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Kyungpook Mathematical Journal 2019; 59(1): 191-202

Published online March 31, 2019

Copyright © Kyungpook Mathematical Journal.

Existence and Uniqueness of Solutions of Fractional Differential Equations with Deviating Arguments under Integral Boundary Conditions

Dnyanoba Dhaigude, and Bakr Rizqan*

Department of Mathematics, Dr. Babasaheb Ambedkar Marathwada University, Aurangabad - 431 004, India
e-mail : dnyanraja@gmail.com and bakeralhaaiti@yahoo.com

Received: April 12, 2017; Revised: August 7, 2018; Accepted: December 19, 2018

The aim of this paper is to develop a monotone iterative technique by introducing upper and lower solutions to Riemann-Liouville fractional differential equations with deviating arguments and integral boundary conditions. As an application of this technique, existence and uniqueness results are obtained.

Keywords: fractional differential equations with deviating arguments, Riemann-Liouville fractional derivatives, existence and uniqueness, monotone iterative technique, integral boundary conditions.

Differential equations with deviating arguments arise in various branches of science, engineering, economics and so on (see [4, 8] and the references therein). Many researchers have studied the existence, uniqueness, continuous dependence, and stability of solutions of nonlinear fractional differential equations (see [1, 2, 3, 5, 6, 7, 10, 11, 12, 15, 16, 17, 19, 20, 25, 30, 33, 34]). The monotone iterative technique [23] combined with the method of upper and lower solutions provides an effective mechanism to prove constructive existence results for nonlinear differential equations. The monotone technique is an interesting and powerful tool to deal with existence results for fractional differential equations. In 2008, the monotone technique for fractional differential equations with initial conditions was first developed by Lakshmikantham and Vatsala [25]. Later, a series of papers appeared in the literature to prove existence and uniqueness of solution of various problems with initial conditions, boundary conditions, integral boundary conditions, nonlinear boundary conditions, and periodic boundary conditions for fractional differential equations, (see, for example [13, 14, 18, 22, 23, 24, 25, 26, 27, 28, 29, 31, 32, 35, 37, 38, 39, 40, 41, 42, 43] and the references therein). However, work on fractional differential equations with deviating argument is rare. In this paper, we study the following problem for the Riemann-Liouville fractional differential equation with a deviating argument and integral boundary conditions:

{D0+αu(t)=f(t,u(t),u(θ(t))),tJ=[0,T],u(0)=λ0Tu(s)ds+d,         d,

where fC (J × ℝ2,ℝ), θC (J, J), θ(t) ≤ t, tJ, λ ≥ 0, 0 < α < 1. The paper is organized as follows. In Section 2, we introduce some useful definitions and basic lemmas. In Section 3, we study the uniqueness of a solution for the problem (1.1) using the Banach fixed point theorem. In Section 4, we develop the monotone method and apply it to obtain existence and uniqueness results for Riemann-Liouville fractional differential equations with deviating arguments and integral boundary conditions.

For the reader’s convenience, we present some necessary definitions and lemmas from the theory of fractional calculus. In addition, we prove some basic results which are useful for further discussion.

Definition 2.1. ([21, 36])

For α > 0, the integral

I0+αu(t)=1Γ(α)0t(t-s)α-1u(s)ds

is called the Riemann-Liouville fractional integral of order α.

Definition 2.2. ([21, 36])

The Riemann-Liouville derivative of order α (n − 1 < αn) can be written as

D0+αu(t)=(ddt)n(I0+n-αu(t))=1Γ(n-α)dndtn0t(t-s)n-α-1u(s)ds,   t>0.

Lemma 2.1. ([21])

Let uCn[0, T], α ∈ (n − 1, n), n ∈ ℕ. Then for tJ,

I0+αD0+αu(t)=u(t)-k=0n-1tkk!u(k)(0).

Consider the space C1−α (J,ℝ) = {uC ((0, T],ℝ) : t1−αuC (J,ℝ)}.

Lemma 2.2. ([9])

Let mC1−α(J,ℝ) where for some t1 ∈ (0, T],

m(t1)=0andm(t)0for0tt1.

Then it follows that

Dαm(t1)0.

Lemma 2.3

Let fC (J × ℝ2,ℝ). A function uC1−α (J,ℝ) is a solution of the problem (1.1) if and only if u is a solution of the integral equation

u(t)=1Γ(α)0t(t-s)α-1f(s,u(s),u(θ(s)))ds+λ0Tu(s)ds+d.
Proof

Assume that u satisfies the problem (1.1). From the first equation of the problem (1.1) and Lemma 2.1, we have

u(t)=1Γ(α)0t(t-s)α-1f(s,u(s),u(θ(s)))ds+λ0Tu(s)ds+d.

