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Normalized solutions for the Kirchhoff equation with combined nonlinearities in ℝ4

An erratum for this article can be found here: https://doi.org/10.1515/anona-2024-3042
  • Xin Qiu , Zeng-Qi Ou , Chun-Lei Tang and Ying Lv EMAIL logo
Published/Copyright: October 16, 2024

Abstract

In this article, we study the following Kirchhoff equation with combined nonlinearities:

a + b R 4 u 2 d x Δ u + λ u = μ u q 2 u + u 2 u , in R 4 , R 4 u 2 d x = c 2 ,

where a , b , c > 0 , μ , λ R , 2 < q < 4 . Under different assumptions on b , c > 0 and μ R , we prove some existence, nonexistence, and asymptotic behavior of the obtained normalized solutions. When μ > 0 :(i) for 2 < q < 3 , we obtain the existence of a local minimizer ground-state solution and a mountain-pass-type solution, (ii) for q = 3 and 3 < q < 4 , we obtain the existence of a mountain-pass type ground-state solution respectively, under different assumptions. When μ < 0 and 2 < q < 4 , we prove the nonexistence result of the aforementioned problem. We also investigate the asymptotic behavior of the normalized ground-state solutions, when μ 0 + and b 0 + , respectively.

MSC 2010: 35J60; 35J50; 35J20

1 Introduction and main results

In this article, we consider the existence, nonexistence, and asymptotic behavior of solutions to the following Kirchhoff problem:

(1.1) a + b R N u 2 d x Δ u + λ u = μ u q 2 u + u p 2 u , in R N ,

with the L 2 -mass constrain

(1.2) S c = { u H 1 ( R N ) : u 2 2 = c 2 > 0 } ,

where N = 4 , λ R , 2 < q < 4 , p = 4 , c > 0 is a priori given, a > 0 is a constant, and b and μ are the positive parameters.

Equation (1.1) is related to the stationary analog of the equation

ρ 2 u t 2 P 0 h + E 2 L 0 L u x 2 d x 2 u x 2 = 0 ,

which is proposed first by Kirchhoff [17] in 1883 to stretch the original wave equation by describing the transversal oscillations of a stretched string. Ma and Munõz Rivera [25] studied a transmission problem concerning a system of two nonlinear elliptic equations of Kirchhoff-type. We can also see [1,2,6,8,11,27,34] and references therein. Mathematically, equation (1.1) is not a pointwise identity as a result of the emergence of the term ( b R N u 2 d x ) Δ u , which causes some mathematical difficulties. Thereafter, there has been a boom in the study of the Kirchhoff-type equation.

In the past few years, many authors have widely studied Kirchhoff-type equations. We point out that the parameter λ R is fixed in the aforementioned articles. Moreover, we are interested in another approach that the prescribed L 2 -norm may represent the number of particles of each component in Bose-Einstein condensates or the power supply in the nonlinear optics framework. In addition, such solutions can give a better insight of the dynamical properties, like orbital stability or instability, and can describe attractive Bose-Einstein condensates. We take ( u , λ ) as normalized solution with u 2 = c and λ is a Lagrange multiplier.

When a = 1 , b = 0 , equations (1.1)–(1.2) reduce to the following Schrödinger equation with combined nonlinearities:

(1.3) Δ u + λ u = μ u q 2 u + u p 2 u , in R N , R N u 2 d x = c 2 .

It is well known that 2 + 4 N is the mass-critical exponent for Schrödinger equation and 2 * is the Sobolev critical exponent ( 2 * = 2 N N 2 if N 3 , and 2 * = + if N = 1 , 2 ). For N 3 and μ R , Soave [29] studied problem (1.3) for the case 2 < q 2 + 4 N p < 2 * and q < p . Soave also showed that nonlinear terms with different power strongly affect existence and asymptotic properties of normalized ground-state solutions, as well as stability/instability results. Moreover, for N 3 , Soave [30] considered the Sobolev critical case and proved the existence and asymptotic properties of normalized ground-state solutions for 2 < q < 2 + 4 N , q = 2 + 4 N and q > 2 + 4 N , respectively. Later, Luo and Zhang [19] studied the fractional Schrödinger equations with combined nonlinearities and obtained existence and nonexistence results of normalized solutions. For more forms, we refer the readers to [23,36] and references therein.

When a > 0 , b > 0 , equation (1.1) is a Kirchhoff-type equation. As far as we know, two mass-critical exponents: 2 + 4 N , 2 + 8 N and Sobolev critical exponent 2 * play a crucial role for the existence of normalized solutions. In particular, when μ = 0 in equation (1.1), Ye [37] studied the following Kirchhoff equation:

(1.4) a + b R N u 2 d x Δ u + λ u = u p 2 u , in R N , R N u 2 d x = c 2 ,

where 1 N 3 , a , b , c > 0 , p ( 2 , 2 * ) . To study normalized solutions of (1.4), it is natural to investigate the energy functional I ( u ) : H 1 ( R N ) R given by

I ( u ) = a 2 R N u 2 d x + b 4 R N u 2 d x 2 1 p R N u p d x .

For p 2 , 2 + 8 N , Ye obtained the sharp existence of global constraint minimizers. When p 2 + 4 N , 2 + 8 N , Ye proved a local minimizer, which is a critical point of I S c . By considering a global minimization problem

(1.5) l c inf S c I ( u ) ,

we have

(1.6) l c ( , 0 ] , if p 2 , 2 + 8 N , l c = , if p 2 + 8 N , 2 * ,

for any given c > 0 , which we can know that (1.5) cannot be attained for the case p 2 + 8 N , 2 * , so immediately after Ye proved the existence of normalized solutions by using the constraint minimization via a suitable submanifold of S c . Moreover, Ye [38,39] proved the existence and mass concentration of critical points for the L 2 -critical case p = 2 + 8 N . After that, using simple energy estimates rather than the concentration-compactness principles introduced in [37], for any c > 0 and p 2 , 2 + 4 N , Zeng and Zhang [40] proved the existence and uniqueness of the minimizer to (1.5), while for p 2 + 4 N , 2 + 8 N , they showed that there exists a threshold mass c * > 0 that demonstrates the existence of a unique minimizer for c > c * and there is no minimizer for any c ( 0 , c * ) . When μ > 0 and N = 3 in equation (1.1), Li et al. [18] proved the existence and asymptotic properties of solutions on S c for the case 2 < q < 14 3 < p 6 or 14 3 < q < p 6 . They also showed a multiplicity result for the case 2 < q < 10 3 and 14 3 < p < 6 and obtained the existence of ground-state solutions for 2 < q < 10 3 < p = 6 or 14 3 < q < p 6 . For the case μ > 0 , 14 3 < q < 6 and p = 6 , Li et al. [20] studied the existence and the asymptotic behavior of solutions on S c with the aid of the Sobolev subcritical approximation method. Besides, for the case μ 0 , Carrião et al. [7] studied the existence of ground-states for 2 < q < 2 * , p = 2 * or 2 < q p ¯ < p < 2 * .

Most results in the literature for the Kirchhoff problem consider 1 N 3 because of 2 + 8 N < 2 * , we can refer to [12,14,24,31,37,38,40] and references therein. Yet there are very few articles that consider N 4 , we can see [16,21,22,26]. Motivated by the results mentioned earlier, the aim of this article is to consider the mass-critical exponent and energy-critical exponent occur simultaneously for Kirchhoff problem in R 4 . In R 4 , three critical exponents: 2 + 4 N , 2 + 8 N , and 2 * may appear simultaneously, which makes the problem more interesting and complicated.

To obtain the solutions of equation (1.1), one usually look for critical points of functional

(1.7) J b , μ ( u ) = a 2 R 4 u 2 d x + b 4 R 4 u 2 d x 2 μ q R 4 u q d x 1 4 R 4 u 4 d x

constrained on the L 2 -spheres S c = { u H 1 ( R N ) : u 2 2 = c 2 > 0 } , where

H 1 ( R 4 ) = { u L 2 ( R 4 ) : u L 2 ( R 4 ) } ,

with the inner product

u , v = R 4 [ u v + u v ] d x

and the norm

u = u , u 1 2 .

We say that u ˜ is a ground-state of (1.1)–(1.2) if it is a solution to (1.1) having minimal energy among all the solutions that belongs to S c :

d J b , μ S c ( u ˜ ) = 0 and J b , μ ( u ˜ ) = inf { J b , μ ( u ) : d J b , μ S c ( u ) = 0 , and u S c } .

In order to obtain the ground-state solution, as in [30], we introduce the Pohozaev manifold

(1.8) P b , μ = { u S c : P b , μ ( u ) = 0 } ,

where

(1.9) P b , μ ( u ) = a u 2 2 + b u 2 4 μ γ q u q q u 4 4 = 0 ,

with γ q = 2 ( q 2 ) q . It is well known that any critical point of J b , μ S c stays in P b , μ , as a consequence of the Pohozaev identity (we refer for instance to [15, Lemma 2.7]). Moreover, P b , μ is a natural constraint.

Due to the L 2 -constraint, it is necessary to define the dilations

( s u ) ( x ) e 2 s u ( e s x ) , for s R and u H 1 ( R 4 ) ,

then u 2 = s u 2 for any s R . As a consequence, given u S c , we have s u S c . We also recall (see [4], Lemma 3.5) that the map ( s , u ) R × H 1 ( R 4 ) ( s u ) H 1 ( R 4 ) is continuous. Hence, it is natural to study the fiber map

(1.10) Ψ u ( s ) J b , μ ( s u ) = a e 2 s 2 u 2 2 + b e 4 s 4 u 2 4 e q γ q s q μ u q q e 4 s 4 u 4 4 .

We see that the monotonicity and convexity properties of Ψ u will strongly affect the structure of P b , μ . We can see that Ψ u ( s ) = P b , μ ( s u ) and

P b , μ = { u S c : Ψ u ( 0 ) = 0 } .

Thus, we can see that s is a critical point of Ψ u ( s ) if and only if s u P b , μ , and in particular u P b , μ if and only if 0 is a critical point of Ψ u ( s ) . In this spirit, we consider the decomposition of P b , μ into the disjoint union P b , μ = P b , μ + P b , μ 0 P b , μ , where

(1.11) P b , μ + { u P b , μ : Ψ u ( 0 ) > 0 } , P b , μ 0 { u P b , μ : Ψ u ( 0 ) = 0 } , P b , μ { u P b , μ : Ψ u ( 0 ) < 0 } ,

for Ψ u ( 0 ) 2 a u 2 2 + 4 b u 2 4 μ q γ q 2 u q q 4 u 4 4 .

Then, we introduce the main results of this article as Theorems 1.11.7.

Theorem 1.1

Let 2 < q < 3 and μ , c > 0 be such that

(1.12) μ c ( 1 γ q ) q < C ,

where

C = a q ( 4 q γ q ) C q q 2 a S 2 ( 2 q γ q ) 4 q γ q 2 q γ q 2 .

(i) If 0 < b < S ¯ (see (2.2)), then (1.1) has a positive ground-state solution u ˜ S c with J b , μ ( u ˜ ) < 0 and u ˜ is an interior local minimizer of J b , μ on the set

A R 0 ( c ) = { u S c : u 2 < R 0 } ,

for a suitable R 0 . Moreover, any ground-state of J b , μ S c is a local minimizer of J b , μ on A R 0 ;

(ii) There exists a number b * = b * ( μ , c ) > 0 such that if b ( 0 , b * ) , then (1.1) also has a positive normalized solution of mountain pass type u ˆ on S c with J b , μ ( u ˆ ) > 0 .

Remark 1.1

There are a few points to say about Theorem 1.1.

