Advances in Pure Mathematics, 2011, 1, 59-62
doi:10.4236/apm.2011.13013 Published Online May 2011 (http://www.scirp.org/journal/apm)
Copyright © 2011 SciRes. APM
An Arbitrary (Fractional) Orders Differential Equation
with Internal Nonlocal and Integral Conditions
Ahmed El-Sayed1, E. O. Bin-Taher2
1Faculty of Science, Alexandr ia University, Alexandria, Egypt
2Faculty of Science, Hadhramout University of Science and Technology , Hadhramout, Yemen
E-mail: amasayed@hotma il.com, ebtsamsam@yahoo.com
Received February 28, 2011; revised March 25, 2011; acce pted Ma rc h 30, 2011
Abstract
In this paper we study the existence of solution for the differential equation of arbitrary ( fractional) orders


d
= ,, 0,1
d
xftDxt
t
, with the general form of internal nonlocal condition


=1 =1
= ,
p
m
kkj j
kj
ax bx
 

 
,0,1, , 0,1,
kj
acdbc d

 . The problem with nonlocal integral condition will be studied.
Keywords: Internal Nonlocal Problem, Integral Condition, Fractional Calculus, Existence of Solution,
Caratheodory Theorem
1. Introduction
Problems with non-local conditions have been exten-
sively studied by several authors in the last two decades.
The reader is referred to ([1-10]), and references therein.
In this work we study the existence of at least one so-
lution for the nonlocal problem of the arbitrary (frac-
tional) or d er different ial equation
 

 
d = ,, 0,1 and 0,1
d
xt ftDxtt
t
 (1)
with the general nonlocal condition


=1 =1
= ,
p
m
kkj j
kj
ax bx
 

(2)
where

 
,0,1, , 0,1,
kj
acdbc d

 
and
0
is parameter.
As an app lication, we deduce the ex istence of solution
for the nonlocal problem of the differential (1) with the
integral cond ition
 
d = d.
cb
ad
x
ss xss

(3)
It must be noticed that the following nonlocal and inte-
gral conditions are special cases of our nonlocal and in-
tegral conditions
 
= , , and ,,
x
xac db
 
 (4)

 
=1 = , , and ,,
m
kk k
k
ax xacdb
 

(5)


=1 = 0, ,,
m
kk k
k
ax ac

(6)
 
d = , ,,
c
a
x
ss xdb

(7)
and
 
d = 0, ,.
c
a
x
ss ac
(8)
2. Preliminaries
Let
1
LI denotes the class of Lebesgue integrable
functions on the interval
=,
I
ab , with the norm
1= d
LI
uutt
and
CI denotes the class of con-
tinuous functions on the interval
, with the norm
= suptI
uut
and
. denotes the gamma func-
tion.
Definition 2.1 The fractional-order integral of the
function
1,
f
Lab of order R
is defined by
(see [11])
 
 
1
= d.
t
aa
ts
I
ftfs s
Definition 2.2 The Caputo fractional-order derivative of
A. EL-SAYED ET AL.
Copyright © 2011 SciRes. APM
60
order
0,1
of the absolutely continuous function

f
t is define d by (se e [1 1] and [ 1 2] )
 
1 d
= .
d
aa
Dft Ift
t

Definition 2.3 The function
:0,1
f
RR is called
1
LCaratheodory if
1)

,tftx is measurable for each
x
R,
2)

,
x
ftx is continuous for almost all
0,1t,
3) there exists

10,1,, mLDD R such that
f
m.
Now we state Caratheodory Theorem ([13] ).
Theorem 2.1 Let
0,1
f
RR be 1
L
Cara-
theodory, then the initial-value problem
 


d=,, for a.e. t>0, and 0=
do
xt
f
txtx x
t (9)
has at least one absolutely continuous solution
0,
x
AC T.
Here we generalize Caratheodory theorem for the
nonlocal problem (1) - (2).
3. Main Results
Consider firstly the fractional-order integral equation
 

1
= ,,ytIf tyt
(10)
Definition 3.1 The function y is called a solution of
the fractional-order integral Equation (10), if
0,1yC and satisfies (10).
Theorem 3.1 Let
:0,1
f
RR be 1
L
Cara
theodory. Then there exists at least one solution of the
fractional-order integral Equation (10).
Proof. Let
 

= Max: 0,1, 0 and 0,1
a
MImtta
 , then



 


 
1
1
, ,d
d , 0.
t
aa
t
a
ts
I
ftytfsys s
ts mssM a

Define the sequence


n
yt by
 
 


10
= , d, 0 ,1
1
t
nn
ts
yt fsysst

which can be written in the operator form


1
1 = ,.
nn
ytI Iftyt
 

Then
 






1
10
1
, d
1
2 2
t
nn
ts
ytIIftytMs
tM
M

 


 







For
12
, 0,1tt such that 12
< tt, then
 
 


 

 

 

 

 

2
1
1
2
1
1
2
1211 0
1
0
2
0
2
1
1
0
1
0
= , d
1
, d
1
= , d
(1 )
, d
(1 )
, d
(1 )
,
(1 )
t
nn n
t
n
t
n
t
n
t
t
n
t
n
ts
yt ytfsyss
ts fsy ss
ts fsy ss
ts fsy ss
ts fsy ss
ts fsy s







