LONG-TIME ASYMPTOTICS FOR THE KORTEWEG DE VRIES EQUATION VIA NONLINEAR STEEPEST DESCENT

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1 LONG-TIME ASYMPTOTICS FOR THE KORTEWEG DE VRIES EQUATION VIA NONLINEAR STEEPEST DESCENT KATRIN GRUNERT AND GERALD TESCHL Abstract We apply the method of nonlinear steepest descent to compute the long-time asymptotics of the Korteweg de Vries equation for decaying initial data in the soliton and similarity region This paper can be viewed as an expository introduction to this method Introduction One of the most famous examples of completely integrable 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, where, as usual, the subscripts denote the differentiation with respect to the corresponding variables Following the seminal work of Gardner, Green, Kruskal, and Miura [7], one can use the inverse scattering transform to establish existence and uniqueness of real-valued classical solutions for the corresponding initial value problem with rapidly decaying initial conditions We refer to, for instance, the monographs by Marchenko [27] or Eckhaus and Van Harten [6] Our concern here are the long-time asymptotics of such solutions The classical result is that an arbitrary short-range solution of the above type will eventually split into a number of solitons travelling to the right plus a decaying radiation part travelling to the left, as illustrated in Figure The first numerical evidence for such a behaviour was found by Zabusky and Kruskal [39] The first mathematical results were given by Ablowitz and Newell [], Manakov [26], and Šabat [3] First rigorous results for the KdV equation were proved by Šabat [3] and Tanaka [34] see also Eckhaus and Schuur [5], where more detailed error bounds are given Precise asymptotics for the radiation part were first formally derived by Zakharov and Manakov [38], by Ablowitz and Segur [2], [32], by Buslaev [6] see also [5], and later on rigorously justified and extended to all orders by Buslaev and Sukhanov [7] A detailed rigorous proof not requiring any a priori information on the asymptotic form of the solution was given by Deift and Zhou [] based on earlier work of Manakov [26] and Its [9] see also [2], [2], [22] For further information on the history of this problem we refer to the survey by Deift, Its, and Zhou [2] To describe the asymptotics in more detail, we recall the well-known fact see eg [9], [27] that qx, t is uniquely determined by the right scattering data of 2 Mathematics Subject Classification Primary 37K4, 35Q53; Secondary 37K45, 35Q5 Key words and phrases Riemann Hilbert problem, KdV equation, solitons Research supported by the Austrian Science Fund FWF under Grant No Y33 Math Phys Anal Geom 2,

2 2 K GRUNERT AND G TESCHL Figure Numerically computed solution qx, t of the KdV equation at time t = 5, with initial condition qx, = sechx sechx the associated Schrödinger operator 2 Ht = d2 + qx, t dx2 The scattering data consist of the right reflection coefficient Rk, t, a finite number of t independent eigenvalues κ 2 j with < κ < κ 2 < < κ N, and norming constants γ j t We will write Rk = Rk, and γ j = γ j for the scattering data of the initial condition Then the long-time asymptotics can be described by distinguishing the following main regions: i The soliton region, x/t > C for some C >, in which the solution is asymptotically given by a sum of one-soliton solutions 3 qx, t 2 N j= where the phase shifts are given by 4 p j = 2 log γ2 j 2κ j κ 2 j cosh 2 κ j x 4κ 3 j t p j, N l=j+ 2 κl κ j κ l + κ j In the case of a pure N-soliton solution ie, Rk, t = this was first established independently by Hirota [8], Tanaka [33], and Wadati and Toda [36] The general case was first established by Šabat [3] and by Tanaka [34] see also [5] and [3] ii The self-similar region, x/3t /3 C for some C >, in which the solution is connected with the Painléve II transcendent This was first established by Segur and Ablowitz [32] iii The collisionless shock region, x < and C < x 3t /3 logt 2/3 < C, for some C >, which only occurs in the generic case ie, when R = Here the asymptotics can be given in terms of elliptic functions as was pointed out by Segur and Ablowitz [32] with further extensions in Deift, Venakides, and Zhou [3]

3 LONG-TIME ASYMPTOTICS FOR THE KDV EQUATION 3 vi The similarity region, x/t < C for some C >, where /2 4νk k 5 qx, t sin6tk 3 νk log92tk 3 3t + δk, 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 Again this was found by Zakharov and Manakov [38] and without a precise expression for δk and assuming absence of solitons by Ablowitz and Segur [2] with further extensions by Buslaev and Sukhanov [7] as discussed before Our aim here is to use the nonlinear steepest descent method for oscillatory Riemann Hilbert problems from Deift and Zhou [] and apply it to rigorously establish the long-time asymptotics in the soliton and similarity regions Theorem 44, respectively, 54, below In fact, our main goal is to give a complete and expository introduction to this method In addition to providing a streamlined and simplified approach, the following items will be different in comparison with [] First of all, in the mkdv case considered in [] there were no solitons present We will add them using the ideas from Deift, Kamvissis, Kriecherbauer, and Zhou [4] following Krüger and Teschl [24] However, in the presence of solitons there is a subtle nonuniqueness issue for the involved Riemann Hilbert problems see eg [4, Chap 38] We will rectify this by imposing an additional symmetry condition and prove that this indeed restores uniqueness Secondly, in the mkdv case the reflection coefficient Rk has always modulus strictly less than one In the KdV case this is generically not true and hence terms of the form Rk/ Rk 2 will become singular and cannot be approximated by analytic functions in the sup norm We will show how to avoid these terms by using left and right instead of just right scattering data on different parts of the jump contour Consequently it will be sufficient to approximate the left and right reflection coefficients Details can be found in Section 6 Moreover, we obtain precise relations between the error terms and the decay of the initial conditions improving the error estimates obtained in Schuur [3] which are stated in terms of smoothness and decay properties of Rk and its derivatives Overall we closely follow the recent review article [25], where Krüger and Teschl applied these methods to compute the long-time asymptotics for the Toda lattice For a general result which applies in the case where Rk has modulus strictly less than one and no solitons are present we refer to Varzugin [35] and for another recent generalization of the nonlinear steepest descent method to McLaughlin and Miller [28] An alternate approach based on the asymptotic theory of pseudodifferential operators was given by Budylin and Buslaev [5] Finally, note that if qx, t solves the KdV equation, then so does q x, t Therefore it suffices to investigate the case t

