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Computational Methods and Function Theory
, Volume 21 (1) – Mar 25, 2021

/lp/springer-journals/solving-the-initial-value-problem-for-the-3-wave-interaction-equations-BYq8sgV0NG

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- Springer Journals
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- Copyright © The Author(s) 2021
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- 1617-9447
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- 2195-3724
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- 10.1007/s40315-020-00357-2
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Starting from the 3-wave interaction equations in 2+1 dimensions (i.e., two space dimensions and one time dimension), we complexify the independent variables, thus doubling the number of real variables, and hence we work in 4+2 dimensions: x , x , y , y and t , t . In this paper we solve the initial value problem of the 3-wave 2 1 2 1 2 interaction equations in 4+2 dimensions. 1 Introduction The modern history of integrable systems begins in the late 1960s when Gardner, Greene, Kruskal and Miura solved the initial value problem of the Korteweg-de Vries (KdV) equation by what was later called the Inverse Scattering Transform [1]. Here “integrable” means that the system in question can be written as the compatibility condition of a set of linear equations, the so-called Lax pair [2,3]. For some years the KdV equation with its striking properties appeared to be a unique case, until Zakharov and Shabat showed that the Non-linear Schrödinger (NLS) equation can also be solved using the Inverse Scattering Transform [4]. The KdV and NLS equations are integrable evolution equations in one spatial dimension. There exist integrable generalizations of the aforementioned equations to two spatial dimensions: the Kadomtsev–Petviashvili (KP) [5] and Davey–Stewartson (DS) [6] equations, respectively. The KP equation, describing non-linear wave motion, has two forms (known as KPI and KPII) which differ in one particular sign appearing Communicated by Elias Wegert. Dedicated to the memory of Stephan Ruscheweyh. B M. C. van der Weele mcv26@cam.ac.uk Department of Applied Mathematics and Theoretical Physics, University of Cambridge, Cambridge CB3 0WA, UK School of Engineering, University of Southern California, Los Angeles, California 90089, USA 123 10 M. C. van der Weele, A. S. Fokas in the equation. The choice of the sign depends on the relative magnitude of the gravitational forces and the surface tension. Progress towards solving these and other equations in 2+1 dimensions (i.e., two spatial dimensions and one time dimension) was made in the works of Zakharov and Manakov [7–9] and Segur [10]. In the early 1980s Ablowitz, one of the authors, and their collaborators solved both the initial value problem of the KPI equation using a non-local Riemann–Hilbert (RH) problem [11] and the analogous problem for the KPII equation (which cannot be written as a RH problem) using the d-bar formalism [12]. This formalism ﬁrst appeared in the works of Beals and Coifman on problems with one spatial dimension [13,14]. By now the above methodologies have been used to solve the initial value problems of a wide variety of integrable evolution equations with two spatial dimensions [15,16]. One of the main current topics in the ﬁeld of integrable systems concerns the existence of non-linear integrable evolution equations in more than two spatial dimen- sions. Several attempts were made to extend the inverse scattering method to non-linear evolution equations in three or more spatial dimensions [17–20] but although these works made important contributions to the solution of inverse problems, they did not result in the desired construction of integrable multidimensional non-linear evolu- tion Partial Differential Equations (PDEs). The difﬁculty they encountered, known as the ‘characterization problem’, was that the associated scattering data must sat- isfy non-linear constraints which seem to be incompatible with the existence of local non-linear integrable evolution PDEs. These non-linear constraints arise in the above approaches because the corresponding eigenvalue equations involve several complex spectral variables. The fact that integrable multidimensional non-linear evolution PDEs exist has been proven by one of the authors, who in 2006 derived equations of this type in four spatial dimensions, which however had the disadvantage of containing two time dimensions [21]. The Cauchy problem of these equations in 4+2 dimensions can be solved by means of a non-local d-bar problem. The main beneﬁt of this approach is that the eigenvalue equations depend on only one complex spectral variable, whereas the sec- ond complex spectral variable only appears in the integration; thus the characterization problem does not appear here. Subsequently, the same author also identiﬁed a large class of integrable evolution equations in any number of spatial dimensions and 1 time dimension, by constructing a non-linear Fourier transform pair which can be used to solve the Cauchy problem of these equations [22,23]. This second class of equations, however, involves a non-local commutator. In this paper we will concentrate on the ﬁrst class, i.e., the one with 2 time dimensions. In general, the choice of the methodology (local RH, local d-bar, non-local RH, non-local d-bar) is closely connected to the dimensionality of the equations one wants to solve [3]. The initial condition q of an integrable evolution equation in 1 and 2 space dimensions depends on 1 and 2 real spatial variables, respectively, hence the associated spectral function should also involve 1 and 2 real spectral variables. In the case of a local RH problem, the spectral function depends on 1 real spectral variable k, which describes the curve on which the “jump” of the RH problem occurs, so this formalism is suitable for solving equations in 1 spatial variable. In the case of a local d-bar problem, the associated spectral function depends on 2 real spectral variables (k , k ) (or equivalently on (k, k), where now k = k + ik and k is the 1 2 1 2 123 Solving the Initial Value Problem... 