# A Note on Fractional Differential Equations with Fractional Separated Boundary Conditions

Authors: Bashir Ahmad; Sotiris K. Ntouyas
Journal: Abstract and Applied Analysis
 (2012)

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## Abstract

We consider a new class of boundary value problems of nonlinear fractional differential equations with fractional separated boundary conditions. A connection between classical separated and fractional separated boundary conditions is developed. Some new existence and uniqueness results are obtained for this class of problems by using standard fixed point theorems. Some illustrative examples are also discussed.


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## Body

## 1. Introduction

In this paper, we investigate the existence of solutions for a fractional boundary value problem with fractional separated boundary conditions given by(1.1)Dcqx(t)=f(t,x(t)),t∈[0,1],1<q≤2,α1x(0)+β1(Dcpx(0))=γ1,α2x(1)+β2(Dcpx(1))=γ2,0<p<1,
where cDq denotes the Caputo fractional derivative of order q, f is a given continuous function, and αi,βi,γi(i=1,2) are real constants, with α1≠0.Fractional calculus has recently gained much momentum as extensive and significant progress on theoretical and practical aspects of the subject can easily be witnessed in the literature. As a matter of fact, the tools of fractional calculus have been effectively applied in the modelling of many physical and engineering problems. The recent development in the theory and methods for fractional calculus indicates its popularity. For some recent work on fractional boundary value problems, See [1–16] and the references therein.


## 2. Linear Problem

Let us recall some basic definitions of fractional calculus [1, 3].Definition 2.1.
For a continuous functiong:[0,∞)→ℝ, the Caputo derivative of fractional order q is defined as
(2.1)cDqg(t)=1Γ(n-q)∫0t(t-s)n-q-1g(n)(s)ds,n-1<q<n,n=[q]+1,
where [q] denotes the integer part of the real number q.Definition 2.2.
The Riemann-Liouville fractional integral of orderq is defined as
(2.2)Iqg(t)=1Γ(q)∫0tg(s)(t-s)1-qds,q>0,
provided the integral exists.
To define the solution of the boundary value problem (1.1) we need the following lemma, which deals with linear variant of the problem (1.1).Lemma 2.3.
For a givenσ∈C([0,1],ℝ), the unique solution of the problem
(2.3)Dcqx(t)=σ(t),t∈[0,1],1<q≤2,α1x(0)+β1(Dpcx(0))=γ1,α2x(1)+β2(Dcpx(1))=γ2,0<p<1,
is given by
(2.4)x(t)=∫0t(t-s)q-1Γ(q)σ(s)ds-tν1(α2∫01(1-s)q-1Γ(q)σ(s)ds+β2∫01(1-s)q-p-1Γ(q-p)σ(s)ds)+α1ν2t+γ1ν1α1ν1,
where
(2.5)ν1=α2Γ(2-p)+β2Γ(2-p),ν2=γ2α1-α2γ1α1.Proof.
It is well known [3] that the solution of fractional differential equation in (2.3) can be written as
(2.6)x(t)=Iqσ(t)-b1-b2t=∫0t(t-s)q-1Γ(q)σ(s)ds-b1-b2t.
Using cDpb=0 (b is a constant), cDpt=t1-p/Γ(2-p), cDpIqσ(t)=Iq-pσ(t), (2.6) gives
(2.7)Dcpx(t)=∫0t(t-s)q-p-1Γ(q-p)σ(s)ds-b2t1-pΓ(2-p).
From the boundary condition α1x(0)+β1(cDpx(0))=γ1, we have
(2.8)α1(-b1)+β1(0)=γ1,whichimpliesthatb1=-γ1α1.
By the boundary condition α2x(1)+β2(Dcpx(1))=γ2, we get
(2.9)α2(∫01(1-s)q-1Γ(q)σ(s)ds-b1-b2)+β2(∫01(1-s)q-p-1Γ(q-p)σ(s)ds-b2Γ(2-p))=γ2,
which, on inserting the value of b1, gives
(2.10)-b2(α2Γ(2-p)+β2Γ(2-p))=γ2α1-α2γ1α1-α2∫01(1-s)q-1Γ(q)σ(s)ds-β2∫01(1-s)q-p-1Γ(q-p)σ(s)ds.
Using (2.5) in the above equation, we obtain
(2.11)b2=-ν2ν1+1ν1(α2∫01(1-s)q-1Γ(q)σ(s)ds+β2∫01(1-s)q-p-1Γ(q-p)σ(s)ds).
Substituting the values of b1 and b2 in (2.6), we get (2.4).Remark 2.4.
In the limitp→1-, it has been observed that the solution (2.4) of problem (2.3) is not reduced to the solution of the resulting problem given by
(2.12)Dcqx(t)=f(t,x(t)),t∈[0,1],1<q≤2,α1x(0)+β1x′(0)=γ1,α2x(1)+β2x′(1)=γ2.
The solution of (2.12) is
(2.13)x(t)=∫0t(t-s)q-1Γ(q)σ(s)ds+(β1-α1t)Δ(α2∫01(1-s)q-1Γ(q)σ(s)ds+β2∫01(1-s)q-2Γ(q-1)σ(s)ds)+1Δ(γ1(α2+β2)-γ2β1+(γ2α1-γ1α2)t),
where Δ=α1(α2+β2)-α2β1≠0. However, we notice that the solution (2.4) of problem (2.3) does not depend on the parameter β1 (appearing in the boundary conditions of (2.3)). Thus we conclude that the parameter β1 is of arbitrary nature. Furthermore, it has been found that the solutions (2.4) and (2.13) coincide by taking β1=0 in (2.13). Hence, for a particular choice of β1=0 in problems (2.3) and (2.12), the two problems have the same solution.


