Existence of positive solutions for a class of quasilinear Schrödinger equations on RN

Accepted Manuscript Existence of positive solutions for a class of quasilinear Schr¨odinger equations on R N Shaoxiong Chen PII: DOI: Reference:

S0022-247X(13)00340-5 http://dx.doi.org/10.1016/j.jmaa.2013.04.031 YJMAA 17529

To appear in:

Journal of Mathematical Analysis and Applications

Received date: 19 October 2012 Please cite this article as: S. Chen, Existence of positive solutions for a class of quasilinear Schr¨odinger equations on R N , J. Math. Anal. Appl. (2013), http://dx.doi.org/10.1016/j.jmaa.2013.04.031 This is a PDF file of an unedited manuscript that has been accepted for publication. As a service to our customers we are providing this early version of the manuscript. The manuscript will undergo copyediting, typesetting, and review of the resulting proof before it is published in its final form. Please note that during the production process errors may be discovered which could affect the content, and all legal disclaimers that apply to the journal pertain.

*Manuscript

Existence of positive solutions for a class of quasilinear Schr¨odinger equations on RN ∗ Shaoxiong Chen

†

Department of Mathematics, Yunnan Normal University, Kunming, Yunnan 650092, P.R.China

Abstract:

In this paper, we study the following quasilinear Schr¨ odinger equation of

the form 1 −△u + V (x)u − △(u2 )u = α|u|p−1 u + β|u|q−1 u, x ∈ RN , 2   N where N ≥ 3, 2 + N2 ≤ q < p < 2 · N2N −2 − 1, α, β ∈ R . Under appropriate assumptions on V (x), we establish the existence of ground state solutions by a minimization argument.

Key Words: Quasilinear Schr¨ odinger equations, Minimization, Implicit function theorem.

1. Introduction and Preliminaries Consider the following quasilinear Schr¨ odinger equation of the form

where 2 + N2

1 −△u + V (x)u − △(u2 )u = α|u|p−1 u + β|u|q−1 u x ∈ RN , (1.1) 2   N ≤ q < p < 2 · N2N −2 − 1, α, β ∈ R, V ∈ C(R , R). Solutions of equation (1.1)

are related to the existence of standing waves solutions for quasilinear Schr¨ odinger equation ∗

This work was supported partially by the National Natural Science Foundation of China (11261070) and

the Natural Science Foundation of Yunnan Province(Grant No:2009CD042). † [email protected].

1

of the form iψt + △ψ − V (x)ψ + k△(h(|ψ|2 ))h′ (|ψ|2 )ψ + g(x, ψ) = 0, x ∈ RN ,

(1.2)

where V (x) is a given potential, k is a real constant, h and g are real functions. The quasilinear schr¨ odinger equations (1.2) are derived as models of several physical phenomena, such as see [2, 3, 4, 9, 10]. It begins with [11] for the studies on mathematics. Several methods can be used to solve the equation (1.1) with β = 0, such as, the existence of a positive ground state solution has been proved in [5, 12] by using a constrained minimization argument; the problem is transformed to a semilinear one in [1, 6, 8] by a change of variables (dual approach); Nehari method is used to get the existence results of ground state solutions in [7, 13]. In this paper, our aim is to search the existence of positive solution of (1.1) by a minimization argument. We need the following several notations. Let C0∞ (RN ) be the collection of smooth functions with compact support. For N ≥ 3, let s := exponent and

2N N −2

be a Sobolev critical

D 1,2 (RN ) := {u ∈ Ls (RN ) : ∇u ∈ L2 (RN )} with the norm kuk2D1,2

=

Z

RN

|∇u|2 dx.

By the Sobolev inequality, D 1,2 (RN ) is continuously embedded into Ls (RN ). Let H 1 (RN ) := {u ∈ L2 (RN ) : ∇u ∈ L2 (RN )} with the inner product hu, viH 1 = and the norm

Z

RN

[∇u · ∇v + uv]dx 1/2

kukH 1 = hu, uiH 1 . In the following, we always assume V ∈ C(RN , R) and inf V (x) ≥ 1. Let us consider RN

the following four assumptions, respectively. 2

(V1 )

lim V (x) = +∞.

|x|→∞

(V2 ) V (x) is radially symmetric. (V3 ) V (x) is periodic in each variable of x1 , · · · , xN . (V4 ) V (x) ≤ V∞ := lim V (y) < ∞ for all x ∈ RN . |y|→∞

The equation (1.1) is the Euler-Lagrange equation of the energy functional 1 J(u) = 2

Z

RN



1+u

2




|∇u| + V (x)u 2

2



dx −

Z

RN



 β α p+1 q+1 |u| + |u| dx. p+1 q+1

But the difficulty is the differentiability of J, in fact under our growth condition of the nonlinearity J is not even defined in H 1 (RN ). Hence we can not directly use the Lagrange multiplier theorem to seek the positive solution (1.1) in H 1 (RN ). To overcome these difficulties, a change of variable is used in [6]. Following the idea, let f be defined by f ′ (t) = p

1 1 + f 2 (t)

on [0, +∞), f (0) = 0 and f (−t) = −f (t) on (−∞, 0]. Then f has following properties (see [14]): (f1 ) f is uniquely defined C ∞ function and invertible. (f2 ) 0 < f ′ (t) ≤ 1 for all t ∈ R. (f3 ) |f (t)| ≤ |t| for all t ∈ R. (f4 ) lim

t→0

(f5 ) (f6 )

f (t) t

lim

t→+∞ 1 2 f (t)

= 1.

f√ (t) t

=

√

√ f (t) 2, lim √ = − 2. t→−∞

|t|

≤ tf ′ (t) ≤ f (t) for all t ≥ 0 and f (t) ≤ tf ′ (t) ≤ 21 f (t) for all t ≤ 0.

(f7 ) |f (t)| ≤

√ p 2 |t| for all t ∈ R.

(f8 ) The function f 2 (t) is strictly convex.

3

(f9 ) There exists a positive constant C such that   C|t|, |t| ≤ 1, |f (t)| ≥ 1  C|t| 2 , |t| ≥ 1.

