Existence and multiplicity of nontrivial solutions for a class of Schrödinger–Kirchhoff-type equations

J. Math. Anal. Appl. 417 (2014) 65–79

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Existence and multiplicity of nontrivial solutions for a class of Schrödinger–Kirchhoff-type equations ✩ Jianjun Nie ∗ School of Mathematics and Computer Science, Guizhou Normal College, Guiyang, Guizhou 550018, PR China

a r t i c l e

i n f o

a b s t r a c t

Article history: Received 4 August 2013 Available online 18 March 2014 Submitted by R. Curto

In the present paper, the following Schrödinger–Kirchhoff-type problem  ⎧   ⎪ ⎨− a + b |∇u|2 dx Δu + λV (x)u = f (x, u),

Keywords: Schrödinger–Kirchhoff-type problem Sobolev embedding Palais–Smale condition

⎪ ⎩

RN

u(x) → 0

in RN ,

(P )

as |x| → ∞,

is studied. When N = 2, 3, 4 and V (x) = 1, the existence theorem of nontrivial weak solutions for problem (P ) is obtained. When N = 1, 2, 3 and V (x) is more general, two existence theorems of nontrivial weak solutions and a sequence of high energy weak solutions for problem (P ) are obtained. © 2014 Elsevier Inc. All rights reserved.

1. Introduction Consider the following Schrödinger–Kirchhoff-type problem   ⎧  ⎪ ⎨ − a + b |∇u|2 dx Δu + λV (x)u = f (x, u), ⎪ ⎩

RN

in RN ,

(P )

u(x) → 0 as |x| → ∞,

where constants a > 0, b  0 and λ > 0. The Kirchhoff-type problem is related to the stationary analogue of the equation ✩ This work was supported by the general foundation of Guizhou Normal College (13YB009), Key Support Subject (Applied Mathematics) and Key project on the reforms of teaching contents and course system of Guizhou Normal College. * Fax: +86 8715516877. E-mail address: [email protected].

http://dx.doi.org/10.1016/j.jmaa.2014.03.027 0022-247X/© 2014 Elsevier Inc. All rights reserved.

J.J. Nie / J. Math. Anal. Appl. 417 (2014) 65–79

66

∂2u ρ 2 − ∂t



E P0 + h 2L

L 2  2

∂u

dx ∂ u = 0

∂x

∂x2

(1.1)

0

which was presented by Kirchhoff [11] as an extension of the classical D’Alembert wave equation for free vibrations of elastic strings. Kirchhoff’s model takes into account the changes in length of the string produced by transverse vibrations. The parameters in Eq. (1.1) have the following meanings: L is the length of the string, h is the area of cross-section, E is the Young modulus of the material, ρ is the mass density and P0 is the initial tension. Some early classical studies of Kirchhoff equations were those of Pohozăev [17] and Bernstein [4]. However, Eq. (1.1) received great attention only after Lions [12] proposed an abstract framework for the problem. Some interesting results can be found in [2,5,7] and the references therein. In the recent years, some interesting studies by variational methods can be found in [1,6,9,13,14,16,21] for Kirchhoff-type problem. Nevertheless, the problems they study are based on a bounded domain of RN . If the problem is set on RN , it is well known that the Sobolev embedding H 1 (RN ) → Ls (RN ) (2  s < 2∗ ) is not compact, hence it is usually difficult to prove Palais–Smale condition if we seek solutions by variational methods. In order to recover the compactness of Sobolev embedding, in [20], the author uses the following assumption: (V ) V ∈ C(RN , R), inf x∈RN V (x)  c > 0 and for each M > 0, meas{x ∈ RN : V (x)  M } < +∞, where meas denotes the Lebesgue measure in RN . Moreover, using the embedding results in [19], we study the following Kirchhoff-type problem (see [15])   ⎧  ⎪ ⎨ − a + b |∇u|2 dx Δu + V |x| u = Q |x| f (u), ⎪ ⎩

in RN ,

RN

(1.2)

u(x) → 0 as |x| → ∞.

