Large deviations of energy transfers in nonequilibrium CFT and asymptotics of non-local Riemann–Hilbert problems

Abstract

A wide class of $1+1$ dimensional unitary conformal field theories allows for an explicit construction of nonequilibrium “profile states” interpolating smoothly between different equilibria on the left and on the right. It has been recently established that the generating function for the full counting statistics of energy transfers in such states may be expressed in terms of the solution to a non-local Riemann–Hilbert problem. Following earlier works on the statistics of energy transfers, in particular the ones of Bernard–Doyon on the “partitioning protocol” in conformal field theory, the full counting statistics of energy transfers in the profile states was conjectured to satisfy a large deviation principle in the limit of long transfer-times. The present paper establishes rigorously this conjecture by carrying out the long-time asymptotic analysis of the underlying non-local Riemann–Hilbert problem.

Appendices

Conformal map of the welded cylinder

In this section we prove Proposition 1.1. Introduce the functions

$$\begin{aligned} \omega ^{(L)}(z) = { \mathchoice{\dfrac{ \widetilde{\gamma }_{+} \cdot z }{ 1 + \text {e}^{- { \mathchoice{\dfrac{2\pi }{ \tau }}{\dfrac{2\pi }{ \tau }}{\frac{2\pi }{ \tau }}{\frac{2\pi }{ \tau }} } z } }}{\dfrac{ \widetilde{\gamma }_{+} \cdot z }{ 1 + \text {e}^{- { \mathchoice{\dfrac{2\pi }{ \tau }}{\dfrac{2\pi }{ \tau }}{\frac{2\pi }{ \tau }}{\frac{2\pi }{ \tau }} } z } }}{\frac{ \widetilde{\gamma }_{+} \cdot z }{ 1 + \text {e}^{- { \mathchoice{\dfrac{2\pi }{ \tau }}{\dfrac{2\pi }{ \tau }}{\frac{2\pi }{ \tau }}{\frac{2\pi }{ \tau }} } z } }}{\frac{ \widetilde{\gamma }_{+} \cdot z }{ 1 + \text {e}^{- { \mathchoice{\dfrac{2\pi }{ \tau }}{\dfrac{2\pi }{ \tau }}{\frac{2\pi }{ \tau }}{\frac{2\pi }{ \tau }} } z } }} } \quad \text {with} \quad \omega ^{(R)}(z) = { \mathchoice{\dfrac{ \widetilde{\gamma }_{-} \cdot z }{ 1 + \text {e}^{ { \mathchoice{\dfrac{2\pi }{ \tau }}{\dfrac{2\pi }{ \tau }}{\frac{2\pi }{ \tau }}{\frac{2\pi }{ \tau }} } z } }}{\dfrac{ \widetilde{\gamma }_{-} \cdot z }{ 1 + \text {e}^{ { \mathchoice{\dfrac{2\pi }{ \tau }}{\dfrac{2\pi }{ \tau }}{\frac{2\pi }{ \tau }}{\frac{2\pi }{ \tau }} } z } }}{\frac{ \widetilde{\gamma }_{-} \cdot z }{ 1 + \text {e}^{ { \mathchoice{\dfrac{2\pi }{ \tau }}{\dfrac{2\pi }{ \tau }}{\frac{2\pi }{ \tau }}{\frac{2\pi }{ \tau }} } z } }}{\frac{ \widetilde{\gamma }_{-} \cdot z }{ 1 + \text {e}^{ { \mathchoice{\dfrac{2\pi }{ \tau }}{\dfrac{2\pi }{ \tau }}{\frac{2\pi }{ \tau }}{\frac{2\pi }{ \tau }} } z } }} } , \end{aligned}$$

(A.1)

with $\widetilde{\gamma }_{\pm }$ as given in the statement of Proposition 1.1 and where $\tau >2\alpha$. These admit the decomposition

$$\begin{aligned} \omega ^{(L)}(z) =\widetilde{\gamma }_{+} \cdot z + \omega ^{(L)}_R(z) \quad \text {and} \quad \omega ^{(R)}(z) =\widetilde{\gamma }_{-} \cdot z + \omega ^{(R)}_L(z) . \end{aligned}$$

(A.2)

Let us set $\Omega (z\mid \varkappa ^{+}, \varkappa ^{-}) = \Upsilon (z) + \omega ^{(L)}(z) + \omega ^{(R)}(z)$. Then, $\Upsilon$ solves the non-local Riemann–Hilbert problem: find $\Upsilon \in \mathcal {O}(\mathcal {S}_{\alpha })$ such that

• $\Upsilon$ has smooth −, resp. $+$, boundary values on ${\mathbb {R}}$, resp. ${\mathbb {R}}-\text {i}\alpha$;

• $\Upsilon _{+}\big (g(x)-\text {i}\alpha \big ) = \Upsilon _{-}\big ( x \big )+ G_{\Upsilon }(x)$, with $x \in {\mathbb {R}}$;

• there exists a constants $C_{\Upsilon }$ and $\eta >0$ such that

  $$\begin{aligned} \Upsilon (z) =\left\{ \begin{array}{ccc} -\omega ^{(L)}_R(z) \, -\, \omega ^{(R)}(z) + \text {O}\Big ( \text {e}^{ - \eta \mathfrak {R}(z) } \Big ) \quad &{} \text {when} \quad &{} \mathfrak {R}(z) \rightarrow + \infty \vspace{2mm}\\ -\omega ^{(L)}(z) \, -\, \omega ^{(R)}_L(z) + C_{\Upsilon } \delta _{\pm ; -} + \text {O}\Big ( \text {e}^{ \eta \mathfrak {R}(z) } \Big ) \quad &{} \text {when} \quad &{} \mathfrak {R}(z) \rightarrow - \infty \end{array} \right. , \end{aligned}$$

  (A.3)

  with an asymptotic expansion that is valid uniformly up to the boundary,

and where

$$\begin{aligned} G_{\Upsilon }(x) = \left\{ \begin{array}{ccc} \omega ^{(L)}(x) - \omega ^{(L)}\big ( g(x) - \text {i}\alpha \big ) + \omega ^{(R)}_L(x) - \omega ^{(R)}_L\big ( g(x) - \text {i}\alpha \big ) &{} \text {for} &{} x \le - M \\ -\text {i}\alpha + \omega ^{(L)}(x) - \omega ^{(L)}\big ( g(x) - \text {i}\alpha \big ) + \omega ^{(R)}(x) - \omega ^{(R)}\big ( g(x) - \text {i}\alpha \big ) &{} \text {for} &{} |x| \le M \\ \omega ^{(L)}_R(x) - \omega ^{(L)}_R\big ( g(x) - \text {i}\alpha \big ) + \omega ^{(R)}(x) - \omega ^{(R)}\big ( g(x) - \text {i}\alpha \big ) &{} \text {for} &{} x \ge M \end{array} \right. \end{aligned}$$

(A.4)

By virtue of Proposition 2.4, this non-local Riemann–Hilbert problem admits a unique solution. Hence, so does the one of $\Omega$.

Since $\Omega \in \mathcal {O}(\mathcal {S}_{\alpha })$, $\Omega$ is open and thus

$$\begin{aligned} \partial _{} \Omega (\mathcal {S}_{\alpha }\mid \varkappa ^{+},\varkappa ^{-}) =\Omega _{+}({\mathbb {R}}-\text {i}\alpha \mid \varkappa ^{+},\varkappa ^{-})\cup \Omega _{-}({\mathbb {R}}\mid \varkappa ^{+},\varkappa ^{-}). \end{aligned}$$

Thus, clearly,

$$\begin{aligned} \Omega (\mathcal {S}_{\alpha }\mid \varkappa ^{+},\varkappa ^{-}) \ni \omega \; \mapsto \; \# \Big \{ \Omega ^{-1}(\omega \mid \varkappa ^{+},\varkappa ^{-}) \Big \} = \int \limits _{ \big \{ {\mathbb {R}}-\text {i}\alpha \} \cup -{\mathbb {R}}}^{} { \mathchoice{\dfrac{ \text {d}s }{2\text {i}\pi }}{\dfrac{ \text {d}s }{2\text {i}\pi }}{\frac{ \text {d}s }{2\text {i}\pi }}{\frac{ \text {d}s }{2\text {i}\pi }} } { \mathchoice{\dfrac{ \Omega ^{\prime }(s\mid \varkappa ^{+},\varkappa ^{-}) }{ \Omega (s\mid \varkappa ^{+}, \varkappa ^{-})- \omega }}{\dfrac{ \Omega ^{\prime }(s\mid \varkappa ^{+},\varkappa ^{-}) }{ \Omega (s\mid \varkappa ^{+}, \varkappa ^{-})- \omega }}{\frac{ \Omega ^{\prime }(s\mid \varkappa ^{+},\varkappa ^{-}) }{ \Omega (s\mid \varkappa ^{+}, \varkappa ^{-})- \omega }}{\frac{ \Omega ^{\prime }(s\mid \varkappa ^{+},\varkappa ^{-}) }{ \Omega (s\mid \varkappa ^{+}, \varkappa ^{-})- \omega }} } \end{aligned}$$

(A.5)

is continuous in $\omega$ on $\Omega (\mathcal {S}_{\alpha }\mid \varkappa ^{+},\varkappa ^{-})$. Being integer valued, it is constant. The asymptotic behaviour of $\Omega$ at infinity entails that $\# \Big \{ \Omega ^{-1}(\omega \mid \varkappa ^{+}, \varkappa ^{-}) \Big \} = 1$ for $\mathfrak {R}(\omega )$ large enough and such that $\omega \in \Omega (\mathcal {S}_{\alpha }\mid \varkappa ^{+},\varkappa ^{-})$. Hence, $\Omega$ is injective and thus a biholomorphism on its image. $\square$

Inversion of the operators $\text {id}- \varvec{ \texttt {L} } ^{\upsilon \upsilon }$ on $L^{2}({\mathbb {R}}^{\upsilon })$

We now discuss the invertibility of $\varvec{ \texttt {id} } - \varvec{ \texttt {L} } ^{\upsilon \upsilon }$ with the help of the Wiener–Hopf technique, see e.g. [8], as will be detailed in the two next subsections. The method builds on the solution of a multiplicative Riemann–Hilbert problem involving the Fourier transform of the kernel $L^{\upsilon }(x)$.

1.1 Inversion of the operators $\text {id}- \varvec{ \texttt {L} } ^{++}$

For the purpose of the present section, we introduce the space

$$\begin{aligned} L^2_{C}({\mathbb {R}}^+) = \bigg \{ f \in L^2_{\mathrm{loc}}({\mathbb {R}}^+) : \; \exists C_f \; \text {and}\, \alpha > 0\quad f(x) = C_f + \text {O}\Big ( \text {e}^{-\alpha x} \Big ) \bigg \} . \end{aligned}$$

(B.1)

Proposition B.1

Let $L^{+}$ be as defined through (2.42) and consider the integral equation

$$\begin{aligned} f(x) - \int \limits _{0}^{+\infty } L^{+}(x-y) f(y) \text {d}y = h(x) , \qquad x \in {\mathbb {R}}^+ \end{aligned}$$

(B.2)

on $L^2_{C}({\mathbb {R}}^+)$ with h such that there exist $\eta >0$ so that

$$\begin{aligned} h(x) = \text {O}\Big ( \text {e}^{-\eta x} \Big ) , \quad when \quad x \rightarrow + \infty . \end{aligned}$$

(B.3)

Then Eq. (B.2) is uniquely solvable on $L^2_{C}({\mathbb {R}}^+)$ and the Fourier transform of the solution takes the form

$$\begin{aligned} \mathcal {F}\big [f](k) = { \mathchoice{\dfrac{1}{ \alpha _{\uparrow }^{(+)}(k) }}{\dfrac{1}{ \alpha _{\uparrow }^{(+)}(k) }}{\frac{1}{ \alpha _{\uparrow }^{(+)}(k) }}{\frac{1}{ \alpha _{\uparrow }^{(+)}(k) }} } \int \limits _{{\mathbb {R}}- \text {i}\eta ^{-} }^{} { \mathchoice{\dfrac{ \text {d}s }{ 2\text {i}\pi }}{\dfrac{ \text {d}s }{ 2\text {i}\pi }}{\frac{ \text {d}s }{ 2\text {i}\pi }}{\frac{ \text {d}s }{ 2\text {i}\pi }} } { \mathchoice{\dfrac{ \alpha _{\downarrow }^{(+)}(s) \cdot \mathcal {F}\big [ h \varvec{1}_{{\mathbb {R}}^+} \big ](s) }{ s-k }}{\dfrac{ \alpha _{\downarrow }^{(+)}(s) \cdot \mathcal {F}\big [ h \varvec{1}_{{\mathbb {R}}^+} \big ](s) }{ s-k }}{\frac{ \alpha _{\downarrow }^{(+)}(s) \cdot \mathcal {F}\big [ h \varvec{1}_{{\mathbb {R}}^+} \big ](s) }{ s-k }}{\frac{ \alpha _{\downarrow }^{(+)}(s) \cdot \mathcal {F}\big [ h \varvec{1}_{{\mathbb {R}}^+} \big ](s) }{ s-k }} } \quad with \quad k \in {\mathbb {R}}+ \text {i}v \end{aligned}$$

(B.4)

for any $0<\eta ^{-}<\eta$ and with $v >0$.

Proof

Following the strategy of the Wiener–Hopf method, one starts by extending f and h to ${\mathbb {R}}$ in such a way that Eq. (B.2) now holds on ${\mathbb {R}}$. We shall make the choice

$$\begin{aligned} h(x)=0 \quad \text {and} \quad f(x)= \int \limits _{0}^{+\infty } L^{+}(x-y) f(y) \text {d}y \quad \text {for} \;\; x <0 . \end{aligned}$$

(B.5)

Given the behaviour of f on ${\mathbb {R}}^+$ and the explicit expression (2.42) for $L^{+}(x)$, it is easy to convince oneself that, upon reducing $\eta$ if need be, these extensions satisfy

$$\begin{aligned} f(x) = \text {O}\Big ( \text {e}^{\eta x} \Big ) \quad \text {and} \quad h(x) =\text {O}\Big ( \text {e}^{\eta x} \Big ) , \end{aligned}$$

(B.6)

when $x \rightarrow - \infty$. Actually, for the purpose of the analysis to come, it is convenient to introduce a specific notation for the restrictions of a function on ${\mathbb {R}}$ to ${\mathbb {R}}^{\pm }$: $f^{\pm }=f\varvec{1}_{{\mathbb {R}}^{\pm }}$. In particular, by construction, we have that $h=h^{+}$. The properties of the extended functions allow one to compute a well-defined Fourier transform provided that $k \in {\mathbb {R}}+ \text {i}v$, $0<v < \eta$. Thus Fourier transforming (B.2) leads to

$$\begin{aligned} \Big ( 1 - \mathcal {F}\big [ L^{+} \big ](k) \Big ) \cdot \mathcal {F}\big [ f^{+} \big ](k) + \mathcal {F}\big [ f^{-} \big ](k) = \mathcal {F}\big [ h^+ \big ](k) \quad \text {with} \quad k \in {\mathbb {R}}+ \text {i}v . \end{aligned}$$

(B.7)

Then, by using the Wiener–Hopf factorisation of $1 - \mathcal {F}\big [ L^{++} \big ]$ given in (2.53), one may recast the equation as

$$\begin{aligned} \alpha _{\uparrow }^{(+)}(k) \cdot \mathcal {F}\big [ f^{+} \big ](k) + \alpha _{\downarrow }^{(+)}(k) \cdot \mathcal {F}\big [ f^{-} \big ](k) = \alpha _{\downarrow }^{(+)}(k) \cdot \mathcal {F}\big [ h^+ \big ](k) . \end{aligned}$$

