Soliton Resolution for the Wadati–Konno–Ichikawa Equation with Weighted Sobolev Initial Data

Abstract

In this work, we employ the ${\bar{\partial }}$-steepest descent method to investigate the Cauchy problem of the Wadati–Konno–Ichikawa (WKI) equation with initial conditions in weighted Sobolev space ${\mathcal {H}}({\mathbb {R}})$. The long time asymptotic behavior of the solution q(x, t) is derived in a fixed space-time cone $S(y_{1},y_{2},v_{1},v_{2})=\{(y,t)\in {\mathbb {R}}^{2}: y=y_{0}+vt, ~y_{0}\in [y_{1},y_{2}], ~v\in [v_{1},v_{2}]\}$. Based on the resulting asymptotic behavior, we prove the soliton resolution conjecture of the WKI equation which includes the soliton term confirmed by $N({\mathcal {I}})$-soliton on discrete spectrum and the $t^{-\frac{1}{2}}$ order term on continuous spectrum with residual error up to $O(t^{-\frac{3}{4}})$.

Appendices

Appendix A: The Parabolic Cylinder Model Problem

Here, we describe the solution of parabolic cylinder model problem [59, 60]. Define the contours $\Sigma ^{pc}=\cup _{j=1}^{4}\Sigma _{j}^{pc}$ where

$$\begin{aligned} \Sigma _{j}^{pc}=\left\{ \lambda \in {\mathbb {C}}|\arg \lambda =\frac{2j-1}{4}\pi \right\} . \end{aligned}$$

(A.1)

For $r_{0}\in {\mathbb {C}}$, let $\nu (r)=-\frac{1}{2\pi }\log (1+|r_{0}|^{2})$, consider the following parabolic cylinder model Riemann–Hilbert problem.

Riemann-Hilbert Problem 9.2

Find a matrix-valued function $M^{(pc)}(\lambda )$ such that (Fig. 5)

$$\begin{aligned}&\bullet \quad M^{(pc)}(\lambda )~ \text {is analytic in}~ {\mathbb {C}}{\setminus }\Sigma ^{pc}, \end{aligned}$$

(A.2)

$$\begin{aligned}&\bullet \quad M_{+}^{(pc)}(\lambda )=M_{-}^{(pc)}(\lambda )V^{(pc)}(\lambda ),\quad \lambda \in \Sigma ^{pc}, \end{aligned}$$

(A.3)

$$\begin{aligned}&\bullet \quad M^{(pc)}(\lambda )={\mathbb {I}}+\frac{M_{1}}{\lambda }+O(\lambda ^{2}),\quad \lambda \rightarrow \infty , \end{aligned}$$

(A.4)

where

$$\begin{aligned} V^{(pc)}(\lambda )=\left\{ \begin{aligned} \lambda ^{-i\nu {\hat{\sigma }}_{3}}e^{\frac{i\lambda ^{2}}{4} {\hat{\sigma }}_{3}}\left( \begin{array}{cc} 1 &{}\quad r_{0} \\ 0 &{}\quad 1 \\ \end{array} \right) ,\quad \lambda \in \Sigma _{1}^{pc},\\ \lambda ^{-i\nu {\hat{\sigma }}_{3}}e^{\frac{i\lambda ^{2}}{4} {\hat{\sigma }}_{3}}\left( \begin{array}{cc} 1 &{}\quad 0 \\ \frac{{\bar{r}}_{0}}{1+|r_{0}|^{2}} &{}\quad 1 \\ \end{array} \right) ,\quad \lambda \in \Sigma _{2}^{pc},\\ \lambda ^{-i\nu {\hat{\sigma }}_{3}}e^{\frac{i\lambda ^{2}}{4} {\hat{\sigma }}_{3}}\left( \begin{array}{cc} 1 &{}\quad \frac{r_{0}}{1+|r_{0}|^{2}} \\ 0 &{}\quad 1 \\ \end{array} \right) ,\quad \lambda \in \Sigma _{3}^{pc},\\ \lambda ^{-i\nu {\hat{\sigma }}_{3}}e^{\frac{i\lambda ^{2}}{4} {\hat{\sigma }}_{3}}\left( \begin{array}{cc} 1 &{}\quad 0 \\ {\bar{r}}_{0} &{}\quad 1 \\ \end{array} \right) ,\quad \lambda \in \Sigma _{4}^{pc}. \end{aligned}\right. \end{aligned}$$

