Large energy entire solutions for the Yamabe type problem of
polyharmonic operator
Yuxia Guo
Department of Mathematics, Tsinghua University, Beijing, 100084, P.R.China
Bo Li
Department of Mathematics, Tsinghua University, Beijing, 100084, P.R.China
Juncheng Wei
Department of Mathematics, The Chinese University of Hong Kong, Shatin, Hong Kong
Abstract: In this paper, we consider the following Yamabe type problem of polyharmonic operator :
(
4m
Dm u
= |u| N −2m u
u
∈ H m (SN ),
on SN
(P )
where N ≥ 2m + 1, m ∈ N+ , SN is the unit sphere with the induced Riemannian
metric g = gSN , and Dm is the elliptic differential operator of 2m order given by
Dm =
m
Y
k=1
1
(−∆g + (N − 2k)(N + 2k − 2))
4
where ∆g is the Laplace-Beltrami operator on SN . We will show that the problem
(P) has infinitely many non-radial sign changing solutions.
Keywords: Polyharmonic equations, Critical exponents, Infinitely many non-radial solutions.
AMS Subject Classification: 35B45, 35J25.
1
Introduction
We consider the following Yamabe type problem for polyharmonic operator:
(
∗
Dm u = |u|m −2 u on SN
u ∈ H m (SN ),
where m∗ =
2N
N −2m , N
(P )
≥ 2m + 1, m ∈ N+ , SN is the unit sphere with the induced Riemannian
metric g = gSN , and Dm is the elliptic differential operator of 2m order given by
Dm =
m
Y
k=1
1
(−∆g + (N − 2k)(N + 2k − 2))
4
1
where ∆g is the Laplace-Beltrami operator on SN (see [7]).
The well known Yamabe problem, which stems from the conformal geometry, is the problem
of finding some scalar curvature K in a compact Riemann manifold (M, g0 ) of dimension n ≥ 2.
More precisely, for a given smooth function K defined on this manifold, we want to find a new
metric g which is conformal to the original metric g0 such that K is actually the scalar curvature
under this new g. In the case of m = 1, the Yamabe problem read as:
(
−∆SN u +
N (N −2)
u
2
N +2
− u N −2 = 0
on SN
u > 0.
(1.1)
see ([3], [4], [31], [10]).
By using the stereo-graphic projection, the problem (1.1) can be reduced to:

N +2

in RN
 −∆u = u N −2




in RN
u >0
u ∈
(1.2)
D1,2 (RN ),
where D1,2 (RN ) is the completion of C0∞ (RN ) under the norm
R
RN
|∇u|2 . It is known that the
only finite energy positive solution to (1.2) are given by the family of the functions ( see [21] ):
µ−
N −2
2
U (µ−1 (x − ξ)), U (x) = (
N −2
2
N
2 , ξ ∈ R , µ > 0.
)
1 + |x|2
(1.3)
Moreover these functions are corresponding to the extremals for the critical Sobolev embedding
(see [29]). And these functions are indeed all positive solutions of (1.2) even without the finite
energy requirement (see [9]). It is natural to ask weather there are finite energy non-radial
sign changing solutions to (1.2). This was first answered by Ding [10]. His proof is variational:
consider the functions of the form
u(x) = u(|x1 |, |x2 |), x = (x1 , x2 ) ∈ S N ⊂ RN +1 = Rk × RN −k , k ≥ 2.
(1.4)
The critical Sobolev embedding becomes compact and hence infinitely many sign changing solutions exist, thanks to the Ljusternik-Schnirelmann theory. See also [17]. Recently, del Pino,
Musso, Pacard and Pistoia [24]-[25] gave another proof of countablely many sign changing nonradial solutions. Their proof is more constructive: they built a sequence of solutions with one
negative bump at the origin and large number of positive bumps in a polygon. This gives more
precise information on such sign changing solutions.
On the other hand, the polyharmonic operator, in particular the biharmonic operator has
found considerable interest due to its geometry roots in recent years. For instance, when m = 2,
the problem (P ) is related to the Paneitz operator, which was introduced by Paneitz [23] for
smooth 4 dimensional Riemannian manifolds and was generalized by [8] to smooth N dimensional Riemannian manifolds. We refer the reader to the papers [2], [5], [6], [11], [12], [14], [15],
2
[16], [26], [27], [29], and the references therein, for various existence and multiplicity results on
the polyharmonic operator and related problems. It is evident from these papers that the polyharmonic operator presents new and challenging features compared with the Laplace operator.
However, few results are known for the Yamabe problem of polyharmonic operator. The purpose
of the present paper is concerned on this topic.
Similar to in the case of m = 1, by using the stereo-graphic projection, the problem (P ) can
be reduced to the following problem in RN , namely
(
4m
(−∆)m u = |u| N −2m u
in RN
(1.5)
u ∈ Dm,2 (RN ),
where N ≥ 2m + 1, m ∈ N+ , and Dm,2 (RN ) is the completion of C0∞ (RN ) with respect to the
norm induced by the scalar product
Z
m
m


∆ 2 u · ∆ 2 v,
if m is even,

RN
Z
(u, v) =
m−1
m−1



∇∆ 2 ∇∆ 2 v, if m is odd.
(1.6)
RN
Moreover it is known that the only finite energy positive solution to equation (1.5) are given by
the family of the functions (see [21],[6]):
µ−
where Pm,N =
m−1
Y
N −2m
2
N −2m
4m
U (µ−1 (x − ξ)), U (x) = Pm,N
(1 + |x|2 )−
N −2m
2
,
(N + 2h).
h=−m
Generalizing the idea of Ding and using variational method, Bartsch and Weth [5] established
an unbounded sequence of sign changing finite energy solutions to (1.5).
In this paper, following the idea in [24]-[25], we will construct a sequence of non-radial sign
changing solutions for problem (1.5). Our result cover the case of Yamabe equations and the
biharmonic equations.
Our main results are:
Let m ≥ 1, N ≥ 2m + 1, and write RN = C × RN −2 . Then for each k large
Theorem 1.1
enough, the problem (1.5) admits a finite energy solution of the form
uk (x) = U (x) −
k
X
− N −2m
2
µk
U (µ−1
k (x − ξj )) + o(1),
j=1
q
where ξj =
2
δkN −2m
k2
1 − µ2k (e
2π(j−1) √
−1
k
N −2m
4m
, 0), j = 1, 2, · · · , k, U (x) = Pm,N
(1 + |x|2 )−
for N ≥ 2m + 2, and µk =
k3
δk2
log2 k
N −2m
2
,
µk =
for N = 2m + 1, and o(1) → 0 uniformly as k → ∞.
δk is a positive number which depends on k only.
As a consequence, we have
3
Theorem 1.2 Suppose that N ≥ 2m + 1, then problem (P ) has infinitely many non-radial sign
changing solutions.
Remark 1.3 The geometry picture of the sign-changing solution u is that it is positive near the
center while negative in the region of the bubbles scattered around the Obata type solution in the
middle.
Remark 1.4 We believe that similar result should also hold for the following Lane-Emden system