(B.8)

Given half-planes $\mathcal {B}_{\uparrow / \downarrow }^{(+)}$ as introduced in (2.52), one may define $U\in \mathcal {O}\Big ( \mathcal {B}_{\uparrow }^{(+)} \cup \mathcal {B}_{\downarrow }^{(+)} {\setminus } \{0\} \Big )$ and having a simple pole at 0 by the piecewise formula

$$\begin{aligned} U(z) = \left\{ \begin{array}{cc} \alpha _{\uparrow }^{(+)}(z) \cdot \mathcal {F}\big [ f^{+} \big ](z) - \varvec{ \texttt {C} } ^{(+)}\Big [ \alpha _{\downarrow }^{(+)} \cdot \mathcal {F}\big [ h^+ \big ] \Big ] (z) \quad &{}, \quad z \in \mathcal {B}_{\uparrow }^{(+)} \vspace{2mm} \\ - \alpha _{\downarrow }^{(+)}(z) \cdot \mathcal {F}\big [ f^{-} \big ](z) - \varvec{ \texttt {C} } ^{(+)}\Big [ \alpha _{\downarrow }^{(+)} \cdot \mathcal {F}\big [ h^+ \big ] \Big ] (z) &{}, \quad z \in \mathcal {B}_{\downarrow }^{(+)} \end{array} \right. \end{aligned}$$

(B.9)

where $\varvec{ \texttt {C} } ^{(+)}$ is the Cauchy transform on $L^2({\mathbb {R}}+\text {i}v )$:

$$\begin{aligned} \varvec{ \texttt {C} } ^{(+)}\big [u \big ] (z) = \int \limits _{ {\mathbb {R}}+ \text {i}v }^{} { \mathchoice{\dfrac{ \text {d}s }{ 2\text {i}\pi }}{\dfrac{ \text {d}s }{ 2\text {i}\pi }}{\frac{ \text {d}s }{ 2\text {i}\pi }}{\frac{ \text {d}s }{ 2\text {i}\pi }} } { \mathchoice{\dfrac{ u(s) }{ s-z }}{\dfrac{ u(s) }{ s-z }}{\frac{ u(s) }{ s-z }}{\frac{ u(s) }{ s-z }} } \quad \text {for} \quad z \in {\mathbb {C}}{\setminus } \big \{ {\mathbb {R}}+\text {i}v \big \}. \end{aligned}$$

(B.10)

Then, by using the relation valid for any $u\in L^p({\mathbb {R}}+\text {i}v)$, $+\infty>p>1$,

$$\begin{aligned} \varvec{ \texttt {C} } _{+}^{(+)}\big [u \big ](k) - \varvec{ \texttt {C} } _{-}^{(+)}\big [u \big ](k) = u(k) , \end{aligned}$$

(B.11)

one gets that $U_+=U_-$ on ${\mathbb {R}}+\text {i}v$ and hence U extends into a meromorphic function on ${\mathbb {C}}$ whose single pole is located at 0 and is simple. Moreover, it follows from (B.9) that

$$\begin{aligned} U(z) = -{ \mathchoice{\dfrac{ \widetilde{\alpha }_{0}^{\,(+)} }{ z }}{\dfrac{ \widetilde{\alpha }_{0}^{\,(+)} }{ z }}{\frac{ \widetilde{\alpha }_{0}^{\,(+)} }{ z }}{\frac{ \widetilde{\alpha }_{0}^{\,(+)} }{ z }} } \mathcal {F}[f^{-}](0) \; + \; \text {O}(1) . \end{aligned}$$

(B.12)

It is easy to see that $\mathcal {F}\big [ f^{+/-} \big ](k) \rightarrow 0$ when $k\rightarrow \infty$ in $\overline{\mathcal {B}_{\uparrow /\downarrow }^{(+)}}$ and this up to the boundary. Hence, $U(k) \rightarrow 0$ as $k\rightarrow \infty$. Since the constant $\mathcal {F}[f^{-}](0)$ is part of the unknowns in the problem, we conclude that there exists a constant $\mathcal {K}^{(+)}$ such that

$$\begin{aligned} U(z) = - { \mathchoice{\dfrac{ \widetilde{\alpha }^{\, (+)}_{0} \mathcal {K}^{(+)} }{ z }}{\dfrac{ \widetilde{\alpha }^{\, (+)}_{0} \mathcal {K}^{(+)} }{ z }}{\frac{ \widetilde{\alpha }^{\, (+)}_{0} \mathcal {K}^{(+)} }{ z }}{\frac{ \widetilde{\alpha }^{\, (+)}_{0} \mathcal {K}^{(+)} }{ z }} } . \end{aligned}$$

(B.13)

This explicit expression for U entails that, for any $k \in \mathcal {B}_{\uparrow }^{(+)}$,

$$\begin{aligned} \mathcal {F}\big [ f^{+} \big ](k) = { \mathchoice{\dfrac{1}{ \alpha _{\uparrow }^{(+)}(k) }}{\dfrac{1}{ \alpha _{\uparrow }^{(+)}(k) }}{\frac{1}{ \alpha _{\uparrow }^{(+)}(k) }}{\frac{1}{ \alpha _{\uparrow }^{(+)}(k) }} } \cdot \bigg \{ \varvec{ \texttt {C} } ^{(+)} \Big [ \alpha _{\downarrow }^{(+)} \cdot \mathcal {F}\big [ h^+ \big ] \Big ] (k) - { \mathchoice{\dfrac{ \widetilde{\alpha }^{\, (+)}_{0} \mathcal {K}^{(+)} }{ k }}{\dfrac{ \widetilde{\alpha }^{\, (+)}_{0} \mathcal {K}^{(+)} }{ k }}{\frac{ \widetilde{\alpha }^{\, (+)}_{0} \mathcal {K}^{(+)} }{ k }}{\frac{ \widetilde{\alpha }^{\, (+)}_{0} \mathcal {K}^{(+)} }{ k }} } \bigg \} . \end{aligned}$$

(B.14)

Note that, owing to (B.3), one may meromorphically continue $\mathcal {F}\big [ f^{+} \big ](k)$ from $\mathcal {B}_{\uparrow }^{(+)}$ up to $\Big \{ z \in {\mathbb {C}}\, : \, \mathfrak {I}(z)> - \eta \Big \}$ by the expression

$$\begin{aligned} \mathcal {F}\big [ f^{+} \big ](k) = { \mathchoice{\dfrac{1}{ \alpha _{\uparrow }^{(+)}(k) }}{\dfrac{1}{ \alpha _{\uparrow }^{(+)}(k) }}{\frac{1}{ \alpha _{\uparrow }^{(+)}(k) }}{\frac{1}{ \alpha _{\uparrow }^{(+)}(k) }} } \cdot \bigg \{ \varvec{ \texttt {C} } ^{(+)} \Big [ \alpha _{\downarrow }^{(+)} \cdot \mathcal {F}\big [ h^+ \big ] \Big ] (k)+ \alpha _{\downarrow }^{(+)}(k) \cdot \mathcal {F}\big [ h^+ \big ](k) - { \mathchoice{\dfrac{ \widetilde{\alpha }^{\, (+)}_{0} \mathcal {K}^{(+)} }{ k }}{\dfrac{ \widetilde{\alpha }^{\, (+)}_{0} \mathcal {K}^{(+)} }{ k }}{\frac{ \widetilde{\alpha }^{\, (+)}_{0} \mathcal {K}^{(+)} }{ k }}{\frac{ \widetilde{\alpha }^{\, (+)}_{0} \mathcal {K}^{(+)} }{ k }} } \bigg \} . \end{aligned}$$

(B.15)

Since $\alpha _{\uparrow }^{(+)}(k)$ has a simple zero at $k=0$, the expression above entails that $\mathcal {F}\big [ f^{+} \big ]$ may have a double pole at $k=0$, and that it is its sole pole in the domain $\mathfrak {I}(k)>-\eta ^{\prime }$, for some $\eta ^{\prime }>0$ and small enough.

Now assume that one is given a meromorphic function w in the tubular neighbourhood $|\mathfrak {I}(z)|<2\eta ^{\prime }$ of ${\mathbb {R}}$ having one pole of order $r+1$ at $k=0$:

$$\begin{aligned} w(k) = \sum \limits _{p=0}^{r} { \mathchoice{\dfrac{ w_p }{ k^{p+1} }}{\dfrac{ w_p }{ k^{p+1} }}{\frac{ w_p }{ k^{p+1} }}{\frac{ w_p }{ k^{p+1} }} } + \text {O}(1) \qquad k \rightarrow 0, \end{aligned}$$

(B.16)

and decaying at least as 1/k at infinity. Then, it is easy to convince oneself that, for $x \not =0$, one has

$$\begin{aligned} \int \limits _{{\mathbb {R}}+ \text {i}\eta ^{\prime } }^{} { \mathchoice{\dfrac{ \text {d}k }{ 2\pi }}{\dfrac{ \text {d}k }{ 2\pi }}{\frac{ \text {d}k }{ 2\pi }}{\frac{ \text {d}k }{ 2\pi }} } \text {e}^{-\text {i}k x } w(k) = - \text {i}\sum \limits _{p=0}^{r} { \mathchoice{\dfrac{ w_p }{ p! }}{\dfrac{ w_p }{ p! }}{\frac{ w_p }{ p! }}{\frac{ w_p }{ p! }} } \cdot (-\text {i}x)^{ p } \; + \; \int \limits _{{\mathbb {R}}- \text {i}\eta ^{\prime } }^{} { \mathchoice{\dfrac{ \text {d}k }{ 2\pi }}{\dfrac{ \text {d}k }{ 2\pi }}{\frac{ \text {d}k }{ 2\pi }}{\frac{ \text {d}k }{ 2\pi }} } \text {e}^{-\text {i}k x } w(k) . \end{aligned}$$

(B.17)

The integral appearing on the rhs of the above identity produces a $\text {O}\big ( \text {e}^{- \eta ^{\prime } x } \big )$ behaviour when $x\rightarrow +\infty$.

$f^{+}$ can be reconstructed from (B.14) by taking the inverse Fourier transform on ${\mathbb {R}}+\text {i}v$. One infers from (B.17) that the only way to give rise to a solution $f^+$ to (B.2) enjoying the asymptotic behaviour that is compatible with $f \in L^2_{C}({\mathbb {R}}^+)$, c.f. (B.1), is that the meromorphic continuation of $\mathcal {F}\big [ f^{+} \big ](k)$ has at most a simple pole at $k=0$. This entails that

$$\begin{aligned} \widetilde{\mathcal {K}}^{(+)} = \mathcal {F}\big [ h^+ \big ](0) . \end{aligned}$$

(B.18)

Thus, if a solution to (B.2) exists in the class (B.1), then it is unique and necessarily takes the form

$$\begin{aligned} \mathcal {F}\big [ f^{+} \big ](k) = { \mathchoice{\dfrac{1}{ \alpha _{\uparrow }^{(+)}(k) }}{\dfrac{1}{ \alpha _{\uparrow }^{(+)}(k) }}{\frac{1}{ \alpha _{\uparrow }^{(+)}(k) }}{\frac{1}{ \alpha _{\uparrow }^{(+)}(k) }} } \cdot \bigg \{ \varvec{ \texttt {C} } ^{(+)} \Big [ \alpha _{\downarrow }^{(+)} \cdot \mathcal {F}\big [ h^+ \big ] \Big ] (k) - { \mathchoice{\dfrac{ \widetilde{\alpha }^{\, (+)}_{0} \mathcal {F}\big [ h^+ \big ](0) }{ k }}{\dfrac{ \widetilde{\alpha }^{\, (+)}_{0} \mathcal {F}\big [ h^+ \big ](0) }{ k }}{\frac{ \widetilde{\alpha }^{\, (+)}_{0} \mathcal {F}\big [ h^+ \big ](0) }{ k }}{\frac{ \widetilde{\alpha }^{\, (+)}_{0} \mathcal {F}\big [ h^+ \big ](0) }{ k }} } \bigg \} , \end{aligned}$$

(B.19)

with $k \in \mathcal {B}^{(+)}_{\uparrow }$. By deforming the contour in the Cauchy transform from ${\mathbb {R}}+\text {i}v$ to ${\mathbb {R}}-\text {i}\eta ^-$ with $0<\eta ^{-}<\eta$ one obtains the representation (B.4).

Reciprocally, it is easy to see that the function f defined as

$$\begin{aligned} f^{\pm }(x) = \int \limits _{ {\mathbb {R}}+ \text {i}v }^{} { \mathchoice{\dfrac{ \text {d}k}{2\pi }}{\dfrac{ \text {d}k}{2\pi }}{\frac{ \text {d}k}{2\pi }}{\frac{ \text {d}k}{2\pi }} } \text {e}^{-\text {i}k x} \gamma ^{\pm }(k) \end{aligned}$$

with

$$\begin{aligned} \gamma ^{+}(k)= & {} { \mathchoice{\dfrac{1}{ \alpha _{\uparrow }^{(+)}(k) }}{\dfrac{1}{ \alpha _{\uparrow }^{(+)}(k) }}{\frac{1}{ \alpha _{\uparrow }^{(+)}(k) }}{\frac{1}{ \alpha _{\uparrow }^{(+)}(k) }} } \cdot \bigg \{ \varvec{ \texttt {C} } ^{(+)}_+ \Big [ \alpha _{\downarrow }^{(+)} \cdot \mathcal {F}\big [ h^+ \big ] \Big ] (k) - { \mathchoice{\dfrac{ \widetilde{\alpha }^{\, (+)}_{0} \mathcal {F}\big [ h^+ \big ](0) }{ k }}{\dfrac{ \widetilde{\alpha }^{\, (+)}_{0} \mathcal {F}\big [ h^+ \big ](0) }{ k }}{\frac{ \widetilde{\alpha }^{\, (+)}_{0} \mathcal {F}\big [ h^+ \big ](0) }{ k }}{\frac{ \widetilde{\alpha }^{\, (+)}_{0} \mathcal {F}\big [ h^+ \big ](0) }{ k }} } \bigg \} \nonumber \\ \gamma ^{-}(k)= & {} { \mathchoice{\dfrac{1}{ \alpha _{\downarrow }^{(+)}(k) }}{\dfrac{1}{ \alpha _{\downarrow }^{(+)}(k) }}{\frac{1}{ \alpha _{\downarrow }^{(+)}(k) }}{\frac{1}{ \alpha _{\downarrow }^{(+)}(k) }} } \cdot \bigg \{ { \mathchoice{\dfrac{ \widetilde{\alpha }^{\, (+)}_{0} \mathcal {F}\big [ h^+ \big ](0) }{ k }}{\dfrac{ \widetilde{\alpha }^{\, (+)}_{0} \mathcal {F}\big [ h^+ \big ](0) }{ k }}{\frac{ \widetilde{\alpha }^{\, (+)}_{0} \mathcal {F}\big [ h^+ \big ](0) }{ k }}{\frac{ \widetilde{\alpha }^{\, (+)}_{0} \mathcal {F}\big [ h^+ \big ](0) }{ k }} } - \varvec{ \texttt {C} } ^{(+)}_- \Big [ \alpha _{\downarrow }^{(+)} \cdot \mathcal {F}\big [ h^+ \big ] \Big ] (k) \bigg \}\nonumber \\ \end{aligned}$$

(B.20)

solves the linear integral equation (B.2) on ${\mathbb {R}}^{+}$.

