with $g(t)$ satisfying orthogonality conditions of the Modulation Lemma and

$\left\lVert(g(t),\partial_{t}g(t))\right\rVert_{H^{1}\times L^{2}}^{2}\leq
C\left[\left\lVert(g(0,x),\partial_{t}g(0,x))\right\rVert_{H^{1}\times
L^{2}}^{2}+\epsilon^{2}\ln{\Big{(}\frac{1}{\epsilon}\Big{)}}^{2}\right]\exp\left(C|t|\left(\frac{\epsilon^{\frac{1}{2}}}{\ln{(\frac{1}{\epsilon})}}\right)\right).$
(188)

###### Proof of Theorem 4.5.

In notation of Theorem 4.1, from Theorem 4.1 and Remark 4.3, there are uniform
positive constants $A_{2},\,A_{1}$ such that for all $t\geq 0$

$A_{2}\left\lVert\overrightarrow{g(t)}\right\rVert^{2}\leq
F(t)+A_{1}\epsilon^{2}\leq
C\Big{(}\left\lVert\overrightarrow{g(t)}\right\rVert^{2}+\epsilon^{2}\Big{)}.$
(189)

From now on, we denote
$G(t)\coloneqq|F(t)|+A_{1}\epsilon^{2}\ln{(\frac{1}{\epsilon})}^{2}.$ From the
inequality (189) and Remark 4.2, there is a constant $C>0$ such that, for all
$t\geq 0,$ $G(t)$ satisfies

$G(t)\leq
G(0)+C\Big{(}\int_{0}^{t}G(s)\frac{\epsilon^{\frac{1}{2}}}{\ln{(\frac{1}{\epsilon})}},ds\Big{)}.$
(190)

In conclusion, from the Fundamental Theorem of Calculus, we obtain that
$G(t)\leq
G(0)\exp\Big{(}\frac{Ct\epsilon^{\frac{1}{2}}}{\ln{(\frac{1}{\epsilon})}}\Big{)}.$
Then, from the definition of $G$ and inequality (189), we verify the
inequality (188). ∎

##  5 Global Dynamics of Modulation Parameters

######  Lemma 5.1.

In notation of Theorem 1.7, $\exists C>0,$ such that if the hypotheses of
Theorem 1.7 are true, then for
$(g_{0}(x),g_{1}(x))=\left(g(0,x),\partial_{t}g(0,x)\right)$ we have that
there are functions $p_{1}(t),\,p_{2}(t)\in C^{1}(R_{\geq 0}),$ such that for
$j\in\\{1,\,2\\},$ we have:

$|\dot{x}_{j}(t)-p_{j}(t)|\lesssim\Big{(}\left\lVert(g_{0},g_{1})\right\rVert_{H^{1}\times
L^{2}}+\epsilon\ln{\Big{(}\frac{1}{\epsilon}\Big{)}}\Big{)}\epsilon^{\frac{1}{2}}\exp\Big{(}\frac{2C\epsilon^{\frac{1}{2}}t}{\ln{(\frac{1}{\epsilon})}}\Big{)},$
(191)
$\Big{|}\dot{p}_{j}(t)-(-1)^{j}8\sqrt{2}e^{-\sqrt{2}z(t)}\Big{|}\lesssim\frac{\Big{(}\left\lVert(g_{0},g_{1})\right\rVert_{H^{1}\times
L^{2}}+\epsilon\ln{\Big{(}\frac{1}{\epsilon}\Big{)}}\Big{)}^{2}}{\ln{\ln{(\frac{1}{\epsilon})}}}\exp\Big{(}\frac{2Ct\epsilon^{\frac{1}{2}}}{\ln{(\frac{1}{\epsilon})}}\Big{)}.$
(192)

###### Proof.

In the notation of Lemma 3.1, we consider the functions $p_{j}(t)$ for
$j\in\\{1,\,2\\}$ and we consider $\theta=\frac{1-\gamma}{2-\gamma},$ the
value of $\gamma$ will be chosen later. From Lemma 3.1, we have that

$|\dot{x}_{j}(t)-p_{j}(t)|\lesssim\Big{[}1+\frac{1}{\gamma
z(t)}\Big{]}\Big{(}\max_{j\in\\{1,2\\}}|\dot{x}_{j}(t)|\left\lVert\overrightarrow{g(t)}\right\rVert+\left\lVert\overrightarrow{g(t)}\right\rVert^{2}\Big{)}+\max_{j\in\\{1,2\\}}|\dot{x}_{j}(t)|z(t)e^{-\sqrt{2}z(t)}.$

We recall from Theorem 2.9 the estimates
$\max_{j\in\\{1,\,2\\}}|\dot{x}_{j}(t)|=O(\epsilon^{\frac{1}{2}}),\,e^{-\sqrt{2}z(t)}=O(\epsilon).$
From Theorem 4.5, we have that

$\left\lVert\overrightarrow{g(t)}\right\rVert\lesssim\left(\left\lVert\overrightarrow{g(0)}\right\rVert+\epsilon\ln{\Big{(}\frac{1}{\epsilon}\Big{)}}\right)\exp\Big{(}\frac{C\epsilon^{\frac{1}{2}}t}{\ln{(\frac{1}{\epsilon})}}\Big{)}.$

To simplify our computations we denote
$c_{0}=\frac{\left\lVert\overrightarrow{g(0)}\right\rVert+\epsilon\ln(\frac{1}{\epsilon})}{\epsilon\ln(\frac{1}{\epsilon})}.$
Then, we obtain for $j\in\\{1,\,2\\}$ that

$|\dot{x}_{j}(t)-p_{j}(t)|\lesssim\left[1+\frac{1}{\gamma\ln{(\frac{1}{\epsilon})}}\right]\Big{(}c_{0}\epsilon^{\frac{3}{2}}\ln{\Big{(}\frac{1}{\epsilon}\Big{)}}\exp\Big{(}\frac{C\epsilon^{\frac{1}{2}}t}{\ln{(\frac{1}{\epsilon})}}\Big{)}+c_{0}^{2}\epsilon^{2}\ln{\Big{(}\frac{1}{\epsilon}\Big{)}}^{2}\exp\Big{(}\frac{2C\epsilon^{\frac{1}{2}}t}{\ln{\frac{1}{\epsilon}}}\Big{)}\Big{)}.$
(193)

Since $e^{-\sqrt{2}z(t)}\lesssim\epsilon,$ we deduce for $\epsilon\ll 1$ that
$z(t)e^{-\sqrt{2}z(t)}\lesssim\epsilon\ln{(\frac{1}{\epsilon})}<\epsilon^{1-\frac{\gamma}{(2-\gamma)2}}\ln{(\frac{1}{\epsilon})}.$
Then, we obtain from the same estimates and the definition (105) of
$\alpha(t),$ that

$\alpha(t)\lesssim
c_{0}^{2}\Big{(}\epsilon\ln{\Big{(}\frac{1}{\epsilon}\Big{)}}\Big{)}^{2}\left[\max_{k\in\\{1,\,2\\}}\Big{(}\frac{1}{\gamma
z(t)}\Big{)}^{k}+\epsilon^{\frac{1-\gamma}{2-\gamma}}\right]\exp\Big{(}2\frac{C\epsilon^{\frac{1}{2}}t}{\ln{(\frac{1}{\epsilon})}}\Big{)}\\\
+c_{0}\epsilon^{2-\frac{\gamma}{(2-\gamma)2}}\ln{\Big{(}\frac{1}{\epsilon}\Big{)}}\exp\Big{(}\frac{C\epsilon^{\frac{1}{2}}t}{\ln{(\frac{1}{\epsilon})}}\Big{)}\left[1+\frac{1}{\gamma
z(t)}+\frac{\epsilon^{\frac{1}{2}}}{(\gamma
z(t))^{2}}\right]+\frac{\epsilon^{1+\frac{2(1-\gamma)}{2-\gamma}}}{z(t)\gamma}.$
(194)

However, if $\gamma\ln{(\frac{1}{\epsilon})}\leq 1$ and
$z(0)\cong\ln{(\frac{1}{\epsilon})},$ which is possible, then the right-hand
side of inequality (194) is greater than or equivalent to
$\epsilon^{2}\ln{(\frac{1}{\epsilon})}^{2}$while
$t\lesssim\frac{\ln{(\frac{1}{\epsilon})}}{\epsilon^{\frac{1}{2}}}.$ But, it
is not difficult to verify that for
$\gamma=\frac{\ln{\ln{(\frac{1}{\epsilon})}}}{\ln{(\frac{1}{\epsilon})}},$ the
right-hand side of inequality (194) is smaller than
$\epsilon^{2}\ln{(\frac{1}{\epsilon})}^{2}$. In conclusion, from now on, we
are going to study the right-hand side of (194) for
$\frac{1}{\ln(\frac{1}{\epsilon})}<\gamma<1$. Since we know that
$\ln{(\frac{1}{\epsilon})}\lesssim z(t)$ from Theorem 2.9, the inequality
(194) implies for $\frac{1}{\ln{(\frac{1}{\epsilon})}}<\gamma<1$ that

$\alpha(t)\lesssim\beta(t)\coloneqq\Big{(}c_{0}\epsilon\ln{\Big{(}\frac{1}{\epsilon}\Big{)}}\Big{)}^{2}\left[\frac{1}{\gamma\ln{(\frac{1}{\epsilon}})}+\epsilon^{\frac{1-\gamma}{2-\gamma}}\right]\exp\Big{(}2\frac{C\epsilon^{\frac{1}{2}}t}{\ln{(\frac{1}{\epsilon})}}\Big{)}+c_{0}\epsilon^{2-\frac{\gamma}{2(2-\gamma)}}\ln{\Big{(}\frac{1}{\epsilon}\Big{)}}\exp\Big{(}\frac{C\epsilon^{\frac{1}{2}}t}{\ln{(\frac{1}{\epsilon})}}\Big{)}+\frac{\epsilon^{1+\frac{2(1-\gamma)}{2-\gamma}}}{\gamma\ln{(\frac{1}{\epsilon})}}\\\
=\beta_{1}(t)+\beta_{2}(t)+\beta_{3}(t),\text{ respectively.}$ (195)

For $\epsilon>0$ small enough, it is not difficult to verify that if
$\beta_{3}(t)\geq\beta_{1}(t),$ then
$\gamma\geq\frac{\ln{\ln{(\frac{1}{\epsilon})}}}{\ln{(\frac{1}{\epsilon})}}.$
Moreover, if we have that
$1>\gamma>8\frac{\ln{\ln{(\frac{1}{\epsilon})}}}{\ln{(\frac{1}{\epsilon})}},$
we obtain from the following estimate

$\beta_{3}(t)=\frac{\epsilon^{2}\epsilon^{\frac{-\gamma}{2-\gamma}}}{\gamma\ln{(\frac{1}{\epsilon})}}>\frac{\epsilon^{2}}{\ln{(\frac{1}{\epsilon})}}\exp\left(\frac{8\ln{\ln\left(\frac{1}{\epsilon}\right)}}{2-\gamma}\right)=\frac{\epsilon^{2}}{\ln{(\frac{1}{\epsilon})}}\ln{\left(\frac{1}{\epsilon}\right)}^{\frac{8}{2-\gamma}},$

that
$\beta_{3}(t)>\frac{\epsilon^{2}\ln{(\frac{1}{\epsilon})}^{2}}{\ln{\ln{(\frac{1}{\epsilon})}}}.$
If
$\gamma\leq\frac{\ln{\ln{(\frac{1}{\epsilon})}}}{\ln{(\frac{1}{\epsilon})}},$
then
$\frac{\epsilon^{2}\ln{(\frac{1}{\epsilon})}^{2}}{\ln{\ln{(\frac{1}{\epsilon})}}}\lesssim\beta_{1}(t).$
In conclusion, for any case we have that
$\frac{\epsilon^{2}\ln{(\frac{1}{\epsilon})}^{2}}{\ln{\ln{(\frac{1}{\epsilon})}}}\lesssim\beta(t),$
so we choose
$\gamma=\frac{\ln{\ln{(\frac{1}{\epsilon})}}}{\ln{(\frac{1}{\epsilon})}}.$ As
a consequence, $\alpha(t)$ is less than or equivalent to

$c_{0}^{2}\epsilon^{2}\frac{\ln{(\frac{1}{\epsilon}})^{2}}{\ln{\ln{(\frac{1}{\epsilon})}}}\exp\Big{(}\frac{2C\epsilon^{\frac{1}{2}}t}{\ln{(\frac{1}{\epsilon})}}\Big{)}.$
(196)

So, the estimates (193), (196), Remark 3.3 and our choice of $\gamma$ imply
the inequalities (191) and (192). ∎

######  Remark 5.2.

If
$\frac{\epsilon^{\frac{1}{2}}}{\ln{(\frac{1}{\epsilon})}^{m}}\lesssim\left\lVert\overrightarrow{g(0)}\right\rVert$
for a constant $m>0,$ then, for $\gamma=\frac{1}{8},$ we have from Lemma 3.1
that there is $p(t)\in C^{2}(\mathbb{R})$ such that for all $t\geq 0$

$\displaystyle|\dot{z}(t)-p(t)|\lesssim\epsilon^{\frac{1}{2}}\left\lVert\overrightarrow{g(0)}\right\rVert,$
(197)
$\displaystyle|\dot{p}(t)-16\sqrt{2}e^{-\sqrt{2}z(t)}|\lesssim\frac{\left\lVert\overrightarrow{g(0)}\right\rVert^{2}}{z(t)}.$
(198)

Then, for the smooth real function $d(t)$ satisfying

$\ddot{d}(t)=16\sqrt{2}e^{-\sqrt{2}d(t)},\,(d(0),\dot{d}(0))=(z(0),\dot{z}(0)),$

and since
$e^{-\sqrt{2}z(t)}\lesssim\epsilon,\,\ln{(\frac{1}{\epsilon})}\lesssim z(t),$
we can deduce that $Y(t)=(z(t)-d(t))$ satisfies the following integral
inequality for a constant $K>0$

$|Y(t)|\leq
K\left(\epsilon^{\frac{1}{2}}\left\lVert\overrightarrow{g(0)}\right\rVert
t+\frac{\left\lVert\overrightarrow{g(0)}\right\rVert^{2}}{\ln{(\frac{1}{\epsilon})}}t^{2}+\int_{0}^{t}\int_{0}^{s}\epsilon|Y(s_{1})|\,ds_{1}\,ds\right),\\\
Y(0)=0,\,\dot{Y}(0)=0.$ (199)

In conclusion, from the Gronwall Lemma, we obtain that $|Y(t)|\lesssim
Q(tK^{\frac{1}{2}}),$ where $Q(t)$ is the solution of the following integral
equation

$Q(t)=\epsilon^{\frac{1}{2}}\left\lVert\overrightarrow{g(0)}\right\rVert
t+\frac{\left\lVert\overrightarrow{g(0)}\right\rVert^{2}}{\ln{(\frac{1}{\epsilon})}}t^{2}+\int_{0}^{t}\int_{0}^{s}\epsilon
Q(s_{1})\,ds_{1}\,ds.$

By standard ordinary differential equation techniques, we deduce that

$|z(t)-d(t)|\lesssim
Q(tK^{\frac{1}{2}})=\left(\frac{\left\lVert\overrightarrow{g(0)}\right\rVert}{2}+\frac{\left\lVert\overrightarrow{g(0)}\right\rVert^{2}}{\epsilon\ln{(\frac{1}{\epsilon})}}\right)e^{\epsilon^{\frac{1}{2}}tK^{\frac{1}{2}}}+\left(\frac{-\left\lVert\overrightarrow{g(0)}\right\rVert}{2}+\frac{\left\lVert\overrightarrow{g(0)}\right\rVert^{2}}{\epsilon\ln{(\frac{1}{\epsilon})}}\right)e^{-\epsilon^{\frac{1}{2}}tK^{\frac{1}{2}}}-2\frac{\left\lVert\overrightarrow{g(0)}\right\rVert^{2}}{\epsilon\ln{(\frac{1}{\epsilon})}},$
(200)

and from $\dot{z}(0)=\dot{d}(0)$ and the estimates (197) and (198), we obtain
that

$|\dot{z}(t)-\dot{d}(t)|\lesssim|p(0)-\dot{z}(0)|+\int_{0}^{t}\epsilon|z(s)-d(s)|\,ds,$
(201)

from which with (200), we obtain that

$|\dot{z}(t)-\dot{d}(t)|\lesssim
e^{\epsilon^{\frac{1}{2}}|t|K^{\frac{1}{2}}}\epsilon^{\frac{1}{2}}\left(\left\lVert\overrightarrow{g(0)}\right\rVert+\frac{\left\lVert\overrightarrow{g(0)}\right\rVert^{2}}{\epsilon\ln{(\frac{1}{\epsilon})}}\right).$
(202)

However, the precision of the estimates (200) and (202) is very bad when
$\epsilon^{-\frac{1}{2}}\ll t,$ which motivate us to apply Lemma (3.1) to
estimate the modulations parameters $x_{1}(t),\,x_{2}(t)$ for
$t\lesssim\frac{\ln{(\frac{1}{\epsilon})}}{\epsilon^{\frac{1}{2}}}.$

We recall from Theorem 1.11 the definitions of the functions
$d_{1}(t),\,d_{2}(t).$ If
$\left\lVert\overrightarrow{g(0)}\right\rVert\geq\frac{\epsilon^{\frac{1}{2}}}{\ln{(\frac{1}{\epsilon})}^{5}},$
because Theorem 2.9 and
$\max_{j\in\\{1,2\\}}|\dot{d}_{j}(0)-\dot{x}_{j}(0)|=0$ imply that
$\max_{j\in\\{1,\,2\\}}|d_{j}(t)-x_{j}(t)|=O(\min(\epsilon
t,\epsilon^{\frac{1}{2}}t)),\,\max_{j\in\\{1,\,2\\}}|\dot{d}_{j}(t)-\dot{x}_{j}(t)|=O(\epsilon
t),$ we deduce for a constant $C>0$ large enough the estimates (9) and (10) of
Theorem 1.11. For the case
$\left\lVert\overrightarrow{g(0)}\right\rVert\leq\frac{\epsilon^{\frac{1}{2}}}{\ln{(\frac{1}{\epsilon})}^{5}}$,
the estimates of
$\max_{j\in\\{1,2\\}}|x_{j}(t)-d_{j}(t)|,\,\max_{j\in\\{1,\,2\\}}|\dot{x}_{j}(t)-\dot{d}_{j}(t)|$
will be done by studying separated cases depending on the initial data
$z(0),\,\dot{z}(0).$

