In this section, we will consider the asymptotically periodic case and prove Theorems 1.2 and 1.3. As in the proof of Theorem 1.1, we first take advantage of the minimizing sequence of Ψ to find a {(\mathit{PS})}_{c} sequence of Φ. In what follows, the important thing is to recover the compactness for the {(\mathit{PS})}_{c} sequence. For this purpose, we need to estimate the functional levels of the problem (NLS) and those of a related periodic problem of (NLS) (roughly speaking, the limit system of (NLS) by (V3))

Hence, first we introduce some definitions and look for solutions for the problem (NLS)_{
p
}. The functional of (NLS)_{
p
} is defined by

{\mathrm{\Phi}}_{p}(u,v)=\frac{1}{2}({\parallel u\parallel}^{2}+{\parallel v\parallel}^{2}+\int {a}_{p}(x){u}^{2}+\int {b}_{p}(x){v}^{2})-\lambda \int uv-\int F(u,v).

The Nehari manifold of (NLS)_{
p
} is

{M}_{p}=\{(u,v)\in H\setminus \{(0,0)\}:\u3008{\mathrm{\Phi}}_{p}^{\mathrm{\prime}}(u,v),(u,v)\u3009=0\},

and {c}_{p}={inf}_{{M}_{p}}{\mathrm{\Phi}}_{p} is the least energy of (NLS)_{
p
} on {M}_{p}. Note that

{\mathrm{\Phi}}_{p}{|}_{{M}_{p}}(u,v)=\int [\frac{1}{2}\mathrm{\nabla}F(u,v)(u,v)-F(u,v)].

(5.1)

As for *c*, we have {c}_{p}\ge 0.

**Lemma 5.1** *Suppose that* {a}_{p}, {b}_{p} *satisfy* (V1) *and* (V2). *Let* (F1)-(F6) *hold*. *Then the problem* (NLS)_{
p
} *has a positive ground state* (u,v)\in {M}_{p} *such that* {\mathrm{\Phi}}_{p}(u,v)={c}_{p}.

*Proof* As a corollary of Theorem 1.1, we infer that the problem (NLS)_{
p
} has a positive ground state. Moreover, from the argument of Theorem 1.1, we find that the ground state of the problem (NLS)_{
p
} we obtained is a minimizer of {\mathrm{\Phi}}_{p} on {M}_{p}. □

The existence of a positive ground state for the problem (NLS)_{
p
} implies that (NLS) has a positive ground state when a={a}_{p} and b={b}_{p}. So, it remains to consider

a\ne {a}_{p}\phantom{\rule{1em}{0ex}}\text{or}\phantom{\rule{1em}{0ex}}b\ne {b}_{p}.

(5.2)

Next, we prove that c<{c}_{p} under some conditions.

**Lemma 5.2** *Suppose that* {a}_{p}, {b}_{p} *satisfy* (V2). *Let* (V1), (V4), (5.2) *and* (F1)-(F6) *hold*. *Then* c<{c}_{p}.

*Proof* Let ({u}_{0},{v}_{0})\in {M}_{p} be a positive ground state of (NLS)_{
p
} such that {\mathrm{\Phi}}_{p}({u}_{0},{v}_{0})={c}_{p}. Assume t>0 satisfies t({u}_{0},{v}_{0})\in M. By (V4), we get

\int [(a(x)-{a}_{p}(x)){u}_{0}^{2}+(b(x)-{b}_{p}(x)){v}_{0}^{2}]\le 0.

Then \mathrm{\Phi}(t{u}_{0},t{v}_{0})\le {\mathrm{\Phi}}_{p}(t{u}_{0},t{v}_{0}).

Replacing Φ and *M* by {\mathrm{\Phi}}_{p} and {M}_{p} respectively, (3.3) also holds. Noting that ({u}_{0},{v}_{0})\in {M}_{p}, we infer that

{\mathrm{\Phi}}_{p}(t{u}_{0},t{v}_{0})\le {\mathrm{\Phi}}_{p}({u}_{0},{v}_{0})\phantom{\rule{1em}{0ex}}\text{and}\phantom{\rule{1em}{0ex}}{\mathrm{\Phi}}_{p}(t{u}_{0},t{v}_{0})={\mathrm{\Phi}}_{p}({u}_{0},{v}_{0})\phantom{\rule{1em}{0ex}}\text{if and only if}t=1.

