Citation: Vernieda B. Vergara, Christy A. Emond, John F. Kalinich. Tissue distribution patterns of solubilized metals from internalized tungsten alloy in the F344 rat[J]. AIMS Environmental Science, 2016, 3(2): 290-304. doi: 10.3934/environsci.2016.2.290
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In recent years, the researchers and modelers have shown a keen interest in the topic of fractional differential equations. In fact, such equations appear in the mathematical models of several real-world phenomena occurring in pure, applied and technical sciences, for instance, see the books [1,2,3]. Unlike the classical derivative, there do exist many definitions of fractional derivatives and integrals. In [4], Hilfer proposed an important definition of fractional derivative (known as Hilfer fractional derivative), which represents both Riemann-Liouville and Caputo fractional derivatives under suitable choice of parameters. Several authors studied initial value problems involving Hilfer fractional derivatives, for example, see [5,6,7,8,9]. Some interesting results on boundary value problems involving Hilfer fractional differential equations can be found in the literature. For example, we refer the reader to works on nonlocal Hilfer problems [10,11], Hilfer Langevin equations [12,13], Hilfer Katugampola operators [14], Hilfer Erdelyi-Kober operators [15], Hilfer inclusion problems [16], Hilfer stochastic differential equations [17], $ \psi $-Hilfer problems [18], $ \psi $-Hilfer coupled systems [19], delay Hilfer fractional differential equations [20], Hilfer equations with variable coefficients [21], Hilfer sequential fractional differential equations [22,23], Hilfer approximate controllability [24] and Hilfer-Hadamard boundary value problems [25]. A variety of recent results on boundary value problems and coupled systems of Hilfer fractional differential equations and inclusions can be found in the survey paper [26].
In [27], the authors introduced and developed the existence and uniqueness of solutions for a new class of coupled systems of Hilfer-type fractional differential equations with nonlocal integral boundary conditions of the form
$ {HDα,βx(t)=f(t,x(t),y(t)),t∈[a,b],HDα1,β1y(t)=g(t,x(t),y(t)),t∈[a,b],x(a)=0,x(b)=m∑i=1θiIφiy(ξi),y(a)=0,y(b)=n∑j=1ζjIψjx(zj), $ | (1.1) |
where $ ^HD^{\alpha, \beta} $, $ ^HD^{\alpha_1, \beta_1} $ are the Hilfer fractional derivatives of orders $ \alpha $, $ \alpha_1 $, $ 1 < \alpha, \alpha_1 < 2 $, and parameters $ \beta $, $ \beta_1 $, $ 0\leq\beta, \beta_1\leq1 $, respectively, and $ I^{\varphi_i} $, $ I^{\psi_j} $ are the Riemann-Liouville fractional integrals of order $ \varphi_i > 0 $ and $ \psi_j > 0 $, respectively, the points $ \xi_i, z_j \in (a, b), a\geq 0, $ $ f, g: [a, b]\times {\mathbb R}\times {\mathbb R}\to {\mathbb R} $ are continuous functions and $ \theta_i $, $ \zeta_j\in \mathbb{R} $, $ i = 1, 2, \ldots, m $, $ j = 1, 2, \ldots, n $ are given real constants.
Recently, in [28], the authors studied a coupled system of $ \psi $-Hilfer fractional order Langevin equations with nonlocal integral boundary conditions given by
$ {HDα1,β1;ψa+(HDp1,q1;ψa++λ1)x(t)=f(t,x(t),y(t)),t∈J:=[a,b],HDα2,β2;ψa+(HDp2,q2;ψa++λ2)y(t)=g(t,x(t),y(t)),t∈J:=[a,b],[0.2cm]x(a)=0,x(b)=m∑i=1ηiIδi;ψa+y(θi),y(a)=0,y(b)=n∑j=1μjIκj;ψa+x(ξj), $ | (1.2) |
where $ {}^{H}\mathfrak{D}_{a^+}^{u, v; \psi } $ is $ \psi $-Hilfer fractional derivatives of order $ u \in \{\alpha_{1}, \alpha_{2}, p_{1}, p_{2}\} $ with $ 0 < u \leq 1 $ and $ v \in \{\beta_{1}, \beta_{2}, q_{1}, q_{2}\} $ with $ 0 \leq v \leq 1 $, $ \mathcal{I}_{a^+}^{w; \psi} $ is $ \psi $-Riemann-Liouville fractional integral of order $ w = \{\delta_{i}, \kappa_{j}\}, $ $ w > 0 $, the points $ \theta_i $, $ \xi_j \in (a, b) $, $ i = 1, 2, \ldots, m $, $ j = 1, 2, \ldots, n $, $ \lambda_{1} $, $ \lambda_{2} \in\mathbb{R} $, $ f $, $ g \in \mathcal{C}([a, b]\times \mathbb{R}^{2}, \mathbb{R}) $ and $ b > a \geq 0 $.
The objective of the present paper is to investigate the existence and uniqueness of solutions for a new class of coupled systems of Langevin type Hilfer fractional differential equations of different orders involving non-integral and autonomous type Riemann-Liouville mixed integral nonlinearities complemented with nonlocal coupled multi-point and Riemann-Liouville integral boundary conditions. This work is motivated by [27] and [28]. In precise terms, we consider the following problem:
$ {HDα1,β1(HDα2,β2+λ1)x(t)=Iζ1a+g1(x(t),y(t))+f1(t,x(t),y(t)),t∈[a,b],HDα3,β3(HDα4,β4+λ2)y(t)=Iζ2b−g2(x(t),y(t))+f2(t,x(t),y(t)),t∈[a,b],x(a)=0,x(b)=m∑i=1μiy(ηi)+n∑k=1νkIqka+y(ξk),qk>0,y(a)=0,y(b)=m∑i=1δix(ηi)+n∑k=1θkIpka+x(ξk),pk>0, $ | (1.3) |
where $ ^HD^{\alpha_j, \beta_j} $ represents Hilfer fractional derivative operator of order $ \alpha_j \in (0, 1) $ with parameter $ \beta_j \in [0, 1], $ $ j = 1, 2, 3, 4 $, $ \lambda_{1}, \lambda_{2}, \mu_{i}, \nu_{k}, \delta_{i} $ and $ \theta_{k}, i = 1, 2, ..., m, k = 1, 2, ..., n $ are constants, $ a < \eta_{i}, \xi_{k} < b, $ where $ a\geq 0 $ and $ m, n \in \mathbb{N}, $ $ I_{a^+}^{\zeta_1}, I_{a^+}^{q_{k}}, I_{a^+}^{p_{k}} $ denote the left Riemann-Liouville fractional integral operators of orders $ \zeta_1 > 0, q_{k} > 0, p_{k} > 0 $ respectively, while $ I^{\zeta_2}_{b^-} $ denotes the right Riemann-Liouville fractional integral operator of order $ \zeta_2 > 0, $ and $ f_{1}, f_{2}: [a, b]\times\mathbb{R}\times\mathbb{R}\rightarrow \mathbb{R}, $ $ g_{1}, g_{2}: \mathbb{R}\times\mathbb{R}\rightarrow \mathbb{R} $ are given continuous functions.
Note that problem (1.3) is more general than problem (1.2), since it contains non-integral as well as Riemann-Liouville mixed integral nonlinearities and nonlocal coupled multi-point and Riemann-Liouville integral boundary conditions.
The rest of the paper is organized as follows. In Section 2, we present some necessary material related to our study and prove an auxiliary lemma to define the solution for the problem at hand. Section 3 contains the main results which rely on Banach contraction mapping principle, Krasnosel'ski${\rm{\mathord{\buildrel{\lower3pt\hbox{$\scriptscriptstyle\smile$}} \over i} }}$'s fixed point theorem and Leray-Schauder alternative. In Section 4, we construct examples for the illustration of the results obtained in Section 3.
We begin this section with some basic concepts used in our study.
Definition 2.1. ([3]) The left and right Riemann–Liouville fractional integrals of order $ \omega > 0 $ for a continuous function $ g, $ existing almost everywhere on $ [a, b], $ are respectively defined by
$ I^{\omega}_{a+} g(t) = \int_{a}^t \frac{(t-s)^{\omega-1}}{\Gamma (\omega)}g(s)ds \, \, \, \, \;{{and}}\; \, \, \, \, I^{\omega}_{b-} g(t) = \int_t^b \frac{(s-t)^{\omega-1}}{\Gamma (\omega)}g(s)ds. $ |
For the sake of simplicity, we write $ I^{\omega}_{a+} $ and $ I^{\omega}_{b-} $ as $ I^{\omega}_{a} $ and $ I^{\omega}_{b} $ respectively.
Definition 2.2. ([4]) For $ n-1 < \alpha < n, \; 0\leq\beta\leq 1, $ the Hilfer fractional derivative of order $ \alpha $ and parameter $ \beta $ for a continuous function $ g $ is defined by
$ ^HD^{\alpha, \beta}g(t) = I_{a}^{\beta(n-\alpha)}D^{n}I_{a}^{(1-\beta)(n-\alpha)}g(t), D = \dfrac{d}{dt}, $ |
where
$ I_{a}^\omega g\left( t \right) = {\rm{ }}\frac{1}{{\Gamma \left( \omega \right)}}\int\limits_a^t {\left( {t - s} \right)^{\omega - 1} g\left( s \right)} ds, \; a\ge 0, $ |
with $ \omega \in \{\beta(n-\alpha), (1-\beta)(n-\alpha)\}. $
Lemma 2.1. ([16]) Let $ h \in L(a, b), \; n-1 < \gamma_{1} \leq n, \; n \in \mathbb{N}, \; 0\leq \gamma_{2} \leq1 $ and $ I_a^{(n-\gamma_{1})(1-\gamma_{2})}h \in AC^{k}[a, b] $. Then
$ Iγ1a(HDγ1,γ2h)(t)=h(t)−n−1∑k=0(t−a)k−(n−γ1)(1−γ2)Γ(k−(n−γ1)(1−γ2)+1)limt⟶a+dkdtk(I(1−γ2)(n−γ1)ah)(t). $ |
In the following lemma, we solve the linear variant of the problem (1.3).
