In this paper, we focus on a class of Hadamard type fractional differential system involving Hadamard type fractional derivatives on an infinite interval. By utilizing the monotone iterative technique and Banach's contraction mapping principle, some explicit monotone iterative sequences for approximating the extreme positive solutions and the unique positive solution for the system are constructed.
Citation: Yaohong Li, Jiafa Xu, Honglin Luo. Approximate iterative sequences for positive solutions of a Hadamard type fractional differential system involving Hadamard type fractional derivatives[J]. AIMS Mathematics, 2021, 6(7): 7229-7250. doi: 10.3934/math.2021424
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In this paper, we focus on a class of Hadamard type fractional differential system involving Hadamard type fractional derivatives on an infinite interval. By utilizing the monotone iterative technique and Banach's contraction mapping principle, some explicit monotone iterative sequences for approximating the extreme positive solutions and the unique positive solution for the system are constructed.
Fractional derivative extends the classical integer order derivative to an arbitrary order case. Fractional order differential equations can better describe various phenomenon than integer order differential equations in many complex and widespread fields of engineering and science such as biology, physics, finance, electrical circuits, signal processing, control theory, and diffusion processes, there has been a rapid growth in the number of fractional differential equations from both theoretical and applied perspectives, see [1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18] and references cited therein.
Note that most of the results on the current works are based on Riemann-Liouville type and Caputo type fractional differential equations in the past ten years. Hadamard type fractional derivative is first introduced in 1892 [19], which contains logarithmic function of arbitrary exponent in the kernel of integral appearing in its definition. Hadamard type integrals arise in the formulation of many problems in mechanics such as in fracture analysis. For details and applications of Hadamard type fractional derivative and integral, see [3,20,21,22]. Recently, more and more scholars pay special attention to Hadamard type fractional differential equations on the finite interval [23,24,25,26,27,28]. For example, by applying Leray-Schauder's alternative and Banach's contraction principle, Ahmad and Ntouyas [29] established the existence and uniqueness of solutions for a coupled system of nonlinear fractional differential equations with a fully Hadamard type integral boundary conditions:
{HDαu(t)=f(t,u(t),v(t)), 1<t<e, 1<α≤2,HDβv(t)=g(t,v(t),u(t)), 1<t<e, 1<β≤2,u(1)=0, u(e)=HIru(σ1)=1Γ(r)∫σ11(logσ1−logs)r−1u(s)dss,v(1)=0, v(e)=HIru(σ2)=1Γ(r)∫σ21(logσ2−logs)r−1v(s)dss, | (1.1) |
where γ>0,1<σ1,σ2<e,HD(⋅) are the Hadamard type fractional derivative and HIr is the Hadamard type fractional integral of order r, f,g:[1,e]×R×R are given continuous functions.
In [30] by means of comparison principle and the monotone iterative technique combined with the method of upper and lower solutions, Yang investigated the extremal iterative solutions for the following coupled system of nonlinear Hadamard type fractional differential equations:
{(HDαa+x)(t)=f(t,x(t),y(t)), 0<α≤1, a<t≤b,(HDαa+y)(t)=g(t,x(t),y(t)), 0<α≤1, a<t≤b,(HJ1−αa+x)(a+)=x∗, (HJ1−αa+y)(a+)=y∗, | (1.2) |
where f,g∈C([a,b]×R×R,R),HDαa+ and HJαa+ are the left-sided Hadamard type fractional derivative and Hadamard type fractional integral of order α, respectively.
On the other hand, some authors have also focused on the existence of solutions for Hadamard type fractional differential equations on the infinite intervals, see[31,32,33,34,35,36] and the references quoted therein. In another study [37], by applying standard fixed point theorems, Tariboon et al. obtained the existence of positive solutions of the Hadamard type fractional differential system with coupled integral boundary conditions:
{HDpx(t)+f(t,x(t),y(t))=0, 1<p≤2,t∈[1,+∞),HDqy(t)+g(t,x(t),y(t))=0, 1<q≤2,t∈[1,+∞),x(0)=0, HDp−1x(+∞)=∑mi=1λiHIαiy(η),y(0)=0, HDq−1y(+∞)=∑nj=1σjHIβjx(ξ), | (1.3) |
In [38] Zhang and Liu focused on a class of Hadamard type fractional differential equation with nonlocal boundary conditions on an infinite interval:
{HDα1+x(t)+a(t)f(t,x(t))=0, 2<α≤3,t∈(1,+∞),x(0)=x′(0)=0, HDα−11+x(+∞)=∑mi=1αiHIβi1+x(η)+b∑nj=1σjx(ξj), | (1.4) |
where HDα1+,HIβi1+ are the Hadamard type fractional derivative of order α and the Hadamard type fractional integral of order βi>0 (i=1,2,3,⋯,m),1<η<ξ1<ξ2<⋯<ξn. b,αi,σj≥0 (i=1,2,3,⋯,m;j=1,2,3,⋯,n) are given constants satisfy certain prior conditions. By using various fixed point methods, the authors not only obtained the existence and uniqueness of solutions, but also the iterative sequences of approximate solutions.
Motivated by the mentioned results above, a nature and meaningful question is if we know the existence of solution for the following Hadamard type fractional differential system (1.5), how can we seek it? This idea lead us to develop the research of approximate sequences of positive solutions for the following Hadamard type fractional differential system with Hadamard type fractional integral boundary conditions:
{HDα1u(t)+f1(t,u(t),v(t),HDα1−1u(t),HDα2−1v(t))=0, 1<α1≤2,t∈J,HDα2v(t)+f2(t,v(t),u(t),HDα1−1u(t),HDα2−1v(t))=0, 1<α2≤2,t∈J,u(0)=0,HDα1−1u(+∞)=∑m1i=1λ1iHIβ1iu(η1),η1∈J,v(0)=0,HDα2−1v(+∞)=∑m2i=1λ2iHIβ2iv(η2),η2∈J | (1.5) |
where J=[1,+∞),HDαj,HIβji are the common Hadamard type fractional derivative of order αj and the Hadamard type fractional integral of order βji>0,fj∈C(J×R×R×R×R,R+),λji>0 are given constants and satisfy Ωj=Γ(αj)−∑mji=1λjiΓ(αj)Γ(αj+βji)(logηj)αj+βji−1>0,j=1,2;i=1,2,⋯,mj,mj∈N+.
In this paper, we emphasize that the nonlinearity terms fj of the system (1.5) involve multiple unknown functions and the lower-order Hadamard type fractional derivative of multiple unknown functions. By utilizing the monotone iterative method, we establish some explicit monotone iterative sequences for approximating the extreme positive solutions and the unique positive solution, which are more valuable and interesting than just constructing the existence of solutions. Further we extend the iterative methods that are often used in a single equation to the system which is different from [31,34,38,39,40,41,42]. Finally we give some examples to verify the application of main results.
