2.1. Integrability and Jost eigenfunctions

Eq. (1) is an integrable system and admits the following Lax pair:

(4a)\begin{equation} \Phi_{x} = X \Phi, \qquad \Phi_{t} = T \Phi, \end{equation}

(4b)\begin{equation} X(x,t;k) = - ik \sigma_3 + Q, \quad T(x,t;k) = -2ik^2 \sigma_3 + i \sigma_3 (Q_x - Q^2 + q_o^2 I_2) + 2k Q, \end{equation}

(4c)\begin{equation} Q(x,t) = \begin{pmatrix} 0 & q(x,t)\\ q^*(x,t) & 0 \end{pmatrix}, \quad \sigma_1 = \begin{pmatrix} 0 & 1\\ 1 & 0 \end{pmatrix}, \quad \sigma_2 = \begin{pmatrix} 0 & -i\\ i & 0 \end{pmatrix}, \quad \sigma_3 = \begin{pmatrix} 1 & 0\\ 0 & -1 \end{pmatrix}. \end{equation}

$^*$

$\sigma_j$

$j=1,2,3$

(5)\begin{equation} \tilde{q}(x)=q_o\left[ \cos \theta +i \sin \theta \tanh x \right] \end{equation}

to generalise $\tanh x$ in [Reference Cuccagna and Jenkins6] to a smooth function which approaches, at exponential rates, the boundary conditions (2). We consider the weighted spaces $L^{p,s}(\mathbb{R})$ with norm defined as

(6)\begin{equation} ||f||_{L^{p,s}(\mathbb{R})}:=\left( \int_\mathbb{R} \langle x \rangle^{2s}|\,f(x)|^p\, dx \right)^{1/p}, \qquad \langle x\rangle:=\sqrt{1+x^2} \end{equation}

(note that $2s=q$, compared to [Reference Cuccagna and Jenkins6]), as well as the Sobolev spaces

(7)\begin{equation} H^\ell(\mathbb{R})=\left\{ f:\mathbb{R} \to \mathbb{C}: f^{(k)}\in L^2(\mathbb{R})\ \text{for } k=0,\dots,\ell \right\} \end{equation}

and

(8)\begin{equation} H^{1,\ell}(\mathbb{R}):=L^{2,\ell}(\mathbb{R})\cap H^\ell(\mathbb{R}), \end{equation}

where the $L^2$-Sobolev bijectivity of the IST for the focusing NLS equation was established in the case of rapidly decaying ICs [Reference Zhou39]. Then, combining results from [Reference Cuccagna and Jenkins6, Reference Demontis, Prinari, van der Mee and Vitale14, Reference Gallo19, Reference Gallo20], we will assume the IC such that $q(x,0)-\tilde{q}(x)\in H^{1,\ell}(\mathbb{R})$ for $\ell=1,3/2,2$ (these are the same as the spaces $\Sigma_m$ in [Reference Cuccagna and Jenkins6] with $m=2,3,4$, respectively), and highlight the corresponding implications on the IST as it becomes relevant.

The Jost eigenfunctions $\Phi_\pm$ are defined as simultaneous solutions of the two linear equations of the Lax pair with boundary conditions

(9a)\begin{equation} \Phi_{\pm}(x,t;k) \sim \left[ Y_{\pm}(k) + o(1)\right] e^{i \Omega(x,t;k) \sigma_3}, \quad x \to \pm \infty, \end{equation}

(9b)\begin{equation} \Omega(x,k;t) = - \lambda x - 2 k \lambda t, \quad \lambda^2 = k^2-q_o^2, \end{equation}

(9c)\begin{equation} Y_{\pm}(k) = I_2 - \frac{i}{k+\lambda} \sigma_3 Q_{\pm}, \quad Q_{\pm} = \begin{pmatrix} 0 & q_{\pm}\\ q_{\pm}^* & 0 \end{pmatrix}. \end{equation}

It is worth remarking that $i\lambda(k)$ with $\lambda(k)$ defined in (9b) is the eigenvalue parameter for the constant coefficient matrices $X_\pm=X-Q_\pm$, where $Q_\pm=\lim_{x\to \pm \infty}Q(x,t)$ (and $Y_\pm$ are the associated matrices of eigenvectors). Then (9a) determines the solutions when $\lambda(k)\in \mathbb{R}$ (purely oscillatory behaviour), and this corresponds to $k^2 \ge q_o^2$, showing that the continuous spectrum for the problem in the $k$-plane is $\mathbb{R}\setminus(-q_o,q_o)$, i.e., there is a gap.

Note that in the expression for $\Omega$ above, the second term has the opposite sign compared to [Reference Biondini and Prinari5, Reference Cuccagna and Jenkins6, Reference Zhaoyu and Fan38]. The correct sign is the one in (9b), but the sign error in [Reference Biondini and Prinari5, Reference Cuccagna and Jenkins6, Reference Zhaoyu and Fan38] does not affect in any significant way the results in those papers, as it ultimately only switches the regions with $t \gt 0$ and those with $t \lt 0$. For convenience, we introduce the uniform variable $z$ defined by the map

(10)\begin{equation} z=k+\lambda \end{equation}

which is inverted via the relations

(11)\begin{equation} k(z) = \frac{1}{2} \left(z+q_o^2/z \right), \quad \lambda(z)=\frac{1}{2}\left(z-q_o^2/z \right). \end{equation}

As mentioned in Section 1, the map $k\to z$ in the context of the IST of the defocusing NLS with nonzero background was first introduced by Faddeev and Takhtajan in [Reference Faddeev and Takhtajan17]. The interested reader can find a discussion on the $2$-to- $1$ nature of the map and other relevant details in [Reference Prinari, Ablowitz and Biondini30].

The asymptotic behaviour of the eigenfunctions can then be written as:

(12a)\begin{equation} \Phi_{\pm}(x,t;z) \sim \left[ Y_{\pm}(z) + o(1)\right] e^{i \Omega(x,t;z) \sigma_3 } , \quad x \to \pm \infty, \end{equation}

(12b)\begin{equation} Y_{\pm}(z) = \begin{pmatrix} 1 & -\frac{i}{z} q_{\pm}\\ \frac{i}{z}q_{\pm}^* & 1 \end{pmatrix} = \begin{pmatrix} 1 & -\frac{i}{z} q_o e^{\pm i \theta}\\ \frac{i}{z} q_o e^{\mp i \theta} & 1 \end{pmatrix}, \end{equation}

(12c)\begin{equation} \Omega(x,t;z) = -\frac{1}{2}(z-q_o^2/z)x-\frac{1}{2}(z^2-q_o^4/z^2)t. \end{equation}

$z$

$\Omega$

$z=0$

$z=\infty$

(13)\begin{equation} M_{\pm}(x,t;z) = \Phi_{\pm}(x,t;z) e^{-i \Omega(x,t;z) \sigma_3} \end{equation}

which satisfy the boundary conditions

(14)\begin{equation} M_{\pm}(x,t;z) \sim Y_{\pm}(z) + o(1), \quad x \to \pm \infty. \end{equation}

If the IC $q(x,0)$ is such that $q(x,0)-\tilde{q}(x)\in H^{1,1}(\mathbb{R})$, then standard Neumann series arguments on the Volterra integral equations for $M_\pm$ can be used to show that the modified eigenfunctions $M_{-,1}(x,t;z)$, $M_{+,2}(x,t;z)$ are analytic in the upper half-plane of $z$ and continuous up to $\mathbb{R}\setminus\left\{0\right\}$, while $M_{-,2}(x,t;z)$, $M_{+,1}(x,t;z)$ are analytic in the lower half-plane of $z$ and continuous up to $\mathbb{R}\setminus\left\{0\right\}$ (see [Reference Cuccagna and Jenkins6, Reference Demontis, Prinari, van der Mee and Vitale14]). Here and in the following, the additional subscript $j=1,2$ in $M_\pm$ denotes the corresponding column of the matrix function. Furthermore, the following lemma holds.

Lemma 2.1. The modified eigenfunctions $M_{\pm}$ satisfy the symmetries

(15)\begin{equation} M_{\pm}^*(z^*) = \sigma_1 \, M_{\pm}(z) \, \sigma_1, \quad M_{\pm} ( q_o^2/z ) = \frac{z}{q_o} \, M_{\pm}(z) \, \sigma_2 \, e^{\mp i \theta \sigma_3} \end{equation}

for any $q_o \gt 0$ and $\theta \in \mathbb{R}$, where each column on both sides of the equations is considered for $z\in \mathbb{C}^+\cup \mathbb{R} \setminus\{0\}$ or $z\in \mathbb{C}^-\cup \mathbb{R} \setminus\{0\}$ depending on the respective domain of analyticity.

Proof. Using the symmetries of the Lax pair

(16a)\begin{equation} X^*(x,t;z^*) = \sigma_1 \, X(x,t;z) \, \sigma_1, \quad X (x,t;q_o^2/z) = X(x,t;z) \end{equation}

(16b)\begin{equation} T^*(x,t;z^*) = \sigma_1 \, T(x,t;z) \, \sigma_1, \quad T(x,t;q_o^2/z) = T(x,t;z) \end{equation}

and the asymptotics of $\Phi_{\pm}$ as $x \to \pm \infty$, one can show that $\Phi_{\pm}$ satisfy the symmetries

(17)\begin{equation} \Phi_{\pm}^*(z^*) = \sigma_1 \, \Phi_{\pm}(z) \, \sigma_1, \quad \Phi_{\pm} ( q_o^2/z ) = \frac{z}{q_o} \, \Phi_{\pm}(z) \, \sigma_2 \, e^{\mp i \theta \sigma_3}. \end{equation}

Symmetries (15) follow directly from Eq. (13) and the symmetries of the function $\Omega$, namely

(18)\begin{equation} \Omega^*(x,t;z^*) = \Omega(x,t;z), \quad \Omega (x,t;q_o^2/z) = - \Omega(x,t;z). \end{equation}

For instance, to prove that the second of (17) holds, one needs to show that the two matrix functions satisfy the same differential equations and have the same asymptotic behaviour as $x\to \pm \infty$. We have:

\begin{align*} \partial_x \Phi_{\pm} ( q_o^2/z ) &= \frac{z}{q_o} \, \partial_x \Phi_{\pm}(z) \, \sigma_2 \, e^{\mp i \theta \sigma_3}= \frac{z}{q_o} X(x,t;z) \, \Phi_{\pm}(z) \, \sigma_2 \, e^{\mp i \theta \sigma_3}\\ &= \frac{z}{q_o} X(x,t;q_o^2/z) \, \Phi_{\pm}(z) \, \sigma_2 \, e^{\mp i \theta \sigma_3}\\ &=\frac{z}{q_o} \, \frac{q_o}{z} X(x,t;q_o^2/z) \, \Phi_{\pm}(q_o^2/z) \, e^{\pm i \theta \sigma_3}\, \sigma_2 \, \sigma_2 \, e^{\mp i \theta \sigma_3} \\ &=X(x,t;q_o^2/z) \, \Phi_{\pm}(q_o^2/z) \end{align*}

and similarly one can show the same for the time-dependence problem. One can also check that the asymptotic behaviour coincides, as well. Indeed, one has:

\begin{equation*} \Phi_{\pm}(z) \sim Y_{\pm}(z) e^{i \Omega(z) \sigma_3}, \qquad \Phi_{\pm}(q_o^2/z) \sim Y_{\pm}(q_o^2/z) e^{i \Omega(q_o^2/z) \sigma_3} \quad \text{as }x \to \pm \infty. \end{equation*}

Moreover, assuming that

\begin{equation*} \Phi_{\pm}(q_o^2/z) = \frac{z}{q_o} \Phi_{\pm}(z) \sigma_2 e^{\mp i \theta \sigma_3} \sim \frac{z}{q_o} \left[ Y_{\pm}(z) e^{i \Omega(z) \sigma_3} \right] \sigma_2 e^{\mp i \theta \sigma_3}, \quad \text{as }x \to \pm \infty \end{equation*}

then one must show that

\begin{equation*} Y_{\pm}(q_o^2/z) e^{i \Omega(q_o^2/z) \sigma_3} \equiv \frac{z}{q_o} \left[ Y_{\pm}(z) e^{i \Omega(z) \sigma_3} \right] \sigma_2 e^{\mp i \theta \sigma_3}, \end{equation*}

which can be easily verified using the definition of $Y_{\pm}$ together with the second of (18).

It is worth mentioning here that the symmetries (15) can be established for $z\in \mathbb{R}$, where all the columns are simultaneously defined, and then extended to $\mathbb{C}^\pm$ column-wise.

2.2. Direct problem: scattering data

The Jost eigenfunctions $\Phi_{\pm}$ are two fundamental solutions of the Lax pair for any $z\in \mathbb{R}\setminus\left\{\pm q_o\right\}$, since

(19)\begin{equation} \det \Phi_{\pm}(x,t;z) \equiv \det Y_\pm (z)= 1 - q_o^2 z^{-2}. \end{equation}

Therefore, there exists a $2 \times 2$ scattering matrix $S(z)$ (independent of $x,t$) such that

(20)\begin{equation} \Phi_{-}(x,t;z) = \Phi_{+}(x,t;z) S(z), \quad S(z) = \begin{pmatrix} a(z) & \bar{b}(z)\\ b(z) & \bar{a}(z) \end{pmatrix}, \quad z \in \mathbb{R} \setminus \{\pm q_o\}. \end{equation}

The time independence of the scattering matrix $S(z)$ is a consequence of having defined the Jost eigenfunctions as simultaneous solutions of the Lax pair, not just of the scattering problem. After replacing the modified eigenfunctions, the last equation becomes

(21a)\begin{equation} M_{-,1}(x,t;z)/a(z) = M_{+,1}(x,t;z) + \rho(z) M_{+,2}(x,t;z) e^{-2i \Omega(x,t;z)}, \end{equation}

(21b)\begin{equation} M_{-,2}(x,t;z)/a^*(z) = M_{+,2}(x,t;z) + \bar{\rho}(z) M_{+,1}(x,t;z) e^{2i \Omega(x,t;z)}, \end{equation}

(22)\begin{equation} \rho(z) = b(z)/a(z), \qquad \bar{\rho}(z)= \bar{b}(z)/\bar{a}(z). \end{equation}

It is then easy to verify that the entries of the scattering matrix $S(z)$ are such that

\begin{equation*} \bar{a}(z)=a^*(z^*), \qquad \bar{b}(z)=b^*(z^*), \end{equation*}

where the symmetry holds for the values of $z$ for which the individual entries of $S(z)$ are defined. In particular, this implies that for all $z\in \mathbb{R}$ one has $\bar{\rho}(z)=\rho^*(z)$, and

(23)\begin{equation} \bar{\rho}(z)=\rho^*(z^*) \end{equation}

when the reflection coefficient $\rho(z)$ can be analytically continued off the real $z$-axis, which will be the case for the box-type ICs considered later on.