In Section 2 of this paper, we study the problem (P ) with V (x) = 1 and f (x, u) = |u|q−1 u. We will find out a nontrivial solution of problem (P ) in Hr1 (RN ) := {u ∈ H 1 (RN ): u is radial}. When 3 < p < 2∗ , this can be obtained very easily by variational methods. Thus we will focus on the case where 1 < q  3. In order to reduce our statement, we give the following notation  1 < q  3, N = 2, 3, (Q) 1 < q < 3, N = 4. In this case, the main difficulty is to prove the boundedness of the Palais–Smale sequences, which cannot be proved directly for 1 < q < 3. We will use an indirect approach which is largely inspired by [10]. Our main result is the following theorem. Theorem 1.1. Assume that V (x) = 1 and (Q) holds. Then one has the following. (i) When N = 2, 3, the problem (P ) has at least one nontrivial weak solution. (ii) When N = 4, the problem (P ) has at least one nontrivial weak solution whenever λ > 0 is sufficiently small. In Section 3, we deal with the problem (P ) when N = 1, 2, 3 and V satisfies the following assumption: (V1 ) V ∈ C(RN , R), V (x)  0. Moreover, there exists a constant d > 0 such that the set {x ∈ RN : V (x)  d} is nonempty and meas{x ∈ RN : V (x)  d} < +∞, where meas denotes the Lebesgue measure in RN .

J.J. Nie / J. Math. Anal. Appl. 417 (2014) 65–79

67

Define the space

Eλ =

   u ∈ H 1 RN : λV (x)u2 dx < +∞ RN

with the inner product and norm  u, vEλ =



∇u∇v + λV (x)uv dx;

  uEλ =

RN



|∇u| + λV (x)u 2

2

 12 dx

.

RN

It is obvious that (V1 ) is weaker than (V ). To the best of our knowledge, it cannot guarantee the compactness of the embedding Eλ → Ls (RN ) (2  s < 2∗ ). This prevents from using the variational techniques in a standard way. In this case, our main difficulty is to verify the compactness property for corresponding energy functional. Our approach is largely inspired by [8]. Fortunately, by (V1 ) and the Poincaré inequality, we deduce that the embedding Eλ → H 1 (RN ) is continuous. Consequently, by the continuity of the embedding Eλ → Ls (RN ) (2  s  2∗ ), there are constants as > 0 (2  s  2∗ ) such that us  as uEλ ,

∀u ∈ Eλ .

(1.3)

In order to state our main results, we first make the following assumptions on f . (f1 ) (f2 ) (f3 ) (f4 ) (f5 )

f ∈ C(RN × R, R) and |f (x, u)|  c(1 + |u|p ) for some 1 < p < 2∗ − 1, where c is a positive constant. f (x, u) = o(|u|) as |u| → 0 uniformly in x ∈ RN . F (x,u) N |u|4 → +∞ as |u| → +∞ uniformly in x ∈ R . uf (x, u)  4F (x, u), ∀x ∈ RN , ∀u ∈ R. There exist μ > 4 and r > 0 such that inf

x∈RN , |u|=r

F (x, u) := β > 0,

and μF (x, u) − f (x, u)u  C0 |u|2 ,

∀x ∈ RN and |u|  r,

where C0 < β(μ−2) . r2 (f6 ) f (x, −u) = −f (x, u) for all x ∈ RN and u ∈ R. Our main results are the following theorems. Theorem 1.2. Assume N = 1, 2, 3. If (V1 ) and (f1 )–(f4 ) hold, then the problem (P ) has at least one nontrivial weak solution whenever λ > 0 is sufficiently large. Furthermore, if the condition (f6 ) is added, then the problem (P ) has a sequence of weak solutions {un } in Eλ with un Eλ → ∞ and I(un ) → +∞. Theorem 1.3. Assume N = 1, 2, 3. If (V1 ), (f1 ), (f2 ) and (f5 ) hold, then the problem (P ) has at least one nontrivial weak solution whenever λ > 0 is sufficiently large. Furthermore, if the condition (f6 ) is added, then the problem (P ) has a sequence of weak solutions {un } in Eλ with un Eλ → ∞ and I(un ) → +∞.

J.J. Nie / J. Math. Anal. Appl. 417 (2014) 65–79

68

Set F (x, u) = functional

u 0

I(u) =

f (x, s) ds. Then a weak solution of problem (P ) is a critical point of the following

a 2

 |∇u|2 dx +

b 4

 

2 |∇u|2 dx

RN

+

1 2

RN



 λV (x)u2 dx − RN

F (x, u) dx.

