CLASSIFICATION OF SOLUTIONS FOR A SYSTEM OF INTEGRAL 2 EQUATIONS WITH NEGATIVE EXPONENTS VIA THE METHOD OF 3 MOVING SPHERES

3

CLASSIFICATION OF SOLUTIONS FOR A SYSTEM OF INTEGRAL
EQUATIONS WITH NEGATIVE EXPONENTS VIA THE METHOD OF
MOVING SPHERES

4

´ˆ C ANH NGO
ˆ
QUO

1
2

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



|x − y|p v(y)−q dy,
 u(x) =
Rn
(0.1)


 v(x) =
|x − y|p u(y)−q dy,
Rn

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.

5

1. I NTRODUCTION

6

Given p, q > 0 and n
1, of interest in the present note is to study non-negative
solutions of the following interesting system of integral equations in Rn



|x − y|p v(y)−q dy,
 u(x) =
Rn
(1.1)

p
−q

 v(x) =
|x − y| u(y) dy.

7

8

Rn

9
10
11

Perhaps, the motivation for studying (1.1) comes from a natural extension from one equation to a system. However, its counterpart when p, q < 0 comes from the study of the sharp
constants C(n, s, p) of the following Hardy-Littlewood-Sobolev inequality

12

Rn
13
14

15

Rn

f (x)g(y)
dxdy
|x − y|p

C(n, s, p) f

Lr

g

Ls

with 1/r + 1/s = 2 − n/p, see [Lieb83, CheLiOu05]. In the special case where u = v,
our interested system (1.1) becomes the following well-known integral equation
|x − y|p u(y)−q dy

u(x) =

(1.2)

Rn
16
17
18
19
20

in Rn where u > 0. To be precise, Eq. (1.2) has its root in geometry as it arises from
studying quantitative properties of geometric curvatures in the conformal geometry. The
classification of the solutions of (1.2) and its variations thus has a long history and is important as it also provides essential ingredients in the study of the Yamabe problem and the
prescribing scalar curvature problem, see [CafGidSpr89, CheLi91, WeiXu99, ZhaHao08]
Date: 11th May, 2015 at 22:11.
2000 Mathematics Subject Classification. 35J60, 58G35, 53C21.
Key words and phrases. Integral equation; Symmetry; Moving sphere.
1

ˆ
Q.A. NGO

2

1
2

for details. As can be observed, the corresponding partial differential equation to (1.2) is
the following well-known semilinear equation
(−∆)

3
4
5
6
7
8

n+p
2

u = u−q

(1.3)

n

with u > 0 in R . Concerning Eq. (1.2), solutions to Eq. (1.2) were first studied by Xu

in [Xu05] when p = 1 and n = 3. He provided the following classification of C 4 entire
solutions of Eq. (1.2).
Theorem 1 (see [Xu05], Thm 1.1). Suppose that u is a C 4 entire positive solution of the
integral equation
|x − y|u(y)−q dy

u(x) =

9

(1.4)

R3
10
11

in R3 with q > 0, then q = 7 and up to a constant multiplication, a translation and a
dilation, u takes the form
u(x) = 1 + |x − x|2

12

13
14
15
16
17
18

1

2

.

Note that we had already dropped the factor 1/8π in (1.4) in the original statement in
[Xu05, Theorem 1.1]. Almost simultaneously, Li studied Eq. (1.2) in its present form and
obtained, among others, the following result in [Li04].
Theorem 2 (see [Li04], Thm 1.5). For n
1, p > 0 and 0 < q
1 + 2n/p, let u be
a nonnegative Lebesgue measurable function in Rn satisfying (1.2). Then q = 1 + 2n/p
and, for some constants a, d > 0 and some x ∈ Rn ,
u(x) = a d + |x − x|2

19

p
2

.

24

Clear, Theorem 2 already includes Theorem 1. We take this chance to mention a recent
paper [XuWuTan14] in which the authors considered a slightly perturbation of (1.2) and
obtained some new phenomena. Let us just go back to the very special case when n = 3,
p = 1. In this setting, Eq. (1.3) is associated with some fourth order partial differential
equation since it simply becomes

25

(−∆)2 u = u−q

20
21
22
23

26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44

(1.5)

3

in R . Although (1.4) and (1.5) are closely related, here we would like to compare the
structure of solutions of (1.4) and (1.5). Clearly, we know from Theorem 1 that it is necessary to have q = 7 in order for (1.4) to admit solutions. In addition, we further know that
all solutions corresponding to the case q = 7 are radial symmetric and asymptotically linear at infinity. However, this is no longer true for (1.5) in the sense that it is not necessary
to have q = 7 for (1.5) to have radial symmetric solutions with asymptotically linearity at
infinity, see [Gue12]; see also [CheFanLi13, FanChe12] for related works concerning the
equivalence of integral and differential equations.
In the literature, to classify solutions of some partial differential equations, people usually use one of the following two methods: moving planes and moving spheres. While the
former method, first invented by Alexandrov in early 1950s, turns out to be powerful when
dealing with partial differential equations such as (1.3) through a series of seminal works
by Serrin [Ser71], Gidas, Ni, and Nirenberg [GidNiNir79], Caffarelli, Gidas, and Spruck
[CafGidSpr89], Chen and Li [CheLi91], the latter method is a variant of the former one.
As far as we know, this method was used by Li and Zhu in [LiZhu95].
A prior to the work by Li and Zhu, to make use of the method of moving planes, one
needs to go through two steps: First to prove solutions are radial symmetric by using the
reflection through some planes, then to classify all radial solutions by solving some transformed ordinary differential equations. In the work by Li and Zhu, the authors successfully


