Zou and Cui Advances in Difference Equations 2013, 2013:233 http://www.advancesindifferenceequations.com/content/2013/1/233
RESEARCH
Open Access
Existence results for a functional boundary value problem of fractional differential equations Yumei Zou1 and Yujun Cui2* *
Correspondence:
[email protected] 2 Department of Mathematics, Shandong University of Science and Technology, Qingdao, 266590, P.R. China Full list of author information is available at the end of the article
Abstract In this paper, a functional boundary value problem of fractional differential equations is studied. Based on Mawhin’s coincidence degree theory, some existence theorems are obtained in the case of nonresonance and the cases of dim Ker L = 1 and dim Ker L = 2 at resonance.
1 Introduction The subject of fractional calculus has gained considerable popularity and importance because of its frequent appearance in various fields such as physics, chemistry, and engineering. In consequence, the subject of fractional differential equations has attracted much attention. Many methods have been introduced to solve fractional differential equations, such as the popular Laplace transform method, the iteration method, the Fourier transform method and the operational method. For details, see [–] and the references therein. Recently, there have been some papers dealing with the basic theory for initial value problems of nonlinear fractional differential equations; for example, see [, ]. Also, there are some articles which deal with the existence and multiplicity of solutions for nonlinear boundary value problems of fractional order differential equations using techniques of topological degree theory. We refer the reader to [–] for some recent results at nonresonance and to [–] at resonance. In [], by making use of the coincidence degree theory of Mawhin, Zhang and Bai discussed the existence results for the following nonlinear nonlocal problem at resonance under the case dim Ker L = : Dα+ u(t) = f t, u(t) ,
< t < ,
u() = ,
βu(η) = u(),
< α ≤ .
Recently, Jiang [] studied the existence of a solution for the following fractional differential equation at resonance under the case dim Ker L = : Dα+ u(t) = f t, u(t), Dα– + u(t) , u() = ,
Dα– + u() =
m i=
ai Dα– + u(ξi ),
Dα– + u() =
n
bj Dα– + u(ηj ).
j=
© 2013 Zou and Cui; licensee Springer. This is an Open Access article distributed under the terms of the Creative Commons Attribution License (http://creativecommons.org/licenses/by/2.0), which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.
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Being directly inspired by [, , ], we intend in this paper to study the following functional boundary value problems (FBVP) of fractional order differential equation: α– Dα+ u(t) = f t, u(t), Dα– + u(t), D+ u(t) , –α I+ u(t)|t= = , Dα– + u(t) = ,
(.)
Dα– + u(t) = ,
(.)
α are the standard Riemann-Liouville differentiation and intewhere < α < , Dα+ and I+ gration, and f ∈ C([, ] × R , R); , : C[, ] → R are continuous linear functionals. In this paper, we shall give some sufficient conditions to construct the existence theorems for FBVP (.), (.) at nonresonance and resonance (both cases of dim Ker L = and dim Ker L = ), respectively. To the best of our knowledge, the method of Mawhin’s theorem has not been developed for fractional order differential equation with functional boundary value problems at resonance. So, it is interesting and important to discuss the existence of a solution for FBVP (.), (.). Many difficulties occur when we deal with them. For example, the construction of the generalized inverse Kp : Im L → dom L ∩ Ker P of L. So, we need to introduce some new tools and methods to investigate the existence of a solution for FBVP (.), (.). The rest of this paper is organized as follows. In Section , we give some notations and lemmas. In Section , we establish the existence results of a solution for functional boundary value problem (.), (.).
2 Preliminaries and lemmas For the convenience of the reader, we present here the necessary definitions from fractional calculus theory. These definitions and properties can be found in the literature. The readers who are unfamiliar with this area can consult, for example, [, , ] for details. Definition . [, ] The Riemann-Liouville fractional integral of order α > of a function u : (, ∞) → R is given by α I+ u(t) =
(α)
t
(t – s)α– u(s) ds,
provided that the right-hand side is pointwise defined on (, ∞). Here (α) is the Gamma +∞ function given by (α) = t α– e–t dt. Definition . [, ] The Riemann-Liouville fractional derivative of order α > of a continuous function u : (, ∞) → R is given by Dα+ u(t) =
n t u(s) d ds, α–n+ (n – α) dt (t – s)
where n – ≤ α < n, provided that the right-hand side is pointwise defined on (, ∞). We use the classical spaces C[, ] with the norm u∞ = maxt∈[,] |u(t)|, L [, ] with the norm u = |u(t)| dt. We also use the space AC n [, ] defined by AC n [, ] = u : [, ] → R | u(n–) are absolutely continuous on [, ]
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and the Banach space C μ [, ] (μ > ) μ C μ [, ] = u(t) | u(t) = I+ x(t) + c t μ– + c t μ– + · · · + cN– t μ–(N–) , x ∈ C[, ], t ∈ [, ], ci ∈ R, i = , , . . . , N = [μ] + μ
μ–(N–)
with the norm uC μ = D+ u∞ + · · · + D+
u∞ + u∞ .
