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Advances in Pure Mathematics, 2013, 3, 153-158
http://dx.doi.org/10.4236/apm.2013.31A021 Published Online January 2013 (http://www.scirp.org/journal/apm)
Regularity of Solutions to an Integral Equation
on a Half-Space n
R
*
Linfen Cao1, Zhaohui Dai2
1College of Mathematics and Information Science, Henan Normal University, Xinxiang, China
2Department of Comp u t e r Science, Henan Normal University, Xinxiang, China
Email: caolf2010@yahoo.com, dzhsoft@sina.com
Received September 18, 2012; revised October 21, 2012; accepted October 29, 2012
ABSTRACT
n
RIn this paper, we discuss the integral equation on a half space
  
11 d,
n
>0, .
p
n
xx R
nn
R
uxuy yu
xy xy







0n
(0.1)
where


11
,, ,
nn
, xxx
n
R is the reflection of th e point x about the
x
. We study the regularity for the
positive solutions of (0.1). A regularity lifting method by contracting operators is used in proving the boundedness of
solutions, and the Lipschitz continuity is d erived by combinations of contracting and shrinking operators introduced by
Ma-Che n-L i ( [ 1 ]).
Keywords: Regularity Lifting; HLS Inequality; Contracting Operators; Shrinking Operators
1. Introduction
Let be the upper half Euclidean space
n
R


12
,,,| 0.
nn
nn
Rxxx xRx
 
n
R
In this paper we consider the regularity of positive
solution of the following integral equation in
 
11 d, .
npn
nn
R
uxuyyx R
xy xy







1pn
R

(1.1)
where . It relates closely to the higher-order PDEs
with Navier boundary conditions in :
 
2in;
0, on.
pn
n
uu R
uuu R
 
  
1
2
,0u
(1.2)
D. Li and R. Zhuo proved the following result:
Proposition 1.1. ([2]) Let
be an even number and
n
pn

ux
2
. If is the smooth solution of the inte-
ux
gral Equation (1.1), then satisfies the PDEs (1.2).
In particular, when
and 2
2
n
pn
, Chen and
Li ([3]) showed the equivalence between the integral
Equation (1.1) and partial differential Equation (1.2). For
more results concerning integral equations, see [4-6].
Firstly, in this paper we have the boundedness for the
positive solutions of (1.1) by using the contracting op-
erators.
Theorem 1.1. Let u be a solution of (1.1). If n
pn
,
and


1np n
uL R

, then u is in
rn n
LR LR

1r
for
any .
Remark 1. In [2], the authors prov ed that Theorem 1.1
is true for the critical case n
pn
. While our result
also covers subcritical case nn
p
nn


and super
critical case n
pn
.
Then we employ the brand new method which is the
combinations of contracting and shrinking operators in-
troduced by Ma-Chen-Li ([1]) to derive the Lipschitz
continuity of solutions.
*This work is supported by grant No.11001076 and 11171091 of NSFC,
N
SF of Henan Provincial Education Committee (No. 2011A110008)
and Foundation for University Key T ea c h e r o f H en a n Prov i n c e.
C
opyright © 2013 SciRes. APM
L. F. CAO, Z. H. DAI
154
Theorem 1.2. Under the same conditions of Theorem
1.1, u is Lipschitz continuous in .
n
R
L
L
2. Estimate by Contracting Operators
In this section, we obtain estimate for positive solu-
tions to the equation (0.1) by using the contracting op-
erators. To prove the Theorem 1.1, we need the follow-
ing equivalent form of Hardy-Littlewood-Sobolev ine-
quality.
Lemma 2.1. Let

nr n
nr
g
LR
for nr
n

.
Define
 
d.
ng yy
1
n
R
Tg xxy
(2.1)
Then



.
nr n
nr
L R
g

rn 1r
,,
rn
LR
TgC nr
Proof of Theorem 1.1: The proof is divided into two
steps.
Step 1. We first show that ,

uxL R

n,
x
R

 
.ax ux
. Define
1p
Then
 
d
ayuy y
11
nnn
R
ux xy xy







For a positive number A, defin e
 
,if
0, elsewher
A
ax ax
ax 
,or
e.
AxA
 
.
BA
ax ax
Let

ax
Obviously,

B, and vanishes outside
the ball ax A

B
ax
0
A
B.
Define
 
d.
yv yy
11
n
AA
nn
R
Tv xa
xy xy







 
d.
11
n
AB
nn
R
F
xa
xy xy







yuyy
 
.
AA
xF x
1r
n
The Equation (0.1) can be rewritten as

ux Tu
We will show that, for any ,
1) TA is a contracting map from to
r
LR
rn
LR
for A large, and
rn
LR
2) A
F
x is in
.
rn
vLR
1) Assume
, then

