From now on, we assume that *Φ* : ℝ^{n} → [0, ∞) satisfy (*G*1)−(*G*5) and (*Δ*_{2}).

Let *Ω* ⊂ℝ be a domain. Following Trudinger [1] we define the space

$$\begin{array}{}{\displaystyle {L}_{\mathit{\Phi}}(\mathit{\Omega})=\left\{u:\mathit{\Omega}\to {\mathbb{R}}^{n}\phantom{\rule{thinmathspace}{0ex}}:\phantom{\rule{thinmathspace}{0ex}}u\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\text{is Lebesgue measurable}\phantom{\rule{thinmathspace}{0ex}}\text{and}\phantom{\rule{thinmathspace}{0ex}}\underset{\mathit{\Omega}}{\int}\mathit{\Phi}(u)dt<\mathrm{\infty}\phantom{\rule{thinmathspace}{0ex}}\right\}.}\end{array}$$

This space equipped with the Luxemburg norm

$$\begin{array}{}{\displaystyle \parallel u{\parallel}_{\mathit{\Phi}}=inf\left\{\nu >0\phantom{\rule{thinmathspace}{0ex}}:\phantom{\rule{thinmathspace}{0ex}}\underset{\mathit{\Omega}}{\int}\mathit{\Phi}\left(\frac{u}{\nu}\right)dt\le 1\right\}}\end{array}$$(2)

is a Banach space. Since *Φ* is *Δ*_{2}-regular, *L*_{Φ}(*Ω*) is also a separable space (c.f. Rem. 8.22 in [8]). Furthermore, *L*_{Φ}(*Ω*) is reflexive if and only if (*Δ*_{2}) is satisfied (c.f. Thm. 8.20 in [8]).

Set *ψ* = *φ*∘∣ ⋅ ∣, i.e. *ψ*(*x*) = *φ*(∣*x*∣) for each *x* ∈ ℝ^{n}. As a consequence of (1), the space *L*_{Φ}(*Ω*) is continuously imbedded in *L*_{ψ}(*Ω*) (c.f. Thm. 8.12 in [8]),

$$\begin{array}{}{\displaystyle {L}_{\mathit{\Phi}}(\mathit{\Omega})\subset {L}_{\psi}(\mathit{\Omega}).}\end{array}$$

Note that ∥*u*∥_{ψ} = ∥∣*u*∣∥_{φ}.

For simplicity of notation, we write *L*_{Φ} instead of *L*_{Φ}(ℝ). Although the norm formula (2) depends on the domain *Ω*, we use the same notation ∥ ⋅ ∥_{Φ} for different subsets of ℝ. It will be clear from the context what *Ω* is.

Let *A**C*_{loc}(ℝ, ℝ^{n}) be the space of locally absolutely continuous functions on ℝ with values in ℝ^{n}. Finally, let *E* denote the Orlicz-Sobolev space

$$\begin{array}{}{\displaystyle E=\left\{u\in A{C}_{loc}(\mathbb{R},{\mathbb{R}}^{n}):\dot{u}\in {L}_{\mathit{\Phi}}(\mathbb{R},{\mathbb{R}}^{n})\right\}}\end{array}$$

with the norm

$$\begin{array}{}{\displaystyle \parallel u\parallel =\parallel \dot{u}{\parallel}_{\mathit{\Phi}}+|u(0)|.}\end{array}$$

We note for later reference that *E* is a separable reflexive Banach space (see [19]).

For every *T* > 0 we define the Banach space *E*_{T} consisting of restrictions of *u* ∈ *E* to the interval [0, *T*] with the induced norm,

$$\begin{array}{}{\displaystyle \parallel u{\parallel}_{{E}_{T}}=|u(0)|+\parallel \dot{u}{\parallel}_{\mathit{\Phi}}.}\end{array}$$

Let *C*([0, *T*], ℝ^{n}) denote the space of continuous functions from [0, *T*] into ℝ^{n} with the standard norm.

