\left\| \int R_1 R_2  - \int \left( \Rtone - \mu_1 \Lambda \Rtone - \omega_1 \partial_y \Rtone \right) \left( \Rttwo- \mu_2 \Lambda \Rttwo -\omega_2 \partial_y \Rttwo \right) \right\|_{L^p} \lesssim e^{-\frac{15}{16} z}\sum_{i=1}^2\left(\mu_i^2 + \omega_i^2\right).
\end{align*}

Therefore, we conclude the proof of <ref> by using (<ref>).

Now we prove (<ref>). By the triangle inequality,

\begin{align*}
&\left\|R_1R_2-\Rtone\Rttwo- \left(\mu_1\Lambda \Rtone+\omega_1 \partial_y \Rtone\right) \Rttwo-\Rtone \left(\mu_2\Lambda \Rttwo+\omega_2 \partial_y \Rttwo\right)\right\|_{H^1} \\
& \lesssim \left\|\left(R_1-\Rtone-\mu_1\Lambda \Rtone-\omega_1\partial_y \Rtone\right) \Rttwo\right\|_{H^1} +\left\|\Rtone \left(R_2-\Rttwo-\mu_2\Lambda \Rttwo-\omega_2\partial_y \Rttwo\right)\right\|_{H^1} \\
& \qquad +\left\|\left(R_1-\Rtone\right)\left(R_2-\Rttwo\right) \right\|_{H^1} .
\end{align*}

Then, we conclude the proof of (<ref>) by using (<ref>) together with Remark <ref>.

Let $Z^{*} \ge 1 $ be chosen large enough. Assume that  $z=z_1-z_2>Z^{*}$, $-z \le z_2 \le -\frac14 z$, $\frac14z \le z_1 \le z$, and recall the definition of $X_i$, $W_i$ and $l$ in (<ref>) and (<ref>). Then, the following estimates hold.

\begin{align}
&\left\|\Rtone X_2\right\|_{H^1}+ \left\|\Rttwo X_1\right\|_{H^1} \lesssim e^{-\frac{15}{16} z} ; \label{est:tildeR_X}\\
& \left\|\Rtone W_2\right\|_{H^1}+ \left\|\Rtone Y_2\right\|_{H^1}\lesssim e^{-\frac{15}{16} z} ;\label{est:tildeR1_W2} \\
& \left\|\Rttwo\left(W_1-l\right)
\right\|_{H^1}+\left\|\Rttwo\left(Y_1-h\right)
\right\|_{H^1} \lesssim e^{-\frac{15}{16} z} . \label{est:tildeR2_W1-l}
\end{align}

Observe that $X_i(x,y)=(-\Delta+1)^{-1}Q(x-z_i,y)$ with $(-\Delta+1)^{-1}Q \in \mathcal{Y}$ (see Lemma <ref>). Thus, estimate (<ref>) follows combining (<ref>) with a change of variable.

Next, we prove estimate (<ref>). We have from the definition of $W_2$ and $Y_2$ in (<ref>), the one of $\Lambda Q$ in (<ref>) and lemma <ref> that

\begin{equation*}
\left|\Rtone W_2 (x,y) \right|+\left|\Rtone Y_2 (x,y) \right| \lesssim e^{-|(x-z_1,y)|} \int_{x-z_2}^{+\infty}e^{-\frac{15}{16}|\tilde{x}|}d\tilde{x}  .
\end{equation*}

In the case where $x>z_1$, we have $|x-z_2|=x-z_2 >z$, so that

\begin{equation*}
\left|\Rtone W_2 (x,y) \right|+\left|\Rtone Y_2 (x,y) \right| \lesssim  e^{-\frac{15}{16}z}e^{-|(x-z_1,y)|}  .
\end{equation*}

In the case where $z_2<x<z_1$, we use that $|(x-z_1,y)| \ge \frac{31}{32}(z_1-x)+\frac1{32}|y|$. Hence,

\begin{equation*}
\left|\Rtone W_2 (x,y) \right| +\left|\Rtone Y_2 (x,y) \right|\lesssim  e^{-\frac{31}{32}z}e^{-\frac1{32}|y|}{\bf 1}_{z_2 <x<z_1}  .
\end{equation*}

In the case where $x<z_2$, we have $|x-z_1|=z_1-x>z$, so that

\begin{equation*}
\left|\Rtone W_2 (x,y) \right|+\left|\Rtone Y_2 (x,y) \right| \lesssim  e^{-\frac{15}{16}z}e^{-\frac1{16}|(x-z_1,y)|} .
\end{equation*}

We deduce combining these estimates and integrating over $\mathbb R^2$ that

\begin{equation*}
\left\|\Rtone W_2\right\|_{L^2}+\left\|\Rtone Y_2\right\|_{L^2} \lesssim e^{-\frac{15}{16} z} .
\end{equation*}

The  estimate

\begin{equation*}
\left\|\nabla \left(\Rtone W_2\right)\right\|_{L^2}+\left\|\nabla \left(\Rtone Y_2\right)\right\|_{L^2} \lesssim e^{-\frac{15}{16} z}
\end{equation*}

is proved in a similar way. This concludes the proof of (<ref>).

Finally, we observe from the definitions of $W_1$, $Y_1$  in (<ref>) and  $l$ and $h$ (<ref>) that

\begin{align*}
&\left|\left(\Rttwo\left(W_1-l\right) \right)(x,y)\right|=\left|Q(x-z_2,y)\int_{-\infty}^{x-z_1}\left(-\Delta+1\right)^{-1}\Lambda Q(\tilde{x},y) d\tilde{x} \right| \\
&\left|\left(\Rttwo\left(Y_1-h\right) \right)(x,y)\right|=\left|Q(x-z_2,y)\int_{-\infty}^{x-z_1}\left(-\Delta+1\right)^{-1}\partial_yQ(\tilde{x},y) d\tilde{x} \right|
\end{align*}

and thus

\begin{align*}
\left|\left(\Rttwo\left(W_1-l\right) \right)(x,y)\right| + \left|\left(\Rttwo\left(Y_1-h\right) \right)(x,y)\right| \lesssim e^{-|(x-z_2,y)|} \int_{-\infty}^{x-z_1}e^{-\frac{15}{16}|\tilde{x}|} d\tilde{x}.
\end{align*}

Hence, the proof of (<ref>) follows arguing as in the proof of (<ref>).

Under the assumptions of Lemmata <ref> and <ref>, we have

\begin{align*}
\MoveEqLeft
\left\|(R_1-\Rtone) X_2\right\|_{H^1}+ \left\|(R_2-\Rttwo) X_1\right\|_{H^1} + \left\|(R_1-\Rtone) W_2\right\|_{H^1}+  \left\|(R_2-\Rttwo)\left(W_1-l\right) \right\|_{H^1} \\
& \qquad + \left\|(R_1-\Rtone) Y_2\right\|_{H^1} + \left\|(R_2-\Rttwo)\left(Y_1-h\right) \right\|_{H^1} \lesssim e^{-\frac{15}{16} z}\sum_{i=1}^2 \left(|\mu_i|+|\omega_i| \right) .
\end{align*}

These estimates are proved by using (<ref>) and arguing as in the proof of Lemma <ref>.

Next, we give the proof of Claim <ref>, which relies on the following estimates.
Let $f\in C^{\infty}(\mathbb R^2)$ be such that $|f(\bx)| \lesssim e^{-\frac78|\bx|}$, for all $\bx \in \mathbb R^2$, and let $Z^{*} \ge 1 $ be chosen large enough. Assume that  $z_1>Z^{*}$ and $z_2<-Z^{*}$ with for any $i$, $\vert z_i \vert > \frac14 z$ and recall the notation $\tilde{\bz_i}=(z_i,0)$, for $i=1,2$. Then, the following estimates hold.
For all $\bx=(x,y) \in \mathbb R^2$,

\begin{align}
& \left| f\left(\bx-\tilde{\bz}_1\right)(1-\psi)(x)\right| \lesssim e^{- \rho z}e^{-\frac1{8}\vert (x-z_1,y)\vert}  ; \label{est:f(1-psi)}\\
& \left| f\left(\bx-\tilde{\bz}_2\right)\psi(x)\right| \lesssim e^{-\rho z}e^{-\frac1{8}\vert (x-z_2,y)\vert}; \label{est:fpsi} \\
& \left| f\left(\bx-\tilde{\bz}_i\right)\psi'(x)\right|\lesssim e^{-\rho z} e^{-\frac1{8}\vert (x-z_i,y)\vert}, \quad i=1,2 . \label{est:fi_psi'}
\end{align}

We only prove (<ref>), since the proofs of (<ref>) and (<ref>) are similar.

In the region $x>z_1/2$, observe from the definition of $\psi$ and the second identity in (<ref>), that $1-\psi(x)=\psi(-x) \lesssim e^{-8\rho x} \lesssim e^{-4 \rho z_1}$. Since $f \in \mathcal{Y}$, it follows that

\begin{equation*}
\left| f\left(\bx-\tilde{\bz}_1\right)(1-\psi)(x)\right| \lesssim e^{-\rho z}e^{-\frac12\vert (x-z_1,y)\vert} .
\end{equation*}

In the region $x<z_1/2$, we have that $|(x-z_1,y)| \ge |x-z_1| \ge \frac{|z_1|}2$. Thus,

\begin{equation*}
\left| f\left(\bx-\tilde{\bz}_1\right)(1-\psi)(x)\right| \lesssim e^{-\frac3{8} z_1}e^{-\frac18\vert (x-z_1,y)\vert} .
\end{equation*}

Combining these two estimates yields (<ref>).

The proofs of estimates (<ref>) and (<ref>) are consequences of Lemma <ref> and Lemma <ref>.

By using the definition of $V$ in (<ref>), we observe that

\begin{equation*}
V \psi'= \sum_{i=1}^2R_i\psi'+ V_A \psi' ,
\end{equation*}

so that the proof of (<ref>) follows from (<ref>), (<ref>), (<ref>) and Lemma <ref>.

By using the definitions of $V$ in (<ref>) and of $\psi_+$ in (<ref>), we decompose $\partial_tV+\partial_xV \psi_+$ as

\begin{align*}
\partial_tV+\partial_xV \psi_+
& = \sum_{i=1}^2 \left( -(\dot{z}_i-\mu_i)\partial_xR_i-\dot{\omega}_i\partial_yR_i+\dot{\mu}_i\Lambda R_i \right) \\
& \qquad + (\mu_1-\mu_2) \left((\psi-1)\partial_xR_1-\psi\partial_xR_2 \right)+\partial_tV_A+\partial_xV_A\psi_+ .
\end{align*}

Thus, we conclude the proof of estimate (<ref>) by gathering (<ref>), (<ref>), (<ref>), (<ref>) with $0<\rho<\frac1{32}$ and Lemma <ref>.

Finally, by using the definitions of $V$ in (<ref>) and of $\psi_{-,e}$ and $\psi_{-,m}$ in (<ref>)-(<ref>), we decompose $\partial_t V \psi_{-,e} + \partial_x V \psi_{-,m}$ as

\begin{align*}
\MoveEqLeft
\partial_t V \psi_{-,e} + \partial_x V \psi_{-,m} \\
&= -(\dot{z}_1-\mu_1)\frac{\psi}{(1+\mu_1)^2}\partial_xR_1-(\dot{z}_2-\mu_2)\frac{1-\psi}{(1+\mu_2)^2}\partial_xR_2 -(\dot{z}_1-\mu_2)\frac{1-\psi}{(1+\mu_2)^2} \partial_xR_1\\ &\quad -(\dot{z}_2-\mu_1)\frac{\psi}{(1+\mu_1)^2} \partial_xR_2 +\sum_{i=1}^2 \left(-\dot{\omega}_i\partial_yR_i+ \dot{\mu}_i\Lambda R_i \right)\psi_{-,e}+\partial_tV_A \psi_{-,e}+\partial_xV_A\psi_{-,m} .
\end{align*}

Hence, we conclude the proof of (<ref>) by using (<ref>), (<ref>), (<ref>), (<ref>) with $0<\rho<\frac1{32}$ and Lemma <ref>, and recalling from
Assumption <ref> that $|\mu_1|+|\mu_2| \le C\nu_0^\star$, for $\nu_0^\star$ small enough.

