Citation: Rajish Kumar P, Sunil Jacob John. On redundancy, separation and connectedness in multiset topological spaces[J]. AIMS Mathematics, 2020, 5(3): 2484-2499. doi: 10.3934/math.2020164
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Differential equations of arbitrary order have been shown to be useful in the study of models of many phenomena in various fields such as: Electrochemistry and material science, they are in fact described by differential equations of fractional order [9,10,15,16,25,26,27,28,29]. For more details, we refer the reader to the books of Hilfer [30], Podlubny [31], Kilbas et al. [34], Miller and Ross [2] and to the following research papers [1,2,3,4,5,6,7,8,11,12,14,16,17,19,20,24,31,35,36,37,38,39,40,41,42]. In this work, we discuss the existence and uniqueness of the solutions for multi-point boundary value problems of nonlinear fractional differential equations with two Riemann-Liouville fractionals:
{Dαx(t)=∑mi=1fi(t,x(t),y(t),φ1x(t),ϕ1y(t)),α∈]1,2],t∈[0,T]Dβy(t)=∑mi=1gi(t,x(t),y(t),φ2x(t),ϕ2y(t)),β∈]1,2],t∈[0,T]I2−αx(0)=0, Dα−2x(T)=θIα−1(x(η)), 0<η<T,I2−βy(0)=0, Dβ−2x(T)=ωIβ−1(x(γ)), 0<γ<T, | (1.1) |
where D(.), I(.) denote the Riemann-Liouville derivative and integral of fractional order (.), respectively, fi, gi:[0,T]×R4→R, i=1,⋯,m are continuous functions on [0,T] and
(φ1x)(t)=∫t0A′1(t,s)x(s)ds, (ϕ1y)(t)=∫t0B′1(t,s)y(s)ds, |
(φ2x)(t)=∫t0A′2(t,s)x(s)ds, (ϕ1y)(t)=∫t0B′2(t,s)y(s)ds, |
with Ai and Bi being continuous functions on [0,1]×[0,1]. However, it is rare to find a work in nonlinear term fi depends on fractional derivative of unknown functions x(t),y(t),φ1x(t),ϕ1y(t) and solutions for multi-order fractional differential equations on the infinite interval [0,T). Motivated by [8,11,12,13,14] and the references therein, we consider the existence and unicity of solution for multi-order fractional differential equations on infinite interval [0,T).
The rest of this paper is organized as follow. In section 2, we present some preliminaries and lemmas. Section 3 is dedicated to showing the existence of a solution for problem (1.1). Finally, section 4 illustrated the proposed results with two examples.
Remark 1.1. This work generalizes the work of Houas and Benbachir [14] on different boundary conditions and for another type of integral.
This section covers the basic concepts of Riemann-Liouville type fractional calculus that will be used throughout this paper.
Definition 2.1. [31,32] The Riemann-Liouville fractional integral operator of order α≥0, of a function f:(0,∞)→R is defined as
{Jαf(t)=1Γ(α)∫t0(t−τ)α−1f(τ)dτ,J0f(t)=f(t), |
where Γ(α):=∫∞0e−uuα−1du.
Definition 2.2. [31,32] The Riemann-Liouville fractional derivative of order α>0, of a continuous function h:(0,∞)→R is defined as
Dαh(t)=1Γ(n−α)(ddt)n∫t0(t−τ)n−α−1h(τ)dτ=(ddt)nIn−αh(τ), |
where n=[α]+1.
