A Cauchy-type problem with a sequential fractional derivative in the space of continuous functions

Boundary Value Problems20122012:58

DOI: 10.1186/1687-2770-2012-58

Received: 12 February 2012

Accepted: 17 May 2012

Published: 17 May 2012

Abstract

A Cauchy-type nonlinear problem for a class of fractional differential equations with sequential derivatives is considered in the space of weighted continuous functions. Some properties and composition identities are derived. The equivalence with the associated integral equation is established. An existence and uniqueness result of the global continuous solution is proved.

AMS Subject Classification: 26A33; 34A08; 34A34; 34A12; 45J08.

Keywords

fractional derivatives Riemann-Liouville fractional derivative sequential fractional derivative fractional differential equation

1 Introduction

We consider a Cauchy-type problem associated with the equation
D a α ( x - a ) r D a β y ( x ) = f ( x , y ) , x > a , 0 < α < 1 , 0 β 1 , r < α , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-58/MediaObjects/13661_2012_Article_152_Equ1_HTML.gif
(1)

where D a α http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-58/MediaObjects/13661_2012_Article_152_IEq1_HTML.gif and D a β http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-58/MediaObjects/13661_2012_Article_152_IEq2_HTML.gif are the Riemann-Liouville fractional derivatives.

In recent years there has been a considerable interest in the theory and applications of fractional differential equations. As for the theory, the investigations include the existence and uniqueness of solutions, asymptotic behavior, stability, etc. See for example the books [13] and the articles [410] and the references therein.

As for the applications, fractional models provide a tool for capturing and understanding complex phenomena in many fields. See for example the surveys in [1, 11] and the collection of applications in [12].

Some recent applications include control systems [13, 14], viscoelasticity [1518], and nanotechnology [19]. Also fractional models are used to model a vibrating string [20], and anomalous transport [21], anomalous diffusion [2224].

Another field of applications is in random walk and stochastic processes [2527] and its applications in financial modeling [2830]. Other physical and engineering processes are given in [31, 32]

In a series of articles, [3335], Glushak studied the uniform well-posedness of a Cauchy-type problem with two fractional derivatives and bounded operator. He also proposed a criterion for the uniform correctness of unbounded operator.

In this article we prove an existence and uniqueness result for a nonlinear Cauchy-type problem associated with the Equation (1) in the space of weighted continuous functions.

We start with some preliminaries in Section 2. In Section 3 we define the sequential derivative and develop some properties and composition identities. In Section 4 we set up the Cauchy-type problem and establish the equivalence with the associated integral equation. Finally, in Section 5 we prove the existence and uniqueness of the solution.

2 Preliminaries

In this section we present some definitions, lemmas, properties and notation which we use later. For more details please see [1].

Let -∞ < a < b < ∞. Let C[a, b] denote the spaces of continuous functions on [a, b]. We denote by L(a, b) the spaces of Lebesgue integrable functions on (a, b). Let CL(a, b) = L(a, b) ⋂ C(a, b].

We introduce the weighted spaces of continuous functions
C γ [ a , b ] = f : ( a , b ] : ( x - a ) γ f ( x ) C [ a , b ] , γ , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-58/MediaObjects/13661_2012_Article_152_Equ2_HTML.gif
(2)
with the norm
f C γ [ a , b ] = ( x - a ) γ f ( x ) C [ a , b ] , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-58/MediaObjects/13661_2012_Article_152_Equ3_HTML.gif
(3)
where
f C [ a , b ] = max x [ a , b ] f ( x ) . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-58/MediaObjects/13661_2012_Article_152_Equ4_HTML.gif
(4)

In the case f is not defined at x = a or γ < 0 we let ( x - a ) γ f ( x ) x = a = lim x - a + ( x - a ) γ f ( x ) . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-58/MediaObjects/13661_2012_Article_152_IEq3_HTML.gif The spaces C γ [a, b] satisfy the following properties.

  • C0[a,b] = C[a,b].

  • C γ 1 [ a , b ] C γ 2 [ a , b ] , γ 1 < γ 2 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-58/MediaObjects/13661_2012_Article_152_IEq4_HTML.gif.

  • C γ [a,b] ⊂CL(a,b),γ < 1.

  • fC γ [a, b] if and only if fC(a, b) and lim x a + ( x - a ) γ f ( x ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-58/MediaObjects/13661_2012_Article_152_IEq5_HTML.gif exists and is finite.

The left-sided Riemann-Liouville fractional integrals and derivatives are defined as follows.

Definition 1 Let fL(a,b). The integral
I a α f ( x ) : = 1 Γ ( α ) a x f ( s ) ( x - s ) 1 - α d s , x > a , α > 0 , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-58/MediaObjects/13661_2012_Article_152_Equ5_HTML.gif
(5)

is called the left-sided Riemann-Liouville fractional integral of order α of the function f.

Definition 2 The expression
D a α f ( x ) : = D I a α - 1 f ( x ) , x > a , 0 < α < 1 , D = d d x , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-58/MediaObjects/13661_2012_Article_152_Equ6_HTML.gif
(6)

is called the left-sided Riemann-Liouville fractional derivative of order α of f provided the right-hand side exists.

For power functions we have the following formulas.

