We investigate the existence of solutions for a Caputo fractional differential equation with periodic boundary condition. Using the positivity of Green's function of the corresponding linear equation, we show the existence of positive solutions by using Krasnosel'skii fixed point theorem. Meanwhile, by using monotone iterative method and lower and upper solutions method, we also discuss the existence of extremal solutions for a special case.
Citation: Xiaoxin Zuo, Weibing Wang. Existence of solutions for fractional differential equation with periodic boundary condition[J]. AIMS Mathematics, 2022, 7(4): 6619-6633. doi: 10.3934/math.2022369
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We investigate the existence of solutions for a Caputo fractional differential equation with periodic boundary condition. Using the positivity of Green's function of the corresponding linear equation, we show the existence of positive solutions by using Krasnosel'skii fixed point theorem. Meanwhile, by using monotone iterative method and lower and upper solutions method, we also discuss the existence of extremal solutions for a special case.
This article is devoted to the following nonlinear fractional differential equation with periodic boundary condition
{cDα0+x(t)−λx(t)=f(t,x(t)), 0<t≤ω,x(0)=x(ω), | (1.1) |
where λ≤0, 0<α≤1 and cDα0+ is Caputo fractional derivative
cDα0+x(t)=1Γ(1−α)∫t0(t−s)−αx′(s)ds. |
Differential equations of fractional order occur more frequently on different research areas and engineering, such as physics, economics, chemistry, control theory, etc. In recent years, boundary value problems for fractional differential equation have become a hot research topic, see [2,3,4,5,6,7,9,10,11,12,13,15,16,17,18,19,20,21,24,25]. In [27], Zhang studied the boundary value problem for nonlinear fractional differential equation
{cDα0+u(t)=f(t,u(t)), 0<t<1,u(0)+u′(0)=0,u(1)+u′(1)=0, | (1.2) |
where 1<α≤2, f:[0,1]×[0,+∞)→[0,+∞) is continuous and cDα0+ is Caputo fractional derivative
cDα0+u(t)=1Γ(2−α)∫t0(t−s)1−αu″(s)ds. |
The author obtained the existence of the positive solutions by using the properties of the Green function, Guo-Krasnosel'skill fixed point theorem and Leggett-Williams fixed point theorem.
Ahmad and Nieto [1] studied the anti-periodic boundary value problem of fractional differential equation
{cDq0+u(t)=f(t,u(t)), 0≤t≤T, 1<q≤2,u(0)=−u(T), cDp0+u(0)=−cDp0+u(T), 0<p<1, | (1.3) |
where f:[0,T]×R→R is continuous. The authors obtained some existence and uniqueness results by applying fixed point principles. The anti-periodic boundary value condition in this article corresponds to the anti-periodic condition u(0)=−u(T),u′(0)=−u′(T) in ordinary differential equation.
In [26], Zhang studied the following fractional differential equation
{Dδ0+u(t)=f(t,u), 0<t≤T,limt→0+t1−αu(t)=u0, | (1.4) |
where 0<δ<1, T>0, u0∈R and Dδ0+ is Riemann-Liouville fractional derivative
Dδ0+u(t)=1Γ(1−δ)ddt∫t0(t−s)−δu(s)ds. |
The author obtained the existence and uniqueness of the solutions by the method of upper and lower solutions and monotone iterative method.
In [7], Belmekki, Nieto and Rodriguez-Lopez studied the following equation
{Dδ0+u(t)−λu(t)=f(t,u(t)),0<t≤1,limt→0+t1−δu(t)=u(1), | (1.5) |
where 0<δ<1, λ∈R, f is continuous. The authors obtained the existence and uniqueness of the solutions by using the fixed point theorem. Cabada and Kisela [8] studied the following equation
{Dδ0+u(t)−λu(t)=f(t,t1−αu(t)),0<t≤1,limt→0+t1−δu(t)=u(1), | (1.6) |
where 0<δ<1, λ≠0(λ∈R), f is continuous. The authors studied the existence and uniqueness of periodic solutions by using Krasnosel'skii fixed point theorem and monotone iterative method. In [7,8], the boundary condition limt→0+t1−δu(t)=u(1) was called as periodic boundary value condition of Riemann-Liouville fractional differential equation, which is different from the periodic condition for ordinary differential equation. The boundary value condition u(0)=u(1) is not suitable for Riemann-Liouville fractional differential equation.