######  Lemma 5.3.

$\exists K>0$ such that if
$\left\lVert\overrightarrow{g(0)}\right\rVert\leq\frac{\epsilon^{\frac{1}{2}}}{\ln{(\frac{1}{\epsilon})}^{5}},\,(g_{0}(x),g_{1}(x))=\left(g(0,x),\partial_{t}g(0,x)\right)$
and all the hypotheses of Theorem 1.11 are true and
$\frac{\epsilon}{\ln{(\frac{1}{\epsilon})}^{8}}\lesssim
e^{-\sqrt{2}z(0)}\lesssim\epsilon,$ then we have for $0\leq t$ that

$\displaystyle\max_{j\in\\{1,\,2\\}}|x_{j}(t)-d_{j}(t)|=O\left(\frac{\max\Big{(}\left\lVert(g_{0},g_{1})\right\rVert,\epsilon\ln{(\frac{1}{\epsilon})}\Big{)}^{2}\ln{(\frac{1}{\epsilon})}^{6}}{\epsilon\ln{\ln(\frac{1}{\epsilon})}}\exp\Big{(}\frac{K\epsilon^{\frac{1}{2}}t}{\ln{(\frac{1}{\epsilon})}}\Big{)}\right),$
(203)
$\displaystyle\max_{j\in\\{1,\,2\\}}|\dot{x}_{j}(t)-\dot{d}_{j}(t)|=O\left(\max\Big{(}\left\lVert(g_{0},g_{1})\right\rVert,\epsilon\ln{\Big{(}\frac{1}{\epsilon}\Big{)}}\Big{)}^{2}\frac{\ln{(\frac{1}{\epsilon})}^{6}}{\epsilon^{\frac{1}{2}}\ln{\ln{(\frac{1}{\epsilon})}}}\exp\Big{(}\frac{K\epsilon^{\frac{1}{2}}t}{\ln{(\frac{1}{\epsilon})}}\Big{)}\right).$
(204)

###### Proof of Lemma 5.3.

First, in notation of Lemma 5.1, we define

$p(t)\coloneqq p_{2}(t)-p_{1}(t),\,z(t)\coloneqq
x_{2}(t)-x_{1}(t),\,\dot{z}(t)\coloneqq\dot{x}_{2}(t)-\dot{x}_{1}(t).$

Also, motivated by Remark 3.3, we consider the smooth function $d(t)$ solution
of the following ordinary differential equation

$\begin{cases*}\ddot{d}(t)=16\sqrt{2}e^{-\sqrt{2}d(t)},\\\
(d(0),\dot{d}(0))=(z(0),\dot{z}(0)).\end{cases*}$

Step 1.(Estimate of $z(t),\,\dot{z}(t)$) From now on, we denote the functions
$W(t)=z(t)-d(t),\,V(t)=p(t)-\dot{d}(t).$ Then, Lemma 5.1 implies that $W,\,V$
satisfy the following ordinary differential estimates

$\displaystyle|\dot{W}(t)-V(t)|=O\left(\max\Big{(}\left\lVert(g_{0},g_{1})\right\rVert,\epsilon\ln{\Big{(}\frac{1}{\epsilon}\Big{)}}\Big{)}\epsilon^{\frac{1}{2}}\exp\Big{(}\frac{2C\epsilon^{\frac{1}{2}}t}{\ln{(\frac{1}{\epsilon})}}\Big{)}\right),$
$\displaystyle\big{|}\dot{V}(t)+16\sqrt{2}e^{-\sqrt{2}d(t)}-16\sqrt{2}e^{-\sqrt{2}z(t)}\big{|}=O\left(\frac{\max\Big{(}\left\lVert(g_{0},g_{1})\right\rVert,\epsilon\ln{\Big{(}\frac{1}{\epsilon}\Big{)}}\Big{)}^{2}}{\ln{\ln{(\frac{1}{\epsilon})}}}\exp\Big{(}\frac{2C\epsilon^{\frac{1}{2}}t}{\ln{(\frac{1}{\epsilon})}}\Big{)}\right).$

From the above estimates and the Taylor’s Expansion Theorem, we deduce the
following almost ordinary differential system of equations, while $|W(t)|<1:$

$\begin{cases}\dot{W}(t)=V(t)+O\Big{(}\max\Big{(}\left\lVert(g_{0},g_{1})\right\rVert,\epsilon\ln{\Big{(}\frac{1}{\epsilon}\Big{)}}\Big{)}\epsilon^{\frac{1}{2}}\exp\Big{(}\frac{2C\epsilon^{\frac{1}{2}}t}{\ln{(\frac{1}{\epsilon})}}\Big{)}\Big{)},\\\
\dot{V}(t)=-32e^{-\sqrt{2}d(t)}W(t)+O\left(e^{-\sqrt{2}d(t)}W(t)^{2}\right)+O\left(\frac{\max\Big{(}\left\lVert(g_{0},g_{1})\right\rVert,\epsilon\ln{(\frac{1}{\epsilon})}\Big{)}^{2}}{\ln{\ln{(\frac{1}{\epsilon})}}}\exp\Big{(}\frac{2C\epsilon^{\frac{1}{2}}t}{\ln{(\frac{1}{\epsilon})}}\Big{)}\right).\end{cases}$

Recalling Remark 3.3, we have that

$d(t)=\frac{1}{\sqrt{2}}\ln{\Big{(}\frac{8}{v^{2}}\cosh{(\sqrt{2}vt+c)}^{2}\Big{)}},$
(205)

where $v>0$ and $c\in\mathbb{R}$ are chosen such that
$(d(0),\dot{d}(0))=(z(0),\dot{z}(0)).$ Moreover, it is not difficult to verify
that

$\displaystyle
v=\Big{(}\frac{\dot{z}(0)^{2}}{4}+8e^{-\sqrt{2}z(0)}\Big{)}^{\frac{1}{2}},\,c=\operatorname{arctanh}{\left(\frac{\dot{z}(0)}{\big{[}32e^{-\sqrt{2}z(0)}+\dot{z}(0)^{2}\big{]}^{\frac{1}{2}}}\right)}.$

Moreover, since $8e^{-\sqrt{2}z(0)}=v^{2}\operatorname{sech}{(c)}^{2}\leq
4v^{2}e^{-2|c|},$ we obtain from the hypothesis for $e^{-\sqrt{2}z(0)}$ that
$\frac{\epsilon^{\frac{1}{2}}}{\ln{(\frac{1}{\epsilon})}^{4}}\lesssim
v\lesssim\epsilon^{\frac{1}{2}}$ and as a consequence the estimate
$|c|\lesssim\ln{(\ln{(\frac{1}{\epsilon})})}.$ Also, it is not difficult to
verify that the functions

$n(t)=(\sqrt{2}vt+c)\tanh{(\sqrt{2}vt+c)}-1,\,m(t)=\tanh{(\sqrt{2}vt+c)}$

generate all solutions of the following ordinary differential equation

$\ddot{y}(t)=-32e^{-\sqrt{2}d(t)}y(t),$ (206)

which is obtained from the linear part of the system (5).

To simplify our computations we use the following notation

$\displaystyle
error_{1}(t)=\max\Big{(}\left\lVert(g_{0},g_{1})\right\rVert,\epsilon\ln{\Big{(}\frac{1}{\epsilon}\Big{)}}\Big{)}\epsilon^{\frac{1}{2}}\exp\Big{(}\frac{2C\epsilon^{\frac{1}{2}}t}{\ln{(\frac{1}{\epsilon})}}\Big{)},$
$\displaystyle
error_{2}(t)=e^{-\sqrt{2}d(t)}(z(t)-d(t))^{2}+\frac{\max\Big{(}\left\lVert(g_{0},g_{1})\right\rVert,\epsilon\ln{\Big{(}\frac{1}{\epsilon}\Big{)}}\Big{)}^{2}}{\ln{\ln{(\frac{1}{\epsilon})}}}\exp\Big{(}\frac{2C\epsilon^{\frac{1}{2}}t}{\ln{(\frac{1}{\epsilon})}}\Big{)}.$

From the variation of parameters technique for ordinary differential equation,
we can write that

$\begin{bmatrix}W(t)\\\ V(t)\end{bmatrix}=c_{1}(t)\begin{bmatrix}m(t)\\\
\dot{m}(t)\end{bmatrix}+c_{2}(t)\begin{bmatrix}n(t)\\\
\dot{n}(t)\end{bmatrix},$ (207)

such that

$\begin{cases}\begin{bmatrix}m(t)&n(t)\\\
\dot{m}(t)&\dot{n}(t)\end{bmatrix}\begin{bmatrix}\dot{c}_{1}(t)\\\
\dot{c}_{2}(t)\end{bmatrix}=\begin{bmatrix}O(error_{1}(t))\\\
O(error_{2}(t))\end{bmatrix},\\\ \\\ \begin{bmatrix}m(0)&n(0)\\\
\dot{m}(0)&\dot{n}(0)\end{bmatrix}\par\begin{bmatrix}c_{1}(0)\\\
c_{2}(0)\end{bmatrix}=\begin{bmatrix}0\\\
O\Big{(}\Big{[}\left\lVert(g_{0},g_{1})\right\rVert+\epsilon\ln{(\frac{1}{\epsilon})}\Big{]}\epsilon^{\frac{1}{2}}\Big{)}\end{bmatrix}.\end{cases}$

The presence of an error in the condition of the initial data
$c_{1}(0),\,c_{2}(0)$ comes from estimate (191) of Lemma 5.1. Since for all
$t\in\mathbb{R}$ $m(t)\dot{n}(t)-\dot{m}(t)n(t)=\sqrt{2}v$, we can verify by
Cramer’s rule and from the fact that
$\frac{\epsilon^{\frac{1}{2}}}{\ln{(\frac{1}{\epsilon})}^{4}}\lesssim v$ that

$\displaystyle
c_{1}(0)=O\left(\max\Big{(}\left\lVert\overrightarrow{g(0)}\right\rVert,\epsilon\ln{\Big{(}\frac{1}{\epsilon}\Big{)}}\Big{)}|c\tanh{(c)}-1|\ln{\Big{(}\frac{1}{\epsilon}\Big{)}}^{4}\right),$
(208) $\displaystyle
c_{2}(0)=O\left(\max\Big{(}\left\lVert\overrightarrow{g(0)}\right\rVert,\epsilon\ln{\Big{(}\frac{1}{\epsilon}\Big{)}}\Big{)}|\tanh{(c)}|\ln{\Big{(}\frac{1}{\epsilon}\Big{)}}^{4}\right),$
(209)

and

$|\dot{c}_{1}(t)|=O\left(\left|m(t)+(\sqrt{2}vt+c)\operatorname{sech}{(\sqrt{2}vt+c)}^{2}\right|\max\left(\left\lVert(g_{0},g_{1})\right\rVert,\epsilon\ln{\Big{(}\frac{1}{\epsilon}\Big{)}}\right)\epsilon^{\frac{1}{2}}\exp\Big{(}\frac{2C\epsilon^{\frac{1}{2}}t}{\ln{(\frac{1}{\epsilon})}}\Big{)}\right)\\\
+O\left(\left[v\operatorname{sech}{(\sqrt{2}vt+c)}^{2}\left|W(t)\right|^{2}+\frac{\max\Big{(}\left\lVert(g_{0},g_{1})\right\rVert,\epsilon\ln{(\frac{1}{\epsilon})}\Big{)}^{2}}{v\ln{\ln{(\frac{1}{\epsilon})}}}\exp\Big{(}\frac{2C\epsilon^{\frac{1}{2}}t}{\ln{(\frac{1}{\epsilon})}}\Big{)}\right]|n(t)|\right),$
(210)
$|\dot{c}_{2}(t)|=O\left(\left[v\operatorname{sech}{(\sqrt{2}vt+c)}^{2}\left|W(t)\right|^{2}+\frac{\max\Big{(}\left\lVert(g_{0},g_{1})\right\rVert,\epsilon\ln{\Big{(}\frac{1}{\epsilon}\Big{)}}\Big{)}^{2}}{v\ln{\ln{(\frac{1}{\epsilon})}}}\exp\Big{(}\frac{2C\epsilon^{\frac{1}{2}}t}{\ln{(\frac{1}{\epsilon})}}\Big{)}\right]|m(t)|\right)\\\
+O\left(\max\left(\left\lVert(g_{0},g_{1})\right\rVert,\epsilon\ln{\Big{(}\frac{1}{\epsilon}\Big{)}}\right)\exp\Big{(}\frac{2C\epsilon^{\frac{1}{2}}t}{\ln{(\frac{1}{\epsilon})}}\Big{)}\epsilon^{\frac{1}{2}}\operatorname{sech}(\sqrt{2}vt+c)^{2}\right).$
(211)

Since we have for all $x\geq 0$ that

$\frac{d}{dx}\left(-\frac{\operatorname{sech}{(x)}^{2}x}{2}+\frac{3\tanh{(x)}}{2}\right)=\frac{\operatorname{sech}{(x)}^{2}}{2}+x\tanh{(x)}\operatorname{sech}{(x)}^{2}\geq\frac{|x\tanh{(x)}-1|\operatorname{sech}{(x)}^{2}}{2},$

we deduce from the Fundamental Theorem of Calculus, the identity
$n(t)=(\sqrt{2}vt+c)\tanh(\sqrt{2}vt+c)-1,$ the inequality
$\frac{\epsilon^{\frac{1}{2}}}{\ln{(\frac{1}{\epsilon})}^{4}}\lesssim v$ and
the estimates (210), (211) that

$|c_{1}(t)-c_{1}(0)|=O\left(\max\left(\left\lVert(g_{0},g_{1})\right\rVert,\epsilon\ln{\Big{(}\frac{1}{\epsilon}\Big{)}}\right)\ln{\Big{(}\frac{1}{\epsilon}\Big{)}}\left[\exp\Big{(}\frac{2Ct\epsilon^{\frac{1}{2}}}{\ln{(\frac{1}{\epsilon})}}\Big{)}-1\right]\right)\\\
+O\left(\left[\exp\Big{(}\frac{2C\epsilon^{\frac{1}{2}}t}{\ln{(\frac{1}{\epsilon})}}\Big{)}-1\right]\left\lVert
n(s)\right\rVert_{L^{\infty}_{s}[0,t]}\max\Big{(}\left\lVert(g_{0},g_{1})\right\rVert,\epsilon\ln{\Big{(}\frac{1}{\epsilon}\Big{)}}\Big{)}^{2}\frac{\ln{(\frac{1}{\epsilon})}^{5}}{\epsilon\ln{\ln(\frac{1}{\epsilon})}}\right)\\\
+O\left(\left|-\frac{\operatorname{sech}{(x)}^{2}x}{2}+\frac{3\tanh{(x)}}{2}\Big{|}_{c}^{\sqrt{2}vt+c}\right|\left\lVert
W(s)\right\rVert_{L^{\infty}_{s}[0,t]}^{2}\right).$ (212)

From a similar argument, we deduce that

$|c_{2}(t)-c_{2}(0)|=O\left(\left\lVert
W(s)\right\rVert_{L^{\infty}_{s}[0,t]}^{2}\Big{[}\tanh{(\sqrt{2}vt+c)}-\tanh{(c)}\Big{]}\right)\\\
+O\left(\max\Big{(}\left\lVert(g_{0},g_{1})\right\rVert,\epsilon\ln{\Big{(}\frac{1}{\epsilon}\Big{)}}\Big{)}^{2}\Big{[}\exp\Big{(}\frac{2C\epsilon^{\frac{1}{2}}t}{\ln{(\frac{1}{\epsilon})}}\Big{)}-1\Big{]}\frac{\ln{(\frac{1}{\epsilon})}^{5}}{\epsilon\ln{\ln{(\frac{1}{\epsilon})}}}+\max\Big{(}\left\lVert(g_{0},g_{1})\right\rVert,\epsilon\ln{\Big{(}\frac{1}{\epsilon}\Big{)}}\Big{)}\ln{\Big{(}\frac{1}{\epsilon}\Big{)}}\exp\Big{(}2Ct\frac{\epsilon^{\frac{1}{2}}}{\ln{(\frac{1}{\epsilon})}}\Big{)}\right).$
(213)

From the estimates
$v\lesssim\epsilon^{\frac{1}{2}},\,|c|\lesssim\ln{\ln{(\frac{1}{\epsilon})}},$
we obtain for $\epsilon\ll 1$ while

$\left\lVert
W(s)\right\rVert_{L^{\infty}_{s}[0,t]}\Big{[}\epsilon^{\frac{1}{2}}t+\ln{\ln{\Big{(}\frac{1}{\epsilon}\Big{)}}}\Big{]}\ln{\ln{\Big{(}\frac{1}{\epsilon}\Big{)}}}\leq
1,$ (214)

that

$\left\lVert
W(s)\right\rVert_{L^{\infty}_{s}[0,t]}^{2}(1+|n(t)|)\lesssim\left\lVert
W(s)\right\rVert_{L^{\infty}_{s}[0,t]}\frac{1}{\ln{\ln{(\frac{1}{\epsilon})}}}.$
(215)

Also, from

$|n(t)|\leq(\sqrt{2}v|t|+|c|),$

we deduce for $t\geq 0$ that

$|n(t)|\lesssim\epsilon^{\frac{1}{2}}t+\ln{\ln{\left(\frac{1}{\epsilon}\right)}}\lesssim\ln{\left(\frac{1}{\epsilon}\right)}\exp\left(\frac{\epsilon^{\frac{1}{2}}t}{\ln{(\frac{1}{\epsilon})}}\right)$
(216)

In conclusion, the estimates (212), (213), (215), (216) and the definition of
$W(t)=z(t)-d(t)$ implies that while the condition (214) is true, then

$|W(t)|\lesssim
f(t)=\frac{\max\Big{(}\left\lVert(g_{0},g_{1})\right\rVert,\epsilon\ln{(\frac{1}{\epsilon})}\Big{)}^{2}\ln{(\frac{1}{\epsilon})}^{6}}{\epsilon\ln{\ln(\frac{1}{\epsilon})}}\exp\left(\frac{(2C+1)\epsilon^{\frac{1}{2}}t}{\ln{(\frac{1}{\epsilon})}}\right)$
(217)

