INFINITELY MANY POSITIVE SOLUTIONS FOR AN NONLINEAR
FIELD EQUATION WITH SUPER-CRITICAL GROWTH
MONICA MUSSO, JUNCHENG WEI, AND SHUSEN YAN
Abstract. We consider the following nonlinear field equation with super critical growth:
(
N +2
−∆u + λu = Q(y)u N −2 , u > 0 in RN +m ,
(∗)
u(y) → 0
as |y| → +∞,
where m ≥ 1, λ ≥ 0 and Q(y) is a bounded positive function. We show that equation (*)
has infinitely many positive solutions under certain symmetry conditions on Q(y).
1. Introduction
In this paper, we consider the following nonlinear field equation with super critical
growth:
(
N +2
−∆u + λu = Q(y)u N −2 , u > 0
u(y) → 0
in RN +m ,
as |y| → +∞,
(1.1)
where m ≥ 1, λ ≥ 0 and Q(y) is a bounded positive function. Note that
N +2
N −2
is a super
critical exponent in RN +m .
Equation (1.1) is a special case of the following problem:
(
−∆u + λu = Q(y)up , u > 0
u(y) → 0
in RN ,
as |y| → +∞,
where p > 1. If λ > 0, (1.2) is the field equation. When p <
(1.2)
N +2
,
N −2
existence of the
ground state solution, or positive solutions with higher energy for the field equations were
+2
considered in [4, 5, 7, 13, 20, 26, 27]. If λ = 0 and p = N
, (1.2) is the prescribed scalar
N −2
N
curvature problem on S , which was studied extensively in the last thirty years. See for
example [2, 3, 6, 8, 9, 10, 11, 12, 21, 22, 23, 24, 25, 38, 40] and the references therein. For
the super-critical case, no result is known for (1.2), except for the case λ = 0 and p is very
N +2
close to N
. See [39].
−2
The aim of this paper is to prove that under some conditions on Q, (1.1) has infinitely
many positive solutions. We will achieve this goal by constructing solutions concentrating
1
2
MONICA MUSSO, JUNCHENG WEI, AND SHUSEN YAN
along large number of m-dimensional manifolds. As far as we know, this seems to be the
first result on the infinite multiplicity for nonlinear field equation with super crtical growth.
To simplify (1.1), we impose the following symmetry condition on Q(y):
(Q): Suppose that Q(y) = Q(r, d), where y = (y ∗ , y ∗∗ ), y ∗ ∈ RN −1 , y ∗∗ ∈ Rm+1 , r = |y ∗ |
and d = |y ∗∗ |.
In this paper, we assume that there is a pair (r0 , d0 ), r0 > 0, d0 > 0, such that (r0 , d0 ) is
dm
. That is, the pair (r0 , d0 ) satisfies
Qr (r0 , d0 ) = 0,
(1.3)
a non-degenerate critical point of the function
Q
N −2
2 (r,d)
N −2
mQ(r0 , d0 )
Qd (r0 , d0 ) =
,
(1.4)
2
d0
dm
and that the 2 × 2 matrix (D2 N −2
) at (r0 , d0 ) is invertible. Moreover, we assume
Q
2
(r,d)
m 3(N − 2) (N − 4)(N − 2) Qrr (r0 , d0 ) + Qdd (r0 , d0 )
−
d20 4(N − 1)
8(N − 1)
Q(r0 , d0 )
2
m
5N − 14
+ 2
> 0.
d0 4(N − 1)(N − 2)
λ−
(1.5)
Our main result in this paper can be stated as follows:
Theorem 1.1. Suppose that N ≥ 5 and Q satisfies the symmetry condition (Q). Assume
dm
has a non-degenerate critical point (r0 , d0 ), satisfying r0 > 0,
that the function N −2
Q
2
(r,d)
d0 > 0, and (1.5). Then problem (1.1) has infinitely many distinct positive solutions.
In the end of the introduction, let us outline the proof of Theorem 1.1. We will construct
solutions which concentrate at large number of spheres and we will use the number of the
sphere as the parameter in the construction. This technique was first developed in [35] to
study the prescribed scalar curvature problem and then was used to study other elliptic
problems [33, 34, 36, 37].
Denote 2∗ = N2N
. It is well-known that the functions
−2
¶ N2−2
µ
¡
¢ N4−2
µ
, µ > 0, x ∈ RN
N (N − 2)
1 + µ2 |y − x|2
are the only solutions to the following problem
∗ −1
−∆u = u2
, u > 0 in RN .
INFINITELY MANY SOLUTIONS
3
For any positive integer k, let
¡
¢
2(j − 1)π
2(j − 1)π
Γj = r0 cos
, r0 sin
, 0, · · · , 0, y ∗∗ , |y ∗∗ | = d0 .
k
k
Then Γj is an m-dimensional sphere in RN +m . Near Γj , we denote
µ
¶ N2−2
¡
¢ N4−2 − N −2
µ
wj (y) = N (N − 2)
Q 4 (r0 , d0 )
,
1 + µ2 |y ∗ − x̃j |2 + µ2 (|y ∗∗ | − d0 )2
where
¡
¢
2(j − 1)π
2(j − 1)π
x̃j = r0 cos
, r0 sin
, 0, · · · , 0 ∈ RN −1 .
k
k
Then, if µ is large, wj is a function concentrating at the sphere Γj . In this paper, we will
prove that for k > 0 large, (1.1) has a solution ui with
ui ≈
k
X
wj ,
j=1
for some large µ.
If m = 0, it was shown in [35] that if the potential Q has a non-degenerate local minimum
point (r0 , 0) with r0 > 0, one can construct positive solutions with large number of bubbles.
These conditions correspond to (1.3), (1.4) and (1.5) in the case m = 0. But there are some
striking differences between the case m = 0 and m > 0. When m = 0, the construction can
be carried out starting from wj as first approximation of the solution. In fact, in this case,
the condition DQ(r0 , 0) = 0 gives that this approximation is good enough to construct
bubbling solutions. On the other hand, when m > 0, we have Qd (r0 , d0 ) 6= 0. This is the
reason why, in this case, if we just consider wj as starting point for our construction, the
error of the approximation is too big, being of order µ−1 . This implies that the contribution
from the perturbation would be of order µ−2 . The calculations show that one needs to find
the exact formula up to order µ−2 in order to be able to determine the concentration rate
µ of the bubbling solutions. This fact forces us to find explicitly the second term in the
expansion of the approximate solution in the case m > 0. Indeed, the second term in the
expansion of the approximate solution is not negligible and it contributes to the constant
in the left hand side of (1.5).
Using the symmetry of the function Q, we will look for solution of the form u =
u(y ∗ , |y ∗∗ |). Another major difference between the case m = 0 and m > 0 is that if
m > 0, though the problem can be reduced to a problem in RN , the reduced equation
4
MONICA MUSSO, JUNCHENG WEI, AND SHUSEN YAN
has singularity at |y ∗∗ | = 0. This will create some difficulty in carrying out the reduction
argument. Note that in [37, 1], to avoid such difficulty, it requires that the domain does
not intersect the hyper-plane y ∗∗ = 0. Arguing in the same way as in this paper, we can
get rid of this condition imposed on the domains in [37, 1].
The existence of positive solutions for elliptic problems involving super-critical nonlinearities is a very delicate problem. See for example [1, 14, 15, 16, 19, 31, 32, 37] for the results
for problems in bounded domains. In [1, 16, 17, 37], existence results were obtained by
constructing solutions concentrating at curves or higher dimensional manifolds at higher
critical Sobolev exponents in bounded domains. The readers can find other results on
solutions concentrating at curves or higher dimensional manifolds in [18, 28, 29, 30].
This paper is arranged as follows. In section 2, we will construct the second approximate
of the solutions. In section 3, we will find the equations which determine the location and
the concentration rate of the bubbles by neglecting the perturbation term. Section 4 is
devoted to the study of a linear problem in a cylinder which plays a crucial role in carrying
out the reduction argument in Section 5. Theorem 1.1 is proved in Section 6. We put all
the other estimates needed in the proof of the main theorem to appendixes.
Acknowledgment. M.Musso is partially supported by Fondecyt Grant 1120151 and
CAPDE-Anillo ACT-125, Chile. J.Wei is partially supported by NSERC of Canada
S.Yan is partially supported by ARC.
2. The approximate solutions near a given curve
In this section, we construct approximate solutions for (1.1) concentrating at the mdimensional sphere Γj , where
¡
¢
2(j − 1)π
2(j − 1)π
Γj = r0 cos
, r0 sin
, 0, · · · , 0, y ∗∗ , |y ∗∗ | = d0 .
k
k
By the rotational invariance of the problem, we just need to construct an approximate
solutions for (1.1) concentrating at Γ1 .
Using the symmetry condition on Q, we can look for solutions of the form u(y ∗ , |y ∗∗ |)
for (1.1). Then u(y ∗ , t), t = |y ∗∗ |, satisfies
N +2
m
ut + λu = Q(|y ∗ |, t)u N −2 ,
t
where ∆ is the Laplace operator for (y1 , · · · , yN −1 , t).
−∆u −
(2.1)
INFINITELY MANY SOLUTIONS
5
Remark 2.1. Equation (2.1) can be written as the following divergent form
N +2
−div(tm Du) + λtm u = tm Q(|y ∗ |, t)u N −2 .
N −2
For large positive integer k, we let ε = k − N −4 .
ε
N −2
2
(2.2)
We make the change of variable
∗
u(εy , εt). Then (2.1) becomes
N +2
m
(2.3)
ut + ε2 λu = Q(ε|y ∗ |, εt)u N −2 .
t
To simplify the notations in the calculations, we use yN to replace t in (2.3). So we have
−∆u −
−∆u −
N +2
m
uyN + ε2 λu = Q(εy ∗ , εyN )u N −2 .
yN
(2.4)
For any constant β > 0, let
α = Q−
N −2
4
(r0 , d0 )β
N −2
2
.
