International Journal of Analysis and Applications
ISSN 2291-8639
Volume 12, Number 2 (2016), 80-86
http://www.etamaths.com

ON EXISTENCE OF SOLUTIONS TO THE CAPUTO TYPE FRACTIONAL

ORDER THREE-POINT BOUNDARY VALUE PROBLEMS

B.M.B. KRUSHNA1,∗ AND K.R. PRASAD2

Abstract. In this paper, we establish the existence of solutions to the fractional order three-point

boundary value problems by utilizing Banach contraction principle and Schaefer’s fixed point theo-

rem.

1. Introduction

Fractional calculus deals with integration and differentiation of an arbitrary real order. Fractional
order derivatives provide a new approach for modeling complex phenomena in physics, mechanics,
control systems, flow in porous media, signal and image processing, aerodynamics, electromagnetics,
viscoelasticity and especially dealing with memory and hereditary properties of various real materials
[5, 6, 9, 11]. Compared to integer order models, the fractional order models offer better description of
underlying processes. In consequence, fractional order differential equations have achieved great deal
of interest and attention of researchers [1, 2, 4, 7, 10, 12, 13, 14, 16, 17, 18].

Zhang [20] obtained the existence and uniqueness of solutions for two-point fractional order bound-
ary value problem

Dα0+u(t) = f
(
t,u(t)

)
, t ∈ (0, 1),

u(0) + u′(0) = 0, u(1) + u′(1) = 0,

where Dα
0+

is the Caputo fractional order derivative and 1 < α ≤ 2.
Shi and Zhang [17] given sufficient conditions for the existence of at least one solution for fractional

order boundary value problem

Dδu(t) + g(t,u) = 0, t ∈ (0, 1),
u(0) = a, u(1) = b,

where 1 < δ ≤ 2, g : [0, 1] ×R → R and Dδ is the Caputo fractional order derivative, using upper and
lower solutions method.

Benchohra, Hamani and Ntouyas [3] derived sufficient conditions for the existence of at least one
solution to the fractional order boundary value problem

CDαu(t) = f
(
t,u(t)

)
, 0 < t < T,

u(0) = g(u), u(T) = uT ,

where 1 < α ≤ 2, CDα is the Caputo fractional order derivative, by using Schaefer’s fixed point
theorem. Also they established criteria for the uniqueness of solutions by virtue of the Banach fixed
point theorem.

In [8], the authors studied the existence and uniqueness of solutions to the boundary value problem

Dαu(t) = f(t,u(t),u′(t)), t ∈ (0, 1),
u(0) = 0, Dpu(1) = δDpu(η),

2010 Mathematics Subject Classification. 26A33, 34A08, 34B15.

Key words and phrases. fractional order derivative; boundary value problem; solution.

c©2016 Authors retain the copyrights of
their papers, and all open access articles are distributed under the terms of the Creative Commons Attribution License.

80



EXISTENCE OF SOLUTIONS TO THE CAPUTO TYPE FBVPS 81

where D is the Caputo fractional order derivative, 1 < α ≤ 2, 0 < δ < p < 1, 0 < η ≤ 1, by using some
fixed point theorem.

In this paper we consider the fractional order three-point boundary value problem

(1.1) CDqy(t) = f
(
t,y(t)

)
, t ∈ (0, 1),

(1.2)



y(0) + y′(0) + y′′(0) =0,

y(η) + y′(η) + y′′(η) =0,

y(1) + y′(1) + y′′(1) =0,

where q ∈ (2, 3],η ∈ (0, 1) and CDq is the standard Caputo fractional order derivative.

We assume that the following conditions hold throughout the paper:

(A1) f : [0, 1] ×R → R is continuous,
(A2) there exists a positive constant M such that

∣∣f(t,y)∣∣ ≤M, for each t ∈ [0, 1] and y ∈ R,
(A3) there exists a positive real constant ξ such that |f(t,y) −f(t,y)| ≤ ξ|y −y| for each t ∈ [0, 1],

for all y, y ∈ R.

1.1. Preliminaries. In this section, we present some definitions and lemmas that are useful in the
proof of our main results.

