CUBO, A Mathematical Journal

Vol. 24, no. 03, pp. 521–539, December 2022

DOI: 10.56754/0719-0646.2403.0521

A derivative-type operator and its application to
the solvability of a nonlinear three point
boundary value problem

René Erlin Castillo1, B

Babar Sultan2

1 Universidad Nacional de Colombia,

Departamento de Matemáticas, Bogotá,

Colombia

recastillo@unal.edu.co B

2 Department of Mathematics,

Quaid-I-Azam University, Islamabad

45320, Pakistan.

babarsultan40@yahoo.com

ABSTRACT

In this paper we introduce an operator that can be thought as

a derivative of variable order, i.e. the order of the derivative

is a function. We prove several properties of this operator,

for instance, we obtain a generalized Leibniz’s formula, Rolle

and Cauchy’s mean theorems and a Taylor type polynomial.

Moreover, we obtain its inverse operator. Also, with this

derivative we analyze the existence of solutions of a nonlinear

three-point boundary value problem of “variable order”.

RESUMEN

En este art́ıculo introducimos un operador que puede ser

pensado como una derivada de orden variable, i.e. el or-

den de la derivada es una función. Demostramos varias

propiedades de este operador, por ejemplo, obtenemos una

fórmula generalizada de Leibniz, teoremas de valor medio de

Rolle y Cauchy y un polinomio de tipo Taylor. Más aún,

obtenemos su operador inverso. También con esta derivada

analizamos la existencia de soluciones de un problema no

lineal de valor en la frontera de tres puntos de “orden vari-

able”.

Keywords and Phrases: Fractional Derivative, boundary value problem, Hammerstein-Volterra integral equation.

2020 AMS Mathematics Subject Classification: 26A33, 34B10, 45D05.

Accepted: 02 December 2022

Received: 17 June, 2022

©2022 R. E. Castillo et al. This open access article is licensed under a Creative Commons

Attribution-NonCommercial 4.0 International License.

http://cubo.ufro.cl/
http://dx.doi.org/10.56754/0719-0646.2403.0521
https://orcid.org/0000-0003-1113-5827
https://orcid.org/0000-0003-2833-4101
mailto:recastillo@unal.edu.co
mailto:babarsultan40@yahoo.com


522 R. E. Castillo & B. Sultan CUBO
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1 Motivation

Derivatives of non-integer order have been studied since the celebrated question of L’Hospital to

Leibniz about the meaning of d
nf

dxn
when n = 1/2. There are several definitions of derivatives of frac-

tional order, e.g., derivative of Riemann-Louville, Caputo, Hadamard, Erdélyi-Kober, Grünwald-

Letnikov and Riesz, among others. Typically, these derivatives are defined using an integral form

of the classical derivative, as a consequence of it, some basic properties of the usual derivative,

as the product rule and chain rule are lost. For a more comprehensible information about these

notions we recommend [17, 20, 30].

Despite of the lack of some properties, derivatives of fractional order appear in many real world

applications as, for instance, in memory effects and future dependence, control theory of dynamical

systems, nanotechnology, viscoelasticity and financial modeling see, e.g., [8, 12, 18, 19, 21, 24, 25,

31, 32]. Thus, due to this development, in the last decades a lot of research has been devoted to

the study of the existence of solutions for several kinds of boundary value problems of fractional

type, see, for instance, [2, 3, 4, 5, 9, 11, 26, 28, 29] and references therein.

In order to overcome the limitations of the classical derivative, in [16] it is introduced a new limit-

based definition of derivative, the so-called conformable fractional derivative, which can be seen as

a natural extension of the fractional derivative, although as it is stated in [7], it is best to consider

the conformable derivative in its own right, independent of fractional derivative theory. Some of

the basic properties, physical interpretation and some boundary value problems for conformable

differential equations can be found in [1, 6, 10, 14, 15, 33, 34] and its references.

In this article, based on the idea of conformable fractional derivative and in ideas from [13], we

consider an extension of the conformable fractional derivative of order α and develop some of

its properties. Additionally, we study the existence and uniqueness of solutions for a nonlinear

three-point boundary value problem in this new setting.

2 Derivative of variable order

We now introduce the notion of (φ, ω)-derivative.

Definition 2.1. Let f : [a, b] −→ R. The (φ, ω)-derivative at the point x ∈ (a, b) (φ(x) ̸= 0) is
defined as

Dφωf(x) = D
φ
ω(f)(x) = D

(φ,1)
ω f(x) = lim

h→0

f(x + hφ(x)) − f(x)
ω(x + h) − ω(x)

. (2.1)

Where ω is a strictly increasing function and φ is a function. At the point x ∈ (a, b) such that



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A derivative-type operator and its application to the solvability... 523

φ(x) = 0 we define the (φ, ω)-derivative as

Dφωf(x) = lim
ξ→x

Dφωf(ξ),

when the limit exists.

