Dynamics of Two Interacting Kinks for the $\phi ^{6}$ Model

Abstract

We consider the nonlinear wave equation known as the $\phi ^{6}$ model in dimension 1+1. We describe the long-time behavior of this model’s solutions close to a sum of two kinks with energy slightly larger than twice the minimum energy of non-constant stationary solutions. We prove orbital stability of two moving kinks. We show for low energy excess $\epsilon$ that these solutions can be described for a long time of order $-\ln {(\epsilon )}\epsilon ^{-\frac{1}{2}}$ as the sum of two moving kinks such that each kink’s center is close to an explicit function which is a solution of an ordinary differential system. We give an optimal estimate in the energy norm of the remainder and we prove that this estimate is achieved during a finite instant t of order $-\ln {(\epsilon )}\epsilon ^{-\frac{1}{2}}.$

Appendices

Appendix A Auxiliary Results

We start the Appendix Section by presenting the following lemma:

Lemma 19

With the same hypothesis as in Theorem 4 and using its notation, we have while $\max _{j\in \{1,2\}}\left|d_{j}(t)-x_{j}(t)\right|<1$ that $\max _{j\in \{1,\,2\}}\left|\ddot{d}_{j}(t)-\ddot{x}_{j}(t)\right|=O\Big (\max _{j\in \{1,\,2\}}\left|d_{j}(t)-x_{j}(t)\right|\epsilon +\epsilon z(t)e^{-\sqrt{2}z(t)} +\left\Vert \overrightarrow{g(t)}\right\Vert \epsilon ^{\frac{1}{2}}\Big ).$

Lemma 20

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

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

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

Proof

By direct computations, we verify that

$$\begin{aligned}{} & {} \dot{U}\left( H^{x_{1}(t)}_{-1,0}+H^{x_{2}(t)}_{0,1}\right) - \dot{U}\left( H^{x_{1}(t)}_{-1,0}\right) -\dot{U}\left( H^{x_{2}(t)}_{0,1}\right) \\{} & {} \quad ={-}24H^{x_{1}(t)}_{-1,0}H^{x_{2}(t)}_{0,1}\left( H^{x_{1}(t)}_{-1,0}+H^{x_{2}(t)}_{0,1}\right) \\ {}{} & {} \qquad {+}\,30H^{x_{1}(t)}_{-1,0}H^{x_{2}(t)}_{0,1}\left[ \left( H^{x_{1}(t)}_{-1,0}\right) ^{3}+\left( H^{x_{2}(t)}_{0,1}\right) ^{3}\right] \\ {}{} & {} \qquad {+}\,60\left( H^{x_{1}(t)}_{-1,0}H^{x_{2}(t)}_{0,1}\right) ^{2}\left[ H^{x_{1}(t)}_{-1,0}+H^{x_{2}(t)}_{0,1}\right] . \end{aligned}$$

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

$$\begin{aligned} 60\left( H^{x_{1}(t)}_{-1,0}H^{x_{2}(t)}_{0,1}\right) ^{2}\left[ H^{x_{1}(t)}_{-1,0}+H^{x_{2}(t)}_{0,1}\right] =\,&\frac{60e^{-2\sqrt{2}z(t)}H^{x_{2}(t)}_{0,1}}{(1+e^{2\sqrt{2}(x-x_{2}(t))})(1+e^{-2\sqrt{2}(x-x_{1}(t))})}\\&{+}\,\frac{60e^{-2\sqrt{2}z(t)}H^{x_{1}(t)}_{-1,0}}{(1+e^{-2\sqrt{2}(x-x_{1}(t))})(1+e^{2\sqrt{2}(x-x_{2}(t))})}. \end{aligned}$$

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

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

(A1)

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 using Lemma 6 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 $H^{k}_{x}({\mathbb {R}})$ and it satisfies for all $k>0$ the following estimate

$$\begin{aligned} \left\Vert \frac{\partial ^{k}}{\partial x^{k}}\left[ (H^{x_{1}(t)}_{-1,0}H^{x_{2}(t)}_{0,1})^{2}(H^{x_{1}(t)}_{-1,0}+H^{x_{2}(t)}_{0,1})\right] \right\Vert _{L^{2}}=O\left( e^{-2\sqrt{2}z(t)}\right) . \end{aligned}$$

(A2)

Next, using the identity

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

(A3)

the identity

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

and Lemma 6, we deduce that

$$\begin{aligned} \left\Vert 24(H^{x_{1}(t)}_{-1,0})^{2}H^{x_{2}(t)}_{0,1}+24e^{-\sqrt{2}z(t)}\frac{H^{x_{1}(t)}_{-1,0}(x)}{(1+e^{-2\sqrt{2}(x-x_{1}(t))})^{\frac{1}{2}}}\right\Vert _{L^{2}}=\,&O\left( e^{-2\sqrt{2}z(t)}\right) , \end{aligned}$$

(A4)

$$\begin{aligned} \left\Vert 30(H^{x_{1}(t)}_{-1,0})^{4}H^{x_{2}(t)}_{0,1}+30e^{-\sqrt{2}z(t)}\left( \frac{(H^{x_{1}(t)}_{-1,0}(x))^{3}}{(1+e^{-2\sqrt{2}(x-x_{1}(t))})^{\frac{1}{2}}}\right) \right\Vert _{L^{2}}=\,&O\left( e^{-3\sqrt{2}z(t)}\right) . \end{aligned}$$

(A5)

The estimate of the remaining terms $-24H^{x_{1}(t)}_{-1,0}\left( H^{x_{2}(t)}_{0,1}\right) ^{2},\,30H^{x_{1}(t)}_{-1,0}\left( H^{x_{2}(t)}_{0,1}\right) ^{4}$ is completely analogous to (A4) and (A5) respectively. In conclusion, all of the estimates above imply the estimate stated in the Lemma 20. $\square$

Proof of Lemma 19

First, we recall the global estimate $e^{-\sqrt{2}z(t)}\lesssim \epsilon .$ We also recall the identity (33)

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

which, by integration by parts, implies that

$$\begin{aligned} \int _{{\mathbb {R}}}24\frac{H_{0,1}(x)\partial _{x}H_{0,1}(x)}{(1+e^{2\sqrt{2}(x)})^{\frac{1}{2}}}-30\frac{(H_{0,1}(x))^{3}\partial _{x}H_{0,1}(x)}{(1+e^{2\sqrt{2}(x)})^{\frac{1}{2}}}\,dx=4. \end{aligned}$$

(A6)

We recall $d_{1}(t),\, d_{2}(t)$ defined in (8) and (9) respectively and $d(t)=d_{2}(t)-d_{1}(t).$ Since $\ddot{d}_{j}(t)=(-1)^{j}8\sqrt{2}e^{-\sqrt{2}d(t)}$ for $j \in \{1,\, 2\},$ we have $\ddot{d}(t)=16\sqrt{2}e^{-\sqrt{2}d(t)},$ which implies clearly with the identities

$$\begin{aligned} \left\Vert \partial _{x}H_{0,1}\right\Vert _{L^{2}}^{2}=\left\Vert \partial ^{2}_{x}H_{0,1}\right\Vert _{L^{2}}^{2}=\frac{1}{2\sqrt{2}} \end{aligned}$$

