A Riemann–Hilbert approach to the modified Camassa–Holm equation with step-like boundary conditions

Abstract

The paper aims at developing the Riemann–Hilbert (RH) approach for the modified Camassa–Holm (mCH) equation on the line with non-zero boundary conditions, in the case when the solution is assumed to approach two different constants at different sides of the line. We present detailed properties of spectral functions associated with the initial data for the Cauchy problem for the mCH equation and obtain a representation for the solution of this problem in terms of the solution of an associated RH problem.

1 Introduction

In the present paper, we consider the initial value problem for the mCH equation (1.1a):

$$\begin{aligned}&m_t+\left( (u^2-u_x^2)m\right) _x=0,\quad m{:}{=}u-u_{xx},\quad t>0,&-\infty<x<+\infty , \end{aligned}$$

(1.1a)

$$\begin{aligned}&u(x,0)=u_0(x),\quad -\infty<x<+\infty , \end{aligned}$$

(1.1b)

assuming that

$$\begin{aligned} u_0(x)\rightarrow {\left\{ \begin{array}{ll} A_1 \text { as } x\rightarrow -\infty \\ A_2 \text { as } x\rightarrow \infty \\ \end{array}\right. }, \end{aligned}$$

(1.2)

where $A_1$ and $A_2$ are some different constants, and that the solution u(x, t) preserves this behavior for all fixed $t>0$.

Equation (1.1a) is an integrable modification, with cubic nonlinearity, of the Camassa–Holm (CH) equation [20, 21]

$$\begin{aligned} m_t+\left( u m\right) _x + u_x m=0,\quad m:= u-u_{xx}. \end{aligned}$$

(1.3)

The Camassa–Holm equation has been studied intensively over more than two decades, due to its rich mathematical structure as well as applications for modeling the unidirectional propagation of shallow water waves over a flat bottom [28, 51]. The CH and mCH equations are both integrable in the sense that they have Lax pair representations, which allows developing the inverse scattering transform (IST) method, in one form or another, to study the properties of solutions of initial (Cauchy) and initial boundary value problems for these equations. In particular, the inverse scattering method in the form of a Riemann–Hilbert (RH) problem developed for the CH equation with linear dispersion [14] allowed studying the large-time behavior of solutions of initial as well as initial boundary value problems for the CH equation [8, 11, 15, 16] using the (appropriately adapted) nonlinear steepest descent method [31].

Over the last few years various modifications and generalizations of the CH equation have been introduced, see, e.g., [70] and references therein. Novikov [62] applied a perturbative symmetry approach in order to classify integrable equations of the form

$$\begin{aligned} \left( 1-\partial _x^2\right) u_t=F(u, u_x, u_{xx}, u_{xxx}, \dots ),\qquad u=u(x,t), \quad \partial _x:=\partial /\partial x, \end{aligned}$$

assuming that F is a homogeneous differential polynomial over $\mathbb {C}$, quadratic or cubic in u and its x-derivatives (see also [59]). In the list of equations presented in [62], equation (32), which was the second equation with cubic nonlinearity, had the form (1.1a). In an equivalent form, this equation was given by Fokas in [41] (see also [63] and [44]). Shiff [67] considered equation (1.1a) as a dual to the modified Korteweg–de Vries (mKdV) equation and introduced a Lax pair for (1.1a) by rescaling the entries of the spatial part of a Lax pair for the mKdV equation. An alternative (in fact, gauge equivalent) Lax pair for (1.1a) was given by Qiao [65], so the mCH equation is also referred to as the Fokas–Olver–Rosenau–Qiao (FORQ) equation [48].

The local well-posedness and wave-breaking mechanisms for the mCH equation and its generalizations, particularly, the mCH equation with linear dispersion, are discussed in [24, 25, 43, 47, 57]. Algebro-geometric quasiperiodic solutions are studied in [48]. The local well-posedness for classical solutions and for global weak solutions to (1.1a) in Lagrangian coordinates are discussed in [45].

The Hamiltonian structure and Liouville integrability of peakon systems are discussed in [2, 22, 47, 63]. In [52], a Liouville-type transformation was presented relating the isospectral problems for the mKdV equation and the mCH equation, and a Miura-type map from the mCH equation to the CH equation was introduced. The Bäcklund transformation for the mCH equation and a related nonlinear superposition formula are presented in [68].

In the case of the Camassa–Holm equation, the inverse scattering transform method (particularly, in the form of a Riemann–Hilbert factorization problem) works for the version of this equation (considered for functions decaying at spatial infinity) that includes an additional linear dispersion term. Equivalently, this problem can be rewritten as a Cauchy problem for equation (1.3) considered on a constant, nonzero background. Indeed, the inverse scattering transform method requires that the spatial equation from the Lax pair associated to the CH equation have continuous spectrum. On the other hand, the asymptotic analysis of the dispersionless CH equation (1.3) on zero background (where the spectrum is purely discrete) requires a different tool (although having a certain analogy with the Riemann–Hilbert method), namely, the analysis of a coupling problem for entire functions [33,34,35].

In the case of the mCH equation, the situation is similar: the inverse scattering method for the Cauchy problem can be developed when equation (1.1a) is considered on a nonzero background. The Riemann–Hilbert formalism for this problem is developed in [9], and the asymptotic analysis of the large-time behavior of the solutions on a uniform nonzero background is presented in [10].

Integrable nonlinear PDE with non-vanishing boundary conditions at infinity has received plenty of attention in the literature, see e.g. [3, 7, 32, 49]. Particularly, initial value problems with initial data approaching different “backgrounds” at different spatial infinities (the so-called step-like initial data) have attracted considerable attention because they can be used as models for studying expanding, oscillatory dispersive shock waves (DSW), which are large scale, coherent excitation in dispersive systems [4, 40]. Large-time evolution of step-like initial data has been studied for models of uni-directional (Korteweg-de Vries equation) wave propagation [1, 36] as well as bi-directional (Nonlinear Schrödinger equation) wave propagation [5, 6, 12, 13, 19, 42, 50].

The RH problem formalism for the step-like initial value problem for the Camassa-Holm equation was presented in [60], and the large-time behavior of the solutions of this problem was discussed in [61].

In the present paper, we develop the Riemann–Hilbert formalism to problem (1.1) with the step-like initial data (1.2) assuming that $0<A_1<A_2$ and that u(x, t) approaches its large-x limits sufficiently fast. We also assume that $m(x,0)=u_0(x)-u_{0xx}(x)>0$ for all x; then it can be shown that $m(x,t)>0$ for all t (see Appendix 1, for the case of the CH equation, see [26, 27]). In Sect. 2, we introduce appropriate transformations of the Lax pair equations and the associated Jost solutions (“eigenfunctions”) and discuss analytic properties of the eigenfunctions and the corresponding spectral functions (scattering coefficients), including the symmetries and the behavior at the branch points. Here the analysis is performed when fixing the branches of the functions $k_j(\lambda ):=\sqrt{\lambda ^2-\frac{1}{A_j^2}}$, $j=1,2$ involved in the Lax pair transformations as having the branch cuts $(-\infty , -\frac{1}{A_j})\cup (\frac{1}{A_j}, \infty )$.

In Sect. 3, the introduced eigenfunctions are used in the construction of the Riemann–Hilbert problems, whose solutions evaluated at $\lambda =0$ (where $\lambda$ is the spectral parameter in the Lax pair equations) give parametric representations of the solution of problem (1.1).

The case $0<A_2<A_1$ is briefly discussed in Appendix 1.

Notations

In what follows, $\sigma _1{:}{=}\left( {\begin{matrix}0&{}1\\ 1&{}0\end{matrix}}\right)$, $\sigma _2{:}{=}\left( {\begin{matrix}0&{}-\mathrm {i}\\ \mathrm {i}&{}0\end{matrix}}\right)$, and $\sigma _3{:}{=}\left( {\begin{matrix}1&{}0\\ 0&{}-1\end{matrix}}\right)$ denote the standard Pauli matrices, ${\mathbb {C}}^+:=\{\lambda \in {\mathbb {C}}|\Im (\lambda )> 0\}$, and ${\mathbb {C}}^-:=\{\lambda \in {\mathbb {C}}|\Im (\lambda )< 0\}$.

2 Lax pairs and eigenfunctions

2.1 Lax pairs

The Lax pair for the mCH equation (1.1a) has the following form [65]:

$$\begin{aligned} \Phi _x(x,t,\lambda )&=U(x,t,\lambda )\Phi (x,t,\lambda ), \end{aligned}$$

(2.1a)

$$\begin{aligned} \Phi _t(x,t,\lambda )&=V(x,t,\lambda )\Phi (x,t,\lambda ), \end{aligned}$$

(2.1b)

where the coefficients U and V are defined by

$$\begin{aligned} U&=\frac{1}{2}\begin{pmatrix} -1 &{}\quad \lambda m \\ -\lambda m &{}\quad 1\end{pmatrix}, \end{aligned}$$

(2.1c)

$$\begin{aligned} V&=\begin{pmatrix}\lambda ^{-2}+\frac{u^2-u_x^2}{2} &{}\quad -\lambda ^{-1}(u-u_x)-\frac{\lambda (u^2-u_x^2)m}{2}\\ \lambda ^{-1}(u+u_x)+\frac{\lambda (u^2- u_x^2)m}{2} &{}\quad -\lambda ^{-2}-\frac{u^2-u_x^2}{2}\end{pmatrix}, \end{aligned}$$

(2.1d)

with $m(x,t)= u(x,t)-u_{xx}(x,t)$. The RH formalism for integrable nonlinear equations is based on using appropriately defined eigenfunctions, i.e., solutions of the Lax pair, whose behavior as functions of the spectral parameter is well-controlled in the extended complex plane. Notice that the coefficient matrices U and V are traceless, which provides that the determinant of a matrix solution to (2.1) (composed of two vector solutions) is independent of x and t.

Also notice that U and V have singularities (in the extended complex $\lambda$-plane) at $\lambda =0$ and $\lambda =\infty$. In particular, U is singular at $\lambda =\infty$, which necessitates a special care when constructing solutions with controlled behavior as $\lambda \rightarrow \infty$. On the other hand, U becomes u-independent at $\lambda =0$ (a property shared by many Camassa–Holm-typed equations, including the CH equation itself), which suggests using the behavior of the constructed solutions as $\lambda \rightarrow 0$ in order to “extract” the solution of the nonlinear equation in question from the solution of an associated Riemann–Hilbert problem (whose construction, in the direct problem, involves the dedicated solutions of the Lax pair equations).

2.1.1 Notations

• We introduce the following notations for various intervals of the real axis:

  $$\begin{aligned} \Sigma _j= \Big ( -\infty ,-\frac{1}{A_j} \Big ] \cup \Big [ \frac{1}{A_j},\infty \Big ), \qquad {\dot{\Sigma }}_j= \Big ( -\infty ,-\frac{1}{A_j} \Big ) \cup \Big ( \frac{1}{A_j},\infty \Big ), \\ \Sigma _0= \Big [-\frac{1}{A_1},-\frac{1}{A_2}\Big ] \cup \Big [\frac{1}{A_2},\frac{1}{A_1}\Big ], \qquad {\dot{\Sigma }}_0= \Big (-\frac{1}{A_1},-\frac{1}{A_2} \Big ) \cup \Big (\frac{1}{A_2},\frac{1}{A_1}\Big ). \end{aligned}$$

  Notice that $\Sigma _1\subset \Sigma _2$ since we assume $A_1<A_2$.

• For $\lambda \in \Sigma _j$ we denote by $\lambda _+$ ($\lambda _-$) the point of the upper (lower) side of $\Sigma _j$ (i.e. $\lambda _\pm =\lim _{\epsilon \downarrow 0}\lambda \pm \mathrm {i}\epsilon$). Then we have $-\lambda _+=(-\lambda )_-$ and $\overline{\lambda _+}=\lambda _-$.

• $k_j(\lambda ):=\sqrt{\lambda ^2-\frac{1}{A_j^2}}$, $j=1,2$ with the branch cut $\Sigma _j$ and the branch is fixed by the condition $k_j(0)=\frac{\mathrm {i}}{A_j}$.

Observe that $\Im k_j(\lambda )\ge 0$ on ${\mathbb {C}}$, and $k_j(\lambda )$ is real valued on the both sides of $\Sigma _j$. Also notice that $k_j(\lambda )=\omega _j^+(\lambda )\omega _j^-(\lambda )$, where $\omega _j^+(\lambda )=\sqrt{\lambda -\frac{1}{A_j}}$ with the branch cut $[\frac{1}{A_j},\infty )$ and $\omega _j^+(0)=\frac{\mathrm {i}}{\sqrt{A_j}}$, and $\omega _j^-(\lambda )=\sqrt{\lambda +\frac{1}{A_j}}$ with the branch cut $(-\infty ,-\frac{1}{A_j}]$ and $\omega _j^-(0)=\frac{1}{\sqrt{A_j}}$.

Observe the following symmetry relations:

$$\begin{aligned} k_j(-\lambda )&=k_j(\lambda ),\quad \lambda \in {\mathbb {C}}{\setminus }\Sigma _j, \end{aligned}$$

(2.2a)

$$\begin{aligned} k_j(\lambda _+)&=-k_j((-\lambda )_+),\quad \lambda \in \Sigma _j, \end{aligned}$$

(2.2b)

$$\begin{aligned} \overline{k_j({\overline{\lambda }})}&=-k_j(\lambda ),\quad \lambda \in {\mathbb {C}}{\setminus }\Sigma _j, \end{aligned}$$

(2.2c)

$$\begin{aligned} \overline{k_j(\lambda _+)}&=k_j(\lambda _+),\quad \lambda \in \Sigma _j \end{aligned}$$

(2.2d)

(here (2.2b) follows from (2.2a) and (2.2c)).

In order to control the large $\lambda$ behavior of solutions of (2.1), we introduce two gauge transformations associated with $x\rightarrow (-1)^j\infty$ and $m\rightarrow A_j$ (in a similar way as it was done in the case of the constant background [9]).

Proposition 2.1

Equation (1.1a) admits Lax pairs of the form ($j=1,2$)

$$\begin{aligned} {\hat{\Phi }}_{jx}+Q_{jx}{\hat{\Phi }}_{{j}}&= {{\hat{U}}}_j{\hat{\Phi }}_{{j}}, \end{aligned}$$

(2.3a)

$$\begin{aligned} {\hat{\Phi }}_{jt}+Q_{jt}{\hat{\Phi }}_{{j}}&= {{\hat{V}}}_j{\hat{\Phi }}_{{j}}, \end{aligned}$$

(2.3b)

whose coefficients $Q_j\equiv Q_j(x,t,\lambda )$, ${{\hat{U}}}_j\equiv {{\hat{U}}}_j(x,t,\lambda )$, and ${{\hat{V}}}_j\equiv {{\hat{V}}}_j(x,t,\lambda )$ are $2\times 2$ matrices given by (2.7) and (2.8), which are characterized by the following properties:

• (i) $Q_j$ is diagonal and is unbounded as $\lambda \rightarrow \infty$.

• (ii) ${{\hat{U}}}_j=\mathrm {O}(1)$ and ${{\hat{V}}}_j=\mathrm {O}(1)$ as $\lambda \rightarrow \infty$.

• (iii) The diagonal parts of ${{\hat{U}}}_j$ and ${{\hat{V}}}_j$ decay as $\lambda \rightarrow \infty$.

• (iv) ${{\hat{U}}}_j\rightarrow 0$ and ${{\hat{V}}}_j\rightarrow 0$ as $x\rightarrow (-1)^j\infty$.

Proof

Notice that U in (2.1c) can be written as

$$\begin{aligned} U(x,t,\lambda )=\frac{ m(x,t)}{2A_j}\begin{pmatrix}-1&{}\quad \lambda A_j\\ -\lambda A_j&{}\quad 1\end{pmatrix}+\frac{ m(x,t)-A_j}{2A_j}\begin{pmatrix}1&{}\quad 0\\ 0&{}\quad -1\end{pmatrix}, \end{aligned}$$

(2.4)

where $m(x,t)-A_j\rightarrow 0$ as $x\rightarrow (-1)^j\infty$. The first (non-decaying, as $x\rightarrow (-1)^j\infty$) term in (2.4) can be diagonalized by introducing

$$\begin{aligned} {\hat{\Phi }}_{{j}}(x,t,\lambda ){:}{=}D_j(\lambda )\Phi (x,t,\lambda ), \end{aligned}$$

(2.5)

where

$$\begin{aligned} D_j(\lambda ){:}{=}\sqrt{\frac{1}{2}}\sqrt{\frac{1}{\mathrm {i}A_jk_j(\lambda )}-1}\begin{pmatrix} \frac{\lambda A_j}{1-\mathrm {i}A_jk_j(\lambda )} &{}\quad - 1 \\ - 1 &{}\quad \frac{\lambda A_j}{1-\mathrm {i}A_jk_j(\lambda )} \\ \end{pmatrix} \end{aligned}$$

(2.6)

with

$$\begin{aligned} D^{-1}_j(\lambda ){:}{=}\sqrt{\frac{1}{2}}\sqrt{\frac{1}{\mathrm {i}A_jk_j(\lambda )}-1}\begin{pmatrix} \frac{\lambda A_j}{1-\mathrm {i}A_jk_j(\lambda )} &{}\quad 1 \\ 1 &{}\quad \frac{\lambda A_j}{1-\mathrm {i}A_jk_j(\lambda )} \\ \end{pmatrix}. \end{aligned}$$

The factor $\sqrt{\frac{1}{2}}\sqrt{\frac{1}{\mathrm {i}A_jk_j(\lambda )}-1}$ provides $\det D_j(\lambda )=1$ for all $\lambda$, and the branch of the square root is chosen so that the branch cut is $[0,\infty )$ and $\sqrt{-1}=\mathrm {i}$; then $\overline{\sqrt{w_j}}=-\sqrt{\overline{w_j}}$. Observe that $\sqrt{\frac{1}{\mathrm {i}A_jk_j(\lambda )}-1}$ is well defined as a function of $\lambda$ on ${\mathbb {C}}{\setminus }\Sigma _j$ as well as on the sides of $\Sigma _j$. Then (2.5) transforms (2.1a) into

$$\begin{aligned} {\hat{\Phi }}_{jx}+\frac{\mathrm {i}k_j(\lambda )m}{2}\sigma _3{\hat{\Phi }}_{{j}}={{\hat{U}}}_j {\hat{\Phi }}_{{j}}, \end{aligned}$$

(2.7a)

where ${{\hat{U}}}_j\equiv {{\hat{U}}}_j(x,t,\lambda )$ is given by

$$\begin{aligned} {{\hat{U}}}_j=\frac{\lambda (m-A_j)}{2 A_j k_j(\lambda )} \sigma _2 +\frac{m-A_j}{2\mathrm {i}A_j^2k_j(\lambda )}\sigma _3. \end{aligned}$$

(2.7b)

In turn, the t-equation (2.1b) of the Lax pair is transformed into

$$\begin{aligned} {\hat{\Phi }}_{jt} +\mathrm {i}A_j k_j(\lambda ) \left( -\frac{1}{2A_j} m(u^2-u_x^2)-\frac{1}{\lambda ^2}\right) \sigma _3{\hat{\Phi }}_{{j}}= {{\hat{V}}}_j {\hat{\Phi }}_{{j}}, \end{aligned}$$

(2.7c)

where ${{\hat{V}}}_j\equiv {{\hat{V}}}_j(x,t,\lambda )$ is given by

$$\begin{aligned} {{\hat{V}}}_j&= -\frac{1}{2 A_j k_j(\lambda )}\left( \lambda (u^2-u_x^2)( m - A_j) +\frac{2 (u-A_j)}{\lambda }\right) \sigma _2 +\frac{ {{\tilde{u}}}_x}{\lambda } \sigma _1 \nonumber \\&\quad -\frac{1}{\mathrm {i}A_j k_j(\lambda )}\left( A_j( u-A_j) +\frac{1}{2A_j}(u^2-u_x^2)( m - A_j)\right) \sigma _3. \end{aligned}$$

(2.7d)

Now notice that equations (2.7a) and (2.7c) have the desired form (2.1), if we define $Q_j$ by

$$\begin{aligned} Q_j(x,t,\lambda ){:}{=}p_j(x,t,\lambda )\sigma _3, \end{aligned}$$

(2.8a)

with

$$\begin{aligned} p_j(x,t,\lambda ){:}{=}\mathrm {i}A_j k_j(\lambda )\left( \frac{1}{2A_j}\int ^x_{(-1)^j\infty } ( m(\xi ,t)-A_j)\mathrm {d}\xi +\frac{x}{2}-t\Big (\frac{1}{\lambda ^2}+\frac{A_j^2}{2}\Big )\right) .\nonumber \\ \end{aligned}$$

(2.8b)

Indeed, we obviously have $p_{jx}=\frac{\mathrm {i}k_j(\lambda )m}{2}$; on the other hand, the equality

$$\begin{aligned} p_{jt}=\mathrm {i}A_j k_j(\lambda ) \left( -\frac{1}{2A_j} m(u^2-u_x^2)-\frac{1}{\lambda ^2}\right) \end{aligned}$$

follows from (1.1a). $\square$

Remark 2.2

In [9], which deals with the mCH equation on a single background, introducing a uniformizing spectral parameter (such that $\lambda$ and the respective $k(\lambda )$ are rational with respect to it) allowed getting rid of square roots and thus avoiding the problem of specifying particular branches. In the present case, since we have to deal with two different functions, $k_1(\lambda )$ and $k_2(\lambda )$, associated with two different backgrounds, we keep the original spectral parameter $\lambda$ as the spectral variable in the RH problem formalism we are going to develop.

