Appendix 1: The coercivity of the quadratic term
In this appendix, we prove Proposition 2.1. The proof of Part (1) is the same as that in Proposition 2.8 (a) [37]. As for Part (2), the proof is divided into several steps.
Step 1: Spectral decomposition. First of all, it follows from the exponential decay of \(\phi _{\omega ,c}\) that \(\mathcal {L}_{+}\) is a relatively compact perturbation of the operator \(-\frac{1}{2} \partial _{x}^{2} + \frac{1}{2}\left( \omega -\frac{c^{2}}{4}\right) .\) By Weyl’s theorem in [30], we obtain that the essential spectrum of \(\mathcal {L}_{+}\) on \(L^{2}\left( {\mathbb {R}}\right) \) is
$$\begin{aligned} \sigma _{\mathrm {ess}}\left( { \mathcal {L}_{+} }\right)&=\sigma _{\mathrm {ess}}\left( { -\frac{1}{2}\partial _{x}^{2} + \frac{1}{2}\left( \omega -\frac{c^{2}}{4}\right) }\right) = \left[ \frac{1}{2}\left( \omega -\frac{c^{2}}{4}\right) ~,~+\infty \right) . \end{aligned}$$
Moreover, all spectrum below the lower bound of the essential spectrum are either an isolated point of \(\sigma \left( { \mathcal {L}_{+} }\right) \) or an eigenvalue of finite multiplicity of \(\mathcal {L}_{+}.\)
Next, since \(\phi _{\omega ,c}\) satisfies
$$\begin{aligned} \left( \omega -\frac{c^2}{4}\right) \phi _{\omega ,c} - \partial ^2_x \phi _{\omega ,c}-\frac{3}{16}\phi _{\omega ,c}^{5} = - \frac{c}{2}\phi _{\omega ,c}^{3}\;, \end{aligned}$$
(6.25)
then by differentiating equation (6.25) with respect to x, we obtain
$$\begin{aligned} \mathcal {L}_{+}\partial _x \phi _{\omega ,c} =0. \end{aligned}$$
(6.26)
Therefore, by \(\partial _x \phi _{\omega ,c}\in L^{2}(\mathbb {R}),\) we obtain from (6.26) that 0 is an eigenvalue of \(\mathcal {L}_{+}.\) By a classical ODE argument as in [37], we obtain
$$\begin{aligned} \ker \mathcal {L}_{+} = \mathrm {span}\left\{ ~\partial _x \phi _{\omega ,c}~\right\} . \end{aligned}$$
(6.27)
Thus, it follows from Sturm-Liouville theory that 0 is the second eigenvalue of \(\mathcal {L}_{+},\) and moreover \(\mathcal {L}_{+}\) enjoys only one negative eigenvalue \(-\lambda ^2_{1}\) with a \(L^{2}(\mathbb {R})\) normalized eigenfunction \(\chi .\) More precisely, we have
$$\begin{aligned} \mathcal {L}_{+}\chi = -\lambda ^2_{1}\chi \quad \text {with }\quad \Vert \chi \Vert _{2}=1. \end{aligned}$$
(6.28)
Now, define
$$\begin{aligned} \mu \triangleq&\inf ~\left\{ { \frac{\left( \,{\mathcal {L}_{+}\psi }\, ,\,{\psi }\,\right) }{\left( \,{ \psi }\, ,\,{\psi }\,\right) } ~:~ \psi \in L^2(\mathbb {R}), ~~ \left( \,{\psi }\, ,\,{\chi }\,\right) = \left( \,{\psi }\, ,\,{\partial _x\phi _{\omega ,c}}\,\right) = 0~ }\right\} , \end{aligned}$$
(6.29)
then by a classical variational argument, it is easy to see that \(\mu >0.\) Therefore, the space \(L^{2}(\mathbb {R})\) can be decomposed as a direct sum as follows
$$\begin{aligned} L^{2}=N\bigoplus \ker \mathcal {L}_{+} \bigoplus P, \end{aligned}$$
(6.30)
where \(N=\mathrm {span}\left\{ ~\chi ~\right\} ,\)
\(\ker \mathcal {L}_{+}\) is defined by (6.27), and P is a closed subspace of \(L^{2}\) such that
$$\begin{aligned} \left( \,{\mathcal {L}_{+}\psi }\, ,\,{\psi }\,\right) \ge \mu \left( \,{\psi }\, ,\,{\psi }\,\right) ,\quad \text { for any } \psi \in P. \end{aligned}$$
(6.31)
Step 2: Nonnegative property. We show
$$\begin{aligned} \inf ~\left\{ { \frac{\left( \,{\mathcal {L}_{+}\psi }\, ,\,{\psi }\,\right) }{\left( \,{ \psi }\, ,\,{\psi }\,\right) } ~:~ \psi \in L^2, ~~ \left( \,{\psi }\, ,\,{\phi _{\omega ,c}}\,\right) = \left( \,{\psi }\, ,\,{\phi _{\omega ,c}^3}\,\right) = \left( \,{\psi }\, ,\,{\partial _x\phi _{\omega ,c}}\,\right) = 0~ }\right\} \ge 0. \end{aligned}$$
In fact, by differentiating equation (6.25) with respect to c and \(\omega ,\) we have
$$\begin{aligned} \mathcal {L}_{+}\partial _{c}\phi _{\omega ,c} = \frac{c}{2}\phi _{\omega ,c}-\frac{1}{2}\phi _{\omega ,c}^{3},\quad \mathcal {L}_{+}\partial _{\omega }\phi _{\omega ,c} =-\phi _{\omega ,c}. \end{aligned}$$
(6.32)
On one hand, (6.30) allows us to decompose \(\partial _{c}\phi _{\omega ,c}\) and \(\partial _{\omega }\phi _{\omega ,c}\) as follows,
$$\begin{aligned} \partial _{c}\phi _{\omega ,c} = a_{1}\chi + b_{1}\partial _{x}\phi _{\omega ,c}+p_{1},\quad \partial _{\omega }\phi _{\omega ,c} = a_{2}\chi +b_{2}\partial _{x}\phi _{\omega ,c}+p_{2}. \end{aligned}$$
(6.33)
where \(a_{1},\)
\(a_{2},\)
\(b_{1}\) and \(b_{2}\) are constants; \(\chi \) is defined by (6.28); \(p_{1}\) and \(p_{2}\) belong to the subspace P defined by (6.31). On the other hand, for any \(\psi \in L^{2}(\mathbb {R})\) with
$$\begin{aligned} \left( \,{\psi }\, ,\,{\phi _{\omega ,c}}\,\right) = \left( \,{\psi }\, ,\,{\phi _{\omega ,c}^3}\,\right) = \left( \,{\psi }\, ,\,{\partial _x\phi _{\omega ,c}}\,\right) = 0, \end{aligned}$$
we decompose \(\psi \) as follows
$$\begin{aligned} \psi =a\chi +p,\quad \text { with } a\in \mathbb {R},\text {~and~} p\in P. \end{aligned}$$
(6.34)
By some straight calculations, we have
$$\begin{aligned} \left( \,{\mathcal {L}_{+} \psi }\, ,\,{\psi }\,\right)&= -\lambda ^{2}_1a^{2}+\left( \,{\mathcal {L}_{+} p}\, ,\,{p}\,\right) . \end{aligned}$$
