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J. Differential Equations 256 (2014) 3463–3495
www.elsevier.com/locate/jde
Entire nonradial solutions for non-cooperative coupledelliptic system with critical exponents in R
3
Yuxia Guo a,∗, Bo Li a, Juncheng Wei b,c
a Department of Mathematics, Tsinghua University, Beijing, 100084, PR Chinab Department of Mathematics, University of British Columbia, Vancouver, BC V6T 1Z2, Canada
c Department of Mathematics, The Chinese University of Hong Kong, Shatin, Hong Kong
Received 15 January 2014
Available online 3 March 2014
Abstract
We consider the following coupled elliptic system:⎧⎪⎪⎨⎪⎪⎩−�u = μ1u
N+2N−2 + βu
2N−2 v
NN−2 in R
N,
−�v = μ2vN+2N−2 + βv
2N−2 u
NN−2 in R
N,
u, v > 0, u, v ∈D1,2(RN),
(SN )
where N � 3, μ1, μ2 are two positive constants and β < 0 is the coupling constant. When N = 3, we provethe existence of infinitely many positive nonradial solutions.© 2014 Elsevier Inc. All rights reserved.
MSC: 35B45; 25J25
Keywords: Non-cooperative coupled systems; Critical exponents; Infinitely many non-radial solutions
* Corresponding author.E-mail address: [email protected] (Y. Guo).
http://dx.doi.org/10.1016/j.jde.2014.02.0070022-0396/© 2014 Elsevier Inc. All rights reserved.
3464 Y. Guo et al. / J. Differential Equations 256 (2014) 3463–3495
1. Introduction
We consider the following coupled elliptic system⎧⎪⎨⎪⎩�u + μ1u
p−1 + βup2 −1v
p2 = 0 in R
N,
�v + μ2vp−1 + βv
p2 −1u
p2 = 0 in R
N,
u > 0, v > 0 in RN,
(PN )
where p = 2NN−2 , (N � 3) is the Sobolev critical exponent, μ1, μ2 are two positive constants, and
β �= 0 is the coupling number. Physically, if β > 0, it means the attractive interaction of the statesu and v. On the other hand, if β < 0, it means the repulsive interaction. More precisely, system(PN ) is closely related to the solitary wave solutions of the time-dependent m coupled nonlinearSchrödinger equations:⎧⎪⎪⎪⎨⎪⎪⎪⎩
−√−1∂
∂tΦj = �Φj +
∑i �=j
βij |Φi | p2 −1Φ
p2j , y ∈R
N, t > 0,
Φj = Φj(y, t) ∈ C, j = 1, . . . ,m,
Φj (y, t) → 0, as |y| → ∞, t > 0,
(1.1)
where μj = βjj > 0’s are positive constants, βij ’s are coupling constants and the exponentp > 2. The above system arises in many physical problems, especially in nonlinear optics (whenp = 4). Physically, Φj denotes the j th component of the beam in Kerr-like photo refractive me-dia; The positive constant μj is for self-focusing in the j th component of the beam; The couplingconstant βij is the interaction between the j th and ith component of the beam. As βij > 0, theinteraction is attractive, and the interaction is repulsive if βij < 0 (see [1] and references therein).In particular, when the spatial dimension is one, the system (1.1) is integrable and there are manyanalytical and numerical results on solitary wave solutions of the general m coupled nonlinearSchrödinger equations (see [6,9,10,12]).
To obtain the solitary wave solutions of system (1.1), one sets Φj(y, t) = e√−1λj tuj (y) and
transforms the system (1.1) to steady state m coupled nonlinear Schrödinger equations:⎧⎪⎪⎪⎨⎪⎪⎪⎩�uj − λjuj + μju
p−1j +
∑i �=j
βij up2i u
p2 −1j = 0, y ∈R
N,
uj (y) > 0, y ∈RN, j = 1, . . . ,m,
uj (y) → 0, as |y| → ∞.
(1.2)
Note that system (1.2) has a gradient structure with respect to the energy functional
E[u1, . . . , um] = 1
2
∫RN
m∑i=1
[|∇ui |2 + λiu2i
]− 1
p
∫RN
∑i,j
βij |ui | p2 |uj | p
2 . (1.3)
In the case of subcritical, i.e. p < 2NN−2 , the existence of ground state solutions for (1.2) may
depend on coupling constants βij ’s. More precisely, when all βij ’s are positive and the matrix∑ = (|βij |) with βjj = μj is positively definite, there exists a ground state solution which is
Y. Guo et al. / J. Differential Equations 256 (2014) 3463–3495 3465
radially symmetric. However, if all βij ’s are negative, or one of βij ’s is negative and the matrix∑= (|βij |) is positively definite, then there is no ground state solutions (see [13] and [2]). Formore related results on the coupled nonlinear Schrödinger equations, we refer the reader to [3–5,14,21] and references therein.
To study the a priori estimates of solutions to (1.2), we have to study the existence and non-existence of the limiting elliptic system (PN ). In the case of p < 2N
N−2 and β12 > −√μ1μ2,
it has been proved that problem (PN ) has no classical solutions [18]. On the other hand, ifβ12 < −√
μ1μ2, non-trivial solutions exist [22].The purpose of the present paper is to study the critical case. So from now on we assume that
N � 3 and p = 2NN−2 . Observe that when β = 0, problem (SN ) decouples to the following (up to
multiplication) well-known Yamabe problem
−�u = uN+2N−2 , u > 0 in R
N. (1.4)
It is well known that all solutions to Yamabe problem (1.4) can be classified
Uε,x0(y) = (N(N − 2)
)N−24
(ε
ε2 + |y − x0|2)N−2
2
.
On the other hand, it is known that when the coupling constant β is positive (the cooperativecase), the only positive solutions to the system (SN ) are radially symmetric around some point
with the form (u, v) = (c1U,c2U), where U(y) = (N(N − 2))N−2
4 ( 11+|y−a|2 )
N−22 is the solution
of Eq. (1.4) and c1, c2 are some positive constants (see [8]). In this paper we consider the caseof non-cooperative, i.e. β < 0. We establish the following result, which seems to exhibit a newphenomena: ∀ fixed μ1,μ2 > 0, β < 0, problem (SN ) admits infinitely many positive nonradialfinite energy solutions, whose energy can be arbitrarily large.
To explain the main ideas of the proof, we have to go back to Eq. (1.4). By remarks before,positive solutions to (1.4) are well classified. It is natural to ask whether or not there are finiteenergy non-radial sign changing solutions to (1.4). This was answered first by Ding [7]. His proofis variational: consider the functions of the form
u(x) = u(|x1|, |x2|
), x = (x1, x2) ∈ SN ⊂R
N+1 =Rk ×R
N−k, k � 2, (1.5)
the critical Sobolev embedding becomes compact and hence infinitely many sign changing so-lutions exist, thanks to the Ljusternik–Schnirelmann theory. See also [11]. Recently, del Pino,Musso, Pacard and Pistoia [15,16] gave another proof of countably many sign changing non-radial solutions. Their proof is more constructive: they built a sequence of solutions with onenegative bump at the origin and large number of positive bumps in a polygon. This gives moreprecise information on such sign changing solutions.
It seems very difficult to apply variational method to obtain non-radial positive solutions to(SN ). So we turn to perturbative method as in [15,16]. First we observe that problem (SN ) is in-variant under rotation, reflection and Kelvin’s transformation. As in [15,16], we build a sequenceof positive solutions with one positive bubble for u at the origin and large number of positivebubbles for v around a polygon. Since our system is coupled to each other, in order to obtain abetter control of the error terms, it is difficult to carry the reduction procedure by using the same
3466 Y. Guo et al. / J. Differential Equations 256 (2014) 3463–3495
norm in [15,16] (see also [23,17,19,20]). We have to modify the norms. Moreover because of thecoupling, the estimates in the reduction procedure is much more complicated than in [15,16].
In this paper, we mainly focus on the problem of dimension N = 3 and prove the infinitelymany positive solutions to (P3) by linearization and perturbation. The existence of solutions isdue to the superlinear behavior of coupling terms −u2v3 and −u3v2 with respect to both u andv components, where contraction holds due to this property.
In sharp contrast with dimension N = 3, the problem in the case of dimension N � 4 is moredifficult. When N = 4, the coupling terms −uv2 and −u2v are linear with u and v respectively.In this case, the linearization method is available while the subsequent nonlinear perturbationmethod fails since the linear structure of coupling terms has become an obstacle to the contractionproperty that we need in the fixed point theory. The case of dimension N � 5 remains an openproblem where the perturbation method here cannot be used either. The nonlinear data terms inthe linearized version become singular when the coupling terms −u
p2 −1v
p2 and −u
p2 v
p2 −1 turn
out to be sublinear with u and v accordingly.Our main result can be presented as follows:
Theorem 1.1. Let N = 3. There exists some sufficiently large k0 ∈ N, such that for any k � k0,the system (P3) has a finite energy solution (uk, vk) of the following form:
{uk(y) = u∗(y) + ψk(y), y ∈R
3,
vk(y) = v∗(y) + ϕk(y), y ∈R3,
where εk ∼ k−15 ln−2 k;
(u∗, v∗
) :=(
μ− 1
41 U1,0,μ
− 14
2
k∑j=1
Uεk,xj
),
xj =(√
1 − ε2k cos
(2(j − 1)π
k
),
√1 − ε2
k sin
(2(j − 1)π
k
),0)
∈R2 ×R,
for j = 1,2, . . . , k, and ‖ψk‖∗ → 0, ‖ϕk‖∗ → 0 as k → ∞, where
‖ξ‖∗ := supy∈R3
{(1 + |y|) 2
q′ · ∣∣ξ(y)∣∣}, 3
2< q <
90
59,
1
q+ 1
q ′ = 1.
2. Approximation and linearization
In this section, we deal with the case when N = 3. Note that in our proofs the values of μ andβ are not essential. Only the sign of them matters. So without loss of generality, we may assumeμ1 = μ2 = 1 and β = −1. Namely we consider the following elliptic system:
⎧⎪⎨⎪⎩−�u = u5 − u2v3 in R
3,
−�v = v5 − v2u3 in R3,
u, v > 0 in R3, u, v ∈D1,2
(R
3).
