# Positive Solutions for Boundary Value Problem of Nonlinear Fractionalq-Difference Equation

Authors: Moustafa El-Shahed; Farah M. Al-Askar
Journal: ISRN Mathematical Analysis
 (2011)

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## Abstract

We investigate the existence of multiple positive solutions to the nonlinearq-fractional boundary value problem cDqau(t)+a(t)f(u(t))=0,
0≤t≤1,2<a≤3,
u(0)=Dq2u(0)=0,γDqu(1)+βDq2u(1)=0, by using a fixed point theorem in a cone.


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## Body

## 1. Introduction

Fractional differential calculus is a discipline to which many researchers are dedicating their time, perhaps because of its demonstrated applications in various fields of science and engineering [1]. Many researchers studied the existence of solutions to fractional boundary value problems, for example, [2–5].Theq-difference calculus or quantum calculus is an old subject that was initially developed by Jackson [6, 7], basic definitions and properties of q-difference calculus can be found in the book mentioned in [8].The fractionalq-difference calculus had its origin in the works by Al-Salam [9] and Agarwal [10]. 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 MittageLeffler function [11], just to mention some.El-Shahed and Hassan [12] studied the existence of positive solutions of the q-difference boundary value problem: (1.1)-Dq2u(t)=a(t)f(u(t)),0≤t≤1,αu(0)-βDqu(0)=0,γu(1)+δDqu(1)=0.Ferreira [13] considered the existence of positive solutions to nonlinear q-difference boundary value problem:
(1.2)(DRLqαu)(t)=-f(t,u(t)),0<t<1,1<α≤2,u(0)=u(1)=0.
In other paper, Ferreira [14] studied the existence of positive solutions to nonlinear q-difference boundary value problem:(1.3)(DRLqαu)(t)=-f(t,u(t)),0<t<1,2<α≤3,u(0)=(Dqu)(0)=0,(Dqy)(1)=β≥0.In this paper, we consider the existence of positive solutions to nonlinearq-difference equation:
(1.4)DCqαu+a(t)f(u(t))=0,0≤t≤1,2<α≤3,
with the boundary conditions
(1.5)u(0)=Dq2u(0)=0,γDqu(1)+βDq2u(1)=0,
where γ,β≥0 and DCqα is the fractional q-derivative of the Caputo type.


