One of the existence of solutions to fractional differential equations

HANOI PEDAGOGICAL UNIVERSITY 2
DEPARTMENT OF MATHEMATICS
——————–o0o———————

NGUYEN THI HAO

ON THE EXISTENCE OF SOLUTIONS TO
FRACTIONAL DIFFERENTIAL EQUATIONS

GRADUATION THESIS
Speciality: Analysis

Hanoi - 2019


HANOI PEDAGOGICAL UNIVERSITY 2
DEPARTMENT OF MATHEMATICS
——————–o0o———————

NGUYEN THI HAO

ON THE EXISTENCE OF SOLUTIONS TO
FRACTIONAL DIFFERENTIAL EQUATIONS

GRADUATION THESIS
Speciality: Analysis

Supervisor
Dr. Hoang The Tuan

Hanoi - 2019

THESIS ASSURANCE

The thesis was written on the basis of my study under the guidance of Dr Hoang The
Tuan and my effort. I have studied and presented the results from bibliographies. The
thesis does not coincide others.
Hanoi, May 2019
Student

Nguyen Thi Hao

1


THESIS ACKNOWLEDGMENTS

This thesis is the final lesson after 4 years I have learned at Hanoi Pedagogical
University 2. I would like to express my sincere gratitude to Dr. Hoang The Tuan for
his guidance and encouragement. He always provides worth opinions and suggestions
through all stages of this thesis. He guided and supported me by valuable knowledge
and explanation. I also would like to express my gratitude to ThS. Nguyen Phuong
Dong for his experience and suggestions that help me to complete my thesis.
I also would like to thank all lecturers of Department of Mathematics - HPU2,
especially Dr Tran Van Bang. He inspires and finds the best opportunities for my class
to study English. It is the most important chance for me to change my mind.
My deepest gratitude goes to my family for their encouragement every time.
They always believe and help me to pursue my dreams.
Due to time, my capacity and conditions are limited, so the thesis can not avoid
errors. Then, I am looking forward to receiving valuable comments from teachers and

friends to complete my thesis. Once more times, thank you very much.
Author

2


Contents

Page
Thesis Assurance

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

1

Thesis Acknowledgements . . . . . . . . . . . . . . . . . . . . . . . . . . . .

2

List of Symbols . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

3

Preface . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

5

1. Fractional Calculus . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

6

1.1. The basic idea . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

6

1.2. Riemann-Liouville Integral . . . . . . . . . . . . . . . . . . . . . . . . .

9

1.3. Riemann - Liouville Derivative . . . . . . . . . . . . . . . . . . . . . . .

15

1.4. Relations Between Riemann-Liouville Integral and Derivative . . . . . .

18

2. Existence and Uniqueness Results for
Riemann-Liouville Fractional Differential Equations . . . . . . . . . . . . .

21

2.1. Main result . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

21

2.2. Existence of Solutions . . . . . . . . . . . . . . . . . . . . . . . . . . . .

22

2.3. Uniqueness of Solutions . . . . . . . . . . . . . . . . . . . . . . . . . . .

25

Bibliography . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

28

3


LIST OF SYMBOLS

N

Set of natural numbers.

R+

Set of strictly positive real numbers.

An , An [a, b]

Set of functions with absolutely continuous derivative of order n − 1.

C, C [a, b]

Set of continuous functions. (cf.Definition 1.1.2)

C k , C k [a, b]

Set of functions with continuous k th derivative.

Hµ , Hµ [a, b]

Holder Space(cf. Definition 1.1.2)

Lp , Lp [a, b]

Lebesgue space (cf. Definition 1.1.2).

.
.

= supa≤x≤b |f (x)|

∞

Chebyshev norm; f

p

Lp norm (1 ≤ p < ∞); f

∞

p

b
|f

a

=

(x)|p dx

.

Ceiling function, x = min {z ∈ Z : z ≥ x}

.

Floor function, x = max {z ∈ Z : z ≤ x}

1/p

Γ

Euler’s Gamma function, (cf. Definition 1.1.1)

o, O

Landau’s symbols

D

Differential operator, Df (x) = f (x)

Dn

n ∈ N : n-fold iterate of the differential operator D.

Dan

n ∈ R+ : Riemann- Liouville fractional differential operator.

I

Identity operator.

Ja

Integral operator, Ja f (x) =

Jan

n ∈ N: n-fold iterate of the integral operator J a
n ∈ R+

x
f
a

(t) dt

N: Riemann- Liouville fractional integral operator.

n = 0: identity operator.

