Positive Solutions of a Four-point Fractional Boundary Value Problem

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Positive Solutions of a Four-point Fractional

Boundary Value Problem

Hongyan Jia

Abstract—In this paper, based on the contraction map prin-ciple and the fixed point index theory, sufficient conditions are established for the uniqueness and existence results of positive solution for four-point boundary value problem of nonlinear differential equation with Caputo’s fractional order derivative. It is interesting to note that the Krein-Rutmann theorem is also used in this paper. Such investigations will provide an important platform for gaining a deeper understanding of our environment.

Index Terms—Caputo’s fractional derivative, Positive solu-tion, Four-point boundary value problem, Fixed point index.

I. INTRODUCTION

F

RACTIONAL derivatives arose in various fields of science and engineering such as control, porous media, electrochemistry, riscoelasticity, electromagnetic and other fields, see [1-5] and the reference therein. As we all known, many physical systems can be represented more accurately using fractional derivative formulations [6,7], thus many works on the basic theory of fractional calculus and fractional order differential equations have been established [6-10].

Recently, there have appeared a very large number of pa-pers, which are devoted to the existence of positive solutions of nonlinear initial value problem or two point boundary value problem, and the solvability of nonlocal boundary value problem, see [11-23] and their references.

However, to the best of our knowledge, there are very few papers published on the positive solution with the nonlocal boundary value problem.

Motivated by the above, in the present paper, we consider the following four-point nonlocal boundary value problems of fractional order

CDα

0+u(t) +f(t, u(t)) = 0, 0< t <1, (1)

u0(0)−βu0(ξ) = 0, u(1) +γu0(η) = 0, (2) where α is a real number with 1 < α 2, 0 < ξ <

η < 1, 0 < β < 1, γ > 0, CDα

0+ is the standard Caputo

fractional derivative. The functionf ∈C([0,1]×[0,+)

[0,+)). By means of contraction map principle and the fixed point index theory, we establish the uniqueness and existence results of positive solution for the problem (1), (2). To the best knowledge of the authors, no work has been done to get positive solution of the problem (1), (2). It is interesting to note that the Krein-Rutmann theorem is also used in this paper.

The work presented in this paper has the following new features. First, the existence of positive solutions for the four-point fractional boundary value problems are considered.

Manuscript received November 2, 2018; revised January 6, 2019. Hongyan Jia is with the School of Mathematics and Statistics, Anyang Normal University, Anyang, Henan 455000, China ( E-mail: jia-hongyanay@163.com).

Second, the uniqueness results of positive solutions are obtained. Third, the Krein-Rutmann theorem is used in this paper.

II. THE PRELIMINARY LEMMAS

For the convenience of readers, we provide some back-ground material in this section.

Definition 2.1 [20] The Riemann-Liouville fractional

in-tegral of orderαfor function y is defined as

0+y(t) = 1 Γ(α)

Z t

0

(t−s)α−1y(s)ds, α >0.

Definition 2.2 [20] The Caputo’s derivative for functiony

is defined as

CDα

0+y(t) = 1 Γ(n−α)

Z t

0

y(n)(s)ds

(t−s)α+1−n, n= [α] + 1,

where[α] denotes the integer part of real numberα. Lemma 2.1[20] Letα >0,then the fractional differential equation

CDα

0+u(t) = 0

has solutions

u(t) =c1+c2t+c3t2+· · ·+cntn−1, ci∈R, i= 1,2,· · ·, n, n= [α] + 1. Lemma 2.2 [9] Letα >0,then

Iα C

0+ 0+u(t) =u(t) +c1+c2t+c3t2+· · ·+cntn−1

for someci∈R, i= 1,2,· · ·, n, n= [α] + 1.

Lemma 2.3 [9] Let P be a cone in a Banach space X, and Ω(P) be a bounded open set in P. Suppose that T :

Ω(P) P is a completely continuous operator. If there existsu0∈P\{θ} such that

u−T u6=µu0, ∀u∈∂Ω(P), µ≥0,

then the fixed point indexi(T,Ω(P), P) = 0.

Lemma 2.4 Let P be a cone in a Banach space X. Suppose that T : P P is a completely continuous operator. If there exists a bounded open setΩ(P)such that each solution of

u=σT u, u∈P, σ∈[0,1]

satisfies u Ω(P), then the fixed point index

i(T,Ω(P), P) = 1.