(A.5)

Fig. 5

Jump matrix $V^{(pc)}$

We have the parabolic cylinder equation expressed as [61]

$$\begin{aligned} \left( \frac{\partial ^{2}}{\partial z^{2}}+\left( \frac{1}{2}-\frac{z^{2}}{2}+a\right) \right) D_{a}=0. \end{aligned}$$

As shown in the literature [28, 62], we obtain the explicit solution $M^{(pc)}(\lambda , r_{0})$:

$$\begin{aligned} M^{(pc)}(\lambda , r_{0})=\Phi (\lambda , r_{0}){\mathcal {P}}(\lambda , r_{0})e^{\frac{i}{4}\lambda ^{2}\sigma _{3}}\lambda ^{-i\nu \sigma _{3}}, \end{aligned}$$

where

$$\begin{aligned} {\mathcal {P}}(\lambda , r_{0})=\left\{ \begin{aligned}&\left( \begin{array}{cc} 1 &{}\quad -r_{0} \\ 0 &{}\quad 1 \\ \end{array} \right) ,\quad&\lambda \in \Omega _{1},\\&\left( \begin{array}{cc} 1 &{}\quad 0\\ -\frac{{\bar{r}}_{0}}{1+|r_{0}|^{2}} &{}\quad 1 \\ \end{array} \right) ,\quad&\lambda \in \Omega _{3},\\&\left( \begin{array}{cc} 1 &{}\quad \frac{r_{0}}{1+|r_{0}|^{2}}\\ 0 &{}\quad 1 \\ \end{array} \right) ,\quad&\lambda \in \Omega _{4},\\&\left( \begin{array}{cc} 1 &{}\quad 0 \\ {\bar{r}}_{0} &{}\quad 1 \\ \end{array} \right) ,\quad&\lambda \in \Omega _{6},\\&~~~I ,\quad&\lambda \in \Omega _{2}\cup \Omega _{5}, \end{aligned}\right. \end{aligned}$$

and

$$\begin{aligned}&\Phi (\lambda , r_{0})\\&=\left\{ \begin{aligned} \left( \begin{array}{cc} e^{-\frac{3\pi \nu }{4}}D_{i\nu }\left( e^{-\frac{3i\pi }{4}}\lambda \right) &{}\quad i\beta _{12}e^{-\frac{3\pi (\nu +i)}{4}}D_{i\nu -1}\left( e^{-\frac{3i\pi }{4}}\lambda \right) \\ -i\beta _{21}e^{-\frac{\pi }{4}(\nu -i)}D_{-i\nu -1}\left( e^{-\frac{i\pi }{4}}\lambda \right) &{}\quad e^{\frac{\pi \nu }{4}}D_{-i\nu }\left( e^{-\frac{i\pi }{4}}\lambda \right) \\ \end{array} \right) ,\quad \lambda \in {\mathbb {C}}^{+},\\ \left( \begin{array}{cc} e^{\frac{\pi \nu }{4}}D_{i\nu }\left( e^{\frac{i\pi }{4}}\lambda \right) &{}\quad i\beta _{12}e^{\frac{\pi }{4}(\nu +i)}D_{i\nu -1}\left( e^{\frac{i\pi }{4}}\lambda \right) \\ -i\beta _{21}e^{-\frac{3\pi (\nu -i)}{4}}D_{-i\nu -1}\left( e^{\frac{3i\pi }{4}}\lambda \right) &{}\quad e^{-\frac{3\pi \nu }{4}}D_{-i\nu }\left( e^{\frac{3i\pi }{4}}\lambda \right) \\ \end{array} \right) ,\quad \lambda \in {\mathbb {C}}^{-}, \end{aligned}\right. \end{aligned}$$

with

$$\begin{aligned}&\beta _{21}=\frac{\sqrt{2\pi }e^{i\pi /4}e^{-\pi \nu /2}}{r_0\Gamma (-i\nu )},\nonumber \\&\beta _{12}=\frac{-\sqrt{2\pi }e^{-i\pi /4}e^{-\pi \nu /2}}{\overline{r_0}\Gamma (i\nu )}=\frac{\nu }{\beta _{21}}. \end{aligned}$$