Eq. (19), combined with (20), yields the following expressions for the scattering coefficients:

(24)\begin{equation} a(z) = \frac{1}{1 - q_o^2 z^{-2}} \mathrm{det} \left( M_{-,1}(z), M_{+,2}(z) \right), \quad b(z)=\frac{1}{1 - q_o^2 z^{-2}} \mathrm{det} \left( M_{+,1}(z), M_{-,1}(z) \right) \end{equation}

where we have omitted the $(x,t)$ dependence in the right-hand side for brevity. The above expressions show that even if the Jost eigenfunctions are continuous at $z=\pm q_o$, in general, the scattering coefficients are singular at those points.

Discrete eigenvalues are the zeros of the scattering coefficient $a(z)$ in $\mathbb{C}^+$ and their complex conjugates (as zeros of $a^*(z^*)$) in $\mathbb{C}^-$. Such zeros are known to be simple, and located on the circle $\mathcal{C}=\left\{z\in \mathbb{C}: |z|=q_o\right\}$ due to the self-adjointness of the scattering problem [Reference Faddeev and Takhtajan17]. Note that the only possible zeros of $a(z)$ embedded in the continuous spectrum, i.e., on the real $z$-axis, are $\pm q_o$, since it was shown in [Reference Faddeev and Takhtajan17] that $a(k)\ne 0$ for any $k\in \mathbb{R} \setminus [-q_o,q_o]$, which corresponds to $z\in \mathbb{R} \setminus \mathbb{C}$. Moreover, in [Reference Demontis, Prinari, van der Mee and Vitale14] it is shown that if $q(x,0)-q_\pm \in L^{1,4}(\mathbb{R}^\pm)$, then $a(z)$ does not vanish at $\pm q_o$ (and hence the number of discrete eigenvalues is necessarily finite, since the only possible accumulation points are $\pm q_o$). Finally, Eq. (24) imply that at each pair of discrete eigenvalues $z_j,z_j^*$:

(25)\begin{equation} M_{-,1}(x,t;z_j) = b_j \, M_{+,2}(x,t;z_j) e^{-2i \Omega(x,t;z_j)}, \quad M_{-,2}(x,t;z_j^*) = b_j^* \, M_{+,1}(x,t;z_j^*) e^{2i \Omega(x,t;z_j^*)} \end{equation}

for some $b_j\in \mathbb{C}$. Additional smoothness and decay properties of the reflection coefficient are established in [Reference Cuccagna and Jenkins6] under suitable assumptions on the IC. Specifically:

• • If $q(x,0)-\tilde{q}(x)\in H^{1,1}(\mathbb{R})$, then $\rho \in L^2(\mathbb{R})$ and $||\log (1-|\rho|^2)||_{L^p} \lt \infty$ for all $p\ge 1$;

• • If $q(x,0)-\tilde{q}(x)\in H^{1,3/2}(\mathbb{R})$, then $\rho\in H^1(\mathbb{R})$ and moreover the discrete eigenvalues are necessarily finite in number.

Even though it will not be relevant for the present work, we mention that in [Reference Cuccagna and Jenkins6] it was also shown that requiring $q(x,0)-\tilde{q}(x)\in H^{1,2}(\mathbb{R})$ allows to obtain bounds of the $\bar{\partial}$-derivatives of the reflection coefficient which were there used for the long-time estimates in the inverse problem.

For completeness, we include the symmetries of the scattering data in the following lemma, whose proof follows straightforwardly from the symmetries (17) of the Jost eigenfunctions.

Lemma 2.2. The scattering matrix $S(z)$ satisfies

(26)\begin{equation} S (q_o^2/z ) = e^{i \theta \sigma_3} \, \sigma_2 \, S(z) \, \sigma_2 \, e^{i \theta \sigma_3}, \qquad \forall z\in \mathbb{R}\setminus \{\pm q_o\}. \end{equation}

Consequently, one has

(27a)\begin{equation} a ( q_o^2/z ) = e^{2 i \theta} a^*(z^*), \quad z\in \mathbb{C}^+ \end{equation}

(27b)\begin{equation} b ( q_o^2/z ) = - b^*(z), \quad \rho (q_o^2/z ) = - e^{-2i \theta} \, \rho^*(z), \quad z\in \mathbb{R}\setminus\{\pm q_o\}. \end{equation}

Moreover, the proportionality constants $b_j$ defined in (25) satisfy

(28)\begin{equation} b_j^*=-b_j. \end{equation}

We emphasise that with a piecewise IC, the direct problem of the IST can be solved exactly. Specifically, as shown in [Reference Biondini and Prinari5], for any piecewise IC (3), one can obtain explicit formulas for the scattering coefficients by expressing, at the boundary of each region, the (fundamental) solution on the left as a linear combination of the (fundamental) solution on the right, which yields:

(29a)\begin{align}a(z) = \frac{1}{\lambda(z) \mu(z)} e^{i(2 L \lambda(z) + \theta)} \Bigg\{\mu(z) \cos(2 L \mu(z)) [ \lambda(z) \cos \theta - i k(z) \sin \theta ] \nonumber \\ + i \sin(2L \mu(z)) \Big[ h q_o \cos \alpha - k(z) ( k(z) \cos \theta - i \lambda(z) \sin \theta ) \Big] \Bigg\}\end{align}

(29b)\begin{align}b(z) = \frac{1}{\lambda(z) \mu(z)} \Bigg\{-q_o \mu(z) \sin \theta \cos(2 L \mu(z)) \nonumber \\ + \Big[ h k(z) \cos \alpha - k(z) q_o \cos \theta -i h \lambda(z) \sin \alpha \Big] \sin(2L \mu(z)) \Bigg\},\end{align}

where

(30)\begin{equation} k(z)=\frac{1}{2}(z+q_o^2/z), \quad \lambda(z)=\sqrt{k(z)^2-q_o^2}\equiv \frac{1}{2}(z-q_o^2/z),\quad \mu(z)=\sqrt{k(z)^2-h^2}\,. \end{equation}

It is important to point out that although $\mu(z)$ is defined in terms of a square root, there is no branching in the IST. Indeed, $\mu$ does not enter in the definition of the Jost eigenfunctions, and equations 29) show that only even functions of $\mu$ appear in the scattering coefficients, which are therefore independent of the choice of sign of the square root.

We also mention that the scattering coefficients when the potential coincides with the background, i.e., for $q(x,t)=q_o\left[e^{i\theta}H(x)+e^{-i\theta}(1-H(x))\right]$, where $H(x)$ is the Heaviside step function, can be obtained from (29) by setting $L=0$. Specifically, we have:

(31)\begin{equation} a(z)=\frac{e^{i\theta}}{\lambda(z)}\left[ \lambda(z)\cos \theta -ik(z)\sin \theta \right], \qquad b(z)=-\frac{q_o \sin \theta}{\lambda(z)}. \end{equation}

When $\theta=0$, i.e., for a constant (in both $x$ and $t$) solution, the above reduce to $a(z)=1$ and $b(z)=0$, so one obtains the trivial reflectionless potential that is the analogue of the solution that is identically zero in the rapidly decaying case.

2.3. Asymptotic behaviour of eigenfunctions and scattering data

To formulate the inverse problem properly, one needs to determine the asymptotic behaviour of eigenfunctions and scattering data as $z \to 0$, $z \to \infty$ and as $z\to\pm q_o$. Specifically, the asymptotic behaviour of the eigenfunctions is obtained from integration by parts on appropriate Volterra integral equations for the eigenfunction (cf., e.g., [Reference Demontis, Prinari, van der Mee and Vitale14]), and it is given by:

(32a)\begin{equation} M_{\pm}(x,t;z) = \begin{pmatrix} 1 + i I_0(x,t)/z & - i q(x,t)/z\\ i q^*(x,t)/z & 1 - i I_0(x,t)/z \end{pmatrix}+\mathcal{O}(z^{-2}) , \quad z \to \infty, \end{equation}

(32b)\begin{equation} M_{\pm}(x,t;z) = \begin{pmatrix} q(x,t)/q_{\pm} & I_0(x,t)/q_{\pm}^* - i q_{\pm}/z\\ I_0(x,t)/q_{\pm} + i q_{\pm}^*/z & q^*(x,t)/q_{\pm}^* \end{pmatrix}+\mathcal{O}(z^2), \quad z \to 0, \end{equation}

(33)\begin{equation} I_0(x,t)= \int_{x}^{+ \infty} (|q(x',t)|^2 - q_o^2) dx'. \end{equation}

Proposition 2.3. The asymptotic behaviour of the scattering coefficients as $z\to \infty$ and as $z\to 0$ is given by:

(34a)\begin{equation} a(z)=1 + \mathcal{O}(z^{-1}), \quad b(z) = \mathcal{O}(z^{-1}), \quad \rho(z) = \mathcal{O}(z^{-1}) \quad z \to \infty, \end{equation}

(34b)\begin{equation} a(z) = q_{+}/q_{-} + \mathcal{O}(z), \quad b(z) = \mathcal{O}(z), \quad \rho(z) = \mathcal{O}(z), \quad z \to 0. \end{equation}

For ICs in $\tilde{q}(x)+H^{1,1}(\mathbb{R})$, the modified eigenfunctions are continuous at $z=\pm q_o$, and Eqs. (24) show that generically the scattering data have simple poles at $\pm q_0$:

(35)\begin{equation} a(z) = \frac{\alpha_{\pm}}{z \mp q_o} + \mathcal{O}(1), \quad b(z) = \frac{\beta_\pm}{z \mp q_o} + \mathcal{O}(1), \qquad \beta_\pm=\mp ie^{-i\theta} \alpha_\pm, \end{equation}

where $\alpha_{\pm} = \mathrm{det} \big( M_{-,1}(x,t;\pm q_o), M_{+,2}(x,t;\pm q_o)\big)$ [and $\alpha_\pm\ne 0$ unless the columns of $M_{\pm}$ become linearly dependent at either $z= q_o$ or $z=-q_o$ or both, in which case the points $\pm q_o$ are called ‘virtual levels’]. Note that [Reference Demontis, Prinari, van der Mee and Vitale14] provides a sufficient condition on the decay of the potential that guarantees $a(\pm q_o) \ne 0$ (even if one between $\alpha_+$ and $\alpha_-$ or both vanish), namely that $q(x,0)-q_\pm \in L^{1,4}(\mathbb{R}^\pm)$, and this will be the case for the ICs considered in this work. Importantly, the reflection coefficient is always bounded at $z=\pm q_o$, and in particular

(36)\begin{equation} \rho(\pm q_o) = \mp i e^{-i \theta}. \end{equation}

Incidentally, we note that in the case of the ‘unperturbed’ background Eqs. (31) give:

(37)\begin{equation} \alpha_\pm =\mp iq_o e^{i\theta}\sin \theta , \qquad \beta_\pm=-q_o\sin \theta. \end{equation}

and $\alpha_\pm=\beta_\pm=0$ when $\theta=0$, i.e., the scattering coefficients are non-singular when $q(x,t)=q_o$, like for any other reflectionless potential, for which $b(z)\equiv 0$ and $a(z)=\prod_{j=1}^n(z-z_j)/(z-z_j^*)$. On the other hand, whenever $b(z)\ne 0$ (or, equivalently, $\rho(z)\ne 0$) the trace formula for $a(z)$ can be used to show that generically $\alpha_\pm \ne 0$ (and hence $\beta_\pm$) [Reference Faddeev and Takhtajan17]. For the purpose of this work, we verify numerically and analytically (cf. Eqs. (136) and the discussion in Appendix A) for all the ICs considered in the examples that indeed $\alpha_\pm,\beta_\pm\ne 0$.

The following result can also be established.

Proposition 2.4. If $\rho(z), \bar{\rho}(z)\in C^1(\mathbb{R})$, the quantity $1-\rho(z) \bar{\rho}(z)$ vanishes quadratically as $z\to \pm q_o$, i.e.,

(38)\begin{equation} 1-\rho(z) \bar{\rho}(z) = O((z\mp q_o)^2), \qquad z\to \pm q_o. \end{equation}

Proof. Clearly, $1-\rho(\pm q_o) \bar{\rho}(\pm q_o)=0$ because of Eq. (36). Differentiating both sides of the second equation in (27b) with respect to $z$ and evaluating at $z=\pm q_o$ we find:

\begin{equation*} \frac{d\rho(z)}{dz} \Bigg|_{z=\pm q_o} = e^{-2i\theta}\frac{d\rho^*(z)}{dz}\Bigg|_{z=\pm q_o}. \end{equation*}

Using the property $\bar{\rho}(z)=\rho^*(z)$ for $z \in \mathbb{R}$, and the above symmetry for $d\rho/dz$ at $z=\pm q_o$, we get:

\begin{equation*} \frac{d}{dz}(1-\rho(z) \bar{\rho}(z)) \Bigg|_{z=\pm q_o} = \frac{d\rho^*(z)}{dz}\Bigg|_{z=\pm q_o} \left( e^{-2i\theta}\rho^*(\pm q_o)+\rho(\pm q_o)\right)\equiv 0 \end{equation*}

where the term in brackets vanishes owing to (36).

In Appendix A, we give the explicit expressions of the asymptotics of the scattering coefficients (29) as $z\to \infty$, $z\to 0$ and $z\to \pm q_o$ in the case of a box IC.

2.4. Inverse problem: Riemann–Hilbert formulation

Within the IST framework, the inverse problem amounts to reconstructing the Jost eigenfunctions in terms of the scattering data, from which one can then recover $q(x,t)$ for any $t \gt 0$. For this purpose, we introduce:

(39)\begin{equation} m(z) := m(x,t;z) = \begin{cases} \Big( M_{-,1}(x,t;z)/a(z), \quad M_{+,2}(x,t;z)\Big), & z \in \mathbb{C}^{+}\\ \Big( M_{+,1}(x,t;z), \quad M_{-,2}(x,t;z)/a^*(z^*)\Big), & z \in \mathbb{C}^{-}. \end{cases} \end{equation}

For a generic IC, the problem might support solitons that arise from the zeros of $a(z)$ in $\mathbb{C}^+\cap\,\mathcal C$, and their complex conjugates, zeros of $a^*(z^*)$ in $\mathbb{C}^-\cap\,\mathcal C$, where, as before, $\mathcal{C}$ denotes the circle of radius $q_o$. Let $N\in \mathbb{N}$ denote the number of solitons (necessarily finite for IC in $\tilde{q}(x)+H^{1,3/2}(\mathbb{R})$, as mentioned above), which implies that $m(z)$ has $N$ poles in $\mathbb{C}^{+}$ with their complex conjugates appearing as poles in $\mathbb{C}^{-}$. The properties established for the eigenfunctions and the scattering coefficient $a(z)$ in the direct problem then yield the following RHP for $m(z)$.

