$U_{n}(t)\in\mathscr{F}^{+}_{r_{0}^{+}/2}$, so we can define the sequence
$(\mathbf{v}^{+}_{n}(t))_{n\in{\mathbb{N}}}$ in $\mathscr{F}^{+}$ as
$\mathbf{v}^{+}_{n}(t):=P^{+}(U_{n}(t))$. We show that such a sequence is
bounded. Indeed, we have

$\forall n\in{\mathbb{N}},\hskip
5.69054pt\lVert\mathbf{v}_{n}^{+}(t)\rVert_{\mathscr{L}}\leq\frac{r_{0}^{+}}{2}+\lVert
U_{n}(t)\rVert_{\mathscr{L}}$ (5.53)

and $(U_{n}(t))_{\mathscr{L}}$ converges weakly in $\mathscr{L}$, so in
particular it is bounded. Therefore, up to an extraction we can assume that
$\mathbf{v}_{n}^{+}(t)\rightharpoonup\mathbf{v}^{+}_{\infty}(t)\in\mathscr{L}$
and by (H3’) we have $\mathbf{v}^{+}_{\infty}(t)\in\mathscr{F}^{\pm}$. Using
now the convergence properties we get the inequality

$\lVert
U_{\infty}(t)-\mathbf{v}^{+}_{\infty}(t)\rVert_{\mathscr{L}}\leq\liminf_{n\to\infty}\lVert
U_{n}(t)-\mathbf{v}_{n}^{+}(t)\rVert_{\mathscr{L}}\leq\frac{r_{0}^{+}}{2},$
(5.54)

so that $U_{n}(t)\in\mathscr{F}^{+}_{r_{0}^{+}/2}$. An identical argument
shows that for all $t\in(-\infty,-T]$ we have
$U_{n}(t)\in\mathscr{F}^{-}_{r_{0}^{-}/2}$. Therefore, we have shown that
$U_{\infty}\in X_{T}$.

  * •

Next, we prove (5.41). We have

$\sup_{n\in{\mathbb{N}}}\int_{\mathbb{R}}\mathcal{E}(U_{n}(t))e^{c_{n}t}dt\leq
M-\sup_{n\in{\mathbb{N}}}\int_{\mathbb{R}}\frac{\lVert
U_{n}^{\prime}(t)\rVert_{\mathscr{L}}^{2}}{2}e^{ct}dt<+\infty,$ (5.55)

by (5.45). Hence, we can apply Fatou’s Lemma to
$(t\in{\mathbb{R}}\to\mathcal{E}(U_{n}(t))e^{c_{n}t})_{n\in{\mathbb{N}}}$ (a
sequence of functions uniformly bounded below by $a$) to show

$\int_{\mathbb{R}}\liminf_{n\to+\infty}\mathcal{E}(U_{n}(t))e^{c_{n}t}dt\leq\liminf_{n\to\infty}\int_{\mathbb{R}}\mathcal{E}(U_{n}(t))e^{c_{n}t}dt,$
(5.56)

which, combined with (5.52) implies

$\int_{\mathbb{R}}\mathcal{E}(U_{\infty}(t))e^{ct}dt\leq\liminf_{n\to\infty}\int_{\mathbb{R}}\mathcal{E}(U_{n}(t))e^{c_{n}t}dt.$
(5.57)

Combining (5.48) and (5.57) we get

$\mathbf{E}_{c}(U_{\infty})\leq\liminf_{n\to\infty}\int_{\mathbb{R}}\frac{\lVert
U^{\prime}_{n}(t)\rVert_{\mathscr{L}}^{2}}{2}e^{c_{n}t}dt+\liminf_{n\to\infty}\int_{\mathbb{R}}\mathcal{E}(U_{n}(t))e^{c_{n}t}dt,$
(5.58)

which, by superadditivity of the limit inferior gives (5.41).

∎

###  5.5 Existence of an infimum for $\mathbf{E}_{c}$ in $X_{T}$

The goal now is to show that, for $T\geq 1$ and $c>0$ fixed, the infimum
$\mathbf{m}_{c,T}$ as defined in (5.15) is attained by a function in $X_{T}$.
This will actually follow easily from Lemma 5.6.

######  Lemma 5.7.

Assume that (H1’), (H2’), (H3’) and (H4’) hold. Let $c>0$, $T\geq 1$ and
$\mathbf{m}_{c,T}$ be as in (5.15). Then, $\mathbf{m}_{c,T}$ is attained for
some $\mathbf{U}_{c,T}\in X_{T}$.

###### Proof.

By (5.18) in Lemma 5.4, we have that there exists a minimizing sequence
$(U_{n})_{n\in{\mathbb{N}}}$ for $\mathbf{E}_{c}$ in $X_{T}$. We apply Lemma
5.6 to $(U_{n})_{n\in{\mathbb{N}}}$ and the sequence of speeds constantly
equal to $c$. We obtain a function $\mathbf{U}_{c,T}\in X_{T}$ such that

$\mathbf{E}_{c}(\mathbf{U}_{c,T})\leq\liminf_{n\to\infty}\mathbf{E}_{c}(U_{n})=\mathbf{m}_{c,T},$
(5.59)

due to (5.41). Therefore, $\mathbf{m}_{c,T}$ is attained by $\mathbf{U}_{c,T}$
in $X_{T}$. ∎

Subsequently, we show that assumption (H6’) implies that the constrained
minimizers are solutions of the equation in a certain set containing $(-T,T)$,
with the proper regularity.

######  Lemma 5.8.

Assume that (H6’) holds. Let $c>0$, $T\geq 1$ and $\mathbf{m}_{c,T}$ be as in
(5.15). Let $\mathbf{U}_{c,T}\in X_{T}$ be such that
$\mathbf{E}_{c}(\mathbf{U}_{c,T})=\mathbf{m}_{c,T}$. Then,
$\mathbf{U}_{c,T}\in\mathcal{A}((-T,T))$, $\mathcal{A}((-T,T))$ as in (4.29)
and

$\mathbf{U}_{c,T}^{\prime\prime}-D_{\mathscr{L}}\mathcal{E}(\mathbf{U}_{c,T})=-c\mathbf{U}_{c,T}^{\prime}\mbox{
in }(-T,T).$ (5.60)

Moreover, if $t\geq T$ is such that

$\mathrm{dist}_{\mathscr{L}}(\mathbf{U}_{c,T}(t),\mathscr{F}^{+})<\frac{r_{0}^{+}}{2},$
(5.61)

then, there exists $\delta^{+}(t)>0$ such that
$\mathbf{U}_{c,T}\in\mathcal{A}((t-\delta^{+}(t),t+\delta^{+}(t)))$ and

$\mathbf{U}_{c,T}^{\prime\prime}-D_{\mathscr{L}}\mathcal{E}(\mathbf{U}_{c,T})=-c\mathbf{U}_{c,T}^{\prime}\mbox{
in }(t-\delta^{+}(t),t+\delta^{+}(t)).$ (5.62)

Similarly, if $t\leq-T$ is such that

$\mathrm{dist}_{\mathscr{L}}(\mathbf{U}_{c,T}(t),\mathscr{F}^{-})<\frac{r_{0}^{-}}{2},$
(5.63)

then, there exists $\delta^{-}(t)>0$ such that
$\mathbf{U}_{c,T}\in\mathcal{A}((t-\delta^{-}(t),t+\delta^{-}(t)))$ and

$\mathbf{U}_{c,T}^{\prime\prime}-D_{\mathscr{L}}\mathcal{E}(\mathbf{U}_{c,T})=-c\mathbf{U}_{c,T}^{\prime}\mbox{
in }(t-\delta^{-}(t),t+\delta^{-}(t)).$ (5.64)

###### Proof.

We first show that

$\forall t\in{\mathbb{R}},\hskip
5.69054pt\mathfrak{P}(\mathbf{U}_{c,T}(t))=\mathbf{U}_{c.T}(t),$ (5.65)

where $\mathfrak{P}$ is the map from (H6’). We claim that the function

$\mathbf{U}_{c,T}^{\mathfrak{P}}:t\in{\mathbb{R}}\to\mathfrak{P}(\mathbf{U}_{c,T}(t))$
(5.66)

belongs to $X_{T}$. Indeed, this follows from (4.27))and (4.28). Property
(4.26) implies that

$\forall t\in{\mathbb{R}},\hskip
5.69054pt\mathcal{E}(\mathbf{U}_{c,T}^{\mathfrak{P}}(t))\leq\mathcal{E}(\mathbf{U}_{c,T}(t)).$
(5.67)

Take now $t\in{\mathbb{R}}$ and $s\in{\mathbb{R}}\setminus\\{t\\}$. Property
(4.27) implies that

$\left\lVert\frac{\mathbf{U}_{c,T}^{\mathfrak{P}}(t)-\mathbf{U}_{c,T}^{\mathfrak{P}}(s)}{t-s}\right\rVert_{\mathscr{L}}\leq\left\lVert\frac{\mathbf{U}_{c,T}(t)-\mathbf{U}_{c,T}(s)}{t-s}\right\rVert_{\mathscr{L}},$
(5.68)

which, by Lebesgue’s differentiation Theorem implies that

$\mbox{for a. e. }t\in{\mathbb{R}},\hskip
5.69054pt\lVert(\mathbf{U}_{c,T}^{\mathfrak{P}})^{\prime}(t)\rVert_{\mathscr{L}}\leq\lVert\mathbf{U}_{c,T}^{\prime}(t)\rVert_{\mathscr{L}}.$
(5.69)

By contradiction, assume now that there exists $t\in{\mathbb{R}}$ such that
$\mathbf{U}_{c,T}(t)\not=\mathfrak{P}(\mathbf{U}_{c,T}(t))=\mathbf{U}_{c,T}^{\mathfrak{P}}(t)$.
Property (4.27) implies that $\mathfrak{P}$ is a $\mathscr{L}$-continuous map.
Therefore, since $\mathbf{U}_{c,T}$ is also $\mathscr{L}$-continuous, we must
have that for some non-empty interval $I_{t}\ni t$, it holds

$\forall s\in I_{t},\hskip
5.69054pt\mathbf{U}_{c,T}(s)\not=\mathfrak{P}(\mathbf{U}_{c,T}(s))=\mathbf{U}_{c,T}^{\mathfrak{P}}(s),$
(5.70)

so that, using (4.26) we get

$\forall s\in I_{t},\hskip
5.69054pt\mathcal{E}(\mathbf{U}_{c,T}^{\mathfrak{P}}(s))<\mathcal{E}(\mathbf{U}_{c,T}(s))$
(5.71)

so that, combining with (5.67) and (5.69) we obtain

$\mathbf{E}_{c}(\mathbf{U}_{c,T}^{\mathfrak{P}})<\mathbf{E}_{c}(\mathbf{U}_{c,T})=\mathbf{m}_{c,T},$
(5.72)

which contradicts the definition of $\mathbf{m}_{c,T}$ (5.15) since
$\mathbf{U}_{c,T}^{\mathfrak{P}}\in X_{T}$. Therefore, we have shown that
(5.65) holds. Next, notice that

$\mathbf{E}_{c}(\mathbf{U}_{c,T};[-T,T])\leq\mathbf{m}_{c,T}-\frac{a}{c}e^{-cT}<+\infty$
(5.73)

and for any
$\phi\in\mathcal{C}^{1}_{c}((-T,T),(\mathscr{H},\lVert\cdot\rVert_{\mathscr{H}}))$
we have $\mathbf{U}_{c,T}+\phi\in X_{T}$, so that
$\mathbf{E}_{c}(\mathbf{U}_{c,T})\leq\mathbf{E}_{c}(\mathbf{U}_{c,T}+\phi)$.
Therefore, the restriction of $\mathbf{U}_{c,T}$ in $(-T,T)$ is a local
minimizer of $\mathbf{E}_{c}(\cdot,[-T,T])$ in the sense of Definition 4.1.
Since $\mathbf{U}_{c,T}$ also verifies (5.65), we can apply the regularity
assumption (H6’). Therefore, $\mathbf{U}_{c,T}\in\mathcal{A}((-T,T))$ and
(5.60) holds. Assume now that there exists $t\geq T$ such that (5.61) holds.
Then, there exists $\mathbf{v}^{+}(t)\in\mathscr{F}^{+}$ such that

$\lVert\mathbf{U}_{c,T}(t)-\mathbf{v}^{+}(t)\rVert_{\mathscr{L}}<\frac{r_{0}^{+}}{2}$
(5.74)

which, since $\mathbf{U}_{c,T}$ is $\mathscr{L}$-continuous, implies that
there exists $\delta^{+}(t)>0$ such that

$\forall s\in(t-\delta^{+}(t),t+\delta^{+}(t)),\hskip
5.69054pt\lVert\mathbf{U}_{c,T}(s)-\mathbf{v}^{+}(t)\rVert_{\mathscr{L}}<\frac{r_{0}^{+}}{2}-d^{+}(t)$
(5.75)

where

$d^{+}(t):=\frac{1}{2}\left(\frac{r_{0}^{+}}{2}-\lVert\mathbf{U}_{c,T}(t)-\mathbf{v}^{+}(t)\rVert_{\mathscr{L}}\right)>0.$
(5.76)

Therefore, if
$\phi\in\mathcal{C}^{1}_{c}((t-\delta^{+}(t),t+\delta^{+}(t)),(\mathscr{H},\lVert\cdot\rVert_{\mathscr{H}}))$
is such that

$\max_{t\in[t-\delta^{+}(t),t+\delta^{+}(t)]}\lVert\phi(t)\rVert_{\mathscr{H}}\leq\frac{d^{+}(t)}{2}$
(5.77)

we have that

$\forall s\in(t-\delta^{+}(t),t+\delta^{+}(t)),\hskip
5.69054pt\lVert\mathbf{U}_{c,T}(s)+\phi(s)-\mathbf{v}^{+}(t)\rVert_{\mathscr{L}}<\frac{r_{0}^{+}}{2}-\frac{d^{+}(t)}{2}$
(5.78)

so that $\mathbf{U}_{c,T}+\phi\in X_{T}$. Meaning that
$\mathbf{E}_{c,T}(\mathbf{U}_{c,T})\leq\mathbf{E}_{c,T}(\mathbf{U}_{c,T}+\phi)$.
Since $\phi$ is supported on $[t-\delta^{+}(t),t+\delta^{+}(t)]$, the previous
implies that

$\mathbf{E}_{c,T}(\mathbf{U}_{c,T};[t-\delta^{+}(t),t+\delta^{+}(t)])\leq\mathbf{E}_{c,T}(\mathbf{U}_{c,T}+\phi;[t-\delta^{+}(t),t+\delta^{+}(t)]),$
(5.79)

so that $\mathbf{U}_{c,T}$ is a local minimizer of
$\mathbf{E}_{c}(\cdot;[t-\delta^{+}(t),t+\delta^{+}(t)])$ in the sense of
Definition 4.1. Since (5.65) also holds, we can apply (H6’) and obtain that
$\mathbf{U}_{c,T}\in\mathcal{A}((t-\delta^{+}(t),t+\delta^{+}(t)))$ and
equation (5.62) holds. If $t\leq-T$ is such that (5.63) holds, the same
reasoning shows that for some $\delta^{-}(t)>0$,
$\mathbf{U}_{c,T}\in\mathcal{A}((t-\delta^{-}(t),t+\delta^{-}(t)))$ and (5.61)
holds, which concludes the proof of the result. ∎

###  5.6 The comparison result

The goal of this subsection is to obtain relevant information on the behavior
of the constrained minimizers. Such information is contained in Corollary 5.1
and it will allow us to remove the constraints later on. In order to carry on
these arguments, assumption (H7’) will become necessary since it will show
that our problem can be somehow dealt with as in the balanced one, which will
allow us to argue in a fashion similar to Smyrnelis [69]. We begin by
introducing some constants. For $0<r\leq r_{0}^{\pm}$, recall the definition
of $\kappa_{r}^{\pm}$ introduced in (4.10), Lemma 4.1. We define

$\eta^{+}_{0}:=\min\left\\{\sqrt{e^{-1}\frac{r_{0}^{+}}{4}\sqrt{2\kappa^{+}_{r_{0}^{+}/4}}},\frac{r_{0}^{+}}{4}\right\\}>0,$
(5.80) $\hat{r}^{+}:=\frac{r_{0}^{+}}{C^{+}+1}>0,$ (5.81)
$\mathcal{E}^{+}_{\max}:=\frac{1}{\left(C^{+}\right)^{2}(C^{+}+1)}\min\left\\{\frac{(\eta_{0}^{+})^{2}}{4},\kappa^{+}_{\eta_{0}^{+}},\beta^{+}(\hat{r}^{+}),\beta^{+}(\eta_{0}^{+})\right\\}>0,$
(5.82)

where the constants $C^{\pm}$,
$\beta^{\pm}(\hat{r}^{\pm}),\beta(\eta_{0}^{\pm})$ were introduced in (H5’).
Recall that in (4.31), (4.32), (4.33) we introduced the analogous constants

$\eta_{0}^{-}:=\min\left\\{\sqrt{e^{-1}\frac{r_{0}^{-}}{4}\sqrt{2(\kappa^{-}_{r_{0}^{-}/4}-a)}},\frac{r_{0}^{-}}{4}\right\\}>0,$
(5.83) $\hat{r}^{-}:=\frac{r_{0}^{-}}{C^{-}+1}>0$ (5.84)

and

$\mathcal{E}_{\max}^{-}:=\frac{1}{(C^{-})^{2}(C^{-}+1)}\min\left\\{\frac{(\eta_{0}^{-})^{2}}{4},\kappa_{\eta_{0}^{-}}^{-}-a,\beta^{-}(\hat{r}^{-}),\beta^{-}(\eta_{0}^{-})\right\\}>0.$
(5.85)

For any $U\in X_{T}$, define

$t^{-}(U,\mathcal{E}_{\max}^{-}):=\sup\left\\{t\in{\mathbb{R}}:\mathcal{E}(U(t))\leq
a+\mathcal{E}_{\max}^{-}\mbox{ and
}\mathrm{dist}_{\mathscr{L}}(U(t),\mathscr{F}^{-})\leq\frac{r_{0}^{-}}{2}\right\\}$
(5.86)

and

$t^{+}(U,\mathcal{E}_{\max}^{+}):=\inf\left\\{t\in{\mathbb{R}}:\mathcal{E}(U(t))\leq\mathcal{E}_{\max}^{+}\mbox{
and
}\mathrm{dist}_{\mathscr{L}}(U(t),\mathscr{F}^{+})\leq\frac{r_{0}^{+}}{2}\right\\}.$
(5.87)

We have the following technical property:

######  Lemma 5.9.

Assume that (H1’) and (H5’) hold. Let $\hat{r}^{\pm}>0$ be as in (5.81),
(5.84) and $\mathcal{E}_{\max}^{\pm}$ be as in (5.82), (5.85). Then, if
$v\in\mathscr{F}^{\pm}_{r_{0}^{\pm}}$ is such that

$\mathcal{E}(v)\leq\min\\{\pm(-a),0\\}+\beta^{\pm}(\hat{r}^{\pm}),$ (5.88)

then

$\forall\lambda\in[0,1],\hskip 5.69054pt\mathcal{E}(\lambda
v+(1-\lambda)P^{\pm}(v))\leq\min\\{\pm(-a),0\\}+\mathsf{C}^{\pm}(\mathcal{E}(v)-\min\\{\pm(-a),0\\}).$
(5.89)

In particular, if

$\mathcal{E}(u)\leq\min\\{\pm(-a),0\\}+\mathcal{E}_{\max}^{\pm},$ (5.90)

then

$\forall\lambda\in[0,1],\hskip 5.69054pt\mathcal{E}(\lambda
v+(1-\lambda)P^{\pm}(v))\leq\min\\{\pm(-a),0\\}+\mathsf{C}^{\pm}\mathcal{E}_{\max}^{\pm}.$
(5.91)

The constants $\mathcal{E}_{\max}^{\pm}$, $\mathsf{C}^{\pm}$ where defined in
(5.85), (5.82) and (4.34) respectively. $P^{\pm}(u)$ is the projection
introduced in (H1’).

