Hindawi Publishing Corporation
Abstract and Applied Analysis
Volume 2013, Article ID 848690, 9 pages
http://dx.doi.org/10.1155/2013/848690
Research Article
Existence of Standing Waves for a Generalized
Davey-Stewartson System
Xiaoxiao Hu,1 Xiao-xun Zhou,2 Wu Tunhua,1 and Min-Bo Yang3
1
School of Information and Engineering, Wenzhou Medical College, Wenzhou, Zhejiang 325035, China
School of Marxism, Tongji University, Shanghai 200092, China
3
Department of Mathematics, Zhejiang Normal University, Jinhua, Zhejiang 321004, China
2
Correspondence should be addressed to Xiaoxiao Hu; smallapple12345@163.com
Received 13 September 2012; Revised 26 December 2012; Accepted 14 January 2013
Academic Editor: Norimichi Hirano
Copyright © 2013 Xiaoxiao Hu et al. This is an open access article distributed under the Creative Commons Attribution License,
which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.
The purpose of this paper is to investigate the existence of standing waves for a generalized Davey-Stewartson system. By reducing
the system to a single Schrödinger equation problem, we are able to establish some existence results for the system by variational
methods.
1. Introduction and Main Results
In this paper, we are going to consider the existence of standing waves for a generalized Davey-Stewartson system in R3
󵄨 󵄨𝑝−2
𝑖𝜓𝑡 + Δ𝜓 = 𝑏 (𝑥) 𝜓𝜑𝑥1 − 𝑎 (𝑥) 󵄨󵄨󵄨𝜓󵄨󵄨󵄨 𝜓,
󵄨 󵄨2
−Δ𝜑 = (𝑏(𝑥)󵄨󵄨󵄨𝜓󵄨󵄨󵄨 )𝑥 .
1
(1)
Here Δ is the Laplacian operator in R3 and 𝑖 is the imaginary
unit, 𝑎(𝑥), 𝑏(𝑥), and 𝑝 satisfy some additional assumptions.
The Davey-Stewartson system is a model for the evolution
of weakly nonlinear packets of water waves that travel
predominantly in one direction, but in which the amplitude
of waves is modulated in two spatial directions. They are given
as
󵄨 󵄨2
𝑖𝜓𝑡 + 𝜓𝑥𝑥 + 𝛿𝜓𝑦𝑦 = 𝑏1 𝜓𝜑𝑥 − 𝑎󵄨󵄨󵄨𝜓󵄨󵄨󵄨 𝜓,
(2)
󵄨 󵄨2
𝑚𝜑𝑥𝑥 + 𝜑𝑦𝑦 = 𝑏2 (󵄨󵄨󵄨𝜓󵄨󵄨󵄨 )𝑥 ,
where 𝑎, 𝑏1 , 𝑏2 ∈ R, 𝜓(𝑡, 𝑥, 𝑦) is the complex amplitude of
the shortwave and 𝜑(𝑡, 𝑥, 𝑦) is the real longwave amplitude
[1]. The physical parameters 𝛿 and 𝑚 play a determining role
in the classification of this system. Depending on their signs,
the system is elliptic-elliptic, elliptic-hyperbolic, hyperbolicelliptic, and hyperbolic-hyperbolic [2], although the last case
does not seem to occur in the context of water waves.
As we know, the system can be reduced to a single
Schrödinger equation by using Fourier transforms. Indeed,
let 𝐸1 be the singular integral operator defined by
F {𝐸1 (𝜓)} (𝜉) = 𝜎1 (𝜉) F (𝜓) (𝜉) ,
(3)
where 𝜎1 (𝜉) = 𝜉12 /|𝜉|2 , 𝜉 ∈ R3 , and F denotes the Fourier
transform:
F (𝜓) (𝜉) = (
1 3/2 −𝑖𝜉𝑥
) ∫ 𝑒 𝜓 (𝑥) 𝑑𝑥.
2𝜋
(4)
Then the generalized Davey-Stewartson system can be
reduced to the following single nonlocal Schrödinger equation
󵄨 󵄨𝑝−2
󵄨 󵄨2
−𝑖𝜓𝑡 − Δ𝜓 = 𝑎 (𝑥) 󵄨󵄨󵄨𝜓󵄨󵄨󵄨 𝜓 + 𝑏 (𝑥) 𝐸1 (𝑏 (𝑥) 󵄨󵄨󵄨𝜓󵄨󵄨󵄨 ) 𝜓. (5)
In this paper, we are interested in the existence of standing
waves for the above equation, that is, solutions in the form of
𝜓 (𝑡, 𝑥) = 𝑒𝑖𝜔𝑡 𝜙 (𝑥) ,
𝜑 (𝑡, 𝑥) = 𝑣 (𝑥) ,
(6)
where 𝜔 > 0, 𝜙, 𝑣 ∈ 𝐻1 (R3 ). Then if (𝜓, 𝜑) is a solution of (1),
then we can see that 𝜙 must satisfy the following Schrödinger
problem:
󵄨 󵄨2
󵄨 󵄨𝑝−2
−Δ𝜙 + 𝜔𝜙 = 𝑏 (𝑥) 𝐸1 (𝑏 (𝑥) 󵄨󵄨󵄨𝜙󵄨󵄨󵄨 ) 𝜙 + 𝑎 (𝑥) 󵄨󵄨󵄨𝜙󵄨󵄨󵄨 𝜙.
(7)
2
Abstract and Applied Analysis
We will consider the generalized Davey-Stewartson system with perturbation. Under suitable assumptions on the
coefficients 𝑎(𝑥), 𝑏(𝑥), the problem can be viewed as the
perturbation of the generalized Davey-Stewartson system
considered in [2, 3]. Here we will not use the critical point
theory or the minimizing methods to establish the existence
results. Moreover, we will not use Lion’s Concentrationcompactness principle to overcome the difficulty of losing
compactness. Instead, we will apply the perturbation method
developed by Ambrosetti and Badiale in [4, 5] to show the
existence of solutions of (8) and (9). In [4, 5], Ambrosetti
and Badiale established an abstract theory to reduce a class of
perturbation problems to a finite dimensional one by some
careful observation on the unperturbed problems and the
Lyapunov-Schmit reduction procedure. This method has also
been successfully applied to many different problems, see
[6] for examples. In this paper we are going to consider the
following two types of perturbed problems for generalized
Davey-Stewartson system. Consider
󵄨 󵄨2
󵄨 󵄨𝑝−2
−Δ𝜙 + 𝜙 = 𝜀𝑏 (𝑥) 𝐸1 (𝑏 (𝑥) 󵄨󵄨󵄨𝜙󵄨󵄨󵄨 ) 𝜙 + (1 + 𝜀𝑎 (𝑥)) 󵄨󵄨󵄨𝜙󵄨󵄨󵄨 𝜙,
(8)
󵄨 󵄨2
󵄨 󵄨𝑝−2
−Δ𝜙 + 𝜔𝜙 = 𝑏 (𝑥) 𝐸1 (𝑏 (𝑥) 󵄨󵄨󵄨𝜙󵄨󵄨󵄨 ) 𝜙 + 𝑎 (𝑥) 󵄨󵄨󵄨𝜙󵄨󵄨󵄨 𝜙.
(9)
and by | ⋅ |𝑠 we denote the usual 𝐿𝑠 -norm; 𝐶, 𝐶𝑖 stand for
different positive constants.
The paper is organized as follows. In Section 2, we outline
the abstract critical point theory for perturbed functionals
and give some properties for the singular operator 𝐸1 . In
Section 3, we prove the main results by some lemmas.
2. The Abstract Theorem
To prove the main results, we need the following known
propositions.
Proposition 4. For any positive constant 𝐴, consider the
following problem, 2 < 𝑝 < 2∗ :
−Δ𝑢 + 𝑢 = 𝐴|𝑢|𝑝−2 𝑢,
𝑢 > 0,
(12)
𝑢 ∈ 𝐻1 (R3 ) .
There is a unique positive radial solution 𝑈, which satisfies the
following decay property:
lim 𝑈 (𝑟) 𝑟𝑒𝑟 = 𝐶 > 0,
𝑟→∞
𝑈󸀠 (𝑟)
= −1,
𝑟 → ∞ 𝑈 (𝑟)
lim
𝑟 = |𝑥| ,
The main results of the paper are the following theorems.
Theorem 1. Assume that 2 < 𝑝 < 6, 𝑎(𝑥) ∈ 𝐿6/(6−𝑝) (R3 )
and 𝑏(𝑥) ∈ 𝐿6 (R3 ). Take the function 𝑈 from Proposition 4 in
Section 2, if there holds
1
1
∫ 𝑏 (𝑥) 𝐸1 (𝑏 (𝑥) |𝑈|2 ) |𝑈|2 + ∫ 𝑎 (𝑥) |𝑈|𝑝 ≠ 0,
4
𝑝
(10)
then for any 𝜀 small, there exists at least one solution of problem
(8).
where 𝐶 > 0 is a constant. The function 𝑈 is a critical point of
𝐶2 functional 𝐼0 : 𝐻1 (R3 ) → R defined by
𝐴
1
𝐼0 (𝑢) = ‖𝑢‖2 − ∫ |𝑢|𝑝 .