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

$$\begin{aligned}{} & {} \dot{U}\left( H^{x_{2}(t)}_{0,1}(x)+H^{x_{1}(t)}_{-1,0}(x)\right) -\dot{U}\left( H^{x_{1}(t)}_{-1,0}(x)\right) -\dot{U}\left( H^{x_{2}(t)}_{0,1}(x)\right) -\ddot{x}_{2}(t)\partial _{x}H^{x_{2}(t)}_{0,1}(x)\nonumber \\{} & {} \quad ={-}\left( \partial ^{2}_{t}g(t,x)-\partial ^{2}_{x}g(t,x)+\ddot{U}\left( H^{x_{2}(t)}_{0,1}(x)+H^{x_{1}(t)}_{-1,0}(x)\right) g(t,x)\right) \nonumber \\{} & {} \qquad {+}\,\sum _{k=3}^{6} U^{(k)}\left( H^{x_{1}(t)}_{-1,0}+H^{x_{2}(t)}_{0,1}\right) \frac{g(t)^{k-1}}{(k-1)!} -\dot{x}_{1}(t)^{2}\partial ^{2}_{x}H^{x_{1}(t)}_{-1,0}(x)\nonumber \\{} & {} \qquad {-}\,\dot{x}_{2}(t)^{2}\partial ^{2}_{x}H^{x_{2}(t)}_{0,1}(x) +\ddot{x}_{1}(t)\partial _{x}H^{x_{1}(t)}_{-1,0}(x). \end{aligned}$$

(A7)

Furthermore, from the estimate (A6), Lemma 20 and Lemma 6, we obtain that

$$\begin{aligned}{} & {} \left\langle \dot{U}\left( H^{x_{1}(t)}_{-1,0}+H^{x_{2}(t)}_{0,1}\right) - \dot{U}\left( H^{x_{1}(t)}_{-1,0}\right) -\dot{U}\left( H^{x_{2}(t)}_{0,1}\right) ,\,\partial _{x}H^{x_{2}(t)}_{0,1}\right\rangle _{L^{2}}\nonumber \\{} & {} \quad =\ddot{x}_{2}(t)\left\Vert \partial _{x}H_{0,1}\right\Vert _{L^{2}}^{2} -(\ddot{x}_{2}(t)-\ddot{d}_{2}(t))\left\Vert \partial _{x}H_{0,1}\right\Vert _{L^{2}}^{2}\nonumber \\{} & {} \qquad {+}\,O\Big (\left|\ddot{x}_{1}(t)\right|z(t)e^{-\sqrt{2}z(t)}\Big )\nonumber \\{} & {} \qquad {+}\,O\Big (e^{-\sqrt{2}z(t)} \max _{j\in \{1,\,2\}}\left|x_{j}(t)-d_{j}(t)\right| +e^{-2\sqrt{2}z(t)}z(t)\Big ). \end{aligned}$$

(A8)

We recall from the proof of Theorem 14 the following estimate

$$\begin{aligned}{} & {} \left|\int _{{\mathbb {R}}}\left[ \ddot{U}\left( H^{x_{2}(t)}_{0,1}(x)\right) -\ddot{U}\left( H^{x_{2}(t)}_{0,1}(x)+H^{x_{1}(t)}_{-1,0}(x)\right) \right] \partial _{x}H^{x_{2}(t)}_{0,1}(x)g(t,x)\,dx\right|\\{} & {} \quad =O\Big (\left\Vert \overrightarrow{g(t)}\right\Vert e^{-\sqrt{2}z(t)}\Big ). \end{aligned}$$

Also, from the Modulation Lemma, we have that

$$\begin{aligned} \langle \partial ^{2}_{t}g(t),\partial _{x}H^{x_{2}(t)}_{0,1}\rangle _{L^{2}}&=\frac{d}{dt}\left[ \langle \partial _{t}g(t),\partial _{x}H^{x_{2}(t)}_{0,1}\rangle _{L^{2}} \right] +\dot{x}_{2}(t)\langle \partial _{t}g(t),\partial _{x}H^{x_{2}(t)}_{0,1}\rangle _{L^{2}} \\ {}&=\frac{d}{dt}\Big [\dot{x}_{2}(t)\langle g(t),\partial ^{2}_{x}H^{x_{2}(t)}_{0,1}\rangle _{L^{2}} \Big ]+\dot{x}_{2}(t)\langle \partial _{t}g(t),\partial _{x}H^{x_{2}(t)}_{0,1}\rangle _{L^{2}}\\ {}&=\ddot{x}_{2}(t)\langle g(t),\partial ^{2}_{x}H^{x_{2}(t)}_{0,1}\rangle _{L^{2}} +2\dot{x}_{2}(t)\langle \partial _{t}g(t),\partial _{x}H^{x_{2}(t)}_{0,1}\rangle _{L^{2}}\\&-\dot{x}_{2}(t)^{2}\langle g(t),\partial ^{3}_{x}H^{x_{2}(t)}_{0,1}\rangle _{L^{2}}. \end{aligned}$$

In conclusion, since $\partial _{x}H^{x_{2}(t)}_{0,1}\in \ker D^{2}E_{pot}\left( H^{x_{2}(t)}_{0,1}\right)$ and $e^{{-}\sqrt{2}z(t)}=O\left( \epsilon ^{\frac{1}{2}}\right) ,$ we obtain from (A8) and (A7) that

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

the estimate of $\left|\ddot{x}_{1}(t)-\ddot{d}_{1}(t)\right|$ is completely analogous, which finishes the proof of Lemma 19. $\square$

Lemma 21

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

$$\begin{aligned} \left\Vert \phi (x) -H_{0,1}(x)\right\Vert _{H^{1}}<+\infty ,\, 0<E_{pot}(\phi (x))-E_{pot}(H_{0,1})<\epsilon (\delta ), \end{aligned}$$

(A9)

then there is a real number y such that

$$\begin{aligned} \left\Vert \phi (x)-H_{0,1}(x-y)\right\Vert _{H^{1}}\le \delta . \end{aligned}$$

Proof of Lemma 21

The proof of Lemma 21 will follow by a contradiction argument. We assume the existence of a sequence of real functions $\left( \phi _{n}(x)\right) _{n}$ satisfying

$$\begin{aligned} \lim _{n\rightarrow +\infty } E_{pot}(\phi _{n})=\,&E_{pot}(H_{0,1}), \end{aligned}$$

(A10)

$$\begin{aligned} \left\Vert \phi _{n}(x)-H_{0,1}(x)\right\Vert _{H^{1}}<&{+}\,\infty , \end{aligned}$$

(A11)

such that

$$\begin{aligned} \lim _{n\rightarrow {+}\infty } \inf _{y\in {\mathbb {R}}} \left\Vert \phi _{n}(x)-H_{0,1}(x+y)\right\Vert _{H^{1}}>0. \end{aligned}$$

(A12)

First, the condition (A10) and the fact that $\lim _{\phi \rightarrow {+\infty }}U(\phi )={+}\infty$ imply the existence of a positive constant c,  which satisfies $\left\Vert \phi _{n}\right\Vert _{L^{\infty }}<c$ if $n\gg 1.$

Next, 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 to verify from the definition of the potential energy functional $E_{pot}$ that if $1\ll n,$ then

$$\begin{aligned} \left\Vert \phi _{n}(x)-1\right\Vert _{L^{2}\left( \{x\vert \phi _{n}(x)>1\}\right) }^{2}+\left\Vert \frac{d\phi _{n}(x)}{dx}\right\Vert _{L^{2}\left( \{x\vert \phi _{n}(x)>1\}\right) }^{2}\lesssim \left|E_{pot}(\phi _{n})-E_{pot}(H_{0,1})\right|. \end{aligned}$$

