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Positive solutions for asymptotically linear Schrödinger equation on hyperbolic space

  • Dongmei Gao , Jun Wang and Zhengping Wang EMAIL logo
Published/Copyright: March 10, 2025
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Advances in Nonlinear Analysis
From the journal Advances in Nonlinear Analysis Volume 14 Issue 1

Abstract

In this article, we study the following stationary Schrödinger equation on hyperbolic space:

− Δ H N u + λ u = f ( u ) , x ∈ H N , N ≥ 3 ,

where Δ H N denotes the Laplace-Beltrami operator on H N , λ ∈ R , and f is locally Lipschitz continuous satisfying asymptotically linear at infinity. After a reduction from hyperbolic case to Euclidean case, using variational methods, we prove the existence of positive solutions for the aforementioned equation under suitable conditions on λ and f .

Keywords: asymptotically linear Schrödinger equation; hyperbolic space; mountain pass theorem; linking theorem
MSC 2010: 35A01; 35A15; 35B09

1 Introduction

In this article, we consider the existence of positive solutions to the following stationary Schrödinger equation on hyperbolic space

(1) − Δ H N u + λ u = f ( u ) , x ∈ H N , N ≥ 3 ,

where Δ H N denotes the Laplace-Beltrami operator on H N , λ is a real parameter, and f is locally Lipschitz continuous satisfying asymptotically linear at infinity. We denote

H N ≔ { x = ( x 0 , x 1 , … , x N ) ∈ R N + 1 : [ x , x ] = 1 , x 0 > 0 } ,

here [ x , y ] = x 0 y 0 − … − x N y N . By a transformation of the polar coordinate,

H N = { ( cosh r , ω sinh r ) ∈ R N + 1 : r > 0 , ω ∈ S N − 1 } ,

we can obtain the following expression of Δ H N :

(2) Δ H N = ∂ r 2 + ( N − 1 ) cosh r sinh r ∂ r + 1 sinh 2 r Δ S N − 1 ,

where S N − 1 ≔ { ω ∈ R N : ∣ ω ∣ = 1 } is the unit sphere on R N .

In recent years, there have been many articles on studying equation (1). For example, Mancini and Sandeep [26] showed the following existence and uniqueness results of positive solutions to (1) with f ( u ) = ∣ u ∣ p u .

Proposition 1.1

(Theorems 1.3 and 1.4 of [26]) Let N ≥ 2 and 0 < p < 4 N − 2 , then (1) has a positive entire solution for λ ≥ − ( N − 1 ) 2 4 . Let λ ≥ − ( N − 1 ) 2 4 if N ≥ 3 and λ > − 2 ( p + 2 ) ( p + 4 ) 2 if N = 2 , then (1) has at most one entire positive solution, up to hyperbolic isometries.

In [26], the non-existence result of positive solutions was also obtained.

Proposition 1.2

(Theorem 1.1 of [26]) Let N ≥ 2 . If λ < − ( N − 1 ) 2 4 , then (1) has no positive solutions. If λ = − ( N − 1 ) 2 4 , there are no positive solutions of (1) in H 1 ( H N ) .

Later, Christianson and Marzuola [14] obtained the existence result by applying the concentration compactness principle after conjugating problem (1) to Euclidean space.

Proposition 1.3

(Theorem 1 and Lemma 6.1 of [14]) Let N ≥ 2 , λ > − ( N − 1 ) 2 4 , and 0 < p < 4 N − 2 , then there exists a positive radial and decreasing solution u ∈ C 2 ( H N ) ∩ H 1 ( H N ) satisfying (1) with f ( u ) = ∣ u ∣ p u .

For the uniqueness of radial solutions to (1) with f ( u ) = ∣ u ∣ p u , Wang [31] give a different proof for the uniqueness result of [26] in a special case, where N ≥ 3 , λ ≥ − ( N − 1 ) 2 4 + 1 4 , and 0 < p < 4 N − 2 are assumed. For more results for nonlinear Schrödinger equations on H N , see [6–8].

The reduction from hyperbolic case to Euclidean case in [14] will also be applied in this article. As follows, let

φ ( r ) ≔ r sinh r N − 1 2 , T R ( r , ω ) ≔ φ ( r ) R ( r , ω ) ,

and then, T is an isometry from L 2 ( r N − 1 d r d ω ) to L 2 ( sinh N − 1 r d r d ω ) . By (2) and a direct computation,

φ − 1 ( − Δ H N ) ( φ R ) = − Δ ˜ N R + V ˜ N ( r ) + ( N − 1 ) 2 4 R ,

where

Δ ˜ N = Δ + r 2 − sinh 2 r r 2 sinh 2 r Δ S N − 1 and V ˜ N ( r ) = ( N − 1 ) ( N − 3 ) 4 r 2 − sinh 2 r r 2 sinh 2 r .

If one considers spherically symmetric solutions in H 1 ( H N ) , the term Δ S N − 1 in Δ ˜ N will be zero; thus, equation (1) can be transformed into the following equation on R N , namely,

(3) − Δ R + V ˜ λ , N ( x ) R − f ˜ ( x , R ) = 0 , x ∈ R N ,

where R = φ − 1 u , potential function as

(4) V ˜ λ , N ( x ) = λ + ( N − 1 ) 2 4 + V ˜ N ( ∣ x ∣ ) , x ∈ R N ,

and nonlinear term as

(5) f ˜ ( x , t ) = φ − 1 ( x ) f ( φ ( x ) t ) , x ∈ R N , t ∈ R .

To the best of the author’s knowledge, there are few articles concerning (1) when f ( u ) is asymptotically linear. While on R N , regarding transformed equation (3) with f ˜ ( x , t ) being asymptotically linear in t , there are numerous research findings. Wang and Zhou [29] were interested in V ˜ λ , N ≡ V λ ( x ) = 1 + λ g ( x ) , λ > 0 and g ∈ L ∞ ( R N ) vanishes on a bounded domain on R N , f ˜ ( x , t ) ≡ f ( t ) satisfies the asymptotically linear condition

1 ≤ α + 1 = lim t → ∞ f ( t ) t < ∞ ,

they obtained the existence and non-existence of positive solutions of (3) for all α and λ by using a variant of the mountain pass theorem under some other assumptions on f and g .

Equation (3) with V ˜ λ , N ≡ V ( x ) was discussed in many articles, for examples, Jeanjean and Tanaka [19,20], and Liu and Wang [23]. When V ( x ) vanishes at infinity, following the work of Ambrosetti, Felli, and Malchiodi [1–3], there are many articles dealing with problem (3), such as [4,5,11,13,27,28,33], where V ( x ) ≥ 0 on R N is assumed. Wang and Zhou [30] considered (3) when V ˜ λ , N ≡ λ V ( x ) changes sign and may vanish at infinity.

Inspired by the aforementioned articles, for λ > − ( N − 1 ) 2 4 , after a reduction (1) to Euclidean case and then using the variational methods, we obtain the existence of positive spherically symmetric solutions for (1). We make further assumptions about the nonlinear term f ,

  1. ( f 0 ) ∀ s > t > 0 , f ( s ) − f ( t ) s − t ≤ C ( s + t ) α , where C > 0 and α ∈ 0 , 4 N − 2 ;

  2. ( f 1 ) f ( t ) ≥ 0 for t ≥ 0 , f ( − t ) = − f ( t ) for t ∈ R ;

  3. ( f 2 ) lim t → 0 + f ( t ) t = 0 , lim t → ∞ f ( t ) t = l < ∞ ;

  4. ( f 3 ) f ( t ) t is nondecreasing in t > 0 ;

  5. ( f 4 ) Let F ( u ) ≔ ∫ 0 u f ( t ) d t , G ( u ) ≔ f ( u ) u 2 − F ( u ) → ∞ as u → ∞ and G ( u ) > 0 for u ≢ 0 ;

  6. ( f 5 ) sup f ( t ) t : t ≠ 0 ≤ β λ + ( N − 1 ) 2 4 , where 0 < β < 1 .

Remark 1.4

In fact, there are indeed many functions that satisfy the aforementioned conditions for nonlinear term. For example, let

(6) f ( t ) = l t 3 1 + t 2 , for some l > 0 ,

and then, we have

(7) F ( u ) = ∫ 0 u f ( t ) d t = l 2 [ u 2 − ln ( 1 + u 2 ) ] ,

(8) G ( u ) = f ( u ) u 2 − F ( u ) = l 2 − u 2 1 + u 2 + ln ( 1 + u 2 ) .

Thus, the functions in (6)–(8) satisfy the conditions ( f 0 ) − ( f 4 ) .