11 d.
n
AA
nn
R
Tvayv yy
xy xy







For any n
rn
, we apply Hardy-Littlewood-Sobo-
lev inequality and Hölder inequality to obtain
 
.
nr n
rnnn rn
nr
AA A
LRLRL RLR
Tv CavCav




Since

nn
axLR
, by the definition of
Axa ,
one can choose a large number A, such that

1.
2
nn
ALR
Ca
and hence arrives at
 
1.
2
rn rn
ALR LR
Tv v


:rn rn
AR LRTL
That is

is a contracting op-
erator.
2) Consider

11 d.
n
AB
nn
R
xayuyy
xy xy







F
For any n
rn
, we apply Hardy-Littlewood-Sobo-
lev inequality and Hölder inequality to obtain
 
.
nr
rnnsn tn
nr
AB B
LRL RLRLR
FCau Cau


We require
11
,, 1.
nr st
nrst

By the bounded-ness of
B
a

, we see that s can be
arbitrary. Since

1np n
uL R
, we take 1np
t
,
and hence
 


11
,as ,
12
2
np np
rs
np p
p

s




we see
1
2
np
p
A
FL
0, for any small . Obviously,

11
2
np np
p

1p
2p2p
since .
If , we are done. If , repeat the above
Copyright © 2013 SciRes. APM
L. F. CAO, Z. H. DAI 155
process and after a few steps, we arrive at


,.
nr
n


rn
ux L R
Step 2. In this step we will show that
.
n
L R
nux
For any point
x
R
, we divide the integral into two
parts









1
1
1
1
\
\
12
11
11
1d
1d
n
n
nn
Bx
nn
RB
x
p
n
Bx
p
n
RBx
ux xy xy
xy xy
uyy
xy
uyy
xy
II









d
d
p
p
uyy
uyy



Consider 2
I
. Since 1
n
xy

n
R
, and by the result in
Step 1, , for

r
ux Lnr
n

, we have
21
I
C.
For 1
I
, we apply Hölder inequality
  

11
1
11q
nq
Bx Bx
Iy
xy
1
1
dd
q
qq
pq
uyy








Choose appropriate q, so that
nqn

, and
hence
 
1
1
Bx xy
1
2
dq
nq
yC




Since


,n
rn
rn
ux L R

,


1
q
pq
Bxuy
1
13
d
q
q
y C





We conclude that
.
n
L R

ux

n
LR


0,1 n
ux CR
ux
3. Lipschitz Continuity by Combinations of
Contracting Operators and Shrinking
Operators
In the previous section we showed that the solution
of (0.1) is in . In this section, we will use
the regularity lifting by combinations of contracting and
shrinking operators to prove , the space
of Lipschitz continuous functions with norm

 

0,1 sup
nn
CRLR xy
vx vy
vv xy







(3.1)
To prove the Theorem 1.2, we need introduce the fol-
lowing definition, property and a more general Regular-
ity Lifting Theorem on the combined use of contracting
and shrinking operators.
Let V be a Hausdorff topological vector space. Sup-
pose there are two extented norms (i.e. the norm of an
element in V might be infinity) defined on V,

::and:: .
XY
XvVvYvVv
 
Definition. (“XY-pair”) Suppose X, Y are two normed
subspaces described above, X and Y are called “XY-pair”,
if whenever the sequence
n
uXn
uu with in X
and nY
uC will imply . uY
Remark 2. The “XY-pair” are quite common, here we
choose
rn
0,1 n
RYC
X
LR
r for 1, and
X
Y:TXX
with the norm defined in (3.1).
Theorem 3.1. (Regularity Lifting Theorem) Suppose
Banach spaces X, Y are an XY-pair”, and let and
be closed subsets of X and Y respectively. Suppose
is
a contraction:
,,forsome0 1;
XX
TfTgfgf g