For the ordinary differential equation, the periodic boundary value problem is closely related to the periodic solution. For the Caputo fractional differential equation, the periodic boundary value condition u(0)=u(w) is meaningful. As far as we know, few work involves the periodic boundary value problem for Caputo fractional. The aim of this paper is to show the existence of positive solutions of (1.1) by using Krasnosel'skii fixed point theorem. Meanwhile, we also use the monotone iterative method to study the extremal solutions problem
{cDα0+u(t)=f(t,u(t)),0<t≤ω,u(0)=u(ω). | (1.7) |
The paper is organized as follows. In Section 2, we recall and derive some results on Mittag-Leffler functions. In Section 3, we use the Laplace transform to obtain the solution of a linear problem and discuss some properties of Green's function. In Section 4, the existence of positive solution is studied by using the Krasnosel'skii fixed point theorem. In Section 5, the existence of extremal solutions is proved by utilizing the monotone iterative technique. Section 6 is conclusion of the paper.
A key role in the theory of linear fractional differential equation is played by the well-known two-parameter Mittag-Leffler function
Eα,β(z)=Σ∞k=0zkΓ(αk+β), z∈R, α, β>0. | (2.1) |
We recall and derive some of their properties and relationships summarized in the following.
Proposition 2.1. Let α∈(0,1],β>0,λ∈R and ξ>0. Then it holds
(C1) limt→0+Eα,β(λtα)=1Γ(β),limt→0+Eα,1(λtα)=1.
(C2) Eα,α+1(λtα)=λ−1t−α(Eα,1(λtα)−1).
(C3) Eα,α(λtα)>0, Eα,1(λtα)>0 for all t≥0.
(C4) Eα,α(λtα) is decreasing in t for λ<0and increasing for λ>0 for all t>0.
(C5) Eα,1(λtα) is decreasing in t for λ<0 and increasing forλ>0 for all t>0.
(C6) ∫ξ0tβ−1Eα,β(λtα)dt=ξβEα,β+1(λξα).
Proof. (C1) It is obtained by an immediate calculation from (2.1).
(C2) By (2.1), we get
Eα,1(λtα)=Σ∞k=0(λtα)kΓ(αk+1)=1+λtαΓ(α+1)+(λtα)2Γ(2α+1)+(λtα)3Γ(3α+1)+⋯, |
Eα,α+1(λtα)=Σ∞k=0(λtα)kΓ(αk+α+1)=1Γ(α+1)+λtαΓ(2α+1)+(λtα)2Γ(3α+1)+⋯. |
Hence,
Eα,α+1(λtα)=λ−1t−α(Eα,1(λtα)−1). |
(C3) It follows from [23,Lemma 2.2].
(C4) It follows from [8,Proposition 1].
(C5) By a direct calculation, we get
ddtEα,1(λtα)=λtα−1Eα,α(λtα), |
since α∈(0,1], t>0 and Eα,α(λtα) is positive by Proposition (C3), the assertion is proved.
(C6) It follows from (1.99) of [20].
In this section, we deal with the linear case that f(t,x)=f(t) is a continuous function by mean of the Laplace transform for caputo fractional derivative
(LcDα0+x)(s)=sαX(s)−sα−1x(0), 0<α≤1, | (3.1) |
where L denotes the Laplace transform operator, X(s) denotes the Laplace transform of x(t).