(f10 ) There exist positive constants C1 and C2 such that |t| ≤ C1 |f (t)| + C2 |f (t)|2 for all t ∈ R. (f11 ) |f (t)f ′ (t)| ≤ 1 for all t ∈ R. (f12 ) For each ξ > 0, there exists C(ξ) > 0 such that f 2 (ξt) ≤ C(ξ)f 2 (t). After the change u = f (v) of variable, J(u) can be reduced to 1 I(v) := 2

Z

RN


|∇v| + V (x)f (v) dx − 2

2

Z

Let E := {v ∈ D

1,2

(R ) : N

RN

Z



RN

 β α p+1 q+1 |f (v)| + |f (v)| dx. (1.3) p+1 q+1

V (x)f 2 (v)dx < +∞}.

Then by (f12 ) we can conclude that E is a linear space. By f (−t) = −f (t) and the convexity of f 2 (t), we can define the norm on E by 

kvkE := kvkD1,2 + inf ξ 1 + ξ>0

Z

RN

V (x)f

2

ξ

−1


v(x) dx .

In the following, we use C or Ci to denote various positive constants. The following Proposition 1.1 appeared in [6], but its proof is omitted there. In order to the completeness, we give also the proof.

Proposition 1.1.

(1) E is a Banach space. R R (2) If kvn − vkE → 0, then RN V (x)f 2 (vn )dx → RN V (x)f 2 (v)dx. 4

Proof. Let {vn } be a Cauchy sequence in E. For every ǫ > 0, there exists N > 0 such that kvn − vn+p kE < ǫ for all n ≥ N and p ∈ N. For ǫ =

1 , 2i+1

i = 1, 2, · · · , we can get a subsequence {vni } of {vn }

such that kvni+1 − vni kE <

1 , i = 1, 2, · · · . 2i+1

Hence there exist ξi ∈ (0, 1), i = 1, 2, · · · , such that   Z

−1 2 ξi 1 + V (x)f ξi vni+1 (x) − vni (x) dx < RN

1 2i+1

(1.4)

.

For every x ∈ RN , we define

k X 

 1 ξi V (x)f 2 ξi−1 vni+1 (x) − vni (x) gk (x) = + V (x)f 2 (4vn1 (x)), 4 i=1

and

∞ X 

g(x) =

i=1



 1 ξi V (x)f 2 ξi−1 vni+1 (x) − vni (x) + V (x)f 2 (4vn1 (x)). 4

We have lim gk (x) = g(x) for all x ∈ RN . By the Fatou Lemma, k→∞

Z

RN

g(x)dx =

Z

lim gk (x)dx =

RN k→∞

Z

lim inf gk (x)dx ≤ lim inf

RN k→∞

k→∞

Z

RN

gk (x)dx.

But lim inf k→∞

Z

RN

gk (x)dx = lim inf k→∞

Z

RN

(

k X  i=1



 ξi V (x)f 2 ξi−1 vni+1 (x) − vni (x)

) 1 + V (x)f 2 (4vn1 (x)) dx 4 k X 1 ≤ lim inf + C1 i+1 k→∞ 2 i=1

≤ C. So, g ∈ L1 (RN ). Furthermore, ∞ X i=1



ξi V (x)f 2 ξi−1 vni+1 (x) − vni (x) < ∞ a.e. on RN . 5

.

If x0 ∈ RN is such that ∞ X i=1



ξi V (x0 )f 2 ξi−1 vni+1 (x0 ) − vni (x0 ) < ∞,

then we assert that {vnk (x0 )} is a Cauchy sequence in R1 . In fact, for any ǫ > 0, by (f1 ), there exists a δ > 0 such that |v − 0| = |f −1 (f (v)) − 0| < ǫ, whenever |f (v)| = |f (v) − f (0)| < δ. For the above δ, there is N > 0 such that k+p X

i=k+1



ξi V (x0 )f 2 ξi−1 vni+1 (x0 ) − vni (x0 ) < V (x0 )δ2 ,

∀p ∈ N,

whenever k > N . By the convexity of f 2 (v) and f (0) = 0, we have V (x0 )f

2

k+p X


ξi V (x0 )f 2 ξi−1 vni+1 (x0 ) − vni (x0 ) . vnk+p (x0 ) − vnk+1 (x0 ) ≤ i=k+1

Hence

It implies that


|f vnk+p (x0 ) − vnk+1 (x0 ) | < δ, f or k > N and p ∈ N. |vnk+p (x0 ) − vnk+1 (x0 )| < ǫ, f or k > N and p ∈ N.

This shows that the sequence {vnk (x0 )} is a Cauchy sequence in R1 . Hence there exists v(x0 ) ∈ R1 such that lim vnk (x0 ) = v(x0 ), and hence vnk (x) → v(x) a.e. on RN . Using k→∞

the convexity of

f 2,

we have

V (x)f (vnk (x)) ≤ 2

k−1 X i=1



1 ξi V (x)f 2 ξi−1 vni+1 (x) − vni (x) + V (x)f 2 (4vn1 (x)) ≤ g(x). 4

Consequently, it follows from Lebesgue dominated convergence theorem that Z

RN

V (x)f 2 (v)dx = lim

Z

k→∞ RN

V (x)f 2 (vnk )dx.

Since {vn } is also a Cauchy sequence in the norm k · kD1,2 , there exists v ′ ∈ D 1,2 (RN ) such that kvn − v ′ kD1,2 → 0, kvn − v ′ kL2∗ → 0. Hence kvnk − v ′ kD1,2 → 0, kvnk − v ′ kL2∗ → 0 and 6

there is a subsequence {vnkl } of {vnk } such that vnkl (x) → v ′ (x) a.e. x ∈ RN . Hence v = v ′ a.e., kvn − vkD1,2 → 0 and kvn − vkLs → 0. We can see Z

V (x)f (v)dx = lim 2

Z

n→∞ RN

RN

V (x)f 2 (vn )dx.

It follows that v ∈ E. We shall show that kvn − vkE → 0. In fact, for every ǫ > 0, there exists N > 0 and ξ > 0 such that, for n > N , 

ξ 1+ Replacing vn1 and

Z

RN

V (x)f

by vn and

1 2i+1

2

ξ

ξ 2i+1

−1


(vn+p − vn ) dx < ǫ, ∀p ∈ N.

