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This article is cited in 2 scientific papers (total in 2 papers) On long-time asymptotics of solution to the non-local Lakshmanan–Porsezian–Daniel equation with step-like initial data Wen-Yu Zhou, Shou-Fu Tian , Xiao-Fan Zhang School of Mathematics, China University of Mining and Technology, Xuzhou, P. R. China
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   § 1. IntroductionAs we know, the research of non-linear partial differential equation is a significant part of non-linear science. Since these equations can describe special physical phenomena, many non-linear integrable models have attracted widespread attention of experts and scholars. For example, the propagation of envelope solitons and optical solitons can be controlled by the non-linear Schrödinger (NLS) equation [1] In optical fibers, the NLS equation can be used to characterize the propagation of picosecond soliton pulses and slow modulation of weakly non-linear wave packet amplitudes in moving medium. However, the NLS equation with only lower-order dispersion term is insufficient to describe some complex phenomena. So, it is necessary to further study the effects of higher-order perturbations. Many modified and generalized NLS equations have been proposed [2]–[5]. One of these equations with higher-order dispersion is the Lakshmanan–Porsezian–Daniel (LPD) equation
$$
\begin{equation}
q_t(x,t)+\frac{1}{2}\, iq_{xx}(x,t)-iq^2(x,t)q(x,t)-\gamma H[q(x,t)]=0,
\end{equation}
\tag{1.2}
$$
with
$$
\begin{equation*}
\begin{aligned} \, H[q(x,t)]&=-iq_{xxxx}(x,t)+6iq(x,t)q_x^2(x,t)+4iq(x,t)q^2_x(x,t) \\ &\qquad+8iq^2(x,t)q_{xx}(x,t)+2iq^2(x,t)q_{xx}(x,t)-6iq^5(x,t), \end{aligned}
\end{equation*}
\notag
$$
where $\gamma$ is an arbitrary positive real parameter and the subscripts represent partial differentiations. The LPD equation was first proposed by Lakshmanan, Porsezian and Daniel [6], who considered the one-dimensional classical Heisenberg ferromagnetic spin system and applied a multiple-scaling method to find the perturbed soliton solution in the non-integrable case. The Lax pair and an infinite number of conservation laws have been given [7], [8]. Besides, the breathers, rogue wave and localized wave solutions and other multi-soliton solutions for the LPD equation have been constructed [9]–[11]. The soliton solution has been obtained by inverse scattering transform (IST) method and the initial-boundary value problem of LPD equation on the half-line has been analyzed [12], [13]. The LPD equation, which has many applications in non-linear optics and physics, can be used as a model for describing the propagation and interaction of the ultrashort pulses in high-speed optical fiber transmission system and has many other extensive applications, and so it has great research value and significance. In this work, to explore the long-time asymptotics different from the local LPD equation, we further study the non-local LPD equation
$$
\begin{equation}
\begin{gathered} \, \begin{cases} q_t(x,t)+\dfrac{1}{2}iq_{xx}(x,t)-iq^2(x,t)r(x,t)-\gamma H[q(x,t)]=0, &x\in\mathbb{R},\ t>0, \\ q(x,0)=q_0(x), &x\in\mathbb{R}, \end{cases} \\ \begin{split} H[q(x,t)]&=-iq_{xxxx}(x,t)+6ir(x,t)q_x^2(x,t)+4iq(x,t)q_x(x,t)r_x(x,t) \\ &\qquad+8ir(x,t)q(x,t)q_{xx}(x,t)+2iq^2(x,t)r_{xx}(x,t)-6ir^2(x,t)q^3(x,t). \end{split} \nonumber \end{gathered}
\end{equation}
\tag{1.3}
$$
The initial function $q_0(x)$ is sufficiently smooth and has the step-like form
$$
\begin{equation}
q_0(x)\to \begin{cases} 0, &x\to-\infty, \\ A, &x\to+\infty, \end{cases}
\end{equation}
\tag{1.4}
$$
where A is a positive arbitrary constant. In addition, we have $r(x,t)=\sigma \overline{q}(-x,t)$, $\sigma=\pm1$. The value $\sigma=1$ corresponds to the defocusing case and $\sigma=-1$ corresponds to the focusing case. We mainly pay close attention to the latter case. The solution $q(x,t)$ satisfies the following boundary conditions: for any positive value of $t$. The potential of non-local equations induced by the non-linear term of the non-local equation is the Parity-Time (PT) symmetry, which was first introduced by Ablowitz and Musslimani to study the non-local NLS equation
$$
\begin{equation}
iq_t(x,t)+q_{xx}(x,t)\pm 2q(x,t)\overline{q}(-x,t)q(x,t)=0,
\end{equation}
\tag{1.6}
$$
which possesses a Lax pair and conservation laws [14]; $\overline{q}(-x,t)$ denotes the complex conjugation of $q(-x,t)$. The PT symmetry is a special reduction from the famous AKNS system as $r(x,t)=\overline{q}(-x,t)$ (see [15]). Compared with local equations, the solutions of non-local equations may exhibit some different properties and phenomena such as non-local NLS equation [16], non-local modified Korteweg–de Vries (mKdV) equation [17], non-local Camassa–Holm (CH) equation [18], and so on. The study of asymptotic solutions to non-linear dispersion equations is a hot topic. It was first brought into forcing with the IST method by Manakov [19]. Later, Deift and Zhou, inspired by this work, developed the non-linear steepest descent method [20]. Subsequently, this method became a powerful tool to studying the long-time asymptotic behavior of integrable systems [21]–[24]. Especially, different from the decaying initial condition, the equations with step-like initial data can generate many new properties and phenomena. They have attracted widespread attention of many experts and scholars. For example, the long-time asymptotics of focusing NLS equation with step-like initial data have been studied [25]–[27]. The long time/zero-dispersion limit of the solution of defocusing NLS equation has been studied for a family of step-like initial data by the steepest descent techniques [28]. Moreover, as different from the long-time asymptotics of the solution $q(x, t)$ for the non-local NLS equation along the rays $x/t = C$, where $C$ is a non-zero constant, Rybalko and Shepelsky [23], [29]–[32] have significantly extended these asymptotic results. In particularly, they found the asymptotics within a one-parameter family of curved wedges that is asymptotically closer to the ray $x = 0$ than any of these rays. In addition, the long-time asymptotics of mKdV equation [33]–[35], the CH equation [36]–[38], and the DNLS equation [39] with step-like initial data were also investigated. Since the LPD equation is a special case of the fifth-order NLS equation, especially for the non-local LPD equation, it has a far-reaching mathematical and physical significance. So far, many experts and scholars have obtained several noteworthy results that can promote a further development of integrable systems. The integrability property, periodic wave, modulation instability, rational soliton solutions and nonsingular solutions have been given by the Darboux transform method or the Bäcklund transformation approach in [40]–[42]. The RH problem based on IST was studied for constructing soliton solutions of non-local LPD equation [12], [43]. On the basis of these integrable properties, we can further study the soliton solutions of this equation, including soliton solutions, rational solutions, breathing solutions and soliton interaction. In addition, these non-linear wave solutions can be used in experiments of physical fields and optical fibers fields. The rich optical solitons and the generated soliton dynamics behaviors contribute to further exploration in the field of non-linear optics. In addition, some new achievements have been made in the long-time asymptotic analysis of the solution to non-local LPD equation. In [44], Wang, Guo, and Liu found the long-time asymptotic behavior of the LPD equation by employing the non-linear steepest descent analysis in the Schwartz space. Besides, in [45], Peng and Chen derived the explicit long-time asymptotic formula of the non-local LPD equation. It is worth noting that the above works study the LPD equation under the condition that the initial data belongs to the Schwartz space. Here, inspired by the above two works and the work on non-local NLS equation [29], we extend these results to derive the asymptotic solutions of non-local LPD equation (1.3) with step-like initial data, and in the asymptotic analysis we will consider the following complex problems. (I) The LPD equation with step-like initial value conditions is investigated. There produces the new singularity point $\xi=0$, and the asymptotic analysis of it needs to be carried out. (II) We make two assumptions (Case 1 and Case 2) about the discrete spectrum. To facilitate subsequent analysis, during a series of deformations of the RH problem, a BP factor is introduced to transform the original RH problem into a regular RH problem without residue conditions. (III) Due to the influence of non-local symmetries $x\to-x$ and $t\to t$, comparing to the solution with rapidly decaying initial data, the results of the long-time asymptotic behaviors for the solution with step-like initial data are more abundant when the value ranges of $x$ are different. (IV) For the space-time region $\mu^2:=(x/t)^2<1/(27\gamma)$, there exist three real saddle points. The asymptotics here have a more complicated error term, which is different from the classical NLS equation. The parabolic cylinder functions are used to solve the leading term. The solution is also compatible with the step-like boundary conditions. (V) For the space-time region $\mu^2:=(x/t)^2>1/(27\gamma)$, there are one real and two conjugate complex saddle points. In this situation, the processing method at the real phase point is same as before. At the complex phase points, we deform the jump contour on the real axis to other contours away from the real axis. Finally, based on the analysis of the two parts, an asymptotic solution is constructed. 1.1. Main resultsThe different space-time regions of the whole plane are decomposed to several parts (see Fig. 1). To find the long-time asymptotic behavior of the solution, we make specific analyses for these space-time regions. In the asymptotic regions $\mathcal{M}_1$ $(\mathcal{M}_2)$ and $\mathcal{N}_1$ $(\mathcal{N}_2)$, which correspond to the case $0<\mu<\sqrt{1/(27\gamma)}$ $(-\sqrt{1/(27\gamma)}<\mu<0)$, the LPD equation has three saddle points. That is to say, in the four regions $\{\mathcal{M}_1, \mathcal{M}_2, \mathcal{N}_1, \mathcal{N}_2\}$, we can construct the same asymptotic solution except the following differences. For the asymptotic formula of solutions, the difference is that the variables $x$ and $t$ have the same signs in the regions $\{\mathcal{M}_1, \mathcal{N}_1\}$ and the opposite signs in the regions $\{\mathcal{M}_2, \mathcal{N}_2\}$. Besides, the signs of the three saddle points are opposite to those in the two parts regions. The main results of this case are as follows. Theorem 1.1. Let the initial data $q_0(x)\in L^1(\mathbb{R}^-)$ and $q_0(x)-A\in L^1(\mathbb{R}^+)$, which is a compact perturbation of the pure step initial data (2.31): $q_0(x)-q_{0A}(x)\,{=}\,0$ for $|x|>\varepsilon$ with some $\varepsilon>0$ in the Cauchy problem (1.3)–(1.5). Moreover, let the scattering data $a_1(\xi)$, $a_2(\xi)$, and $b(\xi)$ determined in (2.19) satisfy the following assumptions: (I) $a_1(\xi)$ has a single simple zero point in $\overline{\mathbb{C}^+}$ at $\xi=i\xi_1$ and $a_2(\xi)$ either has no zero points or has a single simple zero point in $\overline{\mathbb{C}^-}$ at $\xi=0$; (II) $\operatorname{Im} v(\lambda_j)\in(-1/2,1/2)$ for $j=1,2,3$, where $\operatorname{Im} v(\lambda_j)=\frac{1}{2\pi}\int_{-\infty}^{\lambda_j} d \arg(1+r_1(s)r_2(s))$, $r_1(\xi)=\overline{b(-\overline{\xi})}/a_1(\xi)$ and $r_2(\xi)= b(\xi)/a_2(\xi)$. Let $q(x,t)$ be a sufficiently smooth solution of the Cauchy problem (1.3)–(1.5). Then the long-time asymptotics of $q(x,t)$ along any line $\mu^2:=(x/t)^2<1/(27\gamma)$ can be obtained as follows: (i) for $x<0$:
$$
\begin{equation}
\begin{aligned} \, q(x,t)=&-\sum_{s=1}^3t^{-\frac{1}{2}+(-1)^s\operatorname{Im} v(-\lambda_s)}\exp\bigl\{-2[\overline{\chi_s}+\overline{\phi_s}(-\mu,\tau)] \nonumber \\ &\qquad\qquad+(-1)^si\operatorname{Re} v(-\lambda_s)\ln t\bigr\}H_s+R_1(-\mu,t), \end{aligned}
\end{equation}
\tag{1.7}
$$
with
$$
\begin{equation}
H_1 = \frac{\sqrt{2\pi}e^{-\frac{\pi}{2}\overline{v(-\lambda_1)}}e^{\frac{i\pi}{4}}} {\sqrt{48\gamma\lambda^2_1-1}\overline{r_2}(-\lambda_1)\Gamma(-i\overline{v(-\lambda_1)})} \biggl(\frac{1-48\gamma\lambda^2_2}{4(48\gamma\lambda^2_1-1)(48\gamma\lambda^2_3-1)} \biggr)^{i\overline{v(-\lambda_1)}},
\end{equation}
\tag{1.8a}
$$
$$
\begin{equation}
H_2 = \frac{\sqrt{2\pi}e^{-\frac{\pi}{2}v(-\lambda_2)}e^{-\frac{i\pi}{4}}} {\sqrt{1-48\gamma\lambda^2_2}r_2(-\lambda_2)\Gamma(iv(-\lambda_2))} \biggl(\frac{1}{4(48\gamma\lambda^2_3-1)}\biggr)^{-i\overline{v(-\lambda_3)}} \nonumber
\end{equation}
\notag
$$
$$
\begin{equation}
\qquad\times \biggl(\frac{1-48\gamma\lambda^2_2}{48\gamma\lambda^2_1-1}\biggr)^{-i\overline{v(-\lambda_2)}},
\end{equation}
\tag{1.8b}
$$
$$
\begin{equation}
H_3 = \frac{\sqrt{2\pi}e^{-\frac{\pi}{2}\overline{v(-\lambda_3)}}e^{\frac{i\pi}{4}}} {\sqrt{48\gamma\lambda^2_3-1}\overline{r_2}(-\lambda_3)\Gamma(-i\overline{v(-\lambda_3)})} \biggl(\frac{1-48\gamma\lambda^2_2}{4(48\gamma\lambda^2_3-1) (48\gamma\lambda^2_1-1)}\biggr)^{i\overline{v(-\lambda_3)}};
\end{equation}
\tag{1.8c}
$$
(ii) for $x>0$: based on the values of $\operatorname{Im} v(\lambda_j)$, $j=1,2,3$, and assuming that they are in the same interval for all $j=1,2,3$, the three possible types of asymptotics are as follows: (a) $\operatorname{Im} v(\lambda_j)\in I_1=(-1/2,-1/6]$, $j=1,2,3$,
$$
\begin{equation}
\begin{aligned} \, q(x,t)=&\sum_{s=1}^3t^{-\frac{1}{2}+(-1)^s\operatorname{Im} v(\lambda_s)}\exp\bigl\{-2[\chi_s+\phi_s(\mu,\tau)] -(-1)^si\operatorname{Re} v(\lambda_s)\ln t\bigr\}N_s \nonumber \\ &\qquad+A\delta^2(\mu,0)+R_1(\mu,t); \end{aligned}
\end{equation}
\tag{1.9}
$$
(b) $\operatorname{Im} v(\lambda_j)\in I_2=(-1/6,1/6)$, $j=1,2,3$,
$$
\begin{equation}
\begin{aligned} \, q(x,t)&=-\sum_{s=1}^3t^{-\frac{1}{2}-(-1)^s\operatorname{Im} v(\lambda_s)}\exp\bigl\{2[\chi_s+\phi_s(\mu,\tau)] +(-1)^si\operatorname{Re} v(\lambda_s)\ln t\bigr\}L_s \nonumber \\ &\qquad+\sum_{s=1}^3t^{-\frac{1}{2}+(-1)^s\operatorname{Im} v(\lambda_s)}\exp\bigl\{-2[\chi_s+\phi_s(\mu,\tau)] -(-1)^si\operatorname{Re} v(\lambda_s)\ln t\bigr\}N_s \nonumber \\ &\qquad+A\delta^2(\mu,0)+R_3(\mu,t); \end{aligned}
\end{equation}
\tag{1.10}
$$
(c) $\operatorname{Im} v(\lambda_j)\in I_3=[1/6,1/2)$, $j=1,2,3$,
$$
\begin{equation}
\begin{aligned} \, q(x,t)=&-\sum_{s=1}^3t^{-\frac{1}{2}-(-1)^s\operatorname{Im} v(\lambda_s)}\exp\bigl\{2[\chi_s+\phi_s(\mu,\tau)]+(-1)^si\operatorname{Re} v(\lambda_s)\ln t\bigr\}L_s \nonumber \\ &\qquad\qquad+A\delta^2(\mu,0)+R_2(\mu,t), \end{aligned}
\end{equation}
\tag{1.11}
$$
with
$$
\begin{equation}
L_1 = \frac{\sqrt{2\pi}e^{-\frac{\pi}{2}v(\lambda_1)}e^{\frac{i\pi}{4}}} {\sqrt{48\gamma\lambda^2_1-1}r_1(\lambda_1)\Gamma(-iv(\lambda_1))} \biggl(\frac{1-48\gamma\lambda^2_2}{4(48\gamma\lambda^2_1-1)(48\gamma\lambda^2_3-1)}\biggr)^{iv(\lambda_1)},
\end{equation}
\tag{1.12a}
$$
$$
\begin{equation}
L_2 = \frac{\sqrt{2\pi}e^{-\frac{\pi}{2}\overline{v(\lambda_2)}}e^{-\frac{i\pi}{4}}} {\sqrt{1-48\gamma\lambda^2_2}\overline{r_1}(\lambda_2)\Gamma(i\overline{v(\lambda_2)})} \biggl(\frac{1}{4(48\gamma\lambda^2_3-1)}\biggr)^{-iv(\lambda_3)} \biggl(\frac{1-48\gamma\lambda^2_2}{48\gamma\lambda^2_1-1}\biggr)^{-iv(\lambda_2)},
\end{equation}
\tag{1.12b}
$$
$$
\begin{equation}
L_3 = \frac{\sqrt{2\pi}e^{-\frac{\pi}{2}v(\lambda_3)}e^{\frac{i\pi}{4}}} {\sqrt{48\gamma\lambda^2_3-1}r_1(\lambda_3)\Gamma(-iv(\lambda_3))} \biggl(\frac{1-48\gamma\lambda^2_2}{4(48\gamma\lambda^2_3-1)(48\gamma\lambda^2_1-1)}\biggr)^{iv(\lambda_3)},
\end{equation}
\tag{1.12c}
$$
and
$$
\begin{equation}
N_1 = \frac{c^2_0\sqrt{2\pi}e^{-\frac{\pi}{2}v(\lambda_1)}e^{-\frac{i\pi}{4}}} {\sqrt{48\gamma\lambda^2_1-1}r_2(\lambda_1)\Gamma(iv(\lambda_1))\lambda^2_1} \biggl(\frac{1-48\gamma\lambda^2_2}{4(48\gamma\lambda^2_1-1)(48\gamma\lambda^2_3-1)}\biggr)^{-iv(\lambda_1)},
\end{equation}
\tag{1.13a}
$$
$$
\begin{equation}
N_2 = \frac{c^2_0\sqrt{2\pi}e^{-\frac{\pi}{2}\overline{v(\lambda_2)}}e^{\frac{i\pi}{4}}} {\sqrt{1-48\gamma\lambda^2_2}\overline{r_2}(\lambda_2)\Gamma(-i\overline{v(\lambda_2)})\lambda^2_2} \biggl(\frac{1}{4(48\gamma\lambda^2_3-1)}\biggr)^{-iv(\lambda_1)} \biggl(\frac{1-48\gamma\lambda^2_2}{48\gamma\lambda^2_1-1}\biggr)^{iv(\lambda_2)},
\end{equation}
\tag{1.13b}
$$
$$
\begin{equation}
N_3 = \frac{c^2_0\sqrt{2\pi}e^{-\frac{\pi}{2}v(\lambda_3)}e^{-\frac{i\pi}{4}}} {\sqrt{48\gamma\lambda^2_3-1}r_2(\lambda_3)\Gamma(iv(\lambda_3))\lambda^2_3} \biggl(\frac{1-48\gamma\lambda^2_2}{4(48\gamma\lambda^2_3-1)(48\gamma\lambda^2_1-1)}\biggr)^{-iv(\lambda_3)};
\end{equation}
\tag{1.13c}
$$
$\delta(\mu,0)$ is defined by
$$
\begin{equation*}
\delta(\mu,0)=\exp\biggl\{\frac{1}{2\pi i}\biggl(\int_{-\infty}^{\lambda_3}+\int_{\lambda_2}^{\lambda_1}\biggr) \frac{\ln(1+r_1(s)r_2(s))}{s} \, ds \biggr\},
\end{equation*}
\notag
$$
where $v(\lambda_j)$, $j=1,2,3$, are as in (3.6). Moreover, $\chi_j(\xi)$ are given by (3.5); $\phi_j(\mu,\tau(\xi))$ are shown in (3.44), (3.47) and (3.50); $\Gamma(\,{\cdot}\,)$ is the Euler gamma function. The error estimates related by $R_3(\mu,t)=R_1(\mu,t)+R_2(\mu,t)$ and $R_l(\mu,t)$, $l=1,2$, are defined as follows:
$$
\begin{equation}
R_1=\begin{cases} O(t^{-1}), &(-1)^j\operatorname{Im} v(\lambda_j)>0, \\ O(t^{-1}\ln t), &\operatorname{Im} v(\lambda_j)=0, \\ &\quad(-1)^{l}\operatorname{Im} v(\lambda_l)\leqslant0, \, l\neq j, \\ O(t^{-1+2|{\operatorname{Im} v(\lambda_1)}|}), &\operatorname{Im} v(\lambda_1)>0,\, \operatorname{Im} v(\lambda_2)\geqslant0, \\ &\quad\operatorname{Im} v(\lambda_3)\leqslant0, \\ O(t^{-1+2|{\operatorname{Im} v(\lambda_2)}|}), &\operatorname{Im} v(\lambda_1)\leqslant0, \, \operatorname{Im} v(\lambda_2)<0, \\ &\quad\operatorname{Im} v(\lambda_3)\leqslant0, \\ O(t^{-1+2|{\operatorname{Im} v(\lambda_3)}|}), &\operatorname{Im} v(\lambda_1)\leqslant0, \, \operatorname{Im} v(\lambda_2)\geqslant0, \\ &\quad\operatorname{Im} v(\lambda_3)>0, \\ O(t^{-1+2\max\{|{\operatorname{Im} v(\lambda_1)}|,|{\operatorname{Im} v(\lambda_2)}|\}}), &\operatorname{Im} v(\lambda_1)>0, \, \operatorname{Im} v(\lambda_2)<0, \\ &\quad\operatorname{Im} v(\lambda_3)\leqslant0, \\ O(t^{-1+2\max\{|{\operatorname{Im} v(\lambda_2)}|,|{\operatorname{Im} v(\lambda_3)}|\}}), &\operatorname{Im} v(\lambda_1)\leqslant0, \, \operatorname{Im} v(\lambda_2)<0, \\ &\quad \operatorname{Im} v(\lambda_3)>0, \\ O(t^{-1+2\max\{|{\operatorname{Im} v(\lambda_1)}|,|{\operatorname{Im} v(\lambda_3)}|\}}), &\operatorname{Im} v(\lambda_1)>0, \, \operatorname{Im} v(\lambda_2)\geqslant0, \\ &\quad\operatorname{Im} v(\lambda_3)>0, \\ O(t^{-1+2\max\{|{\operatorname{Im} v(\lambda_1)}|,|{\operatorname{Im} v(\lambda_2)}|,|{\operatorname{Im} v(\lambda_3)}|\}}), &(-1)^j\operatorname{Im} v(\lambda_j)<0, \end{cases}
\end{equation}
\tag{1.14}
$$
$$
\begin{equation}
R_2= \begin{cases} O(t^{-1+2\max\{|{\operatorname{Im} v(\lambda_1)}|,|{\operatorname{Im} v(\lambda_2)}|,|{\operatorname{Im} v(\lambda_3)}|\}}), &(-1)^j\operatorname{Im} v(\lambda_j)>0, \\ O(t^{-1+2\max\{|{\operatorname{Im} v(\lambda_1)}|,|{\operatorname{Im} v(\lambda_2)}|\}}), &\operatorname{Im} v(\lambda_1)<0,\, \operatorname{Im} v(\lambda_2)>0, \\ &\quad\operatorname{Im} v(\lambda_3)\geqslant0, \\ O(t^{-1+2\max\{|{\operatorname{Im} v(\lambda_2)}|,|{\operatorname{Im} v(\lambda_3)}|\}}), &\operatorname{Im} v(\lambda_1)\geqslant0,\, \operatorname{Im} v(\lambda_2)>0, \\ &\quad\operatorname{Im} v(\lambda_3)<0, \\ O(t^{-1+2\max\{|{\operatorname{Im} v(\lambda_1)}|,|{\operatorname{Im} v(\lambda_3)}|\}}), &\operatorname{Im} v(\lambda_1)<0,\, \operatorname{Im} v(\lambda_2)\leqslant0, \\ &\quad\operatorname{Im} v(\lambda_3)<0, \\ O(t^{-1+2|{\operatorname{Im} v(\lambda_1)}|}), &\operatorname{Im} v(\lambda_1)<0,\, \operatorname{Im} v(\lambda_2)\leqslant0, \\ &\quad\operatorname{Im} v(\lambda_3)\geqslant0, \\ O(t^{-1+2|{\operatorname{Im} v(\lambda_2)}|}), &\operatorname{Im} v(\lambda_1)\geqslant0,\, \operatorname{Im} v(\lambda_2)>0, \\ &\quad\operatorname{Im} v(\lambda_3)\geqslant0, \\ O(t^{-1+2|{\operatorname{Im} v(\lambda_3)}|}), &\operatorname{Im} v(\lambda_1)\geqslant0, \, \operatorname{Im} v(\lambda_2)\leqslant0, \\ &\quad\operatorname{Im} v(\lambda_3)<0, \\ O(t^{-1}\ln t), &\operatorname{Im} v(\lambda_j)=0, \\ &\quad (-1)^{l}\operatorname{Im} v(\lambda_l)\geqslant0, \, l\neq j, \\ O(t^{-1}), &(-1)^j\operatorname{Im} v(\lambda_j)<0. \end{cases}
\end{equation}
\tag{1.15}
$$
Remark 1.2. For case $x>0$, we only consider all $\operatorname{Im} v(\lambda_j)$ for $j=1,2,3$ that are located in the same interval, otherwise, for instance, $\operatorname{Im} v(\lambda_j)\in I_1(I_3)$, then the solution $q(x,t)$ contains the terms owning $N_j(L_j)$. If $\operatorname{Im} v(\lambda_j)\in I_2$, the solution $q(x,t)$ contains the terms owning $N_j$ and the terms owning $L_j$. Remark 1.3. It can be seen that $A\delta^2(\mu,0)\to A$ as $\lambda_j\to\infty$ for $j=1,2,3$, thus the asymptotic solutions in (1.9)–(1.11) with the boundary condition (1.5b) still hold. Remark 1.4. For the case of pure-step initial data in (2.31), the two assumptions (I) and (II) of Theorem 1.1 can be all satisfied. Furthermore, in this case, we have $1+r_1(\xi)r_2(\xi)=4\xi^2/(4\xi^2+A^2)$, which means that $\operatorname{Im} v(\,{\cdot}\,)=0$. Remark 1.5. We can ensure the existence of a solution $q(x,t)$, since we require that the initial function $q_0(x)$ should be sufficiently smooth and satisfy $q_0(x) \to 0$ as $x\to-\infty$ and $q_0(x)\to A$ as $x\to+\infty$. In fact, we can lower the smoothness requirement in the initial data $q_0(x)$ ensuring only that it should belong to a weighted Sobolev space with certain exponent. Moreover, the reconstruction formula can establish a connection between the RH problem and the solution. Through demonstrating the existence and uniqueness of the solution to RH problem, we can prove the existence of the solution to the LPD equation. A rigorous proof depends on studying the well-posed problem for the equation, and there are currently some relevant works [47] and [48]. In the asymptotic regions $\mathcal{M}_4$ $(\mathcal{M}_3)$ and $\mathcal{N}_4$ $(\mathcal{N}_3)$, which correspond to the case $\sqrt{1/(27\gamma)}<\mu$ $(\mu<-\sqrt{1/(27\gamma)})$, the LPD equation has one real saddle point and a pair of conjugate complex saddle points. Thus, in the regions $\{\mathcal{M}_3, \mathcal{M}_4, \mathcal{N}_3, \mathcal{N}_4\}$, we can construct the same asymptotic solution except the following differences. For the asymptotic formula of solutions, the difference is that the variables $x$ and $t$ have the same signs in the regions $\{\mathcal{M}_4, \mathcal{N}_4\}$ and the opposite signs in the regions $\{\mathcal{M}_3, \mathcal{N}_3\}$. Besides, the signs of the saddle point $\lambda_3$ and the real part of the saddle points $\lambda_1$ and $\lambda_2$ are opposite for the two parts regions. The main results of this case are as follows. Theorem 1.6. Under the conditions of Theorem 1.1, the long-time asymptotics of $q(x,t)$ along any line $\mu^2:=(x/t)^2>1/(27\gamma)$ can be described as follows: (i) for $x<0$:
$$
\begin{equation}
q(x,t)=-t^{-\frac{1}{2}-\operatorname{Im} v(-\lambda_3)} \exp\bigl\{-2[\overline{\pi}+\overline{\phi}(-\mu,\varsigma)]-i\operatorname{Re} v(-\lambda_3)\ln t\bigr\}\alpha +\mathcal{R}_1(-\mu,t),
\end{equation}
\tag{1.16}
$$
with
$$
\begin{equation}
\alpha= \frac{\sqrt{2\pi} \, e^{-\frac{\pi}{2}\overline{v(-\lambda_3)}}e^{\frac{i\pi}{4}}} {\sqrt{48\gamma\lambda^2_1-1}\, \overline{r_2}(-\lambda_3)\Gamma(-i\overline{v(-\lambda_3)})} [4(48\gamma\lambda^2_3-1)]^{-i\overline{v(-\lambda_3)}};
\end{equation}
\tag{1.17}
$$
(ii) for $x>0$: based on the value of $\operatorname{Im} v(\lambda_3)$, the three possible types of asymptotics are as follows: (a) $\operatorname{Im} v(\lambda_3)\in(-1/2,-1/6]$,
$$
\begin{equation}
\begin{aligned} \, q(x,t)=&t^{-\frac{1}{2}-\operatorname{Im} v(\lambda_3)}\exp\bigl\{-2[\pi+\phi(\mu,\varsigma)] +i\operatorname{Re} v(\lambda_3)\ln t\bigr\}\beta_1 \nonumber \\ &\qquad+A\varpi^2(\mu,0)+\mathcal{R}_1(\mu,t); \end{aligned}
\end{equation}
\tag{1.18}
$$
(b) $\operatorname{Im} v(\lambda_3)\in(-1/6,1/6)$,
$$
\begin{equation}
\begin{aligned} \, q(x,t)=&-t^{-\frac{1}{2}+\operatorname{Im} v(\lambda_3)} \exp\bigl\{2[\pi+\phi(\mu,\varsigma)]-i\operatorname{Re} v(\lambda_3)\ln t\bigr\}\beta_2 \nonumber \\ &\qquad+t^{-\frac{1}{2}-\operatorname{Im} v(\lambda_3)} \exp\bigl\{-2[\pi+\phi(\mu,\varsigma)]+i\operatorname{Re} v(\lambda_3)\ln t\bigr\}\beta_1 \nonumber \\ &\qquad+A\varpi^2(\mu,0)+\mathcal{R}_3(\mu,t); \end{aligned}
\end{equation}
\tag{1.19}
$$
(c) $\operatorname{Im} v(\lambda_3)\in[1/6,1/2)$,
$$
\begin{equation}
\begin{aligned} \, q(x,t)=&-t^{-\frac{1}{2}+\operatorname{Im} v(\lambda_3)} \exp\bigl\{2[\pi+\phi(\mu,\varsigma)]-i\operatorname{Re} v(\lambda_3)\ln t\bigr\}\beta_2 \nonumber \\ &\qquad+A\varpi^2(\mu,0)+\mathcal{R}_2(\mu,t), \end{aligned}
\end{equation}
\tag{1.20}
$$
with
$$
\begin{equation}
\beta_1 = \frac{d^2_0\sqrt{2\pi} \, e^{-\frac{\pi}{2} v(\lambda_3)}e^{-\frac{i\pi}{4}}} {\sqrt{48\gamma\lambda^2_3-1} \, r_2(\lambda_3)\Gamma(iv(\lambda_3))\lambda^2_3} [4(48\gamma\lambda^2_3-1)]^{iv(\lambda_3)},
\end{equation}
\tag{1.21a}
$$
$$
\begin{equation}
\beta_2 = \frac{\sqrt{2\pi} \, e^{-\frac{\pi}{2}v(\lambda_3)}e^{\frac{i\pi}{4}}} {\sqrt{48\gamma\lambda^2_1-1} \, r_1(\lambda_3)\Gamma(-iv(\lambda_3))} [4(48\gamma\lambda^2_3-1)]^{-iv(\lambda_3)}.
