# On the Solvability of Caputoq-Fractional Boundary Value Problem Involving p-Laplacian Operator

Authors: Hüseyin Aktuğlu; Mehmet Ali Özarslan
Journal: Abstract and Applied Analysis
 (2013)

---

## Abstract

We consider the model of a Caputoq-fractional boundary value
problem involving p-Laplacian operator. By using the Banach contraction
mapping principle, we prove that, under some conditions, the suggested model
of the Caputo q-fractional boundary value problem involving p-Laplacian
operator has a unique solution for both cases of 0<p<1 and p>2. It is
interesting that in both cases solvability conditions obtained here depend on
q, p, and the order of the Caputo q-fractional differential equation.
Finally, we illustrate our results with some examples.


---

## Body

## 1. Introduction

In this section we will give some basic definitions and results that will be needed in the sequel. For more details about the theory ofq-calculus, fractional calculus, and q-fractional calculus, we refer readers to [1–10].Letq∈(0,1) be a fixed real number. Then for any α∈ℝ,
(1)[α]q:=1-qα1-q.
The  q-binomial function is defined for all  n∈ℕ as
(2)(t-s)qn=∏k=0n-1(t-qks),(t-s)qβ=tβ∏i=0∞(1-(s/t)qi1-(s/t)qi+α),
where β is not a positive integer. It is easy to see that
(3)(at-as)qβ=aβ(t-s)qβ.
The q-analog of Euler's gamma function is denoted by Γq(t) and defined as
(4)Γq(t)=(1-q)qt-1(1-q)t-1,t>0.
The following theorem will be used to compare values of  Γ(t), the usual gamma function, with values of  Γq(t) for a fixed q∈(0,1).Theorem 1 (see [11]).
For0<r<q<1, one has
(5)Γr(t)≤Γq(t)≤Γ(t),for0<t≤1ort≥2,Γ(t)≤Γq(t)≤Γr(t),for1≤t≤2.It is known that for0<q<1,
(6)𝕋q={qn;n∈ℤ}∪{0},𝕋qα={qn+α;n∈ℤ}∪{0},α∈ℝ+∪{0}.
The nabla  q-derivative of the function f:𝕋q→ℝ is defined by
(7)∇qf(s)=f(s)-f(qs)(1-q)s,s∈𝕋q-{0}.
The nabla q-integral of f is defined by
(8)∫0sf(t)∇qt=(1-q)s∑k=0∞‍qkf(sqk).
Jackson in [12] and Thomae in [13] showed that the  q-beta function, which is defined by
(9)Bq(t,s)=Γq(t)Γq(s)Γq(t+s),
has the following q-integral representation:
(10)Bq(t,s)=∫01τt-1(1-qτ)qs-1∇qτ,t,s>0.
The fundamental theorem of  q-calculus states that
(11)∇q∫0sf(t)∇qt=f(s),
and if f is continuous at 0, then
(12)∫0s∇qf(t)∇qt=f(s)-f(0).
Moreover,
(13)∇q∫0tf(t,s)∇qs=∫0t∇qf(t,s)∇qs+f(qt,t),
where the derivative is applied with respect to t.The nablaq-fractional derivative of (t-s)qα with respect to t and for all α∈ℝ is given by
(14)∇q(t-s)qα=1-qα1-q(t-s)qα-1.
Moreover, the q-fractional integral of order α≠0,-1,-2,… is defined by
(15)qI0αf(t)=1Γq(α)∫0t(t-qs)qα-1f(s)∇qs.
The α-order Caputo  q-fractional derivative of a function f is defined by
(16)qC0αf(t)=1Γq(n-α)∫0t(t-qs)n-α-1∇qnf(s)∇qs,
where n=[α]+1 and [α] denotes the greatest integer less than or equal to α.The following lemma enables us to transfer Caputoq-fractional differential equations into an equivalent q-fractional integral equation.Lemma 2 (see [3]).
Assume thatα>0 and f is defined on a suitable domain. Then
(17)qI0qαC0αf(t)=f(t)-∑i=0n-1‍tkΓq(k+1)∇qkf(0)
and if   0<α≤1, then
(18)qI0qαC0αf(t)=f(t)-f(0).On the other hand the operatorφp(s)=|s|p-2s, where p>1 is called the p-Laplacian operator. It is easy to see that φp-1=φr, where (1/p)+(1/r)=1. The following properties of  p-Laplacian operator will be used in the rest of the paper.(P1)
if1<p<2,  xy>0, and |x|,|y|≥m>0, then |φp(x)-φp(y)|≤(p-1)mp-2|x-y|;(P2)
ifp≥2 and |x|,|y|≤M then, |φp(x)-φp(y)|≤(p-1)Mp-2|x-y|.