Lemma . [] Let α > , n = [α] + . Assume that u ∈ L (, ) with a fractional integration of order n – α that belongs to AC n [, ]. Then the equality
α I+ Dα+ u
(t) = u(t) –
n ((I n–α u)(t))(n–i) |t= +
i=
(α – i + )
t α–i
holds almost everywhere on [, ]. –α Remark . If u satisfies Dα+ u = f (t) ∈ L (, ) and I+ u|t= = , then u ∈ C α– [, ]. In fact, with Lemma ., one has α u(t) = I+ f (t) + c t α– + c t α– + c t α– . –α Combine with I+ u|t= = , there is c = . So,
α α– I+ f (t) + c (α) + c t (α–)– . f (t) + c t α– + c t α– = I+ u(t) = I+ In the following lemma, we use the unified notation both for fractional integrals and α fractional derivatives assuming that I+ = D–α + for α < . Lemma . [] Assume α > , then: (i) Let k ∈ N. If Dαa+ u(t) and (Dα+k a+ u)(t) exist, then
Dk Dαa+ u(t) = Dα+k a+ u (t);
(ii) If β > , α + β > , then
α+β α β Ia+ u(t) = Ia+ u (t) Ia+
is satisfied at any point on [a, b] for u ∈ Lp (a, b) and ≤ p ≤ +∞; α )u(t) = u(t) holds on [a, b]; (iii) Let u ∈ C[a, b]. Then (Dαa+ Ia+ (iv) Note that for λ > –, λ = α – , α – , . . . , α – n, we have Dα t λ =
(λ + ) λ–α t , (λ – α + )
Dα t α–i = ,
i = , , . . . , n.
Lemma . [] F ⊂ C μ [, ] is a sequentially compact set if and only if F is uniformly bounded and equicontinuous. Here ‘F is uniformly bounded and equicontinuous’ means that there exists M > such that for every u ∈ F,
u ∞ + u∞ < M uC μ = Dμ+ u ∞ + · · · + Dμ–[μ] +
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and that ∀ > , ∃N > , for all t , t ∈ [, ], |t – t | < δ, u ∈ F, i ∈ {, , . . . , [μ]}, there hold u(t ) – u(t ) < ε,
μ–i D u(t ) – Dμ–i u(t ) < ε, + +
respectively. Next, consider the following conditions: (A) [] [] = . (A) [] = , [] = , [t] = . (A) [] = , [] = , [t] = . (A) [] = , [] = , [t] = . (A) [] = , [] = , [t] = . We shall prove that: If (A) holds, then Ker L = {θ }. It is the so-called nonresonance case. If (A) holds, then Ker L = {at α– : a ∈ R}. If (A) or (A) holds, then Ker L = {at α– : a ∈ R}. If (A) holds, then Ker L = {at α– + bt α– : a, b ∈ R}. In the nonresonance case, FBVP (.), (.) can be transformed into an operator equation. Lemma . Assume that (A) holds. Then functional boundary value problem (.) and (.) has a solution if and only if the operator T : C α– [, ] → C α– [, ], defined by t [ (fu)(s) ds] α– t (t – s) (fu)(s) ds – (α) [] t t [] [ (t – s)(fu)(s) ds] – [ (fu)(s) ds] [t] α– – t , (α – ) [] []
(Tu)(t) = (α)
t
α–
α– has a fixed point, where (fu)(t) = f (t, u(t), Dα– + u(t), D+ u(t)).
Proof If u is a solution to Tu = u, by Lemma ., we get α– Dα+ u(t) = f t, u(t), Dα– + u(t), D+ u(t) , t t [ (fu)(s) ds] α– D+ u(t) = (fu)(s) ds – [] and t [ (fu)(s) ds] t = (t – s)(fu)(s) ds – [] t t [] [ (t – s)(fu)(s) ds] – [ (fu)(s) ds] [t] – . [] []
Dα– + u(t)
t
Considering the linearity of i (i = , ), we have –α I+ u(t)|t= = ,
Dα– + u(t)
=
t
t [ (fu)(s) ds] [] = , (fu)(s) ds – []
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t [ (fu)(s) ds] [t] (t – s)(fu)(s) ds – [] t t [] [ (t – s)(fu)(s) ds] – [ (fu)(s) ds] [t] – [] = . [] []
Dα– + u(t) =
t
So, u is a solution to FBVP (.), (.). If u is a solution to (.), by Lemma ., we can reduce (.) to an equivalent integral equation α (fu)(t) + c t α– + c t α– + c t α– . u(t) = I+
(.)
–α u(t)|t= = , there is c = , and By I+
Dα– + u(t) =
t
Dα– + u(t) =
(fu)(s) ds + c (α),
(.)
(t – s)(fu)(s) ds + c (α)t + c (α – ).
(.)
t
Applying and to (.) and (.), respectively, we obtain = Dα– + u(t) =
= Dα– + u(t) =
t
(fu)(s) ds + c (α) [],
t
(t – s)(fu)(s) ds + c (α) [t] + c (α – ) [].
Thus, t [ (fu)(s) ds] , c = – (α) [] t t [] [ (t – s)(fu)(s) ds] – [ (fu)(s) ds] [t] . c = – (α – ) [] []
(.) (.)
Substituting (.) and (.) into (.), we obtain t [ (fu)(s) ds] α– (t – s) (fu)(s) ds – t (α) [] t t [] [ (t – s)(fu)(s) ds] – [ (fu)(s) ds] [t] α– t . – (α – ) [] []
u(t) = (α)
t
The proof is complete.
α–
The following definitions and lemmas are a preparation for the existence of solutions to (.), (.) at resonance. Definition . Let Y , Z be real Banach spaces, let L : dom L ⊂ Y → Z be a linear operator. L is said to be a Fredholm operator of index zero provided that: (i) Im L is a closed subset of Z, (ii) dim Ker L = codim Im L < +∞.