4

PREFACE

In the seventeenth century, Newton and Leibniz developed the foundations of
differential and integral calculus. In particular, in a letter to de l’Hospital, Leibniz
introduced the symbol. The question of de l’Hospital ”What does

dn
dxn f

(x) mean if

n = 21 ?” was the first occurrence of fractional calculus. In the last few decades, mathematicians have studied and published the enormous numbers of very interesting and
novel applications of fractional differential equations in physics, chemistry, engineering,
finance and psychology. In addition, some applications of fractional calculus within various fields of mathematics itself. It turns out that many of these applications gave rise
to a type of equations, but they had not been covered in the standard mathematical
literature. Leibniz was not able to find the answer of de l’Hospital’s question, except
for the special case f (x) = x. There are many possible generalizations of

dn
dxn f

(x) to

the case n ∈
/ N, while Riemann- Liuoville derivative and Caputo’derivative are popular.
In the first half of the nineteenth centery, in works of Abel, Riemann and Liouville, the
former concepts of fractional calculus was established. Athough it leads to be difficult
when applying it to ”real world” problems, the theories were studied and presented

carefully. The Riemann-Liouville idea is closed to others and is the basis of other results. Based on the monograph by Kai Diethelm, in this thesis, we focuse on equations
with Riemann-Liouville differential operators. The structure of the thesis is arranged
in the following way:
• Chapter 1: Fractional Calculus, we introduce the fundamental concepts and definitions of fractional Riemann-Liouville differential and integral operators;
• Chapter 2: Existence and Uniqueness results for Riemann-Liouville fractional
differential equations.

5


Chapter 1
Fractional Calculus

In this chapter, the basic idea behind fractional calculus is related to a classical standard
result from differential and integral calculus, the fundamental theorem. We also recall
definitions of fractional integral and differential operators.

1.1. The basic idea
Theorem 1.1.1. (Fundamental Theorem of Classical Calculus) Let f : [a, b] →

R be a continuous function, and let F : [a, b] → R be defined by
x

f (t) dt.

F (x) :=
a

Then, F is differentiable and
F = f.

Therefore, we have a closed relation between differential operator and integral
operator. It is one of the goals of fractional calculus to retain this relation in a suitably
generalized sense. Hence there is also a need to deal with fractional integral operators
and actually it turns out to be useful to discuss these first before coming to fractional
differential operators. Thus, it is an answer of de l’Hospital’s question.
It has proven to be convenient to use the notational conventions introduced in
the following definition.
Definition 1.1.1. (i) By D, we denote the operator that maps a differentiable function onto its derivative, i.e
Df (x) := f (x) .
(ii) By Ja , we denote the operator that maps a function f , assumed to be Riemann
integrable on the compact interval [a, b], onto its primitive centered at a, i.e
x

Ja f (x) :=

f (t) dt
a

6


for a ≤ x ≤ b.
(iii) For n ∈ N we use the symbols Dn and Jan to denote the n-fold iterates of D and
Ja , respectively, i.e we set D1 := D, Ja1 := Ja , Dn := DDn−1 and Jan := Ja Jan−1
for n ≥ 2.
Note 1.1.1. The key question is ”How can we extend the concept of Definition 1.1.1 to
n∈
/ N?”. Once we will have provided such an extension, we need to ask for the mapping
properties of the resulting operators and in particular, this includes the question for
their domains and ranges. Note that, from the Fundamental Theorem of Classical

Calculus, we have a notation DJa f = f which implies that
Dn Jan f = f
for n ∈ N, i.e Dn is the left inverse of Jan in a suitable space of functions. We wish to
retain this property. However, the conditions of the theorem are not straightforward
to the fractional case n ∈
/ N. We give a generalization of the concepts in the definition
1.1.1 in next part.
Following the above outline, we begin with the integral operator Jan for n ∈ N.
Two following lemmas are well known in [4, Theorem 2.16].
Lemma 1.1.2. Let f be Riemann integrable on [a, b]. Then, for a ≤ x ≤ b and n ∈ N,
we have
Jan f (x) =

1
(n − 1)!

x

(x − t)n−1 f (t) dt.
a

Moreover, it is an immediate consequence of the fundamental theorem that the
following relation holds for the operators D and Ja :
Lemma 1.1.3. Let m, n ∈ N such that m > n, and let f be a function having the
continuous n-th derivative on the interval [a, b]. Then,
Dn f = Dm Jam−n f.
Proof. We have
f = Dm−n Jam−n f.
Applying the operator Dn to both sides of this relation and using the fact that
Dn Dm−n = Dm , the statement follows.