Lemma 3 For x > a we have
I a α ( t - a ) β - 1 ( x ) = Γ ( β ) Γ ( β + α ) ( x - a ) β + α - 1 , α 0 , β > 0 . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-58/MediaObjects/13661_2012_Article_152_Equ7_HTML.gif
(7)
D a α ( t - a ) α - 1 ( x ) = 0 , 0 < α < 1 . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-58/MediaObjects/13661_2012_Article_152_Equ8_HTML.gif
(8)

Next we present some mapping properties of the operator I a α http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-58/MediaObjects/13661_2012_Article_152_IEq6_HTML.gif.

Lemma 4 For α > 0, I a α http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-58/MediaObjects/13661_2012_Article_152_IEq6_HTML.gif maps L(a, b) into L(a, b).

The proof of Lemma 4 is given in [36]. The following lemma is proved in [37].

Lemma 5 For α > 0, I a α http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-58/MediaObjects/13661_2012_Article_152_IEq7_HTML.gif maps C[a, b] into C[a, b].

The following lemma is proved in [38].

Lemma 6 Let α ≥ 0. If fCL(a, b) then I a α f C L ( a , b ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-58/MediaObjects/13661_2012_Article_152_IEq8_HTML.gif.

The mapping properties of I a α http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-58/MediaObjects/13661_2012_Article_152_IEq7_HTML.gif in the spaces C γ [a, b], 0 ≤ αγ < 1, are given in [1], Lemma 2.8 which is proved in [39] in Russian. For completeness we present here a more general result for α > 0 and γ < 1. First we prove the necessity condition at the left end.

Lemma 7 Let α ≥ 0 and γ < 1. If fC γ [a, b] then
lim x a + ( x - a ) γ - α I a α f ( x ) = c Γ ( 1 - γ ) Γ ( 1 + α - γ ) , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-58/MediaObjects/13661_2012_Article_152_Equ9_HTML.gif
(9)

where c = lim x a + ( x - a ) γ f ( x ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-58/MediaObjects/13661_2012_Article_152_IEq9_HTML.gif.

Proof. Note that from Lemma 3 we have
I a α ( x - a ) - γ = Γ ( 1 - γ ) Γ ( 1 + α - γ ) ( x - a ) α - γ . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-58/MediaObjects/13661_2012_Article_152_Equa_HTML.gif
Thus
( x - a ) γ - α I a α f ( x ) - c Γ ( 1 - γ ) Γ ( 1 + α - γ ) = ( x - a ) γ - α I a α f ( x ) - c ( x - a ) γ - α I a α ( x - a ) - γ = ( x - a ) γ - α I a α f ( x ) - c I a α ( x - a ) - γ = ( x - a ) γ - α I a α ( x - a ) - γ { ( x - a ) γ f ( x ) - c } . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-58/MediaObjects/13661_2012_Article_152_Equb_HTML.gif
Now, given ϵ > 0 there exists δ > 0 such that x - a < δ implies that
( x - a ) γ f ( x ) - c < ε Γ ( 1 + α - γ ) Γ ( 1 - γ ) . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-58/MediaObjects/13661_2012_Article_152_Equc_HTML.gif
Thus
( x - a ) γ - α I a α f ( x ) - c Γ ( 1 - γ ) Γ ( 1 + α - γ ) = ( x - a ) γ - α I a α ( x - a ) - γ { ( x - a ) γ f ( x ) - c } < ε Γ ( 1 + α - γ ) Γ ( 1 - γ ) ( x - a ) γ - α I a α ( x - a ) - γ = ε Γ ( 1 + α - γ ) Γ ( 1 - γ ) ( x - a ) γ - α Γ ( 1 - γ ) Γ ( 1 + α - γ ) ( x - a ) α - γ = ε . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-58/MediaObjects/13661_2012_Article_152_Equd_HTML.gif

This yields the limit (9).

Next we present the mapping properties of I a α http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-58/MediaObjects/13661_2012_Article_152_IEq6_HTML.gif in the spaces C γ [a, b].

Lemma 8 Let α > 0 and γ < 1. If fC γ [a, b] then I a α f C γ - α [ a , b ] http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-58/MediaObjects/13661_2012_Article_152_IEq10_HTML.gif and for x ∈ (a, b] we have
I a α f ( x ) Γ ( 1 - γ ) Γ ( 1 + α - γ ) ( x - a ) α - γ f C γ [ a , b ] . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-58/MediaObjects/13661_2012_Article_152_Equ10_HTML.gif
(10)
Proof. From Lemmas 6 and 7 we have I α fC(a, b) and lim x a + ( x - a ) γ - α I a α f ( x ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-58/MediaObjects/13661_2012_Article_152_IEq11_HTML.gif exists and is finite. Thus I a α f C γ - α [ a , b ] http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-58/MediaObjects/13661_2012_Article_152_IEq12_HTML.gif. Now for x ∈ (a, b] we have
I a α f ( x ) = 1 Γ ( α ) a x ( x - t ) α - 1 ( t - a ) - γ ( t - a ) γ f ( t ) d t 1 Γ ( α ) f C γ [ a , b ] a x ( x - t ) α - 1 ( t - a ) - γ d t . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-58/MediaObjects/13661_2012_Article_152_Eque_HTML.gif

The relation (10) follows by applying Lemma 3.

Consequently, from Lemma 8 we have the following property.

Lemma 9 Let α > 0, γ < 1, and r ∈ ℝ. If fCγ[a, b] then ( x - a ) - r I a α f C γ + r - α [ a , b ] http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-58/MediaObjects/13661_2012_Article_152_IEq13_HTML.gif. In particular, if γ + r - α < 1 then I α fCL(a, b).

