Electronic Journal of Differential Equations, Vol. 2020 (2020), No. 15, pp. 1–16.

ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu

EXISTENCE OF SOLUTIONS TO NONLOCAL BOUNDARY

VALUE PROBLEMS FOR FRACTIONAL DIFFERENTIAL

EQUATIONS WITH IMPULSES

DANIEL CAO LABORA, ROSANA RODRÍGUEZ-LÓPEZ, MOHAMMED BELMEKKI

Abstract. In this work, through the application of fixed point theory, we
consider the properties of the solutions to a nonlocal boundary value problem

for fractional differential equations subject to impulses at fixed times. We

compute the Green’s function related to the problem, which allows us to ob-
tain an integral representation of the solution. This representation gives an

explicit description of the solution when the source term does not depend on

the solution. Nevertheless, when the description of the source term is implicit,
we can not ensure the existence of a solution. In this case, we prove the exis-

tence of a solution for the integral problem via fixed point techniques. To do

this, we develop a slight generalization of Arzelà-Ascoli theorem that makes it
suitable for piecewise uniformly continuous functions.

1. Introduction

The investigation on fractional differential equations has experimented a huge
expansion in the previous decades, and new applications have been proposed since
then. Some examples of applications of fractional order equations can be found,
for instance, in [1], where models of viscoplasticity are considered. The work [3] is
focused on protein dynamics; [7] is devoted to continuum and statistical mechanics;
and [8] in relaxation in filled polymers.

On the other hand, the physical interpretation for fractional differential equations
has been considered in [4] from the point of view of Riemann-Liouville derivatives,
also in [11].

In this article, we consider a boundary value problem with integral conditions
for a class of fractional differential equations subject to impulses. Some existence
results for higher-order fractional differential equations with integral conditions can
be found in [2], while some other results related to Pettis integral are included in
[13].

We represent by Dδ the fractional derivative of Riemann-Liouville type and con-
sider the following impulsive fractional differential equation with nonlocal boundary
conditions:

Dδ
0+u(t) + f(t, u(t)) = 0, t ∈ (0, t1), (1.1)

2010 Mathematics Subject Classification. 26A33, 34B37, 34B05, 34B10, 34B27.
Key words and phrases. Fractional differential equations; nonlocal boundary value problems;

Riemann-Liouville fractional derivative; fixed point results.
c©2020 Texas State University.

Submitted May 9, 2018. Published February 10, 2020.

1



2 D. CAO LABORA, R. RODRÍGUEZ-LÓPEZ, M. BELMEKKI EJDE-2020/15

Dδ
t+1
u(t) + f(t, u(t)) = 0, t ∈ (t1, 1), (1.2)

u(0) = 0, u(t+1 ) = 0, α0u(ξ0) + α1u(ξ1) = u(1), β

∫ 1

0

u(s) ds = u(1), (1.3)

where 1 < δ ≤ 2, 0 = t0 < t1 < 1, ξi ∈ (ti, ti+1), i = 0, 1, α0, α1, β ∈ R \ {0}, and
f : [0, 1]× R→ R continuous on [0, t1]× R. We also require f to be continuous on
(t1, 1] × R and that f |(t1,1]×R can be extended continuously to [t1, 1] × R, that is,
the discontinuity at t1 is of finite jump.

The previous problem is a particular case of the following problem

Dδ
t+k
u(t) + f(t, u(t)) = 0, tk < t < tk+1, k = 0, . . . ,m, (1.4)

u(t+k ) = 0, k = 0, . . . ,m,

m∑
k=0

αku(ξk) = u(T ), β

∫ T

0

u(s) ds = u(T ), (1.5)

where 1 < δ ≤ 2, T > 0, 0 = t0 < t1 < · · · < tk < tk+1 < · · · < tm+1 = T ,
ξi ∈ (ti, ti+1), i = 0, . . . ,m, α0, α1, . . . , αm, β ∈ R, and f : [0, T ]×R→ R continuous
on [0, t1]×R and on (tk, tk+1]×R, for k = 1, . . . ,m, in such a way that f |(tk,tk+1]×R
can be extended continuously to [tk, tk+1]×R, for k = 1, . . . ,m. Here, if we consider
T = 1 and m = 1, that is, 0 = t0 < t1 < t2 = 1, then we obtain problem (1.1)–(1.3).

In Section 2, we present some basic definitions and results. In Section 3, we
consider a linear problem and obtain the corresponding Green’s function, and in
Section 4, we give some existence results for the general nonlinear problem.

2. Basic definitions

We introduce some basic concepts about fractional integrals and derivatives.
Some relevant monographs on fractional calculus are [5, 6, 9, 12, 14].

Definition 2.1. The Riemann-Liouville fractional integral of order δ > 0 of a
function f : (a, b]→ R is given by

Iδa+f(t) =
1

Γ(δ)

∫ t

a

(t− τ)δ−1f(τ) dτ,

provided that the right-hand side is pointwise defined on (a, b], and where Γ denotes
the Gamma function.

Definition 2.2. For a continuous function f : (a, b] → R, the Riemann-Liouville
derivative of fractional order δ > 0 is given by

Dδ
a+f(t) =

1

Γ(n− δ)

( d
dt

)n ∫ t

a

(t− τ)n−δ−1f(τ) dτ,

where n = bδc+ 1, being bδc the integer part of the real number δ.

Lemma 2.3 ([14]). Given δ > 0, the solutions to the fractional differential equation

Dδ
0+u(t) = 0

are the functions

u(t) = c1t
δ−1 + c2t

δ−2 + · · ·+ cnt
δ−n, ci ∈ R, i = 1, . . . , n,

where n = dδe, the largest integer less than equal to δ.