We recall here the definition of Gamma function.
7


Definition 1.1.2. The function Γ : (0, ∞) → R, defined by
∞

tx−1 e−t dt

Γ (x) :=
0

is called Euler’s Gamma function (or Euler’s integral of the second kind).
We have some properties of Gamma function.
Theorem 1.1.4. (i) The functional equation
Γ (x + 1) = xΓ (x)
holds for any x ∈ (0, ∞).
(ii) For n ∈ N, Γ (n + 1) = n!
The proof of this theorem is presented in [2, P. 192].
Before we start the main work, we introduce some function spaces in which we
are going to discuss matters.
Definition 1.1.3. Let 0 < µ ≤ 1, k ∈ N and p ≥ 1. We define
b

f : [a, b] → R, f is measurable on [a, b] and

Lp [a, b] :=

|f (x)|p dx < ∞

a

L∞ [a, b] := {f : [a, b] → R, f is measurable and bounded on [a, b]}
Hµ [a, b] := f : [a, b] → R, ∃c > 0 such that ∀x, y ∈ [a, b] : |f (x) − f (y)| ≤ c|x − y|µ
C k [a, b] := f : [a, b] → R, f has a continuous k th derivative
C [a, b] := C 0 [a, b] and H0 [a, b] := C [a, b].
Definition 1.1.4. By H ∗ or H ∗ [a, b], we denote the set of functions f : [a, b] → R
with the property that there exists some constants L > 0 such that
|f (x + h) − f (x)| ≤ L |h| ln |h|−1
whenever |h| <

1
2

and x, x + h ∈ [a, b].

When working in a Lebesgue space rather than in a space of continuous functions,
we can still retain the main part of the statement of the fundamental theorem. A proof
of this theorem can be found in [3].
8


Theorem 1.1.5. (Fundamental Theorem in Lebesgue Spaces) Let f ∈ L1 [a, b].
Then, Ja f is differentiable almost everywhere in [a, b] and DJa f = f also holds almost
everywhere on [a, b].

1.2. Riemann-Liouville Integral
/ N.
We recall definitions of fractional integral and differential operators Jan and Dn , n ∈
For the detail, see [1].

Definition 1.2.1. Let n ∈ R+ , the operator Jan , defined on L1 [a, b] by
Jan f (x) :=

1
Γ (n)

x

(x − t)n−1 f (t) dt
a

for a ≤ x ≤ b, is called Riemann- Liouville fractional integral operator of order n.
For n = 0, we set Ja0 := I, the identity operator.
Proposition 1.2.1. Let f ∈ L1 [a, b] and n > 0, then the integral Jan f (x) exists for
almost every x ∈ [a, b]. Moreover, the function Jna f itself is also an element of L1 [a, b].
Proof. We write the integral in question as
x

+∞

(x − t)n−1 f (t) dt =

φ1 (x − t) φ2 (t) dt.
−∞

a

where
φ1 (u) =

un−1

f or 0 < u ≤ b − a,

0

otherwise.

and
f or a ≤ u ≤ b,

f (u)

φ2 (u) =

0

otherwise.

By construction, φj ∈ L1 (R) for j ∈ {1, 2} and thus by a classical result on Lebesgue
integration, this proof is completed.
One important property of fractional integral operators is a commutative semigroup with respect to concatenation. The identity operator Ja0 is the neutral element
of this semigroup.
Theorem 1.2.2. Let m, n > 0 and φ ∈ L1 [a, b]. Then,
Jam Jan φ = Jam+n φ
9


and
Jam Jan φ = Jan Jam φ

hold almost everywhere on [a, b]. If additionally φ ∈ C [a, b] and m + n ≥ 1, then the
identity holds everywhere on [a, b].
Proof. We have
Jam Jan φ (x)

x

1
=
Γ (m) Γ (n)

t
m−1

(t − τ )n−1 φ (τ ) dτ dt.

(x − t)
a

a

By Fubini’s Theorem, we interchange the order of integration and obtain
Jam Jan φ (x) =

x

1
Γ (m) Γ (n)

x

(x − t)m−1 (t − τ )n−1 φ (τ ) dtdτ
a

τ

x

1
=
Γ (m) Γ (n)

x

(x − t)m−1 (t − τ )n−1 dtdτ .

φ (τ )
a

τ

Substituting t = τ + s (x − τ ) , we yield
1

x

Jam Jan φ (x)

1
=

Γ (m) Γ (n)

[(x − τ ) (1 − s)]m−1 [s (x − τ )]n−1 (x − τ ) dsdτ

φ (τ )
a

0
x

1
=
φ (τ )(x − τ )m+n−1
Γ (m) Γ (n) a
Applying Euler’s integral of the first kind
1

(1 − s)m−1 sn−1 ds =
0

1

(1 − s)m−1 sn−1 dsdτ .
0

Γ (m) Γ (n)
.
T (m + n)

Thus, we have

Jam Jan φ (x)