Later, the following observation is important.

Lemma 10 Let α > 0 and r < α. If fCL(a, b) then ( x - a ) - r I a α f C L ( a , b ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-58/MediaObjects/13661_2012_Article_152_IEq14_HTML.gif.

Proof. When r ≤ 0 the result follows clearly from Lemma 6. When r > 0 it follows from Lemma 6 that ( x - a ) - r I a α f C ( a , b ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-58/MediaObjects/13661_2012_Article_152_IEq15_HTML.gif and we only need to show that ( x - a ) - r I a α f L ( a , b ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-58/MediaObjects/13661_2012_Article_152_IEq16_HTML.gif.

For any x ∈ (a, b] we have the following inequality.
I a α f ( x ) = 1 Γ ( α ) a x ( x - t ) α - 1 f ( t ) d t = 1 Γ ( α ) a x ( x - t ) α - r - 1 ( x - t ) r f ( t ) d t ( x - a ) r Γ ( α ) a x ( x - t ) α - r - 1 f ( t ) d t = Γ ( α - r ) Γ ( α ) ( x - a ) r I a α - r f ( x ) . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-58/MediaObjects/13661_2012_Article_152_Equf_HTML.gif
Or,
( x - a ) - r I a α f ( x ) Γ ( α - r ) Γ ( α ) I a α - r f ( x ) . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-58/MediaObjects/13661_2012_Article_152_Equg_HTML.gif

From Lemma 4 the right-hand side is in L(a, b) and thus ( x - a ) - r I a α f L ( a , b ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-58/MediaObjects/13661_2012_Article_152_IEq17_HTML.gif. This completes the proof.

The following lemma follows by direct calculations using Dirichlet formula, [36].

Lemma 11 Let α ≥ 0, β ≥ 0, and fCL(a, b). Then
I a α I a β f ( x ) = I a α + β f ( x ) , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-58/MediaObjects/13661_2012_Article_152_Equ11_HTML.gif
(11)

for all x ∈ (a, b].

Lemma 11 leads to the left inverse operator.

Lemma 12 Let α > 0 and fCL(a, b). Then
D a α I a α f ( x ) = f ( x ) , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-58/MediaObjects/13661_2012_Article_152_Equ12_HTML.gif
(12)

for all x ∈ (a, b].

Now we present a version of the fundamental theorem of fractional calculus.

Lemma 13 Let 0 < α < 1. If fC(a, b) and D a α f C L ( a , b ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-58/MediaObjects/13661_2012_Article_152_IEq18_HTML.gif, then f C L ( a , b ) , I a α f ( a + ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-58/MediaObjects/13661_2012_Article_152_IEq19_HTML.gif exists and is finite, and
I a α D a α f ( x ) = f ( x ) - I a 1 - α f ( a + ) Γ ( α ) ( x - a ) α - 1 , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-58/MediaObjects/13661_2012_Article_152_Equ13_HTML.gif
(13)

for all x ∈ (a, b].

Proof. From Lemma 12 we have for all x ∈ (a, b] the relation
D a α I a α D a α f ( x ) = D a α f ( x ) , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-58/MediaObjects/13661_2012_Article_152_Equh_HTML.gif
which we can write as
D a α f - I a α D a α f ( x ) = 0 . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-58/MediaObjects/13661_2012_Article_152_Equi_HTML.gif
This implies that
f ( x ) - I a α D a α f ( x ) = c ( x - a ) α - 1 , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-58/MediaObjects/13661_2012_Article_152_Equ14_HTML.gif
(14)
for some constant c. Since Lemma 6 implies that I a α D a α f C L ( a , b ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-58/MediaObjects/13661_2012_Article_152_IEq20_HTML.gif, we also have fCL(a, b). Also, if we apply I a 1 - α http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-58/MediaObjects/13661_2012_Article_152_IEq21_HTML.gif to both sides of (14) we obtain
I a 1 - α f ( x ) = I a D a α f ( x ) = c Γ ( α ) . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-58/MediaObjects/13661_2012_Article_152_Equj_HTML.gif

Taking the limit yields I a 1 - α f ( a + ) = c Γ ( α ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-58/MediaObjects/13661_2012_Article_152_IEq22_HTML.gif and (13) is obtained.

In the proof of our existence and uniqueness result we will use the following results.

Lemma 14 Let γ ∈ ℝ, a < c < b, gC γ [a, c], gC[c, b] and g is continuous at c. Then gC γ [a, b].

Theorem 15 ([1], Banach Fixed Point Theorem) Let (U, d) be a nonempty complete metric space. Let T : UU be a map such that for every u, vU, the relation
d ( T u , T v ) w d ( u , v ) , 0 w < 1 , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-58/MediaObjects/13661_2012_Article_152_Equk_HTML.gif

holds. Then the operator T has a unique fixed point u* ∈ U.

3 Sequential derivative

In this section we define the sequential derivative and integral that we consider and develop some of their properties. In particular, we derive the composition identities.