For brevity, in the statement of future RHPs, we refer to 1(b) as the condition that $m(z)$ takes continuous boundary values away from the set $\{0\} \cup \{z_j\}_{j=1}^N \cup \{z_j^*\}_{j=1}^N$. Note also that whenever the jump matrix is evaluated on the real $z$-axis, one can use $\bar{\rho}(z)=\rho^*(z)$ and $1+\rho(z)\bar{\rho}(z)=1+|\rho(z)|^2$. However, since in the following we will have to introduce deformations off the real axis, it is preferable to keep both $\rho(z)$ and $\bar{\rho}(z)$, related to each other via symmetry (23). Finally, note that the solution of the above RHP, as well as all the following ones, depends parametrically on $x,t$, but this parametric dependence is usually omitted for brevity unless otherwise specified.

Proof. Let $m$ be a solution of RHP 1. Evaluating the jump condition of $m$ at $z= \pm q_o$, we get

\begin{equation*} m^{+}(x,t;-q_o) = m^{-}(x,t;-q_o) \begin{pmatrix} 0 & i e^{i \theta}\\ i e^{-i \theta} & 1 \end{pmatrix}, \quad m^{+}(x,t;q_o) = m^{-}(x,t;q_o) \begin{pmatrix} 0 & -i e^{i \theta}\\ -i e^{-i \theta} & 1 \end{pmatrix} \end{equation*}

because $\rho(\pm q_o) = \mp i e^{-i \theta}$ (cf. (36)). The first column of the above equations gives

(45)\begin{equation} m^+_{1}(x,t;-q_o) = i e^{-i \theta} \, m^-_{2}(x,t;-q_o), \quad m^+_{1}(x,t;q_o) = - i e^{-i \theta} \, m^-_{2}(x,t;q_o). \end{equation}

Moreover, symmetries (44) at $z= \pm q_o$ yield

(46)\begin{equation} m^+_{1}(x,t;-q_o) = -i e^{-i \theta} \, m^-_{2}(x,t;-q_o), \quad m^+_{1}(x,t;q_o) = i e^{-i \theta} \, m^-_{2}(x,t;q_o), \end{equation}

which together with Eq. (45) gives

(47)\begin{equation} m^+_{1}(x,t;-q_o) = m^-_{2}(x,t;-q_o) =0, \quad m^+_{1}(x,t;q_o) = m^-_{2}(x,t;q_o) =0. \end{equation}

The analyticity of the jump matrix then implies that $m^+_{1}(z)$ and $m^-_{2}(z)$ have analytic extensions in a neighbourhood of $z = \pm q_o$, and therefore the zeros have to be at least of first order.

Remark 2.8. It is important for future reference to establish the behaviour of the columns of $m(x,t;z)$ as $z\to \pm q_o$ from either half-plane. First, note that if $m(x,t;z)$ is defined as in (39), then one can verify that at $t=0$:

(48a)\begin{equation} m_1^+(x,0; z)\sim \frac{z\mp q_o}{\alpha_\pm} \begin{pmatrix} 1 \\ \pm i e^{i\theta} \end{pmatrix}, \quad m_2^+(x,0; z)\sim \begin{pmatrix} \mp i e^{i\theta} \\ 1 \end{pmatrix} \qquad z\to \pm q_o, \ z\in \mathbb{C}^+, \end{equation}

(48b)\begin{equation} m_1^-(x,0; z)\sim \begin{pmatrix} 1 \\ \pm ie^{-i\theta} \end{pmatrix}, \quad m_2^-(x,0; z)\sim \frac{z\mp q_o}{\alpha_\pm^*} \begin{pmatrix} \mp ie^{-i\theta} \\ 1 \end{pmatrix}\qquad z\to \pm q_o, \ z\in \mathbb{C}^-, \end{equation}

where $\alpha_\pm$ are defined in (35) (and in the case of box-type ICs they can be explicitly computed, see Appendix A). The above equations show that the relevant columns of $m^\pm(z)$ have simple zeros as $z\to \pm q_o$, at least at $t=0$. Since $q_o,\alpha_\pm,\theta$ are time-independent, we assume that the above behaviour holds for all $t\ge 0$, similarly to how one would handle pole singularities corresponding to the discrete eigenvalues. The fact that $m_1^+(x,t;z)$ and $m_2^-(x,t;z)$ have simple zeros as $z\to \pm q_o$ for any $t \geq 0$, as it happens at $t=0$ according to Eqs. (48), is consistent with Lemma 2.7, which holds for all $t\ge 0$, since a double zero would contradict this lemma. This is important for the deformed RHPs, which will be defined later on, to establish the precise behaviour of the respective solutions as $z\to \pm q_o$.

Proof. We apply Liouville’s theorem to the matrix $m \, \breve{m}^{-1}$, where $m$ and $\breve{m}$ are two arbitrary solutions to RHP 1. In this case, one of the things to check is whether $m \, \breve{m}^{-1}$ is singular at $z=0$ and/or at the poles from the discrete spectrum, since each of the solutions $m$ and $\breve{m}$ is singular there. Additionally, one must check the points $z=\pm q_o$ where the determinant of both $m$ and $\breve{m}$ vanishes according to Lemma 2.7. We write:

(49)\begin{equation} m(z) \, \breve{m}^{-1}(z) = \frac{1}{\det \breve{m}(z)} \, m(z) \sigma_2 \breve{m}^{T}(z) \sigma_2 \end{equation}

and like in the proof of Lemma 5.3 in [Reference Cuccagna and Jenkins6], we can check from condition 4. of RHP 1, that $m \, \breve{m}^{-1}$ is bounded at $z=0$. Moreover, one can check from the residue conditions 2. of RHP 1 that $m \, \breve{m}^{-1}$ has no poles at $z_j$ and $z_j^*$, for $j=1,2,\cdots N$. These cases were already addressed in the proof of Lemma 5.3 in [Reference Cuccagna and Jenkins6] in a similar manner. In addition to that proof, note that

(50)\begin{equation} \det \breve{m}(z) = \mathcal{O}(z \mp q_o), \quad z\to \pm q_o, \end{equation}

which follows directly from Lemma 2.7. Using Eqs. (49)-(50), Proposition 2.6 and the fact that any solution to RHP 1 takes continuous boundary values away from the set $\{0\} \cup \{z_j\}_{j=1}^N \cup \{z_j^*\}_{j=1}^N$, then we can show that $m \, \breve{m}^{-1}$ is bounded at $z=\pm q_o$, leaving no singularities at these points. Uniqueness of solution of RHP 1 then follows from the large- $z$ asymptotics of $m \, \breve{m}^{-1}$.

Finally, using the asymptotic property of $M_{\pm}(z)$ at infinity and definition (39), one then reconstructs the potential via the following relation

(51)\begin{equation} q(x,t) = \lim_{{\substack{z \to \infty \\ z\in \mathbb{C}^+}}} \Big( i z \, m_{12} (x,t,z) \Big) \end{equation}

where the subscript denotes the $(1,2)$ entry of the matrix function $m$.

Our aim is to solve RHP 1 numerically and use relation (51) to recover $q(x,t)$ at arbitrary values of $x\in \mathbb{R}$, $t \gt 0$. However, RHP 1 presents two numerical challenges. The first arises from the presence of rapidly oscillatory exponential terms in the off-diagonal entries of the jump matrix, which we address by deforming the contour away from the real axis, turning the oscillations into exponential decay. The second challenge arises from Eq. (43b), which indicates that the matrix $m(z)$ has a singularity at $z=0$. The following sections detail our approach to overcoming these challenges.

3.1. Analysis of the phase function

Recall that our aim is to solve RHP 1 numerically, and use it to compute the potential $q(x,t)$. As shown in [Reference Cuccagna and Jenkins6, Reference Zhaoyu and Fan38], the long-time asymptotic analysis identifies three regions in the space-time domain of $x$ and $t$, depending on the value of the parameter $\xi = x/(2t)$, with different characteristics: the solitonic region when $|\xi| \lt q_o$, the solitonless region when $|\xi| \gt q_o$, and the collisionless shock region when $|\xi|=q_o$. While the analysis in [Reference Cuccagna and Jenkins6, Reference Zhaoyu and Fan38] is carried out for the specific case of $q_o=1$, the results can be generalised to any arbitrary value of $q_o$. Therefore, we express the jump matrix $G$ in terms of $\xi$ as follows:

(52)\begin{equation} G(\xi,z) = \begin{pmatrix} 1-\rho(z)\bar{\rho}(z) & -\bar{\rho}(z) e^{-2t \, i \Theta(\xi,z)}\\ \rho(z) e^{2t \, i \Theta(\xi,z)} & 1 \end{pmatrix}, \quad \Theta(\xi,z) = 2 \left( \lambda(z) \xi + k(z) \lambda(z)\right). \end{equation}

Note that $G$ involves the exponentials $e^{\pm 2 t \, i \Theta(\xi,z)}$ which are oscillatory for $z \in \mathbb{R}$ where the jump of the RHP is defined. Thus, solving RHP 1 numerically involves implementing contour deformations away from the real axis following the principles of nonlinear steepest descent, similar to those used for estimating the long-time asymptotics (see [Reference Deift and Zhou9], [Reference Deift, Venakides and Zhou8] and [Reference Deift, Zhou and Venakides13] for more details). We want to emphasise that the deformations are valid for arbitrary $t$, and they have the advantage that, for large $t$, the exponential decay off the real axis makes the numerical scheme more efficient, by reducing the number of terms that need to be computed. Therefore, we implement the deformations based on the sign chart of $\mathop{\rm Re}\nolimits(i \Theta)$ to achieve exponential decay for large $t$ of all the jumps off the real axis.

First, we compute the stationary phase points of the function $\Theta$. We have:

(53a)\begin{equation} \Theta'(z) = \xi \left( 1+q_o^2 z^{-2} \right)+\left( z+q_o^4 z^{-3}\right)= z^{-1}\Big( \xi(z+q_o^2 z^{-1}) + (z+q_o^2 z^{-1})^2-2q_o^2 \Big) \equiv z^{-1} l(s), \end{equation}

(53b)\begin{equation} \Theta''(z) = -z^{-2} l(s)+z^{-1}l'(s)s'(z) \equiv -z^{-2} l(s) + z^{-1} \left( 1-\frac{q_o^2}{z^2}\right)l'(s), \end{equation}

(53c)\begin{equation} l(s) = s^2 + \xi s - 2q_o^2, \qquad s=z+q_o^2 z^{-1}. \end{equation}

$\Theta'(z)$

$|\xi| \geq q_o$

$\Theta'$

$|\xi|\geq q_o$

(54a)\begin{equation} z_k(\xi) = \frac{1}{2} \left| \nu(\xi) + (-1)^{k} \sqrt{\nu^2(\xi) - 4 q_o^2} \right|, \quad k=1,2, \quad \xi \leq -q_o, \end{equation}

(54b)\begin{equation} \hat{z}_k(\xi) = -\frac{1}{2} \left| \nu(\xi) + (-1)^{k} \sqrt{\nu^2(\xi) - 4 q_o^2} \right|, \quad k=1,2, \quad \xi \geq q_o, \end{equation}

(54c)\begin{equation} \nu(\xi) = \frac{1}{2}\left( |\xi|+\sqrt{8 q_o^2 + \xi^2}\right). \end{equation}

(55a)\begin{equation} 0 \lt z_1(\xi) \lt q_o \lt z_2(\xi), \quad \xi \lt -q_o, \end{equation}

(55b)\begin{equation} \hat{z}_2(\xi) \lt -q_o \lt \hat{z}_1(\xi) \lt 0, \quad \xi \gt q_o. \end{equation}

$\xi= \pm q_o$

$z_{k}(-q_o)= q_o$

$\hat{z}_{k}(q_o)= -q_o$

$k=1,2$

• • There is no stationary phase point on $\mathbb{R}$ when $|\xi| \lt q_o$.

• • There is one stationary phase point on $\mathbb{R}$ when $|\xi|=q_o$.

• • There are two stationary phase points on $\mathbb{R}$ when $|\xi| \gt q_o$.

Finally, a direct calculation yields

(56)\begin{equation} \mathop{\rm Re}\nolimits (i \Theta(\xi,z)) = - \xi \, \mathop{\rm Im}\nolimits z \left( 1 + \frac{q_o^2}{ |z|^2}\right) - \mathop{\rm Re}\nolimits z \, \mathop{\rm Im}\nolimits z \left( 1 + \frac{q_o^4}{|z|^4} \right). \end{equation}

Figure 2 shows the sign chart of $\mathop{\rm Re}\nolimits (i \Theta)$ in the three regions for $q_o=1$, along with the location of the stationary phase points in each region. Similar figures can be obtained for arbitrary $q_o$, but the range of $x$-values must be scaled appropriately to account for the location of the stationary phase points.

Figure 2. The sign chart of $\mathop{\rm Re}\nolimits(i \Theta)$, when $|\xi| \lt 1$, $|\xi| \gt 1$ and $|\xi|=1$ for the specific case $q_o=1$. The yellow region corresponds to $\mathop{\rm Re}\nolimits(i \Theta) \gt 0$, and the green region corresponds to $\mathop{\rm Re}\nolimits(i \Theta) \lt 0$.

In the following, we are going to assume that the IC is of box-type, and specifically of the form (3). As shown in [Reference Biondini and Prinari5], if the asymptotic phase difference $\theta=0$ and $0 \lt h \lt q_o$, the scattering problem always has at least one discrete eigenvalue. On the other hand, if $\theta=0$ and $h \gt q_o$, any choice of the box width $L$, and any $\alpha\in[0,\pi] $ satisfying:

(57)\begin{equation} 0\le \alpha \lt \arccos{(q_o/h)} \end{equation}

yields an IC for which no discrete eigenvalue is present. Conversely, if we assume $\alpha=0$, for any choice of $L$ and $h \gt q_o$, any sufficiently small asymptotic phase difference $\theta$, satisfying the condition

(58)\begin{equation} \sin \theta \lt \left( 1-\frac{q_o}{h}\right)\tanh{\left(2q_oL\sqrt{\frac{h^2}{q_o^2}-1}\right)} \end{equation}

(cf. Eq. (45) in [Reference Biondini and Prinari5]) also produces ICs with no discrete eigenvalues.

