Nontrivial Solutions for Schrödinger–Poisson Systems With Doubly Critical Exponents

Abstract

This article investigates the existence and nonexistence of nontrivial solutions for a Schrödinger–Poisson system with doubly critical exponents: $\begin{aligned} {\begin{cases} - Δ u + u + λ ϕ | u |^{3} u = | u |^{4} u + μ | u |^{q - 2} u, & in R^{3}, \\ - Δ ϕ = | u |^{5}, & in R^{3}, \end{cases} \end{aligned}$ where $λ > 0$ , $μ > 0$ , and $q \in (2, 6)$ . Under optimal parameter ranges, we prove both the existence of radial ground state solutions and the nonexistence of nontrivial solutions. This work completely resolves an open problem proposed in [X. Chen, C. Tang, Communications on Pure and Applied Analysis 23 (2024), 1011–1043].

Keywords

Schrödinger–Poisson problem doubly critical exponents existence nonexistence

1. Introduction

In this article, we study the existence and nonexistence of nontrivial solutions for the following Schrödinger–Poission system with doubly critical exponents:

\begin{aligned} {\begin{cases} - Δ u + u + λ ϕ | u |^{3} u = | u |^{4} u + μ | u |^{q - 2} u, & in R^{3}, \\ - Δ ϕ = | u |^{5}, & in R^{3}, \end{cases} \end{aligned}

(1.1)

where

λ > 0

μ > 0

are physical parameters, and

q \in (2, 6)

. The investigation of (1.1) is motivated by recent studies on the Schrödinger–Poisson system

\begin{aligned} {\begin{cases} - Δ u + u + λ ϕ | u |^{p - 2} u = f (u), & in R^{3}, \\ - Δ ϕ = | u |^{p}, & in R^{3}, \end{cases} \end{aligned}

(1.2)

where

p \in [5 / 3, 5]

and the nonlinearity

f

satisfies appropriate growth and regularity conditions. The values

5 / 3

and

5

are referred to as the lower and upper critical exponents, respectively, due to their connection with the Hardy–Littlewood–Sobolev inequality (see Lemma 2.2 below). Over the past decades, system (1.2) has been extensively studied in the case of a subcritical nonlocal term (

5 / 3 < p < 5

), while fewer results are available for the critical case (

p = 5

); see, for instance, Li and Zhang (2013) and Li et al. (2014, 2017). These existing results on the existence or nonexistence of nontrivial solutions reveal significant differences between the critical and subcritical nonlocal cases.

Especially, in the case $p = 2$ , system (1.2) reduces to the following system

\begin{aligned} {\begin{cases} - Δ u + u + λ ϕ u = f (u), & in R^{3}, \\ - Δ ϕ = | u |^{2}, & in R^{3}, \end{cases} \end{aligned}

(1.3)

which was originally introduced by Benci and Fortunato (1998) in 1998 to model the interaction between a charged particle and the electrostatic field in quantum mechanics. Due to its physical relevance, this system has been extensively studied by many researchers in terms of the existence and nonexistence of positive solutions, multiple solutions, least energy solutions, radial solutions, sign-changing solutions and normalized solutions; see, for example, Ambrosetti and Ruiz (2008), Azzollini and Pomponio (2008), Cerami and Vaira (2010), Chen and Tang (2016), Chen and Wu (2022), Coclite (2003), Dou et al. (2024), Seok (2013), Shu et al. (2025) and Tang and Chen (2017).

For (1.1), the Schrödinger–Poisson system with critical nonlinearity and critical nonlocal term, Chen and Tang (2024) proved that the system has no nontrivial solutions when $λ \geq ((q + 2) / 8)^{2}$ and $μ = 1$ . However, establishing the existence of nontrivial solutions for (1.1) is significantly more challenging. Two major difficulties arise: first, the lack of compactness; second, the competition between the critical nonlocal term and the critical nonlinearity. Motivated by this, Chen and Tang (2024) proposed the following open problem:

(Q) Investigate the existence of nontrivial solutions in the case $0 < λ < ((q + 2) / 8)^{2}$ and $μ = 1$ .

In this article, by employing more refined analytical techniques, we prove that (1.1) admits no nontrivial solutions under weaker assumptions on $λ$ for all $μ > 0$ . Moreover, we construct a function $\hat{w} \in H_{rad}^{1} (R^{3})$ with negative energy. Based on this construction, we employ the Ekeland variational principle to obtain a Palais–Smale (PS) sequence in $H_{rad}^{1} (R^{3})$ . Inspired by the work of Ruiz (2006), we develop new analytical methods to prove the boundedness and non-vanishing property of such (PS) sequences. Finally, we show that problem (1.1) admits a nontrivial solution. The following theorem presents our main result and gives a positive resolution to the open problem mentioned above.

