On a three step crisis integro differential equation

R E S E A R C H

Open Access

On a three step crisis integro-differential

equation

Dumitru Baleanu

_{, Khadijeh Ghafarnezhad}

_{and Shahram Rezapour}

*_{Correspondence:} [email protected] 3_{Department of Mathematics,} Azarbaijan Shahid Madani University, Tabriz, Iran Full list of author information is available at the end of the article

Abstract

One of the interesting fractional integro-differential equations is the three step crisis equation which has been reviewed recently. In this paper, we investigate the existence of solutions for a three step crisis fractional integro-differential equation under some boundary conditions.

MSC: Primary 34A08; secondary 34A60

Keywords: Caputo derivation; Pointwise defined equation; Three steps crisis equation; Singularity

1 Preliminaries

It is well known that we can make better exact models for most natural phenomena by using fractional differential equations. Most researchers are working on fractional integro-differential equations (see, for example, [1,2,5–8,10–21]).

In 2010, Agarwal et al. reviewed the existence of solutionsDα_{u(t) +f(t,u(t)) = 0 with}

boundary conditionsu(0) =· · ·=u(n–1)_{= 0 and u(1) =}1

0u(s)dμ(s), wheren≥2, α∈

(n– 1,n),μ(s) is a functional of bounded variation,f may have singularity att= 0 and 1

0 dμ(s) < 1 [3]. In 2012, Agarwal et al. studied positive solutions for the integral value

problemDα_u

i(t) +fi(t,u1(t),u2(t)) = 0 with boundary conditionsui(0) =ui(0) = 0 and

ui(1) =

1

0 ui(t)dη(t) fori= 1, 2, wheret∈(0, 1),α∈(2, 3],D

α _{is the Riemann–Liouville}

fractional derivative of orderα,fi is a real valued continuous map on [0, 1]×R+×R+

and₀1ui(t)dη(t) denotes the Riemann–Stieltjes integral [4]. In 2013, the singular

frac-tional problemDα_u+f(t,u,Dγ_u,Dμ_{u) +g(t,u,D}γ_u,Dμ_{u) = 0 with boundary conditions}

u(0) =u(0) =u(0) =u(0) = 0 was reviewed, where 3 <α< 4, 0 <γ < 1, 1 <μ< 2,Dα_is

the Caputo fractional derivative andf is a Caratheodory function on [0, 1]×(0,∞)3[9]. Recently, the authors introduced a new model for investigating the fractional differential equations called three step crisis integro-differential equations [11]. By using the idea, we investigate the existence of solutions for the three step crisis integro-differential equation

Dαx(t) +f

t,x(t),x(t),Dβx(t), t

0

h(ξ)x(ξ)dξ

= 0 (1)

with boundary conditionsx(0) =x(T0),x(1) =x(T1) andx(0) =x(n)(0) = 0, whereα> 1

withn= [α] – 1,T0,T1,β,λ,μ∈(0, 1),h∈L1[0, 1],Dα is the Caputo fractional

deriva-tive of order α, f(t,x1(t), . . . ,x5(t)) = f1(t,x1(t), . . . ,x4(t)) on [0,λ), f(t,x1(t), . . . ,x5(t)) =

f2(t,x1(t), . . . ,x4(t)) on [λ,μ] andf(t,x1(t), . . . ,x5(t)) =f(t,x1(t), . . . ,x4(t)) on (μ, 1] in which

f1(t,·,·,·,·) andf3(t,·,·,·,·) are continuous on [0,λ) and (μ, 1], respectively, andf2(t,·,·,·,·)

is singular at some pointst∈[λ,μ]. In this case, we use the symbolf= [f1,f2,f3,λ,μ] [11].

