Existence of ground state solutions for an asymptotically 2-linear fractional Schrödinger–Poisson system

Abstract

In this paper, we investigate the following fractional Schrödinger–Poisson system:

$$\left \{ \textstyle\begin{array}{l@{\quad}l} (-\Delta)^{s} u + u + \phi u = f(u), & \text{in } \mathbb{R}^{3}, \\ (-\Delta)^{t} \phi= u^{2}, & \text{in } \mathbb{R}^{3}, \end{array}\displaystyle \right . $$

where $\frac{3}{4} < s < 1$, $\frac{1}{2} < t < 1$, and f is a continuous function, which is superlinear at zero, with $f(\tau) \tau \ge3 F(\tau) \ge0$, $F(\tau) = \int_{0}^{\tau} f(s) \,ds$, $\tau \in\mathbb{R}$. We prove that the system admits a ground state solution under the asymptotically 2-linear condition. The result here extends the existing study.

1 Introduction

In this paper, we study the existence of ground state solutions for the following fractional Schrödinger–Poisson system:

$$ \left \{ \textstyle\begin{array}{l@{\quad}l} (-\Delta)^{s} u + u + \phi u = f(u), & \text{in } \mathbb{R}^{3}, \\ (-\Delta)^{t} \phi= u^{2}, & \text{in } \mathbb{R}^{3}, \end{array}\displaystyle \right . $$

(1.1)

where $\frac{3}{4} < s < 1$, $\frac{1}{2} < t < 1$, $(-\Delta)^{s}$ and $(-\Delta)^{t}$ are the fractional Laplace operators, f satisfies the following conditions:

$(f_{1})$:

$f \in C(\mathbb{R}, \mathbb{R})$, $\lim_{\tau\to0} \frac {f(\tau)}{\tau} = 0$;

$(f_{2})$:

$$\lim_{ \vert \tau \vert \to\infty} \frac{f(\tau)}{ \vert \tau \vert ^{2}} = \mu \quad \text{with } \sqrt{ \frac{54}{\pi} C(3, s)^{-1} S^{2} \biggl(\frac{32 \pi}{3} \biggr)^{\frac{5}{3}}} < \mu< + \infty, $$

where the constant S and the function $C(3, s)$ will be specified in Sect. 2;

$(f_{3})$:

$$f(\tau) \tau \ge3 F(\tau) \ge0, \quad\forall\tau \in\mathbb {R}, \text{where } F(\tau) = \int_{0}^{\tau} f(s) \,ds. $$

In recent years, the nonlinear fractional Schrödinger–Poisson systems have received a lot of attention. In [1], Gao, Tang and Chen studied the existence of ground state solutions of (1.1) in a mild assumption on f with super-quadratic nonlinearity. If u is replaced by $V(x) u$ and $f(u) = \mu|u|^{q-2} u + |u|^{2_{s}^{*} - 2} u$ ($2_{s}^{*} = \frac {6}{3-2s}$) in (1.1), the existence of a nontrivial ground state solution is given by Teng [2]. In [3], based on the symmetric mountain pass theorem, He and Jing investigated a class of fractional Schrödinger–Poisson system with superlinear terms, the existence and multiplicity of nontrivial solutions of such a system are obtained. Wang, Ma and Guan [4] studied the existence of a sign-changing solution of the following nonlinear fractional Schrödinger–Poisson system:

$$\left \{ \textstyle\begin{array}{l@{\quad}l} (-\Delta)^{s} u + V(x) u + \phi u = K(x) f(u), & \text{in } \mathbb{R}^{3}, \\ (-\Delta)^{t} \phi= u^{2}, & \text{in } \mathbb{R}^{3}, \end{array}\displaystyle \right . $$

by means of the constraint variational method and the quantitative deformation lemma.

When $s=t=1$, system (1.1) reduces to the following Schrödinger–Poisson system:

$$ \left \{ \textstyle\begin{array}{l@{\quad}l} -\Delta u + u + \phi u = g(u), & \text{in } \mathbb{R}^{3}, \\ -\Delta\phi= u^{2}, & \text{in } \mathbb{R}^{3}. \end{array}\displaystyle \right . $$

(1.2)

Yin, Wu and Tang [5] proved the existence of ground state solutions of (1.2) by using

$$\lim_{ \vert t \vert \to\infty} \frac{f(t)}{ \vert t \vert ^{2}} = \nu \quad \text{with } \sqrt{ \frac{189}{8 \pi S} \biggl(\frac{32 \pi}{3} \biggr)^{\frac{5}{3}}} < \nu< + \infty, $$

instead of the usual 2-superlinear condition $\lim_{|t| \to\infty} \frac{G(t)}{|t|^{3}} = + \infty$ ($G(t) = \int_{0}^{t} g(s) \,ds$), which relaxed the conditions of nonlinearity in [6–8].