## 3. Main Results

Let𝒞=C([0,1],ℝ) denotes the Banach space of all continuous functions from [0,1]→ℝ endowed with the norm defined by ∥x∥=sup{|x(t)|,t∈[0,1]}.In view of Lemma2.3, we define an operator F:𝒞→𝒞 by(3.1)(Fx)(t)=∫0t(t-s)q-1Γ(q)f(s,x(s))ds-tν1(α2∫01(1-s)q-1Γ(q)f(s,x(s))ds+β2∫01(1-s)q-p-1Γ(q-p)f(s,x(s))ds)+α1ν2t+γ1ν1α1ν1.
Observe that the problem (1.1) has solutions if and only if the operator equation Fx=x has fixed points.Now we are in a position to present our main results. The methods used to prove the existence results are standard; however, their exposition in the framework of problem (1.1) is new.Theorem 3.1.
Suppose thatf:[0,1]×ℝ→ℝ is a continuous function and satisfies the following assumption:(A1)
|f(t,x)-f(t,y)|≤L|x-y|, for all t∈[0,1],L>0, x,y∈ℝ. Then the boundary value problem (1.1) has a unique solution provided
(3.2)LΓ(q+1)(1+|α2||ν1|)+|β2||ν1|LΓ(q-p+1)<1.Proof.
Settingsupt∈[0,1]|f(t,0)|=M<∞ and choosing r≥(ΛM+N)/(1-LΛ), where
(3.3)Λ=1Γ(q+1)(1+|α2||ν1|)+|β2||ν1|1Γ(q-p+1),N=|α1ν2|+|γ1ν1||α1ν1|,
we show that FBr⊂Br, where Br={x∈𝒞:∥x∥≤r}. For x∈Br, we have
(3.4)∥(Fx)(t)∥≤sup⁡t∈[0,1]{∫0t(t-s)q-1Γ(q)|f(s,x(s))|ds+|α2t||ν1|∫01(1-s)q-1Γ(q)|f(s,x(s))|ds+|β2t||ν1|∫01(1-s)q-p-1Γ(q-p)|f(s,x(s))|ds+|α1ν2t+γ1ν1||α1ν1|}≤sup⁡t∈[0,1]{∫0t(t-s)q-1Γ(q)(|f(s,x(s))-f(s,0)|+|f(s,0)|)ds+|α2||ν1|∫01(1-s)q-1Γ(q)(|f(s,x(s))-f(s,0)|+|f(s,0)|)ds+|β2||ν1|∫01(1-s)q-p-1Γ(q-p)(|f(s,x(s))-f(s,0)|+|f(s,0)|)ds+|α1ν2+γ1ν1||α1ν1|}≤(Lr+M)sup⁡t∈[0,1]{∫0t(t-s)q-1Γ(q)ds+|α2||ν1|∫01(1-s)q-1Γ(q)ds+|β2||ν1|∫01(1-s)q-p-1Γ(q-p)ds}+|α1ν2t+γ1ν1||α1ν1|≤(Lr+M){1Γ(q+1)+|α2||ν1|1Γ(q+1)+|β2||ν1|1Γ(q-p+1)}+|α1ν2|+|γ1ν1||α1ν1|=(Lr+M)Λ+N≤r.
Now, for x,y∈𝒞 and for each t∈[0,1], we obtain
(3.5)‖(Fx)(t)-(Fy)(t)‖≤sup⁡t∈[0,1]{∫0t(t-s)q-1Γ(q)|f(s,x(s))-f(s,y(s))|ds+|α2||ν1|∫01(1-s)q-1Γ(q)|f(s,x(s))-f(s,y(s))|ds+|β2||ν1|∫01(1-s)q-p-1Γ(q-p)|f(s,x(s))-f(s,y(s))|ds}≤L{1Γ(q+1)+|α2||ν1|1Γ(q+1)+|β2||ν1|1Γ(q-p+1)}‖x-y‖.
As (L/Γ(q+1))(1+|α2|/|ν1|)+(|β2|/|ν1|)(L/Γ(q-p+1))<1, therefore F is a contraction. Thus, the conclusion of the theorem followed by the contraction mapping principle (Banach fixed point theorem).Example 3.2.
