2.1. Spectral analysis on the lax pair

The mCH (1.1) admits the Lax pair [Reference de Monvel, Karpenko and Shepelsky2]

(2.1) \begin{equation} \Phi _x = X \Phi, \qquad \Phi _t =T \Phi, \end{equation}

where

(2.2) \begin{align} & X=\frac {1}{2}(izm\sigma _2-\sigma _3), \end{align}

(2.3) \begin{align} &T= \left(z^{-2}+\frac {u^2-u_x^2}{2} \right)\sigma _3-i \left(z^{-1} (u-u_x)+\frac {z}{2}\left(u^2-u_x^2 \right)m \right)\sigma _2 \\[6pt] \nonumber \end{align}

and $z\in \mathbb{C}$ is spectrum parameter. Here, we introduce the standard Pauli matrices

\begin{align*} \sigma _1= \left(\begin{array}{l@{\quad}l} 0&1\\[3pt]1&0\\[3pt] \end{array} \right),\quad \sigma _2= \left(\begin{array}{l@{\quad}l} 0&-i\\[3pt]i&0\\[3pt] \end{array} \right),\quad \sigma _3 = \left(\begin{array}{l@{\quad}l} 1&0\\[3pt]0&-1\\[3pt] \end{array} \right). \end{align*}

Since the Lax pair (2.1) admit spectral singularity at $z=\infty$ and $z=0$ , the asymptotic behaviour of the eigenfunction $\Phi$ as $z\to \infty$ and $z\to 0$ need to be controlled.

Case I. $z=\infty$ . For any real constant $C\neq 0$ , we denote a matrix function relying on $C$

(2.4) \begin{equation} D_C(z)=\frac {1}{2}\left (\begin{array}{c@{\quad}c} \phi _C(z)+\phi _C(z)^{-1} & \phi _C(z)^{-1}-\phi _C(z)\\[3pt] \phi _C(z)^{-1}-\phi _C(z) & \phi _C(z)+\phi _C(z)^{-1} \end{array}\right ), \ \phi _C(z)=\left ( \frac {C+z}{C-z}\right ) ^{1/4}, \end{equation}

where $\phi _C(z)$ is analytic on $\mathbb{C}\setminus \left [{-}C,C\right ]$ and the branch is chosen such that as $\ z\to \infty$ , $ \phi _C(z)\sim e^{-\frac {i\pi }{4}}+\mathcal{O}(z^{-1}).$ Denote

(2.5) \begin{align} {\lim _{z\to \infty } D_C(z)}=D_C(\infty )=\frac {\sqrt {2}}{2}(I+i\sigma _1), \end{align}

which is independent of $C$ . For convenience, we use the notation $f_\pm (z)$ of some function $f$ to denote the boundary values of $f$ from the $\pm$ sides of the oriented jump contours. We set the orientation of all curve on the real axis to be directed from the left to the right in this paper. From $\phi _{C,+}(z)=-i\phi _{C,-}(z)$ , it follows that

\begin{align*} D_{C,+}(z)=i\sigma _1D_{C,-}(z), \ \ z\in \Sigma _+, \end{align*}

Under the initial value (1.5), we define two gauge transformations

(2.6) \begin{align} \Psi ^\pm (z;\,x,t)=D_{c_\pm }(z)\Phi (z;\,x,t), \end{align}

which satisfy the following Lax pair

(2.7) \begin{align} & \Psi _{x}^\pm = \left ({-}\frac {i}{2}m\sqrt {z^2-c_\pm ^2}\sigma _3+P^\pm \right )\Psi ^\pm, \end{align}

(2.8) \begin{align} & \Psi _{t}^\pm =\left (i\sqrt {z^2-c_\pm ^2}\left ( \dfrac {m(u^2-u_x^2)}{2}+\dfrac {1}{c_\pm z}\right ) \sigma _3+L^\pm \right )\Psi ^\pm, \end{align}

with

\begin{align*} P^\pm \,:\!=\, &i\dfrac {c_\pm m-1}{2\sqrt {z^2-c_\pm ^2}}\left (c_\pm \sigma _3+iz\sigma _2\right ), \\[3pt] L^\pm \,:\!=\, &i\left ( \dfrac {c_\pm (u^2-u_x^2)(1-c_\pm m)}{2\sqrt {z^2-c_\pm ^2}}-\dfrac {u-1/c_\pm }{\sqrt {z^2-c_\pm ^2}}\right ) \sigma _3+\dfrac {u_x}{c_\pm }\sigma _1\\[3pt] &-\left ( \dfrac {z(u^2-u_x^2)(1-c_\pm m)}{2\sqrt {z^2-c_\pm ^2}}-\dfrac {c_\pm u-1}{z\sqrt {z^2-c_\pm ^2}}\right ) {\sigma _2,} \end{align*}

and the branch of the square root is chosen such that $\sqrt {z^2-c_\pm ^2}\sim ic_\pm$ , $z\to 0$ in $\mathbb{C}^+$ , where $\mathbb{C}^\pm$ denote the upper/lower half complex plane and $c_\pm$ are exactly given in (1.6). For convenience, we denote $\Sigma _\pm =[{-}c_\pm, c_\pm ]$ as the branch cut of $\phi _{c_\pm }(z)$ .

Furthermore, we introduce

(2.9) \begin{align} \mu ^\pm (z;\,x,t)=\Psi ^\pm (z;\,x,t)e^{itp^{(\pm )}(z)\sigma _3}, \end{align}

where $p^{(\pm )}(z)$ are defined by

(2.10) \begin{align} tp^{(\pm )}(z) \,:\!=\, tp^{(\pm )}(z;\,x,t)&=\dfrac {\sqrt {z^2-c_\pm ^2}}{2}\left ( \int _{\pm \infty }^{x}(m(s)-1/c_\pm )ds+\frac {x}{c_\pm }-\frac {2t}{c_\pm z^2}-\frac {t}{c_\pm ^3} \right ). \end{align}

In this paper, whenever convenient, we use $f(z)$ to denote $f(z;\,x,t)$ to emphasise the dependence on $z$ . Then $\mu ^\pm (z;\,x,t)$ solve the two Volterra-type integral equations

(2.11) \begin{equation} \mu ^\pm (z;\,x,t)=I+\int ^{x}_{\pm \infty }e^{\frac {i}{2}\hat {\sigma }_3\sqrt {z^2-c_\pm ^2}\int ^{s}_{x}m(v,t)dv}\left [ P^\pm (z;\,s,t) \mu ^\pm (z;\,s,t) \right ] ds. \end{equation}

It follows from (2.11) that the Jost functions $ \mu ^\pm (z) \,:\!=\, \mu ^\pm (z;\,x,t)$ admit two kinds of symmetries

\begin{align*} \mu ^\pm (z)=\sigma _1\overline {\mu ^\pm (\bar {z})}\sigma _1=\sigma _2\overline {\mu ^\pm ({-}z)}\sigma _2^{-1}. \end{align*}

Again applying (2.11), it is accomplished that $ \det \mu ^\pm (z)=1,$ and

\begin{align*} \mu ^\pm (z)\to I, \ z\to \infty .\end{align*}

Thus it appears that $\mu ^\pm (z)$ are analytical in $\mathbb{C}\setminus \Sigma _\pm$ , respectively. Let

(2.12) \begin{align} \tilde {\mu }^\pm (z;\,x,t) =D_{c_\pm }^{-1}(z)\mu ^\pm (z;\,x,t), \end{align}

then the Volterra-type integrals (2.11) about $\tilde {\mu }^\pm (z) \,:\!=\, \tilde {\mu }^\pm (z;\,x,t)$ are changed into

(2.13) \begin{align} \tilde {\mu }^\pm (z;\,x,t) =\,& D_{c_\pm }^{-1}(z) +\int ^{x}_{\pm \infty }D_{c_\pm }^{-1}(z)e^{\frac {i}{2}\sqrt {z^2-c_\pm ^2}\int ^{s}_{x}m(l,t)dl\hat {\sigma _3}} D_{c_\pm }(z)\nonumber \\[3pt] &\cdot \left (X(z;\,s,t)+ \frac {i}{2}m(s,t)\sqrt {z^2-c_\pm ^2}D_{c_\pm }(z)^{-1}\sigma _3D_{c_\pm }(z)\right ) \tilde {\mu }^\pm (z;\,s,t) ds, \end{align}

where $X$ is defined in (2.2). It follows from (2.13) that the Jost functions $\tilde {\mu }^\pm (z)$ have no more than $-\frac {1}{4}$ -weak singularity at $z=\pm 1$ and $z=\pm c$ as

\begin{align*} \tilde {\mu }^\pm (z)=\mathcal{O}\big ((z\mp 1)^{-1/4}\big ), \quad \tilde {\mu }^\pm (z)=\mathcal{O}\big ((z\mp c)^{-1/4}\big ). \end{align*}

Since $D_{c_\pm }(z)^{-1}\Psi ^\pm (z;\,x,t)$ are two fundamental matrix solutions of the Lax pair (2.1), they are related by a scattering matrix function $S(z)$ independent of $x$ and $t$

(2.14) \begin{equation} D_{c_+}(z)^{-1}\Psi ^+(z;\,x,t)=D_{c_-}(z)^{-1}\Psi ^-(z;\,x,t)S(z), \end{equation}

\begin{align*} S(z) =\left (\begin{array}{c@{\quad}c} s_{11}(z) &s_{12}(z) \\[4pt] s_{21}(z) & s_{22}(z) \end{array}\right ),\qquad \det S(z) =1. \end{align*}

Combining the transformations (2.9), (2.12) with the (2.14), it deduced that

(2.15) \begin{align} S(z)=e^{itp^{(-)}(z;\,x,t)\sigma _3}(\tilde {\mu }^-(z;\,x,t))^{-1}\tilde {\mu }^+(z;\,x,t)e^{-itp^{(+)}(z;\,x,t)\sigma _3}, \end{align}

which is analytic on $\mathbb{C}\setminus \Sigma _+$ . It also implies that the scattering matrix $S(z)$ has no more than $-\frac {1}{4}$ -weak singularity at $z=\pm 1$ and $z=\pm c$ . On the other hand, it is also deduced that

\begin{align*} S(z)\sim e^{\frac {1}{2}Hz\sigma _3 },\ \ z\to \infty, \end{align*}

where $H$ is a constant given by

\begin{align*}H= \left(1-\frac {1}{c}\right)x+\left(\frac {1}{c^3}-1\right)t+\int _{-\infty }^{x} \left(m(s,t)-1 \right)ds+\int _{x}^{+\infty } \left(m(s,t)-\frac {1}{c} \right)ds.\end{align*}

Define two reflecting coefficients by

(2.16) \begin{equation} r_1(z)=\frac {s_{21}(z)}{s_{11}(z)},\qquad r_2(z)=\frac {s_{12}(z)}{s_{22}(z)}. \end{equation}

As $z\to \pm c$ , they then admit asymptotic behaviour $1-r_1(z)r_2(z)=\mathcal{O}((z\mp c)^{1/2}).$

To construct the RH problem, the jump of the Jost functions $\tilde {\mu }^\pm (z)$ on the cut $\Sigma _\pm$ need to be analysed in the following proposition under standard proof.

Proposition 1. The functions $\mu ^\pm$ , $\tilde {\mu }^\pm$ , $S$ and the reflecting coefficients $r_1$ , $r_2$ admit the jump relations

(i) For $z\in \Sigma _\pm$ ,

\begin{align*} \mu ^\pm _+(z)=\sigma _1 \mu ^\pm _-(z)\sigma _1. \end{align*}

(ii) For $z\in \Sigma _-$ ,

\begin{align*} & \tilde {\mu }^\pm _{11,+}(z)=-i \tilde {\mu }^\pm _{12,-}(z), \quad \tilde {\mu }^\pm _{21,+}(z)=-i \tilde {\mu }^\pm _{22,-}(z),\\[3pt] &s_{11,\pm }(z)=s_{22,\mp }(z), \quad s_{12,\pm }(z)=s_{21,\mp }(z), \quad {r_{1,\pm }(z)=r_{2,\mp }(z).} \end{align*}

(iii) For $z\in \Sigma _+\setminus \Sigma _-$ , $\tilde {\mu }^+(z)$ has same jump as above equation while $\tilde {\mu }^-(z)$ has no jump. And

\begin{align*} & s_{11,+}(z)=-is_{12,-}(z), \quad s_{21,+}(z)=-is_{22,-}(z), \quad {r_{1,\pm }(z)r_{2,\mp }(z)=1.} \end{align*}

(iv) For $z\in \mathbb{R}\setminus \Sigma _+$ ,

\begin{align*} r_1(z)=\overline {r_2(z)}. \end{align*}

Under the Assumption1 $r_1(z)$ and $r_2(z)$ are analytic in $\mathbb{C}\setminus \Sigma _+$ .