Then, from the expression for $V(t)$ in the equation (207) and the estimates
(212), (213), (216), we obtain that if inequality (217) is true, then

$|V(t)|\lesssim\max\Big{(}\left\lVert(g_{0},g_{1})\right\rVert,\epsilon\ln{\Big{(}\frac{1}{\epsilon}\Big{)}}\Big{)}^{2}\frac{\ln{(\frac{1}{\epsilon})}^{6}}{\epsilon^{\frac{1}{2}}\ln{\ln{(\frac{1}{\epsilon})}}}\exp\Big{(}\frac{(4C+3)\epsilon^{\frac{1}{2}}t}{\ln{(\frac{1}{\epsilon})}}\Big{)}\\\
+\max\Big{(}\left\lVert(g_{0},g_{1})\right\rVert,\epsilon\ln{\Big{(}\frac{1}{\epsilon}\Big{)}}\Big{)}^{4}\frac{\ln{(\frac{1}{\epsilon})}^{12}}{\epsilon^{\frac{3}{2}}\big{[}\ln{\ln{(\frac{1}{\epsilon})}}\big{]}^{2}}\exp\Big{(}\frac{(4C+3)\epsilon^{\frac{1}{2}}t}{\ln{(\frac{1}{\epsilon})}}\Big{)},$
(218)

which implies the following estimate

$|\dot{W}(t)|\lesssim\max\Big{(}\left\lVert\overrightarrow{g(0)}\right\rVert,\epsilon\ln{\Big{(}\frac{1}{\epsilon}\Big{)}}\Big{)}^{2}\frac{\ln{(\frac{1}{\epsilon})}^{6}}{\epsilon^{\frac{1}{2}}\ln{\ln{(\frac{1}{\epsilon})}}}\exp\Big{(}\frac{(4C+3)\epsilon^{\frac{1}{2}}t}{\ln{(\frac{1}{\epsilon})}}\Big{)}.$
(219)

Indeed, from the bound
$\left\lVert\overrightarrow{g(0)}\right\rVert\lesssim\frac{\epsilon^{\frac{1}{2}}}{\ln{(\frac{1}{\epsilon})}^{4}},$
we deduce that (214) is true for $0\leq
t\leq\frac{\ln{\ln{(\frac{1}{\epsilon})}}\ln{(\frac{1}{\epsilon})}}{(4C+2)\epsilon^{\frac{1}{2}}}.$
As a consequence, the estimates (217) and (219) are true for $0\leq
t\leq\frac{\ln{\ln{(\frac{1}{\epsilon})}}\ln{(\frac{1}{\epsilon})}}{(4C+2)\epsilon^{\frac{1}{2}}}.$
But, for $t\geq 0,$ we have that

$|W(t)|\lesssim\epsilon^{\frac{1}{2}}t\lesssim
3\ln{\left(\frac{1}{\epsilon}\right)}\exp\left(\frac{\epsilon^{\frac{1}{2}}t}{3\ln{(\frac{1}{\epsilon})}}\right),\,|\dot{W}(t)|\lesssim\epsilon
t\lesssim
3\epsilon^{\frac{1}{2}}\ln{\left(\frac{1}{\epsilon}\right)}\exp\left(\frac{\epsilon^{\frac{1}{2}}t}{3\ln{(\frac{1}{\epsilon})}}\right).$
(220)

Since $f(t)$ defined in inequality (217) is strictly increasing and
$f(0)\lesssim\frac{1}{\ln{(\frac{1}{\epsilon})}^{2}\ln{\ln{(\frac{1}{\epsilon})}}}$,
there is an instant $T_{M}>0$ such that

$\exp\left(\frac{\epsilon^{\frac{1}{2}}T_{M}}{\ln{(\frac{1}{\epsilon})}}\right)f(T_{M})=\frac{1}{\ln{(\frac{1}{\epsilon})}\ln{\ln{(\frac{1}{\epsilon})}}^{2}},$
(221)

from which with estimate (217) and condition (214) we deduce that (217) is
true for $0\leq t\leq T_{M}.$ Also, from the identity (221) and the fact that
$\left\lVert\overrightarrow{g(0)}\right\rVert\lesssim\frac{\epsilon^{\frac{1}{2}}}{\ln{(\frac{1}{\epsilon})}^{4}}$
we deduce

$\frac{1}{\ln{(\frac{1}{\epsilon})}\ln{\ln{(\frac{1}{\epsilon})}}^{2}}\lesssim\frac{1}{\ln{(\frac{1}{\epsilon})}^{2}\ln{\ln{(\frac{1}{\epsilon})}}}\exp\left(\frac{(2C+2)\epsilon^{\frac{1}{2}}T_{M}}{\ln{(\frac{1}{\epsilon})}}\right),$

from which we obtain that
$T_{M}\geq\frac{3}{8(C+1)}\frac{\ln{\ln{(\frac{1}{\epsilon})}}\ln{(\frac{1}{\epsilon})}}{\epsilon^{\frac{1}{2}}}$
for $\epsilon\ll 1.$ In conclusion, since $f(t)$ is an increasing function, we
have for $t\geq T_{M}$ and $\epsilon\ll 1$ that

$f(t)\exp\left(\frac{[17(C+1)+4]\epsilon^{\frac{1}{2}}t}{3\ln{(\frac{1}{\epsilon})}}\right)\geq\frac{1}{\ln{(\frac{1}{\epsilon})}\ln{\ln{(\frac{1}{\epsilon})}}^{2}}\exp\left(\frac{[17(C+1)+1]\epsilon^{\frac{1}{2}}t}{3\ln{(\frac{1}{\epsilon})}}\right)\geq\frac{\ln{(\frac{1}{\epsilon})}^{1+\frac{1}{8}}}{\ln{\ln{(\frac{1}{\epsilon})}}^{2}}\exp\left(\frac{\epsilon^{\frac{1}{2}}t}{3\ln{(\frac{1}{\epsilon}})}\right),$

from which with the estimates (220) and (217) we deduce for all $t\geq 0$ that

$|W(t)|\lesssim\frac{\max\Big{(}\left\lVert(g_{0},g_{1})\right\rVert,\epsilon\ln{(\frac{1}{\epsilon})}\Big{)}^{2}\ln{(\frac{1}{\epsilon})}^{6}}{\epsilon\ln{\ln(\frac{1}{\epsilon})}}\exp\left(\frac{(8C+9)\epsilon^{\frac{1}{2}}t}{\ln{(\frac{1}{\epsilon})}}\right).$
(222)

As consequence, we obtain from the estimates (208), (209), (212), (213) and
(222) that

$|\dot{W}(t)|\lesssim\frac{\max\Big{(}\left\lVert(g_{0},g_{1})\right\rVert,\epsilon\ln{(\frac{1}{\epsilon})}\Big{)}^{2}\ln{(\frac{1}{\epsilon})}^{6}}{\epsilon^{\frac{1}{2}}\ln{\ln(\frac{1}{\epsilon})}}\exp\left(\frac{(16C+18)\epsilon^{\frac{1}{2}}t}{\ln{(\frac{1}{\epsilon})}}\right)$
(223)

for all $t\geq 0.$
Step 2.(Estimate of $|x_{1}(t)+x_{2}(t)|,\,|\dot{x}_{1}(t)+\dot{x}_{2}(t)|.$)
First, we define

$M(t)\coloneqq(x_{1}(t)+x_{2}(t))-(d_{1}(t)+d_{2}(t)),\,N(t)\coloneqq(p_{1}(t)+p_{2}(t))-(\dot{d}_{1}(t)+\dot{d}_{2}(t)).$
(224)

From the inequalities (191), (192) of Lemma 5.1, we obtain, respectively:

$|\dot{M}(t)-N(t)|\lesssim\max\Big{(}\left\lVert(g_{0},g_{1})\right\rVert,\epsilon\ln{\left(\frac{1}{\epsilon}\right)}\Big{)}\epsilon^{\frac{1}{2}}\exp\Big{(}\frac{C\epsilon^{\frac{1}{2}}t}{\ln{\left(\frac{1}{\epsilon}\right)}}\Big{)},\,|\dot{N}(t)|\lesssim\frac{\max\Big{(}\left\lVert(g_{0},g_{1})\right\rVert,\epsilon\ln{(\frac{1}{\epsilon})}\Big{)}^{2}}{\ln{\ln(\frac{1}{\epsilon})}}\exp\Big{(}\frac{2C\epsilon^{\frac{1}{2}}t}{\ln{(\frac{1}{\epsilon})}}\Big{)}.$

Also, from inequality (191) and the fact that for
$j\in\\{1,2\\}\,d_{j}(0)=x_{j}(0),\,\dot{d}_{j}(0)=\dot{x}_{j}(0),$ we deduce
that $M(0)=0$ and
$|N(0)|\lesssim\max\left(\left\lVert\overrightarrow{g(0)}\right\rVert,\epsilon\ln{(\frac{1}{\epsilon})}\right)\epsilon^{\frac{1}{2}}.$
Then, from the Fundamental Theorem of Calculus, we obtain that

$\displaystyle
N(t)=O\left(\frac{\max\Big{(}\left\lVert(g_{0},g_{1})\right\rVert,\epsilon\ln{(\frac{1}{\epsilon})}\Big{)}^{2}\ln{(\frac{1}{\epsilon})}}{\epsilon^{\frac{1}{2}}\ln{\ln{(\frac{1}{\epsilon})}}}\exp\Big{(}\frac{4C\epsilon^{\frac{1}{2}}t}{\ln{(\frac{1}{\epsilon})}}\Big{)}\right),$
(225) $\displaystyle
M(t)=O\left(\frac{\max\Big{(}\left\lVert(g_{0},g_{1})\right\rVert,\epsilon\ln{(\frac{1}{\epsilon})}\Big{)}^{2}\ln{(\frac{1}{\epsilon})}^{2}}{\epsilon\ln{\ln{(\frac{1}{\epsilon})}}}\exp\Big{(}\frac{4C\epsilon^{\frac{1}{2}}t}{\ln{(\frac{1}{\epsilon})}}\Big{)}\right).$
(226)

In conclusion, for $K=16C+18,$ we verify from triangle inequality that the
estimates (222) and (226) imply (203) and the estimates (223) and (225) imply
(204). ∎

######  Remark 5.4.

The estimates (226) and (225) are true for any initial data $(g_{0},g_{1})\in
H^{1}(\mathbb{R})\times L^{2}(\mathbb{R})$ such that the hypothesis of Theorem
1.11 are true.

######  Remark 5.5 (Similar Case).

If we add the following conditions

$e^{-\sqrt{2}z(0)}\ll\frac{\epsilon}{\ln{(\frac{1}{\epsilon})}^{8}},\,\frac{\epsilon^{\frac{1}{2}}}{\ln{(\frac{1}{\epsilon})}^{4}}\lesssim
v\lesssim\epsilon^{\frac{1}{2}},\,-\ln{\Big{(}\frac{1}{\epsilon}\Big{)}}^{2}<c<0,$

to the hypotheses of Theorem 1.11, then, by repeating the above proof of Lemma
5.3, we would still obtain (210), (211), (212) and (213).

However, since now $|c|\leq\ln{(\frac{1}{\epsilon})}^{2},$ if $\epsilon\ll 1$
enough, we can verify while

$\left\lVert
W(s)\right\rVert_{L^{\infty}_{s}[0,t]}\left(\epsilon^{\frac{1}{2}}t+\ln{\left(\frac{1}{\epsilon}\right)}^{2}\right)\ln{\ln{\left(\frac{1}{\epsilon}\right)}}\leq
1,$ (227)

that

$\left\lVert
W(s)\right\rVert_{L^{\infty}_{s}[0,t]}^{2}(1+|n(t)|)\lesssim\left\lVert
W(s)\right\rVert_{L^{\infty}_{s}[0,t]}\frac{1}{\ln{\ln{(\frac{1}{\epsilon})}}},$

which implies by a similar reasoning to the proof of Lemma 5.3 for a uniform
constant $C>1$ the following estimates

$\displaystyle|W(t)|\lesssim\frac{\max\Big{(}\left\lVert(g_{0},g_{1})\right\rVert,\epsilon\ln{(\frac{1}{\epsilon})}\Big{)}^{2}\ln{(\frac{1}{\epsilon})}^{7}}{\epsilon\ln{\ln(\frac{1}{\epsilon})}}\exp\Big{(}\frac{C\epsilon^{\frac{1}{2}}t}{\ln{(\frac{1}{\epsilon})}}\Big{)}=f_{1}(t,C),$
(228)
$\displaystyle|\dot{W}(t)|\lesssim\max\Big{(}\left\lVert(g_{0},g_{1})\right\rVert,\epsilon\ln{\Big{(}\frac{1}{\epsilon}\Big{)}}\Big{)}^{2}\frac{\ln{(\frac{1}{\epsilon})}^{7}}{\epsilon^{\frac{1}{2}}\ln{\ln{(\frac{1}{\epsilon})}}}\exp\Big{(}\frac{C\epsilon^{\frac{1}{2}}t}{\ln{(\frac{1}{\epsilon})}}\Big{)}=f_{2}(t,C).$
(229)

From the estimates (228), (229) and
$\left\lVert\overrightarrow{g(0)}\right\rVert\leq\frac{\epsilon^{\frac{1}{2}}}{\ln{(\frac{1}{\epsilon})}^{5}}$,
we deduce that the condition (227) holds while $0\leq
t\leq\frac{\ln{\ln{(\frac{1}{\epsilon})}}\ln{(\frac{1}{\epsilon})}}{4(C+1)\epsilon^{\frac{1}{2}}}.$
Indeed, since
$\left\lVert\overrightarrow{g(0)}\right\rVert^{2}\leq\frac{\epsilon}{\ln{(\frac{1}{\epsilon})}^{10}},$
we can verify that there is an instant
$\frac{\ln{\ln{(\frac{1}{\epsilon})}}\ln{(\frac{1}{\epsilon})}}{4(C+1)\epsilon^{\frac{1}{2}}}\leq
T_{M}$ such that (227) and (228) are true for $0\leq t\leq T_{M}$ and

$f_{1}(T_{M},C)\exp\left(\frac{\epsilon^{\frac{1}{2}}T_{M}}{\ln{(\frac{1}{\epsilon})}}\right)=\frac{1}{\ln{(\frac{1}{\epsilon})}^{2+\frac{1}{2}}\ln{\ln{(\frac{1}{\epsilon})}}}.$

In conclusion, we can repeat the argument in the proof of step $1$ of Lemma
5.3 and deduce that there is $1<K\lesssim C+1$ such that for all $t\geq 0$

$|W(t)|\lesssim f_{1}(t,K),\,|\dot{W}(t)|\lesssim f_{2}(t,K).$ (230)

######  Lemma 5.6.

In notation of Theorem 1.11, $\exists K>1,\,\delta>0$ such that if
$0<\epsilon<\delta,\,0<v\leq\frac{\epsilon^{\frac{1}{2}}}{\ln{(\frac{1}{\epsilon})}^{4}},\,(g_{0}(x),g_{1}(x))=\left(g(0,x),\partial_{t}g(0,x)\right)$
and
$\left\lVert\overrightarrow{g(0)}\right\rVert\leq\frac{\epsilon^{\frac{1}{2}}}{\ln{(\frac{1}{\epsilon})}^{5}},$
then we have for $0\leq t$ that

$\displaystyle\max_{j\in\\{1,\,2\\}}|d_{j}(t)-x_{j}(t)|=O\left(\frac{\max\Big{(}\left\lVert(g_{0},g_{1})\right\rVert,\epsilon\ln{(\frac{1}{\epsilon})}\Big{)}^{2}}{\epsilon\ln{\ln{(\frac{1}{\epsilon})}}}\ln{\Big{(}{\frac{1}{\epsilon}}\Big{)}}^{2}\exp\Big{(}\frac{Kt\epsilon^{\frac{1}{2}}}{\ln{\frac{1}{\epsilon}}}\Big{)}\right),$
(231)
$\displaystyle\max_{j\in\\{1,\,2\\}}|\dot{d}_{j}(t)-\dot{x}_{j}(t)|=O\left(\frac{\max\Big{(}\left\lVert(g_{0},g_{1})\right\rVert,\epsilon\ln{(\frac{1}{\epsilon})}\Big{)}^{2}}{\epsilon^{\frac{1}{2}}\ln{\ln{(\frac{1}{\epsilon})}}}\ln{\Big{(}{\frac{1}{\epsilon}}\Big{)}}\exp\Big{(}\frac{Kt\epsilon^{\frac{1}{2}}}{\ln{\frac{1}{\epsilon}}}\Big{)}\right).$
(232)

###### Proof of Lemma 5.6.

First, we recall that

$d(t)=\frac{1}{\sqrt{2}}\ln{\left(\frac{8}{v^{2}}\cosh{\left(\sqrt{2}vt+c\right)}\right)},$

which implies that

$e^{-\sqrt{2}d(t)}=\frac{v^{2}}{8}\operatorname{sech}{\left(\sqrt{2}vt+c\right)}^{2}.$
(233)

We recall the notation $W(t)=z(t)-d(t),\,V(t)=p(t)-\dot{d}(t).$ From the first
inequality of Lemma 5.1, we have that

$|V(0)|\lesssim\max\left(\left\lVert(g_{0},g_{1})\right\rVert,\epsilon\ln{\left(\frac{1}{\epsilon}\right)}\right)\epsilon^{\frac{1}{2}}.$
(234)

We already verified that $W,\,V$ satisfy the following ordinary differential
system

$\begin{cases}\dot{W}(t)=V(t)+O\Big{(}\max\Big{(}\left\lVert(g_{0},g_{1})\right\rVert,\epsilon\ln{\Big{(}\frac{1}{\epsilon}\Big{)}}\Big{)}\epsilon^{\frac{1}{2}}\exp\Big{(}\frac{C\epsilon^{\frac{1}{2}}t}{\ln{(\frac{1}{\epsilon})}}\Big{)}\Big{)},\\\
\\\
\dot{V}(t)=-32e^{-\sqrt{2}d(t)}W(t)+O(e^{-\sqrt{2}z(t)}(W(t))^{2})+O\Big{(}\frac{\max(\left\lVert(g_{0},g_{1})\right\rVert,\epsilon\ln{(\frac{1}{\epsilon})})^{2}}{\ln{\ln{(\frac{1}{\epsilon})}}}\exp\Big{(}\frac{2C\epsilon^{\frac{1}{2}}t}{\ln{(\frac{1}{\epsilon})}}\Big{)}\Big{)}.\end{cases}$