Lemma 2.2. Let $ h_{1}, h_{2}:[a, b]\rightarrow \mathbb{R} $ be continuous functions and $ \Delta \ne 0. $ Then the unique solution of the following coupled system:
$ {HDα1,β1(HDα2,β2+λ1)x(t)=h1(t),t∈[a,b],HDα3,β3(HDα4,β4+λ2)y(t)=h2(t),t∈[a,b],x(a)=0,x(b)=m∑i=1μiy(ηi)+n∑k=1νkIqkay(ξk),qk>0,y(a)=0,y(b)=m∑i=1δix(ηi)+n∑k=1θkIpkax(ξk),pk>0, $ | (2.1) |
is given by
$ x(t)=Iα1+α2ah1(t)−λ1Iα2ax(t)+(t−a)α2+ϵ1−1ΔΓ(α2+ϵ1){Ω2(λ1Iα2ax(b)−Iα1+α2ah1(b)−λ2m∑i=1μiIα4ay(ηi)+m∑i=1μiIα3+α4ah2(ηi)+n∑k=1νkIqk+α3+α4ah2(ξk)−λ2n∑k=1νkIqk+α4ay(ξk))+Ω4(λ2Iα4ay(b)−Iα3+α4ah2(b)+m∑i=1δiIα1+α2ah1(ηi)−λ1m∑i=1δiIα2ax(ηi)+n∑k=1θkIpk+α1+α2ah1(ξk)−λ1n∑k=1θkIpk+α2ax(ξk))}, $ | (2.2) |
$ y(t)=Iα3+α4ah2(t)−λ2Iα4ay(t)+(t−a)α4+ϵ3−1ΔΓ(α4+ϵ3){Ω3(λ1Iα2ax(b)−Iα1+α2ah1(b)−λ2m∑i=1μiIα4ay(ηi)+m∑i=1μiIα3+α4ah2(ηi)+n∑k=1νkIqk+α3+α4ah2(ξk)−λ2n∑k=1νkIqk+α4ay(ξk))+Ω1(−Iα3+α4ah2(b)+λ2Iα4ay(b)+m∑i=1δiIα1+α2ah1(ηi)−λ1m∑i=1δiIα2ax(ηi)+n∑k=1θkIpk+α1+α2ah1(ξk)−λ1n∑k=1θkIpk+α2ax(ξk))}, $ | (2.3) |
where $ \Delta, \Omega_{i}, \; i = 1, 2, 3, 4 $ are given by
$ Ω1=(b−a)α2+ϵ1−1Γ(α2+ϵ1),Ω2=(b−a)α4+ϵ3−1Γ(α4+ϵ3),Ω3=m∑i=1δi(ηi−a)α2+ϵ1−1Γ(α2+ϵ1)+n∑k=1θk(ξk−a)pk+α2+ϵ1−1Γ(pk+α2+ϵ1),Ω4=m∑i=1μi(ηi−a)α4+ϵ3−1Γ(α4+ϵ3)+n∑k=1νk(ξk−a)qk+α4+ϵ3−1Γ(qk+α4+ϵ3),Δ=Ω1Ω2−Ω3Ω4, $ | (2.4) |
and $ \epsilon_{i} = \alpha_{i}+\beta_{i}-\alpha_{i}\beta_{i}, \; i = 1, 2, 3, 4 $.
Proof. Applying the integral operators $ I_{a}^{\alpha_{1}} $ and $ I_{a}^{\alpha_{3}} $ on the first and second Hilfer fractional differential equations in (2.1) respectively and using Lemma 2.1, we obtain
$ (HDα2,β2+λ1)x(t)−c0(t−a)ϵ1−1Γ(ϵ1)=Iα1ah1(t), $ | (2.5) |
$ (HDα4,β4+λ2)y(t)−d0(t−a)ϵ3−1Γ(ϵ3)=Iα3ah2(t). $ | (2.6) |
Now operating $ I_{a}^{\alpha_{2}} $ and $ I_{a}^{\alpha_{4}} $ respectively to the Eqs (2.5) and (2.6), we get
$ x(t)+λ1Iα2ax(t)−c1(t−a)ϵ2−1Γ(ϵ2)−c0(t−a)α2+ϵ1−1Γ(α2+ϵ1)=Iα1+α2ah1(t), $ | (2.7) |
$ y(t)+λ2Iα4ay(t)−d1(t−a)ϵ4−1Γ(ϵ4)−d0(t−a)α4+ϵ3−1Γ(α4+ϵ3)=Iα3+α4ah2(t). $ | (2.8) |
Using the conditions $ x(a) = 0 $ and $ y(a) = 0 $ in (2.7) and (2.8) respectively, we find that $ c_{1} = d_{1} = 0. $ Thus we have
$ x(t)=Iα1+α2ah1(t)−λ1Iα2ax(t)+c0(t−a)α2+ϵ1−1Γ(α2+ϵ1), $ | (2.9) |
$ y(t)=Iα3+α4ah2(t)−λ2Iα4ay(t)+d0(t−a)α4+ϵ3−1Γ(α4+ϵ3). $ | (2.10) |
Inserting (2.9) and (2.10) in the condition $ x(b) = \sum\limits_{i = 1}^{m}\mu_{i}y(\eta_{i})+\sum\limits_{k = 1}^{n}\nu_{k}I_{a}^{q_{k}}y(\xi_{k}), $ we find that
$ Iα1+α2ah1(b)−λ1Iα2ax(b)+c0(b−a)α2+ϵ1−1Γ(α2+ϵ1)=m∑i=1μi{Iα3+α4ah2(ηi)−λ2Iα4ay(ηi)+d0(ηi−a)α4+ϵ3−1Γ(α4+ϵ3)}+n∑k=1νkIqka{Iα3+α4ah2(ξk)−λ2Iα4ay(ξk)+d0(ξk−a)α4+ϵ3−1Γ(α4+ϵ3)}, $ |
which can alternatively be written as
$ c0(b−a)α2+ϵ1−1Γ(α2+ϵ1)−d0{m∑i=1μi(ηi−a)α4+ϵ3−1Γ(α4+ϵ3)+n∑k=1νk(ξk−a)qk+α4+ϵ3−1Γ(qk+α4+ϵ3)}=λ1Iα2ax(b)−Iα1+α2ah1(b)−λ2m∑i=1μiIα4ay(ηi)+m∑i=1μiIα3+α4ah2(ηi)+n∑k=1νkIqk+α3+α4ah2(ξk)−λ2n∑k=1νkIqk+α4ay(ξk). $ | (2.11) |
In a similar manner, making use of (2.9) and (2.10) in the condition: $ y(b) = \sum_{i = 1}^{m}\delta_{i}x(\eta_{i})+\sum_{k = 1}^{n}\theta_{k}I_{a}^{p_{k}}x(\xi_{k}), $ leads to
$ −c0{m∑i=1δi(ηi−a)α2+ϵ1−1Γ(α2+ϵ1)+n∑k=1θk(ξk−a)pk+α2+ϵ1−1Γ(pk+α2+ϵ1)}+d0(b−a)α4+ϵ3−1Γ(α4+ϵ3)=λ2Iα4ay(b)−Iα3+α4ah2(b)+m∑i=1δiIα1+α2ah1(ηi)−λ1m∑i=1δiIα2ax(ηi)+n∑k=1θkIpk+α1+α2ah1(ξk)−λ1n∑k=1θkIpk+α2ax(ξk). $ | (2.12) |
Making use of the notation in (2.4), we can write (2.11) and (2.12) as
$ Ω1c0−Ω4d0=λ1Iα2ax(b)−Iα1+α2ah1(b)−λ2m∑i=1μiIα4ay(ηi)+m∑i=1μiIα3+α4ah2(ηi)+n∑k=1νkIqk+α3+α4ah2(ξk)−λ2n∑k=1νkIqk+α4ay(ξk),−Ω3c0+Ω2d0=λ2Iα4ay(b)−Iα3+α4ah2(b)+m∑i=1δiIα1+α2ah1(ηi)−λ1m∑i=1δiIα2ax(ηi)+n∑k=1θkIpk+α1+α2ah1(ξk)−λ1n∑k=1θkIpk+α2ax(ξk), $ |
which, on solving for $ c_{0} $ and $ d_{0}, $ yields
$ c0=1Δ{Ω2(λ1Iα2ax(b)−Iα1+α2ah1(b)−λ2m∑i=1μiIα4ay(ηi)+m∑i=1μiIα3+α4ah2(ηi)+n∑k=1νkIqk+α3+α4ah2(ξk)−λ2n∑k=1νkIqk+α4ay(ξk))+Ω4(λ2Iα4ay(b)−Iα3+α4ah2(b)+m∑i=1δiIα1+α2ah1(ηi)−λ1m∑i=1δiIα2ax(ηi)+n∑k=1θkIpk+α1+α2ah1(ξk)−λ1n∑k=1θkIpk+α2ax(ξk))},d0=1Δ{Ω3(λ1Iα2ax(b)−Iα1+α2ah1(b)−λ2m∑i=1μiIα4ay(ηi)+m∑i=1μiIα3+α4ah2(ηi)+n∑k=1νkIqk+α3+α4ah2(ξk)−λ2n∑k=1νkIqk+α4ay(ξk))+Ω1(λ2Iα4ay(b)−Iα3+α4ah2(b)+m∑i=1δiIα1+α2ah1(ηi)−λ1m∑i=1δiIα2ax(ηi)+n∑k=1θkIpk+α1+α2ah1(ξk)−λ1n∑k=1θkIpk+α2ax(ξk))}. $ |
Substituting the values of $ c_0 $ and $ d_0 $ in (2.9) and (2.10) respectively together with (2.4), we get the solution (2.2) and (2.3). By direct computation, one can obtain the converse of this lemma. The proof is finished.
