Indian J. Pure Appl. Math., 43(6): 619-635, December 2012
c Indian National Science Academy
°
MILD AND STRONG SOLUTIONS TO FEW TYPES OF FRACTIONAL
ORDER NONLINEAR EQUATIONS WITH PERIODIC BOUNDARY
CONDITIONS
Mohamed A. E. Herzallah
Faculty of Science, Zagazig University, Zagazig, Egypt.
College of Science in Zulfi, Majmaah University, Saudi Arabia
e-mail: m [email protected]
(Received 13 May 2012; after final revision 25 July 2012;
accepted 26 July 2012)
In this paper, the two fractional periodic boundary value problems
C α
0 D0+ u(t)
− λu(t) = f (t, u(t)), u(0) = u(1),
0 < α < 1,
and
C β
0 D0+ u(t)
− λu(t) = f (t, u(t)), u(0) = u(1), u0 (0) = 0
1 < β < 2,
α
will be studied where C
0 Dt is the ordinary Caputo fractional derivative and
λ ∈ R − {0}. Under some suitable assumptions on the function f , the
existence of at least one mild solution will be proved. Under some conditions,
the uniqueness of this mild solution will be proved to both problems. Finally,
these mild solutions will be strong solutions under certain conditions.
Key words : Mild solution; strong solution; fractional derivative; boundary
value problem; Schaufer fixed point theorem.
620
MOHAMED A. E. HERZALLAH
1. I NTRODUCTION
Fractional Calculus is a field of applied mathematics that deals with derivatives and
integrals of arbitrary orders. Many applications of fractional calculus can be found
in turbulence and fluid dynamics, stochastic dynamical system, plasma physics
and controlled thermonuclear fusion, nonlinear control theory, image processing,
nonlinear biological systems, astrophysics (see [5, 13-16, 18]).
Various phenomena were modelled with fractional differential equations. Recently, there are a large number of papers dealing with the solvability of nonlinear
fractional differential equations by the use of nonlinear analysis techniques (see
[1-4, 8-12, 17, 20-22] and the references therein). Really in some papers the solution is given and is determined if it is a mild or a strong solution, but there is still
a confusion with the given solution in some other papers mainly because of the
vague of these solutions. The vague solutions came from the determination of a
fixed point for the integral equation which is resulted from the fractional one. The
vague solution was neither proved to be mild solution not to be strong solution to
the nonlinear fractional differential equation.
The nonlinear fractional differential equation
R δ
0 Dt u(t)
− λu(t) = f (t, u(t)),
u(1) = lim t1−δ u(t),
t→0+
0 < δ < 1, λ ∈ R − {0},
(1)
δ
where R
0 D0+ is the Riemann-Liouville fractional derivative, has been considered
in [3], and its solution has been given in the form
Z 1
u(t) =
Gλ,δ (t, s)f (s, u(s))ds
(2)
0
where Gλ,δ (t, s) is the Green’s function defined by

Γ(δ)Eδ,δ (λtδ )Eδ,δ (λ(1−s)δ )tδ−1 (1−s)δ−1



1−Γ(δ)Eδ,δ (λ)


 +(t − s)δ−1 E (λ(t − s)δ )
δ,δ
Gλ,δ (t, s) =




δ
δ δ−1
δ−1

 Γ(δ)Eδ,δ (λt )Eδ,δ (λ(1−s) )t (1−s)
1−Γ(δ)Eδ,δ (λ)
0≤s≤t≤1
0≤t≤s≤1
(3)
FRACTIONAL ORDER NONLINEAR EQUATIONS
621
In [11] the existence of periodic mild solution to the problem of a nonlinear
fractional differential equation
R δ
0 Dt u(t)
− λu(t) = f (t, u(t)),
u(0) = u(1) = 0,
1 < δ < 2, λ ∈ R − {0}
has been proved and given in the form
Z 1
u(t) =
Gλ,δ (t, s)y(s)ds,
(4)
(5)
0
where

tδ−1 E (λtδ )


− Eδ,δδ,δ(λ) (1 − s)δ−1 Eδ,δ (λ(1 − s)δ )