Taking φ(x) = x1−α and ω(x) = x we obtain the conformable fractional derivative of order α, cf.

[16].

Theorem 2.2. Let f, g be (φ, ω)-differentiable. Then:

(a) The function f is continuous.

(b) Dφω(a) = 0, a is a constant.

(c) Dφω(af + g) = aD
φ
ω(f) + D

φ
ω(g).

(d) Dφω(fg) = fD
φ
ω(g) + fD

φ
ω(g).

(e) Dφω

(
f

g

)
=

fDφω(g) − fDφω(g)
g2

.

(f) If f and ω are differentiable, we have

Dφω(f)(t) = φ(t)
f′(t)

ω′(t)
. (2.2)

(g) If f, g and ω are differentiable, we have

Dφω(f ◦ g)(t) = f
′(g(t)) · Dφω(g)(t).

Proof. It is a matter of direct calculations.

Formula (2.2) enables us to calculate in a straightforward way some (φ,ω)-derivatives. For example,

letting φ(x) = sin(x), f(x) = cos(x), and ω(x) = x, we have

Dφωφ(x) = sin(x) cos(x), D
φ
ωf(x) = − sin

2(x),

whereas taking φ and f as above with ω(x) = ex − 1 we get

Dφωφ(x) =
sin(x) cos(x)

ex
, Dφωf(x) =

− sin2(x)
ex

.

We now introduce the n-iterated (φ, ω)-derivative.



524 R. E. Castillo & B. Sultan CUBO
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Definition 2.3. By D
(φ,n)
ω f(x) we define the n-iterated (φ, ω)-derivative of the function f, i.e.

D(φ,n)ω f(x) = D
φ
ω

(
D(φ,n−1)ω f

)
(x),

with the convention D
(φ,0)
ω f(x) := f(x).

Theorem 2.4 (Generalized Leibniz’s formula). We have

D(φ,n)ω (f1f2 · · · fm) =
∑

i1+i2+···+im=n
ij=0,n

n!
D

(φ,i1)
ω (f1)D

(φ,i2)
ω (f2) · · · D

(φ,im)
ω (fm)

i1!i2! · · · im!
, (2.3)

where we suppose that all is well-defined.

Proof. For m = 2, equation (2.3) is obtained by induction on n and using the formula for the

(φ, ω)-derivative of the product, in this case we obtain

D(φ,n)ω (f1f2) =

n∑
j=0

(
n

j

)
D(φ,n−j)ω (f1)D

(φ,j)
ω (f2). (2.4)

By the well-known method to prove the multinomial theorem from the binomial theorem we can,

in the same way, obtain (2.3) from (2.4).

Theorem 2.5 (Fermat’s Theorem). Let f : [a, b] −→ R have a local maximum or minimum at
x = c ∈ (a, b) and Dφω(f)(c) exists. Then Dφω(f)(c) = 0.

Proof. Let us suppose, without loss of generality, that x = c is a minimum of f. We have, for

sufficiently small h ̸= 0, that

sgn(hφ(c))
f(c + hφ(c)) − f(c)
ω(c + h) − ω(c)

⩾ 0. (2.5)

From (2.5) and the hypothesis of the existence of Dφω(f)(c) the result follows.

Theorem 2.6 (Rolle’s theorem). Let f : [a, b] −→ R be a continuous function in [a, b] and
(φ, ω)-differentiable in (a, b) such that f(a) = f(b) = 0. Then there exists c ∈ (a, b) such that
Dφω(f)(c) = 0.

Proof. Supposing, without loss of generality, that there exists ξ ∈ (a, b) such that f(ξ) ≥ 0. Then
by Weierstraß theorem, there exists c ∈ (a, b) which is a maximum. Invoking Fermat’s theorem
2.5 we end the proof.



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A derivative-type operator and its application to the solvability... 525

Theorem 2.7 (Cauchy mean-value theorem). Let f, g : [a, b] −→ R be both continuous on the
closed interval [a, b] and (φ, ω)-differentiable in the open interval (a, b). Then there exists a number

ξ ∈ (a, b) such that
[f(b) − f(a)]Dφω(g)(ξ) = [g(b) − g(a)]D

φ
ω(f)(ξ). (2.6)

Proof. The proof follows, as in the classical case, from Rolle’s theorem 2.6 applied to the function

F(x) = f(x)[g(b) − g(a)] − g(x)[f(b) − f(a)].