(2.5)
We define v(x) by the relation
u(y) = αv(x),
(2.6)
where y = (y1 , · · · , yN ),
xi = βyi ,
r0
− f1 ),
ε
are small parameters.
x1 = β(y1 −
and f1 and fN
i = 2, · · · , N − 1,
xN = β(yN −
d0
− fN ),
ε
We expand
µ
¶
xN
Q(εy) =Q(r0 , d0 ) + Qd (r0 , d0 )ε
+ fN
β
¢
¡
x
x
1
+ ε2 D2 Q(r0 , d0 )( + f, + f ) + O ε3 |y|3 ,
2
β
β
since Qr (r0 , d0 ) = 0. Here we let fi = 0, i = 2, · · · , N − 1.
Let
S(u) = ∆u +
m
∗
uyN − ε2 λu + Q(εy)u2 −1 .
yN
(2.7)
6
MONICA MUSSO, JUNCHENG WEI, AND SHUSEN YAN
Then S(u) = Q−
N −2
4
(r0 , d0 )β
N +2
2
S(v), where
∗ −1
S(v) ≡ B1 (v) + B2 (v) + ∆v + v 2
.
(2.8)
Here ∆ denotes the Laplace operator for y, B1 (v) is a linear differential operator defined
by
B1 (v) = −β −2 ε2 λv +
mε
vx ,
βd0 + εxN + εβfN N
(2.9)
and
·
µ
¶
µ
¶¸
Qd (r0 , d0 ) xN
ε2 D2 Q(r0 , d0 ) x
x
∗
B2 (v) = ε
+ fN +
+ f, + f
v 2 −1
Q(r0 , d0 )
β
2 Q(r0 , d0 )
β
β
¡ 3 3 2∗ −1 ¢
+ O ε |x| |v|
.
The major term of an approximate solution concentrating along Γε,1 :=
¡
w(x) = N (N − 2)
¢ N4−2
µ
1
1 + |x|2
¶ N2−2
Γ1
ε
(2.10)
is given by
.
We have
S(w) =B2 (w) + B3 (w)
·
µ
¶¸
ε 2 m xN
−1 εm
=β
− 2
+ fN
wxN − β −2 ε2 λv
d0
d0
β
·
µ
µ
¶¸
¶
Qd (r0 , d0 ) xN
x
ε2 D2 Q(r0 , d0 ) x
∗
+ ε
+ fN +
+ f, + f
w2 −1
Q(r0 , d0 )
β
2 Q(r0 , d0 )
β
β
¡ 3 3 2∗ −1
¢
+ O ε |x| |w|
+ ε3 |x|2 |wyN | .
(2.11)
Write
¡
¢
∗
S(w) = εS1 + εS2 + ε2 S3 + ε2 S4 + O ε3 |x|3 |w|2 −1 + ε3 |x|2 |wyN | ,
where
S1 = β −1
¢
¡ mwxN
Qd (r0 , d0 )
∗
+
xN w2 −1 ,
d0
Q(r0 , d0 )
S3 = −
and
mfN wxN
βd20
Qd (r0 , d0 )
∗
fN w2 −1 ,
Q(r0 , d0 )
¶
µ
D2 Q(r0 , d0 ) x
∗
+
, f w2 −1
Q(r0 , d0 )
β
S2 =
(2.12)
(2.13)
INFINITELY MANY SOLUTIONS
7
mxN wxN
− β −2 λw
β 2 d20
(2.14)
´ ∗
1 ³ D2 Q(r0 , d0 )
D2 Q(r0 , 0, εz)
2 −1
+
(x, x) +
(f, f ) w
.
2 β 2 Q(r0 , d0 )
Q(r0 , d0 )
Note that S1 and S3 are odd, while S2 and S4 are even.
Next, we will make a correction for the first approximate w. We will choose φ, such that
S4 = −
S(w + εφ) is of order ε2 . For this aim, we calculate
S(w + εφ) =B2 (w + εφ) + B3 (w + εφ) + εL0 (φ) + N0 (εφ)
=S(w) + εL0 (φ) + N0 (εφ) + B2 (w + εφ) − B2 (w) + εB3 (φ),
where
∗ −2
L0 (φ) = ∆φ + (2∗ − 1)w2
φ
(2.15)
and
∗ −1
N0 (φ) = (w + φ)2
∗ −1
− w2
∗ −2
− (2∗ − 1)w2
φ.
(2.16)
Consider the following problem:
∗ −2
−∆φ − (2∗ − 1)w2
φ = S1 + S2 .
(2.17)
We can write down the solutions for (2.17). First, we consider
¢
¡ mwxN
Qd (r0 , d0 )
∗
+
xN w2 −1
d0
Q(r0 , d0 )
¢
Qd (r0 , d0 ) ¡ N − 2 1 0
∗
=β −1
xN
w + w2 −1 ,
Q(r0 , d0 )
2 r
∗ −2
− ∆φ − (2∗ − 1)w2
φ = S1 = β −1
(2.18)
d (r0 ,d0 )
where r = |x|. For (2.18), we let φ = β −1 QQ(r
xN ϕ(r). Then
0 ,d0 )
−ϕ00 −
N +1 0
N −21 0
∗
∗
ϕ − (2∗ − 1)w2 −2 ϕ =
w + w2 −1 .
r
2 r
(2.19)
N −2
w
4
(2.20)
N − 2 Qd (r0 , d0 ) −1
β xN w
4 Q(r0 , d0 )
(2.21)
It is easy to check that
ϕ=−
is a solution of (2.19). So,
φ=−
8
MONICA MUSSO, JUNCHENG WEI, AND SHUSEN YAN
is a solution of (2.18).
Now, we consider
∗ −2
−∆φ − (2∗ − 1)w2
φ = S2 =
Qd (r0 , d0 )
∗
fN w2 −1 .
Q(r0 , d0 )
(2.22)
So, it is easy to see
φ=−
N − 2 Qd (r0 , d0 )
fN w
4 Q(r0 , d0 )
(2.23)
is a solution of (2.22).
Combining (2.21) and (2.23), we find that
N − 2 Qd (r0 , d0 ) −1
N − 2 Qd (r0 , d0 )
β xN w −
fN w
4 Q(r0 , d0 )
4 Q(r0 , d0 )
m −1
m
=−
β xN w −
fN w
2d0
2d0
φ=−
(2.24)
is a solution of (2.17).
Let φ be defined in (2.24). Then
S(w + εφ) =ε2 S3 + ε2 S4 + B2 (w + εφ) − B2 (w) + εB3 (φ)
¡
¢
∗
+ N0 (εφ) + O ε3 |x|3 |w|2 −1 + ε3 |x|2 |wyN | .
(2.25)
The approximate solution for S(v) = 0, where S(v) is defined in (2.8), near Γε,1 to be
w + εφ.
To avoid the possible singularity of the term
(2.26)
mε
,
βd0 +εxN +εβfN
we further modify the ap-
proximate solution as follows. Let ξ(s) be a function such that ξ = 1 if 0 ≤ |s| ≤ 12 δ, ξ = 0
if |s| ≥ δ, and 0 ≤ ξ ≤ 1. Define
w(x) = ξ(εxN )(w + εφ) .
For any function ψ, let
¡
¢
r0
d0
ψf,β (y) = αψ β(y1 − − f1 ), βy2 , · · · , βyN −1 , β(yN −
− fN ) .
ε
ε
Define
(2.27)
INFINITELY MANY SOLUTIONS
Wk (y) =
k
X
wf,β ((Rkj )−1 y),
9
(2.28)
j=1
where
Rkj
is the rotation of angle
2jπ
k
in the y1 y2 plane, j = 1, · · · , k. We will use Wk as
an approximate solution for (2.3).
Using the symmetry condition on the function Q, we introduce the following space:
©
Hs = u : u = u(y ∗ , |y ∗∗ |), u(Rk ·) = u(·), u is even in yh , h = 2, · · · , N − 1
Z
¡
¢
ª
|Du|2 + λu2 < +∞ .
RN +m
It is easy to check that Wk ∈ Hs .
Theorem 1.1 is a direct consequence of the following result:
Theorem 2.2. Under the same conditions as in Theorem 1.1, there exists a large constant
k0 > 0, such that for all k ≥ k0 , (1.1) has a solution uk ∈ Hs satisfying
uk = Wk + o(1)
(2.29)
where o(1) → 0 uniformly in RN +m as k → ∞.
3. Key Estimates
Let w be the approximate solution defined in (2.27). In this section, we will find the
equations that determine the parameters in the approximate solutions.