(1.5)

in (1.4), by the convexity of f 2 , we have

k−1 X


ξi V (x)f 2 ξi−1 vni+1 (x) − vni (x) ≤ g(x), a.e. on RN , ξV (x)f 2 ξ −1 (vnk (x) − vn (x)) ≤ ξ ξ i=1

where have

k−1 P i=1

ξ

ξi ξ

≤

Z

RN

k−1 P i=1

ξ 1 2i+1 ξ

V (x)f

2

ξ

≤ 1. Using Lebesgue dominated convergence theorem again, we −1




(v − vn ) dx = lim ξ k→∞

Z

RN


V (x)f 2 ξ −1 (vnk − vn ) dx.

In (1.5), by replacing vn+p by vnk and taking limit k → ∞, we have 

inf ξ 1 +

ξ>0

Z

RN

V (x)f

2

ξ

−1

  Z
(v − vn ) dx ≤ ξ 1 +

RN

V (x)f

2

ξ

−1






(v − vn ) dx ≤ ǫ.

Consequently, we conclude that kvn − vkE → 0. Thus (1) holds. Assume kvn − vkE → 0. By using the similar argument, and noting the previous proof of v ∈ E, we can conclude R R 2 2 RN V (x)f (vn )dx → RN V (x)f (v)dx. Then (2) holds. This completes the proof. Define

ma = inf Ea (u), Ma

where Ma = {u ∈ H 1 (RN ) : kukLp+1 = a}, a > 0, and Ea (u) =

1 2

Z

RN




 1 + u2 |∇u|2 + V (x)u2 dx − 7

β q+1

Z

RN

|u|q+1 dx.

We also define wa = inf Fa (v), Wa

where Wa = {v ∈ E : kf (v)kLp+1 = a}, a > 0, and 1 Fa (v) = 2

Z


|∇v| + V (x)f (v) dx − 2

RN

We have the following fact.

Proposition 1.2.

2

Z

RN

|f (v)|q+1 dx.

ma = wa for every a > 0.

Proof. For any v ∈ Wa , let u = f (v), then Z Z Z 2 2 |∇v| dx < ∞, |∇u| dx ≤ RN

β q+1

RN

RN

2

u dx ≤

Z

RN

V (x)f 2 (v)dx < ∞,

So u ∈ Ma . It follows that Fa (v) = Ea (f (v)) = Ea (u) ≥ ma . Hence wa ≥ ma . Moreover, For any u ∈ Ma , let v = f −1 (u), then u = f (v). We may assume Ea (u) < ∞. Since R u ∈ H 1 (RN ), RN |f (v)|p+1 dx = ap+1 and 2 < q + 1 < p + 1, then u ∈ Lq+1 (RN ) by H¨older inequality. We have Z Z 
 1 β |u|q+1 dx < ∞. 1 + u2 |∇u|2 + V (x)u2 dx ≤ Ea (u) + 2 RN q + 1 RN R R Then RN V (x)f 2 (v)dx = RN V (x)u2 dx < ∞. It shows that v ∈ Wa . It implies that Ea (u) = Ea (f (v)) = Fa (v) ≥ wa , and hence ma ≥ wa . This completes the proof.

Proposition 1.3.

(See Proposition 2.2 in [6]) The map: v → f (v) from E into

Lr (RN ) is continuous for 2 ≤ r ≤ 2s.

Proposition 1.4.

(See Proposition 2.3 in [6]) (1) F is well defined and continuous

on E. (2) F is Gauteaux-differentiable. For v ∈ E, the G-derivative F ′ (v) is a continuous linear functional, and F ′ (v) is continuous in v in the strong-weak topology, that is, if vn → v strongly in E, then F ′ (vn ) ⇀ F ′ (v) weakly. 8

Remark.

The Proposition 1.4 doesn’t show that F (v) is C 1 , so we can not use the

Lagrange multiplier theorem. But we can get our conclusion we want exactly by a similar argument for the Lagrange multiplier theorem.

2. Main results It is well known that if v ∈ E is a critical point of I, i.e. Z Z V (x)f (v)f ′ (v)ϕdx ∇v · ∇ϕdx + hI ′ (v), ϕi = N N R R Z Z p−1 ′ |f (v)|q−1 f (v)f ′ (v)ϕdx |f (v)| f (v)f (v)ϕdx − β −α RN

RN

= 0, for all ϕ ∈ C0∞ (RN ), then u := f (v) is a weak solution of the equation 1 −△u + V (x)u − △(u2 )u = α|u|p−1 u + β|u|q−1 u, 2

x ∈ RN .

(1.1)

Our main results are the following Theorem 2.1 and Theorem 2.2. The idea of our proof is based on the work in [5] and [6].

Theorem 2.1.

Let N ≥ 3, 2 +

2 N

≤ q < p < 2s − 1, β <

N N +2 .

Assume (V1 )

or (V2 ). Then for every a > 0, there exists α(a) ∈ R1 such that the equation (1.1) with α = α(a) has a positive weak solution u ∈ Ma . Furthermore, the following assertions hold: (1) If a → 0+ and 2 +

2 N

≤ q, then α(a) → +∞.

(2) If a → ∞ and 3 ≤ q, then −β

p−q N ≤ lim inf α(a) ≤ lim sup α(a) ≤ 0, for 0 < β < , a→∞ p−1 N+2 a→∞

and for β ≤ 0, α(a) > 0 and

lim α(a) = 0.

a→∞

Proof. Step 1: By the assumptions of (V1 ) and(V2 ), wa is achieved at some 0 ≤ va ∈ Wa with va 6= 0. 9

Let {vn } ∈ Wa be a minimizing sequence for wa . Set un = f (vn ). Then {un } ∈ Ma is a minimizing sequence for ma . We can assume un ≥ 0. It shows that Ea (un ) → ma , so there exists C > 0 such that C ≥ Ea (un ) Z Z 
 1 β 2 2 2 = 1 + un |∇un | + V (x)un dx − |un |q+1 dx 2 RN q + 1 RN  Z Z Z β 1 |un |2 dx − |∇un |2 dx + |un |q+1 dx. ≥ 2 q + 1 N N N R R R By H¨older inequality, Z

RN

|un |q+1 dx ≤

Z

≤λ =λ where λ =

p−q p−1 .