(5.3)

Therefore,

c\le \mathrm{\Phi}(t{u}_{0},t{v}_{0})\le {\mathrm{\Phi}}_{p}(t{u}_{0},t{v}_{0})\le {\mathrm{\Phi}}_{p}({u}_{0},{v}_{0})={c}_{p}.

(5.4)

If c<{c}_{p}, we are done. Otherwise, c={c}_{p}. Then by (5.3) and (5.4), we get t=1 and \mathrm{\Phi}({u}_{0},{v}_{0})=c. Then ({u}_{0},{v}_{0}) is a ground state for (NLS). Note that ({u}_{0},{v}_{0}) is a solution of (NLS)_{
p
}. From the first equations of (NLS) and (NLS)_{
p
}, we infer that a={a}_{p}. Similarly, b={b}_{p} contrary to (5.2). The proof is now complete. □

**Lemma 5.3** *Suppose that* {a}_{p}, {b}_{p} *satisfy* (V1) *and* (V2). *Let* (V1), (V5), (5.2) *and* (F1)-(F7) *hold*. *Then* c<{c}_{p}.

*Proof* Let ({u}_{0},{v}_{0})\in {M}_{p} be a positive ground state of (NLS)_{
p
} such that {\mathrm{\Phi}}_{p}({u}_{0},{v}_{0})={c}_{p}. By (V5) and (F7), we find that ({v}_{0},{u}_{0}) is also a minimizer of {\mathrm{\Phi}}_{p} on {M}_{p}. Let t,\tau >0 be such that t({u}_{0},{v}_{0}),\tau ({v}_{0},{u}_{0})\in M. Using (V5), we have \int (a(x)+b(x)-2V(x))({u}_{0}^{2}+{v}_{0}^{2})\le 0. Then

\begin{array}{c}\int [(a(x)-V(x)){u}_{0}^{2}+(b(x)-V(x)){v}_{0}^{2}]\le 0\phantom{\rule{1em}{0ex}}\text{or}\hfill \\ \int [(a(x)-V(x)){v}_{0}^{2}+(b(x)-V(x)){u}_{0}^{2}]\le 0.\hfill \end{array}

Without loss of generality, we assume that

\int [(a(x)-V(x)){u}_{0}^{2}+(b(x)-V(x)){v}_{0}^{2}]\le 0.

Then \mathrm{\Phi}(t{u}_{0},t{v}_{0})\le {\mathrm{\Phi}}_{p}(t{u}_{0},t{v}_{0}). Below we argue analogously with the proof of Lemma 5.2 to infer that c<{c}_{p}. This ends the proof. □

Now, we are ready to prove Theorems 1.2 and 1.3. The proof is partially inspired by [17], where the authors dealt with Schrödinger-Poisson equations.

*Proof of Theorem 1.2* As the argument of Theorem 1.1, we infer that there exists a sequence ({u}_{n},{v}_{n})\in M such that {\mathrm{\Phi}}^{\prime}({u}_{n},{v}_{n})\to 0 and \mathrm{\Phi}({u}_{n},{v}_{n})\to c.

We claim that \{({u}_{n},{v}_{n})\} is bounded in *H*. Otherwise, suppose {t}_{n}:=\parallel ({u}_{n},{v}_{n})\parallel \to \mathrm{\infty} up to a subsequence. Set ({w}_{n},{z}_{n})=\frac{({u}_{n},{v}_{n})}{\parallel ({u}_{n},{v}_{n})\parallel}. As in the proof of Theorem 1.1, taking a subsequence, we suppose {|{w}_{n}|}_{q}+{|{z}_{n}|}_{q}\to A\in [0,\mathrm{\infty}) and exclude the case that A=0. So, A\ne 0, then we can assume that in {L}^{q}({\mathbb{R}}^{N}). From the Lions compactness lemma, it follows that there exist {\delta}_{0}>0 and {y}_{n}\in {\mathbb{R}}^{N} such that

{\int}_{{B}_{1}({y}_{n})}{|{w}_{n}|}^{2}>{\delta}_{0}.