 

 

2
1
1
2
1
0
d
, d
(1 )
, d.
(1 )
t
n
t
t
n
s
ts fsy ss
ts fsy ss


Therefore
  




 



2
1
22
11
2
12 11
22
1
21
d
1
d d
11
.
2
t
nn t
tt
tt
ts
yt ytmss
tt
mM
tt
M













 


Hence
 
2112 11
< <
nn
ttytyt


  and
n
yt is a sequence of equi-continuous and uni-
formly bounded functions. By Arzela-Ascoli Theorem,
([14] and [15]) there exists a subsequence


k
n
yt of
continuous functions which converges uniformly to a
continuous function y as k.
Now we show that this limit function is the required
solution.
Since


1
, ,
nk
f
sy sms L
and

,nk
fsys is continuous in the second argu-
ment,
A. EL-SAYED ET AL.
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61




.., , as ,
nk
iefsysfsysk
therefore the sequence
 


,,
nk
tsfsys

0,1
satisfies Lebesgue dominated convergence
theorem. Hence

 


 

0
0
, d
lim 1
= , d = (),
1
t
nk
k
t
ts fsy ss
ts fsyssyt



which proves the existence of at least one solution
0,1yC of the fractional-order functional integral
Equation (10).
For the existence of solution for the nonlocal problem
(1) - (2) we have the following theorem.
Theorem 3.2 Let the assumptions of Theorem 3.1 are
satisfied. Then nonlocal problem (1) - (2) has at least one
solution
0,1xAC.
Proof. Consider the nonlocal problem (1) - (2).
Let
 
= , yt Dxt
then
 
1d
= ,
d
x
t
yt It
(11)
 

1
= ,ytIftyt
(12)
and y is the solution of the fractional-order integral
Equation (10).
Operating by
I
on both sides of Equation(11), we
obtain
   
d
= = 0
d
xt
IytIxt x
t
 (13)


= 0 .
x
tx Iyt
(14)
Let =k
t
in Equation (13), we get
 
  
1
0
=1 =1=1
= d0 .
mm m
k
k
kk kk
kk k
s
axayss xa
 
And let =
j
t
in Equation (13), we get
 
  
1
0
=1 =1=1
= d0 .
pp p
j
j
j
jj
jj j
s
bxbjy ssxb
 
From Equation (2), we get

 

 
1
0
=1 =1
1
0
=1 =1
d0
= d0 .
mm
k
k
kk
kk
pp
j
j
j
j
jj
s
ayssxa
s
byssxb



Then we get


 

 
1
0
=1
1
0
=1
0= d
d
mk
k
k
k
pj
j
j
j
s
xAa yss
s
byss
and


 

 

 
1
0
=1
1
0
=1
1
0
= d
d
d
mk
k
k
k
pj
j
j
j
t
s
xtAayss
s
byss
ts ys s
(15)
where
1
=1 =1
=
pm
jk
jk
Aba




which, by Theorem 3.1, has at least one solution
0,1 .xAC
Now, from Equation (15), we have
 

 

 
1
0
=1
0
1
0
=1
0= = d
lim
d
k
j
mk
k
k
t
pj
j
j
s
x
xtAayss
s
Ab yss
and
 

 

 
 
1
0
1=1
11
1
00
=1
1= = d
lim
1
d d
j
mk
k
k
tk
pj
j
j
s
xxtAa yss
ss
A
byssyss




from which we deduce that Equation (15) has at least one
solution
0,1 .xAC
To complete the proof, diffe rentiating (15), we obtain
 

d = = ,.
d
xytftDxt
t
Also from (15) we can prove that the solution satisfies
the nonlocal condition (2).
4. Nonlocal Integral Condition
Let
0,1 .xAC be the solution of the nonlocal prob-
lem (1) - (2).
Let
11 012
=, ,, = <<,<=
kkkkk km
attttattttc


A. EL-SAYED ET AL.
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62
and 11 012
=,(,), = <<,<=
jjjjj jp
bsss sdssssb


then the nonlocal condition (2) will be


11
=1 =1
= .
p
m
kkkj jj
kj
tt xss x




From the continuity of the solution
of the nonlocal
problem (1) - (2) w e can ob tain


11
=1 =1
= .
limlim p
m
kkkj jj
mp
kj
tt xss x


 


and the nonlocal condition (2) transformed to the integral
one
 
d = d .
cb
ad
x
ss xss

(16)
Now, we have the following Theorem
Theorem 4.1 Let the assumptions of Theorem 3.2 are
satisfied. Then there exist at least one solution
0,1 .xAC of the nonlocal problem with integral
condition,



= , , 0, 1,xt ftDxt t
 
d = d , .
cb
ad
x
ssyssbd ca



Letting = 0
in (16), the we can easily prove the
following corollary .
Theorem 4.2 Let the assumptions 1) - 2) are satisfied.
Then the nonlocal problem



= , , 0, 1,xt ftDxt t
 
d = 0, ,0,1
c
axssac
has at least one solution
0,1 .xAC
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