(A.6)

Then, it is not hard to obtain the asymptotic behavior of the solution by using the well-known asymptotic behavior of $D_{a}(z)$,

$$\begin{aligned} M^{(pc)}(r_0,\lambda )={\mathbb {I}}+\frac{M_1^{(pc)}}{i\lambda }+O(\lambda ^{-2}), \end{aligned}$$

(A.7)

where

$$\begin{aligned} M_1^{(pc)}=\begin{pmatrix}0&{}-\beta _{12}\\ \beta _{21}&{}0\end{pmatrix}. \end{aligned}$$

(A.8)

Appendix B: Meromorphic Solutions of the WKI Riemann–Hilbert Problem

Here, we study RHP 3.5 for the reflectionless case, i.e., $r(z)=0$. Under this condition, we know that M(y, t; z) has no jump across the real axis. Then, for given scattering data $\sigma _{d}=\{(z_{k}, c_{k}), z_{k}\in {\mathcal {Z}}\}^{N}_{k=1}$ satisfying $z_{k}\ne z_{j}$ for $k\ne j$, we obtain the following Riemann–Hilbert problem from RHP 3.5.

Riemann–Hilbert Problem B.1 Find a matrix value function $M(y,t;z|\sigma _{d})$ satisfying

• $M(y,t;z|\sigma _{d})$ is analytic in ${\mathbb {C}}{\setminus }({\mathcal {Z}}\bigcup \bar{{\mathcal {Z}}})$;

• $M(y,t;z|\sigma _{d})={\mathbb {I}}+O(z^{-1})$,    $z\rightarrow \infty$;

• $M(y,t;z|\sigma _{d})$ satisfies the following residue conditions at simple poles $z_{k}\in {\mathcal {Z}}$ and ${\bar{z}}_{k}\in \bar{{\mathcal {Z}}}$

  $$\begin{aligned} \begin{aligned}&\mathop {Res}_{z=z_{k}}M(y,t;z|\sigma _{d})=\lim _{z\rightarrow z_{k}}M(y,t;z|\sigma _{d})N_{k},\\&\mathop {Res}_{z={\bar{z}}_{k}}M(y,t;z|\sigma _{d})=\lim _{z\rightarrow {\bar{z}}_{k}}M(y,t;z|\sigma _{d})\sigma _{2}{\bar{N}}_{k}\sigma _{2}, \end{aligned} \end{aligned}$$

  (B.1)

  where

  $$\begin{aligned}&\displaystyle N_{k}=\left( \begin{aligned} \begin{array}{cc} 0 &{}\quad \gamma _{k}(x,t) \\ 0 &{}\quad 0 \end{array} \end{aligned}\right) ,~ \gamma _{k}(x,t)=c_{k}e^{-2it\theta (z_{k})}, \end{aligned}$$

  (B.2)
  $$\begin{aligned}&\displaystyle \theta (z_{k})=2z_{k}^{2}+\frac{y}{t}z_{k}. \end{aligned}$$

  (B.3)

Then, based on the Liouville’s theorem, the uniqueness of the solution is a direct result. Referring to the symmetry shown in Proposition 2.4, we obtain $M(y,t;z|\sigma _{d})=-\sigma _{2} {\bar{M}}(y,t;{\bar{z}}|\sigma _{d})\sigma _{2}$, from which we can derive the following expansion, i.e.,

$$\begin{aligned} M(y,t;z|\sigma _{d})={\mathbb {I}}+\sum _{k=1}^{N}\left[ \frac{1}{z-z_{k}}\left( \begin{aligned} \begin{array}{cc} 0 &{}\quad \zeta _{k}(x,t) \\ 0 &{}\quad \eta _{k}(x,t) \end{array} \end{aligned}\right) +\frac{1}{z-\overline{z_{k}}}\left( \begin{aligned} \begin{array}{cc} \overline{\eta _{k}}(x,t) &{}\quad 0 \\ -\overline{\zeta _{k}}(x,t) &{}\quad 0 \end{array} \end{aligned}\right) \right] , \end{aligned}$$

(B.4)

where $\zeta _{k}(x,t)$ and $\eta _{k}(x,t)$ are unknown coefficients to be determined. Next, in the similar way shown in [42], we obtain the following proposition.

