Electronic Journal of Qualitative Theory of Differential Equations 2008, No. 29, 1-8; http://www.math.u-szeged.hu/ejqtde/

On the stability of a fractional-order differential equation with nonlocal initial condition

El-Sayed A. M. A. & Abd El-Salam Sh. A. E-mail addresses: amasayed@hotmail.com & shrnahmed@yahoo.com Faculty of Science, Alexandria University, Alexandria, Egypt

Abstract

The topic of fractional calculus (integration and differentiation of fractional-order), which concerns singular integral and integro-differential operators, is enjoying interest among mathematicians, physicists and engineers (see [1]-[2] and [5]-[14] and the references therein). In this work, we investigate initial value problem of fractional-order differential equation with nonlocal condition. The stability (and some other properties concerning the existence and uniqueness) of the solution will be proved.

Key words: Fractional calculus; Banach contraction fixed point theorem; Nonlocal condition; Stability.

1 Introduction

Let L_{1}[a, b] denote the space of all Lebesgue integrable functions on the interval [a, b], 0 ≤ a < b < ∞, with the L_{1}-norm ||x||_{L_{1}} = ∫_{0}^{1} |x(t)| dt.

Definition 1.1 The fractional (arbitrary) order integral of the function f ∈ L_{1}[a, b] of order β ∈ R^{+} is defined by (see [11] - [14])

I_{a}^{β} f(t) = ∫_{a}^{t} \frac{(t − s)^{β − 1}}{Γ(β)} f(s) ds,

where Γ(.) is the gamma function.

Definition 1.2 The (Caputo) fractional-order derivative D^{α} of order α ∈ (0, 1] of the function g(t) is defined as (see [12] - [14])

D_{a}^{α} g(t) = I_{a}^{1 − α} \frac{d}{dt} g(t), t ∈ [a, b].

Now the following theorem (some properties of the fractional-order integration and the fractional-order differentiation) can be easily proved.

Theorem 1.1 Let β, γ ∈ R^{+} and α ∈ (0, 1]. Then we have:

(i) I_{a}^{β} : L_{1} → L_{1}, and if f(t) ∈ L_{1}, then I_{a}^{γ} I_{a}^{β} f(t) = I_{a}^{γ+β} f(t). (ii) lim_{β→n} I_{a}^{β} f(t) = I_{a}^{n} f(t) , n = 1, 2, 3, ... uniformly. If f(t) is absolutely continuous on [a, b], then

(iii) lim_{α→1} D_{a}^{α} f(t) = D f(t)

(iv) If f(t) = k ≠ 0, k is a constant, then D_{a}^{α} k = 0. In ([3]) the nonlocal initial value problem for first-order differential inclusions:

\{ \begin{matrix}x′(t) ∈ F (t, x(t)), t ∈ (0, 1], \\ x(0) + ∑_{k=1}^{m} a_{k} x(t_{k}) = x_{0}, \end{matrix}

was studied, where F : J × ℜ → 2^{ℜ} is a set-valued map, J = [0, 1], x_{0} ∈ ℜ is given, 0 < t_{1} < t_{2} < ⋯ < t_{m} < 1, and a_{k} ≠ 0 for all k = 1, 2, ⋯, m. Our objective in this paper is to investigate, by using the Banach contraction fixed point theorem, the existence of a unique solution of the following fractional-order differential equation:

D^{α} x(t) = c(t) f(x(t)) + b(t),  (1)

with the nonlocal condition:

x(0) + ∑_{k=1}^{m} a_{k} x(t_{k}) = x_{0},  (2)

where x_{0} ∈ ℜ and 0 < t_{1} < t_{2} < ⋯ < t_{m} < 1, and a_{k} ≠ 0 for all k = 1, 2, ⋯, m. Then we will prove that this solution is uniformly stable.

2 Existence of solution

Here the space C[0, 1] denotes the space of all continuous functions on the interval [0, 1] with the supremum norm ||y|| = sup_{t∈[0,1]} |y(t)|. To facilitate our discussion, let us first state the following assumptions:

(i) |\frac{∂f}{∂x}|≤ k,

(ii) c(t) is a function which is absolutely continuous, (iii) b(t) is a function which is absolutely continuous.

