Existence of solutions for a functional integro-di erential equation [601904]

Existence of solutions for a functional integro-dierential equation
with innite point and Riemann-Stieltjes integral conditions
A. M. A. El-Sayed1;aand Reda Gamal Ahmed2;b
1Faculty of Science, Alexandria University, Egypt.
2Faculty of Science, Al-Azhar University, Cairo, Egypt.
aEmail [anonimizat]
bEmail [anonimizat]
Abstract
In this article, we study the existence of solutions for two ini-
tial value problems of the functional integro-dierential equation with
nonlocal innite-point and integral Riemann-Stieltjes conditions. we
study the continuous dependenc of the solution.
MSC: 34A12, 34k20, 34k25
keywords: Existence of solutions, continuous dependence, nonlocal condi-
tion, Riemann-Stieltjes condition, innite point condition.
1 Introduction
In the last years, Some investigators have interested studying functional dif-
ferential and integral equation with the nonlocal condition, in order to achieve
various goals; see [[1], [2], [3], [4], [5], [6], [7], [9], [11], [13], [12], [14], [15]],
and the references cited therein.
In this paper, we are concerned with the initial value problem(IVP) for the
functional integro-dierential equation
dx
dt=f(t;x(t);Zt
0g(s;x(s))ds); a:e t2(0;T]; (1.1)
with the nonlocal condition
x(0) +mX
k=1akx(k) =x0; k2(0;T]: (1.2)
The existence of at least one solution, under certain conditions, will be
proved. The continuous dependenc of the solution on the nonlocal data
ak, onx0and on the functional g, will be studied.
1

As applications, the IVP of equation (1.1) with the integral Riemann-Stieltjes
condition
x(0) +ZT
0x(s)dh(s) =x0; (1.3)
will be studied. Also, the IVP of equation (1.1) with innite-point condition
x(0) +1X
k=1akx(k) =x0; (1.4)
will be studied.
2 Integral Representation
Consider the nonlocal problem (1.1)-(1.2) with the assumptions:
1.f: [0;T]R2!Rsatises Caratheodory condition. There exist a
functionc12L1[0;T] and a positive constant b1>0, such that
jf(t;; )jc1(t) +b1jj+b1jj:
2.g: [0;1]R!Rsatises Caratheodory condition. There exist a
functionc22L1[0;T] and a positive constant b2>0, such that
jg(t;)jc2(t) +b2jj:
3.
sup
t2[0;1]Zt
0c1(s)dsM1; sup
t2[0;1]Zt
0Zs
0c2()ddsM2:
4. (1 +EPm
k=1ak)(b1T+1
2b1b2T2)<1
Lemma 2.1. LetPm
k=1ak6=1, the solution of IVP (1.1) -(1.2) , if it exist,
then it can be represented by the integral equation
x(t) =E
x0mX
k=1akZk
0f(s;x(s);Zs
0g(;x())d)ds
+Zt
0f(s;x(s);Zs
0g(;x())d)ds; (2.1)
whereE= (1 +Pm
k=1ak)1:
2

Proof. Letxbe a solution of IVP (1.1)-(1.2), Integrating both sides of (1.1)
we obtain
x(t) =x(0) +Zt
0f(s;x(s);Zs
0g(;x())d)ds: (2.2)
Using the nonlocal condition (1.2), we get
mX
k=1akx(k) =x(0)mX
k=1ak+mX
k=1akZk
0f(s;x(s);Zs
0g(;x())d)ds;
since,Pm
k=1akx(k) =x0x(0), we have
x0x(0) =x(0)mX
k=1ak+mX
k=1akZk
0f(s;x(s);Zs
0g(;x())d)ds;
then
x(0) =1
1 +Pm
k=1ak[x0mX
k=1akZk
0f(s;x(s);Zs
0g(;x())d)ds];(2.3)
Using (2.2) and (2.3), we obtain
x(t) =1
1 +Pm
k=1ak[x0mX
k=1akZk
0f(s;x(s);Zs
0g(;x())d)ds]
+Zt
0f(s;x(s);Zs
0g(;x())d)ds:
3 Existence of solution
Theorem 3.1. Let the assumptions 1{4be satised, then the IVP (1.1) -(1.2)
has at least one solution x2C[0;T].
Proof. Let the operator Fassociated with the integral equation (2.1) by
Fx(t) =E
x0mX
k=1akZk
0f(s;x(s);Zs
0g(;x())d)ds
+Zt
0f(s;x(s);Zs
0g(;x())d)ds:
3