\end{equation}
\tag{1.21b}
$$
The error estimates $\mathcal{R}_1(\mu,t)$ and $\mathcal{R}_2(\mu,t)$ take the forms
$$
\begin{equation}
\mathcal{R}_1(\mu,t)= \begin{cases} O\bigl(\max\bigl\{t^{-1}, t^{-\frac{1}{2}}e^{-\frac{\sqrt{3}}{4}t\mu\zeta_{c}(\mu)}\bigr\}\bigr), &\operatorname{Im} v(\lambda_3)>0, \\ O\bigl(\max\bigl\{t^{-1}\ln t, t^{-\frac{1}{2}}e^{-\frac{\sqrt{3}}{4}t\mu\zeta_{c}(\mu)}\bigr\}\bigr), &\operatorname{Im} v(\lambda_3)=0, \\ O\bigl(\max\bigl\{t^{-1+2|{\operatorname{Im} v(\lambda_3)}|}, t^{-\frac{1}{2}}e^{-\frac{\sqrt{3}}{4}t\mu\zeta_{c}(\mu)}\bigr\}\bigr), &\operatorname{Im} v(\lambda_3)<0, \end{cases}
\end{equation}
\tag{1.22}
$$
$$
\begin{equation}
\mathcal{R}_2(\mu,t)= \begin{cases} O\bigl(\max\bigl\{t^{-1+2|{\operatorname{Im} v(\lambda_3)}|}, t^{-\frac{1}{2}}e^{-\frac{\sqrt{3}}{4}t\mu\zeta_{c}(\mu)}\bigr\}\bigr), &\operatorname{Im} v(\lambda_3)<0, \\ O\bigl(\max\bigl\{t^{-1}\ln t, t^{-\frac{1}{2}}e^{-\frac{\sqrt{3}}{4}t\mu\zeta_{c}(\mu)}\bigr\}\bigr), &\operatorname{Im} v(\lambda_3)=0, \\ O\bigl(\max\bigl\{t^{-1}, t^{-\frac{1}{2}}e^{-\frac{\sqrt{3}}{4}t\mu\zeta_{c}(\mu)}\bigr\}\bigr), &\operatorname{Im} v(\lambda_3)<0, \end{cases}
\end{equation}
\tag{1.23}
$$
and $\mathcal{R}_3(\mu,t)=\mathcal{R}_1(\mu,t)+\mathcal{R}_2(\mu,t)$. In addition,
$$
\begin{equation}
\varpi(\mu,0)=\exp\biggl\{\frac{1}{2\pi i}\int_{-\infty}^{\lambda_3}\frac{\ln(1+r_1(s)r_2(s))}{s} \, ds \biggr\},
\end{equation}
\tag{1.24}
$$
where $\lambda_3$ takes the form in (3.2). $v(\lambda_3)$, $\pi$ and $\phi(\mu,\varsigma)$ are as in (4.5), (4.3) and (4.32); $\Gamma(\,{\cdot}\,)$ is the Euler gamma function, $d_0=A\varpi^2(0,\lambda_3)/(2i)$ is defined by the residue condition (4.18), and $\zeta_{c}$ is given by (4.56). 1.2. Plan of the proofWe give the proofs of Theorem 1.1 and Theorem 1.6 through the non-linear steepest descent method and the RH problem to the long-time asymptotic behavior of the non-local LPD equation. In § 2, we perform the spectral analysis and construct the RH problem of non-local LPD equation. In addition, we also find an expression of the scattering matrix $S(\xi)$ with pure-step initial data. In this case, we classify the zero points of the scattering data $a(\mu)$ and $b(\mu)$ as Case 1 and Case 2. With the reflectionless condition, the soliton solution in Case 2 is obtained. In § 3, through the non-linear steepest descent method, we construct the long-time asymptotics of $q(x,t)$ in regions $\{\mathcal{M}_1, \mathcal{M}_2, \mathcal{N}_1, \mathcal{N}_2\}$ by the following steps. In § 3.1, we decompose the jump matrix $J(x,t,\xi)$ into the upper and lower triangle matrices and introduce the $\delta$ function to get rid of the intermediate matrix. In § 3.2, we perform the second RH deformation to make the jump matrices decline to the identity matrix $I$ for large $t$. In § 3.3, we introduce the BP factor to regularize the RH problem. In § 3.4, we give the local models near saddle points and solve them by the parabolic cylinder functions. In § 3.5, by the Beals–Coifman theory, we conduct the error analysis on the regular RH problem. Finally, the asymptotic solutions of non-local LPD equation are attained. These results are shown in Theorem 1.1. In § 4, we use the technique similar to that of § 3 to obtain the asymptotic solutions in the regions $\{\mathcal{M}_3, \mathcal{M}_4, \mathcal{N}_3, \mathcal{N}_4\}$. In § 4.1, we perform a series of deformations on the RH problem. Here, due to the existence of a pair of conjugate complex saddle points, the symbol table in Fig. 9 indicates that the decomposition forms of jump matrix $\widetilde J_N$ must be treated differently in the left and right half- planes of the complex $\xi$-plane. In § 4.2, the asymptotic solutions of the non-local LPD equation are obtained. The results are shown in Theorem 1.6. § 2. Spectral analysis and the RH problemOur aim here is to conduct a spectral analysis of eigenfunctions and scattering data including their analytic, symmetric and asymptotic properties. After using the Ablowitz–Kaup–Newell–Segur scheme [7], the Lax pair of local LPD equation (1.2) has been derived in [8]. The non-local version of LPD equation is further derived in a local situation, and so the Lax pair of (1.3) constructed in [40] takes the form
$$
\begin{equation}
\begin{alignedat}{2} \phi_x&=M\phi, &\qquad M &=-i\xi\sigma_3+Q, \\ \phi_t&=N\phi, &\qquad N &=i\xi^2\sigma_3-\xi Q+\frac{1}{2}V+\gamma V_P, \end{alignedat}
\end{equation}
\tag{2.1}
$$
with
$$
\begin{equation*}
\begin{gathered} \, Q=\begin{pmatrix} 0 & q(x,t) \\ -\overline{q}(-x,t) & 0 \end{pmatrix},\qquad V=\begin{pmatrix} -iq(x,t)\overline{q}(-x,t) & -iq_x(x,t) \\ -i\overline{q}_x(-x,t) &iq(x,t)\overline{q}(-x,t) \end{pmatrix}, \\ V_P=\begin{pmatrix} iA_P(x,t) & B_P(x,t) \\ -C_P(x,t) & -iA_P(x,t) \end{pmatrix},\qquad\sigma_3=\begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix}, \\ \begin{aligned} \, A_P(x,t) &=-8\xi^4+4\overline{q}(-x,t)q(x,t)\xi^2+2i\overline{q}(-x,t)q_x(x,t)\xi +2iq(x,t)\overline{q}(-x,t)\xi \\ &\qquad-3q^2(x,t)\overline{q}^2(-x,t)+q_x(x,t)\overline{q}_x(-x,t)-q(x,t)\overline{q}(-x,t) \\ &\qquad-\overline{q}(-x,t)q_{xx}(x,t), \\ B_P(x,t) &= 8q(x,t)\xi^3+4iq_x(x,t)\xi^2-2q_{xx}(x,t)\xi-4\overline{q}(-x,t)q^2(x,t) -iq_{xxx}(x,t) \\ &\qquad-6iq(x,t)\overline{q}(-x,t)q_x(x,t), \\ C_P(x,t) &=-8\overline{q}(-x,t)\xi^3-4i\overline{q}_x(-x,t)\xi^2-2\overline{q}_{xx}(-x,t)\xi -4\overline{q}^2(-x,t)q(x,t)\xi \\ &\qquad+i\overline{q}_{xxx}(-x,t)+6i\overline{q}(-x,t)\overline{q}_x(-x,t)q(x,t), \end{aligned} \end{gathered}
\end{equation*}
\notag
$$
where $\phi(x,t,\xi)$ is a $2\times2$ matrix-valued function, the potential function $q(x,t)$ is a complex function and $\xi\in\mathbb{C}$ is a spectral parameter. We can confirm that the matrices $M$ and $N$ of the Lax pair satisfy the zero curve equation $M_t-N_x+[M,N]=0$, which can further leads to the non-local LPD equation (1.3). Considering the asymptotic spectral problem of the Lax pair (2.1) as $x\to\pm\infty$, we have
$$
\begin{equation}
\begin{alignedat}{2} \phi_{\pm x}&=M_{\pm}(\xi)\phi_{\pm}, &\qquad M_{\pm}(\xi) &=-i\xi\sigma_3+Q_{\pm}, \\ \phi_{\pm t}&=N_{\pm}(\xi)\phi_{\pm}, &\qquad N_{\pm}(\xi) &=(-\xi+8\xi^3\gamma)M_{\pm}(\xi), \end{alignedat}
\end{equation}
\tag{2.2}
$$
with
$$
\begin{equation}
Q_+=\begin{pmatrix} 0 & A\\ 0 & 0 \end{pmatrix},\qquad Q_-= \begin{pmatrix} 0 & 0 \\ -A & 0 \end{pmatrix}.
\end{equation}
\tag{2.3}
$$
By calculations, the matrices $M_{\pm}(\xi)$ and $N_{\pm}(\xi)$ can be diagonalized with the same matrices $L_{\pm}(\xi)$ of the form
$$
\begin{equation}
L_+(\xi)= \begin{pmatrix} 1 & \dfrac{A}{2i\xi} \\ 0 & 1 \end{pmatrix},\qquad L_-(\xi)= \begin{pmatrix} 1 & 0 \\ \dfrac{A}{2i\xi} & 1 \end{pmatrix},
\end{equation}
\tag{2.4}
$$
where
$$
\begin{equation}
M_{\pm}L_{\pm}=L_{\pm}(-i\xi\sigma_3), \qquad N_{\pm}L_{\pm}=L_{\pm}(i\xi^2\sigma_3-8i\xi^4\gamma\sigma_3).
\end{equation}
\tag{2.5}
$$
Substituting this into the asymptotic Lax pair (2.2), we derive a system of differential equations. Solving this system directly allows us to determine the asymptotic conditions for the characteristic functions in the Lax pair (2.1). For the convenience to obtain the integral equations along the paths $(-\infty,x)$ and $(x,+\infty)$, we use the notation $\phi_{\pm}$ to define the characteristic functions for each path, which behave asymptotically as
$$
\begin{equation}
\phi_{\pm}(x,t,\xi)\to L_{\pm}(\xi)e^{-i\theta(\xi)\sigma_3}, \qquad x\to\pm\infty,
\end{equation}
\tag{2.6}
$$
where $\theta(\xi)=\xi x-(\xi^2-8\xi^4\gamma)t$. The matrix exponential function satisfies
$$
\begin{equation}
e^{-i\theta(\xi)\sigma_3}= \begin{pmatrix} e^{-i\theta(\xi)} & 0 \\ 0 & e^{i\theta(\xi)} \end{pmatrix},
\end{equation}
\tag{2.7}
$$
because by the definition of matrix power series we have
$$
\begin{equation*}
\begin{aligned} \, e^{\xi\sigma_3} &=\sum^{\infty}_{k=0}\frac{\xi^k\sigma^k_3}{k!} =I+\xi\sigma_3+\frac{1}{2!}(\xi\sigma_3)^2+\cdots \\ &=\begin{pmatrix} {\displaystyle\sum^{\infty}_{k=0}\frac{\xi^k}{k!}} & 0 \\ 0 & {\displaystyle\sum^{\infty}_{k=0}\frac{(-\xi)^k}{k!}} \end{pmatrix} = \begin{pmatrix} e^{\xi} & 0 \\ 0 & e^{-\xi} \end{pmatrix}, \end{aligned}
\end{equation*}
\notag
$$
this definition is consistently used throughout the entire text. Due to the existence of exponential oscillation terms $e^{\pm i\theta(\xi)}$ with respect to the variable $\xi$ in the asymptotic relationship (2.6), in order to satisfy the normalization condition for constructing the RH problem, it is necessary to eliminate this exponential function. Consider the new characteristic functions
$$
\begin{equation}
\phi_{\pm}(x,t,\xi)=\psi_{\pm}(x,t,\xi)e^{-i\theta(\xi)\sigma_3},
\end{equation}
\tag{2.8}
$$
and
$$
\begin{equation}
\psi_{\pm}(x,t,\xi)\to L_{\pm}(\xi), \qquad x\to\pm\infty.
\end{equation}
\tag{2.9}
$$
We next introduce the equivalent Lax pair to (2.1)
$$
\begin{equation}
\begin{gathered} \, (L_{\pm}^{-1}\psi_{\pm})_x-i\xi[L_{\pm}^{-1}\psi_{\pm},\sigma_3] =L_{\pm}^{-1}\Delta Q_{\pm}\psi_{\pm}, \\ (L_{\pm}^{-1}\psi_{\pm})_t+(i\xi^2-8i\xi^4\gamma)[L_{\pm}^{-1}\psi_{\pm},\sigma_3] =L_{\pm}^{-1}\Delta U_{\pm}\psi_{\pm}, \end{gathered}
\end{equation}
\tag{2.10}
$$
where $U=N-(i\xi^2\sigma_3-8i\xi^4\gamma\sigma_3)$, $\Delta X_{\pm}=X-X_{\pm}$, and $[Y,\sigma_3]=Y\sigma_3-\sigma_3Y$. Both $\psi_-(x,t,\xi)$ and $\psi_+(x,t,\xi)$ can be uniquely determined from the Volterra integral equations
$$
\begin{equation}
\begin{aligned} \, \psi_-(x,t,\xi)&=L_-(\xi)+\int_{-\infty}^xG_-(x,y,\xi) \Delta Q_-(y,t)\psi_-(y,t,\xi)e^{i\xi(x-y)\sigma_3} \, dy, \\ \psi_+(x,t,\xi)&=L_+(\xi)+\int_{\infty}^xG_+(x,y,\xi) \Delta Q_+(y,t)\psi_+(y,t,\xi)e^{i\xi(x-y)\sigma_3} \, dy, \end{aligned}
\end{equation}
\tag{2.11}
$$
where $G_{\pm}(x,y,\xi) =L_{\pm}(\xi)e^{-i\xi(x-y)\sigma_3}L_{\pm}^{-1}(\xi)$. Analytic properties of the kernels $G_{\pm}\Delta Q_{\pm}e^{i\xi(x-y)\sigma_3}$ can be obtained via [46]. Since $\phi_{\pm}(x,t,\xi)$ are solutions of the Lax pair (2.1), there exists a matrix $S(\xi)$ independent of $x$ and such that $t$ satisfies
$$
\begin{equation}
\phi_-(x,t,\xi)=\phi_+(x,t,\xi)S(\xi),\qquad \xi\in \mathbb{R}\setminus \{0\}.
\end{equation}
\tag{2.12}
$$
Substituting (2.8) into (2.12), we have
$$
\begin{equation}
\psi_-(x,t,\xi)=\psi_+(x,t,\xi)e^{-i\theta\sigma_3}S(\xi)e^{i\theta\sigma_3},\qquad \xi\in \mathbb{R}\setminus \{0\},
\end{equation}
\tag{2.13}
$$
where the matrix $S(\xi)$ is defined by
$$
\begin{equation}
S(\xi)= \begin{pmatrix} s_{11}(\xi) & s_{12}(\xi)\\ s_{21}(\xi) & s_{22}(\xi) \end{pmatrix}.
\end{equation}
\tag{2.14}
$$
Proposition 2.1. The matrices $\psi_{\pm}(x,t,\xi)$ and the scattering data of $S(\xi)$ satisfy the following symmetry relations: (i) with regard to $\psi_{\pm}(x,t,\xi)$,
$$
\begin{equation}
\sigma_1\overline{\psi_-(-x,t,-\overline{\xi})}\sigma_1=\psi_+(x,t,\xi),\qquad \xi\in\mathbb{R}\setminus \{0\};
\end{equation}
\tag{2.15}
$$
(ii) with regard to $s_{ij}(\xi)$, $i,j=1,2$, where $\sigma_1= \left( \begin{smallmatrix} 0 & 1\\ 1 & 0 \end{smallmatrix} \right)$. Proof. For (i), using the Lax pair (2.10) and since $\sigma_1\overline{Q(-x,t)}\sigma_1{=}\,-Q(x,t)$, we can easily verify that the matrices $\psi_{\pm}(x,t,\xi)$ satisfy the symmetry relation
For (ii), using (2.8), the following symmetry relation for $\phi_{\pm}(x,t,\xi)$ can be easily acquired:
$$
\begin{equation}
\sigma_1\overline{\phi_-(-x,t,-\overline{\xi})}\sigma_1=\phi_+(x,t,\xi),\qquad \xi\in\mathbb{R}\setminus \{0\}.
\end{equation}
\tag{2.17}
$$
Next, using (2.12), we get the symmetry relation for the matrix $S(\xi)$:
$$
\begin{equation}
\sigma_1\overline{{S(-\overline{\xi})}^{-1}}\sigma_1=S(\xi), \qquad \xi\in\mathbb{R}\setminus \{0\}.
\end{equation}
\tag{2.18}
$$
Based on (2.18), we have the symmetry relations $s_{11}(\xi)=\overline{s_{11}(-\overline{\xi})}$, $s_{22}(\xi)=\overline{s_{22}(-\overline{\xi})}$, and $s_{21}(\xi)=-\overline{s_{12}(-\overline{\xi})}$. Now the matrix $S(\xi)$ can be written as $\Box$ Proposition 2.2. Let $q_0(x)\in L^1(\mathbb{R}^-)$, $q_0(x)-A\in L^1(\mathbb{R}^+)$, and let the matrices $\psi_{\pm}(x,t,\xi)$ and the scattering data of $S(\xi)$ satisfy the following analytic relations: (i) $\psi_-^{(1)}$ and $\psi_+^{(2)}$ are analytic with respect to $\xi\in\mathbb{C}^+$ and continuous in $\overline{\mathbb{C}^+}\setminus \{0\}$; $\psi_-^{(2)}$ and $\psi_+^{(1)}$ are analytic with respect to $\xi\in\mathbb{C}^-$ and continuous in $\overline{\mathbb{C}^-}$; (ii) $a_1(\xi)$ is analytic with respect to $\xi\in\mathbb{C}^+$ and continuous in $\overline{\mathbb{C}^+}\setminus \{0\}$; $a_2(\xi)$ is analytic with respect to $\xi\in\mathbb{C}^-$ and continuous in $\overline{\mathbb{C}^-}$; $b(\xi)$ is continuous in $\xi\in\mathbb{R}$; where $\mathbb{C}^+=\{\xi\in\mathbb{C}\mid \operatorname{Im}\xi>0\}$ and $\mathbb{C}^-=\{\xi\in\mathbb{C}\mid \operatorname{Im}\xi<0\}$ stand for the upper and the lower half planes of the complex plane, respectively; $\psi_{\pm}^{(k)}$ is the $k$th column of $\psi_{\pm}$. Proof. For (i), taking $\psi_-^{(1)}$ as an example to prove, the cases $\psi_+^{(2)}$, $\psi_-^{(2)}$, and $\psi_+^{(1)}$ are similar. We define $(A)^{(1)}$ as the first column of the matrix $A$, where $A$ represents the matrix obtained by multiplying several matrices, $\psi^{(ij)}_-$ represents the elements of the $i$th row and $j$th column of matrix $\psi_-$. Now the first column of the Volterra integral equation (2.11) is given by
$$
\begin{equation*}
L^{-1}_-\psi^{(1)}_-(x,t,\xi)= \begin{pmatrix} 1\\ 0 \end{pmatrix} +\int_{-\infty}^x\bigl(e^{-i\xi(x-y)\sigma_3}L^{-1}_- \Delta Q_-(y,t)\psi_-(y,t,\xi)e^{i\xi(x-y)\sigma_3} \bigr)^{(1)} \, dy,
\end{equation*}
\notag
$$
with
$$
\begin{equation*}
\begin{aligned} \, &\bigl(e^{-i\xi(x-y)\sigma_3}L^{-1}_- \Delta Q_-(y,t)\psi_-(y,t,\xi)e^{i\xi(x-y)\sigma_3}\bigr)^{(1)} \\ &\qquad= \begin{pmatrix} 1 & 0 \\ -\dfrac{A}{2i\xi}\, e^{2i\xi(x-y)} & e^{2i\xi(x-y)} \end{pmatrix} \begin{pmatrix} 0 & q(y,t)\\ -\overline{q}(-y,t)+A & 0 \end{pmatrix} \begin{pmatrix} \psi^{(11)}_-(y,t,\xi)\\ \psi^{(21)}_-(y,t,\xi) \end{pmatrix}. \end{aligned}
\end{equation*}
\notag
$$
It can be seen that we only need to consider the analytic properties of $e^{2i\xi(x-y)}$ with respect to the parameter $\xi$ from the definition
$$
\begin{equation*}
e^{2i\xi(x-y)}=e^{2i\operatorname{Re}\xi(x-y)}e^{-2\operatorname{Im}\xi(x-y)},
\end{equation*}
\notag
$$
since the limit of integration implies that $x-y$ is always positive, and so $\psi^{(1)}_-$ is analytic with respect to $\xi\in\mathbb{C}^+$. The functional class required for the solution $q(x,t)$ in the condition is to ensure the existence and uniqueness of Jost solutions defined by Volterra integral equations. For a proof, see [46].
