The Focusing NLS Equation with Step-Like Oscillating Background: The Genus 3 Sector

Abstract

We consider the Cauchy problem for the focusing nonlinear Schrödinger equation with initial data approaching different plane waves $A_j\mathrm {e}^{\mathrm {i}\phi _j}\mathrm {e}^{-2\mathrm {i}B_jx}$, $j=1,2$ as $x\rightarrow \pm \infty$. The goal is to determine the long-time asymptotics of the solution, according to the value of $\xi =x/t$. The general situation is analyzed in a recent paper where we develop the Riemann–Hilbert approach and detect different asymptotic scenarios, depending on the relationships between the parameters $A_1$, $A_2$, $B_1$, and $B_2$. In particular, in the shock case $B_1<B_2$, some scenarios include genus 3 sectors, i.e., ranges of values of $\xi$ where the leading term of the asymptotics is given in terms of hyperelliptic functions attached to a Riemann surface $M(\xi )$ of genus three. The present paper is devoted to the complete asymptotic analysis in such a sector.

Appendices

Appendix A. Exact Solution in Terms of Parabolic Cylinder Functions

Let X denote the cross $X:=X_1\cup \dots \cup X_4\subset \mathbb {C}$, where the rays

$$\begin{aligned}&X_1:=\lbrace s\mathrm {e}^{\frac{\mathrm {i}\pi }{4}}\mid 0\le s<\infty \rbrace ,&\quad&X_2:=\lbrace s\mathrm {e}^{\frac{3\mathrm {i}\pi }{4}}\mid 0\le s<\infty \rbrace ,\nonumber \\&X_3:=\lbrace s\mathrm {e}^{-\frac{3\mathrm {i}\pi }{4}}\mid 0\le s<\infty \rbrace ,&X_4:=\lbrace s\mathrm {e}^{-\frac{\mathrm {i}\pi }{4}}\mid 0\le s<\infty \rbrace \end{aligned}$$

(A.1)

are oriented toward the origin as in Fig. 20.

Fig. 20

The contour $X=X_1\cup \dots \cup X_4$

Define the function $\nu :\mathbb {C}\rightarrow [0,\infty )$ by $\nu (q):=\frac{1}{2\pi }\ln (1+|q|^2)$. Let $\rho (q,z):=\mathrm {e}^{\mathrm {i}\nu (q)\ln _{-\pi /2}z}$, i.e., $\rho (q,z)=z^{\mathrm {i}\nu (q)}$ with the branch cut along the negative imaginary axis.

We consider the following family of RH problems parametrized by $q\in \mathbb {C}$:

$$\begin{aligned} {\left\{ \begin{array}{ll} m^X(q,\,\cdot \,)\in I+\dot{E}^2(\mathbb {C}\setminus X),&{}\\ m_+^X(q,z)=m_-^X(q,z)v^X(q,z)&{}\text {for a.e. }z\in X, \end{array}\right. } \end{aligned}$$

(A.2)

where the jump matrix $v^X(q,z)$ is defined by

$$\begin{aligned} v^X(q,z):={\left\{ \begin{array}{ll} \begin{pmatrix}1&{}0\\ q\rho (q,z)^{-2}\mathrm {e}^{2\mathrm {i}z^2}&{}1\end{pmatrix},&{}k\in X_1,\\ \begin{pmatrix}1&{}{{\bar{q}}}\rho (q,z)^2\mathrm {e}^{-2\mathrm {i}z^2}\\ 0&{}1\end{pmatrix},&{}k\in X_2,\\ \begin{pmatrix}1&{}0\\ -\frac{q}{1+|q|^2}\rho (q,z)^{-2}\mathrm {e}^{2\mathrm {i}z^2}&{}1\end{pmatrix},&{}k\in X_3,\\ \begin{pmatrix}1&{}-\frac{{{\bar{q}}}}{1+|q|^2}\rho (q,z)^2\mathrm {e}^{-2\mathrm {i}z^2}\\ 0&{}1\end{pmatrix},&k\in X_4. \end{array}\right. } \end{aligned}$$

(A.3)

The matrix $v^X$ has entries that oscillate rapidly as $z\rightarrow 0$ and $v^X$ is not continuous at $z=0$; however $v^X(q,\,\cdot \,)-I\in L^2(X)\cap L^{\infty }(X)$. The RH problem (A.2) can be solved explicitly in terms of parabolic cylinder functions [28].