11 complex conjugate of k) and, hence, this problem is used to solve equations in 2 spatial dimensions. Alternatively, sometimes the non-local RH formalism can be used for solving equations in 2 spatial dimensions (as mentioned above for the KPI equation); here the spectral function depends not only on the real variable k (specifying the curve along which the jump occurs) but also on the real variable λ, specifying the integration along a given curve. The non-local RH formalism can be generalized using a limiting procedure [9] to a non-local d-bar formalism. This is the formalism we have to employ when the initial data depend on 4 real spatial variables [21]. In this case, the spectral functions also depend on 4 real spectral variables (k , k ,λ ,λ ) (or equivalently on 1 2 1 2 (k, k,λ, λ) with k = k + ik , λ = λ + i λ ), where λ and λ deﬁne the integration 1 2 1 2 1 2 on the complex λ-plane. As a speciﬁc case study, we choose the N -wave interaction equations [24]: q = α q + (C − J α )q + (α − α )q q , ij ij ij i i ij ij in nj in nj t x y n=1 n=i , j for i = j and q = 0, i , j = 1,..., N , (1.1) ii C −C i j where α = (for i = j ), C , J ∈ R. In particular, we study the case N = 3 ij i i J −J i j which is especially important because it is related to the three-wave resonant interaction equations [25]: u + a u + b u = c u u , (1.2) i i i i i i j n t x y where bar denotes complex conjugation, a , b , c are constants and i , j , n = 1, 2, 3 i i i cyclically permuted. Indeed, Eq. (1.2) can be obtained from (1.1) by assuming that q = σ q for i > j and σ σ =−σ (where the σ ’s are real normalizing ij ij 32 21 31 ij ji constants) and taking q = u , q = u , q = u . The three equations contained in 12 3 23 1 2 Eq. (1.2) describe the non-linear interaction of wave packets and are found in numerous applications such as non-linear optics or internal waves in the ocean [26]. From now on, we will use the indices a and b rather than the more familiar i and j to avoid confusion with the imaginary number i, which appears frequently in the analysis that follows. By complexifying the independent variables of the N -wave interaction equations (1.1)for N = 3, we obtain the following system of non-linear integrable equations in 4+2 dimensions: q = α q + (C − J α )q + (α − α )q q , for a = b, n = a, b, and q = 0, ab ab ab a a ab ab an nb an nb aa x y (1.3) where a, b, n = 1, 2, 3 and x = x + ix , y = y + iy , t = t + it , x , x , y , y , t , t ∈ R. (1.4) 1 2 1 2 1 2 1 2 1 2 1 2 The d-bar derivatives appearing in Eq. (1.3)are givenby ∂ = ∂ + i ∂ and x x x 1 2 analogously for ∂ and ∂ [27]. In this paper we apply various methods outlined in Refs. [21,28,29]. The spectral analysis of the time-independent part of the problem is presented in Sect. 2 via two 123 12 M. C. van der Weele, A. S. Fokas different approaches. Subsequently, the full time-dependent problem in 4+2 dimen- sions is solved in Sect. 3. Finally, in Sect. 4 we discuss the question of reducing the problem to fewer dimensions by explicitly eliminating one of the two time variables. 2 Spectral Analysis of the Time-Independent Part of the Lax Pair In this section we will derive non-linear Fourier transform pairs [30] tailor-made for solving the 3-wave interaction equations (1.3), by performing the spectral analysis of the following eigenvalue equation, which is the time-independent part of the Lax pair associated with the 3-wave interaction equations in 4 + 2 dimensions: μ − J μ − k[J,μ]− Qμ = 0. (2.1) x y The matrix μ is a 3 × 3 matrix valued function which depends on the six real variables (x , x , y , y , k , k ), the eigenvalue k is a complex spectral variable, and J and Q 1 2 1 2 1 2 are deﬁned by ⎛ ⎞ ⎛ ⎞ J 00 0 q q 1 12 13 ⎝ ⎠ ⎝ ⎠ J = 0 J 0 , Q(x , x , y , y ) = q 0 q , 2 1 2 1 2 21 23 00 J q q 0 3 31 32 where J , J , J ∈ R\{0}, with J > J > J and q (x , x , y , y ), a, b = 1, 2, 3, 1 2 3 1 2 3 ab 1 2 1 2 are complex-valued functions which are sufﬁciently smooth and which decay rapidly enough for large values of the spatial variables. We observe that Eq. (2.1) can be written in component form as μ − J μ − k (J − J ) μ − (Qμ) = 0. (2.2) ab a ab a b ab ab x y Let us introduce some notation: x = x + ix , y = y + iy ,ξ = ξ + i ξ , 1 2 1 2 1 2 η = η + i η , k = k + ik ,λ = λ + i λ , 1 2 1 2 1 2 dx = dx dx , dy = dy dy , dξ = dξ dξ , 1 2 1 2 1 2 dη = dη dη , dk = dk dk and dλ = dλ dλ , 1 2 1 2 1 2 where x , x , y , y ,ξ ,ξ ,η ,η , k , k ,λ ,λ ∈ R. 1 2 1 2 1 2 1 2 1 2 1 2 Also, we shall generally write f (k,λ, x , y) instead of f (k , k ,λ ,λ , x , x , y , y ). 1 2 1 2 1 2 1 2 At this point, it is important to emphasise that, although our approach is based on the complexiﬁcation of the Lax pair of the 3-wave interaction equations, the resulting non- linear integrable system (1.3) is not the 3-wave interaction equations with 2+1 complex variables, but a genuine 4+2 dimensional system with six real variables. Namely, it does not depend on x + ix , y + iy , t + it but on x , x , y , y , t , t . This fact 1 2 1 2 1 2 1 2 1 2 1 2 is not surprising. Indeed, in general if an equation involves a complex parameter, the solution depends on both the real and the imaginary parts of this parameter. For example, the time-independent part of the Lax pair Eq. (2.1) or equivalently Eq. (2.2) 123 Solving the Initial Value Problem... 13 depends among other things on the complex variable k = k + ik , but the associated 1 2 eigenfunctions μ depend on the real variables k , k . An explicit demonstration that ab 1 2 the type of equations constructed here are genuinely 4+2 dimensional can be found in [31]; in this paper explicit solutions of the 4+2 Davey–Stewartson equation are analysed which clearly depend on 4 real spatial variables and on 2 real time variables. We shall derive the non-linear Fourier transform pairs for the 4+2 dimensional 3-wave interaction equations in two ways. 2.1 First Method: Fourier Transform Formulation In the above Eq. (2.2), the d-bar derivatives appear with respect to both complex spatial variables x and y. As an important ﬁrst step towards solving the system Eq. (2.2), we will start by disentangling these derivatives. This can be achieved by introducing the (a) (a) (a) (a) following local coordinates v (= v ), v , v and v : 1 2 3 4 (a) (a) (a) (a) x = v , x = v + v , y = v − J v , y =−J v , 1 1 2 1 a 1 2 a 2 4 3 4 for a = 1, 2, 3. 1 1 (a) (a) (a) v = x ,v = x + y ,v = y + J x ,v =− y , 1 1 2 2 1 a 1 2 2 3 4 J J a a (2.3) Hence, 1 ∂ ∂ 1 ∂ ∂ ∂ ∂ = + i = + J + i , (2.4a) x a (a) (a) 2 ∂x ∂x 2 ∂v 1 2 1 ∂v ∂v 3 2 1 ∂ ∂ 1 ∂ 1 ∂ 1 ∂ ∂ = + i = + i − . (2.4b) (a) (a) (a) 2 ∂ y ∂ y 2 J J 1 2 a a ∂v ∂v ∂v 3 2 4 Deﬁning the new complex variables (a) (a) z := v + i v , for a = 1, 2, 3, (2.5) and using Eqs. (2.4a)-(2.4b), we ﬁnd ∂ 1 ∂ ∂ = + i = ∂ − J ∂ . (2.6) x a y (a) (a) 2 ∂v ∂z ∂v Hence, Eq. (2.2) can be written in the desired disentangled form: ∂μ ab − k (J − J ) μ − (Qμ) = 0. (2.7) a b ab ab (a) ∂z Taking into consideration the requirement of boundedness, we rewrite (2.7)inthe form ∂ (a) (a) (a) (a) −k(J −J )z +k(J −J )z −k(J −J )z +k(J −J )z a b a b a b a b μ e = e (Qμ) . (2.8) ab ab (a) ∂z 123 14 M. C. van der Weele, A. S. Fokas In order to solve the direct problem we look for a solution of Eq. (2.8) such that μ → I as the spatial variables tend to inﬁnity, where I denotes the unit matrix. Using the Pompeiu formula [27] and the above boundary condition, we obtain (a) (a) −k(J −J ) ζ −z +k(J −J ) ζ −z a b a b 1 (Qμ) e ab μ = δ − dζ. (2.9) ab ab (a) π 2 ζ − z Differentiating Eq. (2.9) with respect to k we ﬁnd (a) (a) ∂μ (J − J ) −k(J −J ) ζ −z +k(J −J ) ζ −z ab a b a a b b =− (Qμ) e dζ ab π 2 ∂k (a) (a) (2.10) −k(J −J ) ζ −z +k(J −J ) ζ −z a b a b ∂μ Q e ∂k ab − dζ, (a) π 2 ζ − z where (a) (a) (a) (a) μ = μ v ,v + v ,v − J v , −J v , k , k , (2.11) ab ab 1 a 1 a 1 2 2 4 3 4 (a) (a) (a) (a) (Qμ) = q ζ ,v + ζ ,v − J ζ , −J ζ μ ζ ,v + ζ ,v ab an 1 2 a 1 a 2 nb 1 2 2 3 2 3 n=1 n=a −J ζ , −J ζ , k , k ) , (2.12) a 1 a 2 