From Lemma 3.2 of [14], we get
(LEα(λtα))(s)=sα−1sα−λ,Re(s)>0,λ∈C,|λs−α|<1, | (3.2) |
and
(Ltβ−1Eα,β(λtα))(s)=sα−βsα−λ,Re(s)>0,λ∈C,|λs−α|<1. | (3.3) |
We do Laplace transform to the equation
cDα0+x(t)−λx(t)=f(t), x(0)=x(ω). | (3.4) |
By (3.1), we obtain
sαX(s)−λX(s)=F(s)+x(0)⋅sα−1, |
X(s)=F(s)sα−λ+sα−1sα−λ⋅x(0), |
where F denotes the Laplace transform of f. By (3.2) and (3.3), we obtain that
x(t)=∫t0(t−s)α−1Eα,α(λ(t−s)α)f(s)ds+x(0)⋅Eα,1(λtα). | (3.5) |
Hence,
x(ω)=x(0)Eα,1(λwα)+∫ω0(ω−s)α−1Eα,α(λ(ω−s)α)f(s)ds=x(0), |
which implies that
x(0)=∫ω0(ω−s)α−1Eα,α(λ(ω−s)α)f(s)ds1−Eα,1(λωα) |
if Eα,1(λωα)≠1. Therefore, if Eα,1(λωα)≠1, the solution of the problem (3.4) is
x(t)=∫ω0(ω−s)α−1Eα,α(λ(ω−s)α)f(s)ds1−Eα,1(λωα)⋅Eα,1(λtα)+∫t0Eα,α(λ(t−s)α)(t−s)1−αf(s)ds=∫t0Eα,1(λtα)Eα,α(λ(ω−s)α)(1−Eα,1(λωα))(ω−s)1−αf(s)ds+∫t0Eα,α(λ(t−s)α)(t−s)1−αf(s)ds+∫ωtEα,1(λtα)Eα,α(λ(ω−s)α)(1−Eα,1(λωα))(ω−s)1−αf(s)ds. |
Theorem 3.1. Let Eα,1(λωα)≠1, the periodic boundary value problem (3.4) has a unique solutiongiven by
x(t)=∫ω0Gα,λ(t,s)f(s)ds, |
where
Gα,λ(t,s)={Eα,1(λtα)Eα,α(λ(ω−s)α)(1−Eα,1(λωα))(ω−s)1−α+Eα,α(λ(t−s)α)(t−s)1−α, 0≤s<t≤ω,Eα,1(λtα)Eα,α(λ(ω−s)α)(1−Eα,1(λωα))(ω−s)1−α, 0≤t≤s<ω. | (3.6) |
Remark 3.2. The unique solution x of (3.4) is continuous on [0,ω].
Lemma 3.3. Let 0<α≤1,λ≠0 and sign(η) denotes the signumfunction. Then
(F1) limt→0+Gα,λ(t,s)=Eα,α(λ(ω−s)α)(1−Eα,1(λωα))(ω−s)1−α for anyfixed s∈[0,ω),
(F2)lims→ω−Gα,λ(t,s)=sign(−λ)⋅∞ for any fixed t∈[0,ω],
(F3)limt→s+Gα,λ(t,s)=∞ for any fixeds∈[0,ω),
(F4)Gα,λ(t,s)>0 for λ<0 and for all t∈[0,ω] and s∈[0,ω),
(F5)Gα,λ(t,s) changes its sign for λ>0 for t∈[0,ω] and s∈[0,ω).
Proof. (F1) When 0≤t≤s<ω, by Proposition 2.1 (C1) we can get (F1).
(F2) When 0≤t≤s<ω, it follows by Proposition 2.1 (C5) that 1−Eα,1(λωα) is positive for λ<0 and negative for λ>0. The unboundedness is implied by continuity of Mittag-Leffler function, Proposition 2.1 (C1) and the relation limt→0+t−r=∞ for r>0.
(F3) When 0≤s<t≤ω, the first term of (3.6) is finite due to the continuity of the involved functions. And by a similar argument as in the previous point of this proof we have the second term tends to infinity.
(F4) It is obtained by the positivity of all involved functions (Proposition 2.1 (C3)) and the inequation 1−Eα,1(λωα)>0 for λ<0.