2.2 Eigenfunctions

The Lax pair in the form (2.7) allows us to determine, via associated integral equations, dedicated solutions having a well-controlled behavior as functions of the spectral parameter $\lambda$ for large values of $\lambda$. Indeed, introducing

$$\begin{aligned} {{\widetilde{\Phi }}}_j={\hat{\Phi }}_j\mathrm {e}^{Q_j} \end{aligned}$$

(2.9)

(understanding ${{\widetilde{\Phi }}}_j$ as a $2\times 2$ matrix), equations (2.7a) and (2.7c) can be rewritten as

$$\begin{aligned} {\left\{ \begin{array}{ll} {{\widetilde{\Phi }}}_{jx}+[Q_{jx},{{\widetilde{\Phi }}}_j]={{\hat{U}}}_j{{\widetilde{\Phi }}}_j,&{}\\ {{\widetilde{\Phi }}}_{jt}+[Q_{jt},{{\widetilde{\Phi }}}_j]={{\hat{V}}}_j{{\widetilde{\Phi }}}_j,&{} \end{array}\right. } \end{aligned}$$

(2.10)

where $[\,\cdot \,,\,\cdot \,]$ stands for the commutator. We now determine the Jost solutions ${{\widetilde{\Phi }}}_{j}\equiv {{\widetilde{\Phi }}}_{j}(x,t,\lambda )$, $j=1,2$ of (2.10) as the solutions of the associated Volterra integral equations:

$$\begin{aligned} {{\widetilde{\Phi }}}_{j}(x,t,\lambda )= & {} I+\int _{(-1)^j\infty }^x \mathrm {e}^{Q_j(\xi ,t,\lambda )-Q_j(x,t,\lambda )}{{\hat{U}}}_j(\xi ,t,\lambda )\nonumber \\&\quad {{\widetilde{\Phi }}}_{j}(\xi ,t,\lambda ) \mathrm {e}^{Q_j(x,t,\lambda )-Q_j(\xi ,t,\lambda )}\mathrm {d}\xi , \end{aligned}$$

(2.11)

or, taking into account the definition (2.8) of $Q_j$,

$$\begin{aligned} {{\widetilde{\Phi }}}_{j}(x,t,\lambda )= & {} I+\int _{(-1)^j\infty }^x \mathrm {e}^{\frac{\mathrm {i}k_j(\lambda )}{2}\int ^\xi _x m(\tau ,t)\mathrm {d}\tau \sigma _3}{{\hat{U}}}_j(\xi ,t,\lambda )\nonumber \\&\quad {{\widetilde{\Phi }}}_j(\xi ,t,\lambda ) \mathrm {e}^{-\frac{\mathrm {i}k_j(\lambda )}{2}\int ^\xi _x m(\tau ,t)\mathrm {d}\tau \sigma _3}\mathrm {d}\xi , \end{aligned}$$

(2.12)

(I is the $2\times 2$ identity matrix).

Hereafter, ${\hat{\Phi }}_j{:}{=}{{\widetilde{\Phi }}}_j\mathrm {e}^{-Q_j}$, $j=1,2$ denote the corresponding Jost solutions of (2.7) whereas $\Phi _j{:}{=}D_j^{-1}(\lambda ){\hat{\Phi }}_{{j}}$ denote the corresponding Jost solutions of (2.1).

We are now able to analyze the analytic and asymptotic properties of the solutions ${{\widetilde{\Phi }}}_j$ of (2.12) as functions of $\lambda$, using Neumann series expansions. Let $A^{(1)}$ and $A^{(2)}$ denote the columns of a $2\times 2$ matrix $A=\left( A^{(1)}\ \ A^{(2)}\right)$. Using these notations we have the following properties:

• ${{\widetilde{\Phi }}}_j^{(j)}$ is analytic in ${\mathbb {C}}{\setminus }\Sigma _j$ and has a continuous extension on the lower and upper sides of ${\dot{\Sigma }}_j$.

• ${{\widetilde{\Phi }}}_j^{(1)}$ and ${{\widetilde{\Phi }}}_j^{(2)}$ are well defined and continuous on the lower and upper sides of ${\dot{\Sigma }}_j$.

In (2.10) the coefficients are traceless matrices, from which it follows that $\det {{\tilde{\Phi }}}_j$ is independent on x and t, and hence

• $\det {{\widetilde{\Phi }}}_j\equiv 1$.

Regarding the values of ${{\widetilde{\Phi }}}_j$ at particular points in the $\lambda$-plane, (2.12) implies the following:

• $\left( {\begin{matrix} {{\widetilde{\Phi }}}_1^{(1)}&{{\widetilde{\Phi }}}_2^{(2)}\end{matrix}}\right) \rightarrow I$ as $\lambda \rightarrow \infty$ (since the diagonal part of ${{\hat{U}}}_j$ is $\mathrm {O}\left( \frac{1}{\lambda }\right)$ and the off-diagonal part of ${{\hat{U}}}_j$ is bounded).

• ${{\tilde{\Phi }}}_j$ has singularities at $\lambda =\pm \frac{1}{A_j}$ of order $\frac{1}{2}$ (this will be discussed below, see Sect. 2.8).

2.3 “Background” solution

Introduce $\Phi _{0,j}(x,t,\lambda ):=D_j^{-1}(\lambda )\mathrm {e}^{-Q_j(x,t,\lambda )}$. We see that $\Phi _{0,j}$ satisfy the differential equations:

$$\begin{aligned} {\left\{ \begin{array}{ll} \Phi _{0,jx}=\frac{ m(x,t)}{2A_j}\begin{pmatrix}-1&{}\quad \lambda A_j\\ -\lambda A_j&{}\quad 1\end{pmatrix}\Phi _{0,j},&{}\\ \Phi _{0,jt}=\left( -\frac{1}{2A_j} m(u^2-u_x^2)-\frac{1}{\lambda ^2}\right) \begin{pmatrix}-1&{}\quad \lambda A_j\\ -\lambda A_j&{}\quad 1\end{pmatrix}\Phi _{0,j}.&\end{array}\right. } \end{aligned}$$

(2.13)

Comparing this with (2.1), $\Phi _{j}(x,t,\lambda )$ can be characterized as the solutions of the integral equations:

$$\begin{aligned} \Phi _{j}(x,t,\lambda )= & {} \Phi _{0,j}(x,t,\lambda )+\int _{(-1)^j\infty }^x \Phi _{0,j}(x,t,\lambda )\nonumber \\&\Phi _{0,j}^{-1}(y,t,\lambda )\frac{m(y,t)-A_j}{2A_j} \sigma _3\Phi _{j}(y,t,\lambda )dy. \end{aligned}$$

(2.14)

Observe that $\Phi _{0,j}(x,t,\lambda )\Phi _{0,j}^{-1}(y,t,\lambda )$ is entire w.r.t. $\lambda$. Hence the “lack of analyticity” (jumps) of $\Phi _{j}(x,t,\lambda )$ is generated by the “lack of analyticity” of $\Phi _{0,j}(x,t,\lambda )$. Notice that $\det \Phi _j = \det \Phi _{0,j}=1$.

2.4 Spectral functions

Introduce the scattering matrices $s(\lambda _\pm )$ for $\lambda \in {\dot{\Sigma }}_1$ as matrices relating $\Phi _1$ and $\Phi _2$:

$$\begin{aligned} \Phi _1(x,t,\lambda _\pm )=\Phi _2(x,t,\lambda _\pm )s(\lambda _\pm ),\qquad \lambda \in {\dot{\Sigma }}_1 \end{aligned}$$

(2.15)

with $\det s(\lambda _\pm )=1$. In turn, ${{\tilde{\Phi }}}_1$ and ${{\tilde{\Phi }}}_2$ are related by

$$\begin{aligned} D_1^{-1}(\lambda _\pm ){{\tilde{\Phi }}}_1(x,t,\lambda _\pm )=D_2^{-1}(\lambda _\pm ){{\tilde{\Phi }}}_2(x,t,\lambda _\pm )\mathrm {e}^{-Q_2(x,t,\lambda _\pm )}s(\lambda _\pm )\mathrm {e}^{Q_1(x,t,\lambda _\pm )},~\lambda \in {\dot{\Sigma }}_1.\nonumber \\ \end{aligned}$$

(2.16)

Introducing

$$\begin{aligned} \tilde{s}(x,t,\lambda _\pm ):=\mathrm {e}^{-Q_2(x,t,\lambda _\pm )}s(\lambda _\pm )\mathrm {e}^{Q_1(x,t,\lambda _\pm )} \end{aligned}$$

(2.17)

we have

$$\begin{aligned} (D_1^{-1}{{\tilde{\Phi }}}_1)(x,t,\lambda _\pm )=(D_2^{-1}{{\tilde{\Phi }}}_2)(x,t,\lambda _\pm )\tilde{s}(x,t,\lambda _\pm ), \qquad \lambda \in {\dot{\Sigma }}_1. \end{aligned}$$

(2.18)

Notice that the scattering coefficients ($s_{ij}$) can be expressed as follows:

$$\begin{aligned} s_{11}&=\det \Big (\Phi _1^{(1)},\Phi _2^{(2)}\Big ), \end{aligned}$$

(2.19a)

$$\begin{aligned} s_{12}&=\det \Big (\Phi _1^{(2)},\Phi _2^{(2)}\Big ), \end{aligned}$$

(2.19b)

$$\begin{aligned} s_{21}&=\det \Big (\Phi _2^{(1)},\Phi _1^{(1)}\Big ), \end{aligned}$$

(2.19c)

$$\begin{aligned} s_{22}&=\det \Big (\Phi _2^{(1)},\Phi _1^{(2)}\Big ). \end{aligned}$$

(2.19d)

Accordingly,

$$\begin{aligned} \tilde{s}_{1j}&=\det \Big ( \Big (D_1^{-1}{{\tilde{\Phi }}}_1 \Big )^{(j)}, \Big (D_2^{-1}{{\tilde{\Phi }}}_2 \Big )^{(2)} \Big ), \end{aligned}$$

(2.20a)

$$\begin{aligned} \tilde{s}_{2j}&=\det \Big ( \Big (D_2^{-1}{{\tilde{\Phi }}}_2 \Big )^{(1)}, \Big (D_1^{-1}{{\tilde{\Phi }}}_1 \Big )^{(j)} \Big ). \end{aligned}$$

(2.20b)

Then (2.19a) implies that $s_{11}(\lambda )$ can be analytically extended to ${\mathbb {C}}{\setminus }\Sigma _2$ and defined on the upper and lower sides of ${{\dot{\Sigma }}}_2$. On the other hand, since $\Phi _1^{(1)}$ is analytic in ${\mathbb {C}}{\setminus }\Sigma _1$ and $\Phi _2^{(1)}$ is defined on the upper and lower sides of $\Sigma _2$, $s_{21}(\lambda )$ can be extended by (2.19c) to the lower and upper sides of ${\dot{\Sigma }}_2$. It follows that (2.15) and (2.16) restricted to the first column hold also on $\Sigma _0$, namely,

$$\begin{aligned} \Phi _1^{(1)}(x,t,\lambda _\pm )=s_{11}(\lambda _\pm )\Phi _2^{(1)}(x,t,\lambda _\pm )+s_{21}(\lambda _\pm )\Phi _2^{(2)}(x,t,\lambda _\pm ), \quad \lambda \in {{\dot{\Sigma }}}_0,\nonumber \\ \end{aligned}$$

(2.21)

and, respectively,

$$\begin{aligned} \Big (D_1^{-1}{{\tilde{\Phi }}}_1^{(1)}\Big )(\lambda _\pm )=\tilde{s}_{11}\Big (\lambda _\pm \Big ) \Big (D_2^{-1}{{\tilde{\Phi }}}_2^{(1)}\Big ) \Big (\lambda _\pm \Big )+\tilde{s}_{21}\Big (\lambda _\pm \Big ) \Big (D_2^{-1}{{\tilde{\Phi }}}_2^{(2)} \Big ) \Big (\lambda _\pm \Big ), \ \lambda \in {{\dot{\Sigma }}}_0.\nonumber \\ \end{aligned}$$

(2.22)

2.5 Symmetries

Let’s analyse the symmetry relations amongst the eigenfunctions and scattering coefficients. In order to simplify the notations, we will omit the dependence on x and t (e.g., $U(\lambda )\equiv U(x,t,\lambda )$).

First symmetry: $\lambda \longleftrightarrow -\lambda$.

Proposition 2.3

The following symmetries hold:

$$\begin{aligned} \Phi _1^{(1)}(\lambda )&=-\sigma _3\Phi _1^{(1)}(-\lambda ),\quad \lambda \in {\mathbb {C}}{\setminus }\Sigma _1, \end{aligned}$$

(2.23a)

$$\begin{aligned} \Phi _2^{(2)}(\lambda )&=\sigma _3\Phi _2^{(2)}(-\lambda ),\quad \lambda \in {\mathbb {C}}{\setminus }\Sigma _2. \end{aligned}$$

(2.23b)

Proof

observe that $\sigma _3U(\lambda )\sigma _3\equiv U(-\lambda )$ and $\sigma _3V(\lambda )\sigma _3\equiv V(-\lambda )$. Hence $\sigma _3 \Phi _j^{(j)}(-\lambda )$ solves (2.1) together with $\Phi _j^{(j)}(\lambda )$. Comparing their asymptotic behaviour as $x\rightarrow (-1)^j\infty$ and using (2.2a), the symmetries (2.23) follow. $\square$

Corollary 2.4

We have

• (1)
  $$\begin{aligned} s_{11}(-\lambda )=s_{11}(\lambda ), \quad \lambda \in {\mathbb {C}}{\setminus }\Sigma _2. \end{aligned}$$

  (2.24)
• (2)
  $$\begin{aligned} {{\tilde{\Phi }}}_1^{(1)}(\lambda )=\sigma _3\tilde{\Phi }_1^{(1)}(-\lambda ),\quad \lambda \in {\mathbb {C}}{\setminus }\Sigma _1, \end{aligned}$$

  (2.25a)
  $$\begin{aligned} {{\tilde{\Phi }}}_2^{(2)}(\lambda )=-\sigma _3\tilde{\Phi }_2^{(2)}(-\lambda ),\quad \lambda \in {\mathbb {C}}{\setminus }\Sigma _2. \end{aligned}$$

  (2.25b)
• (3)
  $$\begin{aligned} \Big (D_1^{-1}{{\tilde{\Phi }}}_1^{(1)}\Big ) (-\lambda )&=-\sigma _3\Big (D_1^{-1}{{\tilde{\Phi }}}_1^{(1)}\Big )(\lambda ), \quad \lambda \in {\mathbb {C}}{\setminus }\Sigma _1, \end{aligned}$$

  (2.26a)
  $$\begin{aligned} \Big (D_2^{-1}{{\tilde{\Phi }}}_2^{(2)}\Big ) (-\lambda )&=\sigma _3 \Big (D_2^{-1}{{\tilde{\Phi }}}_2^{(2)}\Big )(\lambda ), \quad \lambda \in {\mathbb {C}}{\setminus }\Sigma _2. \end{aligned}$$

  (2.26b)

Proof

• (1) Substitute (2.23) into (2.19a).

• (2) Observe that due to (2.2a), we have $D_j^{-1}(-\lambda )=-\sigma _3D_j^{-1}(\lambda )\sigma _3$ and $Q_j(-\lambda )=Q_j(\lambda )$. Combining this with (2.23) and using the connection between $\Phi _j$ and ${{\tilde{\Phi }}}_j$, we obtain (2.4).

• (3) Combine $D_j^{-1}(-\lambda )=-\sigma _3D_j^{-1}(\lambda )\sigma _3$ and (2.4).

$\square$

Proposition 2.5

The following symmetry holds

$$\begin{aligned} \Phi _j(\lambda _+)=-\sigma _3\Phi _j(-\lambda _+)\sigma _3, \qquad \lambda \in {{\dot{\Sigma }}}_j. \end{aligned}$$

(2.27)

Proof

Since $\sigma _3U(\lambda )\sigma _3\equiv U(-\lambda )$ and $\sigma _3V(\lambda )\sigma _3\equiv V(-\lambda )$ and U and V do not have jumps along $\Sigma _j$, it follows that if $\Phi _j(\lambda _+)$ solves (2.1), so does $\sigma _3 \Phi _j(-\lambda _+)$. Comparing their asymptotic behaviour as $x\rightarrow (-1)^j\infty$ and using (2.2a), the symmetry (2.27) follows. $\square$

Corollary 2.6

We have

• (1)
  $$\begin{aligned} s(\lambda _+)=\sigma _3s(-\lambda _+)\sigma _3,\quad \lambda \in {{\dot{\Sigma }}}_1 \end{aligned}$$

  (2.28)
• (2)
  $$\begin{aligned} {{\tilde{\Phi }}}_j(\lambda _+)=\sigma _3{{\tilde{\Phi }}}_j(-\lambda _+)\sigma _3, \qquad \lambda \in {{\dot{\Sigma }}}_j. \end{aligned}$$

  (2.29)
• (3)
  $$\begin{aligned} \Big (D_j^{-1}{{\tilde{\Phi }}}_j\Big )((-\lambda )_-)=-\sigma _3 \Big (D_j^{-1}{{\tilde{\Phi }}}_j \Big )(\lambda _+)\sigma _3, \quad \lambda _+\in {{\dot{\Sigma }}}_j. \end{aligned}$$

  (2.30)

Proof

• (1) Substitute (2.27) into (2.15).

• (2) Observe that due to (2.2a), we have $D_j^{-1}(-\lambda _+)=-\sigma _3D_j^{-1}(\lambda _+)\sigma _3$ and $Q_j(-\lambda _+)=Q_j(\lambda _+)$. Combining this with (2.27) and using the connection between $\Phi _j$ and ${{\tilde{\Phi }}}_j$, we obtain (2.29).

• (3) Combine $D_j^{-1}(-\lambda _+)=-\sigma _3D_j^{-1}(\lambda _+)\sigma _3$ and (2.29).

$\square$

Second symmetry: $\lambda \longleftrightarrow -{{\overline{\lambda }}}$.

Proposition 2.7

The following symmetry holds

$$\begin{aligned} \Phi _j(\lambda _+)=\sigma _3\Phi _j((-\lambda )_+)\sigma _2, \qquad \lambda \in {{\dot{\Sigma }}}_j. \end{aligned}$$

(2.31)

Proof

Since U and V are single valued functions of $\lambda$, we have $\sigma _3U(\lambda _+)\sigma _3\equiv U((-\lambda )_+)$ and $\sigma _3V(\lambda _+)\sigma _3\equiv V((-\lambda )_+)$ for $\lambda \in \Sigma _j$. Hence, if $\Phi _j(\lambda _+)$ solves (2.1), so does $\sigma _3 \Phi _j((-\lambda _+)$. Comparing their asymptotic behaviour as $x\rightarrow (-1)^j\infty$ and using (2.2b) and the equality $\sqrt{\frac{1}{\mathrm {i}A_jk_j(\lambda _+)}-1}\sqrt{-\frac{1}{\mathrm {i}A_jk_j(\lambda _+)}-1}=-\frac{\lambda _+}{k_j(\lambda _+)}$ for $\lambda _+\in {{\dot{\Sigma }}}_j$, the symmetry (2.31) follows. $\square$

Corollary 2.8

We have

• (1)
  $$\begin{aligned} s(\lambda _+)=\sigma _2s((-\lambda )_+)\sigma _2,\qquad \lambda \in {{\dot{\Sigma }}}_1. \end{aligned}$$

  (2.32)
• (2)
  $$\begin{aligned} s(\lambda _+)=\sigma _1s(\lambda _-)\sigma _1,\qquad \lambda \in {{\dot{\Sigma }}}_1. \end{aligned}$$

  (2.33)
• (3)
  $$\begin{aligned} {{\tilde{\Phi }}}_j(\lambda _+)=\sigma _2{{\tilde{\Phi }}}_j((-\lambda )_+)\sigma _2, \qquad \lambda \in {{\dot{\Sigma }}}_j. \end{aligned}$$

  (2.34)
• (4)
  $$\begin{aligned} \Big (D_j^{-1}{{\tilde{\Phi }}}_j \Big )((-\lambda )_+)=\sigma _3 \Big (D_j^{-1}{{\tilde{\Phi }}}_j \Big )(\lambda _+)\sigma _2, \qquad \lambda \in {{\dot{\Sigma }}}_j. \end{aligned}$$

  (2.35)

Proof

• (1) Substitute (2.31) into (2.15).

• (2) Combine (2.32) with (2.28).

• (3) Observe that $k_j(\lambda _+)\in {\mathbb {R}}$ and that due to (2.2b) and $\sqrt{\frac{1}{\mathrm {i}A_jk_j(\lambda _+)}-1}\sqrt{-\frac{1}{\mathrm {i}A_jk_j(\lambda _+)}-1}=-\frac{\lambda _+}{k_j(\lambda _+)}$, we have $D_j(\lambda _+)\sigma _3D_j^{-1}((-\lambda )_+)=\sigma _2$ and $Q_j((-\lambda )_+)=-Q_j(\lambda _+)$ for $\lambda \in {{\dot{\Sigma }}}_j$. Combining this with (2.31) and using the connection between $\Phi _j$ and ${{\tilde{\Phi }}}_j$, we obtain (2.34).

• (4) Combine $D_j(\lambda _+)\sigma _3D_j^{-1}((-\lambda )_+)=\sigma _2$ and (2.34).

$\square$

Third symmetry: $\lambda \longleftrightarrow {{\overline{\lambda }}}$.

Proposition 2.9

The following symmetries hold

$$\begin{aligned} \overline{\Phi _j^{(j)}({\overline{\lambda }})}=-\Phi _j^{(j)}(\lambda ), \quad \lambda \in {\mathbb {C}}{\setminus }\Sigma _j. \end{aligned}$$

(2.36)

Proof

Since $\overline{U({{\overline{\lambda }}})}\equiv U(\lambda )$ and $\overline{V({{\overline{\lambda }}})}\equiv V(\lambda )$, it follows that $\overline{\Phi _j^{(j)}({{\overline{\lambda }}})}$ solves (2.1a) together with $\Phi _j^{(j)}(\lambda )$. Hence, comparing their asymptotic behaviour as $x\rightarrow (-1)^j\infty$ and using (2.2c) and the equality $\overline{\sqrt{\frac{1}{\mathrm {i}A_j k_j({\overline{\lambda }})}-1}}=-\sqrt{\frac{1}{\mathrm {i}A_j k_j(\lambda )}-1}$, we obtain the symmetries (2.36). $\square$

Corollary 2.10

We have

• (1)
  $$\begin{aligned} \overline{s_{11}({\overline{\lambda }})}=s_{11}(\lambda ),\quad \lambda \in {\mathbb {C}}{\setminus }\Sigma _2. \end{aligned}$$

  (2.37)
• (2)
  $$\begin{aligned} \overline{{{\tilde{\Phi }}}_j^{(j)}({\overline{\lambda }})}={{\tilde{\Phi }}}_j^{(j)}(\lambda ),\quad \lambda \in {\mathbb {C}}{\setminus }\Sigma _j. \end{aligned}$$

  (2.38)
• (3)
  $$\begin{aligned} \overline{(D_j^{-1}{{\tilde{\Phi }}}_j^{(j)})({\overline{\lambda }})}=-(D_j^{-1}{{\tilde{\Phi }}}_j^{(j)})(\lambda ), \quad \lambda \in {\mathbb {C}}{\setminus }\Sigma _j. \end{aligned}$$

  (2.39)

Proof

• (1) Substitute (2.36) into (2.19a).