First we recall some Hadamard type fractional calculus definitions and lemmas that are helpful to the proof of main results.
Definition 2.1 (see [1]). The Hadamard type fractional derivative of order q for a integrable function g:[1,∞)→R is given by
HDqg(t)=1Γ(n−q)(tddt)n∫t1(logt−logs)n−q−1g(s)dss,n−1<q<n, |
where n=[q]+1, [q] denotes the integer part of the real number q and log(⋅)=loge(⋅).
Definition 2.2 (see [1]). The Hadamard type fractional integral of order q for a integrable function g is given by
HIqg(t)=1Γ(q)∫t1(logt−logs)q−1g(s)dss,q>0, |
provided the integral exists.
Lemma 2.1 (see [1,32]). If a,α,β>0, then
(HDαa(logt−loga)β−1)(x)=Γ(β)Γ(β−α)(logx−loga)β−α−1. |
Lemma 2.2 Let hj∈C[1,∞) with 0<∫∞1hj(s)dss<∞ and Ωj>0,j=1,2, then the following Hadamard type fractional differential system with Hadamard type fractional integral boundary conditions
{HDα1u(t)+h1(t)=0,1<α1≤2,t∈J,HDα2v(t)+h2(t)=0,1<α2≤2,t∈J,u(0)=0,HDα1−1u(+∞)=∑m1i=1λ1iHIβ1iu(η1),v(0)=0,HDα2−1v(+∞)=∑m2i=1λ2iHIβ2iv(η2), | (2.1) |
has a unique solution:
{u(t)=∫+∞1G1(t,s)h1(s)dss,v(t)=∫+∞1G2(t,s)h2(s)dss, | (2.2) |
where
Gj(t,s)=gj(t,s)+mj∑i=1λji(logt)αj−1ΩjΓ(αj+βji)gji(ηj,s),j=1,2, | (2.3) |
and
gj(t,s)=1Γ(αj){(logt)αj−1−(logt−logs)αj−1,1≤s≤t<+∞,(logt)αj−1,1≤t≤s<+∞, | (2.4) |
gji(ηj,s)={(logηj)αj+βji−1−(logηj−logs)αj+βji−1,1≤s≤ηj<+∞,(logηj)αj+βji−1,1≤ηj≤s<+∞. | (2.5) |
Proof. Utilizing Lemmas 2.5 of [32], we can derive directly the above results.
Remark 2.1 Applying definition 2.1 of Hadamard type fractional derivative and Lemma 2.1, from (2.2), (2.3), (2.4) and (2.5), by a simple computation, one can obtain
{HDα1−1u(t)=∫+∞1G∗1(t,s)h1(s)dss,HDα2−1v(t)=∫+∞1G∗2(t,s)h2(s)dss, |
where
G∗j(t,s)=k(t,s)+mj∑i=1λjiΓ(αj)ΩjΓ(αj+βji)gji(ηj,s),j=1,2, | (2.6) |
and
k(t,s)={0,1≤s≤t<+∞,1,1≤t≤s<+∞. | (2.7) |
For convenience, we introduce the following notations:
Λj=1Γ(αj)+mj∑i=1λjiΓ(αj)ΩjΓ(αj+βji)(logηj)αj+βji−1, Ξj=1+mj∑i=1λjiΓ(αj)ΩjΓ(αj+βji)(logηj)αj+βji−1,j=1,2. |
Lemma 2.3 (see [32]). The Green's function Gj(t,s) defined by (2.3) has the following properties:
(A1): Gj(t,s)≥0 and Gj(t,s) are continuous for all (t,s)∈J×J,j=1,2;
(A2): Gj(t,s)1+(logt)αj≤Λj for all (t,s)∈J×J,j=1,2.
Remark 2.2 The Green's function Gj(t,s) and G∗j(t,s) defined by (2.3) and (2.6) still have the following properties:
(B1): Gj(t,s)≤Λj(logt)αj−1 for (t,s)∈J×J,j=1,2;
(B2): 0≤G∗j(t,s)≤Ξj for (t,s)∈J×J,j=1,2.
Proof. From (2.4) and (2.5), it is obvious that
gj(t,s)≤(logt)αj−1Γ(αj), gji(ηj,s)≤(logηj)αj+βji−1,(t,s)∈J×J, |
then
Gj(t,s)≤Λj(logt)αj−1,(t,s)∈J×J, |
so (B1) holds. And from (2.6) and (2.7), it is easy to that (B2) holds.
Lemma 2.4 (see [32,33]). Let U⊂X be a bounded set. Then U is a relatively compact in X if the following conditions hold:
(ⅰ) For any u∈U,u(t)1+(logt)α−1 and HDα−1u(t) are equicontinuous on any compact interval of J;
(ⅱ) For any ε>0, there is a constant C=C(ε)>1 such that |u(t1)1+(logt1)α−1−u(t2)1+(logt2)α−1|<ε and |HDα−1u(t1)−Dα−1u(t2)|<ε for any t1,t2≥C and u∈U.
Next we present some assumptions that will play an important role in subsequent discussion.