1
=
Γ (m + n)

x

φ (τ )(x − τ )m+n−1 dτ = Jam+n φ (x)
a

almost everywhere on [a, b].
By the classical theorems on parameter integrals, if φ ∈ C [a, b] then also Jan φ ∈ C [a, b].
Therefore, Jam Jan φ ∈ C [a, b] and Jam+n φ ∈ C [a, b]. It follows that these are two
continuous functions coincide almost everywhere, they must coincide everywhere.
Finally, if φ ∈ C [a, b] and m + n ≥ 1, then we have
Jam Jan φ = Jam+n φ = Jam+n−1 Ja1 φ
almost everywhere. Since Ja1 φ is continuous and Jam+n φ = Jam+n−1 Ja1 φ is also continuous, once again we may conclude that the two functions on either side of the equality
are continuous almost everywhere, that means they must be identical everywhere.
10


Theorem 1.2.3. Let φ ∈ Hµ [a, b] for some µ ∈ [0, 1], and let 0 < n < 1. Then
Jan φ (x) =

φ (a)
(x − a)n + Φ (x)
Γ (n + 1)

with some functions Φ. This function Φ satisfies

Φ (x) = O (x − a)µ+n as x −→ a.
Moreover,

Φ∈




 Hµ+n [a, b]
H ∗ [a, b]



 H [a, b]
1

if µ + n < 1,
if µ + n = 1,
if µ + n > 1.

Proof. We have
Jan φ (x) =

x

φ (a)
Γ (n)

(x − t)n−1 dt +
a

x

1
Γ (n)

a

φ (t) − φ (a)
(x − t)1−n

dt.

This yields the desired representation with
1
Φ (x) =
Γ (n)

x

φ (t) − φ (x)
(x − t)1−n

a

dt.

In the view of φ ∈ Hµ ,
1
|Φ (x)| ≤

Γ (n)

x
a

x

L|t − a|µ

L
dt =
1−n
Γ (n)
(x − t)

L
=
(x − a)µ+n
Γ (n)

(t − a)µ (x − t)n−1 dt
a

1

sµ (1 − s)n−1 ds = O (x − a)µ+n .
0

Now we set g (x) := (φ (x) − φ (a)) /Γ (n) . and let h > 0 and x, x + h ∈ [a, b].
Then,

Φ (x + h) − Φ (x) =

=

x
x+h
g (t) (x + h − t)n−1 dt − a g (t)(x − t)n−1 dt
a
x
g (t)
a

(x + h − t)n−1 − (x − t)n−1 dt +

x+h
g (t) (x + h − t)n−1
x

x

(g (t) − g (x)) (x + h − t)n−1 − (x − 1)n−1 dt

=
a

=:K1
x+h

(g (t) − g (x)) (x + h − t)n−1 dt +K3

+
x

=:K2

11


where K3 contains the remaining terms. An explicit calculation shows that
x

x+h

(x + h − t)n−1 − (x − t)n−1 dt +

K3 = g (x)
a

(x + h − t)n−1

.

x

We estimate the terms K1 , K2 and K3 separately. In the view of our assumption
φ ∈ Hµ , it is clear that φ ∈ Hµ . Hence,
x−a
(g (x − u) − g (x))
0

|K1 | =
≤L

= Lh

x−a µ
u
0

un−1 − (u + h)n−1 du

(x−a)/h
(ht)µ
0

= Lhµ+n

(u + h)n−1 − un−1 du

(th)n−1 − (th + h)n−1 dt

(x−a)/h µ
t
0

tn−1 − (t + 1)n−1 dt

At t → 0, there is no problem with the convergence of the integral since the integral behaves as tµ+n−1 there (the exponent is strictly greater than −1) . In the case
1 µ+n−1
t

dt
0

x − a < h, the integral is bounded by

=

1
(µ+n)

and thus K1 = O hµ+n

in this case. If x − a ≥ h, we find
(x−a)/h µ
t
0

tn−1 − (t + 1)n−1 dt

1

(x−a)/h

tµ tn−1 − (t + 1)n−1 dt +

=
0

<

tµ tn−1 − (t + 1)n−1 dt
1

1
µ+n

(x−a)/h +n−2
t
dt
1

+ (1 − n)

in the view of the mean value theorem of differential calculus. The remaining integral
can easily be calculated explicitly. We find for µ + n < 1 that
µ+n

|K1 | ≤ O h

≤ O hµ+n

1
+ (1 − n)
µ+n
1
+ (1 − n)
µ+n

(x−a)/h

tµ+n−2 dt
1
∞

tµ+n−2 dt O hµ+n
1

because of µ + n < 1 and x − a ≥ h. For µ + n = 1, a similar calculation gives
(x−a)/h

|K1 | ≤ O hµ+n

t−1 dt

1+
1

12

= O h ln h−1 .