Theorem A.1

The RH problem (A.2) has a unique solution $m^X(q,z)$ for each $q\in \mathbb {C}$. This solution satisfies

$$\begin{aligned} m^X(q,z)=I+\frac{\mathrm {i}}{z} \begin{pmatrix} 0&{}-\mathrm {e}^{-\pi \nu }\beta ^X(q)\\ \mathrm {e}^{\pi \nu }\overline{\beta ^X(q)}&{}0\end{pmatrix}+\mathrm {O}\left( \frac{1}{z^2}\right) ,\quad z\rightarrow \infty ,\quad q\in \mathbb {C}, \end{aligned}$$

(A.4)

where the error term is uniform with respect to $\arg z\in [0,2\pi ]$ and q in compact subsets of $\mathbb {C}$, and the function $\beta ^X(q)$ is defined by

$$\begin{aligned} \beta ^X(q):=\frac{\sqrt{\nu (q)}}{2}\mathrm {e}^{\mathrm {i}\left( -\frac{3\pi }{4}-2\nu (q)\ln 2-\arg q+\arg \Gamma (\mathrm {i}\nu (q))\right) },\quad q\in \mathbb {C}. \end{aligned}$$

(A.5)

Moreover, for each compact subset $K\subset \mathbb {C}$,

$$\begin{aligned} \sup _{q\in K}\sup _{z\in \mathbb {C}\setminus X}|m^X(q,z)|<\infty . \end{aligned}$$

(A.6)

Proof

Uniqueness follows because $\det v^X=1$. Define $m^{(X1)}$ by

$$\begin{aligned} m^{(X1)}(q,z):=m^X(q,z)\rho (q,z)^{\sigma _3}\mathrm {e}^{-\mathrm {i}z^2\sigma _3}. \end{aligned}$$

(A.7)

Using that $\rho (q,\,\cdot \,)$ has a branch cut along the negative imaginary axis, we see that $m^X$ satisfies the jump condition in (A.2) iff $m^{(X1)}$ satisfies

$$\begin{aligned} m_+^{(X1)}(q,z)=m_-^{(X1)}(q,z)v^{(X1)}(q,z)\quad \text {for a.e. }z\in X \cup \mathrm {i}\,\mathbb {R}_-, \end{aligned}$$

where

$$\begin{aligned} v^{(X1)}:={\left\{ \begin{array}{ll} \begin{pmatrix}1&{}0\\ q&{}1\end{pmatrix},&{}z\in X_1,\\ \begin{pmatrix}1&{}{{\bar{q}}}\\ 0&{}1\end{pmatrix},&{}z\in X_2,\\ \begin{pmatrix}1&{}0\\ -\frac{q}{1+|q|^2}&{}1\end{pmatrix},&{}z\in X_3,\\ \left( \frac{\rho _+(q,z)}{\rho _-(q,z)}\right) ^{\sigma _3}=\begin{pmatrix}\frac{1}{1+|q|^2}&{}0\\ 0&{}1+|q|^2\end{pmatrix},&{}z\in \mathrm {i}\,\mathbb {R}_-,\\ \begin{pmatrix}1&{}-\frac{{{\bar{q}}}}{1+|q|^2}\\ 0&{}1\end{pmatrix},&z\in X_4, \end{array}\right. } \end{aligned}$$

with all contours oriented toward the origin. We next merge the contours $X_1$ and $X_2$ along $\mathrm {i}\,\mathbb {R}_+$ and the contours $X_3$ and $X_4$ along $\mathrm {i}\,\mathbb {R}_-$. Thus we let

$$\begin{aligned} \psi (q,z)=m^{(X1)}(q,z)B(q,z), \end{aligned}$$

(A.8)

where

$$\begin{aligned} B(z):={\left\{ \begin{array}{ll} \begin{pmatrix}1&{}0\\ q&{}1\end{pmatrix},&{}\arg z\in \left( \frac{\pi }{4},\frac{\pi }{2}\right) ,\\ \begin{pmatrix}1&{}-{{\bar{q}}}\\ 0&{}1\end{pmatrix},&{}\arg z\in \left( \frac{\pi }{2},\frac{3\pi }{4}\right) ,\\ \begin{pmatrix}1&{}0\\ -\frac{q}{1+|q|^2}&{}1\end{pmatrix},&{}\arg z\in \left( \frac{5\pi }{4},\frac{3\pi }{2}\right) ,\\ \begin{pmatrix}1&{}\frac{{{\bar{q}}}}{1+|q|^2}\\ 0&{}1\end{pmatrix},&{}\arg z\in \left( \frac{3\pi }{2},\frac{7\pi }{4}\right) ,\\ \,I,&{}\text {else}. \end{array}\right. } \end{aligned}$$

Then $m^X$ satisfies the jump condition in (A.2) iff $\psi$ satisfies

$$\begin{aligned} \psi _+(q,z)=\psi _-(q,z)v^{\psi }(q)\quad \text {for a.e. }z\in \mathrm {i}\,\mathbb {R}, \end{aligned}$$

(A.9)

where the constant matrix $v^{\psi }$ is defined by

$$\begin{aligned} v^{\psi }(q):=\begin{pmatrix}1+|q|^2&{}\bar{q}\\ q&{}1\end{pmatrix},\quad k\in \mathrm {i}\,\mathbb {R}, \end{aligned}$$

(A.10)

and the contour $\mathrm {i}\,\mathbb {R}$ is oriented downward.

