Applied Mathematics, 2010, 1, 94-100
doi:10.4236/am.2010.12012 Published Online July 2010 (http://www. SciRP.org/journal/am)
Copyright © 2010 SciRes. AM
Liouville-Type Theorems for Some Integral Systems
Zhengce Zhang
College of Science, Xian Jiaotong University, Xian, China
E-mail: zhangzc@mail.xjtu.edu.cn
Received March 20, 2010; revised May 21, 2010; accepted May 27, 2010
Abstract
In this paper, Liouville-type theorems of nonnegative solutions for some elliptic integral systems are consid-
ered. We use a new type of moving plane method introduced by Chen-Li-Ou. Our new ingredient is the use
of Stein-Weiss inequality instead of Maximum Principle.
Keywords: Liouville Theorem, Integral System, Moving Plane Method
1. Introduction
In this paper we consider the nonnegative solutions of
the systems of integral equations in 2
NN,
0
N,
12
12
()
()
N
N
qq
N
pp
N
uy vy
ux dy
xy
vy uy
vx dy
xy
 
 
(1)
and
12
12
()
()
N
N
qq
N
pp
N
uy vy
ux dy
xy
vy uy
vx dy
xy
 
 
(2)
The integral systems are closely related to the follow-
ing systems of differential equations in
N
12
12
/2
/2
()
()
qq
pp
uu v
vv u
 
 
(3)
and
12
12
/2
/2
()
()
qq
pp
uuv
vvu
 
 
(4)
In fact, every positive smooth solution of PDE (3)(or
(4)) multiplied by a constant satisfies (1) (or (2)) respec-
tively. This equivalence between integral and PDE sys-
tems can be verified as in the proof of Theorem 1 in [1].
For single equations, we refer to [2]. Here, in (3) and (4),
we have used the following definition:
/2
()( )uu

 
where
is the Fourier transformation and its in-
verse.
The question is to determine for which values of the
exponents pi and qi the only nonnegative solution (u, v)
of (1) and (2) is trivial, i.e., (u; v) = (0, 0). When 2
,
is the case of the Emden-Fowler equation
0,0  u uuk in N
(5)
When )3)(2/()2(1 
NNNk , it has been pro-
ved in [3,4] that the only solutions of (5) is u = 0. In di-
mension N = 2, a similar conclusion holds for
k0.
It is also well known that in the critical case, /)2(
Nk
)2(
N, problem (5) has a two-parameter family of so-
lutions given by
2
2
2)()(

N
xxd
c
xu , (6)
where 2
1
])2([ dNNc  with d > 0 and
x
N
. If
p1,q1 > 1, p2,q2 0 and

/)2(2,2min 2121
Nqqpp
)2(
N, using Pokhozhaev’s second identity, Chen and
Lu [5] have proved that the problem (4) has no positive
radial solutions with )()( xuxu . Suppose that p1, p
2,
q1 and q2 satisfy 1,,1,0 2211
qp qp and other re-
lated conditions, using the method of integral relations,
Mitidieri [6] has proved that the problem (4) has no posi-
tive solutions of C2(N
). In the present paper, we study
Problem (1) and Problem (2) by virtue of the moving
plane method and obtain the following theorems of non-
Z. C. ZHANG
Copyright © 2010 SciRes. AM
95
existence of positive solutions in the weaker regularity.
Theorem 1.1: Let (u, v) be a nonnegative solution of
Problem (1) and

N
N
qqpp
N
N
2121,,, but not
equal to )/()(
NN at the same time. Suppose
that 1
loc
Lu (N
)2
loc
L(N
) and 2
loc
Lv (N
) 1
loc
L
(N
) with
1
1
1
1
1
q
N
N
q
,
1
1
2
2
2
q
N
N
q
,
1
1
1
1
1
p
N
N
p
and
1
1
2
2
2
p
N
N
p
. Then both u
and v are trivial.
Theorem 1.2: Let (u, v) be a nonnegative solution of
Problem (2) and 1212
,
NN
ppqq
NN
 

but
not both equal to )/()(
NN . Suppose that u, v
()
N
L
()
N
L

with 12
12
1
()1
qq
Nqq
N


and
12
12
1
()1
pp
Npp
N


. Then both u and v are trivial.
Remark 1.1: In the proof of Theorem 1.1 or Theorem
1.2, we only treat the case 
2121 ,,,qqpp
N
N
N
N or

N
N
qqpp
N
N
2121 , respec-
tively. The remaining cases can be handled in the same
way. We leave this to the interested reader.
We shall need the following doubly weighted Hardy-
Littlewood-Sololev inequality of Stein and Weiss (see,
for example, [7])
()
p
N
NqN ,, ,p,NL
L
Vx,yfydyf