• (2) Observe that due to (2.2c) and $\overline{\sqrt{\frac{1}{\mathrm {i}A_jk_j({\overline{\lambda }})}-1}}=-\sqrt{\frac{1}{\mathrm {i}A_jk_j(\lambda )}-1}$, we have $\overline{D_j^{-1}({\overline{\lambda }})}=- D_j^{-1}(\lambda )$ and $\overline{Q_j({\overline{\lambda }})}=Q_j(\lambda )$. Hence combining this with (2.36) and using the connection between $\Phi _j$ and ${{\tilde{\Phi }}}_j$, we obtain (2.38).

• (3) Combine $\overline{D_j^{-1}({\overline{\lambda }})}=- D_j^{-1}(\lambda )$ and (2.38).

$\square$

Proposition 2.11

The following symmetry holds

$$\begin{aligned} \overline{\Phi _j(\overline{\lambda _+})}=-\Phi _j(\lambda _+), \qquad \lambda \in {{\dot{\Sigma }}}_j. \end{aligned}$$

(2.40)

Proof

As above, since $\overline{U({{\overline{\lambda }}})}\equiv U(\lambda )$ and $\overline{V({{\overline{\lambda }}})}\equiv V(\lambda )$ and U and V have no jumps along $\Sigma _j$, we have $\overline{U(\lambda _-)}\equiv U(\lambda _+)$ and $\overline{V(\lambda _-)}\equiv V(\lambda _+)$. It follows that if $\Phi _j(\lambda _+)$ solves (2.1), so does $\overline{ \Phi _j({{\overline{\lambda }}}_+)}$. Comparing their asymptotic behaviour as $x\rightarrow (-1)^j\infty$ and using (2.2c) and the fact that $\overline{\sqrt{\frac{1}{\mathrm {i}A_jk_j({\overline{\lambda }})}-1}}=-\sqrt{\frac{1}{\mathrm {i}A_jk_j(\lambda )}-1}$, the symmetry (2.40) follows. $\square$

Corollary 2.12

We have

• (1)
  $$\begin{aligned} \overline{s(\overline{\lambda _+})}=s(\lambda _+),\quad \lambda \in {{\dot{\Sigma }}}_1. \end{aligned}$$

  (2.41)
• (2)
  $$\begin{aligned} \overline{{{\tilde{\Phi }}}_j(\overline{\lambda _+})}={{\tilde{\Phi }}}_j(\lambda _+), \qquad \lambda \in {{\dot{\Sigma }}}_j. \end{aligned}$$

  (2.42)
• (3)
  $$\begin{aligned} \overline{(D_j^{-1}{{\tilde{\Phi }}}_j)(\overline{\lambda _+})}=-(D_j^{-1}{{\tilde{\Phi }}}_j)(\lambda _+), \quad \lambda \in {{\dot{\Sigma }}}_j. \end{aligned}$$

  (2.43)

Proof

• (1) Substitute (2.40) into (2.15).

• (2) Observe that due to (2.2c) and $\overline{\sqrt{\frac{1}{\mathrm {i}A_jk_j({\overline{\lambda }})}-1}}=-\sqrt{\frac{1}{\mathrm {i}A_jk_j(\lambda )}-1}$, we have $\overline{D_j^{-1}(\lambda _-)}=- D_j^{-1}(\lambda _+)$ and $\overline{Q_j(\lambda _-)}=Q_j(\lambda _+)$ for $\lambda \in {{\dot{\Sigma }}}_j$. Combining this with (2.40) and using the connection between $\Phi _j$ and ${{\tilde{\Phi }}}_j$, we obtain the result.

• (3) Combine $\overline{D_j^{-1}(\lambda _-)}=- D_j^{-1}(\lambda _+)$ and $\overline{Q_j(\lambda _-)}=Q_j(\lambda _+)$ and (2.42).

$\square$

Fourth symmetry $\lambda _+ \longleftrightarrow \lambda _+$.

Proposition 2.13

The following symmetry holds

$$\begin{aligned} \overline{\Phi _j(\lambda _+)}=\mathrm {i}\Phi _j(\lambda _+)\sigma _1, \qquad \lambda \in {{\dot{\Sigma }}}_j. \end{aligned}$$

(2.44)

Proof

Since $\overline{U(\lambda _+)}\equiv U(\lambda _+)$ and $\overline{V(\lambda _+)}\equiv V(\lambda _+)$ for $\lambda \in \Sigma _j$, in follows that if $\Phi _j(\lambda _+)$ solves (2.1), so does $\overline{\Phi _j(\lambda _+)}$. Comparing their asymptotic behaviour as $x\rightarrow (-1)^j\infty$ and using (2.2d) and the equalities $\sqrt{-\frac{1}{\mathrm {i}A_jk_j(\lambda _+)}-1}\cdot \frac{\lambda _+A_j}{1+\mathrm {i}A_jk_j(\lambda _+)}=-\mathrm {i}\sqrt{\frac{1}{\mathrm {i}A_jk_j(\lambda _+)}-1}$ and $\sqrt{\frac{1}{\mathrm {i}A_jk_j(\lambda _+)}-1}\cdot \frac{\lambda _+A_j}{1-\mathrm {i}A_jk_j(\lambda _+)}=\mathrm {i}\sqrt{-\frac{1}{\mathrm {i}A_jk_j(\lambda _+)}-1}$ for $\lambda \in {{\dot{\Sigma }}}_j$, the symmetry (2.44) follows. $\square$

Corollary 2.14

We have

• (1) $s(\lambda _+)=\sigma _1\overline{s(\lambda _+)}\sigma _1, ~ \lambda \in {{\dot{\Sigma }}}_1$, which, in terms of the matrix entries, reads as follows:

  $$\begin{aligned} s_{11}(\lambda _+)&=\overline{s_{22}(\lambda _+)}, \end{aligned}$$

  (2.45a)
  $$\begin{aligned} s_{12}(\lambda _+)&=\overline{s_{21}(\lambda _+)}. \end{aligned}$$

  (2.45b)

• (2) $|s_{11}(\lambda _+)|^2-|s_{21}(\lambda _+)|^2=1$ for $\lambda \in {{\dot{\Sigma }}}_1$.

• (3) $\big | \frac{s_{21}(\lambda _+)}{s_{11}(\lambda _+)} \big |\le 1$ for $\lambda \in {{\dot{\Sigma }}}_1$. Notice that $\big | \frac{s_{21}(\lambda _+)}{s_{11}(\lambda _+)} \big |= 1$ for $\lambda \in {{\dot{\Sigma }}}_1$ iff $s_{11}(\lambda _+)=\infty$.

• (4)
  $$\begin{aligned} s_{11}(\lambda _-)&=\overline{s_{22}(\lambda _-)},\qquad \lambda \in {{\dot{\Sigma }}}_1, \end{aligned}$$

  (2.46a)
  $$\begin{aligned} s_{12}(\lambda _-)&=\overline{s_{21}(\lambda _-)},\qquad \lambda \in {{\dot{\Sigma }}}_1. \end{aligned}$$

  (2.46b)
• (5)
  $$\begin{aligned} \Phi _j(\lambda _+)=\mathrm {i}\Phi _j(\lambda _-) \sigma _1, \qquad \lambda \in {{\dot{\Sigma }}}_j. \end{aligned}$$

  (2.47)
• (6)
  $$\begin{aligned} \Phi _1^{(1)}(\lambda _+)&=\mathrm {i}\Phi _1^{(2)}(\lambda _-), \qquad \lambda \in {{\dot{\Sigma }}}_1, \end{aligned}$$

  (2.48a)
  $$\begin{aligned} \Phi _2^{(2)}(\lambda _+)&=\mathrm {i}\Phi _2^{(1)}(\lambda _-), \qquad \lambda \in {{\dot{\Sigma }}}_2. \end{aligned}$$

  (2.48b)
• (7)
  $$\begin{aligned} s_{11}(\lambda _+)&=s_{22}(\lambda _-),\quad&\lambda \in {{\dot{\Sigma }}}_1, \end{aligned}$$

  (2.49a)
  $$\begin{aligned} s_{11}(\lambda _+)&=-\mathrm {i}s_{21}(\lambda _-),\quad&\lambda \in {{\dot{\Sigma }}}_0, \end{aligned}$$

  (2.49b)
  $$\begin{aligned} s_{11}(\lambda _-)&=\mathrm {i}s_{21}(\lambda _+),\quad&\lambda \in {{\dot{\Sigma }}}_0. \end{aligned}$$

  (2.49c)

• (8) $\big | \frac{s_{21}(\lambda _+)}{s_{11}(\lambda _+)} \big |=1$ for $\lambda \in {{\dot{\Sigma }}}_0$.

• (9)
  $$\begin{aligned} \overline{{{\tilde{\Phi }}}_j(\lambda _+)}=\sigma _1{{\tilde{\Phi }}}_j(\lambda _+)\sigma _1, \qquad \lambda \in {{\dot{\Sigma }}}_j. \end{aligned}$$

  (2.50)
• (10)
  $$\begin{aligned} {{\tilde{\Phi }}}_1^{(1)}(\lambda _-)=\sigma _1{{\tilde{\Phi }}}_1^{(2)}(\lambda _+),\quad \lambda \in \Sigma _1, \end{aligned}$$

  (2.51a)
  $$\begin{aligned} {{\tilde{\Phi }}}_2^{(2)}(\lambda _-)=\sigma _1{{\tilde{\Phi }}}_2^{(1)}(\lambda _+),\quad \lambda \in \Sigma _2. \end{aligned}$$

  (2.51b)
• (11)
  $$\begin{aligned} \overline{(D_j^{-1}{{\tilde{\Phi }}}_j)(\lambda _+)}=\mathrm {i}(D_j^{-1}{{\tilde{\Phi }}}_j)(\lambda _+)\sigma _1, \quad \lambda \in {{\dot{\Sigma }}}_j. \end{aligned}$$

  (2.52)
• (12)
  $$\begin{aligned} D_j^{-1}(\lambda _-){{\tilde{\Phi }}}_j^{(j)}(\lambda _-)&=(-\mathrm {i}D_j^{-1}(\lambda _+){{\tilde{\Phi }}}_j(\lambda _+)\sigma _1)^{(j)},\quad&\lambda \in {{\dot{\Sigma }}}_1, \end{aligned}$$

  (2.53a)
  $$\begin{aligned} D_2^{-1}(\lambda _-){{\tilde{\Phi }}}_2^{(2)}(\lambda _-)&=(-\mathrm {i}D_2^{-1}(\lambda _+){{\tilde{\Phi }}}_2(\lambda _+)\sigma _1)^{(2)},\quad&\lambda \in {{\dot{\Sigma }}}_0, \end{aligned}$$

  (2.53b)
  $$\begin{aligned} D_1^{-1}(\lambda _-){{\tilde{\Phi }}}_1^{(1)}(\lambda _-)&=D_1^{-1}(\lambda _+){{\tilde{\Phi }}}_1^{(1)}(\lambda _+),\quad&\lambda \in {{\dot{\Sigma }}}_0. \end{aligned}$$

  (2.53c)
• (13)
  $$\begin{aligned} (D_1^{-1}{{\tilde{\Phi }}}_1)((-\lambda )_+)&=\sigma _3\overline{(D_1^{-1}{{\tilde{\Phi }}}_1)(\lambda _+)},\quad&\lambda \in {{\dot{\Sigma }}}_1, \end{aligned}$$

  (2.54a)
  $$\begin{aligned} (D_2^{-1}{{\tilde{\Phi }}}_2)((-\lambda )_+)&=-\sigma _3\overline{(D_2^{-1}{{\tilde{\Phi }}}_2)(\lambda _+)},\quad&\lambda \in {{\dot{\Sigma }}}_2. \end{aligned}$$

  (2.54b)
• (14)
  $$\begin{aligned} s_{11}((-\lambda )_+)=\overline{s_{11}(\lambda _+)}, \quad \lambda \in {{\dot{\Sigma }}}_1. \end{aligned}$$

  (2.55)

Proof

• (1) Substitute (2.44) into (2.15).

• (2) This follows from the fact that $\det {s(\lambda _\pm )=1}$ for all $\lambda \in \Sigma _1$ and (2.45).

• (3) Dividing the previous equality by $|s_{11}(\lambda _+)|^2$, we obtain $1-\big |\frac{s_{21}(\lambda _+)}{s_{11}(\lambda _+)}\big |^2=\big |\frac{1}{s_{11}(\lambda _+)}\big |^2\ge 0$. Hence $\big |\frac{s_{21}(\lambda _+)}{s_{11}(\lambda _+)}\big |\le 1$.

• (4) Combine (2.45) and (2.41).

• (5) Combine (2.44) and (2.40).

• (6) Rewrite (2.47) columnwise.

• (7) Substituting (2.47) into (2.19a) leads to (2.49). Notice that in proving (2.49b) and (2.49c) we use the fact that $\Phi _1^{(1)}$ is analytic on $\Sigma _0$.

• (8) Using the previous result for the first equality and (2.37) for the second one, we get $\big | \frac{s_{21}(\lambda _+)}{s_{11}(\lambda _+)} \big |=\big | \frac{-\mathrm {i}s_{11}(\lambda _-)}{s_{11}(\lambda _+)} \big |=\big | \frac{\overline{ s_{11}(\lambda _+)}}{s_{11}(\lambda _+)} \big |=1$.

• (9) Observe that $\sqrt{-\frac{1}{\mathrm {i}A_jk_j(\lambda _+)}-1}\cdot \frac{\lambda _+A_j}{1+\mathrm {i}A_jk_j(\lambda _+)}=-\mathrm {i}\sqrt{\frac{1}{\mathrm {i}A_jk_j(\lambda _+)}-1}$ and $\sqrt{\frac{1}{\mathrm {i}A_jk_j(\lambda _+)}-1}\cdot \frac{\lambda _+A_j}{1-\mathrm {i}A_jk_j(\lambda _+)}=\mathrm {i}\sqrt{-\frac{1}{\mathrm {i}A_jk_j(\lambda _+)}-1}$ imply $\overline{D_j^{-1}(\lambda _+)}=\mathrm {i}D_j^{-1}(\lambda _+)\sigma _1$ and $\overline{D_j(\lambda _+)}=-\mathrm {i}\sigma _1D_j(\lambda _+)$, and (2.2d) imply $\overline{Q_j(\lambda _+)}=-Q_j(\lambda _+)$ for $\lambda \in {{\dot{\Sigma }}}_j$. Combining this with (2.44) and using the connection between $\Phi _j$ and ${{\tilde{\Phi }}}_j$, we obtain (2.50).

• (10) Combine (2.50) and (2.42).

• (11) Combine $\overline{D_j^{-1}(\lambda _+)}=\mathrm {i}D_j^{-1}(\lambda _+)\sigma _1$ and (2.50).

• (12) Use (2.52) combined with (2.43) for the first two equalities and the fact that $k_1(\lambda )$ is analytic on ${{\dot{\Sigma }}}_0$ for the last one.

• (13) Combine (2.52) and (2.35).

• (14) (2.32) implies $s_{22}(\lambda _+)=s_{11}((-\lambda )_+)$. Combine this with (2.45a).

$\square$

2.6 Limits of the eigenfunctions and scattering coefficients from below and above the branch cut

Recall that $k_j(\lambda )$ is analytic in ${\mathbb {C}}{\setminus }\Sigma _j$ and discontinuous across $\Sigma _j$.

Notations

It will be useful in what follows to introduce the following notations (for $\lambda \in \Sigma _j$):

$$\begin{aligned}k_j^+(\lambda ):=k_j(\lambda _+)=\lim _{\epsilon \downarrow 0}k_j(\lambda +\mathrm {i}\epsilon ), \qquad k_j^-(\lambda ):=k_j(\lambda _-)=\lim _{\epsilon \downarrow 0}k_j(\lambda -\mathrm {i}\epsilon ).\end{aligned}$$

Similarly,

$$\begin{aligned}{{\tilde{\Phi }}}_1^{(1)+}(\lambda ):= & {} {{\tilde{\Phi }}}_1^{(1)}(\lambda _+)=\lim _{\epsilon \downarrow 0}{{\tilde{\Phi }}}_1^{(1)}(\lambda +\mathrm {i}\epsilon ),\\ {{\tilde{\Phi }}}_1^{(1)-}(\lambda ):= & {} {{\tilde{\Phi }}}_1^{(1)}(\lambda _-)=\lim _{\epsilon \downarrow 0}{{\tilde{\Phi }}}_1^{(1)}(\lambda -\mathrm {i}\epsilon ).\end{aligned}$$

Observe that

$$\begin{aligned} k_j^-(\lambda )&=-k_j^+(\lambda ),\quad \lambda \in \Sigma _1, \end{aligned}$$

(2.56a)

$$\begin{aligned} k_1^-(\lambda )&=k_1^+(\lambda )=k_1(\lambda ),\quad \lambda \in \Sigma _0, \end{aligned}$$

(2.56b)

$$\begin{aligned} k_2^-(\lambda )&=-k_2^+(\lambda ),\quad \lambda \in \Sigma _0. \end{aligned}$$

(2.56c)

Combining (2.50) and (2.42) we have

$$\begin{aligned} {{\tilde{\Phi }}}_1^{(1)-}(\lambda )=\sigma _1{{\tilde{\Phi }}}_1^{(2)+}(\lambda ),\quad \lambda \in \Sigma _1, \end{aligned}$$

(2.57a)

$$\begin{aligned} {{\tilde{\Phi }}}_2^{(2)-}(\lambda )=\sigma _1{{\tilde{\Phi }}}_2^{(1)+}(\lambda ),\quad \lambda \in \Sigma _2. \end{aligned}$$

(2.57b)

2.7 Discrete spectrum and zeros of scattering coefficients

Multiplying (2.1a) by $\begin{pmatrix} 0&{}\quad -1\\ 1&{}\quad 0 \end{pmatrix}$ we arrive at the spectral problem for a weighted Dirac operator:

$$\begin{aligned} \frac{2}{m}\left( \begin{pmatrix} 0&{}\quad -1\\ 1&{}\quad 0 \end{pmatrix} \Phi _x+\frac{1}{2}\begin{pmatrix} 0&{}\quad 1\\ 1&{}\quad 0 \end{pmatrix}\Phi \right) =\lambda \Phi , \quad x\in (-\infty ,\infty ). \end{aligned}$$

(2.58)

Since $\lim _{x\rightarrow (-1)^j\infty }m(x,t)=A_j\ne 0$, this operator can be viewed as a self-adjoint operator in $L^2(-\infty ,\infty )$ and thus its spectrum in real.

Observe that for $\lambda \in {{\dot{\Sigma }}}_1$, both $k_j(\lambda )$, $j=1,2$ are real-valued and hence the eigenfunctions $\Phi _j$ are bounded but not square integrable near $(-1)^j\infty$. Since they are related by a matrix independent on x and t, $\Phi _j$ are bounded and not square integrable near $\pm \infty$. Hence ${{\dot{\Sigma }}}_1$ comprise the continuous spectrum.

For $\lambda \in (-1/A_2, 1/A_2)$, $\Phi _1^{(1)}$ decays (exponentially fast) as $x\rightarrow -\infty$ and $\Phi _2^{(2)}$ decays (exponentially fast) as $x\rightarrow +\infty$; hence the the eigenvalues in $(-1/A_2, 1/A_2)$ coincides with the zeros of $s_{11}(\lambda )=\det (\Phi _1^{(1)},\Phi _2^{(2)})$.

Note that since $|s_{11}(\lambda _+)|^2-|s_{21}(\lambda _+)|^2=1$ for $\lambda \in {{\dot{\Sigma }}}_1$ (see Corollary 2.14), we have $s_{11}(\lambda _+)\ne 0$ for $\lambda \in {{\dot{\Sigma }}}_1$.

Let’s show that $s_{11}(\lambda _+)\ne 0$ as well as $s_{21}(\lambda _+)\ne 0$ for $\lambda \in {{\dot{\Sigma }}}_0$ (the similar result for $\lambda _-$ will then follow from the symmetry (2.41)). Indeed, we have $|\frac{s_{21}}{s_{11}}(\lambda _\pm )|=1$ for $\lambda \in {{\dot{\Sigma }}}_0$ (see Corollary 2.14). Hence $s_{11}(\lambda _{0+})s_{21}(\lambda _{0+})=0$ iff $s_{11}(\lambda _{0+})=0$ and $s_{21}(\lambda _{0+})=0$ simultaneously. But $s_{11}(\lambda _{0+})=0$ implies that $\Phi _1^{(1)}(\lambda _{0+})$ and $\Phi _2^{(2)}(\lambda _{0+})$ are dependent. Silarly, $s_{21}(\lambda _{0+})=0$ implies that $\Phi _1^{(1)}(\lambda _{0+})$ and $\Phi _2^{(1)}(\lambda _{0+})$ are dependent. Hence $\Phi _2^{(1)}(\lambda _{0+})$ and $\Phi _2^{(2)}(\lambda _{0+})$ are dependent, which contradicts the fact that $\det \Phi _{0,2}\equiv 1$ (the latter follows from evaluating $\det \Phi _{0,2}(x,t,\lambda )$ as $x\rightarrow \infty$ and using the fact that the determinant of a matrix composed by two vector solutions of (2.58) does not depend on x).

Assumption. We will assume that $s_{11}(\lambda )$ has a finite number of zeros on ${\mathbb {R}}{\setminus }\Sigma _2$. Since $s_{11}$ is analytic on ${\mathbb {C}}{\setminus }\Sigma _2$, the uniqueness theorem implies that the sufficient condition is $s_{11}(\pm \frac{1}{A_2})\ne 0$.

Let $\{\lambda _k\}_{k=1}^n$ be the zeros of $s_{11}(\lambda )$. For such $\lambda _k$ we have

$$\begin{aligned}\Phi _1^{(1)}(\lambda _k)=b_k\Phi _2^{(2)}(\lambda _k), \quad b_k{:}{=}\; b(\lambda _k).\end{aligned}$$

Proposition 2.15

The zeros of $s_{11}(\lambda )$ are simple.

Proof

We will denote by $'$ the dirivative w.r.t. $\lambda$.