Lemma 2.5 [9] Suppose that A : C[0,1] C[0,1] is a completely continuous linear operator and A(P)⊂P. If there existψ∈C[0,1]\(−P)and a constantc >0such that

cAψ ψ, then the spectral radius r(A) 6= 0 andA has a positive eigenfunctionφ1corresponding to its first eigenvalue

λ1= (r(A))−1.

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Lemma 2.6 If y ∈C[0,1], 1 < α≤2, then the unique solution of

CDα

0+u(t) +y(t) = 0, 0< t <1, (3)

u0(0)−βu0(ξ) = 0, u(1) +γu0(η) = 0, (4) is

u(t) =

Z 1

0

G(t, s)y(s)ds,

where

G(t, s) =

                                                              

(t−s) α−1

Γ(α) +

(1−s)α−1 Γ(α) +β(1 +γ−t)(ξ−s)

α−2

(1−β)Γ(α−1) +

γ(η−s)α−2 Γ(α−1) ,

s≤ξ, s≤t,

(1−s)α−1 Γ(α) +

β(1 +γ−t)(ξ−s)α−2 (1−β)Γ(α−1) +γ(η−s)α−2

Γ(α−1) , s≤ξ, t≤s,

(t−s) α−1

Γ(α) +

(1−s)α−1 Γ(α) +

γ(η−s)α−2 Γ(α−1) ,

ξ≤s≤η, s≤t,

(1−s)α−1 Γ(α) +

γ(η−s)α−2

Γ(α−1) , ξ≤s≤η, t≤s,

(t−s) α−1

Γ(α) +

(1−s)α−1

Γ(α) , η≤s, s≤t, (1−s)α−1

Γ(α) , η≤s, t≤s.

(5)

Proof We can apply Lemma 2.2 and Definition 2.1 to reduce CDα

0+u(t) +y(t) = 0 to an equivalent integral equation

u(t) = 1

Γ(α)

Z t

0

(t−s)α−1y(s)ds+c1+c2t,

for some c1, c2∈R.By condition (4), one has

c2= β Γ(α−1)

Z ξ

0

(ξ−s)α−2y(s)ds+c 2β,

1

Γ(α)

Z 1

0

(1−s)α−1y(s)ds+c1+c2

γ

Γ(α−1)

Z η

0

(η−s)α−2y(s)ds+c 2γ= 0,

so, we have

c1= 1 Γ(α)

Z 1 0

(1−s)α−1y(s)ds

+ β(1 +γ) (1−β)Γ(α−1)

Z ξ

0

(ξ−s)α−2y(s)ds

+ γ

Γ(α−1)

Z η

0

(η−s)α−2y(s)ds

c2= −β

(1−β)Γ(α−1)

Z ξ

0

(ξ−s)α−2y(s)ds.

Therefore, the unique solution of this problem is

u(t) = 1 Γ(α)

Rt

0(t−s)α−1y(s)ds + 1

Γ(α)

R1

0(1−s)α−1y(s)ds +(1−β(1+β)Γ(γ−α−1)t) R0ξ(ξ−s)α−2y(s)ds +Γ(αγ−1)R0η(η−s)α−2y(s)ds =R01G(t, s)y(s)ds.

The proof is complete.

Lemma 2.7 The function G(t, s) defined by (5) satisfies

G(t, s)0,for t, s∈(0,1). Proof For0≤s≤ξ≤1,

g(t, s)≥ −(t−s) α−1

Γ(α) +

(1−s)α−1 Γ(α) +β(1 +γ−t)(ξ−s)α−2

(1−β)Γ(α−1) +

γ(η−s)α−2 Γ(α−1)

≥ −(1−s) α−1

Γ(α) +

(1−s)α−1 Γ(α) +

βγ(ξ−s)α−2 (1−β)Γ(α−1) +γ(η−s)

α−2

Γ(α−1) 0.

Forξ≤s≤η, g(t, s)≥ −(t−s)

α−1

Γ(α) +

(1−s)α−1 Γ(α) +

γ(η−s)α−2 Γ(α−1)

≥ −(1−s) α−1

Γ(α) +

(1−s)α−1 Γ(α) +

γ(η−s)α−2 Γ(α−1) 0.

Forη≤s≤1,

g(t, s)≥ −(t−s) α−1

Γ(α) +

(1−s)α−1 Γ(α)

≥ −(1−s)α−1

Γ(α) +

(1−s)α−1 Γ(α) = 0.

The proof is complete.

III. THE UNIQUENESS RESULT

Consider the Banach spaceE=C[0,1]with the norm

kuk= max

t∈[0,1]|u(t)|.