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Let Y , Z be real Banach spaces and L : dom L ⊂ Y → Z be a Fredholm operator of index zero. P : Y → Y , Q : Z → Z are continuous projectors such that Im P = Ker L, Ker Q = Im L, Y = Ker L ⊕ Ker P and Z = Im L ⊕ Im Q. It follows that L|dom L∩Ker P : dom L ∩ Ker P → Im L is invertible. We denote the inverse of the mapping by KP (generalized inverse operator of L). If is an open bounded subset of Y such that dom L ∩ = ∅, the mapping N : Y → Z will be called L-compact on , if QN( ) is bounded and KP (I – Q)N : → Y is compact. We need the following known result for the sequel (Theorem . []). Theorem . Let L be a Fredholm operator of index zero, and let N be L-compact on . Assume that the following conditions are satisfied: (i) Lx = λNx for every (x, λ) ∈ [(dom L\ Ker L) ∩ ∂ ] × (, ). (ii) Nx ∈/ Im L for every x ∈ Ker L ∩ ∂ . (iii) deg(QN|Ker L , Ker L ∩ , ) = , where Q : Z → Z is a projector as above with Im L = Ker Q. Then the equation Lx = Nx has at least one solution in dom L ∩ . Let Y = C α– [, ], Z = L [, ]. Let the linear operator L : Y ⊂ dom L → Z with –α u(t)|t= = , dom L = u ∈ C α– [, ] : Dα+ u(t) ∈ Z, I+ α– α– D+ u(t) = , D+ u(t) = be defined by Lu = Dα+ u(t). Let the nonlinear operator N : Y → Z be defined by α– (Nu)(t) = f t, u(t), Dα– + u(t), D+ u(t) . Then (.), (.) can be written as Lu = Nu. Now, we give Ker L, Im L and some necessary operators at dim Ker L = and dim Ker L = , respectively. Lemma . Let L be the linear operator defined as above. If (A) holds, then Ker L = u ∈ dom L : u = at α– , a ∈ R, t ∈ [, ] and t v(s) ds = . Im L = v ∈ Z : –α u(t)|t= = . Considering (A), Proof Let u(t) = at α– . Clearly, Dα+ u(t) = and I+ α– α– [D+ u(t)] = a [(α)] = a(α) [] = and [D+ u(t)] = (α) [t] = . So,
u ∈ dom L : u = at α– , a ∈ R, t ∈ [, ] ⊂ Ker L.
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–α If Lu = Dα+ u(t) = , then u(t) = at α– + bt α– + ct α– . Considering I+ u(t)|t= = and (A), α– we can obtain that b = c = . It yields u(t) = at and Ker L ⊂ {u ∈ dom L : u = at α– , a ∈ R, t ∈ [, ]}. We now show that
t v(s) ds = . Im L = v ∈ Z :
If v ∈ Im L, then there exists u ∈ dom L such that Dα+ u(t) = v(t). Hence, u(t) =
(α)
t
(t – s)α– v(s) ds + at α– + bt α–
for some a, b ∈ R. It yields
Dα– + u(t)
=
t
t
v(s) ds + a(α) [] =
v(s) ds = .
Therefore t v(s) ds = . Im L ⊂ v ∈ Z :
On the other hand, suppose v ∈ Z satisfies
t
v(s) ds = .
Let u(t) = (α)
t
(t – s)
α–
t α– v(s) ds – (α – ) []
t
(t – s)v(s) ds .
–α Obviously, Dα+ u(t) = v(t) and I+ u(t)|t= = . Considering (A) and the linearity of i (i = , ), we have
Dα– + u(t)
t
=
v(s) ds =
and Dα– + u(t) =
t
(t – s)v(s) ds –
[]
t
(t – s)v(s) ds
= .
It yields t v(s) ds = ⊂ Im L. v ∈ Z :
The proof is complete. Lemma . If [t] = , then:
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(i) L is a Fredholm operator of index zero and dim Ker L = codim Im L = . (ii) The linear operator Kp : Im L → dom L ∩ Ker P can be defined by t α– (α – ) []
α v(t) – (Kp v)(t) = I+
t
(t – s)v(s) ds .
(iii) Kp vC α– ≤ v , where = + (α) + (+(α–)) and is the norm of a (α–)| []| continuous linear functional . (iv) The linear operator Kp : Im L → dom L ∩ Ker P ⊂ C α– [, ] is completely continuous.
Proof Firstly, we construct the mapping Q : Z → Z defined by Qy =
[t]
t
y(s) ds .
(.)
Noting that Q y =
[t]
t
(Qy) ds =
[t]
ds (Qy) = Qy,
t
we get Q : Z → Z is a well-defined projector. Now, it is obvious that Im L = Ker Q. Noting that Q is a linear projector, we have Z = Im Q ⊕ Ker Q. Hence, Z = Im Q ⊕ Im L and dim Ker L = codim Im L =. This means L is a Fredholm mapping of index zero. Taking P : Y → Y as (Pu)(t) =
Dα– + u(t)|t= α– t , (α)
then the generalized inverse Kp : Im L → dom L ∩ Ker P of L can be rewritten α (Kp v)(t) = I+ v(t) –
t α– (α – ) []
t
(t – s)v(s) ds .
