Bulletin of Mathematical Analysis and Applications
ISSN: 1821-1291, URL: http://www.bmathaa.org
Volume 3 Issue 4(2011), Pages 59-68.
ON SOLUTIONS OF A SYSTEM OF HIGHER-ORDER
NONLINEAR FRACTIONAL DIFFERENTIAL EQUATIONS
(COMMUNICATED BY DOUGLAS R. ANDERSON)
ZHENYU GUO, MIN LIU
Abstract. A system of higher-order nonlinear fractional differential equations is studied in this article, and some sufficient conditions for existence
and uniqueness of a solution for the system is established by the nonlinear
alternative of Leray-Schauder and Banach contraction principle.
1. Introduction and preliminaries
This article is concerned with the initial value problem for the following system
of fractional order differential equations:
(
)
c ρ
D u(t) = f t, v (n) (t),c Dβ v(t) , u(k) (0) = ηk , 0 < t ≤ T,
(1.1)
(
)
c σ
D v(t) = g t, u(n) (t),c Dα u(t) , v (k) (0) = ξk , 0 < t ≤ T,
(1.2)
where c D denotes the Caputo fractional derivative, f, g : [0, T ] × R2 → R are
given functions, ρ, σ ∈ (m − 1, m), α, β ∈ (n − 1, n), m, n ∈ N, ρ > β, σ > α,
k = 0, 1, 2, · · · , m − 1, T > 0, and ηk , ξk are suitable real constants. In this article,
we consider the case that all of ρ, σ, β and α are non-integer valued.
Recently, fractional order differential equations and systems have been of great
interest. For example, in 2010, Li[9] discussed the existence and uniqueness of mild
solution for
(
)
dq x(t)
= −Ax(t) + f t, x(t), Gx(t) , t ∈ [0, T ],
(1.3)
dtq
x(0) + g(x) = x0 .
Li and Guérékata[10] studied mild solutions of the fractional integrodifferential
equations as follows
∫ t
dq x(t)
+
Ax(t)
=
f
(t,
x(t))
+
a(t − s)g(s, x(s))ds, t ∈ [0, T ], x(0) = x0 .
dtq
0
(1.4)
2000 Mathematics Subject Classification. 34K15, 34C10.
Key words and phrases. System of fractional differential equations; the nonlinear alternative
of Leray-Schauder; Banach contraction principle; fixed point.
c
⃝2011
Universiteti i Prishtinës, Prishtinë, Kosovë.
Submitted August 25, 2011. Published September 13, 2011.
59
60
Z. GUO, M. LIU
In 2011, Anguraj, Karthikeyan and Trujillo[1] investigated the existence and the
uniqueness of the solution for the following fractional integrodifferential equation
∫ t
∫ 1
(
(
)
(
) )
dq x(t)
=
f
t,
x(t),
k
t,
s,
x(s)
ds,
h
t,
s,
x(s)
ds , t ∈ [0, 1],
dtq
0
0
(1.5)
∫ 1
g(s)x(s)ds.
x(0) =
0
Guo and Liu[4] studied the existence of unique solutions of initial value problems
of the following system of fractional order differential equations with infinite delay
Dα y1 (t) = f1 [t, y1t , y2t ], t ∈ [0, b],
y1 (t) = ϕ1 (t),
t ∈ (−∞, 0],
α
D y2 (t) = f2 [t, y1t , y2t ],
y2 (t) = ϕ2 (t),
t ∈ [0, b],
(1.6)
t ∈ (−∞, 0].
For detailed discussion on this topic, refer to the monographs of Kilbas et al.[5],
and the papers by Ahmad and Alsaedi [2], Guo and Liu [3], Kosmatov [6], Lakshmikantham and Vatsala [7], Li and Deng [8], Su [11], Goodrich [12,13], Bonilla
et al. [14], Bai and Fang [15], Kobayashi [16], Wang et al. [17] and the references
therein.
Applying the nonlinear alternative of Leray-Schauder, we obtain a result of existence of a solution for system (1.1)-(1.2). The uniqueness of a solution for the
system is established by Banach contraction principle.
The following notations, definitions, and preliminary facts will be used throughout this paper.
Let X = {u : u ∈ C([0, T ])} and Y = {v : v ∈ C([0, T ])} be normed spaces with
the sup-norm ∥u∥X and ∥v∥Y , respectively, where C([0, T ]) denotes the space of all
continuous functions defined on [0, T ]. Then, (X × Y, ∥ · · · ∥X×Y ) is a normed space
endowed with the sup-norm given by ∥(u, v)∥X×Y := max{∥u∥X , ∥v∥Y }.
Definition 1.1. For a function f ∈ C m ([0, T ]), m ∈ N, where C m ([0, T ]) denotes
the space of all continuous functions with mth order derivative, the Caputo derivative of fractional order α ∈ (m − 1, m) is defined by
∫ t
1
c α
D f (t) =
(t − s)m−α−1 f (m) (s)ds.