Let

$$\begin{aligned} \beta ^X:=-\frac{\mathrm {e}^{\frac{\pi \mathrm {i}}{4}}2^{-2\mathrm {i}\nu }\mathrm {e}^{\frac{\pi \nu }{2}}\sqrt{\pi }}{q\sqrt{2}\,\Gamma (-\mathrm {i}\nu )}. \end{aligned}$$

(A.11)

Then $\beta ^X$ can be written as in (A.5). Define a sectionally analytic function $\psi (q,z)$ by

$$\begin{aligned} \psi (q,z):=\begin{pmatrix} \psi _{11}(q,z)&{}\frac{\left( \frac{\mathrm {d}}{\mathrm {d}z}-2\mathrm {i}z\right) \psi _{22}(q,z)}{4\mathrm {e}^{\pi \nu }\overline{\beta ^X(q)}}\\ \frac{\left( \frac{\mathrm {d}}{\mathrm {d}z}+2\mathrm {i}z\right) \psi _{11}(q,z)}{4\mathrm {e}^{-\pi \nu }\beta ^X(q)}&{}\psi _{22}(q,z)\end{pmatrix},\quad q\in \mathbb {C},\quad z\in \mathbb {C}\setminus \mathrm {i}\,\mathbb {R}, \end{aligned}$$

(A.12)

where the functions $\psi _{11}$ and $\psi _{22}$ are defined by

$$\begin{aligned} \psi _{11}(q,z)&:={\left\{ \begin{array}{ll} 2^{-\mathrm {i}\nu }\mathrm {e}^{-\frac{3\pi \nu }{4}}D_{\mathrm {i}\nu }(2\mathrm {e}^{-\frac{3\pi \mathrm {i}}{4}}z),&{}{\text {Re}}z<0,\\ 2^{-\mathrm {i}\nu }\mathrm {e}^{\frac{\pi \nu }{4}}D_{\mathrm {i}\nu }(2\mathrm {e}^{\frac{\pi \mathrm {i}}{4}}z),&{}{\text {Re}}z>0, \end{array}\right. } \end{aligned}$$

(A.13a)

$$\begin{aligned} \psi _{22}(q,z)&:={\left\{ \begin{array}{ll} 2^{\mathrm {i}\nu }\mathrm {e}^{\frac{5\pi \nu }{4}}D_{-\mathrm {i}\nu }(2\mathrm {e}^{\frac{3\pi \mathrm {i}}{4}}z),&{}{\text {Re}}z<0,\\ 2^{\mathrm {i}\nu }\mathrm {e}^{\frac{\pi \nu }{4}}D_{-\mathrm {i}\nu }(2\mathrm {e}^{-\frac{\pi \mathrm {i}}{4}}z),&{}{\text {Re}}z>0, \end{array}\right. } \end{aligned}$$

(A.13b)

and $D_a(z)$ denotes the parabolic cylinder function.

Since $D_a(z)$ is an entire function of both a and z, $\psi (q,z)$ is analytic in the left and right halves of the complex z-plane with a jump across the imaginary axis. The function $\psi$ satisfies

$$\begin{aligned} \partial _z\psi +2\mathrm {i}z\sigma _3\psi =4\begin{pmatrix}0&{}\beta ^X\mathrm {e}^{-\pi \nu }\\ {{\bar{\beta }}}^X\mathrm {e}^{\pi \nu }&{}0\end{pmatrix}\psi ,\quad q\in \mathbb {C},\quad z\in \mathbb {C}\setminus \mathrm {i}\,\mathbb {R}. \end{aligned}$$

Since $\psi _+$ and $\psi _-$ solve the same second order ODE, there exists a function $v^{\psi }$ independent of z such that (A.9) holds. Setting $z=0$ and using that

$$\begin{aligned} D_{\mathrm {i}\nu }(0)=\frac{\sqrt{\pi }\,2^{\frac{\mathrm {i}\nu }{2}}}{\Gamma \left( \frac{1}{2}(1-\mathrm {i}\nu )\right) },\quad D_{\mathrm {i}\nu }'(0)=-\frac{\sqrt{\pi }\,2^{\frac{1}{2}(1+\mathrm {i}\nu )}}{\Gamma \left( -\frac{\mathrm {i}\nu }{2}\right) }\,, \end{aligned}$$

we find

$$\begin{aligned} v^{\psi }(q)=\psi _-(q,0)^{-1}\psi _+(q,0)=\begin{pmatrix}1+|q|^2&{}\bar{q}\\ q&{}1\end{pmatrix}. \end{aligned}$$

Hence $\psi$ satisfies (A.9). This shows that $m^X$ satisfies the jump condition in (A.2).

For each $\delta >0$, the parabolic cylinder function satisfies the asymptotic formula [39]

$$\begin{aligned} D_a(z)&=z^a\mathrm {e}^{-\frac{z^2}{4}}\left( 1-\frac{a(a-1)}{2z^2}+\mathrm {O}(z^{-4})\right) \\&\quad -\frac{\sqrt{2\pi }\mathrm {e}^{\frac{z^2}{4}}z^{-a-1}}{\Gamma (-a)}\left( 1+\frac{(a+1)(a+2)}{2z^2}+\mathrm {O}(z^{-4})\right) \\&\qquad \times {\left\{ \begin{array}{ll} 0,&{}\arg z\in [-\frac{3\pi }{4}+\delta ,\frac{3\pi }{4}-\delta ],\\ \mathrm {e}^{\mathrm {i}\pi a},&{}\arg z\in [\frac{\pi }{4}+\delta ,\frac{5\pi }{4}-\delta ],\\ \mathrm {e}^{-\mathrm {i}\pi a},&{}\arg z\in [-\frac{5\pi }{4}+\delta ,-\frac{\pi }{4}-\delta ], \end{array}\right. }\quad z\rightarrow \infty ,\quad a\in \mathbb {C}, \end{aligned}$$

