Abstract
A Cauchytype 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; RiemannLiouville fractional derivative; sequential fractional derivative; fractional differential equation1 Introduction
We consider a Cauchytype problem associated with the equation
where and are the RiemannLiouville 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 wellposedness of a Cauchytype 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 Cauchytype 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 Cauchytype 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
with the norm
where
In the case f is not defined at x = a or γ < 0 we let The spaces C_{γ}[a, b] satisfy the following properties.
• C_{0}[a,b] = C[a,b].
• C_{γ}[a,b] ⊂CL(a,b),γ < 1.
• f ∈ C_{γ }[a, b] if and only if f ∈ C(a, b) and exists and is finite.
The leftsided RiemannLiouville fractional integrals and derivatives are defined as follows.
Definition 1 Let f ∈ L(a,b). The integral
is called the leftsided RiemannLiouville fractional integral of order α of the function f.
Definition 2 The expression
is called the leftsided RiemannLiouville fractional derivative of order α of f provided the righthand side exists.
For power functions we have the following formulas.
Lemma 3 For x > a we have
Next we present some mapping properties of the operator .
Lemma 4 For α > 0, 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, maps C[a, b] into C[a, b].
The following lemma is proved in [38].
Lemma 6 Let α ≥ 0. If f ∈ CL(a, b) then .
The mapping properties of in the spaces C_{γ}[a, b], 0 ≤ α ≤ γ < 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 f ∈ C_{γ}[a, b] then
Proof. Note that from Lemma 3 we have
Thus
Now, given ϵ > 0 there exists δ > 0 such that x  a < δ implies that
Thus
This yields the limit (9).
Next we present the mapping properties of in the spaces C_{γ}[a, b].
Lemma 8 Let α > 0 and γ < 1. If f ∈ C_{γ}[a, b] then and for x ∈ (a, b] we have
Proof. From Lemmas 6 and 7 we have I^{α}f ∈ C(a, b) and exists and is finite. Thus . Now for x ∈ (a, b] we have
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 f ∈ C_{γ}[a, b] then . In particular, if γ + r  α < 1 then I^{α}f ∈ CL(a, b).
Later, the following observation is important.
Lemma 10 Let α > 0 and r < α. If f ∈ CL(a, b) then .
Proof. When r ≤ 0 the result follows clearly from Lemma 6. When r > 0 it follows from Lemma 6 that and we only need to show that .
For any x ∈ (a, b] we have the following inequality.
Or,
From Lemma 4 the righthand side is in L(a, b) and thus . This completes the proof.
The following lemma follows by direct calculations using Dirichlet formula, [36].
Lemma 11 Let α ≥ 0, β ≥ 0, and f ∈ CL(a, b). Then
for all x ∈ (a, b].
Lemma 11 leads to the left inverse operator.
Lemma 12 Let α > 0 and f ∈ CL(a, b). Then
for all x ∈ (a, b].
Now we present a version of the fundamental theorem of fractional calculus.
Lemma 13 Let 0 < α < 1. If f ∈ C(a, b) and , then exists and is finite, and
for all x ∈ (a, b].
Proof. From Lemma 12 we have for all x ∈ (a, b] the relation
which we can write as
This implies that
for some constant c. Since Lemma 6 implies that , we also have f ∈ CL(a, b). Also, if we apply to both sides of (14) we obtain
Taking the limit yields 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, g ∈ C_{γ}[a, c], g ∈ C[c, b] and g is continuous at c. Then g ∈ C_{γ}[a, b].
Theorem 15 ([1], Banach Fixed Point Theorem) Let (U, d) be a nonempty complete metric space. Let T : U → U be a map such that for every u, v ∈ U, the relation
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 f ∈ CL(a, b). Define the sequential integral and the sequential derivative by
and
if the righthand sides exist.
From Lemma 3 we have the following formula for the power function.
Lemma 17 Let α > 0, β > 0, r ∈ ℝ. If
then for x > a,
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,
(b) Let 0 < α < 1 and β > 0. Let r ∈ ℝ be such that r < α + β. Then for x > a,
Lemma 19 (Left inverse) Let α > 0, β > 0, and r ∈ ℝ. If f ∈ CL(a, b) such that then
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 .
Lemma 20 Let α > 0, β > 0, and r < 1 + α. Let 0 ≤ γ < min{1, 1 + α  r}. If f ∈ C_{γ}[a, b] then and for x ∈ (a, b] we have
where
Lemma 20 implies the following.
Lemma 21 Let α > 0, β > 0, and r < 1 + α. Let 0 ≤ γ < min{1, 1 + αr}. If r ≤ α + β, then is bounded in C_{γ}[a, b] and
where k is given by (22).
Proof. Since γ + r  α  β ≤ γ, then from Lemma 20 we have
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 , maps C[a, b] into C[a, b] and
where
The following is an analogous result to the result for the RiemannLiouville integral proved in [10].
Lemma 23 Let α > 0, β > 0, and r < α. Let f ∈ CL(a, c). Let
Then
Proof. Since r < α, Lemma 10 implies that . Thus is finite and
where
Since , the limit of the righthand side vanishes and the proof is complete.
The following lemma relates the fractional derivative to the RiemannLiouville derivative .
Lemma 24 Let 0 < α < 1, β ≥ 0, and r ∈ ℝ. If and then exists and finite, and
for all x ∈ (a, b]. If in addition, r < α then .
Proof. Clearly implies that . Thus we can apply Lemma 13 to and obtain
By multiplying both sides by (x  a)^{r }we obtain (26). If r < α then Lemma 10 implies that and thus from (26) we have . This proves the result.
The Next lemma gives an analogous result to the fundamental theorem of calculus in terms of the operators and .