Case II: $z=0$ .

The Lax pair (2.7)–(2.8) is rewritten in the form

\begin{align*} & \Psi ^{\pm }_x = -\frac {i\sqrt {z^2-c_\pm ^2}}{2c_\pm }\sigma _3\Psi ^\pm +P_0^\pm \Psi ^\pm, \\[3pt] & \Psi ^{\pm }_t =i\sqrt {z^2-c_\pm ^2}\left ( \dfrac {1}{2c_\pm ^3}+\dfrac {1}{c_\pm z^2}\right ) \sigma _3\Psi ^\pm +L_0^\pm \Psi ^\pm, \end{align*}

where $c_\pm$ are exactly given in (1.6) and

\begin{align*} P_0^\pm =&iz\dfrac {c_\pm m-1}{2c_\pm \sqrt {z^2-c_\pm ^2}}\left ( z\sigma _3+ic_\pm \sigma _2\right ),\\[3pt] L_0^\pm =&L_\pm +i\sqrt {z^2-c_\pm ^2}\left ( \dfrac {m(u^2-u_x^2)}{2}+\dfrac {1}{c_\pm z}-\dfrac {1}{2c_\pm ^3}+\dfrac {1}{c_\pm z^2}\right ) \sigma _3. \end{align*}

By making transformation

\begin{align*} \mu ^\pm _0(z;\,x,t)&=\Psi _\pm (z;\,x,t) e^{iq^{(\pm )}(z;\,x,t)\sigma _3}, \quad \\[3pt] q^{(\pm )}(z;\,x,t)&=\frac {i\sqrt {z^2-c_\pm ^2}}{2c_\pm }\left [x-\left (\dfrac {1}{c_\pm ^2}+\dfrac {2}{ z^2} \right )t \right ], \end{align*}

$\mu ^\pm _0(z) \,:\!=\, \mu ^\pm _0(z;\,x,t)$ admit a new Lax pair

(2.17) \begin{align} & \mu ^{\pm }_{0,x} = -\frac {i\sqrt {z^2-c_\pm ^2}}{2c_\pm }[\sigma _3,\mu ^\pm _0]+P_0^\pm \mu ^\pm _0, \end{align}

(2.18) \begin{align} & \mu ^{\pm }_{0,t} =i\sqrt {z^2-c_\pm ^2}\left ( \dfrac {1}{2c_\pm ^3}+\dfrac {1}{c_\pm z^2}\right ) [\sigma _3,\mu ^\pm _0(z)]+L_0^\pm \mu ^\pm _0, \\[6pt] \nonumber \end{align}

which can be written into two Volterra type integrals

\begin{align*} \mu ^\pm _0(z)=I+\int ^{x}_{\pm \infty }e^{\frac {i}{2c_\pm }\hat {\sigma }_3\sqrt {z^2-c_\pm ^2}(s-x)}\left [ P_0^\pm (z;\,s,t)\mu ^\pm _0(z;\,s,t) \right ] ds. \end{align*}

Taking $z=0$ in above integral equation implies $\mu ^\pm _0(0)=I$ . Moreover, expanding $\mu ^\pm _0(z)$ at $z=0$ gives that

(2.19) \begin{align} \mu ^\pm _0(z)=I+\frac {z}{2}\left (\begin{array}{c@{\quad}c} 0 & \int _{\pm \infty }^x\left(m-\frac {1}{c_\pm }\right)e^{x-s}ds \\[3pt] -\int _{\pm \infty }^x\left(m-\frac {1}{c_\pm } \right)e^{s-x}ds & 0 \end{array}\right )+\mathcal{O}(z^2), \end{align}

which will be used to reconstruct the potential $u(x, t)$ .

Because $\mu ^\pm _0e^{-iq^{(\pm )}\sigma _3}$ also admit Lax pair (2.7)–(2.8), there exist two matrix functions $C_\pm (z)$ independent of $x$ and $t$ such that

(2.20) \begin{align} \mu ^\pm _0(z;\,x,t)e^{-iq^{(\pm )}(z;\,x,t)\sigma _3}C_\pm (z)=\mu ^\pm (z;\,x,t)e^{-itp^{(\pm )}(z;\,x,t)\sigma _3}. \end{align}

Since $q^{(\pm )}-tp^{(\pm )}=-\dfrac {1}{2}\sqrt {z^2-c_\pm ^2}\int _{\pm \infty }^x(m-1/c_\pm )ds,$ taking the limits $x\to \pm \infty$ , we obtain $C_\pm (z)\equiv I$ . Invoking (2.12) and $D_{c_\pm, +}(0)=i\sigma _1$ , it follows that

(2.21) \begin{align} \tilde {\mu }^\pm _+(0)=-i\left (\begin{array}{c@{\quad}c} 0 & e^{i(q^{(\pm )}_+(0)-tp^{(\pm )}_+ (0))} \\[3pt] e^{-i(q^{(\pm )}_+(0)-tp^{(\pm )}_+ (0))} & 0 \end{array}\right ). \end{align}

Consequently, from (2.15) it follows that as $z\to 0 \in \mathbb{C}^+$ ,

(2.22) \begin{align} s_{11}(z)=e^{i(q^{(-)}_+-q^{(+)}_+) (0)}+\mathcal{O}(z^2),\ s_{22}(z)=e^{-i(q^{(-)}_+-q^{(+)}_+)(0)}+\mathcal{O}(z^2). \end{align}

2.2. Setting up a RH problem with step-like initial data

Define a sectionally analytical matrix

(2.23) \begin{equation} M(z) \,:\!=\, M(z;\,x,t)= {D_{c_-}(\infty )}\times \left \{ \begin{array}{l@{\quad}l} \left ( \tilde {\mu }^-_1 (z), \ \frac {\tilde {\mu }^+_2(z)}{s_{22}(z)}e^{it(p^{(+)}-p^{(-)})}\right ), &\text{as } z\in \mathbb{C}^+,\\[12pt] \left ( \frac {\tilde {\mu }^+_1(z)}{s_{11}(z)}e^{-it(p^{(+)}-p^{(-)})}, \ \tilde {\mu }^-_2(z)\right ), &\text{as }z\in \mathbb{C}^-,\\[3pt] \end{array}\right . \end{equation}

where $\tilde {\mu }^\pm _1 (z)$ and $\tilde {\mu }^\pm _2 (z)$ denote the first and second column of $\tilde {\mu }^\pm (z)$ , respectively and $D_{c_-}(\infty )$ is defined in (2.5).

In order to construct the RH problem only depending explicitly on the scattering data, via the definition of the new scale $y(x, t)$ in (1.7), we define

(2.24) \begin{equation} N(z) \,:\!=\, N(z;\,y,t)= M(z;\,x(y,t),t). \end{equation}

Recall the notation $\xi =y/t$ and $c_-=1$ , then $p^{(-)}$ defined in (2.10) can be rewrite as

\begin{align*}p^{(-)}=\frac {\sqrt {z^2-1}}{2}\left (\xi -1-2z^{-2} \right ). \end{align*}

Then $ N(z)$ is a solution of the following RH problem.

From (2.19), (2.20) and (2.22), it reveals that

(2.28) \begin{align} N(z)=&N_+(0)+N_1z+\mathcal{O}(z^2), \ \ z\to 0\in \mathbb{C}^+, \end{align}

where

\begin{align*} N_+(0)=iD_{c_-}(\infty )\left (\begin{array}{c@{\quad}c} 0 & f_1 \\[3pt] f_1^{-1} & 0 \end{array}\right ),\ N_1=iD_{c_-}(\infty ) \left (\begin{array}{c@{\quad}c} f_2 & 0\\[3pt] 0 & f_3 \end{array}\right ) \end{align*}

with

\begin{align*} f_1&=\exp \left \lbrace -\frac {1}{2}\int _{-\infty }^x(m-1)ds\right \rbrace, \ f_2=\frac {e^{\frac {1}{2}\int _{-\infty }^x(m-1)ds}}{2}\left ( \int _{-\infty }^x(m-1)e^{s-x}ds +1\right ),\\[3pt] f_3&=\frac {e^{-\frac {1}{2}\int _{-\infty }^x(m-1)ds}}{2c}\left (1-\int _{+\infty }^x(cm-1)e^{x-s}ds \right ). \end{align*}

Thus, via the definition of $y$ in (1.7), it follows that

(2.29) \begin{equation} x(y,t)=y-2\ln \left ( i\sqrt {2}N_{21,+}(0)\right ). \end{equation}

Direct calculation shows that

(2.30) \begin{align} &-2\left (N_{12,+}(0)N_{1,11}+N_{21,+}(0)N_{1,22}\right )=f_1f_2+f_1^{-1}f_3\\[3pt] &=\frac {1}{2}\left (\int _{-\infty }^x(m-1)e^{s-x}ds-\int _{+\infty }^x(m-1/c)e^{x-s}ds +(1+c)/c\right ).\nonumber \end{align}

Taking $x\to +\infty$ and $x\to -\infty$ in above equation, respectively, and via the fact that $\underset {x\to -\infty }{\lim }m=1$ , $\underset {x\to +\infty }{\lim }m=1/c$ , we arrive at that

\begin{align*} \underset {x\to +\infty }{\lim }(f_1f_2+f_1^{-1}f_3)=1/c\qquad \underset {x\to -\infty }{\lim }(f_1f_2+f_1^{-1}f_3)=1. \end{align*}

Via taking the derivative with respect to $x$ on (2.30), we obtain

\begin{align*}-2\left (N_{12,+}(0)N_{1,11}+N_{21,+}(0)N_{1,22}\right )+2\partial _x^2(N_{12,+}(0)N_{1,11}+N_{21,+}(0)N_{1,22})=m.\end{align*}

Therefore, we arrive at the following reconstruction formula

(2.31) \begin{align} &u(x,t)=-2(N_{12}(0)N_{1,11}+N_{21,+}(0)N_{1,22}), \end{align}

2.3. An almanac of jump matrix factorisations

The jump matrix $\tilde {V}(z)$ admits the following decomposition from the symmetry of the reflecting coefficients $r_1$ , $r_2$ in Proposition1, which will be used in the asymptotic analysis in the next section.

On the interval $ \mathbb{R}\setminus \Sigma _+$ ,

(2.32) \begin{align} \tilde {V}(z)=&\left (\begin{array}{c@{\quad}c} 1 & 0\\[3pt] -r_1e^{2itp^{(-)}} & 1 \end{array}\right )\left (\begin{array}{c@{\quad}c} 1 & r_2e^{-2itp^{(-)}}\\[3pt] 0 & 1 \end{array}\right )\nonumber \\[3pt] =&\left (\begin{array}{c@{\quad}c} 1 &\frac { r_2e^{-2itp^{(-)}}}{1-r_1r_2}\\[3pt] 0 & 1 \end{array}\right )(1-r_1r_2)^{-\sigma _3}\left (\begin{array}{c@{\quad}c} 1 &0\\[3pt] \frac {-r_1e^{2itp^{(-)}}}{1-r_1r_2} & 1 \end{array}\right ). \\[6pt] \nonumber \end{align}

On the interval $ \Sigma _+\setminus \Sigma _-$ ,

\begin{align*} &\tilde {V}(z) =\left (\begin{array}{c@{\quad}c} 1 & 0\\[3pt] -r_{1,-}(z)e^{2itp^{(-)}} & 1 \end{array}\right )\left (\begin{array}{c@{\quad}c} 1 & r_{2,+}(z)e^{-2itp^{(-)}}\\[3pt] 0 & 1 \end{array}\right )\\[3pt] =&\left (\begin{array}{c@{\quad}c} 1 &\frac { r_{2,-}(z)e^{-2itp^{(-)}}}{1-r_{1,-}(z)r_{2,-}(z)}\\[3pt] 0 & 1 \end{array}\right )\left (\begin{array}{c@{\quad}c} 0 & \frac {r_{2,-}(z)}{e^{2itp^{(-)}}}\\[3pt] \frac {-e^{2itp^{(-)}}}{r_{2,-}(z)} & 0 \end{array}\right )\left (\begin{array}{c@{\quad}c} 1 &0\\[3pt] \frac {-r_{1,+}(z)e^{2itp^{(-)}}}{1-r_{1,+}(z)r_{2,+}(z)} & 1 \end{array}\right ). \end{align*}

On the interval $ \Sigma _-$ ,

\begin{align*} &\tilde {V}(z) =-i\left (\begin{array}{c@{\quad}c} 1 & 0\\[3pt] -r_{1,-}(z)e^{2itp^{(-)}} & 1 \end{array}\right )\sigma _1\left (\begin{array}{c@{\quad}c} 1 & r_{2,+}(z)e^{-2itp^{(-)}}\\[3pt] 0 & 1 \end{array}\right ). \end{align*}