4 4 K GRUNERT AND G TESCHL 2 The Inverse scattering transform and the Riemann Hilbert problem In this section we want to derive the Riemann Hilbert problem for the KdV equation from scattering theory This is essentially classical compare, eg, [4] except for two points The eigenvalues will be added by appropriate pole conditions which are then turned into jumps following Deift, Kamvissis, Kriecherbauer, and Zhou [4] We will impose an additional symmetry conditions to ensure uniqueness later on following Krüger and Teschl [24] For the necessary results from scattering theory respectively the inverse scattering transform for the KdV equation we refer to [27] see also [4] and [9] We consider real-valued classical solutions qx, t of the KdV equation, which decay rapidly, that is 2 max + x qx, t dx <, for all T > t T R Existence of such solutions can for example be established via the inverse scattering transform if one assumes cf [27, Sect 42] that the initial condition satisfies 22 + x qx, + q x x, + q xx x, + q xxx x, dx < R Associated with qx, t is a self-adjoint Schrödinger operator 23 Ht = d2 + q, t, dx2 DH = H2 R L 2 R Here L 2 R denotes the Hilbert space of square integrable complex-valued functions over R and H k R the corresponding Sobolev spaces By our assumption 2 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 24 Htψ ± k, x, t = k 2 ψ ± k, x, t, Imk >, and asymptotically look like the free solutions 25 lim x ± e ikx ψ ± k, x, t = Both ψ ± k, x, t are analytic for Imk > and continuous for Imk The asymptotics of the two Jost solutions are 26 ψ ± k, x, t = e ±ikx + Q ± x, t 2ik + O k 2, as k with Imk >, where 27 Q + x, t = x qy, tdy, Furthermore, one has the scattering relations x Q x, t = qy, tdy 28 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 They have the following well-known properties:

5 LONG-TIME ASYMPTOTICS FOR THE KDV EQUATION 5 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 29 Res iκj T k = iµ j tγ +,j t 2 = iµ j γ 2 +,j, where 2 γ +,j t = ψ + iκ j,, t 2 and ψ + iκ j, x, t = µ j tψ iκ j, x, t Moreover, 2 T kr + k, t + T kr k, t =, T k 2 + R ± k, t 2 = 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 22 The time evolutions of the quantities Rk, t and γ j t are given by Rk, t = Rke 8ik3t, γ j t = γ j e 4κ3 j t, where Rk = Rk, and γ j = γ j We will set up a Riemann Hilbert problem as follows: { T kψ k, x, te 24 mk, x, t = ikx ψ + k, x, te ikx, Imk >, ψ+ k, x, te ikx T kψ k, x, te ikx, Imk < 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 nontangential 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 In general, for an oriented contour Σ, m + k resp m k will denote the limit of mκ as κ k from the positive resp negative side of Σ Here the positive resp negative side is the one which lies to the left resp right as one traverses the contour in the direction of the orientation Theorem 23 Let S + H = {Rk, k ; κ j, γ j, j N} be the right scattering data of the operator H Then mk = mk, x, t defined in 24 is a solution of the following vector Riemann Hilbert problem Find a function mk which is meromorphic away from the real axis with simple poles at ±iκ j and satisfies: i The jump condition Rk 2 Rke 25 m + k = m kvk, vk = tφk Rke tφk, for k R,

6 6 K GRUNERT AND G TESCHL ii the pole conditions Res iκj mk = lim mk k iκ j iγj 2 26 etφiκ j, iγ 2 Res iκj mk = lim mk j e tφiκ j, k iκ j iii the symmetry condition 27 m k = mk iv and the normalization, 28 lim miκ = κ Here the phase is given by 29 Φk = 8ik 3 + 2ik x t Proof The jump condition 25 is a simple calculation using the scattering relations 28 plus 2 The pole conditions follow since T k is meromorphic for Imk > with simple poles at iκ j and residues given by 29 The symmetry condition holds by construction and the normalization 28 is immediate from the following lemma below Observe that the pole condition at iκ j is sufficient since the one at iκ j follows by symmetry Moreover, using qx, t 22 T kψ k, x, tψ + k, x, t = + 2k 2 + O k 4 as k with Imk > observe that the right-hand side is just the diagonal Green s functions of Ht divided by the free one we obtain from 26 Lemma 24 The function mk, x, t defined in 24 satisfies 22 mk, x, t = + Qx, t + O 2ik k 2 Here Qx, t = Q + x, t is defined in 27 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 [4] Choose ε so small 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 iγ2 j etφiκ j, k iκ j < ε, k iκ j 222 mk = iγ 2 j etφiκ j mk k+iκ j, k + iκ j < ε, mk, else

