Existence Theorem for a Nonlinear Functional Integral
Equation and an Initial Value Problem of
Fractional Order in L
1
(R
+
)
Ibrahim Abouelfarag Ibrahim1,2, Tarek S. Amer1,3, Yasser M. Aboessa1,4 1
Mathematics Department, Faculty of Science and Education, Taif University, Al-Khurmah Branch, Taif, KSA
2
Mathematics Department, Faculty of Science, Suez Canal University, Ismailia, Egypt
3
Mathematics Department, Faculty of Science, Tanta University, Tanta, Egypt
4
Mathematics Department, Faculty of Science, Al-Azhar University, Cairo, Egypt Email: [email protected], [email protected], [email protected] Received November 21, 2012; revised December 21, 2012; accepted December 30, 2012
ABSTRACT
The aim of this paper is to study the existence of integrable solutions of a nonlinear functional integral equation in the space of Lebesgue integrable functions on unbounded interval, L1(R+). As an application we deduce the existence of solution of an initial value problem of fractional order that be studied only on a bounded interval. The main tools used are Schauder fixed point theorem, measure of weak noncompactness, superposition operator and fractional calculus.
Keywords: Nonlinear Functional Integral Equation; Volterra Operator; Measure of Weak Noncompactness; Fractional Calculus; Schauder Fixed Point Theorem
1. Introduction
The class of functional integral equations of various types plays very important role in numerous mathemati- cal research areas. An interesting feature of functional integral equations is its role in the study of many prob- lems of functional differential Equations [1-4].
In this work we study the solvability of the following initial value problem
1 0
0 d
, , d
d
, 0,1 , 0 .
t
y t
k t s f s D y s s t
t R y y
,(1)
where D y denotes the fractional derivative of order
of y with
0,1
. Such initial value problem of arbitrary order (1) was investigated in [5-7]. To achieve this goal, let us consider the integral equation
1
2
0 0
, , , d
t s
d ,
x t k t s f s k s x
t R
s (2)
which is different from that studied in [2].
Section 2 contains some basic results. Our main result will be given in Section 3. Solvability of the considered initial value problem will be discussed in Section 4.
2. Basic Concepts
This section is devoted to recall some notations and known results that will be needed in the sequel.
A
If is a Lebesgue measurable subset of the set of real numbers R then we use the symbol meas.
A to denote the Lebesgue measure of A. Let L A1
be the space of all real functions defined and Lebesgue meas- urable on the set A. If xL1
A then the norm of x is defined as:
1 d .
L A A
x x
x t twhen AR
0,
, we will write L1 instead of
1 L R .
2.1. The Superposition Operator
An important operator called the superposition operator can be investigated in the theories of differential integral and functional equations [4,8-10]. It can be defined as follows:
Definition 1. Assume that f I: R R satisfies Carathéodory conditions, that is it is measurable in for any
t x and continuous in x for almost all where
t
,
tI xR. Then for every measurable function x on the interval I we assign the function:
The operator F defined in this way is called the super-position operator generated by the function f .
Carathéodory [11] gave the first contribution to the theory of the superposition operator and proved its measurability according to the measurability of f .
We state the following result giving the necessary and sufficient condition so that the superposition operator F generated by f will map continuously into itself [12].
1 L
Theorem 2. Let f satisfy the conditions in Defini- tion 1. The superposition operator F generated by the function f maps continuously the space into itself if and only if:
1 L
,
,f t x a t b x for all tI and xR
b
, where is a function that belongs to and is a nonnegative constant.
a
1
It is known that a real valued continuous function is measurable and that the converse is not necessarily true. However, for the converse we have the following results due to Dragoni [13].
L
Theorem 3. Let I be a bounded interval and
:
f I R R be a function satisfying Caratheodory conditions. Then for each 0 there exists a closed subset D of the interval I such that meas.
I D
and f D R is continuous.
2.2. Volterra Integral Operator
We proceed by recalling some basic facts concerning the linear Volterra integral operator in the Lebesgue space
1 Suppose is a given function which is measurable with respect to both variables where
.
L k: R
t s, : 0 s t
.
