Filetype PDF | Posted on 28 Jan 2023 | 3 years ago
Authentication
digital resources
266x Filetype PDF File size 0.24 MB Source: core.ac.uk
J. Math. Anal. Appl. 367 (2010) 260–272
Contents lists available at ScienceDirect
Journal of Mathematical Analysis and
Applications
www.elsevier.com/locate/jmaa
Initial value problems for fractional differential equations involving
Riemann–Liouville sequential fractional derivative✩
a,b,∗ b a
Zhongli Wei , Qingdong Li ,JunlingChe
a Department of Mathematics, Shandong Jianzhu University, Jinan, Shandong, 250101, Peoples Republic of China
b School of Mathematics and System Sciences, Shandong University, Jinan, Shandong, 250100, Peoples Republic of China
article info abstract
Article history: In this paper, we shall discuss the properties of the well-known Mittag–Leffler function,
Received 20 September 2009 and consider the existence and uniqueness of solution of the initial value problem for
Available online 18 January 2010 fractional differential equation involving Riemann–Liouville sequential fractional derivative
Submitted by J. Shi by using monotone iterative method.
Keywords: ©2010 Elsevier Inc. All rights reserved.
Initial value problem
Fractional differential equation
Riemann–Liouville sequential fractional
derivatives
Upper solution and lower solution
1. Introduction
Let J =[a,b] be a compact interval on the real axis R,andy be a measurable Lebesgue function, that is, y ∈ L1(a,b).
Let x ∈ J and α ∈R (0<α 1). The Riemann–Liouville fractional integrals Iα and derivative Dα are defined by (see, for
example, [1–3]) a+ a+
x
α 1 α−1 α d 1−α
Ia+y (x)= Γ(α) (x−s) y(s)ds and Da+y (x)= dx Ia+ y (x). (1.1)
a
Wewill work here following the definition of sequentialfractional derivative presented by Miller and Ross in p. 209 of [4],
Dα y=Dα y,
a+ a+ (k−1)α (1.2)
Dkαy=Dα D y (k=2,3,...).
a+ a+ a+
There is a close connection between the sequential fractional derivatives and the nonsequential Riemann–Liouville deriva-
tives. For example, in the case k = 2, 0 < α < 1/2 and the Riemann–Liouville derivatives, the relationship between Dkαy
and Dkα y is given by a+
a+
( α−1
D2α ( 2α 1−α ( t −a) (
a+y x)= Da+ y(t)− Ia+ y a+) Γ(α) x). (1.3)
✩ Research supported by the NNSF-China (10971046) and the NSF of Shandong Province (ZR2009AM004).
* Corresponding author at: Department of Mathematics, Shandong Jianzhu University, Jinan, Shandong, 250101, Peoples Republic of China.
E-mail address: jnwzl@yahoo.com.cn (Z. Wei).
0022-247X/$ – see front matter © 2010 Elsevier Inc. All rights reserved.
doi:10.1016/j.jmaa.2010.01.023
Z. Wei et al. / J. Math. Anal. Appl. 367 (2010) 260–272 261
Weshall consider the existence of minimal and maximal solutions and uniqueness of solution of the initial value problem
for fractional differential equation involving Riemann–Liouville sequential fractional derivative, using the method of upper
and lower solutions and its associated monotone iterative method.
D2αy(x)= fx,y,Dα y, x∈(0,T],
0+ 0+ (1.4)
1−α 1−α α
x y(x)|x=0 = y0, x D0+y (x)|x=0 = y1,
where 04M areconstantsandσ ∈C1−α[0,T],hasthefollowingrepresentationofsolution
u(x)=Γ(α)u e (λ ,x)+Γ(α)(u −λ u ) e (λ ,t)∗e (λ ,t) (x)+ e (λ ,t)∗e (λ ,t)∗σ(t) (x), (1.15)
0 α 2 1 2 0 α 2 α 1 α 2 α 1
where
−N+√N2−4M −N−√N2−4M
λ = ,λ= <0. (1.16)
1 2 2 2
Proof. Let
α
D0+−λ2 u(x)=y(x), x∈(0,T].
Then the problem in (1.14) is equivalent to
α
D −λ y(x)=σ(x), x∈(0,T],
0+ 1 (1.17)
1−α
x y(x)| =y =u −λ u ,
x=0 0 1 2 0
and
α
D −λ u(x)=y(x), x∈(0,T],
0+ 2 (1.18)
1−α
x u(x)|x=0 =u0.
By Lemma 1.1, we have that the linear initial value problems (1.16) and (1.17) have the following representation of solutions
y(x)=Γ(α)y0eα(λ1,x)+ eα(λ1,t)∗σ(t) (x), (1.19)
u(x)=Γ(α)u0eα(λ2,x)+ eα(λ2,t)∗ y(t) (x). (1.20)
Substituting (1.19) into (1.20), we obtain (1.15). The proof of Lemma 1.2 is completed. ✷
Z. Wei et al. / J. Math. Anal. Appl. 367 (2010) 260–272 263
By a direct computation
x ∞ i (i+1)α−1 ∞ j (j+1)α−1
λ(x−t) λt
eα(λ1,t)∗eα(λ2,t) (x)= 1 · 2 dt
0 i=0 Γ((i +1)α) j=0 Γ((j +1)α)
∞ x (i+1)α−1 (j+1)α−1
=λiλj (x−t) · t dt
1 2 Γ((i +1)α) Γ((j+1)α)
n=0i+j=n 0
∞ (i+j+1+1)α−1
=λiλj x
1 2Γ((i+ j+1+1)α)
n=0i+j=n
∞ n+1 n+1 (n+1+1)α−1
λ −λ x
= 1 2 ·
λ −λ Γ((n+1+1)α)
n=0 1 2
∞ n n (n+1)α−1
1 (λ −λ )x
= 1 2
λ −λ Γ((n+1)α)
1 2 n=0
1
=λ −λ eα(λ1,t)−eα(λ2,t) (x), x∈R.
1 2
Hence, we obtain that
Lemma1.3.
1
eα(λ2,t)∗eα(λ1,t) (x)= eα(λ1,t)∗eα(λ2,t) (x) = λ −λ eα(λ1,t)−eα(λ2,t) (x), x∈R. (1.21)
1 2
This paper is organized as follows. In Section 2 we give some preliminaries, including a property of Mittag–Leffler
function which will be used in our main result, a comparison result. The main results are established in Section 3.
2. ApropertyofMittag–Lefflerfunctionandsomelemmas
In the following, we shall use the definition and properties of the Γ function which listed as follows (see [29]):
+∞
Γ(α)= tα−1e−tdt, (2.1)
0
1 = lim 1 α(1+α) 1+α ··· 1+α , (2.2)
n→∞ α
Γ(α) (n+1) 2 n
1 = lim 1 (1+α) 1+α ··· 1+α . (2.3)
n→∞ α
Γ(1+α) (n+1) 2 n
Let
α α
ω (α)=α(1+α) 1+ ··· 1+ . (2.4)
n 2 n
Then
1 = lim 1 ω (α). (2.5)
n→∞ α n
Γ(α) (n+1)
Lemma2.1.For0<α1,thereexistpositiveconstants
n
b0 >0, b1 >0, ..., bn>0, suchthat ω (kα)=biCi+1. (2.6)
n n n n n k+i
i=0
Hence, we have
n
i i+2
(k −1)ω (kα)= (i +2)b C , (2.7)
n n k+i
i=0
no reviews yet
Please Login to review.