## 2. A Model of Caputoq-Fractional Boundary Value Problem Involving  p-Laplacian Operator

In this paper, our main aim is to prove the existence and uniqueness of the solution for the following Caputoq-fractional boundary value problem involving the p-Laplacian operator:
(19)∇q(φp(Cq0αx(t)))=f(t,x(t)),∇qkx(0)=0,fork=2,3,…,n-1,x(0)=a0x(1),∇qx(0)=a1∇qx(1),
where a0,a1≠1,  1<α∈ℝ, and f∈C([0,1]×ℝ,ℝ).Note that, the boundary value problem given in (19) is antiperiodic for a0,a1=-1.In the following lemma we obtain aq-integral equation which is equivalent  to the Caputo q-fractional boundary value problem given in (19).Lemma 3.
Assume thatα>1,a0,a1≠1, and h∈C([0,1]). Then
(20)∇q(φp(Cq0αx(t)))=h(t),∇qkx(0)=0,fork=2,3,…,n-1,x(0)=a0x(1),∇qx(0)=a1∇qx(1)
are equivalent to the following q-integral equation:
(21)x(t)=b0q∫0t(t-qτ)qα-1φr(∫0τh(s)∇qs)∇qτ+b1q∫01(1-qτ)qα-1φr(∫0τh(s)∇qs)∇qτ+b2q(t)∫01(1-qτ)qα-2φr(∫0τh(s)∇qs)∇qτ,
where b0q=1/(Γq(α)),  b1q=a0/(Γq(α)(1-a0)), and b2q(t)=(a1(t+a0(1-t)))/ (Γq(α-1)(1-a0) (1-a1)).Proof.
Using (20) and the fact thatφp(qC0αx(0))=0, we have
(22)φp(C0αqx(t))=∫0th(s)∇qs,
or equivalently,
(23)qC0αx(t)=φr(∫0th(s)∇qs).
Applying  q-fractional integral operatorqI0α to both sides and using Lemma 2, we get
(24)x(t)-∑k=0n-1‍tkΓq(k+1)∇qkx(0)=1Γq(α)∫0t(t-qτ)qα-1φr(∫0τh(s)∇qs)∇qτ.
Using  ∇qkx(0)=0,   for k=2,3,…,[α]-1 in (24), we obtain
(25)x(t)=1Γq(α)∫0t(t-qτ)qα-1φr(∫0τh(s)∇qs)∇qτ+x(0)+t∇qx(0).
According to (13) and (14), we have
(26)∇qx(t)=1Γq(α-1)∫0t(t-qτ)qα-2φr(∫0τh(s)∇qs)∇qτ+∇qx(0).
Taking  t=1 in both sides of (25) and (26), we get
(27)x(1)=1Γq(α)∫01(1-qτ)qα-1φr(∫0τh(s)∇qs)∇qτ+x(0)+∇qx(0),∇qx(1)=1Γq(α-1)∫01(1-qτ)qα-2φr(∫0τh(s)∇qs)∇qτ+∇qx(0).
Solving equations obtained by the given boundary value conditions x(0)=a0x(1) and ∇qx(0)=a1∇qx(1), it follows that
(28)∇qx(0)=a1Γq(α-1)(1-a1)×∫01(1-qτ)qα-2φr(∫0τh(s)∇qs)∇qτ,x(0)=a0Γq(α)(1-a0)∫01(1-qτ)qα-1φr(∫0τh(s)∇qs)∇qτ+a0Γq(α-1)(1-a0)(1-a1)×∫01(1-qτ)qα-2φr(∫0τh(s)∇qs)∇qτ.
Substituting (28) into (25) gives (21) which completes the proof.