(1.7)
Γ(m − α) 0
Definition 1.2. The Riemann-Liouville fractional integral of order α, inversion of
Dα , is defined by
∫ t
1
I α f (t) =
(t − s)α−1 f (s)ds.
(1.8)
Γ(α) 0
Lemma 1.3. [8] If α ∈ (m − 1, m), m ∈ N, f ∈ C m ([0, T ]) and g ∈ C 1 ([0, T ]), then
(1) c Dα I α g(t) = g(t);
(2) I α (c Dα )f (t) = f (t) −
m−1
∑ k
k=0
t (k)
f (0).
k!
Lemma 1.4. [6] If m − 1 < α < β < m and f ∈ C m ([0, T ]), then for all k ∈
{1, 2, · · · , m − 1} and for all t ∈ [0, T ], the following relations hold:
c
Dβ−m+k f m−k (t) =c Dβ f (t),
(1.9)
A SYSTEM OF HIGHER-ORDER FRACTIONAL DIFFERENTIAL EQUATIONS
c
Dβ−α c Dα f (t) =c Dβ f (t).
61
(1.10)
Theorem 1.5. (the nonlinear alternative of Leray-Schauder) Let X be a normed
linear space, S ⊂ X be a convex set, U be open in S with 0 ∈ U , and F : U → S be
a continuous and compact mapping. Then either the mapping F has a fixed point
in U or there exist n ∈ ∂U and λ ∈ (0, 1) with n = λF n.
Now list the following hypotheses for convenience:
(H1) f : [0, T ]×R2 → R are continuously differentiable function with f (0, 0, 0) =
0 and f (t, 0, 0) ̸= 0 on a compact subinterval of (0, T ];
(H2) g : [0, T ] × R2 → R are continuously differentiable function with g(0, 0, 0) =
0 and g(t, 0, 0) ̸= 0 on a compact subinterval of (0, T ];
(H3) there exist nonnegative functions a1 , a2 , a3 , b1 , b2 , b3 ∈ C([0, T ]) such that
|f (t, x, y)| ≤ a1 (t) + a2 (t)|x| + a3 (t)|y|,
t ∈ [0, T ],
|g(t, x, y)| ≤ b1 (t) + b2 (t)|x| + b3 (t)|y|,
t ∈ [0, T ];
(1.11)
(H4) there exist nonnegative functions l1 , l2 , l3 , l4 ∈ C([0, T ]) such that
|f (t, x1 , y1 ) − f (t, x2 , y2 )| ≤ l1 (t)|x1 − x2 | + l2 (t)|y1 − y2 |,
t ∈ [0, T ],
|g(t, x1 , y1 ) − g(t, x2 , y2 )| ≤ l3 (t)|x1 − x2 | + l4 (t)|y1 − y2 |,
t ∈ [0, T ].
(1.12)
2. Existence and uniqueness of a solution
In this section, the theorems of existence and uniqueness of a solution for system
(1.1)-(1.2) will be given.
Lemma 2.1. Let (H1)-(H2) hold and n − 1 < α, β < n ≤ m − 1 < ρ, σ < m. Then,
a function u ∈ C m ([0, T ]) is a solution of the initial value problem (1.1) if and only
if
∫ t
n−1
∑ tk
(t − s)n−1
u(t) =
ηk +
w1 (s)ds, 0 < t ≤ 1,
(2.1)
k!
Γ(n)
0
k=0
(i)
where w1 (t) = u (t) ∈ C m−n ([0, T ]) with u(n+i) (t) = w1 (t), 0 ≤ i ≤ m − n − 1
is a solution of the integral equation
∫ t
∫ s
m−n−1
)
∑ ti
(t − s)ρ−n−1 (
(s − τ )n−β−1
w1 (t) =
ηn+i +
f s, w2 (s),
w2 (τ )dτ ds,
i!
Γ(ρ − n)
Γ(n − β)
0
0
i=0
(2.2)
and a function v ∈ C m ([0, T ]) is a solution of the initial value problem (1.2) if and
only if
∫ t
n−1
∑ tk
(t − s)n−1
v(t) =
ξk +
w2 (s)ds, 0 < t ≤ 1,
(2.3)
k!
Γ(n)
0
(n)
k=0
(n)
where w2 (t) = v
integral equation
w2 (t) =
m−n−1
∑ i
i=0
(i)
(t) ∈ C m−n ([0, T ]) with v (n+i) (t) = w2 (t) is a solution of the
t
ξn+i +
i!
∫
t
0
(t − s)σ−n−1 (
g s, w1 (s),
Γ(σ − n)
∫
0
s
)
(s − τ )n−α−1
w1 (τ )dτ ds.
