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(1)iψt+Δψ=b(x)ψφx1-a(x)|ψ|p-2ψ,-Δφ=(b(x)|ψ|2)x1.
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=b1ψφx-a|ψ|2ψ,mφxx+φyy=b2(|ψ|2)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, hyperbolic-elliptic, 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
(3)ℱ{E1(ψ)}(ξ)=σ1(ξ)ℱ(ψ)(ξ),
where σ1(ξ)=ξ12/|ξ|2, ξ∈ℝ3, and ℱ denotes the Fourier transform:
(4)ℱ(ψ)(ξ)=(12π)3/2∫e-iξxψ(x)dx.
Then the generalized Davey-Stewartson system can be reduced to the following single nonlocal Schrödinger equation
(5)-iψt-Δψ=a(x)|ψ|p-2ψ+b(x)E1(b(x)|ψ|2)ψ.
In this paper, we are interested in the existence of standing waves for the above equation, that is, solutions in the form of
(6)ψ(t,x)=eiωtϕ(x),φ(t,x)=v(x),
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:
(7)-Δϕ+ωϕ=b(x)E1(b(x)|ϕ|2)ϕ+a(x)|ϕ|p-2ϕ.

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 Concentration-compactness 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
(8)-Δϕ+ϕ=εb(x)E1(b(x)|ϕ|2)ϕ+(1+εa(x))|ϕ|p-2ϕ,(9)-Δϕ+ωϕ=b(x)E1(b(x)|ϕ|2)ϕ+a(x)|ϕ|p-2ϕ.

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
(10)14∫b(x)E1(b(x)|U|2)|U|2+1p∫a(x)|U|p≠0,
then for any ε small, there exists at least one solution of problem (8).

Theorem 2.

Let ω=ε2, suppose 2<p<4, and there exists a positive constant A such that a(x),b(x) satisfy

a(x)-A is continuous, bounded and a(x)-A∈L1(ℝ3) with ∫(a(x)-A)≠0;

b(x) is continuous, bounded and b∈L2(ℝ3).

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.

Theorem 3.

Let ω=ε2, suppose 2<p<4 and a(x),b(x) satisfy (b1) and

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.

Throughout this paper, we denote the norm of H1(ℝ3) by
(11)∥u∥=(∫|∇u|2+u2)1/2,
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*:
(12)-Δu+u=A|u|p-2u,u>0,u∈H1(ℝ3).
There is a unique positive radial solution U, which satisfies the following decay property:
(13)limr→∞U(r)rer=C>0,limr→∞U′(r)U(r)=-1,r=|x|,
where C>0 is a constant. The function U is a critical point of C2 functional I0:H1(ℝ3)→ℝ defined by
(14)I0(u)=12∥u∥2-Ap∫ℝ3|u|p.
Moreover, I0 possesses a 3-dimensional manifold of critical points
(15)Z={zθ=U(x+θ),θ∈ℝ3}.
Set
(16)Q(u):=I0′′(U)[u,u]=∫ℝ3[|∇u|2+u2-pAUp-2u2]
and denote X=span{∂U/∂xi,1≤i≤3}. We have

Q(U)=(2-p)A∫ℝ3Updx<0,

KerQ=X,

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
(17)Iε(u)=I0(u)+G(ε,u),
where I0:E→ℝ and G:ℝ×E→ℝ. We need the following hypotheses and assume that

I0 and G are C2 with respect to u;

G is continuous in (ε,u) and G(0,u)=0 for all u;

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.

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.

let TθZ denote the tangent space to Z at zθ, the manifold Z is nondegenerate in the following sense:

Ker(I0′′(z))=TθZ and I0′′(zθ) is an index-0 Fredholm operator for any zθ∈Z.

There exists α>0 and a continuous function Γ:Z→ℝ such that
(18)Γ(z)=limε→0G(ε,z)εα,G′(ε,z)=o(εα/2).

Consider the existence of critical points of the perturbed problem
(19)Iε′(u)=0.
We want to look for solutions of the form u=z+w with z∈Z and w∈W=(TθZ)⊥. 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:
(20)PIε′(z+w)=0,(I-P)Iε′(z+w)=0.
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)(21)Iε(u)=I0(z)+εαΓ(z)+o(εα).
In [4, 5] the following abstract theorem is proved.