3 Integration of variable order

Definition 3.1. Let f : [a, b] −→ R. We define the (φ, ω)-integral of the function f as

Iφω (f)(t) =

∫ t
a

f(ξ)

φ(ξ)
dω(ξ), (3.1)

where the integral is understood in the Lebesgue-Stieltjes sense.

Notice that for f ∈ L∞([a, b]) and 1
φ
∈ L1([a, b], dw) the integral (3.1) is finite.

When f, φ and ω′ are continuous functions, it is straightforward the relation Dφω(I
φ
ω f)(t) = f(t),

since

Dφω(I
φ
ω f)(t) =

φ(t)

ω′(t)
D

(∫ t
a

f(ξ)

φ(ξ)
dω(ξ)

)
(t) = f(t),

using (2.2).

In the case φ and ω′ are continuous functions, the following Lagrange mean-value theorem

Dφω(f)(ξ) =
f(b) − f(a)

I
φ
ω (1)(b) − Iφω (1)(a)

, ξ ∈ (a, b) (3.2)

is valid, when f : [a, b] −→ R is continuous on the closed interval [a, b] and (φ, ω)-differentiable in
the open interval (a, b). The equation (3.2) follows from (2.6) taking g(x) = Iφω (1)(x) (note that

Iφω (1)(a) = 0, but we leave it in (3.2) just for keeping with the parallel in the classical case).

By I
(φ,n)
ω φ(x) we define the n-iterated (φ, ω)-integral of the function f, i.e.

I(φ,n)ω f(x) = I
φ
ω (I

(φ,n−1)
ω f)(x),

with the convention I
(φ,0)
ω f(x) := f(x).



526 R. E. Castillo & B. Sultan CUBO
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4 Taylor formula

In this section we will obtain a Taylor type formula using the (φ,ω)-derivative with a remainder

which generalizes well-know remainders, i.e. Cauchy, Lagrange, Peano, Schlömilch, among others,

cf. [22, 23, 27] for similar remainders for the classical derivative.

Theorem 4.1. Let f : I −→ R be a continuous function in the open interval I and n-times
(φ, ω)-differentiable function in I. We also require that φ, ω′ and I

(φ,j)
ω (1)(x) are continuous

functions, for j = 1, n. Moreover, let g : I −→ R be a n-times (φ, ω)-differentiable function such
that D

(φ,j)
ω g(a) = 0 for j = 1, n − 1 and D

(φ,k)
ω g(y) ̸= 0 for all y different from a and x and

j = 1, n − 1. Then, for all x ∈ I we have

f(x) =

n∑
j=0

D(φ,j)ω (f)(a) I
(φ,j)
ω (1)(x) + Rn(x), (4.1)

with

Rn(x) =
g(x) − g(a)
D

(φ,n)
ω g(ξ)

(
D(φ,n)ω (f)(ξ) − D

(φ,n)
ω (f)(a)

)
, (4.2)

where x ̸= a and ξ is between a and x.

Proof. We first note that, since

D(φ,n)ω (I
(φ,j)
ω 1)(x) =



I
(φ,j−n)
ω (1)(x), j > n,

1, n = j,

0, n > j.

we have

Rn(a) = D
(φ,1)
ω (Rn)(a) = · · · = D

(φ,n−1)
ω (Rn)(a) = 0. (4.3)

By the Cauchy type finite increment formula (2.6), relations (4.3) and the hypothesis on g we have

Rn(x) − Rn(a)
g(x) − g(a)

=
D

(φ,1)
ω (Rn)(θ1) − D

(φ,1)
ω (Rn)(a)

D
(φ,1)
ω (g)(θ1) − D

(φ,1)
ω (g)(a)

= . . . =
D

(φ,n−1)
ω (Rn)(θn−1) − D

(φ,n−1)
ω (Rn)(a)

D
(φ,n−1)
ω (g)(θn−1) − D

(φ,n−1)
ω (g)(a)

=
D

(φ,n)
ω (Rn)(ξ)

D
(φ,n)
ω (g)(ξ)

, (4.4)

where ξ := θn. On the other hand, (φ, ω)-differentiating the equality (4.1) n-times we obtain

D
(φ,n)
ω (f)(x) − D

(φ,n)
ω (f)(a) = D

(n)
ω (Rn)(x) which, together with (4.4), entails (4.2).