The long-time asymptotic of RH problem1 is affected by the growth or decay of the exponential function $e^{\pm 2itp^{(-)}}$ with

\begin{align*}p^{(-)}=\frac {\sqrt {z^2-1}}{2}\left (\xi -1-2z^{-2} \right ),\ \partial _zp^{(-)}=\frac {1}{2\sqrt {z^2-1}z^3}\left [ \left ( \xi -1)z^4+2z^2-4 \right ) \right ]. \end{align*}

Thus, to obtain the long-time asymptotics, we need to analyse the real part of $ 2itp^{(-)}$ . We hope that after appropriately choosing triangular factorisations of the jump matrices and associated deformations of the original RH problem, the jumps remaining on $\mathbb{R}$ can become constant matrices (independent of $z$ , but dependent on $\xi$ and $c$ ) of special structure or a jump matrices of solvable model, whereas the other jumps decay exponentially to the identity matrix. We introduce the g-function mechanism [Reference Deift, Venakides and Zhou17] to problems with step-like background in different regions. This mechanism is relevant when some entries of the jump matrix grow exponentially or oscillate as $t\to \infty$ . The general idea consists in replacing the original phase function in the jump matrix. This new $g$ -function needs to be analytic on $\mathbb{C}$ except some new cut (it is undetermined and do not must be $\Sigma _\pm$ ) and satisfies the above condition. Moreover, it must have same asymptotic properties as $z\to \infty, \ 0\in \mathbb{C}^+$ as $p^{(-)}$ :

(2.33) \begin{align} &p^{(-)}=\frac {1-\xi }{2}z+\mathcal{O}(z^{-1}),\ \partial _zp^{(-)}=\frac {1-\xi }{2}+\mathcal{O}(z^{-2}),\ z\to \infty ; \end{align}

(2.34) \begin{align} &p^{(-)}=\frac {i}{z^2}-\frac {i\xi }{2}+\mathcal{O}(z),\ \partial _zp^{(-)}=\frac {-2i}{z^3}+\mathcal{O}(z),\ z\to 0\in \mathbb{C}^+. \\[6pt] \nonumber \end{align}

The structure of the limiting RH problem is such that the problem can be solved explicitly in terms of Riemann theta functions and Abel integrals on Riemann surfaces associated with the limiting RH problem [Reference Biondini and Mantzavinos1, Reference de Monvel, Kotlyarov and Shepelsky4–Reference Buckingham and Venakides7]. For different ranges of the parameter $\xi = y/t$ , different Riemann surfaces may appear.

Figure 2.

In the white region, $\mathrm{Im}[p^{(-)}]\gt 0$ , while in another region, $\mathrm{Im}[p^{(-)}]\lt 0$ . (a) $\xi \lt 3/4$ ; (b) $3/4\lt \xi \lt 1$ ; (c) $1\leq \xi \lt 3$ .

Figure 3.

Figure of curves $\Sigma _j$ and domains $\Omega _j$ , $j=1,2,$ in the case of $\{(\xi, c)\,\,:\ \xi \lt 3/4\}$ .

3.1. A model RH problem on cuts

The global parametrix $ M^{mod1}(z )$ is given by the following model RH problem:

The solution of this model RH problem is given by

(3.12) \begin{align} M^{mod1}(z)=D_1(\infty )D_1(z)^{-1}, \end{align}

where $D_1$ is defined in (2.4). As $z\to 0\in \mathbb{C}^+$ , it is adduced that

(3.13) \begin{equation} M^{mod1}(z)=\frac {1}{\sqrt {2}}\left (I-i\sigma _1\right )-\frac {zi}{2\sqrt {2}}\left (I+i\sigma _1\right )+\mathcal{O}(z^2). \end{equation}

3.2. The small norm RH problem for error function

In this subsection, we consider the error matrix-function $E(z) \,:\!=\, E(z;\,\xi, c)$ in this region.

Out of $U$ , the jump $V^{ E }$ admits the following estimates

(3.15) \begin{equation} \| V^{ E }-I\|_p\lesssim \exp \left \{-tK_p\right \}, z\in \Sigma ^{ E }\setminus U, \ p\in [1,\infty ], \end{equation}

for positive $K_p$ relying on $p$ . For $z\in \partial U$ , $M^{mod1}(z)$ is bounded, so by (3.10), we find that

(3.16) \begin{align} | V^{ E }(z)-I|&= \mathcal{O}(t^{-1/2}). \end{align}

Therefore, the existence and uniqueness of the RH problem4 is obtained via a small-norm RH problem [Reference Deift and Zhou18, Reference Deift and Zhou19]. According to Beals–Coifman theory, the solution of the RH problem4 can be given by

(3.17) \begin{equation} E(z)=I+\frac {1}{2\pi i}\int _{\Sigma ^{E}}\dfrac {\left ( I+\varpi (s)\right ) (V^{E}(s)-I)}{s-z}ds, \end{equation}

where the $\varpi \in L^\infty (\Sigma ^{E})$ is the unique solution of $(1-C_E)\varpi =C_E\left (I \right ),$ and $C_E$ is a integral operator: $L^\infty (\Sigma ^{E})\to L^2(\Sigma ^{E})$ defined by $ C_E(f)(z)=C_-\left ( \,f(V^{E}(z) -I)\right )$ with the usual Cauchy projection operator $ C_-(f)(s)=\lim _{z\to \Sigma ^{E}_-}\frac {1}{2\pi i}\int _{\Sigma ^{E}}\dfrac {f(s)}{s-z}ds.$

In case (i), under the absence of $\lambda _j$ , $j=1,2$ , it appears that

(3.18) \begin{equation} E(z)=I+\mathcal{O}(e^{-C(\xi, c)t}) \end{equation}

where $C(\xi, c)$ is a positive constant relying on $\xi$ and $c$ . While in the case(ii) and (iii), the stationary points have contribution on $t\to \infty$ . By (3.16), it adduced that

(3.19) \begin{equation} \| C_E\|\leq \| C_-\| \| V^{E}(z)-I\|_2 \lesssim \mathcal{O}(t^{-1/2}), \end{equation}

which implies that $1-C_E$ is invertible for sufficiently large $t$ . So $\varpi$ exists and is unique. Besides,

(3.20) \begin{equation} \| \varpi \|_{L^\infty (\Sigma ^{E})}\lesssim \dfrac {\| C_E\|}{1-\| C_E\|}\lesssim t^{-1/2}. \end{equation}

In order to reconstruct the solution $u(y,t)$ of (1.1), we need the asymptotic behaviour of $E(z)$ as $z\to 0\in \mathbb{C}^+$ and the long-time asymptotic behaviour of $E(0)$ .

Proposition 4. As $z\to 0\in \mathbb{C}^+$ , we have

(3.21) \begin{align} E(z)=E(0)+E_1z+\mathcal{O}(z^2), \end{align}

with long-time asymptotic behaviour

(3.22) \begin{equation} E(0)=I+t^{-1/2}H^{(0)}+\mathcal{O}(t^{-1}), \end{equation}

where

(3.23) \begin{align} H^{(0)}= &\sum _{ p=\pm \lambda _j,j=1,2 }\frac {M^{mod1}(p)A_{j,\pm }(\xi )M^{mod1}(p)^{-1}}{p}. \end{align}

Here $A_{j,\pm }(\xi )$ is given by (3.10). And

\begin{align*} E_1=\frac {1}{2\pi i}\int _{\Sigma ^{E}}\dfrac {\left ( I+\varpi (s)\right ) (V^{E}-I)}{s^2}ds= t^{-1/2}H^{(1)}+\mathcal{O}(t^{-1}), \end{align*}

where

(3.24) \begin{align} H^{(1)}= &\sum _{ p=\pm \lambda _j,j=1,2 }\frac {M^{mod}(p)A_{j,\pm }(\xi )M^{mod}(p)^{-1}}{p^2}. \end{align}

Proof. Substituting the long-time asymptotic behaviour of $V^{E}$ , $\varpi (s)$ and Proposition3 into $2\pi i(E(0)-I)$ , it is found that

(3.25) \begin{align} &\int _{\Sigma ^{E}}\dfrac {\left ( I+\varpi (s)\right ) (V^{E}-I)}{s}ds\nonumber \\[3pt] &=\int _{\partial U}\dfrac { M^{mod1}(s)(M^{j,\pm }(s)-I)M^{mod1}(s)^{-1}}{s}ds+\mathcal{O}(t^{-1})\nonumber \\[3pt] &=t^{-1/2}\int _{\partial U}\dfrac { M^{mod1}(s)A_{j,\pm }(\xi )M^{mod1}(s)^{-1}}{s(z\mp \lambda _j)}ds+\mathcal{O}(t^{-1}). \end{align}

Then by residue theorem we finally arrive at the result.

5.1. Constructing the $g$ -function

To construct the g-function, we first introduce:

(5.1) \begin{align} Y(z) \,:\!=\, Y(z;\,z_0,c)=\left [\dfrac {z^2-z_0^2}{(z^2-1)(z^2-c^2)} \right ]^{1/2}, \end{align}

where $z_0\in (1,c)$ . Its branch cut is

(5.2) \begin{align} \Sigma _{mod}=[{-}c,-z_0]\cup \Sigma _-\cup [z_0,c]. \end{align}

and the branch of the square root is chosen such that $Y_+(z)\in i\mathbb{R}^+$ for $z\in [z_0,c]$ . And d $g$ is the derivative of g-function given as follows

(5.3) \begin{align} \text{d}g=\dfrac { Y(z)}{z^3}\left [ \frac {1-\xi }{2}z^4-\frac {c}{z_0}\left ( 1+\frac {1}{c^2}-\frac {1}{z_0^2}\right ) z^2+\frac {2c}{z_0}\right ] \text{d}z. \end{align}

Here, $\text{d}g$ is a meromorphic differential defined on the 2-genus Riemann surface $\mathcal{M}$ , which has real branch points $\pm 1$ , $\pm c$ and $\pm z_0$ with $1\lt z_0\lt c$ . And the canonical homology basis $\left \lbrace a_j,b_j \right \rbrace _{j=1}^2$ is shown in Figure 7. Simply calculation shows that

\begin{align*} &\partial _zg-\partial _zp^{(-)}=\mathcal{O}( z^{-2}),\text{ as }z\to \infty ;\ \partial _zg-\partial _zp^{(-)}=\mathcal{O}(z),\text{ as }z\to 0\in \mathbb{C}^+. \end{align*}

Thus the g-function is given by

(5.4) \begin{align} g(z) \,:\!=\, g(z;\,\xi, c)=\int _{c}^{z}\text{d}g,\ \ z\in \mathbb{C}\setminus \Sigma _{mod}. \end{align}

Figure 7.

The canonical homology basis $\lbrace a_j,b_j \rbrace _{j=1}^2$ of the genius 2 Riemann surface.

Proof. First, we give the existence of $z_0$ . From the symmetry of d $g$ , it is accomplished that $a_2$ -period of $g(z)$ is zero. Rewrite the function $ Y(z)$ as $Y(z;\,z_0,c)$ . Let $F(s) \,:\!=\, F(s;\,\xi, c)$ be a function defined on $\mathbb{R}$ with

\begin{align*} F(s;\,\xi, c)=\int _{s}^{c}\frac {Y_+(z;s,c)}{z^3}\left [ \frac {\xi -1}{2}z^4+\frac {c}{s}\left ( 1+\frac {1}{c^2}-\frac {1}{s^2}\right ) z^2-\frac {2c}{s}\right ] \text{d}z. \end{align*}

Then it follows that $F(c)=0$ and

\begin{align*}F(1)=\int _{1}^{c}\frac {1}{z^3}\left [ \left (z^2-c^2 \right )^{-1/2}\right ] _+\left ( \frac {\xi -1}{2}z^4+\frac {z^2}{c}-2c\right ) \text{d}z=-\theta ^{(+)}_+(1),\end{align*}

with $\theta ^{(+)}$ defined in (4.1). And we calculate the $s$ -derivative of $F$ at $s=c$ ,

\begin{align*} \partial _sF(c)=-\frac {i}{c^3}\left (c^2-1\right )^{-1/2}\left ( \frac {\xi -1}{2}c^4+c^2-2\right ). \end{align*}

In the case $\xi \lt 1$ , obviously, $F(1)\in i\mathbb{R}^-$ . Thus, when $c^2\gt 2$ , $\xi \gt -\frac {2(c^2-2)}{c^4}+1$ , $\partial _sF(c)\in i\mathbb{R}^-$ . And in the case $1\leq \xi \lt \frac {2}{c}+1$ , from the property of $\theta ^{(+)}$ in above section, we have that $F(1)\in i\mathbb{R}^+$ . While when $c^2\lt 2$ , $1+\frac {2(2-c^2)}{c^4}\lt \xi \lt \frac {2}{c}+1$ and $c^2\gt 2$ , $1\leq \xi \lt \frac {2}{c}+1$ , it is adduced that $\partial _sF(c)\in i\mathbb{R}^+$ . So there must exist $z_0\in (1,c)$ , such that $F(z_0)=0$ .