  1. In the proof of Theorem 1.1, we use the idea of [29] to constrain the functional J b , μ on the Pohozaev set P b , μ and know that J b , μ P b , μ is bounded from below. By decomposing P b , μ manifold and constructing fiber map, then we can obtain a local minimizer u ˜ for J b , μ P b , μ and construct a minimax characterization for J b , μ to obtain the second critical point u ˆ .

  2. Compared with the case 1 N 3 , the mass-energy doubly critical Kirchhoff problem will lead to competition between the nonlocal term and Sobolev critical term in R 4 , which makes the problem more difficult. The compactness analysis and energy estimates involving mass-energy doubly critical exponents are very technical, which are distinct from [29]. To obtain the second solution, we need to assume that b > 0 is suitably small and establish the relationship with the energy of the radial ground-state u * (see Remark 3.2) when b = 0 in (1.1)–(1.2) inspired by [35]. A crucial step of our proof is to give a upper energy estimate of the minimax value m * (see Lemma 3.7) that differs from [16,29].

  3. Theorem 1.1 improves and extends the results of [16]. When 2 < q < 3 and 0 < b < S 2 , Kong and Chen [16] proved the existence of a local minimizer of (1.1)–(1.2) under appropriate assumptions on μ and c . However, not only do we obtain a local minimizer u ˜ for J b , μ P b , μ , but we also obtain the second mountain pass solution u ˆ with J b , μ ( u ˆ ) > 0 by limiting b to a smaller range.

  4. To overcome the compactness, we need to deal with (1.1)–(1.2) in the radial space H r 1 ( R 4 ) , since the embedding H r 1 ( R 4 ) L p ( R 4 ) is compact for p ( 2 , 4 ) . Moreover, we emphasis that (1.12) can be used to ensure that P b , μ is a smooth manifold.

Theorem 1.2

Assume that q = 3 and 0 < b < S ¯ . Moreover, let us suppose that μ , c > 0 satisfy

(1.13) 0 < μ < 3 a 2 C q q c .

Then (1.1) has a positive normalized ground-state solution of mountain pass type on S c for some λ > 0 .

Remark 1.2

For q = 3 , this case has the following characteristics:

  1. The case of Theorem 1.2 is even more special due to the simultaneous occurrence of three critical exponents: 2 + 4 N , 2 + 8 N and 2 * , which becomes more interesting and complicated. In R 4 , double mass-critical exponents and one energy critical exponent occur at the same time, which is a novel phenomenon that has never appeared in previous articles for Kirchhoff equation.

  2. The corresponding fiber map Ψ u ( s ) has four different terms, where exists two different competitions. The first competition is that the nonlocal term e 4 s u 2 4 and energy critical term e 4 s u 4 4 are in competition, and the second competition is that e 2 s u 2 2 and e 2 s u q q are in competition, which will strongly affect the structure of Ψ u ( s ) .

  3. In the previous articles, we usually construct a Palais-Smale sequence of J b , μ satisfying the Pohozaev identity in order to obtain the boundedness of the Palais-Smale sequence, which has been introduced by Jeanjean [15]. This is the key ingredient to prove strong convergence. However, in R 4 , satisfying the Pohozaev identity is not enough to prove the boundedness of the Palais-Smale sequence, and we need to add additional conditions to μ , c , which is due to the special phenomenon of three critical exponents appearing simultaneously.

Theorem 1.3

Let 3 < q < 4 and for any μ , c > 0 , there exists a number b * = b * ( μ , c ) > 0 such that if

(1.14) 0 < b < b * ,

then (1.1) has a positive normalized ground-state solution of mountain pass type on S c for some λ > 0 .

Remark 1.3

For the case 3 < q < 4 , this is also a very special situation.

  1. Theorem 1.3 means that q belongs to 2 + 4 N < q < 2 + 8 N = 2 * ; at this point, the energy functional has a mountain-pass geometry on S c since mass-energy doubly critical appears. In R 4 , the geometry of J b , μ S c is completely different from the structure of N = 3 , which has been proved in [10] for the Kirchhoff equation and in [30] for the Schrödinger equation. Moreover, let us point out that any μ > 0 and c > 0 are admissible while the condition on μ , c > 0 in [10] is more complicated than us. We just need b to be small enough to obtain the geometry of J b , μ S c .

  2. We still look for the critical point on the Pohozaev set P b , μ , so the solution we found is the ground-state solution. The most crucial step of the proof is the ground-state energy m is smaller than a 2 S 2 4 ( 1 b S 2 ) , which aims to recover compactness. Besides, we emphasis that (1.14) can ensure the structure of J b , μ S c and can also be used to prove the boundedness of the Palais-Smale sequence.

Next, we consider the defocusing perturbation case, i.e., μ < 0 , we have:

Theorem 1.4

Let μ < 0 and 2 < q < 4 .

  1. If u is a critical point for J b , μ S c (not necessarily positive, or even real-valued), then the associated Lagrange multiplier λ < 0 and

    J b , μ ( u ) a 2 S 2 4 ( 1 b S 2 ) .

  2. The problem

    a + b R 4 u 2 d x Δ u + λ u = μ u q 2 u + u 2 u , in R 4 ,

    with u > 0 , has no solution u H 1 ( R 4 ) , for any λ < 0 and μ < 0 .

Now, we study the asymptotic behavior of the normalized ground-states as μ 0 + and b 0 + , respectively.

Theorem 1.5

Assume that 2 < q < 4 , and for a fixed c > 0 . Let u c , b , μ S c be the positive ground-state solution of (1.1) with energy level m ( c , b , μ ) .

  1. If 2 < q < 3 , for b ( 0 , S ¯ ) fixed, let μ > 0 satisfy (1.12), then

    m ( c , b , μ ) 0 and u c , b , μ 2 2 0 , as μ 0 + .

  2. If q = 3 , for b ( 0 , S ¯ ) fixed, let μ > 0 satisfy (1.13) and if 3 < q < 4 , let b > 0 satisfy (1.14), then u c , b , μ 0 in H r 1 ( R 4 ) ,

    m ( c , b , μ ) Λ and u c , b , μ 2 2 a S 2 1 b S 2 , as μ 0 + .

Theorem 1.6

Assume that 2 < q < 4 . Let μ , c > 0 satisfy

(1.15) μ c ( 1 γ q ) q < min C , a 2 γ q q γ q q 2 ( a S ) 2 2 ( 2 γ q q ) 2 γ q q 2 4 q C q q ( 4 γ q q )

if 2 < q < 3 and satisfy (1.13) if q = 3 , and any μ , c > 0 are admissible if 3 < q < 4 . Let u c , b , μ S c be the positive ground-state solution of (1.1) with energy level m ( c , b , μ ) . Then, up to a subsequence, we have u c , b , μ u in H r 1 ( R 4 ) as b 0 + , where u S c is a positive ground-state solution of the equation

a Δ u + λ u = μ u q 2 u + u 2 u , in R 4 ,

for some λ > 0 .

Finally, we discuss the asymptotic behavior of the normalized ground-states as b 0 + and μ 0 + at the same time.

Theorem 1.7

Assume that q ( 2 , 4 ) , c > 0 , b n 0 + and μ n 0 + as n . Let u n u c , b n , μ n S c be the positive ground-state solution of (1.1) with energy level m ( c , b n , μ n ) for n large.

  1. If 2 < q < 3 , then

    m ( c , b n , μ n ) 0 and u n 2 2 0 , as n .

  2. If 3 q < 4 , then u n 0 in H r 1 ( R 4 ) ,

    m ( c , b n , μ n ) ( a S ) 2 4 and u n 2 2 a S 2 , as n .

We introduce some notations that will clarify what follows:

  • H r 1 ( R 4 ) = { u H 1 ( R 4 ) : u ( x ) = u ( x ) } .

  • D 1 , 2 ( R 4 ) is the completion of C 0 ( R 4 ) with respect to u D 1 , 2 = u 2 .

  • L p ( R 4 ) with p [ 1 , ) is the Lebesgue space with the norm u p = R 4 u p d x 1 p .

  • The arrows ‘ ’ and ‘ ’ denote the weak convergence and strong convergence, respectively.

  • C , C i denote positive constants that may vary from line to line.

  • S c , r = { u S c : u ( x ) = u ( x ) } .

  • o ( 1 ) means a quantity that tends to 0.

2 Preliminaries

In this section, we show several results that will be often used throughout the rest of this article. We start by introducing the famous Sobolev inequality and Gagliardo-Nirenberg inequality. For N = 4 , there exists an optimal constant S > 0 such that

(2.1) S inf u D 1 , 2 ( R 4 ) \ { 0 } u 2 2 u 4 2 .

It is well known [32] that the optimal constant is achieved by

U ε , y ( x ) = 8 ε ε 2 + x y 2 1 2 , ε > 0 , y R 4 ,

which are the only positive classical solutions to the critical Lane-Emden equation

Δ w = w 2 w , w > 0 in R 4 .

By (2.1), we have u 4 4 S 2 u 2 4 , then 0 u 4 4 u 2 4 S 2 < + . Thus, we denote

(2.2) S ¯ inf u D 1 , 2 ( R 4 ) \ { 0 } u 4 4 u 2 4 .

It is easy to find that S ¯ S 2 . If q ( 2 , 2 * ) , we also recall Gagliardo-Nirenberg inequality [33] that there exists an optimal constant C q depending on q such that

(2.3) u q C q u 2 γ q u 2 1 γ q .

Let η C 0 ( R 4 ) be a radial cut-off function satisfying

0 η 1 , η = 1 , on B 1 ( 0 ) ; η = 0 , on R 4 \ B 2 ( 0 ) .

Define u ε = η U ε , 0 and

(2.4) v ε = c u ε u ε 2 = a ε u ε .

The following estimates can be deduced by standard arguments (see [9,13]): as ε 0 + ,

(2.5) u ε 2 2 = S 2 + O ( ε 2 ) , u ε 6 6 = S 2 + O ( ε 4 )

and

(2.6) u ε s s = O ( ε ( 4 s ) 2 ) , if s > 2 , O ( ε ln ε ) , if s = 2 , O ( ε s 2 ) , if s < 2 ,

where 2 s < 4 .

Lemma 2.1

Let f ε : R + R be defined by

f ε ( t ) = a t 2 2 v ε 2 2 + b t 4 4 v ε 2 4 t 4 4 v ε 4 4 .

Then, as ε 0 + , we have

sup t > 0 f ε ( t ) Λ + O ( ε 1 2 ) ,

where Λ a 2 S 2 4 ( 1 b S 2 ) .

Lemma 2.2

Suppose that 2 < q < 4 . Let μ , c > 0 satisfy (1.13) if q = 3 , and let b > 0 satisfy (1.14) if 3 < q < 4 . If { u n } S c is a Palais-Smale sequence for J b , μ S c at energy level m 0 with the property

P b , μ ( u n ) 0 , as n ,

then { u n } is bounded in H 1 ( R 4 ) and { λ n } is bounded in R .

Proof

If 2 < q < 3 , we have 0 < q γ q < 2 . It follows from (2.3) and

P b , μ ( u n ) = a u n 2 2 + b u n 2 4 μ γ q u n q q u n 4 4 0 , as n ,

we have

(2.7) m + 1 J b , μ ( u n ) = J b , μ ( u n ) 1 4 P b , μ ( u n ) = a 4 u n 2 2 μ q 1 q γ q 4 u n q q + o n ( 1 ) a 4 u n 2 2 μ q 1 q γ q 4 C q q u n 2 q γ q c q ( 1 γ q ) + o n ( 1 ) ,

for n large, from which we see that { u n } is bounded in H 1 ( R 4 ) .