where the error terms are uniform with respect to a in compact subsets and $\arg z$ in the given ranges. Using this formula and the identity

$$\begin{aligned} \frac{\mathrm {d}}{\mathrm {d}z}D_a(z)=\frac{z}{2}D_a(z)-D_{a+1}(z), \end{aligned}$$

the asymptotic equation (A.4) follows from a tedious but straightforward computation. The boundedness in (A.6) is a consequence of (A.4) and the definition (A.12) of $\psi (q,z)$. $\square$

Remark A.2

The definition (A.12) of $\psi$ can be motivated as follows. Since $(\partial _z+2\mathrm {i}z\sigma _3)\psi$ and $\psi$ have the same jump across $\mathrm {i}\,\mathbb {R}$, the function $(\partial _z\psi +2\mathrm {i}z\sigma _3\psi )\psi ^{-1}$ is entire. Suppose $m^X=1+m_1^X(q)z^{-1}+\mathrm {O}(z^{-2})$ as $z\rightarrow \infty$, where the matrix $m_1^X(q)$ is independent of z. Then we expect

$$\begin{aligned} \psi =\left( I+\frac{m_1^X}{z}+\mathrm {O}(z^{-2})\right) \rho ^{\sigma _3}\mathrm {e}^{-\mathrm {i}z^2\sigma _3},\quad z\rightarrow \infty , \end{aligned}$$

(A.14)

which suggests that

$$\begin{aligned} (\partial _z\psi +2\mathrm {i}z\sigma _3\psi )\psi ^{-1}=2\mathrm {i}[\sigma _3,m_1^X]+\mathrm {O}(z^{-1}),\quad z\rightarrow \infty . \end{aligned}$$

(A.15)

Strictly speaking, due to the factor B(q, z) in (A.8), equation (A.14) is not valid for z close to $\mathrm {i}\,\mathbb {R}$. Equation (A.15) implies that $(\partial _z\psi +2\mathrm {i}z\sigma _3\psi )\psi ^{-1}$ is bounded; hence a constant. Thus

$$\begin{aligned} \partial _z\psi +2\mathrm {i}z\sigma _3\psi =2\mathrm {i}[\sigma _3,m_1^X]\psi . \end{aligned}$$

Letting

$$\begin{aligned} \beta _{12}=4\mathrm {i}(m_1^X)_{12},\qquad \beta _{21}=-4\mathrm {i}(m_1^X)_{21}, \end{aligned}$$

we find that the (11) and (22) entries of $\psi$ satisfy the equations

$$\begin{aligned} {\left\{ \begin{array}{ll} \partial _z^2\psi _{11}+(4z^2+2\mathrm {i}-\beta _{12}\beta _{21})\psi _{11}=0,&{}\\ \partial _z^2\psi _{22}+(4z^2-2\mathrm {i}-\beta _{12}\beta _{21})\psi _{22}=0,&{} \end{array}\right. }z\in \mathbb {C}\setminus \mathrm {i}\,\mathbb {R}. \end{aligned}$$

Introducing the new variable $\zeta$ by $\zeta :=2\mathrm {e}^{-\frac{3\pi \mathrm {i}}{4}}z$, we find that $f(\zeta ):=\psi _{11}(z)$ satisfies the parabolic cylinder equation

$$\begin{aligned} \partial _{\zeta }^2f+\left( \frac{1}{2}-\frac{\zeta ^2}{4}+a\right) f=0 \end{aligned}$$

for $a=\frac{\mathrm {i}}{4}\beta _{12}\beta _{21}$. This means that

$$\begin{aligned} \psi _{11}(z)={\left\{ \begin{array}{ll} c_1D_a(2\mathrm {e}^{-\frac{3\pi \mathrm {i}}{4}}z)+c_2D_a(2\mathrm {e}^{\frac{\pi \mathrm {i}}{4}}z),&{}{\text {Re}}z>0,\\ c_3D_a(2\mathrm {e}^{-\frac{3\pi \mathrm {i}}{4}}z)+c_4D_a(2\mathrm {e}^{\frac{\pi \mathrm {i}}{4}}z),&{}{\text {Re}}z<0, \end{array}\right. } \end{aligned}$$

for some constants $\lbrace c_j\rbrace _1^4$. Since

$$\begin{aligned} D_a(\zeta )=\zeta ^a\mathrm {e}^{-\frac{\zeta ^2}{4}}\left( 1+\mathrm {O}(\zeta ^{-2})\right) ,\quad \zeta \rightarrow \infty ,\quad |\arg \zeta |<\frac{3\pi }{4}-\delta , \end{aligned}$$

the sought-after asymptotics (A.14) of $\psi$ can be obtained by choosing

$$\begin{aligned} a=\mathrm {i}\nu ,\quad c_1=0,\quad c_2=2^{-\mathrm {i}\nu }\mathrm {e}^{\frac{\pi \nu }{4}},\quad c_3=2^{-\mathrm {i}\nu }\mathrm {e}^{-\frac{3\pi \nu }{4}},\quad c_4=0. \end{aligned}$$