Proposition B.2

For the given scattering data $\sigma _{d}=\{(z_{k}, c_{k}), z_{k}\in {\mathcal {Z}}\}^{N}_{k=1}$ such that $z_{k}\ne z_{j}$ for $k\ne j$, the solution of RHP B.1 is unique for each $(x,t)\in {\mathbb {R}}^{2}$. Moreover, the solution satisfies

$$\begin{aligned} \Vert M(y,t;z|\sigma _{d})\Vert _{L^{\infty }({\mathbb {C}}{\setminus }({\mathcal {Z}}\cup \bar{{\mathcal {Z}}}))}\lesssim 1. \end{aligned}$$

(B.5)

1.1 B.1 Renormalization of the RHP for Reflectionless Case

For the reflectionless case, following from the trace formula (5.5), we obtain

$$\begin{aligned} s_{22}(z)=\prod _{k=1}^{N}\left( \frac{z-z_{k}}{z-{\bar{z}}_{k}}\right) . \end{aligned}$$

(B.6)

Following from the ideas in [42], we define $\vartriangle \subseteq \{1,2,\ldots ,N\}$, $\bigtriangledown \subseteq \{1,2,\ldots ,N\}{\setminus }\vartriangle$, and

$$\begin{aligned}&s_{22,\vartriangle }=\prod _{k\in \vartriangle }\frac{z-z_{k}}{z-\overline{z_{k}}},\nonumber \\&s_{22,\triangledown }=\frac{s_{11}}{s_{11,\vartriangle }}= \prod _{k\in \triangledown }\frac{z-z_{k}}{z-\overline{z_{k}}}. \end{aligned}$$

(B.7)

Then, the normalized transformation

$$\begin{aligned} M^{\vartriangle }(y,t;z|\sigma _{d}^{\vartriangle })=M(y,t;z|\sigma _{d})s_{22,\vartriangle }(z)^{-\sigma _{3}}, \end{aligned}$$

(B.8)

splits the poles between the columns of $M(y,t;z|\sigma _{d})$ based on the selection of different $\vartriangle$. Then, we can get the modified Riemann–Hilbert problem.

Riemann–Hilbert Problem B.3 Given scattering data $\sigma _{d}=\{(z_{k}, c_{k})\}^{N}_{k=1}$ and $\vartriangle \subseteq \{1,2, \cdots ,N\}$, find a matrix value function $M^{\vartriangle }$ satisfying

• $M^{\vartriangle }(y,t;z|\sigma _{d}^{\vartriangle })$ is analytic in ${\mathbb {C}}{\setminus }({\mathcal {Z}}\bigcup \bar{{\mathcal {Z}}})$;

• $M^{\vartriangle }(y,t;z|\sigma _{d}^{\vartriangle })={\mathbb {I}}+O(z^{-1})$,    $z\rightarrow \infty$;

• $M^{\vartriangle }(y,t;z|\sigma _{d}^{\vartriangle })$ satisfies the following residue conditions at simple poles $z_{k}\in {\mathcal {Z}}$ and $\bar{z_{k}}\in \bar{{\mathcal {Z}}}$

  $$\begin{aligned} \begin{aligned}&\mathop {Res}_{z=z_{k}}M^{\vartriangle }(y,t;z|\sigma _{d}^{\vartriangle })=\mathop {lim}_{z\rightarrow z_{k}}M^{\vartriangle }(y,t;z|\sigma _{d}^{\vartriangle })N^{\vartriangle }_{k},\\&\mathop {Res}_{z={\bar{z}}_{k}}M^{\vartriangle }(y,t;z|\sigma _{d}^{\vartriangle })=\mathop {lim}_{z\rightarrow {\bar{z}}_{k}}M^{\vartriangle }(y,t;z|\sigma _{d}^{\vartriangle })\sigma _{2}\overline{N^{\vartriangle }_{k}}\sigma _{2}, \end{aligned} \end{aligned}$$