Let E be endowed with the ordering u≤v if u(t)≤v(t) for allt∈[0,1]. Denote the coneP ⊂E by

P ={u∈E: u(t)0, t∈[0,1]}.

Foru∈P, define the operatorT : P →P by

(T u)(t) =

Z 1

0

G(t, s)f(s, u(s))ds, (6)

clearly the problem (1), (2) has a solutionuif and only ifu

solves the operator equationu=T u.

Lemma 3.1 The operator T : P →P defined by (6) is completely continuous.

Proof The operator T : P P is continuous in view of the continuity ofG(t, s)andf(t, u). LetBl={u∈P : kuk ≤l}, L= max

0≤t≤1, u∈Bl

|f(t, u)|+ 1, for each u∈Bl,

we have

|(T u)(t)|=

¯ ¯

¯R01G(t, s)(f(s, u(s))ds

¯ ¯

¯≤LR01G(t, s)ds ≤L µ Rξ 0 · 1

Γ(α)(1−s)

α−1

+ β(1 +γ)

(1−β)Γ(α−1)(ξ−s)

α−2

+ γ

Γ(α−1)(η−s)

α−2

¸

ds

+Rξη

·

1

Γ(α)(1−s)

α−1+ γ

Γ(α−1)(η−s)

α−2

¸

ds

+Rη1 1

Γ(α)(1−s)

α−1ds

=L

µ

1 Γ(α+ 1)+

β(1 +γ)ξα−1 (1−β)Γ(α) +

γηα−1 Γ(α)

.

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This shows that T maps bounded sets into bounded sets in

P.

Let Bl ⊂P be a bounded set, t1, t2 [0,1], t1 < t2, for

any u∈Bl,we have

|(T u)(t2)(T u)(t1)|

=|R01(G(t2, s)−G(t1, s))(f(s, u(s))ds|

≤L

µ R1

0 |G(t2, s)−G(t1, s)|ds

.

G(t, s)is uniformly continuous in[0,1]×[0,1]because of the continuity of G(t, s). So, for anyε >0,there exists δ >0, whenever|t2−t1|< δ,we have|G(t2, s)−G(t1, s)|< Lε.

So, |(T u)(t2)(T u)(t1)| < ε, which implies that {T u :

u∈Bτ} is equicontinuous.

Therefore, the operatorT : P→P is completely continuous by the Arzela-Ascoli theorem.

Theorem 3.2 Assume that f(t, u)satisfies

|f(t, u)−f(t, v)|< ω(t)|u−v|, t∈[0,1], u, v∈[0,+).

(7) If

Rξ

0[(1−β)Γ(α−1)(1−s)α−1 +β(1 +γ)Γ(α)(ξ−s)α−2 + (1−β)γΓ(α)(η−s)α−2]ω(s)ds +Rξη[(1−β)Γ(α−1)(1−s)α−1 + (1−β)γΓ(α)(η−s)α−2]ω(s)ds +Rη1[(1−β)Γ(α−1)(1−s)α−1]ω(s)ds

<(1−β)Γ(α)Γ(α−1).

(8)

Then problem (1), (2) has a unique positive solution.

Proof In the following, we will proveTn is a contraction operator for nsufficiently large under the condition (7) and (8). Indeed, for u, v∈P, we have

|(T u−T v)(t)|=R01G(t, s)|f(s, u(s))−f(s, v(s))|ds <ku−vk

µ Rξ

0

·

1

Γ(α)(1−s)

α−1

+ β(1 +γ)

(1−β)Γ(α−1)(ξ−s)

α−2

+ γ

Γ(α−1)(η−s)

α−2

¸

ω(s)ds

+Rξη

·

1

Γ(α)(1−s)

α−1

+ γ

Γ(α−1)(η−s)

α−2

¸

ω(s)ds

+Rη1 1

Γ(α)(1−s)

α−1ω(s)ds

= ku−vk (1−β)Γ(α)Γ(α−1)

µ Rξ

0[(1−β)Γ(α−1)(1−s)α−1 +β(1 +γ)Γ(α)(ξ−s)α−2 + (1−β)γΓ(α)(η−s)α−2]ω(s)ds +Rξη[(1−β)Γ(α−1)(1−s)α−1 + (1−β)γΓ(α)(η−s)α−2]ω(s)ds +Rη1[(1−β)Γ(α−1)(1−s)α−1]ω(s)ds

.