To be continued

For $z>Z_0$, the following estimate holds for $\bx \in \mathbb{R}^2$

\begin{align*}
& Q(\bx) Q(\bx-(z,0)) \lesssim e^{z_2+x}\mathbf{1}_{x \leq z_2} + \\
& \int_{\bx \in \mathbb{R}^2} Q(\bx) Q(\bx-(z,0)) d\bx \lesssim
\end{align*}

For $f,g\in \Y$, with $\vert f(\bx) \vert + \vert g(\bx) \vert \lesssim \langle \bx \rangle^{n}e^{-\vert \bx \vert}$, we have

\begin{align}\label{eq:product_Y}
& \left\vert f\left(\bx+\left(\frac{z}{2},0\right)\right) g\left(\bx-\left(\frac{z}{2},0\right)\right) \right\vert \lesssim \left( \right)  \mathbf{1}_{2 \vert x \vert >z} + \left( \right) \mathbf{1}_{ z- \vert y \vert \leq 2\vert x\vert \leq z} + \left( \right) \mathbf{1}_{\vert y \vert + 2 \vert x \vert \leq z } \\
& \left\vert \partial_x^{-1} f\left( x+\frac{z}{2},y\right) g\left( x-\frac{z}{2},y\right) \right\vert \lesssim  \left( \langle z\rangle +\left\langle \left( x-\frac{z}{2},y \right) \right\rangle\right)^{2n} e^{-z - \left\vert \left( x-\frac{z}{2},y \right) \right\vert}\mathbf{1}_{x>\frac{z}{2}} +\langle z \rangle^{2n} e^{-z} \mathbf{1}_{-\frac{z}{2} < x <\frac{z}{2}- \frac{3}{8}\vert y \vert} \nonumber \\
& \quad + \left( \langle z \rangle^{2n}+ \langle y \rangle^{2n} \right) e^{-z-\frac{y}{2}} \mathbf{1}_{\max(-\frac{z}{2}, \frac{z}{2}- \frac{3}{8} \vert y \vert )\leq x \leq \frac{z}{2}} + \left( \langle z \rangle + \left\langle \left( x-\frac{z}{2},y \right) \right\rangle \right)^{2n} e^{- \left\vert \left( x-\frac{z}{2},y\right)\right\vert} \mathbf{1}_{x<-\frac{z}{2}} \label{eq:product2}
\end{align}

We can replace the last inequality by the following

\begin{align*}
\forall x\in \left( -\frac{z}{2} , \frac{z}{2}\right), \quad \left\vert \partial_x^{-1} f\left(x+\frac{z}{2},y \right) g\left( x- \frac{z}{2},y\right) \right\vert \lesssim \left( \langle z\rangle^{2n} + \langle y \rangle^{2n} \right) \exp \left( - \frac{3}{4} z - \frac{1}{2}y\right).
\end{align*}

For (<ref>),

We continue with the computations for (<ref>). First, we notice that if $f(\bx) \lesssim \langle \bx \rangle^{n} e^{-\vert \bx \vert}$, then

\begin{align*}
\left\vert \partial_x^{-1} f(\bx)\right\vert \lesssim \min \left(1, \left( \langle x \rangle^{n} +\langle y \rangle^{n} \right) e^{-x} \right)
\end{align*}

The multiplication of $\partial_x^{-1}f\left( x+\frac{z}{2},y\right)$ by $g\left( x-\frac{z}{2},y \right)$ with the help of ref gives (<ref>).

The last proof is dealt using the following inequalities.

\begin{align*}
\left\vert \partial_x^{-1} f(\bx) \right\vert \lesssim \partial_x^{-1} \left(\left( \langle x \rangle^{n} + \langle y \rangle^n \right) e^{-\frac{1}{4}\vert y\vert -\frac{3}{4}\vert x \vert} \right) \lesssim \left( \langle x \rangle^{n} + \langle y \rangle^n \right) e^{-\frac{1}{4}\vert y\vert -\frac{3}{4}\vert x \vert} \quad \text{and} \quad \left\vert g(\bx) \right\vert \lesssim \langle \bx \rangle^{n} e^{-\frac{3}{4}\vert x \vert - \frac{1}{4} \vert y \vert}.
\end{align*}

Next, by using the product of the two functions, for $x \in [-\frac{z}{2},\frac{z}{2}]$,

\begin{align*}
\left\vert \partial_x^{-1} f\left( x+\frac{z}{2},y\right) g \left( x-\frac{z}{2},y \right) \right\vert \lesssim \left( \langle z \rangle^{2n} + \langle y \rangle^{2n}\right) \exp \left( -\frac{1}{2}y- \frac{3}{4}\left( \frac{z}{2}- x \right) - \frac{3}{4} \left( \frac{z}{2}+x \right) \right).
\end{align*}

For $z>\red{Z_0}$ The following estimates hold on $\mathbb{R}$

\begin{align}\label{est:non_lin1}
&\left\vert (-\Delta+1)^{-1} \left( Q(\bx) Q(\bx-(z,0))\right) \right\vert \lesssim \frac{e^{-z}}{\sqrt{z}}, \\
& \left\vert (-\Delta+1)^{-1} \left( Q(\bx) Q_{app}^n (\bx,z)\right) \right\vert \lesssim \frac{e^{-z}}{\sqrt{z}} \sum_{k=0}^n \frac{\langle \bx\rangle^{2k}}{z^k}, \label{est:non_lin2} \\
& \left\vert (-\Delta+1)^{-1} \left( Q(\bx)\left(  Q(\bx-(z,0))-Q_{app}^n (\bx,z)\right) \right) \right\vert \lesssim   \label{est:non_lin3}
\end{align}

For (<ref>),

§ SOLUTIONS TO THE SECOND ORDER DYNAMICAL SYSTEM

The goal of this appendix is to study the solutions $Z$ of the autonomous ODE

\begin{align}\label{defi:eq_Z}
\Ddot{Z}(t)= \frac{2}{\langle \Lambda Q, Q \rangle} \int_{\mathbb{R}^2} Q(x + Z(t),y) \partial_x (Q^2) (x,y) dxdy.
\end{align}

The function $Z$ is an approximation of the distance between the two solitary waves. In particular, at $-\infty$ the two solitary waves are far away one from another and are repulsing at a velocity $-2\mu_0$, and thus, we should look for the solutions of (<ref>) satisfying $\lim_{t\rightarrow -\infty} (Z(t),\dot{Z}(t))=(+\infty,-2\mu_0)$. We also expect the collision to be almost symmetric, in the sense that $Z$ should be even.

In the one dimensional case of the gKdV equation, the authors in [41] use the first order asymptotic of (<ref>), $\ddot{Z}=ce^{-Z}$, which admits explicit solutions.

for some positive constant $c$: this equation can be found by replacing in (<ref>) the term $Q(\cdot+Z,\cdot)$ by its first order asymptotic expansion where $\partial_x(Q^2)$ is located, thus around $0$ (see the asymptotic expansion in dimension $2$ in Proposition (<ref>)). In the $1$-dimensional context, the solution $Z$ is explicit and the behaviour of the asymptotic expansion at different points can be given in term of the solution.

In the $2$-dimensional case, the equation and its asymptotics cannot be solved explicitly. Instead, we describe the set of solutions by using ODE arguments.

§.§ Phase portrait of the solutions

In this subsection, we consider for a positive constant $\kappa$, the ODE

\begin{align}\label{defi:eq_Z_general}
\ddot{Z}(t) = \kappa \int_{\mathbb{R}^2} Q(x+Z(t),y) \partial_x \left(Q^2\right)(x,y)dx dy ,
\end{align}

which corresponds to (<ref>) in the case $\kappa=2\langle \Lambda Q, Q \rangle^{-1}$.
We define the associated Hamiltonian function

\begin{align}\label{defi:H_static}
H (Y_0,Y_1) := \frac{Y_1^2}{2} + \kappa \int_{\mathbb{R}^2} Q(x+Y_0,y) Q^2(x,y) dx dy.
\end{align}

The Hamiltonian $H(Z(t),\dot{Z}(t))$ is preserved along the trajectories of (<ref>).

More precisely, let $I$ be an interval containing $0$ and let $Z:I \to \mathbb R$ be a $C^2$ solution to (<ref>) with initial datum $(Z(0),\dot{Z}(0))=(Y_0,Y_1) \in \mathbb R^2$. Then, $H(Z(t),\dot{Z}(t))=H(Y_0,Y_1)$, for all $t \in I$.

We observe that for any solution $Z$ of (<ref>), $\frac{d}{dt}H(Z(t),\dot{Z}(t))=0$. Hence, we conclude the proof by integrating between $0$ and $t$.

Let $(Y_0,Y_1)\in \mathbb{R}^2$. Then, there exists a unique solution $Z \in C^2(\mathbb R : \mathbb R)$ of (<ref>) satisfying $\left(Z(0),\dot{Z}(0)\right)=(Y_0,Y_1)$.
Furthermore, the phase portrait of (<ref>) is given by Figure <ref>.

[->] (-7,0) – (7,0);
[->] (0,-2.75) – (0,2.75);
(7,0) node[right] $Z$;
(0,2.75) node[right] $\dot{Z}$;

[red] (0,0) node $\bullet$;

[blue,->,>=latex,domain=0.3:1,samples=] plot [variable=] (*,rad(atan())*1.1);
[blue,domain=1:2.6,samples=] plot [variable=] (*,rad(atan())*1.1);

[blue,domain=0.3:1,samples=] plot [variable=] (*,-rad(atan())*1.1);
[blue,->,>=latex,domain=2.6:1,samples=] plot [variable=] (*,-rad(atan())*1.1);

[blue,domain=0.3:1,samples=] plot [variable=] (-*,rad(atan())*1.1);
[blue,->,>=latex,domain=2.6:1,samples=] plot [variable=] (-*,rad(atan())*1.1);

[blue,->,>=latex,domain=0.3:1,samples=] plot [variable=] (-*,-rad(atan())*1.1);
[blue,domain=1:2.6,samples=] plot [variable=] (-*,-rad(atan())*1.1);

[->,>=latex,domain=-2.4:1,samples=] plot [variable=] (1+*,rad(atan()));
[domain=1:2.4,samples=] plot [variable=] (1+*,rad(atan()));

[->,>=latex,domain=-2.2:1,samples=] plot [variable=] (2+*,rad(atan())/2);
[domain=1:2.2,samples=] plot [variable=] (2+*,rad(atan())/2);

[->,>=latex,domain=-1.7:1,samples=] plot [variable=] (4+*,rad(atan())/3);
[domain=1:1.7,samples=] plot [variable=] (4+*,rad(atan())/3);

[->,>=latex,domain=-2.4:1,samples=] plot [variable=] (-1-*,-rad(atan()));
[domain=1:2.4,samples=] plot [variable=] (-1-*,-rad(atan()));

[->,>=latex,domain=-2.2:1,samples=] plot [variable=] (-2-*,-rad(atan())/2);
[domain=1:2.2,samples=] plot [variable=] (-2-*,-rad(atan())/2);

[->,>=latex,domain=-1.7:1,samples=] plot [variable=] (-4-*,-rad(atan())/3);
[domain=1:1.7,samples=] plot [variable=] (-4-*,-rad(atan())/3);

[->,>=latex,domain=-6.9:1,samples=] plot [variable=] (,1.5*(*+1)/(*+2));
[domain=1:6.9,samples=] plot [variable=] (,1.5*(*+1)/(*+2));

[->,>=latex,domain=-6.9:1,samples=] plot [variable=] (,2*(*+1)/(*+2));
[domain=1:6.9,samples=] plot [variable=] (,2*(*+1)/(*+2));

[->,>=latex,domain=-6.9:1,samples=] plot [variable=] (,0.25+2.5*(*+2)/(*+4));
[domain=1:6.9,samples=] plot [variable=] (,0.25+2.5*(*+2)/(*+4));

[->,>=latex,domain=-6.9:1,samples=] plot [variable=] (-,-1.5*(*+1)/(*+2));
[domain=1:6.9,samples=] plot [variable=] (-,-1.5*(*+1)/(*+2));

[->,>=latex,domain=-6.9:1,samples=] plot [variable=] (-,-2*(*+1)/(*+2));
[domain=1:6.9,samples=] plot [variable=] (-,-2*(*+1)/(*+2));

[->,>=latex,domain=-6.9:1,samples=] plot [variable=] (-,-0.25-2.5*(*+2)/(*+4));
[domain=1:6.9,samples=] plot [variable=] (-,-0.25-2.5*(*+2)/(*+4));

Phase portrait of equation (<ref>). In red: the constant solution. In blue: the four solutions defined on $\mathbb{R}$ ending at $(0,0)$ when $t$ goes to $-\infty$ or $+\infty$. In black: all the other trajectories.

