Home Mathematics Normalized ground-states for the Sobolev critical Kirchhoff equation with at least mass critical growth
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Normalized ground-states for the Sobolev critical Kirchhoff equation with at least mass critical growth

  • Shiyong Zhang and Qiongfen Zhang EMAIL logo
Published/Copyright: July 25, 2025
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Open Mathematics
From the journal Open Mathematics Volume 23 Issue 1

Abstract

In this article, we investigate the following nonlinear Kirchhoff equation with Sobolev critical growth:

− a + b ∫ R 3 ∣ ∇ u ∣ 2 d x Δ u + λ u = μ f ( u ) + ∣ u ∣ 4 u , in R 3 , u > 0 , ∫ R 3 ∣ u ∣ 2 d x = m 2 , in R 3 , ( P m )

where μ is a positive parameter, a , b > 0 , and the frequency λ appears as a positive Lagrange multiplier. The nonlinearity f is more general and satisfies Sobolev subcritical conditions. With the assistance of the Pohožaev constraints and the Sobolev subcritical approximation method, we have achieved a couple of the normalized ground-state solutions to ( P m ) and the asymptotic behavior of the ground-state is also studied.

Keywords: normalized ground-state solutions; nonlinear Kirchhoff equation; critical growth; Pohozaev manifold
MSC 2010: 34C37; 35A15; 37J45

1 Introduction

In this article, we consider the normalized solutions ( u , λ ) ∈ H 1 ( R N ) × R to the following Kirchhoff equation:

(1) − a + b ∫ R 3 ∣ ∇ u ∣ 2 d x Δ u + λ u = μ f ( u ) + ∣ u ∣ 4 u , in R 3 ,

which has a general nonlinearity, where a , b > 0 , λ appears as a Lagrange multiplier. We will work on the constrained manifold

S m ≔ { u ∈ H 1 ( R 3 ) : ∫ R 3 ∣ u ∣ 2 d x = m 2 } ,

where

H 1 ( R 3 ) ≔ { u ∈ L 2 ( R 3 ) : ∣ ∇ u ∣ ∈ L 2 ( R 3 ) }

is a space with the inner product

⟨ u , v ⟩ = ∫ R 3 [ a ∇ u ⋅ ∇ v + u v ] d x

and equipped with the norm

‖ u ‖ = ⟨ u , u ⟩ 1 2 .

Such ( u , λ ) is called a couple of solutions to the system ( P m ) .

The Kirchhoff equation, as an extension of the classical D’Alembert’s wave equations, was proposed by Kirchhoff [1]. The problem ( P m ) describes the stable state of the following Kirchhoff equation:

(1) u t t − ( a + b ∫ Ω ∣ ∇ u ∣ 2 d x ) Δ u = f ( x , u ) , x ∈ Ω , u = 0 , x ∈ ∂ Ω ,

where Ω ⊂ R 3 is a bounded region. In particular, Kirchhoff’s model is regarded as a nonlocal model, for it takes into account the effects of changes in the length of the strings during the vibrations. From a physical perspective, the prescribed mass represents the energy source in nonlinear optics or the total number of atoms in Bose-Einstein condensation. We can refer to [2] to learn more about the physical context of the Kirchhoff model. Besides its significance in physics, the problem of fixed mass is rather complex, and the study of such problems is particularly valuable and challenging. Li and Ye [3] studied the existence of positive solutions for Kirchhoff-type equations with critical Sobolev exponent, under the following assumptions: f ( t ) t 3 is strictly increasing for t > 0 , and there exists γ > 4 , such that 0 < γ F ( t ) ≤ f ( t ) t , for all t > 0 . Jeanjean [4] studied the following semilinear ellipitic equation:

(2) − Δ u + λ u = g ( u ) , x ∈ R N ,

where N ≥ 1 , λ ∈ R , and g fulfills the following conditions:

(g0) g : R → R is continuous and odd.

(g1) ∃ ( α , β ) ∈ R × R satisfying

2 N + 4 N < α ≤ β < 2 N N − 2 , if N ≥ 3 , 2 N + 4 N < α ≤ β , if N = 1 , 2 ,

such that

α G ( s ) ≤ g ( s ) s ≤ β G ( s ) with G ( s ) = ∫ 0 s g ( t ) d t .

(g2) Let G ˜ : R → R , G ˜ ( s ) = g ( s ) s − 2 G ( s ) . Then, G ˜ ′ exists and

G ˜ ′ ( s ) s > 2 N + 4 N G ˜ ( s ) .

They proved the existence of solutions by searching for critical points of the energy functional

(3) F ( u ) = 1 2 ∫ R N ∣ ∇ u ∣ 2 d x − ∑ i = 1 m a i σ i + 2 ∫ R N ∣ u ∣ σ i + 2 d x .

It is easy to see that the corresponding energy functional (3) is unbounded from below on S m . By making use of a minimax procedure, Jeanjean showed that for each m > 0 , (2) possesses at least one couple ( u n , λ n ) ∈ H 1 ( R N ) × R + of weak solution with ‖ u n ‖ 2 = m , and u n is radial under the conditions (g0) and (g1) for N ≥ 2 . Based on the direct minimization of the energy functional on the linear combination of Nehari and Pohožaev constraints, Bieganowski and Mederski [5] proposed a simple minimization method to prove the existence of normalized ground-states to (2) for the Sobolev subcritical equation. Under conditions similar to hypothesis (g0)–(g1), He et al. [6] considered the general nonlinearities for mass supercritical case in the dimension 1 ≤ N ≤ 3 and obtained a ground-state normalized solution ( λ c , u c ) with λ c > 0 and u c ∈ H rad 1 ( R N ) . Besides, they also studied the asymptotic behavior of u c .

In recent years, a lot of interests are paid on normalized solution to Kirchhoff-type equations, for the nonlinearities f ( u ) = ∣ u ∣ p − 2 u and more general space dimensions. Ye [7] studied the existence of solutions with prescribed L 2 -norm for the following Kirchhoff-type equation:

(4) − a + b ∫ R N ∣ ∇ u ∣ 2 d x Δ u + V ( x ) u = μ ∣ u ∣ p − 2 u + ∣ u ∣ 4 u + λ u ,

where 1 ≤ N ≤ 3 and 2 < p < 2 * . Using exclusion analysis with a minimized sequence dichotomy, Ye proved the existence of solutions to (4). Alternatively, Gao et al. [8] studied the blow-up behavior of the normalized solutions to (4). Later, utilizing the subcritical approximation method, which was used to consider mass-constrained Kirchhoff-type problems, Li et al. [9] studied the existence and asymptotics of normalized ground-states for a Sobolev critical Kirchhoff equation

(5) − a + b ∫ R 3 ∣ ∇ u ∣ 2 d x Δ u + λ u = μ ∣ u ∣ p − 2 u + ∣ u ∣ 4 u ,

where a , b , m , μ > 0 and 14 3 < p < 6 . Lv et al. [10] proved that there exists a positive normalized solution for (5) on R 4 . They also investigated the asymptotic behavior of the normalized ground-state solutions, when μ → 0 + and b → 0 + . When ∣ u ∣ p − 2 u is replaced by ( I α * ∣ u ∣ p ) ∣ u ∣ p − 2 u , where I α is a Riesz potential, Fei and Zhang [11] proved that (5) has a normalized solution by approximation methods and Schwartz symmetrization rearrangements. Using the symmetric mountain pass theorem and considering the Palais-Smale-Pohozaev condition, Xu et al. [12] have dealt the existence and multiplicity of normalized solutions with positive energy for the following equation:

− a + b ∫ R 3 ∣ ∇ u ∣ 2 d x Δ u − λ u = f ( u ) , in R 3 , ∫ R 3 ∣ u ∣ 2 d x = c , in R 3 .

In [13], using appropriate transform, Kong et al. obtained the equivalent system of the following fractional Kirchhoff equation with critical growth:

(6) a + b ∫ R 3 ( − Δ ) s 2 u 2 d x ( − Δ ) s u + λ u = μ ∣ u ∣ q − 2 u + ∣ u ∣ 2 s * − 2 u , in R 3 .