## 3. Solvability of the Caputoq-Fractional Boundary Value Problem

This section is devoted to the solvability of the Caputoq-fractional boundary value problem given in (19). In the first part we shall prove the existence and uniqueness of the solution, and then we shall illustrate our main results with some examples.Recall thatC[0,1] is a Banach space with the norm ∥x∥=maxt∈[0,1]|x(t)|. Now consider Ti:C[0,1]→C[0,1],i=0,1 with
(29)T0x(t):=φr(∫0tf(s,x(s))∇qs),T1x(t)=b0q∫0t(t-qτ)qα-1x(τ)∇qτ+b1q∫01(1-qτ)qα-1x(τ)∇qτ+b2q(t)∫01(1-qτ)qα-2x(τ)∇qτ.
Then T=T1∘T0 is a continuous and compact operator.Theorem 4.
Suppose that1<r<2,  a0,a1≠1,q∈(0,1) is fixed, and the following conditions hold:
∃
λ
>
0
, 0<δ<2/(2-r) and d with
(30)0<d<λ2-rΓq(δ(r-2)+2+α)(r-1)Γq(δ(r-2)+2)×[|1-a0||1-a1|(|1-a0|+|a0|)(|1-a1|+|a1|[δ(r-2)+α+1]q)]
such that
(31)[δ]qλtδ-1≤f(t,x),forany(t,x)∈(0,1]×ℝ,(32)|f(t,x)-f(t,y)|≤d|x-y|,fort∈[0,1],x,y∈ℝ.
Then the boundary value problem (19) has a unique solution.Proof.
Inequality given in (31) implies that
(33)λtδ≤∫0tf(s,x)∇qs,forany(t,x)∈[0,1]×ℝ.
On the other hand using (P1) and (32), we have
(34)|T0x(t)-T0y(t)|=|φr(∫0tf(s,x(s))∇qs)-φr(∫0tf(s,y(s))∇qs)|≤(r-1)(λtδ)r-2|∫0tf(s,x(s))∇qs-∫0tf(s,y(s))∇qs|≤(r-1)(λtδ)r-2∫0t|f(s,x(s))-f(s,y(s))|∇qs≤d(r-1)(λtδ)r-2∫0t|x(s)-y(s)|∇qs≤d(r-1)λr-2tδ(r-2)+1∥x-y∥.
Similarly,(35)|Tx(t)-Ty(t)|=|T1(T0x(t))-T1(T0y(t))|=|b0q∫0t(t-qτ)qα-1((T0x)(τ)-(T0y)(τ))∇qτ+b1q∫01(1-qτ)qα-1((T0x)(τ)-(T0y)(τ))∇q+b2q(t)∫01(1-qτ)qα-2((T0x)(τ)-(T0y)(τ))∇qτ|.
Finally using (34) in (35), we get
(36)|Tx(t)-Ty(t)|≤d(r-1)λr-2∥x-y∥×[b0q∫0t(t-qτ)qα-1τδ(r-2)+1∇qτ××+|b1q|∫01(1-qτ)qα-1τδ(r-2)+1∇qτ××+|b2q(t)|∫01(1-qτ)qα-2τδ(r-2)+1∇qτ].
Since
(37)∫0t(t-qτ)qα-1τδ(r-2)+1∇qτ=∫01(1-qτ)qα-1tδ(r-2)+α+1τδ(r-2)+1∇qτ,
we have
(38)|Tx(t)-Ty(t)|≤d(r-1)λr-2∥x-y∥×[b0q∫01(1-qτ)qα-1tδ(r-2)+α+1τδ(r-2)+1∇qτ+|b1q|∫01(1-qτ)qα-1τδ(r-2)+1∇qτ+|b2q(t)|∫01(1-qτ)qα-2τδ(r-2)+1∇qτ]=d(r-1)λr-2∥x-y∥×[b0qtδ(r-2)+α+1Bq(δ(r-2)+2,α)+|b1q|Bq(δ(r-2)+2,α)+|b2q(t)|Bq(δ(r-2)+2,α-1)b0qtδ(r-2)+α+1]=d(r-1)λr-2∥x-y∥×[[δ(r-2)+α+1]q[α-1]qb0qtδ(r-2)+α+1Bq(δ(r-2)+2,α)+|b1q|Bq(δ(r-2)+2,α)+|b2q(t)|×[δ(r-2)+α+1]q[α-1]qBq(δ(r-2)+2,α)]≤d(r-1)λr-2∥x-y∥Bq(δ(r-2)+2,α)×[b0qtδ(r-2)+α+1+|b1q|+|b2q(t)|[δ(r-2)+α+1]q[α-1]q].