Proposition 3.1. We have the following estimates:
Z
ε
−2
RN
Z
−2
ε
RN
¡N − 2
¢ D
¡
¢
S(w)
w + xDw = 2 + O |f |2 + ε ,
2
β
Z
³ Q (r , d )
∂w
Qrd (r0 , d0 ) ´ 1
∗
rr 0 0
=−
f1 +
fN ∗
w2 + O(ε),
S(w)
∂x1
Q(r0 , d0 )
Q(r0 , d0 )
2 β RN
(3.1)
(3.2)
and
Z
ε
−2
RN
Z
∂w
Qrd (r0 , d0 ) 1 −1
∗
S(w)
= − f1
β
w2
∗
∂xN
Q(r0 , d0 ) 2
RN
Z
¡ m
2
m2 Qdd (r0 , d0 ) 1 ¢ −1
∗
w2 + O(ε)
−
+
β
− fN
2
2
∗
N d0 N (N − 2) d0
Q(r0 , d0 ) 2
RN
(3.3)
10
MONICA MUSSO, JUNCHENG WEI, AND SHUSEN YAN
where
m 3(N − 2)
(N − 4)(N − 2) Qrr (r0 , d0 ) + Qdd (r0 , d0 )
B2 −
B2
2
d0 4(N − 1)
8(N − 1)
Q(r0 , d0 )
m2
5N − 14
+ 2
B2 .
d0 4(N − 1)(N − 2)
D =λB2 −
(3.4)
Proof. Recall that
S(w + εφ)
=ε2 S3 + ε2 S4 + B2 (w + εφ) − B2 (w) + εB3 (φ)
∗
∗
∗
+ (w + εφ)2 −1 − w2 −1 − (2∗ − 1)w2 −2 εφ
¡
¢
∗
+ O ε3 |x|3 |w|2 −1 + ε3 |x|2 |wyN |
¢
Qd (r0 , d0 ) ¡ xN
∗
=ε2 S3 + ε2 S4 + ε2
+ fN (2∗ − 1)w2 −2 φ
Q(r0 , d0 ) β
2
εm
(2∗ − 1)(2∗ − 2) 2∗ −3 2
+
φxN + ε2
w
φ
βd0
2
¡
¢
∗
+ O ε3 |x|3 |w|2 −1 + ε3 |x|2 |wyN | .
Since
N −2
w
2
(3.5)
+ xDw is even, we have
Z
¢
¡N − 2
w + xDw
S(w)
2
N
ZR
¢
¡N − 2
=
w + xDw + O(ε3 )
S(w + εφ)
2
RN
Z
¢
¡N − 2
=ε2
S4
w + xDw
2
RN
Z
¢
¡N − 2
¢
¡ xN
∗
2 Qd (r0 , d0 )
+ε
+ fN (2∗ − 1)w2 −2 φ
w + xDw
Q(r0 , d0 ) RN β
2
Z
¡N − 2
¢
m
φxN
+ ε2
w + xDw
βd0 RN
2
Z
∗
∗
¢
(2 − 1)(2 − 2) 2∗ −3 2 ¡ N − 2
w
φ
w + xDw + O(ε3 ).
+ ε2
2
2
RN
(3.6)
From
S4 = −
´ ∗
mxN wxN
1 ³ D2 Q(r0 , d0 )
D2 Q(r0 , d0 )
+
(x,
x)
+
(f,
f
)
w2 −1 − β −2 λw,
β 2 d20
2 Q(r0 , d0 )β 2
Q(r0 , d0 )
(3.7)
INFINITELY MANY SOLUTIONS
11
we find
Z
S4
RN
¡N − 2
¢ D̄
¡
¢
w + xDw = 2 + O |f |2 ,
2
β
(3.8)
where, in view of (A.1), (A.3) and (A.6),
Z
¡N − 2
¡N − 2
¢ m
¢
xN wxN
D̄ = − λ
w
w + xDw − 2
w + xDw
2
d 0 RN
2
RN
Z
¡N − 2
¢
1
D2 Q(r0 , d0 )
∗
+
(x, x)w2 −1
w + xDw
2 RN Q(r0 , d0 )
2
m 3(N − 2)
(N − 4)(N − 2) ∆Q(r0 , d0 )
=λB2 − 2
B2 −
B2 ,
d0 4(N − 1)
8(N − 1)
Q(r0 , d0 )
R
and B2 = RN w2 .
¿From (1.3) and (1.4), we find
Z
(3.9)
N −2
m
Qr (r0 , d0 ) + Qdd (r0 , d0 ) + Qd (r0 , d0 )
r0
d0
2
2 m
Q(r0 , d0 ).
=Qrr (r0 , d0 ) + Qdd (r0 , d0 ) +
N − 2 d20
∆Q(r0 , d0 ) =Qrr (r0 , d0 ) +
Therefore,
m 3(N − 2)
m2 N − 4
B
−
B2
2
d20 4(N − 1)
d20 4(N − 1)
(N − 4)(N − 2) Qrr (r0 , d0 ) + Qdd (r0 , d0 )
−
B2 .
8(N − 1)
Q(r0 , d0 )
D̄ =λB2 −
(3.10)
On the other hand,
Z
¡N − 2
¢
m
φxN
w + xDw
βd0 RN
2
Z
¡
¢ ¡N − 2
¢
m −1
m
m
−
=
β xN w −
fN w x
w + xDw
N
βd0 RN 2d0
2d0
2
Z
2
¡
¢¡
¢
N −2
m
w + xN w x N
w + xDw
= − β −2 2
2d0 RN
2
2
m N +2
=β −2 2
B2 ,
d0 8(N − 1)
(3.11)
12
MONICA MUSSO, JUNCHENG WEI, AND SHUSEN YAN
Z
¢
¡N − 2
¢
¡ xN
Qd (r0 , d0 )
∗
+ fN (2∗ − 1)w2 −2 φ
w + xDw
Q(r0 , d0 ) RN β
2
Z
¢¡ N − 2
¢
¡
2(N + 2) m −1
1 m −1
2∗ −2
=
w
+
xDw
+ O(|f |2 )
β
x
w
−
β
x
w
N
N
2
(N − 2) d0
2 d0
2
RN
2
N + 2 (N − 2)(N − 4)
m
B2 + O(|f |2 ),
= 2 β −2
d0
(N − 2)2
4(N − 1)
Z
¢
(2∗ − 1)(2∗ − 2) 2∗ −3 2 ¡ N − 2
w
φ
w + xDw
2
2
N
ZR
∗
∗
¢2 ¡ N − 2
¡
¢
(2 − 1)(2 − 2) 2∗ −3 1 m −1
=
β xN w
w
−
w + xDw + O(|f |2 )
2
2 d0
2
RN
2
N + 2 m −2 (N − 4)(N − 2)
=−
β
B2 + O(|f |2 ).
2(N − 2)2 d20
4(N − 1)
(3.12)
(3.13)
So,
Z
¢
¡N − 2
w + xDw
φxN
d0 RN
2
Z
¢
¡N − 2
¢
¡
Qd (r0 , d0 )
xN
∗
+
+ fN (2∗ − 1)w2 −2 φ
w + xDw
Q(r0 , d0 ) RN β
2
Z
∗
∗
¢
(2 − 1)(2 − 2) 2∗ −3 2 ¡ N − 2
+
w
φ
w + xDw
2
2
RN
³
2
m
(N + 2)(N − 4)
N +2 ´
= 2 β −2
B2 + O(|f |2 )
+
d0
8(N − 1)(N − 2) 8(N − 1)
m2
(N + 2)(N − 3)
= 2 β −2
B2 + O(|f |2 ).
d0
4(N − 1)(N − 2)
β
−1 m
(3.14)
Thus, using (3.6), (3.8), (3.10) and (3.14), we find
Z
ε
−2
RN
¢ D
¡N − 2
S(w + εφ)
w + xDw = 2 + O(|f |2 ),
2
β
(3.15)
where
(N − 4)(N − 2) Qrr (r0 , d0 ) + Qdd (r0 , d0 )
m 3(N − 2)
B2 −
B2
2
d0 4(N − 1)
8(N − 1)
Q(r0 , d0 )
m2
5N − 14
+ 2
B2 .
d0 4(N − 1)(N − 2)
D =λB2 −
So (3.1) follows from (3.15) and (3.6).
(3.16)
INFINITELY MANY SOLUTIONS
We now compute
R
RN
S(w + εφ)wxi . Let us compute
mfN wxN
D2 Q(r0 , d0 )
S3 = −
+
βd20
Q(r0 , d0 )
µ
R
RN
x
,f
β
13
S3 wxi . Recall
¶
∗ −1
w2
.
(3.17)
For i = 1,
Z
RN
S3 wx1
Z
³ Q (r , d ) f
Qrd (r0 , d0 ) fN ´
∗
rr 0 0
1
=
x1 wx1 w2 −1
+
Q(r0 , d0 ) β
Q(r0 , d0 ) β
RN
Z
³ Q (r , d )
´
Qrd (r0 , d0 )
1
∗
rr 0 0
=−
f1 +
fN ∗
w2 .
Q(r0 , d0 )
Q(r0 , d0 )
2 β RN
(3.18)
For i = N ,
Z
RN
S3 wxN
¡ m
Qdd (r0 , d0 ) 1 ¢
= − fN
+
N βd20
Q(r0 , d0 ) 2∗ β
Z
Qrd (r0 , d0 ) 1
∗
− f1
w2 .
∗
Q(r0 , d0 ) 2 β RN
Z
∗
w2
RN
(3.19)
On the other hand,
Z
Z
¢
m
m ¡ 1 m −1
1m
−1
β
φxN wxi = β
−
β xN w −
fN w x wxi
N
2 d0
2 d0
RN d0
RN d0
Z
Z
m2
m2 δiN
∗
= − β −1 2 fN
wxN wxi = −fN β −1 2
w2 ,
2d0
2d0 N RN
RN
Z
¢ ∗
¡ xN
Qd (r0 , d0 ) ∗
(2 − 1)
+ fN w2 −2 φwxi
Q(r0 , d0 )
β
N
ZR
¢ ∗
¡ xN 1 m
Qd (r0 , d0 ) ∗
1 m −1
=
(2 − 1)
fN w − fN
β xN w w2 −2 wxi
−
Q(r0 , d0 )
β 2 d0
2 d0
RN
Z
Qd (r0 , d0 ) ∗
m
δiN
∗
=fN
(2 − 1) β −1 ∗
w2
Q(r0 , d0 )
d0
2 RN
Z
2
(N + 2)δiN
m
∗
w2 ,
=fN 2 β −1
d0
N (N − 2) RN
Z
(2∗ − 1)(2∗ − 2) 2∗ −3 2
w
φ wxi
2
RN
Z
(2∗ − 1)(2∗ − 2)
1 m2 −1
∗
w2 −3
β xN w 2 f N w x i
=
2
2
2
d
N
R
Z0
N + 2 m2 −1
∗
= − fN
β δiN
w2 .