However, since $v^{2}\leq\frac{\epsilon}{\ln{(\frac{1}{\epsilon})}^{8}},$ we
deduce from (233) that
$e^{-\sqrt{2}d(t)}\lesssim\frac{\epsilon}{\ln{(\frac{1}{\epsilon})}^{8}}$ for
all $t\geq 0.$ So, while $\left\lVert W(s)\right\rVert_{L^{\infty}[0,t]}<1,$
we have from the differential ordinary system (5) for $t\geq 0$ and some
constant $C>0$ independent of $\epsilon$ that

$|\dot{V}(t)|\lesssim\frac{\epsilon}{\ln{(\frac{1}{\epsilon})}^{8}}\left\lVert
W(s)\right\rVert_{L^{\infty}[0,t]}+\frac{\max\Big{(}\left\lVert(g_{0},g_{1})\right\rVert,\epsilon\ln{(\frac{1}{\epsilon})}\Big{)}^{2}}{\ln{\ln{(\frac{1}{\epsilon})}}}\exp\Big{(}\frac{2C\epsilon^{\frac{1}{2}}t}{\ln{(\frac{1}{\epsilon})}}\Big{)},$

from which we deduce the following estimate

$|V(t)-V(0)|=O\left(\frac{\epsilon
t}{\ln{(\frac{1}{\epsilon})}^{8}}\left\lVert
W(s)\right\rVert_{L^{\infty}[0,t]}\right)+O\left(\frac{\max\Big{(}\left\lVert(g_{0},g_{1})\right\rVert,\epsilon\ln{(\frac{1}{\epsilon})}\Big{)}^{2}\ln{(\frac{1}{\epsilon})}}{\epsilon^{\frac{1}{2}}\ln{\ln{(\frac{1}{\epsilon})}}}\exp\Big{(}\frac{2C\epsilon^{\frac{1}{2}}t}{\ln{(\frac{1}{\epsilon})}}\Big{)}\right).$

In conclusion, while $\left\lVert W(s)\right\rVert_{L^{\infty}[0,t]}<1,$ we
have that

$|\dot{W}(t)|\leq|V(0)|+O\left(\frac{\max\Big{(}\left\lVert(g_{0},g_{1})\right\rVert,\epsilon\ln{(\frac{1}{\epsilon})}\Big{)}^{2}\ln{(\frac{1}{\epsilon})}}{\epsilon^{\frac{1}{2}}\ln{\ln{(\frac{1}{\epsilon})}}}\exp\Big{(}\frac{2C\epsilon^{\frac{1}{2}}t}{\ln{(\frac{1}{\epsilon})}}\Big{)}\right)+O\left(\frac{\epsilon
t}{\ln{(\frac{1}{\epsilon})}^{8}}\left\lVert
W(s)\right\rVert_{L^{\infty}[0,t]}\right).$ (235)

Finally, since $W(0)=0,$ the fundamental theorem of calculus and (235) imply
the following estimate

$\left\lVert
W(s)\right\rVert_{L^{\infty}[0,t]}\leq|V(0)|t+O\left(\frac{\epsilon
t^{2}}{\ln{(\frac{1}{\epsilon})}^{8}}\left\lVert
W(s)\right\rVert_{L^{\infty}[0,t]}+\frac{\max\Big{(}\left\lVert(g_{0},g_{1})\right\rVert,\epsilon\ln{(\frac{1}{\epsilon})}\Big{)}^{2}\ln{(\frac{1}{\epsilon})}^{2}}{\epsilon\ln{\ln{(\frac{1}{\epsilon})}}}\exp\Big{(}\frac{2C\epsilon^{\frac{1}{2}}t}{\ln{(\frac{1}{\epsilon})}}\Big{)}\right).$
(236)

Then, the estimates (234) and (236) imply if $\epsilon\ll 1$ that

$|W(t)|\lesssim\frac{\max\Big{(}\left\lVert(g_{0},g_{1})\right\rVert,\epsilon\ln{(\frac{1}{\epsilon})}\Big{)}^{2}\ln{(\frac{1}{\epsilon})}^{2}}{\epsilon\ln{\ln{(\frac{1}{\epsilon})}}}\exp\Big{(}\frac{(2C+1)\epsilon^{\frac{1}{2}}t}{\ln{(\frac{1}{\epsilon})}}\Big{)},$
(237)

for $0\leq
t\leq\frac{\ln{(\frac{1}{\epsilon})}\ln{\ln(\frac{1}{\epsilon})}}{(8C+4)\epsilon^{\frac{1}{2}}}.$
From (237) and (235), we deduce for $0\leq
t\leq\frac{\ln{(\frac{1}{\epsilon})}\ln{\ln{(\frac{1}{\epsilon})}}}{(8C+4)\epsilon^{\frac{1}{2}}}$
that

$|\dot{W}(t)|\lesssim\frac{\max\Big{(}\left\lVert(g_{0},g_{1})\right\rVert,\epsilon\ln{(\frac{1}{\epsilon})}\Big{)}^{2}\ln{(\frac{1}{\epsilon})}^{2}}{\epsilon^{\frac{1}{2}}\ln{\ln{(\frac{1}{\epsilon})}}}\exp\Big{(}\frac{(2C+1)\epsilon^{\frac{1}{2}}t}{\ln{(\frac{1}{\epsilon})}}\Big{)}.$
(238)

Since $|W(t)|\lesssim\epsilon^{\frac{1}{2}}t,\,|\dot{W}(t)|\lesssim\epsilon t$
for all $t\geq 0,$ we can verify by a similar argument to the proof of Step
$1$ of Lemma 5.3 that for all $t\geq 0$ there is a constant $1<K\lesssim(C+1)$
such that

$\displaystyle|W(t)|\lesssim\frac{\max\Big{(}\left\lVert(g_{0},g_{1})\right\rVert,\epsilon\ln{(\frac{1}{\epsilon})}\Big{)}^{2}\ln{(\frac{1}{\epsilon})}^{2}}{\epsilon\ln{\ln{(\frac{1}{\epsilon})}}}\exp\Big{(}\frac{K\epsilon^{\frac{1}{2}}t}{\ln{(\frac{1}{\epsilon})}}\Big{)},$
(239)
$\displaystyle|\dot{W}(t)|\lesssim\frac{\max\Big{(}\left\lVert(g_{0},g_{1})\right\rVert,\epsilon\ln{(\frac{1}{\epsilon})}\Big{)}^{2}\ln{(\frac{1}{\epsilon})}^{2}}{\epsilon^{\frac{1}{2}}\ln{\ln{(\frac{1}{\epsilon})}}}\exp\Big{(}\frac{K\epsilon^{\frac{1}{2}}t}{\ln{(\frac{1}{\epsilon})}}\Big{)}.$
(240)

In conclusion, estimates (231) and (232) follow from Remark 5.4, inequalities
(239), (240) and triangle inequality. ∎

######  Remark 5.7.

We recall the definition (205) of $d(t).$ It is not difficult to verify that
if
$\left\lVert\overrightarrow{g(0)}\right\rVert\leq\frac{\epsilon^{\frac{1}{2}}}{\ln(\frac{1}{\epsilon})^{5}},\,\frac{\epsilon^{\frac{1}{2}}}{\ln(\frac{1}{\epsilon})^{4}}\lesssim
v$ and one of the following statements

  1. 1.

$e^{-\sqrt{2}z(0)}\ll\frac{\epsilon}{\ln{(\frac{1}{\epsilon})}^{8}}$ and
$c>0,$

  2. 2.

$e^{-\sqrt{2}z(0)}\ll\frac{\epsilon}{\ln{(\frac{1}{\epsilon})}^{8}}$ and
$c\leq-\ln{(\frac{1}{\epsilon})}^{2},$

was true, then we would have that
$e^{-\sqrt{2}d(t)}\ll\frac{\epsilon}{\ln{(\frac{1}{\epsilon})}^{8}}$ for
$0\leq t\lesssim\frac{\ln{(\frac{1}{\epsilon})}^{2}}{\epsilon^{\frac{1}{2}}}.$
Moreover, assuming
$e^{-\sqrt{2}z(0)}\ln{(\frac{1}{\epsilon})}^{8}\ll\epsilon,$ if $c>0,$ then we
have for all $t\geq 0$ that

$e^{-\sqrt{2}d(t)}=\frac{v^{2}}{8}sech{(\sqrt{2}vt+c)}^{2}\leq\frac{v^{2}}{8}sech{(c)}^{2}=e^{-\sqrt{2}z(0)}\ll\frac{\epsilon}{\ln{(\frac{1}{\epsilon})}^{8}},$

otherwise if $c\leq-\ln{(\frac{1}{\epsilon})}^{2},$ since
$v\lesssim\epsilon^{\frac{1}{2}},$ then there is $1\lesssim K$ such that for
$0\leq t\leq\frac{K\ln{(\frac{1}{\epsilon})}^{2}}{\epsilon^{\frac{1}{2}}},$
then $2|\sqrt{2}vt+c|>|c|,$ and so

$e^{-\sqrt{2}d(t)}\leq
v^{2}\operatorname{sech}{\left(-\frac{c}{2}\right)}^{2}\ll\frac{\epsilon}{\ln{(\frac{1}{\epsilon})}^{8}}.$

In conclusion, the result of Lemma 5.6 would be true for these two cases.

From the following inequality

$\max\Big{(}\left\lVert\overrightarrow{g(0)}\right\rVert,\epsilon\ln{\Big{(}\frac{1}{\epsilon}\Big{)}}\Big{)}\leq\ln{\Big{(}\frac{1}{\epsilon}\Big{)}}\max\Big{(}\left\lVert\overrightarrow{g(0)}\right\rVert,\epsilon\Big{)},$

we deduce from Lemmas 5.3, 5.6 and Remarks 5.4, 5.5 and 5.7 the statement of
Theorem 1.11.

##  6 Proof of Theorem 1.7

If
$\left\lVert\overrightarrow{g(0)}\right\rVert\geq\epsilon\ln{(\frac{1}{\epsilon})}$
the result of Theorem 1.7 is a direct consequence of Theorem 4.5. So, from now
on, we assume that
$\left\lVert\overrightarrow{g(0)}\right\rVert<\epsilon\ln{(\frac{1}{\epsilon})}.$
We recall from Theorem 1.11 the notations $v,\,c,\,d_{1}(t),\,d_{2}(t)$ and we
denote $d(t)=d_{2}(t)-d_{1}(t)$ that satisfies

$d(t)=\frac{1}{\sqrt{2}}\ln{\Big{(}\frac{8}{v^{2}}\cosh{(\sqrt{2}vt+c)}^{2}\Big{)}},\,e^{-\sqrt{2}d(t)}=\frac{v^{2}}{8}\operatorname{sech}{(\sqrt{2}vt+c)}^{2}.$

From the definition of $d_{1}(t),\,d_{2}(t),\,d(t)$, we know that
$\max_{j\\{1,\,2\\}}|\ddot{d}_{j}(t)|+e^{-\sqrt{2}d(t)}=O\Big{(}v^{2}\operatorname{sech}{(\sqrt{2}vt+c)}^{2}\Big{)}$
and since $z(0)=d(0),\,\dot{z}(0)=\dot{d}(0),$ we have that $v,\,c$ satisfy
the following identities

$v=\left(e^{-\sqrt{2}z(0)}+\Big{(}\frac{\dot{x}_{2}(0)-\dot{x}_{1}(0)}{2}\Big{)}^{2}\right)^{\frac{1}{2}},\,c=\operatorname{arctanh}{\Big{(}\frac{\dot{x}_{2}(0)-\dot{x}_{1}(0)}{2v}\Big{)}},$

so Theorem 2.9 implies that $v\lesssim\epsilon^{\frac{1}{2}}.$ From the
Corollary 1.13 and the Theorem 1.11, we deduce that $\exists C>0$ such that if
$\epsilon\ll 1$ and $0\leq
t\leq\frac{\ln{\ln{(\frac{1}{\epsilon})}}\ln{(\frac{1}{\epsilon})}}{\epsilon^{\frac{1}{2}}},$
then we have that

$\displaystyle\max_{j\in\\{1,\,2\\}}|\ddot{x}_{j}(t)|=O\Big{(}\max_{j\in\\{1,2\\}}|\ddot{d}_{j}(t)|\Big{)}+O\left(\epsilon^{\frac{3}{2}}\ln{\Big{(}\frac{1}{\epsilon}\Big{)}}^{9}\exp\Big{(}\frac{Ct\epsilon^{\frac{1}{2}}}{\ln{(\frac{1}{\epsilon})}}\Big{)}\right),$
(241) $\displaystyle
e^{-\sqrt{2}z(t)}=e^{-\sqrt{2}d(t)}+O\left(\max\left(e^{-\sqrt{2}d(t)},e^{-\sqrt{2}z(t)}\right)|z(t)-d(t)|\right)=e^{-\sqrt{2}d(t)}+O\left(\epsilon^{2}\ln{\Big{(}\frac{1}{\epsilon}\Big{)}}^{9}\exp\Big{(}\frac{Ct\epsilon^{\frac{1}{2}}}{\ln{(\frac{1}{\epsilon})}}\Big{)}\right).$
(242)

Next, we consider a smooth function $0\leq\chi_{2}(x)\leq 1$ that satisfies

$\chi_{2}(x)=\begin{cases}1,\text{ if $x\leq\frac{9}{20}$,}\\\ 0,\text{ if
$x\geq\frac{1}{2}$.}\end{cases}$

We denote

$\chi_{2}(t,x)=\chi_{2}\Big{(}\frac{x-x_{1}(t)}{x-x_{2}(t)}\Big{)}.$

From Theorem 4.1 and Remark 4.4, the estimates (241) and (242) of the
modulation parameters imply that for the following functional

$L_{1}(t)=\big{\langle}D^{2}E\left(H^{x_{2}(t)}_{0,1}+H^{x_{1}(t)}_{-1,0}\right)\overrightarrow{g(t)},\,\overrightarrow{g(t)}\big{\rangle}_{L^{2}\times
L^{2}}+2\int_{\mathbb{R}}\partial_{t}g(t,x)\partial_{x}g(t,x)\Big{[}\dot{x}_{1}(t)\chi_{2}(t,x)+\dot{x}_{2}(t)(1-\chi_{2}(t,x))\Big{]}\,dx\\\
-2\int_{\mathbb{R}}g(t,x)\Big{(}\dot{U}\left(H^{x_{1}(t)}_{-1,0}(x)\right)+\dot{U}\left(H^{x_{2}(t)}_{0,1}(x)\right)-\dot{U}\left(H^{x_{2}(t)}_{0,1}(x)+H^{x_{1}(t)}_{-1,0}(x)\right)\Big{)}\,dx\\\
+2\int_{\mathbb{R}}g(t,x)\Big{[}(\dot{x}_{1}(t))^{2}\partial^{2}_{x}H^{x_{1}(t)}_{-1,0}(x)+(\dot{x}_{2}(t))^{2}\partial^{2}_{x}H^{x_{2}(t)}_{0,1}(x)\Big{]}\,dx+\frac{1}{3}\int_{\mathbb{R}}U^{(3)}\left(H^{x_{2}(t)}_{0,1}(x)+H^{x_{1}(t)}_{-1,0}(x)\right)g(t,x)^{3}\,dx,$

and the following quantity $\delta_{1}(t)$ denoted by

$\delta_{1}(t)=\left\lVert\overrightarrow{g(t)}\right\rVert\Big{(}e^{-\sqrt{2}z(t)}\max_{j\in\\{1,2\\}}|\dot{x}_{j}(t)|+\max_{j\in\\{1,2\\}}|\dot{x}_{j}(t)|^{3}e^{-\frac{9\sqrt{2}z(t)}{20}}+\max_{j\in\\{1,2\\}}|\dot{x}_{j}(t)||\ddot{x}_{j}(t)|\Big{)}\\\
+\left\lVert\overrightarrow{g(t)}\right\rVert^{2}\Big{(}\frac{\max_{j\in\\{1,\,2\\}}|\dot{x}_{j}(t)|}{z(t)}+\max_{j\in\\{1,\,2\\}}\dot{x}_{j}(t)^{2}+\max_{j\in\\{1,\,2\\}}|\ddot{x}_{j}(t)|\Big{)}+\left\lVert\overrightarrow{g(t)}\right\rVert^{4},$

we have $|\dot{L}_{1}(t)|=O(\delta_{1}(t))$ for $t\geq 0.$ Moreover, estimates
(241), (242) and the bound $\dot{L}_{1}(t)=O(\delta_{1}(t))$ implies that for

$\delta_{2}(t)=\left\lVert\overrightarrow{g(t)}\right\rVert
v^{2}\epsilon^{\frac{1}{2}}\operatorname{sech}{(\sqrt{2}vt+c)}^{2}+\left\lVert\overrightarrow{g(t)}\right\rVert\epsilon^{2}\ln{\Big{(}\frac{1}{\epsilon}\Big{)}}^{9}\exp\Big{(}\frac{Ct\epsilon^{\frac{1}{2}}}{\ln{(\frac{1}{\epsilon})}}\Big{)}\\\
+\epsilon^{\frac{3}{2}}e^{-\frac{9\sqrt{2}z(t)}{20}}\left\lVert\overrightarrow{g(t)}\right\rVert+\max_{j\in\\{1,2\\}}\frac{|\dot{x}_{j}(t)|}{z(t)}\left\lVert\overrightarrow{g(t)}\right\rVert^{2}+\left\lVert\overrightarrow{g(t)}\right\rVert^{4},$

