Existence and uniqueness of positive and nondecreasing solution for nonlocal fractional boundary value problem

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Existence and uniqueness of positive and nondecreasing solution for

nonlocal fractional boundary value problem

Rahmat Darzi∗

Department of Mathematics, Neka Branch, Islamic Azad University, Neka, Iran.

E-mail:r.darzi@iauneka.ac.ir

Bahram Agheli

Department of Mathematics, Qaemshahr Branch, Islamic Azad University, Qaemshahr, Iran.

E-mail:b.agheli@qaemshahriau.ac.ir

Abstract In this article, we verify the existence and uniqueness of a positive and nondecreasing solution for nonlinear boundary value problem of fractional differential equation in the form

Dα

0+x(t) +f(t, x(t)) = 0, 0< t <1, 2< α≤3,

x(0) =x′_{(0) = 0}_, _x′_{(1) =}_βx₍_ξ₎_,

whereDα

0+denotes the standard Riemann-Liouville fractional derivative, 0< ξ <1 and 0< β ξα₋1

< α−1. Our analysis relies on the fixed point theorem in partially ordered sets. An illustrative example is also presented.

Keywords. Boundary value problem, Fixed point theorem, Partially ordered set, Positive solution,

non-decreasing solution.

2010 Mathematics Subject Classification. 26A33, 34A12.

1. Introduction

In recent years, fractional calculus is one of the interesting issues that have at-tracted the attention of many scientists, specially in mathematics and engineering sciences. Many natural phenomena can be presented by boundary value problems of fractional differential equations. Many authors in different fields such as chemical physics, fluid flows, electrical networks, viscoelasticity, try to model the phenomena by boundary value problems of fractional differential equations [1-4]. In order to achieve extra information in fractional calculus, specially boundary value problems, the reader can refer to more valuable papers or books that are written by authors

Received: 14 September 2015 ; Accepted: 2 April 2016. ∗corresponding Author.

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[5-26].

Bai [17] discussed the existence of positive solutions for the BVP

D₀α+u(t) +f(t, u(t)) = 0, 0< t <1, 1< α≤2,

u(0) = 0, Dα−1u(1) =βu(η), 0< η <1,

where α is a real number, 0 < β ηα−1 _< _{1 and} _Dα

0+ denotes Riemann-Liouville

fractional derivative.

In [14], Zhang used some fixed point theorems on cones to verify the existence of a positive solution for the equation

Dα₀+u(t) +f(t, u(t)) = 0, 0< t <1, 1< α≤2,

with the boundary conditionsu(0) +u′_{(0) =}_u_{(1) +}_u′_{(1) = 0.}

El-Shahed [27] studied the existence and nonexistence of positive solutions to the following nonlinear FDE

Dα₀+u(t) +λf(u(t)) = 0, 0< t <1, 2< α≤3,

along with boundary conditions u(0) =u′_{(0) =}_u′_{(1) = 0}_,_where _Dα

0+ denotes

stan-dard Riemann-Liouville fractional derivative.

In this paper, which is motivated by paper [23], we investigate the existence and uniqueness of a positive and nondecreasing solution for a nonlocal boundary value problem for fractional differential equation of the form

Dα₀+x(t) +f(t, x(t)) = 0, 0< t <1, 2< α≤3, (1.1)

x(0) =x′(0) = 0, x′(1) =βx(ξ), (1.2)

where D₀α+ denotes standard Riemann-Liouville fractional derivative, 0 < ξ < 1,

0< βξα−1 _{< α}₋₁_, _and _f _∈_C_([0_,_1]_×_[0_,_∞₎_,_[0_,_∞_{)). The main approach is based} upon a fixed point theorem in partially ordered sets.

2.

Background materials and preliminaries

We now give definitions, lemmas and theorems that will be used in the remainder of this paper.

Definition 2.1. ([7,8]) The Riemann-Liouville fractional integral of orderα >0, of a functionx:R>0_−→_R_{is defined by}

I₀α+x(t) =

1 Γ(α)

Z t

0

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provided that right-hand side is point wise defined onR>0.