In fact, for v ∈ Im L, we have –α I+ (Kp v)(t)|t= = ,
Dα– + (Kp v)(t)
α = Dα– + I+ v(t)
t
=
v(s) ds =
and Dα– + (Kp v)(t) =
t
(t – s)v(s) ds –
[]
t
(t – s)v(s) ds [] = ,
which implies that Kp is well defined on Im L. Moreover, for v ∈ Im L, we have α v(t) = v(t) (LKp )v(t) = Dα+ I+
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and for v ∈ dom L ∩ Ker P, we know α I+ Dα+ v(t) = v(t) –
–α Dα– Dα– v(t)|t= α– I+ v(t)|t= α– + v(t)|t= α– t – + t – t , (α) (α – ) (α)
–α α– v(t)|t= = Dα– v ∈ dom L ∩ Ker P means that I+ + v(t)|t= = [D+ v(t)] = . So,
α Dα+ v(t) – (Kp L)v(t) = I+
t α– α– Dα– + v(t) – D+ v(t)|t= (α – ) []
Dα– t α– + v(t)|t= α– t + Dα– + v(t)|t= = v(t). (α – ) (α – ) []
= v(t) –
That is, Kp = (L|dom L∩Ker P )– . Since Dα– + (Kp v)(t) =
t
v(s) ds,
Dα– + (Kp v)(t) =
t
(t – s)v(s) ds – []
t
(t – s)v(s) ds ,
then Kp v∞ ≤
+ v , (α) (α – )| []|
α–
D (Kp v) ≤ v , + ∞
α–
D (Kp v) ≤ + v . + ∞ | []|
It follows that Kp vC α–
≤ +
( + (α – )) + v . (α) (α – )| []|
Finally, we prove that Kp : Im L → dom L ∩ Ker P ⊂ C α– [, ] is completely continuous. Let V ⊂ Im L ⊂ L [, ] be a bounded set. From the above discussion, we only need to prove that Kp V is equicontinuous on [, ]. For v ∈ V , t , t ∈ [, ] with t < t , we have α– D (Kp v)(t ) – Dα– (Kp v)(t ) = + +
t t
v(s) ds ≤
t
v(s) ds,
t
t α– t D (Kp v)(t ) – Dα– (Kp v)(t ) = (t – s)v(s) ds – (t – s)v(s) ds + + t t ≤ (t – s)v(s) ds + (t – t )v(s) ds ≤
t
t
v(s) ds + (t – t )v
t
and (Kp v)(t ) – (Kp v)(t ) t t α– α– (t – s) v(s) ds – (t – s) v(s) ds ≤ (α)
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t | [ (t – s)v(s) ds]| α– t – t α– (α – )| []| t t α– α– α– ≤ (t – s) – (t – s) v(s) ds (t – s) v(s) ds + (α) t (α) +
v α– t – t α– (α – )| []| t v(s) ds + α – v (t – t ) + v t α– – t α– . ≤ (α) t (α) (α – )| []| +
Therefore, Kp (V ) is equicontinuous. Thus, the operator Kp : Im L → dom L ∩ Ker P is completely continuous. The proof is complete. Similar to Lemmas . and ., we can obtain the following lemma. Lemma . If (A) holds, then Ker L = {at α– : a ∈ R} and t v(s) ds = . Im L = v :
Furthermore, if [t] = also holds, then L is a Fredholm operator of index zero and dim Ker L = codim Im L = . Here, the projectors P : Y → Y , Q : Z → Z can be defined as follows: Dα– + v(t)|t= α– t , (α – ) t [ v(s) ds] (Qv)(t) = . [t] (Pv)(t) =
The generalized inverse operator of L, KP : Im L → dom L ∩ Ker P can be defined by α v(t) – (Kp v)(t) = I+
t [ (t – s)v(s) ds] α– t . (α) [t]
Also, Kp vC α– ≤ v , where = +
(α)
+
(+(α)) . (α)| [t]|
Lemma . If (A) holds, then Ker L = {at α– : a ∈ R} and t Im L = v : (t – s)v(s) ds = .
Furthermore, if [t ] = also holds, then L is a Fredholm operator of index zero and dim Ker L = codim Im L = . Here, the projectors P : Y → Y , Q : Z → Z can be defined as
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follows: Dα– + v|t= α– t , (α – ) t [ (t – s)v(s) ds] . (Qv)(t) = [t ] (Pv)(t) =
The generalized inverse operator of L, KP : Im L → dom L ∩ Ker P can be defined by α (Kp v)(t) = I+ v(t) –
t [ v(s) ds] α– t . (α) []
Also, Kp vC α– ≤ v , where = + tional .
(α)
+
(+(α)) (α)| []|
and is the norm of the continuous linear func-
Lemma . If (A) holds, then Ker L = u ∈ dom L : u = at α– + bt α– , a, b ∈ R, t ∈ [, ] and t t Im L = v ∈ Z : v(s) ds = (t – s)v(s) ds = .
–α u(t)|t= = . Considering Proof Let u(t) = at α– + bt α– . Clearly, Dα+ u(t) = and I+ α– α– (A), [D+ u(t)] = [(α)] = (α) [] = and [D+ u(t)] = a(α) [t] + b(α – ) [] = . So,
u ∈ dom L : u = at α– + bt α– , a, b ∈ R, t ∈ [, ] ⊂ Ker L. If Lu = Dα+ u(t) = , then u(t) = at α– + bt α– + ct α– . Considering Dα+ u(t) = and (A), we can obtain that Ker L ⊂ u ∈ dom L : u = at α– + bt α– , a, b ∈ R, t ∈ [, ] . We now show that t t v(s) ds = (t – s)v(s) ds = . Im L = v ∈ Z :
If v ∈ Im L, then there exists u ∈ dom L such that Dα+ u(t) = v(t). Hence, u(t) =
(α)
t
(t – s)α– v(s) ds + at α– + bt α–
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for some a, b ∈ R. It yields
Dα– + u(t)
t
=
v(s) ds + a(α) [] =
t
v(s) ds =
and Dα– + u(t) =
t
(t – s)v(s) ds + a(α) [t] + b(α – ) []
t
(t – s)v(s) ds = .
=
Therefore, t t v(s) ds = (t – s)v(s) ds = . Im L ⊂ v ∈ Z :
On the other hand, suppose v ∈ Z satisfies
t
t v(s) ds = (t – s)v(s) ds = .
Let α u(t) = I+ v(t) =
(α)
t
(t – s)α– v(s) ds.
–α u(t)|t= = . Considering (A) and the linearity of i Obviously, Dα+ u(t) = v(t) and I+ (i = , ), we have
Dα– + u(t) =
t
v(s) ds =
and Dα– + u(t) =
t
(t – s)v(s) ds = .
It yields t t v ∈ Z : v(s) ds = (t – s)v(s) ds = ⊂ Im L.
The proof is complete.