Indeed, since $\gamma ^{+}$, resp. $\gamma ^{-}$, admits a holomorphic continuation to $\mathcal {B}_{\uparrow }^{(+)}$, resp. $\mathcal {B}_{\downarrow }^{(+)}$, that decays as $\text {O}(1/k)$ at infinity, one readily shows that, indeed, the function

$$\begin{aligned} x \mapsto \int \limits _{ {\mathbb {R}}+ \text {i}v }^{} { \mathchoice{\dfrac{ \text {d}k}{2\pi }}{\dfrac{ \text {d}k}{2\pi }}{\frac{ \text {d}k}{2\pi }}{\frac{ \text {d}k}{2\pi }} } \text {e}^{-\text {i}k x} \gamma ^{\pm }(k) \end{aligned}$$

(B.21)

are supported on ${\mathbb {R}}^{\pm }$ and that they exhibit the required asymptotic behaviour. The previous reasonings taken backwards then ensure that

$$\begin{aligned} \Big ( 1 - \mathcal {F}\big [ L^{+} \big ](k) \Big ) \cdot \gamma ^{+}(k) + \gamma ^{-}(k) = \mathcal {F}\big [ h^+ \big ](k) \quad \text {for} \quad k \in {\mathbb {R}}+ \text {i}v . \end{aligned}$$

(B.22)

Upon taking the inverse Fourier transform, the above relation leads to Eq. (B.2), hence proving the existence of solutions in $L^2_{C}({\mathbb {R}}^+)$. $\square$

1.2 Inversion of the operators $\text {id}- \varvec{ \texttt {L} } ^{--}$

Analogously to the previous setting, we introduce the space

$$\begin{aligned} L^2_{C}({\mathbb {R}}^-) = \bigg \{ f \in L^2({\mathbb {R}}^-) : \; \exists \; C_f \;\; \text {and}\;\; \alpha > 0 \quad f(x) = C_f + \text {O}\big ( \text {e}^{\alpha x} \big ) \bigg \} . \end{aligned}$$

(B.23)

Proposition B.2

Let $L^{-}$ be as defined through (2.42) and consider the integral equation

$$\begin{aligned} f(x) - \int \limits _{-\infty }^{0} L^{-}(x-y) f(y) \text {d}y = h(x) \quad for \quad x \in {\mathbb {R}}^- , \end{aligned}$$

(B.24)

on $L^2_{C}({\mathbb {R}}^-)$ with h such that there exist $\eta >0$ so that

$$\begin{aligned} h(x) = \text {O}\Big ( \text {e}^{\eta x} \Big ) \; \end{aligned}$$

(B.25)

when $x \rightarrow - \infty$.

Then, Eq. (B.2) is uniquely solvable on $L^2_{C}({\mathbb {R}}^-)$ and the Fourier transform of the solution takes the form

$$\begin{aligned} \mathcal {F}\big [f](k) = - \alpha _{\downarrow }^{(-)}(k) \int \limits _{{\mathbb {R}}+ \text {i}\eta ^{-} }^{} { \mathchoice{\dfrac{ \text {d}s }{ 2\text {i}\pi }}{\dfrac{ \text {d}s }{ 2\text {i}\pi }}{\frac{ \text {d}s }{ 2\text {i}\pi }}{\frac{ \text {d}s }{ 2\text {i}\pi }} } { \mathchoice{\dfrac{ \big \{ \alpha _{\uparrow }^{(-)}(s) \big \}^{-1} \cdot \mathcal {F}\big [ h \varvec{1}_{{\mathbb {R}}^-} \big ](s) }{ s-k }}{\dfrac{ \big \{ \alpha _{\uparrow }^{(-)}(s) \big \}^{-1} \cdot \mathcal {F}\big [ h \varvec{1}_{{\mathbb {R}}^-} \big ](s) }{ s-k }}{\frac{ \big \{ \alpha _{\uparrow }^{(-)}(s) \big \}^{-1} \cdot \mathcal {F}\big [ h \varvec{1}_{{\mathbb {R}}^-} \big ](s) }{ s-k }}{\frac{ \big \{ \alpha _{\uparrow }^{(-)}(s) \big \}^{-1} \cdot \mathcal {F}\big [ h \varvec{1}_{{\mathbb {R}}^-} \big ](s) }{ s-k }} } \quad with \quad k \in {\mathbb {R}}- \text {i}v \end{aligned}$$

(B.26)

for any $0<\eta ^{-}<\eta$ and with $v >0$.

Proof

One extends the functions f and h to ${\mathbb {R}}$ as

$$\begin{aligned} f(x) = \int \limits _{-\infty }^{0} L^{-}(x-y) f(y) \text {d}y \quad \text {and} \quad h(x) = 0 \quad \text {for} \quad x>0 , \end{aligned}$$

(B.27)

so that, reducing $\eta >0$ if need be, these extensions possess the $x \rightarrow + \infty$ asymptotic behaviour

$$\begin{aligned} f(x) = \text {O}\Big ( \text {e}^{ - \eta x} \Big ) \quad \text {and} \quad h(x) = \text {O}\Big ( \text {e}^{ - \eta x} \Big ) . \end{aligned}$$

(B.28)

One may then take the Fourier transform of (B.24) extended to ${\mathbb {R}}$, provided that the Fourier variable k satisfies $k \in {\mathbb {R}}-\text {i}v$, with $0<v\ll 1$. This leads to

$$\begin{aligned} \mathcal {F}\big [ f^{+} \big ](k) + \Big ( 1 - \mathcal {F}\big [ L^{-} \big ](k) \Big ) \mathcal {F}\big [ f^{-} \big ](k) = \mathcal {F}\big [ h \big ](k). \end{aligned}$$

(B.29)

Using the Wiener–Hopf factorisation of $1 - \mathcal {F}\big [ L^{-} \big ]$ relatively to ${\mathbb {R}}-\text {i}v$, one may recast the last equation as

$$\begin{aligned} \Big \{ \alpha _{\uparrow }^{(-)}(k) \Big \}^{-1} \cdot \mathcal {F}\big [ f^{+} \big ](k) + \Big \{ \alpha _{\downarrow }^{(-)}(k) \Big \}^{-1} \cdot \mathcal {F}\big [ f^{-} \big ](k) = \Big \{ \alpha _{\uparrow }^{(-)}(k) \Big \}^{-1} \cdot \mathcal {F}\big [ h \big ](k) . \end{aligned}$$

(B.30)

Define $U\in \mathcal {O}\Big ( \mathcal {B}_{\uparrow }^{(-)} \cup \mathcal {B}_{\downarrow }^{(-)} {\setminus } \{0\} \Big )$ by the piecewise formula

$$\begin{aligned} U(k) = \left\{ \begin{array}{cc} \varvec{ \texttt {C} } ^{(-)}\Big [ \big \{ \alpha _{\uparrow }^{(-)} \big \}^{-1} \cdot \mathcal {F}\big [ h \big ] \Big ] (k) - \Big \{ \alpha _{\uparrow }^{(-)}(k) \Big \}^{-1} \cdot \mathcal {F}\big [ f^{+} \big ](k) \qquad &{} , \quad z \in \mathcal {B}_{\uparrow }^{(-)} \vspace{2mm} \\ \Big \{ \alpha _{\downarrow }^{(-)}(k) \Big \}^{-1} \cdot \mathcal {F}\big [ f^{-} \big ](k) + \varvec{ \texttt {C} } ^{(-)}\Big [ \Big \{ \alpha _{\uparrow }^{(-)} \Big \}^{-1} \cdot \mathcal {F}\big [h \big ] \Big ] (k) &{} , \quad z \in \mathcal {B}_{\downarrow }^{(-)} \end{array} \right. \end{aligned}$$

(B.31)

where $\varvec{ \texttt {C} } ^{(-)}$ is the Cauchy transform on $L^2({\mathbb {R}}-\text {i}v)$:

$$\begin{aligned} \varvec{ \texttt {C} } ^{(-)}\big [u \big ] (z) = \int \limits _{ {\mathbb {R}}- \text {i}v }^{} { \mathchoice{\dfrac{ \text {d}s }{ 2\text {i}\pi }}{\dfrac{ \text {d}s }{ 2\text {i}\pi }}{\frac{ \text {d}s }{ 2\text {i}\pi }}{\frac{ \text {d}s }{ 2\text {i}\pi }} } { \mathchoice{\dfrac{ u(s) }{ s-z }}{\dfrac{ u(s) }{ s-z }}{\frac{ u(s) }{ s-z }}{\frac{ u(s) }{ s-z }} } \quad \text {for} \quad z \in {\mathbb {C}}{\setminus } \big \{ {\mathbb {R}}- \text {i}v \big \}. \end{aligned}$$

(B.32)

Since $\alpha _{\uparrow }^{(-)}(k)$ admits a simple zero at $k=0$, one gets that U is meromorphic on $\mathcal {B}_{\uparrow }\cup \mathcal {B}_{\downarrow }$. Its sole pole is located at $k=0$ and is simple. Moreover U vanishes at $\infty$ and satisfies $U_+=U_-$ on ${\mathbb {R}}-\text {i}v$. All of this allows one to infer that

$$\begin{aligned} U(k) = - { \mathchoice{\dfrac{ \mathcal {F}[f^{+}](0) }{ k \alpha _0^{(-)} }}{\dfrac{ \mathcal {F}[f^{+}](0) }{ k \alpha _0^{(-)} }}{\frac{ \mathcal {F}[f^{+}](0) }{ k \alpha _0^{(-)} }}{\frac{ \mathcal {F}[f^{+}](0) }{ k \alpha _0^{(-)} }} } . \end{aligned}$$

(B.33)

However, since $\mathcal {F}[f^{+}](0)$ is part of the unknowns in the problem, it is more convenient to set $\mathcal {K}^{(-)}=\mathcal {F}[f^{+}](0)$.

The expression (B.33) allows one to reconstruct the Fourier transform of $f^{-}$ for $k \in \mathcal {B}_{\downarrow }^{(-)}$ as:

$$\begin{aligned} \mathcal {F}\big [ f^{-} \big ](k) = - \alpha _{\downarrow }^{(-)}(k) \bigg \{ { \mathchoice{\dfrac{ \mathcal {K}^{(-)} }{k \cdot \alpha _0^{(-)} }}{\dfrac{ \mathcal {K}^{(-)} }{k \cdot \alpha _0^{(-)} }}{\frac{ \mathcal {K}^{(-)} }{k \cdot \alpha _0^{(-)} }}{\frac{ \mathcal {K}^{(-)} }{k \cdot \alpha _0^{(-)} }} } + \varvec{ \texttt {C} } ^{(-)} \Big [ \big \{ \alpha _{\uparrow }^{(-)} \big \}^{-1} \cdot \mathcal {F}\big [ h^- \big ] \Big ] (k) \bigg \} . \end{aligned}$$

(B.34)

The meromorphic continuation of $\mathcal {F}\big [ f^{-} \big ](k)$ to $\mathcal {B}_{\uparrow }^{(-)}$ takes the form

$$\begin{aligned} \mathcal {F}\big [ f^{-} \big ](k) =&- \alpha _{\downarrow }^{(-)}(k) \bigg \{ { \mathchoice{\dfrac{ \mathcal {K}^{(-)} }{k \cdot \alpha _0^{(-)} }}{\dfrac{ \mathcal {K}^{(-)} }{k \cdot \alpha _0^{(-)} }}{\frac{ \mathcal {K}^{(-)} }{k \cdot \alpha _0^{(-)} }}{\frac{ \mathcal {K}^{(-)} }{k \cdot \alpha _0^{(-)} }} } - \big \{ \alpha _{\uparrow }^{(-)} (k) \big \}^{-1} \cdot \mathcal {F}\big [ h^- \big ](k) + \varvec{ \texttt {C} } ^{(-)} \Big [ \big \{ \alpha _{\uparrow }^{(-)} \big \}^{-1} \cdot \mathcal {F}\big [ h^- \big ] \Big ] (k) \bigg \} \quad \text {for} \quad k \in \mathcal {B}_{\uparrow }^{(-)} . \end{aligned}$$

(B.35)

The function $\alpha _{\downarrow }^{(-)}(k)$ admits a simple pole at $k=0$. For generic $\mathcal {K}^{(-)}$, the term under the bracket also admits a simple pole at k, so that the meromorphic continuation has a double pole at $k=0$. As in the case of the Wiener–Hopf equation on ${\mathbb {R}}^+$, contour displacements in the inverse Fourier transform ensure that if $f^{-}$ has at most constant asymptotics at $-\infty$ then the meromorphic continuation of $\mathcal {F}\big [ f^{-} \big ](k)$ must have at most a simple pole at $k=0$. This unambiguously fixes the unknown constant as $\mathcal {K}^{(-)}=\mathcal {F}[h^{-}](0)$, leading to

$$\begin{aligned} \mathcal {F}\big [ f^{-} \big ](k) = - \alpha _{\downarrow }^{(-)}(k) \bigg \{ { \mathchoice{\dfrac{\mathcal {F}[h^{-}](0) }{k \cdot \alpha _0^{(-)} }}{\dfrac{\mathcal {F}[h^{-}](0) }{k \cdot \alpha _0^{(-)} }}{\frac{\mathcal {F}[h^{-}](0) }{k \cdot \alpha _0^{(-)} }}{\frac{\mathcal {F}[h^{-}](0) }{k \cdot \alpha _0^{(-)} }} } + \varvec{ \texttt {C} } ^{(-)} \Big [ \big \{ \alpha _{\uparrow }^{(-)} \big \}^{-1} \cdot \mathcal {F}\big [ h^- \big ] \Big ] (k) \bigg \} \end{aligned}$$

(B.36)

for any $k \in \mathcal {B}^{(-)}_{\downarrow }$. Upon deforming the contour in the Cauchy transform $\varvec{ \texttt {C} } ^{(-)}$ up to ${\mathbb {R}}+\text {i}\eta ^{-}$ with $0< \eta ^{-} <\eta$, one arrives to (B.26).

It is easy to see, proceeding similarly as before, that the above expression does give rise to a solution to (B.24). $\square$

Inversion of $\,\text {id}- \varvec{ \texttt {L} } _w^{(0)}$

1.1 Characterisation in terms of a Riemann–Hilbert problem

The operator $\text {id}- \varvec{ \texttt {L} } _w^{(0)}$ on $L^{2}( ] -w \,; w [ )$, as defined through (5.3) and (5.5), is a truncated Wiener–Hopf operator and, as such, can be explicitly inverted in terms of the solution to an auxiliary Riemann–Hilbert problem. Consider the operator $\varvec{ \texttt {V} }$ on $L^2({\mathbb {R}}+\text {i}v)$ with the kernel

$$\begin{aligned} V(k,s) \, =&\, -\mathcal {F}[L^{(0)}](k) \cdot { \mathchoice{\dfrac{ \text {e}^{\text {i}(k-s)w } - \text {e}^{-\text {i}(k-s)w } }{ 2\text {i}\pi (k-s) }}{\dfrac{ \text {e}^{\text {i}(k-s)w } - \text {e}^{-\text {i}(k-s)w } }{ 2\text {i}\pi (k-s) }}{\frac{ \text {e}^{\text {i}(k-s)w } - \text {e}^{-\text {i}(k-s)w } }{ 2\text {i}\pi (k-s) }}{\frac{ \text {e}^{\text {i}(k-s)w } - \text {e}^{-\text {i}(k-s)w } }{ 2\text {i}\pi (k-s) }} }\quad \text {where}\quad \mathcal {F}[L^{(0)}](k) = { \mathchoice{\dfrac{ \cosh \big [k(\tau /2-\alpha -\text {i}\varkappa ) \big ] }{ \cosh \big [k\tau /2 \big ] }}{\dfrac{ \cosh \big [k(\tau /2-\alpha -\text {i}\varkappa ) \big ] }{ \cosh \big [k\tau /2 \big ] }}{\frac{ \cosh \big [k(\tau /2-\alpha -\text {i}\varkappa ) \big ] }{ \cosh \big [k\tau /2 \big ] }}{\frac{ \cosh \big [k(\tau /2-\alpha -\text {i}\varkappa ) \big ] }{ \cosh \big [k\tau /2 \big ] }} } . \end{aligned}$$

(C.1)

Then, it is easy to see that $\mathcal {F}^{-1}( \text {id}+ \varvec{ \texttt {V} } )\mathcal {F} = \text {id}- \varvec{ \texttt {L} } _w^{(0)}$ or, more precisely, if f solves $\big (\text {id}- \varvec{ \texttt {L} } _w^{(0)}\big )[f]=h$ with $h\in L^{2}( ] -w \,; w [ )$, then

$$\begin{aligned} \Big ( \text {id}+ \varvec{ \texttt {V} } \Big )\Big [ \mathcal {F}[f]\Big ](k) = \mathcal {F}[h](k) \end{aligned}$$

(C.2)

for an appropriate extension of f outside $[ -w \,; w ]$.

