In this paper, we study the existence of solutions for non-linear fractional q-difference
equations of order
2 3
involving the p-Laplacian operator with various boundary value conditions. By
using the Banach contraction mapping principle, we prove that, under certain conditions, the suggested
non-linear fractional boundary value problem involving the p-Laplacian operator has a unique solution. Finally,
we illustrate our results with some examples.
chapter 5.pptx: drainage and irrigation engineering
Solvability of Fractionl q -Difference Equations of Order 2 3 Involving The P-Laplacian Operator
1. Invention Journal of Research Technology in Engineering & Management (IJRTEM)
ISSN: 2455-3689
www.ijrtem.com Volume 2 Issue 3 ǁ March. 2018 ǁ PP 17-29
|Volume 2| Issue 3 | www.ijrtem.com | 17 |
Solvability of Fractionlq -Difference Equations of
Order 32 Involving The P-Laplacian Operator
Yawen Yan Chengmin Hou
(Department of mathematics,Yanbian University,Yanji 133002,P.R.China)
ABSTRACT : In this paper, we study the existence of solutions for non-linear fractional q-difference
equations of order 32 involving the p-Laplacian operator with various boundary value conditions. By
using the Banach contraction mapping principle, we prove that, under certain conditions, the suggested
non-linear fractional boundary value problem involving the p-Laplacian operator has a unique solution. Finally,
we illustrate our results with some examples.
KEYWORDS: p-Laplacian operators; Caputo fractional derivative; Caputo fractional boundary value
problem
I. INTRODUCTION
In recent years, the topic of q-calculus has attracted the attention of several researchers and a variety of new
results can be found in the papers [2-10] and the references cited therein. In 2010, Ferreira [1] considered the
existence of nontrivial solutions to the fractional q-difference equation
.0)1()0(
,10)),(,()(0,
xx
ttxtftxDq
In [15], Aktuğlu and Özarslan dealt with the following Caputo q-fractional boundary value problem involving
the p-Laplacian operator:
,)1()0(),1()0(,1,,3,2,0)0(
,10)),(,()))(((
10 xDaxDxaxnkxD
ttxtftxDD
qq
k
q
q
c
pq
where 0, 10 aa , 1 , and ),]1,0([ RRCf . Under some conditions, the authors obtained the
existence and uniqueness of the solution for the above boundary value problem by using the Banach contraction
mapping principle.
In this paper, we focus on the solvability of the following non-linear fractional differential equations of
order 3,2 involving the p-Laplacian operator with boundary conditions:
),1()0(
),1()0(
),1()0(
,0)0(
)),(,()))(((
2
2
2
1
0
xDaxD
xDaxD
xax
xD
txtftxDD
qq
qq
q
c
q
c
pq
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where ),]1,0([ RRCf and ),,]1,0([)( 2
RRCtx ,)(
2
sss
p
p
,1p vp 1
, 1
11
vp
,
q
c
D is the fractional q-derivative of the Caputo type,
2
, qq DD is the fractional q-derivative
and 10 a , 11 a ,
2
1
2
v
a , ]1,0[t ,
D ,
D is the conformable fractional derivative.
2 Preliminaries on fractional q-calculus
In this section, we present basic definitions and results that will be needed in the rest of the paper. More
detailed information about the theory of fractional q-calculus can be found in [6,7,8].
Let )1,0(q and define
q
q
a
a
q
1
1
: , Ra .
The q-analogue of the power function
n
ba )( with ],2,1,0[:0 Nn is
,1:)( 0
ba
1
0
:)(
n
k
kn
bqaba , Rba , .
More generally, if R , then
0
)(
/1
/1
:)(
n
n
n
qab
qab
aba
, 0a .
In particular, if 0b then
aa )(
. We also use the notation
00 )(
for 0 . The q-gamma function
is defined by
1
)1(
)1(
)1(
:)(
x
x
q
q
q
x , ,2,1,0 Rx ,
and satisfies )()1( xxx qqq .
Remark 2 ([14]) If 0 and ,tba then
)()( btat .
It is well known that the beta function ),( stBq has the following integral representation:for any 0, st ,
.)()1()1(
)1(),(
0
)1()1(1
)1(
1
0
)1(
n
tnsnn
q
st
q
qqqq
dqstB
Moreover, ),( stBq can be expressed in terms of )(t , the gamma function, as follows:
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.