For (ii), according to the Lax pair (2.1) and the Abel formula, we have $\operatorname{Tr}(M)=\operatorname{Tr}(N)=0$, and now $(\det\psi_{\pm})_x=(\det\psi_{\pm})_t=0$ can be easily verified. Therefore, $\det(\psi_{\pm})$ have nothing to do with the variables $x$ and $t$, which means that $\det(\phi_{\pm})=\det(\psi_{\pm})=1$. From (2.12), we also have $\det S(\xi)=1$. Due to (2.13), the wronskian representations of the scattering coefficients $a_1(\xi)$, $a_2(\xi)$, and $b(\xi)$ are given by
$$
\begin{equation}
\begin{alignedat}{2} a_1(\xi)&= \operatorname{Wr}\bigl(\psi_-^{(1)}(0,0,\xi),\psi_+^{(2)}(0,0,\xi)\bigr), &\qquad \xi&\in \overline{\mathbb{C}^+}\setminus \{0\}, \\ a_2(\xi)&= \operatorname{Wr}\bigl(\psi_+^{(1)}(0,0,\xi),\psi_-^{(2)}(0,0,\xi)\bigr), &\qquad \xi&\in \overline{\mathbb{C}^-}, \\ b(\xi)&=\operatorname{Wr}\bigl(\psi_+^{(1)}(0,0,\xi),\psi_-^{(1)}(0,0,\xi)\bigr), &\qquad \xi&\in \mathbb{R}, \end{alignedat}
\end{equation}
\tag{2.20}
$$
according to the analytic relations of $\psi_{\pm}(x,t,\xi)$, we get item (ii). $\Box$ Proposition 2.3. Under the conditions in Proposition 2.2, the matrices $\psi_{\pm}(x,t,\xi)$ and the scattering data of $S(\xi)$ satisfy the following asymptotic properties: (i) as $\xi\to\infty$,
$$
\begin{equation}
\begin{alignedat}{3} \psi_-^{(1)}(x,t,\xi) &= \begin{pmatrix} 1\\ 0 \end{pmatrix} +O(\xi^{-1}), &\quad \psi_+^{(2)}(x,t,\xi) &= \begin{pmatrix} 0\\ 1 \end{pmatrix} +O(\xi^{-1}), &\qquad \xi&\in\mathbb{C}^+, \\ \psi_-^{(2)}(x,t,\xi) &= \begin{pmatrix} 0\\ 1 \end{pmatrix} +O(\xi^{-1}), &\quad \psi_+^{(1)}(x,t,\xi) &= \begin{pmatrix} 1\\ 0 \end{pmatrix} +O(\xi^{-1}), &\qquad \xi&\in\mathbb{C}^-, \end{alignedat}
\end{equation}
\tag{2.21}
$$
$$
\begin{equation}
\begin{alignedat}{2} a_j(\xi)&=1+O(\xi^{-1}),\quad j=1,2, &\qquad \xi &\in\overline{\mathbb{C}^{\pm}}, \\ b(\xi)&=O(\xi^{-1}), &\qquad \xi &\in\mathbb{R}; \end{alignedat}
\end{equation}
\tag{2.22}
$$
(ii) as $\xi\to0$,
$$
\begin{equation}
\begin{alignedat}{2} \psi_-^{(1)}(x,t,\xi)&=\frac{1}{\xi} \begin{pmatrix} f_1(x,t)\\ f_2(x,t) \end{pmatrix} +O(1), &\qquad \psi_-^{(2)}(x,t,\xi) &=\frac{2i}{A} \begin{pmatrix} f_1(x,t)\\ f_2(x,t) \end{pmatrix} +O(\xi), \\ \psi_+^{(1)}(x,t,\xi)&=-\frac{2i}{A} \begin{pmatrix} \overline{f_2}(-x,t)\\ \overline{f_1}(-x,t) \end{pmatrix} +O(\xi), &\qquad \psi_+^{(2)}(x,t,\xi) &=-\frac{1}{\xi} \begin{pmatrix} \overline{f_2}(-x,t)\\ \overline{f_1}(-x,t) \end{pmatrix} +O(1), \end{alignedat}
\end{equation}
\tag{2.23}
$$
where $f_1(x,t)$, $f_2(x,t)$ can be obtained from the Volterra integral equations Proof. For (i), because the determinant of $\psi_{\pm}$ is 1, we can get the columns of $\psi_{\pm}$ represented as (2.21). Now by substituting (2.21) into (2.20) we have (2.22).
For (ii), using (2.11), we can assume that
$$
\begin{equation}
\begin{alignedat}{2} \psi_-^{(1)}(x,t,\xi)&=\frac{1}{\xi} \begin{pmatrix} f_1(x,t)\\ f_2(x,t) \end{pmatrix} +O(1), &\qquad \psi_-^{(2)}(x,t,\xi) &= \begin{pmatrix} \widetilde{f}_1(x,t)\\ \widetilde{f}_2(x,t) \end{pmatrix} +O(\xi), \\ \psi_+^{(1)}(x,t,\xi)&= \begin{pmatrix} \widetilde{g}_1(x,t)\\ \widetilde{g}_2(x,t) \end{pmatrix} +O(\xi), &\qquad \psi_+^{(2)}(x,t,\xi) &=\frac{1}{\xi} \begin{pmatrix} g_1(x,t)\\ g_2(x,t) \end{pmatrix} +O(1), \end{alignedat}
\end{equation}
\tag{2.26}
$$
where some $f_j(x,t)$, $\widetilde{f}_j(x,t)$, $g_j(x,t)$, and $\widetilde{g}_j(x,t)$, $j=1,2$, are undetermined.
By the symmetry relation (2.15), we have
$$
\begin{equation}
\begin{pmatrix} g_1(x,t)\\ g_2(x,t) \end{pmatrix}= \begin{pmatrix} -\overline{f_2}(-x,t)\\ -\overline{f_1}(-x,t) \end{pmatrix}, \qquad \begin{pmatrix} \widetilde{g}_1(x,t)\\ \widetilde{g}_2(x,t) \end{pmatrix} = \begin{pmatrix} \overline{\widetilde{f}_2}(-x,t)\\ \overline{\widetilde{f}_1}(-x,t) \end{pmatrix}.
\end{equation}
\tag{2.27}
$$
Substituting
$$
\begin{equation*}
\psi_-(x,t,\xi)= \begin{pmatrix} \dfrac{1}{\xi}f_1(x,t) & \widetilde{f_1}(x,t) \\ \dfrac{1}{\xi}f_2(x,t) & \widetilde{f_2}(x,t) \end{pmatrix}
\end{equation*}
\notag
$$
into the Volterra integral (2.11) and letting $\xi\to0$, these undetermined equations take the forms
$$
\begin{equation}
\widetilde{f_1}(x,t) =\int_{-\infty}^xq(y,t)\widetilde{f_2}(y,t) \, dy,
\end{equation}
\tag{2.28b}
$$
$$
\begin{equation}
f_2(x,t) =\frac{A}{2i}+ \int_{-\infty}^x(-\overline{q(-y,t)}+A)f_1(y,t) \, dy,
\end{equation}
\tag{2.28c}
$$
$$
\begin{equation}
\widetilde{f_2}(x,t) =1+ \int_{-\infty}^x(-\overline{q(-y,t)}+A)\widetilde{f_1}(y,t) \, dy.
\end{equation}
\tag{2.28d}
$$
Hence
$$
\begin{equation}
\begin{pmatrix} \widetilde{f_1}(x,t)\\ \widetilde{f_2}(x,t) \end{pmatrix} =\frac{2i}{A} \begin{pmatrix} f_1(x,t)\\ f_2(x,t) \end{pmatrix}.
\end{equation}
\tag{2.29}
$$
Now the asymptotic properties of (2.23) follow from (2.29) and (2.27). Hence, we can only employ the functions $f_1(x,t)$ and $f_2(x,t)$ to characterize the matrices $\psi_{\pm}(x,t,\xi)$. It is analogous to $(i)$, where we substitute (2.23) into (2.20) to obtain
$$
\begin{equation}
a_1(\xi) =\frac{1}{\xi^2}\bigl(|f_2(0,0)|^2-|f_1(0,0)|^2\bigr)+O(\xi^{-1}),
\end{equation}
\tag{2.30a}
$$
$\Box$ Remark 2.4. In the case of pure-step initial data, that is, when the scattering matrix $S(\xi)$ can be expressed as It is observed that, in this case, $a_1(\xi)$ has a single simple zero $\xi=(A/2)i$ in the upper half-plane, while $a_2(\xi)$ has no zeros in the lower half-plane. Using the scattering relation (2.13) and Proposition 2.2, we define the piece-wise meromorphic matrices
$$
\begin{equation}
M(x,t,\xi)= \begin{cases} \biggl( \dfrac{\psi_-^{(1)}(x,t,\xi)}{a_1(\xi)}, \psi_+^{(2)}(x,t,\xi)\biggr), &\text{as } \xi\in \mathbb{C}^+, \\ \biggl( \psi_+^{(1)}(x,t,\xi),\dfrac{\psi_-^{(2)}(x,t,\xi)}{a_2(\xi)}\biggr), &\text{as }\xi\in \mathbb{C}^-, \end{cases}
\end{equation}
\tag{2.33}
$$
with the boundary values $M_{\pm}(x,t,\xi):=\lim_{\xi_1\to\xi,\, \xi_1\in\mathbb{C}^{\pm}}M(x,t,\xi)$, $\xi\in\mathbb{R}$, satisfying
$$
\begin{equation}
M_+(x,t,\xi)=M_-(x,t,\xi)J(x,t,\xi), \qquad \xi\in\mathbb{R}\setminus \{0\},
\end{equation}
\tag{2.34}
$$
where the jump matrix is defined by
$$
\begin{equation}
J(x,t,\xi)= \begin{pmatrix} 1+\dfrac{b(\xi)\overline{b(-\overline{\xi})}}{a_1(\xi)a_2(\xi)} & -\dfrac{b(\xi)}{a_2(\xi)}e^{-2i\theta} \\ -\dfrac{\overline{b(-\overline{\xi})}}{a_1(\xi)}e^{2i\theta} & 1 \end{pmatrix},
\end{equation}
\tag{2.35}
$$
from this we define the reflection coefficients $r_1$ and $r_2$ satisfying
$$
\begin{equation}
r_1(\xi)=\frac{\overline{b(-\overline{\xi})}}{a_1(\xi)},\qquad r_2(\xi)=\frac{b(\xi)}{a_2(\xi)},
\end{equation}
\tag{2.36}
$$
where $1+r_1(\xi)r_2(\xi)=1/(a_1(\xi)a_2(\xi))$ for $\xi\in\mathbb{R}\setminus\{0\}$. From the symmetry relations of $a_j(\xi)$, $j=1,2$, and $b(\xi)$, we conclude that $\overline{r_1(-\overline{\xi})}=r_1(\xi)$ and $\overline{r_2(-\overline{\xi})}=r_2(\xi)$. Referring back to (2.24), as $\xi\to0$, the distinct behaviors in the two cases $a_2(0)\,{=}\,0$ and $a_2(0)\neq 0$ result in qualitative differences for $M(x,t,\xi)$. The case $a_2(0)\neq 0$ contains pure-step initial data in Remark 2.4, where $a_1(\xi)$ has a single simple zero located on the imaginary axis in $\mathbb{C}^+$, while $a_2(\xi)$ has no zero in $\mathbb{C}^-$. Since small (in the $L^1$ norm) perturbations of the pure-step initial data preserve these properties, we will concentrate on the following two cases. Case 1: $a_1(\xi)$ has a simple and pure imaginary zero in $\overline{\mathbb{C}^+}\setminus \{0\}$ at $\xi=i\xi_1$ with $\xi_1>0$ and $a_2(\xi)$ has no zero in $\overline{\mathbb{C}^-}$. Case 2: $a_1(\xi)$ has a simple and pure imaginary zero in $\overline{\mathbb{C}^+}\setminus \{0\}$ at $\xi=i\xi_1$ with $\xi_1>0$ and $a_2(\xi)$ has a zero in $\overline{\mathbb{C}^-}$ at $\xi=0$. Thus we assume that $\dot{a}_2(0)\neq 0$ and $a_{11}:=\lim_{\xi \to 0}\xi a_1(\xi)\neq 0$. Proposition 2.5. The zero of $a_1(\xi)$ is equal to $i\xi_1$, where: (i) in Case 1,
$$
\begin{equation}
\xi_1=\frac{A}{2}\exp\biggl\{-\frac{1}{2\pi i}\,\textrm{v.p.}\int_{-\infty}^{+\infty} \frac{\ln(\vartheta^2/(\vartheta^2+1)) (1-b(\vartheta)\overline{b(-\vartheta)})}{\vartheta} \, d\vartheta\biggr\};
\end{equation}
\tag{2.37}
$$
(ii) in Case 2,
$$
\begin{equation}
\xi_1=A\frac{\sqrt{(\operatorname{Re} b(0))^2+F_2^2}-\operatorname{Re} b(0)}{2F_1F_2},
\end{equation}
\tag{2.38}
$$
where
$$
\begin{equation}
\begin{aligned} \, F_1 &=\exp\biggl\{\frac{1}{2\pi i}\,\textrm{v.p.}\int_{-\infty}^{+\infty}\frac{\ln(1-b(\vartheta) \overline{b(-\vartheta)})}{\vartheta} \, d\vartheta\biggr\}, \\ F_2 &=\exp\biggl\{\frac{1}{2}\ln(1-|b(0)|^2)\biggr\}. \end{aligned}
\end{equation}
\tag{2.39}
$$
The function $1-b(\vartheta)\overline{b(-\vartheta)}$ assumes complex values, and so, to avoid the difficulties caused by this multivalued function, we define
$$
\begin{equation*}
\ln T(\vartheta)(1-b(\vartheta)\overline{b(-\vartheta)})= \ln\bigl|T(\vartheta)(1-b(\vartheta)\overline{b(-\vartheta)})\bigr|+ i\arg T(\vartheta)(1-b(\vartheta)\overline{b(-\vartheta)}),
\end{equation*}
\notag
$$
and with In this case, $T(\vartheta)(1-b(\vartheta)\overline{b(-\vartheta)})$ is single-valued. Proof. For (i), to construst a scalar RH problem that is analytic and has no zeros in $\overline{\mathbb{C}^+}$ and $\overline{\mathbb{C}^-}$, we perform the following transformations on $a_1(\xi)$ and $a_2(\xi)$: with
Given that $\widehat{a}_1(\xi)$ and $\widehat{a}_2(\xi)$ have no zeros in $\overline{\mathbb{C}^+}$ and $\overline{\mathbb{C}^-}$, and this RH problem is regular, we can use the Sokhotski–Plemelj formula to find a unique solution, that is to say,
$$
\begin{equation}
\widehat{a_1}(\xi)=e^{T(\xi)},\qquad \widehat{a_2}(\xi)=e^{-T(\xi)},
\end{equation}
\tag{2.42}
$$
with
$$
\begin{equation}
T(\xi)=\frac{1}{2\pi i}\int_{-\infty}^{+\infty}\frac{\ln(\vartheta^2/(\vartheta^2+1)) (1-b(\vartheta)\overline{b(-\vartheta)})}{\vartheta-\xi} \, d\vartheta,
\end{equation}
\tag{2.43}
$$
and
$$
\begin{equation}
T(+i0)+T(-i0)=\frac{1}{\pi i}\,\textrm{v.p.}\int_{-\infty}^{+\infty}\frac{\ln(\vartheta^2/(\vartheta^2+1)) (1-b(\vartheta)\overline{b(-\vartheta)})}{\vartheta} \, d\vartheta.
\end{equation}
\tag{2.44}
$$
As $\xi\to0$, we have
$$
\begin{equation}
a_1(\xi)=\frac{\xi_1e^{T(+i0)}}{\xi^2}(1+o(\xi)),\qquad a_2(\xi)=\frac{1}{\xi_1} \, e^{-T(-i0)}.
\end{equation}
\tag{2.45}
$$
On the other hand,
$$
\begin{equation}
a_1(\xi)=\frac{A^2a_2(0)}{4\xi^2}(1+o(\xi)) =\frac{A^2}{4\xi^2\cdot\xi_1} \, e^{-T(-i0)}(1+o(\xi)).
\end{equation}
\tag{2.46}
$$
Combining equations (2.45) and (2.46), we have Now the solution of $\xi_1$ takes the form
$$
\begin{equation}
\xi_1=\frac{A}{2}\exp\biggl\{-\frac{1}{2\pi i} \, \textrm{v.p.}\int_{-\infty}^{+\infty}\frac{\ln(\vartheta^2/(\vartheta^2+1)) (1-b(\vartheta)\overline{b(-\vartheta)})}{\vartheta} \, d\vartheta\biggr\}.
\end{equation}
\tag{2.48}
$$
For (ii), rewriting (2.21) and using the symmetry relation (2.46), we have
$$
\begin{equation}
\psi_-^{(1)}(x,t,\xi) =\frac{1}{\xi} \begin{pmatrix} f_1(x,t)\\ f_2(x,t) \end{pmatrix} + \begin{pmatrix} m_1(x,t)\\ m_2(x,t) \end{pmatrix} +O(\xi),
\end{equation}
\tag{2.49a}
$$
$$
\begin{equation}
\psi_-^{(2)}(x,t,\xi) =\frac{2i}{A} \begin{pmatrix} f_1(x,t)\\ f_2(x,t) \end{pmatrix} +\xi \begin{pmatrix} n_1(x,t)\\ n_2(x,t) \end{pmatrix} +O(\xi^2),
\end{equation}
\tag{2.49b}
$$
$$
\begin{equation}
\psi_+^{(1)}(x,t,\xi) =-\frac{2i}{A} \begin{pmatrix} \overline{f_2}(-x,t)\\ \overline{f_1}(-x,t) \end{pmatrix}-\xi \begin{pmatrix} \overline{n_2}(-x,t)\\ \overline{n_1}(-x,t) \end{pmatrix} +O(\xi^2),
\end{equation}
\tag{2.49c}
$$
$$
\begin{equation}
\psi_+^{(2)}(x,t,\xi) =-\frac{1}{\xi} \begin{pmatrix} \overline{f_2}(-x,t)\\ \overline{f_1}(-x,t) \end{pmatrix} + \begin{pmatrix} \overline{m_2}(-x,t)\\ \overline{m_1}(-x,t) \end{pmatrix} +O(\xi).
\end{equation}
\tag{2.49d}
$$
Now by (2.20) the scattering data take the forms
$$
\begin{equation}
a_1(\xi) =\frac{1}{\xi}(f_1\overline{m_1} -\overline{f_1}m_1-f_2\overline{m_2}+ \overline{f_2}m_2)\big|_{x,t=0}+O(1),
\end{equation}
\tag{2.50a}
$$
$$
\begin{equation}
a_2(\xi) =\frac{2i\xi}{A}(f_1\overline{n_1}+\overline{f_1}n_1 -f_2\overline{n_2}-\overline{f_2}n_2)\big|_{x,t=0}+O(\xi^2),
\end{equation}
\tag{2.50b}
$$
$$
\begin{equation}
b(\xi) = \biggl[\frac{2i}{A}(\overline{f_1}m_1-\overline{f_2}m_2) +(f_1\overline{n_1}-f_2\overline{n_2})\biggr]\bigg|_{x,t=0}+O(\xi),
\end{equation}
\tag{2.50c}
$$
where $\overline{n_1}(0,0)m_1(0,0)-\overline{n_2}(0,0)m_2(0,0)=0$. The functions $f$, $m$, and $n$ satisfy $|f_2(0,0)|^2-|f_1(0,0)|^2=0$, $|m_2(0,0)|^2-|m_1(0,0)|^2=0$ and $|n_2(0,0)|^2-|n_1(0,0)|^2= 0$. We also have
$$
\begin{equation}
a_{11}=\lim_{\xi \to 0}(\xi a_1(\xi))=f_1\overline{m_1} -\overline{f_1}m_1-f_2\overline{m_2}+\overline{f_2}m_2.
\end{equation}
\tag{2.51}
$$
In view of (2.50a), (2.50b) and (2.50c), (2.51) can be rewritten as
$$
\begin{equation}
a_{11}=-\frac{A^2}{4} \, \dot{a_2}(0)+iA\cdot \operatorname{Re} b(0).
\end{equation}
\tag{2.52}
$$
In the case similar to $(i)$, in order to remove the zeros of $a_1(\xi)$ and $a_2(\xi)$, we redefine
$$
\begin{equation}
\widetilde{a}_1(\xi)=a_1(\xi) \, \frac{\xi}{\xi-i\xi_1},\qquad \widetilde{a}_2(\xi)=a_2(\xi) \, \frac{\xi-i\xi_1}{\xi}.
\end{equation}
\tag{2.53}
$$
As a result, we have
$$
\begin{equation}
a_1(\xi) =\frac{\xi-i\xi_1}{\xi}\exp\biggl\{\frac{1}{2\pi i}\int_{-\infty}^{+\infty}\frac{\ln (1-b(\vartheta)\overline{b(-\vartheta)})}{\vartheta-\xi} \, d\vartheta\biggr\},
\end{equation}
\tag{2.54}
$$
$$
\begin{equation}
a_2(\xi) =\frac{\xi}{\xi-i\xi_1}\exp\biggl\{-\frac{1}{2\pi i}\int_{-\infty}^{+\infty}\frac{\ln (1-b(\vartheta)\overline{b(-\vartheta)})}{\vartheta-\xi} \, d\vartheta\biggr\},
\end{equation}
\tag{2.55}
$$
and where $F_1$, $F_2$ are given by (2.39). Since
$$
\begin{equation}
1-|b(0)|^2=\lim_{\xi \to 0}a_1(\xi)a_2(\xi)=a_{11}\dot{a_2}(0),
\end{equation}
\tag{2.57}
$$
we have
$$
\begin{equation}
\dot{a}_2(0)=a_{11}^{-1}(1-|b(0)|^2)=\frac{i}{\xi_1} \, F_1^{-1}F_2.
\end{equation}
\tag{2.58}
$$
Combining this with (2.52), (2.56), and (2.58), we have with the solution $\Box$ Considering the singularity conditions of $a_1(\xi)$ and $\psi_j(x,t,\xi)$ for $j=1,2$ at $\xi=0$, we find the asymptotic behavior of $M(x,t,\xi)$ at $\xi=0$ for both cases. We write $\xi\in\mathbb{C}_{\pm}$ to indicate that the allowed path for $\xi$ lies in the upper or lower half of the complex plane. Now we have the asymptotic estimates of $M_\pm$ as $\xi\to 0$. (i) in Case 1,
$$
\begin{equation}
M_+ = \begin{pmatrix} \dfrac{4}{A^2a_2(0)}f_1(x,t) & -\overline{f_2}(-x,t) \\ \dfrac{4}{A^2a_2(0)}f_2(x,t) & -\overline{f_1}(-x,t) \end{pmatrix} (I+O(\xi)) \begin{pmatrix} \xi & 0 \\ 0 & \dfrac{1}{\xi} \end{pmatrix}, \qquad \xi \in\mathbb{C}_+, \quad \xi \to0,
\end{equation}
\tag{2.61a}
$$
$$
\begin{equation}
M_- =\frac{2i}{A} \begin{pmatrix} -\overline{f_2}(-x,t) & \dfrac{f_1(x,t)}{a_2(0)} \\ -\overline{f_1}(-x,t) & \dfrac{f_2(x,t)}{a_2(0)} \end{pmatrix} +O(\xi), \qquad \xi \in\mathbb{C}_-, \quad \xi \to0;
\end{equation}
\tag{2.61b}
$$
(ii) in Case 2,
$$
\begin{equation}
M_+ = \begin{pmatrix} \dfrac{f_1(x,t)}{a_{11}(\xi)} & -\overline{f_2}(-x,t) \\ \dfrac{f_2(x,t)}{a_{11}(\xi)} & -\overline{f_1}(-x,t) \end{pmatrix} (I+O(\xi)) \begin{pmatrix} 1 & 0 \\ 0 & \dfrac{1}{\xi} \end{pmatrix}, \qquad \xi \in\mathbb{C}_+, \quad \xi \to0,
\end{equation}
\tag{2.62a}
$$
$$
\begin{equation}
M_- =\frac{2i}{A} \begin{pmatrix} -\overline{f_2}(-x,t) & \dfrac{f_1(x,t)}{\dot{a}_2(0)} \\ -\overline{f_1}(-x,t) & \dfrac{f_2(x,t)}{\dot{a}_2(0)} \end{pmatrix} (I+O(\xi)) \begin{pmatrix} 1 & 0 \\ 0 & \dfrac{1}{\xi} \end{pmatrix}, \qquad \xi \in\mathbb{C}_-, \quad \xi \to0.
\end{equation}
\tag{2.62b}
$$
The residue condition of $M(x,t,\xi)$ at the point $\xi=i\xi_1$ takes the form
$$
\begin{equation}
\operatorname*{Res}_{\xi=i\xi_1}[M(x,t,\xi)]_1=\frac{r_1}{\dot{a}_1(i\xi_1)} \, e^{-2\xi_1x+2i\xi_1^2t+16i\xi_1^4\gamma t}[M(x,t,i\xi_1)]_2,
\end{equation}
\tag{2.63}
$$
with
$$
\begin{equation}
\psi_-^{(1)}(x,t,i\xi_1)=r_1\psi_+^{(1)}(x,t,i\xi_1), \qquad |r_1|=1,
\end{equation}
\tag{2.64}
$$
where $r_1$ is a constant. RH problem 2.6. Find a piecewise meromorphic matrix $M(x,t,\xi)$ such that. (i) Jump conditions: The non-tangential limits $M_{\pm}(x,t,\xi)=M(x,t,\xi\pm i0)$ exist a.e. for $\xi\in\mathbb{R}$ such that $M(x,t,{\cdot}\,)-I\in L^2(\mathbb{R}\setminus [-\varepsilon,\varepsilon])$ for any $\varepsilon>0$ and $M_{\pm}(x,t,\xi)$ satisfy the condition
$$
\begin{equation}
M_+(x,t,\xi)=M_-(x,t,\xi)J(x,t,\xi),\qquad \xi\in\mathbb{R}\setminus \{0\},
\end{equation}
\tag{2.65}
$$
where the jump matrix $J(x,t,\xi)$ is given by (2.35) and the jump contour is shown in Fig. 2, with $a_j(\xi)$, $j=1,2$, $b(\xi)$ are given in Case 1 or Case 2. (ii) The normalization condition at $\xi=\infty$:
$$
\begin{equation}
M(x,t,\xi)=I+O\biggl(\frac{1}{\xi}\biggr),\qquad \xi\to\infty.
\end{equation}
\tag{2.66}
$$
(iii) Rhe residue condition (2.63) with $\xi_1$ given in terms of $b(\xi)$ using (2.37) or (2.38). (iv) The singularity conditions at $\xi=0$: $M(x,t,\xi)$ satisfies (2.61a), (2.61b) or (2.62a), (2.62b). Assume that the RH problem (i)–(iv) has a solution $M(x,t,\xi)$. Then the solution of the initial value problem can be expressed in terms of $M_{12}$ and $M_{21}$ as follows:
$$
\begin{equation}
q(x,t) =2i\lim_{\xi \to \infty}\xi M_{12}(x,t,\xi),
\end{equation}
\tag{2.67}
$$
$$
\begin{equation}
q(-x,t) =-2i\lim_{\xi \to \infty}\xi \overline{M_{21}(x,t,\xi)}.