Observe that

$$\begin{aligned} V(\lambda ,\mu ) = { \mathchoice{\dfrac{ \Big ( \varvec{E}_{L}(\lambda ), \varvec{E}_R(\mu ) \Big ) }{ \lambda -\mu }}{\dfrac{ \Big ( \varvec{E}_{L}(\lambda ), \varvec{E}_R(\mu ) \Big ) }{ \lambda -\mu }}{\frac{ \Big ( \varvec{E}_{L}(\lambda ), \varvec{E}_R(\mu ) \Big ) }{ \lambda -\mu }}{\frac{ \Big ( \varvec{E}_{L}(\lambda ), \varvec{E}_R(\mu ) \Big ) }{ \lambda -\mu }} } \end{aligned}$$

(C.3)

where, upon setting $e(\lambda )=\text {e}^{\text {i}w \lambda }$,

$$\begin{aligned} \varvec{E}_{R}(\mu )= { \mathchoice{\dfrac{1}{2\text {i}\pi }}{\dfrac{1}{2\text {i}\pi }}{\frac{1}{2\text {i}\pi }}{\frac{1}{2\text {i}\pi }} }\left( \begin{array}{c} e(\mu ) \\ e^{-1}(\mu ) \end{array} \right) \quad \text {and} \quad \varvec{E}_{L}(\lambda )= -\mathcal {F}[L^{(0)}](\lambda ) \left( \begin{array}{c} -e^{-1}(\lambda ) \\ e(\lambda ) \end{array} \right) , \end{aligned}$$

(C.4)

so that $\Big ( \varvec{E}_{L}(\lambda ), \varvec{E}_R(\lambda ) \Big ) = 0$. This means that $\varvec{ \texttt {V} }$ is an integrable integral operator. As such, it can be studied by means of an associated Riemann–Hilbert problem as first observed in [13].

Assume that $\text {id}+ \varvec{ \texttt {V} }$ is invertible. Then, define the functions $\varvec{F}_{R/L}(\lambda )$ as the solutions to the linear integral equations

$$\begin{aligned} \big [\varvec{F}_{R}\big ] \Big ( \text {id}+ \varvec{ \texttt {V} } \Big )(\lambda ) = \varvec{E}_{R}(\lambda ) \quad \text {and} \quad \Big ( \text {id}+ \varvec{ \texttt {V} } \Big ) \big [\varvec{F}_{L}\big ](\lambda ) = \varvec{E}_{L}(\lambda ) . \end{aligned}$$

(C.5)

The first formula is to be understood as an action of the operator to the left and the second one as its action to the right.

We refer the reader to Sect. 1.4 where the notations used below are introduced.

Theorem C.1

There exists $w_0$ large enough such that the operator $\text {id}+ \varvec{ \texttt {V} }$ acting on $L^2({\mathbb {R}}+\text {i}v)$ with the integral kernel (C.3) is invertible for any $w\ge w_0$ with inverse given by $\text {id}- \varvec{ \texttt {R} }$.  The integral kernel of the resolvent operator $\varvec{ \texttt {R} }$ is expressed as

$$\begin{aligned} R(\lambda ,\mu ) = { \mathchoice{\dfrac{ \Big ( \varvec{F}_{L}(\lambda ), \varvec{F}_R(\mu ) \Big ) }{ \lambda -\mu }}{\dfrac{ \Big ( \varvec{F}_{L}(\lambda ), \varvec{F}_R(\mu ) \Big ) }{ \lambda -\mu }}{\frac{ \Big ( \varvec{F}_{L}(\lambda ), \varvec{F}_R(\mu ) \Big ) }{ \lambda -\mu }}{\frac{ \Big ( \varvec{F}_{L}(\lambda ), \varvec{F}_R(\mu ) \Big ) }{ \lambda -\mu }} } . \end{aligned}$$

(C.6)

The vectors $\varvec{F}_{R/L}(\lambda )$ are given by

$$\begin{aligned} \varvec{F}_{R}(\lambda ) = \chi _{+}(\lambda )\cdot \varvec{E}_{R}(\lambda ) \quad and \quad \varvec{F}_{L}^{ \varvec{ \texttt {t} } }(\lambda ) = \varvec{E}_{L}^{ \varvec{ \texttt {t} } }(\lambda ) \cdot \chi _{+}^{-1}(\lambda ) . \end{aligned}$$

(C.7)

Above, $\varvec{ \texttt {t} }$ is the vector transposition while $\chi$ corresponds to the unique solution to the matrix Riemann–Hilbert problem for²$\chi$:  find $\,\chi \in \mathcal {M}_2\Big ( \mathcal {O}\big ( {\mathbb {C}}{\setminus } \big \{ {\mathbb {R}}+ \text {i}v \big \} \big ) \Big )\,$ such that

• $\chi (\lambda )=I_2+\text {O}\Big ({ \mathchoice{\dfrac{1}{\lambda }}{\dfrac{1}{\lambda }}{\frac{1}{\lambda }}{\frac{1}{\lambda }} }\Big )$ when $\lambda \rightarrow \infty$;

• $\chi$ admits continuous ± boundary values on ${\mathbb {R}}$ such that $\chi _{\pm }-I_2 \in \mathcal {M}_2\big ( L^{2}({\mathbb {R}}+\text {i}v) \big )$. These boundary values are related by

  $$\begin{aligned} \chi _+(\lambda )\, G_{\chi }(\lambda ) = \chi _{-}(\lambda )\, , \end{aligned}$$

  (C.8)

  where the jump matrix takes the form

  $$\begin{aligned} G_{\chi }(\lambda ) = I_2 + 2\text {i}\pi \varvec{E}_{R}(\lambda ) \cdot \varvec{E}_{L}^{ \varvec{ \texttt {t} } } (\lambda ) = \left( \begin{array}{cc} 1+ \mathcal {F}[L^{(0)}](\lambda ) &{} -\mathcal {F}[L^{(0)}](\lambda ) \, e^{2}(\lambda ) \vspace{2mm} \\ \mathcal {F}[L^{(0)}](\lambda ) \, e^{-2}(\lambda ) &{} 1 - \mathcal {F}[L^{(0)}](\lambda ) \end{array} \right) . \end{aligned}$$

  (C.9)

  The unique solution $\chi$ takes the explicit form given in Fig. 9.  It admits the integral representations

  $$\begin{aligned} \chi (\lambda ) \, = \; I_2 - \int \limits _{ {\mathbb {R}}}^{} { \mathchoice{\dfrac{ \varvec{F}_{R}(\mu ) \cdot \varvec{E}_{L}^{ \varvec{ \texttt {t} } }(\mu ) }{ \mu - \lambda }}{\dfrac{ \varvec{F}_{R}(\mu ) \cdot \varvec{E}_{L}^{ \varvec{ \texttt {t} } }(\mu ) }{ \mu - \lambda }}{\frac{ \varvec{F}_{R}(\mu ) \cdot \varvec{E}_{L}^{ \varvec{ \texttt {t} } }(\mu ) }{ \mu - \lambda }}{\frac{ \varvec{F}_{R}(\mu ) \cdot \varvec{E}_{L}^{ \varvec{ \texttt {t} } }(\mu ) }{ \mu - \lambda }} } \text {d}\mu \quad and \quad \chi ^{-1}(\lambda ) \, = \; I_2 + \int \limits _{ {\mathbb {R}}}^{} { \mathchoice{\dfrac{ \varvec{E}_{R}(\mu ) \cdot \varvec{F}_{L}^{ \varvec{ \texttt {t} } }(\mu ) }{ \mu - \lambda }}{\dfrac{ \varvec{E}_{R}(\mu ) \cdot \varvec{F}_{L}^{ \varvec{ \texttt {t} } }(\mu ) }{ \mu - \lambda }}{\frac{ \varvec{E}_{R}(\mu ) \cdot \varvec{F}_{L}^{ \varvec{ \texttt {t} } }(\mu ) }{ \mu - \lambda }}{\frac{ \varvec{E}_{R}(\mu ) \cdot \varvec{F}_{L}^{ \varvec{ \texttt {t} } }(\mu ) }{ \mu - \lambda }} } \text {d}\mu . \end{aligned}$$

  (C.10)

Most results stated in Theorem C.1 are classic and go back to the work [13]. The representation given in Fig. 9 is established throughout Sect. (C.2) to come by a rather standard application of the non-linear steepest descent method [7]. It is a standard fact, which follows from ${\text {det}}G_{\chi }=1$, that the Riemann–Hilbert problem for $\chi$ admits a unique solution, see e.g. [6]. Thus, we will not discuss this question further.

Fig. 9

Piecewise definition of the matrix $\chi$. The curves $\Gamma _{\uparrow /\downarrow }$ separate all poles, other that at 0, of $\lambda \mapsto \mathcal {F}[L^{(0)}](\lambda ) \cdot \Big \{ 1 - \mathcal {F}[L^{(0)}](\lambda ) \Big \}^{-1}$ from ${\mathbb {R}}$ and are such that $\text {dist}(\Gamma _{\uparrow /\downarrow }, {\mathbb {R}}) > \varrho$ for some $\varrho >0$. The piecewise holomorphic matrix $\Pi$ appearing in the above figure is as defined through (C.26)

1.2 Asymptotic resolution of the Riemann–Hilbert problem

1.2.1 Riemann–Hilbert problem for $\Xi$

First, we consider the solution to an auxiliary scalar Riemann–Hilbert problem. Let

$$\begin{aligned} \mathcal {B}_{\uparrow }^{(0)} = \Big \{ z \in {\mathbb {C}}: \; \mathfrak {I}z > v \Big \} \quad \text {and}\quad \mathcal {B}_{\downarrow }^{(0)} = \Big \{ z \in {\mathbb {C}}: \; \mathfrak {I}z < v \Big \} . \end{aligned}$$

(C.11)

One introduces the function

$$\begin{aligned} \alpha ^{(0)} \in \mathcal {O}\big ( {\mathbb {C}}^* {\setminus } \{{\mathbb {R}}+ \text {i}v \} \big ) \quad \text {with} \quad \alpha ^{(0)}(\lambda ) = \left\{ \begin{array}{cc} \alpha ^{(0)}_{\uparrow }(\lambda ) &{} \lambda \in \mathcal {B}_{\uparrow }^{(0)} \vspace{2mm}\\ \alpha ^{(0)}_{\downarrow }(\lambda ) &{} \lambda \in \mathcal {B}_{\downarrow }^{(0)} \end{array} \right. \end{aligned}$$

(C.12)

in which³$\alpha ^{(0)}_{\uparrow }\in \mathcal {O}\Big ( \overline{\mathcal {B}}_{\uparrow }^{(0)} \Big )$, $\,\alpha ^{(0)}_{\downarrow }\in \mathcal {O}\Big ( \overline{\mathcal {B}}_{\downarrow }^{(0)} {\setminus } \{ 0 \} \Big )$, $\,\alpha ^{(0)}_{\uparrow /\downarrow }(\lambda ) \rightarrow 1$ when $\lambda \rightarrow \infty$ in $\overline{\mathcal {B}}_{\uparrow /\downarrow }^{(0)}$ and such that

$$\begin{aligned} { \mathchoice{\dfrac{ \alpha ^{(0)}_{\uparrow }(\lambda ) }{ \alpha ^{(0)}_{\downarrow }(\lambda ) }}{\dfrac{ \alpha ^{(0)}_{\uparrow }(\lambda ) }{ \alpha ^{(0)}_{\downarrow }(\lambda ) }}{\frac{ \alpha ^{(0)}_{\uparrow }(\lambda ) }{ \alpha ^{(0)}_{\downarrow }(\lambda ) }}{\frac{ \alpha ^{(0)}_{\uparrow }(\lambda ) }{ \alpha ^{(0)}_{\downarrow }(\lambda ) }} } = 1 - \mathcal {F}[L^{(0)}](\lambda ) . \end{aligned}$$

(C.13)

$\alpha ^{(0)}_{\uparrow /\downarrow }$ admit meromorphic continuations to $\mathcal {B}_{\downarrow /\uparrow }^{(0)}$ such that

(C.14)

Note that $k=0$ is the only zero and pole of $\alpha _{\uparrow /\downarrow }^{(0)}$ in a fixed v-independent tubular neighbourhood of ${\mathbb {R}}$.

The functions $\alpha _{\uparrow /\downarrow }^{(0)}$ can be read out from Eqs. (2.54)–(2.55) upon the substitution $\varkappa ^{\upsilon } \hookrightarrow \varkappa$.

