Volume 2009, Article ID 421310,10pages doi:10.1155/2009/421310
Research Article
Existence and Uniqueness of Positive and
Nondecreasing Solutions for a Class of Singular Fractional Boundary Value Problems
J. Caballero Mena, J. Harjani, and K. Sadarangani
Departamento de Matem´aticas, Universidad de Las Palmas de Gran Canaria, Campus de Tafira Baja, 35017 Las Palmas de Gran Canaria, Spain
Correspondence should be addressed to K. Sadarangani,[email protected] Received 24 April 2009; Accepted 14 June 2009
Recommended by Juan Jos´e Nieto
We establish the existence and uniqueness of a positive and nondecreasing solution to a singular boundary value problem of a class of nonlinear fractional differential equation. Our analysis relies on a fixed point theorem in partially ordered sets.
Copyrightq2009 J. Caballero Mena et al. This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.
1. Introduction
Many papers and books on fractional differential equations have appeared recently. Most of them are devoted to the solvability of the linear fractional equation in terms of a special functionsee, e.g.,1,2and to problems of analyticity in the complex domain3. Moreover, Delbosco and Rodino4considered the existence of a solution for the nonlinear fractional differential equationDα0u ft, u, where 0 < α < 1 andf : 0, a×R → R, 0 < a≤ ∞ is a given continuous function in 0, a×R. They obtained results for solutions by using the Schauder fixed point theorem and the Banach contraction principle. Recently, Zhang5 considered the existence of positive solution for equationDα0u ft, u, where 0 < α < 1 andf :0,1×0,∞ → 0,∞is a given continuous function by using the sub- and super- solution methods.
In this paper, we discuss the existence and uniqueness of a positive and nondecreasing solution to boundary-value problem of the nonlinear fractional differential equation
Dα0ut ft, ut 0, 0< t <1,
u0 u1 u0 0, 1.1
where 2 < α ≤ 3,Dα0 is the Caputo’s differentiation andf : 0,1×0,∞ → 0,∞with limt→0ft,− ∞i.e.,fis singular att0.
Note that this problem was considered in6where the authors proved the existence of one positive solution for1.1by using Krasnoselskii’s fixed point theorem and nonlinear alternative of Leray-Schauder type in a cone and assuming certain hypotheses on the function f. In6the uniqueness of the solution is not treated.
In this paper we will prove the existence and uniqueness of a positive and nondecreasing solution for the problem 1.1 by using a fixed point theorem in partially ordered sets.
Existence of fixed point in partially ordered sets has been considered recently in7–12.
This work is inspired in the papers6,8.
For existence theorems for fractional differential equation and applications, we refer to the survey13. Concerning the definitions and basic properties we refer the reader to14.
Recently, some existence results for fractional boundary value problem have appeared in the literaturesee, e.g.,15–17.
2. Preliminaries and Previous Results
For the convenience of the reader, we present here some notations and lemmas that will be used in the proofs of our main results.
Definition 2.1. The Riemman-Liouville fractional integral of orderα > 0 of a function f : 0,∞ → Ris given by
I0αft 1 Γα
t
0
t−sα−1fsds 2.1
provided that the right-hand side is pointwise defined on0,∞.
Definition 2.2. The Caputo fractional derivative of orderα > 0 of a continuous function f : 0,∞ → Ris given by
Dα0ft 1 Γn−α
t
0
fns
t−sα−n1ds, 2.2
wheren−1< α≤n, provided that the right-hand side is pointwise defined on0,∞.
The following lemmas appear in14.
Lemma 2.3. Letn−1< α≤n,u∈Cn0,1. Then
I0αD0αut ut−c1−c2t− · · · −cntn−1, 2.3
whereci∈R,i1,2, . . . , n.
Lemma 2.4. The relation
I0αI0βϕI0αβ ϕ 2.4
is valid when Reβ >0, Reαβ>0,ϕx∈L10, b.
The following lemmas appear in6.
