This article dealt with a class of coupled hybrid fractional differential system. It consisted of a mixed type of Caputo and Hilfer fractional derivatives with respect to two different kernel functions, ψ1 and ψ2, respectively, in addition to coupled boundary conditions. The existence of the solution of the system was investigated using the Dhage fixed point theorem. Finally, an illustration was presented to validate our findings.
Citation: M. Latha Maheswari, K. S. Keerthana Shri, Mohammad Sajid. Analysis on existence of system of coupled multifractional nonlinear hybrid differential equations with coupled boundary conditions[J]. AIMS Mathematics, 2024, 9(6): 13642-13658. doi: 10.3934/math.2024666
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This article dealt with a class of coupled hybrid fractional differential system. It consisted of a mixed type of Caputo and Hilfer fractional derivatives with respect to two different kernel functions, ψ1 and ψ2, respectively, in addition to coupled boundary conditions. The existence of the solution of the system was investigated using the Dhage fixed point theorem. Finally, an illustration was presented to validate our findings.
This paper considers a fractional coupled system on an infinite interval involving the Erdélyi-Kober derivative:
{Dγ,δ1βu(x)+F(x,u(x),v(x))=0,x∈(0,+∞),Dγ,δ2βv(x)+G(x,u(x),v(x))=0,x∈(0,+∞),limx→0xβ(2+γ)Iδ1+γ,2−δ1u(x)=0,limx→∞xβ(1+γ)Iδ1+γ,2−δ1u(x)=0,limx→0xβ(2+γ)Iδ2+γ,2−δ2v(x)=0,limx→∞xβ(1+γ)Iδ2+γ,2−δ2v(x)=0, | (1.1) |
where δ1,δ2∈(1,2], γ∈(−2,−1), and β>0. Dγ,δ1β, Dγ,δ2β are Erdélyi-Kober fractional derivatives (EKFDs for short), and Iδ1+γ,2−δ1,Iδ2+γ,2−δ2 are the Erdélyi-Kober fractional integrals. F,G are continuous functions. We discuss the existence of positive solutions for (1.1).
During the past several decades, fractional equations have been studied widely; see [1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20,21,22,23,24,25,26,27,28,29,30,31,32,33,34,35,36] for instance. From the literature, we can see that there are many fractional derivatives used in differential equations. Among these various definitions, the widely used ones are the Riemann-Liouville and Caputo fractional derivatives, in many works. To generalize the Riemann-Liouville fractional derivative, Erdélyi-Kober defined a new fractional derivative, and we call it the Erdélyi-Kober fractional derivative. Moreover, the Erdélyi-Kober operator is very useful; we can refer to [6,9,14,15,16,17] and the references therein. The Erdélyi-Kober operator is a fractional integration operation which was given by Arthur Erdélyi and Hermann Kober in 1940 [23]. Some of these definitions and results were given in Samko et al. [3], Kiryakova [19], and McBride [20].
Nowadays, the theory of fractional operators in the Erdélyi-Kober frame has attracted much interest from researchers. The study of fractional systems is also very important, as these systems appear in various applications, especially in biological sciences. Recently, some problems of Erdélyi-Kober type fractional differential equations on infinite intervals received widespread attention from many scholars; see [8,21,22] for example.
Recently, in [8], the authors investigated the following equation:
{(Dϑ,σθu)(x)+F(u(x))=0,0≤x<∞,limt→0xθ(2−σ)Iσ+ϑ,2−σu(x)=0,limt→+∞xθ(2−σ)Iσ+ϑ,2−σu(x)=0, |
where σ∈(1,2), ϑ∈(1,2), θ>0, and F is a given continuous function, Dϑ,σθ denotes the EKFD, and Iσ+ϑ,2−σ denotes the Erdélyi-Kober fractional integral. The authors studied the existence and nonexistence of positive solutions for this problem by utilizing a fixed point result which uses the strongly positive-like operators and eigenvalue criteria.
In [9], the authors studied a fractional coupled system:
{cDϱu(τ)=F(τ,u(τ),z(τ),cDς1z(τ),Iξz(τ)),τ∈[0,T]:=K,2<ϱ≤3,1<ς1<2,cDςz(τ)=G(τ,u(τ),cDϱ1u(τ),Iζu(τ),z(τ),τ∈[0,T]:=K,2<ς≤3,1<ϱ1<2,u(0)=ϕ1(z),u′(0)=ε1z′(k1),u(T)=γρϑ−ρ(ϖ+v)Γ(ϖ)∫ϑ0σρv+ρ−1z(σ)(ϑρ−σρ)1−ϖdσ:=γJv,ϖρv(ϑ),z(0)=ϕ2(u),z′(0)=ε2z′(k2),z(T)=δvφ−v(θ+ω)Γ(θ)∫φ0σvω+υ−1u(σ)(φv−σv)1−θdσ:=δJω,θvu(φ), |
where cDϱ,cDς1,cDς,cDϱ1 are the Liouville-Caputo fractional derivatives of order 2<ϱ,ς≤3, 1<ς1,ϱ1<2. Iξ,Iζ are the Riemann-Liouville fractional integrals of order 1<ξ,ζ<2. Jυ,ϖρ,Jω,θv are the Erdélyi-Kober fractional integrals of order ϖ,θ>0, with v,ω>0, ρ, ϑ∈(−∞,+∞). F,G:K×(−∞,+∞)4→(−∞,+∞) and ϕ1,ϕ2:C(K,(−∞,+∞))→(−∞,+∞) are continuous functions. γ,δ,ε1,ε2 are positive real constants. The existence result was given by the Leray-Schauder alternative, and the uniqueness result was obtained due to Banach's fixed-point theorem. By the same methods, Arioua and Titraoui [18] studied system (1.1). Moreover, In [10], Arioua and Titraoui also investigated a new fractional problem involving the Erdélyi-Kober derivative. Inspired by the above articles, we use different methods to consider the fractional coupled system involving Erdélyi-Kober derivative (1.1). We employ the Guo-Krasnosel'skii fixed point theorem to discuss (1.1) in a special Banach space, and we also use the monotone iterative technique to study this system. Some existence results of positive solutions for system (1.1) are obtained, including the existence results of at least two positive solutions.
Definition 2.1. (see [2]) Let α∈(−∞,+∞). Cnα, n∈N, denotes a set of all functions f(t),t>0, with f(t)=tpf1(t) with p>α and f1∈Cn[0,∞).
Definition 2.2. (see [1,2]) For a function u∈Cα, the σ-order right-hand Erdélyi-Kober fractional integral is
(Iγ,σβu)(t)=βt−β(γ+σ)Γ(σ)∫t0sβ(γ+1)−1u(s)(tβ−sβ)1−σds,σ,β>0,γ∈(−∞,+∞), |
in which, Γ is the Euler gamma function.