Theorem 1.1

Let $2 < q < 6$ and $μ > 0$ . If

\begin{aligned} \frac{5}{36} < λ < λ_{1} := \frac{5 π}{2 \sqrt[3]{2 π}} {(\frac{8}{q^{4}})}^{\frac{4}{q - 2}} μ^{\frac{4}{q - 2}}, \end{aligned}

(1.4)

then (1.1) has a radial solution

\bar{u}

satisfying

Φ_{λ, μ} (\bar{u}) = inf_{u \in H_{rad}^{1} (R^{3})} Φ_{λ, μ} (u)

, where

Φ_{λ, μ}

is the energy functional associated with (1.1) and defined by (2.16). If

\begin{aligned} λ > λ_{2} := {[\frac{q - 2}{8} {(\frac{6 - q}{4})}^{\frac{6 - q}{q - 2}} μ^{\frac{4}{q - 2}} + \frac{1}{2}]}^{2}, \end{aligned}

(1.5)

then (1.1) does not admit any nontrivial solution.

Remark 1.2

When $μ = 1$ , condition (1.5) reduces to

\begin{aligned} λ > λ_{2} := {[\frac{q - 2}{8} {(\frac{6 - q}{4})}^{\frac{6 - q}{q - 2}} + \frac{1}{2}]}^{2} . \end{aligned}

(1.6)

Since

(6 - q) / 4 < 1

, it follows that

λ_{2} < ((q + 2) / 8)^{2}

. This demonstrates that condition (1.5) is indeed weaker than the one established in Chen and Tang (2024).

The following notations are used throughout the article:

$∙$ $L^{s} (R^{3}) (1 \leq s \leq \infty)$ denotes the Lebesgue space with the norm

‖ u ‖_{s} := {(\int_{R^{3}} | u |^{s} d x)}^{\frac{1}{s}}, 1 \leq s < \infty, ‖ u ‖_{\infty} = ess sup_{x \in R^{3}} | u (x) |;

$∙$ $D^{1, 2} (R^{3}) := {u \in L^{6} (R^{3}) : | \nabla u | \in L^{2} (R^{3})}$ denotes the Hilbert space endowed with the inner product and norm

(u, v)_{D^{1, 2}} := \int_{R^{3}} \nabla u \cdot \nabla v d x, ‖ u ‖_{D^{1, 2}} := {(\int_{R^{3}} | \nabla u |^{2} d x)}^{\frac{1}{2}};

$∙$ $H^{1} (R^{3}) := {u \in L^{2} (R^{3}) : | \nabla u | \in L^{2} (R^{3})}$ denotes the Hilbert space endowed with the inner product and norm

(u, v) := \int_{R^{3}} (\nabla u \cdot \nabla v + u v) d x, ‖ u ‖ := {[\int_{R^{3}} (| \nabla u |^{2} + | u |^{2}) d x]}^{\frac{1}{2}};

$∙$ $H_{rad}^{1} (R^{3}) := {u \in H^{1} (R^{3}) | u (x) = u (| x |) a.e. in R^{3}}$ ;

$∙$ For any $x \in R^{3}$ and $r > 0$ , $B_{r} (x) := {y \in R^{3} : | y - x | < r}$ and $B_{r} = B_{r} (0)$ ;

$∙$ $C_{1}, C_{2}, \dots$ denote positive constants possibly different in different places.

2. Preliminaries

In this section, we present some useful propositions which are crucial for proving our main result.

Lemma 2.1
Agueh (2006) (Soblev inequality) There holds
$\begin{aligned} inf_{D^{1, 2} (R^{3})} \frac{‖ \nabla u ‖_{2}^{2}}{‖ u ‖_{6}^{2}} := S = \frac{3 π}{2} \sqrt[3]{\frac{π}{2}} . \end{aligned}$
(2.1)
Lemma 2.2
Lieb (1983) (Hardy–Littlewood–Sobolev inequality)
$\begin{aligned} \int_{R^{3}} \int_{R^{3}} \frac{| u (x) v (y) |}{| x - y |} d x d y \leq \frac{8 \sqrt[3]{2}}{3 \sqrt[3]{π}} ‖ u ‖_{6 / 5} ‖ v ‖_{6 / 5}, \forall u, v \in L^{6 / 5} (R^{3}) . \end{aligned}$
(2.2)