Definition 2.1 By a solution of the initial value Problem (1) - (2) we mean a function

x ∈ C[0, 1] with \frac{dx}{dt} ∈ L_{1}[0, 1].

Theorem 2.1 If the above assumptions (i) - (iii) are satisfied such that

1 + ∑_{k=1}^{m} a_{k} ≠ 0 and A < \frac{Γ(1 + α)}{k ||c||},

where A = 1 + |a| ∑_{k=1}^{m} |a_{k}| and a = (1 + ∑_{k=1}^{m} a_{k})^{−1} ,

then the initial value Problem (1) - (2) has a unique solution. Proof: For simplicity let c(t)f(x(t)) + b(t) = g(t, x(t)). If x(t) satisfies (1) - (2), then by using the definitions and properties of the fractional-order integration and fractional-order differentiation equation (1) can be written as

I^{1 − α} x′(t) = g(t, x(t)).

Operating by I^{α} on both sides of the last equation, we obtain

x(t) − x(0) = I^{α} g(t, x(t)),

by substituting for the value of x(0) from (2), we get

x(t) = x_{0} − ∑_{k=1}^{m} a_{k} x(t_{k}) + I^{α} g(t, x(t)).  (3)

If we put t = t_{k} in (3), we obtain

x(t_{k}) = x_{0} − ∑_{k=1}^{m} a_{k} x(t_{k}) + I^{α} g(t, x(t))|_{t=t_{k}}.  (4)

Then subtract (3) from (4) to get

x(t_{k}) = x(t) − I^{α} g(t, x(t)) + I^{α} g(t, x(t))|_{t=t_{k}}.  (5)

Substitute from (5) in (3), we get

x(t) = x_{0} + I^{α} g(t, x(t))
− ∑_{k=1}^{m} a_{k} ( x(t) − I^{α} g(t, x(t)) + I^{α} g(t, x(t))|_{t=t_{k}} )
= x_{0} + I^{α} g(t, x(t))
− ∑_{k=1}^{m} a_{k} x(t) + ∑_{k=1}^{m} a_{k} I^{α} g(t, x(t)) − ∑_{k=1}^{m} a_{k} I^{α} g(t, x(t))|_{t=t_{k}},
(1 + ∑_{k=1}^{m} a_{k}) x(t) = x_{0} − ∑_{k=1}^{m} a_{k} I^{α} g(t, x(t))|_{t=t_{k}} + (1 + ∑_{k=1}^{m} a_{k}) I^{α} g(t, x(t)),
x(t) = a ( x_{0} − ∑_{k=1}^{m} a_{k} I^{α} g(t, x(t))|_{t=t_{k}} ) + I^{α} g(t, x(t)).  (6)

Now define the operator T : C → C by

T x(t) = a (x_{0} − ∑_{k=1}^{m} a_{k} ∫_{0}^{t_{k}} \frac{(t_{k} − s)^{α−1}}{Γ(α)} \{c(s)f(x(s)) + b(s)\} ds)+I^{α}\{c(s)f(x(s))+b(s)\}.

(7)

Let x, y ∈ C, then

T x(t) − T y(t) = − a ∑_{k=1}^{m} a_{k} ∫_{0}^{t_{k}} \frac{(t_{k} − s)^{α−1}}{Γ(α)} c(s) f(x(s)) ds

+ a ∑_{k=1}^{m} a_{k} ∫_{0}^{t_{k}} \frac{(t_{k} − s)^{α−1}}{Γ(α)} c(s) f(y(s)) ds

+ ∫_{0}^{t} \frac{(t − s)^{α−1}}{Γ(α)} c(s) \{f(x(s)) − f(y(s))\} ds
= − a ∑_{k=1}^{m} a_{k} ∫_{0}^{t_{k}} \frac{(t_{k} − s)^{α−1}}{Γ(α)} c(s) \{f(x(s)) − f(y(s))\} ds

+ ∫_{0}^{t} \frac{(t − s)^{α−1}}{Γ(α)} c(s) \{f(x(s)) − f(y(s))\} ds,
|T x(t) − T y(t)| ≤ k |a| ∑_{k=1}^{m} |a_{k}| ∫_{0}^{t_{k}} \frac{(t_{k} − s)^{α−1}}{Γ(α)} |c(s)| |x(s) − y(s)| ds