Inspired by [5], the main objective of this paper is to extend the main results of [1], by relaxing the condition of super-quadratic nonlinearity used in [1]. That is, the nonlinearity f is assumed to be asymptotically 2-linear. We deal with the nonlinear fractional Schrödinger–Poisson system (1.1) in view of variational method and some analysis technique. Our result also extends the main results of [5].

2 Preliminaries

The fractional Sobolev space $H^{s}(\mathbb{R}^{3})$ can be described by means of the Fourier transform, i.e.

$$H^{s}\bigl(\mathbb{R}^{3}\bigr) = \biggl\{ u \in L^{2}\bigl(\mathbb{R}^{3}\bigr) : \int_{\mathbb {R}^{3}} \bigl( \vert \xi \vert ^{2s} \bigl\vert \widehat{u}(\xi) \bigr\vert ^{2} + \bigl\vert \widehat{u}( \xi) \bigr\vert ^{2}\bigr) \,d \xi < + \infty \biggr\} $$

endowed with the norm

$$\begin{aligned} \Vert u \Vert & := \Vert u \Vert _{H^{s}} = \biggl( \int_{\mathbb{R}^{3}} \bigl( \vert \xi \vert ^{2s} \bigl\vert \widehat {u}(\xi) \bigr\vert ^{2} + \bigl\vert \widehat{u}( \xi) \bigr\vert ^{2}\bigr) \,d \xi \biggr)^{\frac{1}{2}} \\ &= \biggl( \int_{\mathbb{R}^{3}} \bigl( \bigl\vert (-\Delta)^{\frac{s}{2}} u(x) \bigr\vert ^{2} + \bigl\vert u(x) \bigr\vert ^{2}\bigr) \,dx \biggr)^{\frac{1}{2}}, \quad\forall u \in H^{s}\bigl(\mathbb {R}^{3}\bigr).\end{aligned} $$

Since $4s + 2t > 3$, we have $2 \le\frac{12}{3+2t} \le\frac {6}{3-2t}$, thus $H^{s}(\mathbb{R}^{3}) \hookrightarrow L^{\frac {12}{3+2t}}(\mathbb{R}^{3})$. From [1], we know that there exists a unique $\phi_{u}^{t} \in\mathcal {D}^{t, 2}(\mathbb{R}^{3}) = \{u \in L^{2_{t}^{*}}(\mathbb{R}^{3}) : |\xi|^{t} \widehat{u}(\xi) \in L^{2}(\mathbb{R}^{3})\}$ which is a weak solution of $(-\Delta)^{t} \phi_{u}^{t} = u^{2}$, and it has the following representation:

$$\phi_{u}^{t}(x) = c_{t} \int_{\mathbb{R}^{3}} \frac{u^{2}(y)}{ \vert x-y \vert ^{3-2t}} \,dy, \quad x \in \mathbb{R}^{3}, $$

where $c_{t} = \pi^{-\frac{3}{2}} 2^{-2t} \frac{\varGamma(3-2t)}{\varGamma (t)}$. Substituting $\phi_{u}^{t}$ in (1.1), we obtain the following fractional Schrödinger equation:

$$ (-\Delta)^{s} u + u + \phi_{u}^{t} u = f(u), \quad x \in\mathbb{R}^{3}. $$

(2.1)

For the properties of $\phi_{u}^{t}$, see [2]. By (2.1), we define the functional $\mathcal{I} : H^{s}(\mathbb{R}^{3}) \to\mathbb{R}$ as follows:

$$ \mathcal{I}(u) = \frac{1}{2} \int_{\mathbb{R}^{3}} \bigl( \bigl\vert (-\Delta)^{\frac{s}{2}} u \bigr\vert ^{2} + u^{2} \bigr) \,dx + \frac{1}{4} \int _{\mathbb{R}^{3}} \phi_{u}^{t} u^{2} \,dx - \int_{\mathbb{R}^{3}} F(u) \,dx, $$

(2.2)

where $F(u) = \int_{0}^{u} f(x) \,dx$. It is easy to see that $(f_{1})$ and $(f_{2})$ imply that $\mathcal{I}$ is a well-defined $C^{1}$-functional, and

$$\bigl\langle \mathcal{I}^{\prime}(u), v \bigr\rangle = \int_{\mathbb{R}^{3}} \bigl( (-\Delta)^{\frac{s}{2}} u (- \Delta)^{\frac{s}{2}} v + uv\bigr) \,dx + \int_{\mathbb{R}^{3}} \phi_{u}^{t} u v \,dx - \int_{\mathbb{R}^{3}} f(u) v \, dx, \quad \forall v \in H^{s} \bigl(\mathbb{R}^{3}\bigr). $$

Hence, if u is a critical point of $\mathcal{I}$, then $(u, \phi _{u}^{t})$ is a solution of (1.1).