ZR

N

ZR

N

RN

1−λ |un |p+1 dx RN Z 2 |un |p+1 dx |un | dx + (1 − λ)

|un |2 dx

λ Z

RN

|un |2 dx + (1 − λ)ap+1 ,

Then

C ≥ Ea (un )   Z  Z Z 1 β 2 2 2 p+1 ≥ |un | dx − |∇un | dx + |un | dx + (1 − λ)a λ 2 q+1 N N RN   R RZ Z 1 β(q − 1) β(p − q) 2 2 |un | dx − ≥ |∇un | dx + − ap+1 . 2 (q + 1)(p − 1) (q + 1)(p − 1) RN RN Because of β <

N 1 N +2 , 2

−

β(p−q) (q+1)(p−1)

> 0. It implies that {un } is bounded in H 1 (RN ).

Similarly, we can prove that {∇(u2n )} is bounded in L2 (RN ). By the compact embedding result from H 1 (RN ) into Lr (RN ) for 2 ≤ r < s, we may assume that un ⇀ ua in H 1 (RN ), un → ua in Lr (RN ) for 2 ≤ r < s and un (x) → ua (x) a.e. x ∈ RN . In the other hand, by the Sobolev inequality, it follows from the boundedness of {∇(u2n )} that {u2n } is bounded in Ls (RN ). By interpolation we know un → ua in Lr (RN ) for 2 ≤ r < 2s. Hence ua ∈ Ma . Since again un ≥ 0, ua ≥ 0 and ua 6= 0. Using the same argument as the process of the proof of lemma 2.1 in [5] and noting that

10

un → ua in Lq+1 (RN ), we have m = lim Ea (un ) n→∞   Z Z 
 β 1 2 2 2 q+1 1 + un |∇un | + V (x)un dx − |un | dx ≥ lim inf n→∞ 2 RN q + 1 RN ≥ Ea (ua ).

Hence ma is achieved at ua , and hence va = f −1 (ua ) ∈ Wa , Fa (va ) = Ea (f (va )) = Ea (ua ) = wa and the property (f6 ) implies va ≥ 0 and va 6= 0. Step 2: Set hr (v) =

1 r

R

RN

|f (v)|r dx for 3 +

≤ r < 2s. Then hr (v) ∈ C 1 (E, R1 ).

2 N

For any ϕ ∈ E, we follow from the proof of the Proposition 2.3 in [6] with r instead of p + 1 that |hh′r (v), ϕi|

Z =

RN

|f (v)|

r−2

f (v)f (v)ϕdx ′

≤ Ckf kr−2 (r−2)

2N N+2

kϕkLs

≤ Ckf kr−2 (r−2)

2N N+2

kϕkE ,

L

L

′ ∗ where 2 ≤ (r − 2) N2N +2 < 2s. So hr (v) ∈ E .

Let vn → v in E. Up to a subsequence, we can assume vn → v a.e. in RN and f (vn ) → f (v) in Lr (RN ) for 2 ≤ r < 2s. Hence |hh′r (vn ) − h′r (v), ϕi| Z = (|f (vn )|r−2 f (vn )f ′ (vn ) − |f (v)|r−2 f (v)f ′ (v))ϕdx RN

≤C But

Z

RN

2N |f (vn )|r−2 f (vn )f ′ (vn ) − |f (v)|r−2 f (v)f ′ (v) N+2 dx

 N+2


2N |f (vn )|r−2 f (vn )f ′ (vn ) − |f (v)|r−2 f (v)f ′ (v) N+2   2N 2N ≤ C |f (vn )|(r−2) N+2 + |f (v)|(r−2) N+2 . 11

2N

kϕkE .

Using the Fatou’s lemma, one has Z 2N 2C |f (v)|(r−2) N+2 dx N   ZR  2N 2N lim inf C |f (vn )|(r−2) N+2 + |f (v)|(r−2) N+2 = RN n→∞

′

r−2

′

r−2

 2N
N+2 f (v)f (v) dx ′

− |f (vn )| f (vn )f (vn ) − |f (v)| Z    2N 2N ≤ lim inf C |f (vn )|(r−2) N+2 + |f (v)|(r−2) N+2 r−2

n→∞

RN

RN

n→∞

Hence


2N N+2 f (v)f (v) dx ′

− |f (vn )| f (vn )f (vn ) − |f (v)| Z   2N 2N C |f (vn )|(r−2) N+2 + |f (v)|(r−2) N+2 dx = lim n→∞ RN   Z
2N r−2 ′ r−2 ′ N+2 dx |f (vn )| f (vn )f (vn ) − |f (v)| f (v)f (v) + lim inf − n→∞ RN Z 2N 2C|f (v)|(r−2) N+2 dx = RN Z
2N − lim sup |f (vn )|r−2 f (vn )f ′ (vn ) − |f (v)|r−2 f (v)f ′ (v) N+2 dx. r−2

lim sup n→∞

and hence lim

Z

Z

2N

RN

n→∞ RN

(|f (vn )|r−2 f (vn )f ′ (vn ) − |f (v)|r−2 f (v)f ′ (v)) N+2 dx ≤ 0,


2N |f (vn )|r−2 f (vn )f ′ (vn ) − |f (v)|r−2 f (v)f ′ (v) N+2 dx = 0.

It follows that hr (v) ∈ C 1 (E, R1 ) for 3 +

2 N

≤ r < 2s.

