The main objective of the present note is to study positive solutions of the following interesting system of integral equations in Rn u(x) = Z Rn |x − y| p v(y) −q dy, v(x) = Z Rn |x − y| pu(y) −q dy, (0.1) with p, q > 0 and n > 1. Under the nonnegative Lebesgue measurability condition for solutions (u, v) of (0.1), we prove that pq = p+2n and that u and v are radially symmetric and monotone decreasing about some point. To prove this, we introduce an integral form of the method of moving spheres for systems to tackle (0.1). As far as we know, this is the first attempt to use the method of moving spheres for systems.
Trang 1EQUATIONS WITH NEGATIVE EXPONENTS VIA THE METHOD OF
2
MOVING SPHERES
3
QU ˆ O´C ANH NG ˆ O 4
Dedicated to Professor Hoang Quoc Toan on the occasion of his 70th birthday
A BSTRACT The main objective of the present note is to study positive solutions of the following interesting system of integral equations in Rn
u(x) = Z
R n
|x − y|pv(y)−qdy, v(x) =
Z
R n
|x − y| p u(y)−qdy,
(0.1)
with p, q > 0 and n > 1 Under the nonnegative Lebesgue measurability condition for solutions (u, v) of (0.1), we prove that pq = p+2n and that u and v are radially symmetric and monotone decreasing about some point To prove this, we introduce an integral form
of the method of moving spheres for systems to tackle (0.1) As far as we know, this is the first attempt to use the method of moving spheres for systems.
1 INTRODUCTION 5
Given p, q > 0 and n > 1, of interest in the present note is to study non-negative
6
solutions of the following interesting system of integral equations in Rn
7
u(x) = Z
R n
|x − y|pv(y)−qdy, v(x) =
Z
R n
|x − y|pu(y)−qdy
(1.1)
8
Perhaps, the motivation for studying (1.1) comes from a natural extension from one
equa-9
tion to a system However, its counterpart when p, q < 0 comes from the study of the sharp
10
constants C(n, s, p) of the following Hardy-Littlewood-Sobolev inequality
11
Z
R n
Z
R n
f (x)g(y)
|x − y|p dxdy 6 C(n, s, p)kf kL rkgkLs
12
with 1/r + 1/s = 2 − n/p, see [Lieb83, CheLiOu05] In the special case where u = v,
13
our interested system (1.1) becomes the following well-known integral equation
14
u(x) = Z
R n
15
in Rn where u > 0 To be precise, Eq (1.2) has its root in geometry as it arises from
16
studying quantitative properties of geometric curvatures in the conformal geometry The
17
classification of the solutions of (1.2) and its variations thus has a long history and is
im-18
portant as it also provides essential ingredients in the study of the Yamabe problem and the
19
prescribing scalar curvature problem, see [CafGidSpr89, CheLi91, WeiXu99, ZhaHao08]
20
Date: 11thMay, 2015 at 22:11.
2000 Mathematics Subject Classification 35J60, 58G35, 53C21.
Key words and phrases Integral equation; Symmetry; Moving sphere.
