Gen. Math. Notes, Vol. 6, No. 1, September 2011, pp. 15-24
ISSN 2219-7184; Copyright © ICSRS Publication, 2011
www.i-csrs.org
Available free online at http://www.geman.in
Solution for Boundary Value Problem of
Non-Integer Order in
L2 -Space
Azhaar H. Sallo
Department of Mathematics, College of Science,
University of Duhok, Duhok, Iraq
E-mail: azhaarsallo@yahoo.com
(Received: 12-4-11/Accepted: 15-9-11)
Abstract
In this paper, we shall prove the existence and uniqueness of a square integrable solution in L2 -space, for the boundary value problem of non- integer
order which has the form:.
c
D xα y( x ) = f ( x , y( x ))
, 1<α ≤ 2
y ( a ) = ya ,
y( b ) = y b
Where c D α is the Caputo fractional derivative and
a , b are positive constants
with b≠ a .The contraction mapping principle has been used in establishing our
main results.
Keywords: Fractional differential equation, Caputo’s fractional
2
derivative, L -Space, Boundary Value Problem.
16
1
Azhaar H. Sallo
Introduction
Fractional differential equations have received considerable attention in the recent
years, due to their wide applications in engineering, economy and other fields.
Fractional derivatives provide an excellent instrument for the description of
memory and hereditary properties of various materials and processes. This is
the main advantage of fractional derivatives in comparison with classical integerorder models, in which such effects are in fact neglected. The advantages of
fractional derivatives become apparent in modeling mechanical and electrical
properties of real materials, and in many other fields see ( [3],[4]).
Bai and Lu [1] discussed the existence and the multiplicity of positive solutions
for boundary value problem of nonlinear fractional differential equation
D0α+ u( t ) = f ( t , u( t ))
, 0<t <1
(1)
u( 0 ) = u( 1 ) = 0
Where
1 < α ≤ 2 is a real number, D0α+ is the standard Riemann-Liouville
differentiation, and f : [0 ,1]× [0 ,∞ ) → [0 ,∞ ) is a continuous function.
Hadid [2] studied local and global existence theorems of the nonlinear differential
equation
D α y( t ) = f ( t , y( t ))
0 <α <1
y( t 0 ) = t 0
(2)
by Shauder and Tyconove fixed point theorems.
Zhang [8] considered the existence of solution of nonlinear fractional differential
equation boundary value problems involving Caputo’s derivative.
D0α+ u( t ) + f ( t , u( t )) = 0
u( 0 ) = α ≠ 0 ,
, 0 < t < 1,
1<α ≤ 2
u( 1 ) = β ≠ 0
(3)
Momani [5] studied local and global uniqueness theorems of the fractional
differential equation (1) by using Biharie’s and Gronwell’s inequalities. In this
paper, we consider the existence and uniqueness of a square - integrable solution
of the following boundary value problem of fractional order:
c
D xα y( x ) = f ( x , y( x ))
y ( a ) = ya ,
, 1<α ≤ 2
y( b ) = y b
(4)
Where c Daα the Caputo’s fractional derivative is f (x , y ) is a function on (a ,b ) × E
in to E and E being an Euclidean – space.
Solution for Boundary Value Problem of…
17
Consider the space L2 (a ,b ) , the space of all measurable functions f such that f
2
is Lebesgue integrable on [a ,b ] . For any f ∈ L2 ( a , b ) we define the norm as
b
f
2
2
= ∫ f ( x ) dx
2
(5)
a
Under this norm, it is known that the space L2 (a ,b ) , is a Banach space.
2
Preliminaries
In this section, we mainly demonstrate and study the definitions, lemmas and
some fundamental facts on Caputo’s derivatives on fractional order, which can
been founded in [6].
Definition 2.1. Let f be a function which is defined almost everywhere (a.e) on
[a,b] .for α > 0 ,we define
bα
I
a
b
α −1
1
(b − x )
f =
Γ ( α ) ∫a
f ( x )dt
Provided that this integral (lebsegue) exists, where Γ is gamma function.
Definition 2.2. For a function f defined on the interval [a ,b] , the Caputo’s
fractional derivative of f is defined by
(Da + f )( x ) = Γ (n1− α ) ∫ (x − s )n−α −1 f
a
α
x
(n )
(s , y( s )) ds
Where n = [ α ] + 1 and [α ] denotes the integer part of α .
Definition 2.3. If α ∈ R and a function f ( x ) defined a.e on a ≤ x ≤ b , we define
f
(α )
x
x
a
a
( x ) = I −α f for all x ∈ [a ,b] , provided that I −α f exists.
Lemma 2.1. If α > 0 and f (x ) is continuous on [a ,b] , then D −α f exists and it
is continuous with respect to x on [a , b] .
Lemma 2.2. If α > 0 and f (x ) belongs to L( a ,b ) , then
x ∈ L(a ,b ) if α ≥ 1 and a.e. if α < 1 .
x
a
D −α f exists for all
18
Azhaar H. Sallo
Lemma 2.3. Let α > 0 and n = [α ] + 1 . If y ∈ AC n [a ,b ] or y ∈ C n [a ,b ] ,
then
(I
α
c
)
n −1
D α y ( x ) = y( x ) − ∑
k =0
y (k ) (a )
(x − a )k .
k!
Lemma 2.4. If α , β > 0 and f ( x ) ∈ L(a ,b ) ,then on a ≤ x ≤ b , we have:
x
a.
∫
s
a
D −α f ds = ax D − (α +1) f , or
x
a
D −1 as D −α f = ax D −(α +1 ) f .
a
b. If α ≥ 1 ,then ax D −α f is absolutely continuous in x ∈ [a , b].
d x −(α +1)
f = ax D −α f , everywhere if α ≥ 1 and a.e. if α < 1.
c.
aD
dx
d. ax D − (α + β ) f = ax D −α as D − β f , a.e. if α + β ≤ 1.
Lemma 2.4. (Hölder's inequality)
Let X be a measurable space, let p and q satisfy 1 < p < ∞ , 1 < q < ∞ , and
1 1
+ = 1.
p q
If f ∈ L p ( X ) and g ∈ Lq ( X ), then ( f g ) belongs to L( X ) and satisfies
∫
X
3