Theorem 1.5

Let λ ∈ − ( N − 1 ) 2 4 , ∞ if N = 3 and λ ∈ − N ( N − 1 ) 6 , ∞ if N > 3 . Assume that f is locally Lipschitz continuous on R satisfying ( f 0 )–( f 3 ) and λ + ( N − 1 ) 2 4 > l > μ * , where

μ * ≔ inf σ ( − Δ + V ˜ λ , N ( x ) ) = inf 0 ≢ u ∈ H 1 ( R N ) ∫ R N [ ∣ ∇ u ∣ 2 + V ˜ λ , N ( x ) u 2 ] d x ∫ R N u 2 d x ,

and V ˜ λ , N ( x ) is defined by (4), then (1) has a positive spherically symmetric solution in H 1 ( H N ) .

As in [30], we introduce a weighted Sobolev space E . Set

V ˜ λ , N + ( x ) ≔ max { V ˜ λ , N ( x ) , 0 } , V ˜ λ , N − ( x ) ≔ max { − V ˜ λ , N ( x ) , 0 } ,

and then define

(9) E ≔ u ∈ D 1 , 2 ( R N ) : ∫ R N V ˜ λ , N + ( x ) u 2 d x < ∞ .

Its scalar and norm are given by

⟨ u , w ⟩ E = ∫ R N ( ∇ u ∇ w + V ˜ λ , N + ( x ) u w ) d x , ‖ u ‖ E = ⟨ u , u ⟩ E 1 2 , ∀ u , w ∈ E .

Thus, H 1 ( R N ) ⊂ E and E is a Hilbert space. Set

Ω + ≔ { x ∈ R N : V ˜ λ , N ( x ) ≥ 0 } , F ≔ { u ∈ E : Supp u ⊂ Ω + } .

Then, F is a closed linear subspace of E and F ⊥ denotes the orthogonal complement of F in E . Let

(10) μ j ≔ inf M ⊂ F ⊥ dim M ≥ j sup ∫ R N [ ∣ ∇ u ∣ 2 + V ˜ λ , N ( x ) u 2 ] d x : u ∈ M , ∫ R N V ˜ λ , N − ( x ) u 2 d x = 1 , j = 1 , 2 , … .

Theorem 1.6

Let N > 3 and λ ∈ − ( N − 1 ) 2 4 , − N ( N − 1 ) 6 . Assume that f is locally Lipschitz continuous on R satisfying ( f 0 )–( f 2 ) and ( f 4 )–( f 5 ), l ≥ ( 1 − μ ¯ ) λ + N ( N − 1 ) 6 , where

μ ¯ ≔ inf { μ j : μ j > 1 } ,

and μ j are defined by (10), and then, (1) has a positive spherically symmetric solution in H 1 ( H N ) .

Remark 1.7

The classification of λ ’s range in Theorems 1.5 and 1.6 is according to the sign of potential function V ˜ λ , N ( x ) . By [14], we have V ˜ λ , N ( x ) ∈ λ + N ( N − 1 ) 6 , λ + ( N − 1 ) 2 4 , and V ˜ λ , N ( x ) tends to λ + ( N − 1 ) 2 4 + O ( ∣ x ∣ − 2 ) as ∣ x ∣ → ∞ . Then, in the case of Theorem 3.1, we have V ˜ λ , N ( x ) ≥ 0 for all x ∈ R N , but in the case of Theorem 1.6, V ˜ λ , N ( x ) changes sign.

Remark 1.8

By Remark 1.7, for λ ∈ − ( N − 1 ) 2 4 , − N ( N − 1 ) 6 , V ˜ λ , N ( x ) is increasing to λ + ( N − 1 ) 2 4 > 0 as ∣ x ∣ → ∞ and has the negative part. Therefore by [32, Theorems 4.45 and 4.46], μ j defined by (10) are the positive eigenvalues of

(11) − Δ u ( x ) + V ˜ λ , N + ( x ) u ( x ) = μ V ˜ λ , N − ( x ) u ( x ) , u ∈ F ⊥ .

Moreover, μ 1 ≤ μ 2 ≤ … ≤ μ j → ∞ as j → ∞ and the corresponding eigenfunctions e j , which may be chosen so that ⟨ e i , e j ⟩ E = δ i j , are a basis for F ⊥ . So the infimum μ ¯ defined in Theorem 1.6 is meaningful.

In fact, the existence results obtained earlier are sharp about λ ; by the argument in [26], for λ ≤ − ( N − 1 ) 2 4 , we have the non-existence results.

Theorem 1.9

Let N ≥ 3 and λ < − ( N − 1 ) 2 4 . Assume that f ∈ C ( R ) satisfying ( f 1 ) , then (1) has no positive solutions.

Theorem 1.10

Let N ≥ 3 and λ = − ( N − 1 ) 2 4 . Assume that f ∈ C 1 ( R ) satisfying ( f 0 )–( f 2 ), then (1) has no positive solutions in H 1 ( H N ) .

The rest of this article is organized as follows: in Section 2, we obtain the symmetry property of positive solutions in H 1 ( H N ) ; in Section 3, we give the proof of existence results; in Section 4, we give the proof of non-existence results.

2 Symmetry property of positive solutions

In this section, we show the spherical symmetry of solutions of (1), which prepares for the transformed equation (3).

Theorem 2.1

Let N ≥ 3 and λ ≥ − ( N − 1 ) 2 4 . Assume that u ∈ C 1 ( H N ) ∩ H 1 ( H N ) is a positive solution of (1) and f is locally Lipschitz continuous on R satisfying ( f 0 ) ; then, u has hyperbolic symmetry, i.e., there is x 0 ∈ H N such that u is constant on hyperbolic spheres centered at x 0 .

Proof

We prove the symmetry results using the moving planes method [25] and the sharp Poincare-Sobolev inequality ([26])

(12) S N , p ∫ H N ∣ u ∣ p d V H N 2 p ≤ ∫ H N ∇ H N u 2 − ( N − 1 ) 2 4 ∣ u ∣ 2 d V H N , u ∈ H 1 ( H N ) ,

where p ∈ 2 , 2 N N − 2 if N ≥ 3 , S N , p > 0 .

Assume that there is a family of isometries A t , which generalize translations or rotations in the R N case and belong to C 1 ( R × H N , H N ) . A t satisfying the invariance condition A t I A t = I , where I is a reflection such that I is an isometry and I 2 = Identity , and I fixes a hypersurface U ⊂ H N . Define a one-parameter family of reflections I t = A t I A − t , and let U t be the hypersurface of H N , which is fixed by I t . We assume that H N can be foliated by U t , i.e., ⋃ t ∈ R U t = H N .

For t ∈ R , define

Q t = ⋃ − ∞ < s < t U s and Q t = ⋃ t < s < ∞ U s ,

and then I t ( Q t ) ⊂ Q t and I t ( Q t ) ⊂ Q t . Next, set

u t ( x ) = u ( I t ( x ) ) , x ∈ R N , t ∈ R ,

and

Λ ≔ { t ∈ R : ∀ s > t , u ≥ u s in Q s } .

The proof will be divided into two steps: the first step is to show that Λ is nonempty and bounded from below, and the second step is to derive that u ≡ u t 0 in Q t 0 , where t 0 = inf Λ .

Step 1. By u ∈ H 1 ( H N ) solves (1), λ ≥ − ( N − 1 ) 2 4 and ( f 0 ) , we have

∫ Q t ∣ ∇ H N ( u t − u ) + ∣ 2 d V H N = ∫ Q t [ − Δ H N ( u t − u ) ] ( u t − u ) + d V H N = ∫ Q t [ f ( u t ) − f ( u ) − λ ( u t − u ) ] ( u t − u ) + d V H N ≤ ( N − 1 ) 2 4 ∫ Q t [ ( u t − u ) + ] 2 d V H N + C ∫ Q t ( u t ) α [ ( u t − u ) + ] 2 d V H N .

Since u ∈ H 1 ( H N ) , α ∈ 0 , 4 N − 2 , u ∈ L α + 2 ( H N ) . By Hölder’s inequality and (12), it follows that

∫ Q t ∣ ∇ H N ( u t − u ) + ∣ 2 d V H N ≤ ( N − 1 ) 2 4 ∫ Q t [ ( u t − u ) + ] 2 d V H N + C ∫ Q t ( u t ) α ⋅ α + 2 α d V H N α α + 2 ∫ Q t [ ( u t − u ) + ] 2 ⋅ α + 2 2 d V H N 2 α + 2 ≤ ( N − 1 ) 2 4 ∫ Q t [ ( u t − u ) + ] 2 d V H N + C ∫ Q t ( u t ) α + 2 d V H N α α + 2 ∫ Q t ∣ ∇ H N ( u t − u ) + ∣ 2 − ( N − 1 ) 2 4 [ ( u t − u ) + ] 2 d V H N .

Thus,

1 − C ∫ Q t ( u t ) α + 2 d V H N α α + 2 ∫ Q t ∣ ∇ H N ( u t − u ) + ∣ 2 − ( N − 1 ) 2 4 [ ( u t − u ) + ] 2 d V H N ≤ 0 ,

where C > 0 . We can choose t ˜ ∈ R such that

(13) C ∫ Q t ( u t ) α + 2 d V H N α α + 2 < 1 , ∀ t > t ˜ ,

and then, for any t > t ˜ ,

∫ Q t ∣ ∇ H N ( u t − u ) + ∣ 2 − ( N − 1 ) 2 4 [ ( u t − u ) + ] 2 d V H N ≤ 0 .