)(
)()(
),(
st
st
stB
q
qq
q
Definition 2.1 Let 0 and f be a function defined on T,0 . The fractional q-integral of the
Remann-Liouville type is given by )())(( 0
tftfIq and
.)()1(
)(
)1(
)()(
)(
1
)(
0
)1(1
0
)1(
n
nnn
q
q
t
q
q
tqfqq
qt
sdsfqsttfI
)1.2(
Definition 2.2 The fractional q-derivative of Riemann-Liouville type of order 0 is defined by
)()(0
tftfDq and
),()( tfIDtfD l
q
l
qq
)2.2(,0
where l is the smallest integer greater than or equal to v.
Definition 2.3 ([6]) The fractional q-derivative of the Caputo type of order 0 is defined by
)3.2(,0,)()(
1
)()(
0
1
sdsfDqst
n
tfDItfD q
n
q
t
n
q
qqq
c
where is the smallest integer greater than or equal to .
Recall that )],,([ RbaCn
is the space of all real-valued function )(tx which have continuous derivatives up
to order 1n on ba, . In the following lemmas, we give some auxiliary results which will be used in the
sequel.
Lemma 2.1 ([6]) Let 0 , then the following equality holds:
)4.2().0(
)1(
)()(
1
0
fD
k
t
tftfDI k
q
k q
k
q
c
q
On the other hand, the operator sss
p
p
2
)(
, where 1p , is called the p-Laplacian operator. It is easy to
see that ,1
vp
where .1
11
vp
The following properties of the p-Laplacian operator will play an
important role in the rest of the paper.
Lemma 2.2 ([19])Let p be a p-Lplacian operator.
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(i) If 21 p , 0xy , and 0, myx , then
yxmpyx p
pp 2
1)()( .
(ii) If 2p , and Myx , , then
yxMpyx p
pp 2
1)()( .
Lemma 2.3 Assume that 3,2 , 10 a , 11 a ,
2
1
2
q
a , ]1,0[t and 1,0Ch . Then the
solution )(tx of the boundary value problem
)6.2(
),1()0(
),1()0(
),1()0(
,0)0(
),()))(((
2
2
2
1
0
xDaxD
xDaxD
xax
xD
thtxDD
qq
qq
q
c
q
c
pq
can be represented by the following integral equation:
)7.2(,))(()1()(
))(()1()(
))(()1(
)(
))(()(
)(
1
)(
1
0 0
)3(
1
0 0
)2(
1
0 0
)1(0
0 0
)1(
ddsshqt
ddsshqt
ddsshq
A
ddsshqttx
v
v
v
q
t
v
q
where
0
0
0
1 a
a
A
,
1
1
1
1 a
a
A
,
2
2
2
21
)1(
aq
qa
A
,and
)1(
)( 110
q
tAAA
t ,
.
)2()1(
])1([
)(
2
02021
qq
tqAAtAAA
t
Proof Using (2.4) and the fact that 0))0(( xDq
c
p
, we have
,)())((
0
t
qq
c
p sdshtxD
or equivalently,
)8.2(,))(()(
0
t
qvq
c
sdshtxD
where 1
11
vp
. Applying the fractional integral operator
qI to both sides of (2.8), we get
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II. SOLVABILITY OF THE FRACTIONAL BOUNDARY VALUE PROBLEM
This section is devoted to the solvability of the fractional boundary value problem given in (2.5). First, we
obtain conditions for existence and uniqueness of the solution )(tx of the fractional boundary value problem
given in (2.5). Then, each result obtained here is illustrated by examples. Recall that ,]1,0[C the space of
continuous functions on 1,0 is a Banach space with the norm )(max ]1,0[ txx t . Now
consider ,1,0],1,0[]1,0[: iCCTi with
sdsxsftxT q
t
v
0
0 ))(,(:)(
and
1
0
)2(
1
0
)2(
0
1
0
)1(0)1(
1
)()1()()()1()(
)()1(
)(
)()(
)(
1
)(
qq
t
q
q
q
q
dxqtdxqt
dxq
A
dxqttxT
Then the operator ]1,0[]1,0[: CCT , defined by 01 TTT , is continuous and compact.