Definition 2.3. (see [2]) Let n−1<δ≤n,n∈N, and for u∈Cα, the σ-order right-hand Erdélyi-Kober fractional derivative is
(Dγ,σβu)(t)=n∏j=1(γ+j+tβddt)(Iγ+σ,n−σβu)(t), |
where
n∏j=1(γ+j+tβddt)(Iγ+σ,n−σβu)=(γ+1+tβddt)⋯(γ+n+tβddt)(Iγ+σ,n−σβu). |
Lemma 2.1. (see [10]) Let 1<σ≤2, −2<γ<−1, β>0, and h∈C2α, with ∫∞0sβ(γ+m)−1h(τ)dτ<∞, m=1,2. The fractional problem
{Dγ,σβu(x)+h(x)=0,x>0,limx→0xβ(2+γ)Iσ+γ,2−δu(x)=0,limx→∞xβ(1+γ)Iσ+γ,2−σu(x)=0, |
has a unique solution given by u(x)=∫∞0Gσ(x,s)sβ(γ+1)−1h(s)ds, where
Gσ(x,s)={βΓ(σ)[x−β(γ+1)−x−β(δ+γ)(xβ−sβ)σ−1],0<s≤x<∞,βΓ(σ)x−β(γ+1),0<x≤s<∞. | (2.1) |
Lemma 2.2. (see [10]) For 1<σ≤2, −2<γ<−1, and β>0, the function Gσ, defined in (2.1), has the following properties:
(i) Gσ(x,s)1+x−β(1+γ)>0, for x,s>0;
(ii) Gσ(x,s)1+x−β(1+γ)≤βΓ(σ), for x,s>0;
(iii) for 0<τλ≤x≤τ and s>τλ2, where λ>1,τ>0, we have
Gσ(x,s)1+x−β(1+γ)≥β(σ−1)τ−β(1+γ)Γ(σ)λ−β(1−γ)(1+τ−β(1+γ))=βp(τ)Γ(σ), |
where p(τ)=(σ−1)τ−β(1+γ)λβ(1+γ)(1+τ−β(1+γ)).
Lemma 2.3. (see [18]) Let 0<σ1,σ2≤1 and F,G∈C2α with
∫∞0sβ(γ+m)−1F(s,u(s),v(s))ds<∞,m=1,2, |
∫∞0sβ(γ+m)−1G(s,u(s),v(s))ds<∞,m=1,2. |
Then, (1.1) has a unique solution given by
u(x)=∫∞0Gσ1(x,s)sβ(γ+1)−1F(s,u(s),v(s))ds, |
v(x)=∫∞0Gσ2(x,s)sβ(γ+1)−1G(s,u(s),v(s))ds, |
where
Gσ1(x,s)={βΓ(σ1)[x−β(γ+1)−x−β(σ1+γ)(xβ−sβ)σ1−1],0<s≤x<∞,βΓ(σ1)x−β(γ+1),0<x≤s<∞, | (2.2) |
Gσ2(x,s)={βΓ(σ2)[x−β(γ+1)−x−β(σ2+γ)(xβ−sβ)σ2−1],0<s≤x<∞,βΓ(σ2)x−β(γ+1),0<x≤s<∞. | (2.3) |
The following result is our main tool.
Lemma 2.4. (Guo-Krasnosel'skii fixed point theorem; see [37]) P is a cone in a Banach space E, and D1 and D2 are bounded open sets in E with θ∈D1, ¯D1⊂D2. A:P∩(¯D2∖D1)→P is a completely continuous operator. Consider the following conditions (ⅰ), (ⅱ):
(i) ‖Aw‖≤‖w‖ for w∈P∩∂D1, ‖Aw‖≥‖w‖ for w∈P∩∂D2;
(ii) ‖Aw‖≥‖w‖ for w∈P∩∂D1, ‖Aw‖≤‖w‖ for w∈P∩∂D2.
If one of the preceding conditions (ⅰ), (ⅱ) holds, then A has at least one fixed point in P∩(¯D2∖D1).
Next, we present some hypotheses that will play an important role in the subsequent discussion:
(H1) F,G:(0,+∞)×(−∞,+∞)×(−∞,+∞)→(0,+∞) are continuous and nondecreasing with respect to the second, third variables on (0,+∞).
(H2) For (x,u,v)∈(0,+∞)×(−∞,+∞)×(−∞,+∞),
F1(x,u,v)=xβ(1+γ)−1F(x,(1+x−β(1+γ))u,(1+x−β(1+γ))v), |
F2(x,u,v)=xβ(1+γ)−1G(x,(1+x−β(1+γ))u,(1+x−β(1+γ))v), |
such that
F1(x,u,v)≤φ1(x)ω1(∣u∣)+ψ1(x)ω2(∣v∣), |
F2(x,u,v)≤φ2(x)~ω1(∣u∣)+ψ2(x)~ω2(∣v∣), |
with ωi,~ωi∈C((0,+∞),(0,+∞)) nondecreasing and φi,ψi∈L1(0,+∞), i=1,2.
(H3) There are positive functions qi,˜qi,i=1,2, with
q∗i=∫∞0(1+x−β(1+γ))qi(x)dx<∞, |
˜q∗i=∫∞0(1+x−β(1+γ))˜qi(x)dx<∞, |
such that
xβ(γ+1)−1∣F(x,u,v)−F(x,˜u,˜v)∣≤q1(x)∣u−˜u∣+˜q1(x)∣v−˜v∣, |
xβ(γ+1)−1∣G(x,u,v)−G(x,˜u,˜v)∣≤q2(x)∣u−˜u∣+˜q2(t)∣v−˜v∣, |
for any u,v,˜u,˜v∈(−∞,+∞) and x∈(0,+∞).
(H4) F,G:(0,+∞)×(0,+∞)×(0,+∞)→(0,+∞) are continuous, such that
xβ(1+γ)−1F(x,u,v)=a1(x)F1(x,u,v), |
xβ(1+γ)−1G(x,u,v)=a2(x)G1(x,u,v), |
where a1,a2∈L1((0,+∞),(0,+∞)), F1,G1∈C((0,+∞)×(0,+∞)×(0,+∞),(0,+∞)), 0<∫ττλa1(x)dx<∞, 0<∫ττλa2(x)dx<∞, with τ>0, λ>1. Moreover, xβ(1+γ)−1F(x,u,v), xβ(1+γ)−1G(x,u,v):[0,+∞)×(0,+∞)×(0,+∞)→[0,+∞) also are continuous.
Remark 2.1. These conditions ensure the continuity and integrability of nonlinear terms in an infinite interval, which play a very important role in the proof of completely continuity for the relevant integral operators.
In this section, we use two Banach spaces defined by
X={u∈C((0,+∞),(−∞,+∞))∣limx→0u(x)1+x−β(1+γ) and limt→+∞u(x)1+x−β(1+γ) exist}, |
with the norm
‖u‖X=supx>0∣u(x)1+x−β(1+γ)∣, |
and
Y={v∈C((0,+∞),(−∞,+∞))∣limx→0v(x)1+x−β(1+γ) and limx→+∞v(x)1+x−β(1+γ) exist}, |
with the norm
‖v‖Y=supx>0∣v(x)1+x−β(1+γ)∣. |
So, (X×Y,‖(u,v)‖X×Y) is a Banach space, with the norm ‖(u,v)‖X×Y=‖u‖X+‖v‖Y.
Lemma 3.1. If F,G are continuous, then (u,v)∈X×Y is a solution of system (1.1)⇔(u,v)∈X×Y is a solution of the following equations:
{u(x)=∫∞0Gσ1(x,s)sβ(γ+1)−1F(s,u(s),v(s))ds,v(x)=∫∞0Gσ2(x,s)sβ(γ+1)−1G(s,u(s),v(s))ds. |
For (u,v)∈X×Y, we define an operator A:X×Y→X×Y as follows:
A(u,v)(x)=(A1(u,v)(x),A2(u,v)(x)), |
where
A1(u,v)(x)=∫∞0Gσ1(x,s)sβ(γ+1)−1F(s,u(s),v(s))ds, |
A2(u,v)(x)=∫∞0Gσ2(x,s)sβ(γ+1)−1G(s,u(s),v(s))ds, |
with Gσi(x,s),i=1,2, given by (2.2) and (2.3).