For definiteness, in most examples, we will consider Eq. (1) with NZBCs (2), IC (3) and the values

(59)\begin{equation} L=1, \quad q_o=1, \quad \theta =0, \quad \alpha=0, \quad h = 1.5. \end{equation}

For these box parameter values, the expressions (29) for the scattering coefficients and the reflection coefficient simplify to

(60a)\begin{equation} a(z) = e^{2 i \lambda(z)} \left( \cos(2 \mu(z)) + i \frac{h-k^2(z)}{\lambda(z) \mu(z)} \sin(2 \mu(z)) \right), \end{equation}

(60b)\begin{equation} b(z) = \frac{k(z)(h-1)}{\lambda(z) \mu (z)}\sin(2 \mu(z)), \end{equation}

(60c)\begin{equation} \rho(z) = e^{-2 i \lambda(z)} \left( \frac{i k(z)(h-1)}{k^2(z) - h + i \lambda(z) \mu(z) \cot(2 \mu(z))} \right), \end{equation}

(61)\begin{equation} k(z)=\frac{1}{2}(z+1/z), \qquad \lambda(z)=\frac{1}{2}(z-1/z), \qquad \mu(z)=\sqrt{k(z)^2-h^2}\,. \end{equation}

The function $\mu(z)$ has branching, with real branch points $k(z)=\pm h$, i.e., at $z=h\pm \sqrt{h^2+q_o^2}$ and $z=-h\pm \sqrt{h^2+q_o^2}$. However, as mentioned in Section 2.2, the values of the scattering coefficients are independent of the choice of the branch for $\mu(z)$ since the dependence on $\mu$ is only via $\sin(2\mu(z))/\mu(z)$ or $\mu(z)\cot (2\mu(z))$, i.e., even functions of $\mu$. The box parameters (59) are chosen so that $a(z)$ has no zeros in $\mathbb{C}^{+}$ (and $a^*(z^*)$ has no zeros in $\mathbb{C}^{-}$), indicating no solitons are present. Other choices of parameters which yield a purely radiative (i.e., solitonless) solution, e.g., satisfying (57) or (58), will also be considered for illustrative purposes.

It is worth mentioning that since we are currently restricting ourselves to the solitonless case, $\tilde{q}(x)+H^{1,1}(\mathbb{R})$ provides an appropriate space for the ICs. Although our box-type ICs do not lie in this space, due to their piecewise-smooth, compactly supported (in relation to the background) nature, the scattering theory can nonetheless be implemented, analytically and numerically, without a hitch, giving an appropriately analytic reflection coefficient.

As pointed out in Section 2, the scattering coefficients generically have simple poles at $z=\pm q_o$ unless the columns of $M_{\pm}$ are linearly dependent at $z=q_o$ or $z=-q_o$ or both, in which case $\alpha_{\pm}=0$. However, in this case, it is straightforward to verify that $\alpha_{\pm} \ne 0$.

Moreover, the reflection coefficient can be analytically continued off the real $z$-axis, and it is in fact analytic in the entire complex plane. In this case, RHP 1 simplifies into the following one.

In the following sections, we provide the necessary deformations that lead to a numerical method that is accurate for arbitrary $x\in \mathbb{R}$, $t \gt 0$ in both the solitonic and the solitonless regions.

3.2. Jump matrix factorisations

Direct calculations show that the jump matrix has the following two factorisations:

(66)\begin{equation} G(\xi,z) = M(\xi,z)P(\xi,z) \equiv L(\xi,z) D(z) U(\xi,z), \end{equation}

where

(67a)\begin{equation} M(\xi,z)=\begin{pmatrix} 1 & - \bar{\rho}(z) e^{-2t \, i \Theta(\xi,z)}\\ 0 & 1 \end{pmatrix}, \quad P(\xi,z) = \begin{pmatrix} 1 & 0\\ \rho(z) e^{2t \, i \Theta(\xi,z)} & 1 \end{pmatrix}, \end{equation}

(67b)\begin{equation} L(\xi,z)=\begin{pmatrix} 1 & 0\\ \frac{\rho(z)}{1-\rho(z)\bar{\rho}(z)} e^{2t \, i \Theta(\xi,z)} & 1 \end{pmatrix}, \quad U(\xi,z)=\begin{pmatrix} 1 & -\frac{\bar{\rho}(z)}{1-\rho(z)\bar{\rho}(z)}e^{-2t \, i \Theta(\xi,z)}\\ 0 & 1 \end{pmatrix}, \end{equation}

(67c)\begin{equation} D(z)=\begin{pmatrix} 1-\rho(z)\bar{\rho}(z) & 0\\ 0 & \frac{1}{1-\rho(z)\bar{\rho}(z)} \end{pmatrix}. \end{equation}

$P, M, L$

$U$

$e^{\pm 2 t \, i \Theta(\xi,z)}$

$\mathrm{Re}(i \Theta)$

$G$

3.3. Solitonic region

In this subsection, we provide the required contour deformations in the solitonic region, corresponding to $\xi\in(-q_o,q_o)$. First, we choose a sufficiently small positive angle $\phi$ to ensure the deformed contours remain inside regions with the same sign in the sign chart and open lenses around $z=0$. While $\phi=\pi/4$ is the optimal choice for the direction of the fastest decay in the nonlinear steepest descent method, the primary requirement is simply to select a small positive angle $\phi$ that maintains the consistency of the sign in the sign chart of $\mathrm{Re}(i \Theta)$.

Four new regions are therefore created, which we denote by $\Omega_k$, $k=1,\cdots,4$, given by

(68a)\begin{equation} \Omega_1 = \{z: \mathrm{arg}\, z \in (0,\phi)\}, \quad \Omega_2 = \{z: \mathrm{arg}\, z \in (\pi - \phi,\pi)\}, \end{equation}

(68b)\begin{equation} \Omega_3 = \{z: \mathrm{arg}\, z \in (-\pi, -\pi + \phi)\}, \quad \Omega_4 = \{z: \mathrm{arg}\, z \in (-\phi,0)\}. \end{equation}

(69a)\begin{equation} \Sigma_1 = e^{i \phi} \mathbb{R}_{+}, \quad \Sigma_2 = e^{i(\pi - \phi)}\mathbb{R}_{+}, \quad \Sigma_3 = e^{-i(\pi - \phi)} \mathbb{R}_{+}, \quad \Sigma_4 = e^{- i \phi} \mathbb{R}_{+}, \end{equation}

the left-to-right oriented boundaries of $\Omega = \bigcup_{k=1}^{4} \Omega_{k}$, see Fig. 3. Panels $(a)$ and $(d)$ in Fig. 2 suggest to use the LDU-factorisation in the regions $\Omega_2 \cup \Omega_3$, and the PM-factorisation in the regions $\Omega_1 \cup \Omega_4$.

Figure 3. Panel a: the initial jump for the function $m$. Panel b: the new jump contours after we open lenses in the solitonic region, and the sign of $\mathop{\rm Re}\nolimits(i \Theta)$ when $-1 \lt \xi \lt 1$. $+$ indicates that $\mathop{\rm Re}\nolimits(i \Theta) \gt 0$ in the corresponding regions, and $-$ stands for $\mathop{\rm Re}\nolimits(i \Theta) \lt 0$.

Next, we define the function $\check{m}(z)$:

(70)\begin{equation} \check{m}(z) = \begin{cases} m(z) P^{-1}(\xi,z), & z \in \Omega_1\\ m(z) U^{-1}(\xi,z), & z \in \Omega_2\\ m(z) L(\xi,z), & z \in \Omega_3\\ m(z) M(\xi,z), & z \in \Omega_4\\ m(z), & \text{elsewhere} \end{cases} \end{equation}

which satisfies the following conditions:

1. 1. $\check{m}(z)$ is analytic for $z\in \mathbb{C} \setminus \Sigma$, where $\Sigma = (-\infty,0) \cup \bigcup_{j=1}^{4} \Sigma_j$, taking continuous boundary values away from $z=0$.

2. 2. The boundary values of $\check{m}(z)$ satisfy the following jump relation across $\Sigma$:

   (71a)\begin{equation} \check{m}^+(z) = \check{m}^-(z) \check{G}(\xi, z), \quad z \in \Sigma \end{equation}

   with jump matrix given by

   (71b)\begin{equation} \check{G}(\xi,z) = \begin{cases} P(\xi,z), & z \in \Sigma_1\\[5pt] U(\xi,z), & z \in \Sigma_2\\[5pt] L(\xi,z), & z \in \Sigma_3\\[5pt] M(\xi,z), & z \in \Sigma_4\\[5pt] D(z), & z \in (-\infty,0)\\[5pt] \end{cases} \end{equation}
   Here, $+/-$ are the limiting values of the function $\check{m}(z)$ as $z$ approaches the oriented contours of non-analyticity from the left and right, respectively.

3. 3. $\check{m}(z)$ has the following asymptotic behaviour:

   (72)\begin{equation} \check{m}(z) = I_2 + \mathcal{O}(1/z), \quad z \to \infty \end{equation}

4. 4. For $t \gt 0$, $\check{m}(z)$ satisfies:

   (73)\begin{equation} \check{m}(z) = \frac{q_o}{z} \sigma_2 e^{-i \theta \sigma_3} + \mathcal{O}(1), \quad z \to 0, \quad z \in \mathbb{C}^{\pm}. \end{equation}

5. 5. $\check{m}(z)$ satisfies the symmetries:

   (74)\begin{equation} \check{m}(z) = \sigma_1 \check{m}^*(z^*) \sigma_1, \quad \check{m} (q_o^2/z) = \frac{z}{q_o} \, \check{m}(z) \, \sigma_2 e^{-i \theta \sigma_3}. \end{equation}

   Note that since all the matrices in (70) have determinant equal to 1, according to Lemma 2.7, we also have $\det \check{m}(z)=1-q_o^2/z^2$.

Proof. Most of the properties above follow in a straightforward way from the definition (70) of $\check{m}(z)$ in terms of $m(z)$, and the corresponding properties of $m(z)$. It is worth discussing in some detail the behaviour of $\check{m}(z)$ as $z\to 0$. Direct calculations show that:

\begin{align*} \check{m}_{11}(z) \sim iq_+\frac{\rho(z)}{z}e^{2it\Theta(z)} & \qquad z\to 0,\ z\in \Omega_1, \\ \check{m}_{22}(z) \sim iq_-\frac{\rho^*(z^*)}{z(1-\rho(z)\rho^*(z^*))}e^{-2it\Theta(z)} & \qquad z\to 0, \ z\in \Omega_2, \\ \check{m}_{11}(z) \sim -iq_+\frac{\rho(z)}{z(1-\rho(z)\rho^*(z^*))}e^{2it\Theta(z)} & \qquad z\to 0, \ z\in \Omega_3, \\ \check{m}_{22}(z) \sim -iq_-\frac{\rho^*(z^*)}{z} e^{-2it\Theta(z)} & \qquad z\to 0, \ z\in \Omega_4, \end{align*}

where we recall that $q_\pm =q_oe^{\pm i \theta}$, and the asymptotic behaviour of all the remaining entries of $\check{m}(z)$ is given by (72). In order to verify (72), one then has to show that the non-tangential limits of the above diagonal entries in each sector $\Omega_j$ are zero, which is the case provided $t \gt 0$. This can be shown for instance by letting $z=\zeta e^{i\alpha}$ with $\alpha$ such that $z\in \Omega_j$, and taking the limit $\zeta\to 0^+$ so that $z\to 0$ in $\Omega_j$ along a ray, while using the explicit asymptotics of $\rho(z)=b(z)/a(z)$ as $z\to 0$ which can be obtained from the formulas in Appendix A. For instance, for $z=\zeta e^{i\alpha}\in \Omega_1$ one has:

\begin{equation*} |\rho(z)/z|\sim (h-1) e^{\sin \alpha/\zeta}, \qquad 0 \lt \alpha \lt \phi \lt \pi/4, \end{equation*}

[with box parameters as in (59)] which grows exponentially as $\zeta\to 0^+$, but as long as $t \gt 0$ the necessary decay is provided by $e^{2it\Theta(z)}$, since

\begin{equation*} |e^{2it\Theta(z)}|\sim e^{-t\sin(2\alpha)/\zeta^2}. \end{equation*}

Similarly, for $z=\zeta e^{i\alpha}\in \Omega_3$

\begin{equation*} \left| \frac{\rho(z)}{z(1-\rho(z)\rho^*(z^*))} \right| \sim \frac{1}{h-1} \frac{1}{\zeta^2} e^{-\sin\alpha/\zeta}, \qquad -\pi \lt \alpha \lt -\pi+\phi \end{equation*}

which grows exponentially as $\zeta\to 0^+$, but again for $t \gt 0$ the necessary decay is provided by $e^{2it\Theta(z)}$.

As to the symmetries, taking into account that for $z \in \mathbb{C}^{+}$:

\begin{equation*} \rho(q_o^2/z)= - e^{-2i \theta} \rho^*(z^*), \quad \bar{\rho}(z) = \rho^*(z^*)\end{equation*}

and $\Theta(q_o^2/z) = - \Theta(z)$, one can derive the following symmetries:

(75a)\begin{equation} \sigma_1 M^*(\xi,z^*) \sigma_1 = P^{-1}(\xi,z), \quad \sigma_1 L^*(\xi,z^*) \sigma_1 = U^{-1}(\xi,z) \end{equation}

(75b)\begin{equation} e^{i \theta \sigma_3} \, \sigma_2 \, M(\xi,q_o^2/z) \, \sigma_2 \, e^{-i \theta \sigma_3} = P^{-1}(\xi,z), \quad e^{i \theta \sigma_3} \, \sigma_2 \, L(\xi,q_o^2/z) \, \sigma_2 \, e^{-i \theta \sigma_3} = U^{-1}(\xi,z). \end{equation}

These symmetries, combined with the known symmetries of $m(z)$, then yield the desired symmetry relations for $\check{m}(z)$.

Next, we discuss the behaviour of $\check{m}(z)$ at $z=\pm q_o$.