LetQr=fx2R:jjxjjrg, wherer=Ejx0j+(1+EPm
k=1ak)(M1+b1M2)
1((1+EPm
k=1ak)(b1T+1
2b1b2T2)).
Then we have, for x2Qr
jFx(t)j E
jx0j+mX
k=1akZk
0jf(s;x(s);Zs
0g(;x())d)jds
+Zt
0jf(s;x(s);Zs
0g(;x())d)jds
E
jx0j+mX
k=1akZk
0(c1(s) +b1jx(s)j+b1Zs
0jg(;x())jd)ds
+Zt
0(c1(s) +b1jx(s)j+b1Zs
0g(;x())d)ds
E
jx0j+mX
k=1ak(M1+b1Tr+b1Zk
0Zs
0jc2() +b2jx()jdds)
+M1+b1Tr+b1Zt
0Zs
0(c2() +b2jx()j)dds
Ejx0j+AmX
k=1ak(M1+b1Tr+b1M2+1
2b1b2T2r)
+M1+b1Tr+b1M2+1
2b1b2T2r
=Ejx0j+ (1 +EmX
k=1ak)(M1+b1Tr+b1M2+1
2b1b2T2r) =r:
ThenF:Qr!Qrand the class of function fFxgis uniformly bounded
inQr.
4

Now, lett1;t22(0;1) s. tjt2t1j<, then
jFx(t2)Fx(t1)j=jZt2
0f(s;x(s);Zs
0g(;x())d)ds
Zt1
0f(s;x(s);Zs
0g(;x())d)dsj
Zt2
t1jf(s;x(s);Zs
0g(;x())d)jds
Zt2
t1(c1(s) +b1jx(s)j+b1Zs
0jg(;x())d)jds
Zt2
t1c1(s)ds+ (t2t1)b1r+b1Zt2
t1Zs
0c2()dds
+1
2b1b2r(t2
2t2
1):
Then the class of functions fFxgis equi-continuous in Qr.
Letxn2Qr,xn!x(n!1 ), then from continuity of the functions fand
g, we obtain f(t;xn(t);yn(t))!f(t;x(t);y(t)) andg(t;xn(t))!g(t;x(t)) as
n!1 . Also
lim
n!1Fxn(t) = lim
n!1
E
x0mX
k=1akZk
0f(s;xn(s);Zs
0g(s;xn())d)ds
+Zt
0f(s;xn(s);Zs
0g(;xn())d)ds
: (3.1)
Using assumptions 1{2 and Lebesgue convergence Theorem [10], from (3.1)
we obtain
lim
n!1Fxn(t) =
E
x0mX
k=1akZk
0lim
n!1f(s;xn(s);Zs
0g(;xn())d)ds
+Zt
0lim
n!1f(s;xn(s);Zs
0g(;xn())d)ds
=Fx(t):
ThenFxn!Fxasn!1 . ThenFis continuous.
lim
t!0x(t) =E
x0mX
k=1akZk
0f(s;x(s);Zs
0g(;x())d)ds
2C[0;T]:
5

Then by Theorem [8] there exist at least one solution x2C[0;T] of the
integral equation (2.1).
To complete the proof, dierentiation (2.1) we obtain
dx
dt=d
dt
E
x0mX
k=1akZk
0f(s;x(s);Zs
0g(;x())d)ds
+Zt
0f(s;x(s);Zs
0g(;x())d)ds
=f(s;x(s);Zs
0g(;x())d)ds:
Also, from the integral equation (2.1), we get
x(k) =E
x0mX
k=1akZk
0f(s;x(s);Zs
0g(;x())d)ds
+Zk
0f(s;x(s);Zs
0g(;x())d)ds
x(0) =E
x0mX
k=1akZk
0f(s;x(s);Zs
0g(;x())d)ds
;(3.2)
and
mX
k=1akx(k) =EmX
k=1ak
x0mX
k=1akZk
0f(s;x(s);Zs
0g(;x())d)ds
+mX
k=1akZk
0f(s;x(s);Zs
0g(;x())d)ds
(3.3)
from (3.2) and (3.3) we have
x(0) +mX
k=1akx(k)
=E(1 +mX
k=1ak)
x0mX
k=1akZk
0f(s;x(s);Zs
0g(;x())d)ds
+mX
k=1akZk
0f(s;x(s);Zs
0g(;x())d)ds
6