Set {\tilde{w}}_{n}={w}_{n}(\cdot +{y}_{n}) and {\tilde{z}}_{n}={z}_{n}(\cdot +{y}_{n}). We assume that ({\tilde{w}}_{n},{\tilde{z}}_{n})\rightharpoonup (\tilde{w},\tilde{z}) in *H*, ({\tilde{w}}_{n},{\tilde{z}}_{n})\to (\tilde{w},\tilde{z}) in {L}_{\mathrm{loc}}^{2}({\mathbb{R}}^{N})\times {L}_{\mathrm{loc}}^{2}({\mathbb{R}}^{N}) and ({\tilde{w}}_{n},{\tilde{z}}_{n})\to (\tilde{w},\tilde{z}) a.e. on {\mathbb{R}}^{2N} up to a subsequence. Then by

{\int}_{{B}_{1}({y}_{n})}{|{w}_{n}|}^{2}>{\delta}_{0},

we obtain \tilde{w}\ne 0. So, Lemma 3.2 implies that

\int \frac{F({t}_{n}{\tilde{w}}_{n},{t}_{n}{\tilde{z}}_{n})}{{t}_{n}^{2}}\to \mathrm{\infty}.

Then by (2.1), we get

\begin{array}{rcl}0& \le & \frac{\mathrm{\Phi}({u}_{n},{v}_{n})}{{\parallel ({u}_{n},{v}_{n})\parallel}^{2}}\\ =& \frac{1}{2}\left[\frac{{\parallel {u}_{n}\parallel}^{2}+{\parallel {v}_{n}\parallel}^{2}+\int a(x){u}_{n}^{2}+\int b(x){v}_{n}^{2}-2\lambda \int {u}_{n}{v}_{n}}{{\parallel ({u}_{n},{v}_{n})\parallel}^{2}}\right]-\int \frac{F({t}_{n}{w}_{n},{t}_{n}{z}_{n})}{{t}_{n}^{2}}\\ \le & \frac{\nu}{2}-\int \frac{F({t}_{n}{\tilde{w}}_{n},{t}_{n}{\tilde{z}}_{n})}{{t}_{n}^{2}}\to -\mathrm{\infty}.\end{array}

This is a contradiction.

Hence, \{({u}_{n},{v}_{n})\} is bounded in *H*. Up to a subsequence, we assume that ({u}_{n},{v}_{n})\rightharpoonup (\tilde{u},\tilde{v}) in *H*, ({u}_{n},{v}_{n})\to (\tilde{u},\tilde{v}) in {L}_{\mathrm{loc}}^{2}({\mathbb{R}}^{N})\times {L}_{\mathrm{loc}}^{2}({\mathbb{R}}^{N}) and ({u}_{n},{v}_{n})\to (\tilde{u},\tilde{v}) a.e. on {\mathbb{R}}^{2N}. By Lemma 2.2, we have {\mathrm{\Phi}}^{\prime}(\tilde{u},\tilde{v})=0. Namely, (\tilde{u},\tilde{v}) is a solution of (NLS).

Below we prove that (\tilde{u},\tilde{v})\ne (0,0). We argue by contradiction. Suppose that (\tilde{u},\tilde{v})=(0,0). By (V3), for any \u03f5>0, there exists r>0 such that

|a(x)-{a}_{p}(x)|<\u03f5,\phantom{\rule{2em}{0ex}}|b(x)-{b}_{p}(x)|<\u03f5\phantom{\rule{1em}{0ex}}\text{for all}|x|r.