(C1) Ωj=Γ(αj)−mj∑i=1λjiΓ(αj)Γ(αj+βji)(logηj)αj+βji−1>0 and fj(t,0,0,0,0)≢0,∀t∈J, j=1,2;
(C2) There exist some nonnegative integrable functions aj0(t),ajk(t) defined on J and some constants 0<γjk<1 satisfy
|fj(t,u1,u2,u3,u4)|≤aj0(t)+4∑k=1ajk(t)|uk|γjk,∀t∈J, uk∈R,j=1,2, k=1,2,3,4, |
and
∫+∞1aj0(t)dtt=a∗j0<+∞,∫+∞0aj1(t)[1+(logt)αj−1]γj1dtt=a∗j1<+∞, |
∫+∞1aj2(t)[1+(logt)αj−1]γj2dtt=a∗j2<+∞,∫+∞1aj3(t)dtt=a∗j3<+∞, |
∫+∞1aj4(t)dtt=a∗j4<+∞,j=1,2; |
(C3) Functions fj are nondecreasing with respect to the second, third, fourth and last variables on J,j=1,2;
(C4) There exist some nonnegative integrable functions bjk(t)(j=1,2, k=1,2,3,4) defined on J satisfy
|fj(t,u1,u2,u3,u4)−fj(t,ˉu1,ˉu2,ˉu3,ˉu4)|≤4∑k=1bjk(t)|uk−ˉuk|,∀t∈J,uk,ˉuk∈R, |
and
∫+∞1bj1(t)[1+(logt)αj−1]dtt=b∗j1<+∞,∫+∞1bj2(t)[1+tαj−1]dtt=b∗j2<+∞, |
∫+∞1bj3(t)dtt=b∗j3<+∞,∫+∞1bj4(t)dtt=b∗j4<+∞,∫+∞1|fj(t,0,0,0,0)|dtt=ϱj<+∞. |
In this paper, we will use two Banach spaces which are define by
X={u∈C(J,R),HDα1−1u∈C(J,R)|supt∈J|u(t)|1+(logt)α1−1<+∞,supt∈J|HDα1−1u(t)|<+∞} |
equipped with the norm ‖u‖X=max{‖u‖1,‖HDα1−1u‖}, where ‖u‖1=supt∈J|u(t)|1+(logt)α1−1 and ‖HDα1−1u‖=supt∈J|HDα1−1u(t)|, and
Y={v∈C(J,R),HDα2−1v∈C(J,R)|supt∈J|v(t)|1+(logt)α2−1<+∞,supt∈J|HDα2−1v(t)|<+∞} |
equipped with the norm ‖u‖Y=max{‖v‖2,‖HDα2−1v‖}, where ‖v‖2=supt∈J|v(t)|1+tα2−1 and ‖HDα2−1v‖=supt∈J|HDα2−1v(t)|. Then the space (X,‖⋅‖X) and (Y,‖⋅‖Y) are two Banach spaces which can be shown similarly to Lemma 2.7 of the literature [32]. Moreover, the product space (X×Y,‖⋅‖X×Y) is also a Banach space with the norm
‖⋅‖X×Y=max{‖u‖X,‖v‖Y}. |
Lemma 3.1 If assumption (C2) holds, then for any (u,v)∈X×Y,
∫+∞1|fi(t,u(t),v(t),HDα1−1u(t),HDα2−1v(t))|dtt≤a∗j0+4∑k=1a∗jk||(u,v)||γikX×Y, j=1,2. |
Proof. For any (u,v)∈X×Y, by assumption (C2), one can obtain
|fj(t,u(t),v(t),HDα1−1u(t),HDα2−1v(t))|≤aj0(t)+aj1(t)|u(t)|γj1+aj2(t)|v(t)|γj2+aj3(t)|HDα1−1u(t))|γj3+aj4(t)|HDα2−1v(t))|γj4≤aj0(t)+aj1(t))[1+(logt)α1−1]γj1|u(t)|γj1[1+(logt)α1−1]γj1+aj2(t))[1+(logt)α2−1]γj2|v(t)|γj2[1+(logt)α2−1]γj2+aj3(t)|HDα1−1u(t)|γj3+aj4(t)|HDα2−1v(t)|γj4≤aj0(t)+aj1(t))[1+(logt)α1−1]γj1||u||γj1X+aj2(t))[1+(logt)α2−1]γj2||v||γj2Y+aj3(t)||u||γj3X+aj4(t)||v||γj4Y, j=1,2. |
Thus we have
∫+∞1|fj(t,u(t),v(t),HDα1−1u(t),HDα2−1v(t))|dtt≤∫+∞1(aj0(t)+aj1(t))[1+(logt)α1−1]γj1||u||γj1X+aj2(t))[1+(logt)α2−1]γj2||v||γj2Y+aj3(t)||u||γj3X+aj4(t)||v||γj4Y)dtt≤a∗j0+a∗j1||u||γj1X+a∗i2||v||γj2Y+a∗j3||u||γj3X+a∗j4||v||γj4Y≤a∗j0+4∑k=1a∗jk||(u,v)||γjkX×Y, j=1,2. |
Lemma 3.2 If assumption (C4) holds, then for any (u,v)∈X×Y,
∫+∞1|fj(t,u(t),v(t),HDα1−1u(t),HDα2−1v(t))|dtt≤4∑k=1b∗jk||(u,v)||X×Y+ϱj, j=1,2. |
Proof. For any (u,v)∈X×Y, by assumption (C4), one can obtain
|fj(t,u(t),v(t),HDα1−1u(t),HDα2−1v(t))|=|fj(t,u(t),v(t),HDα1−1u(t),HDα2−1v(t))−fj(t,0,0,0,0)+fi(t,0,0,0,0)|≤|fj(t,u(t),v(t),HDα1−1u(t),HDα2−1v(t))−fj(t,0,0,0,0)|+|fj(t,0,0,0,0)|≤bj1(t)[1+(logt)α1−1]|u(t)|[1+(logt)α1−1]+bj2(t)[1+(logt)α2−1]|v(t)|[1+(logt)α2−1]+bj3(t)|HDα1−1u(t)|+bj4(t)|HDα2−1v(t)|+|fj(t,0,0,0,0)|≤bj1(t)[1+(logt)α1−1]||u||X+bj2(t)[1+(logt)α2−1]||v||Y+bj3(t)||u||X+bj4(t)||v||Y+|fj(t,0,0,0,0)|, j=1,2. |
Thus we have
∫+∞1|fj(t,u(t),v(t),HDα1−1u(t),HDα2−1v(t))|dtt≤b∗j1||u||X+b∗j2||v||Y+b∗j3||u||X+b∗j4||v||Y+ϱj≤4∑k=1b∗jk||(u,v)||X×Y+ϱj, j=1,2. |
Define two cones P1={u∈X|u(t)≥0,HDα1−1u(t)≥0,t∈J} and P2={v∈Y|v(t)≥0,HDα2−1v(t)≥0,t∈J}, then P1×P2⊂X×Y is also a cone by P1×P2={(u,v)∈X×Y|u(t)≥0,v(t)≥0,HDα1−1u(t)≥0,HDα2−1v(t)≥0,t∈J}.
From Lemma 2.2, we can know that the system (2.2) is equivalent to the following system of Hammerstein-type integral equations:
(u(t)v(t))=(∫+∞1G1(t,s)f1(u,v)(s)dss∫+∞1G2(t,s)f2(u,v)(s)dss):=(T1(u,v)(t)T2(u,v)(t)), for (u,v)∈P1×P2,t∈J, | (3.1) |
and for convenience, we set
f1(u,v)(s)=f1(s,u(s),v(s),HDα1−1u(s),HDα2−1v(s)),f2(u,v)(s)=f2(s,u(s),v(s),HDα1−1u(s),HDα2−1v(s)). |
Therefore one can define an operator T:P1×P2→P1×P2 as follows:
T(u,v)(t)=(T1,T2)(u,v)(t), for (u,v)∈P1×P2,t∈J. | (3.2) |
By Remark 2.1, one can also define
(HDα1−1T1(u,v)(t)HDα2−1T2(u,v)(t))=(∫+∞1G∗1(t,s)f1(u,v)(s)dss∫+∞1G∗2(t,s)f2(u,v)(s)dss), for (u,v)∈P1×P2,t∈J. | (3.3) |
Therefore, if (u,v)∈P1×P2/(0,0) is a fixed point of the operator T, then (u,v) is a positive solution for the Hadamard type fractional differential system (1.5). It is obvious that the system (1.5) has a positive solution if and only if the operator equation (u,v)=T(u,v) has a positive fixed point in P1×P2, where T is given as (3.2). Next we will directly consider the existence of fixed points of the operator T.