Moreover, there exists $z_1\in (z_0,c)$ such that $f(z_1^2)=0$ with

\begin{align*}f(x)=\frac {\xi -1}{2}x^2+\frac {c}{z_0}\left ( 1+\frac {1}{c^2}-\frac {1}{z_0^2}\right ) x-\frac {2c}{z_0}.\end{align*}

By simply calculating the $a_1$ -period of $g(z)$ is zero and the both $b$ -period are real. Obviously, $f(0)\lt 0$ . So in the $\xi \gt 1$ case, $f(x)$ only has one zero $z_1$ on $\mathbb{R}^+$ . And in the $\xi \lt 1$ case, we denote another real solution of $f(z)=0$ as $z_2^2$ .

In addition, in the $\xi \lt 1$ case, simple calculation gives that

\begin{align*} &\partial _s F(s;\,\xi )=-\int _{s}^{c}\frac {z(1-\xi )}{2\sqrt {(z^2-1)(z^2-c^2)(z^2-s^2)}s^3}(s^2-z_1^2)(s^2-z_2^2)dz,\\[3pt] &\partial _{ ( 1-\xi )/2} F(s;\,\xi )=\int _{s}^{c}\frac {z\sqrt {z^2-s^2}}{\sqrt {(z^2-1)(z^2-c^2)}}dz. \end{align*}

Thus, when $\xi$ decreases from $1$ , $z_0(\xi )$ increases in $(1,c)$ while $z_2$ as a solution of $f(x)=0$ decreases. When $z_2$ merges with $c$ , we denote this critical condition as $\xi _m$ .

Denote constant

(5.5) \begin{align} g(\infty )=\lim _{z\to \infty }g(z)-p^{(-)}(z). \end{align}

Next, because $g$ will have different sign table in $\xi \gt 1$ and $\xi \lt 1$ , we will discuss $g$ -function separately in these two cases.

5.2. Opening the jump in the region $1\leq \xi \lt +\infty$

In this region, we give the signature table of Im $[g]$ in Figure 8. Let the open domains $\Omega ^\pm _j$ be as in Figure 8. Now we use $g$ to replace $p^{(-)}$ in the exponential function. In this region of $(\xi, c)$ , we introduce a piecewise matrix interpolation function $ G(z) \,:\!=\, G (z;\,\xi, c)$ with

(5.6) \begin{align} \begin{aligned} G(z)=&\left (\begin{array}{c@{\quad}c} 1 &0\\[3pt] \frac {r_1e^{2itg}}{1-r_1r_2} & 1 \end{array}\right ), \text{ } z\in \Omega _2;\quad G(z)= \left (\begin{array}{c@{\quad}c} 1 &\frac { r_2e^{-2itg}}{1-r_1r_2}\\[3pt] 0 & 1 \end{array}\right ), \text{ } z\in \Omega _2^*;\\[3pt] G(z)=&\left (\begin{array}{c@{\quad}c} 1 & -r_2e^{-2itg}\\[3pt] 0 & 1 \end{array}\right ), \text{ } z\in \Omega _1;\quad G(z)= \left (\begin{array}{c@{\quad}c} 1 & 0\\[3pt] -r_1e^{2itg} & 1 \end{array}\right ), \text{ } z\in \Omega _1^*;\\[3pt] G(z)&=I, \text{ } z \text{ in elsewhere}. \end{aligned} \end{align}

Same as above section, $ G (z)$ brings a new $-\frac {1}{4}$ -singularity on $z=\pm c$ . To open the jump contour $\mathbb{R}$ , we define

\begin{align*} \Sigma _2=&\left \lbrace z= -z_1+e^{\frac {3\pi i}{4}}\mathbb{R}^+\right \rbrace \cup \left \lbrace z= z_1+e^{\frac {\pi i}{4}}\mathbb{R}^+\right \rbrace, \\[3pt] \Sigma _1=&\left \lbrace z=-z_0+e^{\psi i}l,\ l\in (0,\frac {z_0-1}{2\cos \psi })\right \rbrace \cup \left \lbrace z=z_0+e^{(\pi -\psi ) i}l,\ l\in (0,\frac {z_0-1}{2\cos \psi })\right \rbrace \\[3pt] &\cup \left \lbrace z=1+e^{\psi i}l,\ l\in (0,\frac {z_0-1}{2\cos \psi })\right \rbrace \cup \left \lbrace z=-1+e^{(\pi -\psi ) i}l,\ l\in (0,\frac {z_0-1}{2\cos \psi })\right \rbrace, \end{align*}

where $ \psi \lt \pi /4$ is chosen as a small enough positive constant such that $\Sigma _1$ is contained in the region of Im $[g]\gt 0$ .

Figure 8.

The opened jump contours $\Sigma ^{(1)}$ and opened domains $\Omega _j\cup \Omega _j^*$ , $j=1,2$ . The yellow region means Im $[g]\lt 0$ , while white region means Im $[g]\gt 0$ .

We define a new matrix-valued function $M^{(1)}(z)$ ,

(5.7) \begin{equation} M^{(1)}(z) \,:\!=\, M^{(1)}(z;\,\xi, c)= e^{itg(\infty )\sigma _3}Ne^{it(p^{(-)}-g)\sigma _3} G, \end{equation}

which satisfies the following RH problem.

To deal with the jump on $\mathbb{R}$ , we define $Y_3(z)= (z^2-1)(z^2-c^2)Y(z),$ and $\delta (z;\,\xi, c) \,:\!=\, \delta (z)$ with

(5.8) \begin{align} \log \delta (z)=&\frac {Y_3(z)}{2\pi i}\left [ \sum _{\pm }\mp \int _{\pm c}^{\pm z_0} \dfrac {\log\!(ir_{2,-}(s))-w_\pm }{(s-z)Y_{3,+}(s)}ds +\int _{\mathbb{R}\setminus \Sigma _+}\dfrac {\log\!(1-r_1(s)r_2(s))}{(s-z)Y_3(s)}ds\right ], \end{align}

where $w_\pm$ satisfy linear system as

\begin{align*} \left(\begin{array}{l@{\quad}l} \int _{-c}^{-z_0}\dfrac {ds}{Y_{3,+}(s)}&\int _{z_0}^{c}\dfrac {ds}{Y_{3,+}(s)}\\[3pt] \int _{-c}^{-z_0}\dfrac {sds}{Y_{3,+}(s)}&\int _{z_0}^{c}\dfrac {sds}{Y_{3,+}(s)}\\[3pt] \end{array} \right)\begin{pmatrix} w_-\\[3pt]w_+\\[3pt] \end{pmatrix}=\begin{pmatrix} \int _{\Sigma _+\setminus [{-}z_0,z_0]} \dfrac {\log\!(ir_{2,-}(s))}{Y_{3,+}(s)}ds+\int _{\mathbb{R}\setminus \Sigma _+}\dfrac {\log\!(1-r_1(s)r_2(s))}{Y_3(s)}ds\\[3pt] \int _{\Sigma _+\setminus [{-}z_0,z_0]} \dfrac {s\log\!(ir_{2,-}(s))}{Y_{3,+}(s)}ds+\int _{\mathbb{R}\setminus \Sigma _+}\dfrac {s\log\!(1-r_1(s)r_2(s))}{Y_3(s)}ds\\[3pt] \end{pmatrix}. \end{align*}

$\delta (z)$ admits the following jump condition:

\begin{align*} &\delta _+(z)=\delta _-(z)(1-r_1(z)r_2(z)), \qquad z\in \mathbb{R}\setminus \Sigma _+;\\[3pt] &\delta _-(z)\delta _+(z)=ir_{2,-}(z)e^{-w_\pm }, \qquad \qquad z\in \mp [\pm c, \pm z_0];\\[3pt] &\delta _-(z)\delta _+(z)=1, \qquad \qquad z\in \Sigma _-. \end{align*}

By a similar way to Proposition5, we obtain the following proposition

By using $\delta (z)$ in (5.8), we define a new matrix function

(5.9) \begin{align} M^{(2)}(z;\,\xi, c) \,:\!=\, \delta _\infty (\xi, c)^{-\sigma _3}M^{(1)}(z;\,\xi, c)\delta (z;\,\xi, c)^{\sigma _3}, \end{align}

which then satisfies the following RH problem.

For $z\in \Sigma ^{(2)}\setminus \mathbb{R}$ , the jump $V^{(2)}(z)$ exponentially approaches the identity matrix as $t\to \infty$ . So we expect to only consider the jump on $\mathbb{R}$ . To arrive at this goal, in this case, we denote $ U \,:\!=\, U(\xi, c)$ as the union set of neighbourhood of $\pm z_0$ :

(5.10) \begin{equation} U=U_{+z_0}\cup U_{-z_0},\ U_{\pm z_0}= \left \lbrace z\,:\,|z\mp z_0|\leq \varrho \right \rbrace, \end{equation}

where $\varrho$ is a small positive constant such that $\varrho \lt \min \left \lbrace \frac {z_0-1}{3}, \frac {z_1-z_0}{3} \right \rbrace$ . The stationary point $\pm z_1$ is on the cut with Im $[g_+](z_1)\lt 0$ , which means that the exponential function in $V^{(2)}(z)$ also decays exponentially on $\pm z_1$ . In fact, it is also decays exponentially on $(z_0,z_1]\cup [{-}z_1,z_0)$ . So the contribution of $\pm z_1$ is small as $t\to \infty$ .

Thus, the jump matrix $V^{(2)}(z)$ uniformly goes to $I$ on $\Sigma ^{(2)}\setminus U$ . So outside $U$ there is only exponentially small error (in $t$ ) by completely ignoring the jump condition of $M^{(2)}(z)$ . It enlightens us to construct the solution $M^{(2)}(z)$ as follow

(5.11) \begin{equation} M^{(2)}(z)=\left \{\begin{array}{l@{\quad}l} E(z;\,\xi, c)M^{mod}(z;\,\xi, c), & z\notin U_{\pm z_0},\\[3pt] E(z;\,\xi, c)M^{lo,+}(z;\,\xi, c), &z\in U_{+z_0},\\[3pt] E(z;\,\xi, c)M^{lo,-}(z;\,\xi, c), &z\in U_{-z_0},\\[3pt] \end{array}\right . \end{equation}

where $M^{mod}(z)$ is the model RH problem on the Riemann surface, which solution is given by theta function in Subsection 5.2.1. $M^{lo,\pm }(z)$ are local model of $\pm z_0$ which solution can be expressed in terms of Airy functions shown in Subsection 5.2.2. And $E(z;\,\xi, c)$ is the error function, which will be discussed in subsection 5.2.3 by the small-norm RH problem theory.

5.2.1. Model RH problem on Riemann surface

We consider the following model RH problem with its jump matrix on $\mathbb{R}$ .

RH problem 7.

1. 1. Analyticity: $M^{mod}(z)$ is analytical in $\mathbb{C}\setminus \Sigma _{mod}$ , where $\Sigma _{mod}$ is given in (5.2);

2. 2. Asymptotic behaviours: $M^{mod}(z) \sim I+\mathcal{O}(z^{-1}),\quad |z| \rightarrow \infty$ ;

3. 3. Jump condition: $M^{mod}(z)$ satisfies the jump relation

   \begin{align*}M^{mod}_+(z)=M^{mod}_-(z)V^{mod}(z), \ \ \ z\in \Sigma _{mod},\end{align*}
   where the jump matrix $V^{mod}(z)$ is given by
   (5.12) \begin{equation} V^{mod}(z)=\left \{ \begin{array}{l@{\quad}l} \left (\begin{array}{c@{\quad}c} 0 & -ie^{-itB_2+w_-}\\[3pt] -ie^{itB_2-w_-} & 0 \end{array}\right ), &\text{as } z\in [{-}c,-z_0], \\[12pt] \left (\begin{array}{c@{\quad}c} 0 & -ie^{-itB_1}\\[3pt] -ie^{itB_1} & 0 \end{array}\right ), &\text{as } z\in \Sigma _-,\\[12pt] \left (\begin{array}{c@{\quad}c} 0 & -ie^{w_+}\\[3pt] -ie^{-w_+} & 0 \end{array}\right ), &\text{as } z\in [z_0,c]. \end{array}\right . \end{equation}

4. 4. Singularity: $M^{mod}(z)$ has at most fourth root singularities at $z=\pm c,\ \pm 1,\ \pm z_0$ .