Then
x(0) +mX
k=1akx(k) =x0:
Then there exist at least one solution x2C[0;T] of the IVP (1.1)-(1.2).
4 Nonlocal Riemann-Stieltjes integral condi-
tion
Letx2C[0;T] be the solution of the IVP (1.1)-(1.2). Let ak=h(tk)
h(tk1),his increasing function, k2(tk1;tk), 0 =t0<t1<t2;:::<tm= 1
then, asm!1 the nonlocal condition (1.2) will be
x(0) +mX
k=1h(tk)h(tk1)x(k) =x0:
And
x(0) + lim
m!1mX
k=1h(tk)h(tk1)x(k) =x(0) +ZT
0x(s)dh(s) =x0:
Theorem 4.1. Let the assumptions 1{4be satised, then the IVP of (1.1) –
(1.3) has at least one solution x2C[0;T].
Proof. Asm!1 , the solution of the IVP (1.1)-(1.2) will be
x(t) = lim
m!11
1 +Pm
k=1ak
x0mX
k=1akZk
0f(s;x(s);Zs
0g(;x())d)ds
+Zt
0f(s;x(s);Zs
0g(;x())d)ds
=1
1 +h(T)h(0)
x0lim
m!1mX
k=1akZk
0f(s;x(s);Zs
0g(;x())d)ds(h(tk)
h(tk1))
+Zt
0f(s;x(s);Zs
0g(;x())d)ds
=1
1 +h(T)h(0)
x0Z1
0Zt
0f(s;x(s);Zs
0g(;x())d)ds:dh (t)
+Zt
0f(s;x(s);Zs
0g(;x())d)ds:
7

5 Innite-point boundary condition
Theorem 5.1. Let the assumptions 1{4be satised, then the IVP of (1.1) –
(1.4) has at least one solution x2C[0;T].
Proof. Let the assumptions of Theorem 3.1 be satised.
LetPm
k=1akbe convergent, then
xm(t) =1
1 +Pm
k=1ak
x0mX
k=1akZk
0f(s;x(s);Zs
0g(;x())d)ds
+Zt
0f(s;xm(s);Zs
0g(;xm())d)ds: (5.1)
Take the limit to (5.1), as m!1 , we have
lim
m!1xm(t) = lim
m!11
1 +Pm
k=1ak
x0mX
k=1akZk
0f(s;x(s);Zs
0g(;x())d)ds
+Zt
0f(s;xm(s);Zs
0g(;xm())d)ds
= lim
m!11
1 +Pm
k=1ak
x0mX
k=1akZk
0f(s;x(s);Zs
0g(;x())d)ds
+ lim
m!1Zt
0f(s;xm(s);Zs
0g(;xm())d)ds: (5.2)
Now
jakx(k)jjakjkxk;
then by comparison testP1
k=1akx(k) is convergent.
Also
jZk
0f(s;x(s);Zs
0g(;x())d)dsjZk
0(c1(s)+b1jx(s)j+b1Zs
0g(;x())d)ds
Zk
0(c1(s) +b1jx(s)j+b1Zs
0(c2(s) +b2jx(s)j)d)ds
M1+b1kxk+b1M2+1
2b1b2kxkM
8