Remark 3.1. Let σ1,σ2,β,γ,λ,τ∈R, such that 1<σ1,σ2≤2,β>0,−2<γ<−1,λ>1,τ>0. If (H2) and (H4) hold, then for (u,v)∈X×Y with u(x),v(x)>0,
∫∞0sβ(γ+1)−1F(s,u(s),v(s))ds≤η∫∞τλ2sβ(γ+1)−1F(s,u(s),v(s))ds, |
∫∞0sβ(γ+1)−1G(s,u(s),v(s))ds≤η∫∞τλ2sβ(γ+1)−1G(s,u(s),v(s))ds, |
where η=max{η1,η2} with η1=1+ιϱ1(λ2−1),η2=1+ι∗ϱ2(λ2−1)>1, ϱ1,ϱ2,ι,ι∗>0.
Proof. By (H4), for x∈[τλ2,τ], we know that there exist two constants ϱ1,ϱ2>0, such that
xβ(γ+1)−1F(s,u,v)≥ϱ1,xβ(γ+1)−1G(s,u,v)≥ϱ2,u,v∈(0,+∞). |
So, for (u,v)∈X×Y with u(x),v(x)>0,
∫∞τλ2sβ(γ+1)−1F(s,u(s),v(s))ds≥∫ττλ2sβ(γ+1)−1F(s,u(s),v(s))ds≥τ(λ2−1)λ2ϱ1, |
∫∞τλ2sβ(γ+1)−1G(s,u(s),v(s))ds≥∫ττλ2sβ(γ+1)−1G(s,u(s),v(s))ds≥τ(λ2−1)λ2ϱ2, |
and hence,
λ2τ(λ2−1)ϱ1∫∞τλ2sβ(γ+1)−1F(s,u(s),v(s))ds≥1, |
λ2τ(λ2−1)ϱ2∫∞τλ2sβ(γ+1)−1G(s,u(s),v(s))ds≥1. |
By (H4), we know that there exist two constants ι,ι∗>0, such that
xβ(γ+1)−1F(x,u(x),v(x))≤ι,xβ(γ+1)−1G(x,u(x),v(x))≤ι∗,for ∀x∈[0,τλ2]. |
Thus,
∫τλ20sβ(γ+1)−1F(s,u(s),v(s))ds≤ιτλ2, |
∫τλ20sβ(γ+1)−1G(s,u(s),v(s))ds≤ι∗τλ2. |
Therefore, we can obtain
∫∞0sβ(γ+1)−1F(s,u(s),v(s))ds=∫τλ20sβ(γ+1)−1F(s,u(s),v(s))ds+∫∞τλ2sβ(γ+1)−1F(s,u(s),v(s))ds≤ιτλ2+∫∞τλ2sβ(γ+1)−1F(s,u(s),v(s))ds≤(1+ιϱ1(λ2−1))∫∞τλ2sβ(γ+1)−1F(s,u(s),v(s))ds=η1∫∞τλ2sβ(γ+1)−1F(s,u(s),v(s))ds. |
Similarly,
∫∞0sβ(γ+1)−1G(s,u(s),v(s))ds≤(1+ιϱ2(λ2−1))∫∞τλ2sβ(γ+1)−1G(s,u(s),v(s))ds=η2∫∞τλ2sβ(γ+1)−1G(s,u(s),v(s))ds. |
Take η=max{η1,η2}, and thus
∫∞0sβ(γ+1)−1F(s,u(s),v(s))ds≤η∫∞τλ2sβ(γ+1)−1F(s,u(s),v(s))ds, |
∫∞0sβ(γ+1)−1G(s,u(s),v(s))ds≤η∫∞τλ2sβ(γ+1)−1G(s,u(s),v(s))ds, |
hold.
Define two cones
K1={u∈X∣u(x)>0,x>0;minx∈[τλ,τ]u(x)1+x−β(1+γ)≥p(τ)η‖u‖X}, |
K2={v∈Y∣v(x)>0,x>0;minx∈[τλ,τ]v(x)1+x−β(1+γ)≥p(τ)η‖v‖Y}. |
Obviously, K1×K2={(u,v)∈X×Y∣u(x)>0,v(x)>0,∀x>0; minx∈[τλ,τ]u(x)1+x−β(1+γ)≥p(τ)η‖u‖X,minx∈[τλ,τ]v(x)1+x−β(1+γ)≥p(τ)η‖v‖Y} is also a cone. For convenience, we first list the following definitions:
F0=lim(u,v)→(0+,0+)supx>0F1(t,(1+x−β(1+γ))u,(1+x−β(1+γ))v)u+v, |
f∞=lim(u,v)→(+∞,+∞)infx>0F1(x,(1+x−β(1+γ))u,(1+x−β(1+γ))v)u+v, |
f0=lim(u,v)→(0+,0+)infx>0F1(x,(1+x−β(1+γ))u,(1+x−β(1+γ))v)u+v, |
F∞=lim(u,v)→(+∞,+∞)supx>0F1(t,(1+x−β(1+γ))u,(1+x−β(1+γ))v)u+v, |
G∗0=lim(u,v)→(0+,0+)supx>0G1(x,(1+x−β(1+γ))u,(1+x−β(1+γ))v)u+v, |
g∗∞=lim(u,v)→(+∞,+∞)infx>0G1(x,(1+x−β(1+γ))u,(1+x−β(1+γ))v)u+v, |
g∗0=lim(u,v)→(0+,0+)infx>0G1(x,(1+x−β(1+γ))u,(1+x−β(1+γ))v)u+v, |
G∗∞=lim(u,v)→(+∞,+∞)supx>0G1(x,(1+x−β(1+γ))u,(1+x−β(1+γ))v)u+v. |
Lemma 4.1. If assumptions (H1) and (H2) hold, then A:K1×K2→K1×K2 is completely continuous.
Proof. First, we show A:K1×K2→K1×K2. By (H1) and (H2), for (u,v)∈K1×K2,
‖A1(u,v)‖X=supt>0|A1(u,v)(x)|1+x−β(1+γ)=supx>0∣∫∞0Gσ1(x,s)1+x−β(1+γ)sβ(γ+1)−1F(s,u(s),v(s))ds∣≤βΓ(σ1)∫∞0∣sβ(γ+1)−1F(s,u(s),v(s))∣ds=βΓ(σ1)∫∞0∣sβ(γ+1)−1F(s,(1+s−β(1+γ))u(s)1+s−β(1+γ),(1+s−β(1+γ))v(s)1+s−β(1+γ))∣ds=βΓ(σ1)∫∞0∣F1(s,u(s)1+s−β(1+γ),v(s)1+s−β(1+γ))∣≤βΓ(σ1)[ω1(‖u‖X)∫∞0φ1(s)ds+ω2(‖v‖Y)∫∞0ψ1(s)ds]<+∞. |
Similarly,
‖A2(u,v)‖Y≤βΓ(σ1)[~ω1(‖u‖X)∫∞0φ2(s)ds+~ω2(‖v‖Y)∫∞0ψ2(s)ds]<+∞. |
By (H1) and Lemma 2.2, for (u,v)∈K1×K2, we have A1(u,v)(x)>0,A2(u,v)(x)>0,x>0. From Lemma 2.2 and Remark 3.1, for x∈[τλ,τ],τ>0, and λ>1,
|A1(u,v)(x)|1+x−β(1+γ)=∫∞0Gσ1(x,s)1+x−β(1+γ)sβ(γ+1)−1F(s,u(s),v(s))ds=∫τλ20Gσ1(x,s)1+x−β(1+γ)sβ(γ+1)−1F(s,u(s),v(s))ds+∫0τλ2Gσ1(x,s)1+x−β(1+γ)sβ(γ+1)−1F(s,u(s),v(s))ds≥∫0τλ2Gσ1(t,s)1+t−β(1+γ)sβ(γ+1)−1F(s,u(s),v(s))ds≥βp(τ)Γ(σ1)∫0τλ2sβ(γ+1)−1F(s,u(s),v(s))ds≥βp(τ)ηΓ(σ1)∫∞0sβ(γ+1)−1F(s,u(s),v(s))ds≥p(τ)η‖A1(u,v)‖X. |
So, A1(u,v)(x)1+x−β(1+γ)≥p(τ)η‖A1(u,v)‖X. Similarly, A2(u,v)(x)1+x−β(1+γ)≥p(τ)η‖A2(u,v)‖Y. Therefore,
minx∈[τλ,τ]A1(u,v)(x)1+x−β(1+γ)≥p(τ)η‖A1(u,v)‖X, |
minx∈[τλ,τ]A2(u,v)(x)1+x−β(1+γ)≥p(τ)η‖A2(u,v)‖Y. |
That is, A:K1×K2→K1×K2 is true.