2
𝑝 R3
(𝑎1 ) 𝑎(𝑥)−𝐴 is continuous, bounded and 𝑎(𝑥)−𝐴 ∈ 𝐿1 (R3 )
with ∫(𝑎(𝑥) − 𝐴) ≠ 0;
(𝑏1 ) 𝑏(𝑥) is continuous, bounded and 𝑏 ∈ 𝐿2 (R3 ).
𝑍 = {𝑧𝜃 = 𝑈 (𝑥 + 𝜃) , 𝜃 ∈ R3 } .
Theorem 3. Let 𝜔 = 𝜀2 , suppose 2 < 𝑝 < 4 and 𝑎(𝑥), 𝑏(𝑥)
satisfy (𝑏1 ) and
(𝑎2 ) 𝑎 − 𝐴 is continuous and there exist 𝐿 ≠ 0 and 0 < 𝛾 < 3
such that |𝑥|𝛾 (𝑎(𝑥) − 𝐴) → 𝐿 as |𝑥| → ∞.
Then for 𝜀 > 0 small enough, there exists a solution 𝜙𝜀 in
𝐻1 (R3 ) for problem (9). Moreover, if 2 < 𝑝 < 2 + 4/3, then
𝜙𝜀 → 0 as 𝜀 → 0.
1
3
Throughout this paper, we denote the norm of 𝐻 (R ) by
2
‖𝑢‖ = (∫ |∇𝑢| + 𝑢 )
1/2
,
(15)
Set
𝑄 (𝑢) := 𝐼0󸀠󸀠 (𝑈) [𝑢, 𝑢] = ∫ [|∇𝑢|2 + 𝑢2 − 𝑝𝐴𝑈𝑝−2 𝑢2 ] (16)
R3
and denote 𝑋 = span{𝜕𝑈/𝜕𝑥𝑖 , 1 ≤ 𝑖 ≤ 3}. We have
Then for 𝜀 > 0 small enough, there exists a solution 𝜙𝜀 in
𝐻1 (R3 ) for problem (9). Moreover, if 2 < 𝑝 < 2 + 4/3, then
𝜙𝜀 → 0 as 𝜀 → 0.
2
(14)
Moreover, 𝐼0 possesses a 3-dimensional manifold of critical
points
2
Theorem 2. Let 𝜔 = 𝜀 , suppose 2 < 𝑝 < 4, and there exists a
positive constant 𝐴 such that 𝑎(𝑥), 𝑏(𝑥) satisfy
(13)
(11)
(1) 𝑄(𝑈) = (2 − 𝑝)𝐴 ∫R3 𝑈𝑝 𝑑𝑥 < 0,
(2) Ker 𝑄 = 𝑋,
(3) 𝑄(𝑤) ≥ 𝐶‖𝑤‖2 , for all 𝑤 ∈ (R𝑈 ⊕ 𝑋)⊥ .
In the following, we outline the abstract theorem of
a variational method to study critical points of perturbed
functionals. Let 𝐸 be a real Hilbert space, we will consider
the perturbed functional defined on it of the form
𝐼𝜀 (𝑢) = 𝐼0 (𝑢) + 𝐺 (𝜀, 𝑢) ,
(17)
where 𝐼0 : 𝐸 → R and 𝐺 : R × 𝐸 → R. We need the
following hypotheses and assume that
(1) 𝐼0 and 𝐺 are 𝐶2 with respect to 𝑢;
(2) 𝐺 is continuous in (𝜀, 𝑢) and 𝐺(0, 𝑢) = 0 for all 𝑢;
Abstract and Applied Analysis
3
(3) 𝐺󸀠 (𝜀, 𝑢) and 𝐺󸀠󸀠 (𝜀, 𝑢) are continuous maps from R ×
𝐸 → 𝐸 and 𝐿(𝐸, 𝐸), respectively, and 𝐿(𝐸, 𝐸) is the
space of linear continuous operators from 𝐸 to 𝐸.
(4) There is a 𝑑-dimensional 𝐶2 manifold 𝑍, 𝑑 ≥ 1,
consisting of critical points of 𝐼0 , and such a 𝑍 will
be called a critical manifold of 𝐼0 .
(5) let 𝑇𝜃 𝑍 denote the tangent space to 𝑍 at 𝑧𝜃 , the
manifold 𝑍 is nondegenerate in the following sense:
Ker(𝐼0󸀠󸀠 (𝑧))
𝐼0󸀠󸀠 (𝑧𝜃 )
= 𝑇𝜃 𝑍 and
is an index-0
Fredholm operator for any 𝑧𝜃 ∈ 𝑍.
(6) There exists 𝛼 > 0 and a continuous function Γ : 𝑍 →
R such that
Γ (𝑧) = lim
𝜀→0
𝐺 (𝜀, 𝑧)
,
𝜀𝛼
(18)
𝐺󸀠 (𝜀, 𝑧) = 𝑜 (𝜀𝛼/2 ) .
(19)
We want to look for solutions of the form 𝑢 = 𝑧+𝑤 with 𝑧 ∈ 𝑍
and 𝑤 ∈ 𝑊 = (𝑇𝜃 𝑍)⊥ . Then we can reduce the problem to a
finite-dimensional one by Lyapunov-Schmit procedure, that
is, it is equivalent to solve the following system:
𝑃𝐼𝜀󸀠 (𝑧 + 𝑤) = 0,
(𝐼 − 𝑃) 𝐼𝜀󸀠 (𝑧 + 𝑤) = 0.
Lemma 5. Suppose assumptions (1)–(6) are satisfied, and there
exists 𝛿 > 0 and 𝑧∗ ∈ 𝑍 such that
min Γ (𝑧) > Γ (𝑧∗ )
‖𝑧−𝑧∗ ‖=𝛿
𝑜𝑟
max Γ (𝑧) < Γ (𝑧∗ ) .
‖𝑧−𝑧∗ ‖=𝛿
(22)
Then for any 𝜀 small, there exists 𝑢𝜀 which is a critical point of
𝐼𝜀 .
We give some facts about the singular integral 𝐸1 in
Cipolatti [2].
Lemma 6. Let 𝐸1 be the singular integral operator defined in
Fourier variable by
F {𝐸1 (𝜓)} (𝜉) = 𝜎1 (𝜉) F (𝜓) (𝜉) ,
(24)
(1) 𝐸1 ∈ L(𝐿𝑝 , 𝐿𝑝 ).
(2) if 𝜓 ∈ 𝐻1 (R3 ), then 𝐸1 (𝜓) ∈ 𝐻1 (R3 ).
(3) 𝐸1 preserves the following operations:
translation: 𝐸1 (𝜓(⋅ + 𝑦))(𝑥) = 𝐸1 (𝜓)(𝑥 + 𝑦), 𝑦 ∈
R3 .
dilation: 𝐸1 (𝜓(𝜆⋅))(𝑥) = 𝐸1 (𝜓)(𝜆𝑥), 𝜆 > 0.
conjugation: 𝐸1 (𝜓) = 𝐸1 (𝜓), 𝜓 is the complex
conjugate of 𝜓.
In this section, we would apply the abstract tools of the previous section to prove the main results. First let us consider
(8), the corresponding energy functional 𝐼𝜀 : 𝐻1 (R3 ) → R
can be defined as
𝐼𝜀 (𝜙) =
(23)
1 󵄩󵄩 󵄩󵄩2
󵄩𝜙󵄩
2󵄩 󵄩
−
𝜀
󵄨 󵄨2 󵄨 󵄨2
∫ 𝑏 (𝑥) 𝐸1 (𝑏 (𝑥) 󵄨󵄨󵄨𝜙󵄨󵄨󵄨 ) 󵄨󵄨󵄨𝜙󵄨󵄨󵄨
4
−
1
󵄨 󵄨𝑝
∫ (1 + 𝜀𝑎 (𝑥)) 󵄨󵄨󵄨𝜙󵄨󵄨󵄨 .
𝑝
(25)
It is easy to see that 𝐼𝜀 : 𝐻1 (R3 ) → R is of 𝐶2 , and thus 𝜙 is a
solution of (8) if and only if 𝜙 is a critical point of the action
functional 𝐼𝜀 (𝜙).