1 2 ζ = ζ + i ζ and dζ = dζ dζ . (2.13) 1 2 1 2 Our aim is to complement the above equations for the μ ’s (which are written in terms ab of the q ’s) with an additional set of equations for the same μ ’s, but written in terms ab ab of appropriate “spectral functions” of the q ’s. In order to obtain this additional set ab of equations for the μ ’s, we shall construct a d-bar problem in which the d-bar ab ∂μ ab derivatives are written in terms of the μ ’s alone (as opposed to Eq. (2.10), ab ∂k ∂μ ab where the expressions on the right hand side contain both the μ ’s and the ’s). ab ∂k To achieve this, we need to relate Eqs. (2.9) and (2.10). In this connection, we ﬁrst introduce appropriate notation for the functions appearing as forcing in (2.10), (J − J ) a b (a) (a) −k(J −J )ζ +k(J −J )ζ a b a b f v ,v , k := − (Qμ) e dζ dζ , 1 2 ab ab 2 3 (2.14) where (Qμ) is as in (2.12). ab (a) (a) We can write f v ,v , k as the two dimensional Fourier transform of the ab 2 3 function H : ab (a) (a) (a) (a) λw −λw f v ,v , k = e H (k,λ)dλ, (2.15) ab ab 2 3 123 Solving the Initial Value Problem... 15 (a) (a) (a) where w = v + i v . 3 2 We next introduce the following new variables (the reasons for this particular choice will become clear later): λ = J λ − J k,λ = λ − k . (2.16) 1 a b 1 2 2 1 2 Using the new variables, Eq. (2.15) takes the form (a) (a) f v ,v , k ab 2 3 (a) (a) 2i [−(J λ −J k )v +(λ −k )v ] a b 1 2 1 2 2 3 =|J | e H (k , k , J λ − J k ,λ − k )dλ dλ a ab 1 2 a b 1 2 1 2 1 2 (a) (a) 2i [−(J λ −J k )v +(λ −k )v ] a 1 b 1 2 2 2 3 =|J | e H (k , k , J λ − J k ,λ − k )dλ dλ . (2.17) a ab 1 2 a 1 b 1 2 2 1 2 Denoting f (k,λ) := H (k , k , J λ − J k ,λ −k ), by inverting equation (2.17) ab ab 1 2 a 1 b 1 2 2 and taking into account equation (2.14), we obtain the following expression which can be considered as the non-linear Fourier transform for the problem under consideration: (J − J ) a b (a) (a) (a) (a) f (k,λ) =− E (k,λ, z ,w ) (Qμ) dz dw , (2.18) ab ab ab π 4 (a) (a) where we have deﬁned E (k,λ, z ,w ) by ab (a) (a) (a) (a) (a) (a) k(J −J )z −k(J −J )z +2i [−(J λ −J k )v +(λ −k )v ] a a a 1 1 2 2 b b b 2 3 E (k,λ, z ,w ) := e , ab (2.19) and where (a) (a) (a) (a) (a) (a) (a) (Qμ) = q v ,v + v ,v − J v , −J v μ v ,v + v ,v an 1 a 1 a nb 1 ab 2 4 3 4 2 4 3 n=1 n=a (a) −J v , −J v , k , k . (2.20) a 1 a 1 2 We next rewrite everything in the original coordinates and make the change of variables ζ = ξ , −J ζ = ξ . (2.21) 1 1 a 2 2 Equation (2.9) then becomes μ (x , y, k) ab i i −k(J −J ) (ξ −x )+ (ξ −y ) +k(J −J ) (ξ −x )− (ξ −y ) a b 1 1 2 2 a b 1 1 2 2 J J a a 1 (Qμ) e ab = δ − dξ dξ . ab 1 2 π 2 |J |[(ξ − x ) + (y − ξ )] a 1 1 2 2 (2.22) 123 16 M. C. van der Weele, A. S. Fokas Taking into account equation (2.17) and the deﬁnition of f (k,λ),Eq. (2.10) becomes ab ∂μ ab (x , y, k) ∂k (Ja −J ) 1 b 2i −(J λ −J k ) x + y +(λ −k )(y +J x )+k (J −J )x +k y a 1 1 2 2 2 2 1 a 1 2 a 1 1 2 b b J J a a =|J | e × f (k,λ) dλ dλ ab 1 2 i i −k(J −J ) (ξ −x )+ (ξ −y ) +k(J −J ) (ξ −x )− (ξ −y ) ∂μ a b 1 1 2 2 a b 1 1 2 2 J J a a Q e ∂k ab − dξ dξ , 1 2 π 2 |J |[(ξ − x ) + (y − ξ )] a 1 1 2 2 (2.23) where in Eqs. (2.22), (2.23)wehave (Qμ) = Q ξ , x + (y − ξ ), y + J (x − ξ ), ξ μ ξ , x 1 2 2 2 1 a 1 1 2 1 2 ab + (y − ξ ), y + J (x − ξ ), ξ , k , k , 2 2 1 a 1 1 2 1 2 ab and ∂μ 1 ∂μ Q = Q ξ , x + (y − ξ ), y + J (x − ξ ), ξ ξ , x 1 2 2 2 1 a 1 1 2 1 2 ∂k a ∂k ab + (y − ξ ), y + J (x − ξ ), ξ , k , k . 2 2 1 a 1 1 2 1 2 ab Also, Eq. (2.18), written in the original coordinates, becomes J − J a b f (k,λ) =− E (k,λ, x , y) (Q(x , y)μ(x , y, k)) dxdy, (2.24) ab ab ab |J | π 4 a R with the function E taking the form ab E (k,λ, x , y) := exp[(J λ − J k)x − (J λ − J k)x + (λ − k)y − (λ − k)y] ab a b a b = exp[2i ((−J λ + J k )x + (J λ − J k )x a 1 b 1 2 a 2 b 2 1 +(k − λ )y + (λ − k )y )]. (2.25) 1 1 2 2 2 1 This expression for E in the original coordinates justiﬁes the speciﬁc transforma- ab tion (2.16) that we used to mix the spectral variables k and λ. The important thing to note here is the approximately symmetric role played by k and λ in this expres- sion (2.25). This will allow us to replace k by λ in several steps of our analysis, which will prove to be very convenient. We will now construct a d-bar problem via Eqs. (2.22) and (2.23). For concreteness, ∂μ we concentrate on the second column of the matrix . Multiplying Eq. (2.23)for ∂k 123 Solving the Initial Value Problem... 17 2i (−J k x +J k x −k y +k y ) 2 1 2 2 2 1 1 2 2 1 a = 1, 2, 3 and b = 2 (second column) by the factor e ,we ﬁnd ∂μ 2i (−J k x +J k x −k y +k y ) 2 1 2 2 2 1 1 2 2 1 e (x , y, k) ∂k 2i (−J λ x +J λ x −λ y +λ y ) 1 1 2 1 2 1 1 2 2 1 =|J | e f (k,λ)dλ 1 12 2i −J k x + (y −ξ ) +J k ξ −k ξ +k (y +J (x −ξ )) 2 1 2 2 2 2 2 1 1 2 2 1 1 1 1 ∂μ ∂μ 22 32 J q + q e 12 13 ∂k ∂k − dξ, π 2 |J |[(ξ − x ) + (y − ξ )] 1 1 1 2 2 (2.26a) ∂μ 2i (−J k x +J k x −k y +k y ) 2 1 2 2 2 1 1 2 2 1 e (x , y, k) ∂k (2.26b) ∂μ ∂μ 12 32 2i (−J k x +J k x −k y +k y ) 2 1 2 2 2 1 1 2 2 1 q + q e 21 23 ∂k ∂k =− dξ, π 2 |J |[(ξ − x ) + (y − ξ )] 2 1 1 2 2 ∂μ 2i (−J k x +J k x −k y +k y ) 2 1 2 2 2 1 1 2 2 1 e (x , y, k) ∂k 2i (−J λ x +J λ x −λ y +λ y ) 3 1 2 3 2 1 1 2 2 1 =|J | e f (k,λ)dλ 3 32 2i −J k x + (y −ξ ) +J k ξ −k ξ +k (y +J (x −ξ )) 2 1 2 2 2 2 2 1 1 2 2 1 3 1 1 ∂μ ∂μ 12 22 J q + q e 31 32 ∂k ∂k − dξ. π 2 |J |[(ξ − x ) + (y − ξ )] R 3 1 1 2 2 (2.26c) 2i (−J k x +J k x −k y +k y ) 1 1 2 1 2 1 1 2 2 1 Multiplying Eq. (2.22)for a = 1, 2, 3 and b = 1by e ,we ﬁnd 2i (−J k x +J k x −k y +k y ) 1 1 2 1 2 1 1 2 2 1 e μ (x , y, k) 2i (−J k x +J k x −k y +k y ) 1 1 2 1 2 1 1 2 2 1 = e (2.27a) 2i (−J k x +J k x −k y +k y ) 1 1 2 1 2 1 1 2 2 1 1 q μ + q μ e ( ) 12 21 13 31 − dξ, π 2 R |J |[(ξ − x ) + (y − ξ )] 1 1 1 2 2 2i −J k x +J k x −k y +k y ( ) 1 1 2 1 2 1 1 2 2 1 e μ (x , y, k) 2i −J k x + (y −ξ ) +J k ξ −k ξ +k (y +J (x −ξ )) 1 1 2 2 2 1 2 1 1 2 2 1 2 1 1 1 (q μ + q μ ) e 21 11 23 31 =− dξ, R |J |[(ξ − x ) + (y − ξ )] 2 1 1 2 2 (2.27b) 2i (−J k x +J k x −k y +k y ) 1 1 2 1 2 1 1 2 2 1 e μ (x , y, k) 2i −J k x + (y −ξ ) +J k ξ −k ξ +k (y +J (x −ξ )) 1 1 2 2 2 1 2 1 1 2 2 1 3 1 1 1 (q μ + q μ ) e 31 11 32 21 =− dξ. π 2 |J |[(ξ − x ) + (y − ξ )] 3 1 1 2 2 (2.27c) 123 18 M. C. van der Weele, A. S. Fokas In each of the above three equations we replace k by λ, multiply by |J | f (k,λ) and 1 12 integrate over dλ. Thus, we obtain the following three equations: 2i (−J1 λ1x2 +J1 λ2x1 −λ1 y2 +λ2 y1 ) |J | e μ (x , y,λ) f (k,λ)dλ 1 11 12 2i (−J λ x +J λ x −λ y +λ y ) 1 1 2 1 2 1 1 2 2 1 =|J | e f (k,λ)dλ 1 12 (2.28a) 2i (−J λ x +J λ x −λ y +λ y ) 1 1 2 1 2 1 1 2 2 1 |J | (q μ + q μ ) e f (k,λ) 1 12 21 13 31 12 − dξdλ, π 2 2 |J |[(ξ − x ) + (y − ξ )] R R 1 1 1 2 2 2i (−J λ x +J λ x −λ y +λ y ) 1 1 2 1 2 1 1 2 2 1 |J | e μ (x , y,λ) f (k,λ)dλ 1 21 12 2i −J λ x + (y −ξ ) +J λ ξ −λ ξ +λ (y +J (x −ξ )) 1 1 2 2 2 1 2 1 1 2 2 1 2 1 1 |J | (q μ + q μ ) e f (k,λ) 1 21 11 23 31 12 =− dξdλ, 2 2 π |J |[(ξ − x ) + (y − ξ )] R R 2 1 1 2 2 (2.28b) 2i (−J λ x +J λ x −λ y +λ y ) 1 1 2 1 2 1 1 2 2 1 |J | e μ (x , y,λ) f (k,λ)dλ 1 31 12 2i −J λ x + (y −ξ ) +J λ ξ −λ ξ +λ (y +J (x −ξ )) 1 1 2 2 2 1 