Remark 3.3. Since in $\Omega_1$ and $\Omega_4$, $\check{m}(z)$ is related to $m(z)$ via the matrices $P^{-1}$ and $M$, which are bounded at $z=q_o$, $\check{m}^\pm(q_o)$ are bounded, but, unlike the corresponding columns of $m(z)$, $\check{m}^+_1(q_o)\ne0$, $\check{m}^-_2(q_o)\ne 0$, although one still has $\det \check{m}^\pm(q_o)=0$ because the two columns are now proportional. Specifically:

(76)\begin{equation} \check{m}_1^\pm (q_o)=ie^{-i\theta}\check{m}_2^\pm(q_o), \quad \check{m}_2^\pm(q_o)=m_2^+(q_o), \end{equation}

where we have used (36). Conversely, for $z \in \Omega_2$, $\check{m}(z) = m(z) U^{-1}(\xi, z)$, and for $z \in \Omega_3$, $\check{m}(z) = m(z) L(\xi,z)$. Proposition 2.4 shows that the $(1,2)$-entry of $U(\xi,z)$ has a double pole at $-q_o$, and the $(2,1)$-entry of $L(\xi,z)$ has a double pole at $-q_o$. Using Proposition 2.6 and Remark 2.8, one can then show that:

(77a)\begin{equation} \check{m}^+_1(z) = \mathcal{O}(z+q_o) \quad \text{as } z\to -q_o, \ z\in \mathbb{C}^+ \end{equation}

(77b)\begin{equation} (z+q_o)\check{m}_2^+(z) = \mathcal{O}(1) \quad \text{as } z\to -q_o, \ z\in \mathbb{C}^+ \end{equation}

(77c)\begin{equation} (z+q_o)\check{m}_1^-(z) = \mathcal{O}(1) \quad \text{as } z\to -q_o, \ z\in \mathbb{C}^- \end{equation}

(77d)\begin{equation} \check{m}^-_2(z) = \mathcal{O}(z+q_o) \quad \text{as } z\to -q_o, \ z\in \mathbb{C}^- \end{equation}

where $\check{m}_j(z)$ denotes the $j$-th column of $\check{m}(z)$. Therefore, $\check{m}(z)$ is singular at $z=-q_o$. However, as we will see below, we can formulate an RHP for a new function $\hat{m}(z)$ which is continuous across $\mathbb{R}^-$, and which we can prove is non-singular at $-q_o$.

It is worth noticing that for large $t$ the jumps across $\Sigma_j$, for $j=1,\cdots,4$ are exponentially close to the identity matrix, as the exponents $e^{\pm 2 t \, i \Theta(\xi,z)}$ decay rapidly. This leads to increased efficiency in the numerical scheme, since fewer terms need to be computed. However, this does not happen with the jump matrix $D(z)$ across $\mathbb{R}^-$. Next, we define the matrix function $\Delta(z)$ explicitly as follows:

(78)\begin{equation} \Delta(z) = \mathrm{diag} \Big( \delta(z),1/\delta(z) \Big), \quad \delta(z) = \mathrm{exp} \left[ \frac{1}{2 \pi i} \int_{-\infty}^{0} \log \left(1-|\rho(s)|^2 \right) \left( \frac{1}{s-z} - \frac{1}{2s} \right)\, ds \right], \end{equation}

where $\delta(z)$ satisfies the following scalar RHP.

In Section 4 and in Appendix B, we provide details on the numerical computation and the analytical derivation of the function $\delta(z)$, respectively. Note that in the numerical code we use:

\begin{equation*} \tilde{\delta}(z) = \mathrm{exp} \left[ \frac{1}{2 \pi i} \int_{-\infty}^{0} \frac{\log \left(1-|\rho(s)|^2 \right)}{s-z} \, ds \right], \end{equation*}

which is simply proportional to $\delta(z)$:

(83)\begin{equation} \tilde{\delta}(z) = \delta_{\infty}^{-1}\, \delta(z), \end{equation}

and such that $\tilde{\delta}(z)\to 1$ as $z\to \infty$. This definition preserves the first symmetry of $\delta(z)$ ( $z \mapsto z^*$) but not the second one ( $z \mapsto q_o^2/z$), since $\tilde{\delta}^*(z^*)=\tilde{\delta}^{-1}(z)$, while $\tilde{\delta}(q_o^2/z)=\delta_\infty^{-2}\tilde{\delta}^{-1}(z)$. However, one can easily go from $\delta(z)$ to $\tilde{\delta}(z)$ via (83), and the use of one versus the other amounts to the conjugation of the solution of the RHP by a constant diagonal matrix.

Using $\Delta(z)$, we can remove the jump across $\mathbb{R}^-$, by letting:

(84)\begin{equation} \hat{m}(z) = \Delta_{\infty} \, \check{m}(z) \Delta^{-1}(z), \quad \Delta_{\infty} = \mathrm{diag} \left(\delta_{\infty}, 1/\delta_{\infty} \right) \end{equation}

which satisfies the following RHP.

RHP 4 (RHP for $\hat{m}$)

Find a $2 \times 2$ matrix-valued function $\hat{m}(z)$ such that:

1. 1. $\hat{m}(z)$ is analytic for $z\in \mathbb{C} \setminus \Sigma'$, where $\Sigma' = \bigcup_{j=1}^{4} \Sigma_j$, taking continuous boundary values away from $z=0$.

2. 2. The boundary values of $\hat{m}(z)$ satisfy the jump relation across $\Sigma'$

   (85a)\begin{equation} \hat{m}^+(z) = \hat{m}^-(z) \hat{G}(\xi,z), \quad z \in \Sigma' \end{equation}

   where the jump matrix is given by

   (85b)\begin{equation} \hat{G}(\xi,z) = \begin{cases} \Delta(z) P(\xi,z) \Delta^{-1}(z), & z \in \Sigma_1\\[5pt] \Delta(z) U(\xi,z) \Delta^{-1}(z), & z \in \Sigma_2\\[5pt] \Delta(z) L(\xi,z) \Delta^{-1}(z), & z \in \Sigma_3\\[5pt] \Delta(z) M(\xi,z) \Delta^{-1}(z), & z \in \Sigma_4 \end{cases} \end{equation}

3. 3. $\hat{m}(z)$ has the asymptotic behaviour:

   (86)\begin{equation} \hat{m}(z) = I_2 + \mathcal{O}(1/z), \quad z \to \infty \end{equation}

4. 4. For $t \gt 0$, $\hat{m}(z)$ is such that:

   (87)\begin{equation} \hat{m}(z) = \frac{q_o \sigma_2 e^{-i \theta \sigma_3}}{z} + \mathcal{O}(1), \quad z \to 0, \quad z \in \mathbb{C}^{\pm} \end{equation}

   where we have used that $\delta(0)=1/\delta_\infty$ (consistently with the second symmetry for $\delta(z)$), and the simplification $\Delta_{\infty} \sigma_2 e^{-i \theta \sigma_3} \Delta_{\infty} \equiv \sigma_2 e^{-i \theta \sigma_3}$.

5. 5. $\hat{m}(z)$ satisfies the symmetries:

   (88)\begin{equation} \hat{m}(z) = \sigma_1 \hat{m}^*(z^*) \sigma_1, \quad \hat{m} (q_o^2/z) = \frac{z}{q_o} \, \hat{m}(z) \, \sigma_2 e^{-i \theta \sigma_3}. \end{equation}

Proof. Most of the above properties follow in a straightforward way from the definition of $\hat{m}(z)$ and the corresponding properties of $m(z)$. As to the symmetries, they can be derived using the known symmetries for $\check{m}(z)$, along with the following symmetries for $\Delta$:

(89)\begin{equation} \Delta(z) = \sigma_1 \Delta^*(z^*) \sigma_1, \quad \Delta(q_o^2/z) \equiv \Delta^{-1}(z) = \sigma_2 \Delta(z) \sigma_2, \quad \Delta_{\infty} = \sigma_1 \Delta_{\infty}^* \sigma_1. \end{equation}

Note that since both $\Delta_\infty$ and $\Delta(z)$ have determinant equal to 1, it follows from (84) that $\det \hat{m}(z)=\det \check{m}(z)=\det m(z)=1-q_o^2/z^2$.

Proof. We apply Liouville’s theorem to the matrix $\hat{m}_1 \, \hat{m}_2^{-1}$, where $\hat{m}_1$ and $\hat{m}_2$ are two arbitrary solutions to RHP 4. $\hat{m}_1 \, \hat{m}_2^{-1}$ is analytic everywhere, since $\hat{m}_1$ and $\hat{m}_2$ have the same jump condition. Moreover, note that at the origin, we have:

\begin{equation*} \lim_{z \to 0} \hat{m}_1 (z) \, \hat{m}_2^{-1}(z) = \lim_{z \to 0} z^2 (z^2 - q_o^2)^{-1} \hat{m}_1(z) \sigma_2 \hat{m}_2^{T}(z) \sigma_2 = I_2\end{equation*}

where we used the property $\hat{m}_2^{-1}(z) = \left( \det \hat{m}_2 (z) \right)^{-1} \sigma_2 \hat{m}_2^{T}(z) \sigma_2$ for any invertible $2 \times 2$ matrix. Uniqueness then follows from the large- $z$ asymptotics of $\hat{m}_1 \, \hat{m}_2^{-1}$, and the fact that $\det \hat{m}_2(z)=1-q_o^2/z^2$.

Next, we discuss the behaviour of $\hat{m}(z)$ at $z=\pm q_o$.

Remark 3.6. Regarding the boundary values of $\hat{m}(z)$ as $z\to q_o$ from $\mathbb{C}^\pm$, we note that since $\check{m}(z)$ is bounded as $z\to q_o$ from $\mathbb{C}^\pm$, it follows that $\hat{m}(z)$, defined via (84), is also bounded in this limit. Specifically, one has:

\begin{equation*} \hat{m}^\pm (q_o)= \begin{pmatrix} \frac{\delta_\infty}{\delta(q_o)}\check{m}_{11}^\pm (q_o) & \delta_\infty \delta(q_o) \check{m}_{12}^\pm (q_o) \\ \frac{1}{\delta_\infty \delta(q_o)}\check{m}_{21}^\pm (q_o) & \frac{\delta(q_o)}{\delta_\infty} \check{m}_{22}^\pm (q_o) \end{pmatrix}, \end{equation*}

and all the entries are non-zero, because all entries of $\check{m}^\pm (q_o)$ are non-zero. On the other hand, $\det \check{m}^\pm(q_o)=0$ on account of (76), which shows that

(90)\begin{equation} \hat{m}_1^\pm (q_o)=\frac{ie^{-i\theta}}{(\delta(q_o))^2}\hat{m}_2^\pm(q_o), \quad \hat{m}_2^\pm (q_o)=\Delta_\infty \delta(q_o)m_2^+(q_o). \end{equation}

Proof. A direct computation from the definition (84) of $\hat{m}(z)$ gives:

(91a)\begin{equation} \lim_{\substack{z \to - q_o\ \\ z \in \mathbb{C}^{+}}} \hat{m}(z) = \Delta_{\infty} \lim_{z \to - q_o} \begin{pmatrix} \frac{m^{+}_{11}(z)}{\delta^{+}(z)} & \quad m^{+}_{11}(z) \delta^{+}(z) \frac{\bar{\rho}(z)}{1-\rho(z) \bar{\rho}(z)} e^{-2t i \Theta(z)} + m^{+}_{12}(z)\delta^{+}(z)\\ \frac{m^{+}_{21}(z)}{\delta^{+}(z)} & \quad m^{+}_{21}(z) \delta^{+}(z) \frac{\bar{\rho}(z)}{1-\rho(z) \bar{\rho}(z)} e^{-2t i \Theta(z)} + m^{+}_{22}(z)\delta^{+}(z) \end{pmatrix} \end{equation}

and

(91b)\begin{equation} \lim_{\substack{z \to - q_o \\ z \in \mathbb{C}^{-}}} \hat{m}(z) = \Delta_{\infty}\lim_{z \to - q_o } \begin{pmatrix} \frac{m^{-}_{11}(z)}{\delta^{-}(z)} + \frac{m^{-}_{12}(z)}{\delta^{-}(z)} \frac{\rho(z)}{1-\rho(z) \bar{\rho}(z)} e^{2t i \Theta(z)} & \quad m^{-}_{12}(z)\delta^{-}(z)\\ \frac{m^{-}_{21}(z)}{\delta^{-}(z)} + \frac{m^{-}_{22}(z)}{\delta^{-}(z)} \frac{\rho(z)}{1-\rho(z) \bar{\rho}(z)} e^{2t i \Theta(z)} & \quad m^{-}_{22}(z)\delta^{-}(z) \end{pmatrix}. \end{equation}

Using Propositions 2.6 and 2.4, along with the known behaviour of $\delta(z)$ as $z \to -q_o$ from $\mathbb{C}^{\pm}$, the fact that $\Delta_{\infty}$ is a constant matrix, and $\Theta(-q_o)=0$, we conclude that $\hat{m}^{\pm}(-q_o)$ exist and are finite.

Moreover, we can express both columns of $\hat{m}^\pm (-q_o)$ as follows:

(92a)\begin{equation} \hat{m}_2^+(-q_o)=-ie^{i\theta}\left( \lim_{\substack{z \to -q_o \\ z \in \mathbb{C}^{+}}} \frac{\delta^2(z)}{1-\rho(z)\bar{\rho}(z)} \right)\hat{m}_1^+(-q_o), \qquad \hat{m}_1^+(-q_o)=\Delta_\infty \lim_{\substack{z \to -q_o \\ z \in \mathbb{C}^{+}}} \frac{m_1(z)}{\delta(z)}, \end{equation}

(92b)\begin{equation} \hat{m}_1^-(-q_o)=ie^{-i\theta}\left( \lim_{\substack{z \to -q_o \\ z \in \mathbb{C}^{-}}} \frac{1}{\delta^2(z)(1-\rho(z)\bar{\rho}(z))} \right)\hat{m}_2^-(-q_o), \quad \hat{m}_2^-(-q_o)=\Delta_\infty \lim_{\substack{z \to -q_o \\ z \in \mathbb{C}^{-}}} \delta(z) m_2(z), \end{equation}

which shows that both columns are non-zero and proportional to each other. This is consistent with $\det \hat{m}^\pm(-q_o)=0$. The jump condition can be used to show that $\hat{m}(z)$ has no jump across $\mathbb{R}^-$, so $\hat{m}_j^+(-q_o)=\hat{m}_j^-(-q_o)$ for $j=1,2$.

