Existence of positive
solutions has been studied by A. Babakhani and V. Daftardar-Gejji (2003) in case
of multiterm nonautonomous fractional differential equations with constant
coefficients. In the present paper we discuss existence of positive solutions in
case of multiterm fractional differential equations of finite delay with polynomial
coefficients.

1. Introduction

In last 30 years, the theory of ordinary differential equations of fractional order has become a new important branch (see, e.g., [1–5] and the references therein). Numerous applications of such equations have been presented [3–10]. Existence of positive solution of fractional ordinary differential equations has been well investigated for fractional functional differential equations [1, 6, 11–14]. Ye et al. [6] have addressed the question of existence of positive solutions for the nonlinear fractional functional differential equation
Dα[x(t)-x(0)]=x(t)f(t,xt),t∈(0,T],x(t)=ϕ(t)≥0,t∈[-w,0],
by using the sub- and supersolution method, where 0<α<1, Dα is the standard Riemann-Liouville fractional derivative, ϕ∈C and f:[0,T]×C→ℝ+ is continuous, as usual, C=C([-w,0],ℝ+) is the space of continuous function from [-w,0] to ℝ+, w>0, equipped with the sup norm:
∥ϕ∥=max-w≤Θ≤0|ϕ(t)|,
and xt denotes the function in C defined by
xt(θ)=x(t+θ),-w≤θ≤0.
They require that the nonlinearity f(t,xt) is increasing in xt for each t∈[0,T].

As a pursuit of this in the present paper, we deal with the existence of positive solutions in the case of multiterm differential equations with polynomial coefficients of the fractional type:
ℒ(D)[x(t)-x(0)]=f(t,xt),t∈(0,T],x(t)=ϕ(t)≥0,t∈[-w,0],
where
ℒ(D)=Dαn-∑j=1n-1pj(t)Dαn-j,0<α1<⋯<αn<1,pj(t)=∑k=0Njajktk,pj(2m)(t)≥0,pj(2m+1)(t)≤0,m=0,1,…,[Nj2],j=1,2,…,n-1,
and Dαj is the standard Riemann-Liouville fractional derivative, T>0, w>0, ϕ∈C=C([-w,0],ℝ+) and f:I×C→ℝ+ is a given continuous function, I=[0,T].

2. Preliminaties

Let E be a real Banach space with a cone K. K introduces a partial order ≤ in E in the following manner [13]:
x≤yify-x∈K.

Definition 2.1 (see [<xref ref-type="bibr" rid="B12">15</xref>]).

For x,y∈E the order interval 〈x,y〉 is defined as
〈x,y〉={z∈E:x≤z≤y}.

Definition 2.2 (see [<xref ref-type="bibr" rid="B12">15</xref>]).

A cone K is called normal, if there exists a positive constant r such that f,g∈K and ϑ≺f≺g implies ∥f∥≤r∥g∥, where ϑ denotes the zero element of K.

Definition 2.3 (see [<xref ref-type="bibr" rid="B13">16</xref>, <xref ref-type="bibr" rid="B14">17</xref>]).

Let f:[a,b]→ℝ, and f∈L1[a,b]. The left-sided Riemann-Liouville fractional integral of f of order α is defined as
Iaαf(x)=1Γ(α)∫ax(x-t)α-1f(t)dt,α>0,x∈[a,b].

Definition 2.4 (see [<xref ref-type="bibr" rid="B13">16</xref>, <xref ref-type="bibr" rid="B14">17</xref>]).

The left-sided Riemann-Liouville fractional derivative of a function f:[a,b]→ℝ is defined as
Daαf(x)=Dm[Iam-αf(x)],x∈[a,b],
where m=[α]+1, Dm=dm/dtm. We denote D0α by Dα and I0α by Iα. If the fractional derivative Daαf(x) is integrable, then [16, page 71]
Iaα(Daβf(x))=Iaα-βf(x)-[Ia1-βf(x)]x=axα-1Γ(α),0<β≤α<1.

