Evolutionary multitasking optimization (EMTO) handles multiple tasks simultaneously by transferring and sharing valuable knowledge from other relevant tasks. How to effectively identify transferred knowledge and reduce negative knowledge transfer are two key issues in EMTO. Many existing EMTO algorithms treat the elite solutions in tasks as transferred knowledge between tasks. However, these algorithms may not be effective enough when the global optimums of the tasks are far apart. In this paper, we study an adaptive evolutionary multitasking optimization algorithm based on population distribution information to find valuable transferred knowledge and weaken the negative transfer between tasks. In this paper, we first divide each task population into K sub-populations based on the fitness values of the individuals, and then the maximum mean discrepancy (MMD) is utilized to calculate the distribution difference between each sub-population in the source task and the sub-population where the best solution of the target task is located. Among the sub-populations of the source task, the sub-population with the smallest MMD value is selected, and the individuals in it are used as transferred individuals. In this way, the solution chosen for the transfer may be an elite solution or some other solution. In addition, an improved randomized interaction probability is also included in the proposed algorithm to adjust the intensity of inter-task interactions. The experimental results on two multitasking test suites demonstrate that the proposed algorithm achieves high solution accuracy and fast convergence for most problems, especially for problems with low relevance.
Citation: Xiaoyu Li, Lei Wang, Qiaoyong Jiang, Qingzheng Xu. An adaptive multitasking optimization algorithm based on population distribution[J]. Mathematical Biosciences and Engineering, 2024, 21(2): 2432-2457. doi: 10.3934/mbe.2024107
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Evolutionary multitasking optimization (EMTO) handles multiple tasks simultaneously by transferring and sharing valuable knowledge from other relevant tasks. How to effectively identify transferred knowledge and reduce negative knowledge transfer are two key issues in EMTO. Many existing EMTO algorithms treat the elite solutions in tasks as transferred knowledge between tasks. However, these algorithms may not be effective enough when the global optimums of the tasks are far apart. In this paper, we study an adaptive evolutionary multitasking optimization algorithm based on population distribution information to find valuable transferred knowledge and weaken the negative transfer between tasks. In this paper, we first divide each task population into K sub-populations based on the fitness values of the individuals, and then the maximum mean discrepancy (MMD) is utilized to calculate the distribution difference between each sub-population in the source task and the sub-population where the best solution of the target task is located. Among the sub-populations of the source task, the sub-population with the smallest MMD value is selected, and the individuals in it are used as transferred individuals. In this way, the solution chosen for the transfer may be an elite solution or some other solution. In addition, an improved randomized interaction probability is also included in the proposed algorithm to adjust the intensity of inter-task interactions. The experimental results on two multitasking test suites demonstrate that the proposed algorithm achieves high solution accuracy and fast convergence for most problems, especially for problems with low relevance.
The topic of fractional differential equations received immense popularity and attraction due to their extensive use in the mathematical modeling of several real world phenomena. Examples include HIV-immune system with memory [1], stabilization of chaotic systems [2], chaotic synchronization [3,4], ecology [5], infectious diseases [6], economic model [7], fractional neural networks [8,9], COVID-19 infection [10], etc. A salient feature distinguishing fractional-order differential and integral operators from the classical ones is their nonlocal nature, which can provide the details about the past history of the phenomena and processes under investigation. In the recent years, many researchers contributed to the development of fractional calculus, for example, see [11,12,13,14,15,16,17,18,19,20,21,22,23,24] and the references cited therein. One can also find a substantial material about fractional order coupled systems in the articles [25,26,27,28,29,30,31,32,33,34].
In this paper, motivated by [30], we consider a Caputo type coupled system of nonlinear fractional differential equations supplemented with a new set of boundary conditions in terms of the sum and difference of the governing functions given by
{CDνφ(t)=f(t,φ(t),ψ(t)),t∈J:=[0,T],CDρψ(t)=g(t,φ(t),ψ(t)),t∈J:=[0,T],P1(φ+ψ)(0)+P2(φ+ψ)(T)=m∑i=1ai(φ+ψ)(σi),∫T0(φ−ψ)(s)ds−∫ζη(φ−ψ)(s)ds=A, | (1.1) |
where CDχ is the Caputo fractional derivative operator of order χ∈{ν,ρ}, ν,ρ∈(0,1], 0<σi<η<ζ<T, i=1,…,m (the case 0<η<ζ<σi<T can be treated in a similar way), P1,P2,ai,A are nonnegative constants, such that P1+P2−m∑i=1ai≠0, T−ζ+η≠0, and f,g:[0,T]×R2→R are continuous functions.
Here it is imperative to notice that the first condition introduced in the problem (1.1) can be interpreted as the sum of the governing functions φ and ψ at the end positions of the interval [0,T] is sum of similar contributions due to arbitrary positions at σi∈(0,T),i=1,...,m, while the second condition describes that the contribution of the difference of the governing functions φ and ψ on the domain [0,T] differs from the one an arbitrary sub-domain (η,ξ) by a constant A.