Definition 2.2. ([7,8]) The Riemann-Liouville fractional derivative of orderα >0, of a functionx:R>0_−→_R_{is given by}

Dα₀+x(t) =

1 Γ(n−α)(

d dt)

n Z t

0

(t−s)n−α−1

x(s)ds, n−1< α≤n. (2.2)

Lemma 2.3. ([7,8]) Let x∈C(0,1)∩L(0,1). Then fractional differential equation Dα

0+x(t) = 0has

x(t) =a1tα−1+a2tα−2+...+antα−n, ai∈R, i= 1, ..., n, n= [α] + 1, (2.3)

as a unique solution.

Lemma 2.4. ([7,8]) Let x ∈ C(0,1)∩L(0,1) with a fractional derivative of order α >0 that belongs to C(0,1)∩L(0,1). Then

I0α+Dα₀+x(t) =x(t) +a1tα−1+a2tα−2+...+antα−n, (2.4)

for someai∈R, i= 1, ..., n, n= [α] + 1.

Theorem 2.5. ([28]) Assume that (K,≤) be a partially ordered set. Suppose that there exists a metric din K so that(K, d) is a complete metric space satisfying the following condition:

if(un)is a nondecreasing sequence in K s.t un→u, then un≤u, ∀n∈N. (2.5)

LetT :K−→K be a nondecreasing mapping so that

d(T u, T v)≤d(u, v)−ϕ(d(u, v)),

where ϕ : R≥0 _−→ R≥0 _{is a continuous and nondecreasing function such that} _ϕ _is positive inR>0_,_ϕ_{(0) = 0}_and_lim

t→∞ϕ(t) =∞.If there existsu0∈Kwithu0≤T u0,

thenT has a fixed point.

Theorem 2.6. ([28]) Assume that hypotheses of Theorem 2.5 are satisfied. Suppose that(K,≤)satisfies the following condition:

f or x, y∈K there exists z∈K which is comparable to x and y. (2.6)

Then, the fixed point is unique.

NOTE. In this paper, we use real Banach space K=C[0,1] with the norm kuk = maxt∈[0,1]{u(t)}. This space can be equipped with a partial order given by

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Nieto and Rodriguez [29] proved (K,≤) with the metric d(u, v) = supt∈[0,1]{|u(t)−

v(t)|}satisfied condition (2.5). Also, foru, v∈C[0,1] if max{u, v} ∈K, then, (K,≤) satisfies condition (2.6).

Lemma 2.7. Assume that 0 < ξ < 1, α−1_β 6= ξα−1 _and _g₍_t₎ _∈ _C_[0_,_1]_{. Then,}

fractional boundary value problem

Dα

0+x(t) +g(t) = 0, 0< t <1, 2< α≤3, (2.7)

x(0) =x′(0) = 0, x′(1) =βx(ξ), (2.8)

has a unique solution

x(t) =

Z 1

0

G1(t, s)g(s)ds+ βt α−1

α−1−βξα−1

Z 1

0

G2(ξ, s)g(s)ds, (2.9)

where

G1(t, s) = 1 Γ(α)

(

tα−1(1−s)α−2₋₍_t₋_s₎α−1_, ₀_≤_s_≤_t_≤₁_,

tα−1₍₁₋_s₎α−2_, ₀_≤_t_≤_s_≤₁_, (2.10)

G2(ξ, s) = 1 Γ(α)

(

ξα−1(1−s)α−2₋₍_ξ₋_s₎α−1_, ₀_≤_s_≤_η_≤₁_,

ξα−1₍₁₋_s₎α−2_, ₀_≤_η_≤_s_≤₁_, (2.11)

Proof. Lemma 2.4 guarantees that

x(t) =a1tα−1+a2tα−2+a3tα−3−I0α+g(t). (2.12)

From boundary conditions (2.8), we obtaina2=a3= 0 and

a1=

α−1

Γ(α)(α−1−βξα−1₎

Z 1

0

(1−s)α−2g(s)ds− β

Γ(α) α−1−βξα−1 Z ξ

0

(ξ−s)α−1g(s)ds.