Lemma 3.3 If assumption (C1), (C2) and (C3) hold, then the operator T:P1×P2→P1×P2 is completely continuous.
Proof. Due to Gj(t,s)≥0,G∗j(t,s)≥0 and fj≥0, we have Tj(u,v)(t)≥0,HDαj−1Tj(u,v)(t)≥0, for any (u,v)∈P1×P2, t∈J, j=1,2, so it is easy to know T:P1×P2→P1×P2.
Next we show in four steps that the operator T:P1×P2→P1×P2 is completely continuous.
Step 1 Take U={(u,v)|(u,v)∈P1×P2,||(u,v)||X×Y≤M}. For any (u,v)∈U, by Lemma 2.3, Lemma 3.1 and Remark 2.2, one can obtain
||T1(u,v)||1=supt∈J|∫+∞1G1(t,s)1+(logt)α1−1f1(u,v)(s)dss|≤Λ1∫+∞1|f1(u,v)(s)|dss≤Λ1(a∗10+4∑k=1a∗1k||(u,v)||γ1kX×Y)<∞ | (3.4) |
and
||HDα1−1T1(u,v)||=supt∈J|∫∞1G∗1(t,s)f1(u,v)(s)dss|≤Ξ1∫+∞1|f1(u,v)(s)|dss≤Ξ1(a∗10+4∑k=1a∗1k||(u,v)||γ1kX×Y)<∞. | (3.5) |
Thus
||T1(u,v)||X≤max{Λ1,Ξ1}(a∗10+4∑k=1a∗1kMγ1k). |
Similarly
||T2(u,v)||Y≤max{Λ2,Ξ2}(a∗20+4∑k=1a∗2kMγ2k). |
Then
||T(u,v)||X×Y=max{‖T1(u,v)‖X,‖T2(u,v)‖Y}≤max{Λ1,Ξ1,Λ2,Ξ2}max(a∗10+4∑k=1a∗1kMγ1k,a∗20+4∑k=1a∗2kMγ2k)<∞. |
which implies that TU is uniformly bounded for any (u,v)∈U.
Step 2 Let I⊂J be any compact interval. Then, for all t1,t2∈I,t2>t1 and (u,v)∈U, we have
|T1(u,v)(t2)1+(logt2)α1−1−T1(u,v)(t1)1+(logt)α1−11|≤|∫+∞1(G1(t2,s)1+(logt2)α1−1−G1(t1,s)1+(logt1)α1−1)f1(u,v)(s)dss|≤∫+∞1|G1(t2,s)1+(logt2)α1−1−G1(t1,s)1+(logt1)α1−1||f1(u,v)(s)|dss. | (3.6) |
Noticing that G1(t,s)/1+(logt)α1−1 is uniformly continuous for any (t,s)∈I×I. Moreover the function G1(t,s)/1+(logt)α1−1 is only associated with t for s≥t, which implies that G1(t,s)/1+(logt)α1−1 is uniformly continuous on I×(J∖I). That is, for all s∈J and t1,t2∈I,∀ϵ>0,∃δ(ϵ)>0 if |t1−t2|<δ such that
|G1(t2,s)1+(logt2)α1−1−G1(t1,s)1+(logt1)α1−1|<ϵ. | (3.7) |
By Lemma 3.1, for all (u,v)∈U, we have
∫+∞1|f1(u,v)(s)|dss≤a∗10+4∑k=1a∗1kMγ1k<∞. | (3.8) |
For all t1,t2∈I,t2>t1 and (u,v)∈U, together (3.6), (3.7) and (3.8) mean that
∀ϵ>0,∃δ(ϵ)>0suchthatif|t1−t2|<δthen|T1(u,v)(t2)1+(logt2)α1−1−T1(u,v)(t1)1+(logt)α1−11|<ϵ. |
That is, T1(u,v)(t)/1+(logt)α1−1 is equicontinuous on I.
Note that
HDα1−1T1(u,v)(t)=∫+∞1G∗1(t,s)f1(u,v)(s)ds |
and function G∗1(t,s)∈C(J×J) is independent of t, which implies that HDα1−1T1(u,v)(t) is equicontinuous on I.
In the same way, one can easily show that T2(u,v)(t)/1+(logt)α2−1 and Dα2−1T2(u,v)(t) are equicontinuous. Hence T1 and T2 are equicontinuous on I. Then the operator T is equicontinuous for all (u,v)∈U on any compact interval I of J.
Step 3 Now we prove the operator T is equiconvergent at +∞. Due to
limt→+∞Gj(t,s)1+(logt)αj−1=1Γ(αj)+mj∑i=1λjiΓ(αj)ΩjΓ(αj+βji)gji(ηj,s)≤1Γ(αj)+mj∑i=1λjiΓ(αj)ΩjΓ(αj+βji)(logηj)αj+βji−1<+∞, j=1,2, |
one can infer that for any ϵ>0, there exists a constant C=C(ϵ)>0, for any t1,t2≥C and s∈J, such that
|Gj(t2,s)1+(logt2)αj−1−Gj(t1,s)1+(logt1)αj−1|<ϵ, j=1,2, |
with the help of Lemma 3.1 and (3.6), which mean that Tj(u,v)(t)/1+(logt)αj−1(j=1,2) are equiconvergent at +∞. Meanwhile function G∗j(t,s)(j=1,2) are independent of t, one can easily show that HDαj−1Tj(u,v)(t)(j=1,2) are equiconvergent at +∞.
From Step 1, Step 2 and Step 3, Lemma 2.4 holds. So the operator T is relatively compact in P1×P2.
Step 4 Finally we prove that the operator T:P1×P2→P1×P2 is continuous. Set (un,vn),(u,v)∈P1×P2 and (un,vn)→(u,v)(n→∞). So ||(un,vn)||X×Y<+∞,||(u,v)||X×Y<+∞. Similar to (3.4) and (3.5), one has
||T1(un,vn)||1=supt∈J|∫+∞0G1(t,s)1+(logt)α1−1f1(un,Vn)(s)dss|≤Λ1[a∗10+4∑k=1a∗1k||(un,vn)||γ1kX×Y], |
and
||HDα1−1T1(un,vn)||=supt∈J|∫+∞0G∗1(t,s)f1(un,vn)(s)ds|≤Ξ1[a∗10+4∑k=1a∗1k||(un,vn)||γ1kX×Y]. |
Via the Lebesgue dominated convergence theorem and continuity of function f1, we know
limn→∞∫+∞1G1(t,s)1+(logt)α1−1f1(un,vn)(s)dss=∫+∞1G1(t,s)1+(logt)α1−1f1(u,v)(s)dss, |
and
limn→∞∫+∞1G∗1(t,s)f1(un,vn)(s)dss=∫∞1G∗1(t,s)f1(u,v)(s)dss. |
Again
‖T1(un,vn)−T1(u,v)‖1≤supt∈J∫+∞1G1(t,s)1+(logt)α1−1|f1(un,vn)(s)−f1(u,v)(s)|dss→0, n→∞, |
and
‖HDα1−1T1(un,vn)−HDα1−1T1(u,v)‖1≤supt∈J∫+∞1K∗1(t,s)|f1(un,vn)(s)−f1(u,v)(s)|dss→0, n→∞. |
Therefore, as n→∞,
‖T1(un,vn)−T1(u,v)‖X=max{‖T1(un,vn)−T1(u,v)‖1,‖HDα1−1T1(un,vn)−HDα1−1T1(u,v)‖}→0. |
This implies that the operator T1 is continuous. At the same way, one can obtain than the operator T2 is continuous. That is, the operator T is continuous.