We will also study the problem (1.1) by replacing A in the last condition with the one containing nonlinear Riemann-Liouville integral term of the form:
1Γ(δ)∫T0(T−s)δ−1h(s,φ(s),ψ(s))ds,δ>0, | (1.2) |
where h:[0,T]×R2→R is a given continuous function.
We organize the rest of the paper as follows. In Section 2, we outline the related concepts of fractional calculus and establish an auxiliary lemma for the linear analogue of the problem (1.1). We apply the standard fixed point theorems to derive the existence and uniqueness results for the problem (1.1) in Section 3. The case of nonlinear Riemann-Liouville integral boundary conditions is discussed in Section 4. The paper concludes with some interesting observations and special cases.
Let us begin this section with some preliminary concepts of fractional calculus [11].
Definition 2.1. The Riemann-Liouville fractional integral of order q>0 of a function h:[0,∞)→R is defined by
Iqh(t)=∫t0(t−s)q−1Γ(q)h(s)ds,t>0, |
provided the right-hand side is point-wise defined on (0,∞), where Γ is the Gamma function.
Definition 2.2. The Caputo fractional derivative of order q for a function h:[0,∞]→R with h(t)∈ACn[0,∞) is defined by
CDqh(t)=1Γ(n−q)∫t0h(n)(s)(t−s)q−n+1ds=In−qh(n)(t), t>0,n−1<q<n. |
Lemma 2.1. Let q>0 and h(t)∈ACn[0,∞) or Cn[0,∞). Then
(IqCDqh)(t)=h(t)−n−1∑k=0h(k)(0)k!tk,t>0,n−1<q<n. | (2.1) |
Now we present an auxiliary lemma related to the linear variant of problem (1.1).
Lemma 2.2. Let F,G∈C[0,T], φ,ψ∈AC[0,T].Then the solution of the following linear coupled system:
{CDνφ(t)=F(t),t∈J:=[0,T],CDρψ(t)=G(t),t∈J:=[0,T],P1(φ+ψ)(0)+P2(φ+ψ)(T)=m∑i=1ai(φ+ψ)(σi),∫T0(φ−ψ)(s)ds−∫ζη(φ−ψ)(s)ds=A, | (2.2) |
is given by
φ(t)=∫t0(t−s)ν−1Γ(ν)F(s)ds+12{AΛ2−1Λ2∫T0(∫s0(s−x)ν−1Γ(ν)F(x)dx−∫s0(s−x)ρ−1Γ(ρ)G(x)dx)ds−P2Λ1(∫T0(T−s)ν−1Γ(ν)F(s)ds+∫T0(T−s)ρ−1Γ(ρ)G(s)ds)+1Λ2∫ξη(∫s0(s−x)ν−1Γ(ν)F(x)dx−∫s0(s−x)ρ−1Γ(ρ)G(x)dx)ds+m∑i=1aiΛ1(∫σi0(σi−s)ν−1Γ(ν)F(s)ds+∫σi0(σi−s)ρ−1Γ(ρ)G(s)ds)}, | (2.3) |
ψ(t)=∫t0(t−s)ρ−1Γ(ρ)G(s)ds+12{−AΛ2+1Λ2∫T0(∫s0(s−x)ν−1Γ(ν)F(x)dx−∫s0(s−x)ρ−1Γ(ρ)G(x)dx)ds−P2Λ1(∫T0(T−s)ν−1Γ(ν)F(s)ds+∫T0(T−s)ρ−1Γ(ρ)G(s)ds)−1Λ2∫ξη(∫s0(s−x)ν−1Γ(ν)F(x)dx−∫s0(s−x)ρ−1Γ(ρ)G(x)dx)ds+m∑i=1aiΛ1(∫σi0(σi−s)ν−1Γ(ν)F(s)ds+∫σi0(σi−s)ρ−1Γ(ρ)G(s)ds)}, | (2.4) |
where
Λ1:=P1+P2−m∑i=1ai≠0, | (2.5) |
Λ2:=T−ζ+η≠0. | (2.6) |
Proof. Applying the operators Iν and Iρ on the first and second fractional differential equations in (2.2) respectively and using Lemma 2.1, we obtain