Set

$$u_{R}(x) = \left \{ \textstyle\begin{array}{l@{\quad}l} \frac{1}{R}, & \vert x \vert \le R, \\ \frac{1}{R} (2 - \frac{ \vert x \vert }{R} ), & R < \vert x \vert \le2R, \\ 0, & \vert x \vert > 2R. \end{array}\displaystyle \right . $$

Hence $u_{R} \in H^{s}(\mathbb{R}^{3})$. By Proposition 3.4 in [9], we have

$$ \Vert u_{R} \Vert _{H^{s}}^{2} = 2 C(3, s)^{-1} \int_{\mathbb{R}^{3}} \vert \xi \vert ^{2s} \bigl\vert \mathcal{F} u_{R}(\xi) \bigr\vert ^{2} \,d \xi, $$

(2.3)

where $\mathcal{F}$ is the usual Fourier transform in $\mathbb{R}^{3}$, and

$$C(3, s) = \biggl( \int_{\mathbb{R}^{3}} \frac{1-\cos(\zeta_{1})}{ \vert \zeta \vert ^{3+2s}} \,d\zeta \biggr)^{-1}, $$

here $\zeta= (\zeta_{1}, \zeta_{2}, \zeta_{3})$. From the inequality $|\xi |^{2s} \le1 + |\xi|^{2}$, $s \in(0, 1]$, together with (2.3), we get

$$ \Vert u_{R} \Vert _{H^{\alpha}}^{2} \le2 C(3, s)^{-1} \int_{\mathbb {R}^{3}} \bigl(1 + \vert \xi \vert ^{2}\bigr) \bigl\vert \mathcal{F} u_{R}(\xi) \bigr\vert ^{2} \,d \xi = 2 C(3, s)^{-1} \Vert u_{R} \Vert _{H^{1}}^{2}. $$

(2.4)

If $t > \frac{1}{2}$, then we have by Lemma 2.3 in [2]

$$\int_{\mathbb{R}^{3}} \phi_{u_{R}(x)}^{t} u_{R}(x)^{2} \,dx \le S_{t}^{2} \vert u_{R} \vert _{\frac{12}{3+2t}}^{4}, $$

where

$$S_{t} = \inf_{u \in\mathcal{D}^{t, 2}(\mathbb{R}^{3}) \setminus\{0\}} \frac{\int_{\mathbb{R}^{3}} \vert (-\Delta)^{\frac{t}{2}} u \vert ^{2} \,dx}{ (\int_{\mathbb{R}^{3}} \vert u(x) \vert ^{2_{t}^{*}} \,dx )^{\frac{2}{2_{t}^{*}}}}. $$

Remark 2.1

If $t = 1$, then the above inequalities modifies to the following inequalities:

$$ \int_{\mathbb{R}^{3}} \phi_{u_{R}(x)} u_{R}(x)^{2} \,dx \le S^{2} \vert u_{R} \vert _{\frac{12}{5}}^{4}, $$

(2.5)

where

$$S = \inf_{u \in\mathcal{D}^{1, 2}(\mathbb{R}^{3}) \setminus\{0\}} \frac{\int_{\mathbb{R}^{3}} \vert \nabla u \vert ^{2} \,dx}{ (\int_{\mathbb{R}^{3}} \vert u(x) \vert ^{6} \,dx )^{\frac{1}{3}}}. $$

From [5], we have

$$ \int_{\mathbb{R}^{3}} \bigl\vert \nabla u_{R}(x) \bigr\vert ^{2} \,dx = \frac{28\pi }{3R}, \qquad \int_{\mathbb{R}^{3}} \bigl\vert u_{R}(x) \bigr\vert ^{3} \,dx \ge\frac{4 \pi}{3}, $$

(2.6)

and

$$ \vert u_{R} \vert _{\frac{12}{5}}^{4} = \biggl( \frac{32 \pi}{3} \biggr)^{\frac{5}{3}} R. $$

(2.7)