  (B.9)

  where

  $$\begin{aligned} N_{k}^{\vartriangle }&=\left\{ \begin{aligned} \left( \begin{array}{cc} 0 &{}\quad \gamma _{k}^{\vartriangle } \\ 0 &{}\quad 0 \\ \end{array} \right) ,\quad k\in \triangledown ,\\ \left( \begin{array}{cc} 0 &{}\quad 0 \\ \gamma _{k}^{\vartriangle } &{}\quad 0 \\ \end{array} \right) ,\quad k\in \vartriangle , \end{aligned}\right. ~~\gamma _{k}^{\vartriangle }=\left\{ \begin{aligned}&c_{k}(s_{22,\vartriangle }(z_{k}))^{2}e^{-2it\theta (z_{k})}\quad k\in \triangledown ,\\&c_{k}^{-1}(s_{22,\vartriangle }^{'}(z_{k}))^{-2}e^{2it\theta (z_{k})}\quad k\in \vartriangle , \end{aligned}\right. \nonumber \\&\theta (z_{k})=2z_{k}^{2}+\frac{y}{t}z_{k}. \end{aligned}$$

  (B.10)

Because $M^{\vartriangle }(y,t;z|\sigma _{d}^{\vartriangle })$ is directly transformed from $M(y,t;z|\sigma _{d})$, it is obvious to find out that RHP B.3 has a unique solution.

For given scattering data $\sigma ^{\triangle }_{d}$, using $q_{sol}(y,t)=q_{sol}(y,t;\sigma ^{\triangle }_{d})$ to denote the unique N-soliton solution of the WKI equation (1.3), by applying (B.8), we can derive that

$$\begin{aligned} q_{sol}(y,t;\sigma ^{\triangle }_{d})=e^{2d}\lim _{z\rightarrow 0}\frac{\partial }{\partial y}\frac{\left( M(0;y,t|\sigma ^{\triangle }_{d})^{-1}M(z;y,t|\sigma ^{\triangle }_{d})\right) _{12}}{z}. \end{aligned}$$

(B.11)

This indicates that each normalization encodes $q_{sol}(y,t)$ in the same way. When the scattering coefficient $s_{22}(z)$ only possesses one zero point $z_{1}$, the one soliton solution can be derived. Taking $z_{1}=\xi +i\eta$, $\xi >0$, $\eta >0$, the one soliton solution of the WKI equation (1.3) is derived as [23]

$$\begin{aligned} q(x,t)=q(y(x,t),t)&=\frac{-2\eta (\xi -\eta i)[\xi \cosh (2\phi )+i\eta \sinh (2\varphi )]e^{2d-2i\varphi }}{\eta [(\xi ^{2}+\eta ^{2})\cosh ^{2}(2\phi )-2\eta ^{2}]},\\ x&=y-\frac{2\eta }{\eta ^{2}(1+e^{4\phi })}, \end{aligned}$$

where $\varphi =\varphi (y,t)$ and $\phi =\phi (y,t)$ are, respectively, defined as

$$\begin{aligned} \varphi (y,t)=\xi y+2(\xi ^{2}-\eta ^{2})t-\frac{1}{2}\arg (c_{1}),\\ \phi (y,t)=4\xi \eta t-\eta y-\frac{1}{2}\log (|c_{1}|). \end{aligned}$$

The constant $c_{1}$ is the norming constant and d is defined in (2.13). However, when the scattering coefficient $s_{22}(z)$ possesses multiple zero point, the exact formula of the solution is too complicated to derive, we do not give them here. In fact, after the elastic collisions, the N-soliton asymptotically separate into N single-soliton solutions as $t\rightarrow \infty$. Of course, the non-generic case, for example, two points of scattering data lie on a vertical line, is an exception. Next, we study the asymptotic behavior of the soliton solutions.