SYSTEM OF INTEGRAL EQUATIONS VIA THE METHOD OF MOVING SPHERES

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2
3
4
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7
8
9

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3

exploited the conformal invariance of the problem which, as a by-product, captures solutions directly rather than going the second step as in the method of moving planes, see
[Li04, Section 1]. It is worth noticing that all these “traditional” methods of moving planes
and spheres can be used to classify solutions to, for example, (1.3) because local properties
of the differential operators of the problem are well-exploited. In case of lack of knowledge
of local properties, it could prevent us from using some known results, see [CheLiOu05,
Section 1].
Now we turn our attention to problem (1.1). In the recent paper [Lei15], Lei studied
positive entire solutions (u, v) of (1.1). Among other things, Lei proved the following
result.

12

Theorem 3 (see [Lei15], Thm 1.2). Assume the positive entire solutions u, v ∈ C 1 (Rn )
of (1.1) are radially symmetric about some point x ∈ Rn , then

13

u(x) ≡ v(x) ≡ a(b2 + |x − x|2 ) 2

11

p

14

with a, b > 0.

15

As can be seen from the proof of [Lei15, Theorem 1.2], the assumption of radial property plays a key role because it allows us to prove that u ≡ v identically. Upon having this,
the form for u and v follows from known results. Also in [Lei15], the author questioned
the assumption of radial property of (u, v) in Theorem 3 can be removed. Motivated by the
question in [Lei15], in the present note we prove Theorem 3 without assuming the radial
property of solutions (u, v). To achieve that goal, we adopt the method of moving spheres
in [Li04] to introduce an integral form of the method of moving spheres for solving systems of integral equations. As far as we know, this is the first attempt to use the method of
moving spheres for systems (with negative exponents).

16
17
18
19
20
21
22
23
24
25
26
27
28

It is worth noticing that in the same paper [Lei15], the author pointed out that J. Xu has
already completed a preprint where an answer for this question was addressed using the
ideas in [LiZhu95] and [Xu07]. Unfortunately, such a preprint has not been made available
yet. Nevertheless, our proof follows [Li04] which should be different from above.
Before discussing further, we state our main result of the present note.

31

Theorem 4. For n
1, p > 0 and q > 0, let (u, v) be a pair of nonnegative Lebesgue
measurable functions in Rn satisfying (1.1). Then q = 1 + 2n/p and, for some constants
a, b > 0 and some x ∈ Rn ,

32

u(x) = v(x) = a(b2 + |x − x|2 ) 2

29
30

p

33

34
35
36
37
38
39
40
41
42
43
44

45

for any x ∈ Rn .
Note that Theorem 4 already includes the case q > 1 + 2n/p which was not considered
in [Li04], see Theorem 2 above.
Nearly a decade ago, Chen, Li, and Ou introduced an integral form of the method of
moving planes in [CheLiOu06] to classify solutions of some integral equations. Since then,
this new method has been used by many mathematicians for different problems under
various contexts. In a continued paper [CheLiOu05], Chen, Li, and Ou also introduced
an integral form of the method of moving planes for system. The basic idea of this new
method that allows the method to go through depends heavily on some new special features
possessed by integral equations. Perhaps, the global form of the integral equations allows
us to overcome the lack of local properties possessed by differential operators. In addition,
we should emphasize that the most important ingredient in this new method is to apply
the Hardy-Littlewood-Sobolev inequality in a very clever way. However, as pointed out


ˆ
Q.A. NGO

4

1
2
3
4
5
6
7
8

9
10
11
12
13
14
15

by Lei in [Lei15], it is inappropriate to apply this new method to (1.1) since the HardyLittlewood-Sobolev inequality does not work due to the presence of the negative exponent
q.
The primary aim of this note is to reformulate the method of moving spheres in an
integral form that suits for our analysis. Following the method of moving spheres by
Li and Zhu, Zhang and Hao first introduced the integral form of the method of moving
spheres in [ZhaHao08] to provide a new proof for [CheLiOu06, Theorem 1.1]. However,
the approach in [ZhaHao08] seems to be narrowed as it still requires the Hardy-LittlewoodSobolev inequality; hence it cannot be applied to our context.
To make our analysis doable, we analyze some conformal invariances found in [Li04]
for one equation and made some necessary changes for systems. For the reader’s convenience and in order to make our note self-contained, we follow the proof of [Li04, Theorem
1.5] closely with a little bit more explanation. Before closing this section, we would like to
mention that our integral system (1.1) is closely related to the following system of partial
differential equations
(−∆)

16
17

n+p
2

u = v −q ,

(−∆)

n+p
2

v = u−q ,

(1.6)

in Rn where u, v > 0. In fact, any C 2 solution (u, v) of (1.1) solves (1.6).

18

2. P RELIMINARIES AND THE METHOD OF MOVING SPHERES FOR SYSTEMS

19

In this section, we setup some preliminaries necessarily for our analysis. The most
important part of this section is the a prior estimates for solutions of (1.1) as stated in
Lemma 1 below. Here and in what follows, by and we mean inequalities up to p, q,
and dimensional constants.