With the equivalent system, they obtained the existence and nonexistence of normalized solutions to (6). In [14], under the normalization condition ∫ R N ∣ u ∣ 2 d x = m 1 and ∫ R N ∣ v ∣ 2 d x = m 2 , Zuo studied the following Kirchhoff-type system:

(7) − a + b ∫ R N ∣ ∇ u ∣ 2 d x Δ u = λ 1 u + μ 1 ∣ u ∣ p − 2 u + β r 1 ∣ u ∣ r 1 − 2 u ∣ v ∣ r 2 , − a + b ∫ R N ∣ ∇ v ∣ 2 d x Δ v = λ 2 v + μ 2 ∣ v ∣ q − 2 v + β r 2 ∣ u ∣ r 1 ∣ v ∣ r 2 − 2 v .

They obtained a ground-state solution for (7) that satisfies the L 2 -supercritical by proving the existence of Palais-Smale sequence and applying minimax methods. Zhang and Han [15] studied similar Schrödinger equations by using Schwartz rearrangement and Ekeland’s variational principle. Zhang et al. [16], have explored high-energy solutions for general Kirchhoff equations without relying on the Ambrosetti-Rabinowitz condition, demonstrating the feasibility of alternative variational approaches of the following equation:

(8) − M ∫ R 3 ∣ ∇ u ∣ 2 d x Δ u + u = a ( x ) f ( u ) , in R 3 .

Recently, Sun et al. [17] studied a class of fractional N s − p Laplace Kirchhoff equations with exponential growth, employing penalization methods and Lyusternik-Schnirelmann theory to establish the existence and concentration of solutions. Their work highlighted the importance of variational methods and the use of fractional Trudinger-Moser inequalities in handling the critical growth of the nonlinearity. Similarly, with the help of the penalization method and Lyusternik-Schnirelmann theory, Liang et al. [18] established the existence, multiplicity, and concentration of solutions for fractional p -Laplace Kirchhoff equations with exponential growth. Sun et al. [19] investigated critical Kirchhoff equations involving the p -sub-Laplacians on the Heisenberg group, and proved the existence and multiplicity of nontrivial solutions using the mountain pass theorem and concentration-compactness principles. Combining the concentration-compactness principle for weighted variable exponent spaces, genus theory, and the Hardy-Littlewood-Sobolev-type inequality, Zhang and Ma [20] obtained the existence of nontrivial solutions for a kind of p ( x ) -Choquard-Kirchhoff problems.

We refer the readers to [21,22] for more results of normalized solutions to the Kirchhoff equation in the whole space R N (see [23,24] for more results of normalized solutions in R 3 , and [25–27] for other equations, such as Schrodinger equations, Choquard equations and Hartree equations).

Inspired by the aforementioned results, in this article, for general nonlinearity f , we discuss the existence and asymptotic of normalized ground-states. To ensure logical proof and for the sake of future discussions, we first suppose that

  1. f ( t ) t is strictly increasing for t ∈ R .

  2. η ≔ limsup ∣ t ∣ → ∞ F ( t ) ∣ t ∣ 14 3 < + ∞ , where F ( t ) = ∫ 0 t f ( s ) d s .

  3. lim ∣ t ∣ → ∞ F ( t ) ∣ t ∣ 6 = 0 .

  4. h ( t ) t ≥ 14 3 H ( t ) for t ∈ R , where H ( t ) = f ( t ) t − 2 F ( t ) and h ≔ H ′ .

  5. 14 3 F ( t ) ≤ f ( t ) t ≤ 6 F ( t ) , for t ∈ R .

  6. H ( ϑ 0 ) > 0 for some ϑ 0 ≠ 0 .

  7. f , h ∈ C ( R , R ) , and there exists a non-negative constant C such that

    ∣ f ( t ) ∣ ≤ C ( ∣ t ∣ + ∣ t ∣ 5 ) , for t ∈ R .

We can deduce from assumption (V1) that the nonlinearity F ( t ) can have a L 2 -critical growth F ( t ) ∼ ∣ t ∣ 14 3 at 0, and from (V2), we know that F ( t ) has Sobolev subcritical growth at ∞ . Under some appropriate assumptions on the nonlinearities f , the normalized solutions to ( P m ) can be obtained by locating critical points of

J f ( u ) = a 2 ∫ R 3 ∣ ∇ u ∣ 2 d x + b 4 ∫ R 3 ∣ ∇ u ∣ 2 d x 2 − μ ∫ R 3 F ( u ) d x − 1 6 ∫ R 3 ∣ u ∣ 6 d x .

Furthermore, u is a ground-state normalized solution to the problem ( P m ) on S m if

d J f ∣ S ( m ) ( u ) = 0 and J f ( u ) = inf { J f ( u ¯ ) : d J f ∣ S ( m ) ( u ¯ ) = 0 and u ¯ ∈ S m } .

We are now in a position to state main results of this article.

Theorem 1.1

Let (V0)–(V6) hold, then there exists non-negative μ ′ such that if μ > μ ′ , problem ( P m ) have a positive and radially symmetric ground-state solution u ∈ H 1 ( R 3 ) for some λ > 0 .

Remark 1.2

Compared with previous works, when proving the existence of normalized solutions to problem ( P m ) , we have to overcome the fundamental difficulties caused by the general nonlinearity function f . To address these difficulties, we have optimized the assumptions made in the literature [4].

To study the asymptotic behavior of the normalized ground-states, we denote the solution u obtained in Theorem 1.1 as u m . We have the following theorem.

Theorem 1.3

Let (V1)–(V6) hold, and if  lim t → 0 F ( t ) ∣ t ∣ 6 = + ∞ , then J ( u m ) → 0 + as m → ∞ .

Remark 1.4

There are some articles that have already studied problem ( P m ), and similar problems for the case f = ∣ u ∣ q − 2 u , where q ∈ ( 14 3 , 6 ) . However, most of them investigated the problem ( P m ) and similar problems on bounded domain, contrary to previous works we consider the problem ( P m ) with a general form of f on the whole space R 3 . Using the fiber map and the concentration-compactness principle, Rădulescu et al. [28] have studied the case f = ∣ u ∣ q − 2 u , but in this article, to the best of our knowledge, the Sobolev subcritical approximation method is used to study more general cases for the first time. This result complements and extends the article by Li et al. [9] regarding the case f = ∣ u ∣ q − 2 u .

The rest of this article is organized as follows. In Section 2, we present some preliminary results and give the proof of Theorem 1.1. In Section 3, we have demonstrated some maximum behavior and prove Theorem 1.3.

2 Preliminaries

Before proving the main result, we first give some key inequalities. First, in this article, we mainly consider the following Kirchhoff equation with Sobolev subcritical growth:

− a + b ∫ R 3 ∣ ∇ u ∣ 2 d x Δ u + λ u = μ f ( u ) + ∣ u ∣ p − 2 u , in R 3 , u > 0 , ∫ R 3 ∣ u ∣ 2 d x = m 2 , in R 3 , ( P )

where a , b , m , μ > 0 and 14 3 < p < 6 . The solutions of (P) correspond to the critical points of the energy functional

(1) ℐ f , p ( u ) = a 2 ∫ R 3 ∣ ∇ u ∣ 2 d x + b 4 ∫ R 3 ∣ ∇ u ∣ 2 d x 2 − μ ∫ R 3 F ( u ) d x − 1 p ∫ R 3 ∣ u ∣ p d x ,

constrained on S m . The Pohožaev identity of the problem (P) is as follows:

P f , p ( u ) ≔ a ∫ R 3 ∣ ∇ u ∣ 2 d x + b ∫ R 3 ∣ ∇ u ∣ 2 d x 2 − 3 2 μ ∫ R 3 [ f ( u ) u − 2 F ( u ) ] d x − 3 p − 6 2 p ∫ R 3 ∣ u ∣ p d x = a ∫ R 3 ∣ ∇ u ∣ 2 d x + b ∫ R 3 ∣ ∇ u ∣ 2 d x 2 − 3 2 μ ∫ R 3 H ( u ) d x − 3 p − 6 2 p ∫ R 3 ∣ u ∣ p d x .

