$D_{\mathtt{h}}=\mathtt{h}-c_{g}^{2}$. The non-degeneracy property
$D_{\mathtt{h}}\neq 0$ corresponds to that in Bridges-Mielke [9, p.183] and
[41, p.409].

######  Lemma 5.3.

For any $\mathtt{h}>0$ it results

$\mathtt{D}_{\mathtt{h}}:=\mathtt{h}-\tfrac{1}{4}\mathtt{e}_{12}^{2}>0\,,\quad\text{and}\quad\lim_{\mathtt{h}\to
0^{+}}\mathtt{D}_{\mathtt{h}}=0\,.$ (5.7)

###### Proof.

We write
$\mathtt{D}_{\mathtt{h}}=(\sqrt{\mathtt{h}}+\frac{1}{2}\mathtt{e}_{12})(\sqrt{\mathtt{h}}-\frac{1}{2}\mathtt{e}_{12})$
whose first factor is positive for $\mathtt{h}>0$. We claim that also the
second factor is positive. In view of (1.2) it is equal to
$\tfrac{1}{2}{\mathtt{c}}_{\mathtt{h}}^{-1}f(\mathtt{h})$ with

$\displaystyle
f(\mathtt{h}):=\big{(}\sqrt{\mathtt{h}}\tanh(\mathtt{h})-\sqrt{\mathtt{h}}+\sqrt{\tanh(\mathtt{h})}\big{)}\big{(}\sqrt{\mathtt{h}}\tanh(\mathtt{h})+\sqrt{\mathtt{h}}-\sqrt{\tanh(\mathtt{h})}\big{)}=:q(\mathtt{h})p(\mathtt{h})\,.$

The function $p(\mathtt{h})$ is positive since $\mathtt{h}>\tanh(\mathtt{h})$
for any $\mathtt{h}>0$. We claim that also the function $q(\mathtt{h})$ is
positive. Indeed its derivative

$q^{\prime}(\mathtt{h})=\frac{1-\tanh(\mathtt{h})}{2\sqrt{\mathtt{h}}\sqrt{\tanh(\mathtt{h})}}\Big{(}-\sqrt{\tanh(\mathtt{h})}+\sqrt{\mathtt{h}}+\sqrt{\mathtt{h}}\,{\tanh(\mathtt{h})}\Big{)}+\sqrt{\mathtt{h}}\big{(}1-\tanh^{2}(\mathtt{h})\big{)}>0$

for any $\mathtt{h}>0$. Since $q(0)=0$ we deduce that $q(\mathtt{h})>0$ for
any $\mathtt{h}>0$. This proves the lemma. ∎

We now state the main result of this section.

######  Lemma 5.4.

(Step of block-decoupling) There exists a $2\times 2$ reversibility-preserving
matrix $X$, analytic in $(\mu,\epsilon)$, of the form

$\displaystyle X$ $\displaystyle:=\begin{pmatrix}x_{11}&\mathrm{i}\,x_{12}\\\
\mathrm{i}\,x_{21}&x_{22}\end{pmatrix}\qquad\qquad\qquad\qquad\text{with}\quad
x_{ij}\in\mathbb{R}\,,\ i,j=1,2\,,$ (5.8)
$\displaystyle=\begin{pmatrix}r_{11}(\epsilon)&\mathrm{i}\,\,r_{12}(\epsilon)\\\
-\mathrm{i}\,\frac{1}{2}\mathtt{D}_{\mathtt{h}}^{-1}(\mathtt{e}_{12}\mathtt{f}_{11}+2{\mathtt{c}}_{\mathtt{h}}^{-\frac{1}{2}})\epsilon+\mathrm{i}\,r_{21}(\epsilon^{2},\mu\epsilon)&\frac{1}{2}\mathtt{D}_{\mathtt{h}}^{-1}({\mathtt{c}}_{\mathtt{h}}^{-\frac{1}{2}}\mathtt{e}_{12}+2\mathtt{h}\mathtt{f}_{11})\epsilon+r_{22}(\epsilon^{2},\mu\epsilon)\end{pmatrix}\,,$

where $\mathtt{e}_{12}$, $\mathtt{f}_{11}$ are defined in (1.2), (4.13) and
$\mathtt{D}_{\mathtt{h}}$ is the positive constant in (5.7), such that the
following holds true. By conjugating the Hamiltonian and reversible matrix
$\mathtt{L}_{\mu,\epsilon}^{(1)}$, defined in (5.2), with the symplectic and
reversibility-preserving $4\times 4$ matrix

$\exp\left(S^{(1)}\right)\,,\quad\text{ where }\qquad
S^{(1)}:=\mathtt{J}_{4}\begin{pmatrix}0&\Sigma\\\
\Sigma^{*}&0\end{pmatrix}\,,\qquad\Sigma:=\mathtt{J}_{2}X\,,$ (5.9)

we get the Hamiltonian and reversible matrix

$\mathtt{L}_{\mu,\epsilon}^{(2)}:=\exp\left(S^{(1)}\right)\mathtt{L}_{\mu,\epsilon}^{(1)}\exp\left(-S^{(1)}\right)=\mathtt{J}_{4}\mathtt{B}_{\mu,\epsilon}^{(2)}=\begin{pmatrix}\mathtt{J}_{2}E^{(2)}&\mathtt{J}_{2}F^{(2)}\\\
\mathtt{J}_{2}[F^{(2)}]^{*}&\mathtt{J}_{2}G^{(2)}\end{pmatrix}\,,$ (5.10)

where the reversibility-preserving $2\times 2$ self-adjoint matrix
$[E^{(2)}]^{*}=E^{(2)}$ has the form

$\displaystyle
E^{(2)}=\begin{pmatrix}\mu\epsilon^{2}\mathtt{e}_{\scriptscriptstyle{\textsc{WB}}}+r_{1}^{\prime}(\mu\epsilon^{3},\mu^{2}\epsilon^{2})-\mathtt{e}_{22}\frac{\mu^{3}}{8}(1+r_{1}^{\prime\prime}(\epsilon,\mu))&\mathrm{i}\,\big{(}\frac{1}{2}\mathtt{e}_{12}\mu+r_{2}(\mu\epsilon^{2},\mu^{2}\epsilon,\mu^{3})\big{)}\\\
-\mathrm{i}\,\big{(}\frac{1}{2}\mathtt{e}_{12}\mu+r_{2}(\mu\epsilon^{2},\mu^{2}\epsilon,\mu^{3})\big{)}&-\mathtt{e}_{22}\frac{\mu}{8}(1+r_{5}(\epsilon,\mu))\end{pmatrix}\,,$
(5.11)

where

$\displaystyle\mathtt{e}_{\scriptscriptstyle{\textsc{WB}}}=\mathtt{e}_{11}-\mathtt{D}_{\mathtt{h}}^{-1}\big{(}{\mathtt{c}}_{\mathtt{h}}^{-1}+\mathtt{h}\mathtt{f}_{11}^{2}+\mathtt{e}_{12}\mathtt{f}_{11}{\mathtt{c}}_{\mathtt{h}}^{-\frac{1}{2}}\big{)}$
(5.12)

(with constants $\mathtt{e}_{11}$, $\mathtt{D}_{\mathtt{h}}$,
$\mathtt{f}_{11}$, $\mathtt{e}_{12}$, defined in (4.13), (5.7), (1.2)), is the
Whitham-Benjamin function defined in (1.1), the reversibility-preserving
$2\times 2$ self-adjoint matrix $[G^{(2)}]^{*}=G^{(2)}$ has the form

$G^{(2)}=\begin{pmatrix}\mu+r_{8}(\mu\epsilon^{2},\mu^{3}\epsilon)&-\mathrm{i}\,r_{9}(\mu\epsilon^{2},\mu^{2}\epsilon)\\\
\mathrm{i}\,r_{9}(\mu\epsilon^{2},\mu^{2}\epsilon)&\tanh(\mathtt{h}\mu)+r_{10}(\mu\epsilon)\end{pmatrix}\,,$
(5.13)

and

$F^{(2)}=\begin{pmatrix}r_{3}(\mu\epsilon^{3})&\mathrm{i}\,r_{4}(\mu\epsilon^{3})\\\
\mathrm{i}\,r_{6}(\mu\epsilon^{3})&r_{7}(\mu\epsilon^{3})\end{pmatrix}\,.$
(5.14)

The rest of the section is devoted to the proof of Lemma 5.4. For simplicity
let $S=S^{(1)}$.

The matrix $\text{exp}(S)$ is symplectic and reversibility-preserving because
the matrix $S$ in (5.9) is Hamiltonian and reversibility-preserving, cfr.
Lemma 3.8 in [6]. Note that $S$ is reversibility preserving since $X$ has the
form (5.8).

We now expand in Lie series the Hamiltonian and reversible matrix
$\mathtt{L}_{\mu,\epsilon}^{(2)}=\exp(S)\mathtt{L}_{\mu,\epsilon}^{(1)}\exp(-S)$.

We split $\mathtt{L}_{\mu,\epsilon}^{(1)}$ into its $2\times 2$-diagonal and
off-diagonal Hamiltonian and reversible matrices

$\displaystyle\qquad\qquad\qquad\qquad\qquad\qquad\mathtt{L}_{\mu,\epsilon}^{(1)}=D^{(1)}+R^{(1)}\,,$
(5.15) $\displaystyle D^{(1)}:=\begin{pmatrix}D_{1}&0\\\
0&D_{0}\end{pmatrix}:=\begin{pmatrix}\mathtt{J}_{2}E^{(1)}&0\\\
0&\mathtt{J}_{2}G^{(1)}\end{pmatrix},\quad
R^{(1)}:=\begin{pmatrix}0&\mathtt{J}_{2}F^{(1)}\\\
\mathtt{J}_{2}[F^{(1)}]^{*}&0\end{pmatrix},$

and we perform the Lie expansion

$\displaystyle\mathtt{L}_{\mu,\epsilon}^{(2)}=\exp(S)\mathtt{L}_{\mu,\epsilon}^{(1)}\exp(-S)=D^{(1)}+\left[S\,,\,D^{(1)}\right]+\frac{1}{2}[S,[S,D^{(1)}]]+R^{(1)}+[S,R^{(1)}]$
(5.16) $\displaystyle+\frac{1}{2}\int_{0}^{1}(1-\tau)^{2}\exp(\tau
S)\text{ad}_{S}^{3}(D^{(1)})\exp(-\tau
S)\,\mathrm{d}\tau+\int_{0}^{1}(1-\tau)\,\exp(\tau
S)\,\text{ad}_{S}^{2}(R^{(1)})\,\exp(-\tau S)\,\mathrm{d}\tau$

where $\text{ad}_{A}(B):=[A,B]:=AB-BA$ denotes the commutator between the
linear operators $A,B$.

