IOSR Journal of Mathematics (IOSR-JM)
e-ISSN: 2278-5728, p-ISSN: 2319-765X. Volume 10, Issue 6 Ver. VI (Nov - Dec. 2014), PP 08-13
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Existence of unique solution for Fractional Differential Equation
by Picard approximation method
MahaAbd Al-Wahab
Department of Applied Science University of Technology Baghdad-Iraq
Abstract: Our work is finding continuous function y on (a, ) which is the unique solution for they()(x)=λ
f(y(x)),x(a, ), 0<≤1 with y(-1)(a) = , where is constant and λ is a real number using the picard
approximation method theorem.
Keywords: Fractional differential equation, picard approximation method.
I.
Introduction
Fractional calculus as well as fractional Differential equations have received increasing attention and
has been a significant development in ordinary and partial fractional differential equations in recent years;
seethe papers by Abbas and Benchohra [1,2,3], Agaiwal et al. [4], monographs of Kilps, Lakshmikatham et al.
[6]. This article studies the existence of the unique solution of fractional differential equationy()(x)=λ f(y(x))
and y(-1)(a)=,  is some constant, 0 < ≤ 1, λ is real number using picard approximation method.
II.
Preliminaries
In this section we introduce notations, definitions and preliminary facts which are used throughout this paper.
Definition (2-1)([7]): Let x={F:F is a real valued function and continuous on [a,)}, for some a(-, ) . Let
the |||| on xbe defined by
F = supx∈[a,) e−γ α F(x)
provided that this norm exist for some constant >0.
Lemma (2-1)([8]):Let 0<≤1 and f,g be continuous functions on (a, ), where aR such that
sup
f( g(x) ) : x  (a, ∞) = M <. Define
x
 (x - a) 1
1
f (x) 

( x  t ) 1  f ( g (t ))dt ,

( )
( ) a
for all x>a and is some constant. Then fc(a,).
Lemma (2-2)([9]): Let us define F(x) = (x-a)-1f(x) on (a,), where fdefine in lemma (2-1) and 0<≤1.Then
Fc[a,).
Lemma (2-3) ([7]):Let ,  R, > -1. If x>a then
 (  a) 

(
t

a
)

, α  γ  negative integer
I
  (    1)
a (  1)

0
, α  γ  negative integer
x
Lemma (2-4) ([9]):suppose G is a banach space and let TL(G) such that T n
n
n
(I − T)−1 = I + ∞
n=1 T , where the series n T converge in L(G).
1
n
< 1. Then I-T is regular and
Definition (2-2) ([9]): Let f be Lebesque- measurable function defined a.e on [a, b], if >0 then we define
b
b
1
I f 
f (t ) (b  t ) 1dt

a
( ) a

provided the integral (Lebesque) exists.
Definition (2-3) ([10]): If R, f is define a.e on the interval [a, b], we define
DOI: 10.9790/5728-10660813
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Existence of unique solution for Fractional Differential Equation by Picard approximation method
x
d f



f
(
x
)

I
f

a
dx
for all x  [a, b]
x
provided that
I  f
exists.
a
Lemma (2-5) ([8]): If 0 < ≤ 1 and f(x) is continuous on [a,b], |f(x)| ≤ M for all x[a,b] (where MR+, M>0).
Then
x
x
I I  f (x)  f ( x) for all x  (a, b).
a