(F5) When 0≤s<t≤ω, the second term of (3.6) is positive due to
lims→t−(t−s)α−1Eα,α(λ(t−s)α)=+∞, |
and by the positivity of all involved functions (Proposition 2.1 (C3)) we get the proof. When 0≤t≤s<ω, it is obtained by 1−Eα,1(λωα)<0 for λ>0 and the positivity of all involved functions (Proposition 2.1 (C3)).
Proposition 3.4. Let α∈(0,1] and λ<0. Then the Green's function (3.6) satisfies
(K1) Gα,λ(t,s)≥m=:Eα,1(λωα)Eα,α(λωα)|λ|ωEα,α+1(λωα)>0,
(K2)∫ω0Gα,λ(t,s)ds=M=:1|λ| for all t∈[0,ω].
Proof. (K1) For 0≤t≤s<ω, we deduce from Proposition 2.1 (C4),(C5) that Gα,λ has the minimum on the line t=s. Hence,
Gα,λ(t,s)≥Gα,λ(t,t)=Eα,1(λtα)Eα,α(λ(ω−t)α)(1−Eα,1(λωα))(ω−t)1−α≥Eα,1(λωα)Eα,α(λωα)(1−Eα,1(λωα))ω1−α=Eα,1(λωα)Eα,α(λωα)[1−(Eα,α+1(λωα)λωα+1)]ω1−α=Eα,1(λωα)Eα,α(λωα)|λ|ωEα,α+1(λωα). |
For 0≤s<t≤ω, we have
Gα,λ(t,s)≥Eα,1(λωα)Eα,α(λωα)|λ|ωEα,α+1(λωα)+Eα,α(λωα)ω1−α≥Eα,1(λωα)Eα,α(λωα)|λ|ωEα,α+1(λωα). |
(K2) Employing Proposition 2.1, we get
∫ω0Gα,λ(t,s)ds=∫ω0Eα,1(λtα)Eα,α(λ(ω−s)α)(ω−s)1−α(1−Eα,1(λωα))ds+∫t0Eα,α(λ(t−s)α)(t−s)1−αds=Eα,1(λtα)1−Eα,1(λωα)∫ω0(ω−s)α−1Eα,α(λ(ω−s)α)ds+∫t0(t−s)α−1Eα,α(λ(t−s)α)ds=Eα,1(λtα)1−Eα,1(λωα)⋅ωαEα,α+1(λωα)+tαEα,α+1(λtα)=Eα,1(λtα)1−Eα,1(λωα)⋅ωα⋅λ−1ω−α(Eα,1(λωα)−1)+tα⋅λ−1t−α(Eα,1(λtα)−1)=1|λ|, |
which completes the proof.
Let C[0,ω] be the space continuous function on [0,ω] with the norm ‖x‖=sup{|x(t)|:t∈[0,ω]}.
In this section, we always assume that λ<0. Clearly, x is a solution of (1.1) if and only if
x(t)=∫ω0Gα,λ(t,s)f(s,x(s))ds, | (4.1) |
where Gα,λ is Green's function defined in Theorem 3.1.
The following famous Krasnosel'skii fixed point theorem, which is main tool of this section.
Theorem 4.1. [22] Let B be a Banach space, and let P⊂B be a cone. AssumeΩ1,Ω2 two open and bounded subsets of B with0∈Ω1,Ω1⊂Ω2 and letA:P∩(¯Ω2∖Ω1)→P bea completely continuous operator such that one of the followingconditions is satisfied:
(L1) ‖Ax‖≤‖x‖, if x∈P∩∂Ω1, and‖Ax‖≥‖x‖, if x∈P∩∂Ω2,
(L2) ‖Ax‖≥‖x‖, if x∈P∩∂Ω1, and‖Ax‖≤‖x‖, if x∈P∩∂Ω2.
Then, A has at least one fixed point inP∩(¯Ω2∖Ω1).