In other words,(39)|Tx(t)-Ty(t)|≤d(r-1)λr-2∥x-y∥Γq(δ(r-2)+2)Γq(α)Γq(δ(r-2)+2+α)×[1Γq(α)tδ(r-2)+α+1+|a0Γq(α)(1-a0)|+|a1(t+a0(1-t))Γq(α-1)(1-a0)(1-a1)|[δ(r-2)+α+1]q[α-1]q]≤d(r-1)λr-2∥x-y∥Γq(δ(r-2)+2)Γq(α)Γq(δ(r-2)+2+α)×[1Γq(α)+|a0|Γq(α)|1-a0|+|a1|(|a0|+|1-a0|)[δ(r-2)+α+1]qΓq(α)|1-a0||1-a1|]≤d(r-1)λr-2∥x-y∥Γq(δ(r-2)+2)Γq(α)Γq(δ(r-2)+2+α)×[(Γq(α)|1-a0||1-a1|)-1([δ(r-2)+α+1]q|1-a0||1-a1|+|a0||1-a1|+|a1|++|a1|(|a0|+|1-a0|)[δ(r-2)+α+1]q)+|a1|×(Γq(α)|1-a0||1-a1|)-1]=d(r-1)λr-2Γq(δ(r-2)+2)Γq(δ(r-2)+2+α)×[(|1-a0|+|a0|)(|1-a1|+|a1|[δ(r-2)+α+1]q)|1-a0||1-a1|]×∥x-y∥=K∥x-y∥,
where  K=d(r-1)λr-2(Γq(δ(r-2)+2)/Γq(δ(r-2)+2+α))[((|1-a0|+|a0|)(|1-a1|+ |a1|[δ(r-2)+α+1]q))/(|1-a0||1-a1|)].
By condition (30), we get 0<K<1, which implies that T is a contraction. As a consequence of the Banach contraction mapping theorem and Lemma 3, the boundary value problem given in (19) has a unique solution.Theorem 5.
Suppose that1<r<2,    a0,a1≠1, and the following conditions hold for a fixed q∈(0,1),
∃
λ
>
0, 0<δ<2/(2-r), and d with
(40)0<d<λ2-rΓq(δ(r-2)+2+α)(r-1)Γq(δ(r-2)+2)×[|1-a0||1-a1|(|1-a0|+|a0|)(|1-a1|+|a1|[δ(r-2)+α+1]q)]
such that
(41)f(t,x)≤-[δ]qλtδ-1,forany(t,x)∈(0,1]×ℝ,|f(t,x)-f(t,y)|≤d|x-y|,fort∈[0,1],x,y∈ℝ.
Then the boundary value problem (19) has a unique solution.Remark 6.
Whenq→1, Theorems 4 and  5 reduce to Theorems 3.1 and  3.2 of  [14].Theorem 7.
Suppose thatr>2,a0,a1≠1, and the following conditions hold for a fixed q∈(0,1). There exists a nonnegative function g(x)∈L[0,1] with M:=∫01g(τ)∇qτ≥0 such that
(42)|f(t,x)|≤g(t),forany(t,x)∈[0,1]×ℝ,
and there exists a constant d with
(43)0<d<Γq(α+2)(r-1)Mr-2×[|1-a0||1-a1|(|1-a0|+|a0|)(|1-a1|+|a1|[α+1]q)],(44)|f(t,x)-f(t,y)|≤d|x-y|,fort∈[0,1],x,y∈ℝ.
Then the boundary value problem (19) has a unique solution.Proof.