Again by (12), we have

∫ Q t [ ( u t − u ) + ] 2 N N − 2 d V H N N − 2 N ≤ ∫ Q t ∣ ∇ H N ( u t − u ) + ∣ 2 − ( N − 1 ) 2 4 [ ( u t − u ) + ] 2 d V H N ≤ 0 .

Therefore, ( u t − u ) + ≡ 0 in Q t , and thus, ( t ˜ , ∞ ) ⊂ Λ , Λ is nonempty.

As the behavior of f is known at zero and at infinity, the strong maximum principle implies that − ∞ does not belong to Λ . Thus, Λ is bounded from below.

Step 2. Since t 0 = inf Λ , by continuity, we see that u ≥ u t 0 in Q t 0 . Assume that u ≢ u t 0 in Q t 0 , set ω t 0 ≔ u − u t 0 , and then ω t 0 satisfies the linear equation

− Δ H N ω t 0 = c t 0 ( x ) ω t 0 , c t 0 ( x ) ≔ f ( u ( x ) ) − f ( u t 0 ( x ) ) u ( x ) − u t 0 ( x ) − λ ∈ L loc ∞ ( H N ) .

Then, the strong maximum principle and connectedness of Q t 0 imply u ≡ u t 0 in Q t 0 (in this case, symmetry is directly obtained) or u > u t 0 in Q t 0 .

Now, suppose u > u t 0 in Q t 0 . By u ∈ L α + 2 ( H N ) , there exist a compact set K ⊂ Q t 0 and δ > 0 such that K ⊂ Q t for t ∈ [ t 0 − δ , t 0 ] and

(14) C ∫ Q t \ K ( u t ) α + 2 d V H N α α + 2 < 1 2 ,

where C is as in (13). Then, same as step 1, we have

(15) ( u t − u ) + ≡ 0 in Q t \ K .

Furthermore, u > u t 0 in Q t 0 implies there exists 0 < δ 1 < δ such that

(16) u > u t in K , ∀ t ∈ [ t 0 − δ 1 , t 0 ] .

Thus, by (15)–(16), we see

u ≥ u t in Q t , ∀ t ∈ [ t 0 − δ 1 , t 0 ] ,

which contradicts with the definition of t 0 . Therefore, u ≡ u t 0 in Q t 0 , the proof is complete.□

3 Proof of the existence results

In this section, we give the proof of Theorems 1.5 and 1.6. By Section 1 and Theorem 2.1, Theorems 1.5 and 1.6 are equivalent to the following theorems.

Theorem 3.1

Let λ ∈ − ( N − 1 ) 2 4 , ∞ if N = 3 and λ ∈ − N ( N − 1 ) 6 , ∞ if N > 3 . Assume that f satisfies ( f 1 )–( f 3 ) and λ + ( N − 1 ) 2 4 > l > μ * , and then (3) has a positive solution in H 1 ( R N ) .

Theorem 3.2

Let N > 3 and λ ∈ − ( N − 1 ) 2 4 , − N ( N − 1 ) 6 . Assume that f satisfies ( f 1 )–( f 2 ) and ( f 4 )–( f 5 ), l ≥ ( 1 − μ ¯ ) λ + N ( N − 1 ) 6 , and then (3) has a positive solution in H 1 ( R N ) .

Remark 3.3

From conditions ( f 1 ) − ( f 5 ) , we can deduce that for any fixed x ∈ R N , there holds

  1. ( g 1 ) f ˜ ( x , t ) ≥ 0 for t ≥ 0 , f ˜ ( x , − t ) = − f ˜ ( x , t ) for t ∈ R ;

  2. ( g 2 ) lim t → 0 + f ˜ ( x , t ) t = 0 , lim t → ∞ f ˜ ( x , t ) t = l < ∞ ;

  3. ( g 3 ) f ˜ ( x , t ) t is nondecreasing in t > 0 ;

  4. ( g 4 ) set

    F ˜ ( x , u ) ≔ ∫ 0 u f ˜ ( x , t ) d t , G ˜ ( x , u ) ≔ f ˜ ( x , u ) u 2 − F ˜ ( x , u ) ,

    then G ˜ ( x , u ) → ∞ as u → ∞ , and G ˜ ( x , u ) > 0 for u ≢ 0 ;

  5. ( g 5 ) sup f ˜ ( x , t ) t : t ≠ 0 ≤ β λ + ( N − 1 ) 2 4 , where 0 < β < 1 .

  6. Indeed,

    0 < φ ( x ) = ∣ x ∣ sinh ∣ x ∣ N − 1 2 < 1 , φ ( x ) → 0 as ∣ x ∣ → ∞ ,

    and by (5), f ˜ ( x , t ) = φ − 1 ( x ) f ( φ ( x ) t ) , and then f ˜ ( x , t ) t = f ( φ t ) φ t , we can easily obtain ( g 1 )–( g 3 ), ( g 5 ) from ( f 1 )–( f 3 ), and ( f 5 ) , respectively. For ( g 4 ) ,

    G ˜ ( x , u ) = 1 2 φ − 1 f ( φ u ) u − φ − 1 ∫ 0 u f ( φ s ) d s = φ − 2 ( x ) 1 2 f ( φ u ) φ u − ∫ 0 φ u f ( s ) d s = φ − 2 ( x ) G ( φ ( x ) u ) → ∞ , as u → ∞ ,

    where the last limit is due to ( f 4 ) .

Define

(17) I ( u ) = 1 2 ∫ R N [ ∣ ∇ u ∣ 2 + V ˜ λ , N ( x ) u 2 ] d x − ∫ R N F ˜ ( x , u ) d x ,

and our aim is to seek critical points of I .

3.1 Proof of Theorem 3.1

To prove Theorem 3.1, we seek critical points of I in H 1 ( R N ) using the following version of mountain pass theorem, which was applied in [16,19].

Proposition 3.4

(Mountain pass theorem) Let B a real Banach space with dual B * , and I ∈ C 1 ( B , R ) with I ( 0 ) = 0 be such that

I ( u ) ≥ 0 , f o r a l l u ∈ B , w i t h ‖ u ‖ B ≤ ρ , inf ‖ u ‖ B = ρ I ( u ) ≥ α , I ( e ) < 0 ,

for some ρ , α > 0 , and e ∈ B with ‖ e ‖ B > ρ . Define

c = inf γ ∈ Γ max 0 ≤ t ≤ 1 I ( γ ( t ) ) ,

where Γ = { γ ∈ C ( [ 0 , 1 ] , B ) : γ ( 0 ) = 0 , I ( γ ( 1 ) ) < 0 } . Then, there exists a sequence { u n } ⊂ B such that

(18) I ( u n ) → c ≥ α , ( 1 + ‖ u n ‖ B ) ‖ I ′ ( u n ) ‖ B * → 0 , as n → ∞ .

A sequence with properties (18) is usually called a Cerami sequence at level c , ( C e ) c sequence for short. Clearly, a ( C e ) c sequence is also a usual ( P S ) c sequence. To obtain a positive solution of (3), it is sufficient to show that { u n } is bounded in H 1 ( R N ) and converges strongly to a nonzero critical point of I .

Next, we prove some lemmas on the energy functional I , which are required when applying the Mountain pass theorem. In the following text, we denote by ‖ u n ‖ and ∣ u n ∣ p , respectively, the standard norms of H 1 ( R N ) and L p ( R N ) , p ≥ 1 .

Lemma 3.5

Let λ ∈ − ( N − 1 ) 2 4 , ∞ if N = 3 and λ ∈ − N ( N − 1 ) 6 , ∞ if N > 3 . Assume that ( g 1 )–( g 2 ) hold and l > μ * , then there exist ρ , β > 0 such that

(19) I ( u ) ≥ 0 , f o r u ∈ H 1 ( R N ) w i t h ‖ u ‖ ≤ ρ , I ( u ) ≥ β > 0 , f o r u ∈ H 1 ( R N ) w i t h ‖ u ‖ = ρ ,

and

(20) I ( e ) < 0 , f o r e ∈ H 1 ( R N ) w i t h ‖ e ‖ > ρ .

Proof

From ( g 1 ) and ( g 2 ) , for any ε > 0 , there exist p ∈ 1 , N + 2 N − 2 and C ( ε , p ) > 0 such that

(21) F ˜ ( x , t ) ≤ ε t 2 + C ( ε , p ) t p + 1 , ∀ t > 0 , x ∈ R N ,

and then by Remark 1.7, Sobolev inequalities, there exist positive constants C 1 , C 2 , and C 3 ( ε ) such that

I ( u ) ≥ C 1 ‖ u ‖ 2 − C 2 ε ‖ u ‖ 2 − C 3 ( ε ) ‖ u ‖ p + 1 .