The solution $M^{mod}$ of the model RH problem can be specifically characterised by $\Theta$ function on the Riemann surface with genus-2. We define

\begin{align*} \mathcal{N}_1(z)=\frac {1}{2} \left (\kappa (z)+\kappa (z)^{-1}\right ),\ \mathcal{N}_2(z)= \frac {1}{2}\left (\kappa (z)^{-1}-\kappa (z)\right ), \end{align*}

where $\kappa (z)$ is analytic function for $z\in \mathbb{C}\setminus \Sigma _{mod},$

\begin{align*} \kappa (z)&=\left [ \frac {(z-c)(z-1)(z+z_0)}{(z-z_0)(z+1)(z+c)}\right ] ^{\frac {1}{4}}, \ \ \end{align*}

and its branch is fixed by requiring that $\kappa (z)=1+\mathcal{O}(z^{-2}), \ \ z\to \infty .$ Let $\omega _i,\ i=1,2$ be the standard holomorphic differentials on the genus 2 Riemann surface $\mathcal{M}$ such that $\int _{a_i}\omega _j=\delta _{ij}$ , $i,j=1,2$ . And denote its $b$ -period matrix $\tilde {B}=[\tilde {B}_{ij}]_{i,j=1,2}\in GL_2(\mathbb{C})$ , with $\tilde {B}_{ij}=\oint _{b_j}\omega _i$ , $i,j=1,2$ . Define the Abel map

\begin{align*} \mathcal{A}\,:\,\mathcal{M}&\to \mathbb{C}^2/\tilde {B}M+N, \ \ M, N\in \mathbb{Z}^2,\\[3pt] P&\mapsto \left (\int _{1}^{P}\omega _1, \ \int _{1}^{P}\omega _2\right )^T. \end{align*}

The theta function $\Theta$ associated to $\tilde {B}$ is defined by

\begin{align*} \Theta (\vec {u})=\sum _{\vec {l}\in \mathbb{Z}^g}\exp (\pi i\lt \tilde {B}\vec {l},\vec {l}\gt +2\pi i\lt \vec {l},\vec {u}\gt ) \end{align*}

with

\begin{align*} \Theta (\vec {u}\pm e_j)&= \Theta (\vec {u}),\ \Theta (\vec {u}\pm \tilde {B}e_j)=\exp (\mp 2\pi iu_j-\pi i\tilde {B}_{jj})\Theta (\vec {u}). \end{align*}

Let $C=(-\frac {tB_1-w_+}{2\pi },-\frac {tB_2+i(w_{-}-w_+)}{2\pi })^T\in \mathbb{C}^2$ be a column vector. According to [Reference Fay23], there exists a constant $\mathcal{K}\in \mathbb{C}^2$ such that $\Theta (\mathcal{A}(P)-\mathcal{K}),\ \Theta (\mathcal{A}(P)+\mathcal{K})$ have the same zeros as $\mathcal{N}_1(z)$ and $\mathcal{N}_2(z)$ , respectively. Define

(5.13) \begin{align} M^{mod}(z)= e^{\frac {w_+}{2}\hat {\sigma }_3} \left (\begin{array}{c@{\quad}c} N^{mod}_{11}(\infty )^{-1}\mathcal{N}_{1}N^{mod}_{11}(z)& N^{mod}_{11}(\infty )^{-1}\mathcal{N}_{2}N^{mod}_{12}(z)\\[10pt] N^{mod}_{22}(\infty )^{-1}\mathcal{N}_{2}N^{mod}_{21}(z)& N^{mod}_{22}(\infty )^{-1}\mathcal{N}_{1}N^{mod}_{22}(z)\\[3pt] \end{array}\right ), \end{align}

which is the solution of RH problem7, where

\begin{align*} N^{mod}(z)= \left (\begin{array}{c@{\quad}c} \frac {\Theta (\mathcal{A}(z)-\mathcal{K}+C)}{\Theta (\mathcal{A}(z)-\mathcal{K})}& \frac {\Theta (-\mathcal{A}(z)-\mathcal{K}+C)}{\Theta (-\mathcal{A}(z)-\mathcal{K})}\\[10pt] \frac {\Theta (\mathcal{A}(z)+\mathcal{K}+C)}{\Theta (\mathcal{A}(z)+\mathcal{K})}& \frac {\Theta (-\mathcal{A}(z)+\mathcal{K}+C)}{\Theta (-\mathcal{A}(z)+\mathcal{K})}\\[3pt] \end{array}\right ),\ N^{mod}(\infty )=\lim _{z\to \infty } N^{mod}(z). \end{align*}

As $z\to 0\in \mathbb{C}^+$ , it follows that

\begin{align*} M^{mod}(z)=M^{mod}_+(0)+M^{mod}_1z+\mathcal{O}(z^2), \end{align*}

where

\begin{align*} &M^{mod}_+(0)= \frac {\sqrt {2}}{2} e^{\frac {w_+}{2}\hat {\sigma }_3}\left (\begin{array}{c@{\quad}c} N^{mod}_{11}(\infty )^{-1}N^{mod}_{11,+}(0)& -iN^{mod}_{11}(\infty )^{-1}N^{mod}_{12,+}(0)\\[10pt] -iN^{mod}_{22}(\infty )^{-1}N^{mod}_{21,+}(0)& N^{mod}_{22}(\infty )^{-1}N^{mod}_{22,+}(0)\\[3pt] \end{array}\right ),\\[3pt] &M^{mod}_1=\frac {\sqrt {2}}{2} \left (\begin{array}{c@{\quad}c}N^{mod}_{11}(\infty )^{-1}& 0\\[10pt] 0& N^{mod}_{22}(\infty )^{-1}\\[3pt] \end{array}\right )\\[3pt] & e^{\frac {w_+}{2}\hat {\sigma }_3}\left [\left (\begin{array}{c@{\quad}c} \partial _zN^{mod}_{11,+}(0)& -i \partial _zN^{mod}_{12,+}(0)\\[10pt] -i \partial _zN^{mod}_{21,+}(0)& \partial _zN^{mod}_{22,+}(0)\\[3pt] \end{array}\right )+ \left(1-\frac {1}{z_0}+\frac {1}{c} \right) \left (\begin{array}{c@{\quad}c} -iN^{mod}_{11,+}(0)& N^{mod}_{12,+}(0)\\[10pt] N^{mod}_{21,+}(0)& -iN^{mod}_{22,+}(0)\\[3pt] \end{array}\right )\right ]. \end{align*}

5.2.2. Localised RH problem near stationary points

Using local approximations, appropriate error estimates as well as higher-order asymptotics beyond the $\mathcal{O}(1)$ term can be derived. In this subsection, we only give the details of the model around $z_0$ . We consider $M^{lo,+}(z)$ here as an example. First, we denote $P(z_0)$ as the neighbourhood of $z_0$ in the Riemann surface $\mathcal{M}$ corresponding to $dg$ . It is an analytic homeomorphism. It follows form the definition of d $g$ in (5.1)–(5.3) that there exist a holomorphic function $f_+$ (we hope the subscript $\pm$ indicates $\pm z_0$ ) on $P(z_0)$ such that

\begin{align*} g=-\frac {2i}{3}f_+^3. \end{align*}

Here, on the complex plane, because $g_+(z) \subset i\mathbb{R}^-$ when $z\in \left ( z_0,c\right )$ , we can choose $f_+\in \mathbb{R}^+$ on $\left ( z_0,c\right ) \cap U_{+z_0}$ . Let

(5.14) \begin{equation} \lambda _+=t^{\frac {2}{3}}f_+^2. \end{equation}

Because of $g_+(z)=-g_-(z)$ , $\lambda _+ \,:\!=\, \lambda _+(z)$ is a holomorphic homeomorphism from $U_{+z_0}$ to a neighbourhood of zero and

\begin{align*} \frac {4}{3}\lambda _+^{\frac {3}{2}}=2itg_+, \ \ z\in \left ( z_0,c\right ) \cap U_{+z_0}, \end{align*}

where the branch cut of $(\cdot )^{\frac {3}{2}}$ is chosen same as in the Airy model in the Appendix A.

From the definition of $\lambda _+$ , it follows that

\begin{eqnarray*} \frac {4}{3}\lambda _+^{\frac {3}{2}}=2itg,\ \ \text{Im}z\gt 0,\ \ \ \frac {4}{3}\lambda _+^{\frac {3}{2}}=-2itg,\ \ \text{Im}z\lt 0. \end{eqnarray*}

Then we explicitly define $M^{lo,+}(z)$ which should have the same jump condition with $M^{(2)}(z)$ locally in $U_{+z_0}$ as follow via the Airy parametrix. Similar to [Reference de Monvel, Lenells and Shepelsky6, Reference Fromm, Lenells and Quirchmayr25, Reference Grava and Minakov29], it is shown that

(5.15) \begin{align} M^{lo,+}(z)=\frac {\sqrt {2}}{2}M^{mod}H(z;\,z_0)^{-1}\left (I-i\sigma _1\right )\lambda _+^{\frac {\sigma _3}{4}}m^{Ai}(\lambda _+)H(z;\,z_0), \end{align}

where

\begin{align*} H(z;\,z_0)=\left \{ \begin{array}{l@{\quad}l} \delta ^{\sigma _3}r_1^{\frac {\sigma _3}{2}},\quad z-z_0\in \mathbb{C}^+\cap U_{+z_0},\\[3pt] \sigma _3\sigma _1\delta ^{\sigma _3}r_2^{-\frac {\sigma _3}{2}}, \ z-z_0\in \mathbb{C}^-\cap U_{+z_0}. \end{array} \right . \end{align*}

Moreover, for convenience, denote

(5.16) \begin{align} F^+(z) \,:\!=\, F^+(z;\,\xi, c)=\frac {\sqrt {2}}{2}M^{mod}(z;\,\xi, c)H(z;\pm z_0)^{-1}\left (I-i\sigma _1\right )f_+^{\frac {1}{2}\sigma _3}, \end{align}

which thereby is an analytic and invertible function in $U_{+z_0}$ . Similarly,

(5.17) \begin{align} M^{lo,-}(z)=\frac {\sqrt {2}}{2}M^{mod}H(z;-z_0)^{-1}\left (I-i\sigma _1\right )\lambda _-^{\frac {\sigma _3}{4}}m^{Ai}(\lambda _-)H(z;-z_0). \end{align}

with

\begin{align*} &H(z;-z_0)=\left \{ \begin{array}{l@{\quad}l} e^{\frac {itB_2}{2}\sigma _3}\sigma _3\delta ^{\sigma _3}r_1^{\frac {\sigma _3}{2}},& z-z_0\in \mathbb{C}^+\cap U_{-z_0},\\[3pt] e^{\frac {itB_2}{2}\sigma _3}\sigma _1\delta ^{\sigma _3}r_2^{-\frac {\sigma _3}{2}},& z-z_0\in \mathbb{C}^-\cap U_{-z_0}, \end{array} \right .\\[3pt] &g=-\frac {2i}{3}f_-^3+\frac {B_2}{2},\qquad \lambda _-=t^{\frac {2}{3}}f_-^2, \end{align*}

where $f_-\in \mathbb{R}+$ on $z\in (-c,-z_0)\cap U_{-z_0}$ and

(5.18) \begin{align} F^- \,:\!=\, M^{mod}H(z;-z_0)^{-1}\left (I-i\sigma _1\right )\lambda _-^{\frac {\sigma _3}{4}} \end{align}

is an analytic and invertible function in $U_{-z_0}$ . Moreover, as $z\to z_0$ in $\mathbb{C}\setminus [z_0,c]$ , $\lambda _+$ has expansion

\begin{align*} \lambda _+=(\tilde {A}_+)^{\frac {2}{3}}(z-z_0)+\frac {2\tilde {B}_+}{3\tilde {A}_+^{\frac {1}{3}}}(z-z_0)^2+\mathcal{O}((z-z_0)^3),\\[3pt] \lambda _-=(\tilde {A}_-)^{\frac {2}{3}}(z+z_0)+\frac {2\tilde {B}_-}{3\tilde {A}_-^{\frac {1}{3}}}(z+z_0)^2+\mathcal{O}((z+z_0)^3), \end{align*}

where

(5.19) \begin{align} \tilde {A}_+=&\frac {t}{2} \left (\frac {2z_0}{(z_0^2-1)(c^2-z_0^2)}\right )^{\frac {1}{2}}\left ( \frac {\xi -1}{2}z_0+\frac {c}{z_0^2}(1+\frac {1}{c^2})-\frac {3c}{z_0^4}\right ),\nonumber \\[6pt]\tilde {B}_+=&\frac {3t}{10}\left [{-}\frac {1}{2}(\frac {2z_0}{(z_0^2-1)(c^2-z_0^2)})^{-\frac {1}{2}}\frac {-7z_0^4+3(1+c^2)z_0^2+c^2}{(z_0^2-1)^2(c^2-z_0^2)^2} \left (\frac {\xi -1}{2}z_0+\frac {c}{z_0^2}\left(1+\frac {1}{c^2} \right)-\frac {3c}{z_0^4}\right )\right .\nonumber \\[4pt]&\left .+\left(\frac {2z_0}{(z_0^2-1)(c^2-z_0^2)}\right)^{\frac {1}{2}}\left(\frac {\xi -1}{2}-\frac {c}{z_0^3}\left(1+\frac {1}{c^2}-\frac {1}{z_0^2}\right)+\frac {6c}{z_0^5}\right)\right ],\nonumber \\[6pt]\tilde {A}_-=&-\frac {it}{2}\left (\frac {2z_0}{(z_0^2-1)(c^2-z_0^2)}\right )^{\frac {1}{2}}\left (-\frac {\xi -1}{2}z_0+\frac {c}{z_0^2}\left(1+\frac {1}{c^2}\right)-\frac {3c}{z_0^4}\right ),\nonumber \\[4pt]\tilde {B}_-=&\frac {3t}{10}\left [{-}\frac {i}{2}\left(\frac {2z_0}{(z_0^2-1)(c^2-z_0^2)}\right)^{-\frac {1}{2}}\frac {-7z_0^4+3(1+c^2)z_0^2+c^2}{(z_0^2-1)^2(c^2-z_0^2)^2} \left(-\frac {\xi -1}{2}z_0+\frac {c}{z_0^2}\left(1+\frac {1}{c^2} \right)-\frac {3c}{z_0^4} \right)\right .\nonumber \\[4pt] &\left .-i\left(\frac {2z_0}{(z_0^2-1)(c^2-z_0^2)}\right)^{\frac {1}{2}}\left(\frac {\xi -1}{2}+\frac {c}{z_0^3}\left(1+\frac {1}{c^2}-\frac {1}{z_0^2}\right)-\frac {6c}{z_0^5}\right)\right ]. \end{align}