Trang 2for details As can be observed, the corresponding partial differential equation to (1.2) is
1
the following well-known semilinear equation
2
3
with u > 0 in Rn Concerning Eq (1.2), solutions to Eq (1.2) were first studied by Xu
4
in [Xu05] when p = 1 and n = 3 He provided the following classification of C4entire
5
solutions of Eq (1.2)
6
Theorem 1 (see [Xu05], Thm 1.1) Suppose that u is a C4entire positive solution of the
7
integral equation
8
u(x) = Z
R 3
9
inR3 withq > 0, then q = 7 and up to a constant multiplication, a translation and a
10
dilation,u takes the form
11
u(x) = 1 + |x − x|21
12
Note that we had already dropped the factor 1/8π in (1.4) in the original statement in
13
[Xu05, Theorem 1.1] Almost simultaneously, Li studied Eq (1.2) in its present form and
14
obtained, among others, the following result in [Li04]
15
Theorem 2 (see [Li04], Thm 1.5) For n > 1, p > 0 and 0 < q 6 1 + 2n/p, let u be
16
a nonnegative Lebesgue measurable function inRn satisfying(1.2) Then q = 1 + 2n/p
17
and, for some constantsa, d > 0 and some x ∈ Rn,
18
u(x) = a d + |x − x|2
p
2
19
Clear, Theorem 2 already includes Theorem 1 We take this chance to mention a recent
20
paper [XuWuTan14] in which the authors considered a slightly perturbation of (1.2) and
21
obtained some new phenomena Let us just go back to the very special case when n = 3,
22
p = 1 In this setting, Eq (1.3) is associated with some fourth order partial differential
23
equation since it simply becomes
24
25
in R3 Although (1.4) and (1.5) are closely related, here we would like to compare the
26
structure of solutions of (1.4) and (1.5) Clearly, we know from Theorem 1 that it is
neces-27
sary to have q = 7 in order for (1.4) to admit solutions In addition, we further know that
28
all solutions corresponding to the case q = 7 are radial symmetric and asymptotically
lin-29
ear at infinity However, this is no longer true for (1.5) in the sense that it is not necessary
30
to have q = 7 for (1.5) to have radial symmetric solutions with asymptotically linearity at
31
infinity, see [Gue12]; see also [CheFanLi13, FanChe12] for related works concerning the
32
equivalence of integral and differential equations
33
In the literature, to classify solutions of some partial differential equations, people
usu-34
ally use one of the following two methods: moving planes and moving spheres While the
35
former method, first invented by Alexandrov in early 1950s, turns out to be powerful when
36
dealing with partial differential equations such as (1.3) through a series of seminal works
37
by Serrin [Ser71], Gidas, Ni, and Nirenberg [GidNiNir79], Caffarelli, Gidas, and Spruck
38
[CafGidSpr89], Chen and Li [CheLi91], the latter method is a variant of the former one
39
As far as we know, this method was used by Li and Zhu in [LiZhu95]
40
A prior to the work by Li and Zhu, to make use of the method of moving planes, one
41
needs to go through two steps: First to prove solutions are radial symmetric by using the
42
reflection through some planes, then to classify all radial solutions by solving some
trans-43
formed ordinary differential equations In the work by Li and Zhu, the authors successfully
Trang 3exploited the conformal invariance of the problem which, as a by-product, captures
solu-1
tions directly rather than going the second step as in the method of moving planes, see
2
[Li04, Section 1] It is worth noticing that all these “traditional” methods of moving planes
3
and spheres can be used to classify solutions to, for example, (1.3) because local properties
4
of the differential operators of the problem are well-exploited In case of lack of knowledge
5
of local properties, it could prevent us from using some known results, see [CheLiOu05,
6
Section 1]
7
Now we turn our attention to problem (1.1) In the recent paper [Lei15], Lei studied
8
positive entire solutions (u, v) of (1.1) Among other things, Lei proved the following
9
result
10
Theorem 3 (see [Lei15], Thm 1.2) Assume the positive entire solutions u, v ∈ C1(Rn)
11
of (1.1) are radially symmetric about some point x ∈ Rn, then
12
u(x) ≡ v(x) ≡ a(b2+ |x − x|2)p2
13
witha, b > 0
14
As can be seen from the proof of [Lei15, Theorem 1.2], the assumption of radial
prop-15
erty plays a key role because it allows us to prove that u ≡ v identically Upon having this,
16
the form for u and v follows from known results Also in [Lei15], the author questioned
