# Positive Solutions for Boundary Value Problems of Singular Fractional Differential Equations

Authors: Zhanbing Bai; Weichen Sun; Weihai Zhang
Journal: Abstract and Applied Analysis
 (2013)

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## Abstract

In this paper, by using a fixed point theorem, we investigate the existence of a positive solution to the singular fractional boundary value problemDC0+αu+ft,u,DC0+νu,DC0+μu+gt,u,DC0+νu,DC0+μu=0,  u0=u′0=u′′0=u′′′0=0, where 3<α<4, 0<ν<1, 1<μ<2, DC0+α is Caputo fractional derivative, ft,x,y,z is singular at the value 0 of its arguments x,y,z, and gt,x,y,z satisfies the Lipschitz condition.


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## Body

## 1. Introduction

In recent years, as an extended concept of integral differential equations, fractional differential equations are widely concerned in various fields of science. For examples, see [1–14]. Many results, such as [1, 2, 6, 15, 16], discuss singular fractional boundary value problems.In [1], the authors discuss positive solutions to the singular Dirichlet problem
(1)D0+αu(t)+f(t,u(t),Dμu(t))=0,u(0)=u(1)=0,
where 1<α<2, 0<μ≤α-1 and f is a Carathéodory function on [0,1]×(0,∞)×ℝ. Here, D0+α is the standard Riemann-Liouville fractional derivative. The existence of positive solutions is obtained by the combination of regularization and sequential techniques with the Guo-Krasnosel’skii fixed point theorem on cone.The singular problem(2)D0+αu(t)+q(t)f(u(t),u′(t),…,u(n-2)(t))=0,n-1<α≤n,n≥2,u(0)=u′(0)=⋯=u(n-2)(0)=0,u(n-2)(1)=0,
was discussed in [16], where f∈C((0,∞)n-1) and q∈Lr[0,1] (r>0) are positive. The existence results of positive solutions are acquired by the use of regularization and sequential techniques with a fixed point theorem for mixed monotone operators on normal cones.Paper [6] investigates positive solutions of singular fractional boundary value problem
(3)D0+αu(t)+f(t,u(t),D0+νu(t),D0+μu(t))=0,u(0)=u′(0)=u′′(0)=u′′(1)=0,
where 3<α≤4, 0<ν≤1, 1<μ≤2, D0+α is the standard Riemann-Liouville fractional derivative, and f is a Carathédory function. The existence and multiplicity of positive solutions are obtained by means of Guo-Krasnosel'skii fixed point theorem on cones.In this paper, we are concerned with the following singular fractional boundary value problem:(4)DC0+αu+f(t,u,DC0+νu,DC0+μu)+g(t,u,DC0+νu,DC0+μu)=0,(5)u(0)=u′(0)=u′′(1)=u′′′(0)=0,
where 3<α<4, 0<ν<1, and 1<μ<2 are real numbers. D0+αC is the Caputo fractional derivative of order α. f satisfies the Carathéodory condition on [0,1]×𝒟, 𝒟⊂ℝ3, f(t,x,y,z) may be singular at the value 0 of all its space variables x, y, z, and g(t,x,y,z) satisfies the Lipschitz condition.A functionu∈C2[0,1] is called a positive solution of problems (4), (5) if u>0 on (0,1], D0+αCu∈L[0,1], and u satisfies boundary condition (5) and equality (4) for a.e. t∈[0,1].Throughout the paper, denote||x||1=∫01‍|x(t)|dt which is the norm of L[0,1], and ∥x∥=max{|x(t)|:t∈[0,1]} is the norm of space C[0,1], while ∥x∥*=max{∥x∥,∥x′∥,∥x′′∥} is the norm of C2[0,1]. AC[0,1] and ACk[0,1] are sets of absolutely continuous functions and functions having absolutely continuous kth derivatives on [0,1], respectively.The following conditions onf and g in (4) will be used.(H1)
f is a Carathéodory function on [0,1]×𝒟, where 𝒟=(0,∞)3, and there exists a positive constant m such that, for a.e. t∈[0,1] and all (x,y,z)∈𝒟,
(6)f(t,x,y,z)≥m.(H2)
g≥0 satisfies the following inequality, for a.e. t∈[0,1] and all (x1,y1,z1), (x2,y2,z2)∈𝒟:
(7)|g(t,x1,y1,z1)-g(t,x2,y2,z2)|≤L1|x1-x2|+L2|y1-y2|+L3|z1-z2|,with(8)1Γ(α-2)[L1+L2Γ(2-ν)+L3Γ(3-μ)]<1.(H3)
For a.e.t∈[0,1] and all (x,y,z)∈𝒟,
(9)f(t,x,y,z)+g(t,x,y,z)≤p(x,y,z)+γ(t)h(x,y,z),whereγ∈L[0,1], p∈C(𝒟), and h∈C([0,∞)3) are positive, p and h are nonincreasing and nondecreasing in all their arguments, respectively,
(10)∫01‍p(2Mtα,M12t4-ν,(2-μ)M6t3-μ)dt<∞,M=mΓ(α+1),limx→∞h(x,x,x)x=0.We will use regularization and sequential techniques to prove the existence of a positive solution of problems (4), (5). Define χn and fn(n∈ℕ) by the following formulas:
(11)χn(t)={t,ift≥1n;1n,ift<1n,
for a.e. t∈[0,1] and all (x,y,z)∈ℝ3,
(12)fn(t,x,y,z)=f(t,χn(x),χn(y),χn(z)).Then, condition (H1) gives that fn is a Carathéodory function on [0,1]×ℝ3,
(13)fn(t,x,y,z)≥m,fora.e.t∈[0,1]andall(x,y,z)∈ℝ3.Condition (H3) gives
(14)fn(t,x,y,z)≤p(1n,1n,1n)+γ(t)h(x+1n,y+1n,z+1n),fora.e.t∈[0,1]andall(x,y,z)∈[0,∞)3,(15)fn(t,x,y,z)≤p(x,y,z)+γ(t)h(x+1n,y+1n,z+1n),fora.e.t∈[0,1]andall(x,y,z)∈𝒟.In Section3, We will firstly investigate the regular fractional differential equation
(16)DC0+αu+fn(t,u,DC0+νu,DC0+μu)+g(t,u,DC0+νu,DC0+μu)=0.