Lemma 2.5. Givenf∈C0,1and 2< α≤3, the unique solution of
Dα0ut ft 0, 0< t <1,
u0 u1 u0 0, 2.5
is given by
ut 1
0
Gt, sfsds, 2.6
where
Gt, s
⎧⎪
⎪⎨
⎪⎪
⎩
α−1t1−sα−2−t−sα−1
Γα , 0≤s≤t≤1, t1−sα−2
Γα−1 , 0≤t≤s≤1.
2.7
Remark 2.6. Note thatGt, s>0 fort /0 andG0, s 0see6.
Lemma 2.7. Let 0 < σ < 1, 2 < α ≤ 3 and F : 0,1 → R is a continuous function with limt→0Ft ∞. Suppose thattσFtis a continuous function on0,1. Then the function defined by
Ht 1
0
Gt, sFsds 2.8
is continuous on [0,1], whereGt, sis the Green function defined inLemma 2.5.
Now, we present some results about the fixed point theorems which we will use later.
These results appear in8.
Theorem 2.8. LetX,≤be a partially ordered set and suppose that there exists a metricdinXsuch thatX, dis a complete metric space. Assume thatX satisfies the following condition: if{xn}is a non decreasing sequence inX such thatxn → xthenxn ≤ xfor alln∈ N. LetT :X → X be a nondecreasing mapping such that
d
Tx, Ty
≤d x, y
−ψ d
x, y
, forx≥y, 2.9
where ψ : 0,∞ → 0,∞ is continuous and nondecreasing function such thatψ is positive in 0,∞,ψ0 0 and limt→ ∞ψt ∞. If there existsx0 ∈X withx0 ≤Tx0thenT has a fixed point.
If we consider thatX,≤satisfies the following condition:
forx, y∈X there existsz∈X which is comparable toxandy, 2.10
then we have the following theorem8.
Theorem 2.9. Adding condition2.10to the hypotheses ofTheorem 2.8one obtains uniqueness of the fixed point off.
In our considerations, we will work in the Banach space C0,1 {x : 0,1 → R, continuous}with the standard normxmax0≤t≤1|xt|.
Note that this space can be equipped with a partial order given by
x, y∈C0,1, x≤y⇐⇒xt≤yt, fort∈0,1. 2.11
In10it is proved thatC0,1,≤with the classic metric given by d
x, y max
0≤t≤1 xt−yt 2.12
satisfies condition2ofTheorem 2.8. Moreover, forx, y∈C0,1, as the function max{x, y}
is continuous in0,1,C0,1,≤satisfies condition2.10.
3. Main Result
Theorem 3.1. Let 0 < σ < 1, 2 < α ≤ 3, f : 0,1×0,∞ → 0,∞ is continuous and limt→0ft,− ∞,tσft, yis a continuous function on0,1×0,∞. Assume that there exists 0< λ≤Γα−σ/Γ1−σsuch that forx, y∈0,∞withy≥xandt∈0,1
0≤tσ f
t, y
−ft, x
≤λ·ln
y−x1
3.1
Then one’s problem1.1has an unique nonnegative solution.
Proof. Consider the cone
P{u∈C0,1:ut≥0}. 3.2
Note that, asPis a closed set ofC0,1,Pis a complete metric space.
Now, foru∈Pwe define the operatorTby
Tut 1
0
Gt, sfs, usds. 3.3
ByLemma 2.7,Tu ∈ C0,1. Moreover, taking into accountRemark 2.6and astσft, y≥ 0 fort, y∈0,1×0,∞by hypothesis, we get
Tut 1
0
Gt, ss−σsσfs, usds≥0. 3.4
Hence,TP⊂P.
In what follows we check that hypotheses in Theorems2.8and2.9are satisfied.
Firstly, the operatorTis nondecreasing since, by hypothesis, foru≥v
Tut 1
0
Gt, sfs, usds
1
0
Gt, ss−σsσfs, usds
≥ 1
0
Gt, ss−σsσfs, vsds Tvt.