then
jakZk
0f(s;x(s);Zs
0g(;x())d)dsjjakj:M:
and by the comparison testP1
k=1akRk
0f(s;x(s);Rs
0g(;x())d)dsis con-
vergent.
Now,jfjjc1(s) +b1kxk+b1M2+b1b2kxk, using assumptions (1)-(2) and
Lebesgue convergence Theorem [10], from (5.2) we obtain
x(t) =1
1 +P1
k=1ak
x01X
k=1akZk
0f(s;x(s);Zs
0g(;x())d)ds
+Zt
0f(s;x(s);Zs
0g(;x())d)ds: (5.3)
The Theorem proved.
6 Uniqueness of the solution
Letfandgsatisfy the following assumptions
5.f: [0;T]R2!Ris measurable in tfor any;2Rand satises the
lipschitz condition
jf(t;; )f(t;u;v )jb1juj+b1jvj; (6.1)
6.g: [0;T]R!Ris measurable in tfor any2Rand satises the
lipschitz condition
jg(t;)g(t;u)jb2juj; (6.2)
7.
sup
t2[0;T]Zt
0jf(s;0;0)jdsL1;sup
t2[0;T]Zt
0Zs
0jg(;0)jddsL2:
Theorem 6.1. Let the assumptions 5{7be satised, then the solution of the
IVP (1.1) -(1.2) is unique.
9

Proof. From assumption 5 we have fis measurable in tfor anyx;y2Rand
satises the lipschitz condition, then it is continuous in ;2R8t2[0;T],
and
jf(t;; )jb1jj+b1jj+jf(t;0;0)j:
Then condition 1 is satised. Also by the same way we can show that as-
sumption 2 satised by assumption 6. Now, from Theorem 3.1 the solution
of the IVP (1.1)-(1.2) exists.
10

Letx;ybe two the solution of (1.1)-(1.2), then
jx(t)y(t)j=jE
mX
k=1akZk
0f(s;x(s);Zs
0g(;x())d)ds
+Zt
0f(s;x(s);Zs
0g(;x())d)ds
E
mX
k=1akZk
0f(s;y(s);Zs
0g(;y())d)ds
Zt
0f(s;y(s);Zs
0g(;y())d)dsj
EmX
k=1akZk
0jf(s;x(s);Zs
0g(;x())d)ds
f(s;y(s);Zs
0g(;y())d)dsj
+Zt
0jf(s;x(s);Zs
0g(;x())d)ds
f(s;y(s);Zs
0g(;y())d)dsj;
EmX
k=1akZk
0j(b1kxyk+b1Zs
0jg(;x())
g(;y())jd)ds+Zt
0j(b1kxyk
+b1Zs
0jg(;x())g(;y())jd)ds
b1TkxykEmX
k=1ak+1
2b1b2T2kxykEmX
k=1ak
+b1Tkxyk+1
2b1b2T2kxyk
= (1 +EmX
k=1ak)(b1T+1
2b1b2T2)kxyk;
11

Hence
(1(1 +EmX
k=1ak)(b1T+1
2b1b2T2))kxyk0:
Since (1 +EPm
k=1ak)(b1T+1
2b1b2T2)<1, thenx(t) =y(t) and the solution
of the IVP (1.1)-(1.2) is unique.
7 Continuous dependence
7.1 Continuous dependenc on x0
Denition 7.1. The solution x2C[0;1] of the IVP (1.1)-(1.2) depends
continuously on x0, if
8>0;9()s:tjx0x
0j<)jjxxjj<;
wherexis the solution of the IVP
dx
dt=f(t;x(t);Zt
0g(s;x(s))ds); a:e t2(0;T]; (7.1)
with the nonlocal condition
x(0) +mX
k=1akx(k) =x
0; k2(0;T]: (7.2)
Theorem 7.1. Let the assumptions of Theorem 6.1 be satised, then the
solution of the IVP (1.1) -(1.2) depends continuously on x0.
Proof. Letx;xbe two solutions of the IVP (1.1)-(1.2) and (7.1)-(7.2) re-
12