RN

It is easy to see that I ∈ C 1 (Eλ , R). In the latter parts of this paper, we use c to denote any positive constant. 2. The radially symmetric case In this section, we consider the problem (P ) when V (x) = 1 and f (x, u) = |u|q−1 u, where q satisfies (Q). In order to give the proof of Theorem 1.1, we will use the following lemma. Lemma 2.1. (See [10, Theorem 1.1].) Let X be a Banach space equipped with the norm . and let J ∈ R+ be an interval. We consider a family (Iμ )μ∈J of C 1 -functionals on X of the form Iμ (u) = A(u) − μB(u),

∀μ ∈ J

where B(u)  0, ∀u ∈ X and such that either A(u) → +∞ or B(u) → +∞ as u → +∞. We assume there are two points (v1 , v2 ) in X such that setting 

 Γ = γ ∈ C [0, 1], X γ(0) = v1 , γ(1) = v2 there hold, ∀μ ∈ J

  βμ := inf max Iμ γ(t) > max Iμ (v1 ), Iμ (v2 ) . γ∈Γ t∈[0,1]

Then, for almost every μ ∈ J, there is a sequence {vn } ⊂ X such that (i) {vn } is bounded,

(ii) Iμ (vn ) → βμ ,

(iii) Iμ (vn ) → 0.

Take X = Hr1 (RN ). Let us apply Lemma 2.1 to the functional Iμ (u) =

a 2

 |∇u|2 dx +

b 4

 

RN

Set

a 2

 RN



|∇u|2 dx + 4b (

RN

2 |∇u|2 dx

+

RN

|∇u|2 dx)2 +

λ 2

 RN

λ 2

 u2 dx − RN

u2 dx = A(u) and

μ q+1

 |u|q+1 dx. RN

1 q+1

 RN

|u|q+1 dx = B(u). Then

Iμ (u) = A(u) − μB(u). It is obvious that Iμ ∈ C 1 (Hr1 (RN ), R), B(u)  0, Iμ (0) = 0 and A(u) → +∞ as u → +∞. Taking μ = 12 , we consider the curve t → vt (x) = t4−N u(t−2 x). One has I 12 (vt ) =

a 4 t 2



RN

b |∇u|2 dx + t8 4

 

2 |∇u|2 dx

RN

+

λ 8 t 2



RN

u2 dx −

1 t(q+1)(4−N )+2N 2(q + 1)

 |u|q+1 dx. RN

J.J. Nie / J. Math. Anal. Appl. 417 (2014) 65–79

69

Note that (q + 1)(4 − N ) + 2N > 8 for 1 < q  3 and N = 2, 3. Consequently, for any 0 = u ∈ Hr1 (RN ), one has I 12 (vt ) → −∞,

as t → +∞.

When N = 4, one has I 12 (vt ) =

a 4 t 2



b |∇u|2 dx + t8 4

 

2 |∇u|2 dx

RN

+

λ 8 t 2

RN

 u2 dx −

1 t8 2(q + 1)

RN

 |u|q+1 dx.

RN

Assume that 1 < q < 3. ∀u = 0, we can choose a small t > 0 such that 

1 2(q + 1)

|tu|q+1 dx >

b 4

 

RN



∇(tu) 2 dx

2 .

RN

Hence there exists a 0 = u0 ∈ Hr1 (RN ) such that 

1 2(q + 1)

0 q+1 b

u

dx > 4

 

RN

Let λ be so small that

λ 2

 RN

|u0 |2 dx <

0 2

∇u dx

2 .

RN

1 2(q+1)

 RN

 |u0 |q+1 dx − 4b ( RN |∇u0 |2 dx)2 . We also have

I 12 vt0 → −∞,

as t → +∞,

where vt0 (x) = u0 (t−2 x). Take a θ > 0 such that I 12 (vθ0 ) < 0. Define γ : [0, 1] → Hr1 (RN ) in the following way  γ(t) :=

0, 0 vθt = (θt)4−N u0 ((θt)−2 x),

if t = 0, if 0 < t  1.

It is easy to see that γ(t) ∈ C([0, 1], Hr1 (RN )) and I 12 (γ(1)) < 0. Note that the map μ → Iμ (u) is decreasing. Then Iμ (γ(1)) < 0 for μ  12 . Set 



 Γ := γ ∈ C [0, 1], Hr1 RN γ(0) = 0, γ(1) = vθ0 . From the preceding consideration, Γ is nonempty. Let

βμ := inf max Iμ γ(t) . γ∈Γ t∈[0,1]