Lemma 5.

Suppose assumptions (1)–(6) are satisfied, and there exists δ>0 and z*∈Z such that
(22)eithermin∥z-z*∥=δΓ(z)>Γ(z*)ormax∥z-z*∥=δΓ(z)<Γ(z*).
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
(23)ℱ{E1(ψ)}(ξ)=σ1(ξ)ℱ(ψ)(ξ),
where σ1(ξ)=ξ12/|ξ|2, ξ∈ℝ3, and ℱ denotes the Fourier transform:
(24)ℱ(ψ)(ξ)=(12π)3/2∫e-iξxψ(x)dx.
For 1<p<∞, E1 satisfies the following properties:

E1∈ℒ(Lp,Lp).

if ψ∈H1(ℝ3), then E1(ψ)∈H1(ℝ3).

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.

3. Proof of the Main Results

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
(25)Iε(ϕ)=12∥ϕ∥2-ε4∫b(x)E1(b(x)|ϕ|2)|ϕ|2-1p∫(1+εa(x))|ϕ|p.
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 <xref ref-type="statement" rid="thm1.1">1</xref>.

Set
(26)I0(ϕ)=12∥ϕ∥2-1p∫|ϕ|p,G(ϕ)=-1p∫a(x)|ϕ|p-14∫b(x)E1(b(x)|ϕ|2)|ϕ|2,
then Iε(u) can be rewritten as
(27)Iε(ϕ)=I0(ϕ)+εG(ϕ).
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
(28)lim|θ|→∞Γ(θ)=0,Γ(θ):=G|Z=-1p∫a(x)|zθ|p-14∫b(x)E1(b(x)|zθ|2)|zθ|2.

From the fact that a∈L6/(6-p)(ℝ3), for any T>0, we have
(29)|∫a(x)|zθ|p|≤|∫|x|≤Ta(x)|zθ|p|+|∫|x|≥Ta(x)|zθ|p|≤(∫|x|≤T|a(x)|6/(6-p))(6-p)/6×(∫|x|≤T|zθ|6)p/2*+(∫|x|≥T|a(x)|6/(6-p))(6-p)/6×(∫|x|≥T|zθ|6)p/6.
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
(30)B1(ϕ)=∫σ1(ξ)|ℱ(b·ϕ)(ξ)|2dξ,
it follows from the Parseval identity that
(31)B1(ϕ)=∫E1(b·ϕ)b·ϕ¯,
and in particular we have
(*)0<B1(ϕ)≤|b(x)ϕ|22.
Then for any T>0, we have
(32)|∫b(x)E1(b(x)|zθ|2)|zθ|2|≤∫|b(x)|zθ|2|2≤(∫|x|≤T|b(x)|6)1/3(∫|x|≤T|zθ|6)2/3+(∫|x|≥T|b(x)|6)1/3(∫|x|≥T|zθ|6)2/3.
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
(33)lim|θ|→∞Γ(θ)=0.
Then by assumption (10) that
(34)14∫b(x)E1(b(x)|U|2)|U|2+1p∫a(x)|U|p≠0,
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).

We are going to consider problem (9). Set
(35)ω=ε2,ϕ(x)=ε2/(p-2)u(εx).
We have
(36)-Δu+u=a(xε)|u|p-2u+ε2(4-p)/(p-2)b(xε)×E1(b(xε)|u|2)u.
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
(37)Iε(u)=12∥u∥2-1p∫a(xε)|u|p-ε2(4-p)/(p-2)4×∫b(xε)E1(b(xε)|u|2)|u|2.
Set
(38)I0(u)=12∥u∥2-Ap∫|u|p.
Then Iε(u) can be rewritten as
(39)Iε(u)=I0(u)+1p∫(A-a(xε))|u|p-ε2(4-p)/(p-2)4×∫b(xε)E1(b(xε)|u|2)|u|2.
Define
(40)G~(ε,u)=1p∫(A-a(xε))|u|p-ε2(4-p)/(p-2)4∫b(xε)E1(b(xε)|u|2)|u|2=G~1(ε,u)+G~2(ε,u)
and for i=1,2(41)Gi(ε,u)={G~i(ε,u),ifε≠0,0,ifε=0.

Lemma 7.

Under assumptions (a1) and (b1), G=G1+G2 is continuous in (ε,u).