Substitutinga1, a2 anda3 into (2.12), we get

x(t) = − 1 Γ(α)

Z t

0

(t−s)α−1g(s)ds+ (α−1)t α−1

Z 1

0

(1−s)α−2g(s)ds

− βt

α−1

Z ξ

0

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Thus, the unique solution of FBVP (2.7)-(2.8) is

x(t) = −1 Γ(α)

Z t

0

(t−s)α−1

g(s)ds+ t α−1

Γ(α)

Z t

0

(1−s)α−2

g(s)ds+

Z 1

t

(1−s)α−2

g(s)ds

+ βt

α−1

Γ(α)(α−1−βξα−1₎ ξ α−1Z

1

0

(1−s)α−2g(s)ds− Z ξ

0

(ξ−s)α−1g(s)ds

! = 1 Γ(α) Z t 0

tα−1₍₁₋_s₎α−2₋₍_t₋_s₎α−1

g(s)ds+ 1 Γ(α)

Z 1

t

tα−1₍₁₋_s₎α−2_g₍_s₎_ds

+ βt

α−1

Z ξ

0

ξα−1₍₁₋_s₎α−2₋₍_ξ₋_s₎α−1

g(s)ds

+ βt

α−1

Z 1

ξ

ξα−1(1−s)α−2_g₍_s₎_ds

=

Z 1

0

G1(t, s)g(s)ds+ βt α−1

α−1−βξα−2

Z 1

0

G2(ξ, s)g(s)ds.

The proof is complete.

Lemma 2.8. The functionG1(t, s) is nonnegative and continuous function.

Proof. It is easy to see thatG1 is continuous. Now, we show thatG1(t, s)≥0. Case 1. s= 1.ObviouslyG1(t,1) = 0

Case 2. For 0≤t≤s≤1, it is clear thatG1(t,1) =tα−1₍₁₋_s₎α−2_≥₀_.

Case 3. For 0≤s≤t≤1, we have

G1(t, s) = 1 Γ(α)

tα−1(1−s)α−2₋₍

t−s)α−1

= 1

Γ(α)

tα−1(1−s)

α−1

(1−s) −(t−s) α−1

≥ 1

Γ(α)

tα−1(1−s)α−1₋₍

t−s)α−1

= 1

Γ(α)

(t−ts)α−1₋₍

t−s)α−1 ≥0.

Thus, the lemma is proved.

Lemma 2.9. For G1 and G2, the following relations are satisfied, respectively,

(i) sup t∈[0,1]

Z 1

0

G1(t, s)ds= 1

(α−1)Γ(α+ 1), (ii)

Z 1

0

G2(ξ, s)ds=ξ α−1

Γ(α) 1

α−1−

ξ α

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Proof. (i) We have

Z 1

0

G1(t, s)ds = 1 Γ(α)

Z t

0

tα−1(1−s)α−2

ds− Z t

0

(t−s)α−1

ds+

Z 1

t

tα−1(1−s)α−2

ds

= 1

Γ(α)

tα

−1

α−1−

tα

α .

Let us set

k(t) = 1 Γ(α)

tα−1

α−1 −

tα

α .

Then, as

k′(t) = 1 Γ(α)

tα−2−tα−1 >0,

k(t) is strictly increasing and this follows that

sup t∈[0,1]

Z 1

0

G1(t, s)ds=k(1) = 1

(α−1)Γ(α+ 1).

(ii)We have

Z 1

0

G2(ξ, s)ds = 1 Γ(α)

Z ξ

0

ξα−1(1−s)α−2

ds− Z ξ

0

(ξ−s)α−1

ds+

Z 1

ξ

ξα−1(1−s)α−2

ds

= ξ

α−1

Γ(α) 1

α−1 −

ξ α

.

Therefore, the proof is completed.

Lemma 2.10. The function G1(t, s) is strictly increasing with respect to the first component.