(2.1)
Y. Guo et al. / J. Differential Equations 256 (2014) 3463–3495 3467
An important observation is the following invariance maps: Let Ti be one of the followingthree maps, i = 1,2,3. Then if (u, v) is a solution of (2.1), (Ti(u), Ti(v)) is also a solution to(2.1). Here maps Ti are given by
(Rotation Invariance Map): for η : R3 →R, (y, y′) ∈ R2 ×R,
T1(η)(y, y′)= η
(e
2jπk
√−1y, y′), j = 1,2,3, . . . , k − 1. (C1)
(Reflection Invariance Map): for η : R3 →R, (y1, y2, y3) ∈R×R×R,
T2(η)(y1, y2, y3) = η(y1,−y2, y3) = η(y1, y2,−y3). (C2)
(Kelvin Invariance Map): for η :R3 → R, y ∈R3,
T3(η)(y) = |y|−5η
(y
|y|2)
. (C3)
Because of the three invariances, we can define a symmetry class as follows
Hs := {(g1, g2) ∈ [D1,2(
R3)]2 ∣∣ Tj (gi) = gi, j = 1,2,3, i = 1,2
}.
As in [15,16], the following approximation solution
(u∗, v∗) =(
U1,0,
k∑j=1
Uεk,xj
)
where
xj =(√
1 − ε2k cos
(2(j − 1)π
k
),
√1 − ε2
k sin
(2(j − 1)π
k
),0)
∈ R2 ×R,
for j = 1,2, . . . , k, belongs to the symmetry class Hs .Let
D(R
3)= {(u, v) ∈ L6(
R3)× L6(
R3) ∣∣∇u ∈ L2(
R3), ∇v ∈ L2(
R3)},
with norm
∥∥(ξ, η)∥∥D := 〈ξ, ξ 〉
12D + 〈η,η〉
12D,
where 〈·,·〉D is defined by:
〈ξ1, ξ2〉D :=∫
3
∇ξ1(y) · ∇ξ2(y) dy, ∀ξ1, ξ2 ∈ D1,2(R
3).
R
3468 Y. Guo et al. / J. Differential Equations 256 (2014) 3463–3495
Linearizing Eqs. (2.1) around (u∗, v∗), we obtain the following two linear operators L0, L00
as follows {L0(ψ,φ) := �ψ + 5u4∗ψ, ∀(ψ,φ) ∈D1,2
(R
3)×D1,2
(R
3),
L00(ψ,φ) := �φ + 5v4∗φ, ∀(ψ,φ) ∈D1,2(R
3)×D1,2
(R
3).
(L)
We rewrite the elliptic system (2.1) in terms of this linear operator L = (L0,L00) as:
{L0(ψ,φ) = [
N5,0(ψ,φ) + N2,3(ψ,φ)]
in R3,
L00(ψ,φ) = [E + N0,5(ψ,φ) + N3,2(ψ,φ)
]in R
3,(LS)
where
E = −v5∗ +k∑
i=1
u5ε,xi
;
N0,5(ψ,φ) = −(v∗ + φ)5 + v5∗ + 5v4∗φ;N3,2(ψ,φ) = (u∗ + ψ)3(v∗ + φ)2;
N5,0(ψ,φ) = −(u∗ + ψ)5 + u5∗ + 5u4∗ψ;N2,3(ψ,φ) = (u∗ + ψ)2(v∗ + φ)3.
In the following, we will focus on the solution to the following problem:
L(ψ,φ) = (h1, h2), (LN )
where (ψ,φ) ∈ Hs .To deal with the particular coupled system (2.1), we need to introduce the following weighted
L∞ norm and the weighted Lq norm:
‖φ‖∗ := supy∈R3
{(1 + |y|) 2
q′ ∣∣φ(y)∣∣}, ∥∥(ψ,φ)
∥∥∗ := ∥∥|ψ | + |φ|∥∥∗;
‖h‖∗∗ := ∥∥(1 + |y|)5− 6q h(y)
∥∥Lq(R3)
,∥∥(h1, h2)
∥∥∗∗ := ∥∥|h1| + |h2|∥∥∗∗,
where 32 < q < 90
59 , and 1q
+ 1q ′ = 1.
In the next section, we need to study the fully nonlinear problem.First, it is well known that (see [15,16,23]) the set of bounded solutions of the decoupled
homogeneous system,
�ψ + 5u4∗ψ = 0, �φ + 5v4∗φ = 0 (2.2)
Y. Guo et al. / J. Differential Equations 256 (2014) 3463–3495 3469
is spanned by 8 functions (Zl,∑k
j=1 Zjs), where l = 1,2,3,4; s = 1,2 and
Zl = ∂ylU1,0(y), l = 1,2,3, y = (y1, y2, y3) ∈ R
3,
Z4(y) = y · ∇U1,0(y) + 1
2U1,0(y), y ∈ R
3;Zj,1(y) = ∂rUεk,xj
, Zj,2(y) = ∂εkUεk,xj
,
with r = |xj | =√
1 − ε2k , j = 1,2, . . . , k.
The following proposition solves (LN ) in the general orthogonal conditions such that,
L(ψ,ϕ) = h +(
4∑l=1
dlZl,
2∑s=1
cs
k∑j=1
U4εk,xj
Zjs
),
where h is orthogonal to the 8 dimensional linear space spanned by Zl ,∑k
j=1 Zjs , l = 1,2,3,4;s = 1,2.
Proposition 2.1. Let h = (h1, h2) be a vector function in (D1,2(R3))2, such that ‖h‖∗∗ < ∞,and satisfies the following orthogonal condition (C0):⎧⎪⎪⎪⎪⎪⎨⎪⎪⎪⎪⎪⎩
∫R3
Zl(y)h1(y) dy = 0, l = 1,2,3,4,
k∑j=1
∫R3
Zjs(y)h2(y) dy = 0; s = 1,2,
(C0)
then the linear problem (LN ) has the unique solution (ψ,φ) = L−1[(h1, h2)] such that‖(ψ,φ)‖∗ < ∞, and⎧⎪⎪⎪⎪⎪⎨⎪⎪⎪⎪⎪⎩
∫R3
U41,0(y)Zl(y)ψ(y)dy = 0, l = 1,2,3,4;
k∑j=1
∫R3
U4εk,xj
(y)Zjs(y)φ(y) dy = 0, s = 1,2.
(PHSI)
Proof. For fixed k, let us consider the subspaces
F = {(h1, h2) ∈ (L 6
5(R
3))2 ∣∣ (h1, h2) satisfies (C0)},
H = {(ψ,φ) ∈ (D1,2(
R3))2 ∣∣ (ψ,φ) satisfies (PHSI)
}.
Then H is a Hilbert space under the induced inner product
⟨(ψ1, φ1), (ψ2, φ2)
⟩H
:=∫
3
∇ψ1(y) · ∇ψ2(y) dy +∫
3
∇φ1(y) · ∇φ2(y) dy.
R R
3470 Y. Guo et al. / J. Differential Equations 256 (2014) 3463–3495
The norm ‖ · ‖H on this space is defined by
∥∥(ψ,φ)∥∥
H:= ⟨
(ψ,φ), (ψ,φ)⟩ 1
2H .
By the Hölder inequality, we have
‖hi‖L
65 (R3)
� C∥∥(1 + |y|)−5+ 6
q∥∥
Lr(R3)
∥∥(1 + |y|)5− 6q hi(y)
∥∥Lq(R3)
� C
[ ∫R3
(1 + |y|)−6 dy
] 1r ‖hi‖∗∗,
where 1q
+ 1r
= 56 , i = 1,2.
The following discussion is focused on the existence of the solution (ψ,ϕ) ∈ H , namely thesolution satisfying the weak form of the system (P ) in the space H , that is, for any testing pair(ξ1, ξ2) ∈ H , it holds that:⎧⎪⎪⎪⎪⎪⎨⎪⎪⎪⎪⎪⎩
∫R3
∇ψ · ∇ξ1 −∫R3
5u4∗ψ · ξ1 +∫R3
h1 · ξ1 = 0,
∫R3
∇φ · ∇ξ2 −∫R3
5v4∗ · ξ2 +∫R3
h2 · ξ2 = 0.
(2.3)
Note that for any (h1, h2) ∈ F , by the uniqueness of Riesz’s theorem, we can define an in-jective, linear and bounded operator A = (A1,A2) : F → H , such that (ψ,φ) = A(h1, h2) =(A1(h1),A2(h2)) and⟨
A1(h1), ξ1⟩D = −(h1, ξ1)L2; ⟨
A2(h2), ξ2⟩D = −(h2, ξ2)L2 .
For convenience, we also define an operator τ = (τ1, τ2) : H → L65 (R3) × L
65 (R3) such that,
τ1(ψ,φ) = −5u4∗ψ; τ2(ψ,φ) = −5v4∗φ.
Then the operator τ is compact due to the fact that uν1∗ (y) · vν2∗ (y) ∼ 1
(1+|y|)4 , for 0 � ν1, ν2 � 4,ν1 + ν2 = 4, and |y| � 2.
Now we define an operator B = A ◦ π ◦ τ : H → H , where π(h) = (π1(h),π2(h)) denotesthe projection operator
π(h) = h −(
4∑l=1
dlZl,
2∑s=1
cs
k∑j=1
U4εk,xj
Zjs
)
and the coefficients dl , cs are such that π(h) satisfies (C0). Then B is also a compact operatorand the system (2.3) is simplified to the following form:
(I − B)(ψ,φ) = A(h1, h2).
Y. Guo et al. / J. Differential Equations 256 (2014) 3463–3495 3471
A direct computation shows that
⟨(ψ,φ),B(ψ, φ)
⟩H
= ⟨(ψ,φ),
(A1 ◦ π1 ◦ τ1(ψ, φ),A2 ◦ π2 ◦ τ2(ψ, φ)
)⟩H
= ⟨ψ,A1 ◦ π1 ◦ τ1(ψ, φ)
⟩D + ⟨
φ,A2 ◦ π2 ◦ τ2(ψ, φ)⟩D
= −(ψ,τ1(ψ, φ))L2 − (
φ, τ2(ψ, φ))L2
=∫R3
ψ · 5u4∗ψ +∫R3
φ · 5v4∗φ
=∫R3
5u4∗ψ · ψ +∫R3
5v4∗φ · φ
= −(τ1(ψ,φ), ψ)L2 − (
τ2(ψ,φ), φ)L2
= ⟨A1 ◦ π1 ◦ τ1(ψ,φ), ψ
⟩D + ⟨
A2 ◦ π2 ◦ τ2(ψ,φ), φ⟩D
= ⟨(A1 ◦ π1 ◦ τ1(ψ,φ),A2 ◦ π2 ◦ τ2(ψ,φ)
), (ψ, φ)
⟩H
= ⟨B(ψ,φ), (ψ, φ)
⟩H
,
which shows that B is also self-adjoint.So the linear problem (LN ) is equivalent to the following equation with self adjoint operators
A, B such that
(I − B)(ψ,φ) = A(h1, h2).