Using the definition of $s_{11}(\lambda )$ we have

$$\begin{aligned}s_{11}'(\lambda )=\det \Big (\Phi _1^{(1)},\Phi _2^{(2)}\Big )'(\lambda )=\det \Big ( \Big (\Phi ')_1^{(1)},\Phi _2^{(2)}\Big )(\lambda )+\det \Big (\Phi _1^{(1)},(\Phi ')_2^{(2)}\Big )(\lambda ).\end{aligned}$$

Since $\Phi _j^{(j)}$ solves (2.1a), we have

$$\begin{aligned}(\Phi ')_{jx}^{(j)}=U(\Phi ')_j^{(j)}+m\begin{pmatrix} 0&{}\quad 1\\ -1&{}\quad 0 \end{pmatrix}\Phi _j^{(j)},\end{aligned}$$

and, using the fact that $\det \Big (U(\Phi ')_1^{(1)},\Phi _2^{(2)}\Big )=-\det ((\Phi ')_1^{(1)},U\Phi _2^{(2)})$, we have

$$\begin{aligned}\frac{d}{dx}\det \Big ((\Phi ')_1^{(1)},\Phi _2^{(2)}\Big )=\det \left( \begin{pmatrix} 0&{}\quad m\\ -m&{}\quad 0 \end{pmatrix}\Phi _1^{(1)},\Phi _2^{(2)}\right) ,\end{aligned}$$

and

$$\begin{aligned}\frac{d}{dx}\det \Big (\Phi _1^{(1)},(\Phi ')_2^{(2)}\Big )=-\det \left( \begin{pmatrix} 0&{}\quad m\\ -m&{}\quad 0 \end{pmatrix}\Phi _2^{(2)},\Phi _1^{(1)}\right) .\end{aligned}$$

Evaluating at $\lambda =\lambda _k$ and using $\Phi _1^{(1)}(\lambda _k)=b_k\Phi _2^{(2)}(\lambda _k)$, we get

$$\begin{aligned} \frac{d}{dx}\det ((\Phi ')_1^{(1)},\Phi _2^{(2)})(\lambda _k)=b_km\det \left( \begin{pmatrix} 0&{}\quad 1\\ -1&{}\quad 0 \end{pmatrix}\Phi _2^{(2)}(\lambda _k),\Phi _2^{(2)}(\lambda _k)\right) ,\end{aligned}$$

$$\begin{aligned}\frac{d}{dx}\det \Big (\Phi _1^{(1)},(\Phi ')_2^{(2)}\Big )(\lambda _k)=-b_km\det \left( \begin{pmatrix} 0&{}\quad 1\\ -1&{}\quad 0 \end{pmatrix}\Phi _2^{(2)}(\lambda _k),\Phi _2^{(2)}(\lambda _k)\right) .\end{aligned}$$

Using the symmetry (2.36) and observing that $\lambda _k\in {\mathbb {R}}$, we have

$$\begin{aligned}\det (\begin{pmatrix} 0&{}\quad 1\\ -1&{}\quad 0 \end{pmatrix}\Phi _2^{(2)}(\lambda _k),\Phi _2^{(2)}(\lambda _k))=-(|(\Phi _2)_{22}|^2+|(\Phi _2)_{12}|^2)(\lambda _k)\end{aligned}$$

and hence

$$\begin{aligned} \frac{d}{dx}\det \Big ((\Phi ')_1^{(1)},\Phi _2^{(2)}\Big )(\lambda _k)=b_k\int _x^{\infty }m \Big (|(\Phi _2)_{22}|^2+|(\Phi _2)_{12}|^2 \Big )d\tau ,\\ \frac{d}{dx}\det \Big (\Phi _1^{(1)},(\Phi ')_2^{(2)}\Big )(\lambda _k)=b_k\int ^x_{-\infty }m \Big (|(\Phi _2)_{22}|^2+|(\Phi _2)_{12}|^2 \Big )dx. \end{aligned}$$

It follows that

$$\begin{aligned}s_{11}'(\lambda _k)=b_k\int ^\infty _{-\infty }m \Big (|(\Phi _2)_{22}|^2+|(\Phi _2)_{12}|^2 \Big )dx,\end{aligned}$$

and thus $s'_{11}(\lambda _k)\ne 0$. $\square$

Observe that due to the symmetry (2.24), if $s_{11}(\lambda _k)=0$, then $s_{11}(-\lambda _k)=0$ as well. Since, according to Proposition 2.15, all zeros of $s_{11}$ are simple, it follows that $s_{11}(0)\ne 0$. This fact will also be discussed in Sect. 3.2.

2.8 Behaviour at the branch points

Observe that $k_j(\pm \frac{1}{A_j})=0$.

Proposition 2.16

${{\tilde{\Phi }}}_j(x,t,\lambda )$ has the following behaviour at the branch points

$$\begin{aligned} {{\tilde{\Phi }}}_j(x,t,\lambda )=\frac{\mathrm {i}\alpha _j(x,t)}{\omega _j^+(\lambda )}\begin{pmatrix} 1&{}\quad 1\\ -1&{}\quad -1 \end{pmatrix}+\begin{pmatrix} a_j(x,t)&{}\quad b_j(x,t)\\ b_j(x,t)&{}\quad a_j(x,t) \end{pmatrix}+\mathrm {O}\Bigg (\sqrt{\lambda -\frac{1}{A_j}}\Bigg ),\quad \lambda \rightarrow \frac{1}{A_j},\\{{\tilde{\Phi }}}_j(x,t,\lambda )=\frac{\alpha _j(x,t)}{\omega _j^-(\lambda )}\begin{pmatrix} 1&{}\quad -1\\ 1&{}\quad -1 \end{pmatrix}+\begin{pmatrix} a_j(x,t)&{}\quad -b_j(x,t)\\ -b_j(x,t)&{}\quad a_j(x,t) \end{pmatrix}+\mathrm {O}\Bigg (\sqrt{\lambda +\frac{1}{A_j}}\Bigg ),\quad \lambda \rightarrow -\frac{1}{A_j},\end{aligned}$$

whith some real-valued $\alpha _j (x,t)$, $a_j(x,t)$, and $b_j(x,t)$, $j=1,2$.

Proof

Recall that $\omega _j^+(\lambda )=\sqrt{\lambda -\frac{1}{A_j}}$ with a branch cut on $[\frac{1}{A_j},\infty )$ and $\omega _j^+(0)=\frac{\mathrm {i}}{\sqrt{A_j}}$, and $\omega _j^-(\lambda )=\sqrt{\lambda +\frac{1}{A_j}}$ with a branch cut on $(-\infty ,-\frac{1}{A_j}]$ and $\omega _j^-(0)=\frac{1}{\sqrt{A_j}}$.

First, consider the behavior of the eigenfunctions near $\frac{1}{A_j}$. Introduce $\tilde{{{\tilde{\Phi }}}}_j(x,t,\lambda )$ such that ${{\tilde{\Phi }}}_j(x,t,\lambda )=W^+\tilde{{{\tilde{\Phi }}}}_j(x,t,\lambda )$ with $W^+=\begin{pmatrix} 1&{}\quad \frac{\mathrm {i}}{\omega _j^+(\lambda )}\\ 1&{}\quad -\frac{\mathrm {i}}{\omega _j^+(\lambda )} \end{pmatrix}$. Then $\tilde{{{\tilde{\Phi }}}}_j(x,t,\lambda )$ solves the following integral equation:

$$\begin{aligned} \tilde{{{\tilde{\Phi }}}}_j(x,t,\lambda )&=\frac{1}{2}\begin{pmatrix} 1&{}\quad 1\\ -\mathrm {i}\omega _j^+(\lambda )&{}\quad \mathrm {i}\omega _j^+(\lambda ) \end{pmatrix} \\&\quad + \int _{(-1)^i\infty }^x A^{-1} \mathrm {e}^{\frac{\mathrm {i}}{2}k_j(\lambda )\int _x^\xi m d\tau \sigma _3}{{\hat{U}}}_j A \tilde{{{\tilde{\Phi }}}}_j \mathrm {e}^{-\frac{\mathrm {i}}{2}k_j(\lambda )\int _x^\xi m d\tau \sigma _3}.\end{aligned}$$

The kernel of this equation and hence $\tilde{{{\tilde{\Phi }}}}_j$ has no singularity at $\frac{1}{A_j}$. Hence

$$\begin{aligned} {{\tilde{\Phi }}}_j(x,t,\lambda )=\frac{\mathrm {i}}{\omega _j^+(\lambda )}\begin{pmatrix} \tilde{{{\tilde{c}}}}_j&{}\quad \tilde{{{\tilde{d}}}}_j\\ -\tilde{\tilde{c}}_j&{}\quad -\tilde{{{\tilde{d}}}}_j \end{pmatrix}+\begin{pmatrix} a_j&{}\quad b_j\\ c_j&{}\quad d_j \end{pmatrix}+\mathrm {O}\left( \sqrt{\lambda -\frac{1}{A_j}}\right) ,\quad \lambda \rightarrow \frac{1}{A_j}.\end{aligned}$$

Using (2.42), we get $\tilde{{{\tilde{c}}}}_j$, $\tilde{{{\tilde{d}}}}_j\in {\mathbb {R}}$ and $a_j,~b_j,~c_j,~d_j\in {\mathbb {R}}$. Then, using (2.50), we get $\tilde{\tilde{c}}_j=\tilde{{{\tilde{d}}}}_j$ and $a_j=d_j,~c_j=b_j$; thus

$$\begin{aligned} {{\tilde{\Phi }}}_j(x,t,\lambda )=\frac{\mathrm {i}\alpha _j(x,t)}{\omega _j^+(\lambda )}\begin{pmatrix} 1&{}\quad 1\\ -1&{}\quad -1 \end{pmatrix}+\begin{pmatrix} a_j&{}\quad b_j\\ b_j&{}\quad a_j \end{pmatrix}+\mathrm {O}\left( \sqrt{\lambda -\frac{1}{A_j}}\right) ,\quad \lambda \rightarrow \frac{1}{A_j}.\end{aligned}$$

In order to get the simiular result for $-\frac{1}{A_j}$, we use $W^-=\begin{pmatrix} \frac{\mathrm {i}}{\omega _j^-(\lambda )}&{}\quad 1\\ \frac{\mathrm {i}}{\omega _j^-(\lambda )}&{}\quad -1 \end{pmatrix}$ instead of $W^+$, which leads to

$$\begin{aligned}{{\tilde{\Phi }}}_j(x,t,\lambda )=\frac{\beta _j(x,t)}{\omega _j^-(\lambda )}\begin{pmatrix} -1&{}\quad 1\\ -1&{}\quad 1 \end{pmatrix}+\begin{pmatrix} {{\hat{a}}}_j&{}\quad {{\hat{b}}}_j\\ {{\hat{b}}}_j&{}\quad {{\hat{a}}}_j \end{pmatrix}+\mathrm {O}(\sqrt{\lambda +\frac{1}{A_j}}),\quad \lambda \rightarrow -\frac{1}{A_j}.\end{aligned}$$

Finally, using (2.29) and (2.34), we get $\alpha _j=-\beta _j$ and $a_j={{\hat{a}}}_j$ and $b_j=-{{\hat{b}}}_j$. $\square$

Evaluating $D_j^{-1}(\lambda )$ near $\pm \frac{1}{A_j}$ gives

Proposition 2.17

$D_j^{-1}(\lambda )$ has the following behaviour at the branch points:

$$\begin{aligned} D_j^{-1}(\lambda )&=\frac{\mathrm {e}^{\frac{3\pi \mathrm {i}}{4}}}{(2A_j)^{\frac{1}{4}}\nu _j^+(\lambda )}\begin{pmatrix} 1&{}\quad 1\\ 1&{}\quad 1 \end{pmatrix}+\frac{\mathrm {i}\mathrm {e}^{\frac{3\pi \mathrm {i}}{4}}(2A_j)^{\frac{1}{4}}\nu _j^+(\lambda )}{2}\begin{pmatrix} 1&{}\quad -1\\ -1&{}\quad 1 \end{pmatrix}\\&\quad +\mathrm {O}\Big ( \Big (\lambda -\frac{1}{A_j}\Big )^{\frac{3}{4}}\Big ),\quad \lambda \rightarrow \frac{1}{A_j}\end{aligned}$$

and

$$\begin{aligned} D_j^{-1}(\lambda )&=\frac{\mathrm {i}}{(2A_j)^{\frac{1}{4}}\nu _j^-(\lambda )}\begin{pmatrix} -1&{}\quad 1\\ 1&{}\quad -1 \end{pmatrix}+\frac{\mathrm {i}(2A_j)^{\frac{1}{4}}\nu _j^-(\lambda )}{2}\begin{pmatrix} 1&{}\quad 1\\ 1&{}\quad 1 \end{pmatrix}\\&\quad +\, \mathrm {O}\Big ( \Big (\lambda +\frac{1}{A_j}\Big )^{\frac{3}{4}}\Big ),\quad \lambda \rightarrow -\frac{1}{A_j}.\end{aligned}$$

Here $\nu _j^+(\lambda )=(\lambda -\frac{1}{A_j})^{\frac{1}{4}}$ with the branch cut $(\frac{1}{A_j},\infty )$ and $\nu _j^+(0)=\frac{\mathrm {e}^{\frac{\pi \mathrm {i}}{4}}}{(A_j)^{\frac{1}{4}}}$, and $\nu _j^-(\lambda )=(\lambda +\frac{1}{A_j})^{\frac{1}{4}}$ with the branch cut $(-\infty ,-\frac{1}{A_j})$ and $\nu _j^-(0)=\frac{1}{(A_j)^{\frac{1}{4}}}$ (observe that $(\nu _j^\pm (\lambda ))^2=\omega _j^\pm (\lambda )$).

3 Riemann–Hilbert problems

3.1 RH problem parametrized by $\varvec{(x,t)}$

Notations

We denote

$$\begin{aligned} \rho (\lambda ):=\frac{s_{21}(\lambda _+)}{s_{11}(\lambda _+)},\quad \lambda \in {{\dot{\Sigma }}}_1\cup {{\dot{\Sigma }}}_0. \end{aligned}$$

(3.1)

Observe that Corollary 2.14 implies that

$$\begin{aligned}&|\rho (\lambda )|\le 1,\quad \lambda \in {{\dot{\Sigma }}}_1, \end{aligned}$$

(3.2a)

$$\begin{aligned}&|\rho (\lambda )|= 1,\quad \lambda \in {{\dot{\Sigma }}}_0. \end{aligned}$$

(3.2b)

Motivated by the analytic properties of eigenfunctions and scattering coefficients, we introduce the matrix-values function

$$\begin{aligned} M(x,t,\lambda )= \left( \frac{(D_1^{-1}{{\tilde{\Phi }}}_1^{(1)})(x,t,\lambda )}{s_{11}(\lambda )\mathrm {e}^{p_1(x,t,\lambda )-p_2(x,t,\lambda )}},(D_2^{-1}{{\tilde{\Phi }}}_2^{(2)})(x,t,\lambda )\right) ,\qquad \lambda \in {\mathbb {C}}{\setminus }\Sigma _2,\nonumber \\ \end{aligned}$$

(3.3a)

meromorphic in ${\mathbb {C}}{\setminus }\Sigma _2$, where $p_j$, $j=1,2$ are defined in (2.8b).

Observe that $D_j^{-1}(\lambda ){{\tilde{\Phi }}}_j(x,t,\lambda )=\Phi _j(x,t,\lambda )\mathrm {e}^{Q_j(x,t,\lambda )}$ and thus $M(x,t,\lambda )$ can be written as

$$\begin{aligned} M(x,t,\lambda )= \left( \frac{\Phi _1^{(1)}(x,t,\lambda )}{s_{11}(\lambda )},\Phi _2^{(2)}(x,t,\lambda )\right) \mathrm {e}^{p_2(x,t,\lambda )\sigma _3}. \end{aligned}$$

(3.3b)

It follows that $\det M\equiv 1$.

3.1.1 Jump matrix

Since $(D_1^{-1}{{\tilde{\Phi }}}_1^{(1)})(\lambda )$ is analytic in ${\mathbb {C}}{\setminus }\Sigma _1$, the limiting values $M^\pm$ of M as $\lambda$ approaches $\Sigma _2$ from ${{\mathbb {C}}}^\pm$ can be expressed as follows:

$$\begin{aligned}&M^\pm (x,t,\lambda ){:}{=}M(x,t,\lambda _\pm )= \left( \frac{(D_1^{-1}{{\tilde{\Phi }}}_1^{(1)})(x,t,\lambda _\pm )}{s_{11}(\lambda _\pm )\mathrm {e}^{p_1(x,t,\lambda _\pm )-p_2(x,t,\lambda _\pm )}},(D_2^{-1}{{\tilde{\Phi }}}_2^{(2)})(x,t,\lambda _\pm )\right) \\&,\quad \lambda \in {{\dot{\Sigma }}}_1, \\&M^\pm (x,t,\lambda ){:}{=}M(x,t,\lambda _\pm )= \left( \frac{(D_1^{-1}{{\tilde{\Phi }}}_1^{(1)})(x,t,\lambda )}{s_{11}(\lambda _\pm )\mathrm {e}^{p_1(x,t,\lambda )-p_2(x,t,\lambda _\pm )}},(D_2^{-1}{{\tilde{\Phi }}}_2^{(2)})(x,t,\lambda _\pm )\right) \\&,\quad \lambda \in {{\dot{\Sigma }}}_0. \end{aligned}$$

Proposition 3.1

$M^+$ and $M^-$ are related as follows:

$$\begin{aligned}M^+(x,t,\lambda ) =M^-(x,t,\lambda ) J(x,t,\lambda ), \quad \lambda \in {{\dot{\Sigma }}}_1\cup {{\dot{\Sigma }}}_0,\end{aligned}$$

where

$$\begin{aligned} J(x,t,\lambda )=\begin{pmatrix} 0&{}\quad \mathrm {i}\\ \mathrm {i}&{}\quad 0 \end{pmatrix}\begin{pmatrix} \mathrm {e}^{-p_2(x,t,\lambda _+)}&{}\quad 0\\ 0&{}\quad \mathrm {e}^{p_2(x,t,\lambda _+)} \end{pmatrix}J_0(\lambda ) \begin{pmatrix} \mathrm {e}^{p_2(x,t,\lambda _+)}&{}\quad 0\\ 0&{}\quad \mathrm {e}^{-p_2(x,t,\lambda _+)} \end{pmatrix} \end{aligned}$$

(3.4a)

with

$$\begin{aligned} J_0(\lambda )={\left\{ \begin{array}{ll} \begin{pmatrix} 1-|\rho (\lambda )|^2&{}\quad -\overline{\rho (\lambda )}\\ \rho (\lambda )&{}\quad 1 \end{pmatrix} , \quad \lambda \in {{\dot{\Sigma }}}_1,\\ \begin{pmatrix} 0&{}\quad -\frac{1}{\rho (\lambda )}\\ \rho (\lambda )&{}\quad 1 \end{pmatrix}, \quad \lambda \in {{\dot{\Sigma }}}_0. \end{array}\right. } \end{aligned}$$

(3.4b)

Proof

(i) $\lambda \in {{\dot{\Sigma }}}_1$. Considering (2.18) columnwise, rearranging the columns and using (2.53a) for $\lambda \in {{\dot{\Sigma }}}_1$, we obtain

$$\begin{aligned} M^+(x,t,\lambda ) =M^-(x,t,\lambda )\mathrm {i}\begin{pmatrix} \frac{{{\tilde{s}}}_{21}(x,t,\lambda _+)\tilde{s}_{11}(x,t,\lambda _-)}{{{\tilde{s}}}_{11}(x,t,\lambda _+)\tilde{s}_{22}(x,t,\lambda _+)}&{}\quad \frac{{{\tilde{s}}}_{11}(x,t,\lambda _-)}{\tilde{s}_{22}(x,t,\lambda _+)}\\ 1-\frac{{{\tilde{s}}}_{21}(x,t,\lambda _+)\tilde{s}_{12}(x,t,\lambda _+)}{{{\tilde{s}}}_{11}(x,t,\lambda _+)\tilde{s}_{22}(x,t,\lambda _+)}&{}\quad -\frac{{{\tilde{s}}}_{12}(x,t,\lambda _+)}{\tilde{s}_{22}(x,t,\lambda _+)} \end{pmatrix}. \end{aligned}$$

(3.5)

Since $\mathrm {e}^{p_1(x,t,\lambda _-)-p_2(x,t,\lambda _-)}=\mathrm {e}^{p_2(x,t,\lambda _+)-p_1(x,t,\lambda _+)}$, from (2.41) and (2.45a) we have $\frac{{{\tilde{s}}}_{11}(\lambda _-)}{{{\tilde{s}}}_{22}(\lambda _+)}=\frac{ s_{11}(\lambda _-)}{ s_{22}(\lambda _+)}=1$. Moreover, using the definition (3.1) of $\rho (\lambda )$ and (2.45), we have $\overline{\rho (\lambda )}=\frac{s_{12}(\lambda _+)}{s_{22}(\lambda _+)}$. Hence we can rewrite the jump condition (3.5) as (3.4a) with (3.4b).

(ii) $\lambda \in {{\dot{\Sigma }}}_0$. Considering (2.22) columnwise, rearranging the columns and using (2.53b) and (2.53c) for $\lambda _+\in {{\dot{\Sigma }}}_0$, we obtain

$$\begin{aligned} M^+(x,t,\lambda ) =M^-(x,t,\lambda )\mathrm {i}\begin{pmatrix} \frac{{{\tilde{s}}}_{21}(x,t,\lambda _+)}{{{\tilde{s}}}_{11}(x,t,\lambda _+)}&{}\quad 1\\ 0&{}\quad -\frac{{{\tilde{s}}}_{11}(x,t,\lambda _+)}{{{\tilde{s}}}_{21}(x,t,\lambda _+)} \end{pmatrix}. \end{aligned}$$

(3.6)

Then, using the definition of $\rho (\lambda )$ together with (2.49c) and (2.49b), we can rewrite the jump condition (3.6) as (3.4a) with (3.4b). $\square$

Remark 3.2

Notice that

$$\begin{aligned} \det J \equiv 1 \end{aligned}$$

(3.7)

and that $J_0(\lambda )$ (and hence J) is continuous at $\pm \frac{1}{A_1}$ if $|\rho (\pm \frac{1}{A_1})|=1$ and $\rho (\pm \frac{1}{A_1}+0)=\rho (\pm \frac{1}{A_1}-0)$, and discontinuous otherwise.