For any $u \in L^{6 / 5} (R^{3})$ , the linear functional $h_{u} : D^{1, 2} (R^{3}) \to R$ defined as
$\begin{aligned} h_{u} (v) := \int_{R^{3}} u v d x, \forall v \in D^{1, 2} (R^{3}) . \end{aligned}$
(2.3)
It is easy to see that
$\begin{aligned} | h_{u} (v) | = | \int_{R^{3}} u v d x | \leq ‖ u ‖_{6 / 5} ‖ v ‖_{6} \leq C ‖ u ‖_{6 / 5} ‖ v ‖, \forall v \in D^{1, 2} (R^{3}) . \end{aligned}$
(2.4)
This shows that $h_{u}$ is bounded in $D^{1, 2} (R^{3})$ . In view of Riesz Representation Theorem, there exists a unique $ψ_{u} \in D^{1, 2} (R^{3})$ such that
$\begin{aligned} \int_{R^{3}} \nabla ψ_{u} \cdot \nabla v d x = \int_{R^{3}} u v d x, \forall v \in D^{1, 2} (R^{3}) . \end{aligned}$
(2.5)
Therefore, for any $u \in L^{6 / 5} (R^{3})$ , there exists a unique $ψ_{u} \in D^{1, 2} (R^{3})$ such that
$\begin{aligned} - Δ ψ_{u} = u \end{aligned}$
(2.6)
in a weak sense. In particular, for any $u \in H^{1} (R^{3})$ , $| u |^{5} \in L^{6 / 5} (R^{3})$ , we set $ϕ_{u} := ψ_{| u |^{5}}$ . Then
$\begin{aligned} \int_{R^{3}} \nabla ϕ_{u} \cdot \nabla v d x = \int_{R^{3}} | u |^{5} v d x, \forall v \in D^{1, 2} (R^{3}) . \end{aligned}$
(2.7)
Moreover, according to Lieb and Loss (2001, Theorem 6.21), we see that
$\begin{aligned} ϕ_{u} (x) = \int_{R^{3}} \frac{| u (y) |^{5}}{4 π | x - y |} d y = \frac{1}{4 π | x |} * | u |^{5} > 0, \forall x \in R^{3} \end{aligned}$
(2.8)
and
$\begin{aligned} \int_{R^{3}} ϕ_{u} | u |^{5} d x = \int_{R^{3}} \int_{R^{3}} \frac{| u (x) |^{5} | u (y) |^{5}}{4 π | x - y |} d x d y . \end{aligned}$
(2.9)

Next, we list some properties of $ϕ_{u}$ , which can directly deduce from Li et al. (2017, Lemma 2.1).
Lemma 2.3
Li et al. (2017) (i)
for all $u \in H^{1} (R^{3})$ ,
$\begin{aligned} ‖ ϕ_{u} ‖_{D^{1, 2}} \leq S^{- \frac{1}{2}} ‖ u ‖_{6}^{5}, \int_{R^{3}} ϕ_{u} | u |^{5} d x \leq S^{- 1} ‖ u ‖_{6}^{10}; \end{aligned}$
(2.10)
(ii)
if $u_{n} ⇀ \bar{u}$ in $L^{6} (R^{3})$ and $u_{n} (x) \to \bar{u} (x)$ a.e. $x \in R^{3}$ , then $ϕ_{u_{n}} ⇀ ϕ_{\bar{u}}$ in $D^{1, 2} (R^{3})$ ;
(iii)
if $u_{n} ⇀ \bar{u}$ in $L^{6} (R^{3})$ and $u_{n} (x) \to \bar{u} (x)$ a.e. $x \in R^{3}$ , then
$\begin{aligned} \int_{R^{3}} ϕ_{u_{n}} (x) | u_{n} |^{5} d x - \int_{R^{3}} ϕ_{u_{n} - \bar{u}} (x) | u_{n} - \bar{u} |^{5} d x - \int_{R^{3}} ϕ_{\bar{u}} | \bar{u} |^{5} d x \to 0. \end{aligned}$
(2.11)