For α<0, we use the convention that Dαh=J−αh. Also for 0≤ρ<α, it is valid that DρJαh=hα−ρ. We note that for ε>−1 and ε≠α−1,α−2,...,α−n, we have
Dαtε=Γ(ε+1)Γ(ε−α+1)tε−α,Dαtα−i=0, i=1,2,...,n. |
In particular, for the constant function h(t)=1, we obtain
Dα1=1Γ(1−α)t−α,α∉N. |
For α∈N, we obtain, of course, Dα1=0 because of the poles of the gamma function at the points 0,−1,−2,... For α>0, the general solution of the homgeneous equation Dαh(t)=0 in C(0,T)∩L(0,T) is
h(t)=c0tα−n+c1tα−n−1+......+cn−2tα−2+cn−1tα−1, |
where ci,i=1,2,....,n−1, are arbitrary real constants. Further, we always have DαIαh=h, and
DαIαh(t)=h(t)+c0tα−n+c1tα−n−1+......+cn−2tα−2+cn−1tα−1. |
Lemma 2.1. [33] Let E be Banach space. Assume that T:E⟶E is a completely continuous operator. If the set V={x∈E:x=μTx, 0<μ<1} is bounded, then T has a fixed point in E.
To define the solution for problem (1.1). We consider the following lemma.
Lemma 2.2. Suppose that (Hi)i=1,…,m⊂C([0,1],R), and consider the problem
Dαh(t)−m∑i=1Hi(t)=0, t∈j, 1<α<2, m∈N∗, | (2.1) |
with the conditions
I2−αh(0)=0, Dα−2h(T)=θIα−1(h(η)), 0<η<T. | (2.2) |
Then we have
h(t)=1Γ(α)m∑i=1∫t0(t−τ)α−1Hi(τ)dτ+tα−1ψ(m∑i=1∫T0(T−τ)Hi(τ)dτ−θΓ(2α)m∑i=1∫η0(η−τ)2α−2Hi(τ)dτ) |
with ψ=θΓ(α)Γ(2α−1)η2α−2−Γ(α)T.
Proof. We have
h(t)=m∑i=1IαHi(t)+c0tα−2+c1tα−1, |
where ci∈R, i=0,1.
We obtain
I2−αh(τ)=m∑i=1I2Hi(τ)+c0I2−ατα−2+c1I2−ατα−1=m∑i=1I2Hi(τ)+c0+c1τ,Iα−1h(τ)=m∑i=1I2α−1Hi(τ)+c0Iα−1τα−2+c1Iα−1τα−1=m∑i=1I2α−1Hi(τ)+c0Γ(α−1)Γ(2α−2)τ2α−3+c1Γ(α)Γ(2α−1)τ2α−2,Dα−2h(τ)=m∑i=1I2Hi(τ)+c0Γ(α−1)+c1Γ(α)τ. |
Using the given conditions: I2−αh(0)=0, we find that c0=0, and since Dα−2h(T)−θIα−1(h(η))=0, we have
m∑i=1I2hi(T)+c1Γ(α)T−θ[m∑i=1I2α−1hi(η)+c1Γ(α)Γ(2α−1)η2α−2]=0, |
then
c1[Γ(α)Γ(2α−1)η2α−2−Γ(α)T]=m∑i=1I2hi(T)−θm∑i=1I2α−1hi(η) |
and
c1=1ψ(m∑i=1I2Hi(T)−θm∑i=1I2α−1Hi(η))=1ψ(m∑i=1∫T0(T−τ)Hi(τ)dτ−θΓ(2α)m∑i=1∫η0(η−τ)2α−2Hi(τ)dτ) |
with
ψ=θΓ(α)Γ(2α−1)η2α−2−Γ(α)T. |
Finally, the solution of (2.1) and (2.2) is
h(t)=1Γ(α)m∑i=1∫t0(t−τ)α−1Hi(τ)dτ+tα−1ψ(m∑i=1∫T0(T−τ)Hi(τ)dτ−θΓ(2α)m∑i=1∫η0(η−τ)2α−2Hi(τ)dτ). |
We denote by
E={x,y∈C([0,T],R);φix,ϕiy∈C([0,T],R) i=1,2}, |
and the Banach space of all continuous functions from [0,T] to R endowed with a topology of uniform convergence with the norm defined by
||(x,y)||E=max(||x||,||y||,||φ1x||,||ϕ1y||,||φ2x||,||ϕ2y||), |
where
||x||=supt∈j|φix(t)|,||y||=supt∈j|y(t)|,||ϕix||=supt∈j|φix(t)|,||ϕiy||=supt∈j|ϕiy(t)|. |
In this section, we prove some existence and uniqueness results to the nonlinear fractional coupled system (1.1).