Step 3: For any a > 0, there exists α(a) ∈ R1 such that 0 < ua = f (va ) ∈ Ma is a weak solution of the equation (1.1) with α = α(a). In fact, by Proposition 1.4, hFa′ (v), ϕi

=

Z

RN

∇v · ∇ϕdx +

Z

RN

′

V (x)f (v)f (v)ϕdx − β

Z

RN

|f (v)|q−1 f (v)f ′ (v)ϕdx

and Fa′ (v) ∈ E ∗ for all v ∈ E. Since h′p+1 (va ) 6= 0, h′p+1 (v) ∈ C(E, R1 ) and va ∈ Wa , the implicit function theorem implies that for all v ∈ N (h′p+1 (va )) (the null space of h′p+1 (va )), there exists a C 1 - map g : [0, 1] → Wa such that g(0) = va and g ′ (0) = v. 12


Now, we prove hFa′ (va ), vi = 0 for all v ∈ N h′p+1 (va ) .

Indeed, for every t > 0, g(t) = va + tv + o(t) ∈ Wa , where

o(t) t

→ 0 as t → 0. Let

Φ(r) = Fa (va + r (tv + o(t))) . By proposition 1.4, Φ(r + λ) − Φ(r) λ→0 λ Fa (va + (r + λ) (tv + o(t))) − Fa (va + r (tv + o(t))) = lim λ→0 λ

Φ′ (r) = lim

= hFa′ (va + r (tv + o(t))) , tv + o(t)i. Hence there exists an θ ∈ (0, 1) such that Fa (va + tv + o(t)) − Fa (va ) = hFa′ (va + θ (tv + o(t))) , tv + o(t)i = thFa′ (va + θ (tv + o(t))) , vi + thFa′ (va + θ (tv + o(t))) ,

o(t) i. t

Take limit t → 0. By proposition 1.4, one has Fa′ (va + θ(tv + o(t))) ⇀ Fa′ (va ) weakly. It follows that hFa′ (va + θ(tv + o(t))), vi → hFa′ (va ), vi and {Fa′ (va + θ(tv + o(t)))} is bounded. Since

o(t) t

→ 0 as t → 0, we have hFa′ (va + θ(tv + o(t))), o(t) t i → 0.

Since Fa (va ) = wa , one has 0 ≤ Fa (va + tv + o(t)) − Fa (va ) = thFa′ (va + θ(tv + o(t))), vi + thFa′ (va + θ(tv + o(t))),

o(t) i. t

Hence 0 ≤ hFa′ (va + θ(tv + o(t))), vi + hFa′ (va + θ(tv + o(t))),

o(t) i. t

Take limit t → 0. We get hFa′ (va ), vi ≥ 0. By arbitrariness of v, one has hFa′ (va ), −vi ≥ 0. It follows that hFa′ (va ), vi = 0, for every v ∈ N (h′p+1 (v0 )). Set v ′ ∈ E be such that hh′p+1 (va ), v ′ i = 1. For every ϕ ∈ E, let ψ = ϕ − hh′p+1 (va ), ϕiv ′ . 13

Then ψ ∈ N (h′p+1 (va )). It means hFa′ (va ), ψi = 0, i.e. hFa′ (va ), ϕi = hFa′ (va ), v ′ ihh′p+1 (va ), ϕi. Put α = α(a) = hFa′ (va ), v ′ i. We have hFa′ (va ), ϕi = αhh′p+1 (va ), ϕi, namely, Z =α

RN

Z

RN

Z

V (x)f (va )f ′ (va )ϕdx Z |f (va )|q−1 f (va )f ′ (va )ϕdx. |f (va )|p−1 f (va )f ′ (va )ϕdx + β

∇va · ∇ϕdx +

RN

(2.1)

RN

It implies that ua = f (va ) ∈ Ma is a weak solution of the equation (1.1). Moreover, the maximum principle implies ua > 0. Step 4: α(a) → +∞ as a → 0+ . Multiplying the equation (2.1) by ua and integrating on RN , we have Z Z 
 2 2 2 |ua |q+1 dx = αap+1 . 1 + 2ua |∇ua | + V (x)ua dx − β RN

RN

Using a similar argument in Step 1, we know Z Z Z 
 1 β(p − q) β(q − 1) p+1 1 2 2 2 2 ua dx− u2a dx− a ≤ αap+1 . 1 + 2ua |∇ua | + V (x)ua dx+ 2 RN 2 RN p−1 p−1 RN

Furthermore, Z  Z 
 β(q − 1) p+1 1 β(p − q) 1 2 2 2 − u2a dx − a ≤ αap+1 . 1 + 2ua |∇ua | + V (x)ua dx + 2 RN 2 p−1 p−1 RN

Since β <

N 1 N +2 , 2

1 2

Z

−

RN



β(p−q) p−1

> 0. Hence


 β(q − 1) p+1 a . 1 + 2u2a |∇ua |2 + V (x)u2a dx ≤ αap+1 + p−1

(2.2)

Now, we claim that α → +∞ as a → 0+ . Indeed, if the conclusion is invalid, then there exists a sequence an → 0+ such that αn := α(an ) ≤ C and Z Z Z 
 |un |q+1 dx = αn 1 + 2u2n |∇un |2 + V (x)u2n dx − β RN

RN

14

RN

|un |p+1 dx,

where un := uan . By the H¨older inequality and the Sobolev inequality, we have Z p+1 |un |p+1 dx an = RN

(1−r)(p+1)

r(p+1)

kun kL2s ≤ kun kL2   p+1 + ku k ≤ C1 kun kp+1 n L2s L2   p+1 p+1 2 2 ≤ C1 kun kL2 + kun kLs " Z  p+1 ≤ C1

≤ C1 ≤ C1 ≤ C1 where

1 p+1

=

r 2

+

1−r 2s ,

2

2

RN

" Z

RN

Z

V

(x)u2n dx

RN



1+

2u2n




+

 p+1

|∇un | dx + 2

RN

Z

|un | dx

2

Z

RN

p+1

ku2n kD21,2 + V

Z

RN

# |∇(u2n )|2 dx

(x)u2n dx

|∇un | + V 2

+

(x)u2n



Z

RN

dx

 p+1 # 2

|∇(u2n )|2 dx

 p+1 2

Z

2

,

and C1 is a constant independent of n. From (2.2) with α = αn ,

taking limit n → ∞, one has Z 
 1 1 + 2u2n |∇un |2 + V (x)u2n dx 2 RN β(q − 1) p+1 a ≤ αn ap+1 + n p−1 n   β(q − 1) ≤ C+ ap+1 → 0. n p−1 Hence

 p+1

(2.3)