Let $ \mathcal{X} = C([a, b], \mathbb{R}) $ denote the Banach space of all continuous functions from $ [a, b] $ to $ \mathbb{R} $ with the norm $ \|x\| = \sup_{t \in [a, b]}|x(t)|. $ Then the product space $ (\mathcal{X} \times \mathcal{X}, \|\cdot\|) $ is also a Banach space endowed with the norm $ \|(x, y)\| = \|x\|+\|y\| $ for $ (x, y) \in \mathcal{X} \times \mathcal{X} $.
In view of Lemma 2.2, we introduce an operator $ \mathcal{T}:\mathcal{X}\times \mathcal{X} \rightarrow \mathcal{X} \times \mathcal{X} $ as
$ T(x,y)(t)=(T1(x,y)(t)T2(x,y)(t)), $ |
where
$ T1(x,y)(t)=Iα1+α2+ζ1ag1(x(t),y(t))+Iα1+α2af1(t,x(t),y(t))−λ1Iα2ax(t)+(t−a)α2+ϵ1−1ΔΓ(α2+ϵ1)×{Ω2(λ1Iα2ax(b)−Iα1+α2+ζ1ag1(x(b),y(b))−Iα1+α2af1(b,x(b),y(b))−λ2m∑i=1μiIα4ay(ηi)+m∑i=1μiIα3+α4a(Iζ2bg2(x(ηi),y(ηi)))+m∑i=1μiIα3+α4af2(ηi,x(ηi),y(ηi))+n∑k=1νkIqk+α3+α4a(Iζ2bg2(x(ξk),y(ξk)))+n∑k=1νkIqk+α3+α4af2(ξk,x(ξk),y(ξk))−λ2n∑k=1νkIqk+α4ay(ξk))+Ω4(λ2Iα4ay(b)−Iα3+α4af2(b,x(b),y(b))+m∑i=1δiIα1+α2+ζ1ag1(x(ηi),y(ηi))+m∑i=1δiIα1+α2af1(ηi,x(ηi),y(ηi))−λ1m∑i=1δiIα2ax(ηi)+n∑k=1θkIpk+α1+α2+ζ1ag1(x(ξk),y(ξk))+n∑k=1θkIpk+α1+α2af1(ξk,x(ξk),y(ξk))−λ1n∑k=1θkIpk+α2ax(ξk))}, $ | (3.1) |
and
$ T2(x,y)(t)=Iα3+α4a(Iζ2bg2(x(t),y(t)))+Iα3+α4af2(t,x(t),y(t))−λ2Iα4ay(t)+(t−a)α4+ϵ3−1ΔΓ(α4+ϵ3)×{Ω3(λ1Iα2ax(b)−Iα1+α2+ζ1ag1(x(b),y(b))−Iα1+α2af1(b,x(b),y(b))−λ2m∑i=1μiIα4ay(ηi)+m∑i=1μiIα3+α4a(Iζ2bg2(x(ηi),y(ηi)))+m∑i=1μiIα3+α4af2(ηi,x(ηi),y(ηi))+n∑k=1νkIqk+α3+α4a(Iζ2bg2(x(ξk),y(ξk)))+n∑k=1νkIqk+α3+α4af2(ξk,x(ξk),y(ξk))−λ2n∑k=1νkIqk+α4ay(ξk))+Ω1(λ2Iα4ay(b)−Iα3+α4af2(b,x(b),y(b))+m∑i=1δiIα1+α2+ζ1ag1(x(ηi),y(ηi))+m∑i=1δiIα1+α2af1(ηi,x(ηi),y(ηi))−λ1m∑i=1δiIα2ax(ηi)+n∑k=1θkIpk+α1+α2+ζ1ag1(x(ξk),y(ξk))+n∑k=1θkIpk+α1+α2af1(ξk,x(ξk),y(ξk))−λ1n∑k=1θkIpk+α2ax(ξk))}. $ | (3.2) |
For computational facilitation, we set
$ σ1=(b−a)α1+α2Γ(α1+α2+1)+Ω1|Δ|{Ω2(b−a)α1+α2Γ(α1+α2+1)+|Ω4|(m∑i=1|δi|(ηi−a)α1+α2Γ(α1+α2+1)+n∑k=1|θk|(ξk−a)pk+α1+α2Γ(pk+α1+α2+1))}, $ | (3.3) |
$ σ2=Ω1|Δ|{Ω2(m∑i=1|μi|(ηi−a)α3+α4Γ(α3+α4+1)+n∑k=1|νk|(ξk−a)qk+α3+α4Γ(qk+α3+α4+1))+|Ω4|(b−a)α3+α4Γ(α3+α4+1)}, $ | (3.4) |
$ σ3=Ω2|Δ|{|Ω3|(b−a)α1+α2Γ(α1+α2+1)+Ω1(m∑i=1|δi|(ηi−a)α1+α2Γ(α1+α2+1)+n∑k=1|θk|(ξk−a)pk+α1+α2Γ(pk+α1+α2+1))}, $ | (3.5) |
$ σ4=(b−a)α3+α4Γ(α3+α4+1)+Ω2|Δ|{|Ω3|(m∑i=1|μi|(ηi−a)α3+α4Γ(α3+α4+1)+n∑k=1|νk|(ξk−a)qk+α3+α4Γ(qk+α3+α4+1))+Ω1(b−a)α3+α4Γ(α3+α4+1)}, $ | (3.6) |
$ σ5=1|Δ|{|λ1|(|Δ|+Ω2Ω1)(b−a)α2Γ(α2+1)+Ω1|λ2Ω4|(b−a)α4Γ(α4+1)+Ω1|λ1Ω4|(m∑i=1|δi|(ηi−a)α2Γ(α2+1)+n∑k=1|θk|(ξk−a)pk+α2Γ(pk+α2+1))+|λ2|Ω2Ω1(n∑k=1|νk|(ξk−a)qk+α4Γ(qk+α4+1)+m∑i=1|μi|(ηi−a)α4Γ(α4+1))}, $ | (3.7) |
$ σ6=1|Δ|{Ω2|λ1Ω3|(b−a)α2Γ(α2+1)+|λ2|(|Δ|+Ω1Ω2)(b−a)α4Γ(α4+1)+|λ1|Ω1Ω2(m∑i=1|δi|(ηi−a)α2Γ(α2+1)+n∑k=1|θk|(ξk−a)pk+α2Γ(pk+α2+1))+Ω2|λ2Ω3|(m∑i=1|μi|(ηi−a)α4Γ(α4+1)+n∑k=1|νk|(ξk−a)qk+α4Γ(qk+α4+1))}, $ | (3.8) |
$ σ7=(b−a)α1+α2+ζ1Γ(α1+α2+ζ1+1)+Ω1|Δ|{Ω2(b−a)α1+α2+ζ1Γ(α1+α2+ζ1+1)+|Ω4|(m∑i=1|δi|(ηi−a)α1+α2+ζ1Γ(α1+α2+ζ1+1)+n∑k=1|θk|(ξk−a)pk+α1+α2+ζ1Γ(pk+α1+α2+ζ1+1))}, $ | (3.9) |
$ σ8=Ω1Ω2(b−a)ζ2|Δ|Γ(ζ2+1)(m∑i=1|μi|(ηi−a)α3+α4Γ(α3+α4+1)+n∑k=1|νk|(ξk−a)qk+α3+α4Γ(qk+α3+α4+1)), $ | (3.10) |
$ σ9=Ω2|Δ|{|Ω3|(b−a)α1+α2+ζ1Γ(α1+α2+ζ1+1)+Ω1(m∑i=1|δi|(ηi−a)α1+α2+ζ1Γ(α1+α2+ζ1+1)+n∑k=1|θk|(ξk−a)pk+α1+α2+ζ1Γ(pk+α1+α2+ζ1+1))}, $ | (3.11) |
$ σ10=(b−a)ζ2Γ(ζ2+1)((b−a)α3+α4Γ(α3+α4+1)+Ω2|Ω3||Δ|(m∑i=1|μi|(ηi−a)α3+α4Γ(α3+α4+1)+n∑k=1|νk|(ξk−a)qk+α3+α4Γ(qk+α3+α4+1))). $ | (3.12) |
In the sequel, we suppose that $ f_{1}, f_{2} :[a, b]\times {\mathbb R}\times {\mathbb R}\to {\mathbb R} $ and $ g_{1}, g_{2}: \mathbb{R}\times\mathbb{R}\rightarrow \mathbb{R} $ are continuous functions satisfying the following assumptions:
$ (H_{1}) $ $ \forall (x_{1}, y_{1}), (x_{2}, y_{2}) \in \mathbb{R}^2, $ there exist positive real constants $ K_{i}, $ i = 1, 2, such that
$ |f1(t,x1,y1)−f1(t,x2,y2)|≤K1(|x1−x2|+|y1−y2|),|f2(t,x1,y1)−f2(t,x2,y2)|≤K2(|x1−x2|+|y1−y2|); $ |
$ (H_{2}) \; \forall (x_{1}, y_{1}), (x_{2}, y_{2}) \in \mathbb{R}^2, $ there exist positive real constants $ L_{i}, $ i = 1, 2, such that
$ |g1(x1,y1)−g1(x2,y2)|≤L1(|x1−x2|+|y1−y2|),|g2(x1,y1)−g2(x2,y2)|≤L2(|x1−x2|+|y1−y2|); $ |
$ (H_{3}) $ We can find real constants $ u_{k}, v_{k}, \omega_{k}, \tau_{k}\geqslant 0, k = 0, 1, 2 $ with $ u_{0}, v_{0}, \omega_{0}, \tau_{0}\neq 0 $ such that
$ |f1(t,x,y)|≤u0+u1|x|+u2|y|,|f2(t,x,y)|≤v0+v1|x|+v2|y|,|g1(x,y)|≤ω0+ω1|x|+ω2|y|,|g2(x,y)|≤τ0+τ1|x|+τ2|y|; $ |
$ (H_{4}) $ There exist nonnegative functions $ \phi _{1}, \phi _{2} \in C([a, b], {\mathbb R}^{+}), $ and positive constants $ \Lambda_{1}, \Lambda_{2} $ such that
$ |f_{1}(t, x, y)|\le \phi_{1}(t) $, $ |f_{2}(t, x, y)|\le \phi_{2}(t) $, $ |g_1(x, y)|\leqslant \Lambda_{1} $, $ |g_2(x, y)|\leqslant \Lambda_{2} $ for all $ (t, x, y)\in [a, b] \times {\mathbb R}\times {\mathbb R}. $
Now we present our first main result dealing with the uniqueness of solutions for the system (1.3), which relies on Banach contraction mapping principle [29].