17
the assumption of radial property of (u, v) in Theorem 3 can be removed Motivated by the
18
question in [Lei15], in the present note we prove Theorem 3 without assuming the radial
19
property of solutions (u, v) To achieve that goal, we adopt the method of moving spheres
20
in [Li04] to introduce an integral form of the method of moving spheres for solving
sys-21
tems of integral equations As far as we know, this is the first attempt to use the method of
22
moving spheres for systems (with negative exponents)
23
It is worth noticing that in the same paper [Lei15], the author pointed out that J Xu has
24
already completed a preprint where an answer for this question was addressed using the
25
ideas in [LiZhu95] and [Xu07] Unfortunately, such a preprint has not been made available
26
yet Nevertheless, our proof follows [Li04] which should be different from above
27
Before discussing further, we state our main result of the present note
28
Theorem 4 For n > 1, p > 0 and q > 0, let (u, v) be a pair of nonnegative Lebesgue
29
measurable functions inRnsatisfying(1.1) Then q = 1 + 2n/p and, for some constants
30
a, b > 0 and some x ∈ Rn,
31
u(x) = v(x) = a(b2+ |x − x|2)p2
32
for anyx ∈ Rn
33
Note that Theorem 4 already includes the case q > 1 + 2n/p which was not considered
34
in [Li04], see Theorem 2 above
35
Nearly a decade ago, Chen, Li, and Ou introduced an integral form of the method of
36
moving planes in [CheLiOu06] to classify solutions of some integral equations Since then,
37
this new method has been used by many mathematicians for different problems under
38
various contexts In a continued paper [CheLiOu05], Chen, Li, and Ou also introduced
39
an integral form of the method of moving planes for system The basic idea of this new
40
method that allows the method to go through depends heavily on some new special features
41
possessed by integral equations Perhaps, the global form of the integral equations allows
42
us to overcome the lack of local properties possessed by differential operators In addition,
43
we should emphasize that the most important ingredient in this new method is to apply
44
the Hardy-Littlewood-Sobolev inequality in a very clever way However, as pointed out
Trang 4by Lei in [Lei15], it is inappropriate to apply this new method to (1.1) since the
Hardy-1
Littlewood-Sobolev inequality does not work due to the presence of the negative exponent
2
q
3
The primary aim of this note is to reformulate the method of moving spheres in an
4
integral form that suits for our analysis Following the method of moving spheres by
5
Li and Zhu, Zhang and Hao first introduced the integral form of the method of moving
6
spheres in [ZhaHao08] to provide a new proof for [CheLiOu06, Theorem 1.1] However,
7
the approach in [ZhaHao08] seems to be narrowed as it still requires the
Hardy-Littlewood-8
Sobolev inequality; hence it cannot be applied to our context
9
To make our analysis doable, we analyze some conformal invariances found in [Li04]
10
for one equation and made some necessary changes for systems For the reader’s
conve-11
nience and in order to make our note self-contained, we follow the proof of [Li04, Theorem
12
1.5] closely with a little bit more explanation Before closing this section, we would like to
13
mention that our integral system (1.1) is closely related to the following system of partial
14
differential equations
15
(−∆)n+p2 u = v−q, (−∆)n+p2 v = u−q, (1.6)
16
in Rnwhere u, v > 0 In fact, any C2solution (u, v) of (1.1) solves (1.6)
17
2 PRELIMINARIES AND THE METHOD OF MOVING SPHERES FOR SYSTEMS 18
In this section, we setup some preliminaries necessarily for our analysis The most
19
important part of this section is the a prior estimates for solutions of (1.1) as stated in
20
Lemma 1 below Here and in what follows, by and & we mean inequalities up to p, q,
21
and dimensional constants
22
Lemma 1 For n> 1 and p, q > 0, let (u, v) be a pair of non-negative Lebesgue
measur-23
able functions inRnsatisfying(1.1) Then there hold
24
Z
R n
(1 + |y|p)u(y)−qdy < ∞,
Z
R n
(1 + |y|p)v(y)−qdy < ∞, (2.1)
25
and
26
lim
|x|→∞
u(x)
|x|p = Z
R n
v(y)−qdy, lim
|x|→∞
v(x)
|x|p = Z
R n
u(y)−qdy, (2.2)
27
andu and v are bounded from below in the following sense
28
29
and above in the following sense
30
31
for allx ∈ Rn In other words, there holds
32
1 + |x|p