Definition 16 Let α > 0, β > 0, r ∈ ℝ. Let fCL(a, b). Define the sequential integral J r , a α , β f http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-58/MediaObjects/13661_2012_Article_152_IEq23_HTML.gif and the sequential derivative D r , a α , β f http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-58/MediaObjects/13661_2012_Article_152_IEq24_HTML.gif by
J r , a α , β f ( x ) = I a α ( x - a ) - r I a β f ( x ) , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-58/MediaObjects/13661_2012_Article_152_Equ15_HTML.gif
(15)
and
D r , a α , β f ( x ) = D a α ( x - a ) r D a β f ( x ) , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-58/MediaObjects/13661_2012_Article_152_Equ16_HTML.gif
(16)

if the right-hand sides exist.

From Lemma 3 we have the following formula for the power function.

Lemma 17 Let α > 0, β > 0, r ∈ ℝ. If
ρ > max { - 1 , β - r - 1 } , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-58/MediaObjects/13661_2012_Article_152_Equl_HTML.gif
then for x > a,
D r , a α , β ( x - a ) ρ = Γ ( ρ + 1 ) Γ ( ρ - β + 1 ) Γ ( ρ + r - β + 1 ) Γ ( ρ + r - β - α + 1 ) ( x - a ) ρ + r - β - α . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-58/MediaObjects/13661_2012_Article_152_Equ17_HTML.gif
(17)

Moreover, from Lemmas 3 and 17 we have the following vanishing derivatives.

Lemma 18

(a) Let α > 0, 0 < β < 1, r ∈ ℝ. Then for x > a,
D r , a α , β ( x - a ) β - 1 = 0 . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-58/MediaObjects/13661_2012_Article_152_Equ18_HTML.gif
(18)
(b) Let 0 < α < 1 and β > 0. Let r ∈ ℝ be such that r < α + β. Then for x > a,
D r , a α , β ( x - a ) α + β - r - 1 = 0 . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-58/MediaObjects/13661_2012_Article_152_Equ19_HTML.gif
(19)
Lemma 19 (Left inverse) Let α > 0, β > 0, and r ∈ ℝ. If fCL(a, b) such that ( x - a ) - r I a α f C L ( a , b ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-58/MediaObjects/13661_2012_Article_152_IEq25_HTML.gif then
D r , a α , β J r , a β , α f ( x ) = f ( x ) , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-58/MediaObjects/13661_2012_Article_152_Equ20_HTML.gif
(20)

for all x ∈ (a, b].

Proof. Relation (20) follows directly by applying Lemma 12 twice.

From Lemmas 8 and 9 we have the following mapping property of the operator J r , a β , α http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-58/MediaObjects/13661_2012_Article_152_IEq26_HTML.gif.

Lemma 20 Let α > 0, β > 0, and r < 1 + α. Let 0 ≤ γ < min{1, 1 + α - r}. If fC γ [a, b] then J r , a β , α f C γ + r - α - β [ a , b ] http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-58/MediaObjects/13661_2012_Article_152_IEq27_HTML.gif and for x ∈ (a, b] we have
J r , a β , α f ( x ) k f C γ [ a , b ] ( x - a ) α + β - r - γ , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-58/MediaObjects/13661_2012_Article_152_Equ21_HTML.gif
(21)
where
k = Γ ( 1 - γ ) Γ ( 1 + α - r - γ ) Γ ( 1 - γ + α ) Γ ( 1 + α + β - r - γ ) . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-58/MediaObjects/13661_2012_Article_152_Equ22_HTML.gif
(22)

Lemma 20 implies the following.

Lemma 21 Let α > 0, β > 0, and r < 1 + α. Let 0 ≤ γ < min{1, 1 + α-r}. If rα + β, then J r , a β , α http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-58/MediaObjects/13661_2012_Article_152_IEq26_HTML.gif is bounded in C γ [a, b] and
J r , a β , α f C γ [ a , b ] k ( b - a ) α + β - r f C γ [ a , b ] , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-58/MediaObjects/13661_2012_Article_152_Equ23_HTML.gif
(23)

where k is given by (22).

Proof. Since γ + r - α - βγ, then from Lemma 20 we have
J r , a β , α f C γ + r - α - β [ a , b ] C γ [ a , b ] . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-58/MediaObjects/13661_2012_Article_152_Equm_HTML.gif

The bound in (23) follows by multiply (21) by (x - a) γ and taking the maximum.

As a special case of Lemma 21, we have

Lemma 22 Let α > 0, β > 0, and r < min{α + 1, α + β}. Then J r , a β , α http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-58/MediaObjects/13661_2012_Article_152_IEq26_HTML.gif, maps C[a, b] into C[a, b] and
J r , a β , α f C [ a , b ] L ( b - a ) α + β - r f C [ a , b ] , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-58/MediaObjects/13661_2012_Article_152_Equ24_HTML.gif
(24)
where
L = Γ ( 1 + α - r ) Γ ( 1 + α ) Γ ( 1 + α + β - r ) . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-58/MediaObjects/13661_2012_Article_152_Equ25_HTML.gif
(25)

The following is an analogous result to the result for the Riemann-Liouville integral proved in [10].