7 LONG-TIME ASYMPTOTICS FOR THE KDV EQUATION 7 Note that for Imk < we redefined mk such that it respects our symmetry 27 Then a straightforward calculation using Res iκ mk = lim k iκ k iκmk shows: Lemma 25 Suppose mk is redefined as in 222 Then mk is holomorphic away from the real axis and the small circles around iκ j and iκ j Furthermore it satisfies 25, 27, 28 and the pole condition is replaced by the jump condition m + k = m k, k iκ j = ε, 223 m + k = m k iγ2 j etφiκ j k iκ j iγ2 j etφiκ j k+iκ j, k + iκ j = ε, where the small circle around iκ j is oriented counterclockwise and the one around iκ j is oriented clockwise 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 26 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 is given by 224 m k = fk f k fk = + 2κ γ 2 e tφiκ + k + iκ k iκ 2κ γ 2 e tφiκ Furthermore, the zero solution is the only solution of the corresponding vanishing problem where the normalization is replaced by lim k mik = In particular, 225 Qx, t = 2γ 2 e tφiκ + 2κ γ 2 e tφiκ 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

8 8 K GRUNERT AND G TESCHL Lemma 27 If mk = for m defined as in 24, then k = Moreover, the zero of at least one component is simple in this case Proof By 24 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 well-known 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 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 where we assume m + k = m kvk, k Σ, m k = mk, lim miκ =, κ Hypothesis 3 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 and vk is symmetric 32 v k = Moreover, suppose detvk = vk, k Σ

9 LONG-TIME ASYMPTOTICS FOR THE KDV EQUATION 9 Now we are ready to show that the symmetry condition in fact guarantees uniqueness Theorem 32 Assume Hypothesis 3 Suppose there exists a solution mk of the Riemann Hilbert problem 3 for which mk = can happen at most k for k = in which case lim sup k m j k is bounded from any direction for j = or j = 2 Then the Riemann Hilbert problem 3 with norming condition replaced by 33 lim miκ = α α κ for given α C, has a unique solution m α k = α mk Proof Let m α k be a solution of 3 normalized according to 33 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 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 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, 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 = normalization we obtain δk = A = α This finishes the proof Now, by the Furthermore, the requirements cannot be relaxed to allow eg second order zeros in stead 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 nontrivial 2 symmetric solution of the vanishing problem ie for α = By Lemma 27 we have Corollary 33 The solution mk = mk, x, t found in Theorem 23 is the only solution of the vector 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 224, 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

10 K GRUNERT AND G TESCHL since our Riemann Hilbert problem is equivalent to a singular integral equation which is Fredholm of index zero see Appendix A 4 Conjugation and Deformation This section demonstrates how to conjugate our Riemann Hilbert problem and how to deform our jump contour, such that the jumps will be exponentially close to the identity away from the stationary phase points Throughout this and the following section, we will assume that the Rk has an analytic extension to a small neighborhood of the real axis This is for example the case if we assume that our solution is exponentially decaying In Section 6 we will show how to remove this assumption For easy reference we note the following result: Lemma 4 Conjugation Assume that Σ Σ Let D be a matrix of the form dk 4 Dk =, dk where d : C\ Σ C is a sectionally analytic function Set 42 mk = mkdk, then the jump matrix transforms according to 43 ṽk = D k vkd + k If d satisfies d k = dk and lim κ diκ =, then the transformation mk = mkdk respects our symmetry, that is, mk satisfies 27 if and only if mk does, and our normalization condition In particular, we obtain v v 44 ṽ = 2 d 2 v 2 d 2, k Σ\ Σ, v 22 respectively 45 ṽ = d d + v v 2 d + d v 2 d + d, k Σ d + d v 22 In order to remove the poles there are two cases to distinguish If ReΦiκ j <, then the corresponding jump is exponentially close to the identity as t and there is nothing to do Otherwise we use conjugation to turn the jumps into one with exponentially decaying off-diagonal entries, again following Deift, Kamvissis, Kriecherbauer, and Zhou [4] 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 42 Assume that the Riemann Hilbert problem for m has jump conditions near iκ and iκ given by m + k = m k, k iκ = ε, 46 m + k = m k iγ2 k iκ iγ2 k+iκ, k + iκ = ε

11 LONG-TIME ASYMPTOTICS FOR THE KDV EQUATION 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 = ε, k iκ2 iγ 2 k+iκ and all remaining data conjugated as in Lemma 4 by k iκ 47 Dk = k+iκ k+iκ 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, and note that Dk is analytic away from the two circles Now set mk = mkdkand note that Dk is also symmetric Therefore the jump conditions can be verified by straightforward calculations and Lemma 4 The jump along the real axis is of oscillatory type and our aim is to apply a contour deformation following [] 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 48 Φk = 8ik 3 + 2ik x t and the stationary phase points, Φ k =, are denoted by ±k, where 49 k = x 2t For x t > we have k ir, and for x t < we have k R For x t > we will also need the value κ defined via ReΦiκ =, that is, x 4 κ = 4t > 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 factorizations of the jump condition 25: 4 vk = b k b + k,