Denote

L=R0ξ[(1−β)Γ(α−1)(1−s)α−1 +β(1 +γ)Γ(α)(ξ−s)α−2 + (1−β)γΓ(α)(η−s)α−2]ω(s)ds +Rξη[(1−β)Γ(α−1)(1−s)α−1 + (1−β)γΓ(α)(η−s)α−2]ω(s)ds +Rη1[(1−β)Γ(α−1)(1−s)α−1]ω(s)ds,

we have

|(T u−T v)(t)|< Lku−vk

(1−β)Γ(α)Γ(α−1),

consequently,

|(T2uT2v)(t)|

=R01G(t, s)|f(s,(T u)(s))−f(s,(T v)(s))|ds < Lku−vk

(1−β)Γ(α)Γ(α−1)

Z 1

0

G(t, s)ω(s)ds

< Lku−vk

((1−β)Γ(α)Γ(α−1))2

µ Rξ

0[(1−β)Γ(α−1)(1−s)α−1 +β(1 +γ)Γ(α)(ξ−s)α−2 + (1−β)γΓ(α)(η−s)α−2]ω(s)ds +Rξη[(1−β)Γ(α−1)(1−s)α−1 + (1−β)γΓ(α)(η−s)α−2]ω(s)ds +Rη1[(1−β)Γ(α−1)(1−s)α−1]ω(s)ds

= L

2kuvk ((1−β)Γ(α)Γ(α−1))2.

By induction, we have

|(TnuTnv)(t)| ≤ Lnku−vk

((1−β)Γ(α)Γ(α−1))n.

Taking into account (8), choosensufficiently large, we have

Ln

((1−β)Γ(α)Γ(α−1))n <

1 2,

and therefore,

kTnuTnvk ≤ 1

2ku−vk,

which gives the proof.

IV. EXISTENCE OF POSITIVE SOLUTIONS

In this section, we impose some growth conditions onf

which allow us to apply some fixed point index lemmas to establish the existence results of positive solutions for problem (1), (2).

Define an operatorA: E→E by

()(t) =

Z 1 0

G(t, s)ψ(s)ds. (9)

Then A is a completely continuous linear operator and

A(P) P. By virtue of the Krein-Rutmann theorem, we have the following lemma.

Lemma 4.1 SupposeAis defined by (9), then the spectral radius r(A) > 0 and A has a positive eigenfunction φ1

corresponding to its first eigenvalueλ1= (r(A))−1.

Proof By Lemma 2.7, we have G(t, s) > 0 for s, t

(0,1). Choose [t1, t2] (0,1) and φ C[0,1] such that

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φ(t) 0, t [0,1], φ(t) > 0, ∀t (t1, t2) and φ(t) = 0, ∀t /∈(t1, t2).Then for t∈[t1, t2],

()(t) =

Z 1 0

G(t, s)φ(s)ds≥

Z t2

t1

G(t, s)φ(s)ds >0,

so there exists a constant c > 0 such that c()(t) φ(t), t [0,1]. In view of Lemma 2.5, we complete the proof.

Theorem 4.2 Suppose that f(t,0) 6≡ 0, t (0,1).

Furthermore,

0limu→+∞maxt∈[0,1]

f(t, u)

u <

½

1 Γ(α+ 1)

+β(1 +γ)ξα−1 (1−β)Γ(α) +

γηα−1 Γ(α)

¾−1 . (10)

Then problem (1), (2) has at least one positive solution.

Proof By (10), there existb >0, 0< N <

µ

1 Γ(α+ 1)+

β(1 +γ)ξα−1 (1−β)Γ(α) +

γηα−1 Γ(α)

−1

, such that

0≤f(t, u)< N u+b, t∈[0,1], u∈[0,+).

Let

Bh={u∈P|ku−b

Z 1 0

G(t, s)dsk ≤h}

be a convex, bounded and closed subset of the Banach Space

E.Foru∈Bh, we have

kuk ≤bk

Z 1

0

G(t, s)dsk+h

≤h+b

µ

1 Γ(α+ 1) +

β(1 +γ)ξα−1 (1−β)Γ(α) +

γηα−1 Γ(α)

and

|T u(t)−bR01G(t, s)ds|

R01G(t, s)|f(s, u(s))−b|ds ≤NkukR01G(t, s)ds

≤Nkuk

µ

1 Γ(α+ 1) +

β(1 +γ)ξα−1 (1−β)Γ(α) +

γηα−1 Γ(α)

≤N

µ

1 Γ(α+ 1)+

β(1 +γ)ξα−1 (1−β)Γ(α) +

γηα−1 Γ(α)

¶ ½

h+b

µ

1 Γ(α+ 1)+

β(1 +γ)ξα−1 (1−β)Γ(α) +

γηα−1 Γ(α)

<(N h+b)

µ

1 Γ(α+ 1) +

β(1 +γ)ξα−1 (1−β)Γ(α) +

γηα−1 Γ(α)

< h

as long as

h > µ b

1 Γ(α+ 1)+

β(1 +γ)ξα−1 (1−β)Γ(α) +

γηα−1 Γ(α)

−1

−N ,

so we have T(Bh) Bh. Using Schauder fixed point

theorem, T has at least one fixed point in Bh, which is a

positive solution of problem (1), (2).