 X
:TYY
and is shrinking:
,,forsome01.
YY
Tggg

 Y
for some.
Sf Tf FF
Define
XY
:.SXY XY
inuTuF
Moreover, assume that
Then there exists a solution u of equation
X
.uY
and more importantly,
The proof and some applications of Theorem 3.1 can
be found in [1,7,8]. n
Proof of Theorem 1.2: For any
x
R


, by elemen-
tary calculus one can verify that


11
0
dd
dd
11 d.
n
t
n
pp
nn
BxR xy
p
n
R
tt
uyy uyy
tt
uyy
nxy


 
 
It follows that the solution of (0.1) only differs by a
constant multiple from the solution of the following
equation
Copyright © 2013 SciRes. APM
L. F. CAO, Z. H. DAI
APM
156

Copyright © 2013 SciRes.




1
d
d.
p
n
t
y yt




and
0d
tt
p
Bx Bx
ux uy yu

  (3.2)
Hence, for convenience of argument, we prove that
every positive solution u of (3.2) is Lipschitz continuous.
Let

2
LL
v u

n
vXLR
X
and

0,1 n
vY CR
Y2.
LL
v u

0

For every , define




1
d
dd.
n
t
yy t
0tt
pp
Bx Bx
Tv xvy yv







 
1
d
dd.
tt
pp
n
Bx Bx
t
Fxuy yuy yt




 
.vTvF
Then obviously , u is a solution of the equation
.Sv Tv F
Write

TX
TY
F
We will show that for suffi-
ciently small,
1) is a contracting operator from to X.
2) is a shrinking operator from to Y.
XY:SXYXY
,fg
and . 3)
1) For any Xn
and for any
x
R
 

 


 

, we have
1
d
dd
tt
pppp n
Bx Bx
t
Tfx Tgxfyg yyfyg yyt





 
 
0

Thus

 

 

 

 



1
0
11
12
1
1
1
0
11
12
1
0
d
dd
d
dd
d
d.
tt
tt
ppppn
Bx Bx
pp
n
Bx Bx
p
tt
n
LL
pp
LLLL
t
TfxTg xfygyyfygyyt
t
pyfygy yyfygy yt
t
Cuf gBxBxt
t
Cu fgCufg
t
















 

 
 

0
and consequently Here we applied the Mean Value Theorem with both
1

y

2
y

f
y and valued between and
g
y

,
and t
B denotes the volume of the ball
t
B
.
Choose sufficiently small such that
1
2,
4
p
L
Cu
1
1.
4
LL
Tf Tgfg



TX
v
Therefore is a contracting operator from to X
for such a small .
Y,n
, then for any
x
2) Assume
zR
 








,

 





1
0
1
0
d
ddd d
d
dd
ttt t
tt
ppp p
n
Bx BzBxBz
pppp n
Bx Bx
t
TvxTvzvyy vyyvyyvyyt
t
vyvyzxyvyvyz xyt







 
















 
 




the last equality above is from the fact

ddd
tt t
pp p
Bz BxBx
vyy vyzxyvyzxy


 
 
Therefore

 





 






0,1 0,1
11
00
1 1
1 2
1 1
00
1 1
1
1
0
dd
dd
dd
dd
d.
t t
t t
pp
nn
Bx Bx
p p
n n
Bx Bx
p p
tt
n
LC LC
tt
TvxTvzzxyvyvyz xy
tt
tt
pvvy vyzxypvvyvyzxy
tt
t
CuvzxB xBxCuvzx
t

 

 

 

pp
vyvy

 

 
 
 
 
 



L. F. CAO, Z. H. DAI 157
Again choosing sufficiently small, we derive
by

0,1
1.
4C
zv
Combining this with the estimate in 1), we arrive at
sup
xz
Tv xTv
zx

0,1 0,1
1,.
2
CC
TvvvY
Hence T is a shrinking rator from Y to Y.
3) To show F is Lipschitz continuous, we split it into
two parts:









ope

tt
Ix Ix
1
1
1
1
12
d
dd
d
dd
tt
pp
n
Bx Bx
pp
n
Bx Bx
t
Fxuy yuy yt
t
uyyuyy
t








 
 