Theorem 3.1. Assume that conditions ($ H_{1}) $ and $ (H_{2} $) hold. Then the system (1.3) has a unique solution on $ [a, b] $ provided that
$ (σ1+σ3)K1+(σ4+σ2)K2+(σ9+σ7)L1+(σ10+σ8)L2+σ5+σ6<1, $ | (3.13) |
where $ \sigma_1, \dots, \sigma_{10} $ are given in (3.3)–(3.12).
Proof. Let us fix $ \sup_{t\in[a, b]}|f_{i}(t, 0, 0)| = M_{i} < \infty, |g_{i}(0, 0)| = 0, \; i = 1, 2. $ In order to satisfy the hypotheses of Banach contraction mapping principle, we first show that $ \mathcal{T}B_{\rho}\subset B_{\rho}, $ where $ B_{\rho} $ is a closed bounded ball $ B_{\rho}\subset \mathcal{X} \times \mathcal{X} $ defined by
$ Bρ={(x,y)∈X×X:‖(x,y)‖≤ρ}, $ |
with
$ ρ≥M1(σ1+σ3)+M2(σ2+σ4)1−[K1(σ1+σ3)+K2(σ2+σ4)+L1(σ7+σ9)+L2(σ8+σ10)+σ5+σ6]. $ | (3.14) |
For an arbitrary element $ (x, y) \in B_{\rho} $ and for each $ t \in [a, b], $ we have
$ |T1(x,y)(t)|≤Iα1+α2+ζ1a(|g1(x(t),y(t))−g1(0,0)|+|g1(0,0)|)+Iα1+α2a(|f1(t,x(t),y(t))−f1(t,0,0)|+|f1(t,0,0)|)+|λ1|Iα2a|x(t)|+(b−a)α2+ϵ1−1|Δ|Γ(α2+ϵ1){Ω2(|λ1|Iα2a|x(b)|+Iα1+α2+ζ1a(|g1(x(b),y(b))−g1(0,0)|+|g1(0,0)|)+Iα1+α2a(|f1(b,x(b),y(b))−f1(b,0,0)|+|f1(b,0,0)|)+|λ2|m∑i=1|μi|Iα4a|y(ηi)|+m∑i=1|μi|Iα3+α4aIζ2b(|g2(x(ηi),y(ηi))−g2(0,0)|+|g2(0,0)|)+m∑i=1|μi|Iα3+α4a(|f2(ηi,x(ηi),y(ηi))−f2(ηi,0,0)|+|f2(ηi,0,0)|)+n∑k=1|νk|Iqk+α3+α4aIζ2b(|g2(x(ξk),y(ξk))−g2(0,0)|+|g2(0,0)|)+n∑k=1|νk|Iqk+α3+α4a(|f2(ξk,x(ξk),y(ξk))−f2(ξk,0,0)|+|f2(ξk,0,0)|)+|λ2|n∑k=1|νk|Iqk+α4a|y(ξk)|)+Ω4(|λ2|Iα4a|y(b)|+Iα3+α4a(|f2(b,x(b),y(b))−f2(b,0,0)|+|f2(b,0,0)|)+m∑i=1|δi|Iα1+α2+ζ1a(|g1(x(ηi),y(ηi))−g1(0,0)|+|g1(0,0)|)+m∑i=1|δi|Iα1+α2a(|f1(ηi,x(ηi),y(ηi))−f1(ηi,0,0)|+|f1(ηi,0,0)|)+n∑k=1|θk|Ipk+α1+α2+ζ1a(|g1(x(ξk),y(ηi))−g1((0,0)|+|g1(0,0)|)+n∑k=1|θk|Ipk+α1+α2a(|f1(ξk,x(ξk),y(ξk))−f1(ξk,0,0)|+|f1(ξk,0,0)|)+|λ1|m∑i=1|δi|Iα2a|x(ηi)|+|λ1|n∑k=1|θk|Ipk+α2a|x(ξk)|)}≤σ1[K1(‖x‖+‖y‖)+M1]+σ2[K2(‖x‖+‖y‖)+M2]+σ7L1(‖x‖+‖y‖)+σ8L2(‖x‖+‖y‖)+σ5(‖x‖+‖y‖). $ |
In a similar manner, one can find that
$ ‖T2(x,y)‖≤σ3[K1(‖x‖+‖y‖)+M1]+σ4[K2(‖x‖+‖y‖)+M2]+σ9L1(‖x‖+‖y‖)+σ10L2(‖x‖+‖y‖)+σ6(‖x‖+‖y‖). $ |
Adding the last two inequalities and using (3.14), we obtain
$ ‖T(x,y)‖≤(σ1+σ3)[K1(‖x‖+‖y‖)+M1]+(σ2+σ4)[K2(‖x‖+‖y‖)+M2]+(σ7+σ9)L1(‖x‖+‖y‖)+(σ8+σ10)L2(‖x‖+‖y‖)+(σ5+σ6)(‖x‖+‖y‖)≤ρ, $ |
which shows that $ \mathcal{T}B_{\rho}\subset B_{\rho} $.
Next, we show that $ \mathcal{T} $ is a contraction on $ \mathcal{X}\times \mathcal{X}. $ For that, let $ (x, y), (x_1, y_1) \in \mathcal{X} \times \mathcal{X}. $ Then we have
$ |T1(x,y)(t)−T1(x1,y1)(t)|≤Iα1+α2+ζ1a|g1(x(t),y(t))−g1(x1(t),y1(t))|+Iα1+α2a|f1(t,x(t),y(t))−f1(t,x1(t),y1(t))|+|λ1|Iα2a|x(t)−x1(t)|+Ω1|Δ|{Ω2(Iα1+α2a|f1(b,x(b),y(b))−f1(b,x1(b),y1(b))|+|λ1|Iα2a|x(b)−x1(b)|+Iα1+α2+ζ1a|g1(x(b),y(b))−g1(x1(b),y1(b))|+|λ2|m∑i=1|μi|Iα4a|y(ηi)−y1(ηi)|+m∑i=1|μi|Iα3+α4aIζ2b(|g2(x(ηi),y(ηi))−g2(x1(ηi),y1(ηi))|)+m∑i=1|μi|Iα3+α4a|f2(ηi,x(ηi),y(ηi))−f2(ηi,x1(ηi),y1(ηi))|+n∑k=1|νk|Iqk+α3+α4a|f2(ξk,x(ξk),y(ξk))−f2(ξk,x1(ξk),y1(ξk))|+n∑k=1|νk|Iqk+α3+α4aIζ2b(|g2(x(ξk),y(ξk))−g2(x1(ξk),y1(ξk))|)+|λ2|n∑k=1|νk|Iqk+α4a|y(ξk)−y1(ξk)|)+|Ω4|(Iα3+α4a|f2(b,x(b),y(b))−f2(b,x1(b),y1(b))|+|λ2|Iα4a|y(b)−y1(b)|+m∑i=1|δi|Iα1+α2a|f1(ηi,x(ηi),y(ηi))−f1(ηi,x1(ηi),y1(ηi))|+m∑i=1|δi|Iα1+α2+ζ1a|g1(x(ηi),y(ηi))−g1(x1(ηi),y1(ηi))|+n∑k=1|θk|Ipk+α1+α2+ζ1a|g1(x(ξk),y(ξk))−g1(x1(ξk),y1(ξk))|+|λ1|n∑k=1|θk|Ipk+α2a|x(ξk)−x1(ξk)|+n∑k=1|θk|Ipk+α1+α2a|f1(ξk,x(ξk),y(ξk))−f1(ξk,x1(ξk),y1(ξk))|+|λ1|m∑i=1|δi|Iα2a|x(ηi)−x1(ηi)|)}, $ |
which leads to
$ ‖T1(x,y)−T1(x1,y1)‖≤[σ1K1+σ2K2+σ7L1+σ8L2+σ5](‖x−x1‖+‖y−y1‖). $ |
Similarly one can obtain
$ ‖T2(x,y)−T2(x1,y1)‖≤[σ3K1+σ4K2+σ9L1+σ10L2+σ6](‖x−x1‖+‖y−y1‖). $ |
It follows from the last two inequalities that
$ ‖T(x,y)−T(x1,y1)‖≤[(σ1+σ3)K1+(σ4+σ2)K2+(σ9+σ7)L1+(σ10+σ8)L2+σ5+σ6][‖x−x1‖+‖y−y1‖]. $ |
Since $ (\sigma_{1}+\sigma_{3})K_{1}+(\sigma_{4}+\sigma_{2})K_{2}+(\sigma_{9}+\sigma_{7})L_{1}+(\sigma_{10}+\sigma_{8})L_{2} +\sigma_{5}+\sigma_{6} < 1 $ by (3.13), therefore $ \mathcal{T} $ is a contraction and hence by Banach's contraction mapping principle, the operator $ \mathcal{T} $ has a unique fixed point. In consequence, the problem (1.3) has a unique solution on $ [a, b]. $ The proof is completed.