Lemma 2.4
If $u_{n} ⇀ \bar{u}$ in $L^{6} (R^{3})$ and $u_{n} (x) \to \bar{u} (x)$ a.e. $x \in R^{3}$ , then $ϕ_{u_{n}} ⇀ ϕ_{\bar{u}}$ in $L^{6} (R^{3})$ .
Proof.
For any $v \in L^{6 / 5} (R^{3})$ , it follows from (2.6), the Green’s formula and Lemma 2.3 (ii) that
$\begin{aligned} \int_{R^{3}} (ϕ_{u_{n}} - ϕ_{\bar{u}}) v d x & = - \int_{R^{3}} (ϕ_{u_{n}} - ϕ_{\bar{u}}) Δ ψ_{v} d x \\ = \int_{R^{3}} (\nabla ϕ_{u_{n}} - \nabla ϕ_{\bar{u}}) \cdot \nabla ψ_{v} d x \\ = (ϕ_{u_{n}} - ϕ_{\bar{u}}, ψ_{v})_{D^{1, 2} (R^{3})} \to 0. \end{aligned}$
This shows that $ϕ_{u_{n}} ⇀ ϕ_{\bar{u}}$ in $L^{6} (R^{3})$ .
Lemma 2.5
Li et al. (2017) Let $q \in [2, 6]$ , $λ > 0$ and $μ > 0$ . If $u \in H^{1} (R^{3})$ is a weak solution of (1.1), then the following Pohozaev type identity holds
$\begin{aligned} \frac{1}{2} ‖ \nabla u ‖_{2}^{2} + \frac{3}{2} \int_{R^{3}} u^{2} d x + \frac{λ}{2} \int_{R^{3}} ϕ_{u} (x) | u |^{5} d x - \frac{3 μ}{q} \int_{R^{3}} | u |^{q} d x - \frac{1}{2} \int_{R^{3}} | u |^{6} d x = 0. \end{aligned}$
(2.12)
Lemma 2.6
Suppose that $u \in H^{1} (R^{3})$ and $- Δ ϕ_{u} = | u |^{5}$ . Then there holds
$\begin{aligned} \int_{R^{3}} (a | \nabla u |^{2} + b ϕ_{u} | u |^{5}) d x \geq 2 \sqrt{a b} \int_{R^{3}} | u |^{6} d x, \forall a, b > 0; u \in H^{1} (R^{3}) . \end{aligned}$
(2.13)
Proof.
By the definition $ϕ_{u}$ and the Green’s formula, we deduce
$\begin{aligned} \int_{R^{3}} ϕ_{u} | u |^{5} d x = \int_{R^{3}} ϕ_{u} (- Δ ϕ_{u}) d x = \int_{R^{3}} | \nabla ϕ_{u} |^{2} d x \end{aligned}$
(2.14)
and
$\begin{aligned} 2 \sqrt{a b} \int_{R^{3}} | u |^{6} d x = 2 \sqrt{a b} \int_{R^{3}} (- Δ ϕ_{u}) | u | d x = 2 \sqrt{a b} \int_{R^{3}} \nabla ϕ_{u} \cdot \nabla | u | d x . \end{aligned}$
(2.15)
From (2.14) and (2.15), we have
$2 \sqrt{a b} \int_{R^{3}} | u |^{6} d x \leq \int_{R^{3}} (a | \nabla u |^{2} + b | \nabla ϕ_{u} |^{2}) d x = \int_{R^{3}} (a | \nabla u |^{2} + b ϕ_{u} | u |^{5}) d x .$
This completes the proof.

System (1.1) is variational, and its solutions are the critical points of the functional defined in $H^{1} (R^{3})$ by
$\begin{aligned} Φ_{λ, μ} (u) = \frac{1}{2} \int_{R^{3}} (| \nabla u |^{2} + u^{2}) d x + \frac{λ}{10} \int_{R^{3}} ϕ_{u} | u |^{5} d x - \frac{μ}{q} \int_{R^{3}} | u |^{q} d x - \frac{1}{6} \int_{R^{3}} | u |^{6} d x . \end{aligned}$
(2.16)
By Lemma 2.3, it is easy to verify that $Φ_{λ, μ}$ is well-defined of classes $C^{1}$ functional, and that
$\begin{aligned} ⟨ Φ_{λ, μ}^{'} (u), v ⟩ = \int_{R^{3}} (\nabla u \cdot \nabla v + u v) d x + \int_{R^{3}} [λ ϕ_{u} | u |^{3} u - μ | u |^{q - 2} u - | u |^{4} u] v d x . \end{aligned}$
(2.17)
Hence if $u \in H^{1} (R^{3})$ is a critical point of $Φ_{λ, μ}$ , then the pair $(u, ϕ_{u})$ is a solution of (1.1).
3. Nonexistence Results

In this section, we give the proof of the second part in Theorem 1.1.

Proof Proof of the second part in Theorems 1.1.

Suppose that $\bar{u}$ is a solution of (1.1). Then it follows that

\begin{aligned} ‖ \nabla \bar{u} ‖_{2}^{2} + ‖ \bar{u} ‖_{2}^{2} + λ \int_{R^{3}} ϕ_{\bar{u}} | \bar{u} |^{5} d x - μ ‖ \bar{u} ‖_{q}^{q} - ‖ \bar{u} ‖_{6}^{6} = 0. \end{aligned}

(3.1)

By using

q \in (2, 6)

and Young’s inequality, we see that

\begin{aligned} μ ‖ \bar{u} ‖_{q}^{q} \leq μ ‖ \bar{u} ‖_{2}^{\frac{6 - q}{2}} ‖ \bar{u} ‖_{6}^{\frac{3 (q - 2)}{2}} \leq ‖ \bar{u} ‖_{2}^{2} + \frac{q - 2}{4} {(\frac{6 - q}{4})}^{\frac{6 - q}{q - 2}} μ^{\frac{4}{q - 2}} ‖ \bar{u} ‖_{6}^{6} . \end{aligned}

(3.2)

By (1.5), (2.13), (3.1), (3.2) and Lemma 2.5, we deduce

\begin{aligned} 0 & = ‖ \nabla \bar{u} ‖_{2}^{2} + ‖ \bar{u} ‖_{2}^{2} + λ \int_{R^{3}} ϕ_{\bar{u}} | \bar{u} |^{5} d x - μ ‖ \bar{u} ‖_{q}^{q} - ‖ \bar{u} ‖_{6}^{6} \\ > ‖ \nabla \bar{u} ‖_{2}^{2} + ‖ \bar{u} ‖_{2}^{2} + λ_{2} \int_{R^{3}} ϕ_{\bar{u}} | \bar{u} |^{5} d x - μ ‖ \bar{u} ‖_{q}^{q} - ‖ \bar{u} ‖_{6}^{6} \\ \geq [2 \sqrt{λ_{2}} - \frac{q - 2}{4} {(\frac{6 - q}{4})}^{\frac{6 - q}{q - 2}} μ^{\frac{4}{q - 2}} - 1] ‖ \bar{u} ‖_{6}^{6}, \end{aligned}

(3.3)

It follows from (1.5) that

\bar{u} = 0

4. Existence Results

In this section, we give the proof of the existence result given in Theorem 1.1.