Definition 1.1. [11] The Riemann–Liouville fractional integral of order p > 0 of a function f :
[0, +∞) → R is given by

I
p
0+
f(t) =

1

Γ(p)

∫ t
0

(t− τ)p−1f(τ)dτ,

provided the right-hand side is defined.

Definition 1.2. [11] The Caputo fractional derivative of order α > 0 of a function f : [a, +∞) → R
is given by

C
a D

α
t f(t) =

1

Γ(α−n)

∫ t
a

f(n)(τ)

(t− τ)α+1−n
dτ, (n− 1 < α < n),

provided the right-hand side is defined.

Lemma 1.1. [11] Let α > 0, then the fractional order differential equation

CDαu(t) = 0

has solution, u(t) = c0 + c1t + c2t
2 + · · · + cn−1tn−1, ci ∈ R, i = 0, 1, 2, · · ·,n− 1, n = [α] + 1.

Lemma 1.2. [11] Let α > 0, then Iα CDαu(t) = u(t) + c0 + c1t + c2t
2 + · · ·+ cn−1tn−1 for ci ∈ R, i =

0, 1, 2, · · ·,n− 1, n = [α] + 1.

The rest of the paper is organized as follows. In Section 2, we establish the conditions for the
existence of solutions to the fractional order boundary value problem (1.1)-(1.2) by using different
fixed point theorems such as Banach contraction principle and Schaefer’s fixed point theorem. In
Section 3, we present examples to illustrate the applicability of the conditions.

2. Main results

In this section, by stating some lemmas, we establish sufficient conditions for the existence of solu-
tions to the fractional order boundary value problem (1.1)-(1.2) using certain fixed point theorems.

Let B = Cq
(
[0, 1],R

)
be the Banach space of all continuous functions from [0, 1] into R equipped

with the norm
‖y‖ = max

t∈[0,1]

∣∣y(t)∣∣.



82 KRUSHNA AND PRASAD

Lemma 2.1. Let ∆ = (η2 − η)Γ(q) 6= 0. Assume that condition (A1) is satisfied. A function y ∈
Cq
(
[0, 1],R

)
is a solution of the fractional integral equation

(2.1)



y(t) =

∫ t
0

(t−s)q−1

Γ(q)
f
(
s,y(s)

)
ds +

[
t2 − 3t + 1

∆

]
×
∫ η
0

Φ(s)f
(
s,y(s)

)
ds

+

[
η2(t− 1) + ηt(2 − t)

∆

]
×
∫ 1
0

Ψ∗(s)f
(
s,y(s)

)
ds,

where

(2.2)

{
Φ(s) =(η −s)q−1 + (q − 1)(η −s)q−2 +

(
q2 − 3q + 2

)
(η −s)q−3,

Ψ∗(s) =(1 −s)q−1 + (q − 1)(1 −s)q−2 +
(
q2 − 3q + 2

)
(1 −s)q−3,

if and only if y is a solution of the fractional order boundary value problem (1.1)-(1.2).

Proof. Let y(t) ∈ Cq[0, 1] be the solution of fractional order boundary value problem (1.1)-(1.2). An
equivalent integral equation for (1.1) and is given by

(2.3) y(t) =

∫ t
0

(t−s)q−1

Γ(q)
f(s,y(s))ds + k1 + k2t + k3t

2.

Using the conditions (1.2) to y(t) in (2.3) and by algebraic calculations, we can get

k1 =
1

∆

[∫ η
0

Φ(s)f
(
s,y(s)

)
ds−η2

∫ 1
0

Ψ∗(s)f
(
s,y(s)

)
ds
]

k2 =
1

∆

[
(η2 + 2η)

∫ 1
0

Ψ∗(s)f
(
s,y(s)

)
ds− 3

∫ 1
0

Φ(s)f
(
s,y(s)

)
ds
]

k3 =
1

∆

[∫ η
0

Φ(s)f
(
s,y(s)

)
ds−η

∫ 1
0

Ψ∗(s)f
(
s,y(s)

)
ds
]
.

By substituting the values of k1,k2,k3 in (2.3), we obtain (2.1). This completes the proof. �

Now we establish the existence of solution to the fractional order boundary value problem (1.1)-(1.2)
by an application of Banach contraction principle [15].