 (−∆)m u = |v|α−1 v;
in RN , m ≥ 1.
(1.7)
 (−∆)m v = |u|β−1 u.
It is known that (see [18]), for N > 2m, α, β ≥ 1 but not equal to 1 such that
1
1
N − 2m
+
>
,
α+1 β+1
N
(1.7) has no any positive solutions. On the other hand, for N > 2m, α, β ≥ 1 such that
1
1
N − 2m
+
≤
.
α+1 β+1
N
(1.7) admits infinitely positive solutions (see [19]), We conjecture that the following is true:
Conjecture 1.1 For N > 2m, α, β ≥ 1 and
1
α+1
+
1
β+1
=
N −2m
N ,
problem (1.7) has infinitely
many sign changing solutions.
The paper is organized as follows: Section 2 contains the construction of an approximation
solution and the estimates of the error. While the Section 3 will devote to the detailed calculus
and further thoughts on the gluing procedures and linearization of the problem. The proof of
the theorem will be also given in this section.
Acknowledgement: The first author was supported by NSFC (11171171) and the third author
was supported from a General Research Grant of RGC of Hong Kong.
2
Approximation solution and the estimate of the error
In this section, we first construct an approximation solution for our problem (1.5). Then we
give the precise estimate of the error.
As we mentioned in the introduction, it is well known that the equation
N +2m
(−∆)m u = u N −2m
has the following radial solution
N −2m
¯ ¯2 N −2m
4m
U (x) = Pm,N
(1 + ¯x¯ )− 2 ,
4
(2.1)
with
m−1
Y
Pm,N =
(N + 2h).
h=−m
Moreover this radial solution U is invariant under the Kelvin type transform :
¯ ¯2m−N
y
u
b(y) = ¯y ¯
u( ¯ ¯2 ).
¯y ¯
(2.2)
b (y) = U (y)(c.f.[6]).
That is, U
We begin with
Lemma 2.1 The equation (2.1) is invariant under the Kelvin transform (2.2), namely,
¯
¯m∗ −2
(−∆)m û(y) = ¯u
b(y)¯
u
b(y),
where
m∗ =
2N
.
N − 2m
Proof. The result is known. For the sake of completeness, we give the proof. We first prove
the case of m = 2. To simplify our proof, we make use of the spherical coordinates, then
∆u(r, θ) = (∂r2 +
N −1
∆θ
∂r + 2 )u(r, θ).
r
r
Iterate the Laplace-Beltrami operators two times, we obtain
N −1
∆θ
∆2 u(r, θ) =(∂r2 +
∂r + 2 )2 u(r, θ)
r
r
h
2(N
−
1)
(N − 1)(N − 3) 2 (N − 1)(N − 3)
= ∂r4 +
∂r3 +
∂r −
∂r
r
r2
r3
8 − 2N
2N − 6
2
1 2i
2
+
∆
+
∆
∂
+
∆
∂
+
∆ u(r, θ),
r
θ
θ
θ
r
r4
r3
r2
r4 θ
which gives the formula of the scalar transform of u as the following:
¯
¯
N −1
∆θ 2
¯
¯
2
2
∆ u(ρ, θ)¯ 1 =(∂ρ +
∂ρ + 2 ) u(ρ, θ)¯ 1
ρ
ρ
ρ= r
ρ= r
h
= ∂r4 + 2(N − 1)r∂r3 + (N − 1)(N − 3)r2 ∂r2 − (N − 1)(N − 3)r3 ∂r
i 1
+ (8 − 2N )r4 ∆θ + (2N − 6)r3 ∆θ ∂r + 2r2 ∆θ ∂r2 + r4 ∆2θ u( , θ).
r
For the same reason, we have for α > 0,
³
i 1
1 ´ h
∆ rα u( , θ) = α(N + α − 2)rα−2 + (3 − N − 2α)rα−3 + rα−4 ∂r2 + rα−2 ∆θ u( , θ),
r
r
(2.3)
(2.4)
and
n
1
∆2 rα u( , θ) = α(α − 2)(N + α − 2)(N + α − 4)rα−4 + rα−8 ∂r4 + (14 − 2N − 4α)rα−7 ∂r3
r
h
i
+ α(α + N − 2) + (3 − N − 2α)(9 − N − 2α) + (α − 4)(N + α − 6) rα−6 ∂r2
h
i
+ α(α + N − 2)(7 − N − 2α) + (3 − N − 2α)(α − 3)(α + N − 5) rα−5 ∂r
h
i
+ α(N + α − 2) + (α − 2)(N + α − 4) rα−4 ∆θ
o 1
+ (10 − 2N − 4α)rα−5 ∆θ ∂r + 2rα−6 ∆θ ∂r2 + rα−4 ∆2θ u( , θ).
r
(2.5)
5
In order to avoid the possible u term and preserve the derivative terms, we set α = 4 − N in
(2.5) above, by comparison with the equation (2.3), we can derive the following formula of the
∆2 operator on the Kelvin type transform, namely,
h
1
∆2 r4−N u( , θ) =r−(N +4) ∂r4 + 2(N − 1)r∂r3 + (N − 1)(N − 3)r2 ∂r2 − (N − 1)(N − 3)r3 ∂r
r
i 1
+ (8 − 2N )r4 ∆θ + (2N − 6)r3 ∆θ ∂r + 2r2 ∆θ ∂r2 + r4 ∆2θ u( , θ)
r
¯
¯
−(N +4) 2
=r
∆ u(ρ, θ)¯ 1 .
ρ= r
(2.6)
By using the formula (2.6), we have,
¯
¯
(−∆)2 u
b(y) =∆2 u
b(y) = r−(N +4) ∆2 u(ρ, θ)¯ 1
ρ= r
¯
¯
=r−(N +4) (−∆)2 u(ρ, θ)¯ 1
ρ=
¯ r
8
¯
=r−(N +4) |u| N −4 u(ρ, θ)¯ 1
ρ= r
=|b
u|
8
N −4
u
b(y).
For any m > 1, to avoid the horrible details and inessential repeats, it is reasonable to give an
induction to reveal the scheme of the proof in the case of m 6= 2. Indeed, for some fixed α > 0,
³
´
1
1
∆m (rα u( , θ)) =∆ ∆m−1 (rα u( , θ))
r
r
n m−2
Y£
¤
1
=∆
(α − 2h)(N + α − (h + 1)) rα−2(m−1) u( , θ)
r
h=0
o
1
+ rα−4(m−1) ∂r2(m−1) u( , θ) + · · ·
r
m−2
Y£
¤ ¡
1 ¢
=
(α − 2h)(N + α − (h + 1)) ∆ rα−2(m−1) u( , θ)
r
h=0
¡
1 ¢
+ ∆ rα−4(m−1) ∂r2(m−1) u( , θ) + ∆ · · ·
r
n m−2
Y£
¤o £
¤
=
(α − 2h)(N + α − (h + 1)) · (α − 2(m − 1))(α + N − 2m) ·
h=0
h
i
1
1
u( , θ) + rα−4m ∂r2m u( , θ) + · · ·
r
r
m−1
¯
Y£
¤
1
¯
=
(α − 2h)(N + α − (h + 1)) rα−2m u( , θ) + rα−4m ∆m u(ρ, θ)¯ 1 .
r
ρ= r
r
α−2m
h=0
By using the same statement as that in the case of m = 2, we set α = 2m−N , and the conformal
invariance under the Kelvin type transform (2.2) holds.
Let
wµ (y − ξ) = µ−
N −2m
2
Then a simple algebra computation shows that:
6
U (µ−1 (y − ξ)).
¤
Lemma 2.2 wµ (y−ξ) is invariant under the Kelvin type transform (2.2) if and only if µ2 +|ξ|2 =
1.
Let k be a large positive integer and µ > 0 be a small concentration parameter such that:

2
 µ = δ N −2m
k −2 , N ≥ 2m + 2,
 µ = δ 2 k −3 log−2 k,
N = 2m + 1,
where δ is a positive parameter that will be fixed later. Let
ξj =
p
1 − µ2 (e
2π(j−1) √
−1
k
),
j = 1, 2, · · · , k,
be the points that are arranged symmetrically as the vertices of a planar regular polygon. Set
Uj (y) = wµ (y − ξj ), j = 1, 2, ..., k
and
U∗ = U −
k
X
Uj .
j=1
In order to find out sign-changing solutions for the problem (1.5). We follow the method of
[24] and use the number of the bubble solutions Uj as a parameter. This was originally developed
by Wei and Yan in [31] for the critical problems with the presence of weights. We will show that
when the bubbles number k is large enough, the problem (1.5) admits a solution of the form:
u(y) = U∗ (y) + φ(y)
where φ is a function which is small when compared with U∗ . With u being this form, the
equation (1.5) can be restated as
(−∆)m φ − p|U∗ |p−1 φ + E − N (φ) = 0
(2.7)
where p = m∗ − 1, and
E = (−∆)m U∗ − |U∗ |p−1 U∗ ,
N (φ) = |U∗ + φ|p−1 (U∗ + φ) − |U∗ |p−1 U∗ − p|U∗ |p−1 φ.
We expect that for k large, the error term E will be controlled small enough so that some
asymptotic estimate holds. In order to get the better control on the error, for a fixed number
N >q>
N
2,
we introduce the following weighted Lq norm:
N +2m− 2N
q
khk∗∗ := k(1 + |y|)
h(y)kLq (RN )
(2.8)
and
kφk∗ := k(1 + |y|N −2m )φ(y)kL∞ (Rn ) .
7
(2.9)
Proposition 2.3 There exists an integer k0 and a positive constant C such that for ∀k ≥ k0 ,
the following estimates for the error term E hold true:
(
1− N
Ck q
if N ≥ 2m + 2;
kEk∗∗ ≤
−1
C log k
if N = 2m + 1.
(2.10)
Proof. We estimate the error in two steps. In the first step, we estimate the error in the exterior
region:
EXT :=
k
\
Bξcj (η/k) :=
j=1
k
\
{|y − ξj | > η/k}.
j=1
In the second step, we estimate the error in the interior:
IN T = EXT c =
k
[
{|y − ξj | ≤ η/k}
where η << 1.
j=1
Step 1: (The estimate for the exterior region EXT ). In order to use mean value theorem
appropriately, we write the formula of E as the following:
E =(−∆)m U∗ − |U∗ |p−1 U∗
k
k
k
X
X
X
¯p−1 ¡
£
¤ ¯
¢
¯
¯
=(−∆) U −
Uj − U −
Uj
U−
Uj
m
j=1
=U p −
k
X
¯
Ujp − ¯U −
j=1
j=1
k
X
¯p−1 ¡
Uj ¯
U−
j=1
j=1
k
X
Uj
¢
j=1
k
P
k
¯U − Uj X
i
h
¯
Ujp
= − |x|p−1 x¯ j=1 +
U
j=1
k
k
k
h¯
i
X
X
¯p−1 ¡
¢ X
= − ¯U − s
Uj ¯
U −s
Uj +
Ujp , for s ∈ (0, 1),
j=1
j=1
j=1
¯U − P Uj
k
k
P
P
Uj |p−1 (U −
Uj ) − |U |p−1 U.
¯ j=1 = |U −
k
where
¯
|x|p−1 x
U
j=1
j=1
We split the exterior region into two parts, namely:
¯
I := {y ¯|y| ≥ 2}
and
II := {|y| < 2}
k
\h \
j=1
8
i
{|y − ξj | > η/k} .
1
|y−ξj |
For y ∈ I, we have
|E(y)| ≤C
n¡
1
1+|y| .
∼
1 + |y|2
Thus
¢−2m
+
k
£X
µ
N −2m
2
¡
µ2 + |y − ξj |2
¢− N −2m ¤
2
4m
N −2m
o
·
j=1
k
£X
µ
N −2m
2
¡
µ2 + |y − ξj |2
¢− N −2m ¤
2
j=1
≤C
h¡
¢
2 −2m
1 + |y|
µ2m k N −2m i X
µ 2
+
·
2
2m
(1 + |y| )
|y − ξj |N −2m
4m
k
N −2m
j=1
≤C
k
X
1
µ
;
2
2m
(1 + |y| )
|y − ξj |N −2m
N −2m
2
j=1
For y ∈ II, we have two cases:
Case 1. There exists i0 ∈ {1, 2, 3, · · · , k} such y is closest to ξi0 , but far away from all the
other ξj ’s (j 6= i0 ) so that
|j − i0 |
1
.
|y − ξj | ≥ |ξj − ξi | ∼
2
k
Case 2. y is far from all ξi ’s , namely, ∃C0 > 0 such that |y − xi | ≥ C0 (1 ≤ i ≤ k).
In both of the cases, we have
|E(y)| ≤C
n¡
1 + |y|2
¢−2m
+
k
£X
µ
N −2m
2
¡
µ2 + |y − ξj |2
¢− N −2m ¤
2
4m
N −2m
o
·
j=1
£
k
X
µ
N −2m
2
¡
µ2 + |y − ξj |2
¢− N −2m ¤
2
j=1
≤C
n¡
1 + |y|2
¢−2m
+
£
N −2m
N −2m
N −2m
k
X
¤ 4m o X
µ 2
µ 2
µ 2
N −2m
·
+
|y − xi0 |N −2m
|y − xj |N −2m
|y − ξj |N −2m
j6=i0
≤C
n¡
1 + |y|2
¢−2m
h
+ µ2m k 4m + max{
X µ2m k 4m
,
|j − i0 |4
j6=xi0
≤C
N −2m
2
µ
(1 + |y|2 )2m
k
X
j=1
j=1
k
io X
4m
k N −2m µ2m } ·
j=1
N −2m
µ 2
|y − ξj |N −2m
1
.
|y − ξj |N −2m
Hence, by combining the results above, we obtain the estimate for E in the exterior region
as:
kEk∗∗(EXT ) =k(1 + |y|)(N +2m)q−2N E q (y)kLq (EXT )
¡
¢(N +2m)q−2N
k hZ
i1/q
1 + |y|
N −2m X
≤Cµ 2
¡
¢4mq
Bξc (η/k) 1 + |y|
|y − ξj |(N −2m)q
j=1
j
Z +∞
i1/q
h Z 1 rN −1 dr
N −2m
−(N +1)
r
dr
+
≤Cµ 2 k
(N −2m)q
1
η/k r

N
1−
1− N
1+2m−N

) ≤ Ck q ,
if N ≥ 2m + 1;
 C(k q + k
≤


C log−1 k,
if N = 2m + 1.
9
Step 2: (For the interior region INT.) In this case, we see that for ∀y ∈ IN T , there
exists j ∈ {1, 2, 3, · · · , k}, such that |y −ξj | ≤ η/k. In order to make the integral region restricted
to the regular region centered at the origin, for this particular j, we define
ej (y) = µ N +2m
2
E
E(ξj + µy).
N −2m
2
Note that µ
¢
ξj ) , where
Uj (ξj + µy) = U (y) and for i 6= j, µ
¯
¯
µ−1 ¯ξj − ξi ¯ ∼ |j−i| .
N −2
2
¡
Ui (ξj + µy) = U y − µ−1 (ξi −
kµ
µ2 (1
|y|4 )
µ2 (1
Since
+
<
+ ηµ−4 k −4 ) ≤ C. We have
¯
¯
ej (y)¯
¯E
¯
¯p−1 ¡ X (kµ)N −2m
X (kµ)N −2m
¢
N −2m
N −2m
¯
¯
2
≤C ¯U (y) +
U
(ξ
+
µy)
·
+
µ
+ µ 2 U (ξj + µy)
¯
j
N
−2m
N
−2m
|j − i|
|j − i|
i6=j
i6=j
X ³ (kµ)N −2m ´p
N +2m
+
+ µ 2 U p (ξj + µy)
|j − i|N −2m
i6=j
¯¡ 1 ¢ N −2m
¯
N −2m
N −2m
N +2m
N −2m ¯p−1
¯
2
2
≤C ¯
·µ 2 +µ 2 p+µ 2
+
µ
¯
2
1 + |y|
¯ µ N −2m
¯
2
N +2m ¯
¯
2
≤C ¯
+
µ
¯
1 + |y|4m
N −2m
µ 2
.
≤C
1 + |y|4m
Hence we get the estimate of the error E in one branch of the interior region as
¯q i1/q
¯ N N +2m
hZ
¯
¯ q− 2 e
Ej (y)¯ dy
kEk∗∗(|x−ξj |<η/k) ≤C
¯µ
|y|≤η/(kµ)
Z
h
≤C µN −2mq
2m
η/(kµ)
0
−N
+4m
q
i1/q
rN −1
dr
1 + r4mq
≤Cµ · k

N
 Ck − q ,
≤
 Ck − Nq · log−4m k,
if N ≥ 2m + 2;
if N = 2m + 1.
At last, by combining the estimates in the exterior region and interior region together, we
get
kEk∗∗ ≤kEk∗∗(EXT ) + kEk∗∗(IN T ) ≤ kEk∗∗(EXT ) +
3
kEk∗∗(|x−ξj |<η/k)
j=1

1− N

 Ck q ,
≤
k
X
if N ≥ 2m + 2;

¢
 ¡ −1
1− N
C log k + k q · log−4m k ≤ C log−1 k,
if N = 2m + 1.
Linearization and gluing
In this section, we focus on the invertibility theory for a linearized equation and the proof of the
main theorem follows from the obtained series of propositions and lemmas.
10
We consider the linear operator L0 defined by
h
i
L0 (φ) := (−∆)m − pU p−1 φ,
with p = m∗ − 1.
We consider the following linear equation
L0 (φ) = h.
(3.1)
Then it is well known that (see [6]) the solution space for the corresponding homogeneous
equation
L0 (φ) = 0
is spanned by the following N + 1 functions,
vi = ∂yi U,
i = 1, 2, 3, · · · , N ;
vN +1 = x · ∇U +
n − 2m
U.
2
We also consider the linear operator L∗ of (2.7), that is
i
h
L∗ (φ) := (−∆)m − p|U∗ |p−1 φ,
with p = m∗ − 1,
Since the region is scattered around the vertices of the regular k-polygonal, the direct calculus
on this L∗ is not convenient. We introduce the following gluing procedure to split the working
space into the respective single branch by some cut-off functions, and the equation (1.5) will be
splitted into k + 1 equations with respective single branches or simple linear operator L0 .
Let ζ(s) be a smooth function satisfying


1, 0 ≤ s < 1/2;






ζ(s) = smooth, 1/2 ≤ s ≤ 1;






