Wengui Yang

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Kyungpook Mathematical Journal 2016; 56(2): 419-430

Published online June 1, 2016

Copyright © Kyungpook Mathematical Journal.

Positive Solutions for Three-point Boundary Value Problem of Nonlinear Fractional q-difference Equation

Wengui Yang

Received: April 16, 2013; Accepted: April 11, 2016

In this paper, we investigate the existence and uniqueness of positive solutions for three-point boundary value problem of nonlinear fractional q-difference equation. Some existence and uniqueness results are obtained by applying some standard fixed point theorems. As applications, two examples are presented to illustrate the main results.

Keywords: Nonlinear fractional $q$-difference equations, Three-point boundary conditions, Existence and uniqueness, Positive solutions, Fixed point theorem.

As a matter of fact, fractional differential equations arise in many engineering and scientific disciplines as the mathematical modeling of systems and processes in the fields of physics, chemistry, aerodynamics, electrodynamics of a complex medium, polymer rheology, etc. involves derivatives of fractional order. Fractional differential equations also serve as an excellent tool for the description of hereditary properties of various materials and processes. In consequence, fractional differential equations have been of great interest; for example, see [2, 3, 4, 7, 16] and the references therein.

The q-difference calculus or quantum calculus is an old subject that was initially developed by Jackson [13, 14], basic definitions and properties of q-difference calculus can be found in the book mentioned in [15].

The fractional q-difference calculus had its origin in the works by Al-Salam [6] and Agarwal [1]. More recently, maybe due to the explosion in research within the fractional differential calculus setting, new developments in this theory of fractional q-difference calculus were made, for example, q-analogues of the integral and differential fractional operators properties such as the q-Laplace transform, q-Taylor’s formula, Mittage-Leffler function [5, 17, 18], just to mention some.

Recently, the question of the existence of solutions for fractional q-difference boundary value problems have aroused considerable attention. There have been some papers dealing with the existence and multiplicity of solutions or positive solutions for boundary value problems involving nonlinear fractional q-difference equations, such as the Krasnosel’skii fixed-point theorem, the Leggett-Williams fixed-point theorem, and the Schauder fixed-point theorem, For examples, see [8, 9] and the references therein.

El-Shahed and Hassan [10] studied the existence of positive solutions of the following q-difference boundary value problem

{-(Dq2u)(t)=a(t)f(u(t)),         0t1,αu(0)-βDqu(0)=0,         γu(1)-δDqu(1)=0.

Ferreira [11] and [12] considered the existence of positive solutions to nonlinear q-difference boundary value problems

{-(Dqαu)(t)=-f(t,u(t)),         0t1,         1<α2,u(0)=u(1)=0,

and

{(Dqαu)(t)=-f(t,u(t)),         0t1,         2<α3,u(0)=(Dqu)(0)=0,         (Dqu)(1)=β0,

respectively.

In this paper, we investigate the existence and uniqueness results for the following nonlinear fractional q-difference equations with three-point boundary conditions

{(Dqαu)(t)+f(t,u(t))=0,         0t1,         1<α2,u(0)=0,         u(1)=βu(ξ),

where 0 < βξα−1 < 1, 0 < ξ < 1, Dqα is the fractional q-derivative of the Riemann-Liouville type of order α, and f : [0, 1] × [0,) → [0,) is continuous function.

For the convenience of the reader, we present some necessary definitions and lemmas of fractional q-calculus theory to facilitate analysis of problem (1.1). These details can be found in the recent literature; see [15] and references therein.

Let q ∈ (0, 1) and define

[a]q=qa-1q-1,a.

The q-analogue of the power (ab)(n) with n ∈ ℕ0 is

(a-b)(0)=1,(a-b)(n)=k=0n-1(a-bqk),n,a,b.

More generally, if α ∈ ℝ, then

(a-b)(α)=aαn=0a-bqna-bqα+n.

Note that, if b = 0 then a(α) = aα. The q-gamma function is defined by

Γq(x)=(1-q)(x-1)(1-q)x-1,x{0,-1,-2,},

and satisfies Γq(x + 1) = [x]qΓq(x).