We look for a $4\times 4$ matrix $S$ as in (5.9) that solves the homological
equation $R^{(1)}+\left[S\,,\,D^{(1)}\right]=0$, which, recalling (5.15),
reads

$\begin{pmatrix}0&\mathtt{J}_{2}F^{(1)}+\mathtt{J}_{2}\Sigma
D_{0}-D_{1}\mathtt{J}_{2}\Sigma\\\
\mathtt{J}_{2}{[F^{(1)}]}^{*}+\mathtt{J}_{2}\Sigma^{*}D_{1}-D_{0}\mathtt{J}_{2}\Sigma^{*}&0\end{pmatrix}=0\,.$
(5.17)

Note that the equation $\mathtt{J}_{2}F^{(1)}+\mathtt{J}_{2}\Sigma
D_{0}-D_{1}\mathtt{J}_{2}\Sigma=0$ implies also
$\mathtt{J}_{2}{[F^{(1)}]}^{*}+\mathtt{J}_{2}\Sigma^{*}D_{1}-D_{0}\mathtt{J}_{2}\Sigma^{*}=0$
and viceversa. Thus, writing $\Sigma=\mathtt{J}_{2}X$, namely
$X=-\mathtt{J}_{2}\Sigma$, the equation (5.17) amounts to solve the
“Sylvester” equation

$D_{1}X-XD_{0}=-\mathtt{J}_{2}F^{(1)}\,.$ (5.18)

We write the matrices $E^{(1)},F^{(1)},G^{(1)}$ in (5.2) as

$E^{(1)}=\begin{pmatrix}E_{11}^{(1)}&\mathrm{i}\,E_{12}^{(1)}\\\
-\mathrm{i}\,E_{12}^{(1)}&E_{22}^{(1)}\end{pmatrix}\,,\quad
F^{(1)}=\begin{pmatrix}F_{11}^{(1)}&\mathrm{i}\,F_{12}^{(1)}\\\
\mathrm{i}\,F_{21}^{(1)}&F_{22}^{(1)}\end{pmatrix}\,,\quad
G^{(1)}=\begin{pmatrix}G_{11}^{(1)}&\mathrm{i}\,G_{12}^{(1)}\\\
-\mathrm{i}\,G_{12}^{(1)}&G_{22}^{(1)}\end{pmatrix}$ (5.19)

where the real numbers $E_{ij}^{(1)},F_{ij}^{(1)},G_{ij}^{(1)}$, $i,j=1,2$,
have the expansion in (5.4), (5.5), (5.6). Thus, by (5.15), (5.8) and (5.19),
the equation (5.18) amounts to solve the $4\times 4$ real linear system

$\displaystyle\underbrace{\begin{pmatrix}G_{12}^{(1)}-E_{12}^{(1)}&G_{11}^{(1)}&E_{22}^{(1)}&0\\\
G_{22}^{(1)}&G_{12}^{(1)}-E_{12}^{(1)}&0&-E_{22}^{(1)}\\\
E_{11}^{(1)}&0&G_{12}^{(1)}-E_{12}^{(1)}&-G_{11}^{(1)}\\\
0&-E_{11}^{(1)}&-G_{22}^{(1)}&G_{12}^{(1)}-E_{12}^{(1)}\end{pmatrix}}_{=:{\mathcal{A}}}\underbrace{\begin{pmatrix}x_{11}\\\
x_{12}\\\ x_{21}\\\
x_{22}\end{pmatrix}}_{=:\vec{x}}=\underbrace{\begin{pmatrix}-F_{21}^{(1)}\\\
F_{22}^{(1)}\\\ -F_{11}^{(1)}\\\ F_{12}^{(1)}\end{pmatrix}}_{=:\vec{f}}.$
(5.20)

We solve this system using the following result, verified by a direct
calculus.

######  Lemma 5.5.

The determinant of the matrix

$A:=\begin{pmatrix}a&b&c&0\\\ d&a&0&-c\\\ e&0&a&-b\\\ 0&-e&-d&a\end{pmatrix}$
(5.21)

where $a,b,c,d,e$ are real numbers, is

$\det A=a^{4}-2a^{2}(bd+ce)+(bd-
ce)^{2}=(bd-a^{2})^{2}-2ce\big{(}a^{2}+bd-\frac{1}{2}ce\big{)}\,.$ (5.22)

If $\det A\neq 0$ then $A$ is invertible and

$\displaystyle A^{-1}=\footnotesize{\frac{1}{\det
A}\left(\begin{array}[]{cccc}\\!a\left(a^{2}-bd-ce\right)&\\!b\left(-a^{2}+bd-
ce\right)&-c\left(a^{2}+bd-ce\right)&\\!-2abc\\\ \\!d\left(-a^{2}+bd-
ce\right)&\\!a\left(a^{2}-bd-ce\right)&2acd&\\!-c\left(-a^{2}-bd+ce\right)\\\
\\!-e\left(a^{2}+bd-ce\right)&\\!2abe&a\left(a^{2}-bd-
ce\right)&\\!b\left(a^{2}-bd+ce\right)\\\
\\!-2ade&\\!-e\left(-a^{2}-bd+ce\right)&d\left(a^{2}-bd+ce\right)&\\!a\left(a^{2}-bd-
ce\right)\end{array}\right)}\,.$ (5.27)

The Sylvester matrix $\mathcal{A}$ in (5.20) has the form (5.21) where, by
(5.4)-(5.6) and since $\tanh(\mathtt{h}\mu)=\mathtt{h}\mu+r(\mu^{3})$,

$\displaystyle
a=G_{12}^{(1)}-E_{12}^{(1)}=-\mathtt{e}_{12}\frac{\mu}{2}\big{(}1+r(\epsilon^{2},\mu\epsilon,\mu^{2})\big{)}\,,\
b=G_{11}^{(1)}=\mu+r_{8}(\mu\epsilon^{2},\mu^{3}\epsilon)\,,$ (5.28)
$\displaystyle
c=E_{22}^{(1)}=-\mathtt{e}_{22}\frac{\mu}{8}(1+r_{5}(\epsilon,\mu))\,,\
d=G_{22}^{(1)}=\mu\mathtt{h}+r(\mu\epsilon,\mu^{3})\,,\
e=E_{11}^{(1)}=r(\mu\epsilon^{2},\mu^{3})\,,$

where $\mathtt{e}_{12}$ and $\mathtt{e}_{22}$, defined respectively in (1.2),
(1.3), are positive for any $\mathtt{h}>0$.

By (5.22), the determinant of the matrix ${\mathcal{A}}$ is

$\displaystyle\det{\mathcal{A}}=(bd-a^{2})^{2}+r(\mu^{4}\epsilon^{2},\mu^{6})=\mu^{4}\mathtt{D}_{\mathtt{h}}^{2}(1+r(\epsilon,\mu^{2}))\,$
(5.29)

where $\mathtt{D}_{\mathtt{h}}$ is defined in (5.7). By (5.27), (5.28), (5.29)
and, since
$\mathtt{D}_{\mathtt{h}}=\mathtt{h}-\frac{1}{4}\mathtt{e}_{12}^{2}$, we obtain

$\displaystyle{\mathcal{A}}^{-1}=(1+r(\epsilon,\mu))\displaystyle{\frac{1}{\mu\mathtt{D}^{2}_{\mathtt{h}}}}\,\begin{pmatrix}\frac{1}{2}{\mathtt{e}_{12}}\mathtt{D}_{\mathtt{h}}&\mathtt{D}_{\mathtt{h}}&\frac{1}{32}\mathtt{e}_{22}(\mathtt{e}_{12}^{2}+4\mathtt{h})&-\frac{1}{8}{\mathtt{e}_{12}}\,\mathtt{e}_{22}\\\
\mathtt{h}\mathtt{D}_{\mathtt{h}}&\frac{1}{2}{\mathtt{e}_{12}}\mathtt{D}_{\mathtt{h}}&\frac{1}{8}\mathtt{e}_{12}\mathtt{e}_{22}\mathtt{h}&-\frac{1}{32}\mathtt{e}_{22}\,(\mathtt{e}_{12}^{2}+4\mathtt{h})\\\
r(\epsilon^{2},\mu^{2})&r(\epsilon^{2},\mu^{2})&\frac{1}{2}{\mathtt{e}_{12}}\mathtt{D}_{\mathtt{h}}&-{\mathtt{D}_{\mathtt{h}}}\\\
r(\epsilon^{2},\mu^{2})&r(\epsilon^{2},\mu^{2})&-\mathtt{h}\mathtt{D}_{\mathtt{h}}&\frac{1}{2}{\mathtt{e}_{12}}\mathtt{D}_{\mathtt{h}}\end{pmatrix}\,.$
(5.30)

Therefore, for any $\mu\neq 0$, there exists a unique solution
$\vec{x}={\mathcal{A}}^{-1}\vec{f}$ of the linear system (5.20), namely a
unique matrix $X$ which solves the Sylvester equation (5.18).

######  Lemma 5.6.

The matrix solution $X$ of the Sylvester equation (5.18) is analytic in
$(\mu,\epsilon)$, and admits an expansion as in (5.8).

###### Proof.

By (5.20), (5.30), (5.19), (5.6) we obtain, for any $\mu\neq 0$

$\displaystyle\footnotesize\begin{pmatrix}x_{11}\\\ x_{12}\\\ x_{21}\\\
x_{22}\end{pmatrix}\footnotesize=\frac{1}{\mathtt{D}^{2}_{\mathtt{h}}}\begin{pmatrix}\frac{1}{2}{\mathtt{e}_{12}}\mathtt{D}_{\mathtt{h}}&\mathtt{D}_{\mathtt{h}}&\frac{1}{32}\mathtt{e}_{22}(\mathtt{e}_{12}^{2}+4\mathtt{h})&-\frac{1}{8}{\mathtt{e}_{12}}\,\mathtt{e}_{22}\\\
\mathtt{h}\mathtt{D}_{\mathtt{h}}&\frac{1}{2}{\mathtt{e}_{12}}\mathtt{D}_{\mathtt{h}}&\frac{1}{8}\mathtt{e}_{12}\mathtt{e}_{22}\mathtt{h}&-\frac{1}{32}\mathtt{e}_{22}\,(\mathtt{e}_{12}^{2}+4\mathtt{h})\\\
r(\epsilon^{2},\mu^{2})\quad&r(\epsilon^{2},\mu^{2})&\frac{1}{2}{\mathtt{e}_{12}}\mathtt{D}_{\mathtt{h}}&-{\mathtt{D}_{\mathtt{h}}}\\\
r(\epsilon^{2},\mu^{2})\quad&r(\epsilon^{2},\mu^{2})&-\mathtt{h}\mathtt{D}_{\mathtt{h}}&\frac{1}{2}\mathtt{e}_{12}\mathtt{D}_{\mathtt{h}}\end{pmatrix}\begin{pmatrix}r(\epsilon)\\\
r(\epsilon)\\\
-\mathtt{f}_{11}\epsilon+r(\epsilon^{3},\mu\epsilon^{2},\mu^{2}\epsilon)\\\
{\mathtt{c}}_{\mathtt{h}}^{-\frac{1}{2}}\epsilon+r(\epsilon^{2},\mu\epsilon)\end{pmatrix}(1+r(\epsilon,\mu))\,,$

which proves (5.8). In particular each $x_{ij}$ admits an analytic extension
at $\mu=0$. Note that, for $\mu=0$, one has $E^{(2)}=G^{(2)}=F^{(2)}=0$ and
the Sylvester equation reduces to tautology. ∎

Since the matrix $S$ solves the homological equation
$\left[S\,,\,D^{(1)}\right]+R^{(1)}=0$, identity (5.16) simplifies to

$\mathtt{L}_{\mu,\epsilon}^{(2)}=D^{(1)}+\frac{1}{2}\left[S\,,\,R^{(1)}\right]+\frac{1}{2}\int_{0}^{1}(1-\tau^{2})\,\exp(\tau
S)\,\text{ad}_{S}^{2}(R^{(1)})\,\exp(-\tau S)\mathrm{d}\tau\,.$ (5.31)

The matrix $\frac{1}{2}\left[S\,,\,R^{(1)}\right]$ is, by (5.9), (5.15), the
block-diagonal Hamiltonian and reversible matrix

$\displaystyle\frac{1}{2}\left[S\,,\,R^{(1)}\right]$ (5.32)
$\displaystyle=\begin{pmatrix}\frac{1}{2}\mathtt{J}_{2}(\Sigma\mathtt{J}_{2}[F^{(1)}]^{*}-F^{(1)}\mathtt{J}_{2}\Sigma^{*})&0\\\
0&\frac{1}{2}\mathtt{J}_{2}(\Sigma^{*}\mathtt{J}_{2}F^{(1)}-[F^{(1)}]^{*}\mathtt{J}_{2}\Sigma)\end{pmatrix}=\begin{pmatrix}\mathtt{J}_{2}\tilde{E}&0\\\
0&\mathtt{J}_{2}\tilde{G}\end{pmatrix},$

where, since $\Sigma=\mathtt{J}_{2}X$,

$\tilde{E}:=\text{{Sym}}\big{(}\mathtt{J}_{2}X\mathtt{J}_{2}[F^{(1)}]^{*}\big{)}\,,\qquad\tilde{G}:=\text{{Sym}}\big{(}X^{*}F^{(1)}\big{)}\,,$
(5.33)

denoting $\text{{Sym}}(A):=\frac{1}{2}(A+A^{*})$.