Finally, for µ + n > 1 and x ≥ b,
(b−a)/h
µ+n

µ+n−2

|K1 | ≤ O h

t

dt

µ+n

=O h

1

b−a
h

µ+n−1

= O (h)

Hence,

K1 =


µ+n


O h

f or µ + n < 1

O h ln h−1

f or µ + n = 1



 O (h)

f or µ + n > 1
Next, we estimate K2 . We use again the fact that g ∈ Hµ to derive (using the substitution s = (t − x) /h)
x+h

(t − x)µ (x + h − t)n−1 dt

|K2 | ≤ L
x

1

sµ (1 − s)n−1 ds = O hµ+n

= Lh+n
0

irrespective of µ and n. Obviously, this bound is stronger than the above bound for
K1 .
Finally, for K3 , we use the Holder assumption on φ which implies that |g (x)| =
L(x − a)µ for some constant L. Evaluating the integrals analytically, we find
|K3 | ≤

L

(x − a)µ (x − a + h)n − (x − a)n
n

In the case x − a ≥ h, this is bounded from above by O hµ+n . If x − a > h, we
estimate the term in brackets by the mean value theorem of differential calculus and
find (taking into account that n < 1)
K3 = O (1) (x − a)µ h(x − a)n−1 = O (h) (x − a)µ+n−1
Once again, we look at the three cases separately and find K3 = O (h) for µ + n = 1,
|K3 | ≤ O (h) hµ+n−1 = O hµ+n for µ + n < 1 and |K3 | ≤ O (h) (b − a)µ+n−1 = O (h)
for µ + n > 1. Again, these estimates are stronger than those we obtained for K1 .
Combining all the estimation, we derive

Φ (x + h) − Φ (x) =


µ+n


O h

O h ln h−1



 O (h)

f or µ + n < 1,
f or µ + n = 1,
f or µ + n > 1.

13


Remark 1.2.1. We can infer a similar result for the integrand

to be in a suitable

Lebesgue class.
Lemma 1.2.4. Let n > 0, p > max {1; 1=n} and

∈ Lp [a; b]. Then,

Jan (x) = o (x − a)n−1=p
as x −→ a+. If additionally n−1=p ∈
= N, then Jan ∈ C
Hn−1=p−

n−1=p

n−1=p

[a; b] and D

n−1=p

Jan ∈

[a; b] :

In the case n − 1=p ∈ N, a sightly modi ed statement may be shown. The Holder

condition on the derivative of Jan must be replaced by a condition similar to the one
de ning the set H ∗ .
Proof. See [1, Theorem 2.6].
Example 1.2.1. Let f (x) = (x − a) for some
Jan f (x) =

> −1 and n > 0. Then,

( + 1)
(x − a)n+
(n + + 1)

In view of the well known corresponding result in the case n ∈ N, this result is precisely
what one would expect from a sensible generalization of the integral operator.
Jan f
=

(x) =

Z

1
(n)

1
(x − a)n+
(n)

x

(t − a) (x − t)n−1 dt:

a
1

Z

s (1 − s)n−1 ds =

0

( + 1)
(x − a)n+ :
(n + + 1)

Next we discuss the interchange of limit operation and fractional integration.
Theorem 1.2.5. Let n > 0. Assume that (fk )∞
k=1 is a uniformly convergent sequence
of continuous functions on [a; b]. Then we may interchange the fractional integral operator and the limit process, i.e.,
Jan lim fk (x) =
k→∞

lim Jan fk (x) :

k→∞

In particular, the sequence of functions (Jan fk )∞
k=1 is uniformly convergent.
Proof. The proof of this theorem is found in [1, P. 21].
Corollary 1.2.6. Let f be analytic in (a − h; a + h) for some h > 0, and n > 0. Then,

Jan f

(x) =

∞
X
(−1)k (x − a)k+n
k=0

k! (n + k) (n)
14

Dk f (x)


for a ≤ x < a + h/2, and
∞

Jan f

(x) =

for a ≤ x < a + h. In particular,

(x − a)k+n k
D f (a)
Γ (k + 1 + n)

k=0
n

Ja f is

analytic in (a, a + h).

Example 1.2.2. Let f (x) = exp (λx) with some λ > 0. Compute J0n f (x) for n > 0.
In the case n ∈ N, we obviously have J0n f (x) = λ−n exp (λx).
For the case n ∈
/ N, we find
∞

J0n f

(x) =

J0n
k=0
∞

=
k=0

(λ·)k
(x) =
k!

