andJessadaTariboon SotirisK.Ntouyas ,SorasakLaoprasittichok HYBRIDFRACTIONALINTEGRO-DIFFERENTIALINCLUSIONS ( 2015 ) 151–164doi:10.7151/dmdico.1174 DiscussionesMathematicaeDifferentialInclusions,ControlandOptimization35

doi:10.7151/dmdico.1174

HYBRID FRACTIONAL INTEGRO-DIFFERENTIAL INCLUSIONS

Sotiris K. Ntouyas

^a,b

, Sorasak Laoprasittichok

^c

and

Jessada Tariboon

^c,1

a

Department of Mathematics, University of Ioannina 451 10 Ioannina, Greece

b

Nonlinear Analysis and Applied Mathematics (NAAM)-Research Group Department of Mathematics, Faculty of Science, King Abdulaziz University

P.O. Box 80203, Jeddah 21589, Saudi Arabia

c

Nonlinear Dynamic Analysis Research Center Department of Mathematics, Faculty of Applied Science King Mongkut’s University of Technology North Bangkok

Bangkok 10800, Thailand e-mail: sntouyas@uoi.gr

sorasak kmutnb@hotmail.com jessada.t@sci.kmutnb.ac.th

Abstract

In this paper we study an existence result for initial value problems for hybrid fractional integro-differential inclusions. A hybrid fixed point theorem for a sum of three operators due to Dhage is used. An example illustrating the obtained result is also presented.

Keywords: fractional differential equations, hybrid differential inclusions, fixed point theorems.

2010 Mathematics Subject Classification: 34A60, 34A08, 34A12.

1

Corresponding author.

1. Introduction

In recent years, initial and boundary value problems of nonlinear fractional dif- ferential equations and inclusions have been studied by many researchers. Frac- tional differential equations appear naturally in various fields of science and en- gineering, uch as physics, chemistry, aerodynamics, electrodynamics of complex medium, polymer rheology, economics, control theory, signal and image process- ing, biophysics, blood flow phenomena, etc. [1]–[6], and constitute an important field of research. For some recent development on the topic, see [7]–[15] and the references therein.

Hybrid fractional differential equations have also been studied by several re- searchers. This class of equations involves the fractional derivative of an unknown function hybrid with the nonlinearity depending on it. Some recent results on hybrid differential equations can be found in a series of papers ([16]–[21]).

In this paper we study existence results for initial value problems for hybrid fractional integro-differential inclusions

(1.1)



 

 



 

 

 D

^α







 x(t) −

m

P

i=1

I

^βi

h

i

(t, x(t)) f (t, x(t), I

^γ

x(t))









∈ F (t, x(t)), t ∈ J := [0, T ],

x(0) = 0,

where D

^α

denotes the Riemann-Liouville fractional derivative of order α, 0 <

α ≤ 1, I

^φ

is the Riemann-Liouville fractional integral of order φ > 0, φ ∈ {β

₁

, β

₂

, . . . , β

_m

, γ}, f ∈ C(J × R

²

, R \ {0}), F : J × R → P(R) is a multi-valued map (P(R) is the family of nonempty subjects of R) and h

i

∈ C(J × R, R) with h

i

(0, 0) = 0, i = 1, 2, . . . , m.

An existence result is obtained for the initial value problem (1.1) by using a hybrid fixed point theorem for three operators in a Banach algebra due to Dhage [22]. In the case of single valued maps, i.e., F = {f }, an initial value problem for hybrid fractional integro-differential equations was studied recently in [23]. Here we extend the results of [23] to cover the multi-valued case.

The rest of the paper is organized as follows: In Section 2 we recall some useful preliminaries. In Section 3 we study the existence of the initial value problem (1.1), while an example illustrating the obtained result is presented in Section 4.

2. Preliminaries

In this section, we introduce some notations and definitions of fractional calculus

[1, 5] and present preliminary results needed in our proofs later.