First, notice for $\tilde{Y}_0, Y_0 \in \mathbb R$ that

\begin{align*}
\left\vert Q(x+Y_0,y)- Q(x+\tilde{Y}_0,y) \right\vert \leq \int_{\tilde{Y}_0}^{Y_0} \left\vert \partial_x Q(x+s,y) \right\vert ds \leq \left\vert Y_0 - \tilde{Y}_0 \right\vert \| \partial_x Q \|_{L^\infty},
\end{align*}

and thus

\begin{align*}
\left\vert \int Q(x+Y_0,y)\partial_x Q^2(x,y)dxdy - \int Q(x+\tilde{Y}_0,y) \partial_x Q^2(x,y)dxdy \right\vert \leq \| \partial_x Q \|_{L^\infty} \| Q \|_{H^1}^2 \left\vert Y_0 -\tilde{Y}_0 \right\vert.
\end{align*}

Therefore, for any initial condition $(Y_0,Y_1)\in \mathbb{R}^2$, we deduce from the Cauchy-Lipschitz-Picard theorem the existence of a unique globally defined solution.

We now describe the phase portrait of the solutions. Let us first focus on initial conditions $(Y_0,Y_1) \in \mathbb{R}_+^*\times \mathbb{R}_+^*$. For such initial conditions, we claim that the curve $\left\{ (Z(t), \dot{Z}(t)) : t \in \mathbb{R}_+\right\}$ remains in the domain $\mathbb{R}_+^*\times \mathbb{R}_+^*$ and that $(Z(t),\dot{Z}(t)) \to (+\infty,l_1)$ when $t \to+\infty$, for some positive $l_1$.

Indeed, it follows form (<ref>) that

\begin{align*}
\ddot{Z}(t) = \kappa \int_{y \in \mathbb{R}} \int_{x \ge 0} \left(Q(x+Z(t),y)-Q(-x+Z(t),y) \right) \partial_x \left(Q^2\right)(x,y)dx dy >0
\end{align*}

since $\partial_x(Q^2)$ is odd in $x$ and $Q$ is positive and decreasing for $x \ge 0$. Thus $Z$ is convex and $\dot{Z}(t)>Y_1>0$ on $\mathbb R_+$. This proves that the curve remains in this domain. Furthermore, $Z$ is bounded by below by $Y_1t$, so that $Z(t)$ tends to $+\infty$ at $+\infty$.

To study the limit of $\dot{Z}$, we use the Hamiltonian defined in (<ref>). Observe that for a fixed $Y_1$ (respectively $Y_0$), the function $H(\cdot, Y_1)$ is strictly decreasing (resp. $H(Y_0,\cdot)$ is strictly increasing). We deduce from Lemma <ref> and (<ref>) that

\begin{align}\label{limit_hamiltonian}
\lim_{t\rightarrow +\infty} \dot{Z}(t)^2 = \lim_{t\rightarrow +\infty} 2H(Z(t), \dot{Z}(t)) = 2H(Y_0,Y_1) .
\end{align}

Observe that each curve of the phase portrait in the region $\mathbb{R}_+^\star \times\mathbb{R}_+^\star$ corresponds to a different Hamiltonian. Indeed, if two initial conditions $(Y_0,Y_1)$ and $(\tilde{Y}_0,\tilde{Y}_1)$ are associated to the same Hamiltonian $H(Y_0,Y_1)=H(\tilde{Y}_0,\tilde{Y}_1)$, let us denote by $Z$ and $\tilde{Z}$ the two corresponding solutions. Assume without loss of generality that $Y_0 < \tilde{Y}_0$. Then, $Z$ is bijective from $[0,+\infty)$ to $[Y_0,+\infty)$, and thus $Z(t_1)=\tilde{Y}_0$ for some $t_1>0$. Since $H(Z(t_1),\dot{Z}(t_1))=H(Y_0,Y_1)=H(\tilde{Z}(0),\dot{\tilde{Z}}(0))$ and $Y_1 \mapsto H(\tilde{Y}_0,Y_1)$ is strictly increasing on $\mathbb{R}_+$, we have $\dot{\tilde{Z}}(0)=\tilde{Y}_1=\dot{Z}(t_1)$. By uniqueness of the Cauchy-Lipschitz-Picard theorem, we conclude that the two solutions are the same up to a time translation.

By (<ref>), we conclude that each curve of the phase portrait has a different limit for $\dot{Z}$ at infinity.

We continue by describing the set of solutions on the boundary of the domain of initial conditions $(Z(0),\dot{Z}(0))=(Y_0,Y_1)\in \mathbb{R}_+^*\times \mathbb{R}_+^*$. First, notice that if $(Y_0,Y_1)=(0,0)$, then the solution is constant $Z(t)=0$. In the case where $(Y_0,Y_1)=(Z_0,0)$ with $Z_0>0$, then by strict convexity of $Z$, the graph of $(Z(t),\dot{Z}(t))$ for any time $t>0$ remains in $\mathbb{R}_+^*\times \mathbb{R}_+^*$. In the case $(Y_0,Y_1)=(0,\mu_1)$ with $\mu_1>0$, we have $Z(t)>0$ close to $0$ and thus the graph remains in the same domain.

We deduce the description of whole set of solutions by symmetry from the set of solutions corresponding to initial conditions in $\mathbb{R}_+\times \mathbb{R}_+$. This concludes the proof of the proposition.

By similar computations, we can also prove that for $Z>Z_0^*$, we have

\begin{align}\label{eq:monotonic}
\int_{\mathbb{R}^2} \partial_x Q(x+Z,y) \partial_x Q^2(x,y) dxdy \leq -\frac{\kappa a_0}{2} \frac{e^{-Z}}{\sqrt{Z}}\int_{\mathbb{R}^2} e^{-x} \partial_x(Q^2)(x,y) dxdy <0.
\end{align}

With the initial conditions at $-\infty$, by multiplying the equation (<ref>) by $\dot{Z}$ and integrating, we have

\begin{align}\label{equation_Z_dot}
2 \mu_0^2 - \frac{1}{2} \dot{Z}(t)^2 = \frac{2}{\langle \Lambda Q, Q \rangle} \int_{\mathbb{R}^2} Q(x+Z(t),y) Q^{2}(x,y) dxdy.
\end{align}

We first prove the existence of a solution for $\mu_0$ small enough. By defining $Z_0^*$ as in the remark above, we define $\mu_0^*$ by the relation (<ref>) with $Z_0^*$ instead of $Z_0$. We now fix $\mu_0\leq \mu_0^*$ and define $Z_0$ satisfying relation (<ref>), and we obtain $Z_0>Z_0^*$. We claim that there exists a unique even solution of (<ref>) with the initial condition $(Z,\dot{Z})(0)=(Z_0,0)$ and that this function is also a solution to the equation of the lemma.

Notice that for $Z>Z_0^*$ the function $Z \mapsto \int Q(x+Z,y) \partial_x (Q^2(x,y))dxdy$ is positive because, by (<ref>), we have

\begin{align*}
\frac{d}{dZ} \int_{\mathbb{R}^2} Q(x+Z,y) \partial_x Q^2(x,y) dxdy <0, \quad \lim_{Z\rightarrow +\infty} \int_{\mathbb{R}^2} Q(x+Z,y) \partial_x Q^2(x,y) dxdy =0.
\end{align*}

Since $\Ddot{Z}>0$, the function $Z$ is strictly convex and the minimum is achieved at $0$. Since $\dot{Z}$ is bounded by (<ref>), the solution $Z$ is globally defined. Note that by local uniqueness, $Z$ is even in $t$. To check the conditions at $-\infty$, we notice that since $Z$ decreases on $\mathbb{R}_-$, for $\epsilon$ small enough we have

\begin{align*}
0<2\mu_0^2- \frac{2}{\langle \Lambda Q, Q \rangle} \int_{\mathbb{R}^2} Q(x+Z_0+\epsilon,y) Q^2(x,y) dx dy < \frac{1}{2} \dot{Z}(t)^2,
\end{align*}

and thus $Z(t)$ goes to $+\infty$ at $-\infty$. Plugging in this condition in the equation satisfied by $\dot{Z}$, we obtain

\begin{align*}
\lim_{t\rightarrow -\infty} \dot{Z}(t)=-2\mu_0.
\end{align*}

As a consequence, we obtain $Z(t)\leq - C t$ for some positive constant $C$ for $t\leq -T_0$ for $T_0$ large enough. We also have $\left( 2\mu_0 -\dot{Z}(t) \right) \left( 2\mu_0 +\dot{Z}(t)\right)>0$, and thus $2\mu_0+\dot{Z}(t)$ is positive on an interval $(-\infty,-T_0)$ increasing $T_0$ if necessary. For the next order term, by using the equation satisfied by $\dot{Z}$, we have

\begin{align*}
\left\vert \left( 2\mu_0 -\dot{Z}(t) \right) \left( 2\mu_0 +\dot{Z}(t)\right) \right\vert \lesssim e^{C t},
\end{align*}

and thus $\dot{Z}(t) +2\mu_0 = O_{-\infty}(e^{Ct})$. An integration from $t<-T_0$ to $-T_0$ for $T_0$ large enough provides

\begin{align*}
\left\vert Z(t) +2\mu_0t -\left( Z(-T_0) +2\mu_0 T_0 \right) \right\vert \lesssim e^{CT_0}.
\end{align*}

By monotonicity of $t\mapsto Z(t)+2\mu_0 t$ close to $-\infty$, we obtain that $Z(t)-2\mu_0t$ converges to a finite limit $l$. Notice that by convexity, $Z$ is above its asymptote at $-\infty$. We obtain the condition of the lemma

\begin{align*}
\lim_{t\rightarrow -\infty} \left( Z(t) + 2\mu_0 t -l \right)=0.
\end{align*}

We now prove the uniqueness of the equation of the lemma for values of $\mu_0$ lower that $\frac{1}{2} \mu_0^*$. The following arguments are due the interpretation of a phase portrait. Suppose that for $\mu_0 < \frac{1}{2} \mu_0^*$, there exists another even solution $Y$ to (<ref>) with the condition

\begin{align*}
\lim_{t\rightarrow -\infty} (Y(t),\dot{Y}(t)) = (+\infty, -2\mu_0).
\end{align*}

By continuity of $Y$ and $\dot{Y}$, there exists $t_0$ close to $-\infty$ such that $\dot{Y}(t_0) \in (-\mu_0^*,0)$ and $Y(t_0)$ large enough such that

\begin{align*}
\frac{1}{\langle \Lambda Q, Q \rangle} \int_{\mathbb{R}^2} Q(x + Y(t_0),y) Q^2(x,y) dxdy \leq\frac{1}{4} (\mu_0^*)^2.
\end{align*}

Now consider the following final velocity

\begin{align*}
\tilde{\mu}^2_0 := \frac{\dot{Y}(t_0)^2}{4} + \frac{1}{\langle \Lambda Q, Q \rangle} \int_{\mathbb{R}^2} Q(x+Y(t_0),y) Q^2(x,y) dx dy \leq \frac{1}{2} (\mu_0^*)^2 < (\mu_0^*)^2,
\end{align*}

and denote by $\tilde{Y}$ the solution constructed at the previous step associated to the velocity $\tilde{\mu}_0$. In particular, since $\dot{\tilde{Y}}$ strictly increases on $(-\infty;0]$ from $-2\tilde{\mu}_0^2$ to $0$, there exists a unique time $t_1$ such that $\dot{\tilde{Y}}(t_1)=\dot{Y}(t_0)$. By the relation (<ref>), we also have $\tilde{Y}(t_1)=Y(t_0)$. Since the equation is autonomous, by uniqueness around one initial condition, $\tilde{Y}$ is a time-translated version of $Y$. Since the two functions are convex and even, they are equal. Thus $Y$ is one of the solution constructed at the previous step, and by the uniqueness of the asymptotic development, we have $\tilde{\mu}_0=\mu_0$. It concludes the proof of the uniqueness of the even solution.