3.1.2 Normalization condition at $\lambda \rightarrow \infty$

Proposition 3.3

As $\lambda \rightarrow \infty$:

$$\begin{aligned} M(x,t,\lambda )={\left\{ \begin{array}{ll}\sqrt{\frac{1}{2}} \begin{pmatrix} -1&{}\quad \mathrm {i}\\ \mathrm {i}&{}\quad -1 \end{pmatrix}+ \mathrm {O}\left( \frac{1}{\lambda }\right) , \quad \lambda \rightarrow \infty ,~ \lambda \in {\mathbb {C}}^+,\\ \sqrt{\frac{1}{2}} \begin{pmatrix} 1&{}\quad \mathrm {i}\\ \mathrm {i}&{}\quad 1 \end{pmatrix}+ \mathrm {O}\left( \frac{1}{\lambda }\right) , \quad \lambda \rightarrow \infty ,~ \lambda \in {\mathbb {C}}^-. \end{array}\right. } \end{aligned}$$

(3.8)

Proof

Expanding $D_j^{-1}(\lambda )$ (2.6) as $\lambda \rightarrow \infty$, we get

$$\begin{aligned}D_j^{-1}(\lambda )={\left\{ \begin{array}{ll}\sqrt{\frac{1}{2}} \begin{pmatrix} -1&{}\quad \mathrm {i}\\ \mathrm {i}&{}\quad -1 \end{pmatrix}+\mathrm {O}\left( \frac{1}{\lambda }\right) , \quad \lambda \rightarrow \infty ,~ \lambda \in {\mathbb {C}}^+,\\ \sqrt{\frac{1}{2}} \begin{pmatrix} 1&{}\quad \mathrm {i}\\ \mathrm {i}&{}\quad 1 \end{pmatrix}+\mathrm {O}\left( \frac{1}{\lambda }\right) , \quad \lambda \rightarrow \infty ,~ \lambda \in {\mathbb {C}}^-. \end{array}\right. }\end{aligned}$$

Recalling that $\left( {\begin{matrix} {{\widetilde{\Phi }}}_1^{(1)}&{{\widetilde{\Phi }}}_2^{(2)}\end{matrix}}\right) \rightarrow I$ as $\lambda \rightarrow \infty$, we have, for $\lambda \in {\mathbb {C}}^+$,

$$\begin{aligned} \Big (D_1^{-1}{{\tilde{\Phi }}}_1^{(1)}\Big )(\lambda )=\sqrt{\frac{1}{2}}\begin{pmatrix} -1\\ \mathrm {i}\end{pmatrix}+\mathrm {O}\Big (\frac{1}{\lambda }\Big ), \quad \lambda \rightarrow \infty ,\\ \Big (D_2^{-1}{{\tilde{\Phi }}}_2^{(2)}\Big )(\lambda )=\sqrt{\frac{1}{2}}\begin{pmatrix} \mathrm {i}\\ -1 \end{pmatrix}+\mathrm {O}\left( \frac{1}{\lambda }\right) , \quad \lambda \rightarrow \infty .\end{aligned}$$

Substituting this into (2.20a), we get $\tilde{s}_{11}(\lambda )=1+\mathrm {O}\left( \frac{1}{\lambda }\right) ,~\lambda \rightarrow \infty .$

Similarly, for $\lambda \in {\mathbb {C}}^-$ we have

$$\begin{aligned}\Big (D_1^{-1}{{\tilde{\Phi }}}_1^{(1)}\Big )(\lambda )=\sqrt{\frac{1}{2}}\begin{pmatrix} 1\\ \mathrm {i}\end{pmatrix}+\mathrm {O}\Big (\frac{1}{\lambda }\Big ), \quad \lambda \rightarrow \infty ,\\(D_2^{-1}{{\tilde{\Phi }}}_2^{(2)}\Big )(\lambda )=\sqrt{\frac{1}{2}}\begin{pmatrix} \mathrm {i}\\ 1 \end{pmatrix}+\mathrm {O}\Big (\frac{1}{\lambda }\Big ), \quad \lambda \rightarrow \infty ,\end{aligned}$$

and ${{\tilde{s}}}_{11}(\lambda )=1+\mathrm {O}\left( \frac{1}{\lambda }\right) ,~\lambda \rightarrow \infty$. Then the claim follows. $\square$

Remark 3.4

In order to have a standard normalisation as $\lambda \rightarrow \infty$, we can introduce

$$\begin{aligned} {{\tilde{M}}}(x,t,\lambda ):={\left\{ \begin{array}{ll}\sqrt{\frac{1}{2}} \begin{pmatrix} -1&{}\quad -\mathrm {i}\\ -\mathrm {i}&{}\quad -1 \end{pmatrix}M(x,t,\lambda ), \quad \lambda \in {\mathbb {C}}^+,\\ \sqrt{\frac{1}{2}} \begin{pmatrix} -1&{}\quad -\mathrm {i}\\ -\mathrm {i}&{}\quad -1 \end{pmatrix}M(x,t,\lambda )\mathrm {i}\sigma _1, \quad \lambda \in {\mathbb {C}}^-. \end{array}\right. } \end{aligned}$$

(3.9)

Then we have ${{\tilde{M}}}\rightarrow I$ at $\lambda \rightarrow \infty$. On the other hand, ${{\tilde{M}}}$ acquires an additional jump across $\lambda \in {\mathbb {R}}{\setminus }\Sigma _2$:

$$\begin{aligned} {{\tilde{M}}}^+(x,t,\lambda ) ={{\tilde{M}}}^-(x,t,\lambda ) \tilde{J}(x,t,\lambda ),~ \lambda \in {\mathbb {R}}{\setminus }\big \{\cup _{j=1,2} \{A_j^{-1}\}\cup \{-A_j^{-1}\}\big \} \end{aligned}$$

with

$$\begin{aligned} {{\tilde{J}}}(x,t,\lambda )={\left\{ \begin{array}{ll} {{\tilde{J}}}_{\Sigma _j}(x,t,\lambda ), \quad \lambda \in {{\dot{\Sigma }}}_j,\quad j=0,1\\ {{\tilde{J}}}_{{\mathbb {R}}{\setminus }\Sigma _2}(x,t,\lambda ), \quad \lambda \in {\mathbb {R}}{\setminus }\Sigma _2, \end{array}\right. } \end{aligned}$$

where $\tilde{J}_{\Sigma _j}(x,t,\lambda )=\mathrm {e}^{-p_2(x,t,\lambda _+)\sigma _3}J_0(\lambda )\mathrm {e}^{p_2(x,t,\lambda _+)\sigma _3}$, $j=0,1$ and ${{\tilde{J}}}_{{\mathbb {R}}{\setminus }\Sigma _2}(x,t,\lambda )=-\mathrm {i}\sigma _1$.

Remark 3.5

Using (2.20b), we obtain $\tilde{s}_{21}(\lambda )=\mathrm {O}\left( \frac{1}{\lambda }\right)$ as $\lambda \rightarrow \infty$. Notice that $\rho (\lambda )=\frac{s_{21}(\lambda _+)}{s_{11}(\lambda _+)}=\frac{\tilde{s}_{21}(\lambda _+)}{\tilde{s}_{11}(\lambda _+)}\mathrm {e}^{-2p_2(x,t,\lambda _+)}$; since $p_2(x,t,\lambda _+)$ is purely imaginary for $\lambda \in \Sigma _2$, $\mathrm {e}^{-2p_2(x,t,\lambda _+)}$ is bounded and thus $\rho (\lambda )=\mathrm {O}\Big (\frac{1}{\lambda }\Big )$ as $\lambda \rightarrow \infty$. Consequently,

$$\begin{aligned} J_{0}(\lambda )=\begin{pmatrix} 1&{}\quad 0\\ 0&{}\quad 1 \end{pmatrix}+\mathrm {O}\Big (\frac{1}{\lambda }\Big ),\quad \lambda \rightarrow \pm \infty \end{aligned}$$

and

$$\begin{aligned} J(x,t,\lambda )=\begin{pmatrix} 0&{}\quad \mathrm {i}\\ \mathrm {i}&{}\quad 0 \end{pmatrix}+\mathrm {O}\Big (\frac{1}{\lambda } \Big ),\quad \lambda \rightarrow \pm \infty . \end{aligned}$$

3.1.3 Symmetries

From the symmetry properties of the eigenfunctions and scattering functions (2.4), (2.39), (2.30), and (2.43) it follows that

$$\begin{aligned} M(-\lambda )&=-\sigma _3 M(\lambda )\sigma _3,\qquad \overline{M({\overline{\lambda }})}=-M(\lambda ),\qquad \lambda \in {\mathbb {C}}{\setminus }\Sigma _2, \end{aligned}$$

(3.10a)

$$\begin{aligned} M((-\lambda )_-)&=-\sigma _3 M(\lambda _+)\sigma _3,\qquad \overline{M(\lambda _-)}=-M(\lambda _+),\qquad \lambda \in {{\dot{\Sigma }}}_1. \end{aligned}$$

(3.10b)

where $M(\lambda )\equiv M(x,t,\lambda )$.

3.1.4 Singularities at $\pm \frac{1}{A_j}$

Let $A^{(ij)}$ denote the elements of a $2\times 2$ matrix $A=\begin{pmatrix} A^{(11)}&{}\quad A^{(12)}\\ A^{(21)}&{}\quad A^{(22)} \end{pmatrix}$.

Proposition 3.6

$M(x,t,\lambda )$ has the following behaviour at the branch points

$$\begin{aligned} M(x,t,\lambda )={\left\{ \begin{array}{ll} \frac{\mathrm {e}^{\frac{3\pi \mathrm {i}}{4}}}{\nu _2^+(\lambda )}\begin{pmatrix} 0&{}\quad \Upsilon _2\\ 0&{}\quad \Lambda _2 \end{pmatrix}+\mathrm {O}(1),\quad \lambda \rightarrow \frac{1}{A_2},\\ \frac{\mathrm {i}}{\nu _2^-(\lambda )}\begin{pmatrix} 0&{}\quad \Upsilon _2\\ 0&{}\quad -\Lambda _2 \end{pmatrix}+\mathrm {O}(1),\quad \lambda \rightarrow -\frac{1}{A_2},\\ \frac{c_+\mathrm {e}^{\frac{3\pi \mathrm {i}}{4}}}{\nu _1^+(\lambda )}\begin{pmatrix} \Upsilon _1&{}\quad 0\\ \Lambda _1&{}\quad 0 \end{pmatrix}+\mathrm {O}(1),\quad \lambda \rightarrow \frac{1}{A_1},~ \lambda \in {\mathbb {C}}_+,\\ \frac{\overline{c_+}\mathrm {e}^{\frac{3\pi \mathrm {i}}{4}}}{\nu _1^+(\lambda )}\begin{pmatrix} \Upsilon _1&{}\quad 0\\ \Lambda _1&{}\quad 0 \end{pmatrix}+\mathrm {O}(1),\quad \lambda \rightarrow \frac{1}{A_1},~ \lambda \in {\mathbb {C}}_-,\\ \frac{\overline{c_+}\mathrm {i}}{\nu _1^-(\lambda )}\begin{pmatrix} -\Upsilon _1&{}\quad 0\\ \Lambda _1&{}\quad 0 \end{pmatrix}+\mathrm {O}(1),\quad \lambda \rightarrow -\frac{1}{A_1},~ \lambda \in {\mathbb {C}}_+,\\ \frac{c_+\mathrm {i}}{\nu _1^-(\lambda )}\begin{pmatrix} -\Upsilon _1&{}\quad 0\\ \Lambda _1&{}\quad 0 \end{pmatrix}+\mathrm {O}(1),\quad \lambda \rightarrow -\frac{1}{A_1},~ \lambda \in {\mathbb {C}}_-, \end{array}\right. } \end{aligned}$$

(3.11)

where $\nu _j^\pm (\lambda )$ are defined in Proposition 2.17, and $\Upsilon _j=-(2A_j)^{\frac{1}{4}}\alpha _j(x,t)+\frac{(a_j(x,t)+b_j(x,t))}{(2A_j)^{\frac{1}{4}}}$, $\Lambda _j=(2A_j)^{\frac{1}{4}}\alpha _j(x,t)+\frac{(a_j(x,t)+b_j(x,t))}{(2A_j)^{\frac{1}{4}}}$ with $\alpha _j(x,t),~a_j(x,t),~b_j(x,t)\in \mathbb {R}$, $j=1,2$ as in Proposition 2.16.

Moreover, $c_+(x,t)=0$ if $\beta _1(x,t)\ne 0$ and $c_+(x,t)=\frac{1}{{{\tilde{s}}}_{11}(x,t,\frac{1}{A_2})}$ if $\beta _1(x,t)= 0$, where $\beta _1(x,t)$ is defined in (3.12b).

Proof

Combining Proposition 2.16 with Proposition 2.17 we get

$$\begin{aligned} D_j^{-1}(\lambda ){{\tilde{\Phi }}}_j(x,t,\lambda )=&\frac{\mathrm {e}^{\frac{3\pi \mathrm {i}}{4}}}{\nu ^+_j(\lambda )}\left( -(2A_j)^{\frac{1}{4}}\alpha _j\begin{pmatrix} 1&{}\quad 1\\ -1&{}\quad -1 \end{pmatrix}+\frac{a_j+b_j}{(2A_j)^{\frac{1}{4}}}\begin{pmatrix} 1&{}\quad 1\\ 1&{}\quad 1 \end{pmatrix}\right) \\&+\mathrm {O}\left( (\lambda -\frac{1}{A_j})^{1/4}\right) \end{aligned}$$

as $\lambda \rightarrow \frac{1}{A_j}$, where $\alpha _j=\alpha _j(x,t)$, $a_j=a_j(x,t)$ and $b_j=b_j(x,t)$.

First, consider the behaviour of M near $\frac{1}{A_2}$. Since $D_1^{-1}(\lambda ){{\tilde{\Phi }}}_1^{(1)}(x,t,\lambda )$ is analytic at $\frac{1}{A_2}$, we have

$$\begin{aligned} D_1^{-1}\left( \frac{1}{A_2}\right) {{\tilde{\Phi }}}_1^{(1)}\left( x,t,\frac{1}{A_2}\right) =\mathrm {i}\begin{pmatrix} a(x,t)\\ c(x,t) \end{pmatrix}\end{aligned}$$

with

$$\begin{aligned}a(x,t)=\left| \sqrt{\frac{A_2+|\sqrt{A_2^2-A_1^2}|}{|\sqrt{A_2^2-A_1^2}|}}\right| \left( \frac{A_1}{A_2+|\sqrt{A_2^2-A_1^2}|}{{\tilde{\Phi }}}_1^{(11)}\left( x,t,\frac{1}{A_2}\right) +{{\tilde{\Phi }}}_1^{(21)}\left( x,t,\frac{1}{A_2}\right) \right) \end{aligned}$$

and

$$\begin{aligned}c(x,t)=\left| \sqrt{\frac{A_2+|\sqrt{A_2^2-A_1^2}|}{|\sqrt{A_2^2-A_1^2}|}}\right| \left( \frac{A_1}{A_2+|\sqrt{A_2^2-A_1^2}|}{{\tilde{\Phi }}}_1^{(21)}\left( x,t,\frac{1}{A_2}\right) +{{\tilde{\Phi }}}_1^{(11)}\left( x,t,\frac{1}{A_2}\right) \right) .\end{aligned}$$

Then, using (2.20a), we get the following expansion of ${{\tilde{s}}}_{11}(x,t,\lambda )$ at $\frac{1}{A_2}$:

$$\begin{aligned} \tilde{s}_{11}(x,t,\lambda )=\frac{\mathrm {i}\mathrm {e}^{\frac{3\pi \mathrm {i}}{4}}}{\nu _2^+(\lambda )}\beta _2(x,t)+\mathrm {O}(1), \quad \lambda \rightarrow \frac{1}{A_2} \end{aligned}$$

with $\beta _2(x,t)=\left( (2A_2)^{\frac{1}{4}}\alpha _2(x,t)(a(x,t)+c(x,t))+\frac{(a_2(x,t)+b_2(x,t))(a(x,t)-c(x,t))}{(2A_2)^{\frac{1}{4}}}\right)$.

Notice that the symmetry (2.38) implies that ${{\tilde{\Phi }}}_1^{(11)}(x,t,\frac{1}{A_2})$ and ${{\tilde{\Phi }}}_1^{(21)}(x,t,\frac{1}{A_2})$ are real-valued and thus $a(x,t)\in {\mathbb {R}}$ and $c(x,t)\in {\mathbb {R}}$.

Recall the assumption $s_{11}(\frac{1}{A_2})\ne 0$, which implies ${{\tilde{s}}}_{11}(\frac{1}{A_2})\ne 0$. Thus there are two possibilities: either $\beta _2(x,t)\ne 0$ or $\beta _2(x,t)=0$ and ${{\tilde{s}}}_{11}(\frac{1}{A_2})=:\gamma \ne 0$. In the both cases,

$$\begin{aligned} M(x,t,\lambda )=\frac{\mathrm {e}^{\frac{3\pi \mathrm {i}}{4}}}{\nu _2^+(\lambda )}\begin{pmatrix} 0&{}\quad -(2A_2)^{\frac{1}{4}}\alpha _2(x,t)+\frac{(a_2(x,t)+b_2(x,t))}{(2A_2)^{\frac{1}{4}}}\\ 0&{}\quad (2A_2)^{\frac{1}{4}}\alpha _2(x,t)+\frac{(a_2(x,t)+b_2(x,t))}{(2A_2)^{\frac{1}{4}}} \end{pmatrix}+\mathrm {O}(1),\quad \lambda \rightarrow \frac{1}{A_2}.\end{aligned}$$

Now consider the behaviour of M as $\lambda$ approaches $\frac{1}{A_1}$ from the upper half-plane. Since $D_2^{-1}(\lambda ){{\tilde{\Phi }}}_2^{(2)}(x,t,\lambda )$ has no singularity at $\frac{1}{A_1}$, we have

$$\begin{aligned} D_2^{-1}\left( \frac{1}{A_1}_+\right) {{\tilde{\Phi }}}_2^{(2)}\left( x,t,\frac{1}{A_1}_+\right) =\begin{pmatrix} b_+(x,t)\\ d_+(x,t) \end{pmatrix}\end{aligned}$$

with

$$\begin{aligned}b_+=\left| \sqrt{\frac{-\mathrm {i}A_1-|\sqrt{A_2^2-A_1^2}|}{|\sqrt{A_2^2-A_1^2}|}}\right| \left( \frac{A_2}{A_1-\mathrm {i}|\sqrt{A_2^2-A_1^2}|}{{\tilde{\Phi }}}_2^{(12)}\left( x,t,\frac{1}{A_1}_+\right) +{{\tilde{\Phi }}}_2^{(22)}\left( x,t,\frac{1}{A_1}_+\right) \right) \end{aligned}$$

and

$$\begin{aligned}d_+=\left| \sqrt{\frac{-\mathrm {i}A_1-|\sqrt{A_2^2-A_1^2}|}{|\sqrt{A_2^2-A_1^2}|}}\right| \left( \frac{A_2}{A_1-\mathrm {i}|\sqrt{A_2^2-A_1^2}|}{{\tilde{\Phi }}}_2^{(22)}\left( x,t,\frac{1}{A_1}_+\right) +{{\tilde{\Phi }}}_2^{(12)}\left( x,t,\frac{1}{A_1}_+\right) \right) .\end{aligned}$$

Then, using (2.20a), we get the following expansion of ${{\tilde{s}}}_{11}(x,t,\lambda )$ at $\frac{1}{A_1}$ in the upper half-plane:

$$\begin{aligned} \tilde{s}_{11}(x,t,\lambda )=\frac{\mathrm {e}^{\frac{3\pi \mathrm {i}}{4}}}{\nu _1^+(\lambda )}\beta _1(x,t)+\mathrm {O}(1), \quad \lambda \rightarrow \frac{1}{A_1},\quad \lambda \in {\mathbb {C}}_+ \end{aligned}$$

(3.12a)

with

$$\begin{aligned} \beta _1(x,t)&=-(2A_2)^{\frac{1}{4}}\alpha _1(x,t)(b_+(x,t)+d_+(x,t))\nonumber \\&\quad +\frac{(a_1(x,t)+b_1(x,t))(d_+(x,t)-b_+(x,t))}{(2A_1)^{\frac{1}{4}}}. \end{aligned}$$

(3.12b)

As above, we have two possibilities: either $\beta _1(x,t)\ne 0$ (generic case) or $\beta _1(x,t)= 0$ and $\tilde{s}_{11}(\frac{1}{A_1}_+)=\gamma _1^+\ne 0$. This gives

$$\begin{aligned} M(x,t,\lambda )&=\frac{c_+\mathrm {e}^{\frac{3\pi \mathrm {i}}{4}}}{\nu _1^+(\lambda )}\begin{pmatrix} -(2A_1)^{\frac{1}{4}}\alpha _1(x,t)+\frac{(a_1(x,t)+b_1(x,t))}{(2A_1)^{\frac{1}{4}}}&{}\quad 0\\ (2A_1)^{\frac{1}{4}}\alpha _1(x,t)+\frac{(a_1(x,t)+b_1(x,t))}{(2A_1)^{\frac{1}{4}}}&{}\quad 0 \end{pmatrix}\\&\quad +\mathrm {O}(1),\quad \lambda \rightarrow \frac{1}{A_1},~ \lambda \in {\mathbb {C}}_+,\end{aligned}$$

where $c_+=0$ if $\beta _1(x,t)\ne 0$, and $c_+=\frac{1}{\tilde{s}_{11}(\frac{1}{A_1}_+)}$ if $\beta _1(x,t)= 0$.

The other the statements follow from the symmetry considerations. $\square$

Remark 3.7

• (1) $\rho (\lambda )=\frac{{{\tilde{s}}}_{21}(\lambda _+)}{\tilde{s}_{11}(\lambda _+)}\mathrm {e}^{-2p_2(x,t,\lambda _+)}=\mathrm {O}(1)$ as $\lambda \rightarrow \frac{1}{A_2}$. Indeed, in the proof of the Proposition 3.6, we have seen that $\tilde{s}_{11}(x,t,\lambda )=\frac{\mathrm {i}\mathrm {e}^{\frac{3\pi \mathrm {i}}{4}}}{\nu _2^+(\lambda )}\beta _2(x,t)+\mathrm {O}(1)$ as $\lambda \rightarrow \frac{1}{A_2}$. Analogously, due to (2.20b), we have $\tilde{s}_{21}(x,t,\lambda )=-\frac{\mathrm {i}\mathrm {e}^{\frac{3\pi \mathrm {i}}{4}}}{\nu _2^+(\lambda )}\beta _2(x,t)+\mathrm {O}(1)$ as $\lambda \rightarrow \frac{1}{A_2}$. Moreover, by our assumptions, $\tilde{s}_{11}(\frac{1}{A_2})\ne 0$, and hence the claim follows.