2 1 1 2 2 1 3 1 1 |J | (q μ + q μ ) e f (k,λ) 1 31 11 32 21 12 =− dξdλ. π 2 2 R R |J |[(ξ − x ) + (y − ξ )] 3 1 1 2 2 (2.28c) We apply a similar procedure to the third column of Eq. (2.22) (with a = 1, 2, 3 and 2i (−J k x +J k x −k y +k y ) 3 1 2 3 2 1 1 2 2 1 b = 3): multiplying these three equations by e , we ﬁnd 2i (−J k x +J k x −k y +k y ) 3 1 2 3 2 1 1 2 2 1 e μ (x , y, k) 2i −J k x + (y −ξ ) +J k ξ −k ξ +k (y +J (x −ξ )) 3 1 2 2 2 3 2 1 1 2 2 1 1 1 1 1 (q μ + q μ ) e 12 23 13 33 =− dξ, R |J |[(ξ − x ) + (y − ξ )] 1 1 1 2 2 (2.29a) 2i (−J k x +J k x −k y +k y ) 3 1 2 3 2 1 1 2 2 1 e μ (x , y, k) 2i −J k x + (y −ξ ) +J k ξ −k ξ +k (y +J (x −ξ )) 3 1 2 2 2 3 2 1 1 2 2 1 2 1 1 1 (q μ + q μ ) e 21 13 23 33 =− dξ, π 2 |J |[(ξ − x ) + (y − ξ )] 2 1 1 2 2 (2.29b) 2i (−J k x +J k x −k y +k y ) 3 1 2 3 2 1 1 2 2 1 e μ (x , y, k) 2i (−J k x +J k x −k y +k y ) 3 1 2 3 2 1 1 2 2 1 = e (2.29c) 2i (−J k x +J k x −k y +k y ) 3 1 2 3 2 1 1 2 2 1 1 (q μ + q μ ) e 31 13 32 23 − dξ. π 2 |J |[(ξ − x ) + (y − ξ )] R 3 1 1 2 2 In Eqs. (2.29a)-(2.29c) we replace k by λ and then multiply by |J | f (k,λ) and 3 32 integrate over dλ. In this way we obtain the following three equations: 123 Solving the Initial Value Problem... 19 2i (−J λ x +J λ x −λ y +λ y ) 3 1 2 3 2 1 1 2 2 1 |J | e μ (x , y,λ) f (k,λ)dλ 3 13 32 2i −J λ x + (y −ξ ) +J λ ξ −λ ξ +λ (y +J (x −ξ )) 3 1 2 2 2 3 2 1 1 2 2 1 1 1 1 |J | (q μ + q μ ) e f (k,λ) 3 12 23 13 33 32 =− dξdλ, π 2 2 |J |[(ξ − x ) + (y − ξ )] R R 1 1 1 2 2 (2.30a) 2i (−J3 λ1x2 +J3 λ2x1 −λ1 y2 +λ2 y1 ) |J | e μ (x , y,λ) f (k,λ)dλ 3 23 32 2i −J λ x + (y −ξ ) +J λ ξ −λ ξ +λ (y +J (x −ξ )) 3 1 2 2 2 3 2 1 1 2 2 1 2 1 1 |J | (q μ + q μ ) e f (k,λ) 3 21 13 23 33 32 =− dξdλ, π 2 2 R R |J |[(ξ − x ) + (y − ξ )] 2 1 1 2 2 (2.30b) 2i (−J λ x +J λ x −λ y +λ y ) 3 1 2 3 2 1 1 2 2 1 |J | e μ (x , y,λ) f (k,λ)dλ 3 33 32 2i (−J λ x +J λ x −λ y +λ y ) 3 1 2 3 2 1 1 2 2 1 =|J | e f (k,λ)dλ 3 32 (2.30c) 2i (−J λ x +J λ x −λ y +λ y ) 3 1 2 3 2 1 1 2 2 1 |J | (q μ + q μ ) e f (k,λ) 3 31 13 32 23 32 − dξdλ. π 2 2 R R |J |[(ξ − x ) + (y − ξ )] 3 1 1 2 2 At this point, we use the similarity of the kernels of Eqs. (2.26), (2.28) and (2.30)to arrive at the desired d-bar problem. Introducing the notations M (x , y, k) a2 ∂μ a2 2i (−J k x +J k x −k y +k y ) 2 1 2 2 2 1 1 2 2 1 = e (x , y, k), a = 1, 2, 3, (2.31) ∂k M (x , y, k) a1 2i (−J λ x +J λ x −λ y +λ y ) 1 1 2 1 2 1 1 2 2 1 =|J | e μ (x , y,λ) f (k,λ)dλ, a = 1, 2, 3, 1 a1 12 (2.32) and M (x , y, k) a3 2i (−J λ x +J λ x −λ y +λ y ) 3 1 2 3 2 1 1 2 2 1 =|J | e μ (x , y,λ) f (k,λ)dλ, a = 1, 2, 3, 3 a3 32 (2.33) equations (2.26) can be rewritten as follows: Component 12 2i (−J λ x +J λ x −λ y +λ y ) 1 1 2 1 2 1 1 2 2 1 M (x , y, k) =|J | e f (k,λ)dλ 12 1 12 (2.34a) ∼ ∼ q M + q M 12 22 13 32 − dξ, π 2 R |J |[(ξ − x ) + (y − ξ )] 1 1 1 2 2 123 20 M. C. van der Weele, A. S. Fokas ∼ ∼ q M + q M 21 12 23 32 Component 22 M (x , y, k) =− dξ, π 2 R |J |[(ξ − x ) + (y − ξ )] 2 1 1 2 2 (2.34b) Component 32 2i (−J λ x +J λ x −λ y +λ y ) 3 1 2 3 2 1 1 2 2 1 M (x , y, k) =|J | e f (k,λ)dλ 32 3 32 (2.34c) ∼ ∼ q M + q M 31 12 32 22 − dξ, R |J |[(ξ − x ) + (y − ξ )] 3 1 1 2 2 where in the integrals of equations (2.34a)–(2.34c) the elements depend on the vari- ables as follows (for a = 1, 2, 3, b = 1, 2, 3 with a = b) y − ξ 2 2 q M = q ξ , x + , y + J (x − ξ ), ξ ab b2 ab 1 2 1 a 1 1 2 y − ξ 2 2 × M ξ , x + , y + J (x − ξ ), ξ , k , k . b2 1 2 1 a 1 1 2 1 2 Adding Eq. (2.28a)to(2.30a), Eq. (2.28b)to(2.30b) and Eq. (2.28c)to(2.30c), and using the notations introduced in (2.32) and (2.33), we obtain the following three expressions: Sum of 11 and 13 2i (−J λ x +J λ x −λ y +λ y ) 1 1 2 1 2 1 1 2 2 1 M (x , y, k) + M (x , y, k) =|J | e f (k,λ)dλ 11 13 1 12 1 (q (M + M ) + q (M + M )) 12 21 23 13 31 33 − dξ, π 2 |J |[(ξ − x ) + (y − ξ )] R 1 1 1 2 2 (2.35a) Sum of 21 and 23 1 (q (M + M ) + q (M + M )) 21 11 13 23 31 33 M (x , y, k) + M (x , y, k) =− dξ, 21 23 π 2 R |J |[(ξ − x ) + (y − ξ )] 2 1 1 2 2 (2.35b) Sum of 31 and 33 2i (−J λ x +J λ x −λ y +λ y ) 3 1 2 3 2 1 1 2 2 1 M (x , y, k) + M (x , y, k) =|J | e f (k,λ)dλ 31 33 3 32 1 (q (M + M ) + q (M + M )) 31 11 13 32 21 23 − dξ, π 2 R |J |[(ξ − x ) + (y − ξ )] 3 1 1 2 2 (2.35c) 123 Solving the Initial Value Problem... 21 where in the integrals of equations (2.35a)–(2.35c) the elements depend on the vari- ables as follows (for a = 1, 2, 3, b = 1, 2, 3 with a = b and j = 1, 3) y − ξ 2 2 q M = q ξ , x + , y + J (x − ξ ), ξ ab bj ab 1 2 1 a 1 1 2 y − ξ 2 2 × M ξ , x + , y + J (x − ξ ), ξ , k , k . bj 1 2 1 a 1 1 2 1 2 Comparing Eqs. (2.34a)-(2.34c) with Eqs. (2.35a)-(2.35c), we ﬁnd ∼ ∼ ∼ M = M + M , M = M + M , M = M + M , 12 11 13 22 21 23 32 31 33 which in the original notation yields the second column of the d-bar problem: ∂μ a2 (x , y, k) = |J | μ (x , y,λ) f (k,λ)E (k,λ, x , y)dλ, a = 1, 2, 3. n an n2 n2 ∂k n=1 n=2 (2.36) In a similar way, following the same procedure for the ﬁrst and third columns of the ∂μ matrix , we arrive at the full d-bar problem for the 3-wave interaction equations in ∂k 4 + 2 dimensions: ∂μ ab (x , y, k) = |J | μ (x , y,λ) f (k,λ)E (k,λ, x , y)dλ, a, b = 1, 2, 3, n an nb nb ∂k n=1 n=b (2.37) where the E ’s are deﬁned in (2.25). nb The above d-bar problem can be written in a more concise form as ∂μ (x , y, k) = μ(x , y,λ)F (k,λ, x , y)dλ, (2.38) ∂k where F (k,λ, x , y) is the 3 × 3 off-diagonal matrix with its ab-th entry equal to F (k,λ, x , y) =|J | f (k,λ)E (k,λ, x , y), for a = b. ab a ab ab Using the Pompeiu formula, and also using the condition that μ → I + O in the limit k →∞,Eq. (2.37) yields 1 dk dλ μ (x , y, k) = δ + |J | μ (x , y,λ) f (k ,λ)E (k ,λ, x , y) . ab ab n an nb nb π k − k n=1 n=b (2.39) In summary, Eq. (2.22) expresses μ in terms of the q ’s, whereas Eq. (2.39) ab ab expresses μ in terms of the f ’s. We can now obtain a relation between the q ’s ab ab ab 123 22 M. C. van der Weele, A. S. Fokas and f ’s by noting that in the limit k →∞ Eqs. (2.39) and (2.2) imply, respectively, ab the following relations: 1 1 μ (x , y, k) ∼ δ + |J | μ (x , y,λ) f (k,λ)E (k,λ, x , y)dkdλ + O ab ab n an nb nb πk k n=1 n=b (2.40) and, for a = b, 1 1 μ (x , y, k) ∼− q (x , y)μ (x , y, k) + O ab ab bb k(J − J ) k a b 1 1 ∼− q (x , y) + O , k →∞. (2.41) ab k(J − J ) k a b The O terms of Eqs. (2.40) and (2.41) yield J − J a b q (x , y) =− |J | μ (x , y,λ) f (k,λ)E (k,λ, x , y)dkdλ. ab n an nb nb π 4 n=1 n=b (2.42) These above expressions for the q ’s in terms of the f ’s, together with the associated ab ab expressions for the f ’s in terms of the q ’s given by (2.24), deﬁne the non-linear ab ab Fourier transform pair needed for the solution of the Cauchy problem of the 4+2 3-wave interaction equations. 