The q-derivative of a function f is here defined by

(Dqf)(x)=f(x)-f(qx)(1-q)x,         (Dqf)(0)=limx0(Dqf)(x),

and q-derivatives of higher order by

(Dq0f)(x)=f(x)and (Dqnf)(x)=Dq(Dqn-1f)(x),n.

The q-integral of a function f defined in the interval [0, b] is given by

(Iqf)(x)=0xf(t)dqt=x(1-q)n=0f(xqn)qn,x[0,b].

If a ∈ [0, b] and f is defined in the interval [0, b], its integral from a to b is defined by

abf(t)dqt=0bf(t)dqt-0af(t)dqt.

Similarly as done for derivatives, an operator Iqn can be defined, namely,

(Iq0f)(x)=f(x)and (Iqnf)(x)=Iq(Iqn-1f)(x),n.

The fundamental theorem of calculus applies to these operators Iq and Dq, i.e.,

(DqIqf)(x)=f(x),

and if f is continuous at x = 0, then

(IqDqf)(x)=f(x)-f(0).

Basic properties of the two operators can be found in the book [15]. We now point out three formulas that will be used later (iDq denotes the derivative with respect to variable i)

[a(t-s)](α)=aα(t-s)(α),Dtq(t-s)(α)=[α]q(t-s)(α-1),(Dxq0xf(x,t)dqt)(x)=0xDxqf(x,t)dqt+f(qx,x).

Denote that if α > 0 and abt, then (ta)(α) ≥ (tb)(α) [11].

Definition 2.1

([19]) Let α ≥ 0 and f be function defined on [0, 1]. The fractional q-integral of the Riemann-Liouville type is Iq0f(x)=f(x) and

(Iqαf)(x)=1Γq(α)0x(x-qt)(α-1)f(t)dqt,         α>0,         x[0,1].

Definition 2.2

([19]) The fractional q-derivative of the Riemann-Liouville type of order α ≥ 0 is defined by Dq0f(x)=f(x) and

(Dqαf)(x)=(DqmIqm-αf)(x),         α>0,

where m is the smallest integer greater than or equal to α.

Lemma 2.1

([19]) Let α β ≥ 0 and f be a function defined on [0, 1]. Then the next formulas hold:

  • (a) (IqβIqαf)(x)=Iqα+βf(x),

  • (b) (DqαIqαf)(x)=f(x).

Lemma 2.2

([11]) Let α > 0 and p be a positive integer. Then the following equality holds:

(IqαDqpf)(x)=(DqpIqαf)(x)-k=0p-1xα-p+kΓq(α+k-p+1)(Dqkf)(0).

Lemma 2.3

Let yC[0, 1] and 1 < α ≤ 2, the unique solution of

(Dqαu)(t)+y(t)=0,         0t1,u(0)=0,         u(1)=βu(ξ),

is given by

u(t)=01G(t,qs)y(s)dqs,

where

G(t,s)={tα-1(1-s)(α-1)-βtα-1(ξ-s)(α-1)-(t-s)(α-1)(1-βξα-1)(1-βξα-1)Γq(α),0st1,sξ,tα-1(1-s)(α-1)-(t-s)(α-1)(1-βξα-1)(1-βξα-1)Γq(α),0<ξst1,tα-1(1-s)(α-1)-βtα-1(ξ-s)(α-1)(1-βξα-1)Γq(α),0tsξ1,tα-1(1-s)(α-1)(1-βξα-1)Γq(α),0ts1,ξs.
Proof

At first, by Lemma 2.1 and Lemma 2.2, the equation (2.1) is equivalent to the integral equation

u(t)=-Iqαy(t)+B1tα-1+B2tα-2,         B1,B2,

that is,

u(t)=-0t(t-qs)(α-1)Γq(α)y(s)dqs+B1tα-1+B2tα-2.

By the boundary conditions u(0) = 0 and u(1) = βu(η), we have

B1=01(1-qs)(α-1)(1-βξα-1)Γq(α)y(s)dqs-0ξβ(ξ-qs)(α-1)(1-βξα-1)Γq(α)y(s)dqs,         B2=0.

Therefore, the solution u(t) of boundary value problem (2.1) satisfies

u(t)=-0t(t-qs)(α-1)Γq(α)y(s)dqs+01tα-1(1-qs)(α-1)(1-βξα-1)Γq(α)y(s)dqs-0ξβtα-1(ξ-qs)(α-1)(1-βξα-1)Γq(α)y(s)dqs=01G(t,qs)y(s)dqs,

where G(t, s) is given by (2.2). The proof is completed.