\end{equation}
\tag{2.68}
$$
Proposition 2.7. Suppose that $a_1(\xi)$, $a_2(\xi)$, and $b(\xi)$ satisfy the following conditions: (i) $a_1(\xi)$ and $a_2(\xi)$ are given by Case 2; (ii) $b(\xi)=0$ for all $\xi\in\mathbb{R}$. Then $\xi_1$ is uniquely determined by $\xi_1=A/2$. The exact solution $q(x,t)$ of problem (1.3) and (1.5) is given by where $r_1=e^{i\alpha}$ with $\alpha\in\mathbb{R}$. Proof. Since $b(\xi)=0$, we have $b(0)=0$. From Case 2 of Proposition 2.5, it follows that $\xi_1=A/2$. Next, and
Since $b(\xi)=0$, we see that $M(x,t,\xi)$ is a meromorphic function with the only pole at $\xi=i\xi_1$. Now comparing (2.62a) and (2.62b), we conclude that $f_1(x,t)=-\overline{f_2}(-x,t)$. Therefore, the singularity conditions (2.62) convert into the residue condition
$$
\begin{equation}
\operatorname*{Res}_{\xi=0}[M(x,t,\xi)]_2=\frac{A}{2i} [M(x,t,0)]_1.
\end{equation}
\tag{2.72}
$$
Now considering item (ii) of RH problem 2.6 at $\xi=\infty$, we derive
$$
\begin{equation}
M(x,t,\xi)=\begin{pmatrix} \dfrac{\xi+f_1(x,t)}{\xi-iA/2} & \dfrac{f_1(x,t)}{\xi} \\ \dfrac{-\overline{f_1}(-x,t)}{\xi-iA/2} &\dfrac{\xi-\overline{f_1}(-x,t)}{\xi} \end{pmatrix}.
\end{equation}
\tag{2.73}
$$
Using the residue condition (2.63) at $\xi=(A/2)i$, we obtain
$$
\begin{equation}
f_1(x,t)=\frac{A}{2i} \, \frac{1}{1-e^{-Ax+\frac{i}{2}A^2t+iA^4\gamma t+i\alpha}}.
\end{equation}
\tag{2.74}
$$
Finally, by (2.67), we get the one-soliton solution (2.69). $\Box$ § 3. Long-time asymptotics for $\mu^2:=(x/t)^2<1/(27\gamma)$In this section, we consider the long-time asymptotic behavior of the solution $q(x,t)$ to the non-local LPD equation. The original RH problem 2.6 can be transformed to an explicit problem via the non-linear steepest descent method. To solve the problem of oscillations in exponential functions $e^{\pm i\theta}$ at the large time, we need to identify the exponential decay regions. First, it is necessary to investigate the steady-state phase points of the phase function and give the signature table of $\operatorname{Re}(i\theta(\xi))$. In view of equations (2.67) and (2.68), it suffices to focus on studying the RH problem for $x>0$. Let $\mu=x/t$, then the phase function can be expressed by The exponential function takes the form $e^{2it\theta(\xi)}=e^{2t\varphi(\xi)}$. There we have the signature table of $\operatorname{Re}(i\theta(\xi))$ and the distribution of phase points shown in Fig. 3, from which we can determine the decaying regions of the oscillating factor $e^{\pm 2it\theta(\xi)}$. By calculation, it can be concluded that we have the following possibilities for the roots of $\varphi'(\xi)$: (i) when $\mu^2>1/(27\gamma)$, there exists one real stationary point; (ii) when $\mu^2=1/(27\gamma)$, there exist three real stationary points, of which two are equal; (iii) when $\mu^2<1/(27\gamma)$, there exist three different real stationary points. Let us consider the case $\mu^2<1/(27\gamma)$ with three different real roots
$$
\begin{equation}
\begin{aligned} \, \lambda_1 &=\frac{w^2}{4}\sqrt[3]{-\frac{\mu}{\gamma}+\frac{1}{\gamma}\sqrt{\mu^2-\frac{1}{27\gamma}}} +\frac{w}{4}\sqrt[3]{-\frac{\mu}{\gamma}-\frac{1}{\gamma}\sqrt{\mu^2-\frac{1}{27\gamma}}}, \\ \lambda_2 &=\frac{w}{4}\sqrt[3]{-\frac{\mu}{\gamma}+\frac{1}{\gamma}\sqrt{\mu^2-\frac{1}{27\gamma}}} +\frac{w^2}{4}\sqrt[3]{-\frac{\mu}{\gamma}-\frac{1}{\gamma}\sqrt{\mu^2-\frac{1}{27\gamma}}}, \\ \lambda_3 &=\frac{1}{4}\sqrt[3]{-\frac{\mu}{\gamma}+\frac{1}{\gamma}\sqrt{\mu^2-\frac{1}{27\gamma}}} +\frac{1}{4}\sqrt[3]{-\frac{\mu}{\gamma}-\frac{1}{\gamma}\sqrt{\mu^2-\frac{1}{27\gamma}}}, \end{aligned}
\end{equation}
\tag{3.2}
$$
where $w=(-1+\sqrt{3}\, i)/2$. When $\mu\in D_1=(\epsilon,\sqrt{1/(27\gamma)}-\epsilon)$ for any positive constant $\epsilon$, we have $\lambda_1>0$, $\lambda_2>0$ and $\lambda_3<0$. The signature table for the distribution of $\operatorname{Re}\varphi(\xi)$ in the complex $\xi$-plane is shown in Fig. 2, which corresponds to Fig. 3, (a). Remark 3.1. For the case $\mu\in D_2=(-\sqrt{1/(27\gamma)}+\epsilon,-\epsilon)$ with arbitrary positive constant $\epsilon$, which corresponds to Fig. 3, (b), a similar discussion can be conduced. The differences are manifested as follows. First, the positive or negative signs of these three real roots are exactly opposite, which will lead to the singularity point zero located to the right of $\lambda_1$. This difference will be reflected in the discussion of singularity conditions at zero points. Besides, in the asymptotic formula, the variables $x$ and $t$ have the opposite signs for the case $\mu\in D_2$ and the same signs for the case $\mu\in D_1$. 3.1. Factorization of the jump matrixFirst, there are two types of triangular factorizations of the jump matrix $J(x,t,\xi)$:
$$
\begin{equation*}
J(x,t,\xi)= \begin{cases} {\displaystyle \begin{pmatrix} 1 & -r_2(\xi)e^{-2it\theta} \\ 0 & 1 \end{pmatrix} \begin{pmatrix} 1 & 0 \\ -r_1(\xi)e^{2it\theta} & 1 \end{pmatrix}}, \\ {\displaystyle \begin{pmatrix} 1 & 0 \\ -\dfrac{r_1(\xi)e^{2it\theta}}{1+r_1(\xi)r_2(\xi)} & 1 \end{pmatrix} \begin{pmatrix} 1+r_1(\xi)r_2(\xi) & 0 \\ 0 & \dfrac{1}{1+r_1(\xi)r_2(\xi)} \end{pmatrix}} \\ \quad\times{\displaystyle \begin{pmatrix} 1 & -\dfrac{r_2(\xi)e^{-2it\theta}}{1+r_1(\xi)r_2(\xi)} \\ 0 & 1 \end{pmatrix}}. \end{cases}
\end{equation*}
\notag
$$
To get rid of the intermediate matrix of the second factorization, it is necessary to introduce a function $\delta(\xi,\mu)$ as the solution of the scalar RH problem: (i) $\delta(\xi,\mu)$ is holomorphic for $\xi\in\mathbb{C}\setminus ((-\infty,\lambda_3]\cup[\lambda_2,\lambda_1])$; (ii) $\delta_+(\xi,\mu)=\delta_-(\xi,\mu)(1+r_1(\xi)r_2(\xi))$, $\xi\in(-\infty,\lambda_3)\cup(\lambda_2,\lambda_1)$; (iii) $\delta(\xi,\mu)\to 1$, $\xi\to\infty$. Using the Plemelj formula, the solution can be formulated as a Cauchy-type integral
$$
\begin{equation}
\delta(\xi,\mu)=\exp\biggl\{\frac{1}{2\pi i} \biggl(\int_{-\infty}^{\lambda_3}+\int_{\lambda_2}^{\lambda_1}\biggr) \frac{\ln(1+r_1(\zeta)r_2(\zeta))}{\zeta-\xi} \, d\zeta \biggr\}.
\end{equation}
\tag{3.3}
$$
From the symmetry relations of $r_1(\xi)$ and $r_2(\xi)$, we have $\delta(\xi,\mu)=\overline{\delta(-\overline{\xi},\mu)}$. Next, $\delta(\xi,\mu)$ can be written as
$$
\begin{equation}
\begin{aligned} \, \delta(\xi,\mu)&=(\xi-\lambda_1)^{iv(\lambda_1)}\biggl(\frac{\xi-\lambda_2}{\xi-\lambda_3} \biggr)^{-iv(\lambda_1)}e^{\chi_1(\xi)} \nonumber \\ &=(\xi-\lambda_3)^{iv(\lambda_3)}\biggl(\frac{\xi-\lambda_2}{\xi-\lambda_1} \biggr)^{-iv(\lambda_2)}e^{\chi_2(\xi)} \nonumber \\ &=(\xi-\lambda_3)^{iv(\lambda_3)}\biggl(\frac{\xi-\lambda_2}{\xi-\lambda_1} \biggr)^{-iv(\lambda_3)}e^{\chi_3(\xi)}, \end{aligned}
\end{equation}
\tag{3.4}
$$
with
$$
\begin{equation}
\begin{aligned} \, \chi_1(\xi)&\,{=}\,\frac{1}{2\pi i} \biggl[\int_{\lambda_2}^{\lambda_3}\ln\biggl(\frac{1+r_1(\zeta)r_2(\zeta)}{1\,{+}\,r_1(\lambda_1)r_2(\lambda_1)} \biggr) \, \frac{d\zeta}{\zeta\,{-}\,\xi}-\!\int_{-\infty}^{\lambda_1}\!\ln(\xi\,{-}\,\zeta) \, d\ln(1\,{+}\,r_1(\zeta)r_2(\zeta)) \biggr], \\ \chi_2(\xi)&\,{=}\,\frac{1}{2\pi i} \biggl[\int_{\lambda_2}^{\lambda_1}\ln\biggl(\frac{1+r_1(\zeta)r_2(\zeta)}{1\,{+}\,r_1(\lambda_2)r_2(\lambda_2)} \biggr) \, \frac{d\zeta}{\zeta\,{-}\,\xi}-\!\int_{-\infty}^{\lambda_3}\!\ln(\xi\,{-}\,\zeta) \, d\ln(1\,{+}\,r_1(\zeta)r_2(\zeta)) \biggr], \\ \chi_3(\xi)&\,{=}\,\frac{1}{2\pi i} \biggl[\int_{\lambda_2}^{\lambda_1}\ln\biggl(\frac{1+r_1(\zeta)r_2(\zeta)}{1\,{+}\,r_1(\lambda_3)r_2(\lambda_3)} \biggr) \, \frac{d\zeta}{\zeta\,{-}\,\xi}-\!\int_{-\infty}^{\lambda_3}\!\ln(\xi\,{-}\,\zeta) \, d\ln(1\,{+}\,r_1(\zeta)r_2(\zeta)) \biggr], \end{aligned}
\end{equation}
\tag{3.5}
$$
where $v(\lambda_l)$ $(l=1,2,3)$ can be expressed as
$$
\begin{equation}
\begin{gathered} \, v(\lambda_l) =-\frac{1}{2\pi}\ln|1+r_1(\lambda_l)r_2(\lambda_l)|-\frac{i}{2\pi}\Delta(\lambda_l), \qquad l=1,2,3, \\ \Delta(\lambda_l)=\int_{-\infty}^{\lambda_l}d \arg(1+r_1(\zeta)r_2(\zeta)). \nonumber \end{gathered}
\end{equation}
\tag{3.6}
$$
Assuming that $\Delta(\xi)\in(-\pi,\pi)$ for $\xi\in\mathbb{R}$, we have $|{\operatorname{Im} v(\xi)}|<1/2$. Under this assumption, $\ln(1+r_1(\xi)r_2(\xi))$ is single-valued and the singularities $\xi=\lambda_l$, $l=1,2,3$, of $\delta(\xi,\mu)$ are square integrable. Theorem 3.2. Let a $2\times2$ matrix function $\widetilde{M}(x,t,\xi)$ satisfy Proof. Based on definition (2.34), we decompose the jump matrix $J(x,t,\xi)$ into $J_1(x,t,\xi)$, $J_2(x,t,\xi)$ and $J_3(x,t,\xi)$ for $\xi\in(-\infty,\lambda_3)\cup(\lambda_2,\lambda_1)$. We next transform
$$
\begin{equation}
\widetilde{M}(x,t,\xi)=M(x,t,\xi)\delta^{a\sigma_3}(\xi,\mu),
\end{equation}
\tag{3.8}
$$
and now combining with (2.33) this gives
$$
\begin{equation*}
\begin{aligned} \, \widetilde{M}_+(x,t,\xi)&=M_+(x,t,\xi)\delta^{a\sigma_3}_+(\xi,\mu) =M_-(x,t,\xi)J(x,t,\xi)\delta^{a\sigma_3}_+(\xi,\mu) \\ &=\widetilde{M}_-(x,t,\xi)\delta^{-a\sigma_3}_-(\xi,\mu)J(x,t,\xi)\delta^{a\sigma_3}_+(\xi,\mu) \triangleq\widetilde{M}_-(x,t,\xi)\widetilde{J}(x,t,\xi). \end{aligned}
\end{equation*}
\notag
$$
For convenience, we omit the above parameters. According to the assumption, we obtain the decomposition
$$
\begin{equation*}
\delta^{-a\sigma_3}_-J\delta^{a\sigma_3}_+= \delta^{-a\sigma_3}_-J_1\cdot J_2\cdot J_3\delta^{a\sigma_3}_+ =\delta^{-a\sigma_3}_-J_1\delta^{a\sigma_3}_-\cdot\delta^{-a\sigma_3}_- J_2\delta^{a\sigma_3}_+\cdot\delta^{-a\sigma_3}_+J_3\delta^{a\sigma_3}_+.
\end{equation*}
\notag
$$
Since the middle part is the identity matrix, we obtain the undetermined coefficient $a$, and we proceed as
$$
\begin{equation*}
\begin{aligned} \, \delta^{-a\sigma_3}_-J_2\delta^{a\sigma_3}_+&= \begin{pmatrix} \delta^{-a}_- & 0 \\ 0 & \delta^{a}_- \end{pmatrix} \begin{pmatrix} 1+r_1r_2 & 0 \\ 0 & \dfrac{1}{1+r_1r_2} \end{pmatrix} \begin{pmatrix} \delta^{a}_+ & 0 \\ 0 & \delta^{-a}_+ \end{pmatrix} \\ &=\begin{pmatrix} \delta^{-a}_-(1+r_1r_2)\delta^{a}_+ & 0 \\ 0 & \delta^{a}_-\biggl(\dfrac{1}{1+r_1r_2}\biggr)\delta^{-a}_+ \end{pmatrix} =I. \end{aligned}
\end{equation*}
\notag
$$
As a result,
$$
\begin{equation*}
\delta^{-a}_-(1+r_1r_2)\delta^{a}_+=1,\qquad \delta^{a}_-\biggl(\frac{1}{1+r_1r_2}\biggr)\delta^{-a}_+=1;
\end{equation*}
\notag
$$
for $a=-1$, this relations is equivalent to item (ii) of the scalar RH problem. This completes the proof of (3.7). $\Box$ Now let us formulate the RH problem of the function $\widetilde{M}(x,t,\xi)$. RH problem 3.3. Find a matrix-valued function $\widetilde{M}(x,t,\xi)$ that id piecewise meromorphic and satisfies the following conditions: (i) $\widetilde{M}(x,t,\xi)$ is meromorphic for $\xi\in\mathbb{C}\setminus\mathbb{R}$ and has a simple pole located at $\xi=i\xi_1$, $\xi_1>0$; (ii) the jump conditions: the non-tangential limits $\widetilde{M}(x,t,\xi)=\widetilde{M}(x,t,\xi\pm i0)$ exist a.e. for $\xi\in\mathbb{R}$ such that $\widetilde{M}(x,t,{\cdot}\,)-I\in L^2(\mathbb{R}\setminus [-\varepsilon,\varepsilon])$ for any $\varepsilon>0$ and $\widetilde{M}_{\pm}(x,t,\xi)$ satisfy
$$
\begin{equation}
\widetilde{M}_+(x,t,\xi)=\widetilde{M}_-(x,t,\xi)\widetilde{J}(x,t,\xi),\qquad \xi\in\mathbb{R}\setminus \{0\},
\end{equation}
\tag{3.9}
$$
where
$$
\begin{equation}
\widetilde{J}= \begin{cases} \begin{pmatrix} 1 & 0 \\ -\dfrac{r_1(\xi)\delta_-^{-2}(\xi,\mu)}{1+r_1(\xi)r_2(\xi)}e^{2it\theta} & 1 \end{pmatrix} \\ \quad\times \begin{pmatrix} 1 & -\dfrac{r_2(\xi)\delta_+^2(\xi,\mu)}{1+r_1(\xi)r_2(\xi)}e^{-2it\theta} \\ 0 & 1 \end{pmatrix}, &\xi<\lambda_3, \, \lambda_2<\xi<\lambda_1, \\ \begin{pmatrix} 1 & -r_2(\xi)\delta^2(\xi,\mu)e^{-2it\theta} \\ 0 & 1 \end{pmatrix} \\ \quad\times \begin{pmatrix} 1 & 0 \\ -r_1(\xi)\delta^{-2}(\xi,\mu)e^{2it\theta} & 1 \end{pmatrix}, &\xi>\lambda_1, \, \lambda_3<\xi<\lambda_2; \end{cases}
\end{equation}
\tag{3.10}
$$
(iii) the normalization condition at $\xi=\infty$:
$$
\begin{equation}
\widetilde{M}(x,t,\xi)=I+O\biggl(\frac{1}{\xi}\biggr),\qquad \xi\to\infty;
\end{equation}
\tag{3.11}
$$
(iv) the residue condition:
$$
\begin{equation}
\operatorname*{Res}_{\xi=i\xi_1}\bigl[\widetilde{M}(x,t,\xi)\bigr]_1=\frac{r_1}{\dot{a}_1(i\xi_1)\delta^2(i\xi_1)} \, e^{-2\xi_1x+2i\xi_1^2t+16i\xi_1^4\gamma t}\bigl[\widetilde{M}(x,t,i\xi_1)\bigr]_2;
\end{equation}
\tag{3.12}
$$
(v) the singularity conditions at $\xi=0$: in Case 1,
$$
\begin{equation}
\widetilde{M}_+ = \begin{pmatrix} \dfrac{4 f_1(x,t)}{A^2a_2(0)\delta(0,\mu)} & -\delta(0,\mu)\overline{f_2}(-x,t) \\ \dfrac{4 f_2(x,t)}{A^2a_2(0)\delta(0,\mu)} & -\delta(0,\mu)\overline{f_1}(-x,t) \end{pmatrix} \nonumber
\end{equation}
\notag
$$
$$
\begin{equation}
\qquad\times (I+O(\xi)) \begin{pmatrix} \xi & 0 \\ 0 & \dfrac{1}{\xi} \end{pmatrix}, \qquad \xi \in\mathbb{C}_+, \quad \xi \to0,
\end{equation}
\tag{3.13a}
$$
$$
\begin{equation}
\widetilde{M}_- =\frac{2i}{A} \begin{pmatrix} -\dfrac{\overline{f_2}(-x,t)}{\delta(0,\mu)} & \dfrac{\delta(0,\mu)f_1(x,t)}{a_2(0)} \\ -\dfrac{\overline{f_1}(-x,t)}{\delta(0,\mu)} & \dfrac{\delta(0,\mu)f_2(x,t)}{a_2(0)} \end{pmatrix}+O(\xi), \qquad \xi \in\mathbb{C}_-, \quad \xi \to0;
\end{equation}
\tag{3.13b}
$$
in Case 2,
$$
\begin{equation}
\widetilde{M}_+ = \begin{pmatrix} \dfrac{f_1(x,t)}{a_{11}\delta(0,\mu)} & -\overline{f_2}(-x,t)\delta(0,\mu) \\ \dfrac{f_2(x,t)}{a_{11}\delta(0,\mu)} & -\overline{f_1}(-x,t)\delta(0,\mu) \end{pmatrix} \nonumber
\end{equation}
\notag
$$
$$
\begin{equation}
\qquad\times (I+O(\xi)) \begin{pmatrix} 1 & 0 \\ 0 & \dfrac{1}{\xi} \end{pmatrix},\qquad \xi\in\mathbb{C}_+,\quad\xi\to0,
\end{equation}
\tag{3.14a}
$$
3.2. RH problem transformationIt is necessary to carry out the second transformation to transform the contour, which can make the jump matrices decline to the identity matrix $I$ for large $t$. Generally speaking, the new RH transformation $\widetilde{M}(x,t,\xi)$ depends on the reflection coefficients $r_j(\xi)$, $r_j(\xi)/(1+r_1(\xi)r_2(\xi))$, $j=1,2$. Now by the classical Deift–Zhou method, they can be approximated by some rational functions with good error control. In the following analysis, for clarity, we will assume that the initial data $q_0(x)$ permit a compact perturbation of the pure-step initial data $q_{0A}(x)$ as depicted in (2.31), which makes certain that the eigenfunctions $\psi_{\pm}^{(s)}(x,0,\xi)$, $s=1,2$, and thus $r_j(\xi)$, $j=1,2$, are meromorphic in $\mathbb{C}$. Next, we define the function $\widehat{M}(x,t,\xi)$ as follows (see Fig. 5):
$$
\begin{equation}
\widehat{M}(x,t,\xi)= \begin{cases} \widetilde{M}(x,t,\xi) \begin{pmatrix} 1 & \dfrac{r_2(\xi)\delta^2(\xi,\mu)}{1+r_1(\xi)r_2(\xi)}e^{-2it\theta} \\ 0 & 1 \end{pmatrix}, &\xi\in \Omega_2, \\ \widetilde{M}(x,t,\xi) \begin{pmatrix} 1 & 0 \\ r_1(\xi)\delta^{-2}(\xi,\mu)e^{2it\theta} & 1 \end{pmatrix}, &\xi\in \Omega_1, \\ \widetilde{M}(x,t,\xi) \begin{pmatrix} 1 & -r_2(\xi)\delta^2(\xi,\mu)e^{-2it\theta} \\ 0 & 1 \end{pmatrix}, &\xi\in \Omega^*_1, \\ \widetilde{M}(x,t,\xi) \begin{pmatrix} 1 & 0 \\ -\dfrac{r_1(\xi)\delta^{-2}(\xi,\mu)}{1+r_1(\xi)r_2(\xi)}e^{2it\theta} & 1 \end{pmatrix}, &\xi\in \Omega^*_2, \\ \widetilde{M}(x,t,\xi), &\xi\in \Omega_0\cup \Omega^*_0. \end{cases}
\end{equation}
\tag{3.15}
$$
Here, we choose an appropriate angle between the real axis and the rays $\Upsilon_j$, $\Upsilon^*_j$ to ensure that the discrete spectrum $i\xi_1$ is located in the sector $\Omega_0$. We thus arrive at the following RH problem $\widehat{M}(x,t,\xi)$ on the contour $\Upsilon$. RH problem 3. Find a matrix-valued function $\widehat{M}(x,t,\xi)$ that is piecewise meromorphic and satisfies the following conditions: (i) $\widehat{M}(x,t,\xi)$ is meromorphic for $\xi\in\mathbb{C}\setminus\Upsilon$ and has a simple pole located at $\xi=i\xi_1$, $\xi_1>0$; (ii) the jump conditions: the non-tangential limits $\widehat{M}(x,t,\xi)=\widehat{M}(x,t,\xi\pm i0)$ exist a.e. for $\xi\in\mathbb{R}$ such that $\widehat{M}(x,t,{\cdot}\,)-I\in L^2(\mathbb{R}\setminus [-\varepsilon,\varepsilon])$ for any $\varepsilon>0$ and $\widehat{M}_{\pm}(x,t,\xi)$ satisfy
$$
\begin{equation}
\widehat{M}_+(x,t,\xi)=\widehat{M}_-(x,t,\xi)\widehat{J}(x,t,\xi),\qquad \xi\in \Upsilon,
\end{equation}
\tag{3.16}
$$
where
$$
\begin{equation}
\widehat{J}(x,t,\xi)= \begin{cases} \begin{pmatrix} 1 & -\dfrac{r_2(\xi)\delta^2(\xi,\mu)}{1+r_1(\xi)r_2(\xi)}\, e^{-2it\theta} \\ 0 & 1 \end{pmatrix}, &\xi\in \Upsilon_2, \\ \begin{pmatrix} 1 & 0 \\ -r_1(\xi)\delta^{-2}(\xi,\mu)e^{2it\theta} & 1 \end{pmatrix}, &\xi\in \Upsilon_1, \\ \begin{pmatrix} 1 & r_2(\xi)\delta^2(\xi,\mu)e^{-2it\theta} \\ 0 & 1 \end{pmatrix}, &\xi\in \Upsilon^*_1, \\ \begin{pmatrix} 1 & 0 \\ \dfrac{r_1(\xi)\delta^{-2}(\xi,\mu)}{1+r_1(\xi)r_2(\xi)} \, e^{2it\theta} & 1 \end{pmatrix}, &\xi\in \Upsilon^*_2; \end{cases}
\end{equation}
\tag{3.17}
$$
(iii) the normalization condition at $\xi=\infty$:
$$
\begin{equation}
\widehat{M}(x,t,\xi)=I+O\biggl(\frac{1}{\xi}\biggr),\qquad \xi\to\infty;
\end{equation}
\tag{3.18}
$$
(iv) the residue condition:
$$
\begin{equation}
\operatorname*{Res}_{\xi=i\xi_1}\bigl[\widehat{M}(x,t,\xi)\bigr]_1=c_1(x,t)\bigl[\widehat{M}(x,t,i\xi_1)\bigr]_2,
\end{equation}
\tag{3.19}
$$
where
$$
\begin{equation*}
c_1(x,t)=\frac{r_1}{\dot{a}_1(i\xi_1)\delta^2(i\xi_1)}\, e^{-2\xi_1x+2i\xi_1^2t+16i\xi_1^4\gamma t};
\end{equation*}
\notag
$$
(v) the singularity conditions at $\xi=0$, $\widehat{M}(x,t,\xi)$ satisfies the following relations in both Cases 1 and 2:
$$
\begin{equation}
\widehat{M}_+ {=}\, \begin{pmatrix} -\dfrac{2i \overline{f}_2(-x,t)}{A \delta(0,\mu)}+O(\xi) & -\dfrac{1}{\xi}\overline{f}_2(-x,t)\delta(0,\mu)+O(1) \\ -\dfrac{2i \overline{f}_1(-x,t)}{A \delta(0,\mu)}+O(\xi) & -\dfrac{1}{\xi}\overline{f}_1(-x,t)\delta(0,\mu)+O(1) \end{pmatrix},\qquad \xi\in\mathbb{C}_+, \quad\xi\to0,
\end{equation}
\tag{3.20a}
$$
$$
\begin{equation}
\widehat{M}_- {=}\,\frac{2i}{A} \begin{pmatrix} -\dfrac{\overline{f}_2(-x,t)}{\delta(0,\mu)}\,{+}\,O(\xi) & -\dfrac{A}{2i\xi}\delta(0,\mu)\overline{f}_2(-x,t)\,{+}\,O(\xi) \\ -\dfrac{\overline{f}_1(-x,t)}{\delta(0,\mu)}+O(\xi) & -\dfrac{A}{2i\xi}\delta(0,\mu)\overline{f}_1(-x,t)+O(\xi) \end{pmatrix},\qquad\!\! \xi\,{\in}\,\mathbb{C}_-,\quad\xi \to 0.