Let $l_1\in \mathbb{R}_+^*$. There exists a unique solution, up to time translation, to (<ref>) with the condition

\begin{align} \label{coro:initial_condition.1}
\lim_{t\rightarrow -\infty} ( Z(t), \dot{Z}(t) ) = (+\infty,-l_1).
\end{align}

Let us split the different possibilities of $l_1$, and compare with the hamiltonian (<ref>). We begin with the existence of a solution.

* $l_1^2 < 2\kappa \| Q \|_{L^3}^3$. In this case, define $Y_0$ as the unique positive real coefficient such that

\begin{align}\label{defi:Z_0_l_1}
l_1^2 = 2\kappa \int_{\mathbb{R}^2} Q(x+Y_0,y) Q^2(x,y) dxdy.
\end{align}

The solution $Z$ associated to the initial condition $(Y_0,0)$ has the hamiltonian $H(Y_0,0)=\frac{l_1^2}{2}$, and thus is a solution to the problem by (<ref>).

* $l_1^2 > 2\kappa \| Q \|_{L^3}^3$. In this case, define $Y_1$ as the unique negative real coefficient such that to

\begin{align*}
Y_1^2 =l_1^2 -2\kappa \| Q\|_{L^3}^3.
\end{align*}

The solution $Z$ associated to the initial condition $(0,Y_1)$ has the hamiltonian $H(0,Y_1)=\frac{l_1^2}{2}$, and is a solution to the problem by (<ref>).

* $l_1^2 = 2\kappa \| Q \|_{L^3}^3$. This case corresponds to the blue curve on the phase portrait, ending at $(+\infty,-l_1)$ as $t\rightarrow -\infty$ and to $(0,0)$ as $t\rightarrow+\infty$.

By the previous discussion on the phase portrait, each of the solution is uniquely determined by the hamiltonian, thus the uniqueness of the solution up to time translation.

Let $0<l_1$ with $l_1^2 < 2\kappa \| Q \|_{L^3}^3$. Then the unique solution $Z$ of (<ref>) with the initial condition $(Z(0),\dot{Z}(0))=(Y_0,0)$, with $Y_0$ defined in (<ref>), is even and has an asymptote at $+\infty$. More precisely, there exist constants $l = l(l_1)\in \mathbb{R}$ and $c=c(l_1)>0$ such that

\begin{align*}
\left\vert Z(t) -l_1 t -l \right\vert \lesssim e^{-ct}.
\end{align*}

If $Z(t)$ is a solution to (<ref>) with the initial condition $(Y_0,0)$, then $t\mapsto Z(-t)$ is also a solution, and by uniqueness we have $Z(t)=Z(-t)$. We can thus focus on the case $t\geq 0$.

Since $Z(t)>0$, for all $t >0$ (from Proposition <ref>),

\begin{align*}
\ddot{Z}(t) \geq \kappa \int_{\mathbb{R}^2} Q(x+Z,y) \partial_x (Q^2)(x,y) dx dy>0 .
\end{align*}

Hence, for a fixed $t_0>0$ and any $t>t_0$, we have $Z(t)>\dot{Z}(t_0)(t-t_0)$ with $\dot{Z}(t_0)>0$. Moreover, it follows from Lemma <ref> that

\begin{align}\label{equation_Z_dot}
\frac{l_1^2}{2} = H(Z(t),\dot{Z}(t)) = \frac{\dot{Z}(t)^2}{2} + \kappa \int_{\mathbb{R}^2} Q(x+Z(t),y) Q^2(x,y) dx dy.
\end{align}

As underlined in the proof of Proposition <ref>, the function $Z \mapsto \int Q(x+Z,y) Q^2(x,y)dx dy$ is strictly decreasing on $\mathbb{R}_+$. Thus for $t\geq t_0$

\begin{align*}
l_1^2 - 2\kappa \int_{\mathbb{R}^2} Q(x+\dot{Z}(t_0)(t-t_0),y) Q^2(x,y) dx dy \leq \dot{Z}(t)^2 .
\end{align*}

From Lemma <ref>, there exist two positive constants $c$ and $C$ such that, for $t$ large enough,
$ l_1^2- Ce^{-ct} \leq \dot{Z}(t)^2$, and thus $  0<l_1-\dot{Z}(t) \leq \frac{C}{l_1}e^{-ct}$.
An integration from $t$ to $+\infty$ yields

\begin{align*}
0 \leq \lim_{s\rightarrow +\infty} \left( l_1 s -Z(s)\right) - l_1t +Z(t) \lesssim e^{-ct}.
\end{align*}

Since $l_1 t-Z(t)$ is concave and bounded on $\mathbb{R}_+$, it has a finite limit $l$ at $+\infty$, which concludes the proof of the lemma.

The following proposition gives an approximation of the right-hand-side of (<ref>) for $Z$ large.

There exist two positive constants $Z^\star$ and $c$ such that it holds, for any $Z>Z^\star$ and $i\in \{0,1,2\}$,

\begin{align}
& \left\vert \int_{\mathbb{R}^2} Q (x+Z,y) \partial_x^i(Q^2)(x,y) dx dy -(-1)^i c\frac{e^{-Z}}{\sqrt{Z}} \right\vert \lesssim \frac{e^{-Z}}{Z},  \label{propo:approx_e^-Z.1}\\
& \left\vert \int_{\mathbb{R}^2} \left( \partial_x Q + Q \right)(x+Z,y) Q^2(x,y) dx dy \right\vert \lesssim \frac{e^{-Z}}{Z}. \label{propo:approx_e^-Z.2}
\end{align}

Proposition <ref> echoes Remark <ref>. Indeed, it follows from (<ref>) that the asymptotic of (<ref>) has the form $\ddot{Z}=ce^{-Z}Z^{-\frac12}$ which does not possess explicit solutions, unlike its $1$-dimensional equivalent $\ddot{Z}=ce^{-Z}$ (see [41]).

Notice that this proposition echoes Remark <ref> with the equation $\ddot{Z}=ce^{-Z}$ found in [41]. Indeed, the assumption to have $Z$ large makes sense in the situation that the two solitary waves remain far one from another. The differences between the $1$ and the $2$-dimensional cases are the constant coefficient, but also the term $Z^{-\frac{1}{2}}$ that comes from the dimension $2$.

We only prove estimate (<ref>) in the case $i=0$, since the proof for the other cases is similar. On the one hand, it follows by using (<ref>) and taking $Z^\star$ large enough that

\begin{align*}
\MoveEqLeft
\int_{\vert(x,y)\vert \geq Z^\frac14} Q(x+ Z,y) Q^2(x,y)dxdy \\
& \leq C \sup_{(x,y)} \left(Q(x+Z,y) Q(x,y)\right) \int_{\vert(x,y)\vert>Z^\frac14} \vert Q(x,y) \vert dx dy \lesssim e^{- Z-Z^\frac14}.
\end{align*}

On the other hand, we infer from the asymptotic expansion of Proposition <ref> that

\begin{align*}
\left\vert \int_{\vert (x,y)\vert \leq Z^\frac14} Q(x+Z,y) Q^2(x,y) dx dy - \frac{e^{-Z}}{Z^\frac{1}{2}} \left( \kappa a_0 \int_{\vert (x,y)\vert \leq Z^\frac14} e^{-x} Q^2(x,y) dx dy \right) \right\vert \lesssim \frac{e^{-Z}}{Z}.
\end{align*}

Therefore, we conclude the proof of (<ref>) in the case $i=0$ by gathering these two estimates.

Estimate (<ref>) follows by combining (<ref>) in the case $i=0$ with (<ref>) in the case $i=1$ and observing that, after integration by parts, the terms of order $e^{-Z}Z^{-\frac12}$ cancel each other out. Note that it is important that the constant $c$ appearing in (<ref>) is independent of $i \in \{0,1\}$.

§.§ Quantified evolution of a function z close to Z

While the previous subsection dealt with the general equation (<ref>) for any arbitrary positive constant $\kappa$, this subsection focuses on the ODEs (<ref>) which are close to (<ref>). More precisely, for a parameter $\nu\in(-\frac12,\frac12)$, we are interested in the following equation and its associated Hamiltonian

\begin{align}\label{defi:eq_Z_nu}
& \Ddot{Z}(t)= \frac{2(1+\nu)}{\langle \Lambda Q, Q \rangle} \int_{\mathbb{R}^2} Q(x + Z(t),y) \partial_x (Q^2) (x,y) dxdy \\
& H_\nu (Y_0,Y_1) := \frac{Y_1^2}{2} + \frac{2(1+\nu)}{\langle \Lambda Q, Q \rangle} \int_{\mathbb{R}^2} Q(x+Y_0,y) Q^2(x,y) dx dy. \label{defi:H_nu}
\end{align}

Recall from (<ref>) that $\langle \Lambda Q, Q \rangle>0$. The case $\nu=0$ corresponds to (<ref>), where $Z$ is a good approximation of the distance $z$ between the $2$ solitary waves during the collision. However, due to the error terms, $z$ is not an exact solution of (<ref>). This is why we study the set of equations (<ref>) with a small parameter $\nu$.

[Equivalent of $l_1$ for $Y_0$ large]
There exist $Y^\star>0$ large enough and two positive constants $c$ and $C$ such that the following holds. For any $\nu \in (-\frac12,\frac12)$ and $Y_0>Y^\star$, define $l_1>0$ by $H_\nu(Y_0,0) = \frac{l_1^2}2$. Then the following holds

\begin{align}\label{eq:comparison_l_1_e^Z_0}
c\frac{e^{-Y_0}}{\sqrt{Y_0}} \leq l_1^2 \leq C \frac{e^{-Y_0}}{\sqrt{Y_0}}.
\end{align}

From the definition of $H_\nu$ in (<ref>), we have

\begin{align*}
l_1^2=\frac{4(1+\nu)}{\langle \Lambda Q, Q \rangle} \int_{\mathbb{R}^2} Q(x+Y_0,y) Q^2(x,y) dx dy ,
\end{align*}

which implies (<ref>) by using (<ref>) with $i=0$ and choosing $Y^\star$ large enough. Note that the constants $c$ and $C$ can be chosen uniformly in $\nu \in (-\frac12,\frac12)$.

[Uniform lower bound of $\dot{Z}$ close to $0$]
There exist $Y^\star>0$, $\eta^\star>0$ small and positive constants $c$ and $C$ such that the following holds. For any $Y_0>Y^\star$, $\nu \in (-\frac12, \frac12)$ and $\eta\in (0, \eta^\star)$, consider $Z$ the unique even $\mathcal{C}^2$-function solving $H_\nu(Z,\dot{Z})=H_\nu(Y_0,0)$ and $T$ the unique positive time such that $Z(T) = Y_0+\eta^2$. Denote $l_1$ as in (<ref>). Then

\begin{align*}
c \eta l_1 \leq \dot{Z}(T) \leq C \eta l_1.
\end{align*}

Recall that the existence and uniqueness of the even function $Z$ are given in Proposition <ref> and Corollary <ref>. The conservation of the Hamiltonian in (<ref>) associated to $Z$ implies that

\begin{align*}
\dot{Z}(T)^2 = -\frac{4(1+\nu)}{\langle \Lambda Q, Q \rangle} \int_{0}^{Z(T)-Y_0} \left(\int_{\mathbb{R}^2} \partial_x Q(x+Y_0+s,y)  Q^2(x,y) dx dy \right) ds.
\end{align*}

Let us find an equivalent on the inner integral for $Y_0$ large enough and a fixed value $s>0$.