######  Lemma 5.7.

The self-adjoint and reversibility-preserving matrices $\tilde{E},\ \tilde{G}$
in (5.33) have the form

$\displaystyle\tilde{E}=\begin{pmatrix}\tilde{\mathtt{e}}_{11}\mu\epsilon^{2}+\tilde{r}_{1}(\mu\epsilon^{3},\mu^{2}\epsilon^{2})&\mathrm{i}\,\tilde{r}_{2}(\mu\epsilon^{2})\\\
-\mathrm{i}\,\tilde{r}_{2}(\mu\epsilon^{2})&\tilde{r}_{5}(\mu\epsilon^{2})\end{pmatrix}\,,\quad\tilde{G}=\begin{pmatrix}\tilde{r}_{8}(\mu\epsilon^{2})&\mathrm{i}\,\tilde{r}_{9}(\mu\epsilon^{2})\\\
-\mathrm{i}\,\tilde{r}_{9}(\mu\epsilon^{2})&\tilde{r}_{10}(\mu\epsilon^{2})\end{pmatrix}\,,$
(5.34)
$\displaystyle\tilde{\mathtt{e}}_{11}:=-\mathtt{D}_{\mathtt{h}}^{-1}\big{(}{\mathtt{c}}_{\mathtt{h}}^{-1}+\mathtt{h}\mathtt{f}_{11}^{2}+\mathtt{e}_{12}\mathtt{f}_{11}{\mathtt{c}}_{\mathtt{h}}^{-\frac{1}{2}}\big{)}\,.$

###### Proof.

For simplicity we set $F=F^{(1)}$. By (5.8), (5.6), one has

$\displaystyle\mathtt{J}_{2}X\mathtt{J}_{2}F^{*}$
$\displaystyle=\begin{pmatrix}x_{21}F_{12}-x_{22}F_{11}&\mathrm{i}\,(x_{21}F_{22}+x_{22}F_{21})\\\
\mathrm{i}\,(x_{11}F_{12}+x_{12}F_{11})&-x_{11}F_{22}+x_{12}F_{21}\end{pmatrix}=\begin{pmatrix}\tilde{\mathtt{e}}_{11}\mu\epsilon^{2}+r(\mu\epsilon^{3},\mu^{2}\epsilon^{2})&\mathrm{i}\,r(\mu\epsilon^{2})\\\
\mathrm{i}\,r(\mu\epsilon^{2})&r(\mu\epsilon^{2})\end{pmatrix}$

with $\tilde{\mathtt{e}}_{11}$ defined in (5.34). The expansion of $\tilde{E}$
in (5.34) follows in view of (5.33). Since $X=\mathcal{O}(\epsilon)$ by (5.8)
and $F=O(\mu\epsilon)$ by (5.6) we deduce that
$X^{*}F=\mathcal{O}(\mu\epsilon^{2})$ and the expansion of $\tilde{G}$ in
(5.34) follows. ∎

Note that the term $\tilde{\mathtt{e}}_{11}\mu\epsilon^{2}$ in the matrix
$\tilde{E}$ in (5.33)-(5.34), has the same order of the $(1,1)$-entry of
$E^{(1)}$ in (5.4), thus will contribute to the Whitham-Benjamin function
$\mathtt{e}_{\scriptscriptstyle{\textsc{WB}}}$ in the $(1,1)$-entry of
$E^{(2)}$ in (5.11). Finally we show that the last term in (5.31) is small.

######  Lemma 5.8.

The $4\times 4$ Hamiltonian and reversibility matrix

$\frac{1}{2}\int_{0}^{1}(1-\tau^{2})\,\exp(\tau
S)\,\textup{ad}_{S}^{2}(R^{(1)})\,\exp(-\tau
S)\,\mathrm{d}\tau=\begin{pmatrix}\mathtt{J}_{2}\widehat{E}&\mathtt{J}_{2}F^{(2)}\\\
\mathtt{J}_{2}[F^{(2)}]^{*}&\mathtt{J}_{2}\widehat{G}\end{pmatrix}$ (5.35)

where the $2\times 2$ self-adjoint and reversible matrices $\widehat{E}$,
$\widehat{G}$ have entries

$\widehat{E}_{ij}\ ,\widehat{G}_{ij}=r(\mu\epsilon^{3})\,,\quad i,j=1,2\,,$
(5.36)

and the $2\times 2$ reversible matrix $F^{(2)}$ admits an expansion as in
(5.14).

###### Proof.

Since $S$ and $R^{(1)}$ are Hamiltonian and reversibility-preserving then
$\textup{ad}_{S}R^{(1)}=[S,R^{(1)}]$ is Hamiltonian and reversibility-
preserving as well. Thus each $\exp(\tau
S)\,\textup{ad}_{S}^{2}(R^{(1)})\,\exp(-\tau S)$ is Hamiltonian and
reversibility-preserving, and formula (5.35) holds. In order to estimate its
entries we first compute $\textup{ad}_{S}^{2}(R^{(1)})$. Using the form of $S$
in (5.9) and $[S,R^{(1)}]$ in (5.32) one gets

$\textup{ad}_{S}^{2}(R^{(1)})=\begin{pmatrix}0&\mathtt{J}_{2}\tilde{F}\\\
\mathtt{J}_{2}\tilde{F}^{*}&0\end{pmatrix}\qquad\text{where}\qquad\tilde{F}:=2\left(\Sigma\mathtt{J}_{2}\tilde{G}-\tilde{E}\mathtt{J}_{2}\Sigma\right)$
(5.37)

and $\tilde{E}$, $\tilde{G}$ are defined in (5.33). Since
$\tilde{E},\tilde{G}=\mathcal{O}(\mu\epsilon^{2})$ by (5.34), and
$\Sigma=\mathtt{J}_{2}X=\mathcal{O}(\epsilon)$ by (5.8), we deduce that
$\tilde{F}=\mathcal{O}(\mu\epsilon^{3})$. Then, for any $\tau\in[0,1]$, the
matrix $\exp(\tau S)\,\textup{ad}_{S}^{2}(R^{(1)})\,\exp(-\tau
S)=\textup{ad}_{S}^{2}(R^{(1)})(1+\mathcal{O}(\mu,\epsilon))$. In particular
the matrix $F^{(2)}$ in (5.35) has the same expansion of $\tilde{F}$, namely
$F^{(2)}=\mathcal{O}(\mu\epsilon^{3})$, and the matrices $\widehat{E}$,
$\widehat{G}$ have entries as in (5.36). ∎

###### Proof of Lemma 5.4..

It follows by (5.31)-(5.32), (5.15) and Lemmata 5.7 and 5.8. The matrix
$E^{(2)}:=E^{(1)}+\tilde{E}+\widehat{E}$ has the expansion in (5.11), with
$\mathtt{e}_{\scriptscriptstyle{\textsc{WB}}}=\mathtt{e}_{11}+\tilde{\mathtt{e}}_{11}$
as in (5.12). Similarly $G^{(2)}:=G^{(1)}+\tilde{G}+\widehat{G}$ has the
expansion in (5.13). ∎

###  5.2 Complete block-decoupling and proof of the main results

We now block-diagonalize the $4\times 4$ Hamiltonian and reversible matrix
$\mathtt{L}_{\mu,\epsilon}^{(2)}$ in (5.10). First we split it into its
$2\times 2$-diagonal and off-diagonal Hamiltonian and reversible matrices

$\displaystyle\qquad\qquad\qquad\qquad\quad\mathtt{L}_{\mu,\epsilon}^{(2)}=D^{(2)}+R^{(2)}\,,$
$\displaystyle D^{(2)}:=\begin{pmatrix}\mathtt{J}_{2}E^{(2)}&0\\\
0&\mathtt{J}_{2}G^{(2)}\end{pmatrix},\quad
R^{(2)}:=\begin{pmatrix}0&\mathtt{J}_{2}F^{(2)}\\\
\mathtt{J}_{2}[F^{(2)}]^{*}&0\end{pmatrix}.$ (5.38)

######  Lemma 5.9.

There exist a $4\times 4$ reversibility-preserving Hamiltonian matrix
$S^{(2)}:=S^{(2)}(\mu,\epsilon)$ of the form (5.9), analytic in
$(\mu,\epsilon)$, of size $\mathcal{O}(\epsilon^{3})$, and a $4\times 4$
block-diagonal reversible Hamiltonian matrix $P:=P(\mu,\epsilon)$, analytic in
$(\mu,\epsilon)$, of size ${\mathcal{O}(\mu\epsilon^{6})}$ such that

$\exp(S^{(2)})(D^{(2)}+R^{(2)})\exp(-S^{(2)})=D^{(2)}+P\,.$ (5.39)

###### Proof.

We set for brevity $S=S^{(2)}$. The equation (5.39) is equivalent to the
system

$\begin{cases}\Pi_{D}\big{(}e^{S}\big{(}D^{(2)}+R^{(2)}\big{)}e^{-S}\big{)}-D^{(2)}=P\\\
\Pi_{\varnothing}\big{(}e^{S}\big{(}D^{(2)}+R^{(2)}\big{)}e^{-S}\big{)}=0\,,\end{cases}$
(5.40)

where $\Pi_{D}$ is the projector onto the block-diagonal matrices and
$\Pi_{\varnothing}$ onto the block-off-diagonal ones. The second equation in
(5.40) is equivalent, by a Lie expansion, and since $[S,R^{(2)}]$ is block-
diagonal, to

$R^{(2)}+\left[S\,,\,D^{(2)}\right]+\underbrace{\Pi_{\varnothing}\int_{0}^{1}(1-\tau)e^{\tau
S}\text{ad}_{S}^{2}\big{(}D^{(2)}+R^{(2)}\big{)}e^{-\tau
S}\mathrm{d}\tau}_{=:\mathcal{R}(S)}=0\,.$ (5.41)

The “nonlinear homological equation” (5.41),

$[S,D^{(2)}]=-R^{(2)}-\mathcal{R}(S)\,,$ (5.42)

is equivalent to solve the $4\times 4$ real linear system

${\mathcal{A}}\vec{x}=\vec{f}(\mu,\epsilon,\vec{x})\,,\quad\vec{f}(\mu,\epsilon,\vec{x})=\mu\vec{v}(\mu,\epsilon)+\mu\vec{g}(\mu,\epsilon,\vec{x})$
(5.43)

associated, as in (5.20), to (5.42). The vector $\mu\vec{v}(\mu,\epsilon)$ is
associated with $-R^{(2)}$ where $R^{(2)}$ is in (5.38). The vector
$\mu\vec{g}(\mu,\epsilon,\vec{x})$ is associated with the matrix
$-\mathcal{R}(S)$, which is a Hamiltonian and reversible block-off-diagonal
matrix (i.e of the form (5.15)). The factor $\mu$ is present in $D^{(2)}$ and
$R^{(2)}$, see (5.11), (5.13), (5.14) and the analytic function
$\vec{g}(\mu,\epsilon,\vec{x})$ is quadratic in $\vec{x}$ (for the presence of
$\text{ad}_{S}^{2}$ in $\mathcal{R}(S)$). In view of (5.14) one has

$\mu\vec{v}(\mu,\epsilon):=(-F^{(2)}_{21},F^{(2)}_{22},-F^{(2)}_{11},F^{(2)}_{12})^{\top},\quad
F^{(2)}_{ij}=\,{r(\mu\epsilon^{3})}\,.$ (5.44)

System (5.43) is equivalent to
$\vec{x}={\mathcal{A}}^{-1}\vec{f}(\mu,\epsilon,\vec{x})$ and, writing
${\mathcal{A}}^{-1}=\frac{1}{\mu}{\mathcal{B}}(\mu,\epsilon)$ (cfr. (5.30)),
to

$\vec{x}={\mathcal{B}}(\mu,\epsilon)\vec{v}(\mu,\epsilon)+{\mathcal{B}}(\mu,\epsilon)\vec{g}(\mu,\epsilon,\vec{x})\,.$