By the Sobolev embedding, we have Iμ (u) =

a 2

 |∇u|2 dx + RN



b 4

2

  |∇u|2 dx RN

 a λ  min , u2 − cuq+1 2 2     a λ 2 q−1  u min , − cu . 2 2

+

λ 2

 u2 dx − RN

μ q+1

 |u|q+1 dx RN

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70

Hence, there exists a ρ > 0 such that Iμ |Bρ (0)  0

and Iμ |∂Bρ (0) > c > 0,

where Bρ (0) ⊂ Hr1 (RN ). This implies that βμ > max{Iμ (0), Iμ (vθ0 )}. Consequently, we obtain the following lemma. Lemma 2.2. Assume that λ > 0 is sufficiently small for N = 4 and (Q) holds. Then, for almost every μ ∈ J = [ 12 , +∞), there exists a bounded Palais–Smale sequence of Iμ at the level βμ in Hr1 (RN ). Furthermore, by the compactness of the embedding Hr1 (RN ) → Ls (RN ) (N  2 and 2 < s < 2∗ ), we have the following lemma. Lemma 2.3. Assume that λ > 0 is sufficiently small for N = 4. Then, for almost every μ ∈ J = [ 12 , +∞), there exists uμ ∈ Hr1 (RN ) such that Iμ (uμ ) = 0 and Iμ (uμ ) = βμ . Proof. By Lemma 2.2, there exists a bounded sequence {un } ⊂ Hr1 (RN ) such that Iμ (un ) → βμ

and Iμ (un ) → 0.

By the Sobolev embedding, passing to a subsequence, we can assume that un uμ in Hr1 (RN ), un → uμ in Ls (RN ) for all s ∈ (2, 2∗ ). Note that 

Iμ (un )

−

Iμ (uμ ), un

− uμ



    2 = a + b |∇un | dx ∇un ∇(un − uμ ) dx + λ |un − uμ |2 dx RN

RN

   − a + b |∇uμ |2 dx ∇uμ ∇(un − uμ ) dx  −μ

RN

RN

RN



|un |q − |uμ |q (un − uμ ) dx

RN

   

2

2



∇(un − uμ ) dx + λ |un − uμ |2 dx = a + b |∇un | dx RN

RN

RN

    −b |∇uμ |2 dx − |∇un |2 dx ∇uμ ∇(un − uμ ) dx RN

 −μ

RN

RN



|un |q − |uμ |q (un − uμ ) dx

RN

 min{a, λ}un − uμ 2     2 2 −b |∇uμ | dx − |∇un | dx ∇uμ ∇(un − uμ ) dx RN

 −μ

RN

RN



|un |q − |uμ |q (un − uμ ) dx.

RN

(2.1)

J.J. Nie / J. Math. Anal. Appl. 417 (2014) 65–79

71

One has   min{a, λ}un − uμ 2  Iμ (un ) − Iμ (uμ ), un − uμ + μ





|un |q − |uμ |q (un − uμ ) dx

RN

    2 2 +b |∇uμ | dx − |∇un | dx ∇uμ ∇(un − uμ ) dx. RN

RN

RN

 1 Set E = {u ∈ L2 (RN ): ∇u ∈ L2 (RN )} with the norm uE = ( RN |∇u|2 ) 2 . Then the embedding Hr (RN ) → Er := {u ∈ E: u is radial} is continuous. Hence, by the boundedness of {un }, we have     2 2 b |∇uμ | dx − |∇un | dx ∇uμ ∇(un − uμ ) dx → 0, RN

RN

RN

as n → +∞. Moreover,

 





q q



| − |u | − u ) dx |u (u |un |q + |uμ |q |un − uμ | dx n μ n μ



RN

RN

 un qq+1 + uμ qq+1 un − uμ q+1 → 0,

as n → +∞. Consequently, by Iμ (un ) → 0, we know that un − uμ  → 0. Hence, one has Iμ (uμ ) = 0 and Iμ (uμ ) = βμ . This completes the proof. 2 Proof of Theorem 1.1. By Lemma 2.3, we can take μn → 1 and choose {un } ⊂ Hr1 (RN ) such that Iμ n (un ) = 0 and Iμn (un ) = βμn . Take advantage from the fact that {un } is a sequence of solutions of Iμ n (u) = 0. Hence it satisfies the Pohozaev equality (the computations are the same as in [3]) 

(N − 2)a 2

N |∇un | dx + λ 2



2

RN

(N − 2)b |un | dx + 2

 

RN

2 |∇un | dx

2

2

N μn − q+1

RN

 |un |q+1 dx = 0. RN

Set  An = a

 |∇un |2 dx,

RN

Bn = λ RN

|un |2 dx,

  2 Cn = b |∇un |2 dx , RN

 Dn = μn

|un |q+1 dx.

