# Results of Positive Solutions for the Fractional Differential System on an Infinite Interval

Authors: Ying Wang; Jun Yang; Yumei Zi
Journal: Journal of Function Spaces
 (2020)

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## Abstract

The chief topic of this paper is to investigate the fractional differential system on an infinite interval. By introducing an appropriate compactness criterion in a special function space and applying the Schauder fixed-point theorem and the Banach contraction mapping principle, we established the results for the existence and uniqueness of positive solutions. An example is then given to show the utilization of the main results.


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## Body

## 1. Introduction

In this paper, we investigate the following fractional differential system on an infinite interval:(1)D0+α1ut+f1t,ut,vt=0,D0+α2vt+f2t,ut,vt=0,0<t<+∞,u0=u′0=u″0=⋯=un−20=0,D0+α1−1u+∞=μ1∫0∞a1tutdA1t,v0=v′0=v″0=⋯=vm−20=0,D0+α2−1v+∞=μ2∫0∞a2tvtdA2t,where 1 ≤ n − 1 < α1 ≤ n, 1 ≤ m − 1 < α2 ≤ m, and n, m ≥ 2, D0+αi is the Riemann–Liouville derivative operator. μi > 0 is a constant, ∫0∞a1tutdA1t and ∫0∞a2tvtdA2t denote the Riemann–Stieltjes integral, and Ai is a function of bounded positive variation. ai ∈ L[0, +∞), μi∫0∞aittαi−1dAit<Γαi, ∫0∞aitdAit<+∞, and fi: [0, +∞) × [0, +∞) × [0, +∞) ⟶ [0, +∞) is continuous, i = 1, 2.Numerous models in physics, chemistry, biology, medicine, and other fields have promoted the research of differential equations, for instance, evaluation of water quality on receiving water [1], and the advection-dispersion equation can be formulated as shown for the case of one-dimensional flow:(2)∂C∂t+u∂C∂x=DL∂2C∂t2−fC,where C is the concentration of a generic pollutant, t is the time, x is the longitudinal displacement, u is the velocity, DL is the diffusion coefficient, and f(C) is a generic term for reactions involving the pollutant C. Westerlund [2] established a one-dimensional model to describe the transmission of the electromagnetic wave:(3)με∂2Ex,t∂x2+μεζDtνEx,t+∂2Ex,t∂t2=0,where μ, ɛ, and ζ are constants and DtνEx,t=∂νEx,t/∂tν is a fractional derivative. In the process of establishing the model, k-Hessian equations [3], Sobolev equations [4], and Schrödinger elliptic equations [5, 6], there are also huge applications.Under the proper initial or boundary conditions, to study the positive solution of the above models is very necessary; especially, for the boundary value problems on the infinite interval, many authors put their interest in it [7–16]. Liang and Zhang [17] applied the fixed-point theorem to obtain the existence of positive solutions for the following fractional differential equation:(4)D0+αut+atfut=0,0<t<+∞,u0=u′0=0,limt⟶+∞D0+α−1ut=∑i=1m−2βiuξi,where 2 < α ≤ 3, D0+α is the Riemann–Liouville fractional derivative, 0 < ξ1 < ξ2 < ⋯ < ξm − 2 < +∞, βi > 0, 0<∑i=1m−2βiuξi<Γα, and f : [0, +∞) ⟶ [0, +∞)is continuous.For all we know, there are few studies on fractional differential systems of infinite intervals, although it is necessary to do so. In this paper, we aimed at getting the existence and uniqueness of positive solutions for system (1) on infinite interval. Compared with the existing literature, the innovations of this paper are as follows. Firstly, the paper which we discuss is the system rather than a single equation. Secondly, we study the system with integral boundary value conditions on infinite intervals, which are more general than those of two-point, three-point, and multipoint boundary value condition. At last, we use two different techniques: the Schauder fixed-point theorem and the Banach contraction mapping principle, for system (1), not only to obtain the existence of positive solutions but also the uniqueness of positive solutions.