Thus, (19) is obtained by choosing ε = C 1 2 C 2 and ρ > 0 small enough.

Recall that

μ * ≔ inf σ ( − Δ + V ˜ λ , N ( x ) ) = inf 0 ≢ u ∈ H 1 ( R N ) ∫ R N [ ∣ ∇ u ∣ 2 + V ˜ λ , N ( x ) u 2 ] d x ∫ R N u 2 d x

and l > μ * imply that there exists ψ ∈ H 1 ( R N ) \ { 0 } such that

(22) ∫ R N [ ∣ ∇ ψ ∣ 2 + V ˜ λ , N ( x ) ψ 2 ] d x < l ∫ R N ψ 2 d x .

We can assume that ψ ≥ 0 on R N , and by Fatou’s lemma, there holds

(23) lim t → ∞ I ( t ψ ) t 2 = 1 2 ∫ R N [ ∣ ∇ ψ ∣ 2 + V ˜ λ , N ( x ) ψ 2 ] d x − lim t → ∞ ∫ R N F ˜ ( x , ψ ) t 2 ψ 2 ψ 2 d x ≤ 1 2 ∫ R N [ ∣ ∇ ψ ∣ 2 + V ˜ λ , N ( x ) ψ 2 − l ψ 2 ] d x < 0 .

So (20) is obtained by choosing e = t ψ for t large.□

By Lemma 3.5, we see that I satisfies all the conditions of Proposition 3.4 with B = H 1 ( R N ) ; thus, we have

Lemma 3.6

Under the assumptions of Lemma 3.5, there exists a ( C e ) c sequence { u n } ∈ H 1 ( R N ) , i.e.,

(24) I ( u n ) → c > 0 , ‖ I ′ ( u n ) ‖ ( 1 + ‖ u n ‖ ) → 0 , a s n → ∞ ,

where c is defined in Proposition 3.4.

Next, we show that { u n } is bounded in H 1 ( R N ) by applying the following variant of the concentration-compactness lemma [34, Lemma 2.1], essentially based on [22].

Lemma 3.7

(Lemma 2.1 of [34]) For R > 0 , set B R ≔ { x ∈ R N : ∣ x ∣ ≤ R } . Let { ρ n } be a sequence in L 1 ( R N ) satisfying

ρ n ≥ 0 o n R N , lim n → ∞ ∫ R N ρ n d x = η , f o r s o m e ρ > 0 .

Then, there exists a subsequence of { ρ n } , still denoted by { ρ n } , satisfying one of the following two conditions:

  1. (Vanishing) lim n → ∞ sup y ∈ R N ∫ B R ( y ) ρ n d x = 0 for all R > 0 .

  2. (Nonvanishing) There exists γ ∈ ( o , η ) , R < ∞ , and { y n } ⊂ R N such that

    lim n → ∞ ∫ B R ( y n ) ρ n d x ≥ γ > 0 .

Lemma 3.8

Let λ ∈ − ( N − 1 ) 2 4 , ∞ if N = 3 and λ ∈ − N ( N − 1 ) 6 , ∞ if N > 3 . Assume that ( g 1 )–( g 3 ) hold and l > μ * , then, the sequence { u n } obtained in Lemma 3.6  is bounded in H 1 ( R N ) .

Proof

Let

(25) ω n = β u n ‖ u n ‖ = β n u n , β n = β ‖ u n ‖ ,

where β > 0 will be determined below, and then, { ω n } is bounded in H 1 ( R N ) . By Lemma 3.7, we may assume that { ω n } satisfies

Vanishing: lim n → ∞ sup y ∈ R N ∫ B R ( y ) ∣ ω n ∣ 2 d x = 0 for all R > 0 ,

or

Nonvanishing: there exist γ > 0 , R < ∞ and { y n } ⊂ R N such that

lim n → ∞ ∫ B R ( y n ) ∣ ω n ∣ 2 d x ≥ γ > 0 .

We will show { u n } is bounded in H 1 ( R N ) by contradiction, i.e., if ‖ u n ‖ → ∞ , then { u n } would satisfy neither vanishing nor nonvanishing, which contradicts with Lemma 3.7.

First, we rule out vanishing case. By ( g 1 ) and ( g 2 ) , for any ε > 0 , there exist p ∈ 1 , N + 2 N − 2 and C ε > 0 such that for all x ∈ R N ,

∫ R N F ˜ ( x , ω n ) d x ≤ ε ∣ ω n ∣ 2 2 + C ε ∣ ω n ∣ p + 1 p + 1 .

So if vanishing occurs, by the well-known vanishing lemma ([22, Lemma I.1 of Part 2]), there holds

∫ R N F ˜ ( x , ω n ) d x → 0 , n → ∞ .

Hence, by Remark 1.7, there exists η > 0 such that

(26) I ( ω n ) = I β ‖ u n ‖ u n = I ( β n u n ) ≥ η β 2 + o ( 1 ) ,

here, and in what follows, o ( 1 ) denotes a quantity, which tends to zero as n → ∞ . Note that condition ( g 3 ) implies for any D ≥ 1 , t ∈ [ 0 , 1 ] and s ≥ 0 , x ∈ R N , there holds

(27) 0 ≤ 1 2 t 2 f ˜ ( x , s ) s − F ˜ ( x , t s ) ≤ D 1 2 f ˜ ( x , s ) s − F ˜ ( x , s ) ,

or

(28) 0 ≤ 1 2 f ˜ ( x , t s ) t s − F ˜ ( x , t s ) ≤ D 1 2 f ˜ ( x , s ) s − F ˜ ( x , s ) .

If (27) occurs, from (24) and ω n = β n u n , we have

I ( ω n ) + o ( 1 ) = I ( β n u n ) − ( β n ) 2 2 ⟨ I ′ ( u n ) , u n ⟩ = ∫ R N ( β n ) 2 2 f ˜ ( x , u n ) u n − F ˜ ( x , β n u n ) d x ≤ D ∫ R N 1 2 f ˜ ( x , u n ) u n − F ˜ ( x , u n ) d x = D I ( u n ) − 1 2 ⟨ I ′ ( u n ) , u n ⟩ → D c .

Taking β 2 > D c η , then the aforementioned inequality contradicts with (26) and vanishing cannot occur.

If (28) occurs, we apply the argument as in [19]. Since ‖ u n ‖ → ∞ , for sufficiently large n , we have β n = β ‖ u n ‖ ∈ ( 0 , 1 ) , and from (26), we obtain

(29) max t ∈ [ 0 , 1 ] I ( t u n ) ≥ I β ‖ u n ‖ u n ≥ η 2 β 2 , for n large enough .

By (24), I ( u n ) → c > 0 , then I ( u n ) ≤ M for some M > 0 . Taking β > 0 so large that M < η 2 β 2 . Clearly, the maximum in (29) cannot be attained at t = 0 or 1. Thus, for each sufficiently large n , there exists t n ∈ ( 0 , 1 ) such that I ( t n u n ) = max t ∈ ( 0 , 1 ) I ( t u n ) , then ⟨ I ′ ( t n u n ) , t n u n ⟩ = 0 . Hence, by (28), we have

I ( t n u n ) = I ( t n u n ) − 1 2 ⟨ I ′ ( t n u n ) , t n u n ⟩ = ∫ R N 1 2 f ˜ ( x , t n u n ) t n u n − F ˜ ( x , t n u n ) d x ≤ D ∫ R N 1 2 f ˜ ( x , u n ) u n − F ˜ ( x , u n ) d x = D I ( u n ) − 1 2 ⟨ I ′ ( u n ) , u n ⟩ → D c ,

which contradicts with (29) if we choose β 2 > max 2 η M , 2 η D c . So vanishing cannot occur.

Next we rule out nonvanishing case. We claim that if nonvanishing occurs, then { y n } must be bounded. In fact, if { y n } is unbounded, passing to a subsequence, we may assume that ∣ y n ∣ → ∞ . For any φ ∈ C 0 ∞ ( R N ) , let φ n ( x ) = φ ( x − y n ) , it follows from (24) that

∣ ⟨ I ′ ( u n ) , φ n ⟩ ∣ ≤ ‖ I ′ ( u n ) ‖ H − 1 ( R N ) ‖ φ n ‖ = ‖ I ′ ( u n ) ‖ H − 1 ( R N ) ‖ φ ‖ → 0 , n → ∞ .

Set u ˆ n ( x ) = u n ( x + y n ) and ω ˆ n ( x ) = ω n ( x + y n ) , where ω n is defined by (25). Then,

(30) ∫ R N ( ∇ ω ˆ n ∇ φ + V ˜ λ , N ( x + y n ) ω ˆ n φ − f ˜ ( x , u ˆ n ) u n ˆ ω ˆ n φ ) d x = o ( 1 ) .