Summarize all of the above discussions, one can infer that the operator T:P1×P2→P1×P2 is completely continuous. So the proof of Lemma 3.3 is completed.
For convenience, we set
Υ=max{Λ1,Λ2,Ξ1,Ξ2}. |
Define a partial order over the product space:
(u1v1) ≥ (u2v2) |
if u1(t)≥u2(t),v1(t)≥v2(t),HDα1−1u1(t)≥HDα1−1u2(t),HDα2−1v1(t)≥HDα2−1v2(t),t∈J.
Theorem 3.1 If assumption (C1), (C2) and (C3) hold, then the system (1.5) exist two positive solutions (u∗,v∗) and (w∗,z∗) satisfying 0≤‖(u∗,v∗)‖X×Y≤R and 0≤‖(w∗,z∗)‖X×Y≤R, where R is a positive preset constant. Moreover, there exist limn→∞(un,vn)=(u∗,v∗) and limn→∞(wn,zn)=(w∗,z∗), where (un,vn) and (wn,zn) are given by the following monotone iterative sequences
(un(t)vn(t))=(T1(un−1,vn−1)(t)T2(un−1,vn−1)(t)),n=1,2,…, with (u0(t)v0(t))=(R(logt)α1−1R(logt)α2−1) | (3.9) |
and
(wn(t)zn(t))=(T1(wn−1,zn−1)(t)T2(wn−1,zn−1)(t)),n=1,2,…, with (w0(t)z0(t))=(00). | (3.10) |
In addition
(w0(t)z0(t))≤(w1(t)z1(t))≤⋯≤(wn(t)zn(t))≤⋯≤(w∗z∗)≤⋯≤(u∗v∗)≤⋯≤(un(t)vn(t)) |
≤⋯≤(u1(t)v1(t))≤(u0(t)v0(t)) | (3.11) |
and
(HDα1−1w0(t)HDα2−1z0(t))≤(HDα1−1w1(t)HDα2−1z1(t))≤⋯≤(HDα1−1wn(t)HDα2−1zn(t))≤⋯≤(HDα1−1w∗HDα2−1z∗)≤⋯≤ |
(HDα1−1u∗HDα2−1v∗)≤⋯≤(HDα1−1un(t)HDα2−1vn(t))≤⋯≤(HDα1−1u1(t)HDα2−1v1(t))≤(HDα1−1u0(t)HDα2−1v0(t)). | (3.12) |
Proof. First, Lemma 3.3 means the fact that T(P1×P2)⊂P1×P2 for any (u,v)∈P1×P2,t∈J.
Next, for 0≤γ1k,γ2k<1(k=1,2,3,4), set
R≥max{5a∗10Υ,5a∗20Υ,(5Υa∗1k)1/(1−γ1k),(5Υa∗2k)1/(1−γ2k),k=1,2,3,4}, |
and UR={(u,v)∈P1×P2:||(u,v)||X×Y≤R}. For any (u,v)∈UR, similar to (3.4) and (3.5), one can obtain
||T1(u,v)||1≤Λ1[a∗10+4∑k=1a∗1k||(u,v)||γ1kX×Y]≤Υ[a∗10+4∑k=1a∗1kRγ1k]≤R |
and
|HDα1−1T1(u,v)||≤Ξ1[a∗10+4∑k=1a∗1k||(u,v)||γ1kX×Y]≤Υ[a∗10+4∑k=1a∗1kRγ1k]≤R. |
This implies that ||T1(u,v)||X≤R for all (u,v)∈UR. In the same way, ||T2(u,v)||Y≤R. Consequently one has
||T(u,v)||X×Y=max{‖T1(u,v)‖X,‖T2(u,v)‖Y}≤R. |
That is, T(UR)⊂UR.
Via the complete continuity of the operator T, we present the sequences (un,vn) and (wn,zn) by (un,vn)=T(un−1,vn−1),(wn,zn) = T(wn−1,zn−1) for n=1,2,⋯. In virtue of (3.9) and (3.10), it is obvious that (u0(t),v0(t)), (w0(t),z0(t))∈UR.
Due to T(UR)⊂UR, it is easy to see that (un,vn),(wn,zn)∈T(UR) for n=1,2,⋯. Thus we just 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 (1.5).
For t∈J,(un,vn)∈UR, from Lemma 2.2 and (3.9), one has
u1(t)=T1(u0,v0)(t)=∫+∞1G1(t,s)f1(u0,v0)(s)dss≤Λ1[a∗10+4∑k=1a∗1kRγ1k](logt)α1−1≤R(logt)α1−1=u0(t) |
and
v1(t)=T2(u0,v0)(t)=∫+∞1G2(t,s)f2(u0,v0)(s)dss≤Λ2[a∗20+4∑k=1a∗2kRγ2k](logt)α2−1≤R(logt)α2−1=v0(t), |
that is
(u1(t)v1(t))=(T1(u0,v0)(t)T2(u0,v0)(t))≤(R(logt)α1−1R(logt)α2−1)=(u0(t)v0(t)). | (3.13) |
Next we consider the monotonicity of the Hadamard type fractional derivative of (u,v). By (3.13) we have
HDα1−1u1(t)=HDα1−1T1(u0,v0)(t)=∫+∞1G∗1(t,s)f1(u0,v0)(s)dss≤Ξ1[a∗10+4∑k=1a∗1kRγ1k]≤R=HDα1−1u0(t),HDα2−1v1(t)=HDα2−1T2(u0,v0)(t)=∫+∞1G∗2(t,s)f2(u0,v0)(s)dss≤Ξ2[a∗20+4∑k=1a∗2kRγ2k]≤R=HDα2−1v0(t), |
that is
(HDα1−1u1(t)HDα2−1v1(t))=(HDα1−1T1(u0,v0)(t)HDα2−1T2(u0,v0)(t))≤(RR)=(HDα1−1u0(t)HDα2−1v0(t)) | (3.14) |
Then, by (3.13) and (3.14), for any t∈J, via the monotonicity conditions (C3) of functions fj(j=1,2), we do the second iteration
(u2(t)v2(t)) = (T1(u1,v1)(t)T2(u1,v1)(t)) ≤ (T1(u0,v0)(t)T2(u0,v0)(t)) = (u1(t)v1(t)), |
(HDα1−1u2(t)HDα2−1v2(t)) = (HDα1−1T1(u1,v1)(t)HDα2−1T2(u1,v1)(t)) ≤ (HDα1−1T1(u0,v0)(t)HDα2−1T2(u0,v0)(t)) = (HDα1−1u1(t)HDα2−1v1(t)). |
For t∈J, by method of induction, the sequences {(un,vn)}∞n=0 satisfy
(un+1(t)vn+1(t)) ≤ (un(t)vn(t)), (HDα1−1un+1(t)HDα2−1vn+1(t)) ≤ (HDα1−1un(t)HDα2−1vn(t)). |
With the help of iterative sequences (un+1,vn+1)=T(un,vn) and the complete continuity of the operator T, one can easily infer that (un,vn)→(u∗,v∗) and T(u∗,v∗)=(u∗,v∗).