If q = 3 , we have q γ q = 2 , then (2.7) becomes

(2.8) m + 1 J b , μ ( u n ) 1 2 a 2 μ 3 C q q c u n 2 2 ,

for n large and (1.13); thus, { u n } is bounded in H 1 ( R 4 ) .

If 3 < q < 4 , we have 2 < q γ q < 4 , and

(2.9) m + 1 J b , μ ( u n ) = J b , μ ( u n ) 1 γ q q P b , μ ( u n ) a 1 2 1 γ q q u n 2 2 b 1 γ q q 1 4 u n 4 4 S ¯ + 1 γ q q 1 4 u n 4 4 = a 1 2 1 γ q q u n 2 2 + 1 b S ¯ 1 γ q q 1 4 u n 4 4 a 1 2 1 γ q q u n 2 2 ,

for n large and (1.14); thus, { u n } is bounded in H 1 ( R 4 ) .

Since { u n } is a bounded Palais-Smale sequence of J b , μ S c , by the Lagrange multiplier rule, there exists λ n R such that

(2.10) ( a + b u n 2 2 ) R 4 u n φ d x μ R 4 u n q 2 u n φ d x R 4 u n 2 u n φ d x + λ n R 4 u n φ d x = o n ( 1 ) ,

for every φ H 1 ( R 4 ) . In particular, choosing φ = u n , then we have

(2.11) λ n c 2 = a u n 2 2 + b u n 2 4 μ u n q q u n 4 4 + o n ( 1 ) ,

which implies that { λ n } is bounded due to the boundedness of { u n } in H 1 ( R 4 ) .□

Proposition 2.1

Suppose that 2 < q < 4 . Let μ , c > 0 satisfy (1.13) if q = 3 , and let b > 0 satisfy (1.14) if 3 < q < 4 . If { u n } S c , r is a Palais-Smale sequence for J b , μ S c at energy level M 0 , with

M < Λ and P b , μ ( u n ) 0 , as n ,

then, up to a subsequence, one of the following alternatives holds:

(i) either u n u 0 in H r 1 ( R 4 ) , where u solves

(2.12) ( a + B b ) Δ u + λ u = u 2 u + μ u q 2 u , in R 4 ,

for some λ > 0 , and M Λ I μ ( u ) a 2 + B b 4 u 2 2 μ q u q q 1 4 u 4 4 for B lim n u n 2 2 > 0 .

(ii) or u n u in H r 1 ( R 4 ) for some u H r 1 ( R 4 ) , which implies that u S c , J b , μ ( u ) = M , and u solves (1.1) for some λ > 0 .

Proof

The proof is divided into two main steps.

Step 1: λ > 0 and u 0 . By Lemma 2.2, we know the sequences { u n } and { λ n } are bounded. Hence, there exist u H r 1 ( R 4 ) and λ R such that, up to a subsequence, u n u in H r 1 ( R 4 ) , u n u in L q ( R 4 ) and u n u almost everywhere in R 4 and up to a subsequence, λ n λ as n . Using again, P b , μ ( u n ) 0 , as n , we have

λ c 2 = lim n λ n c 2 = lim n μ ( 1 γ q ) u n q q = μ ( 1 γ q ) u q q .

Since μ > 0 and 0 < γ q < 1 , we deduce that λ 0 , with equality if and only if u 0 . If λ n 0 , we have u n 0 in L q ( R 4 ) . By (2.11), then we obtain

lim n ( a u n 2 2 + b u n 2 4 ) = lim n u n 4 4 = R .

It follows that lim n u n 2 2 = b + a 2 4 b 2 a 2 b , and by (2.1), there holds b S 2 + a S 1 2 . Note that

(2.13) M = lim n J b , μ ( u n ) = lim n a 2 u n 2 2 + b 4 u n 2 4 1 4 u n 4 4 = a 4 lim n u n 2 2 = a 4 b + a 2 4 b 2 a 2 8 b

due to P b , μ ( u n ) 0 , as n . Since M 0 , we have 0 and

(2.14) a S 1 b S 2 2 .

Substituting (2.14) into (2.13), we have M = lim n J b , μ ( u n ) Λ , which contradicts with our assumptions M < Λ . Hence, we have λ > 0 and u 0 .

Step 2: One of the alternatives ( i ) and ( i i ) holds. From u n u 0 in H r 1 ( R 4 ) , we have B lim n u n 2 2 u 2 2 > 0 . Then, passing to the limit in (2.10), we have

(2.15) ( a + B b ) R 4 u φ d x μ R 4 u q 2 u φ d x R 4 u 2 u φ d x + λ R 4 u φ d x = 0 ,

for every φ H 1 ( R 4 ) , which means that u satisfies

( a + B b ) Δ u + λ u = u 2 u + μ u q 2 u .

Then, we can obtain the corresponding Pohozaev identity

Q b , μ ( u ) ( a + B b ) u 2 2 μ γ q u q q u 4 4 = 0 .

Letting v n = u n u , this means v n 0 in H r 1 ( R 4 ) , v n 0 in L q ( R 4 ) , v n 0 a.e. in R 4 , and

u n 2 2 = u 2 2 + v n 2 2 + o n ( 1 ) .

By the Brézis-Lieb lemma in [3], we have

u n 4 4 = u 4 4 + v n 4 4 + o n ( 1 ) , u n q q = u q q + o n ( 1 ) .

Rewrite P b , μ ( u n ) = o n ( 1 ) as

P b , μ ( u n ) = ( a + B b ) u n 2 2 μ γ q u q q u n 4 4 + o n ( 1 ) .

From Q b , μ ( u ) = 0 , we have that up to a subsequence

(2.16) = lim n v n 4 4 = lim n ( a + B b ) v n 2 2 lim n ( a v n 2 2 + b v n 2 4 ) .

The Sobolev inequality (2.1) implies that

a S 1 2 + b S 2 and lim n ( a v n 2 2 + b v n 2 4 ) lim n v n 4 4 1 S 2 lim n v n 2 4 .

We deduce that

either a S 1 b S 2 2 and lim n v n 2 2 a S 2 1 b S 2 , or = 0 = lim n v n 2 2 .

Let us suppose at first that a S 1 b S 2 2 and lim n v n 2 2 a S 2 1 b S 2 . From (2.16), we deduce that

M = lim n J b , μ ( u n ) = I μ ( u ) + lim n a 2 v n 2 2 + B b 4 v n 2 2 v n 4 4 4 = I μ ( u ) + lim n a 4 v n 2 2 I μ ( u ) + Λ ,

where I μ ( u ) ( a 2 + B b 4 ) u 2 2 1 4 u 4 4 μ q u q q . Hence, alternative ( i ) of the proposition follows. If instead = 0 = lim n v n 2 2 , we have u n u in D 1 , 2 ( R 4 ) and L 4 ( R 4 ) . We test (2.10) and (2.15) with φ = u n u in order to prove that u n u in L 2 ( R 4 ) , and subtract, obtaining

( a + B b ) ( u n u ) 2 2 + λ u n u 2 2 0 , as n ,

which implies that alternative ( i i ) of the proposition follows. The proof is complete.□

3 Proof of Theorem 1.1

We first consider the existence of the local minimizer for J b , μ S c . For the L 2 -subcritical perturbation case: 2 < q < 3 , there holds 0 < γ q q < 2 . By (2.1) and (2.3), we have

(3.1) J b , μ ( u ) = a 2 u 2 2 + b 4 u 2 4 1 4 u 4 4 μ q u q q a 2 u 2 2 S 2 4 u 2 4 μ q C q q c ( 1 γ q ) q u 2 γ q q ,

for every u S c . Therefore, we consider the function h : R + R

h ( t ) a 2 t 2 S 2 4 t 4 μ q C q q c ( 1 γ q ) q t γ q q , t > 0 .

Since μ > 0 and 0 < γ q q < 2 , we have that h ( 0 + ) = 0 and h ( + ) = . Moreover,

(3.2) Ψ u ( s ) = a e 2 s 2 u 2 2 + b e 4 s 4 u 2 4 e q γ q s q μ u q q e 4 s 4 u 4 4 a e 2 s 2 u 2 2 + b e 4 s 4 S ¯ u 4 4 e q γ q s q μ u q q e 4 s 4 u 4 4 = a e 2 s 2 u 2 2 + e 4 s 4 b S ¯ 1 u 4 4 e q γ q s q μ u q q .

Since 0 < b < S ¯ and 0 < γ q q < 2 , Ψ u ( + ) = and Ψ u ( ) = 0 . In order to study the properties of Ψ u , we need the following technical lemma.

Lemma 3.1

Under assumption (1.12), then the function h ( t ) has a strict local minimum with negative value and a strict global maximum with positive value. Moreover, there exist two positive numbers R 0 < R 1 depending on μ and c such that h ( R 0 ) = h ( R 1 ) = 0 and h ( t ) > 0 if and only if t ( R 0 , R 1 ) .

Proof

Define φ : R + R by

φ ( t ) φ c ( t ) = h ( t ) t q γ q = a 2 t 2 γ q q S 2 4 t 4 γ q q μ q C q q c ( 1 γ q ) q .

Direct computation shows that φ achieves its maximum at t * = 2 a S 2 ( 2 γ q q ) 4 γ q q 1 2 with

φ ( t * ) = 2 a S 2 ( 2 γ q q ) 4 γ q q 2 γ q q 2 a 4 γ q q μ q C q q c ( 1 γ q ) q > 0

due to assumption (1.12). Then, h ( t * ) = ( t * ) γ q q φ ( t * ) > 0 , which combined with h ( 0 + ) = 0 and h ( + ) = implies that h has at least two zero points 0 < R 0 < R 1 . Note that h ( t ) = 0 is equivalent to φ ( t ) = 0 , and clearly, φ has only one critical point. Therefore, h has exactly two zero points 0 < R 0 < R 1 .

As a direct consequence, h ( t ) > 0 if and only if t ( R 0 , R 1 ) , h achieves its global maximum with positive value in ( R 0 , R 1 ) and h achieves a local minimum with negative value in ( 0 , R 0 ) . Note also that h ( t ) = 0 is equivalent to

ψ ( t ) a t 2 γ q q S 2 t 4 γ q q μ γ q C q q c ( 1 γ q ) q = 0 .

Clearly, ψ ( t ) has only one critical point, which is a strict maximum, and hence, the aforementioned equation has at most two solutions, which necessarily are the local minimum and the global maximum of h previously found.□

Using again (1.12), we can also prove as in [29, Lemma 5.2] that P b , μ 0 = , and P is a smooth manifold of codimension 2 in H 1 ( R 4 ) . Recalling the decomposition P b , μ = P b , μ + P b , μ 0 P b , μ , then manifold P b , μ is divided into two components P b , μ + and P b , μ .

Lemma 3.2

Let a , c > 0 , 2 < q < 3 , μ c ( 1 γ q ) q < C , and 0 < b < S ¯ . For every u S c , the function Ψ u has exactly two critical points s u < t u R and two zeros c u < d u R , with s u < c u < t u < d u . Moreover,

  1. s u u P b , μ + and t u u P b , μ , and if s u P b , μ , then either s = s u or s = t u ;

  2. ( s u ) 2 R 0 for every s c u , and

    J b , μ ( s u u ) = min { J b , μ ( s u ) : s R and ( s u ) 2 < R 0 } < 0 ;

  3. there holds

    J b , μ ( t u u ) = max s R J b , μ ( s u ) > 0

    and Ψ u ( s ) is strictly decreasing and concave on ( t u , + ) ; hence, if t u < 0 , then P b , μ ( u ) < 0 ;

  4. the maps u S c s u R and u S c t u R are of class C 1 .