By the injectivity of the operator A, for ∀(v1, v2) ∈ Ker(I − B),
(I − B)(v1, v2) = (0,0) = A(0,0),
which means that (v1, v2) is actually the solution to the decoupled homogeneous system (2.2).Therefore, there exist some constants ap , bt , p = 1,2,3,4; t = 1,2, such that
⎧⎪⎪⎪⎪⎪⎨⎪⎪⎪⎪⎪⎩v1(y) =
4∑p=1
ap · Zp(y), y ∈ R3;
v2(y) =2∑
t=1
bt
k∑i=1
Zit (y), y ∈ R3.
We put this formula into the restriction system (PHSI), then it yields that ap = 0, bt = 0, and(v1, v2) = (0,0), hence Ker(I − B) = {(0,0)}, so that
R(I − B) = Ker⊥(I − B∗)= Ker⊥(I − B) = H.
Hence we get the unique existence of the solution for the system (LN ). �
3472 Y. Guo et al. / J. Differential Equations 256 (2014) 3463–3495
Proposition 2.2. Under the same assumption of Proposition 2.1, there exists a large k0 ∈ N,and a constant C independent of k, such that for any k � k0, the solution (ψk,φk) to the linearproblem (LN ) is equivalent to the equation (ψk,φk) = L−1[(h1k, h2k)], with∥∥L−1[(h1k, h2k)
]∥∥∗ = ∥∥(ψk,φk)∥∥∗ � C‖hk‖∗∗,
which shows that L is invertible and L−1 is a bounded linear operator.
Proof. We prove the results by contradiction. Suppose that the conclusion does not hold true,then we can find a series of ((ψk,φk))k∈N and ((h1k, h2k))k∈N satisfying∥∥(ψk,φk)
∥∥∗ = ∥∥|ψk| + |φk|∥∥∗ ≡ 1; ‖hk‖∗∗ = ∥∥|h1k| + |h2k|
∥∥∗∗ → 0, as k → ∞.
We rewrite the linear problem (LN ) into the following form:{−�ψk = 5u4∗ψk − h1k,
−�φk = 5v4∗φk − h2k.(2.4)
Multiply the 3-dimensional kernels Γy(z) = |z−y|−1 on both sides of the system (2.4) above,and integrate them on the whole space R
3, we obtain
C3ψk(y) =∫R3
5u4∗(z)ψk(z) − h1k(z)
|z − y| dz, (2.5)
C3φk(y) =∫R3
5v4∗(z)φk(z) − h1k(z)
|z − y| dz. (2.6)
Using Hölder’s inequality and the formula (2.5), we estimate ψk(y) pointwise that
∣∣ψk(y)∣∣� C
[ ∫R3
u4∗(z) dz
|z − y|(1 + |z|) 2q′
· ‖ψk‖∗ +∫R3
h1k(z)(1 + |z|)5− 6q dz
|z − y|(1 + |z|)5− 6q
]
� C
[ ∫R3
dz
|z − y|(1 + |z|)4+ 2q′
‖ψk‖∗ +( ∫R3
dz
|z − y|q ′(1 + |z|)6−q ′
) 1q′
‖h1k‖∗∗]. (2.7)
By the standard argument as Appendix B in [23], it is direct to give the following estimates,∫R3
dz
|z − y|(1 + |z|)4+ 2q′
� C(1 + |y|)−1
, (2.8)
and ∫3
dz
|z − y|q ′(1 + |z|)6−q ′ � C
(1 + |y|)−2
. (2.9)
R
Y. Guo et al. / J. Differential Equations 256 (2014) 3463–3495 3473
For completeness, the proof of (2.8) and (2.9) will be provided in Lemma A and Lemma B ofAppendix A.
Put (2.8) and (2.9) in (2.7) and multiply both sides of the inequality by the term (1 + |y|) 2q′ ,
we get,
(1 + |y|) 2
q′ ∣∣ψk(y)∣∣� C
(1 + |y|) 2
q′ [(1 + |y|)−1‖ψk‖∗ + (1 + |y|)− 2
q′ ‖h1k‖∗∗]
� C[(
1 + |y|) 2q′ −1‖ψk‖∗ + ‖h1k‖∗∗
], ∀y ∈ R
3. (2.10)
Similarly, on the case of φk , we also make use of the Hölder inequality and estimate (2.9)such that,
(1 + |y|) 2
q′ ∣∣φk(y)∣∣� C
[(1 + |y|) 2
q′∫R3
v4∗(z) dz
|z − y|(1 + |z|) 2q′
‖φk‖ + ‖h2k‖∗∗]. (2.11)
The crucial point here in (2.11) is the estimate of integral∫R3
v4∗(z) dz
|z−y|(1+|z|)2q′
. We divide
this integral by two parts, one is the integral on the interior region INT := {x ∈ R3 | ∃j ∈
{1,2,3, . . . , k}, s.t. |x − xj | <ηk} where 0 < η � 1, while the other integral is on the exterior
region EXT := R3\INT .
I. Consider the interior region INT .For z ∈ INT , ∃j ∈ {1,2,3, . . . , k} such that z ∈ Bxj
(ηk), the small ball with radius η
kand
center at xj . By changing of variables, i.e. z �→ w := z−xj
εk, then w ∈ B0(
ηkεk
), the ball with largeradius η
kεkand center at the origin. Meanwhile, for any different indices i, j (i �= j), we have
|xj −xi |εk
∼ |j−i|kεk
which dominates ηkεk
, so
v∗(z) = v∗(εkw + xj ) � C
k∑i=1
ε12k
(ε2k + |εkw + xj − xi |2) 1
2
� C
(∑i �=j
ε− 1
2k
(1 + |w + xj −xi
εk|) + ε
− 12
k
(1 + |w|))
� Ckε− 1
2k
(1 + |w|)−1
, ∀w ∈ B0
(η
kεk
). (2.12)
II. Consider the exterior region EXT .For z ∈ EXT , we observe that |z − xj | � η
kfor all j = 1,2, . . . , k.
(II.-1). If z ∈ EXT ∩ B0(2), which is a bounded region in EXT , we have only two alternativechoices here.
3474 Y. Guo et al. / J. Differential Equations 256 (2014) 3463–3495
(II.-1.-i). ∃ some i0 ∈ {1,2,3, . . . , k}, s.t. z is closest to xi0 but relatively far from all theother xj ’s (j �= i0), then |z − xj | � 1
2 |xj − xi0 | ∼ j−i0k
, j �= i0. So we obtain that
v∗(z) � C
k∑i=1
ε12k
εk + |z − xi |
� C
(∑i �=i0
kε12k
|i − i0| + kε12k
η
)� Ck ln k · ε
12k � k ln k · ε
12k
(1 + |z|)−1
, (2.13)
since z ∈ EXT ∩ B0(2) ⊂ B0(2).(II.-1.-ii). If z is far from all xi ’s, i = 1,2,3, . . . , k, such that there exists some fixed con-
stant c0 > 0, |z − xj | � c0 for all j = 1,2,3, . . . , k, then for the same reason that z islocated in the bounded region, it yields,
v∗(z) � C
k∑j=1
ε12k
εk + |z − xj | � Ckε12k � Ckε
12k
(1 + |z|)−1
. (2.14)
(II.-2). If z ∈ EXT\B0(2), then |z − xj | ∼ (1 + |z|) for all j = 1,2, . . . , k. So we have,
v∗(z) � C
k∑j=1
ε12k
εk + |z − xj | � C
k∑j=1
ε12k
(1 + |z|)−1 � Ckε
12k
(1 + |z|)−1
. (2.15)
Conclude (II.-1) and (II.-2), we get
v∗(z) � Ck lnkε12k
(1 + |z|)−1
, z ∈ EXT. (2.16)
By the estimates (2.8), (2.12) and (2.16) above, we can compute that,
∫R3
v4∗(z) dz
|z − y|(1 + |z|) 3q′
=∫
INT
+∫
EXT
v4∗(z) dz
|z − y|(1 + |z|) 3q′
=k∑
j=1
∫Bxj
(ηk)
v4∗(z) dz
|z − y|(1 + |z|) 3q′
+∫
EXT
v4∗(z) dz
|z − y|(1 + |z|) 3q′
� Cε3k
k∑j=1
∫Bx (
η)
v4∗(εkw + xj ) dw
|εkw + xj − y| + Ck4 ln4 kε2k
∫EXT
dz
|z − y|(1 + |z|)4+ 2q′
j kεk
Y. Guo et al. / J. Differential Equations 256 (2014) 3463–3495 3475
� Ck4εk
∫B0(
ηkεk
)
k∑j=1
1
|εkw + xj − y|dw
(1 + |w|)4+ Ck4 ln4 kε2
k
∫R3
dz
|z − y|(1 + |z|)4+ 2q′
� Ck4εk
∫B0(
ηkεk
)
k∑j=1
1
|εkw + xj − y|dw
(1 + |w|)4+ Ck4 ln4 kε2
k
(1 + |y|)−1
. (2.17)
III. Consider the outer region R3\B0(2).
For y ∈ R3\B0(2) = EXT\B0(2), from (II.-2) we know that |y − xj | ∼ 1 + |y| for all j =
1,2,3, . . . , k. And for all w ∈ B0(η
kεk), |εkw| is dominated by 1 + |y|, hence |y − xj |, then
k∑j=1
1
|εkw + xj − y| � Ck(1 + |y|)−1
. (2.18)
Putting (2.17) and (2.18) back into the estimate (2.11) for φk , it gives
(1 + |y|) 2
q′ ∣∣φk(y)∣∣
� C
[(1 + |y|) 2
q′∫R3
v4∗(z) dz
|z − y|(1 + |z|) 3q′
· ‖φk‖∗ + ‖h2k‖∗∗]
� C
[(1 + |y|) 2
q′ −1(
k5εk
∫B0(
ηkεk
)
dw
(1 + |w|)4+ k4 ln4 kε2
k
)‖φk‖∗ + ‖h2k‖∗∗
]
� C[(
k5εk + k4 ln4 kε2k
)(1 + |y|) 2
q′ −1‖φk‖∗ + ‖h2k‖∗∗]
� C[(
1 + |y|) 2q′ −1‖φk‖∗ + ‖h2k‖∗∗
], ∀y ∈R
3\B0(2), since εk ∼ k−15 ln−2 k. (2.19)
Combining the estimate (2.10) for ψk and the estimate (2.19) for φk , we have
(1 + |y|) 2
q′ (∣∣ψk(y)∣∣+ ∣∣φk(y)
∣∣)� C
[(1 + |y|) 2
q′ −1(‖ψk‖∗ + ‖φk‖∗)+ (‖h1k‖∗∗ + ‖h2k‖∗∗
)], ∀y ∈R
3\B0(2). (2.20)
Notice that the assumption 32 < q < 90
59 implies that 2 < 9031 < q ′ < 3, therefore 2
q ′ − 1 < 0 and
(1 + |y|) 2
q′ −1 → 0 as |y| → +∞.