 +(t − s)δ−1 E (λ(t − s)δ )
0≤s≤t
δ,δ
Gλ,δ (t, s) =




δ−1
δ

 − t Eδ,δ (λt ) (1 − s)δ−1 Eδ,δ (λ(1 − s)δ ) t ≤ s ≤ 1.
Eδ,δ (λ)
(6)
and the conditions for uniqueness has been given.
In this paper, consider the two periodic problems:
C α
0 Dt u(t)
− λu(t) = f (t, u(t)), u(0) = u(1),
0 < α < 1,
λ ∈ R − {0}
(7)
and
C β
0 D0+ u(t)
− λu(t) = f (t, u(t)), u(0) = u(1), u0 (0) = 0
1 < β < 2,
λ ∈ R.
(8)
The existence and uniqueness of solution to the two problems in linear case are
given in [6, 7, 10] as a special case. In both problems the existence and uniqueness
of a solution in the linear case will be proved, its formula in terms of the MittagLeffler function will be given, the existence of a mild solution to the nonlinear
one will be proved, the conditions to have uniqueness of this mild solution will be
given, and these mild solutions will be strong solutions under certain conditions.
622
MOHAMED A. E. HERZALLAH
This paper takes the example of fractional Langevin’s type of equation with
fractional order of differentiation (α, β) in one case fractional order as 0 < α < 1,
and in second case of fractional order as 1 < β < 2.
The paper is organized as follows:
In section 2, we give the principal definitions and theorems used in this paper.
In section 3 we present the main results. Our conclusion are presented in section 4.
2. P RELIMINARIES
The following classes of functions u from the interval J = [a, b], − ∞ < a <
b < ∞, into R will be used in this paper. Let L(J, R) be the space of all integrable functions defined on the interval J with values in R, C(J, R) be the space
of continuous functions from J into R, C m (J, R) = {u ∈ C(J, X) : u(k) (t) ∈
C(J, X), k ≤ m}, AC(J, R) be the space of all absolutely continuous functions
which can be defined by f ∈ AC(J, R) if and only if there exists a Lebesgue
Rx
integrable function g on J such that f (x) = f (a) + a g(t)dt for all x ∈ J),
AC m (J, R) = {u ∈ AC(J, R) : u(k) (t) ∈ AC(J, R), k ≤ m} and W m,1 (J, R)
be the Sobolev space (class of all functions whose derivative up to order m belong
to L(J, R)). For simplicity we write L(J) instead of L(J, R) and so on (see [19,
23]).
Here we give some definitions and theorems which are used in the sequel of
this paper. Let f ∈ L(J), J = [a, b] and let α be a positive real number.
Definition 2.1 — The fractional integral of order α of the function f ∈ L(J) is
defined by (see [14, 16, 18, 19]).
Z
α
a It f (t)
=
a
t
(t − s)α−1
f (s)ds
Γ(α)
Definition 2.2 — The (Riemman-Liouvile) fractional derivative of the function
FRACTIONAL ORDER NONLINEAR EQUATIONS
623
f (t) of order α ∈ (n − 1, n) is defined by (see [14, 16, 18, 19])
R α
a Dt f (t)
=
n−α
Dn Ia+
f (t)
dn
= n
dt
Z
t
a
(t − s)n−α−1
f (s)ds.
Γ(n − α)