$|\dot{L}_{1}(t)|=O(\delta_{2}(t))$ if $0\leq
t\leq\frac{\ln{\ln{(\frac{1}{\epsilon})}}\ln{(\frac{1}{\epsilon})}}{\epsilon^{\frac{1}{2}}}.$
Now, similarly to the proof of Theorem 4.5, we denote
$G(s)=\max\big{(}\left\lVert\overrightarrow{g(s)}\right\rVert,\epsilon\big{)}.$
From Theorem 4.1 and Remark 4.4, we have that there are positive constants
$K,k>0$ independent of $\epsilon$ such that

$k\left\lVert\overrightarrow{g(t)}\right\rVert^{2}\leq
L_{1}(t)+K\epsilon^{2}.$

We recall that Theorem 2.9 implies that

$\ln{\Big{(}\frac{1}{\epsilon}\Big{)}}\lesssim
z(t),\,e^{-\sqrt{2}z(t)}+\max_{j}|\dot{x}_{j}(t)|^{2}+\max_{j\in\\{1,\,2\\}}|\ddot{x}_{j}(t)|=O(\epsilon),$

from which with the definition of $G(s)$ and estimates (241) and (242) we
deduce that

$\delta_{1}(t)\lesssim
G(t)v^{2}\operatorname{sech}{(\sqrt{2}vt+c)}^{2}\epsilon^{\frac{1}{2}}+G(t)\epsilon^{\frac{39}{20}}+G(t)^{2}\frac{\epsilon^{\frac{1}{2}}}{\ln{(\frac{1}{\epsilon})}},$

while $0\leq
t\leq\frac{\ln{\ln{(\frac{1}{\epsilon})}}\ln{(\frac{1}{\epsilon})}}{\epsilon^{\frac{1}{2}}}.$
In conclusion, the Fundamental Theorem of Calculus implies that $\exists K>0$
independent of $\epsilon$ such that

$G(t)^{2}\leq
K\left(G(0)^{2}+\int_{0}^{t}G(s)v^{2}\operatorname{sech}{(\sqrt{2}vs+c)}^{2}\epsilon^{\frac{1}{2}}+G(s)\epsilon^{\frac{39}{20}}+G(s)^{2}\frac{\epsilon^{\frac{1}{2}}}{\ln{(\frac{1}{\epsilon})}}\,ds\right),$
(243)

while $0\leq
t\leq\frac{\ln{\ln{(\frac{1}{\epsilon})}}\ln{(\frac{1}{\epsilon})}}{\epsilon^{\frac{1}{2}}}.$
Since
$\frac{d}{dt}[\tanh{(\sqrt{2}vt+c)}]=\sqrt{2}v\operatorname{sech}{(\sqrt{2}vt+c)}^{2},$
we verify that while the term
$G(s)v^{2}\operatorname{sech}{(\sqrt{2}vt+c)}^{2}\epsilon^{\frac{1}{2}}$ is
the dominant in the integral of the estimate (243), then $G(t)\lesssim G(0).$
The remaining case corresponds when
$G(s)^{2}\frac{\epsilon^{\frac{1}{2}}}{\ln{(\frac{1}{\epsilon})}}$ is the
dominant term in the integral of (243) from an instant $0\leq
t_{0}\leq\frac{\ln{\ln{(\frac{1}{\epsilon})}}\ln{(\frac{1}{\epsilon})}}{\epsilon^{\frac{1}{2}}}.$
Similarly to the proof of 4.5, we have for $t_{0}\leq
t\leq\frac{\ln{\ln{(\frac{1}{\epsilon})}}\ln{(\frac{1}{\epsilon})}}{\epsilon^{\frac{1}{2}}}$
that $G(t)\lesssim
G(t_{0})\exp\Big{(}C\frac{(t-t_{0})\epsilon^{\frac{1}{2}}}{\ln{(\frac{1}{\epsilon})}}\Big{)}.$
In conclusion, in any case we have for $0\leq
t\leq\frac{\ln{\ln{(\frac{1}{\epsilon})}}\ln{(\frac{1}{\epsilon})}}{\epsilon^{\frac{1}{2}}}$
that

$G(t)\lesssim
G(0)\exp\Big{(}C\frac{t\epsilon^{\frac{1}{2}}}{\ln{(\frac{1}{\epsilon})}}\Big{)}.$
(244)

However, for
$T\geq\frac{\ln{\ln{(\frac{1}{\epsilon})}}\ln{(\frac{1}{\epsilon})}}{\epsilon^{\frac{1}{2}}}$
and $K>2$ we have that

$\epsilon\ln{\Big{(}\frac{1}{\epsilon}\Big{)}}\exp\left(K\frac{\epsilon^{\frac{1}{2}}T}{\ln{(\frac{1}{2})}}\right)\leq\epsilon\exp\left(\frac{2K\epsilon^{\frac{1}{2}}T}{\ln{(\frac{1}{2})}}\right).$

In conclusion, from the result of Theorem 4.5, we can exchange the constant
$C>0$ by a larger constant such that estimate (244) is true for all $t\geq 0.$

##  Appendix A Auxiliary Results

We start the Appendix Section by presenting the following lemma:

######  Lemma A.1.

With the same hypothesis as in Theorem 1.11 and using its notation, we have
while $\max_{j\in\\{1\,,2\\}}|d_{j}(t)-x_{j}(t)|<1$ that
$\max_{j\in\\{1,\,2\\}}|\ddot{d}_{j}(t)-\ddot{x}_{j}(t)|=O\Big{(}\max_{j\in\\{1,\,2\\}}|d_{j}(t)-x_{j}(t)|\epsilon+\epsilon
z(t)e^{-\sqrt{2}z(t)}+\left\lVert\overrightarrow{g(t)}\right\rVert\epsilon^{\frac{1}{2}}\Big{)}.$

######  Lemma A.2.

For $U(\phi)=\phi^{2}(1-\phi^{2})^{2},$ we have that

$\dot{U}\left(H^{x_{1}(t)}_{-1,0}(x)+H^{x_{2}(t)}_{0,1}(x)\right)-\dot{U}\left(H^{x_{1}(t)}_{-1,0}(x)\right)-\dot{U}\left(H^{x_{2}(t)}_{0,1}(x)\right)=24e^{-\sqrt{2}z(t)}\left(\frac{H^{x_{1}(t)}_{-1,0}(x)}{(1+e^{-2\sqrt{2}(x-x_{1}(t))})^{\frac{1}{2}}}+\frac{H^{x_{2}(t)}_{0,1}(x)}{(1+e^{2\sqrt{2}(x-x_{2}(t))})^{\frac{1}{2}}}\right)\\\
-30e^{-\sqrt{2}z(t)}\left(\frac{H^{x_{1}(t)}_{-1,0}(x)^{3}}{(1+e^{-2\sqrt{2}(x-x_{1}(t))})^{\frac{1}{2}}}+\frac{H^{x_{2}(t)}_{0,1}(x)^{3}}{(1+e^{2\sqrt{2}(x-x_{2}(t))})^{\frac{1}{2}}}\right)+r(t,x),$

such that $\left\lVert
r(t)\right\rVert_{L^{2}_{x}(\mathbb{R})}=O(e^{-2\sqrt{2}z(t)}).$

###### Proof.

By directly computations, we verify that

$\dot{U}\left(H^{x_{1}(t)}_{-1,0}(x)+H^{x_{2}(t)}_{0,1}(x)\right)-\dot{U}\left(H^{x_{1}(t)}_{-1,0}(x)\right)-\dot{U}\left(H^{x_{2}(t)}_{0,1}(x)\right)=-24H^{x_{1}(t)}_{-1,0}H^{x_{2}(t)}_{0,1}(H^{x_{1}(t)}_{-1,0}+H^{x_{2}(t)}_{0,1})\\\
+30H^{x_{1}(t)}_{-1,0}H^{x_{2}(t)}_{0,1}((H^{x_{1}(t)}_{-1,0})^{3}+(H^{x_{2}(t)}_{0,1})^{3})+60(H^{x_{1}(t)}_{-1,0}H^{x_{2}(t)}_{0,1})^{2}(H^{x_{1}(t)}_{-1,0}+H^{x_{2}(t)}_{0,1}).$

First, from the definition of $H_{0,1}(x),$ we verify that

$\displaystyle
60(H^{x_{1}(t)}_{-1,0}H^{x_{2}(t)}_{0,1})^{2}(H^{x_{1}(t)}_{-1,0}+H^{x_{2}(t)}_{0,1})=60e^{-2\sqrt{2}z(t)}\left(\frac{H^{x_{2}(t)}_{0,1}}{(1+e^{2\sqrt{2}(x-x_{2}(t))})(1+e^{-2\sqrt{2}(x-x_{1}(t))})}\right)$
$\displaystyle+60e^{-2\sqrt{2}z(t)}\left(\frac{H^{x_{1}(t)}_{-1,0}}{(1+e^{-2\sqrt{2}(x-x_{1}(t))})(1+e^{2\sqrt{2}(x-x_{2}(t))})}\right).$

Using (4), we can verify using by induction for any $k\in\mathbb{N}$ that

$\left|\frac{d^{k}}{dx^{k}}\left[\frac{1}{(1+e^{2\sqrt{2}x})}\right]\right|=\left|\frac{d^{k}}{dx^{k}}\left[1-\frac{e^{2\sqrt{2}x}}{(1+e^{2\sqrt{2}x})}\right]\right|=\left|\frac{d^{k}}{dx^{k}}\left[H_{0,1}(x)^{2}\right]\right|=O(1),$
(245)

and since
$\frac{H_{0,1}(x)}{(1+e^{2\sqrt{2}x})}=\frac{e^{\sqrt{2}x}}{(1+e^{2\sqrt{2}x})^{\frac{3}{2}}}$
is a Schwartz function, we deduce that
$60(H^{x_{1}(t)}_{-1,0}H^{x_{2}(t)}_{0,1})^{2}(H^{x_{1}(t)}_{-1,0}+H^{x_{2}(t)}_{0,1})$
is in $H^{k}_{x}(\mathbb{R})$ for all $k>0$ and

$\left\lVert(H^{x_{1}(t)}_{-1,0}H^{x_{2}(t)}_{0,1})^{2}(H^{x_{1}(t)}_{-1,0}+H^{x_{2}(t)}_{0,1})\right\rVert_{H^{k}(\mathbb{R})}=O(e^{-2\sqrt{2}z(t)}).$
(246)

Next, using the identity

$H^{x_{1}(t)}_{-1,0}(x)H^{x_{2}(t)}_{0,1}(x)=-\frac{e^{-\sqrt{2}z(t)}}{(1+e^{2\sqrt{2}(x-x_{2}(t))})^{\frac{1}{2}}(1+e^{-2\sqrt{2}(x-x_{1}(t))})^{\frac{1}{2}}},$
(247)

the identity

$1-\frac{1}{(1+e^{2\sqrt{2}x})^{\frac{1}{2}}}=\frac{e^{2\sqrt{2}x}}{(1+e^{2\sqrt{2}x})^{\frac{1}{2}}+(1+e^{2\sqrt{2}x})},$

and Lemma 2.3, we deduce that

$\displaystyle\left\lVert
24(H^{x_{1}(t)}_{-1,0})^{2}H^{x_{2}(t)}_{0,1}+24e^{-\sqrt{2}z(t)}\frac{H^{x_{1}(t)}_{-1,0}(x)}{(1+e^{-2\sqrt{2}(x-x_{1}(t))})^{\frac{1}{2}}}\right\rVert_{L^{2}_{x}(\mathbb{R})}=O(e^{-2\sqrt{2}z(t)}),$
(248) $\displaystyle\left\lVert
30(H^{x_{1}(t)}_{-1,0})^{4}H^{x_{2}(t)}_{0,1}+30e^{-\sqrt{2}z(t)}\left(\frac{(H^{x_{1}(t)}_{-1,0}(x))^{3}}{(1+e^{-2\sqrt{2}(x-x_{1}(t))})^{\frac{1}{2}}}\right)\right\rVert_{L^{2}_{x}(\mathbb{R})}=O(e^{-3\sqrt{2}z(t)}).$
(249)

The estimate of the remaining terms
$-24H^{x_{1}(t)}_{-1,0}(H^{x_{2}(t)}_{0,1})^{2},\,30H^{x_{1}(t)}_{-1,0}(H^{x_{2}(t)}_{0,1})^{4}$
is completely analogous to (248) and (249) respectively. In conclusion, all of
the estimates above imply the estimate stated in the Lemma A.2. ∎

###### Proof of Lemma A.1.

First, we recall the global estimate $e^{-\sqrt{2}z(t)}\lesssim\epsilon.$ We
also recall the identity (27)

$\int_{\mathbb{R}}\big{(}8(H_{0,1}(x))^{3}-6(H_{0,1}(x))^{5}\big{)}e^{-\sqrt{2}x}\,dx=2\sqrt{2},$

and the global estimate $e^{-\sqrt{2}z(t)}\lesssim\epsilon.$ which, by
integration by parts, implies that

$\int_{\mathbb{R}}24\frac{H_{0,1}(x)\partial_{x}H_{0,1}(x)}{(1+e^{2\sqrt{2}(x)})^{\frac{1}{2}}}-30\frac{(H_{0,1}(x))^{3}\partial_{x}H_{0,1}(x)}{(1+e^{2\sqrt{2}(x)})^{\frac{1}{2}}}\,dx=4.$
(250)

We recall $d_{1}(t),\,d_{2}(t)$ defined in (7) and (8) respectively and
$d(t)=d_{2}(t)-d_{1}(t).$ Since, we have for $j\in\\{1,\,2\\}$ that
$\ddot{d}_{j}(t)=(-1)^{j}8\sqrt{2}e^{-\sqrt{2}d(t)},$ we have
$\ddot{d}(t)=16\sqrt{2}e^{-\sqrt{2}d(t)},$ which clearly with the fact that

$\left\lVert\partial_{x}H_{0,1}\right\rVert_{L^{2}}^{2}=\left\lVert\partial^{2}_{x}H_{0,1}\right\rVert_{L^{2}}^{2}=\frac{1}{2\sqrt{2}},$

imply that
$\ddot{d}_{j}(t)\left\lVert\partial_{x}H_{0,1}\right\rVert_{L^{2}}^{2}=(-1)^{j}4e^{-\sqrt{2}d(t)}.$
We also recall the partial differential equation satisfied by the remainder
$g(t,x)$ (155), which can be rewritten as

$\dot{U}\left(H^{x_{2}(t)}_{0,1}(x)+H^{x_{1}(t)}_{-1,0}(x)\right)-\dot{U}\left(H^{x_{1}(t)}_{-1,0}(x)\right)-\dot{U}\left(H^{x_{2}(t)}_{0,1}(x)\right)-\ddot{x}_{2}(t)\partial_{x}H^{x_{2}(t)}_{0,1}(x)=\\\
-\left(\partial^{2}_{t}g(t,x)-\partial^{2}_{x}g(t,x)+\ddot{U}\left(H^{x_{2}(t)}_{0,1}(x)+H^{x_{1}(t)}_{-1,0}(x)\right)g(t,x)\right)+\sum_{k=3}^{6}U^{(k)}\left(H^{x_{1}(t)}_{-1,0}+H^{x_{2}(t)}_{0,1}\right)\frac{g(t)^{k-1}}{(k-1)!}\\\
-\dot{x}_{1}(t)^{2}\partial^{2}_{x}H^{x_{1}(t)}_{-1,0}(x)-\dot{x}_{2}(t)^{2}\partial^{2}_{x}H^{x_{2}(t)}_{0,1}(x)+\ddot{x}_{1}(t)\partial_{x}H^{x_{1}(t)}_{-1,0}(x).$
(251)

In conclusion, from the estimate (250), Lemma A.2 and Lemma 2.3, we obtain
that

$\left\langle\dot{U}\left(H^{x_{1}(t)}_{-1,0}+H^{x_{2}(t)}_{0,1}\right)-\dot{U}\left(H^{x_{1}(t)}_{-1,0}\right)-\dot{U}\left(H^{x_{2}(t)}_{0,1}\right),\,\partial_{x}H^{x_{2}(t)}_{0,1}\right\rangle_{L^{2}(\mathbb{R})}-\ddot{x}_{2}(t)\left\lVert\partial_{x}H_{0,1}\right\rVert_{L^{2}}^{2}=-(\ddot{x}_{2}(t)-\ddot{d}_{2}(t))\left\lVert\partial_{x}H_{0,1}\right\rVert_{L^{2}}^{2}\\\
+O\Big{(}|\ddot{x}_{1}(t)|z(t)e^{-\sqrt{2}z(t)}+e^{-\sqrt{2}z(t)}\max_{j\in\\{1,\,2\\}}|x_{j}(t)-d_{j}(t)|+e^{-2\sqrt{2}z(t)}z(t)\Big{)}.$
(252)

We recall from the proof of Theorem 4.1 the following estimate

$\left|\int_{\mathbb{R}}\left[\ddot{U}\left(H^{x_{2}(t)}_{0,1}(x)\right)-\ddot{U}\left(H^{x_{2}(t)}_{0,1}(x)+H^{x_{1}(t)}_{-1,0}(x)\right)\right]\partial_{x}H^{x_{2}(t)}_{0,1}(x)g(t,x)\,dx\right|=O\Big{(}\left\lVert\overrightarrow{g(t)}\right\rVert
e^{-\sqrt{2}z(t)}\Big{)}.$

Also, from the Modulation Lemma, we have that

$\displaystyle\langle\partial^{2}_{t}g(t),\partial_{x}H^{x_{2}(t)}_{0,1}\rangle_{L^{2}}$
$\displaystyle=\frac{d}{dt}\left[\langle\partial_{t}g(t),\partial_{x}H^{x_{2}(t)}_{0,1}\rangle_{L^{2}}\right]+\dot{x}_{2}(t)\langle\partial_{t}g(t),\partial_{x}H^{x_{2}(t)}_{0,1}\rangle_{L^{2}}$
$\displaystyle=\frac{d}{dt}\Big{[}\dot{x}_{2}(t)\langle
g(t),\partial^{2}_{x}H^{x_{2}(t)}_{0,1}\rangle_{L^{2}}\Big{]}+\dot{x}_{2}(t)\langle\partial_{t}g(t),\partial_{x}H^{x_{2}(t)}_{0,1}\rangle_{L^{2}}$
$\displaystyle=\ddot{x}_{2}(t)\langle
g(t),\partial^{2}_{x}H^{x_{2}(t)}_{0,1}\rangle_{L^{2}}+2\dot{x}_{2}(t)\langle\partial_{t}g(t),\partial_{x}H^{x_{2}(t)}_{0,1}\rangle_{L^{2}}.$