As is well known, the Caputo fractional derivative of order α > 0 of a function f : (0,∞)→Ris defined byc_Dα_{f(t) =} 1

Γ(n–α)

t 0

fn_(s)

(t–s)α+1–nds, wheren= [α] + 1 (see, for

ex-ample, [13]). LetΨ be the family of nondecreasing functionsψ : [0,∞)→[0,∞) such that∞_n=1ψn(t) <∞for allt> 0 [22]. One can check thatψ(t) <tfor allt> 0 [22]. Let T :X→Xandα:X×X→[0,∞) be two maps. ThenT is called anα-admissible map whenever α(x,y)≥1 impliesα(Tx,Ty)≥1 [22]. Let (X,d) be a complete metric space, ψ∈Ψ andα:X×X→[0,∞) a map. A self-mapT:X→Xis called anα-ψ-contraction wheneverα(x,y)d(Tx,Ty)≤ψ(d(x,y)) for allx,y∈X[22]. We need the following results. Lemma 1([23]) Let0 <n– 1≤α<n.Then Iα_Dα_{x(t) =x(t) +}n–1

i=0 citifor some constants

c0, . . . ,cn–1.

Lemma 2 ([24]) If E is a closed,bounded and convex subset of a Banach space X and

T:E→E is completely continuous,then T has a fixed point in E.

Lemma 3([22]) Let(X,d)be a complete metric space,ψ∈Ψ,α:X×X→[0,∞)a map

and T:X→X anα-admissibleα-ψ-contraction.If T is continuous and there exists x0∈X

such thatα(x0,Tx0)≥1,then T has a fixed point.

2 Main results

Now, we are ready to state and prove our main results.

Lemma 4 Letα> 1,n= [α] + 1,T0,T1∈(0, 1)and f ∈L1[0, 1].Then x(t) =

1

0 G(t,s)f(s)ds

is the solution of the pointwise defined equation Dα_{x(t) +f(t) = 0with boundary}

condi-tions x(0) =x(T0),x(1) =x(T1)and x(0) =· · ·=x(n–1)(0) = 0,where G(t,s) = –(t–s)

α–1 Γ(α) + (1+t)(1–s)α–1

Γ(α) –

(1+t)(T1–s)α–2 Γ(α–1) –

t(T0–s)α–2

Γ(α–1) whenever 0 ≤ s ≤ t, s ≤ T0, G(t,s) =

–(t–s)α–1 Γ(α) + (1+t)(1–s)α–1

Γ(α) –

(1+t)(T1–s)α–2

Γ(α–1) whenever0≤T0≤s≤t,s≤T1,G(t,s) =

(1+t)(1–s)α–1 Γ(α) –

(1+t)(T1–s)α–2 Γ(α–1)

whenever 0≤t≤s≤T1,s≥T0,G(t,s) = (1+t)(1–s)

α–1 Γ(α) –

(1+t)(T1–s)α–2 Γ(α–1) –

t(T0–s)α–2

Γ(α–1) whenever

0≤t≤s≤T0≤T1, G(t,s) = –(t–s)

α–1 Γ(α) +

(1+t)(1–s)α–1

Γ(α) whenever 0≤T0≤T1≤s≤t and

G(t,s) =(1+t)(1–s)_Γ(α)α–1 whenever0≤t≤s,s≥T1.

Proof Suppose that the equationDα_{x(t) +f(t) = 0 holds for allt∈E⊂[0, 1], wherem(E}c_{) =}

0 andmis the Lebesgue measure onR. Letf0be a function such thatf0=f onE. It is easy

to check thatIα_{(f(t)) =I}α_(f

0(t)) for allt∈[0, 1]. This implies thatIα(Dαx(t)) =Iα(–f0(t))

and by using Lemma1we getx(t) = – 1 Γ(α)

t 0(t–s)

α–1_f(s)ds+c

0+c1tfor some constants

c0andc1. By using the boundary conditions, we obtainx(0) =c0and

x(T0) = –

1 Γ(α– 1)

T0

0

(T0–s)α–1f(s)ds+c1.

Thus,c1–c0= –_Γ(α1_–1)

T0 0 (T0–s)

α–1_{f(s)ds. Sincex(1) =x(T}

1), we get

c0=

1 Γ(α– 1)

1

0

(1 –s)α–1f(s)ds– 1 Γ(α– 1)

T1

0

and so

c1=

1 Γ(α)

1

0

(1 –s)α–1f(s)ds

– 1 Γ(α– 1)

T1

0

(T1–s)α–2f(s)ds–

1 Γ(α– 1)

T0

0

(T0–s)α–2f(s)ds.