(6.35)
Then, it follows from \(\left( \,{\psi }\, ,\,{\phi _{\omega ,c}}\,\right) = 0\) and (6.32) that
$$\begin{aligned} \left( \,{\psi }\, ,\,{\mathcal {L}_{+}\partial _{\omega }\phi _{\omega ,c}}\,\right) =0, \end{aligned}$$
which together with (6.33) and (6.34) implies that
$$\begin{aligned} -aa_{2}\lambda ^{2}_1+\left( \,{\mathcal {L}_{+} p}\, ,\,{p_{2}}\,\right) =0. \end{aligned}$$
(6.36)
By a similar argument as above, we have
$$\begin{aligned} -aa_{1}\lambda ^{2}_1+\left( \,{\mathcal {L}_{+} p}\, ,\,{p_{1}}\,\right) =0. \end{aligned}$$
(6.37)
Next, since
$$\begin{aligned} \det d''\left( {\omega }\, ,\,{c}\right) <0, \end{aligned}$$
there exists \(\left( {\xi _{1}~,~\xi _{2}}\right) \in \mathbb {R}^{2}\) such that
$$\begin{aligned} \begin{pmatrix} \xi _{1} &{} \xi _{2} \\ \end{pmatrix} d''\left( {\omega }\, ,\,{c}\right) \begin{pmatrix} \xi _{1} \\ \xi _{2} \\ \end{pmatrix}>0. \end{aligned}$$
(6.38)
Now, let \(\left( {\xi _{1}~,~\xi _{2}}\right) \in \mathbb {R}^{2}\) satisfy (6.38), and
$$\begin{aligned} p_{0}\triangleq ~&\xi _{1}p_{1}+\xi _{2}p_{2}, \end{aligned}$$
then by a straight calculation, we have
$$\begin{aligned} \left( \,{\mathcal {L}_{+}p_{0}}\, ,\,{p_{0}}\,\right) =&\; \xi _{1}^{2}\left( \,{\mathcal {L}_{+}p_{1}}\, ,\,{p_{1}}\,\right) + 2\xi _{1}\xi _{2}\left( \,{\mathcal {L}_{+}p_{1}}\, ,\,{p_{2}}\,\right) + \xi _{2}^{2}\left( \,{\mathcal {L}_{+}p_{2}}\, ,\,{p_{2}}\,\right) \nonumber \\ =&\; \xi _{1}^{2}\left( \,{\mathcal {L}_{+}\partial _{c}\phi _{\omega ,c}}\, ,\,{\partial _{c}\phi _{\omega ,c}}\,\right) + 2\xi _{1}\xi _{2} \left( \,{\mathcal {L}_{+}\partial _{c}\phi _{\omega ,c}}\, ,\,{\partial _{\omega }\phi _{\omega ,c}}\,\right) \nonumber \\&\; + \xi _{2}^{2} \left( \,{\mathcal {L}_{+}\partial _{\omega }\phi _{\omega ,c}}\, ,\,{\partial _{\omega }\phi _{\omega ,c}}\,\right) + \xi _{1}^{2}a_{1}^{2}\lambda ^{2}_1 + 2\xi _{1}\xi _{2}a_{1}a_{2}\lambda ^{2}_1 + \xi _{2}^{2}a_{2}^{2}\lambda ^{2}_1 \nonumber \\ =&\; - \begin{pmatrix} \xi _{1} &{} \xi _{2} \\ \end{pmatrix} d''\left( {\omega }\, ,\,{c}\right) \begin{pmatrix} \xi _{1} \\ \xi _{2} \\ \end{pmatrix} + \left( {a_1\xi _{1}+a_2\xi _{2}}\right) ^{2}\lambda ^{2}_1 \nonumber \\ <&\; \left( {a_{1}\xi _{1}+a_{2}\xi _{2}}\right) ^{2}\lambda ^{2}_1. \end{aligned}$$
(6.39)
Next, by (6.37), (6.36), (6.39) and the Cauchy-Schwarz inequality, it is easy to see that,
$$\begin{aligned} \left( \,{ \mathcal {L}_{+}p }\, ,\,{p}\,\right) \ge \frac{ {\left( \,{ \mathcal {L}_{+}p }\, ,\,{p_{0}}\,\right) }^{2} }{ \left( \,{ \mathcal {L}_{+}p_{0} }\, ,\,{p_{0}}\,\right) }&\ge \frac{a^{2}\lambda ^{2}_1\left( { a_{1}\xi _{1}+a_{2}\xi _{2} }\right) ^{2} }{ \left( { a_{1}\xi _{1}+a_{2}\xi _{2} }\right) ^{2} } =a^{2}\lambda ^{2}_1, \end{aligned}$$
which, together (6.35), implies that,
$$\begin{aligned} \left( \,{ \mathcal {L}_{+}\psi }\, ,\,{ \psi }\,\right) \ge \;0. \end{aligned}$$
Step 3: Positive property. Last we show
$$\begin{aligned} \inf ~\left\{ { \frac{\left( \,{\mathcal {L}_{+}\psi }\, ,\,{\psi }\,\right) }{\left( \,{ \psi }\, ,\,{\psi }\,\right) } ~:~ \psi \in L^2(\mathbb {R}), ~ \left( \,{\psi }\, ,\,{\phi _{\omega ,c}}\,\right) = \left( \,{\psi }\, ,\,{\phi _{\omega ,c}^3}\,\right) = \left( \,{\psi }\, ,\,{\partial _x\phi _{\omega ,c}}\,\right) = 0~ }\right\} > 0. \end{aligned}$$
We argue by contradiction. Suppose that there exists a sequence \(\psi _{n}\in L^2(\mathbb {R})\) such that
$$\begin{aligned} \left( \,{ \mathcal {L}_{+}\psi _{n}}\, ,\,{ \psi _{n} }\,\right) \rightarrow 0, \end{aligned}$$
with
$$\begin{aligned} \left( \,{\psi _{n}}\, ,\,{\phi _{\omega ,c}}\,\right) = \left( \,{\psi _{n}}\, ,\,{\phi _{\omega ,c}^3}\,\right) = \left( \,{\psi _{n}}\, ,\,{\partial _x\phi _{\omega ,c}}\,\right) = 0 \text {~~and~~} \left( \,{\psi _{n}}\, ,\,{\psi _{n}}\,\right) =1. \end{aligned}$$
By a decomposition similar as (6.34), we have for any n
$$\begin{aligned} \psi _{n}=a_{n}\chi +p_{n}, \quad \text { with ~} a_{n}\in \mathbb {R}, \text {~and~} p_{n}\in P, \end{aligned}$$
moreover, \(\left( \,{ \mathcal {L}_{+}p_{n} }\, ,\,{ p_{0} }\,\right) =\left( a_1\xi _1+a_2\xi _2\right) a_{n}\lambda ^{2}_1.\) Therefore, by the similar arguments as in Step 2, we have
$$\begin{aligned} 0\leftarrow \left( \,{ \mathcal {L}_{+}\psi _{n}}\, ,\,{ \psi _{n} }\,\right)&\ge -a^2_{n}\lambda ^{2}_1 + \frac{ a_{n}^{2}\left( { \xi _{1}a_{1}+\xi _{2}a_{2} }\right) ^{2}\lambda ^{4}_1 }{ \left( \,{ \mathcal {L}_{+}p_{0} }\, ,\,{ p_{0} }\,\right) } \\&=a_{n}^{2}\lambda ^{2}_1 \left( { \frac{ \lambda ^{2}_1\left( { \xi _{1}a_{1}+\xi _{2}a_{2} }\right) ^{2} }{ \left( \,{ \mathcal {L}_{+}p_{0} }\, ,\,{ p_{0} }\,\right) } -1 }\right) . \end{aligned}$$
Thus, it follows from (6.39) that,
$$\begin{aligned} a_{n}\rightarrow 0,\qquad \text {as } n\rightarrow \infty , \end{aligned}$$
which implies that
$$\begin{aligned} \left( \,{ \mathcal {L}_{+}p_{n} }\, ,\,{ p_{n} }\,\right) \rightarrow 0. \end{aligned}$$
Thus \(p_{n}\rightarrow 0\) in \(L^2(\mathbb {R})\), which is in contradiction with \(\left( \,{\psi _{n}}\, ,\,{\psi _{n}}\,\right) =1.\) This ends the proof.