3.3.1. Removing the singularity at $z=0$

The contour deformations described in the previous subsection shift the original jump away from the real axis, transforming the oscillatory behaviour into exponential decay. As a result, the jump matrices are now along rays where, for large $t$, they approach the identity exponentially fast. However, the matrix function $\hat{m}(z)$ in the last (so far) deformed RHP is still singular at $z=0$, which is numerically challenging. To circumvent this problem, we introduce a function $\widetilde{m}(z)$ that satisfies the same RHP as $\hat{m}(z)$ but is non-singular at $z=0$. Specifically, we define $\widetilde{m}(z)$ via the relation:

(93)\begin{equation} \hat{m}(z)=E(z)\widetilde{m}(z), \qquad E(z)= I_2+\frac{q_o\sigma_2e^{-i\theta \sigma_3}}{z}\widetilde{m}^{-1}(0). \end{equation}

We emphasise that the singularity at $z=0$ is not being neglected. We show that there exists a transformation, specifically (93), that takes a solution of RHP 4, singular at $z=0$, into a solution of a new RHP, defined below, which is not singular at $z=0$. This is accomplished via multiplication by $E(z)$, which carries the $1/z$ term. This multiplication ensures that $\widetilde{m}(z)$ is regular at $z=0$, eliminating the singularity from RHP (5) and allowing one to, in effect, ignore condition (87). The transformation (93) could, in principle, introduce singularities at $z=\pm q_o$. As shown in Appendix D, if $\hat{m}(z)$ satisfies the solvability condition

(94)\begin{equation} W\left(\hat{m}_2(q_o),\hat{m}_1(-q_o)\right)\ne 0, \end{equation}

that is, if $\hat{m}_2(q_o)$ and $\hat{m}_1(-q_o)$ are linearly independent, then the function $\widetilde{m}(z)$ defined via relation (93) is bounded at $z=\pm q_o$ (for further details, see Remark 3.10). Moreover, it satisfies the following RHP.

RHP 5 (RHP for $\widetilde{m}$)

Find a $2 \times 2$ matrix-valued function $\widetilde{m}(z)$ such that:

1. 1. $\widetilde{m}(z)$ is analytic for $z\in \mathbb{C} \setminus \Sigma' $, where $\Sigma' = \bigcup_{j=1}^{4} \Sigma_j$, taking continuous boundary values onto $\Sigma^\prime$.

2. 2. The boundary values of $\widetilde{m}(z)$ satisfy the jump relation across $\Sigma'$

   (95a)\begin{equation} \widetilde{m}^+(z) = \widetilde{m}^-(z) \widetilde{G}(\xi,z), \quad z \in \Sigma' \end{equation}
   where $\widetilde{G}(\xi,z)$ is the same jump matrix as in Eq. (85b) (i.e., $\widetilde{G}(\xi,z) = \hat{G}(\xi,z)$).

3. 3. $\widetilde{m}(z)$ has the following asymptotic behaviour:

   (96a)\begin{equation} \widetilde{m}(z) = I_2 + \mathcal{O}(1/z), \quad z \to \infty. \end{equation}

4. 4. The non-tangential limits of $\widetilde{m}(z)$ as $z \to 0$ from $\mathbb{C}^{\pm}$ exist and are equal, and we let

   \begin{equation*} \widetilde{m}(x,t;0) := \lim_{\substack{\zeta \to 0 \\ \zeta \in \mathbb{C}^{\pm}}} \widetilde{m}(x,t;\zeta).\end{equation*}

5. 5. $\widetilde{m}(z)$ satisfies the symmetries

   (97)\begin{equation} \widetilde{m}(z) = \sigma_1 \widetilde{m}^*(z^*) \sigma_1, \quad \widetilde{m}(q_o^2/z)=\widetilde{m}(0)\sigma_2 e^{-i\theta \sigma_3}\widetilde{m}(z)\sigma_2 e^{-i\theta \sigma_3}. \end{equation}

Proof. This follows directly from symmetries (88), the definition of $E(z)$ and relation (93).

Lemma 3.9. Any solution $\widetilde{m}(z)$ of RHP 5 is such that $\det \widetilde{m}(z) = 1$, for all $z \in \mathbb{C}$. Consequently, standard RHP theory implies that if RHP 5 has a solution, then the solution is unique.

Proof. Let $\widetilde{m}(z)$ be a solution to RHP 5. Then, for all $z \in \Sigma'$, $\det \widetilde{m}^{+}(z) = \det \widetilde{m}^{-}(z)$, which implies that $\det \widetilde{m}(z)$ is entire. Moreover, $\det \widetilde{m}(z)=1+\mathcal{O}(1/z)$ as $z \to \infty$, which completes the proof.

Remark 3.10. Here, we want to emphasise that although $\widetilde{m}(z)$ defined in Eq. (93) as $\widetilde{m}(z) = E^{-1}(z) \hat{m}(z)$ may appear to have poles at the points where $\det E(z)=0$, namely at $z = \pm q_o$, this is not actually the case. As shown in Appendix D, for any $x,t$ for which $\hat{m}(x,t;z)$ satisfies the solvability condition (94), there exists a unique $\widetilde{m}(x,t;0)$, consistent with the symmetries, that ensures absence of singularities for $\widetilde{m}(x,t;z)$ at $z=\pm q_o$.

The question of the existence of a solution of RHP 5 remains an open problem. We note that by analytic Fredholm theory a solution can only fail on a set of measure zero in the $(x,t)$-plane. This fact does not appear to adversely affect our numerical scheme.

3.3.2. Reconstructing the solution of RHP 4

The main idea is that we solve numerically the RHP for $\widetilde{m}$ and we recover $\hat{m}$ for any $z$ using relation (93). The following proposition holds.

Proposition 3.11. If $\widetilde{m}(z)$ solves RHP 5, then $\hat{m}(z)$ defined in (93), solves RHP 4.

Proof.

1. 1. $\hat{m}(z)$ and $\widetilde{m}(z)$ share the same jump across $\Sigma'$, and $E(z)$ is analytic there, so the jump condition in RHP 4 follows immediately.

2. 2. The asymptotic behaviour of $\hat{m}(z)$ as $z \to 0$ and $z \to \infty$ in RHP 4 follows directly from (93) and RHP 5.

3. 3. The first symmetry in (88) holds similarly. For the second symmetry, observe that:

   \begin{equation*} \hat{m}(q_o^2/z) = E(q_o^2/z) \widetilde{m}(0) \sigma_2 e^{-i \theta \sigma_3} E^{-1}(z) \hat{m}(z) \sigma_2 e^{-i \theta \sigma_3}.\end{equation*}

   This simplifies to the desired one after using the property:

   \begin{equation*} \frac{z}{q_o}E^{-1}(q_o^2/z)E(z)=\widetilde{m}(0)\sigma_2e^{-i\theta \sigma_3} \end{equation*}

   which is obtained directly from the definition of $E(z)$, and Lemma 3.9.

Remark 3.12. Note that $\hat{m}(z)$ cannot be singular at $z=\pm q_o$, since $E(z)$ is regular at these points and $\widetilde{m}(z)$ is also non-singular there.

The question of which constraint/constraints are needed on the IC (or, correspondingly, on the scattering data) so that the solvability condition (D.10) is satisfied, and whether/how these constraints depend on $x$ and or $t$ is currently an open one. Remark D.5 shows that at $t=0$ the solvability condition is satisfied for any $x\in \mathbb{R}$ by any box-type IC provided $\theta\ne \pi/2$, but a proof that this is the case also for $t \gt 0$ is currently not available. It is reasonable to conjecture that this constraint be related to the existence of solutions of RHP 5. Indeed, if such a solution $\widetilde{m}(z)$ exists, then, on one hand, it is necessarily non-singular at $z=\pm q_o$, and, on the other hand, by virtue of Lemma 3.9 this solution is unique. In turn, this unique solution, regular at $z=\pm q_o$, allows one to obtain via (93) the (unique, by Lemma 3.5) solution $\hat{m}(z)$ of RHP 4. And, based on the calculations in Appendix D, this implies that: (i) this $\hat{m}(z)$ satisfies the solvability condition (otherwise, the corresponding $\widetilde{m}(z)$ defined via (93) would not be bounded at both $z=\pm q_o$); and (ii) $\widetilde{m}(0)$ obtained from RHP 5 must be related to $\hat{m}(\pm q_o)$ via Eqs. (D.19), because this is the unique value of $\widetilde{m}(0)$ which guarantees absence of singularities of $\widetilde{m}(z)$ at $z=\pm q_o$. Arguably, the situations when the solvability condition fails to be satisfied are the same as those in which RHP 5 does not admit a solution, and this is consistent with the fact that in this case one would have to modify RHP 5 so as to include a singularity at either $z=q_o$ or $z=-q_o$.

In our approach, once we choose the IC, we directly assess, at any given $x\in\mathbb{R}$ and $t \gt 0$, whether a numerical solution $\widetilde{m}(x,t;z)$ to RHP 5 exists, and build from it the solution $\hat{m}(x,t;z)$ of RHP 4, and, in turn, the potential $q(x,t)$ (see below). The solvability condition, as well as the fact that Eqs. (D.19) hold, can then be checked a posteriori. This is a strong indication that in all these cases one can expect to be able to rigorously prove that indeed RHP 5 has a solution, but this is left for future work.

Now, Proposition 3.11 implies that the function $\hat{m}(z)$ we recover from relation (93), which is the one obtained after solving numerically for $\widetilde{m}(z)$, is the unique solution to RHP 4. Moreover, by tracking back the sequence of transformations leading to the original function $m(z)$, we obtain the following reconstruction formula for $m(z)$ in terms of $\hat{m}(z)$:

(98)\begin{equation} m(z) = \begin{cases} \Delta_{\infty}^{-1} \hat{m}(z) \Delta(z) P(z), & z \in \Omega_1\\ \Delta_{\infty}^{-1} \hat{m}(z) \Delta(z) U(z), & z \in \Omega_2\\ \Delta_{\infty}^{-1} \hat{m}(z) \Delta(z) L^{-1}(z), & z \in \Omega_3\\ \Delta_{\infty}^{-1} \hat{m}(z) \Delta(z) M^{-1}(z),& z \in \Omega_4\\ \Delta_{\infty}^{-1} \hat{m}(z) \Delta(z), & z \in \mathbb{C} \setminus \Omega_j, \quad j=1,\cdots,4 \end{cases}. \end{equation}

Using the conditions of RHP 4, we can verify that $m(z)$, defined as in (98), satisfies the conditions of RHP 2. Hence, by uniqueness, the solution constructed through (98), is the unique solution to RHP 2.

3.3.3. Reconstruction formula for the potential

Using the above deformations, one can compute $\widetilde{m}(z)$ numerically for any $z$. However, the potential $q(x,t)$ is reconstructed via (51) in terms of the solution of the initial RHP formulated for $m(z)$. Therefore, we need to relate $m(z)$ to $\widetilde{m}(z)$. Tracking the subsequent transformations $m(z) \mapsto \hat{m}(z) \mapsto \widetilde{m}(z)$, we have:

(99)\begin{equation} m(z) = \Delta_{\infty}^{-1} \left( I_2 + \frac{q_o \sigma_2 \, e^{-i \theta \sigma_3}}{z} \widetilde{m}^{-1}(0)\right) \widetilde{m}(z) \Delta(z), \quad z \in \mathbb{C} \setminus \bigcup_{j=1}^4 \Omega_j. \end{equation}

For $z \in \Omega_j$, one can still write the equivalent expression of $m$ in terms of $\widetilde{m}$ using the respective definition of $\check{m}$ in each region. However, the potential $q(x,t)$ is determined by the large- $z$ behaviour of $m(x,t,z)$, and $\check{m}$ coincides with $m$ as $z \to \infty$. Thus, Eq. (99) is sufficient for this purpose. Using Eqs. (51) and (99), one can recover the potential $q(x,t)$ for any $(x,t)$ via the relation

(100)\begin{equation} q(x,t) = i \left[ \Delta_{\infty}^{-1} \left( \widetilde{m}^{(-1)}(x,t) + q_o \sigma_2 \, e^{-i \theta \sigma_3}\widetilde{m}^{-1}(0) \right) \Delta_{\infty} \right]_{12} \end{equation}

where $\widetilde{m}^{(-1)}(x,t)$ is the $\mathcal{O}(1/z)$-order term in the asymptotics of the function $\widetilde{m}$ as $z \to \infty$, and the subscript denotes the $(1,2)$ entry of the prescribed matrix.

In Section 4, we will discuss how to implement a numerical scheme to compute the solution to RHP 4 in the solitonic region numerically by solving the corresponding integral equations for $\widetilde{m}$ using the RHPackage [Reference Olver26] and ISTPackage packages [Reference Trogdon31].

3.4. Solitonless region

In this section, we describe the required contour deformations in the solitonless region.

3.4.1. The case ξ<-q_o

We first consider the case $\xi \lt -q_o$, see panel $(f)$ in Fig. 2. When $\xi \gt q_o$ (panel $(c)$ in Fig. 2), the sign chart changes, but the approach for the deformations is similar, and we will discuss it briefly in the next section. In the solitonless region, unlike the solitonic region, the phase function $\Theta$ has two real stationary phase points. Thus, we fix a sufficiently small positive angle $\phi$ to ensure the deformed contours remain inside regions with the same sign in the sign chart, and open lenses around the origin and the two stationary phase points, $z_1$ and $z_2$. Let us consider the real intervals

(101)\begin{equation} l_1 = \left( 0, \frac{|z_1|}{2} \, \sec \phi \right), \quad l_2 = \left( 0, \frac{|z_2 - z_1|}{2} \, \sec \phi \right) \end{equation}

and denote the boundaries produced after opening lenses by $\Sigma_{kj}$:

(102a)\begin{equation} \Sigma_{01}=e^{i \phi} l_1, \quad \Sigma_{02} = e^{i (\pi - \phi)} \mathbb{R}^{+}, \quad \Sigma_{11} = z_1 + e^{i (\pi - \phi)} l_1, \quad\Sigma_{12} = z_1 + e^{i \phi} l_2 \end{equation}

(102b)\begin{equation} \Sigma_{21} = z_2 + e^{i \phi} \mathbb{R}^{+}, \quad \Sigma_{22} = z_2 + e^{i (\pi - \phi)} l_2 \end{equation}

(102c)\begin{equation} \Sigma_{k4} = \Sigma_{k1}^*, \quad \Sigma_{k3} = \Sigma_{k2}^*, \quad k=0,1,2. \end{equation}

We then denote by $\Omega_{kj}$, $k=0,1,2$ and $j=1,\cdots,4$ the 12 new regions enclosed by the above boundaries, see Fig. 4. Based on panel $(f)$ in Fig. 2, we have

(103a)\begin{equation} \mathrm{Re}(i \Theta(z)) \gt 0, \quad \text{for} \quad z \in \Omega_{kj}, \quad k=0,1,2, \quad j=2,4 \end{equation}

(103b)\begin{equation} \mathrm{Re}(i \Theta(z)) \lt 0, \quad \text{for} \quad z \in \Omega_{kj}, \quad k=0,1,2, \quad j=1,3, \end{equation}

which suggests to use the $\mathrm{LDU}$ factorisation for the jump matrix $G$ in the regions $\Omega_{kj}$, for $k=0,1,2$ and $j=2,3$, and the $\mathrm{PM}$ factorisation in the regions $\Omega_{kj}$, for $k=0,1,2$ and $j=1,4$. Accordingly, we define the function $\check{m}$ as follows:

(104)\begin{equation} \check{m}(z) := \check{m}(\xi,z) = \begin{cases} m(z) P^{-1}(\xi,z), & z \in \Omega_{01} \cup \Omega_{11} \cup \Omega_{21}\\ m(z) U^{-1}(\xi,z), & z \in \Omega_{02} \cup \Omega_{12} \cup \Omega_{22}\\ m(z) L(\xi,z), & z \in \Omega_{03} \cup \Omega_{13} \cup \Omega_{23}\\ m(z) M(\xi,z), & z \in \Omega_{04} \cup \Omega_{14} \cup \Omega_{24}\\ m(z), & \text{elsewhere} \end{cases} \end{equation}

Figure 4. Panel a: the initial jump for the function $m$. Panel b: the new jump contours after we open lenses in the solitonless region when $\xi \lt -q_o$, and the sign of $\mathop{\rm Re}\nolimits(i \Theta)$, where $+$ indicates the real part $\mathop{\rm Re}\nolimits(i \Theta) \gt 0$ in the corresponding regions, and $-$ stands for $\mathop{\rm Re}\nolimits(i \Theta) \lt 0$.

which satisfies the following conditions:

1. 1. $\check{m}(z)$ is analytic for $z\in \mathbb{C} \setminus \Sigma$, where $\Sigma = (z_1,z_2) \cup (-\infty,0) \cup \bigcup_{k=0}^{2} \bigcup_{j=1}^{4} \Sigma_{k j}$

2. 2. $\check{m}(z)$ satisfies the jump relation across $\Sigma$

   (105a)\begin{equation} \check{m}^+(z) = \check{m}^-(z) \check{G}(\xi,z), \quad z \in \Sigma \end{equation}

   with $\check{G}$ given by

   (105b)\begin{equation} \check{G}(\xi,z) = \begin{cases} P(\xi,z), & z \in \Sigma_{01} \cup \Sigma_{11} \cup \Sigma_{21}\\[5pt] U(\xi,z), & z \in \Sigma_{02} \cup \Sigma_{12} \cup \Sigma_{22}\\[5pt] L(\xi,z), & z \in \Sigma_{03} \cup \Sigma_{13} \cup \Sigma_{23}\\[5pt] M(\xi,z), & z \in \Sigma_{04} \cup \Sigma_{14} \cup \Sigma_{24}\\[5pt] D(z), & z \in (-\infty,0) \cup (z_1,z_2)\\[5pt] \end{cases} \end{equation}

3. 3. $\check{m}(z)$ has the following asymptotic behaviour:

   (106)\begin{equation} \check{m}(z) = I_2 + \mathcal{O}(1/z), \quad z \to \infty \end{equation}

4. 4. For $t \gt 0$, $\check{m}(z)$ satisfies:

   (107)\begin{equation} \check{m}(z) = \frac{q_o}{z} \sigma_2 e^{-i \theta \sigma_3} + \mathcal{O}(1), \quad z \to 0, \quad z \in \mathbb{C}^{\pm} \end{equation}

5. 5. $\check{m}(z)$ satisfies the symmetries:

   (108)\begin{equation} \check{m}(z) = \sigma_1 \check{m}^*(z^*) \sigma_1, \quad \check{m} (q_o^2/z) = \frac{z}{q_o} \, \check{m}(z) \, \sigma_2 e^{-i \theta \sigma_3}. \end{equation}

Note that in order for $\check{m}(z)$ to satisfy the second symmetry for all $z$, the contour deformations have to be chosen appropriately for $|z| \lt q_o$ based on the contours for $|z| \gt q_o$. While we believe this can in general be done, it is not practical to enforce it numerically. For this reason, we will not include the symmetries explicitly in some of the deformed RHPs considered below.

Remark 3.13. Since $\Sigma_{02}$ and $\Sigma_{03}$ open around $(-\infty,0)$, and $\Sigma_{12} \cup \Sigma_{22}$ and $\Sigma_{13} \cup \Sigma_{23}$ open around $(z_1,z_2)$, and given that $q_o \in (z_1,z_2)$ and $-q_o \in (-\infty,0)$ (see Eq. (55)), together with the fact that the definition of $\check{m}(z)$ in the respective domains involves the matrices $U(\xi,z)$ and $L(\xi,z)$ (whose entries have double poles at $z=\pm q_o$), Remark 3.3 applies here as well. The key difference compared to the solitonic region is that we must now examine $\check{m}(z)$ as $z \to \pm q_o$, as it may exhibit singular behaviour at both points. In fact, from the definition of $\check{m}(z)$ and Proposition 2.6, one can show that:

• • $\check{m}^+_1(z) = \mathcal{O}(z \mp q_o)$, as $z\to \pm q_o$ from $\mathbb{C}^+$

• • $\check{m}^+_2(z)$ is singular at $\pm q_o$, with $(z \mp q_o)\check{m}_2^+(z) = \mathcal{O}(1)$, as $z\to \pm q_o$ from $\mathbb{C}^+$

• • $\check{m}^-_1(z)$ is singular at $\pm q_o$, with $(z \mp q_o)\check{m}_1^-(z) = \mathcal{O}(1)$, as $z\to \pm q_o$ from $\mathbb{C}^-$

• • $\check{m}^-_2(z) = \mathcal{O}(z \mp q_o)$, as $z\to \pm q_o$ from $\mathbb{C}^-$.

Like in the solitonic region, we need to introduce an appropriate function $\hat{\Delta}$ to remove the jump matrix $D(z)$ across $(-\infty,0) \cup (z_1,z_2)$. Note that in general, $\hat{\Delta}$ is singular at the stationary phase points. To properly handle the singularities, we introduce circles around both $z_1$ and $z_2$, see Fig. 5.

Figure 5. Panel a: circles introduced to avoid the singularities of $\hat{\Delta}$ at the two stationary phase points. Panel b: new regions of non-analyticity for the function $\hat{m}$ near $z_1$. Panel c: jump contours for $\hat{m}$ near $z_1$.

This introduces new regions, denoted by $\omega_j, \omega_j'$, for $j=1,\cdots,6$, see Figs. 5 and 6.

We define the function $\hat{m}$ as follows:

(109)\begin{equation} \hat{m}(z) := \hat{m}(\xi,z) = \begin{cases} \check{m}(z) U^{-1}(\xi,z) \hat{\Delta}(z), & z \in \omega_1 \cup \omega_1'\\ \check{m}(z) P(\xi,z) U^{-1}(\xi,z) \hat{\Delta}(z), & z \in \omega_2 \cup \omega_3 \cup \omega_5' \cup \omega_6'\\ \check{m}(z) M(\xi,z) P(\xi,z) U^{-1}(\xi,z) \hat{\Delta}(z), & z \in \omega_4 \cup \omega_4'\\ \check{m}(z) D(z) \hat{\Delta}(z), & z \in \omega_5 \cup \omega_3'\\ \check{m}(z) \hat{\Delta}(z), & z \in \omega_6 \cup \omega_2'\\ \check{m}(z), & \text{elsewhere} \end{cases} \end{equation}

which satisfies the following conditions:

1. 1. $\hat{m}(z)$ is analytic for $z\in\mathbb{C} \setminus \Sigma'$, taking continuous boundary values, where

   (110)\begin{equation} \Sigma' = \Sigma \cup \bigcup_{j=1}^{6} \sigma_{j} \cup \bigcup_{j=1}^{6} \sigma'_{j}, \quad \Sigma = (-\infty,0) \cup (z_1,z_2) \cup \bigcup_{j=1}^{4} \Sigma_{0 j} \cup \bigcup_{j=1}^4 \hat{\Sigma}_{1 j} \cup \bigcup_{j=1}^{4} \hat{\Sigma}_{2 j} \end{equation}

2. 2. $\hat{m}(z)$ satisfies the jump relation across $\Sigma'$

   (111a)\begin{equation} \hat{m}^+(z) = \hat{m}^-(z) \hat{G}(\xi,z), \quad z \in \Sigma' \end{equation}

   where the jump matrix $\hat{G}$ is given by

   (111b)\begin{equation} \hat{G}(\xi,z) = \begin{cases} U^{-1}(\xi,z) \hat{\Delta}(z), & z \in \sigma_1 \cup \sigma_1'\\[5pt] P(\xi,z) U^{-1}(\xi,z) \hat{\Delta}(z), & z \in \sigma_2 \cup \sigma_3 \cup \sigma_5' \cup \sigma_6'\\[5pt] M(\xi,z) P(\xi,z) U^{-1}(\xi,z) \hat{\Delta}(z), & z \in \sigma_4 \cup \sigma_4'\\[5pt] D(z) \hat{\Delta}(z), & z \in \sigma_5 \cup \sigma_3'\\[5pt] \hat{\Delta}(z), & z \in \sigma_6 \cup \sigma_2'\\[5pt] D(z), & z \in (-\infty,0) \cup (z_1,z_2) \end{cases}\end{equation}
   and the jumps across the segments $\Sigma_{0j}$, for $j=1,\cdots,4$, and $\hat{\Sigma}_{kj}$, for $k=1,2$ and $j=1,\dots,4$ are the same as the jumps for the function $\check{m}$. Here, $\hat{\Sigma}_{kj}$ denote the portions of the segments $\Sigma_{kj}$ which lie outside the circular regions $\omega_j$ and $\omega_j'$. Notice that $\hat{m}(z)$ is analytic inside the circular regions.

3. 3. $\hat{m}(z) = I_2 + \mathcal{O}(1/z), \quad z \to \infty$.

4. 4. Since for $z \in \mathbb{C} \setminus \bigcup_{j=1}^{6} (\omega_j \cup \omega'_j)$, $\hat{m}(z)$ and $\check{m}(z)$ coincide, their limiting behaviour at $z=0$ must also coincide, i.e., for $t \gt 0$

   (112)\begin{equation} \hat{m}(z) = \frac{q_o}{z} \sigma_2 e^{-i \theta \sigma_3} + \mathcal{O}(1), \quad z \to 0, \quad z \in \mathbb{C}^{\pm} \end{equation}

5. 5. For all $z \in \mathbb{C} \setminus \bigcup_{j=1}^{6}( \omega_j \cup \omega'_j)$, $\hat{m}(z)$ satisfies the same symmetries as $\check{m}(z)$ supposing that the contours have been chosen such that they remain invariant under the symmetry $z \mapsto q_o^2/z^2$.

Remark 3.14. Note that since $\hat{m}(z)$ coincides with $\check{m}(z)$ outside $\bigcup_{j=1}^{6} \left(\omega_j \cup \omega'_j\right)$, $\hat{m}(z)$ and $\check{m}(z)$ have the same behaviour at $z = \pm q_o$, i.e.,

• • $\hat{m}^+_1(z) = \mathcal{O}(z \mp q_o)$, as $z\to \pm q_o$ from $\mathbb{C}^+$

• • $\hat{m}^+_2(z)$ is singular at $\pm q_o$, with $(z \mp q_o)\hat{m}_2^+(z) = \mathcal{O}(1)$, as $z\to \pm q_o$ from $\mathbb{C}^+$

• • $\hat{m}^-_1(z)$ is singular at $\pm q_o$, with $(z \mp q_o)\hat{m}_1^-(z) = \mathcal{O}(1)$, as $z\to \pm q_o$ from $\mathbb{C}^-$

• • $\hat{m}^-_2(z) = \mathcal{O}(z \mp q_o)$, as $z\to \pm q_o$ from $\mathbb{C}^-$.

Next, we define the matrix function $\hat{\Delta}(z)$ explicitly as follows:

(113a)\begin{equation} \hat{\Delta}(z) = \mathrm{diag} \Big( \hat{\delta}(z),1/\hat{\delta}(z) \Big), \end{equation}

(113b)\begin{equation}\hat{\delta}(z) = \mathrm{exp} \left[ \frac{1}{2 \pi i} \int_{(-\infty,0) \cup (z_1,z_2)} \log \left(1-|\rho(s)|^2 \right) \left( \frac{1}{s-z} - \frac{1}{2s} \right)\, ds \right] \end{equation}

and $\hat{\delta}(z)$ satisfies the following scalar RHP.

RHP 6 (RHP for $\hat{\delta}$)

Find a scalar function $\hat{\delta}(z)$ such that:

1. 1. $\hat{\delta}(z)$ is analytic in $\mathbb{C} \setminus (-\infty,0) \cup (z_1,z_2)$, taking continuous boundary values away from $\pm q_o$

2. 2. $\hat{\delta}(z)$ satisfies the jump relation

   (114)\begin{equation} \hat{\delta}^{+}(z) = \hat{\delta}^{-}(z) \left(1-|\rho(z)|^2 \right) , \quad z \in (-\infty,0) \cup (z_1,z_2) \end{equation}

3. 3. $\hat{\delta}(z) = \hat{\delta}_{\infty} + \mathcal{O}(1/z)$, as $z \to \infty$, where $\hat{\delta}_{\infty} = \mathrm{exp} \left[ - \frac{1}{4 \pi i} \int_{(-\infty,0) \cup (z_1,z_2)} \frac{\log \left(1-|\rho(s)|^2 \right) }{s} \, ds \right]$.

4. 4. $\hat{\delta}(z)$ has the following asymptotic behaviour as $z\to \pm q_o$:

   (115a)\begin{equation} \hat{\delta}(z) (z \mp q_o)^{-1} = \mathcal{O}(1), \quad z \to \pm q_o, \quad z \in \mathbb{C}^{+} \end{equation}

   (115b)\begin{equation} \hat{\delta}(z) (z \mp q_o) = \mathcal{O}(1), \quad z \to \pm q_o, \quad z \in \mathbb{C}^{-} \end{equation}

5. 5. $\hat{\delta}(z)$ satisfies the symmetries:

   \begin{equation*} \hat{\delta}^*(z^*) = \hat{\delta}^{-1}(z) = \hat{\delta}(q_o^2/z).\end{equation*}

Proof. Using the definition of $\hat{\delta}(z)$ and observing that the two stationary phase points satisfy the symmetry $\frac{1}{z_1} = \frac{z_2}{q_o^2}$, we can establish the second symmetry above.

Using $\hat{\Delta}(z)$, we can remove the jump across $(-\infty,0) \cup (z_1,z_2)$ by letting:

(116)\begin{equation} \breve{m}(z) = \hat{\Delta}_{\infty} \, \hat{m}(z) \hat{\Delta}^{-1}(z), \quad \hat{\Delta}_{\infty} = \mathrm{diag} \left(\hat{\delta}_{\infty}, 1/\hat{\delta}_{\infty} \right) \end{equation}

which satisfies the following RHP.