By the implicit function theorem this equation admits a unique small solution
$\vec{x}=\vec{x}(\mu,\epsilon)$, analytic in $(\mu,\epsilon)$, with size
${\mathcal{O}(\epsilon^{3})}$ as $\vec{v}$ in (5.44). Then the first equation
of (5.40) gives $P=[S,R^{(2)}]+\Pi_{D}\int_{0}^{1}(1-\tau)e^{\tau
S}\text{ad}_{S}^{2}\big{(}D^{(2)}+R^{(2)}\big{)}e^{-\tau S}\mathrm{d}\tau$,
and its estimate follows from those of $S$ and $R^{(2)}$ (see (5.14)). ∎

Proof of Theorems 2.5 and 1.1.  By Lemma 5.9 and recalling (3.1) the operator
$\mathcal{L}_{\mu,\epsilon}:\mathcal{V}_{\mu,\epsilon}\to\mathcal{V}_{\mu,\epsilon}$
is represented by the $4\times 4$ Hamiltonian and reversible matrix

$\mathrm{i}\,{\mathtt{c}}_{\mathtt{h}}\mu+\exp(S^{(2)})\mathtt{L}_{\mu,\epsilon}^{(2)}\exp(-S^{(2)})=\mathrm{i}\,{\mathtt{c}}_{\mathtt{h}}\mu+\begin{pmatrix}\mathtt{J}_{2}E^{(3)}&0\\\
0&\mathtt{J}_{2}G^{(3)}\end{pmatrix}=:\begin{pmatrix}\mathtt{U}&0\\\
0&\mathtt{S}\end{pmatrix}\,,$

where the matrices $E^{(3)}$ and $G^{(3)}$ expand as in (5.11), (5.13).
Consequently the matrices $\mathtt{U}$ and $\mathtt{S}$ expand as in (2.40).
Theorem 2.5 is proved. Theorem 1.1 is a straight-forward corollary. The
function $\underline{\mu}(\epsilon)$ in (1.4) is defined as the implicit
solution of the function
$\Delta_{\scriptscriptstyle{\textsc{BF}}}(\mathtt{h};\mu,\epsilon)$ in (1.6)
for $\epsilon$ small enough, depending on $\mathtt{h}$.

##  Appendix A Expansion of the Kato basis

In this appendix we prove Lemma 4.2. We provide the expansion of the basis
$f_{k}^{\pm}(\mu,\epsilon)=U_{\mu,\epsilon}f_{k}^{\pm}$, $k=0,1$, in (4.1),
where $f_{k}^{\pm}$ defined in (4.2) belong to the subspace
$\mathcal{V}_{0,0}:=\text{Rg}(P_{0,0})$. We first Taylor-expand the
transformation operators $U_{\mu,\epsilon}$ defined in (3.7). We denote
$\partial_{\epsilon}$ with an apex and $\partial_{\mu}$ with a dot.

######  Lemma A.1.

The first jets of $U_{\mu,\epsilon}P_{0,0}$ are

$\displaystyle U_{0,0}P_{0,0}$ $\displaystyle=P_{0,0}\,,\quad
U_{0,0}^{\prime}P_{0,0}=P_{0,0}^{\prime}P_{0,0}\,,\quad\dot{U}_{0,0}P_{0,0}=\dot{P}_{0,0}P_{0,0}\,,$
(A.1) $\displaystyle\dot{U}_{0,0}^{\prime}P_{0,0}$
$\displaystyle=\big{(}\dot{P}_{0,0}^{\prime}-\tfrac{1}{2}P_{0,0}\dot{P}_{0,0}^{\prime}\big{)}P_{0,0}\,,$
(A.2)

where

$\displaystyle P_{0,0}^{\prime}$
$\displaystyle=\frac{1}{2\pi\mathrm{i}\,}\oint_{\Gamma}(\mathscr{L}_{0,0}-\lambda)^{-1}\mathscr{L}_{0,0}^{\prime}(\mathscr{L}_{0,0}-\lambda)^{-1}\mathrm{d}\lambda\,,$
(A.3) $\displaystyle\dot{P}_{0,0}$
$\displaystyle=\frac{1}{2\pi\mathrm{i}\,}\oint_{\Gamma}(\mathscr{L}_{0,0}-\lambda)^{-1}\dot{\mathscr{L}}_{0,0}(\mathscr{L}_{0,0}-\lambda)^{-1}\mathrm{d}\lambda\,,$
(A.4)

and

$\displaystyle\dot{P}_{0,0}^{\prime}$
$\displaystyle=-\frac{1}{2\pi\mathrm{i}\,}\oint_{\Gamma}(\mathscr{L}_{0,0}-\lambda)^{-1}\dot{\mathscr{L}}_{0,0}(\mathscr{L}_{0,0}-\lambda)^{-1}\mathscr{L}_{0,0}^{\prime}(\mathscr{L}_{0,0}-\lambda)^{-1}\mathrm{d}\lambda$
(A.5a)
$\displaystyle\qquad-\frac{1}{2\pi\mathrm{i}\,}\oint_{\Gamma}(\mathscr{L}_{0,0}-\lambda)^{-1}\mathscr{L}_{0,0}^{\prime}(\mathscr{L}_{0,0}-\lambda)^{-1}\dot{\mathscr{L}}_{0,0}(\mathscr{L}_{0,0}-\lambda)^{-1}\mathrm{d}\lambda$
(A.5b)
$\displaystyle\qquad+\frac{1}{2\pi\mathrm{i}\,}\oint_{\Gamma}(\mathscr{L}_{0,0}-\lambda)^{-1}\dot{\mathscr{L}}_{0,0}^{\prime}(\mathscr{L}_{0,0}-\lambda)^{-1}\mathrm{d}\lambda\,.$
(A.5c)

The operators $\mathscr{L}_{0,0}^{\prime}$ and $\dot{\mathscr{L}}_{0,0}$ are

$\mathscr{L}_{0,0}^{\prime}=\begin{bmatrix}\partial_{x}\circ p_{1}(x)&0\\\
-a_{1}(x)&p_{1}(x)\circ\partial_{x}\end{bmatrix},\qquad\dot{\mathscr{L}}_{0,0}=\begin{bmatrix}0&\textup{sgn}(D)m(D)\\\
0&0\end{bmatrix},$ (A.6)

where $\operatorname*{sgn}(D)$ is defined in (2.31) and $m(D)$ is the real,
even operator

$m(D):=\tanh(\mathtt{h}|D|)+\mathtt{h}|D|(1-\tanh^{2}(\mathtt{h}|D|))$ (A.7)

and $a_{1}(x)$ and $p_{1}(x)$ are given in Lemma 2.2.

The operator $\dot{\mathscr{L}}_{0,0}^{\prime}$ is

$\dot{\mathscr{L}}_{0,0}^{\prime}=\begin{bmatrix}\mathrm{i}\,p_{1}(x)&0\\\
0&\mathrm{i}\,p_{1}(x)\end{bmatrix}\,.$ (A.8)

###### Proof.

By (3.7) and (3.6) one has the Taylor expansion in $\mathcal{L}(Y)$

$U_{\mu,\epsilon}P_{0,0}=P_{\mu,\epsilon}P_{0,0}+\frac{1}{2}(P_{\mu,\epsilon}-P_{0,0})^{2}P_{\mu,\epsilon}P_{0,0}+\mathcal{O}(P_{\mu,\epsilon}-P_{0,0})^{4}\,,$

where
$\mathcal{O}(P_{\mu,\epsilon}-P_{0,0})^{4}=\mathcal{O}(\epsilon^{4},\epsilon^{3}\mu,\epsilon^{2}\mu^{2},\epsilon\mu^{3},\mu^{4})\in\mathcal{L}(Y)$.
Consequently one derives (A.1), (A.2), using also the identity
$\dot{P}_{0,0}P_{0,0}^{\prime}P_{0,0}+P_{0,0}^{\prime}\dot{P}_{0,0}P_{0,0}=-P_{0,0}\dot{P}_{0,0}^{\prime}P_{0,0}$,
which follows differentiating $P_{\mu,\epsilon}^{2}=P_{\mu,\epsilon}$.
Differentiating (3.5) one gets (A.3)-(A.5c). Formulas (A.6)-(A.8) follow by
(3.2) using also that the Fourier multiplier
$\Pi_{0}\big{(}\tanh(\mathtt{h}|D|)+\mathtt{h}|D|\big{(}1-\tanh^{2}(\mathtt{h}|D|)\big{)}\big{)}=0$.
∎

By the previous lemma we have the Taylor expansion

$f_{k}^{\sigma}(\mu,\epsilon)=f_{k}^{\sigma}+\epsilon
P_{0,0}^{\prime}f_{k}^{\sigma}+\mu\dot{P}_{0,0}f_{k}^{\sigma}+\mu\epsilon\big{(}\dot{P}_{0,0}^{\prime}-\frac{1}{2}P_{0,0}\dot{P}_{0,0}^{\prime}\big{)}f_{k}^{\sigma}+\mathcal{O}(\mu^{2},\epsilon^{2})\,.$
(A.9)

In order to compute the vectors $P_{0,0}^{\prime}f_{k}^{\sigma}$ and
$\dot{P}_{0,0}f_{k}^{\sigma}$ using (A.3) and (A.4), it is useful to know the
action of $(\mathscr{L}_{0,0}-\lambda)^{-1}$ on the vectors

$\displaystyle
f_{k}^{+}:=\begin{bmatrix}{\mathtt{c}}_{\mathtt{h}}^{1/2}\cos(kx)\\\
{\mathtt{c}}_{\mathtt{h}}^{-1/2}\sin(kx)\end{bmatrix}\,,\quad
f_{k}^{-}:=\begin{bmatrix}-{\mathtt{c}}_{\mathtt{h}}^{1/2}\sin(kx)\\\
{\mathtt{c}}_{\mathtt{h}}^{-1/2}\cos(kx)\end{bmatrix}\,,$ (A.10)
$\displaystyle
f_{-k}^{+}:=\begin{bmatrix}{\mathtt{c}}_{\mathtt{h}}^{1/2}\cos(kx)\\\
-{\mathtt{c}}_{\mathtt{h}}^{-1/2}\sin(kx)\end{bmatrix}\,,\quad
f_{-k}^{-}:=\begin{bmatrix}{\mathtt{c}}_{\mathtt{h}}^{1/2}\sin(kx)\\\
{\mathtt{c}}_{\mathtt{h}}^{-1/2}\cos(kx)\end{bmatrix}\,,\quad
k\in\mathbb{N}\,.$

######  Lemma A.2.

The space $H^{1}(\mathbb{T})$ decomposes as
$H^{1}(\mathbb{T})=\mathcal{V}_{0,0}\oplus\mathcal{U}\oplus{\mathcal{W}_{H^{1}}}$,
with
$\mathcal{W}_{H^{1}}=\overline{\bigoplus\limits_{k=2}^{\infty}\mathcal{W}_{k}}^{H^{1}}$
where the subspaces $\mathcal{V}_{0,0},\mathcal{U}$ and $\mathcal{W}_{k}$,
defined below, are invariant under $\mathscr{L}_{0,0}$ and the following
properties hold:

  * (i)

$\mathcal{V}_{0,0}=\text{span}\\{f^{+}_{1},f^{-}_{1},f^{+}_{0},f^{-}_{0}\\}$
is the generalized kernel of $\mathscr{L}_{0,0}$. For any $\lambda\neq 0$ the
operator $\mathscr{L}_{0,0}-\lambda:\mathcal{V}_{0,0}\to\mathcal{V}_{0,0}$ is
invertible and