In conclusion, since $\partial_{x}H^{x_{2}(t)}_{0,1}\in
kerD^{2}E_{pot}\left(H^{x_{2}(t)}_{0,1}\right),$ we obtain from (252) and
(251) that

$|\ddot{x}_{2}(t)-\ddot{d}_{2}(t)|=O\Big{(}\max_{j\in\\{1,\,2\\}}|d_{j}(t)-x_{j}(t)|\epsilon+\epsilon
z(t)e^{-\sqrt{2}z(t)}\\\ +\left\lVert\overrightarrow{g(t)}\right\rVert
e^{-\sqrt{2}z(t)}+\left\lVert\overrightarrow{g(t)}\right\rVert\epsilon^{\frac{1}{2}}\Big{)},$

the estimate of $|\ddot{x}_{1}(t)-\ddot{d}_{1}(t)|$ is completely analogous,
which finishes the demonstration. ∎

######  Lemma A.3.

For any $\delta>0$ there is a $\epsilon(\delta)>0$ such that if

$\left\lVert\phi(x)-H_{0,1}(x)\right\rVert_{H^{1}(\mathbb{R})}<+\infty,\,0<E_{pot}(\phi(x))-E_{pot}(H_{0,1})<\epsilon(\delta),$
(253)

then there is a real number $y$ such that

$\left\lVert\phi(x)-H_{0,1}(x-y)\right\rVert_{H^{1}}\leq\delta.$

###### Proof of Lemma A.3..

The proof of Lemma 2.6 will follow by a contradiction argument. We assume that
there is a $c>0$ and sequence of real functions $\left(\phi_{n}(x)\right)_{n}$
satisfying

$\displaystyle\lim_{n\to+\infty}E_{pot}(\phi_{n})=E_{pot}(H_{0,1}),$ (254)
$\displaystyle\left\lVert\phi_{n}(x)-H_{0,1}(x)\right\rVert_{H^{1}(\mathbb{R})}<+\infty,$
(255)

such that

$\inf_{y\in\mathbb{R}}\left\lVert\phi_{n}(x)-H_{0,1}(x+y)\right\rVert_{H^{1}(\mathbb{R})}>c.$
(256)

Since $U(\phi)=\phi^{2}(1-\phi^{2})^{2}$ and
$\left|E_{pot}(\phi_{n})-E_{pot}(H_{0,1})\right|\ll 1$ for $1\ll n,$ it is not
difficult from the definition of the potential energy functional $E_{pot}$
that if $1\ll n,$ then

$\left\lVert\phi_{n}(x)-1\right\rVert_{L^{2}\left(\\{x|\phi_{n}(x)>1\\}\right)}^{2}+\left\lVert\frac{d\phi_{n}(x)}{dx}\right\rVert_{L^{2}\left(\\{x|\phi_{n}(x)>1\\}\right)}^{2}\lesssim\left|E_{pot}(\phi_{n})-E_{pot}(H_{0,1})\right|.$

By an analogous argument, we can verify that

$\left\lVert\phi_{n}(x)\right\rVert_{L^{2}\left(\\{x|-\frac{1}{2}<\phi_{n}(x)<0\\}\right)}^{2}+\left\lVert\frac{d\phi_{n}(x)}{dx}\right\rVert_{L^{2}\left(\\{x|-\frac{1}{2}<\phi_{n}(x)<0\\}\right)}^{2}\lesssim\left|E_{pot}(\phi_{n})-E_{pot}(H_{0,1})\right|,$

and if there is $x_{0}\in\mathbb{R}$ such that
$\phi_{n}(x_{0})\leq-\frac{1}{2},$ we would obtain that

$\displaystyle\int_{x_{0}}^{+\infty}\frac{1}{2}\frac{d\phi_{n}(x)}{dx}^{2}+U(\phi_{n}(x))\,dx=\int_{x_{0}}^{+\infty}\sqrt{2U(\phi_{n}(x))}\left|\frac{d\phi_{n}(x)}{dx}\right|\,dx+\frac{1}{2}\int_{x_{0}}^{+\infty}\left(\left|\frac{d\phi_{n}(x)}{dx}\right|-\sqrt{2U(\phi_{n}(x))}\right)^{2}\,dx$
$\displaystyle\geq\int_{-\frac{1}{2}}^{1}\sqrt{2U(\phi)}\,d\phi=E_{pot}(H_{0,1})+\int_{-\frac{1}{2}}^{0}\sqrt{2U(\phi)}\,d\phi>E_{pot}(H_{0,1}),$

which contradicts (254) if $n\gg 1.$ In conclusion, we can restrict the proof
to the case where $0\leq\phi_{n}(x)\leq 1$ and $n\gg 1.$ Now, from the density
of $H^{2}(\mathbb{R})$ in $H^{1}(\mathbb{R}),$ we can also restrict the
contradiction hypotheses to the situation where $\frac{d\phi_{n}}{dx}(x)$ is a
continuous function for all $n\in\mathbb{N}.$ Also, we have that if
$\left\lVert\phi(x)-H_{0,1}(x)\right\rVert_{H^{1}(\mathbb{R})}<+\infty,$ then
$E_{pot}(\phi(x))\geq E_{pot}(H_{0,1}(x)).$ In conclusion, there is a sequence
of positive numbers $\left(\epsilon_{n}\right)_{n}$ such that

$E_{pot}(\phi_{n})=E_{pot}(H_{0,1})+\epsilon_{n},\,\lim_{n\to+\infty}\epsilon_{n}=0.$

Also, $\tau_{y}\phi(x)=\phi(x-y)$ satisfies
$E_{pot}(\phi(x))=E_{pot}(\tau_{y}\phi(x))$ for any $y\in\mathbb{R}.$ In
conclusion, since for all $n\in\mathbb{N},$ $\lim_{x\to+\infty}\phi_{n}(x)=1$
and $\lim_{x\to-\infty}\phi_{n}(x)=0,$ we can restrict to the case where

$\phi_{n}(0)=\frac{1}{\sqrt{2}},$

for all $n\in\mathbb{N}.$ For $(v)_{+}=\max(v,0)$ and
$(v)_{-}=-\left(v-(v)_{+}\right),$ since $\frac{d\phi_{n}(x)}{dx}$ is a
continuous function on $x,$ we deduce that
$\left(\frac{d\phi_{n}(x)}{dx}\right)_{+}$ and
$\left(\frac{d\phi_{n}(x)}{dx}\right)_{-}$ are also continuous functions on
$x$ for all $n\in\mathbb{N}.$ In conclusion, for any $n\in\mathbb{N},$ we have
that the set

$U=\left\\{x\in\mathbb{R}|\,\frac{d\phi_{n}(x)}{dx}<0\right\\}$ (257)

is an enumerable union of disjoint open intervals
$(a_{k,n},b_{k,n})_{k\in\mathbb{N}}$, which are bounded, since
$\lim_{x\to+\infty}\phi_{n}(x)=1,\,\lim_{x\to-\infty}\phi_{n}(x)=0$ and
$0\leq\phi_{n}(x)\leq 1.$ Also, we observe that for every collection $E$ of
disjoint open bounded intervals $(h_{i,n},l_{i,n})\subset\mathbb{R}$
satisfying the conditions

$\phi_{n}(h_{i,n})=\phi_{n}(l_{i,n}),$ (258)

$i\in I\subset\mathbb{Z}$ and $l_{i,n}<h_{i+1,n}$ that the following function

$f_{i,n}(x)=\begin{cases}\phi_{n}(x)\text{ if $x\leq h_{i,n}$},\\\
\phi_{n}(x+l_{i,n}-h_{i,n})\text{ if $x>h_{i,n}$},\end{cases}$

satisfies $E_{pot}(H_{0,1})\leq E_{pot}(f_{k,n})\leq
E_{pot}(\phi_{n})=E_{pot}(H_{0,1})+\epsilon_{n},$ which implies that

$\int_{h_{i,n}}^{l_{i,n}}\frac{1}{2}\frac{d\phi_{n}(x)}{dx}^{2}+U(\phi_{n}(x))\leq\epsilon_{n}.$

Furthermore, we can deduce from Lebesgue’s dominated convergence theorem that

$\sum_{i\in
I}\int_{h_{i,n}}^{l_{i,n}}\frac{1}{2}\frac{d\phi_{n}(x)}{dx}^{2}+U(\phi_{n}(x))\leq\epsilon_{n},$
(259)

for every finite or enumerable collection $E$ of disjoint open bounded
intervals $(h_{i,n},l_{i,n})\subset\mathbb{R},\,i\in I\subset\mathbb{Z}$ such
that $\phi_{n}(h_{i,n})=\phi_{n}(l_{i,n}).$ In conclusion, we can deduce from
(259) that

$\int_{\mathbb{R}}\left(\frac{d\phi_{n}(x)}{dx}\right)_{-}^{2}\,dx\leq
2\epsilon_{n},$ (260)

and so for $1\ll n$ we have that

$\left\lVert\frac{d\phi_{n}(x)}{dx}-\left|\frac{d\phi_{n}(x)}{dx}\right|\right\rVert_{L^{2}(\mathbb{R})}^{2}\leq
8\epsilon_{n},\,\phi_{n}(0)=\frac{1}{\sqrt{2}}.$ (261)

Moreover, we can verify that

$E_{pot}(\phi_{n})=\frac{1}{2}\left[\int_{\mathbb{R}}\left(\left|\frac{d\phi_{n}(x)}{dx}\right|-\sqrt{2U(\phi_{n}(x))}\right)^{2}\,dx\right]+\int_{\mathbb{R}}\sqrt{2U(\phi_{n}(x))}\left|\frac{d\phi_{n}(x)}{dx}\right|\,dx,$

from which we deduce with $\lim_{x\to-\infty}\phi_{n}(x)=0$ and
$\lim_{x\to+\infty}\phi_{n}(x)=1$ that

$\displaystyle E_{pot}(H_{0,1})+\epsilon_{n}=E_{pot}(\phi_{n})$
$\displaystyle\geq\frac{1}{2}\left[\int_{\mathbb{R}}\left(\left|\frac{d\phi_{n}(x)}{d(x)}\right|-\sqrt{2U(\phi_{n}(x))}\right)^{2}\,dx\right]+\int_{0}^{1}\sqrt{2U(\phi)}\,d\phi$
$\displaystyle=\frac{1}{2}\left[\int_{\mathbb{R}}\left(\left|\frac{d\phi_{n}(x)}{d(x)}\right|-\sqrt{2U(\phi_{n}(x))}\right)^{2}\,dx\right]+E_{pot}(H_{0,1}).$

Then, from estimate (261), we have that

$\frac{d\phi_{n}(x)}{dx}=\sqrt{2U(\phi_{n}(x))}+r_{n}(x),\,\phi_{n}(0)=\frac{1}{\sqrt{2}},$
(262)

with $\left\lVert r_{n}\right\rVert_{L^{2}(\mathbb{R})}\lesssim\epsilon_{n}$
for all $1\ll n.$ We recall that $U(\phi)=\phi^{2}(1-\phi^{2})^{2}$ is a
Lipschitz function in the set $\\{\phi|\,0\leq\phi\leq 1\\}.$ Then, because
$H_{0,1}(x)$ is the unique solution of the following ordinary differential
equation

$\begin{cases}\frac{d\phi(x)}{dx}=\sqrt{2U(\phi(x))},\\\
\phi(0)=\frac{1}{\sqrt{2}},\end{cases}$

we deduce from Gronwall Lemma that for any $K>0$ we have

$\lim_{n\to+\infty}\left\lVert\phi_{n}(x)-H_{0,1}(x)\right\rVert_{L^{\infty}[-K,K]}=0,\,\lim_{n\to+\infty}\left\lVert\frac{d\phi_{n}(x)}{dx}-\dot{H}_{0,1}(x)\right\rVert_{L^{2}[-K,K]}=0.$
(263)

Also, if $1\ll n,$ then
$\left\lVert\frac{d\phi_{n}(x)}{dx}\right\rVert_{L^{2}_{x}(\mathbb{R})}^{2}<2E_{pot}(H_{0,1})+1,$
and so we obtain from Cauchy-Schwartz inequality that

$\left|\phi_{n}(x)-\phi_{n}(y)\right|\leq|x-y|^{\frac{1}{2}}\left\lVert\frac{d\phi_{n}}{dx}\right\rVert_{L^{2}(\mathbb{R})}^{2}<M|x-y|^{\frac{1}{2}},$
(264)

for a constant $M>0.$ The inequality (264) implies that for any $1>\omega>0$
there is a number $h(\omega)\in\mathbb{N}$ such that if $n\geq h(\omega)$ then

$|\phi_{n}(x)-H_{0,1}(x)|_{L^{\infty}\\{x|\,\frac{1}{\omega}<|x|\\}}<\omega,$
(265)

otherwise we would obtain that there is $0<\theta<\frac{1}{4}$ and a
subsequence $(m_{n})_{n\in\mathbb{N}}$ and a sequence of real numbers
$(x_{n})_{n\in\mathbb{N}}$ with
$\lim_{n\to+\infty}m_{n}=+\infty,\,|x_{n}|>n+1$ such that

$\displaystyle\left|\phi_{m_{n}}(x_{n})-1\right|>\theta\text{ if $x_{n}>0,$}$
(266) $\displaystyle\left|\phi_{m_{n}}(x_{n})\right|>\theta\text{ if
$x_{n}<0.$}$ (267)

However, since we are considering $\phi_{n}(x)\in H^{2}(\mathbb{R})\subset
C^{1}(\mathbb{R})$ and $0\leq\phi_{n}\leq 1,$ we would obtain from the mean
value theorem that there would exist a sequence $(y_{n})_{n}$ with
$y_{n}>x_{n}>n+1$ or $y_{n}<x_{n}<-n-1$ such that

$\displaystyle 1-\theta\leq\phi_{m_{n}}(y_{n})\leq 1+\theta,\text{ if
$y_{n}>0,$}$ (268) $\displaystyle\phi_{m_{n}}(y_{n})=\theta\text{ otherwise.}$
(269)

But, estimates (264), (268), (269) and identity
$U(\phi)=\phi^{2}(1-\phi^{2})^{2}$ would imply that

$1\lesssim\int_{|x|\geq n-2}U(\phi_{m_{n}}(x))\,dx\text{ for all $n\gg 1,$}$
(270)

and because of estimate (263) and the following identity

$\lim_{K\to+\infty}\int_{-K}^{K}\frac{1}{2}\dot{H}_{0,1}(x)^{2}+U(H_{0,1}(x))=E_{pot}(H_{0,1}(x)),$
(271)

estimate (270) would imply that
$\lim_{n\to+\infty}E_{pot}(\phi_{m_{n}})>E_{pot}(H_{0,1})$ which contradicts
our hypotheses. In conclusion, for any $1>\omega>0$ there is a number
$h(\omega)$ such that if $n\geq h(\omega)$ then (265) holds. So we deduce for
any $0<\omega<1$ that there is a number $h_{1}(\omega)$ such that

$\text{if $n\geq h_{1}(\omega),$ then
}|\phi_{n}(x)-H_{0,1}(x)|\leq\omega\text{ for all $x\in\mathbb{R}.$}$ (272)

Then, if $\omega\leq\frac{1}{100},\,n\geq h(\omega)$ and $K\geq 200,$
estimates (272) and (263) imply that

$\displaystyle\int_{K}^{+\infty}U(\phi_{n}(x))+\frac{1}{2}\frac{d\phi_{n}(x)}{dx}^{2}\,dx\geq\frac{1}{2}\int_{K}^{+\infty}\left(1-\phi_{n}(x)\right)^{2}+\frac{d\phi_{n}(x)}{dx}^{2}\,dx,$
(273)
$\displaystyle\int_{-\infty}^{-K}U(\phi_{n}(x))+\frac{1}{2}\frac{d\phi_{n}(x)}{dx}^{2}\,dx\geq\frac{1}{2}\int_{-\infty}^{-K}\phi_{n}(x)^{2}+\frac{d\phi_{n}(x)}{dx}^{2}\,dx.$
(274)

In conclusion, from estimates (272), (273), (274) and

$\lim_{K\to+\infty}\int_{|x|\geq
K}\frac{1}{2}\dot{H}_{0,1}(x)^{2}+U(H_{0,1}(x))\,dx=0,$

we obtain that
$\lim_{n\to+\infty}\left\lVert\phi_{n}-H_{0,1}(x)\right\rVert_{L^{2}(\mathbb{R})}=0$
and, from the equation in (262) is satisfied for each $\phi_{n},$ we conclude
that
$\lim_{n\to+\infty}\left\lVert\frac{d\phi_{n}}{dx}-\dot{H}_{0,1}(x)\right\rVert_{L^{2}(\mathbb{R})}=0.$
In conclusion, if $1\ll n,$ inequality (256) is false. ∎

From Lemma A.3, we obtain the following corollary:

######  Corollary A.4.

For any $\delta>0$ there is a $\epsilon_{0}>0$ such that if
$\epsilon\leq\epsilon_{0},\,\left\lVert\phi(x)-H_{0,1}(x)-H_{-1,0}(x)\right\rVert_{H^{1}(\mathbb{R})}<+\infty$
and $E_{pot}(\phi)=2E_{pot}(H_{0,1})+\epsilon,$ then there are
$x_{2},x_{1}\in\mathbb{R}$ such that

$x_{2}-x_{1}\geq\frac{1}{\delta},\,\left\lVert\phi(x)-H_{0,1}(x-x_{2})+H_{-1,0}(x-x_{1})\right\rVert_{H^{1}_{x}(\mathbb{R})}\leq\delta.$
(275)

###### proof of Corollary A.4..

First, from a similar reasoning to the proof of Lemma A.3 we can assume by
density that $\frac{d\phi(x)}{dx}\in H^{1}_{x}(\mathbb{R}).$ Next, from
hypothesis
$\left\lVert\phi(x)-H_{0,1}(x)-H_{-1,0}(x)\right\rVert_{H^{1}(\mathbb{R})}<+\infty,$
we deduce using the mean value theorem that there is an $y\in\mathbb{R}$ such
that $\phi(y)=0.$ Now, we consider the functions