2
2(N − 2)N d0
RN
−1
(3.20)
(3.21)
(3.22)
14
MONICA MUSSO, JUNCHENG WEI, AND SHUSEN YAN
So, we obtain
Z
−2
ε
RN
S(w + εφ)wx1
Z
³ Q (r , d )
Qrd (r0 , d0 ) ´ 1
∗
rr 0 0
f1 +
fN ∗
w2 + O(ε),
=−
Q(r0 , d0 )
Q(r0 , d0 )
2 β RN
(3.23)
and
Z
ε
−2
RN
S(w + εφ)wxN
Z
Qrd (r0 , d0 ) 1 −1
∗
w2
β
= − f1
∗
Q(r0 , d0 ) 2
RN
Z
¡ m
2
m2 Qdd (r0 , d0 ) 1 ¢ −1
2∗
− fN
−
+
β
w
+ O(ε).
2
2
N d0 N (N − 2) d0
Q(r0 , d0 ) 2∗
RN
(3.24)
¤
Recall that
N −2
N +2
m
∗
uyN − ε2 λu + Q(εy)u2 −1 = Q− 4 (r0 , d0 )β 2 S(v),
yN
and for any function ψf,β , we use the notation
S(u) = ∆u +
(3.25)
¡
¢
r0
d0
ψf,β (y) = αψ β(y1 − − f1 ), βy2 , · · · , βyN −1 , β(yN −
− fN ) ,
ε
ε
and ξ(s) is a function such that ξ = 1 if 0 ≤ |s| ≤ 12 δ, ξ = 0 if |s| ≥ δ, and 0 ≤ ξ ≤ 1.
To find the equations that determine the parameters f and β, we need the following
result.
Proposition 3.2. We have the following estimates:
Z
−2
ε (Q(r0 , d0 ))
N −2
2
S(Wm )
RN
D
B
= 2 − N −2
β
β
∂(ξ(εxN )w)f,β
∂β
¡
¢
+ O |f |2 + ε ,
(3.26)
Z
∂(ξ(εxN )w)f,β
∂x1
RN
Z
³ Q (r , d )
´
Qrd (r0 , d0 )
1
∗
rr 0 0
=−
f1 +
fN ∗
w2 + O(ε),
Q(r0 , d0 )
Q(r0 , d0 )
2 β RN
−2
ε (Q(r0 , d0 ))
and
N −2
2
S(Wm )
(3.27)
INFINITELY MANY SOLUTIONS
15
Z
∂(ξ(εxN )w)f,β
S(Wm )
∂xN
RN
Z
Qrd (r0 , d0 ) 1 −1
∗
= − f1
β
w2
∗
Q(r0 , d0 ) 2
RN
Z
¡ m
2
m2 Qdd (r0 , d0 ) 1 ¢ −1
2∗
−
+
+ O(ε),
− fN
β
w
2
2
N d0 N (N − 2) d0
Q(r0 , d0 ) 2∗
RN
−2
ε (Q(r0 , d0 ))
N −2
2
(3.28)
where the constant D is the same as in Proposition 3.1, and B is a positive constant.
Proof. It follows from (2.8),
S(Wk ) =
k
X
S(αwf,β ((Rkj )−1 y))
j=1
k
³ ∗
´
X
∗
2 −1
+ Q Wk
−
(αwf,β ((Rkj )−1 y))2 −1 .
(3.29)
j=1
To prove (3.26), we use (2.25) to find that
Z
k
X
RN j=2
=ε2 O
S(αwf,β ((Rkj )−1 y))
k
³X
j=2
∂(ξ(εxN )w)f,β
∂β
´
¡ N −2 N −2 ¢
εN −2
2
=
ε
O
k
ε
,
|xj − x1 |N −2
(3.30)
On the other hand, it is standard to prove
Z
RN
=
k
³ ∗
´ ∂(ξ(εx )w)
X
∗
N
f,β
2 −1
Q Wk
−
(αwf,β ((Rkj )−1 y, z))2 −1
∂β
j=1
B
(Q(r0 , d) ))
N −2
2
β N −2
k N −2 εN −2 ,
where B > 0 is a constant.
It is easy to see that (3.1), (3.30) and (3.31) imply (3.26).
To prove (3.27), we first note that
(3.31)
16
MONICA MUSSO, JUNCHENG WEI, AND SHUSEN YAN
Z
k
X
S(αwf,β ((Rkj )−1 y))
RN j=2
=ε2 O
∂(ξ(εxN )w)f,β
∂xi
´
¡
¢
εN −2
= ε2 O k N −2 εN −2 ,
N
−2
|xj − x1 |
k
³X
j=2
(3.32)
and
Z
k
´
³ ∗
X
j −1
2 −1
2∗ −1 ∂(ξ(εxN )w)f,β
Q Wk
−
(αwf,β ((Rk ) y, z))
∂xi
RN
j=1
¡ N −1 N −1 ¢
=O k
ε
(3.33)
which, together with (3.2), imply (3.27). We can prove (3.28) in a similar way.
¤
4. A linear Problem
Let Ω = {r0 − 4δ < |y ∗ | < r0 + 4δ} × {d0 − 4δ < yN < d0 + 4δ}, y ∗ = (y1 , · · · , yN −1 ).
Define Ωε = ε−1 Ω. For j = 1, · · · , k, denote
¡
¢
2(j − 1)π −1
2(j − 1)π
xj = (ε−1 r0 − f1 ) cos
, (ε r0 − f1 ) sin
, 0, · · · , 0, ε−1 d0 − fN ∈ RN .
k
k
Let
kuk∗ = sup
k
³X
y∈Ωε
j=1
+ sup
y∈Ωε
´−1
1
(1 + |y − xj |)
k
³X
N −2
+τ
2
´−1
1
N
j=1
|u(y)|
(1 + |y − xj |) 2 +τ
(4.1)
|Du(y)|,
and
kf k∗∗ = sup
y∈Ωε
where τ =
N −4
.
N −2
k
³X
j=1
´−1
1
(1 + |y − xj |)
For this choice of τ , we find that
N +2
+τ
2
|f (y)|,
(4.2)
INFINITELY MANY SOLUTIONS
k
X
j=2
17
k
X 1
1
τ τ
≤
Ck
ε
≤ Cετ k ≤ C 0 .
τ
|xj − x1 |τ
j
j=2
Lε u = −∆u −
m
uy + λε2 u.
yN N
(4.3)
Consider

2∗ −2
∗
∗

Lε u − (2 − 1)Q(ε|y |, εyN )Wk u = h in Ωε ,
u = 0, on ∂Ωε ,

u ∈ H ,
s
(4.4)
Lemma 4.1. Let u be the solution of (4.4). Then there is a constant C > 0 and θ > 0,
such that
k
³X
´−1
1
N −2
j=1
(1 + |y − xj |) 2
Pk
+τ
³
j=1
≤C khk k∗∗ + kuk∗ Pk
|u(y, z)| +
(1+|y−xj |)
1
N −2 +τ +θ
2
´−1
1
N
j=1
1
j=1
k
³X
(1 + |y − xj |) 2 +τ
|Du(y, z)|
´
.
N −2
(1+|y−xj |) 2 +τ
Proof. To use Lemma C.2, we first make the change of variable ū(·) = ε−
N −2
2
u(ε−1 ·). Then,
ū satisfies

2∗ −2
− N +2
∗
∗
−1

L̃ε ū − (2 − 1)Q(|y |, yN )W̄k ũ = ε 2 h(ε ·) in Ω,
u = 0, on ∂Ω,

u ∈ H ,
s
where W̄k (·) = ε−
N −2
2
Wk (ε−1 ·)
It is easy to check that
L̃ε u =
m
m
−div(yN
Du) + λyN
u
,
m
yN
m
m
and −div(yN
Du) + λyN
u is uniformly elliptic in Ω. It follows from Lemma C.2 that
18
MONICA MUSSO, JUNCHENG WEI, AND SHUSEN YAN
Z
¯
¡
¢ m
N +2
∗
¯
|ū(εx)| =
G(εx, y)) (2∗ − 1)Q(|y ∗ |, yN )W̄k2 −2 ū + ε− 2 h(ε−1 y) yN
dy
Ω
Z
¡ 2∗ −2
¢
1
− N2+2
−1
W̄
|ū|
+
ε
≤C
|h(ε
y)|
dy
k
N −2
Ω |εx − y|
Z
¡ 2∗ −2
¢
1
− N 2−2
=Cε
W
|u|
+
|h(y)|
dy
k
N −2
Ωε |x − y|
Z
k
h³X
´ ∗
1
1
− N 2−2
≤Cε
Wk2 −2 kuk∗
N −2
N
−2
+τ
|x
−
y|
2
RN
j=1 (1 + |y − xj |)
+
k
³X
´
1
(1 + |y − xj |)
j=1
N +2
+τ
2
(4.5)
i
khk∗∗ .
So, we obtain
Z
h³X
´ ∗
1
1
Wk2 −2 kuk∗
|x − y|N −2 j=1 (1 + |y − xj |) N2−2 +τ
k
|u(x)| ≤C
RN
+
k
³X
j=1
´
1
(1 + |y − xj |)
N +2
+τ
2
i
khk∗∗ .
(4.6)
Using Lemma B.3, we have
Z
k
RN
≤C
³X
´ ∗
1
1
Wm2 −2
|x − y|N −2 j=1 (1 + |y − xj |) N2−2 +τ
k
X
j=1
1
(1 + |x − xj |)
N −2
+τ +θ
2
(4.7)
.