Definition 2.1. The Riemann-Liouville fractional derivative of order q > 0 of a continuous function f : (0, ∞) → R is defined by

D

^q

f (t) = 1 Γ(n − q)

 d dt



n

Z

t 0

(t − s)

^n−q−1

f (s)ds, n − 1 < q < n,

where n = [q] + 1, [q] denotes the integer part of a real number q, provided the right-hand side is point-wise defined on (a, ∞), where Γ is the gamma function defined by Γ(q) = R

∞

0

e

^−s

s

^q−1

ds.

Definition 2.2. The Riemann-Liouville fractional integral of order p > 0 of a continuous function f : (0, ∞) → R is defined by

I

^p

f (t) = 1 Γ(p)

Z

t 0

(t − s)

^p−1

f (s)ds, provided the right-hand side is point-wise defined on (0, ∞).

Lemma 2.1 [1]. Let q > 0 and x ∈ C(0, T ) ∩ L(0, T ). Then the fractional differential equation

D

^q

x(t) = 0 has a unique solution

x(t) = k

₁

t

^q−1

+ k

₂

t

^q−2

+ · · · + k

_n

t

^q−n

, where k

i

∈ R, i = 1, 2, . . . , n, and n − 1 < q < n.

Lemma 2.2 [1]. Let q > 0. Then for x ∈ C(0, T ) ∩ L(0, T ) it holds

I

^q

D

^q

x(t) = x(t) −

n

X

j=1

(I

^n−q

x)

^(n−j)

(0) Γ(q − j + 1) t

^q−j

, where n − 1 < q < n.

Let E = C(J, R) be the space of continuous real-valued functions defined on J = [0, T ]. Define a norm k · k and a multiplication in E by

kxk = sup

t∈J

|x(t)| and (xy)(t) = x(t)y(t), ∀t ∈ J.

Clearly E is a Banach algebra with respect to above supremum norm and the multiplication in it.

Next we recall some basic definitions of multivalued analysis.

For a normed space (A, k · k), let P

cl

(A) = {A

1

∈ P(A) : A

₁

is closed}, P

b

(A) = {A

₁

∈ P(A) : A

₁

is bounded}, P

cp

(A) = {A

1

∈ P(A) : A

₁

is compact}, and P

_cp,c

(A) = {A

₁

∈ P(A) : A

₁

is compact and convex}. A multi-valued map G : A → P(A) is convex (closed) valued if G(a) is convex (closed) for all a ∈ A.

The map G is bounded on bounded sets if G(B) = ∪

x∈B

G(x) is bounded in A for all B ∈ P

b

(A) (i.e., sup

_x∈B

{sup{|y| : y ∈ G(x)}} < ∞). G is called upper semi-continuous (u.s.c.) on A if for each a

₀

∈ A, the set G(a

₀

) is a nonempty closed subset of A, and if for each open set N of A containing G(a

0

), there exists an open neighborhood N

₀

of a

₀

such that G(N

₀

) ⊆ N. G is said to be completely continuous if G(B) is relatively compact for every B ∈ P

b

(A). If the multi-valued map G is completely continuous with nonempty compact values, then G is u.s.c.

if and only if G has a closed graph, i.e., a

_n

→ a

_∗

, b

_n

→ b

_∗

, b

_n

∈ G(a

_n

) imply b

∗

∈ G(a

∗

). G has a fixed point if there is a ∈ A such that a ∈ G(a). The fixed point set of the multivalued operator G will be denoted by Fix G. A multivalued map G : J → P

_cl

(R) is said to be measurable if for every b ∈ R, the function t 7−→ d(b, G(t)) = inf{|b − c| : c ∈ G(t)} is measurable.

Let C(J, R) denote the Banach space of continuous functions from J into R with the norm kxk = sup

_t∈J

|x(t)|. Let L

¹

(J, R) be the Banach space of mea- surable functions x : J → R which are Lebesgue integrable and normed by kxk

_L1

= R

T

0

|x(t)|dt.