Lemma . If [t] [t ] – [t ] [t ] = , then L is a Fredholm operator of index zero and dim Ker L = codim Im L = . Furthermore, the linear operator Kp : Im L → dom L ∩ Ker P can be defined by α v(t). (Kp v)(t) = I+
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Also, Kp vC α– ≤ +
v . (α)
Proof Firstly, we construct the mapping Q : Z → Z defined by t t [t ] [ v(s) ds] – [t ] [ (t – s)v(s) ds] (Qv)(t) = [t] [t ] – [t ] [t ] t t [t ] [ v(s) ds] – [t] [ (t – s)v(s) ds] t. – [t] [t ] – [t ] [t ] Let T v =
t t [t ] [ v(s) ds] – [t ] [ (t – s)v(s) ds] [t] [t ] – [t ] [t ]
and t t [t ] [ v(s) ds] – [t] [ (t – s)v(s) ds] T v = – . [t] [t ] – [t ] [t ] We have Qv = T v + (T v)t.
(.)
Noting that t t [t ] [ (T v) ds] – [t ] [ (t – s)(T v) ds] T (T v) = [t] [t ] – [t ] [t ] t t [t ] [ ds] – [t ] [ (t – s) ds] (T v) = [t] [t ] – [t ] [t ]
=
[t ] [t] – [t ] [ t ] (T v) [t] [t ] – [t ] [t ]
= T v,
t t [t ] [ (T v)s ds] – [t ] [ (t – s)(T v)s ds] T (T v) = [t] [t ] – [t ] [t ] t t [t ] [ s ds] – [t ] [ (t – s)s ds] (T v) = [t] [t ] – [t ] [t ]
=
[t ] [ t ] – [t ] [ t ] (T v) [t] [t ] – [t ] [t ]
= ,
t t [t ] [ (T v) ds] – [t] [ (t – s)(T v) ds] [t] [t ] – [t ] [t ] t t [t ] [ ds] – [t] [ (t – s) ds] (T v) =– [t] [t ] – [t ] [t ]
T (T v) = –
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[t ] [t] – [t] [ t ] (T v) =– [t] [t ] – [t ] [t ] = and t t [t ] [ (T v)s ds] – [t] [ (t – s)(T v)s ds] T (T v) = – [t] [t ] – [t ] [t ] t t [t ] [ s ds] – [t] [ (t – s)s ds] (T v) =– [t] [t ] – [t ] [t ]
[t ] [ t ] – [t] [ t ] =– (T v) [t] [t ] – [t ] [t ] = T v, we have, for each v ∈ Z, that Q v = T T v + (T v)t + T T v + (T v)t t = T v + (T v)t = Qv. So, Q : Z → Z is a well-defined projector. Now we will show that Ker Q = Im L. If v ∈ Ker Q, from Qv = , we have T v = and T v = . Considering the definitions of T and T , we have ⎧ ⎨ [t ] [ t v(s) ds] – [t ] [ t (t – s)v(s) ds] = , t ⎩ [t ] [ v(s) ds] – [t] [ t (t – s)v(s) ds] = .
Since [t ] [t ]
– [t ] = – [t] t + t t = , – [t]
t t so [ v(s) ds] = [ (t – s)v(s) ds] = , which yields v ∈ Im L. On the other hand, if t t v ∈ Im L, from [ v(s) ds] = [ (t – s)v(s) ds] = and the definition of Q, it is obvious that Qv = , thus v ∈ Ker Q. Hence, Ker Q = Im L. For v ∈ Z, from v = (v – Qv) + Qv, v – Qv ∈ Ker Q = Im L, Qv ∈ Im Q, we have Z = Im L + Im Q. And for any v ∈ Im L ∩ Im Q, from v ∈ Im Q, there exist constants a, b ∈ R such that v(t) = a + bt. From v ∈ Im L, we obtain ⎧ ⎨ [t] · a + [ t ] · b = , ⎩ [ t ] · a + [ t ] · b = .
In view of [t] [ t ]
[ t ] [t] t – t t = , = [ t ]
(.)
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therefore (.) has a unique solution a = b = , which implies Im L ∩ Im Q = {θ } and Z = Im L ⊕ Im Q. Since dim Ker L = dim Im Q = codim Im L = , thus L is a Fredholm map of index zero. Let P : Y → Y be defined by (Pv)(t) =
Dα– u(t)|t= α– Dα– + u(t)|t= α– t + + t . (α) (α – )
Then the generalized inverse Kp : Im L → dom L ∩ Ker P of L can be rewritten α v(t). (Kp v)(t) = I+
In fact, for v ∈ Im L, we have –α (Kp v)(t)|t= = , I+
α– α Dα– + (Kp v)(t) = D+ I+ v(t) =
t
v(s) ds =
and Dα– + (Kp v)(t) =
t
(t – s)v(s) ds = ,
which implies that Kp is well defined on Im L. Moreover, for v ∈ Im L, we have α v(t) = v(t) (LKp )v(t) = Dα+ I+
and for v ∈ dom L ∩ Ker P, we know α I+ Dα+ v(t) = v(t) –
–α Dα– Dα– v(t)|t= α– I+ v(t)|t= α– + v(t)|t= α– t – + t – t , (α) (α – ) (α)
–α α– v ∈ dom L ∩ Ker P means that I+ v(t)|t= = Dα– + v(t)|t= = D+ v(t)|t= = . So, α Dα+ v(t) = v(t). (Kp L)v(t) = I+
That is, Kp = (L|dom L∩Ker P )– . Since Dα– + (Kp v)(t) =
t
v(s) ds,
Dα– + (Kp v)(t) =
t
(t – s)v(s) ds,
then v , (α)
α–
D (Kp v) ≤ v , + ∞ Kp v∞ ≤
α–
D (Kp v) ≤ v . + ∞
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It follows that Kp vC α–
≤ +
v . (α)
The proof is complete.