#### Proposition 2.1

*The inclusion map* *E*_{T} → *C*([0, *T*], ℝ^{n}) *is continuous*, *i.e. there is* *C*_{T} > 0 *such that for each* *u* ∈ *E*_{T} *one has*

$$\begin{array}{}{\displaystyle \underset{t\in [0,T]}{max}|u(t)|\le {C}_{T}\parallel u{\parallel}_{{E}_{T}}.}\end{array}$$

#### Proof

One has

$$\begin{array}{}{\displaystyle |u(t)|=\left|u(0)+\underset{0}{\overset{t}{\int}}\dot{u}(s)ds\right|\le |u(0)|+\underset{0}{\overset{t}{\int}}|\dot{u}(s)|ds}\\ {\displaystyle \phantom{\rule{2em}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\le |u(0)|+\underset{0}{\overset{T}{\int}}|\dot{u}(s)|ds\le |u(0)|+2\parallel 1{\parallel}_{{\phi}^{\ast}}\parallel |\dot{u}|{\parallel}_{\phi}}\\ {\displaystyle \phantom{\rule{2em}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\le (1+2\parallel 1{\parallel}_{{\phi}^{\ast}})\left(|u(0)|+\parallel |\dot{u}|{\parallel}_{\phi}\right)}\\ {\displaystyle \phantom{\rule{2em}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\le {C}_{T}\left(|u(0)|+\parallel \dot{u}{\parallel}_{\mathit{\Phi}}\right)={C}_{T}\parallel u{\parallel}_{{E}_{T}}.}\end{array}$$□

#### Proposition 2.2

*If a sequence* {*u*_{k}}_{k∈ℕ} ⊂ *E*_{T} *converges weakly to* *u*_{0} ∈ *E*_{T} *then it converges uniformly to* *u*_{0} *in* *C*([0, *T*], ℝ^{n}).

#### Proof

Since {*u*_{k}}_{k∈ℕ} converges to *u*_{0} weakly in *E*_{T} then, by Proposition 2.1, it also converges to *u*_{0} weakly in *C*([0, *T*], ℝ^{n}). Furthermore, ∥*u*_{k}∥_{ET} ≤ *M* for some *M* > 0 and every *k* ∈ ℕ.

Let 0 ≤ *s* ≤ *t* ≤ *T*. Then

$$\begin{array}{}{\displaystyle |{u}_{k}(t)-{u}_{k}(s)|=\left|\underset{s}{\overset{t}{\int}}{\dot{u}}_{k}(\tau )d\tau \right|\le \underset{s}{\overset{t}{\int}}|{\dot{u}}_{k}(\tau )|d\tau}\\ {\displaystyle \phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\le 2\parallel 1{\parallel}_{{\phi}^{\ast}}\parallel |{\dot{u}}_{k}|{\parallel}_{\phi}\le 2\parallel 1{\parallel}_{{\phi}^{\ast}}\parallel {u}_{k}{\parallel}_{{E}_{T}}}\\ {\displaystyle \phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\le 2M{\left(({\phi}^{\ast}{)}^{-1}\left(\frac{1}{t-s}\right)\right)}^{-1}.}\end{array}$$

Thus {*u*_{k}}_{k∈ℕ} is a sequence of equicontinuous functions. By the Arzela-Ascoli Theorem, every sequence {*u*_{k}_{i}}_{i∈ℕ} contains a subsequence converging to a certain *û* in *C*([0, *T*], ℝ^{n}). By the uniqueness of the weak limit, *û* = *u*_{0}, which completes the proof.□

In what follows, *Φ* : ℝ^{2} → ℝ and *V* : ℝ × (ℝ^{2} ∖ {*ξ*}) → ℝ satisfy the assumptions of Theorem 1.2.

For each *u* ∈ *E*, we define a functional *I* by setting

$$\begin{array}{}{\displaystyle I(u)=\underset{-\mathrm{\infty}}{\overset{\mathrm{\infty}}{\int}}\left(\mathit{\Phi}(\dot{u}(t))-V(t,u(t))\right)dt.}\end{array}$$(3)

Let

$$\begin{array}{}{\displaystyle {\alpha}_{\epsilon}=inf\{-V(t,x):x\notin {B}_{\epsilon}(0)\},}\end{array}$$(4)

where 0 < *ε* ≤ $\begin{array}{}\frac{1}{2}\end{array}$∣*ξ*∣ and *B*_{ε}(0) denotes the ball of radius *ε* centered at the origin. By (*V*_{1})−(*V*_{3}) we have *α*_{ε} > 0.