12 2 K GRUNERT AND G TESCHL where Rke tφk 42 b k =, b + k = Rke tφk for k > Rek and 43 vk = B k Rk 2 B + k, Rk 2 where 44 B k =, B RketΦk + k = Rk 2 Rke tφk Rk 2 for k < Rek 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 T k, k We define the partial transmission coefficient with respect to k by k+iκ j k iκ j, k ir +, 45 T k, k = N κ j κ, j= k+iκ j k iκ j exp k 2πi k log T ζ 2 ζ k dζ, k R +, for k C\Σk, where Σk = [ Rek, Rek ] oriented from left to right Thus T k, k is meromorphic for k C\Σk Note that T k, k can be computed in terms of the scattering data since T k 2 = R + k, t 2 Moreover, we set 2κ j, k ir +, 46 Thus 47 T k = N κ j κ, j= 2κ j + 2π k k log T ζ 2 dζ, k R + T k, k = + T k i k + O k 2, as k Theorem 43 The partial transmission coefficient T k, k is meromorphic in C\Σk, where 48 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 49 T + k, k = T k, k Rk 2, for k Σk Moreover, i T k, k = T k, k, k C\Σk, ii T k, k = T k, k, k C, in particular T k, k is real for k ir, and iii if k R + the behaviour near k = is given by T k, k = T kc + o with C for Imk

13 LONG-TIME ASYMPTOTICS FOR THE KDV EQUATION 3 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 s formulas i, ii, and iii are straightforward to check Now we are ready to perform our conjugation step Introduce k iκ j iγ j 2etΦiκ j D iγ 2 j etφiκ j k, k iκ j < ε, κ < κ j, k iκ j 42 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 D k = T k, k T k, k Observe that Dk respects our symmetry, D k = Dk Now we conjugate our problem using Dk and set 42 mk = mkdk Note that even though Dk might be singular at k = if k > and R =, mk is nonsingular since the possible singular behaviour of T k, k from D k cancels with T k in mk by virtue of Theorem 43 iii Then using Lemma 4 and Lemma 42 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, 422 ṽk = k+iκ j, k + iκ j = ε, iγ 2 j etφiκ j T k,k 2 and corresponding to κ > κ j if any by ṽk =, k iκ j = ε, 423 ṽk = iγ2 j etφiκ j T k,k 2 k iκ j iγ2 j etφiκ j T k,k 2 k+iκ j, k + iκ j = ε In particular, all jumps corresponding to poles, except for possibly one if κ j = κ, are exponentially close to the identity for t In the latter case we will keep the pole condition for κ j = κ which now reads 424 Res iκj mk = lim mk k iκ j Res iκj mk = lim mk k iκ j iγj 2etΦiκj T iκ j, k 2, iγ 2 j e tφiκj T iκ j, k 2

14 4 K GRUNERT AND G TESCHL k ir k R k k + Figure 2 Sign of ReΦk for different values of k Furthermore, the jump along R is given by { b k b + k, k R\Σk, 425 ṽk = B k B+ k, k Σk, where 426 b k = and R ke tφk T k,k 2, b+ k = Rke tφk, T k,k 2 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 R k = Rk, k R, T ± k, k = T k, k, k Σk, and the jump condition 49 for the partial transmission coefficient T k, k along Σk in the last step This also shows that the matrix entries are bounded for k R near k = since T ± k, k = T ± k, k Since we have assumed that Rk has an analytic continuation to a neighborhood of the real axis, we can now deform the jump along R to move the oscillatory terms into regions where they are decaying According to Figure 2 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 ± ReΦk < and such that the circles around ±iκ j lie outside the region in between Σ and Σ + see Figure 3 Then we can split our jump by redefining mk according to mk b + k, < Imk < ε, 427 mk = mk b k, ε < Imk <, mk, else

15 LONG-TIME ASYMPTOTICS FOR THE KDV EQUATION 5 ReΦ > ReΦ < Σ + ReΦ > R ReΦ < Σ Figure 3 Deformed contour for k ir + ReΦ< Σ + Σ + k k Σ ReΦ> ReΦ> Σ 2 + Σ 2 ReΦ< ReΦ< R Σ ReΦ> Figure 4 Deformed contour for k R + Thus the jump along the real axis disappears and the jump along Σ ± is given by { b+ k, k Σ vk = b k, k Σ All other jumps are unchanged Note that the resulting Riemann Hilbert problem still satisfies our symmetry condition 27, since we have 429 b± k = b k By construction the jump along Σ ± is exponentially close to the identity as t Case 2: k R, k : We set Σ ± = Σ ± Σ 2 ± according to Figure 4 chosen such that the circles around ±iκ j lie outside the region in between Σ and Σ + Again note that Σ ± respectively Σ 2 ± lie in the region with ± ReΦk < Then we can split our jump by redefining

