We first consider the following equation

− Δ u + a u = b 1 | u | p − 2 u + b 2 | u | q − 2 u ,   u ∈ H 1 ( ℝ N ) , (2.1)

where p , q ∈ ( 2,2 * ) , p > q , a > 0 , b j > 0 , j = 1,2 .

For each u ∈ H 1 ( ℝ N ) , the energy functional associated to Equation (2.1) is J a b → ( u ) = 1 2 ∫ ℝ N ( | ∇ u | 2 + a u 2 ) d x − b 1 p ∫ ℝ N | u | p d x − b 2 q ∫ ℝ N | u | q d x .

The weak solutions of Equation (2.1) are critical points of J a b → ∈ C 1 ( H 1 ( ℝ N ) , ℝ ) . We denote the least energy by Θ a b → = inf N a b → J a b → , where N a b → : = { u ∈ H 1 ( ℝ N ) \ { 0 } : 〈 ( J a b → ) ′ ( u ) , u 〉 = 0 } is the Nehari manifold. The set of least energy solutions can be denoted by S a b → = { u ∈ H 1 ( ℝ N ) : J a b → ( u ) = Θ a b → , ( J a b → ) ′ ( u ) = 0 } . In particular, we set J ∞ : = J τ ∞ ς → ∞ , N ∞ : = N τ ∞ ς → ∞ and Θ ∞ : = Θ τ ∞ ς → ∞ .

Lemma 2.1. The functional J a b → satisfies that

1) There exist ρ > 0 and κ > 0 such that J a b → ( u ) > κ for all ‖ u ‖ = ρ ;

2) For u ≠ 0 , J a b → ( t u ) → − ∞ as t → + ∞ .

Similar to the proof of Lemma 2.4 in [20] , we have the following result.

Lemma 2.2. Let ϒ a b → : = { γ ∈ C ( [ 0 , 1 ] , H 1 ( ℝ N ) ) : γ ( 0 ) = 0 ,   J a b → ( γ ( 1 ) ) < 0 } , then

Θ a b → = inf u ∈ H 1 ( ℝ N ) \ { 0 } max t ≥ 0 J a b → ( t u ) = inf γ ∈ ϒ a b → max t ∈ [ 0 , 1 ] J a b → ( γ ( t ) ) > 0.

Lemma 2.3. Θ a b → is achieved and S a b → is compact in H 1 ( ℝ N ) .

Proof. For any u ∈ H 1 ( ℝ N ) , we choose the equivalent norm ‖ u ‖ 1 2 = ∫ ℝ N ( | ∇ u | 2 + a u 2 ) d x . Clearly, N a b → is not empty. Let u n ∈ N a b → with u n ≥ 0 and J a b → ( u n ) → Θ a b → as n → ∞ , By Lemma 1.4, there is u n * ∈ H r 1 ( ℝ N )

with u n * ≥ 0 such that ‖ u n * ‖ 1 ≤ ‖ u n ‖ 1 , | u n * | p = | u n | p , | u n * | q = | u n | q . Observe that ‖ u n * ‖ 1 2 ≤ b 1 | u n * | p p + b 2 | u n * | q q . If ‖ u n * ‖ 1 2 = b 1 | u n * | p p + b 2 | u n * | q q , then u n * ∈ N a b → . If ‖ u n * ‖ 1 2 < b 1 | u n * | p p + b 2 | u n * | q q , then there exists t n ∈ ( 0,1 ) such that t n u n * ∈ N a b → and Θ a b → ≤ J a b → ( t n u n * ) < q − 2 2 q ‖ u n ‖ 1 2 + p − q p q b 1 | u n | p p = J a b → ( u n ) → Θ a b → as n → ∞ . Hence J a b → ( t n u n * ) → Θ a b → as n → ∞ . Define w n : = t n u n * , then w n ∈ H r 1 ( ℝ N ) ∩ N a b → , w n ≥ 0 , and J a b → ( w n ) → Θ a b → as n → ∞ .

Clearly, { w n } ⊂ H 1 ( ℝ N ) is bounded. We assume w n ⇀ w in H r 1 ( ℝ N ) as n → ∞ up to a subsequence if necessary. By Lemma 1.5 and w n ∈ N a b → , we have ‖ w n ‖ 1 2 ≤ C ( ‖ w n ‖ 1 p + ‖ w n ‖ 1 q ) , which ensures that w ≠ 0 by letting n → ∞ . Due to the weakly lower semi-continuity of norm, we obtain ‖ w ‖ 1 2 ≤ b 1 | w | p p + b 2 | w | q q . Thus we have w ∈ N a b → .

In the end, we can obtain ( J a b → ) ′ ( w ) = 0 , where w ∈ S a b → is positive and radially symmetric. With similar arguments as aboves, S a b → is compact in H 1 ( ℝ N ) .

Lemma 2.4. Let a i > 0 , b i 1 , b i 2 > 0 , i = 1 , 2 .

1) If min { a 2 − a 1 , b 1 1 − b 2 1 , b 1 2 − b 2 2 } ≥ 0 , then Θ a 1 b → 1 ≤ Θ a 2 b → 2 ;

2) If min { a 2 − a 1 , b 1 1 − b 2 1 , b 1 2 − b 2 2 } ≥ 0 and max { a 2 − a 1 , b 1 1 − b 2 1 , b 1 2 − b 2 2 } > 0 , then Θ a 1 b → 1 < Θ a 2 b → 2 .

Lemma 2.5. If u is a ground state solution of

− Δ u + τ ∞ u = ς 1 ∞ | u | p − 2 u + ς 2 ∞ | u | q − 2 u ,   u ∈ H 1 ( ℝ N ) (2.2)

with the energy Θ ∞ . Setting z ( x ) = λ u ( a τ ∞ x ) , then z is a ground state solution of

− Δ z + a z = ( a ς 1 ∞ τ ∞ b 1 λ 2 − p ) b 1 | z | p − 2 z + ( a ς 2 ∞ τ ∞ b 2 λ 2 − q ) b 2 | z | q − 2 z ,   z ∈ H 1 ( ℝ N ) (2.3)

with the energy Θ ( λ ) = λ 2 ( a τ ∞ ) 1 − N 2 Θ ∞ .

Proof. Observe that u is a ground state solution of Equation (2.2) if and only if z is a ground state solution of Equation (2.3). Indeed,

− Δ z + a z = λ a τ ∞ ( − Δ u ( a τ ∞ x ) + τ ∞ u ( a τ ∞ x ) ) = ( a ς 1 ∞ τ ∞ λ 2 − p ) | z | p − 2 z + ( a ς 2 ∞ τ ∞ λ 2 − q ) | z | q − 2 z .

Furthermore, u ∈ N ∞ if and only if z ∈ N ( λ ) . Hence Θ ( λ ) = λ 2 ( a τ ∞ ) 1 − N 2 Θ ∞ .

Lemma 2.6. Assume that a ≤ τ ∞ , b → ≥ ς → ∞ . Then m ( a , b → ) Θ a b → ≤ Θ ∞ .

Proof. Note that if λ > 0 satisfies max { a ς 1 ∞ τ ∞ b 1 λ 2 − p , a ς 2 ∞ τ ∞ b 2 λ 2 − q } ≤ 1 , we have Θ a b → ≤ Θ ( λ ) . By the definition of m ( a , b → ) , we can find two cases:

( ς 2 ∞ b 2 ) ≤ ( a τ ∞ ) q − p p − 2 ( ς 1 ∞ b 1 ) q − 2 p − 2 (2.4)

or

( ς 1 ∞ b 1 ) < ( a τ ∞ ) p − q q − 2 ( ς 2 ∞ b 2 ) p − 2 q − 2 . (2.5)

If (2.4) holds, we choose λ = ( a ς 1 ∞ τ ∞ b 1 ) 1 p − 2 , then Θ ( λ ) = ( a τ ∞ ) p p − 2 − N 2 ⋅ ( ς 1 ∞ b 1 ) 2 p − 2 Θ ∞ . Thus we get m ( a , b → ) Θ a b → ≤ Θ ∞ .