Consider the following fractional boundary value problem(3.6)Dc3/2x(t)=1(t+2)2|x|1+|x|,t∈[0,1],x(0)+β1(Dc1/2x(0))=12,12x(1)+13(D1/2cx(1))=2.
Here, q=3/2, p=1/2, α1=1, α2=1/2, β2=1/3, γ1=1/2, γ2=2, β1 is arbitrary, and f(t,x)=(1/(t+2)2)(|x|/1+|x|). As |f(t,x)-f(t,y)|≤(1/4)|x-y|, therefore, (A1) is satisfied with L=1/4. Further, ν1=1/2+2/3π, ν2=7/4 and
(3.7)LΓ(q+1)(1+|α2||ν1|)+|β2||ν1|LΓ(q-p+1)=13π+2+π2(3π+4)≃0.390505<1.
Thus, by the conclusion of Theorem 3.1, the boundary value problem (3.6) has a unique solution on [0,1].
Now, we prove the existence of solutions of (1.1) by applying Krasnoselskii’s fixed point theorem [17].Theorem 3.3 (Krasnoselskii’s fixed point theorem).
LetM be a closed, bounded, convex, and nonempty subset of a Banach space  X. Let  A,B be the operators such that (i)  Ax+By∈M whenever  x,y∈M; (ii)  A is compact and continuous; (iii)  B is a contraction mapping. Then there exists  z∈M such that  z=Az+Bz.Theorem 3.4.
Letf:[0,1]×ℝ→ℝ be a jointly continuous function satisfying the assumption  (A1). In addition one assumes that (A2)
|f(t,x)|≤μ(t), for all (t,x)∈[0,1]×ℝ, and μ∈C([0,1],ℝ+). Then the problem (1.1) has at least one solution on [0,1] if
(3.8)|α2||ν1|LΓ(q+1)+|β2||ν1|LΓ(q-p+1)<1.Proof.
Lettingsupt∈[0,1]|μ(t)|=∥μ∥, we choose a real number r¯ satisfying the inequality
(3.9)r¯≥‖μ‖{1Γ(q+1)+|α2||ν1|1Γ(q+1)+|β2||ν1|1Γ(q-p+1)+|α1ν2|+|γ1ν1||α1ν1|},
and consider Br¯={x∈C:∥x∥≤r¯}. We define the operators 𝒫 and 𝒬 on Br¯ as
(3.10)(Px)(t)=∫0t(t-s)q-1Γ(q)f(s,u(s))ds,(Qx)(t)=-tν1(α2∫01(1-s)q-1Γ(q)f(s,x(s))ds+β2∫01(1-s)q-p-1Γ(q-p)f(s,x(s))ds)+α1ν2t+γ1ν1α1ν1.
For x,y∈Br¯, we find that
(3.11)‖Px+Qy‖≤‖μ‖{1Γ(q+1)+|α2||ν1|1Γ(q+1)+|β2||ν1|1Γ(q-p+1)+|α1ν2|+|γ1ν1||α1ν1|}≤r¯.
Thus, 𝒫x+𝒬y∈Br¯. It follows from the assumption (A1) together with (3.8) that 𝒬 is a contraction mapping. Continuity of f implies that the operator 𝒫 is continuous. Also, 𝒫 is uniformly bounded on Br¯ as
(3.12)‖Px‖≤‖μ‖Γ(q+1).
Now we prove the compactness of the operator𝒫.