For the first part, we have






 



1
11 d
tt
pp
Bx Bz
Ix Izu yyu


 
1
111
1
1
d
dd d
d1.
t t
p p
n
Bx Bz
pp
nn
LL
t
yyuyyu yy t
t
CutxzxzCuxz
t



 


 



 
useHere wed the fact that

1
the volume of\\.
n
tt tt
Bx BzBz BxCtxz


It follows that

11 .
sup IxIzC
(3.3)
xz xz
For the second part, we use a different approach. Write
x
z
, then
 










11
1
33
d
ddd d
,,
ttt t
ppp p
n
Bx BzBzBx
t
Izuyyu yyu yyu yy t
Ixz Izx






 







 
1
Ix






 


 



 
 
11
11 1
111
1
d
d
dd d
d1dd11
1d1
t
t
ttt
p
n
B zBz
n n
ppp
nn n
BzBz Bz
p
n
yz
t
u uyy
tt t
uyyuyy uyy
tt t
uyy
yz
 
 



 
 






 








 
 

12
d.
np
R
Cuyy C


(3.4)
Similarly,





1
n
1
31
11
d
,dd d
tt
pp p
n
Bx Bz
t
Ixzuyyuyyyy
tt


  
 





1
311
11
dd
,dd dd
tt t
t
pp pp
nn
BzBxB xBx
tt
Izxuyy uyyuyy uyy
tt

 



 



 


11
11
1
11
dd
d1 d
dd
d11 d
tt
t t
n
pp
nn
Bx Bx
n
pp
nn
Bx Bx
tt
uy
y uyy
tt
tt
uyy uyy
tt




 
 
 
 




 
 
 

 

 




1
23
11nC






(3.5)
Note that here we applied the Mean Value Theorem
with both 1
fore,
and 2 valued between 0 and
.
Combining (3.4) and (3.5), we have

 

223 3
,,
I
xIz IxzIzxC

 
The same inequality holds for
 
22
I
zIx. There-
22 .
sup
xz
Ix IzC
xz
Also from the definition of

(3.6)
F
x, we immediately
have
Copyright © 2013 SciRes. APM
L. F. CAO, Z. H. DAI
158
.Fu (3.7)
y (
LL

Obviousl (3.3), 3.6) and (3.7) imply that
F
x is
Li ith (3 ipschitz continuous, and this together w.7)mply
.FXY
Finally, to see that S
maps XY tself, we
only need to verify that
into i
if2, then
LL .
L
vu Tvu

 (3.8)
In fact,
L








1
0tt
n
Bx Bx
p
vt
vBxBx
1
0
.
tt
n
L
p
L
t
Cu
d
dd
pp
t
Tvxvy yy y
dt



 
Choosing



sufficiently smallut independent of v),
w 1), 2) and 3), by the The
3.1 and Remark 2, we conclude that the solutio
is Lipschitz continuous. This completes the pro
Th.
Usually, contracting operators are used to lift reg
ties. For a linear operator, if it is “shrinking”, then it is
contracting. While for nonlinear problems, as were seen
in Section 3, sometimes it is very difficult or even im-
possible to prove that it is contracting in a given functio
space. However, one can show that it is “shrinking”, a
can still lift the regularity of solutions in many cases. The
ge la
4. Acknowledgements
enxiong Chen for his
discussions.
Society, Vol.
138, 2010, pp. 2779-2791.
doi:10.1090/S
(b
e can guarantee (3.8).
So far we have verifiedorem
n u of (0.1)
of of the
eorem 1.2ulari-
n
nd
neral Regurity Lifting Theorem is applied for inte-
gral equations and system of integral equations associ-
ated with Bessel potentials and Wolff potentials (see [1]
and [7]), and therefore arrive at higher regularity as
Lipschitz continuity of solutions.
Most of this work was completed when the first author
was visiting the Department of Mathematics, Yeshiva
University, and she would like to thank the hospitality of
the Department. Besides, the authors would like to ex-
press their gratitude to Professor W
hospitality and many valuable
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