Assume that one is given a solution $\chi$ to the Riemann–Hilbert problem for $\chi$, and define

$$\begin{aligned} \Xi (\lambda ) \, = \, \chi (\lambda ) \cdot \Big ( \alpha ^{(0)}(\lambda ) \Big )^{-\sigma _3} . \end{aligned}$$

(C.15)

It is clear that the Riemann–Hilbert problem for $\chi$ is in one-to-one correspondence with the Riemann–Hilbert problem for $\Xi$. The latter consists in finding $\Xi \in \mathcal {M}_2\Big ( \mathcal {O}\big ( {\mathbb {C}}^{*} {\setminus } \big \{ {\mathbb {R}}+\text {i}v \big \} \big ) \Big )$ such that

• $\Xi$ admits a simple pole at 0;

• $\Xi (\lambda ) = I_2 + \text {O}\Big ({ \mathchoice{\dfrac{1}{\lambda }}{\dfrac{1}{\lambda }}{\frac{1}{\lambda }}{\frac{1}{\lambda }} }\Big )$ when $\lambda \rightarrow \infty$;

• $\Xi (\lambda ) \cdot \Big ( \alpha ^{(0)}_{\downarrow }(\lambda ) \Big )^{\sigma _3}$ is regular at $\lambda =0$;

• $\Xi$ admits continuous ± boundary values on ${\mathbb {R}}+\text {i}v$ such that $\Xi _{\pm }-I_2 \in \mathcal {M}_2\big ( L^{2}({\mathbb {R}}+\text {i}v) \big )$. These boundary values are related as

  $$\begin{aligned} \Xi _+(\lambda )\, G_{\Xi }(\lambda ) = \Xi _{-}(\lambda ), \quad \text {where} \quad G_{\Xi }(\lambda ) = \left( \begin{array}{cc} 1+ P(\lambda )Q(\lambda ) &{} P(\lambda ) e^{2}(\lambda ) \\ Q(\lambda ) e^{-2}(\lambda ) &{} 1 \end{array} \right) . \end{aligned}$$

  (C.16)

Note that the jump matrix factorises as $\,G_{\Xi }(\lambda ) = M_{\uparrow }(\lambda ) \cdot M_{\downarrow }(\lambda )$ in which

$$\begin{aligned} M_{\uparrow }(\lambda ) = \left( \begin{array}{cc} 1 &{} P(\lambda ) e^{2}(\lambda ) \\ 0 &{} 1 \end{array} \right) \quad \text {and} \quad M_{\downarrow }(\lambda ) = \left( \begin{array}{cc} 1 &{} 0 \\ Q(\lambda ) e^{-2}(\lambda ) &{} 1 \end{array} \right) . \end{aligned}$$

(C.17)

The expression for these matrices involve the functions

$$\begin{aligned} P(\lambda ) = - \alpha ^{(0)}_{\uparrow }(\lambda )\cdot \alpha ^{(0)}_{\downarrow }(\lambda ) \cdot \mathcal {F}[L^{(0)}](\lambda ) \quad \text {and} \quad Q(\lambda ) = { \mathchoice{\dfrac{ \mathcal {F}[L^{(0)}](\lambda ) }{ \alpha ^{(0)}_{\uparrow }(\lambda )\cdot \alpha ^{(0)}_{\downarrow }(\lambda ) }}{\dfrac{ \mathcal {F}[L^{(0)}](\lambda ) }{ \alpha ^{(0)}_{\uparrow }(\lambda )\cdot \alpha ^{(0)}_{\downarrow }(\lambda ) }}{\frac{ \mathcal {F}[L^{(0)}](\lambda ) }{ \alpha ^{(0)}_{\uparrow }(\lambda )\cdot \alpha ^{(0)}_{\downarrow }(\lambda ) }}{\frac{ \mathcal {F}[L^{(0)}](\lambda ) }{ \alpha ^{(0)}_{\uparrow }(\lambda )\cdot \alpha ^{(0)}_{\downarrow }(\lambda ) }} } . \end{aligned}$$

(C.18)

In particular, Q is analytic on a tubular neighbourhood of ${\mathbb {R}}$ and satisfies

$$\begin{aligned} Q(0)= { \mathchoice{\dfrac{1}{\alpha _0^{(0)}\cdot \widetilde{\alpha }_{0}^{(0)} }}{\dfrac{1}{\alpha _0^{(0)}\cdot \widetilde{\alpha }_{0}^{(0)} }}{\frac{1}{\alpha _0^{(0)}\cdot \widetilde{\alpha }_{0}^{(0)} }}{\frac{1}{\alpha _0^{(0)}\cdot \widetilde{\alpha }_{0}^{(0)} }} } = -1, \end{aligned}$$

(C.19)

see (2.58).

The matrices $M_{\uparrow /\downarrow }$ are such that their off-diagonal entries are exponentially small in w for $\lambda$ belonging to ${\mathbb {H}}^{\pm }$ and uniformly away from ${\mathbb {R}}$.

1.2.2 Riemann–Hilbert problem for $\Upsilon$

Next, one defines $\Upsilon$ as in Fig. 10. The contours $\Gamma _{\uparrow /\downarrow }$ are chosen such that it holds $\Gamma _{\uparrow }=-\Gamma _{\downarrow }$. It is clear that the Riemann–Hilbert problems for $\Xi$ is in one-to-one correspondence with the one for $\Upsilon$.

Find $\Upsilon \in \mathcal {M}_2\Big ( \mathcal {O}\Big ( {\mathbb {C}}^{*} {\setminus } \big \{ \Gamma _{\uparrow } \cup \Gamma _{\downarrow } \big \} \Big ) \Big )$ such that

• $\Upsilon$ admits a simple pole at 0;

• $\Upsilon (\lambda )=I_2+\text {O}\Big ({ \mathchoice{\dfrac{1}{\lambda }}{\dfrac{1}{\lambda }}{\frac{1}{\lambda }}{\frac{1}{\lambda }} }\Big )$ when $\lambda \rightarrow \infty$;

• $\Upsilon (\lambda )\cdot M_{\downarrow }(\lambda ) \cdot \Big ( \alpha ^{(0)}_{\downarrow }(\lambda ) \Big )^{\sigma _3}$ is regular at $\lambda =0$;

• $\Upsilon$ admits continuous ± boundary values on $\Gamma _{\uparrow } \cup \Gamma _{\downarrow }$ such that $\Upsilon _{\pm }-I_2 \in \mathcal {M}_2\big ( L^{2}(\Gamma _{\uparrow } \cup \Gamma _{\downarrow }) \big )$. These boundary values are related by

  $$\begin{aligned} \Upsilon _+(\lambda )\, G_{\Upsilon }(\lambda ) = \Upsilon _{-}(\lambda )\, \quad \text {with} \quad G_{\Upsilon }(\lambda ) = M_{\downarrow }(\lambda )\cdot \varvec{1}_{ \Gamma _{\downarrow } }(\lambda ) + M_{\uparrow }(\lambda )\cdot \varvec{1}_{ \Gamma _{\uparrow } }(\lambda ) . \end{aligned}$$

  (C.20)

Fig. 10

Piecewise definition of the matrix $\Upsilon$ in terms of the matrix $\Xi$. The curves $\Gamma _{\uparrow /\downarrow }$ separate all poles, other than the one at 0, of $\lambda \mapsto \mathcal {F}[L^{(0)}](\lambda ) \cdot \Big \{ 1 - \mathcal {F}[L^{(0)}](\lambda ) \Big \}^{-1}$ from ${\mathbb {R}}$ and are such that $\text {dist}(\Gamma _{\uparrow /\downarrow }, {\mathbb {R}}) > \varrho$ for some $\varrho >0$

1.2.3 Auxiliary Riemann–Hilbert problem for $\Pi$

To continue further, one first introduces $\Pi$ as the unique solution to the below Riemann–Hilbert problem for $\Pi$. Find $\Pi \in \mathcal {M}_2\Big ( \mathcal {O}\big ( {\mathbb {C}}{\setminus } \big \{ \Gamma _{\uparrow } \cup \Gamma _{\downarrow } \big \} \big ) \Big )$ such that:

• $\Pi (\lambda )=I_2+\text {O}\Big ({ \mathchoice{\dfrac{1}{\lambda }}{\dfrac{1}{\lambda }}{\frac{1}{\lambda }}{\frac{1}{\lambda }} }\Big )$ when $\lambda \rightarrow \infty$;

• $\Pi$ admits continuous ± boundary values on $\Gamma _{\uparrow } \cup \Gamma _{\downarrow }$ such that $\Pi _{\pm }-I_2 \in \mathcal {M}_2\big ( L^{2}(\Gamma _{\uparrow } \cup \Gamma _{\downarrow } ) \big )$. These boundary values are related by

  $$\begin{aligned} \Pi _+(\lambda )\, G_{\Upsilon }(\lambda ) = \Pi _{-}(\lambda ) . \end{aligned}$$

  (C.21)

Again, there exists at most a one solution to the Riemann–Hilbert problem for $\Pi$. Existence may be established by the singular integral equation method introduced in [1].

Indeed, introduce the singular integral operator on the space $\mathcal {M}_2\big ( L^2( \Gamma _{\uparrow } \cup \Gamma _{\downarrow } ) \big )$ of $2\times 2$ matrix-valued $L^2\big ( \Gamma _{\uparrow } \cup \Gamma _{\downarrow } \big )$ functions by

$$\begin{aligned} \mathcal {C}^{(+)}_{ \Gamma _{\uparrow } \cup \Gamma _{\downarrow } }\big [ \Psi \big ] (\lambda ) = \lim _{ \begin{array}{c} z \rightarrow \lambda \\ z \in + \text {side} \, \text {of} \, \Gamma _{\uparrow } \cup \Gamma _{\downarrow } \end{array} } \int _{ \Gamma _{\uparrow } \cup \Gamma _{\downarrow } }{} { \mathchoice{\dfrac{ \Psi (t)\cdot (G_{\Upsilon }-I_2)(t) }{t-z}}{\dfrac{ \Psi (t)\cdot (G_{\Upsilon }-I_2)(t) }{t-z}}{\frac{ \Psi (t)\cdot (G_{\Upsilon }-I_2)(t) }{t-z}}{\frac{ \Psi (t)\cdot (G_{\Upsilon }-I_2)(t) }{t-z}} } \cdot { \mathchoice{\dfrac{ \text {d}t}{ 2 \text {i}\pi }}{\dfrac{ \text {d}t}{ 2 \text {i}\pi }}{\frac{ \text {d}t}{ 2 \text {i}\pi }}{\frac{ \text {d}t}{ 2 \text {i}\pi }} } . \end{aligned}$$

(C.22)

Since $G_{\Upsilon }-I_2 \in \mathcal {M}_2\Big ( \big (L^{\infty } \cap L^2 \big ) \big ( \Gamma _{\uparrow } \cup \Gamma _{\downarrow } \big ) \Big )$ and $\Gamma _{\uparrow } \cup \Gamma _{\downarrow }$ is a Lipschitz curve, it follows from [4] that $\mathcal{C}^{(+)}_{ \Gamma _{\uparrow } \cup \Gamma _{\downarrow } }$ is continuous on $\mathcal {M}_2\Big (L^2\big ( \Gamma _{\uparrow } \cup \Gamma _{\downarrow } ) \Big )$ and fulfils:

$$\begin{aligned} \big | \big | \big | \mathcal {C}^{(+)}_{ \Gamma _{\uparrow } \cup \Gamma _{\downarrow } } \big | \big | \big |_{ \mathcal {M}_2(L^2( \Gamma _{\uparrow } \cup \Gamma _{\downarrow } )) } \; \le \; C \text {e}^{- \varrho w} . \end{aligned}$$

(C.23)

Hence, since

$$\begin{aligned} G_{\Upsilon }-I_2 \in \mathcal {M}_2\Big ( L^2\big ( \Gamma _{\uparrow } \cup \Gamma _{\downarrow } \big ) \Big ) \quad \text {and} \quad \mathcal {C}^{(+)}_{ \Gamma _{\uparrow } \cup \Gamma _{\downarrow } }[I_2] \in \mathcal {M}_2\Big ( L^2\big ( \Gamma _{\uparrow } \cup \Gamma _{\downarrow } \big ) \Big ) , \end{aligned}$$

(C.24)

provided that w is large enough, it follows that the singular integral equation

$$\begin{aligned} \Big (I_2 + \mathcal {C}^{(+)}_{ \Gamma _{\uparrow } \cup \Gamma _{\downarrow } } \Big ) \big [ \Pi _+ \big ] = I_2 \end{aligned}$$

(C.25)

admits a unique solution $\Pi _{+}$ such that $\Pi _{+} - I_2 \in \mathcal {M}_2\Big ( L^2 \big ( \Gamma _{\uparrow } \cup \Gamma _{\downarrow } ) \Big )$. It is then a standard fact [1] in the theory of Riemann–Hilbert problems that the matrix

$$\begin{aligned} \Pi (\lambda ) = I_2 \; - \; \int \limits _{ \Gamma _{\uparrow } \cup \Gamma _{\downarrow } }^{} { \mathchoice{\dfrac{ \Pi _+(t) (G_{\Upsilon }-I_2)(t) }{ t-\lambda }}{\dfrac{ \Pi _+(t) (G_{\Upsilon }-I_2)(t) }{ t-\lambda }}{\frac{ \Pi _+(t) (G_{\Upsilon }-I_2)(t) }{ t-\lambda }}{\frac{ \Pi _+(t) (G_{\Upsilon }-I_2)(t) }{ t-\lambda }} } \cdot { \mathchoice{\dfrac{ \text {d}t }{ 2\text {i}\pi }}{\dfrac{ \text {d}t }{ 2\text {i}\pi }}{\frac{ \text {d}t }{ 2\text {i}\pi }}{\frac{ \text {d}t }{ 2\text {i}\pi }} } \end{aligned}$$

(C.26)

is the unique solution to the Riemann–Hilbert problem for $\Pi$. It is a direct consequence of the Neumann expansion of the solution to the singular integral equation (C.25) for $\Pi _+$ and of the local holomorphicity of the jump matrices that, for some $\varrho >0$,

$$\begin{aligned} \Pi (\lambda ) = I_2 + \text {O}\bigg ( { \mathchoice{\dfrac{ \text {e}^{-\varrho w} }{ 1+|\lambda | }}{\dfrac{ \text {e}^{-\varrho w} }{ 1+|\lambda | }}{\frac{ \text {e}^{-\varrho w} }{ 1+|\lambda | }}{\frac{ \text {e}^{-\varrho w} }{ 1+|\lambda | }} } \bigg ) \end{aligned}$$

(C.27)

uniformly on ${\mathbb {C}}$ and with a differentiable remainder.

The piecewise holomorphic matrix $\Pi$ thus constructed enjoys a few properties that will be useful below. Indeed, one readily infers from the identity $M_{\uparrow }(-\lambda ) = \sigma ^x \cdot M_{\downarrow }^{-1}(\lambda ) \cdot \sigma ^x$, adjoined to the contour symmetry $\Gamma _{\uparrow }= \big \{ -z \,: \, z \in \Gamma _{\downarrow } \big \}$ and the uniqueness of the Riemann–Hilbert problem for $\Pi$ that the relation $\Pi (\lambda ) = \sigma ^x \cdot \Pi (-\lambda ) \cdot \sigma ^x$ holds. In particular,

$$\begin{aligned} \Pi (0) = \sigma ^x \cdot \Pi (0) \cdot \sigma ^x \quad \text {and} \quad \Pi ^{\prime }(0) = -\sigma ^x \cdot \Pi ^{\prime }(0) \cdot \sigma ^x . \end{aligned}$$

(C.28)

These properties lead to the the $\lambda \rightarrow 0$ expansion

$$\begin{aligned} \Pi (\lambda ) = \left( \begin{array}{cc} \Pi _{11}(0) &{} \Pi _{21}(0) \\ \Pi _{21}(0) &{} \Pi _{11}(0) \end{array} \right) + \lambda \left( \begin{array}{cc} \Pi _{11}^{\prime }(0) &{}- \Pi _{21}^{\prime }(0) \\ \Pi _{21}^{\prime }(0) &{} -\Pi _{11}^{\prime }(0) \end{array} \right) \; + \; \text {O}(\lambda ^2) . \end{aligned}$$

(C.29)

In other words, by setting

$$\begin{aligned} c_1=\Pi _{11}(0)\Pi _{11}^{\prime }(0) - \Pi _{21}(0)\Pi _{21}^{\prime }(0) \quad \text {and} \quad c_2=\Pi _{11}(0)\Pi _{21}^{\prime }(0) - \Pi _{21}(0)\Pi _{11}^{\prime }(0), \end{aligned}$$

(C.30)

since $\,{\text {det}}\Pi (\lambda )=1$,  one infers that

$$\begin{aligned} \Pi ^{-1}(0) \Pi (\lambda ) = I_2 + \lambda \left( \begin{array}{cc} c_1 &{} -c_2 \\ c_2 &{} - c_1 \end{array} \right) + \text {O}(\lambda ^2) . \end{aligned}$$

(C.31)

1.2.4 Solution of the Riemann–Hilbert problem for $\Upsilon$

With $\Pi$ defined, the solution to the Riemann–Hilbert problem for $\Upsilon$ can be constructed as $\Upsilon (\lambda ) = \mathcal {P}(\lambda ) \cdot \Pi (\lambda )$, where $\mathcal {P}(\lambda )$ is a meromorphic matrix on ${\mathbb {C}}$ whose only pole is located at $\lambda =0$. Below, we establish that this meromorphic matrix takes the form

$$\begin{aligned} \mathcal {P}(\lambda ) = \Pi (0) \cdot \bigg (I_2 + { \mathchoice{\dfrac{ \theta }{ \lambda }}{\dfrac{ \theta }{ \lambda }}{\frac{ \theta }{ \lambda }}{\frac{ \theta }{ \lambda }} } \varvec{ \texttt {D} } \bigg ) \cdot \Pi ^{-1}(0) \end{aligned}$$

(C.32)

where

$$\begin{aligned} \varvec{ \texttt {D} } = \left( \begin{array}{cc} -1 &{} - 1 \\ 1 &{} 1 \end{array} \right) \quad \text {and} \quad \theta = { \mathchoice{\dfrac{ 1 }{ (e^{-2}Q)^{\prime }(0) +2(c_1 +c_2) }}{\dfrac{ 1 }{ (e^{-2}Q)^{\prime }(0) +2(c_1 +c_2) }}{\frac{ 1 }{ (e^{-2}Q)^{\prime }(0) +2(c_1 +c_2) }}{\frac{ 1 }{ (e^{-2}Q)^{\prime }(0) +2(c_1 +c_2) }} } , \end{aligned}$$

(C.33)

with $c_1,c_2$ as introduced in (C.30). The matrix $\mathcal {P}$ is constructed so that

$$\begin{aligned} \lambda \; \mapsto \; \mathcal {P}(\lambda ) \Pi (\lambda ) M_{\downarrow }(\lambda ) \Big ( \alpha ^{(0)}(\lambda ) \Big )^{\sigma _3} \end{aligned}$$

(C.34)

is regular at $\lambda =0$.