For an arbitrary function xL1 define Volterra inte-gral operator as follows:
0
, d ,
t
Kx t
k t s x s s tR.It is well known that if K L: 1L1 then it is con-tinuous [4,9].
In general, it is rather difficult to find necessary and sufficient conditions for the function k t s
, guarantee-ing that the integral operator K transforms the space1 into itself. Some special cases of this problem were discussed in [4,14]. In this direction we state the next result [15]:
L
Theorem 4. Let be measurable on and such that
k
0
ess sup , d .
s s
k t s s
Then the Volterra integral operator generated by
maps (continuously) the space
K
k L1L R1
intoitself and the norm K of this operator is majorized by the number
0
ess sup , d
s s
k t s s
.Observe that if is a nonempty and measurable subset of
D
R then we can also consider the linear Volterra integral operator associated with the Lebesgue space L D1
.Namely,
1 xL D
R
If where is a nonempty and measur-able subset of
D
then we extend x
to the whole half
axis R by putting x t
0 for tR D. Then we can treat K in the usual way. When the operator K transforms L D1
into itself its norm will be denoted by K D.2.3. Measures of Weak Noncompactness
Let us assume that is an infinite dimensional Banach space with the norm
E
and the zero element . De-note by mE
E
the family of all nonempty and bounded subsets of and by W
E
n its subfamily consisting of all relatively weakly compact sets. The symbol W
X stands for the weak closure of a set X and the symbol
will denote the convex closed hull (with respect to the norm topology) of a set
ConvX
X . We denote by
,B x r the ball centered at x and of radius . We write r instead of
r B B
,r . In what follows we ac-cept the following definition [16]Definition 5. A function is said to be a measure of weak noncompactness if it satisfies the fol-lowing conditions: The Family
: E
µ m R
1) The family kerµ
XmE :
X 0
is non- .W empty and is nonempty and ker µnE
2) X Y
X
Y . 3) µ ConvX
X .4) µ
X
1
Y
X 1
Y , for
0,1 .
5) , and 1 for 1, 2,
W
n E n n n n
X m X X X X n
If
And if limµ X
n 0n then the intersection is non-empty
1
n n
X X
The family kerµ is said to be the kernel of the meas-ure of weak noncompactness µ. Let us observe that the intersection set X from 5) belongs to kerµ. Indeed, since µ
X µ
Xn for every n then we have that
0µ X .
We can construct a useful measure of weak noncom-pactness in the space 1 that based on the following criterion for weak noncompactness due to Dieudonné
We state here some results concerning the above men-tioned operators:
[17,18].
Theorem 6. A bounded set X is relatively weakly compact in 1 if and only if the following two condi-tions are satisfied:
L
a) for any 0 there exists 0 such that if meas.
D Then
dD
x t t
for all, xX . b) for any 0 there is T 0 such that
d Tx t t
for any ,xX .Now, for a nonempty and bounded subset X of the space L1 let us define:
X c X d X , (3) where
0
lim sup sup d : , meas ,
x X D
c X x t t D R D
and
lim sup
d : .T
T
d X x t t x X
It can be shown [17] that the function is a measure of weak noncompactness in the space L1 such that
X
X 2
X , for any XmL1, where denotes the De Blasi measure of weak noncompactness in L1. Moreover, µ B
r 2r.In our approach we will need the following fixed point theorem due to Schauder.
Theorem 7. Let be a nonempty, convex, closed, and bounded subset of a Banach space . Let
C
E
:
H CC be a completely continuous mapping. Then H has at least one fixed point in C.
2.4. Fractional Calculus
The definitions of both differential operator and the inte-gral operator of fractional order are stated as follows [19,20].
Definition 8. Let f L1,R. The Riemman- Liouville (R-L) fractional integral of the function f t
of order is defined as
1
d ,0, .
t a
a
t s
I f t f s s
a t b
Definition 9. Let g t
be an absolutely continuous function on
a b, . Then the fractional derivative of or-der
0,1 of
g t
is defined as
1 ,
d d .
a a
D g t I Dg t
D t
1) Let f Df, L1and ,
0,1 , then
i) thenI I f ta a
Ia f t
.ii) DI f ta
I Df ta
, when f a
0.
I 2) The operator a
maps L1 into itself continuously.
