Existence results for fractional q-differential equations with integral and multi-point boundary conditions

Invention Journal of Research Technology in Engineering & Management (IJRTEM)
ISSN: 2455-3689
www.ijrtem.com Volume 2 Issue 9 ǁ September 2018 ǁ PP 11-19
|Volume 2| Issue 9 | www.ijrtem.com | 11 |
Existence results for fractional q-differential equations with
integral and multi-point boundary conditions
Yawen Yan1,
Chengmin Hou1*
(Department of Mathematics, Yanbian University, Yanji, 133002, P.R. China)
ABSTRACT: This paper concerns a new kind of fractional q-differential equation of arbitrary order by
combining a multi-point boundary condition with an integral boundary condition. By solving the equation which
is equivalent to the problem we are going to investigate, the Green’s functions are obtained. By defining a
continuous operator on a Banach space and taking advantage of the cone theory and some fixed-point theorems,
the existence of multiple positive solutions for the BVPs is proved based on some properties of Green’s functions
and under the circumstance that the continuous functions f satisfy certain hypothesis. Finally, examples are
provided to illustrate the results.
KEYWORDS: fractional q-differential equation; boundary value problem; Green’s function; fixed point
theorems
I. INTRODUCTION
Fractional calculus has attracted many researchers’ interests because of its wide application in solving practical
problems that arise in fields like viscoelasticity, biological science, ecology, aerodynamics, etc. Numerous
writings have showed that fractional order differential equations could provide more methods to deal with
complex problems in statistical physics and environmental issues. Especially, writers introduced the development
history of fractional calculus in [1], and authors in [2] stated some pioneering applications of fractional calculus.
For specific applications, see [3, 4] and the references therein. Fractional-order differential equations with
boundary value problems sprung up dramatically. Multi-point boundary conditions and integral boundary
conditions become hot spots of research among different types of boundary value problems, and the studies in [5–
9] are excellent. However, most researchers tend to investigate either integral conditions or multi-point conditions.
For instance, the authors explored the fractional-order equation with integral boundary conditions as follows in
[9]




=−=
=+

1
0
,)()()1(,20,0)0(
),1,0(,0))(,()()(
sdsusgunjuD
ttutfthtuD
q
j
q
q


(1.1)
where 10  q ,

qD denotes the Riemann-Liouville type fractional q-derivative of order  ,
],1( nn − and 3n is an integer, 0 is a constant, the functions g , h , f are continuous. They
devoted themselves to finding the existence of solutions by making use the comparison theorem, monotone
iterative technique and lower-upper solution method. In [8], Yang and zhao investigated the following fractional
q-difference equation by using the Banach contraction principle, Krasnoselskii’s fixed point theorem and
Schauder fixed point theorem:




=+=
=

−
−
2
1
21 ,10),()()(),()0(
,21],1,0[)),(,()(
m
i
iiq
c
q
c
q
c
uuDbuDauu
ttutftuD




(1.2)
where )2,,2,1(10 2111 −= − mim   , 
qD represents the Caputo
type fractional q-derivative of order  . The nonlinear function Rf ]1,0[: R→ is a
given continuous function and  , a , b , Ri  .In consideration of the fact that integral boundary
conditions and multi-point boundary conditions have been investigated in a variety of papers (see [10-12]), in
this paper we are dedicated to considering fractional differential equations that contain both the integral
boundary condition and the multi-point boundary condition:

Existence results for fractional q-differential equations with…








+=
−==
=+
 
−
=
−
=
i
i
m
i iq
m
i i
i
q
q
xsdsxx
nixD
txtftxD



0
2
1
2
1
),()()1(
,2,2,1,0,0)0(
,0))(,()(
 (1.3)
where

qD represents the standard Riemann-Liouville fractional derivative of order  satisfying
nn − 1 with 3n and
+
 Nn . In addition, 0, ii  and 10 221  −m 
with 21 − mi , where m is an integer satisfying 3m . RRf →]1,0[: is a given continuous
function?
To ensure that readers can easily understand the results, the rest of the paper is planned as follows. Section 2 is
aimed to recall certain basic definitions and lemmas to obtain the Green’s functions. Section 3 is devoted to
reviewing Krasnoselkii’s fixed point theorem, Schauder type fixed point theorem, Banach’s contraction mapping
principle and nonlinear alternative for single-valued maps and to applying them to analyze the problem in order
to show the main results. In the last section, some examples are given to verify that the results are practical.
II. PRELIMINARIES
Definition 2.1 [13] Let f be a function defined on [0,1]. The fractional q-integral of the
Riemann-Liouville type of order 0 is )())(( 0
tftfIq = and
sdsfqsttfI q
t
q
q )()(
)(
1
)(
0
)1(