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
(42)4|G2(ε,u)-G2(ε0,u0)|=|ε2(4-p)/(p-2)∫b(xε)E1(b(xε)|u|2)|u|2-ε02(4-p)/(p-2)∫b(xε0)E1(b(xε0)|u0|2)|u0|2|≤|ε|2(4-p)/(p-2)|∫b(xε)E1(b(xε)|u|2)|u|2-∫b(xε0)E1(b(xε0)|u0|2)|u0|2|+|ε2(4-p)/(p-2)-ε02(4-p)/(p-2)|×|∫b(xε0)E1(b(xε0)|u0|2)|u0|2|≔|ε|2(4-p)/(p-2)I1+I2.
It is obvious that I2→0, as ε→ε0. At the same time, we know
(43)I1≤|∫[b(xε)-b(xε0)]E1(b(xε)|u|2)|u|2|+|∫b(xε0)E1([b(xε)-b(xε0)]|u|2)|u|2|+|∫b(xε0)E1(b(xε0)[|u|2-|u0|2])|u|2|+|∫b(xε0)E1(b(xε0)|u0|2)(|u|2-|u0|2)|≔Π1+Π2+Π3+Π4.
Estimating the first term Π1, by Hölder inequality, we know
(44)Π1=|∫[b(xε)-b(xε0)]E1(b(xε)|u|2)|u|2|≤(∫(|b(xε)-b(xε0)||u|2)2)1/2×(∫|E1(b(xε)|u|2)|2)1/2.
Since b(x) is bounded and continuous, the operator E1∈ℒ(L2,L2), the dominated convergence theorem implies that
(45)Π1→0,asε→ε0.
Similarly, we can deduce that Π2,Π3,Π4 vanishes, as (ε,u)→(ε0,u0). Hence
(46)|G2(ε,u)-G2(ε0,u0)|→0
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
(47)4|G2(ε,u)|≤|ε|2(4-p)/(p-2)∫|b(xε)E1(b(xε)|u|2)|u|2|≤|ε|2(4-p)/(p-2)∫|b(xε)|2|u(x)|4≤|ε|2(4-p)/(p-2)(∫|b(xε)|6)1/3(∫|u|6)2/3,
therefore G2(ε,u)→0, as (ε,u)→(0,u). Hence G=G1+G2 is continuous and the lemma is proved.

Lemma 8.

Under assumptions (a1) and (b1), G′ and G′′ are continuous in (ε,u).

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
(48)∥G2′(ε,u)-G2′(ε0,u0)∥=sup∥v∥=1{(b(xε0)|u0|2)ε2(4-p)/(p-2)×∫b(xε)E1(b(xε)|u|2)uv-ε02(4-p)/(p-2)×∫b(xε0)E1(b(xε0)|u0|2)u0v}≤sup∥v∥=1{(b(xε0)|u0|2)|ε|2(4-p)/(p-2)×∫(b(xε)E1(b(xε)|u|2)uv-b(xε0)E1(b(xε0)|u0|2)u0v)+|ε2(4-p)/(p-2)-ε02(4-p)/(p-2)|×|∫b(xε0)E1(b(xε0)|u0|2)u0v|}≔|ε|2(4-p)/(p-2)sup∥v∥=1I1+sup∥v∥=1I2.
Estimating the second term, since E1∈ℒ(L2,L2), by Hölder inequality, we know
(49)sup∥v∥=1I2=sup∥v∥=1|ε2(4-p)/(p-2)-ε02(4-p)/(p-2)|×|∫b(xε0)E1(b(xε0)|u0|2)u0v|≤sup∥v∥=1C0|ε2(4-p)/(p-2)-ε02(4-p)/(p-2)|×(∫|b(xε0)|u0|2|2)1/2(∫|b(xε0)u0v|2)1/2≤C1|ε2(4-p)/(p-2)-ε02(4-p)/(p-2)|(∫|u0|6)1/2.
Thus sup∥v∥=1I2→0 as ε→ε0. Estimating the first term I1, we know
(50)sup∥v∥=1I1=sup∥v∥=1{(xε0)∫(b(xε)E1(b(xε)|u|2)uv-b(xε0)E1(b(xε0)|u0|2)u0v)}≤sup∥v∥=1{|∫[b(xε)-b(xε0)]E1(b(xε)|u|2)uv|+|∫b(xε0)E1×([b(xε)-b(xε0)]|u|2)uv(xε0)|+|∫b(xε0)E1×(b(xε0)[|u|2-|u0|2])uv|+|∫b(xε0)E1×(b(xε0)|u0|2)(uv-u0v)|}≔A1+A2+A3+A4.
As in Lemma 7, by Hölder inequality again, we can prove that Ai→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
(51)∥G2′(ε,u)∥=sup∥v∥=1|ε2(4-p)/(p-2)∫b(xε)E1(b(xε)|u|2)uv|≤sup∥v∥=1{|ε|2(4-p)/(p-2)(∫|b(xε)|6)1/3×(∫|u|6)1/2(∫|v|6)1/6}.
And we know ||G2′(ε,u)||→0, as ε→0. From the above arguments, we know G′=G1′+G2′ is continuous in (ε,u).