This yields the expression (A.13a) for $\psi _{11}$; the expression (A.13b) for $\psi _{22}$ is derived in a similar way. The functions $\psi _{12}$ and $\psi _{21}$ can be obtained from the equations

$$\begin{aligned} \psi _{21}=\beta _{12}^{-1}(\psi _{11}'(z)+2\mathrm {i}z\psi _{11}(z)),\qquad \psi _{12}=\beta _{21}^{-1}(\psi _{22}'(z)-2\mathrm {i}z\psi _{22}(z)). \end{aligned}$$

Using that $\det \psi =1$, the (21) and (12) entries of the jump condition $(\psi _-)^{-1}\psi _+=v^{\psi }$ evaluated at $z=0$ then yield

$$\begin{aligned} v_{21}^{\psi }(q)=-\frac{\sqrt{2\pi }\mathrm {e}^{\frac{\pi \mathrm {i}}{4}} 2^{-2\mathrm {i}\nu +1}\mathrm {e}^{-\frac{\pi \nu }{2}}}{\beta _{12}\Gamma (-\mathrm {i}\nu )},\qquad v_{12}^{\psi }(q)=-\frac{\sqrt{2\pi }\mathrm {e}^{-\frac{\pi \mathrm {i}}{4}} 2^{2\mathrm {i}\nu +1}\mathrm {e}^{\frac{3\pi \nu }{2}}}{\beta _{21}\Gamma (\mathrm {i}\nu )}. \end{aligned}$$

Thus $v^{\psi }$ has the form (A.10) provided that $\beta _{12}=4\mathrm {e}^{-\pi \nu }\beta ^X$ and $\beta _{21}=4\mathrm {e}^{\pi \nu }{{\bar{\beta }}}^X$ with $\beta ^X$ given by (A.11). This motivates the form of equation (A.12).

Appendix B. Exact Solution in Terms of Airy Functions

Let $Y:=Y_1\cup \dots \cup Y_4\subset \mathbb {C}$ denote the union of the four rays

$$\begin{aligned}&Y_1:=\lbrace s\mid 0\le s<\infty \rbrace ,&\qquad&Y_2:=\lbrace s\mathrm {e}^{\frac{2\mathrm {i}\pi }{3}}\mid 0\le s<\infty \rbrace ,\nonumber \\&Y_3:=\lbrace -s\mid 0\le s<\infty \rbrace ,&Y_4:=\lbrace s\mathrm {e}^{-\frac{2\mathrm {i}\pi }{3}}\mid 0\le s<\infty \rbrace , \end{aligned}$$

(B.1)

oriented toward the origin as in Fig. 21.

Fig. 21

The sectors $S_j$, $j=1,\dots ,4$

Define the open sectors $\lbrace S_j\rbrace _1^4$ by

$$\begin{aligned}&S_1:=\lbrace \arg k\in (0,\tfrac{2\pi }{3})\rbrace ,&\qquad&S_2:=\lbrace \arg k\in (\tfrac{2\pi }{3},\pi )\rbrace ,\\&S_3:=\lbrace \arg k\in (\pi ,\tfrac{4\pi }{3})\rbrace ,&S_4:=\lbrace \arg k\in (\tfrac{4\pi }{3},2\pi )\rbrace . \end{aligned}$$

Let $\omega :=\mathrm {e}^{\frac{2\pi \mathrm {i}}{3}}$. Define the function $m^{\mathrm {Ai}}(\zeta )$ for $\zeta \in \mathbb {C}\setminus Y$ by

$$\begin{aligned} m^{\mathrm {Ai}}(\zeta ):=\Psi (\zeta )\times {\left\{ \begin{array}{ll} \,\mathrm {e}^{\frac{2}{3}\zeta ^{3/2}\sigma _3},&{}\zeta \in S_1\cup S_4,\\ \begin{pmatrix}1&{}0\\ -1&{}1\end{pmatrix}\mathrm {e}^{\frac{2}{3}\zeta ^{3/2}\sigma _3},&{}\zeta \in S_2,\\ \begin{pmatrix}1&{}0\\ 1&{}1\end{pmatrix}\mathrm {e}^{\frac{2}{3}\zeta ^{3/2}\sigma _3},&\zeta \in S_3, \end{array}\right. } \end{aligned}$$

(B.2)

where

$$\begin{aligned} \Psi (\zeta ):={\left\{ \begin{array}{ll} \begin{pmatrix}\mathrm {Ai}(\zeta )&{}\mathrm {Ai}(\omega ^2\zeta )\\ \mathrm {Ai}'(\zeta )&{}\omega ^2\mathrm {Ai}'(\omega ^2\zeta )\end{pmatrix}\mathrm {e}^{-\frac{\pi \mathrm {i}}{6}\sigma _3},&{}\zeta \in \mathbb {C}^+,\\ \begin{pmatrix}\mathrm {Ai}(\zeta )&{}-\omega ^2\mathrm {Ai}(\omega \zeta )\\ \mathrm {Ai}'(\zeta )&{}-\mathrm {Ai}'(\omega \zeta )\end{pmatrix}\mathrm {e}^{-\frac{\pi \mathrm {i}}{6}\sigma _3},&\zeta \in \mathbb {C}^-. \end{array}\right. } \end{aligned}$$