## 2. Preliminaries of Fractionalq-Calculus

Letq∈(0,1) and define [8] (2.1)[α]q=qa-1q-1=qa-1+⋯+1,a∈R.
The q-analogue of the power (a-b)n is (2.2)(a-b)(0)=1,(a-b)(n)=∏k=0n-1(a-bqk),a,b∈R,n∈N.
If  α  is not a positive integer, then  (2.3)(a-b)(α)=aα∏i=0∞(1-(b/a)qi)(1-(b/a)qα+i).
Note that if b=0, then a(α)=aα. The q-gamma function is defined by(2.4)Γq(x)=(1-q)(x-1)(1-q)x-1,x∈R∖{0,-1,-2,…},0<q<1
and satisfies Γq(x+1)=[x]qΓq(x).Theq-derivative of a function f is here defined by (2.5)Dqf(x)=dqf(x)dqx=f(qx)-f(x)(q-1)x
and q-derivatives of higher order by(2.6)Dqnf(x)={f(x)ifn=0,DqDqn-1f(x)ifn∈N.
The q-integral of a function f defined in the interval [0,b] is given by(2.7)∫0xf(t)dqt=x(1-q)∑n=0∞f(xqn)qn,0≤|q|<1,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 (2.8)∫abf(t)dqt=∫0bf(t)dqt-∫0af(t)dqt.
Similarly as done for derivatives, an operator Iqn can be defined, namely,(2.9)(Iq0f)(x)=f(x),(Iqnf)(x)=Iq(Iqn-1f)(x),n∈N.
The fundamental theorem of calculus applies to these operators Iq and Dq, that is,
(2.10)(DqIqf)(x)=f(x),
and if f is continuous at x=0, then (2.11)(IqDqf)(x)=f(x)-f(0).
Basic properties of the two operators can be found in the book mentioned in [8]. We now point out three formulas that will be used later (Diq denotes the derivative with respect to variable i) [13] (2.12)[a(t-s)](α)=aα(t-s)(α),(2.13)Dtq(t-s)(α)=[α]q(t-s)(α-1),(2.14)(Dxq∫0xf(x,t)dqt)(x)=∫0xDxqf(x,t)dqt+f(qx,x).Remark 2.1.
We note that ifα>0 and a≤b≤t, then (t-a)(α)≥(t-b)(α) [13].
The following definition was considered first in [10].Definition 2.2.
Letα≥0 and f be a function defined on [0,1]. The fractional q-integral of the Riemann-Liouville type is (IRLq0f)(x)=f(x) and
(2.15)(IRLqαf)(x)=1Γq(α)∫ax(x-qt)(α-1)f(t)dqt,α∈R+,x∈[0,1].Definition 2.3 (see [15]).
The fractionalq-derivative of the Riemann-Liouville type of order α≥0 is defined by (DRLq0f)(x)=f(x) and
(2.16)(DRLqαf)(x)=(Dq[α]Iq[α]-αf)(x),α>0,
where [α] is the smallest integer greater than or equal to α.Definition 2.4 (see [15]).
The fractionalq-derivative of the Caputo type of order α≥0 is defined by
(2.17)(DCqαf)(x)=(Iq[α]-αDq[α]f)(x),α>0,
where [α] is the smallest integer greater than or equal to α.Lemma 2.5.
Letα,β≥0 and let f be a function defined on [0,1]. Then, the next formulas hold:(1)
(IqβIqαf)(x)=(Iqα+βf)(x),(2)
(DqαIqαf)(x)=f(x).The next result is important in the sequel. It was proved in a recent work by the author of [13].Theorem 2.6.
Letα>0 and n∈N. Then, the following equality holds:
(2.18)(IRLqαDRLqnf)(x)=DRLqnIRLqαf(x)-∑k=0α-1xα-n+kΓq(α+k-n+1)(Dqkf)(0).Theorem 2.7 (see [15]).
Letα>0 and n∈ℜ+∖N. Then, the following equality holds:
(2.19)(IqαDCqαf)(x)=f(x)-∑k=0[α]-1xkΓq(k+1)(Dqkf)(0).