Lemma 2.1

If$(f_{1})$and$(f_{2})$hold, then

• (i) there exists a$v \in H^{s}(\mathbb{R}^{3}) \setminus\{0\}$such that$\mathcal{I}(v) \le0$;

• (ii) $c := \inf_{\gamma\in\varGamma} \max_{t \in[0, 1]} \mathcal {I}(\gamma(t)) > 0$, where

  $$\varGamma= \bigl\{ \gamma\in C\bigl([0, 1], H^{s}\bigl( \mathbb{R}^{3}\bigr)\bigr) : \gamma(0) = 0, \gamma(1) = v\bigr\} . $$

Proof

Set $R = \frac{8 \pi\mu_{0}}{3 S^{2} (\frac{32 \pi}{3} )^{\frac {5}{3}}}$, where

$$\mu_{0} = \sqrt{ \frac{45}{8 \pi} C(3, s)^{-1} S^{2} \biggl(\frac{32 \pi }{3} \biggr)^{\frac{5}{3}}} < \sqrt{ \frac{6}{\pi} C(3, s)^{-1} S^{2} \biggl(\frac{32 \pi}{3} \biggr)^{\frac{5}{3}}}. $$

Denote $u_{R, \theta} = \theta^{2} u_{R}(\theta x)$, from (2.2), (2.4), Fatou’s lemma, $(f_{2})$ and (2.5)–(2.7), we obtain

$$\begin{aligned} \lim_{\theta\to+ \infty} \frac{\mathcal{I}(u_{R, \theta})}{\theta ^{3}} &= \lim_{\theta\to+ \infty} \frac{1}{\theta^{3}} \biggl(\frac{1}{2} \int_{\mathbb{R}^{3}} \bigl( \bigl\vert (-\Delta)^{\frac{s}{2}} u_{R, \theta} \bigr\vert ^{2} + u_{R, \theta}^{2} \bigr) \,dx \\ &\quad+ \frac{1}{4} \int_{\mathbb{R}^{3}} \phi_{u_{R, \theta}}^{t} u_{R, \theta}^{2} \,dx - \int_{\mathbb{R}^{3}} F(u_{R, \theta}) \,dx \biggr) \\ &\le\lim_{\theta\to+ \infty} \frac{1}{\theta^{3}} \biggl( C(3, s)^{-1} \int_{\mathbb{R}^{3}} \bigl( \vert \nabla u_{R, \theta} \vert ^{2} + u_{R, \theta}^{2} \bigr) \,dx \\ &\quad+ \frac{1}{4} \int_{\mathbb{R}^{3}} \phi_{u_{R, \theta}}^{t} u_{R, \theta}^{2} \,dx - \int_{\mathbb{R}^{3}} F(u_{R, \theta}) \,dx \biggr) \\ &= \lim_{\theta\to+ \infty} \frac{1}{\theta^{3}} \biggl( C(3, s)^{-1} \biggl[\theta^{3} \int_{\mathbb{R}^{3}} \bigl\vert \nabla u_{R}(x) \bigr\vert ^{2} \,dx + \theta \int_{\mathbb{R}^{3}} u_{R}(x)^{2} \,dx \biggr] \\ &\quad + \frac{\theta^{1+2t}}{4} \int_{\mathbb{R}^{3}} \phi_{u_{R}(x)}^{t} u_{R}(x)^{2} \,dx - \int_{\mathbb{R}^{3}} F\bigl(\theta^{2} u_{R}(\theta x)\bigr) \,dx \biggr) \\ &= C(3, s)^{-1} \int_{\mathbb{R}^{3}} \bigl\vert \nabla u_{R}(x) \bigr\vert ^{2} \,dx + \frac{1}{4} \int_{\mathbb{R}^{3}} \phi_{u_{R}(x)}^{t} u_{R}(x)^{2} \,dx \cdot\lim_{\theta\to+ \infty} \frac{1}{\theta ^{2(1-t)}} \\ &\quad- \lim_{\theta\to+ \infty} \int_{\mathbb{R}^{3}} \frac{F(\theta^{2} u_{R})}{ \vert \theta^{2} u_{R} \vert ^{3}} \vert u_{R} \vert ^{3} \,dx \\ &\le \left\{ \textstyle\begin{array}{l@{\quad}l} \frac{28 \pi}{3R} C(3, s)^{-1} + \frac{S^{2}}{4} \vert u_{R} \vert _{\frac{12}{5}}^{4} - \frac{\mu}{3} \int_{\mathbb{R}^{3}} \vert u_{R} \vert ^{3} \,dx, & t = 1, \\ \frac{28 \pi}{3R} C(3, s)^{-1} - \frac{\mu}{3} \int_{\mathbb{R}^{3}} \vert u_{R} \vert ^{3} \,dx, & t < 1 \end{array}\displaystyle \right . \\ &< \frac{28 \pi}{3R} C(3, s)^{-1} + \frac{S^{2}}{4} \vert u_{R} \vert _{\frac{12}{5}}^{4} - \sqrt{ \frac{6}{\pi} C(3, s)^{-1} S^{2} \biggl(\frac{32 \pi}{3} \biggr)^{\frac{5}{3}}} \int_{\mathbb{R}^{3}} \vert u_{R} \vert ^{3} \,dx \\ &< \frac{28 \pi}{3R} C(3, s)^{-1} + \frac{S^{2}}{4} \biggl( \frac{32 \pi }{3} \biggr)^{\frac{5}{3}} R - \frac{4 \pi}{3} \mu_{0} \\ &= - \frac{2 \pi}{3R} C(3, s)^{-1} < 0.\end{aligned} $$