For each y ∈ C(J, R), define the set of selections of F by S

F,x

:= {v ∈ L

¹

(J, R) : v(t) ∈ F (t, x(t)) for a.e. t ∈ J}.

For the forthcoming analysis, we need the following lemma.

Lemma 2.3 ([24]). Let X be a Banach space. Let F : J × X × X → P

cp,c

(X) be an L

¹

− Carath´ eodory multivalued map and let Θ be a linear continuous mapping from L

¹

(J, X) to C(J, X). Then the operator

Θ ◦ S

_F

: C(J, X) → P

_cp,c

(C(J, X)), x 7→ (Θ ◦ S

_F

)(x) = Θ(S

_F,x

) is a closed graph operator in C(J, X) × C(J, X).

3. Main result

In this section we prove our main result for the initial value problem (1.1). The following hybrid fixed point theorem for three operators in a Banach algebra E, due to Dhage [22], will be used to prove the existence result for the initial value problem (1.1).

Lemma 3.1. Let X be a Banach algebra and let A, C : X → X and B : X →

P

_cp,c

(X) be three operators satisfying:

(a

1

) A and C are Lipschitzian with Lipschitz constants δ and ρ, respectively, (b

₁

) B is compact and upper semi-continuous,

(c

₁

) δM + ρ < 1/2, where M = k ∪ B(X)k

P

. Then either

(i) the operator inclusion x ∈ AxBx + Cx has a solution, or (ii) the set E = {u ∈ X : λu ∈ AuBu + Cu, λ > 1} is unbounded.

Lemma 3.2. Let y : J → R be a continuous function. The unique solution of the hybrid fractional integro-differential problem

(3.1)



 

 



 

 

 D

^α







 x(t) −

P

m i=1

I

^βi

h

_i

(t, x(t)) f (t, x(t), I

^γ

x(t))









= y(t), t ∈ J := [0, T ], x(0) = 0,

is given by

(3.2) x(t) = f (t, x(t), I

^γ

x(t)) Z

t

0

(t − s)

^α−1

Γ(α) y(s)ds +

m

X

i=1

I

^βi

h

i

(t, x(t)), t ∈ J.

Proof. Applying the Riemann-Liouville fractional integral of order α to both sides of (3.1) and using Lemma 2.2, we have







 x(t) −

m

P

i=1

I

^βi

h

i

(t, x(t)) f (t, x(t), I

^γ

x(t))









− t

^α−1

Γ(α) I

^1−α

 x(t) − P

m

i=1

I

^βi

h

_i

(t, x(t)) f (t, x(t)), I

^γ

x(t)



t=0

= I

^α

y(t).

Since x(0) = 0, h

_i

(0, 0) = 0, i = 1, 2, . . . , m and f (0, 0, 0) 6= 0, it follows that x(t) = f (t, x(t), I

^γ

x(t))I

^α

y(t) +

m

X

i=1

I

^βi

h

_i

(t, x(t)).

Thus (3.2) holds. The proof is completed.

Definition 3.1. A function x ∈ C

¹

(J, R) is called a solution of problem (1.1) if there exists a function v ∈ L

¹

(J, R) with v(t) ∈ F (t, x(t)), a.e. J such that x(0) = 0 and

x(t) = f (t, x(t), I

^γ

x(t)) Z

t

0

(t − s)

^α−1

Γ(α) v(s)ds +

m

X

i=1

I

^βi

h

i

(t, x(t)), t ∈ J.

Theorem 3.1. Assume that:

(H

1

) The functions f : J × R

²

→ R \ {0} and h

i

: J × R → R, h

i

(0, 0) = 0, i = 1, 2, . . . , m, are continuous and there exist two positive functions φ, ψ

i

, i = 1, 2, . . . , m with bounds kφk and kψ

_i

k, i = 1, 2, . . . , m, respectively, such that (3.3) |f (t, x

₁

, x

₂

) − f (t, y

₁

, y

₂

)| ≤ φ(t)(|x

₁

− x

₂

| + |y

₁

− y

₂

|),

and

(3.4) |h

_i

(t, x) − h

_i

(t, y)| ≤ ψ

_i

(t)|x − y|, i = 1, 2, . . . , m, for t ∈ J and x, y, x

₁

, x

₂

, y

₁

, y

₂

∈ R.