If (2.5) holds, we choose λ = ( a ς 2 ∞ τ ∞ b 2 ) 1 q − 2 , then Θ ( λ ) = ( a τ ∞ ) q q − 2 − N 2 ⋅ ( ς 2 ∞ b 2 ) 2 q − 2 Θ ∞ . Thus we get m ( a , b → ) Θ a b → ≤ Θ ∞ .

Lemma 2.7. If τ < τ ∞ ,   ς → v ≥ ς → ∞ , then m ( τ , ς → v ) > 1 and Θ τ ς → v < Θ ∞ . If τ w ≤ τ ∞ ,   ς → > ς → ∞ , then m ( τ w , ς → ) ≥ 1 and Θ τ w ς → < Θ ∞ .

Proof. Choose a = τ ,   b j = ς j v ,   j = 1 , 2 in Equation (2.1), Equations (2.3), (2.4) and (2.5), respectively. By the definition of m ( τ , ς → v ) , we have m ( τ , ς → v ) > 1 . By Lemma 2.6, we have Θ τ ς → v < Θ ∞ .

Similarly, we choose a = τ w ,   b j = ς j ,   j = 1 , 2 in Equation (2.1), Equations (2.3), (2.4) and (2.5), respectively. By the definition of m ( τ w , ς → ) , we have m ( τ w , ς → ) ≥ 1 . If (2.4) holds, we choose λ = ( τ w ς 1 ∞ τ ∞ ς 1 ) 1 p − 2 , then Θ τ w ς → ≤ Θ ( λ ) ≤ Θ ∞ by Lemma 2.4 and Lemma 2.5. If ς 1 > ς 1 ∞ , then Θ τ w ς → ≤ Θ ( λ ) < Θ ∞ by Lemma 2.5. If ς 2 > ς 2 ∞ , then Θ τ w ς → < Θ ( λ ) ≤ Θ ∞ by Lemma 2.4. Hence Θ τ w ς → < Θ ∞ . If (2.5) holds, we choose λ = ( τ w ς 2 ∞ τ ∞ ς 2 ) 1 q − 2 , then Θ τ w ς → ≤ Θ ( λ ) ≤ Θ ∞ . If ς 1 > ς 1 ∞ , then Θ τ w ς → < Θ ( λ ) ≤ Θ ∞ . If ς 2 > ς 2 ∞ , then Θ τ w ς → ≤ Θ ( λ ) < Θ ∞ . Thus Θ τ w ς → < Θ ∞ . Lemma 2.8. There exist constants C , c > 0 such that for every u ∈ S a b → , u ( x ) ≤ C e − c | x | for all x ∈ ℝ N .

Proof. Let a ′ : = a − b 1 u p − 2 − b 2 u q − 2 , we obtain − Δ u + a ′ u = 0 . For R large enough, we get 2 a ′ ≥ a for | x | ≥ R . Define ϕ ( x ) = C 1 e − c 2 | x | or ϕ ( s ) = C 1 e − c 2 s , where C 1 > 0 , s = | x | . Choose C₁ large enough such that ϕ ( x ) ≥ u ( x ) for | x | = R . Since − Δ ϕ ( x ) + a ′ ϕ ( x ) ≥ ( a 2 − c 2 2 ) ϕ ( s ) , we choose 0 < c 2 ≤ a 2 such that − Δ ϕ ( x ) + a ′ ϕ ( x ) ≥ 0 for | x | ≥ R . Therefore,

( − Δ ϕ ( x ) + a ′ ϕ ( x ) ≥ − Δ u + a ′ u ∀ | x | ≥ R ϕ ( x ) ≥ u ( x ) ∀ | x | = R

By comparison principle, ϕ ( x ) ≥ u ( x ) for all | x | ≥ R , then u ( x ) ≤ C 1 e − c 2 | x | for all | x | ≥ R . For C₁ large enough, we get that u ( x ) ≤ C 1 e − c 2 | x | for all | x | < R . Thus u ( x ) ≤ C 1 e − c 2 | x | for all x ∈ ℝ N .

In this subsection, we consider the following equation for p , q ∈ ( 2,2 * ) and p > q ,

− Δ u + V ε a ( x ) u = W 1 ε b 1 ( x ) | u | p − 2 u + W 2 ε b 2 ( x ) | u | q − 2 u ,   u ∈ H 1 ( ℝ N ) , (2.6)

where τ ≤ a ≤ τ ∞ , ς → ≥ b → ≥ ς → ∞ , V ε a ( x ) : = V a ( ε x ) : = max { a , V ( x ) } and W j ε b j ( x ) : = W j b j ( ε x ) : = min { b j , W j ( x ) } , j = 1 , 2 .

For each u ∈ H 1 ( ℝ N ) , the energy functional associated to Equation (2.6) is

J ε a b → ( u ) = 1 2 ∫ ℝ N ( | ∇ u | 2 + V ε a ( x ) u 2 ) d x − 1 p ∫ ℝ N   W 1 ε b 1 ( x ) | u | p d x       − 1 q ∫ ℝ N   W 2 ε b 2 ( x ) | u | q d x .

The weak solutions of Equation (2.6) are critical points of J ε a b → ∈ C 1 ( H 1 ( ℝ N ) , ℝ ) . We denote the least energy by Θ ε a b → = inf N ε a b → J ε a b → , where N ε a b → : = { u ∈ H 1 ( ℝ N ) \ { 0 } : 〈 ( J ε a b → ) ′ ( u ) , u 〉 = 0 } is the Nehari manifold. The set of least energy solutions was denoted by S ε a b → = { u ε ∈ H 1 ( ℝ N ) : J ε a b → ( u ε ) = Θ ε a b → ,   ( J ε a b → ) ′ ( u ε ) = 0 } . In particular, we set J ε ∞ : = J ε τ ∞ ς → ∞ ,     N ε ∞ : = N ε τ ∞ ς → ∞ ,     Θ ε ∞ : = Θ ε τ ∞ ς → ∞ ;

V ε ∞ : = V ε τ ∞ ,     W j ε ∞ : = W j ε ς j ∞ ,     j = 1 , 2.

Lemma 2.9. The functional J ε a b → satisfies that

1) There exist ρ > 0 and κ > 0 both dependent on N , p , q , τ , ς → and independent of a , b → such that J ε a b → ( u ) > κ for all ‖ u ‖ = ρ ;

2) For u ≠ 0 , J ε a b → ( t u ) → − ∞ as t → + ∞ .

Lemma 2.10. Let ϒ ε a b → : = { γ ∈ C ( [ 0 , 1 ] , H 1 ( ℝ N ) ) : γ ( 0 ) = 0 ,   J ε a b → ( γ ( 1 ) ) < 0 } , then

Θ ε a b → = inf u ∈ H 1 ( ℝ N ) \ { 0 } max t ≥ 0 J ε a b → ( t u ) = inf γ ∈ ϒ ε a b → max t ∈ [ 0 , 1 ] J ε a b → ( γ ( t ) ) > 0.

Lemma 2.11. If J ε ∞ has a ( P S ) c sequence, then either c = 0 or c ≥ Θ ε ∞ . Furthermore, Θ ε ∞ ≥ Θ ∞ .