5.2.3. The small norm RH problem for error function

In this subsection, we consider the error matrix-function $E(z) \,:\!=\, E(z;\,\xi, c)$ in this region.

RH problem 8.

1. 1. Analyticity: $E(z)$ is analytical in $\mathbb{C}\setminus \Sigma ^{ E }$ , where

   \begin{align*}\Sigma ^{ E }= \partial U\cup \left [ \Sigma ^{(2)}\setminus \left ( U\cup [{-}c,-z_1]\cup \Sigma _-\cup [z_1,c]\right ) \right ] \end{align*}
   with $U$ defined in (5.10);

2. 2. Asymptotic behaviours: $E(z) \sim I+\mathcal{O}(z^{-1}),\qquad |z| \rightarrow \infty$ ;

3. 3. Jump condition: $E(z)$ has continuous boundary values $E_\pm (z)$ on $\Sigma ^{E}$ satisfying $E_+(z)=E_-(z)V^{ E }(z),$ where the jump matrix $V^{E}(z)$ is given by

   (5.20) \begin{equation} V^{ E }(z)=\left \{\begin{array}{l@{\quad}l@{\quad}l@{\quad}l} M^{mod}(z)V^{(2)}(z)M^{mod}(z)^{-1}, & z\in \Sigma ^{ E }\setminus \partial U,\\[4pt] M^{lo,\pm }(z)M^{mod}(z)^{-1}, & z\in \partial U_{\pm z_0}, \end{array}\right . \end{equation}
   which is shown in Figure 9 .

By (5.15), (5.17) and (A.5), the jump matrix of above RH problem satisfies

\begin{align*} \parallel V^{E}(z)-I\parallel _2 \lesssim \mathcal{O}(t^{-1}). \end{align*}

Similar to the discussion in Subsection 3.2, the RH problem8 admits a unique solution given by

(5.21) \begin{equation} E(z)=I+\frac {1}{2\pi i}\int _{\Sigma ^{ E }}\dfrac {\left ( I+\varpi (s)\right ) (V^{E}(s)-I)}{s-z}ds, \end{equation}

where the $\varpi \in L^\infty (\Sigma ^{ E })$ is the unique solution of $ (1-C_E)\varpi =C_E\left (I \right ).$

Figure 9.

The jump contour $\Sigma ^{E}$ for the $E(z)$ . The red circles are $\partial U$ .

In order to reconstruct the solution $u(y,t)$ of (1.1), we need the asymptotic behaviour of $E(z)$ as $z\to 0\in \mathbb{C}^+$ and the long-time asymptotic behaviour of $E(0)$ .

Proposition 8. As $z\to 0\in \mathbb{C}^+$ , we have

\begin{align*} E(z)=E(0)+E_1z+\mathcal{O}(z^2), \end{align*}

with long-time asymptotic behaviour

(5.22) \begin{equation} E(0)=I+t^{-1}H^{(0)}+\mathcal{O}(t^{-2}). \end{equation}

And

\begin{align*} H^{(0)} \,:\!=\, H^{(0)}(\xi, c)=&\sum _{ p=\pm z_0 }\partial _z\left (\frac {1 }{z}F^{\pm }\left (\begin{array}{c@{\quad}c} 0&-\frac {5i}{48}\tilde {A}_\pm ^{-\frac {4}{3}}\\[3pt] 0&0\\[3pt] \end{array}\right )(F^\pm )^{-1}\right )(p)\\[3pt] &+\sum _{ p=\pm z_0 }\frac {1 }{p}\left ( F^{\pm }\left (\begin{array}{c@{\quad}c} 0&\frac {5}{36}\tilde {B}_\pm \tilde {A}^{-\frac {7}{3}}_\pm \\[3pt] -\frac {7i}{48}\tilde {A}^{-\frac {2}{3}}_\pm &0\\[3pt] \end{array}\right )(F^\pm )^{-1}\right )(p). \end{align*}

Here, $\tilde {B}_\pm$ , $\tilde {A}_\pm$ are given in (5.19). And

\begin{align*} E_1 \,:\!=\, E_1(\xi, c)=\frac {1}{2\pi i}\int _{\Sigma ^{E}}\dfrac {\left ( I+\varpi (s)\right ) (V^{ E }-I)}{s^2}ds, \end{align*}

satisfies long-time asymptotic behaviour condition

(5.23) \begin{equation} E_1=t^{-1}H^{(1)}+\mathcal{O}(t^{-2}), \end{equation}

where

\begin{align*} H^{(1)} \,:\!=\, H^{(1)}(\xi, c)=&\sum _{ p=\pm z_0 }\partial _z\left (\frac {1 }{z^2}F^{\pm }\left (\begin{array}{c@{\quad}c} 0&-\frac {5i}{48}\tilde {A}_\pm ^{-\frac {4}{3}}\\[3pt] 0&0\\[3pt] \end{array}\right )(F^\pm )^{-1}\right )(p)\\[3pt] &+\sum _{ p=\pm z_0 }\frac {1 }{p^2}\left ( F^{\pm }\left (\begin{array}{c@{\quad}c} 0&\frac {5}{36}\tilde {B}_\pm \tilde {A}^{-\frac {7}{3}}_\pm \\[3pt] -\frac {7i}{48}\tilde {A}^{-\frac {2}{3}}_\pm &0\\[3pt] \end{array}\right )(F^\pm )^{-1}\right )(p). \end{align*}

Proof. By using expansion of $V^{ E }$ and $\varpi (s)=\mathcal{O}(t^{-1})$ , we have

\begin{align*} &\int _{\Sigma ^{E}}\dfrac {\left ( I+\varpi (s)\right ) (V^{E}-I)}{s}ds=\frac {1}{t}\int _{\partial U_{\pm z_0}}\dfrac { M^{mod}H(s;\pm z_0)^{-1} m^{Ai}_1H(s;\pm z_0)(M^{mod})^{-1}}{sf_\pm ^{3}}ds+\mathcal{O}(t^{-2}). \end{align*}

From the definition of $F^\pm$ in (5.16) and (5.18), we rewrite

\begin{align*} M^{mod}H(s;\pm z_0)^{-1}& m^{Ai}_1H(s;\pm z_0)(M^{mod})^{-1}\\[3pt] &=\frac {1}{2}F^\pm f_\pm ^{-\frac {\sigma _3}{2}}\left (I+i\sigma _1\right )m^{Ai}_1\left (I-i\sigma _1\right )f_\pm ^{\frac {\sigma _3}{2}}(F^\pm )^{-1}. \end{align*}

Here, from (5.19) and $m^{Ai}_1$ in Appendix A, as $z\to z_0\in \mathbb{C}^+$ , we have

\begin{align*} &\frac {1}{2}f_\pm ^{-3}f_\pm ^{-\frac {\sigma _3}{2}}\left (I+i\sigma _1\right )m^{Ai}_1\left (I-i\sigma _1\right )f_\pm ^{\frac {\sigma _3}{2}}=\\[3pt] &\frac {1}{(z\mp z_0)^2}\left (\begin{array}{c@{\quad}c} 0&-\frac {5i}{48}\tilde {A}_\pm ^{-\frac {4}{3}}\\[3pt] 0&0\\[3pt] \end{array}\right )+\frac {1}{(z\mp z_0)} \left (\begin{array}{c@{\quad}c} 0&\frac {5}{36}\tilde {B}_\pm \tilde {A}^{-\frac {7}{3}}_\pm \\[3pt] -\frac {7i}{48}\tilde {A}^{-\frac {2}{3}}_\pm &0\\[3pt] \end{array}\right )+\mathcal{O}(1). \end{align*}

Then by residue theorem, we finally arrive at the result.

5.3. Opening the jump in the region $3/4\lt \xi \lt 1$

This region includes two cases:

1. (1) $\left \{ (\xi, c)\,:\,2\lt c^2\lt 4, 1-\frac {2(c^2-2)}{c^4} \lt \xi \lt 1\right \}$ ;

2. (2) $\left \{(\xi, c)\,:\,c^2\gt 4, \xi _m\lt \xi \lt 1\right \}$ .

In both cases we introduce the same $g$ function by

\begin{align*} \text{d}g=\dfrac { Y(z)}{z^3}\left [ \frac {1-\xi }{2}z^4-\frac {c}{z_0}\left ( 1+\frac {1}{c^2}-\frac {1}{z_0^2}\right ) z^2+\frac {2c}{z_0}\right ] \text{d}z \end{align*}

which will have another zero on $\mathbb{R}$ except on cut. It means that $g$ has three pairs of stationary points on $\mathbb{R}$ , which will gives additional contribution as $t\to \infty$ . The stationary points of $g$ are $\pm z_0$ and the zeros of the equation

\begin{align*} \frac {1-\xi }{2}z^4-\frac {c}{z_0}\left ( 1+\frac {1}{c^2}-\frac {1}{z_0^2}\right ) z^2+\frac {2c}{z_0}=0. \end{align*}

It has two pairs of zeros on $\mathbb{R}$ : $\pm z_1\in (z_0,c)$ , $\pm z_2$ . Note that, $z_1^2z_2^2=\frac {4c}{z_0(1-\xi )}$ . For a given $c$ , when $\xi$ decreases from $1$ , $z_2$ as a function of $\xi$ decreases from $+\infty$ . We denote $\xi _m$ as the critical condition of $\xi$ that stationary point $z_2$ merge $c$ . Under this case, the sign table of Im $g$ is shown in Figure 10.

Similar to the above section, we define the contour $\Sigma _j$ and closed region $\Omega _j$ relying on $(\xi, c)$ as in Figure 10. Here the angle of $\Sigma _j$ is a small enough positive constant such that $\Sigma _j,\ j=1,2$ are contained in the region of Im $g\gt 0$ .

Figure 10.

The domains $\Omega _j$ and curves $\Sigma _j$ , $j=1,2$ . The yellow region means Im $[g](z)\lt 0$ while white region means Im $[g](z)\gt 0$ .