1.2 B.2 Long-Time Behavior of Soliton Solutions

Define a distance

$$\begin{aligned} \mu ({\mathcal {I}})=\min _{z_{k}\in {\mathcal {Z}}{\setminus } {\mathcal {Z}}({\mathcal {I}})}\{Im(z_{k})dist(Rez_{k},I)\}, \end{aligned}$$

(B.12)

and a space-time cone

$$\begin{aligned} S(y_{1},y_{2},v_{1},v_{2})=\{(y,t),y=y_{0}+vt ~with ~y_{0}\in [y_{1},y_{2}],v\in [v_{1},v_{2}]\}, \end{aligned}$$

(B.13)

where $v_{1}\le v_{2}\in {\mathbb {R}}$ are given velocities (Fig. 6).

Fig. 6

Space-time $S(y_{1},y_{2},v_{1},v_{2})$

Proposition B.4

Given scattering data $\sigma _{d}^{\vartriangle _{z_{0}}^{-}}=\{(z_{k},c_{k})\},$ fix $y_{1},y_{2},v_{1},v_{2}\in {\mathbb {R}}$ and $y_{1}<y_{2}$, $v_{1}<v_{2}$. Let ${\mathcal {I}}=\left[ -\frac{v_{2}}{4},-\frac{v_{1}}{4}\right]$. Then as $t\rightarrow \infty$ and $(y,t)\in S(y_{1},y_{2},v_{1},v_{2})$, we have

$$\begin{aligned} M^{\vartriangle _{z_{0}}^{\mp }}(z|\sigma _{d}^{\vartriangle _{z_{0}}^{\pm }})= ({\mathbb {I}}+O(e^{-8\mu |t|}))M^{\vartriangle ^{\mp }_{{\mathcal {I}}}} (z|{\hat{\sigma }}_{d}({\mathcal {I}})), \end{aligned}$$

(B.14)

where $M^{\vartriangle ^{\mp }_{{\mathcal {I}}}} (z|{\hat{\sigma }}_{d}({\mathcal {I}}))$ is $N({\mathcal {I}})=|{\mathcal {Z}}({\mathcal {I}})|$-soliton solutions corresponding to scattering data (Fig. 7)

$$\begin{aligned}&{\hat{\sigma }}_{d}({\mathcal {I}})=\{(z_{k},c_{k}({\mathcal {I}})),z_{k}\in {\mathcal {Z}}({\mathcal {I}})\},\nonumber \\&c_{k}({\mathcal {I}})=c_{k}\prod _{z_{j}\in {\mathcal {Z}}{\setminus } {\mathcal {Z}}({\mathcal {I}})}\left( \frac{z_{k}-z_{j}}{z_{k}-{\bar{z}}_{j}}\right) ^{2}. \end{aligned}$$

(B.15)

Fig. 7

For fixed $v_{1}<v_{2}$, ${\mathcal {I}}=\left[ -\frac{v_{2}}{4},-\frac{v_{1}}{4}\right]$

Proof

We first consider the case of $M^{\vartriangle _{z_{0}}^{-}}(z|\sigma _{d}^{\vartriangle _{z_{0}}^{-}})$. Define

$$\begin{aligned}&\triangle ^{-}({\mathcal {I}})=\{k: Rez_{k}<-\frac{v_{2}}{4}\},\\&\triangle ^{+}({\mathcal {I}})=\{k: Rez_{k}>-\frac{v_{1}}{4}\}. \end{aligned}$$

Then, if we choose $\vartriangle =\vartriangle ^{-}({\mathcal {I}})$ in RHP B.3, it is easy to check that

$$\begin{aligned} \big |\big |N_{k}^{\vartriangle ^{-}({\mathcal {I}})}\big |\big |=\left\{ \begin{aligned}&o(1) \quad k\in {\mathcal {Z}}({\mathcal {I}}),\\&o(e^{-8\mu ({\mathcal {I}})|t|})\quad k\in {\mathcal {Z}}{\setminus }{\mathcal {Z}}({\mathcal {I}}), \end{aligned}\right. ~~t\rightarrow -\infty , \end{aligned}$$

(B.16)

which implies that the residues with $z_{k}\in {\mathcal {Z}}{\setminus }{\mathcal {Z}}({\mathcal {I}})$ have little contribution to the solution $M^{\vartriangle _{z_{0}}^{\pm }}$.