Thus, $\mathcal{I}(u_{R, \theta}) \le0$ if θ is sufficiently large.

(ii) By $(f_{1})$ and $(f_{2})$, for $\varepsilon= \frac{1}{4} > 0$, there exists $C > 0$ such that

$$ f(\theta) \le\frac{1}{4} \theta+ C \theta^{2}. $$

(2.8)

From (2.8) and by using the Sobolev inequality, we obtain

$$\mathcal{I}(u) \ge\frac{1}{2} \Vert u \Vert ^{2} - \frac{1}{4} \vert u \vert _{2}^{2} - C \vert u \vert _{3}^{3} \ge\frac{1}{4} \Vert u \Vert ^{2} - C S_{s,3}^{-\frac{3}{2}} \Vert u \Vert ^{3}, $$

where

$$S_{s, 3} = \inf_{u \in\mathcal{D}^{s, 2}} \frac{\int_{\mathbb{R}^{3}} ( \vert (-\Delta)^{\frac{s}{2}} u \vert ^{2} + u^{2}) \,dx}{ (\int_{\mathbb{R}^{3}} \vert u(x) \vert ^{3} \,dx )^{\frac{2}{3}} }. $$

For sufficiently small $\rho> 0$, we have $I(u) > 0$ with $\|u\| = \rho$. □

Similar to the proof of Lemma 2.2 in [5], we have the following lemma.

Lemma 2.2

Suppose that$(f_{1})$, $(f_{2})$hold. If$\{u_{n}\} \subset H^{s}(\mathbb{R}^{3})$is a bounded$(PS)_{c\neq0}$sequence of$\mathcal{I}$, then there exists$u_{0} \neq 0$such that$\mathcal{I}^{\prime}(u_{0}) = 0$.

3 Main result

Theorem 3.1

Assume the conditions$(f_{1})$–$(f_{3})$are satisfied, then system (1.1) has at least a ground state solution.

Proof

For convenience, we introduce a functional on $H^{s}(\mathbb{R}^{3})$ as follows:

$$\begin{aligned}[b] \mathcal{J}(u) ={}& \frac{1+2s}{2} \int_{\mathbb{R}^{3}} \bigl\vert (-\Delta)^{\frac {s}{2}} u \bigr\vert ^{2} \,dx + \frac{1}{2} \int_{\mathbb{R}^{3}} u^{2} \,dx + \frac{5-2t}{4} \int_{\mathbb {R}^{3}} \phi_{u}^{t} u^{2} \,dx \\ &+ \int_{\mathbb{R}^{3}} \bigl[(1+2t) F(u) - 2 f(u) u\bigr] \,dx.\end{aligned} $$

(3.1)

Inspired by the idea of Jeanjean [10], we define the map $\varPsi: \mathbb{R} \times H^{s} (\mathbb{R}^{3}) \to H^{s} (\mathbb{R}^{3})$ by $\varOmega(\lambda, w)(x) = e^{2 \lambda} w(e^{\lambda} x)$. For each λ and $w \in H^{s} (\mathbb{R}^{3})$, we can compute the functional $\mathcal{I} \circ\varOmega$ as follows:

$$\begin{aligned}[b] \mathcal{I}\bigl(\varOmega(\lambda, w)\bigr) = {}&\frac{e^{(1+2s)}\lambda}{2} \int _{\mathbb{R}^{3}} \bigl\vert (-\Delta)^{\frac{s}{2}} w \bigr\vert ^{2} + \frac{e^{\lambda}}{2} \int_{\mathbb{R}^{3}} w^{2} \\ &+ \frac{e^{(5-2t)\lambda}}{4} \int_{\mathbb{R}^{3}} \phi _{w}^{t} w^{2} - \frac{1}{e^{3\lambda}} \int_{\mathbb{R}^{3}} F\bigl(e^{2\lambda} w\bigr).\end{aligned} $$