16 6 K GRUNERT AND G TESCHL mk according to mk b + k, k between R and Σ +, mk b k, k between R and Σ, 43 mk = mk B + k, k between R and Σ 2 +, 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 given by b+ k, k Σ +, b k, k Σ 43 vk =, B + k, k Σ 2 +, B k, k Σ 2 All other jumps are unchanged Again the resulting Riemann Hilbert problem still satisfies our symmetry condition 27 and the jump along Σ ± \{k, k } is exponentially decreasing as t Theorem 44 Assume 432 R + x +l qx, dx < for some integer l 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 433 respectively 434 where x qy, tdy = 4 qx, t = N i=j+ κ i 2γj 2 x, t + 2κ j γj 2x, t + Ot l, 4κ j γj 2 x, t + 2κ j γj 2x, + t2 Ot l, 435 γ 2 j x, t = γ 2 j e 2κ jx+8κ 3 j t N If x t c j ε, for all j, one has 436 respectively 437 x qy, tdy = 4 κ i κ, i=j+ qx, t = Ot l 2 κi κ j κ i + κ j κ i + Ot l, κ = x 4t,

17 LONG-TIME ASYMPTOTICS FOR THE KDV EQUATION 7 Proof Since mk = mk for k sufficiently far away from R equations 22, 42, and 47 imply the following asymptotics 438 mk = + 2T k + Qx, t + O 2ik k 2 By construction, the jump along Σ ± is exponentially decreasing as t Hence we can apply Theorem A6 as follows: If x t c j > ε resp κ 2 κ 2 j > ε for all j we can choose γt = and w t = ŵ in Theorem A6 Since ŵ is exponentially small as t, the solutions of the associated Riemann Hilbert problems only differ by Ot l for any l Comparing m = with the above asymptotics shows Q + x, t = 2T k + Ot l If x t c j < ε resp κ 2 κ 2 j < ε for some j, we choose γt = γ k x, t and w t = ŵ in Theorem A6, where γ 2 j x, t = γ 2 j e tφiκ j T iκ j, i κ j 3 2 = γ 2 j e 2κ jx+8κ 3 j t N i=j+ 2 κi κ j κ i + κ j As before we conclude that ŵ is exponentially small and so the associated solutions of the Riemann Hilbert problems only differ by Ot l From Lemma 26, we have the one-soliton solution m k = fk f k with fk = + 2κ j γ 2 j x, t + k + iκ j k iκ j 2κ j γ 2 j x, t As before, comparing with the above asymptotics shows Qx, t = 2T i κ j 2γj 2 x, t κ j γj 2x, t + Ot l To see the second part just use 22 in place of 22 This finishes the proof in the case where Rk has an analytic extensions We will remove this assumption in Section 6 thereby completing the proof Since the one-soliton solution is exponentially decaying away from its minimum, we also obtain the form stated in the introduction: Corollary 45 Assume 432, then the asymptotic in the soliton region, x/t C for some C >, is given by 439 qx, t = 2 where N j= 44 p j = 2 log κ 2 j cosh 2 κ j x 4κ 3 j t p j + Ot l, γ2 j 2κ j N i=j+ 2 κi κ j κ i + κ j

18 8 K GRUNERT AND G TESCHL Σ 3 R3 ζ R 4ζ Σ 4 Rζ Σ 2 R 2ζ Figure 5 Contours of a cross 5 Reduction to a Riemann Hilbert problem on a small cross 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 close to the identity except in small neighborhoods of the stationary phase points k and k Hence we need to continue our investigation of this case in this section 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 In fact, the solution of both crosses can be reduced to the following model problem: Introduce the cross Σ = Σ Σ 4 see Figure 5 by 5 Σ = {ue iπ/4, u [, } Σ 3 = {ue 3iπ/4, u [, } Σ Σ 2 = {ue iπ/4, u [, } Σ 4 = {ue 3iπ/4, u [, } Orient Σ such that the real part of k increases in the positive direction Denote by D = {ζ, ζ < } the open unit disc Throughout this section ζ iν will denote the function e iν logζ, where the branch cut of the logarithm is chosen along the negative real axis, Introduce the following jump matrices v j for ζ Σ j R ζζ v = 2iν e tφζ, v 2 =, 52 R3 ζζ v 3 = 2iν e tφζ and consider the RHP given by 53, v 4 = R 2 ζζ 2iν e tφζ R 4 ζζ 2iν e tφζ M + ζ = M ζv j ζ, ζ Σ j, j =, 2, 3, 4, Mζ I, ζ The solution is given in the following theorem of Deift and Zhou [] for a proof of the version stated below see Krüger and Teschl [25] Theorem 5 [] Assume there is some ρ > such that v j ζ = I for ζ > ρ Moreover, suppose that within ζ ρ the following estimates hold:

19 LONG-TIME ASYMPTOTICS FOR THE KDV EQUATION 9 i The phase satisfies Φ = iφ ir, Φ =, Φ = i and ± Re Φζ { + for ζ Σ Σ 3, 54 4 ζ 2, else, Φζ Φ iζ2 2 C ζ 3 ii There is some r D and constants α, L, ], such that R j, j =,, 4, satisfy Hölder conditions of the form R ζ r L ζ α, R 2 ζ r L ζ α, r R 3 ζ r 2 r L ζ α, R 4 ζ r 2 L ζ α Then the solution of the RHP 53 satisfies 57 Mζ = I + ζ for ζ > ρ, where i t /2 β β + Ot +α 2, 58 β = νe iπ/4 argr+argγiν e itφ t iν, ν = 2π log r 2 Furthermore, if R j ζ and Φζ 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 Theorem 52 Decoupling Consider the Riemann Hilbert problem 59 m + k = m kvk, k Σ, lim miκ =, κ with detv 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 I are Ot β away from some ε neighborhoods of some points k j Σ, j n Moreover, suppose that the solution of the matrix problem with jump vk restricted to the ε neighborhood of k j has a solution which satisfies 5 M j k = I + M j ρt α + Oρt β, k k j > ε k k j Then the solution mk is given by 5 mk = + ρt α n j= where the error term depends on the distance of k to Σ M j k k j + Oρt β, Proof In this proof we will use the theory developed in Appendix A with m k = and the usual Cauchy kernel Ω s, k = I ds s k Moreover, since symmetry is not important, we will consider C w on L 2 Σ rather than restricting it to the symmetric subspace L 2 sσ Here w ± = ±b ± I correspond to some factorization

20 2 K GRUNERT AND G TESCHL v = b b + of v eg, b = I and b + = v Assume that mk exists, then the same arguments as in the appendix show that mk = + µsw + s + w sω s, k, 2πi where µ solves Introduce ˆmk by 52 ˆmk = Σ I C w µ = C w { mkm j k, k k j 2ε, mk, else The Riemann Hilbert problem for ˆmk has jumps given by M j k, k k j = 2ε, M j kvkm j k, k Σ, ε < k k j < 2ε, 53 ˆvk = I, k Σ, k k j < ε, vk, else By assumption the jumps are I + Oρt α on the circles k k j = 2ε and even I + Oρt β on the rest both in the L 2 and L norms In particular, we infer that I Cŵ exists for sufficiently large t and using the Neumann series to estimate ˆµ = I Cŵ Cŵ cf the proof of Theorem A6 we obtain 54 ˆµ 2 c ŵ 2 c ŵ = Oρt α Thus we conclude 55 mk = + ˆµsŵs ds 2πi ˆΣ s k n = + 2πi j= s k j =2ε = ρt α 2πi = + ρt α n and hence the claim is proven j= ˆµsM j s I ds s k + Oρt β n M j j= s k j =2ε M j k k j + Oρt β, s k j ds s k + Oρt β Now let us turn to the solution of the problem on Σ c k = Σ + Σ {k k k < ε} for some small ε > Without loss we can also deform our contour slightly such that Σ c k consists of two straight lines Next, note Φk = 6ik 3, Φ k = 48ik As a first step we make a change of coordinates 56 ζ = 48k k k, k = k + ζ 48k such that the phase reads Φk = Φk + i 2 ζ2 + Oζ 3

21 LONG-TIME ASYMPTOTICS FOR THE KDV EQUATION 2 Next we need the behavior of our jump matrix near k, that is, the behavior of T k, k near k Lemma 53 Let k R, then iν k k 57 T k, k = T k, k, 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 58 T k, k = exp k T ζ 2 log k iκ j 2πi T k 2 ζ k dζ j= is Hölder continuous of any exponent less than at the stationary phase point k = k and satisfies T k, k = Proof First of all observe that k iν 59 exp log T k 2 k 2πi k ζ k dζ k = k + k k Hölder continuity of any exponent less than is well-known cf [29] If kζ is defined as in 56 and < α <, then there is an L > such that 52 T kζ, k ζ iν T k, k e iν log2k 48k L ζ α, where the branch cut of ζ iν is chosen along the negative real axis We also have 52 Rkζ Rk L ζ α and thus the assumptions of Theorem 5 are satisfied with 522 r = Rk T k, k 2 e 2iν log2k 48k and ν = 2π log Rz 2 since r = Rz Therefore we can conclude that the solution on Σ c k is given by 523 where β 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 524 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 and /2 < α <

22 22 K GRUNERT AND G TESCHL 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: 525 M c 2k = M k c = I 48k k + k i t /2 Applying Theorem 52 yields the following result: Theorem 54 Assume 526 R + x 6 qx, dx <, β + Ot α β then the asymptotics in the similarity region, x/t C for some C >, are given by x 527 respectively k qy, tdy = 4 κ j log T ζ 2 dζ π κ j κ, k νk 3k t cos6tk3 νk log92tk 3 + δk + Ot α 528 qx, t = 4νk k 3t for any /2 < α < Here k = x 2t and sin6tk 3 νk log92tk 3 + δk + Ot α 529 νk = π log T k, 53 δk = π 4 argrk + argγiνk + 4 π T ζ 2 log k T k 2 dζ ζ k k Proof By Theorem 52 we have mk = + i β 48k t /2 k k β = + i 48k t /2 k which leads to + Ot α, l= k k N j= l β β arctan κ j k β β k + k l= Qx, t = 2T k k t /2 Reβ + Ot α + Ot α k l β β k upon comparison with 438 Using the fact that β/ ν = proves the first claim