+ k ∫_{0}^{t} \frac{(t − s)^{α−1}}{Γ(α)} |c(s)| |x(s) − y(s)| ds
≤ k |a| ∑_{k=1}^{m} |a_{k}| sup_{t} |c(t)| sup_{t} |x(t) − y(t)| ∫_{0}^{t_{k}} \frac{(t_{k} − s)^{α−1}}{Γ(α)} ds

+ k sup_{t} |c(t)| sup_{t} |x(t) − y(t)| ∫_{0}^{t} \frac{(t − s)^{α−1}}{Γ(α)} ds
≤ k |a| ∑_{k=1}^{m} |a_{k}| ∥c∥ ∥x − y∥ \frac{t_{k}^{α}}{Γ(1 + α)}
+ k ∥c∥ ∥x − y∥ \frac{t^{α}}{Γ(1 + α)}
≤ \frac{k}{Γ(1 + α)} (1 + |a| ∑_{k=1}^{m} |a_{k}|) ∥c∥ ∥x − y∥
≤ \frac{k A ∥c∥}{Γ(1 + α)} ∥x − y∥ = K||x − y||.

but since K = \frac{kA∥c∥}{Γ(1+α)} < 1, then we get

||T x − T y|| < K ||x − y||,

which proves that the map T : C → C is contraction. Applying the Banach contraction fixed point theorem we deduce that (7) has a unique fixed point x ∈ C[0, 1].

Now, differentiate (6) to obtain

x′(t) = \frac{d}{dt} I^{α} ( c(t) f(x(t)) + b(t) )
= (c(t) f(x(t)) + b(t))|_{t=0} \frac{t^{α−1}}{Γ(α)} + I^{α} \frac{d}{dt} (c(t) f(x(t)) + b(t))
= K_{1} \frac{t^{α−1}}{Γ(α)} + I^{α} (c′(t) f(x(t)) + \frac{∂f}{∂x} x′(t) c(t) + b′(t)),

∫_{0}^{1} |x′(t)| dt ≤ \frac{K_{1}}{Γ(1 + α)} t^{α}|_{0}^{1}
+ ∫_{0}^{1} ∫_{0}^{t} \frac{(t − s)^{α−1}}{Γ(α)} |c′(s) f(x(s)) + \frac{∂f}{∂x} x′(s) c(s) + b′(s)| ds dt
= \frac{K_{1}}{Γ(1 + α)} + ∫_{0}^{1} |c′(s) f(x(s)) + \frac{∂f}{∂x} x′(s) c(s) + b′(s)| ∫_{s}^{1} \frac{(t − s)^{α−1}}{Γ(α)} dt ds
≤ \frac{K_{1}}{Γ(1 + α)} + \frac{1}{Γ(1 + α)} ∫_{0}^{1} |c′(s) f(x(s)) + \frac{∂f}{∂x} x′(s) c(s) + b′(s)| ds,
∥x′∥_{L_{1}} ≤ \frac{K_{1}}{Γ(1 + α)} + \frac{1}{Γ(1 + α)} (∥c′∥_{L_{1}} ∥f∥ + k ∥x′∥_{L_{1}} ∥c∥ + ∥b′∥_{L_{1}}) ,

(1 − \frac{k∥c∥}{Γ(1 + α)}) ∥x′∥_{L_{1}} ≤ \frac{K_{1}}{Γ(1 + α)} + \frac{1}{Γ(1 + α)} (∥c′∥_{L_{1}} ∥f∥ + ∥b′∥_{L_{1}}) ,
∥x′∥_{L_{1}} ≤ (1 − \frac{k∥c∥}{Γ(1 + α)})^{−1} (\frac{K_{1}}{Γ(1 + α)} + \frac{1}{Γ(1 + α)} (∥c′∥_{L_{1}} ∥f∥ + ∥b′∥_{L_{1}})) .

Therefore we obtain that x′ ∈ L_{1}[0, 1]. To complete the equivalence of equation (6) with the initial value problem (1) - (2), let x(t) be a solution of (6), differentiate both sides, and get

x′(t) = \frac{d}{dt} I^{α} g(t, x(t))
= g(t, x(t))|_{t=0} \frac{t^{α−1}}{Γ(α)} + I^{α} \frac{d}{dt} g(t, x(t)).