(H

₂

) There exists a continuous function p : J → (0, ∞) such that t → F (t, x) is measurable and

kF (t, x)k ≤ p(t), for almost all t ∈ J and x ∈ R.

(H

3

)

(3.5) kφk



1 + T

^γ

Γ(γ + 1)



kpk T

^α

Γ(α + 1) +

m

X

i=1

kψk

_i

T

^βi

Γ(β

_i

+ 1) < 1.

Then the problem (1.1) has at least one solution on J .

Proof. Set X = C(J, R) and define three operators A : X → X by Ax(t) = f (t, x(t), I

^γ

x(t)), t ∈ J,

(3.6)

B : X → P(X) by Bx =



u ∈ X : u(t) = Z

t

0

(t − s)

^α−1

Γ(α) v(s)ds, v ∈ S

_F,x

 , (3.7)

and C : X → X by Cx(t) =

m

X

i=1

I

^βi

h

i

(t, x(t))

=

m

X

i=1

Z

t 0

(t − s)

^βi−1

Γ(β

_i

) h

_i

(s, x(s))ds, t ∈ J.

(3.8)

Then the problem (1.1) is transformed into an operator inclusion as x ∈ AxBx + Cx.

(3.9)

We shall show that the operators A, B and C satisfy all the conditions of Lemma 3.1. This will be achieved in the series of the following steps.

First we show that the operators A and C define single-valued operators A, C : X → X and B : X → P

_cp,c

(X). The claim concerning A and C is obvious, because the functions f and h

i

, i = 1, 2, . . . , m are continuous on J × R

²

and J × R respectively. We only prove the claim for the multi-valued operator B on X. Note that the operator B is equivalent to the composition L ◦ S

F

, where L is the continuous linear operator on L

¹

(J, R) into C(J, R), defined by

L(v)(t) = Z

t

0

(t − s)

^α−1

Γ(α) v(s)ds.

Suppose that x ∈ X is arbitrary and let {v

n

} be a sequence in S

_F,x

. Then, by definition of S

F,x

, we have v

n

(t) ∈ F (t, x(t)) for almost all t ∈ J . Since F (t, x(t)) is compact for all t ∈ J , there is a convergent subsequence of {v

_n

(t)} (we denote it by {v

_n

(t)} again) that converges in measure to some v(t) ∈ S

_F,x

for almost all t ∈ J . On the other hand, L is continuous, so L(v

n

)(t) → L(v)(t) pointwise on J . In order to show that the convergence is uniform, we have to show that {L(v

_n

)} is an equi-continuous sequence. Let t

₁

, t

₂

∈ [0, 1] with t

₁

< t

₂

. Then, we have

|L(v

_n

)(t

₂

) − L(v

_n

)(t

₁

)| ≤     

Z

t1

0

[(t

₂

− s)

^α−1

− (t

₁

− s)

^α−1

]

Γ(α) v

_n

(s)ds

+ Z

t2

t1

(t

₂

− s)

^α−1

Γ(α) v

n

(s)ds      .

We see that the right hand of the above inequality tends to zero as t

₂

→ t

₁

. Thus, the sequence {L(v

n

)} is equi-continuous and by the Arzel´ a-Ascoli theorem, we get that there is a uniformly convergent subsequence. So, there is a subsequence of {v

_n

} (we denote it again by {v

_n

}) such that L(v

_n

) → L(v). Note that, L(v) ∈ L(S

_F,x

). Hence, B(x) = L(S

F,x

) is compact for all x ∈ X. So B(x) is compact.