Theorem 4.3 Assume that

lim

u→0+tmin∈[0,1]

f(t, u)

u > λ1, (11)

lim

u→+∞tmax∈[0,1]

f(t, u)

u < λ1, (12)

whichλ1 is the first eigenvalue of the operator defined by

(9). Then problem (1), (2) has at least one positive solution.

Proof By (11), there existsl1>0 such that

f(t, u)≥λ1u, 0≤t≤1, 0≤u≤r1.

Letψ∗ be the positive eigenfunction ofA corresponding to

λ1, thusλ1Aψ∗=ψ∗.For eachψ∈Bl1∩P, one has

(T ψ)(t)≥λ1

Z 1 0

G(t, s)ψ(s)ds=λ1()(t), t∈[0,1].

(13) Suppose that T has no fixed point on ∂Bl1 ∩P, now we

prove that

ψ−T ψ6=µψ∗, ψ∂B

l1∩P, µ≥0. (14)

If (14) is not true, there existψ1 ∂Bl1∩P andµ0 >0

such thatψ1−T ψ1=µ0ψ∗,so

ψ1=T ψ1+µ0ψ∗≥µ0ψ∗. (15)

Set µ∗= sup{µ|ψ1 ≥µψ∗}, clearly0< µ0 ≤µ∗ <+∞,

andψ1≥µ∗ψ∗ and therefore,λ11≥µ∗λ1Aψ∗=µ∗ψ∗.

Therefore, by (13) and (15),

ψ1=T ψ1+µ0ψ∗≥λ11+µ0ψ∗≥µ∗ψ∗+µ0ψ∗,

which contradicts the definition of µ. Hence, (14) holds. Lemma 2.3 implies that

i(T, Bl1∩P, P) = 0. (16)

On the other hand, by (12), there exists 0 < σ < 1 and

l2> l1 such that

f(t, u)≤σλ1u, 0≤t≤1, u≥l2.

Let A1ψ = σλ1Aψ, ψ C[0,1]. Then A1 : C[0,1]

C[0,1]is a bounded linear operator andA1(P)⊂P.Let

N = sup

u∈Bl2∩P, s∈[0,1]

f(s, u(s)) max

t∈[0,1]

Z 1

0

G(t, s)ds,

clearly0< N <+∞. Let

S={ψ∈P|ψ=µAψ, 0≤µ≤1}.

In the following, we prove thatSis bounded. For anyψ∈S,

set ψ(t) = min(t), l2} and E(ψ) = {t [0,1](t) >

l2},then

ψ(t) =µAψ(t)(T ψ)(t)R01G(t, s)f(s, ψ(s))ds

=RE(ψ)G(t, s)f(s, ψ(s))ds

+R[0,1]\E(ψ)G(t, s)f(s, ψ(s))ds R01G(t, s)σλ1ψ(s)ds+

R1

0 G(t, s)f(s, ψ(s))ds

(A1ψ)(t) +N , t∈[0,1].

Thus ((I −A1)ψ)(t) N , t [0,1]. Since λ1 is the

first eigenvalue of A and 0 < σ < 1, so (r(A1))−1 > 1.

Therefore,(I−A1)−1exists and

(I−A1)−1=I+A1+A21+· · ·+An1+· · ·.

It follows from A1(P) P that (I −A1)−1 P. So,

ψ(t)(I−A1)−1N , t [0,1]. Therefore, S is bounded.

Choosel3>max{l2, k(I−A1)−1Nk}.By Lemma 2.4, we

have

i(T, Bl3∩P, P) = 1. (17)

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In view of (16),(17), one has

i(T,(Bl3∩P)\(Bl1∩P), P)

=i(T, Bl3∩P, P)−i(T, Bl1∩P, P) = 1.

ThenT has at least one fixed point on(Bl3∩P)\(Bl1∩P),

which means that problem (1), (2) has at least one positive solution.

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IAENG International Journal of Applied Mathematics, 49:2, IJAM_49_2_09

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