For each discrete spectrum point $z_{k}\in {\mathcal {Z}}{\setminus } {\mathcal {Z}}({\mathcal {I}})$, we make a small disk $D_{k}$ corresponding to each spectrum point $z_{k}$. And the radius of the disk $D_{k}$ is sufficiently small to guarantee that they are non-overlapping. Denote $\partial D_{k}$ as the boundary of $D_{k}$. Then, we introduce that

$$\begin{aligned} \Xi (z)=\left\{ \begin{aligned}&{\mathbb {I}}-\frac{N_{k}^{\vartriangle ^{-}({\mathcal {I}})}}{z-z_{k}} \quad z\in D_{k},\\&{\mathbb {I}}-\frac{\sigma _{2} (\overline{N_{k}})^{\vartriangle ^{-}({\mathcal {I}})}\sigma _{2}}{z-{\bar{z}}_{k}} \quad z\in \overline{D_{k}},\\&{\mathbb {I}},\quad elsewhere. \end{aligned}\right. \end{aligned}$$

(B.17)

By introducing a transformation that ${\hat{M}}^{\vartriangle _{z_{0}}^{-}}(z)=M^{\vartriangle _{z_{0}}^{-}}(z)\Xi (z)$, we can derive that ${\hat{M}}^{\vartriangle _{z_{0}}^{-}}(z)$ has a new jump in $\partial D_{k}$. Then, ${\hat{M}}^{\vartriangle _{z_{0}}^{-}}(z)$ satisfied the following jump relationship

$$\begin{aligned} {\hat{M}}^{\vartriangle _{z_{0}}^{-}}_{+}(z)={\hat{M}}^{\vartriangle _{z_{0}}^{-}}_{-}(z){\hat{V}},~~ z\in \partial D_{k}\cup \partial \overline{D_{k}}. \end{aligned}$$

(B.18)

By using the estimate (B.16), the jump matrix ${\hat{V}}$ satisfies that

$$\begin{aligned} ||{\hat{V}}-I||=O(e^{-8\mu ({\mathcal {I}})|t|}), ~~z\in \partial D_{k}\cup \partial \overline{D_{k}},~~ t\rightarrow -\infty . \end{aligned}$$

(B.19)

Observing a fact that ${\hat{M}}^{\vartriangle _{z_{0}}^{-}}(z|\sigma _{d})$ and $M^{\vartriangle ^{-}_{{\mathcal {I}}}}(z|{\hat{\sigma }}_{d}({\mathcal {I}}))$ possess the same poles and residue conditions. Therefore, we can show that

$$\begin{aligned} {\hat{M}}^{\vartriangle _{z_{0}}^{-}}(z|\sigma _{d}) [M^{\vartriangle ^{-}_{{\mathcal {I}}}}(z|{\hat{\sigma }}_{d}({\mathcal {I}}))]^{-1}\triangleq \varepsilon (z) \end{aligned}$$

(B.20)

has no poles. And, its jumps across the $\partial D_{k}\cup \partial {\bar{D}}_{k}$ satisfy the same estimates with (B.19). Then, with the application of the theory of small-norm Riemann–Hilbert problems, one can easily derive that

$$\begin{aligned} \varepsilon (z)={\mathbb {I}}+O(e^{-8\mu ({\mathcal {I}})|t|}), ~~t\rightarrow \infty , \end{aligned}$$

which together with ${\hat{M}}^{\vartriangle _{z_{0}}^{-}}(z)=M^{\vartriangle _{z_{0}}^{-}}(z)\Xi (z)$ gives the formula (B.14). The other case of $M^{\vartriangle _{z_{0}}^{+}}(z|\sigma _{d}^{\vartriangle _{z_{0}}^{+}})$ can be proved similarly. $\square$