$\displaystyle(\mathscr{L}_{0,0}-\lambda)^{-1}f_{1}^{+}=-\frac{1}{\lambda}f_{1}^{+}\,,\quad(\mathscr{L}_{0,0}-\lambda)^{-1}f_{1}^{-}=-\frac{1}{\lambda}f_{1}^{-},\quad(\mathscr{L}_{0,0}-\lambda)^{-1}f_{0}^{-}=-\frac{1}{\lambda}f_{0}^{-}\,,$
(A.11)
$\displaystyle(\mathscr{L}_{0,0}-\lambda)^{-1}f_{0}^{+}=-\frac{1}{\lambda}f_{0}^{+}+\frac{1}{\lambda^{2}}f_{0}^{-}\,.$
(A.12)

  * (ii)

$\mathcal{U}:=\text{span}\left\\{f_{-1}^{+},f_{-1}^{-}\right\\}$. For any
$\lambda\neq\pm 2\mathrm{i}\,$ the operator
$\mathscr{L}_{0,0}-\lambda:\mathcal{U}\to\mathcal{U}$ is invertible and

$\displaystyle(\mathscr{L}_{0,0}-\lambda)^{-1}f_{-1}^{+}=\frac{1}{\lambda^{2}+4{\mathtt{c}}_{\mathtt{h}}^{2}}\left(-\lambda
f_{-1}^{+}+2{\mathtt{c}}_{\mathtt{h}}f_{-1}^{-}\right),$ (A.13)
$\displaystyle(\mathscr{L}_{0,0}-\lambda)^{-1}f_{-1}^{-}=\frac{1}{\lambda^{2}+4{\mathtt{c}}_{\mathtt{h}}^{2}}\left(-2{\mathtt{c}}_{\mathtt{h}}f_{-1}^{+}-\lambda
f_{-1}^{-}\right)\,.$

  * (iii)

Each subspace $\mathcal{W}_{k}:=\text{span}\left\\{f_{k}^{+},\
f_{k}^{-},f_{-k}^{+},\ f_{-k}^{-}\right\\}$ is invariant under
$\mathscr{L}_{0,0}$. Let
$\mathcal{W}_{L^{2}}=\overline{\bigoplus\limits_{k=2}^{\infty}\mathcal{W}_{k}}^{L^{2}}$.
For any $|\lambda|<\delta(\mathtt{h})$ small enough, the operator
${\mathscr{L}_{0,0}-\lambda:\mathcal{W}_{H^{1}}\to\mathcal{W}_{L^{2}}}$ is
invertible and for any $f\in{\mathcal{W}_{L^{2}}}$

$(\mathscr{L}_{0,0}-\lambda)^{-1}f=\big{(}{\mathtt{c}}_{\mathtt{h}}^{2}\partial_{x}^{2}+|D|\tanh(\mathtt{h}|D|)\big{)}^{-1}\begin{bmatrix}{\mathtt{c}}_{\mathtt{h}}\partial_{x}&-|D|\tanh(\mathtt{h}|D|)\\\
1&{\mathtt{c}}_{\mathtt{h}}\partial_{x}\end{bmatrix}f+\lambda\varphi_{f}(\lambda,x)\,,$
(A.14)

for some analytic function $\lambda\mapsto\varphi_{f}(\lambda,\cdot)\in
H^{1}(\mathbb{T},\mathbb{C}^{2})$.

###### Proof.

By inspection the spaces $\mathcal{V}_{0,0}$, $\mathcal{U}$ and
${\mathcal{W}_{k}}$ are invariant under $\mathscr{L}_{0,0}$ and, by Fourier
series, they decompose $H^{1}(\mathbb{T},\mathbb{C}^{2})$. Formulas
(A.11)-(A.12) follow using that $f_{1}^{+},f_{1}^{-},f_{0}^{-}$ are in the
kernel of $\mathscr{L}_{0,0}$, and $\mathscr{L}_{0,0}f_{0}^{+}=-f_{0}^{-}$.
Formula (A.13) follows using that
$\mathscr{L}_{0,0}f^{+}_{-1}=-2{\mathtt{c}}_{\mathtt{h}}f^{-}_{-1}$ and
$\mathscr{L}_{0,0}f^{-}_{-1}=2{\mathtt{c}}_{\mathtt{h}}f^{+}_{-1}$. Let us
prove item $(iii)$. Let $\mathcal{W}:=\mathcal{W}_{H^{1}}$. The operator
${{\left.\kern-1.2pt(\mathscr{L}_{0,0}-\lambda\mathrm{Id})\vphantom{\big{|}}\right|_{\mathcal{W}}}}$
is invertible for any
$\lambda\notin\\{\pm\mathrm{i}\,\sqrt{|k|\tanh{(\mathtt{h}|k|)}}\pm\mathrm{i}\,k{\mathtt{c}}_{\mathtt{h}},k\geq
2,k\in{\mathbb{N}}\\}$ and

$\footnotesize({\left.\kern-1.2pt\mathscr{L}_{0,0}\vphantom{\big{|}}\right|_{\mathcal{W}}})^{-1}=\left({\mathtt{c}}_{\mathtt{h}}^{2}\partial_{x}^{2}+|D|\tanh(\mathtt{h}|D|)\right)^{-1}\begin{bmatrix}{\mathtt{c}}_{\mathtt{h}}\partial_{x}&-|D|\tanh(\mathtt{h}|D|))\\\
1&{\mathtt{c}}_{\mathtt{h}}\partial_{x}\end{bmatrix}_{|\mathcal{W}}\,.$

By Neumann series, for any $\lambda$ such that
$|\lambda|\|({\left.\kern-1.2pt\mathscr{L}_{0,0}\vphantom{\big{|}}\right|_{\mathcal{W}}})^{-1}\|_{{\mathcal{L}(\mathcal{W},H^{1}(\mathbb{T}))}}<1$
we have

$({\left.\kern-1.2pt\mathscr{L}_{0,0}\vphantom{\big{|}}\right|_{\mathcal{W}}}-\lambda)^{-1}=({\left.\kern-1.2pt\mathscr{L}_{0,0}\vphantom{\big{|}}\right|_{\mathcal{W}}})^{-1}\big{(}\mathrm{Id}-\lambda({\left.\kern-1.2pt\mathscr{L}_{0,0}\vphantom{\big{|}}\right|_{\mathcal{W}}})^{-1}\big{)}^{-1}=({\left.\kern-1.2pt\mathscr{L}_{0,0}\vphantom{\big{|}}\right|_{\mathcal{W}}})^{-1}\sum_{k\geq
0}(({\left.\kern-1.2pt\mathscr{L}_{0,0}\vphantom{\big{|}}\right|_{\mathcal{W}}})^{-1}\lambda)^{k}\,.$

Formula (A.14) follows with
$\varphi_{f}(\lambda,x):=({\left.\kern-1.2pt\mathscr{L}_{0,0}\vphantom{\big{|}}\right|_{\mathcal{W}}})^{-1}\sum_{k\geq
1}\lambda^{k-1}[({\left.\kern-1.2pt\mathscr{L}_{0,0}\vphantom{\big{|}}\right|_{\mathcal{W}}})^{-1}]^{k}f$.
∎

We shall also use the following formulas obtained by (A.6), (A.7) and (4.2):

$\displaystyle\mathscr{L}_{0,0}^{\prime}f_{1}^{+}=\begin{bmatrix}2{\mathtt{c}}_{\mathtt{h}}^{-1/2}\,\sin(2x)\\\
\frac{1}{2}{\mathtt{c}}_{\mathtt{h}}^{5/2}(1-{\mathtt{c}}_{\mathtt{h}}^{-4})(1+\cos(2x))\end{bmatrix}\,,\qquad\mathscr{L}_{0,0}^{\prime}f_{1}^{-}=\begin{bmatrix}2\,{\mathtt{c}}_{\mathtt{h}}^{-1/2}\,\cos(2x)\\\
-\frac{1}{2}{\mathtt{c}}_{\mathtt{h}}^{5/2}(1-{\mathtt{c}}_{\mathtt{h}}^{-4})\sin(2x)\end{bmatrix}\,,$
(A.15)
$\displaystyle\mathscr{L}_{0,0}^{\prime}f_{0}^{+}=\begin{bmatrix}2{\mathtt{c}}_{\mathtt{h}}^{-1}\sin(x)\\\
\left({\mathtt{c}}_{\mathtt{h}}^{2}+{\mathtt{c}}_{\mathtt{h}}^{-2}\right)\cos(x)\end{bmatrix}\,,\qquad\mathscr{L}_{0,0}^{\prime}f_{0}^{-}=0\,,$
$\displaystyle\dot{\mathscr{L}}_{0,0}f_{1}^{+}=-\mathrm{i}\,b(\mathtt{h})\begin{bmatrix}\cos(x)\\\
0\end{bmatrix}\,,\qquad\dot{\mathscr{L}}_{0,0}f_{1}^{-}=\mathrm{i}\,b(\mathtt{h})\begin{bmatrix}\sin(x)\\\
0\end{bmatrix}\,,\quad
b(\mathtt{h}):={\mathtt{c}}_{\mathtt{h}}^{-1/2}\big{(}{\mathtt{c}}_{\mathtt{h}}^{2}+\mathtt{h}(1-{\mathtt{c}}_{\mathtt{h}}^{4})\big{)}\,,$
$\displaystyle\dot{\mathscr{L}}_{0,0}f_{0}^{+}=0\,,\qquad\dot{\mathscr{L}}_{0,0}f_{0}^{-}=0\,.$

######  Remark.

In deep water we have $\dot{\mathscr{L}}_{0,0}f_{0}^{-}=f_{0}^{+}$ (cfr.
formula (A.14) in [6]). In finite depth instead
$\dot{\mathscr{L}}_{0,0}f_{0}^{-}=0$ because the Fourier multiplier
$\operatorname*{sgn}(D)m(D)$ in (A.7) vanishes on the constants.

We finally compute $P_{0,0}^{\prime}f_{k}^{\sigma}$ and
$\dot{P}_{0,0}f_{k}^{\sigma}$.

######  Lemma A.3.

One has

$\displaystyle
P_{0,0}^{\prime}f^{+}_{1}={\begin{bmatrix}\frac{1}{2}{\mathtt{c}}_{\mathtt{h}}^{-\frac{11}{2}}(3+{\mathtt{c}}_{\mathtt{h}}^{4})\,\cos(2x)\\\
\frac{1}{4}{\mathtt{c}}_{\mathtt{h}}^{-\frac{13}{2}}(1+{\mathtt{c}}_{\mathtt{h}}^{4})(3-{\mathtt{c}}_{\mathtt{h}}^{4})\sin(2x)\end{bmatrix}}\,,\quad
P_{0,0}^{\prime}f^{-}_{1}={\begin{bmatrix}-\frac{1}{2}{\mathtt{c}}_{\mathtt{h}}^{-\frac{11}{2}}(3+{\mathtt{c}}_{\mathtt{h}}^{4})\,\sin(2x)\\\
\frac{1}{4}{\mathtt{c}}_{\mathtt{h}}^{-\frac{13}{2}}(1+{\mathtt{c}}_{\mathtt{h}}^{4})(3-{\mathtt{c}}_{\mathtt{h}}^{4})\cos(2x)\end{bmatrix}}\,,$
(A.16) $\displaystyle
P_{0,0}^{\prime}f^{+}_{0}=\tfrac{1}{4}{\mathtt{c}}_{\mathtt{h}}^{-\frac{5}{2}}(3+{\mathtt{c}}_{\mathtt{h}}^{4})f^{+}_{-1}\,,\quad
P_{0,0}^{\prime}f^{-}_{0}=0\,,\quad\dot{P}_{0,0}f_{0}^{+}=0\,,\quad\dot{P}_{0,0}f_{0}^{-}=0\,,$
$\displaystyle\dot{P}_{0,0}f_{1}^{+}=\frac{\mathrm{i}\,}{4}\big{(}1+{\mathtt{c}}_{\mathtt{h}}^{-2}\mathtt{h}(1-{\mathtt{c}}_{\mathtt{h}}^{4})\big{)}f^{-}_{-1}\,,\quad\dot{P}_{0,0}f_{1}^{-}=\frac{\mathrm{i}\,}{4}\big{(}1+{\mathtt{c}}_{\mathtt{h}}^{-2}\mathtt{h}(1-{\mathtt{c}}_{\mathtt{h}}^{4})\big{)}f^{+}_{-1}\,.$