Let $κ > 0$ and $w = κ e^{- | x |}$ . Then $w \in H_{rad}^{1} (R^{3})$ , and

\begin{aligned} ‖ \nabla w ‖_{2}^{2} = \int_{R^{3}} | \nabla w |^{2} d x = 4 π κ^{2} \int_{0}^{+ \infty} r^{2} e^{- 2 r} d r = π κ^{2} \end{aligned}

(4.1)

and

\begin{aligned} ‖ w ‖_{s}^{s} = \int_{R^{3}} | w |^{s} d x = 4 π κ^{s} \int_{0}^{+ \infty} r^{2} e^{- s r} d r = \frac{8 π κ^{s}}{s^{3}}, \forall s \in [2, 6) . \end{aligned}

(4.2)

Both (2.2) and (4.2) imply

\begin{aligned} \int_{R^{3}} ϕ_{w} (x) | w |^{5} d x \leq \frac{2 \sqrt[3]{2}}{3 π \sqrt[3]{π}} ‖ w ‖_{6}^{10} = \frac{2}{3^{6}} \sqrt[3]{2 π} κ^{10} . \end{aligned}

(4.3)

Lemma 4.1

Let $q \in (2, 6)$ and $μ > 0$ . Assume that (1.4) holds. Then $Φ_{λ, μ}$ is coercive in $H_{rad}^{1} (R^{3})$ .

Proof.

Arguing by contradiction, suppose there exists a sequence ${u_{n}} \subset H_{rad}^{1} (R^{3})$ such that $‖ u_{n} ‖ \to \infty$ and $Φ_{λ, μ} (u_{n}) \leq d$ for some $d \in (0, \infty)$ . Then it follows from (1.4), (2.13) and (2.16) that

\begin{aligned} Φ_{λ, μ} (u) & = \frac{1}{2} (‖ \nabla u ‖_{2}^{2} + ‖ u ‖_{2}^{2}) + \frac{λ}{10} \int_{R^{3}} ϕ_{u} | u |^{5} d x - \frac{μ}{q} ‖ u ‖_{q}^{q} - \frac{1}{6} ‖ u ‖_{6}^{6} \\ \geq \frac{36 λ - 5}{2 (36 λ + 5)} ‖ \nabla u ‖_{2}^{2} + \frac{1}{2} ‖ u ‖_{2}^{2} + \frac{36 λ - 5}{720} \int_{R^{3}} ϕ_{u} | u |^{5} d x - \frac{μ}{q} ‖ u ‖_{q}^{q} \\ \geq \frac{36 λ - 5}{4 (36 λ + 5)} ‖ \nabla u ‖_{2}^{2} + \frac{1}{4} ‖ u ‖_{2}^{2} + \frac{36 λ - 5}{1440} \int_{R^{3}} ϕ_{u} | u |^{5} d x \\ + \int_{R^{3}} [\frac{1}{4} u^{2} + \frac{36 λ - 5}{12 \sqrt{10 (36 λ + 5)}} | u |^{6} - \frac{μ}{q} | u |^{q}] d x \\ = \frac{36 λ - 5}{4 (36 λ + 5)} ‖ \nabla u ‖_{2}^{2} + \frac{1}{4} ‖ u ‖_{2}^{2} + \frac{36 λ - 5}{1440} \int_{R^{3}} ϕ_{u} | u |^{5} d x \\ + \int_{R^{3}} [\frac{1}{4} + \frac{36 λ - 5}{12 \sqrt{10 (36 λ + 5)}} | u |^{4} - \frac{μ}{q} | u |^{q - 2}] u^{2} d x . \end{aligned}

(4.4)

Set

\begin{aligned} h (s) := \frac{1}{4} + \frac{36 λ - 5}{12 \sqrt{10 (36 λ + 5)}} s^{4} - \frac{μ}{q} s^{q - 2}, s \in [0, + \infty) . \end{aligned}

(4.5)

Since

q \in (2, 6), μ > 0

and

36 λ - 5 > 0

h (0) = 1 / 4

and

h (s) \to \infty

s \to \infty

, and then the infimum

h_{0} := inf_{s \in [0, \infty)} h (s) = h (s_{0})

is attained at some

s_{0} \in (0, \infty)