\end{equation}
\tag{3.20b}
$$
Moreover, it can be noted that the singularity conditions at $\xi=0$ in both cases can be reduced to the same residue condition
$$
\begin{equation}
\operatorname*{Res}_{\xi=0}\bigl[\widehat{M}(x,t,\xi)\bigr]_2=c_0(\mu)\bigl[\widehat{M}(x,t,0)\bigr]_1,
\end{equation}
\tag{3.21}
$$
with $c_0(\mu)=A\delta^2(0,\mu)/(2i)$. 3.3. Regular RH problemIn the discussion of this subsection, we will employ the BP factor to convert the RH problem 3, which includes two residue conditions (3.19) and (3.21) into a regular RH problem devoid of residue conditions. After making the transformation
$$
\begin{equation}
\widehat{M}(x,t,\xi)=B(x,t,\xi)\widehat{M}^{\mathrm{reg}}(x,t,\xi) \begin{pmatrix} 1 & 0 \\ 0 & \dfrac{\xi-i\xi_1}{\xi} \end{pmatrix},\qquad \xi\in\mathbb{C},
\end{equation}
\tag{3.22}
$$
finding the solution $\widehat{M}(x,t,\xi)$ of RH problem 3 can be reduced that of the solution $\widehat{M}^{\mathrm{reg}}(x,t,\xi)$ of the regular RH problem. Here, the function $B(x,t,\xi)$ takes the form $B(x,t,\xi)=I+(i\xi_1/(\xi-i\xi_1))P(x,t)$; $B(x,t,\xi)$ and $P(x,t)$ are the BP factors. Now we introduce the regular RH problem as follows. RH problem 3.5. Find a matrix-valued function $\widehat{M}^{\mathrm{reg}}(x,t,\xi)$ that is piecewise meromorphic and satisfies the following conditions: (i) $\widehat{M}^{\mathrm{reg}}(x,t,\xi)$ is analytic for $\xi\in\mathbb{C}\setminus\Upsilon$; (ii) the jump conditions:
$$
\begin{equation}
\widehat{M}^{\mathrm{reg}}_+(x,t,\xi)=\widehat{M}^{\mathrm{reg}}_-(x,t,\xi)\widehat{J}^{\mathrm{reg}}(x,t,\xi),
\end{equation}
\tag{3.23}
$$
where
$$
\begin{equation}
\widehat{J}^{\mathrm{reg}}(x,t,\xi)= \begin{pmatrix} 1 & 0 \\ 0 & \dfrac{\xi-i\xi_1}{\xi} \end{pmatrix} \widehat{J}(x,t,\xi) \begin{pmatrix} 1 & 0 \\ 0 & \dfrac{\xi}{\xi-i\xi_1} \end{pmatrix},
\end{equation}
\tag{3.24}
$$
$$
\begin{equation}
\widehat{J}^{\mathrm{reg}}(x,t,\xi)= \begin{cases} \begin{pmatrix} 1 & -\dfrac{r_2^{\mathrm{reg}}(\xi)\delta^2(\xi,\mu)}{1+r_1^{\mathrm{reg}}(\xi)r_2^{\mathrm{reg}}(\xi)} \, e^{-2it\theta} \\ 0 & 1 \end{pmatrix}, &\xi\in \Upsilon_2, \\ \begin{pmatrix} 1 & 0 \\ -r_1^{\mathrm{reg}}(\xi)\delta^{-2}(\xi,\mu)e^{2it\theta} & 1 \end{pmatrix}, &\xi\in \Upsilon_1, \\ \begin{pmatrix} 1 & r_2^{\mathrm{reg}}(\xi)\delta^2(\xi,\mu)e^{-2it\theta} \\ 0 & 1 \end{pmatrix}, &\xi\in \Upsilon^*_1, \\ \begin{pmatrix} 1 & 0 \\ \dfrac{r_1^{\mathrm{reg}}(\xi)\delta^{-2}(\xi,\mu)}{1+r_1^{\mathrm{reg}}(\xi)r_2^{\mathrm{reg}}(\xi)} \, e^{2it\theta} & 1 \end{pmatrix}, &\xi\in \Upsilon^*_2, \end{cases}
\end{equation}
\tag{3.25}
$$
$$
\begin{equation}
r_1^{\mathrm{reg}}(\xi)=\frac{\xi-i\xi_1}{\xi} \, r_1(\xi),\qquad r_2^{\mathrm{reg}}(\xi)=\frac{\xi}{\xi-i\xi_1} \, r_2(\xi);
\end{equation}
\tag{3.26}
$$
(iii) the normalization condition at $\xi=\infty$:
$$
\begin{equation}
\widehat{M}^{\mathrm{reg}}(x,t,\xi)=I+O\biggl(\frac{1}{\xi}\biggr),\qquad \xi\to\infty;
\end{equation}
\tag{3.27}
$$
(iv) the elements of the matrix-valued factor $P(x,t)$ determined by $\widehat{M}^{\mathrm{reg}}(x,t,\xi)$ take the form
$$
\begin{equation}
\begin{aligned} \, P_{12}(x,t)&=\frac{u_1(x,t)v_1(x,t)}{u_1(x,t)v_2(x,t)-u_2(x,t)v_1(x,t)}, \\ P_{21}(x,t)&=-\frac{u_2(x,t)v_2(x,t)}{u_1(x,t)v_2(x,t)-u_2(x,t)v_1(x,t)}, \end{aligned}
\end{equation}
\tag{3.28}
$$
where $P_{ij}(x,t)$ $(i,j=1,2)$ represent the position of the $i$s row and the $j$s column in the matrix $P(x,t)$. $u(x,t)=(u_1(x,t),u_2(x,t))^\top$ and $v(x,t)=(v_1(x,t),v_2(x,t))^\top$ are given by
$$
\begin{equation}
\begin{aligned} \, u(x,t)&=i\xi_1\bigl[\widehat{M}^{\mathrm{reg}}(x,t,i\xi_1)\bigr]_1 -c_1(x,t)\bigl[\widehat{M}^{\mathrm{reg}}(x,t,i\xi_1)\bigr]_2, \\ v(x,t)&=i\xi_1\bigl[\widehat{M}^{\mathrm{reg}}(x,t,0)\bigr]_2 +c_0(\mu)\bigl[\widehat{M}^{\mathrm{reg}}(x,t,0)\bigr]_1. \end{aligned}
\end{equation}
\tag{3.29}
$$
Proof. Items (i)–(iii) are easy to check. As for item (iv), by writing the terms of transformation (3.22) in matrix form and recalling the residue conditions (3.19) and (3.21), we can derive (3.28) by a straightforward computation. $\Box$ Remark 3.6. Since $r_1(\xi)=\overline{r_1(-\overline{\xi})}$ and $r_2(\xi)=\overline{r_2(-\overline{\xi})}$, we have Next, a relation between $q(x,t)$ and $\widehat{M}^{\mathrm{reg}}(x,t,\xi)$ can be established in the following proposition. From the rough approximation $\widehat{M}^{\mathrm{reg}}(x,t,\xi)\approx I$ as $t\to\infty$, we derive the approximate error estimate for the large $t$ asymptotics of $q(x,t)$. Proposition 3.7. The solution $q(x,t)$ of the Cauchy problem (1.3) and (1.5) can be written as Furthermore, as $t\to\infty$, the rough estimate for $q(x,t)$ is defined by along any ray $\mu=x/t=\mathrm{const}\in(\epsilon,\sqrt{1/(27\gamma)}+\epsilon)$ or $\mu\in(-\sqrt{1/(27\gamma)}+\epsilon,\epsilon)$. Proof. Using transformation (3.22) and since $B(x,t,\xi)=I+(i\xi_1/(\xi-i\xi_1))P(x,t)$, we get the asymptotic expansion of $\widehat{M}(x,t,\xi)$ as $\xi\to\infty$:
$$
\begin{equation}
\widehat{M}(x,t,\xi)= \begin{pmatrix} 1 & 0 \\ 0 & 1-\dfrac{i\xi_1}{\xi} \end{pmatrix} +\frac{i\xi_1}{\xi-i\xi_1}P(x,t) +\frac{\widehat{M}^{\mathrm{reg}}_1(x,t)}{\xi}+O(\xi^{-2}), \qquad \xi\to\infty,
\end{equation}
\tag{3.34}
$$
where $\widehat{M}^{\mathrm{reg}}(x,t,\xi)=I+\widehat{M}^{\mathrm{reg}}_1(x,t)/\xi+O(\xi^{-2})$ as $\xi\to\infty$. Now by combining (2.67), (2.68) and using transformations (3.7) and (3.15), we arrive at equation (3.32).
The rough approximation $\widehat{M}^{\mathrm{reg}}(x,t,\xi)\approx I$ as $t\to\infty$ indicates that, for $x>0$,
$$
\begin{equation}
\begin{pmatrix} u_1(x,t) \\ u_2(x,t) \end{pmatrix} \approx \begin{pmatrix} i\xi_1 \\ 0 \end{pmatrix},\qquad \begin{pmatrix} v_1(x,t) \\ v_2(x,t) \end{pmatrix} \approx \begin{pmatrix} c_0(\mu) \\ i\xi_1 \end{pmatrix},
\end{equation}
\tag{3.35}
$$
hence,
$$
\begin{equation}
q(x,t)\approx -2\xi_1\frac{i\xi_1c_0(\mu)}{-\xi^2_1+c_0(\mu)c_1(x,t)}\approx 2ic_0(\mu)=A\delta^2(0,\mu).
\end{equation}
\tag{3.36}
$$
For $x<0$,
$$
\begin{equation}
\begin{pmatrix} \overline{u_1(-x,t)} \\ \overline{u_2(-x,t)} \end{pmatrix} \approx \begin{pmatrix} i\xi_1 \\ 0 \end{pmatrix},\qquad \begin{pmatrix} \overline{v_1(-x,t)} \\ \overline{v_2(-x,t)} \end{pmatrix} \approx \begin{pmatrix} c_0(\mu) \\ i\xi_1 \end{pmatrix},
\end{equation}
\tag{3.37}
$$
therefore, $\Box$ 3.4. Local models near the saddle pointsIt is evident that we can show that the jump matrix $\widehat{J}^{\mathrm{reg}}(x,t,\xi)$ approaches the identity matrix as $t\to\infty$, but the neighborhood of the saddle points $\lambda_1$, $\lambda_2$, and $\lambda_3$ require an additional analysis. The objective of this subsection is to achieve a precise approximation of the function $\widehat{M}^{\mathrm{reg}}(x,t,\xi)$ near these saddle points by parabolic cylinder functions, thereby obtaining the long-time asymptotics of $\widehat{M}^{\mathrm{reg}}(x,t,\xi)$. In the following discussion, we take $r_j(\xi)\neq0$, $j=1,2$. If as least one of $r_j(\xi)$, $j=1,2$, is zero, then $v(\lambda_j)=0$. It is enough to estimate the long-time asymptotic solution of $\widehat{J}^{\mathrm{reg}}(x,t,\xi)$ at $\xi=0$, $\xi=i\xi_1$, and $\xi=\infty$ using (3.32). Furthermore, we observe that this RH problem is analogous to the case with zero boundary conditions, thus we will adopt the approach of [44] for a subsequent analysis. The scaling transformations are defined as follows:
$$
\begin{equation}
\xi \to\lambda_1+\frac{\tau}{\sqrt{4t(48\gamma\lambda_1^2-1)}},
\end{equation}
\tag{3.39a}
$$
$$
\begin{equation}
\xi \to\lambda_2+\frac{\tau}{\sqrt{4t(1-48\gamma\lambda_2^2)}},
\end{equation}
\tag{3.39b}
$$
$$
\begin{equation}
\xi \to\lambda_3+\frac{\tau}{\sqrt{4t(48\gamma\lambda_3^2-1)}}.
\end{equation}
\tag{3.39c}
$$
Besides, the function $\delta$ can be written as
$$
\begin{equation}
\begin{aligned} \, \delta(\xi(\tau),\mu)&=\biggl(\frac{\tau}{\sqrt{4t(48\gamma\lambda_1^2-1)}}\biggr)^{iv(\lambda_1)} \Biggl(\sqrt{\frac{48\gamma\lambda_3^2-1}{1-48\gamma\lambda_2^2}} \Biggr)^{-iv(\lambda_1)}e^{\chi_1(\xi)} \\ &=\biggl(\frac{\tau}{\sqrt{4t(48\gamma\lambda_3^2-1)}}\biggr)^{iv(\lambda_3)} \Biggl(\sqrt{\frac{48\gamma\lambda_1^2-1}{1-48\gamma\lambda_2^2}} \Biggr)^{-iv(\lambda_2)}e^{\chi_2(\xi)} \\ &=\biggl(\frac{\tau}{\sqrt{4t(48\gamma\lambda_3^2-1)}}\biggr)^{iv(\lambda_3)} \Biggl(\sqrt{\frac{48\gamma\lambda_1^2-1}{1-48\gamma\lambda_2^2}} \Biggr)^{-iv(\lambda_3)}e^{\chi_3(\xi)}, \end{aligned}
\end{equation}
\tag{3.40}
$$
where $v(\lambda_j)$ and $\chi_j(\xi)$, $j=1,2,3$, are as in (3.5) and (3.6). We define $D_{\varepsilon}(\lambda_j)$ to express the open disk centered at $\lambda_j$ of radius $\varepsilon$. We also consider the contours
$$
\begin{equation}
\Upsilon_{\lambda_j}=\Upsilon\cap D_{\varepsilon}(\lambda_j) =\Upsilon_{1,\varepsilon_j}\cup\Upsilon_{2,\varepsilon_j} \cup\Upsilon^*_{1,\varepsilon_j}\cup\Upsilon^*_{2,\varepsilon_j}
\end{equation}
\tag{3.41}
$$
for $j=1,2,3$, and define $\Upsilon=\Upsilon_1\cup\Upsilon_2\cup\Upsilon^*_1\cup\Upsilon^*_2$. At the next step, we introduce the $2\times2$ matrices $\widehat{M}^{\mathrm{reg}}_{\lambda_j}(x,t,\xi)$ and $\Lambda_j(\mu,t)$ for $j=1,2,3$. The matrix exponential function is defined by $N^{\sigma_3}=\left(\begin{smallmatrix} N & 0 \\ 0 & N^{-1} \end{smallmatrix}\right)$ for any function $N$:
$$
\begin{equation}
\widehat{M}^{\mathrm{reg}}_{\lambda_1}(x,t,\xi)=\Lambda_1(\mu,t) \widehat{m}^{\mathrm{pc}}_{\lambda_1}(\lambda_1,\tau(\xi))\Lambda^{-1}_1(\mu,t),
\end{equation}
\tag{3.42}
$$
with
$$
\begin{equation}
\Lambda_1(\mu,t)=e^{[\chi_1(\lambda_1)+\phi_1(\mu,\tau(\lambda_1))]\sigma_3} \biggl(\frac{1-48\gamma\lambda_2^2}{4t(48\gamma\lambda_1^2-1)(48\gamma\lambda_3^2-1)}\biggr) ^{\frac{i}{2}v(\lambda_1)\sigma_3},
\end{equation}
\tag{3.43}
$$
$$
\begin{equation}
\begin{split} \phi_1(\mu,\tau(\lambda_1))=&-\frac{i\gamma\tau^4}{2t(48\gamma\lambda_1^2-1)^2} -\frac{4i\gamma\lambda_1\tau^3}{\sqrt{t(48\gamma\lambda_1^2-1)^3}} \\ &\qquad+\frac{i\tau^2}{4t(48\gamma\lambda_1^2-1)} -\frac{i(16\gamma\lambda^2_1-t)\lambda_1\tau}{\sqrt{t(48\gamma\lambda_1^2-1)}} -4\gamma\lambda_1^4+\frac{1}{2}\lambda_1^2, \end{split}
\end{equation}
\tag{3.44}
$$
$$
\begin{equation}
\widehat{M}^{\mathrm{reg}}_{\lambda_2}(x,t,\xi)=\Lambda_2(\mu,t) \widehat{m}^{\mathrm{pc}}_{\lambda_2}(\lambda_2,\tau(\xi))\Lambda^{-1}_2(\mu,t),
\end{equation}
\tag{3.45}
$$
with
$$
\begin{equation}
\Lambda_2(\mu,t)=e^{[\chi_2(\lambda_2)+\phi_2(\mu,\tau(\lambda_2))]\sigma_3} \biggl(\frac{1}{4t(48\gamma\lambda_3^2-1)}\biggr)^{-\frac{i}{2}v(\lambda_3)\sigma_3} \biggl(\frac{1-48\gamma\lambda_2^2}{48\gamma\lambda_1^2-1}\biggr)^{-\frac{i}{2}v(\lambda_2)\sigma_3},
\end{equation}
\tag{3.46}
$$
$$
\begin{equation}
\begin{split} \phi_2(\mu,\tau(\lambda_2))=&-\frac{i\gamma\tau^4}{2t(1-48\gamma\lambda_2^2)^2} -\frac{4i\gamma\lambda_2\tau^3}{\sqrt{t(1-48\gamma\lambda_2^2)^3}} \\ &\qquad+\frac{i\tau^2}{4t(1-48\gamma\lambda_2^2)} -\frac{i(16\gamma\lambda^2_2-t)\lambda_2\tau}{\sqrt{t(1-48\gamma\lambda_2^2)}} -4\gamma\lambda_2^4+\frac{1}{2}\lambda_2^2, \end{split}
\end{equation}
\tag{3.47}
$$
$$
\begin{equation}
\widehat{M}^{\mathrm{reg}}_{\lambda_3}(x,t,\xi)=\Lambda_3(\mu,t) \widehat{m}^{\mathrm{pc}}_{\lambda_3}(\lambda_3,\tau(\xi))\Lambda^{-1}_3(\mu,t),
\end{equation}
\tag{3.48}
$$
with
$$
\begin{equation}
\Lambda_3(\mu,t)=e^{[\chi_3(\lambda_3)+\phi_3(\mu,\tau(\lambda_3))]\sigma_3} \biggl(\frac{1-48\gamma\lambda_2^2}{4t(48\gamma\lambda_3^2-1)(48\gamma\lambda_1^2-1)}\biggr) ^{\frac{i}{2}v(\lambda_3)\sigma_3},
\end{equation}
\tag{3.49}
$$
$$
\begin{equation}
\begin{split} \phi_3(\mu,\tau(\lambda_3))=&-\frac{i\gamma\tau^4}{2t(48\gamma\lambda_3^2-1)^2} -\frac{4i\gamma\lambda_3\tau^3}{\sqrt{t(48\gamma\lambda_3^2-1)^3}} \\ &\qquad+\frac{i\tau^2}{4t(48\gamma\lambda_3^2-1)} -\frac{i(16\gamma\lambda^2_3-t)\lambda_3\tau}{\sqrt{t(48\gamma\lambda_3^2-1)}} -4\gamma\lambda_3^4+\frac{1}{2}\lambda_3^2. \end{split}
\end{equation}
\tag{3.50}
$$
Here, $\widehat{m}^{\mathrm{pc}}_{\lambda_j}(\lambda_j,\tau(\xi))$, $j=1,2,3$, satisfy the parameterized RH problems
$$
\begin{equation}
\begin{cases} \widehat{m}^{\mathrm{pc}}_{\lambda_j,+}(\lambda_j,\tau)= \widehat{m}^{\mathrm{pc}}_{\lambda_j,-}(\lambda_j,\tau)J^{\mathrm{pc}}_{\lambda_j}(\tau), \quad \tau\in\Sigma^{\mathrm{pc}}_{\lambda_j}, \\ \widehat{m}^{\mathrm{pc}}_{\lambda_j}(\lambda_j,\tau)\to I,\quad \tau\to\infty. \end{cases}
\end{equation}
\tag{3.51}
$$
These RH problems can be explicitly solved by employing the parabolic cylinder functions. The method for solving this problem is well-established; see [23] for a detailed procedure. The jump contours $\Upsilon_{\lambda_s}$, $s=1,3$, and $\Upsilon_{\lambda_2}$ of $\widehat{M}^{\mathrm{reg}}_{\lambda_j}(x,t,\xi)$, $j=1,2,3$, are defined in Figs. 6 and 7:
$$
\begin{equation}
J^{\mathrm{pc}}_{\lambda_s}(\tau) = \begin{cases} \begin{pmatrix} 1 & -\dfrac{r_2^{\mathrm{reg}}(\lambda_s)}{1+r_1^{\mathrm{reg}}(\lambda_s)r_2^{\mathrm{reg}}(\lambda_s)} \, e^{-\frac{i}{2}\tau^2}\tau^{2iv(\lambda_s)}\\ 0 & 1 \end{pmatrix}, &\tau\in \Sigma_{2,\varepsilon_s}, \\ \begin{pmatrix} 1 & 0 \\ -r_1^{\mathrm{reg}}(\lambda_s)e^{\frac{i}{2}\tau^2}\tau^{-2iv(\lambda_s)} & 1 \end{pmatrix}, &\tau\in \Sigma_{1,\varepsilon_s}, \\ \begin{pmatrix} 1 & r_2^{\mathrm{reg}}(\lambda_s)e^{-\frac{i}{2}\tau^2}\tau^{2iv(\lambda_s)} \\ 0 & 1 \end{pmatrix}, &\tau\in \Sigma^*_{1,\varepsilon_s}, \\ \begin{pmatrix} 1 & 0 \\ \dfrac{r_1^{\mathrm{reg}}(\lambda_s)}{1+r_1^{\mathrm{reg}}(\lambda_s)r_2^{\mathrm{reg}}(\lambda_s)} \, e^{\frac{i}{2}\tau^2}\tau^{-2iv(\lambda_s)} & 1 \end{pmatrix}, &\tau\in \Sigma^*_{2,\varepsilon_s}, \end{cases}
\end{equation}
\tag{3.52}
$$
$$
\begin{equation}
J^{\mathrm{pc}}_{\lambda_2}(\tau) = \begin{cases} \begin{pmatrix} 1 & 0 \\ -r_1^{\mathrm{reg}}(\lambda_2)e^{-\frac{i}{2}\tau^2}\tau^{2iv(\lambda_2)} & 1 \end{pmatrix}, &\tau\in \Sigma_{2,\varepsilon_2}, \\ \begin{pmatrix} 1 & -\dfrac{r_2^{\mathrm{reg}}(\lambda_2)}{1+r_1^{\mathrm{reg}}(\lambda_2)r_2^{\mathrm{reg}}(\lambda_2)} \, e^{\frac{i}{2}\tau^2}\tau^{-2iv(\lambda_2)} \\ 0 & 1 \end{pmatrix}, &\tau\in \Sigma_{1,\varepsilon_2}, \\ \begin{pmatrix} 1 & 0 \\ \dfrac{r_1^{\mathrm{reg}}(\lambda_2)}{1+r_1^{\mathrm{reg}}(\lambda_2)r_2^{\mathrm{reg}}(\lambda_2)} \, e^{-\frac{i}{2}\tau^2}\tau^{2iv(\lambda_2)} & 1 \end{pmatrix}, &\tau\in \Sigma^*_{1,\varepsilon_2}, \\ \begin{pmatrix} 1 & r_2^{\mathrm{reg}}(\lambda_2)e^{\frac{i}{2}\tau^2}\tau^{-2iv(\lambda_2)} \\ 0 & 1 \end{pmatrix}, &\tau\in \Sigma^*_{2,\varepsilon_2}. \end{cases}
\end{equation}
\tag{3.53}
$$
The function $\widehat{m}^{\mathrm{pc}}_{\lambda_j}(\lambda_j,\tau(\xi))$, $j=1,2,3$, which can be directly solved using parabolic cylindrical function, is defined by
$$
\begin{equation}
\widehat{m}^{\mathrm{pc}}_{\lambda_j}(\lambda_j,\tau)= \begin{cases} m_{\lambda_j}(\lambda_j,\tau)(G^{l}_{\lambda_j})^{-1}(\lambda_j,\tau), &\tau\in\Omega_{l,\varepsilon_j}, \, l=0,1,2, \\ m_{\lambda_j}(\lambda_j,\tau)(G^{l*}_{\lambda_j})^{-1}(\lambda_j,\tau), &\tau\in\Omega^*_{l,\varepsilon_j}, \, l=0,1,2. \end{cases}
\end{equation}
\tag{3.54}
$$
We assume that $G^0_{\lambda_s}=G^{0*}_{\lambda_s}=e^{-\frac{i}{4}\tau^2\sigma_3}\tau^{iv(\lambda_s)\sigma_3}$, $s=1,3$, and $G^0_{\lambda_2}=G^{0*}_{\lambda_2}=e^{\frac{i}{4}\tau^2\sigma_3}\tau^{-iv(\lambda_2)\sigma_3}$,
$$
\begin{equation*}
\begin{alignedat}{2} G^1_{\lambda_s} &=G^0_{\lambda_s} \begin{pmatrix} 1 & 0 \\ -r^{\mathrm{reg}}_1(\lambda_s) & 1 \end{pmatrix}, &\quad G^{1*}_{\lambda_s} &=G^{0*}_{\lambda_s} \begin{pmatrix} 1 & r^{\mathrm{reg}}_2(\lambda_s) \\ 0 & 1 \end{pmatrix}, \\ G^2_{\lambda_s} &=G^0_{\lambda_s} \begin{pmatrix} 1 & -\dfrac{r^{\mathrm{reg}}_2(\lambda_s)}{1\,{+}\,r^{\mathrm{reg}}_1(\lambda_s)r^{\mathrm{reg}}_2(\lambda_s)} \\ 0 & 1 \end{pmatrix}, &\quad G^{2*}_{\lambda_s} &=G^{0*}_{\lambda_s} \begin{pmatrix} 1 & 0 \\ \dfrac{r^{\mathrm{reg}}_1(\lambda_s)}{1\,{+}\,r^{\mathrm{reg}}_1(\lambda_s)r^{\mathrm{reg}}_2(\lambda_s)} & 1 \end{pmatrix}, \\ G^1_{\lambda_2} &=G^0_{\lambda_2} \begin{pmatrix} 1 & -\dfrac{r^{\mathrm{reg}}_2(\lambda_2)}{1\,{+}\,r^{\mathrm{reg}}_1(\lambda_2)r^{\mathrm{reg}}_2(\lambda_2)} \\ 0 & 1 \end{pmatrix}, &\quad G^{1*}_{\lambda_s} &=G^{0*}_{\lambda_s} \begin{pmatrix} 1 & 0 \\ \dfrac{r^{\mathrm{reg}}_1(\lambda_2)}{1\,{+}\,r^{\mathrm{reg}}_1(\lambda_2)r^{\mathrm{reg}}_2(\lambda_2)} & 1 \end{pmatrix}, \\ G^2_{\lambda_2} &=G^0_{\lambda_2} \begin{pmatrix} 1 & 0 \\ -r^{\mathrm{reg}}_1(\lambda_2) & 1 \end{pmatrix}, &\quad G^{2*}_{\lambda_2} &=G^{0*}_{\lambda_2} \begin{pmatrix} 1 & r^{\mathrm{reg}}_2(\lambda_2) \\ 0 & 1 \end{pmatrix}. \end{alignedat}
\end{equation*}
\notag
$$
The functions $m_{\lambda_j}(\lambda_j,\tau)$, $j=1,2,3$, satisfies the RH problem
$$
\begin{equation}
\begin{alignedat}{2} m_{\lambda_j,+}(\lambda_j,\tau) &=m_{\lambda_j,-}(\lambda_j,\tau)J_j(\lambda_j), &\qquad \tau&\in\mathbb{R}, \\ m_{\lambda_j}(\lambda_j,\tau)&=(I+O(\tau^{-1})) e^{(-1)^j\frac{i}{4}\tau^2\sigma_3}\tau^{(-1)^{j+1}iv(\lambda_j)\sigma_3},&\qquad\tau&\to\infty, \end{alignedat}
\end{equation}
\tag{3.55}
$$
with
$$
\begin{equation}
J_j(\lambda_j)= \begin{pmatrix} 1+r^{\mathrm{reg}}_1(\lambda_j)r^{\mathrm{reg}}_2(\lambda_j) & -r^{\mathrm{reg}}_2(\lambda_j) \\ -r^{\mathrm{reg}}_1(\lambda_j) & 1 \end{pmatrix}.