By (42), we can get that
(45)∫0t|f(τ,x(τ))|∇qτ≤∫01g(τ)∇qτ=M
for all t∈[0,1]. By the definition of  T0, we have
(46)|T0x(t)-T0y(t)|=|φr(∫0tf(s,x(s))∇qs)-φr(∫0tf(s,y(s))∇qs)|.
Using (P2) and (45) gives
(47)|T0x(t)-T0y(t)|≤(r-1)Mr-2×|∫0tf(s,x(s))∇qs-∫0tf(s,y(s))∇qs|≤(r-1)Mr-2×∫0t|f(s,x(s))-f(s,y(s))|∇qs≤d(r-1)Mr-2∫0t|x(s)-y(s)|∇qs≤d(r-1)Mr-2t∥x-y∥.
Therefore,(48)|Tx(t)-Ty(t)|=|T1(T0x(t))-T1(T0y(t))|=|b0q∫0t(t-qτ)qα-1((T0x)(τ)-(T0y)(τ))∇qτ+b1q∫01(1-qτ)qα-1((T0x)(τ)-(T0y)(τ))∇q+b2q(t)∫01(1-qτ)qα-2((T0x)(τ)-(T0y)(τ))∇qτ|≤d(r-1)Mr-2∥x-y∥×[1Γq(α)∫0t(t-qτ)qα-1τ∇qτ+|a0|Γq(α)|1-a0|∫01(1-qτ)qα-1τ∇qτ+|a1(t+a0(1-t))|Γq(α-1)|1-a0||1-a1|∫01(1-qτ)qα-2τ∇qτ].
Since
(49)∫0t(t-qτ)qα-1τ∇qτ=∫01tα+1(1-qτ)qα-1τ∇qτ
we have
(50)|Tx(t)-Ty(t)|≤d(r-1)Mr-2∥x-y∥×[|a1|(|a0|+t|1-a0|)Γq(α-1)|1-a0||1-a1|1Γq(α)∫01tα+1(1-qτ)qα-1τ∇qτ+|a0|Γq(α)|1-a0|∫01(1-qτ)qα-1τ∇qτ+|a1|(|a0|+t|1-a0|)Γq(α-1)|1-a0||1-a1|∫01(1-qτ)qα-2τ∇qτ].
Using  q-Beta function and the fact that t∈[0,1], we get
(51)|Tx(t)-Ty(t)|≤d(r-1)Mr-2∥x-y∥×[1Γq(α)Bq(2,α)+|a0|Γq(α)|1-a0|Bq(2,α)+|a1|(|a0|+|1-a0|)Γq(α-1)|1-a0||1-a1|[α+1]q[α-1]qBq(2,α)]≤d(r-1)Mr-2Bq(2,α)∥x-y∥×[1Γq(α)+|a0|Γq(α)|1-a0|+|a1|(|a0|+|1-a0|)[α+1]qΓq(α)|1-a0||1-a1|]≤d(r-1)Mr-2Bq(2,α)∥x-y∥×[(Γq(α)|1-a0||1-a1|)-1([α+1]q|1-a0||1-a1|+|a0||1-a1|+|a1|(|a0|+|1-a0|)[α+1]q)×(Γq(α)|1-a0||1-a1|)-1]≤d(r-1)Mr-2Γq(α+2)×[(|1-a0|+|a0|)(|1-a1|+|a1|[α+1]q)|1-a0||1-a1|]×∥x-y∥≤K∥x-y∥,
where
(52)K=d(r-1)Mr-2Γq(α+2)×[(|1-a0|+|a0|)(|1-a1|+|a1|[α+1]q)|1-a0||1-a1|].
By condition (43), we get K<1 which implies that T is a contraction; therefore boundary value problem given in (19) has a unique solution.Next, we give some examples to illustrate our results.Example 8.
Consider the following Boundary value problem(53)∇q(φ7/3(Cq03/2x(t)))=4t2(2+cos(πx24+ω)),t∈(0,1),∇qkx(0)=0,fork=2,3,…,n-1,x(0)=12x(1),∇qx(0)=12∇qx(1),
where
(54)p=73,α=32,δ=4,a0=12,a1=12.
Then r=7/4, and take δ=4,  λ=1, and d=π/6. Using  Theorem 1  and the fact that [3/2]q>1 for any fixed q∈(0,1), we have