3 Main results From Lemma ., we can obtain the existence theorem for FBVP (.), (.). Theorem . Assume that (A) and the following conditions hold: f (t, x , x , x ) – f (t, y , y , y ) ≤ β |x – y | + |x – y | + |x – y | . Then FBVP (.), (.) has a unique solution in C α– [, ] provided that
β + + + + < . | []| (α) (α – )| []| | []| Proof We shall prove that Tx = x has a unique solution in C α– [, ]. For each u, v ∈ C α– [, ], considering the linearity of i (i = , ), we have (Tu)(t) – (Tv)(t) = (α)
t
(t – s)α– (fu)(s) – (fu)(s) ds
t
[ ((fu)(s) – (fu)(s)) ds] α– t (α) [] t [ (t – s)((fu)(s) – (fu)(s)) ds] α– t – (α – ) [] t [ ((fu)(s) – (fu)(s)) ds] [t] α– t . + (α – ) [] []
–
Then (Tu)(t) – (Tv)(t) ≤ βu – vC α–
+ (α) (α – )| []|
α– D (Tu)(t) – Dα– (Tv)(t) ≤ βu – vC α– + + + | []|
+ , | []|
and
α– D (Tu)(t) – Dα– (Tv)(t) ≤ βu – vC α– + + . + + | []| | []| So,
Tu – TvC α– ≤ βu – vC α– + + + + . | []| (α) (α – )| []| | []| The above inequality implies that T is a contraction. By using Banach’s contraction principle, Tx = x has a unique solution in C α– [, ]. From Lemma ., FBVP (.), (.) has a unique solution in C α– [, ]. The proof is complete.
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From Lemmas .-. and Theorem ., we can obtain the existence theorem for FBVP (.), (.) in the case of dim Ker L = . Theorem . Let f : [, ] × R → R be a continuous function. Assume that [t] = , (A) and the following conditions (H)-(H) hold: (H) There exist functions α, β, γ , ω ∈ L [, ] such that for all (x, y, z) ∈ R , t ∈ [, ], f (t, x, y) ≤ ω(t) + α(t)|x| + β(t)|y| + γ (t)|z|. (H) There exists a constant A > such that for u ∈ dom L, if |Dα– + u(t)| > A for all t α– α– t ∈ [, ], then [ f (s, u(s), D+ u(s), D+ u(s)) ds] = . (H) There exists a constant B > such that either for each a ∈ R : |a| > B,
t
a
f s, asα– , a(α), ds >
or for each a ∈ R : |a| > B,
t
a
f s, asα– , a(α), ds < .
Then FBVP (.), (.) has at least one solution in C α– [, ] provided
+ + α + β + γ < , (α)
where is the same as in Lemma .. Proof Set
= u ∈ dom L\ Ker L : Lu = λNu for some λ ∈ [, ] . Then, for u ∈ , since Lu = λNu, so λ = , Nu ∈ Im L = Ker Q, hence
t
α– f s, u(s), Dα– u(s), D u(s) ds = . + +
Thus, from (H), there exists t ∈ [, ] such that α– D u(t ) ≤ A. + Now, α– Dα– + u(t) = D+ u(t ) +
t t
Dα+ u(s) ds,
and so α–
D u() ≤ Dα– u(t) ≤ Dα– u(t ) + Dα u(t) + + + + ∞ ≤ A + Lu ≤ A + Nu .
(.)
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Again, for u ∈ , u ∈ dom L\Ker L, then (I – P)u ∈ dom L ∩ Ker P and LPu = . Thus, from Lemma ., we have
(I – P)u α– = KP L(I – P)u α– ≤ L(I – P)u C C ≤ Nu .
(.)
From (.), (.), we have
uC α– ≤ PuC α– + (I – P)u C α–
+ Dα– = + u() + (I – P)u C α– (α)
≤A + + + + Nu . (α) (α) By this and (H), we have uC α–
+ + + + ω ≤A (α) (α)
+ + + α + β + γ uC α– (α)
and uC α– ≤
A( (α) + ) + ( (α) + + )ω – ( (α) + + )(α + β + γ )
.
Therefore, is bounded. Let
= {u ∈ Ker L : Nu ∈ Im L}. For u ∈ , there is u ∈ Ker L = {u ∈ dom L | u = at α– , t ∈ [, ], a ∈ R}, and Nu ∈ Im L, thus t f s, at α– , a(α), a(α)s ds = . A From (H), we get |a| ≤ (α) , thus is bounded. Next, according to the condition (H), for any a ∈ R, if |a| > B, then either
t
a
f s, asα– , a(α), ds <
(.)
or else a
t
f s, as
α–
, a(α), ds > .
If (.) holds, set
= u ∈ Ker L : –λJu + ( – λ)QNu = , λ ∈ [, ] ,
(.)
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here Q is given by (.) and J : Ker L → Im Q is the linear isomorphism given by J(at α– ) = a , ∀a ∈ R, t ∈ [, ]. For u = at α– ∈ , [t]
t
λa = ( – λ)
f s, asα– , a(α), ds .