−
−

= 

, 0 , ]1,0[t .
Definition 2.2 [14] The fractional q-derivative of the Riemann-Liouville type of order  is
   







=

=
−
−
+
+
,0),(
;0),(
;0),(
)(
0,
0,






xfID
xf
xfI
xfD
qq
q
q
where   denotes the smallest integer greater or equal to  .
Lemma 2.1 [13] Let 0,  and n be a positive integer. Then the following equality holds:

−
=
−−
+−+
+=
1
0
1
)0(
)1(
)()(
n
i
i
q
q
n
q
n
q
n
qq fD
ni
t
tfIDtfDI



.
Lemma 2.2 Assume that ]1,0[1
Lh  , ]1,0[n
ACx  and nn − 1 with 3n , then the
solution to the fractional differential equation
]1,0[,0)()( =+ tthtxDq

(2.1)
with multi-point and integral boundary conditions





+=
−==
 
−
=
−
=
i
i
m
i iq
m
i i
i
q
xsdsxx
nixD


0
2
1
2
1
),()()1(
,2,2,1,0,0)0( 
(2.2)
is given by

   
−
=
−−
=
−
++=
2
1
1
0
1
1
0
2
1
1
0
1
)(),()(),()(),()(
m
i
qii
m
i
qiiq sdshsG
t
sdshsH
t
sdshstGtx 




,
where 0
1
1
2
1
1
2
1
−−= 
−
=
−
−
=
m
i
ii
m
i
ii



 and



−
−−−

= −−
−−−
,10,)1(
,10,)()1(
)(
1
),( 11
111
stst
tsstststG
q



(2.3)



−
−−−
+
= −
−
.10,)1(
,10,)()1(
)1(
1
),( 1
1
stst
tsstststH
q



(2.4)
Proof By Lemma 2.1, the following equality holds:
=
−
+−=
n
k
k
kq tCthItx
1
)()( 
.
In view of the boundary conditions, the parameters 032 ==== nCCC  are
concluded and
),)(())((
)()(
)1()1(
1
2
1
11
2
1
2
1
0
2
1
1
1







iiq
m
i
iiiq
m
i
i
i
m
i
iq
m
i
i
q
ChI
C
hI
xsdsx
tChIx
i
+−++−=
+=
+−=


−
=
+
−
=
−
=
−
=
−
i.e.,
.)()(
)(
)()(
)1(
)()1(
)(
11
1
0
1
2
1
0
2
1
1
0
1
1
2
1
1
2
1
sdshs
sdshs
sdshsC
qi
m
i q
i
qi
m
i q
i
q
q
m
i
ii
m
i
ii
i
i



−
−
=
−
=
−
−
=
−
−
=
−

−
−
+
−
−

=






−−














So,
.)()(
)(
)()(
)1(
)()1(
)(
11
0
1
2
1
0
2
1
1
0
1
1




−

−




−
+
−−

=


−
−
=
−
=
−
sdshs
sdshssdshsC
qi
m
i q
i
qi
m
i q
i
q
q
i
i











Hence, the solution is

.)(),()(),()(),(
)()(
)(
)()(
)1(
)()1(
)1(
)1(
)()1(
)(
)()(
)(
1
)()(
)(
)()(
)1(
)()1(
)(
)()(
)(
1
)(
1
0
2
1
1
1
0
2
1
1
1
0
0
)1(
2
1
1
0
)(
2
1
1
1
0
)1(
1
1
0
)1(
0
1
)1(
0
)1(
2
1
1
0
)(
2
1
1
1
0
)1(
0
1
)1(
sdshsG
t
sdshsH
t
sdshstG
sdshqs
t
sdshqs
t
sdshqs
t
sdshqs
t
sdshqst
sdshqs
t
sdshqs
t
sdshqs
t
sdshqsttx
qi
m
i
iqi
m
i
iq
qi
m
i q
i
qi
m
i q
i
q
q
q
t
q
q
q
qi
m
i q
i
qi
m
i q
i
q
t
q
q
q
i
i
ii






−
=
−−
=
−
−
−
=
−
−
=
−
−
−
−
−
−
−
−
=
−−
=
−
−
−
−
++=
−

−
−
+
−−
+
−
+
−

+−

−=
−

−−
+
−
−

+−

−=























 