spectively. Then
jx(t)x(t)j=jE
x0mX
k=1akZk
0f(s;x(s);Zs
0g(;x())d)ds
+Zt
0f(s;x(s);Zs
0g(;x())d)ds
E
x
0mX
k=1akZk
0f(s;x(s);Zs
0g(;x())d)ds
+Zt
0f(s;x(s);Zs
0g(;x())d)dsj
Ejx0x
0j+EmX
k=1akZk
0jf(s;x(s);Zs
0g(;x())d)
f(s;x(s);Zs
0g(;x())d)dsj+Zt
0jf(s;x(s);Zs
0g(;x())d)
f(s;x(s);Zs
0g(;x())d)dsj;
Ejx0x
0j+EmX
k=1akZk
0(b1kxxk
+b1Zs
0jg(;x())g(;x())jd)ds+Zt
0(b1kxxk
+b1Zs
0jg(;x())g(;x())jd)ds
Ejx0x
0j+b1TkxykEmX
k=1ak+1
2b1b2T2kxxkEmX
k=1ak
+b1Tkxxk+1
2T2b1b2kxxk
E+ (1 +EmX
k=1ak)(b1T+1
2b1b2T2)kxxk:
Hence
kxxkE
[1(1 +EPm
k=1ak)(b1T+1
2b1b2T2)]=:
Then the solution of the IVP (1.1)-(1.2) depends continuously on x0. The
proof is completed.
13

7.2 Continuous dependenc on the nonlocal data ak
Denition 7.2. The solution x2C[0;1] of the IVP (1.1)-(1.2) depends
continuously on the nonlocal data ak, if
8>0;9()s:tjaka
kj<)jjxxjj<;
wherexis the solution of the IVP
dx
dt=f(t;x(t);Zt
0g(s;x(s))ds); a:e t2(0;1); (7.3)
with the nonlocal condition
x(0) +mX
k=1a
kx(k) =x0; ak0; k2(0;1): (7.4)
Theorem 7.2. Let the assumptions of Theorem 6.1 be satised, then the
solution of the IVP (1.1) -(1.2) depends continuously on the nonlocal data ak.
Proof. LetPm
k=1a
k6=1 andx;xbe two solutions of the IVP (1.1)-(1.2)
and (7.3)-(7.4) respectively. Then
jx(t)x(t)j=jE
x0mX
k=1akZk
0f(s;x(s);Zs
0g(;x())d)ds
+Zt
0f(s;x(s);Zs
0g(;x())d)dsE
x0mX
k=1a
kZk
0f(s;x(s);Zs
0g(;x())d)ds
Zt
0f(s;x(s);Zs
0g(s;x())d)dsj
14

EEmjx0j+jEmX
k=1a
kZk
0f(s;x(s);Zs
0g(;x())d)ds
EmX
k=1a
kZk
0f(s;x(s);Zs
0g(;x())d)ds
+E[mX
k=1a
kZk
0f(s;x(s);Zs
0g(;x())d)ds
mX
k=1akZk
0f(s;x(s);Zs
0g(;x())d)ds]j
+b1Tkxxk+1
2T2b1b2kxxk
EEmjx0j+m[(b1T+1
2b1b2T2)kxk+TL1+1
2b1T2L2]mX
k=1a
k
+Ej[mX
k=1a
kZk
0f(s;x(s);Zs
0g(;x())d)dsmX
k=1akZk
0f(s;x(s);Zs
0g(;x())d)ds
+mX
k=1akZk
0f(s;x(s);Zs
0g(;x())d)dsmX
k=1akZk
0f(s;x(s);Zs
0g(;x())d)ds]j
+b1Tkxxk+1
2T2b1b2kxxk
EEmx 0+m[(b1T+1
2b1b2T2)kxk+TL1+1
2b1T2L2]mX
k=1a
k
+E[m[(b1T+1
2b1b2T2)kxk+TL1+1
2b1T2L2]
+mX
k=1ak(b1Tkxxk+1
2T2b1b2kxxk)]
+b1Tkxxk+1
2T2b1b2kxxk
EEmx 0+m[(b1T+1
2b1b2T2)kxk+TL1+1
2b1T2L2](E+mX
k=1a
k)
+ (EmX
k=1ak+ 1)(b1T+1
2T2b1b2)kxxk
15