Furthermore, we always assume that there exists a constant C c such that the following inequalities hold:

(2) γ η C c m 8 3 < 1 ,

where γ > 20 3 . By (V1)–(V2), (V4), and (V6), for any ε > 0 , there exists a non-negative constant C ε such that

(3) H ( x ) ≤ γ F ( x ) ≤ γ ( ε ∣ x ∣ 6 + ( ε + η ) ∣ x ∣ 14 3 + C ε ∣ x ∣ q ) ,

for any x ∈ R , where q ∈ 14 3 , 6 . As an alternative, from (V4), we have

(4) F ( x ) , H ( x ) ≥ 0 , for any x ∈ R .

D 1 , 2 ( R 3 ) denotes the closure of the function space C ∞ ( R 3 ) with the norm

‖ u ‖ D 1 , 2 ( R 3 ) = ‖ ∇ u ‖ 2 ,

and the optimal embedding constant S of D 1 , 2 ( R 3 ) ↪ L 6 ( R 3 ) is given by

(5) S ≔ inf ∀ u ∈ D 1 , 2 ( R 3 ) \ { 0 } ‖ ∇ u ‖ 2 2 ‖ u ‖ 6 2 .

Set

(6) P f , p ≔ { u ∈ S m : P f , p ( u ) = 0 } .

We introduce the following lemma.

Lemma 2.1

Assume that (V0)–(V2) hold. For arbitrary u ∈ S m , there exists a unique s 0 ∈ R such that u s 0 ∈ P f , q . Moreover, s 0 is the unique critical point of ℐ f , q ( u s ) and

(7) ℐ f , p ( u s 0 ) = max s ∈ R ℐ f , p ( u s ) .

In addition, if P f , p ( u ) < 0 , then s 0 ≤ 0 .

Proof

For arbitrary u ∈ S m and s > 0 , define

(8) u s ( x ) ≔ e 3 2 s u ( e s x ) ,

then,

(9) ℐ f , p ( u s ) = a 2 ∫ R 3 ∣ ∇ u s ∣ 2 d x + b 4 ∫ R 3 ∣ ∇ u s ∣ 2 d x 2 − μ ∫ R 3 F ( u s ) d x − 1 p ∫ R 3 ∣ u s ∣ p d x = 1 2 a e 2 s ∫ R 3 ∣ ∇ u ∣ 2 d x + 1 4 b e 4 s ∫ R 3 ∣ ∇ u ∣ 2 d x 2 − μ e − 3 s ∫ R 3 F ( e 3 2 s u ) d x − 1 p e 3 p − 6 2 s ∫ R 3 ∣ u ∣ p d x

and

(10) P f , p ( u s ) = a e 2 s ∫ R 3 ∣ ∇ u ∣ 2 d x + b e 4 s ∫ R 3 ∣ ∇ u ∣ 2 d x 2 − 3 p − 6 2 p e 3 p − 6 2 s ∫ R 3 ∣ u ∣ p d x − 3 2 μ e − 3 s ∫ R 3 f e 3 2 s u e 3 2 s u − 2 F e 3 2 s u d x = a e 2 s ∫ R 3 ∣ ∇ u ∣ 2 d x + b e 4 s ∫ R 3 ∣ ∇ u ∣ 2 d x 2 − 3 2 μ e − 3 s ∫ R 3 H e 3 2 s u d x − 3 p − 6 2 p e 3 p − 6 2 s ∫ R 3 ∣ u ∣ p d x .

Define

(11) ψ ( s ) = a e 2 s ∫ R 3 ∣ ∇ u ∣ 2 d x + b ∫ R 3 ∣ ∇ u ∣ 2 d x 2 − 3 p − 6 2 p e 3 p − 14 2 s ∫ R 3 ∣ u ∣ p d x − 3 2 μ e − 7 s ∫ R 3 f e 3 2 s u e 3 2 s u − 2 F e 3 2 s u d x .

It is simple to understand that ψ ( s ) is continuous for all s ∈ R . Since 14 3 < p < 6 , we obtain that 3 q − 14 2 > 0 . Then, we can prove that ψ ( s ) → + ∞ as s → − ∞ and ψ ( s ) → − ∞ as s → + ∞ . According to the intermediate value theorem, there exists s 0 ∈ R such that ψ ( s 0 ) = 0 . Next, we proceed to prove that s 0 is unique. Assume that there exists s 1 < s 2 such that ψ ( s 1 ) = ψ ( s 2 ) = 0 , then

0 < a 1 e 2 s 1 − 1 e 2 s 2 ∫ R 3 ∣ ∇ u ∣ 2 d x = μ 1 e 7 s 1 − 1 e 7 s 2 ∫ R 3 3 2 f e 3 2 s 1 u e 3 2 s 1 u − 2 ∫ R 3 F e 3 2 s 1 u − 3 2 f e 3 2 s 2 u e 3 2 s 2 u − 2 ∫ R 3 F e 3 2 s 2 u d x + 3 p − 6 2 p e 3 p − 14 2 s 1 − e 3 p − 14 2 s 2 ∫ R 3 ∣ u ∣ p d x < 0 ,

which is a contraction. Hence, there exists a unique s 0 ∈ R such that ψ ( s 0 ) = 0 . Since P f , p ( u s ) = e 4 s ψ ( s ) , P f , p ( u s 0 ) = 0 . So, we derive u s 0 ∈ P f , p . Let ξ f , p ( s ) = ℐ f , p ( u s ) , then

ξ f , p ′ ( s ) = a e 2 s ∫ R 3 ∣ ∇ u ∣ 2 d x + e 4 s b ∫ R 3 ∣ ∇ u ∣ 2 2 d x − 3 p − 6 2 p e 3 p − 6 2 s ∫ R 3 ∣ u ∣ p d x − 3 2 μ e − 3 s ∫ R 3 f e 3 2 s u e 3 2 s u − 2 F e 3 2 s u d x = P f , p ( u s ) .

It is easy to verify that ℐ f , p ( u s ) > 0 as τ → − ∞ and ℐ f , p ( u s ) → − ∞ as s → + ∞ , so s 0 is a unique critical point of ℐ f , p ( u s ) and

ℐ f , p ( u s 0 ) = max s ∈ R ℐ f , p ( u s ) .

Alternatively, if P f , p ( u ) ≤ 0 , then s 0 ≤ 0 . Otherwise,

(12) 0 = P f , p ( u s 0 ) = a e 2 s 0 ∫ R 3 ∣ ∇ u ∣ 2 d x + e 4 s 0 b ∫ R 3 ∣ ∇ u ∣ 2 2 d x − 3 2 μ e − 3 s 0 ∫ R 3 f e 3 2 s 0 u e 3 2 s 0 u − 2 F e 3 2 s 0 u d x − 3 p − 6 2 p e 3 p − 6 2 s 0 ∫ R 3 ∣ u ∣ p d x = e 4 s 0 a e 2 s 0 ∫ R 3 ∣ ∇ u ∣ 2 d x + b ∫ R 3 ∣ ∇ u ∣ 2 2 d x − 3 2 μ e − 7 s 0 ∫ R 3 H e 3 2 s 0 u d x − 3 p − 6 2 p e 3 p − 14 2 s 0 ∫ R 3 ∣ u ∣ p d x = e 4 s 0 P f , p ( u ) ≤ 0 ,

which is contradictory. We have completed the proof.□

Lemma 2.2

Assume that (V1), (V2), (V4), (V5), (V6) and (2) hold, then

Ψ f , p = inf P f , p ℐ f , p > 0 .