 






 

 




Therefore,
we complete the proof. 
Lemma 2.3 The functions ),( stG and ),( stH obtained in Lemma 2.5 are continu-
ous and nonnegative on ]1,0[]1,0[  . It is easy to figure out that
)(
1
),(0
q
stG

 and
)1(
1
),(0
+

q
stH hold for all ]1,0[, st .
III. MAIN RESULTS
Based on the lemmas mentioned in the previous section, we define an operator CCS →: as
follows:
,))(,(),(
))(,(),())(,(),())((
1
0
2
1
1
1
0
2
1
1
1
0
sdsxsfsG
t
sdsxsfsH
t
sdsxsfstGtSx
qi
m
i
i
qi
m
i
iq


−
=
−
−
=
−
+
+=






(3.1)
where )],1,0([ RCC = denotes the Banach space of all continuous functions defined on ]1,0[ that are
mapped into R with the norm defined as )(sup
]1,0[
txx
t
= . And G and H are provided in Lemma 2.6.
Solutions to the problem exist if and only if the operator S has fixed points. In order to make the analysis clear,
we introduce some fixed-point theorems that play a role in our proof.
Lemma 3.1 (Krasnoselskii [19]) Let Q be a closed, convex, bounded and nonempty subset of
a Banach space Y . Let 21,  be operators such that
(i) Qvv + 2211  whenever Qvv 21, ;
(ii) 1 is compact and continuous;
(iii) 2 is a contraction mapping.
Then there exists Qv  such that 2211 vvv  += .
Lemma 3.2 ([14]) Let X be a Banach space. Assume that XXT →: is a completely
continuous operator and the set }10,:{ == TuuXuV is bounded. Then T has a

fixed point in X .
Lemma 3.3 ([15]) Let E be a Banach space, 1E be a closed, convex subset of E , V be an open subset of
1E and V0 . Suppose that 1: EVU → is a continuous, compact (that is, )(VU is a relatively compact
subset of 1E ) map. Then either
(i) U has a fixed point in V , or
(ii) there are Vx  (the boundary of V in 1E and )1,0(k with )(xkUx = .
For convenience, we set some notations:
)1(
1
)1()1(
)1(1
)(
1
12
1
12
1
1
+
−
+
++
−+
+

=
−−
=
+−
=
 








q
ii
m
i
i
q
ii
m
i
i
q
,
.
)(
1
12


q
−= (3.2)
Theorem 3.4 Let RRf →]1,0[: be a continuous function that satisfies the conditions:
(H1) yxlytfxtf −− ),(),( for all ]1,0[t and Ryx , ;
(H2) )(),( txtf  for all Rxt  ]1,0[),( and )],1,0([ +
 RC ,
then BVP (1.3) has at least one solution on ]1,0[ when 12 l with 2 given in (3.2).
Proof Define a set }:{  = xCxB , where 1  with 1 defined in (3.2) and
)(sup [0,1]t t = . Let the operators 1S and 2S on B be defined as
,))(,(),())((
1
0
1 sdsxsfstGtxS q=
.))(,(),())(,(),())((
1
0
2
1
1
1
0
2
1
1
2 sdsxsfsG
t
sdsxsfsH
t
txS qi
m
i
iqi
m
i
i 
−
=
−−
=
−
+= 




It is easy to understand 121 + ySxS for any Byx , that means BySxS + 21 .
By assumption (H1),
.
)1()1()1(
)1(
),(),(),(
1
sup
),(),(),(sup
sup
2
2
1
112
1
1
0
2
1]1,0[
1
0
2
1
1
]1,0[
22
]1,0[
22
yxl
yx
l
sdysfxsfsG
sdysfxsfsH
t
ySxSySxS
m
i q
ii
i
q
ii
m
i
i
qi
m
i
i
t
qi
m
i
i
t
t
−
−








+
−
+
++
−+

−+
−
−=−



−
=
−+−
=
−
=
−
=
−
















The operator 2S is a contraction because of 12 l . As we all know, 1S is a continuous result from
the continuity of f . Moreover,
  ,
)()(
))(,(
sup))(,(),(sup
1
0]1,0[
1
0]1,0[
1


 q
q
qt
q
t
sd
sxsf
sdsxsfstGxS











=  
which implies that 1S is uniformly bounded on B . Apart from that, the following inequalities hold with
+= mBxt fxtf ),(sup ]1,0[),( 
and 10 21  tt :