(3.2)

By (3.2), $(f_{1})$ and $(f_{2})$, we see that $\mathcal{I} \circ\varOmega$ is continuously Fréchet-differentiable on $\mathbb{R} \times H^{s}(\mathbb{R}^{3})$. By virtue of Lemma 2.1, there exists $\lambda^{*} \in\mathbb{R}$ such that $(\mathcal{I} \circ\varOmega)(\lambda^{*}, u_{R}) < 0$. The mountain pass level of $\mathcal{I} \circ\varOmega$ is given as follows:

$$ \bar{c} = \inf_{\bar{\gamma} \in\bar{\varGamma}} \sup_{t \in[0, 1]} (\mathcal{I} \circ\varOmega) \bigl(\bar{\gamma}(t)\bigr), $$

(3.3)

where the family of paths is denoted by

$$\bar{\varGamma} = \bigl\{ \bar{\gamma} \in C\bigl([0, 1]; \mathbb{R} \times H^{s}(\mathbb{R})\bigr) : \bar{\gamma}(0) = (0, 0), \ (\mathcal{I} \circ\varOmega) \bigl(\bar{\gamma}(1)\bigr) < 0\bigr\} . $$

For $\varGamma= \{\varOmega\circ\bar{\gamma}: \bar{\gamma} \in\bar{\varGamma }\}$, we have $c \le\bar{c}$. Obviously, $\{0\} \times\varGamma\subset\bar{\varGamma}$ and then $\bar{c} \le c$. Thus, $\bar{c} = c$. It follows for each $(\eta, u) \in\mathbb {R} \times H^{s}(\mathbb{R}^{3})$ that

$$\begin{gathered}\begin{aligned}[b] \mathcal{I}^{\prime}\bigl(\varOmega(\lambda_{n}, w_{n})\bigr) \bigl[\varOmega(\lambda_{n}, u)\bigr] ={}& e^{(1+2s) \lambda_{n}} \int_{\mathbb{R}^{3}} (-\Delta)^{\frac{s}{2}} w_{n} (- \Delta)^{\frac{s}{2}}u \\ &+ e^{\lambda_{n}} \int_{\mathbb{R}^{3}} w_{n} u + e^{(5-2t)\lambda_{n}} \int _{\mathbb{R}^{3}} \phi_{w_{n}}^{t} w_{n} u - \frac{1}{e^{\lambda_{n}}} \int_{\mathbb{R}^{3}} f\bigl(e^{2 \lambda_{n}} w_{n}\bigr) u,\end{aligned} \\ (\mathcal{I} \circ\varOmega)^{\prime}(\lambda_{n}, w_{n})[\eta, u] = \mathcal {I}^{\prime}\bigl(\varOmega( \lambda_{n}, w_{n})\bigr)\bigl[\varOmega( \lambda_{n}, u)\bigr] + \mathcal{J}\bigl(\varOmega( \lambda_{n}, w_{n})\bigr)\eta.\end{gathered} $$

From Theorem 2.9 of [11], (3.3) and setting $u_{n} = \varOmega(\lambda_{n}, w_{n})$, one has

$$ \mathcal{I}(u_{n}) \to c > 0, \qquad \mathcal{I}^{\prime }(u_{n}) \to0,\qquad \mathcal{J}(u_{n}) \to0. $$

(3.4)

By (3.4) and $(f_{3})$, we derive that

$$\begin{aligned}[b] &c \ge\mathcal{I}(u_{n}) - \frac{1}{5-2t} \mathcal{J}(u_{n}) + o(1) \\ &\quad= \frac{2-(s+t)}{5-2t} \int_{\mathbb{R}^{3}} \bigl\vert (-\Delta)^{\frac{s}{2}} u_{n} \bigr\vert ^{2} \,dx + \frac{2-t}{5-2t} \int_{\mathbb{R}^{3}} u_{n}^{2} \,dx \\ &\qquad+ \frac{2}{5-2t} \int_{\mathbb{R}^{3}} \bigl[f(u_{n}) u_{n} - 3 F(u_{n})\bigr] \,dx + o(1) \\ &\quad\ge\frac{2-t}{5-2t} \int_{\mathbb{R}^{3}} u_{n}^{2} \,dx + o(1),\end{aligned} $$