For the sequences {(wn,zn)}∞n=0, we employ a similar discussion. For t∈J, we have
(w1(t)z1(t)) = (T1(w0,z0)(t)T2(w0,z0)(t)) = (∫+∞1G1(t,s)f1(w0,z0)(s)dss∫+∞1G2(t,s)f2(w0,z0)(s)dss) ≥ (00) = (w0(t)z0(t)), |
(HDα1−1w1(t)HDα2−1z1(t)) = (HDα1−1T1(w0,z0)(t)HDα2−1T2(w0,z0)(t)) = (∫+∞1G∗1(t,s)f1(w0,z0)(s)dss∫+∞1G∗2(t,s)f1(w0,z0)(s)dss) ≥ (00) = (HDα1−1w0(t)HDα2−1z0(t)). |
Using the the monotonicity condition (C3) of functions fj, one has
(w2(t)z2(t)) = (T1(w1,z1)(t)T2(w1,z1)(t)) ≥ (T1(w0,z0)(t)T2(w0,z0)(t)) = (w1(t)z1(t)), |
(HDα1−1w2(t)HDα2−1z2(t)) = (HDα1−1T1(w1,z1)(t)HDα2−1T2(w1,z1)(t)) ≥ (HDα1−1T1(w0,z0)(t)HDα2−1T2(w0,z0)(t)) = (HDα1−1w1(t)HDα2−1z1(t)). |
Analogously, for n=0,1,2,… and t∈J, one has
(wn+1(t)zn+1(t)) ≥ (wn(t)zn(t)), (HDα1−1wn+1(t)HDα2−1zn+1(t)) ≥ (HDα1−1wn(t)HDα2−1zn(t)). |
In virtue of the iterative sequences (wn+1,zn+1)=T(wn,zn) and the complete continuity of the operator T, it is also easy to conclude that (wn,zn)→(w∗,z∗) and T(w∗,z∗)=(w∗,z∗). Finally we demonstrate that (u∗,v∗) and (w∗,z∗) are the minimal and maximal positive solutions of the system (1.5). Suppose that (ξ(t),η(t)) is any positive solution of the Hadamard type fractional differential system (1.5), then T(ξ(t),η(t))=(ξ(t),η(t)) and
(w0(t)z0(t)) = (00) ≤ (ξ(t)η(t)) ≤ (Rtα1−1Rtα2−1) = (u0(t)v0(t)), |
(HDα1−1w0(t)HDα2−1z0(t)) ≤ (HDα1−1ξ(t)HDα2−1η(t)) ≤ (HDα1−1u0(t)HDα2−1v0(t)). |
Using the monotone conditions (C3) of the operator T, we obtain
(w1(t)z1(t)) = (T1(w0,z0)(t)T2(w0,z0)(t)) ≤ (ξ(t)η(t)) ≤ (T1(u0,v0)(t)T2(u0,v0)(t)) = (u1(t)v1(t)), |
(HDα1−1w1(t)HDα2−1z1(t)) ≤ (HDα1−1ξ(t)HDα2−1η(t)) ≤ (HDα1−1u1(t)HDα2−1v1(t)). |
Repeating the above process, we have
(wn(t)zn(t)) ≤ (ξ(t)η(t)) ≤ (un(t)vn(t)), |
(HDα1−1wn(t)HDα2−1zn(t)) ≤ (HDα1−1ξ(t)HDα2−1η(t)) ≤ (HDα1−1un(t)HDα2−1vn(t)), |
which combine limn→∞(wn,zn)=(w∗,z∗) and limn→∞(un,un)=(u∗,v∗), we gain the results (3.11) and (3.12).
On the other hand, due to f(t,0,0,0,0)≠0 for all t∈J, we know that (0,0) isn't a solution of the Hadamard type fractional differential system (1.5). From (3.11) and (3.12), it is clear that (w∗,z∗) and (u∗,v∗) are two extreme positive solutions of the system (1.5), which can be constructed via limit of two monotone iterative sequences in (3.9) and (3.10).
Theorem 3.2 If assumption (C1), (C4) and
m=Υmax{4∑k=1b1k,4∑k=1b2k}<1 | (3.15) |
hold, then the system (1.5) has a unique positive solution (x,y) in P1×P2. Further there exists a iterative sequence (un,vn) such that limn→∞(un,vn)=(x,y) is satisfied uniformly on any finite interval of J, where
(un(t)vn(t))=(T1(un−1,vn−1)(t)T2(un−1,vn−1)(t)),n=1,2,⋯. | (3.16) |
Moreover there exists an error estimate for the approximation sequence
||(un,vn)−(x,y)||X×Y=mn1−m||(u1,v1)−(u0,v0)||X×Y,n=1,2,⋯. | (3.17) |
Proof. Take
r≥Υϱ/(1−m), |
where m is defined by (3.15) and ϱ=max{ϱ1,ϱ2}, ϱj (j=1,2) are defined by the aussumption (C4).