φ(t)=∫t0(t−s)ν−1Γ(ν)F(s)ds+c1, | (2.7) |
ψ(t)=∫t0(t−s)ρ−1Γ(ρ)G(s)ds+c2, | (2.8) |
where c1,c2∈R. Inserting (2.7) and (2.8) in the condition P1(φ+ψ)(0)+P2(φ+ψ)(T)=m∑i=1ai(φ+ψ)(σi), we get
c1+c2=1Λ1{m∑i=1ai(∫σi0(σi−s)ν−1Γ(ν)F(s)ds+∫σi0(σi−s)ρ−1Γ(ρ)G(s)ds)−P2(∫T0(T−s)ν−1Γ(ν)F(s)ds+∫T0(T−s)ρ−1Γ(ρ)G(s)ds)}. | (2.9) |
Using (2.7) and (2.8) in the condition ∫T0(φ−ψ)(s)ds−∫ζη(φ−ψ)(s)ds=A, we obtain
c1−c2=1Λ2{A−∫T0(∫s0(s−x)ν−1Γ(ν)F(x)dx−∫s0(s−x)ρ−1Γ(ρ)G(x)dx)ds+∫ξη(∫s0(s−x)ν−1Γ(ν)F(x)dx−∫s0(s−x)ρ−1Γ(ρ)G(x)dx)ds}. | (2.10) |
Solving (2.9) and (2.10) for c1 and c2, yields
c1=12{AΛ2−1Λ2∫T0(∫s0(s−x)ν−1Γ(ν)F(x)dx−∫s0(s−x)ρ−1Γ(ρ)G(x)dx)ds−P2Λ1(∫T0(T−s)ν−1Γ(ν)F(s)ds+∫T0(T−s)ρ−1Γ(ρ)G(s)ds)+1Λ2∫ξη(∫s0(s−x)ν−1Γ(ν)F(x)dx−∫s0(s−x)ρ−1Γ(ρ)G(x)dx)ds+1Λ1m∑i=1ai(∫σi0(σi−s)ν−1Γ(ν)F(s)ds+∫σi0(σi−s)ρ−1Γ(ρ)G(s)ds)}, |
and
c2=12{−AΛ2+1Λ2∫T0(∫s0(s−x)ν−1Γ(ν)F(x)dx−∫s0(s−x)ρ−1Γ(ρ)G(x)dx)ds−P2Λ1(∫T0(T−s)ν−1Γ(ν)F(s)ds+∫T0(T−s)ρ−1Γ(ρ)G(s)ds)−1Λ2∫ξη(∫s0(s−x)ν−1Γ(ν)F(x)dx−∫s0(s−x)ρ−1Γ(ρ)G(x)dx)ds+1Λ1m∑i=1ai(∫σi0(σi−s)ν−1Γ(ν)F(s)ds+∫σi0(σi−s)ρ−1Γ(ρ)G(s)ds)}. |
Substituting the values of c1 and c2 in (2.7) and (2.8) respectively, we get the solution (2.3) and (2.4). By direct computation, one can obtain the converse of this lemma. The proof is complete.
Let X=C([0,T],R)×C([0,T],R) denote the Banach space endowed with the norm ‖(φ,ψ)‖=‖φ‖+‖ψ‖=supt∈[0,T]|φ(t)|+supt∈[0,T]|ψ(t)|, (φ,ψ)∈X. In view of Lemma 2.2, we define an operator Φ:X→X in relation to the problem (1.1) as
Φ(φ,ψ)(t):=(Φ1(φ,ψ)(t),Φ2(φ,ψ)(t)), | (3.1) |
where
Φ1(φ,ψ)(t)=1Γ(ν)∫t0(t−s)ν−1f(s,φ(s),ψ(s))ds+12{AΛ2−1Λ2∫T0(∫s0(s−x)ν−1Γ(ν)f(x,φ(x),ψ(x))dx−∫s0(s−x)ρ−1Γ(ρ)g(x,φ(x),ψ(x))dx)ds−P2Λ1(∫T0(T−s)ν−1Γ(ν)f(s,φ(s),ψ(s))ds+∫T0(T−s)ρ−1Γ(ρ)g(s,φ(s),ψ(s))ds)+1Λ2∫ξη(∫s0(s−x)ν−1Γ(ν)f(x,φ(x),ψ(x))dx−∫s0(s−x)ρ−1Γ(ρ)g(x,φ(x),ψ(x))dx)ds+1Λ1m∑i=1ai(∫σi0(σi−s)ν−1Γ(ν)f(s,φ(s),ψ(s))ds+∫σi0(σi−s)ρ−1Γ(ρ)g(s,φ(s),ψ(s))ds)}, | (3.2) |
and
Φ2(φ,ψ)(t)=1Γ(ρ)∫t0(t−s)ρ−1g(s,φ(s),ψ(s))ds+12{−AΛ2+1Λ2∫T0(∫s0(s−x)ν−1Γ(ν)f(x,φ(x),ψ(x))dx−∫s0(s−x)ρ−1Γ(ρ)g(x,φ(x),ψ(x))dx)ds−P2Λ1(∫T0(T−s)ν−1Γ(ν)f(s,φ(s),ψ(s))ds+∫T0(T−s)ρ−1Γ(ρ)g(s,φ(s),ψ(s))ds)−1Λ2∫ξη(∫s0(s−x)ν−1Γ(ν)f(x,φ(x),ψ(x))dx−∫s0(s−x)ρ−1Γ(ρ)g(x,φ(x),ψ(x))dx)ds+1Λ1m∑i=1ai(∫σi0(σi−s)ν−1Γ(ν)f(s,φ(s),ψ(s))ds+∫σi0(σi−s)ρ−1Γ(ρ)g(s,φ(s),ψ(s))ds)}. | (3.3) |
In the forthcoming analysis, we need the following assumptions.