RN

Then we have ⎧ 1 1 1 1 ⎪ ⎪ A + B + C − D = βμn , (i) ⎪ ⎪2 n 2 n 4 n q+1 n ⎨ (ii) An + Bn + Cn − Dn = 0, ⎪ ⎪ N N − 2 N N − 2 ⎪ ⎪ ⎩ An + Bn + Cn − Dn = 0. (iii) 2 2 2 q+1 An 2 By (i) − N1 (iii), we have N1 An + 4−N 4N Cn = βμn . Since βμn is bounded and Cn = b( a ) , An and Cn are −2)q bounded. And by (iii) − N 2−2 (ii), we have Bn = (N +2)−(N Dn . Consequently, we deduce that Bn is 2(q+1) bounded and hence un  is bounded. Then by the Sobolev embedding, passing to a subsequence, we can assume that un u in Hr1 (RN ), un → u in Ls (RN ) for all s ∈ (2, 2∗ ). Similar to the proof of Lemma 2.3, one has un → u in Hr1 (RN ). Hence I(u) = β1 = 0, I  (u) = 0. This completes the proof. 2

J.J. Nie / J. Math. Anal. Appl. 417 (2014) 65–79

72

3. The non-radially symmetric case In this section we consider the problem (P ) when V (x) satisfies (V1 ). We will give the proofs of Theorem 1.2 and Theorem 1.3. To complete the proofs, we need the following lemma. Lemma 3.1. (See [8, Theorem A.2].) Let Ω be an open set in RN and f ∈ C(Ω × R, R) a function such that |f (x, u)|  c(|u|r + |u|s ) for some c > 0 and 1  r < s < ∞. Suppose that s  p + 1 < ∞, r  t < ∞, t > 1, {un } is a bounded sequence in Lp+1 (Ω) ∩ Lt (Ω), un → u a.e. in Ω and in Lp+1 (Ω ∩ BR ) ∩ Lt (Ω ∩ BR ) for all R > 0. Then, passing to a subsequence, there exists a sequence vn → u in Lp+1 (Ω) ∩ Lt (Ω) such that f (x, un ) − f (x, un − vn ) − f (x, u) → 0

in Lt/r (Ω) + L(p+1)/s (Ω),

where vn (x) = χ(2|x|/Rn )u(x), χ ∈ C ∞ (R, [0, 1]) be such that χ(t) = 1 for t  1, χ(t) = 0 for t  2, Rn > 0 is a sequence of constants with Rn → ∞ as n → ∞, the space Lp+1 (Ω) ∩ Lt (Ω) with the norm u(p+1)∧t := up+1 + ut and the space Lp+1 (Ω) + Lt (Ω) with the norm   u(p+1)∨t := inf vp+1 + wt : v ∈ Lp+1 (Ω), w ∈ Lt (Ω), u = v + w . Lemma 3.2. Assume that (V1 ), (f1 ), (f2 ) and (f4 ) hold. Then there exists Λ > 0 such that I satisfies the (PS)c condition for all λ  Λ. Proof. Let {un } be a (PS)c sequence. By (f4 ), for large n, we have  1  I (un ), un 4    1 1 2 f (x, un )un − F (x, un ) dx = un Eλ + 4 4

c + un Eλ + o(1)  I(un ) −

RN



1 un 2Eλ . 4

This implies that {un } is bounded in Eλ . Hence, passing to a subsequence, we can assume that un u in Eλ , un → u in Lsloc (RN ) for 2  s < 2∗ and un (x) → u(x) a.e. in RN . Take vn (x) = χ(2|x|/Rn )u(x). We claim that vn → u in Eλ . Indeed, u ∈ Eλ implies that to any ε > 0 there is a ρ = ρ(ε) such that 

 λV (x)u2  ε and RN \Bρ (0)

|∇u|2  ε. RN \Bρ (0)

It follows from (3.1) that  vn − u2Eλ = RN



= RN



∇(vn − u) 2 dx +

 λV (x)|vn − u|2 dx

RN





∇ χ 2|x|/Rn u(x) − u(x) 2 dx +



RN



2

λV (x) χ 2|x|/Rn u(x) − u(x) dx

(3.1)

J.J. Nie / J. Math. Anal. Appl. 417 (2014) 65–79





2 4 χ 2|x|/Rn − 1 |∇u|2 dx + 2 Rn

 RN







2 χ 2|x|/Rn u2 dx

RN



2 λV (x) χ 2|x|/Rn − 1 u2 dx

+ RN

 

Bρ (0)

73



2 4 χ 2|x|/Rn − 1 |∇u|2 dx + 2 Rn



+





2 χ 2|x|/Rn u2 dx

RN

2 λV (x) χ 2|x|/Rn − 1 u2 dx + cε.