###### Proof.

We first compute $P_{0,0}^{\prime}f_{1}^{+}$. By (A.3), (A.11) and (A.15) we
deduce

$P_{0,0}^{\prime}f_{1}^{+}=-\frac{1}{2\pi\mathrm{i}\,}\oint_{\Gamma}\frac{1}{\lambda}(\mathscr{L}_{0,0}-\lambda)^{-1}\begin{bmatrix}2{\mathtt{c}}_{\mathtt{h}}^{-1/2}\,\sin(2x)\\\
\frac{1}{2}{\mathtt{c}}_{\mathtt{h}}^{5/2}(1-{\mathtt{c}}_{\mathtt{h}}^{-4})(1+\cos(2x))\end{bmatrix}\mathrm{d}\lambda\,.$

We note that
$\footnotesize\begin{bmatrix}2{\mathtt{c}}_{\mathtt{h}}^{-1/2}\,\sin(2x)\\\
\frac{1}{2}{\mathtt{c}}_{\mathtt{h}}^{5/2}(1-{\mathtt{c}}_{\mathtt{h}}^{-4})(1+\cos(2x))\end{bmatrix}=\frac{1}{2}{\mathtt{c}}_{\mathtt{h}}^{5/2}(1-{\mathtt{c}}_{\mathtt{h}}^{-4})f_{0}^{-}+\mathcal{W}$.
Therefore by (A.11) and (A.14) there is an analytic function
$\lambda\mapsto\varphi(\lambda,\cdot)\in H^{1}(\mathbb{T},\mathbb{C}^{2})$ so
that

$P_{0,0}^{\prime}f_{1}^{+}=-\frac{1}{2\pi\mathrm{i}\,}\oint_{\Gamma}\frac{1}{\lambda}\Big{(}-\dfrac{{\mathtt{c}}_{\mathtt{h}}^{5/2}(1-{\mathtt{c}}_{\mathtt{h}}^{-4})}{2\lambda}f_{0}^{-}{-\frac{1+{\mathtt{c}}_{\mathtt{h}}^{4}}{4{\mathtt{c}}_{\mathtt{h}}^{6}}\begin{bmatrix}2{\mathtt{c}}_{\mathtt{h}}\frac{{\mathtt{c}}_{\mathtt{h}}^{-\frac{1}{2}}(3+{\mathtt{c}}_{\mathtt{h}}^{4})}{1+{\mathtt{c}}_{\mathtt{h}}^{4}}\cos(2x)\\\
{\mathtt{c}}_{\mathtt{h}}^{-\frac{1}{2}}(3-{\mathtt{c}}_{\mathtt{h}}^{4})\sin(2x)\end{bmatrix}}+\lambda\varphi(\lambda)\Big{)}\,\mathrm{d}\lambda\,,$

where we exploited the identity
$\tanh(2\mathtt{h})=\frac{2{\mathtt{c}}_{\mathtt{h}}^{2}}{1+{\mathtt{c}}_{\mathtt{h}}^{4}}$
in applying (A.14). Thus, by means of residue Theorem we obtain the first
identity in (A.16). Similarly one computes $P_{0,0}^{\prime}f_{1}^{-}$. By
(A.3), (A.11) and (A.15), one has $P_{0,0}^{\prime}f_{0}^{-}=0$. Next we
compute $P_{0,0}^{\prime}f_{0}^{+}$. By (A.3), (A.11), (A.12) and (A.15) we
get

$P_{0,0}^{\prime}f_{0}^{+}=-\frac{1}{2\pi\mathrm{i}\,}\oint_{\Gamma}\frac{1}{\lambda}(\mathscr{L}_{0,0}-\lambda)^{-1}\begin{bmatrix}2{\mathtt{c}}_{\mathtt{h}}^{-1}\sin(x)\\\
({\mathtt{c}}_{\mathtt{h}}^{2}+{\mathtt{c}}_{\mathtt{h}}^{-2})\cos(x)\end{bmatrix}\mathrm{d}\lambda\,.$

Next we decompose
$\footnotesize\begin{bmatrix}2{\mathtt{c}}_{\mathtt{h}}^{-1}\sin(x)\\\
({\mathtt{c}}_{\mathtt{h}}^{2}+{\mathtt{c}}_{\mathtt{h}}^{-2})\cos(x)\end{bmatrix}=\underbrace{{\frac{1}{2}{\mathtt{c}}_{\mathtt{h}}^{-\frac{3}{2}}({\mathtt{c}}_{\mathtt{h}}^{4}+3)}}_{=:\alpha}f^{-}_{-1}+\underbrace{{\frac{1}{2}{\mathtt{c}}_{\mathtt{h}}^{-\frac{3}{2}}({\mathtt{c}}_{\mathtt{h}}^{4}-1)}}_{=:\beta}f^{-}_{1}$.
By (A.15) and (A.13) we get

$P_{0,0}^{\prime}f_{0}^{+}=-\frac{1}{2\pi\mathrm{i}\,}\oint_{\Gamma}\Big{(}-\frac{2\alpha{\mathtt{c}}_{\mathtt{h}}}{\lambda(\lambda^{2}+4{\mathtt{c}}_{\mathtt{h}}^{2})}f_{-1}^{+}-\frac{\alpha}{\lambda^{2}+4{\mathtt{c}}_{\mathtt{h}}^{2}}f_{-1}^{-}+\frac{\beta}{\lambda^{2}}f^{-}_{1}\Big{)}\mathrm{d}\lambda=\frac{\alpha}{2{\mathtt{c}}_{\mathtt{h}}}f^{+}_{-1}\,,$

where in the last step we used the residue theorem. We compute now
$\dot{P}_{0,0}f^{+}_{1}$. First we have $\dot{P}_{0,0}f_{1}^{+}=\
\frac{\mathrm{i}\,}{2\pi\mathrm{i}\,}b({\mathtt{h}})\oint_{\Gamma}\frac{1}{\lambda}(\mathscr{L}_{0,0}-\lambda)^{-1}\footnotesize\begin{bmatrix}\cos(x)\\\
0\end{bmatrix}\mathrm{d}\lambda$, where $b(\mathtt{h})$ is in (A.15), and
then, writing $\footnotesize\begin{bmatrix}\cos(x)\\\
0\end{bmatrix}=\frac{1}{2}{\mathtt{c}}_{\mathtt{h}}^{-\frac{1}{2}}(f_{1}^{+}+f_{-1}^{+})$
and using (A.13), we conclude using again the residue theorem
$\dot{P}_{0,0}f_{1}^{+}=\frac{\mathrm{i}\,}{4}\big{(}1+{\mathtt{h}}(1-{\mathtt{c}}_{\mathtt{h}}^{4}){\mathtt{c}}_{\mathtt{h}}^{-2}\big{)}f^{-}_{-1}$.
The computation of $\dot{P}_{0,0}f^{-}_{1}$ is analogous. Finally, in view of
(A.15), we have

$\displaystyle\dot{P}_{0,0}f^{+}_{0}=\frac{1}{2\pi\mathrm{i}\,}\oint_{\Gamma}(\mathcal{L}_{0,0}-\lambda)^{-1}\dot{\mathcal{L}}_{0,0}\big{(}\frac{1}{\lambda^{2}}f_{0}^{-}-\frac{1}{\lambda}f_{0}^{+}\big{)}\mathrm{d}\lambda=0\,,$
$\displaystyle\dot{P}_{0,0}f^{-}_{0}=-\frac{1}{2\pi\mathrm{i}\,}\oint_{\Gamma}\frac{1}{\lambda}(\mathcal{L}_{0,0}-\lambda)^{-1}\dot{\mathcal{L}}_{0,0}f_{0}^{-}\mathrm{d}\lambda=0\,.$

In conclusion all the formulas in (A.16) are proved. ∎

So far we have obtained the linear terms of the expansions (4.3), (4.4),
(4.5), (4.6). We now provide further information about the expansion of the
basis at $\mu=0$. The proof of the next lemma follows as that of Lemma A.4 in
[6].

######  Lemma A.4.

The basis $\\{f_{k}^{\sigma}(0,\epsilon),\,k=0,1,\sigma=\pm\\}$ is real. For
any $\epsilon$ it results $f_{0}^{-}(0,\epsilon)\equiv f_{0}^{-}$. The
property (4.8) holds.

We now provide further information about the expansion of the basis at
$\epsilon=0$. The following lemma follows as Lemma A.5 in [6]. The key
observation is that the operator
${\left.\kern-1.2pt\mathscr{L}_{\mu,0}\vphantom{\big{|}}\right|_{\mathcal{Z}}}$,
where $\mathcal{Z}$ is the invariant subspace
$\mathcal{Z}:=\text{span}\\{f_{0}^{+},\,f_{0}^{-}\\}$, has the two eigenvalues
$\pm\mathrm{i}\,\sqrt{\mu\tanh(\mathtt{h}\mu)}$, which, for small $\mu$, lie
inside the loop $\Gamma$ around $0$ in (3.5).

######  Lemma A.5.

For any small $\mu$, we have $f_{0}^{+}(\mu,0)\equiv f_{0}^{+}$ and
$f_{0}^{-}(\mu,0)\equiv f_{0}^{-}$. Moreover the vectors $f_{1}^{+}(\mu,0)$
and $f_{1}^{-}(\mu,0)$ have both components with zero space average.

We finally consider the $\mu\epsilon$ term in the expansion (A.9).

######  Lemma A.6.

The derivatives
$(\partial_{\mu}\partial_{\epsilon}f_{k}^{\sigma})(0,0)=\left(\dot{P}_{0,0}^{\prime}-\frac{1}{2}P_{0,0}\dot{P}_{0,0}^{\prime}\right)f_{k}^{\sigma}$
satisfy

$\displaystyle(\partial_{\mu}\partial_{\epsilon}f_{1}^{+})(0,0)=\mathrm{i}\,\begin{bmatrix}odd(x)\\\
even(x)\end{bmatrix},\qquad(\partial_{\mu}\partial_{\epsilon}f_{1}^{-})(0,0)-=\mathrm{i}\,\begin{bmatrix}even(x)\\\
odd(x)\end{bmatrix},$ (A.17)
$\displaystyle(\partial_{\mu}\partial_{\epsilon}f_{0}^{+})(0,0)=\mathrm{i}\,\begin{bmatrix}odd(x)\\\
even_{0}(x)\end{bmatrix},\qquad(\partial_{\mu}\partial_{\epsilon}f_{0}^{-})(0,0)=\mathrm{i}\,\begin{bmatrix}even_{0}(x)\\\
odd(x)\end{bmatrix}\,.$

###### Proof.

We prove that
$\dot{P}^{\prime}_{0,0}=\eqref{Pmisto1}+\eqref{Pmisto2}+\eqref{Pmisto3}$ is
purely imaginary, see footnote 4. This follows since the operators in
$\eqref{Pmisto1}$, $\eqref{Pmisto2}$ and $\eqref{Pmisto3}$ are purely
imaginary because $\dot{\mathscr{L}}_{0,0}$ is purely imaginary,
$\mathscr{L}_{0,0}^{\prime}$ in (A.6) is real and
$\dot{\mathscr{L}}_{0,0}^{\prime}$ in (A.8) is purely imaginary (argue as in
Lemma 3.2-$(iii)$ of [6]). Then, applied to the real vectors $f^{\sigma}_{k}$,
$k=0,1$, $\sigma=\pm$, give purely imaginary vectors.