Proof of Theorem 1. Set
1 󵄩 󵄩2 1 󵄨 󵄨 𝑝
𝐼0 (𝜙) = 󵄩󵄩󵄩𝜙󵄩󵄩󵄩 − ∫ 󵄨󵄨󵄨𝜙󵄨󵄨󵄨 ,
2
𝑝
(21)
In [4, 5] the following abstract theorem is proved.
𝑒𝑖𝑡ℎ𝑒𝑟
1 3/2 −𝑖𝜉𝑥
) ∫ 𝑒 𝜓 (𝑥) 𝑑𝑥.
2𝜋
For 1 < 𝑝 < ∞, 𝐸1 satisfies the following properties:
(20)
Here 𝑃 is the orthogonal projection onto 𝑊. Under the
conditions above, the first equation in this system can be
solved by implicit function theorem, and then by using the
Taylor expansion, we obtain for 𝑢 = 𝑧 + 𝑤(𝜀, 𝑧)
𝐼𝜀 (𝑢) = 𝐼0 (𝑧) + 𝜀𝛼 Γ (𝑧) + 𝑜 (𝜀𝛼 ) .
F (𝜓) (𝜉) = (
3. Proof of the Main Results
Consider the existence of critical points of the perturbed
problem
𝐼𝜀󸀠 (𝑢) = 0.
where 𝜎1 (𝜉) = 𝜉12 /|𝜉|2 , 𝜉 ∈ R3 , and F denotes the Fourier
transform:
𝐺 (𝜙) = −
1
󵄨 󵄨2 󵄨 󵄨2
󵄨 󵄨𝑝 1
∫ 𝑎 (𝑥) 󵄨󵄨󵄨𝜙󵄨󵄨󵄨 − ∫ 𝑏 (𝑥) 𝐸1 (𝑏 (𝑥) 󵄨󵄨󵄨𝜙󵄨󵄨󵄨 ) 󵄨󵄨󵄨𝜙󵄨󵄨󵄨 ,
𝑝
4
(26)
then 𝐼𝜀 (𝑢) can be rewritten as
𝐼𝜀 (𝜙) = 𝐼0 (𝜙) + 𝜀𝐺 (𝜙) .
(27)
Thus 𝐼0 (𝜙) and 𝐺(𝜙) are both 𝐶2 with respect to 𝜙. To apply
Lemma 5, by Proposition 4, we need only to check that
lim Γ (𝜃) = 0,
|𝜃| → ∞
Γ (𝜃) := 𝐺|𝑍 = −
−
1
󵄨 󵄨𝑝
∫ 𝑎 (𝑥) 󵄨󵄨󵄨𝑧𝜃 󵄨󵄨󵄨
𝑝
1
󵄨 󵄨2 󵄨 󵄨2
∫ 𝑏 (𝑥) 𝐸1 (𝑏 (𝑥) 󵄨󵄨󵄨𝑧𝜃 󵄨󵄨󵄨 ) 󵄨󵄨󵄨𝑧𝜃 󵄨󵄨󵄨 .
4
(28)
4
Abstract and Applied Analysis
From the fact that 𝑎 ∈ 𝐿6/(6−𝑝) (R3 ), for any 𝑇 > 0, we have
󵄨
󵄨󵄨
󵄨󵄨∫ 𝑎 (𝑥) 󵄨󵄨󵄨𝑧 󵄨󵄨󵄨𝑝 󵄨󵄨󵄨 ≤
󵄨󵄨
𝜃 󵄨 󵄨󵄨
󵄨
󵄨
󵄨
We are going to consider problem (9). Set
󵄨󵄨
󵄨 󵄨
󵄨
󵄨󵄨
󵄨 󵄨𝑝 󵄨󵄨 󵄨󵄨
󵄨 󵄨𝑝 󵄨󵄨
𝑎 (𝑥) 󵄨󵄨󵄨𝑧𝜃 󵄨󵄨󵄨 󵄨󵄨󵄨 + 󵄨󵄨󵄨∫
𝑎 (𝑥) 󵄨󵄨󵄨𝑧𝜃 󵄨󵄨󵄨 󵄨󵄨󵄨
󵄨󵄨∫
󵄨󵄨 󵄨󵄨 |𝑥|≥𝑇
󵄨󵄨
󵄨󵄨 |𝑥|≤𝑇
𝜔 = 𝜀2 ,
𝜙 (𝑥) = 𝜀2/(𝑝−2) 𝑢 (𝜀𝑥) .
(6−𝑝)/6
≤ (∫
|𝑥|≤𝑇
|𝑎 (𝑥)|6/(6−𝑝) )
We have
𝑥
−Δ𝑢 + 𝑢 = 𝑎 ( ) |𝑢|𝑝−2 𝑢
𝜀
𝑝/2∗
× (∫
|𝑥|≤𝑇
󵄨󵄨 󵄨󵄨6
󵄨󵄨𝑧𝜃 󵄨󵄨 )
(29)
6/(6−𝑝)
+ (∫
|𝑥|≥𝑇
|𝑎 (𝑥)|
𝑝/6
󵄨󵄨 󵄨󵄨6
󵄨󵄨𝑧𝜃 󵄨󵄨 )
|𝑥|≥𝑇
× (∫
)
𝑥
+ 𝜀2(4−𝑝)/(𝑝−2) 𝑏 ( )
𝜀
(6−𝑝)/6
.
󵄨2
󵄨
𝐵1 (𝜙) = ∫ 𝜎1 (𝜉) 󵄨󵄨󵄨F (𝑏 ⋅ 𝜙) (𝜉)󵄨󵄨󵄨 𝑑𝜉,
It can be proved that 𝜙(𝑥) = 𝜀2/(𝑝−2) 𝑢(𝜀𝑥) ∈ 𝐻1 (R3 ) is a
solution of system (9) if and only if 𝑢 ∈ 𝐻1 (R3 ) is a critical
point of the functional 𝐼𝜀 : 𝐻1 (R3 ) → R defined by
𝐼𝜀 (𝑢) =
−
𝐵1 (𝜙) = ∫ 𝐸1 (𝑏 ⋅ 𝜙) 𝑏 ⋅ 𝜙,
(37)
Set
󵄨2
󵄨
0 < 𝐵1 (𝜙) ≤ 󵄨󵄨󵄨𝑏 (𝑥) 𝜙󵄨󵄨󵄨2 .
1
𝐴
𝐼0 (𝑢) = ‖𝑢‖2 − ∫ |𝑢|𝑝 .
2
𝑝
(∗)
󵄨󵄨
󵄨
󵄨󵄨∫ 𝑏 (𝑥) 𝐸 (𝑏 (𝑥) 󵄨󵄨󵄨𝑧 󵄨󵄨󵄨2 ) 󵄨󵄨󵄨𝑧 󵄨󵄨󵄨2 󵄨󵄨󵄨
󵄨󵄨
1
𝜃 󵄨 󵄨 𝜃 󵄨 󵄨󵄨
󵄨
󵄨
󵄨
𝐼𝜀 (𝑢) = 𝐼0 (𝑢) +
󵄨
󵄨 󵄨2 󵄨2
≤ ∫ 󵄨󵄨󵄨󵄨𝑏 (𝑥) 󵄨󵄨󵄨𝑧𝜃 󵄨󵄨󵄨 󵄨󵄨󵄨󵄨
6
|𝑏 (𝑥)| )
(38)
Then 𝐼𝜀 (𝑢) can be rewritten as
Then for any 𝑇 > 0, we have
|𝑥|≥𝑇
𝜀2(4−𝑝)/(𝑝−2)
4
𝑥
𝑥
× ∫ 𝑏 ( ) 𝐸1 (𝑏 ( ) |𝑢|2 ) |𝑢|2 .
𝜀
𝜀
(31)
and in particular we have
+ (∫
𝑥
1 2 1
‖𝑢‖ − ∫ 𝑎 ( ) |𝑢|𝑝
2
𝑝
𝜀
(30)
it follows from the Parseval identity that
|𝑥|≤𝑇
(36)
𝑥
× 𝐸1 (𝑏 ( ) |𝑢|2 ) 𝑢.
𝜀
Since 𝑈 exponentially decays at infinity, we know the right
side of the equality goes to 0, if 𝜃 → ∞.
Let 𝐵1 be the quadratic functional on 𝐿2 defined by
≤ (∫
(35)
−
1/3
1/3
|𝑏 (𝑥)|6 )
(32)
2/3
󵄨󵄨 󵄨󵄨6
(∫
󵄨󵄨𝑧𝜃 󵄨󵄨 )
|𝑥|≤𝑇
2/3
󵄨󵄨 󵄨󵄨6
󵄨󵄨𝑧𝜃 󵄨󵄨 )
|𝑥|≥𝑇
(∫
Since 𝑈 exponentially decays at infinity, the right side of the
inequality (32) goes to 0. Thus from (29) and (32) above we
soon get
lim Γ (𝜃) = 0.