20
21
22
23
24

Lemma 1. For n 1 and p, q > 0, let (u, v) be a pair of non-negative Lebesgue measurable functions in Rn satisfying (1.1). Then there hold
(1 + |y|p )u(y)

25

−q

26

27

28

31
32

33

34

dy < ∞,

(2.1)

Rn

and
lim

|x|→∞

u(x)

=
|x|p

−q

v(y)

dy,

Rn

lim

|x|→∞

v(x)
=
|x|p

−q

u(y)

dy,

(2.2)

Rn

and u and v are bounded from below in the following sense

29
30

−q

(1 + |y|p )v(y)

dy < ∞,

Rn

u(x), v(x)

1 + |x|p

(2.3)

u(x), v(x)

1 + |x|p

(2.4)

and above in the following sense
for all x ∈ Rn . In other words, there holds
1 + |x|p
C
n
in R for some constant C 1.

u(x), v(x)

C(1 + |x|p )

37

Proof. To prove our lemma, we first observe from (1.1) that both u and v are strictly
positive everywhere in Rn and are finite within a set of positive measure. Hence there
exist some R > 1 sufficiently large and some Lebesgue measurable set E ⊂ Rn such that

38

E ⊂ {y : u(y) < R, v(y) < R} ∩ B(0, R)

35
36

(2.5)


SYSTEM OF INTEGRAL EQUATIONS VIA THE METHOD OF MOVING SPHERES

1

with meas(E)

2

v(x)

1/R. Using this, we can easily bound v from below as follows
1
Rq

|x − y|p u(y)−q dy
E

3
4

5

|x − y|p dy =
E

1
Rq

|y|p dy
E+x

for any x ∈ Rn . Choose ε > 0 small enough and then fix it in such a way that vol(B(0, ε)) <
|E|/2. Then we can estimate
|y|p dy

|y|p dy

E+x

E+x\B(0,ε)

εp

5

dy
E+x\B(0,ε)

= εp |E + x| − vol(B(0, ε)) .
6
7

From this, it is clear that v is bounded from below by some positive constant. The same
reason applied to u shows that there exists some constant C0 > 0 such that
u(x), v(x) > C0

8

9
10
11

14
15
16
17
18
19

(2.6)

Proof of (2.3). To prove this, we first consider |x| 2R where R is defined through (2.5).
Note that for every y ∈ E ⊂ B(0, R), there holds |x − y| |x| − |y| |x|/2 thanks to
|x| 2R. Using this we can estimate
1
Rq

v(x)

12

13

∀x ∈ Rn .

vol(E) p
|x|
(2R)p

|x − y|p dy
E

for any |x| 2R. A similar argument also shows u(x) vol(E)(2R)−p |x|p in the region
{|x|
2R}. Hence, it is easy to select a large constant C > 1 in such a way that (2.3)
holds in the region {|x| 2R}. Thanks to (2.6), we can further decrease C, if necessary,
to obtain the estimate (2.3) in the ball {|x| 2R}; hence the proof of (2.3) follows.
Proof of (2.1). We only need to estimate v since u can be estimated similarly. To this
purpose, we first show that u−q ∈ L1 (Rn ). Clearly for some x satisfying 1

|x|
2,
there holds
|x − y|p u(y)

20

−q

dy = v(x) ∈ (0, +∞).

Rn
21

Observer that for any y ∈ Rn \B(0, 4), |x − y|
Rn \B(0,4)

23

|x − y|p u(y)

dy < +∞.

Rn

In the small ball B(0, 4), thanks to (2.3), it is obvious to verify that
u(y)−q dy

24

(1 + |y|p ))−q dy < +∞.

B(0,4)
25

−q

u(y)−q dy <

22

|y| − |x| > 1; hence

B(0,4)
−q

Thus, we have just shown that u

∈ L (Rn ). In view of (2.1), it suffices to prove that
1

|y|p u(y)−q dy < +∞.

26

(2.7)

Rn
27

To see this, we again observe that |y|
|y|p u(y)−q dy

28

Rn
29

|x − y|p u(y)
Rn

\B(0,4)

−q

dy < +∞.

\B(0,4)

In the small ball B(0, 4), it is obvious to see that
|y|p u(y)−q dy

30

B(0,4)
31

2|x − y| for all y ∈ Rn \B(0, 4). Therefore,

u(y)−q dy < +∞,

B(0,4)

thanks to u−q ∈ L1 (Rn ). From this, (2.7) follows, so does (2.1).


ˆ
Q.A. NGO

6

1
2

3

4
5
6
7
8
9

Proof of (2.2). We only consider the limit |x|−p v(x) as |x| → ∞ since the limit |x|−p u(x)
can be proved similarly. Indeed, using (1.1), we first obtain
v(x)
= lim
|x|→∞ |x|p
|x|→∞
lim

12
13
14

15

16
17
18
19
20
21

22

(2.8)

Observe that as |x| → +∞, (|x − y|/|x|) u(y)−q → u(y)−q almost everywhere y in
Rn . Hence we can apply the Lebesgue dominated convergence theorem to pass (2.8) to
the limit to conclude (2.1) provided we can show that |x − y|p |x|−p u(y)−q is bounded by
some integrable function. To this end, for each |x| 1 fixed we first split
Rn = {y ∈ Rn : |x − y|

2|x|} ∪ {y ∈ Rn : |x − y| > 2|x|} = D1 ∪ D2 .