When looking for $\mathcal {P}$, it is convenient to parameterise

$$\begin{aligned} \mathcal {P}(\lambda ) = \Pi (0) \mathcal {G}(\lambda ) \Pi ^{-1}(0) \quad \text {with} \quad \mathcal {G}(\lambda ) = I_2 + { \mathchoice{\dfrac{1}{\lambda }}{\dfrac{1}{\lambda }}{\frac{1}{\lambda }}{\frac{1}{\lambda }} } \left( \begin{array}{cc} g_{11} &{} g_{12} \\ g_{21} &{} g_{22} \end{array} \right) \end{aligned}$$

(C.35)

Then, one has $\mathcal {P}(\lambda ) \Pi (\lambda ) M_{\downarrow }(\lambda ) \Big ( \alpha ^{(0)}_{\downarrow }(\lambda ) \Big )^{\sigma _3}= \Pi (0) H(\lambda )$, with

$$\begin{aligned} H_{11}(\lambda ) \;= & {} \; \alpha ^{(0)}_{\downarrow }(\lambda ) \Big [ \mathcal {G}_{11}(\lambda )+Q(\lambda )e^{-2}(\lambda ) \mathcal {G}_{12}(\lambda ) \Big ] \, \nonumber \\&+ \, c_1\lambda \alpha ^{(0)}_{\downarrow }(\lambda ) \Big [ \mathcal {G}_{11}(\lambda )-Q(\lambda )e^{-2}(\lambda ) \mathcal {G}_{12}(\lambda ) \Big ] \nonumber \\&+ c_2 \lambda \alpha ^{(0)}_{\downarrow }(\lambda ) \Big [ \mathcal {G}_{12}(\lambda )-Q(\lambda )e^{-2}(\lambda ) \mathcal {G}_{11}(\lambda ) \Big ] + \text {O}(1) , \end{aligned}$$

(C.36)

as well as $H_{21}=[H_{11}]_{\mid \mathcal {G}_{1a} \hookrightarrow \mathcal {G}_{2a} }$ and $H_{a2}(\lambda ) = \text {O}(1)$, when $\lambda \rightarrow 0$.

In principle, $H_{11}$ admits a second order pole at $\lambda =0$. By imposing that $H_{11}$ is regular at $\lambda =0$, one obtains the system of equations on the coefficients $g_{1a}$:

$$\begin{aligned} g_{11}= -Q(0) g_{12} \quad \text {and} \quad g_{12}\cdot \bigg [(e^{-2}Q)^{\prime }(0)-2 c_1 Q(0) + c_2 \big [1+Q^2(0) \big ] \bigg ] = -1 . \end{aligned}$$

(C.37)

These equations are solvable owing to $|c_1|+|c_2| =\text {O}( \text {e}^{-\varrho w } )$, what is in itself a consequence of (C.27).

Likewise, by requiring that $H_{21}$ is regular at $\lambda =0$, one obtains the system of equations on the coefficients $g_{2a}$:

$$\begin{aligned} g_{21}= -Q(0) g_{22} \quad \text {and} \quad g_{22}\cdot \bigg [(e^{-2}Q)^{\prime }(0)-2 c_1 Q(0)+ c_2 \big [1+Q^2(0) \big ] \bigg ] = -Q(0) . \end{aligned}$$

(C.38)

All in all, this yields that

$$\begin{aligned} \left( \begin{array}{cc} g_{11} &{} g_{12} \\ g_{21} &{} g_{22} \end{array} \right) = { \mathchoice{\dfrac{ 1 }{ (e^{-2}Q)^{\prime }(0)-2 c_1 Q(0)+ c_2 \big [1+Q^2(0) \big ] }}{\dfrac{ 1 }{ (e^{-2}Q)^{\prime }(0)-2 c_1 Q(0)+ c_2 \big [1+Q^2(0) \big ] }}{\frac{ 1 }{ (e^{-2}Q)^{\prime }(0)-2 c_1 Q(0)+ c_2 \big [1+Q^2(0) \big ] }}{\frac{ 1 }{ (e^{-2}Q)^{\prime }(0)-2 c_1 Q(0)+ c_2 \big [1+Q^2(0) \big ] }} } \cdot \left( \begin{array}{cc} Q(0) &{} -1 \\ Q^2(0) &{} -Q(0) \end{array} \right) . \end{aligned}$$

(C.39)

The form of $\mathcal {P}(\lambda )$ then follows upon recalling that $Q(0)=-1$.

By tracing backwards the various transformations, one gets that the unique solution $\chi$ to the Riemann–Hilbert problem for $\chi$ takes the piecewise form as depicted in Fig. 9.

1.3 Resolvent kernel of $\,\text {id}+ \varvec{ \texttt {V} }$

It follows from the results of Theorem C.1 that the solution to (C.2) takes the form

$$\begin{aligned} \mathcal {F}[f](k)= \mathcal {F}[h](k)\; - \; \int \limits _{ {\mathbb {R}}+ \text {i}v }^{} \text {d}\mu \, R(k,\mu ) \mathcal {F}[h](\mu ) . \end{aligned}$$

(C.40)

Since $\chi _{+}(\lambda ) \varvec{E}_{R}(\lambda )=\chi _{-}(\lambda ) \varvec{E}_{R}(\lambda )$, and since the vectors $\varvec{E}_{L/R}$ are analytic in a tubular neighbourhood of ${\mathbb {R}}$, it follows that $R(\lambda ,\mu )$ is also analytic in some open neighbourhood of ${\mathbb {R}}^2$.

1.3.1 Support restrictions

One may explicitly check that the integral term only involves the values of h inside of $[ -w \,; w ]$. Indeed, one has

$$\begin{aligned} \int \limits _{ {\mathbb {R}}+ \text {i}v }^{}\text {d}\mu \, \text {e}^{\text {i}\mu x} R(k,\mu ) = \int \limits _{ {\mathbb {R}}+ \text {i}v }^{} \text {d}\mu \, \text {e}^{\text {i}\mu x} { \mathchoice{\dfrac{ \Big ( \varvec{E}_{L}(k), \chi _{+}^{-1}(k) \cdot \chi _+(\mu ) \varvec{E}_R(\mu ) \Big ) }{ k - \mu }}{\dfrac{ \Big ( \varvec{E}_{L}(k), \chi _{+}^{-1}(k) \cdot \chi _+(\mu ) \varvec{E}_R(\mu ) \Big ) }{ k - \mu }}{\frac{ \Big ( \varvec{E}_{L}(k), \chi _{+}^{-1}(k) \cdot \chi _+(\mu ) \varvec{E}_R(\mu ) \Big ) }{ k - \mu }}{\frac{ \Big ( \varvec{E}_{L}(k), \chi _{+}^{-1}(k) \cdot \chi _+(\mu ) \varvec{E}_R(\mu ) \Big ) }{ k - \mu }} } \end{aligned}$$

(C.41)

If $x>w$, then $\text {e}^{\text {i}\mu x} \varvec{E}_{R}(\mu )$ is bounded on ${\mathbb {H}}^{+}_{v}={\mathbb {H}}^+ + \text {i}v$, and so, since the integrand vanishes at $\infty$ in ${\mathbb {H}}^+_{v}$, one obtains zero by deforming the integration contour to $+\text {i}\infty$. One arrives to the same conclusion when $x<-w$ upon using $\chi _{+}(\mu ) \varvec{E}_{R}(\mu )=\chi _{-}(\mu ) \varvec{E}_{R}(\mu )$. Hence, for any function h on ${\mathbb {R}}$ with exponential decay at $\pm \infty$, one gets, for $0<v$ small enough, that

$$\begin{aligned} \int \limits _{ {\mathbb {R}}+ \text {i}v }^{}\text {d}\mu \, R(k,\mu ) \mathcal {F}[h](\mu ) = \int \limits _{ {\mathbb {R}}+ \text {i}v }^{} \text {d}\mu \, R(k,\mu ) \mathcal {F}[h \varvec{1}_{ [ -w \,; w ] }](\mu ) . \end{aligned}$$

(C.42)

1.3.2 Leading asymptotic form of the resolvent

The resolvent may be approximated, in the large-w limit, by inserting the leading behaviour of the matrix $\chi$ into the expression for the vectors $\varvec{F}_{R/L}$ (C.7), and then inserting the latter into the formula for the resolvent kernel (C.6).

For further convenience, given $\varvec{ \texttt {D} }$ as in (C.33), set

$$\begin{aligned} \mathcal {P}_{\infty }(\lambda ) = I_2 \; + \, { \mathchoice{\dfrac{1}{\lambda \mathfrak {b}^{\prime }(0)}}{\dfrac{1}{\lambda \mathfrak {b}^{\prime }(0)}}{\frac{1}{\lambda \mathfrak {b}^{\prime }(0)}}{\frac{1}{\lambda \mathfrak {b}^{\prime }(0)}} } \varvec{ \texttt {D} } \quad \text {with} \quad \mathfrak {b}(\lambda ) = { \mathchoice{\dfrac{ e^{-2}(\lambda ) }{ \alpha _{\downarrow }^{(0)}(\lambda ) \alpha _{\uparrow }^{(0)}(\lambda ) }}{\dfrac{ e^{-2}(\lambda ) }{ \alpha _{\downarrow }^{(0)}(\lambda ) \alpha _{\uparrow }^{(0)}(\lambda ) }}{\frac{ e^{-2}(\lambda ) }{ \alpha _{\downarrow }^{(0)}(\lambda ) \alpha _{\uparrow }^{(0)}(\lambda ) }}{\frac{ e^{-2}(\lambda ) }{ \alpha _{\downarrow }^{(0)}(\lambda ) \alpha _{\uparrow }^{(0)}(\lambda ) }} } , \end{aligned}$$

(C.43)

so that, by using that $(e^{-2}Q)^{\prime }(0) = \mathfrak {b}^{\prime }(0)$, one may decompose

$$\begin{aligned} \mathcal {P}(\lambda ) = \mathcal {P}_{\infty }(\lambda ) + \delta \mathcal {P}(\lambda ) \quad \text {with} \quad \delta \mathcal {P}(\lambda ) = \text {O}\bigg ( { \mathchoice{\dfrac{ \text {e}^{-\varrho w} }{ |\lambda | }}{\dfrac{ \text {e}^{-\varrho w} }{ |\lambda | }}{\frac{ \text {e}^{-\varrho w} }{ |\lambda | }}{\frac{ \text {e}^{-\varrho w} }{ |\lambda | }} } \bigg ) . \end{aligned}$$

(C.44)

It is as well convenient to introduce an analogous parameterisation gathering the exponentially small corrections to $\Pi (\lambda )=I_2 + \delta \Pi (\lambda )$, where, by virtue of (C.27), one has

$$\begin{aligned} \delta \Pi (\lambda ) = \text {O}\bigg ( { \mathchoice{\dfrac{ \text {e}^{-\varrho w} }{ 1+|\lambda | }}{\dfrac{ \text {e}^{-\varrho w} }{ 1+|\lambda | }}{\frac{ \text {e}^{-\varrho w} }{ 1+|\lambda | }}{\frac{ \text {e}^{-\varrho w} }{ 1+|\lambda | }} } \bigg ) \end{aligned}$$

(C.45)

uniformly on ${\mathbb {C}}$.

From there, one obtains that, uniformly throughout the region $\mathcal {D}_{II}$, as defined in Fig. 9, one has

$$\begin{aligned} \chi (\lambda ) = \chi _{\infty }^{(II)}(\lambda ) + \delta \chi ^{(II)}(\lambda ) \quad \text {with} \quad \chi _{\infty }^{(II)}(\lambda ) = \mathcal {P}_{\infty }(\lambda ) M_{\uparrow }^{-1}(\lambda ) \big [ \alpha _{\uparrow }^{(0)}(\lambda ) \big ]^{\sigma _3} , \end{aligned}$$

(C.46)

and

$$\begin{aligned} \delta \chi ^{(II)}(\lambda )= \delta \mathcal {P}(\lambda ) \Pi (\lambda ) M_{\uparrow }^{-1}(\lambda ) \big [ \alpha _{\uparrow }^{(0)}(\lambda ) \big ]^{\sigma _3} \; + \; \mathcal {P}_{\infty }(\lambda ) \delta \Pi (\lambda ) M_{\uparrow }^{-1}(\lambda ) \big [ \alpha _{\uparrow }^{(0)}(\lambda ) \big ]^{\sigma _3} . \end{aligned}$$

(C.47)

By direct inspection, one obtains that uniformly in $\lambda \in \overline{\mathcal {D}}_{II}$,

$$\begin{aligned} \delta \chi ^{(II)}(\lambda ) = \text {O}\bigg ( { \mathchoice{\dfrac{ \text {e}^{- \varrho w } }{ 1 + |\lambda | }}{\dfrac{ \text {e}^{- \varrho w } }{ 1 + |\lambda | }}{\frac{ \text {e}^{- \varrho w } }{ 1 + |\lambda | }}{\frac{ \text {e}^{- \varrho w } }{ 1 + |\lambda | }} } \bigg ). \end{aligned}$$

(C.48)

Likewise, uniformly throughout the region $\mathcal {D}_{III}$, one has the decomposition

$$\begin{aligned} \chi (\lambda ) = \chi _{\infty }^{(III)}(\lambda ) + \delta \chi ^{(III)}(\lambda ) \quad \text {with} \quad \chi _{\infty }^{(III)}(\lambda ) = \mathcal {P}_{\infty }(\lambda ) M_{\downarrow }(\lambda ) \big [ \alpha _{\downarrow }^{(0)}(\lambda ) \big ]^{\sigma _3} , \end{aligned}$$

(C.49)

and

$$\begin{aligned} \delta \chi ^{(III)}(\lambda )= \delta \mathcal {P}(\lambda ) \Pi (\lambda ) M_{\downarrow }(\lambda ) \big [ \alpha _{\downarrow }^{(0)}(\lambda ) \big ]^{\sigma _3} \; + \; \mathcal {P}_{\infty }(\lambda ) \delta \Pi (\lambda ) M_{\downarrow }(\lambda ) \big [ \alpha _{\downarrow }^{(0)}(\lambda ) \big ]^{\sigma _3} . \end{aligned}$$

(C.50)

Again, a direct analysis shows that for $\lambda \in \mathcal {D}_{III}$ and uniformly away from 0, one has

$$\begin{aligned} \delta \chi ^{(III)}(\lambda ) = \text {O}\bigg ( { \mathchoice{\dfrac{ \text {e}^{- \varrho w + 2 v w } }{ 1 + |\lambda | }}{\dfrac{ \text {e}^{- \varrho w + 2 v w } }{ 1 + |\lambda | }}{\frac{ \text {e}^{- \varrho w + 2 v w } }{ 1 + |\lambda | }}{\frac{ \text {e}^{- \varrho w + 2 v w } }{ 1 + |\lambda | }} } \bigg ). \end{aligned}$$

(C.51)

Note that the additional term $\text {e}^{ 2 v w }$ present in the estimates on the remainder is due to the presence of $e^{-2}$ in the off-diagonal entry of $M_{\downarrow }$ and the fact that $\mathcal {D}_{III}\cap {\mathbb {H}}^{+} = \Big \{ \lambda \in {\mathbb {C}}\, : \, 0< \mathfrak {I}(\lambda ) < v \Big \}$.