If f is continuous on [a,b], then [Ia1-βf(x)]x=a=0 and (2.5) reduces to
Iaα(Daβf(x))=Iaα-βf(x),0<β≤α<1.

Proposition 2.5.

Let y be continuous on [0,T], T>0 and let n be a nonnegative integer, then
Iα(tnx(t))=∑k=0n(-αk)[Dktn][Iα+kx(t)]=∑k=0n(-αk)n!tn-k(n-k)!Iα+kx(t),
where
(-αk)=(-1)kΓ(α+1)k!Γ(α)=(-1)k(αk)=Γ(1-α)Γ(k+1)Γ(1-α-k).

The proof of the above proposition can be found in [17, page 53].

Corollary 2.6.

Let x∈C[0,T], T>0 and pj(t)=∑k=0Njajktk, Nj∈ℕ∪{0}, j=1,2,…,n, then
Iα(∑j=1npj(t)x(t))=∑j=1n∑k=0Nj∑r=0kajk(-αr)k!tk-r(k-r)![Iα+rx(t)].

Theorem 2.7 (see [<xref ref-type="bibr" rid="B19">10</xref>]).

Let E be a Banach space with C⊆E closed and convex. Assume that U is a relatively open subset of C with 0∈U and F:U¯→C is a continuous and compact map. Then either

F has a fixed point in U¯, or

there exists u∈∂U and λ∈(0,1) with u=λF(u).

3. Existence of Positive Solution

In this section, we discuss the existence of positive solutions for (1.4). Using (2.5), (2.6), and Corollary 2.6, (1.4) is equivalent to the integral equation
x(t)={x(0)+ℐ[x(t)-x(0)]+Iαnf(t,xt),t∈(0,T],ϕ(t),t∈[-w,0],
where
ℐ=∑j=1n-1∑k=0Nj∑r=0kajk(-αnr)k!tk-r(k-r)!Iαn-αn-j+r.
Let y(·):[-w,T]→[0,+∞) be the function defined by
y(t)={ϕ(0),t∈I,ϕ(t)≥0,t∈[-w,0],
then y0=ϕ, for each z∈C(I,ℝ) with z(0)=0, we denote by z̅ the function define by
z̅(t)={z(t),t∈I,0,t∈[-w,0].
We can decompose x(·) as x(t)=z̅(t)+y(t), t∈[-w,T], which implies xt=z̅t+yt, for t∈I. Therefore, (3.1) is equivalent to the integral equation
z(t)=ℐz(t)+Iαnf(t,z̅t+yt),t∈I,
where ℐ is defined (3.2). Set A0={z∈C(I,ℝ):z0=0} and let ∥z∥T be the seminorm in A0 defined by
∥z∥T=∥z0∥+∥z∥=∥z∥=:sup{|z(t)|:t∈I},z∈A0,
and A0 is a Banach space with norm ∥·∥T. Let K be a cone of A0, K={z∈A0;z(t)≥0,t∈I} and
K*={x∈C([-w,T],ℝ+);x(t)=ϕ(t)≥0,t∈[-w,0]}.
Define the operator F:K→K by
Fz(t)=ℐz(t)+Iαnf(t,z̅t+yt),t∈I.

Theorem 3.1.

Suppose that the following conditions hold:

there exist p,q∈C(I,ℝ+) such that f(t,xt)≤p(t)+q(t)∥xt∥, for t∈I, xt∈C, and ∥Iαnp∥=supt∈[0,T]Iαnp(t)<∞, ∥Iαnq∥=supt∈[0,T]Iαnq(t)<∞,

1-ℐ(T)-∥Iαnq∥>0, where
ℐ(T)=∑j=1n-1∑k=0Nj∑r=0k|ajk(-αnr)|k!Tαn-αn-j+k(k-r)!Γ(αn-αn-j+r).

Then (1.4) has at least a positive solution x*∈K*, satisfying ∥x*∥≤max{∥ϕ∥,h}, where
h=λ∥ϕ∥∥Iαnq∥+λ∥Iαnp∥1-λℐ(T)-λ∥Iαnq∥+1.Proof.