Proof. Let { u n } ⊂ H 1 ( ℝ N ) is a ( P S ) c sequence of J ε ∞ , then J ε ∞ ( u n ) → c and ( J ε ∞ ) ′ ( u n ) → 0 in H − 1 ( ℝ N ) as n → ∞ . We will show that c ≥ Θ ε ∞ when c ≠ 0 . Since { u n } ⊂ H 1 ( ℝ N ) is bounded, we may assume u n ⇀ u in H 1 ( ℝ N ) as n → ∞ . Hence ( J ε ∞ ) ′ ( u ) = 0 . Set y n : = u n − u . By Lemma 1.32 in [23] , we obtain

J ε ∞ ( y n ) → c − J ε ∞ ( u )   as   n → ∞ . (2.7)

For all φ ∈ H 1 ( ℝ N ) , we have

〈 ( J ε ∞ ) ′ ( y n ) , φ 〉 − 〈 ( J ε ∞ ) ′ ( u n ) , φ 〉 = ∫ ℝ N     W 1 ε ∞ ( x ) ( | u n | p − 2 u n − | y n | p − 2 y n − | u | p − 2 u ) φ d x         + ∫ ℝ N     W 2 ε ∞ ( x ) ( | u n | q − 2 u n − | y n | q − 2 y n − | u | q − 2 u ) φ d x . (2.8)

For any σ > 0 , there is R > 0 such that ∫ | x | > R | u | p d x < σ p and ∫ | x | > R | u | q d x < σ q . By mean value theorem and Hölder inequality, we obtain ∫ | x | > R | ( | u n | p − 2 u n − | y n | p − 2 y n ) φ | d x ≤ C σ ‖ φ ‖ . Furthermore, by Hölder inequality again, we get that ∫ | x | > R | | u | p − 2 u φ | d x ≤ C σ ‖ φ ‖ . Thus

| ∫ | x | > R     W 1 ε ∞ ( x ) ( | u n | p − 2 u n − | y n | p − 2 y n − | u | p − 2 u ) φ d x | ≤ C σ ‖ φ ‖ . (2.9)

Similarly,

| ∫ | x | > R     W 2 ε ∞ ( x ) ( | u n | q − 2 u n − | y n | q − 2 y n − | u | q − 2 u ) φ d x | ≤ C σ ‖ φ ‖ . (2.10)

By Lemma 1.5, we obtain | u n | μ − 2 u n − | u | μ − 2 u → 0 in L l o c μ μ − 1 ( ℝ N ) as n → ∞ with μ = p , q , respectively. Hence

| ∫ | x | ≤ R     W 1 ε ∞ ( x ) ( | u n | p − 2 u n − | y n | p − 2 y n − | u | p − 2 u ) φ d x | = o ( 1 ) ‖ φ ‖ . (2.11)

Similarly,

| ∫ | x | ≤ R     W 2 ε ∞ ( x ) ( | u n | q − 2 u n − | y n | q − 2 y n − | u | q − 2 u ) φ d x | = o ( 1 ) ‖ φ ‖ . (2.12)

By (2.8) - (2.12), we get that ( J ε ∞ ) ′ ( y n ) → 0 in H − 1 ( ℝ N ) as n → ∞ .

For all n ∈ ℤ + , if y n ≠ 0 , there is t n > 0 such that t n y n ∈ N ε ∞ . Thus

J ε ∞ ( t n y n ) ≥ Θ ε ∞ (2.13)

and 0 = 〈 ( J ε ∞ ) ′ ( t n y n ) , t n y n 〉 . By o ( 1 ) = 〈 ( J ε ∞ ) ′ ( y n ) , y n 〉 , then we obtain

o ( 1 ) = ( 1 − t n p − 2 ) ∫ ℝ N     W 1 ε ∞ ( x ) | y n | p d x + ( 1 − t n q − 2 ) ∫ ℝ N     W 2 ε ∞ ( x ) | y n | q d x . (2.14)

Moreover, ‖ y n ‖ 2 ≤ C ∫ ℝ N ( | ∇ y n | 2 + V ε ∞ ( x ) y n 2 ) d x ≤ C ( | y n | p p + | y n | q q ) + o ( 1 ) . If | y n | p → 0 and | y n | q → 0 as n → ∞ , we can get u n → u in H 1 ( ℝ N ) as n → ∞ and c ≥ Θ ε ∞ . If | y n | p ≥ σ > 0 or | y n | q ≥ σ > 0 in (2.14), we obtain t n → 1 as n → ∞ . Hence by (2.7), we have J ε ∞ ( t n y n ) → c − J ε ∞ ( u ) as n → ∞ . By (2.13), we get that c ≥ Θ ε ∞ . If there is y n k ≡ 0 , then J ε ∞ ( u ) = c ≠ 0 and u ∈ N ε ∞ . Hence c ≥ Θ ε ∞ .

Observe that J ε ∞ ( u ) ≥ J ∞ ( u ) for all u ∈ H 1 ( ℝ N ) . According to Lemma 2.2 and Lemma 2.10, we obtain Θ ε ∞ ≥ Θ ∞ .

Similar to the proof of Lemma 2.11, we also have the following result.

Lemma 2.12. If J ε a b → has a ( P S ) c sequence, then either c = 0 or c ≥ Θ ε a b → .

Lemma 2.13. For all c < Θ ε ∞ , J ε a b → satisfies ( P S ) c condition.

Proof. Let { u n } ⊂ H 1 ( ℝ N ) is a ( P S ) c sequence of J ε a b → , then J ε a b → ( u n ) → c and ( J ε a b → ) ′ ( u n ) → 0 in H − 1 ( ℝ N ) as n → ∞ . We assume u n ⇀ u in H 1 ( ℝ N ) as n → ∞ . Hence ( J ε a b → ) ′ ( u ) = 0 . Set y n : = u n − u . Due to the proof of Lemma 2.11, we obtain

J ε a b → ( y n ) → c − J ε a b → ( u ) ,     ( J ε a b → ) ′ ( y n ) → 0     in   H − 1 ( ℝ N )   as   n → ∞ . (2.15)

Next, we will show that J ε ∞ ( y n ) → c − J ε a b → ( u ) and ( J ε ∞ ) ′ ( y n ) → 0 in H − 1 ( ℝ N ) as n → ∞ . By definition, we get that for all σ > 0 , there is R ¯ > 0 such that for all | x | > R ¯ ,

| V ε ∞ ( x ) − V ε a ( x ) | ≤ σ ,   | W j ε ∞ ( x ) − W j ε b j ( x ) | ≤ σ ,     j = 1,2. (2.16)

Thus by (2.16), we have

| J ε ∞ ( y n ) − J ε a b → ( y n ) | ≤ σ ( 1 2 | y n | 2 2 + 1 p | y n | p p + 1 q | y n | q q ) + C ( | y n | L 2 ( B R ¯ ) 2 + | y n | L p ( B R ¯ ) p + | y n | L q ( B R ¯ ) q ) ,

which together with Lemma 1.5 and (2.15) imply that

J ε ∞ ( y n ) → c − J ε a b → ( u )   as   n → ∞ . (2.17)

Similarly, by Lemma 1.5 and (2.15) again, we have

( J ε ∞ ) ′ ( y n ) → 0   in   H − 1 ( ℝ N )   as   n → ∞ . (2.18)

By (2.17) and (2.18), we obtain { y n } is a ( P S ) c − J ε a b → ( u ) sequence of J ε ∞ . By Lemma 2.11, either c = J ε a b → ( u ) or c ≥ J ε a b → ( u ) + Θ ε ∞ . The latter contradicts our assumption c < Θ ε ∞ . Hence c = J ε a b → ( u ) and J ε a b → ( u n ) → J ε a b → ( u )   as   n → ∞ . (2.19)

Now we prove that u n → u in H 1 ( ℝ N ) as n → ∞ . Since

o ( 1 ) = 〈 ( J ε a b → ) ′ ( u n ) , u n 〉 , we get J ε a b → ( u n ) = q − 2 2 q ∫ ℝ N ( | ∇ u n | 2 + V ε a ( x ) u n 2 ) d x + p − q p q ∫ ℝ N     W 1 ε b 1 ( x ) | u n | p d x + o ( 1 ) . By 0 = 〈 ( J ε a b → ) ′ ( u ) , u 〉 , we obtain J ε a b → ( u ) = q − 2 2 q ∫ ℝ N ( | ∇ u | 2 + V ε a ( x ) u 2 ) d x + p − q p q ∫ ℝ N     W 1 ε b 1 ( x ) | u | p d x . By Lemma 1.6, we assume there exist R > 0 and δ > 0 such that ∫ B R ( x n ) | y n | p d x ≥ δ > 0 for some x n ∈ ℝ N . Moreover,

l i m n → ∞ J ε a b → ( y n ) ≥ l i m n → ∞ ( p − q p q ∫ B R ( x n )     W 1 ε b 1 ( x ) | y n | p d x + o ( 1 ) ) ≥ l i m n → ∞ ( C ∫ B R ( x n ) | y n | p d x + o ( 1 ) ) ≥ C δ > 0,

which is impossible. By (2.19), we conclude that ‖ u n ‖ → ‖ u ‖ as n → ∞ . Thus u n → u in H 1 ( ℝ N ) as n → ∞ .