Definition 2.3 — The (Caputo) fractional derivative of order α ∈ (n − 1, n) of
the function g(t) is defined by (see [14, 18])
C β
a Dt g(t)
n−β n
= Ia+
D g(t),
Dn =
dn
.
dtn
Lemma 2.4 — Let α ∈ (n, n + 1) (see [14]):
(a) If f ∈ C[a, b], then
C D α I α f (t)
a t a t
= f (t).
α
(b) If f ∈ AC n [a, b], then a Itα C
a Dt f (t) = f (t) −
Pn−1
k=0
f (k) (a)
k! (t
− a)k
The Mittag-Leffler function with two parameters plays a very important role in
fractional calculus as the the role played by exponential function in the theory of
integer-order differential equation.
Mittag-Leffler function in two parameters is defined by (see [14, 18]),
Eα,β (z) =
∞
X
k=0
zk
,
Γ(αk + β)
which gives
E1,1 (z) = ez , E1,2 (z) =
√
sinh( z)
= √z .
ez −1
z ,
α, β > 0,
(9)
√
E2,1 (z) = cosh( z) and E2,2 (z)
The Laplace transform, with variable ξ, of the Mittag-Leffler function in two
parameters is given by:
(k)
L(tαk+β−1 Eα,β (±atα )) =
(k)
where Eα,β (y) =
k!ξ α−β
,
(ξ α ∓ a)k+1
1
Re(ξ) > |a| α ,
(10)
dk
E (y).
dy k α,β
In proving the existence of the solution of our problems we used the following
fixed point theorem (see [23]):
624
MOHAMED A. E. HERZALLAH
Theorem 2.5 — Schaefer fixed point theorem
Let X be a normed linear space, and let the operator T : X → X be compact.
Then either
(i) the operator T has a fixed point in X, or
(ii) the set B = {u ∈ X : u = µT (u), µ ∈ (0, 1)} is unbounded.
3. M AIN R ESULTS
Firstly, the mild solution and strong solution to problems (7) and (8) are defined
as:
Definition 3.1 — Let J = [0, 1], the function u ∈ C(J) is called:
(1) a strong solution of problem (7) on J if u ∈ AC(J) and (7) holds on J;
(2) a strong solution of problem (8) on J if u ∈ AC 1 (J) and (8) holds on J;
(3) a mild solution of (7) on J if
u(t) = u(0) + λ 0 Itα u(t) + 0 Itα f (t, u(t))
(11)
holds on J;
(4) a mild solution of (7) on J if
u(t) = u(0) + λ 0 Itβ u(t) + 0 Itβ f (t, u(t))
(12)
holds on J;
3.1 Linear problem
In this subsection we consider the two linear problems of (7) and (8) in the forms
C α
0 Dt u(t)
− λu(t) = y(t),
0 < α < 1, u(0) = u(1), λ ∈ R − {0}
(13)
FRACTIONAL ORDER NONLINEAR EQUATIONS
625
and
C β
0 Dt u(t) − λu(t)
1 < β < 2, u(0) = u(1), u0 (0) = 0, λ ∈ R − {0}
(14)
where y is a known function. By the results given in [6, 7, 10] one can prove the
existence and uniqueness of a strong solution to (13) if y ∈ W 1,1 [0, 1] and to (14)
1
π
if y ∈ W 1,1 [0, 1] and |arg( (ξ+1)
β | ≤ θ < 2 for Re(ξ) ≥ 0).
= y(t),
Here we discuss these problems by using Laplace transform with getting the
form of this solution. Taking Laplace transform of (13) we get
ξ α û(ξ) − ξ α−1 u(0) − λû(ξ) = ŷ(ξ)
which gives
û(ξ) =
1
ξ α−1
u(0) + α
ŷ(ξ).
α
ξ −λ
ξ −λ
Taking the inverse Laplace transform we get
Z
u(t) = Eα,1 (λtα )u(0) +
0
t
(t − s)α−1 Eα,α (λ(t − s)α )y(s)ds.