We will show that the operator F:K→K is continuous and completely continuous.Step 1.

The operator F:K→K is continuous in view of the continuity of f.

Step 2.

F maps bounded sets into bounded sets in K.

Let G⊂K be bounded; that is, there exists a positive constant l such that ∥z∥T≤l, for all z∈G. For each z∈G, we have
|Fz(t)|≤∑j=1n-1∑k=0Nj∑r=0k|ajk(-αnr)|k!tk-r(k-r)!Iαn-αn-j+r|z(t)|+Iαnf(t,z̅t+yt)≤∑j=1n-1∑k=0Nj∑r=0k|ajk(-αnr)|lk!tαn-αn-j+k(k-r)!Γ(αn-αn-j+r)+Iαnf(t,z̅t+yt)≤lℐ(T)+Iαn{p(t)+q(t)∥z̅t+yt∥},
where ℐ(T) is defined in (3.9). It follows that
∥Fz∥T≤lℐ(T)+∥Iαnp∥+l∥Iαnq∥+∥ϕ(t)∥∥Iαnq∥.
Hence FG is bounded.

Step 3.

F maps bounded sets into equicontinuous sets of K.

We will show that FG is equicontinuous. For each z∈G, t1,t2∈I and t1<t2, then for given ϵ>0, choose
δ=min{[ϵC(j,k,r)4]1/(αn-αn-j+r),[ϵΓ(αn+1)4(∥p∥+∥q∥(l+∥ϕ∥))]1/αn},
where j=1,2,…,n-1, k=0,1,…,Nj, r=0,1,…,k,
C(j,k,r)=(k-r)!∑i=1n-1(Ni+1)(Ni+2)×Γ(αn-αn-j+r+1)|ajk(-αnr)|lηk!
and η=max{1,TNj,j=1,2,…,n-1}. If |t1-t2|<δ,
|Fz(t1)-Fz(t2)|=|∑j=1n-1∑k=0Nj∑r=0k|ajk(-αnr)|k!t1k-r(k-r)!Γ(αn-αn-j+r)∫0t1(t1-s)αn-αn-j+r-1z(s)ds-∑j=1n-1∑k=0Nj∑r=0k|ajk(-αnr)|k!t2k-r(k-r)!Γ(αn-αn-j+r)∫0t2(t2-s)αn-αn-j+r-1z(s)ds|+1Γ(αn)|∫0t1(t1-s)αn-1f(s,z̅s+ys)ds-∫0t2(t2-s)αn-1f(s,z̅s+ys)ds|≤|∑j=1n-1∑k=0Nj∑r=0k|ajk(-αnr)|k!t2k-r(k-r)!Γ(αn-αn-j+r)∫0t1(t2-s)αn-αn-j+r-1z(s)ds-∑j=1n-1∑k=0Nj∑r=0k|ajk(-αnr)|k!t1k-r(k-r)!Γ(αn-αn-j+r)∫0t1(t1-s)αn-αn-j+r-1z(s)ds|+∑j=1n-1∑k=0Nj∑r=0k|ajk(-αnr)|k!t2k-r(k-r)!Γ(αn-αn-j+r)∫t1t2(t2-s)αn-αn-j+r-1|z(s)|ds+∥f∥∞Γ(αn){∫0t1[(t2-s)αn-1-(t1-s)αn-1]ds+∫t1t2(t2-s)αn-1ds}≤∑j=1n-1∑k=0Nj∑r=0k|ajk(-αnr)|lk!Tk-r(k-r)!Γ(αn-αn-j+r)×∫0t2{(t2-s)αn-αn-j+r-1-(t1-s)αn-αn-j+r-1}ds+∥p∥+∥q∥(l+∥ϕ∥)Γ(αn){∫0t1[(t2-s)αn-1-(t1-s)αn-1]ds+∫t1t2(t2-s)αn-1ds}≤∑j=1n-1∑k=0Nj∑r=0k|ajk(-αnr)|2lk!η(t2-t1)αn-αn-j+r(k-r)!Γ(αn-αn-j+r+1)+2(∥p∥+∥q∥(l+∥ϕ∥))(t2-t1)αnΓ(αn+1)=∑j=1n-1∑k=0Nj∑r=0k|ajk(-αnr)|2lk!ηδαn-αn-j+r(k-r)!Γ(αn-αn-j+r+1)+2(∥p∥+∥q∥(l+∥ϕ∥))δαnΓ(αn+1)≤ϵ2+ϵ2=ϵ.
Hence FG is equicontiuous. The Arzela-Ascoli theorem implies that F(G)¯ is compact and F:K→K is continuous and completely continuous.