Proof. We assume that

g1(t) = 1 Γ(α)

tα−1(1−s)α−2₋₍

t−s)α−1

, s≤t,

g2(t) = 1 Γ(α)t

α−1₍₁₋

s)α−2

, t≤s,

wheresis fixed. Fors≤t,we have

g′1(t) = 1 Γ(α)

(α−1)tα−2(1−s)α−2₋₍

α−1)(t−s)α−2

= 1

Γ(α−1)

(t−ts)α−2₋₍

t−s)α−2 ≥0.

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whent1< t2≤s,ands≤t1< t2, we haveG1(t1, s)< G1(t2, s). Now, we show that in the caset1≤s≤t2 witht1 < t2, G1(t1, s)< G1(t2, s). We haveg2(t1)≤g2(s) =

g1(s)≤g1(t2). Ifg2(t1) = g1(t2), then g2(t1) =g2(s) =g1(s) =g1(t2) and since g1 andg2 are monotone, we conclude thatt1=t2 =s, which contradicts with t1 < t2. Theng2(t1)< g1(t2) and this shows thatG1(t1, s)< G1(t2, s).

3.

Main results

For convenience of presentation, we now present the following hypothesis to be used in the rest of the paper.

(H1)f : [0,1]×R≥0_−→R≥0_{is a continuous function and nondecreasing with respect}

to second variable.

(H2) There existH ⊂[0,1] withµ(H)>0 so thatf(t, x(t))6= 0 fort∈H andx∈R_.

(H3) There exist 0< γ < ₍_α_−1)Γ(1_α₊₁₎+ βξα₋1

(α−1−βξα

−1)Γ(α)(

1 α−1−

ξ α)

−1

such that for

x, y∈R≥0_{, with}_x_≤_y

f(t, x)−f(t, y)≤γκ(x−y), 0≤t≤1,

whereκ(x) = x x+2.

Theorem 3.1. Assume that the hypotheses (H1)-(H3) hold. Then, the boundary

value problem (1.1)-(1.2) has a unique positive and strictly increasing solutionx(t).

Proof. Consider the cone

K={x∈C[0,1] :x(t)≥0}.

SinceKis a closed set,K is a complete metric space with the distance given by

d(x, y) = sup t∈[0,1]

|x(t)−y(t)|.

Let us define the operatorA_:_C_[0_,_1]_−→_C_[0_,_1]_,_by

Ax(t) =

Z 1

0

G1(t, s)f(s, x(s))ds+ βt α−1

α−1−βξα−1

Z 1

0

G2(ξ, s)f(s, x(s))ds. (3.1)

By Lemma 2.8 and (H1), it follows thatA₍_K₎_⊆_K_.

Now, using Theorems 2.5 and 2.6 and in three steps, we will prove to theorem.

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By (H1), forx, y∈K withx≥y, we have

A_x₍_t_{) =} Z 1

0

G1(t, s)f(s, x(s))ds+ βt α−1

1−βξα−2

Z 1

0

G2(ξ, s)f(s, x(s))ds

≥ Z 1

0

G1(t, s)f(s, y(s))ds+ βt α−1

1−βξα−2

Z 1

0

G2(ξ, s)f(s, y(s))ds

= A_y₍_t₎_.

This shows thatA_{is nondecreasing operator. Also, for}_x_≥_y_{, we have}

d(Ax,Ay) = sup t∈[0,1]

|(Ax)(t)−(Ay)(t)|= sup t∈[0,1]

(Ax)(t)−(Ay)(t)

≤ sup

t∈[0,1]

Z 1

0

G1(t, s)

f(s, x(s))−f(s, y(s))

ds

+ βt

α−1

α−1−βξα−1

Z 1

0

G2(ξ, s)

f(s, x(s))−f(s, y(s))

ds

≤ sup

t∈[0,1]

Z 1

0

G1(t, s)γ

x(s)−y(s)

x(s)−y(s) + 2ds

+ β

α−1−βξα−1

Z 1

0

G2(ξ, s)γ

x(s)−y(s)

x(s)−y(s) + 2

ds.