(15)
Using the boundary conditions we get the solution in the form
Z 1
Eα,1 (λtα )
u(t) =
(1 − s)α−1 Eα,α (λ(1 − s)α )y(s)ds
1 − Eα,1 (λ) 0
Z t
+ (t − s)α−1 Eα,α (λ(t − s)α )y(s)ds.
(16)
0
By the definition and properties of Mittag-Leffler function one can prove that
α
α
0 It (Eα,1 (λt ))
= tα Eα,α+1 (λtα ) and 0 Itα (tα−1 Eα,1 (λtα )) = t2α−1 Eα,2α (λtα )
which for y ∈ C(J) gives ( where * denotes the convolution operator (f ∗ g)(t) =
626
Rt
0
MOHAMED A. E. HERZALLAH
f (t − s)g(s)ds)
u(0) + λ 0 Itα u(t) + 0 Itα y(t) = u(0) + λ 0 Itα {Eα,1 (λtα )u(0)
Z t
+ (t − s)α−1 Eα,α (λ(t − s)α )y(s)ds}
0
+ 0 Itα y(t)
= u(0) + λ{tα Eα,α+1 (λtα )u(0)
+(t2α−1 Eα,2α (λtα )) ∗ y(t)} + 0 Itα y(t)
∞
X
λk tαk
α
= u(0) + λt u(0)
Γ(αk + α + 1)
k=0
Ã
!
∞
k tαk
X
λ
+ λt2α−1
∗ y(t)
Γ(αk + 2α)
k=0
+
tα−1
Γ(α)
= u(0) +
Ã
∗ y(t)
∞
X
k=1
+ λtα−1
λk tαk
Γ(αk + 1)
∞
X
k=1
λk tαk
Γ(αk + α)
!
∗ y(t)
tα−1
∗ y(t)
Γ(α)
= u(0) + (Eα,1 (λtα ) − 1)u(0)
tα−1
+(tα−1 Eα,α (λtα ) −
) ∗ y(t)
Γ(α)
tα−1
+
∗ y(t)
Γ(α)
= u(t)
+
Thus u(t) given by (16) is a mild solution of (13).
Also for y ∈ W 1,1 (J) with
(16) we get
d
α
dt (Eα,1 (λt ))
= λtα−1 Eα,α (λtα ), differentiate
u0 (t) = λtα−1 Eα,α (λtα )u(0) + tα−1 Eα,α (λtα )y(0) + (tα−1 Eα,α (λtα )) ∗ y 0 (t)
(17)
FRACTIONAL ORDER NONLINEAR EQUATIONS
627
which gives that u0 (t) is integrable thus u ∈ AC(J). Now we want to prove that
u(t) given by (16) satisfies (13) we have u(0) = u(1) and from the definition of
Mittag-Liffler function one gets
1−α α−1
(t
Eα,α (λtα ))
0 It
= Eα,1 (λtα )
(18)
which gives
C
α
D0+
u =
1−α 0
u (t)
0 It
α
Z
α
= λEα,1 (λt )u(0) + Eα,1 (λt )y(0) +
0
t
Eα,1 (λ(t − s)α )y 0 (s)ds
= λEα,1 (λtα )u(0) + Eα,1 (λtα )y(0) + (Eα,1 (λ(t − s)α )y(s))|s=t
s=0
Z t
+
λ(t − s)α−1 Eα,α (λ(t − s)α )y(s)ds
0
= λu(t) + y(t).
Thus we have proved that.
Theorem 3.2 — If y is a continuous function and λ 6= 0 is a real parameter
then the linear problem (13) has a continuous mild solution given by (16). This
solution becomes a strong solution if y ∈ W 1,1 (J).
As proving Theorem 3.2 one can prove the following theorem for linear problem
(14).
Theorem 3.3 — If y is a continuous function and λ 6= 0 is a real parameter
then the linear problem (14) has a continuous mild solution given by
Z 1
Eβ,1 (λtβ )
(1 − s)β−1 Eβ,β (λ(1 − s)β )y(s)ds
u(t) =
1 − Eβ,1 (λ) 0
Z t
+ (t − s)β−1 Eβ,β (λ(t − s)β )y(s)ds.
(19)
0
This solution becomes a strong solution if y ∈ W 1,1 (J).
Remark 3.4 :
(1) If y ∈ C(J) in (16) we get the mild solution u ∈ C(J).
628
MOHAMED A. E. HERZALLAH
(2) If y ∈ C(J) in (19) we get the mild solution u ∈ C 1 (J).
(3) The solution of the linear problem (13) and (14) can be written in the form
Z
1
u(t) =
0
Gλ,γ (t, s)y(s)ds
(20)
where Gλ,γ (t, s) is the Green’s function given by:
 E (λtγ )
γ,1
γ−1 E
γ