f g dx ≤  ∫ f
X
p
1
p
 
dx   ∫ g
 X
q

dx 

1
q
The Main Results
In this section we shall prove existence and uniqueness of a square-integrable
solution in L2 (a ,b ) space for the boundary value problem of non-integer order
of the form (4).
Lemma 3.1
Let 1 < α ≤ 2 and let h : [a ,b] → R be continuous. A function y is a solution of
fractional integral equation
(x − a) (b −t )α−1 h(t )dt− (x −b) y + (x − a) y
1
( x −t )α−1h(t )dt−
∫
(b − a) Γ(α ) ∫a
(b − a) a (b − a) b
Γ(α ) a
x
y( x ) =
b
if and only if y is a solution of the fractional boundary value problem
(6 )
Solution for Boundary Value Problem of…
c
Dxα y( x) = f ( x, y( x ))
y( a) = ya ,
19
, 1<α ≤ 2
(7 )
y(b) = yb
(8)
Proof: By using definition (2.3), equation (7) can be written as
x
I −α y( x ) = f ( x, y( x ))
a
, 1 <α ≤ 2
(9)
x
Operating on both sides of equation (9) by the operator I −α , we obtain :
a
x
t
−α
a
a
Iα I
x
y( x ) = I α f ( x , y( x ))
a
by using Lemma (2.3), we get:
x
y( x ) = y( a ) + y ′( a ) ( x − a ) + I α f ( x , y( x ))
a
( 10 )
Using the boundary condition (10), we obtain:
y( a ) = ya
( 11)
b
y( b ) = ya + y′( a ) (b − a) + I α f ( x, y( x ))
( 12 )
a
y′( a ) =
yb
y
1 bα
− a −
I f ( x, y( x ))
(b − a) (b − a) (b − a) a
( 13 )
Substituting equation (13) and (11) in equation (10) , we obtain the final form of
y ( x ) is :
1
(x −a) (b−t )α−1h(t )dt− (x −b) y + (x −a) y
( x −t )α−1h(t )dt−
∫
(b−a)Γ(α ) ∫a
(b−a) a (b−a) b
Γ(α ) a
x
y( x) =
b
Theorem 3.2 Let the right hand side
equation (4) be such that:
(14)
f (x, y) of the fractional differential
i. it satisfies the Lipschitz condition in y with Lipschitz constant K, that is,
f ( x, y2 ) − f ( x, y1 ) ≤ K y2 − y1
on the domain
D , where:
( 15 )
20
Azhaar H. Sallo
D = { ( x , y ):
x − x0 ≤ a , y − y 0 ≤ b }
( 16 )
ii. it is a square –intgerable ( f ∈ L2 ( a ,b ) ) as a function of x ∈ ( a ,b ) . Then
If
2 k ( b − a )α
1 + 2α
<1
( 17 )
Γ (α )
2α( 2α − 1 )
There exists a square –integrable solution for the fractional differential equation
(7) with the boundary condition (8) on (a, b) .
Proof. Let the mapping T on L 2 (a , b) be defined as:
(x−a) (b−t )α−1 f(t,y(t ))dt−(x−b) y +(x−a) y
1
T( y(x))=
(x−t )α−1 f (t,y(t ))dt−
∫
(b−a)Γ(α)∫a
(b−a) a (b−a) b
Γ(α) a
x
b
(18)
every function g ∈ L 2 (a, b) into a function which
we claim that T takes
belongs to L 2 (a, b) . Let:
h( x)=
x
(x−a) b (b−t )α−1 f (t, y(t ))dt−(x−b) y +(x−a) y
1
α−1
(
x
−
t
)
f
(
t
,
y
(
t
))
dt
−
(b−a)Γ(α) ∫a
(b−a) a (b−a) b
Γ(α) ∫a
(19)
Then
(x −a) (b −t )α−1 f (t, y(t ))dt− (x −b) y + (x −a) y
1
h( x ) =
( x −t )α−1 f ( t, y( t ))dt−
∫
(b −a)Γ(α ) ∫a
(b −a) a (b −a) b
Γ (α ) a
x
b
2
2
( 20)
By the lemma (2.4(b)) , , the first term of equation (19) exists and it is absolutely
continuous so it is continuous for all x ∈ [a , b ] .also the other terms are continuous
for all x ∈ [a , b ] .
2
Thus h(x) is continuous and measurable in D . Hence h( x ) is measurable.
Now, we have to show that
f +g
p
(
≤ 2p f
p
+g
p
)
h( x )
2
is Lebsegue integrable. Since
and from equation (20), we have
1 x
(
x −a) b
(x −b) y + (x −a) y
α−1
h(x) ≤4
(x − t) f (t, y(t))dt + 4
(b − t)α−1f (t, y(t))dt−
∫
∫
Γ(α) a
(b −a)Γ(α) a
(b −a) a (b −a) b
2
2
2
Solution for Boundary Value Problem of…
the term
21
(x − a ) b (b − t ) α−1 f ( t, y(t )) dt − (x − a ) y + (x − a ) y
(b − a )Γ(α) ∫a
(b − a ) a (b − a ) b
is Lebsegue integrable.
Let
 1 x