(3.5)

which implies that $\{u_{n}\}$ is bounded in $L^{2}(\mathbb{R}^{3})$. According to (3.4), we obtain

$$\begin{aligned}[b] \int_{\mathbb{R}^{3}} f(u_{n}) u_{n} \,dx + o\bigl( \Vert u_{n} \Vert \bigr) &= \int_{\mathbb{R}^{3}} \bigl( \bigl\vert (-\Delta)^{\frac{s}{2}} u_{n} \bigr\vert ^{2} + u_{n}^{2} \bigr) \,dx + \int_{\mathbb{R}^{3}} \phi_{u_{n}}^{t} u_{n}^{2} \,dx \\ &\ge \int_{\mathbb{R}^{3}} \bigl\vert (-\Delta)^{\frac{s}{2}} u_{n} \bigr\vert ^{2} \,dx.\end{aligned} $$

(3.6)

Combining $(f_{1})$ and $(f_{2})$, we have

$$ f(\tau) \tau\le C \vert \tau \vert ^{3} + \varepsilon \tau^{2}. $$

(3.7)

By means of (3.7), the interpolation and Sobolev inequalities, we get

$$\begin{aligned}[b] \int_{\mathbb{R}^{3}} f(u_{n}) u_{n} \,dx &\le C \int_{\mathbb{R}^{3}} \vert u_{n} \vert ^{3} \,dx + \varepsilon \int_{\mathbb{R}^{3}} u_{n}^{2} \,dx \\ &\le C \biggl( \int_{\mathbb{R}^{3}} \vert u_{n} \vert ^{2} \,dx \biggr)^{\frac{6s-3}{4s}} \cdot \biggl( \int_{\mathbb{R}^{3}} \vert u_{n} \vert ^{2_{s}^{*}} \,dx \biggr)^{\frac{3-2s}{4s}} + C \varepsilon \\ &\le C \biggl( \int_{\mathbb{R}^{3}} \vert u_{n} \vert ^{2} \,dx \biggr)^{\frac{6s-3}{4s}} \cdot \biggl( \int_{\mathbb{R}^{3}} \bigl\vert (-\Delta)^{\frac {s}{2}} u_{n} \bigr\vert ^{2} \,dx \biggr)^{\frac{3}{4s}} + C \varepsilon.\end{aligned} $$

(3.8)

Since $s > \frac{3}{4}$, by (3.6) and (3.8), we know that $\{(-\Delta )^{\frac{s}{2}} u_{n}\}$ is bounded in $L^{2}(\mathbb{R}^{3})$. Hence, $\{ u_{n}\}$ is bounded in $H^{s}(\mathbb{R}^{3})$.

Define

$$m = \inf_{M} \mathcal{I}(u), \quad M = \bigl\{ u \in H^{s}\bigl(\mathbb{R}^{3}\bigr) \setminus\{ 0\} | \mathcal{I}^{\prime}(u) = 0\bigr\} . $$

In view of Lemma 2.2 and $\{u_{n}\}$ being bounded, we obtain $u_{0} \neq 0$ and $\mathcal{I}^{\prime}(u_{0}) = 0$. Thus M is not empty and $0 \le m \le\mathcal{I}(u_{0})$. In the following we will prove m can be achieved in M. Suppose that $\{ u_{n}\}$ is a sequence of nontrivial critical points of $\mathcal{I}$ satisfying $\mathcal{I}(u_{n}) \to m$. Similar to the proofs of (3.5), (3.6) and (3.8), we find that $\{u_{n}\}$ is bounded in $H^{s}(\mathbb {R}^{3})$. By $\mathcal{I}^{\prime}(u_{n}) u_{n} = 0$, (2.8) and the Sobolev inequality, we have

$$\begin{aligned}[b] \Vert u_{n} \Vert ^{2}& = \int_{\mathbb{R}^{3}} f(u_{n}) u_{n} \,dx - \int_{\mathbb{R}^{3}} \phi_{u_{n}}^{t} u_{n}^{2} \,dx \\ &\le\frac{1}{4} \vert u_{n} \vert _{2}^{2} + C \vert u_{n} \vert _{3}^{3} \le \frac{1}{4} \Vert u_{n} \Vert ^{2} + C S_{s,3}^{-\frac{3}{2}} \Vert u_{n} \Vert ^{3}.\end{aligned} $$

(3.9)

From (3.9), there exists a positive $\rho> 0$ such that

$$ \lim\inf_{n \to\infty} \Vert u_{n} \Vert \ge\rho> 0. $$

(3.10)

