k j k + k j Furthermore we can choose α j, β j R for j =,, n, such that we can match the values of R and its first four derivatives at k, k at f n k Thus we will set R k = f 4 k, with α = ReRk, β = ImRk and so on Note that R k is integrable Hence Hk C 4 R and vanishes together with its first three derivatives at k, k Note that Φk/i = 8k 3 3kk 2 is a polynomial of order three which has a maximum at k and a minimum at k Thus the phase Φk/i restricted to Σk gives a one to one coordinate transform Σk [Φk /i, Φ k /i] = [ 6k, 3 6k] 3 and we can hence express Hk in this new coordinate setting it equal to zero outside this interval The coordinate transform locally looks like a cube root near k and k, however, due to our assumption that H vanishes there, H is still C 2 in this new coordinate and the Fourier transform with respect to this new coordinates exists and has the required properties Moreover, as in Lemma 5 we obtain: Lemma 53 Let Hk be as in the previous lemma Then we can split Hk according to Hk = H a,t k+h r,t k such that H a,t k is analytic in the region ReΦk < and H a,t ke Φkt/2 = O, ReΦk <, Imk, H r,t k = Ot, k R 5 28

34 Chapter 5 Analytic Approximation Proof We choose H a,t k = Kt ĤxexΦk dx with Kt = t/2 Then we can conclude as in Lemma 5: H a,t ke Φkt/2 and H r,t k Kt ĤxexΦk+Φkt/2 dx Ĥx e KtΦk+Φkt/2 Kt Ĥx e Φkt/2+Φkt/2 = Ĥx const Ĥx dx const Kt x 4 dx const Kt const 3/2 t By construction R a,t k = R k+k k k +k H a,t k will satisfy the required Lipschitz estimate in a vicinity of the stationary phase points uniformly in t and all jumps will be I + Ot Hence we can proceed as in Chapter 4 29

35 Appendix A The Riemann-Hilbert problem on a small cross In this chapter, which is taken from Krüger and Teschl [3], we will solve the Riemann Hilbert problem on a small cross Introduce the cross Σ = Σ Σ 4 see Figure A by Σ = {ue iπ/4, u [, } Σ 2 = {ue iπ/4, u [, } Σ 3 = {ue 3iπ/4, u [, } Σ 4 = {ue 3iπ/4, u [, } A Orient Σ such that the real part of z increases in the positive direction Denote by D = {z, z < } the open unit disc Throughout this section z iν will denote the function e iν logz, where the branch cut of the logarithm is chosen along the negative real axis, Now consider the Riemann Hilbert problem given by m + z = m zv j z, z Σ j, j =, 2, 3, 4, A2 mz I, z, Σ 3 R3z Σ 2 Rz Σ 4 R 4z R 2z Figure A: Contours of a cross Σ 3

36 Chapter A The Riemann-Hilbert problem on a small cross where the jump matrices are given as follows: v j for z Σ j R zz v = 2iν e tφz, v 2 = R 2 zz 2iν e tφz, R3 zz v 3 = 2iν e tφz, v 4 = R 4 zz 2iν e tφz A3 We can now state the next theorem, which gives us the solution of the Riemann Hilbert problem A2 In the proof we follow the computations of section 3 and 4 in Deift and Zhou [9] We will allow some variation, in all parameters as indicated Theorem A There is some ρ > such that v j z = I for z > ρ Moreover, suppose that within z ρ the following estimates hold: i The phase satisfies Φ ir, Φ =, Φ = i and ±Re Φz Φ { + for z Σ Σ 3, 4 z 2, else, A4 Φz Φ iz2 2 C z 3 A5 ii There is some r D and constants α, L, ], such that R j, j =,, 4, satisfy Hölder conditions of the form R z r L z α, R 2 z r L z α, r R 3 z r 2 r L z α, R 4 z r 2 L z α A6 Then the solution of the Riemann Hilbert problem A2 satisfies for z > ρ, where mz = I + z i t /2 β + Ot +α 2, A7 β β = νe iπ/4 argr+argγiν e tφ t iν, ν = 2π log r 2 A8 Furthermore, if R j z and Φz depend on some parameter, the error term is uniform with respect to this parameter as long as r remains within a compact subset of D and the constants in the above estimates can be chosen independent of the parameters Remark A2 Note that the solution of the Riemann Hilbert problem A2 is unique This follows from the usual Liouville argument [5, Lem 78] since detv j = The proof will be given in the rest of this chapter, but split into a few parts for a better overview 3

37 Chapter A The Riemann-Hilbert problem on a small cross A Approximation A close look at the stated theorem shows, that the actual value of ρ is of no importance In fact, if we choose < ρ < ρ, then the solution m of the problem with jump ṽ, where ṽ is equal to v for z < ρ and I otherwise, differs from m only by an exponentially small error This already indicates, that we should be able to replace R j z by their respective values at z = To see this we start by rewriting our Riemann Hilbert problem as a singular integral equation We will use the theory developed in Appendix B for the case of 2 2 matrix valued functions with m z = I and the usual Cauchy kernel since we won t require symmetry in this section Ω s, z = I ds s z Moreover, since our contour is unbounded, we will again assume w L Σ L 2 Σ All results from Appendix B still hold in this case with some straightforward modifications, as the only difference is that µ is now a matrix and no longer a vector Indeed, as in Theorem B2, in the special case b + z = v j z and b z = I for z Σ j, we obtain mz = I + 2πi Σ µsws ds s z, where µ I is the solution of the singular integral equation that is, A9 I C w µ I = C w I, A µ = I + I C w C w I, C w f = C wf A Here C denotes the usual Cauchy operator and we set wz = w + z since w z = As our first step we will get rid of some constants and rescale the entire problem by setting ˆmz = Dt mzt /2 Dt, A2 where Dt = dt, dt = e tφ/2 t iν/2 A3 dt Then one easily checks that ˆmz solves the Riemann Hilbert problem ˆm + z = ˆm zˆv j z, z Σ j, j =, 2, 3, 4, A4 ˆmz I, z, z / Σ, 32