In view of (A1), we define sup(t,x)∈[0,1]×Br¯|f(t,x)|=f¯, and consequently we have
(3.13)|(Px)(t1)-(Px)(t2)|=|1Γ(q)∫0t1[(t2-s)q-1-(t1-s)q-1]f(s,x(s))ds+∫t1t2(t2-s)q-1f(s,x(s))ds|≤f¯Γ(q+1)|2(t2-t1)q+t1q-t2q|,
which is independent of x. Thus,𝒫 is equicontinuous. Hence, by the Arzelá-Ascoli Theorem, 𝒫 is compact on Br¯. Thus, all the assumptions of Theorem 3.3 are satisfied. So the conclusion of Theorem 3.3 implies that the boundary value problem (1.1) has at least one solution on [0,1].Our next existence result is based on Leray-Schauder nonlinear alternative [18].Lemma 3.5 (nonlinear alternative for single-valued maps).
LetE be a Banach space,  C a closed, convex subset of  E,U an open subset of  C and  0∈U. Suppose that  F:U¯→C is a continuous, compact   (i.e., F(U¯) is a relatively compact subset of  C) map. Then either (i)
F has a fixed point in U¯, or(ii)
there is au∈∂U (the boundary of U in C) and λ∈(0,1) with u=λF(u).Theorem 3.6.
Letf:[0,1]×ℝ→ℝ be a jointly continuous function. Assume that: (A3)
there exist a functionp∈C([0,1],ℝ+) and a nondecreasing function ψ:ℝ+→ℝ+ such that |f(t,x)|≤p(t)ψ(∥x∥), for all (t,x)∈[0,1]×ℝ;(A4)
there exists a constantM>0 such that(3.14)Mψ(M){(1/Γ(q+1))(1+|α2|/|ν1|)+(|β2|/|ν1|Γ(q-p+1))}‖p‖+(|α1ν2|+|γ1ν1|)/|α1ν1|>1.
Then the boundary value problem (1.1) has at least one solution on [0,1].Proof.
Consider the operatorF:𝒞→𝒞 defined by (3.1). We show that Fmaps bounded sets into bounded sets inC([0,1],ℝ). For a positive number r, let Br={x∈C([0,1],ℝ):∥x∥≤r} be a bounded set in C([0,1],ℝ). Then
(3.15)|(Fx)(t)|≤∫0t(t-s)q-1Γ(q)p(s)ψ(‖x‖)ds+|α2||ν1|∫01(1-s)q-1Γ(q)p(s)ψ(‖x‖)ds+|β2||ν1|∫01(1-s)q-p-1Γ(q-p)p(s)ψ(‖x‖)ds+|α1ν2+γ1ν1||α1ν1|≤ψ(‖x‖){∫0t(t-s)q-1Γ(q)p(s)ds+|α2||ν1|∫01(1-s)q-1Γ(q)p(s)ds+|β2||ν1|∫01(1-s)q-p-1Γ(q-p)p(s)ds}+|α1ν2|+|γ1ν1||α1ν1|≤ψ(‖x‖){1Γ(q+1)(1+|α2||ν1|)+|β2||ν1|Γ(q-p+1)}‖p‖+|α1|ν2+|γ1ν1||α1ν1|.
Thus
(3.16)‖Fx‖≤ψ(r){1Γ(q+1)(1+|α2||ν1|)+|β2||ν1|Γ(q-p+1)}‖p‖+|α1ν2+γ1ν1||α1ν1|.
Next we show that Fmaps bounded sets into equicontinuous sets ofC([0,1],ℝ). Let t′,t′′∈[0,1] with t′<t′′ and x∈Br, where Br is a bounded set of C([0,1],ℝ). Then we obtain