Second, it will give a simply prove that A is continuous. Let D={(u,v)|(u,v)∈K1×K2,‖(u,v)‖X×Y≤K,K>0}, a bounded subset in K1×K2. Let (un,vn)∈D be a sequence that converges to (u,v) in ∈K1×K2. Then ‖(un,vn)‖X×Y≤K. From Lemma 2.2,
‖A1(un,vn)−A1(u,v)‖X=supx>0∣A1(un,vn)(x)−A1(u,v)(x)1+x−β(1+γ)∣≤βΓ(σ1)∣∫∞0sβ(γ+1)−1F(s,un(s),vn(s))ds−∫∞0sβ(γ+1)−1F(s,u(s),v(s))ds∣≤βΓ(σ1)∫∞0∣sβ(γ+1)−1(F(s,un(s),vn(s))−F(s,u(s),v(s)))∣ds. |
By (H2),
∣sβ(γ+1)−1F(s,un(s),vn(s))∣=∣sβ(γ+1)−1F(s,(1+s−β(1+γ))un(s)1+s−β(1+γ),(1+s−β(1+γ))vn(s)1+s−β(1+γ))∣=F1(s,un(s)1+s−β(1+γ),vn(s)1+s−β(1+γ))≤φ1(s)ω1(‖un‖X)+ψ1(s)ω2(‖vn‖Y)∈L1(0,∞). |
By the continuity of sβ(γ+1)−1F(s,u(s),v(s)) and the Lebesgue dominated convergence theorem,
∫∞0sβ(γ+1)−1F(s,un(s),vn(s))ds→∫∞0sβ(γ+1)−1F(s,u(s),v(s))ds,n→∞. |
Therefore, ‖A1(un,vn)−A1(u,v)‖X→0,n→∞. Similarly, ‖A2(un,vn)−A2(u,v)‖Y→0,n→∞.
So, ‖A(un,vn)−A(u,v)‖X×Y→0,n→∞. That is, A is continuous in D. In the end, we know that A(D) is relatively compact on (0,∞) and is equi-convergent at ∞ by [18]. Therefore, A:K1×K2→K1×K2 is completely continuous.
Theorem 4.1. Assume that (H2) and (H4) hold. If F0=0,G∗0=0,f∞=∞,g∗∞=∞, then the system (1.1) has at least one positive solution.
Proof. We divide the proof into several steps.
Step 1. A:K1×K2→K1×K2 is completely continuous. This result easily follows from Lemma 4.1.
Step 2. We show that there exist R1>0 and D1={(u,v)∈X×Y,‖(u,v)‖X×Y<R1} such that ‖A(u,v)‖X×Y≤‖(u,v)‖X×Y, (u,v)∈(K1×K2)∩∂D1.
Because F0=0,G∗0=0, we choose R1>0, such that
F1(x,(1+x−β(1+γ))u,(1+x−β(1+γ))v)≤ϵ1(u+v), |
G1(x,(1+x−β(1+γ))u,(1+x−β(1+γ))v)≤ϵ2(u+v), |
for 0<u+v≤R1,x>0, where ϵ1,ϵ2>0 satisfy
ϵ1≤12Γ(σ1)β∫∞0a1(s)ds,ϵ2≤12Γ(σ2)β∫∞0a2(s)ds. |
So, for (u,v)∈K1×K2 and ‖(u,v)‖X×Y=R1, by Lemma 2.2,
A1(u,v)(x)1+x−β(1+γ)=∫∞0Gσ1(x,s)1+x−β(1+γ)sβ(γ+1)−1F(s,u(s),v(s))ds≤βΓ(σ1)∫∞0sβ(γ+1)−1F(s,u(s),v(s))ds, |
A2(u,v)(x)1+x−β(1+γ)=∫∞0Gσ2(x,s)1+x−β(1+γ)sβ(γ+1)−1G(s,u(s),v(s))ds≤βΓ(σ2)∫∞0sβ(γ+1)−1G(s,u(s),v(s))ds. |
By (H4),
A1(u,v)(x)1+x−β(1+γ)≤βΓ(σ1)∫∞0a1(s)F1(s,u(s),v(s))ds=βΓ(σ1)∫∞0a1(s)F1(s,(1+s−β(1+γ))u(s)1+s−β(1+γ),(1+s−β(1+γ))v(s)1+s−β(1+γ))ds≤βΓ(σ1)∫∞0a1(s)ϵ1u(s)+v(s)1+s−β(1+γ)ds≤βΓ(σ1)ϵ1‖(u,v)‖X×Y∫∞0a1(s)ds≤12‖(u,v)‖X×Y. |
Similarly,
A2(u,v)(x)1+x−β(1+γ)≤βΓ(σ2)ϵ2‖(u,v)‖X×Y∫∞0a2(s)ds≤12‖(u,v)‖X×Y. |
Therefore,
‖A(u,v)‖X×Y≤‖(u,v)‖X×Y, for (u,v)∈K1×K2, and ‖(u,v)‖X×Y=R1. |
Let D1={(u,v)∈X×Y,‖(u,v)‖X×Y<R1}. Then,
‖A(u,v)‖X×Y≤‖(u,v)‖X×Y, for (u,v)∈(K1×K2)∩∂D1. |
Step 3. We show that there exist R2>0 and D2={(u,v)∈X×Y,‖(u,v)‖X×Y<R2} such that
‖A(u,v)‖X×Y≥‖(u,v)‖X×Y, for (u,v)∈(K1×K2)∩∂D2. |
Because f∞=∞,g∗∞=∞, there exists R>0, such that
F1(x,(1+x−β(1+γ))u,(1+x−β(1+γ))v)≥m1(u+v), |
G1(x,(1+x−β(1+γ))u,(1+x−β(1+γ))v)≥m2(u+v), |
for u+v≥R,x>0, where m1,m2>0 satisfy
m1≥12η1ηΓ(σ1)βp2(τ)∫τλτa1(s)ds,m2≥12η2ηΓ(σ2)βp2(τ)∫τλτa2(s)ds,η=max{η1,η2}. |
Let R2≥max{R1,ηRp(τ)}, and D2={(u,v)∈X×Y,‖(u,v)‖X×Y<R2}. Then, D1⊂D2.