 
(7)
where

 yyxxyxV ),( , pN
 /0
, 0/Nq
,
1/1/1
pp and qNp /11/)(/1 

.
There are some related works about this paper. For
2
, 0
11 qp and 2
1p, )2/()2(
2
 NNq
(but p2 and q2 are not both equal to )2/()2(
NN ),
Figueiredo and Felmer (see [8]) proved the similar Liou-
ville-type theorems to Theorem 1.1 and Theorem 1.2 on
nonnegative positive of C2(N
) using the moving plane
method and Maximum Principle. Busca and Manásevich
obtained a new result (see [9]) using the same method as
in [8]. It allows p2 and q2 to reach regions where one of
the two exponents is supercritical. In [10], Zhang, Wang
and Li first introduced the Kelvin transforms and gave a
different proof using the method of moving spheres. This
approach was suggested in [11], while Li and Zhang had
made significant simplifications in the proof of some
Liouville theorems for a single equation in [12]. In this
paper, by virtue of Hardy-Littlewood-Sololev inequality
instead of Maximum Principle, we consider the integral
systems (1) and (2) with N
0 and the general
nonlinearities. Therefore, it is a generalization of Liou-
ville-type theorems in [5,6,8,9,13-15].
Let us emphasize that considerable attention has been
drawn to Liouville-type results and existence of positive
solutions for general nonlinear elliptic equations and
systems, and that numerous related works are devoted to
some of its variants, such as more general quasilinear
operators, and domains. We refer the interested reader to
[16-24], and some of the references therein.
In the following, we shall use C to denote different
constants which depend only on N,
, i
q, i
p(i = 1,2)
and the solutions u and v in varying places.
2. Kelvin-Type Transform and Proof of
Theorem 1.1
To prove Theorem 1.1, we shall use the method of mov-
ing planes. We first introduce the Kelvin-type transforms
u and v as follows,
)()( 2
x
x
uxxuN
and )()( 2
x
x
vxxv N
which are defined for
x
N

0\ . Then by elementary
calculations, one verifies that (1) and (2) are transformed
into the following form:


 
12
12
1212
N
N
Ns s
qq
Nr rpp
u( x)xyyuyyvydy
v( x)xyyvyyuydy
 
 
 
 
(8)
and
12
12
12
12
/2
/2
()
()
ss
qq
rr
pp
uxu xv
vxv xu




(9)
with
0
)()(
ii qNNs
, )()(
 NNr
i
)2,1(0
i pi, and
 
 
12
12
N
N
Ns qq
Nr pp
u( x)xyyuyvydy
v( x)xyyvyuydy




(10)
Z. C. ZHANG
Copyright © 2010 SciRes. AM
96
and
12
12
/2
/2
()
()
sqq
rpp
uxuv
vxvu
 
 
(11)
where 0))(()( 21
 qqNNs
and
)(
Nr
0))(( 21  ppN
. Obviously, both )(xu and )(xv
have singularities at origin. Since u is locally 21

LL
and v is locally 12

LL (in Theorem 1.1), it is easy to
see that )(xu and )(xv have no singularity at infinity,
i.e., for any domain that is a positive distance away
from the origin,
dyyu )(
1
and dyyu )(
2
. (12)
dyyv)(
2
and dyyv )(
1
. (13)
For
, define
12 1
,,
N
xxxxx
 
.
Let

N
xxxx ,,221

and define
)()(
xuxu and )()(
xvxv
Lemma 2.1: For any solution (u,v) of (8), we have
dyv
y
v
yyx
yx
dyu
y
u
y
yx
yx
xuxu
q
s
q
sN
N
q
s
q
s
NN
)
11
)(
1
1
()
11
(
)
11
()()(
2
2
2
2
1
1
1
1



(14)
and
dyu
y
u
yyx
yx
dyv
y
v
y
yx
yx
xvxv
p
r
p
rN
N
p
r
p
r
NN
)
11
)(
1
1
()
11
(
)
11
()()(
2
2
2
2
1
1
1
1



(15)
Proof. Let


c
Nxxxx
11 ),(. Then it is
easy to see that
dyyv
yyx
dyyu
yyx
xu
q
sN
q
sN
2
2
1
1
)(
11
)(
11
)(
dyyu
yyx
dyyv
yyx
dyyu
yyx
yv
yyx
dyyu
yyx
q
sN
q
sN
q
sN
q
sN
q
sN
c
c
1
1
2
2
1
1
2
2
1
1
)(
11
)(
11
)(
11
)(
11
)(
11
2
2
1
1
2
2
1
1
2
2
)(
11
)(
11
)(
11
)(
11
)(
11
q
sN
q
sN
q
sN
q
sN
q
sN
yv
yyx
dyyu
yyx
dyyv
yyx
dyyu
yyx
yv
yyx
Here we have used the fact that yxyx