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5 Three-point boundary value problems of variable order

Inspired in [10], we are interested in the use of the (φ, ω)-derivative to study the solutions of the

following nonlinear boundary value problem

Dφω(D + λ)x(t) = f(t, x(t)), t ∈ [0, 1] (5.1)

x(0) = 0, x′(0) = α, x(1) = βx(η), (5.2)

where Dφω is the derivative of variable order, D is the ordinary derivative, f : [0, 1] × R −→ R
is a known function, β, λ and α are real numbers, λ ̸= 0 and η ∈ (0, 1). Notice that in virtue
of Theorem 2.2, a sufficient condition for the well posedness of equation (5.1) is, by considering

ω ∈ C1[0, 1], and x ∈ C2[0, 1]. Thus, in the sequel we consider these conditions on the functions ω
and x. In addition, in order to use the (φ, ω)-integral, we are going to assume that φ is continuous

and bounded away from zero.

From the conditions on the functions ω and φ we conclude that the following non negative numbers

are finite

Ω := sup
t∈[0,1]

ω′(t) < ∞, M := sup
t∈[0,1]

∣∣∣∣ 1φ(t)
∣∣∣∣ < ∞.

We will use these numbers in the sequel to establish the existence results.

First, as usual, we will consider the linear boundary value problem:

Dφω(D + λ)x(t) = g(t), t ∈ [0, 1], g ∈ C[0, 1] (5.3)

x(0) = 0, x′(0) = α, x(1) = βx(η), α, β, λ ∈ R, λ ̸= 0, η ∈ (0, 1). (5.4)

To obtain a solution for the boundary value problem, we apply the (φ, ω)-integral to equation (5.3):

(D + λ)x(t) + (D + λ)x(0) = Iφω (g)(t), (5.5)

where, using the boundary condition (5.4), (D + λ)x(0) = α. Then, equation (5.5) simplifies as

(D + λ)x(t) + α = Iφω (g)(t). (5.6)

Let y(t) = eλtx(t), Then we rewrite (5.6) as

Dy(t) = eλtIφω (g)(t) − αe
λt.

Integrating from 0 to t we obtain

y(t) − y(0) =
∫ t
0

eλsIφω (g)(s) ds −
α

λ
(eλt − 1)



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y(t) =

∫ t
0

eλs
∫ s
0

g(r)

φ(r)
ω′(r) dr ds −

α

λ
(eλt − 1), (y(0) = x(0) = 0).

Now, notice that

∫ t
0

eλs
∫ s
0

g(r)

φ(r)
ω′(r) dr ds =

eλt

λ

∫ t
0

g(s)

φ(s)
ω′(s) ds −

1

λ

∫ t
0

eλsg(s)

φ(s)
ω′(s) ds, 0 ≤ s ≤ t ≤ 1.

From here we have that

y(t) =
eλt

λ

∫ t
0

g(s)

φ(s)
ω′(s) ds −

1

λ

∫ t
0

eλsg(s)

φ(s)
ω′(s) ds −

α

λ
(eλt − 1), 0 ≤ s ≤ t ≤ 1.

Thus,

x(t) =
1

λ

∫ t
0

g(s)

φ(s)
ω′(s) ds −

e−λt

λ

∫ t
0

eλsg(s)

φ(s)
ω′(s) ds +

α

λ
(e−λt − 1).

Finally, from the condition βx(η) = x(1) we get

β

λ

∫ η
0

g(s)

φ(s)
(1 − eλ(s−η))ω′(s) ds +

αβ

λ
(e−λη − 1) −

1

λ

∫ 1
0

g(s)

φ(s)
(1 − eλ(s−1)) ds −

α

λ
(e−λ − 1) = 0.

Therefore, introducing this equality into the formula of function x above, we obtain the following

expression for x satisfying boundary value problem (5.3)–(5.4)

x(t) =
1

λ

∫ t
0

g(s)

φ(s)
(1 − eλ(s−t))ω′(s) ds +

β

λ

∫ η
0

g(s)

φ(s)

(
1 − eλ(s−η)

)
ω′(s) ds

−
1

λ

∫ 1
0

g(s)

φ(s)

(
1 − eλ(s−1)

)
ω′(s) ds +

α

λ

(
e−λt − e−λ + βe−λη − β

)
.

Notice that actually we just proved the following result.

Theorem 5.1. The linear boundary value problem (5.3)–(5.4) has a unique solution given by

x(t) =
1

λ

∫ t
0

g(s)

φ(s)
ω′(s)k(s, t) ds +

β

λ

∫ η
0

g(s)

φ(s)
ω′(s)k(s, η) ds −

1

λ

∫ 1
0

g(s)

φ(s)
k(s, 1)ω′(s) ds

+
α

λ

(
e−λt − e−λ + βe−λη − β

)
.

where, k(s, t) = 1 − eλ(s−t).