23 LONG-TIME ASYMPTOTICS FOR THE KDV EQUATION 23 To see the second part, as in the proof of Theorem 44, just use 22 in place of 22, which shows 4k qx, t = 3t Imβ + Ot α This finishes the proof in the case where Rk has an analytic extensions We will remove this assumption in Section 6 thereby completing the proof Equivalence of the formula for δk given in the previous theorem with the one given in the introduction follows after a simple integration by parts Remark 55 Formally the equation 528 for q can be obtained by differentiating the equation 527 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 56 Note that Theorem 52 does not require uniform boundedness of the associated integral operator in contradistinction to Theorem A6 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 6 Analytic Approximation In this section 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 6 Rk = e ikx ˆRxdx R will be sufficient It is well-known that our decay assumption 432 implies ˆR L R and the estimate cf [27, Sect 32] 62 ˆR 2x const implies x l ˆR x L, x qrdr, x, Lemma 6 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 63 R a,t ke βt = Ot l, < Imk < ε, R r,t k = Ot l, k R Proof We choose R a,t k = Kt eikx ˆRxdx with Kt = β ε t for some positive β < β Then, for < Imk < ε, R a,t ke βt e βt ˆRx e Imkx dx e βt e Ktε ˆR = ˆR e β βt, Kt which proves the first claim Similarly, for Imk =, R r,t k Kt x l ˆR x x l dx xl ˆR x L, Kt l const t l

24 24 K GRUNERT AND G TESCHL To apply this lemma in the soliton region k ir + we choose 64 β = min ReΦk > Imk=ε and split Rk = R a,t k + R r,t k according to Lemma 6 to obtain 65 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 426 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 Section 4 while leaving the rest on the real axis Hence, rather then 428, the jump now reads ba,t,+ k, k Σ +, 66 ˆvk = ba,t, k, k Σ, 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 4 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 To remedy this we will rewrite them in terms of left rather then right scattering data For this purpose let us use the notation 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 be the right and T l k, k T k/t k, k be the left partial transmission coefficient With this notation we have { b k b + k, Rek < k, 67 ṽk = B k B+ k, Rek > k, where and b k = R r ke tφk T r k,k 2, b+ k = R r ke tφk, T rk,k 2 B k = T r, k,k 2 T k R 2 r ke tφk, B T r,+k,k 2 + k = T k R 2 r ke tφk Using 2 we can further write 68 B k = R l ke tφk, B+ k = T l k,k 2 R l ke tφk T l k,k 2 Now we can proceed as before with B ± k as with b ± k by splitting R l k rather than R r k 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

25 LONG-TIME ASYMPTOTICS FOR THE KDV EQUATION 25 real line 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 In this respect note that 432 implies Rk C l R cf [23] We begin with Lemma 62 Suppose Rk C 5 R Then we can split Rk according to 69 Rk = R k + k k k + k Hk, k Σk, 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 6 Hk = Ĥxe xφk dx, 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 Since Rk is C 5 we infer that Hk C 4 R and it vanishes together with its first three derivatives at k, k Note that Φk/i = 8k 3 3k 2 k 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 6 we obtain: Lemma 63 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 6 H a,t ke Φkt/2 = O, ReΦk <, Imk, H r,t k = Ot, k R Proof We choose H a,t k = Kt ĤxexΦk dx with Kt = t/2 Then we can conclude as in Lemma 6: and H r,t k H a,t ke Φkt/2 Ĥx e KtΦk+Φkt/2 Ĥx const Kt Ĥx dx const Kt x 4 dx const Kt 3/2 const t

26 26 K GRUNERT AND G TESCHL 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 The remaining two parts of the real line, k ] and [k, can be handled analogously and hence we can proceed as in Section 5 Appendix A Singular integral equations In this section we show how to transform a meromorphic vector Riemann Hilbert problem with simple poles at iκ, iκ, m + k = m kvk, k Σ, A iγ 2, Res iκ mk = lim mk k iκ iγ 2 Res iκ mk = lim mk k iκ m k = mk, lim mi k = k into a singular integral equation Since we require the symmetry condition for our Riemann Hilbert problem we need to adapt the usual Cauchy kernel to preserve this symmetry Moreover, we keep the single soliton as an inhomogeneous term which will play the role of the leading asymptotics in our applications The classical Cauchy-transform of a function f : Σ C which is square integrable is the analytic function Cf : C\Σ C given by A2 Cfk = 2πi Σ fs s k ds, k C\Σ Denote the tangential boundary values from both sides taken possibly in the L 2 - sense see eg [8, eq 72] by C + f respectively C f Then it is well-known that C + and C are bounded operators L 2 Σ L 2 Σ, which satisfy C + C = I see eg [8] Moreover, one has the Plemelj Sokhotsky formula [29], A3 C ± = ih ± I, 2 where A4 Hfk = π fs ds, k Σ, Σ k 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 A5 f k = fk Clearly this will only be possible if we require our jump data to be symmetric as well: Hypothesis A Suppose the jump data Σ, v satisfy the following assumptions: i 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