Proof

For any u ∈ P f , p , the Gagliardo-Nirenberg inequality is the following:

(13) ‖ u ‖ t ≤ C t ‖ u ‖ 2 1 − δ t ‖ ∇ u ‖ 2 δ t ,

where δ t = 3 ( t − 2 2 t ) . Consequently, using (3)–(4), and (13), we derive that

(14) a ∫ R 3 ∣ ∇ u ∣ 2 d x + b ∫ R 3 ∣ ∇ u ∣ 2 2 d x = 3 2 μ ∫ R 3 [ f ( u ) u − 2 F ( u ) ] d x + 3 p − 6 2 p ∫ R 3 ∣ u ∣ p d x = 3 2 μ ∫ R 3 H ( u ) d x + 3 p − 6 2 p ∫ R 3 ∣ u ∣ p d x ≤ 3 2 μ γ ∫ R 3 ε ∣ u ∣ 6 + ( ε + η ) ∣ u ∣ 14 3 + C ε ∣ u ∣ q d x + 3 p − 6 2 p ∫ R 3 ∣ u ∣ p d x = 3 2 μ γ ε ‖ u ‖ 6 6 + 3 2 μ γ ( ε + η ) ‖ u ‖ 14 3 14 3 + 3 2 μ γ C ε ‖ u ‖ q q + 3 p − 6 2 p ‖ u ‖ p p ≤ 3 2 μ γ ε S − 3 ‖ ∇ u ‖ 2 6 + 3 2 μ γ ( ε + η ) C c m 8 3 ‖ ∇ u ‖ 2 2 + 3 2 μ γ C ε C q q m q ( 1 − δ p ) ‖ ∇ u ‖ 2 q δ q + 3 p − 6 2 p C p p m p ( 1 − δ p ) ‖ ∇ u ‖ 2 p δ p ,

where γ is determined in (2). Taking ε < 1 − γ η C c m 8 3 γ C c m 8 3 and using (2), we derive that there exists a constant ℳ > 0 such that ‖ ∇ u ‖ 2 2 ≥ ℳ > 0 , owing to the fact p δ p > q δ q > 4 . Hence,

ℐ f , p ( u ) = ℐ f , p ( u ) − 1 4 P f , p ( u ) = a 4 ∫ R 3 ∣ ∇ u ∣ 2 d x + 3 8 μ ∫ R 3 f ( u ) u − 14 3 F ( u ) d x + 3 p − 14 8 p ∫ R 3 ∣ u ∣ p d x ≥ 1 4 a ℳ > 0 .

Therefore, Ψ f , p > 0 .□

Corollary 2.3

Assume that (V1), (V2), (V4), (V5), (V6), and (2) hold, then Ψ f , 6 > 0 .

Lemma 2.4

Assume that (V4) and (V6) hold and f is odd, then Ψ f , q is attainted by a nonnegative and radially symmetric function.

Proof

Take a minimizing sequence { u n } ⊂ P f , p for Ψ f , p . Then, u n ∈ P f , p , ℐ f , p ( u n ) → Ψ f , p as n → ∞ . In other words, u n ∈ S m , P f , p ( u n ) = 0 , ℐ f , p ( u n ) → Ψ f , p as n → ∞ . Let u n * be the symmetric decreasing rearrangement of u n , then we have u n * = ∣ u n ∣ * . Without loss of generality, we may suppose that u n is nonnegative and apply Fubini’s theorem, then we can see that

∫ R 3 F ( u n ) d x = ∫ R 3 ∫ 0 u n f ( s ) d s d x = ∫ 0 ∞ f ( s ) ∣ { x : u n ( x ) > s } ∣ d s = ∫ 0 ∞ f ( s ) ∣ { x : u n * ( x ) > s } ∣ d s = ∫ R 3 F ( u n * ) d x .

Recalling the Polya-Szego inequality, we have

∫ R 3 ∣ ∇ u n * ∣ 2 d x ≤ ∫ R 3 ∣ ∇ u n ∣ 2 d x .

Hence, P f , p ( u n * ) ≤ P f , p ( u n ) = 0 . Set ( u n * ) s = e 3 2 s u n * ( e s x ) . From Lemma 2.1, there exists a unique s n ≤ 0 such that

( u n * ) s n ∈ P f , p .

Consequently,

(15) Ψ f , p ≤ ℐ f , p ( ( u n * ) s n ) = ℐ f , p ( ( u n * ) s n ) − 1 4 P f , p ( ( u n * ) s n ) = a 4 ∫ R 3 ∣ ∇ ( u n * ) s n ∣ 2 d x + 3 8 μ ∫ R 3 f ( ( u n * ) s n ) ( u n * ) s n − 14 3 F ( ( u n * ) s n ) d x + 3 p − 14 8 p ∫ R 3 ∣ ( u n * ) s n ∣ p d x = a 4 e 2 s n ∫ R 3 ∣ ∇ u n * ∣ 2 d x + 3 p − 14 8 p e 3 p − 6 2 s n ∫ R 3 ( u n * ) p d x + 3 8 μ e − 3 s n ∫ R 3 f e 3 2 s n ( u n * ) e 3 2 s n ( u n * ) − 14 3 F e 3 2 s n ( u n * ) d x ≤ a 4 e 2 s n ∫ R 3 ∣ ∇ u n ∣ 2 d x + 3 p − 14 8 p e 3 p − 6 2 s n ∫ R 3 ∣ u n ∣ p d x + 3 8 μ e − 3 s n ∫ R 3 f e 3 2 s n ( ∣ u n ∣ ) e 3 2 s n ( ∣ u n ∣ ) − 14 3 F e 3 2 s n ( ∣ u n ∣ ) d x ≤ a 4 ∫ R 3 ∣ ∇ u n ∣ 2 d x + 3 8 μ ∫ R 3 f ( u n ) ( u n ) − 14 3 F ( u n ) d x + 3 p − 14 8 p ∫ R 3 u n p d x = ℐ f , p ( u n ) − 1 4 P f , p ( u n ) = ℐ f , p ( u n ) → Ψ f , p , as n → ∞ .

Obviously, we obtain

(16) ℐ f , p ( ( u n * ) s n ) → Ψ f , p , as n → ∞ .

We know that { ( u n * ) s n } is a minimizing sequence for ℐ f , p , and together with the proof of Lemma 2.2, { ( u n * ) s n } is bounded in H 1 ( R 3 ) . Up to a subsequence, we may assume that ( u n * ) s n ⇀ u in H 1 ( R 3 ) , ( u n * ) s n → u in L r ( R 3 ) for 2 < r < 6 , and ( u n * ) s n ( x ) → u ( x ) a.e. on R 3 . Consequently, we have

∫ R 3 ∣ ∇ u ∣ 2 d x ≤ liminf n → ∞ ∫ R 3 ∣ ∇ ( u n * ) s n ∣ 2 d x , ∫ R 3 ∣ u ∣ 2 d x ≤ liminf n → ∞ ∫ R 3 ∣ ( u n * ) s n ∣ 2 d x = m 2 .

Furthermore, together with the uniqueness of the existence of limit, we have

(17) P f , p ( u ) = a ∫ R 3 ∣ ∇ u ∣ 2 d x + b ∫ R 3 ∣ ∇ u ∣ 2 d x 2 − 3 2 μ ∫ R 3 [ f ( u ) u − 2 F ( u ) ] d x − 3 p − 6 2 p ∫ R 3 ∣ u ∣ p d x ≤ liminf n → ∞ ∫ R 3 ∣ ∇ ( u n * ) s n ∣ 2 d x + b liminf n → ∞ ∫ R 3 ∣ ∇ ( u n * ) s n ∣ 2 d x 2 − 3 2 μ lim n → ∞ ∫ R 3 [ f ( ∣ u ∣ * ) s n ( ∣ u ∣ * ) s n − 2 F ( ( ∣ u ∣ * ) s n ) ] d x − 3 p − 6 2 p lim n → ∞ ∫ R 3 ∣ ( ∣ u ∣ * ) s n ∣ p d x ≤ liminf n → ∞ P f , p ( ( u n * ) s n ) = 0 .

Set ∫ R 3 u 2 d x ≔ θ 2 , then θ 2 ≤ m 2 . Define

u ˜ ( x ) ≔ ( θ m − 1 ) 2 p − 2 u ( θ m − 1 ) 2 p 3 ( p − 2 ) x ,

then we obtain that

∫ R 3 f ( u ˜ ) u ˜ d x = ( θ m − 1 ) − 6 p ( 3 p − 2 ) ∫ R 3 f ( θ m − 1 ) 2 q − 2 u ( θ m − 1 ) 2 p − 2 u d x ≤ ∫ R 3 f ( u ) u d x ,

∫ R 3 ∣ u ˜ ∣ 2 d x = ( θ m − 1 ) 4 p − 2 ∫ R 3 u 2 ( θ m − 1 ) 2 p 3 ( p − 2 ) x d x = m 2 , ∫ R 3 ∣ u ˜ ∣ p d x = ∫ R 3 ∣ u ∣ p d x ,

∫ R 3 ∣ ∇ u ˜ ∣ 2 d x = ∫ R 3 ∇ ( θ m − 1 ) 2 p − 2 u ( θ m − 1 ) 2 p 3 ( p − 2 ) x 2 d x = ( θ m − 1 ) 2 ( 6 − p ) 3 ( p − 2 ) ∫ R 3 ∣ ∇ u ∣ 2 d x ≤ ∫ R 3 ∣ ∇ u ∣ 2 d x .