Now, we are going to analyze the existence of solutions for the nonlinear boundary value problem:

Dφω(D + λ)x(t) = f(t, x(t)), t ∈ [0, 1], λ ∈ (−1, ∞) \ {0} (5.7)

x(0) = 0, x′(0) = α, x(1) = βx(η). (5.8)

As in Theorem 5.1, we can transform boundary value problem (5.7)–(5.8) into the nonlinear



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A derivative-type operator and its application to the solvability... 529

Hammerstein-Volterra integral equation

x(t) =
1

λ

∫ t
0

f(s, x(s))

φ(s)
k(s, t)ω′(s) ds +

β

λ

∫ η
0

f(s, x(s))

φ(s)
k(s, η)ω′(s) ds

−
1

λ

∫ 1
0

f(s, x(s))

φ(s)
k(s, 1)ω′(s) ds +

α

λ

(
e−λt − e−λ + βe−λη − β

)
.

where, k(s, t) = 1 − eλ(s−t).

In order to investigate the existence of a solution for this integral equation, we analyze it as a fixed

point problem; that is, letting

T : (C2[0, 1], ∥ · ∥∞) −→ (C2[0, 1], ∥ · ∥∞)
x(t) 7−→ Tx(t)

Tx(t) :=
1

λ

∫ t
0

f(s, x(s))

φ(s)
k(s, t)ω′(s) ds +

1

λ

∫ 1
0

f(s, x(s))

φ(s)

(
χ(0,η)(s)βk(s, η) − k(s, 1)

)
ω′(s) ds

+
α

λ

(
e−λt − e−λ + βe−λη − β

)
, (5.9)

(with χ(0,η)(s) the characteristic function of the interval (0, η)), we have that the existence of the

solution of the integral equation is equivalent to the existence of a fixed point of the operator T.

To assure that the operator T applies C2[0, 1] into itself, we assume that f(t, x(t)) is continuous

and differentiable in the first variable.

We are going to use metric fixed point theory (Banach’s contraction principle) to provide conditions

to guarantee that the boundary value problem (5.7)–(5.8) has a unique solution.

Theorem 5.2. Let f : [0, 1] × R −→ R be a continuous and differentiable in the first variable
function satisfying that

|f(t, x) − f(t, y)| ≤ K|x − y|, K > 0, for all t ∈ [0, 1], x, y ∈ R.

Then, the nonlinear boundary value problem (5.7)–(5.8) has a unique solution provide that

(|β| + 1)MKΩ
|λ|

<
1

4
,

where M := supt∈[0,1]
1

|φ(t)|
and Ω := supt∈[0,1] w

′(t).

Proof. As we saw, it is sufficient to show that the operator T defined by the formula (5.9) has a

unique fixed point. Let x and y be two functions in C2[0, 1]. Then,



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|Tx(t) − Ty(t)| =

∣∣∣∣∣1λ
∫ t
0

(f(s, x(s)) − f(s, y(s))
φ(s)

k(s, t)ω′(s) ds

+
1

λ

∫ 1
0

(f(s, x(s)) − f(s, y(s))
φ(s)

(
χ(0,η)(s)βk(s, η) − k(s, 1)

)
ω′(s) ds

∣∣∣∣∣
≤

1

|λ|

∫ t
0

|k(s, t)|
|φ(s)|

|f(s, x(s)) − f(s, y(s))|ω′(s) ds

+
1

|λ|

∫ 1
0

∣∣∣χ(0,η)(s)βk(s, η) − k(s, 1)∣∣∣
|φ(s)|

|f(s, x(s)) − f(s, y(s))|ω′(s) ds.

On the other hand,

|k(s, t)| = |1 − eλ(s−t)| ≤ 1 + eλ(s−t).

Notice that for −1 < λ < 0, the inequality |ex − 1| < 7/4|x|, for 0 < |x| < 1, gives us the estimate

|1 − eλ(s−t)| <
7

4
λ(s − t) <

7

4
< 2.

Then,

sup
s∈[0,t]

(1 + eλ(s−t)) ≤ 2, −1 < λ < 0.

Now, for λ > 0,

sup
s∈[0,t]

(1 + eλ(s−t)) = 1 + e−λt ≤ 2, for any t ∈ [0, 1].