Lemma 23 Let α > 0, β > 0, and r < α. Let fCL(a, c). Let
g ( x ) = 1 Γ ( β ) a c ( x - t ) β - 1 ( t - a ) - r I a α f ( t ) d t . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-58/MediaObjects/13661_2012_Article_152_Equn_HTML.gif
Then
lim x c + g ( x ) = J r , a β , α f ( c ) . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-58/MediaObjects/13661_2012_Article_152_Equo_HTML.gif
Proof. Since r < α, Lemma 10 implies that ( x - a ) - r I a α f ( x ) C L ( a , c ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-58/MediaObjects/13661_2012_Article_152_IEq28_HTML.gif. Thus J r , a β , α f ( c ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-58/MediaObjects/13661_2012_Article_152_IEq29_HTML.gif is finite and
g ( x ) - J r , a β , c f ( c ) 1 Γ ( β ) a c k ( x , t ) ( t - a ) - r I a α f ( t ) d t , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-58/MediaObjects/13661_2012_Article_152_Equp_HTML.gif
where
k ( x , t ) = ( c - t ) β - 1 - ( x - t ) β - 1 . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-58/MediaObjects/13661_2012_Article_152_Equq_HTML.gif

Since lim x c + k ( x , t ) = 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-58/MediaObjects/13661_2012_Article_152_IEq30_HTML.gif, the limit of the right-hand side vanishes and the proof is complete.

The following lemma relates the fractional derivative D r , a α , β http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-58/MediaObjects/13661_2012_Article_152_IEq31_HTML.gif to the Riemann-Liouville derivative D a β http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-58/MediaObjects/13661_2012_Article_152_IEq32_HTML.gif.

Lemma 24 Let 0 < α < 1, β ≥ 0, and r ∈ ℝ. If D a β y ( x ) C ( a , b ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-58/MediaObjects/13661_2012_Article_152_IEq33_HTML.gif and D r , a α , β y C L ( a , b ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-58/MediaObjects/13661_2012_Article_152_IEq34_HTML.gif then ( x - a ) r D a β y C L ( a , b ) , I a 1 - α ( x - a ) r D a β y ( a + ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-58/MediaObjects/13661_2012_Article_152_IEq35_HTML.gif exists and finite, and
D a β y ( x ) = ( x - a ) - r I a α D r , a α , β y ( x ) + I 1 - α ( x - a ) r D a β y ( a + ) Γ ( α ) ( x - a ) α - r - 1 , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-58/MediaObjects/13661_2012_Article_152_Equ26_HTML.gif
(26)

for all x ∈ (a, b]. If in addition, r < α then D a β y C L ( a , b ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-58/MediaObjects/13661_2012_Article_152_IEq36_HTML.gif.

Proof. Clearly D a β y C ( a , b ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-58/MediaObjects/13661_2012_Article_152_IEq37_HTML.gif implies that ( x - a ) r D a β y C ( a , b ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-58/MediaObjects/13661_2012_Article_152_IEq38_HTML.gif. Thus we can apply Lemma 13 to ( x - a ) r D a β y http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-58/MediaObjects/13661_2012_Article_152_IEq39_HTML.gif and obtain
I a α D r , a α , β y ( x ) = I a α D a α ( x - a ) r D a β y ( x ) = ( x - a ) r D a β y ( x ) - I a 1 - α ( x - a ) r D a β y ( a + ) Γ ( α ) ( x - a ) α - 1 . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-58/MediaObjects/13661_2012_Article_152_Equr_HTML.gif

By multiplying both sides by (x - a) r we obtain (26). If r < α then Lemma 10 implies that ( x - a ) - r I a α D r , a α , β y ( x ) C L ( a , b ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-58/MediaObjects/13661_2012_Article_152_IEq40_HTML.gif and thus from (26) we have D a β y C L ( a , b ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-58/MediaObjects/13661_2012_Article_152_IEq41_HTML.gif. This proves the result.

The Next lemma gives an analogous result to the fundamental theorem of calculus in terms of the operators D r , a α , β http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-58/MediaObjects/13661_2012_Article_152_IEq31_HTML.gif and J r , a α , β http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-58/MediaObjects/13661_2012_Article_152_IEq42_HTML.gif.

Lemma 25 Let 0 < α < 1 and 0 < β < 1. Let yC(a, b) be such that D r , a α , β y C L ( a , b ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-58/MediaObjects/13661_2012_Article_152_IEq43_HTML.gif and D a β y C L ( a , b ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-58/MediaObjects/13661_2012_Article_152_IEq41_HTML.gif. Then both I a 1 - α ( x - a ) r D a β y ( a + ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-58/MediaObjects/13661_2012_Article_152_IEq44_HTML.gif and I a 1 - β y ( a + ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-58/MediaObjects/13661_2012_Article_152_IEq45_HTML.gif exist, yCL(a, b), and
J r , a β , α D r , a α , β y ( x ) = y ( x ) - I a 1 - β y ( a + ) Γ ( β ) ( x - a ) β - 1 - I a 1 - α ( x - a ) r D a β y ( a + ) Γ ( α ) Γ ( α + r ) Γ ( α + β - r ) ( x - a ) α + β - r - 1 , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-58/MediaObjects/13661_2012_Article_152_Equ27_HTML.gif
(27)

for all x ∈ (a, b].