Note that $\det m^{\mathrm {Ai}}(\zeta )=\mathrm {e}^{\pi \mathrm {i}/6}/(2\pi )$ is constant and nonzero. For each integer $N\ge 0$, define the asymptotic approximations $m_{\mathrm {as},N}^{\mathrm {Ai}}(\zeta )$ and $m_{\mathrm {as},N}^{\mathrm {Ai},\mathrm {inv}}(\zeta )$ of $m^{\mathrm {Ai}}(\zeta )$ and $m^{\mathrm {Ai}}(\zeta )^{-1}$, respectively, by

$$\begin{aligned} m_{\mathrm {as},N}^{\mathrm {Ai}}(\zeta )&:=\frac{\mathrm {e}^{\frac{\mathrm {i}\pi }{12}}}{2\sqrt{\pi }}\sum _{k=0}^N\frac{1}{\left( \frac{2}{3}\zeta ^{3/2}\right) ^k}\,\zeta ^{-\frac{1}{4}\sigma _3}\!\begin{pmatrix}(-1)^ku_k&{}u_k\\ -(-1)^k\nu _k&{}\nu _k\end{pmatrix}\mathrm {e}^{-\frac{\pi \mathrm {i}}{4}\sigma _3},&\quad&\zeta \in \mathbb {C}\setminus Y, \end{aligned}$$

(B.3a)

$$\begin{aligned} m_{\mathrm {as},N}^{\mathrm {Ai},\mathrm {inv}}(\zeta )&:=\sqrt{\pi }\mathrm {e}^{-\frac{\mathrm {i}\pi }{12}}\sum _{k=0}^N\frac{1}{\left( \frac{2}{3}\zeta ^{3/2}\right) ^k}\mathrm {e}^{\frac{\pi \mathrm {i}}{4}\sigma _3}\begin{pmatrix}\nu _k &{} -u_k\\ (-1)^k\nu _k &{}(-1)^k u_k\end{pmatrix}\zeta ^{\frac{1}{4}\sigma _3},&\zeta \in \mathbb {C}\setminus Y, \end{aligned}$$

(B.3b)

where the real constants $\lbrace u_j,\nu _j\rbrace _0^{\infty }$ are defined by $u_0:=\nu _0:=1$ and

$$\begin{aligned} u_k:=\frac{(2k+1)(2k+3)\dots (6k-1)}{(216)^kk!}\,,\quad \nu _k:=\frac{6k+1}{1-6k}u_k,\quad k=1,2,\dots \end{aligned}$$

For $N=0$, we have

$$\begin{aligned} m_{\mathrm {as},0}^{\mathrm {Ai}}(\zeta )=m_{\mathrm {as},0}^{\mathrm {Ai},\mathrm {inv}}(\zeta )^{-1}=\frac{\mathrm {e}^{\frac{\mathrm {i}\pi }{12}}}{2\sqrt{\pi }}\begin{pmatrix}\zeta ^{-\frac{1}{4}}&{}0\\ 0&{}\zeta ^{\frac{1}{4}}\end{pmatrix}\begin{pmatrix}1&{}1\\ -1&{}1\end{pmatrix}\mathrm {e}^{-\frac{\pi \mathrm {i}}{4}\sigma _3}. \end{aligned}$$

Theorem B.1

(a) The function $m^{\mathrm {Ai}}(\zeta )$ defined in (B.2) is analytic for $\zeta \in \mathbb {C}\setminus Y$ and satisfies the jump condition

$$\begin{aligned} m_+^{\mathrm {Ai}}(\zeta )=m_-^{\mathrm {Ai}}(\zeta )v^{\mathrm {Ai}}(\zeta ),\quad \zeta \in Y\setminus \lbrace 0\rbrace , \end{aligned}$$

(B.4)

where the jump matrix $v^{\mathrm {Ai}}$ is defined by

$$\begin{aligned} v^{\mathrm {Ai}}(\zeta ):={\left\{ \begin{array}{ll} \begin{pmatrix}1&{}-\mathrm {e}^{-\frac{4}{3}\zeta ^{3/2}}\\ 0&{}1\end{pmatrix},&{}\zeta \in Y_1,\\ \begin{pmatrix}1&{}0\\ \mathrm {e}^{\frac{4}{3}\zeta ^{3/2}}&{}1\end{pmatrix},&{}\zeta \in Y_2\cup Y_4,\\ \begin{pmatrix}0&{}1\\ -1&{}0\end{pmatrix},&\zeta \in Y_3. \end{array}\right. } \end{aligned}$$