 
 .)()(
)1(
))(,(),(),(
))(,(),())(,(),())(())((
12
1
1
1
2
1
0
12
1
0
1
1
0
21222


tttt
f
sdsxsfstGstG
sdsxsfstGsdsxsfstGtxStxS
q
m
q
qq
−+−
+

−
−=−
−−


The operator 1S is compact due to the Arzela-Ascoli theorem. Since three conditions are satisfied, BVP (1.3)
has at least one solution on ]1,0[ by application of Krasnoselskii’s fixed point theorem. 
Theorem 3.5 Assume that there exists a constant L such that Lxtf ),( for any ]1,0[t and
]1,0[Cx  . Then there exists at least one solution to BVP (1.3).
Proof Firstly, we set out to verify that the operator S given in (3.1) is completely continuous. Define a
bounded set )],1,0([ +
 RCU , then 1))(( LtSx  holds when we take Ux . On the top of that,
 
.)(
)1(
)()(
))(,(),(
))(,(),(
))(,(),(),())(())((
1
1
1
22
12
1
1
1
2
1
0
2
1
1
1
1
2
1
0
2
1
1
1
1
2
1
0
1212








−+
+
−+−

−
+
−
+
−=−
−−
−−
−
=
−−
−
=
−−













tt
tttt
L
sdsxsfsG
tt
sdsxsfsH
tt
sdsxsfstGstGtSytSx
q
qi
m
i
i
qi
m
i
i
q
Hence, S is equicontinuous on ]1,0[ in view of

t and
1−
t is equicontinuous on ]1,0[ . The operator S
is deduced to be completely continuous by the Arzela
-Ascoli theorem along with the continuity of S decided by f .
Secondly, we consider the set  10,: == SxxCxV and prove that V is bounded. In fact, for
each Vx and ]1,0[t ,
1
2
1
112
1 )1(
1
)1()1(
)1(1
)(
1
))((










L
L
tSxx
m
i q
ii
i
q
ii
m
i
i
q
=








+
−
+
++
−+
+


=

−
=
−+−
=
Consequently, the set V is bounded by definition.
Finally, we conclude that BVP (1.3) has at least one solution according to Lemma 3.5 and the proof is completed.
Theorem 3.6 Assume that RRf →]1,0[: is a continuous function and satisfies condition (H1) with
11 l , where 1 is defined in (3.2). Then the BVP has a unique solution on ]1,0[ .
Proof Let }:{ rxCxPr = be a bounded set. To show rr PSP  with the operator S defined in
(3.1), + lrsxsf ))(,( holds when we take rPx  for ]1,0[t and with the condition provided by
= )0,(sup ]1,0[ tft and
1
1
1 

l
r
−
 . Beyond that,

.)(
,))(,(),(
))(,(),())(,(),(sup)(
1
1
0
2
1
1
1
0
2
1
1
1
0]1,0[
rlr
sdsxsfsG
t
sdsxsfsH
t
sdsxsfstGSx
qi
m
i
i
qi
m
i
iq
t
+



+



+


−
=
−
−
=
−








rr PSP  is strictly proved. Next, choosing Cyx , with ]1,0[t , we get

.
))(,())(,(),(
))(,())(,(),(
))(,())(,(),(sup
1
1
0
2
1
1
1
0
2
1
1
1
0]1,0[
yxl
sdsysfsxsfsG
t
sdsysfsxsfsH
t
sdsysfsxsfstGSySx
qi
m
i
i
qi
m
i
i
q
t
−



−+
−+
−−



−
=
−
−
=
−








The operator S is a contraction with the assumption 11 l . Therefore, BVP (1.3) has a unique solution by
Banach’s contraction mapping principle. 
Theorem 3.7 Let RRf →]1,0[: be a continuous function, and assume that
(H3) there exist a function )],1,0([ +
 RCp and a nondecreasing function
+
Rq : +
→ R such that
)()(),( xqtpxtf  for all Rxt  ]1,0[),( ;
(H4) there exists a constant 0N such that
1
)1(
1
)1()1(
)1(1
)(
1
)(
1
2
1
112
1