Hence,

x(t) = – 1 Γ(α)

t

0

(t–s)α–1_f(s)ds– 1

Γ(α) 1

0

(1 –s)α–1_f(s)ds

– 1 Γ(α– 1)

T1

0

(T1–s)α–2f(s)ds+

t Γ(α)

1

0

(1 –s)α–1f(s)ds

– t Γ(α– 1)

T1

0

(T1–s)α–2f(s)ds–

t Γ(α– 1)

T0

0

(T0–s)α–2f(s)ds

= – 1 Γ(α)

t

0

(t–s)α–1f(s)ds– 1 +t Γ(α)

1

0

(1 –s)α–1f(s)ds

– 1 +t Γ(α– 1)

T1

0

(T1–s)α–2f(s)ds–

t Γ(α– 1)

T0

0

(T0–s)α–2f(s)ds.

Now it is easy to check thatx(t) =₀1G(t,s)f(s)ds, whereGis the given Green function.

By using some usual calculations, we find that|G(t,s)| ≤2+_Γα₍+T0_α) (1 –s)α–2_{for allt,s∈[0, 1]}

and|∂_∂G_t(t,s)| ≤ _Γ3₍α_α)(1 –s)α–2_{for allt,s∈[0, 1]. Also, it is easy to see thatD}μ_{x∈C[0, 1] and}

|Dμ_{x| ≤} x

Γ(2–μ) wheneverx∈C

1_{[0, 1]. Here, 0≤μ≤1. Now, consider the Banach space}

X=C1[0, 1] with the norm x ∗=max{ x , x }, · is thesupnorm onC[0, 1]. Assume thatf = [f1,f2,f3,λ,μ]. DefineT:X→Xby

Tx(t) =

1

0

G(t,s)f

s,x(s),x(s),Dβx(s), s

0

h(ξ)x(ξ)dξ,φx(s)ds

= – 1 Γ(α)

t

0

(t–s)α–1_{f(s,x(s),x(s),D}β_x(s),

s

0

h(ξ)x(ξ)dξ,φx(s)ds

– 1 +t Γ(α)

1

0

(1 –s)α–1_{f(s,x(s),x(s),D}β

x(s), s

0

h(ξ)x(ξ)dξ,φx(s)ds

– 1 +t Γ(α– 1)

T1

0

(T1–s)α–2f(s,x(s),x(s),Dβx(s),

s

0

h(ξ)x(ξ)dξ,φx(s)ds

– t Γ(α– 1)

T0

0

(T0–s)α–2f(s,x(s),x(s),Dβx(s),

s

0

h(ξ)x(ξ)dξ,φx(s)ds

for allx∈Xandt∈[0, 1]. Note that the singular pointwise defined problem (1) has a solution if and only ifT has a fixed point inX. We are going to investigate the singular pointwise defined problem (1) under two different conditions. Here. we present first one. In our second result we denote the mapTbyF.

Theorem 5 Let f = [f1,f2,f3,λ,μ],f1(t, 0, 0, 0, 0, 0) = 0,f2(s, 0, 0, 0, 0, 0) = 0and f3(u, 0, 0, 0,

Λ,Λ:X→[0,∞)and mappings a1,a2,a3,a4: (λ,μ)→[0,∞)such thatlimz→0+Λ(z)

then the pointwise defined equation(1)with boundary conditions has a solution,where

+Λ Dβ(x1–x2)(s) +Λ

₀sh(ξ)x1(ξ) –x2(ξ)

dξ

ds

≤

λ

0

G(t,s)

Λ x1–x2

+Λx₁–x₂

+ΛDβ(x1–x2)+Λ

s

0

|h(ξ) x1–x2 dξ

ds

+

μ

λ

_{G(t,s) a1(s) x1–x2 +a2(s)x} 1–x2

+a3(s)Dβ(x1–x2)+a4(s)

s

0 |

h(ξ) x1–x2 dξ

ds

+ λ

μ

G(t,s)