. If

h_{0} \geq 0

, then it follows from (4.4) that

{‖ u_{n} ‖^{2}}

is bounded, which is a contradiction. If

h_{0} := - m < 0

, then we can obtain that

{s > 0 : h (s) < 0} = (a, b)

for some

b > a > 0

. Set

\begin{aligned} D_{n} := {x \in R^{3} : | u_{n} (x) | \in (a, b)}, A_{n} := \int_{D_{n}} u_{n}^{2} d x . \end{aligned}

(4.6)

Then it follows from (4.4), (4.5) and (4.6) that

\begin{aligned} d \geq \frac{36 λ - 5}{4 (36 λ + 5)} ‖ u ‖^{2} + \frac{36 λ - 5}{1440} \int_{R^{3}} ϕ_{u} | u |^{5} d x - m A_{n} . \end{aligned}

(4.7)

In view of the Strauss inequality Willem (1996, Lemma 4.5), for any

u \in H_{rad}^{1} (R^{3})

, we have that

\begin{aligned} | u (x) | \leq c_{0} | x |^{- 1} ‖ u ‖, a . e . x \in R^{3}, \end{aligned}

(4.8)

for some

c_{0} > 0

. Therefore, from (2.9), (4.6) and (4.8), we have

\begin{aligned} \int_{R^{3}} ϕ_{u_{n}} | u_{n} |^{5} d x & = \int_{R^{3}} \int_{R^{3}} \frac{| u_{n} (x) |^{5} | u_{n} (y) |^{5}}{4 π | x - y |} d x d \\ \geq \int_{R^{3}} \int_{R^{3}} \frac{| u_{n} (x) |^{5} | u_{n} (y) |^{5}}{4 π (| x | + | y |)} d x d y \\ \geq \frac{1}{4 π c_{0} ‖ u_{n} ‖} \int_{R^{3}} \int_{R^{3}} \frac{| u_{n} (x) |^{5} | u_{n} (y) |^{5}}{| u_{n} (x) |^{- 1} + | u_{n} (y) |^{- 1}} d x d y \\ \geq \frac{1}{4 π c_{0} ‖ u_{n} ‖} \int_{D_{n}} \int_{D_{n}} \frac{| u_{n} (x) |^{5} | u_{n} (y) |^{5}}{| u_{n} (x) |^{- 1} + | u_{n} (y) |^{- 1}} d x d y \\ \geq \frac{a^{7}}{8 π c_{0} ‖ u_{n} ‖} {(\int_{D_{n}} | u_{n} |^{2} d x)}^{2} \\ \geq \frac{a^{7} A_{n}^{2}}{8 π c_{0} ‖ u_{n} ‖}, \end{aligned}

(4.9)

which, together with (4.7), implies that

\begin{aligned} d \geq \frac{36 λ - 5}{4 (36 λ + 5)} ‖ u_{n} ‖^{2} + \frac{(36 λ - 5) a^{7} A_{n}^{2}}{11520 π c_{0} ‖ u_{n} ‖} - m A_{n} . \end{aligned}

(4.10)

Thus, the sequence

{‖ u_{n} ‖}

is bounded. This contradiction shows that

Φ_{λ, μ}

is coercive in

H_{rad}^{1} (R^{3})

Set

\begin{aligned} m_{λ, μ} := inf_{u \in H_{rad}^{1} (R^{3})} Φ_{λ, μ} (u) . \end{aligned}

(4.11)

Lemma 4.2

Let $q \in (2, 6)$ and $μ > 0$ . Assume that (1.4) holds. Then $m_{λ, μ}$ is well-defined and $m_{λ, μ} \in (- \infty, 0)$ .

Proof.

By of Lemma 4.1, $Φ_{λ, μ}$ is coercive in $H_{rad}^{1} (R^{3})$ , so $m_{λ}$ is well-defined. Next, we prove that $m_{λ} < 0$ . Let

\hat{κ} := {(\frac{q^{4}}{8 μ})}^{\frac{1}{q - 2}}, \hat{w} := \hat{κ} e^{- | x |} .

Then

\hat{w} \in H_{rad}^{1} (R^{3})

, and from (1.4), (2.16), (4.1), (4.2) and (4.3), we have

\begin{aligned} Φ_{λ, μ} (\hat{w}) & = \frac{1}{2} \int_{R^{3}} (| \nabla \hat{w} |^{2} + {\hat{w}}^{2}) d x + \frac{λ}{10} \int_{R^{3}} ϕ_{\hat{w}} (x) | \hat{w} |^{5} d x - \frac{μ}{q} \int_{R^{3}} | \hat{w} |^{q} d x - \frac{1}{6} \int_{R^{3}} | \hat{w} |^{6} d x \\ < \frac{1}{2} (‖ \nabla \hat{w} ‖_{2}^{2} + ‖ \hat{w} ‖_{2}^{2}) + \frac{λ_{1}}{10} \int_{R^{3}} ϕ_{\hat{w}} (x) | \hat{w} |^{5} d x - \frac{μ}{q} ‖ \hat{w} ‖_{q}^{q} - \frac{1}{6} ‖ \hat{w} ‖_{6}^{6} \\ \leq [1 + \frac{\sqrt[3]{2 π} λ_{1}}{5 \cdot 3^{6} π} {\hat{κ}}^{8} - \frac{8 μ {\hat{κ}}^{q - 2}}{q^{4}} - \frac{8 {\hat{κ}}^{4}}{6^{4}}] π {\hat{κ}}^{2} \\ = [\frac{\sqrt[3]{2 π} λ_{1}}{5 \cdot 3^{6} π} {(\frac{q^{4}}{8 μ})}^{\frac{4}{q - 2}} - \frac{8}{6^{4}}] π {\hat{κ}}^{6} = 0, \forall 0 < λ < λ_{1} . \end{aligned}