• (2) $\rho (\lambda )=\mathrm {O}(1)$ as $\lambda \rightarrow \frac{1}{A_1}$. Indeed, we already know that $\tilde{s}_{11}(\lambda )=\frac{\mathrm {e}^{\frac{3\pi \mathrm {i}}{4}}}{\nu _1^+(\lambda )}\beta _1(x,t)+\mathrm {O}(1)$ as $\lambda \rightarrow \frac{1}{A_1}$, $\lambda \in {\mathbb {C}}_+$. Analogously, (2.20b) together with (2.52) implies that if $\beta _1\ne 0$ we have $\tilde{s}_{21}(\lambda )=\frac{\mathrm {i}\mathrm {e}^{\frac{3\pi \mathrm {i}}{4}}}{\nu _1^+(\lambda )}\overline{\beta _1(x,t)}+\mathrm {O}(1)$, $\lambda \rightarrow \frac{1}{A_1}$. Moreover, by our assumptions, ${{\tilde{s}}}_{11}(\frac{1}{A_1}_+)\ne 0$, and hence the claim follows.

3.1.5 Residue conditions

By (2.17), zeros of ${{\tilde{s}}}_{11}(\lambda )$ coincide with zeros $s_{11}(\lambda )$; hence, by Proposition 2.15, they are real and simple. Moreover, the symmetry (2.24) implies that $-\lambda _k$ is a zero of ${{\tilde{s}}}_{11}(\lambda )$ together with $\lambda _k$; we will denote the set of zeros of $s_{11}(\lambda )$ by $\{\lambda _k,-\lambda _k\}_1^n$, where $\lambda _k\in (0,\frac{1}{A_2})$.

Proposition 3.8

$M^{(1)}$ has simple poles at $\{\lambda _k,-\lambda _k\}_1^n$. Moreover,

$$\begin{aligned} {{\,\mathrm{Res}\,}}_{\pm \lambda _k}M^{(1)}(x,t,\lambda )=\frac{b_k}{s'_{11}(\lambda _k)}\mathrm {e}^{2p_2(\lambda _k)}M^{(2)}(x,t,\pm \lambda _k), \end{aligned}$$

(3.13)

Moreover, $\frac{b_k}{s'_{11}(\lambda _k)}\mathrm {e}^{2p_2(\lambda _k)}\in {\mathbb {R}}$.

Proof

Recall that $\Phi _1^{(1)}(\lambda _k)=b_k\Phi _2^{(2)}(\lambda _k)$ with $b_k=b(\lambda _k)\in {\mathbb {R}}$ due the symmetry (2.36). Then $(D_1^{-1}{{\tilde{\Phi }}}_1^{(1)})(\lambda _k)$ and $(D_2^{-1}{{\tilde{\Phi }}}_2^{(2)})(\lambda _k)$ are related as

$$\begin{aligned} \frac{(D_1^{-1}\tilde{\Phi }_1^{(1)})(\lambda _k)}{s_{11}(\lambda _k) \mathrm {e}^{p_1(\lambda _k)-p_2(\lambda _k)}}=\frac{b_k}{s_{11}(\lambda _k)} \mathrm {e}^{2p_2(\lambda _k)}(D_2^{-1}\tilde{\Phi }_2^{(2)})(\lambda _k), \end{aligned}$$

and hence (3.13) follows. Moreover, differentiating (2.37) and using the fact that $\lambda _k\in {\mathbb {R}}$, we get $s'_{11}(\lambda _k)\in {\mathbb {R}}$, and thus $\frac{b_k}{s'_{11}(\lambda _k)}\mathrm {e}^{2p_2(\lambda _k)}\in {\mathbb {R}}$.

Differentiating (2.24), we get $s'_{11}(\lambda _k)=-s'_{11}(-\lambda _k)$. On the other hand, (2.23) implies that $b(-\lambda _k)=-b(\lambda _k)$. Combining these facts, we obtain (3.13) with the minus sign. $\square$

Remark 3.9

In terms of ${{\tilde{M}}}$ (3.9), the residue conditions take the following form:

$$\begin{aligned}&{{\tilde{M}}}^{(1)}(x,t,\lambda )=\frac{1}{\lambda -\lambda _{k}} \frac{b_k}{s'_{11}(\lambda _k)}\mathrm {e}^{2p_2(\lambda _{k})}{{\tilde{M}}}^{(2)}(x,t,\lambda _{k+})+\mathrm {O}(1), ~\lambda \rightarrow \lambda _k,~\lambda \in {\mathbb {C}}_+, \end{aligned}$$

(3.14a)

$$\begin{aligned}&{{\tilde{M}}}^{(2)}(x,t,\lambda )=\frac{1}{\lambda -\lambda _{k}} \frac{b_k}{s'_{11}(\lambda _k)}\mathrm {e}^{2p_2(\lambda _{k})}\tilde{M}^{(1)}(x,t,\lambda _{k-})+\mathrm {O}(1),~\lambda \rightarrow \lambda _k,~\lambda \in {\mathbb {C}}_-. \end{aligned}$$

(3.14b)

3.1.6 RH problem parametrized by $\varvec{(x,t)}$

In the framework of the Riemann–Hilbert approach to nonlinear evolution equations, one interprets the jump relation, normalization condition, singularity conditions, and residue conditions as a Riemann–Hilbert problem, with the jump matrix and residue parameters determined by the initial data for the nonlinear problem in question. The considerations above imply that $M(x,t,\lambda )$ can be characterized as the solution of the following Riemann–Hilbert problem:

Find a $2\times 2$ meromorphic matrix $M(x,t,\lambda )$ that satisfies the following conditions:

• Jump condition (3.1).

• Normalization condition (3.8).

• Singularity conditions: the singularities of $M(x,t,\lambda )$ at $\pm \frac{1}{A_j}$ are of order not bigger than $\frac{1}{4}$.

• Residue conditions (if any): given $\lbrace \lambda _k, \kappa _k\rbrace _1^N$ with $\lambda _k\in (0,\frac{1}{A_2})$ and $\kappa _k\in {\mathbb {R}}{\setminus }\{0\}$, $M^{(1)}(x,t,\lambda )$ has simple poles at $\lbrace \lambda _k, -\lambda _k\rbrace _1^N$, with the residues satisfying the equations

  $$\begin{aligned} {{\,\mathrm{Res}\,}}_{\pm \lambda _k}M^{(1)}(x,t,\lambda )=\kappa _k\mathrm {e}^{2p_2(\lambda _k)}M^{(2)}(x,t,\pm \lambda _k). \end{aligned}$$

  (3.15)

Remark 3.10

The solution of the RH problem above, if exists, satisfies the following properties:

• 1. $\det M\equiv 1$ (follows from the fact that $\det J\equiv 1$).

• 2. Symmetries

  $$\begin{aligned} M(-\lambda )&=-\sigma _3 M(\lambda )\sigma _3,\qquad \overline{M({\overline{\lambda }})}=-M(\lambda ),\qquad \lambda \in {\mathbb {C}}{\setminus }\Sigma _2, \end{aligned}$$

  (3.16a)
  $$\begin{aligned} M((-\lambda )_-)&=-\sigma _3 M(\lambda _+)\sigma _3,\qquad \overline{M(\lambda _-)}=-M(\lambda _+),\qquad \lambda \in {{\dot{\Sigma }}}_1. \end{aligned}$$

  (3.16b)

  where $M(\lambda )\equiv M(x,t,\lambda )$ (follows from the respective symmetries of the jump matrix and the residue conditions, assuming the uniqueness of the solution).

Remark 3.11

We do not need to specify the singularities at the branch points $\pm \frac{1}{A_j}$ in order to formulate RH problem. It is enough to require them to be of order not bigger than $\frac{1}{4}$.

As for other Camassa–Holm-type equations, a principal drawback of the RH formalism presented above is that the jump condition (3.1) involves not only the scattering functions uniquely determined by the initial data for problem (1.1), but the solution itself, via $p_2(x,t,\lambda )$ involving m(x, t) (2.8b). In order to have the data for a RH problem to be explicitly determined by the initial data only, we introduce the space variable $y(x,t){:}{=}x-\frac{1}{A_2}\int _x^{+\infty }( m(\xi ,t)-A_2)\mathrm {d}\xi -A_2^2 t$, which will play the role of a parameter (together with t) for the RH problem, see Sect. 3.3 below.

In order to determine an efficient way for retrieving the solution of the mCH equation from the solution of the RH problem, we will use the behavior of the Jost solutions of the Lax pair equations evaluated at $\lambda =0$, for which the x-equation (2.1a) of the Lax pair becomes trivial (independent of the solution of the mCH equation).

3.2 Eigenfunctions near 0

In the case of the Camassa–Holm equation [14] as well as other CH-type nonlinear integrable equations studied so far, see, e.g., [17], the analysis of the behavior of the respective Jost solutions at a dedicated point in the complex plane of the spectral parameter (in our case, at $\lambda =0$) requires a dedicated gauge transformation of the Lax pair equations.

It is remarkable that in the case of the mCH equation, in order to control the behavior of the eigenfunctions at $\lambda =0$, we don’t need to introduce an additional transformation; all we need is to regroup the terms in the Lax pair (2.7).

Namely, we rewrite (2.7) as follows:

$$\begin{aligned} {\hat{\Phi }}_{jx}+\frac{\mathrm {i}A_j k_j(\lambda )}{2}\sigma _3{\hat{\Phi }}_{{j}}={{\hat{U}}}^0_j {\hat{\Phi }}_{{j}}, \end{aligned}$$

(3.17a)

where ${{\hat{U}}}^0_j\equiv {{\hat{U}}}^0_j(x,t,\lambda )$ is given by

$$\begin{aligned} {{\hat{U}}}^0_j=\frac{(m-A_j)}{2}\frac{\lambda }{\mathrm {i}k_j(\lambda )} \begin{pmatrix} \lambda &{}\quad \frac{1}{A_j} \\ -\frac{1}{A_j} &{}\quad -\lambda \\ \end{pmatrix}, \end{aligned}$$

(3.17b)

and

$$\begin{aligned} {\hat{\Phi }}_{jt} +\mathrm {i}A_j k_j(\lambda ) \left( -\frac{A_j^2}{2} -\frac{1}{\lambda ^2}\right) \sigma _3{\hat{\Phi }}_{j}= {{\hat{V}}}^0_j {\hat{\Phi }}_{j}, \end{aligned}$$

(3.17c)

where ${{\hat{V}}}^0_j\equiv {{\hat{V}}}^0_j(x,t,\lambda )$ is given by

$$\begin{aligned} {{\hat{V}}}^0_{j}&={{\hat{V}}}_j+\mathrm {i}A_j k_j(\lambda )\left( \frac{(u^2-u_x^2)m}{2A_j}-\frac{A_j^2}{2}\right) \sigma _3. \end{aligned}$$

(3.17d)

Further, introduce (compare with (2.8b))

$$\begin{aligned} p^0_j(x,t,\lambda ){:}{=}\frac{\mathrm {i}A_j k_j(\lambda )}{2}\left( x-2\left( \frac{A_j^2}{2} +\frac{1}{\lambda ^2}\right) t\right) . \end{aligned}$$

(3.18)

Then, introducing $Q^0_j{:}{=}p^0_j\sigma _3$ and ${{\widetilde{\Phi }}}^0_j{:}{=}{\hat{\Phi }}_{{j}} \mathrm {e}^{Q_j^0}$, equations (3.17a) and (3.17c) reduce to

$$\begin{aligned} {\left\{ \begin{array}{ll} {{\widetilde{\Phi }}}^0_{jx}+[Q^0_{jx},{{\widetilde{\Phi }}}_j^0]={{\hat{U}}}^0_j{{\widetilde{\Phi }}}_j^0,&{}\\ {{\widetilde{\Phi }}}^0_{jt}+[Q^0_{jt},{{\widetilde{\Phi }}}_j^0]={{\hat{V}}}^j_0{{\widetilde{\Phi }}}_j^0.&{} \end{array}\right. } \end{aligned}$$

(3.19)

Define the Jost solutions ${{\widetilde{\Phi }}}_{j}^0$ of (3.19) as the solutions of the integral equations

$$\begin{aligned} {{\widetilde{\Phi }}}_{j}^0(x,t,\lambda )=I+\int _{(-1)^j\infty }^x\mathrm {e}^{\frac{-\mathrm {i}A_j k_j(\lambda )}{2}(x-\xi )\sigma _3}{{\hat{U}}}_j^0(\xi ,t,\lambda ){{\widetilde{\Phi }}}_j^{0}(\xi ,t,\lambda )\mathrm {e}^{\frac{\mathrm {i}A_j k_j(\lambda )}{2}(x-\xi )\sigma _33}\mathrm {d}\xi .\nonumber \\ \end{aligned}$$

(3.20)

Further, defining ${\hat{\Phi }}_{j}^0{:}{=}{{\widetilde{\Phi }}}^0_{j}\mathrm {e}^{-p_j^0 \sigma _3}$, we observe that ${\hat{\Phi }}_{j}^0(x,t,\lambda )$ and ${\hat{\Phi }}_{{j}}(x,t,\lambda )$ satisfy the same differential equations (2.7) and thus they are related by matrices $C_j(\lambda )$ independent of x and t:

$$\begin{aligned} {\hat{\Phi }}_{j}={\hat{\Phi }}_{j}^{0}C_j(\lambda ). \end{aligned}$$

Consequently,

$$\begin{aligned} {{\widetilde{\Phi }}}_{j}(x,t,\lambda )={{\widetilde{\Phi }}}_j^{0}(x,t,\lambda )\mathrm {e}^{-p_j^0(x,t,\lambda ) \sigma _3}C_j(\lambda )\mathrm {e}^{p_j(x,t,\lambda )\sigma _3}. \end{aligned}$$

(3.21)

Since $p_j(x,t,\lambda )-p_j^0(x,t,\lambda )=\frac{\mathrm {i}k_j(\lambda )}{2 }\int _x^{(-1)^j\infty }( m(\xi ,t)-A_j)\mathrm {d}\xi$ and

$$\begin{aligned} {{\widetilde{\Phi }}}_{j}(x,t,\lambda )={{\widetilde{\Phi }}}_j^{0}(x,t,\lambda )\mathrm {e}^{\frac{\mathrm {i}k_j(\lambda )}{2 }\int ^x_{(-1)^j\infty }( m(\xi ,t)-A_j)\mathrm {d}\xi \sigma _3}, \end{aligned}$$

passing to the limits $x\rightarrow (-1)^j\infty$, we get $C_j(\lambda )=I$.

Noticing that ${{\hat{U}}}_j^0(x,t,0)\equiv 0$, it follows from (3.20) that ${{\widetilde{\Phi }}}_j^{0}(x,t,0)\equiv I$ and thus ${{\widetilde{\Phi }}}_j(x,t,0)=\mathrm {e}^{-\frac{1}{2A_j }\int ^x_{(-1)^j\infty }( m(\xi ,t)-A_j)\mathrm {d}\xi \sigma _3}$. Combining this with $D_j^{-1}(0)=\begin{pmatrix} 0&{}\quad i\\ i&{}\quad 0 \end{pmatrix}$ gives

$$\begin{aligned} (D_j^{-1}{{\widetilde{\Phi }}}_j)(x,t,0)=\mathrm {i}\begin{pmatrix} 0&{}\quad \mathrm {e}^{\frac{1}{2A_j }\int ^x_{(-1)^j\infty }( m(\xi ,t)-A_j)\mathrm {d}\xi }\\ \mathrm {e}^{-\frac{1}{2A_j }\int ^x_{(-1)^j\infty }( m(\xi ,t)-A_j)\mathrm {d}\xi }&{}\quad 0 \end{pmatrix} \end{aligned}$$

Consequently,

$$\begin{aligned} {{\tilde{s}}}_{11}(0)=\mathrm {e}^{-\frac{1}{2A_1 }\int ^x_{-\infty }( m(\xi ,t)-A_1)\mathrm {d}\xi -\frac{1}{2A_2 }\int _x^{\infty }( m(\xi ,t)-A_2)\mathrm {d}\xi } \end{aligned}$$

(hence ${{\tilde{s}}}_{11}(0)\ne 0$) and

$$\begin{aligned} M(x,t,0) = \mathrm {i}\begin{pmatrix} 0&{}\quad \mathrm {e}^{-\frac{1}{2A_2 }\int _x^{\infty }( m(\xi ,t)-A_2)\mathrm {d}\xi }\\ \mathrm {e}^{\frac{1}{2A_2 }\int _x^{\infty }( m(\xi ,t)-A_2)\mathrm {d}\xi }&{}\quad 0 \end{pmatrix}. \end{aligned}$$

(3.22)

Remark 3.12

Considering $M(x,t,\lambda )$ as the solution of the RH problem in Sect. 3.1.6, the matrix structure of M(x, t, 0) as in (3.22), i.e.,

$$\begin{aligned} M(x,t,0) = \mathrm {i}\begin{pmatrix} 0 &{}\quad a_1(x,t) \\ a_1^{-1}(x,t) &{}\quad 0 \end{pmatrix} \end{aligned}$$

(3.23)

with some $a(x,t)\in \mathbb {R}$, which follows from the symmetry properties (3.16a) of the solution taking into account that $\det M\equiv 1$ (provided the solution is unique).

In order to extract the solution of the mCH equation from the solution of the associated RH problem, it turns to be useful to find the next term in the expansion of $M(x,t,\lambda )$ at $\lambda =0$.

First, expanding $D_j^{-1}(\lambda )$ near 0, we have

$$\begin{aligned}D_j^{-1}(\lambda )=\begin{pmatrix} 0&{}\quad i\\ i&{}\quad 0 \end{pmatrix}+\lambda \begin{pmatrix} \mathrm {i}\frac{A_j}{2}&{}\quad 0\\ 0&{}\quad \mathrm {i}\frac{A_j}{2} \end{pmatrix}+\mathrm {O}(\lambda ^2).\end{aligned}$$

On the other hand, $\mathrm {e}^{\frac{\mathrm {i}k_j(\lambda )}{2 }\int ^x_{(-1)^j\infty }( m(\xi ,t)-A_j)\mathrm {d}\xi \sigma _3}=\mathrm {e}^{-\frac{ 1}{2A_j }\int ^x_{(-1)^j\infty }( m(\xi ,t)-A_j)\mathrm {d}\xi \sigma _3}+\mathrm {O}(\lambda ^2),~\lambda \rightarrow 0.$ Then, expanding ${{\widetilde{\Phi }}}_{j}^0(x,t,\lambda )$ at 0 using the Neumann series, we have

$$\begin{aligned}{{\widetilde{\Phi }}}_{j}^0(x,t,\lambda )=I+\lambda \begin{pmatrix} 0&{}\quad -\int _{(-1)^j\infty }^x e^{x-\xi }\frac{m-A_j}{2}\mathrm {d}\xi \\ \int _{(-1)^j\infty }^x e^{-(x-\xi )}\frac{m-A_j}{2}\mathrm {d}\xi &{}\quad 0 \end{pmatrix}+\mathrm {O}(\lambda ^2).\end{aligned}$$

In particular,

$$\begin{aligned} {{\tilde{s}}}_{11}(\lambda )=\mathrm {e}^{-\frac{1}{2A_1 }\int ^x_{-\infty }( m(\xi ,t)-A_1)\mathrm {d}\xi -\frac{1}{2A_2 }\int _x^{\infty }( m(\xi ,t)-A_2)\mathrm {d}\xi }+\mathrm {O}(\lambda ^2). \end{aligned}$$

Finally, we have

$$\begin{aligned} M(x,t,\lambda )=\mathrm {i}\begin{pmatrix} 0&{}\quad a_1(x,t)\\ a_1^{-1}(x,t)&{}\quad 0 \end{pmatrix}+\mathrm {i}\lambda \begin{pmatrix} a_2(x,t)&{}\quad 0\\ 0&{}\quad a_3(x,t) \end{pmatrix}+\mathrm {O}(\lambda ^2), \end{aligned}$$

(3.24)

where

$$\begin{aligned} a_1(x,t)&= \mathrm {e}^{-\frac{1}{2A_2 }\int _x^{\infty }( m(\xi ,t)-A_2)\mathrm {d}\xi }, \end{aligned}$$

(3.25a)

$$\begin{aligned} a_2(x,t)&= \left( \int _{-\infty }^x e^{-(x-\xi )}\frac{m-A_1}{2}\mathrm {d}\xi +\frac{A_1}{2}\right) \mathrm {e}^{\frac{1}{2A_2 }\int _x^{\infty }( m(\xi ,t)-A_2)\mathrm {d}\xi }, \end{aligned}$$

(3.25b)

$$\begin{aligned} a_3(x,t)&= \left( \int ^{\infty }_x e^{(x-\xi )}\frac{m-A_2}{2}\mathrm {d}\xi +\frac{A_2}{2}\right) \mathrm {e}^{-\frac{1}{2A_2 }\int _x^{\infty }( m(\xi ,t)-A_2)\mathrm {d}\xi }. \end{aligned}$$

(3.25c)

Notice that the matrix structure of terms in the r.h.s. of (3.24) is consistent with the symmetry properties (3.16a) of M.