For the sake of convenience, we impose the following hypotheses:
(H1) For each i=1,2,⋯,m, the functions fi and gi :[0,T]×R4⟶R are continuous.
(H2) There exist nonnegative real numbers ξik,φik,k=1,2,3,4,i=1,2,⋯,m, such that for all t∈[0,T] and all (x1,x2,x3,x4), (y1,y2,y3,y4)∈R4, we have
|fi(t,x1,x2,x3,x4)−fi(t,y1,y2,y3,y4)|≤4∑k=1 ξik|xk−yk|, |
and
|gi(t,x1,x2,x3,x4)−gi(t,y1,y2,y3,y4)|≤4∑k=1 χik|xk−yk|. |
(H3) There exist nonnegative constants (Li) and (Ki) i=1,...,m, such that: For each t∈[0,T] and all (x1,x2,x3,x4)∈R4,
|fi(t,x1,x2,x3,x4)|≤Li,|gi(t,x1,x2,x3,x4)|≤Ki,i=1,...,m. |
We also consider the following quantities:
A1=TαΓ(α+1)m∑i=1(ξi1+ξi2+ξi3+ξi4),A2=TβΓ(β+1)m∑i=1(χi1+χi2+χi3+χi4),A3=maxt,s∈[0,1]||A′1(t,s)||×A1,A4=maxt,s∈[0,1]||A′2(t,s)||×A1,A5=maxt,s∈[0,1]||B′1(t,s)||×A2,A6=maxt,s∈[0,1]||B′2(t,s)||×A2,ν1=[TαΓ(α+1)+1ψ(Tα+12+θT3α−2(2α−1)2Γ(2α−1))],ν2=[TβΓ(β+1)+1ψ′(Tβ+12+ωT3β−2(2β−1)2Γ(2β−1))],ν3=maxt,s∈[0,1]|A′1(t,s)|ν1,ν4=maxt,s∈[0,1]|A′2(t,s)|ν1,ν5=maxt,s∈[0,1]|B′1(t,s)|ν2,ν6=maxt,s∈[0,1]|B′2(t,s)|ν2. |
The first result is based on Banach contraction principle. We have
Theorem 3.1. Assume that (H2) holds. If the inequality
max(A1,A2,A3,A4,A5,A6)<1, | (3.1) |
is valid, then the system (1.1) has a unique solution on [0,T].
Proof. We define the operator T:E⟶E by
T(x,y)(t)=(T1(x,y)(t),T2(x,y)(t)),t∈[0,T], |
such that
T1(x,y)(t)=1Γ(α)m∑i=1∫t0(t−τ)α−1Hi(τ)dτ+tα−1ψ(m∑i=1∫T0(T−τ)Hi(τ)dτ−θΓ(2α)m∑i=1∫η0(η−τ)2α−2Hi(τ)dτ) | (3.2) |
and
T2(x,y)(t)=1Γ(β)m∑i=1∫t0(t−τ)β−1Gi(τ)dτ+tβ−1ψ′(m∑i=1∫T0(T−τ)Gi(τ)dτ−ωΓ(2β)m∑i=1∫γ0(γ−τ)2β−2Gi(τ)dτ) | (3.3) |
where
Hi(τ)=fi(τ,x(τ),y(τ),φ1x(τ),ϕ1y(τ)) |
and
Gi(τ)=gi(τ,x(τ),y(τ),φ2x(τ),ϕ2y(τ)). |
We obtain
φiT1(x,y)(t)=∫t0Ai(t,s)T1(x,y)(s)ds, ϕiT2(x,y)(t)=∫t0Bi(t,s)T2(x,y)(s)ds |
where i=1,2.