∞

k=0

λk n

J (.)k (x)
k! 0

λk
xk+n = λ−n
Γ (k + n + 1)

∞

k=0

(λx)k+n
.
Γ (k + n + 1)

1.3. Riemann - Liouville Derivative
Having established these fundamental properties of Riemann- Liouville integral operators, we come to the corresponding differential operators.
Definition 1.3.1 (See [1]). Let n ∈ R+ and m = n . The operator Dan defined by
Dan f := Dm Jam−n f
is called the Riemann - Liouville fractional differential operator of order n.
For n = 0, we set Da0 := I, the identity operator.
Lemma 1.3.1. Let n ∈ R+ and let m ∈ N such that m > n. Then,
Dan = Dm Jam−n .
Proof. Our assumptions on m imply that m ≥ n . Thus,
Dm Jam−n = D

n

Dm−

n

m− n

Ja

Ja

n −n

= Dan

in the view of the semigroup property of fractional integration.

Next, we present a very simple sufficient condition for the existence of Dan f .
Lemma 1.3.2. Let f ∈ A1 [a, b] and 0 < n < 1. Then, Dan f exists almost everywhere
in [a, b]. Moreover, Dan f ∈ Lp [a, b] for 0 ≤ p < 1 and
Dan f

1
(x) =
Γ (1 − n)

f (a)
+
(x − a)n
15

x
a

f (t) (x − t)−n dt .


Proof. We use the definition of the Rienmann - Liouville differential operator and the
fact that f ∈ A1 . It follows that
Dan f (x) =
=
=
=

x

1
d
Γ (1 − n) dx

a
x

d
1
Γ (1 − n) dx

t

f (a) +
a

f (u) du (x − t)−n dt

a
x

1
d
Γ (1 − n) dx
1
Γ (1 − n)

f (t) (x − t)−n dt

f (a)
a

x

dt
+
(x − t)n

f (a)
d
n +
dx
(x − a)

x
a

t

a

t

f (u)(x − t)−n dudt

a

f (u)(x − t)−n dudt .

a

By Fubini’s theorem, we may interchange the order of integration in the double integral.
This implies that
Dan f (x) =

1
Γ (1 − n)

f (a)
d
n +
dx
(x − a)

x

f (u)

a

(x − u)1−n
du .
1−n

The standard rules on the differentiation of parameter integrals give the desired representation. The integrability statement is an immediate consequence of the representation using classical results from Lebesgue integration theory.
Example 1.3.1. Let f (x) = (x − a)β for some β > −1 and n > 0. Then,
Dan f (x) = D

n

n

Ja−n f (x) =

Γ (β + 1)
D (. − a)
Γ ( n − n + β + 1)

n −n−β

(x) .

Specifically, if n − β ∈ N, the right-hand side is the n -th derivative of a classical
polynomial of degree n −(n − β) ∈ {0, 1, ..., n − 1}, and so the expression vanishes,
i.e,
Dan (. − a)n−m (x) = 0
for all n > 0, m ∈ {1, 2, .., n }.
On the other hand, if n − β ∈

/ N, we find
Dan (. − a)β (x) =

Γ (β + 1)
(x − a)β−n .
Γ (β + 1 − n)

Both these relations are straightforward generalization of what we know for integerorder derivatives. Note that in the last expression the argument of Gamma function
may be negative.
Theorem 1.3.3. Assume that n1 , n2 ≥ 0. Moreover let φ ∈ L1 [a, b] and f = Jan1 +n2 φ,
Dan1 Dan2 f = Dan1 +n2 f.
16


Proof. By our assumption on f and the definition of the Riemann-Liouville differential
operator, we have
Dan1 Dan2 f = Dan1 Dan2 Jan1 +n2 φ = D

n1

J

n1 −n1

D

n2

Ja

n2 −n2 n1 +n2
φ.
Ja

The semigroup property of the integral operators allows us to rewrite this expression as
Dan1 Dan2 f = D

n1

J

n1 −n1 D n2

Ja

n2 +n1

φ

.
= D n1 J n1 −n1 D n2 Ja J n1 φ
The orders of the integral and differential operators involved are natural numbers, we
n2

find that this is equivalent to
Dan1 Dan2 f = D

n1

J

n1 −n1 n1
Ja φ

=D

n1

Ja

n1

φ

where we have once again the semigroup property of fractional integration. We imply
that
Dan1 Dan2 f = φ.
The proof that Dan1 +n2 f = φ goes along similar lines.
The smoothness and zero condition in this theorem is not just a technicality. The
following examples show some cases where the condition is not satisfied. They prove
that an unconditional semigroup property of fractional differentiation in the RiemannLiouville sense does not hold.
Example 1.3.2. Consider example 1.3.1, let f (x) = x−1/2 and n1 = n2 = 1/2. Then,
D0n1 f (x) = D0n2 f (x) = 0
and
D0n1 D0n2 f (x) = 0.
However,
D0n1 +n2 f (x) = D1 f (x) = − 2x2/3

−1

.