Appendix 2: The linearization of the action functional
In this part, we show Lemma 6.1. First of all, we show the following claim,
Claim 1
Let \(\mathcal {R}_{k}\) be one of the expression \(R_{k},\) \(\partial _{x}R_{k}\) and \(\partial _{x}^{2}R_{k}\), and \(\mathfrak {g}\) and \(\mathfrak {h}\) be defined by (5.2), then
$$\begin{aligned}&\displaystyle \int \left| { \mathcal {R}_{1}\left( {t\,,\,x}\right) ~\mathcal {R}_{2}\left( {t\,,\,x}\right) }\right| \;\mathrm {d}x\le C~ e^{ -8\theta _{2}\left( { \frac{L}{2} + 8\theta _{2}\;t }\right) }.\end{aligned}$$
(6.40)
$$\begin{aligned}&\displaystyle \int \left| { \mathcal {R}_{1}\left( {t\,,\,x}\right) ~\mathfrak {h}\left( {t\,,\,x}\right) }\right| \;\mathrm {d}x+ \int \left| { \mathcal {R}_{2}\left( {t\,,\,x}\right) ~\mathfrak {g}\left( {t\,,\,x}\right) }\right| \;\mathrm {d}x\le C~ e^{ -8\theta _{2} \left( { \frac{L}{2} + 8\theta _{2}t }\right) }, \end{aligned}$$
(6.41)
Proof
Firstly, by Lemma 4.3, we have
$$\begin{aligned} \dot{x}_{2}\left( {t}\right) -\dot{x}_{1}\left( {t}\right)&= \left( {\; c_{2}^{0}-c_{1}^{0}\;}\right) + \left( {\; \dot{x}_{2}\left( {t}\right) - c_{2}\left( {t}\right) \;}\right) - \left( {\; \dot{x}_{1}\left( {t}\right) - c_{1}\left( {t}\right) \;}\right) \\&\quad + \left( {\; c_{2}\left( {t}\right) - c_{2}\left( {0}\right) \;}\right) - \left( {\; c_{1}\left( {t}\right) - c_{1}\left( {0}\right) \; }\right) + \left( {\; c_{2}\left( {0}\right) - c_{2}^{0}\; }\right) - \left( {\; c_{1}\left( {0}\right) - c_{1}^{0} \; }\right) \\&\ge \frac{ c_{2}^{0}-c_{1}^{0} }{ 4 }, \end{aligned}$$
therefore, integrating in \(t>0\) gives us that for \(t>0\)
$$\begin{aligned} x_{2}\left( {t}\right) -x_{1}\left( {t}\right) \ge \frac{L}{2} + \frac{ c_{2}^{0}-c_{1}^{0} }{ 4 }\; t \ge \frac{L}{2} + 8\theta _{2}\;t. \end{aligned}$$
Thus,
$$\begin{aligned} \int \left| { \mathcal {R}_{1}\left( {t\,,\,x}\right) ~\mathcal {R}_{2}\left( {t\,,\,x}\right) }\right| \;\mathrm {d}x&\le C \int e^{ -\frac{ \sqrt{ 4\omega _{1}\left( {t}\right) -c_{1}^{2}\left( {t}\right) } }{2}\left| { x-x_{1}\left( {t}\right) }\right| } ~ e^{ -\frac{ \sqrt{4\omega _{2}\left( {t}\right) -c_{2}^{2}\left( {t}\right) } }{2}\left| { x-x_{2}\left( {t}\right) }\right| } \;\mathrm {d}x\\&\le C \int e^{ -\frac{ \sqrt{ 4\omega _{1}^{0}-\left( {c_{1}^{0}}\right) ^{2} } }{4}\left| { x-x_{1}\left( {t}\right) }\right| } ~ e^{ -\frac{ \sqrt{ 4\omega _{2}^{0}-\left( {c_{2}^{0}}\right) ^{2} } }{4}\left| { x-x_{2}\left( {t}\right) }\right| } \;\mathrm {d}x\\&\le C e^{ -\frac{ \sqrt{ 4\omega _{1}^{0}-\left( {c_{1}^{0}}\right) ^{2} } }{8}\left| { x_{2}\left( {t}\right) -x_{1}\left( {t}\right) }\right| }\\&\le C e^{ -8\theta _{2}\left( { \frac{L}{2} + 8\theta _{2}\;t }\right) }. \end{aligned}$$
Secondly, as for (6.41), we only estimate the former term since the later term can be proved in the same way. By Lemma 4.3, we have for sufficiently small \(\alpha _0\) and sufficiently large \(L_0\)
$$\begin{aligned} \frac{d}{dt}\left( \overline{x}^{0} + \sigma t - \sqrt{t+a} - x_{1}\left( {t}\right) \right) =&\; \sigma - \frac{1}{ 2\sqrt{t+a}}-\dot{x}_{1}\left( {t}\right) \nonumber \\ \ge&\left( { \sigma -c_{1}^{0} }\right) - \frac{4}{L} - \left( {\; \dot{x}_{1}\left( {t}\right) - c_{1}\left( {t}\right) \;}\right) + \left( {\; c_{1}^{0} - c_{1}\left( {t}\right) \; }\right) \nonumber \\ \ge&\frac{\sigma -c_{1}^{0}}{4}, \end{aligned}$$
(6.42)
By integrating with respect to t, we obtain
$$\begin{aligned} \overline{x}^{0} + \sigma t - \sqrt{t+a} - x_{1}\left( {t}\right) \ge&\; \overline{x}^{0} - \frac{L}{8} - x_{1}\left( {0}\right) + \frac{\sigma -c_{1}^{0}}{4}\; t \nonumber \\ \ge&\frac{L}{4} + 4\theta _{2}t. \end{aligned}$$
(6.43)
This implies that
$$\begin{aligned} \int \left| { \mathcal {R}_{1}\left( {t\,,\,x}\right) ~\mathfrak {h}\left( {t\,,\,x}\right) }\right| \;\mathrm {d}x\le&\; C \int _{x> \overline{x}^{0} + \sigma t - \sqrt{t+a}} e^{ -\frac{ \sqrt{ 4\omega _{1}\left( {t}\right) -c_{1}^{2}\left( {t}\right) } }{2}\left| { x-x_{1}\left( {t}\right) }\right| } \;\mathrm {d}x\\ \le&\; C \int _{x > \overline{x}^{0} + \sigma t - \sqrt{t+a}} e^{ -\frac{ \sqrt{ 4\omega _{1}^{0}-\left( {c_{1}^{0}}\right) ^{2} } }{4}\left| { x-x_{1}\left( {t}\right) }\right| } \;\mathrm {d}x\\ \le&\; C e^{ -16\theta _{2} \left( { \frac{L}{4} + 4\theta _{2}t }\right) }. \end{aligned}$$
This ends the proof. \(\square \)
Proof of Lemma 6.1
We now expand \(\mathfrak {E}\left( {u\left( {t}\right) }\right) \) one by one.