+
−
+




++
−+
+

−
−
=
−+−
=

m
i q
ii
i
q
ii
m
i
i
q
pvqN








.
Then BVP (1.3) has at least one solution on [0,1].
Proof The first step is to show that the operator S given in (3.1) maps bounded sets into bounded sets in C .
Let v be a positive number and  vxCxBv = : be a bounded set in C . For each vBx  and by
(H3), the following equalities are obtained:
sdsxsfsG
t
sdsxsfsH
t
sdsxsfstGtSx
qi
m
i
i
qi
m
i
iq
))(,(),(
))(,(),())(,(),())((
1
0
2
1
1
1
0
2
1
1
1
0


−
=
−
−
=
−
+
+






.)(
)1(
1
)1()1(
)1(1
)(
1
)(
1
2
1
112
1









pvq
pvq
m
i q
ii
i
q
ii
m
i
i
q
=




+
−
+




++
−+
+

 
−
=
−+−
=
The next step is to verify that the operator S maps bounded sets into equicontinuous sets of C .
Choose ]1,0[, 21 tt with 21 tt  and take vBx  ,

 








−+
+
−+−

−
+
−
+
−=−
−−
−−
−
=
−−
−
=
−−



)(
)1(
)()(
)(
))(,(),(
))(,(),(
))(,(),(),()()(
1
1
1
22
12
1
1
1
2
1
0
2
1
1
1
1
2
1
0
2
1
1
1
1
2
1
0
1212










tt
tttt
pvq
sdsxsfsG
tt
sdsxsfsH
tt
sdsxsfstGstGtSytSx
q
qi
m
i
i
qi
m
i
i
q
By the Arzela-Ascoli theorem, the operator S is completely continuous since the right-hand
side tends to zero independent of vBx  as 12 tt → . Let x be a solution of BVP (1.3), then
for )1,0( and by the same method applied to show the boundedness of S , the results hold:
,
)1(
1
)1()1(
)1(1
)(
1
)(
))(()(
2
1
112
1 



+
−
+




++
−+
+


=

−
=
−+−
=
m
i q
ii
i
q
ii
m
i
i
q
pxq
tSxtx









i.e.,
1
)1(
1
)1()1(
)1(1
)(
1
)(
1
2
1
112
1













+
−
+




++
−+
+

−
−
=
−+−
=

m
i q
ii
i
q
ii
m
i
i
q
pvqx








.
The final step is to select a set  1: += NxCxQ in order to take (H4) into consideration where there
exists N such that Nx  . Even though CQS →: is completely continuous can be verified, there is
no Nx  that can satisfy )(xSx = for )1,0( decided by the selection of N .
Above all, we complete the proof that the operator S has a fixed point in Q which is a solution to BVP (1.3).

IV. EXAMPLE
Example 4.1 Consider the fractional differential equations with boundary value as follows:









+++=
==
=+++

−−
)
3
1
(2)
4
1
(
2
3
)(4)(
2
5
)1(
,0)0(,0)0(
],1,0[,0
2
1
tansin)(
3
1
0
4
1
0
2
12
7
xxsdsxsdsxx
xDxD
txexttxD
qq
qq
t
q
.
(4.1)
From the equation above, it is clear that
2
7
= , 2=m ,
2
5
1 = , 42 = ,
2
3
1 = ,
22 = ,
4
1
1 = ,
3
1
2 = . Consequently, we can get 026.0= , 7918.01 = , 4785.02 =
by computation.
Since
2
1
cossin),( ++= −
xextxtf t
,
,2
tantansinsin
tantansinsin),(),(
11
11
yx
yxeyxt
yexeytxtytfxtf
t
tt
−
−+−
−+−=−
−−−
−−−−

that implies 2=l and both 11 l and 12 l hold. Problem (4.1) has a unique solution
on [0,1] because all the conditions of Theorem 3.6 are satisfied.
V. CONCLUSION
We have proved the existence of solutions for fractional differential equations with integral and multi-point
boundary conditions. The problem is issued by applying some fixed point theorems and the properties of Green’s
function. We also provide examples to make our results clear.
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Calc. Appl. Anal. 17, 552-578 (2014)
[3] Konjik, S, Oparnica, L, Zorica, D: Waves in viscoelastic media described by a linear fractional model,
Integral Transforms Spec. Funct. 22, 283-291 (2011)
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Multi-Point Conditions.
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Theorem[J]. International Journal of Nonlinear Sciences & Numerical Simulation, 2017, 18(7).
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Yawen Yan, Existence results for fractional q-differential equations with integral and multi-point
boundary conditions. Invention Journal of Research Technology in Engineering & Management
(IJRTEM), 2(9), 11-19. Retrieved September 4, 2018, from www.ijrtem.com.

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