Theorem 3.1 Suppose 21 v , 10 a , 11 a , ,
2
1
2
v
a )1,0(q and the following conditions
hold: 0 ,
v
2
2
0 and d with
)1.3(,)2(1)2()2)1()1((2
)1)2(1)(1()2)2(()1(/
)1)(2)2((
0012
10
2
qq
qq
q
v
vvqAAAA
vAAvv
qvd
such that
),(1
xtftq
for any )2.3(,1,0),( Rxt
and
yxdytfxtf ),(),( for 1,0t and )3.3(., Ryx
Then boundary value problem (2.5) has a unique solution.
Proof Using inequality (3.3), we get
dsxsft
t
0
),(
for any .1,0),( Rxt
By Lemma 2.2 (i) and (3.3), we have
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)4.3(.)1(
)()()1(
))(,())(,()1(
))(,())(,())(1(
))(,())(,(
)()(
1)2(2
0
)2(2
0
)2(2
00
2
00
00
yxtvd
sdsysxtvd
sdsysfsxsftv
sdsysfsdsxsftv
sdsysfsdsxsf
tyTtxT
vv
q
t
vv
q
t
vv
q
t
q
t
v
q
t
vq
t
v
Moreover,
)5.3(.)()()1()(
)()()1()(
)()()1(
)(
)()()(
)(
1
))(())((
)()(
1
0
00
)2(
1
0
00
)2(
1
0
00
)1(0
0
00
)1(
0101
q
q
q
q
t
q
q
dyTxTqt
dyTxTqt
dyTxTq
A
dyTxTqt
tyTTtxTT
tTytTx
Finally, substituting (3.4) in (3.5), we get
)6.3(.)1()(
)1()(
)1(
)(
)(
)(
1
1
)()(
1
0
1)2()3(
1
0
1)2()2(
1
0
1)2()1(0
0
1)2()1(2
q
v
q
v
q
v
q
t
q
v
q
v
dqt
dqt
dq
A
dqtyxvd
tTytTx
Using the equality
,
)2)2((
)()2)2((
),2)2(( 1)2(1)2(
0
1)2(1
v
v
tvBt
dqt
q
qqv
q
v
t
q
v
in (3.6) one can write that
)2,2)2(()()1,2)2(()(
),2)2((
)(
),2)2((
)(
1
)()(
0
1)2(
2
vBtvBt
vB
A
vB
t
yxvd
tTytTx
q
q
v
v
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,
12
1)2()2(
)2()1(
2)1(2
1
1)2(
)1(
)(
)()(
),2)2((1
2
02021
010
1)2(
2
qq
qq
q
q
q
qqq
v
v
vv
q
tqAAtAAA
vtAAAt
vByxvd
using (1), we get that
,
)2(1)2(
)1(
2)1(12
1)2(11
)2)2((
)2)2((
1
)2(1)2(
)()1(
2)1(12
1)2(
)(
)1(
)()(
1
)2)2((
)()2)2((
1
)()(
0012
10
2
02021
010
2
yxK
yxvv
q
qAAAA
vAA
v
v
vd
vv
q
qAAAAA
v
AAA
v
v
yxvd
tTytTx
qq
q
q
qv
qq
q
q
qqq
q
qqv
where
)7.3(,)2(1)2(
)1(
2)1(12
1)2(11
)2)2((
)2)2((
1
0012
10
2
qq
q
q
qv
vv
q
qAAAA
vAA
v
v
vdK
Combining (3.7) with (3.1) implies that ,10 K thereforeT is a contraction. As a consequence of the
Banach contraction mapping theorem [22] the boundary value problem given in (2.5) has a unique solution.
□
Theorem 3.2 Suppose 21 v , 0a , 11 a , ,
2
1
2
v
a )1,0(q and the following conditions
hold: 0 ,
v
2
2
0 , and d satisfies (3.1).
such that
1
),(
txtf q , for any .1,0),( Rxt
and
yxdytfxtf ),(),( , for 1,0t and ., Ryx
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Then boundary value problem (2.5) has a unique solution.