###### Proof.

Assume that (5.88) holds for $v\in\mathscr{F}^{\pm}_{r_{0}^{\pm}}$. Then,
invoking (H5’) we have that $v\in\mathscr{F}_{\mathscr{H},\hat{r}^{\pm}}$, so
in particular the projection $P^{\pm}_{\mathscr{H}}(u)$ is well defined. Fix
$\lambda\in[0,1]$. Since $v\in\mathscr{F}^{\pm}_{r_{0}^{\pm}}$, the projection
$P^{\pm}(v)$ is well defined by (H1’). Using (4.20) we obtain

$\displaystyle\lVert\lambda
v+(1-\lambda)P^{\pm}(v)-P^{\pm}(v)\rVert_{\mathscr{H}}$
$\displaystyle=\lambda\lVert v-P^{\pm}(v)\rVert_{\mathscr{H}}$ (5.92)
$\displaystyle\leq(C^{\pm}+1)\lVert
v-P^{\pm}_{\mathscr{H}}(v)\rVert_{\mathscr{H}}\leq(C^{\pm}+1)\hat{r}^{\pm}$
(5.93)

so that $\lambda
v+(1-\lambda)P^{\pm}(v)\in\mathscr{F}^{\pm}_{\mathscr{H},r_{0}^{\pm}}$ by the
definition of $\hat{r}^{\pm}$ in (5.81), (5.84). Using now again (4.20) along
with the estimate (4.22) in (H5’), we get

$\lVert
P^{\pm}(v)-P^{\pm}_{\mathscr{H}}(v)\rVert_{\mathscr{H}}^{2}\leq(C^{\pm})^{2}(\mathcal{E}(v)-\min\\{\pm(-a),0\\})$
(5.94)

which, plugging into (5.92), gives

$\displaystyle\lVert\lambda
v+(1-\lambda)P^{\pm}(v)-P_{\mathscr{H}}^{\pm}(v)\rVert_{\mathscr{H}}^{2}\leq\frac{1}{2}((C^{\pm})^{2}+(C^{\pm}+1)^{2})(\mathcal{E}(v)-\min\\{\pm(-a),0\\}),$
(5.95)

that, using again (4.22), implies exactly (5.89). Assuming now that (5.90)
holds, we have by (5.85), (5.82) that in particular (5.88) holds. Therefore,
(5.91) follows from (5.89). ∎

Next, we have the following property:

######  Lemma 5.10.

Assume that (H5’) and (H7’) hold. Let $c>0$ and $T\geq 1$. Assume that $U\in
X_{T}$ is such that $\mathbf{E}_{c}(U)\leq 0$. Then the quantities
$t^{-}(U,\mathcal{E}_{\max}^{-})$ and $t^{+}(U,\mathcal{E}_{\max}^{+})$
defined in (5.86) and (5.87), respectively, are well defined as real numbers.
Moreover, it holds that

$\mathcal{E}(U(t^{-}(U,\mathcal{E}_{\max}^{-}))\leq
a+\mathcal{E}_{\max}^{-},\hskip
2.84526pt\mathrm{dist}_{\mathscr{L}}(U(t^{-}(U,\mathcal{E}_{\max}^{-})),\mathscr{F}^{-})\leq\frac{r_{0}^{-}}{2}$
(5.96)

and

$\mathcal{E}(U(t^{+}(U,\mathcal{E}_{\max}^{+}))\leq\mathcal{E}_{\max}^{+},\hskip
2.84526pt\mathrm{dist}_{\mathscr{L}}(U(t^{+}(U,\mathcal{E}_{\max}^{+})),\mathscr{F}^{+})\leq\frac{r_{0}^{+}}{2}.$
(5.97)

###### Proof.

Using that $\mathbf{E}_{c}(U)\leq 0$ and the fact that
$\\{t\in{\mathbb{R}}:\mathcal{E}(U(t))>0\\}$ is nonempty since $U\in X_{T}$,
we must have that

$\\{t\in{\mathbb{R}}:\mathcal{E}(U(t))<0\\}\not=\emptyset$ (5.98)

and if $v\in\mathscr{L}$ is such that $\mathcal{E}(v)<0$, then we have
$\mathrm{dist}_{\mathscr{L}}(v,\mathscr{F}^{-})\leq r_{0}^{-}/2$ by (4.38) in
(H7’) and $\mathcal{E}(v)<a+\mathcal{E}_{\max}^{-}$ since we assume
$a+\mathcal{E}_{\max}^{-}>0$. Therefore, $t^{-}(U,\mathcal{E}_{\max}^{-})$ is
well defined, as we have shown that

$\left\\{t\in{\mathbb{R}}:\mathcal{E}(U(t))\leq a+\mathcal{E}^{-}_{\max}\mbox{
and
}\mathrm{dist}_{\mathscr{L}}(U(t),\mathscr{F}^{-})\leq\frac{r_{0}^{-}}{2}\right\\}\not=\emptyset$
(5.99)

and such set is bounded above by $T$, because
$\mathscr{F}^{-}_{r_{0}^{-}/2}\cap\mathscr{F}^{+}_{r_{0}^{+}/2}=\emptyset$.
Using Lemma 5.3, we have that $t^{+}(U,\mathcal{E}_{\max}^{+})$ is well
defined. Finally, inequalities (5.96) and (5.97) follow because
$t\in{\mathbb{R}}\to\mathcal{E}(U(t))\in{\mathbb{R}}$ is lower semicontinuous
by (H1’) and $t\to\mathrm{dist}_{\mathscr{L}}(U(t),\mathscr{F}^{\pm})$ is
continuous whenever $U(t)\in\mathscr{F}^{\pm}_{r_{0}^{\pm}/2}$ by (H1’)
(recall that $t\in{\mathbb{R}}\to U(t)\in\mathscr{L}$ is continuous because
$U\in H^{1}_{\mathrm{loc}}({\mathbb{R}},\mathscr{L})$). ∎

The main work is done by the following result:

######  Proposition 5.1.

Assume that (H5’) and (H7’) hold. Let $c>0$ and $T\geq 1$. Consider $U\in
X_{T}$ with $\mathbf{E}_{c}(U)\leq 0$. Let
$t^{\pm}:=t^{\pm}(U,\mathcal{E}_{\max}^{\pm},\eta_{0}^{\pm})$ be as in (5.86)
and (5.87). Then, $t^{\pm}$ are well defined by Lemma 5.10. Moreover, if there
exists $\tilde{t}^{-}<t^{-}$ such that

$r_{0}^{-}\geq\mathrm{dist}_{\mathscr{L}}(U(\tilde{t}^{-}),\mathscr{F}^{-})\geq\frac{r_{0}^{-}}{2}$
(5.100)

then, we can find $\tilde{U}^{-}\in X_{T}$ such that

$\forall t\leq t^{-},\hskip
5.69054pt\mathrm{dist}_{\mathscr{L}}(\tilde{U}^{-}(t),\mathscr{F}^{-})<\frac{r_{0}^{-}}{2},$
(5.101)

and

$\mathbf{E}_{c}(\tilde{U}^{-})<\mathbf{E}_{c}(U).$ (5.102)

Analogously, if there exists $\tilde{t}^{+}>t^{+}$ such that

$r_{0}^{+}\geq\mathrm{dist}_{\mathscr{L}}(U(\tilde{t}^{+}),\mathscr{F}^{+})\geq\frac{r_{0}^{+}}{2},$
(5.103)

then we can find $\tilde{U}^{+}\in X_{T}$ such that

$\forall t\geq t^{+},\hskip
5.69054pt\mathrm{dist}_{\mathscr{L}}(\tilde{U}^{+}(t),\mathscr{F}^{+})<\frac{r_{0}^{+}}{2}$
(5.104)

and

$\mathbf{E}_{c}(\tilde{U}^{+})<\mathbf{E}_{c}(U).$ (5.105)

Furthermore, we have that

$0<t^{+}-t^{-}\leq\mathsf{T}_{\star}(c),$ (5.106)

where

$\mathsf{T}_{\star}(c):=\frac{1}{c}\ln\left(\frac{-a}{\alpha_{\star}}+1\right),$
(5.107)

with $\alpha_{\star}>0$ a constant which is independent from $c$, $T$ and $U$.

The idea of the proof of Proposition 5.1 is pictured in Figure 5.2.

###### Proof of Proposition 5.1.

We begin by proving the first part of the result for $\mathscr{F}^{-}$. Recall
that Lemma 5.10 gives

$\mathcal{E}(U(t^{-}))\leq a+\mathcal{E}_{\max}^{-}.$ (5.108)

and $U(t^{-})\in\mathscr{F}^{-}_{r_{0}^{-}/2}$. Since
$a+\mathcal{E}_{\max}^{-}\leq\beta(\mathcal{E}_{\max}^{-})$ by the definition
of $\mathcal{E}_{\max}^{-}$, (5.85), we have by (H5’) that

$\mathrm{dist}_{\mathscr{L}}(U(t^{-}),\mathscr{F}^{-})\leq\eta_{0}^{-}.$
(5.109)

Assume that there exists $\tilde{t}^{-}<t^{-}$ such that (5.100) is satisfied.
Moreover, we assume, as we can, that

$\tilde{t}^{-}:=\max\left\\{t\leq
t^{-}:\mathrm{dist}_{\mathscr{L}}(U(t),\mathscr{F}^{-})\geq\frac{r_{0}^{-}}{2}\right\\}$
(5.110)

(the $\sup$ can be replaced by a $\max$ by continuity). Define

$t^{-}_{0}:=\inf\\{t\in[\tilde{t}^{-},t^{-}]:\mathcal{E}(U(t))\leq
a+\mathcal{E}_{\max}^{-}\mbox{ and
}\mathrm{dist}_{\mathscr{L}}(U(t))\leq\eta_{0}^{-}\\}.$ (5.111)

Let $\mathbf{v}^{-}:=P^{-}(U(t^{-}_{0}))\in\mathscr{F}^{-}$, with $P^{-}$ as
in (H1’). Notice that due to (5.110), we have

$\forall t\in[t^{-}_{0},t^{-}],\hskip
5.69054pt\mathrm{dist}(U(t),\mathscr{F}^{-})<\frac{r_{0}^{-}}{2}.$ (5.112)

The proof now bifurcates according to two possible cases:

  * •

$t^{-}_{0}\leq\tilde{t}^{-}+1$. In that case, set

$\tilde{U}^{-}(t):=\begin{cases}\mathbf{v}^{-}&\mbox{ if }t\leq
t_{0}^{-}-1,\\\ (t_{0}^{-}-t)\mathbf{v}^{-}+(t-t_{0}^{-}+1)U(t_{0}^{-})&\mbox{
if }t^{-}_{0}-1\leq t\leq t_{0}^{-},\\\ U(t)&\mbox{ if }t\geq
t^{-}_{0},\end{cases}$ (5.113)

which belongs to $X_{T}$. Due to the definition of $\tilde{U}^{-}$ and
(5.112), we have that $\tilde{U}^{-}$ satisfies (5.101). It remains to check
(5.102). We have

$\int_{t_{0}^{-}-1}^{t_{0}^{-}}\mathbf{e}_{c}(\tilde{U}^{-}(t))dt\leq\int_{t_{0}^{-}-1}^{t_{0}^{-}}\left[\frac{\lVert
U(t_{0}^{-})-\mathbf{v}^{-}\rVert_{\mathscr{L}}^{2}}{2}+\mathcal{E}(\tilde{U}^{-}(t))\right]e^{ct}dt.$
(5.114)

Fix $t\in[t^{-}_{0}-1,t^{-}_{0}]$. Choosing $\lambda=t-t_{0}^{-}+1\in[0,1]$
and applying (5.91) in Lemma 5.9 and (5.108), we have that

$\mathcal{E}(\tilde{U}^{-}(t))\leq a+\mathsf{C}^{-}\mathcal{E}_{\max}^{-}.$
(5.115)

The previous fact combined with (5.109), (5.108) and 5.114, gives

$\int_{t_{0}^{-}-1}^{t_{0}^{-}}\mathbf{e}_{c}(\tilde{U}^{-}(t))dt\leq\left(\frac{(\eta_{0}^{-})^{2}}{2}+\mathsf{C}^{-}\mathcal{E}_{\max}^{-}\right)e^{ct^{-}_{0}}+\frac{a(e^{ct^{-}_{0}}-e^{c(t_{0}^{-}-1)})}{c}.$
(5.116)

The continuity of $U$ and (5.100) implies that there exists
$\tilde{t}^{-}_{2}\in(\tilde{t}^{-},t^{-}_{0})$ such that

$\mathrm{dist}(U(\tilde{t}_{2}^{-}),\mathscr{F}^{-})=\frac{r_{0}^{-}}{4}\mbox{
and }\forall t\in[\tilde{t}^{-},\tilde{t}^{-}_{2}],\hskip
5.69054pt\mathrm{dist}_{\mathscr{L}}(U(t),\mathscr{F}^{-})\geq\frac{r_{0}^{-}}{4}.$
(5.117)

Using (5.117), we get

$\int_{\tilde{t}^{-}}^{\tilde{t}_{2}^{-}}\lVert
U^{\prime}(t)\rVert_{\mathscr{L}}e^{ct}dt\geq\frac{r_{0}^{-}e^{c\tilde{t}^{-}}}{4}$
(5.118)

and (5.117) also implies

$\forall t\in[\tilde{t}^{-},\tilde{t}_{2}^{-}],\hskip
5.69054pt\mathcal{E}(U(t))\geq\kappa^{-}_{r_{0}^{-}/4}.$ (5.119)

Inequalities (5.118) and (5.119) along with the definition of $\eta_{0}^{-}$
in (5.83) and Young’s inequality give

$\displaystyle\int_{\tilde{t}^{-}}^{\tilde{t}_{2}^{-}}\mathbf{e}_{c}(U(t))dt$
$\displaystyle\geq\frac{r_{0}^{-}e^{c\tilde{t}^{-}_{0}}}{4}\sqrt{2(\kappa^{-}_{r_{0}^{-}/4}-a)}+a\frac{e^{c\tilde{t}_{2}^{-}}-e^{c\tilde{t}^{-}}}{c}$
(5.120)
$\displaystyle=e(\eta_{0}^{-})^{2}e^{c\tilde{t}^{-}}+a\frac{e^{c\tilde{t}_{2}^{-}}-e^{c\tilde{t}^{-}}}{c},$
(5.121)

which gets to, using also that $\tilde{t}^{-}\geq t^{-}_{0}-1$,

$\displaystyle\int_{-\infty}^{t^{-}_{0}}\mathbf{e}_{c}(U(t))dt$
$\displaystyle=\int_{(-\infty,t^{-}_{0}]\setminus[\tilde{t}^{-},\tilde{t}_{2}^{-}]}\mathbf{e}_{c}(U(t))dt+\int_{\tilde{t}^{-}}^{\tilde{t}_{2}^{-}}\mathbf{e}_{c}(U(t))dt$
(5.122) $\displaystyle\geq
e(\eta_{0}^{-})^{2}e^{c\tilde{t}^{-}}+a\frac{e^{ct^{-}_{0}}}{c}\geq(\eta_{0}^{-})^{2}e^{ct_{0}^{-}}+a\frac{e^{ct^{-}_{0}}}{c}.$
(5.123)

Using now (5.116) we get

$\displaystyle\int_{-\infty}^{t^{-}_{0}}\mathbf{e}_{c}(\tilde{U}^{-}(t))dt$
$\displaystyle=\int_{-\infty}^{t^{-}_{0}-1}\mathbf{e}_{c}(\tilde{U}^{-}(t))dt+\int_{t^{-}_{0}-1}^{t^{-}_{0}}\mathbf{e}_{c}(\tilde{U}^{-}(t))dt$
(5.125)
$\displaystyle\leq\left(\frac{(\eta_{0}^{-})^{2}}{2}+\mathsf{C}^{-}\mathcal{E}_{\max}^{-}\right)e^{ct^{-}_{0}}+\frac{ae^{ct^{-}_{0}}}{c}.$
(5.126)

Therefore, subtracting (5.125) from (5.122), we get

$\displaystyle\int_{-\infty}^{t^{-}_{0}}\mathbf{e}_{c}(U(t))dt-\int_{-\infty}^{t^{-}_{0}}\mathbf{e}_{c}(\tilde{U}^{-}(t))dt$
(5.127)
$\displaystyle\geq\left(\frac{(\eta_{0}^{-})^{2}}{2}+\mathsf{C}^{-}\mathcal{E}_{\max}^{-}\right)e^{ct^{-}_{0}},$
(5.128)

which is positive because (5.85) implies

$\mathsf{C}^{-}\mathcal{E}_{\max}^{-}\leq\frac{(\eta_{0}^{-})^{2}}{4}.$
(5.130)

Since $\tilde{U}^{-}$ and $U$ coincide in $[t^{-}_{0},+\infty)$, the proof of
the first case is concluded.

  * •

$t_{0}^{-}>\tilde{t}+1$. In such a case, set

$\tilde{U}^{-}(t):=\begin{cases}\mathbf{v}^{-}&\mbox{ if
}t\leq\tilde{t}^{-},\\\
(t-\tilde{t}^{-})U(t_{0}^{-})+(\tilde{t}^{-}+1-t)\mathbf{v}^{-}&\mbox{ if
}\tilde{t}^{-}\leq t\leq\tilde{t}^{-}+1,\\\ U(t^{-}_{0})&\mbox{ if
}\tilde{t}^{-}+1\leq t\leq t^{-}_{0},\\\ U(t)&\mbox{ if }t^{-}_{0}\leq
t,\end{cases}$ (5.131)

which clearly belongs to $X_{T}$ and for all $t\leq t^{-}$,
$U(t)\in\mathscr{F}^{-}_{r_{0}^{-}/2}$ by (5.112). We have that
$\tilde{U}^{-}$ is constant in $[\tilde{t}^{-}+1,t_{0}^{-}]$, therefore

$\int_{\tilde{t}^{-}+1}^{t^{-}_{0}}\mathbf{e}_{c}(\tilde{U}^{-}(t))dt\leq(a+\mathcal{E}_{\max}^{-})\frac{e^{ct_{0}^{-}}-e^{c\tilde{t}^{-}+1}}{c}$
(5.132)

and, due to the definitions of $\mathcal{E}_{\max}^{-}$ in (5.85) and
$t_{0}^{-}$ in (5.111)

$\int_{\tilde{t}^{-}+1}^{t_{0}^{-}}\mathbf{e}_{c}(U(t))dt\geq\min\\{a+\mathcal{E}_{\max}^{-},\kappa_{\eta_{0}^{-}}\\}\frac{e^{ct_{0}^{-}}-e^{c\tilde{t}^{-}+1}}{c}\geq\int_{\tilde{t}^{-}+1}^{t^{-}_{0}}\mathbf{e}_{c}(\tilde{U}^{-}(t))dt,$
(5.133)

because $\mathcal{E}_{\max}^{-}+a>0$ by (H7’) and $t_{0}\geq\tilde{t}^{-}+1$
by assumption. Hence

$\int_{\tilde{t}^{-}+1}^{+\infty}\mathbf{e}_{c}(\tilde{U}(t))dt\leq\int_{\tilde{t}^{-}+1}^{\infty}\mathbf{e}_{c}(U(t))dt.$
(5.134)

Arguing as in the first case scenario, we can prove that

$\int_{-\infty}^{\tilde{t}^{-}+1}\mathbf{e}_{c}(\tilde{U}(t))dt<\int_{-\infty}^{\tilde{t}^{-}+1}\mathbf{e}_{c}(U(t))dt,$
(5.135)

which concludes the proof of the second case.