By an analogous argument, we can verify that

$$\begin{aligned}{} & {} \left\Vert \phi _{n}(x)\right\Vert _{L^{2}(\{x\vert {-}\frac{1}{2}<\phi _{n}(x)<0\})}^{2}+\left\Vert \frac{d\phi _{n}(x)}{dx}\right\Vert _{L^{2}(\{x\vert {-}\frac{1}{2}<\phi _{n}(x)<0\})}^{2}\\{} & {} \quad \lesssim \left|E_{pot}(\phi _{n})-E_{pot}(H_{0,1})\right|, \end{aligned}$$

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

$$\begin{aligned}{} & {} \int _{x_{0}}^{+\infty }\frac{1}{2}\frac{d\phi _{n}(x)}{dx}^{2}+U(\phi _{n}(x))\,dx\\{} & {} \quad =\int _{x_{0}}^{+\infty }\sqrt{2U(\phi _{n}(x))}\left|\frac{d\phi _{n}(x)}{dx}\right|\,dx+\frac{1}{2}\int _{x_{0}}^{+\infty }\left( \left|\frac{d\phi _{n}(x)}{dx}\right|-\sqrt{2U(\phi _{n}(x))}\right) ^{2}\,dx\\{} & {} \quad \ge \int _{-\frac{1}{2}}^{1}\sqrt{2U(\phi )}\,d\phi = E_{pot}(H_{0,1})+\int _{-\frac{1}{2}}^{0}\sqrt{2U(\phi )}\,d\phi >E_{pot}(H_{0,1}), \end{aligned}$$

which contradicts (A10) if $n\gg 1.$ Thus, if we consider the following function

$$\begin{aligned} \varphi _{n}(x)=\min \left( \max \left( \phi _{n}(x),0\right) ,1\right) , \end{aligned}$$

which satisfies $E_{pot}\left( \varphi _{n}\right) \ge E_{pot}\left( H_{0,1}\right)$ and

$$\begin{aligned} \frac{d\varphi _{n}(x)}{dx}={\left\{ \begin{array}{ll} \frac{d \phi _{n}(x)}{dx} \text {, if } 0<\phi _{n}(x)<1,\\ 0 \text {, for almost every } x\in {\mathbb {R}} \text { satisfying either } \phi _{n}(x)\le 0 \text { or } \phi _{n}(x)\ge 1, \end{array}\right. } \end{aligned}$$

we can deduce with the estimates above and inequality $\limsup _{n\rightarrow {+}\infty }\left\Vert \phi _{n}\right\Vert _{L^{\infty }}<c$ that if $n\gg 1,$ then

$$\begin{aligned} \left\Vert \phi _{n}(x)-\varphi _{n}(x)\right\Vert _{L^{2}}^{2}+\left\Vert \frac{d\phi _{n}(x)}{dx}-\frac{d\varphi _{n}(x)}{dx}\right\Vert _{L^{2}}^{2}&\lesssim \left| E_{pot}\left( \phi _{n}\right) -E_{pot}\left( H_{0,1}\right) \right| ,\\ \left| E_{pot}\left( \phi _{n}\right) -E_{pot}\left( \varphi _{n}\right) \right|&\lesssim \left| E_{pot}\left( \phi _{n}\right) -E_{pot}\left( H_{0,1}\right) \right| . \end{aligned}$$

Consequently, using triangle inequality and conditions (A10), (A12), we would obtain that

$$\begin{aligned} \lim _{n\rightarrow {+}\infty }\inf _{y\in {\mathbb {R}}}\left\Vert \varphi _{n}(x)-H_{0,1}(x+y)\right\Vert _{H^{1}}>0. \end{aligned}$$

In conclusion, we can restrict the proof to the case where $0\le \phi _{n}(x)\le 1$ 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 $\left\Vert \phi (x)-H_{0,1}(x)\right\Vert _{H^{1}}<+\infty ,$ then $E_{pot}(\phi (x))\ge E_{pot}(H_{0,1}(x)).$ In conclusion, there is a sequence of positive numbers $\left( \epsilon _{n}\right) _{n}$ such that

$$\begin{aligned} E_{pot}(\phi _{n})=E_{pot}(H_{0,1})+\epsilon _{n},\,\lim _{n\rightarrow +\infty }\epsilon _{n}=0. \end{aligned}$$

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 $n\in {\mathbb {N}},$ $\lim _{x\rightarrow +\infty }\phi _{n}(x)=1$ and $\lim _{x\rightarrow -\infty }\phi _{n}(x)=0,$ we can restrict to the case where

$$\begin{aligned} \phi _{n}(0)=\frac{1}{\sqrt{2}}, \end{aligned}$$

for all $n\in {\mathbb {N}}.$

Next, we consider the notations $(v)_{+}=\max (v,0)$ 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

$$\begin{aligned} U=\left\{ x\in {\mathbb {R}}\vert \,\frac{d\phi _{n}(x)}{dx}<0\right\} \end{aligned}$$

(A13)

is an enumerable union of disjoint open intervals $(a_{k,n},b_{k,n})_{k \in {\mathbb {N}}}$, which are bounded, since $\lim _{x\rightarrow +\infty }\phi _{n}(x)=1,\,\lim _{x\rightarrow -\infty }\phi _{n}(x)=0$ and $0\le \phi _{n}(x)\le 1.$

Now, let E be a set of disjoint open bounded intervals $(h_{i,n},l_{i,n})\subset {\mathbb {R}}$ satisfying the conditions

$$\begin{aligned} \phi _{n}(h_{i,n})=\phi _{n}(l_{i,n}), \end{aligned}$$

(A14)

and $\{i\vert \,(h_{i,n},l_{i,n})\in E \}=I\subset {\mathbb {Z}}.$ For any $i\in I,$ the following function

$$\begin{aligned} f_{i,n}(x)={\left\{ \begin{array}{ll} \phi _{n}(x)\text { if } x\le h_{i,n},\\ \phi _{n}(x+l_{i,n}-h_{i,n}) \text { if } x>h_{i,n},\\ \end{array}\right. } \end{aligned}$$

satisfies $E_{pot}(H_{0,1})\le E_{pot}(f_{i,n})\le E_{pot}(\phi _{n})=E_{pot}(H_{0,1})+\epsilon _{n},$ which implies that

$$\begin{aligned} \int _{h_{i,n}}^{l_{i,n}}\frac{1}{2}\frac{d\phi _{n}(x)}{dx}^{2}+U(\phi _{n}(x))\le \epsilon _{n}. \end{aligned}$$

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

$$\begin{aligned} \sum _{i\in I} \int _{h_{i,n}}^{l_{i,n}}\frac{1}{2}\frac{d\phi _{n}(x)}{dx}^{2}+U(\phi _{n}(x))\le \epsilon _{n}, \end{aligned}$$

(A15)

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 (A15) that

$$\begin{aligned} \int _{{\mathbb {R}}}\left( \frac{d\phi _{n}(x)}{dx}\right) _{-}^{2}\,dx\le 2\epsilon _{n}, \end{aligned}$$

(A16)