3. Existence Theorem
Consider the integral Equation (2) and let H denotes the operator determined by the right hand side of this equation, i.e.,
1
2
0 0
, , , d
t s
d ,
Hx t k t s f s k s x s
(4)where tR In fact the operator H can be written as the product HK F K F1 1 2 2 of the linear Volterra opera-tor
0
, d , 1
t
i i
K x t
k t s x s s i , 2.and the superposition operator
Fx t f t x t
,
,tR.Therefore Equation (4) can be written as:
1 1 2 2 .
xHxK F K F x (5) To establish our main result concerning existence of an integrable solution of Equation (2) we impose suitable conditions on the functions involved in that equation. Namely we assume
1) The functions f R: R R satisfy the Cara- theodory conditions and there exist functions aL1 and constants b0 such that
,
f t x a t b x holds for all
t x, RR.2) The functions satisfy the Cara- theodory conditions and the linear Volterra operators
:
i
k RRR 1, 2
K K associated with 1 2 map 1into itself. 3)
,
k k L
:R R
is increasing, absolutely continuous
and there exists a constant such that
0
M
t M a.e. on R.4) b K1 K2 M 1 1
.
Now we can state our main result in the next theorem.
Theorem 10. Under the above assumptions the Equa-tion (2) has at least one soluEqua-tion xL1.
Proof. Since H is a nonlinear operator defined by Equation (5), then based on assumptions i) and ii) if
1
indeed bounded, we have
1 1 2 2 1 1 2 2
1 2
0 0
1 2
0 0
1 1 2
1 1 2
0
1 1 2
0 1
1 1 2
0 1
1 1 2
, , d d
, d d d d . s s
Hx K F K F x K F K F x
K f s k s x s
d
K a s b k s x s
K a b K K x
K a b K K x
K a b K K x
M
K a b K K M x u u
K a b K K M x
The above estimate shows that the operator H maps into itself, where
r B
11
1 1 1 2 .
r K a b K K M
Moreover, according to Theorem 2, we deduce that the operator H is continuous on the space L₁.
Next, to prove that H is a contraction, let X be a nonempty subset of Br. Fix 0
e
and take a measur-able subset DR such that m as.D. Then for any xX , we get
1 1 11 2 1 1 2
1 2
0
1 2
0
1 1 2
1 1 2
1 1 2
1 2
1 d
d
, , d d ,
, d d
d d d d d D D d L D D s D D s D D D D D D
D D L D
D
D D D L D
D
D D D
D
Hx t t
K FK x t t K F K x
K f s k s x s
K a s b k s x s
D
K a t t b K K x t t
K a t t b K K x
K a t t b K K x
b K K
K a t t x u u
M
where the symbol Ddenotes the operator norm acting from the space L D1
into itself. Also in the above calculation we used the fact that a t
0 for tR. From the absolute continuity of the function and theobvious equality
0
lim sup d : , meas. 0
D
a t t D R D
.and using Theorem 6 we obtain
1 2
.
D D
b K K
c HX c X
M
(6)
Furthermore, fixing T 0 we can deduce that
1
1 2 , 1 2
1 2 1 d d d d T L T T T T T T T
Hx t t
,
K FK x K FK x t t
bK K
K a t x u u
M
where the symbol T denotes the operator norm acting from the space L T1
,
into itself. Now according to the fact that the set consisting of one element is weakly compact, by using Theorem 6 and the formula
lim sup
d : .T
T
d X x t t x X
lim ,T T we get and since
1 2
.
b K K
d HX d X
M
(7)
According to Equation (3), combining (6) and (7), we get
b K1 K2
HX X
M
(8)
q b K₁ ₂K M
Put . Clearly, according to as-
sumption iv) q1. Consider the sequence of sets
nr
B , where B1r Conv GB
r
,Br2 Conv GB
r1
B
and
so on. Obviously this sequence is decreasing i.e.
r for
1
n n
r
B n1, 2,. Moreover, r. Apart from this, all sets belonging to this sequence are closed and convex, so weakly closed. On the other hand in view of inequality (8) we have
1
r B B
Brn qn
Br ,
which yields that lim
rn 0.n B
Consequently, by axiom 5) of Definition 5 we infer that the set
is nonempty, closed, convex and weakly compact (in view of µ Y
0). Moreover, GYY.GY
In the sequel we show that the set is relatively compact in the set 1.