If λ = , then a = . Otherwise, if |a| > B, in view of (.), one has
t
a( – λ)
f s, asα– , a(α), ds < ,
which contradicts λa ≥ . Thus, ⊂ {u ∈ Ker L | u = at α– , |a| ≤ B} is bounded. If (.) holds, then define the set
= x ∈ Ker L : λJu + ( – λ)QNu = , λ ∈ [, ] , here J is as above. Similar to the above argument, we can show that is bounded too. In the following, we shall prove that all the conditions of Theorem . are satisfied. Let
be a bounded open subset of Y such that i= i ⊂ . By Lemma . and standard arguments, we can prove that KP (I – Q)N : → Y is compact, thus N is L-compact on . Then, by the above argument, we have (i) Lu = λNu, for every (u, λ) ∈ [(dom L\ Ker L) ∩ ∂ ] × (, ), (ii) Nu ∈/ Im L for u ∈ Ker L ∩ ∂ . Finally, we will prove that (iii) of Theorem . is satisfied. Let H(u, λ) = ±λJu + ( – λ)QNu. According to the above argument, we know H(u, λ) = for u ∈ Ker L ∩ ∂ . Thus, by the homotopy property of degree, we have deg(QN|Ker L , Ker L ∩ , ) = deg H(·, ), Ker L ∩ , = deg H(·, ), Ker L ∩ , = deg(J, Ker L ∩ , ) = . Then, by Theorem ., Lu = Nu has at least one solution in dom L ∩ , so that FBVP (.), (.) has a solution in C α– [, ]. The proof is complete. Theorem . Let f : [, ] × R → R be a continuous function. Assume that [t] = , (A), (H) and the following conditions (H), (H) hold: α– (H) There exists a constant A > such that for u ∈ dom L, if |Dα– + u(t)| + |D+ u(t)| > A t α– α– for all t ∈ [, ], then [ f (s, u(s), D+ u(s), D+ u(s)) ds] = . (H) There exists a constant B > such that either for each a ∈ R : |a| > B,
t
a
f s, asα– , , a(α – ) ds >
or for each a ∈ R : |a| > B,
t
a
f s, as
α–
, , a(α – ) ds < .
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Then FBVP (.), (.) has at least one solution in C α– [, ] provided
+ + α + β + γ < , (α – )
where is the same as in Lemma .. Theorem . Let f : [, ] × R → R be a continuous function. Assume that [t ] = , (A), (H) and the following conditions (H), (H) hold: α– (H) There exists a constant A > such that for u ∈ dom L, if |Dα– + u(t)| + |D+ u(t)| > A t α– α– for all t ∈ [, ], then [ (t – s)f (s, u(s), D+ u(s), D+ u(s)) ds] = . (H) There exists a constant B > such that either for each a ∈ R : |a| > B,
t
a
(t – s)f s, asα– , , a(α – ) ds >
or for each a ∈ R : |a| > B,
t
a
(t – s)f s, asα– , , a(α – ) ds < .
Then FBVP (.), (.) has at least one solution in C α– [, ] provided
+ + α + β + γ < , (α – )
where is the same as in Lemma .. The proofs of Theorem . and Theorem . are similar to that of Theorem .. So, we omit them. The above Theorem ., Theorem . and Theorem . are the existence of solutions to FBVP (.), (.) in the case of dim Ker L = . By making use of Theorem ., Lemma . and Lemma ., we obtain the existence of solutions for FBVP (.), (.) in the case of dim Ker L = . Theorem . Let f : [, ] × R → R be a continuous function. Assume that [t] [t ] – [t ] [t ] = , (A), (H) and the following conditions (H), (H) hold: α– (H) There exists a constant A > such that for u ∈ dom L, if |Dα– + u(t)| + |D+ u(t)| > A for all t ∈ [, ], then
t
α– α– f s, u(s), D+ u(s), D+ u(s) ds = or
t
α– (t – s)f s, u(s), Dα– u(s), D u(s) ds . = + +
(H) There exists a constant B > such that for a , a ∈ R satisfying |a | + |a | > B, either
t
α– α– ds > , N a t + a t
t
(t – s)N a t α– + a t α– ds >
a
a
(.)
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or
t
α– α– ds < , N a t + a t
t
(t – s)N a t α– + a t α– ds < .
a
a
(.)
Then FBVP (.), (.) has at least one solution in C α– [, ] provided
++ α + β + γ < . (α) (α – )
Proof Set
= u ∈ dom L\ Ker L : Lu = λNu for some λ ∈ [, ] . Then, for u ∈ , since Lu = λNu, so λ = , Nu ∈ Im L = Ker Q, hence
t
α– f s, u(s), Dα– u(s), D u(s) ds = + +
t
α– α– (t – s)f s, u(s), D+ u(s), D+ u(s) ds = .
and
Thus, from (H), there exists t ∈ [, ] such that α– D u(t ) + Dα– u(t ) ≤ A. + + Now, α– Dα– + u(t) = D+ u(t ) +
t
α– Dα– + u(t) = D+ u(t ) +
Dα+ u(s) ds,
t t t
Dα– + u(s) ds,
and so
α– D u() ≤ Dα– u(t) ≤ Dα– u(t ) + Dα u(t) + + + + ∞ ≤ A + Lu ≤ A + Nu ,
α– D u() ≤ Dα– u(t) ≤ Dα– u(t ) + Dα– u(t) + + + + ∞ α– α–
≤ D+ u(t ) + D+ u(t) ∞ α– α
≤ Dα– + u(t ) + D+ u(t ) + D+ u(t) ≤ A + Nu .
(.)
(.)
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Again, for u ∈ , u ∈ dom L\Ker L, then (I – P)u ∈ dom L ∩ Ker P and LPu = . Thus, from Lemma ., we have
(I – P)u α– = KP L(I – P)u α– ≤ + C C ≤ +
Nu . (α)
L(I – P)u (α) (.)
From (.), (.) and (.), we have
uC α– ≤ PuC α– + (I – P)u C α–
α–
+ D+ u() + + Dα– = + u() + (I – P)u C α– (α) (α – )
+ + + ++ Nu . ≤A (α) (α – ) (α) (α – ) By this and (H), we have uC α–
+ + + ++ ω ≤A (α) (α – ) (α) (α – )
+ ++ α + β + γ uC α– (α) (α – )
and uC α– ≤
A( (α) +
(α–)
– ( (α) ++
+ ) + ( (α) ++ )(α (α–)
)ω (α–)
+ β + γ )
.