Lemma 2.4

The function G(t, s) defined by (2.2) satisfies G(t, qs) ≥ 0 for all 0 ≤ s, t ≤ 1.

Proof

We start by defining four functions as follows

g1(t,s)=tα-1(1-s)(α-1)-βtα-1(ξ-s)(α-1)-(t-s)(α-1)(1-βξα-1),         0st1,sξ,g2(t,s)=tα-1(1-s)(α-1)-(t-s)(α-1)(1-βξα-1),         0<ξst1,g3(t,s)=tα-1(1-s)(α-1)-βtα-1(ξ-s)(α-1),         0tsξ1,g4(t,s)=tα-1(1-s)(α-1),         0ts1,ξs.

Firstly, we prove g1(t, qs) ≥ 0, 0 ≤ st ≤ 1, sξ. In view of the fact that if α > 0 and abt, then (ta)(α) ≥ (tb)(α) [11], we get

g1(t,qs)=tα-1(1-qs)(α-1)-βtα-1(ξ-qs)(α-1)-(t-qs)(α-1)(1-βξα-1)=tα-1((1-qs)(α-1)-β(ξ-qs)(α-1)-(1-qst)(α-1)(1-βξα-1))tα-1((1-qs)(α-1)-β(ξ-qs)(α-1)-(1-qs)(α-1)(1-βξα-1))=tα-1(βξα-1(1-qs)(α-1)-β(ξ-qs)(α-1))=tα-1(βξα-1(1-qs)(α-1)-βξα-1(1-qsξ)(α-1))βξα-1tα-1((1-qs)(α-1)-(1-qs)(α-1))=0.

Therefore, g1(t, qs) ≥ 0, 0 ≤ st ≤ 1, sξ. Similarly, with 0 < βξα−1 < 1, 0 < 1 − βξα−1 < 1, we have

g2(t,qs)=tα-1(1-qs)(α-1)-(t-qs)(α-1)(1-βξα-1)tα-1(1-qs)(α-1)-(t-qs)(α-1)=tα-1((1-qs)(α-1)-(1-qst)(α-1))tα-1((1-qs)(α-1)-(1-qs)(α-1))=0,g3(t,qs)=tα-1(1-qs)(α-1)-βtα-1(ξ-qs)(α-1)tα-1((1-qs)(α-1)-βξα-1(1-qsξ)(α-1))tα-1(1-qs)(α-1)(1-βξα-1)0.

It is obvious that g4(t, qs) = tα−1(1 − qs)(α−1) ≥ 0, 0 ≤ ts ≤ 1, ξs. Hence, G(t, qs) ≥ 0 for all 0 ≤ s, t ≤ 1. The proof is completed.

Let ℂ = C([0, 1], ℝ) denote the Banach space of all continuous functions from [0, 1] → ℝ endowed with the norm defined by ||u|| = maxt∈[0,1] |u(t)|. Define the cone ℙ ℂ by

={uu(t)0,for t[0,1]}.

Lemma 2.5

Let ℐ : ℙ → ℂ be the operator defined by

Tu(t):=01G(t,qs)f(s,u(s))dqs=-0t(t-qs)(α-1)Γq(α)f(s,u(s))dqs+01tα-1(1-qs)(α-1)(1-βξα-1)Γq(α)f(s,u(s))dqs-0ξβtα-1(ξ-qs)(α-1)(1-βξα-1)Γq(α)f(s,u(s))dqs.

Then ℐ : ℙ → ℙ is completely continuous.

Proof

The operator : ℙ → ℙ is continuous in view of nonnegativeness and continuity of G and f. Let Ω ℙ be bounded, i.e., there exists a positive constant M > 0 such that ||u|| ≤ M, for all u ∈ Ω. Let K = max0≤t≤1,0≤uM |f(t, u)| + 1, then, for all u ∈ Ω, we have

Tu(t)=|01G(t,qs)f(s,u(s))dqs|K01max0t1G(t,qs)dqs.

Hence, (Ω) is bounded.