We shall now prove that T is contractive.
Let T1(x1,y1),T2(x2,y2)∈E. Then, for each t∈[0,T], we have
|T1(x1,y1)−T1(x2,y2)|≤[1Γ(α)m∑i=1∫t0(t−τ)α−1dτ+tα−1ψ(m∑i=1∫T0(T−τ)dτ−θΓ(2α)m∑i=1∫η0(η−τ)2α−2dτ)]×maxτ∈[0,T]m∑i=1|(fi(τ,x1(τ),y1(τ),φ1x1(τ),ϕ1y1(τ))−fi(τ,x2(τ),y2(τ),φ1x2(τ),ϕ1y2(τ)))|≤TαΓ(α+1)maxτ∈[0,T]m∑i=1|(fi(τ,x1(τ),y1(τ),φ1x1(τ),ϕ1y1(τ))−fi(τ,x2(τ),y2(τ),φ1x2(τ),ϕ1y2(τ)))|. |
By (H2), it follows that
||T1(x1,y1)−T1(x2,y2)||≤TαΓ(α+1)m∑i=1(ξi1+ξi2+ξi3+ξi4)×max(||x1−x2||,||y1−y2||,||φ1(x1−x2)||,||φ2(x1−x2)||,||ϕ1(y1−y2)||,||ϕ2(y1−y2)||). |
Hence,
||T1(x1,y1)−T1(x2,y2)||≤A1||x1−x2,y1−y2||E. | (3.4) |
With the same arguments as before, we can show that
||T2(x1,y1)−T2(x2,y2)||≤A2||x1−x2,y1−y2||E. | (3.5) |
On the other hand, we have
||φ1(T1(x1,y1)−T1(x2,y2))||≤∫t0||A′1(t,s)||||T1(x1,y1)−T1(x2,y2)||ds≤maxt,s∈[0,1]||A′1(t,s)||×A1||x1−x2,y1−y2||E. |
Hence,
||φ1(T1(x1,y1)−T1(x2,y2))||≤A3||x1−x2,y1−y2||E | (3.6) |
and
||φ2(T1(x1,y1)−T1(x2,y2))||≤A4||x1−x2,y1−y2||E. | (3.7) |
Also, we have
||ϕ1(T2(x1,y1)−T2(x2,y2))||≤A5||x1−x2,y1−y2||E | (3.8) |
and
||ϕ2(T2(x1,y1)−T2(x2,y2))||≤A6||x1−x2,y1−y2||E. | (3.9) |
Thanks to (3.4)–(3.9), we get
||T(x1,y1)−T(x2,y2)||≤max(A1,A2,A3,A4,A5,A6)×||(x1−x2,y1−y2)||E. | (3.10) |
Thanks to (3.10), we conclude that T is a contractive operator. Therefore, by Banach fixed point theorem, T has a unique fixed point which is the solution of the system (1.1).
Our second main result is based on Lemma 2.1. We have
Theorem 3.2. Assume that the hypotheses (H1) and (H3) are satisfied. Then, system (1.1) has at least a solution on [0,T].
Proof. The operator T is continuous on E in view of the continuity of fi and gi (hypothesis (H1)).