EJDE-2020/15 EXISTENCE OF SOLUTIONS 3

Lemma 2.4. Given δ > 0, the solutions to the fractional differential equation

Dδ
a+u(t) = 0

are the functions

u(t) = c1(t− a)δ−1 + c2(t− a)δ−2 + · · ·+ cn(t− a)δ−n, ci ∈ R, i = 1, . . . , n,

where n = dδe.

Proof. The result is well known for δ ∈ Z+. Consequently, we can assume that
δ 6∈ Z+. Then, the equation

Dδ
a+u(t) =

1

Γ(n− δ)

( d
dt

)n ∫ t

a

(t− τ)n−δ−1u(τ) dτ = 0

can be written, via the change of variable τ = a+ s, as

1

Γ(n− δ)

( d
dt

)n ∫ t−a

0

(t− a− s)n−δ−1u(a+ s) ds = 0,

or
1

Γ(n− δ)

( d
dt

)n ∫ t−a

0

(t− a− s)n−δ−1v(s) ds = 0,

where v(s) := u(a+ s). Now, if we consider z = t− a, we obtain that

1

Γ(n− δ)

( d
dz

)n ∫ z

0

(z − s)n−δ−1v(s) ds = 0.

Hence, the function v satisfies the equation Dδ
0+v(z) = 0. From Lemma 2.3, we

know that

u(a+ z) = v(z) = c1z
δ−1 + c2z

δ−2 + · · ·+ cnz
δ−n, ci ∈ R, i = 1, . . . , n,

or, equivalently,

u(t) = v(t−a) = c1 (t−a)δ−1 +c2 (t−a)δ−2 +· · ·+cn (t−a)δ−n, ci ∈ R, i = 1, . . . , n.

�

It is straightforward to derive the following result, after a direct application of
Lemma 2.4.

Lemma 2.5. Given δ > 0, we have

Iδa+(Dδ
a+u(t)) = u(t) + c1(t− a)δ−1 + c2(t− a)δ−2 + · · ·+ cn(t− a)δ−n,

for ci ∈ R, i = 1, . . . , n.

Proof. We write Iδa+(Dδ
a+u(t)) = u(t)+f(t) and we apply Dδ

a+ to both sides of this

expression. Since Dδ
a ◦ Iδa = Id, we have that Dδ

a+f = 0 and the conclusion follows
from Lemma 2.4. �

Remark 2.6. From Lemma 2.5, for a = tk, k = 0, 1, . . . ,m, and 1 < δ ≤ 2, we
obtain

Iδ
t+k

(Dδ
t+k
u(t)) = u(t) + c1,k(t− tk)δ−1 + c2,k(t− tk)δ−2, c1,k, c2,k ∈ R.



4 D. CAO LABORA, R. RODRÍGUEZ-LÓPEZ, M. BELMEKKI EJDE-2020/15

3. Green’s function for a linear fractional differential equation

In this section, we consider a related linear fractional differential equation with
the same boundary conditions, to obtain the explicit expression of the Green’s
function. First, we need to define the following space of functions.

Definition 3.1. Given a partition 0 = t0 < t1 < t2 = 1, the space of piecewise
continuous functions from [0, 1] to R, is defined as

PC([0, 1],R) =
{
u : [0, 1]→ R : u ∈ C([0, t1],R) and u ∈ C((t1, 1],R)

with lim
t→t+1

u(t) finite
}
.

Lemma 3.2. Consider 1 < δ ≤ 2, α0, α1, β ∈ R \ {0}, 0 = t0 < t1 < t2 = 1,
and ξi ∈ (ti, ti+1) for i = 0, 1. Assume also that σ ∈ PC([0, 1],R). A function
u ∈ PC([0, 1],R) is a solution to the boundary value problem

Dδ
0+u(t) + σ(t) = 0, t ∈ (0, t1), (3.1)

Dδ
t1+u(t) + σ(t) = 0, t ∈ (t1, 1), (3.2)

u(0) = 0, u(t+1 ) = 0, α0u(ξ0) + α1u(ξ1) = u(1), β

∫ 1

0

u(s) ds = u(1) (3.3)

if and only if it satisfies the integral equation

u(t) =

∫ 1

0

H(t, s)σ(s) ds,

where H(t, s) is the Green’s function given in the proof, provided that

ϕ(1)−
1∑
k=0

αkϕ(ξk) =
tδ1β(1− t1)1−δ(1− t1)δ−1

δ − β(1− t1)
− α0ξ

δ−1
0

− α1
tδ1β(1− t1)1−δ(ξ1 − t1)δ−1

δ − β(1− t1)
6= 0,

where

ϕ(t) =

t
δ−1, t ∈ (0, t1),

tδ1β(1−t1)1−δ(t−t1)δ−1

δ−β(1−t1) , t ∈ (t1, 1].

Proof. Integrating (3.1) and (3.2), we obtain

Iδ
t+k
Dδ
t+k
u(t) = −Iδ

t+k
σ(t), t ∈ (tk, tk+1), k = 0, 1.

From Remark 2.6, equations (3.1) and (3.2) can be rewritten as

u(t) + c1,k(t− tk)δ−1 + c2,k(t− tk)δ−2 = − 1

Γ(δ)

∫ t

tk

(t− τ)δ−1σ(τ) dτ,

for t ∈ (tk, tk+1), where c1,k, c2,k ∈ R, k = 0, 1, or equivalently, by renaming the
constants used,

u(t) = − 1

Γ(δ)

∫ t

tk

(t− τ)δ−1σ(τ) dτ + c1,k(t− tk)δ−1 + c2,k(t− tk)δ−2,

for t ∈ (tk, tk+1), where c1,k, c2,k ∈ R, for k = 0, 1.



EJDE-2020/15 EXISTENCE OF SOLUTIONS 5

Note that the conditions u(t+k ) = 0, for k = 0, 1, imply that c2,k = 0, for k = 0, 1,
so that

u(t) = − 1

Γ(δ)

∫ t

tk

(t− τ)δ−1σ(τ) dτ + c1,k(t− tk)δ−1, t ∈ (tk, tk+1),

where c1,k ∈ R, for k = 0, 1.