Hence
kxxkEEmx 0+m[(b1T+1
2b1b2T2)kxk+TL1+1
2b1T2L2](E+Pm
k=1a
k)
1(1 +EPm
k=1ak)(b1T+1
2T2b1b2)=;
whereE= (1+Pm
k=1a
k)1:Then the solution of the IVP (1.1)-(1.2) depends
continuously on the nonlocal data ak. The proof is completed.
7.3 Continuous dependenc on the functional g
Denition 7.3. The solution x2C[0;T] of the IVP (1.1)-(1.2) depends
continuously on the functional g, if
8>0;9()s:tjggj<)jjxxjj<;
wherexis the solution of the IVP
dx
dt=f(t;x(t);Zt
0g(s;x(s))ds); a:e t2(0;T]; (7.5)
with the nonlocal condition
x(0) +mX
k=1akx(k) =x0; k2(0;T]: (7.6)
Theorem 7.3. Let the assumptions of Theorem 6.1 be satised, then the
solution of the IVP (1.1) -(1.2) depends continuously on the functional g.
Proof. Letx;xbe two solutions of the IVP (1.1)-(1.2) and (7.5)-(7.6) re-
spectively. Then
jx(t)x(t)j
EmX
k=1akZk
0j(b1kxxk+b1Zs
0jg(;x())g(;x())jd)ds
+Zt
0j(b1kxxk+b1Zs
0jg(;x())g(;x())jd)ds
b1TkxxkEmX
k=1ak+1
2b1T2EmX
k=1ak+1
2b1b2T2kxxkEmX
k=1ak
+1
2b1T2+b1T2kxxk+1
2b1b2T2kxxk
(1 +EmX
k=1ak)1
2b1T2+ (1 +EmX
k=1ak)(b1T+1
2b1b2T2)kxxk:
16

Hence
kxxkb1
[1(1 +EPm
k=1ak)(b1T+1
2b1b2T2)]Pm
k=1ak=:
Then the solution of the IVP (1.1)-(1.2) depends continuously on the func-
tionalg. The proof is completed.
8 Examples:
Example 8.1. Consider the following nonlinear integro{dierential equation
dx
dt=t4et+ln(1 +x(t))
4 +t3+Zt
01
9(sin(3s+ 3)
+s4cosx(s)
ejx(s)j)dt; a:e t2(0;1]; (8.1)
with innite point boundary condition
x(0) +1X
k=11
k2x(k1
k) =x0: (8.2)
Set
f(t;x(t);Zt
0g(s;x(s))ds) =t4et+ln(1 +x(t))
4 +t3+Zt
01
9(sin(3s+ 3)
+s4cosx(s)
ejx(s)j)dt:
Then
jf(t;x(t);Zt
0g(s;x(s))ds)j t4et+1
4(jxj+1
4Zt
04
9j(cos(3s+ 3)
+s4cosx(s)
ejx(s)j)dtj);
and also
jg(s;x(s))j4
9jcos(3s+ 3)j+4
9jx(s)j:
17

It is clear that the assumptions 1{4 of Theorem 3.1 are satised with c1(t) =
t3et2L1[0;1],c2(t) =1
2jcos(3t+ 3)j 2L1[0;1],b1=1
3,b2=4
9;
(1 +P1
k=11
k2
1+P1
k=11
k2)(b1+1
2b1b2) = (1 +2
6
1+2
6)(1
3+2
27)<1;
and the series:1X
k=11
k2;
is convergent. Therefore, by applying to Theorem 3.1, the given IVP (8.1)-
(8.2) has a continuous solution.
Example 8.2. Consider the following nonlinear integro{dierential equation
dx
dt=t5+t2+ 1 +x(t)pt+ 9+Zt
01
4(sin2(3s+ 3)
+sx(s)
2s(1 +x(s)))dt; a:e t2(0;1]; (8.3)
with innite point boundary condition
x(0) +1X
k=11
k4x(k2+k1
k2+k) =x0: (8.4)
Set
f(t;x(t);Zt
0g(s;x(s))ds) =t5+t2+ 1 +x(t)pt+ 9+1
4Zt
0(sin2(3s+ 3)
+sx(s)
2s(1 +x(s)))dt:
Then
jf(t;x(t);Zt
0g(s;x(s))ds)j t5+t2+ 1 +1
3jxj+1
3Zt
03
4j(sin2(3s+ 3)
+sx(s)
2s(1 +x(s))jdt;
and also
jg(s;x(s))j3
4j(sin2(3s+ 3)j+3
8jx(s)j:
18