Taking θ = 12 , we can find some k0 > 0 large enough such that for all k � k0,
C(‖h1k‖∗∗ + ‖h2k‖∗∗
)<
θ
2,
where we assume ‖hk‖∗∗ = ‖|h1k| + |h2k|‖∗∗ → 0 in our hypothesis.
3476 Y. Guo et al. / J. Differential Equations 256 (2014) 3463–3495
Then we choose some R > 2 large enough such that for all y satisfying |y| > R, then it yields
C(1 + |y|) 2q′ −1
< θ4 .
Therefore, for any k � k0, we have
(1 + |y|) 2
q′ (∣∣ψk(y)∣∣+ ∣∣φk(y)
∣∣)� θ
4
(‖ψk‖∗ + ‖φk‖∗)+ θ
2� θ = 1
2, ∀y ∈R
3\B0(R),
and
1 ≡ ∥∥(ψk,φk)∥∥∗ = sup
y∈R3
{(1 + |y|) 2
q′ (∣∣ψk(y)∣∣+ ∣∣φk(y)
∣∣)}= sup
y∈R3
{(1 + |y|) 2
q′ (∣∣ψk(y)∣∣+ ∣∣φk(y)
∣∣)}= sup
y∈B0(R)
{(1 + |y|) 2
q′ (∣∣ψk(y)∣∣+ ∣∣φk(y)
∣∣)}� (1 + R)
2q′ ∥∥|ψk| + |φk|
∥∥L∞(B0(R))
. (2.21)
Hence, we can find η := (1 + R)− 2
q′ > 0, which is independent of k, such that
∥∥|ψk| + |φk|∥∥
L∞(B0(R))� (1 + R)
− 2q′ := η > 0. (2.22)
Letting k → ∞ in the (LN ) problem: L(ψk,φk) = (h1k, h2k), as ‖(h1k, h2k)‖∗∗ → 0 in ourhypothesis, and we can find a weak limit (ψ0, φ0) to the decoupled homogeneous system (2.2):
L(ψ0, φ0) = (0,0),∥∥|ψ0| + |φ0|
∥∥L∞(B0(R))
� η > 0. (2.23)
It is known that (ψ0, φ0) lies in Ker(L) while the condition (PHSI) restricted that (ψ0, φ0)
orthogonal to Ker(L), so (ψ0, φ0) = (0,0), which contradicts to the conclusion (2.23). �Proposition 2.3. Let h1k , h2k be such that
Tj (h1k) = h1k, Tj (h2k) = h2k, j = 1,2,3,
where Tj ’s are the three invariance maps defined at the beginning of this section, then there existsa bounded linear operator L−1 as that in Proposition 2.2, such that for any k � k0, the problem(LN ) admits a unique weak solution (ψk,φk) = L−1[(h1k, h2k)] such that ‖(ψk,φk)‖∗ < ∞,and ⎧⎪⎪⎪⎪⎪⎨⎪⎪⎪⎪⎪⎩
∫R3
U41,0(y)Zl(y)ψ(y)dy = 0, l = 1,2,3,4,
k∑j=1
∫R3
U4εk,xj
(y)Zjs(y)φ(y) dy = 0, s = 1,2.
(PHSI)
Y. Guo et al. / J. Differential Equations 256 (2014) 3463–3495 3477
Proof. In the proof of Proposition 2.1, it is sufficient to check the condition (H )⎧⎪⎪⎪⎪⎪⎨⎪⎪⎪⎪⎪⎩
∫R3
Zp(y)h1k(y) dy = 0, p = 1,2,3,4;
k∑j=1
∫R3
Zjs(y)h2k(y) dy = 0, s = 1,2.
(H )
By the oddness of Z3 and the condition (C2), it is easy to verify that∫R3
Z3(y)h1k(y) dy = 0.
For Z1, Z2, we consider the vector integral
I =∫R3
h1k(y)
[Z1(y)
Z2(y)
]dy = −
∫R3
h1k(y)
(1 + |y|2) 32
[y1y2
]dy,
then by the condition (C1), we calculate that,
e2πk
√−1I = −∫R3
h1k(y)
(1 + |y|2) 32
e2πk
√−1[
y1y2
]dy
= −∫R3
h1(e− 2π
k
√−1(z1, z2), z3)
(1 + |(e− 2πk
√−1(z1, z2), z3)|2) 32
[z1z2
]dz
= I,
which yields that I = [ 00
]. Hence we get that
∫R3
Z1(y)h1k(y) dy =∫R3
Z2(y)h1k(y) dy = 0.
For Z4, observe that Z4(y) = ∂λ|λ=1[λ 12 U1,0(λy)], we define the function I (λ) by
I (λ) = λ12
∫R3
U1,0(λy)h1k(y) dy.
By changing variables y �→ y
|y|2 and the condition (C3), we have that I (λ) = I ( 1λ). Thus
∂λI (λ)∣∣ = − 1 · ∂sI (s)
∣∣ 1 = −∂λI (λ)∣∣ ,
λ=1 λ2 s=λ,λ=1 λ=1
3478 Y. Guo et al. / J. Differential Equations 256 (2014) 3463–3495
and
∂λI (λ)∣∣λ=1 =
∫R3
Z4(y)h1k(y) dy = 0.
Hence ∫R3
Zi(y) · h1k(y) dy = 0, i = 1,2,3,4.
Similarly, ∫R3
Zi(y) · h2k(y) dy = 0, i = 1,2,3,4.
For the functions Zjs , we define the unit vectors as ej = (cos( 2(j−1)πk
), sin(2(j−1)π
k),0) ∈ R
3,j = 1,2, . . . , k. Then a direct computation shows that
∫R3
Zj,1(y)h2k(y) dy =∫R3
∂
∂r
(ε
ε2 + |y − xj |2) 1
2 · h2k(y) dy
= ε12 ·∫R3
y · ej − r
(ε2 + |y − xj |2) 32
· h2k(y) dy,
and
∫R3
Zj,2(y)h2k(y) dy =∫R3
∂
∂ε
(ε
ε2 + |y − xj |2) 1
2 · h2k(y) dy
= 1
2· ε− 1
2 ·∫R3
|y − xj |2 − ε2
(ε2 + |y − xj |2) 32
· h2k(y) dy
= 1
2· ε− 1
2 ·∫R3
(|y|2 − 1)h2k(y)
(ε2 + |y − xj |2) 32
dy − r
ε
∫R3
Zj,1(y) · h2k(y) dy.
It follows from the condition (C3) that
1
2· ε− 1
2 ·∫R3
(|y|2 − 1)h2k(y)
(ε2 + |y − xj |2) 32
dy = 1
2· ε− 1
2 ·∫R3
(| y
|y|2 |2 − 1)h2k(y
|y|2 )
(ε2 + | y
|y|2 − xj |2) 32
d
(y
|y|2)
= 1
2· ε− 1
2 ·∫
3
(1 − |y|2)h2k(y)
(ε2 + |y − xj |2) 32
dy
R
Y. Guo et al. / J. Differential Equations 256 (2014) 3463–3495 3479
= −1
2· ε− 1
2 ·∫R3
(|y|2 − 1)h2k(y)
(ε2 + |y − xj |2) 32
dy = 0.
Therefore ∫R3
Zj,2(y)h2k(y) dy = − r
ε
∫R3
Zj,1(y) · h2k(y) dy.
And it is sufficient to prove that∫R3
Zj,1(y) · h2k(y) dy = 0, j = 1,2, . . . , k.