## 2. Preliminaries and Lemmas

Definition 1.
(see [18, 19]). Let α > 0 and u be piecewise continuous on (0, +∞) and integrable on any finite subinterval of [0, +∞). Then, for t > 0, we call(5)I0+αut=1Γα∫0tt−sα−1usdsthe Riemann–Liouville fractional integral of u of order α.Definition 2.
(see [18, 19]). The Riemann–Liouville fractional derivative of order α > 0, n − 1 ≤ α < n, n∈ℕ, is defined as(6)D0+αut=1Γn−αddtn∫0tt−sn−α−1usds,where ℕ denotes the natural number set and the function u(t) is n times continuously differentiable on [0, +∞).Lemma 1 (see [18, 19]).
Letα > 0, and if the fractional derivative D0+α−1ut and D0+αut are continuous on [0, +∞), then,(7)I0+αD0+αut=ut+c1tα−1+c2tα−2+⋯+cntα−n,where c1, c2, …, cn ∈ (−∞, +∞), n is the smallest integer greater than or equal to α.Lemma 2.
Letyi ∈ C(0, +∞) ∩ L[0, +∞); then, the fractional system(8)D0+α1ut+y1t=0,D0+α1vt+y2t=0,0<t<+∞,u0=u′0=u″0=···=un−20=0,D0+α1−1u+∞=μ1∫0∞a1tutdA1t,v0=v′0=v″0=···=vm−20=0,D0+α2−1v+∞=μ2∫0∞a2tvtdA2t,has a unique integral representation(9)ut=∫0∞G1t,sy1sds,vt=∫0∞G2t,sy2sds,where(10)Git,s=Gi1=t,s+Gi2t,s,i=1,2,Gi1t,s=1Γαitαi−1−t−sαi−1,0≤s≤t≤+∞,tαi−1,0≤t≤s≤+∞,Gi2t,s=μitαi−1Γαi−μi∫0∞aittαi−1dAit∫0∞aitGi1t,sdAit.Proof.
By Lemma1, the equations in system (8) can be transformed into the equivalent integral equations(11)ut=−I0+α1y1t+c1tα1−1+c2tα1−2+⋯+cntα1−n,cj∈−∞,+∞,j=1,2,…,n,vt=−I0+α2y2t+c¯1tα2−1+c¯2tα2−2+⋯+cmtα2−m,c¯j∈−∞,+∞,j¯=1,2,…,m,that is,(12)ut=−1Γα1∫0tt−sα1−1y1sds+c1tα1−1+c2tα1−2+⋯+cntα1−n,cj∈−∞,+∞,j=1,2,…,n,vt=−1Γα2∫0tt−sα2−1y2sds+c¯1tα2−1+c¯2tα2−2+⋯+c¯mtα2−m,c¯j∈−∞,+∞,j=1,2,…,m.
Since(13)u0=u′0=⋯=un−2=0,v0=v′0=⋯=vm−20=0,we have(14)c2=c3=⋯=cn=0,c¯2=c¯3=⋯=c¯m=0.
So,(15)ut=−1Γα1∫0tt−sα1−1y1sds+c1tα1−1,vt=−1Γα2∫0tt−sα2−1y2sds+c¯1tα2−1.
We also have(16)D0+α1−1ut=c1Γα1−∫0ty1sds,D0+α2−1vt=c¯1Γα2−∫0ty2sds.
For(17)D0+α1−1u+∞=μ1∫0∞a1susdA1s,D0+α2−1v+∞=μ2∫0∞a2svsdA2s,we obtain(18)c1=1Γα1μ1∫0∞a1susdA1s+∫0+∞y1sds,c¯1=1Γα2μ2∫0∞a2svsdA2s+∫0+∞y2sds.