Appendix C: Detailed Calculations for the Pure ${\bar{\partial }}$-Problem

Proposition C.1

For large t, there exist constants $c_{j}(j=1,2,3)$ such that $I_{j}(j=1,2,3)$ defined in (8.7) and (8.8) possess the following estimate

$$\begin{aligned} I_{j}\le c_{j}t^{-\frac{1}{4}},~~ j=1,2,3. \end{aligned}$$

(C.1)

Proof

Let $s=p+iq$ and $z=\xi +i\eta$. Considering the fact that

$$\begin{aligned} \Big |\Big |\frac{1}{s-z}\Big |\Big |_{L^{2}}(q+z_{0})=\left( \int _{q+z_{0}}^{\infty }\frac{1}{|s-z|^{2}}\mathrm{d}p\right) ^{\frac{1}{2}} \le \frac{\pi }{q-\eta }, \end{aligned}$$

we can derive that

$$\begin{aligned} \begin{aligned} |I_{1}|&\le \int _{0}^{+\infty }\int _{q+z_{0}}^{+\infty } \frac{|{\bar{\partial }}\chi _{{\mathcal {Z}}}(s)|e^{-4|t|q(p-z_{0})}}{|s-z|}\mathrm{d}p\mathrm{d}q\\&\le \int _{0}^{+\infty }e^{-4|t|q^{2}}\big |\big |{\bar{\partial }}\chi _{{\mathcal {Z}}}(s)\big |\big |_{L^{2}(q+z_{0})} \Big |\Big |\frac{1}{s-z}\Big |\Big |_{L^{2}(q+z_{0})}\mathrm{d}q \\&\le c_{1}\int _{0}^{+\infty }\frac{e^{-4|t|q^{2}}}{\sqrt{|q-\eta |}}\mathrm{d}q \le c_{1}|t|^{-\frac{1}{4}}. \end{aligned} \end{aligned}$$

(C.2)

Similarly, considering that $r\in H^{1,1}({\mathbb {R}})$, we obtain the estimate

$$\begin{aligned} |I_{2}|\le \int _{0}^{+\infty }\int _{q+z_{0}}^{+\infty } \frac{|r'(p)|e^{-4|t|q^{2}}}{|s-z|}\mathrm{d}p\mathrm{d}q \le c_{2}|t|^{-\frac{1}{4}}. \end{aligned}$$

(C.3)

To obtain the estimate of $I_{3}$, we consider the following $L^{k}(k>2)$ norm

$$\begin{aligned} \bigg |\bigg |\frac{1}{\sqrt{|s-z_{0}|}}\bigg |\bigg |_{L^{k}} \le \left( \int _{q+z_{0}}^{+\infty } \frac{1}{|p-z_{0}+iq|^{\frac{k}{2}}}\mathrm{d}p\right) ^{\frac{1}{k}} \le cq^{\frac{1}{k}-\frac{1}{2}}. \end{aligned}$$

(C.4)

Similarly, we can derive that

$$\begin{aligned} \bigg |\bigg |\frac{1}{|s-z|}\bigg |\bigg |_{L^{k}}\le c|q-\eta |^{\frac{1}{k}-\frac{1}{2}}. \end{aligned}$$

(C.5)

By applying (C.4) and (C.5), it is not hard to check that

$$\begin{aligned} \begin{aligned} |I_{3}|&\le \int _{0}^{+\infty }\int _{q}^{+\infty } \frac{|z-z_{0}|^{-\frac{1}{2}}e^{-4|t|q(p-z_{0})}}{|s-z|}\mathrm{d}p\mathrm{d}q\\&\le \int _{0}^{+\infty }e^{-4|t|q^{2}}\bigg |\bigg |\frac{1}{\sqrt{|s-z_{0}|}}\bigg |\bigg |_{L^{k}} \bigg |\bigg |\frac{1}{|s-z|}\bigg |\bigg |_{L^{k}}\mathrm{d}q \le c_{3}t^{-\frac{1}{4}}. \end{aligned} \end{aligned}$$

(C.6)

Now, we complete the estimates of $I_{j}(j=1,2,3)$. $\square$