Again, the proof is completely analogous to the one in [29, Lemma 5.3].

Now, for k > 0 , we set

A k { u S c : u 2 < k } , and m ( c ) inf u A R 0 J b , μ ( u ) .

From Lemma 3.2, it follows that P + A R 0 and sup P + J b , μ 0 inf P J b , μ . Moreover, as in [29, Lemma 5.5], we have that m ( c ) ( , 0 ) ,

(3.3) m ( c ) = inf P J b , μ = inf P + J b , μ and m ( c ) < inf A R 0 \ A R 0 ρ J b , μ ,

for ρ > 0 small enough.

Recall that in the proof of Lemma 3.1, we define

φ c ( t ) = a 2 t 2 γ q q S 2 4 t 4 γ q q μ q C q q c ( 1 γ q ) q .

Next, we show the monotonicity of φ c ( t ) in order to obtain the sub-additivity of m ( c ) .

Lemma 3.3

Let c 1 > 0 and t 1 > 0 be such that φ c 1 ( t 1 ) 0 . If c 2 ( 0 , c 1 ] , then

φ c 2 ( t ) 0 , for t c 2 c 1 t 1 , t 1 .

Proof

It is easy to show that φ c 2 ( t 1 ) φ c 1 ( t 1 ) 0 and

φ c 2 c 2 c 1 t 1 c 2 c 1 2 γ q q and φ c 1 ( t 1 ) 0 .

We assume by contradiction that φ c 2 ( t ) < 0 for some t ( c 2 c 1 t 1 , t 1 ) , which implies φ c 2 has a local minimum point in ( c 2 c 1 t 1 , t 1 ) . By Lemma 3.1, we would find a contradiction since the function φ c 2 has a unique critical point at t * , which φ c 2 achieves its global maximum. Thus, the proof is complete.□

By the assumption μ c ( 1 γ q ) q < C and for μ > 0 , we denote

c < c * c * ( μ ) = C μ 1 ( 1 γ q ) q .

Lemma 3.4

  1. The map c ( 0 , c * ) m ( c ) R is continuous.

  2. If c ( 0 , c * ) , then

    m ( c ) m ( c 0 ) + m ( c 2 c 0 2 ) , for c 0 ( 0 , c ) .

    Moreover, if m ( c 0 ) or m ( c 2 c 0 2 ) is achieved, then the aforementioned inequality is strict.

Proof

( i ) Let { c n } ( 0 , c * ) be such that lim n c n = c ( 0 , c * ) . For ε > 0 small, there exists u A t * ( c ) such that J b , μ ( u ) m ( c ) + ε < m ( c ) + 3 ε < 0 . Setting u n = c n c u , we have u n 2 = c n and u n 2 < t * for n large, which means that u n A t * ( c n ) for n large. Then, since J b , μ ( u n ) J ( u ) as n , we have

(3.4) m ( c n ) J b , μ ( u n ) = J b , μ ( u ) + ε m ( c ) + 2 ε ,

for n large. For above ε > 0 , we see from (3.4) that there exists u n A t * ( c n ) such that

J b , μ ( u n ) m ( c n ) + ε m ( c ) + 3 ε < 0 ,

for n large. Moreover, setting v n = c c n u n , it follows immediately that v n 2 = c . We wish to obtain

(3.5) v n 2 < t * ,

and hence, v n A t * ( c ) . If this is valid, there holds

(3.6) m ( c ) J b , μ ( v n ) J b , μ ( u n ) + ε m ( c n ) + 2 ε ,

for n large. Combining (3.4) and (3.6), we conclude that m ( c n ) m ( c ) as n . It remains to prove (3.5). In the case c n c , we can infer that

v n 2 = c c n u n 2 u n 2 < t * .

When c n < c , it follows from Lemma 3.3 and φ c ( t * ) > 0 that

φ c n ( t ) 0 , for t c n c t * , t * .

Consequently, on the basis of J b , μ ( u n ) < 0 and u n A t * ( c n ) , we have u n 2 < c n c t * , and hence,

v n 2 = c c n u n 2 < t * .

( i i ) Letting c ( 0 , c * ) and c 0 ( 0 , c ) , we first prove that

(3.7) m ( θ c 0 ) θ 2 m ( c 0 ) , for θ 1 , c c 0 ,

and if, in addition, m ( c 0 ) is achieved, then the inequality is strict. Indeed, for ε > 0 small, there exists u A t * ( c 0 ) such that

(3.8) J b , μ ( u ) m ( c 0 ) + ε < 0 .

Since φ c ( t * ) 0 , we see from Lemma 3.3 that φ c 0 ( t ) 0 for t c 0 c t * , t * . Since J b , μ ( u ) < 0 and u A t * ( c 0 ) , there must be u 2 < c 0 c t * . Setting v ( x ) = u ( θ 1 2 x ) , we have v 2 = θ u 2 = θ c 0 and v 2 = θ 1 2 u 2 < t * , which means that v A t * ( θ c 0 ) . Then, by (3.8),

m ( θ c 0 ) J b , μ ( v ) = a θ 2 u 2 2 + b θ 2 4 u 2 4 θ 2 4 u 4 4 μ q θ 2 u q q < θ 2 J b , μ ( u ) θ 2 ( m ( c 0 ) + ε ) .

Since ε > 0 is arbitrary, we have m ( θ c 0 ) θ 2 m ( c 0 ) . If m ( c 0 ) is achieved, then one can choose ε = 0 in (3.8), and hence, the strict inequality holds. Now, we use (3.7) to deduce that

m ( c ) = c 0 2 c 2 m c c 0 c 0 + c 2 c 0 2 c 2 m c c 2 c 0 2 c 2 c 0 2 m ( c 0 ) + m ( c 2 c 0 2 ) ,

with a strict inequality if m ( c 0 ) or m ( c 2 c 0 2 ) is achieved.□

Remark 3.1

As a consequence of Lemma 3.4 (ii) combined with m ( c ) < 0 , we can see c ( 0 , c * ) m ( c ) R is decreasing.

Proof of Theorem 1.1(i) for the local minimizer

Let { v n } S c , r be a minimizing sequence for inf A R 0 J b , μ . Moreover, for every n , we can take s v n v n P + , observing that then ( s v n v n ) 2 < R 0 by Lemma 3.2, and

J b , μ ( s v n v n ) = min { J b , μ ( s v n ) : s R and ( s v n ) 2 < R 0 } J b , μ ( v n ) ,

in this way, we obtain a new minimizing sequence { w n = s v n v n } , with w n S c , r P + radially decreasing for every n . By (3.3), w n 2 < R 0 ρ for every n , and hence, Ekeland’s variational principle yields in a standard way the existence of a new minimizing sequence { u n } A R 0 for m ( c ) , with the property that u n w n 0 as n , which is also a Palais-Smale sequence for J b , μ on S c , r . The condition u n w n 0 , together with the boundedness of { w n } (each w n stays in A R 0 ), implies that P ( u n ) = P ( w n ) + o ( 1 ) 0 as n .

By Lemma 2.2, { u n } is bounded in H r 1 ( R 4 ) . Therefore, there exists u H r 1 ( R 4 ) such that, up to a subsequence, u n u in H r 1 ( R 4 ) for some u 0 . Setting u ˜ n = u n u , we have

(3.9) u ˜ n 2 2 = u n 2 2 u 2 2 + o ( 1 ) , u ˜ n 2 2 = u n 2 2 u 2 2 + o ( 1 ) = c 2 u 2 2 + o ( 1 )

and

(3.10) m ( c ) = J b , μ ( u n ) + o ( 1 ) = J b , μ ( u ˜ n ) + J b , μ ( u ) + b 2 u ˜ n 2 2 u 2 2 + o ( 1 ) .

We shall prove that u ˜ n 0 in H r 1 ( R 4 ) by the following two steps.

Step 1: u ˜ n 2 0 as n , i.e., u 2 = c . Suppose by contradiction that u 2 c 0 < c . By (3.9), we have u ˜ n 2 c and u ˜ n 2 u n 2 < t * for n large, which means u ˜ n A t * ( u ˜ n 2 ) , and thus, J b , μ ( u ˜ n ) m ( u ˜ n 2 ) . Combining this with (3.10) yields

m ( c ) = J b , μ ( u ˜ n ) + J b , μ ( u ) + b 2 u ˜ n 2 2 u 2 2 + o ( 1 ) m ( u ˜ n 2 ) + J b , μ ( u ) + o ( 1 ) .

By Lemma 3.4(i),

(3.11) m ( c ) m ( c 2 c 0 2 ) + J b , μ ( u ) .

We also have u A t * ( c 0 ) , which implies J b , μ ( u ) m ( c 0 ) . If J b , μ ( u ) > m ( c 0 ) , then it follows from (3.11) and Lemma 3.4(ii) that

m ( c ) > m ( c 2 c 0 2 ) + m ( c 0 ) m ( c ) ,

which is absurd. If J b , μ ( u ) = m ( c 0 ) , then the strict inequality in Lemma 3.4(ii) holds and

m ( c ) m ( c 2 c 0 2 ) + J b , μ ( u ) = m ( c 2 c 0 2 ) + m ( c 0 ) > m ( c ) ,

which is also a contradiction. Therefore, we have u 2 = c .

Step 2: u ˜ n 2 0 as n . Note that u ˜ n q 0 as n and

u ˜ n 2 < t * = 2 a S 2 ( 2 γ q q ) 4 γ q q 1 2 ,

for n large. Then,

J b , μ ( u ˜ n ) = a 2 u ˜ n 2 2 + b 4 u ˜ n 2 4 1 4 u ˜ n 4 4 μ q u ˜ n q q a 2 u ˜ n 2 2 S 2 4 u ˜ n 2 4 + o ( 1 ) = u ˜ n 2 2 a 2 S 2 4 u ˜ n 2 2 + o ( 1 ) a 4 γ q q u ˜ n 2 2 + o ( 1 ) ,

which combined with (3.10) and J b , μ m ( c ) since u A t * ( c ) implies that u ˜ n 2 0 as n .

Thus, we have u n u in H r 1 ( R 4 ) for some u , which implies J b , μ ( u ) = m ( c ) and u solves (1.1)–(1.2) for some λ > 0 . Moreover, we see from (3.3) that u is an interior local minimizer of J b , μ on the set A R 0 ( c ) and, by u n 2 0 as n and the maximum principle, u is a positive ground-state solution. The proof is complete.□

Next, we focus now on the existence of the second critical point for J b , μ S c . To construct a minimax structure, we give some lemmas.

Lemma 3.5

For u S c with J b , μ ( u ) < m ( c ) , the value t u given in Lemma 3.2is negative.

The first part of Theorem 1.1 states that there exists u P + such that J b , μ ( u ) = m ( c ) < 0 . It is also clear that J b , μ ( t u ) < 2 m ( c ) for t > 0 sufficiently large. This implies the set

Γ Γ ( c , b , μ ) = { γ C ( [ 0 , 1 ] , S c , r ) : γ ( 0 ) P + , J b , μ ( γ ( 1 ) ) 2 m ( c ) }

is not empty. Now, we consider the minimax value

m * m * ( c , b , μ ) = inf γ Γ sup t [ 0 , 1 ] J b , μ ( γ ( t ) ) .

Lemma 3.6

There holds m * = inf P J b , μ ( u ) = inf u S c max s R J b , μ ( s u ) > 0 .

The proof can argue in the same way as that of [10, Lemmas 4.9 and 4.10] and hence omit the details.