To sum up, we have shown that if (5.100) is satisfied, then there exists
$\tilde{U}^{-}$ such that (5.101) and (5.102) hold, as we wanted.

Assume now that there exists $\tilde{t}^{+}>t^{+}$ such that (5.103) holds. As
before, Lemma 5.10 and the definition of $\mathcal{E}_{\max}^{+}$, (5.82),
imply that $t^{+}:=t^{+}(U,\mathcal{E}_{\max}^{+})$ is such that

$\mathrm{dist}_{\mathscr{L}}(U(t^{+}),\mathscr{F}^{+})\leq\eta_{0}^{+}$
(5.136)

and

$\mathcal{E}(U(t^{+}))\leq\mathcal{E}_{\max}^{+}.$ (5.137)

We claim that we can assume without loss of generality that

$\forall t\in[t^{+},+\infty),\hskip 5.69054pt\mathcal{E}(U(t))\geq 0.$ (5.138)

Indeed, if we can find $t_{0}\in(t^{+},+\infty)$ such that
$\mathcal{E}(U(t))<0$, then by (H7’) we have have that
$\mathcal{E}(U(t_{0}))\leq a+\mathcal{E}_{\max}^{-}$ and by (4.38) in (H7’) we
also have that $\mathrm{dist}_{\mathscr{L}}(U(t_{0}),\mathscr{F}^{-})\leq
r_{0}^{-}/2$. Therefore, we have by the definitions (5.86) and (5.87) that
$t^{-}\geq t_{0}$ and $t^{+}>t^{-}$, a contradiction since we assume
$t_{0}>t^{+}$.

For the positive case, the proof is simpler as it suffices to define
$\mathbf{v}^{+}:=P^{+}(U(t^{+}))$ and

$\tilde{U}^{+}(t):=\begin{cases}\mathbf{v}^{+}&\mbox{ if }t\geq t^{+}+1\\\
(t-t^{+})\mathbf{v}^{+}+(t^{+}+1-t)U(t^{+})&\mbox{ if }t^{+}+1\geq t\geq
t^{+},\\\ U(t)&\mbox{ if }t^{+}\geq t,\end{cases}$ (5.139)

which is such that $U\in X_{T}$. Moreover, it holds that for all $t\geq
t^{+}$, we have $\tilde{U}^{+}(t)\in\mathscr{F}^{+}_{r_{0}^{+}/2}$. Therefore,
the requirements (5.104) and (5.105) hold for $\tilde{U}^{+}$. It remains to
check that (5.105) is also fulfilled. We have that

$\int_{t^{+}}^{t^{+}+1}\mathbf{e}_{c}(\tilde{U}^{+}(t))dt=\int_{t^{+}}^{t^{+}+1}\left[\frac{\left\lVert
U(t^{+})-\mathbf{v}^{+}\right\rVert_{\mathscr{L}}^{2}}{2}+\mathcal{E}(\tilde{U}^{+}(t))\right]e^{ct}dt.$
(5.140)

Using (5.91) in Lemma 5.9 and (5.137), we get

$\int_{t^{+}}^{t^{+}+1}\mathcal{E}(\tilde{U}^{+}(t))e^{ct}dt\leq\mathsf{C}^{+}\mathcal{E}_{\max}^{+}e^{c(t^{+}+1)}.$
(5.141)

Using now (5.136), we get

$\int_{t^{+}}^{t^{+}+1}\frac{\left\lVert
U(t^{+})-\mathbf{v}^{+}\right\rVert_{\mathscr{L}}^{2}}{2}e^{ct}dt\leq\frac{(\eta_{0}^{+})^{2}}{2}e^{c(t^{+}+1)}.$
(5.142)

Plugging (5.141) and (5.142) into (5.140), we get

$\int_{t^{+}}^{t^{+}+1}\mathbf{e}_{c}(\tilde{U}^{+}(t))dt\leq\left(\frac{(\eta_{0}^{+})^{2}}{2}+\mathsf{C}^{+}\mathcal{E}_{\max}^{+}\right)e^{ct^{+}+1}.$
(5.143)

Since for all $t\geq t^{+}+1$ we have $\tilde{U}^{+}(t)=\mathbf{v}^{+}$, we
obtain from (5.143)

$\int_{t^{+}}^{+\infty}\mathbf{e}_{c}(\tilde{U}^{+}(t))dt\leq\left(\frac{(\eta_{0}^{+})^{2}}{2}+\mathsf{C}^{+}\mathcal{E}_{\max}^{+}\right)e^{ct^{+}+1}.$
(5.144)

Next, notice that by continuity we can find
$\tilde{t}_{2}^{+}\in(t^{+},\tilde{t}^{+})$ such that

$\mathrm{dist}(U(\tilde{t}_{2}^{+}),\mathscr{F}^{+})=\frac{r_{0}^{+}}{4}\mbox{
and }\forall t\in[\tilde{t}^{+},\tilde{t}_{2}^{+}],\hskip
5.69054pt\frac{r_{0}^{+}}{2}\geq\mathrm{dist}_{\mathscr{L}}(U(t),\mathscr{F}^{+})\geq\frac{r_{0}^{+}}{4}.$
(5.145)

Therefore, using (5.103) and (5.145), we obtain

$\int_{\tilde{t_{2}}^{+}}^{\tilde{t}^{+}}\lVert
U^{\prime}(t)\rVert_{\mathscr{L}}e^{ct}dt\geq\frac{r_{0}^{+}}{4}e^{ct^{+}+1}e^{-1}$
(5.146)

and (5.145), (4.7) in (H1’) imply

$\forall t\in[\tilde{t}^{+},\tilde{t}_{2}^{+}],\hskip
5.69054pt\mathcal{E}(U(t))\geq\kappa_{r_{0}^{+}/4}^{+}.$ (5.147)

Inequalities (5.146), (5.147) yield by Young’s inequality

$\int_{\tilde{t}_{2}^{+}}^{\tilde{t}^{+}}\mathbf{e}_{c}(U(t))dt\geq\frac{r_{0}^{+}}{4}e^{ct^{+}+1}e^{-1}\sqrt{2\kappa_{r_{0}^{+}/4}^{+}}=(\eta_{0}^{+})^{2}e^{t^{+}+1},$
(5.148)

where the last equality is due to the definition of $\eta_{0}^{+}$ in (5.80).
Combining with (5.138), we get

$\int_{t^{+}}^{+\infty}\mathbf{e}_{c}(U(t))dt\geq(\eta_{0}^{+})^{2}e^{t^{+}+1}.$
(5.149)

The definition of $\mathcal{E}_{\max}^{+}$ in (5.82) and (5.144) imply then
that

$\int_{t^{+}}^{+\infty}\mathbf{e}_{c}(U(t))dt>\int_{t^{+}}^{+\infty}\mathbf{e}_{c}(\tilde{U}^{-}(t))dt,$
(5.150)

which establishes (5.105).

We now show the last part of the proof: we show that (5.106) holds with the
constant $T_{\star}(c)$ defined in (5.107). The argument is the same as in
[6], Lemma 2.10. Assume by contradiction that there exists
$t\in(t^{-},+\infty)$ such that $\mathcal{E}(U(t))<0$. Then, arguing as above
we must have $t<t^{-}$ by the definition of $t^{-}$ in (5.86), a
contradiction. Therefore, we can write

$\displaystyle\mathbf{E}_{c}(U)$
$\displaystyle=\frac{1}{2}\int_{\mathbb{R}}\lVert
U^{\prime}(t)\rVert_{\mathscr{L}}^{2}e^{ct}dt-\int_{-\infty}^{t^{-}}\mathcal{E}^{-}(U(t))e^{ct}dt$
(5.151) $\displaystyle+\int_{{\mathbb{R}}}\mathcal{E}^{+}(U(t))e^{ct}dt,$
(5.152)

where $\mathcal{E}^{-}$ and $\mathcal{E}^{+}$ stand for the positive and the
negative part of $\mathcal{E}$, respectively. We have that

$\int_{-\infty}^{t^{-}}\mathcal{E}^{-}(U(t))e^{ct}dt\leq\frac{-a}{c}e^{ct^{-}}.$
(5.153)

Set
$\alpha_{\star}:=\min\\{\mathcal{E}_{\max}^{+},\mathcal{E}_{\max}^{-}+a\\}>0$,
which is independent on $U$, $c$ and $T$. Notice that for all
$t\in(t^{-},t^{+})$ we have that $\mathcal{E}(U(t))\geq\alpha_{\star}$.
Indeed, if $\mathcal{E}(U(t))<\alpha_{\star}$, then by the definition of
$t^{-}$ and $t^{+}$ in (5.86) and (5.87) we get

$\mathrm{dist}_{\mathscr{L}}(U(t),\mathscr{F}^{\pm})\geq\frac{r_{0}^{\pm}}{2},$
(5.154)

which implies that

$\mathcal{E}(U(t))\geq\min\\{\kappa^{+}_{r_{0}^{+}/2},\kappa^{-}_{r_{0}^{-}/2}+a\\}\geq\alpha_{\star},$
(5.155)

by (5.85) and (5.82), a contradiction. Therefore,

$\int_{{\mathbb{R}}}\mathcal{E}^{+}(U(t))e^{ct}dt\geq\int_{t^{-}}^{t^{+}}\mathcal{E}^{+}(U(t))e^{ct}dt\geq\frac{\alpha_{\star}}{c}\left(e^{ct^{+}}-e^{ct^{-}}\right).$
(5.156)

Plugging (5.153) and (5.156) into (5.151) and using that
$\mathbf{E}_{c}(U)\leq 0$, we obtain

$0\geq\frac{a}{c}e^{ct^{-}}+\frac{\alpha_{\star}}{c}\left(e^{ct^{+}}-e^{ct^{-}}\right)\geq\left(\frac{a}{c}+\frac{\alpha_{\star}}{c}\left(e^{c(t^{+}-t^{-})}-1\right)\right)e^{ct^{-}},$
(5.157)

that is,

$0\geq-\left(\frac{-a}{\alpha_{\star}}+1\right)+e^{c(t^{+}-t^{-})},$ (5.158)

which implies

$0<t^{+}-t^{-}\leq\frac{1}{c}\ln\left(\frac{-a}{\alpha_{\star}}+1\right)=\mathsf{T}_{\star}(c),$
(5.159)

which is exactly (5.106) according to the definition (5.107). ∎

Figure 5.2: As it has been shown, the proof of Proposition 5.1 consists on
showing that if the function $U$ gets too far from $\mathscr{F}^{\pm}$ after
getting too close, then we can find a suitable competitor with strictly less
energy. Above, we see a design for the positive case (the competitor
$\tilde{U}^{+}$ is represented in blue). The second and third picture
correspond to the two possible scenarios for the negative case (the competitor
$\tilde{U}^{-}$ is represented in blue).

The importance of Proposition 5.1 is summarized by the following result, which
gives important information on the behavior of the constrained minimizers:

######  Corollary 5.1.

Assume that (H3’), (H4’), (H5’) and (H7’) hold. Let $c>0$ and $T\geq 1$. Let
$\mathbf{U}_{c,T}$ be an associated minimizer of $\mathbf{E}_{c}$ in $X_{T}$
given by Lemma 5.7. Then, if
$t^{\pm}:=t^{\pm}(\mathbf{U}_{c,T},\mathcal{E}_{\max}^{\pm})$ are as in
(5.86), (5.87) it holds that

$\forall t\leq t^{-},\hskip
5.69054pt\mathrm{dist}_{\mathscr{L}}(\mathbf{U}_{c,T}(t),\mathscr{F}^{-})<\frac{r_{0}^{-}}{2}$
(5.160)

and

$\forall t\geq t^{+},\hskip
5.69054pt\mathrm{dist}_{\mathscr{L}}(\mathbf{U}_{c,T}(t),\mathscr{F}^{+})<\frac{r_{0}^{+}}{2},$
(5.161)

Moreover, we have

$\forall t\geq t^{-},\hskip 5.69054pt\mathcal{E}(\mathbf{U}_{c,T}(t))\geq 0.$
(5.162)

Finally, we have that if $\mathbf{E}_{c}(\mathbf{U}_{c,T})\leq 0$, then

$0<t^{+}-t^{-}\leq\mathsf{T}_{\star}(c),$ (5.163)

where $\mathsf{T}_{\star}(c)$ is as in (5.107). In particular, the function

$c\in(0,+\infty)\to\mathsf{T}_{\star}(c)$ (5.164)

is continuous.

###### Proof.

If we assume by contradiction that (5.161) does not hold, then we necessarily
have that there exists $\tilde{t}^{-}<t^{-}$ such that (5.100) holds.
Proposition 5.1 implies then the existence of $\tilde{U}\in X_{T}$ such that
$\mathbf{E}_{c}(\tilde{U})<\mathbf{E}_{c}(\mathbf{U}_{c,T})=\mathbf{m}_{c,T}$,
a contradiction. Therefore, (5.161) holds. Similarly, we can show that (5.160)
also holds. Finally, in order to establish (5.162), we argue as in the proof
Proposition 5.1. Indeed, due to the definition of $t^{-}$, we have that for
$t\geq t^{-}$ it holds that either

$\mathcal{E}(\mathbf{U}_{c,T}(t))\geq a+\mathcal{E}_{\max}^{-}>0$ (5.165)

(which is (H7’)) or

$\mathrm{dist}_{\mathscr{L}}(\mathbf{U}_{c,T}(t),\mathscr{F}^{-})\geq\frac{r_{0}^{-}}{2}$
(5.166)

which by (4.38) in (H7’) implies that $\mathcal{E}(\mathbf{U}_{c,T}(t))\geq
0$. Therefore, (5.162) holds and the proof is concluded. ∎

Moreover, Lemma 5.8 applies to $\mathbf{V}_{c,T}$ as follows:

######  Corollary 5.2.

Assume that (H3’), (H4’), (H5’) and (H7’) hold. Let $c>0$ and $T\geq 1$. Let
$\mathbf{U}_{c,T}$ be an associated minimizer of $\mathbf{E}_{c}$ in $X_{T}$
given by Lemma 5.7. Then, if
$t^{\pm}:=t^{\pm}(\mathbf{U}_{c,T},\mathcal{E}_{\max}^{\pm})$ are as in
(5.86), (5.87) it holds that there exists $\delta_{c,T}>0$ such that the set

$S_{c,T}:=(-\infty,t^{-}+\delta_{c,T})\cup(-T,T)\cup(t^{+}-\delta_{c,T},+\infty)$
(5.167)

is such that $\mathbf{U}_{c,T}\in\mathcal{A}(S_{c,T})$ (see (4.29)) and

$\mathbf{U}_{c,T}^{\prime\prime}-D_{\mathscr{L}}\mathcal{E}(\mathbf{U}_{c,T})=-c\mathbf{U}_{c,T}^{\prime}\mbox{
in }S_{c,T}.$ (5.168)

The proof of 5.2 is obtained in a straightforward manner by combining Lemma
5.8 with the informations given by Corollary 5.1. Notice that 5.2 implies that
constrained solutions are picewise solutions and, in particular, they solve
the equation for times with large absolute value.

###  5.7 Existence of the unconstrained solutions

We now establish the existence of the unconstrained solutions making use of
the previous comparison results. As in [6] and [7], we define the set

$\mathcal{C}:=\\{c>0:\exists T\geq 1\mbox{ and }U\in X_{T}\mbox{ such that
}\mathbf{E}_{c}(U)<0\\}.$ (5.169)

We first prove some important properties for $\mathcal{C}$ which are the same
to Lemma 2.12 in [6] and Lemma 27 in [7]:

######  Lemma 5.11.

Assume that (H3’), (H4’) and (H5’) hold. Let $\mathcal{C}$ be the set defined
in (5.169). Then, $\mathcal{C}$ is open and non-empty. Moreover, if we assume
that (H7’) holds, then $\mathcal{C}$ is also bounded with

$\sup\mathcal{C}\leq\frac{\sqrt{-2a}}{d_{0}},$ (5.170)

where $d_{0}$ was defined in (4.36).

###### Proof.

Firstly, we show that $\mathcal{C}\not=\emptyset$. For that purpose, consider
the function $\Psi$ introduced in (5.16). Consider the function

$f:c\in(0,+\infty)\to
e^{-c}\left(\frac{a}{c}+e^{2c}\int_{-1}^{1}\left(\frac{\lVert\Psi^{\prime}(t)\rVert^{2}_{\mathscr{L}}}{2}+\mathcal{E}(\Psi(t))\right)dt\right)\in{\mathbb{R}},$
(5.171)

which is well defined by Lemma 5.4. We have that for all $c>0$

$\mathbf{E}_{c}(\Psi)=\frac{-a}{c}e^{-c}+\int_{-1}^{1}\left(\frac{\lVert\Psi^{\prime}(t)\rVert^{2}_{\mathscr{L}}}{2}+\mathcal{E}(\Psi(t))\right)e^{ct}dt\leq
f(c)$ (5.172)

and $f$ is a continuous function such that $\lim_{c\to 0}f(c)=-\infty$ because
$a<0$. Moreover, we have that for all $c>0$,

$f^{\prime}(c)=-e^{-c}a+ce^{2c}\int_{-1}^{1}\left(\frac{\lVert\Psi^{\prime}(t)\rVert^{2}_{\mathscr{L}}}{2}+\mathcal{E}(\Psi(t))\right)dt>0$
(5.173)

and $\lim_{c\to+\infty}f(c)=+\infty$. Therefore, there exists a unique
$c_{\Psi}>0$ such that $f(c_{\Psi})=0$ and for all $c<c_{\Psi}$ we have
$\mathbf{E}_{c}(\Psi)<0$ by (5.172). Therefore,
$(0,c_{\Psi})\subset\mathcal{C}$, meaning that $\mathcal{C}\not=\emptyset$ as
we wanted to show.