Proposition 4.2. Assume that there exist 0<r<R,0<c1<c2 such that
f:[0,ω]×[mc1ωMc2r,R]→R is continuous, | (4.2) |
c1≤f(t,u)≤c2, ∀(t,u)∈[0,ω]×[mc1ωMc2r,R]. | (4.3) |
Let P⊂C[0,ω] be the cone
P={x∈C[0,ω]:mint∈[0,ω]x(t)≥mc1ωMc2‖x‖}. |
Then the operator A:¯PR∖Pr→P givenby
Ax(t)=∫ω0Gα,λ(t,s)f(s,x(s))ds | (4.4) |
is completely continuous, where Pl={u∈P:‖u‖<l}.
Proof. Let x∈¯PR∖Pr, then
mc1ωMc2r≤x(t)≤R for all t∈[0,ω]. | (4.5) |
We first show that A is well-defined, i.e. that A:¯PR∖Pr→P. Note that
Ax(t)=∫ω0Gα,λ(t,s)f(s,x(s))ds=∫ω0Eα,1(λtα)Eα,α(λ(ω−s)α)(ω−s)1−α(1−Eα,1(λωα))f(s,x(s))ds +∫t0Eα,α(λ(t−s)α)(t−s)1−αf(s,x(s))ds=kq(ω)Eα,1(λtα)+q(t), | (4.6) |
where k=11−Eα,1(λωα) and
q(t)={∫t0Eα,α(λ(t−s)α)(t−s)1−αf(s,x(s))ds,0<t≤ω,0,t=0. | (4.7) |
Clearly, for t∈(0,ω]
0<q(t)≤c2∫t0Eα,α(λ(t−s)α)(t−s)1−αds≤c2Γ(α)∫t0(t−s)α−1ds=c2Γ(α+1)tα, | (4.8) |
which implies that q is continuous at t=0. On the other hand,
q(t)=∑k<1α−1λk∫t0(t−s)αk+α−1Γ(αk+α)f(s,x(s))ds+∫t0∑k≥1α−1λkΓ(αk+α)(t−s)αk+α−1f(s,x(s))ds=∑k<1α−1λkΓ(αk+α)∫t0uαk+α−1f(t−u,x(t−u))du+∫t0∑k≥1α−1λkΓ(αk+α)(t−s)αk+α−1f(s,x(s))ds=:H1(t)+H2(t), |
where
H1(t)=∑k<1α−1λkΓ(αk+α)∫t0uαk+α−1f(t−u,x(t−u))du,H2(t)=∫t0∑k≥1α−1λkΓ(αk+α)(t−s)αk+α−1f(s,x(s))ds. |
Since
|uαk+α−1f(t−u,x(t−u))|≤c2uαk+α−1,u>0, t∈(0,ω], x∈¯PR/Pr,|λk(t−s)αk+α−1Γ(αk+α)f(s,x(s))|≤λkΓ(αk+α)tαk+α−1c2, 0≤s≤t, x∈¯PR/Pr,∫t0uαk+α−1du<+∞, t∈(0,ω],∑k≥1α−1λktαk+α−1Γ(αk+α)<+∞, t∈(0,ω], |
we obtain that H1∈C[0,ω] and
H2(t)=∑k≥1α−1λkΓ(αk+α)∫t0(t−s)αk+α−1f(s,x(s))ds=:∑k≥1α−1λkΓ(αk+α)uk(t). |
Noting that uk∈C(0,ω]
|uk(t)|≤c2ωαk+ααk+α, t∈(0,ω], ∑λkΓ(αk+α)ωαk+ααk+α<+∞, |
we have H2∈C(0,ω]. Hence, q∈C[0,ω].
Moreover,
mc1ωMc2‖Ax‖=mc1ωMc2sup∫ω0Gα,λ(t,s)f(s,x(s))ds≤mc1ωMsup∫ω0Gα,λ(t,s)ds=mc1ω≤min∫ω0Gα,λ(t,s)f(s,x(s))ds=mint∈[0,ω]Ax(t), | (4.9) |
which means that A:¯PR∖Pr→P.