C 6 u(x), v(x) 6 C(1 + |x|p)
33
inRnfor some constantC > 1
34
Proof To prove our lemma, we first observe from (1.1) that both u and v are strictly
35
positive everywhere in Rn and are finite within a set of positive measure Hence there
36
exist some R > 1 sufficiently large and some Lebesgue measurable set E ⊂ Rnsuch that
37
E ⊂ {y : u(y) < R, v(y) < R} ∩ B(0, R) (2.5)
Trang 5with meas(E)> 1/R Using this, we can easily bound v from below as follows
1
v(x) >
Z
E
|x − y|pu(y)−qdy > 1
Rq
Z
E
|x − y|pdy = 1
Rq
Z
E+x
|y|pdy
2
for any x ∈ Rn Choose ε > 0 small enough and then fix it in such a way that vol(B(0, ε)) <
3
|E|/2 Then we can estimate
4
Z
E+x
|y|p
dy >
Z
E+x\B(0,ε)
|y|pdy
> εp Z
E+x\B(0,ε)
dy
= εp |E + x| − vol(B(0, ε))
5
From this, it is clear that v is bounded from below by some positive constant The same
6
reason applied to u shows that there exists some constant C0> 0 such that
7
8
Proof of (2.3) To prove this, we first consider |x|> 2R where R is defined through (2.5)
9
Note that for every y ∈ E ⊂ B(0, R), there holds |x − y|> |x| − |y| > |x|/2 thanks to
10
|x| > 2R Using this we can estimate
11
v(x) > 1
Rq
Z
E
|x − y|pdy > vol(E)
(2R)p|x|p
12
for any |x|> 2R A similar argument also shows u(x) > vol(E)(2R)−p|x|pin the region
13
{|x| > 2R} Hence, it is easy to select a large constant C > 1 in such a way that (2.3)
14
holds in the region {|x|> 2R} Thanks to (2.6), we can further decrease C, if necessary,
15
to obtain the estimate (2.3) in the ball {|x|6 2R}; hence the proof of (2.3) follows
16
Proof of (2.1) We only need to estimate v since u can be estimated similarly To this
17
purpose, we first show that u−q ∈ L1(Rn) Clearly for some x satisfying 1 6 |x| 6 2,
18
there holds
19
Z
R n
|x − y|pu(y)−qdy = v(x) ∈ (0, +∞)
20
Observer that for any y ∈ Rn\B(0, 4), |x − y| > |y| − |x| > 1; hence
21
Z
R n \B(0,4)
u(y)−qdy <
Z
R n
|x − y|pu(y)−qdy < +∞
22
In the small ball B(0, 4), thanks to (2.3), it is obvious to verify that
23
Z
B(0,4)
u(y)−qdy
Z
B(0,4)
(1 + |y|p))−qdy < +∞
24
Thus, we have just shown that u−q∈ L1(Rn) In view of (2.1), it suffices to prove that
25
Z
R n
26
To see this, we again observe that |y|6 2|x − y| for all y ∈ Rn\B(0, 4) Therefore,
27
Z
R n \B(0,4)
|y|pu(y)−qdy
Z
R n \B(0,4)
|x − y|pu(y)−qdy < +∞
28
In the small ball B(0, 4), it is obvious to see that
29
Z
B(0,4)
|y|pu(y)−qdy
Z
B(0,4)
u(y)−qdy < +∞,
30
thanks to u−q∈ L1(Rn) From this, (2.7) follows, so does (2.1)
Trang 6Proof of (2.2) We only consider the limit |x|−pv(x) as |x| → ∞ since the limit |x|−pu(x)
1
can be proved similarly Indeed, using (1.1), we first obtain
2
lim
|x|→∞
v(x)
|x|p = lim
|x|→∞
Z
R n
|x − y|p
3
Observe that as |x| → +∞, (|x − y|/|x|)pu(y)−q → u(y)−q almost everywhere y in
4
Rn Hence we can apply the Lebesgue dominated convergence theorem to pass (2.8) to
5
the limit to conclude (2.1) provided we can show that |x − y|p|x|−pu(y)−qis bounded by
6
some integrable function To this end, for each |x|> 1 fixed we first split
7
Rn = {y ∈ Rn: |x − y| 6 2|x|} ∪ {y ∈ Rn : |x − y| > 2|x|} = D1∪ D2
8
Then, in D1, we immediately obtain
9
|x − y|
|x|
p
6 2p (1 + |y|p),
10
for any y ∈ D1while in D2we realize that |y|> |x − y| − |x| > |x − y|/2 which helps
11
us to estimate
12
|x − y|
|x|
p
6 |x − y|p |y|p 1 + |y|p
13
thanks to |x|> 1 Thus, we have just proved that for any |x| > 1, the following estimate
14
|x − y|
|x|
p
u(y)−q (1 + |y|p)u(y)−q
15
holds Our proof now follows by observing (1 + |y|p)u(y)−q ∈ L1(Rn) by (2.1)
16
Proof of (2.4) To see this, we first observe from (2.2) that there exists some large number
17
k > 1/R such that
18
u(x)
|x|p < 1 +
Z
R n
v(y)−qdy
19
in Rn\B(0, kR) In the ball B(0, kR), it is easy to estimate |x − y|p
|x|p+ |y|pwhich
20
helps us to conclude that
21
u(x) (kR)p
Z
R n
(1 + |y|p)v(y)−qdy
22
in the ball B(0, kR) From this and our estimate for u outside B(0, kR), we obtain the
23
desired estimate Our estimate for v follows the same lines; hence we obtain (2.4) as
24
25
In the next result, we prove a regularity result similar to that obtained by Li in [Li04,
26
Lemma 5.2]
27
Lemma 2 For n> 1 and p, q > 0, let (u, v) be a pair of non-negative Lebesgue
measur-28
able functions inRnsatisfying(1.1) Then u and v are smooth
29
Proof Our proof is similar to that of [Li04, Lemma 5.2] Let R > 0 be arbitrary, first we
30
decompose u and v into the following way
31
u(x) =u1R(x) + u2R(x) =
Z
|y|62R
+ Z
|y|>2R
!