(H1) There exist continuous nonnegative functions μi,κi∈C([0,1],R+),i=1,2,3, such that
|f(t,φ,ψ)|≤μ1(t)+μ2(t)|φ|+μ3(t)|ψ|∀(t,φ,ψ)∈J×R2; |
|g(t,φ,ψ)|≤κ1(t)+κ2(t)|φ|+κ3(t)|ψ|∀(t,φ,ψ)∈J×R2. |
(H2) There exist positive constants αi,βi,i=1,2, such that
|f(t,φ1,ψ1)−f(t,φ2,ψ2)|≤α1|φ1−φ2|+α2|ψ1−ψ2|,∀t∈J,φi,ψi∈R,i=1,2; |
|g(t,φ1,ψ1)−g(t,φ2,ψ2)|≤β1|φ1−φ2|+β2|ψ1−ψ2|,∀t∈J,φi,ψi∈R,i=1,2. |
For computational convenience, we introduce the notation:
ϱ1=12|Λ1|[m∑i=1aiσνiΓ(ν+1)+P2TνΓ(ν+1)]+12|Λ2|[ζν+1−ην+1Γ(ν+2)+Tν+1Γ(ν+2)], | (3.4) |
ϱ2=12|Λ1|[m∑i=1aiσρiΓ(ρ+1)+P2TρΓ(ρ+1)]+12|Λ2|[ζρ+1−ηρ+1Γ(ρ+2)+Tρ+1Γ(ρ+2)], | (3.5) |
and
M0=min{1−[‖μ2‖(2ϱ1+TνΓ(ν+1))+‖κ2‖(2ϱ2+TρΓ(ρ+1))],1−[‖μ3‖(2ϱ1+TνΓ(ν+1))+‖κ3‖(2ϱ2+TρΓ(ρ+1))]}. |
We make use of the following fixed point theorem [35] to prove the existence of solutions for the problem (1.1).
Lemma 3.1. Let E be the Banach space and Q:E→E be a completely continuous operator. If the set Ω={x∈E|x=μQx,0<μ<1} is bounded, then Q has a fixed point in E.
Theorem 3.1. Suppose that f,g:J×R2→R are continuousfunctions and the condition (H1) holds. Then there exists at least one solution for the problem (1.1) on J if
‖μ2‖(2ϱ1+TνΓ(ν+1))+‖κ2‖(2ϱ2+TρΓ(ρ+1))<1,‖μ3‖(2ϱ1+TνΓ(ν+1))+‖κ3‖(2ϱ2+TρΓ(ρ+1))<1, | (3.6) |
where ϱi(i=1,2) are defined in (3.4)–(3.5).
Proof. Observe that continuity of Φ:X→X follows from that of the functions f and g. Now we show that the operator Φ maps any bounded subset of X into a relatively compact subset of X. For that, let Ωˉr⊂X be bounded. Then, for the positive real constants Lf and Lg, we have
|f(t,φ(t),ψ(t))|≤Lf,|g(t,φ(t),ψ(t))|≤Lg,∀(φ,ψ)∈Ωˉr. |
So, for any (φ,ψ)∈Ωˉr, t∈J, we get
|Φ1(φ,ψ)(t)|≤LfΓ(ν)∫t0(t−s)ν−1ds+12{A|Λ2|+1Λ2∫T0(Lf∫s0(s−x)ν−1Γ(ν)dx+Lg∫s0(s−x)ρ−1Γ(ρ)dx)ds+P2|Λ1|(Lf∫T0(T−s)ν−1Γ(ν)ds+Lg∫T0(T−s)ρ−1Γ(ρ)ds)+1|Λ2|∫ξη(Lf∫s0(s−x)ν−1Γ(ν)dx+Lg∫s0(s−x)ρ−1Γ(ρ)dx)ds+1|Λ1|m∑i=1ai(Lf∫σi0(σi−s)ν−1Γ(ν)ds+Lg∫σi0(σi−s)ρ−1Γ(ρ)ds)}≤LfTνΓ(ν+1)+Lf2|Λ1|[m∑i=1aiσνiΓ(ν+1)+P2TνΓ(ν+1)]+Lf2|Λ2|[ζν+1−ην+1Γ(ν+2)+Tν+1Γ(ν+2)]+Lg2|Λ1|[m∑i=1aiσρiΓ(ρ+1)+P2TρΓ(ρ+1)]+Lg2|Λ2|[ζρ+1−ηρ+1Γ(ρ+2)+Tρ+1Γ(ρ+2)]+A2|Λ2|, |
which, in view of (3.4) and (3.5), takes the form:
|Φ1(φ,ψ)(t)|≤Lf(TνΓ(ν+1)+ϱ1)+Lgϱ2+A2|Λ2|. | (3.7) |
In a similar fashion, one can obtain
|Φ2(φ,ψ)(t)|≤Lfϱ1+Lg(TρΓ(ρ+1)+ϱ2)+A2|Λ2|. | (3.8) |
From (3.7) and (3.8), we get
‖Φ(φ,ψ)‖=‖Φ1(φ,ψ)‖+‖Φ2(φ,ψ)‖≤Lf(TνΓ(ν+1)+2ϱ1)+Lg(TρΓ(ρ+1)+2ϱ2)+A|Λ2|. |
From the foregoing inequality, we deduce that the operator Φ is uniformly bounded.