Thus, for (u,v)∈K1×K2, ‖(u,v)‖X×Y=R2, we have
u(x)1+x−β(1+γ)≥minx∈[τλ,τ]u(x)1+x−β(1+γ)≥p(τ)η1‖u‖X, |
v(x)1+x−β(1+γ)≥minx∈[τλ,τ]v(x)1+x−β(1+γ)≥p(τ)η2‖v‖Y. |
So,
u(x)+v(x)1+x−β(1+γ)≥p(τ)η1‖u‖X+p(τ)η2‖v‖Y≥p(τ)η(‖u‖X+‖v‖Y)=p(τ)η‖(u,v)‖X×Y=p(τ)ηR2≥R. |
By (H4), for x∈[τλ,τ], we can obtain
A1(u,v)(x)1+x−β(1+γ)≥βp(τ)η1Γ(σ1)∫∞0sβ(γ+1)−1F(s,u(s),v(s))ds=βp(τ)η1Γ(σ1)∫∞0a1(s)F1(s,u(s),v(s))ds=βp(τ)η1Γ(σ1)∫∞0a1(s)F1(s,(1+s−β(1+γ))u(s)1+s−β(1+γ),(1+s−β(1+γ))v(s)1+s−β(1+γ))ds≥βp(τ)η1Γ(σ1)m1∫∞0a1(s)u(s)+v(s)1+s−β(1+γ)ds≥βp(τ)η1Γ(σ1)m1∫∞0a1(s)dsp(τ)η1‖u‖X+βp(τ)η1Γ(σ1)m1∫∞0a1(s)dsp(τ)η2‖v‖Y≥βp(τ)η1Γ(σ1)m1∫τλτa1(s)dsp(τ)η1‖u‖X+βp(τ)η1Γ(σ1)m1∫τλτa1(s)dsp(τ)η2‖v‖Y=βp2(τ)η1Γ(σ1)m1∫τλτa1(s)ds(1η1‖u‖X+1η2‖v‖Y)≥βp2(τ)η1Γ(σ1)m1∫τλτa1(s)ds1η‖(u,v)‖X×Y≥12‖(u,v)‖X×Y. |
Similarly, A2(u,v)(x)1+x−β(1+γ)≥12‖(u,v)‖X×Y. Therefore,
‖A(u,v)‖X×Y≥‖(u,v)‖X×Y, for (u,v)∈(K1×K2)∩∂D2. |
Finally, by Lemma 2.4, A has a fixed point in (K1×K1)∩∂(¯D2∖D1). So, (1.1) has at least one positive solution.
Theorem 4.2. Assume that (H2) and (H4) hold. If f0=∞,g∗0=∞,F∞=0,G∗∞=0, then (1.1) has at least one positive solution.
Proof. We divide the proof into several steps.
Step 1. A:K1×K2→K1×K2 is completely continuous. This result easily follows from Lemma 4.1.
Step 2. We show that there exist r1>0 and D1={(u,v)∈X×Y,‖(u,v)‖X×Y<r1} such that
‖A(u,v)‖X×Y≥‖(u,v)‖X×Y, for (u,v)∈(K1×K2)∩∂D1. |
Because f0=∞,g∗0=∞, there exists r1>0 such that
F1(x,(1+x−β(1+γ))u,(1+x−β(1+γ))v)≥M1(u+v), |
G1(x,(1+x−β(1+γ))u,(1+x−β(1+γ))v)≥M2(u+v), |
for 0<u+v≤r1,x>0, where M1,M2>0, satisfy
M1≥12η1ηΓ(σ1)βp2(τ)∫τλτa1(s)ds,M2≥12η2ηΓ(σ2)βp2(τ)∫τλτa2(s)ds,η=max{η1,η2}. |
Let D1={(u,v)∈X×Y,‖(u,v)‖X×Y<r1}. So, for (u,v)∈K1×K2 with ‖(u,v)‖X×Y=r1, and x∈[τλ,τ], then by (H4),
A1(u,v)(x)1+x−β(1+γ)≥βp(τ)η1Γ(σ1)∫∞0sβ(γ+1)−1F(s,u(s),v(s))ds=βp(τ)η1Γ(σ1)∫∞0a1(s)F1(s,u(s),v(s))ds=βp(τ)η1Γ(σ1)∫∞0a1(s)F1(s,(1+s−β(1+γ))u(s)1+s−β(1+γ),(1+s−β(1+γ))v(s)1+s−β(1+γ))ds≥βp(τ)η1Γ(σ1)M1∫∞0a1(s)u(s)+v(s)1+s−β(1+γ)ds≥βp(τ)η1Γ(σ1)M1∫∞0a1(s)dsp(τ)η1‖u‖X+βp(τ)η1Γ(σ1)M1∫∞0a1(s)dsp(τ)η2‖v‖Y≥βp(τ)η1Γ(σ1)M1∫τλτa1(s)dsp(τ)η1‖u‖X+βp(τ)η1Γ(σ1)M1∫τλτa1(s)dsp(τ)η2‖v‖Y=βp2(τ)η1Γ(σ1)M1∫τλτa1(s)ds(1η1‖u‖X+1η2‖v‖Y)≥βp2(τ)η1Γ(σ1)M1∫τλτa1(s)ds1η‖(u,v)‖X×Y≥12‖(u,v)‖X×Y. |
Similarly, A2(u,v)(x)1+x−β(1+γ)≥12‖(u,v)‖X×Y. Thus,
‖A(u,v)‖X×Y≥‖(u,v)‖X×Y, for (u,v)∈(K1×K2)∩∂D1. |
Step 3. We show that there exist r2>0 and D2={(u,v)∈X×Y,‖(u,v)‖X×Y<r2} such that
‖A(u,v)‖X×Y≤‖(u,v)‖X×Y for (u,v)∈(K1×K2)∩∂D2. |
Because F∞=0,G∗∞=0, there exists r>0, such that
F1(x,(1+x−β(1+γ))u,(1+x−β(1+γ))v)≤ϵ1(u+v), |
G1(x,(1+x−β(1+γ))u,(1+x−β(1+γ))v)≤ϵ2(u+v), |
for u+v>r,x>0, where ϵ1,ϵ2>0 satisfy
ϵ1≤12Γ(σ1)β∫∞0a1(s)ds,ϵ2≤12Γ(σ2)β∫∞0a2(s)ds. |
Let D2={(u,v)∈X×Y,‖(u,v)‖X×Y<r2}, where r2>max{r1,r}. Then D1⊂D1. We define two functions U1,U2 as follows:
U1:(−∞,+∞)→(−∞,+∞),U1(a)=sup0<u+v≤asupx>0F1(x,(1+x−β(1+γ))u,(1+x−β(1+γ))v), |
U2:(−∞,+∞)→(−∞,+∞),U2(a)=sup0<u+v≤asupx>0G1(x,(1+x−β(1+γ))u,(1+x−β(1+γ))v). |
For (u,v)∈K1×K2 and ‖(u,v)‖X×Y=r2,
U1(r2)=sup0<u+v≤r2supx>0F1(x,(1+x−β(1+γ))u,(1+x−β(1+γ))v)≤ϵ1sup0<u+v≤r2(u+v)=ϵ1r2=ϵ1‖(u,v)‖X×Y, |
U2(r2)=sup0<u+v≤r2supx>0G1(x,(1+x−β(1+γ))u,(1+x−β(1+γ))v)≤ϵ2sup0<u+v≤r2(u+v)=ϵ2r2=ϵ2‖(u,v)‖X×Y. |
By Lemma 2.2 and (H4),
A1(u,v)(x)1+x−β(1+γ)≤βΓ(σ1)∫∞0sβ(γ+1)−1F(s,u(s),v(s))ds=βΓ(σ1)∫∞0a1(s)F1(s,u(s),v(s))ds=βΓ(σ1)∫∞0a1(s)F1(s,(1+s−β(1+γ))u(s)1+s−β(1+γ),(1+s−β(1+γ))v(s)1+s−β(1+γ))ds≤βΓ(σ1)∫∞0a1(s)sup0<u+v≤r2supx>0F1(x,(1+x−β(1+γ))u,(1+x−β(1+γ))v)ds=βΓ(σ1)∫∞0a1(s)U1(r2)ds≤βΓ(σ1)∫∞0a1(s)dsϵ1‖(u,v)‖X×Y≤12‖(u,v)‖X×Y. |
Similarly, A2(u,v)(x)1+x−β(1+γ)≤12‖(u,v)‖X×Y. Therefore, ‖A(u,v)‖X×Y≤‖(u,v)‖X×Y, for (u,v)∈(K1×K2)∩∂D2. Finally, by Lemma 2.4, A has a fixed point in (K1×K1)∩∂(¯D2∖D1). So, the system (1.1) has at least one positive solution.