.
Substituting x by
x, we get
2
2
1
1
2
2
1
1
)(
11
)(
11
)(
11
)(
11
)(
q
sN
q
sN
q
sN
q
sN
yv
y
yx
dyyu
y
yx
dyyv
y
yx
dyyu
y
yx
xu
Thus
dyv
y
v
yyx
yx
dyu
y
u
y
yx
yx
xuxu
q
s
q
sN
N
q
s
q
s
NN
)
11
)(
1
1
()
11
(
)
11
()()(
2
2
2
2
1
1
1
1



Z. C. ZHANG
Copyright © 2010 SciRes. AM
97
This implies (14). Similarly, we can get (15). So
Lemma 2.1 is proved.
Proof of Theorem 1.1.
Outline: Let x1 and x2 be any two points in N
. We
shall show that
)()( 21 xuxu and )()(21 xvxv
and therefore u and v must be constants. This is impossi-
ble unless u = v = 0. To obtain this, we show that u and v
are symmetric about the midpoint (x1 + x2) / 2. Since the
integral equations are invariant under translation, we may
assume that the midpoint is at the origin. Let uand v
be the Kelvin-type transformations of u and v respec-
tively. Then what left to prove is that u and v are sym-
metric about the origin. We shall carry this out in the
following three steps.
Step 1. Define
)}()( ,{xuxuxx
u


and
)}()( ,{xvxvxx
v


We want to show that for sufficiently negative values
of
¸ both u
and v
¸ must be empty. We note that
yxyx 
¸ x, y
. Moreover, since
<
0, yy
for any
y. Then by (14), for any
u
x
,
111
222
111
222
111
1
()()
()
(()())
()
(()())
(() ())
(()())
[( )]
u
v
N
N
sqq
N
N
sqq
Ns qq
Ns qq
Ns
q
ux ux
xyx y
yuyuydy
xyxy
yvyvydy
xyyuyuydy
x
yyvyvydy
qxyyuuu








 
 

221
2
()
[()]()
u
v
Ns
q
ydy
qxyyvvvydy



It follows from (7) and then the Hölder inequality that,
for any )}/(),/(max{ 21
 sNNsNN ,
)(
1
)(
1
1
))](([



u
u
u
L
q
sN
L
dyyuuu
yyxC
uu

2
2
1
1
2
2
1
()
1
() ()
1
() ()
[()]()
v
u
uu
vv
Ns
q
L
q
LL
q
LL
xy y
vvvydy
Cuu u
vvv











(16)
where
1
1
1
1
1
q
N
N
q
and
1
1
2
2
2
q
N
N
q
.
Similarly, one can show that
1
1
2
2
()
1
() ()
1
() ()
v
vv
uu
L
p
LL
p
LL
vv
Cvv v
uuu






(17)
where
1
1
1
1
1
p
N
N
p
and
1
1
2
2
2
p
N
N
p
.
Combining (16) and (17), we have


12
12
12
12
()
1
11
()()
1
11
() ()
()
1
1
u
uu
vv
u
L
qp
LL
pq
LL
L
uu
CCu u
Cv v
uu


















(18)
By conditions (12) and (13), we can choose N suffi-
ciently large, such that for N
, we have

12
12
12
12
1
11
() ()
1
11
() ()
1
1
1 .
2
uu
vv
qp
LL
pq
LL
CCu u
Cv v














 


Then (18) implies that
0
)(  u
L
uu
,
and therefore
u
must be measure zero, and hence
empty. Similarly, one can show that v
is empty.
Step 2. Now we have that for N
,
)()( xuxu
and )()( xvxv

x. (19)
So the plane can start moving continuously from
N
to the right as long as (19) holds. Suppose that
there exists 0
0
such that, on 0
,
)()( 0xuxu
and )()( 0xvxv
,
Z. C. ZHANG
Copyright © 2010 SciRes. AM
98
but meas
00
()() 0xuxux


or meas
0
x
0
()() 0vxv x
.
We shall show that the plane can move further to the
right, i.e., there exists an
depending on N, pi, qi (i = 1,
2), and the solutions (vu,) such that (19) holds for all
),[00
 .
Once either meas
00
()()0xuxux


or
meas
00
()()0xvxvx


holds, we see by (14)
that in )()( 0xuxu
the interior of 0
. Let

)()(, 00
0
~
xuxu xx
u


and

000
, ()()
~
v
x
xvxvx

 
 .
It is easy to see that ~
0
u
has measure zero, and
~
00
0
lim 
uu


. The similar results also hold for v.
Let
be the reflection of the set about the plane
1
x. From (16) and (17), we deduce
12
12
12
12
()
1
11
(( )*)(( )*)
1
11
(( )*)(( )*)
()
(1 )
(1)
u
uu
vv
u
L
qp
LL
pq
LL
L
uu
CCuu
Cv v
uu

