To do this let us take an arbitrary sequence L
yn Y and fix arbitrarily a number 0. Since is weakly compact, in view of Theorem 6 we deduce that there ex-ists such that for any natural number the fol-lowing inequality is satisfiedY
0
T n
d .4
n T
y t t
(9)To apply the classical Schauder fixed point theorem, we need to prove that the set HY is relatively compact in L1. For this aim let us consider the functions f t x
, on the set
0,T and the functions k t si
, on the set
0,T 0,T i1,2
.In view of Theorem 3 we can find a closed subset D of the interval
0,T such that meas.
Dc (where
0,c
D T D) and such that the functions f D R
and i D 0,
1, 2 Tk i
are continuous. Hence we infer that ki D 0,T
i 1, 2
are uniformly continuous. In what follows we show that
yn is an equicon-tinuous on D, for that let us take arbitrarily t t1, 2D. Without loss of generality we can assume that t1t2. Then, keeping in mind our assumptions, for an arbitrary fixed nN we obtain:
2 1 1 1 2 11 2 2
0 0
1 1 2
0 0
1 2 1 2
0 0
1 2
0
, , , d d
, , , d d
, d d d n n t s n t s n T s T n t s n t
Hy t Hy t
k t s f s k s y s
k t s f s k s y s
k t t a s bk y s
k a s bk y s
d
where denotes the modulus of continuity of the function on the set
1,T k
1
k D
0,T and
2 2 1 1 2 11 2 1 2
0 0 0
1 1 2
0
max , : , 0, , 1, 2 .
, d
d d d
i i
n n
T T s T
n
t t s
n
t t
k k t s t s D T i
Hy t Hy t
k t t a s s bk y s
k a s s bk k y s
By rearranging the order of double integrations, we get
2 1 2
1 1
2 2 1
2 1
1
2 1 1 2 1
2
0 0
1 1 2
0
1 1 2
1 1
2 2
0 0
1 1 2 2 1
0
1 2 2 1
,
d d d
d d d
d d , d d d d T n n
T T T
n
t t t
n t t t t n t T T T n t t n t n
Hy t Hy t k t t
a s s bk y s
k a s s bk k y s
b k k y s
k t t a s s bk T y
k a s s bk k t t y
bk k t t y
d
2 1 t t
From the above estimate and the consideration of the fact that YBr we obtain
2 1 1 2 1 2 12 1 1 2 1
2
0 0
1 1 2 2 1
0 1 2 2 1
2
1 2 1 1
1 2 2 1
, d d d d d , d . T n n n t t n t t n t t T t
Hy t Hy t k t t
a s s bk T y
k a s s bk k t t y
bk k t t y
bTk
k t t a r k a s s
M
bk k t t r M
yn Now, utilizing the fact that the sequence is weakly compact and taking into account Theorem 6 we can show that the number
2 1 d t ta s s
d d
is arbitrarily small provided the number
2 1 t t is taken to be sufficiently small (it is a consequence of the fact that a one element set is weakly compact in L1).
1 2
0 0
1 2
0 0
1 1 2
0 0
1 1 2
0 0
1 1 2
0 0
1 1 2
0 0
1 0
, , , d d
, ,
d d d
d d d
d d
d d
d
n
t s
n
t s
n
t t t
n
t t t
n
t t
n
t t
n t
Hy t
k t s f s k s y s
k t s a s b k s y s
k a s s bk k y s
k a s s bk k y s
k a s s bk k y t
k a s s bk k T y
k a s s bk
d d
1 2 0 1 2 1
0 0
d
d d
t n t t
n
k T y
M
bk k T
k a s s y u u
M
Hence
11 1 2 .
n
Hy t k a bk k TM r
Hence consequently the sequence
Hyn is a se-quence of uniformly bounded and equicontinuous func-tions on D. Hence, in view of Ascoli-Arzela theorem we deduce that the sequence
Hyn is relatively com-pact subset in the space C D
.Further observe that the above reasoning does not de-pend on the choice of . Thus we can construct a se-quence
Dp of closed subsets of the interval
0,T such that meas.