$\phi_{-}(x)=\begin{cases}\phi(x)\text{ if $x\leq y,$}\\\ 0\text{
otherwise,}\end{cases}$

and

$\phi_{+}(x)=\begin{cases}0\text{ if $x\leq y,$}\\\ \phi(x)\text{
otherwise.}\end{cases}$

Clearly, $\phi(x)=\phi_{-}(x)$ for $x<y$ and $\phi(x)=\phi_{+}(x)$ for $x>y.$
From identity $U(0)=0,$ we deduce that

$E_{pot}(\phi)=E_{pot}(\phi_{-})+E_{pot}(\phi_{+}),$

also we have that

$E_{pot}(H_{-1,0})<E_{pot}(\phi_{-}),\,E_{pot}(H_{0,1})<E_{pot}(\phi_{+}).$

In conclusion, since $E_{pot}(\phi)=2E_{pot}(H_{0,1})+\epsilon,$ Lemma A.3
implies that if $\epsilon<\epsilon_{0}\ll 1,$ then there is
$x_{2},\,x_{1}\in\mathbb{R}$ such that

$\left\lVert\phi(x)-H_{0,1}(x-y_{2})-H_{-1,0}(x-y_{1})\right\rVert_{H^{1}}\leq\left\lVert\phi_{+}-H_{0,1}(x-y_{2})\right\rVert_{H^{1}}+\left\lVert\phi_{-}-H_{-1,0}(x-y_{1})\right\rVert_{H^{1}}\leq\delta.$
(276)

So, to finish the proof of Corollary A.4, we need only to verify that we have
$x_{2}-x_{1}\geq\frac{1}{\delta}$ if $0<\epsilon_{0}\ll 1.$ But, we recall
that $H_{0,1}(0)=\frac{1}{\sqrt{2}},$ from which with estimate (276) we deduce
that

$\left|\phi_{+}(x_{2})-\frac{1}{\sqrt{2}}\right|\lesssim\delta,\left|\phi_{-}(x_{1})+\frac{1}{\sqrt{2}}\right|\lesssim\delta,$
(277)

so if $\epsilon_{0}\ll 1,$ then $x_{1}<y<x_{2}.$ From Lemma 2.8, we can verify
that $f(z)=\left\lVert
DE_{pot}(H_{0,1}^{z}(x)+H_{-1,0}(x))\right\rVert_{L^{2}}$ is a bounded
function in $\mathbb{R}_{+},$ from which with estimate (276) we deduce that if
$0<\epsilon_{0}\ll 1,$ then

$\left|E_{pot}(\phi)-E_{pot}\left(H_{0,1}(x-x_{2})+H_{-1,0}(x-x_{1})\right)\right|<e^{-2\sqrt{2}\frac{1}{\delta}}.$

In conclusion, we obtain from Lemma 2.4 and the estimate above that
$x_{2}-x_{1}\geq\frac{1}{\delta}$ if $0<\epsilon_{0}\ll 1$ and
$\epsilon<\epsilon_{0}.$ ∎

Now, we complement our material by presenting the proof of Identity (27) and
the proof of The Modulation Lemma.

###### Proof of Identity (27).

From the definition of the function $H_{0,1}(x),$ we have

$\int_{\mathbb{R}}\big{(}8(H_{0,1}(x))^{3}-6(H_{0,1}(x))^{5}\big{)}e^{-\sqrt{2}x}\,dx=\int_{\mathbb{R}}\frac{8e^{2\sqrt{2}x}+2e^{4\sqrt{2}x}}{(1+e^{2\sqrt{2}x})^{\frac{5}{2}}}\,dx,$

by the change of variable $y(x)=(1+e^{2\sqrt{2}x}),$ we obtain

$\displaystyle\int_{\mathbb{R}}\big{(}8(H_{0,1}(x))^{3}-6(H_{0,1}(x))^{5}\big{)}e^{-\sqrt{2}x}\,dx$
$\displaystyle=\frac{1}{2\sqrt{2}}\int_{1}^{\infty}\frac{8}{y^{\frac{5}{2}}}+\frac{2(y-1)}{y^{\frac{5}{2}}}\,dy=\frac{1}{2\sqrt{2}}\int_{1}^{\infty}\frac{6}{y^{\frac{5}{2}}}+\frac{2}{y^{\frac{3}{2}}}\,dy,$
$\displaystyle=\frac{1}{2\sqrt{2}}(-4y^{-\frac{3}{2}}-4y^{-\frac{1}{2}})\Big{|}_{1}^{\infty}=2\sqrt{2}.$

∎

###### Proof of the Modulation Lemma.

First, let $x_{2},\,x_{1}\in\mathbb{R}$ and $g\in H^{1}(\mathbb{R})$ such that
$x_{2}-x_{1}\geq\frac{1}{\delta_{0}}$ with $\delta_{0}>0$ small enough to be
chosen later. Then, we define the following map $F:\mathbb{R}^{2}\times
H^{1}(\mathbb{R})\to\mathbb{R}^{2}$ by

$F((h_{2},h_{1}),g(x))=\begin{bmatrix}\langle\partial_{x}H^{x_{2}+h_{2}}_{0,1},\,H^{x_{2}}_{0,1}+H^{x_{1}}_{-1,0}-H^{x_{1}+h_{1}}_{-1,0}+g\rangle_{L^{2}}\\\
\langle\partial_{x}H^{x_{1}+h_{1}}_{-1,0},\,H^{x_{2}}_{0,1}+H^{x_{1}}_{-1,0}-H^{x_{2}+h_{2}}_{0,1}+g\rangle_{L^{2}}\\\
\end{bmatrix}$

for any $((h_{1},h_{2}),g)\in\mathbb{R}^{2}\times H^{1}(\mathbb{R}).$ Clearly,
$F(0,0,0)=(0,0),$ also, we can verify that the Derivative
$DF_{h_{2},h_{1}}((0,0),g)$ is given by

$\begin{bmatrix}\left\lVert\partial_{x}H_{0,1}\right\rVert_{L^{2}}^{2}+\langle\partial_{x}H^{x_{2}}_{0,1},\,\frac{dg}{dx}\rangle&\langle\partial_{x}H^{x_{2}}_{0,1},\,\partial_{x}H^{x_{1}}_{-1,0}\rangle\\\
\langle\partial_{x}H^{x_{2}}_{0,1},\,\partial_{x}H^{x_{1}}_{-1,0}\rangle&\left\lVert\partial_{x}H_{0,1}\right\rVert_{L^{2}}^{2}+\langle\partial_{x}H^{x_{1}}_{-1,0},\,\frac{dg}{dx}\rangle\end{bmatrix}.$

Then,
$R_{g}(h_{2},h_{1})=F(h_{2},h_{1},g)-F(0,0,g)-DF_{h_{2},h_{1}}(0,0,g)(h_{2},h_{1})$
satisfies the following identity

$R_{g}(h_{2},h_{1})=\begin{bmatrix}\langle\partial_{x}H^{x_{2}+h_{2}}_{0,1}-\partial_{x}H^{x_{2}}_{0,1}+h_{2}\partial^{2}_{x}H^{x_{2}}_{0,1},\,g\rangle\\\
\langle\partial_{x}H^{x_{1}+h_{1}}_{-1,0}-\partial_{x}H^{x_{1}}_{-1,0}+h_{1}\partial^{2}_{x}H^{x_{1}}_{-1,0},\,g\rangle\end{bmatrix}+\begin{bmatrix}\langle\partial_{x}H^{x_{2}+h_{2}}_{0,1},\,H^{x_{2}}_{0,1}-H^{x_{2}+h_{2}}_{0,1}-h_{2}\partial_{x}H^{x_{2}+h_{2}}_{0,1}\rangle\\\
\langle\partial_{x}H^{x_{1}+h_{1}}_{-1,0},\,H^{x_{1}}_{-1,0}-H^{x_{1}+h_{1}}_{-1,0}-h_{1}\partial_{x}H^{x_{1}+h_{1}}_{-1,0}\rangle\end{bmatrix}\\\
+\begin{bmatrix}\langle\partial_{x}H^{x_{2}+h_{2}}_{0,1},\,H^{x_{1}}_{-1,0}-H^{x_{1}+h_{1}}_{-1,0}\rangle-
h_{1}\langle\partial_{x}H^{x_{2}}_{0,1},\,\partial_{x}H^{x_{1}}_{-1,0}\rangle\\\
\langle\partial_{x}H^{x_{1}+h_{1}}_{-1,0},\,H^{x_{2}}_{0,1}-H^{x_{2}+h_{2}}_{0,1}\rangle-
h_{2}\langle\partial_{x}H^{x_{2}}_{0,1},\,\partial_{x}H^{x_{1}}_{-1,0}\rangle\end{bmatrix}$
(278)

for all $(h_{2},h_{1})\in\mathbb{R}^{2},$ also it is not difficult to verify
that $R_{g}(0,0)=(0,0).$ Also for $\delta_{0}>0$ small enough, if
$\max\\{|h_{1}|,|h_{2}|\\}=O(\delta_{0}),\,\left\lVert
g\right\rVert_{H^{1}(\mathbb{R})}\leq\delta_{0},$ it is not difficult to
verify from Lemma 2.3 that

$|R_{g}(h_{2},h_{1})|\lesssim\left\lVert
g\right\rVert_{H^{1}}\max\\{|h_{1}|,|h_{2}|\\}^{2}+\max\\{|h_{1}|,|h_{2}|\\}^{3}+\max\\{|h_{1}|,|h_{2}|\\}^{2}(x_{2}-x_{1})e^{-\sqrt{2}(x_{2}-x_{1})},$

from which we deduce that

$|R_{g}(h_{2},h_{1})|=O\Big{(}(\delta_{0}^{2}+e^{-\frac{\sqrt{2}}{2\delta_{0}}})(|h_{1}|+|h_{2}|)\Big{)},$
(279)

for any $((h_{2},h_{1}),g)\in\mathbb{R}^{2}\times H^{1}(\mathbb{R})$ such that
$\max\\{|h_{2}|,|h_{1}|\\}=O(\delta_{0})$ and $\left\lVert
g\right\rVert_{H^{1}}\leq\delta_{0}.$ In particular, estimate (279) implies
that $DF_{h_{2},h_{1}}((h_{2},h_{1}),g)$ is an uniformly non-degenerate
matrix, for any $(h_{2},h_{1}),\,(x_{2},x_{1})\in\mathbb{R}^{2}$ and $g\in
H^{1}(\mathbb{R})$ such $x_{2}-x_{1}\geq\frac{1}{\delta_{0}},\left\lVert
g\right\rVert_{H^{1}}\leq\delta_{0}$ and
$\max\\{|h_{2}|,|h_{1}|\\}=O(\delta_{0}).$ As a consequence, the result of the
Modulation Lemma follows from the Implicit Function Theorem for Banach Spaces
with the fact that $F((0,0),0)=(0,0)$. ∎

##  Appendix B Optimality of Theorem 1.7

######  Theorem B.1.

In notation of Theorem 1.7, for any constant $C>0$ and any function
$s:\mathbb{R}_{+}\to\mathbb{R}_{+}$ with $\lim_{h\to 0}s(h)=0,$ we can find a
positive value $\delta(s)$ such that if $0<\epsilon\leq\delta(s),$ then for
any $\left\lVert\overrightarrow{g(0)}\right\rVert\leq\epsilon s(\epsilon)$
there is a
$0<T\lesssim\frac{\ln{(\frac{1}{\epsilon})}}{\epsilon^{\frac{1}{2}}}$ such
that $\epsilon\lesssim\left\lVert\overrightarrow{g(T)}\right\rVert.$

###### Proof of Optimality of Theorem 1.7.

We use the notations of Theorem 1.7 and Theorem 1.11. Clearly, if the result
of Theorem B.1 is false, then by contradiction there is a function
$q:\mathbb{R}_{+}\to\mathbb{R}_{+}$ with $\lim_{h\to 0}q(h)=0$ such that for
any $1\ll N\in\mathbb{N}$ is possible to have

$\left\lVert\overrightarrow{g(t)}\right\rVert\leq q(\epsilon)\epsilon$ (280)

for all $0\leq t\leq
N\frac{\ln{(\frac{1}{\epsilon})}}{\epsilon^{\frac{1}{2}}}=T$ if $\epsilon\ll
1$ enough. From Modulation Lemma, we can denote the solution $\phi(t,x)$ as

$\phi(t,x)=H^{x_{1}(t)}_{-1,0}(x)+H^{x_{2}(t)}_{0,1}(x)+g(t,x),$

such that

$\langle
g(t,x),\,\partial_{x}H^{x_{1}(t)}_{-1,0}(x)\rangle_{L^{2}(\mathbb{R})}=0,\,\langle
g(t,x),\,\partial_{x}H^{x_{2}(t)}_{0,1}(x)\rangle_{L^{2}(\mathbb{R})}=0.$

Also, for all $t\geq 0,$ we have that $g(t,x)$ has a unique representation as

$g(t,x)=P_{1}(t)\partial^{2}_{x}H^{x_{1}(t)}_{-1,0}(x)+P_{2}(t)\partial^{2}_{x}H^{x_{2}(t)}_{0,1}(x)+r(t,x),$
(281)

such that $r(t)$ satisfies the following new orthogonality conditions

$\langle
r(t),\,\partial^{2}_{x}H^{x_{1}(t)}_{-1,0}\rangle_{L^{2}(\mathbb{R})}=0,\,\langle
r(t),\,\partial^{2}_{x}H^{x_{2}(t)}_{0,1}\rangle_{L^{2}(\mathbb{R})}=0.$ (282)

In conclusion, we deduce that

$\left\lVert
g(t,x)\right\rVert_{L^{2}(\mathbb{R})}^{2}=\left\lVert\ddot{H}_{0,1}(x)\right\rVert_{L^{2}}^{2}(P_{1}^{2}+P_{2}^{2})+\left\lVert
r(t)\right\rVert_{L^{2}_{x}(\mathbb{R}}^{2}+2P_{1}P_{2}\langle\ddot{H}^{z(t)}_{0,1}(x),\,\ddot{H}_{-1,0}(x)\rangle_{L^{2}(\mathbb{R})}.$
(283)

We recall from Theorem 2.9 that
$\frac{1}{\sqrt{2}}\ln{(\frac{1}{\epsilon})}<z(t)$ for all $t\geq 0.$ Since,
from Lemma 2.3, we have that
$\langle\partial^{2}_{x}H^{x_{1}(t)}_{-1,0},\,\partial^{2}_{x}H^{x_{2}(t)}_{0,1}\rangle\lesssim
z(t)e^{-\sqrt{2}z(t)}$ and
$z(t)e^{-\sqrt{2}z(t)}\lesssim\epsilon\ln{(\frac{1}{\epsilon})}$ if
$0<\epsilon\ll 1,$ we deduce from the equation (283) that there is a uniform
constant $K>1$ such that for all $t\geq 0$ we have the following estimate

$\frac{\left\lVert
g(t)\right\rVert_{L^{2}}}{K}\leq|P_{1}(t)|+|P_{2}(t)|+\left\lVert
r(t)\right\rVert_{L^{2}(\mathbb{R})}\leq
K\left\lVert\overrightarrow{g(t)}\right\rVert.$ (284)

From Theorem 2.9 and orthogonality condition (282), we deduce that

$\left\langle\partial_{t}r(t,x),\,\partial^{2}_{x}H^{x_{2}(t)}_{0,1}(x)\right\rangle_{L^{2}}=\dot{x}_{2}(t)\left\langle
r(t,x),\,\partial^{3}_{x}H^{x_{2}(t)}_{0,1}(x)\right\rangle_{L^{2}}=O\Big{(}\left\lVert
r(t)\right\rVert_{L^{2}}\epsilon^{\frac{1}{2}}\Big{)}.$

In conclusion, estimate (284) and Lemma 2.3 imply that there is a $K>1$ such
that

$|\dot{P}_{1}(t)|+|\dot{P}_{2}(t)|+\left\lVert\partial_{t}r(t)\right\rVert_{L^{2}(\mathbb{R})}\leq
K\left\lVert\overrightarrow{g(t)}\right\rVert$ (285)

for all $t\geq 0.$ Finally, Minkowski inequality and estimate (284) imply that
there is a uniform constant $K>1$ such that

$\left\lVert\partial_{x}r(t,x)\right\rVert_{L^{2}(\mathbb{R})}\leq
K\left\lVert\overrightarrow{g(t)}\right\rVert.$ (286)

We recall from Theorem 2.10 the following estimate

$\frac{\epsilon}{K}\leq\left\lVert\overrightarrow{g(t)}\right\rVert^{2}+\dot{x}_{1}(t)^{2}+\dot{x}_{2}(t)^{2}+e^{-\sqrt{2}z(t)}\leq
K\epsilon$ (287)

for some uniform constant $K>1.$ Now, from hypothesis (280), we obtain from
Theorem 1.11 and Corollary 1.13 that there are constants $M\in\mathbb{N}$ and
$C>0$ such that for all $t\geq 0$ the following inequalities are true

$\displaystyle\max_{j\in\\{1,\,2\\}}|x_{j}(t)-d_{j}(t)|\leq\epsilon\ln{\Big{(}\frac{1}{\epsilon}\Big{)}}^{M+1}\exp\Big{(}\frac{10C\epsilon^{\frac{1}{2}}t}{\ln{(\frac{1}{\epsilon})}}\Big{)},$
(288)
$\displaystyle\max_{j\in\\{1,\,2\\}}|\dot{x}_{j}(t)-\dot{d}_{j}(t)|\leq\epsilon^{\frac{3}{2}}\ln{\Big{(}\frac{1}{\epsilon}\Big{)}}^{M}\exp\Big{(}\frac{10C\epsilon^{\frac{1}{2}}t}{\ln{(\frac{1}{\epsilon})}}\Big{)},$
(289)
$\displaystyle\max_{j\in\\{1,\,2\\}}|\ddot{x}_{j}(t)-\ddot{d}_{j}(t)|\leq\epsilon^{\frac{3}{2}}\ln{\Big{(}\frac{1}{\epsilon}\Big{)}}\exp\Big{(}\frac{10C\epsilon^{\frac{1}{2}}t}{\ln{(\frac{1}{\epsilon})}}\Big{)},$
(290)

for a uniform constant $C>0.$ From the partial differential equation (1)
satisfied by $\phi(t,x)$ and the representation (281) of $g(t,x)$, we deduce
in the distributional sense that for any $h(x)\in H^{1}(\mathbb{R})$ that