Using the integral condition β
∫ 1

0
u(s)ds = u(1), we obtain

c1,1(1− t1)δ−1 = β

∫ 1

0

u(s) ds+
1

Γ(δ)

∫ 1

t1

(1− s)δ−1σ(s) ds.

The previous calculations lead, for k = 0, to

u(t) = − 1

Γ(δ)

∫ t

0

(t− τ)δ−1σ(τ) dτ + c1,0 t
δ−1, t ∈ (0, t1), (3.4)

and, for k = 1, to

u(t) =− 1

Γ(δ)

∫ t

t1

(t− τ)δ−1σ(τ) dτ + β(1− t1)1−δ(t− t1)δ−1

∫ 1

0

u(s) ds

+
(1− t1)1−δ

Γ(δ)
(t− t1)δ−1

∫ 1

t1

(1− s)δ−1σ(s) ds, t ∈ (t1, 1).

(3.5)

From these expressions, we obtain∫ 1

0

u(s) ds =

∫ t1

0

u(s) ds+

∫ 1

t1

u(s) ds

= − 1

Γ(δ)

∫ t1

0

∫ t

0

(t− s)δ−1σ(s) ds dt+

∫ t1

0

c1,0 t
δ−1 dt

− 1

Γ(δ)

∫ 1

t1

∫ t

t1

(t− s)δ−1σ(s) ds dt

+ β(1− t1)1−δ
∫ 1

t1

(t− t1)δ−1 dt

∫ 1

0

u(s) ds

+
(1− t1)1−δ

Γ(δ)

∫ 1

t1

(t− t1)δ−1 dt

∫ 1

t1

(1− s)δ−1σ(s) ds

= − 1

δΓ(δ)

∫ t1

0

(t1 − s)δσ(s) ds+
c1,0
δ
tδ1

− 1

δΓ(δ)

∫ 1

t1

(1− s)δσ(s) ds+ β
1− t1
δ

∫ 1

0

u(s) ds

+
1− t1
δΓ(δ)

∫ 1

t1

(1− s)δ−1σ(s) ds.

Then∫ 1

0

u(s) ds = − 1

(δ − β(1− t1))Γ(δ)

∫ t1

0

(t1 − s)δσ(s) ds+
c1,0

(δ − β(1− t1))
tδ1

− 1

(δ − β(1− t1))Γ(δ)

∫ 1

t1

(1− s)δσ(s) ds

+
(1− t1)

(δ − β(1− t1))Γ(δ)

∫ 1

t1

(1− s)δ−1σ(s) ds.

(3.6)



6 D. CAO LABORA, R. RODRÍGUEZ-LÓPEZ, M. BELMEKKI EJDE-2020/15

Therefore, we can find a piecewise defined integral kernel G(t, s) that allows us
to express u in a simple way. On the one hand, for t ∈ (0, t1), it follows from (3.4)
that

u(t) = − 1

Γ(δ)

∫ t

0

(t− s)δ−1σ(s) ds+ c1,0t
δ−1

=

∫ 1

0

G(t, s)σ(s) ds+ c1,0t
δ−1

=

∫ 1

0

G(t, s)σ(s) ds+ c1,0ϕ(t), t ∈ (0, t1).

On the other hand, if we replace expression (3.6) in (3.5), we can describe u(t), for
t ∈ (t1, 1), as

u(t) =− 1

Γ(δ)

∫ t

t1

(t− s)δ−1σ(s) ds

− β(1− t1)1−δ(t− t1)δ−1

(δ − β(1− t1))Γ(δ)

∫ t1

0

(t1 − s)δσ(s) ds

+
c1,0

δ − β(1− t1)
tδ1β(1− t1)1−δ(t− t1)δ−1

− β(1− t1)1−δ(t− t1)δ−1

(δ − β(1− t1))Γ(δ)

∫ 1

t1

(1− s)δσ(s) ds

+
β(1− t1)2−δ(t− t1)δ−1

(δ − β(1− t1))Γ(δ)

∫ 1

t1

(1− s)δ−1σ(s) ds

+
(1− t1)1−δ

Γ(δ)
(t− t1)δ−1

∫ 1

t1

(1− s)δ−1σ(s) ds

=− 1

Γ(δ)

∫ t1

0

β(1− t1)1−δ(t− t1)δ−1

δ − β(1− t1)
(t1 − s)δσ(s) ds

+
1

Γ(δ)

∫ t

t1

[
−(t− s)δ−1 + (1− t1)1−δ(t− t1)δ−1(1− s)δ−1

]
σ(s) ds

− 1

Γ(δ)

∫ t

t1

β(1− t1)1−δ(t− t1)δ−1

δ − β(1− t1)
(1− s)δσ(s) ds

+
1

Γ(δ)

∫ t

t1

β(1− t1)2−δ(t− t1)δ−1

δ − β(1− t1)
(1− s)δ−1σ(s) ds

− 1

Γ(δ)

∫ 1

t

β(1− t1)1−δ(t− t1)δ−1

δ − β(1− t1)
(1− s)δσ(s) ds

+
1

Γ(δ)

∫ 1

t

β(1− t1)2−δ(t− t1)δ−1

δ − β(1− t1)
(1− s)δ−1σ(s) ds

+
1

Γ(δ)

∫ 1

t

(1− t1)1−δ(t− t1)δ−1(1− s)δ−1σ(s) ds

+
c1,0

δ − β(1− t1)
tδ1β(1− t1)1−δ(t− t1)δ−1

=

∫ 1

0

G(t, s)σ(s) ds+
c1,0

δ − β(1− t1)
tδ1 β(1− t1)1−δ(t− t1)δ−1



EJDE-2020/15 EXISTENCE OF SOLUTIONS 7

=

∫ 1

0

G(t, s)σ(s) ds+ c1,0ϕ(t), t ∈ (t1, 1].