Let Ij (t) = ∫R3 wε(y − t · xj ) · h2k(y) dy with wε(y) = ε− 1
2 U1,0(ε−1y), then
∂t |t=1Ij (t) =∫R3
∂t |t=1wε(y − t · xj ) · h2k(y) dy = r ·∫R3
Zj,1(y) · h2k(y) dy. (2.24)
By changing the variables y �→ y
|y|2 and the condition (C2), we obtain
Ij (t) =∫R3
wε
(y
|y|2 − t · xj
)· h2k
(y
|y|2)
d
(y
|y|2)
=∫R3
wε
(y
|y|2 − t · xj
)· |y|−1 · h2k(y) dy
=∫R3
(ε
ε2 + t2r2
) 12 ·(
1
|y − t ·xj
ε2+t2r2 |2 + ε2
(ε2+t2r2)2
) 12 · h2k(y) dy
=∫R3
wε(t)
(y − r(t) · xj
) · h2k(y) dy,
where ε(t) = ε
ε2+t2r2 , r(t) = t
ε2+r2t2 . Notice that ε(1) = ε; r(1) = 1, we have,
∂t |t=1Ij (t) =[ ∫R3
∂ε(t)wε(t)
(y − r(t)xj
)h2k(y) dy · ε′(t)
−∫R3
∂r(t)wε(t)
(y − r(t)xj
)h2k(y) dy · r ′(t)
]∣∣∣∣t=1
3480 Y. Guo et al. / J. Differential Equations 256 (2014) 3463–3495
= (−2εr2) ·∫R3
Zj,2(y)h2k(y) dy − (1 − 2r2) ·
∫R3
Zj,1(y)h2k(y) dy
= (2r3 + 2r2 − 1
) ·∫R3
Zj,1(y)h2k(y) dy. (2.25)
Compare the identity (2.24) and the identity (2.25), we obtain
r ·∫R3
Zj,1(y)h2k(y) dy = (2r3 + 2r2 − 1
) ·∫R3
Zj,1(y)h2k(y) dy, ∀0 < r < 1,
hence ∫R3
Zj,2(y)h2k(y) dy = − r
ε
∫R3
Zj,1(y) · h2k(y) dy = 0,
and the condition (H ) holds as desired. �3. Estimates of the data terms
In this section, we go back to the original system (P3). Note that the problem (P3) is closelyrelated to the problem (LN ), if we rewrite (P3) in the following form,{
L0(ψk,φk) = [N5,0(ψk,φk) + N2,3(ψk,φk)
] := h1k(ψk,φk) in R3,
L00(ψk,φk) = [E + N0,5(ψk,φk) + N3,2(ψk,φk)
] := h2k(ψk,φk) in R3,
(LS)
which share the same linear operator L in both of the (LN ) and (LS) systems.However, (LS) is essentially more difficult to deal with since the nonlinear data term
(h1k, h2k) in (LS) depends on the solution (ψk,φk) itself, where (h1k, h2k) is decomposed into(N5,0(ψk,φk) + N2,3(ψk,φk),E + N0,5(ψk,φk) + N3,2(ψk,φk)), where
N5,0(ψ,φ) = −(u∗ + ψ)5 + u5∗ + 5u4∗ψ;N2,3(ψ,φ) = (u∗ + ψ)2(v∗ + φ)3;
E = −v5∗ +k∑
i=1
u5εk,xi
;
N0,5(ψ,φ) = −(v∗ + φ)5 + v5∗ + 5v4∗φ;N3,2(ψ,φ) = (u∗ + ψ)3(v∗ + φ)2,
while the data term in (LN ) is independently given. We will present the estimates of those 5nonlinear data terms separately.
Y. Guo et al. / J. Differential Equations 256 (2014) 3463–3495 3481
Lemma 3.1.
‖E‖∗∗ � Ck592 − 45
q ln3− 3q k.
Proof. Since the term E can be written in a polynomial as the following:
E =(
k∑i=1
uεk,xi
)5
−k∑
i=1
u5εk,xi
=∑
i1+i2+···+ik=5i1,i2,...,ik∈Ni1,i2,...,ik �=5
k∏l=1
uilεk,xl
,
which is a sum of (k5 − k) terms. Without loss of generality, we only consider the termuεk,x1 · u4
εk,x2(i1 = 1, i2 = 4, i3 = i4 = · · · = ik = 0).
IV. Consider the interior region INT for E.From I we have known that for any y ∈ INT , ∃j ∈ {1,2,3, . . . , k}, s.t. y ∈ Bxj
(ηk) and by
changing of the variable: y �→ w = y−xj
εk, then w ∈ B0(
ηkεk
), |i−j |kεk
∼ |i−j |kεk
which dominates |w|and for different indices i �= j , so
uεk,x1(y) · u4εk,x2
(y) = uεk,x1(εkw + xj ) · u4εk,x2
(εkw + xj )
�Cε
− 52
k
(1 + |w + xj −x1k
|)(1 + |w + xj −x2k
|)4
�
⎧⎪⎪⎪⎪⎪⎪⎪⎪⎪⎨⎪⎪⎪⎪⎪⎪⎪⎪⎪⎩
Cε− 5
2k
(1+|w|)(1+|w+ x1−x2εk
|)4, j = 1,
Cε− 5
2k
(1+|w+ x2−x1εk
|)(1+|w|)4, j = 2,
Cε− 5
2k
(1+|w+ xj −x1εk
|)(1+|w+ xj −x2εk
|)4, j �= 1,2
� Ckε− 3
2k
(1 + |w|)−4
. (3.1)
Therefore, we can estimate the ‖ · ‖∗∗ norm restricted on INT region by using (3.1),
∥∥uεk,x1u4εk,x2
∥∥∗∗(INT)�
k∑j=1
∥∥uεk,x1u4εk,x2
∥∥∗∗(Bxj(
ηk))
� Cε3q
k∑j=1
( ∫B0(
ηkεk
)
uqεk,x1(εkw + xj )u
4qεk,x2(εkw + xj ) dw
) 1q
3482 Y. Guo et al. / J. Differential Equations 256 (2014) 3463–3495
� Ck2ε3q− 3
2k
( ∫B0(
ηkεk
)
dw
(1 + |x|)4q
) 1q
� Ck2ε3q− 3
2k . (3.2)
Since E contains k5 − k such terms as uεk,x1u4εk,x2
, then the interior estimate of E followsfrom the sum of those terms as
‖E‖∗∗(INT) � Ck7ε3q− 3
2k . (3.3)
V. Consider the exterior region EXT for E.Similar to the case of II, we get |y − x1| > η
kand |y − x2| > η
k, so
∣∣uεk,x1(y)u4εk,x2
(y)∣∣� C
ε12
εk + |y − x1| · ε2k
ε4k + |y − x2|4
� Cε
12k
η/k· ε−2
k
1 + | y−x2εk
|4 � Ckε− 3
2k
1
1 + | y−x2εk
|4 . (3.4)
Letting y �→ w := y−x2εk
, then
∥∥uεk,x1u4εk,x2
∥∥∗∗(EXT)� Ckε
− 32
k
[ ∫EXT
(1 + |y|)5q−6 dy
(1 + | y−x2εk
|4)q] 1
q
� Ckε3q− 3
2k
[ ∫R3
(1 + |εkw + x2|)5q−6 dw
1 + |w|4q
] 1q
� Ckε3q− 3
2k
[ ∫|w|� 2
εk
dw
1 + |w|4q+
∫|w|> 2
εk
ε5q−6k |w|5q−6 dw
1 + |w|4q
] 1q
� Ckε3q− 3
2k
[1 + ε
4q−3k
] 1q � Ckε
3q− 3
2k . (3.5)
Then by summing k5 − k such terms, we get the ‖ · ‖∗∗ estimate of E on EXT that
‖E‖∗∗(EXT) � Ck6ε3q− 3
2k .
In all, conclude the cases of IV and V, since εk ∼ k−15 ln−2 k and 32 < q < 90
59 , we have
‖E‖∗∗ � ‖E‖∗∗(EXT) + ‖E‖∗∗(INT) � Ck7ε3q− 3
2k � Ck
592 − 45
q ln3− 6q k → 0,
as k → ∞. � (3.6)
The following Lemma 3.2 is due to Lemma 3.5 of [24].
Y. Guo et al. / J. Differential Equations 256 (2014) 3463–3495 3483
Lemma 3.2. For any t ∈ R and q > 1, we have
∣∣[(1 + t)+]q − 1 − qt
∣∣�⎧⎪⎨⎪⎩
C min{|t |q, |t |2}, if 1 < q � 2, |t | > 12 ,
C|t |2, if 1 < q � 2, |t |� 12 ,
C(|t |2 + |t |q), if q > 2
(3.7)
where (1 + t)+ = max{1 + t,0}.
Lemma 3.3.∥∥N5,0(ψk,ϕk)∥∥∗∗ � C‖ψk‖2∗
(1 + ‖ψk‖3∗
); ∥∥N0,5(ψk,φk)∥∥∗∗ � C‖φk‖2∗
(k
532 − 45
q + ‖φk‖3∗).
Proof. Notice that (uk, vk) = (u∗ + ψk, v∗ + φk) > (0,0) in the problem (P3), so we can applyLemma 3.2 that ∀y ∈R
3,∣∣N5,0(ψk(y),φk(y)
)∣∣= ∣∣−(u∗(y) + ψk(y))5 + 5u5∗ + 5u4∗(y)ψk(y)
∣∣= u5∗(y)
∣∣∣∣(1 + ψk(y)
u∗(y)
)5
− 1 − 5ψk(y)
u∗(y)
∣∣∣∣� Cu5∗(y)
[∣∣∣∣ψk(y)
u∗(y)
∣∣∣∣2 +∣∣∣∣ψk(y)
u∗(y)
∣∣∣∣5]� C[∣∣ψk(y)
∣∣2u3∗(y) + ∣∣ψk(y)∣∣5]
� C(1 + |y|)−3− 4
q′ ‖ψk‖2∗ + C(1 + |y|)− 10
q′ ‖ψk‖5∗. (3.8)
Since 32 < q < 90
59 in our assumption, it implies that −2 − 2q < −5 and 4 − 5q < −3, so wehave
∥∥(1 + |y|)−3− 4q′ ∥∥∗∗ =
( ∫R3
(1 + |y|)−2q−2
dy
) 1q
< ∞, (3.9)
and
∥∥(1 + |y|)− 10q′ ∥∥∗∗ =
( ∫R3
(1 + |y|)4−5q
dy
) 1q
< ∞. (3.10)
From (3.8), (3.9) and (3.10), we obtain the weighted estimate for N5,0(ψk,φk) as the follow-ing,
∥∥N5,0(ψk,φk)∥∥∗∗ � C
[∥∥(1 + |y|)−3− 4q′ ∥∥∗∗ · ‖ψk‖2∗ + ∥∥(1 + |y|)− 10
q′ ∥∥∗∗ · ‖ψk‖5∗]
� C
[ ∫R3
(1 + |y|)−2−2q
dy
] 1q · ‖ψk‖2∗ + C
[ ∫R3
(1 + |y|)4−5q
dy
] 1q · ‖ψk‖5∗
� C‖ψk‖2(1 + ‖ψk‖3). (3.11)
∗ ∗
3484 Y. Guo et al. / J. Differential Equations 256 (2014) 3463–3495
Next for the term N0,5(ψk,φk), we can also make use of Lemma 3.2 that, for any y ∈ R3, we
have
∣∣N0,5(ψk(y),φk(y)
)∣∣= v4∗(y)
∣∣∣∣(1 + φk(y)
v∗(y)
)5
− 1 − 5φk(y)
v∗(y)
∣∣∣∣� Cv5∗(y)
[∣∣∣∣φk(y)
v∗(y)
∣∣∣∣2 +∣∣∣∣φk(y)
v∗(y)
∣∣∣∣5]� C[∣∣φk(y)
∣∣2v3∗(y) + ∣∣φk(y)∣∣5]
� C[(
1 + |y|)− 4q′ v3∗(y) · ∥∥φk(y)
∥∥2∗ + (
1 + |y|)− 10q′ ‖φk‖5∗
]. (3.12)
By the estimates (2.12) and (2.16) of v∗ in I and II, and the weighted estimates (3.10) alongwith (3.12) above, we compute that,∥∥N0,5(ψk,φk)
∥∥∗∗
� C[∥∥(1 + |y|)− 4
q′ v3∗(y)∥∥∗∗ · ‖φk‖2∗ + ∥∥(1 + |y|)− 10
q′ ∥∥∗∗ · ‖φk‖5∗]
� C[∥∥(1 + |y|)− 4
q′ v3∗(y)∥∥∗∗(INT)
+ ∥∥(1 + |y|)− 4q′ v3q∗ (y)
∥∥∗∗(EXT)
] · ‖φk‖2∗ + C‖φk‖5∗
� C
[ε
3q
k
k∑j=1
( ∫B0(
ηkεk
)
v3q∗ (εkw + xj ) dw
) 1q +
( ∫EXT
(1 + |y|)q−2
v3q∗ (y) dy
) 1q
]· ‖φk‖2∗
+ C‖φk‖5∗
� C
[k4ε
q3 − 3
2k
( ∫B0(
ηkεk
)
dw
(1 + |w|)3q
) 1q + k3 ln3 kε
32k
( ∫EXT
(1 + |y|)−2−2q
dy
) 1q]
· ‖φk‖2∗
+ C‖φk‖5∗
� C(k4ε
q3 − 3
2k + k3 ln3 kε
32k
) · ‖φk‖2∗ + C‖φk‖5∗
� Ck532 − 45
q ‖φk‖2∗ + C‖φk‖5∗ � C(k
532 − 45
q + ‖φk‖3∗)‖φk‖2∗. � (3.13)
Lemma 3.4. ∥∥N2,3(ψk,φk)∥∥∗∗ � C
(k
532 − 45
q + ‖φk‖3∗) · (1 + ‖ψk‖2∗
);∥∥N3,2(ψk,φk)∥∥∗∗ � C
(k
18− 45q + ‖φk‖2∗
) · (1 + ‖ψk‖3∗).