Λ x1–x2

+Λx₁–x₂

+Λ|Dβ x1–x2

+Λ

s

0

|h(ξ) x1–x2 dξ

ds

≤ λ

0

G(t,s)

Λ x1–x2

+Λx₁–x₂

+Λ

x₁–x₂ Γ(2 –β)

+Λm0 x1–x2

ds

+

μ

λ

G(t,s)

a1(s) x1–x2 +a2(s)x1–x2

+a3(s)

x₁–x₂

Γ(2 –β) +a4(s)m0 x1–x2

ds

+ λ

μ

G(t,s)

Λ x1–x2

+Λx₁–x₂

+Λ

x1–x2

Γ(2 –β)

+Λm0 x1–x2

ds

≤

λ

0

G(t,s)

Λl x1–x2 ∗+Λl x1–x2 ∗

+Λ

l x1–x2 ∗

Γ(2 –β)

+Λm0l x1–x2 ∗

ds

+

μ

λ

G(t,s)

a1(s)l x1–x2 ∗+a2(s)l x1–x2 ∗

+a3(s)

l x1–x2 ∗

Γ(2 –β) +a4(s)m0l x1–x2 ∗

ds

+ λ

μ

G(t,s)

Λl x1–x2 ∗+Λl x1–x2 ∗

+Λ

l x1–x2 ∗

Γ(2 –β)

+Λm0l x1–x2 ∗

ds, (*)

where m0 =

1

0 |h(ξ)|dξ and l = max{1, 1

Γ(2–β),m0,θ0 + θ1}. On the other hand,

lim_z→0+Λ(z)

z =qand so for each> 0 there exists 0 <δΛ=δ(,Λ) such that| Λ(z)

z –q|<for

By choosing 0 <z≤δ1:=min{δΛ,}, we have

Λ(z) < (+q)|z|< (+q). (2)

ForΛwe have similar conclusion, that is,

Λ(z) <+q (3)

for all 0 <z≤δ1:=min{δΛ,}. Let> 0 be given,l x1–x2 ∗<min{δ1,δ2}andx1→x2. By

using (2) and (3), we getΛ(l x1–x2 ∗) < (+q)andΛ(l x1–x2 ∗) < (+q). Now by

using (*), we obtain

Tx1(t) –Tx2(t)

≤4(q+)

λ

0

G(t,s) ds+

μ

λ

a1(s) +· · ·+a4(s) G(t,s) ds

+ 4q+ 1

μ

G(t,s) ds≤4(q+)2 +α+T0 Γ(α)

λ

0

(1 –s)α–2_ds

+(2 +α+T0) Γ(α)

4

i=1

μ

λ

ai(s)(1 –s)α–2ds+ 4

q+2 +α+T0

Γ(α) 1

μ

(1 –s)α–2_ds

=2 +α+T0 Γ(α)

4(q+)· 1 α– 1

1 – (1 –λ)α–1

+

4

i=1

ˆai [λ,μ]+ 4

q+· 1

α– 1(1 –μ)

α–1

.

Hence,

Tx1–Tx2 ≤

2 +α+T0

Γ(α)

4(q+)· 1 α– 1

1 – (1 –λ)α–1

+

4

i=1

ˆai [λ,μ]+ 4

q+· 1

α– 1(1 –μ)

α–1

.

In a similar way, we get

T_x1 (t) –T_x2 (t)

≤

1

0

∂_∂G_t(t,s) f

s,x1(s),x1(s),D β

x1(s),

s

0

h(ξ)x1(ξ)dξ

–f

s,x2(s),x2(s),D β

x2(s),

s

0

h(ξ)x2(ξ)dξ

ds

≤

λ

0

∂G

∂t(t,s)

f1

s,x1(s),x1(s),D β

x1(s),

s

0

h(ξ)x1(ξ)dξ

–f1

s,x2(s),x2(s),Dβx2(s),

s

0

h(ξ)x2(ξ)dξ

≤4Λl x ∗

Hence,T is equi-continuous onEand soT:E→Eis completely continuous. Now by using Lemma2,Thas a fixed point onEand so the problem (1) has a solution.