Conversely, assume that uC1−α (J,ℝ) satisfies the integral equation (2.1). Applying the Riemann-Liouville operator D0+α to both sides of the integral equation (2.1), we have

D0+αu(t)=D0+α(1Γ(α)0t(t-s)α-1f(s,u(s),u(θ(s)))ds+λ0Tu(s)ds+d)D0+αu(t)=f(t,u(t),u(θ(t))).

In addition, we have u(0)=λ0Tu(s)ds+d from the integral equation (2.1). The proof is complete.

Lemma 2.4

Suppose that {uε} is a family of continuous functions defined on J, for each ε > 0, which satisfies

{D0+αuɛ(t)=f(t,uɛ(t),uɛ(θ(t))),uɛ(0)=λ0Tuɛ(s)ds+d,

where |f (t, uε(t), uε(θ(t)))| ≤ M for tJ. Then the family {uε} is equicontinuous on J.

Proof

For 0 ≤ t1 < t2T, consider

uɛ(t1)-uɛ(t2)=1Γ(α)|0t1(t1-s)α-1f(s,uɛ(s),uɛ(θ(s)))ds-0t2(t2-s)α-1f(s,uɛ(s),uɛ(θ(s)))ds|MΓ(α)(0t1[(t1-s)α-1-(t2-s)α-1]ds+t1t2(t2-s)α-1ds)MΓ(α+1)[t1α-t2α+2(t2-t1)α]2MΓ(α+1)(t2-t1)α<ɛ,

provided that t2-t1<δ=[ɛΓ(α+1)2M]1α, proving the result.

In this section, we obtain the uniqueness of solution of the problem (1.1) for Riemann-Liouville fractional differential equations with deviating argument and integral boundary conditions.

Theorem 3.1

Assume that

  • fC (J × ℝ2,ℝ), θ(t) ∈ C (J, J), θt, tJ,

  • there exists nonnegative constants M and N such that function f satisfiesf(t,u1,u2)-f(t,υ1,υ2)Mu1-υ1+Nu2-υ2,

    for all tJ, ui, υi ∈ ℝ, i = 1, 2. Ifλ<Γ(α+1)-Tα(M+N)TΓ(α+1), then the problem (1.1) has a unique solution.

Proof

Consider the operator T defined by

(Tu)(t)=λ0Tu(s)ds+d+1Γ(α)0t(t-s)α-1f(s,u(s),u(θ(s))ds.

Now, we show that T : C1−α(J,ℝ) → C1−α(J,ℝ) is a contraction operator. For any u, υC1−α(J,ℝ), we have

Tu-TυC=maxtJ(Tu)(t)-(Tυ)(t)maxtJλ0Tu(s)-υ(s)ds+maxtJ1Γ(α)0t(t-s)α-1×f(s,u(s),u(θ(s)))-f(s,υ(s),υ(θ(s)))dsλ0Tdsu-υC+maxtJ1Γ(α)0t(t-s)α-1×[M(u(s)-υ(s))+N(u(θ(s))-υ(θ(s)))]dsλTu-υC+maxtJ(M+N)Γ(α)0t(t-s)α-1u-υCdsλTu-υC+maxtJ(M+N)tαΓ(α)01(1-η)α-1dηu-υC[λT+TαΓ(α+1)(M+N)]u-υC.

Therefore, ||Tu||C < ||uυ||C. By the Banach fixed point theorem, the operator T has a unique fixed point, i.e. the problem (1.1) has a unique solution. The proof is complete.

Corollary 3.1

Let M, N be constants, σC1−α(J,ℝ). The linear problem

{D0+αu(t)+Mu(t)+Nu(θ(t))=σ(t),   0<α<1,   tJ,u(0)=λ0Tu(s)ds+d,   d,

has a unique solution.

Proof

It follows from the Theorem 3.1.

In this section, we prove the existence and uniqueness of solution for the problem (1.1) by monotone iterative technique combined with the method of upper and lower solutions. Now we define the functional interval as follows:

[υ0,w0]={uC1-α(J,):υ0(t)u(t)w0(t)tJ}.

First, we prove the following comparison result, which plays an important role in our further discussion.

Lemma 4.1

Let θC (J, J) where θ(t) ≤ t on J. Suppose that pC1−α(J,ℝ) satisfies the inequalities

{D0+αp(t)-Mp(t)-Np(θ(t))Fp(t),         tJ,   p(0)0,

where M and N are constants. If

-(1+Tα)[M+N]<Γ(1+α),

then p(t) ≤ 0 for all tJ.