We can obtain

P f , p ( u ˜ ) = a ∫ R 3 ∣ ∇ u ˜ ∣ 2 d x + b ∫ R 3 ∣ ∇ u ˜ ∣ 2 d x 2 − 3 2 μ ∫ R 3 [ f ( u ˜ ) u ˜ − 2 F ( u ˜ ) ] d x − 3 p − 6 2 p ∫ R 3 ∣ u ˜ ∣ p d x ≤ a ∫ R 3 ∣ ∇ u ∣ 2 d x + b ∫ R 3 ∣ ∇ u ∣ 2 d x 2 − 3 2 μ ∫ R 3 [ f ( u ) u − 2 F ( u ) ] d x − 3 p − 6 2 p ∫ R 3 ∣ u ∣ p d x ≤ P f , p ( u ) ≤ 0 .

By Lemma 2.1, we obtain that there exists s ≤ 0 such that u ˜ s ∈ P f , p . And then, we have

Ψ f , p ≤ ℐ f , p ( u ˜ s ) = ℐ f , p ( u ˜ s ) − 2 3 p − 6 P f , p ( u ˜ s ) = 3 p − 10 2 ( 3 p − 6 ) a ∫ R 3 ∣ ∇ u ˜ s ∣ 2 d x + 3 p − 14 4 ( 3 p − 6 ) b ∫ R 3 ∣ ∇ u ˜ s ∣ 2 d x 2 + 1 p − 2 μ ∫ R 3 [ f ( u ˜ s ) u ˜ s − p F ( u ˜ s ) ] d x = 3 p − 10 2 ( 3 p − 6 ) a e 2 s ∫ R 3 ∣ ∇ u ˜ ∣ 2 d x + 3 p − 14 4 ( 3 p − 6 ) b e 4 s ∫ R 3 ∣ ∇ u ˜ ∣ 2 d x 2 + 1 p − 2 e − 3 s μ ∫ R 3 ( f ( u ˜ ) u ˜ − p F ( u ˜ ) ) d x ≤ 3 p − 10 2 ( 3 p − 6 ) a e 2 s ∫ R 3 ∣ ∇ u ∣ 2 d x + 3 p − 14 4 ( 3 p − 6 ) b e 4 s ∫ R 3 ∣ ∇ u ∣ 2 d x 2 + 1 p − 2 e − 3 s μ ∫ R 3 [ f ( u ) u − p F ( u ) ] d x ≤ 3 p − 10 2 ( 3 p − 6 ) a ∫ R 3 ∣ ∇ u ∣ 2 d x + 3 p − 14 4 ( 3 p − 6 ) b ∫ R 3 ∣ ∇ u ∣ 2 d x 2 + 1 p − 2 μ ∫ R 3 [ f ( u ) u − p F ( u ) ] d x ≤ 3 p − 10 2 ( 3 p − 6 ) a liminf n → ∞ ∫ R 3 ∣ ∇ ( ∣ u n ∣ ∗ ) s n ∣ 2 d x + 3 p − 14 4 ( 3 p − 6 ) b liminf n → ∞ ∫ R 3 ∣ ∇ ( ∣ u n ∣ ∗ ) s n ∣ 2 d x 2 + 1 p − 2 μ lim n → ∞ ∫ R 3 [ f ( ∣ u * ∣ s n ) ( ∣ u * ∣ s n ) − p F ( ∣ u * ∣ s n ) ] d x ≤ liminf n → ∞ ℐ f , p ( ∣ u * ∣ s n ) − 2 3 p − 6 P f , p ( ∣ u * ∣ s n ) = Ψ f , p .

This shows that

ℐ f , p ( u ˜ τ ) = Ψ f , p , s = 0 , and θ = m .

Consequently, u ∈ P f , p is a minimizer of ℐ f , p , and it is nonnegative and radially symmetric. By the maximum principle, we conclude that u > 0 .□

Set

ρ f , p ≔ { u ∈ P f , p : ℐ f , p = Ψ f , p } .

Lemma 2.5

Assume that (V3), (V4) and ρ f , p ≠ ∅ hold, then for any u ∈ ρ f , p , there exists λ > 0 such that u fulfills

− a + b ∫ R 3 ∣ ∇ u ∣ 2 d x Δ u + λ u = μ f ( u ) + ∣ u ∣ p − 2 u .

Proof

For arbitrary u ∈ ρ f , p , we can obtain that u ∈ P f , p and

ℐ f , p = Ψ f , p = inf P f , p ℐ f , p .

We apply the Lagrange multiplier rule to (2.12), and there exist λ and η such that

η − 2 a △ u − 4 b ∫ R 3 ∣ ∇ u ∣ 2 d x Δ u − 3 2 μ h ( u ) − 3 p − 6 2 ∣ u ∣ p − 2 u = − a + b ∫ R 3 ∣ ∇ u ∣ 2 d x Δ u + λ u − μ f ( u ) − ∣ u ∣ p − 2 u .

That is to say,

(18) − ( 1 − 2 η ) a − ( 1 − 4 η ) b ∫ R 3 ∣ ∇ u ∣ 2 d x Δ u + λ u = μ f ( u ) − 3 2 η μ h ( u ) + 1 − η ⋅ 3 p − 6 2 ∣ u ∣ p − 2 u .

Next, we prove that η = 0 . To calculate the Pohožaev identity for (18), we obtain

(19) ( 1 − 2 η ) a ∫ R 3 ∣ ∇ u ∣ 2 d x + ( 1 − 4 η ) b ∫ R 3 ∣ ∇ u ∣ 2 d x 2 − 3 2 μ ∫ R 3 [ f ( u ) u − 2 F ( u ) ] d x + 9 4 η μ ∫ R 3 h ( u ) − 4 3 H ( u ) d x − 3 p − 6 2 p 1 − η 3 p − 6 2 ∫ R 3 ∣ u ∣ p d x = 0 .

Together with P f , p = 0 , subtracting P f , p from (19) gives

(20) 0 = − 2 η a ∫ R 3 ∣ ∇ u ∣ 2 d x − 4 b η ∫ R 3 ∣ ∇ u ∣ 2 d x 2 + 9 4 η μ ∫ R 3 h ( u ) u − 4 3 H ( u ) d x + 3 p − 6 2 p η ⋅ 3 p − 6 2 ∫ R 3 ∣ u ∣ p d x .

If η ≠ 0 , then dividing both sides of (20) by η gives

(21) 0 = 2 a ∫ R 3 ∣ ∇ u ∣ 2 d x + 4 b ∫ R 3 ∣ ∇ u ∣ 2 d x 2 − 9 4 μ ∫ R 3 h ( u ) u − 4 3 H ( u ) d x − 3 p − 6 2 p ⋅ 3 p − 6 2 ∫ R 3 ∣ u ∣ p d x .

Then, computing 4 P f , p –(21), we obtain

0 = 2 a ∫ R 3 ∣ ∇ u ∣ 2 d x + 3 p − 6 2 p ⋅ 3 p − 14 2 ∫ R 3 ∣ u ∣ p d x + 9 4 μ ∫ R 3 h ( u ) u − 4 3 H ( u ) d x + 6 μ ∫ R 3 [ f ( u ) u − 2 F ( u ) ] d x .

Hence, from assumptions (V3) and (V4), we deduce that u = 0 , which is a contradiction. Therefore, we have η = 0 . Then,

− a + b ∫ R 3 ∣ ∇ u ∣ 2 d x Δ u + λ u − μ f ( u ) − ∣ u ∣ p − 2 u = 0 .

Based on P f , p = 0 , we can also conclude that

λ m 2 = 3 2 ∫ R 3 f ( u ) u − 4 3 F ( u ) d x + 6 − p 2 p ∫ R 3 ∣ u ∣ p d x .