Therefore, we obtain the following bound

|k(s, t)| ≤ 2. (5.10)

Moreover,

∣∣∣χ(0,η)(s)βk(s, η) − k(s, 1)∣∣∣ = ∣∣∣−χ(0,η)(s)βeλ(s−η) + eλ(s−1) + χ(0,η)(s)β − 1∣∣∣
≤ | − χ(0,η)(s)βeλ(s−η)| + |eλ(s−1)| + |β| + 1,

where, for −1 < λ < 0, we have that

sup
s∈[0,η]

eλ(s−η) = 1, sup
s∈[0,1]

eλ(s−1) = 1.

In the case λ > 0, we get

sup
s∈[0,η]

eλ(s−η) = e−λη ≤ 1, sup
s∈[0,1]

eλ(s−1) = e−λ ≤ 1.



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With these bounds we obtain the following estimation

∣∣∣χ(0,η)(s)βk(s, η) − k(s, 1)∣∣∣ ≤ 2(|β| + 1). (5.11)
We introduce the bounds (5.10) and (5.11) into the difference |Tx(t) − Ty(t)|:

|Tx(t) − Ty(t)| ≤
2

|λ|

∫ t
0

1

|φ(s)|
|f(s, x(s)) − f(s, y(s))|ω′(s) ds

+
2(|β| + 1)

|λ|

∫ 1
0

1

|φ(s)|
|f(s, x(s)) − f(s, y(s))|ω′(s) ds. (5.12)

Since f(s, x(s)) is Lipschitz in the second variable, then

|Tx(t) − Ty(t)| ≤
2

|λ|

∫ t
0

K

|φ(s)|
|x(s) − y(s)|ω′(s) ds +

2(|β| + 1)
|λ|

∫ 1
0

K

|φ(s)|
|x(s) − y(s)|ω′(s) ds

≤
2(|β| + 1)

|λ|

∫ t
0

K

|φ(s)|
|x(s) − y(s)|ω′(s) ds

+
2(|β| + 1)

|λ|

∫ 1
0

K

|φ(s)|
|x(s) − y(s)|ω′(s) ds.

Taking the maximum over t ∈ [0, 1] we obtain

∥Tx − Ty∥∞ ≤ 2
2(|β| + 1)

|λ|
KMΩ∥x − y∥∞.

Therefore, T is a contraction operator, since µ = 2
2(|β|+1)

|λ| KMΩ < 1. Thus from the Banach

contraction principle, T has a unique fixed point as desired.

Now, we are going to use topological fixed point theory, more precisely Schaefer’s fixed point

theorem, to establish the existence of at least one solution of boundary value problem (5.7)–(5.8),

dropping the Lipschitzianity of the function f.

First, we prove that the operator T is compact.

Theorem 5.3. The operator T : (C2[0, 1], ∥ · ∥∞) −→ (C2[0, 1], ∥ · ∥∞) is compact.

Proof. We start by proving the continuity of T . Let (xn) ⊂ C2[0, 1], x ∈ C2[0, 1] be such that
∥xn − x∥∞ → 0. We have to show that ∥Txn − Tx∥∞ → 0. Fixed ε > 0, there exists K ≥ 0 such
that

∥xn∥∞ ≤ K, ∀n ∈ N

∥x∥∞ ≤ K.

Since f : [0, 1] × [−K, K] −→ R is continuous, then it is uniformly continuous on [0, 1] × [−K, K].



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Thus there exists δ(ε) > 0 such that

|f(s1, x(s1)) − f(s2, y(s2))| ≤ ε,

for every (s1, x(s1)), (s2, y(s2)) ∈ [0, 1] × [−K, K] such that ∥(s1 − s2, x(s1) − y(s2))∥2 < δ(ε).

From the fact that ∥xn − x∥∞ → 0, it follow that there exists N(ε) ∈ N such that

sup
t∈[0,1]

|xn(t) − x(t)| < δ,

for every n ≥ N(ε). Consequently, from (5.12),

∥Txn − Tx∥∞ = sup
t∈[0,1]

|Txn(t) − Tx(t)|

≤ sup
t∈[0,1]

{
2

|λ|

∫ t
0

|f(s, xn(s)) − f(s, x(s))|
|φ(s)|

ω′(s) ds

+
2(|β| + 1)

|λ|

∫ 1
0

|f(s, xn(s)) − f(s, x(s))|
|φ(s)|

ω′(s) ds

}

<
2|β| + 4

|λ|
MΩε, M := sup

t∈[0,1]

1

|φ(t)|
, Ω := sup

t∈[0,1]
ω′(t).

Therefore, the operator T is continuous. To prove the compactness we consider a bounded set

X ⊂ C2[0, 1] and we will show that T(X) is relatively compact in (C2[0, 1], ∥ · ∥∞) by using the
Arzela-Ascoli theorem. Let K ≥ 0 be such that

∥x∥∞ ≤ K,

for every x ∈ X.