## 2. Preliminaries

Definition 1.
The Caputo fractional derivative of orderβ>0 of a function υ∈C[0,1] is defined by
(17)DC0+βυ(t)=1Γ(n-β)∫0t‍(t-s)n-β-1υ(n)(s)ds,
provided that the right-hand side is pointwise defined on [0,1], where n=[β]+1 and [β] means the integer part of the number β. Γ is the Euler function.Definition 2.
The fractional integral of orderα>0 of a function y:[0,1]→ℝ is defined by
(18)I0+αy(t)=1Γ(α)∫0t‍(t-s)α-1y(s)ds,
provided the right-hand side is pointwise defined on [0,1].Lemma 3 (see [10]).
One has(19)I0+α:L1[0,1]⟶{L1[0,1],ifα∈(0,1),AC[α]-1[0,1],ifα≥1,
where [α] means the integral part of α and AC0[0,1]=AC[0,1].Lemma 4 (see [10]).
Suppose thatα>0, α∉ℕ. If x∈C(0,1] and CD0+αx∈L1[0,1], then
(20)x(t)=I0+αDC0+αx(t)+∑k=0n-1‍cktk,fort∈(0,1],
where n=[α]+1 and ck∈ℝ, k=0,1,…,n-1.Lemma 5.
Givenρ∈L[0,1], then for t∈[0,1],
(21)u(t)=∫01‍G(t,s)ρ(s)ds
is the unique solution in C2[0,1] of the equation
(22)D0+αu(t)+ρ(t)=0,
satisfying the boundary condition (5), where α∈(3,4) and
(23)G(t,s)={t2(1-s)α-32Γ(α-2)-(t-s)α-1Γ(α),if0≤s≤t≤1,t2(1-s)α-32Γ(α-2),if0≤t≤s≤1.Proof.
By Lemma4,
(24)u(t)=-I0+αρ(t)+c0+c1t1+c2t2+c3t3,for3<α<4,
are all solutions of (22) in C[0,1], where cj∈ℝ. Lemma 3 guarantees that I0+αρ∈AC2[0,1], for 3<α<4; therefore,
(25)u(t)=-I0+αρ(t)+c2t2,for3<α<4,
are all solutions of (22) in C2[0,1], where c2,c4∈ℝ. Considering that the solutions should satisfy u(0)=u′(0)=u′′(1)=u′′′(0)=0, we get that c2=(1/2Γ(α-2))∫01‍(1-s)α-3ρ(s)ds. Consequently,
(26)u(t)=t2∫01‍(1-s)α-3ρ(s)ds2Γ(α-2)-∫0t‍(t-s)α-1ρ(s)dsΓ(α)=∫01‍G(t,s)ρ(s)ds
is the unique solution of problems (22), (5).Lemma 6.
LetG be as defined in (2.3). Then, (1)
G(t,s)∈C([0,1]×[0,1]) and G(t,s)>0 on (0,1)×(0,1),(2)
G(t,s)≤1/Γ(α-1) for (t,s)∈[0,1]×[0,1],(3)
∫01‍G(t,s)ds≥(α2-α-2)tα/2Γ(α+1) for t∈[0,1],(4)
(∂/∂t)G(t,s)∈C([0,1]×[0,1]) and (∂/∂t)G(t,s)>0 on (0,1)×(0,1),(5)
(∂/∂t)G(t,s)≤1/Γ(α-2) for (t,s)∈[0,1]×[0,1],(6)
∫01‍(∂/∂t)G(t,s)ds≥(α-2)tα-1/Γ(α) for t∈[0,1],(7)
(∂2/∂t2)G(t,s)∈C([0,1]×[0,1]) and (∂2/∂t2)G(t,s)>0 on (0,1)×(0,1),(8)
(∂2/∂t2)G(t,s)≤1/Γ(α-2) for (t,s)∈[0,1]×[0,1],(9)
∫01‍(∂2/∂t2)G(t,s)ds≥t(1-tα-2)/Γ(α-1) for t∈[0,1].Proof.
(1), (4), and (7) are as follows(27)G(t,s)={t2(1-s)α-32Γ(α-2)-(t-s)α-1Γ(α),if0≤s≤t≤1,t2(1-s)α-32Γ(α-2),if0≤t≤s≤1,∂∂tG(t,s)={t(1-s)α-3Γ(α-2)-(t-s)α-2Γ(α-1),if0≤s≤t≤1,t(1-s)α-3Γ(α-2),if0≤t≤s≤1,∂2∂t2G(t,s)={(1-s)α-3Γ(α-2)-(t-s)α-3Γ(α-2),if0≤s≤t≤1,(1-s)α-3Γ(α-2),if0≤t≤s≤1.
Because(∂2/∂t2)G(t,s)≥0, therefore (∂/∂t)G(t,s)≥(∂/∂t)G(0,s)≥0 and G(t,s)≥G(0,s)≥0.
It is obvious that (2), (3), (5), (6), (8), and (9) hold.