$\left\langle
h(x),\,(\ddot{P}_{1}(t)+\dot{x}_{1}(t)^{2})\partial^{2}_{x}H^{x_{1}(t)}_{-1,0}+(\ddot{P}_{2}(t)+\dot{x}_{2}(t)^{2})\partial^{2}_{x}H^{x_{2}(t)}_{0,1}\right\rangle_{L^{2}_{x}(\mathbb{R})}=\left\langle
h(x),-P_{1}(t)\Big{[}\Big{(}-\partial^{2}_{x}+\ddot{U}(H^{x_{1}(t)}_{-1,0})\Big{)}\partial^{2}_{x}H^{x_{1}(t)}_{-1,0}\Big{]}\right\rangle\\\
\left\langle
h(x),\,-P_{2}(t)\Big{[}\Big{(}-\partial^{2}_{x}+\ddot{U}(H^{x_{2}(t)}_{0,1})\Big{)}\partial^{2}_{x}H^{x_{2}(t)}_{0,1}\Big{]}-\Big{[}\partial^{2}_{t}r(t)-\partial^{2}_{x}r(t)+\ddot{U}(H^{x_{2}(t)}_{0,1}+H^{x_{1}(t)}_{-1,0})r(t)\Big{]}\right\rangle_{L^{2}_{x}(\mathbb{R})}\\\
-\left\langle
h(x),\,\left[\dot{U}(H^{x_{2}(t)}_{0,1}+H^{x_{1}(t)}_{-1,0})-\dot{U}(H_{0,1}^{x_{2}(t)})-\dot{U}(H_{-1,0}^{x_{1}(t)})\right]-\ddot{x}_{1}(t)\partial_{x}H^{x_{1}(t)}_{-1,0}(x)-\ddot{x}_{2}(t)\partial_{x}H^{x_{2}(t)}_{0,1}(x)\right\rangle_{L^{2}_{x}(\mathbb{R})}\\\
\left\langle
h(x),\,-P_{1}(t)\Big{[}\Big{(}\ddot{U}(H^{x_{2}(t)}_{0,1}+H^{x_{1}(t)}_{-1,0})-\ddot{U}(H^{x_{1}(t)}_{-1,0})\Big{)}\partial^{2}_{x}H^{x_{1}(t)}_{-1,0}\Big{]}-P_{2}(t)\Big{[}\Big{(}\ddot{U}(H^{x_{2}(t)}_{0,1}+H^{x_{1}(t)}_{-1,0})-\ddot{U}(H^{x_{2}(t)}_{0,1})\Big{)}\partial^{2}_{x}H^{x_{2}(t)}_{0,1}\Big{]}\right\rangle\\\
+O\left(\left\lVert h\right\rVert_{L^{2}}\left[\left\lVert
g(t)\right\rVert_{H^{1}}^{2}+\max_{j\in\\{1,\,2\\}}|\ddot{x}_{j}(t)|+\max_{j\in\\{1,\,2\\}}|\dot{P}_{j}(x)\dot{x}_{j}(t)|+\max_{j\in\\{1,\,2\\}}|P_{j}(t)|e^{-\sqrt{2}z(t)}+|P_{j}(t)\ddot{x}_{j}(t)|+|P_{j}(t)\dot{x}_{j}(t)^{2}|\right]\right).$
(291)

From Lemma A.2 and estimates (288) and (290), we obtain from (291) that

$\left\langle
h(x),\,(\ddot{P}_{1}(t)+\dot{x}_{1}(t)^{2})\partial^{2}_{x}H^{x_{1}(t)}_{-1,0}+(\ddot{P}_{2}(t)+\dot{x}_{2}(t)^{2})\partial^{2}_{x}H^{x_{2}(t)}_{0,1}\right\rangle_{L^{2}_{x}(\mathbb{R})}=\left\langle
h(x),-P_{1}(t)\Big{[}\Big{(}-\partial^{2}_{x}+\ddot{U}(H^{x_{1}(t)}_{-1,0})\Big{)}\partial^{2}_{x}H^{x_{1}(t)}_{-1,0}\Big{]}\right\rangle_{L^{2}(\mathbb{R})}\\\
\left\langle
h(x),\,-P_{2}(t)\Big{[}\Big{(}-\partial^{2}_{x}+\ddot{U}(H^{x_{2}(t)}_{0,1})\Big{)}\partial^{2}_{x}H^{x_{2}(t)}_{0,1}\Big{]}-\Big{[}\partial^{2}_{t}r(t)-\partial^{2}_{x}r(t)+\ddot{U}(H^{x_{2}(t)}_{0,1}+H^{x_{1}(t)}_{-1,0})r(t)\Big{]}\right\rangle_{L^{2}_{x}(\mathbb{R})}\\\
+O\left(\left\lVert
h\right\rVert_{L^{2}}\left[\max_{j\in\\{1,\,2\\}}|\ddot{x}_{j}(t)-\ddot{d}_{j}(t)|+e^{-\sqrt{2}d(t)}+|z(t)-d(t)|e^{-\sqrt{2}z(t)}+e^{-2\sqrt{2}z(t)}\right]\right)\\\
+O\left(\left\lVert h\right\rVert_{L^{2}}\left[\left\lVert
g(t)\right\rVert_{H^{1}}^{2}+\max_{j\in\\{1,\,2\\}}|\ddot{x}_{j}(t)|+\max_{j\in\\{1,\,2\\}}|\dot{P}_{j}(x)\dot{x}_{j}(t)|+\max_{j\in\\{1,\,2\\}}|P_{j}(t)|e^{-\sqrt{2}z(t)}+|P_{j}(t)\ddot{x}_{j}(t)|+|P_{j}(t)\dot{x}_{j}(t)^{2}|\right]\right).$
(292)

From the condition (282), we deduce that

$\displaystyle\left\langle\partial^{2}_{t}r(t),\,\partial^{2}_{x}H^{x_{2}(t)}_{0,1}\right\rangle_{L^{2}}=\frac{d}{dt}\left[\dot{x}_{2}(t)\left\langle
r(t),\,\partial^{3}_{x}H^{x_{2}(t)}_{0,1}\right\rangle_{L^{2}}\right]+\dot{x}_{2}(t)\left\langle\partial_{t}r(t),\,\partial^{3}_{x}H^{x_{2}(t)}_{0,1}\right\rangle_{L^{2}},$
$\displaystyle\left\langle\partial^{2}_{t}r(t),\,\partial^{2}_{x}H^{x_{1}(t)}_{-1,0}\right\rangle_{L^{2}}=\frac{d}{dt}\left[\dot{x}_{1}(t)\left\langle
r(t),\,\partial^{3}_{x}H^{x_{1}(t)}_{-1,0}\right\rangle_{L^{2}}\right]+\dot{x}_{1}(t)\left\langle\partial_{t}r(t),\,\partial^{3}_{x}H^{x_{1}(t)}_{-1,0}\right\rangle_{L^{2}},$

which with the Theorem 2.9 imply that there is a uniform constant $C>0$ such
that

$\left|\left\langle\partial^{2}_{t}r(t),\,\partial^{2}_{x}H^{x_{2}(t)}_{0,1}\right\rangle_{L^{2}}\right|\leq
C\epsilon^{\frac{1}{2}}\left\lVert\overrightarrow{r(t)}\right\rVert,\,\left|\left\langle\partial^{2}_{t}r(t),\,\partial^{2}_{x}H^{x_{1}(t)}_{-1,0}\right\rangle_{L^{2}}\right|\leq
C\epsilon^{\frac{1}{2}}\left\lVert\overrightarrow{r(t)}\right\rVert.$ (293)

From (284), (285) and (286), we obtain that
$\left\lVert\overrightarrow{r(t)}\right\rVert\lesssim\left\lVert\overrightarrow{g(t)}\right\rVert.$
In conclusion, after we apply the partial differential equation (292) in
distributional sense to
$\partial^{2}_{x}H^{x_{2}(t)}_{0,1},\,\partial^{2}_{x}H^{x_{1}(t)}_{-1,0},$
the estimates (284), (285), (286), (288), (290) and (293) imply that there is
a uniform constant $K_{1}>0$ such that if $\epsilon\ll 1$ enough, then for
$j\in\\{1,\,2\\}$ we have that for $0\leq
t\leq\frac{N\ln{(\frac{1}{\epsilon})}}{\epsilon^{\frac{1}{2}}}$

$\big{|}\ddot{P}_{j}(t)+\dot{x}_{j}(t)^{2}\big{|}\leq
K_{1}\Big{(}e^{-\sqrt{2}d(t)}+\epsilon^{\frac{3}{2}}\ln{\Big{(}\frac{1}{\epsilon}\Big{)}}^{M+1}\exp\Big{(}\frac{10C\epsilon^{\frac{1}{2}}t}{\ln{(\frac{1}{\epsilon})}}\Big{)}+\epsilon
q(\epsilon)\Big{)},$

from which we deduce for all $0\leq t\leq
N\frac{\ln{(\frac{1}{\epsilon})}}{\epsilon^{\frac{1}{2}}}$ that

$\Big{|}\sum_{j=1}^{2}\ddot{P}_{j}(t)+\dot{x}_{j}(t)^{2}\Big{|}\leq
2K_{1}\Big{(}e^{-\sqrt{2}d(t)}+\epsilon^{\frac{3}{2}}\ln{\Big{(}\frac{1}{\epsilon}\Big{)}}^{M+1}\exp\Big{(}\frac{10C\epsilon^{\frac{1}{2}}t}{\ln{(\frac{1}{\epsilon})}}\Big{)}+\epsilon
q(\epsilon)\Big{)}.$ (294)

Since
$\Big{|}\sum_{j=1}^{2}\ddot{P}_{j}(t)\Big{|}\geq-\Big{|}\sum_{j=1}^{2}\ddot{P}_{j}(t)+\dot{x}_{j}(t)^{2}\Big{|}+\sum_{j=1}^{2}\dot{x}_{j}(t)^{2},$
we deduce from the estimates (294) and (287) that

$\Big{|}\sum_{j=1}^{2}\ddot{P}_{j}(t)\Big{|}\geq\frac{\epsilon}{K}-\Big{[}e^{-\sqrt{2}z(t)}+\left\lVert\overrightarrow{g(t)}\right\rVert^{2}\Big{]}-2K_{1}\Big{[}e^{-\sqrt{2}d(t)}+\epsilon^{\frac{3}{2}}\ln{\Big{(}\frac{1}{\epsilon}\Big{)}}^{M+1}\exp\Big{(}\frac{10C\epsilon^{\frac{1}{2}}t}{\ln{(\frac{1}{\epsilon})}}\Big{)}\Big{]}-2K_{1}\epsilon
q(\epsilon).$ (295)

We recall that from the statement of Theorem 1.11 that
$e^{-\sqrt{2}d(t)}=\frac{v^{2}}{8}\operatorname{sech}{(\sqrt{2}vt+c)}^{2},$
with
$v=\Big{(}\frac{\dot{z}(0)^{2}}{4}+8e^{-\sqrt{2}z(0)}\Big{)}^{\frac{1}{2}},$
which implies that $v\lesssim\epsilon^{\frac{1}{2}}.$ Since we have verified
in Theorem 2.9 that $e^{-\sqrt{2}z(t)}\lesssim\epsilon,$ the mean value
theorem implies that
$|e^{-\sqrt{2}z(t)}-e^{-\sqrt{2}d(t)}|=O(\epsilon|z(t)-d(t)|),$ from which we
deduce from 288 that

$|e^{-\sqrt{2}z(t)}-e^{-\sqrt{2}d(t)}|=O\Big{(}\epsilon^{2}\ln{\Big{(}\frac{1}{\epsilon}\Big{)}}^{M+1}\exp\Big{(}\frac{10C\epsilon^{\frac{1}{2}}t}{\ln{(\frac{1}{\epsilon})}}\Big{)}\Big{)}.$

In conclusion, if $\epsilon\ll 1$ enough, we obtain for $0\leq
t\leq\frac{N\ln{(\frac{1}{\epsilon})}}{\epsilon^{\frac{1}{2}}}$ from (295)
that

$\Big{|}\sum_{j=1}^{2}\ddot{P}_{j}(t)\Big{|}\geq\frac{\epsilon}{K}-\Big{[}e^{-\sqrt{2}d(t)}+\left\lVert\overrightarrow{g(t)}\right\rVert^{2}\Big{]}-4K_{1}\Big{[}e^{-\sqrt{2}d(t)}+\epsilon^{\frac{3}{2}}\ln{\Big{(}\frac{1}{\epsilon}\Big{)}}^{M+1}\exp\Big{(}\frac{10C\epsilon^{\frac{1}{2}}t}{\ln{(\frac{1}{\epsilon})}}\Big{)}\Big{]}-2K_{1}\epsilon
q(\epsilon).$ (296)

The conclusion of the demonstration will follow from studying separate cases
in the choice of $0<v,\,c.$ We also observe that $K,\,K_{1}$ are uniform
constants and the value of $N\in\mathbb{N}_{>0}$ can be chosen in the
beginning of the proof to be as much large as we need.
Case 1.($v^{2}\leq\frac{8\epsilon}{(1+4K_{1})2K}.$) From inequality (296), we
deduce that

$\Big{|}\sum_{j=1}^{2}\ddot{P}_{j}(t)\Big{|}\geq\frac{\epsilon}{2K}-\left\lVert\overrightarrow{g(t)}\right\rVert^{2}-4K_{1}\Big{(}\epsilon^{\frac{3}{2}}\ln{\Big{(}\frac{1}{\epsilon}\Big{)}}^{M+1}\exp\Big{(}\frac{10C\epsilon^{\frac{1}{2}}t}{\ln{(\frac{1}{\epsilon})}}\Big{)}\Big{)}-2K_{1}\epsilon
q(\epsilon),$

then, from (280) we deduce for $0\leq
t\leq\frac{\ln{(\frac{1}{\epsilon})}}{\epsilon^{\frac{1}{2}}}$ that if
$\epsilon$ is small enough, then
$\Big{|}\sum_{j=1}^{2}\ddot{P}_{j}(t)\Big{|}\geq\frac{\epsilon}{4K},$ and so,

$\Big{|}\sum_{j=1}^{2}\dot{P}_{j}(t)\Big{|}\geq\frac{\epsilon
t}{4K}-\Big{|}\sum_{j=1}^{2}\dot{P}_{j}(0)\Big{|},$

which contradicts the fact that (285) and (280) should be true for
$\epsilon\ll 1.$
Case
2.($v^{2}\geq\frac{8\epsilon}{(1+4K_{1})2K},\,|c|>2\ln{(\frac{1}{\epsilon})}.$)
It is not difficult to verify that for $0\leq
t\leq\min(\frac{|c|}{2\sqrt{2}v},N\frac{\ln{(\frac{1}{\epsilon})}}{\epsilon^{\frac{1}{2}}}),$
we have that
$e^{-\sqrt{2}d(t)}\leq\frac{v^{2}}{8}\operatorname{sech}{(\frac{c}{2})}^{2}\lesssim\epsilon^{3},$
then estimate (296) implies that
$\Big{|}\sum_{j=1}^{2}\ddot{P}_{j}(t)\Big{|}\geq\frac{\epsilon}{4K}$ is true
in this time interval. Also, since now $v\cong\epsilon^{\frac{1}{2}},$ we have
that

$\frac{\ln{(\frac{1}{\epsilon})}}{\epsilon^{\frac{1}{2}}}\lesssim\frac{|c|}{2\sqrt{2}v},$

so we obtain a contradiction by similar argument to the Case $1.$
Case 3.($v^{2}\geq\frac{8\epsilon}{(1+4K_{1})2K},\,|c|\leq
2\ln{(\frac{1}{\epsilon})}.$) For $1\ll N$ enough and
$t_{0}=\frac{(1+4K_{1})^{\frac{1}{2}}K^{\frac{1}{2}}\sqrt{2}\ln{(\frac{1}{\epsilon})}}{\epsilon^{\frac{1}{2}}},$
we have during the time interval $\left\\{t_{0}\leq t\leq
2\frac{(1+4K_{1})^{\frac{1}{2}}K^{\frac{1}{2}}\sqrt{2}\ln{(\frac{1}{\epsilon})}}{\epsilon^{\frac{1}{2}}}\right\\}$
that
$e^{-\sqrt{2}d(t)}\leq\frac{v^{2}}{8}\operatorname{sech}{\Big{(}2\ln{\Big{(}\frac{1}{\epsilon}\Big{)}}\Big{)}}^{2}\lesssim\epsilon^{5}.$
In conclusion, estimate (295) implies that
$\Big{|}\sum_{j=1}^{2}\ddot{P}_{j}(t)\Big{|}\geq\frac{\epsilon}{4K}$ is true
in this time interval. From the Fundamental Calculus Theorem, we have that

$\Big{|}\sum_{j=1}^{2}\dot{P}_{j}(t)\Big{|}\geq\frac{\epsilon(t-t_{0})}{4K}-\Big{|}\sum_{j=1}^{2}\dot{P}_{j}(t_{0})\Big{|}.$

In conclusion, hypothesis (280) and estimate (285) imply for
$T=2\frac{(1+2K_{1})^{\frac{1}{2}}K^{\frac{1}{2}}\sqrt{2}\ln{(\frac{1}{\epsilon})}}{\epsilon^{\frac{1}{2}}}$
that

$\Big{|}\sum_{j=1}^{2}\dot{P}_{j}(T)\Big{|}\geq\frac{\epsilon^{\frac{1}{2}}(1+2K_{1})^{\frac{1}{2}}\sqrt{2}\ln{(\frac{1}{\epsilon})}}{8K^{\frac{1}{2}}},$

which contradicts the fact that (280) and (285) should be true, which finishes
our proof. ∎

######  Remark B.2.

Clearly, we can use Theorem B.1 to verify that there is a sequence
$(t_{n})_{n\in\mathbb{R}}$ such that for all
$n\in\mathbb{N},\,\epsilon\lesssim\left\lVert\overrightarrow{g(t_{n})}\right\rVert_{H^{1}\times
L^{2}}.$

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