In this region of $(\xi, c)$ , we introduce a piecewise matrix interpolation function

(5.24) \begin{equation} G(z) \,:\!=\, G(z;\,\xi, c)=\left \{ \begin{array}{l@{\quad}l} \left (\begin{array}{c@{\quad}c} 1 & -r_2e^{-2itg}\\[3pt] 0 & 1 \end{array}\right ), &\text{as } z\in \Omega _1;\\[12pt] \left (\begin{array}{c@{\quad}c} 1 & 0\\[3pt] -r_1e^{2itg} & 1 \end{array}\right ), &\text{as } z\in \Omega _1^*;\\[3pt] \left (\begin{array}{c@{\quad}c} 1 &0\\[3pt] \frac {r_1e^{2itg}}{1-r_1r_2} & 1 \end{array}\right ), &\text{as } z\in \Omega _2;\\[12pt] \left (\begin{array}{c@{\quad}c} 1 &\frac { r_2e^{-2itg}}{1-r_1r_2}\\[3pt] 0 & 1 \end{array}\right ), &\text{as } z\in \Omega _2^*;\\[12pt] I &\text{as } z \text{ in elsewhere}, \end{array}\right .. \end{equation}

Same as above section, $ G (z)$ brings a new singularity. To deal with the jump on $\mathbb{R}$ , we introduce an auxiliary function $\delta (z) \,:\!=\, \delta (z;\,\xi, c)$ . Define $Y_3(z)= (z^2-1)(z^2-c^2)Y(z),$ and

\begin{align*} \log \delta (z)=&\frac {Y_3(z)}{2\pi i}\sum _{\pm }\mp \int _{\pm c}^{\pm z_0} \dfrac {\log\!(ir_{2,-}(s))-w_\pm }{(s-z)Y_{3,+}(s)}ds +\frac {Y_3(z)}{2\pi i}\left ( \int _{-z_2}^{-c}+\int _{c}^{z_2}\right ) \dfrac {\log\!(1-r_1(s)r_2(s))}{(s-z)Y_3(s)}ds, \end{align*}

where

\begin{align*} \begin{pmatrix} \int _{-c}^{-z_0}\dfrac {ds}{Y_{3,+}(s)}&\int _{z_0}^{c}\dfrac {ds}{Y_{3,+}(s)}\\[3pt] \int _{-c}^{-z_0}\dfrac {sds}{Y_{3,+}(s)}&\int _{z_0}^{c}\dfrac {sds}{Y_{3,+}(s)}\\[3pt] \end{pmatrix}\begin{pmatrix} w_-\\[3pt]w_+\\[3pt] \end{pmatrix}=\begin{pmatrix} \int _{\Sigma _+\setminus [{-}z_0,z_0]} \dfrac {\log\!(ir_{2,-}(s))}{Y_{3,+}(s)}ds+\int _{[{-}z_2,z_2]\setminus \Sigma _+}\dfrac {\log\!(1-r_1(s)r_2(s))}{Y_3(s)}ds\\[3pt] \int _{\Sigma _+\setminus [{-}z_0,z_0]} \dfrac {s\log\!(ir_{2,-}(s))}{Y_{3,+}(s)}ds+\int _{[{-}z_2,z_2]\setminus \Sigma _+}\dfrac {s\log\!(1-r_1(s)r_2(s))}{Y_3(s)}ds\\[3pt] \end{pmatrix}. \end{align*}

$\delta (z)$ admits the following jump condition

\begin{align*} &\delta _+(z)=\delta _-(z)(1-r_1r_2), \qquad z\in [{-}z_2,z_2]\setminus \Sigma _+;\\[3pt] &\delta _-(z)\delta _+(z)=ir_{2,-}(z)e^{-w_\pm }, \quad z\in \mp [\pm c, \pm z_0];\\[3pt] &\delta _-(z)\delta _+(z)=1, \qquad \qquad z\in \Sigma _-. \end{align*}

and the following proposition

Proposition 9. The scalar function $\delta (z)$ satisfies the following properties

1. (a) $\delta (z)$ is analytic on $\mathbb{C}\setminus \left ( ({-}z_2,-z_0)\cup \Sigma _-\cup (z_0,z_2)\right )$ ;

2. (b) $\delta (z)$ has singularity at $z=\pm c$ with

   (5.25) \begin{align} \delta (z)=\mathcal{O}(z-p)^{\mp 1/4}, \quad z\in \mathbb{C}^\pm \setminus \mathbb{R} \to p, \quad p=c,-c. \end{align}

3. (c) As $z\to \infty \in \mathbb{C}\setminus \mathbb{R}$ , $\delta (z)\sim \delta _\infty$ with

   \begin{align*} \log \delta _\infty =&-\frac {1}{2\pi i}\int _{\Sigma _+\setminus [{-}z_0,z_0]} \dfrac {s^2\log\!(ir_{2,-}(s))}{Y_{3,+}(s)} ds-\frac {1}{2\pi i}\int _{[{-}z_2,z_2]\setminus \Sigma _+}\dfrac {s^2\log\!(1-r_1(s)r_2(s))}{Y_3(s)}ds. \end{align*}

4. (d) As $z\to 0\in \mathbb{C}^+$ ,

   (5.26) \begin{align} \delta (z)=&\delta _+(0)\left ( 1+\delta ^{(1)}z\right ) +\mathcal{O}(z^2), \end{align}
   where
   \begin{align*} & \delta ^{(1)}= \frac {cz_0}{2\pi }\left (\int _{-c}^{-z_0}+\int _{z_0}^{c} \right ) \dfrac {\log\!(ir_{2,-}(s))}{s^2Y_{3,+}(s)}ds +\frac {cz_0}{2\pi }\left ( \int _{-z_2}^{-c}+\int _{c}^{z_2}\right )\dfrac {\log\!(1-r_1(s)r_2(s))}{s^2Y_3(s)}ds; \end{align*}

5. (e) For $z_2$ , there exists an analytic function $\delta _{+z_2}(z)$ on $z\in U_{+z_2}\setminus (c,z_2)$ which is continuous to the boundary such that for $\nu (z_2)= \frac {1}{2\pi }\log\!(1-r_1(z_2)r_2(z_2) )$ ,

(5.27) \begin{align} \delta (z)= \delta _{\pm z_2}(z)(z-z_2)^{-i\nu (z_2)}, \quad \arg\!(z-z_2)\in (-\pi, \pi ), \end{align}

and

\begin{align*} \left |\delta _{+z_2}(z)-\delta _{+z_2}(z_2)\right |\lesssim \left |z-z_2\right |. \end{align*}

Through $\delta (z)$ and $G(z)$ , in this region of $(\xi, c)$ , same as above subsection, we give series of transformations:

\begin{align*} N\overset {(5.7)}{\longrightarrow } M^{(1)} \overset {(5.9)}{\longrightarrow }M^{(2)}=\delta _\infty ^{-\sigma _3}e^{itg(\infty )\sigma _3}Ne^{it(p-_-g)\sigma _3}\delta ^{\sigma _3}, \end{align*}

which then satisfies the following RH problem.

Except the cut away from $\mathbb{R}$ , the jump $ {V}^{(2)}(z)$ exponentially approaches the identity matrix as $t\to \infty$ . So we expect to only consider the jump on $\mathbb{R}$ . However, different from above section, in this region, $g$ has another pair of stationary points on $\mathbb{R}$ . So in this case, we denote $ U \,:\!=\, U(\xi, c)$ as the union set of neighbourhood of $\pm z_0$ and $\pm z_2$

(5.29) \begin{equation} U=U_{\pm z_0}\cup U_{\pm z_2},\ U_{\pm z_j}= \left \lbrace z\,:\,|z\mp z_j|\leq \varrho \right \rbrace, \ j=0,2. \end{equation}

Here, $\varrho$ is a small positive constant such that $\varrho \lt \min \left \lbrace \frac {z_0-1}{3}, \frac {z_1-z_0}{3},\frac {z_2-c}{3},\epsilon \right \rbrace$ . Outside $U$ there is only exponentially small error (in $t$ ) by completely ignoring the jump condition of $M^{(2)}(z)$ . This proposition enlightens us to construct the solution $M^{(2)}(z)$ as follow

\begin{align*} M^{(2)}(z)= \left \{\begin{array}{l@{\quad}l} E(z;\,\xi, c)M^{mod}(z;\,\xi, c), & z\notin U,\\[4pt] E(z;\,\xi, c)M^{0,\pm }(z;\,\xi, c), &z\in U_{\pm z_0},\\[4pt] E(z;\,\xi, c)M^{mod}(z;\,\xi, c)M^{2,\pm }(z;\,\xi, c), &z\in U_{\pm z_2},\\[3pt] \end{array}\right . \end{align*}

where same as $M^{mod}(z;\,\xi, c)$ is the model RH problem on the Riemann surface, which solution is given by theta function in Subsection 5.2.1. The difference is $M^{j,\pm }(z;\,\xi, c)$ are local model of $\pm z_j$ , $j=0,2$ . When $j=2$ , its solution can be expressed in terms of parabolic cylinder parametrix. When $j=0$ , same as above subsection, its solution can be expressed in terms of Airy parametrix shown in Subsection 5.2.2. Its contribution will be higher-order term of the local parametrix of $M^{2,\pm }(z;\,\xi, c)$

Similar to discussion of the Proposition3 in the Section 3, let

\begin{align*}\zeta ^{2,+}=2\sqrt {g''(\pm z_2)}(z\mp z_2),\ r_{+z_2}=r_2(+ z_2)\delta ^2_{+ z_2}(\pm z_2)e^{-2itg(\pm z_2)}(4tg''(\pm z_2))^{2i\nu (z_2)}.\end{align*}

One can find that $M^{2,+}(z)$ is well approximated by $P_1(\zeta ^{2,+};r_{z_2}^+)$ and $M^{2,+}(z)=\sigma _2M^{2,-}({-}z)\sigma _2^{-1}$ . Therefore, similar to the Proposition3, we give the approximation below.

Proposition 10. $M^{2,\pm }(z)$ admits the following asymptotic expansion

(5.30) \begin{align} M^{2,\pm }(z)=I+t^{-1/2} \frac {A_\pm (\xi )}{z\mp z_2} +\mathcal{O}(t^{-1}), \ \ t\to +\infty, \end{align}

where

(5.31) \begin{align} &A_\pm (\xi, c)=\frac {1}{2\sqrt {|g''(\pm z_2)|}} \left (\begin{array}{c@{\quad}c} 0 & \tilde {\beta }^\pm _{12}\\[3pt] \tilde {\beta }^\pm _{21} & 0 \end{array}\right ),\\[3pt] &\tilde {\beta }^-_{12}=\tilde {\beta }^+_{21}=\frac {\sqrt {2\pi }e^{\frac {\pi \nu (z_2)}{2}}e^{\frac {\pi i}{4} }}{r_{+z_2}\Gamma (i\nu (z_2))},\quad \tilde {\beta }^\pm _{21}\tilde {\beta }^\pm _{12}=-\nu (z_2),\nonumber \end{align}

where $\delta _{+z_2}$ and $\nu (z_2)$ are defined in (5.27).

5.3.1. The small norm RH problem for error function

In this subsection, we consider the error matrix-function $E(z) \,:\!=\, E(z;\,\xi, c)$ in this region.

RH problem 10.

1. 1. Analyticity: $E(z)$ is analytical in $\mathbb{C}\setminus \Sigma ^{E}$ , where

   \begin{align*}\Sigma ^{E}= \partial U\cup \left [ \Sigma ^{(2)}\setminus \left ( U\cup [{-}c,-z_1]\cup \Sigma _-\cup [z_1,c]\right ) \right ] ;\end{align*}

2. 2. Asymptotic behaviours: $E(z) \sim I+\mathcal{O}(z^{-1}),\qquad |z| \rightarrow \infty$ ;

3. 3. Jump condition: $E(z)$ has continuous limit $E_\pm (z)$ on $\Sigma ^{E}$ satisfying $E_+(z)=E_-(z)V^{E}(z),$ where the jump matrix $V^{E}(z)$ is given by

   (5.32) \begin{equation} V^{E}(z)=\left \{\begin{array}{l@{\quad}l@{\quad}l@{\quad}l} M^{mod}(z)V^{(2)}(z)M^{mod}(z)^{-1}, & z\in \Sigma ^{E}\setminus \partial U,\\[4pt] M^{lo,\pm }(z)M^{mod}(z)^{-1}, & z\in \partial U_{\pm z_0}, \\[4pt] M^{mod}(z)M^{2,\pm }(z)M^{mod}(z)^{-1}, & z\in \partial U_{\pm z_2}, \end{array}\right . \end{equation}
   which is shown in Figure 11 .

Figure 11.

The jump contour $\Sigma ^{E}$ for the $E(z)$ . The red circles are $\partial U$ .

Similar to the discussion in Section 3.2, the RH problem10 satisfies

(5.33) \begin{align} | V^{E}(z)-I|&= \mathcal{O}(t^{-1/2}). \end{align}

and admits a unique solution, which can be given by

(5.34) \begin{equation} E(z)=I+\frac {1}{2\pi i}\int _{\Sigma ^{ E }}\dfrac {\left ( I+\varpi (s)\right ) (V^{E}(s)-I)}{s-z}ds, \end{equation}

where the $\varpi \in L^\infty (\Sigma ^{ E })$ is the unique solution of the equation $ (1-C_E)\varpi =C_E\left (I \right ).$

In order to reconstruct the solution $u(y,t)$ of (1.1), we need the asymptotic behaviour of $E(z)$ as $z\to 0\in \mathbb{C}^+$ and the long-time asymptotic behaviour of $E(0)$ , which is obtained from Proposition4.