As a direct conclusion of Lemma 3.6, there holds the function b ( 0 , b * ) m * ( c , b , μ ) is monotone non-decreasing.

Remark 3.2

Using the same arguments as the first part (i) of Theorem 1.1, we can also conclude that (1.1) with b = 0 has a positive ground-state solution u * S c for some λ = λ * > 0 . Therefore,

J 0 , μ ( u * ) = m ( c , 0 , μ ) inf u A R 0 ( c ) J 0 , μ ( u ) < 0 ,

where

J 0 , μ ( u ) = a 2 R 4 u 2 d x 1 4 R 4 u 4 d x μ q R 4 u q d x .

Lemma 3.7

There exists a number b * = b * ( μ , c ) > 0 such that if b ( 0 , b * ) , then

m * < m ( c , 0 , μ ) + ( a S ) 2 4 .

Proof

Let u ε be as in Section 2. Choose ε = b and denote

v b , τ ( x ) = u * ( x ) + τ u b ( x ) and w b , τ ( x ) = s v b , τ ( s x ) ,

where τ 0 . Then,

(3.12) w b , τ 2 2 = v b , τ 2 2 and w b , τ 4 4 = v b , τ 4 4

and

(3.13) w b , τ 2 2 = s 2 v b , τ 2 2 , w b , τ q q = s γ q q q v b , τ q q .

Choosing s = v b , τ 2 c , we have w b , τ S c , and then by Lemma 3.2, there exists t b , τ such that t b , τ w b , τ P . Since u * P c , 0 , μ + , we have t b , 0 > 0 . It follows from (2.5) and

a e 2 t b , τ w b , τ 2 2 + b e 4 t b , τ w b , τ 2 4 = e 4 t b , τ w b , τ 4 4 + μ γ q e γ q q t b , τ w b , τ q q

that t b , τ as τ + uniformly for b > 0 small. Note that t b , τ is continuous with respect to τ . Then, for b > 0 small, there exists τ b > 0 such that t b , τ b = 0 , which means w b , τ b P . Hence, by Lemma 3.6,

(3.14) m * J b , μ ( w b , τ b ) sup τ 0 J b , μ ( w b , τ ) .

In view of (3.12) and (3.13), we rewrite J b , μ ( w b , τ ) as

(3.15) J b , μ ( w b , τ ) = a 2 v b , τ 2 2 + b 4 v b , τ 2 4 1 4 v b , τ 4 4 μ q s γ q q q v b , τ q q = a 2 u * 2 2 + a τ R 4 u * u b d x + a τ 2 2 u b 2 2 + b 4 u * 2 4 + b τ 2 R 4 u * u b d x 2 + b τ 4 4 u b 2 4 + b τ u * 2 2 R 4 u * u b d x + b τ 2 2 u * 2 2 u b 2 2 + b τ 3 u b 2 2 R 4 u * u b d x 1 4 u * + τ u b 4 4 μ q s γ q q q u * + τ u b q q .

Note that

s 2 = v b , τ 2 2 c 2 = 1 + 2 τ c 2 R 4 u * u b d x + τ 2 c 2 u b 2 2 .

Then, there exists τ 0 > 0 such that

(3.16) J b , μ ( w b , τ ) < m ( c , 0 , μ ) + ( a S ) 2 4 , for b > 0 small and τ ( 0 , τ 0 ) .

We also see from (3.15) that

J b , μ ( w b , τ ) J 0 , μ ( u * ) + a τ 2 2 u b 2 2 + b τ 4 4 u b 2 4 τ 4 4 u b 4 4 + b 4 u * 2 4 + a τ R 4 u * u b + b τ u * 2 2 R 4 u * u b d x + b τ 2 R 4 u * u b d x 2 + b τ 2 2 u * 2 2 u b 2 2 + b τ 3 u b 2 2 R 4 u * u b d x 1 4 R 4 ( u * + τ u b 4 u * 4 ( τ u b ) 4 ) d x + μ q u * q q J 0 , μ ( u * ) + a τ 2 2 u b 2 2 + b τ 4 4 u b 2 4 τ 4 4 u b 4 4 + b 4 u * 2 4 + a τ R 4 u * u b d x + b τ u * 2 2 R 4 u * u b d x + b τ 2 R 4 u * u b d x 2 + b τ 2 2 u * 2 2 u b 2 2 + b τ 3 u b 2 2 R 4 u * u b d x + μ q u * q q .

Hence, by (2.5), there exists τ 1 > 0 such that

(3.17) J b , μ ( w b , τ ) < m ( c , 0 , μ ) + ( a S ) 2 4 , for b > 0 small and τ ( τ 1 , + ) .

Finally, we consider τ 0 τ τ 1 . Since u * S c is a positive ground-state solution of (1.1) with b = 0 for some λ = λ * > 0 , there holds

a R 4 u * u b d x + λ * R 4 u * u b d x = R 4 u * 3 u b d x + μ R 4 u * q 1 u b d x

and

λ * c 2 = λ * u * 2 2 = μ ( γ q 1 ) u * q q .

Then, by (3.15), (2.5), and (2.6), we have

(3.18) J b , μ ( w b , τ ) J 0 , μ ( u * ) + a τ 2 2 u b 2 2 + b τ 4 4 u b 2 4 τ 4 4 u b 4 4 + b 4 u * 2 4 + b τ 2 R 4 u * u b d x 2 + b τ 2 2 u * 2 2 u b 2 2 + b τ 3 u b 2 2 R 4 u * u b d x + b τ u * 2 2 R 4 u * u b d x 1 4 R 4 ( u * + τ u b 4 u * 4 ( τ u b ) 4 4 τ u * 3 u b ) d x μ q R 4 ( u * + τ u b q u * q q τ u * q 1 u b ) d x + τ λ * R 4 u * u b d x τ μ c 2 ( γ q 1 ) u * + τ u b q q R 4 u * u b d x τ 2 μ 2 c 2 ( γ q 1 ) u * + τ u b q q u b 2 2 J 0 , μ ( u * ) + ( a S ) 2 4 + O ( b 1 2 ) τ 2 μ 2 c 2 ( γ q 1 ) u * + τ u b q q u b 2 2 τ μ c 2 ( γ q 1 ) ( u * + τ u b q q u * q q ) R 4 u * u b d x R 4 u * ( τ u b ) 5 d x m ( c , 0 , μ ) + ( a S ) 2 4 + o ( b 1 4 ) C b 1 4 < m ( c , 0 , μ ) + ( a S ) 2 4 ,

for b > 0 small, where we used the inequality ( r + s ) 4 r 4 s 4 4 r 3 s 4 r s 3 for r , s 0 . Therefore, combining (3.14) with (3.16), (3.17), and (3.18), we conclude that there exists b * = b * ( μ , c ) > 0 such that

m * sup τ 0 J b , μ ( w b , τ ) < m ( c , 0 , μ ) + ( a S ) 2 4 , for b ( 0 , b * ) .

The proof is complete.□

Lemma 3.8

Suppose that 2 < q < 3 and μ c ( 1 γ q ) q < C . Then, there exists a Palais-Smale sequence { u n } S c , r for J b , μ S c at the level m * with the properties

u n 2 0 and P ( u n ) 0 , as n ,

where u = min { u , 0 } .

Proof

We introduce the augmented functional J ˜ b , μ : H r 1 ( R 4 ) × R R in the spirit of [15] as

J ˜ b , μ ( u , s ) = J b , μ ( s u ) = a e 2 s 2 u 2 2 + b e 4 s 4 u 2 4 e 4 s 4 u 4 4 μ e γ q q s q u q q .

Set

Γ ˜ = { γ ˜ C ( [ 0 , 1 ] , S c , r × R ) : γ ˜ ( 0 ) ( P + , 0 ) , γ ˜ ( 1 ) ( J b , μ 2 m ( c ) , 0 ) } .

It is easy to see that if γ Γ , then γ ˜ ( γ , 0 ) Γ ˜ and J ˜ b , μ ( γ ˜ ( t ) ) = J b , μ ( γ ( t ) ) for t [ 0 , 1 ] ; however, if γ ˜ = ( γ ˜ 1 , γ ˜ 2 ) Γ ˜ , then γ ( ) γ ˜ 2 ( ) γ ˜ 1 ( ) Γ and J b , μ ( γ ( t ) ) = J ˜ b , μ ( γ ˜ ( t ) ) for t [ 0 , 1 ] . Therefore, we have

m ˜ * inf γ ˜ Γ ˜ sup t [ 0 , 1 ] J ˜ b , μ ( γ ˜ ( t ) ) = m * .

By the definition of m * , for ε n = 1 n 2 , there exists γ n Γ such that

sup t [ 0 , 1 ] J b , μ ( γ n ( t ) ) m * + 1 n 2 .

Replacing γ n by γ n if necessary, we may assume that γ n ( t ) 0 in R 4 for all t [ 0 , 1 ] . Setting γ ˜ n = ( γ n , 0 ) Γ ˜ , we have

sup t [ 0 , 1 ] J ˜ b , μ ( γ ˜ n ( t ) ) m * + 1 n 2 .

Using Ekeland’s variational principle as in [15, Lemma 2.3] yields the existence of a sequence { ( v n , s n ) } S c , r × R such that, as n + ,

(3.19) s n + dist ( v n , γ n ( [ 0 , 1 ] ) ) 0 ,

(3.20) J ˜ b , μ ( v n , s n ) m * ,

(3.21) ( J ˜ b , μ S c , r × R ) ( v n , s n ) 0 .

Note that J ˜ b , μ ( v n , s n ) = J ˜ b , μ ( s n v n , 0 ) and

(3.22) ( J ˜ b , μ S c , r × R ) ( v n , s n ) , ( φ , s ) = ( J ˜ b , μ S c , r × R ) ( s n v n , 0 ) , ( s n φ , s ) ,

for ( φ , s ) H r 1 ( R 4 ) × R with R 4 v n φ = 0 . Setting u n = s n v n S c , r , we see from (3.20) that

J b , μ ( u n ) = J ˜ b , μ ( s n v n , 0 ) = J ˜ b , μ ( v n , s n ) m * , as n .

Taking ( 0 , 1 ) as a test function in (3.22), we deduce from (3.21) that

P b , μ ( u n ) = s J ˜ b , μ ( u n , 0 ) 0 , as n .

For w H r 1 ( R 4 ) with R 4 ( s n v n ) w = 0 , we take ( s n w , 0 ) as a test function in (3.22) and then deduce from (3.19) and (3.21) that ( J b , μ S c , r ) ( u n ) 0 as n . By (3.19) again, u n 2 2 = s n ( v n ) 2 2 = v n 2 2 0 as n . The proof is complete.□

Proof of Theorem 1.1(ii) for the mountain pass type

By Lemma 3.8, there exists a sequence { u n } S c , r with the following properties

J b , μ ( u n ) m * , ( J b , μ S c , r ) ( u n ) 0 , u n 2 0 , P ( u n ) 0 , as n .

If alternative (i) in Proposition 2.1 occurs, then u n u in H r 1 ( R 4 ) for some u 0 and due to u 2 2 lim n u n 2 2 = B ; thus,

m * Λ I μ ( u ) = a 2 + B b 4 u 2 2 μ q u q q 1 4 u 4 4 a 2 u 2 2 + b 4 u 2 4 μ q u q q 1 4 u 4 4 = J b , μ ( u ) .