a
Theorem (2-1)([9]):Let 0<≤1 and  be positive constant. Let g(x)=(g1(x), g2(x), …, gn(x))T, x[a,), wheregi
1
are continuous on [a,), i= 1, 2, …, n and g(x) =
n
2 2
i=1 gi
and |g(x)|≤x+c, where c is positive constant.
fi=(f1, f2, …, fn)T such that fi[a,) andsup{|f(x)|: x[a, )}=M<. Choose λ such that λ < eα
T
−1
c α
α
. Then
()
there exists continuous vector function y(x)=(y1(x), y2(x), …, yn(x)) , x(a,) such that y (x)=λf(y(x)),
x(a,) with y(-1)(a)=, where =(1, 2, …, n)T is some constant vector and satisfied
|y(x)|<exp (c-1|x|).constant.
III.
Main Results
In this section we prove the existence of a continuous function y on (a, ) which is the unique solution
for y()(x)=λf(y(x)), and y(-1)(a)=, where  is some constant,0 < ≤ 1, λ is real number using picard
approximation method.
Theorem
(3):Let
0<≤1,
g(x)
is
continuous
function
on
[a,
)
and
|g(x)|≤|x|
...(3.1) where x[a,). Let f(y(x)) be a continuous function on [a, ) such that
sup f(y(x)) :x[a,∞) =M<. Then there exist a continuous function y on (a, ) which is the unique solution
for
y()(x)=λ f(y(x))
x(a,) with
y(-1)(a)=, where  is some constantand λ is real number.
Proof:Let[a, a+h]be any compact subinterval of [a,) and let (X, ||||) be the space defined in definition (2-1).
Consider
x
1
y(x)  y 0 (x) 
( x  t ) 1 f ( y(t ))dt , x  (a, ) ...(3.2)

( ) a
where
y0 
(x - a)
1
 (x - a) 1
, it follows from lemma (2-1) that y(a,).Then
( )
(x - a)1

y(x)  b 
( x  t ) 1 f ( y(t ))dt , x(a, ). Where b 

( ) a
( )
x
Let F(x)=(x-a)1-y(x) , x(a,) ...(3.3)
Where y is given in (3.2) and define
F(x, y(t))=(x-a)1-f(y(x)), a≤t<x< ...(3.4)
Thus from Lemma (2-2) we have Fc[a, ).
Now define a linear operator K on [a, a+h] as:
x
KF(x)  1  ( x  t ) 1 F ( x, y(t ))dt
( ) a
, x[a, a+h]
...(3.5),
and consider the equation
DOI: 10.9790/5728-10660813
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Existence of unique solution for Fractional Differential Equation by Picard approximation method
 x
F(x)  b 
( x  t ) 1 F ( x, y (t ))d

( ) a

where b 
and  is some constant.
( )
, x[a, a+h]
...(3.6)
Now we prove
1
n
limx→∞ K n
= 0, from (3.3) we have
x
KF (x) =
1
( x  t ) 1 F ( x, y (t )) dt