Proof. By applying Lemma 13 twice we obtain
J r , a β , α D r , a α , β y ( x ) = I a β ( x - a ) - r I a α D a α ( x - a ) r D a β y ( x ) = I β ( x - a ) - r ( x - a ) r D a β y ( x ) - I a 1 - α ( x - a ) r D a β y ( a + ) Γ ( α ) ( x - a ) α - 1 = I a β D a β y - I a 1 - α ( x - a ) r D a β y ( a + ) Γ ( α ) I a β ( x - a ) α - 1 - r = y ( x ) - I a 1 - β y ( a + ) Γ β ( x - a ) β - 1 - I a 1 - α ( x - a ) r D a β y ( a + ) Γ ( α ) Γ ( α + r ) Γ ( α + β - r ) ( x - a ) α + β - r - 1 . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-58/MediaObjects/13661_2012_Article_152_Equs_HTML.gif

4 Cauchy-type problem and equivalency

Consider the Cauchy-type problem
D r , a α , β y ( x ) = f ( x , y ( x ) ) , a < x b , 0 < α < 1 , 0 < β < 1 , r , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-58/MediaObjects/13661_2012_Article_152_Equ28_HTML.gif
(28)
lim x a + I a 1 - α ( x - a ) r D a β y ( x ) = c 1 , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-58/MediaObjects/13661_2012_Article_152_Equ29_HTML.gif
(29)
lim x a + I a 1 - β y ( x ) = c 0 , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-58/MediaObjects/13661_2012_Article_152_Equ30_HTML.gif
(30)

where c0 and c1 are real numbers.

In this problem there are two conditions even when 0 < α + β < 1. The two initial conditions are based on the composition (27). The condition (29) is of one order less than that in the differential Equation (28) while the condition (30) is one order less than the equation for D a β y http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-58/MediaObjects/13661_2012_Article_152_IEq46_HTML.gif.

In addition, from [1, Lemma 3.2], the condition (30) follows from the condition
lim x a + ( x - a ) 1 - β y ( x ) = c 0 Γ ( β ) , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-58/MediaObjects/13661_2012_Article_152_Equ31_HTML.gif
(31)
and if 0 < α - r < 1 then (29) follows from the condition
lim x a + ( x - a ) 1 - α + r D a β y ( x ) = c 1 Γ ( α ) . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-58/MediaObjects/13661_2012_Article_152_Equ32_HTML.gif
(32)

Consequently, the results below hold under conditions of the type (31) and (32).

Now, Based on the composition in Lemma 24, in the next theorem we establish an equivalence with the following fractional integro-differential equation:
D a β y = ( x - a ) - r I a α f [ x , y ( x ) ] + c 1 Γ ( α ) ( x - a ) α - r - 1 , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-58/MediaObjects/13661_2012_Article_152_Equ33_HTML.gif
(33)
lim x a + I a 1 - β y ( x ) = c 0 . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-58/MediaObjects/13661_2012_Article_152_Equ34_HTML.gif
(34)

Theorem 26 Let 0 < α < 1, β > 0 and r ∈ ℝ. Let f : (a, b] × ℝ → ℝ be a function such that f(.,y(.)) ∈ C1-α[a, b] for any yC1-α[a, b]. Then we have the following.

(a) If yC1-α[a, b] satisfies (33) and (34) then y(x) satisfies (28-30).
  1. (b)

    If yC1-α[a, b] with D a β y C ( a , b ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-58/MediaObjects/13661_2012_Article_152_IEq37_HTML.gif satisfy (28-30), then y(x) satisfies (33-34).

     

Proof.

For assertion (a), let yC1-α[a, b] satisfy (33-34). We multiply (33) by (x - a) r to obtain
( x - a ) r D a β y = I a α f ( x , y ( x ) ) + c 1 Γ ( α ) ( x - a ) α - 1 . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-58/MediaObjects/13661_2012_Article_152_Equ35_HTML.gif
(35)

Next we apply D a α http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-58/MediaObjects/13661_2012_Article_152_IEq1_HTML.gif to both sides of (35) to obtain (28). As for the initial condition, apply I a 1 - α http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-58/MediaObjects/13661_2012_Article_152_IEq21_HTML.gif to both sides of (35) and then take the limit to obtain (29).

For assertion (b), let yC1-α[a, b] satisfy (28-30). Since f(x, y(x)) ∈ C1-α[a, b], then from (28) we have D r , a α , β y C 1 - α [ a , b ] http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-58/MediaObjects/13661_2012_Article_152_IEq47_HTML.gif. Since also by hypothesis D a β y C ( a , b ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-58/MediaObjects/13661_2012_Article_152_IEq37_HTML.gif, we can apply Lemma 24 and the formula (26) holds. By substituting the initial condition we obtain (33). This completes the proof.

The composition in Lemma 25 leads to the nonlinear integral equation,
y ( x ) = J r , a β , α f ( x , y ( x ) ) + c 0 Γ ( β ) ( x - a ) β - 1 + c 1 Γ ( α - r ) Γ ( α ) Γ ( α + β - r ) ( x - a ) α + β - r - 1 . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-58/MediaObjects/13661_2012_Article_152_Equ36_HTML.gif
(36)

The following theorem establishes an equivalence with this equation.

Theorem 27 Let 0 < α < 1, 0 < β < 1 and r < α. Let f : (a, b] × ℝ → ℝ be a function such that f(.,y(.)) ∈ C1-β[a, b] for any yC1-β[a, b]. Then the following statements hold.

(a) If yC1-β[a, b] satisfies the integral Equation (36) then y(x) satisfies the Cauchy-type problem (28-30).

(b) If yC1-β[a, b] with D a β y C ( a , b ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-58/MediaObjects/13661_2012_Article_152_IEq37_HTML.gif satisfies the Cauchy-type problem (28-30), then y(x) satisfies the integral Equation (36).