Combining (15) and (18), we have(19)ut=−1Γα1∫0tt−sα1−1y1sds+c1tα1−1=∫0∞G11t,sy1sds+μ1tα1−1Γα1∫0∞a1tutdA1t,(20)vt=−1Γα2∫0tt−sα2−1y2sds+c¯1tα2−1=∫0∞G21t,sy2sds+μ2tα2−1Γα2∫0∞a2tvtdA2t.
(19) and (20) are multiplied by a1(t) and a2(t), respectively, and then solved the integral from 0 to +∞ with respect to A1(t) and A2(t); then, we have(21)∫0∞a1tutdA1t=Γα1Γα1−μ1∫0∞a1tdA1t∫0∞a1t∫0∞G11t,sy1sdsdA1t,∫0∞a2tvtdA2t=Γα2Γα2−μ2∫0∞a2tdA2t∫0∞a2t∫0∞G21t,sy2sdsdA2t.
Therefore,(22)ut=∫0∞G11t,sy1sds+μ1tα1−1Γα1∫0∞a1tutdA1t=∫0∞G11t,sy1sds+μ1tα1−1Γα1−μ1∫0∞a1tdA1t∫0∞a1t∫0∞G11t,sy1sdsdA1t=∫0∞G11t,sy1sds+∫0∞G12t,sy1sds,vt=∫0∞G21t,sy2sds+μ2tα2−1Γα2∫0∞a2tvtdA2t=∫0∞G21t,sy2sds+μ2tα2−1Γα2−μ2∫0∞a2tdA2t∫0∞a2t∫0∞G21t,sy2sdsdA2t=∫0∞G21t,sy2sds+∫0∞G22t,sy2sds.
So, (9) holds. The proof is completed.Lemma 3.
The Green function in Lemma2 has the following properties:(1)
Gi1(t, s) is continuous and Gi1(t, s) ≥ 0, (t, s) ∈ [0, +∞) × [0, +∞).(2)
Gi1t,s/1+tαi−1≤1/Γαi,Git,s/1+tαi−1≤ϖ,t,s∈0,+∞×0,+∞,where(23)ϖ=max1Γα1+Γα1+μ1∫0∞a1tdA1tΓα1Γα1−μ1∫0∞a1ttα1−1dA1t,1Γα2+Γα2+μ2∫0∞a2tdA2tΓα2Γα2−μ2∫0∞a2ttα2−1dA2t.
The spaceX = E1 × E2 will be used in the study of system (1), where(24)E1=u∈C0,+∞:supt∈0,+∞ut1+tα1−1<+∞,E2=v∈C0,+∞:supt∈0,+∞vt1+tα2−1<+∞.
Then, (E1, ‖·‖) and (E2, ‖·‖) are the Banach space with the norm(25)u=supt∈0,+∞ut1+tα1−1,v=supt∈0,+∞vt1+tα2−1.
Clearly, (X, ‖·‖) is a Banach space with the norm u,v = ‖u‖ + v. Define nonlinear integral operators Ti : X ⟶ Ei and T : X ⟶ X by(26)Tiu,vt=∫0∞Git,sfis,us,vsds,i=1,2,(27)Tu,v=T1u,v,T2u,v.
Thus, the existence of solutions to system (1) is equivalent to the existence of solutions in X for operator equation Tu,v=u,v defined by (27).Lemma 4 (see [20, 21]).
LetE be defined as (24) and M be any bounded subset of E. Then, M is relatively compact in E if xt/1+tα−1:x∈M is equicontinuous on any finite subinterval of J, and for any given ɛ > 0, there exists N > 0 such that xt1/1+t1α−1−xt2/1+t2α−1<ε uniformly with respect to all x ∈ M, and t1, t2 > N.