We next prove that $\mathcal{C}$ is open. Let $c\in\mathcal{C}$, we have that
$\mathbf{E}_{c}(\mathbf{U}_{c,T})<0$, where $\mathbf{U}_{c,T}$ is a minimizer
of $\mathbf{E}_{c}$ in $X_{T}$ given by Lemma 5.7. By (5.11) in Lemma 5.3,
there exists a sequence $(t_{n})_{n\in{\mathbb{N}}}$ in $[T,+\infty)$ such
that $t_{n}\to+\infty$ and

$\lim_{n\to\infty}\mathcal{E}(\mathbf{U}_{c,T}(t_{n}))=0.$ (5.174)

Up to subsequences, we have that for all $n\in{\mathbb{N}}$,
$\mathbf{U}_{c,T}(t_{n})\in\mathscr{F}^{+}_{r_{0}^{+}}$. Hence, we can define

$\mathbf{U}_{c,T}^{n}(s):=\begin{cases}\mathbf{U}_{c,T}(s)&\mbox{ if }s\leq
t,\\\
(1+t_{n}-s)\mathbf{U}_{c,T}(t_{n})+(s-t_{n})P^{+}(\mathbf{U}_{c,T}(t_{n}))&\mbox{
if }t_{n}\leq s\leq t_{n}+1,\\\ P^{+}(\mathbf{U}_{c,T}(t_{n}))&\mbox{ if
}t_{n}+1\leq s.\end{cases}$ (5.175)

We have that for all $n\in{\mathbb{N}}$,

$\displaystyle\mathbf{E}_{c}(\mathbf{U}^{n}_{c,T}(s))$
$\displaystyle=\int_{-\infty}^{t_{n}}\mathbf{e}_{c}(\mathbf{U}_{c,T}(s))ds+\frac{\lVert\mathbf{U}_{c,T}(t_{n})-P^{+}(\mathbf{U}_{c,T}(t_{n}))\rVert_{\mathscr{L}}^{2}}{2}$
(5.176)
$\displaystyle+\int_{t_{n}}^{t_{n}+1}\mathcal{E}(\mathbf{U}^{n}_{c,T}(s))ds$
(5.177)
$\displaystyle\leq\mathbf{E}_{c}(\mathbf{U}_{c,T})+\frac{\lVert\mathbf{U}_{c,T}(t_{n})-P^{+}(\mathbf{U}_{c,T}(t_{n}))\rVert_{\mathscr{L}}^{2}}{2}$
(5.178)
$\displaystyle+\int_{t_{n}}^{t_{n}+1}\mathcal{E}((1+t_{n}-s)\mathbf{U}_{c,T}(t_{n})+(s-t_{n})P^{+}(\mathbf{U}_{c,T}(t_{n})))ds,$
(5.179)

where we have used that $t_{n}\geq T$ in order to obtain the inequality. Let
$\beta^{+}(\hat{r}^{+})$ be as in Lemma 5.9. Up to a subsequence, we have that
for all $n\in{\mathbb{N}}$ it holds
$\mathcal{E}(\mathbf{U}_{c,T}(t_{n}))\leq\beta^{+}(\hat{r}^{+})$. Therefore,
by Lemma 5.9 we have that for all $\lambda\in[0,1]$ and $n\in{\mathbb{N}}$ it
holds

$\mathcal{E}(\lambda\mathbf{U}_{c,T}(t_{n})+(1-\lambda)P^{+}(\mathbf{U}_{c,T}(t_{n})))\leq\mathsf{C}^{+}\mathcal{E}(\mathbf{U}_{c,T}(t_{n})),$
(5.180)

where $\mathsf{C}^{+}>0$ is independent on $n$ (see (4.34)). Plugging into
(5.176), we obtain that for all $n\in{\mathbb{N}}$ it holds

$\mathbf{E}_{c}(\mathbf{U}^{n}_{c,T}(s))\leq\mathbf{E}_{c}(\mathbf{U}_{c,T})+\frac{\lVert\mathbf{U}_{c,T}(t_{n})-P^{+}(\mathbf{U}_{c,T}(t_{n}))\rVert_{\mathscr{L}}^{2}}{2}+\mathsf{C}^{+}\mathcal{E}(\mathbf{U}_{c,T}(t_{n})).$
(5.181)

Recall that we assume Notice also that (5.12) implies in particular that

$\lim_{n\to+\infty}\lVert\mathbf{U}_{c,T}(t_{n})-P^{+}(\mathbf{U}_{c,T}(t_{n}))\rVert_{\mathscr{L}}^{2}=0$
(5.182)

which, in combination with inequalities (5.174) and (5.181) together with the
fact that $\mathbf{E}_{c}(\mathbf{U}_{c,T})<0$, gives that there exists
$N\in{\mathbb{N}}$ such that $\mathbf{E}_{c}(\mathbf{U}_{c,T}^{N})<0$. Since
$\mathbf{U}_{c,T}^{N}$ is constant in $[t_{N}+1,+\infty)$, we have that for
all $\tilde{c}>0$, $\mathbf{E}_{\tilde{c}}(\mathbf{U}_{c,T}^{N})<+\infty$.
Therefore, by Lemma 5.5 we have that

$\tilde{c}\in(0,+\infty)\to\mathbf{E}_{\tilde{c}}(\mathbf{U}_{c,T}^{N})\in{\mathbb{R}}$
(5.183)

is well defined and continuous. Therefore, we can find some $\delta>0$ such
that for all $\tilde{c}\in(c-\delta,c+\delta)$,
$\mathbf{E}_{\tilde{c}}(\mathbf{U}_{c,T}^{N})<0$. As a consequence, we have
that $(c-\delta,c+\delta)\subset\mathcal{C}$, which shows that $\mathcal{C}$
is open.

We now assume that (H7’) holds and we use it to establish the bound (5.170).
In particular, we can apply Proposition 5.1. Let $c>0$ and $T\geq 1$ be such
that $\mathbf{E}_{c}(\mathbf{U}_{c,T})<0$ with $\mathbf{U}_{c,T}\in X_{T}$ a
minimizing solution given by Lemma 5.7. Let
$t^{\pm}:=t^{\pm}(\mathbf{U}_{c,T},\mathcal{E}_{\max}^{\pm})$ be as in (5.86),
(5.87). Inequality (5.163) in Proposition 5.1 implies that $t^{-}<t^{+}$.
Recall the definition of $d_{0}$ in (4.36) and the fact that
$\mathbf{U}_{c,T}(t^{\pm})\in\mathscr{F}^{\pm}_{r_{0}^{\pm}/2}$. Those facts
imply

$d_{0}\leq\lVert\mathbf{U}_{c,T}(t^{+})-\mathbf{U}_{c,T}(t^{-})\rVert_{\mathscr{L}}.$
(5.184)

Since (H7’) holds, we can use (5.162) in Corollary 5.1 to obtain

$\displaystyle\lVert\mathbf{U}_{c,T}(t^{+})-\mathbf{U}_{c,T}(t^{-})\rVert_{\mathscr{L}}^{2}$
$\displaystyle\leq
2\int_{{\mathbb{R}}}\frac{\lVert\mathbf{U}_{c,T}^{\prime}(t)\rVert^{2}}{2}e^{ct}dt\left(\frac{e^{-ct^{-}}-e^{-ct^{+}}}{c}\right)$
(5.185) $\displaystyle\leq
2\left(\mathbf{E}_{c}(\mathbf{U}_{c,T})-\frac{a}{c}e^{ct^{-}}\right)\left(\frac{e^{-ct^{-}}-e^{-ct^{+}}}{c}\right).$
(5.186)

Using now that $\mathbf{E}_{c}(\mathbf{U}_{c,T})\leq 0$, the fact that
$t^{-}<t^{+}$ and (5.184), the inequality above becomes

$d_{0}^{2}\leq-2a\frac{1-e^{c(t^{-}-t^{+})}}{c^{2}}\leq\frac{-2a}{c^{2}},$
(5.187)

so that (5.170) follows. ∎

We now have all the ingredients for establishing the existence of the
unconstrained solutions:

######  Proposition 5.2.

Assume that (H3’), (H4’), (H5’), (H6’) and (H7’) hold. Let
$\overline{c}\in\partial(\mathcal{C})\cap(0,+\infty)$, where
$\partial(\mathcal{C})$ stands for the boundary of the set $\mathcal{C}$
defined in (5.169). Then, there exists $\overline{T}\geq 1$ such that
$\mathbf{m}_{\overline{c},\overline{T}}=0$
($\mathbf{m}_{\overline{c},\overline{T}}$ as in (5.15)) and
$\overline{\mathbf{U}}\in X_{\overline{T}}$ an associated minimizer of
$\mathbf{E}_{\overline{c}}$ in $X_{\overline{T}}$ which does not saturate the
constraints, i. e.

$\forall t\geq\overline{T},\hskip
5.69054pt\mathrm{dist}_{\mathscr{L}}(\overline{\mathbf{U}}(t),\mathscr{F}^{+})<\frac{r_{0}^{+}}{2}$
(5.188)

and

$\forall t\leq-\overline{T},\hskip
5.69054pt\mathrm{dist}_{\mathscr{L}}(\overline{\mathbf{U}}(t),\mathscr{F}^{-})<\frac{r_{0}^{-}}{2}.$
(5.189)

Moreover, $\overline{\mathbf{U}}\in\mathcal{A}({\mathbb{R}})$ and the pair
$(\overline{c},\overline{\mathbf{U}})$ solves (4.1).

######  Remark 5.3.

Notice that Lemma 5.11 implies that (under the necessary assumptions) the set
$\mathcal{C}$ is bounded, meaning that
$\partial(\mathcal{C})\cap(0,+\infty)\not=\emptyset$. Such a fact, in
combination with Proposition 5.2 shows the existence of the unconstrained
solutions.

###### Proof of Proposition 5.2.

By Lemma 5.11, we have that $\mathcal{C}\not=\emptyset$ is open, which implies
that $\partial(\mathcal{C})\subset{\mathbb{R}}\setminus\mathcal{C}$.
Therefore, we have $\overline{c}\not\in\mathcal{C}$. Recall that due to the
definition of $\mathcal{C}$ in (5.169), we have that

$\forall T\geq 1,\hskip 5.69054pt\mathbf{m}_{\overline{c},T}\geq 0.$ (5.190)

The definition of the boundary allows to consider a sequence
$(c_{n})_{n\in{\mathbb{N}}}$ contained in $\mathcal{C}$ such that
$c_{n}\to\overline{c}$. Then, for each $n\in{\mathbb{N}}$ there exists
$T_{n}\geq 1$ such that $\mathbf{E}_{c_{n}}(\mathbf{U}_{c_{n},T_{n}})<0$,
where, for each $n\in{\mathbb{N}}$, $\mathbf{U}_{c_{n},T_{n}}$ is a minimizer
of $\mathbf{E}_{c_{n}}$ in $X_{T_{n}}$. For each $n\in{\mathbb{N}}$, set
$t_{n}^{\pm}:=t^{+}(\mathbf{U}_{c_{n},T_{n}},\mathcal{E}_{\max}^{\pm})$ as in
(5.86), (5.87). Using (5.163) in Corollary 5.1 we have that that

$\forall n\in{\mathbb{N}},\hskip
5.69054pt0<t^{+}_{n}-t^{-}_{n}\leq\mathsf{T}_{\star}(c_{n}),$ (5.191)

and the function

$c\in(0,+\infty)\to\mathsf{T}_{\star}(c)\in(0,+\infty)$ (5.192)

is continuous. Since the sequence $(c_{n})_{n\in{\mathbb{N}}}$ is bounded, we
have that

$\mathsf{T}_{\star}:=\max\left\\{1,\sup_{n\in{\mathbb{N}}}\mathsf{T}_{\star}(c_{n})\right\\}<+\infty$
(5.193)

and

$\forall n\in{\mathbb{N}},\hskip
5.69054pt0<t^{+}_{n}-t^{-}_{n}\leq\mathsf{T}_{\star},$ (5.194)

so that we have a bound on $(t^{+}_{n}-t^{-}_{n})_{n\in{\mathbb{N}}}$.
Moreover, (5.160) and (5.161) in Corollary 5.1 imply

$\forall n\in{\mathbb{N}},\forall t\geq t^{+}_{n},\hskip
5.69054pt\mathrm{dist}_{\mathscr{L}}(\mathbf{U}_{c_{n},T_{n}}(t),\mathscr{F}^{+})<\frac{r_{0}^{+}}{2}$
(5.195)

and

$\forall n\in{\mathbb{N}},\forall t\leq t^{-}_{n},\hskip
5.69054pt\mathrm{dist}_{\mathscr{L}}(\mathbf{U}_{c_{n},T_{n}}(t),\mathscr{F}^{-})<\frac{r_{0}^{-}}{2}.$
(5.196)

For each $n\in{\mathbb{N}}$, define the function
$\mathbf{U}_{c_{n},T_{n}}^{t_{n}^{+}}:=\mathbf{U}_{c_{n},T_{n}}(\cdot+t_{n}^{+})$.
Then, (5.194) implies that (5.195) and (5.196) write as

$\forall n\in{\mathbb{N}},\forall t\geq 0,\hskip
5.69054pt\mathrm{dist}_{\mathscr{L}}(\mathbf{U}_{c_{n},T_{n}}^{t_{n}^{+}}(t),\mathscr{F}^{-})<\frac{r_{0}^{-}}{2}.$
(5.197)

and

$\forall n\in{\mathbb{N}},\forall t\leq-\mathsf{T}_{\star},\hskip
5.69054pt\mathrm{dist}_{\mathscr{L}}(\mathbf{U}_{c_{n},T_{n}}(t),\mathscr{F}^{-})<\frac{r_{0}^{-}}{2}.$
(5.198)

so that for all $n\in{\mathbb{N}}$ we have
$\mathbf{U}_{c_{n},T_{n}}^{t_{n}^{+}}\in X_{\mathsf{T}_{\star}}$. Moreover, a
computation shows

$\forall n\in{\mathbb{N}},\hskip
5.69054pt\mathbf{E}_{c_{n}}(\mathbf{U}_{c_{n},T_{n}}^{t_{n}^{+}})=e^{-c_{n}t_{n}^{+}}\mathbf{E}_{c_{n}}(\mathbf{U}_{c_{n},T_{n}})<0.$
(5.199)

Therefore, if we apply Lemma 5.6 with sequence of speeds
$(c_{n})_{n\in{\mathbb{N}}}$ and the sequence
$(\mathbf{U}_{c_{n},T_{n}}^{t_{n}^{+}})_{n\in{\mathbb{N}}}$ in
$X_{\mathsf{T}_{\star}}$, we obtain $\overline{\mathbf{U}}\in
X_{\mathsf{T}_{\star}}$ such that

$\mathbf{E}_{\overline{c}}(\overline{\mathbf{U}})\leq\liminf_{n\to\infty}\mathbf{E}_{c_{n}}(\mathbf{U}_{c_{n},T_{n}}^{t_{n}^{+}})\leq
0,$ (5.200)

which in combination with (5.190) implies that
$\mathbf{m}_{\overline{c},\mathsf{T}_{\star}}=0$. Therefore, we have
$\mathbf{E}_{\overline{c}}(\overline{\mathbf{U}})=0$, so that
$\overline{\mathbf{U}}$ is a minimizer of $\mathbf{E}_{\overline{c}}$ in
$X_{\mathsf{T}_{\star}}$. Set
$t^{\pm}_{\star}:=t^{\pm}(\overline{\mathbf{U}},\mathcal{E}_{\max}^{\pm})$ as
in (5.86), (5.87). Invoking (5.190) and Corollary 5.2, we obtain that for all
$T\geq 1$ such that $\mathbf{m}_{\overline{c},T}=0$ and
$\overline{\mathbf{U}}\in X_{T}$, we have
$\overline{\mathbf{U}}\in\mathcal{A}(S_{\overline{c},T})$ with

$S_{\overline{c},T}:=(-\infty,t^{-}_{\star}+\delta_{\star}(T))\cup(-T,T)\cup(t^{+}_{\star}-\delta_{\star}(T),+\infty)$
(5.201)

for some $\delta_{\star}(T)>0$ and

$\overline{\mathbf{U}}^{\prime\prime}-D_{\mathscr{L}}\mathcal{E}(\overline{\mathbf{U}})=-\overline{c}\overline{\mathbf{U}}^{\prime}\mbox{
in }S_{\overline{c},T}.$ (5.202)

Moreover, using (5.160) and (5.161) in Corollary 5.1, we obtain as before that

$\forall t\geq t^{+}_{\star},\hskip
5.69054pt\mathrm{dist}_{\mathscr{L}}(\overline{\mathbf{U}}(t),\mathscr{F}^{+})<\frac{r_{0}^{+}}{2}$
(5.203)

and

$\forall t\leq t^{-}_{\star},\hskip
5.69054pt\mathrm{dist}_{\mathscr{L}}(\overline{\mathbf{U}}(t),\mathscr{F}^{-})<\frac{r_{0}^{-}}{2}.$
(5.204)

Therefore, if we set $\overline{T}=\max\\{1,t^{+}_{\star},-t^{-}_{\star}\\}$,
then (5.203) and (5.204) imply that $\overline{\mathbf{U}}\in
X_{\overline{T}}$ and that (5.188), (5.189) hold. Moreover, we have that
$\mathbf{E}_{\overline{c}}(\overline{\mathbf{U}})=0$, so that
$\overline{\mathbf{U}}$ is a minimizer of $\mathbf{E}_{\overline{c}}$ in
$X_{\overline{T}}$ by (5.190). Therefore, we obtain that
$\overline{\mathbf{U}}\in\mathcal{A}(S_{\overline{c},\overline{T}})$ and

$\overline{\mathbf{U}}^{\prime\prime}-D_{\mathscr{L}}\mathcal{E}(\overline{\mathbf{U}})=-\overline{c}\overline{\mathbf{U}}^{\prime}\mbox{
in }S_{\overline{c},\overline{T}},$ (5.205)

with $S_{\overline{c},\overline{T}}$ as in (5.201). The choice of
$\overline{T}$ implies that $S_{\overline{c},\overline{T}}={\mathbb{R}}$.
Therefore, $\overline{\mathbf{U}}\in\mathcal{A}({\mathbb{R}})$ and
$(\overline{c},\overline{\mathbf{U}})$ solves (4.1), which finishes the proof.
∎

Notice that our Proposition 5.2 follows very similar lines than the analogous
results in [6] and [7].

###  5.8 Uniqueness of the speed

The precise statement of the uniqueness result is as follows:

######  Proposition 5.3.

Assume that (H6’) holds. Let $X$ be the set defined in (4.39). Let
$(c_{1},c_{2})\in(0,+\infty)^{2}$ be such that there exist $\mathbf{U}_{1}$
and $\mathbf{U}_{2}$ in $X\cap\mathcal{A}({\mathbb{R}})$ such that
$(c_{1},\mathbf{U}_{1})$ and $(c_{2},\mathbf{U}_{2})$ solve (4.1) and for each
$i\in\\{1,2\\}$, $\mathbf{E}_{c_{i}}(\mathbf{U}_{i})<+\infty$. Assume moreover
that

$\forall i\in\\{1,2\\},\forall j\in\\{1,2\\}\setminus\\{i\\},\hskip
5.69054pt\mathbf{E}_{c_{i}}(\mathbf{U}_{j})\geq 0.$ (5.206)

Then, we have $c_{1}=c_{2}$.