Since ‖ ω n ‖ = β and { ω n } is nonvanishing, there exists ω ˆ ∈ H 1 ( R N ) \ { 0 } such that

ω ˆ n ⇀ ω ˆ weakly  in  H 1 ( R N ) , ω ˆ n → ω ˆ a.e.  in R N .

From Remark 1.7, V ˜ λ , N ( x ) ∈ L ∞ ( R N ) , V ˜ λ , N ( x ) → λ + ( N − 1 ) 2 4 as ∣ x ∣ → ∞ , then V ˜ λ , N ( x + y n ) ω ˆ n is bounded in L 2 ( R N ) , and by ∣ y n ∣ → ∞ , we have

V ˜ λ , N ( x + y n ) ω ˆ n ( x ) → λ + ( N − 1 ) 2 4 ω ˆ ( x ) , a.e. in R N , n → ∞ .

Hence, V ˜ λ , N ( x + y n ) ω ˆ n ⇀ λ + ( N − 1 ) 2 4 ω ˆ weakly in L 2 ( R N ) , i.e.,

(31) ∫ R N V ( x + y n ) ω ˆ n φ d x → ∫ R N λ + ( N − 1 ) 2 4 ω ˆ φ d x , n → ∞ .

Moreover,

(32) ∫ R N f ˜ ( x , u ˆ n ) u n ˆ ω ˆ n φ d x → l ∫ R N ω ˆ φ d x , n → ∞ .

Indeed, set

Ω 0 ≔ { x ∈ R N : ω ˆ ( x ) = 0 } , Ω 1 ≔ { x ∈ R N : ω ˆ ( x ) ≠ 0 } .

For x ∈ Ω 0 , we have ω ˆ n ( x ) → ω ˆ ( x ) ≡ 0 a.e. in Ω 0 . By ( g 1 ) and ( g 2 ) , there exists C > 0 such that f ˜ ( x , t ) t ≤ C for all t ∈ R , x ∈ R N , then

f ˜ ( x , u ˆ n ( x ) ) u ˆ n ( x ) ω ˆ n ( x ) φ → 0 ≡ l ω ˆ ( x ) , a.e. in Ω 0 , n → ∞ .

For x ∈ Ω 1 , we have ω ˆ n ( x ) → ω ˆ ( x ) ≠ 0 a.e. in Ω 1 , and since ‖ u ˆ n ‖ = ‖ u n ‖ → ∞ , we see that ∣ u ˆ n ( x ) ∣ → ∞ . Then, ( g 1 ) and ( g 2 ) yield

f ˜ ( x , u ˆ n ( x ) ) u ˆ n ( x ) → l , a.e. in Ω 1 , n → ∞ .

Hence,

f ˜ ( x , u ˆ n ( x ) ) u ˆ n ( x ) ω ˆ n ( x ) → l ω ˆ ( x ) a.e. in R N , n → ∞ .

Since f ˜ ( x , t ) t ≤ C for all t ∈ R , x ∈ R N , we have f ˜ ( x , u ˆ n ( x ) ) u ˆ n ( x ) ω ˆ n ( x ) is bounded in L 2 ( R N ) ; thus,

f ˜ ( x , u ˆ n ( x ) ) u ˆ n ( x ) ω ˆ n ( x ) ⇀ l ω ˆ ( x ) , weakly in L 2 ( R N ) , n → ∞ ,

which implies (32).

Using (30)–(32), we see that ω ˆ satisfies

− Δ ω ˆ + λ + ( N − 1 ) 2 4 ω ˆ = l ω ˆ , in H 1 ( R N ) .

However, by the Pohozaev identity in [9], the aforementioned equation has only the zero solution in H 1 ( R N ) for any λ and l , which is a contradiction.

Therefore, if nonvanishing occurs, { y n } is bounded and there exists 0 ≢ ω ∈ H 1 ( R N ) such that ω n ⇀ ω weakly in H 1 ( R N ) . By (24), for any φ ∈ C 0 ∞ ( R N ) , we have

⟨ I ′ ( u n ) , φ ⟩ ‖ u n ‖ = o ( 1 ) ,

i.e.,

(33) ∫ R N [ ∇ ω n ∇ φ + V ˜ λ , N ( x ) ω n φ ] d x − ∫ R N f ˜ ( x , u n ) u n ω n φ d x = o ( 1 ) .

Since ω n ⇀ ω weakly in H 1 ( R N ) , we see

(34) ∫ R N [ ∇ ω n ∇ φ + V ˜ λ , N ( x ) ω n φ ] d x → ∫ R N [ ∇ ω ∇ φ + V ˜ λ , N ( x ) ω φ ] d x , n → ∞ .

As (32), we have

(35) ∫ R N f ˜ ( x , u n ) u n ω n φ d x → l ∫ R N ω φ d x , n → ∞ .

From (33)–(35), we obtain that ω ≢ 0 and satisfies

(36) − Δ ω + V ˜ λ , N ( x ) ω = l ω ,

and then, by l > μ * and [29, Proposition 2.1(i)], there is no nontrivial solution of (36), which is a contradiction. So nonvanishing cannot occur, the lemma is proved.□

From Lemma 3.8, we obtain a bounded ( C e ) c sequence { u n } . One way to prove Theorem 3.1 is to show that I satisfies the so-called Cerima condition (or PS condition), ( C ) c condition for short, i.e., { u n } has a strongly convergent subsequence in H 1 ( R N ) . By the proof of [21, Theorem 1.2], we need only to show that for any ε > 0 , there exist R ( ε ) and n ( ε ) such that

(37) ∫ ∣ x ∣ ≥ R [ ∣ ∇ u n ∣ 2 + ∣ u n ∣ 2 ] d x ≤ ε , for all R ≥ R ( ε ) and n ≥ n ( ε ) .

In fact, we have the following lemma.

Lemma 3.9

Let N ≥ 3 and λ ∈ − ( N − 1 ) 2 4 , ∞ . Assume that ( g 1 )–( g 3 ) hold, l < λ + ( N − 1 ) 2 4 and { u n } ⊂ H 1 ( R N ) be a bounded ( C e ) c sequence for I at level c > 0 , then for any ε > 0 , there exist R ( ε ) and n ( ε ) such that (37) holds.

Proof

The proof is motivated by [17]. For any fixed R > 0 , we take a cut-off function ξ R ∈ C ∞ ( R N , R ) such that

(38) ξ R ( x ) = 0 , ∣ x ∣ ≤ R , 1 , ∣ x ∣ ≥ 2 R ,

and

(39) ∣ ∇ ξ R ( x ) ∣ ≤ C 0 R , for some C 0 > 0 (independent of  R ) .

Then, for u ∈ H 1 ( R N ) and R ≥ 1 ,

‖ ξ R u ‖ ≤ C 1 ‖ u ‖ , for some C 1 > 0 (independent of  R ) .

Since { u n } ⊂ H 1 ( R N ) is a bounded ( C e ) c sequence, for n large, there holds

(40) ∣ ⟨ I ′ ( u n ) , ξ R u n ⟩ ∣ ≤ ‖ I ′ ( u n ) ‖ H − 1 ( R N ) ‖ ξ R u n ‖ ≤ o ( 1 ) ,

i.e.,

(41) ∫ R N ( ∣ ∇ u n ∣ 2 + V ˜ λ , N ( x ) ∣ u n ∣ 2 ) ξ R d x + ∫ R N u n ∇ u n ∇ ξ R d x ≤ ∫ R N f ˜ ( x , u n ) u n ξ R d x + o ( 1 ) .

Since l < λ + ( N − 1 ) 2 4 , taking δ ≔ 1 2 λ + ( N − 1 ) 2 4 − l > 0 , there exists R 1 > 0 such that

(42) V ˜ λ , N ( x ) ≥ l + δ , for ∣ x ∣ ≥ R 1 .

By ( g 1 ) – ( g 3 ) , we have

(43) f ˜ ( x , u n ) u n ≤ l ∣ u n ∣ 2 , for any x ∈ R N .

Thus, for R ≥ 2 R 1 , (41)–(43) yield

(44) ∫ R N ( ∣ ∇ u n ∣ 2 + δ ∣ u n ∣ 2 ) ξ R d x ≤ ∫ R N u n ∇ u n ∇ ξ R d x + o ( 1 ) ≤ C 0 R ‖ u n ‖ 2 + o ( 1 ) .

As { u n } ⊂ H 1 ( R N ) is bounded, choosing R and n large enough in (44), we have that (38) holds.□

Proof of Theorem 3.1

By Lemmas 3.8–3.9, we obtain that I ( u ) has a nontrivial critical point in H 1 ( R N ) . Then, by ( g 1 ) and the strong maximum principle, we know that (3) has positive solutions in H 1 ( R N ) .□

3.2 Proof of Theorem 3.2

To prove Theorem 3.2, we first recall the following variant version of classical Rabinowitz’s linking theorem.