(33)
(39)
𝑥
𝑥
× ∫ 𝑏 ( ) 𝐸1 (𝑏 ( ) |𝑢|2 ) |𝑢|2 .
𝜀
𝜀
̃ (𝜀, 𝑢) = 1 ∫ (𝐴 − 𝑎 ( 𝑥 )) |𝑢|𝑝
𝐺
𝑝
𝜀
−
𝜀2(4−𝑝)/(𝑝−2)
𝑥
𝑥
∫ 𝑏 ( ) 𝐸1 (𝑏 ( ) |𝑢|2 ) |𝑢|2
4
𝜀
𝜀
̃2 (𝜀, 𝑢)
̃1 (𝜀, 𝑢) + 𝐺
=𝐺
(40)
Then by assumption (10) that
1
1
∫ 𝑏 (𝑥) 𝐸1 (𝑏 (𝑥) |𝑈|2 ) |𝑈|2 + ∫ 𝑎 (𝑥) |𝑈|𝑝 ≠ 0,
4
𝑝
𝜀2(4−𝑝)/(𝑝−2)
4
Define
.
|𝜃| → ∞
𝑥
1
∫ (𝐴 − 𝑎 ( )) |𝑢|𝑝
𝑝
𝜀
and for 𝑖 = 1, 2
(34)
we know Γ(0) ≠ 0. Thus, the conclusion follows from
Lemma 5 that any strict maximum or minimum of Γ gives
rise to a critical point of the perturbed functional and hence
to a solution of (8).
̃ (𝜀, 𝑢) , if 𝜀 ≠ 0,
𝐺
𝐺𝑖 (𝜀, 𝑢) = { 𝑖
0,
if 𝜀 = 0.
(41)
Lemma 7. Under assumptions (𝑎1 ) and (𝑏1 ), 𝐺 = 𝐺1 + 𝐺2 is
continuous in (𝜀, 𝑢).
Abstract and Applied Analysis
5
Proof. From the proof of Lemma 4.1 in [7], we know 𝐺1 is
continuous in (𝜀, 𝑢) ∈ R × 𝐻1 (R3 ), and hence we only need
to prove that 𝐺2 is continuous in (𝜀, 𝑢).
If (𝜀, 𝑢) → (𝜀0 , 𝑢0 ), with 𝜀0 ≠ 0. Then we can estimate that
󵄨
󵄨
󵄨
4 󵄨󵄨𝐺2 (𝜀, 𝑢) − 𝐺2 (𝜀0 , 𝑢0 )󵄨󵄨󵄨
󵄨󵄨
𝑥
𝑥
= 󵄨󵄨󵄨󵄨𝜀2(4−𝑝)/(𝑝−2) ∫ 𝑏 ( ) 𝐸1 (𝑏 ( ) |𝑢|2 ) |𝑢|2
𝜀
𝜀
󵄨
2(4−𝑝)/(𝑝−2)
−𝜀0
∫𝑏(
𝑥 󵄨 󵄨2 󵄨 󵄨2 󵄨󵄨󵄨
𝑥
) 𝐸1 (𝑏 ( ) 󵄨󵄨󵄨𝑢0 󵄨󵄨󵄨 ) 󵄨󵄨󵄨𝑢0 󵄨󵄨󵄨 󵄨󵄨󵄨
󵄨󵄨
𝜀0
𝜀0
󵄨󵄨
𝑥
𝑥
≤ |𝜀|2(4−𝑝)/(𝑝−2) 󵄨󵄨󵄨󵄨∫ 𝑏 ( ) 𝐸1 (𝑏 ( ) |𝑢|2 ) |𝑢|2
𝜀
𝜀
󵄨
−∫𝑏(
𝑥
𝑥 󵄨 󵄨2 󵄨 󵄨2 󵄨󵄨󵄨
) 𝐸1 (𝑏 ( ) 󵄨󵄨󵄨𝑢0 󵄨󵄨󵄨 ) 󵄨󵄨󵄨𝑢0 󵄨󵄨󵄨 󵄨󵄨󵄨
󵄨󵄨
𝜀0
𝜀0
󵄨
2(4−𝑝)/(𝑝−2) 󵄨󵄨
󵄨󵄨
+ 󵄨󵄨󵄨󵄨𝜀2(4−𝑝)/(𝑝−2) − 𝜀0
󵄨
:= |𝜀|2(4−𝑝)/(𝑝−2) 𝐼1 + 𝐼2 .
(42)
It is obvious that 𝐼2 → 0, as 𝜀 → 𝜀0 . At the same time, we
know
󵄨󵄨
󵄨󵄨
𝑥
𝑥
𝑥
󵄨
󵄨
𝐼1 ≤ 󵄨󵄨󵄨∫ [𝑏 ( ) − 𝑏 ( )] 𝐸1 (𝑏 ( ) |𝑢|2 ) |𝑢|2 󵄨󵄨󵄨
󵄨󵄨
𝜀
𝜀0
𝜀
󵄨󵄨
󵄨󵄨
󵄨󵄨󵄨
𝑥
𝑥
𝑥
󵄨
+ 󵄨󵄨󵄨∫ 𝑏 ( ) 𝐸1 ([𝑏 ( ) − 𝑏 ( )] |𝑢|2 ) |𝑢|2 󵄨󵄨󵄨
󵄨󵄨
𝜀0
𝜀
𝜀0
󵄨󵄨
󵄨󵄨󵄨
󵄨󵄨󵄨
𝑥
𝑥
(43)
󵄨 󵄨2
+ 󵄨󵄨󵄨∫ 𝑏 ( ) 𝐸1 (𝑏 ( ) [|𝑢|2 − 󵄨󵄨󵄨𝑢0 󵄨󵄨󵄨 ]) |𝑢|2 󵄨󵄨󵄨
󵄨󵄨
󵄨󵄨
𝜀0
𝜀0
󵄨󵄨
󵄨
𝑥
𝑥 󵄨 󵄨2
󵄨
󵄨 󵄨2 󵄨󵄨
+ 󵄨󵄨󵄨∫ 𝑏 ( ) 𝐸1 (𝑏 ( ) 󵄨󵄨󵄨𝑢0 󵄨󵄨󵄨 ) (|𝑢|2 − 󵄨󵄨󵄨𝑢0 󵄨󵄨󵄨 )󵄨󵄨󵄨
󵄨󵄨
󵄨󵄨
𝜀0
𝜀0
:= Π1 + Π2 + Π3 + Π4 .
Estimating the first term Π1 , by Hölder inequality, we know
󵄨󵄨
󵄨󵄨
𝑥
𝑥
𝑥
󵄨
󵄨
Π1 = 󵄨󵄨󵄨∫ [𝑏 ( ) − 𝑏 ( )] 𝐸1 (𝑏 ( ) |𝑢|2 ) |𝑢|2 󵄨󵄨󵄨
󵄨󵄨
󵄨󵄨
𝜀
𝜀0
𝜀
(44)
󵄨󵄨
󵄨󵄨2 1/2
𝑥
× (∫ 󵄨󵄨󵄨󵄨𝐸1 (𝑏 ( ) |𝑢|2 )󵄨󵄨󵄨󵄨 ) .
𝜀
󵄨
󵄨
as 𝜀 󳨀→ 𝜀0 .
therefore 𝐺2 (𝜀, 𝑢) → 0, as (𝜀, 𝑢) → (0, 𝑢). Hence 𝐺 = 𝐺1 +
𝐺2 is continuous and the lemma is proved.
Proof. 𝐺1󸀠 and 𝐺1󸀠󸀠 are continuous in (𝜀, 𝑢), see [7, Lemma
4.2] for the details. Here we only prove that 𝐺2󸀠 and 𝐺2󸀠󸀠 are
continuous in (𝜀, 𝑢).