Example1 Consider the pointwise defined equation

D72_{x(t) +f}

with the boundary conditions in the last result, where

x5) =t2

4

i=1xi. Then we havef1(t, 0, 0, 0, 0) =f2(t, 0, 0, 0, 0) =f3(t, 0, 0, 0, 0) = 0,

f1(t,x1,x2,x3,x4) –f1(t,y1,y2,y3,y4) ≤t 4

i=1

_|xi|–|yi|

≤t

4

i=1

|xi–yi| ≤0.1 4

i=1

Λ|xi–yi|

,

f2(t,x1,x2,x3,x4) –f2(t,y1,y2,y3,y4) ≤d(t) 4

i=1

_|xi|–|yi|

≤d(t)

4

i=1

|xi–yi|,

and

f3(t,x1,x2,x3,x4) –f3(t,y1,y2,y3,y4) ≤

1 2t

4

i=1

_|xi|–|yi|

≤1

2t

4

i=1

|xi–yi| ≤

1 2

4

i=1

Λ|xi–yi|

,

whereΛ(x) =|x|andΛ(x) = 1₂|x|. Hence,limz→0+Λ _(z)

z = 0.1 :=q,limz→0+ Λ(z)

z = 1 2 :=q,

ˆ

ai=dˆ∈L1[0.1, 0.8],

4

i=1 ˆai [λ,μ]< 0.092 and

4q(1 – (1 –λ)α–1₎

α– 1 +

4

i=1

ˆai +

4q

α– 1(1 –μ)

α–1

<

4×0.1(1 – (1 – 0.1)52 5

2

+ 0.092 +4×0.5(1 – 0.9)

5 2 5 2

<Γ(α)

3lα .

Now by using Theorem5, the problem has a solution.

Now, we present our second result by using different conditions.

Theorem 6 Suppose that f = [f1,f2,f3,λ,μ],f is nonnegative on[0, 1]and there exist

non-negative functions a1,a2,a3,a4: [0,λ]→R+,maps b1, . . . ,bk0: [λ,μ]→R+for some k0≥1,

and functions c1,c2,c3,c4: [μ, 1]→R+ such thataˆi∈L1[0,λ],bˆj∈L1[λ,μ],cˆi∈L1[μ, 1]

andaˆ1(s) = (1 –s)α–2ai(s).Assuming that there are nonnegative and nondecreasing

func-tionsφi,Φi:R+→R+and Hj:R+4→R+such thatlimz→0+ φi(z)

zμi :=lμi<∞,limz→0+ Φi(z)

zγi :=

lγi<∞andlimz→0+ Hj(z,z,z,z)

zm :=qj<∞for someμi,γi,m∈[1,∞)and Hjare nonnegative

and nondecreasing with respect to all their components(1≤i≤4, 1≤j≤k0),

f1(t,x1, . . . ,x4) –f1(t,y1, . . . ,y4) ≤ 4

i=1

ai(t)φi

|xi–yi|

,

_{f2(t,x1, . . . ,x4) –f2(t,y1, . . . ,y4) ≤}5 i=1

bj(t)Hj

|x1–y1|, . . . ,|x4–y4|

and|f3(t,x1, . . . ,x4)–f3(t,y1, . . . ,y4)| ≤

4

i=1ci(t)Φi(|xi–yi|).Suppose that|f2(t,x1, . . . ,x4)| ≤

Θ(t)Λ(x1, . . . ,x4), whereΛare nonnegative and nondecreasing with respect to all their

components,lim_x→0+ Λ(x,x,x,x)

x :=P2, <∞,Θˆ ∈L

1_[λ,μ],lim

max|xi|→0|f1(t,x1,...,x4)max|xi| |=P1(t)and

limmax|xi|→0|f3(t,x1,...,x4)max|xi| |=P3(t),wherePˆ1∈L

1_[0,λ],Pˆ

3∈L1[μ, 1].If

max

3α Γ(α),

2 +α+T0

Γ(α)

max

₅

i=1

ˆai [0,λ](lμi) + k0

j=1

ˆbj [λ,μ]qj

+

4

i=1

ˆci [μ,1]](lγi),max

1, 1

Γ(2 –β),m0

ˆP1 [0,λ]+P2 Θ [λ,μ]+ ˆP3 [μ,1]

< 1,

then the pointwise defined equation(1)with boundary conditions has a solution.