These formulae allow to compute the leading behaviour of the vector $\varvec{F}_R(\lambda )$ inside each of the domains. One infers that

$$\begin{aligned} \varvec{F}_R(\lambda )=&\varvec{F}_{R;\infty }(\lambda )+ \delta \varvec{F}_R^{(A)}(\lambda ) \quad \text {with} \quad \delta \varvec{F}_R^{(A)}(\lambda ) = \delta \chi ^{(A)}(\lambda ) \varvec{E}_R(\lambda ) \nonumber \\&\quad \quad \text {for} \quad \lambda \in \mathcal {D}_{A} , \; A \in \{II, III\}. \end{aligned}$$

(C.52)

We stress that the expression for $\varvec{F}_{R;\infty }(\lambda )$ does not depend on whether $\lambda \in \mathcal {D}_{II}$ or $\lambda \in \mathcal {D}_{III}$. A direct calculation shows that

$$\begin{aligned} \varvec{F}_{R;\infty }(\lambda )= & {} { \mathchoice{\dfrac{ 1 }{ 2 \text {i}\pi }}{\dfrac{ 1 }{ 2 \text {i}\pi }}{\frac{ 1 }{ 2 \text {i}\pi }}{\frac{ 1 }{ 2 \text {i}\pi }} } \mathcal {P}_{\infty }(\lambda ) \cdot \left( \begin{array}{c} \alpha _{\downarrow }^{(0)}(\lambda ) e(\lambda ) \\ \big \{ \alpha _{\uparrow }^{(0)}(\lambda ) e(\lambda ) \big \}^{-1} \end{array} \right) \nonumber \\ \;= & {} \; { \mathchoice{\dfrac{ 1 }{ 2 \text {i}\pi }}{\dfrac{ 1 }{ 2 \text {i}\pi }}{\frac{ 1 }{ 2 \text {i}\pi }}{\frac{ 1 }{ 2 \text {i}\pi }} } \left( \begin{array}{c} \alpha _{\downarrow }^{(0)}(\lambda ) e(\lambda ) - { \mathchoice{\dfrac{ 1 }{ \lambda \mathfrak {b}^{\prime }(0) }}{\dfrac{ 1 }{ \lambda \mathfrak {b}^{\prime }(0) }}{\frac{ 1 }{ \lambda \mathfrak {b}^{\prime }(0) }}{\frac{ 1 }{ \lambda \mathfrak {b}^{\prime }(0) }} } \Big [ \alpha _{\downarrow }^{(0)}(\lambda ) e(\lambda ) + \big \{ \alpha _{\uparrow }^{(0)}(\lambda ) e(\lambda ) \big \}^{-1} \Big ] \\ \big \{ \alpha _{\uparrow }^{(0)}(\lambda ) e(\lambda ) \big \}^{-1} + { \mathchoice{\dfrac{ 1 }{ \lambda \mathfrak {b}^{\prime }(0) }}{\dfrac{ 1 }{ \lambda \mathfrak {b}^{\prime }(0) }}{\frac{ 1 }{ \lambda \mathfrak {b}^{\prime }(0) }}{\frac{ 1 }{ \lambda \mathfrak {b}^{\prime }(0) }} } \Big [ \alpha _{\downarrow }^{(0)}(\lambda ) e(\lambda ) + \big \{ \alpha _{\uparrow }^{(0)}(\lambda ) e(\lambda ) \big \}^{-1} \Big ] \end{array} \right) \nonumber \\ \;= & {} \; { \mathchoice{\dfrac{ 1 }{ 2 \text {i}\pi }}{\dfrac{ 1 }{ 2 \text {i}\pi }}{\frac{ 1 }{ 2 \text {i}\pi }}{\frac{ 1 }{ 2 \text {i}\pi }} } \left( \begin{array}{c} \alpha _{\downarrow }^{(0)}(\lambda ) e(\lambda ) \cdot \Big [ 1 - { \mathchoice{\dfrac{ 1 }{ \lambda \mathfrak {b}^{\prime }(0) }}{\dfrac{ 1 }{ \lambda \mathfrak {b}^{\prime }(0) }}{\frac{ 1 }{ \lambda \mathfrak {b}^{\prime }(0) }}{\frac{ 1 }{ \lambda \mathfrak {b}^{\prime }(0) }} } \big ( 1+\mathfrak {b}(\lambda ) \big ) \Big ] \\ \big \{ \alpha _{\uparrow }^{(0)}(\lambda ) e(\lambda ) \big \}^{-1} \cdot \Big [ 1 + \, { \mathchoice{\dfrac{ 1 }{ \lambda \mathfrak {b}^{\prime }(0) }}{\dfrac{ 1 }{ \lambda \mathfrak {b}^{\prime }(0) }}{\frac{ 1 }{ \lambda \mathfrak {b}^{\prime }(0) }}{\frac{ 1 }{ \lambda \mathfrak {b}^{\prime }(0) }} } \big ( 1 + \big \{ \mathfrak {b}(\lambda )\big \}^{-1} \big ) \Big ] \end{array} \right) \nonumber \\ \;= & {} \; { \mathchoice{\dfrac{ 1 }{ 2 \text {i}\pi }}{\dfrac{ 1 }{ 2 \text {i}\pi }}{\frac{ 1 }{ 2 \text {i}\pi }}{\frac{ 1 }{ 2 \text {i}\pi }} } \left( \begin{array}{c} f_{+;\infty }(\lambda ) \\ f_{-;\infty }(\lambda ) \end{array} \right) . \end{aligned}$$

(C.53)

Above, $\mathfrak {b}$ is as introduced in (C.43). It is easy to see that the above expression for $f_{\pm ;\infty }(\lambda )$ is analytic in a tubular neighbourhood of ${\mathbb {R}}$. In particular, there is no pole at $\lambda =0$ as follows from $\mathfrak {b}(0)=-1$.

Similarly, using the relation ${\text {det}}\chi (\lambda )=1$, one infers that

$$\begin{aligned}&\varvec{F}_L(\lambda )=\varvec{F}_{L;\infty }(\lambda )+ \delta \varvec{F}_L^{(A)}(\lambda ) \quad \text {with} \quad \delta \varvec{F}_L^{(A)}(\lambda ) \nonumber \\&\quad = \text {CoMat}\Big (\delta \chi ^{(A)}(\lambda )\Big ) \varvec{E}_L(\lambda ) \quad \text {for} \quad \lambda \in \mathcal {D}_{A} , \; A \in \{II, III\} \end{aligned}$$

(C.54)

where

$$\begin{aligned} \varvec{F}_{L;\infty }(\lambda )=- \mathcal {F}\big [ L^{(0)} \big ](\lambda ) \left( \begin{array}{c} - f_{-;\infty }(\lambda ) \\ f_{+;\infty }(\lambda ) \end{array} \right) \end{aligned}$$

(C.55)

and $\text {CoMat}(M)$ stands for the Comatrix of M.

From the above one infers that the resolvent admits the following expansion

$$\begin{aligned} R(\lambda ,\mu ) = R_{\infty }(\lambda ,\mu ) \, + \, \delta R(\lambda ,\mu ) \quad \text {uniformly}\, \text {in} \quad \mathcal {D}_{II}\cup \mathcal {D}_{III} , \end{aligned}$$

(C.56)

where

$$\begin{aligned} R_{\infty }(\lambda ,\mu ) = { \mathchoice{\dfrac{ - \mathcal {F}\big [ L^{(0)} \big ](\lambda ) }{ 2\text {i}\pi (\lambda -\mu ) }}{\dfrac{ - \mathcal {F}\big [ L^{(0)} \big ](\lambda ) }{ 2\text {i}\pi (\lambda -\mu ) }}{\frac{ - \mathcal {F}\big [ L^{(0)} \big ](\lambda ) }{ 2\text {i}\pi (\lambda -\mu ) }}{\frac{ - \mathcal {F}\big [ L^{(0)} \big ](\lambda ) }{ 2\text {i}\pi (\lambda -\mu ) }} } \cdot \Big ( -f_{-;\infty }(\lambda ) \; \; f_{+;\infty }(\lambda ) \Big )\cdot \left( \begin{array}{c} f_{+;\infty }(\mu ) \\ f_{-;\infty }(\mu ) \end{array} \right) \end{aligned}$$

(C.57)

while, for $(\lambda ,\mu ) \in \mathcal {D}_{A}\times \mathcal {D}_{B}$ with $A,B \in \{II, III\}$,

$$\begin{aligned} \delta R(\lambda ,\mu ) \;= & {} \; { \mathchoice{\dfrac{1}{\lambda -\mu }}{\dfrac{1}{\lambda -\mu }}{\frac{1}{\lambda -\mu }}{\frac{1}{\lambda -\mu }} } \Bigg \{ \Big (\varvec{E}_L(\lambda ), ^{ \varvec{ \texttt {t} } }\text {CoMat}\Big ( \delta \chi ^{(A)}(\lambda ) \Big ) \varvec{F}_{R;\infty }(\mu ) \Big ) \; \nonumber \\&+ \; \Big (\varvec{F}_{L;\infty }(\lambda ), \delta \chi ^{(B)}(\mu ) \varvec{E}_{R}(\mu ) \Big ) \nonumber \\&\; + \; \Big (\varvec{E}_L(\lambda ), ^{ \varvec{ \texttt {t} } }\text {CoMat}\Big ( \delta \chi ^{(A)}(\lambda ) \Big ) \cdot \delta \chi ^{(B)}(\mu ) \varvec{E}_{R}(\mu ) \Big ) \Bigg \} . \end{aligned}$$

(C.58)

The leading resolvent may be explicitly cast as

$$\begin{aligned} R_{\infty }(\lambda ,\mu ) =&{ \mathchoice{\dfrac{ - \mathcal {F}[L^{(0)}](\lambda ) }{ 2\text {i}\pi (\lambda - \mu ) }}{\dfrac{ - \mathcal {F}[L^{(0)}](\lambda ) }{ 2\text {i}\pi (\lambda - \mu ) }}{\frac{ - \mathcal {F}[L^{(0)}](\lambda ) }{ 2\text {i}\pi (\lambda - \mu ) }}{\frac{ - \mathcal {F}[L^{(0)}](\lambda ) }{ 2\text {i}\pi (\lambda - \mu ) }} } \Big ( - \{ \alpha _{\uparrow }^{(0)}(\lambda ) e (\lambda ) \}^{-1} , \; \alpha _{\downarrow }^{(0)}(\lambda ) e(\lambda ) \Big ) \nonumber \\&\cdot \bigg (I_{2}+ { \mathchoice{\dfrac{\lambda -\mu }{\lambda \mu \mathfrak {b}^{\prime }(0) }}{\dfrac{\lambda -\mu }{\lambda \mu \mathfrak {b}^{\prime }(0) }}{\frac{\lambda -\mu }{\lambda \mu \mathfrak {b}^{\prime }(0) }}{\frac{\lambda -\mu }{\lambda \mu \mathfrak {b}^{\prime }(0) }} } \varvec{ \texttt {D} } \bigg ) \left( \begin{array}{c} \alpha _{\downarrow }^{(0)}(\mu ) \; e(\mu ) \vspace{2mm} \\ \{ \alpha _{\uparrow }^{(0)}(\mu ) e (\mu ) \}^{-1} \end{array} \right) \end{aligned}$$

(C.59)

as follows from $\varvec{ \texttt {D} } ^2=0$.

Obviously, $R_{\infty }(\lambda ,\mu )$ is analytic in a tubular neighbourhood of ${\mathbb {R}}^2$ and satisfies, for some $\alpha >0$, the bounds

$$\begin{aligned} \big | R_{\infty }(\lambda ,\mu ) \big | = \text {O}\Bigg ( { \mathchoice{\dfrac{ \text {e}^{-\alpha |\lambda |} }{ |\lambda -\mu | }}{\dfrac{ \text {e}^{-\alpha |\lambda |} }{ |\lambda -\mu | }}{\frac{ \text {e}^{-\alpha |\lambda |} }{ |\lambda -\mu | }}{\frac{ \text {e}^{-\alpha |\lambda |} }{ |\lambda -\mu | }} } \text {e}^{ w ( |\mathfrak {I}(\lambda )| + |\mathfrak {I}(\mu )| ) } \Bigg ) , \end{aligned}$$

(C.60)

which is valid throughout $\Big \{ \mathcal {D}_{II} \cup \mathcal {D}_{III} \cup \big \{ {\mathbb {R}}+ \text {i}v \big \} \Big \}^2$,  provided that $\lambda , \mu$ are both uniformly away from 0.

Since $\delta R= R-R_{\infty }$, one infers that $\delta R$ is analytic in a tubular neighbourhood of ${\mathbb {R}}^2$. One can bound $\delta R$, globally on $\Big \{ \mathcal {D}_{II} \cup \mathcal {D}_{III} \cup \big \{ {\mathbb {R}}+ \text {i}v \big \} \Big \}^2$ by using its patch-wise valid decomposition. This yields that

$$\begin{aligned} \big | \delta R(\lambda ,\mu ) \big | = \text {O}\Bigg ( { \mathchoice{\dfrac{ \text {e}^{-\alpha |\lambda |- \varrho w + 4 v w } }{ | \lambda - \mu | }}{\dfrac{ \text {e}^{-\alpha |\lambda |- \varrho w + 4 v w } }{ | \lambda - \mu | }}{\frac{ \text {e}^{-\alpha |\lambda |- \varrho w + 4 v w } }{ | \lambda - \mu | }}{\frac{ \text {e}^{-\alpha |\lambda |- \varrho w + 4 v w } }{ | \lambda - \mu | }} } \text {e}^{w(|\mathfrak {I}(\lambda )|+|\mathfrak {I}(\mu )|)} \Bigg ) . \end{aligned}$$

(C.61)

Upon putting these two bounds together, one concludes that for $\lambda , \mu$ throughout $\Big \{ \mathcal {D}_{II} \cup \mathcal {D}_{III} \cup \big \{ {\mathbb {R}}+ \text {i}v \big \} \Big \}$ but both uniformly away from 0,

$$\begin{aligned} \big | R(\lambda ,\mu ) \big | \, \le \, { \mathchoice{\dfrac{ C \text {e}^{-\alpha |\lambda |} }{ |\lambda -\mu | }}{\dfrac{ C \text {e}^{-\alpha |\lambda |} }{ |\lambda -\mu | }}{\frac{ C \text {e}^{-\alpha |\lambda |} }{ |\lambda -\mu | }}{\frac{ C \text {e}^{-\alpha |\lambda |} }{ |\lambda -\mu | }} } \text {e}^{w(|\mathfrak {I}(\lambda )|+|\mathfrak {I}(\mu )|)} , \end{aligned}$$

(C.62)

for some $\alpha >0$.