Lemma 2.14. lim sup ε → 0 Θ ε a b → ≤ Θ a ˜ b ˜ → , where a ˜ = V a ( 0 ) , b ˜ j = W j b j ( 0 ) , j = 1 , 2 , b ˜ → = ( b ˜ 1 , b ˜ 2 ) . Meanwhile, if V ( 0 ) ≤ a , W j ( 0 ) ≥ b j , j = 1 , 2 , then lim ε → 0 Θ ε a b → = Θ a b → .

Proof. Setting V ˜ ε ( x ) : = V ε a ( x ) − a ˜ and W ˜ j ε ( x ) : = b ˜ j − W j ε b j ( x ) , j = 1 , 2 , we have

V ˜ ε ( x ) → 0,     W ˜ j ε ( x ) → 0,   j = 1,2     a .e .   on   ℝ N   as   ε → 0. (2.20)

Furthermore,

J ε a b → ( u ) − J a ˜ b ˜ → ( u ) = ∫ ℝ N     V ˜ ε ( x ) u 2 d x 2 + ∫ ℝ N     W ˜ 1 ε ( x ) | u | p d x p + ∫ ℝ N     W ˜ 2 ε ( x ) | u | q d x q . (2.21)

Due to Lemma 2.3, there is α ∈ S a ˜ b ˜ → satisfying J a ˜ b ˜ → ( α ) = Θ a ˜ b ˜ → for α ∈ N a ˜ b ˜ → . Let t ε > 0 such that t ε α ∈ N ε a b → , we obtain

max t ≥ 0 J ε a b → ( t α ) = J ε a b → ( t ε α ) ≥ Θ ε a b → . (2.22)

Observe that lim t → + ∞ J ε a b → ( t α ) = − ∞ , there is T > 0 such that

J ε a b → ( t α ) < 0 ,   ∀ t > T . (2.23)

Combining (2.22) with (2.23), we have t ε ≤ T . Let t ε → t 0 as ε → 0 . By applying (2.20) - (2.22) and the Lebesgue dominated convergence theorem, we obtain Θ ε a b → ≤ J ε a b → ( t ε α ) → J a ˜ b ˜ → ( t 0 α ) ≤ J a ˜ b ˜ → ( α ) = Θ a ˜ b ˜ → as ε → 0 . Hence lim sup ε → 0 Θ ε a b → ≤ Θ a ˜ b ˜ → .

In the end, a ˜ = a and b ˜ j = b j , j = 1 , 2 when V ( 0 ) ≤ a and W j ( 0 ) ≥ b j , j = 1 , 2 , namely, for all x ∈ ℝ N , we obtain V ˜ ε ( x ) ≥ 0 , W ˜ j ε ( x ) ≥ 0 , j = 1 , 2 . By Lemma 2.2, Lemma 2.10 and (2.21), we have Θ ε a b → ≥ Θ a ˜ b ˜ → . According to Θ a ˜ b ˜ → ≤ lim inf ε → 0 Θ ε a b → ≤ lim sup ε → 0 Θ ε a b → ≤ Θ a ˜ b ˜ → , we obtain lim ε → 0 Θ ε a b → = Θ a ˜ b ˜ → = Θ a b → .

Lemma 2.15. If τ ≤ a < τ ∞ ,   ς → ≥ b → ≥ ς → ∞ or τ ≤ a ≤ τ ∞ ,   ς → > b → ≥ ς → ∞ , then there is ε a b → > 0 such that Θ ε a b → is achieved at u ε a b → > 0 for all ε ≤ ε a b → .

Proof. By Lemma 2.7, we have Θ a ˜ b ˜ → < Θ ∞ , where a ˜ = V a ( 0 ) , b ˜ j = W j b j ( 0 ) , j = 1 , 2 . By Lemma 2.11 and Lemma 2.14, there is ε a b → > 0 such that Θ ε a b → < Θ ∞ ≤ Θ ε ∞ for all ε ≤ ε a b → . By Lemma 2.13, J ε a b → satisfies the ( P S ) Θ ε a b → condition for all ε ≤ ε a b → , which combined Lemma 2.9 with Lemma 2.10, we have Θ ε a b → is achieved at u ε a b → ∈ H 1 ( ℝ N ) . We set u ε a b → is a ground state solution of Equation (2.6). If ( u ε a b → ) ± ≠ 0 , by

0 = 〈 ( J ε a b → ) ′ ( u ε a b → ) , ( u ε a b → ) ± 〉 = 〈 ( J ε a b → ) ′ ( ( u ε a b → ) ± ) , ( u ε a b → ) ± 〉 implies that

( u ε a b → ) ± ∈ N ε a b → . Thus Θ ε a b → = J ε a b → ( u ε a b → ) = J ε a b → ( ( u ε a b → ) + ) + J ε a b → ( ( u ε a b → ) − ) ≥ 2 Θ ε a b → ,

which is impossible. Hence u ε a b → does not change the sign. Then we may assume u ε a b → ≥ 0 . By the elliptic regularity theory, u ε a b → ∈ C 2 ( ℝ N ) . By strong maximum principle, we have u ε a b → > 0 .

Without loss of generality, we assume x j v = 0 . Then V ( 0 ) = τ , W j ( 0 ) = ς j v , j = 1 , 2 .

Lemma 3.1. Equation (3.1) has at least m pairs of solutions.

Proof. We choose a = τ ,   b j = ς j v ,   j = 1,2 in Equation (2.1) and by Lemma 2.3 and Lemma 2.8, there are u ∈ S τ ς → v and u > 0 . Let s > 0 , ζ s ∈ C 0 ∞ ( ℝ + ) satisfies ζ s ( t ) = 0 if t ≥ s + 1 and ζ s ( t ) = 1 if t ≤ s with | ζ ′ s ( t ) | ≤ 1 . Assume u s ( x ) : = ζ s ( | x | ) u ( x ) for x ∈ ℝ N . By ‖ u s − u ‖ → 0 as s → ∞ , we get that u s → u in H 1 ( ℝ N ) as s → ∞ and u s → u in L μ ( ℝ N ) for μ ∈ [ 2,2 * ] as s → ∞ . There is a unique t s > 0 such that t s u s ∈ N τ ς → . Therefore,

max t ≥ 0 J τ ς → v ( t u s ) = p − 2 2 p t s p ∫ ℝ N     ς 1 v | u s | p d x + q − 2 2 q t s q ∫ ℝ N     ς 2 v | u s | q d x → p − 2 2 p ∫ ℝ N     ς 1 v | u | p d x + q − 2 2 q ∫ ℝ N     ς 2 v | u | q d x   ( s → ∞ ) = max t ≥ 0 J τ ς → v ( t u ) = J τ ς → v ( u ) = Θ τ ς → v . (3.2)