Then, in D1 , we immediately obtain
|x − y|
|x|

10

11

Rn
p

|x − y|p
−q
u(y) dy.
|x|p

p

2p

(1 + |y|p ),

for any y ∈ D1 while in D2 we realize that |y| |x − y| − |x| |x − y|/2 which helps
us to estimate
p
|x − y|
|x − y|p |y|p 1 + |y|p
|x|
thanks to |x| 1. Thus, we have just proved that for any |x| 1, the following estimate
|x − y|
|x|

p

u(y)−q

(1 + |y|p )u(y)−q

holds. Our proof now follows by observing (1 + |y|p )u(y)−q ∈ L1 (Rn ) by (2.1).
Proof of (2.4). To see this, we first observe from (2.2) that there exists some large number
k > 1/R such that
u(x)
v(y)−q dy
<1+
n
|x|p
R
in Rn \B(0, kR). In the ball B(0, kR), it is easy to estimate |x − y|p |x|p + |y|p which
helps us to conclude that
u(x)

(1 + |y|p )v(y)−q dy

(kR)p
Rn

23
24
25

26
27
28
29

30
31

in the ball B(0, kR). From this and our estimate for u outside B(0, kR), we obtain the
desired estimate. Our estimate for v follows the same lines; hence we obtain (2.4) as
claimed.
In the next result, we prove a regularity result similar to that obtained by Li in [Li04,
Lemma 5.2].
Lemma 2. For n 1 and p, q > 0, let (u, v) be a pair of non-negative Lebesgue measurable functions in Rn satisfying (1.1). Then u and v are smooth.
Proof. Our proof is similar to that of [Li04, Lemma 5.2]. Let R > 0 be arbitrary, first we
decompose u and v into the following way
u(x) =u1R (x) + u2R (x) =

|x − y|p v(y)−q dy,

+
|y| 2R

|y|>2R

32

1
2
v(x) =vR
(x) + vR
(x) =

33
34

|x − y|p u(y)−q dy.

+
|y| 2R

|y|>2R

2
Thanks to (2.1), we immediately see that we can continuously differentiate u2R and vR
2
∞
2
under the integral sign for any |x| < R. Consequently, uR ∈ C (B(0, R)) and vR ∈


SYSTEM OF INTEGRAL EQUATIONS VIA THE METHOD OF MOVING SPHERES

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2
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C ∞ (B(0, R)). In view of (2.3) and (2.4), we know that u−q ∈ L∞ (B(0, 2R)) which
1
implies that vR
is at least H¨older continuous in B(0, R). Similar reasons tell us that u1R
is also at least H¨older continuous in B(0, R). Hence, we have just proved that u and v
are at least H¨older continuous in B(0, R), so are at least H¨older continuous in Rn since
R > 0 is arbitrary. Standard bootstrap argument shows u ∈ C ∞ (Rn ) and at the same time
v ∈ C ∞ (Rn ) follows the same lines.
Once we have the smoothness for solutions of (1.1), we can narrow the range for q as
follows.
Proposition 1. For n
1 and p, q > 0. Then in order for (1.1) to have solutions, it is
necessary to have q 1 + 2n/p.

12

Proof. The proof is just a direct consequence of [HuaYu13, Theorem 1] and Lemma 2
above, see also [Lei15, Theorem 1.1]; hence we omit its details.

13

3. T HE METHOD OF MOVING SPHERES FOR SYSTEMS

11

14
15

As a consequence of Proposition 1, from now on, we only consider the case q
2n/p. Let w be a positive function on Rn . For x ∈ Rn and λ > 0 we define

17
18

19

∀y ∈ Rn

w(ξ x,λ ),

where
ξ x,λ = x + λ2

dy =

ξ−x

λ
|z − x|

|y−x| λ

2n

dz.

(3.3)

|ξ x,λ − z x,λ |p v(z x,λ )−q
|z−x| λ

22

|ξ x,λ − z x,λ |p

=
|z−x| λ

λ
|z − x|

λ
|z − x|

2n

vx,λ (z)−q dz.

−p

|ξ x,λ − y|p v(y)−q dy
|y−x| λ

=

24

|y−x| λ

|ξ − z|p

=

−p

λ |ξ x,λ − z x,λ |
|z − x| |ξ − z|

|z−x| λ

λ
|z − x|

|ξ x,λ − y|p v(y)−q dy

2n−pq+p

vx,λ (z)−q dz.

Similarly, we obtain
λ
|ξ − x|

−p

|ξ x,λ − y|p v(y)−q dy
|y−x| λ

26

|ξ − z|p

=
|z−x| λ

λ
|z − x|

dz

2n−pq

Then, using the relation |z − x||ξ − x||ξ x,λ − z x,λ | = λ2 |ξ − z|, we obtain
λ
|ξ − x|

25

(3.2)

Note that if y = z x,λ , then z = y x,λ . Therefore, we have
|ξ x,λ − y|p v(y)−q dy =

23

(3.1)

2.

|ξ − x|
Clearly, upon the change of variable y = z x,λ , we then have

20

21

p

|ξ − x|
λ

wx,λ (ξ) =

16

1+

2n−pq+p

vx,λ (z)−q dz.


ˆ
Q.A. NGO

8

1

Lemma 3. For any solutions (u, v) of (1.1), we have
ux,λ (ξ) =

2

λ
|z − x|

2n−pq+p

|ξ − z|p

λ
|z − x|

2n−pq+p

|ξ − z|p

Rn
3

and
vx,λ (ξ) =

4

Rn
5

for any ξ ∈ Rn .