In the case J b , μ ( u ) 0 , we have

m * Λ 0 > m ( c , 0 , μ ) ,

yielding a contradiction with Lemma 3.7; however, in the case J b , μ ( u ) < 0 , we have u 2 < t * , and then, by u 2 c and Remark 3.1 that

J b , μ ( u ) m ( u 2 ) m ( c ) m ( c , 0 , μ ) ,

which also contradicts the result of Lemma 3.7. Therefore, alternative (ii) in Proposition 2.1 holds true, and hence, u n u in H r 1 ( R 4 ) with u being a ground-state solution of (1.1)–(1.2) for some λ > 0 . Moreover, by u n 2 0 as n and the maximum principle, we can see that u is positive. The proof is complete.□

4 Proof of Theorem 1.2

Next we prove Theorem 1.2 for the case q = 3 . Let us keep in mind that γ q q = 2 throughout this subsection. Assuming 0 < b < S ¯ and (1.13) holds, the energy functional J b , μ possesses a mountain-pass geometric structure on S c . Specifically speaking, we begin by recalling the fiber mapping

(4.1) Ψ u ( s ) = a e 2 s 2 u 2 2 + b e 4 s 4 u 2 4 e 2 s 3 μ u 3 3 e 4 s 4 u 4 4 = e 2 s a u 2 2 2 μ u 3 3 3 + e 4 s 4 ( b u 2 4 u 4 4 ) .

We have

μ u 3 3 3 μ 3 C q q c u 2 2 < a 2 u 2 2 and b u 2 4 < u 4 4

due to assumption (1.13) and 0 < b < S ¯ . Thus, the fiber mapping satisfies Ψ u ( ) = 0 + and Ψ u ( + ) = . So it is easy to see that J b , μ S c has a mountain-pass geometric structure. We consider once again the Pohozaev manifold P b , μ , and the decomposition P b , μ = P b , μ + P b , μ 0 P b , μ defined in (1.11).

Lemma 4.1

P b , μ 0 = P b , μ + = .

Proof

For u P b , μ , we have

(4.2) 0 = a u 2 2 + b u 2 4 u 4 4 μ γ q u 3 3 a u 2 2 + ( b S 2 ) u 2 4 μ γ q C q q c u 2 2 = ( a μ γ q C q q c ) u 2 2 + ( b S 2 ) u 2 4 .

Since γ q = 2 3 , 0 < b < S ¯ S 2 , and (4.2), we obtain

(4.3) inf u P b , μ u 2 > 0 .

Moreover, for u P b , μ , using (4.3) and 0 < b < S ¯ , we can see

( Ψ u ) ( 0 ) = ( Ψ u ) ( 0 ) q γ q P b , μ ( u ) = 2 ( b u 2 4 u 4 4 ) < 0 ,

which means u P b , μ . Therefore, P b , μ = P b , μ , and P b , μ 0 = P b , μ + = .□

Lemma 4.2

For every u S c , there exists a unique t u R such that t u u P b , μ . t u is the unique critical point of Ψ u ( s ) and is a strict maximum point at the positive level. Moreover,

  1. P b , μ = P b , μ ;

  2. P b , μ ( u ) < 0 if and only if t u < 0 ;

  3. The map u S c t u R is of class C 1 .

Proof

For every u S c , from (4.1), we have Ψ u ( ) = 0 + and Ψ u ( + ) = . Therefore, Ψ u has a global maximum point t u at the positive level, which means Ψ u ( t u ) = 0 and t u u P b , μ . Moreover, it is easy to see that Ψ u ( s ) has only one critical point. Therefore we know that t u is the unique critical point of Ψ u . By Lemma 4.1, we know P b , μ = P b , μ and we deduce that t u u P b , μ . Observing that Ψ u ( t ) < 0 if and only if t > t u , we have P b , μ ( u ) = Ψ u ( 0 ) < 0 if and only if t u < 0 . Define Φ : H 1 ( R 4 ) × R R by Φ ( u , t ) = Ψ u ( t ) . Clearly, Φ is of class C 1 , Φ ( u , t u ) = 0 and t Φ ( u , t u ) = Ψ u ( t u ) < 0 . Applying the Implicit Function Theorem, we see that the map u S c t u R is of class C 1 .□

Lemma 4.3

m m ( c , b , μ ) = inf u P b , μ J b , μ ( u ) > 0 .

Proof

Lemma 4.2 indicates that P b , μ . For u P b , μ , combined with assumptions (1.13) and (4.3), there holds

(4.4) J b , μ ( u ) = J b , μ ( u ) 1 4 P b , μ ( u ) = a 4 u 2 2 μ 6 u 3 3 1 2 a 2 μ C q q c 3 u 2 2 > 0 .

Thus, we have the desired result.□

Lemma 4.4

There exists k > 0 sufficiently small such that

0 < sup A k ¯ J b , μ < m and u A k ¯ J b , μ ( u ) , P b , μ ( u ) > 0 ,

where A k { u S c : u 2 2 < k } .

Proof

By the Sobolev inequality (2.1), the Gagliardo-Nirenberg inequality (2.3), and assumption (1.13), we have

J b , μ ( u ) = a 2 u 2 2 + b 4 u 2 4 1 4 u 4 4 μ 3 u 3 3 a 2 u 2 2 + b 4 u 2 4 S 2 4 u 2 4 μ 3 C q q c u 2 2 = a 2 μ 3 C q q c u 2 2 + 1 4 b 1 S 2 u 2 4 > 0 , P b , μ ( u ) = a u 2 2 + b u 2 4 u 4 4 2 μ 3 u 3 3 a u 2 2 + b u 2 4 S 2 u 2 4 2 μ C q q c 3 u 2 2 = a 2 μ C q q c 3 u 2 2 + ( b S 2 ) u 2 4 > 0 ,

if u A k ¯ with k small enough. If necessary replacing k with a smaller quantity, we also have J b , μ ( u ) u 2 2 2 < m for every u A k ¯ .□

We introduce the minimax class

Γ ¯ { γ ¯ C ( [ 0 , 1 ] , S c , r ) : γ ¯ ( 0 ) A k ¯ , J b , μ ( γ ¯ ( 1 ) ) < 0 } ,

then Γ ¯ . In fact, for every u S c , r , there exists s 0 1 and s 1 1 , such that s 0 u A k ¯ and J b , μ ( s 1 u ) < 0 , and s s u is continuous. Thus, we can define the minimax value

(4.5) m ¯ m ¯ ( c , b , μ ) = inf γ ¯ Γ ¯ max t [ 0 , 1 ] J b , μ ( γ ¯ ( t ) ) .

Lemma 4.5

There holds m = m ¯ .

Proof

For any u P b , μ , we have

lim t ( t u ) 2 2 = 0 , lim t + ( t u ) 2 2 = + , and lim t + J b , μ ( t u ) = .

Then, there exists t 1 = t 1 ( u ) < 0 and t 2 = t 2 ( u ) > 0 such that t 1 u A k 1 , t 2 u S c \ A k 2 , and J b , μ ( t 2 u ) < 0 . Setting γ ¯ ( t ) = u ( 1 t ) t 1 + t t 2 for t [ 0 , 1 ] , we have γ ¯ Γ ¯ and sup t [ 0 , 1 ] J b , μ ( γ ¯ ( t ) ) = J b , μ ( u ) , from which it follows that m m ¯ . To prove m m ¯ , it suffices to verify that γ ¯ ( [ 0 , 1 ] ) P b , μ for any γ ¯ Γ ¯ . Due to (4.4), we have P b , μ ( γ ¯ ( 1 ) ) < γ q q J b , μ ( γ ¯ ( 1 ) ) 0 for any γ ¯ Γ ¯ . Note that P b , μ ( γ ¯ ( 0 ) ) > 0 . Let us define t 1 = inf { t [ 0 , 1 ) : P b , μ ( γ ¯ ( s ) ) < 0 for s ( t , 1 ] } . Then, there must be P b , μ ( γ ¯ ( t 1 ) ) = 0 , which implies that γ ¯ ( t 1 ) γ ¯ ( [ 0 , 1 ] ) P b , μ . The proof is complete.□

Lemma 4.6

0 < m < Λ .

Proof

Let v ε be as in (2.4). By Lemmas 4.2 and 4.3, there exists a unique t ε R such that t ε v ε P b , μ and 0 < m J b , μ ( t ε v ε ) . To conclude the proof, it suffices to show that J b , μ ( t ε v ε ) < Λ for ε > 0 sufficiently small. In view of (2.5), (2.6), and (4.3), there exists a constant C > 0 such that, as ε 0 + ,

e 2 t ε v ε 3 3 = a ε 2 e 2 t ε a ε u ε 3 3 = ( t ε v ε ) 2 2 u ε 2 2 c u ε 3 3 u ε 2 C ε 1 4 ,

which combined with Lemma 2.1 leads to

J b , μ ( t ε v ε ) = a e 2 t ε 2 v ε 2 2 + b e 4 t ε 4 v ε 2 4 e 4 t ε 4 v ε 4 4 μ e 2 t ε 3 v ε 3 3 Λ + O ( ε 1 2 ) μ 3 C ε 1 4 < Λ .

Letting ε > 0 small, we obtain the desired conclusion.□

Proof of Theorem 1.2

Similar to 3.8 and by Lemma 4.5, there exists a sequence { u n } S c , r with the following properties:

J b , μ ( u n ) m , ( J b , μ S c , r ) ( u n ) 0 , u n 2 0 , P b , μ ( u n ) 0 , as n .

According to Proposition 2.1, one of the two alternatives in Proposition 2.1 occurs. Suppose that alternative (i) occurs, then u n u in H r 1 ( R 4 ) for some u 0 , which means that u satisfies

( a + B b ) Δ u + λ u = u 2 u + μ u u .

Then, we can obtain the corresponding Pohozaev identity

Q b , μ ( u ) = ( a + B b ) u 2 2 2 μ 3 u q q u 4 4 = 0 .

Moreover,

m Λ I μ ( u ) = I μ ( u ) 1 4 Q b , μ ( u ) = a 4 u 2 2 μ 6 u 3 3 a 4 u 2 2 μ 6 C q q c u 2 2 = 1 2 a 2 μ 3 C q q c u 2 2 .

By assumption (1.13), we have m Λ 0 , which contradicts the result of Lemma 4.6. Therefore, alternative (ii) in Proposition 2.1 holds true, and hence, u n u in H r 1 ( R 4 ) with u being a ground-state solution of (1.1)–(1.2) in H r 1 ( R 4 ) for some λ > 0 . Moreover, from u n 2 0 as n and the maximum principle, we can see that u is positive.□

5 Proof of Theorem 1.3

Next, we prove Theorem 1.3 for the case 3 < q < 4 . At this point, there holds 2 < γ q q < 4 , so it is easy to see that Ψ u ( s ) possesses a mountain-pass geometric structure on S c . We consider once again the Pohozaev manifold P b , μ , and the decomposition P b , μ = P b , μ + P b , μ 0 P b , μ defined in (1.11).

Lemma 5.1

P b , μ 0 = P b , μ + = .

Proof

For u P b , μ , since 0 < b < b * is small enough and 2 < γ q q < 4 , we have

0 = a u 2 2 + b u 2 4 u 4 4 μ γ q u q q a u 2 2 + ( b S 2 ) u 2 4 μ γ q C q q c ( 1 γ q ) q u 2 γ q q ,

which implies that

(5.1) inf u P b , μ u 2 > 0 .