γ,γ (λ(1 − s) )

1−Eγ,1 (λ) (1 − s)


 +(t − s)γ−1 E (λ(t − s)γ )
0≤s<t
γ,γ
Gλ,γ (t, s) =




 Eγ,1 (λtγ ) (1 − s)γ−1 E (λ(1 − s)γ ) t ≤ s < 1
γ,γ
1−Eγ,1 (λ)
(21)
where we put γ = α for (13) and γ = β for (14).
(4) As a special case of (21) for γ = 1 we get



Gλ,1 (t, s) =


1+eλ λ(t−s)
e
1−eλ
0≤s<t
(22)
eλ
eλ(t−s)
1−eλ
t≤s≤1
(5) for γ = 2 we get
Gλ,2 (t, s) =
1
√
√
λ(1 − cosh( λ))

√
√

cosh( λs) sinh( λ(1 − t))



 + sinh(√λ(t − s))
0≤s≤t




 cosh(√λt) sinh(√λ(1 − s)) t ≤ s ≤ 1
(23)
3.2 Nonlinear problem
Now for the nonlinear problem (7). If u is a mild solution of (7), then it is given by
Z
u(t) =
0
1
Gλ,α (t, s)f (s, u(s))ds,
(24)
FRACTIONAL ORDER NONLINEAR EQUATIONS
629
where Gλ,α (t, s) is the Green function given by (21).
Define the operator T : C[0, 1] → C[0, 1] by
Z
T u(t) =
0
1
Gλ,α (t, s)f (s, u(s))ds
(25)
Theorem 3.5 — Let the function f : [0, 1] × R → R satisfies the following
(i) f (t, u) is bounded, that is there exists M > 0 such that
|f (t, u)| ≤ M, ∀ t ∈ [0, 1], u ∈ R,
(26)
(ii) f (t, u) satisfies Lipschitz condition, that is there exists a constant K > 0
such that
|f (t, u) − f (t, v)| ≤ K|u − v|, t ∈ [0, 1], u, v ∈ R
(27)
Then the operator T given by (25) is compact.
P ROOF : Let t1 < t2 ∈ (0, 1) and u ∈ C[0, 1], then we have
¯
Z 1
¯ Eα,1 (λtα1 )
|T u(t1 ) − T u(t2 )| = ¯¯
(1 − s)α−1 Eα,α (λ(1 − s)α )f (s, u(s))ds
1 − Eα,1 (λ) 0
Z t1
+
(t1 − s)α−1 Eα,α (λ(t1 − s)α )
0
f (s, u(s))ds
Z 1
Eα,1 (λtα2 )
+
(1 − s)α−1 Eα,α (λ(1 − s)α )f (s, u(s))ds
1 − Eα,1 (λ) 0
¯
Z t2
¯
α−1
α
−
(t2 − s)
Eα,α (λ(t2 − s) )f (s, u(s))ds¯¯
0
630
MOHAMED A. E. HERZALLAH
¯µ
¶Z 1
¯ Eα,1 (λtα1 )
Eα,1 (λtα2 )
¯
= ¯
−
(1 − s)α−1
1 − Eα,1 (λ) 1 − Eα,1 (λ)
0
Eα,α (λ(1 − s)α )f (s, u(s))ds
Z t1
+
((t1 − s)α−1 Eα,α (λ(t1 − s)α ) − (t2 − s)α−1
0
Eα,α (λ(t2 − s)α ))f (s, u(s))ds
¯
Z t2
¯
α−1
α
−
(t2 − s)
Eα,α (λ(t2 − s) )f (s, u(s))ds¯¯
t1
¯
·¯
¯
¯
1
α
α
(Eα,1 (λt1 ) − Eα,1 (λt2 )¯¯
≤ M ¯¯
1 − Eα,1 (λ)
Z 1
|(1 − s)α−1 Eα,α (λ(1 − s)α )|ds
0
Z t1
+
|((t1 − s)α−1 Eα,α (λ(t1 − s)α )
0
−(t2 − s)α−1 Eα,α (λ(t2 − s)α ))|ds
¸
Z t2
α−1
α
+
|(t2 − s)
Eα,α (λ(t2 − s) )|ds
t1
using the definition and properties of Mittag Liffler function to find the values of
the three integrations in the last inequality we get:
For the first integration:
Z
Z
1
α−1
|(1 − s)
0
α
Eα,α (λ(1 − s) )|ds ≤
=
1
(1 − s)
0
∞
X
k=0
=
α−1
∞
X
k=0
∞
X
|λ|k (1 − s)αk
k=0
Z 1
|λ|k
Γ(αk + α)
Γ(αk + α)
ds
(1 − s)αk+α−1 ds
0
|λ|k
Γ(αk + α + 1)
= Eα,α+1 (|λ|).
Similar for the third integration we get
Z t2
|(t2 − s)α−1 Eα,α (λ(t2 − s)α )|ds ≤ (t2 − t1 )α Eα,α+1 (|λ|(t2 − t1 )α )
t1
FRACTIONAL ORDER NONLINEAR EQUATIONS
631
Now for the second integration we get
Z t1
|((t1 − s)α−1 Eα,α (λ(t1 − s)α ) − (t2 − s)α−1 Eα,α (λ(t2 − s)α ))|ds
0 ¯
¯
Z t1 ¯X
∞
h
i¯
λk
¯
¯
=
(t1 − s)αk+α−1 − (t2 − s)αk+α−1 ¯ ds
¯
¯
0 ¯k=0 Γ(αk + α)
Z
∞
i
t1 h
X
|λ|k
≤
(t2 − s)αk+α−1 − (t1 − s)αk+α−1 ds
Γ(αk + α) 0
≤
k=0
∞
X
k=0
|λ|k
[−(t2 − t1 )αk+α + tαk+α
− tαk+α
]
2
1
Γ(αk + α + 1)
≤ −(t2 − t1 )α Eα,α+1 (|λ|(t2 − t1 )α ) + tα2 Eα,α+1 (|λ|tα2 ) − tα1 Eα,α+1 (|λ|tα1 ).
From these results we get that
|T u(t1 ) − T u(t2 )| → 0 ∀ |t2 − t1 | → 0.
(28)
Now we prove that T is continuous. For u, v ∈ C[0, 1] we get