The property (3.10) implies that
$(\partial_{\mu}\partial_{\epsilon}f_{k}^{\sigma})(0,0)$ have the claimed
parity structure in (A.17). We shall now prove that
$(\partial_{\mu}\partial_{\epsilon}f_{0}^{\pm})(0,0)$ have zero average. We
have, by (A.12) and (A.15)

$\displaystyle\eqref{Pmisto1}f_{0}^{+}:=\frac{1}{2\pi\mathrm{i}\,}\oint_{\Gamma}(\mathscr{L}_{0,0}-\lambda)^{-1}\dot{\mathscr{L}}_{0,0}(\mathscr{L}_{0,0}-\lambda)^{-1}\frac{1}{\lambda}\begin{bmatrix}2{\mathtt{c}}_{\mathtt{h}}^{-1}\sin(x)\\\
\left({\mathtt{c}}_{\mathtt{h}}^{2}+{\mathtt{c}}_{\mathtt{h}}^{-2}\right)\cos(x)\end{bmatrix}\,\mathrm{d}\lambda$

and since the operators $(\mathscr{L}_{0,0}-\lambda)^{-1}$ and
$\dot{\mathscr{L}}_{0,0}$ are both Fourier multipliers, hence they preserve
the absence of average of the vectors, then $\eqref{Pmisto1}f_{0}^{+}$ has
zero average. Next $\eqref{Pmisto2}f_{0}^{+}=0$ since
$\dot{\mathscr{L}}_{0,0}f_{0}^{\pm}=0$, cfr. (2.31). Finally, by (A.12) and
(A.8) where $p_{1}(x)=p_{1}^{[1]}\cos(x)$,

$\eqref{Pmisto3}f_{0}^{+}=\frac{\mathrm{i}\,p_{1}^{[1]}}{2\pi\mathrm{i}\,}\oint_{\Gamma}(\mathscr{L}_{0,0}-\lambda)^{-1}\Big{(}-\frac{1}{\lambda}\begin{bmatrix}\cos(x)\\\
0\end{bmatrix}+\frac{1}{\lambda^{2}}\begin{bmatrix}0\\\
\cos(x)\end{bmatrix}\Big{)}\,\mathrm{d}\lambda$

is a vector with zero average. We conclude that
$\dot{P}_{0,0}^{\prime}f_{0}^{+}$ is an imaginary vector with zero average, as
well as $(\partial_{\mu}\partial_{\epsilon}f_{0}^{+})(0,0)$ since $P_{0,0}$
sends zero average functions in zero average functions. Finally, by (3.10),
$(\partial_{\mu}\partial_{\epsilon}f_{0}^{+})(0,0)$ has the claimed structure
in (A.17).

We finally consider $(\partial_{\mu}\partial_{\epsilon}f_{0}^{-})(0,0)$. By
(A.11) and $\mathscr{L}_{0,0}^{\prime}f_{0}^{-}=0$ (cfr. (A.15)), it results

$\eqref{Pmisto1}f_{0}^{-}=-\frac{1}{2\pi\mathrm{i}\,}\oint_{\Gamma}\frac{(\mathscr{L}_{0,0}-\lambda)^{-1}}{\lambda}\dot{\mathscr{L}}_{0,0}(\mathscr{L}_{0,0}-\lambda)^{-1}\mathscr{L}_{0,0}^{\prime}f_{0}^{-}\mathrm{d}\lambda=0\,.$

Next by (A.11) and $\dot{\mathscr{L}}_{0,0}f_{0}^{-}=0$ we get
$\eqref{Pmisto2}f_{0}^{-}=0$. Finally by (A.11) and (A.8)

$\displaystyle\eqref{Pmisto3}f_{0}^{-}=-\frac{1}{2\pi\mathrm{i}\,}\oint_{\Gamma}(\mathscr{L}_{0,0}-\lambda)^{-1}\frac{1}{\lambda}\begin{bmatrix}0\\\
\mathrm{i}\,p_{1}^{[1]}\cos(x)\end{bmatrix}\mathrm{d}\lambda$

has zero average since $(\mathscr{L}_{0,0}-\lambda)^{-1}$ is a Fourier
multiplier (and thus preserves average absence). ∎

This completes the proof of Lemma 4.2.

##  Appendix B Expansion of the Stokes waves in finite depth

In this Appendix we provide the expansions (2.6)-(2.7), (2.15), (2.20)-(2.23).
Proof of (2.6)-(2.7). Writing

$\begin{aligned}
&\eta_{\epsilon}(x)=\epsilon\eta_{1}(x)+\epsilon^{2}\eta_{2}(x)+\mathcal{O}(\epsilon^{3})\,,\\\
&\psi_{\epsilon}(x)=\epsilon\psi_{1}(x)+\epsilon^{2}\psi_{2}(x)+\mathcal{O}(\epsilon^{3})\,,\end{aligned}\qquad\quad
c_{\epsilon}={\mathtt{c}}_{\mathtt{h}}+\epsilon
c_{1}+\epsilon^{2}c_{2}+\mathcal{O}(\epsilon^{3})\,,$ (B.1)

where $\eta_{i}$ is $even(x)$ and $\psi_{i}$ is $odd(x)$ for $i=1,2$, we solve
order by order in $\epsilon$ the equations (2.5), that we rewrite as

$\begin{cases}-c\,\psi_{x}+\eta+\dfrac{\psi_{x}^{2}}{2}-\dfrac{\eta_{x}^{2}}{2(1+\eta_{x}^{2})}(c-\psi_{x})^{2}=0\\\
c\,\eta_{x}+G(\eta)\psi=0\,,\end{cases}$ (B.2)

having substituted $G(\eta)\psi$ with $-c\,\eta_{x}$ in the first equation. We
expand the Dirichlet-Neumann operator
$G(\eta)=G_{0}+G_{1}(\eta)+G_{2}(\eta)+\mathcal{O}(\eta^{3})$ where, according
to [13][formula (2.14)],

$\displaystyle G_{0}$
$\displaystyle:=D\tanh(\mathtt{h}D)=|D|\tanh(\mathtt{h}|D|)\,,$ (B.3)
$\displaystyle G_{1}(\eta)$
$\displaystyle:=D\big{(}\eta-\tanh(\mathtt{h}D)\eta\tanh(\mathtt{h}D)\big{)}D=-\partial_{x}\eta\partial_{x}-|D|\tanh(\mathtt{h}|D|)\eta|D|\tanh(\mathtt{h}|D|),$
$\displaystyle G_{2}(\eta)$
$\displaystyle:=-\frac{1}{2}D\Big{(}D{\eta}^{2}\tanh(\mathtt{h}D)+\tanh(\mathtt{h}D){\eta}^{2}D-2\tanh(\mathtt{h}D)\eta
D\tanh(\mathtt{h}D)\eta\tanh(\mathtt{h}D)\Big{)}D\,.$

First order in $\epsilon$. Substituting in (B.2) the expansions in (B.1), we
get the linear system

$\left\\{\begin{matrix}-{\mathtt{c}}_{\mathtt{h}}(\psi_{1})_{x}+\eta_{1}=0\\\
{\mathtt{c}}_{\mathtt{h}}(\eta_{1})_{x}+G_{0}\psi_{1}=0\,,\end{matrix}\right.\quad\text{i.e.}\,\begin{bmatrix}\eta_{1}\\\
\psi_{1}\end{bmatrix}\in\text{Ker }\mathcal{B}_{0}\text{ with
}\mathcal{B}_{0}:=\begin{bmatrix}1&-{\mathtt{c}}_{\mathtt{h}}\partial_{x}\\\
{\mathtt{c}}_{\mathtt{h}}\partial_{x}&G_{0}\end{bmatrix},$ (B.4)

where $\eta_{1}$ is $even(x)$ and $\psi_{1}$ is $odd(x)$.

######  Lemma B.1.

The kernel of the linear operator $\mathcal{B}_{0}$ in (B.4) is

$\text{Ker }\mathcal{B}_{0}=\text{span}\,\Big{\\{}\begin{bmatrix}\cos(x)\\\
{\mathtt{c}}_{\mathtt{h}}^{-1}\sin(x)\end{bmatrix}\Big{\\}}.$ (B.5)

###### Proof.

The action of $\mathcal{B}_{0}$ on each subspace
span$\footnotesize{\,\Big{\\{}\begin{bmatrix}\cos(kx)\\\
0\end{bmatrix},\begin{bmatrix}0\\\ \sin(kx)\end{bmatrix}\Big{\\}}}$,
$k\in\mathbb{N}$, is represented by the $2\times 2$ matrix
$\footnotesize{\begin{bmatrix}1&-{\mathtt{c}}_{\mathtt{h}}k\\\
-{\mathtt{c}}_{\mathtt{h}}k&k\tanh(\mathtt{h}k)\end{bmatrix}}$. Its
determinant
$k\tanh(\mathtt{h}k)-{\mathtt{c}}_{\mathtt{h}}^{2}k^{2}=k^{2}\Big{(}\frac{\tanh(\mathtt{h}k)}{k}-\tanh(\mathtt{h})\Big{)}$
vanishes if and only if $k=1$. Indeed the function
$x\mapsto\frac{\tanh(\mathtt{h}x)}{x}$ is monotonically decreasing for $x>0$,
since its derivative
$\frac{2x\mathtt{h}-\sinh(2\mathtt{h}x)}{2\cosh^{2}(\mathtt{h}x)x^{2}}$ is
negative for $x>0$. For $k=1$ we obtain the kernel of $\mathcal{B}_{0}$ given
in (B.5). For $k=0$ it has no kernel since $\psi_{1}(x)$ is odd. ∎

We set $\eta_{1}(x):=\cos(x)$,
$\psi_{1}(x):={\mathtt{c}}_{\mathtt{h}}^{-1}\sin(x)$ in agreement with (2.6).
Second order in $\epsilon$. By (B.2), and since
${\mathtt{c}}_{\mathtt{h}}^{2}(\eta_{1})_{x}^{2}=(G_{0}\psi_{1})^{2}$, we get
the linear system

$\mathcal{B}_{0}\begin{bmatrix}\eta_{2}\\\
\psi_{2}\end{bmatrix}=\begin{bmatrix}c_{1}(\psi_{1})_{x}-\frac{1}{2}(\psi_{1})_{x}^{2}+\frac{1}{2}(G_{0}\psi_{1})^{2}\\\
-c_{1}(\eta_{1})_{x}-G_{1}(\eta_{1})\psi_{1}\end{bmatrix}\,,$ (B.6)

where $\mathcal{B}_{0}$ is the self-adjoint operator in (B.4). System (B.6)
admits a solution if and only if its right-hand term is orthogonal to the
Kernel of $\mathcal{B}_{0}$ in (B.5), namely

$\Big{(}\begin{bmatrix}c_{1}(\psi_{1})_{x}-\frac{1}{2}(\psi_{1})_{x}^{2}+\frac{1}{2}(G_{0}\psi_{1})^{2}\\\
-c_{1}(\eta_{1})_{x}-G_{1}(\eta_{1})\psi_{1}\end{bmatrix}\;,\;\begin{bmatrix}\cos(x)\\\
{\mathtt{c}}_{\mathtt{h}}^{-1}\sin(x)\end{bmatrix}\Big{)}=0\,.$ (B.7)

In view of the first order expansion (2.6), (B.3) and the identity
$\tanh(2\mathtt{h})=\displaystyle{\frac{2{\mathtt{c}}_{\mathtt{h}}^{2}}{1+{\mathtt{c}}_{\mathtt{h}}^{4}}}$,
it results $[G_{0}\psi_{1}](x)={\mathtt{c}}_{\mathtt{h}}\sin(x)$,
$\big{[}G_{1}(\eta_{1})\psi_{1}\big{]}(x)=\frac{1-{\mathtt{c}}_{\mathtt{h}}^{4}}{{\mathtt{c}}_{\mathtt{h}}(1+{\mathtt{c}}_{\mathtt{h}}^{4})}\sin(2x)$
so that (B.7) implies $c_{1}=0$, in agrement with (2.6). Equation (B.6)
reduces to

$\displaystyle\begin{bmatrix}1&-{\mathtt{c}}_{\mathtt{h}}\partial_{x}\\\
{\mathtt{c}}_{\mathtt{h}}\partial_{x}&G_{0}\end{bmatrix}\begin{bmatrix}\eta_{2}\\\
\psi_{2}\end{bmatrix}=\begin{bmatrix}-\frac{1}{4}({\mathtt{c}}_{\mathtt{h}}^{-2}-{\mathtt{c}}_{\mathtt{h}}^{2})-\frac{1}{4}({\mathtt{c}}_{\mathtt{h}}^{-2}+{\mathtt{c}}_{\mathtt{h}}^{2})\cos(2x)\\\
-\frac{1-{\mathtt{c}}_{\mathtt{h}}^{4}}{{\mathtt{c}}_{\mathtt{h}}(1+{\mathtt{c}}_{\mathtt{h}}^{4})}\sin(2x)\end{bmatrix}.$
(B.8)