Observe that $0 \le Z(T)-Y_0 = \eta^2$. Inspired from the computation of (<ref>), we adjust $\eta^\star$ so that for $s\leq \eta^2 <(\eta^\star)^2$, it holds on $\mathbb{R}^2$

\begin{align*}
\left\vert \partial_x Q(x+ Y_0+s,y)- \partial_x Q (x+Y_0,y) \right\vert \lesssim s \frac{e^{-\left\vert (x+Y_0,y)\right\vert}}{\langle (x+Y_0,y) \rangle^\frac{1}{2}} .
\end{align*}

Thus, it follows arguing as in (<ref>) that

\begin{align*}
\MoveEqLeft
\left\vert \int_{s=0}^{Z(T)-Y_0} \int_{\mathbb{R}^2} \left( \partial_x Q \left( x+ Y_0+ s,y\right) - \partial_x Q(x+ Y_0,y) \right) Q^2(x,y) dxdy ds \right\vert \\
& \lesssim \eta^4 \sup \left( \frac{e^{-\left\vert (x+Y_0,y) \right\vert }}{\langle (x+Y_0,y) \rangle^\frac12} Q(x,y)\right) \| Q \|_{L^1} \lesssim \eta^4 \frac{e^{-Y_0}}{Y_0^\frac12}.
\end{align*}

Then, we deduce from (<ref>) that

\begin{align*}
\left\vert \dot{Z}(T) ^2 - \frac{4(1+\nu)}{\langle \Lambda Q, Q \rangle} \int_{s=0}^{Z(T)-Y_0} \int_{\mathbb{R}^2} Q(x+ Y_0 ,y) Q^2(x,y) dxdy ds \right\vert  \leq \int_{s=0}^{Z(T)-Y_0} C \frac{e^{-Y_0}}{Y_0}  ds + C\eta^4 \frac{e^{-Y_0}}{Y_0^\frac12},
\end{align*}

which combined with (<ref>) yields

\begin{align*}
\left\vert \dot{Z}(T)^2 - \eta^2 l_1^2 \right\vert
\leq \eta^2 \left( C \frac{e^{-Y_0}}{Y_0} +C \eta^2 \frac{e^{-Y_0}}{Y_0^\frac12} \right).
\end{align*}

By using Lemma <ref>, where the constants can be chosen uniformly in $\nu$, we obtain that for $l_1$ small enough

\begin{align*}
\eta^2 l_1^2 \left( 1- \frac{C}{\vert \ln(l_1) \vert^\frac12 } -C \eta^2\right) \leq \dot{Z}(T)^2 \leq \eta^2 l_1^2 \left( 1+ \frac{C}{\vert \ln(l_1) \vert^\frac12 } +C \eta^2\right) .
\end{align*}

We conclude the proof of the lemma by choosing $l_1^\star$ and $\eta^\star$ smaller if necessary and by taking the square root of the previous estimate.

The next proposition quantifies the defect of closeness of a function to a solution to the equation (<ref>). More precisely, let us denote by $Z$ the even solution to (<ref>) with $\nu=0$. If a function $z$ defined on a time interval $[t_0,t_1]$ is close to $Z$ at time $t_0$ and the flows $H_{\pm \nu}(z,\dot{z})$ are well controlled along the time, then $z$ is close to $Z$ at any time $t\in [t_0,t_1]$.

[Evolution of $(z,\dot{z})$ with $H_0(z,\dot{z})$ almost constant]
Let $\eta\in (0,1)$. There exist some constants $C>0$, $\varepsilon_0^\star$ and $\nu_6^\star \leq \frac{1}{2}$ such that the following holds. Let $\nu\in (0, \nu_6^\star)$ and $\varepsilon_0 \in (0, \varepsilon_0^\star)$. Let us define a constant $h < H_{-\nu_6^\star}(0,0)$ small and the three hamiltonian $H:=H_0$, $H_+:= H_\nu$ and $H_-:=H_{-\nu}$. We set $Z$ the unique even $\mathcal{C}^1$-function such that $H(Z,\dot{Z})=h$ and $t_0>0$ satisfying $Z(t_0)\geq Z(0)+\eta^2$. Let a $\mathcal{C}^1$-function $z$ defined on a time interval $[t_0,t_1]$ that satisfies the initial condition

\begin{align*}
\left\vert z(t_0) -Z(t_0) \right\vert \leq \varepsilon_0,
\end{align*}

and for any $t\in [t_0,t_1]$

\begin{align}\label{condition:H_+(z)_H_-(z)}
H_-(z(t),\dot{z}(t)) \leq h \leq H_+(z(t),\dot{z}(t)).
\end{align}

Then for any $t$ in $[t_0,t_1]$ we have

\begin{align*}
\left\vert z(t) -Z(t) \right\vert \leq C \nu + C\varepsilon_0.
\end{align*}

[->,,samples=] (-8,0) – (-4,0);
[->,samples=] (-7.5,-1.5) – (-7.5,1.5);
(-3.7,0) node[below] $Z_-$;
(-7.5,1.5) node[left] $\dot{Z}_-$;

[->,>=latex,domain=-1.6:1,samples=] plot [variable=] (-6.6+*,rad(atan())/1.5);
[domain=1:1.6,samples=] plot [variable=] (-6.6+*,rad(atan())/1.5);
(-4,0.75) node[right] $h_0$;

[->,blue,>=latex,domain=-1.75:1,samples=] plot [variable=] (-7.1+*,rad(atan())*1.2);
[blue,domain=1:1.75,samples=] plot [variable=] (-7.1+*,rad(atan())*1.2);
(-4,1.25) node[right] $h$;

[->,>=latex,domain=-1.85:1,samples=] plot [variable=] (-7.4+*,rad(atan())*1.5);
[domain=1:1.85,samples=] plot [variable=] (-7.4+*,rad(atan())*1.5);
(-4,1.75) node[right] $h_1$;

[red,domain=1.2:1.75,samples=] plot [variable=] (-7.25+*,rad(atan())*(1+*sin(*1200)/15);

[->,samples=] (-2,0) – (2,0);
[->,samples=] (-1.5,-1.5) – (-1.5,1.5);
(2.3,0) node[below] $Z$;
(-1.5,1.5) node[left] $\dot{Z}$;

[->,>=latex,domain=-1.5:1,samples=] plot [variable=] (-0.3+*,rad(atan())/2);
[domain=1:1.5,samples=] plot [variable=] (-0.3+*,rad(atan())/2);
(2,0.5) node[right] $h_0$;

[->,>=latex,domain=-1.65:1,samples=] plot [variable=] (-0.75+*,rad(atan()));
[domain=1:1.65,samples=] plot [variable=] (-0.75+*,rad(atan()));
(2,1) node[right] $h$;

[->,>=latex,domain=-1.75:1,samples=] plot [variable=] (-1.1+*,rad(atan())*1.4);
[domain=1:1.75,samples=] plot [variable=] (-1.1+*,rad(atan())*1.4);
(2,1.5) node[right] $h_1$;

[red,domain=1.2:1.75,samples=] plot [variable=] (-1.25+*,rad(atan())*(1+*sin(*1200)/15);

[->,samples=] (4,0) – (8,0);
[->,samples=] (4.5,-1.5) – (4.5,1.5);
(8.5,0) node[below] $Z_+$;
(4.5,1.5) node[left] $\dot{Z}_+$;

[->,>=latex,domain=-1.45:1,samples=] plot [variable=] (5.9+*,rad(atan())/2.5);
[domain=1:1.45,samples=] plot [variable=] (5.9+*,rad(atan())/2.5);
(8,0.25) node[right] $h_0$;

[->,blue,>=latex,domain=-1.6:1,samples=] plot [variable=] (5.5+*,rad(atan())*0.8);
[blue,domain=1:1.6,samples=] plot [variable=] (5.5+*,rad(atan())*0.8);
(8,0.75) node[right] $h$;

[->,>=latex,domain=-1.75:1,samples=] plot [variable=] (5+*,rad(atan())*1.3);
[domain=1:1.75,samples=] plot [variable=] (5+*,rad(atan())*1.3);
(8,1.25) node[right] $h_1$;

[red,domain=1.2:1.75,samples=] plot [variable=] (4.75+*,rad(atan())*(1+*sin(*1200)/15);

Comparison of a trajectory $(z,\dot{z})$ in red with the phase portrait associated to the equations related to $H_-$ (left), $H$ (middle) and $H_+$ (right). On each graph, the three lines are associated to three values $h_0<h<h_1$.

In the proof of the collision, we will set $h=2\mu_0^2$.

For a constant $C_\star$ to define later, define the inequality

\begin{align}\label{bootstrap_estimate_z_Z}
\left\vert z(t) -Z(t) \right\vert \leq C_\star \nu + C_\star \varepsilon_0.
\end{align}

The coefficients $\nu_6^\star$ and $\varepsilon_0^\star$ are adjusted so that

\begin{align}\label{restriction_e_0_nu}
C_\star \nu_6^\star + C_\star \varepsilon_0^\star \leq \frac{\eta^2}{4}.
\end{align}

Define the time

\begin{align*}
T^\star := \sup \left\{ t \in [t_0;t_1]; \, \forall t \in [t_0;T^\star), \text{ estimate } \eqref{bootstrap_estimate_z_Z} \text{ holds} \right\}.
\end{align*}

$T^\star$ is well-defined by assumption and continuity of $z$ and $Z$. By a bootstrap argument, we claim that $T^\star=t_1$. We suppose that $T^\star<t_1$ and show that this statement leads to a contradiction. To do so, we compare the two functions $z$ and $Z$ along the flow of two well-chosen functions $Z_-$ and $Z_+$ associated to the two hamiltonian $H_+$ and $H_-$. We thus define $Z_-$ and $Z_+$ as the unique even functions satisfying

\begin{align}\label{defi:Z_-_Z_+}
H_-(Z_-,\dot{Z}_-)=h, \quad \text{and} \quad H_+(Z_+,\dot{Z}_+)=h.
\end{align}

Those two functions are represented in blue in Figure <ref>. Remark that the constraint $h<H_{-\nu_6^\star}(0,0)$ justifies that the function $Z_-$ is associated to an initial condition under the form $(Y_0,0)$ for $Y_0>0$, and not under the form $(0,Y_1)$ as explained in Proposition <ref>. $Z$ (respectively $Z_-$, $Z_+$) goes from $\mathbb{R}$ to $(Z(0);+\infty)$ (resp. $Z_-(0)$, $Z_+(0)$). By abuse of notations, we denote by $Z^{-1}$ (resp. $Z_-^{-1}$, $Z_+^{-1}$) the bijection from $(Z(0);+\infty)$ (resp. $Z_-(0)$, $Z_+(0)$) to $\mathbb{R}_+^*$.

We will make use of the following claims, whose proofs are postponed.
[Comparison of the three functions]
At the initial time, we have

\begin{align}\label{eq:Z_Z_+_Z_-(0)}
\end{align}

In particular, the function $\nu \mapsto Z_+(0)$ for $\nu$ close to $0$ is a decreasing function and

\begin{align}\label{eq:limit_nu}
\lim_{\nu \rightarrow 0} Z_+(0) = Z(0).
\end{align}

For any $s>Z_+(0)$ it holds

\begin{align}\label{eq:Z_Z_+_Z_-(s)}
\dot{Z}_+\circ Z_+^{-1}(s) \leq \dot{Z} \circ Z^{-1}(s) \leq \dot{Z}_-\circ Z_-^{-1} (s).
\end{align}

[Comparison of the derivatives of the inverse functions]
The two following inequalities hold

\begin{align*}
& \frac{1}{\dot{Z} \circ Z^{-1}(s)} - \frac{1}{\dot{Z}_- \circ Z_-^{-1}(s)} \leq \frac{2\nu}{\langle \Lambda Q, Q \rangle} \frac{\int Q(x+s,y) Q^2(x,y)dx dy}{ \left( \dot{Z} \circ Z^{-1}(s)\right)^3}, \\
& \frac{1}{\dot{Z} \circ Z^{-1}(s)} - \frac{1}{\dot{Z}_+\circ Z_+^{-1}(s)} \geq -\frac{2\nu}{\langle \Lambda Q,Q \rangle} \frac{\int Q(x+s,y)Q^2(x,y)dxdy}{\left(\dot{Z}_+ \circ Z_+^{-1} (s)\right)^3}.
\end{align*}

To continue the proof of (<ref>), we split the inequality into the lower and the upper bound.