From the bounds (5.10) and (5.11) we have

|Tx(t)| ≤
1

|λ|

∫ t
0

|f(s, x(s))|
|φ(s)|

|k(s, t)|ω′(s) ds +
1

|λ|

∫ 1
0

|f(s, x(s))|
|φ(s)|

|χ(0,η)βk(s, η) − k(s, 1)|ω′(s) ds

+

∣∣∣∣αλ
∣∣∣∣ |e−λt − e−λ + βe−λη − β|

≤
2MΩ

|λ|

∫ t
0

|f(s, x(s))| ds +
2(|β| + 1)

|λ|
MΩ

∫ 1
0

|f(s, x(s))| ds

+

∣∣∣∣αλ
∣∣∣∣ |e−λt − e−λ + βe−λη − β|

≤
2|β| + 4

|λ|
MΩ

∫ 1
0

|f(s, x(s))| ds +
∣∣∣∣αλ
∣∣∣∣ |e−λt − e−λ + βe−λη − β|.



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We obtain a upper bound for |e−λt − e−λ + βe−λη − β|, namely

|e−λt − e−λ + βe−λη − β| ≤ ∆, t ∈ [0, 1],

where

∆ :=



(|β| + 1)e−λ, −1 < λ < 0

2(|β| + 1), λ > 0.

On the other hand, since the function f is uniformly continuous on the compact set [0, 1]×[−K, K],
then there exists, and it is finite, the positive number

RK = ∥f∥∞ = sup
x∈X

sup
s∈[0,1]

|f(s, x(s))| < ∞, (s, x(s)) ∈ [0, 1] × [−K, K].

Thus, we have

∥Tx∥∞ ≤
2|β| + 4

|λ|
MΩRK +

∣∣∣∣αλ
∣∣∣∣∆, (5.13)

for every x ∈ X. That means, the set T(X) is bounded in C2[0, 1]. Now, if t1, t2 ∈ [0, 1], are such
that t1 ≤ t2 and satisfy |t1 − t2| < δ, then

|Tx(t1) − Tx(t2)| =

∣∣∣∣∣1λ
∫ t1
0

f(s, x(s))

φ(s)
ω′(s) ds −

1

λ

∫ t2
0

f(s, x(s))

φ(s)
ω′(s) ds −

α

λ
e−λt1 +

α

λ
e−λt2

∣∣∣∣∣
=

∣∣∣∣∣1λ
∫ t2
t1

f(s, x(s))

φ(s)
ω′(s) ds +

α

λ
(e−λt2 − e−λt1)

∣∣∣∣∣
→ 0, as |t1 − t2| → 0

for every x ∈ X, so the set T(X) ⊂ C2[0, 1] satisfies the hypotheses of Arzela-Ascoli’s theorem, so
T(X) is relatively compact in C2[0, 1]. Therefore, the operator T is compact.

Now, we establish the following existence result.

Theorem 5.4. Let f : [0, 1] × R −→ R be a continuous and differentiable in the first variable
function, and let us assume that there exist C, D ≥ 0 and q ∈ (0, 1) such that

|f(s, r)| ≤ C|r|q + D.

For every (s, r) ∈ [0, 1] × R. Then, the nonlinear boundary value problem (5.7)–(5.8) has at least
one solution.

Proof. The theorem is proved once we assure the existence of at least a fixed point of the operator

T . Let

S = {x ∈ C2[0, 1] : ∃σ ∈ [0, 1] such that x = σTx}.



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To apply Schaefer’s fixed point theorem we should show that S is bounded. Let x ∈ S,

∥x∥∞ =σ∥Tx∥∞.

Now, from (5.13) we have

|Tx(t)| ≤
2|β| + 4

|γ|

∫ 1
0

|f(s, x(s))|
|φ(s)|

ω′(s) ds +

∣∣∣∣αβ
∣∣∣∣∆ ≤ 2|β| + 4|γ| MΩ(C∥x∥q∞ + D) +

∣∣∣∣αβ
∣∣∣∣∆ < ∞.

Then

∥x∥∞ = σ∥Tx∥∞ ≤ σ
2|β| + 4

|γ|
MΩ(C∥x∥q∞ + D) +

∣∣∣∣αβ
∣∣∣∣∆σ < ∞.

This inequality and the fact q ∈ (0, 1) shows that S is bounded. Thus, from Schaefer’s fixed point
theorem, the operator T has a fixed point, which implies that boundary value problem (5.7)–(5.8)

has a solution.