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

$$\begin{aligned} \left\Vert \frac{d\phi _{n}(x)}{dx}-\left|\frac{d\phi _{n}(x)}{dx}\right|\right\Vert _{L^{2}}^{2}\le 8\epsilon _{n},\,\phi _{n}(0)=\frac{1}{\sqrt{2}}. \end{aligned}$$

(A17)

Moreover, we can verify that

$$\begin{aligned} E_{pot}(\phi _{n})=\frac{1}{2}\left[ \int _{{\mathbb {R}}}\left( \left|\frac{d\phi _{n}(x)}{dx}\right|-\sqrt{2U(\phi _{n}(x))}\right) ^{2}\,dx\right] +\int _{{\mathbb {R}}}\sqrt{2U(\phi _{n}(x))}\left|\frac{d\phi _{n}(x)}{dx}\right|\,dx, \end{aligned}$$

from which we deduce with $\lim _{x\rightarrow -\infty }\phi _{n}(x)=0$ and $\lim _{x\rightarrow +\infty }\phi _{n}(x)=1$ that

$$\begin{aligned} E_{pot}(H_{0,1})+\epsilon _{n}\ge&\frac{1}{2}\left[ \int _{{\mathbb {R}}}\left( \left|\frac{d\phi _{n}(x)}{dx}\right|-\sqrt{2U(\phi _{n}(x))}\right) ^{2}\,dx\right] +\int _{0}^{1}\sqrt{2 U(\phi )}\,d\phi \\ =\,&\frac{1}{2}\left[ \int _{{\mathbb {R}}}\left( \left|\frac{d\phi _{n}(x)}{dx}\right|-\sqrt{2U(\phi _{n}(x))}\right) ^{2}\,dx\right] +E_{pot}(H_{0,1}). \end{aligned}$$

Then, from estimate (A17), we have that

$$\begin{aligned} \frac{d\phi _{n}(x)}{dx}=\sqrt{2U(\phi _{n}(x))}+r_{n}(x),\,\phi _{n}(0)=\frac{1}{\sqrt{2}}, \end{aligned}$$

(A18)

with $\left\Vert r_{n}\right\Vert _{L^{2}}^{2}\lesssim \epsilon _{n}$ for all $1\ll n.$

We recall that $U(\phi )=\phi ^{2}(1-\phi ^{2})^{2}$ is a Lipschitz function in the set $\{\phi \vert \,0\le \phi \le 1\}.$ Then, because $H_{0,1}(x)$ is the unique solution of the following ordinary differential equation

$$\begin{aligned} {\left\{ \begin{array}{ll} \begin{aligned} \frac{d\phi (x)}{dx}=\,&{}\sqrt{2U(\phi (x))},\\ \phi (0)=\,&{}\frac{1}{\sqrt{2}}, \end{aligned} \end{array}\right. } \end{aligned}$$

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

$$\begin{aligned} \lim _{n\rightarrow +\infty } \left\Vert \phi _{n}(x)-H_{0,1}(x)\right\Vert _{L^{\infty }[-K,K]}=0,\,\lim _{n\rightarrow +\infty }\left\Vert \frac{d\phi _{n}(x)}{dx}-\dot{H}_{0,1}(x)\right\Vert _{L^{2}[-K,K]}=0.\nonumber \\ \end{aligned}$$

(A19)

Also, if $1\ll n,$ then $\left\Vert \frac{d\phi _{n}(x)}{dx}\right\Vert _{L^{2}}^{2}<2E_{pot}(H_{0,1})+1,$ and so we obtain from Cauchy–Schwarz inequality that

$$\begin{aligned} \left|\phi _{n}(x)-\phi _{n}(y)\right|\le \left|x-y\right|^{\frac{1}{2}}\left\Vert \frac{d\phi _{n}}{dx}\right\Vert _{L^{2}}^{2}<M\left|x-y\right|^{\frac{1}{2}}, \end{aligned}$$

(A20)

for a constant $M>0.$ The inequality (A20) implies that for any $1>\omega >0$ there is a number $h(\omega )\in {\mathbb {N}}$ such that if $n\ge h(\omega )$ then

$$\begin{aligned} \left\Vert \phi _{n}(x)-H_{0,1}(x)\right\Vert _{L^{\infty }\{x\vert \,\frac{1}{\omega }<\left|x\right|\}}<\omega , \end{aligned}$$

(A21)

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

$$\begin{aligned} \left|\phi _{m_{n}}(x_{n})-1\right|&>\theta \text { if } x_{n}>0, \end{aligned}$$

(A22)

$$\begin{aligned} \left|\phi _{m_{n}}(x_{n})\right|&>\theta \text { if } x_{n}<0. \end{aligned}$$

(A23)

However, since we are considering $\phi _{n}(x)\in C^{1}({\mathbb {R}})$ and $0\le \phi _{n}\le 1,$ we would obtain from the intermediate 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

$$\begin{aligned} 1-\theta&\le \phi _{m_{n}}(y_{n})\le 1+\theta , \text { if } y_{n}>0, \end{aligned}$$

(A24)

$$\begin{aligned} \phi _{m_{n}}(y_{n})&= \theta \text { otherwise.} \end{aligned}$$

(A25)

But, estimates (A20), (A24), (A25) and identity $U(\phi )=\phi ^{2}(1-\phi ^{2})^{2}$ would imply that

$$\begin{aligned} 1\lesssim \int _{\left|x\right|\ge n-2}U(\phi _{m_{n}}(x))\,dx \text { for all } n\gg 1, \end{aligned}$$

(A26)

and because of estimate (A19) and the following identity

$$\begin{aligned} \lim _{K\rightarrow +\infty }\int _{-K}^{K}\frac{1}{2}\dot{H}_{0,1}(x)^{2}+U(H_{0,1}(x))=E_{pot}(H_{0,1}(x)), \end{aligned}$$

(A27)

estimate (A26) would imply that $\lim _{n\rightarrow +\infty }E_{pot}(\phi _{m_{n}})>E_{pot}(H_{0,1})$ which contradicts our hypotheses.

In conclusion, for any $1>\omega >0$ there is a number $h(\omega )$ such that if $n\ge h(\omega )$ then (A21) holds. So we deduce for any $0<\omega <1$ that there is a number $h_{1}(\omega )$ such that

$$\begin{aligned} \text {if } n\ge h_{1}(\omega ), \text { then } \left|\phi _{n}(x)-H_{0,1}(x)\right|\le \omega \text { for all } x\in {\mathbb {R}}. \end{aligned}$$

(A28)

Then, if $\omega \le \frac{1}{100},\,n\ge h(\omega )$ and $K\ge 200,$ estimates (A28) and (A19) imply that

$$\begin{aligned} \int _{K}^{+\infty }U(\phi _{n}(x))+\frac{1}{2}\frac{d\phi _{n}(x)}{dx}^{2}\, dx&\ge \frac{1}{2}\int _{K}^{+\infty }\left( 1-\phi _{n}(x)\right) ^{2}+\frac{d\phi _{n}(x)}{dx}^{2}\,dx, \end{aligned}$$

(A29)

$$\begin{aligned} \int _{-\infty }^{-K}U(\phi _{n}(x))+\frac{1}{2}\frac{d\phi _{n}(x)}{dx}^{2}\, dx&\ge \frac{1}{2}\int _{-\infty }^{-K}\phi _{n}(x)^{2}+\frac{d\phi _{n}(x)}{dx}^{2}\,dx. \end{aligned}$$

(A30)