Therefore, is bounded. Let
= {u ∈ Ker L : Nu ∈ Im L}. For u ∈ , there is u ∈ Ker L = {u ∈ dom L | u = at α– + bt α– , t ∈ [, ], a, b ∈ R}, and Nu ∈ Im L, thus
t
f s, asα– + bsα– , a(α), a(α)s + b(α – ) ds =
t
(t – s)f s, asα– + bsα– , a(α), a(α)s + b(α – ) ds = .
and
From (H), we get (α)|a| + (α – )|b| ≤ A. Then, for u ∈ , we have uC α– ≤ (α) + |a| + (α – ) + |b| ≤ + thus is bounded.
A, (α – )
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Next, for any a , a ∈ R, define a linear isomorphism J : Ker L → Im Q by a [t ] – a [t ] – (a [t ] – a [t])t J a t α– + a t α– = . [t] [t ] – [t ] [t ] If (.) holds, set
= u ∈ Ker L : λJu + ( – λ)QNu = , λ ∈ [, ] , where Q is given by (.). For u = a t α– + a t α– ∈ , from λJu + ( – λ)QNu = , we obtain t
Nu(s) ds t a λ + ( – λ)
t (t – s)Nu(s) ds = – t a λ + ( – λ)
and t
a λ + ( – λ) Nu(s) ds t
t (t – s)Nu(s) ds = . – [t] a λ + ( – λ)
By [t] [t ] – [t ] [t ] = , it yields ⎧ ⎨a λ + ( – λ) [ t Nu(s) ds] = , ⎩a λ + ( – λ) [ t (t – s)Nu(s) ds] = .
If λ = , then a = a = . Otherwise, if |a | + |a | > B, considering the above equalities and (.), we have t t λ a + a = –( – λ) a Nu(s) ds + a (t – s)Nu(s) ds < ,
which contradicts λ(a + a ) ≥ . If (.) holds, then we take
= u ∈ Ker L : –λJu + ( – λ)QNu = , λ ∈ [, ] and, again, obtain a contradiction. Thus, in either case,
uC α– = a t α– + a t α– C α– ≤ |a | + (α) + |a | + (α – ) ≤ B + (α) + (α – ) for all u ∈ , that is, is bounded.
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In the following, we shall prove that all the conditions of Theorem . are satisfied. Let
be a bounded open subset of Y such that i= i ⊂ . By Lemma . and standard arguments, we can prove that KP (I – Q)N : → Y is compact, thus N is L-compact on . Then, by the above argument, we have (i) Lu = λNu for every (u, λ) ∈ [(dom L\ Ker L) ∩ ∂ ] × (, ), (ii) Nu ∈/ Im L for u ∈ Ker L ∩ ∂ . Finally, we will prove that (iii) of Theorem . is satisfied. Let H(u, λ) = ±λJu + ( – λ)QNu. According to the above argument, we know H(u, λ) = for u ∈ Ker L ∩ ∂ . Thus, by the homotopy property of degree, we have deg(QN|Ker L , Ker L ∩ , ) = deg H(·, ), Ker L ∩ , = deg H(·, ), Ker L ∩ , = deg(±J, Ker L ∩ , ) = . Then, by Theorem ., Lu = Nu has at least one solution in dom L ∩ , so that FBVP (.), (.) has a solution in C α– [, ]. The proof is complete.
Competing interests The authors declare that they have no competing interests. Authors’ contributions All authors contributed equally and significantly in this research work. All authors read and approved the final manuscript. Author details 1 Department of Statistics and Finance, Shandong University of Science and Technology, Qingdao, 266590, P.R. China. 2 Department of Mathematics, Shandong University of Science and Technology, Qingdao, 266590, P.R. China. Acknowledgements The authors would like to thank the referees for carefully reading this article and making valuable comments and suggestions. This work is supported by the Foundation items: NSFC (10971179, 11071141), NSF (BS2010SF023, BS2012SF022) of Shandong Province. Received: 11 January 2013 Accepted: 16 July 2013 Published: 7 August 2013 References 1. Podlubny, I: Fractional Differential Equations. Mathematics in Science and Engineering. Academic Press, New York (1999) 2. Kilbas, AA, Srivastava, HM, Trujillo, JJ: Theory and Applications of Fractional Differential Equations. North-Holland Mathematics Studies, vol. 204. Elsevier, Amsterdam (2006) 3. Babakhani, A, Gejji, VD: Existence of positive solutions of nonlinear fractional differential equations. J. Math. Anal. Appl. 278, 434-442 (2003) 4. Lakshmikantham, V, Leela, S, Devi, JV: Theory of Fractional Dynamic Systems. Cambridge Academic Publishers, Cambridge (2009) 5. Lakshmikantham, V, Leela, S: Nagumo-type uniqueness result for fractional differential equations. Nonlinear Anal. TMA 71, 2886-2889 (2009) 6. Bai, Z: On positive solutions of a nonlocal fractional boundary value problem. Nonlinear Anal. TMA 72, 916-924 (2010) 7. Bai, Z, Lü, H: Positive solutions of boundary value problems of nonlinear fractional differential equation. J. Math. Anal. Appl. 311, 495-505 (2005) 8. Su, X: Boundary value problem for a coupled system of nonlinear fractional differential equations. Appl. Math. Lett. 22, 64-69 (2009) 9. Jiang, D, Yuan, C: The positive properties of the Green function for Dirichlet-type boundary value problems of nonlinear fractional differential equations and its application. Nonlinear Anal. TMA 72(2), 710-719 (2010) 10. Zhang, S: The existence of a positive solution for a nonlinear fractional differential equation. J. Math. Anal. Appl. 252, 804-812 (2000)
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doi:10.1186/1687-1847-2013-233 Cite this article as: Zou and Cui: Existence results for a functional boundary value problem of fractional differential equations. Advances in Difference Equations 2013 2013:233.
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