3.5
Besides, foru≥v
dTu, Tv max
t∈0,1|Tut−Tvt|
max
t∈0,1Tut−Tvt max
t∈0,1
1 0
Gt, s
fs, us−fs, vs ds
max
t∈0,1
1
0
Gt, ss−σsσ
fs, us−fs, vs ds
≤max
t∈0,1
1
0
Gt, ss−σλ·lnus−vs 1ds
3.6
As the functionϕx lnx1is nondecreasing then, foru≥v,
lnus−vs 1≤lnu−v1 3.7
and from last inequality we get
dTu, Tv≤max
t∈0,1
1
0
Gt, ss−σλ·lnus−vs 1ds
≤λ·lnu−v1·max
t∈0,1
1
0
Gt, ss−σds
λ·lnu−v1
·max
t∈0,1
t
0
α−1t1−sα−2−t−sα−1
Γα s−σds 1
t
t1−sα−2 Γα−1 s−σds
≤λ·lnu−v1
·max
t∈0,1
t
0
α−1t1−sα−2
Γα s−σds 1
t
t1−sα−2·s−σ Γα−1 ds
≤λ·lnu−v1
·max
t∈0,1
t
0
α−11−sα−2
Γα s−σds 1
t
1−sα−2·s−σ Γα−1 ds
λ·lnu−v1·max
t∈0,1
t
0
1−sα−2s−σ Γα−1 ds
1
t
1−sα−2s−σ Γα−1 ds
λ·lnu−v1 Γα−1 ·max
t∈0,1
1
0
1−sα−2s−σds
λ·lnu−v1 Γα−1 ·
1
0
1−sα−2s−σds λ·lnu−v1
Γα−1 ·β1−σ, α−1 λ·lnu−v1
Γα−1 ·Γ1−σ·Γα−1 Γα−σ λ·lnu−v1·Γ1−σ
Γα−σ ≤ Γα−σ
Γ1−σ ·λ·lnu−v1· Γ1−σ Γα−σ lnu−v1 u−v −u−v −lnu−v1.
3.8
Putψx x−lnx1. Obviously,ψ :0,∞ → 0,∞is continuous, nondecreasing, positive in0,∞,ψ0 0 and limx→ ∞ψx ∞.
Thus, foru≥v
dTu, Tv≤du, v−ψdu, v. 3.9
Finally, take into account that for the zero function, 0≤T0, byTheorem 2.8our problem1.1 has at least one nonnegative solution. Moreover, this solution is unique sinceP,≤satisfies condition2.10 see comments at the beginning of this sectionandTheorem 2.9.
Remark 3.2. In6, lemma 3.2it is proved that T : P → P is completely continuous and Schauder fixed point theorem gives us the existence of a solution to our problem1.1.
In the sequel we present an example which illustratesTheorem 3.1.
Example 3.3. Consider the fractional differential equationthis example is inspired in6
D5/20 ut t−1/22√ln2ut
t 0, 0< t <1 u0 u1 u0 0
3.10
In this case,ft, u t−1/22ln2ut/√
t for t, u ∈ 0,1×0,∞. Note that f is continuous in0,1×0,∞and limt→0ft,− ∞. Moreover, foru≥ vandt ∈ 0,1we have
0≤√ t
t−1
2 2
ln2u−
t−1 2
2
ln2v
3.11
becausegx lnx2is nondecreasing on0,∞, and
√t
t−1 2
2
ln2u−
t−1 2
2
ln2v
√ t·
t−1
2 2
ln2u−ln2v
t
t−1
2 2
ln 2u
2v
√ t
t−1
2 2
ln
2vu−v 2v
≤ 1
2 2
ln1u−v.
3.12
Note thatΓα−σ/Γ1−σ Γ5/2−1/2/Γ1−1/2 Γ2/Γ1/2 1/√
π≥1/4.
Theorem 3.1give us that our fractional differential3.10has an unique nonnegative solution.