Step 4.

We now show that there exists an open set U⊆K with z≠λF(z) for λ∈(0,1) and z∈∂U. Let z∈K be any solution of z=λFz, λ∈(0,1), where F is given by (3.8); since F:K→K is continuous and completely continuous, we have
z(t)=λFz(t)≤λ∑j=1n-1∑k=0Nj∑r=0k|ajk(-αnr)|k!tk-r(k-r)!Γ(αn-αn-j+r)×∫0t(t-s)αn-αn-j+r-1z(s)ds+λΓ(αn)∫0t(t-s)αn-1f(s,z̅s+ys)ds≤λ∑j=1n-1∑k=0Nj∑r=0k|ajk(-αnr)|∥z∥k!tk-r(k-r)!Γ(αn-αn-j+r)∫0t(t-s)αn-αn-j+r-1ds+λΓ(αn)∫0t(t-s)αn-1[p(s)+q(s)∥z̅s+ys∥]ds≤λ{∥z∥ℐ(T)+∥z∥Iαnq(t)+∥ϕ∥Iαnq(t)+Iαnp(t)}.
So
∥z∥(1-λℐ(T)-λ∥Iαnp∥)≤λ∥ϕ∥∥Iαnq∥+λ∥Iαnp∥.
Now, by (3.10) and (3.17), we know that any solution z of (3.8) satisfies ∥z∥≠h; let
U={z∈K;∥z∥<h}.
Therefore, Theorem 2.7 guarantees that (3.1) has at least a positive solution z∈U̅. Hence, (1.4) has at least a positive solution x*∈K*, satisfying ∥x*∥≤max{∥ϕ∥,h} and the proof is complete.

Note that we can complete the above mentioned procedure by using only the continuity of f without condition (1), but with our procedure and details of condition (1) in Theorem 3.1 answers all the questions exist in the following remark.

Remark 3.2.

When f is continuous on (0,T]×C, limt→0+f(t,·)=+∞, (i.e., f is singular at t=0) in (1.4). Suppose ∃σ∈(0,αn], such that tσf(t,xt) is a continuous function on [0,T]×C, then Iαnf(t,xt)=Iαnt-σtσf(t,xt) is continuous on I×C by Lemma 2.1 in [12, page 613]. We also obtain results about the existence to (1.4) by using a nonlinear alternative of Leray-Schauder type. The proof is similar to that of Theorem 3.1 as long as we let

tσf(t,xt)≤p(t)+q(t)∥xt∥, for t∈I, xt∈C, and ∥Iαnt-σp∥<∞, ∥Iαnt-σq∥<∞,

1-ℐ(T)-∥Iαnt-σq∥>0, then (1.4) has at least a positive solution x*∈K*, satisfying ∥x*∥≤max{∥ϕ∥,h}, where
h=λ∥Iαnt-σp∥+λ∥ϕ∥∥Iαnt-σq∥1-λℐ(T)-λ∥Iαnt-σq∥+1.

4. Unique Existence of Solution

In this section, we will give uniqueness of positive solution to (1.4).

Theorem 4.1.

Let f:I×C→ℝ+ be continuous and λ∈L1([0,T],ℝ+) with ∥Iαnλ∥<∞. Further assume

|f(t,u̅t+yt)-f(t,v̅t+yt)|≤λ(t)∥u̅t-v̅t∥, for all u,v∈K, t∈[0,T],

ℐ(T)+∥Iαnλ∥<1.