In order to show that Φ maps bounded sets into equicontinuous sets of X, let t1,t2∈[0,T],t1<t2, and (φ,ψ)∈Ωˉr. Then
|Φ1(φ,ψ)(t2)−Φ1(φ,ψ)(t1)|≤|1Γ(ν)(∫t10[(t2−s)ν−1−(t1−s)ν−1]f(s,φ(s),ψ(s))ds+∫t2t1(t2−s)ν−1f(s,φ(s),ψ(s))ds)|≤Lf(2(t2−t1)ν+tν2−tν1Γ(ν+1)). |
Analogously, we can obtain
|Φ2(φ,ψ)(t2)−Φ2(u,v)(t1)|≤Lg(2(t2−t1)ρ+tρ2−tρ1Γ(ρ+1)). |
Clearly the right-hand sides of the above inequalities tend to zero when t1→t2, independently of (φ,ψ)∈Ωˉr. Thus it follows by the Arzelá-Ascoli theorem that the operator Φ:X→X is completely continuous.
Next we consider the set E={(φ,ψ)∈X|(φ,ψ)=λΦ(φ,ψ),0<λ<1} and show that it is bounded. Let (φ,ψ)∈E, then (φ,ψ)=λΦ(φ,ψ),0<λ<1. For any t∈J, we have
φ(t)=λΦ1(φ,ψ)(t),ψ(t)=λΦ2(φ,ψ)(t). |
As in the previous step, using ϱi(i=1,2) given by (3.4)-(3.5), we find that
|φ(t)|=λ|Φ1(φ,ψ)(t)|≤(‖μ1‖+‖μ2‖‖φ‖+‖μ3‖‖ψ‖)(TνΓ(ν+1)+ϱ1)+(‖κ1‖+‖κ2‖‖φ‖+‖κ3‖‖ψ‖)ϱ2+A2|Λ2|, |
|ψ(t)|=λ|Φ2(φ,ψ)(t)|≤(‖μ1‖+‖μ2‖‖φ‖+‖μ3‖‖ψ‖)ϱ1+(‖κ1‖+‖κ2‖‖φ‖+‖κ3‖‖ψ‖)(TρΓ(ρ+1)+ϱ2)+A2|Λ2|. |
In consequence, we get
‖φ‖+‖ψ‖≤‖μ1‖(2ϱ1+TνΓ(ν+1))+‖κ1‖(2ϱ2+TρΓ(ρ+1))+A|Λ2|+[‖μ2‖(2ϱ1+TνΓ(ν+1))+‖κ2‖(2ϱ2+TρΓ(ρ+1))]‖φ‖+[‖μ3‖(2ϱ1+TνΓ(ν+1))+‖κ3‖(2ϱ2+TνΓ(ν+1))]‖ψ‖. |
Thus, by the condition (3.6), we have
‖(φ,ψ)‖≤1M0{‖μ1‖(2ϱ1+TνΓ(ν+1))+‖κ1‖(2ϱ2+TρΓ(ρ+1))+A|Λ2|}, |
which shows that ‖(φ,ψ)‖ is bounded for t∈J. In consequence, the set E is bounded. Thus it follows by the conclusion of Lemma 3.1 that the operator Φ has at least one fixed point, which is indeed a solution of the problem (1.1).
Letting μ2(t)=μ3(t)≡0 and κ2(t)=κ3(t)≡0, the statement of Theorem 3.1 takes the following form.
Corollary 3.1. Let f,g:J×R2→R be continuousfunctions such that
|f(t,φ,ψ)|≤μ1(t),|g(t,φ,ψ)|≤κ1(t),∀(t,φ,ψ)∈J×R2, |
where μ1,κ1∈C([0,T],R+). Then there exists at least one solution for the problem (1.1) on J.
Corollary 3.2. If μi(t)=λi,κi(t)=εi,i=1,2,3, then the condition (H1) becomes:
(H′1) there exist real constants λi,εi>0,i=1,2, such that
|f(t,φ,ψ)|≤λ1+λ2|φ|+λ3|ψ|∀(t,φ,ψ)∈J×R2; |
|f(t,φ,ψ)|≤ε1+ε2|φ|+ε3|ψ|∀(t,φ,ψ)∈J×R2; |
and (3.6) takes the form:
λ2(2ϱ1+TνΓ(ν+1))+ε2(2ϱ2+TρΓ(ρ+1))<1,λ3(2ϱ1+TνΓ(ν+1))+ε3(2ϱ2+TρΓ(ρ+1))<1. |
Then there exists at least one solution for the problem (1.1) on J.
The next result is concerned with the existence of a unique solution for the problem (1.1) and is reliant on the contraction mapping principle due to Banach.
Theorem 3.2. Let f,g:[0,1]×R2→R be continuous functions and the assumption (H2) holds.Then the problem (1.1) has a unique solution on J if
α(TνΓ(ν+1)+2ϱ1)+β(TρΓ(ρ+1)+2ϱ2)<1, | (3.9) |
where α=max{α1,α2},β=max{β1,β2} and ϱi,i=1,2, are defined in (3.4)-(3.5).