2.2 Second Method: Green’s Function Formulation In order to solve the direct problem we look for a solution of Eq. (2.2) such that μ → I as the spatial variables tend to inﬁnity, where I denotes the unit matrix. A solution of Eq. (2.2) statisfying the above condition is given by the following equation μ (x , y, k) = δ + G (x − x , y − y , k) Q(x , y )μ(x , y , k) dx dy , ab ab ab ab (2.43) where we require that the Green’s function G(x , y, k) satisﬁes G (x , y, k) − J G (x , y, k) − k (J − J ) G (x , y, k) ab a ab a b ab x y ρx −ρx +σ y−σ y = δ(x )δ(y) = e dρdσ, (2.44) where ρ = ρ + i ρ , σ = σ + i σ ,dρ = dρ dρ ,dσ = dσ dσ and we have used 1 2 1 2 1 2 1 2 the notation δ(x ) = δ(x )δ(x ), δ(y) = δ(y )δ(y ). Hence, from Eq. (2.44)wehave 1 2 1 2 ρx −ρx +σ y−σ y 1 e G (x , y, k) = dρdσ. (2.45) ab π 4 −ρ + J σ − k(J − J ) R a a b 123 Solving the Initial Value Problem... 23 Our aim again is to complement the above equations for the μ ’s with another set ab of equations for the same μ ’s, but now written in terms of appropriate “spectral ab functions” of the q ’s, which we will denote by q . In order to obtain this additional ab ab set of equations for the μ ’s, we shall construct a d-bar problem. Differentiating ab Eq. (2.45) with respect to k we ﬁnd, for a = b ρx −ρx +σ y−σ y ∂G 1 e ∂ 1 ab (x , y, k) = dρdσ. (2.46) 4 −ρ+J σ π −(J − J ) ∂k a b ∂k − + k J −J a b Using the formula ∂ 1 = πδ(z − ζ) = πδ(z − ζ )δ(z − ζ ), 1 1 2 2 ∂z z − ζ where z = z + iz and ζ = ζ + i ζ , with z , z ,ζ ,ζ ∈ R, we obtain 1 2 1 2 1 2 1 2 ∂ 1 −ρ + J σ ρ − J σ 1 a 1 2 a 2 = πδ k − δ k − . (2.47) 1 2 −ρ+J σ J − J J − J ∂k a b a b k − J −J a b So combining Eqs. (2.46) and (2.47) we get ∂G ab (x , y, k) ∂k ρx −ρx +σ y−σ y 1 e −ρ + J σ ρ − J σ 1 a 1 2 a 2 = πδ k − δ k − dρdσ 1 2 π 4 −(J − J ) J − J J − J a b a b a b ρx −ρx +σ y−σ y 1 e J − J J − J 1 a b a b = δ k + ρ − σ 1 1 1 π −(J − J ) J J J a b a a a J − J 1 a b × δ − k + ρ − σ dρdσ 2 2 2 J J a a 1 J − J a b 2i (ρ x +ρ x +σ y +σ y ) 1 2 2 1 1 2 2 1 =− e dρ 3 2 σ = [(J − J )k + ρ ] π 2 1 a b 1 1 (J ) R J a a σ =− [(J − J )k − ρ ] 2 a b 2 2 1 1 1 J − J 2i ρ x +ρ x + [(J −J )k +ρ ]y − [(J −J )k −ρ ]y a b 1 2 2 1 a b 1 1 2 a b 2 2 1 J J a a =− e dρ, 3 2 π 2 (J ) δ(x ) where in the second step we have used the identity δ(αx ) = , α ∈ R\{0}. |α| With the change of variables ρ =−J λ + J k ,ρ = J λ − J k , 1 a 1 b 1 2 a 2 b 2 we ﬁnd 123 24 M. C. van der Weele, A. S. Fokas ∂G J − J ab a b 2i[(−J λ +J k )x +(J λ −J k )x +(k −λ )y +(λ −k )y ] a 1 b 1 2 a 2 b 2 1 1 1 2 2 2 1 (x , y, k) =− e dλ π 2 ∂k J − J a b (J λ−J k)x −(J λ−J k)x +(λ−k)y−(λ−k)y a b a b =− e dλ J − J a b =− E (k,λ, x , y)dλ, (2.48) ab π 2 where E is deﬁned in (2.25). Equation (2.48) was found for a = b, but this equation ab is also valid for a = b. Indeed, since Eq. (2.45) does not depend on k or k when a = b, ∂G aa we have that (x , y, k) = 0, a = 1, 2, 3, which is consistent with Eq. (2.48). ∂k Differentiating Eq. (2.43) with respect to k we get ∂μ ab (x , y, k) ∂k ∂G ab = (x − x , y − y , k) Q(x , y )μ(x , y , k) dx dy ab ∂k ∂μ + G (x − x , y − y , k) Q(x , y ) (x , y , k) dx dy ab ∂k ab J − J (2.48) a b =− E (k,λ, x − x , y − y ) Q(x , y )μ(x , y , k) dλ dx dy ab 3 ab π 4 2 R R ∂μ + G (x − x , y − y , k) Q(x , y ) (x , y , k) dx dy ab ∂k R ab J − J a b =− E (k,λ, x , y)E (k,λ, x , y ) Q(x , y )μ(x , y , k) dx dy dλ ab ab 3 ab 2 4 R R ∂μ + G (x − x , y − y , k) Q(x , y ) (x , y , k) dx dy . ab ∂k ab By deﬁning ∧ J − J a b q (k,λ) := − E (k,λ, x , y) (Q(x , y)μ(x , y, k)) dxdy, (2.49) ab ab ab π 4 we have ∂μ ∧ ab (x , y, k) = E (k,λ, x , y)q (k,λ)dλ ab ab ∂k ∂μ + G (x − x , y − y , k) Q(x , y ) (x , y , k) dx dy . ab R ∂k ab (2.50) Now, we shall again construct a d-bar problem, this time via equations (2.43) and (2.50). Multiplying equations (2.43), for a = 1, 2, 3 and b = 1 with k replaced by λ,by q (k,λ)E (k,λ, x , y) and integrating over dλ, we ﬁnd 123 Solving the Initial Value Problem... 25 μ (x , y,λ)q (k,λ)E (k,λ, x , y)dλ 11 12 = E (k,λ, x , y)q (k,λ)dλ + G (x − x , y − y ,λ)E (k,λ, x , y)q (k,λ) q (x , y )μ (x , y ,λ) 11 12 12 21 2 4 R R + q (x , y )μ (x , y ,λ) dx dy dλ, (2.51a) 13 31 μ (x , y,λ)q (k,λ)E (k,λ, x , y)dλ 21 12 = G (x − x , y − y ,λ)E (k,λ, x , y)q (k,λ) q (x , y )μ (x , y ,λ) 21 12 21 11 2 4 R R + q (x , y )μ (x , y ,λ) dx dy dλ, (2.51b) 23 31 μ (x , y,λ)q (k,λ)E (k,λ, x , y)dλ 31 12 = G (x − x , y − y ,λ)E (k,λ, x , y)q (k,λ) q (x , y )μ (x , y ,λ) 31 12 31 11 2 4 R R + q (x , y )μ (x , y ,λ) dx dy dλ. (2.51c) 32 21 Now, multiplying (2.43), for a = 1, 2, 3 and b = 3 with k replaced by λ,by q (k,λ)E (k,λ, x , y) and integrating over dλ, we ﬁnd μ (x , y,λ)q (k,λ)E (k,λ, x , y)dλ 13 32 = G (x − x , y − y ,λ)E (k,λ, x , y)q (k,λ) q (x , y )μ (x , y ,λ) 13 32 12 23 2 4 R R + q (x , y )μ (x , y ,λ) dx dy dλ, (2.52a) 13 33 μ (x , y,λ)q (k,λ)E (k,λ, x , y)dλ 23 32 = G (x − x , y − y ,λ)E (k,λ, x , y)q (k,λ) q (x , y )μ (x , y ,λ) 23 32 21 13 2 4 R R + q (x , y )μ (x , y ,λ) dx dy dλ, (2.52b) 23 33 μ (x , y,λ)q (k,λ)E (k,λ, x , y)dλ 33 32 32 = E (k,λ, x , y)q (k,λ)dλ + G (x − x , y − y ,λ)E (k,λ, x , y)q (k,λ) q (x , y )μ (x , y ,λ) 33 32 31 13 2 4 R R + q (x , y )μ (x , y ,λ) dx dy dλ. (2.52c) 32 23 Adding equation (2.51a)to(2.52a), equation (2.51b)to(2.52b) and equation (2.51c) to (2.52c), we obtain the following three expressions: 123 26 M. C. van der Weele, A. S. Fokas ∧ ∧ μ (x , y,λ)E (k,λ, x , y)q (k,λ) + μ (x , y,λ)E (k,λ, x , y)q (k,λ) dλ 11 12 12 13 32 32 = E (k,λ, x , y)q (k,λ)dλ + q (x , y ) G (x − x , y − y ,λ)E (k,λ, x , y)q (k,λ)μ (x , y ,λ) 12 11 12 21 4 2 R R + G (x − x , y − y ,λ)E (k,λ, x , y)q (k,λ)μ (x , y ,λ)dλ 13 32 23 + q (x , y ) G (x − x , y − y ,λ)E (k,λ, x , y)q (k,λ)μ (x , y ,λ) 13 11 12 31 + G (x − x , y − y ,λ)E (k,λ, x , y)q (k,λ)μ (x , y ,λ)dλ dx dy , (2.53a) 13 32 33 ∧ ∧ μ (x , y,λ)E (k,λ, x , y)q (k,λ) + μ (x , y,λ)E (k,λ, x , y)q (k,λ) dλ 21 12 23 32 12 32 = q (x , y ) G (x − x , y − y ,λ)E (k,λ, x , y)q (k,λ)μ (x , y ,λ) 21 21 12 11 4 2 R R + G (x − x , y − y ,λ)E (k,λ, x , y)q (k,λ)μ (x , y ,λ)dλ 23 32 13 + q (x , y ) G (x − x , y − y ,λ)E (k,λ, x , y)q (k,λ)μ (x , y ,λ) 23 21 12 31 + G (x − x , y − y ,λ)E (k,λ, x , y)q (k,λ)μ (x , y ,λ)dλ dx dy , (2.53b) 23 32 33 ∧ ∧ μ (x , y,λ)E (k,λ, x , y)q (k,λ) + μ (x , y,λ)E (k,λ, x , y)q (k,λ) dλ 31 12 33 32 12 32 = E (k,λ, x , y)q (k,λ)dλ + q (x , y ) G (x − x , y − y ,λ)E (k,λ, x , y)q (k,λ)μ (x , y ,λ) 31 31 12 11 4 2 R R + G (x − x , y − y ,λ)E (k,λ, x , y)q (k,λ)μ (x , y ,λ)dλ 33 32 32 13 + q (x , y ) G (x − x , y − y ,λ)E (k,λ, x , y)q (k,λ)μ (x , y ,λ) 32 31 12 12 21 + G (x − x , y − y ,λ)E (k,λ, x , y)q (k,λ)μ (x , y ,λ)dλ dx dy . (2.53c) 33 32 32 23 ∂μ We now consider the second column of the matrix , i.e., the equations obtained ∂k from (2.50)for b = 2: ∂μ (x , y, k) = E (k,λ, x , y)q (k,λ)dλ ∂k R ∂μ + G (x − x , y − y , k) q (x , y ) (x , y , k) 12 12 ∂k ∂μ +q (x , y ) (x , y , k) dx dy , (2.54a) ∂k 123 Solving the Initial Value Problem... 