Furthermore,

V ( ε x ) → V ( 0 ) = τ ,   W j ( ε x ) → W j ( 0 ) = ς j v ,   j = 1,2   as   ε → 0 (3.3)

uniformly of x on any bounded set. There is a unique t s ε > 0 such that t s ε u s ∈ N ε . Observe that t s ε → t s as ε → 0 . Hence (3.2) and (3.3) imply that

max t ≥ 0 J ε ( t u s ) = p − 2 2 p t s ε p ∫ | x | ≤ s + 1     W 1 ( ε x ) | u s | p d x + q − 2 2 q t s ε q ∫ | x | ≤ s + 1     W 2 ( ε x ) | u s | q d x → p − 2 2 p t s p ∫ | x | ≤ s + 1     ς 1 v | u s | p d x + q − 2 2 q t s q ∫ | x | ≤ s + 1     ς 2 v | u s | q d x   ( ε → 0 ) = max t ≥ 0 J τ ς → v ( t u s ) → Θ τ ς → v   ( s → ∞ ) . (3.4)

By Lemma 2.7, m ( τ , ς → v ) > 1 . We choose m v = m ( τ , ς → v ) . For the maximal integer m ∈ ℤ + with m < m v , we have m ≥ 1 . Define ξ s l ( x ) : = u s ( x 1 − 2 l ( s + 1 ) , x 2 , ⋯ , x N ) for l = 0,1, ⋯ , m − 1 , and set E s m : = span { ξ s l ( x ) : l = 0,1, ⋯ , m − 1 } . Clearly, ( ξ s i , ξ s j ) = 0 if i ≠ j . Hence dim E s m = m . Combining (3.2) with (3.3) again, for all l = 1,2, ⋯ , m − 1 , we have

max t ≥ 0 J ε ( t ξ s l ) = p − 2 2 p t l ε p ∫ | x | ≤ s + 1     W 1 ( ε x ) | ξ s l | p d x + q − 2 2 q t l ε q ∫ | x | ≤ s + 1     W 2 ( ε x ) | ξ s l | q d x → p − 2 2 p t l p ∫ | x | ≤ s + 1     ς 1 v | ξ s l | p d x + q − 2 2 q t l q ∫ | x | ≤ s + 1 ς 2 v | ξ s l | q d x   ( ε → 0 ) = max t ≥ 0 J τ ς → v ( t u s ) → Θ τ ς → v   ( s → ∞ ) ,

where t l ε and t l are the unique constants satisfying t l ε ξ s l ∈ N ε and t l ξ s l ∈ N τ ς → v , respectively, and t l ε → t l as ε → 0 . Therefore, for all δ > 0 , there are s δ > 0 and ε δ > 0 such that for all l = 0,1, ⋯ , m − 1 , we get

max t ≥ 0 J ε ( t ξ s l ) ≤ Θ τ ς → v + δ ,   ∀ s ≥ s δ ,   ∀ ε ≤ ε δ . (3.5)

Let u = t 0 ξ s 0 + t 1 ξ s 1 + ⋯ + t m − 1 ξ s ( m − 1 ) for any u ∈ E s m , where t 0 , ⋯ , t m − 1 ∈ ℝ . According to (3.5), for all s ≥ s δ and ε ≤ ε δ , we obtain J ε ( u ) = J ε ( t 0 ξ s 0 ) + J ε ( t 1 ξ s 1 ) + ⋯ + J ε ( t m − 1 ξ s ( m − 1 ) ) ≤ m ( Θ τ ς → v + δ ) . Thus sup u ∈ E s m J ε ( u ) ≤ m ( Θ τ ς → v + δ ) for all s ≥ s δ and ε ≤ ε δ . By Lemma 2.6,

m Θ τ ς → v < Θ ∞ . We choose 0 < δ < Θ ∞ m − Θ τ ς → v , then there exist s m > 0 and ε m > 0 such that

sup u ∈ E s m J ε ( u ) < Θ ∞ ,   ∀ s ≥ s m ,   ∀ ε ≤ ε m . (3.6)

Next, we shall define constants c 1 , c 2 , ⋯ , c m and prove that they are critical values of J ε . Consider the symmetric group ℤ 2 = { id , − id } and we denote by Σ : = { A ⊂ H 1 ( ℝ N ) : A   is   closed   and   A = − A } and

H : = { h ∈ C ( H 1 ( ℝ N ) , H 1 ( ℝ N ) ) : h   is   an   odd   homeomorphism } .

For any A ∈ Σ , we define a version of Benci pseudo-index of A as follows, i ( A ) : = min h ∈ H gen ( h ( A ) ∩ ∂ B ρ ) , where gen ( A ) : = inf { n : ∃ g ∈ C ( A , ℝ n \ { 0 } )   and   g   is   odd } is the Krasnoselskii genus of A, and ρ > 0 is a constant given in Lemma 2.9. Let c l : = inf i ( A ) ≥ l sup u ∈ A J ε ( u ) , l = 1,2, ⋯ , m . Observe that c 1 ≤ c 2 ≤ ⋯ ≤ c m . For any A ∈ Σ and i ( A ) ≥ 1 , we have gen ( A ∩ ∂ B ρ ) ≥ 1 , then A ∩ ∂ B ρ is not empty. By Lemma 2.9, it follows from sup u ∈ A J ε ( u ) > κ that c 1 ≥ κ , where κ is defined in Lemma 2.9.

Noticing that gen ( A ) satisfies dimension property in [24] , for all h ∈ H , we have gen ( h ( E s m ) ∩ ∂ B ρ ) = dim E s m = m . Hence i ( E s m ) = m , then we obtain c m ≤ sup u ∈ E s m J ε ( u ) . Combining (3.6) with Lemma 2.11, we have

κ ≤ c 1 ≤ c 2 ≤ ⋯ ≤ c m ≤ sup u ∈ E s m J ε ( u ) < Θ ∞ ≤ Θ ε ∞ . (3.7)

Let c 0 : = κ , c ∞ : = sup u ∈ E s m J ε ( u ) , J ε c : = { u ∈ H 1 ( ℝ N ) : J ε ( u ) ≤ c } , and Ψ c : = { u ∈ H 1 ( ℝ N ) : J ε ( u ) = c ,   J ′ ε ( u ) = 0 } . Clearly, J ε is an even functional. For all c ∈ [ c 0 , c ∞ ] , we obtain

J ε c ∈ Σ   and   Ψ c ∈ Σ . (3.8)

By using (3.7) and Lemma 2.13, for all c ∈ [ c 0 , c ∞ ] , J ε satisfies ( P S ) c condition and

Ψ c   is   compact   in   H 1 ( ℝ N ) . (3.9)

Set ( Ψ c ) ι : = { u ∈ H 1 ( ℝ N ) : dist ( u , Ψ c ) < ι } , where ι > 0 for any c ∈ [ c 0 , c ∞ ] , then we choose δ = ι 4 , we have there is ε ˜ > 0 such that

‖ J ′ ε ( u ) ‖ ≥ 8 ε ˜ δ ,   ∀ u ∈ J ε − 1 ( [ c − 2 ε ˜ , c + 2 ε ˜ ] ) \ ( Ψ c ) ι 2 ¯ . (3.10)

Let P : = H 1 ( ℝ N ) \ ( Ψ c ) ι , then P 2 δ = H 1 ( ℝ N ) \ ( Ψ c ) ι 2 ¯ . By (3.10), we have ‖ J ′ ε ( u ) ‖ ≥ 8 ε ˜ δ for all u ∈ J ε − 1 ( [ c − 2 ε ˜ , c + 2 ε ˜ ] ) ∩ P 2 δ . By Lemma 2.3 in [23] , there is η ¯ ∈ C ( [ 0,1 ] × H 1 ( ℝ N ) , H 1 ( ℝ N ) ) such that for all t ∈ [ 0,1 ] , η ¯ ( t , ⋅ ) is

an odd homeomorphism of H 1 ( ℝ N ) and η ¯ ( 1, J ε c + ε ˜ ∩ P ) ⊂ J ε c − ε ˜ . Set η : = η ¯ ( 1, ⋅ ) , then η is an odd homeomorphism of H 1 ( ℝ N ) and