Upper bound of (<ref>). We prove in this part an upper bound of the quantity $z(t)-Z(t)$ in terms of the initial condition $z(t_0)-Z(t_0)$ and $\nu$. Notice first that $Z_-$ is a bijection from $(0;+\infty)$ to $(Z_-(0);+\infty)$. From (<ref>), the bootstrap assumption (<ref>) and the restriction (<ref>) on $\epsilon_0$ and $\nu$, we have for any $t\in [t_0,T^\star)$ and the assumption on $t_0$

\begin{align*}
Z_-(0) < Z(0) < Z(0) +\frac34 \eta^2 \leq z(t).
\end{align*}

The functions $Z_-^{-1}\circ z$ and $Z_-^{-1}\circ Z$ are well-defined on $[t_0;T^\star)$.
We now study the derivative of the difference of those functions for $t\in[t_0,T^\star)$. The assumption (<ref>) implies the bound

\begin{align*}
\dot{Z}_- \circ Z_-^{-1} \circ z(t) = \left( 2h - \frac{4(1-\nu)}{\langle \Lambda Q, Q \rangle} \int_{\mathbb{R}^2} Q(x+z(t),y) Q^2(x,y) dx dy \right)^{\frac12} \geq \dot{z}(t)
\end{align*}

and thus

\begin{align}\label{est:Z_-^-1}
\frac{d}{dt} \left( Z_-^{-1} \circ z(t) - Z_-^{-1}\circ Z(t) \right) \leq \dot{Z}(t) \left( \frac{1}{\dot{Z}\circ Z^{-1}} - \frac{1}{\dot{Z}_- \circ Z_-^{-1}}\right) \circ Z(t).
\end{align}

Plugging the estimate of Claim <ref> in (<ref>) with Proposition <ref> with the assumption of $h$ small, we obtain

\begin{align*}
\frac{d}{dt} \left( Z_-^{-1}\circ z(t) - Z_-^{-1} \circ Z(t) \right) \leq \frac{4\nu}{\dot{Z}(t)^2} \ddot{Z}(t).
\end{align*}

An integration from $t_0$ to $t\in (t_0,T^\star)$ with the lower bound provided by Lemma <ref> yields

\begin{align*}
\left( Z_-^{-1} \circ z(t) - Z_-^{-1} \circ Z(t) \right) - \left( Z_-^{-1} \circ z(t_0) - Z_-^{-1} \circ Z (t_0) \right) \leq \frac{4\nu}{c\eta l_1}.
\end{align*}

We continue using the mean value theorem which states that for any $t\in [t_0,T^\star)$ there exists a coefficient $\lambda_t \in \left[ Z(t)- C_\star \left( \nu+ \varepsilon_0\right),  Z(t)+ C_\star \left( \nu+ \varepsilon_0\right)\right]$ such that

\begin{align*}
Z_-\circ z(t) - Z_-^{-1} \circ Z(t)= \frac{z(t)-Z(t)}{\dot{Z}_- \circ Z_-^{-1}(\lambda_t)}.
\end{align*}

Notice that the previous quantity is well-defined since $\lambda_t >Z_0 +\frac34 \eta^2 >Z_-(0)$, and we have

\begin{align*}
z(t)-Z(t) \leq \frac{\dot{Z}_-\circ Z_-^{-1}(\lambda_t)}{\dot{Z}_-\circ Z_-^{-1}(\lambda_{t_0})} \left( z(t_0) -Z(t_0) \right) + \frac{4\nu}{c\eta l_1} \dot{Z}_-\circ Z_-^{-1}(\lambda_t).
\end{align*}

In particular, Lemma <ref> applied to $\lambda_t >Z_-(0) +\frac12 \eta^2$ yields $\dot{Z}_-\circ Z_-^{-1}(\lambda_t) \geq c2^{-\frac12}\eta l_1$. We thus obtain with $\dot{Z}_-\leq l_1$

\begin{align}\label{est:z_Z_upper_bound}
z(t)-Z(t) \leq \frac{\dot{Z}_-\circ Z_-^{-1}(\lambda_t)}{\dot{Z}_-\circ Z_-^{-1}(\lambda_{t_0})} \left( z(t_0) -Z(t_0) \right) + \frac{4\nu}{c\eta} \quad \text{with} \quad \frac{\dot{Z}_-\circ Z_-^{-1}(\lambda_t)}{\dot{Z}_-\circ Z_-^{-1}(\lambda_{t_0})} \in \left[ \frac{c}{\sqrt{2}} , \frac{\sqrt{2}}{c\eta}\right].
\end{align}

Lower bound of (<ref>) The proof of the lower bound on $z(t)-Z(t)$ follows similar ideas, let us emphasize the main differences. First, the functions $Z_+^{-1}\circ Z$ and $Z_+^{-1}\circ Z$ are well-defined on $[t_0,T^\star)$. Indeed, by the limit in (<ref>), decreasing $\nu^\star$ if necessary and the assumption (<ref>), we have

\begin{align*}
Z(0) < Z_+(0) < Z(0)+\frac{\eta^2}{4} < Z(0)+\frac34 \eta^2 <z(t).
\end{align*}

These inequalities justify the functions $Z_+^{-1}\circ Z$ and $Z_+^{-1}\circ z$. Since Proposition <ref> provides the inequality

\begin{align*}
\frac{2 \nu}{\langle \Lambda Q, Q \rangle} \int Q(x+Z(t),y) Q^2(x,y) dx dy \leq 2 \nu \ddot{Z}_+ \circ Z_+^{-1} \circ Z(t)
\end{align*}

we can compute, as for $Z_-$, the time derivative of the difference of the functions and using assumption (<ref>) and Claim (<ref>)

\begin{align*}
\frac{d}{dt} \left( Z_+^{-1}\circ z- Z_+^{-1} \circ Z \right) (t) \geq -2 \nu \frac{\ddot{Z}_+\circ Z_+^{-1} \circ Z(t)}{\left( \dot{Z}_+ \circ Z_+^{-1} \circ Z (t) \right)^3} \dot{Z}(t).
\end{align*}

An integration from $t_0$ to $t$ yields, by the change of variable $s'=Z_+^{-1}\circ Z(s)$

\begin{align*}
\MoveEqLeft
\left(Z_+^{-1} \circ z - Z_+^{-1} \circ Z \right) (t) - \left( Z_+^{-1} \circ z - Z_+^{-1} \circ Z \right) (t_0) \\
& \qquad \geq 2 \nu \left( \frac{1}{\dot{Z}_+\circ Z_+^{-1} \circ Z(t)}- \frac{1}{\dot{Z}_+ \circ Z_+^{-1} \circ Z(t_0)} \right).
\end{align*}

By the bootstrap assumption (<ref>) and the mean value theorem, for any $t\in [t_0, T^\star)$, there exists $\lambda_t$ such that $\left\vert \lambda_t - Z(t) \right\vert \leq C_\star \nu + C_\star \varepsilon_0$ and

\begin{align*}
Z_+^{-1} \circ z(t) - Z_+^{-1} \circ Z (t) = \frac{1}{\dot{Z}_+ \circ Z_+^{-1}(\lambda_t)} (z(t)-Z(t)).
\end{align*}

Since $Z_+$ increases on $[0,+\infty)$, we obtain the inequality

\begin{align*}
z(t)-Z(t) \geq \frac{\dot{Z}_+\circ Z_+^{-1}(\lambda_t)}{\dot{Z}_+\circ Z_+^{-1}(\lambda_{t_0})} (z(t_0)-Z(T_0)) -2 \nu \frac{\dot{Z}_+\circ Z_+^{-1}(\lambda_t)}{\dot{Z}_+ \circ Z_+^{-1}\circ Z(t_0)}.
\end{align*}

Since $Z(t)>Z(0)+\frac34 \eta^2 >Z_+(0)+\frac12 \eta^2$ for any $t\in [t_0,T^\star)$, by applying Lemma <ref>, we have $\dot{Z}_+ \circ Z_+^{-1} \circ Z(t_0) \geq c\eta l_1$. We thus obtain the lower bound

\begin{align}\label{est:z_Z_lower_bound}
z(t)-Z(t) \geq \frac{\dot{Z}_+\circ Z_+^{-1}(\lambda_t)}{\dot{Z}_+\circ Z_+^{-1}(\lambda_{t_0})} (z(t_0)-Z(T_0)) -\frac{2 \nu}{c\eta} \quad \text{with} \quad \frac{\dot{Z}_+\circ Z_+^{-1}(\lambda_t)}{\dot{Z}_+\circ Z_+^{-1}(\lambda_{t_0})} \in \left[ \frac{c}{\sqrt{2}}, \frac{\sqrt{2}}{c\eta} \right].
\end{align}

Gathering the upper bound (<ref>) and lower bound (<ref>), we obtain that

\begin{align*}
\left\vert z(t)-Z(t) \right\vert \leq \frac{\sqrt{2}}{c\eta} \varepsilon_0 + \frac{4}{c\eta} \nu.
\end{align*}

By choosing $C_\star > 4(c\eta)^{-1}$, we strictly improve the bound on $z(t)-Z(t)$, which contradicts the maximality of $T^\star$. Thus $T^\star=t_1$, and the proposition is proved.

By the definition of the Hamiltonian we have

\begin{align*}
\frac{2}{\langle \Lambda Q, Q\rangle} \int Q(\cdot +Z(0),\cdot) Q^2 & = \frac{2(1-\nu)}{\langle \Lambda Q,Q \rangle} \int Q (\cdot +Z_-(0),\cdot )Q^2= \frac{2(1+\nu)}{\langle \Lambda Q,Q \rangle} \int Q (\cdot +Z_+(0),\cdot )Q^2\\
& =h.
\end{align*}

Since $\nu$ is positive, we obtain <ref>. Under the previous form, the statement on the limit of $\nu$ is straightforward.

Concerning the second inequality, all the derivatives are positive and by the definition of the Hamiltonian

\begin{align*}
\frac{1}{1-\nu}\left( 2h -\dot{Z}_- \circ Z_-^{-1}(s)^2 \right)= 2h -\dot{Z} \circ Z^{-1}(s)^2 = \frac{1}{1+\nu} \left(2h - \dot{Z}_+ \circ Z_+^{-1}(s)^2\right)>0 ,
\end{align*}

which proves (<ref>) since $\nu>0$.

Let us begin with the proof of the inequality with $Z_-$. By developing, we have

\begin{align*}
\frac{1}{\dot{Z} \circ Z^{-1}(s)} - \frac{1}{\dot{Z}_- \circ Z_-^{-1}(s)} = \frac{4\nu}{\langle \Lambda Q, Q \rangle} \frac{\int Q(x+s,y) Q^2(x,y)dx dy}{\dot{Z}\circ Z^{-1}(s) \dot{Z}_-\circ Z_-^{-1}(s) \left( \dot{Z}_- \circ Z_-^{-1}(s) + \dot{Z}\circ Z^{-1}(s)\right)}.
\end{align*}

Then, estimate (<ref>) concludes the proof of the first inequality. The second inequality holds with the same computations with the other side of (<ref>).

In the previous proposition, the constraint on the time interval $[t_0,t_1]$ is that the time interval is far enough from $0$ so that $Z(t_0) > Z(0) + \eta^2$. The coefficient $\epsilon_0^\star$ and $\nu_6^\star$ are thus adapted in terms of $\eta$. A similar statement thus holds on a time interval $t_0<t_1<0$.