Example 1.3.3. Let f (x) = x1/2 , n1 = 1/2 and n2 = 3/2. Consider Example 1.3.1
again, we have D0n1 f (x) =

√
π
2

and D0n2 f (x) = 0. This implies D0n1 D0n2 f (x) = 0, but

D0n2 D0n1 f (x) = −x−3/2 /4 = D2 f (x) = Dn1 +n2 f (x) .
In other words, the first of these two examples shows that it is possible to have
D0n1 D0n2 = D0n2 D0n1 f (x) = Dn1 +n2 f (x) ,
17


and in the second example, the case
D0n1 D0n2 = D0n2 D0n1 f (x) = Dn1 +n2 f (x)
holds.

1.4. Relations Between Riemann-Liouville Integral and Derivative
In this section we discuss on the relationship between the Riemann-Liouvill integral
and derivative.
Theorem 1.4.1. Let n ≥ 0. Then for f ∈ L1 [a, b], the following statement holds
Dan Jan f = f
almost everywhere.
Proof. See [1, Theorem 2.14].
By Theorem 1.4.1, we imply a very important result that Dan is the left inverse of
Jan . However, we can not claim that it is the right inverse. We consider the following

situation.
Theorem 1.4.2. Let n > 0. If there exists some φ ∈ L1 [a, b] such that f = Jan φ, then
Jan Dan f = f
almost everywhere.
Proof. This is an immediate consequence of Theorem 1.4.1, that
Jan Dan f = Jan [Dan Jan φ] = Jan φ = f

If f is not required in Theorem 1.4.2, then we consider one of different representation for Jan Dan f .
Proof. The case n = 0 is trivial for then Dan and Jan are both the identity operator.
For n > 0, let m = n
Dan Jan f (x) = Dm Jam−n Jan f (x) = Dm Jam f (x) = f (x) .

18


Theorem 1.4.3. Let n > 0 and m = n + 1. Assume that f is satisfied Jam−n f ∈
Am [a, b]. Then,
m−1

Jan Dan f

(x) = f (x) −
k=0

(x − a)n−k−1
lim Dm−k−1 Jam−n f (z).
Γ (n − k) z→a+

Specifically, for 0 < n < 1, we have
Jan Dan f (x) = f (x) −

(x − a)n−1
lim J 1−n f (z) .
Γ (n) z→a+ a

Proof. It is easy to imply that the right hand side exists (since assumption on f and the
continuity of Dm−a Jam−n f ). Moreover, because of this assumption, there exists some
φ ∈ L1 such that Dm−a Jam−n f = Dm−1 Jam−n f (a) + Ja1 φ. This is a classical differential
equation of order m − 1 for Jam−n f . Its solution is in the form
m−1

Jam−n f

(x) =
k=0

(x − a)k
lim Dk Jam−n f (z) + Jam φ (x).
z→a+
k!

Thus, by definition of Dan ,
Jan Dan f (x) = Jan Dm Jam−n f (x)
= Jan Dm

m−1

(.−a)k
lim Dk Jam−n f
k! z→a+

k=0

= Jan Dm Jam φ (x) +

(z) + Jam φ (x)

m−1 Jan Dm (.−a)k (x)
k!

k=0

lim Dk Jam−n f (z)

k→a+

= Jan φ (x) .
Next, we apply the operator Dam−n to both side and find
m−1

Jan Dan f (x) =

Jan Dm (. − a)k (x)
k!

k=0
m−1

Jan Dm (. − a)k (x)

=

k=0

lim Dk Jam−n f (z) + Dm−n Jam φ (x)

k→a+

k!

lim Dk Jam−n f (z) + Da1 Ja1−m+n Jam φ (x) .

k→a+

Moreover, we have
m−1

f (x) =
k=0

(x − a)k+n−m
lim Dk Jam−n f (z) + Jan φ (x) .
Γ (k + n − m + 1) k→a+
19


Finally, we substitute k in the sum by m − k − 1, solve for Jan φ (x) and from above
results, we imply that
m−1

Jan Dan f

(x) =

Jan φ (x)

= f (x) −
k=0

(x − a)n−k−1
lim Dm−k−1 Jam−n f (z).
Γ (n − k) k→a+

20


Chapter 2
Existence and Uniqueness Results for
Riemann-Liouville Fractional Differential Equations

The question of existence and uniqueness of solutions, i.e, the classical questions concerning ordinary differential equations involving fractional derivative is discussed by
many mathematicians under other types of conditions. The thesis focuses on equations with Riemann-Liouville differential operators. The fundamental result presented
and proved in this chapter is an existence and uniqueness theorem.
Without loss of generality, we assume in this result and in the ensuing developments
that the fractional derivatives are developed at the point 0.