η ( J ε c + ε ˜ \ ( Ψ c ) ι ) ⊂ J ε c − ε ˜ . (3.11)

For any A ∈ Σ and A ⊂ J ε c 0 , it follows from J ε ( u ) > κ for all u ∈ ∂ B ρ that A ∩ ∂ B ρ = ∅ . Hence gen ( A ∩ ∂ B ρ ) = 0 and

i ( A ) = min h ∈ H gen ( h ( A ) ∩ ∂ B ρ ) = 0. (3.12)

Moreover,

E s m ⊂ J ε c ∞   and   i ( E s m ) = m ≥ 1. (3.13)

By applying the Theorem 1.4 in [24] , (3.8), (3.9) and (3.11) - (3.13), we have c 1 , ⋯ , c m are critical values of J ε , and gen ( Ψ c ) ≥ s + 1 , if c : = c k = c k + 1 = ⋯ = c k + s with k ≥ 1 and k + s ≤ m . Since J ε is even, then J ε has at least m pairs of critical points being solutions of Equation (3.1).

Lemma 3.2. Equation (3.1) has at least one positive and one negative ground state solutions for m ≥ 1 and has at least a pair of sign-changing solutions for m ≥ 2 .

Proof. If a = τ , b j = ς j , j = 1 , 2 in Equation (2.1), then a ˜ = V τ ( 0 ) = V ( 0 ) = τ , b ˜ j = W j ς j ( 0 ) = W j ( 0 ) = ς j v , j = 1 , 2 . By Lemma 2.9 and Lemma 2.13, J ε τ ς → has a ( P S ) Θ ε sequence and satisfies ( P S ) Θ ε condition. By Lemma 2.15, there exists ε 0 > 0 such that Θ ε is achieved at u ε > 0 for all ε ≤ ε 0 . Thus u ε and − u ε are positive and negative ground state solutions of Equation (3.1), respectively.

Let α ± ∈ S τ ς → v with α + > 0 . Define α s ± ( x ) : = ζ s ( | x | ) α ± ( x ) for x ∈ ℝ N , where ζ s is given in Lemma 3.1. Then α s ± → α ± in H 1 ( ℝ N ) as s → ∞ .

Choose s > 0 , x s ∈ ℝ with | x s | large enough and dist ( B s + 1 ( 0 ) ¯ , B s + 1 ( x s ) ¯ ) > 0 .

Let t s ± ∈ ℝ such that u s + : = t s + α s + ∈ N ε and u s − : = t s − α s − ( ⋅ − x s ) ∈ N ε . Then u s + ≥ 0 and u s − ≤ 0 , supp u s + ∩ supp u s − is empty and u s : = u s + + u s − ∈ N ε . Define L ε : = { u ∈ N ε : u ± ∈ N ε } , then we have u s ∈ L ε . Define l ε : = inf u ∈ L ε J ε ( u ) , then l ε ≥ 2 Θ ε > 0 .

Next, we will prove l ε < Θ ε ∞ for ε small enough. Due to l ε ≤ J ε ( u s ) , we get

l i m ε → 0 l ε ≤ l i m s → ∞ l i m ε → 0 J ε ( u s ) . (3.14)

Observe that t s ± → 1 as s → ∞ and

J ε ( u s ) = J ε ( u s + ) + J ε ( u s − ) → J τ ς → v ( u s + ) + J τ ς → v ( u s − )   ( ε → 0 ) = J τ ς → v ( t s + α s + ) + J τ ς → v ( t s − α s − ( x − x s ) ) → 2 J τ ς → v ( α )   ( s → ∞ ) = 2 Θ τ ς → v . (3.15)

By m ≥ 2 and combining Lemma 2.6 with Lemma 2.11, we have

2 Θ τ ς → v < Θ ∞ ≤ Θ ε ∞ . (3.16)

By (3.14) - (3.16), we get that l ε < Θ ε ∞ for ε small enough, which implies J ε satisfies ( P S ) l ε condition for ε small enough.

Now we show that there is a ( P S ) l ε sequence of J ε . Since { u n } ⊂ N ε , then u n ± ∈ N ε . We assume u n ± ⇀ u ± in H 1 ( ℝ N ) with u ± ≠ 0 . There exist t + > 0 and t − < 0 such that t ± u ± ∈ N ε , u = t + u + + t − u − ∈ L ε , we get J ε ( u ) = l ε . Assume by contradiction that if u ˜ is not a sign-changing solution of Equation (3.1), there exists φ ∈ H 1 ( ℝ N ) such that 〈 J ′ ε ( u ˜ ) , φ 〉 ≤ − 1 / 2 . We choose ε ^ > 0 small enough, satisfying 〈 J ′ ε ( t u + + h u − + ρ φ ) , φ 〉 ≤ − 1 / 2 for all | t − 1 | + | h − 1 | + | ρ | ≤ ε ^ . Let η be a cut off function such that

η ( t , h ) = ( 1 | t − 1 | ≤ 1 / 2 ε ^     and   | h − 1 | ≤ 1 / 2 ε ^ 0 | t − 1 | ≥ ε ^     or     | h − 1 | ≥ ε ^ .

Then J ε ( t u + + h u − + ε ^ η ( t , h ) φ ) ≤ J ε ( u ˜ ) = l ε . Hence max 0 ≤ t , h ≤ 2 J ε ( t u + + h u − + ε ^ η ( t , h ) φ ) < l ε . By a degree theory argument in [25] , we find a , b ∈ ( 0,2 ) such that u ˜ : = a u + + b u − + ε ^ η ( a , b ) φ ∈ L ε and J ε ( u ˜ ) < l ε , which contradits that the defination of l ε .

In the end, we prove l ε is achieved at some u ε ∈ L ε . Let { u n } ⊂ L ε and J ε ( u n ) → l ε as n → ∞ . By Ekeland vainational principle there is { u ¯ n } ⊂ L ε such that J ε ( u ¯ n ) → l ε and J ′ ε ( u ¯ n ) → 0 as n → ∞ , then ‖ u ¯ n − u n ‖ → 0 as n → ∞ . Hence { u ¯ n } is a ( P S ) l ε sequence of J ε . Going of necessary to a subsequence, for ε small enough we may assume u ¯ n → u ε in H 1 ( ℝ N ) as n → ∞ . Hence J ε ( u ε ) = l ε and J ′ ε ( u ε ) = 0 . Then u ε ∈ N ε , we have u ε ± ≠ 0 , u ε ± ∈ N ε . Thus u ε ∈ L ε and ± u ε are a pair of sign-changing solutions of

Equation (3.1). Let v ε ( x ) = u ε ( x ε ) , then ± v ε are a pair of sign-changing solutions of Equation (1.2).

This completes the proof of Theorem 1.1.

We can assume without loss of generality that x w = 0 . Then V ( 0 ) = τ w , W j ( 0 ) = ς j , j = 1,2 . Letting a = τ w , b j = ς j , j = 1,2 in Equation (2.1), there is u ∈ S τ w ς → by Lemma 2.3. Due to Lemma 2.7, m ( τ w , ς → ) ≥ 1 , we choose

m w = ( m ( τ w , ς → ) if   m ( τ w , ς → ) > 1 3 2 if   m ( τ w , ς → ) = 1 .

For the maximal integer m < m w , then m ≥ 1 . By Lemma 2.6 and Lemma 2.7, we have m Θ τ w ς → < Θ ∞ . The following proof of Theorem 1.2 is similar to that of Theorem 1.1 and so is omitted.