#### Lemma 2.3

*Suppose that* *u* ∈ *E* *and* *u*(*t*) ∉ *B*_{ε}(0) *for each* *t* ∈ [*a*, *b*]. *Then*, *there is* *C* > 0 *such that*

$$\begin{array}{}{\displaystyle (I(u)+1{)}^{2}\ge C\cdot length\phantom{\rule{thinmathspace}{0ex}}({u}_{|[a,b]})\ge C|u(b)-u(a)|.}\end{array}$$(5)

#### Proof

One has

$$\begin{array}{}{\displaystyle |u(b)-u(a)|=\left|\underset{a}{\overset{b}{\int}}\dot{u}(t)dt\right|\le \underset{a}{\overset{b}{\int}}|\dot{u}(t)|dt\le 2\parallel |\dot{u}|{\parallel}_{\phi}\parallel 1{\parallel}_{{\phi}^{\ast}}.}\end{array}$$

The last estimation follows from Hölder’s inequality in Orlicz spaces (c.f. [5], Par. 8.11). Directly from the definition, one has

$$\begin{array}{}{\displaystyle \parallel 1{\parallel}_{{\phi}^{\ast}}={\left[({\phi}^{\ast}{)}^{-1}\left(\frac{1}{b-a}\right)\right]}^{-1}.}\end{array}$$

Set *δ* = length (*u*_{∣[a, b]}) and *τ* = *b* − *a*. Then

$$\begin{array}{}{\displaystyle \parallel |\dot{u}|{\parallel}_{\phi}\ge \frac{1}{2}\delta \parallel 1{\parallel}_{{\phi}^{\ast}}^{-1}=\frac{1}{2}\delta \cdot ({\phi}^{\ast}{)}^{-1}\left(\frac{1}{\tau}\right).}\end{array}$$

Consequently,

$$\begin{array}{}{\displaystyle I(u)\ge \underset{a}{\overset{b}{\int}}\left(\mathit{\Phi}(\dot{u}(t))-V(t,u(t))\right)dt=\underset{a}{\overset{b}{\int}}\mathit{\Phi}(\dot{u}(t))dt+\underset{a}{\overset{b}{\int}}-V(t,u(t))dt}\\ {\displaystyle \phantom{\rule{2em}{0ex}}\phantom{\rule{1em}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\ge \underset{a}{\overset{b}{\int}}\phi (|\dot{u}(t)|)dt+{\alpha}_{\epsilon}\tau \ge \parallel |\dot{u}|{\parallel}_{\phi}-1+{\alpha}_{\epsilon}\tau}\\ {\displaystyle \phantom{\rule{2em}{0ex}}\phantom{\rule{1em}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\ge \frac{1}{2}\delta \cdot ({\phi}^{\ast}{)}^{-1}\left(\frac{1}{\tau}\right)-1+{\alpha}_{\epsilon}\tau .}\end{array}$$(6)

Hence

$$\begin{array}{}{\displaystyle I(u)+1\ge \frac{1}{2}\delta \cdot ({\phi}^{\ast}{)}^{-1}\left(\frac{1}{\tau}\right)+{\alpha}_{\epsilon}\tau \ge \frac{1}{2}\frac{\delta}{\tau k}\cdot ({\phi}^{\ast}{)}^{-1}(k)+{\alpha}_{\epsilon}\tau ,}\end{array}$$

where the natural number *k* satisfies *τ* *k* ≥ 1 and the last inequality follows from the fact that (*φ*^{*})^{−1} is concave. We choose the smallest *k* with the property *τ* *k* ≥ 1. In particular, we set *k* = 1 if *τ* ≥ 1. Now, if *τ* ≥ 1 then

$$\begin{array}{}{\displaystyle f(\tau )=\frac{1}{2}\frac{\delta}{\tau}\cdot ({\phi}^{\ast}{)}^{-1}(1)+{\alpha}_{\epsilon}\tau}\end{array}$$

achieves its minimum at the point

$$\begin{array}{}{\displaystyle {\tau}_{min}={\left(\frac{\delta \cdot ({\phi}^{\ast}{)}^{-1}(1)}{2{\alpha}_{\epsilon}}\right)}^{\frac{1}{2}},}\end{array}$$

which is equal to $\begin{array}{}{f}_{min}=(2\delta {\alpha}_{\epsilon}({\phi}^{\ast}{)}^{-1}(1){)}^{\frac{1}{2}}.\end{array}$ If *τ* < 1 then

$$\begin{array}{}{\displaystyle \frac{1}{2}\frac{\delta}{\tau k}\cdot ({\phi}^{\ast}{)}^{-1}(k)+{\alpha}_{\epsilon}\tau \ge \frac{1}{4}\delta \cdot ({\phi}^{\ast}{)}^{-1}(k)+{\alpha}_{\epsilon}\tau \ge \frac{1}{4}\delta \cdot ({\phi}^{\ast}{)}^{-1}(1).}\end{array}$$

Finally, set

$$\begin{array}{}{\displaystyle C=min\left\{2{\alpha}_{\epsilon}({\phi}^{\ast}{)}^{-1}(1),\phantom{\rule{thinmathspace}{0ex}}\frac{1}{4}({\phi}^{\ast}{)}^{-1}(1)\right\}.}\end{array}$$□

#### Corollary 2.5

*If* *u* ∈ *E* *and* *I*(*u*) < ∞ *then* *u* ∈ *L*^{∞}(ℝ, ℝ^{2}).