It follows from Lemma B.2 that
Z
k
RN
≤C
³X
´
1
1
dy
|x − y|N −2 j=1 (1 + |y − xj |) N2+2 +τ
k
X
j=1
1
(1 + |x − xj |)
So, from (4.7), (4.8) and (4.6), we find
N −2
+τ
2
,
(4.8)
INFINITELY MANY SOLUTIONS
|u(x)| ≤C
k
X
j=1
+C
1
(1 + |x − xj |)
k
X
N −2
+τ +θ
2
19
kuk∗
(4.9)
1
(1 + |x − xj |)
j=1
N −2
+τ
2
khk∗∗
For the estimate of |Du|, we use
|Dū(εx)|
Z
¯
¡
¢ m ¯
N +2
∗
¯
dy ¯
=
DG(εx, y)) (2∗ − 1)Q(|y ∗ |, yN )W̄k2 −2 ū + ε− 2 h(ε−1 y) yN
Ω
≤Cε
Z
RN
+
h³X
´ ∗
1
1
Wk2 −2 kuk∗
|x − y|N −1 j=1 (1 + |y − xj |) N2−2 +τ
k
−N
2
k
³X
j=1
´
1
(1 + |y − xj |)
N +2
+τ
2
(4.10)
i
khk∗∗ .
So, we obtain
|Du(x)|
Z
k
h³X
´ ∗
1
1
≤C
Wk2 −2 kuk∗
N −2
N
−1
+τ
2
RN |x − y|
j=1 (1 + |y − xj |)
+
k
³X
j=1
≤C
k
X
j=1
´
1
(1 + |y − xj |)
1
(1 + |x − xj |)
N
2
+τ +θ
N +2
+τ
2
i
(4.11)
khk∗∗
kuk∗ + C
k
X
1
j=1
(1 + |x − xj |) 2 +τ
N
khk∗∗ ,
and the result follows.
¤
Recall that for any function ψ, we use the notation
¢
¡
d0
r0
− fN ) ,
ψf,β (x) = αψ β(x1 − − f1 ), βx2 , · · · , βxN −1 , β(xN −
ε
ε
and ξ(s) is a function such that ξ = 1 if 0 ≤ |s| ≤ 12 δ, ξ = 0 if |s| ≥ δ, and 0 ≤ ξ ≤ 1. In
the following, we will use that notations:
20
MONICA MUSSO, JUNCHENG WEI, AND SHUSEN YAN
Z1,1 = ∂x1 (ξ(εxN )w)f,β , Z1,2 = ∂xN (ξ(εxN )w)f,β , Z1,3 = ∂β (ξ(εxN )w)f,β ,
and Zi,t (·) = Z1,t (Ri−1 ·), where Ri is the rotation of angle
2iπ
k
in the x1 x2 plane.
Now we consider the a priori estimates for the following problem:

P
P
∗
Lε ϕ − (2∗ − 1)Q(ε|y ∗ |, εyN )Wk2 −2 ϕ = h + 3t=1 ct ki=1 Zi,t in Ωε ,



ϕ = 0 on ∂Ω ,
ε
R
Pk

N ϕ

i=1 Zi,j = 0, j = 1, 2, 3,

 R
φ ∈ Hs .
(4.12)
Lemma 4.2. Assume that ϕk solves (4.12). Then kϕk k∗ ≤ Ckhk∗∗ .
Proof. We argue by contradiction. Suppose that there are k → +∞, and ϕk solving (4.12)
with khk k∗∗ → 0, and kϕk k∗ = 1. For simplicity, we drop the subscript k.
We estimate cl , l = 1, 2, 3. Multiplying (4.12) by Z1,l and integrating over RN , we see
that ct satisfies
3 X
k Z
X
t=1 i=1
Z
¡
=
RN
Zi,t Z1,l ct
RN
2∗ −2
Lε ϕ − (2∗ − 1)Wk
¢
ϕ Z1,l −
(4.13)
Z
hZ1,l .
RN
It follows from Lemma B.1 that
¯­
®¯
¯ h, Z1,l ¯
Z
≤Ckhk∗∗
RN
k
X
1
1
dy
N +2
N
−2
+τ
(1 + |y − x1 |)
2
j=1 (1 + |y − xj |)
(4.14)
≤Ckhk∗∗ .
On the other hand, we have
Z
ZR
N
=
RN
¢
¡
∗
Lε ϕ − (2∗ − 1)Q(ε|y ∗ |, εyN )Wk2 −2 ϕ Z1,l
¢
¡
∗
−∆Z1j − (2∗ − 1)Q(ε|y ∗ |, εyN )Wk2 −2 Z1j ϕ
Z
¡
+
RN
−
¢
m
ϕxN + ε2 λϕ Z1,l = o(1)kϕk∗ .
xN
(4.15)
INFINITELY MANY SOLUTIONS
21
Thus we obtain from (4.13), (4.14) and (4.15) that
cl = o(kϕk∗ ) + O(khk∗∗ ).
(4.16)
So, it follows from Lemma 4.1,
k
³X
j=1
´−1
1
(1 + |y − xj |)
Pk
N −2
+τ
2
³
j=1
≤ khk∗∗ + kϕk∗ Pk
|ϕ(y)| +
1
N −2 +τ +θ
2
1
´−1
1
N
(1 + |y − xj |) 2 +τ
j=1
(1+|y−xj |)
j=1
k
³X
|Dϕ(y)|
(4.17)
´
.
N −2
(1+|y−xj |) 2 +τ
Next, we show that for any R > 0,
|ϕk | + |Dϕk | → 0,
uniformly in BR (xi ).
(4.18)
Suppose that there is (xk ) ∈ BR (xi ), such that
|ϕk (xk )| + |Dϕk (xk )| ≥ c0 > 0.
(4.19)
1
Then, ϕ̄k (·) = ϕk (· − xk ) → ϕ̄ in Cloc
(RN ) and ϕ̄ satisfies
∗ −2
−∆ϕ − (2∗ − 1)w2
N −2
w + xDw.
2
Thus, ϕ̄ is a linear combination of ∂xi w,
Z
ϕ = 0.
By the assumption, we can deduce
Z
¢
¡N − 2
w + xDw ϕ̄ = 0.
2
RN
RN
This implies ϕ̄ = 0. So we obtain a contradiction.
∂xi wϕ̄ = 0,
¿From (4.18) and (4.17), we find
kϕk k∗ = max
k
h³X
x∈RN
j=1
+
´−1
1
(1 + |y − xj |)
k
³X
N −2
+τ
2
´−1
1
N
j=1
(1 + |y − xj |) 2 +τ
|ϕk (y)|
i
|Dϕ(y)|
≤o(1) + Ckhk k∗∗ + oR (1)kϕk k∗ .
This is a contradiction to kϕk k∗ = 1.
¤
22
MONICA MUSSO, JUNCHENG WEI, AND SHUSEN YAN
¿From Lemma 4.2, using the same argument as in the proof of Proposition 4.1 in [14],
we can prove the following result :
Proposition 4.3. There exists k0 > 0 and a constant C > 0, independent of k, such that
for all k ≥ k0 , problem (4.12) has a unique solution ϕk satisfying
kϕk k∗ ≤ Ckhk∗∗ .
(4.20)
5. Finite-dimensional Reduction
Our objective is to construct a solution of the form
u = Wk + ϕ,
for
(
N +2
−∆u + ε2 λu = Q(ε|y ∗ |, ε|y ∗∗ |)u N −2 , u > 0, in RN +m ,
u(y) → 0
as |y| → +∞,
where φ ∈ Hs is some small perturbation. Then φ satisfies
Aϕ = E + N (ϕ),
(5.1)
(5.2)
where
∗ −2
Aϕ = −∆ϕ + ε2 λϕ − (2∗ − 1)Q(ε|y ∗ |, ε|y ∗∗ |)Wk2
∗ −1
E = Q(ε|y ∗ |, ε|y ∗∗ |)Wk2
ϕ,
+ ∆Wk − ε2 λWk2 ,
(5.3)
(5.4)
and
³¡
´
¢2∗ −1
∗
∗
N (ϕ) = Q(ε|y ∗ |, ε|y ∗∗ |) Wk + ϕ
− Wk2 −1 − (2∗ − 1)Wk2 −2 ϕ .
(5.5)
The solutions we will construct are radially symmetric in y ∗∗ . So they satisfy (2.4). To
overcome the difficulty caused by the singularity at yN = 0 in (2.4), we need to separate
this problem into two problems.
Let
Dε,δ
k
[
ª
©
=
x : d(y, Γj,ε ) ≤ δε−1 ,
j=1
INFINITELY MANY SOLUTIONS
23
where
¡
¢
2(j − 1)π −1
2(j − 1)π
Γj,ε = ε−1 r0 cos
, ε r0 sin
, 0, · · · , 0, y ∗∗ ,
k
k
Similar to Section 4, we define the norms
kuk∗,Γ = sup
k
³X
y∈Dε,2δ
j=1
y∈Dε,2δ
+
(1 + d(y, Γj,ε ))
k
³X
+ sup
j=1
sup
´−1
1
N −2
+τ
2
|u(y)|
´−1
1
(1 + d(y, Γj,ε ))
N
2
+τ
|y ∗∗ | = ε−1 d0 .
|Du(y)|
(5.6)
|u(y)|,
RN +m \Dε,δ
and
kf k∗∗,Γ = sup
k
³X
y∈Dε,2δ
+
j=1
sup
´−1
1
(1 + d(y, Γi,ε ))
N +2
+τ
2
|f (y)|
(5.7)
|f (y)|,
RN +m \Dε,δ
where τ =
N −4
.
N −2
In this section, we discuss the solvability of

P
P

= E + N (ϕ) + 3t=1 ct ki=1 Zi,t in RN +m ,
Aϕ
Pk R
i=1 RN +m ϕZi,j = 0, j = 1, 2, 3,

ϕ ∈ H ,
s
(5.8)
for some constants ct .