In this subsection, we will consider the case (P3) - (1), the other case can be handled similarly. Without loss of generality, we assume x j v = 0 . Then V ( 0 ) = τ , W j ( 0 ) = ς j v , j = 1,2 .

Lemma 3.3. u ε → u as ε → 0 up to a sequence after translations.

Proof. Let ε k → 0 as k → ∞ , u k : = u ε k ∈ S ε k with u k > 0 . By Lemma 2.14, we obtain lim k → ∞ Θ ε k = Θ τ ς → , which together with

Θ ε k = J ε k ( u k ) ≥ p − 2 2 p ∫ ℝ N ( | ∇ u k | 2 + V ( ε k x ) u k 2 ) d x ≥ C ‖ u k ‖ 2 , implies that { u k } ⊂ H 1 ( ℝ N ) is bounded. By Lemma 2.8, there exist σ > 0 , R > 0 and z ′ k ∈ ℝ N such that

∫ B R ( z ′ k )     u k 2 d x ≥ σ . (3.17)

Let u ^ k ( x ) : = u k ( x + z ′ k ) , V ^ ε k ( x ) : = V ( ε k ( x + z ′ k ) ) , W ^ j ε k ( x ) : = W j ( ε k ( x + z ′ k ) ) , j = 1,2 . Then u ^ k is a solution of

− Δ u ^ k + V ^ ε k ( x ) u ^ k = W ^ 1 ε k ( x ) u ^ k p − 1 + W ^ 2 ε k ( x ) u ^ k q − 1 . (3.18)

Furthermore,

Θ ^ ε k = J ^ ε k ( u ^ k ) = J ε k ( u k ) = Θ ε k . (3.19)

Since { u ^ k } is bounded, we can assume that u ^ k ⇀ u in H 1 ( ℝ N ) as k → ∞ . Then u ^ k → u in L l o c μ ( ℝ N ) for μ ∈ [ 2,2 * ) as k → ∞ . By (3.17), u ≠ 0 .

Since V and W j , j = 1,2 are bounded, up to a subsequence if necessary, we can assume

V ( ε k z ′ k ) → V 0 ,   W j ( ε k z ′ k ) → W j 0 ,   j = 1,2   as   k → ∞ , (3.20)

and W → 0 : = ( W 10 , W 20 ) . For all x ∈ ℝ N , by the boundedness of ∇ V : | ∇ V ( x ) | ≤ C , for given arbitrarily R > 0 , we obtain | V ( ε k x + ε k z ′ k ) − V ( ε k z ′ k ) | ≤ ε k C R for all x ∈ B R ( 0 ) . Hence V ^ ε k ( x ) → V 0 as k → ∞ uniformly on any bounded set of x. Similarly, W ^ j ε k ( x ) → W j 0 , j = 1,2 as k → ∞ uniformly on any bounded set of x. Similar to the proof of Lemma 2.14, we have

lim sup k → ∞ Θ ^ ε k ≤ Θ V 0 W → 0 . (3.21)

By (3.18), for any φ ∈ C 0 ∞ ( ℝ N ) , we obtain

0 = l i m k → ∞ ∫ ℝ N ( ∇ u ^ k ∇ φ + V ^ ε k ( x ) u ^ k φ − W ^ 1 ε k ( x ) u ^ k p − 1 φ − W ^ 2 ε k ( x ) u ^ k q − 1 φ ) d x = ∫ ℝ N ( ∇ u ∇ φ + V 0 u φ − W 10 u p − 1 φ − W 20 u q − 1 φ ) d x ,

which implies that u is a ground state solution of

− Δ u + V 0 u = W 10 u p − 1 + W 20 u q − 1 (3.22)

with the energy functional

J V 0 W → 0 ( u ) = p − 2 2 p ∫ ℝ N     W 10 u p d x + q − 2 2 q ∫ ℝ N     W 20 u q d x ≥ Θ V 0 W → 0 . (3.23)

By Fatou’s Lemma,

p − 2 2 p ∫ ℝ N     W 10 u p d x + q − 2 2 q ∫ ℝ N     W 20 u q d x ≤ lim inf t → 0 ∫ ℝ N ( p − 2 2 p W ^ 1 ε k ( x ) u ^ k p + q − 2 2 q W ^ 2 ε k ( x ) u ^ k q ) d x , (3.24)

Combining (3.19) with (3.22) - (3.24), we have

Θ V 0 W → 0 ≤ J V 0 W → 0 ( u ) ≤ lim inf k → ∞ J ^ ε k ( u ^ k ) ≤ lim sup k → ∞ Θ ^ ε k ≤ Θ V 0 W → 0 .

Hence

lim k → ∞ Θ ^ ε k = Θ V 0 W → 0 = J V 0 W → 0 ( u ) . (3.25)

Set η ∈ C 0 ∞ ( ℝ ) satisfies η ( t ) = 0 if t ≥ 2 and η ( t ) = 1 if t ≤ 1 . Define u ˜ k ( x ) : = η ( | 2 x | k ) u ( x ) and z k ( x ) = u ^ k ( x ) − u ˜ k ( x ) for x ∈ ℝ N . Then u ˜ k → u

and z k ⇀ 0 in H 1 ( ℝ N ) as k → ∞ , u ˜ k → u in L μ ( ℝ N ) for μ ∈ [ 2,2 * ] and z k → 0 in L l o c μ ( ℝ N ) for μ ∈ [ 2,2 * ) as k → ∞ , u ˜ k → u and z k → 0 a.e. on ℝ N as k → ∞ . We define

J ^ ε k ( z k ) : = 1 2 ∫ ℝ N ( | ∇ z k | 2 d x + V ^ ε k ( x ) z k 2 ) d x                             − 1 p ∫ ℝ N     W ^ 1 ε k ( x ) | z k | p d x − 1 q ∫ ℝ N     W ^ 2 ε k ( x ) | z k | q d x . Now we show that

J ^ ε k ( z k ) → 0 and 〈 J ^ ′ ε k ( z k ) , z k 〉 → 0 as k → ∞ . By Remark 1.33 in [23] , we have

‖ z k ‖ 2 = ‖ u ^ k ‖ 2 − ‖ u ˜ k ‖ 2 + o ( 1 ) . (3.26)

For any σ > 0 , there exists k 0 > 0 such that

∫ ℝ N | | u ^ k | μ − | z k | μ − | u ˜ k | μ | d x ≤ C σ ,   ∀ k > k 0 . (3.27)

By choosing μ = 2 , p , q in (3.27), respectively, we obtain

∫ ℝ N     V ^ ε k ( x ) u ^ k 2 d x = ∫ ℝ N     V ^ ε k ( x ) z k 2 d x + ∫ ℝ N     V ^ ε k ( x ) u ˜ k 2 d x + o ( 1 ) , (3.28)

∫ ℝ N     W ^ 1 ε k ( x ) | u ^ k | p d x = ∫ ℝ N     W ^ 1 ε k ( x ) | z k | p d x + ∫ ℝ N     W ^ 1 ε k ( x ) | u ˜ k | p d x + o ( 1 ) , (3.29)

∫ ℝ N     W ^ 2 ε k ( x ) | u ^ k | q d x = ∫ ℝ N     W ^ 2 ε k ( x ) | z k | q d x + ∫ ℝ N     W ^ 2 ε k ( x ) | u ˜ k | q d x + o ( 1 ) . (3.30)

By using the Lebesgue dominated convergence theorem,

∫ ℝ N     V ^ ε k ( x ) u ˜ k 2 d x = ∫ ℝ N     V 0 u 2 d x + o ( 1 ) , (3.31)

∫ ℝ N     W ^ 1 ε k ( x ) | u ˜ k | p d x = ∫ ℝ N     W 10 | u | p d x + o ( 1 ) , (3.32)

∫ ℝ N     W ^ 2 ε k ( x ) | u ˜ k | q d x = ∫ ℝ N     W 20 | u | q d x + o ( 1 ) . (3.33)