Lemma 2.2. (Banach contraction principle) Let (X,d) be a complete metric space. Let T : X →
X be a contraction. That is, there exists λ ∈ [0, 1) such that d(Tx,Ty) ≤ λd(x,y), for all x,y ∈ X.
Then there exists a unique fixed point of T.

Theorem 2.3. Assume that the conditions (A1) and (A3) are satisfied. If

(2.4) ζ =

∣∣∣∣ξ(η + 1)(ηq−2 + 1)Γ(q + 1)(η − 1)
∣∣∣∣ < 1,

then the fractional order boundary value problem (1.1)-(1.2) has a solution on [0, 1].

Proof. We transform the fractional order boundary value problem (1.1)-(1.2) in to a fixed point problem
by defining an operator T.

Let T : B →B be the operator defined by

(2.5)



Ty(t) =

∫ t
0

(t−s)q−1

Γ(q)
f
(
s,y(s)

)
ds−

[t2 − 3t + 1
∆

]
×
∫ η
0

Φ(s)f
(
s,y(s)

)
ds

−
[η2(t− 1) + η(2 − t)t

∆

]
×
∫ 1
0

Ψ∗(s)f
(
s,y(s)

)
ds.

Clearly the fixed points of the operator T are the solutions of the fractional order boundary value
problem (1.1)-(1.2). Now we show that the operator T is a contraction mapping. Let y, y ∈ B and



EXISTENCE OF SOLUTIONS TO THE CAPUTO TYPE FBVPS 83

using (2.2), then for each t ∈ [0, 1], we have

∣∣Ty(t) −Ty(t)∣∣ ≤∫ t
0

(t−s)q−1

Γ(q)

∣∣∣f(s,y(s))−f(s,y(s))∣∣∣ds
+

∣∣∣∣t2 − 3t + 1∆
∣∣∣∣×
∫ η
0

Φ(s)
∣∣∣f(s,y(s))−f(s,y(s))∣∣∣ds

+

∣∣∣∣η2(1 − t) + η(t− 2)t∆
∣∣∣∣×
∫ 1
0

Ψ∗(s)
∣∣∣f(s,y(s))−f(s,y(s))∣∣∣ds

≤

{
(t−s)q

qΓ(q)
+

1

|∆|
×
∫ η
0

Φ(s)ds +
∣∣∣ η
∆

∣∣∣×∫ 1
0

Ψ∗(s)ds

}[
ξ‖y −y‖∞

]
≤
∣∣∣∣ξ(ηq−2 + 1)(η + 1)(η − 1)Γ(q + 1)

∣∣∣∣∥∥y −y∥∥∞.
Hence ‖Ty(t)−Ty(t)‖∞ ≤ ζ‖y−y‖. Therefore T is a contraction mapping. By the contraction mapping
principle, the operator T has a fixed point and is the solution of the fractional order boundary value
problem (1.1)-(1.2). �

Next we establish the result by applying Schaefer’s fixed point theorem [19].

Lemma 2.4. (Schaefer′s fixed point theorem) Let F : X → X be a completely continuous op-
erator. If the set E(F) =

{
x ∈ X : x = λ∗Fx, for some λ∗ ∈ [0, 1]

}
is bounded. Then F has fixed

points.

Theorem 2.5. Assume that the conditions (A1)-(A2) are satisfied. Then the fractional order boundary
value problem (1.1)-(1.2) has at least one solution on [0, 1].

Proof. We prove that the operator T is defined by (2.5) has a fixed point by utilizing Schaefer’s fixed
point theorem. Now we establish the result in 4 steps.
Step 1. The operator T given by (2.5) is continuous. Let {yn} be a sequence such that yn → y in B.
Using (2.2) and for each t ∈ [0, 1], we have

∣∣Tyn(t) −Ty(t)∣∣ ≤∫ t
0

(t−s)q−1

Γ(q)

∣∣∣f(s,yn(s))−f(s,y(s))∣∣∣ds + ∣∣∣t2 − 3t + 1
∆

∣∣∣×∫ η
0

Φ(s)
∣∣∣f(s,yn(s))−f(s,y(s))∣∣∣ds + ∣∣∣∣η2(1 − t) + ηt(t− 2)∆

∣∣∣∣×∫ 1
0

Ψ∗(s)
∣∣∣f(s,yn(s))−f(s,y(s))∣∣∣ds

≤
∫ t
0

(t−s)q−1

Γ(q)
max
s∈[0,1]