Bρ (0)

Furthermore, by the Lebesgue dominated convergence theorem, we obtain vn − uEλ → 0,

as n → +∞.

Moreover, similar to (2.1), one has 

I  (un ) − I  (vn ), un − vn



    |∇vn |2 dx − |∇un |2 dx ∇vn ∇(un − vn ) dx  min{a, 1}un − vn 2Eλ − b  −

RN

RN

RN

  f (x, un ) − f (x, vn ) (un − vn ) dx.

RN

Then  min{a, 1}un − vn 2Eλ −

  f (x, un ) − f (x, vn ) (un − vn ) dx

RN







 I (un ) − I (vn ), un − vn



    2 2 +b |∇vn | dx − |∇un | dx ∇vn ∇(un − vn ) dx. RN

RN

(3.2)

RN

Sine un u in Eλ and I  (un ) → 0, we have I  (un ) − I  (u), un − u → 0 as n → ∞. By vn − uEλ → 0, I ∈ C 1 (Eλ , R) and the boundedness of {un } in Eλ , one has

  

I (un ) − I  (vn ), un − vn

   

 I  (un ) − I  (u), un − vn + I  (u) − I  (vn ), un − vn

     

 I  (un ) − I  (u), un − u + I  (un ) − I  (u), u − vn + I  (u) − I  (vn ), un − vn

→ 0,

(3.3)

  = {u ∈ L2 (RN ): ∇u ∈ L2 (RN )} with the norm u  = ( N |∇u|2 dx) 12 . Then the as n → ∞. Set E E R  is continuous. Hence un u and vn → u in E.  Consequently, by the boundedness of embedding Eλ → E {un } and {vn } in Eλ , one has

J.J. Nie / J. Math. Anal. Appl. 417 (2014) 65–79

74



   



2 2

b |∇vn | dx − |∇un | dx ∇vn ∇(un − vn ) dx



RN

RN

RN

 

 





b |∇vn |2 dx − |∇un |2 dx ∇u∇(un − vn ) dx

RN

RN

RN

 

 



+

b |∇vn |2 dx − |∇un |2 dx ∇(vn − u)∇(un − vn ) dx

RN

RN

RN

 

 



2 2

 b |∇vn | dx − |∇un | dx ∇u∇(un − u) dx

RN

RN

RN

 

 



2 2

+ b |∇vn | dx − |∇un | dx ∇u∇(u − vn ) dx

RN

RN

RN

RN

RN

RN

 

 



+

b |∇vn |2 dx − |∇un |2 dx ∇(vn − u)∇(un − vn ) dx

→ 0,

(3.4)

 as n → ∞. Now, we prove | RN [f (x, un ) − f (x, vn ) − f (x, un − vn )](un − vn ) dx| → 0, as n → +∞. Indeed, by (f1 ) and (f2 ) we know that for any ε > 0, there is c(ε) > 0 such that



f (x, u)  ε|u| + c(ε)|u|p

(3.5)



F (x, u)  ε |u|2 + c(ε) |u|p+1 . 2 p+1

(3.6)

and

Take r = 1, s = p and t = 2. It follows from Lemma 3.1 that

p+1 gn (x) → 0 in L2 RN + L p RN , where gn (x) = f (x, un ) − f (x, u) − f (x, un − vn ). Then 



f (x, un ) − f (x, u) − f (x, un − vn ) |un − vn | dx  gn 2∨(p+1) un − vn 2∧(p+1) → 0,

RN

as n → +∞, where (p + 1) =

p+1 p .

Take un = vn for all n > 0 in Lemma 3.1. Then

f (x, vn ) − f (x, u) → 0 Consequently, we also obtain that

 RN



p+1 in L2 RN + L p RN .

|f (x, vn ) − f (x, u)||un − vn | dx → 0 as n → +∞. Then, one has







f (x, un ) − f (x, vn ) − f (x, un − vn ) (un − vn ) dx



RN



 RN

→ 0,



f (x, un ) − f (x, u) − f (x, un − vn ) |un − vn | dx +





f (x, u) − f (x, vn ) |un − vn | dx

RN

(3.7)

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as n → +∞. It follows from (3.2), (3.3), (3.4) and (3.7) that  min{a, 1}un − vn 2Eλ −

f (x, un − vn )(un − vn ) dx  o(1),

(3.8)

RN

for large n. Set wn = un − vn . Then, by (V1 ) and wn 0, one has  wn 22

 wn2

= V (x)d

∗

Take 0 < α < min{ (N −2)(22

−p−1)

wn2 dx 

dx +

1 wn 2Eλ + o(1). λd

(3.9)