|x − y|pv(y)−qdy,
v(x) =v1R(x) + vR2(x) =
Z
|y|62R
+ Z
|y|>2R
!
|x − y|pu(y)−qdy
32
Thanks to (2.1), we immediately see that we can continuously differentiate u2R and v2R
33
under the integral sign for any |x| < R Consequently, u2 ∈ C∞(B(0, R)) and v2 ∈
34
Trang 7C∞(B(0, R)) In view of (2.3) and (2.4), we know that u−q ∈ L∞(B(0, 2R)) which
1
implies that v1
R is at least H¨older continuous in B(0, R) Similar reasons tell us that u1
R 2
is also at least H¨older continuous in B(0, R) Hence, we have just proved that u and v
3
are at least H¨older continuous in B(0, R), so are at least H¨older continuous in Rn since
4
R > 0 is arbitrary Standard bootstrap argument shows u ∈ C∞(Rn) and at the same time
5
6
Once we have the smoothness for solutions of (1.1), we can narrow the range for q as
7
follows
8
Proposition 1 For n > 1 and p, q > 0 Then in order for (1.1) to have solutions, it is
9
necessary to haveq 6 1 + 2n/p
10
Proof The proof is just a direct consequence of [HuaYu13, Theorem 1] and Lemma 2
11
above, see also [Lei15, Theorem 1.1]; hence we omit its details
12
3 THE METHOD OF MOVING SPHERES FOR SYSTEMS 13
As a consequence of Proposition 1, from now on, we only consider the case q 6 1 +
14
2n/p Let w be a positive function on Rn For x ∈ Rnand λ > 0 we define
15
wx,λ(ξ) = |ξ − x|
λ
p
16
where
17
ξx,λ= x + λ2 ξ − x
18
Clearly, upon the change of variable y = zx,λ, we then have
19
dy =
|z − x|
2n
20
Note that if y = zx,λ, then z = yx,λ Therefore, we have
21
Z
|y−x|>λ
|ξx,λ− y|pv(y)−qdy =
Z
|z−x|6λ
|ξx,λ− zx,λ|pv(zx,λ)−q
|z − x|
2n
dz
= Z
|z−x|6λ
|ξx,λ− zx,λ|p
|z − x|
2n−pq
vx,λ(z)−qdz
22
Then, using the relation |z − x||ξ − x||ξx,λ− zx,λ| = λ2|ξ − z|, we obtain
23
|ξ − x|
−pZ
|y−x|>λ
|ξx,λ− y|pv(y)−qdy
= Z
|y−x|>λ
|z − x|
|ξx,λ− zx,λ|
|ξ − z|
−p
|ξx,λ− y|pv(y)−qdy
= Z
|z−x|6λ
|ξ − z|p
|z − x|
2n−pq+p
vx,λ(z)−qdz
24
Similarly, we obtain
25
|ξ − x|
−pZ
|y−x|6λ
|ξx,λ− y|pv(y)−qdy
= Z
|z−x|>λ
|ξ − z|p
|z − x|
2n−pq+p
vx,λ(z)−qdz
26
Trang 8Lemma 3 For any solutions (u, v) of (1.1), we have
1
ux,λ(ξ) =
Z
R n
|ξ − z|p
|z − x|
2n−pq+p
vx,λ(z)−qdz
2
and
3
vx,λ(ξ) =
Z
R n
|ξ − z|p
|z − x|
2n−pq+p
ux,λ(z)−qdz
4
for anyξ ∈ Rn
5
Proof Using our system (1.1), we obtain
6
ux,λ(ξ) = |ξ − x|
λ
p
u(ξx,λ)
= |ξ − x|
λ
pZ
R n
|ξx,λ− y|pv(y)−qdy
= Z
R n
|ξ − z|p
|z − x|
2n−pq+p
vx,λ(z)−qdz
7
8