## 3. Fractional Boundary Value Problem

We will consider now the question of existence of positive solutions to the following problem:(3.1)DCqαu(t)+a(t)f(u(t))=0,0≤t≤1,2<α≤3,u(0)=Dq2u(0)=0,γDqu(1)+βDq2u(1)=0,
where γ,β≥0,DCqα is the fractional q-derivative of the Caputo type, and f∈C([0,∞),[0,∞)),a∈C([0,1],[0,∞)) such that ∫01a(s)dqs>0. To that end, we need the following theorem.Theorem 3.1 (see [16, 17]).
LetX be a Banach space, and P⊂X is a cone in X. Assume that Ω1 and Ω2 are open subsets of X with 0∈Ω1 and Ω¯1⊂Ω2.
LetT:P∩(Ω¯2∖Ω1)→P be completely continuous operator. In addition suppose either:P1:
∥Tu∥≤∥u∥,u∈P∩∂Ω1 and  ∥Tu∥≥∥u∥,u∈P∩∂Ω2 orP2:
∥Tu∥≤∥u∥,u∈P∩∂Ω2 and ∥Tu∥≥∥u∥,u∈P∩∂Ω1,holds. Then, T has a fixed point in P∩(Ω¯2∖Ω1).Lemma 3.2.
Lety∈C[0,1]; then the boundary value problem
(3.2)DCqαu(t)+y(t)=0,0≤t≤1,2<α≤3,u(0)=Dq2u(0)=0,γDqu(1)+βDq2u(1)=0
has a unique solution
(3.3)u(t)=∫01G(t,qs)y(s)dqs,
where
(3.4)G(t,s)={t(1-s)(α-2)Γq(α-1)+βγt(1-s)(α-3)Γq(α-2)0≤t≤s≤1,t(1-s)(α-2)Γq(α-1)+βγt(1-s)(α-3)Γq(α-2)-(t-s)(α-1)Γq(α)0≤s≤t≤1.Proof.
We may apply Lemma2.5 and Theorem 2.7; we see that
(3.5)u(t)=u(0)+Dqu(0)Γq(2)t+Dq2u(0)Γq(3)t2-Iqαy(t).
By using the boundary conditions u(0)=Dq2u(0)=0, we get
(3.6)u(t)=At-∫0t(t-qs)(α-1)Γq(α)y(s)dqs.
Differentiating both sides of (3.6), one obtains, with the help of (2.13), and (2.14)
(3.7)(Dqu)(t)=A-∫0t[α-1]q(t-qs)(α-2)Γq(α)y(s)dqs,(Dq2u)(t)=-∫0t[α-1]q[α-2]q(t-qs)(α-3)Γq(α)y(s)dqs.
Then, by the condition γDqu(1)+βDq2u(1)=0, we have
(3.8)A=∫01(1-qs)(α-2)Γq(α-1)y(s)dqs+βγ∫01(1-qs)(α-3)Γq(α-2)y(s)dqs.
The proof is complete.Lemma 3.3.
FunctionG defined above satisfies the following conditions:
(3.9)G(t,qs)≥0,G(1,qs)≥G(t,qs),0≤t,s≤1,(3.10)G(t,qs)≥g(t)G(1,qs),0≤t,s≤1withg(t)=t.Proof.
We start by defining two functions(3.11)g1(t,s)=t(1-s)(α-2)Γq(α-1)+βγt(1-s)(α-3)Γq(α-2)-(t-s)(α-1)Γq(α),0≤s≤t≤1,g2(t,s)=t(1-s)(α-2)Γq(α-1)+βγt(1-s)(α-3)Γq(α-2),0≤t≤s≤1.
It is clear that g2(t,qs)≥0. Now, g1(0,qs)=0 and, in view of Remark 2.1, for t≠0(3.12)g1(t,qs)=t(1-qs)(α-2)Γq(α-1)+βγt(1-qs)(α-3)Γq(α-2)-(t-qs)(α-1)Γq(α)≥t(1-qs)(α-2)Γq(α-1)+βγt(1-qs)(α-3)Γq(α-2)-tα-1(1-qs)(α-1)Γq(α)≥t(1-qs)(α-2)Γq(α-1)+βγt(1-qs)(α-3)Γq(α-2)-t(1-qs)(α-1)Γq(α)=tΓq(α)[[α-1]q(1-qs)(α-2)-(1-qs)(α-1)]≥0.
Therefore, G(t,qs)≥0. Moreover, for fixed s∈[0,1],
(3.13)Dtqg1(t,qs)=(1-qs)(α-2)Γq(α-1)+βγ(1-qs)(α-3)Γq(α-2)-[α-1]q(t-qs)(α-2)Γq(α)≥1Γq(α)[[α-1]q(1-qs)(α-2)-[α-1]q(t-qs)(α-2)]=1Γq(α-1)[(1-qs)(α-2)-(t-qs)(α-2)]≥0,
that is, g1(t,qs) is an increasing function of t. Obviously, g2(t,qs) is increasing in t, therefore; G(t,qs) is an increasing function of t for fixed s∈[0,1]. This concludes the proof of (3.9).
Suppose now thatt≥qs. Then,
(3.14)G(t,qs)G(1,qs)=γ[α-1]qt(1-qs)(α-2)+β[α-1]q[α-2]qt(1-qs)(α-3)-γ(t-qs)(α-1)γ[α-1]q(1-qs)(α-2)+β[α-1]q[α-2]q(1-qs)(α-3)-γ(1-qs)(α-1)≥γ[α-1]qt(1-qs)(α-2)+β[α-1]q[α-2]qt(1-qs)(α-3)-γt(1-qs)(α-1)γ[α-1]q(1-qs)(α-2)+β[α-1]q[α-2]q(1-qs)(α-3)-γ(1-qs)(α-1)=t.