(4.12)

(4.12) shows that

m_{λ, μ} < 0

Proof Proof of the second part in Theorem 1.1.

In view of the Ekeland variational principle, there exists a sequences of ${u_{n}} \subset H_{rad}^{1} (R^{3})$ such that

\begin{aligned} Φ_{λ, μ} (u_{n}) \to m_{λ, μ}, Φ_{λ, μ}^{'} (u_{n}) \to 0. \end{aligned}

(4.13)

By Lemma 4.1,

{u_{n}}

is bounded in

H_{rad}^{1} (R^{3})

. Therefore, there exists

\bar{u} \in H_{rad}^{1} (R^{3})

such that, passing to a subsequence,

\begin{aligned} {\begin{cases} u_{n} ⇀ \bar{u}, & in H_{rad}^{1} (R^{3}); \\ u_{n} \to \bar{u}, & in L^{s} (R^{3}), \forall s \in (2, 6); \\ u_{n} \to \bar{u}, & a.e. on R^{3} . \end{cases} \end{aligned}

(4.14)

First, we prove that

\bar{u} \neq 0

. Otherwise, if

\bar{u} = 0

, then

u_{n} ⇀ 0

H_{rad}^{1} (R^{3})

and

u_{n} \to 0

L^{s} (R^{3})

for all

s \in (2, 6)

. From (2.17), (4.13) and (4.14), we have

\begin{aligned} ‖ \nabla u_{n} ‖_{2}^{2} + ‖ u_{n} ‖_{2}^{2} + λ \int_{R^{3}} ϕ_{u_{n}} | u_{n} |^{5} d x = ‖ u_{n} ‖_{6}^{6} + o (1) . \end{aligned}

(4.15)

lim_{n \to \infty} ‖ u_{n} ‖_{6}^{6} = 0

, then it follows from (4.15) that

lim_{n \to \infty} ‖ \nabla u_{n} ‖_{2}^{2} = 0

lim_{n \to \infty} ‖ u_{n} ‖_{2}^{2} = 0

and

lim_{n \to \infty} \int_{R^{3}} ϕ_{u_{n}} | u_{n} |^{5} d x = 0

. Hence, from (2.16), (4.13) and (4.14), we have

\begin{aligned} m_{λ, μ} + o (1) = \frac{1}{2} (‖ \nabla u_{n} ‖_{2}^{2} + ‖ u_{n} ‖_{2}^{2}) + \frac{λ}{10} \int_{R^{3}} ϕ_{u_{n}} | u_{n} |^{5} d x - \frac{μ}{q} ‖ u_{n} ‖_{q}^{q} - \frac{1}{6} ‖ u_{n} ‖_{6}^{6} = o (1), \end{aligned}

(4.16)

which is a contradiction. If

lim_{n \to \infty} ‖ u_{n} ‖_{6}^{6} > 0

, then it follows from (2.1) and (4.15) that

\begin{aligned} lim_{n \to \infty} ‖ u_{n} ‖_{6}^{6} \geq S^{\frac{3}{2}} . \end{aligned}

(4.17)

Hence, from (1.4), (2.13), (2.16), (4.13), (4.14), (4.15) and (4.17), we have

\begin{aligned} m_{λ, μ} + o (1) & = \frac{1}{2} (‖ \nabla u_{n} ‖_{2}^{2} + ‖ u_{n} ‖_{2}^{2}) + \frac{λ}{10} \int_{R^{3}} ϕ_{u_{n}} | u_{n} |^{5} d x - \frac{μ}{q} ‖ u_{n} ‖_{q}^{q} - \frac{1}{6} ‖ u_{n} ‖_{6}^{6} \\ = \frac{1}{2} (‖ \nabla u_{n} ‖_{2}^{2} + ‖ u_{n} ‖_{2}^{2}) + \frac{λ}{10} \int_{R^{3}} ϕ_{u_{n}} | u_{n} |^{5} d x - \frac{1}{6} ‖ u_{n} ‖_{6}^{6} + o (1) \\ \geq \frac{1}{2} ‖ u_{n} ‖_{2}^{2} + (\sqrt{\frac{λ}{5}} - \frac{1}{6}) ‖ u_{n} ‖_{6}^{6} + o (1) \\ \geq (\sqrt{\frac{λ}{5}} - \frac{1}{6}) S^{\frac{3}{2}} + o (1), \end{aligned}