27 ∂μ ∂μ 22 12 (x , y, k) = G (x − x , y − y , k) q (x , y ) (x , y , k) 22 21 ∂k ∂k ∂μ +q (x , y ) (x , y , k) dx dy , (2.54b) ∂k ∂μ ∧ (x , y, k) = E (k,λ, x , y)q (k,λ)dλ ∂k ∂μ + G (x − x , y − y , k) q (x , y ) (x , y , k) 32 31 ∂k ∂μ +q (x , y ) (x , y , k) dx dy . (2.54c) ∂k Comparing equations (2.53) to equations (2.54) we ﬁnd ∂μ ∧ (x , y, k) = μ (x , y,λ)E (k,λ, x , y)q (k,λ) 11 12 ∂k +μ (x , y,λ)E (k,λ, x , y)q (k,λ) dλ, (2.55a) 13 32 ∂μ ∧ (x , y, k) = μ (x , y,λ)E (k,λ, x , y)q (k,λ) 21 12 ∂k +μ (x , y,λ)E (k,λ, x , y)q (k,λ) dλ, (2.55b) 23 32 ∂μ ∧ (x , y, k) = μ (x , y,λ)E (k,λ, x , y)q (k,λ) 31 12 ∂k +μ (x , y,λ)E (k,λ, x , y)q (k,λ) dλ, (2.55c) 33 32 provided that G (x − x , y − y , k)E (k,λ, x , y ) = G (x − x , y − y ,λ)E (k,λ, x , y), 12 12 11 12 (2.56a) G (x − x , y − y , k)E (k,λ, x , y ) = G (x − x , y − y ,λ)E (k,λ, x , y), 12 32 13 32 (2.56b) G (x − x , y − y , k)E (k,λ, x , y ) = G (x − x , y − y ,λ)E (k,λ, x , y), 22 12 21 12 (2.56c) G (x − x , y − y , k)E (k,λ, x , y ) = G (x − x , y − y ,λ)E (k,λ, x , y), 22 32 23 32 (2.56d) G (x − x , y − y , k)E (k,λ, x , y ) = G (x − x , y − y ,λ)E (k,λ, x , y), 32 12 31 12 (2.56e) G (x − x , y − y , k)E (k,λ, x , y ) = G (x − x , y − y ,λ)E (k,λ, x , y). 32 32 33 32 (2.56f) 123 28 M. C. van der Weele, A. S. Fokas Thus, as a consistency check, we will now verify equations (2.56a) and (2.56b)(the other equations can be veriﬁed in a similar way). Equivalently, we want to show that G (x , y, k) = G (x , y,λ)E (k,λ, x , y), (2.57a) 12 11 12 G (x , y, k) = G (x , y,λ)E (k,λ, x , y). (2.57b) 12 13 32 These conditions hold, since (i) G (x , y, k) ρx −ρx +σ y−σ y 1 e (2.45) = dρdσ π 4 −ρ + J σ − k(J − J ) 1 1 2 ρx −ρx +σ y−σ y 1 e = dρdσ π (−ρ + J σ − k (J − J )) + i (ρ − J σ − k (J − J )) 1 1 1 1 1 2 2 1 2 2 1 2 (2.58) We replace the variables (ρ ,ρ ,σ ,σ ) with (ρ + J k − J λ ,ρ − J k + 1 2 1 2 1 2 1 1 1 2 2 2 J λ ,σ + k − λ ,σ − k + λ ). Then equation (2.58) becomes 1 2 1 1 1 2 2 2 2i (ρ x +ρ x +σ y +σ y ) 1 2 2 1 1 2 2 1 1 e G (x , y, k) = dρdσ E (k,λ, x , y) 12 12 π (−ρ + J σ ) + i (ρ − J σ ) 1 1 1 2 1 2 = G (x , y,λ)E (k,λ, x , y) 11 12 which veriﬁes equation (2.57a) and consequently equation (2.56a). (ii) We now replace the variables (ρ ,ρ ,σ ,σ ) with (ρ + J k − J λ ,ρ − J k + 1 2 1 2 1 2 1 3 1 2 2 2 J λ ,σ + k − λ ,σ − k + λ ) in equation (2.58). Then (2.58) becomes 3 2 1 1 1 2 2 2 G (x , y, k) 2i (ρ x +ρ x +σ y +σ y ) 1 2 2 1 1 2 2 1 1 e = dρdσ π 4 (−ρ + J σ − λ (J − J )) + i (ρ − J σ − λ (J − J )) R 1 1 1 1 1 3 2 1 2 2 1 3 × E (k,λ, x , y) = G (x , y,λ)E (k,λ, x , y) 13 32 which veriﬁes equation (2.57b) and consequently equation (2.56b). Hence, equations (2.55) hold and yield the second column of the d-bar problem ∂μ a2 (x , y, k) = μ (x , y,λ)q (k,λ)E (k,λ, x , y)dλ, a = 1, 2, 3. an n2 n2 ∂k n=1 n=2 (2.59) 123 Solving the Initial Value Problem... 29 In a similar way, we arrive at the full d-bar problem for the 3-wave interaction equations in 4 + 2 dimensions: ∂μ ∧ ab (x , y, k) = μ (x , y,λ)q (k,λ)E (k,λ, x , y)dλ. (2.60) an nb nb ∂k n=1 n=b Comparing equations (2.24) and (2.49), we observe that q (k,λ) =|J | f (k,λ). a ab ab Hence, the above equation is exactly the same as equation (2.37), i.e., the d-bar problem we found in Sect. 2.1. Thus, we ﬁnd again J − J a b q (x , y) =− |J | μ (x , y,λ) f (k,λ)E (k,λ, x , y)dkdλ ab n an nb nb π 4 n=1 n=b J − J ∧ a b =− μ (x , y,λ)q (k,λ)E (k,λ, x , y)dkdλ. (2.61) an nb nb π 4 n=1 n=b Hence, equations (2.61) and (2.49) comprise again a non-linear Fourier transform pair tailor-made for the solution of the Cauchy problem of the 4+2 3-wave interaction equations. This completes the analysis of the time-independent part of the Lax pair. 3 The Time-dependent Problem Let us use the non-linear Fourier transform pair q and f , given by equations (2.42) ab ab and (2.24) respectively. Suppose that f (k,λ), a, b = 1, 2, 3 are allowed to depend ab on the complex variable t, where t = t + it , with t , t ∈ R. To avoid confusion, let 1 2 1 2 h (k,λ, t ) denote these functions. Then q (x , y) for a, b = 1, 2, 3 will also depend ab ab on the time variable t and we denote these functions by g (x , y, t ). ab Let us consider the following non-local d-bar problem: ∂μ (x , y, t , k) = μ(x , y, t,λ)F (k,λ, x , y, t )dλ, (3.1a) ∂k (1) μ (x , y, t ) 1 μ(x , y, t , k) = I + + O , k →∞, (3.1b) k k where F (k,λ, x , y, t ) ab =|J | h (k,λ, t )E (k,λ, x , y) a ab ab =|J | f (k,λ) exp[(J λ − J k)x − (J λ − J k)x + (λ − k)y − (λ − k)y a ab a b a b + (C λ − C k) t − C λ − C k t ], for a, b = 1, 2, 3 with a = b, a b a b and F (k,λ, x , y, t ) = 0, for a = 1, 2, 3, (3.2) aa 123 30 M. C. van der Weele, A. S. Fokas where C , C , C ∈ R\{0} (with C > C > C ), E ’s are deﬁned in (2.25), f ’s 1 2 3 1 2 3 ab ab in (2.24) and (C λ−C k)t − C λ−C k t a b a b h (k,λ, t ) = f (k,λ)e . (3.3) ab ab We assume that the d-bar problem characterized by (3.1) has a unique solution. Which means that if we ﬁnd two operators L and M such that (i) Lμ and M μ satisfy Eq. (3.1a) and (ii) Lμ and M μ are of O as k →∞, then we have Lμ = 0 and M μ = 0. The above argument is the main idea of the dressing method (see e.g. [9,32]). Using the dressing method we will show that μ(x , y, t , k) satisﬁes the following eigenvalue equations: Lμ = μ − J μ − k[J,μ]− Qμ = 0, (3.4) x y M μ = μ − C μ − k[C,μ]− Aμ = 0, (3.5) where Q and A are off-diagonal 3 × 3 matrices deﬁned in terms of μ by (1) (1) Q = μ , J , A = μ , C , (3.6) where C is the diagonal 3 × 3 matrix with entries C , C , C . We note that Eqs. (3.6), 1 2 3 (1) by eliminating μ , give that the ab-th entry of the 3 ×3matrix A is equal to α Q , ab ab C −C a b where α := for a = b. ab J −J a b Let us now introduce the following operators: D μ := μ + kμJ , (3.7a) x x D μ := μ + kμ, (3.7b) y y D μ := μ + kμC . (3.7c) t t We shall show that D − JD − Q μ = 0. (3.8) x y We do this in two steps: (i) First we show that D − JD − Q μ satisﬁes (3.1a). We observe that the oper- x y ators D , D and D , commute with the operator . Hence, also using (3.2), we x y ∂k ﬁnd ∼ ∼ ∂ ∂μ D − JD − Q μ(k) = D − JD − Q (k) x y x y ∂k ∂k = D − JD − Q μ(λ)F (k,λ)dλ x y = μ (λ)F (k,λ) + μ(λ)F (k,λ) + kμ(λ)F (k,λ)J − J μ (λ)F (k,λ) x x y 123 Solving the Initial Value Problem... 31 − J μ(λ)F (k,λ) − kJ μ(λ)F (k,λ) − Qμ(λ)F (k,λ)dλ = μ (λ)F (k,λ) + μ(λ) (λJF (k,λ) − kF (k,λ)J ) + kμ(λ)F (k,λ)J − J μ (λ)F (k,λ) + J μ(λ)(k − λ)F (k,λ) − kJ μ(λ)F (k,λ) − Qμ(λ)F (k,λ)dλ = μ (λ) − J μ (λ) + λμ(λ)J − λJ μ(λ) − Qμ(λ) F (k,λ)dλ x y = D − JD − Q μ(λ) F (k,λ)dλ, (3.9) x y where for convenience we have surpressed the dependence of μ and F on {x , y, t }. (ii) The second step is to recognize that D − JD − Q μ = O as k →∞. (3.10) x y Indeed, by taking into account the asymptotic expansion (3.1b)of μ, we obtain (1) D − JD μ = μ − J μ −k[J,μ]= μ , J +O , k →∞. (3.11) x y x y From the deﬁnition (3.6)of Q in terms of μ, we conclude that (3.10) holds. Therefore, by virtue of (3.9) and (3.10), we have indeed D − JD − Q μ = 0. x y As a consequence, μ(x , y, t , k) satisﬁes (3.4). In a similar way we can show that D − CD − A μ = 0. (3.12) t y After expanding the operators, equation (3.12) takes the form (3.5). Let us denote the ab-th entry of the matrix Q by g (x , y, t ). Then by the note we ab made under (3.6), we have that the ab-th entry of A is equal to α g (x , y, t ), where ab ab C −C a b α := (a = b). In the Appendix we show that (3.4) and (3.5) provide a Lax ab J −J a b pair for the following 3-wave system in 4+2, i.e., in four spatial and two temporal dimensions g = α g + (C − J α )g + (α − α )g g , ab ab ab a a ab ab an nb an nb x y for a = b and n = a, b, and g = 0, (3.13) aa where a, b, n = 1, 2, 