\end{equation}
\tag{3.56}
$$
In addition, the RH problems $\widehat{m}^{\mathrm{pc}}_{\lambda_j}(\lambda_j,\tau)$, $j=1,2,3$, behave asymptotically as
$$
\begin{equation}
\widehat{m}^{\mathrm{pc}}_{\lambda_j}(\lambda_j,\tau)=I+\frac{i}{\tau} \begin{pmatrix} 0 & \beta^{\mathrm{reg}}_j(\lambda_j) \\ -\gamma^{\mathrm{reg}}_j(\lambda_j) & 0 \end{pmatrix}+O(\tau^{-2}),\qquad \tau\to\infty,
\end{equation}
\tag{3.57}
$$
where
$$
\begin{equation}
\beta^{\mathrm{reg}}_s(\lambda_s) =-\frac{\sqrt{2\pi}\, e^{-\frac{\pi}{2}v(\lambda_s)}e^{\frac{i\pi}{4}}} {r^{\mathrm{reg}}_1(\lambda_s)\Gamma(-iv(\lambda_s))}, \qquad s =1,3,
\end{equation}
\tag{3.58a}
$$
$$
\begin{equation}
\gamma^{\mathrm{reg}}_s(\lambda_s) =-\frac{\sqrt{2\pi} \, e^{-\frac{\pi}{2}v(\lambda_s)}e^{-\frac{i\pi}{4}}} {r^{\mathrm{reg}}_2(\lambda_s)\Gamma(iv(\lambda_s))}, \qquad s =1,3.
\end{equation}
\tag{3.58b}
$$
By symmetry (see (3.31)), we have
$$
\begin{equation}
\widehat{m}^{\mathrm{pc}}_{\lambda_2}(\lambda_2,\tau) =\overline{\widehat{m}^{\mathrm{pc}}_{\lambda_1}(\lambda_1,-\overline{\tau})},
\end{equation}
\tag{3.59}
$$
which implies that
$$
\begin{equation}
\beta^{\mathrm{reg}}_2(\lambda_2)=\overline{\beta^{\mathrm{reg}}_1(\lambda_2)},\qquad \gamma^{\mathrm{reg}}_2(\lambda_2)=\overline{\gamma^{\mathrm{reg}}_1(\lambda_2)}.
\end{equation}
\tag{3.60}
$$
3.5. The long-time asymptotic behaviorThe purpose of this subsection is to find am explicit long-time asymptotic formula for the non-local LPD equation. After acquiring the local parametrix $\widehat{M}^{\mathrm{reg}}_{\lambda_j}(x,t,\xi)$, $j=1,2,3$, we introduce
$$
\begin{equation}
\breve{M}^{\mathrm{reg}}(x,t,\xi)= \begin{cases} \widehat{M}^{\mathrm{reg}}(x,t,\xi)(\widehat{M}^{\mathrm{reg}}_{\lambda_1})^{-1}(x,t,\xi), &|\xi-\lambda_1|<\varepsilon, \\ \widehat{M}^{\mathrm{reg}}(x,t,\xi)(\widehat{M}^{\mathrm{reg}}_{\lambda_2})^{-1}(x,t,\xi), &|\xi-\lambda_2|<\varepsilon, \\ \widehat{M}^{\mathrm{reg}}(x,t,\xi)(\widehat{M}^{\mathrm{reg}}_{\lambda_3})^{-1}(x,t,\xi), &|\xi-\lambda_3|<\varepsilon, \\ \widehat{M}^{\mathrm{reg}}(x,t,\xi), &\text{elsewhere}, \end{cases}
\end{equation}
\tag{3.61}
$$
here $\varepsilon$ is small enough so that $|\lambda_j|>\varepsilon$ and $|i\xi_1-\lambda_j|>\varepsilon$. Consider the jump contour $\breve{\Upsilon}=\Upsilon\cup\partial D_{\varepsilon}(\lambda_1)\cup\partial D_{\varepsilon}(\lambda_2)\cup\partial D_{\varepsilon}(\lambda_3)$ of $\breve{M}(x,t,\xi)$ (see Fig. 8). We also set $\Upsilon_{\varepsilon}=\Upsilon\cap D_{\varepsilon}(\lambda_j)$, $j=1,2,3$, as the jump contours of $\widehat{M}^{\mathrm{reg}}_{\lambda_j}(x,t,\xi)$. The function $\breve{M}^{\mathrm{reg}}(x,t,\xi)$ satisfies the following RH problem. RH problem 3.8. Find a matrix-valued function $\breve{M}^{\mathrm{reg}}(x,t,\xi)$ that is piecewise meromorphic and satisfies the following conditions: (i) $\breve{M}^{\mathrm{reg}}(x,t,\xi)$ is analytic for $\xi\in\mathbb{C}\setminus\breve{\Upsilon}$; (ii) the jump conditions:
$$
\begin{equation}
\breve{M}^{\mathrm{reg}}_+(x,t,\xi)=\breve{M}^{\mathrm{reg}}_-(x,t,\xi)\breve{J}(x,t,\xi),
\end{equation}
\tag{3.62}
$$
where
$$
\begin{equation}
\breve{J}(x,t,\xi)= \begin{cases} \widehat{M}^{\mathrm{reg}}_{\lambda_1}(x,t,\xi) \widehat{J}^{\mathrm{reg}}(x,t,\xi)(\widehat{M}^{\mathrm{reg}}_{\lambda_1})^{-1}(x,t,\xi), &\xi\in \Upsilon\cap D_{\varepsilon}(\lambda_1), \\ \widehat{M}^{\mathrm{reg}}_{\lambda_2}(x,t,\xi) \widehat{J}^{\mathrm{reg}}(x,t,\xi)(\widehat{M}^{\mathrm{reg}}_{\lambda_2})^{-1}(x,t,\xi), &\xi\in \Upsilon\cap D_{\varepsilon}(\lambda_2), \\ \widehat{M}^{\mathrm{reg}}_{\lambda_3}(x,t,\xi) \widehat{J}^{\mathrm{reg}}(x,t,\xi)(\widehat{M}^{\mathrm{reg}}_{\lambda_3})^{-1}(x,t,\xi), &\xi\in \Upsilon\cap D_{\varepsilon}(\lambda_3), \\ (\widehat{M}^{\mathrm{reg}}_{\lambda_1})^{-1}(x,t,\xi), &\xi\in \partial D_{\varepsilon}(\lambda_1), \\ (\widehat{M}^{\mathrm{reg}}_{\lambda_2})^{-1}(x,t,\xi), &\xi\in \partial D_{\varepsilon}(\lambda_2), \\ (\widehat{M}^{\mathrm{reg}}_{\lambda_3})^{-1}(x,t,\xi), &\xi\in \partial D_{\varepsilon}(\lambda_3), \\ \widehat{J}^{\mathrm{reg}}(x,t,\xi), &\xi\in \Upsilon\setminus \Upsilon_{\varepsilon}; \end{cases}
\end{equation}
\tag{3.63}
$$
(iii) the normalization condition at $\xi=\infty$: The next step is to obtain the large $t$ estimate of $\breve{M}^{\mathrm{reg}}(x,t,\xi)$. We now set $w(x,t,\xi)=\breve{J}(x,t,\xi)-I$. From the symmetry relations $\widehat{J}^{\mathrm{reg}}(x,t,\xi)=\overline{\widehat{J}^{\mathrm{reg}}(x,t,-\overline{\xi})}$ and $\widehat{M}^{\mathrm{reg}}(x,t,\xi)=\overline{\widehat{M}^{\mathrm{reg}}(x,t,-\overline{\xi})}$, we have The function $w(x,t,\xi)$ is estimated as follows (see [44]) with $1\leqslant n\leqslant \infty$:
$$
\begin{equation}
\begin{gathered} \, \| w(x,t,\xi) \|_{(L^1\cap L^2\cap L^{\infty})(\Upsilon\setminus \Upsilon_{\varepsilon})}=O(e^{-ct}), \\ \| w(x,t,\xi) \|_{L^{n}(\Upsilon_{\varepsilon})} =O\bigl(t^{-\frac{1}{2}-\frac{1}{2n}+\max\{|{\operatorname{Im} v(\lambda_1)}|,|{\operatorname{Im} v(\lambda_2)}|,|{\operatorname{Im} v(\lambda_3)}|\}}\ln t). \end{gathered}
\end{equation}
\tag{3.66}
$$
We now consider equations (3.43), (3.46), and (3.49). We assume that the $2\times 1$ column vectors $\Lambda^{(i)}_j(\mu,t)$, $j=1,2,3$, represent the $i$th column of the $2\times2$ matrix-valued functions $\Lambda_j(\mu,t)$. We have the following estimates for each column:
$$
\begin{equation}
\begin{alignedat}{2} \Lambda_1(\mu,t) &=\bigl(\Lambda^{(1)}_1(\mu,t),\Lambda^{(2)}_1(\mu,t)\bigr)=\bigl(O(t^{\frac{1}{2}\operatorname{Im} v(\lambda_1)}),O(t^{-\frac{1}{2}\operatorname{Im} v(\lambda_1)})\bigr), &\qquad t &\to\infty, \\ \Lambda_2(\mu,t) &=\bigl(\Lambda^{(1)}_2(\mu,t),\Lambda^{(2)}_2(\mu,t)\bigr)=\bigl(O(t^{-\frac{1}{2}\operatorname{Im} v(\lambda_2)}),O(t^{\frac{1}{2}\operatorname{Im} v(\lambda_2)})\bigr), &\qquad t&\to\infty, \\ \Lambda_3(\mu,t) &=\bigl(\Lambda^{(1)}_3(\mu,t),\Lambda^{(2)}_3(\mu,t)\bigr)=\bigl(O(t^{\frac{1}{2}\operatorname{Im} v(\lambda_3)}),O(t^{-\frac{1}{2}\operatorname{Im} v(\lambda_3)})\bigr), &\qquad t&\to\infty. \end{alignedat}
\end{equation}
\tag{3.67}
$$
The vectors $\Lambda^{(i)}_j(\mu,t)$ in the $i$th column ($i=1,2$) are evaluated in the $i$th column on the right-hand side of the above equations. We will use this notation to define the estimates for the first and second column vectors of a matrix below. The condition $|{\operatorname{Im} v(\xi)}|< 1/2$ needs to be satisfied, and $\lambda_j$ are as in (3.2). For $\xi\in\partial D_{\varepsilon}(\lambda_1)$, we have
$$
\begin{equation}
\begin{aligned} \, w&=(\widehat{M}^{\mathrm{reg}}_{\lambda_1})^{-1}(x,t,\xi)-I =\Lambda_1(\mu,t)\bigl((\widehat{m}^{\Upsilon}_{\lambda_1})^{-1}(\lambda_1,\tau(\xi))-I\bigr) \Lambda^{-1}_1(\mu,t) \nonumber \\ &=\Lambda_1(\mu,t)\biggl(-\frac{i}{\tau} \begin{pmatrix} 0 & \beta^{\mathrm{reg}}_1(\lambda_1) \\ -\gamma^{\mathrm{reg}}_1(\lambda_1) & 0 \end{pmatrix}+O(\tau^{-2})\biggr) \Lambda^{-1}_1(\mu,t) \nonumber \\ &=\frac{\Xi_1(\mu,t)}{\sqrt{t}(\xi-\lambda_1)}+\widehat{R}^1_1(\lambda_1,t), \end{aligned}
\end{equation}
\tag{3.68}
$$
with
$$
\begin{equation}
\Xi_1=-\frac{i}{2\sqrt{48\gamma\lambda^2_1-1}} \begin{pmatrix} 0 & \beta^{\mathrm{reg}}_1(\lambda_1)e^{2[\chi_1+\phi_1]} F_1^{iv(\lambda_1)} \\ -\gamma^{\mathrm{reg}}_1(\lambda_1)e^{-2[\chi_1+\phi_1]} F_1^{-iv(\lambda_1)} & 0 \end{pmatrix},
\end{equation}
\tag{3.69}
$$
and
$$
\begin{equation}
\begin{gathered} \, \widehat{R}^1_1(\lambda_1,t)=\bigl(\widehat{R}^{1(1)}_1(\lambda_1,t),\widehat{R}^{1(2)}_1(\lambda_1,t)\bigr) =\bigl(O(t^{-1-\operatorname{Im} v(\lambda_1)}),O(t^{-1+\operatorname{Im} v(\lambda_1)})\bigr), \\ F_1=\frac{1-48\gamma\lambda^2_2}{4t(48\gamma\lambda^2_1-1)(48\gamma\lambda^2_3-1)}, \nonumber \end{gathered}
\end{equation}
\tag{3.70}
$$
where $\widehat{R}^{1(i)}_1(\lambda_1,t)$, $i=1,2$, is the $i$th column vector of the matrix function $\widehat{R}^1_1(\lambda_1,t)$. Using the same method, it can be shown that, for $\xi\in\partial D_{\varepsilon}(\lambda_2)$,
$$
\begin{equation}
w=\frac{\Xi_2(\mu,t)}{\sqrt{t}(\xi-\lambda_2)}+\widehat{R}^2_1(\lambda_2,t),
\end{equation}
\tag{3.71}
$$
with
$$
\begin{equation}
\Xi_2=-\frac{i}{2\sqrt{1-48\gamma\lambda^2_2}} \begin{pmatrix} 0 & \beta^{\mathrm{reg}}_2(\lambda_2)e^{2[\chi_2+\phi_2]} \dfrac{F_2^{-iv(\lambda_2)}}{\widetilde{F}_2} \\ -\gamma^{\mathrm{reg}}_2(\lambda_2)e^{-2[\chi_2+\phi_2]} \dfrac{F_2^{iv(\lambda_2)}}{\widetilde{F}_2^{-1}} & 0 \end{pmatrix},
\end{equation}
\tag{3.72}
$$
and
$$
\begin{equation}
\begin{gathered} \, \widehat{R}^2_1(\lambda_2,t)=\bigl(\widehat{R}^{2(1)}_1(\lambda_2,t),\widehat{R}^{2(2)}_1(\lambda_2,t)\bigr) =\bigl(O(t^{-1+\operatorname{Im} v(\lambda_2)}),O(t^{-1-\operatorname{Im} v(\lambda_2)})\bigr), \\ F_2=\frac{1}{4t(1-48\gamma\lambda^2_2)},\qquad \widetilde{F}_2=\bigl(4t(48\gamma\lambda^2_1-1)\bigr)^{iv(\lambda_2)} \biggl(\frac{1}{4t(48\gamma\lambda^2_3-1)}\biggr)^{-iv(\lambda_3)}, \nonumber \end{gathered}
\end{equation}
\tag{3.73}
$$
where $\widehat{R}^{2(i)}_1(\lambda_2,t)$, $i=1,2$, is the $i$th column vector of the matrix function $\widehat{R}^2_1(\lambda_2,t)$. For $\xi\in\partial D_{\varepsilon}(\lambda_3)$, we have
$$
\begin{equation}
w=\frac{\Xi_3(\mu,t)}{\sqrt{t}(\xi-\lambda_3)}+\widehat{R}^3_1(\lambda_3,t),
\end{equation}
\tag{3.74}
$$
with
$$
\begin{equation}
\Xi_3=-\frac{i}{2\sqrt{48\gamma\lambda^2_3-1}} \begin{pmatrix} 0 & \beta^{\mathrm{reg}}_3(\lambda_3)e^{2[\chi_3+\phi_3]} F_3^{iv(\lambda_3)} \\ -\gamma^{\mathrm{reg}}_3(\lambda_3)e^{-2[\chi_3+\phi_3]} F_3^{-iv(\lambda_3)} & 0 \end{pmatrix},
\end{equation}
\tag{3.75}
$$
and
$$
\begin{equation}
\begin{gathered} \, \widehat{R}^3_1(\lambda_3,t)=\bigl(\widehat{R}^{3(1)}_1(\lambda_3,t),\widehat{R}^{3(2)}_1(\lambda_3,t)\bigr) =\bigl(O(t^{-1-\operatorname{Im} v(\lambda_3)}),O(t^{-1+\operatorname{Im} v(\lambda_3)})\bigr), \\ F_3=\frac{1-48\gamma\lambda^2_2}{4t(48\gamma\lambda^2_3-1)(48\gamma\lambda^2_1-1)}, \nonumber \end{gathered}
\end{equation}
\tag{3.76}
$$
where $\widehat{R}^{3(i)}_1(\lambda_3,t)$, $i=1,2$, is the $i$th column vector of the matrix function $\widehat{R}^3_1(\lambda_3,t)$. Furthermore, we have the following estimates for $w(x,t,\xi)$ as $t\to\infty$:
$$
\begin{equation}
\begin{aligned} \, \| w(x,t,\xi) \|_{(L^1\cap L^2)(\breve{\Upsilon})} &=O\bigl(t^{-\frac{1}{2}+\max\{|{\operatorname{Im} v(\lambda_1)}|,|{\operatorname{Im} v(\lambda_2)}|,|{\operatorname{Im} v(\lambda_3)}|\}}\bigr), \\ \| w(x,t,\xi) \|_{L^{\infty}(\breve{\Upsilon})} &=O\bigl(t^{-\frac{1}{2}+\max\{|{\operatorname{Im} v(\lambda_1)}|,|{\operatorname{Im} v(\lambda_2)}|,|{\operatorname{Im} v(\lambda_3)}|\}}\ln t\bigr), \\ \| w^{(l)}(x,t,\xi) \|_{(L^1\cap L^2)(\breve{\Upsilon})} &=O\bigl(t^{-\frac{1}{2}+(-1)^{l+j}\max\{\operatorname{Im} v(\lambda_1),\operatorname{Im} v(\lambda_2),\operatorname{Im} v(\lambda_3)\}}\bigr), \\ \| w^{(l)}(x,t,\xi) \|_{L^{\infty}(\breve{\Upsilon})} &=O\bigl(t^{-\frac{1}{2}+(-1)^{l+j}\max\{\operatorname{Im} v(\lambda_1),\operatorname{Im} v(\lambda_2),\operatorname{Im} v(\lambda_3)\}}\ln t\bigr), \end{aligned}
\end{equation}
\tag{3.77}
$$
where $\xi\in\breve{\Upsilon}$ and $w^{(l)}$ is the $l$th column of $w$. Consider the Cauchy operator $C_w\colon L^2(\breve{\Upsilon})+L^{\infty}(\breve{\Upsilon})\to L^2(\breve{\Upsilon})$, where $C_w(f)=C_-(fw)$ and $C_-$ acts from $L^2(\breve{\Upsilon})$ to $L^2(\breve{\Upsilon})$. Since $(C_- f)(\xi)$, $\xi\in \breve{\Upsilon}$, is the negative (according to the orientation of $\breve{\Upsilon}$) non-tangential boundary value of
$$
\begin{equation*}
(Cf)(\xi')=\frac{1}{2\pi i}\int_{\breve{\Upsilon}}\frac{f(s)}{s-\xi'} \, ds,\qquad \xi'\in\mathbb{C}\setminus\breve{\Upsilon},
\end{equation*}
\notag
$$
we find that
$$
\begin{equation}
\| C_w \|\leqslant T\| w \|_{L^{\infty}(\breve{\Upsilon})} =O\bigl(t^{-\frac{1}{2}+\max\{|{\operatorname{Im} v(\lambda_1)}|,|{\operatorname{Im} v(\lambda_2)}|,|{\operatorname{Im} v(\lambda_3)}|\}}\ln t\bigr),\qquad t\to\infty,
\end{equation}
\tag{3.78}
$$
where $T$ is a constant. It can be shown that $\| C_w \|$ degenerates to zero as $t\to \infty$, which means that $I-C_w$ is reversible for the large $t$. We now introduce $\breve{v}(x,t,\xi)-I\in L^2(\breve{\Upsilon})$. The function $\breve{v}(x,t,\xi)$ is the solution of the Fredholm-type equation We also have
$$
\begin{equation}
\|\breve{v}(x,t,\xi)-I\|_{L^2(\breve{\Upsilon})}\leqslant T\| w(x,t,\xi) \|_{(L^1\cap L^2)(\breve{\Upsilon})}.
\end{equation}
\tag{3.80}
$$
Hence
$$
\begin{equation}
\|\breve{v}(x,t,\xi)-I\|_{L^2(\breve{\Upsilon})}=O\bigl(t^{-\frac{1}{2}+\max\{|{\operatorname{Im} v(\lambda_1)}|,|{\operatorname{Im} v(\lambda_2)}|,|{\operatorname{Im} v(\lambda_3)}|\}}\bigr),\qquad t\to\infty.
\end{equation}
\tag{3.81}
$$
According to the Beals–Coifman theory [49], the RH problem $\breve{M}^{\mathrm{reg}}(x,t,\xi)$ is solvable if and only if so is the singular integral equation (3.79). From the uniqueness of the solution to the RH problem, it can be inferred that the solution to RH problem 3.8 is independent of the decomposition of the jump matrix $\breve{J}(x; t; \xi)$. After considering the trivial decomposition of $\breve{J}(x; t; \xi)$, the solution to the RH problem 3.8 can be further expressed as
$$
\begin{equation}
\breve{M}^{\mathrm{reg}}(x,t,\xi)=I+C(\breve{v}w)=I+\frac{1}{2\pi i}\int_{\breve{\Upsilon}} \breve{v}(x,t,s)w(x,t,s) \, \frac{ds}{s-\xi}.
\end{equation}
\tag{3.82}
$$
We have
$$
\begin{equation}
\lim_{\xi\to\infty}\xi(\breve{M}^{\mathrm{reg}}(x,t,\xi)-I)=-\frac{1}{2\pi i}\int_{\breve{\Upsilon}}\breve{v}(x,t,s)w(x,t,s) \, ds.