α−1
(Z(x)) = 
(
x
−
t
)
f
(
t
,
y
(
t
))
dt

∫
Γ ( α ) a

2
2
( 21)
We have to show that (Z(x)) 2 is Lebesgue integrable. Since ( x − t )α −1 ∈ L2 (a , b )
and by hypothesis (ii) f ( t , g ( t )) ∈ L2 ( a , b ) as a function of t, then by HÖlder
inequality ( p = q = 2) and from equation (18) we obtain:
x
1 x
2(α−1)
(Z(x))≤
( x −t )
dt ∫ f 2( t, y( t ))dt
∫
Γ(α ) a
a
2
( x − a )2α−1 x 2
=
f ( t, y( t ))dt
( 2α −1) Γ (α) ∫a
( 22)
For x ∈ [a , b ] and by using definition (2.1) and Lemma (2.4(a)) we have:
x t
t 2





 ∫ ( t − s )1−1 f 2 ( t , y( t )) ds  dt
f
(
t
,
y
(
t
))
ds
dt
=
∫∫
∫



aa
aa


x
x
= ∫(
x 1
a
I f 2 ) dt
a
x 2
=a
I f2
x
Thus by Lemma (2.2), it follows that
∫f
2
dt is Lebesgue integrable for all
a
x ∈ [a ,b ] .
If there exist a Lebesgue integrable function g ( x ) on
[a ,b]
f ( x ) ≤ g( x ), a.e on [a ,b] , where f ( x ) is measurable then
such that
f ( x ) is
Lebesgue integrable function. Hence from inequality (22), (Z(x)) 2 is Lebesgue
2
integrable. Thus h(x) Lebesgue integrable and therefore T maps L 2 ( a , b )
into itself.
To prove that T is a contraction mapping, Let g 1 , g 1 be any two function that
belong to L 2 ( a , b ) , and consider:
22
Azhaar H. Sallo
T ( y 2 ( t )) − T ( y 1 ( t ))
2
2
1 x
( x − t )α −1 [ f ( t , y 2 ( t )) − f ( t , y 1 ( t ))]dt −
∫
Γ (α ) a
=
(
x − a) b
−
( b − t )α −1 [ f ( t , y 2 ( t )) −
∫
(b − a )Γ ( α ) a
x
1
2
(
x−a) b
[ f (t,y2(t ))− f (t,y1(t ))]dt−
(b−t )α−1[ f (t,y2(t ))− f (t,y1(t ))]dt dx
∫
(b−a)Γ(α) a
2
x
α−1
∫ Γ(α) ∫( x−t )
a
f ( t , y 1 ( t ))] dt
2
a
2