Moreover,

| ∇ u ˜ k | 2 2 = | ∇ u | 2 2 + o ( 1 ) . (3.34)

Combining (3.25) - (3.34) and (3.18) with (3.22), we have

J ^ ε k ( z k ) = J ^ ε k ( u ^ k ) − J ^ ε k ( u ˜ k ) + o ( 1 ) = Θ ^ ε k − J V 0 W → 0 ( u ) + o ( 1 ) = o ( 1 ) (3.35)

and

〈 J ^ ′ ε k ( z k ) , z k 〉 = 〈 J ^ ′ ε k ( u ^ k ) , u ^ k 〉 − 〈 J ^ ′ ε k ( u ˜ k ) , u ˜ k 〉 + o ( 1 ) = 〈 J ^ ′ ε k ( u ^ k ) , u ^ k 〉 − 〈 ( J V 0 W → 0 ) ′ ( u ) , u 〉 + o ( 1 ) = o ( 1 ) . (3.36)

In the end, by (3.35) and (3.36), we have o ( 1 ) = J ^ ε k ( z k ) − 1 p 〈 J ^ ′ ε k ( z k ) , z k 〉 ≥ C ‖ z k ‖ 2 , which implies that z k → 0 in H 1 ( ℝ N ) as k → ∞ . Thus u ^ k → u in H 1 ( ℝ N ) as k → ∞ .

Lemma 3.4. u ^ k ( x ) → 0 as | x | → ∞ uniformly in k ∈ ℤ + .

Proof. We use the contradiction method to obtain that there are σ > 0 for x n ∈ ℝ N , | x n | → ∞ as n → ∞ such that u ^ k n ( x n ) ≥ σ . Moreover, there exists

C > 0 (independent of k) such that u ^ k n ( x n ) ≤ C ( ∫ B 1 ( x n )     u ^ k n 2 d x ) 1 2 . Thus by the Minkowski inequality, we have

u ^ k n ( x n ) ≤ C ( ∫ ℝ N | u ^ k n − u | 2 d x ) 1 2 + C ( ∫ B 1 ( x n )     u 2 d x ) 1 2 → 0 as n → ∞ , which is impossible.

Lemma 3.5. { ε k z ′ k } k is bounded on ℝ N .

Proof. Assume by contradiction that there is | ε k z ′ k | → ∞ as k → ∞ up to a subsequence. Hence V 0 ≥ τ ∞ > τ and W j 0 ≤ ς j ∞ < ς j v , j = 1 , 2 . By Lemma 2.4, we have Θ V 0 W → 0 > Θ τ ς → v . According to (3.19), (3.25) and Lemma 2.14, Θ V 0 W → 0 = lim k → ∞ Θ ^ ε k = lim k → ∞ Θ ε k ≤ lim sup k → ∞ Θ ε k ≤ Θ τ ς → v , which is impossible.

By Lemma 3.5, we may assume ε k z ′ k → x 0 as k → ∞ . By (3.20), we obtain V 0 = V ( x 0 ) and W j 0 = W j ( x 0 ) , j = 1 , 2 . Applying (3.22), we get that u is a ground state solution of Equation (1.4).

Lemma 3.6. { ε z ε } ε is bounded, where z ε ∈ ℝ N is a maximum point of u ε .

Proof. If the thesis were not true, there were ε k → 0 with | ε k z k | → ∞ , where z k : = z ε k is a maximum point of u k : = u ε k . Repeating Lemma 3.3 - Lemma 3.5, we can get that there exists z ′ k ∈ ℝ N such that u ^ k = u k ( ⋅ + z ′ k ) → u ≠ 0 in H 1 ( ℝ N ) as k → ∞ , u ^ k ( x ) → 0 as | x | → ∞ uniformly in k ∈ ℤ + , { ε k z ′ k } k is bounded on ℝ N . Thus | ε k z k − ε k z ′ k | → ∞ as k → ∞ , then | z k − z ′ k | → ∞ as k → ∞ . Since max ℝ N u k = u k ( z k ) = u ^ k ( z k − z ′ k ) → 0 as k → ∞ , then u ^ k ( x ) → 0 as k → ∞ uniformly in x ∈ ℝ N , which contradicts with u ≠ 0 .

Lemma 3.7. lim ε → 0 dist ( ε z ε , A v ) = 0 .

Proof. By Lemma 3.5 and Lemma 3.6, there exists ε k → 0 as k → ∞ with

ε k z ′ k → x 0 ,   ε k z k → z 0   as   k → ∞ , (3.37)

where z k = z ε k is a maximum point of u k = u ε k . By Lemma 3.3 and Lemma 3.5, there exists z ′ k ∈ ℝ N such that u ^ k ( x ) = u k ( x + z ′ k ) . By Lemma 3.4, we may assume u ^ k ( x ′ k ) = max ℝ N u ^ k and { x ′ k } k is bounded on ℝ N . Hence z k = x ′ k + z ′ k and ε k x ′ k → 0 as k → ∞ . By (3.32) and (3.34), which imply that

z 0 = x 0 ,     V ( z 0 ) = V 0 ,     W j ( z 0 ) = W j 0 ,   j = 1 , 2. (3.38)

Assume indirectly that z 0 ∉ A v , then V ( z 0 ) > τ , W j ( z 0 ) ≤ ς j v , j = 1,2 or V ( z 0 ) = τ , W 1 ( z 0 ) < ς 1 v , W 2 ( z 0 ) = ς 2 v or V ( z 0 ) = τ , W 1 ( z 0 ) = ς 1 v , W 2 ( z 0 ) < ς 2 v . By Lemma 2.4,

Θ V ( z 0 ) W → ( z 0 ) > Θ τ ς → v . (3.39)

Combining (3.19), (3.25), (3.38) and (3.39) with Lemma 2.14, we have

lim k → ∞ Θ ε k = lim k → ∞ Θ ^ ε k = Θ V 0 W → 0 = Θ V ( z 0 ) W → ( z 0 ) > Θ τ ς → v ≥ lim sup k → ∞ Θ ε k ,

which is impossible. Hence x 0 = z 0 ∈ A v .

By Lemma 3.6, if V ∩ ( W 1 ∩ W 2 ) is not empty, we assume x 0 ∈ A v = V ∩ ( W 1 ∩ W 2 ) , which implies that

lim ε → 0 dist ( ε z ε , V ∩ ( W 1 ∩ W 2 ) ) = 0     and     V ( x 0 ) = τ ,   W j ( x 0 ) = ς j ,   j = 1,2.

Hence u is a groundstate solution of Equation (1.5). This completes the proof of Theorem 1.3.

Similar to the proof of Step 6 in [18] , we have the following result.

Lemma 3.8. There exists C > 0 such that for small ε > 0 , u ε ( x ) ≤ C e − τ 2 | x − z ε | for all x ∈ ℝ N .

Now we prove Theorem 1.3 by Lemma 3.3 - Lemma 3.8. Set x ε = ε z ε , then v ε ( x ε ) = u ε ( z ε ) . By Lemma 3.6, x ε is a maximum point of v ε and { x ε } ε is bounded on ℝ N . By Lemma 3.7, lim ε → 0 dist ( x ε , A v ) = 0 . By Lemma 3.3 and Lemma 3.4, u ^ ε ( x ) = u ε ( x + z ′ ε ) = v ε ( ε x + x ε − ε x ′ ε ) , where x ′ ε = z ε − z ′ ε is a maximum point of u ^ ε with ε x ′ ε → 0 as ε → 0 . By Lemma 3.8, we obtain

v ε ( x ) ≤ C e − c ε | x − x ε | , where C depends on N , τ .

Consequently, we establish the multiplicity of the semi-classical solutions for Equation (1.2), and we obtain the existence, concentration, convergence, exponential decay estimates of the positive ground state solution. We also prove the existence of sign-changing solutions of Equation (1.2).