The next existence result is based on Krasnosel'ski${\rm{\mathord{\buildrel{\lower3pt\hbox{$\scriptscriptstyle\smile$}} \over i} }}$'s fixed point theorem.
Lemma 3.1. (Krasnosel'ski${{\mathord{\buildrel{\lower3pt\hbox{$\scriptscriptstyle\smile$}} \over i} }}$'s fixed point theorem). [30] Let $ B $ be a closed, convex, bounded and nonempty subset of a Banach space $ X. $ Let $ E_1 $ and $ E_2 $ be the operators such that (i) $ E_1x+E_2y \in B $ whenever $ x, y \in B; $ (ii) $ E_1 $ is compact and continuous; (iii) $ E_2 $ is a contractionmapping. Then there exists $ z \in B $ such that $ z = E_1z+E_2z. $
Theorem 3.2. Assume that $ (H_1), $ $ (H_2) $ and $ (H_{4}) $ hold and
$ σ5+σ6<1, $ | (3.15) |
where $ \sigma_{5} $ and $ \sigma_{6} $ are given by (3.7) and (3.8) respectively. Then the problem (1.3) has at least one solution on $ [a, b]. $
Proof. Let us split the operators $ \mathcal{T}_{1} $ and $ \mathcal{T}_{2} $ defined by (3.1) and (3.2) respectively into four operators as follows
$ \mathcal{T}_{1}(x, y)(t) = \mathcal{T}_{1, 1}(x, y)(t)+\mathcal{T}_{1, 2}(x, y)(t), \, \, \mathcal{T}_{2}(x, y)(t) = \mathcal{T}_{2, 1}(x, y)(t)+\mathcal{T}_{2, 2}(x, y)(t), $ |
where
$ T1,1(x,y)(t)=Iα1+α2+ζ1ag1(x(t),y(t))+Iα1+α2af1(t,x(t),y(t))+(t−a)α2+ϵ1−1ΔΓ(α2+ϵ1)×{Ω2(−Iα1+α2+ζ1ag1(x(b),y(b))−Iα1+α2af1(b,x(b),y(b))+m∑i=1μiIα3+α4a(Iζ2bg2(x(ηi),y(ηi)))+m∑i=1μiIα3+α4af2(ηi,x(ηi),y(ηi))+n∑k=1νkIqk+α3+α4a(Iζ2bg2(x(ξk),y(ξk)))+n∑k=1νkIqk+α3+α4af2(ξk,x(ξk),y(ξk)))+Ω4(−Iα3+α4af2(b,x(b),y(b))+m∑i=1δiIα1+α2+ζ1ag1(x(ηi),y(ηi))+m∑i=1δiIα1+α2af1(ηi,x(ηi),y(ηi))+n∑k=1θkIpk+α1+α2+ζ1ag1(x(ξk),y(ξk))+n∑k=1θkIpk+α1+α2af1(ξk,x(ξk),y(ξk)))},T1,2(x,y)(t)=−λ1Iα2ax(t)+(t−a)α2+ϵ1−1ΔΓ(α2+ϵ1){Ω2(−λ1Iα2ax(b)−λ2m∑i=1μiIα4ay(ηi)−λ2n∑k=1νkIqk+α4ay(ξk))+Ω4(λ2Iα4ay(b)−λ1m∑i=1δiIα2ax(ηi)−λ1n∑k=1θkIpk+α2ax(ξk))}, $ |
$ T2,1(x,y)(t)=Iα3+α4a(Iζ2bg2(x(t),y(t)))+Iα3+α4af2(t,x(t),y(t))+(t−a)α4+ϵ3−1ΔΓ(α4+ϵ3)×{Ω3(−Iα1+α2+ζ1ag1(x(b),y(b))−Iα1+α2af1(b,x(b),y(b))+m∑i=1μiIα3+α4a(Iζ2bg2(x(ηi),y(ηi)))+m∑i=1μiIα3+α4af2(ηi,x(ηi),y(ηi))+n∑k=1νkIqk+α3+α4a(Iζ2bg2(x(ξk),y(ξk)))+n∑k=1νkIqk+α3+α4af2(ξk,x(ξk),y(ξk)))+Ω1(−Iα3+α4af2(b,x(b),y(b))+m∑i=1δiIα1+α2+ζ1ag1(x(ηi),y(ηi))+m∑i=1δiIα1+α2af1(ηi,x(ηi),y(ηi))+n∑k=1θkIpk+α1+α2+ζ1ag1(x(ξk),y(ξk))+n∑k=1θkIpk+α1+α2af1(ξk,x(ξk),y(ξk)))},T2,2(x,y)(t)=−λ2Iα4ay(t)+(t−a)α4+ϵ3−1ΔΓ(α4+ϵ3)×{Ω3(λ1Iα2ax(b)−λ2m∑i=1μiIα4ay(ηi)−λ2n∑k=1νkIqk+α4ay(ξk))+Ω1(λ2Iα4ay(b)−λ1m∑i=1δiIα2ax(ηi)−λ1n∑k=1θkIpk+α2ax(ξk))}. $ |
Now we verify the hypotheses of Krasnosel'ski${\rm{\mathord{\buildrel{\lower3pt\hbox{$\scriptscriptstyle\smile$}} \over i} }}$'s fixed point theorem (Lemma 3.1) in three steps.
(ⅰ) In this step, it will be shown that $ \mathcal{T}_{1}(x, y)+\mathcal{T}_{2}(\widehat{x}, \widehat{y}) \in B_{r} $ for all $ (x, y), (\widehat{x}, \widehat{y}) \in B_{r}, $ where $ B_{r}\subset \mathcal{X}\times \mathcal{X} $ is a bounded closed ball with radius
$ r \geqslant \dfrac{ (\sigma_{1}+\sigma_{3})\|\phi_{1}\|+(\sigma_{2}+\sigma_{4})\|\phi_{2}\|+(\sigma_{7}+\sigma_{9})\Lambda_{1}+(\sigma_{8}+\sigma_{10})\Lambda_{2}}{1-\sigma_{5}-\sigma_{6}}. $ |
As in the proof of Theorem 3.1, we can find that
$ |T1,1(x,y)(t)+T1,2(x,y)(t)|≤σ1‖ϕ1‖+σ2‖ϕ2‖+σ7Λ1+σ8Λ2+σ5r, $ |
and
$ |T2,1(ˆx,ˆy)(t)+T2,2(ˆx,ˆy)(t)|≤σ3‖ϕ1‖+σ4‖ϕ2‖+σ9Λ1+σ10Λ2+σ6r, $ |
which lead to the inequality
$ ‖T1(x,y)+T2(ˆx,ˆy)‖≤(σ1+σ3)‖ϕ1‖+(σ2+σ4)‖ϕ2‖+(σ7+σ9)Λ1+(σ8+σ10)Λ2+(σ5+σ6)r≤r. $ |
Thus $ \mathcal{T}_{1}(x, y)+\mathcal{T}_{2}(\widehat{x}, \widehat{y}) \in B_{r} $.
(ⅱ) Here we establish that $ (\mathcal{T}_{1, 2}, \mathcal{T}_{2, 2}) $ is a contraction mapping. Let $ (x, y), (\widehat{x}, \widehat{y}) \in B_{r}. $ Then it is easy to find that
$ |\mathcal{T}_{1, 2}(x, y)(t)-\mathcal{T}_{1, 2}(\widehat{x}, \widehat{y})(t)| \leq \sigma_{5}[\|x-\widehat{x}\|+\|y-\widehat{y}\|], $ |
$ |\mathcal{T}_{2, 2}(x, y)(t)-\mathcal{T}_{2, 2}(\widehat{x}, \widehat{y})| \leq \sigma_{6}[\|x-\widehat{x}\|+\|y-\widehat{y}\|]. $ |
Consequently, we get
$ ‖(T1,2,T2,2)(x,y)−(T1,2,T2,2)(ˆx,ˆy)‖≤(σ5+σ6)[‖x−ˆx‖+‖y−ˆy‖], $ |
which, by (3.15), implies that $ (\mathcal{T}_{1, 2}, \mathcal{T}_{2, 2}) $ is a contraction.
(ⅲ) We show that $ (\mathcal{T}_{1, 1}, \mathcal{T}_{2, 1}) $ is compact and continuous.