Now, we show that B(x) is convex for all x ∈ X. Let z

₁

, z

₂

∈ B(x). We select f

₁

, f

₂

∈ S

_F,x

such that

z

_i

(t) = Z

t

0

(t − s)

^α−1

Γ(α) f

_i

(s)ds, i = 1, 2, for almost all t ∈ [0, 1]. Let 0 ≤ λ ≤ 1. Then, we have

[λz

1

+ (1 − λ)z

2

](t) = Z

t

0

(t − s)

^α−1

Γ(α) [λf

1

(s) + (1 − λ)f

2

(s)]ds.

Since F has convex values, so S

F,u

is convex and λf

1

(s) + (1 − λ)f

2

(s) ∈ S

F,x

. Thus

λz

1

+ (1 − λ)z

2

∈ B(x).

Consequently, B is convex-valued.

Step 1. We show that A and C are Lipschitz on X.

Let x, y ∈ X. Then by (H

1

), for t ∈ J we have

|Ax(t) − Ay(t)| = |f (t, x(t), I

^γ

x(t)) − f (t, y(t), I

^γ

y(t))|

≤ φ(t)(|x(t) − y(t)| + |I

^γ

x(t) − I

^γ

y(t)|)

≤ kφk



1 + T

^γ

Γ(γ + 1)



kx − yk,

which implies kAx − Ayk ≤ kφk 

1 +

_Γ(γ+1)^Tγ



kx − yk for all x, y ∈ X. Therefore, A is a Lipschitz on X with Lipschitz constant kφk 

1 +

_Γ(γ+1)^Tγ

 . Analogously, for any x, y ∈ X, we have

|Cx(t) − Cy(t)| =     

m

X

i=1

I

^βi

h

_i

(t, x(t)) −

m

X

i=1

I

^βi

h

_i

(t, y(t))     

≤

m

X

i=1

Z

t 0

(t − s)

^βi−1

Γ(β

_i

) ψ

_i

(s)|x(s) − y(s)|ds

≤ kx − yk

m

X

i=1

kψ

_i

kT

^βi

Γ(β

i

+ 1) . This means that

kCx − Cyk ≤

m

X

i=1

kψ

_i

kT

^βi

Γ(β

_i

+ 1) kx − yk.

Thus, C is a Lipschitz on X with Lipschitz constant P

m i=1

kψ_ikTβi

Γ(βi+1)

. Step 2. The operator B is compact and upper semi-continuous.

We will show that B maps bounded sets into bounded sets in X. Let B

_r

= {x ∈ X : kxk ≤ r}. Then for h ∈ B, x ∈ B

_r

there exists v ∈ S

F,x

such that

h(t) = Z

t

0

(t − s)

^α−1

Γ(α) v(s)ds.

Then for t ∈ J we have

|h(t)| ≤ Z

t

0

(t − s)

^α−1

Γ(α) |v(s)|ds

≤ Z

t

0

(t − s)

^α−1

Γ(α) p(s)ds ≤ kpk T

^α

Γ(α + 1) := K

₁

,

for all t ∈ J . Therefore, khk ≤ K

₁

which shows that B is uniformly bounded on X.

Now, we will show that B(X) is an equicontinuous set in X. Let τ

₁

, τ

₂

∈ J with τ

₁

< τ

₂

and x ∈ S. Then we have

|h(τ

₂

) − h(τ

1

)|

=     

Z

_τ2

0

(τ

₂

− s)

^α−1

Γ(α) v(s)ds − Z

_τ1

0

(τ

₁

− s)

^α−1

Γ(α) v(s)ds

    

≤ Z

τ1

0

   

(τ

₂

− s)

^α−1

− (τ

₁

− s)

^α−1

Γ(α)

   

|v(s)|ds + Z

τ2

τ1

(τ

₂

− s)

^α−1

Γ(α) |v(s)|ds

≤

"

Z

τ1

0

   

(τ

2

− s)

^α−1

− (τ

₁

− s)

^α−1

Γ(α)

   

kpkds + Z

τ2

τ1

(τ

2

− s)

^α−1

Γ(α) kpkds,

which is independent of x. As τ

₁

→ τ

₂

, the right-hand side of the above inequality tends to zero. Therefore, it follows from the Arzel´ a-Ascoli theorem that B is a completely continuous operator on B

r

.