\end{equation}
\tag{3.83}
$$
Taking into account (3.68)–(3.75) and (3.81), we have, for $j=1,2,3$,
$$
\begin{equation}
\begin{aligned} \, &\oint_{|s-\lambda_j|=\varepsilon}\breve{v}(x,t,s)w(x,t,s) \, ds \nonumber \\ &\qquad=\oint_{|s-\lambda_j|=\varepsilon}w(x,t,s) \, ds +\oint_{|s-\lambda_j|=\varepsilon}(\breve{v}(x,t,s)-I)w(x,t,s) \, ds \nonumber \\ &\qquad=\frac{\Xi_j(\mu,t)}{\sqrt{t}}\oint_{|s-\lambda_j|=\varepsilon}\frac{1}{s-\lambda_j} \, ds +\widehat{R}^j_1(\lambda_j,t)+\widehat{R}^j_2(\lambda_j,t) \nonumber \\ &\qquad=-2\pi i\widetilde{\Xi}_j(\mu,t)+\widehat{R}^j_1(\lambda_j,t)+\widehat{R}^j_2(\lambda_j,t), \end{aligned}
\end{equation}
\tag{3.84}
$$
where $\widetilde{\Xi}_j(\mu,t)=-\Xi_j(\mu,t)/\sqrt{t}$. In addition, $\Xi_j(\mu,t)$, $j=1,2,3$, are given by (3.69), (3.72), and (3.75), $\widehat{R}^j_1(\lambda_j,t)$, $j=1,2,3$, are given by (3.70), (3.73), and (3.76). We also have
$$
\begin{equation}
\widehat{R}^1_2(\lambda_1,t) =\| \breve{v}(x,t,s)-I \|_{L^2(\partial D_{\varepsilon}(\lambda_1))}O(\widetilde{\Xi}_1(\mu,t)) \nonumber
\end{equation}
\notag
$$
$$
\begin{equation}
=\bigl(O\bigl(t^{-1+|{\operatorname{Im} v(\lambda_1)}|-\operatorname{Im} v(\lambda_1)}\bigr), O\bigl(t^{-1+|{\operatorname{Im} v(\lambda_1)}|+\operatorname{Im} v(\lambda_1)}\bigr)\bigr),
\end{equation}
\tag{3.85}
$$
$$
\begin{equation}
\widehat{R}^2_2(\lambda_2,t) =\| \breve{v}(x,t,s)-I \|_{L^2(\partial D_{\varepsilon}(\lambda_2))}O(\widetilde{\Xi}_2(\mu,t)) \nonumber
\end{equation}
\notag
$$
$$
\begin{equation}
=\bigl(O\bigl(t^{-1+|{\operatorname{Im} v(\lambda_2)}|+\operatorname{Im} v(\lambda_2)}\bigr), O\bigl(t^{-1+|{\operatorname{Im} v(\lambda_2)}|-\operatorname{Im} v(\lambda_2)}\bigr)\bigr),
\end{equation}
\tag{3.86}
$$
$$
\begin{equation}
\begin{aligned} \, \widehat{R}^3_2(\lambda_3,t)&=\| \breve{v}(x,t,s)-I \|_{L^2(\partial D_{\varepsilon}(\lambda_3))}O(\widetilde{\Xi}_3(\mu,t)) \nonumber \\ &=\bigl(O\bigl(t^{-1+|{\operatorname{Im} v(\lambda_3)}|-\operatorname{Im} v(\lambda_3)}\bigr), O\bigl(t^{-1+|{\operatorname{Im} v(\lambda_3)}|+\operatorname{Im} v(\lambda_3)}\bigr)\bigr). \end{aligned}
\end{equation}
\tag{3.87}
$$
In equations (3.85)–(3.87), the vectors $\widehat{R}^j_2(\lambda_j,t)$, $j=1,2,3$, for the $i$th $(i=1,2)$ column on the left-hand side are evaluated in the $i$th (i=1,2) column vector of the matrix function after the second equality sign on the right-hand side. Setting $R(\mu,t)=\sum_{j=1}^3\widehat{R}_j(\mu,t)$ with $\widehat{R}_1(\mu,t)=\sum_{q=1}^3\widehat{R}^q_1(\lambda_q,t)$, $\widehat{R}_2(\mu,t)=\sum_{p=1}^3\widehat{R}^p_2(\lambda_p,t)$, and $\widehat{R}_3(\mu,t)=\widehat{R}_1(\mu,t)+\widehat{R}_2(\mu,t)$, we have
$$
\begin{equation}
\widehat{R}_3(\mu,t)=\bigl(\widehat{R}^{(1)}_3(\mu,t),\widehat{R}^{(2)}_3(\mu,t)\bigr) =\bigl(O\bigl(t^{-1+m^{(1)}-m^{(2)}}\bigr),O\bigl(t^{-1+m^{(1)}+m^{(2)}}\bigr)\bigr),
\end{equation}
\tag{3.88}
$$
with
$$
\begin{equation*}
\begin{aligned} \, m^{(1)} &=\max\{|{\operatorname{Im} v(\lambda_1)}|,|{\operatorname{Im} v(\lambda_2)}|,|{\operatorname{Im} v(\lambda_3)}|\}, \\ m^{(2)} &=\max\{\operatorname{Im} v(\lambda_1),\operatorname{Im} v(\lambda_2),\operatorname{Im} v(\lambda_3)\}, \end{aligned}
\end{equation*}
\notag
$$
where $\widehat{R}^{(i)}_3(\mu,t)$ is the $i$th column vector of the matrix function $\widehat{R}_3(\mu,t)$. Let $R(\mu,t)=\left( \begin{smallmatrix} R_1(\mu,t) & R_2(\mu,t) \\ R_1(\mu,t) & R_2(\mu,t) \end{smallmatrix} \right)$, where $R_1(\mu,t)$ and $R_2(\mu,t)$ are estimated in (1.14) and (1.15). In view of (3.84), we can write (3.82) as
$$
\begin{equation}
\begin{aligned} \, &\breve{M}^{\mathrm{reg}} =I-\frac{1}{2\pi i}\oint_{|s-\lambda_1| =\varepsilon}\frac{\widetilde{\Xi}_1(\mu,t)}{(s-\lambda_1)(s-\xi)} \, ds- \frac{1}{2\pi i}\oint_{|s-\lambda_2|=\varepsilon}\frac{\widetilde{\Xi}_2(\mu,t)}{(s-\lambda_2)(s-\xi)} \, ds \nonumber \\ &\qquad-\frac{1}{2\pi i} \oint_{|s-\lambda_3|=\varepsilon}\frac{\widetilde{\Xi}_3(\mu,t)}{(s-\lambda_3)(s-\xi)} \, ds +R(\mu,t),\qquad |\xi-\lambda_j|>\varepsilon,\quad j=1,2,3. \end{aligned}
\end{equation}
\tag{3.89}
$$
Using (3.61), we have $\widehat{M}^{\mathrm{reg}}(x,t,\xi)=\breve{M}^{\mathrm{reg}}(x,t,\xi)$. Hence
$$
\begin{equation}
\lim_{\xi\to\infty}\xi(\widehat{M}^{\mathrm{reg}}(x,t,\xi)-I) =\widetilde{\Xi}_1(\mu,t)+\widetilde{\Xi}_2(\mu,t)+\widetilde{\Xi}_3(\mu,t)+R(\mu,t),
\end{equation}
\tag{3.90}
$$
and
$$
\begin{equation}
\widehat{M}^{\mathrm{reg}}(x,t,0) =I-\frac{\widetilde{\Xi}_1(\mu,t)}{\lambda_1}- \frac{\widetilde{\Xi}_2(\mu,t)}{\lambda_2}- \frac{\widetilde{\Xi}_3(\mu,t)}{\lambda_3}+R(\mu,t),
\end{equation}
\tag{3.91a}
$$
$$
\begin{equation}
\widehat{M}^{\mathrm{reg}}(x,t,i\xi_1) =I-\frac{\widetilde{\Xi}_1(\mu,t)}{\lambda_1-i\xi_1}- \frac{\widetilde{\Xi}_2(\mu,t)}{\lambda_2-i\xi_1}- \frac{\widetilde{\Xi}_3(\mu,t)}{\lambda_3-i\xi_1}+R(\mu,t).
\end{equation}
\tag{3.91b}
$$
Next, we estimate the elements $P_{12}(x,t)$ and $P_{21}(x,t)$ of the matrix-valued factor $P(x,t)$ defined in (3.28). By the definitions of $u(x,t)$ and $v(x,t)$ in (3.29), we have
$$
\begin{equation}
\begin{aligned} \, u_1(x,t) &=i\xi_1+R_1(\mu,t), \\ u_2(x,t) &=-i\xi_1\biggl(\frac{(\widetilde{\Xi}_1)_{21}(\mu,t)}{\lambda_1-i\xi_1} +\frac{(\widetilde{\Xi}_2)_{21}(\mu,t)}{\lambda_2-i\xi_1}+ \frac{(\widetilde{\Xi}_3)_{21}(\mu,t)}{\lambda_3-i\xi_1}\biggr)+R_1(\mu,t), \end{aligned}
\end{equation}
\tag{3.92}
$$
$$
\begin{equation}
\begin{aligned} \, v_1(x,t) &=c_0(\mu)-i\xi_1\biggl(\frac{(\widetilde{\Xi}_1)_{12}(\mu,t)}{\lambda_1} +\frac{(\widetilde{\Xi}_2)_{12}(\mu,t)}{\lambda_2}+ \frac{(\widetilde{\Xi}_3)_{12}(\mu,t)}{\lambda_3}\biggr) \\ &\qquad+R_3(\mu,t), \\ v_2(x,t) &=i\xi_1-c_0(\mu)\biggl(\frac{(\widetilde{\Xi}_1)_{21}(\mu,t)}{\lambda_1} +\frac{(\widetilde{\Xi}_2)_{21}(\mu,t)}{\lambda_2}+ \frac{(\widetilde{\Xi}_3)_{21}(\mu,t)}{\lambda_3}\biggr) \\ &\qquad+R_3(\mu,t), \end{aligned}
\end{equation}
\tag{3.93}
$$
with $R_3(\mu,t)=R_1(\mu,t)+R_2(\mu,t)$. We next have
$$
\begin{equation}
\begin{aligned} \, u_1v_1&=i\xi_1c_0(\mu)+\xi^2_1\biggl(\frac{(\widetilde{\Xi}_1)_{12}(\mu,t)}{\lambda_1} +\frac{(\widetilde{\Xi}_2)_{12}(\mu,t)}{\lambda_2}+ \frac{(\widetilde{\Xi}_3)_{12}(\mu,t)}{\lambda_3}\biggr) \\ &\qquad+R_3(\mu,t), \\ u_1v_2&=-\xi^2_1-i\xi_1c_0(\mu)\biggl(\frac{(\widetilde{\Xi}_1)_{21}(\mu,t)}{\lambda_1} +\frac{(\widetilde{\Xi}_2)_{21}(\mu,t)}{\lambda_2}+ \frac{(\widetilde{\Xi}_3)_{21}(\mu,t)}{\lambda_3}\biggr) \\ &\qquad+R_3(\mu,t), \\ u_2v_1&=-i\xi_1c_0(\mu)\biggl(\frac{(\widetilde{\Xi}_1)_{21}(\mu,t)}{\lambda_1-i\xi_1} +\frac{(\widetilde{\Xi}_2)_{21}(\mu,t)}{\lambda_2-i\xi_1}+ \frac{(\widetilde{\Xi}_3)_{21}(\mu,t)}{\lambda_3-i\xi_1}\biggr) \\ &\qquad+R_1(\mu,t), \\ u_2v_2&=\xi^2_1\biggl(\frac{(\widetilde{\Xi}_1)_{21}(\mu,t)}{\lambda_1-i\xi_1} +\frac{(\widetilde{\Xi}_2)_{21}(\mu,t)}{\lambda_2-i\xi_1}+ \frac{(\widetilde{\Xi}_3)_{21}(\mu,t)}{\lambda_3-i\xi_1}\biggr)+R_1(\mu,t). \end{aligned}
\end{equation}
\tag{3.94}
$$
Plugging this into (3.28), we have, after direct calculations,
$$
\begin{equation}
\begin{aligned} \, P_{12}(x,t) &=-\frac{ic_0}{\xi_1}-\biggl(\frac{(\widetilde{\Xi}_1)_{12}(\mu,t)}{\lambda_1} +\frac{(\widetilde{\Xi}_2)_{12}(\mu,t)}{\lambda_2}+ \frac{(\widetilde{\Xi}_3)_{12}(\mu,t)}{\lambda_3}\biggr) \nonumber \\ &\qquad+\frac{ic^2_0}{\xi_1}\biggl(\frac{(\widetilde{\Xi}_1)_{21}(\mu,t)}{\lambda_1(\lambda_1-i\xi_1)} +\frac{(\widetilde{\Xi}_2)_{21}(\mu,t)}{\lambda_2(\lambda_2-i\xi_1)}+ \frac{(\widetilde{\Xi}_3)_{21}(\mu,t)}{\lambda_3(\lambda_3-i\xi_1)}\biggr)+R_3(\mu,t), \end{aligned}
\end{equation}
\tag{3.95}
$$
$$
\begin{equation}
\begin{aligned} \, P_{21}(x,t) &=\biggl(\frac{(\widetilde{\Xi}_1)_{21}(\mu,t)}{\lambda_1-i\xi_1} +\frac{(\widetilde{\Xi}_2)_{21}(\mu,t)}{\lambda_2-i\xi_1}+ \frac{(\widetilde{\Xi}_3)_{21}(\mu,t)}{\lambda_3-i\xi_1}\biggr)+R_1(\mu,t). \end{aligned}
\end{equation}
\tag{3.96}
$$
The expressions of $P_{12}(x,t)$ and $P_{21}(x,t)$ contain explicitly the parameter $\xi_1$. We next set
$$
\begin{equation}
(\widetilde{\Xi}_j)_{12}=\frac{\lambda_j}{\lambda_j-i\xi_1}(\Xi^{\mathrm{reg}}_j)_{12},\qquad (\widetilde{\Xi}_j)_{21}=\frac{\lambda_j-i\xi_1}{\lambda_j}(\Xi^{\mathrm{reg}}_j)_{21},
\end{equation}
\tag{3.97}
$$
where $r^{\mathrm{reg}}_l(\lambda_j)$ is substituted for $r_l(\lambda_j)$, $l=1,2$, $j=1,2,3$, in $\widetilde{\Xi}_j$. Hence $\Xi^{\mathrm{reg}}_j$. Consequently, we find that the terms with explicit expressions involving $\xi_1$ have decayed in the main asymptotic terms. Now by substituting (3.95)–(3.97) and (3.90) into (3.32), we find the long-time asymptotic behavior of the solutions of the LPD equation for $x>0$ and $x<0$. The main results are collected in Theorem 1.1 above.
§ 4. Long-time asymptotics for $\mu^2:=(x/t)^2>1/(27\gamma)$In this section, we find the long-time asymptotics of the solution in the case $\mu^2>1/(27\gamma)$. Three different roots are shown in (3.2), where $\lambda_1$, $\lambda_2$ is a pair of conjugate complex numbers, and $\lambda_3$ is a real number. We next analyze the asymptotic behavior of the solution under the condition $\mu\in D_3=(\sqrt{1/(27\gamma)}+\epsilon,+\infty)$, which corresponds to the case (e) in Fig. 3. The signature table illustrating the distribution of $\operatorname{Re}\varphi(\xi)$ in the complex $\xi$-plane is shown in Fig. 9. Remark 4.1. For $\mu\in D_3=(\sqrt{1/(27\gamma)}+\epsilon,+\infty)$, the distribution of the phase points corresponds to case (f) in Fig. 3. The asymptotic behavior of the solution obtained in this case is same as in case (e) in Fig. 3, except that $\lambda_3$ is now a positive real number, and the real parts of $\lambda_1$ and $\lambda_2$ are negative. 4.1. The deformations of RH problemLet $\varpi(\xi,\lambda_3)$ be the solution to the following scalar RH problem: (i) $\varpi(\xi,\lambda_3)$ is holomorphic for $\xi\in\mathbb{C}\setminus (-\infty,\lambda_3]$, (ii) $\varpi_+(\xi,\lambda_3)=\varpi_-(\xi,\lambda_3)(1+r_1(\xi)r_2(\xi))$, $\xi\in(-\infty,\lambda_3)$, (iii) $\varpi(\xi,\lambda_3)\to 1$, $\xi\to\infty$. By the Plemelj formula, its solution can be written as the Cauchy-type integral
$$
\begin{equation}
\varpi(\xi,\lambda_3)=\exp\biggl\{\frac{1}{2\pi i}\int_{-\infty}^{\lambda_3}\frac{\ln(1+r_1(s)r_2(s))}{s-\xi} \, ds \biggr\}.
\end{equation}
\tag{4.1}
$$
An integration by parts gives
$$
\begin{equation}
\varpi(\xi,\lambda_3)=(\xi-\lambda_3)^{iv(\lambda_3)}e^{\pi(\lambda_3,\xi)},
\end{equation}
\tag{4.2}
$$
where
$$
\begin{equation}
\pi(\lambda_3,\xi)=-\frac{1}{2\pi i} \int_{-\infty}^{\lambda_3}\ln(\xi-\zeta) \, d_{\zeta}\ln(1+r_1(\zeta)r_2(\zeta))
\end{equation}
\tag{4.3}
$$
is uniformly bounded, and $v(\lambda_3)$ can be expressed as
$$
\begin{equation}
\begin{gathered} \, v(\lambda_3) =-\frac{1}{2\pi}\ln|1+r_1(\lambda_3)r_2(\lambda_3)|-\frac{i}{2\pi}\Delta(\lambda_3), \\ \Delta(\lambda_3)=\int_{-\infty}^{\lambda_3}d \arg(1+r_1(\zeta)r_2(\zeta)). \nonumber \end{gathered}
\end{equation}
\tag{4.4}
$$
Assuming that $\Delta(\xi)\in(-\pi,\pi)$ for $\xi\in\mathbb{R}$, we have $|{\operatorname{Im} v(\xi)}|< 1/2$. Under this assumption, $\ln(1+r_1(\xi)r_2(\xi))$ is single-valued, and the singularities $\xi=\lambda_3$ of $\varpi(\xi,\lambda_3)$ are square integrable. We can now redefine the function $M(x,t,\xi)$ in terms of $\varpi(\xi,\lambda_3)$:
$$
\begin{equation}
\widetilde{N}(x,t,\xi)= \begin{cases} M(x,t,\xi)\varpi^{-\sigma_3}(\xi,\lambda_3), &\operatorname{Re}\xi\leqslant0, \\ M(x,t,\xi), &\operatorname{Re}\xi>0. \end{cases}
\end{equation}
\tag{4.5}
$$
We now present a matrix $\widetilde{N}(x,t,\xi)$ satisfying the following RH problem. RH problem 4.2. Find a matrix-valued function $\widetilde{N}(x,t,\xi)$ that is piecewise meromorphic and satisfies the following conditions: (i) $\widetilde{N}(x,t,\xi)$ is meromorphic for $\xi\in\mathbb{C}\setminus\mathbb{R}$ and has a simple pole located at $\xi=i\xi_1$, $\xi_1>0$; (ii) the jump conditions: the non-tangential limits $\widetilde{N}(x,t,\xi)=\widetilde{N}(x,t,\xi\pm i0)$ exist a.e. for $\xi\in\mathbb{R}$ such that $\widetilde{N}(x,t,{\cdot}\,)-I\in L^2(\mathbb{R}\setminus [-\varepsilon,\varepsilon])$ satisfies
$$
\begin{equation}
\widetilde{N}_+(x,t,\xi)=\widetilde{N}_-(x,t,\xi)\widetilde{J}_N(x,t,\xi),\qquad \xi\in\mathbb{R}\setminus \{0 \},
\end{equation}
\tag{4.6}
$$
for any $\varepsilon>0$ and $\widetilde{N}_{\pm}(x,t,\xi)$, where
$$
\begin{equation}
\widetilde{J}_N= \begin{cases} \begin{pmatrix} 1 & 0 \\ -\dfrac{r_1(\xi)\varpi_-^{-2}(\xi,\lambda_3)}{1+r_1(\xi)r_2(\xi)}e^{2it\theta} & 1 \end{pmatrix} \\ \quad\times \begin{pmatrix} 1 & -\dfrac{r_2(\xi)\varpi_+^2(\xi,\lambda_3)}{1+r_1(\xi)r_2(\xi)}e^{-2it\theta} \\ 0 & 1 \end{pmatrix}, &-\infty<\xi<\lambda_3, \\ \begin{pmatrix} 1 & -r_2(\xi)\varpi^2(\xi,\lambda_3)e^{-2it\theta} \\ 0 & 1 \end{pmatrix} \\ \quad\times \begin{pmatrix} 1 & 0 \\ -r_1(\xi)\varpi^{-2}(\xi,\lambda_3)e^{2it\theta} & 1 \end{pmatrix}, &\lambda_3<\xi<0, \\ J, &0<\xi<+\infty; \end{cases}
\end{equation}
\tag{4.7}
$$
(iii) the normalization condition at $\xi=\infty$:
$$
\begin{equation}
\widetilde{N}(x,t,\xi)=I+O\biggl(\frac{1}{\xi}\biggr),\qquad \xi\to\infty;
\end{equation}
\tag{4.8}
$$
(iv) the residue condition:
$$
\begin{equation}
\operatorname*{Res}_{\xi=i\xi_1}\bigl[\widetilde{N}(x,t,\xi)\bigr]_1=\frac{r_1}{\dot{a}_1(i\xi_1)\varpi^2(i\xi_1)} \, e^{-2\xi_1x+2i\xi_1^2t+16i\xi_1^4\gamma t}\bigl[\widetilde{N}(x,t,i\xi_1)\bigr]_2;
\end{equation}
\tag{4.9}
$$
(v) the singularity conditions at $\xi=0$: in Case 1,
$$
\begin{equation}
\widetilde{N}_+ = \begin{pmatrix} \dfrac{4 f_1(x,t)}{A^2a_2(0)\varpi(0,\lambda_3)} & -\varpi(0,\lambda_3)\overline{f_2}(-x,t) \\ \dfrac{4 f_2(x,t)}{A^2a_2(0)\varpi(0,\lambda_3)} & -\varpi(0,\lambda_3)\overline{f_1}(-x,t) \end{pmatrix} (I\,{+}\,O(\xi)) \begin{pmatrix} \xi & 0 \\ 0 & \dfrac{1}{\xi} \end{pmatrix}, \!\!\qquad \xi \to {+}i0,
\end{equation}
\tag{4.10a}
$$
$$
\begin{equation}
\widetilde{N}_- =\frac{2i}{A} \begin{pmatrix} -\dfrac{\overline{f_2}(-x,t)}{\varpi(0,\lambda_3)} & \dfrac{\varpi(0,\lambda_3)f_1(x,t)}{a_2(0)} \\ -\dfrac{\overline{f_1}(-x,t)}{\varpi(0,\lambda_3)} & \dfrac{\varpi(0,\lambda_3)f_2(x,t)}{a_2(0)} \end{pmatrix}+O(\xi), \!\!\qquad \xi \to {-}i0;
\end{equation}
\tag{4.10b}
$$
in Case 2,
$$
\begin{equation}
\widetilde{N}_+ = \begin{pmatrix} \dfrac{f_1(x,t)}{a_{11}\varpi(0,\lambda_3)} & -\overline{f_2}(-x,t)\varpi(0,\lambda_3) \\ \dfrac{f_2(x,t)}{a_{11}\varpi(0,\lambda_3)} & -\overline{f_1}(-x,t)\varpi(0,\lambda_3) \end{pmatrix}(I+O(\xi)) \begin{pmatrix} 1 & 0 \\ 0 & \dfrac{1}{\xi} \end{pmatrix}, \qquad \xi \to+i0,
\end{equation}
\tag{4.11a}
$$
$$
\begin{equation}
\widetilde{N}_- =\frac{2i}{A} \begin{pmatrix} -\dfrac{\overline{f_2}(-x,t)}{\varpi(0,\lambda_3)} & \dfrac{\varpi(0,\lambda_3)f_1(x,t)}{\dot{a}_2(0)} \\ -\dfrac{\overline{f_1}(-x,t)}{\varpi(0,\lambda_3)} & \dfrac{\varpi(0,\lambda_3)f_2(x,t)}{\dot{a}_2(0)} \end{pmatrix}(I+O(\xi)) \begin{pmatrix} 1 & 0 \\ 0 & \dfrac{1}{\xi} \end{pmatrix}, \qquad \xi \to-i0.
\end{equation}
\tag{4.11b}
$$
In order to make the jump matrices decay to the identity matrix as $t\to\infty$, we use the standard procedure (see [19]) to make factorizations of the jump matrices along the real axis to deform the contours onto those on which the oscillatory jump on the left half of the real $\xi$-axis is traded for exponential decay, then perform translation transformations to move them to the areas parallelling to the real axis in the right half of the real $\xi$-axis. As shown in Fig. 10, the contours
$$
\begin{equation*}
\Gamma_3=\biggl\{\xi\biggm|\zeta_{a}+\frac{\sqrt{3}}{8}\,i\zeta_b\biggr\}
\end{equation*}
\notag
$$
and
$$
\begin{equation*}
\Gamma^*_3=\biggl\{\xi\biggm|\zeta_{a}-\frac{\sqrt{3}}{8}\,i\zeta_{b}\biggr\}
\end{equation*}
\notag
$$
are parallel to the real axis with $\zeta_b=\operatorname{Im} \lambda_2=-\operatorname{Im} \lambda_1$ and $\zeta_a>0$ for $\zeta_a,\zeta_b\in\mathbb{R}$. Next, we define the new function $\widehat{N}(x,t,\xi)$ as follows:
$$
\begin{equation}
\widehat{N}(x,t,\xi)= \begin{cases} \widetilde{N}(x,t,\xi) \begin{pmatrix} 1 & \dfrac{r_2(\xi)\varpi^2(\xi,\lambda_3)}{1+r_1(\xi)r_2(\xi)}e^{-2it\theta} \\ 0 & 1 \end{pmatrix}, &\xi\in G_2, \\ \widetilde{N}(x,t,\xi) \begin{pmatrix} 1 & 0 \\ r_1(\xi)\varpi^{-2}(\xi,\lambda_3)e^{2it\theta} & 1 \end{pmatrix}, &\xi\in G_1, \\ \widetilde{N}(x,t,\xi) \begin{pmatrix} 1 & -r_2(\xi)\varpi^2(\xi,\lambda_3)e^{-2it\theta} \\ 0 & 1 \end{pmatrix}, &\xi\in G^*_1, \\ \widetilde{N}(x,t,\xi) \begin{pmatrix} 1 & 0 \\ -\dfrac{r_1(\xi)\varpi^{-2}(\xi,\lambda_3)}{1+r_1(\xi)r_2(\xi)}e^{2it\theta} & 1 \end{pmatrix}, &\xi\in G^*_2, \\ \widetilde{N}(x,t,\xi) \begin{pmatrix} 1 & 0 \\ r_1(\xi)e^{2it\theta} & 1 \end{pmatrix}, &\xi\in G_3, \\ \widetilde{N}(x,t,\xi) \begin{pmatrix} 1 & -r_2(\xi)e^{-2it\theta} \\ 0 & 1 \end{pmatrix}, &\xi\in G^*_3, \\ \widetilde{N}(x,t,\xi), &\xi\in G_{0,4}\cup G^*_{0,4}. \end{cases}
\end{equation}
\tag{4.12}
$$
The above function $\widehat{N}(x,t,\xi)$ satisfies the following RH problem. RH problem 4.3. Find a matrix-valued function $\widehat{N}(x,t,\xi)$ that is piecewise meromorphic and satisfies the following conditions: (i) $\widehat{N}(x,t,\xi)$ is meromorphic for $\xi\in\mathbb{C}\setminus\Gamma$ and has a simple pole located at $\xi=i\xi_1$, $\xi_1>0$; (ii) the jump conditions: the non-tangential limits $\widehat{N}(x,t,\xi)=\widehat{N}(x,t,\xi\pm i0)$ exist a.e. for $\xi\in\mathbb{R}$ such that $\widehat{N}(x,t,\cdot)-I\in L^2(\mathbb{R}\setminus [-\varepsilon,\varepsilon])$ satisfies
$$
\begin{equation}
\widehat{N}_+(x,t,\xi)=\widehat{N}_-(x,t,\xi)\widehat{J}_N(x,t,\xi),\qquad \xi\in \Gamma,
\end{equation}
\tag{4.13}
$$
for any $\varepsilon>0$ and $\widehat{N}_{\pm}(x,t,\xi)$, where
$$
\begin{equation}
\widehat{J}_N(x,t,\xi)= \begin{cases} \begin{pmatrix} 1 & -\dfrac{r_2(\xi)\varpi^2(\xi,\lambda_3)}{1+r_1(\xi)r_2(\xi)}e^{-2it\theta} \\ 0 & 1 \end{pmatrix}, &\xi\in \Gamma_2, \\ \begin{pmatrix} 1 & 0 \\ -r_1(\xi)\varpi^{-2}(\xi,\lambda_3)e^{2it\theta} & 1 \end{pmatrix}, &\xi\in \Gamma_1, \\ \begin{pmatrix} 1 & r_2(\xi)\varpi^2(\xi,\lambda_3)e^{-2it\theta} \\ 0 & 1 \end{pmatrix}, &\xi\in \Gamma^*_1, \\ \begin{pmatrix} 1 & 0 \\ \dfrac{r_1(\xi)\varpi^{-2}(\xi,\lambda_3)}{1+r_1(\xi)r_2(\xi)}e^{2it\theta} & 1 \end{pmatrix}, &\xi\in \Gamma^*_2, \\ \begin{pmatrix} 1 & 0 \\ -r_1(\xi)e^{2it\theta} & 1 \end{pmatrix}, &\xi\in \Gamma_3, \\ \begin{pmatrix} 1 & -r_2(\xi)e^{-2it\theta}\\ 0 & 1 \end{pmatrix}, &\xi\in \Gamma^*_3; \end{cases}
\end{equation}
\tag{4.14}
$$
(iii) the normalization condition at $\xi=\infty$:
$$
\begin{equation}
\widehat{N}(x,t,\xi)=I+O\biggl(\frac{1}{\xi}\biggr),\qquad \xi\to\infty;
\end{equation}
\tag{4.15}
$$
(iv) the residue condition:
$$
\begin{equation}
\operatorname*{Res}_{\xi=i\xi_1}\bigl[\widehat{N}(x,t,\xi)\bigr]_1=d_1(x,t)\bigl[\widehat{N}(x,t,i\xi_1)\bigr]_2,
\end{equation}
\tag{4.16}
$$
where
$$
\begin{equation*}
d_1(x,t)=\frac{r_1}{\dot{a}_1(i\xi_1)\varpi^2(i\xi_1)}\, e^{-2\xi_1x+2i\xi_1^2t+16i\xi_1^4\gamma t};
\end{equation*}
\notag
$$
(v) the singularity conditions at $\xi=0$, $\widehat{N}(x,t,\xi)$ satisfies, in Cases 1 and 2,
$$
\begin{equation}
\widehat{N}_+ = \begin{pmatrix} -\dfrac{2i \overline{f}_2(-x,t)}{A \varpi(0,\lambda_3)}+O(\xi) & -\dfrac{1}{\xi}\overline{f}_2(-x,t)\varpi(0,\lambda_3)+O(1) \\ -\dfrac{2i \overline{f}_1(-x,t)}{A \varpi(0,\lambda_3)}+O(\xi) & -\dfrac{1}{\xi}\overline{f}_1(-x,t)\varpi(0,\lambda_3)+O(1) \end{pmatrix}, \qquad \xi \to+i0,
\end{equation}
\tag{4.17a}
$$
$$
\begin{equation}
\widehat{N}_- =\frac{2i}{A} \begin{pmatrix} -\dfrac{\overline{f}_2(-x,t)}{\varpi(0,\lambda_3)}+O(\xi) & -\dfrac{A}{2i\xi}\varpi(0,\lambda_3)\overline{f}_2(-x,t)+O(\xi) \\ -\dfrac{\overline{f}_1(-x,t)}{\varpi(0,\lambda_3)}+O(\xi) & -\dfrac{A}{2i\xi}\varpi(0,\lambda_3)\overline{f}_1(-x,t)+O(\xi) \end{pmatrix}, \qquad \xi \to-i0.