With the notations of Proposition <ref>, consider the same assumptions and definitions on $\eta$, $C$, $\nu_6^\star$, $\varepsilon_0^\star$, $h$, $H$, $H_+$, $H_-$ and $Z$. Let $t_1<0$ satisfying $Z(t_1) \geq Z(0)+\eta^2$. Let $z$ be a $\mathcal{C}^1$-function defined on a time interval $[t_0,t_1]$ that satisfies the conditions

\begin{align*}
\left\vert z(t_0) -Z(t_0) \right\vert\leq \varepsilon_0,
\end{align*}

and, for any $t\in [t_0,t_1]$,

\begin{align*}
H_-(z(t),\dot{z}(t)) \leq h \leq H_+(z(t),\dot{z}(t)).
\end{align*}

Then for any $t\in [t_0,t_1]$, we have

\begin{align*}
\left\vert z(t)-Z(t) \right\vert \leq C\nu + C\varepsilon_0.
\end{align*}

The proof is exactly the same as the one of Proposition <ref>. We emphasize that at the final time $t_1$, $z(t_1)$ and $Z(t_1)$ have to be far enough from $Z(0)$. From (<ref>), we indeed have

\begin{align*}
z(t_1) \geq Z(t_1) - C_\star \left( \nu_6^\star + \varepsilon_0^\star \right) \geq Z(0) + \frac{3}{4} \eta^2,
\end{align*}

and the rest of the arguments are similar.

§ ACKNOWLEDGEMENTS

The authors were supported by a Trond Mohn Forskningsstiftelse (TMF) grant. The authors would like to warmly thank Yvan Martel for many insightful conversations and for his constant encouragements.

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corresponds to the CP transformation $X_{0}$ that is of the following form in
the representations ${\bf 1}$ and ${\bf 3}$

$X_{0}({\bf 1})=1\;\;\mbox{and}\;\;X_{0}({\bf
3})=\left(\begin{array}[]{ccc}1&0&0\\\ 0&0&1\\\ 0&1&0\end{array}\right)\;.$
(150)

All other CP transformations $X$ of interest correspond to the composition of
the automorphism in eq. (149) and a group transformation $g$. The CP
transformation $X(\mathrm{{\bf r}})$ in the representation $\mathrm{{\bf r}}$
is of the form

$X(\mathrm{{\bf r}})=g(\mathrm{{\bf r}})\,X_{0}(\mathrm{{\bf
r}})\;\;\mbox{with}\;\;g(\mathrm{{\bf r}})=a(\mathrm{{\bf
r}})^{\alpha}\,c(\mathrm{{\bf r}})^{\gamma}\,d(\mathrm{{\bf
r}})^{\delta}\;\;\mbox{and}\;\;g(\mathrm{{\bf r}})=a(\mathrm{{\bf
r}})^{\alpha}\,b(\mathrm{{\bf r}})^{\beta}\,c(\mathrm{{\bf
r}})^{\gamma}\,d(\mathrm{{\bf r}})^{\delta}$ (151)

for $G_{f}=\Delta(3\,n^{2})$ and $G_{f}=\Delta(6\,n^{2})$, respectively, as
long as $X(\mathrm{{\bf r}})$ represents a symmetric matrix in flavour space,
see eq. (1). The CP symmetries and transformations relevant for the different
cases, Case 1) through Case 3 b.1), were given in section 3.

##  Appendix B Conventions of mixing angles, CP invariants and neutrino masses

We follow the conventions of the PDG in the parametrisation of a unitary
mixing matrix ($W$) in terms of the lepton mixing angles and the Dirac phase
$\delta$ [56]

$W=\begin{pmatrix}c_{12}c_{13}&s_{12}c_{13}&s_{13}e^{-i\delta}\\\
-s_{12}c_{23}-c_{12}s_{23}s_{13}e^{i\delta}&c_{12}c_{23}-s_{12}s_{23}s_{13}e^{i\delta}&s_{23}c_{13}\\\
s_{12}s_{23}-c_{12}c_{23}s_{13}e^{i\delta}&-c_{12}s_{23}-s_{12}c_{23}s_{13}e^{i\delta}&c_{23}c_{13}\end{pmatrix}$
(152)

with $s_{ij}=\sin\theta_{ij}$ and $c_{ij}=\cos\theta_{ij}$, while we define
the Majorana phases $\alpha$ and $\beta$ through

$P=\left(\begin{array}[]{ccc}1&0&0\\\ 0&e^{i\alpha/2}&0\\\
0&0&e^{i(\beta/2+\delta)}\end{array}\right)$ (153)

so that

$U_{\mathrm{PMNS}}=\left(\begin{array}[]{ccc}U_{e1}&U_{e2}&U_{e3}\\\ U_{\mu
1}&U_{\mu 2}&U_{\mu 3}\\\ U_{\tau 1}&U_{\tau 2}&U_{\tau
3}\end{array}\right)=W\,P$ (154)

with $0\leq\theta_{ij}\leq\pi/2$ and $0\leq\alpha,\beta,\delta\leq 2\,\pi$. We
extract the sine squares of the lepton mixing angles as follows

$\sin^{2}\theta_{13}=|U_{e3}|^{2}\;,\;\;\sin^{2}\theta_{12}=\frac{|U_{e2}|^{2}}{1-|U_{e3}|^{2}}\;,\;\;\sin^{2}\theta_{23}=\frac{|U_{\mu
3}|^{2}}{1-|U_{e3}|^{2}}\,.$ (155)

The CP phases are most conveniently extracted with the help of the CP
invariants $J_{\mathrm{CP}}$ [57], $I_{1}$ and $I_{2}$ [58]

$J_{\mathrm{CP}}={\rm Im}\left[U_{e1}U_{e3}^{\star}U_{\tau 1}^{\star}U_{\tau
3}\right]=\frac{1}{8}\,\sin 2\theta_{12}\,\sin 2\theta_{23}\,\sin
2\theta_{13}\,c_{13}\,\sin\delta$ (156)

and

$I_{1}={\rm
Im}[U_{e2}^{2}(U_{e1}^{\star})^{2}]=s^{2}_{12}\,c^{2}_{12}\,c^{4}_{13}\,\sin\alpha\;,\;\;I_{2}={\rm
Im}[U_{e3}^{2}(U_{e1}^{\star})^{2}]=s^{2}_{13}\,c^{2}_{12}\,c^{2}_{13}\,\sin\beta\;.$
(157)

From these, $\sin\delta$, $\sin\alpha$ and $\sin\beta$ can be computed.

The light neutrino masses $m_{i}$, $i=1,2,3$, can be expressed in terms of the
lightest neutrino mass $m_{0}$ and the two measured mass squared differences
$\Delta m_{\mathrm{sol}}^{2}$ and $\Delta m_{\mathrm{atm}}^{2}$, which are
defined as

$\Delta m_{\mathrm{sol}}^{2}=m_{2}^{2}-m_{1}^{2}\;\;\;\mbox{and}\;\;\;\Delta
m_{\mathrm{atm}}^{2}=\left\\{\begin{array}[]{cc}m_{3}^{2}-m_{1}^{2}&\mbox{for
NO}\\\ m_{3}^{2}-m_{2}^{2}&\mbox{for IO}\end{array}\right.\;,$ (158)

depending on the light neutrino mass ordering, NO or IO. For NO, the light
neutrino masses $m_{i}$ read

$m_{1}=m_{0}\;\;,\;\;\;m_{2}=\sqrt{m_{0}^{2}+\Delta
m_{\mathrm{sol}}^{2}}\;\;,\;\;\;m_{3}=\sqrt{m_{0}^{2}+\Delta
m_{\mathrm{atm}}^{2}}\;,$ (159)

while for IO we have

$m_{1}=\sqrt{m_{0}^{2}+|\Delta m_{\mathrm{atm}}^{2}|-\Delta
m_{\mathrm{sol}}^{2}}\,,\;\;m_{2}=\sqrt{m_{0}^{2}+|\Delta
m_{\mathrm{atm}}^{2}|}\,,\;\;m_{3}=m_{0}\;.$ (160)

Furthermore, we define the sum of the light neutrino masses

$\Sigma_{\nu}=m_{1}+m_{2}+m_{3}\;,$ (161)

which is constrained by cosmological measurements.

##  Appendix C Data on lepton mixing parameters and neutrino masses

We use the latest global fit results from the NuFIT Collaboration, NuFIT 5.0
(July 2020) [36] (without SK atmospheric data). For the lepton mixing angles
we have
for NO

$\displaystyle\sin^{2}\theta_{13}=0.02221^{+0.00068}_{-0.00062}\;\;\;\;\;\;\;\;\,\,\mbox{and}\;\;\;0.02034\leq\sin^{2}\theta_{13}\leq
0.02430\;,$ (162)
$\displaystyle\sin^{2}\theta_{12}=0.304^{+0.013}_{-0.012}\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\,\mbox{and}\;\;\;0.269\leq\sin^{2}\theta_{12}\leq
0.343\;,$
$\displaystyle\sin^{2}\theta_{23}=0.570^{+0.018}_{-0.024}\;\;\;\,\;\;\;\;\;\;\;\;\;\;\;\;\mbox{and}\;\;\;0.407\leq\sin^{2}\theta_{23}\leq
0.618\;,$

and for IO

$\displaystyle\sin^{2}\theta_{13}=0.02240^{+0.00062}_{-0.00062}\;\;\;\;\;\;\;\;\,\,\mbox{and}\;\;\;0.02053\leq\sin^{2}\theta_{13}\leq
0.02436\;,$ (163)
$\displaystyle\sin^{2}\theta_{12}=0.304^{+0.013}_{-0.012}\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\,\mbox{and}\;\;\;0.269\leq\sin^{2}\theta_{12}\leq
0.343\;,$
$\displaystyle\sin^{2}\theta_{23}=0.575^{+0.017}_{-0.021}\;\;\;\,\;\;\;\;\;\;\;\;\;\;\;\;\mbox{and}\;\;\;0.411\leq\sin^{2}\theta_{23}\leq
0.621\;.$

The experimental constraint on the Dirac phase $\delta$ reads
for NO

$\delta=\left(195^{+51}_{-25}\right)^{\circ}\;\;\;\mbox{and}\;\;\;107^{\circ}\leq\delta\leq
403^{\circ}\;,$ (164)

and for IO

$\delta=\left(286^{+27}_{-32}\right)^{\circ}\;\;\;\mbox{and}\;\;\;192^{\circ}\leq\delta\leq
360^{\circ}\;\;\;\mbox{at the}\;\;3\,\sigma\;\;\mbox{level}.$ (165)

For NO the experimental results for the mass squared differences are

$\displaystyle\Delta
m_{\mathrm{sol}}^{2}=\left(7.42^{+0.21}_{-0.20}\right)\times
10^{-5}\,\mathrm{eV}^{2}\;\;\;\;\;\;\;\;\;\mbox{and}\;\;\;6.82\leq\frac{\Delta
m_{\mathrm{sol}}^{2}}{10^{-5}\,\mathrm{eV}^{2}}\leq 8.04\;,$ (166)
$\displaystyle\Delta
m_{\mathrm{atm}}^{2}=\left(2.514^{+0.028}_{-0.027}\right)\times
10^{-3}\,\mathrm{eV}^{2}\;\;\;\;\,\mbox{and}\;\;\;2.431\leq\frac{\Delta
m_{\mathrm{atm}}^{2}}{10^{-3}\,\mathrm{eV}^{2}}\leq 2.598\;,$

and for IO

$\displaystyle\Delta
m_{\mathrm{sol}}^{2}=\left(7.42^{+0.21}_{-0.20}\right)\times
10^{-5}\,\mathrm{eV}^{2}\;\;\;\;\;\;\;\;\;\;\,\,\mbox{and}\;\;\;6.82\leq\frac{\Delta
m_{\mathrm{sol}}^{2}}{10^{-5}\,\mathrm{eV}^{2}}\leq 8.04\;,$ (167)
$\displaystyle\Delta
m_{\mathrm{atm}}^{2}=\left(-2.497^{+0.028}_{-0.028}\right)\times
10^{-3}\,\mathrm{eV}^{2}\;\;\;\,\,\mbox{and}\;\;\;-2.583\leq\frac{\Delta
m_{\mathrm{atm}}^{2}}{10^{-3}\,\mathrm{eV}^{2}}\leq-2.412\;.$

(Note that NO is currently slightly preferred over IO by experimental data
with $\Delta\chi^{2}=\chi^{2}_{\mathrm{IO}}-\chi^{2}_{\mathrm{NO}}=2.7$.)