In the following we prove that G′′ is continuous in (ε,u). As we know
(52)G2′′(ε,u)[w,v]=ε2(4-p)/(p-2)×∫(b(xε)E1(b(xε)|u|2)wv+2b(xε)E1(b(xε)wu)uv).

If (ε,u)→(ε0,u0) with ε0≠0, then
(53)∥G2′′(ε,u)-G2′′(ε0,u0)∥=sup∥w∥=∥v∥=1|G2′′(ε,u)[w,v]-G2′′(ε0,u0)[w,v]|≤sup∥w∥=∥v∥=1{I1+I2},
where
(54)I1≔|ε2(4-p)/(p-2)∫b(xε)E1(b(xε)|u|2)wv-ε02(4-p)/(p-2)∫b(xε0)E1(b(xε0)|u0|2)wv|,I2≔2|ε2(4-p)/(p-2)∫b(xε)E1(b(xε)uw)uv-ε02(4-p)/(p-2)∫b(xε0)E1(b(xε0)u0w)u0v|.
We estimate I1 only, and I2 can be estimated in a similar way, indeed
(55)I1≤|ε2(4-p)/(p-2)-ε02(4-p)/(p-2)|×|∫b(xε)E1(b(xε)|u|2)wv-∫b(xε0)E1(b(xε0)|u0|2)wv|+|ε0|2(4-p)/(p-2)|∫b(xε)E1(b(xε)|u|2)wv-∫b(xε0)E1(b(xε0)|u0|2)wv|.
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).

If (ε,u)→(0,u0), then from the definition of G2, we know
(56)∥G2′′(ε,u)∥=sup∥w∥=∥v∥=1|G2′′(ε,u)[w,v]|≤sup∥w∥=∥v∥=1{(b(xε)wu)ε2(4-p)/(p-2)×∫(b(xε)E1(b(xε)|u|2)wv+2b(xε)E1(b(xε)wu)uv)}≔sup∥w∥=∥v∥=1{I3+I4}.
Using Hölder inequality, we know
(57)I3≤C0|ε|2(4-p)/(p-2)(∫|b(xε)|2|wv|2)1/2×(∫|b(xε)|2|u|4)1/2≤C0|ε|2(4-p)/(p-2)(∫|b(xε)|6)1/3(∫|u|6)1/3×(∫|w|6)1/6(∫|v|6)1/6,I4≤C1|ε|2(4-p)/(p-2)(∫|b(xε)|6)1/3(∫|u|6)1/3×(∫|w|6)1/6(∫|v|6)1/6.
Therefore
(58)∥G2′′(ε,u)∥→0asε→0.
From the above arguments, we know G′′ is continuous in (ε,u) and the proof is complete.

Lemma 9.

Assume (a1) and (b1) are satisfied. Define
(59)Γ(θ)=-1pUp(θ)∫(a(x)-A).
Then
(60)limε→0G(ε,zθ)ε3=Γ(θ),G′(ε,zθ)=o(ε3/2).

Proof.