(20)
Again conditions (12) and (13) ensure that one can
choose
small enough such that, for all
00
[, )

,
12
12
12
12
1
11
(( )*)(( )*)
1
11
(( )*)(( )*)
(1 )
1
(1) .
2
uu
vv
qp
LL
pq
LL
CCuu
Cv v













 

Now by (20), we have
() 0
u
L
uu

and therefore u
must be empty. Similarly,
v
must be empty, too.
Step 3. If the plane stops at 0
0
, then )(xu and
)(xv must be symmetric and monotone about the plane
x1 = 0
. This implies that )(xu and )(xv have no
singularity at the origin. But Equation (8) tells us that
this is impossible if )(xu and )(xv are nontrivial.
Hence we can move the plane to x1 = 0. Then )(xu and
)(xv must be symmetric about the origin. Thus u = v =
0. This completes the proof.
3. Proof of Theorem 1.2
In this section we establish Theorem 1.2. The proof is
along the same line of the proof of Theorem 1.1. First
establishing Lemma 3.1 which is similar to Lemma 2.1,
we omit its proof.
Lemma 3.1: For any solution (vu,) of (10), we have
dyvu
y
vu
y
yxyx
xuxu
qq
s
qq
s
NN
)
11
(
)
11
()()(
21
21 


(21)
and
dyuv
y
uv
y
yxyx
xvxv
pp
r
pp
r
NN
)
11
(
)
11
()()(
21
21 


(22)
Proof of Theorem 1.2.
Here we only provide necessary changes in Step 1, the
rest is same as the proof of Theorem 1.1. u
and
v
are defined as above. We want to show that for suffi-
ciently negative values of
, both u
and
v
must
be empty. We note that yxyx 
x,
y.
Moreover, since
< 0, yy
for any
y.
Then by (21), for any
u
x
,
12 12
21 1
122
12
1
1
2
()()
()
()()()()
()[()()]
()[()()]
[()]()
u
v
u
N
Ns
qq qq
Ns qq q
Ns qq q
Ns
qq
ux ux
xyx yy
uyvyuyvydy
x
yyvyuyuydy
x
yyuyvyvydy
qxyyuvuuydy
qx











 


21
1
[()]().
v
Ns
qq
yyvuvvydy



It follows from (7) and then the Hölder inequality that,
for any )/(
sNN ,
Z. C. ZHANG
Copyright © 2010 SciRes. AM
99






)(
1
)(
1
)(
))](([
))](([
12
2
1
u
v
u
u
u
L
qq
sN
L
q
q
sN
L
dyyvvuv
yyx
dyyuuvu
yyxC
uu


12
12
1
()() ()
1
() ()()
u
uu
vu v
qq
LLL
qq
LL L
Cuvuu
uv vv


 
 

 




(23)
Similarly, one can show that
12
12
()
11
() ()()
1
()
() ()
v
uu u
v
vv
L
pp
LL L
pp
L
LL
vv
Cvuu u
vuvv
 
 








(24)
where
1)(
1
21
21


pp
N
N
pp
and
1)(
1
21
21

qq
N
N
qq
.
Combining (23) and (24), we have
12
12
12
12
()
1
1
()()
1
()()
1
1
()
()
1
() ()()
(1
(1
u
u
u
vv
v
v
uu u
L
qq
LL
qq
LL
pp
L
L
pp
LLL
uu
CCuv
uv
Cv u
vu uu



 


 










(25)
By the hypotheses in Theorem 1.2, we can choose N
sufficiently large, such that for N
,
12
12
12
12
1
1
()
()
1
()()
1
1
()
()
1
() ()
(1
(1
1
.
2
u
u
vv
v
v
uu
qq
L
L
qq
LL
pp
L
L
pp
LL
CCuv
uv
Cv u
vu














Then (25) implies that
()
0
u
L
uu

,
and therefore u
must be measure zero, and hence
empty. Similarly
v
is also empty.
4. Acknowledgements
This work is supported by Youth Foundation of NSFC
(No. 10701061), and it was completed while the author
was visiting University of Notre Dame.
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