## 3. Main Results

We list the conditions to be used later:H1 there exist nonnegative functions pit,git,hit∈L10,+∞ and tα1−1git, tα2−1hit∈L10,+∞ such that(28)fit,u,v≤pit+gitu+hitv,t,u,v∈0,+∞×0,+∞×0,+∞.(H2) |fi(t, 0, 0)| ∈ L1[0, +∞), there exist nonnegative functions git, hi(t) ∈ L1[0, +∞) and tα1−1git,tα2−1hit∈L10,+∞ such that(29)fit,u1,v1−fit,u2,v2≤gitu1−u2+hitv1−v2,t,u1,v1,t,u1,v1∈0,+∞×0,+∞×0,+∞.Remark 1.
If (H1) holds, then(30)∫0∞fit,ut,vtdt≤pi∗+gi∗+hi∗u,v,u,v∈X,where(31)pi∗=∫0∞pitdt,gi∗=∫0∞1+tα1−1gitdt,hi∗=∫0∞1+tα2−1hitdt.
In fact, by (H1), for any u,v ∈ X, we have(32)∫0∞fit,ut,vtdt≤∫0∞pit+gitut+hitvtdt=∫0∞pit+1+tα1−1gitut1+tα1−1+1+tα2−1hitvt1+tα2−1dt≤∫0∞pit+1+tα1−1gitu+1+tα2−1hitvdt=∫0∞pitdt+∫0∞1+tα1−1gitdtu+∫0∞1+tα2−1hitdtv=pi∗+gi∗u+hi∗v≤pi∗+gi∗+hi∗u,v.Theorem 1.
Assume that (H1) holds; then, T : X ⟶ X is a completely continuous operator.Proof.
First, we show thatT : X ⟶ X is continuous. Suppose {(un,vn)} ⊂ X, (u,v) ∈ X with ‖(un,vn) − (u,v)‖ ⟶ 0 (n ⟶ +∞), and there exists a constant r > 0 such that ‖(un,vn)‖ ≤ r and ‖(u,v)‖ ≤ r. By (H1) and (30), we have(33)∫0∞G1t,sf1s,uns,vnsds−∫0∞G1t,sf1s,us,vsds≤∫0∞G1t,sf1s,uns,vnsds+∫0∞G1t,sf1s,us,vsds≤ϖ∫0∞f1s,uns,vnsds+ϖ∫0∞f1s,us,vsds≤2ϖp1∗+g1∗u+h1∗v≤2ϖp1∗+g1∗+h1∗u,v<+∞.
From (H1) and (33), for any ɛ > 0, there exists sufficiently large M0 such that(34)∫M0∞G1t,sf1s,uns,vnsds+∫M0∞G1t,sf1s,us,vsds<ε2.
On the contrary, by the continuity off1(t, u,v) on 0,M0×0,1+M0α1−1r×0,1+M0α1−1r, there exists N > 0 such that when n > N and t ∈ [0, M0], we have(35)f1s,uns,vns−f1s,us,vs<ε2ϖM0.
Hence, for anyt ∈ [0, +∞) and n > N, we obtain(36)T1un,vnt1+tα1−1−T1u,vt1+tα1−1=∫0∞G1t,s1+tα1−1f1s,uns,vnsds−∫0∞G1t,s1+tα1−1f1s,us,vsds≤∫0M0G1t,s1+tα1−1f1s,uns,vns−f1s,us,vsds+∫M0∞G1t,s1+tα1−1f1s,uns,vnsds+∫M0∞G1t,s1+tα1−1f1s,us,vsds≤∫0M0G1t,s1+tα1−1f1s,uns,vns−f1s,us,vsds+∫M0∞G1t,sf1s,uns,vnsds+∫M0∞G1t,sf1s,us,vsds<ϖ∫0M0f1s,uns,vnsds−f1s,us,vsds+ε2<ε2+ε2=ε.
Thus, we know that ‖T1(un,vn) − T1(u,v)‖ ⟶ 0 (n ⟶ +∞). By the similar proof as (33)–(36), we know ‖T2(un,vn) − T2(u,v)‖ ⟶ 0 (n ⟶ +∞). So, T : X ⟶ X is continuous.
Next, we show thatT : X ⟶ X is relatively compact. Let Ω be a bounded subset of X; then, there exists constant M > 0 such that ‖(u,v)‖ ≤ M, (u,v) ∈ Ω. For any (u,v) ∈ Ω, t ∈ [0, +∞), and by (32), we obtain(37)T1un,vnt1+tα1−1=∫0∞G1t,s1+tα1−1f1s,uns,vnsds≤ϖ∫0∞f1s,uns,vnsds≤ϖp1∗+g1∗+h1∗u,v≤ϖp1∗+g1∗+h1∗M<+∞.
Similarly, we have(38)T2un,vnt1+tα2−1≤ϖp1∗+g1∗+h1∗u,v≤ϖp1∗+g1∗+h1∗M<+∞.
Consequently,T(Ω) is uniformly bounded.