Lemma 4 For any solutions (u, v) of (1.1), we have
9
ux,λ(ξ) − u(ξ) =
Z
|z−x|>λ
k(x, λ; ξ, z)
v(z)−q−
|z − x|
2n−pq+p
vx,λ(z)−q
dz,
10
and
11
vx,λ(ξ) − v(ξ) =
Z
|z−x|>λ
k(x, λ; ξ, z)
u(z)−q−
|z − x|
2n−pq+p
ux,λ(z)−q
dz,
12
for anyξ ∈ Rnwhere
13
k(x, λ; ξ, z) = |ξ − x|
λ
p
|ξx,λ− z|p− |ξ − z|p
14
Moreover,k(x, λ; ξ, z) > 0 for any |ξ − x| > λ > 0 and |z − x| > λ > 0
15
Proof First, we observe that
16
ux,λ(ξ) =
Z
|z−x|>λ
|ξ − z|p
|z − x|
2n−pq+p
vx,λ(z)−qdz
+ |ξ − x|
λ
pZ
|z−x|>λ
|ξx,λ− z|pv(z)−qdz
17
and
18
u(ξ) = Z
|z−x|>λ
|ξx,λ− z|pv(z)−qdz
+ |ξ − x|
λ
pZ
|z−x|>λ
|ξx,λ− z|p
|z − x|
2n−pq+p
vx,λ(z)−qdz
19
Therefore,
20
ux,λ(ξ) − u(ξ) =
Z
|z−x|>λ
k(x, λ; ξ, z)
v(z)−q−
λ
|z − x|
2n−pq+p
vx,λ(z)−q
dz,
21
where
22
k(x, λ; ξ, z) = |ξ − x|
λ
p
|ξx,λ− z|p− |ξ − z|p
23
Trang 9The representation of vx,λ(ξ) − v(ξ) can be found similarly Finally, the positivity of the
1
kernel k for any |ξ − x| > λ and |z − x| > λ is clear due to the following magic formula
2
|ξ − x|
λ
2
|ξx,λ− z|2− |ξ − z|2= 1
λ2 λ2− |z − x|2
λ2− |ξ − x|2
3
4
For future usage, we note that |ξ − x||ξx,λ− z| = |z − x||zx,λ− ξ|; hence we can rewrite
5
the kernel k as follows
6
k(x, λ; ξ, z) = |z − x|
λ
p
|ξ − zx,λ|p− |ξ − z|p
7
Therefore, each component of ∇ξk(x, λ; ξ, z) can be easily calculated as the following
8
∂ξik(x, λ; ξ, z) =p |z − x|
λ
p
|ξ − zx,λ|p−2 ξi− (zx,λ)i
− p|ξ − z|p−2(ξi− zi)
=p |z − x|
λ
p
|ξ − x|
|z − x||ξ
x,λ− z|
p−2
ξi− (zx,λ)i
− p|ξ − z|p−2(ξi− zi)
(3.4)
9
In particular, we would have
10
∇k(x, λ; ξ, z) · ξ =p|z − x|
2|ξ − x|p−2
λp |ξx,λ− z|p−2 |ξ|2− zx,λ· ξ
− p|ξ − z|p−2(|ξ|2− z · ξ)
(3.5)
11
In the following lemma, we prove that the method of moving spheres can get started
12
starting from a very small radius
13
Lemma 5 For each x ∈ Rn, there existsλ0(x) > 0 such that
14
ux,λ(y) > u(y), vx,λ(y) > v(y)
15
for any pointy and any λ such that |y − x| > λ with 0 < λ < λ0(x)
16
Proof Since u is a positive C1function and p > 0, there exists some r0> 0 small enough
17
such that
18
∇y
|y − x|−p2u(y)· (y − x) < 0
19
for all 0 < |y − x| < r0 Consequently, by definition
20
ux,λ(y) = |y − x|
λ
p
u(yx,λ)
=|y − x|p2|yx,λ− x|−p2u(yx,λ)
>u(y)
21
for all 0 < λ < |y − x| < r0 Note that in the previous estimates, we made use of the fact
22
that if |y − x| > λ then |yx,λ− x| < λ For small λ0∈ (0, r0) and for 0 < λ < λ0, we
23
have
24
ux,λ(y) > |y − x|