If (𝜀, 𝑢) → (𝜀0 , 𝑢0 ) with 𝜀0 ≠ 0, then
󵄩󵄩 󸀠
󵄩
󵄩󵄩𝐺2 (𝜀, 𝑢) − 𝐺2󸀠 (𝜀0 , 𝑢0 )󵄩󵄩󵄩
󵄩
󵄩
= sup {𝜀2(4−𝑝)/(𝑝−2)
‖𝑣‖=1
𝑥
𝑥
× ∫ 𝑏 ( ) 𝐸1 (𝑏 ( ) |𝑢|2 ) 𝑢𝑣
𝜀
𝜀
2(4−𝑝)/(𝑝−2)
− 𝜀0
×∫𝑏(
𝑥
𝑥 󵄨 󵄨2
) 𝐸1 (𝑏 ( ) 󵄨󵄨󵄨𝑢0 󵄨󵄨󵄨 ) 𝑢0 𝑣}
𝜀0
𝜀0
≤ sup {|𝜀|2(4−𝑝)/(𝑝−2)
(48)
‖𝑣‖=1
𝑥
𝑥
× ∫ (𝑏 ( ) 𝐸1 (𝑏 ( ) |𝑢|2 ) 𝑢𝑣
𝜀
𝜀
𝑥
𝑥 󵄨 󵄨2
−𝑏 ( ) 𝐸1 (𝑏 ( ) 󵄨󵄨󵄨𝑢0 󵄨󵄨󵄨 ) 𝑢0 𝑣)
𝜀0
𝜀0
󵄨
2(4−𝑝)/(𝑝−2) 󵄨󵄨
󵄨󵄨
+ 󵄨󵄨󵄨󵄨𝜀2(4−𝑝)/(𝑝−2) − 𝜀0
󵄨
󵄨󵄨
󵄨󵄨
𝑥
𝑥
󵄨
󵄨
󵄨 󵄨2
× 󵄨󵄨󵄨∫ 𝑏 ( ) 𝐸1 (𝑏 ( ) 󵄨󵄨󵄨𝑢0 󵄨󵄨󵄨 ) 𝑢0 𝑣󵄨󵄨󵄨}
󵄨󵄨
󵄨󵄨
𝜀0
𝜀0
:= |𝜀|2(4−𝑝)/(𝑝−2) sup 𝐼1 + sup 𝐼2 .
‖𝑣‖=1
Since 𝑏(𝑥) is bounded and continuous, the operator 𝐸1 ∈
L(𝐿2 , 𝐿2 ), the dominated convergence theorem implies that
Π1 󳨀→ 0,
󵄨󵄨 𝑥
󵄨󵄨
𝑥
󵄨
󵄨
4 󵄨󵄨󵄨𝐺2 (𝜀, 𝑢)󵄨󵄨󵄨 ≤ |𝜀|2(4−𝑝)/(𝑝−2) ∫ 󵄨󵄨󵄨󵄨𝑏 ( ) 𝐸1 (𝑏 ( ) |𝑢|2 ) |𝑢|2 󵄨󵄨󵄨󵄨
𝜀
󵄨 𝜀
󵄨
2
󵄨
󵄨
󵄨 𝑥 󵄨
≤ |𝜀|2(4−𝑝)/(𝑝−2) ∫ 󵄨󵄨󵄨󵄨𝑏 ( )󵄨󵄨󵄨󵄨 |𝑢 (𝑥)|4
󵄨 𝜀 󵄨
2/3
󵄨󵄨 𝑥 󵄨󵄨6 1/3
≤ |𝜀|2(4−𝑝)/(𝑝−2) (∫ 󵄨󵄨󵄨󵄨𝑏 ( )󵄨󵄨󵄨󵄨 ) (∫ |𝑢|6 ) ,
󵄨 𝜀 󵄨
(47)
Lemma 8. Under assumptions (𝑎1 ) and (𝑏1 ), 𝐺󸀠 and 𝐺󸀠󸀠 are
continuous in (𝜀, 𝑢).
󵄨󵄨
𝑥
𝑥 󵄨 󵄨2 󵄨 󵄨2 󵄨󵄨󵄨
󵄨
× 󵄨󵄨󵄨∫ 𝑏 ( ) 𝐸1 (𝑏 ( ) 󵄨󵄨󵄨𝑢0 󵄨󵄨󵄨 ) 󵄨󵄨󵄨𝑢0 󵄨󵄨󵄨 󵄨󵄨󵄨
󵄨󵄨
󵄨󵄨
𝜀0
𝜀0
1/2
󵄨󵄨 𝑥
𝑥 󵄨󵄨󵄨󵄨 2 2
󵄨󵄨
≤ (∫ (󵄨󵄨𝑏 ( ) − 𝑏 ( )󵄨󵄨 |𝑢| ) )
󵄨󵄨 𝜀
𝜀0 󵄨󵄨
as (𝜀, 𝑢) → (𝜀0 , 𝑢0 ).
If (𝜀, 𝑢) → (0, 𝑢0 ), by definition, 𝐺2 (0, 𝑢) = 0. Since
𝑏(𝑥) ∈ 𝐿2 (R3 ) is also bounded, we know 𝑏(𝑥) ∈ 𝐿6 (R3 ),
applying Parseval identity and Hölder inequality, we get
(45)
Similarly, we can deduce that Π2 , Π3 , Π4 vanishes, as (𝜀, 𝑢) →
(𝜀0 , 𝑢0 ). Hence
󵄨󵄨
󵄨
(46)
󵄨󵄨𝐺2 (𝜀, 𝑢) − 𝐺2 (𝜀0 , 𝑢0 )󵄨󵄨󵄨 󳨀→ 0
‖𝑣‖=1
Estimating the second term, since 𝐸1 ∈ L(𝐿2 , 𝐿2 ), by Hölder
inequality, we know
󵄨
2(4−𝑝)/(𝑝−2) 󵄨󵄨
󵄨󵄨
sup 𝐼2 = sup 󵄨󵄨󵄨󵄨𝜀2(4−𝑝)/(𝑝−2) − 𝜀0
󵄨
‖𝑣‖=1
‖𝑣‖=1
󵄨󵄨
󵄨󵄨
𝑥
𝑥 󵄨 󵄨2
󵄨
󵄨
× 󵄨󵄨󵄨∫ 𝑏 ( ) 𝐸1 (𝑏 ( ) 󵄨󵄨󵄨𝑢0 󵄨󵄨󵄨 ) 𝑢0 𝑣󵄨󵄨󵄨
󵄨󵄨
󵄨󵄨
𝜀0
𝜀0
6
Abstract and Applied Analysis
󵄨
2(4−𝑝)/(𝑝−2) 󵄨󵄨
󵄨󵄨
≤ sup 𝐶0 󵄨󵄨󵄨󵄨𝜀2(4−𝑝)/(𝑝−2) − 𝜀0
󵄨
In the following we prove that 𝐺󸀠󸀠 is continuous in (𝜀, 𝑢).
As we know
‖𝑣‖=1
󵄨󵄨 𝑥
󵄨󵄨 𝑥
󵄨󵄨2 1/2
󵄨2 1/2
󵄨
󵄨
󵄨
󵄨 󵄨2 󵄨󵄨
× (∫ 󵄨󵄨󵄨𝑏 ( ) 󵄨󵄨󵄨𝑢0 󵄨󵄨󵄨 󵄨󵄨󵄨 ) (∫ 󵄨󵄨󵄨𝑏 ( ) 𝑢0 𝑣󵄨󵄨󵄨 )
󵄨󵄨 𝜀0
󵄨󵄨 𝜀0
󵄨󵄨
󵄨󵄨
𝐺2󸀠󸀠 (𝜀, 𝑢) [𝑤, 𝑣] = 𝜀2(4−𝑝)/(𝑝−2)
𝑥
𝑥
× ∫ (𝑏 ( ) 𝐸1 (𝑏 ( ) |𝑢|2 ) 𝑤𝑣
𝜀
𝜀
1/2
󵄨
2(4−𝑝)/(𝑝−2) 󵄨󵄨
󵄨󵄨 (∫ 󵄨󵄨󵄨󵄨𝑢0 󵄨󵄨󵄨󵄨6 ) .
≤ 𝐶1 󵄨󵄨󵄨󵄨𝜀2(4−𝑝)/(𝑝−2) − 𝜀0
󵄨
𝑥
𝑥
+2𝑏 ( ) 𝐸1 (𝑏 ( ) 𝑤𝑢) 𝑢𝑣) .
𝜀
𝜀
(52)
(49)
Thus sup‖𝑣‖=1 𝐼2 → 0 as 𝜀 → 𝜀0 . Estimating the first term 𝐼1 ,
we know
𝑥
𝑥
sup 𝐼1 = sup { ∫ (𝑏 ( ) 𝐸1 (𝑏 ( ) |𝑢|2 ) 𝑢𝑣
𝜀
𝜀
‖𝑣‖=1
‖𝑣‖=1
−𝑏 (
𝑥
𝑥 󵄨 󵄨2
) 𝐸1 (𝑏 ( ) 󵄨󵄨󵄨𝑢0 󵄨󵄨󵄨 ) 𝑢0 𝑣)}
𝜀0
𝜀0
󵄨󵄨
󵄨󵄨
𝑥
𝑥
𝑥
󵄨
󵄨
≤ sup {󵄨󵄨󵄨∫ [𝑏 ( ) − 𝑏 ( )] 𝐸1 (𝑏 ( ) |𝑢|2 ) 𝑢𝑣󵄨󵄨󵄨
󵄨
󵄨󵄨
𝜀
𝜀0
𝜀
‖𝑣‖=1 󵄨
󵄨󵄨󵄨
𝑥
+ 󵄨󵄨󵄨∫ 𝑏 ( ) 𝐸1
𝜀0
󵄨󵄨
󵄨󵄨󵄨
𝑥
𝑥
× ([𝑏 ( ) − 𝑏 ( )] |𝑢|2 ) 𝑢𝑣󵄨󵄨󵄨
󵄨󵄨
𝜀
𝜀0
󵄨󵄨
𝑥
󵄨
+ 󵄨󵄨󵄨∫ 𝑏 ( ) 𝐸1
󵄨󵄨
𝜀0
󵄨󵄨
𝑥
󵄨
󵄨 󵄨2
× (𝑏 ( ) [|𝑢|2 − 󵄨󵄨󵄨𝑢0 󵄨󵄨󵄨 ]) 𝑢𝑣󵄨󵄨󵄨
󵄨󵄨
𝜀0
󵄨󵄨
𝑥
󵄨
+ 󵄨󵄨󵄨∫ 𝑏 ( ) 𝐸1
󵄨󵄨
𝜀0
󵄨󵄨
𝑥 󵄨 󵄨2
󵄨
× (𝑏 ( ) 󵄨󵄨󵄨𝑢0 󵄨󵄨󵄨 ) (𝑢𝑣 − 𝑢0 𝑣)󵄨󵄨󵄨}
󵄨󵄨
𝜀0
:= 𝐴 1 + 𝐴 2 + 𝐴 3 + 𝐴 4 .