 0, s > 1.
We define the cut-off functions as
 ³
¯
¯´
 ζ kη −1 |y|−2 · ¯y − |y|2 ξk ¯ , if |y| ≥ 1;
´
³
ζj (y) =
 ζ kη −1 ¯¯y − ξ ¯¯ , if |y| < 1,
j
such that
ζj (y) = ζj (
y
),
|y|2
¯
supp{ζj } ⊂ {y ¯|y − ξj | ≤ η/k}, j = 1, 2, · · · , k.
By means of the cut-off functions, we can split the equation (2.7) into a system comprised
of k + 1 equations.
k
P
Let φ =
φej + ψ, y = (y1 , y2 );
j=1
y 0 = (y3 , · · · , yN ), we assume
φej (y, y 0 ) = φe1 (e−
2π(j−1) √
−1
k
11
y, y 0 ), j = 1, · · · , k,
(3.2)
¯ ¯2m−N
y
φe1 (y) = ¯y ¯
φe1 ( 2 ),
|y|
φe1 (y1 , · · · , ys , · · · , yN ) = φe1 (y1 , · · · , −ys , · · · , yN ),
(3.3)
s = 2, 3, · · · , N,
(3.4)
and
kφ1 k∗ ≤ ρ with ρ << 1,
where φ1 (y) := µ
N −2m
2
(3.5)
φe1 (ξ1 + µy).
Then the equation (2.7) can be splitted into the following system:
h
i

X

(−∆)m φej − p|U∗ |p−1 ζj φej + ζj − p|U∗ |p−1 ψ + E − N (φej +
φei + ψ) = 0,




i6=j




k
k
h
i

X
X

m
p−1
p−1
p−1
p−1
(−∆) ψ − pU
ψ + − p(|U∗ |
−U
)(1 −
ζj ) + pU
(
ζj ) ψ

j=1
j=1




k
k
k

X
X ¡
X

¢

p−1
ej + (1 −

−
p|U
|
(1
−
ζ
)
φ
ζ
)
E
−
N
(
φej + ψ) = 0.

∗
j
j

j=1
j=1
j = 1, · · · , k;
j=1
(3.6)
3.1
The existence of ψ
In this subsection, we will focus on the existence of ψ in the second equation of the system (3.6).
For this purpose, we first prove the following
Proposition 3.1 Assume that
Z
N
2
RN
Then the equation
< q < N , let h(y) be a function such that khk∗∗ < +∞, and
vl h = 0,
∀l = 1, 2, · · · , N + 1.
i
h
L0 (φ) = (−∆)m − pU p−1 φ = h,
(3.7)
has a unique solution φ satisfying kφk∗ < ∞ and
Z
U p−1 Zl φ = 0, ∀l = 1, 2, · · · , N + 1.
RN
Moreover, there is a constant C depending only on q and N such that
kφk∗ ≤ Ckhk∗∗ .
Proof. Let
H = {φ ∈ D
m,2
¯Z
¯
(R )¯
N
Rn
U p−1 vl φ = 0,
∀l = 1, 2, · · · , N + 1}.
Then H is a Hilbert space endowed with the inner product:
Z


∆m/2 u(y) · ∆m/2 v(y)dy

n
< u, v >H := ZR h¡
¢ ¡
¢i
m−1
m−1



∇∆ 2 u(y) · ∇∆ 2 v(y) dy
Rn
12
if m is even;
if m is odd.
Moreover, for any φe ∈ H
¡
¢ ¡
¢
¡
¢
L0 φ, φe = (−∆)m φ, φe − p U p−1 φ, φe
¡
¢
=< φ, φe >H −p U p−1 φ, φe
¡
¢
e φ >H − φ,
e pU p−1 φ
=< φ,
¡
¢
= φ, L0 φe .
By the Sobolev inequality, Lp estimates ( Caldéron-Zygmund inequality)(c.f.[13]), and the
iteration by m times, we get
i
 h
m
 C k(−∆)m φk22 + k∆ 2 φk22 + kφk22 ,
m is even;
h
i
kL0 φk22 ≤
 C k(−∆)m φk2 + k∇∆ m−1
2
2
2 φk + kφk
m is odd,
2
2
2 ,
h
i

m
 C k(−∆)m φk22 + k∆ 2 φk22 + kφk22 ,
m is even;
h
i
≤
 C k(−∆)m φk2 + k∆ m+1
2
2
2 φk + kφk
m is odd,
2
2
2 ,
i
 h
m
 C k(−∆)m−1 φk22 + k∆ 2 −1 φk22 + kφk22 ,
m is even;
i
h
≤
 C k(−∆)m−1 φk2 + k∆ m−1
2
2
2 φk + kφk
m is odd,
2
2 ,
2
≤···
≤Ckφk22 .
³
´
Hence L0 is a bounded, linear and self-adjoint operator in H, (·, ·) . The F redholm alternative in Hilbert space tells us that the closure of the range of the operator L0 is the orthogonal
complement of the null space of L∗0 = L0 . We have known from the beginning of this section,
the null space is actually spanned by {v1 , v2 , · · · , vn+1 }, therefore, the invertibility problem (3.1)
has a weak solution if and only if
(h, vi ) = 0,
for i = 1, 2, 3, · · · , N + 1,
which are exactly the assumption required in Proposition 3.1. It admits a weak solution φ.
Since khk∗∗ < ∞, we choose the pair r =
we have
hZ
khkr ≤
2N
0
N +2m , r
|h|q (1 + |y|)(N +2m)q−2N dy
=
2N
N −2m
i1/q h Z
·
= p + 1, by Hölder inequality,
(1 + |y|)−2N dy
i1−1
r
RN
RN
q
(3.8)
≤Ckhk∗∗ < ∞,
and
´ N −2m ³ Z
´ 4m
N +2m
N +2m
(N +2m)r
(N +2m)r
U (p−1)r· 4m
|φ|r· N −2m
N
N
R
R
³Z
´ 2m
2N
N
U N −2m
=kφkp+1 ·
³Z
kU p−1 φkr ≤
RN
≤Ckφkp+1 = Ckφkm∗ ≤ Ck(∇)m φk = CkφkH < ∞.
13
(3.9)
By (3.8) and (3.9), we have f = pU p−1 φ + h ∈ Lr , and the weak solution can be represented by
the following equations
Z
Z
m
m

2 φ · ∆ 2 ψ +

∆
f ψ = 0,

N
N
R
R
Z h
i Z
m−1
m−1


2
2

(∇∆
φ) · (∇∆
ψ) +
RN
m is even;
for ∀ψ ∈ H.
f ψ = 0,
(3.10)
m is odd,
RN
Now we define the functional Af : H → R by
Z
Af (ψ) = −
then
f ψ,
RN
Z
m
m


∆ 2 φ · ∆ 2 ψ = Af (ψ),

N
ZR h
i
m−1
m−1



(∇∆ 2 φ) · (∇∆ 2 ψ) = Af (ψ),
m is even;
m is odd.
RN
Moreover, by Hölder inequality, we know that
¯
¯
¯Af (ψ)¯ ≤ kf kr kψkp+1 ≤ Ckf kr kψkH .
³ ¡ ¢´
Thus Af is a bounded linear functional on the Hilbert space H, ·, · , by the Riesz’s representation theorem, there exists a unique φ ∈ H such that
Z
m
m


∆2φ·∆2ψ

N
ZR h
Af (ψ) =
i
m−1
m−1



(∇∆ 2 φ) · (∇∆ 2 ψ)
m is even,
m is odd.
RN
Consequently, we can define an operator A : Lr → H, through the functional Af , by
A(f ) = φ,
and
¡
¢
< A(f ), ψ >H = f, ψ ,
∀ψ ∈ H.
As a result, the equation (3.1) can be equivalent to
φ = A(h) + A(pU p−1 φ) = A(h) + A(τ (φ)),
where τ : H → Lr ,
φ 7→ pU p−1 φ, is a compact mapping by the rapid decreasing rate of U p−1 .
Set B = A ◦ τ , then B is the operator from H to H. Moreover, it is easy to see that B is also
compact since it is the composition of the bounded linear operator and the compact operator,
hence the equation (3.1) can be rewritten as
(I − B)φ = A(h).
Also it is natural to verify that B is also self-adjoint, and the Fredholm alternative applies.
Hence (I − B)φ = A(h) has a solution if and only if
∀v ∈ Ker(I − B),
(I − B)v = 0 = A(0),
14
since A is injective.
Then, we obtain h ≡ 0 with
A(0) ∈ R(I − B) = (Ker(I − B ∗ ))⊥ = (Ker(I − B))⊥ .
Therefore the equation (3.1) is reduced to the homogeneous version, that is
L0 (v) = 0,
where v can be written as the sum of vi ’s,
v(y) =
N
+1
X
al · vl (y),
l=1
with constants a1 , a2 , · · · , aN +1 .
Recall the constraint of the H is such that
Z
Z
0=
U p−1 vl · v = al
RN
RN
U p−1 (y)vl2 (y)dy,
which yields the vanishing components
al = 0,
l = 1, 2, · · · , N + 1,
and v ≡ 0,
Ker(I − B) = {0}.
Hence, the orthogonal complement R(I − B) = H,
this shows the existence of φ by
(I − B)φ = A(h),
and the uniqueness of φ by
Ker(I − B) = {0}.
In the following , we will show that
kφk∗ ≤ Ckhk∗∗ .
Set φ0 = φ, the linearized equation (3.1) is equivalent to the following system:


(−∆)φ = φ1 ,






(−∆)φ1 = φ2 ,


···




 (−∆)φ
m−1
= pU p−1 φ + h.
By the elliptic regularity, we know that {φl , l = 0, 1, 2, · · · , m − 1} are all bounded in L∞ norm,
and h is also bounded in the L∞ norm. Moreover by the Local elliptic estimates, we have
kD2 φm−1 kLq (B1 ) + kDφm−1 kLq (B1 ) + kφm−1 kL∞ (B1 ) ≤ CkhkL2 (B2 ) ≤ Ckhkr ≤ Ckhk∗∗ ;
15
kD2 φl kLq (B l ) + kDφl kLq (B l ) + kφl kL∞ (B l ) ≤ Ckφl+1 kL2 (B l+1 ) ≤ Ckφl+1 kL∞ (B l+1 ) ,
m
m
m
m
m
l = 0, 1, · · · , m − 2.
Hence
kφkL∞ (B 1 ) ≤ Ckφm−1 k ≤ Ckhk∗∗ .
m
Without loss of generality, we can write
kφk∗(B1 ) = k(1 + |y|N −2m )φ(y)kL∞ (B1 ) ≤ CkφkL∞ (B1 ) ≤ Ckφk∗∗ .
To complete the estimate outside the unit ball, we make use of the Kelvin type transform
¯ ¯2m−N
e
φ(y)
= ¯y ¯
φ( y 2 ). And a simple algebra shows,
|y|
e − pU p−1 (y)φ(y)
e
(−∆)m φ(y)
¯ ¯2m−N
¯ ¯−(2m+N )
y
y
=¯y ¯
(−∆)m φ( 2 ) − pU p−1 (y) · ¯y ¯
φ( 2 )
|y|
|y|
¯ ¯(p−1)(2m−N ) p−1 y
¯ ¯2m−N
¯ ¯−(2m+N )
y
y
U
( 2 ) · ¯y ¯
φ( 2 )
=¯y ¯
(−∆)m φ( 2 ) − p¯y ¯
|y|
|y|
|y|
i
¯ ¯−(2m+N ) h
y
=¯y ¯
(−∆)m − pU p−1 φ( 2 )
|y|
¯ ¯−(2m+N )
y
=¯y ¯
· h( 2 ) = e
h(y).
|y|
It turns out that
e
(−∆)m φe − pup−1 φ(y)
=e
h(y).
Similarly, by using the Hölder inequality and the local elliptic estimates, we have
³Z ¯ ¯
¯
¯ ´1/q
N +2m− 2N
¯y ¯(N +2m)q−2N ¯hq (y)¯dy
q hk = Ckhk ,
ke
hkLq (B2 ) ≤ C
≤ Ck(1 + |y|)
q
∗∗
B c1
2
and
e L∞ (B ) ≤ Cke
kφk∗(Bc ) = k(1 + |y|N −2m )φkL∞ (B1c ) ≤ Ckφk
hkLq (B2 ) ≤ Ckhk∗∗ .
1
1
Therefore, we get the estimate for φ
kφk∗ ≤ kφk∗B1 + kφk∗Bc ≤ Ckhk∗∗ .
1
This completes the proof of proposition 3.1.
¤
Now we return to the existence and uniqueness of solution ψ for the equation in (3.6), which
can be simplified to
(−∆)m ψ − pU p−1 ψ + (V1 + V2 ) · ψ + M (ψ) = 0,
where
k
³¯ ¯
´³
´
X
p−1
V1 = −p ¯U∗ ¯
− U p−1 1 −
ζj ,
j=1
16
V2 = pU p−1
k
³X
j=1
(3.11)
´
ζj ,
k
k
k
i
X
X
¯ ¯p−1 X
¡
¡
¢
¢h
M (ψ) = −p¯U∗ ¯
1 − ζj φj + 1 −
ζj E − N (
φej + ψ) ,
j=1
and
j=1
(3.12)
j=1
¯
¯p−1 ¡
¯ ¯p−1
¢ ¯ ¯p−1
N (φ) = ¯U∗ + φ¯
U∗ + φ − ¯U∗ ¯ U∗ − p¯U∗ ¯ φ.
(3.13)
Proposition 3.2 There exists some positive constants k0 , C, ρ0 , such that for ∀k ≥ k0 , and φej
satisfying (3.2)−(3.5), with ρ < ρ0 . Then there exists a unique solution ψ = Ψ(φ1 ) to (3.11)
satisfying the symmetries:
ψ(y, y3 , · · · , yl , · · · , yn ) =ψ(y, y3 , · · · , −yl , · · · , yn );
2πj √
ψ(y, y 0 ) =ψ(e k −1 y, y 0 ), j = 1, 2, · · · , k − 1;
¯ ¯2m−N ¡ y ¢
ψ(y) =¯y ¯
ψ
.
|y|2
Moreover
i
h