Using Lemma 2.9 and (H3) and since we know the function κ(x) = _x₊₂x is

nonde-creasing, we get

d(A_x,A_y₎ _≤ _γ_· kx−yk

kx−yk+ 1 t∈[0sup,1]

Z 1

0

G1(t, s)ds+ β

α−1−βξα−1

Z 1

0

G2(ξ, s)ds

= γ· kx−yk

kx−yk+ 1

1

(α−1)Γ(α+ 1)+

βξα−1

(α−1−βξα−1_)Γ(_α₎( 1

α−1−

ξ α)

<

kx−yk − kx−yk − kx−yk kx−yk+ 1

.

Let us setϕ(u) = u−_u₊₁u . It is easy to see that ϕ : R≥0 _−→ _R≥0 _{is continuous,} nondecreasing, positive inR>0_{, ϕ}_{(0) = 0 and lim}_u_→∞_ϕ₍_u_{) =}_∞_._{So, for} _x_≥_y_{, we}

have

d(A_x,A_y₎_≤_d₍_{x, y}₎₋_ϕ₍_d₍_{x, y}₎₎_.

On the other hand,G1(t, s)≥0 andf ≥0 yields

(A₀₎₍_t_{) =} Z 1

0

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Now, all the conditions of Theorem 2.5 are satisfied. Consequently, the BVP (1.1)-(1.2) has at least one nonnegative the solution.

Step 2. Uniqueness of the solution.

It is clear that (K,≤) satisfies condition 2.6 and so by Theorem 2.6, we conclude the uniqueness of solution.

Step 3. The solution is strictly increasing.

As x(0) = R1

0 G1(0, s)f(s, x(s))ds and G1(0, s) = 0, we havex(0) = 0. We assume thatt1, t2∈[0,1] witht1< t2. We distinguish the following cases.

Case 1. Ift1= 0, thenx(t1) = 0. We know thatx(t)≥0. Assumingx(t2) = 0, we get

G1(t2, s)f(s, x(s)) = 0, a.e.(s) and sinceG1(t2, s)6= 0 we havef(s, x(s)) = 0, a.e.(s).

On the other hand, f is nondecreasing with respect to the second variable, and so

we have f(s,0) ≤ f(s, x(s)) = 0, a.e.(s), which contradicts the hypothesis (H2). Consequently,x(t1) = 0< x(t2).

Case 2. Ift1>0,then

x(t2)−x(t1) = (A_x₎₍_t₂₎₋₍A_x₎₍_t₁₎

+

Z 1

0

G1(t2, s)−G1(t1, s)f(s, x(s))ds

+β(t α−1 2 −tα

−1

1 )

α−1−βξα−1

Z 1

0

G2(η, s)f(s, x(s))ds.

Now, by Lemma 2.10 and sincef is nonnegative, we getx(t1) =x(t2) and orx(t1)< x(t2).

Assume thatx(t1) =x(t2).Then

Z 1

0

G1(t2, s)−G1(t1, s)f(s, x(s))ds= 0,

and this follows that

G1(t2, s)−G1(t1, s)f(s, x(s))ds= 0, a.e.(s),

and Lemma 2.10 yields

f(s, x(s))ds= 0, a.e.(s),

which again contradicts the hypothesis (H2). Thus, x(t1) < x(t2). Therefore, the

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4.

An example

Example 4.1. Consider the boundary value problem

(

D 5 2

0+x(t) +tx

+x

4x+8 = 0, 0< t <1,

x(0) =x′_{(0) = 0}_, _x′_{(1) =}_x₍1 2),

(4.1)

in whichβ= 1,ξ= 1

2,f(t, x) = t+1

4 · x x+2 with

f(t, x)−f(t, y) = t+ 1 4

x x+ 2 −

y y+ 2

< 1

2·

x−y

x−y+ 2, x≥y.

Furthermore,

1

(α−1)Γ(α+ 1)+

βξα−1

(α−1−βξα−1_)Γ(_α₎( 1

α−1 −

ξ α)

−1

≃3.2373> 1

2 =γ.

Thus, the conclusion of Theorem 3.1 applies to problem (4.1).

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