Continuity of $ (\mathcal{T}_{1, 1}, \mathcal{T}_{2, 1}) $ is obvious. For $ (x, y) \in B_{r}, $ we have
$ |T1,1(x,y)(t)|≤Iα1+α2+ζ1a(|g1(x(t),y(t))|)+Iα1+α2a(|f1(t,x(t),y(t))|)+(b−a)α2+ϵ1−1|Δ|Γ(α2+ϵ1){Ω2(Iα1+α2+ζ1a(|g1(x(b),y(b))|)+Iα1+α2a(|f1(b,x(b),y(b))|)+m∑i=1|μi|Iα3+α4a(Iζ2b(|g2(x(ηi),y(ηi))|)+m∑i=1|μi|Iα3+α4a(|f2(ηi,x(ηi),y(ηi))|)+n∑k=1|νk|Iqk+α3+α4a(Iζ2b(|g2(x(ξk),y(ξk))|)+n∑k=1|νk|Iqk+α3+α4a(|f2(ξk,x(ξk),y(ξk))|))+Ω4(Iα3+α4a(|f2(b,x(b),y(b))|)+m∑i=1δiIα1+α2+ζ1a(|g1(x(ηi),y(ηi))|)+m∑i=1|δi|Iα1+α2a(|f1(ηi,x(ηi),y(ηi))|)+n∑k=1θkIpk+α1+α2+ζ1a(|g1(x(ξk),y(ηi))|)+n∑k=1|θk|Ipk+α1+α2a(|f1(ξk,x(ξk),y(ξk))|))}≤σ1‖ϕ1‖+σ2‖ϕ2‖+σ7Λ1+σ8Λ2. $ |
In a similar manner, we can get $ |\mathcal{T}_{2, 1}(x, y)(t)| \leq \sigma_{3}\|\phi_{1}\|+\sigma_{4}\|\phi_{2}\|+ \sigma_{9}\Lambda_{1}+ \sigma_{10}\Lambda_{2}. $ Thus
$ \|(\mathcal{T}_{1, 1}, \mathcal{T}_{2, 1})(x, y)\|\leq (\sigma_{1}+\sigma_{3})\|\phi_{1}\|+(\sigma_{2}+\sigma_{4})\|\phi_{2}\|+(\sigma_{7}+\sigma_{9})\Lambda_{1} + (\sigma_{8}+\sigma_{10})\Lambda_{2}, $ |
which means that $ (\mathcal{T}_{1, 1}, \mathcal{T}_{2, 1}) $ is uniformly bounded on $ B_{r}. $
In order to show the equicontinuity of $ (\mathcal{T}_{1, 1}, \mathcal{T}_{2, 1}), $ we take $ t_{1}, t_{2} \in [a, b] $ with $ t_{1} < t_{2}. $ Then, for arbitrary $ (x, y) \in B_{r}, $ we obtain
$ |T1,1(x,y)(t2)−T1,1(x,y)(t1)|≤|∫t2a(t2−s)α1+α2−1Γ(α1+α2)f1(s,x(s),y(s))ds−∫t1a(t1−s)α1+α2−1Γ(α1+α2)f1(s,x(s),y(s))ds|+|∫t2a(t2−s)α1+α2+ζ1−1Γ(α1+α2+ζ1)g1(x(s),y(s))ds−∫t1a(t1−s)α1+α2+ζ1−1Γ(α1+α2)+ζ1g1(x(s),y(s))ds|+|(t2−a)α2+ϵ1−1−(t1−a)α2+ϵ1−1||Δ|Γ(α2+ϵ1){Ω2(Iα1+α2+ζ1a(|g1(x(b),y(b))|)+Iα1+α2a(|f1(b,x(b),y(b))|)+m∑i=1|μi|Iα3+α4a(Iζ2b(|g2(x(ηi),y(ηi))|)+m∑i=1|μi|Iα3+α4a(|f2(ηi,x(ηi),y(ηi))|)+n∑k=1|νk|Iqk+α3+α4a(Iζ2b(|g2(x(ξk),y(ξk))|)+n∑k=1|νk|Iqk+α3+α4a(|f2(ξk,x(ξk),y(ξk))|))+Ω4(Iα3+α4a(|f2(b,x(b),y(b))|)+m∑i=1|δi|Iα1+α2+ζ1a(|g1(x(ηi),y(ηi))|)+m∑i=1|δi|Iα1+α2a(|f1(ηi,x(ηi),y(ηi))|)+n∑k=1|θk|Ipk+α1+α2+ζ1a(|g1(x(ξk),y(ηi))|)+n∑k=1|θk|Ipk+α1+α2a(|f1(ξk,x(ξk),y(ξk))|))}⩽‖ϕ1‖Γ(α1+α2+1){2(t2−t1)α1+α2+|(t2−a)α1+α2−(t1−a)α1+α2|}+Λ1Γ(α1+α2+ζ1+1){2(t2−t1)α1+α2+ζ1+|(t2−a)α1+α2+ζ1−(t1−a)α1+α2+ζ1|}+|(t2−a)α2+ϵ1−1−(t1−a)α2+ϵ1−1||Δ|Γ(α2+ϵ1){Ω2(‖ϕ1‖(b−a)α1+α2Γ(α1+α2+1)+Λ1(b−a)α1+α2+ζ1Γ(α1+α2+ζ1+1)+m∑i=1|μi|‖ϕ2‖(ηi−a)α3+α4Γ(α3+α4+1)+m∑i=1|μi|Λ2(b−a)ζ2(ηi−a)α3+α4Γ(α3+α4+1)Γ(ζ2+1)+n∑k=1|νk|‖ϕ2‖(ξk−a)qk+α3+α4Γ(qk+α3+α4+1))+n∑k=1|νk|Λ2(b−a)ζ2(ξk−a)qk+α3+α4Γ(ζ2+1)Γ(qk+α3+α4+1))+|Ω4|(‖ϕ2‖(b−a)α3+α4Γ(α3+α4+1)+m∑i=1|δi|‖ϕ1‖(ηi−a)α1+α2Γ(α1+α2+1)+m∑i=1|δi|Λ1(ηi−a)α1+α2+ζ1Γ(α1+α2+ζ1+1)+n∑k=1|θk|‖ϕ1‖(ξk−a)pk+α1+α2Γ(pk+α1+α2+1)+n∑k=1|θk|Λ1(ξk−a)pk+α1+α2+ζ1Γ(pk+α1+α2+ζ1+1))}→0, $ |
as $ t_{2}\rightarrow t_{1} $ independently of $ (x, y) \in B_{r}. $ Also
$ |T2,1(x,y)(t2)−T2,1(x,y)(t1)|≤‖ϕ2‖Γ(α3+α4+1){2(t2−t1)α3+α4+|(t2−a)α3+α4−(t1−a)α3+α4|}+Λ2(b−a)ζ2Γ(ζ2+1)Γ(α3+α4+1){2(t2−t1)α3+α4+|(t2−a)α3+α4−(t1−a)α3+α4|}+|(t2−a)α4+ϵ3−1−(t1−a)α4+ϵ3−1||Δ|Γ(α4+ϵ3){|Ω3|(‖ϕ1‖(b−a)α1+α2Γ(α1+α2+1)+Λ1(b−a)α1+α2+ζ1Γ(α1+α2+ζ1+1)+m∑i=1|μi|‖ϕ2‖(ηi−a)α3+α4Γ(α3+α4+1)+m∑i=1|μi|Λ2(b−a)ζ2(ηi−a)α3+α4Γ(ζ2+1)Γ(α3+α4+1)+n∑k=1|νk|‖ϕ2‖(ξk−a)qk+α3+α4Γ(qk+α3+α4+1)+n∑k=1|νk|Λ2(b−a)ζ2(ξk−a)qk+α3+α4Γ(ζ2+1)Γ(qk+α3+α4+1))+Ω1(‖ϕ2‖(b−a)α3+α4Γ(α3+α4+1)+m∑i=1|δi|‖ϕ1‖(ηi−a)α1+α2Γ(α1+α2+1)+m∑i=1|δi|Λ1(ηi−a)α1+α2+ζ1Γ(α1+α2+ζ1+1)+n∑k=1|θk|‖ϕ1‖(ξk−a)pk+α1+α2Γ(pk+α1+α2+1)+n∑k=1|θk|Λ1(ξk−a)pk+α1+α2+ζ1Γ(pk+α1+α2+ζ1+1))}→0, $ |
as $ t_{2}\rightarrow t_{1} $ independently of $ (x, y) \in B_{r}. $ Thus $ |(\mathcal{T}_{1, 1}, \mathcal{T}_{2, 1})(x, y)(t_{2})-(\mathcal{T}_{1, 1}, \mathcal{T}_{2, 1})(x, y)(t_{1})| $ vanishes as $ t_{2}\rightarrow t_{1} $ independently of $ (x, y) \in B_{r}, $ which shows that $ (\mathcal{T}_{1, 1}, \mathcal{T}_{2, 1}) $ is equicontinuous. So we deduce by the Arzelá-Ascoli theorem that $ (\mathcal{T}_{1, 1}, \mathcal{T}_{2, 1}) $ is compact on $ B_{r}. $
It follows from the steps (i)-(iii) that the hypotheses of Krasnosel'ski${\rm{\mathord{\buildrel{\lower3pt\hbox{$\scriptscriptstyle\smile$}} \over i} }}$'s fixed point theorem are satisfied, so its conclusion implies that the problem (1.3) has at least one solution on $ [a, b]. $ This finishes the proof.
Remark 3.1. The conclusion of Theorem 3.2 can also be achieved by assuming $ (H_{1}), (H_{2}), (H_4) $ and the condition: $ (\sigma_{1}+\sigma_{3})K_{1}+(\sigma_{2}+\sigma_{4})K_{2}+(\sigma_{7}+\sigma_{9})L_{1}+(\sigma_{8}+\sigma_{10})L_{2} < 1, $ where $ \sigma_1, \dots, \sigma_4 $ are given in (3.3)–(3.6) and $ \sigma_7, \dots, \sigma_{10} $ are given in (3.9)–(3.12).