3. Using the non-linear Fourier transform pairs in four dimensions f and q , which are ab ab deﬁned in equations (2.24) and (2.42), respectively, we can solve the Cauchy problem of equations (3.13) supplemented with the initial condition g (x , x , y , y , 0, 0) = q (x , x , y , y ), ∀a, b = 1, 2, 3. (3.14) ab 1 2 1 2 ab 1 2 1 2 123 32 M. C. van der Weele, A. S. Fokas Indeed, suppose that the functions q are Schwartz functions in four dimensions which ab satisfy appropriate small norm conditions such that equations (2.22) and (2.39)have unique solutions. We deﬁne the f ’s by equations (2.24), then the h ’s are given ab ab by equations (3.3). We deﬁne g (x , y, t ) by equations (2.42) where the f ’s are ab ab replaced by the corresponding h ’s. Then, the functions g satisfy equation (3.13) ab ab and condition (3.14) for all a, b = 1, 2, 3. The above discussion implies that the solution of the Cauchy problem of the 3-wave interaction equations in 4+2 dimensions is given by J − J a b g (x , y, t ) =− |J | μ (x , y, t,λ) f (k,λ)E (k,λ, x , y) ab n an nb nb n=1 n=b (C λ−C k)t − C λ−C k t n b n b ×e dkdλ, (3.15) where the E ’s are deﬁned in (2.25) and where the μ ’s are the entries of the time- nb an dependent 3 × 3 matrix-valued function μ which satisﬁes the non-local d-bar problem (3.1). Hence, these time-dependent μ ’s are given by Eq. (2.9) with the f ’s replaced an ab by the corresponding h ’s of Eq. (3.3). ab 4 On the Reduction to Fewer Dimensions Having solved the problem in 4+2 dimensions, one would like to be able to reduce this to 3+1 dimensions in order to comply with the physical world. This is not a simple matter, as we will discuss in this section. A logical ﬁrst step is to eliminate one of the two time variables, and this step at least can be accomplished in a straightforward manner. We show this by eliminating the t variable, focusing in ﬁrst instance on the relatively simple case of the linear limit, i.e., for q → q + O( ), with ab ab → 0 and a, b = 1, 2, 3. The t variable may be eliminated in analogous fashion. The present analysis may be compared with that in [29], where the reduction of the 4+2 dimensional Davey–Stewartson system to 3+1 dimensions was discussed. In the linear limit we have, as can be inferred from Eq. (2.39), μ → 1,μ → 0, for a = b, (4.1) aa ab implying that in Eq. (3.15) we now have only to deal with a single term out of the sum-over-n, i.e. only for n = a. Noting that the exponent appearing in Eq. (3.15)is of the following form (with n = a): (C λ − C k) t − C λ − C k t = 2i [(C λ − C k ) t − (C λ − C k ) t ] , a b a b a 2 b 2 1 a 1 b 1 2 (4.2) 123 Solving the Initial Value Problem... 33 we see that the t variable can be eliminated by imposing the condition C λ − C k = 0, i.e., λ = k . (4.3) a 2 b 2 2 2 The t variable may be eliminated in an analogous manner by imposing C λ − 2 a 1 C k = 0. In what follows, however, we focus on the elimination of t . b 1 1 In the linear limit, the 3-wave interaction equations decouple, so we only have to consider one of the equations (e.g. the one for a = 1 and b = 2) for a dependent variable denoted by u: u(x , x , y , y , t ). For the choice a = 1 and b = 2, the 1 2 1 2 2 condition (4.3) becomes λ = k . (4.4) 2 2 Making in equations (2.42) and (2.24) the substitutions μ = 1,μ = 0,μ = 1,μ = 0, f (k,λ) = f (k,λ), q (x , y) = u(x , y; 0), 11 13 22 32 12 12 (4.5) we obtain the Fourier transform pair in four dimensions: |J |(J − J ) 1 1 2 2i[(−J λ +J k )x +(J λ −J k )x +(k −λ )y +(λ −k )y )] 1 1 2 1 2 1 2 2 2 1 1 1 2 2 2 1 u(x , y; 0) =− e π 4 × f (k,λ) dkdλ, (4.6) (J − J ) 1 2 −2i[(−J λ +J k )x +(J λ −J k )x +(k −λ )y +(λ −k )y )] 1 1 2 1 2 1 2 2 2 1 1 1 2 2 2 1 f (k,λ) =− e |J | π 4 ×u(x , y; 0) dxdy. (4.7) In order to impose the reduction (4.4) in the above pair, we make in equation (4.6)the substitution f (k,λ) = f (k,λ) δ λ − k . (4.8) 2 2 Then, (4.6) becomes u(x , y; 0) C C 2 2 |J |(J − J ) 2i (−J λ +J k )x + J −J k x +(k −λ )y + −1 k y ) 1 1 2 1 1 2 1 2 1 2 2 1 1 1 2 2 1 C C 1 1 =− e π 3 × f (k , k ,λ ) dk dk dλ , 1 2 1 1 2 1 (4.9) where f (k , k ,λ ) denotes f k , k ,λ , k . Introducing the notation 1 2 1 1 2 1 2 2i J −J k x 1 2 2 1 (k , k ,λ ; x ) = f (k , k ,λ )e , (4.10) 1 2 1 1 1 2 1 123 34 M. C. van der Weele, A. S. Fokas equation (4.9) can be rewritten in the form |J |(J − J ) 1 1 2 2i (−J λ +J k )x +(k −λ )y + −1 k y ) 1 1 2 1 2 1 1 2 2 1 u(x , y; 0) =− e × (k , k ,λ ; x ) dk dk dλ 1 2 1 1 1 2 1 |J |(J − J ) 1 1 2 2i (J x +y )k + −1 y k +(−J x −y )λ 2 2 2 1 1 2 1 2 2 1 =− e π 3 × (k , k ,λ ; x ) dk dk dλ . (4.11) 1 2 1 1 1 2 1 Deﬁning p := (k , k ,λ ) ≡ (p , p , p ) (4.12) 1 2 1 1 2 3 and → C r := 2 (J x + y ) , 2 − 1 y , 2 (−J x − y ) ≡ (r , r , r ) . (4.13) 2 2 2 1 1 2 2 1 2 3 So we can write u(x , y; 0) as a function of r , r , r and x , which we will denote 1 2 3 1 by h(r , r , r ; x ). 1 2 3 1 Hence, we can write equation (4.11)as → → 1 → → i p · r 2 3 u(x , y; 0) ≡ h(r , r , r ; x ) = e −8π |J |(J − J ) ( p ; x ) d p . 1 2 3 1 1 1 2 1 (2π) (4.14) Employing the inverse Fourier transform in the variables (r , r , r ), the above equa- 1 2 3 tion yields → → → → 2 −i p · r 3 − 8π |J |(J − J ) ( p ; x ) = e h(r , r , r ; x )d r . (4.15) 1 1 2 1 1 2 3 1 We want to change variables from d r to dx dy dy , hence we compute the Jacobian 2 1 2 ∂(r , r , r ) C 1 2 3 2 = 8 − 1 (J − J ). (4.16) 1 2 ∂(x , y , y ) C 2 1 2 1 Thus, ( p ; x ) − 1 C 1 −2i (J x +y )k + −1 y k +(−J x −y )λ 2 2 2 1 1 2 1 2 2 1 =− e u(x , y; 0)dx dy dy . 2 1 2 π |J | 3 1 R (4.17) 123 Solving the Initial Value Problem... 35 Hence, recalling the deﬁnition of from (4.10), the above equation becomes − 1 C C 2 2 1 −2i (−J λ +J k )x + J −J k x +(k −λ )y + −1 k y ) 1 1 2 1 2 1 2 2 1 1 1 2 2 1 C C 1 1 f (k , k ,λ ) =− e 1 2 1 π |J | 3 × u(x , y; 0) dx dy dy . (4.18) 2 1 2 The reduction (4.4) imposes the constraint J C − J C 1 2 2 1 ∂ u = ∂ u. (4.19) x y 1 1 C − C 2 1 The “reduced” Fourier transform pair (4.9) and (4.18) can be used for the solution of the t -independent linearized version of equation (1.3)for a = 1 and b = 2, i.e. of the equation u = α u + (C − J α )u , (4.20) t 12 x 1 1 12 y supplemented with the constraint (4.19). Indeed, using in (3.15) the substitutions (4.5) and (4.8) we ﬁnd the equation u(x , y, t ) C C 2 2 |J |(J − J ) 2i (−J λ +J k )x + J −J k x +(k −λ )y + −1 k y ) 1 1 2 1 1 2 1 2 1 2 2 1 1 1 2 2 1 C C 1 1 =− e π 3 −2i (C λ −C k )t 1 1 2 1 2 ×e f (k , k ,λ ) dk dk dλ . (4.21) 1 2 1 1 2 1 In summary, there exist two different ways to solve the system of equations (4.19) and (4.20). One way is to solve equations (4.19) and (4.20) using the novel Fourier transform pair (4.9) and (4.18): consider the Cauchy problem of equations (4.19) and (4.20), where u(x , x , y , y , 0) = u (x , x , y , y ), x , x , y , y ∈ R, (4.22) 1 2 1 2 0 1 2 1 2 1 2 1 2 with u a scalar function. Then, equation (4.21) with f deﬁned in terms of u by 0 0 equation (4.18) where u(x , x , y , y , 0) is replaced by u , provides the solution of 1 2 1 2 0 the above Cauchy problem. Indeed, it is straightforward to verify that if u is deﬁned by the right hand side of (4.21), this function satisﬁes (4.19) and (4.20). Furthermore, evaluating (4.21)at t = 0, we ﬁnd (4.22) in lieu of the validity of the Fourier transform pair (4.9) and (4.18). Hence, the above deﬁned function u(x , x , y , y , t ) solves the 1 2 1 2 2 Cauchy problem. The second way of solving Eqs. (4.19) and (4.20)istouse (4.19) to eliminate one of the space variables from (4.20) and then to use the standard Fourier transform, or alternatively the method of characteristics. For example, eliminating ∂ from (4.20) 123 36 M. C. van der Weele, A. S. Fokas we ﬁnd J C − J C 1 2 2 1 iu = α u + iu + (C − J α ) u + iu (4.23) t 12 y x 1 1 12 y y 2 1 2 1 2 C − C 2 1 C −C 1 2 and recalling that α = , we see that Eq. (4.23) simpliﬁes further to J −J 1 2 u = α u + (C − J α )u , (4.24) t 12 x 1 1 12 y 2 2 2 which has the form of the 2+1 dimensional linearized 3-wave interaction equations. This shows that, in the linear limit, the elimination of t has brought about a greater reduction than intended; instead of a problem in 3+1 dimensions we are left with a 2+1 dimensional equation. That is to say, the evolution involving the variables x , y 2 2 and t is described by Eq. (4.24), while the relation involving the x and y variables is 2 1 1 governed by Eq. (4.19). This separation can also be inferred directly from the way in which the general solution (4.21), which satisﬁes both (4.19) and (4.24), depends on the four spatial variables x , x , y , y and the time variable t . Namely, the exponential 1 2 1 2 2 expression can be separated into one part containing x and y that involves only 1 1 the spectral variable k , and a second part containing x , y and t that involves the 2 2 2 2 remaining spectral variables k and λ . In the end, then, Eqs. (4.19) and (4.24) can be 1 1 solved independently of each other. It is readily checked that the elimination of the t variable leads to a similar conclu- sion. So the method presented above exceeds its original aim, giving an over-reduced set of equations, at least in the linear limit. This over-reduction appears to be a pecu- liarity of the 3-wave interaction equations, since the same method applied to the linear limit of the 4+2 dimensional Davey–Stewartson system yields a 3+1 dimensional result [29]. The non-linear problem is technically more difﬁcult. In this case, in contrast to the linear limit, the μ ’s do not simplify and hence, in the function g of Eq. (3.15), we an ab have to deal with both terms of the sum-over-n (n = 1, 2, 3, n = b). To eliminate t (or t ) from the exponential appearing in this function, we thus have to impose two conditions for each value of b, instead of the single condition (4.3) we had in the linear limit. Furthermore, since the equations remain coupled in the non-linear case, all three values b = 1, 2, 3 must be considered, giving six conditions in total. The implementation of these conditions, and their implications for the desired reduction of the full non-linear problem to 3+1 dimensions, are currently under investigation. As a ﬁnal remark, we note that the original 2+1 dimensional problem is easily recovered from the 4+2 dimensional version. Indeed, to accomplish this reduction, it is sufﬁcient to assume that the functions g appearing in Eq. (3.13) are independent ab of x , y and t .Then Eq.(3.13) reduces to a non-linear problem in 2+1 dimensions, 2 2 2 equivalent to the original N -wave interaction equations (1.1)for N = 3. Acknowledgements M. C. van der Weele gratefully acknowledges support from the Cambridge Trust and Christ’s College via the “Vice-Chancellor’s and Christ’s College Warwick Scholarship”, from the Foundation for Education and European Culture (Founders Nicos and Lydia Tricha), from the Onassis Foundation — Scholarship ID: F ZQ 004-1/2020-2021, and from The A. G. Leventis Foundation. A. S. 123 Solving the Initial Value Problem... 37 Fokas is supported by EPSRC via a senior fellowship. Both authors want to thank the anonymous referee for very insightful comments and suggestions. Open Access This article is licensed under a Creative Commons Attribution 4.0 International License, which permits use, sharing, adaptation, distribution and reproduction in any medium or format, as long as you give appropriate credit to the original author(s) and the source, provide a link to the Creative Commons licence, and indicate if changes were made. The images or other third party material in this article are included in the article’s Creative Commons licence, unless indicated otherwise in a credit line to the material. If material is not included in the article’s Creative Commons licence and your intended use is not permitted by statutory regulation or exceeds the permitted use, you will need to obtain permission directly from the copyright holder. To view a copy of this licence, visit http://creativecommons.org/licenses/by/4.0/. Appendix Here we shall show that equations (3.4) and (3.5) indeed provide a Lax pair for (3.13). In order to see this explicitly, let us prove that the compatibility condition μ = μ (5.1) tx xt yields equation (3.13). Differentiating equation (3.4) with respect to t we get ∼ ∼ μ − J μ − k J,μ − Q μ − Qμ = 0. (5.2) xt yt t t t Likewise differentiating equation (3.5) with respect to x we get μ − C μ − k [C,μ ] − A μ − Aμ = 0. (5.3) yx x x x tx Hence, combining equations (5.1), (5.2) and (5.3) we obtain ∼ ∼ J μ +kJ μ − kμ J + Q μ + Qμ = C μ +kC μ − kμ C + A μ + Aμ . (5.4) yx x x x x yt t t t t We will use the following relations: • μ = C μ + k C,μ + A μ + Aμ , this equation is found by differentiating yt yy y y y equation (3.5) with respect to y and using the compatibility condition μ = μ . yt t y • μ = C μ + k[C,μ]+ Aμ, which is equation (3.5). t y ∼ ∼ • μ = J μ +k J,μ + Q μ+ Qμ , this equation was found by differentiating yx yy y y equation (3.4) with respect to y and using the compatibility condition μ = μ . x y yx • μ = J μ + k [J,μ] + Qμ, which is equation (3.4). x y Substituting the above in equation (5.4)wearriveat J C μ + k C,μ + A μ + Aμ + kJ C μ + k[C,μ]+ Aμ yy y y y y ∼ ∼ − k C μ + k[C,μ]+ Aμ J + Q μ + Q C μ + k[C,μ]+ Aμ y y ∼ ∼ = C J μ + k J,μ + Q μ + Qμ yy y y 123 38 M. C. van der Weele, A. S. Fokas ∼ ∼ + kC J μ + k [J,μ] + Qμ − k J μ + k [J,μ] + Qμ C y y + A μ + A J μ + k [J,μ] + Qμ . x y Thus, ∼ ∼ ∼ Q = A + C Q − JA +[A, Q] (5.5) x y t y where we have used that [J , A]=[C , Q], i.e., (J − J )A = (C − C )g which a b ab a b ab C −C a b is true because of the deﬁnitions of the matrix A and the coefﬁcient α := . ab J −J a b We observe that the component form of equation (5.5) gives us the 3-wave sys- tem (3.13) and, hence, equations (3.4) and (3.5) indeed provide a Lax pair for (3.13). References 1. Gardner, C.S., Greene, J.M., Kruskal, M.D., Miura, R.M.: Method for solving the Korteweg-deVries equation. Phys. Rev. Lett. 19, 1095–1097 (1967) 2. Lax, P.D.: Integrals of nonlinear equations of evolution and solitary waves. Commun. Pure Appl. Math. 21(5), 467–490 (1968) 3. Fokas, A.S.: Lax pairs: a novel type of separability. Inverse Probl. 25, 123007 (2009) 4. 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