Proof. These two terms N2,3(ψk,φk) and N3,2(ψk,φk) are coupled terms in the system, we canrewrite that for any y ∈R3,
∣∣N2,3(ψk(y),φk(y)
)∣∣= ∣∣(u∗(y) + ψk(y))2 · (v∗(y) + φk(y)
)3∣∣� C
(u2(y) + ∣∣ψk(y)
∣∣2)(v3(y) + ∣∣φk(y)∣∣3)
∗ ∗
Y. Guo et al. / J. Differential Equations 256 (2014) 3463–3495 3485
� C[(
1 + |y|)−2 + (1 + |y|)− 4
q′ ‖ψk‖2∗] · [v3∗(y) + (
1 + |y|)− 6q′ ‖φk‖3∗
].
(3.14)
Notice that 32 < q < 90
59 and 9031 < q ′ < 3, so the weighted estimate of (3.14) above can be
simplified by (2.12), (2.16), (3.10) and (3.13) that
∥∥N2,3(ψk,φk)∥∥∗∗ �
∥∥(1 + |y|)− 4q′ v3∗(y) + (
1 + |y|)− 10q′ ∥∥φ‖3∗‖∗∗
(1 + ‖ψk‖2∗
)� C
(k
532 − 45
q + ‖φk‖3∗)(
1 + ‖ψk‖2∗). (3.15)
Next, for the other coupled term N3,2(ψk,φk), we can also proceed in the same manner that,for any y ∈R
3,
∣∣N3,2(ψk(y),φk(y)
)∣∣= ∣∣u∗(y) + ψk(y)∣∣3∣∣v∗(y) + φk(y)
∣∣2� C
(u3∗(y) + ∣∣ψk(y)
∣∣3)(v2∗(y) + ∣∣φk(y)∣∣2)
� C[(
1 + |y|)−3 + (1 + |y|)− 6
q′ ‖ψk‖3∗] · [v2∗(y) + (
1 + |y|)− 4q′ ‖φk‖2∗
]� C
[(1 + |y|)− 6
q′ v2∗(y) + (1 + |y|)− 10
q′ ‖φk‖2∗] · (1 + ‖ψk‖3∗
). (3.16)
For the same reason of the estimates (2.12) and (2.16), we obtain that
∥∥N3,2(ψk,φk)∥∥∗∗ � C
[∥∥(1 + |y|)− 6q′ v2∗(y)
∥∥∗∗ + ‖φk‖2∗](
1 + ‖ψk‖3∗)
� C
[k∑
j=1
∥∥(1 + |y|)− 6q′ v2∗(y)
∥∥∗∗(Bxj(
ηk))
+ ∥∥(1 + |y|)− 6q′ v2∗(y)
∥∥∗∗(EXT)
+ ‖φk‖2∗
](1 + ‖ψk‖3∗
)
� C
[ε
3q
k
k∑j=1
( ∫B0(
ηkεk
)
v2q∗ (εkw + xj ) dw
) 1q +
( ∫EXT
(1 + |y|)−q
v2q∗ (y) dy
) 1q
+ ‖φk‖2∗
](1 + ‖ψk‖3∗
)� C
[k3ε
3q−1
k
( ∫B0(
ηkεk
)
dw
(1 + |w|)2q
) 1q + k2(ln k)2εk
( ∫EXT
(1 + |y|)−3q
dy
) 1q
+ ‖φk‖2∗](
1 + ‖ψk‖3∗)
3486 Y. Guo et al. / J. Differential Equations 256 (2014) 3463–3495
� C[k3ε
3q−1
k + k2(lnk)2εk + ‖φk‖2∗](
1 + ‖ψk‖3∗)
� C(k
18− 45q + ‖φk‖2∗
)(1 + ‖ψk‖3∗
). � (3.17)
4. Proof of the Theorem 1.1
Recall that the solution (uk, vk) := (u∗ +ψk, v∗ +φk) to the problem (P3) is equivalent to thesolution (ψk,φk) to the linearized equation
L(ψk,φk) = (N5,0(ψk,φk) + N2,3(ψk,φk),E + N0,5(ψk.φk) + N3,2(ψk,φk)
). (4.1)
To solve this linearized problem, we construct two Banach spaces, X and Y ,
X := {(ψ,φ) ∈ (D1,2(
R3))2 ∣∣
(ψ,φ) satisfies (C1), (C2), (C3) symmetries and∥∥(ψ,φ)
∥∥∗ < ∞}; (4.2)
Y := {(h1, h2) ∈ (D1,2(
R3))2 ∣∣
(h1, h2) satisfies (C1), (C2), (C3) symmetries and∥∥(h1, h2)
∥∥∗∗ < ∞}. (4.3)
Since Proposition 2.2 implies that the linear operator L : X �→ Y is an invertible operator, wecan also define an operator M : X → X such that
M(ψ,φ) = L−1[(N5,0(ψ,φ) + N2,3(ψ,φ),E + N0,5(ψ,φ) + N3,2(ψ,φ))]
, (4.4)
then the solution (uk, vk) to the problem (P3) is further reduced to the fixed point problem(ψk,φk) under the mapping M, i.e. M(ψk,φk) = (ψk,φk). It is sufficient to show two propertiesVI and VII of the mapping M:
VI. M : Xρ → Xρ ; where Xρ := {(ψ,φ) ∈ X | ‖(ψ,φ)‖∗ � C(ρ + k592 − 45
q ln3− 6q k),
0 < ρ � 1}.For VI, we combine the results of Proposition 2.2, Lemma 3.1, Lemma 3.3 and Lemma 3.4,
we know that for any (ψ,φ) ∈ Xρ ,∥∥M(ψ,φ)∥∥∗ = ∥∥L−1[(N5,0(ψ,φ) + N2,3(ψ,φ),E + N0,5(ψ,φ) + N3,2(ψ,φ)
)]∥∥∗� C
∥∥(N5,0(ψ,φ) + N2,3(ψ,φ),E + N0,5(ψ,φ) + N3,2(ψ,φ))∥∥∗∗
� C[‖E‖∗∗ + ∥∥N5,0(ψ,φ)
∥∥∗∗ + ∥∥N0,5(ψ,φ)∥∥∗∗ + ∥∥N2,3(ψ,φ)
∥∥∗∗+ ∥∥N3,2(ψ,φ)
∥∥∗∗]
� C[k
592 − 45
q ln3− 6q k + ‖ψk‖2∗
(1 + ‖ψk‖3∗
)+ ‖φk‖2∗(k
532 − 45
q + ‖φk‖3∗)
+ (k
532 − 45
q + ‖φk‖3∗)(
1 + ‖ψk‖2∗)+ (
k18− 45
q + ‖φk‖2∗)(
1 + ‖ψk‖3∗)]
� C(ρ2 + k
592 − 45
q ln3− 6q k)� C
(ρ + k
592 − 45
q ln3− 6q k). (4.5)
Therefore, M(ψ,φ) ∈ Xρ and M : Xρ → Xρ .
Y. Guo et al. / J. Differential Equations 256 (2014) 3463–3495 3487
VII. M is a contraction mapping in Xρ .For VII, we choose any two different elements (ψ1, φ1), (ψ2, φ2) ∈ Xρ , then we make a
difference form of the operator M as,
∥∥M(ψ1, φ1) −M(ψ2, φ2)∥∥∗
= ∥∥L−1[(N5,0(ψ1, φ1) − N5,0(ψ2, φ2) + N2,3(ψ1, φ1) − N2,3(ψ2, φ2),
N0,5(ψ1, φ1) − N0,5(ψ2, φ2) + N3,2(ψ1, φ1) − N3,2(ψ2, φ2))]∥∥∗
� C[∥∥N5,0(ψ1, φ1) − N5,0(ψ2,φ2)
∥∥∗∗ + ∥∥N2,3(ψ1, φ1) − N2,3(ψ2, φ2)∥∥∗∗∥∥N0,5(ψ1, φ1) − N0,5(ψ2, φ2)
∥∥∗∗ + ∥∥N3,2(ψ1, φ1) − N3,2(ψ2, φ2)∥∥∗∗
]. (4.6)
For the rest of VII, we give the estimates of those weighted norms ‖N5,0(ψ1, φ1) −N5,0(ψ2, φ2)‖∗∗, ‖N2,3(ψ1, φ1) − N2,3(ψ2, φ2)‖∗∗, ‖N0,5(ψ1, φ1) − N0,5(ψ2, φ2)‖∗∗,‖N3,2(ψ1, φ1) − N3,2(ψ2, φ2)‖∗∗ to show the contraction property as k large enough and ρ
small enough.