It is clear that the assumptions 1{4 of Theorem 3.1 are satised with c1(t) =
t5+t2+ 12L1[0;1],c2(t) =3
4j(sin2(3s+ 3)j2L1[0;1],b1=1
3,b2=3
8;
(1 +P1
k=11
k4
1+P1
k=11
k4)(b1+1
2b1b2) = (1 +4
90
1+4
90)(1
3+1
16)<1;
and the series:1X
k=11
k4;
is convergent. Therefore, by applying to Theorem 3.1, the given IVP (8.3)-
(8.4) has a continuous solution.
References
[1] K. Deimling, Nonlinear functional analysis , Springer-Verlag, 1985.
[2] J. Dugundji and A. Granas, Fixed point theory , Monograe Mathematy-
czne. PWN. Warsaw, 1982.
[3] Kh. W. El-Kadeky, A nonlocal proplem of the dierential equation x0=
f(t;x;x0), Journal of Fractional Calculus and Applications 3(7) (2012),
1{8.
[4] H. El-Owaidy, A. M. A. El-Sayed, and Reda Gamal Ahmed, Existence of
solutions of a coupled system of functional integro-dierential equations
of arbitrary (fractional) orders , Malaya J. Mat 6(2018), no. 4, 774{780.
[5] A. M. A. El-Sayed and RG Ahmed, Innite point and riemann-stieltjes
integral conditions for an integro-dierential equation , Nonlinear Anal-
ysis: Modelling and Control (In press).
[6] A. M. A. El-Sayed and Sh. A. Abd El-Salam, On the stability of a frac-
tional order dierential equation with nonlocal initial condtion , EJQTDE
(2008), no. 29, 1{8.
[7] F. Ge, H. Zhou, and C. Kou, Existence of solutions for a coupled frac-
tional dierential equations with innitely many points boundary condi-
tions at resonance on an unbounded domain , Dier. Equ. Dyn. Syst 24
(2016), 1{17.
19

[8] K. Goeble and W.A. Kirk, Topics in metric xed point theory , Cam-
birdge Universty Press, 1990.
[9] L. Guo, L. Liu, and Y. Wu, Existence of positive solutions for singular
fractional dierential equations with innite-point boundary conditions ,
Nonlinear Analysis: Modelling and Control 21(2016), 635{650.
[10] A. N. Kolomogorov and S. V. Fomin, Inroductory real analysis , Dover
Puble. Inc, 1975.
[11] Bingmei Liu, Junling Li, Lishan Liu, and Yongqing Wang, Existence
and uniqueness of nontrivial solutions to a system of fractional dier-
ential equations with riemann{stieltjes integral conditions , Advances in
Dierence Equations 2018 (2018), no. 1, 306.
[12] Xiaoqian Liu, Lishan Liu, and Yonghong Wu, Existence of positive
solutions for a singular nonlinear fractional dierential equation with
integral boundary conditions involving fractional derivatives , Boundary
Value Problems 2018 (2018), no. 1, 24.
[13] Dandan Min, Lishan Liu, and Yonghong Wu, Uniqueness of positive
solutions for the singular fractional dierential equations involving inte-
gral boundary value conditions , Boundary Value Problems 2018 (2018),
no. 1, 23.
[14] Hari M. Srivastava, Ahmed M. A. El-Sayed, and Fatma M. Gaafar, A
class of nonlinear boundary value problems for an arbitrary fractional-
order dierential equation with the riemann-stieltjes functional integral
and innite-point boundary conditions , Symmetry 10(2018), no. 10.
[15] Xinqiu Zhang, Lishan Liu, Yonghong Wu, and Yumei Zou, Existence
and uniqueness of solutions for systems of fractional dierential equa-
tions with riemann{stieltjes integral boundary condition , Advances in
Dierence Equations 2018 (2018), no. 1, 204.
20