In the section, we obtain the multiplicity of positive solution of (1.1) by using the monotone iterative technique.
Theorem 5.1. If (H1) and (H2) hold, then (1.1) has two positive solutions (u∗,v∗) and (w∗,z∗) satisfying 0≤‖(u∗,v∗)‖X×Y≤Υ and 0≤‖(w∗,z∗)‖X×Y≤Υ, where Υ is a positive preset constant. Moreover, limn→∞(un,vn)=(u∗,v∗) and limn→∞(wn,zn)=(w∗,z∗), where (un,vn) and (wn,zn) are given by
(un(x),vn(x))=(A1(un−1,vn−1)(x),A2(un−1,vn−1)(x)),n=1,2,…, | (5.1) |
with
(u0(x),v0(x))=(Υ1[1+x−β(γ+1)],Υ2[1+x−β(γ+1)]),Υ1,Υ2>0,Υ1+Υ2≤Υ, |
and
(wn(x),zn(x))=(A1(wn−1,zn−1)(x),A2(wn−1,zn−1)(x)),n=1,2,…, | (5.2) |
with (w0(x),z0(x))=(0,0). In addition,
(w0(x),z0(x))≤(w1(x),z1(x))≤⋯≤(wn(x),zn(x))≤⋯≤(w∗,z∗)≤(u∗,v∗)≤⋯≤(un(x),vn(x))≤⋯≤(u1(x),v1(x))≤(u0(x),v0(x)). | (5.3) |
Proof. First, from Lemma 4.1, A(K1×K2)⊂K1×K2 for (u,v)∈K1×K2. Let
Υ1=βΓ(σ1)[ω1(Υ)∫∞0φ1(s)ds+ω2(Υ)∫∞0ψ1(s)ds]<∞, |
Υ2=βΓ(σ2)[~ω1(Υ)∫∞0φ2(s)ds+~ω2(Υ)∫∞0ψ2(s)ds]<∞, |
and Υ≥Υ1+Υ2 with DΥ={(u,v)∈K1×K2:‖(u,v)‖X×Y≤Υ}. For any (u,v)∈DΥ, from (H2) and Lemma 2.2,
‖A1(u,v)‖X=supx>0|A1(u,v)(x)|1+x−β(1+γ)=supx>0∣∫∞0Gσ1(x,s)1+t−β(1+γ)sβ(γ+1)−1F(s,u(s),v(s))ds∣≤βΓ(σ1)∫∞0∣sβ(γ+1)−1F(s,u(s),v(s))ds∣≤βΓ(σ1)[ω1(∣u(s)∣1+s−β(1+γ))∫∞0φ1(s)ds+ω2(∣v(s)∣1+s−β(1+γ))∫∞0ψ1(s)ds]≤βΓ(σ1)[ω1(‖u‖X)∫∞0φ1(s)ds+ω2(‖v‖Y)∫∞0ψ1(s)ds]≤βΓ(σ1)[ω1(Υ)∫∞0φ1(s)ds+ω2(Υ)∫∞0ψ1(s)ds]=Υ1. |
Similarly, ‖A2(u,v)‖Y≤Υ2 for (u,v)∈DΥ. Thus,
‖A(u,v)‖X×Y=‖A1(u,v)‖X+‖A2(u,v)‖Y≤Υ1+Υ2≤Υ. |
That is, A(DΥ)⊂DΥ. We construct two sequences as follows:
(un,vn)=A(un−1,vn−1),(wn,zn)=A(wn−1,zn−1), n=1,2,3,…. |
Obviously, (u0(x),v0(x)),(w0(x),z0(x))∈DΥ. Because A(DΥ)⊂DΥ, (un,vn),(wn,zn)∈DΥ,n=1,2,…. We need to show that there exist (u∗,v∗) and (w∗,z∗) satisfying limn→∞(un,vn)=(u∗,v∗) and limn→∞(wn,zn)=(w∗,z∗) which are two monotone sequences for approximating positive solutions of the system (1.1).
For x∈(0,+∞),(un,vn)∈DΥ, from Lemma 2.2 and (5.1),
u1(x)=A1(u0,v0)(x)=∫∞0Gσ1(x,s)sβ(γ+1)−1F(s,u0(s),v0(s))ds≤βΓ(σ1)∫∞0(1+t−β(1+γ))sβ(γ+1)−1F(s,u0(s),v0(s))ds≤βΓ(σ1)(1+x−β(1+γ))[ω1(∣u0(s)∣1+s−β(1+γ))∫∞0φ1(s)ds+ω2(∣v0(s)∣1+s−β(1+γ))∫∞0ψ1(s)ds]≤βΓ(σ1)(1+x−β(1+γ))[ω1(‖u0‖X)∫∞0φ1(s)ds+ω2(‖v0‖Y)∫∞0ψ1(s)ds]≤βΓ(σ1)(1+x−β(1+γ))[ω1(Υ)∫∞0φ1(s)ds+ω2(Υ)∫∞0ψ1(s)ds]=(1+x−β(1+γ))Υ1=u0(x) |
and
v1(x)=A2(u0,v0)(x)=∫∞0Gσ2(x,s)sβ(γ+1)−1G(s,u0(s),v0(s))ds≤βΓ(σ2)∫∞0(1+x−β(1+γ))sβ(γ+1)−1G(s,u0(s),v0(s))ds≤βΓ(σ2)(1+t−β(1+γ))[~ω1(∣u0(s)∣1+s−β(1+γ))∫∞0φ2(s)ds+~ω2(∣v0(s)∣1+s−β(1+γ))∫∞0ψ2(s)ds]≤βΓ(σ2)(1+t−β(1+γ))[~ω1(‖u0‖X)∫∞0φ2(s)ds+~ω2(‖v0‖Y)∫∞0ψ2(s)ds]≤βΓ(σ2)(1+x−β(1+γ))[~ω1(Υ)∫∞0φ2(s)ds+~ω2(Υ)∫∞0ψ2(s)ds]=(1+x−β(1+γ))Υ2=v0(x), |
that is,
(u1(x),v1(x))=(A1(u0,v0)(x),A2(u0,v0)(x))≤((1+x−β(1+γ))Υ1,(1+x−β(1+γ))Υ2)=(u0(x),v0(x)). |
So, by the condition (H1),
(u2(x),v2(x))=(A1(u1,v1)(x),A2(u1,v1)(x))≤(A1(u0,v0)(x),A2(u0,v0)(x))=(u1(x),v1(x)). |
For x∈(0,+∞), the sequences {(un,vn)}∞n=0 satisfy (un+1(x),vn+1(x))≤(un(x),vn(x)). By the iterative sequences (un+1,vn+1)=A(un,vn) and the complete continuity of the operator A, (un,vn)→(u∗,v∗), and A(u∗,v∗)=(u∗,v∗).