4(x−a) b b
α−1
α−1



 dx
≤
(
x
−
t
)
f
(
t
,
y
(
t
)
)
−
f
(
t
,
y
(
t
)
)
dt
dx
+
(
b
−
t
)
f
(
t
,
y
(
t
)
)
−
f
(
t
,
y
(
t
)
)
dt
2
1
2
1
∫
∫
∫
∫
2
2 


(Γ(α))2 aa
b
−
a
Γ
α
(
)
(
(
)
)
a a


4
2
b x
2
Using lipschitz condition ( 15 ) ,we have :
T ( y 2 ( t )) − T ( y 1 ( t ))
2
2
4K2
≤
2
x

α −1
∫  ∫ ( x − t ) y 2 (t ) − y1 (t ) dt  dx +
aa

b
(Γ (α ))2
2

4(x −a) K2 b b
α−1

 dx
(
)
(
)
+
(
b
−
t
)
y
t
−
y
t
dt
2
1

(b−a)2(Γ(α))2 ∫a ∫a

2
Since
y 2 , y 1 ∈ L2 ( a ,b )
L2 ( a ,b )
and
is
a
(23)
linear
space,
then y 2 − y1 ∈ L2 ( a , b ) , ( x − t )α −1 ∈ L2 ( a ,b ) and ( b − t )α −1 ∈ L2 ( a , b ) by
using HÖlder inequality, where ( p = q = 1 ) , inequality (23) becomes
T ( y 2 ( t )) − T ( y 1 ( t ))
2
2
≤
4K2
(Γ (α ))2
 x
 x

2
2 (α − 1 )

 dx +


(
x
−
t
)
dt
y
(
t
)
−
y
(
t
)
dt

2
1
∫a  ∫a

∫

 a

b
4(x −a) K2
b
 x

2(α−1)

∫ y2(t) − y1(t) 2dtdx
(
b
−
t
)
dt
2
2 ∫∫


(b−a) (Γ(α)) a a
a

2
+
b
To integrate the right hand side of inequality (21):
x
Let
r ( x ) = ∫ y 2 (t ) − y1 (t ) dt
2
a
and
r ' ( x ) = y 2 (t ) − y 1 (t )
2
a .e
(24)
Solution for Boundary Value Problem of…
2
T( g2(t ))−T( g1(t )) 2 ≤
b
4 K2
2α−1
( x − a)
(2α −1)(Γ(α)) ∫
2
23
r(x)dx+
a
4K2
2
T( y2(t ))−T( y1(t )) 2 ≤
≤
2α(2α−1)(Γ(α))
2
b
2α−1
( b − a)
(b −a) (2α −1) (Γ(α)) ∫
2
2
r(x)dx
a
[r(b)(b−a) −∫(x−a)
b
2α
2
4(x −a) K2
r (x)dx+
2α '
a
4(x−a) K2 (b−a)2α−1 b
2
(b−a)2(2α−1) (Γ(α))2 ∫a
r(x)dx
b
b

 4(x−a)2 K2 (b−a)2α−1 b b

2
2
2α
2α
 y −y1 2dtdx(25)
(
b
−
a
)
y
−
y
dx
−
(
x
−
a
)
y
−
y
dx
+

2
1
2
1
∫
∫
2
2
2 ∫ ∫ 2

2α(2α−1)(Γ(α)) 
a
a
 (b−a) (2α−1) (Γ(α)) a a

4K2
b
Since ∫ ( x − a ) 2α y 2 − y 1 dx , a ≤ x ≤ b , then inequality (25) gives
2
a
T ( y 2 ( t )) − T ( y 1 ( t ))
2
2
≤
4 K 2 ( b − a ) 2α
b
∫
2α (2α − 1)(Γ (α ))
2
y 2 − y 1 dx +
2
a
+
4 K 2 ( b − a ) 2α
b
(2α − 1)(Γ (α )) ∫
2
y 2 − y 1 dx
2
a
T ( y 2 ( t )) − T ( y 1 ( t ))
2
2
≤
2 K ( b − a )α
Γ (α )
1 + 2α
2α ( 2α − 1 )
y 2 − y1
2
2
Consequently by (17) T is a contraction mapping, as a consequence of Banach
fixed point theorem, we deduce that T has a unique fixed point y( x ) ∈ L2 ( a , b ) ,
that is T y ( x ) = y( x ) .Therefore from equation (18) we have
(x − a) (b −t )α−1 f (t,g(t ))dt − (x −b) y + (x − a) y
1
y( x ) =
( x − t )α−1 f (t, g(t ))dt −
∫
Γ(α ) a
(b − a)Γ(α ) ∫a
(b − a) a (b − a) b
x
b
This completes the proof.
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24
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