In the following result, we prove the existence of solutions for the problem (1.3) by applying the Leray-Schauder alternative [29].
Lemma 3.2. (Leray-Schauder alternative [29]) Let $ E $ be a Banach space, $ M $ be closed, convex subset of $ E $, $ U $ is an open subset of $ C $ and $ 0 \in U. $ Suppose that $ F:\overline{U}\to C $ is continuous, compact map (that is, $ F(U) $ is a relatively compact subset of $ C $). Then either (i) $ F $ has a fixed point in $ \overline{U} $, or (ii) there are $ u \in \partial U, $ and $ \lambda \in (0, 1) $ with $ u = \lambda F(U) $.
Theorem 3.3. Assume that ($ H_{3}) $ holds. Then there exists at least one solution for the problem (1.3) on $ [a, b] $ provided that
$ (σ1+σ3)ui+(σ2+σ4)vi+(σ7+σ9)ωi+(σ8+σ10)τi+(σ5+σ6)<1,i=1,2, $ | (3.16) |
where $ \sigma_1, \dots, \sigma_{10} $ are given in (3.3)–(3.12).
Proof. For all $ (x, y) \in B_{\rho}\subset \mathcal{X} \times \mathcal{X}, $ where $ B_{\rho} $ defined by (3.14), there exist positive constants $ N_{1}, \dots, N_4 $ such that $ |f_{1}(t, x, y)|\leq N_{1}, |f_{2}(t, x, y)|\leq N_{2}, |g_{1}(x, y)|\leq N_{3}, |g_{2}(x, y)|\leq N_{4}, $ Then we show that $ \mathcal{T}: \mathcal{X} \times \mathcal{X} \rightarrow \mathcal{X} \times \mathcal{X} $ is completely continuous. Observe that continuity of $ f_{1}, f_{2}, g_{1}, g_2 $ implies that of the operator $ \mathcal{T}. $ For $ (x, y) \in B_{\rho} $, as in the proof of Theorem 3.1, we have
$ ‖T1(x,y)‖≤σ1N1+σ2N2+σ7N3+σ8N4+ρσ5,‖T2(x,y)‖≤σ3N1+σ4N2+σ9N3+σ10N4+ρσ6. $ |
From the preceding inequalities, we get
$ \| \mathcal{T}(x, y)\| \leq (\sigma_{1}+ \sigma_{3})N_{1}+(\sigma_{2}+\sigma_{4})N_{2}+(\sigma_{7}+\sigma_{9})N_{3}+(\sigma_{8}+\sigma_{10})N_{4}+\rho (\sigma_{5}+\sigma_{6}), $ |
which implies that $ \mathcal{T}B_{\rho} $ is uniformly bounded.
Next we show that $ \mathcal{T}B_{\rho} $ is equicontinues. Let $ t_{1}, t_{2} \in [a, b] $ with $ t_{2} > t_{1}. $ Then, for arbitrary $ (x, y) \in B_{\rho}, $ we obtain
$ |T1(x,y)(t2)−T1(x,y)(t1)|≤|∫t2a(t2−s)α1+α2−1Γ(α1+α2)f1(s,x(s),y(s))ds−∫t1a(t1−s)α1+α2−1Γ(α1+α2)f1(s,x(s),y(s))ds|+|∫t2a(t2−s)α1+α2+ζ1−1Γ(α1+α2+ζ1)g1(x(s),y(s))ds−∫t1a(t1−s)α1+α2+ζ1−1Γ(α1+α2+ζ1)g1(x(s),y(s))ds|+|λ1|Γ(α2)|∫t1a[(t2−s)α2−1−(t1−s)α2−1]x(s)ds+∫t2t1(t2−s)α2−1x(s)ds|+|(t2−a)α2+ϵ1−1−(t1−a)α2+ϵ1−1||Δ|Γ(α2+ϵ1){Ω2(Iα1+α2+ζ1a(|g1(x(b),y(b))|)+Iα1+α2a(|f1(b,x(b),y(b))|)+|λ1|Iα2a|x(b)|+|λ2|m∑i=1|μi|Iα4a|y(ηi)|+m∑i=1|μi|Iα3+α4a(Iζ2b(|g2(x(ηi),y(ηi))|)+m∑i=1|μi|Iα3+α4a(|f2(ηi,x(ηi),y(ηi))|)+n∑k=1|νk|Iqk+α3+α4a(Iζ2b(|g2(x(ξk),y(ξk))|)+n∑k=1|νk|Iqk+α3+α4a(|f2(ξk,x(ξk),y(ξk))|)+|λ2|n∑k=1|νk|Iqk+α4a|y(ξk)|)+Ω4(|λ2|Iα4a|y(b)|+Iα3+α4a(|f2(b,x(b),y(b))|)+m∑i=1|δi|Iα1+α2+ζ1a(|g1(x(ηi),y(ηi))|)+m∑i=1|δi|Iα1+α2a(|f1(ηi,x(ηi),y(ηi))|)+|λ1|m∑i=1|δi|Iα2a|x(ηi)|+n∑k=1|θk|Ipk+α1+α2+ζ1a(|g1(x(ξk),y(ηi))|)+n∑k=1|θk|Ipk+α1+α2a(|f1(ξk,x(ξk),y(ξk))|)+|λ1|n∑k=1|θk|Ipk+α2a|x(ξk)|)}⩽N1Γ(α1+α2+1){2(t2−t1)α1+α2+|(t2−a)α1+α2−(t1−a)α1+α2|}+N3Γ(α1+α2+ζ1+1){2(t2−t1)α1+α2+ζ1+|(t2−a)α1+α2+ζ1−(t1−a)α1+α2+ζ1|}+|λ1|ρΓ(α2+1){2(t2−t1)α2+|(t2−a)α2−(t1−a)α2|}+|(t2−a)α2+ϵ1−1−(t1−a)α2+ϵ1−1||Δ|Γ(α2+ϵ1){Ω2(|λ1|ρ(b−a)α2Γ(α2+1)+m∑i=1|λ2μi|ρ(ηi−a)α4Γ(α4+1)+n∑k=1|λ2νk|ρ(ξk−a)qk+α4Γ(qk+α4+1)+N1(b−a)α1+α2Γ(α1+α2+1)+N3(b−a)α1+α2+ζ1Γ(α1+α2+ζ1+1)+m∑i=1|μi|N2(ηi−a)α3+α4Γ(α3+α4+1)+m∑i=1|μi|N4(b−a)ζ2(ηi−a)α3+α4Γ(α3+α4+1)Γ(ζ2+1)+n∑k=1|νk|N2(ξk−a)qk+α3+α4Γ(qk+α3+α4+1)+n∑k=1|νk|N4(b−a)ζ2(ξk−a)qk+α3+α4Γ(ζ2+1)Γ(qk+α3+α4+1))+|Ω4|(N2(b−a)α3+α4Γ(α3+α4+1)+m∑i=1|δi|N1(ηi−a)α1+α2Γ(α1+α2+1)+m∑i=1|δi|N3(ηi−a)α1+α2+ζ1Γ(α1+α2+ζ1+1)+n∑k=1|θk|N1(ξk−a)pk+α1+α2Γ(pk+α1+α2+1)+n∑k=1|θk|N3(ξk−a)pk+α1+α2+ζ1Γ(pk+α1+α2+ζ1+1)+|λ2|ρ(b−a)α4Γ(α4+1)+m∑i=1|λ1δi|ρ(ηi−a)α2Γ(α2+1)+n∑k=1|λ1θk|ρ(ξk−a)pk+α2Γ(pk+α2+1))}, $ |
which tends to zero as $ t_{1}\rightarrow t_{2} $ independent of $ (x, y) \in B_{\rho}. $ Similarly, it can be established that $ | \mathcal{T}_{2}(x, y)(t_{2})-\mathcal{T}_{2}(x, y)(t_{1})|\rightarrow 0 $ as $ t_{2}\rightarrow t_{1} $ independently of $ (x, y) \in B_{\rho}. $ Thus the operator $ \mathcal{T} $ is equicontinuous. Hence, by the Arzelá-Ascoli theorem, the operator $ \mathcal{T} $ is completely continuous.
Next we consider the set
$ \Omega = \{(x, y) \in \mathcal{X}\times \mathcal{X}: (x, y) = r \mathcal{T}(x, y), \; \; 0\leq r \leq 1\}, $ |
and show that it is bounded. Let $ (x, y) \in \Omega $, then $ (x, y) = r \mathcal{T}(x, y) $ implies that $ x(t) = r \mathcal{T}_{1}(x, y)(t), $ and $ y(t) = r \mathcal{T}_{2}(x, y)(t), \forall t \in [a, b]. $ By the condition ($ H_{3} $), we obtain
$ ‖x‖≤σ1(u0+u1|x|+u2|y|)+σ2(v0+v1|x|+v2|y|)+σ7(ω0+ω1|x|+ω2|y|)+σ8(τ0+τ1|x|+τ2|y|)+σ5(‖x‖+‖y‖),‖y‖≤σ3(u0+u1|x|+u2|y|)+σ4(v0+v1|x|+v2|y|)+σ9(ω0+ω1|x|+ω2|y|)+σ10(τ0+τ1|x|+τ2|y|)+σ6(‖x‖+‖y‖). $ |
Adding the above inequalities, we get
$ ‖x‖+‖y‖≤(σ1+σ3)(u0+u1|x|+u2|y|)+(σ2+σ4)(v0+v1|x|+v2|y|)+(σ7+σ9)(ω0+ω1|x|+ω2|y|)+(σ8+σ10)(τ0+τ1|x|+τ2|y|)+(σ5+σ6)(‖x‖+‖y‖), $ |
which leads to
$ \|(x, y)\| \leq \dfrac{(\sigma_{1}+\sigma_{3})u_{0}+(\sigma_{2}+\sigma_{4})v_{0}+(\sigma_{7}+\sigma_{9})\omega_{0}+(\sigma_{8}+\sigma_{10}) \tau_{0}}{\sigma^{*}}, $ |
where
$ σ∗=min{1−(σ1+σ3)u1−(σ2+σ4)v1−(σ7+σ9)ω1−(σ8+σ10)τ1−(σ5+σ7),1−(σ1+σ3)u2−(σ2+σ4)v2−(σ7+σ9)ω2−(σ8+σ10)τ2−(σ5+σ7)}>0 $ |
by condition (3.16). Thus the set $ \Omega $ is bounded. Hence it follows by the Leray-Schauder alternative for single-valued maps [29] that the problem (1.3) has at least one solution on $ [a, b], $ which completes the proof.