6

Proof. Using our system (1.1), we obtain
u(ξ x,λ )
p

|ξ x,λ − y|p v(y)−q dy
Rn

|ξ − z|p

=

ux,λ (z)−q dz

p

|ξ − x|
λ
|ξ − x|
=
λ

ux,λ (ξ) =
7

vx,λ (z)−q dz

Rn

λ
|z − x|

2n−pq+p

vx,λ (z)−q dz.

8

The formula for v follows the same line as above.

9

Lemma 4. For any solutions (u, v) of (1.1), we have

10

ux,λ (ξ) − u(ξ) =

λ
|z − x|

2n−pq+p

k(x, λ; ξ, z) v(z)−q −

λ
|z − x|

2n−pq+p

k(x, λ; ξ, z) u(z)−q −

|z−x| λ
11

12

and
vx,λ (ξ) − v(ξ) =
|z−x| λ

13

vx,λ (z)−q dz,

ux,λ (z)−q dz,

for any ξ ∈ Rn where
k(x, λ; ξ, z) =

14

|ξ − x|
λ

p

|ξ x,λ − z|p − |ξ − z|p .

15

Moreover, k(x, λ; ξ, z) > 0 for any |ξ − x| > λ > 0 and |z − x| > λ > 0.

16

Proof. First, we observe that
|ξ − z|p

ux,λ (ξ) =
|z−x| λ
17

18

vx,λ (z)−q dz

p

|ξ − x|
λ

+

2n−pq+p

λ
|z − x|

|ξ x,λ − z|p v(z)−q dz
|z−x| λ

and
|ξ x,λ − z|p v(z)−q dz

u(ξ) =
|z−x| λ
19

+
20

21

|ξ − x|
λ

p

|ξ x,λ − z|p
|z−x| λ

23

2n−pq+p

vx,λ (z)−q dz.

Therefore,
k(x, λ; ξ, z) v(z)−q −

ux,λ (ξ) − u(ξ) =
|z−x| λ

22

λ
|z − x|

where
k(x, λ; ξ, z) =

|ξ − x|
λ

λ
|z − x|

2n−pq+p

p

|ξ x,λ − z|p − |ξ − z|p .

vx,λ (z)−q dz,


SYSTEM OF INTEGRAL EQUATIONS VIA THE METHOD OF MOVING SPHERES

1
2

9

The representation of vx,λ (ξ) − v(ξ) can be found similarly. Finally, the positivity of the
kernel k for any |ξ − x| > λ and |z − x| > λ is clear due to the following magic formula
2

3

4

5
6

7

8

|ξ − x|
1
|ξ x,λ − z|2 − |ξ − z|2 = 2 λ2 − |z − x|2 λ2 − |ξ − x|2 .
λ
λ
Thus the proof follows.
For future usage, we note that |ξ −x||ξ x,λ −z| = |z −x||z x,λ −ξ|; hence we can rewrite
the kernel k as follows
p

|z − x|
k(x, λ; ξ, z) =
|ξ − z x,λ |p − |ξ − z|p .
λ
Therefore, each component of ∇ξ k(x, λ; ξ, z) can be easily calculated as the following
∂ξi k(x, λ; ξ, z) =p

|z − x|
λ

p

|ξ − z x,λ |p−2 ξi − (z x,λ )i

− p|ξ − z|p−2 (ξi − zi )
9

|z − x|
=p
λ

p

|ξ − x| x,λ
|ξ − z|
|z − x|

(3.4)

p−2

ξi − (z

x,λ

)i

− p|ξ − z|p−2 (ξi − zi ).
10

In particular, we would have
|z − x|2 |ξ − x|p−2 x,λ
|ξ − z|p−2 |ξ|2 − z x,λ · ξ
λp
− p|ξ − z|p−2 (|ξ|2 − z · ξ).

∇k(x, λ; ξ, z) · ξ =p
11

(3.5)

13

In the following lemma, we prove that the method of moving spheres can get started
starting from a very small radius.

14

Lemma 5. For each x ∈ Rn , there exists λ0 (x) > 0 such that

15

ux,λ (y)

12

16

u(y),

for any point y and any λ such that |y − x|

vx,λ (y)

v(y)

λ with 0 < λ < λ0 (x).

19

Proof. Since u is a positive C 1 function and p > 0, there exists some r0 > 0 small enough
such that
p
∇y |y − x|− 2 u(y) · (y − x) < 0

20

for all 0 < |y − x| < r0 . Consequently, by definition

17

18

ux,λ (y) =
21

|y − x|
λ
p

p

u(y x,λ )
p

=|y − x| 2 |y x,λ − x|− 2 u(y x,λ )
>u(y)

22
23
24
25
26
27
28
29

for all 0 < λ < |y − x| < r0 . Note that in the previous estimates, we made use of the fact
that if |y − x| > λ then |y x,λ − x| < λ. For small λ0 ∈ (0, r0 ) and for 0 < λ < λ0 , we
have
p

|y − x|
ux,λ (y)
inf u u(y)
λ
B(x,r0 )
for all |y − x| r0 . Hence, we have just shown that ux,λ (y) u(y) for all point y and any
λ such that |y − x| λ with 0 < λ < λ0 . A similar argument shows that vx,λ (y) v(y)
for all point y and any λ such that |y − x| λ with 0 < λ < λ1 for some λ1 ∈ (0, r1 ).
Simply setting λ0 (x) = min{λ0 , λ1 } we obtain the desired result.