If t≤qs, Then
(3.15)G(t,qs)G(1,qs)=t,
and this finishes the proof of (3.10).Remark 3.4.
If we let0<τ<1, then
(3.16)min⁡t∈[τ,1]G(t,qs)≥τG(1,qs),fors∈[0,1].
Let X=C[0,1] be the Banach space endowed with norm ∥u∥=supt∈[τ,1]|u(t)|. Let τ=qn [18] for a given n∈N and define the cone P⊂X by
(3.17)P={u∈X:u(t)≥0,min⁡t∈[τ,1]u(t)≥τ‖u‖},
and the operator  T:P→X is defined by
(3.18)Tu(t)=∫01G(t,qs)a(s)f(u(s))dqs.Remark 3.5.
It follows from the nonnegativeness and continuity ofG,a, and f that the operator T:P→X is completely continuous [3]. Moreover, foru∈P,Tu(t)≥0 on [0,1] and
(3.19)min⁡t∈[τ,1]⁡Tu(t)=min⁡t∈[τ,1]∫01G(t,qs)a(s)f(u(s))dqs≥τ∫01G(1,qs)a(s)f(u(s))dqs=τ‖Tu‖
that is, T(P)⊂P.
Throughout this section, we will use the following notations:(3.20)Λ1=(∫01G(1,qs)a(s)dqs)-1,Λ2=(τ2∫τ1G(1,qs)a(s)dqs)-1.
It is obvious that Λ1,Λ2>0. Also, we define
(3.21)f0=lim⁡u→0+max⁡0≤t≤1f(u)u,f∞=lim⁡u→+∞max⁡0≤t≤1f(u)u.Theorem 3.6.
Letτ=qn with n∈N. And assume that the following assumptions are satisfied:(H1)
f0=f∞=0;(H2)
There exist constantk>0 and M∈(Λ2,∞) such that(3.22)f(u)≥Mk,foru∈[τk,k].
Then, the BVP (3.1) has at least two positive solutions u1,u2, such that
(3.23)r<‖u1‖<k<‖u2‖<R.Proof.
First, sincef0=0, for any ε∈(0,Λ1) there exists r∈(0,k) such that
(3.24)f(u)≤εu,foru∈[0,r].
Letting Ω1={u∈P:∥u∥<r}, for any u∈∂Ω1, we get
(3.25)‖Tu‖=max⁡t∈[0,1]∫01G(t,qs)a(s)f(u(s))dqs≤εr∫01G(1,qs)a(s)dqs<r=‖u‖,
which yields
(3.26)‖Tu‖≤‖u‖,∀u∈∂Ω1.
Thus,
(3.27)i(T,Ω1,P)=1.
Second, in view of f∞=0, then for any ε∈(0,Λ1), there exists l>k such that
(3.28)f(u)≤εu,foru∈[l,∞)
and we consider two cases.Case 1 (2.12).
Suppose thatf(u) is unbounded; then from f∈C([0,∞),[0,∞)), there is R>l such that
(3.29)f(u)≤f(R),foru∈[0,R],
then, from (3.28) and (3.29), one has
(3.30)f(u)≤f(R)≤εR,foru∈[0,R].
For u∈∂Ω2, we get
(3.31)‖Tu‖=max⁡t∈[0,1]∫01G(t,qs)a(s)f(u(s))dqs≤∫01G(1,qs)a(s)f(R)dqs≤εR∫01G(1,qs)a(s)dqs<R=‖u‖.Case 2 (2.13).
Suppose thatf(u) is bounded and L=max0≤t≤1,0≤u≤R|f(t,u(t))|. Taking R=max{L/ε,k}, for u∈∂Ω2, we get
(3.32)‖Tu‖=max⁡t∈[0,1]∫01G(t,qs)a(s)f(u(s))dqs≤L∫01G(1,qs)a(s)dqs≤εR∫01G(1,qs)a(s)dqs<R=‖u‖.
Hence, in either case, we always may set Ω2={u∈P:∥u∥<R} such that
(3.33)‖Tu‖≤‖u‖,∀u∈∂Ω2.
Thus,
(3.34)i(T,Ω2,P)=1.
Finally, set Ω={u∈P:∥u∥<k}, for u∈∂Ω, since mint∈[τ,1]u(t)≥τ∥u∥=τk for  u∈P, and hence, for any u∈∂Ω, from (H2), we can get
(3.35)‖Tu‖=max⁡t∈[0,1]∫01G(t,qs)a(s)f(u(s))dqs≥τ∫01G(1,qs)a(s)f(u(s))dqs≥τ2Mk∫τ1G(1,qs)a(s)dqs>k=‖u‖,
which implies
(3.36)‖Tu‖≥‖u‖,∀u∈∂Ω.
Thus,
(3.37)i(T,Ω,P)=0.
Hence, since r<k<R and from (3.27), (3.34), and (3.37), it follows from the additivity of the fixed point index that
(3.38)i(T,Ω2∖Ω̅,P)=1,i(T,Ω∖Ω̅1,P)=-1.
Thus, T has two  positive solutions u1,u2 such that r<∥u1∥<k<∥u2∥<R for t∈(0,1]. So, the proof is complete.



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