#### Proof

Assume that *u* ∉ *L*^{∞}(ℝ, ℝ^{2}). Then for every *n* ∈ ℕ there exists *t*_{n} ∈ ℝ such that ∣*u*(*t*_{n})∣ > *n*. Consequently, by Lemma 2.3 we get

$$\begin{array}{}{\displaystyle (I(u)+1{)}^{2}\ge C|u({t}_{n})-u({t}_{1})|\ge C(|u({t}_{n})|-|u({t}_{1})|)\ge C(n-|u({t}_{1})|)}\end{array}$$

for *n* ∈ ℕ, contrary to *I*(*u*) < ∞.□

#### Lemma 2.6

*If* *u* ∈ *E* *and* *I*(*u*) < ∞ *then* $\begin{array}{}\underset{t\to \pm \mathrm{\infty}}{lim}\end{array}$ *u*(*t*) = 0.

Lemma 2.6 is analogous to Proposition 3.11 of [20] and Lemma 2.4 of [21]. In spite of different assumptions on the potential *V*, the claims are similar.

#### Proof

Let *A*(*u*) denote the set of limit points of *u*(*t*), as *t* → −∞. From Corollary 2.5 we conclude that *A*(*u*) ≠ ∅. Assume that there are *ε* > 0 and *ρ* ∈ ℝ such that if *t* < *ρ* then *u*(*t*) ∉ *B*_{ε}(0). By (4) we obtain,

$$\begin{array}{}{\displaystyle I(u)\ge \underset{-\mathrm{\infty}}{\overset{\rho}{\int}}-V(t,u(t))dt=\mathrm{\infty},}\end{array}$$

a contradiction. Thus *A*(*u*) contains 0. It is sufficient to note that *A*(*u*) consists of a point. If not, there is *ε* > 0 such that *u*(*t*) intersects $\begin{array}{}\mathrm{\partial}{B}_{\frac{\epsilon}{2}}(0)\end{array}$ and *∂* *B*_{ε}(0) infinitely many times. Let *τ*_{0} ≥ 0 be the smallest number such that

$$\begin{array}{}{\displaystyle I(u)+1\ge \frac{1}{2}\frac{\epsilon}{2}\cdot ({\phi}^{\ast}{)}^{-1}\left(\frac{1}{{\tau}_{0}}\right)+{\alpha}_{\frac{\epsilon}{2}}{\tau}_{0}.}\end{array}$$

Since lim_{τ→∞}(*φ*^{*})^{−1}(*τ*) = ∞, one has *τ*_{0} > 0. By Remark 2.4, we obtain

$$\begin{array}{}{\displaystyle I(u)+1\ge n{\alpha}_{\frac{\epsilon}{2}}{\tau}_{0}}\end{array}$$

for each *n* ∈ ℕ, and hence *I*(*u*) = ∞, a contradiction.

In the same manner we can see that $\begin{array}{}\underset{t\to \mathrm{\infty}}{lim}\end{array}$ *u*(*t*) = 0.□

#### Lemma 2.7

*If* [*a*, *b*] *is an interval such that* *u*([*a*, *b*]) ⊂𝓝 ∖ {*ξ*} *then it holds*

$$\begin{array}{}{\displaystyle |U(u(b))|-|U(u(a))|\le 2(I(u)+1{)}^{2}.}\end{array}$$(7)

#### Proof

We first note that

$$\begin{array}{}{\displaystyle |U(u(b))|\le |U(u(a))|+\left|\underset{a}{\overset{b}{\int}}\frac{d}{dt}U(u(t))dt\right|}\\ {\displaystyle \phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\le |U(u(a))|+\left|\underset{a}{\overset{b}{\int}}\left(\mathrm{\nabla}U(u(t)),\dot{u}(t)\right)dt\right|}\\ {\displaystyle \phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\le |U(u(a))|+\underset{a}{\overset{b}{\int}}|\mathrm{\nabla}U(u(t))||\dot{u}(t)|dt}\\ {\displaystyle \phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\le |U(u(a))|+2\parallel |\mathrm{\nabla}U(u)|{\parallel}_{{\phi}^{\ast}}\parallel |\dot{u}|{\parallel}_{\phi}}\end{array}$$