Moreover, for u P b , μ , using (5.1), we can see

( Ψ u ) ( 0 ) = ( Ψ u ) ( 0 ) q γ q P b , μ ( u ) = a ( 2 q γ q ) u 2 2 + b ( 4 q γ q ) u 2 4 ( 4 q γ q ) u 4 4 a ( 2 q γ q ) u 2 2 ( 4 q γ q ) ( S ¯ b ) u 2 4 < 0 ,

which means u P b , μ . Therefore, P b , μ = P b , μ , and P b , μ 0 = P b , μ + = .□

Lemma 5.2

For every u S c , there exists a unique t u R such that t u u P b , μ . t u is the unique critical point of Ψ u ( s ) and is a strict maximum point at positive level. Moreover,

  1. P b , μ = P b , μ ;

  2. P b , μ ( u ) < 0 if and only if t u < 0 ;

  3. The map u S c t u R is of class C 1 .

Proof

For every u S c , since 2 < γ q q < 4 , (3.2) and 0 < b < b * is small enough, we have lim s Ψ u ( s ) = 0 + and lim s + Ψ u ( s ) = . Therefore, Ψ u has a global maximum point t u at positive level. Indeed, Ψ u ( s ) = 0 implies that φ ( s ) = μ γ q u q q with

φ ( s ) = a u 2 2 e ( 2 q γ q ) s ( u 4 4 b u 2 4 ) e ( 4 q γ q ) s .

We can see that the function φ ( s ) is continuous and monotonically decreasing, with lim s φ ( s ) = + and lim s + φ ( s ) = , respectively. So, equation φ ( s ) = μ γ q u q q has exactly one solution. Therefore, we know that t u is the unique critical point of Ψ u . Since Ψ u ( t u ) = 0 , t u u P b , μ . By maximality Ψ u ( t u ) 0 , and since P b , μ 0 = , we deduce that t u u P b , μ , which implies that P b , μ = P b , μ . The rest of the proofs are the same as the proofs of Lemma 4.2.□

Lemma 5.3

m m ( c , b , μ ) = inf u P b , μ J b , μ ( u ) > 0 .

Proof

Lemma 5.2 indicates that P b , μ . For u P b , μ , since 0 < b < b * is small enough, there holds

(5.2) J b , μ ( u ) = J b , μ ( u ) 1 γ q q P b , μ ( u ) = a 1 2 1 γ q q u 2 2 b 1 γ q q 1 4 u 2 4 + 1 γ q q 1 4 u 4 4 a 1 2 1 γ q q u 2 2 b 1 γ q q 1 4 u 4 4 S ¯ + 1 γ q q 1 4 u 4 4 = a 1 2 1 γ q q u 2 2 + 1 b S ¯ 1 γ q q 1 4 u 4 4 a 1 2 1 γ q q u 2 2 ,

which combined with (5.1) concludes the proof.□

Lemma 5.4

There exists k > 0 sufficiently small such that

0 < sup A k ¯ J b , μ < m and u A k ¯ J b , μ ( u ) , P b , μ ( u ) > 0 ,

where A k { u S c : u 2 2 < k } .

Proof

By the Sobolev inequality (2.1), the Gagliardo-Nirenberg inequality (2.3), and 2 < γ q q < 4 , we have

J b , μ ( u ) = a 2 u 2 2 + b 4 u 2 4 1 4 u 4 4 μ q u q q a 2 u 2 2 S 2 4 u 2 4 μ q C q q c ( 1 γ q ) q u 2 γ q q > 0 , P b , μ ( u ) = a u 2 2 + b u 2 4 μ γ q u q q u 4 4 a u 2 2 μ γ q C q q c ( 1 γ q ) q u 2 γ q q S 2 u 2 4 > 0 ,

if u A k ¯ with k small enough. If necessary replacing k with a smaller quantity, we also have J b , μ ( u ) u 2 2 2 < m for every u A k ¯ .□

Lemma 5.5

There holds m = m ¯ , where m ¯ is defined in (4.5).

Proof

The proof is same as Lemma 4.5, and we omit it.□

Lemma 5.6

0 < m < Λ .

Proof

Let v ε be as in (2.4). By Lemmas 5.2 and 5.3, there exists a unique t ε R such that t ε v ε P b , μ and 0 < m J b , μ ( t ε v ε ) . To conclude the proof, it suffices to show that J b , μ ( t ε v ε ) < Λ for ε > 0 sufficiently small. In view of (2.5), (2.6), and (5.1), there exists a constant C > 0 such that, as ε 0 + ,

e γ q q t ε v ε q q = a ε γ q q e γ q q t ε a ε ( 1 γ q ) q u ε q q = ( t ε v ε ) 2 γ q q u ε 2 γ q q c ( 1 γ q ) q u ε q q u ε 2 ( 1 γ q ) q C ε 4 q 4 ,

which combined with Lemma 2.1 leads to

J b , μ ( t ε v ε ) = a e 2 t ε 2 v ε 2 2 + b e 4 t ε 4 v ε 2 4 e 4 t ε 4 v ε 4 4 μ e γ q q t ε q v ε q q Λ + O ( ε 1 2 ) μ q C ε 4 q 4 .

Letting ε > 0 small yields the desired conclusion.□

Proof of Theorem 1.3

Similar to 3.8 and by Lemma 5.5, there exists a sequence { u n } S c , r with the following properties:

J b , μ ( u n ) m , ( J b , μ S c , r ) ( u n ) 0 , u n 2 0 , P b , μ ( u n ) 0 , as n .

According to Proposition 2.1, one of the two alternatives in Proposition 2.1 occurs. Suppose that alternative (i) occurs, then u n u in H r 1 ( R 4 ) for some u 0 , which means that u satisfies

( a + B b ) Δ u + λ u = u 2 u + μ u q 2 u .

Then, we can obtain the corresponding Pohozaev identity:

Q b , μ ( u ) = ( a + B b ) u 2 2 μ γ q u q q u 4 4 = 0 .

Moreover,

m Λ I μ ( u ) = I μ ( u ) 1 q γ q Q b , μ ( u ) = a 1 2 1 q γ q u 2 2 B b 1 q γ q 1 4 u 2 2 + 1 q γ q 1 4 u 4 4 a 1 2 1 q γ q u 2 2 B b 1 q γ q 1 4 u 2 2 + 1 q γ q 1 4 S ¯ u 2 4 = a 1 2 1 q γ q u 2 2 + ( S ¯ u 2 2 B b ) u 2 2 1 q γ q 1 4 .

Since 0 < b < b * is small enough and 2 < γ q q < 4 , we have m Λ 0 , which contradicts the result of Lemma 5.6. Therefore, alternative (ii) in Proposition 2.1 holds true, and hence, u n u in H r 1 ( R 4 ) with u being a ground-state solution of (1.1)–(1.2) in H r 1 ( R 4 ) for some λ > 0 . Moreover, we see from u n 2 0 as n and the maximum principle that u is positive.□

6 Nonexistence result

In this section, we prove Theorem 1.4. The main tool is the Hadamard three spheres theorem [28].

Proof of Theorem 1.4(i)

Let u be a critical point of the energy functional J b , μ ( u ) on S c . Then, u solves (1.1)–(1.2) for some λ R ; testing (1.1) by u and integrating over R 4 , we have

a u 2 2 + b u 2 4 + λ u 2 2 = μ u q q + u 4 4 = 0 .

Combined with P b , μ ( u ) = 0 , i.e.,

P b , μ ( u ) = a u 2 2 + b u 2 4 μ γ q u q q u 4 4 = 0 ,

we have

λ u 2 2 = μ ( 1 γ q ) u q q .

Since u S c (thus u 0 ), μ < 0 , and 0 < γ q < 1 , we infer that λ < 0 . Now, we are going to prove J b , μ ( u ) a 2 S 2 4 ( 1 b S 2 ) . Again using P b , μ ( u ) = 0 , we have

(6.1) ( a + b u 2 2 ) u 2 2 = μ γ q u q q + u 4 4 u 4 4 ,

where we used that μ γ q < 0 . From (6.1) and (2.1), we obtain

a u 2 2 + b u 2 4 S 2 u 2 4 .

Thus, we have

(6.2) u 2 2 a S 2 1 b S 2 .

Moreover,

J b , μ ( u ) = J b , μ ( u ) 1 4 P b , μ ( u ) = a 4 u 2 2 + μ q q γ q 4 1 u q q a 4 u 2 2 a 2 S 2 4 ( 1 b S 2 ) ,

where we used that 0 < q γ q < 4 , μ < 0 , and (6.2).□

Proof of Theorem 1.4(ii)

Suppose that u is a solution of (1.1)–(1.2). Using the Brezis-Kato argument [5], we have that u , u , Δ u L ( R 4 ) . This and the fact that u L 2 ( R 4 ) allow us to conclude that

(6.3) u ( x ) 0 , as x .

Thus, there exists R 0 > 0 such that

u 2 2 1 , for all x R 0 .

Therefore,

1 a + b 1 a + b u 2 ,

for all x R 0 . This implies that

Δ u = 1 a + b u 2 2 ( u 2 u + μ u q 2 u λ u ) λ a + b u > 0 ,

for all x R 0 . Hence, u is superharmonic at infinity. Now, we define

m ( r ) = min x = r { u ( x ) } .

Since u > 0 , we have that m ( r ) > 0 . The Hadamard three spheres theorem (see [28]) gives us that

m ( r ) m ( r 1 ) 1 r 2 1 r 2 2 + m ( r 2 ) 1 r 1 2 1 r 2 1 r 1 2 1 r 2 2 ,

for all R 0 < r 1 < r < r 2 . As (6.3) holds, we have that m ( r 2 ) 0 , as r 2 . Moreover, the function r r 2 m ( r ) is monotone non-decreasing for r > R 0 . Therefore,

m ( r ) m ( R 0 ) R 0 2 r 2 ,

for all x R 0 . Therefore, we have

u H 1 ( R 4 ) 2 C u 2 2 C R 0 m ( r ) 2 r 3 d r C R 0 1 r d r = ,

where we used that H 1 ( R 4 ) is continuously embedding into L 2 ( R 4 ) . This is enough to conclude that the problem does not have solution.□

7 Asymptotic behavior of normalized solutions

In this section, we investigate the asymptotic behavior of normalized ground states for (1.1)–(1.2) and give the proofs of Theorems 1.51.7. Let us start by considering μ 0 + .

Proof of Theorem 1.5(i)

Let c > 0 and 0 < b < S ¯ be fixed. We recall that u b , μ u c , b , μ is an interior local minimizer of J b , μ in A R 0 , where R 0 ( c , μ ) is defined by Lemma 3.1. It is easy to check that R 0 ( c , μ ) 0 as μ 0 + , and then, u b , μ 2 < R 0 ( c , μ ) 0 as well. Moreover, from the Gagliardo-Nirenberg inequality (2.1) and the Sobolev inequality (2.3), we have

0 > m ( b , μ ) = J b , μ ( u b , μ ) = a 2 u b , μ 2 2 + b 4 u b , μ 2 4 1 4 u b , μ 4 4 μ q u b , μ q q a 2 u b , μ 2 2 + b 4 u b , μ 2 4 S 2 4 u b , μ 2 4 μ q C q q c ( 1 γ q ) q u b , μ 2 γ q q 0 ,

as μ 0 + , which implies that m ( b , μ ) 0 .□

Next, we consider the case where 3 q < 4 .

Lemma 7.1

There holds

inf u P b , μ J b , μ ( u ) = inf u S c max t R J b , μ ( t u ) .

Proof

We see from Lemmas 4.2 and 5.2 that

J b , μ ( u ) = max t R J b , μ ( t u ) inf u S c max t R J b , μ ( t u ) , for u P b , μ ,

and if u S c , then there is a unique t u , b , μ t u , c , b , μ R such that t u , b , μ u P b , μ and

max t R J b , μ ( t u ) = J b , μ ( t u , b , μ u ) inf u P b , μ J b , μ ( u ) .