First we show that TUr⊂Ur, where Ur={(u,v)∈P1×P2,||(u,v)||X×Y≤r}. For any (u,v)∈Ur, by Lemma 3.2 and Remark 2.2, we have
||T1(u,v)||1≤Λ1(4∑k=1b∗1kr+ϱ1) |
and
||HDα1−1T1(u,v)||≤Ξ1(4∑k=1b∗1kr+ϱ1), |
which implies
||T1(u,v)||X≤Υ(4∑k=1b∗1kr+ϱ1)≤mr+Υϱ1, ∀(u,v)∈Ur. |
Similar
||T2(u,v)||Y≤Υ(4∑k=1b∗2kr+ϱ2)≤mr+Υϱ2, ∀(u,v)∈Ur. |
So one has
||T(u,v)||X×Y≤mr+Υϱ≤r,∀(u,v)∈Ur. |
Now we demonstrate that operator T is a contraction. For any (u1,v1),(u2,v2)∈Ur, by assumption (C4), we have
||T1(u1,v1)−T1(u2,v2)||1≤supt∈J∫+∞1G1(t,s)1+(logt)α1−1|f1(u1,v1)(s)−f1(u2,v2)(s)|dss≤Λ1∫+∞1[b11(s)(1+(logs)α1−1)|u1(s)−u2(s)|1+(logs)α1−1+b12(s)(1+(logs)α2−1)|v1(s)−v2(s)|1+(logs)α2−1+b13(s)|HDα1−1u1(s)−HDα1−1u2(s)|]+b14(s)|HDα2−1v1(s)−HDα2−1v2(s)|]dss≤Λ14∑k=1b∗1k||(u1,v1)−(u2,v2)||X×Y |
and
||HDα1−1T1(u1,v1)−HDα1−1T1(u2,v2)||≤supt∈J∫+∞0G∗1(t,s)|f1(u1,v1)(s)−f1(u2,v2)(s)|ds≤Ξ14∑k=1b∗1k||(u1,v1)−(u2,v2)||X×Y, |
which implies
||T1(u1,v1)−T1(u2,v2)||X≤Υ4∑k=1b∗1k||(u1,v1)−(u2,v2)||X×Y. | (3.18) |
In the same way, one can obtain
||T2(u1,v1)−T2(u2,v2)||Y≤Υ4∑k=2b∗2k||(u1,v1)−(u2,v2)||X×Y. | (3.19) |
By (3.18) and (3.19), we gain
||T(u1,v1)−T(u2,v2)||X×Y≤m||(u1,v1)−(u2,v2)||X×Y,∀(u1,v1),(u2,v2)∈Ur. | (3.20) |
Due to m<1, then operator T is a contraction. With the help of the Banach fixed-point theorem, T has a unique fixed point (x,y) in Ur. That is, the system (1.5) has a unique positive solution (x,y).
Further, for any (u0,v0)∈Ur,‖(un,vn)−(x,y)‖X×Y→0 as n→∞, where un=T1(un−1,vn−1),vn=T2(un−1,vn−1),n=1,2,⋯. From (3.20), we have
||(un,vn)−(un−1,vn−1)||X×Y≤mn−1||(u1,v1)−(u0,v0)||X×Y, |
and
||(un,vn)−(uj,vj)||X×Y≤||(un,vn)−(un−1,vn−1)||X×Y+||(un−1,vn−1)−(un−2,vn−2)||X×Y+⋯+||(uj+1,vj+1)−(uj,vj)||X×Y≤mn(1−mj−n)1−m||(u1,v1)−(u0,v0)||X×Y. | (3.21) |
Taking j→+∞ on both sides of (3.21), one can obtain
||(un,vn)−(x,y)||X×Y≤mn1−m||u1−u0||X×Y. |
So the proof of Theorem 3.2 is completed.
Example 4.1 Consider the following Hadamard type fractional differential system
{−HD1.8u(t)=e−2t+e−t|u(t)|0.1[1+(logt)0.8]0.1+e−2t|v(t)|0.3[1+(logt)0.5]0.3+t|HD0.8u(t)|0.22(4+t)2+t|HD0.5v(t)|0.45(1+t2),−HD1.5v(t)=t−5+e−3t|u(t)|0.2[1+(logt)0.8]0.2+e−4t|v(t)|0.4[1+(logt)0.5]0.4+t|HD1.5u(t)|0.25(1+t2)+t|HD0.5v(t)|0.6(9+t)2,u(1)=0, HD0.8u(+∞)=0.2HI1.8u(2.5)+0.1HI2.8u(2.5),v(1)=0,HD0.5v(+∞)=0.1HI1.5u(1.5)+0.3HI2.5u(1.5)+2HI3.5u(1.5), | (4.1) |
where α1=1.8, α2=1.5, β11=1.8, β12=2.8, λ11=0.2, λ12=0.1, β21=1.5, β22=2.5, β23=3.5, λ21=0.1, λ22=0.3, λ23=2, η1=2.5, η2=1.5 and
f1(t,u1,u2,u3,u4)=e−2t+e−t|u1|0.1[1+(logt)0.8]0.1+e−2t|u2|0.3[1+(logt)0.5]0.3+t|u3|0.22(4+t)2+t|u4|0.45(1+t2), |
f2(t,u1,u2,u3,u4)=t−5+e−3t|u1|0.2[1+(logt)0.8]0.2+e−4t|u2)|0.4[1+(logt)0.5]0.4+t|u3|0.25(1+t2)+t|u4|0.6(9+t)2. |
Here γ11=0.1, γ12=0.3, γ13=0.2, γ14=0.4, γ21=0.2, γ22=0.4, γ23=0.2, γ23=0.2, γ24=0.6.
We find that f1(t,0,0,0,0)≢0,f1(t,0,0,0,0)≢0, for ∀t∈J and Ω1=Γ(α1)−2∑i=1λ1iΓ(α1)Γ(α1+β1i)(logη1)α1+β1i−1≈0.885609>0, Ω2=Γ(α2)−3∑i=1λ2iΓ(α2)Γ(α2+β2i)(logη2)α2+β2i−1≈0.872852>0. So assumption (C1) holds.
Noting that
|f1(t,u1,u2,u3,u4)|≤e−2t+e−t|u1|0.1[1+(logt)0.8]0.1+e−2t|u2|0.3[1+(logt)0.5]0.3+t|u3|0.22(4+t)2+t|u4|0.45(1+t2)=a10(t)+a11(t)|u1|0.1+a12(t)|u2|0.3+a13(t)|u3|0.2+a14(t)|u4|0.4, |
|f2(t,u1,u2,u3,u4)|≤t−5+e−3t|u1|0.2[1+(logt)0.8]0.2+e−4t|u2)|0.4[1+(logt)0.5]0.4+t|u3|0.25(1+t2)+t|u4|0.6(9+t)2=a20(t)+a21(t)|u1|0.2+a22(t)|u2|0.2+a23(t)|u3|0.2+a24(t)|u4|0.6 |
and
a∗10=∫+∞1a10(t)dtt=∫+∞1e−2tdtt≤∫+∞1e−2tdt=12e2<∞,a∗11=∫+∞1a11(t)[1+(logt)0.8]0.1dtt=∫+∞1e−t[1+(logt)0.8]0.1[1+(logt)0.8]0.1dtt≤1e<∞,a∗12=∫+∞1a12(t)[1+(logt)0.5]0.3dtt=∫+∞1e−2t[1+(logt)0.5]0.3[1+(logt)0.5]0.3dtt≤12e2<∞,a∗13=∫+∞1a13(t)dtt=∫+∞1t2(4+t)2dtt=110<∞,a∗14=∫+∞1a14(t)dtt=∫+∞1t5(1+t2)dtt=π10<∞,a∗20=∫+∞1a20(t)dtt=∫+∞1t−5dtt=15<∞, |
a∗21=∫+∞1a11(t)[1+(logt)0.8]0.2dtt=∫+∞1e−3t[1+(logt)0.8]0.2[1+(logt)0.8]0.2dtt≤13e3<∞,a∗22=∫+∞1a12(t)[1+(logt)0.5]0.4dtt=∫+∞1e−4t[1+(logt)0.5]0.3[1+(logt)0.5]0.4dtt≤14e4<∞,a∗23=∫+∞1a13(t)dtt=∫+∞1t5(1+t2)dtt≤π10<∞,a∗24=∫+∞1a14(t)dtt=∫+∞1t(9+t)2dtt≤110<∞, |
which imply that assumption (C2) holds.