## 3. Auxiliary Regular Problems (16), (5)

LetX=C2[0,1], and let
(28)P={x∈X:x(t)≥0,x′(t)≥0,x′′(t)≥0,fort∈[0,1]}.Forx∈P, we can obtain that
(29)DC0+νx∈C[0,1],DC0+μx∈C[0,1],DC0+νx(t)≥0,DC0+μx(t)≥0,forx∈P,t∈[0,1].We define the operatorsΦn and Ψ on P as
(30)(Φnx)(t)=∫01‍G(t,s)fn(s,x(s),DC0+νx(s),DC0+μx(s))ds,n=1,2,…,(Ψx)(t)=∫01‍G(t,s)g(s,x(s),DC0+νx(s),DC0+μx(s))ds.Lemma 7.
Φ
n
:
P
→
P is a completely continuous operator.The proof is similar to Lemma 3.1 of [6], so we omit it.Lemma 8 (see [17]).
LetM be a closed convex and nonempty subset of a Banach space X. Let A, B be the operators such that (i) Ax+By∈M wherever x,y∈M, (ii) A is compact and continuous, and (iii) B is a contraction mapping. Then, there exists z∈M such that z=Az+Bz.Theorem 9.
Let (H1) and (H2) hold. Then, problems (16), (5) have a solution un∈P such that
(31)un(t)≥mtα(α2-α-2)2Γ(α+1),fort∈[0,1].Proof.
By Lemma7, Φn:P→P is a completely continuous operator. Now, for x,y∈X, we obtain that
(32)∥(Ψx)-(Ψy)∥=max0≤t≤1|∫01‍G(t,s)×[g(s,x(s),DC0+νx(s),DC0+μx(s))-g(s,y(s),DC0+νy(s),DC0+μy(s))]ds{∫01‍}|≤|max0≤t≤1∫01‍G(t,s)ds|×(L1∥x-y∥+L2∥DC0+νx-DC0+νy∥+L3∥DC0+μx-DC0+μy∥)≤|max0≤t≤1∫01‍G(t,s)ds|×({L3Γ(3-μ)}L1∥x-y∥*+L2Γ(2-ν)∥x-y∥*+L3Γ(3-μ)∥x-y∥*)≤∥x-y∥*Γ(α-2)(L1+L2Γ(2-ν)+L3Γ(3-μ)).
Therefore,Ψ is a contraction mapping, and it is obvious that Φnx+Ψy∈X, for x,y∈X. Thus, all the assumptions of Lemma 8 are satisfied, and the conclusion of Lemma 8 implies that the boundary value problems (16), (5) have at least one solution.Lemma 10.
Suppose that (H1), (H2), and (H3) hold and un be a solution of problems (16), (5). Then, the sequence {un} is relatively compact in X.Proof.
Note that, fort∈[0,1] and n∈ℕ,
(33)un(t)=∫01G(t,s)fn(s,xn(s),DC0+νxn(s),DC0+μxn(s))ds+∫01G(t,s)g(s,xn(s),DC0+νxn(s),DC0+μxn(s))ds.
And un fulfills (31).
Lemma6 and (13) imply that
(34)un′(t)≥m∫01‍∂∂tG(t,s)ds≥m(α-2)tα-1Γ(α),fort∈[0,1],n∈ℕ,un′′(t)≥m∫01‍∂2∂t2G(t,s)ds≥m(1-tα-2)tΓ(α-1),fort∈[0,1],n∈ℕ.
So,(35)CD0+μun(t)=1Γ(2-μ)∫0t‍(t-s)1-μun′′(s)ds≥mΓ(2-μ)Γ(α-1)∫0t‍(t-s)1-μs(1-sα-2)ds,DC0+νun(t)=1Γ(1-ν)∫0t‍(t-s)-νun′(s)ds≥m(α-1)Γ(1-ν)Γ(α)∫0t‍(t-s)-νsα-1ds.
Since(36)∫0t‍(t-s)1-μs(1-sα-2)ds>∫0t‍(t-s)1-μs(1-s)ds=12-μ∫0t‍(t-s)2-μ(1-2s)ds=12-μ(t3-μ3-μ-2t4-μ(3-μ)(4-μ))=t3-μ2-μ(4-μ-2t(3-μ)(4-μ))≥t3-μ2-μ(2-μ(3-μ)(4-μ))=t3-μ(3-μ)(4-μ),∫0t‍(t-s)-νsα-1ds≥∫0t‍(t-s)-νs3ds=∫0t‍(t-s)1-ν×3s2(1-ν)ds=∫0t‍(t-s)2-ν×6s(1-ν)(2-ν)ds=∫0t‍(t-s)3-ν×6(1-ν)(2-ν)(3-ν)ds=6t4-ν(1-ν)(2-ν)(3-ν)(4-ν)≥t4-ν(1-ν)(2-ν)(3-ν)(4-ν),