In conclusion, from estimates (A28), (A29), (A30) and

$$\begin{aligned} \lim _{K\rightarrow +\infty }\int _{\vert x\vert \ge K}\frac{1}{2}\dot{H}_{0,1}(x)^{2}+U(H_{0,1}(x))\,dx=0, \end{aligned}$$

we obtain that $\lim _{n\rightarrow +\infty }\left\Vert \phi _{n}(x)-H_{0,1}(x)\right\Vert _{L^{2}}=0$ and, from the initial value problem (A18) satisfied for each $\phi _{n},$ we conclude that $\lim _{n\rightarrow +\infty }\left\Vert \frac{d\phi _{n}}{dx}(x)-\dot{H}_{0,1}(x)\right\Vert _{L^{2}}=0.$ In conclusion, inequality (A12) is false. $\square$

From Lemma 21, we obtain the following corollary:

Corollary 22

For any $\delta >0$ there exists $\epsilon _{0}>0$ such that if $0<\epsilon \le \epsilon _{0},\left\Vert \phi (x)-H_{0,1}(x)-H_{-1,0}(x)\right\Vert _{H^{1}}<+\infty$ and $E_{pot}(\phi )=2E_{pot}(H_{0,1})+\epsilon ,$ then there exist $x_{2},x_{1} \in {\mathbb {R}}$ such that

$$\begin{aligned} x_{2}-x_{1}\ge \frac{1}{\delta },\,\left\Vert \phi (x)-H_{0,1}(x-x_{2})+H_{-1,0}(x-x_{1})\right\Vert _{H^{1}}\le \delta . \end{aligned}$$

(A31)

Proof of Corollary 22

First, from a similar reasoning to the proof of Lemma 21 we can assume by density that $\frac{d\phi (x)}{dx}\in H^{1}_{x}({\mathbb {R}}).$ Next, from hypothesis $\left\Vert \phi (x)-H_{0,1}(x)-H_{-1,0}(x)\right\Vert _{H^{1}({\mathbb {R}})}<+\infty ,$ we deduce using the intermediate value theorem that there is a $y\in {\mathbb {R}}$ such that $\phi (y)=0.$ Now, we consider the functions

$$\begin{aligned} \phi _{-}(x)={\left\{ \begin{array}{ll} \phi (x)&{} \text { if } x\le y,\\ 0&{} \text { otherwise,} \end{array}\right. } \end{aligned}$$

and

$$\begin{aligned} \phi _{+}(x)={\left\{ \begin{array}{ll} 0 &{} \text { if } x\le y,\\ \phi (x) &{} \text { otherwise.} \end{array}\right. } \end{aligned}$$

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

$$\begin{aligned} E_{pot}(\phi )=E_{pot}(\phi _{-})+E_{pot}(\phi _{+}), \end{aligned}$$

also, we have that

$$\begin{aligned} E_{pot}(H_{-1,0})< E_{pot}(\phi _{-}),\,E_{pot}(H_{0,1})<E_{pot}(\phi _{+}). \end{aligned}$$

In conclusion, since $E_{pot}(\phi )=2E_{pot}(H_{0,1})+\epsilon ,$ Lemma 21 implies that if $\epsilon <\epsilon _{0}\ll 1,$ then there exist $x_{2},\,x_{1} \in {\mathbb {R}}$ such that

$$\begin{aligned}{} & {} \left\Vert \phi (x)-H_{0,1}(x-x_{2})-H_{-1,0}(x-x_{1})\right\Vert _{H^{1}}\nonumber \\{} & {} \quad \le \left\Vert \phi _{+}-H_{0,1}(x-x_{2})\right\Vert _{H^{1}}+\left\Vert \phi _{-}-H_{-1,0}(x-x_{1})\right\Vert _{H^{1}}\le e^{{-}\frac{4}{\delta }}\ll \delta . \end{aligned}$$

(A32)

So, to finish the proof of Corollary 22, we need only to verify that we have $x_{2}-x_{1}\ge \frac{1}{\delta }$ if $0<\epsilon _{0}\ll 1.$ But, we recall that $H_{0,1}(0)=\frac{1}{\sqrt{2}},$ from which with estimate (A32) we deduce that

$$\begin{aligned} \left|\phi _{+}(x_{2})-\frac{1}{\sqrt{2}}\right|\lesssim \delta , \left|\phi _{-}(x_{1})+\frac{1}{\sqrt{2}}\right|\lesssim \delta , \end{aligned}$$

(A33)

so if $\epsilon _{0}\ll 1,$ then $x_{1}<y<x_{2}.$ Using the fact that U is a smooth function, Lemma 10 and identity (35), we can verify the existence of a constant $C>0$ satisfying the following inequality

$$\begin{aligned} \left| DE_{pot}\left( H_{0,1}(x-x_{2})+H_{-1,0}(x-x_{1})+u\right) (v)\right| \le C \left\Vert v\right\Vert _{H^{1}}. \end{aligned}$$

for any $u,\,v\in H^{1}({\mathbb {R}})$ such that $\left\Vert u\right\Vert _{H^{1}}\le 1.$ Therefore, using estimate (A32) and the Fundamental Theorem of Calculus, we deduce that if $0<\epsilon _{0}\ll 1,$ then

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

(A34)

Furthermore, since the function $A(z)=E_{pot}\left( H^{z}_{0,1}(x)+H_{{-}1,0}(x)\right)$ is a continuous function on ${\mathbb {R}}_{\ge 0}$ and $A(z)>2E_{pot}\left( H_{0,1}\right)$ for any $z\ge 0$, we have for any $k>0$ that there exists $\delta _{k}>0$ satisfying

$$\begin{aligned} \sup _{\{z\in [0,k]\}}A(z)>2E_{pot}\left( H_{0,1}\right) +\delta _{k}. \end{aligned}$$

In conclusion, we obtain from Lemma 7 and the estimate (A34) that $x_{2}-x_{1}\ge \frac{1}{\delta }$ if $0<\epsilon _{0}\ll 1$ and $\epsilon <\epsilon _{0}.$ $\square$

Now, we complement our manuscript by presenting the proof of identity (33).

Proof of Identity (33)

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

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

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

$$\begin{aligned}{} & {} \int _{{\mathbb {R}}}\big (8(H_{0,1}(x))^{3}-6(H_{0,1}(x))^{5}\big )e^{-\sqrt{2}x}\,dx\\{} & {} \quad =\frac{1}{2\sqrt{2}}\int _{1}^{\infty }\frac{8}{y^{\frac{5}{2}}}+\frac{2(y-1)}{y^{\frac{5}{2}}}\,dy\\{} & {} \quad =\frac{1}{2\sqrt{2}}\int _{1}^{\infty }\frac{6}{y^{\frac{5}{2}}}+\frac{2}{y^{\frac{3}{2}}}\,dy, =\frac{1}{2\sqrt{2}}(-4y^{-\frac{3}{2}}-4y^{-\frac{1}{2}})\Big \vert _1^\infty =2\sqrt{2}. \end{aligned}$$

$\square$

Appendix B Proof of Theorem 3

Proof of Theorem 3

We use the notations of Theorems 2 and 4. Clearly, if the result of Theorem 3 is false, then by contradiction for any $N\gg 1$ the inequality

$$\begin{aligned} \left\Vert \overrightarrow{g(t)}\right\Vert \le \frac{\epsilon }{N} \end{aligned}$$

(B35)

could be possible for all $0\le t\le N\frac{\ln {\frac{1}{\epsilon }}}{\epsilon ^{\frac{1}{2}}}=T$ if $\epsilon \ll 1$ enough.