Next, we show that A is continuous on ¯PR∖Pr. Let xn,x∈¯PR∖Pr and ‖xn−x‖→0. From (4.2), we have ‖f(t,xn(t))−f(t,x(t))‖→0,
‖Axn−Ax‖=supt∈[0,ω]|∫ω0Gα,λ(t,s)(f(s,x(s))−f(s,y(s)))ds|≤supt∈[0,ω]∫ω0Gα,λ(t,s)ds‖f(t,xn(t))−f(t,x(t))‖≤M‖f(t,xn(t))−f(t,x(t))‖→0, |
which implies that A is continuous. From (4.6), we get that Ax(t) is uniformly bounded. Finally, we show that {Ax|x∈¯PR/Pr} is an equicontinuity in C[0,ω]. By (4.6), we have
|Ax(t1)−Ax(t2)|=|∫ω0(Gα,λ(t1,s)−Gα,λ(t2,s))f(s,x(s))ds|≤kq(ω)|Eα,1(λtα1)−Eα,1(λtα2)|+|q(t1)−q(t2)|. |
Since Eα,1(λtα)∈C[0,ω], q(t)∈C[0,ω] are uniformly continuous, |Eα,1(λtα1)−Eα,1(λtα2)| and |q(t1)−q(t2)| tend to zero as |t1−t2|→0. Hence, {Ax(t)|x∈¯PR∖Pr} is equicontinuous in C[0,ω].
Finally, by Arzela-Ascoli theorem, we can obtain that A is compact. Hence, it is completely continuous.
Theorem 4.3. Assume that there exist 0<r<R,0<c1<c2 such that (4.2)and (4.3) hold. Further suppose one of the followingconditions is satisfied
(i)f(t,u)≥Mc2m2ω2c1u, ∀(t,u)∈[0,ω]×[mc1ωMc2r,r],
f(t,u)≤|λ|u, ∀(t,u)∈[0,ω]×[mc1ωMc2R,R],
(ii)f(t,u)≤|λ|u, ∀(t,u)∈[0,ω]×[mc1ωMc2r,r],
f(t,u)≥Mc2m2ω2c1u, ∀(t,u)∈[0,ω]×[mc1ωMc2R,R].
Then (1.1) has at least a positive solution x with r≤‖x‖≤R.
Proof. Here we only consider the case (i). By Proposition 4.2, A:¯PR∖Pr→P is completely continuous. For x∈∂Pr, we have
‖x‖=r, mc1ωMc2r≤x(t)≤r, ∀t∈[0,ω] |
and
Ax(t)≥m∫ω0f(s,x(s))ds≥Mc2mω2c1∫ω0x(s)ds≥r=‖x‖. |
Similarly, if x∈∂PR,
mc1ωMc2R≤x(t)≤R, t∈[0,ω], |
0≤Ax(t)≤∫ω0Gα,λ(t,s)|λ|x(s)ds≤|λ|R∫ω0Gα,λ(t,s)ds=R=‖x‖. |
By Theorem 4.1, there exists x∈¯PR∖Pr such that Ax=x and x is a solution of (1.1). Moreover,
mc1ωMc2r≤x(t)≤R. |
Corollary 4.4. Let c1<c2 be positive reals and f(t,x) satisfy the conditions
(i) c1≤f(t,x)≤c2 for all x≥0,
(ii) f:[0,w]×(0,+∞)→R is a continuous function.
Then problem (1.1) has a positive solution.
Proof. Let 0<r<c21m2ω2Mc2, R>c22M|λ|m1c1ω, then (4.2) and (4.3) are satisfied. Clearly, for (t,u)∈[0,ω]×[mc1ωMc2r,r],
f(t,u)≥c1≥Mc2m2ω2c1r≥Mc2m2ω2c1u |
and for (t,u)∈[0,ω]×[mc1ωMc2R,R],
f(t,u)≤c2≤|λ|mc1ωMc2R≤|λ|u. |
Hence, by Theorem 4.3 (1.1) has at least a positive solution.