27 LONG-TIME ASYMPTOTICS FOR THE KDV EQUATION 27 A6 A7 ii The distance between Σ and {iy y y } is positive for some y > and ±iκ Σ iii Σ is invariant under k k and is oriented such that under the mapping k k sequences converging from the positive sided to Σ are mapped to sequences converging to the negative side iv The jump matrix v is invertible and can be factorized according to v = b b + = I w I + w +, where w ± = ±b ± I satisfy w ± k = v The jump matrix satisfies w k, k Σ w = w + L Σ + w L Σ <, w 2 = w + L2 Σ + w L2 Σ < Next we introduce the Cauchy operator A8 Cfk = 2πi k+iκ s+iκ s k k iκ s iκ s k Σ fsω κ s, k acting on vector-valued functions f : Σ C 2 Here the Cauchy kernel is given by A9 Ω κ s, k = ds = ds, for some fixed iκ / Σ In the case κ = we set A Ω s, k = s k ds s k s k s+iκ s k s iκ and one easily checks the symmetry property: A Ω κ s, k = Ω κ s, k The properties of C are summarized in the next lemma Lemma A2 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 A2 and C + C = I A3 Cf iκ =, Cfiκ = Furthermore, C restricts to L 2 sσ, that is A4 Cf k = Cfk, k C\Σ for f L 2 sσ or L s Σ and by A6 we also have A5 C ± fw k = C fw ± k, k Σ Proof Everything follows from A and the fact that C inherits all properties from the classical Cauchy operator

28 28 K GRUNERT AND G TESCHL We have thus obtained a Cauchy transform with the required properties Following Section 7 and 8 of [3], we can solve our Riemann Hilbert problem using this Cauchy operator Introduce the operator C w : L 2 sσ L 2 sσ by A6 C w f = C + fw + C fw +, f L 2 sσ By our hypothesis A7 C w is also well-defined as operator from L s Σ L 2 sσ and we have A7 C w L 2 s L 2 s const w respectively C w L s L 2 s const w 2 Furthermore recall from Lemma 26 that the unique solution corresponding to v I is given by A8 m k = fk f k, fk = + 2κ γ 2 e tφiκ + k + iκ k iκ 2κ γ 2 e tφiκ Observe that for γ = we have fk = and for γ = we have fk = k+iκ k iκ In particular, fk is uniformly bounded for all γ [, ] if k iκ > ε Then we have the next result Theorem A3 Assume Hypothesis A Suppose m solves the Riemann Hilbert problem A Then A9 mk = c m k + µsw + s + w sω κ s, k, 2πi where µ = m + b + = m b and c = Σ µsw + s + w sω κ s, 2πi Σ Here m j denotes the j th component of a vector Furthermore, µ solves A2 I C w µk c m k = c C w m k Conversely, suppose µ solves A2 I C w µk m k = C w m k, and c = µsw + s + w sω κ s,, 2πi Σ then m defined via A9, with c = + c and µ = + c µ, solves the Riemann Hilbert problem A and µ = m ± b ± Proof If m solves A and we set µ = m ± b ±, then m satisfies an additive jump given by m + m = µw + + w Hence, if we denote the left hand side of A9 by m, both functions satisfy the same additive jump Furthermore, Hypothesis 3 implies that µ is symmetric and hence so is m Using A3 we also see that m satisfies the same pole conditions as m In summary, m m has no jump and solves A with v I except for the normalization which is given by lim k mik mik = Hence Lemma 26 implies m m =

29 LONG-TIME ASYMPTOTICS FOR THE KDV EQUATION 29 Moreover, if m is given by A9, then A2 implies A22 m ± = c m + C ± µw + C ± µw + = c m + C w µ ± µw ± = c m I C w µ + µb ± From this we conclude that µ = m ± b ± solves A2 Conversely, if µ solves A2, then set mk = m k + 2πi Σ µsw + s + w sω ζ s, k, and the same calculation as in A22 implies m ± = µb ±, which shows that m = + c m solves the Riemann Hilbert problem A Remark A4 In our case m k L s Σ, but m k is not square integrable and so µ L 2 sσ + L s Σ in general Note also that in the special case γ = we have m k = and we can choose κ as we please, say κ = such that c = c = in the above theorem Hence we have a formula for the solution of our Riemann Hilbert problem mk in terms of m + I C w C w m and this clearly raises the question of bounded invertibility of I C w as a map from L 2 sσ L 2 sσ This follows from Fredholm theory cf eg [37]: Lemma A5 Assume Hypothesis A The operator I C w is Fredholm of index zero, A23 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 keri C w = We are interested in comparing a Riemann Hilbert problem for which w and w 2 is small with the one-soliton problem For such a situation we have the following result: Theorem A6 Fix a contour Σ and choose κ, γ = γ t, v t depending on some parameter t R such that Hypothesis A holds Assume that w t satisfies A24 w t ρt and w t 2 ρt for some function ρt as t Then I C w t : L 2 sσ L 2 sσ exists for sufficiently large t and the solution mk of the Riemann Hilbert problem A differs from the one-soliton solution m t k only by Oρt, where the error term depends on the distance of k to Σ {±iκ} Proof By A7 we conclude that C w t L 2 s L 2 s = Oρt respectively C w t L s L2 s = Oρt Thus, by the Neumann series, we infer that I C w t exists for sufficiently large t and I C w t I L 2 s L 2 = Oρt s Next we observe that µ t m t = I C w t C w tm t L 2 sσ