The term:
\(\displaystyle \int \left| { \partial _{x} u\left( {t}\right) }\right| ^{2}\;\mathrm {d}x.\) By (6.40) and integration by parts, we have
$$\begin{aligned} \int \left| { \partial _{x} u\left( {t}\right) }\right| ^{2}\;\mathrm {d}x=&\sum _{k=1}^{2}\int \left| { \partial _{x} R_{k}\left( {t}\right) }\right| ^{2}\;\mathrm {d}x- \sum _{k=1}^{2} 2 \mathfrak {R}\int \partial _{x}^{2}R_{k}\left( {t}\right) \; \overline{\varepsilon }\left( {t}\right) \;\mathrm {d}x+ \int \left| { \partial _{x} \varepsilon \left( {t}\right) }\right| ^{2}\;\mathrm {d}x\\&+ \mathrm {O}\left( ~{ e^{ -8\theta _{2}\left( { \frac{L}{2} + 8\theta _{2}\;t }\right) } }~\right) . \end{aligned}$$
The term:
\(\displaystyle \int \left| {u\left( {t}\right) }\right| ^{6}\;\mathrm {d}x.\) By (6.40) and Gagliardo-Nirenberg inequality, we have
$$\begin{aligned} \int \left| {u\left( {t}\right) }\right| ^{6}\;\mathrm {d}x=&\sum _{k=1}^{2}\int \left| { R_{k}\left( {t}\right) }\right| ^{6} \;\mathrm {d}x+ \sum _{k=1}^{2} \int 6 \left| { R_{k}\left( {t}\right) }\right| ^{4}\mathfrak {R}\left( { R_{k} \;\overline{ \varepsilon }}\right) \left( {t}\right) \;\mathrm {d}x\\&+ \sum _{k=1}^{2} \int 3 \left| { R_{k}\left( {t}\right) }\right| ^{4} \left| { \varepsilon \left( {t}\right) }\right| ^{2} + 12 \left| { R_{k}\left( {t}\right) }\right| ^{2} \left[ \mathfrak {R}\left( { R_{k} \;\overline{ \varepsilon }}\right) \left( {t}\right) \right] ^{2} \;\mathrm {d}x\\&+ \left\| \varepsilon \left( {t}\right) \right\| _{H^1(\mathbb {R})}^{2}\beta \left( { \left\| \varepsilon \left( {t}\right) \right\| _{H^1(\mathbb {R})} }\right) + \mathrm {O}\left( ~{ e^{ -8\theta _{2}\left( { \frac{L}{2} + 8\theta _{2}\;t }\right) } }~\right) . \end{aligned}$$
The term:
\(\displaystyle \frac{\omega _{1}\left( {0}\right) }{2}\int \left| {u\left( {t}\right) }\right| ^{2}\mathfrak {g}\left( {t}\right) \;\mathrm {d}x\) and \(\displaystyle \frac{\omega _{2}\left( {0}\right) }{2}\int \left| {u\left( {t}\right) }\right| ^{2}\mathfrak {h}\left( {t}\right) \;\mathrm {d}x.\) By (6.41) and the Cauchy-Schwarz inequality, we have
$$\begin{aligned}&\frac{\omega _{1}\left( {0}\right) }{2}\int \left| {u\left( {t}\right) }\right| ^{2}\mathfrak {g}\left( {t}\right) \;\mathrm {d}x\\&\quad = \frac{\omega _{1}\left( {0}\right) }{2} \int \left| {\sum _{k=1}^{2} R_{k}\left( {t}\right) + \varepsilon \left( {t}\right) }\right| ^{2}\mathfrak {g}\left( {t}\right) \;\mathrm {d}x\\&\quad = \frac{\omega _{1}\left( {0}\right) }{2} \int \left| { R_{1}\left( {t}\right) }\right| ^{2} - \left| { R_{1}\left( {t}\right) }\right| ^{2} \mathfrak {h}\left( {t}\right) + \left| { R_{2}\left( {t}\right) }\right| ^{2} \mathfrak {g}\left( {t}\right) + 2 \mathfrak {R}\left( { R_{1}\;\overline{ R_{2} } }\right) \left( {t}\right) \mathfrak {g}\left( {t}\right) \;\mathrm {d}x\\&\qquad + \frac{\omega _{1}\left( {0}\right) }{2} \int \mathfrak {R}\left( { R_{1}\;\overline{ \varepsilon } }\right) - \mathfrak {R}\left( { R_{1}\;\overline{ \varepsilon } }\right) \left( {t}\right) \mathfrak {h}\left( {t}\right) + \mathfrak {R}\left( { R_{2}\;\overline{ \varepsilon } }\right) \left( {t}\right) \mathfrak {g}\left( {t}\right) \;\mathrm {d}x\\&\qquad + \frac{\omega _{1}\left( {0}\right) }{2} \int \left| { \varepsilon \left( {t}\right) }\right| ^{2}\mathfrak {g}\left( {t}\right) \;\mathrm {d}x\\&\quad = \frac{\omega _{1}\left( {0}\right) }{2} \int \left| { R_{1}\left( {t}\right) }\right| ^{2} \;\mathrm {d}x+ \frac{\omega _{1}\left( {0}\right) }{2} \int \mathfrak {R}\left( { R_{1}\;\overline{ \varepsilon } }\right) \;\mathrm {d}x+ \frac{\omega _{1}\left( {0}\right) }{2} \int \left| { \varepsilon \left( {t}\right) }\right| ^{2}\mathfrak {g}\left( {t}\right) \;\mathrm {d}x\\&\qquad + \mathrm {O}\left( ~{ \int \left| { R_{1}\left( {t}\right) }\right| ^{2} \mathfrak {h}\left( {t}\right) + \left| { R_{2}\left( {t}\right) }\right| ^{2} \mathfrak {g}\left( {t}\right) + \left| { R_{1}\left( {t}\right) }\right| \;\left| { R_{2}\left( {t}\right) }\right| \mathfrak {g}\left( {t}\right) \;\mathrm {d}x}~\right) \\&\qquad + \mathrm {O}\left( ~{ \int \left| { R_{1}\left( {t}\right) }\right| \;\left| { \varepsilon \left( {t}\right) }\right| \mathfrak {h}\left( {t}\right) + \left| { R_{2}\left( {t}\right) }\right| \;\left| { \varepsilon \left( {t}\right) }\right| \mathfrak {g}\left( {t}\right) \;\mathrm {d}x}~\right) \\&\quad = \frac{\omega _{1}\left( {0}\right) }{2} \int \left| { R_{1}\left( {t}\right) }\right| ^{2} \;\mathrm {d}x+ \frac{\omega _{1}\left( {0}\right) }{2} \int \mathfrak {R}\left( { R_{1}\;\overline{ \varepsilon } }\right) \;\mathrm {d}x+ \frac{\omega _{1}\left( {t}\right) }{2} \int \left| { \varepsilon \left( {t}\right) }\right| ^{2}\mathfrak {g}\left( {t}\right) \;\mathrm {d}x\\&\qquad + \frac{ \omega _{1}\left( {0}\right) - \omega _{1}\left( {t}\right) }{ 2 } \int \left| { \varepsilon \left( {t}\right) }\right| ^{2}\mathfrak {g}\left( {t}\right) \;\mathrm {d}x\\&\qquad + \mathrm {O}\left( ~{ \int \left| { R_{1}\left( {t}\right) }\right| ^{2} \mathfrak {h}\left( {t}\right) + \left| { R_{2}\left( {t}\right) }\right| ^{2} \mathfrak {g}\left( {t}\right) + \left| { R_{1}\left( {t}\right) }\right| \;\left| { R_{2}\left( {t}\right) }\right| \mathfrak {g}\left( {t}\right) \;\mathrm {d}x}~\right) \\&\qquad + \mathrm {O}\left( ~{ \int \left| { R_{1}\left( {t}\right) }\right| \;\left| { \varepsilon \left( {t}\right) }\right| \mathfrak {h}\left( {t}\right) + \left| { R_{2}\left( {t}\right) }\right| \;\left| { \varepsilon \left( {t}\right) }\right| \mathfrak {g}\left( {t}\right) \;\mathrm {d}x}~\right) \\&\quad = \frac{\omega _{1}\left( {0}\right) }{2} \int \left| { R_{1}\left( {t}\right) }\right| ^{2} \;\mathrm {d}x+ \frac{\omega _{1}\left( {0}\right) }{2} \int \mathfrak {R}\left( { R_{1}\;\overline{ \varepsilon } }\right) \;\mathrm {d}x+ \frac{\omega _{1}\left( {t}\right) }{2} \int \left| { \varepsilon \left( {t}\right) }\right| ^{2}\mathfrak {g}\left( {t}\right) \;\mathrm {d}x\\&\qquad + \left\| \varepsilon \left( {t}\right) \right\| _{H^1(\mathbb {R})}^{2} \beta \left( { \left\| \varepsilon \left( {t}\right) \right\| _{H^1(\mathbb {R})} }\right) + \mathrm {O}\left( ~{ \left| { \omega _{1}\left( {t}\right) - \omega _{1}\left( {0}\right) }\right| ^{2} }~\right) + \mathrm {O}\left( ~{ e^{ -8\theta _{2} \left( { \frac{L}{2} + 8\theta _{2}t }\right) } }~\right) , \end{aligned}$$
and
$$\begin{aligned}&\frac{\omega _{2}\left( {0}\right) }{2}\int \left| {u\left( {t}\right) }\right| ^{2}\mathfrak {h}\left( {t}\right) \;\mathrm {d}x\\&\quad = \frac{\omega _{2}\left( {0}\right) }{2} \int \left| { R_{2}\left( {t}\right) }\right| ^{2} \;\mathrm {d}x+ \frac{\omega _{2}\left( {0}\right) }{2} \int \mathfrak {R}\left( { R_{2}\;\overline{ \varepsilon } }\right) \;\mathrm {d}x+ \frac{\omega _{2}\left( {t}\right) }{2} \int \left| { \varepsilon \left( {t}\right) }\right| ^{2}\mathfrak {g}\left( {t}\right) \;\mathrm {d}x\\&\qquad + \left\| \varepsilon \left( {t}\right) \right\| _{H^1(\mathbb {R})}^{2} \beta \left( { \left\| \varepsilon \left( {t}\right) \right\| _{H^1(\mathbb {R})} }\right) + \mathrm {O}\left( ~{ \left| { \omega _{2}\left( {t}\right) - \omega _{2}\left( {0}\right) }\right| ^{2} }~\right) + \mathrm {O}\left( ~{ e^{ -8\theta _{2} \left( { \frac{L}{2} + 8\theta _{2}t }\right) } }~\right) . \end{aligned}$$
The term
\(\displaystyle - \frac{c_{1}\left( {0}\right) }{2}\; \mathfrak {I}\int \left( { \overline{u}\partial _{x}u }\right) \left( {t}\right) \mathfrak {g}\left( {t}\right) \;\mathrm {d}x\)
and
\(\displaystyle - \frac{c_{2}\left( {0}\right) }{2}\; \mathfrak {I}\int \left( { \overline{u}\partial _{x}u }\right) \left( {t}\right) \mathfrak {h}\left( {t}\right) \;\mathrm {d}x.\) Similarly, we have
$$\begin{aligned}&- \frac{c_{1}\left( {0}\right) }{2}\; \mathfrak {I}\int \left( { \overline{u}\partial _{x}u }\right) \left( {t}\right) \mathfrak {g}\left( {t}\right) \;\mathrm {d}x\\&\quad = - \frac{c_{1}\left( {0}\right) }{2}\; \mathfrak {I}\int \overline{R}_{1}\; \partial _{x} R_{1}\left( {t}\right) \;\mathrm {d}x- c_{1}\left( {0}\right) \; \mathfrak {I}\int \partial _{x} R_{1}\left( {t}\right) \;\overline{\varepsilon } \;\mathrm {d}x\\&\qquad - \frac{c_{1}\left( {t}\right) }{2}\; \mathfrak {I}\int \overline{\varepsilon }\; \partial _{x} \varepsilon \left( {t}\right) \mathfrak {g}\left( {t}\right) \;\mathrm {d}x\\&\qquad + \left\| \varepsilon \left( {t}\right) \right\| _{H^1(\mathbb {R})}^{2} \beta \left( { \left\| \varepsilon \left( {t}\right) \right\| _{H^1(\mathbb {R})} }\right) + \mathrm {O}\left( ~{ \left| { c_{1}\left( {t}\right) - c_{1}\left( {0}\right) }\right| ^{2} }~\right) + \mathrm {O}\left( ~{ e^{ -8\theta _{2} \left( { \frac{L}{2} + 8\theta _{2}t }\right) } }~\right) , \end{aligned}$$
and
$$\begin{aligned}&- \frac{c_{2}\left( {0}\right) }{2}\; \mathfrak {I}\int \left( { \overline{u}\partial _{x}u }\right) \left( {t}\right) \left( {t}\right) \mathfrak {h}\left( {t}\right) \;\mathrm {d}x\\&\quad = - \frac{c_{2}\left( {0}\right) }{2}\; \mathfrak {I}\int \overline{R}_{2}\; \partial _{x} R_{2}\left( {t}\right) \;\mathrm {d}x- c_{2}\left( {0}\right) \; \mathfrak {I}\int \partial _{x} R_{2}\left( {t}\right) \;\overline{\varepsilon } \;\mathrm {d}x\\&\qquad - \frac{c_{2}\left( {t}\right) }{2}\; \mathfrak {I}\int \overline{\varepsilon }\; \partial _{x} \varepsilon \left( {t}\right) \mathfrak {h}\left( {t}\right) \;\mathrm {d}x\\&\qquad + \left\| \varepsilon \left( {t}\right) \right\| _{H^1(\mathbb {R})}^{2} \beta \left( { \left\| \varepsilon \left( {t}\right) \right\| _{H^1(\mathbb {R})} }\right) + \mathrm {O}\left( ~{ \left| { c_{2}\left( {t}\right) - c_{2}\left( {0}\right) }\right| ^{2} }~\right) + \mathrm {O}\left( ~{ e^{ -8\theta _{2} \left( { \frac{L}{2} + 8\theta _{2}t }\right) } }~\right) . \end{aligned}$$
The term
\(\displaystyle \frac{c_{1}\left( {0}\right) }{8}\int \left| {u\left( {t}\right) }\right| ^{4}\mathfrak {g}\left( {t}\right) \;\mathrm {d}x\)
and
\(\displaystyle \frac{c_{2}\left( {0}\right) }{8}\int \left| {u\left( {t}\right) }\right| ^{4}\mathfrak {h}\left( {t}\right) \;\mathrm {d}x.\)
$$\begin{aligned}&\frac{c_{1}\left( {0}\right) }{8}\int \left| {u\left( {t}\right) }\right| ^{4}\mathfrak {g}\left( {t}\right) \;\mathrm {d}x\\&\quad = \frac{c_{1}\left( {0}\right) }{8} \int \left| {R_{1}\left( {t}\right) }\right| ^{4} \;\mathrm {d}x+ \frac{c_{1}\left( {0}\right) }{4} \int \left| {R_{1}\left( {t}\right) }\right| ^{2}\mathfrak {R}\left( {R_{1}\;\overline{\varepsilon }}\right) \left( {t}\right) \;\mathrm {d}x\\&\qquad + \frac{c_{1}\left( {t}\right) }{8} \int 2 \left| {R_{1}\left( {t}\right) }\right| ^{2}\left| {\varepsilon \left( {t}\right) }\right| ^{2} + \left[ \mathfrak {R}\left( {R_{1}\;\overline{\varepsilon }}\right) \left( {t}\right) \right] ^{2} \;\mathrm {d}x\\&\qquad + \left\| \varepsilon \left( {t}\right) \right\| _{H^1(\mathbb {R})}^{2} \beta \left( { \left\| \varepsilon \left( {t}\right) \right\| _{H^1(\mathbb {R})} }\right) + \mathrm {O}\left( ~{ \left| { c_{1}\left( {t}\right) - c_{1}\left( {0}\right) }\right| ^{2} }~\right) + \mathrm {O}\left( ~{ e^{ -8\theta _{2} \left( { \frac{L}{2} + 8\theta _{2}t }\right) } }~\right) , \end{aligned}$$