###### Proof.

We prove the result by contradiction. Hence, we can assume without loss of
generality that $c_{1}<c_{2}$. A direct computation shows that for every
$(c,U)\in(0,+\infty)\times(X\cap\mathcal{A}({\mathbb{R}}))$ a solution to
(4.1), we have

$\forall t\in{\mathbb{R}},\hskip 5.69054pt\frac{\lVert
U^{\prime}(t)\rVert_{\mathscr{L}}^{2}}{2}+\mathcal{E}(U(t))=e^{-ct}\left(\frac{e^{ct}}{c}\left(\mathcal{E}(U(t))-\frac{\lVert
U^{\prime}(t)\rVert_{\mathscr{L}}^{2}}{2}\right)\right)^{\prime}.$ (5.207)

Replacing $(c_{2},U_{2})$ in (5.207) and multiplying for each
$t\in{\mathbb{R}}$ by $e^{c_{1}t}$, computations show that

$\displaystyle\forall t_{1}<t_{2},\hskip
5.69054ptc_{1}\mathbf{E}_{c_{1}}(\mathbf{U}_{2};(t_{1},t_{2}))$
$\displaystyle=(c_{1}-c_{2})\int_{t_{1}}^{t_{2}}\lVert\mathbf{U}_{2}^{\prime}(t)\rVert_{\mathscr{L}}^{2}e^{c_{1}t}dt$
(5.208)
$\displaystyle+\left[e^{c_{1}t}\left(\mathcal{E}(\mathbf{U}_{2}(t))-\frac{\lVert\mathbf{U}_{2}^{\prime}(t)\rVert_{\mathscr{L}}^{2}}{2}\right)\right]^{t_{2}}_{t_{1}}.$
(5.209)

Notice now that the definition of $X$ in (4.39) implies that

$X=\bigcup_{T\geq 1}X_{T},$ (5.210)

which means that there exists $T\geq 1$ such that $\mathbf{U}_{2}\in X_{T}$.
Combining then Lemma 5.1 and the fact that
$\mathbf{E}_{c_{2}}(\mathbf{U}_{2})<+\infty$, we get that
$\mathbf{e}_{c_{2}}(\mathbf{U}_{2}(\cdot))\in L^{1}({\mathbb{R}})$. Therefore,
we can find two sequences $(t_{n}^{+})_{n\in{\mathbb{N}}}$ and
$(t_{n}^{-})_{n\in{\mathbb{N}}}$ such that $t_{n}^{\pm}\to\pm\infty$ and

$\lim_{n\to\infty}\mathbf{e}_{c_{2}}(\mathbf{U}_{2}(t_{n}^{\pm}))=0.$ (5.211)

Since we have $c_{1}<c_{2}$, it holds

$\forall t\in{\mathbb{R}},\hskip
5.69054pte^{c_{1}t}\left\lvert\mathcal{E}(\mathbf{U}_{2}(t))-\frac{\lVert\mathbf{U}_{2}^{\prime}(t)\rVert_{\mathscr{L}}^{2}}{2}\right\rvert\leq\lvert\mathbf{e}_{c_{2}}(\mathbf{U}_{2}(t))\rvert$
(5.212)

which in combination with (5.211) implies

$\lim_{n\to\infty}e^{c_{1}t_{n}^{\pm}}\left(\mathcal{E}(\mathbf{U}_{2}(t_{n}^{\pm}))-\frac{\lVert\mathbf{U}_{2}^{\prime}(t_{n}^{\pm})\rVert_{\mathscr{L}}^{2}}{2}\right)=0.$
(5.213)

Therefore, if we replace for each $n\in{\mathbb{N}}$ in (5.208) with
$a=t_{n}^{-}$ and $b=t_{n}^{+}$, we can then pass to the limit (5.213) and
obtain

$c_{1}\mathbf{E}_{c_{1}}(\mathbf{U}_{2})=(c_{1}-c_{2})\int_{{\mathbb{R}}}\lVert\mathbf{U}_{2}^{\prime}(t)\rVert_{\mathscr{L}}^{2}e^{c_{1}t}dt<0$
(5.214)

because we assume $c_{1}<c_{2}$. However, by (5.206) we have
$\mathbf{E}_{c_{1}}(\mathbf{U}_{2})\geq 0$, which is a contradiction. ∎

######  Remark 5.4.

Again, the proof of Proposition 5.3 is essentially a direct adaptation of that
given in [6] and [7]. Our hypothesis are slightly weaker, since we only assume
that the solutions have finite energies and (5.206), while in [6] and [7] it
is assumed that the solutions are global minimizers of the corresponding
energy functionals. Notice also that (H7’) is not needed for proving
Proposition 5.3, which holds in a more general setting.

Proposition 5.3 along with Proposition 5.2 allows to show that the set
$\mathcal{C}$ defined in (5.169) is in fact an open interval:

######  Corollary 5.3.

Assume that (H3’), (H4’), (H5’), (H6’) and (H7’) hold. Let

$c(\mathcal{C}):=\sup\mathcal{C}.$ (5.215)

Then, we have $\mathcal{C}=(0,c(\mathcal{C}))$.

###### Proof.

The statement of the result is equivalent to show that

$\partial(\mathcal{C})\cap(0,+\infty)=\\{c(\mathcal{C})\\}$ (5.216)

The quantity $c(\mathcal{C})$ is well defined in $(0,+\infty)$ because
$\mathcal{C}$ is non-empty and bounded by Lemma 5.11. Therefore, we have
$c(\mathcal{C})\in\partial(\mathcal{C})\cap(0,+\infty)$ because $\mathcal{C}$
is open, so it does not contain its limit points. By Proposition 5.2, we find
$\mathbf{U}^{\mathcal{C}}\in X$ such that
$(c(\mathcal{C}),\mathbf{U}^{\mathcal{C}})$ solves (4.1). Let now
$\overline{c}\in\partial(\mathcal{C})\cap(0,+\infty)$. If we show that
$\overline{c}=c(\mathcal{C})$, the proof will be finished. Applying
Proposition 5.2 with $\overline{c}$, we find $\overline{\mathbf{U}}\in X$ such
that $(\overline{c},\overline{\mathbf{U}})$ solves (4.1). Proposition 5.2,
along with the fact that $\overline{c}$ and $c(\mathcal{C})$ do not belong to
$\mathcal{C}$, also implies that

$\inf_{U\in
X}\mathbf{E}_{\overline{c}}(U)=\mathbf{E}_{\overline{c}}(\overline{\mathbf{U}})=0=\mathbf{E}_{c(\mathcal{C})}(\mathbf{U}^{\mathcal{C}})=\inf_{U\in
X}\mathbf{E}_{c(\mathcal{C})}(U)$ (5.217)

so that

$\mathbf{E}_{c(\mathcal{C})}(\overline{\mathbf{U}})\geq 0\mbox{ and
}\mathbf{E}_{\overline{c}}(\mathbf{U}^{\mathcal{C}})\geq 0$ (5.218)

meaning that we can apply Proposition 5.3 to
$(c(\mathcal{C}),\mathbf{U}^{\mathcal{C}})$,
$(\overline{c},\overline{\mathbf{U}})$. As a consequence, we have
$\overline{c}=c(\mathcal{C})$, which concludes the proof. ∎

###  5.9 Proof of Theorem 5 completed

All the elements of the proof of Theorem 5 are already present in previous
result. Indeed, Proposition 5.2 along with Corollary 5.3 implies the existence
of $(c^{\star},\mathbf{U})\in(0,+\infty)\times X_{T^{\star}}$ a solution to
(4.1) with $c^{\star}=c(\mathcal{C})$. Conditions (4.2) and (4.3) are
satisfied due to the fact that $\mathbf{U}\in X_{T^{\star}}$. The statement
regarding the uniqueness of the speed $c^{\star}$ follows from Proposition
5.3. Finally, we have that (5.12) is exactly the exponential rate of
convergence (4.42), which completes the proof.

∎

###  5.10 Asymptotic behavior of the constrained solutions at $-\infty$

As it has been pointed out before, almost nothing can be said about the
behavior of arbitrary function in $X_{T}$ at $-\infty$. However, it turns out
that constrained minimizers converge exponentially at $-\infty$ with respect
to the $\mathscr{L}$-norm provided that the speed fulfills an explicit upper
bound, see Proposition 5.4. Such an upper bound also allows to establish some
other properties. Once Proposition 5.4 will have been established, we will be
able to complete the proofs of Theorems 6 and 7. The results of this
subsection are obtained by combining ideas from Smyrnelis [69], Alikakos and
Katzourakis [7], Alikakos, Fusco and Smyrnelis [6]. It is worth to point out
that the arguments we present here strongly rely on the fact that the
solutions considered are minimizers and that we do not expect them to hold for
more general critical points.

We begin by showing the following preliminary result, which follows by a
direct computation:

######  Lemma 5.12.

Assume that (H6’) holds. Let $c>0$, $t_{1}<t_{2}$ and
$U\in\mathcal{A}((t_{1},t_{2}))$ such that

$U^{\prime\prime}-D_{\mathscr{L}}\mathcal{E}(U)=-cU^{\prime}\mbox{ in
}(t_{1},t_{2}).$ (5.219)

Then, we have the formula

$\forall t\in(t_{1},t_{2}),\hskip
5.69054pt\frac{d}{dt}\left(\mathcal{E}(U(t))-\frac{\lVert
U^{\prime}(t)\rVert_{\mathscr{L}}^{2}}{2}\right)=c\lVert
U^{\prime}(t)\rVert^{2}_{\mathscr{L}}.$ (5.220)

Lemma 5.12 gives the following pointwise bounds for constrained solutions:

######  Lemma 5.13.

Assume that (H3’), (H4’), (H5’), (H6’) and (H7’) hold. Let $\mathbf{U}_{c,T}$
be a constrained solution given by Lemma 5.7 and
$t^{-}:=t^{-}(\mathbf{U}_{c,T},\mathcal{E}_{\max}^{-})$ be as in (5.86). Then
for all $t<t^{-}$ we have the inequality

$\frac{\lVert\mathbf{U}_{c,T}^{\prime}(t)\rVert_{\mathscr{L}}^{2}}{2}\leq\mathcal{E}(\mathbf{U}_{c,T}(t))-a.$
(5.221)

Similarly, it holds that for all $t>t^{+}$

$\mathcal{E}(\mathbf{U}_{c,T}(t))\leq\frac{\lVert\mathbf{U}_{c,T}^{\prime}(t)\rVert_{\mathscr{L}}^{2}}{2},$
(5.222)

where $t^{+}:=t^{+}(\mathbf{U}_{c,T},\mathcal{E}_{\max}^{+})$ is as in (5.87).

###### Proof.

Notice that (5.168) in Corollary 5.2 implies that $\mathbf{U}_{c,T}$ solves

$\mathbf{U}_{c,T}^{\prime\prime}-D_{\mathscr{L}}\mathcal{E}(\mathbf{U}_{c,T})=-c\mathbf{U}_{c,T}^{\prime}\mbox{
in }(-\infty,t^{-}).$ (5.223)

Therefore, the function

$f_{c,T}:t\in(-\infty,t^{-}]\to
e^{ct}\left(\mathcal{E}(\mathbf{U}_{c,T}(t))-a-\frac{\lVert\mathbf{U}_{c,T}^{\prime}(t)\rVert_{\mathscr{L}}^{2}}{2}\right),$
(5.224)

is $C^{1}$ and we clearly have that $f_{c,T}\in L^{1}((-\infty,t^{-}])$. By
(5.220) in Lemma 5.12, we have

$\forall t\in(-\infty,t^{-}),\hskip
5.69054ptf_{c,T}^{\prime}(t)=cf_{c,T}(t)+ce^{ct}\lVert\mathbf{U}^{\prime}_{c,T}(t)\rVert_{\mathscr{L}}^{2}\geq
0,$ (5.225)

and we also have $f_{c,T}^{\prime}\in L^{1}((-\infty,t^{-}))$. Therefore, it
holds that

$\lim_{t\to-\infty}f_{c,T}(t)=0.$ (5.226)

Fix $t_{1}<t_{2}\leq t^{-}$. Integrating (5.225) in $[t_{1},t_{2}]$ we get

$f_{c,T}(t_{2})\geq f_{c,T}(t_{1}),$ (5.227)

which in combination with (5.226) gives

$\forall t<t^{-},\hskip 5.69054ptf_{c,T}(t)\geq 0,$ (5.228)

which is (5.221). Inequality (5.222) is obtained in an identical fashion. ∎

We conclude this subsection by proving the exponential convergence result,
which is inspired by the ideas in Proof of (28) in Smyrnelis [69].

######  Proposition 5.4.

Assume that (H3’), (H4’), (H5’), (H6’) and (H7’) hold. Let $c>0$ and $T\geq
1$. Assume moreover that $c<\gamma^{-}$, where $\gamma^{-}$ is defined in
(4.35). Let $\mathbf{U}_{c,T}$ be a constrained solution given by Lemma 5.7.
Then, there exists $\overline{M}^{-}>0$ such that for all
$\varepsilon\in(0,\gamma^{-}-c)$ and $t\in{\mathbb{R}}$ it holds

$\int_{-\infty}^{t}(\mathcal{E}(\mathbf{U}_{c,T}(s))-a)e^{-\varepsilon
s}ds\leq\overline{M}^{-}e^{(\gamma^{-}-c-\varepsilon)t}.$ (5.229)

Furthermore, there exist $M^{-}>0$ and
$\mathbf{v}_{c,T}^{-}\in\mathscr{F}^{-}$ such that for all $t\in{\mathbb{R}}$

$\lVert\mathbf{U}_{c,T}(t)-\mathbf{v}_{c,T}^{-}\rVert_{\mathscr{L}}^{2}\leq
M^{-}e^{(\gamma^{-}-c)t}.$ (5.230)

###### Proof.

Let $t^{-}:=t^{-}(\mathbf{U}_{c,T},\mathcal{E}_{\max}^{-})$ be as in (5.86).
By applying (5.160) in Corollary 5.1, we obtain that for all $t\leq t^{-}$,
$\mathbf{U}_{c,T}(t)\in\mathscr{F}^{-}_{r_{0}^{-}/2}$. For all $t\leq t^{-}$,
define $\mathbf{v}^{-}(t):=P^{-}(\mathbf{U}_{c,T}(t))$. Consider the function

$\tilde{U}^{-}_{t}(s):=\begin{cases}\mathbf{v}^{-}(t)&\mbox{ if }s\leq t-1,\\\
(t-s)\mathbf{v}^{-}(t)+(s-t+1)\mathbf{U}_{c,T}(t)&\mbox{ if }t-1\leq s\leq
t,\\\ \mathbf{U}_{c,T}(s)&\mbox{ if }t\leq s,\end{cases}$ (5.231)

which belongs to $X_{T}$. Therefore,

$\mathbf{E}_{c}(\mathbf{U}_{c,T})\leq\mathbf{E}_{c}(\tilde{U}_{t}^{-})$
(5.232)

and, equivalently

$\int_{-\infty}^{t}\mathbf{e}_{c}(\mathbf{U}_{c,T}(s))ds\leq\frac{a}{c}e^{ct}+\int_{t-1}^{t}\left(\frac{\lVert\mathbf{U}_{c,T}(t)-\mathbf{v}^{-}(t)\rVert_{\mathscr{L}}^{2}}{2}+(\mathcal{E}(\tilde{U}^{-}_{t}(s))-a)\right)e^{cs}ds.$
(5.233)

Using Lemma 5.9 and (4.7) in (H1’), (5.233) becomes

$\int_{-\infty}^{t}\mathbf{e}_{c}(\mathbf{U}_{c,T}(s))ds\leq\frac{a}{c}e^{ct}+(C^{-}+\mathsf{C}^{-})(\mathcal{E}(\mathbf{U}_{c,T}(t)-a)e^{ct},$
(5.234)

which gives

$\int_{-\infty}^{t}(\mathcal{E}(\mathbf{U}_{c,T}(s)-a)e^{cs}ds\leq\frac{1}{\gamma^{-}}(\mathcal{E}(\mathbf{U}_{c,T}(t)-a)e^{ct},$
(5.235)

where $\gamma^{-}$ was defined in (4.35). Define the function

$\theta^{-}_{c,T}:t\in(-\infty,t^{-}]\to\int_{-\infty}^{t}(\mathcal{E}(\mathbf{U}_{c,T}(s)-a)e^{cs}ds\in{\mathbb{R}}.$
(5.236)

By (H6’), the function $\theta_{c,T}^{-}$ defined in (5.236) verifies that for
all $t\in(-\infty,t^{-})$

$(\theta^{-}_{c,T})^{\prime}(t)=(\mathcal{E}(\mathbf{U}_{c,T}(t))-a)e^{ct}$
(5.237)

which, by (5.235) implies

$\forall t\leq t^{-},\hskip
5.69054pt\gamma^{-}\theta_{c,T}^{-}(t)\leq(\theta^{-}_{c,T})^{\prime}(t).$
(5.238)

Fix now $t\in(-\infty,t^{-})$ and assume that $\theta^{-}_{c,T}(t)>0$. The
previous inequality is equivalent to

$\gamma^{-}\leq(\ln(\theta^{-}_{c,T}(t)))^{\prime}.$ (5.239)

which, by integrating in $[t,t^{-}]$ becomes

$\gamma^{-}(t^{-}-t)\leq\ln(\theta^{-}_{c,T}(t^{-}))-\ln(\theta^{-}_{c,T}(t)),$
(5.240)

hence

$e^{\gamma^{-}(t^{-}-t)}\leq\frac{\theta^{-}_{c,T}(t^{-})}{\theta^{-}_{c,T}(t)},$
(5.241)

that is

$\theta^{-}_{c,T}(t)e^{\gamma^{-}(t^{-}-t)}\leq\theta^{-}_{c,T}(t^{-}),$
(5.242)

which clearly also holds if we drop the assumption $\theta^{-}_{c,T}(t)>0$, as
$\theta^{-}_{c,T}$ is a non-negative function. Thus, we have shown that

$\forall t\leq t^{-},\hskip
5.69054pt\theta^{-}_{c,T}(t)\leq\theta^{-}_{c,T}(t^{-})e^{-\gamma^{-}(t^{-}-t)}.$
(5.243)

Now, we have that using (5.243) we get for any fixed $t\leq t^{-}-1$,
$\varepsilon>0$ and $i\in{\mathbb{N}}$

$\displaystyle\int_{t-i-1}^{t-i}(\mathcal{E}(\mathbf{U}_{c,T}(s))-a)e^{-\varepsilon
s}ds$ $\displaystyle\leq
e^{-(c+\varepsilon)(t-i-1)}\int_{t-i-1}^{t-i}(\mathcal{E}(\mathbf{U}_{c,T}(s))-a)e^{cs}ds$
(5.244) $\displaystyle\leq
e^{-(c+\varepsilon)(t-i-1)}\theta^{-}_{c,T}(t^{-})e^{-\gamma^{-}(t^{-}-t+i)}$
(5.245)
$\displaystyle=e^{(c+\varepsilon)(1-t^{-})}\theta^{-}_{c,T}(t^{-})e^{(\gamma^{-}-c-\varepsilon)(t-t^{-})}e^{(c+\varepsilon-\gamma^{-})i}.$
(5.246)

Since we assume that $c<\gamma^{-}$, we have that by choosing any
$\varepsilon\in(0,\gamma^{-}-c)$ it holds

$\sum_{i\in{\mathbb{N}}}e^{(c+\varepsilon-\gamma^{-})i}=\frac{1}{1-e^{(c+\varepsilon-\gamma^{-})}},$
(5.247)

which, in combination with (5.244) gives (5.229) (notice that the case
$t>t^{-}-1$ presents no problem, as $e^{(\gamma^{-}-c-\varepsilon)t}$ is then
large). Therefore, by (5.221) in Lemma 5.13 we have that for all
$\varepsilon\in(0,\gamma^{-}-c)$ and $t\in{\mathbb{R}}$

$\int_{-\infty}^{t}\frac{\lVert
U^{\prime}(s)\rVert_{\mathscr{L}}^{2}}{2}e^{-\varepsilon
s}ds\leq\overline{M}^{-}e^{(\gamma^{-}-c-\varepsilon)t},$ (5.248)

which, by Cauchy-Schwarz inequality, means that

$\int_{-\infty}^{t}\lVert\mathbf{U}_{c,T}^{\prime}(s)\rVert_{\mathscr{L}}ds\leq\left(\frac{e^{\varepsilon
t}}{\varepsilon}\int_{-\infty}^{t}\lVert\mathbf{U}_{c,T}^{\prime}(s)\rVert_{\mathscr{L}}^{2}e^{-\varepsilon
s}ds\right)^{\frac{1}{2}}\leq\frac{2\overline{M}^{-}}{\varepsilon}e^{(\gamma^{-}-c)t},$
(5.249)

where we have used that $\lim_{s\to-\infty}e^{\varepsilon s}=0$, because
$\varepsilon>0$. Since $c<\gamma^{-}$, in particular inequality (5.249)
implies the existence of some $\tilde{v}^{-}\in\mathscr{L}$ such that

$\lim_{t\to-\infty}\lVert\mathbf{U}_{c,T}(t)-\tilde{v}^{-}\rVert_{\mathscr{L}}=0.$
(5.250)

Inequality (5.249) also implies that for all $\tilde{t}<t\in{\mathbb{R}}$ we
have

$\lVert\mathbf{U}_{c,T}(t)-\mathbf{U}_{c,T}(\tilde{t}^{-})\rVert_{\mathscr{L}}\leq\frac{2\overline{M}^{-}}{\varepsilon}e^{(\gamma^{-}-c)t},$
(5.251)

which by taking the limit $\tilde{t}\to-\infty$ and using (5.250) gives
(5.230), by choosing for instance
$\varepsilon=(\gamma^{-}-c)/2\in(0,\gamma^{-}-c)$ and
$M^{-}=\frac{2\overline{M}^{-}}{\varepsilon}>0$.