Consider a coupled system of Hilfer fractional differential equations with boundary conditions:
$ {HD1/2,3/4(HD1/6,4/5+1/90)x(t)=I1/20+g1(x,y)+f1(t,x,y),t∈[0,1],HD1/2,3/4(HD1/2,1/7+1/100)y(t)=I1/31−g2(x,y)+f2(t,x,y),t∈[0,1],x(0)=y(0)=0,x(1)=1100y(1/10)+1200y(1/5)+1300y(3/10)+1400y(2/5)+1500y(1/2)+190I1/20+y(3/5)+170I1/20+y(7/10)+120I1/20+y(4/5),y(1)=135x(1/10)+1100x(1/5)+121x(3/10)+170x(2/5)+1500x(1/2)+1100I1/30+x(3/5)+1200I1/30+x(7/10)+1300I1/30+x(4/5). $ | (4.1) |
Here $ \alpha_{1} = 1/2, \alpha_{2} = 1/6, \alpha_{3} = 1/2, \alpha_{4} = 1/2,$ $ \beta_{1} = 3/4, \beta_{2} = 4/5, \beta_{3} = 3/4, \beta_{4} = 1/7, \lambda_{1} = 1/90, \lambda_{2} = 1/100,$ $\epsilon_{1} = 7/8 = \epsilon_{3}, q_{k} = 1/2, p_{k} = 1/3,$ $k = 1, 2, 3, m = 5, n = 3, \eta_{1} = 1/10, \eta_{2} = 1/5, \eta_{3} = 3/10,$ $\eta_{4} = 2/5, \eta_{5} = 1/2, \xi_{1} = 3/5, \xi_{2} = 7/10, \xi_{3} = 4/5,$ $\mu_{1} = 1/100, \mu_{2} = 1/200,$ $\mu_{3} = 1/300, \mu_{4} = 1/400, \mu_{5} = 1/500,$ $ \nu_{1} = 1/90, \nu_{2} = 1/70, \nu_{3} = 1/20,$ $\delta_{1} = 1/35, \delta_{2} = 1/100, \delta_{3} = 1/21, \delta_{4} = 1/70, \delta_{5} = 1/500, \theta_{1} = 1/100,$ $ \theta_{2} = 1/200, \theta_{3} = 1/300, \zeta_{1} = 1/2, \zeta_{2} = 1/3 $.
With the given data, it is found that $ |\Delta| = 1.1419, \sigma_{1} = 2.2278, \sigma_{3} = 0.1836, \sigma_{2} = 0.1096,$ $\sigma_{4} = 2.0132, \sigma_{5} = 0.0009, \sigma_{6} = 0.0025, \sigma_{7} = 1.8560, \sigma_{8} = 0.0475, \sigma_{9} = 0.1328,$ $\sigma_{10} = 1.1252. $
$ ({\bf a)} $ For illustrating Theorem 3.1, we take
$ {f1(t,x,y)=2arctanx+π14π(1+t)+17(t+π)sin|y|,f2(t,x,y)=17arctanx+3(21+t)|y|(1+|y|)+t3(1+t2),g1(x,y)=112(|x|(1+|x|)+|y|),g2(x,y)=117(sin|x|+arctan|y|). $ | (4.2) |
It can easily be verified that $ f_{1}, f_{2} $ satisfy the condition $ (H_1) $ with $ K_{1} = 1/7\pi, K_{2} = 1/7, $ respectively and $ g_{1}, g_{2} $ satisfy the condition $ (H_2) $ with $ L_{1} = 1/12, L_{2} = 1/17, $ respectively. Furthermore
$ (\sigma_{1}+\sigma_{3})K_{1}+(\sigma_{4}+\sigma_{2})K_{2}+(\sigma_{9}+\sigma_{7})L_{1}+(\sigma_{10}+\sigma_{8})L_{2} +\sigma_{5}+\sigma_{6}\approx 0.65102 < 1. $ |
Clearly the hypotheses of Theorem 3.1 are satisfied and hence it follows by its conclusion that the system (4.1) with $ f_{1}(t, x, y), f_{2}(t, x, y), g_{1}(x, y) $ and $ g_{2}(x, y) $ given by (4.2) has a unique solution on $ [0, 1]. $ On the other hand, one can deduce that the system (4.1) with (4.2) has at least one solution on $ [0, 1] $ by the application of Remark 3.1 with $ (\sigma_{1}+\sigma_{3})K_{1}+(\sigma_{4}+\sigma_{2})K_{2}+(\sigma_{9}+\sigma_{7})L_{1}+(\sigma_{10}+\sigma_{8})L_{2}\approx 0.6476 < 1. $
$ ({\bf b)} $ As an application of Theorem 3.2, consider
$ {f1(t,x,y)=arctanx10(t2+1)+sin|y|17(1+t),f2(t,x,y)=2√t2+2+2|x|5π(8+t)(1+|x|),g1(x,y)=|x|2(1+|x|)+16arctany,g2(x,y)=13e−|x|+17cos|x|. $ | (4.3) |
Using the given values, we find that the assumption ($ H_{4} $) is satisfied since $ |f_{1}(t, x, y)| $ $ \leqslant \dfrac{\pi}{20(1+t^2)}+\dfrac{1}{17(t+1)} = \phi_{1}(t) $ and $ |f_{2}(t, x, y)|\leqslant \dfrac{2}{\sqrt{t^{2}+2}} + \dfrac{2}{5\pi(8+t)} = \phi_{2}(t), $ $ |g_{1}(x, y)|\leqslant (6+\pi)/12 = \Lambda_{1}, $ $ |g_{2}(x, y)|\leqslant 10/21 = \Lambda_{2}. $ Also $ (\sigma_{5}+\sigma_{6})\approx 0.0034 < 1 $ holds true. As all the assumptions of Theorem 3.2 are satisfied, so its conclusion implies that the system (4.1) with the nonlinearities (4.3) has at least one solution on $ [0, 1]. $
$ ({\bf c)} $ In order to demonstrate the application of Theorem 3.3, let us choose
$ {f1(t,x,y)=arctanx+e−t|x|217(1+|x|)+126ycosx,f2(t,x,y)=2√t2+2+2π(8+t)xarctany+|x||y|5(1+|x|),g1(x,y)=ln7+121xsin|y|+113y,g2(x,y)=3e−|x|+17xcosy+111yarctanx. $ | (4.4) |
Obviously ($ H_{3} $) holds true with positive values of $ u_{0}, v_{0}, \omega_{0}, \tau_{0} $ and $ u_{1} = 1/17, u_{2} = 1/26, v_1 = 1/8, v_2 = 1/5, \omega_{1} = 1/21, \omega_{2} = 1/13, \tau_{1} = 1/7, \tau_{2} = \pi/22. $ Also, $ (\sigma_{1}+\sigma_{3})u_{1}+(\sigma_{2}+\sigma_{4})v_{1}+(\sigma_{7}+\sigma_{9})\omega_{1}+(\sigma_{8}+\sigma_{10})\tau_{1}+(\sigma_{5}+\sigma_{6})\approx 0.6728 < 1, $ and $ (\sigma_{1}+\sigma_{3})u_{2}+(\sigma_{2}+\sigma_{4})v_{2}+(\sigma_{7}+\sigma_{9})\omega_{2}+(\sigma_{8}+\sigma_{10})\tau_{2}+(\sigma_{5}+\sigma_{6})\approx 0.8412 < 1. $ As the hypothesis of Theorem 3.3 is verified, therefore we deduce by its conclusion that there exists at least one solution of the system (4.1) with $ f_{1}, f_{2}, g_{1} $ and $ g_2 $ given by (4.4).
In the present research work, we investigated the existence and uniqueness of solutions for a new coupled system of multi-term Hilfer fractional differential equations of different orders involving non-integral and autonomous type Riemann-Liouville mixed integral nonlinearities equipped with nonlocal coupled multi-point and Riemann-Liouville integral boundary conditions. Firstly, we proved an auxiliary result concerning the linear variant of the given problem, helping us to transform the problem at hand into a fixed point problem. Then we proved the existence of a unique solution for the given problem by applying Banach's contraction mapping principle and derived the existence results by means of Krasnosel'ski${\rm{\mathord{\buildrel{\lower3pt\hbox{$\scriptscriptstyle\smile$}} \over i} }}$'s fixed point theorem and Leray-Schauder nonlinear alternative. All the obtained results are well illustrated by numerical examples. Our results are new and enrich the literature on nonloacl nonlinear integral boundary value problems for Hilfer fractional differential equations.
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 declare no conflict of interest.
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