Now, we show that T is completely continuous:
(i) First, we prove that T maps bounded sets of E into bounded sets of E. Taking λ>0, and (x,y)∈Ωλ={(x,y)∈E;||(x,y)||≤λ}, then for each t∈[0,T], we have:
|T1(x,y)|≤[1Γ(α)∫t0(t−τ)α−1dτ+tα−1ψ(∫T0(T−τ)dτ−θΓ(2α)∫η0(η−τ)2α−2dτ)]×supt∈[0,T]m∑i=1|fi(t,x(t),y(t),φ1x(t),ϕ1y(t))|≤[TαΓ(α+1)+1ψ(Tα+12+θT3α−2(2α−1)2Γ(2α−1))]×supt∈[0,T]m∑i=1|fi(t,x(t),y(t),φ1x(t),ϕ1y(t))|, |
Thanks to (H3), we can write
||T1(x,y)||≤[TαΓ(α+1)+1ψ(Tα+12+θT3α−2(2α−1)2Γ(2α−1))]m∑i=1Li. |
Thus,
||T1(x,y)||≤ν1m∑i=1Li. | (3.11) |
As before, we have
||T2(x,y)||≤ν2m∑i=1Ki. | (3.12) |
On the other hand, for all j=1,2, we get
|ϕjT1(x,y)(t)|=|∫t0A′j(t,s)T1(x,y)(s)ds|≤maxt,s∈[0,1]|A′j(t,s)|ν1m∑i=1Li. |
This implies that
||ϕ1T1(x,y)(t)||≤ν3m∑i=1Li, | (3.13) |
||ϕ2T1(x,y)(t)||≤ν4m∑i=1Li. | (3.14) |
Similarly, we have
||φ1T2(x,y)(t)||≤ν5m∑i=1Ki, | (3.15) |
||φ2T2(x,y)(t)||≤ν6m∑i=1Ki. | (3.16) |
It follows from (3.11)–(3.16) that:
||T(x,y)||E≤max(ν1m∑i=1Li,ν2m∑i=1Ki,ν3m∑i=1Li,ν4m∑i=1Li,,ν5m∑i=1,ν6m∑i=1). |
Thus,
||T(x,y)||E<∞. |
(ii) Second, we prove that T is equi-continuous:
For any 0≤t1<t2≤T and (x,y)∈Ωλ, we have
|T1(x,y)(t2)−T1(x,y)(t1)|≤[1Γ(α)∫t10(t2−τ)α−1−(t1−τ)α−1dτ+1Γ(α)∫t2t1(t2−τ)α−1dτ+tα−12−tα−11ψ(T22−θη2α−1Γ(2α−1)2Γ(2α−1))]×supt∈[0,T]m∑i=1|fi(t,x(t),y(t),φ1x(t),ϕ1y(t))|≤[2Γ(α+1)(t2−t1)α−1+(tα−12−tα−11)[T22ψ−θη2α−1ψΓ(2α−1)2Γ(2α−1)+1Γ(α+1)]]×m∑i=1Li. |
Therefore,
||T1(x,y)(t2)−T1(x,y)(t1)||E[2Γ(α+1)(t2−t1)α−1+(tα−12−tα−11)[T22ψ+1Γ(α+1)]]×m∑i=1Li. | (3.17) |
We also have
||T2(x,y)(t2)−T2(x,y)(t1)||E[2Γ(β+1)(t2−t1)β−1+(tβ−12−tβ−11)[T22ψ′+1Γ(β+1)]]×m∑i=1Ki. | (3.18) |
On the other hand,
|ϕiT1(x,y)(t2)−ϕiT1(x,y)(t1)|≤[maxs∈[0,1]|A′i(t2,s)−A′i(t1,s)|+(t2−t1)maxs∈[0,1]|A′i(t1,s)|]×sups∈[0,1]|T1(x,y)(s)|. |
Consequently, for all i=1,2, we obtain
||ϕiT1(x,y)(t2)−ϕiT1(x,y)(t1)||≤[maxs∈[0,1]|A′i(t2,s)−A′i(t1,s)|+(t2−t1)maxs∈[0,1]|A′i(t1,s)|]ν1m∑i=1Li. | (3.19) |
Similarly,
||φiT1(x,y)(t2)−φiT1(x,y)(t1)||≤[maxs∈[0,1]|B′i(t2,s)−B′i(t1,s)|+(t2−t1)maxs∈[0,1]|B′i(t1,s)|]ν2m∑i=1Ki. | (3.20) |
where i=1,2. Using (3.17)–(3.20), we deduce that
||T(x,y)(t2)−T(x,y)(t1)||E⟶0 |
as t2→t1.