(b) For each integer $N\ge 0$, the functions $m_{\mathrm {as},N}^{\mathrm {Ai}}(\zeta )$ and $m_{\mathrm {as},N}^{\mathrm {Ai},\mathrm {inv}}(\zeta )$ are analytic for $\zeta \in \mathbb {C}\setminus (-\infty ,0]$ and satisfy the following jump relations on the negative real axis:

$$\begin{aligned} m_{\mathrm {as},N+}^{\mathrm {Ai}}(\zeta )&=m_{\mathrm {as},N-}^{\mathrm {Ai}}(\zeta )\begin{pmatrix}0&{}1\\ -1&{}0\end{pmatrix},&\quad&\zeta <0, \end{aligned}$$

(B.5a)

$$\begin{aligned} m_{\mathrm {as},N+}^{\mathrm {Ai},\mathrm {inv}}(\zeta )&=\begin{pmatrix}0&{}-1\\ 1&{}0\end{pmatrix}m_{\mathrm {as},N-}^{\mathrm {Ai},\mathrm {inv}}(\zeta ),&\zeta <0. \end{aligned}$$

(B.5b)

(c) The functions $m_{\mathrm {as},N}^{\mathrm {Ai}}$ and $m_{\mathrm {as},N}^{\mathrm {Ai},\mathrm {inv}}$ approximate $m^{\mathrm {Ai}}$ and its inverse as $\zeta \rightarrow \infty$ in the sense that

$$\begin{aligned} m^{\mathrm {Ai}}(\zeta )^{-1} m_{\mathrm {as},N}^{\mathrm {Ai}}(\zeta )&=I+\mathrm {O}(\zeta ^{-\frac{3(N+1)}{2}}),&\quad&\zeta \rightarrow \infty , \end{aligned}$$

(B.6a)

$$\begin{aligned} m_{\mathrm {as},N}^{\mathrm {Ai},\mathrm {inv}}(\zeta )m^{\mathrm {Ai}}(\zeta )&=I+\mathrm {O}(\zeta ^{-\frac{3(N+1)}{2}}),&\zeta \rightarrow \infty , \end{aligned}$$

(B.6b)

where the error terms are uniform with respect to $\arg \zeta \in [0,2\pi ]$.

Proof

The analyticity of $m^{\mathrm {Ai}}$ is a direct consequence of the Airy function $\mathrm {Ai}(\zeta )$ being entire. The jump condition (B.4) can be verified by means of the identity

$$\begin{aligned} \mathrm {Ai}(\zeta )+\omega \mathrm {Ai}(\omega \zeta )+\omega ^2\mathrm {Ai}(\omega ^2\zeta )=0,\quad \zeta \in \mathbb {C}. \end{aligned}$$

On the other hand, the analyticity of $m_{\mathrm {as},N}^{\mathrm {Ai}}(\zeta )$ and $m_{\mathrm {as},N}^{\mathrm {Ai},\mathrm {inv}}(\zeta )$ is immediate from (B.3). Using that $(\zeta ^{3/2})_{\pm }=\mp \mathrm {i}|\zeta |^{3/2}$ and $(\zeta ^{1/4})_{\pm }=\mathrm {e}^{\pm \frac{\pi \mathrm {i}}{4}}|\zeta |^{1/4}$ for $\zeta \in (-\infty ,0)$, a straightforward computation gives (B.5). This proves (a) and (b).

In order to prove (c), we note that, for each small number $\delta >0$, the Airy function satisfies the following asymptotic expansions uniformly in the stated sectors, see [39]:

$$\begin{aligned} {\left\{ \begin{array}{ll} \mathrm {Ai}(z)=\frac{\mathrm {e}^{-\frac{2}{3}z^{3/2}}}{2\sqrt{\pi }z^{1/4}}\sum \limits _{k=0}^{\infty }\frac{(-1)^ku_k}{\left( \frac{2}{3}z^{3/2}\right) ^k},&{}\\ \mathrm {Ai}'(z)=-\frac{z^{1/4}\mathrm {e}^{-\frac{2}{3}z^{3/2}}}{2\sqrt{\pi }}\sum \limits _{k=0}^{\infty }\frac{(-1)^k(6k+1)u_k}{(1-6k)\left( \frac{2}{3}z^{3/2}\right) ^k},&{} \end{array}\right. }\qquad z\rightarrow \infty ,\quad |\arg z|\le \pi -\delta , \end{aligned}$$

(B.7)

and

$$\begin{aligned}&{\left\{ \begin{array}{ll} \mathrm {Ai}(z)=\frac{(-z)^{-1/4}}{\sqrt{\pi }}\sum \limits _{k=0}^{\infty }(-1)^k\left( \frac{u_{2k}\cos \left( \frac{2}{3}(-z)^{3/2}-\frac{\pi }{4}\right) }{\left( \frac{2}{3}(-z)^{3/2}\right) ^{2k}}+\frac{u_{2k+1}\sin \left( \frac{2}{3}(-z)^{3/2}-\frac{\pi }{4}\right) }{\left( \frac{2}{3}(-z)^{3/2}\right) ^{2k+1}}\right) ,&{}\\ \mathrm {Ai}'(z)=\frac{(-z)^{1/4}}{\sqrt{\pi }}\sum \limits _{k=0}^{\infty }(-1)^k\left( \frac{\nu _{2k}\sin \left( \frac{2}{3}(-z)^{3/2}-\frac{\pi }{4}\right) }{\left( \frac{2}{3}(-z)^{3/2}\right) ^{2k}}-\frac{\nu _{2k+1}\cos \left( \frac{2}{3}(-z)^{3/2}-\frac{\pi }{4}\right) }{\left( \frac{2}{3}(-z)^{3/2}\right) ^{2k+1}}\right) ,&{} \end{array}\right. }\nonumber \\& z\rightarrow \infty ,\quad \frac{\pi }{3}+\delta \le \arg z\le \frac{5\pi }{3}-\delta . \end{aligned}$$