∎

######  Remark 5.5.

Notice that combining (5.229) in Proposition 5.4 with (5.221) in Lemma 5.13,
we obtain in particular that $\mathbf{U}_{c,T}^{\prime}\in
L^{2}({\mathbb{R}},\mathscr{L})$ provided that $c<\gamma^{-}$ (see the
statements of the results for the notations).

###  5.11 Proof of Theorem 6 completed

Assume first that (H7’) holds. Let $(c^{\star},\mathbf{U})$ be the solution to
(4.1) with conditions at infinity (4.2) and (4.42) given by Proposition 5.2
and Lemma 5.3. Since we took $c^{\star}=\sup\mathcal{C}$ with $\mathcal{C}$ as
in (5.169), inequality (5.170) in Lemma 5.11 implies that

$c^{\star}\leq\frac{\sqrt{-2a}}{d_{0}}$ (5.252)

which by (4.44) in (H8’) implies that

$c^{\star}<\gamma^{-}$ (5.253)

so that we can apply (5.230) in Proposition 5.4 to $\mathbf{U}$, as it is a
minimizer of $\mathbf{E}_{c^{\star}}$ in $X_{T^{\star}}$ for some
$T^{\star}\geq 1$. Therefore, (4.45) holds for $\mathbf{U}$, which completes
the proof.

∎

###  5.12 Proof of Theorem 7 completed

Since we assume that (H8’) holds and $\tilde{\mathbf{U}}$ is such that
$\tilde{\mathbf{U}}\in X_{T}$ for some $T\geq 1$ and
$\mathbf{E}_{c^{\star}}(\tilde{\mathbf{U}})=0$, then by Proposition 5.2, we
can apply Proposition 5.4 to $\mathbf{U}$. We recall that by Remark 5.5 we
have that $\mathbf{U}^{\prime}\in L^{2}({\mathbb{R}},\mathscr{L})$ and by
(5.229) in Proposition 5.4 we have that $\mathcal{E}\circ\mathbf{U}\in
L^{1}((-\infty,t])$ for all $t\in{\mathbb{R}}$. Therefore, we can find a
sequence $(t_{n}^{-})_{n\in{\mathbb{N}}}$ in ${\mathbb{R}}$ such that

$\lim_{n\to\infty}t_{n}^{-}=-\infty$ (5.254)

and

$\lim_{n\to\infty}\left(\mathcal{E}(\mathbf{U}(t_{n}^{-}))-a-\frac{\lVert\mathbf{U}^{\prime}(t_{n}^{-})\rVert_{\mathscr{L}}^{2}}{2}\right)=0.$
(5.255)

Similarly, since $\mathbf{E}_{c}(\mathbf{U})=0<+\infty$, we have that
$\mathcal{E}\circ\mathbf{U}\in L^{1}([t,+\infty))$ for all $t\in{\mathbb{R}}$,
which means that we can find $(t_{n}^{+})_{n\in{\mathbb{N}}}$ a sequence of
real numbers such that

$\lim_{n\to\infty}t_{n}^{+}=+\infty$ (5.256)

and

$\lim_{n\to\infty}\left(\frac{\lVert\mathbf{U}^{\prime}(t_{n}^{+})\rVert_{\mathscr{L}}^{2}}{2}-\mathcal{E}(\mathbf{U}(t_{n}^{+}))\right)=0.$
(5.257)

Taking the scalar product in $\mathscr{L}$ between equation (4.1) and
$\mathbf{U}^{\prime}$, we obtain

$\forall t\in{\mathbb{R}},\hskip
5.69054pt\langle\mathbf{U}^{\prime\prime}(t),\mathbf{U}^{\prime}(t)\rangle_{\mathscr{L}}-\langle
D_{\mathscr{L}}\mathcal{E}(\mathbf{U}(t)),\mathbf{U}^{\prime}(t)\rangle_{\mathscr{L}}=-c\lVert\mathbf{U}^{\prime}(t)\rVert_{\mathscr{L}}^{2}$
(5.258)

so that

$\forall t\in{\mathbb{R}},\hskip
5.69054pt\langle\mathbf{U}^{\prime\prime}(t),\mathbf{U}^{\prime}(t)\rangle_{\mathscr{L}}-(\mathcal{E}(\mathbf{U}(t)))^{\prime}=-c^{\star}\lVert\mathbf{U}^{\prime}(t)\rVert_{\mathscr{L}}^{2}.$
(5.259)

Fix $n\in{\mathbb{N}}$. Integrating above in $[t_{n}^{-},t_{n}^{+}]$ (which is
non-empty up to an extraction) we obtain

$\int_{t_{n}^{-}}^{t_{n}^{+}}\langle\mathbf{U}^{\prime\prime}(t),\mathbf{U}^{\prime}(t)\rangle_{\mathscr{L}}dt-\mathcal{E}(\mathbf{U}(t_{n}^{+}))+\mathcal{E}(\mathbf{U}(t_{n}^{-}))=-c^{\star}\int_{t_{n}^{-}}^{t_{n}^{+}}\lVert\mathbf{U}^{\prime}(t)\rVert_{\mathscr{L}}^{2}dt.$
(5.260)

Integrating by parts we obtain

$\int_{t_{n}^{-}}^{t_{n}^{+}}\langle\mathbf{U}^{\prime\prime}(t),\mathbf{U}^{\prime}(t)\rangle_{\mathscr{L}}dt=\lVert\mathbf{U}^{\prime}(t_{n}^{+})\rVert_{\mathscr{L}}^{2}-\lVert\mathbf{U}^{\prime}(t_{n}^{-})\rVert_{\mathscr{L}}^{2}-\int_{t_{n}^{-}}^{t_{n}^{+}}\langle\mathbf{U}^{\prime}(t),\mathbf{U}^{\prime\prime}(t)\rangle_{\mathscr{L}}dt,$
(5.261)

which means

$\int_{t_{n}^{-}}^{t_{n}^{+}}\langle\mathbf{U}^{\prime\prime}(t),\mathbf{U}^{\prime}(t)\rangle_{\mathscr{L}}dt=\frac{1}{2}(\lVert\mathbf{U}^{\prime}(t_{n}^{+})\rVert_{\mathscr{L}}^{2}-\lVert\mathbf{U}^{\prime}(t_{n}^{-})\rVert_{\mathscr{L}}^{2}).$
(5.262)

Plugging into (5.260) we obtain

$a+\left(\mathcal{E}(\mathbf{U}(t_{n}^{-}))-a-\frac{\lVert\mathbf{U}^{\prime}(t_{n}^{-})\rVert_{\mathscr{L}}^{2}}{2}\right)+\left(\frac{\lVert\mathbf{U}^{\prime}(t_{n}^{+})\rVert_{\mathscr{L}}^{2}}{2}-\mathcal{E}(\mathbf{U}(t_{n}^{+}))\right)=-c^{\star}\int_{t_{n}^{-}}^{t_{n}^{+}}\lVert\mathbf{U}^{\prime}(t)\rVert_{\mathscr{L}}^{2}dt.$
(5.263)

Using (5.254), (5.255), (5.256) and (5.257), along with the fact that
$\mathbf{U}^{\prime}\in L^{2}({\mathbb{R}},\mathscr{L})$, we can pass to the
limit $n\to\infty$ and we get that

$a=-c^{\star}\int_{\mathbb{R}}\lVert\mathbf{U}^{\prime}(t)\rVert_{\mathscr{L}}^{2}dt,$
(5.264)

which shows (4.47). We now show that (4.48) holds. Inspecting again the proof
of Theorem 5, we have that $c^{\star}$ is equal to $c(\mathcal{C})$ as in
Corollary 5.3. Take $c<c^{\star}$, then by Corollary 5.3 we have that
$c\in\mathcal{C}$. The definitions of $\mathcal{C}$ in (5.169) implies then
that

$\exists T\geq 1,\hskip 5.69054pt\inf_{U\in X_{T}}\mathbf{E}_{c}(U)<0$ (5.265)

which, by considering $\tilde{U}\in X_{T}$ such that
$\mathbf{E}_{c}(\tilde{U})<0$ and then the sequence
$(\tilde{U}(\cdot+n))_{n\in{\mathbb{N}}}$ which is contained in $X$, implies
that $\inf_{U\in X}\mathbf{E}_{c}(U)=-\infty$. If we now take $c>c^{\star}$,
we have again by Corollary 5.3 that

$\forall T\geq 1,\hskip 5.69054pt\inf_{U\in X_{T}}\mathbf{E}_{c}(U)\geq 0$
(5.266)

which means

$\inf_{U\in X}\mathbf{E}_{c}(U)=0.$ (5.267)

Therefore, (4.48) follows. Finally, we have that (4.49) is exactly (5.170) in
Lemma 5.11. ∎

##  6 Proofs of the main results completed

Once we have proved the abstract results, we are ready to prove the main ones.
In order to do such a thing, we need to show that the main problem can be put
into the abstract framework. This is shown in Lemma 6.1 which is in subsection
6.1. The next subsections are then devoted to the conclusion of the proofs of
the main results, which are Theorems 1, 2 and 4. However, as pointed out
before, we do not have a counterpart of Theorem 3 in the abstract setting,
which means that we prove it using arguments relative to the main setting.

###  6.1 Proving the link between the main setting and the abstract setting

The following result establishes the link between the main assumptions and the
abstract ones. As a consequence, the main results can be deduced from the
abstract framework, which we have already established.

######  Lemma 6.1.

Assume that (H5) holds. Set

$\mathscr{L}:=L^{2}({\mathbb{R}},{\mathbb{R}}^{k}),\hskip
5.69054pt\mathscr{H}:=H^{1}({\mathbb{R}},{\mathbb{R}}^{k}),\hskip
5.69054pt\tilde{\mathscr{H}}:=H^{2}({\mathbb{R}},{\mathbb{R}}^{k}),$ (6.1)
$r_{0}^{\pm}:=\rho_{0}^{\pm}$ (6.2)

and

$\forall v\in\mathscr{L},\hskip
5.69054pt\mathcal{E}(v):=\begin{cases}E(\psi+v)-\mathfrak{m}^{+}&\mbox{ if
}v\in\mathscr{H},\\\ +\infty&\mbox{ otherwise},\end{cases}$ (6.3)

where $\mathfrak{m}^{+}$ was introduced in (H5), the constants
$\rho_{0}^{\pm}$ are those from (2.9) and the function $\psi$ is any smooth
function in $X(\sigma^{-},\sigma^{+})$ converging to $\sigma^{\pm}$ at
$\pm\infty$ at an exponential rate and such that $\psi^{\prime}\in
H^{2}({\mathbb{R}},{\mathbb{R}}^{k})$. Under this choice, assumptions (H1’),
(H2’), (H3’), (H4’), (H5’) and (H6’) hold. Moreover, we have:

  * •

If (H6) holds, then (H7’) holds.

  * •

If (H7) holds, then (H8’) holds.

###### Proof.

The fact that the functional

$v\in\mathscr{H}\to E(\psi+v)$ (6.4)

is well defined and, moreover, is a $C^{1}$ functional on
$(\mathscr{H},\lVert\cdot\rVert_{\mathscr{H}})$ is proven by classical
arguments. See for instance Bisgard [28], Montecchiari and Rabinowitz [56].
See also [60] for the precise statement in this setting. We obviously have
$\mathscr{F}^{\pm}=\mathcal{F}^{\pm}-\psi$. We now pass to prove that the
assumptions are satisfied.

Assumption (H1’) is satisfied:

The fact that $\mathcal{E}$ is weakly lower semicontinuous in $\mathscr{L}$ is
standard, see Lemma 3.1 in [69]. We already invoked Lemma 2.1 in [67] in
Schatzmann so that (2.9) and (2.11) hold. That is, due to (6.2) we have that
if

$\inf_{\tau\in{\mathbb{R}}}\lVert
v+\psi-\mathfrak{q}^{\pm}(\cdot+\tau)\rVert_{\mathscr{L}}\leq r_{0}^{\pm}$
(6.5)

there is a unique $\tau(v)\in{\mathbb{R}}$ which attains the infimum in (6.5).
Moreover, the correspondence $v\to\tau(v)$ defined on the subset of
$\mathscr{L}$ composed of functions that verify (6.5) is of class $C^{2}$.
Therefore, the applications

$P^{\pm}:v\in\mathscr{F}^{-}_{r_{0}^{\pm}/2}\to\mathfrak{q}^{\pm}(\cdot+\tau(v))-\psi\in\mathscr{F}^{\pm},$
(6.6)

satisfy the properties required. Finally, we have that estimate (4.7) follows
by Lemma 3.2 in [57], up to modifying the choice of the constants
$\rho_{0}^{\pm},\beta_{0}^{\pm}$.

Assumption (H2’) is satisfied:

By (6.1), we have that
$\tilde{\mathscr{H}}\subset\mathscr{H}\subset\mathscr{L}$ and the associated
norms verify

$\lVert\cdot\rVert_{\mathscr{L}}\leq\lVert\cdot\rVert_{\mathscr{H}}\leq\lVert\cdot\rVert_{\tilde{\mathscr{H}}}.$
(6.7)

As we pointed out before, $\mathcal{E}$ restricted to
$(\mathscr{H},\lVert\cdot\rVert_{\mathscr{H}})$ is a $C^{1}$ functional.
Moreover, as shown in [28, 56], we have that the differential is given by

$\forall v\in\mathscr{H},\hskip
5.69054ptD\mathcal{E}(v):w\in\mathscr{H}\to\int_{\mathbb{R}}\left(\langle\psi^{\prime}+v^{\prime},w^{\prime}\rangle+\langle\nabla
V(\psi+v),w\rangle\right)\in{\mathbb{R}}.$ (6.8)

Let now $v\in\tilde{\mathscr{H}}$, since $\psi$ is smooth with good behavior
at infinity we can integrate by parts to get

$\displaystyle\forall w\in\mathscr{H},\hskip 5.69054ptD\mathcal{E}(v)(w)$
$\displaystyle=\int_{\mathbb{R}}\langle-(\psi^{\prime\prime}+v^{\prime\prime})+\nabla
V(\psi+v),w\rangle$ (6.9) $\displaystyle=\langle
D_{\mathscr{L}}\mathcal{E}(v),w\rangle_{\mathscr{L}},$ (6.10)

where we have set

$D_{\mathscr{L}}\mathcal{E}:v\in(\tilde{\mathscr{H}},\lVert\cdot\rVert_{\tilde{\mathscr{H}}})\to-(\psi^{\prime\prime}+v^{\prime\prime})+\nabla
V(\psi+v)\in(\mathscr{L},\lVert\cdot\rVert_{\mathscr{L}}),$ (6.11)

which, by standard arguments, can be shown to be continuous. Notice that
(4.13) in (H2’) is exactly (6.9) above, which concludes this part of the
proof.

Assumption (H3’) is satisfied:

Let $(\mathbf{v}_{n}^{-})_{n\in{\mathbb{N}}}$ be a $\mathscr{L}$-bounded
sequence in $\mathscr{F}^{-}$. We want to show the existence of a subsequence
of $(\mathbf{v}_{n}^{-})_{n\in{\mathbb{N}}}$ strongly convergent in
$\mathscr{H}$. Since

$\mathscr{F}^{-}=\mathcal{F}^{-}-\psi=\\{\mathfrak{q}^{-}(\cdot+\tau)-\psi:\tau\in{\mathbb{R}}\\},$
(6.12)

we have that
$(\mathbf{v}_{n}^{-})_{n\in{\mathbb{N}}}=(\mathfrak{q}^{-}(\cdot+\tau_{n})-\psi)_{n\in{\mathbb{N}}}$
with $(\tau_{n})_{n\in{\mathbb{N}}}$ a bounded sequence of real numbers. Since
such a sequence is bounded in $\mathscr{L}$, we know that, up to an
extraction, there exists $\tilde{v}\in\mathscr{L}$ such that
$\mathfrak{q}^{-}(\cdot+\tau_{n})-\psi\rightharpoonup\tilde{v}$ weakly in
$\mathscr{L}$. Due to the weak lower semicontinuity of $\mathcal{E}$, we have
that

$\mathcal{E}(\tilde{v})\leq\liminf_{n\to+\infty}\mathcal{E}(\mathfrak{q}^{-}(\cdot+\tau_{n})-\psi)=a,$
(6.13)

which, by minimality, implies that $\mathcal{E}(v)=0$, that is,
$\tilde{v}\in\mathscr{F}^{-}$. We can then write
$\tilde{v}=\mathfrak{q}^{-}(\cdot+\tau)-\psi$ for some $\tau\in{\mathbb{R}}$.
Now, notice that, by the compactness of minimizing sequences (2.6), there
exists a sequence $(\tau_{n}^{\prime})_{n\in{\mathbb{N}}}$ of real numbers
such that, up to an extraction

$\mathfrak{q}^{-}(\cdot+\tau_{n}+\tau_{n}^{\prime})-\mathfrak{q}^{-}\to
0\mbox{ strongly in }\mathscr{H}$ (6.14)

which necessarily implies that

$\tau_{n}+\tau_{n}^{\prime}\to 0$ (6.15)

and, since $(\tau_{n})_{n\in{\mathbb{N}}}$ is bounded, we have that
$(\tau_{n}^{\prime})_{n\in{\mathbb{N}}}$ is a bounded sequence as well.
Therefore, we can assume, up to an extraction, that
$\tau_{n}^{\prime}\to\tau$. Combining this information with (6.14), we obtain

$\mathfrak{q}^{-}(\cdot+\tau_{n})-\mathfrak{q}^{-}(\cdot-\tau^{-})\to 0\mbox{
strongly in }\mathscr{H},$ (6.16)

which establishes the claim.

We need to show the same for $\mathscr{F}^{+}$. The argument is identical to
the one above, except for the fact that the compactness of minimizing
sequences is replaced by 3. in assumption (H5), which is in fact stronger, and
we use that the elements in $\mathscr{F}^{+}$ are local minimizers (instead of
global ones), which does not require any modification of the reasoning.