In summary,

u(t) =

∫ 1

0

G(t, s)σ(s) ds+ c1,0ϕ(t), t ∈ (0, 1], (3.7)

where

G(t, s) =
1

Γ(δ)



−(t− s)δ−1 if 0 ≤ s ≤ t < t1,

0 if 0 ≤ t ≤ s ≤ 1, 0 ≤ t < t1,

−β(1−t1)1−δ(t−t1)δ−1

δ−β(1−t1) (t1 − s)δ if 0 ≤ s ≤ t1 ≤ t,

−(t− s)δ−1 + (1− t1)1−δ(t− t1)δ−1(1− s)δ−1

−β(1−t1)1−δ(t−t1)δ−1

δ−β(1−t1) (1− s)δ

+β(1−t1)2−δ(t−t1)δ−1

δ−β(1−t1) (1− s)δ−1 if t1 ≤ s ≤ t ≤ 1,

−β(1−t1)1−δ(t−t1)δ−1

δ−β(1−t1) (1− s)δ

+β(1−t1)2−δ(t−t1)δ−1

δ−β(1−t1) (1− s)δ−1

+(1− t1)1−δ(t− t1)δ−1(1− s)δ−1 if t1 ≤ t ≤ s ≤ 1;

that is,

G(t, s) =
1

Γ(δ)



−(t− s)δ−1 if 0 ≤ s ≤ t < t1,

0 if 0 ≤ t ≤ s ≤ 1, 0 ≤ t < t1,

−β(1−t1)1−δ(t−t1)δ−1

δ−β(1−t1) (t1 − s)δ if 0 ≤ s ≤ t1 ≤ t,

−(t− s)δ−1 + (1− t1)1−δ(t− t1)δ−1(1− s)δ−1

×
[
1− β(1−s)

δ−β(1−t1) + β(1−t1)
δ−β(1−t1)

]
if t1 ≤ s ≤ t ≤ 1,

(1− t1)1−δ(t− t1)δ−1(1− s)δ−1

×
[
1− β(1−s)

δ−β(1−t1) + β(1−t1)
δ−β(1−t1)

]
if t1 ≤ t ≤ s ≤ 1,

or, equivalently,

G(t, s)

=
1

Γ(δ)(δ − β(1− t1))

×



−(t− s)δ−1(δ − β(1− t1)) if 0 ≤ s ≤ t < t1,

0 if 0 ≤ t ≤ s ≤ 1, 0 ≤ t < t1,

−β(1− t1)1−δ(t− t1)δ−1(t1 − s)δ if 0 ≤ s ≤ t1 ≤ t,

−(t− s)δ−1(δ − β(1− t1))

+(1− t1)1−δ(t− t1)δ−1(1− s)δ−1(δ − β(1− s)) if t1 ≤ s ≤ t ≤ 1,

(1− t1)1−δ(t− t1)δ−1(1− s)δ−1(δ − β(1− s)) if t1 ≤ t ≤ s ≤ 1.

(3.8)

Finally, if we consider the condition α0u(ξ0) + α1u(ξ1) = u(1), we have

u(1) =

∫ 1

0

G(1, s)σ(s) ds+ c1,0ϕ(1)



8 D. CAO LABORA, R. RODRÍGUEZ-LÓPEZ, M. BELMEKKI EJDE-2020/15

= α0

∫ 1

0

G(ξ0, s)σ(s) ds+ α0 c1,0ϕ(ξ0) + α1

∫ 1

0

G(ξ1, s)σ(s) ds+ α1 c1,0ϕ(ξ1)

=

∫ 1

0

[α0G(ξ0, s) + α1G(ξ1, s)]σ(s) ds+ (α0ϕ(ξ0) + α1ϕ(ξ1))c1,0

=

∫ 1

0

1∑
k=0

αkG(ξk, s)σ(s) ds+ c1,0

1∑
k=0

αkϕ(ξk);

that is, ∫ 1

0

[ 1∑
k=0

αkG(ξk, s)−G(1, s)
]
σ(s) ds = c1,0

[
ϕ(1)−

1∑
k=0

αkϕ(ξk)
]
.

Since, by hypothesis, ϕ(1)−
∑1
k=0 αkϕ(ξk) 6= 0, we have

c1,0 =
1

ϕ(1)−
∑1
k=0 αkϕ(ξk)

∫ 1

0

[ 1∑
k=0

αkG(ξk, s)−G(1, s)
]
σ(s) ds.

We know that

G(1, s) =
1

Γ(δ)(δ − β(1− t1))

{
−β(t1 − s)δ, 0 ≤ s ≤ t1,
(1− s)δ−1β(s− t1), t1 ≤ s ≤ 1.

Moreover, for ξ0 ∈ (0, t1), we have

G(ξ0, s) =
1

Γ(δ)(δ − β(1− t1))

{
−(ξ0 − s)δ−1(δ − β(1− t1)), 0 ≤ s ≤ ξ0,
0, ξ0 ≤ s ≤ 1,

and, for ξ1 ∈ (t1, 1),

G(ξ1, s) =
1

Γ(δ)(δ − β(1− t1))

×


−β(1− t1)1−δ(ξ1 − t1)δ−1(t1 − s)δ, 0 ≤ s ≤ t1,

−(ξ1 − s)δ−1(δ − β(1− t1))

+(1− t1)1−δ(ξ1 − t1)δ−1(1− s)δ−1(δ − β(1− s)), t1 ≤ s ≤ ξ1,

(1− t1)1−δ(ξ1 − t1)δ−1(1− s)δ−1(δ − β(1− s)), ξ1 ≤ s ≤ 1.