Since

$$\begin{array}{}{\displaystyle \parallel |\mathrm{\nabla}U(u)|{\parallel}_{{\phi}^{\ast}}\le 1+\underset{a}{\overset{b}{\int}}{\phi}^{\ast}(|\mathrm{\nabla}U(u(t))|)dt\le 1+\underset{a}{\overset{b}{\int}}-V(t,u(t))dt}\end{array}$$

and

$$\begin{array}{}{\displaystyle \parallel |\dot{u}|{\parallel}_{\phi}\le 1+\underset{a}{\overset{b}{\int}}\phi (|\dot{u}(t)|)dt}\end{array}$$

we obtain

$$\begin{array}{}{\displaystyle |U(u(b))|\le |U(u(a))|+2(I(u)+1{)}^{2}.}\end{array}$$

□

As an immediate consequence of (7) one has that *u*(*t*) ≠ *ξ* for *t* ∈ ℝ provided that *I*(*u*) < ∞ (c.f. [7], Eq. (2.21)). In fact, we obtain the following

#### Corollary 2.8

*(c.f. [17]) If the action functional* *I* *is bounded on some set* *W* ⊂ *E*, *say* *I*(*W*) ⊂[0, *β*] *then there is* *ρ* > 0 *depending on* *β* *such that for every* *u* ∈ *W* *and* *t* ∈ ℝ *one has* ∣*u*(*t*) − *ξ*∣ ≥ *ρ*.

Set

$$\begin{array}{}{\displaystyle \mathit{\Lambda}=\left\{u\in E:\underset{t\to \pm \mathrm{\infty}}{lim}u(t)=0,\phantom{\rule{thinmathspace}{0ex}}u(\mathbb{R})\subset {\mathbb{R}}^{2}\setminus \{\xi \}\right\}.}\end{array}$$

If *I*(*u*) < ∞ then *u* ∈ *Λ*. Consequently, *u* describes a closed curve in ℝ^{2} ∖ {*ξ*} that starts and ends at 0. Hence its homotopy class [*u*] represents an element of the fundamental group *π*_{1}(ℝ^{2} ∖ {*ξ*}).

Let us remind that two functions *u*_{0}, *u*_{1} ∈ *Λ* are homotopic if and only if there exists a continuous map *h* : [0, 1] → *Λ* such that *h*(0) = *u*_{0} and *h*(1) = *u*_{1}. The rotation number (or winding number) *rot*_{ξ}(*u*) of *u* around *ξ* is constant on every connected component of *Λ* and induces an isomorphism *rot*_{*} : *π*_{1}(ℝ^{2} ∖ {*ξ*}) → ℤ,

$$\begin{array}{}{\displaystyle ro{t}_{\ast}([u])=ro{t}_{\xi}(u).}\end{array}$$

Equivalently, *Λ* is a sum of its path connected components labeled by the integers.

Similarily to [17] one can prove the following result.

#### Proposition 2.9

*Let* *W* ⊂*Λ* *be a set such that the functional* *I* *restricted to* *W* *is bounded*. *Then there exists* *D* ∈ ℕ *such that* ∣*rot*_{ξ}(*u*)∣ ≤ *D* *for all* *u* ∈ *W*.

Let

$$\begin{array}{}{\displaystyle {\mathit{\Lambda}}^{\pm}=\{u\in \mathit{\Lambda}:\pm \phantom{\rule{thinmathspace}{0ex}}{rot}_{\xi}(u)>0\},}\end{array}$$

and

$$\begin{array}{}{\displaystyle {\lambda}^{\pm}=\underset{u\in {\mathit{\Lambda}}^{\pm}}{inf}I(u).}\end{array}$$(8)

Our main result is an immediate consequence of the following.

#### Theorem 2.10

*If the assumptions of Theorem 1.2 are satisfied then there exists* *u*^{±} ∈ *Λ*^{±} *such that* *I*(*u*^{±}) = *λ*^{±} > 0. *Moreover*, *u*^{±} *is a classical homoclinic solution of* (LS).

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