(B.8)

The function $m_{\mathrm {as},N}^{\mathrm {Ai}}(\zeta )$ is defined by the expression obtained by substituting the asymptotic sums in (B.7) from $k=0$ to $k=N$ into the definition of $m^{\mathrm {Ai}}(\zeta )$ for $\zeta \in S_1$. Similarly, $m_{\mathrm {as},N}^{\mathrm {Ai},\mathrm {inv}}(\zeta )$ is defined by the expression obtained by substituting the asymptotic sums in (B.7) from $k=0$ to $k=N$ into the following expression for $m^{\mathrm {Ai}}(\zeta )^{-1}$ which is valid for $\zeta \in S_1$:

$$\begin{aligned} m^{\mathrm {Ai}}(\zeta )^{-1}=2\pi \mathrm {i}\omega ^2\mathrm {e}^{-\frac{2}{3}\zeta ^{3/2} \sigma _3}\mathrm {e}^{\frac{\pi \mathrm {i}}{6}\sigma _3} \begin{pmatrix}\omega ^2 \mathrm {Ai}'(\omega ^2\zeta )&{}-\mathrm {Ai}(\omega ^2\zeta )\\ -\mathrm {Ai}'(\zeta ) &{} \mathrm {Ai}(\zeta )\end{pmatrix}. \end{aligned}$$

It follows that the estimates in (B.6) hold as $\zeta \rightarrow \infty$ in ${{\bar{S}}}_1$. Long but straightforward computations using (B.7) and (B.8) show that the estimates in (B.6) hold as $\zeta \rightarrow \infty$ in $\bar{S}_2\cup {{\bar{S}}}_3\cup {{\bar{S}}}_4$ as well. $\square$

Appendix C. Endpoint Behavior of a Cauchy Integral

Let $\gamma :[\alpha ,\beta ]\rightarrow \mathbb {C}$ be a smooth simple contour from $a=\gamma (\alpha )$ to $b=\gamma (\beta )$ with $a\ne b$. By a slight abuse of notation, we let $\gamma$ denote both the map $[\alpha ,\beta ]\rightarrow \mathbb {C}$ and its image $\gamma :=\gamma ([\alpha ,\beta ])$ as a subset of $\mathbb {C}$.

Lemma C.1

Let $f:\gamma \rightarrow \mathbb {C}$ be such that $t\mapsto f(\gamma (t))$ is $C^1$ on $[\alpha ,\beta ]$. Suppose g(k) is a $C^1$-function of $k\in \mathbb {C}$ and define the function h(k) by

$$\begin{aligned} h(k):=\frac{g(k)}{2\pi \mathrm {i}}\int _{\gamma }\frac{f(s)\mathrm {d}s}{s-k},\quad k\in \mathbb {C}\setminus \gamma . \end{aligned}$$

Then the functions $\mathrm {e}^{\mathrm {i}h(k)}$ and $\mathrm {e}^{-\mathrm {i}h(k)}$ are bounded as $k\in \mathbb {C}\setminus \gamma$ approaches b if and only if the product g(b)f(b) is purely imaginary.

Proof

We know (see [37, Appendix A2]) that the endpoint behavior of the Cauchy integral (here near $k=b$) is given by

$$\begin{aligned} \frac{1}{2\pi \mathrm {i}}\int _a^b\frac{f(s)}{s-k}\,\mathrm {d}s=\frac{f(b)}{2\pi \mathrm {i}}\ln (k-b)+\Phi _b(k), \end{aligned}$$

where $\Phi _b(k)$ is bounded near $k=b$. We thus have $h(k)=\frac{g(k)}{2\pi \mathrm {i}}f(b)\ln (k-b)+\Psi _b(k)$ where $\Psi _b(k)$ is bounded near $k=b$. Moreover,

$$\begin{aligned} \frac{g(k)}{2\pi \mathrm {i}}f(b)\ln (k-b)=\frac{g(k)-g(b)}{2\pi \mathrm {i}}f(b)\ln (k-b)+\frac{g(b)}{2\pi \mathrm {i}}f(b)\ln (k-b), \end{aligned}$$

where the first term on the right-hand side is clearly bounded as $k \rightarrow b$. The term $\frac{g(b)}{2\pi \mathrm {i}}f(b)\ln (k-b)$ is unbounded as $k\rightarrow b$, but its contribution to $\mathrm {e}^{\pm \mathrm {i}h(k)}$ is bounded iff the product g(b)f(b) is purely imaginary. $\square$