and
$$\begin{aligned}&\frac{c_{2}\left( {0}\right) }{8}\int \left| {u\left( {t}\right) }\right| ^{4}\mathfrak {h}\left( {t}\right) \;\mathrm {d}x\\&\quad = \frac{c_{2}\left( {0}\right) }{8} \int \left| {R_{2}\left( {t}\right) }\right| ^{4} \;\mathrm {d}x+ \frac{c_{2}\left( {0}\right) }{4} \int \left| {R_{2}\left( {t}\right) }\right| ^{2}\mathfrak {R}\left( {R_{2}\;\overline{\varepsilon }}\right) \left( {t}\right) \;\mathrm {d}x\\&\qquad + \frac{c_{2}\left( {t}\right) }{8} \int 2 \left| {R_{2}\left( {t}\right) }\right| ^{2}\left| {\varepsilon \left( {t}\right) }\right| ^{2} + \left[ \mathfrak {R}\left( {R_{2}\;\overline{\varepsilon }}\right) \left( {t}\right) \right] ^{2} \;\mathrm {d}x\\&\qquad + \left\| \varepsilon \left( {t}\right) \right\| _{H^1(\mathbb {R})}^{2} \beta \left( { \left\| \varepsilon \left( {t}\right) \right\| _{H^1(\mathbb {R})} }\right) + \mathrm {O}\left( ~{ \left| { c_{2}\left( {t}\right) - c_{2}\left( {0}\right) }\right| ^{2} }~\right) + \mathrm {O}\left( ~{ e^{ -8\theta _{2} \left( { \frac{L}{2} + 8\theta _{2}t }\right) } }~\right) . \end{aligned}$$
Summing up the above terms, we can conclude the proof by (3.9) for \(k=1, 2\) and the orthogonal conditions (4.11). \(\square \)
Appendix 3: The coercivity of the localized quadratic term
Let L be large enough, \(x_{1}\) and \(x_{2}\in \mathbb {R}\) with \(x_{2}-x_{1}>\frac{L}{2}.\) Now, define
$$\begin{aligned} g\left( {x}\right) = \left\{ \begin{array}{ll} 1, &{} x\le x_{1}+\frac{L}{8}, \\ 0<\cdot<1, &{} x_{1}+\frac{L}{8}< x < x_{2}-\frac{L}{8}, \\ 0, &{} x\ge x_{2}-\frac{L}{8}, \end{array} \right. \quad \text {and}\; h\left( {x}\right) = 1 - g\left( {x}\right) . \end{aligned}$$
(6.44)
In order to prove Lemma 6.2, it suffices to show the following result.
Lemma 6.4
Let \(L>1\) be large enough, g, h be given by (6.44). Then there exists \(C_1>0\) such that
$$\begin{aligned} \mathcal {H}_{2}\left( {~\varepsilon ~,~\varepsilon ~}\right) \ge C_1 \left\| \varepsilon \right\| _{H^1(\mathbb {R})}, \end{aligned}$$
where
$$\begin{aligned} \mathcal {H}_{2}\left( {~\varepsilon ~,~\varepsilon ~}\right) =\,&\frac{1}{2}\int \left| {\varepsilon _{x}}\right| ^{2} - \frac{1}{32} \left( 3 \int \left| {R_{1}}\right| ^{4}\left| {\varepsilon }\right| ^{2} + 12 \int \left| {R_{1}}\right| ^{2}\left[ \mathfrak {R}\left( { \overline{R}_{1}\varepsilon }\right) \right] ^{2} \right) \\&- \frac{1}{32} \left( 3 \int \left| {R_{2}}\right| ^{4}\left| {\varepsilon }\right| ^{2} + 12 \int \left| {R_{2}}\right| ^{2}\left[ \mathfrak {R}\left( { \overline{R}_{2}\varepsilon }\right) \right] ^{2} \right) \\&+ \frac{ \omega _{1} }{ 2 }\int \left| { \varepsilon }\right| ^{2}g - \frac{c_{1}}{2}\mathfrak {I}\int \overline{\varepsilon }\varepsilon _{x}g + \frac{c_{1}}{8} \left( { 2 \int \left| {R_{1}}\right| ^{2}\left| {\varepsilon }\right| ^{2} + 4 \int \left[ \mathfrak {R}\left( { \overline{R}_{1}\varepsilon }\right) \right] ^{2} }\right) \\&+ \frac{ \omega _{2} }{ 2 }\int \left| { \varepsilon }\right| ^{2}h - \frac{c_{2}}{2}\mathfrak {I}\int \overline{\varepsilon }\varepsilon _{x}h + \frac{c_{2}}{8} \left( { 2 \int \left| {R_{2}}\right| ^{2}\left| {\varepsilon }\right| ^{2} + 4 \int \left[ \mathfrak {R}\left( { \overline{R}_{2}\varepsilon }\right) \right] ^{2} }\right) , \end{aligned}$$
with \(R_{k}\left( {x}\right) =\varphi _{\omega _{k},c_{k}}\left( {x - x_{k}}\right) e^{i\gamma _{k}}\) \(\left( { k=1,\,2}\right) .\)
First, we give a localized version of the ‘single solitary’ coercive result. For the convenience of notation, we denote
$$\begin{aligned} R\left( {x}\right) = \varphi _{\omega ,c}\left( {x-y_0}\right) e^{i\gamma }, \end{aligned}$$
with \(4\omega >c^{2},\)
\(y_{0},\,\theta \in \mathbb {R}.\) Let \(\Phi \,:\,\mathbb {R}\mapsto \mathbb {R}\) be an even \(C^{2}\) function with
$$\begin{aligned} \Phi \left( {x}\right) = \left\{ \begin{array}{ll} 1, &{} \left| {x}\right| \le 1, \\ e^{-|x|}\le \cdot \le 3 e^{-|x|}, &{} 1<\left| {x}\right| <2, \\ e^{-|x|}, &{} \left| {x}\right| \ge 2, \end{array} \right. \end{aligned}$$
and \(\Phi '\left( {x}\right) \le 0\) for \(x>0.\)
Lemma 6.5
Let \(B>1\) be large enough. If \(\varepsilon \in H^{1}\left( {\mathbb {R}}\right) \) satisfies the following orthogonality condition,
$$\begin{aligned}&\mathfrak {R}\int R(x)\; \overline{\varepsilon (x)} \;\mathrm {d}x=0, \quad \mathfrak {R}\int \left( {i\partial _x R + \frac{1}{2} \left| {R}\right| ^2R}\right) (x)\; \overline{\varepsilon (x)} \;\mathrm {d}x=0,\\&\mathfrak {R}\int \partial _x R(x)\; \overline{\varepsilon (x)} \;\mathrm {d}x=0, \quad \mathfrak {R}\int iR(x)\; \overline{ \varepsilon (x)} \;\mathrm {d}x=0. \end{aligned}$$
Then, we have
$$\begin{aligned} \mathcal {H}_{ B, y_{0} }\left( \,{ \varepsilon }\, ,\,{ \varepsilon }\,\right) \ge \frac{C_{0}}{4}\int \left( { \left| {\varepsilon _{x}}\right| ^{2} + \left| {\varepsilon }\right| ^{2} }\right) \Phi _{B,y_{0}}\, \mathrm {d}x, \end{aligned}$$
where