Notice from the proof of the theorem above, that we can use the functions φ and ω given in the

definition of the (φ, ω)-derivative to rewrite Theorem 5.4 as:

Theorem 5.5. Let f : [0, 1] × R −→ R be a continuous and differentiable in the first variable
function, and let us assume that there exist C, D ≥ 0 and q ∈ (0, 1) such that

|f(s, r)|
|φ(s)|

ω′(s) ≤ C|r|q + D.

For every (s, r) ∈ [0, 1] × R. Then, the nonlinear boundary value problem (5.7)–(5.8) has at least
one solution.

Schaefer’s theorem is a consequence of the Schauder fixed point theorem, which is a localization

fixed point result. We will use Schauder’s theorem to give a localization result for the solutions of

boundary value problem (5.7)–(5.8).

Theorem 5.6. Let f : [0, 1] × R −→ R be a continuous and differentiable in the first variable
function and, in addition, let us assume that f ∈ L1([0, 1] × R). Let B(r) be the closed ball with
radius r. Then, the nonlinear boundary value problem (5.7)–(5.8) has at least one solution for

every closed ball B(r) such that

r ≥
2|β| + 4

|λ|
MΩ∥f∥1 +

∣∣∣∣αλ
∣∣∣∣∆, (5.14)

with,

∆ :=



(|β| + 2)e−λ, −1 < λ < 0

2(|β| + 1), λ > 0
, M := sup

t∈[0,1]

∣∣∣∣ 1φ(t)
∣∣∣∣ , Ω = sup

t∈[0,1]
ω′(t).



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A derivative-type operator and its application to the solvability... 535

Proof. Since the operator T is continuous and compact, we can apply Schauder’s fixed point

theorem, once we prove that T(B(r)) ⊂ B(r).

From (5.13) and the hypotheses, we have

|Tx(t)| ≤
2|β| + 4

|γ|

∫ 1
0

|f(s, x(s))|
|φ(s)|

ω′(s) ds +

∣∣∣∣αβ
∣∣∣∣∆ (5.15)

≤
2|β| + 4

|γ|
MΩ∥f∥1 +

∣∣∣∣αβ
∣∣∣∣∆

≤ r.

Thus, ∥Tx∥∞ ≤ r. Consequently T(B(r)) ⊂ B(r). Finally, from Schauder’s theorem, T has a
fixed point, as so boundary value problem (5.7)–(5.8) has at least one solution, for every closed

ball B(r) with radius r as in (5.14).

We can control the growth behavior of the nonlinear function f and still guarantee the existence of

solutions for BVP (5.7)–(5.8). Some of these behaviors, as we will see, can be controlled in terms

of the functions φ and ω given in the definition of the (φ, ω)-derivative, which can be interpreted

as behaviors scaled for the (φ, ω)-derivative.

The main idea is to replace the integral term (5.15) with some condition which allows found a

bound for it. For instance if we assume that f is uniformly bounded by A > 0 on [0, 1] × R, then
use the estimate |f(s, x(s))| ≤ A in (5.15) and obtain the radius r ≥ 2|β|+4|λ| MΩAK +

∣∣α
λ

∣∣∆.
If we assume that |f(s, y)| ≤ A |φ(s)|

w′(s)
, for some A > 0, for all (s, x) ∈ [0, 1] × R, the integral term

is less or equal to A and the radius is r ≥ 2|β|+4|λ| A +
∣∣α
λ

∣∣∆.
Finally, if |f(s, x(s))| ≤

|γ|
2(|β| + 2)MΩ

s|x(s)|, and we assume that

∣∣∣∣αλ
∣∣∣∣∆ ≤ r2, for each r > 0 given.

Then, estimate (5.15) is rewrite as

|Tx(t)| ≤
∥x∥∞
2

+

∣∣∣∣αβ
∣∣∣∣∆ ≤ r2 + r2 = r.

This proves that T applies any ball of radius r into itself. Therefore, we conclude that BVP

(5.7)–(5.8) as at leat one solution on each ball of radius r.

In similar fashion it can be proved that for

|f(s, y)| ≤
|γ|

4(|β| + 2)MΩ( |an|
n+1

+ · · · + |a0|)
|ansn + · · · + a0|)|x(s)|,



536 R. E. Castillo & B. Sultan CUBO
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and ∣∣∣∣αλ
∣∣∣∣∆ ≤ r2, for each r > 0,

the same conclusion holds.



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A derivative-type operator and its application to the solvability... 537

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	Motivation
	Derivative of variable order
	Integration of variable order
	Taylor formula
	Three-point boundary value problems of variable order