 1 + 2u2n |∇un |2 + V (x)u2n dx RN   β(q − 1) ap+1 ≤2 C+ n p−1  Z β(q − 1) |un |p+1 dx =2 C+ p−1 RN  Z  p+1  2 
 β(q − 1) 2 2 2 . 1 + 2un |∇un | + V (x)un dx ≤ 2C1 C + p−1 RN

It follows that 1 ≤ 2C1



β(q − 1) C+ p−1

 Z

RN



1+

2u2n




|∇un | + V 2

15

(x)u2n



dx

 p+1 −1 2

→ 0, (n → ∞)

This is a contradiction. This shows that (1) holds. Step 5: The proof of (2). For q ≥ 3 and a ≥ 1, let u ∈ Ma be such that Ea (u) = ma and Z

RN




 1 + 2u2 |∇u|2 + V (x)u2 dx − β

Z

RN

Then u1 = a1 u ∈ M1 and Z

RN




 1 + 2(au1 )2 |∇(au1 )|2 + V (x)(au1 )2 dx − β

|u|q+1 dx = αap+1 .

Z

RN

|au1 |q+1 dx = αap+1 .

For large a, we have Z Z 
 |au1 |q+1 dx 1 + u21 |∇(au1 )|2 + V (x)(au1 )2 dx − β αap+1 ≥ N N R R   Z Z 
 β q+1 2 2 2 2 1 |u1 | dx 1 + u1 |∇u1 | + V (x)u1 dx − ≥a 2 RN q + 1 RN Z Z Z 1 a2 β + |au1 |q+1 dx |au1 |2 dx + |u1 |q+1 dx − β 2 RN q + 1 RN RN  Z Z 1 1 ≥ a2 m 1 + |au1 |q+1 dx |au1 |2 dx − β 1 − 2 RN (q + 1)aq−1 N R Z 1 2 2 |au1 | dx ≥ a m1 + 2 RN    Z 1 p−q q − 1 p+1 2 −β 1− |au1 | dx + a (q + 1)aq−1 p − 1 RN p−1   Z p − q (q + 1)aq−1 − 1 1 |au1 |2 dx −β · ≥ a2 m 1 + 2 p−1 (q + 1)aq−1 N R   1 q − 1 p+1 −β 1− a (q + 1)aq−1 p − 1   1 q − 1 p+1 ≥ a2 m 1 − β 1 − a q−1 (q + 1)a p−1 q − 1 p+1 ≥ a2 m 1 − β a , p−1

(2.4)

and hence lim inf α ≥ −β a→∞

In (2.4),

1 2

p−q − β p−1 ·

(q+1)aq−1 −1 (q+1)aq−1

q−1 . p−1

> 0 for large a. Indeed, set g(t) =

1 2

p−q − β p−1 t. Hence g(t)

is continuous on R1 and g(1) > 0. Hence there exists δ > 0 such that g(t) > 0 for all 16

t ∈ (1 − δ, 1 + δ). Taking limit a → +∞, one has a0 > 0 such that 1 − δ < 1 2

−

β p−q p−1

·

(q+1)aq−1 −1 (q+1)aq−1

(q+1)aq−1 −1 (q+1)aq−1

> 0 for a > a0 .

(q+1)aq−1 −1 (q+1)aq−1

→ 1. Consequently, there is   q−1 −1 < 1 + δ for a > a0 . So g (q+1)a > 0, that is, (q+1)aq−1

If β ≤ 0, then (2.4) implies that α > 0 and lim inf α ≥ 0. On the other hand, For a→∞

every u ∈ M1 , au ∈ Ma . Hence ma ≤ Ea (au) ≤ aq+1 E1 (u) for a ≥ 1 and q ≥ 3. It shows ma ≤ aq+1 m1 . Let u0 ∈ Ma be such that Z

RN




 1 + 2u20 |∇u0 |2 + V (x)u20 dx − β

Then Ea (u0 ) = ma and Z p+1 αa =






RN

|u0 |q+1 dx = αap+1 .



Z

|u0 |q+1 dx 1+ |∇u0 | + V dx − β RN  Z 
 1 ≤ (q + 1) 1 + u20 |∇u0 |2 + V (x)u20 dx 2 RN  Z β q+1 − |u0 | dx q + 1 RN RN

2u20

Z

2

(x)u20

= (q + 1)ma ≤ (q + 1)aq+1 m1 . Since p > q, lim sup α ≤ 0. Hence a→∞

−β

N p−q ≤ lim inf α(a) ≤ lim sup α(a) ≤ 0, for 0 < β < , a→∞ p−1 N +2 a→∞

and lim α = 0 whenever β ≤ 0. This completes the proof. a→∞

Theorem 2.2.

Let N ≥ 3, 3 ≤ q < p < 2s − 1, β <

N +2 8 .

Assume that (V3 ) or

(V4 ) hold. Then for every a > 0, there exists α(a) ∈ RN such that the equation (1.1) has a positive weak solution ua ∈ Ma . Furthermore, the following assertions hold: (1) If a → 0+ , then α(a) → +∞. (2) If a → ∞, then −β

p−q N+2 ≤ lim inf α(a) ≤ lim sup α(a) ≤ 0, for 0 < β < , a→∞ p−1 8 a→∞ 17

and for β ≤ 0,

α(a) > 0 and

lim α(a) = 0.

a→∞

Proof. By the proof of Theorem 2.1, we know that it is sufficient to prove that for any a > 0, ma is achieved at some 0 < ua ∈ Ma . We first consider the case (V3 ). We need to consider the case a = 1 only. Let {un } ⊂ M1 be a minimizing sequence for m1 . From the Step 1 in the proof of Theorem 2.1, we know C ≥ E1 (un ) Z Z 
 1 β = |un |q+1 dx 1 + u2n |∇un |2 + V (x)u2n dx − 2 RN q + 1 RN  Z Z
1 ≥ |un |2 dx 1 + u2n |∇un |2 dx + 2 RN RN   Z Z q−1 p−q β 2 p+1 |un | dx + |un | dx − q + 1 p − 1 RN p − 1 RN    Z Z Z 1 1 β(p − q) 2 2 2 2 ≥ |un | dx |∇un | dx + u |∇un | dx + − 2 RN n 2 (q + 1)(p − 1) RN RN β(q − 1) − , (q + 1)(p − 1) p−q where 12 − β (q+1)(p−1) > 0 for 3 ≤ q < p < 2s − 1 and β < N 8+2 . It follows that both R R 2 2 2 RN un |∇un | dx and RN |un | dx are bounded. By the concentration-compactness principle,

we know that there is γ ∈ (0, 1] and xn ∈ RN such that for any 0 < ǫ < γ, there exists

R > 0 such that for every R′ ≥ R, lim inf n→∞

and lim inf n→∞

Z

Z

BR (xn )

RN \BR′ (xn )

|un |p+1 dx ≥ γ − ǫ,

|un |p+1 dx ≥ (1 − γ) − ǫ.