Proposition 3.13

u(x, t) and $u_x(x,t)$ can be algebraically expressed in terms of the coefficients $a_j(x,t)$, $j=1,3$ in the development (3.24) of $M(x,t,\lambda )$ as follows:

$$\begin{aligned} u(x,t)&= a_1(x,t)a_2(x,t)+ a_1^{-1}(x,t) a_3(x,t), \end{aligned}$$

(3.26a)

$$\begin{aligned} u_x(x,t)&= -a_1(x,t)a_2(x,t)+ a_1^{-1}(x,t)a_3(x,t). \end{aligned}$$

(3.26b)

Proof

Introduce $v(x,t) := a_1(x,t)a_2(x,t)+ a_1^{-1}(x,t) a_3(x,t)$. Using (3.25) it follows that

$$\begin{aligned} v(x,t) = \frac{A_1+A_2}{2}+\int _{-\infty }^x e^{-(x-\xi )}\frac{m-A_1}{2}\mathrm {d}\xi +\int ^{\infty }_x e^{(x-\xi )}\frac{m-A_2}{2}\mathrm {d}\xi \qquad \end{aligned}$$

(3.27)

and thus, differentiating w.r.t. x,

$$\begin{aligned} v_x(x,t) = \frac{A_2-A_1}{2}-\int _{-\infty }^x e^{-(x-\xi )}\frac{m-A_1}{2}\mathrm {d}\xi +\int ^{\infty }_x e^{(x-\xi )}\frac{m-A_2}{2}\mathrm {d}\xi .\qquad \end{aligned}$$

(3.28)

Since we assume that $\lim _{x\rightarrow (-1)^j\infty }m(x,t)=A_i$, from (3.27) it follows that $v-v_{xx}=m$ and that

$$\begin{aligned} \lim _{x\rightarrow (-1)^j\infty }v(x,t)=A_i, \qquad \lim _{x\rightarrow (-1)^i\infty }v_x(x,t)=0; \end{aligned}$$

thus $v\equiv u$. Finally, we notice that the expression in the r.h.s. of (3.28) can be written as the r.h.s. of (3.26b) taking into account (3.25). $\square$

3.3 RH problem in the $\varvec{(y,t)}$ scale

As we already mentioned, the jump condition (3.1) involves not only the scattering functions uniquely determined by the initial data for problem (1.1), but the solution itself, via m(x, t), which enters the definition of $p_2(x,t,\lambda )$ (2.8b). In order to have the data for the RH problem to be explicitly determined by the initial data only, we introduce the new space variable y(x, t) by

$$\begin{aligned} y(x,t)=x-\frac{1}{A_2}\int _x^{+\infty }(m(\xi ,t)-A_2)\mathrm {d}\xi -A_2^2 t, \end{aligned}$$

(3.29)

Then, introducing ${{\hat{M}}}(y,t,\lambda )$ so that $M(x,t,\lambda )={{\hat{M}}} (y(x,t),t,\lambda )$, the dependence of the jump matrix in (3.1) on y and t as parameters becomes explicit: the jump condition for ${{\hat{M}}}(y,t,\lambda )$ has the form

$$\begin{aligned} {{\hat{M}}}^+(y,t,\lambda )={{\hat{M}}}^-(y,t,\lambda ){{\hat{J}}}(y,t,\lambda ),\quad \lambda \in {{\dot{\Sigma }}}_1\cup {{\dot{\Sigma }}}_0. \end{aligned}$$

(3.30a)

Here

$$\begin{aligned} {{\hat{J}}}(y,t,\lambda ){:}{=}\begin{pmatrix} 0&{}\quad \mathrm {i}\\ \mathrm {i}&{}\quad 0 \end{pmatrix}\begin{pmatrix} \mathrm {e}^{-{{\hat{p}}}_2(y,t,\lambda _+)}&{}\quad 0\\ 0&{}\quad \mathrm {e}^{{{\hat{p}}}_2(y,t,\lambda _+)} \end{pmatrix}J_0(\lambda ) \begin{pmatrix} \mathrm {e}^{{{\hat{p}}}_2(y,t,\lambda _+)}&{}\quad 0\\ 0&{}\quad \mathrm {e}^{-{{\hat{p}}}_2(y,t,\lambda _+)} \end{pmatrix},\nonumber \\ \end{aligned}$$

(3.30b)

where $J_0(\lambda )$ is defined by (3.4b) and $p_2$ is explicitly given in terms of y and t:

$$\begin{aligned} {{\hat{p}}}_2(y,t,\lambda ) {:}{=}\frac{\mathrm {i}A_2 k_2(\lambda )}{2}\left( y-\frac{2t}{\lambda ^2}\right) . \end{aligned}$$

(3.30c)

Similarly, the residue conditions (3.15) become explicit as well:

$$\begin{aligned} {{\,\mathrm{Res}\,}}_{\pm \lambda _k}{{\hat{M}}}^{(1)}(y,t,\lambda )=\kappa _k\mathrm {e}^{2{{\hat{p}}}_2(y,t,\lambda _k)}{{\hat{M}}}^{(2)}(y,t,\pm \lambda _k), \end{aligned}$$

(3.31)

with $\kappa _k=\frac{b_k}{s'_{11}(\lambda _k)}$.

Noticing that the normalization condition (3.8) and the singularity conditions at $\lambda =\pm \frac{1}{A_j}$ hold in the new scale (y, t), we arrive at the basic RH problem characterizing problem (1.1a).

BasicRHproblem

Given $\rho (\lambda )$ for $\lambda \in {{\dot{\Sigma }}}_1\cup {{\dot{\Sigma }}}_0$, and $\lbrace \lambda _k,\kappa _k\rbrace _1^N$ with $\lambda _k\in (0,\frac{1}{A_2})$ and $\kappa _k\in {\mathbb {R}}{\setminus }\{0\}$, associated with the initial data $u_0(x)$ in (1.1), find a piece-wise (w.r.t. ${{\dot{\Sigma }}}_2$) meromorphic, $2\times 2$-matrix valued function ${{\hat{M}}}(y,t,\lambda )$ satisfying the following conditions:

• Jump condition (3.30) across ${{\dot{\Sigma }}}_1\cup {{\dot{\Sigma }}}_0$ (with $J_0(\lambda )$ defined by (3.4b)).

• Residue conditions (3.31).

• Normalization condition:

  $$\begin{aligned} {{\hat{M}}}(y,t,\lambda )={\left\{ \begin{array}{ll}\sqrt{\frac{1}{2}} \begin{pmatrix} -1&{}\quad \mathrm {i}\\ \mathrm {i}&{}\quad -1 \end{pmatrix}+\mathrm {O}\left( \frac{1}{\lambda }\right) , \quad \lambda \rightarrow \infty ,~ \lambda \in {\mathbb {C}}^+,\\ \sqrt{\frac{1}{2}} \begin{pmatrix} 1&{}\quad \mathrm {i}\\ \mathrm {i}&{}\quad 1 \end{pmatrix}+\mathrm {O}\left( \frac{1}{\lambda }\right) , \quad \lambda \rightarrow \infty ,~ \lambda \in {\mathbb {C}}^-. \end{array}\right. } \end{aligned}$$

  (3.32)

• Singularity conditions: the singularities of ${{\hat{M}}}(y,t,\lambda )$ at $\pm \frac{1}{A_j}$ are of order not bigger than $\frac{1}{4}$.

Evaluating the solution of this problem as $\lambda \rightarrow 0$, we are able to present the solution u to the initial value problem (1.1) in a parametric form, see below. As for the data for the RH problem, the scattering matrix $s(\lambda )$ (and hence $s_{11}(\lambda )$, $s_{21}(\lambda )$, and $\rho (\lambda )$) as well as the discrete data $\left\{ \lambda _k, \kappa _k\right\} _1^n$ are determined by $u_0(x)$ via the solutions of (2.11) considered for $t=0$.

The uniqueness of the solution of the basic RH problem follows using standard arguments based on the application of Liouville’s theorem to the ratio ${{\hat{M}}}_1({{\hat{M}}}_2)^{-1}$ of two potential solutions, ${{\hat{M}}}_1$ and ${{\hat{M}}}_2$. Particularly, the singularity condition implies that the possible singularities of ${{\hat{M}}}_1({{\hat{M}}}_2)^{-1}$ are of order no bigger that 1/2 and that these singularities, being isolated, are removable.

The uniqueness, in particular, implies the symmetries

$$\begin{aligned} {{\hat{M}}}(-\lambda )&=-\sigma _3 {{\hat{M}}}(\lambda )\sigma _3,\qquad \overline{{{\hat{M}}}({\overline{\lambda }})}=-M(\lambda ),\qquad \lambda \in {\mathbb {C}}{\setminus }\Sigma _2, \end{aligned}$$

(3.33a)

$$\begin{aligned} {{\hat{M}}}((-\lambda )_-)&=-\sigma _3 {{\hat{M}}}(\lambda _+)\sigma _3,\qquad \overline{{{\hat{M}}}(\lambda _-)}=-{{\hat{M}}}(\lambda _+),\qquad \lambda \in {{\dot{\Sigma }}}_1. \end{aligned}$$

(3.33b)

where ${{\hat{M}}}(\lambda )\equiv {{\hat{M}}}(y,t,\lambda )$, which follows from the corresponding symmetries of ${{\hat{J}}}(y,t,\lambda )$.

3.4 Recovering $\varvec{u(x,t)}$ from the solution of the basic RH problem

Comparing the RH problem (3.1), (3.8), (3.15) parametrized by x and t with the RH problem (3.30)–(3.32) parametrized by y and t and using (3.25)–(3.29) we arrive at our main representation result.

Theorem 3.14

Assume that u(x, t) is the solution of the Cauchy problem (1.1) and let ${{\hat{M}}}(y,t,x)$ be the solution of the associated RH problem (3.30)–(3.32), whose data are determined by $u_0(x)$. Let

$$\begin{aligned} {{\hat{M}}}(y,t,\lambda )=\mathrm {i}\begin{pmatrix} 0&{}\quad {{\hat{a}}}_1(y,t)\\ {{\hat{a}}}_1^{-1}(y,t)&{}\quad 0 \end{pmatrix}+\mathrm {i}\lambda \begin{pmatrix} {{\hat{a}}}_2(y,t)&{}\quad 0\\ 0&{}\quad {{\hat{a}}}_3(y,t) \end{pmatrix}+\mathrm {O}(\lambda ^2) \end{aligned}$$

(3.34)

be the development of ${{\hat{M}}}(y,t,x)$ at $\lambda =0$. Then the solution u(x, t) of the Cauchy problem (1.1) can be expressed, in a parametric form, in terms of ${{\hat{a}}}_j(y,t)$, $j=1,2,3$: $u(x,t)={{\hat{u}}}(y(x,t),t)$, where

$$\begin{aligned}&{{\hat{u}}}(y,t)={{\hat{a}}}_1(y,t){{\hat{a}}}_2(y,t)+{{\hat{a}}}_1^{-1}(y,t){{\hat{a}}}_3(y,t), \end{aligned}$$

(3.35a)

$$\begin{aligned}&x(y,t)=y-2\ln {{\hat{a}}}_1(y,t)+A_2^2 t. \end{aligned}$$

(3.35b)

Additionally, ${{\hat{u}}}_x(y,t)$ can also be algebraically expressed in terms of ${{\hat{a}}}_j(y,t)$, $j=1,2,3$: $u_x(x,t)={{\hat{u}}}_x(y(x,t),t)$, where

$$\begin{aligned} {{\hat{u}}}_x(y,t)=-{{\hat{a}}}_1(y,t){{\hat{a}}}_2(y,t)+{{\hat{a}}}_1^{-1}(y,t){{\hat{a}}}_3(y,t). \end{aligned}$$

(3.35c)

Alternatively, one can express ${{\hat{u}}}_x(y,t)$ in terms of the first term in (3.34) only. The price to pay is that this expression involves the derivatives of this term.

Proposition 3.15

The x-derivative of the solution u(x, t) of the Cauchy problem (1.1) has the parametric representation

$$\begin{aligned}&{{\hat{u}}}_x(y,t)=-\frac{1}{A_2}\partial _{ty}\ln {{\hat{a}}}_1(y,t), \end{aligned}$$

(3.36a)

$$\begin{aligned}&x(y,t)=y-2\ln {{\hat{a}}}_1(y,t)+A_2^2 t. \end{aligned}$$

(3.36b)

Proof

Differentiating the identity $x(y(x,t),t)=x$ w.r.t. t gives

$$\begin{aligned} 0=\frac{d}{dt}\left( x(y(x,t),t)\right) =x_y(y,t)y_t(x,t)+x_t(y,t). \end{aligned}$$

(3.37)

From (3.29) it follows that

$$\begin{aligned} x_y(y,t)=\frac{A_2}{{{\hat{m}}}(y,t)}, \end{aligned}$$

(3.38)

where ${{\hat{m}}}(y,t)= m(x(y,t),t)$, and

$$\begin{aligned} y_t(x,t)=-\frac{1}{A_2}( u^2- u_x^2)m. \end{aligned}$$

Substituting this and (3.38) into (3.37) we obtain

$$\begin{aligned} x_t(y,t)={{\hat{u}}}^2(y,t)-{{\hat{u}}}_x^2(y,t). \end{aligned}$$

(3.39)

Further, differentiating (3.39) w.r.t. y we get

$$\begin{aligned} x_{ty}(y,t)=({{\hat{u}}}^2(y,t)-{{\hat{u}}}_x^2(y,t))_x x_y(y,t)=2A_2{{\hat{u}}}_x(y,t) \end{aligned}$$

(3.40)

and thus

$$\begin{aligned} u_x(x(y,t),t)\equiv {{\hat{u}}}_x(y,t)=\frac{1}{2A_2}\partial _{ty}x(y,t) =-\frac{1}{A_2}\partial _{ty}\ln {{\hat{a}}}_1(y,t). \end{aligned}$$

$\square$

4 Concluding remarks

We have presented the Riemann–Hilbert problem approach for the modified Camassa–Holm equation on the line with step-like boundary conditions. In the proposed formalism, we have taken the branch cut of $k_j(\lambda )$ along the half-lines $\Sigma _j$ (outer cuts), which is convenient since we extract the solution of the mCH equation exploiting the development of the solution of the RH problem at a point laying in the domain of analyticity. Notice that it is possible to formulate RH problem taking the branch cut of $k_j(\lambda )$ to be the segments $(-\frac{1}{A_j},\frac{1}{A_j})$ (inner cuts). In the case with inner cuts, the properties of Jost solutions are more conventional (two of the columns are analytic in the upper half-plane and other two in the lower half-plane), but, on the other hand, possible eigenvalues are located on the jump.

The present paper is focused on the representation results while assuming the existence of a solution of problem (1.1) in certain functional classes. To the best of our knowledge, the question of existence is still open. One of the ways to answering it is to appeal to functional analytic PDE techniques to obtain well-posedness in appropriate functional classes. However, very little is known for the cases of nonzero boundary conditions, particularly, for backgrounds having different behavior at different infinities. Since 1980s, existence problems for integrable nonlinear PDE with step-like initial conditions have been addressed using the classical Inverse Scattering Transform method [53]. A more recent progress in this direction (in the case of the Korteweg-de Vries equation) has been reported in [37, 39, 46] (see also [38]). Another way to show existence is to infer it from the RH problem formalism (see, e.g., [42] for the case of defocusing nonlinear Schrödinger equation), where a key point consists in establishing a solution of the associated RH problem and controlling its behavior w.r.t. the spatial parameter. For Camassa-Holm-type equations, where the RH problem formalism involves the change of the spatial variable, it is natural to study the existence of solution in both (x, t) and (y, t) scales. More precisely, the solvability problem splits into two problems: (i) the solvability of the RH problem parametrized by (y, t) and (ii) the bijectivity of the change of the spatial variable. Particularly, it is possible that it is the change of variables that can be responsible of the wave breaking [9, 18]. The solvability problem for problem (1.1) in the current setting will be addressed elsewhere.

Another interesting and important problem that can be addressed using the developed approach is the investigation of the large-time behavior of the solutions of the Cauchy problem (1.1) adapting the nonlinear steepest descent method.

Appendices

Appendix A: Sign-preserving property of m

Assume that $u(x,t)-A_1\in H^{3}(-\infty , a)$ and $u(x,t)-A_2\in H^{3}(a,\infty )$ for any real a and for any $t\in (0,T)$, where $T\le +\infty$ is the maximal existing time. Then Morrey’s inequality implies that $(mu_x)(s,x)$ is uniformly bounded for $0<s<t<T$, $x\in {\mathbb {R}}$.

Consider the Cauchy problem for q(t, x):

$$\begin{aligned}&\frac{\mathrm {d}q}{\mathrm {d}t}=(u^2-u_x^2)(q(t,x),t),\quad t\in (0,T), ~x\in {\mathbb {R}}, \end{aligned}$$

(A.1a)

$$\begin{aligned}&q(0,x)=x,\quad x\in {\mathbb {R}}, \end{aligned}$$

(A.1b)

where u(x, t) solves (1.1). Differentiating (A.1) with respect to x leads to

$$\begin{aligned}&\frac{\mathrm {d}}{\mathrm {d}t} q_x(t,x)=(2mu_x)(q(t,x),t)q_x(t,x), \end{aligned}$$

(A.2a)

$$\begin{aligned}&q_x(0,x)=1,\quad x\in {\mathbb {R}}. \end{aligned}$$

(A.2b)

It follows that

$$\begin{aligned} q_x(t,x)=\mathrm {e}^{2\int _{0}^{t}(mu_x)(q(s,x),s)\mathrm {d}s}>0 \end{aligned}$$

(A.3)

and, moreover,

$$\begin{aligned} \mathrm {e}^{k(t)}\le q_x(t,x)\le \mathrm {e}^{K(t)},\quad t\in [0,T) \end{aligned}$$

(A.4)

for some k(t) and K(t).

Now observe that from (1.1a) and (A.1) it follows that $\frac{\mathrm {d}}{\mathrm {d}t} \left[ m(q(t,x),t)q_x(t,x)\right] =0$. Indeed,

$$\begin{aligned} \frac{d}{dt}&\left[ m(q(t,x),t)q_x(t,x) \right] \\&= \left[ m_t(q(t,x),t)+m_x(q(t,x),t)q_t(t,x) \right] (q(t,x),t)q_x(t,x) + m(q(t,x),t)q_{tx}(t,x) \\&= \left[ -(u^2-u_x^2)_xm-(u^2-u_x^2)m_x +m_x(u^2-u_x^2) \right] (q(t,x),t)q_x(t,x)\\&\qquad +2(m^2u_x)(q(t,x),t)q_x(t,x) =0. \end{aligned}$$

Thus, due to (A.1b) and (A.2b), we have

$$\begin{aligned} m(t,q(t,x))q_x(t,x)=m(0,q(0,x))q_x(0,x)=m(0,x). \end{aligned}$$

Hence, if $m(x,0)>0$, then $m(q(t,x),t)>0$ for all $t\in [0, T)$, $x\in {\mathbb {R}}$. Since $q_x(t,x)>0$, we have that q(t, x) is strictly increasing function. Moreover, integrating (A.4) w.r.t. x, we also have $\lim _{x\rightarrow \pm \infty }q(t,x)=\pm \infty$. Hence q(x, t) is one-to-one from ${\mathbb {R}}$ onto ${\mathbb {R}}$ and thus $m(t,x)>0$ for all $t\in [0, T)$, $x\in {\mathbb {R}}$.

Appendix B: The case $A_2<A_1$

Notice that in this case $\Sigma _2\subset \Sigma _1$ and $\Sigma _0=[-\frac{1}{A_2},-\frac{1}{A_1}]\cup [\frac{1}{A_1},\frac{1}{A_2}]$.

We define $\Phi _i$ and ${{\tilde{\Phi }}}_i$ as in (2.14) and (2.12), and introduce the scattering matrices $s(\lambda _\pm )$, this time for $\lambda \in {\dot{\Sigma }}_2$, as matrices relating $\Phi _1$ and $\Phi _2$ (for brevity we keep for it the same notation s):

$$\begin{aligned}&\Phi _1(x,t,\lambda _\pm )=\Phi _2(x,t,\lambda _\pm )s(\lambda _\pm ),\qquad \lambda \in {\dot{\Sigma }}_2\end{aligned}$$

(B.1a)

with $\det s(\lambda _\pm )=1$. In turn, ${{\tilde{\Phi }}}_1$ and ${{\tilde{\Phi }}}_2$ are related by

$$\begin{aligned}&D_1^{-1}(\lambda _\pm ){{\tilde{\Phi }}}_1(x,t,\lambda _\pm )=D_2^{-1}(\lambda _\pm ){{\tilde{\Phi }}}_2(x,t,\lambda _\pm )\mathrm {e}^{-Q_2(x,t,\lambda _\pm )}s(\lambda _+)\mathrm {e}^{Q_1(x,t,\lambda _\pm )},\qquad \lambda \in {\dot{\Sigma }}_2. \end{aligned}$$

(B.2a)

The scattering coefficients $s_{ij}$ can be expressed as in (2.19). However, in this case, (2.19a) implies that $s_{11}(\lambda )$ can be analytically extended to ${\mathbb {C}}{\setminus }\Sigma _1$ and defined on the upper and lower parts of ${{\dot{\Sigma }}}_1$, and, since $\Phi _2^{(2)}$ is analytic in ${\mathbb {C}}{\setminus }\Sigma _2$ and $\Phi _1^{(2)}$ is defined on the upper and lower sides of $\Sigma _1$, $s_{12}(\lambda )$ can be extended by (2.19c) to the lower and upper sides of ${\dot{\Sigma }}_1$. Thus the following relations hold also on ${{\dot{\Sigma }}}_0$:

$$\begin{aligned} \Phi _2^{(2)}(x,t,\lambda _\pm )=s_{11}(\lambda _\pm )\Phi _1^{(2)}(x,t,\lambda _\pm )-s_{12}(\lambda _\pm )\Phi _1^{(1)}(x,t,\lambda _\pm ), \quad \lambda \in {{\dot{\Sigma }}}_0. \end{aligned}$$

(B.3a)

and, respectively,

$$\begin{aligned} (D_2^{-1}\Phi _2^{(2)})(x,t,\lambda _\pm )=\tilde{s}_{11}(x,t,\lambda _\pm )(D_1^{-1}\Phi _1^{(2)})(x,t,\lambda _\pm )-\tilde{s}_{12}(x,t,\lambda _\pm )(D_1^{-1}\Phi _1^{(1)})(x,t,\lambda _\pm ), ~\lambda \in {{\dot{\Sigma }}}_0, \end{aligned}$$

(B.4a)

where ${{\tilde{s}}}(x,t,\lambda _\pm ):=\mathrm {e}^{-Q_2(x,t,\lambda _\pm )}s(\lambda _\pm )\mathrm {e}^{Q_1(x,t,\lambda _\pm )}$.