λ
p
inf
B(x,r 0 )u > u(y)
25
for all |y − x|> r0 Hence, we have just shown that ux,λ(y) > u(y) for all point y and any
26
λ such that |y − x| > λ with 0 < λ < λ0 A similar argument shows that vx,λ(y) > v(y)
27
for all point y and any λ such that |y − x| > λ with 0 < λ < λ1for some λ1 ∈ (0, r1)
28
Simply setting λ0(x) = min{λ0, λ1} we obtain the desired result
29
Trang 10For each x ∈ Rnwe define
1
λ(x) = sup {µ > 0 : ux,λ(y) > u(y), vx,λ(y) > v(y), ∀0 < λ < µ, |y − x| > λ}
2
In view of Lemma 5 above, we get 0 < λ(x) 6 +∞ In the next few lemmas, we show
3
that whenever λ(x) is finite for some point x, we can write down precisely the form of
4
solutions (u, v)
5
Lemma 6 If λ(x0) < ∞ for some point x0∈ Rn
then
6
ux
0 ,λ(x0)≡ u, vx
0 ,λ(x0)≡ v
7
inRn In addition, we obtainq = 1 + 2n/p
8
Proof By the definition of λ(x0), we know that
9
ux
0 ,λ(x0)(y) > u(y), vx
0 ,λ(x0)(y) > v(y) (3.6)
10
for any |y − x0| > λ(x) In view of Lemma 4, we obtain
11
ux
0 ,λ(x0)(y) − u(y)
= Z
|z−x 0 |>λ(x 0 )
k(x0, λ(x0); y, z)
v(z)−q−
λ(x0)
|z − x0|
2n−pq+p
vx
0 ,λ(x0)(z)−q
dz, (3.7)
12
and
13
vx
0 ,λ(x0)(y) − v(y)
= Z
|z−x 0 |>λ(x 0 )
k(x0, λ(x0); y, z)
u(z)−q−
λ(x0)
|z − x0|
2n−pq+p
ux
0 ,λ(x0)(z)−q
dz, (3.8)
14
for any y ∈ Rn Keep in mind that 2n − pq + p> 0, there are two possible cases:
15
Case 1 Either ux0,λ(x0)(y) = u(y) or vx
0 ,λ(x0)(y) = v(y) for any |y − x0| > λ(x0)
16
Without loss of generality, we assume that the formal case occurs Using (3.7) and the
17
positivity of the kernel k, we get that 2n − pq + p = 0 and that vx
0 ,λ(x0)(y) = v(y) for
18
any |y − x0| > λ(x0) Hence again by (3.7) we conclude that ux0,λ(x0)(y) = u(y) in the
19
whole Rn A similar argument also shows that vx
0 ,λ(x0)(y) = v(y) in Rn and we are
20
done
21
Case 2 Or ux
0 ,λ(x0)(y) > u(y) and vx
0 ,λ(x0)(y) > v(y) for any |y − x0| > λ(x0) In
22
this case, we derive a contradiction by showing that in can slightly move spheres a little bit
23
over λ(x0) which then violates the definition of λ(x0)
24
To this purpose, still using (3.7), in the region |z − x0| > λ(x0), first we know from
25
(3.6) that
26
v(z)−q−
λ(x0)
|z − x0|
2n−pq+p
vx
0 ,λ(x 0 )(z)−q> v(z)−q− vx
0 ,λ(x 0 )(z)−q
27
Hence, by the positivity of the kernel k, we can estimate
28
ux
0 ,λ(x0)(y) − u(y) >
Z
|z−x 0 |>λ(x 0 )
k(x0, λ(x0); y, z)hv(z)−q− vx
0 ,λ(x0)(z)−qidz
(3.9)
29