(55)λ2-rΓq(δ(r-2)+2+α)(r-1)Γq(δ(r-2)+2)×[(|1-a1|+|a1|[δ(r-2)+α+1]q))-1(|1-a0||1-a1|)×([1]q(|1-a0|+|a0|)×××(|1-a1|+|a1|[δ(r-2)+α+1]q))-1]=23[[3/2]qΓq(3/2)(1+[3/2]q)]>23[[3/2]qΓq(3/2)([3/2]q+[3/2]q)]>13Γ(32)=π6=d>0,K=d(r-1)λr-2Γq(δ(r-2)+2)Γq(δ(r-2)+2+α)×[(|1-a0|+|a0|)(|1-a1|+|a1|[δ(r-2)+α+1]q)|1-a0||1-a1|]=π4Γq(5/2)[(1+[32]q)]<π4[3/2]qΓq(3/2)[([32]q+[32]q)]≤π2Γ(3/2)=1.
Moreover, it can be easily seen that
(56)[δ]qλtδ-1=[4]qt3≤4t2(2+cos(πx24+ω))=f(t,x).
Finally,
(57)|f(t,x)-f(t,y)|=|4t2(2+cos(πx24+ω))=-4t2(2+cos(πy24+ω))|=4t2|cos(πx24+ω)-cos(πy24+ω)|≤4|(πx24+ω)-(πy24+ω)|=π6|x-y|.
Therefore as a consequence of Theorem 4, boundary value problem given in (53) has a unique solution.Example 9.
Consider the following boundary value problem:(58)∇q(φ9/4(φ31/15(Cq03/2x(t))))=4t2(2+cos(πx24+ω)),t∈(0,1),∇qkx(0)=0,fork=2,3,…,n-1,x(0)=12x(1),∇qx(0)=12∇qx(1).
Then
(59)φ9/4(φ31/15(s))=φ9/4(|s|1/15s)=||s|1/15s|(9/4)-2|s|1/15s=|s|(1/15)(1/4)|s|1/4|s|1/15s=|s|(1/60)+(1/4)+(1/15)s=|s|1/3s=|s|(7/3)-2s=φ7/3(s).
Therefore boundary value problem given in (58) reduces to the boundary value problem given in (53), and it has a unique solution.Example 10.
Now consider  the following antiperiodic boundary value problem:(60)∇q(φ7/4(Cq03/2x(t)))=(sin2(πx40+ω)),t∈(0,1),∇qkx(0)=0,fork=2,3,…,n-1,x(0)=-x(1),∇qx(0)=-∇qx(1),
where
(61)p=74,α=32,a0=-1,a1=-1.
Then  r=7/3, and take d=π/20. Using  Theorem 1 and taking g(t)=1, we get that
(62)M=1,Γq(α+2)(r-1)Mr-2×[|1-a0||1-a1|(|1-a0|+|a0|)(|1-a1|+|a1|[α+1]q)]=[Γq(7/2)(2+[5/2]q)]>[[5/2]q[3/2]qΓq(3/2)(2[5/2]q+[5/2]q)]=[3/2]qΓq(3/2)3>Γ(3/2)3=π6>π20=d.
On the other hand,
(63)|f(t,x)-f(t,y)|≤|sin2(πx40+ω)-sin2(πy40+ω)|≤π20|x-y|,fort∈[0,1],x,y∈ℝ,K=d(r-1)Mr-2Γq(α+2)×[(|1-a0|+|a0|)(|1-a1|+|a1|[α+1]q)|1-a0||1-a1|]=π20Γq(7/2)[2+[52]q]<π20[5/2]q[3/2]qΓq(3/2)[3[52]q]=3π20[3/2]qΓq(3/2)<3π20Γ(3/2)=310<1.
Therefore by Theorem 7, the antiperiodic boundary value problem given in (60) has a unique solution.



---

*Source: 658617-2013-06-27.xml*