RHP 7 (RHP for $\breve{m}$)

Find a $2 \times 2$ matrix-valued function $\breve{m}(z)$ such that:

1. 1. $\breve{m}(z)$ is analytic for $z\in \mathbb{C} \setminus \Sigma^{\prime\prime}$, where

   (117)\begin{equation} \Sigma^{\prime\prime} = \bigcup_{j=1}^{6} \sigma_{j} \cup \bigcup_{j=1}^{6} \sigma'_{j} \cup \bigcup_{j=1}^{4} \Sigma_{0 j} \cup \bigcup_{j=1}^4 \hat{\Sigma}_{1 j} \cup \bigcup_{j=1}^{4} \hat{\Sigma}_{2 j} \end{equation}

2. 2. $\breve{m}(z)$ satisfies the jump relation across $\Sigma^{\prime\prime}$

   (118a)\begin{equation} \breve{m}^+(z) = \breve{m}^-(z) \breve{G}(\xi,z), \quad z \in \Sigma^{\prime\prime} \end{equation}

   where the jump matrix $\breve{G}$ is given by

   (118b)\begin{equation} \breve{G}(\xi,z) = \begin{cases} \hat{\Delta}(z) P(\xi,z) \hat{\Delta}^{-1}(z), & z \in \Sigma_{01} \cup \hat{\Sigma}_{11} \cup \hat{\Sigma}_{21}\\[5pt] \hat{\Delta}(z) U(\xi,z) \hat{\Delta}^{-1}(z), & z \in \Sigma_{02} \cup \hat{\Sigma}_{12} \cup \hat{\Sigma}_{22}\\[5pt] \hat{\Delta}(z) L(\xi,z) \hat{\Delta}^{-1}(z), & z \in \Sigma_{03} \cup \hat{\Sigma}_{13} \cup \hat{\Sigma}_{23}\\[5pt] \hat{\Delta}(z) M(\xi,z) \hat{\Delta}^{-1}(z), & z \in \Sigma_{04} \cup \hat{\Sigma}_{14} \cup \hat{\Sigma}_{24}\\[5pt] \hat{\Delta}(z) U^{-1}(\xi,z), & z \in \sigma_{1} \cup \sigma_{1}'\\[5pt] \hat{\Delta}(z) P(\xi,z) U^{-1}(\xi,z), & z \in \sigma_{2} \cup \sigma_{3} \cup \sigma_{5}' \cup \sigma_{6}'\\[5pt] \hat{\Delta}(z) M(\xi,z) P(\xi,z) U^{-1}(\xi,z), & z \in \sigma_{4} \cup \sigma_{4}'\\[5pt] \hat{\Delta}(z) D(z), & z \in \sigma_{5} \cup \sigma_{3}'\\[5pt] \hat{\Delta}(z), & z \in \sigma_{6} \cup \sigma_{2}' \end{cases}\end{equation}
   where again by $\hat{\Sigma}_{kj}$, for $k=1,2$ and $j=1,\cdots,4$, we denote the portions of the segments $\Sigma_{kj}$ which lie outside the circular regions $\omega_j$ and $\omega_j'$.

3. 3. $\breve{m}(z)$ has the following asymptotic behaviour:

   (119)\begin{equation} \breve{m}(z) = I_2 + \mathcal{O}(1/z), \quad z \to \infty \end{equation}

4. 4. For $t \gt 0$, $\breve{m}(z)$ satisfies

   (120)\begin{equation} \breve{m}(z) = \frac{q_o \sigma_2 \, e^{-i \theta \sigma_3}}{z} + \mathcal{O}(1), \quad z \to 0, \quad z \in \mathbb{C}^{\pm}. \end{equation}

Next, we discuss the behaviour of $\breve{m}(z)$ as $z \to \pm q_o$.

Proposition 3.15. The boundary values of $\breve{m}(z)$ as $z \to \pm q_o$ from $\mathbb{C}^\pm$ are bounded.

Proof. For all $z \in \mathbb{C} \setminus \bigcup_{j=1}^6 (\omega_j \cup \omega_j')$, $\hat{m}(z)=\check{m}(z)$, and thus Eqs. (84) and (116) are equivalent. Furthermore, the functions denoted by $\check{m}(z)$ in the two (solitonic and solitonless) regions are defined identically in a neighbourhood of $-q_o$. Consequently, we may use relations (91) for the limiting behaviour of $\breve{m}(z)$ at $z=-q_o$. Moreover, from (104), observe that $\check{m}(z)$ is defined identically in a neighbourhood of both $q_o$ and $-q_o$, and thus the behaviours of $\breve{m}(z)$ at $z=\pm q_o$ coincide. We then write:

(121a)\begin{equation} \lim_{\substack{z \to \pm q_o\ \\ z \in \mathbb{C}^{+}}} \breve{m}(z) = \hat{\Delta}_{\infty} \lim_{z \to \pm q_o} \begin{pmatrix} \frac{m^{+}_{11}(z)}{\hat{\delta}^{+}(z)} & \quad m^{+}_{11}(z) \hat{\delta}^{+}(z) \frac{\bar{\rho}(z)}{1-\rho(z) \bar{\rho}(z)} e^{-2t i \Theta(z)} + m^{+}_{12}(z) \hat{\delta}^{+}(z)\\ \frac{m^{+}_{21}(z)}{\hat{\delta}^{+}(z)} & \quad m^{+}_{21}(z) \hat{\delta}^{+}(z) \frac{\bar{\rho}(z)}{1-\rho(z) \bar{\rho}(z)} e^{-2t i \Theta(z)} + m^{+}_{22}(z) \hat{\delta}^{+}(z) \end{pmatrix} \end{equation}

and

(121b)\begin{equation} \lim_{\substack{z \to \pm q_o \\ z \in \mathbb{C}^{-}}} \breve{m}(z) = \hat{\Delta}_{\infty}\lim_{z \to \pm q_o } \begin{pmatrix} \frac{m^{-}_{11}(z)}{\hat{\delta}^{-}(z)} + \frac{m^{-}_{12}(z)}{\hat{\delta}^{-}(z)} \frac{\rho(z)}{1-\rho(z) \bar{\rho}(z)} e^{2t i \Theta(z)} & \quad m^{-}_{12}(z) \hat{\delta}^{-}(z)\\ \frac{m^{-}_{21}(z)}{\hat{\delta}^{-}(z)} + \frac{m^{-}_{22}(z)}{\hat{\delta}^{-}(z)} \frac{\rho(z)}{1-\rho(z) \bar{\rho}(z)} e^{2t i \Theta(z)} & \quad m^{-}_{22}(z) \hat{\delta}^{-}(z) \end{pmatrix}. \end{equation}

Using Propositions 2.4 and 2.6, along with the known behaviour of $\hat{\delta}(z)$ as $z \to \pm q_o$ from $\mathbb{C}^{\pm}$, the fact that $\hat{\Delta}_{\infty}$ is a constant matrix, and $\Theta(-q_o)=0$, we conclude that $\breve{m}^{\pm}(\pm q_o)$ exist and are finite.

Notice that $\breve{m}(z)$ is singular at $z=0$. Proposition 3.15 shows that in the region under consideration, opening lenses does not introduce singularities at $\pm q_o$. Therefore, we can use the same approach as in Section 3.3.1 to remove the singularity at $z=0$. In this case, we denote the analogue of the function $\widetilde{m}(z)$ as $\tilde{m}(z)$, defined by $\breve{m}(z) = E(z) \tilde{m}(z)$, with $E(z)$ given as in (93). Like in the solitonic region, using the above deformations, we can compute $\tilde{m}(z)$ numerically for any $z \in \mathbb{C}$. Since in recovering the potential $q(x,t)$, we use the large- $z$ behaviour of the respective transformations rather than their local properties, then $q(x,t)$ is still obtained by means of (100) with $\Delta_{\infty}$ replaced by $\hat{\Delta}_{\infty}$.

Remark 3.16. In performing the deformations depicted in Fig. 6, two things are important: (i) the angle at which the black contours leave each circle, and (ii) the radius of the circle. For a generic second-order stationary phase point, the second derivative of the phase function $\Theta(z)$ is non-zero, and these contours should all be separated by angles of $\pi/2$, making angles of $\pi/4$ with the real axis, in order to match the local path of steepest descent. As far as (ii) is concerned, this means that the circle intersects the path of steepest ascent, and we may have growth on the circle. To avoid this issue, we choose the radius proportional to $1/\sqrt{\Theta''(z_j)}$ where $z_j$ is the stationary phase point under consideration. This is feasible because while $\delta$ has a singularity inside the circle, it is a bounded singularity – it behaves as $(z- z_j)^{\pm i \nu}$ for $\nu \gt 0$. This scaling is discussed further in Section 4 below. Additionally, in our numerical implementations, we find it more convenient to use squares instead of circles for the local deformations near the stationary phase points, which of course can be done without loss of generality.

Figure 6. Panel a: new regions of non-analyticity for the function $\hat{m}$ near $z_2$. Panel b: jump contours for $\hat{m}$ near $z_2$.

3.4.2. The case $\xi \gt q_o$

We now briefly discuss the appropriate deformations for the solitonless region when $\xi \gt q_o$. Similarly to the case $\xi \lt -q_o$, we open lenses around the origin and the two stationary phase points, $\hat{z}_1$ and $\hat{z}_2$, creating 12 new regions $\Omega_{kj}$ enclosed by the boundaries $\Sigma_{kj}$, as shown in Fig. 7. The boundaries $\Sigma_{kj}$ are now defined as follows:

(122a)\begin{equation} \Sigma_{01} = \hat{z}_2 + e^{i \phi} l_2, \quad \Sigma_{02} = \hat{z}_2 + e^{i (\pi - \phi)} \mathbb{R}^{+}, \quad \Sigma_{11} = \hat{z}_1 + e^{i (\pi - \phi)} l_2, \quad \Sigma_{12} = \hat{z}_1 + e^{i \phi} l_1 \end{equation}

(122b)\begin{equation} \Sigma_{21} = e^{i \phi} \mathbb{R}^{+}, \quad \Sigma_{22} = e^{i (\pi - \phi)} l_1 \end{equation}

(122c)\begin{equation} \Sigma_{k4} = \Sigma_{k1}^*, \quad \Sigma_{k3} = \Sigma_{k2}^*, \quad k=0,1,2 \end{equation}

and the real intervals $l_1$ and $l_2$ are still given by Eq. (101). Based on panel $(c)$ in Fig. 2, Eqs. (103) also hold in this case, suggesting the use of the $\mathrm{LDU}$ factorisation for the jump matrix $G$ in the regions $\Omega_{kj}$, for $k=0,1,2$ and $j=2,3$, and the $\mathrm{PM}$ factorisation in the regions $\Omega_{kj}$, for $k=0,1,2$ and $j=1,4$. Consequently, the definition of the first deformation $\check{m}(z)$ is identical to (104), and it satisfies the same conditions as in the solitonless region with $\xi \lt -q_o$, with the contour of non-analyticity now given by $\Sigma = (-\infty,\hat{z}_2) \cup (\hat{z}_1,0) \cup \bigcup_{k=0}^{2} \bigcup_{j=1}^{4} \Sigma_{k j} $. Thus, the jump matrix $D(z)$ is across the contour $(-\infty,\hat{z}_2) \cup (\hat{z}_1,0)$. To eliminate this jump, we define

(123a)\begin{equation} \check{\Delta}(z) = \mathrm{diag} \Big( \check{\delta}(z),1/\check{\delta}(z) \Big), \end{equation}

(123b)\begin{equation}\check{\delta}(z) = \mathrm{exp} \left[ \frac{1}{2 \pi i} \int_{(-\infty,\hat{z}_2) \cup (\hat{z}_1,0)} \log \left(1-|\rho(s)|^2 \right) \left( \frac{1}{s-z} - \frac{1}{2s} \right) \, ds \right]. \end{equation}

Figure 7. Panel a: The initial jump for the function $m(z)$. Panel b: The new jump contours after we open lenses in the solitonless region when $\xi \gt q_o$, and the sign of $\mathop{\rm Re}\nolimits(i \Theta)$, where $+$ indicates the real part $\mathop{\rm Re}\nolimits(i \Theta) \gt 0$ in the corresponding regions, and $-$ stands for $\mathop{\rm Re}\nolimits(i \Theta) \lt 0$.

The second deformation $\hat{m}(z)$ is defined as in Eq. (109), except that in this case $\hat{\Delta}(z)$ is replaced by $\check{\Delta}(z)$, $\omega_j$’s are the new circular regions of non-analyticity of $\hat{m}(z)$ near $\hat{z}_1$ and $\omega_j'$’s are the new circular regions of non-analyticity of $\hat{m}(z)$ near $\hat{z}_2$, for $j=1 \cdots 4$. The function $\hat{m}(z)$ satisfies conditions analogous to those for $\hat{m}(z)$ in the solitonless region with $\xi \lt -q_o$, except that $\Sigma$ is now replaced by

\begin{equation*} \Sigma = (-\infty,\hat{z}_2) \cup (\hat{z}_1,0) \cup \bigcup_{j=1}^{4} \hat{\Sigma}_{0j} \cup \bigcup_{j=1}^{4} \hat{\Sigma}_{1j} \cup \bigcup_{j=1}^{4} \Sigma_{2j}\end{equation*}

and again $\hat{\Sigma}_{kj}$ denote the portions of the segments $\Sigma_{kj}$, for $k=0,1$ and $j=1 \cdots 4$, which lie outside the circular regions $\omega_j$ and $\omega_j'$. The jump of $\hat{m}(z)$ across $\Sigma$ is given by Eq. (111b) where $\hat{\Delta}(z)$ is replaced by $\check{\Delta}(z)$, and $D(z)$ is now a jump across $(-\infty,\hat{z}_2) \cup (\hat{z}_1,0)$. Note that $\hat{m}(z)$ is analytic inside the circular regions. To eliminate the jump $D(z)$, we introduce the function $\breve{m}(z)$ given by

\begin{equation*} \breve{m}(z) = \check{\Delta}_{\infty} \, \hat{m}(z) \, \check{\Delta}^{-1}(z)\end{equation*}

where $\check{\Delta}(z)$ is as in (123).

Notice that since $\Sigma_{01}$ and $\Sigma_{11}$ (and likewise $\Sigma_{04}$ and $\Sigma_{14}$) open around $(\hat{z}_2,\hat{z}_1)$, and $\Sigma_{21}$ and $\Sigma_{24}$ open around $(0,\infty)$, and given that $-q_o \in (\hat{z}_2,\hat{z}_1)$ and $q_o \gt 0$ (see Eq. (55)), together with the fact that $\check{m}(z)$ in the respective domains is defined via the matrices $P(\xi,z)$ and $M(\xi,z)$, which are bounded at $\pm q_o$, then $\check{m}(z)$ is also bounded at these points.