The proof is complete.□

Lemma 7.2

We have

inf u P b , 0 J b , 0 ( u ) = inf u S c max t R J b , 0 ( t u ) = Λ .

Proof

Similar to Lemma 7.1, we have

inf u P b , 0 J b , 0 ( u ) = inf u S c max t R J b , 0 ( t u ) .

Lemma 2.1 indicates that inf u S c max t R J b , 0 ( t u ) Λ . Let u S c and denote

a 1 = a u 2 2 , b 1 = b u 2 4 , and c 1 = u 4 4 .

It follows from the definition of S that

max t R J b , 0 ( t u ) = max t R a 1 t 2 2 + b 1 t 4 4 c 1 t 4 4 = a 1 2 4 ( c 1 b 1 ) Λ .

Hence, we have inf u S c max t R J b , 0 ( t u ) Λ , which completes the proof.□

Lemma 7.3

Let μ ¯ , c > 0 be such that 0 < μ < 3 a 2 C q q c if q = 3 and any μ , c > 0 are admissible if 3 < q < 4 . Then, the function μ m ( b , μ ) is non-increasing on [ 0 , μ ¯ ] .

Proof

Let 0 μ 1 < μ 2 μ ¯ . By Lemmas 7.1 and 7.2, we have

m ( b , μ 2 ) = inf u S c max t R J b , μ 2 ( t u ) inf u S c max t R J b , μ 1 ( t u ) = m ( b , μ 1 ) .

The proof is complete.□

Proof of Theorem 1.5(ii) and (iii)

Let c > 0 , and let μ ¯ satisfy the assumption of Lemma 7.3 for this choice of c . First, we know that the family of positive radial ground-states { u b , μ , μ : 0 < μ < μ ¯ } is bounded in H r 1 ( R 4 ) from (2.8) and (2.9).

Assume up to a subsequence that u b , μ u in H r 1 ( R 4 ) , u b , μ u in L q ( R 4 ) , u b , μ u a.e. in R 4 and u b , μ 2 2 B 0 as μ 0 + . Then, by P b , μ ( u b , μ ) = 0 again and the Sobolev inequality (2.1), we have

a B + b B 2 S 2 B 2 ,

from which it follows that either B = 0 or B a S 2 1 b S 2 . If B = 0 , then u b , μ 0 in D 1 , 2 ( R 4 ) , and hence, J b , μ ( u b , μ ) 0 as μ 0 + . On the other hand, Lemma 7.3 indicates that J b , μ ( u b , μ ) = m ( b , μ ) m ( b , μ ¯ ) > 0 for 0 < μ < μ ¯ , which comes to a contradiction. Therefore, we have B a S 2 1 b S 2 . Combining Lemmas 7.2 and 7.3 yields that

a 4 B = lim μ 0 + J b , μ ( u b , μ ) 1 4 P b , μ ( u b , μ ) = lim μ 0 + m ( b , μ ) m ( b , 0 ) = Λ .

Then, there must be B = a S 2 1 b S 2 and lim μ 0 + m ( b , μ ) = Λ . Finally, we prove u = 0 , i.e., u b , μ 0 in H r 1 ( R 4 ) as μ 0 + . Let λ μ > 0 be the Lagrange multiplier associated with the ground-state solution u b , μ . By P b , μ ( u b , μ ) = 0 , we have

λ μ c 2 = λ μ u b , μ 2 2 = μ ( 1 γ q ) u b , μ q q ,

and then λ μ 0 as μ 0 + . Combining this with u b , μ u in H r 1 ( R 4 ) implies that u weakly solves

( a + b B ) Δ u = u 3 , u 0 , in R 4 , u H r 1 ( R 4 ) ,

which has only trivial solution in H r 1 ( R 4 ) . Therefore, we have u = 0 .□

Next, we consider the case when b 0 + . We first consider the case 2 < q < 3 and recall the following lemma proved in [30].

Lemma 7.4

There holds m ( 0 , μ ) = inf u P 0 , μ J 0 , μ ( u ) < 0 .

Lemma 7.5

If 0 < b < 1 , then m ( 0 , μ ) m ( b , μ ) m ( 1 , μ ) .

Proof

We see from Lemma 3.1 that R 0 is independent of b . Then,

m ( 0 , μ ) = inf u A R 0 ( c ) J 0 , μ ( u ) m ( b , μ ) = inf u A R 0 ( c ) J b , μ ( u ) inf u A R 0 ( c ) J 1 , μ ( u ) = m ( 1 , μ ) .

The proof is complete.□

Lemma 7.6

m ( b , μ ) m ( 0 , μ ) as b 0 + .

Proof

By (1.15) and [30, Theorem 1.1], there exists u P 0 , μ + such that J 0 , μ ( u ) = m ( 0 , μ ) . Then, a u 2 2 u 4 4 μ γ q u q q = 0 and 2 a u 2 2 4 u 4 4 μ γ q 2 q u q q > 0 . According to Lemma 3.2, there is a unique s u , b , μ s u , c , b , μ R such that u s u , b , μ P b , μ + , which implies

a e 2 s u , b , μ u 2 2 + b e 4 s u , b , μ u 2 4 e 4 s u , b , μ u 4 4 μ γ q e γ q q s u , b , μ u q q = 0

and

2 a e 2 s u , b , μ u 2 2 + 4 b e 4 s u , b , μ u 2 4 4 e 4 s u , b , μ u 4 4 μ γ q 2 q e γ q q s u , b , μ u q q > 0 .

Hence, s u , b , μ 0 as b 0 + , and thus,

m ( 0 , μ ) m ( b , μ ) J b , μ ( u s u , b , μ ) J 0 , μ ( u ) = m ( 0 , μ ) .

Thus, the proof is complete.□

Proof of Theorem 1.6 for 2 < q < 3

Let b n 0 + as n and u n u c , b n , μ be the ground-state solution of 1.51.7 obtained by Theorem 1.1. Then, there exists λ n R such that for φ H r 1 ( R 4 ) ,

(7.1) a R 4 u n φ d x + b n R 4 u n 2 R 4 u n φ d x + R 4 λ n u n φ d x = R 4 u n 2 u n φ d x + μ R 4 u n q 2 u n φ d x .

By Lemma 7.5, it is easy to verify that { u n } is bounded in H r 1 ( R 4 ) and { λ n } is bounded in R . Assume u n u in H r 1 ( R 4 ) , u n u in L q ( R 4 ) , u n u a.e. in R 4 , u n 2 2 B 0 and λ n λ as n . In view of P b n , μ ( u n ) = 0 and γ q < 1 , we have

λ c 2 = lim n λ u n 2 2 = μ ( 1 γ q ) u q q .

If u = 0 , then a u n 2 2 = u n 4 4 + o ( 1 ) , which combined with Lemma 7.6 implies

m ( 0 , μ ) = lim n m ( b n , μ ) = lim n J b n , μ ( u n ) = a 4 B 0 .

This contradicts the result of Lemma 7.4. Thus, u 0 , and henceforth, λ > 0 . Letting n in (7.1), we have

(7.2) a R 4 u φ d x + R 4 λ u φ d x = R 4 u 2 u φ d x + μ R 4 u q 2 u φ d x ,

which means that u is a weak solution of

(7.3) a Δ u + λ u = μ u q 2 u + u 2 u , in R 4 .

It is standard to show that u satisfies the following Pohozaev identity:

(7.4) a u 2 2 u 4 4 μ γ q u q q = 0 .

Setting v n = u n u , we have

u n 2 2 = v n 2 2 + u 2 2 + o ( 1 ) and u n 4 4 = v n 4 4 + u 4 4 + o ( 1 ) ,

from which it follows that

P b n , μ ( u n ) = a u n 2 2 u n 4 4 μ γ q u n q q + o ( 1 ) = a v n 2 2 v n 4 4 + a u 2 2 u 4 4 μ γ q u q q + o ( 1 ) .

Combining this with P b n , μ ( u n ) = 0 and (7.4) yields a v n 2 2 = v n 4 4 + o ( 1 ) S 2 v n 2 4 + o ( 1 ) . Assuming lim n v n 2 2 = w , we have a w S 2 w 2 . Then, either w = 0 or w a S 2 . If w a S 2 , there holds

m ( 0 , μ ) = J b n , μ ( u n ) + o ( 1 ) = a 2 u 2 2 1 4 u 4 4 μ q u q q + a 2 v n 2 2 1 4 v n 4 4 + o ( 1 ) = a 4 u 2 2 μ q 1 γ q q 4 u q q + a 4 w + o ( 1 ) a 4 u 2 2 μ q 1 γ q q 4 C q q c ( 1 γ q ) q u 2 γ q q + ( a S ) 2 4 + o ( 1 ) ,

where we used (7.4) and Lemma (2.3). Defining ξ : R + R by

ξ ( t ) = a 4 t 2 μ q 1 γ q q 4 C q q c ( 1 γ q ) q t γ q q ,

we see from (1.15) that

min t > 0 ξ ( t ) = 2 γ q q 2 2 γ q q a γ q q 2 γ q q μ q C q q 1 γ q q 4 c ( 1 γ q ) q 2 2 γ q q > ( a S ) 2 4 .

Then m ( 0 , μ ) > 0 , which contradicts the result of Lemma 7.4. Therefore, w = 0 , and thus, u n u in D 1 , 2 ( R 4 ) and in L 4 ( R 4 ) . Choosing u n u as a test function in (7.1) and (7.2), respectively, subtracting them, and taking limit, we see that u n u in L 2 ( R 4 ) . The proof is complete.□

Next, we consider the case where 3 q < 4 . The following lemma is taken from [30].

Lemma 7.7

We have

m ( 0 , μ ) = inf u P 0 , μ J 0 , μ ( u ) = inf u S c max t R J 0 , μ ( t u )

and

0 < m ( 0 , μ ) < ( a S ) 2 4 .

Lemma 7.8

The function b m ( b , μ ) is non-decreasing on [ 0 , 1 ] .

Proof

Let 0 b 1 < b 2 1 . By Lemmas 7.1 and 7.7, there holds

m ( b 1 , μ ) = inf u S c max t R J b 1 , μ ( t u ) inf u S c max t R J b 2 , μ ( t u ) = m ( b 2 , μ ) ,

which completes the proof.□

Lemma 7.9

m ( b , μ ) m ( 0 , μ ) as b 0 + .

Proof

It is similar to that of Lemma 7.6 and now is based on Lemmas 4.2 and 5.2.□

Proof of Theorem 1.6 for 3 ≤ q < 4

In view of Lemmas 7.77.9, we can repeat the arguments in the proof of Theorem 1.6 for 2 < q < 3 to conclude the proof.□

Proof of Theorem 1.7

The proof is similar to that of Theorem 1.5, so we omit it.□

Acknowledgement

The authors are very grateful for the anonymous reviewers for their careful reading of the manuscript and valuable comments.

  1. Funding information: This research was supported by Innovation Research 2035 Pilot Plan of Southwest University (SWU-XDPY22015), National Natural Science Foundation of China (No. 12371120) and Chongqing Postgraduate Research and Innovation Programme (CYS240130).

  2. Author contributions: Xin Qiu: Writing-original draft, Writing-review & editing; Zeng-Qi Ou: Supervision, Formal Analysis; Chun-Lei Tang: Methodology, Supervision; Ying Lv: Writing-review & editing.

  3. Conflict of interest: The authors state no conflict of interest.

  4. Data availability statement: Data sharing not applicable to this article as no data sets were generated or analyzed during the current study.

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Received: 2024-03-12
Revised: 2024-07-29
Accepted: 2024-08-26
Published Online: 2024-10-16

© 2024 the author(s), published by De Gruyter

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

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