(5.5)

Note that (\tilde{u},\tilde{v})=(0,0), after passing to a subsequence, we assume ({u}_{n},{v}_{n})\to (0,0) in {L}^{2}({B}_{r}(0))\times {L}^{2}({B}_{r}(0)). So, for the above *ϵ*, there exists {J}_{1}\in \mathbb{N} such that for n>{J}_{1}, we have

{\int}_{{B}_{r}(0)}{u}_{n}^{2}<\u03f5,\phantom{\rule{2em}{0ex}}{\int}_{{B}_{r}(0)}{v}_{n}^{2}<\u03f5.

Combining with (5.5), for n>{J}_{1}, we get

|\int (a(x)-{a}_{p}(x)){u}_{n}^{2}|\le {\int}_{{B}_{r}(0)}|a(x)-{a}_{p}(x)|{u}_{n}^{2}+\u03f5{\int}_{{\mathbb{R}}^{N}\setminus {B}_{r}(0)}{u}_{n}^{2}<({|a|}_{\mathrm{\infty}}+{|{a}_{p}|}_{\mathrm{\infty}})\u03f5+C\u03f5.

Then \int (a(x)-{a}_{p}(x)){u}_{n}^{2}\to 0. Similarly, \int (b(x)-{b}_{p}(x)){v}_{n}^{2}\to 0. Therefore,

{\mathrm{\Phi}}_{p}({u}_{n},{v}_{n})=\mathrm{\Phi}({u}_{n},{v}_{n})+{o}_{n}(1),\phantom{\rule{2em}{0ex}}\u3008{\mathrm{\Phi}}_{p}^{\mathrm{\prime}}({u}_{n},{v}_{n}),({u}_{n},{v}_{n})\u3009=\u3008{\mathrm{\Phi}}^{\prime}({u}_{n},{v}_{n}),({u}_{n},{v}_{n})\u3009+{o}_{n}(1).

Hence,

{\mathrm{\Phi}}_{p}({u}_{n},{v}_{n})=c+{o}_{n}(1),\phantom{\rule{2em}{0ex}}\u3008{\mathrm{\Phi}}_{p}^{\mathrm{\prime}}({u}_{n},{v}_{n}),({u}_{n},{v}_{n})\u3009={o}_{n}(1).

(5.6)

Let {s}_{n}>0 be such that {s}_{n}({u}_{n},{v}_{n})\in {M}_{p}. We claim that {s}_{n}\ge 1 for large *n* and {s}_{n}\to 1.

First, we prove that

\underset{n\to \mathrm{\infty}}{lim\hspace{0.17em}sup}{s}_{n}\le 1.

(5.7)

Otherwise, there exist \delta >0 and a subsequence of {s}_{n}, still denoted by {s}_{n}, such that {s}_{n}\ge 1+\delta for all n\in \mathbb{N}. From (5.6) we have

{\parallel {u}_{n}\parallel}^{2}+{\parallel {v}_{n}\parallel}^{2}+\int {a}_{p}(x){u}_{n}^{2}+\int {b}_{p}(x){v}_{n}^{2}-2\lambda \int {u}_{n}{v}_{n}=\int \mathrm{\nabla}F({u}_{n},{v}_{n})({u}_{n},{v}_{n})+{o}_{n}(1).

Moreover, by {s}_{n}({u}_{n},{v}_{n})\in {M}_{p}, we get

\begin{array}{c}{s}_{n}^{2}[{\parallel {u}_{n}\parallel}^{2}+{\parallel {v}_{n}\parallel}^{2}+\int {a}_{p}(x){u}_{n}^{2}+\int {b}_{p}(x){v}_{n}^{2}-2\lambda \int {u}_{n}{v}_{n}]\hfill \\ \phantom{\rule{1em}{0ex}}=\int \mathrm{\nabla}F({s}_{n}{u}_{n},{s}_{n}{v}_{n})({s}_{n}{u}_{n},{s}_{n}{v}_{n}).\hfill \end{array}

Hence,

\int [\frac{\mathrm{\nabla}F({s}_{n}{u}_{n},{s}_{n}{v}_{n})({u}_{n},{v}_{n})}{{s}_{n}}-\mathrm{\nabla}F({u}_{n},{v}_{n})({u}_{n},{v}_{n})]={o}_{n}(1).