Proof

Consider pε(t) = p(t) − ε(1 + tα), ε > 0. Then

D0+αpɛ(t)=D0+αp(t)-D0+αɛ(1+tα)Fp(t)-ɛtαΓ(1-α)-ɛΓ(1+α)=Fpɛ(t)+ɛ[-M(1+tα)-N(1+tα)-1tαΓ(1-α)-Γ(1+α)]<Fpɛ(t)+ɛ[-(1+tα)(M+N)-Γ(1+α)]<Fpɛ(t)

and

pɛ(0)=p(0)-ɛ(1+tα)<0.

We prove that pε(t) < 0 on J. Assume that pε(t) ≮ 0 on J. Thus there exists a t1 ∈ (0, T] such that pε(t1) = 0 and pε(t) < 0, t ∈ (0, t1). In view of Lemma 2.2, we have D0+αpɛ(t1)0. It follows that

0<Fpɛ(t)=-Npɛ(θ(t1)).

If N = 0, then 0 < 0, which is a contradiction. If −N >0, then pε(θ(t1)) > 0, which is again a contradiction. This proves that pε(t) < 0 on J. So p(t) − ε(1 + tα) < 0 on J. Taking ε → 0, we obtain required result.

Definition 4.1

A pair of functions [υ0, w0] in C1−α(J,ℝ) are called lower and upper solutions of the problem (1.1) if

D0+αυ0(t)f(t,υ0(t),υ0(θ(t))),         υ0(0)0Tυ0(s)ds+d

and

D0+αw0(t)f(t,w0(t),w0(θ(t))),         w0(0)0Tw0(s)ds+d.

Theorem 4.1

Assume that

  • fC (J × ℝ2,ℝ), θC (J, J), θ(t) ≤ t, tJ,

  • functions υ0and w0in C1−α(J,ℝ) are lower and upper solutions of the problem (1.1) such that υ0(t) ≤ w0(t) on J,

  • there exists nonnegative constants M, N such that function f satisfies the conditionf(t,u1,u2)-f(t,υ1,υ2)-M(u1-υ1)-N(u2-υ2),

    for υ0(t) ≤ υ1u1w0(t), υ0(θ(t)) ≤ υ2u2w0(θ(t)).

Then there exists monotone sequences {υn(t)} and {wn(t)} in C1−α(J,ℝ) such that

{υn(t)}υ(t)   and   {wn(t)}w(t)   asn

for all tJ, where υ and w are minimal and maximal solutions of the problem (1.1) respectively and υ(t) ≤ u(t) ≤ w(t) on J.

Proof

For any ηC1−α(J,ℝ) such that η ∈ [υ0, w0], we consider the following linear problem:

{D0+αu(t)=f(t,η(t),η(θ(t)))+M[η(t)-u(t)]+N[η(θ(t))-u(θ(t))],u(0)=0Tu(s)ds+d,

By Corollary 3.1, the linear problem (4.6) has a unique solution u(t).

Next, we define the iterates as follows and construct the sequences {υn}, {un}

{D0+αυn+1(t)=f(t,υn(t),υn(θ(t)))-M[υn+1(t)-υn(t)]-N[υn+1(θ(t))-υn(θ(t))],υn+1(0)=0Tυn(s)ds+d,

and

{D0+αwn+1(t)=f(t,wn(t),wn(θ(t)))-M[wn+1(t)-wn(t)]-N[wn+1(θ(t))-wn(θ(t))],wn+1(0)=0Twn(s)ds+d,

Clearly, the existence of solutions υn+1 and wn+1 of the problems (4.7) and (4.8), respectively, follows from the above arguments. Further, by setting n = 0 in the problems (4.7), (4.8), we get the existence of solutions υ1 and w1, respectively. We show that υ0(t) ≤ υ1(t) ≤ w1(t) ≤ w0(t). Set p(t) = υ1(t) − υ0(t). Since υ0 is the lower solution of the problem (4.7), we have

D0+αp(t)=D0+αυ1(t)-D0+αυ0(t)f(t,υ0(t),υ0(θ(t)))-f(t,υ0(t),υ0(θ(t)))-M[υ1(t)-υ0(t)]-N[υ1(θ(t))-υ0(θ(t))]-Mp(t)-Np(θ(t))

and

p(0)=υ1(0)-υ0(0)0Tυ0(s)ds+d-0Tυ0(s)ds-d=0.

From Lemma 4.1, we obtain p(t) ≥ 0, which implies that υ1(t) ≥ υ0(t) on J. Similarly, we can prove υ1(t) ≤ w1(t) and w1(t) ≤ w0(t) on J. Thus υ0(t) ≤ υ1(t) ≤ w1(t) ≤ w0(t). Assume that for some k > 1,

υk-1(t)υk(t)wk(t)wk-1(t)on J.