(50)
As in Lemma 7, by Hölder inequality again, we can prove
that 𝐴 𝑖 → 0, as (𝜀, 𝑢) → (𝜀0 , 𝑢0 ), 𝑖 = 1, 2, 3, 4. Therefore
‖𝐺2󸀠 (𝜀, 𝑢) − 𝐺2󸀠 (𝜀0 , 𝑢0 )‖ → 0 as (𝜀, 𝑢) → (𝜀0 , 𝑢0 ).
If 𝜀0 = 0, from the definition of 𝐺2 , we know ‖𝐺2󸀠 (0, 𝑢0 )‖ =
0. Hence
󵄩
󵄩󵄩 󸀠
󵄩󵄩𝐺2 (𝜀, 𝑢)󵄩󵄩󵄩
󵄩
󵄩
𝑥
𝑥
󵄨󵄨
󵄨󵄨
= sup 󵄨󵄨󵄨󵄨𝜀2(4−𝑝)/(𝑝−2) ∫ 𝑏 ( ) 𝐸1 (𝑏 ( ) |𝑢|2 ) 𝑢𝑣󵄨󵄨󵄨󵄨
𝜀
𝜀
󵄨
‖𝑣‖=1 󵄨
󵄨󵄨 𝑥 󵄨󵄨6 1/3
≤ sup {|𝜀|2(4−𝑝)/(𝑝−2) (∫ 󵄨󵄨󵄨󵄨𝑏 ( )󵄨󵄨󵄨󵄨 )
󵄨 𝜀 󵄨
‖𝑣‖=1
×(∫ |𝑢|6 )
1/2
(∫ |𝑣|6 )
1/6
(51)
If (𝜀, 𝑢) → (𝜀0 , 𝑢0 ) with 𝜀0 ≠ 0, then
󵄩󵄩 󸀠󸀠
󵄩
󵄩󵄩𝐺2 (𝜀, 𝑢) − 𝐺2󸀠󸀠 (𝜀0 , 𝑢0 )󵄩󵄩󵄩
󵄩
󵄩
󵄨󵄨 󸀠󸀠
󵄨
= sup 󵄨󵄨󵄨𝐺2 (𝜀, 𝑢) [𝑤, 𝑣] − 𝐺2󸀠󸀠 (𝜀0 , 𝑢0 ) [𝑤, 𝑣]󵄨󵄨󵄨󵄨
‖𝑤‖=‖𝑣‖=1
≤
sup
‖𝑤‖=‖𝑣‖=1
(53)
{𝐼1 + 𝐼2 } ,
where
󵄨󵄨
𝑥
𝑥
𝐼1 := 󵄨󵄨󵄨󵄨𝜀2(4−𝑝)/(𝑝−2) ∫ 𝑏 ( ) 𝐸1 (𝑏 ( ) |𝑢|2 ) 𝑤𝑣
𝜀
𝜀
󵄨
2(4−𝑝)/(𝑝−2)
−𝜀0
∫𝑏(
󵄨󵄨
𝑥 󵄨 󵄨2
𝑥
󵄨
) 𝐸1 (𝑏 ( ) 󵄨󵄨󵄨𝑢0 󵄨󵄨󵄨 ) 𝑤𝑣󵄨󵄨󵄨 ,
󵄨󵄨
𝜀0
𝜀0
󵄨󵄨
𝑥
𝑥
𝐼2 := 2 󵄨󵄨󵄨󵄨𝜀2(4−𝑝)/(𝑝−2) ∫ 𝑏 ( ) 𝐸1 (𝑏 ( ) 𝑢𝑤) 𝑢𝑣
𝜀
𝜀
󵄨
2(4−𝑝)/(𝑝−2)
−𝜀0
∫𝑏(
󵄨󵄨
𝑥
𝑥
󵄨
) 𝐸1 (𝑏 ( ) 𝑢0 𝑤) 𝑢0 𝑣󵄨󵄨󵄨 .
󵄨󵄨
𝜀0
𝜀0
(54)
We estimate 𝐼1 only, and 𝐼2 can be estimated in a similar way,
indeed
󵄨
2(4−𝑝)/(𝑝−2) 󵄨󵄨
󵄨󵄨
𝐼1 ≤ 󵄨󵄨󵄨󵄨𝜀2(4−𝑝)/(𝑝−2) − 𝜀0
󵄨
𝑥
𝑥
󵄨󵄨󵄨
× 󵄨󵄨󵄨∫ 𝑏 ( ) 𝐸1 (𝑏 ( ) |𝑢|2 ) 𝑤𝑣
𝜀
𝜀
󵄨
−∫𝑏(
󵄨󵄨
𝑥
𝑥 󵄨 󵄨2
󵄨
) 𝐸1 (𝑏 ( ) 󵄨󵄨󵄨𝑢0 󵄨󵄨󵄨 ) 𝑤𝑣󵄨󵄨󵄨
󵄨󵄨
𝜀0
𝜀0
𝑥
𝑥
󵄨 󵄨2(4−𝑝)/(𝑝−2) 󵄨󵄨󵄨󵄨
2
+ 󵄨󵄨󵄨𝜀0 󵄨󵄨󵄨
󵄨󵄨∫ 𝑏 ( ) 𝐸1 (𝑏 ( ) |𝑢| ) 𝑤𝑣
𝜀
𝜀
󵄨
−∫𝑏(
󵄨󵄨
𝑥 󵄨 󵄨2
𝑥
󵄨
) 𝐸1 (𝑏 ( ) 󵄨󵄨󵄨𝑢0 󵄨󵄨󵄨 ) 𝑤𝑣󵄨󵄨󵄨 .
󵄨󵄨
𝜀0
𝜀0
(55)
}.
And we know ||𝐺2󸀠 (𝜀, 𝑢)|| → 0, as 𝜀 → 0. From the above
arguments, we know 𝐺󸀠 = 𝐺1󸀠 + 𝐺2󸀠 is continuous in (𝜀, 𝑢).
Similar to the proof in Lemma 7, we know 𝐼1 → 0 as 𝜀 → 𝜀0
and 𝑢 → 𝑢0 . Thus we know ‖𝐺2󸀠󸀠 (𝜀, 𝑢) − 𝐺2󸀠󸀠 (𝜀0 , 𝑢0 )‖ → 0 as
(𝜀, 𝑢) → (𝜀0 , 𝑢0 ).