Proof. (a). Let yC1-β[a, b] satisfy the integral Equation (36). By hypothesis we have fC1-β[a, b]. Moreover, from Lemma 9 and the hypothesis r < α, we have
( x - a ) - r I a α f C 1 + r - α - β [ a , b ] C L ( a , b ) . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-58/MediaObjects/13661_2012_Article_152_Equt_HTML.gif
Thus the hypothesis of Lemmas 18 and 19 are satisfied. Applying the operator D r , a α , β http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-58/MediaObjects/13661_2012_Article_152_IEq31_HTML.gif to both sides of (36) and using Lemmas 18 and 19 yields (28) as follows.
D r , a α , β y ( x ) = D r , a α , β J r , a β , α f ( x , y ( x ) ) + D r , a α , β c 0 Γ ( β ) ( x - a ) β - 1 + c 1 Γ ( α - r ) Γ ( α ) Γ ( α + β - r ) ( x - a ) α + β - r - 1 = f ( x , y ( x ) ) . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-58/MediaObjects/13661_2012_Article_152_Equu_HTML.gif
Next, applying I a 1 - β http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-58/MediaObjects/13661_2012_Article_152_IEq48_HTML.gif to both sides of (36) yields
I a 1 - β y ( x ) = J r , a 1 , α f ( x , y ( x ) ) + c 0 + c 1 Γ ( α ) Γ ( α - r ) Γ ( α + β - r ) Γ ( α + β - r ) Γ ( α - r + 1 ) ( x - a ) α - r . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-58/MediaObjects/13661_2012_Article_152_Equ37_HTML.gif
(37)

Since r < α, taking the limit we obtain the initial condition (30).

Applying I a 1 - α ( x - a ) r D a β http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-58/MediaObjects/13661_2012_Article_152_IEq49_HTML.gif to both sides of (36) and using Lemmas 3, 11, and 12 yields
I a 1 - α ( x - a ) r D a β y ( x ) = I a 1 - α ( x - a ) r D a β J r , a β , α f ( x , y ( x ) ) + I a 1 - α ( x - a ) r D a β c 1 Γ ( α - r ) Γ ( α ) Γ ( α + β - r ) ( x - a ) α + β - r - 1 = I a 1 f ( x , y ( x ) ) + c 1 . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-58/MediaObjects/13661_2012_Article_152_Equv_HTML.gif

Again, taking the limit we obtain the initial condition (29).

(b). Let yC1-β[a, b] satisfy (28-30). Since f(x, y(x)) ∈ CL(a, b) then from (28), D r , a α , β y C L ( a , b ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-58/MediaObjects/13661_2012_Article_152_IEq43_HTML.gif. Since r < α then from Lemma 24 we have D a β y C L ( a , b ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-58/MediaObjects/13661_2012_Article_152_IEq41_HTML.gif. Thus we can apply Lemma 25 and the formula (27) holds. By using the initial conditions we obtain (36). This completes the proof.

In the next section we use this equivalence to prove the existence and uniqueness of solutions.

5 Existence and uniqueness of the solution of the Cauchy-type problem

In this section we prove an existence and uniqueness result for the Cauchy-type problem (28-30) using the integral Equation (36). For this purpose we introduce the following lemma.

Lemma 28 Let 0 < r < α < 1, 0 < β < 1, then the fractional differentiation operator J r , a β , α http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-58/MediaObjects/13661_2012_Article_152_IEq26_HTML.gif is bounded in C1-β[a, b] and
J r , a β , α f C 1 - β [ a , b ] K ( b - a ) α + β - r f C 1 - β [ a , b ] , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-58/MediaObjects/13661_2012_Article_152_Equ38_HTML.gif
(38)
where
K = Γ ( β ) Γ ( α + β - r ) Γ ( α + β ) Γ ( α + 2 β - r ) . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-58/MediaObjects/13661_2012_Article_152_Equ39_HTML.gif
(39)

Proof. Clearly from the hypothesis we have r < α + β and 0 < 1 - β < min{1, 1 + α - r}. Thus the result follows by taking γ = 1 - β in Lemma 21.

Theorem 29 Let 0 < r < α < 1, 0 ≤ β < 1. Let f : (a, b] × ℝ → ℝ be a function such that f(.,y(.)) ∈ C1-β[a, b] for any yC1-β[a, b] and the condition:
f ( x , y 1 ) - f ( x , y 2 ) A y 1 - y 2 , A > 0 , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-58/MediaObjects/13661_2012_Article_152_Equ40_HTML.gif
(40)

is satisfied for all x ∈ (a, b] and for all y1, y2 ∈ ℝ.

Then the Cauchy-type problem (28-30) has a solution yC1-β[a, b]. Furthermore, if for this solution D a β y C ( a , b ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-58/MediaObjects/13661_2012_Article_152_IEq37_HTML.gif, then this solution is unique.

Proof.