(4.18)

which is also a contradiction. Therefore,

\bar{u} \neq 0

. By (4.14) and Lemma 2.3, we deduce

ϕ_{u_{n}} ⇀ ϕ_{\bar{u}}

D^{1, 2} (R^{3})

, and so Lemma 2.4 implies that

ϕ_{u_{n}} ⇀ ϕ_{\bar{u}}

L^{6} (R^{3})

. Hence, for any

ψ \in C_{0}^{\infty} (R^{3}) \cap H_{rad}^{1} (R^{3})

, we have

\begin{aligned} \int_{R^{3}} (ϕ_{u_{n}} - ϕ_{\bar{u}}) | \bar{u} |^{3} \bar{u} ψ d x = o (1) \end{aligned}

(4.19)

and

\begin{aligned} | \int_{R^{3}} ϕ_{u_{n}} (| u_{n} |^{3} u_{n} - | \bar{u} |^{3} \bar{u}) ψ d x | & \leq \int_{R^{3}} \int_{R^{3}} \frac{| u_{n} (y) |^{5} | | u_{n} (x) |^{3} u_{n} (x) - | \bar{u} (x) |^{3} \bar{u} (x) | | ψ (x) |}{4 π | x - y |} d x d y \\ \leq C_{1} ‖ u_{n} ‖_{6}^{5} \int_{R^{3}} {[{| | u_{n} (x) |^{3} u_{n} (x) - | \bar{u} (x) |^{3} \bar{u} (x) |}^{\frac{6}{5}} | ψ (x) |^{\frac{6}{5}} d x]}^{\frac{5}{6}} \\ \leq C_{1} ‖ ψ ‖_{\infty} ‖ u_{n} ‖_{6}^{5} \int_{R^{3}} {[{| | u_{n} (x) |^{3} u_{n} (x) - | \bar{u} (x) |^{3} \bar{u} (x) |}^{\frac{6}{5}} d x]}^{\frac{5}{6}} \\ = o (1) . \end{aligned}

(4.20)

Combining (4.19) and (4.20), we have

\begin{aligned} \int_{R^{3}} ϕ_{u_{n}} | u_{n} |^{3} u_{n} v d x \to \int_{R^{3}} ϕ_{\bar{u}} | \bar{u} |^{3} \bar{u} v d x, \forall v \in H_{rad}^{1} (R^{3}) . \end{aligned}

(4.21)

By a standard argument, we can deduce from (2.17), (4.13), (4.14) and (4.21),

Φ_{λ, μ}^{'} (\bar{u}) = 0

. Let

v_{n} = u_{n} - \bar{u}

. Then

v_{n} ⇀ 0

H_{rad}^{1} (R^{3})

and

v_{n} \to 0

L^{q} (R^{3})

, and

v_{n} \to 0

a.e. on

R^{3}

. Hence, it follows from (2.11), (2.13), (2.16), (4.11), (4.13), (4.14) and the Brézis–Lieb lemma that

\begin{aligned} m_{λ, μ} & = lim_{n \to \infty} Φ_{λ, μ} (u_{n}) \\ = lim_{n \to \infty} [\frac{1}{2} (‖ \nabla v_{n} ‖_{2}^{2} + ‖ v_{n} ‖_{2}^{2}) + \frac{λ}{10} \int_{R^{3}} ϕ_{v_{n}} (x) | v_{n} |^{5} d x - \frac{1}{6} ‖ v_{n} ‖_{6}^{6}] + Φ_{λ, μ} (\bar{u}) \\ \geq lim_{n \to \infty} [\frac{1}{2} ‖ v_{n} ‖_{2}^{2} + (\sqrt{\frac{λ}{5}} - \frac{1}{6}) ‖ v_{n} ‖_{6}^{6}] + Φ_{λ, μ} (\bar{u}) \\ \geq lim_{n \to \infty} [\frac{1}{2} ‖ v_{n} ‖_{2}^{2} + (\sqrt{\frac{λ}{5}} - \frac{1}{6}) ‖ v_{n} ‖_{6}^{6}] + m_{λ, μ} . \end{aligned}

(4.22)

(4.22) show that

lim_{n \to \infty} ‖ v_{n} ‖_{2}^{2} = 0

lim_{n \to \infty} ‖ v_{n} ‖_{6}^{6} = 0

and

lim_{n \to \infty} ‖ v_{n} ‖ = 0

. It follows that

Φ_{λ, μ} (\bar{u}) = m_{λ, μ}

and

Φ_{λ, μ}^{'} (\bar{u}) = 0

, which implies that

\bar{u}

is a ground solution of (1.1).

The proof is now complete.

Footnotes

Funding

The authors received no financial support for the research, authorship, and/or publication of this article.

Declaration of Conflicting Interests

The authors declared no potential conflicts of interest with respect to the research, authorship, and/or publication of this article.

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