If we take A = 1
ϕ(1)−

∑1
k=0 αkϕ(ξk)

and B = A
Γ(δ)(δ−β(1−t1)) , we have

c1,0 =A

∫ 1

0

[ 1∑
k=0

αkG(ξk, s)−G(1, s)
]
σ(s) ds

=A

∫ 1

0

[α0G(ξ0, s) + α1G(ξ1, s)−G(1, s)]σ(s) ds

=B

∫ ξ0

0

[
− α0(ξ0 − s)δ−1(δ − β(1− t1))

− α1β(1− t1)1−δ(ξ1 − t1)δ−1(t1 − s)δ + β(t1 − s)δ
]
σ(s) ds

+B

∫ t1

ξ0

[−α1β(1− t1)1−δ(ξ1 − t1)δ−1(t1 − s)δ + β(t1 − s)δ]σ(s) ds



EJDE-2020/15 EXISTENCE OF SOLUTIONS 9

+B

∫ ξ1

t1

−α1(ξ1 − s)δ−1(δ − β(1− t1))σ(s) ds

+B

∫ ξ1

t1

α1(1− t1)1−δ(ξ1 − t1)δ−1(1− s)δ−1(δ − β(1− s))σ(s) ds

+B

∫ ξ1

t1

[−(1− s)δ−1β(s− t1)]σ(s) ds

+B

∫ 1

ξ1

[
α1(1− t1)1−δ(ξ1 − t1)δ−1(1− s)δ−1(δ − β(1− s))

− (1− s)δ−1β(s− t1)
]
σ(s) ds

=

∫ 1

0

K(s) ds,

where

K(s) =



B
[
− α0(ξ0 − s)δ−1(δ − β(1− t1))

−α1β(1− t1)1−δ(ξ1 − t1)δ−1(t1 − s)δ
]

+Bβ(t1 − s)δ, 0 ≤ s ≤ ξ0,

B[−α1β(1− t1)1−δ(ξ1 − t1)δ−1(t1 − s)δ + β(t1 − s)δ], 0 ≤ ξ0 ≤ s ≤ t1,

−Bα1(ξ1 − s)δ−1(δ − β(1− t1))

+Bα1(1− t1)1−δ(ξ1 − t1)δ−1(1− s)δ−1(δ − β(1− s))
+B[−(1− s)δ−1β(s− t1)], t1 ≤ s ≤ ξ1,

Bα1(1− t1)1−δ(ξ1 − t1)δ−1(1− s)δ−1(δ − β(1− s))
−B(1− s)δ−1β(s− t1), ξ1 < s ≤ 1.

Then, for t ∈ (0, t1], we can write

u(t) = − 1

Γ(δ)

∫ t

0

(t− s)δ−1σ(s) ds+ c1,0 t
δ−1 =

∫ 1

0

H(t, s)σ(s) ds,

where H is defined in the following way. Denoting C = A
δ−β(1−t1) , for t ∈ (0, ξ0],

we have

H(t, s) =
tδ−1

Γ(δ)



−(t− s)δ−1t1−δ + C
[
− α0(ξ0 − s)δ−1

−α1β(1− t1)1−δ(ξ1 − t1)δ−1(t1 − s)δ
]

+ Cβ(t1 − s)δ

if 0 ≤ s ≤ t ≤ ξ0,

C
[
− α0(ξ0 − s)δ−1 − α1β(1− t1)1−δ(ξ1 − t1)δ−1(t1 − s)δ

]
+Cβ(t1 − s)δ if 0 ≤ t < s ≤ ξ0,

C
[
− α1β(1− t1)1−δ(ξ1 − t1)δ−1(t1 − s)δ + β(t1 − s)δ

]
if 0 ≤ t ≤ ξ0 < s ≤ t1,

−Cα1(ξ1 − s)δ−1(δ − β(1− t1))

+Cα1(1− t1)1−δ(ξ1 − t1)δ−1(1− s)δ−1(δ − β(1− s))
+C
[
− (1− s)δ−1β(s− t1)

]
if 0 ≤ t ≤ ξ0, t1 ≤ s ≤ ξ1,

Cα1(1− t1)1−δ(ξ1 − t1)δ−1(1− s)δ−1(δ − β(1− s))
−C(1− s)δ−1β(s− t1) if 0 ≤ t ≤ ξ0, ξ1 < s ≤ 1,



10 D. CAO LABORA, R. RODRÍGUEZ-LÓPEZ, M. BELMEKKI EJDE-2020/15

and, for t ∈ (ξ0, t1],

H(t, s) =
tδ−1

Γ(δ)



−(t− s)δ−1t1−δ + C
[
− α0(ξ0 − s)δ−1

−α1β(1− t1)1−δ(ξ1 − t1)δ−1(t1 − s)δ
]

+ Cβ(t1 − s)δ

if 0 ≤ s ≤ ξ0 < t ≤ t1,

−(t− s)δ−1t1−δ

+C
[
−α1β(1− t1)1−δ(ξ1 − t1)δ−1(t1 − s)δ + β(t1 − s)δ

]
if 0 ≤ ξ0 ≤ s ≤ t ≤ t1,

C[−α1β(1− t1)1−δ(ξ1 − t1)δ−1(t1 − s)δ + β(t1 − s)δ]
if 0 ≤ ξ0 ≤ t < s ≤ t1
−Cα1(ξ1 − s)δ−1(δ − β(1− t1))

+Cα1(1− t1)1−δ(ξ1 − t1)δ−1(1− s)δ−1(δ − β(1− s))
+C[−(1− s)δ−1β(s− t1)] if ξ0 < t ≤ t1, t1 ≤ s ≤ ξ1,

Cα1(1− t1)1−δ(ξ1 − t1)δ−1(1− s)δ−1(δ − β(1− s))
−C(1− s)δ−1β(s− t1) if ξ0 < t ≤ t1, ξ1 < s ≤ 1.