GivenI ⊂ [0, +∞) be any compact interval. For any t1, t2 ∈ I, t1 < t2 and (u,v) ∈ Ω, we deduce(39)T1u,vt21+t2α1−1−T1u,vt11+t1α1−1=∫0∞G1t2,s1+t2α1−1f1s,us,vsds−∫0∞G1t1,s1+t1α1−1f1s,us,vsds≤∫0∞G11t2,s1+t2α1−1−G11t1,s1+t1α1−1f1s,us,vsds+∫0∞G12t2,s1+t2α1−1−G12t1,s1+t1α1−1f1s,us,vsds≤∫0∞G11t2,s1+t2α1−1−G11t1,s1+t1α1−1f1s,us,vsds+t2α1−11+t2α1−1−t1α1−11+t1α1−1∫0∞μ1∫0∞a1tG11t,sdA1tΓα1−μ1∫0∞a1ttα1−1dA1tf1s,us,vsds≤∫0∞G11t2,s1+t2α1−1−G11t1,s1+t2α1−1f1s,us,vsds+∫0∞G11t1,s1+t2α1−1−G11t1,s1+t1α1−1f1s,us,vsds+t2α1−11+t2α1−1−t1α1−11+t1α1−1∫0∞μ1∫0∞a1tG11t,sdA1tΓα1−μ1∫0∞a1ttα1−1dA1tf1s,us,vsds.
Since(40)∫0∞G11t2,s1+t2α1−1−G11t1,s1+t2α1−1f1s,us,vsds≤∫0t1G11t2,s1+t2α1−1−G11t1,s1+t2α1−1f1s,us,vsds+∫t1t2G11t2,s1+t2α1−1−G11t1,s1+t2α1−1f1s,us,vsds+∫t2+∞G11t2,s1+t2α1−1−G11t1,s1+t2α1−1f1s,us,vsds≤1Γα1∫0t1t2α1−1−t1α1−1+t2−sα1−1−t1−sα1−11+t2α1−1f1s,us,vsds+1Γα1∫t1t2t2α1−1−t1α1−1+t2−sα1−11+t2α1−1f1s,us,vsds+1Γα1∫t2+∞t2α1−1−t1α1−11+t2α1−1f1s,us,vsds⟶0,ast1⟶t2.
In the same way, we can know(41)∫0∞G11t1,s1+t2α1−1−G11t1,s1+t1α1−1f1s,us,vsds⟶0,ast1⟶t2.
So,(42)T1u,vt21+t2α1−1−T1u,vt11+t1α1−1⟶0,ast1⟶t2.
Similar to (39)–(41), we have(43)T2u,vt21+t2α2−1−T2u,vt11+t1α2−1⟶0,ast1⟶t2.
Therefore,T(Ω) is equicontinuous.
By (H1) and (30), for any ɛ > 0, there exists κ > 0 such that(44)∫κ∞fs,uns,vnsds1<ε.
Due tolimt⟶+∞tα1−1/1+tα1−1=1, there exists sufficiently large N1 > 0 such that, for any t1, t2 > N1, we have(45)t2α1−11+t2α1−1−t1α1−11+t1α1−1<ε.
Also because oflimt⟶+∞t−κα1−1/1+tα1−1=1, there exists sufficiently large N2 > κ such that, for any t1, t2 > N2 and 0 ≤ s ≤ κ, we have(46)t2−sα1−11+t2α1−1−t1−sα1−11+t1α1−1<ε.
ChooseN > max{N1, N2}; for any t1, t2 > N, we get(47)T1u,vt21+t2α1−1−T1u,vt11+t1α1−1≤∫0∞G11t2,s1+t2α1−1−G11t1,s1+t1α1−1f1s,us,vsds+t2α1−11+t2α1−1−t1α1−11+t1α1−1∫0∞μ1∫0∞a1tG11t,sdA1tΓα1−μ1∫0∞a1ttα1−1dA1tf1s,us,vsds.
In (47),(48)∫0∞G11t2,s1+t2α1−1−G11t1,s1+t1α1−1f1s,us,vsds≤∫0t1G11t2,s1+t2α1−1−G11t1,s1+t1α1−1f1s,us,vsds+∫t1t2G11t2,s1+t2α1−1−G11t1,s1+t1α1−1f1s,us,vsds+∫t2∞G11t2,s1+t2α1−1−G11t1,s1+t1α1−1f1s,us,vsds≤1Γα1∫0t1t2α1−11+t2α1−1−t1α1−11+t1α1−1+t2−sα1−11+t2α1−1−t1−sα1−11+t1α1−1f1s,us,vsds+1Γα1∫t1t2t2α1−11+t2α1−1−t1α1−11+t1α1−1+t2−sα1−11+t2α1−1f1s,us,vsds+1Γα1∫t2∞t2α1−11+t2α1−1−t1α1−11+t1α1−1f1s,us,vsds≤1Γα1∫0κt2α1−11+t2α1−1−t1α1−11+t1α1−1+t2−sα1−11+t2α1−1−t1−sα1−11+t1α1−1f1s,us,vsds+1Γα1∫κt1t2α1−11+t2α1−1−t1α1−11+t1α1−1+t2−sα1−11+t2α1−1−t1−sα1−11+t1α1−1f1s,us,vsds+1Γα1∫κ∞t2α1−11+t2α1−1−t1α1−11+t1α1−1+1f1s,us,vsds+1Γα1∫κ∞t2α1−11+t2α1−1−t1α1−11+t1α1−1f1s,us,vsds≤s∈0,κ,u,v∈Ωmaxf1s,us,vsΓα1∫0κt2α1−11+t2α1−1−t1α1−11+t1α1−1+t2−sα1−11+t2α1−1−t1−sα1−11+t1α1−1ds+1Γα1∫κ∞t2α1−11+t2α1−1−t1α1−11+t1α1−1+t2−sα1−11+t2α1−1−t1−sα1−11+t1α1−1f1s,us,vsds+1Γα1∫κ∞t2α1−11+t2α1−1−t1α1−11+t1α1−1+1f1s,us,vsds+1Γα1∫κ∞t2α1−11+t2α1−1−t1α1−11+t1α1−1f1s,us,vsds≤s∈0,κ,u,v∈Ωmaxf1s,us,vsΓα1∫0κt2α1−11+t2α1−1−t1α1−11+t1α1−1+t2−sα1−11+t2α1−1−t1−sα1−11+t1α1−1ds+9Γα1∫κ∞f1s,us,vsds≤s∈0,κ,u,v∈Ωmaxf1s,us,vsΓα12κε+9Γα1ε.
By (47) and (48), we have(49)T1u,vt21+t2α1−1−T1u,vt11+t1α1−1≤maxs∈0,κ,u,v∈Ωf1s,us,vsΓα12κε+9Γα1ε+ε∫0∞μ1∫0∞a1tG11t,sdA1tΓα1−μ1∫0∞a1ttα1−1dA1tf1s,us,vsds.
So,T1(u,v) is equicontinuous on +∞, proof similar to (49), and we know T2(u,v) is equicontinuous on +∞; thus, T(u,v) is equicontinuous on +∞. It follows from Lemma 4 that T : X ⟶ X is relatively compact. Therefore, T : X ⟶ X is completely continuous. The proof is completed.Theorem 2.
Assume that (H1) holds; then, system (1) has at least one positive solution if ϖg1∗+g2∗+h1∗+h2∗<1.Proof.
Let(50)r≥ϖp1∗+p2∗1−ϖg1∗+g2∗+h1∗+h2∗,K=u,v∈X,u,v≤r.
Now, we illustrateT(K) ⊂ K for any (u,v) ∈ K and t ∈ [0, +∞); by Lemma 3 and Remark 1, we have(51)T1u,vt1+tα1−1=∫0∞G1t,s1+tα1−1f1s,us,vsds≤ϖ∫0∞f1s,us,vsds≤ϖp1∗+g1∗+h1∗u,v≤ϖp1∗+g1∗+h1∗r≤r.
Similarly, we have(52)T2u,vt1+tα2−1≤ϖp2∗+g2∗+h2∗r≤r.
By (51) and (52),(53)Tu,v=T1u,v+T2u,v≤ϖp1∗+g1∗+h1∗r+p2∗+g2∗+h2∗r≤r.
Therefore,T(K) ⊂ K. By Theorem 1, we know that T : K ⟶ K is completely continuous. So, by the Schauder fixed-point theorem, system (1) has at least one positive solution. The proof is completed.Theorem 3.
Assume that (H2) holds; then, system (1) has a unique positive solution if ϖg1∗+g2∗+h1∗+h2∗<1.Proof.
From (H2), we know(54)fit,u,v≤gitu+hitv+fit,0,0.
So, for any (u,v) ∈ X, we have(55)∫0∞fit,ut,vtdt≤∫0∞gitut+hitvt+fit,0,0dt=∫0∞1+tα1−1gitut1+tα1−1+1+tα2−1hitvt1+tα2−1+fit,0,0dt≤∫0∞1+tα1−1gitut+1+tα2−1hitvt+fit,0,0dt=∫0∞1+tα1−1gitdtut+∫0∞1+tα2−1hitdtvt+∫0∞fit,0,0dt=gi∗u+hi∗v+∫0∞fit,0,0dt≤gi∗+hi∗u,v+∫0∞fit,0,0dt.
For any (u1,v1), (u2,v2) ∈ X and t ∈ [0, +∞), by Lemma 3, we have(56)T1u1,v1t−T1u2,v2t1+tα1−1=∫0∞G1t,s1+tα1−1f1s,u1s,v1s−f1s,u2s,v2sds≤ϖ∫0∞g1su1s−u2s+h1sv1s−v2sds≤ϖ∫0∞1+sα1−1g1su1s−u2s1+sα1−1+1+sα1−1h1sv1s−v2s1+sα1−1ds≤ϖg1∗u1−u2+h1∗v1−v2.
By the similar proof, we have(57)T2u1,v1t−T2u2,v2t1+tα2−1≤ϖg2∗u1−u2+h2∗v1−v2.
Now, inequalities (56) and (57) can show that(58)Tu,v≤ϖg1∗+g2∗+h1∗+h2∗u1−u2+v1−v2.
Thus, by the Banach contraction mapping theorem thatT has a unique fixed point in X, system (1) has a unique positive solution. The proof is completed.