Combining (i) and (ii), we conclude that T is completely continuous.
(iii) Finally, we shall prove that the set F defined by
F={(x,y)∈E,(x,y)=ρT(x,y), 0<ρ<1} |
is bounded.
Let (x,y)∈F, then (x,y)=ρT(x,y), for some 0<ρ<1. Thus, for each t∈[0,T], we have:
x(t)=ρT1(x,y)(t), y(t)=ρT2(x,y)(t). | (3.21) |
Thanks to (H3) and using (3.11) and (3.12), we deduce that
||x||≤ρν1m∑i=1Li, ||y||≤ρν2m∑i=1Ki. | (3.22) |
Using (3.13)–(3.16), it yields that
{||ϕ1x||≤ρν3∑mi=1Li||ϕ2x||≤ρν4∑mi=1Li||φ1y||≤ρν5∑mi=1Ki||φ2y||≤ρν6∑mi=1Ki. | (3.23) |
It follows from (3.22) and (3.23) that
||T(x,y)||E≤ρmax(ν1∑mi=1Li,ν2∑mi=1Ki,ν3∑mi=1Li,ν4∑mi=1Li,,ν5∑mi=1,ν6∑mi=1). |
Consequently,
||(x,y)||E<∞. |
This shows that F is bounded. By Lemma (2.1), we deduce that T has a fixed point, which is a solution of (1.1).
To illustrate our main results, we treat the following examples.
Example 4.1. Consider the following system:
{D32x(t)=cos(πt)(x+y+φ1x(t)+ϕ1y(t))10π(x+y+φ1x(t)+ϕ1y(t))+132π2e(cosx(t)+cosy(t)+φ1x(t)+ϕ1y(t)4π),D32y(t)=18π3(t+1)(x+y+φ2x(t)+ϕ2y(t)3+x+y+φ2x(t)+ϕ2y(t))+1(10π+et)e(t+1)(sinx(t)+siny(t)+cosφ2x(t)+cosϕ2y(t)2+sinx(t)+siny(t)+cosφ2x(t)+cosϕ2y(t)),I12x(0)=0, D−12x(T)=I12(x(1)),I12y(0)=0, D−12y(T)=I12(y(1)). | (4.1) |
We have
α=32, β=32, T=1, θ=1, ω=1, γ=1, m=2, η=1. |
Also,
f1(t,x(t),y(t),φ1x(t),ϕ1y(t))=cos(πt)(x+y+φ1x(t)+ϕ1y(t))10π(1+x+y+φ1x(t)+ϕ1y(t)), | (4.2) |
f2(t,x(t),y(t),φ1x(t),ϕ1y(t))=132π2e(cosx(t)+cosy(t)+φ1x(t)+ϕ1y(t)4π). | (4.3) |
For t∈[0,1] and (x1,y1,φ1x1,ϕ1y1),(x2,y2,φ1x2,ϕ1y2)∈R4, we have
|f1(t,x1,y1,φ1x1,ϕ1y1)−f1(t,x2,y2,φ1x2,ϕ1y2)|≤|cos(πt)|10π|x1+y1+φ1x1+ϕ1y11+x1+y1+φ1x1+ϕ1y1−x2+y2+φ1x2+ϕ1y2)1+x2+y2+φ1x2+ϕ1y2)|≤110π(|x1−x2|+|y1−y2|+|φ1x1−φ1x2|+|ϕ1y1−ϕ1y2|) | (4.4) |
and
|f2(t,x1,y1,φ1x1,ϕ1y1)−f2(t,x2,y2,φ1x2,ϕ1y2)|≤132πe(|x1−x2|+|y1−y2|+|φ1x1−φ1x2|+|ϕ1y1−ϕ1y2|). | (4.5) |
So, we can take
ξ11=ξ12=ξ13=ξ14=110π, |
ξ21=ξ22=ξ23=ξ24=132πe. |
We also have
g1(t,x(t),y(t),φ2x(t),ϕ2y(t))=18π3(t+1)(x+y+φ2x(t)+ϕ2y(t)3+x+y+φ2x(t)+ϕ2y(t)) |