If t ∈ (t1, 1], we have

u(t) =

∫ 1

0

G(t, s)σ(s) ds+
tδ1β(1− t1)1−δ(t− t1)δ−1

δ − β(1− t1)
c1,0

=

∫ 1

0

H(t, s)σ(s) ds,

where G is given by (3.8). To find an explicit expression for H when t ∈ (t1, 1], we
distinguish two cases. When t ∈ (t1, ξ1], we have

H(t, s) =
tδ1β(1− t1)1−δ(t− t1)δ−1

Γ(δ)(δ − β(1− t1))



EJDE-2020/15 EXISTENCE OF SOLUTIONS 11

×



− (t1−s)δ
tδ1

+ C
[
− α0(ξ0 − s)δ−1(δ − β(1− t1))

−α1β(1− t1)1−δ(ξ1 − t1)δ−1(t1 − s)δ
]

+Cβ(t1 − s)δ

if 0 ≤ s ≤ ξ0, t1 < t ≤ ξ1,

− (t1−s)δ
tδ1

+ C[−α1β(1− t1)1−δ(ξ1 − t1)δ−1(t1 − s)δ + β(t1 − s)δ]
if 0 ≤ ξ0 ≤ s ≤ t1 < t ≤ ξ1,

− (δ−β(1−t1))

tδ1β(1−t1)1−δ(t−t1)δ−1 (t− s)δ−1 + 1
tδ1β

(1− s)δ−1(δ − β(1− s))
−Cα1(ξ1 − s)δ−1(δ − β(1− t1))

+Cα1(1− t1)1−δ(ξ1 − t1)δ−1(1− s)δ−1(δ − β(1− s))
+C[−(1− s)δ−1β(s− t1)] if t1 ≤ s ≤ t ≤ ξ1,
(δ−β(1−s))

tδ1β
(1− s)δ−1 − Cα1(ξ1 − s)δ−1(δ − β(1− t1))

+Cα1(1− t1)1−δ(ξ1 − t1)δ−1(1− s)δ−1(δ − β(1− s))
+C[−(1− s)δ−1β(s− t1)] if t1 ≤ t < s ≤ ξ1,
(δ−β(1−s))

tδ1β
(1− s)δ−1

+Cα1(1− t1)1−δ(ξ1 − t1)δ−1(1− s)δ−1(δ − β(1− s))
−C(1− s)δ−1β(s− t1) if t1 < t ≤ ξ1 < s ≤ 1,

and, for t ∈ (ξ1, 1],

H(t, s) =
tδ1β(1− t1)1−δ(t− t1)δ−1

Γ(δ)(δ − β(1− t1))

− (t1−s)δ
tδ1

+ C
[
− α0(ξ0 − s)δ−1(δ − β(1− t1))

−α1β(1− t1)1−δ(ξ1 − t1)δ−1(t1 − s)δ
]

+ Cβ(t1 − s)δ

if 0 ≤ s ≤ ξ0, ξ1 < t ≤ 1,

− (t1−s)δ
tδ1

+ C[−α1β(1− t1)1−δ(ξ1 − t1)δ−1(t1 − s)δ + β(t1 − s)δ]
if 0 ≤ ξ0 ≤ s ≤ t1, ξ1 < t ≤ 1,

− (δ−β(1−t1))

tδ1β(1−t1)1−δ(t−t1)δ−1 (t− s)δ−1 + 1
tδ1β

(1− s)δ−1(δ − β(1− s))
−Cα1(ξ1 − s)δ−1(δ − β(1− t1))

+Cα1(1− t1)1−δ(ξ1 − t1)δ−1(1− s)δ−1(δ − β(1− s))
+C

[
−(1− s)δ−1β(s− t1)

]
if t1 ≤ s ≤ ξ1 < t ≤ 1,

− (δ−β(1−t1))

tδ1β(1−t1)1−δ(t−t1)δ−1 (t− s)δ−1 + 1
tδ1β

(1− s)δ−1(δ − β(1− s))
Cα1(1− t1)1−δ(ξ1 − t1)δ−1(1− s)δ−1(δ − β(1− s))
−C(1− s)δ−1β(s− t1) if ξ1 < s ≤ t < 1,

(δ−β(1−s))
tδ1β

(1− s)δ−1

+Cα1(1− t1)1−δ(ξ1 − t1)δ−1(1− s)δ−1(δ − β(1− s))
−C(1− s)δ−1β(s− t1) if ξ1 < t < s < 1.

The proof is complete. �



12 D. CAO LABORA, R. RODRÍGUEZ-LÓPEZ, M. BELMEKKI EJDE-2020/15

4. Existence results

We denote by X the set of piecewise uniformly continuous functions on [0, 1].
Let tj for j ∈ {0, 1, . . . , n}, with 0 = t0 < t1 < · · · < tn−1 < tn = 1 be a fixed
collection of finitely many points in [0, 1]. We will say that u ∈ X if and only if
u|(tj−1,tj) is uniformly continuous for each j from 1 to n. We endow the set X with
the supremum norm

‖u‖∞ := sup
t∈[0,1]

|u(t)|, u ∈ X,

which makes X a Banach space.
The main goal of this section is to show that the integral equation

u(t) =

∫ 1

0

H(t, s) f(s, u(s)) ds (4.1)

has a solution in X.
The kernel H associated to the problem of interest has been already computed,

and it is trivial to check that there exists a finite collection of (sixteen) open rect-
angular regions such that H is uniformly continuous on each of them, with their
closures covering [0, 1] × [0, 1]. In particular, we can conclude that H is bounded,
in absolute value, by some constant A ∈ R+.