We claim that υk(t) ≤ υk+1(t) ≤ wk+1(t) ≤ wk(t) on J. To prove our claim, set p(t) = υk+1(t) − υk(t). Then we have

D0+αp(t)=D0+αυk+1(t)-D0+αυk(t)=f(t,υk(t),υk(θ(t)))-M[υk+1(t)-υk(t)]-N[υk+1(θ(t))-υk(θ(t))]-f(t,υk-1(t),υk-1(θ(t)))+M[υk(t)-υk-1(t)]+N[υk(θ(t))-υk-1(θ(t))]-M[υk+1(t)-υk(t)]-N[υk+1(θ(t))-υk(θ(t))]-Mp(t)-Np(θ(t)),

and

p(0)=υk+1(0)-υk(0)=0Tυk(s)ds+d-0Tυk-1(s)ds-d0T[υk(s)-υk(s)]ds=0.

By Lemma 4.1, we obtain p(t) ≥ 0, implying that υk+1(t) ≥ υk(t) for all tJ. Similarly, we can prove υk+1(t) ≤ wk+1(t) and wk+1(t) ≤ wk(t) for all tJ. From the principle of mathematical induction, we have

υ0υ1υ1υkwkw2w1w0on J.

Clearly, the sequences {υn}, {wn} are monotonic and uniformly bounded. Further we observe that { D0+αυn } and {D0+αwn } are also uniformly bounded on J, in view of the relations (4.7), (4.8). Applying Lemma 2.4 we can conclude that sequences {υn}, {wn} are equicontinuous. Hence by the Ascoli-Arzela theorem the sequences {n}, {wn} converge uniformly to υ and w on J respectively.

Now, we prove that υ and w are the minimal and maximal solutions of the problem (1.1). Let u be any solution of the problem (1.1) different from υ and w. So there exists a k such that υk(t) ≤ u(t) ≤ wk(t) on J. Set p(t) = u(t) − υk+1(t). Then we have

D0+αp(t)=D0+αu(t)-D0+αυk+1(t)=f(t,u(t),u(θ(t)))-f(t,υk(t),υk(θ(t)))+M[υk+1(t)-υk(t)]+N[υk+1(θ(t))-υk(θ(t))]-M[u(t)-υk+1(t)]-N[u(θ(t))-υk+1(θ(t))]-Mp(t)-Np(θ(t)),

and

p(0)=u(0)-υk+1(0)=0T[u(s)-υk(s)]ds0.

By Lemma 4.1, we obtain p(t) ≥ 0, implying that u(t) ≥ υk+1(t) for all k on J. Similarly, we can prove u(t) ≤ wk+1(t) for all k on J. Since υ0(t) ≤ u(t) ≤ u0(t) on J. By induction it follows that υk(t) ≤ u(t) and u(t) ≤ wk(t) for all k. Thus υk(t) ≤ u(t) ≤ wk(t) on J. Taking the limit as k→∞, we obtain υ(t) ≤ u(t) ≤ w(t) on J. Thus the functions υ(t), w(t) are the minimal and maximal solutions of the problem (1.1). The proof is complete.

Next we prove the uniqueness of solution of the problem (1.1) as follows.

Theorem 4.2

Assume that

  • all the conditions of the Theorem 4.1 hold,

  • there exists nonnegative constants M, N such that the function f satisfies the conditionf(t,u1,u2)-f(t,υ1,υ2)M(u1-υ1)+N(u2-υ2),

    for υ0(t) ≤ υ1u1w0(t), υ0(θ(t)) ≤ υ2u2w0(θ(t)).

Then the problem (1.1) has a unique solution.

Proof

We know υ(t) ≤ w(t) on J. It is sufficient to prove that υ(t) ≥ w(t) on J. Consider p(t) = w(t) − υ(t). Then we have

D0+αp(t)=D0+αw(t)-D0+αυ(t)=f(t,w(t),w(θ(t)))-f(t,υ(t),υ(θ(t)))-M[υ(t)-w(t)]-N[υ(θ(t))-w(θ(t))]=-Mp(t)-Np(θ(t))

and

p(0)=w(0)-υ(0)=0T[w(s)-υ(s)]ds0.

By Lemma 4.1, we know p(t) ≤ 0, implying that υ(t) ≥ w(t), and the result follows.

The authors are grateful to the respected anonymous referee and the English language expert Prof. J. T. Neugebauer for their valuable comments and suggestions that greatly improved the presentation of the paper.

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