From Modulation Lemma, we can denote the solution $\phi (t,x)$ as

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

such that

$$\begin{aligned} \langle g(t,x),\,\partial _{x}H^{x_{1}(t)}_{-1,0}(x)\rangle _{L^{2}}=0,\,\langle g(t,x),\,\partial _{x}H^{x_{2}(t)}_{0,1}(x)\rangle _{L^{2}}=0. \end{aligned}$$

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

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

(B36)

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

$$\begin{aligned} \left\langle r(t),\,\partial ^{2}_{x}H^{x_{1}(t)}_{-1,0}\right\rangle _{L^{2}}=0,\, \left\langle r(t),\,\partial ^{2}_{x}H^{x_{2}(t)}_{0,1}\right\rangle _{L^{2}}=0. \end{aligned}$$

(B37)

In conclusion, we deduce that

$$\begin{aligned} \left\Vert g(t)\right\Vert _{L^{2}}^{2}=\left\Vert \partial ^{2}_{x} H_{0,1}\right\Vert _{L^{2}}^{2}(P_{1}^{2}+P_{2}^{2})+\left\Vert r(t)\right\Vert _{L^{2}}^{2}+2P_{1} P_{2}\left\langle \partial ^{2}_{x} H^{z(t)}_{0,1},\,\partial ^{2}_{x} H_{-1,0}\right\rangle _{L^{2}}. \nonumber \\ \end{aligned}$$

(B38)

We recall from Theorem 11 that $\frac{1}{\sqrt{2}}\ln {\frac{1}{\epsilon }}<z(t)$ for all $t\ge 0.$ Since, from Lemma 6, we have that $\left\langle \partial ^{2}_{x}H^{x_{1}(t)}_{-1,0}, \,\partial ^{2}_{x}H^{x_{2}(t)}_{0,1}\right\rangle _{L^{2}} \lesssim z(t) e^{-\sqrt{2}z(t)}$ and $z(t)e^{-\sqrt{2}z(t)}\lesssim \epsilon \ln {\frac{1}{\epsilon }}$ if $0<\epsilon \ll 1,$ we deduce from the Eq. (B38) that there is a uniform constant $K>1$ such that for all $t\ge 0$ we have the following estimate

$$\begin{aligned} \frac{\left\Vert g(t)\right\Vert _{L^{2}}}{K} \le \left|P_{1}(t)\right|+\left|P_{2}(t)\right|+\left\Vert r(t)\right\Vert _{L^{2}}\le K \left\Vert \overrightarrow{g(t)}\right\Vert . \end{aligned}$$

(B39)

From Theorem 11 and the orthogonality conditions (B37), we deduce that

$$\begin{aligned} \left\langle \partial _{t}r(t),\,\partial ^{2}_{x}H^{x_{2}(t)}_{0,1}\right\rangle _{L^{2}}=\,&\dot{x}_{2}(t)\left\langle r(t),\,\partial ^{3}_{x}H^{x_{2}(t)}_{0,1}\right\rangle _{L^{2}}=O\Big (\left\Vert r(t)\right\Vert _{L^{2}}\epsilon ^{\frac{1}{2}}\Big ),\\ \left\langle \partial _{t}r(t),\,\partial ^{2}_{x}H^{x_{1}(t)}_{{-}1,0}\right\rangle _{L^{2}}=\,&\dot{x}_{2}(t)\left\langle r(t),\,\partial ^{3}_{x}H^{x_{1}(t)}_{{-}1,0}\right\rangle _{L^{2}}=O\Big (\left\Vert r(t)\right\Vert _{L^{2}}\epsilon ^{\frac{1}{2}}\Big ). \end{aligned}$$

In conclusion, estimate (B39) and Lemma 6 imply that there is a $K>1$ such that

$$\begin{aligned} \left|\dot{P}_{1}(t)\right|+\left|\dot{P}_{2}(t)\right|+\left\Vert \partial _{t}r(t)\right\Vert _{L^{2}}\le K\left\Vert \overrightarrow{g(t)}\right\Vert \end{aligned}$$

(B40)

for all $t\ge 0.$ Finally, Minkowski inequality and estimate (B39) imply that there is a uniform constant $K>1$ such that

$$\begin{aligned} \left\Vert \partial _{x}r(t,x)\right\Vert _{L^{2}}\le K\left\Vert \overrightarrow{g(t)}\right\Vert . \end{aligned}$$

(B41)

We recall from Theorem 12 the following estimate

$$\begin{aligned} \frac{\epsilon }{K} \le \left\Vert \overrightarrow{g(t)}\right\Vert ^{2}+\dot{x}_{1}(t)^{2}+\dot{x}_{2}(t)^{2}+ e^{-\sqrt{2}z(t)}\le K \epsilon \end{aligned}$$

(B42)

for some uniform constant $K>1.$ Now, from hypothesis (B35), we obtain from Theorem 4 and Corollary 5 that there are constants $M\in {\mathbb {N}}$ and $C>0$ such that for all $t\ge 0$ the following inequalities are true

$$\begin{aligned} \max _{j\in \{1,\,2\}}\left|x_{j}(t)-d_{j}(t)\right|&\le \epsilon \left( \ln {\frac{1}{\epsilon }}\right) ^{M+1}\exp \Big (\frac{10C\epsilon ^{\frac{1}{2}}t}{\ln {\frac{1}{\epsilon }}}\Big ), \end{aligned}$$

(B43)

$$\begin{aligned} \max _{j\in \{1,\,2\}}\left|\dot{x}_{j}(t)-\dot{d}_{j}(t)\right|&\le \epsilon ^{\frac{3}{2}}\left( \ln {\frac{1}{\epsilon }}\right) ^{M}\exp \Big (\frac{10C\epsilon ^{\frac{1}{2}}t}{\ln {\frac{1}{\epsilon }}}\Big ), \end{aligned}$$

(B44)

$$\begin{aligned} \max _{j\in \{1,\,2\}}\left|\ddot{x}_{j}(t)-\ddot{d}_{j}(t)\right|&\le \epsilon ^{\frac{3}{2}}\left( \ln {\frac{1}{\epsilon }}\right) \exp \Big (\frac{10C\epsilon ^{\frac{1}{2}}t}{\ln {(\frac{1}{\epsilon })}}\Big ), \end{aligned}$$

(B45)

for a uniform constant $C>0.$

From the partial differential equation (1) satisfied by $\phi (t,x)$ and the representation (B36) of g(t, x), we deduce in the distributional sense that for any $h(x)\in H^{1}({\mathbb {R}})$ that

$$\begin{aligned}{} & {} \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}}\nonumber \\{} & {} \quad ={-}\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}}\nonumber \\{} & {} \qquad {-}\,\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 ]\right\rangle _{L^{2}}\nonumber \\{} & {} \qquad {-}\,\left\langle h(x),\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}}\nonumber \\{} & {} \qquad {-}\,\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] \right\rangle _{L^{2}}\nonumber \\{} & {} \qquad {+}\,\left\langle h(x),\,\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}} \nonumber \\{} & {} \qquad {-}\,\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 ]\right\rangle _{L^{2}}\nonumber \\{} & {} \qquad {-}\,\left\langle h(x), 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 _{L^{2}}\nonumber \\{} & {} \qquad {+}\,O\left( \left\Vert h\right\Vert _{L^{2}}\left[ \left\Vert g(t)\right\Vert _{H^{1}}^{2}+\max _{j\in \{1,\,2\}}\left|\ddot{x}_{j}(t)\right|\right] \right) \nonumber \\{} & {} \qquad {+}\,O\left( \left\Vert h\right\Vert _{L^{2}}\left[ \max _{j\in \{1,\,2\}}\left|\dot{P}_{j}(t)\dot{x}_{j}(t)\right|+\max _{j \in \{1,\,2\}} \left|P_{j}(t)\right|e^{-\sqrt{2}z(t)}\right] \right) \nonumber \\{} & {} \qquad {+}\,O\left( \left|P_{j}(t)\ddot{x}_{j}(t)\right|+\left|P_{j}(t)\dot{x}_{j}(t)^{2}\right|\right) . \end{aligned}$$