Observe that, until now, we have not imposed hypotheses over the function f ,
since we were focused on the study of linear problems. Now, we assume the following
conditions:

(H1) There exists α ∈ [0, 1) and B ∈ R+ such that |f(t, x)tα| ≤ B for all
(t, x) ∈ (0, 1]× R.

(H2) H is uniformly continuous on each (ti, ti+1)×(tj , tj+1), for i, j ∈ {0, . . . , n−
1}.

(H3) There exists A ∈ R+ such that |H(t, s)| ≤ A, for all (t, s) ∈ [0, 1]× [0, 1].
(H4) f(t, x)tα is uniformly continuous on each (ti, ti+1) × [− AB

1−α ,
AB
1−α ], for i ∈

{0, . . . , n− 1}.
Equation (4.1) is of Hammerstein-type, and it is not very difficult to prove (in

fact, it is well known) that it has continuous solutions under the assumption of
continuity on the functions H and f . In this case, the essential tool to prove the
existence of continuous solutions is Schauder’s fixed point theorem. In our case, we
will reproduce a similar proof. However, since we only have hypotheses concerning
piecewise uniform continuity, we will only seek for piecewise uniformly continuous
solutions.

To apply Schauder’s theorem to our problem, we need to find a closed and convex
set S ⊂ X such that T : S → S is continuous, and T (S) is a relatively compact
subset of X, where

[Tu](t) =

∫ 1

0

H(t, s) f(s, u(s)) ds, t ∈ [0, 1].

Lemma 4.1. Under the hypotheses (H1) and (H2), the map T : X → X is well
defined.

Proof. From (H1), we know that there exist B > 0 and α ∈ [0, 1) such that

|f(s, u)| ≤ Bs−α, for all (s, u) ∈ (0, 1]× R.



EJDE-2020/15 EXISTENCE OF SOLUTIONS 13

Furthermore, we know that (ti, ti+1)× (tj , tj+1), for i, j ∈ {0, . . . , n− 1}, is a finite
collection of open rectangles in [0, 1] × [0, 1] such that H is uniformly continuous
on each of them, and the union of their closures covers [0, 1]× [0, 1]. Thus, we can
ensure that, for each ε > 0 and each i ∈ {0, . . . , n−1} fixed, there exists δ > 0 such
that

|H(t, s)−H(t̂, s)| < 1− α
B

ε, for every t, t̂ ∈ (ti, ti+1) with |t− t̂| < δ.

Note that the pair of points (t, s) and (t̂, s) lie in the same rectangle for each
s ∈ [0, 1], but that common rectangle changes with the value of s. However, the
conclusion holds because of the finiteness of the number of rectangles.

The argument to check that T maps X into itself can be developed in the fol-
lowing way. We consider a fixed u ∈ X and a fixed i ∈ {0, . . . , n − 1}. Then, we
deduce that

|[Tu](t)− [Tu](t̂)| =
∣∣ ∫ 1

0

(H(t, s)−H(t̂, s))f(s, u(s))ds
∣∣

≤
∫ 1

0

|H(t, s)−H(t̂, s)||f(s, u(s))|ds

≤ B 1− α
B

ε

∫ 1

0

s−αds = ε,

(4.2)

for t, t̂ ∈ (ti, ti+1) such that |t− t̂| < δ. Hence, we have shown that Tu is uniformly
continuous on any interval (ti, ti+1), for i ∈ {0, . . . , n − 1}. By definition, this
implies that Tu ∈ X. �

As we mentioned above, the idea is to show the existence of a fixed point for the
equation u = Tu, with a suitable application of the following theorem.

Theorem 4.2 (Schauder’s Theorem [15]). Let S be a closed and convex set in a
Banach space and assume that T : S → S is a continuous mapping such that T (S)
is a relatively compact set of S. Then T has, at least, one fixed point.

The first step in our study is to make an adequate choice for S, the set in which
we will look for a fixed point of the mapping T . Under conditions (H1) and (H3),
we consider

S =
{
u ∈ X : ‖u‖∞ ≤

AB

1− α

}
,

where A,B are bounds for H(t, s) and f(t, x)tα, respectively, with α given by (H1).

Lemma 4.3. Under hypotheses (H1) and (H3), T (S) ⊆ S.

Proof. Certainly, T maps the set S into itself since, for every t ∈ [0, 1], we have

|[Tu](t)| =
∣∣ ∫ 1

0

H(t, s) f(s, u(s)) ds
∣∣ ≤ ∫ 1

0

|H(t, s)| |f(s, u(s))| ds ≤ AB

1− α
.

Furthermore, S is closed since it is the pre-image of a closed interval via a continuous
function (the norm function). Note that, by the triangle inequality, S is also convex.

�

By Theorem 4.2, we only need to prove that T (S) is relatively compact, and that
T is continuous on S. The relative compactness will be an immediate consequence
of previous checks and a general version of Arzelà-Ascoli theorem. First, we state
the original version of Arzelà-Ascoli theorem with a trivial remark.



14 D. CAO LABORA, R. RODRÍGUEZ-LÓPEZ, M. BELMEKKI EJDE-2020/15

Theorem 4.4 (Arzelà-Ascoli Theorem [16]). Consider a subset U .of the set of
continuous functions on a compact interval I. Then U is relatively compact if and
only if the functions in U are uniformly bounded and they are equicontinuous.

Remark 4.5. The assumption concerning the compactness of the interval I can
be replaced by relative compactness, provided that the functions in U are uni-
formly continuous. The reason is the well known theorem of extension of uniformly
continuous functions on a relatively compact set.