N
 kψk∗ ≤ C k 1− q + kφ1 k2∗ , if N ≥ 2m + 2;
h
i
 kψk ≤ C log−1 k + kφ k2 , if N = 2m + 1.
∗
1 ∗
And the operator Ψ satisfies the Lipschitz condition
kΨ(φ11 ) − Ψ(φ21 )k∗ ≤ Ckφ11 − φ21 k∗ .
Proof. Notice that
h¡
¢
¡ ¢i¡ y ¢
¢
¡ ¢i¡ ¢ ¯ ¯−(N +2m) h¡
V 1 + V2 · ψ + M ψ
V1 + V 2 · ψ + M ψ y = ¯y ¯
,
|y|2
we digress to a general problem for (3.7) in Proposition 3.1, where h satisfies
h(y, y3 , · · · , yl , · · · , yN ) =h(y, y3 , · · · , −yl , · · · , yN );
2πj √
h(y, y 0 ) =h(e k −1 y, y 0 ), j = 1, 2, · · · , k − 1;
¯ ¯−(N +2m) ¡ y ¢
h
h(y) =¯y ¯
.
|y|2
We claim that (3.7) has a unique bounded solution ψ = T (h) such that there is a constant
C, depending on q and N satisfying
kφk∗ ≤ Ckhk∗ .
Thanks to the results in Proposition 3.1, it is sufficient to check
Z
(h, vl ) =
hvl = 0, ∀ l = 1, 2, · · · , N + 1.
Rn
From the assumption that h is even with respect to y3 , y4 , · · · , yn , and the oddness of vl =
it is natural (h, vl ) = 0 for l = 3, · · · , N.
17
∂U
∂yl ,
For l = 1, 2, we consider the vector integral
#
#
"
"
Z
Z
v1
y1
h(y)
I=
h
= cN
·
dy.
¡
¢N
v2
y2
RN
RN 1 + |y|2 2 −1+m
Let
(z, z 0 ) = (e
2πj √
−1
k
y, y 0 ),
by the invariance of h and the integral I under this change of variables, we know that
"
#
Z
2πj √
2πj √
y1
h(y)
−1
e k
· I =cN
·
· e k −1 dy
N
¡
¢
y2
RN 1 + |y|2 2 −1+m
"
#
Z
z1
h(z)
=cN
·
dz
¡
¢N
z2
RN 1 + |z|2 2 −1+m
=I,
which yields I = 0, since e
2πj √
−1
k
6= 0 for k ≥ 2.
For l = N + 1, we define a function I(λ), for λ > 0, by
Z
N −2m
U (λy)h(y)dy.
I(λ) = λ 2
RN
By changing the variables y 7→ z =
y
,
|y|2
I(λ) =λ
we have
N −2m
2
Z
U (λy)h(y)dy
RN
Z
λy
y
y
U ( 2 )h( 2 )d( 2 )
|y|
|y|
|y|
Z
¡ −1 ¢ N −2m
2
U (λ−1 y)h(y)dy
= λ
=λ
N −2m
2
RN
RN
=I(λ
−1
) := g(λ),
which shows
I 0 (1) = g 0 (1) = −
1 0 1 ¯¯
I( )
= −I 0 (1).
λ2 λ λ=1
Thus
0 = I 0 (1) = (h, vN +1 ),
Up to now we have verified that all conditions of proposition 3.1 are satisfied, hence
T (h) = kψk∗ ≤ Ckhk∗∗ ,
and T is a bounded linear operator.
Now we return to our problem, take h = (V1 + V2 )ψ + M (ψ), then the unique existence of ψ
is reduced to the survey of the fixed point of an operator M from the complete space X to itself,
where X denotes the linear space with bounded norm k · k∗ and all symmetries in Proposition
3.2. For this purpose, we consider the following fixed point problem:
h¡
i
¢
ψ = −T V1 + V2 + M (ψ) := M(ψ), ψ ∈ X.
18
By the uniqueness result of Proposition 3.1 and fact that
ψl (y) = ψ(y, y3 , · · · , −yl , · · · , yN ),
ψj2 (y) = ψ(e
2πj √
−1
k
l = 3, 4, · · · , N ;
¯ ¯2m−N ¡ y ¢
ψN +1 (y) = ¯y ¯
ψ
|y|2
y, y 0 );
satisfy the ψ− equation in (3.7), we obtain that
ψ = ψl = ψj2 = ψN +1 ,
which are exactly the symmetries required in Proposition 3.2.
Throughout the last part of this section, we will prove that M is a contraction mapping.
This crucial conclusion is derived from a series of estimates of V1 ,
Recall
V2 ,
M respectively.
k
³¯ ¯
´³
´
X
p−1
p−1
¯
¯
V1 = −p U∗
−U
1−
ζj ,
j=1
³
then the multiplier 1 −
k
P
j=1
´
ζj
shows that suppV1 ⊂ EXT.
By using the similar arguments as in the discussion of Step 1 in Proposition 2.3, for y ∈
EXT, there exists s ∈ (0, 1) such that
¯
¯
¯ ¯
¡
¢
1
¯
¯V1 (y)ψ(y)¯ =¯¯V1 (y)ψ(y) 1 + |y|N −2m ·
¯
1 + |y|N −2m
¯
¯ ¯
¯
¯ ¯N −2m
¯
¯ ¯
¯
≤C ¯V1 (y)U (y)¯ · ¯1 + ¯y ¯
ψ(y)¯
¯
¯
¯
¯
≤Ckψk∗ ¯V1 (y)U (y)¯
¯
¯ ¯p−1 ¯¯
¯ p−1
≤Ckψk∗ ¯U
(y) − ¯U∗ ¯ (y)¯U (y)
k
k
¯
¯p−2 h X
i
X
¯
¯
=Ckψk∗ U (y)¯U (y) − s
Uj (y)¯
Uj (y) .
j=1
Note that EXT = I
k
X
S
j=1
II. For y ∈ I, we have µ2 + |y − ξj |2 ∼ 1 + |y|2 , hence
Ui (y) ≤Ckµ
i=1
≤
N −2m
2
U (y)

 Ck 2m+1−N U (y) ≤ CU (y),
N ≥ 2m + 2;
 Ck − 21 log−1 k · U (y) ≤ CU (y),
N = 2m + 1.
For y ∈ II, we have
k
X
i=1
Ui (y) =
k
X
µ
N −2m
2
³
i=1
N −2m
´ N −2m
1
2
2
2
µ + |y − ξi |
N −2m
≤Ck · k 2 µ 2

 Ck 2m+2−N
2
≤ C ≤ CU (2) ≤ CU (y),
≤
1
 C(log k)− 2 ≤ CU (2) ≤ CU (y),
19
N ≥ 2m + 2;
N = 2m + 1.
Combine the obtained results for I and II, we have
k
X
Ui (y) ≤ CU (y),
∀y ∈ EXT.
(3.14)
i=1
Put (3.14) into the estimate of |V1 ψ|, we obtain, for y ∈ EXT,
k
¯
¯
¡X
¢
¯V1 (y)ψ(y)¯ ≤Ckψk∗ U p−1 (y) ·
Ui (y)
i=1
≤Ckφk∗
¡
k
N −2m
¢2m X
1
µ 2
.
2
(1 + |y|)
|y − ξi |N −2m
i=1
At last by using the similar arguments as in the the Step 2 of Proposition 2.3, we know that

 Ck 1− Nq ,
N ≥ 2m + 2;
kV1 · ψk∗∗ ≤
 C log−1 k,
N = 2m + 1.
Now we turn to the estimate of V2 · ψ. Recall
V2 = pU
p−1
k
³X
´
ζj .
j=1
The multiplier
³P
k
j=1
´
ζj
shows that the support of V2 · ψ lies in the annular region, namely,
¯¯
p
¯ η
¯
suppV2 ⊂ IN T ⊂ {y ¯¯|y| − 1 − µ2 ¯ ≤ } := AN N.
k
By the argument of measures, we obtain
kV2 ψk∗∗ ≤Ckψk∗
k
X
kU p ζj k∗∗(|y−ξj |≤η/k)
j=1
¡
¢
≤Ckψk∗ · meas AN N
≤Ckψk∗ · k 1−N < Ckψk∗ k
1− N
q
.
The discussion of M (ψ) will be more technical, since the operator M is not linear as V1 and
V2 are. We introduce the following notations to simplify the presentation of M (ψ).
Define
k
¯ ¯p−1 X
¯
¯
M1 = −p U∗
(1 − ζj )φj ;
M2 = (1 −
j=1
M3 (ψ) = −(1 −
k
X
j=1
k
X
k
X
ζj )N (
φej + ψ).
j=1
j=1
Then the nonlinear operator M (ψ) can be rewritten as
M (ψ) = M1 + M2 + M3 (ψ).
20
ζj )E;
For M1 , apply the estimate of the exterior region, we have
kM1 k∗∗ ≤C
k
X
¯ ¯p−1
k¯U∗ ¯ φj k∗∗
j=1
≤
(|y−ξj |>η/k)

 Ckψk∗ k 1− Nq ,
N ≥ 2m + 2;
 Ckψk log−1 k,
∗
N = 2m + 1.
The estimate for M2 is the same as that for the error term E.
For M3 (ψ), we have
suppM3 (ψ) ⊂ EXT.
Recall the formula of N , by means of the mean value theorem, there exist s, t ∈ (0, 1) such
that
k
k
k
k
X
X
¯ X
¯ ¯¯¯
¯p−1 ¡
¯ ¯p−1 ¡ X
¢ ¯ ¯p−1
¢¯¯
¯N (
φej + ψ)¯ =¯¯U∗ +
φej + ψ ¯
· U∗ +
φej + ψ − ¯U∗ ¯
· U∗ − p¯U∗ ¯
φej + ψ ¯
j=1
j=1
j=1
j=1
k
k
¯¯
X
¯p−1 ¯ ¯p−1 ¡ X
¢¯¯
¯
=p¯¯U∗ + s(
− ¯U∗ ¯
φej + ψ ¯
φej + ψ)¯
j=1
j=1
k
X
¯
=p¯U∗ + ts(
¯p−2 ¯
φej + ψ)¯ ¯
j=1
k
X
¯2
φej + ψ ¯
j=1
We restricted N to the exterior region, by the proof of Step 1 in the Proposition 2.3, we
have



 Ckφ1 k∗ U (y);
k
¯X
¯ 
N −2
k
¯
ψej (y)¯ ≤ X
µ 2

.

j=1


|y − ξ |N −2
j=1
j
Hence
k
X
kM3 (ψ)k∗∗ =k(1 −
k
X
ζj )N (
φej + ψ)k∗∗
j=1
k
X
≤CkN (
j=1
φej + ψ)k∗∗(EXT )
j=1
k
k
X
£
¤p−1 £ X
¤
≤Ck |U∗ | + |
φej | + |ψ|
· |
φej |2 + |ψ|2 k∗∗(EXT )
j=1
≤Ckφ1 k2∗ kU p−1 ·
j=1
k
X
j=1
N −2
2
µ
k∗∗
+ Ckψk2∗ kU p k∗∗(EXT )
|y − ξj |N −2 (EXT )