Step 3. B has a closed graph.

Let x

n

→ x

∗

, h

n

∈ B(x

_n

) and h

n

→ h

∗

. We need to show that h

∗

∈ B(x

∗

).

Associated with h

n

∈ T (x

_n

), there exists v

n

∈ S

_F,xn

such that for each t ∈ J, h

_n

(t) =

Z

t 0

(t − s)

^α−1

Γ(α) v

_n

(s)ds.

Thus it suffices to show that there exists v

∗

∈ S

_F,x∗

such that for each t ∈ J, h

∗

(t) =

Z

t 0

(t − s)

^α−1

Γ(α) v

∗

(s)ds.

Let us consider the linear operator Θ : L

¹

(J, R) → C(J, R) given by f 7→ Θ(f )(t) =

Z

t 0

(t − s)

^α−1

Γ(α) v(s)ds.

Observe that

kh

_n

(t) − h

∗

(t)k =

Z

t 0

(t − s)

^α−1

Γ(α) (v

n

(s) − v

∗

(s))ds

→ 0, as n → ∞.

Thus, it follows by Lemma 2.3 that Θ◦S

F

is a closed graph operator. Further, we have h

_n

(t) ∈ Θ(S

_F,xn

). Since x

_n

→ x

_∗

, then we have

h

∗

(t) = Z

t

0

(t − s)

^α−1

Γ(α) v

∗

(s)ds,

for some v

∗

∈ S

_F,x∗

. Hence B has a closed graph (and therefore it has closed values). In consequence, the operator B is upper semicontinuous.

Step 4. Finally we show that δM + ρ < 1, that is, (c

₁

) of Lemma 3.1 holds.

Since

(3.10) M = kB(S)k = sup

x∈S

 sup

t∈J

|Bx(t)|



≤ kpk T

^α

Γ(α + 1) , then by (H

3

) we have

kφk



1 + T

^γ

Γ(γ + 1)



kpk T

^α

Γ(α + 1) +

m

X

i=1

kψ

_i

kT

^βi

Γ(β

i

+ 1) < 1, with δ = kφk



1 +

_Γ(γ+1)^Tγ



and ρ = P

m i=1

kψk_iT^βi Γ(βi+1)

.

Thus all the conditions of Lemma 3.1 are satisfied and therefore, the con- clusion (i) or (ii) holds. We show that the conclusion (ii) of Lemma 3.1 is not possible. Let x be any solution of the (1.1) such that µx ∈ AxBx + Cx for some µ > 1. Then there exists v ∈ S

_F,x

such that

x(t) = λf (t, x(t), I

^γ

x(t)) Z

t

0

(t − s)

^α−1

Γ(α) v(s)ds + λ

m

X

i=1

I

^βi

h

i

(t, x(t)), t ∈ J

where λ =

_µ¹

< 1. Therefore, we have

|x(t)| ≤ |f (t, x(t), I

^γ

x(t))|

Z

t 0

(t − s)

^α−1

Γ(α) |v(s)|ds +

m

X

i=1

Z

t 0

(t − s)

^βi−1

Γ(β

_i

) |h

_i

(s, x(s))|ds

≤ (|f (t, x(t), I

^γ

x(t)) − f (t, 0, 0))| + |f (t, 0, 0)|) kpk Z

t

0

(t − s)

^α−1

Γ(α) ds +

m

X

i=1

Z

t 0

(t − s)

^βi−1

Γ(β

_i

) (|h

i

(s, x(s)) − h

i

(s, 0)| + |h

i

(s, 0)|) ds

≤

"

kφk 1 + T

^γ

Γ(γ + 1)