By {s}_{n}\ge 1+\delta and (F3), we obtain

\int [\frac{\mathrm{\nabla}F((1+\delta ){u}_{n},(1+\delta ){v}_{n})({u}_{n},{v}_{n})}{1+\delta}-\mathrm{\nabla}F({u}_{n},{v}_{n})({u}_{n},{v}_{n})]\le {o}_{n}(1).

(5.8)

Similar to the proof of Theorem 1.1, if ({u}_{n},{v}_{n})\to (0,0) in {L}^{q}({\mathbb{R}}^{N})\times {L}^{q}({\mathbb{R}}^{N}), then ({u}_{n},{v}_{n})\to (0,0) in *H*. Contrary to (3.2), since ({u}_{n},{v}_{n})\in M, therefore, in {L}^{q}({\mathbb{R}}^{N})\times {L}^{q}({\mathbb{R}}^{N}). Suppose in {L}^{q}({\mathbb{R}}^{N}). Then from the Lions compactness lemma, it follows that there exist {x}_{n}\in {\mathbb{R}}^{N} and {\delta}_{1}>0 such that

{\int}_{{B}_{1}({x}_{n})}{u}_{n}^{2}>{\delta}_{1}.

(5.9)

We denote {\overline{u}}_{n} and {\overline{v}}_{n} by {\overline{u}}_{n}={u}_{n}(\cdot +{x}_{n}) and {\overline{v}}_{n}={v}_{n}(\cdot +{x}_{n}). Similarly, we assume that ({\overline{u}}_{n},{\overline{v}}_{n})\rightharpoonup (\overline{u},\overline{v}) in *H*, ({\overline{u}}_{n},{\overline{v}}_{n})\to (\overline{u},\overline{v}) in {L}_{\mathrm{loc}}^{2}({\mathbb{R}}^{N})\times {L}_{\mathrm{loc}}^{2}({\mathbb{R}}^{N}) and ({\overline{u}}_{n},{\overline{u}}_{n})\to (\overline{u},\overline{v}) a.e. on {\mathbb{R}}^{2N} up to a subsequence. By (5.9), we have

{\int}_{{B}_{1}(0)}{\overline{u}}_{n}^{2}>{\delta}_{1}.

So, \overline{u}\ne 0. From (5.8), (F3) and the Fatou lemma, we obtain

0<\int [\frac{\mathrm{\nabla}F((1+\delta )\overline{u},(1+\delta )\overline{v})(\overline{u},\overline{v})}{1+\delta}-\mathrm{\nabla}F(\overline{u},\overline{v})(\overline{u},\overline{v})]\le 0,

which is impossible. Consequently, (5.7) holds.

Now, we show that {s}_{n}\ge 1 for large *n*. Indeed, on the contrary, passing to a subsequence, we assume that {s}_{n}<1. Using (3.1) and (5.1), we have

\begin{array}{rcl}{c}_{p}& \le & {\mathrm{\Phi}}_{p}({s}_{n}{u}_{n},{s}_{n}{v}_{n})=\int [\frac{1}{2}\mathrm{\nabla}F({s}_{n}{u}_{n},{s}_{n}{v}_{n})({s}_{n}{u}_{n},{s}_{n}{v}_{n})-F({s}_{n}{u}_{n},{s}_{n}{v}_{n})]\\ \le & \int [\frac{1}{2}\mathrm{\nabla}F({u}_{n},{v}_{n})({u}_{n},{v}_{n})-F({u}_{n},{v}_{n})]\\ =& \mathrm{\Phi}({u}_{n},{v}_{n})+{o}_{n}(1)=c+{o}_{n}(1),\end{array}

(5.10)

where (5.10) follows from the fact that *α* is increasing in (0,\mathrm{\infty}) by Lemma 2.1. Then {c}_{p}\le c, contrary to Lemma 5.2. Therefore, combining with (5.7), we may assume that

{s}_{n}\ge 1,\phantom{\rule{1em}{0ex}}\text{for large}n\phantom{\rule{1em}{0ex}}\text{and}\phantom{\rule{1em}{0ex}}\underset{n\to \mathrm{\infty}}{lim}{s}_{n}=1.