(B46)

From Lemma 20 and estimates (B43) and (B45), we obtain from (B46) that

$$\begin{aligned}{} & {} \left\langle h(x),\,({\ddot{P}}_{1}(t)+\dot{x}_{1}(t)^{2})\partial ^{2}_{x}H^{x_{1}(t)}_{-1,0}+ ({\ddot{P}}_{2}(t)+\dot{x}_{2}(t)^{2})\partial ^{2}_{x}H^{x_{2}(t)}_{0,1}\right\rangle _{L^{2}}\nonumber \\{} & {} \quad ={-}\left\langle h(x), P_{1}(t)\Big [\Big (-\partial ^{2}_{x}+\ddot{U}(H^{x_{1}(t)}_{-1,0})\Big )\partial ^{2}_{x}H^{x_{1}(t)}_{-1,0}\Big ]\right\rangle _{L^{2}}\nonumber \\{} & {} \qquad {-}\,\left\langle h(x),P_{2}(t)\Big [\Big (-\partial ^{2}_{x}+\ddot{U}(H^{x_{2}(t)}_{0,1})\Big )\partial ^{2}_{x}H^{x_{2}(t)}_{0,1}\Big ]\right\rangle _{L^{2}}\nonumber \\{} & {} \qquad {-}\,\left\langle h(x),\Big [\partial ^{2}_{t}r(t) -\partial ^{2}_{x}r(t)+{\ddot{U}}(H^{x_{2}(t)}_{0,1}+H^{x_{1}(t)}_{-1,0})r(t)\Big ]\right\rangle _{L^{2}}\nonumber \\{} & {} \qquad {+}\,O\left( \left\Vert h\right\Vert _{L^{2}}\left[ \max _{j\in \{1,\,2\}}\left|\ddot{x}_{j}(t)-\ddot{d}_{j}(t)\right|+e^{-\sqrt{2}d(t)}\right] \right) \nonumber \\{} & {} \qquad {+}\,O\left( \left\Vert h\right\Vert _{L^{2}}\left[ \left|z(t)-d(t)\right|e^{-\sqrt{2}z(t)}+e^{-2\sqrt{2}z(t)}\right] \right) \nonumber \\{} & {} \qquad {+}\,O\left( \left\Vert h\right\Vert _{L^{2}}\left[ \left\Vert g(t)\right\Vert _{H^{1}}^{2}+\max _{j\in \{1,\,2\}}\left|\ddot{x}_{j}(t)\right|\right] \right) \nonumber \\{} & {} \qquad {+}\,O\left( \left\Vert h\right\Vert _{L^{2}}\left[ \max _{j\in \{1,\,2\}}\left|\dot{P}_{j}(t)\dot{x}_{j}(t)\right|+\max _{j \in \{1,\,2\}} \left|P_{j}(t)\right|e^{-\sqrt{2}z(t)}+\left|P_{j}(t)\ddot{x}_{j}(t)\right|\right] \right) \nonumber \\{} & {} \qquad {+}\,O\left( \left\Vert h\right\Vert _{L^{2}}\left|P_{j}(t)\dot{x}_{j}(t)^{2}\right|\right) . \end{aligned}$$

(B47)

From the condition (B37), we deduce that

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

which imply with Theorem 11 the existence of a uniform constant $C>0$ such that

$$\begin{aligned} \left|\left\langle \partial ^{2}_{t}r(t),\,\partial ^{2}_{x}H^{x_{2}(t)}_{0,1}\right\rangle _{L^{2}}\right|\le C\epsilon ^{\frac{1}{2}}\left\Vert \overrightarrow{r(t)}\right\Vert ,\, \left|\left\langle \partial ^{2}_{t}r(t),\,\partial ^{2}_{x}H^{x_{1}(t)}_{-1,0}\right\rangle _{L^{2}}\right|\le C\epsilon ^{\frac{1}{2}}\left\Vert \overrightarrow{r(t)}\right\Vert . \nonumber \\ \end{aligned}$$

(B48)

From (B39), (B40) and (B41), we obtain that $\left\Vert \overrightarrow{r(t)}\right\Vert \lesssim \left\Vert \overrightarrow{g(t)}\right\Vert .$

In conclusion, after we apply the partial differential equation (B47) in the distributional sense to $\partial ^{2}_{x}H^{x_{2}(t)}_{0,1},\,\partial ^{2}_{x}H^{x_{1}(t)}_{-1,0},$ the estimates (B39), (B40), (B41), (B43), (B45) and (B48) 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 $0\le t\le \frac{N\ln {\frac{1}{\epsilon }}}{\epsilon ^{\frac{1}{2}}}$

$$\begin{aligned} \left|{\ddot{P}}_{j}(t)+\dot{x}_{j}(t)^{2}\right|\le K_{1}\left( e^{-\sqrt{2}d(t)}+\epsilon ^{\frac{3}{2}}\left( \ln {\frac{1}{\epsilon }}\right) ^{M+1}\exp \Big (\frac{10C\epsilon ^{\frac{1}{2}}t}{\ln {\frac{1}{\epsilon }}}\Big )+\frac{\epsilon }{N} \right) , \end{aligned}$$

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

$$\begin{aligned} \left|\sum _{j=1}^{2}{\ddot{P}}_{j}(t)+\dot{x}_{j}(t)^{2}\right|\le 2K_{1}\left( e^{-\sqrt{2}d(t)}+\epsilon ^{\frac{3}{2}}\left( \ln {\frac{1}{\epsilon }}\right) ^{M+1}\exp \Big (\frac{10C\epsilon ^{\frac{1}{2}}t}{\ln {\frac{1}{\epsilon }}}\Big )+\frac{\epsilon }{N}\right) . \nonumber \\ \end{aligned}$$

(B49)

Since $\left|\sum _{j=1}^{2}{\ddot{P}}_{j}(t)\right|\ge -\left|\sum _{j=1}^{2}{\ddot{P}}_{j}(t)+\dot{x}_{j}(t)^{2}\right|+\sum _{j=1}^{2} \dot{x}_{j}(t)^{2},$ we deduce from the estimates (B49) and (B42) that

$$\begin{aligned} \left|\sum _{j=1}^{2}{\ddot{P}}_{j}(t)\right|\ge & {} \frac{\epsilon }{K}-\Big [e^{-\sqrt{2}z(t)}+\left\Vert \overrightarrow{g(t)}\right\Vert ^{2}\Big ]\nonumber \\{} & {} {-}\,2K_{1}\Big [e^{-\sqrt{2}d(t)}+\epsilon ^{\frac{3}{2}}\left( \ln {\frac{1}{\epsilon }}\right) ^{M+1}\exp \Big (\frac{10C\epsilon ^{\frac{1}{2}}t}{\ln {\frac{1}{\epsilon }}}\Big )\Big ]-\frac{2K_{1}\epsilon }{N}.\nonumber \\ \end{aligned}$$