Proof Letx,y∈Xandt∈[0, 1]. Then we have

Fx(t) –Fy(t)

≤

λ

0

_{G(t,s) f1s,x(s),x(s),D}β

x(s), s

0

h(ξ)x(ξ)dξ

–f1

s,y(s),y(s),Dβy(s), s

0

h(ξ)y(ξ)dξ ds

+

μ

λ

G(t,s) f2

s,x(s),x(s),Dβx(s), s

0

h(ξ)x(ξ)dξ

–f2

s,y(s),y(s),Dβ_y(s),

s

0

h(ξ)y(ξ)dξ ds

+ 1

μ

G(t,s) f3

s,x(s),x(s),Dβ_x(s),

s

0

h(ξ)x(ξ)dξ

–f3

s,y(s),y(s),Dβy(s), s

0

h(ξ)y(ξ)dξ ds

≤

λ

0

G(t,s) [a1(s)φ x(s) –y(s) +a2(s)φ x(s) –y(s)

+a3(s)φ Dβx(s) –Dβy(s) +a4(s)φ

₀sh(ξ)x(ξ) –y(ξ)dξ

ds

+

μ

λ

G(t,s)

k0

i=1

bi(s)Hi( x(s) –y(s) , x(s) –y(s) ,

Dβ_{x(s) –D}β_{y(s) ,}

s

0

h(ξ)x(ξ) –y(ξ)dξ ds

+ 1

μ

G(t,s)|

c1(s)Φ x(s) –y(s) +c2(s)Φ x(s) –y(s)

+c3(s)Φ Dβx(s) –Dβy(s) +c4(s)Φ(

₀sh(ξ)x(ξ) –y(ξ)dξ

ds

≤2 +α+T0

Γ(α) λ

0

(1 –s)α–2a1(s)φ

+a3(s)φ

ilar way, we get

y. This implies thatFis continuous onX. Sincelimx→0+Λ(x,x,x,x)

x =P2,limx→0+

Λ(lx,lx,lx,lx) lx =

P2, where l=max{1,_Γ(2–1_β),m0}. Thus for each > 0 there exists δ1=δ1() such that Λ(lx,lx,lx,lx)

lx <P2+for allx∈(0,δ1]. Hence,

Λ(lx,lx,lx,lx) < (P2+)lx (4)

forx∈(0,δ1]. Also,lim|xi|→0|f1(t,x1,...,x4)min|xi| |=P1(t). Thus, there existsδ2=δ2() such that

f1(t,x1, . . . ,x4) <

P1(t) +

min|xi| (5)

for allt∈[0, 1] and|xi| ∈(0,δ2] for 1≤i≤4. Similarly, there existsδ3=δ3() such that

f3(t,x1, . . . ,x4) <

P3(t) +

min|xi| (6)

for allt∈[0, 1] and|xi| ∈(0,δ3] for 1≤i≤4. Since ˆP1 [0,λ]+P2 Θ [λ,μ]+ ˆP3 [μ,1]<Γ_lθ(α0),

we can choose0> 0 such that ˆP1 [0,λ]+α–10 (1 – (1 –λ) α–1_{) + (P}

2+0) Θ [λ,μ]+ ˆP3 [μ,1]+ 0

α–1(1 –μ)

α–1_<Γ(α) lθ0 . Since

max

3α Γ(α),

2 +α+T0

Γ(α)

5

i=1

ˆai [0,λ](lμi) + k0

j=1

ˆbj [λ,μ]qj+ 4

i=1

ˆci [μ,1]

(lγi)] < 1,

pick1∈(0, 1) such that

max

3α Γ(α),

2 +α+T0

Γ(α)

5

i=1

ˆai [0,λ](lμi+1)

+

k0

j=1

ˆbj [λ,μ](qj+1) + 4

i=1

ˆci [μ,1]

(lγi+1)] < 1. (7)

Letr0=min{δ1(0),δ2(0),δ3(0),₂1}, andC={x∈X: x ∗<r0}. Define the mapαonX×

Xby

α(x,y) = ⎧ ⎨ ⎩

1, x,y∈C, 0, otherwise.