 Ckφ1 k2 k 1− Nq + Ckψk2 , N ≥ 2m + 2
∗
∗
≤
2
−1
 Ckφ k log k + Ckψk2 , N = 2m + 1.
1 ∗
∗
21
Summing together the obtained estimates above, we have
kM (ψ)k∗∗ ≤kM1 k∗∗ + kM2 k∗∗ + kM3 (ψ)k∗∗
 ¡
¢
 C k 1− Nq + kφ1 k2 k 1− Nq + kψk2 ,
∗
∗
≤
 C ¡ log−1 k + kφk2 log−1 k + kψk2 ¢,
∗
∗
N ≥ 2m + 2
N = 2m + 1.
On the other hand, for any ψ1 , ψ2 ∈ Bρ/2 ⊂ X, by the mean value theorem, there exists some
s, t ∈ (0, 1) such that
kM (ψ1 ) − M (ψ2 )k∗∗ =kM3 (ψ1 ) − M3 (ψ2 )k∗∗
k
k
X
X
e
φej + ψ2 )k∗∗(EXT )
≤CkN (
φj + ψ1 ) − N (
j=1
¯
=Ckp¯U∗ +
j=1
k
X
¯p−1
¯ ¯p−1
φej + s(ψ1 − ψ2 )¯ (ψ1 − ψ2 ) − p¯U∗ ¯ (ψ1 − ψ2 )k∗∗(EXT )
j=1
k
k
X
¯
¯ ¯
¯p−2 ¯ X
¯
e
¯
¯
¯
·
φej + s(ψ1 ψ2 )¯ · ¯ψ1 − ψ2 ¯k∗∗(EXT )
φj + ts(ψ1 − ψ2 )
≤Ck U∗ + t
¡
j=1
j=1
¢
≤C kφ1 k∗ + kψ1 − ψ2 k∗ kψ1 − ψ2 k∗ · kU p k∗∗(EXT )
≤Cρkψ1 − ψ2 k∗ .
More generally, we have
£
¡
¢¤
kM(ψ1 − ψ2 )k∗ =k − T (V1 + V2 ) · (ψ1 − ψ2 ) + M (ψ1 − ψ2 ) k∗
h ¡
i
¢
≤C k V1 + V2 · (ψ1 − ψ2 )k∗∗ + kM3 (ψ1 ) − M3 (ψ2 )k∗∗

 C(k 1− Nq + ρ)kψ1 − ψ2 k∗ , N ≥ 2m + 2;
≤
 C(log−1 k + ρ)kψ − ψ k , N = 2m + 1.
1
2 ∗
Choosing k0 large enough and ρ0 small enough, then for any k ≥ k0 and ρ ≤ ρ0 , it holds:

 C(k 1− Nq + ρ) < 1,
 C(log−1 k + ρ) < 1.
Hence M is a contraction mapping from the small ball in X to the ball itself. The Banach fixed
point theorem gives the unique existence of ψ.
3.2
¤
The existence of φ̃j in (3.6)
In this subsection, we will turn to study the first series of equations (3.6):
h
i
X
(−∆)m φej − p|U∗ |p−1 ζj φej + ζj − p|U∗ |p−1 ψ + E − N (φej +
φei + ψ) = 0,
j = 1, · · · , k;
i6=j
Indeed, these equations can all be reduced to the φ1 − equation by means of the changing of the
variables, that is the equation:
22
h
i
X
(−∆)m φe1 − p|U∗ |p−1 ζ1 φe1 + ζ1 − p|U∗ |p−1 ψ + E − N (φe1 +
φei + ψ) = 0.
i6=1
In order to simplify the horrible formula above, we introduce the following new notation N , e
h:
h
i
X
¡
¢
N (φ1 ) := p |U1 |p−1 − |U∗ |p−1 ζ1 φe1 + ζ1 − p|U∗ |p−1 ψ + E − N (φe1 +
φei + ψ) ;
i6=1
e
h := ζ1 E + N (φ1 );
where e
h is even with respect to each of the variables y2 , y3 , · · · , yN and satisfies
h̃(y) = |y|−(N +2m)e
h(
y
).
|y|2
Then the φ1 − equation can be simplified as
h
Recall the definition of µ,
¯ ¯p−1 i
(−∆)m − p¯U1 ¯ ζ1 φe1 + e
h = 0.
(3.15)

2
 µ = δ N −2m
k −2 ,
N ≥ 2m + 2,
 µ = δ 2 k −3 log−2 k,
N = 2m + 1.
We know that µ is relevant to δ, hence
R
cN +1 (δ) :=
RN (ζ1 E
R
RN
+ N (φ1 ))e
vN +1
2
U1p−1 veN
+1
R
=R
RN
e
he
vN +1
p−1 2
U1 veN +1
RN
is also a variable relevant to δ.
By the argument of changing variables through translating and scaling, we can get the
equivalence between equation (3.15) and equation (3.7).
Consider the result of Proposition 3.1, it is evident to assert that, the unique existence of φe1
is equivalent to the verification of the following series of conditions
Z
Z
e
he
vl =
hvl = 0, l = 1, 2, 3, · · · , N + 1.
RN
RN
From the formula of cN +1 (δ), we know one of the conditions holds
Z
Z
e
hvN +1 = 0 ⇔ cN +1 (δ) = 0,
he
vN +1 = 0 ⇔
RN
RN
under a selected positive number δ.
And the existence of this particular δ is granted by the following lemma (see [24])
R
Lemma 3.3 We can write the δ related RN e
he
vN +1 in the following form as
Z
RN
e
he
vN +1 =



 AN


 A3
δ
£
¤
δaN − 1 +
1
Θk (δ),
N ≥ 2m + 2;
¤
δ £
1
δa3 − 1 + 2
Θk (δ),
k log k
k log2 k
N = 2m + 1,
k N −2m
23
k N −m
where Θk (δ) is continuous w.r.t δ and uniformly bounded as k → ∞, AN = p
R
RN
U p−1 vN +1 ,
with the positive number

k
X

N −2m
1
1


2
2
, N ≥ 2m + 2;
lim N −2m


k→∞ k
|ξ − ξj |N −2m

j=2 1
aN =
k

√

1 X
1


lim
, N = 2m + 1.