Example 4.5. Consider the equation
{cDα0+x(t)−λx(t)=1+x1β(t), 0<x≤ω,x(0)=x(ω), | (4.10) |
where 0<α≤1,β>1 and
Λ={λ<0,|λ|Eα,1(λωα)Eα,α(λωα)≥4Eα,α+1(λωα)}≠∅. |
Choosing c1=1,c2=2,r=110min{1,m2ω22M},R=1. It is easy to check that (4.2) and (4.3) hold. For λ∈Λ,
f(t,u)=1+u1β≥1≥Mc2m2ωc1r≥Mc2m2ωc1u,∀(t,u)∈[0,ω]×[mc1ωMc2r,r], |
f(t,u)≤2≤|λ|⋅mc1ωMc2=|λ|⋅mω2M≤|λ|u,∀(t,u)∈[0,ω]×[mc1ωMc2R,R]. |
Hence, (4.10) has at least one positive solution for λ∈Λ.
In this section, by using the monotone iterative method, we discuss the existence of solutions when λ=0 in (1.1). Firstly, we give the definition of the upper and lower solutions and get monotone iterative sequences with the help of the corresponding linear equation. Finally, we prove the limits of the monotone iterative sequences are solutions of (1.7).
Definition 5.1. Let h,k∈C1[0,ω]. h and k are called lower solution and upper solution of problem (1.7), respectively if h and k satisfy
cDα0+h(t)≤f(t,h(t)), 0<t≤ω, h(0)≤h(ω), | (5.1) |
cDα0+k(t)≥f(t,k(t)), 0<t≤ω, k(0)≥k(ω), | (5.2) |
Clearly, if g the lower solution or upper solution of (1.7), then cDα0+g is continuous on [0,ω].
Lemma 5.2. Let δ∈C[0,ω] with δ≥0 and p∈R with p≤0. Then
{cDα0+z(t)−λz(t)=δ(t), 0<t≤ω,z(ω)−z(0)=p, | (5.3) |
has a unique solution z(t)≥0 for t∈[0,ω], where 0<α≤1,λ<0.
Proof. Let z1, z2 are two solutions of (5.3) and v=z1−z2, then
{cDα0+v(t)−λv(t)=0, 0<t≤ω,v(ω)=v(0). | (5.4) |
Using Theorem 3.1, (5.4) has trivial solution v=0.
By (3.5), we can verify that problem (5.3) has a unique solution
z=∫ω0Gα,λ(t,s)δ(t)ds+p⋅Eα,1(λωα)Eα,1(λωα)−1. |
As consequence, by Proposition 2.1 (C3), (C5) and Lemma 3.3 (F4), we conclude that z(t)≥0. This completes the proof.
Theorem 5.3. Assume that h,k are the lower and upper solutions ofproblem (1.7) and h≤k. Moreover, supposethat f satisfies the following properties:
(M) there is λ<0 such that for all fixed t∈[0,ω], f(t,x)−λx is nondecreasing in h(t)≤x≤k(t),
(J) f:[0,ω]×[h(t),k(t)]→R is a continuous function.
Then there are two monotone sequences {hn} and {kn} are nonincreasingand nondecreasing, respectively with h0=h and k0=ksuch that limn→∞hn=¯h(t), limn→∞kn=¯k(t)uniformly on [0,ω], and ¯h,¯k are the minimal and the maximalsolutions of (1.7) respectively, such that
h0≤h1≤h2≤...≤hn≤¯h≤x≤¯k≤kn≤...≤k2≤k1≤k0 |
on [0,ω], where x is any solution of (1.7) such that h(t)≤x(t)≤k(t)on [0,ω].
Proof. Let [h,k]={u∈C[0,ω]:h(t)≤u(t)≤k(t),t∈[0,ω]}. For any η∈[h,k], we consider the equation
{cDα0+x(t)−λx(t)=f(t,η(t))−λη(t), 0<t≤ω,x(0)=x(ω), |
Theorem 3.1 implies the above problem has a unique solution
x(t)=∫ω0Gα,λ(t,s)(f(s,η(s))−λη(s))ds. | (5.5) |
Define an operator B by x=Bη, we shall show that
(a) k≥Bk,Bh≥h,
(b) B is nondecreasing on [h,k].