The main result of this section is the following:
Proposition 5.1. There is an integer k0 > 0, such that for each k ≥ k0 , (5.8) has a unique
solution ϕ, satisfying
kϕk∗,Γ ≤ Cε1+σ ,
where σ > 0 is a small constant.
Proof. Step 1. Decomposition of the problem.
24
MONICA MUSSO, JUNCHENG WEI, AND SHUSEN YAN
To use the results obtained in section 4, we need to decompose (5.2) into two problems.
Let η1,δ (|y ∗ |) be a function satisfying η1,δ = 1 if 0 ≤ ||y ∗ | − ε−1 r0 | < ε−1 δ, η1,δ = 0 if
||y ∗ |−ε−1 r0 | > 2ε−1 δ, and 0 ≤ η ≤ 1, where y ∗ = (y1 , · · · , yN −1 ). We take another function
η2,δ (|y ∗∗ |), satisfying η2,δ = 1 if 0 ≤ ||y ∗∗ | − ε−1 d0 | < ε−1 δ, η2,δ = 0 if ||y ∗∗ | − ε−1 d0 | >
2ε−1 δ, and 0 ≤ η ≤ 1, where y ∗∗ = (yN , · · · , yN +m ). Define ηδ (y) = η(|y ∗ |, |y ∗∗ |) =
η1,δ (y ∗ )η2,δ (y ∗∗ ). Then, ϕ = η2ε−1 δ ω + ψ satisfies (5.8), if ω and ψ satisfy
(
¡
¢ P
P
∗
Aω = η2δ E + N (ηδ ω + ψ) − (2∗ − 1)Q(ε|y ∗ |, ε|y ∗∗ |)Wk2 −2 ψ + 3t=1 ct ki=1 Zi,t ,
ω = 0, on ∂Ωε ,
(5.9)
−1
−1
∗
−1
−1
−1
−1
∗∗
−1
where Ωε = {ε r0 −4δε < |y | < ε r0 +4δε }×{ε d0 −4δε < |y | < ε d0 +4δε−1 },
and
− ∆ψ + ε2 λψ − (2∗ − 1)(1 − η2δ )Q(ε|y ∗ |, ε|y ∗∗ |)Wk2
=(1 −
¡
2
η2δ
)
+ (1 −
E+
3
X
ct
t=1
2
η2δ )N (η2δ ω
k
X
Zi,t
∗ −2
ψ
¢
(5.10)
i=1
+ ψ) + 2Dη2δ Dω + ω∆η2δ
2
=(1 − η2δ
)N (η2δ ω + ψ) + 2Dη2δ Dω + ω∆η2δ .
Step 2. Solving (5.10).
Given any ω satisfying kωk∗,Γ ≤ C (see (5.6) for the definition of the norm k · k∗,Γ ), we
find
|2Dη2δ Dω + ω∆η2δ | ≤ Ckωk∗,Γ ε2 kε
N −2
+τ
2
≤ Ckωk∗,Γ ε
N +2
2
.
(5.11)
So,
2
|(1 − η2δ
)N (η2δ ω + ψ) + 2Dη2δ Dω + ω∆η2δ |
¡
¢
N +2
∗
∗
2
≤C(1 − η2δ
) |η2δ ω|2 −1 + |ψ|2 −1 + Ckωk∗,Γ ε 2
∗
(2∗ −1)(N −2)
∗
2
≤Ckωk2∗,Γ−1 ε
+ C|ψ|2 −1 + Ckωk∗,Γ ε
¡
¢ N +2
∗
∗
=C kωk2∗,Γ−1 + kωk∗,Γ ε 2 + C|ψ|2 −1 .
N +2
2
(5.12)
∗
On the other hand, the operator −∆ψ + ε2 λψ − (2∗ − 1)(1 − η2δ )Q(ε|y ∗ |, ε|y ∗∗ |)Wk2 −2 ψ
is invertible in L∞ (RN +m ). So using the contraction mapping theorem, we can prove that
(5.10) has a solution ψ = ψ(ω), satisfying
INFINITELY MANY SOLUTIONS
25
¡
¢ N +2
∗
kψkL∞ (RN +m ) ≤ C kωk2∗,Γ−1 + kωk∗,Γ ε 2 .
(5.13)
Moreover, we can also prove
¡
¢
N +2
∗
∗
kψ(ω1 ) − ψ(ω2 )kL∞ (RN +m ) ≤ C kω1 k2∗,Γ−2 + kω2 k2∗,Γ−2 + 1 kω1 − ω2 k∗,Γ ε 2 .
Step 3.
(5.14)
Solving (5.9).
We insert ψ = ψ(ω) into (5.9). Given ω, it follows from Proposition 4.3 that there exists
B(ω), satisfying
(
P
P
AB(ω) = hε + 3t=1 ct ki=1 Zi,t ,
B(ω) = 0, on ∂Ωε ,
(5.15)
kB(ω)k∗,Γ ≤ Ckhε k∗∗,Γ ,
(5.16)
and
where
¡
¢
∗
hε = η2δ E + N (ηδ ω + ψ) − (2∗ − 2)Q(ε|y ∗ |, ε|y ∗∗ |)Wk2 −2 ψ(ω) .
It is easy to check that
∗ −2
|η2δ Wk2
≤C
¡
ψ|
∗
kωk2∗,Γ−1
k
¢ N +2 X
+ kωk∗,Γ ε 2
j=1
≤C
¡
∗
kωk2∗,Γ−1
η2δ
(1 + d(y, Γj,ε ))4
k
¢ 4−τ X
+ Ckωk∗,Γ ε
j=1
(5.17)
1
(1 + d(y, Γj,ε ))
N +2
+τ
2
,
and
¡
¢
∗
∗
|η2δ N (η2δ ω + ψ)| ≤ Cη2δ |ω|2 −1 + |ψ|2 −1
¢2∗ −1 (2∗ −1)(N +2)
¡
∗
∗
2
ε
≤Cη2δ |ω|2 −1 + Cη2δ kωk2∗,Γ−1 + kωk∗,Γ
2∗ −1
≤Ckωk∗,Γ
k
X
j=1
So, we obtain
1
(1 + d(y, Γj,ε ))
N +2
+τ
2
.
(5.18)
26
MONICA MUSSO, JUNCHENG WEI, AND SHUSEN YAN
∗
khε k∗∗,Γ ≤ Ckωk2∗,Γ−1 + Ckωk∗,Γ ε4−τ .
(5.19)
Let
k Z
X
©
F = ω : kωk∗,Γ ≤ ε, ω ∈ Hs ,
i=1
ª
ωZi,j = 0, j = 1, 2, 3 .
RN +m
¿From (5.19) and (5.16), we find that B maps F to F .
On the other hand, ω̃ =: B(ω1 ) − B(ω2 ) satisfying
¡
¢
Aω̃ =η2δ N (ηδ ω1 + ψ(ω1 )) − N (ηδ ω2 + ψ(ω2 ))
∗
∗
− (2 − 1)Q(ε|y |, ε|y
∗∗
∗
|)Wk2 −2
¡
3
k
X
¢ X
ψ(ω1 )) − ψ(ω2 ) +
ct
Zi,t ,
t=1
(5.20)
i=1
which, together with (5.14), gives
¡
¢
∗
∗
kω̃k∗,Γ ≤ C kω1 k2∗,Γ−2 + kω2 k2∗,Γ−2 kω1 − ω2 k∗,Γ + Ckω1 − ω2 k∗,Γ ε4−τ .
(5.21)
Thus, B is a contraction map.
Using the contraction mapping theorem, we find that (5.9) has a solution ω, satisfying
ω ∈ F , and
kωk∗,Γ ≤ CkEk∗∗,Γ .
So, the result follows from Lemma 5.3.
¤
Lemma 5.2. If N ≥ 5, then
min(2∗ −1,2)
kN (ϕ)k∗∗,Γ ≤ Ckϕk∗,Γ
.
Proof. If N ≥ 6, we have
∗ −1
|N (ϕ)| ≤ C|ϕ|2
.
Using
k
X
j=1
aj bj ≤
k
¡X
j=1
apj
k
¢1 1 1
¢ p1 ¡X
+ = 1, aj , bj ≥ 0,
bqj q ,
p q
j=1
INFINITELY MANY SOLUTIONS
27
we obtain that for y ∈ Dε,δ ,
|N (ϕ)| ≤
∗
Ckϕk2∗ −1
k
³X
(1 + d(y, Γj,ε ))
j=1
∗
≤Ckϕk2∗ −1
k
X
j=1
∗
≤Ckϕk2∗ −1
j=1
k
³X
1
(1 + d(y, Γj,ε ))
k
X
´2∗ −1
1
N +2
+τ
2
1
(1 + d(y, Γj,ε ))
N +2
+τ
2
j=1
N −2
+τ
2
´ N4−2
1
(1 + d(y, Γj,ε ))τ
(5.22)
.
Thus, the result follows for N ≥ 6.
If N = 5, then
∗ −3
|N (ϕ)| ≤ CWk2
∗ −1
|ϕ|2 + C|ϕ|2
.
So, for y ∈ Dε,δ ,
|N (ϕ)|
∗
≤Ckϕk2∗,Γ Wk2 −3
k
³X
j=1
∗ −1
+ Ckϕk2∗
(1 + d(y, Γj,ε ))
k
³X
j=1
j=1
´2∗ −1
(1 + d(y, Γj,ε ))
j=1
k
X
N −2
+τ
2
1
k
¡
¢³X
2
2∗ −1
=C kϕk∗,Γ + kϕk∗
≤Ckϕk2∗
´2
1
´2∗ −1
1
(1 + d(y, Γj,ε ))
1
(1 + d(y, Γj,ε ))
N −2
+τ
2
N +2
+τ
2
N −2
+τ
2
.