Assumption (H4’) is satisfied:

More precisely, we show that 2. in (H4’) holds. Notice that since the results
are local in nature and (H1’) implies that locally the situation does not
change between $\mathscr{F}^{-}$ and $\mathscr{F}^{+}$, we can treat both
cases together. Let $(v,\mathbf{v}^{\pm})\in\mathscr{F}^{\pm}_{r_{0}^{\pm}}$.
Let $\tau(v)$ be given by the projection map defined in (6.6). We have that
$\mathbf{v}^{\pm}=\mathfrak{q}^{\pm}(\cdot+\tau)-\psi$ for some
$\tau\in{\mathbb{R}}$. Define

$\hat{P}^{\pm}_{(v,\mathbf{v}^{\pm})}:w\in\mathscr{L}\to
w(\cdot-\tau(v)+\tau)-\psi+\psi(\cdot-\tau(v)+\tau)\in\mathscr{L}.$ (6.17)

Clearly, using the definition of the projection in (6.6) and $\tau(v)$

$\displaystyle\lVert\hat{P}^{\pm}_{(v,\mathbf{v}^{\pm})}(v)-\mathbf{v}^{\pm}\rVert_{\mathscr{L}}$
$\displaystyle=\lVert
v(\cdot-\tau(v)+\tau)-(\mathfrak{q}^{\pm}(\cdot+\tau)-\psi(\cdot-\tau(v)+\tau))\rVert_{\mathscr{L}}$
(6.18) $\displaystyle=\lVert
v-(\mathfrak{q}^{\pm}(\cdot+\tau(v))-\psi)\rVert_{\mathscr{L}}=\inf_{\tilde{\tau}\in{\mathbb{R}}}\lVert
v-(\mathfrak{q}^{\pm}(\cdot+\tilde{\tau})-\psi)\rVert_{\mathscr{L}},$ (6.19)

meaning that

$P^{\pm}(\hat{P}^{\pm}_{(v,\mathbf{v}^{\pm})}(v))=\mathbf{v}^{\pm}$ (6.20)

and

$\mathrm{dist}_{\mathscr{L}}(\hat{P}^{\pm}_{(v,\mathbf{v}^{\pm})}(v),\mathscr{F}^{\pm})=\mathrm{dist}_{\mathscr{L}}(v,\mathscr{F}^{\pm})$
(6.21)

which are (4.14) and (4.15) respectively. Next, notice that for
$(w_{1},w_{2})\in\mathscr{L}^{2}$ and $h\in{\mathbb{R}}$ we have

$\hat{P}^{\pm}_{(v,\mathbf{v}^{\pm})}(w_{1}+hw_{2})=\hat{P}^{\pm}_{(v,\mathbf{v}^{\pm})}(w_{1})+hw_{2}(\cdot-\tau(v)+\tau)$
(6.22)

so that $\hat{P}^{\pm}_{(v,\mathbf{v}^{\pm})}$ is differentiable and

$\forall(w_{1},w_{2})\in\mathscr{L}^{2},\hskip
5.69054ptD(\hat{P}^{\pm}_{(v,\mathbf{v}^{\pm})})(w_{1},w_{2})=w_{2}(\cdot-\tau(v)+\tau)$
(6.23)

so that

$\forall(w_{1},w_{2})\in\mathscr{L}^{2},\hskip 5.69054pt\lVert
D(\hat{P}^{\pm}_{(v,\mathbf{v}^{\pm})})(w_{1},w_{2})\rVert_{\mathscr{L}}=\lVert
w_{2}\rVert_{\mathscr{L}},$ (6.24)

which is (4.16). Finally, notice that $v\in\mathscr{H}$ if and only if
$\hat{P}^{\pm}_{(v,\mathbf{v}^{\pm})}\in\mathscr{H}$. Assuming that
$v\in\mathscr{H}$ we have

$\displaystyle\mathcal{E}(\hat{P}^{\pm}_{(v,\mathbf{v}^{\pm})}(v))$
$\displaystyle=\mathcal{E}(v(\cdot-\tau(v)+\tau)-\psi+\psi(\cdot-\tau(v)+\tau))$
(6.25)
$\displaystyle=E(v(\cdot-\tau(v)+\tau)+\psi(\cdot-\tau(v)+\tau))=E(\psi+v)=\mathcal{E}(v)$
(6.26)

and if $v\in\mathscr{L}\setminus\mathscr{H}$, we have
$\mathcal{E}(\hat{P}^{\pm}_{(v,\mathbf{v}^{\pm})}(v))=+\infty=\mathcal{E}(v)$.
Therefore, (4.17) holds. We have then showed that (H4’) holds.

Assumption (H5’) is satisfied:

Lemma 2.1 in Schatzmann states that for
$v\in\mathscr{F}_{\mathscr{H},r_{0}^{\pm}}^{\pm}$ the problem

$\inf_{\tau\in{\mathbb{R}}}\lVert
v+\psi-\mathfrak{q}^{\pm}(\cdot+\tau)\rVert_{\mathscr{H}}$ (6.27)

has a unique solution $\tau_{\mathscr{H}}(v)\in{\mathbb{R}}$ and the
projection map

$P^{\pm}_{\mathscr{H}}:v\in\mathscr{F}^{\pm}_{\mathscr{H},r_{0}^{\pm}}\to\mathfrak{q}^{\pm}(\cdot+\tau_{\mathscr{H}}(v))\in\mathscr{F}^{\pm}$
(6.28)

is $C^{1}$ with respect to the $\mathscr{H}$-norm. Next, we have that (4.20)
is Corollary 2.3 in [67]. Finally, the fact that (4.21) implies (4.22) for the
constants $C^{\pm}$ (up to possibly increasing) is a consequence of the
compactness of the minimizing sequences. See for example Corollary 3.2 in
[67].

Assumption (H6’) is satisfied:

We show the existence of the map $\mathfrak{P}$. We follow Lemma 3.3 in [67].
Let $R_{0}>0$ be the constant from (H2). For $R\geq R_{0}$ define in
${\mathbb{R}}^{k}$

$f_{R}(u):=\begin{cases}u&\mbox{ if }\lvert u\rvert\leq R,\\\ R\frac{u}{\lvert
u\rvert}&\mbox{ otherwise},\end{cases}$ (6.29)

where $R_{0}$ is the constant from (H2). For $u\in R^{k}$ such that $\lvert
u\rvert\leq R$, we have $f_{R}(u)=u$. Assume that $u\in{\mathbb{R}}^{k}$ is
such that $\lvert u\rvert>R$. In that case, there exists
$\xi\in\left(\frac{R}{\lvert u\rvert},1\right)$ such that

$V(u)=V(f_{R}(u))+\langle\nabla_{u}V(\xi
u),u-f_{R}(u)\rangle=V(f_{R}(u))+\frac{1}{\xi}\left(1-\frac{R}{\lvert
u\rvert}\right)\langle\nabla_{u}V(\xi u),\xi u\rangle$ (6.30)

which, by (H2) implies

$\forall R\geq R_{0},\forall u\in{\mathbb{R}}^{k}:\lvert u\rvert>R,\hskip
5.69054ptV(u)\geq V(f_{R}(u))+\frac{1}{\xi}\left(1-\frac{R}{\lvert
u\rvert}\right)\nu_{0}\lvert\xi u\rvert^{2}>V(f_{R}(u)).$ (6.31)

In particular, we have shown

$\forall R\geq R_{0},\forall u\in{\mathbb{R}}^{k},\hskip 5.69054ptV(u)\geq
V(f_{R}(u)).$ (6.32)

Next, let $J\subset{\mathbb{R}}$ be a compact interval and $v\in
H^{1}(J,{\mathbb{R}}^{k})$. For $R\geq R_{0}$, consider the function
$v_{R}:=f_{R}\circ v$. Since we clearly have that for all
$u\in{\mathbb{R}}^{k}$, $\lvert f_{R}(u)\rvert\leq\lvert u\rvert$, it holds
that $v_{R}\in L^{2}(J,{\mathbb{R}}^{k})$. Next, we have that $f_{R}$ is the
projection onto the closed ball of center 0 and radius $R$, so that it is non-
expansive. As a consequence, we have

$\forall R\geq R_{0},\forall u\in{\mathbb{R}}^{k},\hskip 5.69054pt\lvert
Df_{R}(u)\rvert\leq 1.$ (6.33)

Therefore, applying the chain rule we obtain

$\mbox{for a. e. }t\in J,\hskip 5.69054pt\lvert
v_{R}^{\prime}(t)\rvert\leq\lvert v^{\prime}(t)\rvert,$ (6.34)

which means that $v_{R}\in H^{1}(J,{\mathbb{R}}^{k})$ and, combining with
(6.32) we obtain

$E(v_{R};J)\leq E(v;J)$ (6.35)

and, by (6.31) the equality above holds if and only if $v_{R}=v$. Let now

$R_{\max}:=2\max\\{R_{0},\lVert\mathfrak{q}^{-}\rVert_{L^{\infty}({\mathbb{R}},{\mathbb{R}}^{k})},\lVert\mathfrak{q}^{+}\rVert_{L^{\infty}({\mathbb{R}},{\mathbb{R}}^{k})}\\}.$
(6.36)

Consider now the application

$\mathfrak{P}:v\in\mathscr{L}\to f_{R_{\max}}\circ(v+\psi)-\psi\in\mathscr{L}$
(6.37)

which is well-defined due to the previous considerations. Moreover, the choice
of $R_{\max}$ implies that $\mathfrak{P}$ equals the identity on
$\\{\mathfrak{q}^{-}(\cdot+\tau)-\psi:\tau\in{\mathbb{R}}\\}$ and
$\\{\mathfrak{q}^{+}(\cdot+\tau)-\psi:\tau\in{\mathbb{R}}\\}$, which is
exactly (4.28). Inequality (6.35) gives (4.26). Finally, using (6.33) we have
that

$\displaystyle\forall(v_{1},v_{2})\in\mathscr{L}^{2},\hskip
5.69054pt\lVert\mathfrak{P}(v_{1})-\mathfrak{P}(v_{2})\rVert_{\mathscr{L}}^{2}$
$\displaystyle=\int_{\mathbb{R}}\lvert
f_{R_{\max}}\circ(v_{1}+\psi)-f_{R_{\max}}\circ(v_{2}+\psi)\rvert^{2}$ (6.38)
$\displaystyle\leq\int_{\mathbb{R}}\sup_{u\in{\mathbb{R}}^{k}}\lvert
Df_{R_{\max}}(u)\rvert^{2}\lvert v_{1}-v_{2}\rvert^{2}$ (6.39)
$\displaystyle\leq\int_{\mathbb{R}}\lvert v_{1}-v_{2}\rvert^{2}=\lVert
v_{1}-v_{2}\rVert_{\mathscr{L}}^{2},$ (6.40)

which is (4.27). Therefore, our map $\mathfrak{P}$ satisfies the required
properties. Next, let $\mathbf{W}$ be a local minimizer of $\mathbf{E}_{c}$.
We show that $\mathbf{W}$ satisfies the desired regularity properties, that
is, $\mathbf{W}\in\mathcal{A}(I)$ with $\mathcal{A}(I)$ as in (4.29). Write
$\overline{\mathbf{W}}:=\mathbf{W}+\psi$. We assume that for all $t\in I$,
$\mathbf{W}(t)=\mathfrak{P}(\mathbf{W})$. The definition of $\mathfrak{P}$ in
(6.37) implies that

$\forall(x_{1},x_{2})\in I\times{\mathbb{R}},\hskip
5.69054pt\overline{\mathbf{W}}(x_{1},x_{2})=f_{R_{\max}}(\overline{\mathbf{W}}(x_{1},x_{2}))$
(6.41)

so that

$\lVert\overline{\mathbf{W}}\rVert_{L^{\infty}(I\times{\mathbb{R}},{\mathbb{R}}^{k})}\leq
R_{\max}.$ (6.42)

Therefore, by classical elliptic regularity arguments, we have that, with the
obvious identifications, $\overline{\mathbf{W}}$ solves

$-c\partial_{x_{1}}\overline{\mathbf{W}}-\Delta\overline{\mathbf{W}}=-\nabla_{u}V(\overline{\mathbf{W}})\mbox{
in }I\times{\mathbb{R}}$ (6.43)

and for all $\alpha\in(0,1)$ we have that
$\overline{\mathbf{W}}\in\mathcal{C}^{3,\alpha}(I_{C}\times{\mathbb{R}},{\mathbb{R}}^{k})$
for any compact $I_{C}\subset I$. It is then clear that

$\mathbf{W}\in\mathcal{C}^{2}(I_{C},L^{2}({\mathbb{R}},{\mathbb{R}}^{k}))\cap\mathcal{C}^{1}(I_{C},H^{1}({\mathbb{R}},{\mathbb{R}}^{k})\cap\mathcal{C}^{0}(I_{C},H^{2}({\mathbb{R}},{\mathbb{R}}^{k}))$
(6.44)

for any $I_{C}\subset I$ compact, which means that
$\mathbf{W}\in\mathcal{A}(I)$.

Assumption (H6) implies (H7’): Immediate.

Assumption (H7) implies (H8’): Immediate. ∎

Once Lemma 6.1 has been established, the main results are easily obtained by
rephrasing the abstract ones.

###  6.2 Proof of Theorem 1 completed

Assume that (H6) holds. Notice that (H6) implies that (H1), (H2), (H3), (H4)
and (H5) hold. Therefore, applying Lemma 6.1 we have that, choosing the
objects as in its statement, we get that (H3’), (H4’), (H5’), (H6’) and (H7’)
hold. Those are exactly the assumptions which are needed for Theorem 5 to
hold, meaning that we obtain $(c^{\star},\mathbf{U})$ with $c^{\star}>0$ and
$\mathbf{U}\in\mathcal{A}({\mathbb{R}})\cap X$, with
$\mathcal{A}({\mathbb{R}})$ as in (4.29) and $X$ as in (4.39), which solves

$\mathbf{U}^{\prime\prime}-D_{\mathscr{L}}\mathcal{E}(\mathbf{U})=-c\mathbf{U}^{\prime}\mbox{
in }{\mathbb{R}}$ (6.45)

and satisfies the conditions at infinity

$\exists T^{-}\leq 0:\forall t\leq T^{-},\hskip
5.69054pt\mathbf{U}(t)\in\mathscr{F}^{-}_{r_{0}^{-}/2}\hskip 2.84526pt\mbox{
and }\hskip
2.84526pt\exists\mathbf{v}^{+}(\mathbf{U})\in\mathscr{F}^{+}:\lim_{t\to+\infty}\lVert\mathbf{U}(t)-\mathbf{v}^{+}(\mathbf{U})\rVert_{\mathscr{H}}=0.$
(6.46)

We now pass to prove each of the three statements of Theorem 1 separately:

  1. 1.

Existence. Recall that for all $t\in{\mathbb{R}}$ we have
$\mathbf{U}(t)\in\mathscr{L}=L^{2}({\mathbb{R}},{\mathbb{R}}^{k})$. Let us
then define

$\mathfrak{U}:(x_{1},x_{2})\in{\mathbb{R}}^{2}\to\mathbf{U}(x_{1})(x_{2})\in{\mathbb{R}}^{k}.$
(6.47)

It is clear then that since $\mathbf{U}\in\mathcal{A}({\mathbb{R}})$ we have
that
$\mathfrak{U}\in\mathcal{C}^{2}_{\mathrm{loc}}({\mathbb{R}},{\mathbb{R}}^{k})$
and, moreover for all $(x_{1},x_{2})$ and any pair of index
$(i,j)\in\\{0,1,2\\}^{2}$ such that $i+j\leq 2$ we have

$\partial_{x_{1}}^{i}\partial_{x_{2}}^{j}\mathfrak{U}(x_{1},x_{2})=(\mathbf{U}^{(i)}(x_{1}))^{(j)}(x_{2})$
(6.48)

where for a curve $f$ taking values in a Hilbert space we denote by $f^{(i)}$
its $i$-th derivative, $i\in{\mathbb{N}}$. As a consequence of (6.45), (6.48)
and the formula for $D\mathcal{E}$ when we make
$\mathcal{E}=E-\mathfrak{m}^{+}$ (see (6.8)) we obtain that

$-c\partial_{x_{1}}\mathfrak{U}-\Delta\mathfrak{U}=-\nabla_{u}V(\mathfrak{U})\mbox{
in }{\mathbb{R}}^{2}$ (6.49)

and by (6.46) we obtain that for some $L\in{\mathbb{R}}$ we have for some
$x_{1}\leq L$ that $\mathfrak{U}(x_{1},\cdot)\in\mathcal{F}^{-}_{\rho^{-}/2}$,
since we choose $r_{0}^{\pm}=\rho^{\pm}$, so that
$\mathcal{F}^{\pm}_{\rho^{\pm}/2}=\mathscr{F}^{\pm}_{r_{0}^{\pm}/2}$. The
variational characterization (2.23) follows directly from Theorem 1, using the
fact that we have $X=S$ and $\mathbf{E}_{c}=E_{2,c}$ for all $c>0$ (again we
implicitly identify $\mathfrak{U}$ with $\mathbf{U}$ via (6.47). Finally, we
have that for all $t\in{\mathbb{R}}$,
$\mathbf{U}(t)=\mathfrak{P}(\mathbf{U}(t))$. According to the choice of
$\mathfrak{P}$ made in Lemma 6.1, this implies that
$\lVert\mathfrak{U}\rVert_{L^{\infty}({\mathbb{R}},{\mathbb{R}}^{k})}<+\infty$,
which by classical Schauder theory and the smoothness properties of $V$
implies that for all $\alpha\in(0,1)$,
$\mathfrak{U}\in\mathcal{C}^{2,\alpha}({\mathbb{R}}^{2},{\mathbb{R}}^{k})$.
The proof of the existence part of Theorem 1 is hence completed.

  2. 2.

Uniqueness of the speed. Again, we have that $X=S$ and
$\mathbf{E}_{c}=E_{2,c}$ for all $c>0$, meaning that the proof of this
statement follows from the analogous one in Theorem 5.

  3. 3.

Exponential convergence. Using the exponential rate of convergence of
$\mathbf{U}$ given by Theorem 5, which is (4.42), we obtain that for some
$b>c^{\star}/2$ it holds

$\lim_{x_{1}\to+\infty}\lVert\mathfrak{U}(x_{1},\cdot)-\mathfrak{q}^{+}(\cdot+\tau^{+})\rVert_{H^{1}({\mathbb{R}},{\mathbb{R}}^{k})}e^{bx_{1}}=0$
(6.50)

for some $\tau^{+}\in{\mathbb{R}}$. This concludes the proof of the statement.