V (x)
, 1}. Then 2 <

2(p+1−α) 2−α

< 2∗ . By (3.9) and the Höder inequality, one has

p+1−α α −2 wn p+1 wn p+1 p+1  wn 2 wn  2(p+1−α)  c(λd) Eλ + o(1). α

(3.10)

2−α

Consequently, by (3.5), (3.9) and (3.10), we have







f (x, wn )wn dx  εwn 22 + c(ε)wn p+1  ε wn 2E + c1 (ε)α wn p+1 + o(1). p+1 Eλ



λ λd (λd) 2 RN

Hence, by (3.8) and the boundedness of {wn }, we have  o(1)  min{a, 1}un − vn 2Eλ −

  f (x, un − vn ) (un − vn ) dx 

 min{a, 1} −

RN

c2 (ε) ε − α λd (λd) 2

 wn 2Eλ .

Letting Λ > 0 be so large that the term in the brackets above is positive when λ  Λ, we get wn → 0 in Eλ . Since wn = un − vn and vn → u, it follows that un → u in Eλ . This completes the proof. 2 Proof of Theorem 1.2. For any 0 < ε < there is a constant c(ε) > 0 such that

min{a,1} a22

(a2 appears in (1.3)), by (3.5) and (3.6) we know that



f (x, u)  ε|u| + c(ε)|u|p and



F (x, u)  ε |u|2 + c(ε) |u|p+1 . 2 p+1 Note that (f3 ) implies p + 1 > 4. Hence, for small ρ > 0, 1 ε c(ε) min{a, 1}u2Eλ − u22 − up+1 p+1 2 2 p+1

1 c(ε) p+1 a up+1  min{a, 1} − a22 ε u2Eλ − Eλ 2 p + 1 p+1

1  min{a, 1} − a22 ε u2Eλ 4

I(u) 

for all u ∈ Bρ , where Bρ = {u ∈ Eλ : uEλ < ρ}. Therefore, I|∂Bρ 

1 min{a, 1} − a22 ε ρ2 := α0 > 0. 4

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Since Eλ → L2 (RN ) and L2 (RN ) is a separable Hilbert space, Eλ has a countable orthogonal basis {ei }. Set Eλk = span{e1 , . . . , ek } and Zλk = (Eλk )⊥ . Then Eλ = Eλk ⊕ Zλk . Therefore, I|∂Bρ ∩Zλk  α0 > 0. λ ⊂ Eλ , there is a positive integral number m such that Moreover, for any finite dimensional subspace E m  Eλ ⊂ Eλ . Since all norms are equivalent in a finite dimensional space, there is a constant b4 > 0 such that u4  b4 uEλ , By (f1 ), (f2 ) and (f3 ) we know that for any M >

b , 4b44

∀u ∈ Eλm . there is a constant C(M ) > 0 such that

F (x, u)  M |u|4 − C(M )|u|2 ,

∀(x, u) ∈ RN × R.

Hence 1 b max{a, 1}u2Eλ + u4Eλ − M u44 + C(M )u22 2 4   1 b 2 4 u4Eλ + C(M )a22 u2Eλ  max{a, 1}uEλ − M b4 − 2 4

I(u) 

λ \ Br . for all u ∈ Eλm . Consequently, there is a large r > 0 such that I < 0 on E By Lemma 3.2, there exists Λ > 0 such that I satisfies the (PS)c condition for all λ  Λ. Moreover, it is obvious that I(0) = 0. Hence I possesses a critical value η  α0 by Theorem 2.2 in [18], i.e. the problem (P ) has a nontrivial solution in Eλ . Moreover, obviously, I is bounded on each bounded subset of Eλ and (f6 ) implies that I is even. Hence the second conclusion follows from Theorem 9.12 in [18]. This completes the proof. 2 Proof of Theorem 1.3. From the proof of first segment in Theorem 1.2, we know that there exist constants ρ > 0 and δ > 0 such that I|∂Bρ  δ > 0. Moreover, for any (x, z) ∈ RN × R, set

h(t) := F x, t−1 z tμ ,

∀t ∈ [1, +∞).