Dcp as and such that the
0 p sequence Hyn
is relatively compact in every space
pC D . Passing to subsequences if necessary we can assume that
Hyn is a Cauchy sequence in each space
pC D , for p1, 2,.
In what follows, utilizing the fact that the set HY
0
is weakly compact, let us choose a number such that for each closed subset D of the interval
0,T such that meas.
Dc we have
d4
c
D
Hy t t
(10)for any yY.
Keeping in mind the fact that the sequence
Hyn is a Cauchy sequence in each space C D
p we canchoose a natural number p0 such that meas.
p0c
D and for arbitrary natural numbers the following inequality holds
0
,
n m p
0 4meas.n m
p
Hy t Hy t
D
for any tDp0. Obviously without loss of generality we can assume that meas.
Dp 00
Now, using the above facts and (10) we obtain .
0
0
0 0
0 0
d
d
d
meas.
4meas.
3
d .
4
p
c p
c p
T
n m
m m
D
m m
D
p p
m m
D
Hy t Hy t t
Hy t Hy t t
Hy t Hy t t
D
D
Hy t Hy t t
(11)
Finally, from (10) and (11) we get
0
d ,
m n n m
Hy Hy Hy t Hy t t
Hynwhich means that is a Cauchy sequence in the space L1L R1
. Hence we conclude that the setHY is relativelycompact in this space.
In the last step of the proof let us consider the set
0 .
Y Conv HY Y
In view of the Mazur theorem we infer that the set 0 is compact in the space 1. Moreover, we have that the operator
L
H transforms continuously the set 0 into itself. Thus the classical Schauder fixed point principle gives that
Y
H has at least one fixed point. This proves that there exists at least one xL1 that solves Equation (4).
4. Nonlinear Equation of Convolution Type
Assume that is an integrable function. For an arbitrary function:
k R R 1 xL set
0
d , .
t
Kx t
k ts x s s tRThis operator K is a linear integral operator of con-volution type and maps 1 into itself continuously. Now, consider the following condition
L
v k R: RandkL1.,
Then we have the following Corollary
Corollary 11. Let the hypotheses i)-v) are satisfied. Then a nonlinear equation of convolution type
1
2
0 0
, , d d
t s
x t k t s f s k s x
t R
shas at least one integrable solution xL1.
In the next subsection, we prove an existence theorem for integral equation of fractional order as a special form of Equation (12).
Initial Value Problems of Fractional Order As a special case of Equation (14), we consider
1
0 0
, , d d ,
1
t s s
x t k t s f s x s t R
(13)where
2 , 0,
1
s k s
1
and
. Equation (13) is an integral equation of fractional order that can be written in the form
1
1 0
, , d ,
t
x t
k t s f s I x s s tR (14)Obviously, Equation (14) has at least one integrable solution xL1.
Definition 12. By a solution of the initial value prob-lem (1) we mean an absolutely continuous function x satisfies the initial value problem (1).
Theorem 13. Let b K2
2
1 and
0,1
. If assumptions i)-iii) and v) are satisfied, then the initial value problem (1) has at least one solution 1.Proof. Let
yL
x be a solution of the integral Equation (14). Putting
0
d .
t
y t
x Since x is integrable, then
0
d .
t
Dy t D x
a e.where d
d D
t
. Moreover, the integral
0 d
t
x
ofintegrable functionx is absolutely continuous then
1
Dy t DI x t Then we have,
. .Dy t x t a e
Furthermore, we obtain
1 1
1 Dy t x t
I Dy t I x t
D y t I x t
Consequently, Equation (14) gives
1
0 d
, , d
d
t
,
y t Dy t k t s f s D y s s t
Since x is integrable and absolutely continuous, then
1 1
0 0
0 0 d
d d
d
d
t t
t
Dy x
I Dy I x
y x
t
y t y x
Clearly, y
0 y0. Hence we deduce that is an absolutely continuous function satisfies the initial value problem (1). Hence the proof is complete.y
5. Conclusion
The existence theorem of functional integrable equation in the space of Lebesgue integrable functions on un-bounded interval L₁
0,
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