\end{equation}
\tag{4.17b}
$$
Similarly, we also have the singularity conditions at $\xi=0$ in both cases, which can be reduced to the same residue condition
$$
\begin{equation}
\operatorname*{Res}_{\xi=0}\bigl[\widehat{N}(x,t,\xi)\bigr]_2=d_0(\mu)\bigl[\widehat{N}(x,t,0)\bigr]_1,
\end{equation}
\tag{4.18}
$$
with the condition $d_0(\mu)=A\varpi^2(0,\lambda_3)/(2i)$. Proceeding now as in § 3.3, we introduce the BP factor to transform $\widehat{N}(x,t,\xi)$ into a regular RH problem with
$$
\begin{equation}
\widehat{N}(x,t,\xi)= \begin{cases} B(x,t,\xi)\widehat{N}^{\mathrm{reg}}(x,t,\xi) \begin{pmatrix} 1 & 0 \\ 0 & \dfrac{\xi-i\xi_1}{\xi} \end{pmatrix}, &\operatorname{Re}\xi\leqslant0, \\ \widehat{N}^{\mathrm{reg}}(x,t,\xi), &\operatorname{Re}\xi>0, \end{cases}
\end{equation}
\tag{4.19}
$$
where $B(x,t,\xi)=I+(i\xi_1/(\xi-i\xi_1))P(x,t)$, and $B(x,t,\xi)$, $P(x,t)$ are BP factors. Now we introduce the regular RH problem as follows. RH problem 4.4. Find a matrix-valued function $\widehat{N}^{\mathrm{reg}}(x,t,\xi)$ that is piecewise meromorphic and satisfies the following conditions: (i) $\widehat{N}^{\mathrm{reg}}(x,t,\xi)$ is analytic for $\xi\in\mathbb{C}\setminus\Gamma$; (ii) the jump conditions:
$$
\begin{equation}
\widehat{N}^{\mathrm{reg}}_+(x,t,\xi)=\widehat{N}^{\mathrm{reg}}_-(x,t,\xi)\widehat{J}^{\mathrm{reg}}_N(x,t,\xi),
\end{equation}
\tag{4.20}
$$
where
$$
\begin{equation}
\widehat{J}^{\mathrm{reg}}_N(x,t,\xi) = \begin{cases} \begin{pmatrix} 1 & 0 \\ 0 & \dfrac{\xi-i\xi_1}{\xi} \end{pmatrix}\widehat{J}_N(x,t,\xi) \begin{pmatrix} 1 & 0 \\ 0 & \dfrac{\xi}{\xi-i\xi_1} \end{pmatrix}, &\operatorname{Re}\xi\leqslant0, \\ \widehat{J}_N(x,t,\xi), &\operatorname{Re}\xi>0, \end{cases}
\end{equation}
\tag{4.21}
$$
$$
\begin{equation}
\widehat{J}^{\mathrm{reg}}_N(x,t,\xi) = \begin{cases} \begin{pmatrix} 1 & -\dfrac{r_2^{\mathrm{reg}}(\xi)\varpi^2(\xi,\lambda_3)}{1+r_1^{\mathrm{reg}}(\xi)r_2^{\mathrm{reg}}(\xi)} e^{-2it\theta}\\ 0 & 1 \end{pmatrix}, &\xi\in \Gamma_2, \\ \begin{pmatrix} 1 & 0 \\ -r_1^{\mathrm{reg}}(\xi)\varpi^{-2}(\xi,\lambda_3)e^{2it\theta} & 1 \end{pmatrix}, &\xi\in \Gamma_1, \\ \begin{pmatrix} 1 & r_2^{\mathrm{reg}}(\xi)\varpi^2(\xi,\lambda_3)e^{-2it\theta} \\ 0 & 1 \end{pmatrix}, &\xi\in \Gamma^*_1, \\ \begin{pmatrix} 1 & 0 \\ \dfrac{r_1^{\mathrm{reg}}(\xi)\varpi^{-2}(\xi,\lambda_3)}{1+r_1^{\mathrm{reg}}(\xi)r_2^{\mathrm{reg}}(\xi)} e^{2it\theta} & 1 \end{pmatrix}, &\xi\in \Gamma^*_2, \\ \begin{pmatrix} 1 & 0 \\ -r_1(\xi)e^{2it\theta} & 1 \end{pmatrix}, &\xi\in \Gamma_3, \\ \begin{pmatrix} 1 & -r_2(\xi)e^{-2it\theta}\\ 0 & 1 \end{pmatrix}, &\xi\in \Gamma^*_3, \end{cases}
\end{equation}
\tag{4.22}
$$
with $r_1^{\mathrm{reg}}(\xi)$ and $r_2^{\mathrm{reg}}(\xi)$ are defined in (3.26); (iii) the normalization condition at $\xi=\infty$:
$$
\begin{equation}
\widehat{N}^{\mathrm{reg}}(x,t,\xi)=I+O\biggl(\frac{1}{\xi}\biggr),\qquad \xi\to\infty;
\end{equation}
\tag{4.23}
$$
(iv) the elements of the matrix-valued factor $P(x,t)$ are determined in terms of $\widehat{N}^{\mathrm{reg}}(x,t,\xi)$,
$$
\begin{equation}
P_{12}(x,t) =\frac{f_1(x,t)g_1(x,t)}{f_1(x,t)g_2(x,t)-f_2(x,t)g_1(x,t)},
\end{equation}
\tag{4.24a}
$$
$$
\begin{equation}
P_{21}(x,t) =-\frac{f_2(x,t)g_2(x,t)}{f_1(x,t)g_2(x,t)-f_2(x,t)g_1(x,t)},
\end{equation}
\tag{4.24b}
$$
where $f(x,t)=(f_1(x,t),f_2(x,t))^\top$, $g(x,t)=(g_1(x,t),g_2(x,t))^\top$ are given by We now have the potential formula, which is proved similarly to Proposition 3.7. Proposition 4.5. The solution $q(x,t)$ of the Cauchy problem (1.3) and (1.5) can be written in terms of: (i) for $\operatorname{Re}\xi\leqslant0$,
$$
\begin{equation}
q(x,t)= \begin{cases} {\displaystyle-2\xi_1P_{12}(x,t)+2i\lim_{\xi\to\infty}\xi\widehat{N}^{\mathrm{reg}}_{12}(x,t,\xi)}, &x>0, \\ {\displaystyle-2\xi_1\overline{P_{21}(-x,t)}-2i\lim_{\xi\to\infty}\xi\overline{\widehat{N}^{\mathrm{reg}}_{21} (-x,t,\xi)}}, & x<0; \end{cases}
\end{equation}
\tag{4.26}
$$
(ii) for $\operatorname{Re}\xi>0$, 4.2. Long-time asymptotic behaviorWe first address the local model at the point $\lambda_3$, following the discussion in § 3.4. We also introduce the new variable $\varsigma$ by
$$
\begin{equation}
\xi\to\lambda_1+\frac{\varsigma}{\sqrt{4t(48\gamma\lambda_3^2-1)}}.
\end{equation}
\tag{4.28}
$$
Now we can write the function $\varpi$ as
$$
\begin{equation}
\varpi(\xi(\varsigma),\lambda_3)=\biggl(\frac{\varsigma}{\sqrt{4t(48\gamma\lambda_3^2-1)}}\biggr)^{iv(\lambda_3)} e^{\pi(\lambda_3,\xi)},
\end{equation}
\tag{4.29}
$$
where $v(\lambda_3)$ and $\pi(\lambda_3,\xi)$ are as in (4.4). We next introduce the local parametrix $\widehat{N}^{\mathrm{reg}}_{\lambda_3}(x,t,\xi)$ by
$$
\begin{equation}
\widehat{N}^{\mathrm{reg}}_{\lambda_3}(x,t,\xi)=\Delta(\lambda_3,t) \widehat{n}^{\mathrm{pc}}_{\lambda_3}(\lambda_3,\varsigma)\Delta^{-1}(\lambda_3,t),
\end{equation}
\tag{4.30}
$$
with
$$
\begin{equation}
\Delta(\lambda_3,t)=e^{[\pi(\lambda_3)+\phi(\lambda_3,\varsigma)]\sigma_3} [4t(48\gamma\lambda_3^2-1)]^{-\frac{i}{2}v(\lambda_3)\sigma_3},
\end{equation}
\tag{4.31}
$$
$$
\begin{equation}
\begin{split} \phi(\lambda_3,\varsigma) &=-\frac{i\gamma\varsigma^4}{2t(48\gamma\lambda_3^2-1)^2} -\frac{4i\gamma\lambda_3\varsigma^3}{\sqrt{t(48\gamma\lambda_3^2-1)^3}}+ \frac{i\varsigma^2}{4t(48\gamma\lambda_3^2-1)} \\ &\qquad -\frac{i(16\gamma\lambda_3^2-t)\lambda_3\varsigma}{\sqrt{t(48\gamma\lambda_3^2-1)}} -4\gamma\lambda_3^4+\frac{1}{2}\lambda_3^2. \end{split}
\end{equation}
\tag{4.32}
$$
Now the parameterized RH problem $\widehat{n}^{\mathrm{pc}}_{\lambda_3}(\lambda_3,\varsigma)$ can be solved by using parabolic cylinder functions, and the following asymptotic behavior takes place:
$$
\begin{equation}
\widehat{n}^{\mathrm{pc}}_{\lambda_3}(\lambda_3,\varsigma)=I+\frac{i}{\varsigma} \begin{pmatrix} 0 & \beta^{\mathrm{reg}}(\lambda_3) \\ -\gamma^{\mathrm{reg}}(\lambda_3) & 0 \end{pmatrix}+O(\varsigma^{-2}),\qquad \varsigma\to\infty,
\end{equation}
\tag{4.33}
$$
where
$$
\begin{equation}
\beta^{\mathrm{reg}}(\lambda_3) =-\frac{\sqrt{2\pi}e^{-\frac{\pi}{2}\, v(\lambda_3)}e^{\frac{i\pi}{4}}} {r^{\mathrm{reg}}_1(\lambda_3)\Gamma(-iv(\lambda_3))},
\end{equation}
\tag{4.34a}
$$
$$
\begin{equation}
\gamma^{\mathrm{reg}}(\lambda_3) =-\frac{\sqrt{2\pi} \, e^{-\frac{\pi}{2}v(\lambda_3)}e^{-\frac{i\pi}{4}}} {r^{\mathrm{reg}}_2(\lambda_3)\Gamma(iv(\lambda_3))}.
\end{equation}
\tag{4.34b}
$$
Once the local parametrix $\widehat{N}^{\mathrm{reg}}_{\lambda_3}(x,t,\xi)$ is introduced, we define the error matrix by
$$
\begin{equation}
\breve{N}^{\mathrm{reg}}(x,t,\xi)= \begin{cases} \widehat{N}^{\mathrm{reg}}(x,t,\xi)(\widehat{N}^{\mathrm{reg}}_{\lambda_3})^{-1}(x,t,\xi), &\lambda_3-\varepsilon<\operatorname{Re}\xi<\lambda_3+\varepsilon, \\ \widehat{N}^{\mathrm{reg}}(x,t,\xi), &\lambda_3+\varepsilon<\operatorname{Re}\xi<0, \\ &\quad\operatorname{Re}\xi<\lambda_3-\varepsilon, \\ \widehat{N}(x,t,\xi), &\operatorname{Re}\xi>0, \end{cases}
\end{equation}
\tag{4.35}
$$
where $\varepsilon$ is small enough to make $|\lambda_3|>\varepsilon$ and $|i\xi_1-\lambda_3|>\varepsilon$. Let $\Gamma_{\varepsilon}=\Gamma_-\cap D_{\varepsilon}(\lambda_3)$ (see Fig. 11) be the jump contour of the local parametrix $\widehat{N}^{\mathrm{reg}}_{\lambda_3}(x,t,\xi)$ with $\Gamma_-=\Gamma_1\cup\Gamma_2\cup\Gamma^*_1\cup\Gamma^*_2$. Next, $\breve{\Gamma}=\Gamma_-\cup\partial D_{\varepsilon}(\lambda_3)$ (see Fig. 12) is the jump contour of $\breve{N}^{\mathrm{reg}}(x,t,\xi)$, and we redefine $\breve{\Gamma}=\Gamma_{1,\varepsilon}\cup\Gamma_{2,\varepsilon} \cup\Gamma^*_{1,\varepsilon}\cup\Gamma^*_{2,\varepsilon}$. Now the function $\breve{N}^{\mathrm{reg}}(x,t,\xi)$ satisfies the following RH problem. RH problem 4.6. Find a matrix-valued function $\breve{N}^{\mathrm{reg}}(x,t,\xi)$ that is piecewise meromorphic and satisfies the following conditions: (i) $\breve{N}^{\mathrm{reg}}(x,t,\xi)$ is analytic for $\xi\in\mathbb{C}\setminus\breve{\Gamma}$; (ii) the jump conditions:
$$
\begin{equation}
\breve{N}^{\mathrm{reg}}_+(x,t,\xi)=\breve{N}^{\mathrm{reg}}_-(x,t,\xi)\breve{J}_N(x,t,\xi),
\end{equation}
\tag{4.36}
$$
where
$$
\begin{equation}
\breve{J}_N(x,t,\xi)= \begin{cases} \widehat{N}^{\mathrm{reg}}_{\lambda_3}(x,t,\xi) \widehat{J}^{\mathrm{reg}}_N(x,t,\xi)(\widehat{N}^{\mathrm{reg}}_{\lambda_3})^{-1}(x,t,\xi), &\xi\in \Gamma_-\cap D_{\varepsilon}(\lambda_3), \\ (\widehat{N}^{\mathrm{reg}}_{\lambda_3})^{-1}(x,t,\xi), &\xi\in \partial D_{\varepsilon}(\lambda_3), \\ \widehat{J}^{\mathrm{reg}}_N(x,t,\xi), &\xi\in \Gamma_-\setminus \Gamma_{\varepsilon}; \end{cases}
\end{equation}
\tag{4.37}
$$
(iii) the normalization condition at $\xi=\infty$: The next step is to obtain the large $t$ estimate of the solution $q(x,t)$. It can be seen that the processing procedures are different for $\xi\in E$ and $\xi\in \mathbb{C}\setminus E$ with $E=\{\xi\in\mathbb{C}\mid\operatorname{Re}\xi\leqslant0\}$. For $\xi\in E$, we proceed similarly to the error analysis in § 3.5. We define $w_N(x,t,\xi)=\breve{J}_N(x,t,\xi)-I$ and consider the Fredholm integral function We also consider the Cauchy integral operator
$$
\begin{equation}
(Cf)(\xi')=\frac{1}{2\pi i}\int_{\breve{\Upsilon}}\frac{f(s)}{s-\xi'} \, ds, \qquad\xi'\in \mathbb{C}\setminus\breve{\Gamma}.
\end{equation}
\tag{4.40}
$$
We have
$$
\begin{equation}
\breve{N}^{\mathrm{reg}}(x,t,\xi)=I+\frac{1}{2\pi i}\int_{\breve{\Gamma}} \breve{v}_N(x,t,s)w_N(x,t,s) \, \frac{ds}{s-\xi}.
\end{equation}
\tag{4.41}
$$
After some calculations, we find that
$$
\begin{equation}
\breve{N}^{\mathrm{reg}}(x,t,\xi)=I-\frac{1}{2\pi i}\oint_{|s-\lambda_3|=\varepsilon} \frac{\Xi^{\mathrm{reg}}_N(\mu,t)}{(s-\lambda_3)(s-\xi)} \, ds +R^N(\mu,t),\qquad |\xi-\lambda_3|>\varepsilon,
\end{equation}
\tag{4.42}
$$
where
$$
\begin{equation}
\Xi^{\mathrm{reg}}_N(\mu,t)=\frac{i}{2\sqrt{48\gamma\lambda^2_3-1}} \begin{pmatrix} 0 & \dfrac{\beta^{\mathrm{reg}}(\lambda_3)e^{2[\pi+\phi]}}{[4t(48\gamma\lambda^2_3-1)]^{\frac{1}{2}+iv(\lambda_3)}} \\ \dfrac{-\gamma^{\mathrm{reg}}(\lambda_3)e^{-2[\pi+\phi]}}{[4t(48\gamma\lambda^2_3-1)]^{\frac{1}{2}-iv(\lambda_3)}} & 0 \end{pmatrix}.
\end{equation}
\tag{4.43}
$$
Let
$$
\begin{equation*}
R^N(\mu,t)= \begin{pmatrix} R_1^N(\mu,t) & R_2^N(\mu,t) \\ R_1^N(\mu,t) & R_2^N(\mu,t) \end{pmatrix},
\end{equation*}
\notag
$$
where $R_1^N(\mu,t)$ and $R_2^N(\mu,t)$ behave as
$$
\begin{equation}
R_1^N(\mu,t) = \begin{cases} O(t^{-1}), &\operatorname{Im} v(\lambda_3)>0, \\ O(t^{-1}\ln t), &\operatorname{Im} v(\lambda_3)=0, \\ O(t^{-1+2|{\operatorname{Im} v(\lambda_3)}|}), &\operatorname{Im} v(\lambda_3)<0, \end{cases}
\end{equation}
\tag{4.44}
$$
$$
\begin{equation}
R_2^N(\mu,t) = \begin{cases} O(t^{-1+2|{\operatorname{Im} v(\lambda_3)}|}), &\operatorname{Im} v(\lambda_3)<0, \\ O(t^{-1}\ln t), &\operatorname{Im} v(\lambda_3)=0, \\ O(t^{-1}), &\operatorname{Im} v(\lambda_3)<0, \end{cases}
\end{equation}
\tag{4.45}
$$
and $R_3^N(\mu,t)=R_1^N(\mu,t)+R_2^N(\mu,t)$. Now the estimates follow after some calculations. (i) For $x<0$, we have
$$
\begin{equation}
q(x,t)=-t^{-\frac{1}{2}-\operatorname{Im} v(-\lambda_3)} \exp\bigl\{-2[\overline{\pi}+\overline{\phi}(-\mu,\varsigma)]-i\operatorname{Re} v(-\lambda_3)\ln t\bigr\} \alpha+R_1^N(-\mu,t),
\end{equation}
\tag{4.46}
$$
where
$$
\begin{equation}
\alpha= \frac{\sqrt{2\pi} \, e^{-\frac{\pi}{2}\overline{v(-\lambda_3)}}e^{\frac{i\pi}{4}}} {\sqrt{48\gamma\lambda^2_1-1} \, \overline{r_2}(-\lambda_3)\Gamma(-i\overline{v(-\lambda_3)})}\, [4(48\gamma\lambda^2_3-1)]^{-i\overline{v(-\lambda_3)}}.
\end{equation}
\tag{4.47}
$$
(ii) For $x>0$, based on the value of $\operatorname{Im} v(\lambda_3)$, there are three possible types of asymptotics to consider:
where
$$
\begin{equation}
\beta_1 = \frac{d^2_0\sqrt{2\pi} \, e^{-\frac{\pi}{2}v(\lambda_3)}e^{-\frac{i\pi}{4}}} {\sqrt{48\gamma\lambda^2_3-1}\, r_2(\lambda_3)\Gamma(iv(\lambda_3))\lambda^2_3}\, [4(48\gamma\lambda^2_3-1)]^{iv(\lambda_3)},
\end{equation}
\tag{4.51a}
$$
$$
\begin{equation}
\beta_2 = \frac{\sqrt{2\pi} \, e^{-\frac{\pi}{2}v(\lambda_3)}e^{\frac{i\pi}{4}}} {\sqrt{48\gamma\lambda^2_1-1} \, r_1(\lambda_3)\Gamma(-iv(\lambda_3))} \, [4(48\gamma\lambda^2_3-1)]^{-iv(\lambda_3)}.
\end{equation}
\tag{4.51b}
$$
For $\xi\in \mathbb{C}\setminus E$, using again the Fredholm integral function (4.39) and the Cauchy integral operator (4.40), we have
$$
\begin{equation}
w_N(x,t,\xi)+I=\breve{J}_N(x,t,\xi)=\widehat{J}_N(x,t,\xi).
\end{equation}
\tag{4.52}
$$
We have the estimate, as $t\to\infty$,
$$
\begin{equation}
\widehat{J}_N(x,t,\xi)=I+O(e^{2it\theta})=I+O(e^{-2t\operatorname{Im}\theta}),
\end{equation}
\tag{4.53}
$$
where
$$
\begin{equation}
\operatorname{Im}\theta=\operatorname{Im}\xi\bigl(32\gamma(\operatorname{Re}\xi)^3 -32\gamma\operatorname{Re}\xi(\operatorname{Im}\xi)^2-2\operatorname{Re}\xi+\mu\bigr).
\end{equation}
\tag{4.54}
$$
As $t\to\infty$, $\operatorname{Im}\xi$ remains unchanged on the contour $\xi\in\Gamma_3$, and the maximum value of $e^{-2t\operatorname{Im}\theta}$ is obtained with zero $\operatorname{Re}\xi$. We now have the asymptotic formula
$$
\begin{equation}
\widehat{J}_N(x,t,\xi)=I+O\bigl(e^{-\frac{\sqrt{3}}{4}t\mu\zeta_{c}(\mu)}\bigr),
\end{equation}
\tag{4.55}
$$
where
$$
\begin{equation}
\zeta_{c}(\mu)=\sqrt[3]{-\frac{\mu}{\gamma}+\frac{1}{\gamma}\sqrt{\mu^2-\frac{1}{27\gamma}}} -\sqrt[3]{-\frac{\mu}{\gamma}-\frac{1}{\gamma}\sqrt{\mu^2-\frac{1}{27\gamma}}}.
\end{equation}
\tag{4.56}
$$
According to the recovery formulas (4.27) of the potential $q(x,t)$ and the error analysis in § 3.5, we have the long-time asymptotic formula for the solution $q(x,t)$
$$
\begin{equation}
q(x,t)=O\bigl(t^{-\frac{1}{2}}e^{-\frac{\sqrt{3}}{4}t\mu\zeta_{c}(\mu)}\bigr).
\end{equation}
\tag{4.57}
$$
Now the long-time asymptotics of the solution $q(x,t)$, for $\xi\in\mathbb{C}$, as required in Theorem 1.6, follows from the above two parts. § 5. Conclusions and discussionsIn this work, the non-linear steepest descent method of Deift and Zhou is developed to study the long-time asymptotic behavior of non-local Lakshmanan–Porsezian–Daniel (LPD) equation with step-like initial data $q_0(x)=o(1)$ as $x\to-\infty$ and $q_0(x)=A+o(1)$ as $x\to+\infty$, where $A$ is an arbitrary positive constant. Comparing to the existing results in [44], [45], we consider the non-decaying initial value condition with step-like structure. Additionally, the non-local symmetries $x\to-x$ and $t\to t$ are taken into account, which enrich the analysis of the asymptotic solution $q(x,t)$, especially when the range of $x$ is varied. In addition, we have derived the long-time asymptotics of the solution $q(x,t)$ for the different space-time sectors on the whole $(x,t)$-plane. Due to the presence of multiple phase points, the regions become richer and the asymptotic analysis of the leading-order term and error term will be different from that of the classical NLS equation. In the direct scattering part, we make spectral analysis of the Lax pair. For the asymptotic properties at the zero singularity point, we make asymptotic expansions of the eigenfunctions as $\xi\to 0$. Here, we set the parts independent of the spectral parameter $\xi$ as undetermined functions, since they do not affect our asymptotic analysis. Then, since the scattering matrix $S (\xi)$ admits a representation in terms of the matrix eigenfunctions $\psi_\pm (x, t, \xi$), the asymptotic properties of the scattering data $\{a_1(\xi), a_2(\xi), b(\xi)\}$ can be obtained based on the Volterra integral equations and the asymptotic behavior of $\psi_\pm (x, t, \xi)$. Given the pure-step initial data condition, which guarantees the absence of zero points in $a_2(\xi)$, we classify our analysis into two distinct cases (Cases 1 and 2) based on the zero point conditions of $a_2(\xi)$. In the inverse scattering part, we construct the RH problem and obtain one soliton solution. In the part of asymptotic solution construction, for the regions $\mu^2:=(x/t)^2<1/(27\gamma)$, we decompose the jump matrix $J(x,t,\xi)$ into triangular matrices. We next perform the second RH deformation to make the jump matrices decay to the identity matrix $I$ for large $t$. After that, a BP factor is introduced to transform the RH problem into a regular one, which can be solved using parabolic cylinder functions. The Beals–Coifman theory is applied for the error analysis of the regular RH problem. Since the solution $q(x, t)$ can be expressed via the solution to the RH problem, where the oscillations terms $e^{\pm i \theta}$ in the jump matrix significantly influence the asymptotic analysis, we implement the aforementioned sequence of transformations to the original RH problem. These transformations are designed to ensure the exponential decay of $e^{\pm i\theta}$ as $t\to \infty$ and yield a solvable RH problem. Following the methodology outlined in [22], we systematically examine the asymptotic properties of the solvable RH problem at infinity, which ultimately allows us to characterize the long-time behavior of solutions to the nonlocal LPD equation. For the regions $\mu^2:=(x/t)^2>1/(27\gamma)$, the LPD equation has one real saddle point and a pair of conjugate complex saddle points. Similarly to the previous analysis processes, we give the long-time asymptotics of the solution in this case. AcknowledgementsThe authors express their sincere thanks to the editors and the anonymous referees for their careful reading and corrections of this paper, constructive comments, and suggestions for improving the quality of this paper.
Citation in format AMSBIB
\Bibitem{ZhoTiaZha25} |
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