Auxiliary lemma

Lemma D.1

Given $\sigma , v>0$ and $r \in {\mathbb {N}}$ there exists $C>0$ such that one has the upper bound

$$\begin{aligned} \int \limits _{ {\mathbb {R}}\pm \text {i}(\sigma +v) }^{} \text {d}t \cdot { \mathchoice{\dfrac{ \big [ \ln (1+|t|) \big ]^{r} }{ (1+|t|)\cdot |k-t|}}{\dfrac{ \big [ \ln (1+|t|) \big ]^{r} }{ (1+|t|)\cdot |k-t|}}{\frac{ \big [ \ln (1+|t|) \big ]^{r} }{ (1+|t|)\cdot |k-t|}}{\frac{ \big [ \ln (1+|t|) \big ]^{r} }{ (1+|t|)\cdot |k-t|}} } \; \le \; C \cdot { \mathchoice{\dfrac{ \big [ \ln (1+|k|) \big ]^{r + 1} }{ 1+|k| }}{\dfrac{ \big [ \ln (1+|k|) \big ]^{r + 1} }{ 1+|k| }}{\frac{ \big [ \ln (1+|k|) \big ]^{r + 1} }{ 1+|k| }}{\frac{ \big [ \ln (1+|k|) \big ]^{r + 1} }{ 1+|k| }} } , \end{aligned}$$

(D.1)

for any $k\in {\mathbb {C}}$ satisfying $|\mathfrak {I}k| \le v$.

Proof

First of all, by changing $\mathfrak {R}(t) \hookrightarrow - \mathfrak {R}(t)$ under the integral, one may always assume that $\mathfrak {R}(k)>0$. Furthermore, for $|\mathfrak {R}(k)|<M$ for some fixed M, the integral is well-defined and the bound (D.1) is obvious. Hence, from now on, one may assume $\mathfrak {R}(k)$ to be large enough.

Given $t=u\pm \text {i}(\sigma + v)$, one has

$$\begin{aligned} \big [ \ln (1+|t|) \big ]^{r} \; \le \; \big [ \ln (1+|u|+\sigma +v ) \big ]^{r} \, \le \, \sum \limits _{\ell =0}^{r} C^{r}_{\ell } \, \big [ \ln (1+|u|) \big ]^{\ell } \cdot \big [ \ln (1+\sigma +v ) \big ]^{r-\ell } , \end{aligned}$$

(D.2)

with $C^{r}_{\ell }$ being the binomial coefficients.

Given the same parameterisation for t, since

$$\begin{aligned} |k-t| \, \ge \, { \mathchoice{\dfrac{1}{3}}{\dfrac{1}{3}}{\frac{1}{3}}{\frac{1}{3}} } \Big \{ \sigma + |x-u| \Big \} \quad \text {where} \quad k = x+\text {i}\mathfrak {I}(k) \quad \text {as}\, \text {well}\, \text {as} \quad 1 + |t|\ge 1+|u| \, , \end{aligned}$$

(D.3)

one gets the upper bound

$$\begin{aligned}&\int \limits _{ {\mathbb {R}}\pm \text {i}(\sigma +v) }^{} \text {d}t \cdot { \mathchoice{\dfrac{ \big [ \ln (1+|t|) \big ]^{r} }{ (1+|t|)\cdot |k-t|}}{\dfrac{ \big [ \ln (1+|t|) \big ]^{r} }{ (1+|t|)\cdot |k-t|}}{\frac{ \big [ \ln (1+|t|) \big ]^{r} }{ (1+|t|)\cdot |k-t|}}{\frac{ \big [ \ln (1+|t|) \big ]^{r} }{ (1+|t|)\cdot |k-t|}} } \, \le \, \sum \limits _{\ell =0}^{r} 3 C^{r}_{\ell } \, \big [ \ln (1+\sigma +v ) \big ]^{r-\ell } \, \mathcal {I}_{\ell } \nonumber \\&\quad \quad \text {with} \quad \mathcal {I}_{\ell }= \int \limits _{ {\mathbb {R}}}^{} { \mathchoice{\dfrac{ \text {d}u \cdot \big [ \ln (1+|u|) \big ]^{\ell } }{ (1+|u|)\cdot (\sigma + |x-u|) }}{\dfrac{ \text {d}u \cdot \big [ \ln (1+|u|) \big ]^{\ell } }{ (1+|u|)\cdot (\sigma + |x-u|) }}{\frac{ \text {d}u \cdot \big [ \ln (1+|u|) \big ]^{\ell } }{ (1+|u|)\cdot (\sigma + |x-u|) }}{\frac{ \text {d}u \cdot \big [ \ln (1+|u|) \big ]^{\ell } }{ (1+|u|)\cdot (\sigma + |x-u|) }} } . \end{aligned}$$

(D.4)

Then, one may decompose $\mathcal {I}_{\ell }$ as

$$\begin{aligned} \mathcal {I}_{\ell } = \underbrace{ \int \limits _{ -\infty }^{0} { \mathchoice{\dfrac{ \text {d}u \cdot \big [ \ln (1-u) \big ]^{\ell } }{ (1-u)\cdot (\sigma + x-u) }}{\dfrac{ \text {d}u \cdot \big [ \ln (1-u) \big ]^{\ell } }{ (1-u)\cdot (\sigma + x-u) }}{\frac{ \text {d}u \cdot \big [ \ln (1-u) \big ]^{\ell } }{ (1-u)\cdot (\sigma + x-u) }}{\frac{ \text {d}u \cdot \big [ \ln (1-u) \big ]^{\ell } }{ (1-u)\cdot (\sigma + x-u) }} } }_{ = \mathcal {I}_{\ell }^{(1)} } + \underbrace{ \int \limits _{ 0}^{x} { \mathchoice{\dfrac{ \text {d}u \cdot \big [ \ln (1+u) \big ]^{\ell } }{ (1+u)\cdot (\sigma + x-u) }}{\dfrac{ \text {d}u \cdot \big [ \ln (1+u) \big ]^{\ell } }{ (1+u)\cdot (\sigma + x-u) }}{\frac{ \text {d}u \cdot \big [ \ln (1+u) \big ]^{\ell } }{ (1+u)\cdot (\sigma + x-u) }}{\frac{ \text {d}u \cdot \big [ \ln (1+u) \big ]^{\ell } }{ (1+u)\cdot (\sigma + x-u) }} } }_{ = \mathcal {I}_{\ell }^{(2)} } + \underbrace{ \int \limits _{x}^{ +\infty } { \mathchoice{\dfrac{ \text {d}u \cdot \big [ \ln (1+u) \big ]^{\ell } }{ (1+u)\cdot (\sigma +u- x) }}{\dfrac{ \text {d}u \cdot \big [ \ln (1+u) \big ]^{\ell } }{ (1+u)\cdot (\sigma +u- x) }}{\frac{ \text {d}u \cdot \big [ \ln (1+u) \big ]^{\ell } }{ (1+u)\cdot (\sigma +u- x) }}{\frac{ \text {d}u \cdot \big [ \ln (1+u) \big ]^{\ell } }{ (1+u)\cdot (\sigma +u- x) }} } }_{ = \mathcal {I}_{\ell }^{(3)} } . \end{aligned}$$

(D.5)

$\mathcal {I}_{\ell }^{(2)}$ may be estimated by direct bounds

$$\begin{aligned} \mathcal {I}_{\ell }^{(2)} \,&\le \, { \mathchoice{\dfrac{ \big [ \ln (1+x) \big ]^{\ell } }{ 1 + x + \sigma }}{\dfrac{ \big [ \ln (1+x) \big ]^{\ell } }{ 1 + x + \sigma }}{\frac{ \big [ \ln (1+x) \big ]^{\ell } }{ 1 + x + \sigma }}{\frac{ \big [ \ln (1+x) \big ]^{\ell } }{ 1 + x + \sigma }} }\int \limits _{ 0}^{x} \text {d}u \cdot \bigg ( { \mathchoice{\dfrac{ 1 }{ (1+u) }}{\dfrac{ 1 }{ (1+u) }}{\frac{ 1 }{ (1+u) }}{\frac{ 1 }{ (1+u) }} } + { \mathchoice{\dfrac{1}{ (\sigma + x-u) }}{\dfrac{1}{ (\sigma + x-u) }}{\frac{1}{ (\sigma + x-u) }}{\frac{1}{ (\sigma + x-u) }} } \bigg ) \nonumber \\ \,&= \, { \mathchoice{\dfrac{ \big [ \ln (1+x) \big ]^{\ell } }{ 1 + x + \sigma }}{\dfrac{ \big [ \ln (1+x) \big ]^{\ell } }{ 1 + x + \sigma }}{\frac{ \big [ \ln (1+x) \big ]^{\ell } }{ 1 + x + \sigma }}{\frac{ \big [ \ln (1+x) \big ]^{\ell } }{ 1 + x + \sigma }} } \cdot \Big ( \ln (1+x) - \ln \sigma + \ln (\sigma + x) \Big ) . \end{aligned}$$

(D.6)

To estimate $\mathcal {I}_{\ell }^{(3)}$ one first decomposes it as

$$\begin{aligned} \mathcal {I}_{\ell }^{(3)} \, =&\, \int \limits _{0}^{ +\infty } { \mathchoice{\dfrac{ \text {d}u \cdot \big [ \ln (1+u+x) \big ]^{\ell } }{ (1+u+x) \cdot (\sigma +u) }}{\dfrac{ \text {d}u \cdot \big [ \ln (1+u+x) \big ]^{\ell } }{ (1+u+x) \cdot (\sigma +u) }}{\frac{ \text {d}u \cdot \big [ \ln (1+u+x) \big ]^{\ell } }{ (1+u+x) \cdot (\sigma +u) }}{\frac{ \text {d}u \cdot \big [ \ln (1+u+x) \big ]^{\ell } }{ (1+u+x) \cdot (\sigma +u) }} } = \int \limits _{0}^{ x } \text {d}u { \mathchoice{\dfrac{ \big [ \ln (1+u+x) \big ]^{\ell } }{ 1+x-\sigma }}{\dfrac{ \big [ \ln (1+u+x) \big ]^{\ell } }{ 1+x-\sigma }}{\frac{ \big [ \ln (1+u+x) \big ]^{\ell } }{ 1+x-\sigma }}{\frac{ \big [ \ln (1+u+x) \big ]^{\ell } }{ 1+x-\sigma }} } \bigg ( { \mathchoice{\dfrac{ 1 }{ \sigma + u }}{\dfrac{ 1 }{ \sigma + u }}{\frac{ 1 }{ \sigma + u }}{\frac{ 1 }{ \sigma + u }} } - { \mathchoice{\dfrac{1}{ 1 + u + x }}{\dfrac{1}{ 1 + u + x }}{\frac{1}{ 1 + u + x }}{\frac{1}{ 1 + u + x }} } \bigg ) \nonumber \\&+ \int \limits _{0}^{ +\infty } { \mathchoice{\dfrac{ \text {d}u \cdot \big [ \ln (1 + u + 2 x) \big ]^{\ell } }{ (1 + u + 2x)\cdot (\sigma +u+x) }}{\dfrac{ \text {d}u \cdot \big [ \ln (1 + u + 2 x) \big ]^{\ell } }{ (1 + u + 2x)\cdot (\sigma +u+x) }}{\frac{ \text {d}u \cdot \big [ \ln (1 + u + 2 x) \big ]^{\ell } }{ (1 + u + 2x)\cdot (\sigma +u+x) }}{\frac{ \text {d}u \cdot \big [ \ln (1 + u + 2 x) \big ]^{\ell } }{ (1 + u + 2x)\cdot (\sigma +u+x) }} }. \end{aligned}$$

(D.7)

The logarithmic term in the first integral may be bounded by $\big [ \ln (1+2x) \big ]^{\ell }$ while, in the second integral one uses the bound valid for x large enough

$$\begin{aligned} { \mathchoice{\dfrac{ 1 + 2 x + u }{ \sigma + x + u }}{\dfrac{ 1 + 2 x + u }{ \sigma + x + u }}{\frac{ 1 + 2 x + u }{ \sigma + x + u }}{\frac{ 1 + 2 x + u }{ \sigma + x + u }} } \, \le \, { \mathchoice{\dfrac{ 1+2x}{ \sigma + x}}{\dfrac{ 1+2x}{ \sigma + x}}{\frac{ 1+2x}{ \sigma + x}}{\frac{ 1+2x}{ \sigma + x}} } , \end{aligned}$$

(D.8)

so as to integrate only a function of the variable $1 + 2 x + u$. Then, the identity

$$\begin{aligned} \int \limits _{0}^{+\infty } \text {d}u { \mathchoice{\dfrac{ \big [ \ln (A+u)\big ]^{\ell } }{ (A+u)^2}}{\dfrac{ \big [ \ln (A+u)\big ]^{\ell } }{ (A+u)^2}}{\frac{ \big [ \ln (A+u)\big ]^{\ell } }{ (A+u)^2}}{\frac{ \big [ \ln (A+u)\big ]^{\ell } }{ (A+u)^2}} } = { \mathchoice{\dfrac{1}{A}}{\dfrac{1}{A}}{\frac{1}{A}}{\frac{1}{A}} } \sum \limits _{p=0}^{\ell } { \mathchoice{\dfrac{ \ell ! }{ p! }}{\dfrac{ \ell ! }{ p! }}{\frac{ \ell ! }{ p! }}{\frac{ \ell ! }{ p! }} } \big [ \ln A\big ]^{p} \end{aligned}$$

(D.9)

leads to

$$\begin{aligned} \mathcal {I}_{\ell }^{(3)} \, \le \, { \mathchoice{\dfrac{ \big [ \ln (1+u+x) \big ]^{\ell } }{ 1+x-\sigma }}{\dfrac{ \big [ \ln (1+u+x) \big ]^{\ell } }{ 1+x-\sigma }}{\frac{ \big [ \ln (1+u+x) \big ]^{\ell } }{ 1+x-\sigma }}{\frac{ \big [ \ln (1+u+x) \big ]^{\ell } }{ 1+x-\sigma }} } \Big \{ \ln (\sigma +x)-\ln (1+2 x) - \ln \sigma + \ln (1+x) \Big \} + \sum \limits _{p=0}^{\ell } { \mathchoice{\dfrac{ \ell ! }{ p! }}{\dfrac{ \ell ! }{ p! }}{\frac{ \ell ! }{ p! }}{\frac{ \ell ! }{ p! }} } { \mathchoice{\dfrac{ \big [ \ln (1+2x) \big ]^{ p } }{ \sigma + x }}{\dfrac{ \big [ \ln (1+2x) \big ]^{ p } }{ \sigma + x }}{\frac{ \big [ \ln (1+2x) \big ]^{ p } }{ \sigma + x }}{\frac{ \big [ \ln (1+2x) \big ]^{ p } }{ \sigma + x }} } . \end{aligned}$$

(D.10)

By using analogous techniques, one may bound $\mathcal {I}_{\ell }^{(1)}$ concluding that

$$\begin{aligned} \mathcal {I}_{\ell }^{(1)}=\text {O}\Big ( { \mathchoice{\dfrac{ \big [ \ln (1+x) \big ]^{\ell +1} }{ 1 + x }}{\dfrac{ \big [ \ln (1+x) \big ]^{\ell +1} }{ 1 + x }}{\frac{ \big [ \ln (1+x) \big ]^{\ell +1} }{ 1 + x }}{\frac{ \big [ \ln (1+x) \big ]^{\ell +1} }{ 1 + x }} } \Big )\, . \end{aligned}$$

(D.11)

All in all, this entails the claim. $\square$