Abstract and Applied Analysis
7
If (𝜀, 𝑢) → (0, 𝑢0 ), then from the definition of 𝐺2 , we
know
󵄩
󵄩󵄩 󸀠󸀠
󵄩󵄩𝐺2 (𝜀, 𝑢)󵄩󵄩󵄩
󵄩
󵄩
󵄨
󵄨
= sup 󵄨󵄨󵄨󵄨𝐺2󸀠󸀠 (𝜀, 𝑢) [𝑤, 𝑣]󵄨󵄨󵄨󵄨
‖𝑤‖=‖𝑣‖=1
≤
sup
‖𝑤‖=‖𝑣‖=1
{𝜀2(4−𝑝)/(𝑝−2)
𝑥
𝑥
× ∫ (𝑏 ( ) 𝐸1 (𝑏 ( ) |𝑢|2 ) 𝑤𝑣
𝜀
𝜀
(56)
sup
‖𝑤‖=‖𝑣‖=1
1/2
󵄨󵄨 𝑥 󵄨󵄨2
𝐼3 ≤ 𝐶0 |𝜀|2(4−𝑝)/(𝑝−2) (∫ 󵄨󵄨󵄨󵄨𝑏 ( )󵄨󵄨󵄨󵄨 |𝑤𝑣|2 )
󵄨 𝜀 󵄨
= 𝐶1 |𝜀|2(4−𝑝)/(𝑝−2) ∫ |𝑏 (𝑥)|2 󳨀→ 0,
1/2
󵄨󵄨 𝑥 󵄨󵄨2
× (∫ 󵄨󵄨󵄨󵄨𝑏 ( )󵄨󵄨󵄨󵄨 |𝑢|4 )
󵄨 𝜀 󵄨
since 4 > 𝑝 > 2. Thus we obtain
󵄨󵄨 𝑥
≤ 𝐶0 |𝜀|2(4−𝑝)/(𝑝−2) (∫ 󵄨󵄨󵄨󵄨𝑏 ( )󵄨󵄨󵄨󵄨 )
󵄨 𝜀 󵄨
× (∫ |𝑤| )
(∫ |𝑣| )
1/6
(∫ |𝑢|6 )
1/3
lim
(57)
× (∫ |𝑤| )
Therefore
1/6
6
(∫ |𝑣| )
1/6
󵄩
󵄩󵄩 󸀠󸀠
󵄩󵄩𝐺2 (𝜀, 𝑢)󵄩󵄩󵄩 󳨀→ 0
󵄩
󵄩
(64)
𝐺1󸀠 (𝜀, 𝑧𝜃 ) = 𝑜 (𝜀3/2 ) .
.
as 𝜀 󳨀→ 0.
1
Γ (𝜃) = − 𝑈𝑝 (𝜃) ∫ (𝑎 (𝑥) − 𝐴) .
𝑝
(58)
(59)
Then
𝐺 (𝜀, 𝑧𝜃 )
lim
= Γ (𝜃) ,
𝜀→0
𝜀3
3/2
𝐺 (𝜀, 𝑧𝜃 ) = 𝑜 (𝜀
(66)
Also, since 𝑧 is bounded, it is easy to check that
󵄩󵄩 󸀠
󵄩
󵄩󵄩𝐺2 (𝜀, 𝑧𝜃 )󵄩󵄩󵄩
󵄩
󵄩 = |𝜀|−3/2 sup 󵄨󵄨󵄨(𝐺󸀠 (𝜀, 𝑧 ) , 𝑣)󵄨󵄨󵄨
󵄨󵄨 2
󵄨󵄨
𝜃
3/2
|𝜀|
‖𝑣‖=1
= |𝜀|2(4−𝑝)/(𝑝−2)−3/2
󵄨󵄨
󵄨󵄨
𝑥
𝑥 󵄨 󵄨2
× sup 󵄨󵄨󵄨󵄨∫ 𝑏 ( ) 𝐸1 (𝑏 ( ) 󵄨󵄨󵄨𝑧𝜃 󵄨󵄨󵄨 ) 𝑧𝜃 𝑣󵄨󵄨󵄨󵄨
𝜀
𝜀
󵄨
‖𝑣‖=1 󵄨
≤ 𝐶0 |𝜀|2(4−𝑝)/(𝑝−2)−3/2
󵄨󵄨 𝑥 󵄨󵄨2 󵄨 󵄨4 1/2
× sup {(∫ 󵄨󵄨󵄨󵄨𝑏 ( )󵄨󵄨󵄨󵄨 󵄨󵄨󵄨𝑧𝜃 󵄨󵄨󵄨 )
󵄨 𝜀 󵄨
‖𝑣‖=1
(60)
).
Proof. By changing of variable, we know
1/2
󵄨󵄨 𝑥 󵄨󵄨2 󵄨 󵄨2
×(∫ 󵄨󵄨󵄨󵄨𝑏 ( )󵄨󵄨󵄨󵄨 󵄨󵄨󵄨𝑧𝜃 󵄨󵄨󵄨 |𝑣|2 ) }
󵄨 𝜀 󵄨
≤ 𝐶1 |𝜀|2(4−𝑝)/(𝑝−2)−3/2
1
𝑥
𝐺1 (𝜀, 𝑧𝜃 ) = − ∫ (𝑎 ( ) − 𝐴) 𝑈𝑝 (𝑥 + 𝜃)
𝑝
𝜀
𝜀3
∫ (𝑎 (𝑥) − 𝐴) 𝑈𝑝 (𝜀𝑥 + 𝜃) .
𝑝
(65)
From the proof of [7], we first know
Lemma 9. Assume (𝑎1 ) and (𝑏1 ) are satisfied. Define
= −
𝐺 (𝜀, 𝑧𝜃 )
= Γ (𝜃) .
𝜀3
𝐺󸀠 (𝜀, 𝑧𝜃 ) = 𝑜 (𝜀3/2 ) .
From the above arguments, we know 𝐺󸀠󸀠 is continuous in
(𝜀, 𝑢) and the proof is complete.
󸀠
𝜀→0
Now we are ready to prove
,
1/3
󵄨󵄨 𝑥 󵄨󵄨6 1/3
𝐼4 ≤ 𝐶1 |𝜀|2(4−𝑝)/(𝑝−2) (∫ 󵄨󵄨󵄨󵄨𝑏 ( )󵄨󵄨󵄨󵄨 ) (∫ |𝑢|6 )
󵄨 𝜀 󵄨
6
as 𝜀 󳨀→ 0,
(63)
󵄨󵄨6 1/3
6
On the other hand, since 𝑧𝜃 is bounded and 𝑏(𝑥) ∈ 𝐿2 (R3 ),
then, changing of variable, we know
󵄨󵄨 𝐺 (𝜀, 𝑧 ) 󵄨󵄨 󵄨󵄨 2(4−𝑝)/(𝑝−2)−3
󵄨𝜀
󵄨󵄨 2
𝜃 󵄨󵄨
󵄨󵄨 = 󵄨󵄨󵄨
󵄨󵄨
3
󵄨
󵄨󵄨
𝜀
4
󵄨󵄨 󵄨󵄨󵄨
󵄨
󵄨󵄨 𝑥 󵄨󵄨2
≤ 𝐶1 |𝜀|2(4−𝑝)/(𝑝−2)−3 ∫ 󵄨󵄨󵄨󵄨𝑏 ( )󵄨󵄨󵄨󵄨
󵄨 𝜀 󵄨
Using Hölder inequality, we know
1/6
(62)
󵄨󵄨 𝑥 󵄨󵄨2 󵄨 󵄨4
≤ |𝜀|2(4−𝑝)/(𝑝−2)−3 (∫ 󵄨󵄨󵄨󵄨𝑏 ( )󵄨󵄨󵄨󵄨 󵄨󵄨󵄨𝑧𝜃 󵄨󵄨󵄨 )
󵄨 𝜀 󵄨
{𝐼3 + 𝐼4 } .
6
𝐺1 (𝜀, 𝑧𝜃 )
= Γ (𝜃) .
𝜀→0
𝜀3
lim
󵄨
𝑥
𝑥 󵄨 󵄨2 󵄨 󵄨2 󵄨󵄨
× ∫ 𝑏 ( ) 𝐸1 (𝑏 ( ) 󵄨󵄨󵄨𝑧𝜃 󵄨󵄨󵄨 ) 󵄨󵄨󵄨𝑧𝜃 󵄨󵄨󵄨 󵄨󵄨󵄨󵄨
𝜀
𝜀
󵄨󵄨
𝑥
𝑥
+2𝑏 ( ) 𝐸1 (𝑏 ( ) 𝑤𝑢) 𝑢𝑣)}
𝜀
𝜀
:=
Since 𝑎(𝑥) is continuous and bounded, the dominated convergence theorem implies that
(61)
󵄨󵄨 󵄨
󵄨󵄨 𝑥 󵄨󵄨2 1/2 󵄨󵄨󵄨󵄨
󵄨󵄨 󵄨󵄨 𝑥 󵄨󵄨󵄨2 1/2
󵄨
󵄨
󵄨
× sup 󵄨󵄨(∫ 󵄨󵄨𝑏 ( )󵄨󵄨 ) (∫ 󵄨󵄨󵄨󵄨𝑏 ( ) 𝑣󵄨󵄨󵄨󵄨 ) 󵄨󵄨󵄨 .
󵄨󵄨
󵄨 𝜀 󵄨
󵄨 𝜀 󵄨
‖𝑣‖=1 󵄨󵄨󵄨
󵄨
(67)
8
Abstract and Applied Analysis
Moreover, recall that 𝑏(𝑥) ∈ 𝐿2 is bounded and use Hölder
inequality, we get
󵄩
󵄩󵄩 󸀠
󵄩󵄩𝐺2 (𝜀, 𝑧𝜃 )󵄩󵄩󵄩
󵄩 ≤ 𝐶 |𝜀|2(4−𝑝)/(𝑝−2) ,
󵄩
(68)
1
3/2
|𝜀|
since 4 > 𝑝, we have
𝐺2󸀠
(𝜀, 𝑧𝜃 )
= 0.