(VII.-1). ‖N5,0(ψ1, φ1) − N5,0(ψ2, φ2)‖∗∗.
Since N5,0(ψ1, φ1) − N5,0(ψ2, φ2) = [(u∗ + ψ2)5 − (u∗ + ψ2)
5] − 5u4∗(ψ2 − ψ1), we makeuse of the Mean Value Theorem twice such that ∃θ0, τ0 ∈ [0,1], with
N5,0(ψ1, φ1) − N5,0(ψ2, φ2)
= 5(u∗ + θ0ψ1 + (1 − θ0)ψ2
)4(ψ2 − ψ1) − 5u4∗(ψ2 − ψ1)
= 5[(
u∗ + θ0ψ1 + (1 − θ0)ψ2)4 − u4∗
](ψ2 − ψ1)
= 20(u∗ + τ0θ0ψ1 + τ0(1 − θ0)ψ2
)3(θ0ψ1 + (1 − θ0)ψ2
)(ψ2 − ψ1). (4.7)
Therefore,
∥∥N5,0(ψ1, φ1) − N5,0(ψ2, φ2)∥∥∗∗
� C∥∥(u∗ + |ψ1| + |ψ2|
)3(|ψ1| + |ψ2|)|ψ2 − ψ1|
∥∥∗∗
� C∥∥[(1 + |y|)−3− 4
q′ + (1 + |y|)− 10
q′ (‖ψ1‖∗ + ‖ψ2‖∗)](‖ψ1‖∗ + ‖ψ2‖∗
)∥∥∗∗‖ψ2 − ψ1‖∗∗
� C(ρ + k
592 − 45
q)∥∥[(1 + |y|)−3− 4
q′ + (ρ + k
592 ln
3− 6q′ k
)(1 + |y|)− 10
q′ ]∥∥∗∗ · ‖ψ2 − ψ1‖∗
� C(ρ + k
592 − 45
q)∥∥(1 + |y|)− 10
q′ ∥∥∗∗‖ψ2 − ψ1‖∗
� C(ρ + k
592 − 45
q)‖ψ2 − ψ1‖∗. (4.8)
(VII.-2). ‖N2,3(ψ1, φ1) − N2,3(ψ2, φ2)‖.
3488 Y. Guo et al. / J. Differential Equations 256 (2014) 3463–3495
Notice that
N2,3(ψ1, φ1) − N2,3(ψ2, φ2)
= (u∗ + ψ1)2(v∗ + φ1)
3 − (u∗ + ψ2)2(v∗ + φ2)
3
= [(u∗ + ψ1)
2 − (u∗ + ψ2)2] · (v∗ + φ1)
3 + (u∗ + ψ2)2[(v∗ + φ1)
3 − (v∗ + φ2)3]. (4.9)
By using the Mean Value Theorem in two square brackets above, the for y ∈ R3, we can find
some θ0, τ0 ∈ [0,1], s.t.
N2,3(ψ1(y),φ1(y)
)− N2,3(ψ2(y),φ2(y)
)= 2
(u∗(y) + θ0ψ1(y) + (1 − θ0)ψ2(y)
) · (ψ1(y) − ψ2(y)) · (v∗(y) + φ1(y)
)3
+ 3(u∗(y) + ψ2(y)
)2(v∗(y) + τ0φ1(y)(1 − τ0)φ2(y)
)2(φ1(y) − φ2(y)
). (4.10)
Hence, by (2.12) in I and (2.16) in II, along with (4.10), the weighted norm of N2,3 can beestimated that,∥∥N2,3(ψ1, φ1) − N2,3(ψ2, φ2)
∥∥∗∗� C
∥∥(u∗ + |ψ1| + |ψ2|)|ψ1 − ψ2|
(v∗ + |φ1|
)3∥∥∗∗+ C
∥∥(u∗ + |ψ2|)2(
v∗ + |φ1| + |φ2|)2|φ1 − φ2|
∥∥∗∗
� C∥∥[(1 + |y|)−1 + (
1 + |y|)− 2q′ (‖ψ1‖∗ + ‖ψ2‖∗
)]· [(1 + |y|)− 2
q′ v3∗(y) + (1 + |y|)− 8
q′ ‖φ1‖3∗]∥∥∗∗ · ‖ψ1 − ψ2‖∗
+ C∥∥[(1 + |y|)−2 + (
1 + |y|)− 4q′ ‖ψ2‖2∗
]· [(1 + |y|)− 2
q′ v2∗(y) + (1 + |y|)− 6
q′ (‖φ1‖∗ + ‖φ2‖∗)2]∥∥ · ‖φ1 − φ2‖∗
� C(∥∥(1 + |y|)− 4
q′ v3∗(y)∥∥∗∗ + ∥∥(1 + |y|)− 10
q′ ∥∥∗∗ · ‖φ1‖3∗)‖ψ1 − ψ2‖∗
+ C(∥∥(1 + |y|)− 6
q′ v2∗(y)∥∥∗∗ + ∥∥(1 + |y|)− 10
q′ ∥∥∗∗ · (‖φ1‖∗ + ‖φ2‖∗)2)‖φ1 − φ2‖∗
� C[k
532 − 45
q + (ρ + k
592 − 45
q)3]‖ψ1 − ψ2‖∗ + C
[k
18− 45q + (
ρ + k592 − 45
q)2]‖φ1 − φ2‖∗
� C(ρ + k
592 − 45
q)(‖ψ1 − ψ2‖∗ + ‖φ1 − φ2‖∗
). (4.11)
(VII.-3). ‖N0,5(ψ1, φ1) − N0,5(ψ2, φ2)‖∗∗.
Similar to the case of (VII.-1), we can use the Mean Value Theorem twice so that ∃θ0, τ0 ∈[0,1], s.t.
N0,5(ψ1, φ1) − N0,5(ψ2, φ2)
= [(v∗ + φ2)
5 − (v∗ + φ1)5]− 5u4(φ2 − φ1)
∗
Y. Guo et al. / J. Differential Equations 256 (2014) 3463–3495 3489
= 5(v∗ + θ0φ1 + (1 − θ0)φ2
)4(φ2 − φ1) − 5u4∗(φ2 − φ1)
= 5[(
v∗ + θ0φ1 + (1 − θ0)φ2)4 − v4∗
] · (φ2 − φ1)
= 20(v∗ + τ0θ0φ1 + τ0(1 − θ0)φ2
)3(θ0φ1 + (1 − θ0)φ2
)(φ2 − φ1). (4.12)
From (2.12) and (2.16), we obtain that,∥∥N0,5(ψ1, φ1) − N0,5(ψ2, φ2)∥∥∗∗
� C∥∥(v∗ + |φ1| + |φ2|
)3 · (|φ1| + |φ2|)|φ1 − φ2|
∥∥∗∗� C
∥∥[v3∗ + (|φ1| + |φ2|)3] · (|φ1| + |φ2|
)|φ1 − φ2|∥∥∗∗
� C∥∥(1 + |y|)− 4
q′ v3∗(y) + (1 + |y|)− 10
q′ (‖φ1‖∗ + ‖φ2‖∗)3∥∥∗∗
(‖φ1‖∗ + ‖φ2‖∗)‖φ2 − φ1‖∗
� C(ρ + k
592 − 45
q)[∥∥(1 + |y|)− 4
q′ v3∗(y)∥∥∗∗ + ∥∥(1 + |y|)− 10
q′ ∥∥∗∗(ρ + k
592 − 45
q)3]‖φ2 − φ1‖∗
� C(ρ + k
592 − 45
q)[
k532 − 45
q + (ρ + k
592 − 45
q)3]‖φ2 − φ1‖∗
� C(ρ + k
592 − 45
q)2‖φ1 − φ2‖∗. (4.13)
(VII.-4). ‖N3,2(ψ1, φ1) − N3,2(ψ2, φ2)‖∗∗.
For the same reason as the case of (VII.-2), we can write that, by the Mean Value Theorem,∃θ0, τ0 ∈ [0,1], s.t.
N3,2(ψ1, φ1) − N3,2(ψ2, φ2)
= [(u∗ + ψ1)
3 − (u∗ + ψ2)3] · (v∗ + φ1)
2 + (u∗ + ψ2)3[(v∗ + φ1)
2 − (v∗ + φ2)2]
= 3(u∗ + θ0ψ1 + (1 − θ0)ψ2
)2(ψ1 − ψ2)(v∗ + φ1)
2
+ 2(u∗ + ψ2)3(v∗ + τ0φ1 + (1 − τ0)φ2
)(φ2 − φ1), (4.14)
so the last term ‖N3,2(ψ1, φ1) − N3,2(ψ2, φ2)‖∗∗ can also be evaluated by the interior estimate(2.12) and the exterior estimate (2.16) of v∗ as the following,∥∥N3,2(ψ1, φ1) − N3,2(ψ2, φ2)
∥∥∗∗� C
∥∥(u2∗ + |ψ1|2 + |ψ2|2)|ψ1 − ψ2|
(v2∗ + |φ1|2
)∥∥∗∗+ C
∥∥(u3∗ + |ψ2|3)(
v∗ + |φ1| + |φ2|)|φ1 − φ2|
∥∥∗∗
� C∥∥[(1 + |y|)−2 + (
1 + |y|)− 4q′ (‖ψ1‖∗ + ‖ψ2‖∗
)2]· [(1 + |y|)− 2
q′ v2∗(y) + (1 + |y|)− 6
q′ ‖φ1‖2∗]∥∥∗∗ · ‖ψ1 − ψ2‖∗
+ C∥∥[(1 + |y|)−3 + (
1 + |y|)− 6q′ ‖ψ2‖3∗
]· [(1 + |y|)− 2
q′ v∗(y) + (1 + |y|)− 4
q′ (‖φ1‖∗ + ‖φ2‖∗)]∥∥ · ‖φ1 − φ2‖∗
∗∗
3490 Y. Guo et al. / J. Differential Equations 256 (2014) 3463–3495
� C[∥∥(1 + |y|)− 6
q′ v2∗(y)∥∥∗∗ + ∥∥(1 + |y|)− 10
q′ ∥∥∗∗‖φ1‖2∗] · ‖ψ1 − ψ2‖∗
+ C[∥∥(1 + |y|)− 8
q′ v∗(y)∥∥∗∗ + ∥∥(1 + |y|)− 10
q′ ∥∥∗∗(‖φ1‖∗ + ‖φ2‖∗
)] · ‖φ1 − φ2‖∗
� C[k
18− 45q + (
ρ + k592 − 45
q)2] · ‖ψ1 − ψ2‖∗ + C
[k− 9
2 − 3q + ρ + k
592 − 45
q] · ‖φ1 − φ2‖∗.