From the expression of function fj, we can infer that fj is increasing respect to the variables u1,u2,u3,u4,∀t∈J,j=1,2. Hence assumption (C3) is also satisfied. By Theorem 3.1, it follows that the system (4.1) have two pairs of positive solutions (u∗,v∗) and (w∗,z∗), which can be constructed via the limit of two explicit monotone iterative sequences in (3.11) and (3.12).
Example 4.2 Consider the following Hadamard type fractional differential system
{−HD1.8u(t)=e−2t+e−t|u(t)|[1+(logt)0.8]+e−2t|v(t)|0.3[1+(logt)0.5]+t|HD0.8u(t)|2(4+t)2+t|HD0.5v(t)|5(1+t2),−HD1.5v(t)=t−5+e−3t|u(t)|[1+(logt)0.8]+e−4t|v(t)|0.4[1+(logt)0.5]+t|HD1.5u(t)|5(1+t2)+t|HD0.5v(t)|(9+t)2,u(1)=0, HD0.8u(+∞)=0.2HI1.8u(2.5)+0.1HI2.8u(2.5),v(1)=0,HD0.5v(+∞)=0.1HI1.5u(1.5)+0.3HI2.5u(1.5)+2HI3.5u(1.5), | (4.2) |
where α1=1.8, α2=1.5, β11=1.8, β12=2.8, λ11=0.2, λ12=0.1, β21=1.5, β22=2.5, β23=3.5, λ21=0.1, λ22=0.3, λ23=2, η1=2.5, η2=1.5 and
f1(t,u1,u2,u3,u4)=e−2t+e−t|u1|1+(logt)0.8+e−2t|u2|[1+(logt)0.5]+t|u3|2(4+t)2+t|u4|5(1+t2), |
f2(t,u1,u2,u3,u4)=t−5+e−3t|u1|1+(logt)0.8+e−4t|u2)|[1+(logt)0.5]+t|u3|5(1+t2)+t|u4|(9+t)2. |
Same to example (4.1), it is easy to verify that assumption (C1) hods.
Observing that
|f1(t,u1,u2,u3,u4)−f1(t,¯u1,¯u2,¯u3,¯u4)|≤e−t1+(logt)0.8|u1−¯u1|+e−2t1+(logt)0.5|u2−¯u2|+t2(4+t)2|u3−¯u3|+t5(1+t2)|u4−¯u4|=b11(t)|u1−¯u1|+b12(t)|u2−¯u2|+b13(t)|u3−¯u3|+b14(t)|u4−¯u4|,|f2(t,u1,u2,u3,u4)−f2(t,¯u1,¯u2,¯u3,¯u4)|≤e−3t1+(logt)0.8|u1−¯u1|+e−4t1+(logt)0.5|u2−¯u2|+t5(1+t2)|u3−¯u3|+t(9+t)2|u4−¯u4|=b21(t)|u1−¯u1|+b22(t)|u2−¯u2|+b23(t)|u3−¯u3|+b24(t)|u4−¯u4|, |
by a same computation as example (4.1), one can obtain
b∗11=∫+∞1b11(t)[1+(logt)0.8]dtt≤1e<∞, b∗12=∫+∞1b12(t)[1+(logt)0.5]dtt≤12e2<∞,b∗13=∫+∞1b13(t)dtt=110<∞,b∗14=∫+∞1a14(t)dtt=π10<∞,b∗21=∫+∞1b21(t)[1+(logt)0.8]dtt≤13e3<∞,b∗22=∫+∞1b22(t)[1+(logt)0.5]dtt≤14e4<∞,b∗23=∫+∞1b13(t)dtt≤π10<∞,b∗24=∫+∞1b14(t)dtt≤110<∞,λ1=∫+∞1f1(t,0,0,0,0)dt=∫+∞1e−2tdtt≤12e2<∞,λ2=∫+∞1f2(t,0,0,0,0)dt=∫+∞1t−5dtt=15<∞, |
which show that assumption (C4) holds. By direct computation, one can obtain that Λ1=1.125161, Λ2=1.143496, Ξ1=1.051490, Ξ2=1.015251, Υ=1.143496,
m=Υmax{4∑k=1b∗1k,4∑k=1b∗2k}≤1.143496×max{0.849706,0.435334}=0.971635<1. |
Hence all presupposed conditions of Theorem 3.2 are satisfied. Then the system (4.2) has a unique positive solution (x,y), which can be constructed via the limit of the iterative sequence in (3.16).
In this paper, we consider a class of Hadamard type fractional differential system. By the aid of monotone iterative technique and Banach's contraction mapping principle, under certain nonlinear and linear increasing conditions, we construct some explicit monotone iterative sequences for approximating the extreme positive solutions and the unique positive solutions. Our results generalize iterative solution of a single equation to the case of a system, and the nonlinear term contains Hadamard type fractional derivative which can be used more widely. Further work is still needed including discussions on iterative solution for Hadamard type fractional differential system with coupling integral condition and additional studies on iterative solution for impulsive Hadamard type fractional differential system.
This work is supported by Foundation of Anhui Provincial Education Department (Grant No. KJ2020A0735 and 2020jxtd286), the Foundation of Suzhou University (Grant No. 2019XJZY02, 2019XJSN03 and szxy2020xxkc03), the China Postdoctoral Science Foundation (Grant No. 2019M652348), Natural Science Foundation of Chongqing (Grant No. cstc2020jcyj-msxmX0123), Technology Research Foundation of Chongqing Educational Committee (Grant No. KJQN201900539 and KJQN202000528) and the Key Laboratory Open Issue of School of Mathematical Science, Chongqing Normal University (Grant No. CSSXKFKTM202003).
The authors declare no conflict of interest.
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