∣∣∣f(s,yn(s))−f(s,y(s))∣∣∣ds+
1

|∆|
×
∫ η
0

Φ(s) max
s∈[0,1]

∣∣∣f(s,yn(s))−f(s,y(s))∣∣∣ds+
|η|
|∆|
×
∫ 1
0

Ψ∗(s) max
s∈[0,1]

∣∣∣f(s,yn(s)) −f(s,y(s))∣∣∣ds
≤
‖f
(
·,yn(·)

)
−f

(
·,y(·)

)
‖

Γ(q)
×
∫ t
0

(t−s)q−1ds+

1

|η −η2|
×
∫ η
0

Φ(s)ds +

∣∣∣∣ ηη −η2
∣∣∣∣×
∫ 1
0

Ψ∗(s)ds

≤
∣∣∣∣(η + 1)(ηq−2 + 1)Γ(q + 1)

∣∣∣∣∥∥∥f(·,yn(·))−f(·,y(·))∥∥∥.



84 KRUSHNA AND PRASAD

Since f is continuous, we have

∥∥∥Tyn(t) −Ty(t)∥∥∥ ≤ ∣∣∣∣(η + 1)(ηq−2 + 1)Γ(q + 1)(n− 1)
∣∣∣∣∥∥∥f(·,yn(·))−f(·,y(·))∥∥∥ → 0

as n →∞.

Therefore the operator T is continuous.
Step 2. The operator T maps bounded sets in to bounded sets in B. Now we will show that for any
ϑ > 0, there exits a positive constant m∗ such that for each y ∈ Bϑ =

{
y ∈ B : ‖y‖∞ ≤ ϑ

}
, we have

‖Ty‖∞ ≤ m∗. By the condition (A2) and (2.2), for each t ∈ [0, 1], we have

|Ty(t)| ≤
∫ t
0

(t−s)q−1

Γ(q)

∣∣∣f(s,y(s))∣∣∣ds + ∣∣∣∣t2 − 3t + 1∆
∣∣∣∣×
∫ η
0

Φ(s)
∣∣∣f(s,y(s))∣∣∣ds

+

∣∣∣∣η(1 − t) + ηt(t− 2)∆
∣∣∣∣×
∫ 1
0

Ψ∗(s)f
(
s,y(s)

)
ds

≤M

{∫ t
0

(t−s)q−1

Γ(q)
ds +

1

|∆|
×
∫ η
0

Φ(s)ds +
∣∣∣ η
∆

∣∣∣×∫ 1
0

Ψ∗(s)ds

}

≤M
∣∣∣∣(η + q)(1 + ηq−2)Γ(q + 1)(η − 1)

∣∣∣∣ .
Thus

‖Ty‖∞ ≤M
∣∣∣∣(η + q)(1 + ηq−2)Γ(q + 1)(η − 1)

∣∣∣∣ = m∗.
Step 3. The operator T maps bounded sets into equicontinuous sets B. Let t1, t2 ∈ (0, 1], t1 < t2, Bϑ
be a bounded set of B as in Step 2 and y ∈ Bϑ. Then

|Ty(t2) −Ty(t1)| ≤
∫ t1
0

[(t2 −s)q−1 − (t1 −s)q−1]
Γ(q)

f
(
s,y(s)

)
ds−

∫ t2
t1

(t2 −s)q−1

Γ(q)
f
(
s,y(s)

)
ds

−
[

(t22 − 3t2 + 1) − (t21 − 3t1 + 1)
∆

]
×
∫ η
0

f
(
s,y(s)

)
Φ(s)ds

−
[
η2(t22 − t21) + η(t21 − t22 + 2t2 − 2t1)

∆

]
×
∫ 1
0

f
(
s,y(s)

)
ds

≤M

{[(t2 − t1)q − (t2 − t1)q
Γ(q)

]
−

(t22 − t21) − 3(t2 − t1)
∆
[
ηq + qηq−1 + q(q − 1)ηq−2

]
−

(1 + q2)η2
[
(η − 1)(t22 − t21) + (2t2 − 2t1)

]
∆

}
.