Proof. Consider the operator Φ:X→X defined by (3.1) and take
r>M1(TνΓ(ν+1)+2ϱ1)+M2(TρΓ(ρ+1)+2ϱ2)+A|Λ2|1−(α(TνΓ(ν+1)+2ϱ1)+β(TρΓ(ρ+1)+2ϱ2)), |
where M1=supt∈[0,T]|f(t,0,0)|, and M2=supt∈[0,T]|g(t,0,0)|. Then we show that ΦBr⊂Br, where Br={(φ,ψ)∈X:‖(φ,ψ)‖≤r}. By the assumption (H1), for (φ,ψ)∈Br,t∈[0,T], we have
|f(t,φ(t),ψ(t))|≤|f(t,φ(t),ψ(t))−f(t,0,0)|+|f(t,0,0)|≤α(|φ(t)|+|ψ(t)|)+M1≤α(‖φ‖+‖ψ‖)+M1. |
In a similar manner, one can find that
|g(t,φ(t),ψ(t))|≤β(‖φ‖+‖ψ‖)+M2. |
In consequence, for (φ,ψ)∈Br, we obtain
|Φ1(φ,ψ)(t)|≤TνΓ(ν+1)(α(‖φ‖+‖ψ‖)+M1)+12[A|Λ2|+1|Λ2|(Tν+1Γ(ν+2)(α(‖φ‖+‖ψ‖)+M1)+Tρ+1Γ(ρ+2)(β(‖φ‖+‖ψ‖)+M2))+P2|Λ1|(TνΓ(ν+1)(α(‖φ‖+‖ψ‖)+M1)+TρΓ(ρ+1)(β(‖φ‖+‖ψ‖)+M2))+1|Λ2|(ζν+1−ην+1Γ(ν+2)(α(‖φ‖+‖ψ‖)+M1)+ζρ+1−ηρ+1Γ(ρ+2)(β(‖φ‖+‖ψ‖)+M2))+1|Λ1|m∑i=1ai(σνiΓ(ν+1)(α(‖φ‖+‖ψ‖)+M1)+σρiΓ(ρ+1)(β(‖φ‖+‖ψ‖)+M2))], |
which, on taking the norm for t∈J, yields
‖Φ1(φ,ψ)‖≤(α(TνΓ(ν+1)+ϱ1)+βϱ2)(‖φ‖+‖ψ‖)+M1(TνΓ(ν+1)+ϱ1)+M2ϱ2+A2|Λ2|. |
In the same way, for (φ,ψ)∈Br, one can obtain
‖Φ2(φ,ψ)‖≤(αϱ1+β(TρΓ(ρ+1)+ϱ2))(‖φ‖+‖ψ‖)+M1ϱ1+M2(TρΓ(ρ+1)+ϱ2)+A2|Λ2|. |
Therefore, for any (φ,ψ)∈Br, we have
‖Φ(φ,ψ))‖=‖Φ1(φ,ψ)‖+‖Φ2(φ,ψ)‖≤(α(TνΓ(ν+1)+2ϱ1)+β(TρΓ(ρ+1)+2ϱ2))(‖φ‖+‖ψ‖)+M1(TνΓ(ν+1)+2ϱ1)+M2(TρΓ(ρ+1)+2ϱ2)+A|Λ2|<r, |
which shows that Φ maps Br into itself.
Next it will be shown that the operator Φ is a contraction. For (φ1,ψ1),(φ2,ψ2)∈E,t∈[0,T], it follows by (H2) that
|Φ1(φ1,ψ1)(t)−Φ1(φ2,ψ2)(t)|≤∫t0(t−s)ν−1Γ(ν)|f(s,φ1(s),ψ1(s))−f(s,φ2(s),ψ2(s))|ds+12{1|Λ2|∫T0(∫s0(s−x)ν−1Γ(ν)|f(x,φ1(x),ψ1(x))−f(x,φ2(x),ψ2(x))|dx+∫s0(s−x)ρ−1Γ(ρ)|g(x,φ1(x),ψ1(x))−g(x,φ2(x),ψ2(x))|dx)ds+P2|Λ1|(∫T0(T−s)ν−1Γ(ν)|f(s,φ1(s),ψ1(s))−f(s,φ2(s),ψ2(s))|ds+∫T0(T−s)ρ−1Γ(ρ)|g(s,φ1(s),ψ1(s))−g(s,φ2(s),ψ2(s))|ds)+1|Λ2|∫ξη(∫s0(s−x)ν−1Γ(ν)|f(x,φ1(x),ψ1(x))−f(x,φ2(x),ψ2(x))|dx+∫s0(s−x)ρ−1Γ(ρ)|g(x,φ1(x),ψ1(x))−g(x,φ2(x),ψ2(x))|dx)ds+1|Λ1|m∑i=1ai(∫σi0(σi−s)ν−1Γ(ν)|f(s,φ1(s),ψ1(s))−f(s,φ2(s),ψ2(s))|ds+∫σi0(σi−s)ρ−1Γ(ρ)|g(s,φ1(s),ψ1(s))−g(s,φ2(s),ψ2(s))|ds)}≤{α(TνΓ(ν+1)+ϱ1)+βϱ2}(‖φ1−φ2‖+‖ψ1−ψ2‖), |
and
|Φ2(φ1,ψ1)(t)−Φ2(φ2,ψ2)(t)|≤∫t0(t−s)ρ−1Γ(ρ)|g(s,φ1(s),ψ1(s))−g(s,φ2(s),ψ2(s))|ds+12{1|Λ2|∫T0(∫s0(s−x)ν−1Γ(ν)|f(x,φ1(x),ψ1(x))−f(x,φ2(x),ψ2(x))|dx+∫s0(s−x)ρ−1Γ(ρ)|g(x,φ1(x),ψ1(x))−g(x,φ2(x),ψ2(x))|dx)ds+P2|Λ1|(∫T0(T−s)ν−1Γ(ν)|f(s,φ1(s),ψ1(s))−f(s,φ2(s),ψ2(s))|ds+∫T0(T−s)ρ−1Γ(ρ)|g(s,φ1(s),ψ1(s))−g(s,φ2(s),ψ2(s))|ds)+1|Λ2|∫ξη(∫s0(s−x)ν−1Γ(ν)|f(x,φ1(x),ψ1(x))−f(x,φ2(x),ψ2(x))|dx+∫s0(s−x)ρ−1Γ(ρ)|g(x,φ1(x),ψ1(x))−g(x,φ2(x),ψ2(x))|dx)ds+1|Λ1|m∑i=1ai(∫σi0(σi−s)ν−1Γ(ν)|f(s,φ1(s),ψ1(s))−f(s,φ2(s),ψ2(s))|ds+∫σi0(σi−s)ρ−1Γ(ρ)|g(s,φ1(s),ψ1(s))−g(s,φ2(s),ψ2(s))|ds)}≤{αϱ1+β(TρΓ(ρ+1)+ϱ2)}(‖φ1−φ2‖+‖ψ1−ψ2‖). |