$$\begin{aligned} \mathcal {H}_{ B, y_{0} }\left( \,{ \varepsilon }\, ,\,{ \varepsilon }\,\right) =&\frac{1}{2}\int \left| {\varepsilon _{x}}\right| ^{2}\Phi _{B,y_{0}}\,\mathrm {d}x+ \frac{\omega }{2}\int \left| {\varepsilon }\right| ^{2}\Phi _{B,y_{0}}\,\mathrm {d}x- \frac{c}{2}\mathfrak {I}\int \overline{\varepsilon }\varepsilon _{x}\Phi _{B,y_{0}}\,\mathrm {d}x\\&+ \frac{c}{8} \left( { 2 \int \left| { R }\right| ^{2}\left| {\varepsilon }\right| ^{2}\Phi _{B,y_{0}}\,\mathrm {d}x+ 4 \int \left[ \mathfrak {R}\left( {\overline{R}\varepsilon }\right) \right] ^{2}\Phi _{B,y_{0}}\,\mathrm {d}x}\right) \\&- \frac{1}{32} \left( { 3 \int \left| { R }\right| ^{4}\left| {\varepsilon }\right| ^{2}\Phi _{B,y_{0}}\,\mathrm {d}x+ 12 \int \left| { R }\right| ^{2}\left[ \mathfrak {R}\left( {\overline{R}\varepsilon }\right) \right] ^{2}\Phi _{B,y_{0}}\,\mathrm {d}x}\right) . \end{aligned}$$
Proof
By setting \(\zeta \left( {x}\right) =\sqrt{\Phi _{B,y_{0}}\left( {x}\right) }\varepsilon \left( {x}\right) ,\) we have
$$\begin{aligned}&\left| {\varepsilon _{x}}\right| ^{2}\Phi _{B,y_{0}} = \left| {\zeta _{x}}\right| ^{2} - \frac{ \Phi _{B,y_{0}}' }{ \Phi _{B,y_{0}} }\mathfrak {R}\left( { \overline{\zeta }\zeta _{x} }\right) + \frac{1}{4} \left( { \frac{ \Phi _{B,y_{0}}' }{ \Phi _{B,y_{0}} } }\right) ^{2}\left| {\zeta }\right| ^{2}.\\&\mathfrak {I}\left( { \overline{\varepsilon }\varepsilon _{x} }\right) \Phi _{B,y_{0}} = \mathfrak {I}\left( { \overline{\zeta }\zeta _{x} }\right) , \quad \text { and }\quad \left| {\varepsilon }\right| ^{2}\Phi _{B,y_{0}} = \left| {\zeta }\right| ^{2}. \end{aligned}$$
Now, we rewrite the quadratic form \(\mathcal {H}_{ B, y_{0} }\) as a quadratic form with respect to \(\zeta ,\) which means
$$\begin{aligned} \mathcal {H}_{ B, y_{0} }\left( \,{\varepsilon }\, ,\,{\varepsilon }\,\right) =&\mathcal {H}_{\omega ,c}\left( \,{ \zeta }\, ,\,{ \zeta }\,\right) - \frac{1}{2}\mathfrak {R}\int \frac{ \Phi _{B,y_{0}}' }{ \Phi _{B,y_{0}} }\overline{\zeta }\zeta _{x}\,\mathrm {d}x- \frac{1}{4} \int \left( { \frac{ \Phi _{B,y_{0}}' }{ \Phi _{B,y_{0}} } }\right) ^{2}\left| {\zeta }\right| ^{2}\,\mathrm {d}x. \end{aligned}$$
On the one hand, as a consequence of Lemma 3.2, we obtain
$$\begin{aligned} \mathcal {H}_{\omega ,c}\left( \,{ \zeta }\, ,\,{ \zeta }\,\right)\ge & {} \frac{C_{0}}{2}\left\| \zeta \right\| _{H^1(\mathbb {R})}^{2} \\&- \frac{ 2 }{ C_{0} } \left[ \left( { \mathfrak {R}\int R\; \overline{\zeta } \,\mathrm {d}x}\right) ^{2} + \left( { \mathfrak {R}\int \left( {i\partial _x R + \frac{1}{2} \left| {R}\right| ^2R}\right) \; \overline{\zeta } \,\mathrm {d}x}\right) ^{2} \right] \\&- \frac{ 2 }{C_{0} } \left[ \left( { \mathfrak {R}\int \partial _x R\; \overline{\zeta } \,\mathrm {d}x}\right) ^{2} + \left( { \mathfrak {R}\int iR\; \overline{ \zeta } \,\mathrm {d}x}\right) ^{2} \right] . \end{aligned}$$
On the other hand, a straight calculation implies that
$$\begin{aligned} \left| { \mathfrak {R}\int R\; \overline{\zeta } \,\mathrm {d}x}\right|&= \left| { \mathfrak {R}\int R\; \overline{\varepsilon }\left( { 1-\sqrt{\Phi _{B,y_{0}}} }\right) \,\mathrm {d}x}\right| \\&= \left| { \mathfrak {R}\int _{ \left| { x-y_{0} }\right|>B }R\; \overline{\varepsilon }\left( { 1-\sqrt{\Phi _{B,y_{0}}} }\right) \,\mathrm {d}x}\right| \\&\le \Vert \varepsilon \Vert _{2}\left( { \int _{ \left| { x-y_{0} }\right| >B }\left| {R}\right| ^{2} \,\mathrm {d}x}\right) ^{\frac{1}{2}}\\&\le \frac{ C }{ e^{ \sqrt{ 4\omega -c^{2} }\frac{B}{2} } }\Vert \varepsilon \Vert _{2}, \end{aligned}$$
moreover, applying the similar argument to \(\mathfrak {R}\int \left( {i\partial _x R + \frac{1}{2} \left| {R}\right| ^2R}\right) \; \overline{\zeta }, \)
\(\mathfrak {R}\int \partial _x R\; \overline{\zeta }\) and \(\mathfrak {R}\int i R\;\overline{ \zeta }\) gives us that
$$\begin{aligned} \mathcal {H}_{\omega ,c}\left( \,{ \zeta }\, ,\,{ \zeta }\,\right) \ge \frac{C_{0}}{2}\left\| \zeta \right\| _{H^1(\mathbb {R})}^{2} - \frac{ C }{ e^{ \sqrt{ 4\omega -c^{2} }B } }\Vert \zeta \Vert _{2}^{2}. \end{aligned}$$
Now it follows from \(\left| { \Phi _{B,y_{0}}' }\right| \le \frac{C }{B}\Phi _{B,y_{0}}\) that, for \(B>1\) large enough,
$$\begin{aligned} \mathcal {H}_{ B,y_{0} }\left( \,{\varepsilon }\, ,\,{\varepsilon }\,\right) \ge \frac{C_{0}}{2}\left\| \zeta \right\| _{H^1(\mathbb {R})}^{2} - \frac{ C }{ e^{ \sqrt{ 4\omega -c^{2} }B } }\Vert \zeta \Vert _{2}^{2} - \frac{C}{B^{2}} \left\| \zeta \right\| _{H^1(\mathbb {R})}^{2} \ge \frac{3C_{0}}{8}\left\| \zeta \right\| _{H^1(\mathbb {R})}^{2}. \end{aligned}$$
Since
$$\begin{aligned} \left| {\zeta _{x}}\right| ^{2} =&\left| {\varepsilon _{x}}\right| ^{2}\Phi _{B,y_{0}} + \frac{ \Phi _{B,y_{0}}' }{ \Phi _{B,y_{0}} }\mathfrak {R}\left( { \overline{\varepsilon }\varepsilon _{x} }\right) \Phi _{B,y_{0}} + \frac{1}{4} \left( { \frac{ \Phi _{B,y_{0}}' }{ \Phi _{B,y_{0}} } }\right) ^{2} \left| {\varepsilon }\right| ^{2}\Phi _{B,y_{0}}\\ \ge&\left( { 1-\frac{C}{B^{2}} }\right) \left| {\varepsilon _{x}}\right| ^{2}\Phi _{B,y_{0}} - \frac{C}{B^{2}}\left| {\varepsilon }\right| ^{2}\Phi _{B,y_{0}} \end{aligned}$$
we obtain, for B large enough,
$$\begin{aligned} \mathcal {H}_{ B, y_{0} }\left( \,{\varepsilon }\, ,\,{\varepsilon }\,\right) \ge \frac{C_{0}}{4}\int \left( { \left| {\varepsilon _{x}}\right| ^{2} + \left| {\varepsilon }\right| ^{2} }\right) \Phi _{B,y_{0}} \,\mathrm {d}x. \end{aligned}$$
This ends the proof. \(\square \)
Proof of Lemma 6.4
Since \(L>1\) is sufficiently large enough, we can take \(B\in \left( {1,~\frac{L}{4}~}\right) \) such that Claim 6.5 holds.