Then operate by I^{1−α} on both sides to obtain

D^{α} x(t) = g(t, x(t)).

And if t = 0 we find that the nonlocal condition (2) is satisfied. Which proves the equivalence.

3 Stability

In this section we study the uniform stability (see [1], [4] and [6]) of the solution of the initial-value problem (1) - (2). Theorem 3.1 The solution of the initial-value problem (1) - (2) is uniformly stable Proof: Let x(t) be a solution of

x(t) = a (x_{0} − ∑_{k=1}^{m} a_{k} ∫_{0}^{t_{k}} \frac{(t_{k} − s)^{α−1}}{Γ(α)} \{c(s)f(x(s)) + b(s)\} ds) + I^{α}\{c(s)f(x(s)) + b(s)\}

(8)

and let x̃(t) be a solution of equation (8) such that x̃(0) = x̃_{0} − ∑_{k=1}^{m} a_{k} x̃(t_{k}). Then

x(t) − x̃(t) = a (x_{0} − x̃_{0}) − a ∑_{k=1}^{m} a_{k} ∫_{0}^{t_{k}} \frac{(t_{k} − s)^{α−1}}{Γ(α)} c(s) f(x(s)) ds
+ a ∑_{k=1}^{m} a_{k} ∫_{0}^{t_{k}} \frac{(t_{k} − s)^{α−1}}{Γ(α)} c(s) f(x̃(s)) ds
+ ∫_{0}^{t} \frac{(t − s)^{α−1}}{Γ(α)} c(s) \{f(x(s)) − f(x̃(s))\} ds,
|x(t) − x̃(t)| ≤ |a| |x_{0} − x̃_{0}|
+ |a| ∑_{k=1}^{m} |a_{k}| ∫_{0}^{t_{k}} \frac{(t_{k} − s)^{α−1}}{Γ(α)} |c(s)| |f(x(s)) − f(x̃(s))| ds
+ ∫_{0}^{t} \frac{(t − s)^{α−1}}{Γ(α)} |c(s)| |f(x(s)) − f(x̃(s))| ds
≤ |a| |x_{0} − x̃_{0}|
+ k |a| ∑_{k=1}^{m} |a_{k}| sup_{t} |c(t)| ∫_{0}^{t_{k}} \frac{(t_{k} − s)^{α−1}}{Γ(α)} |(x(s) − x̃(s)| ds
+ k sup_{t} |c(t)| ∫_{0}^{t} \frac{(t − s)^{α−1}}{Γ(α)} |x(s) − x̃(s)| ds
≤ |a| |x_{0} − x̃_{0}|
+ k |a| ∥c∥ ∑_{k=1}^{m} |a_{k}| sup_{t} |x(t) − x̃(t)| ∫_{0}^{t_{k}} \frac{(t_{k} − s)^{α−1}}{Γ(α)} ds
+ k ∥c∥ sup_{t} |x(t) − x̃(t)| ∫_{0}^{t} \frac{(t − s)^{α−1}}{Γ(α)} ds,
∥x − x̃∥ ≤ |a| |x_{0} − x̃_{0}| + k |a| ∥c∥ ∑_{k=1}^{m} |a_{k}| ∥x − x̃∥ \frac{t_{k}^{α}}{Γ(1 + α)}
+ k ∥c∥ ∥x − x̃∥ \frac{t^{α}}{Γ(1 + α)}
≤ |a| |x_{0} − x̃_{0}| + \frac{k ∥c∥}{Γ(1 + α)} (1 + |a| ∑_{k=1}^{m} |a_{k}|) ∥x − x̃∥

= |a| |x_{0} − x̃_{0}| + \frac{k A ∥c∥}{Γ(1 + α)} ∥x − x̃∥,
(1 − \frac{k A ∥c∥}{Γ(1 + α)}) ||x − x̃|| ≤ |a| |x_{0} − x̃_{0}|,
||x − x̃|| ≤ (1 − \frac{k A ∥c∥}{Γ(1 + α)})^{−1} |a| |x_{0} − x̃_{0}|.

Therefore, if | x_{0} − x̃_{0} | < δ(ε), then || x − x̃ || < ε, which complete the proof of the theorem.

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(Received February 7, 2008)