As limit on the sum of the light neutrino masses, we use the one given by the
Planck Collaboration in [41],

$\Sigma_{\nu}\leq 0.12\;\mathrm{eV}\;\;\mbox{corresponding
to}\;\;m_{0}\lesssim 0.04\,\mathrm{eV}$ (168)

for the lightest neutrino mass. The experimental limits from KATRIN [59] and
from the searches for $0\nu\beta\beta$ decay [42, 43, 44, 45, 46] do not pose
stronger constraints on the lightest neutrino mass $m_{0}$ than the ones
obtained from cosmology.

##  Appendix D Numerical treatment and fit procedure

Predicting the CP phases for the different choices of the CP symmetry requires
identifying the values of the free angle $\theta_{S}$ which lead to a set of
lepton mixing parameters in agreement with experimental data. The free angle
$\theta_{S}$ is fit by maximising the joint likelihood of the predictions for
$\sin^{2}\theta_{ij}$. As mentioned in appendix C, we refer to the combination
of experimental data (latest update on global fits) provided by NuFIT 5.0
[36]. In order to fit $\theta_{S}$ with high accuracy, and especially to take
into account the “double well” structure in $\sin^{2}\theta_{23}$, we
interpolate the numerical $\Delta\chi^{2}$ values (available at [36]), and
linearly extrapolate beyond the provided ranges to ensure a smooth behaviour
for arbitrary input values. The interpolated $\chi^{2}$ functions are then
transformed into probability distributions so that a global joint likelihood
of all relevant parameters ($\sin^{2}\theta_{ij}$ and, in the case of the
(3,3) ISS framework, also $\Delta m_{ij}^{2}$) can be constructed. To ensure
that the cosmological bound on the sum of light neutrino masses is respected,
a half-normal distribution (as a gaussian upper limit) is further included. We
first fit predictions for $U_{S}$ and therefore $\theta_{S}$ on the data for
$\sin^{2}\theta_{ij}$, from which we proceed to consider the model-dependent,
i.e. (3,3) ISS framework. This is done by maximising the joint likelihood
function using the migrad algorithm of the iminuit library [60]. Local maximum
likelihood estimators lying outside of the global $3\,\sigma$ region around
the (experimental) best-fit point are rejected.

Due to the peculiar structure and almost degenerate heavy states in the ISS
mass matrix, a large numerical precision ($\sim 100$ digits) is needed for a
reliable matrix diagonalisation. This is achieved using the mpmath python
library [61] and algorithms within. To study the effects of the heavy sterile
states on the predictions for lepton mixing parameters, we use “effective”
mixing angles (and phases) which we define as in appendix B (eqs.
(155,156,157), with $U_{\text{PMNS}}$ replaced by $\widetilde{U}_{\nu}$). The
free angle $\theta_{S}$ then needs to be re-fitted, using the results obtained
within the model-independent approach as starting values for the fit, thus
allowing to study deviations from those predictions.

Keeping the lightest neutrino mass $m_{0}$ fixed - and thus the lightest
Majorana mass - ($\mu_{1}$ or $\mu_{3}$ depending on the ordering of the light
neutrino mass spectrum), the remaining two Majorana masses $\mu_{i}$ are
treated as free parameters to be determined by a fit to $\sin^{2}\theta_{ij}$
and $\Delta m_{ij}^{2}$ data. The starting values for $\mu_{i}$ are determined
by inverting the leading order expression given in eq. (17) and a modified
Casas-Ibarra parametrisation [62]

$\mu_{S}\,\simeq\,M_{NS}^{T}\,m_{D}^{-1}\,U^{\star}_{S}\,m^{\text{diag}}_{\nu}\,U^{\dagger}_{S}\,{m_{D}^{T}}^{-1}\,M_{NS}\,,$
(169)

where $U_{S}$, the matrix which diagonalises $\mu_{S}$ and at leading order
also the light neutrino mass matrix, is determined by the flavour symmetry
$G_{f}$ and CP and the residual symmetry $G_{\nu}$. For choices of $y_{0}$ and
$M_{0}$ in conflict with bounds on the unitarity of $\widetilde{U}_{\nu}$,
sizeable departures from the model-independent results are observed, as
described in the main body of the paper.

##  Appendix E Loop functions

The loop functions and their relevant limits212121Note that in Ref. [49] the
loop function $F_{\text{Xbox}}$ is named $F_{\text{box}}$ and has an opposite
global sign compared to Ref. [47], which also reflects in the form factor
$F_{\text{box}}^{\beta 3\alpha}$. are taken from Refs. [47, 49]. The photon
dipole and anapole functions, as well as some relevant limits, are given by

$\displaystyle F_{\gamma}(x)$ $\displaystyle=$
$\displaystyle\frac{7x^{3}-x^{2}-12x}{12(1-x)^{3}}-\frac{x^{4}-10x^{3}+12x^{2}}{6(1-x)^{4}}\log
x\,,$ $\displaystyle F_{\gamma}(x)$ $\displaystyle\xrightarrow[x\gg 1]{}$
$\displaystyle-\frac{7}{12}-\frac{1}{6}\log x\,,$ $\displaystyle
F_{\gamma}(0)$ $\displaystyle=$ $\displaystyle 0\,,$ (170) $\displaystyle
G_{\gamma}(x)$ $\displaystyle=$
$\displaystyle-\frac{x(2x^{2}+5x-1)}{4(1-x)^{3}}-\frac{3x^{3}}{2(1-x)^{4}}\log
x\,,$ $\displaystyle G_{\gamma}(x)$ $\displaystyle\xrightarrow[x\gg 1]{}$
$\displaystyle\frac{1}{2}\,,$ $\displaystyle G_{\gamma}(0)$ $\displaystyle=$
$\displaystyle 0\,.$ (171)

The functions associated with the $Z$ penguins are given by a two-point
function

$\displaystyle F_{Z}(x)$ $\displaystyle=$
$\displaystyle-\frac{5x}{2(1-x)}-\frac{5x^{2}}{2(1-x)^{2}}\log x\,,$
$\displaystyle F_{Z}(x)$ $\displaystyle\xrightarrow[x\gg 1]{}$
$\displaystyle\frac{5}{2}-\frac{5}{2}\log x\,,$ $\displaystyle F_{Z}(0)$
$\displaystyle=$ $\displaystyle 0\,,$ (172)

and two three-point functions which are symmetric under interchange of the
arguments.

$\displaystyle G_{Z}(x,y)$ $\displaystyle=$
$\displaystyle-\frac{1}{2(x-y)}\left[\frac{x^{2}(1-y)}{1-x}\log
x-\frac{y^{2}(1-x)}{1-y}\log y\right]\,,$ $\displaystyle G_{Z}(x,x)$
$\displaystyle=$ $\displaystyle-\frac{x}{2}-\frac{x\log x}{1-x}\,,$
$\displaystyle G_{Z}(0,x)$ $\displaystyle=$ $\displaystyle-\frac{x\log
x}{2(1-x)}\,,$ $\displaystyle G_{Z}(0,x)$ $\displaystyle\xrightarrow[x\gg
1]{}$ $\displaystyle\frac{1}{2}\log x\,,$ $\displaystyle G_{Z}(0,0)$
$\displaystyle=$ $\displaystyle 0\,,$ (173) $\displaystyle H_{Z}(x,y)$
$\displaystyle=$
$\displaystyle\frac{\sqrt{xy}}{4(x-y)}\left[\frac{x^{2}-4x}{1-x}\log
x-\frac{y^{2}-4y}{1-y}\log y\ \right]\,,$ $\displaystyle H_{Z}(x,x)$
$\displaystyle=$
$\displaystyle\frac{(3-x)(1-x)-3}{4(1-x)}-\frac{x^{3}-2x^{2}+4x}{4(1-x)^{2}}\log
x\,,$ $\displaystyle H_{Z}(0,x)$ $\displaystyle=$ $\displaystyle 0\,.$ (174)

The (symmetric) box-loop-functions and their limits are given by

$\displaystyle F_{\text{box}}(x,y)$ $\displaystyle=$
$\displaystyle\frac{1}{x-y}\left\\{\left(4+\frac{xy}{4}\right)\left[\frac{1}{1-x}+\frac{x^{2}}{(1-x)^{2}}\log
x-\frac{1}{1-y}-\frac{y^{2}}{(1-y)^{2}}\log y\right]\right.$
$\displaystyle\left.-2xy\left[\frac{1}{1-x}+\frac{x}{(1-x)^{2}}\log
x-\frac{1}{1-y}-\frac{y}{(1-y)^{2}}\log y\right]\right\\}\,,$ $\displaystyle
F_{\text{box}}(x,x)$ $\displaystyle=$
$\displaystyle-\frac{1}{4(1-x)^{3}}\left[x^{4}-16x^{3}+31x^{2}-16+2x\left(3x^{2}+4x-16\right)\log
x\right]\,,$ $\displaystyle F_{\text{box}}(0,x)$ $\displaystyle=$
$\displaystyle\frac{4}{1-x}+\frac{4x}{(1-x)^{2}}\log x\,,$ $\displaystyle
F_{\text{box}}(0,x)$ $\displaystyle\xrightarrow[x\gg 1]{}$ $\displaystyle
0\,,$ $\displaystyle F_{\text{box}}(0,0)$ $\displaystyle=$ $\displaystyle
4\,,$ (175) $\displaystyle F_{\text{Xbox}}(x,y)$ $\displaystyle=$
$\displaystyle-\frac{1}{x-y}\left\\{\left(1+\frac{xy}{4}\right)\left[\frac{1}{1-x}+\frac{x^{2}}{(1-x)^{2}}\log
x-\frac{1}{1-y}-\frac{y^{2}}{(1-y)^{2}}\log y\right]\right.$
$\displaystyle\left.-2xy\left[\frac{1}{1-x}+\frac{x}{(1-x)^{2}}\log
x-\frac{1}{1-y}-\frac{y}{(1-y)^{2}}\log y\right]\right\\}\,,$ $\displaystyle
F_{\text{Xbox}}(x,x)$ $\displaystyle=$
$\displaystyle\frac{x^{4}-16x^{3}+19x^{2}-4}{4(1-x)^{3}}+\frac{3x^{3}+4x^{2}-4x}{2(1-x)^{3}}\log
x\,,$ $\displaystyle F_{\text{Xbox}}(0,x)$ $\displaystyle=$
$\displaystyle-\frac{1}{1-x}-\frac{x}{(1-x)^{2}}\log x\,,$ $\displaystyle
F_{\text{Xbox}}(0,x)$ $\displaystyle\xrightarrow[x\gg 1]{}$ $\displaystyle
0\,,$ $\displaystyle F_{\text{Xbox}}(0,0)$ $\displaystyle=$
$\displaystyle-1\,,$ (176) $\displaystyle G_{\text{box}}(x,y)$
$\displaystyle=$
$\displaystyle-\frac{\sqrt{xy}}{x-y}\left\\{(4+xy)\left[\frac{1}{1-x}+\frac{x}{(1-x)^{2}}\log
x-\frac{1}{1-y}-\frac{y}{(1-y)^{2}}\log y\right]\right.$
$\displaystyle\left.-2\left[\frac{1}{1-x}+\frac{x^{2}}{(1-x)^{2}}\log
x-\frac{1}{1-y}-\frac{y^{2}}{(1-y)^{2}}\log y\right]\right\\}\,,$
$\displaystyle G_{\text{box}}(x,x)$ $\displaystyle=$
$\displaystyle\frac{2x^{4}-4x^{3}+8x^{2}-6x}{(1-x)^{3}}-\frac{x^{4}+x^{3}+4x}{(1-x)^{3}}\log
x\,,$ $\displaystyle G_{\text{box}}(0,x)$ $\displaystyle=$ $\displaystyle
0\,.$ (177)

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