Similarly, for the sequences {(wn,zn)}∞n=0, we have
(w1(x),z1(x))=(A1(w0,z0)(x),A2(w0,z0)(x))=(∫∞0Gσ1(x,s)sβ(γ+1)−1F(s,w0(s),z0(s))ds,∫∞0Gσ2(x,s)sβ(γ+1)−1G(s,w0(s),z0(s))ds)≥(0,0)=(w0(x),z0(x)). |
Then, by the condition (H1),
(w2(x),z2(x))=(A1(w1,z1)(x),A2(w1,z1)(x))≥(A1(w0,z0)(x),A2(w0,z0)(x))=(w1(x),z1(x)). |
Analogously, for x∈(0,+∞), we have (wn+1(x),zn+1(x))≥(wn(x),zn(x)). By the iterative sequences (wn+1,zn+1)=A(wn,zn) and the complete continuity of the operator A, (wn,zn)→(w∗,z∗), and A(w∗,z∗)=(w∗,z∗).
Finally, we prove that (u∗,v∗) and (w∗,z∗) are the minimal and maximal positive solutions of (1.1). Assume that (ς(x),μ(x)) is any positive solution of (1.1). Then, A(ς(x),μ(x))=(ς(x),μ(x)), and
(w0(x),z0(x))=(0,0)≤(ς(x),μ(x))≤((1+x−β(1+γ))Υ1,(1+x−β(1+γ))Υ2)=(u0(x),v0(x)). |
Therefore,
(w1(x),z1(x))=(A1(w0,z0)(x),A2(w0,z0)(x))≤(ς(x),μ(x))≤(A1(u0,v0)(x),A2(u0,v0)(x))=(u1(x),v1(x)). |
That is, (w1(x),z1(x))≤(ς(x),μ(x))≤(un(x),vn(x)). So, (5.3) holds. By (H1), (0,0) is not a solution of (1.1). From (5.1), (w∗,z∗) and (u∗,v∗) are two extreme positive solutions of (1.1), which can be constructed via limitS of two monotone iterative sequences in (5.1) and (5.2).
Example 6.1. We consider the following system:
{D−32,531u(x)+x32(u1+x12)2e−x+x32(v1+x12)2e−x=0,t∈(0,+∞),D−32,321v(x)+x52e−2x2(u1+x12)2ln(1+(u1+x12)2)+x52e−2x2(u1+x12)2ln(1+(u1+x12)2),x∈(0,+∞),limx→0x12I16,13u(x)=0,limx→∞x−12I16,13u(x)=0,limx→0x12I0,12v(x)=0,limx→∞x−12I0,12v(x)=0, | (6.1) |
where σ1=53,σ2=32,γ=−32,β=1,
F(x,u,v)=x32e−x[(u1+x12)2+(v1+x12)2], |
G(x,u,v)=x52e−2x2[(u1+x12)2ln(1+(u1+x12)2)+(u1+x12)2ln(1+(u1+x12)2)]. |
First, for F1(x,u,v)=xβ(1+γ)−1F(x,(1+x−β(1+γ))u,(1+x−β(1+γ))v)=e−x(u2,v2), we choose ω1(u)=u2∈C((0,+∞),(0,+∞)), ω2(v)=v2∈C((0,+∞),(0,+∞)), and φ1(x)=ψ1(x)=e−x∈L1(0,+∞). Then,
∣F1(x,u,v)∣≤φ1(x)ω1(∣u∣)+ψ1(t)ω2(∣v∣), (0,+∞)×(−∞,+∞)×(−∞,+∞). |
Similarly, for F2(x,u,v)=xβ(1+γ)−1G(x,(1+x−β(1+γ))u,(1+x−β(1+γ))v)=xe−2x2[u2ln(u2+1)+v2ln(v2+1)], we choose ~ω1(u)=u2ln(u2+1)∈C((0,+∞),(0,+∞)), ~ω2(v)=v2ln(v2+1)∈C((0,+∞),(0,+∞)), and φ2(x)=ψ2(x)=xe−2x2∈L1(0,+∞). Then,
∣F2(x,u,v)∣≤φ2(x)~ω1(∣u∣)+ψ2(x)~ω2(∣v∣), (0,+∞)×(−∞,+∞)×(−∞,+∞). |
So, the condition (H2) holds. Obviously, F,G:(0,+∞)×(0,+∞)×(0,+∞)→(0,+∞) are continuous.