We may assume the components of xn are integer multiples of the periods of V (x). Then {un (· + xn )} is still a minimizing sequence. We claim that γ = 1. Otherwise, γ < 1. Let ηR : [0, +∞) → [0, 1] be a smooth function satisfying ηR (t) = 1 for 0 ≤ t ≤ R and ηR (t) = 0 18

′ (t) ≤ for t > R and ηR

2 R.

Define

yn (x) = ηR (|x − xn |)un (x) ,

zn (x) = (1 − ηR (|x − xn |))un (x).

Then, for large n, Z

p+1

RN

|yn |

dx − γ ≤ ǫ,

Z

p+1

RN

|zn |

dx − (1 − γ) ≤ ǫ.

Notice that Z Z 
 β 1 |un |q+1 dx 1 + u2n |∇un |2 + V (x)u2n dx − 2 RN q + 1 RN Z Z  

  1 1 ≥ 1 + yn2 |∇yn |2 + V (x)yn2 dx + 1 + zn2 |∇zn |2 + V (x)zn2 dx 2 RN 2 RN Z C β |un |q+1 dx − , − q + 1 RN R

19

where C > 0 is a constant independent of n, ǫ and R. We have m1 + o(1) = E1 (un )     yn zn q+1 ≥ kyn kq+1 E + kz k E n Lp+1 1 Lp+1 1 ky k p+1 kzn kLp+1 n L Z     1 2 2 kyn k2Lp+1 − kyn kq+1 + |∇y | + V (x)y dx n p+1 n L 2 RN Z    1 2 2 + kzn k2Lp+1 − kzn kq+1 |∇z | + V (x)z dx n p+1 n L 2 RN Z  1 yn2 |∇yn |2 dx kyn k4Lp+1 − kyn kq+1 + p+1 L 2 RN Z 1 q+1 4 + zn2 |∇zn |2 dx kzn kLp+1 − kzn kLp+1 2 RN  Z Z Z C β q+1 q+1 q+1 |un | dx − |zn | dx − |yn | dx + + q+1 R RN RN RN   ≥ kyn kq+1 + kzn kq+1 m1 Lp+1 Lp+1 Z   1 |∇yn |2 + V (x)yn2 dx kyn k2Lp+1 − kyn kq+1 + p+1 L 2 RN ! Z Z Z C β q+1 q+1 q+1 |un | dx − + |zn | dx − |yn | dx + q+1 R B2R (xn ) B2R (xn ) B2R (xn ) h i q+1 q+1 ≥ (γ − ǫ) p+1 + (1 − γ − ǫ) p+1 m1  Z  q−1  2 1 p+1 p+1 1 − (γ + ǫ) |∇yn |2 + V (x)yn2 dx + (γ − ǫ) 2 RN ! Z Z Z C β |un |q+1 dx − + |zn |q+1 dx − |yn |q+1 dx + q+1 R B2R (xn ) B2R (xn ) B2R (xn ) h i q+1 q+1 C ≥ (γ − ǫ) p+1 + (1 − γ − ǫ) p+1 m1 − R i h  Z q−1 2 1 |∇un (x + xn )|2 + V (x + xn ) (un (x + xn ))2 dx + (γ − ǫ) p+1 1 − (γ + ǫ) p+1 2 B2R (0) Z β + |yn (x + xn )|q+1 dx q+1 B2R (0)  Z Z q+1 q+1 |un (x + xn )| dx . |zn (x + xn )| dx − + B2R (0)

B2R (0)

20

Because of q < p and |ar − br | ≤ |a − b|r (for a > 0, b > 0, r ∈ (0, 1]), it shows that m1 + o(1) q+1 C ≥ (1 − 2ǫ) p+1 m1 − R i h  Z q−1 2 1 |∇un (x + xn )|2 + V (x + xn ) (un (x + xn ))2 dx + (γ − ǫ) p+1 1 − (γ + ǫ) p+1 2 B2R (0) Z Z β q+1 + |zn (x + xn )|q+1 dx |yn (x + xn )| dx + q+1 B2R (0) B2R (0)  Z q+1 |un (x + xn )| dx . −

B2R (0)

Since {un (· + xn )} is a minimizing sequence, {un (· + xn )} and {u2n (· + xn )} are bounded in H 1 (RN ) (see the proof of lemma 3.1.in [5]). Up to a subsequence, we may assume un (· + xn ) ⇀ u in H 1 (RN ) and un (x + xn ) → u(x) a.e. x ∈ RN . Note that {u2n (· + xn )} is bounded in H 1 (B2R (0)). We may assume u2n (· + xn )|B2R (0) → u2 |B2R (0) in L

q+1 2

(B2R (0))

for 4 ≤ q + 1 < 2s. Set u2R = u|B2R (0) . Then u2R 6= 0 for large R because of γ > 0. It is clear that yn (x + xn ) → η2R (|x|)u2R (x), zn (x + xn ) → (1 − η2R (|x|))u2R (x) in Lq+1 (B2R (0)). Take limit n → ∞. We have q+1

m1 ≥ (1 − 2ǫ) p+1 m1

 Z q−1   2 1 p+1 p+1 + (γ − ǫ) |∇u2R |2 + V (x)u22R dx 1 − (γ + ǫ) 2 B2R (0) Z Z β |(1 − η2R (|x|))u2R (x)|q+1 dx + |η2R (|x|)u2R (x)|q+1 dx + q+1 B2R (0) B2R (0)  Z C |u2R (x)|q+1 dx − . (2.5) − R B2R (0)