This example give us uniqueness of the solution for the fractional differential equation appearing in6in the particular caseσ1/2 andα5/2
Remark 3.4. Note that ourTheorem 3.1works if the condition3.1is changed by, forx, y ∈ 0,∞withy≥xandt∈0,1
0≤tσ f
t, y
−ft, x
≤λ·ψ y−x
3.13
whereψ:0,∞ → 0,∞is continuous andϕx x−ψxsatisfies aϕ:0,∞ → 0,∞and nondecreasing;
bϕ0 0;
cϕis positive in0,∞;
dlimx→ ∞ϕx ∞.
Examples of such functions areψx arctgxandψx x/1x.
Remark 3.5. Note that the Green functionGt, sis strictly increasing in the first variable in the interval0,1. In fact, forsfixed we have the following cases
Case 1. Fort1, t2≤sandt1 < t2as, in this case,
Gt, s t1−sα−2
Γα−1 . 3.14 It is trivial that
Gt1, s t11−sα−2
Γα−1 < t21−sα−2
Γα−1 Gt2, s. 3.15 Case 2. Fort1≤s≤t2andt1< t2, we have
Gt2, s−Gt1, s
α−1t21−sα−2
Γα −t2−sα−1 Γα
−
t11−sα−2 Γα−1
t21−sα−2−t11−sα−2
Γα−1 −t2−sα−1 Γα
> t2−t11−sα−2
Γα−1 −t2−sα−1 Γα−1 t2−t11−sα−2
Γα−1 −t2−st2−sα−2 Γα−1 .
3.16
Now,t2−t1≥t2−sand1−s≥t2−sthen t2−t11−sα−2
Γα−1 > t2−st2−sα−2
Γα−1 . 3.17 Hence, taking into account the last inequality and3.16, we obtainGt1, s< Gt2, s.
Case 3. Fors≤t1, t2andt1 < t2<1, we have
∂G
∂t α−11−sα−2−α−11−sα−2
Γα α−1 Γα
1−sα−2−t−sα−2
, 3.18
and, as1−sα−2 >t−sα−2fort∈0,1, it can be deduced that∂G/∂t >0 and consequently, Gt2, s> Gt1, s.
This completes the proof.
Remark 3.5 gives us the following theorem which is a better result than that 6, Theorem 3.3 because the solution of our problem 1.1 is positive in 0,1 and strictly increasing.
Theorem 3.6. Under assumptions ofTheorem 3.1, our problem1.1has a unique nonnegative and strictly increasing solution.
Proof. ByTheorem 3.1we obtain that the problem1.1has an unique solutionut∈C0,1 withut ≥ 0. Now, we will prove that this solution is a strictly increasing function. Let us taket2, t1∈0,1witht1< t2, then
ut2−ut1 Tut2−Tut1
1
0
Gt2, s−Gt1, sfs, usds. 3.19
Taking into accountRemark 3.4and the fact thatf≥0, we getut2−ut1≥0.
Now, if we suppose thatut2−ut1 0 then1
0Gt2, s−Gt1, sfs, usds0 and as,Gt2, s−Gt1, s>0 we deduce thatfs, us 0 a.e.
On the other hand, iffs, us 0 a.e. then
ut 1
0
Gt, sfs, usds0 fort∈0,1. 3.20
Now, as limt→0ft,0 ∞, then forM > 0 there existsδ > 0 such that fors ∈0,1with 0< s < δwe getfs,0> M. Observe that0, δ⊂ {s∈0,1:fs, us> M}, consequently,
δμ0, δ≤μ s∈0,1:fs, us> M
3.21 and this contradicts thatfs, us 0 a.e.
Thus, ut2 − ut1 > 0 for t2, t1 ∈ 0,1 with t2 > t1. Finally, as u0 1
0G0, sfs, usds0 we have that 0< utfort /0.
Acknowledgment
This research was partially supported by ”Ministerio de Educaci ´on y Ciencia” Project MTM 2007/65706.
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