Letx,y∈Xandα(x,y)≥1. Thenx,y∈Cand so _{Fx(t) ≤} λ

0

_{G(t,s) f1s,x(s),x(s),D}β

x(s), s

0

h(ξ)x(ξ)dξ ds

+

μ

λ

G(t,s) f2

s,x(s),x(s),Dβx(s), s

0

h(ξ)x(ξ)dξ ds

+ 1

μ

G(t,s) f3

s,x(s),x(s),Dβx(s), s

0

h(ξ)x(ξ)dξ ds

≤2 +α+T0

Γ(α)

λ

0

(1 –s)α–2 _f 1

s,x(s),x(s),Dβ_x(s),

s

0

= 2 +α+T0 Γ(α) l x ∗

λ

0

(1 –s)α–2P1(s)ds+0

λ

0

(1 –s)α–2ds

+ (P2+0) ˆΘ [λ,μ]+

1

μ

(1 –s)α–2P3(s)ds+0

1

μ

(1 –s)α–2ds

= 2 +α+T0 Γ(α) l x ∗

ˆP1 [0,λ]+

0

α– 1

1 – (1 –λ)α–1

+ (P2+0) ˆΘ [λ,μ]+ ˆP3 [μ,1]+

0

α– 1(1 –μ)

α–1

≤θ0l

ˆP1 [0,λ]+

0

α– 1

1 – (1 –λ)α–1

+ (P2+0) ˆΘ [λ,μ]+ ˆP3 [μ,1]+

0

α– 1(1 –μ)

α–1

x ∗

≤ x ∗

and so Fx ≤ x ∗<r0. Also, we can conclude that Fx ≤ x ∗<r0. Hence, Fx <r0

and soFx∈C. For the same reason,Fy∈C. Similar to (7), we conclude that

Fx–Fy ∗≤max

3α Γ(α),

2 +α+T0

Γ(α)

5

i=1

ˆai [0,λ](lμi+1) x–y μi ∗

+

k0

j=1

ˆbj [λ,μ](qj+1) x–y m∗ + 4

i=1

ˆci [μ,1]

(lγi+1) x–y γi ∗]

≤max

3α Γ(α),

2 +α+T0

Γ(α)

5

i=1

ˆai [0,λ](lμi+1) x–y ∗

+

k0

j=1

ˆbj [λ,μ](qj+1) x–y ∗+ 4

i=1

ˆci [μ,1]

(lγi+1) x–y ∗]

=max

3α Γ(α),

2 +α+T0

Γ(α)

5

i=1

ˆai [0,λ](lμi+1)

+

k0

j=1

ˆbj [λ,μ](qj+1) + 4

i=1

ˆci [μ,1]

(lγi+1)] x–y ∗

whenever x–y ∗<1. Thus, x–y ∗≤ x ∗+ y ∗≤0wheneverx,y∈C. Hence,

Fx–Fy ∗≤max

3α Γ(α),

2 +α+T0

Γ(α)

5

i=1

ˆai [0,λ](lμi+1)

+

k0

j=1

ˆbj [λ,μ](qj+1) + 4

i=1

ˆci [μ,1]

(lγi+1)] x–y ∗

where

ψ(t) =max

3α Γ(α),

2 +α+T0

Γ(α)

5

i=1

ˆai [0,λ](lμi+1)

+

k0

j=1

ˆbj [λ,μ](qj+1)

+

4

i=1

ˆci [μ,1]

(lγi+1)]t.

Note thatψ ∈Ψ. Now by using Theorem6,F has a fixed point and so the pointwise defined problem (1) has a solution.