$$\begin{aligned}&\mathcal {H}_{2}\left( {~\varepsilon ~,~\varepsilon ~}\right) \\&\quad = \mathcal {H}_{ B,x_{1} }\left( \,{ \varepsilon }\, ,\,{ \varepsilon }\,\right) + \mathcal {H}_{ B,x_{2} }\left( \,{ \varepsilon }\, ,\,{ \varepsilon }\,\right) \\&\qquad + \frac{1}{2} \int \left[ \left| {\varepsilon _{x}}\right| ^{2} + \omega _{1}\left| {\varepsilon }\right| - c_{1}\mathfrak {I}\left( { \overline{\varepsilon }\varepsilon _{x} }\right) \right] \left( { g - \Phi _{B,x_{1}} }\right) \,\mathrm {d}x\\&\qquad + \frac{1}{2} \int \left[ \left| {\varepsilon _{x}}\right| ^{2} + \omega _{2}\left| {\varepsilon }\right| - c_{2}\mathfrak {I}\left( { \overline{\varepsilon }\varepsilon _{x} }\right) \right] \left( { h - \Phi _{B,x_{2}} }\right) \,\mathrm {d}x\\&\qquad + \frac{c_{1}}{8} \left( { 2 \int \left| {R_{1}}\right| ^{2}\left| {\varepsilon }\right| ^{2} \left( { 1 - \Phi _{B,x_{1}} }\right) \,\mathrm {d}x+ 4 \int \left[ \mathfrak {R}\left( { \overline{R}_{2}\varepsilon }\right) \right] ^{2} \left( { 1 - \Phi _{B,x_{1}} }\right) \,\mathrm {d}x}\right) \\&\qquad + \frac{c_{2}}{8} \left( { 2 \int \left| {R_{2}}\right| ^{2}\left| {\varepsilon }\right| ^{2} \left( { 1 - \Phi _{B,x_{2}} }\right) \,\mathrm {d}x+ 4 \int \left[ \mathfrak {R}\left( { \overline{R}_{1}\varepsilon }\right) \right] ^{2} \left( { 1 - \Phi _{B,x_{2}} }\right) \,\mathrm {d}x}\right) \\&\qquad - \frac{1}{32} \left( 3 \int \left| {R_{1}}\right| ^{4}\left| {\varepsilon }\right| ^{2} \left( { 1 - \Phi _{B,x_{1}} }\right) \,\mathrm {d}x+ 12 \int \left| {R_{1}}\right| ^{2}\left[ \mathfrak {R}\left( { \overline{R}_{1}\varepsilon }\right) \right] ^{2} \left( { 1 - \Phi _{B,x_{1}}}\right) \,\mathrm {d}x\right) \\&\qquad - \frac{1}{32} \left( 3 \int \left| {R_{2}}\right| ^{4}\left| {\varepsilon }\right| ^{2} \left( { 1 - \Phi _{B,x_{2}} }\right) \,\mathrm {d}x+ 12 \int \left| {R_{2}}\right| ^{2}\left[ \mathfrak {R}\left( { \overline{R}_{2}\varepsilon }\right) \right] ^{2} \left( { 1 - \Phi _{B,x_{2}} }\right) \,\mathrm {d}x\right) . \end{aligned}$$
It follows from a direct computation that
$$\begin{aligned} g\left( {x}\right) -\Phi _{B,x_{1}}\left( {x}\right) \left\{ \begin{array}{ll} = 0, &{} \left| { x-x_{1} }\right|<\frac{L}{8}, \\ \ge -e^{-\frac{L}{8B}}, &{} \text { else }, \end{array} \right. \\ h\left( {x}\right) -\Phi _{B,x_{2}}\left( {x}\right) \left\{ \begin{array}{ll} = 0, &{} \left| { x-x_{2} }\right|<\frac{L}{8}, \\ \ge -e^{-\frac{L}{8B}}, &{} \text { else }, \end{array} \right. \\ 1-\Phi _{B,x_{1}}\left( {x}\right) \left\{ \begin{array}{ll} = 0, &{} \left| { x-x_{1} }\right| <\frac{L}{8}, \\ \ge -e^{-\frac{L}{8B}}, &{} \text { else }, \end{array} \right. \end{aligned}$$
and
$$\begin{aligned} 1-\Phi _{B,x_{2}}\left( {x}\right) \left\{ \begin{array}{ll} = 0, &{} \left| { x-x_{2} }\right| <\frac{L}{8}, \\ \ge -e^{-\frac{L}{8B}}, &{} \text { else }. \end{array} \right. \end{aligned}$$
Moreover, since, for \(k=1,2,\)
\(c_{k}^{2}<4\omega _{k},\) there exists \(\delta _{k}>0\) such that
$$\begin{aligned} \left| {\varepsilon _{x}}\right| ^{2} + \omega _{k}\left| {\varepsilon }\right| ^{2} - c_{k}\mathfrak {I}\left( { \overline{\varepsilon }\varepsilon _{x} }\right) \ge \delta _{k}\left( { \left| {\varepsilon _{x}}\right| ^{2} + \left| {\varepsilon }\right| ^{2} }\right) . \end{aligned}$$
Thus, taking L large enough, we obtain
$$\begin{aligned} \mathcal {H}_{2}\left( {~\varepsilon ~,~\varepsilon ~}\right)&\ge \frac{C_{0}}{4} \int \left( { \left| {\varepsilon _{x}}\right| ^{2} + \left| {\varepsilon }\right| ^{2} }\right) \Phi _{B,x_{1}}\,\mathrm {d}x+ \frac{C_{0}}{4} \int \left( { \left| {\varepsilon _{x}}\right| ^{2} + \left| {\varepsilon }\right| ^{2} }\right) \Phi _{B,x_{2}}\,\mathrm {d}x\\&\quad + \delta _{1} \int \left( { \left| {\varepsilon _{x}}\right| ^{2} + \left| {\varepsilon }\right| ^{2} }\right) \left( { g - \Phi _{B,x_{1}} }\right) \,\mathrm {d}x+ \delta _{2} \int \left( { \left| {\varepsilon _{x}}\right| ^{2} + \left| {\varepsilon }\right| ^{2} }\right) \left( { h - \Phi _{B,x_{2}} }\right) \,\mathrm {d}x\\&\quad - C e^{-\frac{L}{4B}} \int \left( { \left| {\varepsilon _{x}}\right| ^{2} + \left| {\varepsilon }\right| ^{2} }\right) \,\mathrm {d}x- C e^{ -\sqrt{ 4\omega _{1}-c_{1}^{2} }\frac{L}{4} }\int \left| {\varepsilon }\right| ^{2}\,\mathrm {d}x\\&\quad - C e^{ -\sqrt{ 4\omega _{2}-c_{2}^{2} }\frac{L}{4} }\int \left| {\varepsilon }\right| ^{2}\,\mathrm {d}x\\&\ge C_1 \left\| \varepsilon \right\| _{H^1(\mathbb {R})}^{2}, \end{aligned}$$
where \(C_1 = \frac{1}{2}\min ~\left\{ {\frac{C_{0}}{4},\delta _{1},\delta _{2}}\right\} .\) This concludes the proof. \(\square \)