Hence, combining (V4), we obtain that λ > 0 .□

Lemma 2.6

We believe that the following inequality is valid:

limsup p → 6 Ψ f , p ≤ Ψ f , 6 .

Proof

Let

g ( p ) ≔ 1 p e 3 p − 6 2 s ∫ R 3 ∣ u ∣ p d x .

Then, g ( p ) is continuous on 14 3 , 6 . By the expression of ℐ f , 6 ( u s ) , for any ζ ∈ ( 0 , 1 ) , there exists δ > 0 such that

∣ ℐ f , p ( u s ) − ℐ f , 6 ( u s ) ∣ < ζ ,

for 6 − δ ≤ p ≤ 6 . Consequently, ℐ f , p ( u s ) ≤ ℐ f , 6 ( u s ) + ζ for any 6 − δ ≤ p ≤ 6 . Because u ∈ P f , p , combining Lemma 2.1, we have

ℐ f , 6 ( u ) = max s ∈ R ℐ f , 6 ( u s ) ,

and there exists u ∈ P f , 6 such that ℐ f , p < Ψ f , 6 + ζ . Hence, we obtain

Ψ f , p ≤ ℐ f , p ( u s ) < ℐ f , 6 ( u ) s + ζ ≤ ℐ f , 6 ( u ) + ζ < Ψ f , 6 + 2 ζ ,

for arbitrary p ∈ ( 6 − δ , 6 ) . The conclusion holds.□

Lemma 2.7

Assume that (V0), (V1), (V2), (V4), (V5), and (2) hold. Then, we have

liminf p → 6 Ψ f , p > 0 .

Proof

From Lemma 2.4, we can take a minimizing sequence { u f , p } p ⊂ P f , p satisfying ℐ f , p ( u f , p ) = Ψ f , p . Considering the Young inequality, we obtain

(22) ∣ u ∣ p = ∣ u ∣ 6 − p 6 − q ⋅ q ⋅ ∣ u ∣ p − q 6 − q ⋅ 6 ≤ 6 − p 6 − q ∣ u ∣ q + p − q 6 − q ∣ u ∣ 6 .

Using (3) and (22), we obtain

(23) a ∫ R 3 ∣ ∇ u f , p ∣ 2 d x + b ∫ R 3 ∣ ∇ u f , p ∣ 2 d x 2 = 3 2 μ ∫ R 3 H ( u f , p ) d x + 3 q − 6 2 q ∫ R 3 ∣ u f , p ∣ p d x ≤ γ μ ∫ R 3 [ ε ∣ u f , p ∣ 6 + ( ε + η ) ∣ u f , p ∣ 14 3 + C ε ∣ u f , p ∣ q ] d x + 3 p − 6 2 p ∫ R 3 ∣ u f , p ∣ p d x ≤ γ μ [ ε ‖ u f , p ‖ 6 6 + ( ε + η ) ‖ u f , p ‖ 14 3 14 3 + C ε ‖ u f , p ‖ q q ] + 3 p − 6 2 p ⋅ 6 − p 6 − q ‖ u f , p ‖ q q + 3 p − 6 2 p ⋅ p − q 6 − q ‖ u f , p ‖ 6 6 ≤ γ μ C ε + 3 p − 6 2 p ⋅ 6 − p 6 − q C q q m q ( 1 − δ q ) ‖ ∇ u f , p ‖ 2 q δ q + γ μ ε + 3 p − 6 2 p ⋅ p − q 6 − q S − 3 ‖ ∇ u f , p ‖ 2 6 + γ μ ( ε + η ) C c m 8 3 ‖ ∇ u f , q ‖ 2 2 .

Then, from (23), we know that there exists a constant M > 0 that satisfies ‖ ∇ u f , p ‖ 2 2 ≥ M . Hence,

ℐ f , p ( u ) = ℐ f , p ( u ) − 1 4 P f , p ( u ) = a 4 ∫ R 3 ∣ ∇ u ∣ 2 d x + 3 8 μ ∫ R 3 f ( u ) u − 14 3 F ( u ) d x + 3 p − 14 8 p ∫ R 3 ∣ u ∣ p d x ≥ 1 4 a M > 0 .

Consequently, we obtain

□ liminf p → 6 Ψ f , p > 0 .

Lemma 2.8

There exists μ ′ > 0 such that

0 < Ψ f , p < a b S 3 + a b 2 S 6 + 4 a S 3 6 , f o r a n y μ > μ ′ .

Proof

Assume that there exists a sequence { u m } with u m → ∞ such that

Ψ f , p ≥ a b S 3 + a b 2 S 6 + 4 a S 3 6 .

For u ∈ S m and combining Lemma 2.1, there exists s μ m ∈ R such that

u s μ m ∈ P f , 6 and ℐ f , 6 ( u s μ m ) = max s ∈ R ℐ f , 6 ( u s ) .

Then, it follows that

(24) a e 2 s μ m ∫ R 3 ∣ ∇ u ∣ 2 d x + b e 4 s μ m ∫ R 3 ∣ ∇ u ∣ 2 d x 2 = 3 2 μ e − 3 s μ m ∫ R 3 f e 3 2 s u e 3 2 s u − 2 F e 3 2 s u d x + e 6 s μ m ∫ R 3 ∣ u ∣ 6 d x .

We derive that { e s μ m } is bounded in R . So, there exists α ≥ 0 such that e s μ m → α as m → ∞ . We claim that α = 0 . If not, there exists α 0 > 0 . Evidently,

lim m → ∞ 3 2 μ e − 3 s μ m ∫ R 3 f e 3 2 s μ m u e 3 2 s μ m u − 2 F e 3 2 s μ m u d x + e 6 s μ m ∫ R 3 ∣ u ∣ 6 d x = + ∞ .

However,

a e 2 s μ m ∫ R 3 ∣ ∇ u ∣ 2 d x + b e 4 s μ m ∫ R 3 ∣ ∇ u ∣ 2 d x 2 → a α 0 2 ∫ R 3 ∣ ∇ u ∣ 2 d x + b α 0 4 ∫ R 3 ∣ ∇ u ∣ 2 d x 2 ,

which contradicts with (24). Therefore, α = 0 . So

max s ∈ R ℐ f , 6 ( u s μ m ) = ℐ f , 6 ( u s μ m ) = 1 2 a e 2 s μ m ∫ R 3 ∣ ∇ u ∣ 2 d x + 1 4 b e 4 s μ m ∫ R 3 ∣ ∇ u ∣ 2 d x 2 − μ e − 3 s μ m ∫ R 3 F e 3 2 s μ m u d x − 1 6 e 6 s μ m ∫ R 3 ∣ u ∣ 6 d x → 0 , as m → ∞ .

As a result,

0 < a b S 3 + a b 2 S 6 + 4 a S 3 6 ≤ Ψ f , q ≤ ℐ f , q ( u s μ m ) → 0 ,

as m → ∞ , which is a contradiction.□

In the sequel, we assume: μ ≥ μ ′ .

Lemma 2.9

Assume that (V1), (V2), (V4), (V5), (V6), and (2)  hold and f is odd, then Ψ f , 6 is attainted by a nonnegative and radially symmetric function.

Proof

Let p m → 6 − as m → ∞ . From Lemma 2.4, there exists a sequence of nonnegative and radially symmetric function { u m } ≔ { u f , p m } ⊂ P f , p m such that

ℐ f , p m ( u m ) = Ψ f , p m ≤ Ψ f , 6 + 1 .

Together with Lemma 2.6, we can obtain

Ψ f , 6 + 1 ≥ ℐ f , p m ( u m ) = ℐ f , p m ( u m ) − 1 4 P f , p m ( u m ) = a 4 ∫ R 3 ∣ ∇ u m ∣ 2 d x + 3 8 μ ∫ R 3 f ( u m ) u m − 14 3 F ( u m ) d x + 3 p m − 14 8 p m ∫ R 3 ∣ u m ∣ p m d x ≥ a 4 ∫ R 3 ∣ ∇ u m ∣ 2 d x .