On the other hand, given ε > 0, setting

δ=min {1,((1-βξα-1)Γq(α+1)ɛK(2-βξα-1+βξα)2α)1α-1},

then, for each u ∈ Ω, t1, t2 ∈ [0, 1], t1 < t2 and t2t1 < δ, one has |ℐu(t2) − (t1)| < ε. That is to say, (Ω) is equicontinuity. In fact,

Tu(t2)-Tu(t1)=|-0t2(t2-qs)(α-1)Γq(α)f(s,u(s))dqs+01t2α-1(1-qs)(α-1)(1-βξα-1)Γq(α)f(s,u(s))dqs-0ξβt2α-1(ξ-qs)(α-1)(1-βξα-1)Γq(α)f(s,u(s))dqs+0t1(t1-qs)(α-1)Γq(α)f(s,u(s))dqs-01t1α-1(1-qs)(α-1)(1-βξα-1)Γq(α)f(s,u(s))dqs+0ξβt1α-1(ξ-qs)(α-1)(1-βξα-1)Γq(α)f(s,u(s))dqs||0t2(t2-qs)(α-1)Γq(α)f(s,u(s))dqs-0t1(t1-qτ)(α-1)Γq(α)f(s,u(s))dqs|+|01t2α-1(1-qs)(α-1)(1-βξα-1)Γq(α)f(s,u(s))dqs-01t1α-1(1-qs)(α-1)(1-βξα-1)Γq(α)f(s,u(s))dqs|+|0ξβt2α-1(ξ-qs)(α-1)(1-βξα-1)Γq(α)f(s,u(s))dqs-0ξβt1α-1(ξ-qs)(α-1)(1-βξα-1)Γq(α)f(s,u(s))dqs|K(t2α-t1α)Γq(α+1)+K(1+βξα)(t2α-1-t1α-1)(1-βξα-1)Γq(α+1).

In the following, we divide the proof into two cases.

Case 1

δt1 < t2 < 1, with the use of mean value theorem,

t2α-1-t1α-1δα-2(α-1)(t2-t1)(α-1)δα-1.
Case 2

0 ≤ t1 < δ, t2 < 2δ. Then we have

t2α-1-t1α-1t2α-1<(2δ)α-1.

Consequently, we have

max{t2α-1-t1α-1,t2α-t1α}2αδα-1

and

Tu(t2)-Tu(t1)K(2-βξα+βξα)2α(1-βξα-1)Γq(α+1)δα-1ɛ.

By means of the Arzela-Ascoli theorem, we have that : ℙ → ℙ is completely continuous. The proof is complete.

In this section, our objective is to give and prove our main results.

Theorem 3.1

Suppose f(t, u) satisfies

0lim supu+maxt[0,1]f(t,u)u<(1-βξα-1)Γq(α+1).

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

Proof

By (3.1), taking into account the nonnegativity and continuity of f, there exist C > 0, 0<M < (1 − βξα−1q(α + 1) such that

0f(t,u)<Mu+C,         for         t[0,1],         u[0,+).

Let

BR={uu-C01G(t,qs)dqsR}

be a convex, bounded, and closed subset of the Banach space E. For uBR, we have

uC01G(t,qs)dqs+RR+C(1-βξα-1)Γq(α+1)

and

|Tu(t)-C01G(t,qs)dqs|01G(t,qs)f(t,u(s))-Cdqsmax {Mu1-βξα-1)Γq(α+1),C1-βξα-1)Γq(α+1)}max {M1-βξα-1)Γq(α+1)(R+C(1-βξα-1)Γq(α+1)),C(1-βξα-1)Γq(α+1)}R

as long as

RC(1-βξα-1)Γq(α+1)-M.

So, we have (BR) ⊂ BR. Then, combining with Lemma 2.5, the Schauder fixed point theorem assures that operator has at least one fixed point in BR and then the problem (1.1) has at least one positive solution. The proof is complete.

Theorem 3.2

Suppose that f : [0, 1]×ℝ → [0,+) is a jointly continuous function satisfying the condition

f(t,u)-f(t,v)Lu-v,         for         t[0,1],         u,v[0,+).