Example2 Consider the problem

D72_{x(t) +f}

t,x(t),x(t),D12_x(t),

t

0

x(ξ)dξ

= 0

with boundary conditionsx(0) =x(1₃),x(1) =x(1₂) andx(0) = 0, where

f(t,x1, . . . ,x4) =

⎧ ⎪ ⎪ ⎨ ⎪ ⎪ ⎩

f1(t,x1, . . . ,x4) :=t2(

4

i=1xi(s)), t∈[0, 0.7),

f2(t,x1, . . . ,x4) :=_p(t)0.1

4 i=1 |

xi(t)|

1+|xi(t)|, 3t ∈[0.7, 0.7],

f3(t,x1, . . . ,x4) :=t(

4

i=1xi(s)), t∈[0.9, 1],

and

p(t) = ⎧ ⎨ ⎩

0, t∈[0.2, 0.9]∩Q, t, t∈[0.2, 0.9]∩Qc_.

Putai(t) =a(t) =t2,bj(t) =b(t) = _p(t)0.1 andci(t) =c(t) =t(for 1≤i≤4,k0= 1). Then we

have|f1(t,x1, . . . ,x4) –f1(t,y1, . . . ,y4)| ≤t24i=1|xi–yi|=a(t)4i=1φ(|xi–yi|),

f3(t,x1, . . . ,x4) –f3(t,y1, . . . ,y4) ≤t 4

i=1

|xi–yi|=c(t) 4

i=1

Φ|xi–yi|

,

and

f2(t,x1, . . . ,x4) –f2(t,y1, . . . ,y4) ≤t2 4

i=1

|xi–yi|=b(t) 4

i=1

H|x–y|, . . . ,|x–y|,

whereφ(z) =z,Φ(z) =zandH(z1, . . . ,z4) =z1+· · ·+z4. Putμi=γi=m= 1. Then we have

limz→0+φ(z)

z = 1,limz→0+ Φ(z)

z = 1 andlimz→0+ Hj(z,z,z,z)

z = 1. Also,limmax|xi|→0

|f1(t,x1,...,x4)|

max|xi| =

4t2 _=P

1(t), limmax|xi|→0|f3(t,x1,...,x4)max|xi| | = 4t=P3(t) and |f2(t,x1, . . . ,x4)| ≤Θ(t)Λ(x1, . . . ,x4),

whereΘ(t) =p(t) andΛ(x1, . . . ,x4) =

5 i=1 |

xi|

the conditions of Theorem6andlimx→0+Λ(x,x,x,x)

x = 4 :=P2. Also, we have

max

3α Γ(α),

2 +α+T0

Γ(α)

·max

5

i=1

ˆai [0,λ](lμi) + k0

j=1

ˆbj [λ,μ]qj

+

4

i=1

ˆci [μ,1]](lγi),max

1, 1

Γ(2 –β),m0

ˆP1 [0,λ]+P2 Θ [λ,μ]+ ˆP3 [μ,1]

≤ 212 15√π

8

max0.19 + 0.005 + 0.9, 1.13[0.19 + 0.02 + 0.9]< 1.

By using Theorem6, the pointwise defined problem has a solution.

3 Conclusion

It is very important that we increase our abilities of natural phenomenon modeling. In this way, it is better we investigate different types of high order fractional integro-differential equations or new type model ones. One of the new models is described by the three step crisis fractional integro-differential equations which have been introduced recently. In this work, we reviewed the existence of solutions for a three step crisis fractional integro-differential equation under some boundary conditions.

Acknowledgements

The second and third authors were supported by Azarbaijan Shahid Madani University. The authors express their gratitude to the unknown referees for their helpful suggestions, which improved the final version of this paper. Funding

Not applicable.

Availability of data and materials

Data sharing not applicable to this paper as no datasets were generated or analyzed during the current study. Competing interests

The authors declare that they have no competing interests. Authors’ contributions

Each of the authors contributed to each part of this study equally and approved of the final version of the manuscript. Author details

1_{Department of Mathematics, Cankaya University, Ankara, Turkey.}2_{Institute of Space Sciences, Magurele-Bucharest,} Romania. 3_{Department of Mathematics, Azarbaijan Shahid Madani University, Tabriz, Iran.}

Publisher’s Note

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Received: 5 January 2019 Accepted: 9 April 2019

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