(3.17)|(Fx)(t′′)-(Fx)(t′)|=|1Γ(q)∫0t′′(t′′-s)q-1f(s,x(s))ds-1Γ(q)∫0t′(t′-s)q-1f(s,x(s))ds|+|ν2||ν1||t′′-t′|≤|1Γ(q)∫0t′[(t′′-s)q-1-(t′-s)q-1]ψ(r)p(s)ds|+|1Γ(q)∫t′t′′(t′′-s)q-1ψ(r)p(s)ds|+|ν2||ν1||t′′-t′|.
Obviously the right hand side of the above inequality tends to zero independently of x∈Br as t′′-t′→0. As F satisfies the above assumptions, therefore it follows by the Arzelá-Ascoli theorem that F:C([0,1],ℝ)→C([0,1],ℝ) is completely continuous.
Letx be a solution. Then, for t∈[0,1] and using the computations in proving that F is bounded, we have
(3.18)|x(t)|=|λ(Fx)(t)|≤ψ(‖x‖){1Γ(q+1)(1+|α2||ν1|)+|β2||ν1|Γ(q-p+1)}‖p‖+|α1ν2|+|γ1ν1||α1ν1|.
Consequently, we have(3.19)‖x‖ψ(‖x‖){(1/Γ(q+1))(1+|α2|/|ν1|)+(|β2|/|ν1|Γ(q-p+1))}‖p‖+(|α1ν2|+|γ1ν1|)/|α1ν1|≤1.
In view of (A4), there exists M such that ∥x∥≠M. Let us set
(3.20)U={x∈C([0,1],R):‖x‖<M+1}.
Note that the operator F:U¯→C([0,1],ℝ) is continuous and completely continuous. From the choice of U, there is no x∈∂U such that x=λF(x) for some λ∈(0,1). Consequently, by the nonlinear alternative of Leray-Schauder type (Lemma 3.5), we deduce that F has a fixed point x∈U¯ which is a solution of the problem (1.1). This completes the proof.In the special case whenp(t)=1 and ψ(|x|)=κ|x|+N we have the following corollary.Corollary 3.7.
Letf:[0,1]×ℝ→ℝ be a continuous function. Assume that there exist constants  0≤κ<1/Λ1, where  Λ1=(1/Γ(q+1)(1+|α2|/|ν1|))+(|β2|/|ν1|Γ(q-p+1)) and  N1>0 such that  |f(t,x)|≤κ|x|+N1 for all  t∈[0,1],  x∈C[0,1]. Then the boundary value problem (1.1) has at least one solution.Example 3.8.
Consider the following boundary value problem:(3.21)D3/2cx(t)=1(6π)sin⁡(2πx)+|x|1+|x|,t∈[0,1],x(0)+β1(Dc1/2x(0))=12,12x(1)+13(Dc1/2x(1))=2.
Here, q=3/2, p=1/2, α1=1, α2=1/2, β2=1/3, γ1=1/2, γ2=2, β1 is arbitrary, and
(3.22)|f(t,x)|=|1(6π)sin⁡(2πx)+|x|1+|x||≤13|x|+1.
Clearly N1=1 and
(3.23)κ=13<1Λ1=0.640196.
Thus, all the conditions of Corollary 3.7 are satisfied, and consequently the problem (3.21) has at least one solution.



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