According to Theorem 27(a), we can consider the existence of an C1-β[a, b] solution for the integral Equation (36). This equation holds in any interval (a, x1] ⊂ (a, b], a < x1 < b. Choose x1 such that
w 1 : = A K ( x 1 - a ) α + β - r < 1 , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-58/MediaObjects/13661_2012_Article_152_Equw_HTML.gif
where K is given by (39). We rewrite the integral equation in the form y(x) = Ty(x), where
T y ( x ) = v 0 ( x ) + J r , a β , α f ( x , y ( x ) ) , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-58/MediaObjects/13661_2012_Article_152_Equx_HTML.gif
and
v 0 ( x ) = c 0 Γ ( β ) ( x - a ) β - 1 + c 1 Γ ( α - r ) Γ ( α ) Γ ( α + β - r ) ( x - a ) α + β - r - 1 . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-58/MediaObjects/13661_2012_Article_152_Equy_HTML.gif
Since r < α then v0C1-β[a, b]. Thus, it follows from Lemma 28 that if yC1-β[a, x1] then TyC1-β[a, x1]. Also, for any y1, y2 in C1-β[a, x1], we have
T y 1 - T y 2 C 1 - β [ a , x 1 ] J r , a β , α f ( x , y 1 ( x ) ) - f ( x , y 2 ( x ) ) C 1 - β [ a , x 1 ] A J r , a β , α y 1 ( x ) - y 2 ( x ) C 1 - β [ a , x 1 ] A K ( x 1 - a ) α + β - r y 1 - y 2 C 1 - β [ a , x 1 ] w 1 y 1 - y 2 C 1 - β [ a , x 1 ] , 0 < w 1 < 1 . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-58/MediaObjects/13661_2012_Article_152_Equz_HTML.gif

Hence by Theorem 15 there exists a unique solution y* ∈ C1-β[a, x1] to the Equation (36) on the interval (a, x1].

If x1b then we consider the interval [x1, b]. On this interval we consider solutions yC[x1, b] for the equation
y ( x ) = T y ( x ) : = v 01 ( x ) + J r , x 1 β , α f ( x , y ( x ) ) , x [ x 1 , b ] , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-58/MediaObjects/13661_2012_Article_152_Equ41_HTML.gif
(41)
where
v 01 ( x ) = y 0 Γ ( β ) ( x - a ) β - 1 + 1 Γ ( α ) Γ ( β ) a x 1 ( x - t ) β - 1 ( t - a ) - r a t ( t - s ) α - 1 f ( s , y ( s ) ) d s d t . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-58/MediaObjects/13661_2012_Article_152_Equaa_HTML.gif
Now we select x2 ∈ (x1, b] such that
w 2 : = A L ( x 2 - x 1 ) α + β - r < 1 , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-58/MediaObjects/13661_2012_Article_152_Equab_HTML.gif
where L is given by (25). Since the solution is uniquely defined on the interval (a, x1], we can consider v01(x) to be a known function. For y1, y2C[x1, x2], it follows from the Lipschitz condition and Lemma 22 that
T y 1 - T y 2 C [ x 1 , x 2 ] J r , x 1 β , α f ( x , y 1 ( x ) ) - f ( x , y 2 ( x ) ) C [ x 1 , x 2 ] A J r , x 1 β , α y 1 ( x ) - y 2 ( x ) C [ x 1 , x 2 ] A L ( x 2 - x 1 ) α + β - r y 1 - y 2 C [ x 1 , x 2 ] w 2 y 1 - y 2 C [ x 1 , x 2 ] . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-58/MediaObjects/13661_2012_Article_152_Equac_HTML.gif
Since 0 < w2 < 1, T is a contraction. Since f(x, y(x)) ∈ C[x1, x2] for any yC[x1, x2], then J r , x 1 β , α f C [ x 1 , x 2 ] http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-58/MediaObjects/13661_2012_Article_152_IEq50_HTML.gif. Moreover, clearly v01(x) is in C[x1, x2]. Thus the right-hand side of (41) is in C[x1, x2]. Therefore T maps C[x1, x2] into itself. By Theorem 15, there exists a unique solution y 1 * C [ x 1 , x 2 ] http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-58/MediaObjects/13661_2012_Article_152_IEq51_HTML.gif to the equation on the interval [x1, x2]. Moreover, it follows from Lemma 23 that y 1 * ( x 1 ) = y 0 * ( x 1 ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-58/MediaObjects/13661_2012_Article_152_IEq52_HTML.gif. Therefore if
y * ( x ) = y 0 * ( x ) , a < x x 1 , y 1 * ( x ) , x 1 < x x 2 , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-58/MediaObjects/13661_2012_Article_152_Equad_HTML.gif

then by Lemma 14, y* ∈ C1-β[a, x2]. So y* is the unique solution of (36) in C1-β[a, x2] on the interval (a, x2].

If x2b, we repeat the process as necessary, say M - 2 times, to obtain the unique solutions y k * C 1 - β [ x k , x k + 1 ] , k = 2 , 3 , , M http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-58/MediaObjects/13661_2012_Article_152_IEq53_HTML.gif, where a = x0 < x1 < ··· < x M = b, such that
w k + 1 = A L ( x k + 1 - x k ) α + β - r < 1 . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-58/MediaObjects/13661_2012_Article_152_Equae_HTML.gif
As a result we have the unique solution y* ∈ C1-β[a, b] of (36) given by
y * ( x ) = y k * ( x ) , x ( x k , x k + 1 ] , k = 0 , 1 , , M - 1 . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-58/MediaObjects/13661_2012_Article_152_Equ42_HTML.gif
(42)

This solution is also a solution for (28-30).

If D a β y C ( a , b ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2012-58/MediaObjects/13661_2012_Article_152_IEq37_HTML.gif then the uniqueness follows from part (b) of Theorem 27. This completes the proof.

Declarations

Acknowledgements

The author was grateful for the support provided by the King Fahd University of Petroleum & Minerals and the financial support by the BAE Systems through the PDSR program by the British Council in Saudi Arabia.

Authors’ Affiliations

(1)
Department of Mathematics & Statistics, King Fahd University of Petroleum & Minerals

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