Then (1.4) has unique solution which is positive, where ℐ(T) is given in (3.9).Proof.

Let u,v∈K. Then we obtain
|Fu(t)-Fv(t)|≤∑j=1n-1∑k=0Nj∑r=0k|ajk(-αnr)|k!tk-r(k-r)!Iαn-αn-j+r|u(t)-v(t)|+Iαn|f(t,u̅t+y+t)-f(t,v̅t+yt)|≤∥u-v∥T{∑j=1n-1∑k=0Nj∑r=0k|ajk(-αnr)|k!tαn-αn-j+k(k-r)!Γ(αn-αn-j+r+1)+Iαnλ(t)}≤∥u-v∥T{∑j=1n-1∑k=0Nj∑r=0k|ajk(-αnr)|k!Tαn-αn-j+k(k-r)!Γ(αn-αn-j+r+1)+Iαnλ(t)},
where F is given in (3.8). Hence,
∥Fu-Fv∥T≤(ℐ(T)+∥Iαnλ∥)∥u-v∥T.
In view of Banach fixed point theorem F has unique fixed point in K, which is the unique positive solution of (2.7) and (1.4) has a unique positive solution in K*.

Remark 4.2.

When λ(t)=L>0, then condition (i) reduces to the Lipschitz condition.

Example 4.3.

Let λ(t)=L>0 and f(t,xt)=Lxt+et=Lx(t-ω)+et, ω>0. Consider the equation
(D1/2-at2D1/4-btD1/6-cD1/8)x=Lx(t-w)+et,t∈(0,64],x(t)=0,t∈[-w,0].
Then (4.3) is equivalent to the integral equation,
x(t)=∑j=13∑k=0Nj∑r=0kajk(-12r)k!tk-r(k-r)!I1/2-α3-j+rx(t)+I1/2(Lx(t-ω)+et).
Here α3=1/2, p1(t)=∑k=02a1ktk=at2, then N1=2, a10=a11=0, a12=a, p2(t)=∑k=01a2ktk=bt, then N2=1, a20=0,a21=b and p3(t)=∑k=00a3ktk=c, then N3=0, a30=c. Hence
x(t)=a10(-120)I1/2-1/4x+a11[(-120)tI1/2-1/4x+(-121)I1/2-1/4+1]+a12[(-120)t2I1/2-1/4x+2(-121)tI1/2-1/4+1x+2(-122)I1/2-1/4+2x]+a20(-120)I1/2-1/6x+a21[(-120)tI1/2-1/6x+(-121)I1/2-1/6+1x]+a30(-120)I1/2-1/8x+LI1/2x(t-ω)+I1/2et.
In view of (2.8) and that Γ(1/2)=π, Γ(-1/2)=-2π and Γ(-3/4)=4π/3 we obtain
x(t)=a[t2I1/4x(t)-tI5/4x(t)+34I9/4x(t)]+b[(1+t)I1/3x(t)-12I4/3x(t)]+cI3/8+LI1/2x(t-ω)+I1/2et.
If |a|≤3/5, |b|≤2/5, |c|≤1/5, 0<L<4/5 in the above equation satisfy the conditions required in Theorem 4.1, the iterated sequence is
x1(t)=I1/2et=t1/2E1,3/2(t),x2(t)=[a(t2I1/4-tI5/4+34I9/4)+b((1+t)I1/3-12I4/3)+cI3/8+LI1/2]x1(t)+x1(t),xn+1(t)=∑k=0n[a(t2I1/4-tI5/4+34I9/4)+b((1+t)I1/3-12I4/3)+cI3/8+LI1/2]n-kx1(t),
for n=1,2,3,…, where Iαx1=tα+1/2E1,α+3/2(t), α>0, x(t)=limn→∞xn(t) is the unique solution, which may not be positive, where Eα,β(t)=∑k=0∞(tk/Γ(αk+β)) is Mittag-Leffler function.

Acknowledgment

The authors thank the referee’s efforts for their remarks and Professor Ferhan Merdivenci Atici, Western Kentucky University, USA, for her regular contact.

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