then,
(37)DC0+μun(t)≥m(2-μ)Γ(5-μ)Γ(α-1)t3-μ,fort∈[0,1],n∈ℕ,DC0+νun(t)≥m(α-1)Γ(5-ν)Γ(α)t4-ν,fort∈[0,1],n∈ℕ.
Let(38)m·min{1Γ(α+1),1Γ(α),1(α-1)}:=M.
It follows from (31) and (37) that, for t∈[0,1], n∈ℕ,
(39)un(t)≥2Mtα,DC0+νun(t)≥M12t4-ν,DC0+μun(t)≥(2-μ)M6t3-μ.
Therefore,(40)p(un(t),DC0+νun(t),DC0+μun(t))≤p(2Mtα,M12t4-ν,(2-μ)M6t3-μ).
By Lemma6, (15), and (39), there hold that
(41)0≤un′′(t)=∫01‍∂2∂t2G(t,s)fn(s,un(s),DC0+νun(s),DC0+μun(s))ds≤1Γ(α-2)∫01‍p(2Msα,M12s4-ν,(2-μ)M6s3-μ)ds+1Γ(α-2)h(∥un∥*+1n,∥un∥*Γ(3-ν)+1n,∥un∥*Γ(3-μ)+1n)×∫01‍γ(s)ds≤1Γ(α-2)×(Λ+h(∥un∥*+1n,∥un∥*Γ(3-ν)+1n,∥un∥*Γ(3-μ)+1n)∥γ∥q),0≤un′(t)=∫0t‍un′′(s)ds≤1Γ(α-2)×(Λ+h(∥un∥*+1n,∥un∥*Γ(3-ν)+1n,∥un∥*Γ(3-μ)+1n)∥γ∥q),0≤un(t)=∫0t‍un′(s)ds≤1Γ(α-2)×(Λ+h(∥un∥*+1n,∥un∥*Γ(3-ν)+1n,∥un∥*Γ(3-μ)+1n)∥γ∥q),
where t∈[0,1], n∈ℕ, and Λ=∫01‍p(2Msα,(M/12)s4-ν,((2-μ)M/6)s3-μ)ds.
It follows from (H2) and the assumption that Λ<∞. Hence,
(42)∥un∥*≤1Γ(α-2)×(Λ,+h(∥un∥*+1n,∥un∥*Γ(3-ν)+1n,∥un∥*Γ(3-μ)+1n)∥γ∥q),
where n∈ℕ. Since limx→∞h(x,x,x)/x=0, there exists L>0 such that, for υ≥L,
(43)1Γ(α-2)×(Λ+h(υ+1n,υΓ(3-ν)+1n,υΓ(3-μ)+1n)∥γ∥q)<υ.
Consequently,∥un∥*<L for n∈ℕ, so that {un} is bounded in X. We are now in a position to prove that {un′′} is equicontinuous on [0,1]. Let
(44)V1=h(L+1n,LΓ(3-ν)+1n,LΓ(3-μ)+1n),(45)Θ(t)=p(2Mtα,M12t4-ν,(2-μ)M6t3-μ),fort∈(0,1].
Then,Λ=∫01‍Θ(t)dt and, for a.e. t∈[0,1], all n∈ℕ,
(46)Θ(t)+V1γ(t)≥fn(t,un(t),DC0+νun(t),DC0+μun(t))+g(t,un(t),DC0+νun(t),DC0+μun(t))
holds. Suppose that 0≤t1<t2≤1, then
(47)|un′′(t2)-un′′(t1)|=|∫01‍(∂2∂t2G(t2,s)-∂2∂t2G(t1,s))×(fn(s,un(s),DC0+νun(s),DC0+μun(s))+g(s,un(s),DC0+νun(s),DC0+μun(s)))ds{∫01‍}|≤1Γ(α-2)[∫t1t2‍(t2-s)α-3(Θ(s)+V1γ(s))ds+∫0t1‍((t2-s)α-3-(t1-s)α-3)×(Θ(s)+V1γ(s))ds{∫0t1‍}]≤1Γ(α-2)[{∫0t1‍}(t2-t1)α-3(Λ+V1∥γ∥q)+∫0t1‍((t2-s)α-3-(t1-s)α-3)×(Θ(s)+V1γ(s))ds{∫0t1‍}].
The proof is similar to that of Lemma7. We choose ε>0. Then, there exists δ0>0 such that (t2-s)α-3-(t1-s)α-3<ε, for any 0≤t1<t2≤1, t2-t1<δ0, and 0≤s≤t1. Suppose that 0<δ<min{δ0,εα-3}. Then, for t1,t2∈[0,1], 0<t2-t1<δ, n∈ℕ, we have
(48)|un′′(t2)-un′′(t1)|≤2εΓ(α-2)(Λ+V1∥γ∥q).
Thus,{un′′} is equicontinuous on [0,1].