## 4. An Example

Consider the following fractional differential system:(59)D0+5/2ut+f1t,ut,vt=0,D0+7/2vt+f2t,ut,vt=0,0<t<+∞,u0=u′0=0,D0+3/2u+∞=15∫0∞e−tutdt,v0=v′0=v″0=0,D0+5/2v+∞=14∫0∞e−tvtdt,where α1=5/2,α2=7/2,μ1=1/5,μ2=1/4,A1t=A2t=t, a1(t) = e−t = a2(t) = e−t. Then, we have(60)∫0∞a1tdA1t=∫0∞a2tdA2t=∫0∞e−tdt=1<+∞,∫0∞a1ttα1−1dA1t=∫0∞e−tt3/2dt=2.3562<+∞,∫0∞a2ttα2−1dA2t=∫0∞e−tt5/2dt=3.3233<+∞,ϖ=2.0929.Take(61)f1t,ut,vt=11+et+utcost1+t3/2e7t+vtsint71+t5/2e3t,f2t,ut,vt=121+et+ut1+t3/2e6t+vt31+t5/2e4t.Let(62)g1t=11+t3/2e7t,g2t=11+t3/2e6t,h1t=171+t5/2e3t,h2t=131+t5/2e4t.Through calculation, we get(63)g1∗=∫0∞1+tα1−1g1tdt=∫0∞e7tdt=17,g2∗=∫0∞1+tα1−1g2tdt=∫0∞e6tdt=16,h1∗=∫0∞1+tα2−1h1tdt=17∫0∞e3tdt=121,h2∗=∫0∞1+tα2−1h2tdt=13∫0∞e4tdt=112,ϖg1∗+g2∗+h1∗+h2∗=2.092917+16+121+112=0.9219<1.Then, by Theorem2, system (59) has at least one positive solution.



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*Source: 5174529-2020-05-05.xml*