ˆ
Q.A. NGO

10

1

For each x ∈ Rn we define

2

λ(x) = sup {µ > 0 : ux,λ (y)

u(y), vx,λ (y)

v(y),

∀0 < λ < µ, |y − x|

λ} .

5

In view of Lemma 5 above, we get 0 < λ(x) +∞. In the next few lemmas, we show
that whenever λ(x) is finite for some point x, we can write down precisely the form of
solutions (u, v).

6

Lemma 6. If λ(x0 ) < ∞ for some point x0 ∈ Rn then

3
4

ux0 ,λ(x0 ) ≡ u,

7

vx0 ,λ(x0 ) ≡ v

8

in Rn . In addition, we obtain q = 1 + 2n/p.

9

Proof. By the definition of λ(x0 ), we know that
ux0 ,λ(x0 ) (y)

10

11

for any |y − x0 |

u(y),

vx0 ,λ(x0 ) (y)

v(y)

(3.6)

λ(x). In view of Lemma 4, we obtain

ux0 ,λ(x0 ) (y) − u(y)
k(x0 , λ(x0 ); y, z) v(z)

=

12

−q

−

|z−x0 | λ(x0 )

λ(x0 )

|z − x0 |

2n−pq+p

vx0 ,λ(x0 ) (z)−q dz,
(3.7)

13

and
vx0 ,λ(x0 ) (y) − v(y)

14

k(x0 , λ(x0 ); y, z) u(z)−q −

=
|z−x0 | λ(x0 )

λ(x0 )
|z − x0 |

2n−pq+p

ux0 ,λ(x0 ) (z)−q dz,
(3.8)

15
16
17

18
19
20
21
22
23
24
25
26

27

28

29

for any y ∈ Rn . Keep in mind that 2n − pq + p

0, there are two possible cases:

Case 1. Either ux0 ,λ(x0 ) (y) = u(y) or vx0 ,λ(x0 ) (y) = v(y) for any |y − x0 |
λ(x0 ).
Without loss of generality, we assume that the formal case occurs. Using (3.7) and the
positivity of the kernel k, we get that 2n − pq + p = 0 and that vx0 ,λ(x0 ) (y) = v(y) for
any |y − x0 | λ(x0 ). Hence again by (3.7) we conclude that ux0 ,λ(x0 ) (y) = u(y) in the
whole Rn . A similar argument also shows that vx0 ,λ(x0 ) (y) = v(y) in Rn and we are
done.
Case 2. Or ux0 ,λ(x0 ) (y) > u(y) and vx0 ,λ(x0 ) (y) > v(y) for any |y − x0 |
λ(x0 ). In
this case, we derive a contradiction by showing that in can slightly move spheres a little bit

over λ(x0 ) which then violates the definition of λ(x0 ).
To this purpose, still using (3.7), in the region |z − x0 |
(3.6) that
v(z)−q −

λ(x0 )
|z − x0 |

λ(x0 ), first we know from

2n−pq+p

vx0 ,λ(x0 ) (z)−q

v(z)−q − vx0 ,λ(x0 ) (z)−q .

Hence, by the positivity of the kernel k, we can estimate
ux0 ,λ(x0 ) (y) − u(y)

|z−x0 | λ(x0 )

k(x0 , λ(x0 ); y, z) v(z)−q − vx0 ,λ(x0 ) (z)−q dz.
(3.9)


SYSTEM OF INTEGRAL EQUATIONS VIA THE METHOD OF MOVING SPHERES

1
2

11

Estimate of ux0 ,λ − u outside B(x0 , λ(x0 ) + 1). Using the Fatou lemma, from (3.9) we
obtain
lim inf |y|−p (ux0 ,λ(x0 ) − u)(y)
|y|→∞

lim inf
|y|→∞

3

|z−x0 | λ(x0 )

|y|−p k(x0 , λ(x0 ); y, z) v(z)−q − vx0 ,λ(x0 ) (z)−q dz

|z|
λ(x0 )

|z−x0 | λ(x0 )

p

−1

v(z)−q −

2n−pq+p

λ(x0 )

|z − x0 |

vx0 ,λ(x0 ) (z)−q dz

>0,
4

5

6

where we have used the following formula
k(x0 , λ(x0 ); y, z)
=
|y|p

lim inf
|y|→∞

9
10
11

14

15

16
17

18
19
20
21
22

|y x0 ,λ − z|p
−
λ(x0 )

|y − z|
|y|

p

|z|
λ(x0 )

k(x0 , λ(x0 ); y, z)
|y|p

p

− 1.

|y|p while
As a consequence, outside a large ball, we would have (ux0 ,λ(x0 ) − u)(y)
in that ball and outside of B(x0 , λ(x0 ) + 1) we would also have (ux0 ,λ(x0 ) − u)(y)
|y|p thanks to the smoothness of ux0 ,λ(x0 ) − u and our assumption ux0 ,λ(x0 ) (y) > u(y).
Therefore, there exists some ε1 > 0 such that

(ux0 ,λ(x0 ) − u)(y)

12
13

p

to obtain the following estimate

7

8

|y − x0 |
|y|

ε1 |y|p

for all |y − x0 | λ(x0 ) + 1. Recall that ux0 ,λ (y) = (|x0 − y|/λ)p u(y x0 ,λ ); hence there
exists some ε2 ∈ (0, ε1 ) such that
(ux0 ,λ − u)(y) =(ux0 ,λ(x0 ) − u)(y) + (ux0 ,λ − ux0 ,λ(x0 ) )(y)
ε1 p
ε1 |y|p + (ux0 ,λ − ux0 ,λ(x0 ) )(y)
|y|
2