Proposition 3.10

[18, Proposition 2.3] Let H be a real Hilbert space. Suppose that I ∈ C 1 ( H , R ) , H = H 1 ⊕ H 2 , where dim H 2 < ∞ , and there exist R > ρ > 0 , α > 0 , and 0 ≠ e 0 ∈ H 1 such that

inf I ( H 1 ∩ S ρ ) ≥ α , sup I ( ∂ Q ) ≤ 0 ,

where S ρ = { u ∈ H : ‖ u ‖ H = ρ } and Q = { u = ν + t e 0 : ν ∈ H 2 , t ≥ 0 , ‖ u ‖ H ≤ R } . If I satisfies the ( C ) c condition for all c ∈ [ α , sup I ( Q ) ] , then I has a critical value in [ α , sup I ( Q ) ] .

We want to apply the aforementioned linking theorem with H = E , which is defined by (9), then we need some decomposition of the weighted Sobolev space E . By Remark 1.8, set

E − ≔ span { e j : μ j < μ ¯ } , E + ≔ span { e j : μ j ≥ μ ¯ } ,

and if μ 1 > 1 , we set E − ≔ { 0 } . Then, we have the orthogonal decomposition

E = E − ⊕ E + ⊕ F ,

and we know that dim E − < ∞ , since μ j → ∞ .

Lemma 3.11

Let N > 3 and λ ∈ − ( N − 1 ) 2 4 , − N ( N − 1 ) 6 . Assume that ( g 1 )–( g 2 ) hold, then there exist ρ , α > 0 such that I ( u ) ≥ α for all u ∈ E + ⊕ F with ‖ u ‖ E = ρ .

Proof

For u ∈ E + ⊕ F , then u = u 1 + u 2 with u 1 ∈ E + and u 2 ∈ F . Since ⟨ u 1 , u 2 ⟩ E = 0 and u 2 ∈ F , we have

(45) ∫ R N ( ∣ ∇ u ∣ 2 + V ˜ λ , N ( x ) u 2 ) d x = ∫ R N ( ∣ ∇ u 1 ∣ 2 + V ˜ λ , N ( x ) ∣ u 1 ∣ 2 ) d x + ∫ R N ( ∣ ∇ u 2 ∣ 2 + V ˜ λ , N + ( x ) ∣ u 2 ∣ 2 ) d x .

For u 1 ∈ E + , we can write u 1 = ∑ j : μ j ≥ μ ¯ a j e j , and then,

(46) ∫ R N ( ∣ ∇ u 1 ∣ 2 + V ˜ λ , N ( x ) ∣ u 1 ∣ 2 ) d x = ‖ u 1 ‖ E 2 − ∫ R N V ˜ λ , N − ( x ) ∣ u 1 ∣ 2 d x ≥ 1 − 1 μ ¯ + 1 ‖ u 1 ‖ E 2 .

Combining (45) and (46), we have, for u ∈ E + ⊕ F ,

(47) I ( u ) ≥ C ‖ u ‖ E 2 − ∫ R N F ˜ ( x , u ) d x , C > 0 .

By ( g 1 ) and ( g 2 ) , for any ε > 0 , there exists C ε > 0 such that

F ˜ ( x , u ) ≤ ε u 2 + C ε ∣ u ∣ 2 N N − 2 , for all u ∈ R , x ∈ R N .

Thus

(48) ∫ R N F ˜ ( x , u ) d x ≤ ε ∫ R N u 2 d x + C ε ∫ R N ∣ u ∣ 2 N N − 2 d x ≤ ε ‖ u ‖ E 2 + C ε ‖ u ‖ E 2 N N − 2 .

Combining (47) and (48), by choosing ε > 0 small enough, then the lemma holds.□

Lemma 3.12

Let N > 3 and λ ∈ − ( N − 1 ) 2 4 , − N ( N − 1 ) 6 . Assume that ( g 1 )–( g 2 ) hold and l ≥ ( 1 − μ ¯ ) λ + N ( N − 1 ) 6 , then there exist R > ρ and e ¯ ∈ E + ⊕ F with ‖ e ¯ ‖ E ≥ R such that I ( u ) ≤ 0 for all u ∈ E − ⊕ R e ¯ with ‖ u ‖ E ≥ R , where ρ is given by Lemma 3.11.

Proof

Let e ¯ be an eigenfunction of μ ¯ , since μ ¯ > 1 , we have e ¯ ∈ E + . Define

W ( x , u ) = F ˜ ( x , u ) − l 2 u 2 ,

and then, for u ∈ E − ⊕ R e ¯ , we have

(49) I ( u ) = 1 2 ∫ R N [ ∣ ∇ u ∣ 2 + ( V ˜ λ , N ( x ) − l ) ∣ u ∣ 2 ] d x − ∫ R N W ( x , u ) d x .

By ( g 1 ) and ( g 2 ) , we deduce that

(50) lim t → ∞ 1 t 2 ∫ R N W ( x , t u ) d x = 0 , for u ∈ E − ⊕ R e ¯ with ∣ u ∣ 2 = 1 .

Noting that dim ( E − ⊕ R e ¯ ) < ∞ , then for any ε > 0 , there exists R ( ε ) > 0 such that

(51) ∫ R N W ( x , u ) d x ≤ ε ∫ R N u 2 d x , for u ∈ E − ⊕ R e ¯ with ∣ u ∣ 2 ≥ R ( ε ) ,

and then by Remark 1.7, there holds

(52) I ( u ) ≤ 1 2 ( μ ¯ − 1 ) ∫ R N V ˜ λ , N − ( x ) u 2 d x − l ∫ R N u 2 d x + ∫ R N W ( x , u ) d x ≤ 1 2 ( μ ¯ − 1 ) ∫ R N V ˜ λ , N − ( x ) u 2 d x − 1 2 ( l − ε ) ∫ R N u 2 d x ≤ 1 2 ( 1 − μ ¯ ) λ + N ( N − 1 ) 6 − ( l − ε ) ∫ R N u 2 d x .

Thus, I ( u ) ≤ 0 by choosing ε small enough.

Since dim ( E − ⊕ R e ¯ ) < ∞ , we have ∣ u ∣ 2 ≥ C ‖ u ‖ E for all u ∈ E − ⊕ R e ¯ , and then, the lemma holds by (52).□

Lemma 3.13

Under the assumptions of Theorem 3.2, if { u n } ⊂ E satisfies

(53) I ( u n ) → c > 0 , ( ‖ 1 + ‖ u n ‖ E ) ‖ I ′ ( u n ) ‖ E * → 0 , a s n → ∞ ,

then { u n } has a strongly convergent subsequence.

Proof

First, we claim that { ‖ u n ‖ E } is bounded. Argue by contradiction, we assume that ‖ u n ‖ E → ∞ . Setting v n = u n ‖ u n ‖ E , we have ‖ v n ‖ E = 1 , and there exists v ∈ E such that v n ⇀ v weakly in E . By (53), we have

⟨ I ′ ( u n ) , u n ⟩ ‖ u n ‖ E 2 = o ( 1 ) ,

i.e.,

(54) o ( 1 ) = ‖ v n ‖ E 2 − ∫ R N V ˜ λ , N − ( x ) ∣ v n ∣ 2 d x − ∫ R N f ˜ ( x , u n ) u n ∣ v n ∣ 2 d x .

Then, v ≢ 0 , otherwise, by Remark 1.7, we have ∫ R N V ˜ λ , N − ( x ) ∣ v n ∣ 2 d x = o ( 1 ) . Moreover, by ( g 1 ) , ( g 2 ) , and ( g 5 ) , there exists R > 0 large enough such that

∫ ∣ x ∣ ≥ R f ˜ ( x , u n ) u n ∣ v n ∣ 2 d x ≤ β ∫ ∣ x ∣ ≥ R V ˜ λ , N ( x ) ∣ v n ∣ 2 d x < ‖ v n ‖ E 2 = 1

and

∫ ∣ x ∣ ≤ R f ˜ ( x , u n ) u n ∣ v n ∣ 2 d x ≤ C ∫ ∣ x ∣ ≤ R ∣ v n ∣ 2 d x = o ( 1 ) .

Then, by (54), we have 1 < 1 , which is a contradiction; thus, v ≢ 0 .

Let D ≔ { x ∈ R N : V ˜ λ , N ( x ) ≠ 0 } , then ∣ D ∣ > 0 , where ∣ D ∣ denotes the Lebesgue measure of D . By ( g 4 ) and ‖ u n ‖ E → ∞ , we have 0 < G ˜ ( x , u n ( x ) ) → ∞ for a.e. x ∈ D . Therefore, from (53), we have

C + o ( 1 ) = I ( u n ) − 1 2 ⟨ I ′ ( u n ) , u n ⟩ = ∫ R N G ˜ ( x , u n ( x ) ) d x ≥ ∫ D G ˜ ( x , u n ( x ) ) d x → ∞ ,

which is a contradiction; thus, { ‖ u n ‖ E } is bounded.