Lemma 25 Let 0 < α < 1 and 0 < β < 1. Let y ∈ C(a, b) be such that and . Then both and exist, y ∈ CL(a, b), and
for all x ∈ (a, b].
Proof. By applying Lemma 13 twice we obtain
4 Cauchytype problem and equivalency
Consider the Cauchytype problem
where c_{0 }and c_{1 }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 .
In addition, from [1, Lemma 3.2], the condition (30) follows from the condition
and if 0 < α  r < 1 then (29) follows from the condition
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 integrodifferential equation:
Theorem 26 Let 0 < α < 1, β > 0 and r ∈ ℝ. Let f : (a, b] × ℝ → ℝ be a function such that f(.,y(.)) ∈ C_{1α }[a, b] for any y ∈ C_{1α }[a, b]. Then we have the following.
(a) If y ∈ C_{1α}[a, b] satisfies (33) and (34) then y(x) satisfies (2830).
(b) If y ∈ C_{1α}[a, b] with satisfy (2830), then y(x) satisfies (3334).
Proof.
For assertion (a), let y ∈ C_{1α}[a, b] satisfy (3334). We multiply (33) by (x  a)^{r }to obtain
Next we apply to both sides of (35) to obtain (28). As for the initial condition, apply to both sides of (35) and then take the limit to obtain (29).
For assertion (b), let y ∈ C_{1α}[a, b] satisfy (2830). Since f(x, y(x)) ∈ C_{1α}[a, b], then from (28) we have . Since also by hypothesis , 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,
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(.)) ∈ C_{1β}[a, b] for any y ∈ C_{1β}[a, b]. Then the following statements hold.
(a) If y ∈ C_{1β}[a, b] satisfies the integral Equation (36) then y(x) satisfies the Cauchytype problem (2830).
(b) If y ∈ C_{1β}[a, b] with satisfies the Cauchytype problem (2830), then y(x) satisfies the integral Equation (36).
Proof. (a). Let y ∈ C_{1β}[a, b] satisfy the integral Equation (36). By hypothesis we have f ∈ C_{1β}[a, b]. Moreover, from Lemma 9 and the hypothesis r < α, we have
Thus the hypothesis of Lemmas 18 and 19 are satisfied. Applying the operator to both sides of (36) and using Lemmas 18 and 19 yields (28) as follows.
Next, applying to both sides of (36) yields
Since r < α, taking the limit we obtain the initial condition (30).
Applying to both sides of (36) and using Lemmas 3, 11, and 12 yields
Again, taking the limit we obtain the initial condition (29).
(b). Let y ∈ C_{1β}[a, b] satisfy (2830). Since f(x, y(x)) ∈ CL(a, b) then from (28), . Since r < α then from Lemma 24 we have . 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 Cauchytype problem
In this section we prove an existence and uniqueness result for the Cauchytype problem (2830) 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 is bounded in C_{1β}[a, b] and
where
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(.)) ∈ C_{1β}[a, b] for any y ∈ C_{1β}[a, b] and the condition:
is satisfied for all x ∈ (a, b] and for all y_{1}, y_{2 }∈ ℝ.
Then the Cauchytype problem (2830) has a solution y ∈ C_{1β}[a, b]. Furthermore, if for this solution , then this solution is unique.
Proof.
According to Theorem 27(a), we can consider the existence of an C_{1β}[a, b] solution for the integral Equation (36). This equation holds in any interval (a, x_{1}] ⊂ (a, b], a < x_{1 }< b. Choose x_{1 }such that
where K is given by (39). We rewrite the integral equation in the form y(x) = Ty(x), where
and
Since r < α then v_{0 }∈ C_{1β}[a, b]. Thus, it follows from Lemma 28 that if y ∈ C_{1β}[a, x_{1}] then Ty ∈ C_{1β}[a, x_{1}]. Also, for any y_{1}, y_{2 }in C_{1β}[a, x_{1}], we have
Hence by Theorem 15 there exists a unique solution y* ∈ C_{1β}[a, x_{1}] to the Equation (36) on the interval (a, x_{1}].
If x_{1 }≠ b then we consider the interval [x_{1}, b]. On this interval we consider solutions y ∈ C[x_{1}, b] for the equation
where
Now we select x_{2 }∈ (x_{1}, b] such that
where L is given by (25). Since the solution is uniquely defined on the interval (a, x_{1}], we can consider v_{01}(x) to be a known function. For y_{1}, y_{2 }∈ C[x_{1}, x_{2}], it follows from the Lipschitz condition and Lemma 22 that
Since 0 < w_{2 }< 1, T is a contraction. Since f(x, y(x)) ∈ C[x_{1}, x_{2}] for any y ∈ C[x_{1}, x_{2}], then . Moreover, clearly v_{01}(x) is in C[x_{1}, x_{2}]. Thus the righthand side of (41) is in C[x_{1}, x_{2}]. Therefore T maps C[x_{1}, x_{2}] into itself. By Theorem 15, there exists a unique solution to the equation on the interval [x_{1}, x_{2}]. Moreover, it follows from Lemma 23 that . Therefore if
then by Lemma 14, y* ∈ C_{1β}[a, x_{2}]. So y* is the unique solution of (36) in C_{1β}[a, x_{2}] on the interval (a, x_{2}].
If x_{2 }≠ b, we repeat the process as necessary, say M  2 times, to obtain the unique solutions , where a = x_{0 }< x_{1 }< ··· < x_{M }= b, such that
As a result we have the unique solution y* ∈ C_{1β}[a, b] of (36) given by
This solution is also a solution for (2830).
If then the uniqueness follows from part (b) of Theorem 27. This completes the proof.
Competing interests
The author declares that they have no competing interests.
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.
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