Proof The inequality
1
),(
txtf implies that ),(1
xtft
. Therefore replace
),( xtf by ),( xtf in the proof of Theorem 1, we can get the proof of this theorem. □
Theorem 3.3 Suppose 2v , 0a , 11 a ,
2
1
2
v
a . There exists a non-negative function
1,0)( Lxg , with 0)(:
1
0
dgM such that
)(),( tgxtf for any )8.3(,1,0),( Rxt
and there exists a constant d with
)9.3(,1)2)1()1((2
)11)(1()1(/
)1)(2(
0012
10
2
qq
q
q
v
qAAAA
AAv
qMd
and
yxdytfxtf ),(),( for 1,0t and ., Ryx
Then boundary value problem (2.5) has a unique solution.
Proof Using (3.8), we get that
)10.3(.)())(,(
1
00
Mdgdxf q
t
For all 1,0t . By the definition of the operator 0T , one can write that
)11.3(.))(,())(,(
)()(
00
00
sdsysfsdsxsf
tyTtxT
q
t
vq
t
v
As a consequence of (2.4), (3.10) and (3.11), we have
.)1(
)()()1(
))(,())(,()1(
))(,())(,()1(
)()(
2
0
2
0
2
00
2
00
yxtMv
sdsysxMvd
sdsysfsxsfMv
sdsysfsdsxsfMv
tyTtxT
v
q
t
v
q
t
v
q
t
q
t
v
Moreover,
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By (3.9), we get 1K , which implies thatT is a contraction,therefore the boundary value problem given in
(2.5) has a unique solution. □
In the present part, we illustrate our results by examples. We will use inequation in [18], one has
),()( ttq for 10 t or 2t ; ),()( tt q for 21 t .
Example 1 Consider the following anti-periodic boundary value problem:
)12.3(
),1(
10
3
)0(),1()0(
),1()0(
,0)0(
1,0)),
32
cos(2(4)))(((
22
2
5
22
5
3
7
xDxDxDxD
xx
xD
t
x
ttxDD
qqqq
q
c
q
c
q
where
3
7
p ,
2
5
, and
10
3
,1 210 aaa .
Then
4
7
v ,
2
1
210 AAA , and take 4 , 1 and
8
d . because of )1,0(q ,
so we let
2
1
q , obviously,
,
8
)
2
3
(
4
1
)
2
3
(
4
1
)2(1)2(2)1(12
2)2(11
)2)2(()1()1)(2)2((
0012
10
2
d
vvqAAAA
vAA
vvqv
q
qq
q
qq
v
.1
64
55
)23(128
55
)23(128
55
)2(1)2(
)1(
2)1(12
1)2(11
)2)2((
)2)2((
1
0012
10
2
q
qq
q
q
qv
vv
q
qAAAA
vAA
v
v
vdK
Moreover, it
can be easy see that
).,())
32
cos(2(44 231
xtf
x
ttt qq
Finally,
.
8
)
32
()
32
(4
)
32
cos()
32
cos(4
))
32
cos(2(4))
32
cos(2(4
),(),(
2
22
yx
yx
yxt
ytxt
ytfxtf
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There fore as a consequence of Theorem 3.1, boundary value problem given in (3.12) has a unique solution.
Example 2 Consider the following boundary value problem:
)13.3(
),1(
10
3
)0(),1()0(
),1()0(
,.0)0(
1,0),
40
(sin)))(((
22
2
5
22
5
4
7
xDxDxDxD
xx
xD
t
x
txDD
qqqq
q
c
q
c
q
where
4
7
p ,
2
5
, and
10
3
,1 210 aaa .
Then
3
7
v ,
2
1
210 AAA , and take 1 and
20
d . Because of 1,0q , so
we let
2
1
q , taking 1)( tg , we get 1M ,
,
20
1
12)1(12
2111)1)(2(
0012
10
2
d
qAAAA
AAvqM
qq
qq
v
.1
27
11
1
)1(
2)1(12
111
)2(
1
0012
10
2
qq
q
q
v
q
qAAAA
AA
Mvd
K
On the other hand,
.
20
))
40
(s))
40
(s
),(),(
22
yx
yinxin
ytfxtf
For 1,0t and ., Ryx
Therefore by Theorem 3.3, the antiperiodic boundary value problem given in (3.13) has a unique solution.
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