 2 k→∞
k log k
|ξ1 − ξj |
j=2
R
he
vN +1 < 0, while for δ large
In fact, by Lemma 3.3, we can see that for δ small enough, RN e
R
enough, RN e
he
vN +1 > 0. By the continuity arguments with respect to δ, we can always find some
R
he
vN +1 = 0.
δ0 > 0 such that RN e
For the simplified version (3.15), and for this particular δ0 , we give the following theorem to
complete the unique existence problem of splitting system (3.6).
h given above, assume in addition that
Proposition 3.4 For e
h(y) := µ
N +2m
2
e
h(ξ1 + µy)
satisfying
khk∗∗ < ∞.
Then the equation (3.15) has a unique solution φe := Te(e
h) that is even with respect to each of
the variables y2 , y3 , · · · , yN , and is invariant under the Kelvin type transform:
¯ ¯2m−N ¡ y ¢
e
φe
,
φ(y)
= ¯y ¯
|y|2
and
Z
RN
where φ(y) = µ
N −2m
2
φU p−1 vN +1 = 0,
with
kφk∗ ≤ Ckhk∗∗ .
e 1 + µy).
φ(ξ
Proof. By Lemma 3.3, we have
Z
RN
hvN +1 = 0.
It follows from the oddness of v2 , v3 , · · · , vN and the evenness of h that
Z
hvj = 0, j = 2, 3, · · · , N.
RN
In the following, we only need to prove that
R
RN
hv1 = 0. However, this needs some toiling work,
since we do not have any symmetry of h with respect to the first component y1 .
Define an integral I(t) and wµ (y) as
Z
wµ (y − tξ1 )e
h(y)dy,
I(t) :=
RN
24
wµ (y) = µ−
N −2m
2
U (µ−1 y).
Then the direct calculus gives,
Z
¯
N −2m
N +2m
ξ1
y − tξ1
y − ξ1
0
I (1) = −
· U(
) · h(
) · µ− 2 · µ− 2 dy ¯t=1
µ
µ
RN µ
p
Z
2
1−µ
U1 (y)h(y)dy
=−
µ
RN
p
Z
1 − µ2
=−
hv1 .
µ
RN
Changing the variable y 7→ z =
y
,
|y|2
we obtain
Z
I(t) =
ZR
N
=
RN
Z
=
ZR
N
=
RN
(3.16)
wµ (y − tξ1 )|y|−(N +2m)e
h(
y
)dy
|y|2
¯ ¯2m−N
y
e
− tξ1 )¯z ¯
h(z)dz
2
|y|
³
´ N −2m N −2m h¯
i− N −2m
¯2
tξ1
µ2
µ
2
2
4m
e
¯y −
¯ +
P
·
h(y)dy
m,N
2
2
2
2
2
2
2
2
2
2
µ + t |ξ1 |
µ + t |ξ1 |
(µ + t |ξ1 | )
wµ (
wµ(t) (y − s(t)ξ1 )e
h(y)dy,
where
µ(t) :=
µ2
µ
,
+ t2 |ξ1 |2
s(t) =
µ2
t
.
+ t2 |ξ1 |2
Take the derivative on both sides of the formula of I(t), we get
Z
hZ
i¯
¡
¢
¯
0
0
1
e
I (1) =
∂µ wµ (y − s(t)ξ) · h(y)dy · µ (t) − ξ1
∂y1 wµ(t) (y − s(t)ξ1 ) e
h(y)dys0 (t) ¯
t=1
N
N
R
R
p
Z
Z
1 − µ2
(2µ2 − 1)
hv1
=2µ2
veN +1 (y)e
h(y)dy −
µ
N
RN
p R
Z
1 − µ2
=
(2µ2 − 1)
hv1 .
µ
RN
(3.17)
Compare (3.16) and (3.17), we have
p
p
Z
Z
1 − µ2
1 − µ2
0
2
hv1 = I (1) =
hv1 ,
−
(2µ − 1)
µ
µ
RN
RN
R
and the equality holds if and only if RN hv1 = 0, this is what we need at last.
Apply Proposition 3.1, we see that if e
h is a general function satisfying all the symmetries in
Proposition 3.2, then there exists some unique solution φe := Te(e
h) that is even with respect to
each of the variables y2 , y3 , · · · , yN and
e ∗ = kTee
kφk
hk∗ ≤ Cke
hk∗∗ .
However, this e
h in our problem (3.15) is not general, but relevant to φ1 itself, which tells us
the direct application of the Proposition 3.1 can not work. We need to emulate what we have
done in Proposition 3.2, that is to construct a contraction mapping, then the Banach fixed point
25
theorem will give the answer to our problem. Since e
h = ζ1 E + N (φ1 ), we define an operator M
by
¡
¢
M(φ1 ) := Te ζ1 E + N (φ1 ) .
Then the unique existence of the solution φ1 is reduced to the existence of a fixed point of a
contraction mapping M.
In the following, we split ζ1 E + N (φ1 ) into several shorter terms and estimate these terms
one by one. Define
¡
¢
f1 := pζ1 U1p−1 − |U∗ |p−1 · φe1 ;
¯ ¯p−1
f3 := −pζ1 ¯U∗ ¯ ψ(φ1 );
f4 := ζ1 N
f2 := (1 − ζ1 )U1p−1 φe1 ;
k
¡X
¢
φej + ψ(φ1 ) ;
f5 := ζ1 E,
j=1
and
n+2m
fei (y) = µ 2 fi (ξ1 + µy), i = 1, 2, 3, 4, 5.
Set
e
h=
5
X
fi .
i=1
Thanks to the multiplier ζ1 , we know that
¯
suppfj ⊂ {y ¯|y − ξ1 | < η/k} =: IN T1 ⊂ IN T,
j = 1, 3, 4, 5.
For f1 , we have
k
¯ ¯
¯ ¯
X
¯p−1
¯
¡
¢
N −2m
¯
¯
|fe1 (y)| ≤¯p¯U (y) +
U y + µ−1 (ξ1 − ξj ) − µ 2 U (ξ1 + µy)¯
− pU p−1 (y)¯ · ¯φ1 (y)¯
j=2
k
¯X
¯p−2
¡
¢
N −2m
≤C ¯
U y + µ−1 (ξ1 − ξj ) + µ 2 U (ξ1 + µy) + U (y)¯
j=2
¯ N −2m
¯
· ¯µ 2 U (ξ1 + µy) + U (y)¯ · U (y) · kφ1 k∗
N −2m
≤Ckφ1 k∗ U
p−1
(y) · µ
N −2m
2
µ 2
≤ Ckφ1 k∗
,
1 + |y|4m
Thus we can proceed the same discussion of Step 2 in Proposition 2.3 and obtain

 Ckφ1 k∗ k − Nq ,
N ≥ 2m + 2,
kf1 k∗∗ = kf1 k∗∗(IN T1 ) ≤
 Ckφ k log−1 k,
N = 2m + 1.
1 ∗
(3.18)
Similarly, with the application of the estimate of ψ in Proposition 3.2, we can get the estimate
of f3 , f4 , for y ∈ IN T1 :
¯
¯
¯fe3 (y)¯ ≤CU p−1 µ N −2m
2
kψ(φ1 )k∞
N −2m
≤CU p−1 µ 2 kψ(φ1 )k∗

¢
N −2m ¡ 1− N
1
2


 Cµ 2 k q + kφ1 k∗ · 1 + |y|4m , N ≥ 2m + 2;
≤
¢
N −2m ¡
1


 Cµ 2
log−1 k + kφ1 k2∗ ·
, N ≥ 2m + 1,
1 + |y|4m
26
and
 ¡
¢
 C k 1− Nq + kφ1 k2 , N ≥ 2m + 2;
∗
≤
 C ¡ log−1 k + kφ k2 ¢, N = 2m + 1,
1 ∗
kf3 k∗∗
kf4 k∗∗

 Ckφ1 k2 k 1− Nq + Ckψk2 , N ≥ 2m + 2;
∗
∗
≤
 Ckφ k2 log−1 k + Ckψk2 , N ≥ 2m + 1,
1 ∗
∗

¡
¢
N
1−
 Ckφ1 k2 k q + C kφ1 k2 + k 1− Nq 2 , N ≥ 2m + 2;
∗
∗
≤
 Ckφ k2 log−1 k + C ¡kφ k2 + log−1 k ¢2 , N ≥ 2m + 1.
1 ∗
(3.19)
(3.20)
1 ∗
The estimate of f5 can be directly derived from the error term E

n
 Ck 1− q , N ≥ 2m + 2;
kf5 k∗∗ ≤
 C log−1 k, N = 2m + 1.
(3.21)
For f2 , we know that
¯
¯
¯ ¯
¯fe2 (y)¯ = ¯ζ1 (µy + ξ1 ) − 1¯ · U p−1 · |φ1 | ≤ CU p kφ1 k∗ ,
hence
hZ
kf2 k∗∗ ≤C
|y−ξ1 |>η/k
h
¯ q n − ξ1 ¯ i1/q
¯fe (
)¯dy
2
µ
i1/q
r(N +2)q−2N −(N −2)pq dr
(1 + |y|)(N +2m)q−2n µ−
− N +2m
q (N +2m)q−2N
2
≤C µ
µ
Z
+∞
·
N +2m
q
2
(3.22)
η/(kµ)
≤Cµ
N +2
1
− 2q
2
< Ck
−N
q
.
b φb1 , φb2 ∈ Bρ (0) ⊂ X,
Sum all the estimates obtained above, we can see that for φ,
 ¡
¢
5
 C k 1− Nq + kφk
b ∗ , N ≥ 2m + 2;
X
b ∗ ≤C
b ∗∗ ≤
kM(φ)k
kfi (φ)k
¢
 C ¡ log−1 k + kφk
b , N ≥ 2m + 1.
∗
i=1
and
kM(φb1 ) − M(φb2 )k∗ ≤C
4
X
kfi (φb1 ) − fi (φb2 )k∗∗
i=1
 ¡ N
¢
 C k − q + kφb1 k∗ + kφb2 k∗ kφb1 − φb2 k∗ , N ≥ 2m + 2;
≤
 C ¡k − Nq + k −1 log−1 k + kφb k + kφb k ¢kφb − φb k , N = 2m + 1,
1 ∗
2 ∗
1
2 ∗
=:J kφb1 − φb2 k∗ ,
with J < 1.
Hence M is a contraction mapping from Bρ (0) to Bρ (0), for k is large enough and ρ is small
enough. By the Banach fixed point theorem, there exists a unique solution φe1 of the equation
(3.15).
¤
27
3.3
The Proof of the Main theorem
Since we are looking for the solution with the form:
u(y) = U∗ (y) + φ(y),
with u being this form, our equation (1.5) can be restated as (2.7). Then for φ =
Pk
j=1 φ̃j
+ ψ,
by means of the cut-off functions, we split the equation (2.7) into a system equation of φ̃j , j =
1, 2, ..., k and ψ (see (3.6)). In this way, the original problem is reduced to prove the existence of
ψ and φ̃j , j = 1, 2, ..., k. These are done in Subsection 3.1 and Subsection 3.2 , respectively. Thus
k
P
for any k ≥ k0 , we get the sign-changing solution uk = U∗ +
φj + ψ for the poly-harmonic
j=1
equation (1.5). This completes the proof of the main theorem.
¤
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