( ) a
x
1
( x  t ) 1 F ( x, y (t )) dt
≤

( ) a
x
1
=
(x − t)α−1 eγ y(t) dt
Γ(α)
a
Then from (3.1) we have
x
F
KF (x) ≤
(x − t)α−1 eγ t dt
Γ(α)
≤
F e
Γ(α)
a
x
γx
(x − t)α−1 dt
a
F eγ x −(x − t)α x
=
Γ(α)
α
a
F eγ x (x − a)α
=
Γ(α)
α
F eγ x (x − a)α
=
… (3.7)
Γ(α + 1)
Now by induction we prove the following inequality
F eγ x (x − a)nα
K n F (x) ≤
Γ(nα + 1)
F eγ x hnα
≤
, x ∈ a, a + h and n = 1, 2, …
Γ(nα + 1)
by using (3.7), it is obvious that (3.8) holds for n=1.
Next suppose that (3.8) is true for positive integer n, then we have
K n+1 F (x) = K(K n F) x
1
=
Γ α
… (3,8)
x
x−t
α −1
K n F x, y t dt
a
1
≤
Γ α
x
x−t
K
1
F (x) ≤
Γ(α)
F
≤
Γ(α)
F eγ x
≤
Γ(α)
x
a
K n F x, y t
dt
a
It follows from (3.8) that
n+1
α−1
x
a
x
a
(x − t)α−1 F (y(t) − a)nα eγ y(t)
Γ(nα + 1)
(x − t)α−1 (t − a)nα eγ t
Γ(nα + 1)
dt
dt
(x − t)α−1 (t − a)nα
dt
Γ(nα + 1)
DOI: 10.9790/5728-10660813
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Existence of unique solution for Fractional Differential Equation by Picard approximation method
x
I
(t − a)nα
a
Γ(nα + 1)
Then by lemma (2-3) we have
γx
= F e
x − a nα+α
,
Γ nα + α + 1
K n+1 F x ≤ F eγ x
γx
x ∈ a, a + h
α(n+1)
F e
x−a
Γ(α n + 1 + 1)
F eγ x hα(n+1)
≤
Γ(α n + 1 + 1)
Thus (3.8) hold for all n=1, 2, 3, ...
Hence
=
e−γ x
Kn F x ≤
F hnα
, x ∈ a, a + h
Γ nα + 1
And so by definition (2-1) we get
F hnα
Kn F ≤
Γ nα + 1
and it follows from definition (2-2) that
hnα
Kn ≤
Γ nα + 1
Now
1
n
hnα
lim
= lim
n→∞
n→∞ Γ nα + 1
1
n
1
α
= h lim
n→∞ nα Γ nα
1
= hα lim
1
1
n→∞
(nα)n [Γ nα ]n
Since n= 1, 2, 3, ... then
1
Kn n
1
n
1
lim (nα)n = lim n
n→∞
n→∞
1
(α)n ≥ 1
And also we have
 nα = 2π nα
1
nα −2 (−nα+θ) 12nα
e
, 0 < θ < 1, 𝑛ϵ Z+
seeArtine (1964) and so
 nα
1
n
= 2π
1
n
nα
1
α
nα
1
2n
e−α eθ
12n 2 α
And hence
lim  nα
1
n
n→∞
= lim
n→∞
2π
1
n
nα
α
1
nα
1
2n
e−α eθ
12n 2 α
= 1 . ∞. 1 . e−∞ . 1 = ∞
Consequently we have
limn→∞ K n
lim (λK)n
n→∞
1
n
1
n
= 0 and this implies that
1
= λ lim K n n = 0
n→∞
Then by lemma (2-4), (I − λK)−1 = I + n λn K n
and then series is convergent.
From (3.5) and (3.6) we have
F x = I − λK −1 (b) , therefore F is exists and is the unique solution of
λ
F x =b+
Γ α
x
x−t
α−1
F x, y t dt.
a
Then from (3.3) we get
μ
λ x−a
F x =
+
Γ α
Γ α
x
1−α
x−t
α−1
f y t dt
a
for allx(a, a+h] and by using (3.3) it follows that
DOI: 10.9790/5728-10660813
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Existence of unique solution for Fractional Differential Equation by Picard approximation method
1−α
x−a
for allx(a, a+h]
μ x−a
y x =
Γ α
α−1
λ
+
Γ α
x
1−α
μ
λ x−a
y x =
+
Γ α
Γ α
x−t
α−1
f y t dt
a
x
x−t
α−1
f y t dt
a
Therefore by definition (2-2) we get
x
 (x - a) 1
y(x) 
   f
a
( )
x  (a, a  h] ...(3.9)
,
andso
x
I

a
x
yI

a
x
t
 (t - a) -1
  I  I  f
a
a
( )
x
But from lemma (2-3) we have
x
t
a
a
Thus
I  y   f(y(x)) ,
I

a
(t - a) -1
 0 and by lemma(2-5) we get
( )
I  I  f  f(y(x)) for all x(a, a+h]
x
a
x(a, a+h]
Then by using definition (2-5) we get
x
y ( ) (x)  I  y   f(y(x)) x(a, a+h]
a
Furthermore from (3.9) we have
x
x
a
a
x
t
 (t - a)  1
   1   f
a
a
( )
 1 y   1
It follows from lemma (2-3) that
x
x
t
x
a
a
a
a
 1 y      1   f      1 f
 
x

f ( y (t ))dt
a
x
and so 
a
1
y
exists for all x[a, a+h]
since by definition (2-3)
x
y ( 1) ( x)   1 y therefore
a
y
( 1)
(a )   .
Now from theorem (2-1) equation (3.11) we have
F(x) ≤ b + λ kF(x)
then from theorem (2-1) equation (3.8) we have
eγc
F(x) ≤ b + λ F c eγ x
γ
< b + F eγ x sincec=
= eγ x (be−γ x + F )
< eγ x (b + F )
thus by using theorem (2-1) equation (3.3) we obtain
DOI: 10.9790/5728-10660813
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Existence of unique solution for Fractional Differential Equation by Picard approximation method
x − a 1−α y(x) < eγ x b + F
h1−α y(x) < eγ x b + F
y(x) < eγ x hα−1 b + F
and so the solution function satisfied
|y(x)|<exp (c-1|x|).constant
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