2.1. Main result
Our task in this chapter is to prove the following theorem.
Theorem 2.1.1. Let n > 0, n ∈
/ N and m = n . Moreover let K > 0, h∗ > 0 and
b1 , ..., bm ∈ R. Define
G :=



 (x; y) ∈ R2 : 0 ≤ x ≤ h∗ , y ∈ R x = 0 and
m

xm−n y −




k=1





bk xm−k /Γ (n − k + 1) < K 


,
and assume that the function f : G −→ R is continuous and bounded in G. It fulfills a
Lipschitz condition with respect to the second variable, i.e there exists a constant L > 0
such that for all (x, y1 ) and (x, y2 ) ∈ G,
|f (x, y1 ) − f (x, y2 )| < L |y1 − y2 | .
Then, the differential equation
D0n y (x) = f (x, y (x))
equipped with the initial conditions
D0n−k y (0) = bk

(k = 1, 2, ..., m − 1) ,
21

lim J0m−n y (z) = bm

z→0+


has a uniquely solution y ∈ C (0, h] where
Γ (n + 1) K
M

∗

h := min

h , h,

1/m

with M := sup(x,z)∈G |f (x, z)| and h being an arbitrary positive number satisfying the
constraint
Γ (2n − m + 1)
Γ (n − m + 1) L

h<

1/n

.

The result is very similar to the known classical results for first-order equations.
We shall first transform the initial value problem into an equivalent Volterra integral
equation. Then we are going to prove the existence and uniqueness of the solution of
the fractional differential equation by a Picard -type iteration process. The proof of
Theorem 2.1.1 is the consequence of two lemmas below.

2.2. Existence of Solutions
Lemma 2.2.1. Assume the hypotheses of Theorem 2.1.1 and let h > 0. The function
y ∈ C (0, h] is a solution of the differential equation
D0n y (x) = f (x, y (x)) ,
equipped with the initial conditions
D0n−k y (0) = bk (k = 1, 2, ..., m − 1) , lim J0m−n y (z) = bm ,
z→0+

if and only id it is a solution of the Volterra integral equation
m

y (x) =
k=1

x

bk xn−k
1
+
Γ (n − k + 1) Γ (n)

(x − t)n−1 f (t, y (t)) dt.
0

Proof. If M = 0 then f (x, y) = 0 for all (x, y) ∈ G. In this case, it is evident that
m−1

the function y : [0, h] −→ R with y (x) =
k=0

(k)

y0 .xk
k!

is a solution of the initial value

problem. Hence we conclude, as required, that is a solution exists in this case.
Otherwise, we consider the Volterra equation. We introduce the polynomial T that
satisfies the initial conditions, as follow
m−1

T (x) :=
k=0

and the set U := {y ∈ C [0, h] : y − T

∞

xk (k)
y
k! 0

≤ K}. It implies that U is closed and convex

subset of the Banach space of all continuous functions on [0, h], equipped with the
22


Chebyshev norm. Hence, U is a Banach space too. Since the polynomial T is an
element of U, we also see that U is non-empty. On this set U, we define the operator
A by

x

1
(Ay) (x) := T (x) +
Γ (n)

(x − t)n−1 f (t, y (t)) dt
0

Using this operator, the equation whose solvability we need to prove, viz the Volterra
equation, can be rewritten as
y = Ay
and thus, in order to prove our desired existence result, we have to show that A has a
fixed point. We therefore proceed by investigating the properties of the operator more
closed.
Our first goal in this context is to show that Ay ∈ U for y ∈ U . To this end we begin
by notice that, for 0 ≤ x1 ≤ x2 ≤ h,
|(Ay) (x1 ) − (Ay) (x2 )|
=

1
Γ(n)

x1
0

=

1
Γ(n)

x1
0

≤

M
Γ(n)

(x1 − t)n−1 f (t, y (t)) dt −

x1
0

x2
0

(x2 − t)n−1 f (t, y (t)) dt

(x1 − t)n−1 − (x2 − t)n−1 f (t, y (t)) dt −

(x1 − t)n−1 − (x2 − t)n−1 dt −

x2
x1

x2
x1

(x2 − t)n−1 f (t, y (t)) dt

(x2 − t)n−1 dt

The second integral in the right hand side of the above equation has the value (x2 − x1 )n /n.
For the first integral, we look at the three cases n = 1, n < 1 and n > 1 separately. In
the first case n = 1, the integrand vanishes identically and hence the integral has the
value zero. Secondly, for n < 1, we have n−1 < 0 and hence (x1 − t)n−1 ≥ (x2 − t)n−1 .
Thus,
x1

x1

(x1 − t)n−1 − (x2 − t)n−1 dt =

0

(x1 − t)n−1 − (x2 − t)n−1 dt

0

=

1 n
x − xn2 + (x2 − x1 )n
n 1

≤

1
(x2 − x1 )n
n

23