Setting $\eta_{2}=\eta_{2}^{[0]}+\eta_{2}^{[2]}\cos(2x)$ and
$\psi_{2}=\psi_{2}^{[2]}\sin(2x)$, system (B.8) amounts to

$\displaystyle\left\\{\begin{matrix}\eta_{2}^{[0]}+\big{(}\eta_{2}^{[2]}-2{\mathtt{c}}_{\mathtt{h}}\psi_{2}^{[2]}\big{)}\cos(2x)=-\frac{1}{4}\left({\mathtt{c}}_{\mathtt{h}}^{-2}-{\mathtt{c}}_{\mathtt{h}}^{2}\right)-\frac{1}{4}\left({\mathtt{c}}_{\mathtt{h}}^{-2}+{\mathtt{c}}_{\mathtt{h}}^{2}\right)\cos(2x)\\\
(-2{\mathtt{c}}_{\mathtt{h}}\eta_{2}^{[2]}+2\psi_{2}^{[2]}\tanh(2\mathtt{h}))\sin(2x)=-\frac{1-{\mathtt{c}}_{\mathtt{h}}^{4}}{{\mathtt{c}}_{\mathtt{h}}(1+{\mathtt{c}}_{\mathtt{h}}^{4})}\sin(2x)\,,\end{matrix}\right.$

which leads to the expansions of $\eta_{2}^{[0]}$, $\eta_{2}^{[2]}$,
$\psi_{2}^{[2]}$ given in (2.6)-(2.7).

Third order in $\epsilon$. It remains to determine $c_{2}$ in (2.8). We get
the linear system

$\mathcal{B}_{0}\begin{bmatrix}\eta_{3}\\\
\psi_{3}\end{bmatrix}=\begin{bmatrix}c_{2}(\psi_{1})_{x}-(\psi_{1})_{x}(\psi_{2})_{x}-(\eta_{1})_{x}^{2}(\psi_{1})_{x}{\mathtt{c}}_{\mathtt{h}}+(\eta_{1})_{x}(\eta_{2})_{x}{\mathtt{c}}_{\mathtt{h}}^{2}\\\
-c_{2}(\eta_{1})_{x}-G_{1}(\eta_{1})\psi_{2}-G_{1}(\eta_{2})\psi_{1}-G_{2}(\eta_{1})\psi_{1}\end{bmatrix}\,.$
(B.9)

System (B.9) has a solution if and only if the right hand side is orthogonal
to the Kernel of $\mathcal{B}_{0}$ given in (B.5). This condition determines
uniquely $c_{2}$. Denoting $\Pi_{1}$ the $L^{2}$-orthogonal projector on
span$\,\\{\cos(x),\sin(x)\\}$, it results

$\displaystyle
c_{2}(\psi_{1})_{x}=c_{2}{\mathtt{c}}_{\mathtt{h}}^{-1}\cos(x)\,,\quad
c_{2}(\eta_{1})_{x}=-c_{2}\sin(x)\,,\quad\Pi_{1}[(\psi_{1})_{x}(\psi_{2})_{x}]=\psi_{2}^{[2]}{\mathtt{c}}_{\mathtt{h}}^{-1}\cos(x)\,,$
$\displaystyle\Pi_{1}[{\mathtt{c}}_{\mathtt{h}}(\eta_{1})_{x}^{2}(\psi_{1})_{x}]=\tfrac{1}{4}\cos(x)\,,\quad\Pi_{1}[{\mathtt{c}}_{\mathtt{h}}^{2}(\eta_{1})_{x}(\eta_{2})_{x}]=\eta_{2}^{[2]}{\mathtt{c}}_{\mathtt{h}}^{2}\cos(x)\,,$

and, in view of (B.3), and (2.6), (2.7),

$\displaystyle\Pi_{1}[G_{1}(\eta_{1})\psi_{2}]$
$\displaystyle=\psi_{2}^{[2]}\frac{1-{\mathtt{c}}_{\mathtt{h}}^{4}}{1+{\mathtt{c}}_{\mathtt{h}}^{4}}\sin(x)\,,\quad\Pi_{1}[G_{2}(\eta_{1})\psi_{1}]={\mathtt{c}}_{\mathtt{h}}\frac{3{\mathtt{c}}_{\mathtt{h}}^{4}-1}{4(1+{\mathtt{c}}_{\mathtt{h}}^{4})}\sin(x)\,,$
$\displaystyle\Pi_{1}[G_{1}(\eta_{2})\psi_{1}]$
$\displaystyle={\mathtt{c}}_{\mathtt{h}}^{-1}\Big{(}\eta_{2}^{[0]}(1-{\mathtt{c}}_{\mathtt{h}}^{4})+\tfrac{1}{2}\eta_{2}^{[2]}(1+{\mathtt{c}}_{\mathtt{h}}^{4})\Big{)}\sin(x)\,.$

Therefore the orthogonality condition proves (2.8).

Proof of (2.15). We expand the function
$\mathfrak{p}(x)=\epsilon\mathfrak{p}_{1}(x)+\epsilon^{2}\mathfrak{p}_{2}(x)+\mathcal{O}(\epsilon^{3})$
defined by the fixed point equation (2.14). We first note that the constant
$\mathtt{f}_{\epsilon}=\mathcal{O}(\epsilon^{2})$ because
$\eta_{1}(x)=\cos(x)$ has zero average. Then
$\mathfrak{p}(x)=\frac{\mathcal{H}}{\tanh(\mathtt{h}|D|)}\big{[}\epsilon\eta_{1}+\epsilon^{2}\big{(}\eta_{2}+(\eta_{1})_{x}\mathfrak{p}_{1}\big{)}+\mathcal{O}(\epsilon^{3})\big{]}$,
and, using that $\mathcal{H}\cos(kx)=\sin(kx)$, for any $k\in\mathbb{N}$, we
get

$\displaystyle\mathfrak{p}_{1}(x)$
$\displaystyle=\frac{\mathcal{H}}{\tanh(\mathtt{h}|D|)}\cos(x)={\mathtt{c}}_{\mathtt{h}}^{-2}\sin(x)\,,$
(B.10) $\displaystyle\mathfrak{p}_{2}(x)$
$\displaystyle=\frac{\mathcal{H}}{\tanh(\mathtt{h}|D|)}((\eta_{1})_{x}\mathfrak{p}_{1}+\eta_{2})=\frac{(1+{\mathtt{c}}_{\mathtt{h}}^{4})({\mathtt{c}}_{\mathtt{h}}^{4}+3)}{8{\mathtt{c}}_{\mathtt{h}}^{8}}\sin(2x)\,.$
(B.11)

Finally

$\displaystyle\mathtt{f}_{\epsilon}$
$\displaystyle=\frac{\epsilon^{2}}{2\pi}\int_{\mathbb{T}}\big{(}\eta_{2}+(\eta_{1})_{x}\mathfrak{p}_{1}\big{)}\mathrm{d}x+\mathcal{O}(\epsilon^{3})=\epsilon^{2}\big{(}\eta_{2}^{[0]}-\tfrac{1}{2}{\mathtt{c}}_{\mathtt{h}}^{-2}\big{)}+\mathcal{O}(\epsilon^{3})\stackrel{{\scriptstyle\eqref{expcoef}}}{{=}}\epsilon^{2}\frac{{\mathtt{c}}_{\mathtt{h}}^{4}-3}{4{\mathtt{c}}_{\mathtt{h}}^{2}}+\mathcal{O}(\epsilon^{3})\,.$

The expansion (2.15) is proved.
Proof of Lemma 2.2. In view of (2.6)-(2.7), the expansions of the functions
$B$, $V$ in (2.10) are

$\displaystyle B=:\epsilon
B_{1}(x)+\epsilon^{2}B_{2}(x)+\mathcal{O}(\epsilon^{3})=\epsilon{\mathtt{c}}_{\mathtt{h}}\sin(x)+\epsilon^{2}\frac{3-2{\mathtt{c}}_{\mathtt{h}}^{4}}{2{\mathtt{c}}_{\mathtt{h}}^{5}}\sin(2x)+\mathcal{O}(\epsilon^{3})$
(B.12)

and

$\displaystyle V=:\epsilon
V_{1}(x)+\epsilon^{2}V_{2}(x)+\mathcal{O}(\epsilon^{3})=\epsilon{\mathtt{c}}_{\mathtt{h}}^{-1}\cos(x)+\epsilon^{2}\Big{[}\frac{{\mathtt{c}}_{\mathtt{h}}}{2}+\frac{3-{\mathtt{c}}_{\mathtt{h}}^{8}}{4{\mathtt{c}}_{\mathtt{h}}^{7}}\cos(2x)\Big{]}+\mathcal{O}(\epsilon^{3})\,.$
(B.13)

In view of (2.18), denoting derivatives w.r.t $x$ with an apex and suppressing
dependence on $x$ when trivial, we have

$\displaystyle{\mathtt{c}}_{\mathtt{h}}+p_{\epsilon}(x)$
$\displaystyle=({\mathtt{c}}_{\mathtt{h}}+\epsilon^{2}c_{2}-V(x)-V^{\prime}(x)\mathfrak{p}(x)+\mathcal{O}(\epsilon^{3}))(1-\mathfrak{p}^{\prime}(x)+(\mathfrak{p}^{\prime}(x))^{2}+\mathcal{O}(\epsilon^{3}))$
$\displaystyle={\mathtt{c}}_{\mathtt{h}}+\epsilon\underbrace{(-V_{1}-{\mathtt{c}}_{\mathtt{h}}\mathfrak{p}_{1}^{\prime})}_{=:p_{1}}+\epsilon^{2}\underbrace{\big{(}c_{2}+V_{1}\mathfrak{p}_{1}^{\prime}-V_{2}-V_{1}^{\prime}\mathfrak{p}_{1}-{\mathtt{c}}_{\mathtt{h}}\mathfrak{p}_{2}^{\prime}+{\mathtt{c}}_{\mathtt{h}}(\mathfrak{p}_{1}^{\prime})^{2}\big{)}}_{=:p_{2}}+\,\mathcal{O}(\epsilon^{3})\,.$
(B.14)

Similarly by (2.18)

$\displaystyle 1+a_{\epsilon}($ $\displaystyle
x):=\frac{1}{1+\mathfrak{p}_{x}(x)}-({\mathtt{c}}_{\mathtt{h}}+p_{\epsilon}(x))B_{x}(x+\mathfrak{p}(x))$
$\displaystyle=$ $\displaystyle
1+\epsilon\underbrace{\big{(}-\mathfrak{p}_{1}^{\prime}-{\mathtt{c}}_{\mathtt{h}}B_{1}^{\prime}\big{)}}_{=:a_{1}}+\epsilon^{2}\underbrace{\big{(}(\mathfrak{p}_{1}^{\prime})^{2}-\mathfrak{p}_{2}^{\prime}-{\mathtt{c}}_{\mathtt{h}}B_{2}^{\prime}-{\mathtt{c}}_{\mathtt{h}}B_{1}^{\prime\prime}\mathfrak{p}_{1}(x)+B_{1}^{\prime}V_{1}+{\mathtt{c}}_{\mathtt{h}}B_{1}^{\prime}\mathfrak{p}_{1}^{\prime}\big{)}}_{=:a_{2}}+\mathcal{O}(\epsilon^{3})\,.$
(B.15)

By (B.13), (B.10), (2.6), (B.11), (B.12) we deduce that the functions $p_{1}$,
$p_{2}$, $a_{1}$, $a_{2}$ in (B.14) and (B.15) have an expansion as in
(2.20)-(2.23).∎

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