The proof of Theorem 1 is concluded. ∎

###  6.3 Proof of Theorem 2 completed

Assume that (H6) and (H7) hold. Arguing as in the proof of Theorem 1, we have
that the assumptions of Theorem 6 are fulfilled if we choose as in Lemma 6.1.
Therefore, Theorem 2 is readily obtained by a straightforward rewriting of
Theorem 6. ∎

###  6.4 Proof of Theorem 3

We now provide the proof of Theorem 3, which is a consequence of the following
results, which are more general as required by Theorem 3 and might be of
independent interest:

######  Lemma 6.2.

Assume that (H1), (H2) and (H3) hold. Let
$(\hat{\sigma}_{-},\hat{\sigma}_{+})\in\Sigma^{2}$ (possibly equal) and $q\in
X(\hat{\sigma}_{-},\hat{\sigma}_{+})$. Assume moreover that there exists
$L^{+}\in{\mathbb{R}}$ and $U\in
H^{1}_{\mathrm{loc}}([L^{+},+\infty)\times{\mathbb{R}},{\mathbb{R}}^{k})$
uniformly continuous and such that

$\int_{L^{+}}^{+\infty}\lvert E(U(x_{1},\cdot))-E(q)\rvert dx_{1}<+\infty,$
(6.51) $\lim_{x_{1}\to+\infty}\lVert
U(x_{1},\cdot)-q\rVert_{L^{2}({\mathbb{R}},{\mathbb{R}}^{k})}=0.$ (6.52)

Then, it holds that

$\lim_{x_{1}\to+\infty}\lVert
U(x_{1},\cdot)-q\rVert_{L^{\infty}({\mathbb{R}},{\mathbb{R}}^{k})}=0$ (6.53)

and

$\lim_{x_{2}\to\pm\infty}\lVert
U(\cdot,x_{2})-\hat{\sigma}_{\pm}\rVert_{L^{\infty}([L^{+},+\infty),{\mathbb{R}}^{k})}=0.$
(6.54)

Similarly, we have the following

######  Lemma 6.3.

Assume that (H1), (H2) and (H3) hold. Let
$(\hat{\sigma}_{-},\hat{\sigma}_{+})\in\Sigma^{2}$ (possibly equal) and $q\in
X(\hat{\sigma}_{-},\hat{\sigma}_{+})$. Assume moreover that there exists
$L^{-}\in{\mathbb{R}}$ and $U\in
H^{1}_{\mathrm{loc}}((-\infty,L^{-}]\times{\mathbb{R}},{\mathbb{R}}^{k})$
uniformly continuous and such that

$\int_{-\infty}^{L^{-}}lvertE(U(x_{1},\cdot))-E(q)\rvert dx_{1}<+\infty,$
(6.55) $\lim_{x_{1}\to-\infty}\lVert
U(x_{1},\cdot)-q\rVert_{L^{2}({\mathbb{R}},{\mathbb{R}}^{k})}=0.$ (6.56)

Then, it holds that

$\lim_{x_{1}\to-\infty}\lVert
U(x_{1},\cdot)-q\rVert_{L^{\infty}({\mathbb{R}},{\mathbb{R}}^{k})}=0$ (6.57)

and

$\lim_{x_{2}\to\pm\infty}\lVert
U(\cdot,x_{2})-\hat{\sigma}_{\pm}\rVert_{L^{\infty}((-\infty,L^{-}],{\mathbb{R}}^{k})}=0.$
(6.58)

In the proof of Lemmas 6.2 and 6.3, we will need to use the following fact:

######  Lemma 6.4.

Assume that (H1), (H2) and (H3) hold. Let
$(\hat{\sigma}_{-},\hat{\sigma}_{+})\in\Sigma^{2}$ (possibly equal) and $q\in
X(\hat{\sigma}_{-},\hat{\sigma}_{+})$. Assume that
$(q_{n})_{n\in{\mathbb{N}}}$ is a sequence in
$X(\hat{\sigma}_{-},\hat{\sigma}_{+})$ such that

$\lim_{n\to\infty}\lVert
q_{n}-q\rVert_{L^{2}({\mathbb{R}},{\mathbb{R}}^{k})}=0$ (6.59)

and

$\lim_{n\to\infty}E(q_{n})=E(q),$ (6.60)

then, it holds that

$\lim_{n\to\infty}\lVert
q_{n}-q\rVert_{H^{1}({\mathbb{R}},{\mathbb{R}}^{k})}=0.$ (6.61)

###### Proof.

First, notice that

$\sup_{n\in{\mathbb{N}}}\lVert
q_{n}\rVert_{L^{\infty}({\mathbb{R}},{\mathbb{R}}^{k})}<+\infty.$ (6.62)

Indeed, (6.60) implies that $(q_{n}^{\prime})_{n\in{\mathbb{N}}}$ is bounded
in $L^{2}({\mathbb{R}},{\mathbb{R}}^{k})$ which, in combination with (6.59))
means that $(q_{n})_{n\in{\mathbb{N}}}$ is bounded in
$H^{1}({\mathbb{R}},{\mathbb{R}}^{k})$, hence in
$L^{\infty}({\mathbb{R}},{\mathbb{R}}^{k})$. We also have that

$\nabla V(q)\in L^{2}({\mathbb{R}},{\mathbb{R}}^{k}),$ (6.63)

which follows easily from the fact that $V$ is smooth and quadratic near the
wells. For all $n\in{\mathbb{N}}$, we write the following expansion

$V(q_{n})=V(q)+\langle\nabla
V(q),q_{n}-q\rangle+\int_{0}^{1}D^{2}V(q+\lambda(q_{n}-q))(q_{n}-q)(q_{n}-q)d\lambda,$
(6.64)

which holds pointwise in ${\mathbb{R}}$. Therefore, Cauchy-Schwartz inequality
implies

$\left(\int_{{\mathbb{R}}}\lvert
V(q_{n})-V(q)\rvert\right)^{2}\leq\left(\int_{{\mathbb{R}}}\lvert\nabla
V(q)\rvert^{2}+\sup_{\begin{subarray}{c}u\in{\mathbb{R}}^{k}\\\ \lvert
u\rvert\leq\lVert q_{n}-q\rVert_{L^{\infty}}\end{subarray}}\lvert
D^{2}V(u)\rvert\right)\lVert
q_{n}-q\rVert_{L^{2}({\mathbb{R}},{\mathbb{R}}^{k})}^{2},$ (6.65)

hence, by (6.62) and (6.63) we find a constant $C>0$ such that for all
$n\in{\mathbb{N}}$

$\int_{{\mathbb{R}}}\lvert V(q_{n})-V(q)\rvert\leq C\lVert
q_{n}-q\rVert_{L^{2}({\mathbb{R}},{\mathbb{R}}^{k})}$ (6.66)

which by (6.59) means that $V(q_{n})-V(q)\to 0$ in
$L^{1}({\mathbb{R}},{\mathbb{R}}^{k})$. As a consequence, (6.60) implies that

$\lim_{n\to\infty}\lVert
q_{n}^{\prime}\rVert_{L^{2}({\mathbb{R}},{\mathbb{R}}^{k})}=\lVert
q^{\prime}\rVert_{L^{2}({\mathbb{R}},{\mathbb{R}}^{k})}.$ (6.67)

Suppose now by contradiction that (6.61) does not hold. Then, we can find a
subsequence $(q_{n_{m}})_{m\in{\mathbb{N}}}$ and $\hat{\delta}>0$ such that
for all $m\in{\mathbb{N}}$

$\lVert
q_{n_{m}}-q\rVert_{H^{1}({\mathbb{R}},{\mathbb{R}}^{k})}\geq\hat{\delta}.$
(6.68)

Since $(q_{n_{m}}^{\prime})_{m\in{\mathbb{N}}}$ is bounded in
$L^{2}({\mathbb{R}},{\mathbb{R}}^{k})$, it converges weakly in $L^{2}$ up to
an extraction, and the limit is $q^{\prime}$ by uniqueness of the limit in the
sense of distributions. By (6.67), we have that such a subsequence also
converges strongly in $L^{2}({\mathbb{R}},{\mathbb{R}}^{k})$, which combining
with (6.59) contradicts (6.68). ∎

We now prove Lemma 6.2. The proof of Lemma 6.3 being analogous, we skip it.

###### Proof of Lemma 6.2.

Assume by contradiction that (6.53) does not hold. Then, we can find a
sequence $(x_{1,n})_{n\in{\mathbb{N}}}$ in $[L^{+},+\infty)\times{\mathbb{R}}$
such that $x_{1,n}\to+\infty$ as $n\to\infty$ as well as $\hat{\delta}>0$ such
that for all $n\in{\mathbb{N}}$

$\lVert
U(x_{1,n},\cdot)-q\rVert_{L^{\infty}({\mathbb{R}},{\mathbb{R}}^{k})}\geq\hat{\delta}.$
(6.69)

By uniform continuity, there exists $\nu>0$ such that for all
$n\in{\mathbb{N}}$ we have

$\max_{x_{1}\in[x_{1,n}-\nu,x_{1,n}+\nu]}\lVert
U(x_{1},\cdot)-q\rVert_{L^{\infty}({\mathbb{R}},{\mathbb{R}}^{k})}\geq\frac{\hat{\delta}}{2}$
(6.70)

Let $A:=\cup_{n\in{\mathbb{N}}}[x_{1,n}-\nu,x_{1,n}+\nu]$. By (6.51) we have
that

$\int_{A}(E(U(x_{1},\cdot))-E(q))dx_{1}<+\infty,$ (6.71)

and since $A$ has positive measure and it is unbounded by above, we find a
sequence $(y_{1,n})_{n\in{\mathbb{N}}}$ in $A$ such that $y_{1,n}\to+\infty$
as $n\to\infty$ and $\lim_{n\to\infty}E(U(y_{1,n},\cdot))=E(q)$. Combining
this fact with (6.52), we have that assumptions (6.59) and (6.60) in Lemma 6.4
hold, which means that

$\lim_{n\to\infty}\lVert
U(y_{1,n},\cdot)-q\rVert_{H^{1}({\mathbb{R}},{\mathbb{R}}^{k})}=0$ (6.72)

which contradicts (6.70). Therefore, we have shown that (6.53) holds. In order
to prove (6.54), we first show that there exists $\underline{L}^{+}\leq L^{+}$
such that

$\lim_{x_{2}\to\pm\infty}\lVert
U(\cdot,x_{2})-\hat{\sigma}_{\pm}\rVert_{L^{\infty}([\underline{L}^{+},+\infty),{\mathbb{R}}^{k})}=0.$
(6.73)

We prove (6.73) by contradiction. The other case being handled in an analogous
fashion, assume that there exists a sequence $(x_{2,n})_{n\in{\mathbb{N}}}$ in
${\mathbb{R}}$ such that $x_{2,n}\to+\infty$ as $n\to\infty$, a sequence
$(x_{1,n})_{n\in{\mathbb{N}}}$ in $[L^{+},+\infty)$ tending to $+\infty$ and
$\hat{\delta}>0$ such that for all $n\in{\mathbb{N}}$

$\lvert U(x_{1,n},x_{2,n})-\hat{\sigma}_{+}\rvert\geq\hat{\delta}.$ (6.74)

Since we already proved that (6.53) holds, there exists $N_{1}\in{\mathbb{N}}$
such that for all $n\geq{\mathbb{N}}$ we have

$\lVert
U(x_{1,n},\cdot)-q\rVert_{L^{\infty}({\mathbb{R}},{\mathbb{R}}^{k})}\leq\frac{\hat{\delta}}{4}$
(6.75)

and, since $q\in X(\hat{\sigma}_{-},\hat{\sigma}_{+})$, there exists
$\hat{t}\in{\mathbb{R}}$ such that for all $t\geq\hat{t}$ we have

$\lvert q(t)-\hat{\sigma}_{+}\rvert\leq\frac{\hat{\delta}}{4}.$ (6.76)

Let $N_{2}\in{\mathbb{N}}$ be such that for all $n\in{\mathbb{N}}$,
$x_{2,n}\geq\hat{t}$. Taking any $n\geq\max\\{N_{1},N_{2}\\}$, we obtain by
(6.75) and (6.76) that

$\lvert U(x_{1,n},x_{2,n})-\hat{\sigma}_{+}\rvert\leq\frac{\delta}{2},$ (6.77)

which contradicts (6.74) and establishes (6.73). In order to establish (6.54),
we handle the limit $x_{2}\to+\infty$, as the other one is treated
identically. Let
$\rho_{\Sigma}^{+}:=\mathrm{dist}(\sigma^{+},\Sigma\setminus\\{\sigma^{+}\\})>0$.
We claim that for every $\tilde{L}\geq L^{+}$ we have that if

$\lim_{x_{2}\to\pm\infty}\lVert
U(\cdot,x_{2})-\hat{\sigma}_{\pm}\rVert_{L^{\infty}([\tilde{L},+\infty),{\mathbb{R}}^{k})}=0,$
(6.78)

then

$\lim_{x_{2}\to\pm\infty}\lVert
U(\cdot,x_{2})-\hat{\sigma}_{\pm}\rVert_{L^{\infty}([\tilde{L}-\eta_{\Sigma}^{+},+\infty),{\mathbb{R}}^{k})}=0,$
(6.79)

where

$\eta_{\Sigma}^{+}:=\min\left\\{\tilde{L}-L^{+},\frac{\rho_{\Sigma}}{4\lVert
D\mathfrak{U}\rVert_{L^{\infty}({\mathbb{R}}^{2},{\mathbb{R}}^{k})}}\right\\}.$
(6.80)

Such a claim allows to easily complete the proof of (6.54) by a finite
induction process, due to the fact that (6.73) holds. ∎

It remains to establish one claim in the proof of Lemma 6.2.

###### Proof that (6.78) implies (6.79).

Assume that (6.78) holds. We show that for every
$\varepsilon\in(0,\eta_{\Sigma}^{+})$ we have

$\lim_{x_{2}\to\pm\infty}\lVert
U(\cdot,x_{2})-\hat{\sigma}_{\pm}\rVert_{L^{\infty}([\tilde{L}-\eta_{\Sigma}^{+}+\varepsilon,+\infty),{\mathbb{R}}^{k})}=0,$
(6.81)

which clearly implies (6.79) by uniform continuity. Fix then
$\varepsilon\in(0,\eta_{\Sigma}^{+})$. By assumption, there exists
$\overline{x}_{2}^{+}\in{\mathbb{R}}$ such that for all
$x_{2}\geq\overline{x}_{2}^{+}$ we have

$\lvert U(\tilde{L},x_{2})-\sigma^{+}\rvert\leq\frac{\rho_{\Sigma}^{+}}{4},$
(6.82)

which, by (6.80) implies that for all
$(x_{1},x_{2})\in[\tilde{L}-\eta^{+}_{\Sigma}+\varepsilon,\tilde{L}]\times[\overline{x}_{2},+\infty)$,
it holds

$\lvert U(x_{1},x_{2})-\sigma^{+}\rvert\leq\frac{\rho_{\Sigma}^{+}}{2}$ (6.83)

and the definition of $\rho_{\Sigma}^{+}$ gives in turn that for all such
$(x_{1},x_{2})$ and $\sigma\in\Sigma\setminus\\{\sigma^{+}\\}$ we have

$\lvert U(x_{1},x_{2})-\sigma\rvert\geq\frac{\rho_{\Sigma}^{+}}{2}.$ (6.84)

Assume now that (6.81) does not hold. Then, inequalities (6.83) and (6.84)
imply that we can find a sequence $(x_{1,n},x_{2,n})_{n\in{\mathbb{N}}}$
contained in
$[\tilde{L}-\eta^{+}_{\Sigma}+\varepsilon,\tilde{L}]\times[\overline{x}_{2},+\infty)$,
such that $x_{2,n}\to+\infty$ as $n\to\infty$ and $\hat{\delta}>0$ such that
for all $n\in{\mathbb{N}}$ and $\sigma\in\Sigma$

$\lvert U(x_{1,n},x_{2,n})-\sigma\rvert\geq\hat{\delta}.$ (6.85)

By uniform continuity, we can find $\nu\in(0,\varepsilon)$ such that for all
$n\in{\mathbb{N}}$ and

$(x_{1},x_{2})\in
B((x_{1,n},x_{2,n}),\nu)\subset[\tilde{L}-\eta^{+}_{\Sigma},\tilde{L}]\times[\overline{x}_{2},+\infty)$
(6.86)

we have for all $\sigma\in\Sigma$

$\lvert U(x_{1},x_{2})-\sigma\rvert\geq\frac{\hat{\delta}}{2}$ (6.87)

or, equivalently

$V(U(x_{1},x_{2}))\geq
V_{\hat{\delta}/2}:=\min\left\\{V(u):u\in{\mathbb{R}}^{k},\mathrm{dist}(u,\Sigma)\geq\frac{\hat{\delta}}{2}\right\\}$
(6.88)

which is positive by (H1) and (H3). Up to an extraction and since
$x_{2,n}\to+\infty$ as $n\to\infty$, we can assume that whenever $n\not=m$ we
have

$B((x_{1,n},x_{2,n}),\nu)\cap B((x_{1,m},x_{2,m}),\nu)=\emptyset,$ (6.89)

which, due to the definition of $\eta_{\Sigma}^{+}$ in (6.80) and (6.88)
implies that

$\displaystyle\int_{L^{+}}^{+\infty}\lvert E(U(x_{1}))-E(q)\rvert dx_{1}$
$\displaystyle\geq\int_{\tilde{L}-\eta^{+}_{\Sigma}}^{\tilde{L}}E(U(x_{1}))dx_{1}-\eta_{\Sigma}^{+}E(q)$
(6.90)
$\displaystyle\geq\sum_{n\in{\mathbb{N}}}\left(\int_{B((x_{1,n},x_{2,n}),\nu)}V(U(x_{1},x_{2}))dx_{1}dx_{2}\right)-\eta_{\Sigma}^{+}E(q)$
(6.91)
$\displaystyle\geq\sum_{n\in{\mathbb{N}}}(\pi\nu^{2}V_{\hat{\delta}/2})-\eta_{\Sigma}^{+}E(q)=+\infty,$
(6.92)

which enters in contradiction with (6.51). Therefore, the claim has been
proven. ∎

We have now all the necessary ingredients for completing the proof of Theorem
3:

###### Proof of Theorem 3 completed.

Let $(c^{\star},\mathfrak{U})$ be the solution given by Theorem 1, interpreted
via the choices made in Lemma 6.1. We will invoke Lemma 6.2. The $L^{2}$
exponential convergence (2.25) given by Theorem 1 implies in particular that
assumption (6.52) in Lemma 6.2 holds with $U=\mathfrak{U}$,
$q=\mathfrak{q}^{+}(\cdot+\tau^{+})$. Moreover, since
$E_{2,c^{\star}}(\mathfrak{U})=0<+\infty$, assumption (6.51) in Lemma 6.2
holds for all $L\in{\mathbb{R}}$ in view of the definition of
$E_{2,c^{\star}}$ (recall that $c^{\star}>0$). Finally, we have by Theorem 1
that
$\mathfrak{U}\in\mathcal{C}^{2,\alpha}({\mathbb{R}}^{2},{\mathbb{R}}^{k})$,
$\alpha\in(0,1)$, so that $\mathfrak{U}$ is uniformly continuous. As a
consequence, Lemma 6.2 applies and we have (6.53) and (6.54) for all
$L\in{\mathbb{R}}$, and this is exactly (2.29) and (2.30) for all
$L\in{\mathbb{R}}$.

Assume now that (H7) holds, so that Theorem 2 applies. We will show that we
can invoke Lemma 6.3. We have that (2.28) in Theorem 2 implies that (6.56) in
Lemma 6.3 holds with $U=\mathfrak{U}$ and
$q=\mathfrak{q}^{-}(\cdot+\tau^{-})$. Moreover, the abstract result
Proposition 5.4 in combination with Lemma 6.1 implies in particular that for
all $L\in{\mathbb{R}}$, (6.56) in Lemma 6.3 holds. Since $\mathfrak{U}$ is
uniformly continuous, Lemma 6.3 applies, which means that (6.57) holds, so
that we have proven (2.31) in Theorem 3. Moreover, for all $L\in{\mathbb{R}}$
we have that (6.58) holds, which combined with (2.30) (which also holds for
all $L\in{\mathbb{R}}$) gives (2.32) and completes the proof. ∎

###  6.5 Proof of Theorem 4 completed

Assume that (H6) and (H7) hold. Arguing as in the proof of Theorem 1, we have
that the assumptions of Theorem 7 are fulfilled if we choose as in Lemma 6.1.
Notice that Theorem 4 is exactly Theorem 7 if we choose the abstract objects
as in Lemma 6.1. Therefore, Theorem 4 is established. ∎

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