(3.10)

λ(x0 ) + 1 and all λ ∈ (λ(x0 ), λ(x0 ) + ε2 ). Repeating the above
for all |y − x0 |

arguments shows that (3.10) is also valid for vx0 ,λ − v, that is
ε1 p
(vx0 ,λ − v)(y)
|y|
(3.11)
2
for a possibly new constant ε1 > 0.
Estimate of ux0 ,λ − u inside B(x0 , λ(x0 ) + 1). Now for ε ∈ (0, ε2 ) to be determined later
and for λ ∈ (λ(x0 ), λ(x0 ) + ε) ⊂ (λ(x0 ), λ(x0 ) + ε2 ) and for λ |y − x0 | λ(x0 ) + 1,
from (3.9), we estimate
k(x0 , λ; y, z)[v(z)−q − vx0 ,λ (z)−q ]dz

(ux0 ,λ − u)(y)
|z−x0 | λ(x0 )

k(x0 , λ; y, z)[v(z)−q − vx0 ,λ (z)−q ]dz
λ(x0 )+1 |z−x0 | λ

k(x0 , λ; y, z)[v(z)−q − vx0 ,λ (z)−q ]dz

+
λ(x0 )+3 |z−x0 | λ(x0 )+2

23

λ(x0 )+1 |z−x0 | λ

k(x0 , λ; y, z)[vx0 ,λ(x0 ) (z)−q − vx0 ,λ (z)−q ]dz
k(x0 , λ; y, z)[v(z)−q − vx0 ,λ (z)−q ]dz

+
λ(x0 )+3 |z−x0 | λ(x0 )+2

=I + II.


ˆ
Q.A. NGO

12

1
2

As we shall see later, I + II
by term.

0 provided ε > 0 is small. We now estimate I and II term

4

(z) δ1 for
Estimate of II. Thanks to (3.11), there exists δ1 > 0 such that v −q − vx−q
0 ,λ
any λ(x0 ) + 2 |z − x0 | λ(x0 ) + 3. Note by the definition of k given in Lemma 4 that

5

k(x0 , λ; y, z) = k(0, λ; y − x0 , z − x0 )

3

6

and from (3.5) there holds
∇y k(0, λ; y, z) · y

7

8
9

|z|
for all λ(x0 ) + 2
independent of ε such that

12
13

16

δ2 (|y| − λ)

k(0, λ; y, z)

for all λ(x0 ) λ |y| λ(x0 )+1 and all λ(x0 )+2 |z| λ(x0 )+3. Simply replacing
y by y − x0 and z by z − z0 and making use of the rule k(x0 , λ; y, z) = k(0, λ; y − x0 , z −
x0 ), we obtain with the same constant δ2 > 0 as above the following estimate
k(x0 , λ; y, z)

14
15

= p|y − z|p−2 |z|2 − |y|2 > 0

λ(x0 ) + 3. Hence, there exists some constant δ2 > 0 is

10
11

|y|=λ

for all λ(x0 ) λ |y − x0 |
Thus, we have just proved that

|z − x0 |

λ(x0 ) + 1 and all λ(x0 ) + 2

δ1 δ2 (|y − x0 | − λ)

II

17

δ2 (|y − x0 | − λ)
λ(x0 ) + 3.

dz.

(3.12)

λ(x0 )+3 |z−x0 | λ(x0 )+2
18

Estimate of I. To estimate I, we first observe that
|vx0 ,λ(x0 ) −q − v −q |(z)

19

20

for all λ(x0 )

|z − x0 |

λ

λ − λ(x0 )

λ(x0 ) + 1 and all λ(x0 )

λ(x0 ) + ε and that

k(0, λ; y − x0 , z)dz

k(x0 , λ; y, z)dz =
λ |z−x0 | λ+1

λ

ε

λ |z| λ+1

λ |z| λ+1
21

|y − x0 |
λ

p

− 1 |(y − x0 )0,λ − z|p dz

(|(y − x0 )0,λ − z|p − |(y − x0 ) − z|p )dz

+
λ |z| λ+1

C(|y − x0 | − λ) + C|(y − x0 )0,λ − (y − x0 )|
C(|y − x0 | − λ).
22

where C > 0 is constant independent of ε. Thus, we obtain
I

23

−Cε

k(x0 , λ; y, z)dz.

(3.13)

λ(x0 )+1 |z−x0 | λ
24

25

Thus, by combining (3.13) and (3.12), it follows that for some small ε > 0 we have
(ux0 ,λ − u)(y)

dz − Cε (|y − x0 | − λ)

δ1 δ2

0

λ(x0 )+3 |z−x0 | λ(x0 )+2
26
27
28

29

for λ(x0 )

λ

λ(x0 ) + ε and λ

|y − x0 |

λ(x0 ) + 1.