Together with u m ∈ S m , we know that { u m } is bounded in H 1 ( R 3 ) . Consequently, there exists a sequence of nonnegative and radially symmetric function u ∈ H 1 ( R 3 ) such that u m ⇀ u in H 1 ( R 3 ) , u m → u in L r ( R 3 ) for 2 < r < 6 , and u m → u ( x ) a.e. on R 3 . According to Lemma 2.5, there exists λ m > 0 such that u m satisfies

− ( a + b ∫ R 3 ∣ ∇ u m ∣ 2 d x ) Δ u m + λ m u m = μ f ( u m ) + ∣ u m ∣ p m − 2 u m .

Together with u m ∈ P f , p , the same as Lemma 2.5, we have

λ m m 2 = 3 2 ∫ R 3 f ( u m ) u m − 4 3 F ( u m ) d x + 6 − p m 2 p m ∫ R 3 ∣ u m ∣ p m d x .

We derive that { λ m } is bounded in R . So, there exists λ ≥ 0 such that λ m → λ as m → ∞ . In the following proof, we will divide it into the following steps.

First, u solves

− a + b ∫ R 3 ∣ ∇ u ∣ 2 d x Δ u + λ u = μ f ( u ) + ∣ u ∣ 4 u .

Based on the previous discussions, we only need to prove that

(25) ∫ R 3 ∣ u m ∣ p m − 2 u m d x → ∫ R 3 ∣ u ∣ 4 u d x .

In fact, for arbitrary ω ∈ C 0 ∞ ( R 3 ) , using Young’s inequality and Lemma 2.2 in [29], we have

∣ ∣ u m ∣ q m − 2 u m ω ∣ ≤ C ( ∣ u m ∣ ∣ ω ∣ + ∣ u m ∣ 5 ∣ ω ∣ ) ≤ C ( ∣ x ∣ − 1 2 ∣ ω ∣ + ∣ x ∣ − 5 2 ∣ ω ∣ ) ∈ L 1 ( R 3 ) ,

where q m → 6 − as m → ∞ , C is nonnegative and independent of m . Then,

∫ R 3 ∣ u m ∣ p m − 2 u m ω d x → ∫ R 3 ∣ u ∣ 4 u ω d x , as m → ∞ .

Next, we shall prove that u ≢ 0 . We assume that u ≡ 0 . Then,

∫ R 3 [ f ( u m ) u m − 2 F ( u m ) ] d x → 0 , ∫ R 3 ∣ u m ∣ q d x → 0 ,

as m → ∞ . From u m ∈ P f , p m and (22), we derive

a ∫ R 3 ∣ ∇ u m ∣ 2 d x + b ∫ R 3 ∣ ∇ u m ∣ 2 d x 2 = 3 2 μ ∫ R 3 H ( u m ) d x + 3 p m − 6 2 p m ∫ R 3 ∣ u m ∣ p m d x ≤ 6 μ ∫ R 3 [ ε ∣ u m ∣ 6 + ( ε + η ) ∣ u m ∣ 14 3 + C ε ∣ u m ∣ q ] d x + 3 p m − 6 2 p m ∫ R 3 ∣ u m ∣ p m d x = 3 p m − 6 2 p m ∫ R 3 ∣ u m ∣ p m d x ≤ 3 p m − 6 2 p m 6 − p m 6 − q ∫ R 3 ∣ u m ∣ q d x + p m − q 6 − q ∫ R 3 ∣ u m ∣ 6 d x + o ( 1 ) = 3 p m − 6 2 p n ⋅ p m − q 6 − q ∫ R 3 ∣ u m ∣ 6 d x + o ( 1 ) ≤ 3 p m − 6 2 p m ⋅ p m − q 6 − q ⋅ S − 3 ‖ ∇ u m ‖ 6 2 + o ( 1 ) .

It implies that

liminf n → ∞ ‖ ∇ u m ‖ 2 2 ≥ b S 3 + b 2 S 6 + 4 a S 3 2 .

By Lemmas 2.6 and 2.4, we have

Ψ f , 6 ≥ limsup p m → 6 Ψ f , p m = limsup p m → 6 ℐ f , p m

= limsup p m → 6 ℐ f , p m ( u m ) − 2 3 p m − 6 P f , p m ( u m ) ≥ a b S 3 + a b 2 S 6 + 4 a S 3 6 .

This is in conflict with Ψ f , p < a b S 3 + a b 2 S 6 + 4 a S 3 6 .

Furthermore, we define u ˆ ( x ) ≔ ( c m − 1 ) 1 2 u ( c m − 1 x ) , where ∫ R 3 u 2 d x ≔ c 2 such that c 2 ≤ m 2 . It is evident that

(26) P f , p ( u ˆ ) = a ∫ R 3 ∣ ∇ u ˆ ∣ 2 d x + b ∫ R 3 ∣ ∇ u ˆ ∣ 2 d x 2 − 3 2 μ ∫ R 3 [ f ( u ˆ ) u ˆ − 2 F ( u ˆ ) ] d x − ∫ R 3 ∣ u ˆ ∣ 6 d x ≤ a ∫ R 3 ∣ ∇ u ∣ 2 d x + b ∫ R 3 ∣ ∇ u ∣ 2 d x 2 − 3 2 μ ∫ R 3 [ f ( u ) u − 2 F ( u ) ] d x − ∫ R 3 ∣ u ∣ 6 d x = 0 .

According to Lemma 2.1, there exists a unique s 0 ≤ 0 such that

u ˆ s 0 = e 3 2 s 0 u ˆ ( e s 0 x ) ∈ P f , 6 .

Then, based on the weak lower semi-continuity of the norm, (V4), Fatou’s lemma, and Lemma 2.6, we derive

(27) Ψ f , 6 ≤ ℐ f , 6 ( u ˆ s 0 ) = ℐ f , 6 ( u ˆ s 0 ) − 1 4 P f , 6 ( u ˆ s 0 ) = 1 4 a ∫ R 3 ∣ ∇ u ˆ s 0 ∣ 2 d x + 3 8 μ ∫ R 3 f ( u ˆ s 0 ) u ˆ s 0 − 14 3 F ( u ˆ s 0 ) d x + 1 12 ∫ R 3 ∣ u ˆ s 0 ∣ 6 d x = 1 4 a e 2 s 0 ∫ R 3 ∣ ∇ u ˆ ∣ 2 d x + 3 8 μ e − 3 s 0 ∫ R 3 f ( e 3 2 s 0 u ˆ ) e 3 2 s 0 u ˆ − 14 3 F ( e 3 2 s 0 u ˆ ) d x + 1 12 e 6 s 0 ∫ R 3 ∣ u ˆ ∣ 6 d x ≤ 1 4 a e 2 s 0 ∫ R 3 ∣ ∇ u ∣ 2 d x + 3 8 μ e − 3 s 0 ∫ R 3 f ( u ) u − 14 3 F ( u ) d x + 1 12 e 6 s 0 ∫ R 3 ∣ u ∣ 6 d x ≤ 1 4 a ∫ R 3 ∣ ∇ u ∣ 2 d x + 3 8 μ ∫ R 3 f ( u ) u − 14 3 F ( u ) d x + 1 12 ∫ R 3 ∣ u ∣ 6 d x ≤ 1 4 a liminf m → ∞ ∫ R 3 ∣ ∇ u m ∣ 2 d x + liminf m → ∞ 3 8 μ ∫ R 3 f ( u m ) u m − 14 3 F ( u ) d x + liminf m → ∞ 3 p m − 14 8 p m ∫ R 3 ∣ u m ∣ p m d x ≤ liminf m → ∞ ℐ f , p m ( u ˆ s 0 ) − 1 4 P f , p m ( u ˆ s 0 ) = liminf m → ∞ ℐ f , p m ( u m ) = liminf m → ∞ Ψ f , p m ≤ limsup m → ∞ Ψ f , p m ≤ Ψ f , 6 .

As a result, ℐ f , 6 ( u ˆ s 0 ) = Ψ f , 6 . Namely, u ˆ s 0 is a minimizer of Ψ f , 6 and is nonnegative and radially symmetric. By the strong maximum principle, u ˆ s 0 > 0 in R N .□

Proof of Theorem 1.1

Taking into account of Lemma 2.9, Ψ f , 6 = ℐ f , p is attained at u ∈ P f , 6 and u is a nonnegative and radially symmetric function. Then, by Lemma 2.5, there exists λ > 0 such that u fulfills the following equation:

− a + b ∫ R 3 ∣ ∇ u ∣ 2 d x Δ u + λ u = μ f ( u ) + ∣ u ∣ 4 u , in R 3 .