Applying the Lions lemma in [11], if $\{u_{n}\}$ vanishes, one has $u_{n} \to0$ in $L^{q}(\mathbb{R}^{3})$ for all $q \in(2, 6)$. Thus, if $t > \frac{1}{2}$, then it follows (12) in [2] that

$$\int_{\mathbb{R}^{3}} \phi_{u_{n}}^{t} u_{n}^{2} \,dx \le S_{t}^{\frac{1}{2}} \vert u_{n} \vert _{\frac{12}{3+2t}}^{4} \to0. $$

By (3.7) and $u_{n} \to0$ in $L^{q}(\mathbb{R}^{3})$, we get $\int_{\mathbb {R}^{3}} f(u_{n}) u_{n} \,dx \to0$. Combining with (3.9), we can easily deduce that $\|u_{n}\| \to0$ in a contradiction with (3.10). Hence there exist $r, \delta> 0$ and a sequence $\{y_{n}\} \subset\mathbb{R}^{3}$ such that $\lim_{n \to\infty} \sup_{y_{n} \in\mathbb{R}^{3}} \int_{B_{r}(y_{n})} |u_{n}|^{2} \ge\delta> 0$. Set $\bar{u}_{n} = u_{n}(x+y_{n})$, then we have $\bar{u}_{n} \rightharpoonup u \neq 0$ in $H^{s}(\mathbb{R}^{3})$, $\mathcal {I}(\bar{u}_{n}) \to m$ and $\mathcal{I}^{\prime}(\bar{u}_{n}) = 0$. Thus, we get $\mathcal{I}^{\prime}(u) = 0$ and $\mathcal{I}(u) \ge m$. Since $\mathcal{I}^{\prime}(u) = 0$, one has

$$ \int_{\mathbb{R}^{3}} \bigl( \bigl\vert (-\Delta)^{\frac{s}{2}} u \bigr\vert ^{2} + u^{2} \bigr) \,dx + \int_{\mathbb{R}^{3}} \phi_{u}^{t} u^{2} \,dx - \int_{\mathbb{R}^{3}} f(u) u \,dx = 0, $$

(3.11)

and by the Pohožaev identity [2, 12], we have

$$ \frac{3-2s}{2} \int_{\mathbb{R}^{3}} \bigl\vert (-\Delta)^{\frac {s}{2}} u \bigr\vert ^{2} \,dx + \frac{3}{2} \int_{\mathbb{R}^{3}} u^{2} \,dx + \frac{3+2t}{4} \int_{\mathbb{R}^{3}} \phi_{u}^{t} u^{2} \,dx = 3 \int_{\mathbb {R}^{3}} F(u) \,dx. $$

(3.12)

According to (3.11)–(3.12), one has $\mathcal{J}(u) = 0$. As $\mathcal {I}^{\prime}(\bar{u}_{n}) = 0$, $\mathcal{I}^{\prime}(u) = 0$ and by Fatou’s lemma we have

$$\begin{aligned} m &= \lim_{n \to\infty} \biggl(\frac{2-(s+t)}{5-2t} \int_{\mathbb {R}^{3}} \bigl\vert (-\Delta)^{\frac{s}{2}} \bar{u}_{n} \bigr\vert ^{2} \,dx + \frac{2-t}{5-2t} \int_{\mathbb{R}^{3}} \bar{u}_{n}^{2} \,dx \\ &\quad + \frac{2}{5-2t} \int_{\mathbb{R}^{3}} \bigl[f(\bar{u}_{n}) \bar{u}_{n} - 3 F(\bar{u}_{n})\bigr] \,dx \biggr) \\ &\ge\frac{2-(s+t)}{5-2t} \int_{\mathbb{R}^{3}} \bigl\vert (-\Delta)^{\frac{s}{2}} u \bigr\vert ^{2} \,dx + \frac{2-t}{5-2t} \int_{\mathbb{R}^{3}} u^{2} \,dx \\ &\quad+ \frac{2}{5-2t} \int_{\mathbb{R}^{3}} \bigl[f(u) u - 3 F(u)\bigr] \,dx \\ &= \mathcal{I}(u) - \frac{1}{5-2t} \mathcal{J}(u) = \mathcal{I}(u) \ge m.\end{aligned} $$

Hence $\mathcal{I}(u) = m$. The proof is complete. □

Remark 3.1

If $s = t =1$, Our main result Theorem 3.1 reduces to Theorem 1.1 in [5]. On the other hand, Theorem 3.1 in this paper relaxes the condition of super-quadratic nonlinearity in [1] to being asymptotically 2-linear.