By changing of variable, we know
(61)G1(ε,zθ)=-1p∫(a(xε)-A)Up(x+θ)=-ε3p∫(a(x)-A)Up(εx+θ).
Since a(x) is continuous and bounded, the dominated convergence theorem implies that
(62)limε→0G1(ε,zθ)ε3=Γ(θ).
On the other hand, since zθ is bounded and b(x)∈L2(ℝ3), then, changing of variable, we know
(63)|G2(ε,zθ)ε3|=|ε2(4-p)/(p-2)-34×∫b(xε)E1(b(xε)|zθ|2)|zθ|2ε2(4-p)/(p-2)-34|≤|ε|2(4-p)/(p-2)-3(∫|b(xε)|2|zθ|4)≤C1|ε|2(4-p)/(p-2)-3∫|b(xε)|2=C1|ε|2(4-p)/(p-2)∫|b(x)|2→0,asε→0,
since 4>p>2. Thus we obtain
(64)limε→0G(ε,zθ)ε3=Γ(θ).

Now we are ready to prove
(65)G′(ε,zθ)=o(ε3/2).
From the proof of [7], we first know
(66)G1′(ε,zθ)=o(ε3/2).
Also, since z is bounded, it is easy to check that
(67)∥G2′(ε,zθ)∥|ε|3/2=|ε|-3/2sup∥v∥=1|(G2′(ε,zθ),v)|=|ε|2(4-p)/(p-2)-3/2×sup∥v∥=1|∫b(xε)E1(b(xε)|zθ|2)zθv|≤C0|ε|2(4-p)/(p-2)-3/2×sup∥v∥=1{(∫|b(xε)|2|zθ|4)1/2×(∫|b(xε)|2|zθ|2|v|2)1/2}≤C1|ε|2(4-p)/(p-2)-3/2×sup∥v∥=1|(∫|b(xε)|2)1/2(∫|b(xε)v|2)1/2|.
Moreover, recall that b(x)∈L2 is bounded and use Hölder inequality, we get
(68)∥G2′(ε,zθ)∥|ε|3/2≤C1|ε|2(4-p)/(p-2),
since 4>p, we have
(69)limε→0G2′(ε,zθ)ε3/2=0.
From the above arguments, we know
(70)limε→0G′(ε,zθ)ε3/2=0,
and the proof is completed.

Proof of Theorem <xref ref-type="statement" rid="thm1.2">2</xref>.

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 index-0 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.

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
(71)Γ(θ)=-Lp+1∫|x|-γUp+1(x+θ).
Then for all θ∈R3, we have
(72)limε→0G(ε,zθ)εγ=Γ(θ),lim|θ|→∞Γ(θ)=0,limε→0G′(ε,zθ)εγ/2=0.

Proof.

As we know
(73)G1(ε,zθ)=-εγp+1∫(a(xε)-A)|x|γεγUp+1(x+θ)|x|γ.
By assumption (a2) and the decay property of U,
(74)limε→0G1(ε,zθ)εγ=Γ(θ).
Moreover, by the boundedness of zθ, we know
(75)|G2(ε,zθ)|=ε2(4-p)/(p-2)4×∫b(x)E1(b(x)|zθ|2)|zθ|2≤ε2(4-p)/(p-2)4(∫|b(xε)|zθ|2|2)≤C0|ε|2(4-p)/(p-2)+3.
Since 3>γ we obtain
(76)limε→0G(ε,zθ)εγ=Γ(θ).

To study the property of G′(ε,zθ), since γ<3 and U exponentially decays at infinity, from the proof in [7], we know
(77)limε→0G1′(ε,zθ)εγ/2=0.
On the other hand, from the boundedness of Zθ and b(x), we have
(78)∥G2′(ε,zθ)∥=|ε|2(4-p)/(p-2)×sup∥v∥=1|∫b(xε)E1(b(xε)|zθ|2)zθv|≤C0|ε|2(4-p)/(p-2)×sup∥v∥=1{(∫|b(xε)|2|zθ|4)1/2×(∫|b(xε)|2|zθ|2|v|2)1/2}≤C1|ε|2(4-p)/(p-2)×sup∥v∥=1|(∫|b(xε)|2)1/2(∫|v|2)1/2|≤C2|ε|2(4-p)/(p-2)+3/2.
Since γ<3, we get
(79)limε→0G2′(ε,zθ)εγ/2=0.
From the above arguments, we know
(80)limε→0G′(ε,zθ)εγ/2=0.

Proof of Theorem <xref ref-type="statement" rid="thm1.3">3</xref>.

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
(81)min|θ|=RΓ(θ)>Γ(0)ormax|θ|=RΓ(θ)<Γ(0).
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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