(5.11)

For \u03f5>0 and 1\le s\le {s}_{n}, using (2.2) we get

|\int \mathrm{\nabla}F(s{u}_{n},s{v}_{n})({u}_{n},{v}_{n})|\le \u03f5{s}_{n}({|{u}_{n}|}_{2}^{2}+{|{v}_{n}|}_{2}^{2})+{C}_{\u03f5}{s}_{n}^{q-1}({|{u}_{n}|}_{q}^{q}+{|{v}_{n}|}_{q}^{q}).

(5.12)

Combining (5.11) with (5.12), one easily has that

\int [F({s}_{n}{u}_{n},{s}_{n}{v}_{n})-F({u}_{n},{v}_{n})]={\int}_{1}^{{s}_{n}}[\int \mathrm{\nabla}F(s{u}_{n},s{v}_{n})({u}_{n},{v}_{n})\phantom{\rule{0.2em}{0ex}}dx]\phantom{\rule{0.2em}{0ex}}ds={o}_{n}(1).

Since {a}_{p},{b}_{p}\in {L}^{\mathrm{\infty}}({\mathbb{R}}^{N}) and \{({u}_{n},{v}_{n})\} is bounded, we get

\frac{{s}_{n}^{2}-1}{2}[{\parallel {u}_{n}\parallel}^{2}+{\parallel {v}_{n}\parallel}^{2}+\int {a}_{p}(x){u}_{n}^{2}+\int {b}_{p}(x){v}_{n}^{2}-2\lambda \int {u}_{n}{v}_{n}]={o}_{n}(1).

Hence, {\mathrm{\Phi}}_{p}({s}_{n}{u}_{n},{s}_{n}{v}_{n})={\mathrm{\Phi}}_{p}({u}_{n},{v}_{n})+{o}_{n}(1). Then using (5.6), we have {c}_{p}\le {\mathrm{\Phi}}_{p}({s}_{n}{u}_{n},{s}_{n}{v}_{n})=c+{o}_{n}(1). Then {c}_{p}\le c. However, Lemma 5.2 implies that c<{c}_{p}. This is a contradiction. Note that this contradiction follows from the hypothesis that (\tilde{u},\tilde{v})=(0,0). So, (\tilde{u},\tilde{v})\ne (0,0). Then (\tilde{u},\tilde{v})\in M.

It suffices to show that \mathrm{\Phi}(\tilde{u},\tilde{v})=c. By (3.1) we have

\begin{array}{rcl}c+{o}_{n}(1)& =& \mathrm{\Phi}({u}_{n},{v}_{n})=\int [\frac{1}{2}\mathrm{\nabla}F({u}_{n},{v}_{n})({u}_{n},{v}_{n})-F({u}_{n},{v}_{n})]\\ \ge & \int [\frac{1}{2}\mathrm{\nabla}F(\tilde{u},\tilde{v})(\tilde{u},\tilde{v})-F(\tilde{u},\tilde{v})]+{o}_{n}(1)\\ =& \mathrm{\Phi}(\tilde{u},\tilde{v})+{o}_{n}(1),\end{array}

(5.13)

where the inequality (5.13) holds by (2.3) and the Fatou lemma. Then \mathrm{\Phi}(\tilde{u},\tilde{v})\le c. According to (\tilde{u},\tilde{v})\in M, we have \mathrm{\Phi}(\tilde{u},\tilde{v})=c. Then (\tilde{u},\tilde{v}) is a ground state for (NLS). Below we argue analogously with the proof of Theorem 1.1 to get a positive ground state for (NLS). The proof is complete. □

*Proof of Theorem 1.3* By Lemma 5.3, repeating the argument of Theorem 1.2, we show the existence of a ground state for (NLS) and then look for a positive ground state as the argument of Theorem 1.1. □