(4.15)
Combining the results of (VII.-1), (VII.-2), (VII.-3), (VII.-4), we have that∥∥M(ψ1, φ1) −M(ψ2, φ2)∥∥∗
� C[∥∥N5,0(ψ1, φ2) − N5,0(ψ2, φ2)
∥∥∗∗ + ∥∥N2,3(ψ1, φ2) − N2,3(ψ2, φ2)∥∥∗∗
+ ∥∥N0,5(ψ1, φ2) − N0,5(ψ2, φ2)∥∥∗∗ + ∥∥N3,2(ψ1, φ2) − N3,2(ψ2, φ2)
∥∥∗∗]
� C[(
ρ + k592 − 45
q)(‖ψ1 − ψ2‖∗ + ‖φ1 − φ2‖∗
)+ (ρ + k
592 − 45
q)2‖φ1 − φ2‖∗
]� C
(ρ + k
592 − 45
q)∥∥(ψ1, φ1) − (ψ2, φ2)
∥∥∗. (4.16)
Choose ρ0 small enough and k0 large enough such that Cρ0 < 14 , Ck
592 − 45
q
0 < 14 , then for any
0 < ρ < ρ0 and k > k0, M is a contraction map. Therefore, by the Banach fixed point theorem,we can find (ψk,φk) as fixed point, and (uk, vk) = (u∗ + ψk, v∗ + φk) the positive solution to(P3) for any k � k0, which means (P3) possesses infinitely many positive solutions.
Remark 4.1. When N � 4, the perturbation method (uk, vk) := (u∗ + ψk, v∗ + φk) and theweighted norms of type
‖ξ‖∗ := supy∈RN
{(1 + |y|)N−1
q′ ∣∣ξ(y)∣∣}, ‖h‖∗∗ := ∥∥(1 + |y|)N+2− 2N
q h(y)∥∥
Lq(RN)
present mathematical difficulties in obtaining contraction maps for the fixed point theory. Webelieve this problem is only due to our specific choice of the weighted norms here and (PN ),(N � 4) can still be solved by other methods.
Acknowledgments
Y. Guo was supported by NSFC (11171171, 11331010) and J. Wei was supported by a GRFgrant from RGC of Hong Kong and an NSERC of Canada.
Appendix A
Let N = 3, our assumption of 32 < q < 90
59 implies 2 < 9031 < q ′ < 3, so the integrals∫
R3dz
|z−y|(1+|z|)4+ 2q′
in Lemma A and∫R3
dz
|z−y|q′(1+|z|)6−q′ in Lemma B are finite when |y| is close
to 0.Without loss of generality, we always set |y| � 2 in the following proofs, and define a distance
d := |y| ∼ 1 + |y|.
2
Y. Guo et al. / J. Differential Equations 256 (2014) 3463–3495 3491
The whole region R3 is divided by three parts, i.e.,
R3 = B0(d) ∪ By(d) ∪ (
R3\(B0(d) ∪ By(d)
))where B0(d) denotes the ball of radius d and center at 0, while By(d) means the ball of radius d
and center at y.
Lemma A. ∫R3
dz
|z − y|(1 + |z|)4+ 2q′
� C(1 + |y|)−1
. (A.1)
Proof. Instead of estimating the integral on R3 directly, we divert to the integrals on smaller
regions B0(d), By(d) and R3\(B0(d) ∪ By(d)) respectively.
(A.-1). For z ∈ Bd(0), we have |z − y|� d2 and∫
B0(d)
dz
|z − y|(1 + |z|)4+ 2q′
� Cd−1∫
B0(d)
dz
(1 + |z|)4+ 2q′
� Cd−1(1 + d−1− 2
q′ )� Cd−1 � C(1 + |y|)−1
. (A.2)
(A.-2). For z ∈ Bd(y), we have |z| + 1 � d2 and∫
By(d)
dz
|z − y|(1 + |z|)4+ 2q′
� Cd−4− 2
q′∫
By(d)
dz
|z − y|
� Cd−4− 2
q′∫
B0(d)
dz
|z| � Cd−4− 2
q′d∫
0
r dr
� Cd−2− 2
q′ � C(1 + |y|)−2− 2
q′ . (A.3)
(A.-3). For z ∈ R3\(B0(d) ∪ By(d)), we have two choices:
(A.-3.-i). If |z| � 2|y|, then |z − y| � |z| − |y| > |z|2 , so
1
|z − y|(1 + |z|)4+ 2q′
� 2
|z|(1 + |z|)4+ 2q′
;
(A.-3.-ii). If |z| < 2|y|, then |z − y| � d >|z|4 , so
1
|z − y|(1 + |z|)4+ 2q′
� 4
|z|(1 + |z|)4+ 2q′
.
3492 Y. Guo et al. / J. Differential Equations 256 (2014) 3463–3495
In all, for all z ∈ R3\(B0(d) ∪ By(d)), we have
1
|z − y|(1 + |z|)4+ 2q′
� 4
|z|(1 + |z|)4+ 2q′
,
therefore, ∫R3\(B0(d)∪By(d))
dz
|z − y|(1 + |z|)4+ 2q′
� C
∫R3\(B0(d)∪By(d))
dz
|z|(1 + |z|)4+ 2q′
� C
∫R3\B0(d)
dz
|z|(1 + |z|)4+ 2q′
� C
+∞∫d
r−3− 2
q′ dr � C(1 + |y|)−2− 2
q′ . (A.4)
Summing the integrals (A.2), (A.3) and (A.4) together, we get the estimate of the originalintegral on R
3 that∫R3
dz
|z − y|(1 + |z|)4+ 2q′
=∫
B0(d)
+∫
By(d)
+∫
R3\(B0(d)∪By(d))
dz
|z − y|(1 + |z|)4+ 2q′
� C(1 + |y|)−1 + C
(1 + |y|)−2− 2
q′ � C(1 + |y|)−1
. � (A.5)
Lemma B. ∫R3
dz
|z − y|q ′(1 + |z|)6−q ′ � C
(1 + |y|)−2
. (A.6)
Proof. Similar to the procedure in Lemma A, we deal with the corresponding integrals onsmaller regions B0(d), By(d) and R
3\(B0(d) ∪ By(d)).
(B.-1). For z ∈ B0(d), we have |z − y|� d2 and
∫B0(d)
dz
|z − y|q ′(1 + |z|)6−q ′ �
∫B0(d)
dz
|z − y|q ′(1 + |z|)5−q ′
� Cd−q ′∫
B0(d)
dz
(1 + |z|)5−q ′ � Cd−q ′(1 + dq ′−2)
� Cd−2 � C(1 + |y|)−2
. (A.7)
Y. Guo et al. / J. Differential Equations 256 (2014) 3463–3495 3493
The fourth inequality above is due to the assumption that 9031 < q ′ < 3, so q ′ − 2 > 0 and
the term dq ′−2 dominates 1.(B.-2). For z ∈ By(d), we have |z|+ 1 � d
2 + 1. Moreover, notice the assumption 9031 < q ′ < 3,
so∫ d
0 r2−q ′dr = d3−q ′
, and the singular estimate is evaluated that,
∫By(d)
dz
|z − y|q ′(1 + |z|)6−q ′ �
∫By(d)
dz
|z − y|q ′(1 + |z|)5−q ′
� Cdq ′−5∫
By(d)
dz
|z − y|q ′ � Cdq ′−5∫
B0(d)
dz
|z|q ′
� Cdq ′−5
d∫0
r2−q ′dr � Cd−2 � C
(1 + |y|)−2
. (A.8)
(B.-3). For z ∈R3\(B0(d) ∪ By(d)), we have two choices:
(B.-3.-i). If |z| � 2|y|, then |z − y|� |z| − |y| > |z|2 , so
1
|z − y|q ′(1 + |z|)6−q ′ �
2q ′
|z|q ′(1 + |z|)6−q ′ �
2q ′
|z|q ′(1 + |z|)5−q ′ ;
(B.-3.-ii). If |z| < 2|y|, then |z − y| � d >|z|4 , so
1
|z − y|q ′(1 + |z|)6−q ′ �
4q ′
|z|q ′(1 + |z|)6−q ′ �
4q ′
|z|q ′(1 + |z|)5−q ′ .
In all, for all z ∈ R3\(B0(d) ∪ By(d)), we have
1
|z − y|q ′(1 + |z|)6−q ′ �
4q ′
|z|q ′(1 + |z|)5−q ′ ,
therefore, ∫R3\(B0(d)∪By(d))
dz
|z − y|q ′(1 + |z|)6−q ′
� C
∫R3\(B0(d)∪By(d))
dz
|z|q ′(1 + |z|)5−q ′
� C
∫3
dz
|z|q ′(1 + |z|)5−q ′ � C
+∞∫d
r−3 dr � C(1 + |y|)−2
. (A.9)
R \B0(d)
3494 Y. Guo et al. / J. Differential Equations 256 (2014) 3463–3495
Summing the integrals (A.7), (A.8) and (A.9) together, we get the estimate of the originalintegral on R3 that
∫R3
dz
|z − y|q ′(1 + |z|)6−q ′
=∫
B0(d)
+∫
By(d)
+∫
R3\(B0(d)∪By(d))
dz
|z − y|q ′(1 + |z|)6−q ′ � C
(1 + |y|)−2
. � (A.10)
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