¤
Next, we estimate E.
Lemma 5.3. If N ≥ 5, then
kEk∗∗,Γ ≤ Cε1+σ ,
where σ > 0 is a small constant.
28
MONICA MUSSO, JUNCHENG WEI, AND SHUSEN YAN
Proof. Define
©
­ y 0 xj ®
πª
Ωj = y : y ∈ Ω,
,
≥
cos
, y 0 = (y1 , y2 , , 0, · · · , 0).
0
|y | |xj |
k
Using the estimates in Section 3, we find for y ∈ Dε,2δ ,
k
¯ 2∗ −1 X
¡
¢2∗ −1 ¯
j −1
¯
¯
|E| ≤C W
−
αwf,β ((Rm
) y)
k
j=1
+ Cε
2
k
X
j=1
1
(1 + d(y, Γj,ε ))N −2
=:J0 + J1 .
¿From the symmetry, we can assume that y ∈ Ω1 . Then,
|y − xj | ≥ |y − x1 |,
∀ y ∈ Ω1 .
Firstly, we claim
1
C
≤
, ∀ y ∈ Ω1 , j 6= 1.
(5.23)
1 + |y − xj |
|xj − x1 |
In fact, if |y − x1 | ≤ 21 |x1 − xj |, then |y − xj | ≥ 12 |x1 − xj |. If |y − x1 | ≥ 12 |x1 − xj |, then
|y − xj | ≥ |y − x1 | ≥ 12 |x1 − xj |, since y ∈ Ω1 .
For the estimate of J0 , we have for y ∈ Dε,2δ ∩ Ω1 ,
k
k
³X
´2∗ −1
X
1
1
C
|J0 | ≤
+
C
. (5.24)
(1 + d(y, Γ1,ε ))4 j=2 (1 + d(y, Γj,ε ))N −2
(1 + d(y, Γj,ε ))N −2
j=2
Using (5.23), we obtain for any y ∈ Ω1 ,
k
X
1
1
(1 + d(y, Γ1,ε ))4 j=2 (1 + d(y, Γj,ε ))N −2
≤
≤
k
X
C
(1 + d(y, Γ1,ε ))
N +2
+τ
2
C
(1 + d(y, Γ1,ε ))
N +2
+τ
2
j=2
(εk)
Using the Hölder inequality, we obtain
1
(1 + |xj − x1 |)
N +2
−τ
2
≤
(5.25)
N +2
−τ
2
Cε1+σ
(1 + d(y, Γ1,ε ))
N +2
+τ
2
.
INFINITELY MANY SOLUTIONS
k
³X
j=2
≤
≤
´2∗ −1
1
(1 + d(y, Γj,ε ))N −2
k
X
j=2
k
³X
1
(1 + d(y, Γj,ε ))
k
X
j=1
29
N +2
+τ
2
(1 + d(y, Γj,ε ))
N +2
+τ
2
(N +2)( 12 −τ N1+2 )
≤C(εk)
k
X
j=2
(1 + |x1 − xj |)
´ N4−2
N +2 N −2
−2
( 2 −τ N
)
4
N +2
1
k
X
j=1
j=1
(1 + d(y, Γj,ε ))
k
³X
1
≤Cε1+σ
j=2
1
´ N4−2
N +2 N −2
−2
( 2 −τ N
)
4
N +2
(5.26)
1
(1 + d(y, Γj,ε ))
1
(1 + d(y, Γj,ε ))
N +2
+τ
2
N +2
+τ
2
.
So, we obtain
kJ0 k∗∗ ≤ Cε1+σ .
For the estimate of J1 ,
|J1 | ≤ Cε
1+σ
k
X
j=1
k
X
1
1
1+σ
≤
Cε
,
N +2
+τ
(1 + d(y, Γi,ε ))N −1−σ
2
j=1 (1 + d(y, Γi,ε ))
(5.27)
which gives
kJ1 k∗∗ ≤ Cε1+σ .
¤
6. Proof of Theorem 2.2
In this section, we will choose f and β, such that the constants ct in (5.8) is zero. For
this purpose, we only need to solve the following problem:
Z
¡
RN +m
and
Z
¡
RN +m
¢ ∂(ξ(εxN )w)f,β
= 0,
∂β
(6.1)
¢ ∂(ξ(εxN )w)f,β
= 0; i = 1, N.
∂xi
(6.2)
Aϕ − E − N (ϕ)
Aϕ − E − N (ϕ)
30
MONICA MUSSO, JUNCHENG WEI, AND SHUSEN YAN
Using Proposition 5.1, we can prove the following result.
Proposition 6.1. We have
Z
¢ ∂(ξ(εxN )w)f,β
Aϕ − E − N (ϕ)
∂β
RN +m
¡
¢
+ O |f |2 + εσ ,
−2
ε (Q(r0 , 0, z))
D
B
= 2 − N −2
β
β
¡
N −2
2
Z
¢ ∂(ξ(εxN )w)f,β
Aϕ − E − N (ϕ)
∂x1
RN +m
Z
³ Q (r , d )
´
Qrd (r0 , d0 )
1
∗
rr 0 0
w2 + O(εσ ),
=−
f1 +
fN ∗
Q(r0 , d0 )
Q(r0 , d0 )
2 β RN
−2
ε (Q(r0 , 0, z))
(6.3)
¡
N −2
2
(6.4)
and
Z
−2
ε (Q(r0 , 0, z))
¡
N −2
2
RN +m
¢ ∂(ξ(εxN )w)f,β
Aϕ − E − N (ϕ)
∂xN
Z
Qrd (r0 , d0 ) 1 −1
∗
= − f1
β
w2
∗
Q(r0 , d0 ) 2
RN
Z
¡ m
m2 Qdd (r0 , d0 ) 1 ¢ −1
2
∗
− fN
−
+
β
w2 + O(εσ ),
2
2
∗
N d0 N (N − 2) d0
Q(r0 , d0 ) 2
RN
where B and D are the same as in Proposition 3.2.
(6.5)
Proof. We use ∂ to denote either ∂xi , or ∂β . It is standard to show that
Z
¡
¢
Aϕ − E − N (ϕ) ∂(ξ(εxN )w)f,β
N +m
ZR
=
N
ZR
¡
¢
S(Wm )∂(ξ(εxN )w)f,β + O εkφk∗,Γ + kφk2∗,Γ
S(Wm )∂(ξ(εxN )w)f,β + O(ε2+σ ).
=
RN
So (6.3), (6.4) and (6.5) follow from Propositions 3.2 and 5.1. We omit the details of the
proof here.
¤
Proof of Theorem 2.2. By Proposition 6.1, we need to solve the following equations:
¡ 2
¢
D
B
σ
−
=
O
|f
|
+
ε
= 0,
β 2 β N −2
(6.6)
INFINITELY MANY SOLUTIONS
31
Z
³ Q (r , d )
Qrd (r0 , d0 ) ´ 1
∗
rr 0 0
f1 +
fN ∗
w2 = O(εσ ),
Q(r0 , d0 )
Q(r0 , d0 )
2 β RN
(6.7)
Z
Qrd (r0 , d0 ) 1 −1
∗
f1
β
w2
∗
Q(r0 , d0 ) 2
RN
Z
¡ m
2
m2 Qdd (r0 , d0 ) 1 ¢ −1
∗
+ fN
w2 = O(εσ ).
−
+
β
2
2
∗
N d0 N (N − 2) d0
Q(r0 , d0 ) 2
RN
(6.8)
and
At a critical point (r0 , d0 ), direct calculations show
´
∂2 ³
dm
N − 2 dm
0 Qrr (r0 , d0 )
=
−
,
N
−2
N
2
∂r Q 2 (r, d)
2
Q 2 (r0 , d0 )
´
∂2 ³
dm
N − 2 dm
0 Qrd (r0 , d0 )
,
=−
N −2
N
∂r∂d Q 2 (r, d)
2
Q 2 (r0 , d0 )
and using (1.4),
´
∂2 ³
dm
∂d2 Q N2−2 (r, d)
=
m(m − 1)dm−2
0
N −2
− (N − 2)m
dm−1
Qd (r0 , d0 )
0
N
+
2
N (N − 2) dm
0 Qd (r0 , d0 )
N
4
Q 2 +1 (r0 , d0 )
Q 2 (r0 , d0 )
Q 2 (r0 , d0 )
N − 2 dm
0 Qdd (r0 , d0 )
−
N
2
Q 2 (r0 , d0 )
³
N − 2 dm
N m2 ´
dm−2
0
0 Qdd (r0 , d0 )
2
−
= m(m − 1) − 2m +
N −2
N
N − 2 Q 2 (r0 , d0 )
2
Q 2 (r0 , d0 )
³ 2m2
´
N − 2 dm
dm−2
0
0 Qdd (r0 , d0 )
=
−
−m
.
N −2
N
N −2
2
Q 2 (r0 , d0 )
Q 2 (r0 , d0 )
So, we find that the matrix in (6.7) and (6.8) equals
−
Q
N −2
2
(r0 , d0 ) 1
N dm
β
0
Z
³
∗
w2 D2
dm
´
N −2
Q 2 (r, d) (r0 ,d0 )
By our assumptions, we can prove easily that (6.6), (6.7) and (6.8) have solution:
βk =
RN
¡ B ¢ N1−4
+ O(εσ ),
D
|f1,k | |fN,k | = O(εσ ).
¤
32
MONICA MUSSO, JUNCHENG WEI, AND SHUSEN YAN
Appendix A. Some Estimates
Recall that
w(x) =
αN
(1 + |x|2 )
N −2
2
,
αN = (N (N − 2))
N −2
4
.
Let
Z
w2 .