Estimates of ux0 ,λ −u and vx0 ,λ −v when |y−x0 | λ(x0 )+1. Combining the preceding
estimate for ux0 ,λ − u in the ball B(x0 , λ(x0 ) + 1) and (3.10) above gives
(ux0 ,λ − u)(y)

0


SYSTEM OF INTEGRAL EQUATIONS VIA THE METHOD OF MOVING SPHERES

13

for λ(x0 ) λ λ(x0 ) + ε and λ |y − x0 |. Again by repeating the whole procedure
above for the difference vx0 ,λ − v, we can conclude that
(vx0 ,λ − v)(y)
1
2

3
4
5
6
7
8

0

for λ(x0 ) λ λ(x0 ) + ε and λ |y − x0 | where ε could be smaller if necessary; thus
giving us a contradiction to the definition of λ(x0 ).
In the last lemma of the current section, we prove that whenever λ(x0 ) < ∞ for some
point x0 ∈ Rn , there must hold λ(x) < ∞ for any point x ∈ Rn .
Lemma 7. If λ(x0 ) < ∞ for some point x0 ∈ Rn then λ(x) < ∞ for any point x ∈ Rn ;
hence
ux,λ(x) ≡ u, vx,λ(x) ≡ v
for all x ∈ Rn in Rn .
Proof. Suppose that there exists some x0 ∈ Rn such that λ(x0 ) < ∞, then by Lemma 6
and for |y| sufficiently large, we have
−p

|y|

u(y) = |y|−p ux0 ,λ(x0 ) (y)
−p

= |y|

−p

lim |y|

|y|→∞

lim |y|

ux,λ (y)

2

|y − x0 |

(3.14)

v(y) = λ(x0 )−p v(x0 ).

u(y),

vx,λ (y)

for all 0 < λ < λ(x) and all x, y such that |y − x|
thanks to (3.14), one can easily see that
−p

lim inf |y|
|y|→∞

17

y − x0

2

(3.15)

Let x ∈ Rn be arbitrary. By the definition of λ(x) we get

14

16

u x0 + λ(x0 )

u(y) = λ(x0 )−p u(x0 ).

−p

|y|→∞

15

p

2

Repeating the above argument then gives

12

13

|y − x0 |

which implies

10

11

y − x0

2

u x0 + λ(x0 )

|y − x0 |
|y|

= λ(x0 )−p
9

−p

λ(x0 )
|y − x0 |

u(y)

−p

lim inf |y|
|y|→∞

−p

= lim inf |y|
|y|→∞

v(y)

λ. Then by a direct computation and

ux,λ (y)
λ
|y − x|

−p

u x + λ2

y−x
2

(3.16)

|y − x|

= λ−p u(x)

19

for all 0 < λ < λ(x). Combining (3.14) and (3.16) gives λ(x0 )−p u(x0 ) λ−p u(x) for
all 0 < λ < λ(x). From this, we obtain λ(x) < ∞ for all x ∈ Rn as claimed.

20

4. P ROOF OF T HEOREM 4

21

To conclude Theorem 4, we first recall the following two calculus lemmas from [Li04].
These two lemmas have been used repeatedly in many works

18

22


ˆ
Q.A. NGO

14

1

Lemma 8. For ν ∈ R and f a function defined on Rn and valued in [−∞, +∞] satisfying
λ
|y − x|

2

3
4
5

6

7

8

9
10
11
12

ν

x + λ2

f

y−x
|y − x|

f (y)

2

for all x, y satisfying |x − y| > λ > 0. Then f is constant or is identical to infinity.
Lemma 9. For ν ∈ R and f a continuous function in Rn . Suppose that for every x ∈ Rn ,
there exists λ(x) > 0 such that
λ(x)
|y − x|
Then for some a

ν

f

x + λ(x)

2

y−x

∀y ∈ Rn \ {x}.

= f (y),

2

|y − x|

0, d > 0 and x ∈ Rn
f (x) = ±a

ν
2

1

.

2

d + |x − x|

Now, to prove Theorem 4, we consider the two possible cases:
Case 1. If λ(x) = ∞ for any x ∈ Rn then ux,λ (y) u(y) for all λ > 0 and for any x, y
satisfying |y − x|
λ. By Lemma 8, u must be constant. Similarly, v is also constant.
However, this is not the case since constant solutions do not solve (1.1).

14

Case 2. If there exists some x0 ∈ Rn such that λ(x0 ) < ∞, then by Lemma 7, we deduce
that λ(x) < ∞ for any point x ∈ Rn . Then Lemma 9 tells us that u is of the form

15

u(x) = a1 (b21 + |x − x1 |2 ) 2

13

p

16

(4.1)

n

for some a1 , d1 > 0 and some point x1 ∈ R . Similarly v is of the following form
p

17

18
19

20

v(x) = a2 (b22 + |x − x2 |2 ) 2

(4.2)

n

for some a2 , d2 > 0 and some point x2 ∈ R . To realize that u ≡ v, we observe that u
given in (4.1) solves the following equation
|x − y|p u(y)−(1+2n)/p dy,

u(x) =
Rn

21

see [Li04, Appendix A]. From this and (1.1), we must have u ≡ v in Rn and hence
p

22
23

u(x) = v(x) = a(b2 + |x − x|2 ) 2
for some constants a, b > 0 and some x ∈ Rn as claimed.

24

ACKNOWLEDGMENTS

25

27

The author wants to thank NINH Van Thu and DO Duc Thuan for useful discussion.
Thanks also go to the Vietnam Institute for Advanced Study in Mathematics (VIASM) for
hosting and support where this note were done.

28

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SYSTEM OF INTEGRAL EQUATIONS VIA THE METHOD OF MOVING SPHERES

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(Q.A. Ngˆo) D EPARTMENT OF M ATHEMATICS , C OLLEGE OF S CIENCE , V I Eˆ T NAM NATIONAL U NIVER ˆ , V I Eˆ T NAM .
H A` N OI

SITY,

33

E-mail address:

34

E-mail address: bookworm