Next, we show that { u n } has a strongly convergent subsequence. By (53), for R > 0 , we have

(55) o ( 1 ) = ⟨ I ′ ( u n ) , u n ξ R ⟩ = ∫ R N ∣ ∇ u n ∣ 2 ξ R d x + ∫ R N V ˜ λ , N ( x ) ∣ u n ∣ 2 ξ R d x + ∫ R N u n ∇ u n ∇ ξ R d x − ∫ R N f ˜ ( x , u n ) u n ξ R d x ,

where ξ R is defined by (38) with (39). From Remark 1.7, we know that V ˜ λ , N ( x ) → λ + ( N − 1 ) 2 4 + O ( ∣ x ∣ − 2 ) as ∣ x ∣ → ∞ ; this together with λ > − ( N − 1 ) 2 4 , for any ε > 0 , there exists R ( ε ) > 0 such that

(56) C 0 2 R 2 ≤ 4 ε 2 V ˜ λ , N ( x ) = 4 ε 2 V ˜ λ , N + ( x ) for R ( ε ) ≤ R ≤ ∣ x ∣ ≤ 2 R .

Thus, for any integer n ≥ 1 and R ≥ R ( ε ) , there holds

(57) ∫ R N ∣ u n ∇ u n ∇ ξ R ∣ d x ≤ ε ∫ R N ∣ ∇ u n ∣ 2 d x + 1 4 ε ∫ R ≤ ∣ x ∣ ≤ 2 R C 0 2 R 2 ∣ u n ∣ 2 d x ≤ ε ∫ R N ∣ ∇ u n ∣ 2 d x + ε ∫ R ≤ ∣ x ∣ ≤ 2 R V ˜ λ , N + ( x ) ∣ u n ∣ 2 d x ≤ ε ‖ u n ‖ E 2 .

Recall ( g 5 ) , for any integer n ≥ 1 , there exists R 1 > 0 such that for R ≥ R 1 , there holds

(58) ∫ R N ∣ f ˜ ( x , u n ) u n ξ R ∣ d x = ∫ ∣ x ∣ ≥ R ∣ f ˜ ( x , u n ) u n ξ R ∣ d x ≤ β ∫ ∣ x ∣ ≥ R V ˜ λ , N ( x ) ∣ u n ∣ 2 ξ R d x = β ∫ R N V ˜ λ , N + ( x ) ∣ u n ∣ 2 ξ R d x .

Combining (55), (57), and (58), we obtain, for any integer n ≥ 1 and R ≥ max { R ( ε ) , R 1 } ,

(59) o ( 1 ) = ∣ ⟨ I ′ ( u n ) , u n ξ R ⟩ ∣ ≥ ∫ R N ∣ ∇ u n ∣ 2 ξ R d x + ( 1 − β ) ∫ R N V ˜ λ , N + ( x ) ∣ u n ∣ 2 ξ R d x − ε ‖ u n ‖ E 2 .

Since { ‖ u n ‖ E } is bounded and from (59), we obtain for R ≥ max { R ( ε ) , R 1 } ,

(60) ∫ ∣ x ∣ ≥ R [ ∣ ∇ u n ∣ 2 + ( 1 − β ) V ˜ λ , N + ( x ) ∣ u n ∣ 2 ] d x ≤ o ( 1 ) ε .

Thus, for any ε > 0 , there exists n ( ε ) > 0 and R ˜ ( ε ) > 0 such that for all n ≥ n ( ε ) and R ≥ R ˜ ( ε ) ,

(61) ∫ ∣ x ∣ ≥ R [ ∣ ∇ u n ∣ 2 + V ˜ λ , N + ( x ) ∣ u n ∣ 2 ] d x ≤ C ε .

Then, by a standard procedure of [21, Theorem 1.2], we can show that { u n } has a strongly convergent subsequence.□

Proof of Theorem 3.2

By Proposition 3.10 and Lemmas 3.11–3.13, we obtain that I ( u ) has a nonzero critical point u in E . Then, similar to the proof of [24, Theorem 1.1], we know that u ∈ L 2 ( R N ) . Thus, by ( g 1 ) and the strong maximum principle, we know that (3) has positive solutions in H 1 ( R N ) .□

4 Proof of the non-existence results

In this section, we give the proof of Theorems 1.9–1.10. A basic fact about the spectrum of − Δ H N on H N is that

(62) λ 1 ( − Δ H N ) ≔ inf 0 ≢ u ∈ H 1 ( H N ) ∫ H N ∣ ∇ H N u ∣ 2 d V H N ∫ H N u 2 d V H N = ( N − 1 ) 2 4 , N ≥ 3 ,

which gives the non-existence result for λ < − ( N − 1 ) 2 4 in Theorem 1.9.

Proof of Theorem 1.9

Argue by contradiction, assume that (1) has a positive solution u . Recall that in [10], λ 1 ( Ω ) , the first eigenvalue of − Δ H N can be characterized as

λ 1 ( Ω ) = sup { t : Δ H N ψ + t ψ ≤ 0 , for some ψ > 0 in Ω } ,

where Ω ⊂ H N is any smooth bounded domain. Therefore, by equation (1) and ( f 1 ) , we have λ 1 ( Ω ) ≥ − λ > ( N − 1 ) 2 4 when Ω is the geodesic ball. This is impossible since the first eigenvalue of the geodesic ball converges to ( N − 1 ) 2 4 as the radius goes to infinity [12].□

Before proving Theorem 1.10, we consider the monotonicity and exponential decay of positive solutions for (1).

Let u be a positive symmetric solution of (1), defined on B N ≔ { ξ ∈ R N : ∣ ξ ∣ < 1 } , the hyperbolic gradient ∇ H N and the hyperbolic laplacian Δ H N can be written as

(63) ∇ H N = 1 − ∣ ξ ∣ 2 2 , Δ H N = 1 − ∣ ξ ∣ 2 2 2 Δ + ( N − 2 ) 1 − ∣ ξ ∣ 2 2 ⟨ ξ , ∇ ⟩ .

Then, (1) can be rewritten as

(64) 1 − ∣ ξ ∣ 2 2 2 Δ u + ( N − 2 ) 1 − ∣ ξ ∣ 2 2 ⟨ ∇ u , ξ ⟩ − λ u + f ( u ) = 0 .

Set ∣ ξ ∣ ≔ tanh t 2 , u ( t ) ≔ u tanh t 2 , q ≔ ( sinh t ) N − 1 , then

∫ B N u 2 d V B N = ω N ∫ 0 ∞ q u 2 d t , ∫ B N ∣ ∇ B N u ∣ 2 d V B N = ω N ∫ 0 ∞ q ∣ u ′ ∣ 2 d t ,

where ω N is the volume of the N dimensional unit sphere and ω 1 ≔ 2 . Therefore,

(65) u ∈ H 1 ( B N ) ⇔ ∫ B N [ u 2 + ∣ ∇ B N u ∣ 2 ] d V B N = ω N ∫ 0 ∞ q [ u 2 + ∣ u ′ ∣ 2 ] d t < ∞ .

In addition, (64) can be rewritten as

(66) u ″ + N − 1 tanh t u ′ − λ u + f ( u ) = 0 , u ′ ( 0 ) = 0 ,

namely,

(67) ( q u ′ ) ′ − λ q u + f ( u ) = 0 , u ′ ( 0 ) = 0 .

The following lemma shows that positive solutions of (66) is decreasing and vanishing at infinity in the case of λ ≤ 0 .

Lemma 4.1

Let N ≥ 3 and λ ≤ 0 . Assume that f satisfies ( f 1 ) and u > 0 be a solution of (66), and then, u ′ ( t ) < 0 for every t > 0 and lim t → ∞ u ( t ) = lim t → ∞ u ′ ( t ) = 0 .

Proof

From equation (67), λ ≤ 0 and ( f 1 ) , we obtain

(68) ( q u ′ ) ′ ( t ) < 0 , ∀ t > 0 .

Integrating (68) from 0 to t and u ′ ( 0 ) = 0 , we see that u ′ ( t ) < 0 for t > 0 , and thus, u ( ∞ ) ≔ lim t → ∞ u ( t ) exists.

Define

J u ( t ) ≔ ( u ′ ) 2 2 − λ 2 u 2 + ∫ 0 u f ( s ) d s .

Since u solves (66), we obtain

d d t J u ( t ) = − N − 1 tanh t ( u ′ ) 2 ≤ 0 , ∀ t > 0 ,

and then, by u > 0 , we have u ′ ( ∞ ) ≔ lim t → ∞ u ′ ( t ) = 0 . Thus, u ″ ( ∞ ) ≔ lim t → ∞ u ″ ( t ) = 0 , which with (66) and ( f 1 ) gives

0 = u ″ ( ∞ ) = λ u ( ∞ ) − f ( u ( ∞ ) ) ≤ 0 ,

and hence, u ( ∞ ) = 0 .□

Next, we show the exponential decay of positive solutions for (66), which plays a key role in the proof of Theorem 1.10.