Z 1
|T u(t) − T v(t)| ≤
|Gλ,α (t, s)||f (s, u(s)) − f (s, v(s))|ds
0
Z 1
≤ K
|Gλ,α (t, s)||u(s) − v(s)|ds
0
Z 1
⇒ kT u − T vk ≤ Kku − vk
|Gλ,α (t, s)|ds
0
About the Green function as in proving (28) we have
µ
¶
Z 1
Eα,1 (|λ|)
|Gλ,α (t, s)|ds ≤
+ 1 Eα,α+1 (|λ|)
|1 − Eα,1 (λ)|
0
thus we get
kT u − T vk ≤ K
µ
¶
Eα,1 (|λ|)
+ 1 Eα,α+1 (|λ|)ku − vk
|1 − Eα,1 (λ)|
(29)
(30)
which prove that T is a continuous operator.
Now let D be a bounded set in C[0, 1]. By continuity of the operator T we get
that T D = {T u : u ∈ D} is bounded, and by (28) we have T D is an equicontinuous set in the space C[0, 1]. By the Arzela-Ascoli theorem we get that T D is
relatively compact set, which prove that T is a compact operator.
632
MOHAMED A. E. HERZALLAH
Theorem 3.6 — If the hypothesis in Theorem 3.5 satisfied such that,
µ
¶
Eα,1 (|λ|)
K
+ 1 Eα,α+1 (|λ|) < 1
|1 − Eα,1 (λ)|
(31)
then there is a unique mild solution of problem (7).
P ROOF : From (30) if we have (31) then T is a contraction mapping. which by
Banach fixed point theorem there is a unique fixed point of the operator T which is
the unique mild solution of (7).
Theorem 3.7 — Assume that the hypothesis of Theorem 3.5 satisfied. Then
there is at least one mild solution in C[0, 1] to problem (7).
P ROOF : Consider the set B = {u ∈ C[0, 1] : u = µT (u), µ ∈ (0, 1)}. Let
u ∈ B then u = µT u which gives
Z 1
|Gλ,α (t, s)||f (s, u(s))|ds
|u(t)| ≤ µ
0
µ
¶
Eα,1 (|λ|)
≤ µM
+ 1 Eα,α+1 (|λ|) < ∞
|1 − Eα,1 (λ)|
Thus kuk < ∞ which proves that the set B is bounded. By Theorem 2.4 we
get that T has a fixed point in C[0, 1] which is the required mild solution.
Similar to proving theorems 3.5, 3.6 and 3.7 with replacing α ∈ (0, 1) by
β ∈ (1, 2) we can prove:
Theorem 3.8 — Assume that the hypothesis of Theorem 3.5 satisfied. Then
there is at least one mild solution in C[0, 1] to the problem (8). And if
µ
¶
Eβ,1 (|λ|)
K
+ 1 Eβ,β+1 (|λ|) < 1
|1 − Eβ,1 (λ)|
satisfied then this mild solution is unique.
Using Remark 3.4(ii) and Theorem 3.3 one can prove that:
Theorem 3.9 — Assume that the hypothesis of Theorem 3.5 satisfied with ∂f
∂t
are
continuous
then
the
mild
solution
given
in
Theorem
3.8
is
a
strong
and ∂f
∂u
solution.
FRACTIONAL ORDER NONLINEAR EQUATIONS
633
Example 3.10 : Consider the nonlinear fractional differential equation
C β
0 Dt u(t)
− λu(t) =
A
t sin(u(t)) + t4 , u(0) = u(1), u0 (0) = 0,
2
1 < β < 2, λ ∈ R − {0}, A = constant. (32)
It is obvious that f (t, u(t)) = A2 tsin(u(t)) + t4 satisfies the conditions of
Theorem 3.5 with M = A/2 + 1 and K = A/2. Using the results of Theorem 3.8
and Theorem 3.9 we get that (32) has at least one strong solution given by
Z 1
u(t) =
Gλ,β (t, s)f (s, u(s))ds,
(33)
0
where Gλ,β (t, s) is the Green function given by (21).
4. C ONCLUSION
In this paper, the existence of at least one mild solution for two fractional nonlinear
α
periodic boundary value problems with Caputo fractional derivative C
0 Dt with
α ∈ (0, 1) and (1, 2) were discussed, the conditions to get a unique mild solution
were determined. Finally our mild solution becomes strong under some conditions
in linear and nonlinear cases.
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