and
g2(t,x(t),y(t),φ2x(t),ϕ2y(t))=1(10π+et)et+1(sinx(t)+siny(t)+cosφ2x(t)+cosϕ2y(t)2+sinx(t)+siny(t)+cosφ2x(t)+cosϕ2y(t)) | (4.6) |
For t∈[0,1] and (x1,y1,φ2x1,ϕ2y1),(x2,y2,φ2x2,ϕ2y2)∈R4, we can write
|g1(t,x1,y1,φ2x1,ϕ2y1)−g1(t,x2,y2,φ2x2,ϕ2y2)|≤18π3(|x1−x2|+|y1−y2|+|φ2x1−φ2x2|+|ϕ2y1−ϕ2y2|), | (4.7) |
and
|g2(t,x1,y1,φ2x1,ϕ2y1)−g2(t,x2,y2,φ2x2,ϕ2y2)|≤110πe2(|x1−x2|+|y1−y2|+|φ2x1−φ2x2|+|ϕ2y1−ϕ2y2|). | (4.8) |
Hence,
χ11=χ12=χ13=χ14=18π3, |
χ21=χ22=χ23=χ24=110πe2. |
Therefore,
A1=0.0589009676,A2=0.0250930393. |
Suppose
A′i=B′i=1, i=1,2, |
so,
A1=A3=A4,A2=A5=A6. |
Thus,
max(A1,A2,A3,A4,A5,A6)<1, | (4.9) |
and by Theorem 3.1, we conclude that the system (4.1) has a unique solution on [0,1].
Example 4.2.
{D32x(t)=π(t+1)sin(φ1x(t)+ϕ1y(t))2−cos(x(t)+y(t))+et2π+cos(x(t)+φ1x(t))+sin(sin(y(t)+ϕ1y(t)), t∈[0,1],D43y(t)=e2sin(x(t)+y(t))2π+cos(φ2x(t)+ϕ2y(t))+3t2cosy(t)et3+1−cos(x(t)+y(t)−φ2x(t)−ϕ2y(t)), t∈[0,1],I12x(0)=0, D−12x(T)=I12(x(1)),I23y(0)=0, D−23y(T)=I13(y(1)). | (4.10) |
We have
α=32, β=43, T=1, θ=1, ω=1, γ=1, m=2, η=1. |
Since
|f1(t,x(t),y(t),φ1x(t),ϕ1y(t))|=|π(t+1)sin(φ1x(t)+ϕ1y(t))2−cos(x(t)+y(t))|≤2π,|f2(t,x(t),y(t),φ1x(t),ϕ1y(t))|=|et2π+cos(x(t)+φ1x(t))+sin(sin(y(t)+ϕ1y(t))|≤e2π+2,|g1(t,x(t),y(t),φ2x(t),ϕ2y(t))|=|e2sin(x(t)+y(t))2π+cos(φ2x(t)+ϕ2y(t))|≤e22π+1,|g2(t,x(t),y(t),φ2x(t),ϕ2y(t))|=|3t2cosy(t)et3+1−cos(x(t)+y(t)−φ2x(t)−ϕ2y(t))|≤3e−1. |
The functions f1, f2, g1 and g2 are continuous and bounded on [0,1]×R4. So, by Theorem 3.2, the system (4.10) has at least one solution on [0,1].
We have proved the existence of solutions for fractional differential equations with integral and multi-point boundary conditions. The problem is solved by applying some fixed point theorems. We also provide examples to make our results clear.
The authors declare that they have no conflicts of interest in this paper.
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