1.1 Symmetries

The symmetries are similar to the case $A_1<A_2$. In particular,

• (1)
  $$\begin{aligned} |s_{11}(\lambda _+)|^2-|s_{12}(\lambda _+)|^2=1,\quad \lambda \in {{\dot{\Sigma }}}_2. \end{aligned}$$

  (B.5)
• (2)
  $$\begin{aligned} \big | \frac{s_{12}(\lambda _+)}{s_{11}(\lambda _+)} \big |\le 1,\quad \lambda \in {{\dot{\Sigma }}}_2 \end{aligned}$$

  (B.6)
• (3)
  $$\begin{aligned} s_{11}(\lambda _+)&=s_{22}(\lambda _-),\quad \lambda \in {{\dot{\Sigma }}}_2,\end{aligned}$$

  (B.7a)
  $$\begin{aligned} s_{11}(\lambda _+)&=\mathrm {i}s_{12}(\lambda _-),\quad \lambda \in {{\dot{\Sigma }}}_0,\end{aligned}$$

  (B.7b)
  $$\begin{aligned} s_{11}(\lambda _-)&=-\mathrm {i}s_{12}(\lambda _+),\quad \lambda \in {{\dot{\Sigma }}}_0. \end{aligned}$$

  (B.7c)
• (4)
  $$\begin{aligned} \big | \frac{s_{12}(\lambda _+)}{s_{11}(\lambda _+)} \big |=1,\quad \lambda \in {{\dot{\Sigma }}}_0 \end{aligned}$$

  (B.8)
• (5)
  $$\begin{aligned} (D_j^{-1}{{\tilde{\Phi }}}_j)((-\lambda )_-)=-\sigma _3(D_j^{-1}{{\tilde{\Phi }}}_j)(\lambda _+)\sigma _3, \quad \lambda _+\in {{\dot{\Sigma }}}_j. \end{aligned}$$

  (B.9)
• (6)
  $$\begin{aligned} \overline{(D_j^{-1}{{\tilde{\Phi }}}_j^{(j)})({\overline{\lambda }})}=-(D_j^{-1}{{\tilde{\Phi }}}_j^{(j)})(\lambda ), \quad \lambda \in {\mathbb {C}}{\setminus }\Sigma _j, \end{aligned}$$

  (B.10)
• (7)
  $$\begin{aligned} (D_1^{-1}{{\tilde{\Phi }}}_1^{(1)})(-\lambda )&=-\sigma _3(D_1^{-1}{{\tilde{\Phi }}}_1^{(1)})(\lambda ), \quad \lambda \in {\mathbb {C}}{\setminus }\Sigma _1,\end{aligned}$$

  (B.11a)
  $$\begin{aligned} (D_2^{-1}{{\tilde{\Phi }}}_2^{(2)})(-\lambda )&=\sigma _3(D_2^{-1}{{\tilde{\Phi }}}_2^{(2)})(\lambda ), \quad \lambda \in {\mathbb {C}}{\setminus }\Sigma _2. \end{aligned}$$

  (B.11b)
• (8)
  $$\begin{aligned} D_j^{-1}(\lambda _-){{\tilde{\Phi }}}_j^{(j)}(\lambda _-)&=(-\mathrm {i}D_j^{-1}(\lambda _+){{\tilde{\Phi }}}_j(\lambda _+)\sigma _1)^{(j)},\quad \lambda \in {{\dot{\Sigma }}}_2,\end{aligned}$$

  (B.12a)
  $$\begin{aligned} D_1^{-1}(\lambda _-){{\tilde{\Phi }}}_1^{(1)}(\lambda _-)&=(-\mathrm {i}D_1^{-1}(\lambda _+){{\tilde{\Phi }}}_1(\lambda _+)\sigma _1)^{(1)},\quad \lambda \in {{\dot{\Sigma }}}_0,\end{aligned}$$

  (B.12b)
  $$\begin{aligned} D_2^{-1}(\lambda _-){{\tilde{\Phi }}}_2^{(2)}(\lambda _-)&=D_2^{-1}(\lambda _+){{\tilde{\Phi }}}_2^{(2)}(\lambda _+),\quad \lambda \in {{\dot{\Sigma }}}_0. \end{aligned}$$

  (B.12c)

1.2 Discrete spectrum

It can be shown in a similar way as for the case $A_1<A_2$ that discrete spectrum is located on $(-\frac{1}{A_1},\frac{1}{A_1})$ (assuming that spectral singularities do not arise in the branch points).

1.3 RH problem parametrized by $\varvec{(x,t)}$

Notations

In this case it is convenient to introduce ${{\check{\rho }}}$ as

$$\begin{aligned} {{\check{\rho }}}(\lambda )=\frac{s_{12}(\lambda _+)}{s_{11}(\lambda _+)},\quad \lambda \in {{\dot{\Sigma }}}_2\cup {{\dot{\Sigma }}}_0. \end{aligned}$$

(B.13)

Observe that (B.6) and (B.8) imply that

$$\begin{aligned} |{{\check{\rho }}}(\lambda )|\le 1,\quad \lambda \in {{\dot{\Sigma }}}_2, \end{aligned}$$

(B.14a)

$$\begin{aligned} |{{\check{\rho }}}(\lambda )|= 1,\quad \lambda \in {{\dot{\Sigma }}}_0. \end{aligned}$$

(B.14b)

Recalling the analytic properties of eigenfunctions and scattering coefficients, we introduce the matrix-valued function

$$\begin{aligned} N(x,t,\lambda )= \left( (D_1^{-1}{{\tilde{\Phi }}}_1^{(1)})(x,t,\lambda ),\frac{(D_2^{-1}{{\tilde{\Phi }}}_2^{(2)})(x,t,\lambda )}{s_{11}(\lambda )\mathrm {e}^{p_1(x,t,\lambda )-p_2(x,t,\lambda )}}\right) ,\quad \lambda \in {\mathbb {C}}{\setminus }\Sigma _2,\nonumber \\ \end{aligned}$$

(B.15)

meromorphic in ${\mathbb {C}}{\setminus }\Sigma _2$, where $p_j$, $j=1,2$, are defined in (2.8b). Since $D_j^{-1}(\lambda ){{\tilde{\Phi }}}_j(x,t,\lambda )=\Phi _j(x,t,\lambda )\mathrm {e}^{Q_j(x,t,\lambda )}$, $N(x,t,\lambda )$ can be written as

$$\begin{aligned} N(x,t,\lambda )= \left( \Phi _1^{(1)}(x,t,\lambda ),\frac{\Phi _2^{(2)}(x,t,\lambda )}{s_{11}(\lambda )}\right) \mathrm {e}^{p_1(x,t,\lambda )\sigma _3}. \end{aligned}$$

Proceeding as in case $A_1<A_2$, we conclude that $N(x,t,\lambda )$ can be characterized as the solution of the following Riemann–Hilbert problem:

Find a $2\times 2$ meromorphic matrix $N(x,t,\lambda )$ that satisfies the following conditions:

• The jump condition

  $$\begin{aligned} N^+(x,t,\lambda )=N^-(x,t,\lambda )G(x,t,\lambda ),\quad \lambda \in {{\dot{\Sigma }}}_2\cup {{\dot{\Sigma }}}_0, \end{aligned}$$

  (B.16a)

  where

  $$\begin{aligned} G(x,t,\lambda )= \begin{pmatrix} 0&{}\quad \mathrm {i}\\ \mathrm {i}&{}\quad 0 \end{pmatrix}\begin{pmatrix} \mathrm {e}^{-p_1(\lambda _+)}&{}\quad 0\\ 0&{}\quad \mathrm {e}^{p_1(\lambda _+)} \end{pmatrix}G_0(\lambda ) \begin{pmatrix} \mathrm {e}^{p_1(\lambda _+)}&{}\quad 0\\ 0&{}\quad \mathrm {e}^{-p_1(\lambda _+)} \end{pmatrix}\nonumber \\ \end{aligned}$$

  (B.16b)

  with

  $$\begin{aligned} G_0(\lambda )={\left\{ \begin{array}{ll} \begin{pmatrix} 1&{}\quad -{{\check{\rho }}}(\lambda )\\ \overline{{{\check{\rho }}}(\lambda )}&{}\quad 1-|{{\check{\rho }}}(\lambda )|^2 \end{pmatrix} , \quad \lambda \in {{\dot{\Sigma }}}_2,\\ \begin{pmatrix} 1&{}\quad -{{\check{\rho }}}(\lambda )\\ \frac{1}{{{\check{\rho }}}(\lambda )}&{}\quad 0 \end{pmatrix}, \quad \lambda \in {{\dot{\Sigma }}}_0. \end{array}\right. } \end{aligned}$$

  (B.16c)

• The normalization condition:

  $$\begin{aligned} N(x,t,\lambda )={\left\{ \begin{array}{ll}\sqrt{\frac{1}{2}} \begin{pmatrix} -1&{}\quad \mathrm {i}\\ \mathrm {i}&{}\quad -1 \end{pmatrix}+\mathrm {O}\left( \frac{1}{\lambda }\right) ,\quad \lambda \rightarrow \infty , \quad \lambda \in {\mathbb {C}}^+,\\ \sqrt{\frac{1}{2}} \begin{pmatrix} 1&{}\quad \mathrm {i}\\ \mathrm {i}&{}\quad 1 \end{pmatrix} +\mathrm {O}\left( \frac{1}{\lambda }\right) ,\quad \lambda \rightarrow \infty , \quad \lambda \in {\mathbb {C}}^-, \end{array}\right. } \end{aligned}$$

  (B.17)

• Singularity conditions: the singularities of $N(x,t,\lambda )$ at $\pm \frac{1}{A_j}$ are of order not bigger than $\frac{1}{4}$.

• Residue conditions (if any): given $\lbrace {{\check{\lambda }}}_k, {{\check{\kappa }}}_k\rbrace _1^{\check{N}}$ with ${{\check{\lambda }}}_k\in (0,\frac{1}{A_1})$ and ${{\check{\kappa }}}_k\in {\mathbb {R}}{\setminus }\{0\}$, $N^{(2)}(x,t,\lambda )$ has simple poles at $\lbrace {{\check{\lambda }}}_k, -{{\check{\lambda }}}_k\rbrace _1^{\check{N}}$, with the residues satisfying the equations

  $$\begin{aligned} {{\,\mathrm{Res}\,}}_{\pm {{\check{\lambda }}}_k}N^{(2)}(x,t,\lambda )={{\check{\kappa }}}_k\mathrm {e}^{-2p_1({{\check{\lambda }}}_k)}N^{(2)}(x,t,\pm {{\check{\lambda }}}_k). \end{aligned}$$

  (B.18)

Remark B.1

The solution of the RH problem above, if exists, satisfies the following properties:

• (1) $\det N\equiv 1$.

• (2) Symmetries:

  $$\begin{aligned} N(-\lambda )&=-\sigma _3 N(\lambda )\sigma _3,\qquad \overline{N({\overline{\lambda }})}=-N(\lambda ),\qquad \lambda \in {\mathbb {C}}{\setminus }\Sigma _1, \end{aligned}$$

  (B.19a)
  $$\begin{aligned} N((-\lambda )_-)&=-\sigma _3 N(\lambda _+)\sigma _3,\qquad \overline{N(\lambda _-)}=-N(\lambda _+),\qquad \lambda \in {{\dot{\Sigma }}}_2. \end{aligned}$$

  (B.19b)

  where $N(\lambda )\equiv N(x,t,\lambda )$ (follows from the respective symmetries of the jump matrix and the residue conditions, assuming the uniqueness of the solution).

1.4 Eigenfunctions near $\lambda =0$

Introducing ${{\tilde{\Phi }}}_{0,j}$ as in (3.20) and proceeding as in case $A_1<A_2$, the following development of $N(x,t,\lambda )$ near $\lambda =0$ holds:

$$\begin{aligned} N(x,t,\lambda )=\mathrm {i}\begin{pmatrix} 0&{}\quad b_1(x,t)\\ b_1^{-1}(x,t)&{}\quad 0 \end{pmatrix}+\mathrm {i}\lambda \begin{pmatrix} b_2(x,t)&{}\quad 0\\ 0&{}\quad b_3(x,t) \end{pmatrix}+\mathrm {O}(\lambda ^2),\qquad \quad \end{aligned}$$

(B.20)

where

$$\begin{aligned} b_1(x,t)&= \mathrm {e}^{\frac{1}{2A_1 }\int ^x_{-\infty }( m(\xi ,t)-A_1)\mathrm {d}\xi }, \end{aligned}$$

(B.21a)

$$\begin{aligned} b_2(x,t)&= \left( \int _{-\infty }^x e^{-(x-\xi )}\frac{m-A_1}{2}\mathrm {d}\xi +\frac{A_1}{2}\right) \mathrm {e}^{-\frac{1}{2A_1 }\int ^x_{-\infty }( m(\xi ,t)-A_1)\mathrm {d}\xi }, \end{aligned}$$

(B.21b)

$$\begin{aligned} b_3(x,t)&= \left( \int ^{\infty }_x e^{(x-\xi )}\frac{m-A_2}{2}\mathrm {d}\xi +\frac{A_2}{2}\right) \mathrm {e}^{\frac{1}{2A_1 }\int ^x_{-\infty }( m(\xi ,t)-A_1)\mathrm {d}\xi }. \end{aligned}$$

(B.21c)

Proposition B.2

u(x, t) and $u_x(x,t)$ can be algebraically expressed in terms of the coefficients $b_j(x,t)$, $j=1,3$ in the development (B.20) of $N(x,t,\lambda )$ as follows:

$$\begin{aligned} u(x,t)&= b_1(x,t)b_2(x,t)+ b_1^{-1}(x,t) b_3(x,t), \end{aligned}$$

(B.22a)

$$\begin{aligned} u_x(x,t)&= -b_1(x,t)b_2(x,t)+ b_1^{-1}(x,t)b_3(x,t). \end{aligned}$$

(B.22b)

1.5 RH problem in the $\varvec{(y,t)}$ scale

Introducing the new space variable $\check{y}(x,t)$ by

$$\begin{aligned} {\check{y}}(x,t)=x+\frac{1}{A_1}\int ^x_{-\infty }(m(\xi ,t)-A_1)\mathrm {d}\xi -A_1^2 t \end{aligned}$$

(B.23)

and introducing ${{\hat{N}}}({\check{y}},t,\lambda )$ so that $N(x,t,\lambda )={{\hat{N}}} ({\check{y}}(x,t),t,\lambda )$, the jump condition (B.16a) becomes

$$\begin{aligned} {{\hat{N}}}^+({\check{y}},t,\lambda )={{\hat{N}}}^-({\check{y}},t,\lambda ){{\hat{G}}}({\check{y}},t,\lambda ),\quad \lambda \in {{\dot{\Sigma }}}_2\cup {{\dot{\Sigma }}}_0, \end{aligned}$$

(B.24a)

where

$$\begin{aligned} {{\hat{G}}}({\check{y}},t,\lambda ){:}{=}\begin{pmatrix} 0&{}\quad \mathrm {i}\\ \mathrm {i}&{}\quad 0 \end{pmatrix}\begin{pmatrix} \mathrm {e}^{-{{\hat{p}}}_1({\check{y}},t,\lambda _+)}&{}\quad 0\\ 0&{}\quad \mathrm {e}^{{{\hat{p}}}_1({\check{y}},t,\lambda _+)} \end{pmatrix}G_0(\lambda ) \begin{pmatrix} \mathrm {e}^{{{\hat{p}}}_1({\check{y}},t,\lambda _+)}&{}\quad 0\\ 0&{}\quad \mathrm {e}^{-{{\hat{p}}}_1({\check{y}},t,\lambda _+)} \end{pmatrix},\nonumber \\ \end{aligned}$$

(B.24b)

$G_0(\lambda )$ is defined by (B.16c),

$$\begin{aligned} {{\hat{p}}}_1({\check{y}},t,\lambda ) {:}{=}\frac{\mathrm {i}A_1 k_1(\lambda )}{2}\left( {\check{y}}-\frac{2t}{\lambda ^2}\right) . \end{aligned}$$

(B.24c)

Thus $G(x,t,\lambda )={{\hat{G}}}({\check{y}}(x,t),t,\lambda )$ and $p_1(x,t,\lambda )={{\hat{p}}}_1({\check{y}}(x,t),t,\lambda )$, where the jump $G(x,t,\lambda )$ and the phase $p_1(x,t,\lambda )$ are defined in (B.16b) and (2.8b), respectively.

Accordingly, the residue conditions (B.18) become

$$\begin{aligned} \begin{aligned} {{\,\mathrm{Res}\,}}_{\pm {{\check{\lambda }}}_k}{{\hat{N}}}^{(2)}({\check{y}},t,\lambda )&={{\check{\kappa }}}_k\mathrm {e}^{-2{{\hat{p}}}_1({\check{y}},t,\lambda _k)}{{\hat{N}}}^{(1)}({\check{y}},t,\pm {{\check{\lambda }}}_k), \end{aligned} \end{aligned}$$

(B.25)

with ${{\check{\kappa }}}_k=\frac{1}{{\check{b}}_k s'_{11}({{\check{\lambda }}}_k)}$.

Noticing that the normalization condition (B.17), the symmetries (3.10), and the singularity conditions at $\lambda =\pm \frac{1}{A_j}$ hold in the new scale $({\check{y}},t)$, we arrive at the basic RH problem.

BasicRHproblem

Given ${{\check{\rho }}}(\lambda )$ for $\lambda \in {{\dot{\Sigma }}}_2\cup {{\dot{\Sigma }}}_0$, and $\lbrace {{\check{\lambda }}}_k,{{\check{\kappa }}}_k\rbrace _1^{{\check{N}}}$ with ${{\check{\lambda }}}_k\in (0,\frac{1}{A_1})$ and ${{\check{\kappa }}}_k\in {\mathbb {R}}{\setminus }\{0\}$, associated with the initial data $u_0(x)$ in (1.1), find a piece-wise (w.r.t. ${{\dot{\Sigma }}}_1$) meromorphic, $2\times 2$-matrix valued function ${{\hat{N}}}({\check{y}},t,\lambda )$ satisfying the following conditions:

• The jump condition (B.24) across ${{\dot{\Sigma }}}_2\cup {{\dot{\Sigma }}}_0$ (with $G_0(\lambda )$ defined by (B.16c)).

• The residue conditions (B.25).

• The normalization condition:

  $$\begin{aligned} {{\hat{N}}}({\check{y}},t,\lambda )={\left\{ \begin{array}{ll}\sqrt{\frac{1}{2}} \begin{pmatrix} -1&{}\quad \mathrm {i}\\ \mathrm {i}&{}\quad -1 \end{pmatrix}+\mathrm {O}\left( \frac{1}{\lambda }\right) , \quad \lambda \rightarrow \infty ,~ \lambda \in {\mathbb {C}}^+,\\ \sqrt{\frac{1}{2}} \begin{pmatrix} 1&{}\quad \mathrm {i}\\ \mathrm {i}&{}\quad 1 \end{pmatrix}+\mathrm {O}\left( \frac{1}{\lambda }\right) , \quad \lambda \rightarrow \infty ,~ \lambda \in {\mathbb {C}}^-. \end{array}\right. } \end{aligned}$$

  (B.26)

• Singularity conditions: ${{\hat{N}}}({\check{y}},t,\lambda )$ may have singularities at $\pm \frac{1}{A_j}$ of order $\frac{1}{4}$.

• Symmetries:

  $$\begin{aligned} {{\hat{N}}}(-\lambda )&=-\sigma _3 {{\hat{N}}}(\lambda )\sigma _3,\qquad \overline{{{\hat{N}}}({\overline{\lambda }})}=-N(\lambda ),\qquad \lambda \in {\mathbb {C}}{\setminus }\Sigma _2, \end{aligned}$$

  (B.27a)
  $$\begin{aligned} {{\hat{N}}}((-\lambda )_-)&=-\sigma _3 {{\hat{N}}}(\lambda _+)\sigma _3,\qquad \overline{{{\hat{N}}}(\lambda _-)}=-{{\hat{N}}}(\lambda _+),\qquad \lambda \in {{\dot{\Sigma }}}_1. \end{aligned}$$

  (B.27b)

  where ${{\hat{N}}}(\lambda )\equiv {{\hat{N}}}({\check{y}},t,\lambda )$.

1.6 Recovering $\varvec{u(x,t)}$ from the solution of the RH problem

Theorem B.3

Assume that u(x, t) is the solution of the Cauchy problem (1.1) and let ${{\hat{N}}}({\check{y}},t,x)$ be the solution of the associated RH problem (B.24)–(B.26), whose data are determined by $u_0(x)$. Let

$$\begin{aligned} {{\hat{N}}}({\check{y}},t,\lambda )=\mathrm {i}\begin{pmatrix} 0&{}\quad {{\hat{b}}}_1({\check{y}},t)\\ {{\hat{b}}}_1^{-1}({\check{y}},t)&{}\quad 0 \end{pmatrix}+\mathrm {i}\lambda \begin{pmatrix} {{\hat{b}}}_2({\check{y}},t)&{}\quad 0\\ 0&{}\quad {{\hat{b}}}_3({\check{y}},t) \end{pmatrix}+\mathrm {O}(\lambda ^2)\qquad \end{aligned}$$

(B.28)

be the development of ${{\hat{N}}}({\check{y}},t,x)$ at $\lambda =0$. Then the solution u(x, t) of the Cauchy problem (1.1) can be expressed, in a parametric form, in terms of ${{\hat{b}}}_j({\check{y}},t)$, $j=1,2,3$: $u(x,t)={{\hat{u}}}({\check{y}}(x,t),t)$, where

$$\begin{aligned}&{{\hat{u}}}({\check{y}},t)={{\hat{b}}}_1({\check{y}},t){{\hat{b}}}_2({\check{y}},t)+{{\hat{b}}}_1^{-1}({\check{y}},t){{\hat{b}}}_3({\check{y}},t),\end{aligned}$$

(B.29a)

$$\begin{aligned}&x({\check{y}},t)={\check{y}}-2\ln {{\hat{b}}}_1({\check{y}},t)+A_2^2 t. \end{aligned}$$

(B.29b)

Additionally, ${{\hat{u}}}_x({\check{y}},t)$ can also be algebraically expressed in terms of ${{\hat{b}}}_j({\check{y}},t)$, $j=1,2,3$: $u_x(x,t)={{\hat{u}}}_x({\check{y}}(x,t),t)$, where

$$\begin{aligned} {{\hat{u}}}_x({\check{y}},t)=-{{\hat{b}}}_1({\check{y}},t){{\hat{b}}}_2({\check{y}},t)+{{\hat{b}}}_1^{-1}({\check{y}},t){{\hat{b}}}_3({\check{y}},t). \end{aligned}$$

(B.29c)