Lemma 4.2

Let N ≥ 3 and λ = − ( N − 1 ) 2 4 . Assume that f ∈ C 1 ( R ) satisfying ( f 0 )–( f 2 ) and u > 0 be a solution of (66), and then

(69) lim t → ∞ log u 2 t = lim t → ∞ log ( u ′ ) 2 t = lim t → ∞ log [ u 2 + ( u ′ ) 2 ] t = − ( N − 1 ) .

Proof

For any ε > 0 , there exists t ε > 0 such that

coth t ≤ 1 + ε , ∀ t ≥ t ε .

By Lemma 4.1, we have u ′ ( t ) < 0 , together with ( f 1 ) , there holds

(70) p ( u ) ≔ u ″ + ( N − 1 ) ( 1 + ε ) u ′ − λ u ≤ u ″ + ( N − 1 ) ( coth t ) u ′ − λ u + f ( u ) = 0 , ∀ t ≥ t ε .

Set

χ ± ( ε ) ≔ − ( N − 1 ) ( 1 + ε ) ± ( N − 1 ) 2 ( 1 + ε ) 2 + 4 λ 2 = − ( N − 1 ) ( 1 + ε ) ± ( ε 2 + 2 ε ) ( N − 1 ) 2 2 ,

and then, χ ± ( ε ) < 0 are the characteristic roots of p ( u ) in (70).

Define

(71) w = u ′ − χ + ( ε ) u ,

and then from (70) we have

(72) w ′ − χ − ( ε ) w = p ( u ) ≤ 0 , ∀ t ≥ t ε .

Therefore,

(73) ( e − χ − ( ε ) t w ) ′ ≤ 0 , ∀ t ≥ t ε ,

then integrating inequality (73) on [ t , ∞ ] , and by Lemma 4.1, we obtain

(74) w ( t ) ≥ e χ − ( ε ) t lim s → ∞ ( e − χ − ( ε ) s w ( s ) ) ≥ 0 , ∀ t ≥ t ε .

Again integrating inequality (73) on [ τ , t ] , we obtain

(75) u ′ ( t ) − χ + ( ε ) u ( t ) = w ( t ) ≤ ( e − χ − ( ε ) τ w ( τ ) ) e χ − ( ε ) t ≔ C ε ( τ ) e χ − ( ε ) t , ∀ t ε ≤ τ ≤ t ,

where C ε ( τ ) ≔ e − χ − ( ε ) τ w ( τ ) > 0 by (74). Finally, by integration, we obtain

(76) u ( t ) ≤ e − χ + ( ε ) τ u ( τ ) − C ε ( τ ) χ − ( ε ) − χ + ( ε ) e ( χ − ( ε ) − χ + ( ε ) ) τ e χ + ( ε ) t + C ε ( τ ) e χ − ( ε ) t χ − ( ε ) − χ + ( ε ) , ∀ t ε ≤ τ ≤ t .

Then, u > 0 , χ − ( ε ) < χ + ( ε ) and (76) imply

e − χ + ( ε ) τ u ( τ ) − C ε ( τ ) χ − ( ε ) − χ + ( ε ) e ( χ − ( ε ) − χ + ( ε ) ) τ ≥ 0 , ∀ τ ≥ t ε ,

i.e.,

u ( τ ) ≥ C ε ( τ ) e χ − ( ε ) τ χ − ( ε ) − χ + ( ε ) = w ( τ ) χ − ( ε ) − χ + ( ε ) = u ′ ( τ ) − χ + ( ε ) u ( τ ) χ − ( ε ) − χ + ( ε ) , ∀ τ ≥ t ε .

Hence,

(77) u ′ ( τ ) ≥ χ − ( ε ) u ( τ ) , ∀ τ ≥ t ε ,

and integrating inequality (77) on [ t ε , t ] we find

(78) u ( t ) ≥ ( u ( t ε ) e − χ − ( ε ) t ε ) e χ − ( ε ) t , ∀ t ≥ t ε .

Next, we observe that u ″ ( t ) > 0 for t large. If not, there exists a sequence t j → ∞ such that d ′ ( t j ) = 0 , where d ≔ u ′ < 0 . Taking the derivative of (66), we have

d ″ + ( N − 1 ) coth ( t ) d ′ + − λ − N − 1 sinh 2 t + f ′ ( u ) d = 0 ,

by lim t → ∞ u ( t ) = 0 , f ∈ C 1 ( R ) and ( f 0 ) , we see that there is t ¯ such that d ″ ( t j ) > 0 for t j ≥ t ¯ , while d ″ has to change sign at two adjacent zeros of d ′ , which is a contradiction. Thus, u ″ > 0 for t large.

Then, by (66), there exists t ˆ large such that

( N − 1 ) coth ( t ) u ′ − λ u < 0 , ∀ t ≥ t ˆ ,

i.e.,

u ′ u ≤ λ tanh t N − 1 = − ( N − 1 ) tanh t 4 , ∀ t ≥ t ˆ .

Thus,

(79) u ( t ) ≤ u ( t ˆ ) e − ( N − 1 ) tanh t 4 ( t − t ˆ ) , ∀ t ≥ t ˆ .

From (77), we can obtain the similar upper bound in (79) for − u ′ . Now by (67) and (78), for any ε > 0 , there exists C 1 ε > 0 such that

(80) − ( q u ′ ) ( t ) ≥ − ( q u ′ ) ( t ε ) − λ ∫ t ε t q e − N − 1 2 + ε τ d τ ≥ C 1 ε e N − 1 2 − ε t , ∀ t ≥ t ε .

So by (77), (79), and (80), we obtain for any ε > 0 , there exist C 2 ε , C 3 ε > 0 such that

(81) C 2 ε e − N − 1 2 + ε t ≤ u ≤ C 3 ε e − N − 1 4 − ε t , C 2 ε e − N − 1 2 + ε t ≤ − u ′ ≤ C 3 ε e − N − 1 4 − ε t , ∀ t ≥ t ˆ .

Set

U ¯ ≔ ( u , u ′ ) ⊥ , F ¯ ( U ¯ , t ) ≔ ( 0 , ( N − 1 ) u ′ ( 1 − coth t ) − f ( u ) ) ⊥ , A ¯ ≔ 0 1 λ − ( N − 1 ) .

We can write (67) as U ¯ ′ = A ¯ U ¯ + F ¯ ( U ¯ , t ) , then (81) gives κ ≔ lim t → ∞ log ∣ U ¯ ∣ t < 0 . By ( f 1 ) and ( f 2 ) ,

∣ F ¯ ( U ¯ , t ) ∣ ≤ ε ∣ U ¯ ∣ , for t large and ∣ U ¯ ∣ small .

Then, κ must be a non-positive characteristic root of A ¯ (see [15, Theorem 4.3]), i.e.,

κ = − ( N − 1 ) ± ( N − 1 ) 2 − 4 λ 2 = − N − 1 2 ,

which proves (69).□

Proof of Theorem 1.10

Because of the asymptotic behavior (69) obtained in Lemma 4.2, and recall that (65), we know that positive solutions of (1) cannot be in H 1 ( B N ) if λ = − ( N − 1 ) 2 4 .□

Acknowledgement

The authors wish to thank the referees for their very helpful comments and suggestions that improved the exposition of this manuscript.

  1. Funding information: Dongmei Gao and Zhengping Wang were supported by NNSF of China (Grant Nos 11931012, 12071482, and 12371118). Jun Wang was supported by NNSF of China (Grant No. 12371114) and National Key R&D Program of China (Grant No. 2022YFA1005601).

  2. Author contributions: The authors contributed equally to the preparation, the revision, and the writing of the manuscript.

  3. Conflict of interest: The authors declare that they have no known competing financial interests or personal relationships that could have appeared to influence the work reported in this article.

  4. Data availability statement: No data were used for the research described in the article.

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Received: 2024-02-20
Revised: 2024-08-19
Accepted: 2025-01-23
Published Online: 2025-03-10

© 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/anona-2025-0070
Keywords for this article
asymptotically linear Schrödinger equation; hyperbolic space; mountain pass theorem; linking theorem
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  69. Linear stability and bifurcation analysis for a free boundary problem arising in a double-layered tumor model
  70. Hopf's lemma, asymptotic radial symmetry, and monotonicity of solutions to the logarithmic Laplacian parabolic system
  71. Generalized quasi-linear fractional Wentzell problems
  72. Existence, symmetry, and regularity of ground states of a nonlinear Choquard equation in the hyperbolic space
  73. Normalized solutions for NLS equations with general nonlinearity on compact metric graphs
  74. Boundedness and global stability in a predator-prey chemotaxis system with indirect pursuit-evasion interaction and nonlocal kinetics
  75. Review Article
  76. Existence and stability of contact discontinuities to piston problems
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