!

kxk + F

₀

#

kpk T

^α

Γ(α + 1) + (kxk + K

0

)

m

X

i=1

kψ

_i

kT

^βi

Γ(β

i

+ 1) , where F

₀

= sup

_t∈J

|f (t, 0)| and K

₀

= sup

_t∈J

|h

_i

(t, 0)|, i = 1, 2, . . . , m, which leads to

kxk ≤

F

0

kpk T

^α

Γ(α + 1) + K

0 m

X

i=1

T

^βi

Γ(β

_i

+ 1) 1 −

"

kφk 1 + T

^γ

Γ(γ + 1)

!

kpk T

^α

Γ(α + 1) +

m

X

i=1

kψ

_i

kT

^βi

Γ(β

i

+ 1)

# .

As a result, the conclusion (ii) of Lemma 3.1 does not hold. Hence, the conclusion (i) holds and consequently the problem (1.1) has a solution on J . This completes the proof.

4. An example

In this section, we present an example to illustrate our result.

Example 4.1. Consider the following hybrid fractional integro-differential equa- tion

(4.1)



 

 

 

 

 

 

 

  D

¹²













 x(t) −

4

X

i=1

I

²ⁱ⁺¹²

h

i

(t, x(t)) f (t, x(t), I

¹¹²

x(t))















∈ F (t, x(t)), t ∈

 0, 3

2

 ,

x(0) = 0, where

h

_i

(t, x(t)) = x

²

(t) + |x(t)|

2(4 + i + t)(1 + |x(t)|) ,

f (t, x(t), I

¹¹²

x(t)) = cos

²

t

1 + e

^t

sin |x(t)| + 1 2(2 + t)

1 1 +

 

 I

¹¹²

x(t)   

+ 1

!  

 I

¹¹²

x(t)    + 2

3 , and the multi-valued map F : [0, 3/2] → P(R) is given by

x → F (t, x) =

"

e

^−x2

+ sin |x|

8 + 3t

²

+ 5t

⁴

, |x|

(3 + t

²

)(1 + |x|)

#

.

Here α = 1/2, T = 3/2, m = 4, β

1

= 3/2, β

2

= 5/2, β

3

= 7/2, β

4

= 9/2 and γ = 11/2. With the given data, we find that

|f (t, x

₁

, x

₂

) − f (t, y

₁

, y

₂

)| ≤ 1

2 + t (|x

₁

− x

₂

| + |y

₁

− y

₂

|), and

|h

_i

(t, x) − h

_i

(t, y)| ≤ 1

4 + i + t |x − y|, i = 1, 2, 3, 4,

for x, y, x

j

, y

j

∈ R, j = 1, 2. Choosing φ(t) = 1/(2 + t) and ψ

i

(t) = 1/(4 + i + t), it follows that kφk = 1/2 and kψ

_i

k = 1/(4 + i), i = 1, 2, 3, 4. For f ∈ F , we have

|f | ≤ max e

^−x2

+ sin |x|

8 + 3t

²

+ 5t

⁴

, |x|

(3 + t

²

)(1 + |x|)

!

≤ 1

3 + t

²

, x ∈ R.

Thus,

kF (t, x)k

_P

:= sup{|f | : f ∈ F (t, x)} ≤ 1 3 + t

²

,

for all t ∈ [0, 3/2], x ∈ R. Setting p(t) = 1/(3 + t

²

), we get kpk = 1/3. By direct computation, we have

kφk 1 + T

^γ

Γ(γ + 1)

!

kpk T

^α

Γ(α + 1) +

m

X

i=1

kψk

_i

T

^βi

Γ(β

_i

+ 1) = 0.7179368020 < 1.

Consequently all conditions in Theorem 3.1 are satisfied. Therefore, the problem (4.1) has at least one solution on [0, 3/2].

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Received 24 September 2015