In view of the foregoing inequalities, it follows that
‖Φ(φ1,ψ1)−Φ(φ2,ψ2)‖=‖Φ1(φ1,ψ1)−Φ1(φ2,ψ2)‖+‖Φ2(φ1,ψ1)−Φ2(φ2,ψ2)‖≤{α(TνΓ(ν+1)+2ϱ1)+β(TρΓ(ρ+1)+2ϱ2)}‖(φ1−φ2,ψ1−ψ2)‖. |
Using the condition (3.9), we deduce from the above inequality that Φ is a contraction mapping. Consequently Φ has a unique fixed point by the application of contraction mapping principle. Hence there exists a unique solution for the problem (1.1) on J. The proof is finished.
Example 3.1. Consider the following problem
{CD1/2φ(t)=f(t,φ(t),ψ(t)),t∈J:=[0,2],CD4/5ψ(t)=g(t,φ(t),ψ(t)),t∈J:=[0,2],(φ+ψ)(0)+5/2(φ+ψ)(2)=1/2(φ+ψ)(1/4)+3/2(φ+ψ)(1/2),∫20(φ−ψ)(s)ds−∫3/42/3(φ−ψ)(s)ds=1, | (3.10) |
where ν=1/2,ρ=4/5,η=2/3,ζ=3/4,a1=1/2,a2=3/2,P1=1,P2=5/2,σ1=1/4,σ2=1/2,A=1,T=2, and f(t,φ,ψ) and g(t,φ,ψ) will be fixed later.
Using the given data, we find that Λ1=1.5,Λ2=1.91666667, ϱ1=2.110627579,ϱ2=2.494392906, where Λ1,Λ2,ϱ1 and ϱ2 are respectively given by (2.5), (2.6), (3.4) and (3.5). For illustrating theorem 3.1, we take
f(t,φ,ψ)=e−t5√16+t2(tan−1φ+ψ+cost)andg(t,φ,ψ)=1(t+2)6(|φ|1+|ψ|+tψ+e−t). | (3.11) |
Clearly f and g are continuous and satisfy the condition (H1) with μ1(t)=e−tcost5√16+t2,μ2(t)=e−t5√16+t2,μ3(t)=e−t10√16+t2,κ1(t)=e−t(t+2)6,κ2(t)=1(t+2)6, and κ3(t)=12(t+2)6. Also
‖μ2‖(2ϱ1+TνΓ(ν+1))+‖κ2‖(2ϱ2+TρΓ(ρ+1))≈0.398009902, |
and
‖μ3‖(2ϱ1+TνΓ(ν+1))+‖κ3‖(2ϱ2+TρΓ(ρ+1))≈0.199004951<1. |
Thus all the conditions of theorem 3.1 hold true and hence the problem (3.10) with f(t,φ,ψ) and g(t,φ,ψ) given by (3.11) has at least one solution on [0,2].
Next we demonstrate the application of Theorem 3.2. Let us choose
f(t,φ,ψ)=e−ttan−1φ+cosψ5√16+t2andg(t,φ,ψ)=1(2+t)6(|φ|2+|φ|+sinψ). | (3.12) |
It is easy to show that the condition (H2) is satisfied with α1=α2=1/20=α and β1=1/64,β2=1/128 and so, β=1/64. Also α(TνΓ(ν+1)+2ϱ1)+β(TρΓ(ρ+1)+2ϱ2)≈0.39800990<1. Thus the hypothesis of Theorem 3.2 holds and hence its conclusion implies that the problem (3.10) with f(t,φ,ψ) and g(t,φ,ψ) given by (3.12) has a unique solution on [0,2].