(B50)

We recall that from the statement of Theorem 4 that $e^{-\sqrt{2}d(t)}=\frac{v^{2}}{8}{{\,\textrm{sech}\,}}{(\sqrt{2}vt+c)}^{2},$ with $v=\Big (\frac{\dot{z}(0)^{2}}{4}+8 e^{-\sqrt{2}z(0)}\Big )^{\frac{1}{2}},$ which implies that $v\lesssim \epsilon ^{\frac{1}{2}}.$ Since we have verified in Theorem 11 that $e^{-\sqrt{2}z(t)}\lesssim \epsilon ,$ the mean value theorem implies that $\left|e^{-\sqrt{2}z(t)}-e^{-\sqrt{2}d(t)}\right|=O(\epsilon \left|z(t)-d(t)\right|),$ from which we deduce from (B43) that

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

In conclusion, if $\epsilon \ll 1$ enough, we obtain for $0\le t \le \frac{N\ln {(\frac{1}{\epsilon })}}{\epsilon ^{\frac{1}{2}}}$ from (B50) that

$$\begin{aligned} \left|\sum _{j=1}^{2}{\ddot{P}}_{j}(t)\right|\ge & {} \frac{\epsilon }{K}-\Big [e^{-\sqrt{2}d(t)}+\left\Vert \overrightarrow{g(t)}\right\Vert ^{2}\Big ]\nonumber \\{} & {} {-}\,4K_{1}\Big [e^{-\sqrt{2}d(t)}+\epsilon ^{\frac{3}{2}}\left( \ln {\frac{1}{\epsilon }}\right) ^{M+1}\exp \Big (\frac{10C\epsilon ^{\frac{1}{2}}t}{\ln {\frac{1}{\epsilon }}}\Big )\Big ]-\frac{2K_{1}\epsilon }{N}.\nonumber \\ \end{aligned}$$

(B51)

The conclusion of the demonstration will follow from studying separate cases in the choice of $v>0,\,c.$ We also observe that $K,\, K_{1}$ are uniform constants and the value of $N \in {\mathbb {N}}_{>0}$ can be chosen at the beginning of the proof to be as much large as we need.

Case 1. ($v^{2}\le \frac{8\epsilon }{(1+4K_{1})2K}.$) From inequality (B51), we deduce that

$$\begin{aligned} \left|\sum _{j=1}^{2}{\ddot{P}}_{j}(t)\right|\ge \frac{\epsilon }{2K}-\left\Vert \overrightarrow{g(t)}\right\Vert ^{2}-4K_{1}\Big (\epsilon ^{\frac{3}{2}}\left( \ln {\frac{1}{\epsilon }}\right) ^{M+1}\exp \Big (\frac{10C\epsilon ^{\frac{1}{2}}t}{\ln {\frac{1}{\epsilon }}}\Big )\Big )-\frac{2K_{1}\epsilon }{N}, \end{aligned}$$

then, from (B35) we deduce for $0\le t \le \frac{\ln {\frac{1}{\epsilon }}}{\epsilon ^{\frac{1}{2}}}$ that if $\epsilon$ is small enough and $N>10K K_{1},$ then $\left|\sum _{j=1}^{2}{\ddot{P}}_{j}(t)\right|\ge \frac{\epsilon }{4K},$ and so,

$$\begin{aligned} \left|\sum _{j=1}^{2}\dot{P}_{j}(t)\right|\ge \frac{\epsilon t}{4K}-\left|\sum _{j=1}^{2}\dot{P}_{j}(0)\right|, \end{aligned}$$

which contradicts the fact that (B40) and (B35) should be true for $\epsilon \ll 1.$

Case 2. ($v^{2}\ge \frac{8\epsilon }{ (1+4K_{1}) 2K},\,\left|c\right|>2\ln {(\frac{1}{\epsilon })}.$) It is not difficult to verify that for $0\le t\le \min ( \frac{\left|c\right|}{2\sqrt{2}v},N\frac{\ln {\frac{1}{\epsilon }}}{\epsilon ^{\frac{1}{2}}}),$ we have that $e^{-\sqrt{2}d(t)}\le \frac{v^{2}}{8}{{\,\textrm{sech}\,}}{(\frac{c}{2})}^{2}\lesssim \epsilon ^{3}.$ Therefore, if $N>10 K K_{1}$ and $\epsilon >0$ is small enough, estimate (B51) would imply that $\left|\sum _{j=1}^{2}\ddot{P}_{j}(t)\right|\ge \frac{\epsilon }{4K}$ is true in this time interval. Also, since now $v\cong \epsilon ^{\frac{1}{2}},$ we have that

$$\begin{aligned} \frac{\ln {\frac{1}{\epsilon }}}{\epsilon ^{\frac{1}{2}}} \lesssim \frac{\left|c\right|}{2\sqrt{2}v}, \end{aligned}$$

so we obtain a contradiction by a similar argument to the Case 1.

Case 3. ($v^{2}\ge \frac{8\epsilon }{ (1+4K_{1}) 2K}$ and $\left|c\right|\le 2\ln {\frac{1}{\epsilon }}.$) For $N\gg 1$ 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 $\left\{ t_{0}\le t \le 2\frac{(1+4K_{1})^{\frac{1}{2}} K^{\frac{1}{2}} \sqrt{2} \ln {\frac{1}{\epsilon }}}{\epsilon ^{\frac{1}{2}}}\right\}$ that $e^{-\sqrt{2}d(t)}\le \frac{v^{2}}{8}{{\,\textrm{sech}\,}}{\Big (2\ln {\frac{1}{\epsilon }}\Big )}^{2}\lesssim \epsilon ^{5}$ and $\frac{\epsilon }{N}<\frac{\epsilon }{20K}.$ In conclusion, estimate (B50) implies that $\left|\sum _{j=1}^{2}{\ddot{P}}_{j}(t)\right|\ge \frac{\epsilon }{4K}$ is true in this time interval. From the Fundamental Calculus Theorem, we have that

$$\begin{aligned} \left|\sum _{j=1}^{2}\dot{P}_{j}(t)\right|\ge \frac{\epsilon (t-t_{0})}{4K}-\left|\sum _{j=1}^{2}\dot{P}_{j}(t_{0})\right|. \end{aligned}$$

In conclusion, hypothesis (B35) and estimate (B40) imply for $T=2\frac{(1+2K_{1})^{\frac{1}{2}} K^{\frac{1}{2}} \sqrt{2} \ln {\frac{1}{\epsilon }}}{\epsilon ^{\frac{1}{2}}}$ and $N\gg 1$ that

$$\begin{aligned} \left|\sum _{j=1}^{2}\dot{P}_{j}(T)\right|\ge \frac{\epsilon ^{\frac{1}{2}} (1+2K_{1})^{\frac{1}{2}} \sqrt{2} \ln {\frac{1}{\epsilon }} }{8K^{\frac{1}{2}}}, \end{aligned}$$

which contradicts the fact that (B35) and (B40) should be true, which finishes our proof. $\square$