𝜀3/2
From the above arguments, we know
lim
(69)
𝜀→0
󸀠
𝐺 (𝜀, 𝑧𝜃 )
= 0,
𝜀3/2
and the proof is completed.
(70)
lim
𝜀→0
Remark 10. The hypothesis ∫(𝑎(𝑥) − 𝐴) ≠ 0 is used to apply
Lemma 5 and has been already used in [4, 7]. If ∫(𝑎(𝑥) − 𝐴) is
identically zero, we can not conclude that there exist critical
points of 𝐼𝜀 .
In the following we prove Theorem 3.
Lemma 11. Assume (𝑎2 ) and (𝑏1 ) are satisfied. Then 𝐺, 𝐺󸀠 , and
𝐺󸀠󸀠 are continuous in (𝜀, 𝑢).
Proof . Keeping the exponentially decay property of 𝑈 in
mind, the continuity of 𝐺1 , 𝐺1󸀠 , and 𝐺1󸀠󸀠 in (𝜀, 𝑢) can be proved
similarly as in [7]. We can also repeat the proof in Lemma 7 to
know the continuity of 𝐺2 . Thus the lemma is concluded.
Lemma 12. Assume (𝑎2 ) and (𝑏1 ) are satisfied. Define
𝐿
∫ |𝑥|−𝛾 𝑈𝑝+1 (𝑥 + 𝜃) .
𝑝+1
(71)
(74)
Moreover, by the boundedness of 𝑧𝜃 , we know
2(4−𝑝)/(𝑝−2)
󵄨 𝜀
󵄨󵄨
󵄨󵄨𝐺2 (𝜀, 𝑧𝜃 )󵄨󵄨󵄨 =
4
󵄨 󵄨2 󵄨 󵄨2
× ∫ 𝑏 (𝑥) 𝐸1 (𝑏 (𝑥) 󵄨󵄨󵄨𝑧𝜃 󵄨󵄨󵄨 ) 󵄨󵄨󵄨𝑧𝜃 󵄨󵄨󵄨
≤
𝜀
2(4−𝑝)/(𝑝−2)
4
󵄨󵄨 𝑥 󵄨 󵄨2 󵄨󵄨2
(∫ 󵄨󵄨󵄨󵄨𝑏 ( ) 󵄨󵄨󵄨𝑧𝜃 󵄨󵄨󵄨 󵄨󵄨󵄨󵄨 )
󵄨
󵄨 𝜀
(75)
(72)
𝐺 (𝜀, 𝑧𝜃 )
= 0.
𝜀→0
𝜀𝛾/2
Since 3 > 𝛾 we obtain
lim
𝜀→0
𝐺 (𝜀, 𝑧𝜃 )
= Γ (𝜃) .
𝜀𝛾
(76)
To study the property of 𝐺󸀠 (𝜀, 𝑧𝜃 ), since 𝛾 < 3 and 𝑈
exponentially decays at infinity, from the proof in [7], we
know
𝐺1󸀠 (𝜀, 𝑧𝜃 )
= 0.
𝜀→0
𝜀𝛾/2
lim
(77)
On the other hand, from the boundedness of 𝑍𝜃 and 𝑏(𝑥), we
have
󵄩󵄩 󸀠
󵄩󵄩
2(4−𝑝)/(𝑝−2)
󵄩󵄩󵄩𝐺2 (𝜀, 𝑧𝜃 )󵄩󵄩󵄩 = |𝜀|
󵄨󵄨
󵄨󵄨
𝑥
𝑥 󵄨 󵄨2
× sup 󵄨󵄨󵄨󵄨∫ 𝑏 ( ) 𝐸1 (𝑏 ( ) 󵄨󵄨󵄨𝑧𝜃 󵄨󵄨󵄨 ) 𝑧𝜃 𝑣󵄨󵄨󵄨󵄨
𝜀
𝜀
󵄨
‖𝑣‖=1 󵄨
≤ 𝐶0 |𝜀|2(4−𝑝)/(𝑝−2)
󵄨󵄨 𝑥 󵄨󵄨2 󵄨 󵄨4 1/2
× sup {(∫ 󵄨󵄨󵄨󵄨𝑏 ( )󵄨󵄨󵄨󵄨 󵄨󵄨󵄨𝑧𝜃 󵄨󵄨󵄨 )
󵄨 𝜀 󵄨
‖𝑣‖=1
1/2
󵄨󵄨 𝑥 󵄨󵄨2 󵄨 󵄨2
×(∫ 󵄨󵄨󵄨󵄨𝑏 ( )󵄨󵄨󵄨󵄨 󵄨󵄨󵄨𝑧𝜃 󵄨󵄨󵄨 |𝑣|2 ) }
󵄨 𝜀 󵄨
(78)
󵄨󵄨 󵄨
1/2 󵄨󵄨󵄨
󵄨󵄨 󵄨󵄨 𝑥 󵄨󵄨󵄨2 1/2
󵄨
2
󵄨
󵄨
󵄨
× sup 󵄨󵄨(∫ 󵄨󵄨𝑏 ( )󵄨󵄨 ) (∫ |𝑣| ) 󵄨󵄨󵄨
󵄨
󵄨󵄨
𝜀
󵄨
󵄨
‖𝑣‖=1 󵄨󵄨
󵄨
≤ 𝐶2 |𝜀|2(4−𝑝)/(𝑝−2)+3/2 .
lim Γ (𝜃) = 0,
|𝜃| → ∞
󸀠
lim
Since 𝛾 < 3, we get
𝐺2󸀠 (𝜀, 𝑧𝜃 )
= 0.
𝜀→0
𝜀𝛾/2
lim
Proof. As we know
𝐺1 (𝜀, 𝑧𝜃 ) = −
𝐺1 (𝜀, 𝑧𝜃 )
= Γ (𝜃) .
𝜀𝛾
≤ 𝐶1 |𝜀|2(4−𝑝)/(𝑝−2)
Then for all 𝜃 ∈ 𝑅3 , we have
𝐺 (𝜀, 𝑧𝜃 )
lim
= Γ (𝜃) ,
𝜀→0
𝜀𝛾
lim
𝜀→0
≤ 𝐶0 |𝜀|2(4−𝑝)/(𝑝−2)+3 .
Proof of Theorem 2. By the exponential decay property of
proposition 𝑈, it is easy to check that 𝐼0󸀠󸀠 is a compact
perturbation of the identity map, and so it is an index0 Fredholm operator. By Proposition 4, we know that 𝑍
is a nondegenerate 3-dimensional critical manifold. From
Lemmas 7 to 9, we know all the assumptions of Lemma 5 are
satisfied. Since 𝑈 has a strict (global) maximum at 𝑥 = 0, Γ has
a strict (global) maximum or minimum at 𝜃 = 0 depending
on the sign of ∫(𝑎(𝑥) − 𝐴). By the abstract theorem, we know
the existence of family solutions {(𝜀, 𝑢𝜀 )} ⊂ R × 𝐻1 (R3 ). If
2 < 𝑝 < 2+4/3, it is easy to check that 𝜓𝜀 → 0 as 𝜀 → 0.
Γ (𝜃) = −
By assumption (𝑎2 ) and the decay property of 𝑈,
(79)
From the above arguments, we know
𝛾
𝛾
𝑝+1
𝜀
𝑥
|𝑥| 𝑈 (𝑥 + 𝜃)
.
∫ (𝑎 ( ) − 𝐴) 𝛾
𝑝+1
𝜀
𝜀
|𝑥|𝛾
(73)
𝐺󸀠 (𝜀, 𝑧𝜃 )
= 0.
𝜀→0
𝜀𝛾/2
lim
(80)
Abstract and Applied Analysis
9
Proof of Theorem 3. From Lemmas 11 and 12, we know
that all the assumptions of Lemma 5 are satisfied. Since
lim|𝜃| → ∞ Γ(𝜃) = 0 and Γ(0) ≠ 0, we know that there is 𝑅 > 0
such that either
min Γ (𝜃) > Γ (0)
|𝜃|=𝑅
or max Γ (𝜃) < Γ (0) .
|𝜃|=𝑅
(81)
By the abstract Theorem 2, we know the existence of family
solutions {(𝜀, 𝑢𝜀 )} ⊂ R × 𝐻1 (R3 ). If 2 < 𝑝 < 2 + 4/3, it is easy
to check that (𝜙𝜀 , 𝜓𝜀 ) → 0 as 𝜀 → 0.
Acknowledgments
This work is supported by ZJNSF (Y7080008, R6090109,
LQ12A01015,
Y201016244,
2012C31025),
SRPWZ
(G20110004), and NSFC (10971194, 11005081, 21207103).
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