In this section, we consider a variant of the problem (1.1) involving a nonlinear Riemann-Liouville integral term in the last boundary condition given by
{CDνφ(t)=f(t,φ(t),ψ(t)),t∈J:=[0,T],CDρψ(t)=g(t,φ(t),ψ(t)),t∈J:=[0,T],P1(φ+ψ)(0)+P2(φ+ψ)(T)=m∑i=1ai(φ+ψ)(σi),∫T0(φ−ψ)(s)ds−∫ζη(φ−ψ)(s)ds=1Γ(δ)∫T0(T−s)δ−1h(s,φ(s),ψ(s))ds,δ>0. | (4.1) |
Now we state a uniqueness result for the problem (4.1). We do not provide the proof of this result as it is similar to that of Theorem 3.2.
Theorem 4.1. Let f,g,h:[0,1]×R2→R be continuous functions and the following assumption holds:
(¯H2) There exist positive constants αi,βi,γi,i=1,2, such that
|f(t,φ1,ψ1)−f(t,φ2,ψ2)|≤α1|φ1−φ2|+α2|ψ1−ψ2|,∀t∈J,φi,ψi∈R,i=1,2; |
|g(t,φ1,ψ1)−g(t,φ2,ψ2)|≤β1|φ1−φ2|+β2|ψ1−ψ2|,∀t∈J,φi,ψi∈R,i=1,2; |
|h(t,φ1,ψ1)−h(t,φ2,ψ2)|≤γ1|φ1−φ2|+γ2|ψ1−ψ2|,∀t∈J,φi,ψi∈R,i=1,2. |
Then the problem (4.1) has a unique solution on J if
\begin{equation} \frac{\gamma T^\delta}{|\Lambda_2|\Gamma(\delta+1)}+\alpha\Big(\frac{T^\nu}{\Gamma(\nu+1)}+2\varrho_1\Big)+\beta\Big(\frac{T^\rho}{\Gamma(\rho+1)}+2\varrho_2\Big) < 1, \end{equation} | (4.2) |
where \alpha = \max \{\alpha_1, \alpha_2\}, \, \beta = \max \{\beta_1, \beta_2\}, \gamma = \max \{\gamma_1, \gamma_2\}, and \varrho_i, \, i = 1, 2 are defined in (3.4)-(3.5).
Example 4.1. Let us consider the data given in Example 3.1 for the problem (4.1) with (3.12), h(t, \varphi, \psi) = (\sin \varphi + \cos \psi+1/2)/\sqrt{t^2+49} and \delta = 3/2 . Then \gamma = 1/7 and
\frac{\gamma T^\delta}{|\Lambda_2|\Gamma(\delta+1)}+\alpha\Big(\frac{T^\nu}{\Gamma(\nu+1)}+2\varrho_1\Big)+\beta\Big(\frac{T^\rho}{\Gamma(\rho+1)}+2\varrho_2\Big)\approx 0.5565956 < 1. |
Clearly the assumptions of Theorem 4.1 are satisfied. Hence, by the conclusion of Theorem 4.1, the problem (4.1) with the given data has a unique solution on [0, 2].
We have studied a coupled system of nonlinear Caputo fractional differential equations supplemented with a new class of nonlocal multipoint-integral boundary conditions with respect to the sum and difference of the governing functions by applying the standard fixed point theorems. The existence and uniqueness results presented in this paper are not only new in the given configuration but also provide certain new results by fixing the parameters involved in the given problem. For example, our results correspond to the ones with initial-multipoint-integral and terminal-multipoint-integral boundary conditions by fixing P_2 = 0 and P_1 = 0 respectively in the present results. By taking A = 0 in the present study, we obtain the results for the given coupled system of fractional differential equations with the boundary conditions of the form:
P_1(\varphi+\psi)(0)+P_2(\varphi+\psi)(T) = \sum\limits_{i = 1}^{m} a_{i}(\varphi+\psi)(\sigma_{i}), \, \int_0^T(\varphi-\psi)(s)ds = \int_{\eta}^{\zeta}(\varphi-\psi)(s)ds, |
where the second (integral) condition means that the contribution of the difference of the unknown functions (\varphi-\psi) on the domain (0, T) is equal to that on the sub-domain (\eta, \zeta). Such a situation arises in heat conduction problems with sink and source. In the last section, we discussed the uniqueness of solutions for a variant of the problem (1.1) involving nonlinear Riemann-Liouville integral term in the last boundary condition of (1.1). This consideration further enhances the scope of the problem at hand. As a special case, the uniqueness result (Theorem 4.1) for the problem (4.1) corresponds to nonlinear integral boundary conditions for \delta = 1.
This project was funded by the Deanship of Scientific Research (DSR), King Abdulaziz University, Jeddah, Saudi Arabia under grant no. (KEP-PhD-80-130-42). The authors, therefore, acknowledge with thanks DSR technical and financial support. The authors also thank the reviewers for their useful remarks that led to the improvement of the original manuscript.
The authors declare that they have no competing interests.
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