To prove (a). Denote θ=k−Bk, we have
cDα0+θ(t)−λθ(t)= cDα0+k(t)− cDα0+Bk(t)−λ(k(t)−Bk(t))≥f(t,k(t))−((f(t,k(t))−λk(t))−λk(t)=0, |
and θ(w)−θ(0)≤0. Since k∈C1[0,ω],
cDα0+k∈C[0,ω], cDα0+Bk∈C[0,ω]. |
By Lemma 5.2, θ≥0, i.e. k≤Bk. In an analogous way, we can show that Bh≥h.
To prove (b). We show that Bη1≤Bη2 if h≤η1≤η2≤k. Let z1=Bη1, z2=Bη2 and z=z2−z1, then by (M), we have
cDα0+z(t)−λz(t)= cDα0+z2(t)− cDα0+z1(t)−λ(z2(t)−z1(t))=f(t,η2(t))−λη2(t)−(f(t,η1(t))−λη1(t))≥0, |
and v(ω)=v(0). By Lemma 5.2, z(t)≥0, which implies Bη1≤Bη2.
Define the sequence {hn}, {kn} with h0=h,k0=k such that hn+1=Bhn,kn+1=Bkn for n=0,1,2,.... From (a) and (b), we have
h0≤h1≤h2≤...≤hn≤kn≤...≤k2≤k1≤k0 |
on t∈[0,ω], and
hn(t)=∫ω0Gα,λ(t,s)(f(s,hn−1(s))−λhn−1(s))ds, |
kn(t)=∫ω0Gα,λ(t,s)(f(s,kn−1(s))−λkn−1(s))ds. |
Therefore, there exist ¯h,¯k such that limn→∞hn=¯h, limn→∞kn=¯k.
Similar to the proof of Proposition 4.2, we can show that B:[h,k]→[h,k] is a completely continuous operator. Therefore, ¯h,¯k are solutions of (1.7).
Finally, we prove that if x∈[h0,k0] is one solution of (1.7), then ¯h(t)≤x(t)≤¯k(t) on [0,ω]. To this end, we assume, without loss of generality, that hn(t)≤x(t)≤kn(t) for some n. From property (b), we can get that hn+1(t)≤x(t)≤kn+1(t),t∈[0,ω]. Since h0(t)≤x(t)≤k0(t), we can conclude that
hn(t)≤x(t)≤kn(t), forall n. |
Passing the limit as n→∞, we obtain ¯h(t)≤x(t)≤¯k(t), t∈[0,ω]. This completes the proof.
Example 5.4. Consider the equation
{cDα0+x(t)=t+1−x2(t), 0<x≤1,x(0)=x(1). | (5.6) |
It easy to check that h=1,k=2 are the low solution and upper solution of (5.6), respectively. Let λ=−10. For all t∈[0,ω],
f(t,u)−λu=t2+1−u2+10u |
is nondecreasing on u∈[1,2] and
f(t,u)=t+1−u2 |
is continuous on [0,ω]×[1,2].
Hence, there exist two monotone sequences {hn} and {kn}, nonincreasing and nondecreasing respectively, that converge uniformly to the extremal solutions of (5.6) on [h,k].
This paper focuses on the existence of solutions for the Caputo fractional differential equation with periodic boundary value condition. We use Green's function to transform the problem into the existence of the fixed points of some operator, and we prove the existence of positive solutions by using the Krasnosel'skii fixed point theorem. On the other hand, the existence of the extremal solutions for the special case of the problem is obtained from monotone iterative technique and lower and upper solutions method. Since the fractional differential equation is nonlocal equation, the process of verifying the compactness of operator is very tedious, and we will search for some better conditions to prove the compactness of the operator A in the follow-up research. Meanwhile, since the existence result for 0<α≤1 is obtained in present paper, we will discuss the existence of solutions for the Caputo fractional differential equation when n−1<α≤n in follow-up research.
The authors declare that there are no conflicts of interest regarding the publication of this paper.
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