Proposition 11. As $z\to 0\in \mathbb{C}^+$ , we have

(5.35) \begin{align} E(z)=E(0)+E_1z+\mathcal{O}(z^2), \end{align}

with long-time asymptotic behaviour

(5.36) \begin{equation} E(0)=I+t^{-1/2}H^{(0)}+\mathcal{O}(t^{-2}). \end{equation}

And

(5.37) \begin{align} H^{(0)} \,:\!=\, H^{(0)}(\xi, c)=&\sum _{ p=\pm z_2 }\frac {M^{mod}(p)A_\pm (\xi )M^{mod}(p)^{-1}}{p}. \end{align}

Here, $\tilde {B}_\pm$ , $\tilde {A}_\pm$ are shown in (5.19). And

(5.38) \begin{equation} E_1=\frac {1}{2\pi i}\int _{\Sigma ^{E}}\dfrac {\left ( I+\varpi (s)\right ) (V^{E}-I)}{s^2}ds, \end{equation}

satisfies long-time asymptotic behaviour condition

(5.39) \begin{equation} E_1=t^{-1/2}H^{(1)}+\mathcal{O}(t^{-1}), \end{equation}

where

(5.40) \begin{align} H^{(1)} \,:\!=\, H^{(1)}(\xi, c)=&\sum _{ p=\pm z_2 }\frac {M^{mod}(p)A_\pm (\xi )M^{mod}(p)^{-1}}{p^2}. \end{align}

6.1. Constructing the $g$ -function

Figure 12.

The canonical homology basis $\lbrace a_j,b_j \rbrace _{j=1}^2$ of the genius 2 Riemann surface.

To construct the g-function, we introduce

(6.1) \begin{align} Z(z)=\left [\dfrac {(z^2-z_1^2)(z^2-z_2^2)}{(z^2-1)} \right ]^{1/2}, \end{align}

whose branch cut is

(6.2) \begin{align} \Sigma _{mod} \,:\!=\, \Sigma _{mod} (\xi, c)=[{-}z_2,-z_1]\cup \Sigma _-\cup [z_1,z_2], \end{align}

and the branch of the square root is chosen such that $Z_+(z)\in i\mathbb{R}^+$ for $z\in [z_1,z_2]$ . And $z_1$ , $z_2$ admit:

(6.3) \begin{align} \frac {1-\xi }{2}=\frac {1}{z_1z_2}\left (1-\frac {1}{z_1^2} -\frac {1}{z_2^2}\right ) . \end{align}

Denote

(6.4) \begin{align} \text{d}g=\dfrac {Z(z)}{z^3}\left [ \frac {1-\xi }{2}z^2-\frac {2}{z_1z_2}\right ] \text{d}z. \end{align}

$\text{d}g$ is a meromorphic differential defined on the 2-genus Riemann surface, with d $g$ on the upper sheet and $-$ d $g$ on the lower sheet. Similarly, the g-function is given by

(6.5) \begin{align} g(z)=\int _{z_2}^{z}\text{d}g,\ z\in \mathbb{C}\setminus \Sigma _{mod}. \end{align}

Figure 13.

The region of $\Omega _1\cup \Omega _1^*$ and the contour $\Sigma _1\cup \Sigma _1^*$ . The shaded region means Im $[g](z)\lt 0$ , while white region means Im $[g](z)\gt 0$ .

Proof. Denote $\eta =-\frac {2}{z_1z_2}$ . Thus, (6.3) gives

\begin{align*} z_1^2+z_2^2=\frac {4}{\eta ^2}\left ( 1+\frac {1-\xi }{\eta }\right ) . \end{align*}

Then the $a_2$ -period of $g$ equals to zero if and only if $F(\eta, \xi )=0$ with

\begin{align*} F(\eta, \xi )=\int _{z_1}^{z_2}\dfrac {Z(z)}{z^3}\left [ \frac {1-\xi }{2}z^2-\frac {2}{z_1z_2}\right ] \text{d}z. \end{align*}

When $\xi =\frac {3}{4}$ , $F(\eta, \xi )=0$ has solution $(-\frac {1}{2},\frac {3}{4})$ , and on the other end $\xi =\xi _m$ , $F(\eta, \xi )=0$ has solution as shown in Proposition6: $(-\frac {2}{cz_0(\xi _m)},\xi _m)$ . And

\begin{align*} \partial _ \eta F(\eta, \xi )=\int _{z_1}^{z_2}\dfrac {z}{\sqrt {(z^2-1)(z^2-z_1^2)(z^2-z_2^2)}}\left [ 1+(1-\xi )\left (\frac {2}{\eta ^3}+\frac {2}{\eta ^4}(1-\xi ) \right ) \right ] \text{d}z. \end{align*}

Consider the function $f(x)=x^4+2(1-\xi )x+3(1-\xi )^2$ . It is noticed that (6.3) implies $-\eta \gt 1-\xi$ , so simple calculation gives that $f(\eta )\gt f(\xi -1)\gt 0$ . So $ \partial _ \eta F\neq 0$ , which gives the existence of solution $\eta$ .

6.2. Opening the jump contour

Similar to the above section, we define the following contour $\Sigma _1$ and closed region $\Omega _1$ relying on $(\xi, c)$ as in Figure 13 to open the jump on $\mathbb{R}$ . In this region of $(\xi, c)$ , we introduce a piecewise matrix interpolation function

(6.6) \begin{equation} G(z) \,:\!=\, G (z;\,\xi, c)=\left \{\begin{array}{l@{\quad}l} {\left ({\begin{array}{c@{\quad}c}1 & -r_2e^{-2itg}\\[3pt] 0 & 1\end{array}}\right )},&\text{as } z\in \Omega _1;\\[12pt] {\left ({\begin{array}{c@{\quad}c}1 & 0\\[3pt]-r_1e^{2itg} & 1\end{array}}\right )}, &\text{as } z\in \Omega _1^*;\\[3pt] I &\text{as } z \text{ in elsewhere}, \end{array}\right . \end{equation}

Same as above sections, $ G (z)$ brings a new singularity. To deal with the jump on $\mathbb{R}$ , we give a introduction of an auxiliary function $\delta (z) \,:\!=\, \delta (z;\,\xi, c)$ . Define

(6.7) \begin{align} Z_3(z)=&(z^2-1)Z(z),\quad \log \delta (z)=\frac {Z_3(z)}{2\pi i}\sum _{\pm }\mp \int _{\pm z_2}^{\pm z_0} \dfrac {\log\!(ir_{2,-}(s))-w_\pm }{(s-z)Z_{3,+}(s)}ds. \end{align}

where $w_\pm$ satisfy linear system as

\begin{align*} \begin{pmatrix} \int _{-z_2}^{-z_0}\dfrac {ds}{Z_{3,+}(s)}&\int _{z_0}^{ z_2}\dfrac {ds}{Z_{3,+}(s)}\\[3pt] \int _{- z_2}^{-z_0}\dfrac {sds}{Z_{3,+}(s)}&\int _{z_0}^{ z_2}\dfrac {sds}{Z_{3,+}(s)}\\[3pt] \end{pmatrix}\begin{pmatrix} w_-\\[3pt]w_+\\[3pt] \end{pmatrix}=\begin{pmatrix} \int _{[{-}z_2,z_2]\setminus [{-}z_0,z_0]} \dfrac {\log\!(ir_{2,-}(s))}{Z_{3,+}(s)}ds\\[3pt] \int _{[{-}z_2,z_2]\setminus [{-}z_0,z_0]} \dfrac {s\log\!(ir_{2,-}(s))}{Z_{3,+}(s)}ds\\[3pt] \end{pmatrix}. \end{align*}

$\delta (z)$ admits the following jump condition

\begin{align*} \delta _-(z)\delta _+(z) & = ir_{2,-}e^{-w_-}, \qquad z\in [{-}z_2,z_2] \setminus [{-}z_1,z_1];\\[3pt] \delta _-(z)\delta _+(z) & = 1, \qquad \qquad\quad\ \ z \in \Sigma _-, \end{align*}

and the following proposition

Through $\delta (z)$ and $G(z)$ , in this region of $(\xi, c)$ , same as above subsection, we give the same transformation in this case:

\begin{align*} N(z)\overset {(5.7)}{\longrightarrow } M^{(1)}(z)\overset {(5.9)}{\longrightarrow } M^{(2)}(z)=\left \{\begin{array}{l@{\quad}l} E(z;\,\xi, c)M^{mod}(z;\,\xi, c) & z\notin U\\[3pt] E(z;\,\xi, c)M^{j,pm}(z;\,\xi, c) &z\in U_{\pm z_j},\ j=1,2,\\[3pt] \end{array}\right . \end{align*}

where $M^{mod}(z;\,\xi, c)$ is the model RH problem on the Riemann surface, which solution is given by theta function in Subsection 6.3. For $j=1,2$ , $M^{j,pm}(z;\,\xi, c)$ are local model of $\pm z_j$ which solution can be expressed in terms of Airy functions similarly in Subsection 5.2.2. And $E(z;\,\xi, c)$ is the error function, which has jump contour in Figure 14, and it is similarly in subsection 5.2.3. We obtain its asymptotic property directly as follow

Figure 14.

The jump contour $\Sigma ^{E}$ for the $E(z;\,\xi, c)$ . The red circles are $\partial U$ .

Proposition 14. As $z\to 0\in \mathbb{C}^+$ , we have

(6.9) \begin{align} E(z;\,\xi, c)=E(0)+E_1z+\mathcal{O}(z^2), \end{align}

with long-time asymptotic behaviour

(6.10) \begin{equation} E(0)=I+t^{-1}H^{(0)}+\mathcal{O}(t^{-2}). \end{equation}

And

\begin{align*} H^{(0)} \,:\!=\, H^{(0)}(\xi, c)=&\sum _{ p=\pm z_j,j=1,2 }\partial _z\left (\frac {1 }{z}F^{\pm }\left (\begin{array}{c@{\quad}c} 0&-\frac {5i}{48}[\tilde {A}^j_\pm ]^{-\frac {4}{3}}\\[3pt] 0&0\\[3pt] \end{array}\right )(F^\pm )^{-1}\right )(p) \\[3pt] &+\sum _{ p=\pm z_j,j=1,2}\frac {1 }{p}\left ( F^{\pm }\left (\begin{array}{c@{\quad}c} 0&\frac {5}{36}\tilde {B}^{\kern1pt j}_\pm [\tilde {A}_\pm ^j]^{-\frac {7}{3}}\\[3pt] -\frac {7i}{48}[\tilde {A}_\pm ^j]^{-\frac {2}{3}}&0\\[3pt] \end{array}\right )(F^\pm )^{-1}\right )(p), \end{align*}

where $\tilde {B}^{\kern1pt j}_\pm$ and $\tilde {A}^j_\pm$ are shown in (6.14). Further, $E_1$ admits the following asymptotic expansion

(6.11) \begin{equation} E_1=t^{-1}H^{(1)}+\mathcal{O}(t^{-2}), \end{equation}

where

\begin{align*} H^{(1)} \,:\!=\, H^{(1)}(\xi, c)=&\sum _{ p=\pm z_j,j=1,2 }\partial _z\left (\frac {1 }{z^2}F^{\pm }\left (\begin{array}{c@{\quad}c} 0&-\frac {5i}{48}[\tilde {A}_\pm ^j]^{-\frac {4}{3}}\\[3pt] 0&0\\[3pt] \end{array}\right )(F^\pm )^{-1}\right )(p)\\[3pt] &+\sum _{ p=\pm z_j,j=1,2}\frac {1 }{p^2}\left ( F^{\pm }\left (\begin{array}{c@{\quad}c} 0&\frac {5}{36}\tilde {B}^j_\pm [\tilde {A}_\pm ^j]^{-\frac {7}{3}}\\[3pt] -\frac {7i}{48}[\tilde {A}^j_\pm ]^{-\frac {2}{3}}&0\\[3pt] \end{array}\right )(F^\pm )^{-1}\right )(p). \end{align*}

Here we denote

(6.12) \begin{align} &F^\pm (z) \,:\!=\, M^{mod}H(z;\pm z_0)^{-1}N^{-1}f_\pm ^{\frac {\sigma _3}{2}}, \end{align}

(6.13) \begin{align} &\frac {3}{2}itg(z)=\tilde {A}_{+}^1(z-z_1)^{3/2}+\tilde {B}_{+}^1(z-z_1)^2+\mathcal{O}((z-z_1)^{5/2}), \\[6pt] \nonumber \end{align}

where the definition of the square root is mapping $\mathbb{R}^+$ to $\mathbb{R}^+$ , and $\tilde {A}^1_-,\tilde {B}_-^1,\tilde {A}^2_+,\tilde {B}_+^2,\tilde {A}^2_-,\tilde {B}_-^2$ are as the similar definition on $-z_1,z_2,-z_2$ , respectively.