We have completed the proof of Theorem 1.1.□

3 Asymptotic behavior

Before proceeding with this section, we first assume that the following inequality holds:

(28) 2 η C c ∫ R 3 u 2 d x 2 3 < 1 ,

where u ∈ H 1 ( R 3 ) \ { 0 } .

Set

(29) χ ( λ ) ≔ ℐ f , p ( λ 3 2 u ( λ x ) ) .

Lemma 3.1

Assume that (V1)–(V6) hold. Then, there exists an interval [ a , b ] ⊂ ( 0 , + ∞ ) and λ ∈ [ a , b ] , such that λ 3 2 u ( λ x ) ∈ P f , p .

Proof

By (V1), we have

χ ( λ ) = ℐ f , p ( λ 3 2 u ( λ x ) ) = a 2 ∫ R 3 ∣ ∇ λ 3 2 u ( λ x ) ∣ 2 d x + b 4 ∫ R 3 ∣ ∇ λ 3 2 u ( λ x ) ∣ 2 d x 2 − μ ∫ R 3 F ( λ 3 2 u ( λ x ) ) d x − 1 p ∫ R 3 λ 3 2 u ( λ x ) p d x = a 2 λ 2 ∫ R 3 ∣ ∇ u ∣ 2 d x + b 4 λ 4 ∫ R 3 ∣ ∇ u ∣ 2 d x 2 − μ λ − 3 ∫ R 3 F ( λ 3 2 u ) d x − 1 p λ 3 p − 6 2 ∫ R 3 ∣ u ∣ p d x → 0 ,

as λ → 0 + , and from (4), we obtain that χ ( λ ) → ∞ as λ → + ∞ . Let

L ≔ ‖ u ‖ 2 2 = ‖ λ 3 2 u ( λ x ) ‖ 2 2 > 0 .

Using (3), for arbitrary ε > 0 , there exists C ε > 0 such that

∫ R 3 F ( u ) d x ≤ ( ε + η ) ‖ u ‖ 14 3 14 3 + C ε ‖ u ‖ 6 6 ≤ ( ε + η ) C c L 2 3 ‖ ∇ u ‖ 2 2 + S − 3 C ε ‖ ∇ u ‖ 2 6 .

Then,

χ ( λ ) λ 2 = a 2 ∫ R 3 ∣ ∇ u ∣ 2 d x + b 4 λ 2 ∫ R 3 ∣ ∇ u ∣ 2 d x 2 − μ λ − 5 ∫ R 3 F ( λ 3 2 u ) d x − 1 p λ 3 p − 10 2 ∫ R 3 ∣ u ∣ p d x ≥ 1 2 a ‖ ∇ u ‖ 2 2 + 1 4 b λ 2 ‖ u ‖ 2 4 − ( ε + η ) C c L 2 3 ‖ ∇ u ‖ 2 2 − S − 3 C ε λ 4 ∣ ∇ u ‖ 2 6 − 1 p λ 3 p − 10 2 ∣ u ∣ p p = 1 2 a − ( ε + η ) C c L 2 3 ‖ ∇ u ‖ 2 2 + 1 4 b λ 2 ‖ u ‖ 2 4 − S − 3 C ε λ 4 ∣ ∇ u ‖ 2 6 − 1 p λ 3 p − 10 2 ‖ u ‖ ∣ u ∣ p p .

By combining with (28), we obtain that there exists a small enough ε > and λ > 0 such that χ ( λ ) > 0 . Therefore, there exists an interval [ a , b ] ⊂ ( 0 , + ∞ ) such that χ ( λ ) has at least one extreme value in [ a , b ] . Note that

(30) λ χ ′ ( λ ) = λ 2 a ∫ R 3 ∣ ∇ u ∣ 2 d x + λ 4 b ∫ R 3 ∣ ∇ u ∣ 2 d x 2 − 3 2 μ λ − 3 ∫ R 3 H λ 3 2 u d x − 3 p − 6 2 p λ 3 p − 6 2 p ∫ R 3 ∣ u ∣ p d x = P f , p λ 3 2 u ( λ ⋅ ) .

Particularly, there exists λ n ∈ [ a , b ] such that λ n 3 2 u ( λ n x ) ∈ P f , p .□

Lemma 3.2

lim m → ∞ Ψ f , 6 → 0 + .

Proof

Set u m = m n u , then u m ∈ S m . Let m n → ∞ as n → ∞ . From Lemma 3.1, there exists λ n > 0 , such that κ ≔ λ n 3 2 u ( λ n x ) ∈ P f , 6 . Consequently, according to Corollary 2.3, we have

(31) 0 < Ψ f , 6 ( κ ) ≤ ℐ f , 6 ( κ ) ≤ a 2 ∫ R 3 ∣ ∇ κ ∣ 2 d x + b 4 ∫ R 3 ∣ ∇ κ ∣ 2 d x 2 = a 2 λ n 2 m n 2 ∫ R 3 ∣ ∇ u ∣ 2 d x + b 4 λ n 4 m n 4 ∫ R 3 ∣ ∇ u ∣ 2 d x 2 .

Since κ ∈ P f , 6 , from the last equals in (31), we have

a 2 λ n 2 m n 2 ∫ R 3 ∣ ∇ u ∣ 2 d x + b 4 λ n 4 m n 4 ∫ R 3 ∣ ∇ u ∣ 2 d x 2 = 3 2 μ ∫ R 3 H ( κ ) d x + ∫ R 3 ∣ κ ∣ 6 d x = 3 2 μ λ n − 3 ∫ R 3 H λ n 3 2 m n u d x + λ n 6 m n 6 ∫ R 3 ∣ u ∣ 6 d x .

Therefore,

(32) a 2 ∫ R 3 ∣ ∇ u ∣ 2 d x + b 4 λ n 2 m n 2 ∫ R 3 ∣ ∇ u ∣ 2 d x 2 = 3 2 μ λ n − 5 m n − 2 ∫ R 3 H λ n 3 2 m n u d x + λ n 4 m n 4 ∫ R 3 ∣ u ∣ 6 d x .

Based on hypothesis (V4) and the assumpion lim t → 0 F ( t ) ∣ t ∣ 6 = + ∞ , we obtain, for arbitrary σ > 0 , there exists δ > 0 such that

(33) H ( s ) ≥ 14 3 F ( s ) ≥ σ ∣ s ∣ 6 , s ∈ ( − δ , 0 ) ∪ ( 0 , δ ) .

Therefore, by combining (32) and (33), we have

a 2 ∫ R 3 ∣ ∇ u ∣ 2 d x + b 4 λ n 2 m n 2 ∫ R 3 ∣ ∇ u ∣ 2 d x 2 ≥ 3 2 μ λ n − 5 m n − 2 σ ∫ R 3 λ n 3 2 m n u 6 d x = 3 2 μ σ ( λ n m n ) 4 ∫ R 3 ∣ u ∣ 6 d x .

We can derive that λ n m n → 0 as n → ∞ , then from (31), we can see that Ψ f , 6 → 0 as n → ∞ . Therefore, we conclude that Ψ f , 6 → 0 + as m → ∞ .□

Proof of Theorem 1.3

Evidently, from Lemma 3.2, we can directly deduce that Theorem 1.3 holds. We have completed the proof.□

Acknowledgments

The authors would like to thank the referees for their useful suggestions that have significantly improved the manuscript.

  1. Funding information: This work was supported by the National Natural Science Foundation of China (No. 11961014) and Guangxi Natural Science Foundation (2021GXNSFAA196040).

  2. Author contributions: S.Z. and Q.Z. provided equal contribution to this research article. All authors read and approved the final manuscript.

  3. Conflict of interest: Authors state no conflict of interest.

  4. Data availability statement: Not applicable.

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Received: 2024-11-28
Revised: 2025-05-19
Accepted: 2025-06-16
Published Online: 2025-07-25

© 2025 the author(s), published by De Gruyter

This work is licensed under the Creative Commons Attribution 4.0 International License.

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https://doi.org/10.1515/math-2025-0182
Keywords for this article
normalized ground-state solutions; nonlinear Kirchhoff equation; critical growth; Pohozaev manifold
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