x−32F(x,u,v)=e−x[(u1+x12)2+(v1+x12)2]=a1(x)F1(x,u,v), |
x−32G(x,u,v)=xe−2x2[(u1+x12)2ln(1+(u1+x12)2)+(u1+x12)2ln(1+(u1+x12)2)]=a2(x)G1(x,u,v), |
where a1(x)=e−x,a2(x)=xe−2x2, F1(x,u,v)=(u1+x12)2+(v1+x12)2, G1(t,u,v)=(u1+x12)2ln(1+(u1+x12)2)+(u1+x12)2ln(1+(u1+x12)2). So, x−32f(x,u,v),x−32G(x,u,v):[0,+∞)×(0,+∞)×(0,+∞)→[0,+∞) are continuous. Hence, the condition (H4) holds. Finally,
F0=lim(u,v)→(0+,0+)u2+v2u+v=0,G∗0=lim(u,v)→(0+,0+)u2ln(u2+1)+v2ln(v2+1)u+v=0, |
f∞=lim(u,v)→(+∞,+∞)u2+v2u+v=∞,g∗∞=lim(u,v)→(+∞,+∞)u2ln(u2+1)+v2ln(v2+1)u+v=∞. |
Therefore, from Theorem 4.1, (6.1) has at least one positive solution (u(x),v(x)). Further,
{u(x)=32Γ(23)[x12∫∞0s−32F(s,u(s),v(s))ds−x−83∫∞x(x−s)23s−32F(s,u(s),v(s))ds],v(x)=2√π[x12∫∞0s−32G(s,u(s),v(s))ds−∫∞x(x−s)12s−32G(s,u(s),v(s))ds]. |
Example 6.2. We consider the following system:
{D−32,321u(x)+x52e−2x2+1[arctan(u1+x12)2+1π]+x52e−2x2+1[arctan(u1+x12)2+π]=0,x∈(0,+∞),D−32,761v(x)+x32e−x[arctan(ln((u1+x12)2+1))+32π]+x32e−x[arctan(ln((v1+x12)2+1))+1],x∈(0,+∞),limx→0x12I012u(x)=0,limx→∞x−12I0,12u(x)=0,limx→0x12I−13,56v(x)=0,limx→∞x−12I−13,56v(x)=0, | (6.2) |
where σ1=32,σ2=76,γ=−32,β=1,
F(x,u,v)=x52e−2x2+1[arctan(u1+x12)2+1π]+x52e−2x2+1[arctan(u1+x12)2+1π], |
G(x,u,v)=x32e−x[arctan(ln((u1+x12)2+1))+32π]+x32e−x[arctan(ln((v1+x12)2+1))+1]. |
First, for
F1(x,u,v)=xβ(1+γ)−1F(x,(1+x−β(1+γ))u,(1+x−β(1+γ))v)=xe−2x2+1[arctanu2+1π+arctanv2+π], |
we choose ω1(u)=arctanu2+1π∈C((0,+∞),(0,+∞)),ω2(v)=arctanv2+π∈C((0,+∞),(0,+∞)), and φ1(x)=ψ1(x)=xe−2x2+1∈L1(0,+∞). Then,
∣F1(x,u,v)∣≤φ1(x)ω1(∣u∣)+ψ1(x)ω2(∣v∣), (0,+∞)×(−∞,+∞)×(−∞,+∞). |
Similarly, for
F2(x,u,v)=xβ(1+γ)−1g(x,(1+x−β(1+γ))u,(1+x−β(1+γ))v)=e−x[arctan(ln(u2+1))+32π+arctan(ln(v2+1))+1], |
we choose ~ω1(u)=arctan(ln(u2+1))+32π∈C((0,+∞),(0,+∞)), ~ω2(v)=arctan(ln(v2+1))+1∈C((0,+∞),(0,+∞)), and φ2(x)=ψ2(x)=e−x∈L1(0,+∞). Then,
∣F2(x,u,v)∣≤φ2(x)~ω1(∣u∣)+ψ2(x)~ω2(∣v∣), (0,+∞)×(−∞,+∞)×(−∞,+∞). |
That is, (H2) holds. Second, F,G:(0,+∞)×(0,+∞)×(0,+∞)→(0,+∞) are continuous. And
x−32F(x,u,v)=xe−2x2+1[arctan(u1+x12)2+1π+arctan(v1+x12)2+π]=a1(x)F1(x,u,v), |
x−32G(x,u,v)=e−x[arctan(ln((u1+x12)2+1))+32π+arctan(ln((v1+x12)2+1))+1]=a2(x)G1(x,u,v), |
where a1(x)=xe−2x2+1,a2(x)=e−x, F1(x,u,v)=arctan(u1+x12)2+1π+arctan(v1+x12)2+π, G1(x,u,v)=arctan(ln((u1+x12)2+1))+32π+arctan(ln((v1+x12)2+1))+1. So, x−32F(x,u,v),x−32G(x,u,v):[0,+∞)×(0,+∞)×(0,+∞)→[0,+∞) are continuous. That is, (H4) holds. In addition,
f0=lim(u,v)→(0+,0+)arctanu2+1π+arctanv2+πu+v=∞, |
g∗0=lim(u,v)→(0+,0+)arctan(ln(u2+1))+32π+arctan(ln(v2+1))+1u+v=∞, |
F∞=lim(u,v)→(+∞,+∞)arctanu2+1π+arctanv2+πu+v=0, |
G∗∞=lim(u,v)→(+∞,+∞)arctan(ln(u2+1))+32π+arctan(ln(v2+1))+1u+v=0. |
Therefore, from Theorem 4.2, (6.2) has at least one positive solution (u(x),v(x)). Further,
{u(x)=2√π[x12∫∞0s−32F(s,u(s),v(s))ds−∫∞x(x−s)12s−32F(s,u(s),v(s))ds],v(x)=6Γ(16)[x12∫∞0s−32G(s,u(s),v(s))ds−x13∫∞x(x−s)16s−32G(s,u(s),v(s))ds]. |
Example 6.3. We consider the following system:
{D−32,531u(x)+x32e−x3∣u1+x12∣+x52ln(∣v1+x12∣+1)e−2x2+110=0,x∈(0,+∞),D−32,321v(x)+x52e−2x2+1arctan(∣u1+x12∣+1√π)+x52e−2x2+15∣v1+x12∣=0,x∈(0,+∞),limx→0x12I16,13u(x)=0,limx→∞x−12I16,13u(x)=0,limx→0x12I0,12v(x)=0,limx→∞x−12I0,12v(x)=0, | (6.3) |
where σ1=53,σ2=32,γ=−32,β=1,
F(x,u,v)=x32e−x3∣u1+x12∣+x52ln(∣v1+x12∣+1)e−2x2+110, |
G(x,u,v)=x52e−2x2+1arctan(∣u1+x12∣+1√π)+x52e−2x2+15∣v1+x12∣. |
Obviously, F,G:(0,+∞)×(−∞,+∞)×(−∞,+∞)→(0,+∞) are continuous and nondecreasing with respect to the second and the third variables on (0,+∞). That is, (H1) holds. Next,
F1(x,u,v)=xβ(1+γ)−1F(x,(1+x−β(1+γ))u,(1+x−β(1+γ))v)=e−x3∣u∣+xe−2x2+110ln(∣v∣+1). |
We choose ω1(u)=∣u∣∈C((0,+∞),(0,+∞)), ω2(v)=ln(∣v∣+1)∈C((0,+∞),(0,+∞)), and φ1(x)=e−x3,ψ1(x)=xe−2x2+110∈L1(0,+∞). Then,
∣F1(x,u,v)∣≤φ1(x)ω1(∣u∣)+ψ1(x)ω2(∣v∣), (0,+∞)×(−∞,+∞)×(−∞,+∞). |
Similarly, for
F2(x,u,v)=xβ(1+γ)−1G(x,(1+x−β(1+γ))u,(1+x−β(1+γ))v)=xe−2x2+1arctan(∣u∣+1√π)+xe−2x2+15∣v∣, |
we choose ~ω1(u)=arctan(∣u∣+1√π)∈C((0,+∞),(0,+∞)), ~ω2(v)=∣v∣∈C((0,+∞),(0,+∞)), and φ2(x)=xe−2x2+1,ψ2(x)=xe−2x2+15∈L1(0,+∞). Then,
∣F2(x,u,v)∣≤φ2(x)~ω1(∣u∣)+ψ2(x)~ω2(∣v∣), (0,+∞)×(−∞,+∞)×(−∞,+∞). |
That is, (H2) holds. Therefore, from Theorem 5.1, (6.3) has two positive solutions (u∗,v∗) and (w∗,z∗) with (0,0)≤(u∗(x),v∗(x)),(w∗(x),z∗(x))≤((1+x12)Υ1,(1+x12)Υ2), where Υ1+Υ2≤Υ, and Υ satisfies
95.58191.86Υ−0.69arctan(Υ+0.56)≥136. |
This paper studies the Erdélyi-Kober fractional coupled system (1.1), where the variable is in an infinite interval. We give some proper conditions and set a special Banach space. We obtain the existence of at least one positive solution for (1.1) by using the Guo-Krasnosel'skii fixed point theorem, and we get the existence of at least two positive solutions for (1.1) by using the monotone iterative technique. Our methods and results are different from ones in [18]. Moreover, we give three examples to show the plausibility of our main results. For future work, we intend to use other fixed point theorems to solve some Erdélyi-Kober fractional differential equations.
The authors declare they have not used artificial intelligence (AI) tools in the creation of this article.
This paper is supported by the Fundamental Research Program of Shanxi Province (202303021221068).
The authors declare that they have no competing interests.
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