## 4. Main Result

Theorem 11.
Suppose that (H1), (H2), and (H3) hold. Then, problems (4), (5) has a positive solution u and, for t∈[0,1],
(49)u(t)≥2Mtα,DC0+νu(t)≥M24t4-ν,DC0+μu(t)≥(2-μ)M6t3-μ.Proof.
Theorem9 shows that problems (16), (5) have a solution un∈P. In addition, Lemma 10 gives that {un} is relatively compact in X and satisfies inequality (39) for t∈[0,1],n∈ℕ. Assume that {un} itself is convergent in X and limn→∞un=u. Then, u∈P satisfies the boundary condition (5), and limn→∞DC0+μun=DC0+μu and limn→∞DC0+νun=DC0+νu in C[0,1]. Consequently, u satisfies (49). Furthermore,
(50)limn→∞fn(t,un(t),DC0+νun(t),DC0+μun(t))=f(t,u(t),DC0+νu(t),DC0+μu(t)).
LetK=sup{∥un∥*:n∈ℕ}. Then, it follows from 0<ν<1 and 1<μ<2(51)∥DC0+μun∥≤KΓ(3-μ),∥DC0+νun∥≤KΓ(2-ν)forn∈ℕ.
Hence, for a.e.(t,s)∈[0,1]×[0,1] and all un∈ℕ, we have
(52)0≤G(t,s)(fn(s,un(s),DC0+νun(s),DC0+μun(s))+g(s,un(s),DC0+νun(s),DC0+μun(s)))≤1Γ(α-1)×(Θ(s)+h(K+1n,KΓ(2-ν)+1n,KΓ(3-μ)+1n)γ(s)),
where Θ is defined by (45). Putting n→∞, by the Lebesgue dominated convergence theorem, we have, for t∈[0,1],
(53)u(t)=∫01‍G(t,s)×(f(s,u(s),DC0+νu(s),DC0+μu(s))+g(s,u(s),DC0+νu(s),DC0+μu(s)))ds.
Consequently,u is a positive solution of problems (4), (5) and satisfies inequality (49). The proof is complete.



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