Recall that, given the sequence of points 0 = t0 < t1 < t2 < · · · < tn−1 < tn = 1,
S consists of the functions u such that the restrictions u|(0,t1), u|(t1,t2), . . . , u|(tn−1,tn)

are uniformly continuous, and such that ‖u‖∞ ≤ AB
1−α . Define Xj as the set of

uniformly continuous functions on (tj−1, tj), for j from 1 to n. Then, there is a
natural isometry between X and

P := R×X1 × R×X2 × · · · ×Xn−1 × R×Xn × R,

where each copy of R reflects the value of u ∈ X at ti, for i = 0, 1, . . . , n. The
distance between two elements in the previous cartesian product is the maximum
distance between their respective coordinates, and the distance between each pair
of respective coordinates is obtained with ‖ · ‖∞.

Theorem 4.6 (General version of Arzelà-Ascoli Theorem). Consider a subset R ⊂
X that is isometric to a cartesian product

C0 ×R1 × C1 ×R2 × · · · ×Rn−1 × Cn−1 ×Rn × Cn ⊂ P. (4.3)

Then R is relatively compact if and only if the functions in the subset R are uni-
formly bounded and piecewise uniformly equicontinuous.

Proof. Assume that R is relatively compact, that is, R̄ is compact. By the isometry
between X and P , we deduce that

C0 ×R1 × C1 ×R2 × · · · ×Rn−1 × Cn−1 ×Rn × Cn

is compact. However, the product of the closures is the closure of the product, so

C0 ×R1 × C1 ×R2 × · · · ×Rn−1 × Cn−1 ×Rn × Cn

is compact [10]. We also know that a cartesian product is compact if and only if
each factor is compact, so each Ri and each Cj is compact, meaning that each Ri
and each Cj is relatively compact. By the original version of Arzelà-Ascoli theorem
and Remark 4.5, we know that each Ri and each Cj is uniformly bounded and
equicontinuous. However, this is clearly equivalent to say that

C0 ×R1 × C1 ×R2 × · · · ×Rn−1 × Cn−1 ×Rn × Cn

is uniformly bounded and equicontinuous in each coordinate. By the isometry,
we conclude that R is uniformly bounded and piecewise uniformly equicontinuous.
Note that every argument used above is reversible, so the converse deduction also
holds. �

Remark 4.7. In our case, the condition of equicontinuity of Cj is trivial, since Cj
consists of a family of functions whose domain is {tj}.



EJDE-2020/15 EXISTENCE OF SOLUTIONS 15

We will use Theorem 4.6 to deduce that S is relatively compact. Note that
S is isometric to a cartesian product, like the one stated in (4.3), after choosing
C0 = · · · = Cn = [− AB

1−α ,
AB
1−α ], and Ri, for each i ∈ {0, . . . , n − 1}, as the set of

uniformly continuous functions on (ti−1, ti) that are bounded, in absolute value, by
AB
1−α .

Lemma 4.8. Operator T : S → S is continuous and T (S) is relatively compact,
provided that (H1)–(H4) hold.

Proof. By the construction of S and Lemma 4.3, it is trivial to show that T (S) ⊂ S
is uniformly bounded. Furthermore, in (4.2), we have proved that the functions
in T (S) are piecewise uniformly equicontinuous. Hence, Theorem 4.6 implies that
T (S) is relatively compact.

To conclude, we only need to prove that T : S → S is continuous. If u ∈ S, we
know that |u(t)| ≤ AB

1−α for every t ∈ [0, 1]. Moreover, because of (H4), we know

that f(t, x)tα is uniformly continuous on each (ti, ti+1) × [− AB
1−α ,

AB
1−α ]. Hence, if

we fix ε > 0 and an index i ∈ {0, . . . , n − 1}, we can find δi > 0 ensuring that, if

ti < t < ti+1 and x, x̂ ∈
[
− AB

1−α ,
AB
1−α

]
are such that |x− x̂| < δi, we have

|f(t, x)− f(t, x̂)| < 1− α
A

εt−α.

Hence, after fixing u0 ∈ S and ε > 0, given any u ∈ S such that

‖u− u0‖ < δ := min{δi : i = 0, . . . n− 1},

we have |u(t)− u0(t)| < δ, for all t ∈ [0, 1]. Thus,

‖Tu− Tu0‖ = sup
t∈[0,1]

∣∣ ∫ 1

0

H(t, s)(f(s, u(s))− f(s, u0(s))) ds
∣∣

≤ sup
t∈[0,1]

∫ 1

0

|H(t, s)| |f(s, u(s))− f(s, u0(s))| ds

≤ A1− α
A

ε

∫ 1

0

s−αds = ε.

This proves the continuity of T on S so, due to Schauder’s fixed point theorem,
we conclude the existence of at least one solution u for the equation u = Tu, with
u ∈ S. �

Remark 4.9. Note that, in the previous results, the term tα can be replaced by

any integrable function g on [0, 1], replacing 1− α by the value
∫ 1

0
g(s)ds.

Acknowledgments. R. Rodŕıguez-López was partially supported by grant num-
ber MTM2016-75140-P (AEI/FEDER, UE), and by grant MTM2013-43014-P [Min-
isterio de Economı́a y Competitividad, co-financed by the European Community
fund FEDER].

References

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16 D. CAO LABORA, R. RODRÍGUEZ-LÓPEZ, M. BELMEKKI EJDE-2020/15

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Daniel Cao Labora

Departamento de Estad́ıstica, Análisis Matemático y Optimización, Universidade de San-
tiago de Compostela, 15782, Spain

Email address: daniel.cao@usc.es

Rosana Rodŕıguez-López

Departamento de Estad́ıstica, Análisis Matemático y Optimización, Universidade de San-

tiago de Compostela, 15782, Spain
Email address: rosana.rodriguez.lopez@usc.es

Mohammed Belmekki
Ecole Supérieure en Sciences Appliquées, BP. 165 RP, Bel Horizon, Tlemcen,13000, Al-
geria

Email address: m.belmekki@yahoo.fr


	1. Introduction
	2. Basic definitions
	3. Green's function for a linear fractional differential equation
	4. Existence results
	Acknowledgments

	References