Then the problem (1.1) has a unique positive solution if

L(1-βξα-1)Γq(α+1)2(2-βξα-1+βξα).
Proof

Defining supt∈[0,1] |f(t, 0)| =K < and selecting

r2K(2-βξα-1+βξα)(1-βξα-1)Γq(α+1),

we show that TBrBr, where Br = {u ∈ ℂ : ||u|| ≤ r}. For uBr, we have

Tu(t)0t(t-qs)(α-1)Γq(α)f(s,u(s))dqs+01tα-1(1-qs)(α-1)(1-βξα-1)Γq(α)f(s,u(s))dqs+0ξβtα-1(ξ-qs)(α-1)(1-βξα-1)Γq(α)f(s,u(s))dqs0t(t-qs)(α-1)Γq(α)(f(s,u(s))-f(s,0)+f(s,0))dqs+01tα-1(1-qs)(α-1)(1-βξα-1)Γq(α)(f(s,u(s))-f(s,0)+f(s,0))dqs+0ξβtα-1(ξ-qs)(α-1)(1-βξα-1)Γq(α)(f(s,u(s))-f(s,0)+f(s,0))dqs(Lr+K)(0t(t-qs)(α-1)Γq(α)dqs+01tα-1(1-qs)(α-1)(1-βξα-1)Γq(α)dqs+0ξβtα-1(ξ-qs)(α-1)(1-βξα-1)Γq(α)dqs)(Lr+K)2-βξα-1+βξα(1-βξα-1)Γq(α+1)r.

Taking the maximum over the interval [0, 1], we get ||ℐu(t)|| ≤ r. Now, for u, v ∈ ℂ and for each t ∈ [0, 1], we obtain

T(u)(t)-Tv(t)0t(t-qs)(α-1)Γq(α)f(s,u(s))-f(s,v(s))dqs+01tα-1(1-qs)(α-1)(1-βξα-1)Γq(α)f(s,u(s))-f(s,v(s))dqs+0ξβtα-1(ξ-qs)(α-1)(1-βξα-1)Γq(α)f(s,u(s))-f(s,v(s))dqsLu-v(0t(t-qs)(α-1)Γq(α)dqs+01tα-1(1-qs)(α-1)(1-βξα-1)Γq(α)dqs+0ξβtα-1(ξ-qs)(α-1)(1-βξ(α-1))Γq(α)dqs)L(2-βξα-1+βξα)(1-βξα-1)Γq(α+1)u-v=ΛL,α,β,ξu-v,

where

ΛL,α,β,ξ=L(2-βξα-1+βξα)(1-βξα-1)Γq(α+1),

which depends only on the parameters involved in the problem. As ΛL,α,β,ξ < 1, Then, combining with Lemma 2.5, the Banach fixed point theorem assures that operator has a unique fixed point in ℂ and then the problem (1.1) has a unique positive solution. The proof is complete.

In this section, we will present some examples to illustrate the main results.

Example 4.1

Consider the following q-fractional three-point boundary value problem

{(Dqαu)(t)=(2u2+u)(2+sin u)9u+1,         t[0,1],u(0)=0,         u(1)=βu(ξ).

where α = 1.5 and β = ξ = q = 0.5. By simple computation, we can easily have

0lim supu+maxt[0,1]f(t,u)u0.666667<(1-βξα-1)Γq(α+1)0.769655.

Thus, all the assumptions of Theorem 3.1 holds. Consequently, the conclusion of Theorem 3.1 implies that the problem (4.1) has at least one positive solution.

Example 4.2

Consider the following q-fractional three-point boundary value problem

{(Dqαu)(t)=e-πtu(t)(6+e-πt)(1+u(t)),         t[0,1],u(0)=0,         u(1)=βu(ξ),

where α = 1.5 and β = ξ = q = 0.5. Let

f(t,u)=e-πtu(6+e-πt)(1+u).

Clearly, L = 1/6 as |f(t, u) − f(t, v)| ≤ 1/6|uv|. Further,

2L(2-βξα-1+βξα)(1-βξα-1)Γq(α+1)=2(2-0.5*0.50.5+0.5*0.51.5)6(1-0.5*0.50.5)Γ0.5(2.5)0.789628<1.

Thus, all the assumptions of Theorem 3.2 are satisfied. Therefore, the conclusion of Theorem 3.2 implies that the problem (4.2) has a unique positive solution.

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