For |z|  r and t ∈ [1, |z| r ], by (f5 ), one has  



z h (t) = f x, t−1 z − 2 tμ + F x, t−1 z μtμ−1 t  −1

 μ−1 =t μF x, t z − t−1 zf x, t−1 z  C0 tμ−3 |z|2 . Then 

|z| h r

|z|



r − h(1) =

h (t) dt

1 |z|

r 

C0 tμ−3 |z|2 dt 1

=

C0 |z|2 C0 |z|μ . − (μ − 2)rμ−2 μ−2

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Hence, we have 

|z| r



C0 |z|μ (μ − 2)rμ−2   C0 β |z|μ .  − rμ (μ − 2)rμ−2

F (x, z) = h(1)  h

By C0 < β(μ−2) , we have r2 θ < μ, and hence

β rμ

−

C0 (μ−2)r μ−2

−

> 0. Since μ > 4, there exists a constant 4 < θ < 2∗ such that

lim

|u|→∞

F (x, u) = +∞. |u|θ

(3.11)

λ ⊂ Eλ , by the equivalence of norms in the finite dimensional space, For any finite dimensional subspace E there is a constant C(θ) > 0 such that uθ  C(θ)uEλ ,

λ . ∀u ∈ E

By (f1 ), (f2 ) and (3.11), we know that for any M > 0, there is a constant C(M ) > 0 such that F (x, u)  M |u|θ − C(M )|u|2 ,

∀(x, u) ∈ RN × R.

(3.12)

Hence 1 b max{a, 1}u2Eλ + u4Eλ − M uθθ + C(M )u22 2 4

θ 1 b  max{a, 1}u2Eλ + u4Eλ − M C(θ) uθEλ + C1 (M )u2Eλ 2 4

I(u) 

λ . Consequently, there is a large r1 > 0 such that I < 0 on Eλ \ Br . for all u ∈ E 1 Now, we prove that I satisfies the (PS)c condition for large λ. Indeed, if a sequence {un } ⊂ Eλ is such that I(un ) → c and I  (un ) → 0, we need to prove that {un } possesses a convergent subsequence. From the proof of Lemma 3.2, it is sufficient to prove that {un } is bounded in Eλ . If {un } is unbounded in Eλ , we can assume un Eλ → +∞. Set vn = ununE . Then vn Eλ = 1 and vn s  as vn Eλ = as for s ∈ [2, 2∗ ]. λ Note that 1 I  (un ), un  = un θEλ un θEλ





a|∇un | + λV (x)|un | 2

2

RN

b dx + un θEλ

 

2 |∇un | dx 2

RN

 − RN

f (x, un )un dx. un θEλ

Since θ > 4, we deduce that  lim

n→+∞ RN

f (x, un )un dx = 0. un θEλ

Since Eλ is a Hilbert space and vn Eλ = 1, passing to a subsequence, there is a point v ∈ Eλ such that vn v in Eλ , vn → v in Lsloc (RN ) (2  s < 2∗ ) and vn (x) → v(x) a.e. in RN . Set Ω = {x ∈ RN : v(x) = 0}. If meas(Ω) > 0, then |un (x)| → +∞ a.e. in Ω. By (f1 ), (f2 ), (f5 ) and (3.12), there are two positive constants c1 and c2 such that f (x, u)u  c1 |u|θ − c2 |u|2

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for all (x, u) ∈ RN × R. Hence  RN

f (x, un )un vn 22 θ dx  c v  − c . 1 n 2 θ un θEλ un θ−2 Eλ

Consequently,   0 = lim inf

n→+∞ RN

f (x, un )un vn 22 dx + c 2 un θEλ un θ−2 Eλ



  lim inf

n→+∞

c1 vn θθ



c1 vθθ

|v|θ dx > 0.

= c1 Ω

This is a contradiction. Hence meas(Ω) = 0. Therefore, v(x) = 0 a.e. x ∈ RN . Then, by (V1 ), one has  vn 22 =

 vn2 dx 

vn2 dx + V (x)d

1 2 vn 2Eλ + o(1)  , λd λd

V (x)
for large n. By (f1 ), (f2 ) and (f5 ), we know that there is a constant c > 0 such that μF (x, u) − uf (x, u)  c|u|2 for all (x, u) ∈ RN × R. Consequently, for large n,   1  0← μI(un ) − I  (un ), un  2 un Eλ

 

  μ μF (x, un ) − f (x, un )un − 1 min{a, 1} − dx 2 un 2Eλ 

RN

μ − 1 min{a, 1} − cvn 22 2   c μ − 1 min{a, 1} − .  2 λd 

c Letting λ > 0 be so large that the term ( μ2 − 1) min{a, 1} − λd is positive, we get a contradiction. Therefore, {un } is bounded in Eλ for large λ. Finally, the conclusions follow from Theorem 2.2 and Theorem 9.12 in [18]. This completes the proof. 2

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