B2 =
RN
Let ωN be the area of the unit sphere S N −1 . We compute
Z
Z
Z
¡N − 2
¢ N −2
1
2
w
w + xDw =
w +
xDw2
2
2
2
N
N
N
R
R
Z
Z
Z R
N
N −2
w2 −
w2 = −
=
w2 = −B2 ,
2
2 RN
RN
RN
Z
Z
¢
¡N − 2
N (N − 2)
2
w + xDw = −
xDw
w +
|x|2 |Dw|2
2
4
RN
RN
RN
Z ∞
Z ∞
´
³
N −1
N +3
r
r
N (N − 2)
2
2
−
=ωN αN (N − 2)
(1 + r2 )N
4
(1 + r2 )N −2
0
0
Z
Z
N
N
³
∞
1 2
t 2 +1
N (N − 2) ∞ t 2 −1 ´
2
= αN ωN (N − 2)
−
2
(1 + t)N
4
(1 + t)N −2
0
0
Z ∞
N
t 2 −1
3(N − 2)N 1
2
=
ωN α N
4(N − 1) 2
(1 + t)N −2
0
3(N − 2)N
=
B2 ,
4(N − 1)
(A.1)
Z
Z
¢
¡N − 2
1
w + xDw =
xN wxN
2
N
RN
3(N − 2)
=
B2 .
4(N − 1)
We also need
Z
xDw
RN
¡N − 2
¢
w + xDw
2
(A.2)
(A.3)
INFINITELY MANY SOLUTIONS
Z
¡N − 2
¢
2
|x| w
w + xDw = − ∗
2
2
RN
Z
2
N − 2 2∗
|x|
=−
αN
2 N
N
RN (1 + |x| )
Z ∞
N
N − 2 2∗ −2 1
t2
2
αN
ωN α N
=−
N
2
(1 + t)N
0
Z ∞
N
t2
21
2
= − (N − 2) ωN αN
.
2
(1 + t)N
0
2
33
Z
2∗ −1
|x|2 w2
∗
RN
(A.4)
But
Z
Z ∞
Z ∞ N +1
N
N
t2
t2
t2
=
−
N
N
−1
(1 + t)
(1 + t)
(1 + t)N
0
0
0
Z ∞
Z ∞
N
N
N
t 2 −1
(N + 2)N
t 2 −1
=
−
2(N − 2) 0 (1 + t)N −2 4(N − 1)(N − 2) 0 (1 + t)N −2
Z ∞
N
N (N − 4)
t 2 −1
=
.
4(N − 2)(N − 1) 0 (1 + t)N −2
∞
(A.5)
So, we obtain
Z
∗ −1
|x|2 w2
RN
¢
¡N − 2
(N − 4)N (N − 2)
w + xDw = −
B2 .
2
4(N − 1)
(A.6)
Appendix B. Basic Estimates
In this section, we list some lemmas, whose proof can be found in [35].
For each fixed i and j, i 6= j, consider the following function
1
1
,
α
(1 + |y − xj |) (1 + |y − xi |)β
where α ≥ 1 and β ≥ 1 are two constants.
gij (y) =
(B.1)
Lemma B.1. For any constant 0 < σ ≤ min(α, β), there is a constant C > 0, such that
gij (y) ≤
³
´
1
1
C
+
.
|xi − xj |σ (1 + |y − xi |)α+β−σ (1 + |y − xj |)α+β−σ
Lemma B.2. For any constant 0 < σ < N − 2, there is a constant C > 0, such that
34
MONICA MUSSO, JUNCHENG WEI, AND SHUSEN YAN
Z
RN
1
1
C
dz ≤
.
N
−2
2+σ
|y − z|
(1 + |z|)
(1 + |y|)σ
Let recall that Wk is defined in (2.28).
Lemma B.3. Suppose that N ≥ 4. Then there is a small θ > 0, such that
Z
k
X
4
1
1
N −2
(z)
dz
W
N −2
k
N
−2
+τ
2
(1
+
|z
−
x
|)
RN |y − z|
j
j=1
≤C
k
X
1
(1 + |y − xj |)
j=1
N −2
+τ +θ
2
.
Proof. The proof can be found in [35].
¤
Appendix C. Fundamental solutions for linear elliptic operators
In this section, we discuss the fundamental solutions for some elliptic linear operators
in divergent form. Firstly, we consider the existence of fundamental solution in RN for
N ≥ 3. The following result may be known. But we can not find the references. So we
will give the proof of it.
Lemma C.1. Suppose that N ≥ 3. The following elliptic problem
−
2
λ0 |ξ| ≤
N
X
Di (aij (x)Dj u) + c(x)u = δp
in RN
j=1
PN
j=1
aij (x)ξi ξj ≤ λ1 |ξ|2 , λ1 ≥ λ0 > 0, c(x) ≥ 0, has a solution Γ(x, p) satisfying
0 < Γ(x, p) ≤
C
C
; |DΓ(x, p)| ≤
,
N
−2
|x − p|
|x − p|N −1
as x → p.
Proof. It is well known that
−
N
X
Di (aij (p)Dj u) + c(p)u = δp
in RN
j=1
has a solution Γ̄(x, p) satisfying
0 < Γ̄(x, p) ≤
C
C
, |DΓ̄(x, p)| ≤
.
N
−2
|x − p|
|x − p|N −1
INFINITELY MANY SOLUTIONS
35
where C > 0 is independent of p.
Consider
−
N
X
Di (aij (x)Dj v) + c(x)v = f (x, p)
j=1
=: −
N
X
¡
¢
¡
¢
Di (aij (x) − aij (p) Dj (ξp Γ̄(x, p))) + c(x) − c(p) (ξp Γ̄(x, p)) in RN ,
(C.1)
j=1
where ξp = 1 in B1 (p), ξp = 0 in RN \ B2 (p) and 0 ≤ ξp ≤ 1. Since
|f (x, p)| ∈ Lr (RN ),
2,r
for r ∈ (1, NN−1 ), (C.1) has a solution v ∈ Wloc
(RN ). So, Γ(x, p) = ξp Γ̄(x, p) + v is the
solution. It is easy to show that Γ(x, p) > 0.
On the other hand, let w be a the solution of
−
N
X
Di (aij (x)Dj v) + c(x)v = |f (x, p)|,
in RN .
(C.2)
j=1
Then |v| ≤ w.
Now we estimate the blow up rate of w at the singularity x = p. By the Harnack
inequality, we find for x 6= p,
³
¡
1
w(y) ≤C
inf
w(y) + |x − p|
sup
y∈B 1 |x−p| (x)
|x − p|N
y∈B 1 |x−p| (x)
2
2
³
´
1
≤C
inf
w(y) +
,
y∈B 1 |x−p| (x)
|x − p|N −3+η1
2
Z
|f (x, p)|r
¢ r1 ´
B|x−p| (x)
2
if
N
N −1
Nr
− r > 0 is small enough, where η1 > 0 is a small constant. From w ∈ W 2,r ⊂ L N −2r ,
we find that for a small η > 0,
inf
y∈B 1 |x−p| (x)
w(y) ≤
2
C
,
|x − p|N −2−η
∀ x close to p.
Otherwise, we have a sequence of xn → p, such that
inf
y∈B 1 |x
2
n −p|
(xn )
w(y) ≥
C
.
|xn − p|N −2−η
36
MONICA MUSSO, JUNCHENG WEI, AND SHUSEN YAN
Then for q =
Nr
,
N −2r
r ∈ (1, NN−1 ),
Z
Z
q
C≥
wq ≥
w ≥
B1 (p)
B 1 |x
2
n −p|
(xn )
|xn −
c0
(N
p| −2−η)q−N
→ +∞,
as xn → p,
since (N − 2 − η)q − N > 0 if r > 1 and η > 0 is small. This is a contradiction. So, we
have proved that
|v| ≤ w(x) ≤
C
.
|x − p|N −2−η
Using the Lq estimate, we also have
sup
|Dv(y)|
y∈B 1 |x−p| (x)
4
¡
C ³
1
≤
sup
|v(y)| + |x − p|2
|x − p| y∈B 1 |x−p| (x)
|x − p|N
2
≤
Z
|f (x, p)|q
¢ 1q ´
B 1 |x−p| (x)
2
C
.
|x − p|N −1−η
Since Γ(x, p) = Γ̄(x, p) + v, we find that for x ∈ B1 (p),
0 < Γ(x, p) ≤
C
C
;
|DΓ(x,
p)|
≤
.
|x − p|N −2
|x − p|N −1
¤
Lemma C.2. Let Ω be a bounded domain in RN . The following elliptic problem
( P
− N
j=1 Di (aij (x))Dj u) + c(x))u = δp ,
u = 0,
in Ω
on ∂Ω,
(C.3)
where aij and c satisfy the same condition as in Lemma C.1 in Ω, has a solution G(x, p)
satisfying
0 < G(x, p) ≤
C
C
, |DG(x, p)| ≤
.
N
−2
|x − p|
|x − p|N −1
Proof. Let γ be the solution of
( P
− N
j=1 Di (aij (x)Dj γ) + c(x)γ = 0, in Ω,
γ(x) = −Γ(x, p)
on ∂Ω,
(C.4)
where Γ(x, p) is the function obtained in Lemma C.2. Then γ < 0. So, G(x, p) = Γ + γ is
the solution satisfying the condition.
INFINITELY MANY SOLUTIONS
37
¤
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Departamentto de Matemática, Pontificia Universidad Catolica de Chile, Avda. Vicuna
Mackenna 4860, Macul, Chile
E-mail address: mmusso@mat.puc.cl
Department of Mathematics, University of British Columbia, Vancouver, BC V6T 1Z2
Canada and Department of Mathematics, Chinese University of Hong Kong, Shatin, Hong
Kong
E-mail address: jcwei@math.ubc.ca
Department of Mathematics, The University of New England, Armidale, NSW 2351,
Australia
E-mail address: syan@turing.une.edu.au