As t2 → t1, the right hand side of the above inequality tends to zero. From Steps 1 to 3, we can
conclude that T : B →B is completely continuous.
Step 4. The set A =

{
y ∈B : y = λTy for some 0 < λ < 1

}
is bounded. Let y ∈ A, then y = λTy

for some 0 < λ < 1. Therefore, for each t ∈ [0, 1], we have

y(t) =λ

{∫ t
0

(t−s)q−1

Γ(q)
f
(
s,y(s)

)
ds−

[
t2 − 3t + 1

∆

]
×
∫ η
0

Φ(s)f
(
s,y(s)

)
ds

−
[
η2(t− 1) + ηt(2 − t)

∆

]
×
∫ 1
0

Ψ∗(s)f
(
s,y(s)

)
ds

}



EXISTENCE OF SOLUTIONS TO THE CAPUTO TYPE FBVPS 85

and ∣∣∣Ty(t)∣∣∣ ≤Mλ
{∫ t

0

(t−s)q−1

Γ(q)
ds +

∣∣∣∣t2 − 3t + 1∆
∣∣∣∣×
∫ η
0

Φ(s)ds

+

∣∣∣∣η2(t− 1) + ηt(2 − t)∆
∣∣∣∣×
∫ 1
0

Ψ∗(s)ds

}

≤Mλ

{∫ t
0

(t−s)q−1

Γ(q)
ds +

∫ η
0

Φ(s)

∆
ds + η

∫ 1
0

Ψ∗(s)

∆
ds

}

≤
M(1 + η)(1 + ηq−1)

Γ(q + 1)
.

Thus for every t ∈ [0, 1], we have ∥∥Ty∥∥∞ ≤M
∣∣∣∣(1 + η)(1 + ηq−1)Γ(q + 1)

∣∣∣∣ .
This shows that the set A is bounded. By Schaefer’s fixed point theorem, the operator T has a fixed
point which is a solution of the fractional order boundary value problem (1.1)-(1.2). �

3. Examples

In this section, we give two examples to illustrate the usefulness of our main results.

Example 3.1 Consider the fractional order three-point boundary value problem,

(3.1) CD2.6y(t) =
3e−2ty

(19 + 5e2t)(3 + y)
, t ∈ (0, 1),

(3.2)




y(0) + y′(0) + y′′(0) =0,

y
(2

5

)
+ y′

(2
5

)
+ y′′

(2
5

)
=0,

y(1) + y′(1) + y′′(1) =0.

For y,y ∈ [0,∞) and t ∈ [0, 1], we have

|f(t,y) −f(t,y)| =
3e−2t

19 + 5e2t

∣∣∣∣ y3 + y − y3 + y
∣∣∣∣ ≤ 924∣∣y −y∣∣.

Now we check the condition (2.4) with ξ = 9
24
,

ζ =

∣∣∣∣ 924
(
2
5

+ 1
)[(

2
5

)0.6
+ 1
)]

Γ(3.6)
(−3

5

) ∣∣∣∣ = 0.42 < 1.
Then all conditions of Theorem 2.3 are satisfied. Thus by Theorem 2.3 the fractional order boundary
value problem (3.1)-(3.2) has unique solution on [0, 1].
Example 3.2 Consider the fractional order three-point boundary value problem,

(3.3) CD2.5y(t) =
e−ty

(1 + et)(1 + y)
, t ∈ (0, 1),

(3.4)




y(0) + y′(0) + y′′(0) =0,

y
(1

2

)
+ y′

(1
2

)
+ y′′

(1
2

)
=0,

y(1) + y′(1) + y′′(1) =0.

By simple algebraic calculations, one can determine m∗ =

∣∣∣∣ 32
(
1 + 0.50.5

)]
Γ(3.5)

(−1
2

) ∣∣∣∣ = 1.6 and M = 12. Then
the fractional order boundary value problem (3.3)-(3.4) satisfies all conditions of Theorem 2.5. Hence
it has unique solution on [0, 1].



86 KRUSHNA AND PRASAD

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1Department of Mathematics, MVGR College of Engineering (Autonomous), Vizianagaram, 535 005, India

2Department of Applied Mathematics, Andhra University, Visakhapatnam, 530 003, India

∗Corresponding author: muraleebalu@yahoo.com