Choose ǫk → 0+ and 12 Rk+1 ≥ Rk → +∞ be such that q+1

m1 ≥ (1 − 2ǫk ) p+1 m1

 Z q−1   2 1 (γ − ǫk ) p+1 1 − (γ + ǫk ) p+1 |∇u2Rk |2 + V (x)u22Rk dx 2 B2Rk (0) Z Z β |(1 − η2Rk (|x|))u2Rk (x)|q+1 dx |η2Rk (|x|)u2Rk (x)|q+1 dx + + q+1 B2Rk (0) B2Rk (0)  Z C |u2Rk (x)|q+1 dx − − . (2.6) R k B2R (0) +

k

21

Notice that |η2Rk (|x|)u2Rk (x)|q+1 → |u(x)|q+1 a.e. x ∈ RN , 0 ≤ |η2Rk (|x|)u2Rk (x)|q+1 ≤ |η2Rk+1 (|x|)u2Rk+1 (x)|q+1 , |(1 − η2Rk (|x|))u2Rk (x)|q+1 → 0 a.e. x ∈ RN and |(1 − η2Rk (|x|))u2Rk (x)|q+1 ≥ |(1 − η2Rk+1 (|x|))u2Rk+1 (x)|q+1 . It implies that

Z

lim

k→∞ B2R (0) k

Z

|η2Rk (|x|)u2Rk (x)|q+1 dx

|η2Rk (|x|)u2Rk (x)|q+1 dx = lim k→∞ RN Z |u(x)|q+1 dx, = RN

and

lim

Z

k→∞ B2R (0) k

≤ lim

Z

k→∞ RN

|(1 − η2Rk (|x|))u2Rk (x)|q+1 dx

|(1 − η2Rk (|x|))u2Rk (x)|q+1 dx

= 0. Furthermore, lim

Z

k→∞ B2R (0) k

Z

= lim

k→∞ B2R (0) k

= lim

k→∞

=

Z

RN

|u2Rk (x)|q+1 dx |u(x)|q+1 dx

k Z X i=1

B2Rk (0)\B2Rk−1 (0)

|u(x)|q+1 dx

|u(x)|q+1 dx.

From (2.6) taking limit k → ∞, we have

 Z  q−1  2 1 p+1 m1 ≥ m1 + γ |∇u|2 + V (x)u2 dx. 1 − γ p+1 2 RN

This is a contradiction. Hence γ = 1. It means that un → u in Lp+1 (RN ). Next, using a argument similar to Theorem 2.1, we get that u is a minimizer for m1 . 22

For the case (V4 ). By checking the details of the proofs of both the case of (V3 ) above and the case of (V4 ) in [5], we know that it is sufficient to deal with the following term: β q+1

Z

RN

q+1

|yn |

dx +

Z

RN

q+1

|zn |

dx −

Z

RN

q+1

|un |

dx



→ 0.

as n → ∞, ǫ → 0. Let {un } ⊂ M1 be a minimizing sequence for m1 . Then {un } and {u2n } are bounded in H 1 (RN ). By translation invariant property, we know that {un (· + xn )} and {u2n (· + xn )} are also bounded in H 1 (RN ). Using the same method of the proof of the case of (V3 ), we get that β q+1

Z

RN

q+1

|yn |

dx +

Z

RN

q+1

|zn |

dx −

Z

RN

q+1

|un |

dx



→ 0.

as n → ∞, ǫ → 0. This completes the proof.

References [1] M. Colin, L. Jeanjean, Solutions for a quasilinear Schr¨ odinger equation: a dual approach, Nonlinear Anal. 56(2004), 213-226. [2] S. Kurihura, Large-amplitude quasi-solitons in superfluid films, J. Phys. Soc. Jpn. 50(1981), 3262-3267. [3] E.W. Laedke, K.H. Spatschek, L. Stenflo, Evolution theorem for a class of perturbed envelope soliton solutions, J. Math. Phys. 24(1983), 2764-2769. [4] A.G. Litvak, A.M. Sergeev, One dimensional collapse of plasma waves, JETP Lett. 27(1978), 517-520. [5] J. Liu, Z. Q. Wang, Soliton solutions for quasilinear Schr¨ odinger equations:I, Proc. Am. Math. Soc. 131(2003), 441-448. [6] J. Liu, Y. Wang, Z. Q. Wang, Soliton solutions for quasilinear Schr¨ odinger equations: II, J. Diff. Eqns 187(2003), 473-493. 23

[7] J. Liu, Y. Wang, Z. Q. Wang, Solutions for quasilinear Schr¨ odinger equations via the Nehari method, Commun. Partial Diff. Eqns. 29(2004), 879-901. ´ J M, Miyagaki O H, Soares S H M, Soliton solutions for quasilinear Schr¨ [8] do O odinger equations: the critical exponential case, Nonlinear Anal, 67(2007), 3357-72. [9] A. Nakamura, Damping and modification of exciton solitary waves, J. Phys. Soc. Jpn 42(1977), 1824-1835. [10] M. Porkolab, M.V. Goldman, Upper hybrid solitons and oscillating two-stream instabilities, Phys. Fluids, 19(1976), 872-881. [11] M. Poppenberg, On the local well posedness of quasi-linear Schr¨ odinger equations in arbitrary space dimension, J. Diff. Eqns. 172(2001), 83-115. [12] M. Poppenberg, K. Schmitt, Z. Q. Wang, On the existence of soliton solutions to quasilinear Schr¨ odinger equations, Calc. Var. 14(2002), 14 329-344. [13] David Ruiz, Gaetano Siciliano, Existence of ground states for a modified nonlinear Schr¨ odinger equation, Nonlinearity, 23(2010), 1221-1233. [14] Y. Chen, X. Wu, Existence of Nontrivial Solutions and High Energy Solutions for a Class of Quasilinear Schr¨ odinger Equations on RN , to appear.

24