Γ(n − α)
(2.4)
62
Z. GUO, M. LIU
Proof. Since the two parts of the Lemma is similar, we only give the proof of the
first part briefly. Lemma 1.4 ensures that
(
)
c ρ−n (n)
D
u (t) =c Dρ u(t) = f t, v (n) (t),c Dβ v(t) .
(2.5)
By Definition 1.1, we obtain
c
∫
(
Dρ−n u(n) (t) = f t, v (n) (t),
)
(t − s)n−β−1 (n)
v (s)ds .
Γ(n − β)
t
0
(2.6)
It follows from Definition 1.2, Lemma 1.3 (2) and the substitutions u(n) (t) =
w1 (t), v (n) (t) = w2 (t) that
w1 (t) = u(n) (t) =
m−n−1
∑ i
i=0
=
m−n−1
∑ i
t (i)
w (0)
i! 1
i=0
∫
t
+
=
t (n+i)
u
(0) + I ρ−n (c Dρ−n u(n) (t))
i!
(t − s)ρ−n−1 ( (n)
f s, v (s),
Γ(ρ − n)
∫
s
0
0
m−n−1
∑ i
)
(s − τ )n−β−1 (n)
v (τ )dτ ds
Γ(n − β)
(2.7)
t
ηn+i
i!
i=0
∫
t
+
0
(t − s)ρ−n−1 (
f s, w2 (s),
Γ(ρ − n)
Conversely, suppose that w1 ∈ C
u(n) (t) = w1 (t) =
m−n
∫
s
0
)
(s − τ )n−β−1
w2 (τ )dτ ds.
Γ(n − β)
([0, T ]) is a solution of (2.2). Then,
m−n−1
∑ i
i=0
t
ηn+i
i!
∫ s
)
(s − τ )n−β−1
(t − s)ρ−n−1 (
f s, w2 (s),
w2 (τ )dτ ds
+
Γ(ρ − n)
Γ(n − β)
0
0
m−n−1
(
)
∑ ti
=
ηn+i + I ρ−n f t, v (n) (t),c Dβ v(t) .
i!
i=0
∫
t
(2.8)
Since ρ − n ∈ (m − n − 1, m − n), by Lemma 1.3 (1) and Lemma 1.4, we have
c
Dρ u(t) =c Dρ−n u(n) (t)
=c Dρ−n
( m−n−1
∑ ti
i=0
i!
)
(
)
ηn+i +c Dρ−n I ρ−n f t, v (n) (t),c Dβ v(t)
(2.9)
(
)
= f t, v (n) (t),c Dβ v(t) , 0 < t ≤ 1.
Differentiating (2.2), we get
(k)
w1
=
∫ t
k
∏
(t − s)ρ−n−1−k
t
ηn+i+k +
(ρ − n − j)
i!
Γ(ρ − n)
0
j=1
∫
s
)
(
(s − τ )n−β−1
w2 (τ )dτ ds
f s, w2 (s),
Γ(n − β)
0
m−n−k−1
i
∑
i=0
(2.10)
A SYSTEM OF HIGHER-ORDER FRACTIONAL DIFFERENTIAL EQUATIONS
63
for each k = 0, 1, · · · , m − n − 1. As ρ − n − 1 − k ∈ (−1, m − n − 1), the second
term in (2.10) goes to zero as t → 0. Thus, we have
(k)
u(n+k) (0) = w1 (0) = ηn+k , k = 0, 1, · · · , m − n − 1,
(m−n)
which means that u(k) (0) = ηk , k = 0, 1, · · · , m − 1. Clearly, w1
C([0, T ]). Therefore, u is a solution of (1.1).
(2.11)
= u(m) ∈
For the sake of simplicity, Lemma 2.1 can be rewritten as
Lemma 2.2. Let f, g : [0, T ]×R → R be continuous functions. Then (u, v) ∈ X ×Y
is a solution of (1.1)-(1.2) if and only if (u, v) ∈ X × Y is a solution of (2.1)-(2.4).
Theorem 2.3. Assume (H1)-(H3) hold, and
∫ t
)
(t − s)ρ−n−1 (
sn−β
B1 = sup
a2 (s) +
a3 (s) ds < 1,
Γ(ρ − n)
Γ(n − β + 1)
t∈[0,T ] 0
∫ t
)
σ−n−1 (
(t − s)
sn−α
B2 = sup
b2 (s) +
b3 (s) ds < 1,
Γ(σ − n)
Γ(n − α + 1)
t∈[0,T ] 0
∫ t
(
)
ρ−n−1
(t − s)
0 < C1 = sup |η(t)| +
a1 (s)ds < +∞,
Γ(ρ − n)
t∈[0,T ]
0
∫
t
(
)
(t − s)σ−n−1
0 < C2 = sup |ξ(t)| +
b1 (s)ds < +∞,
Γ(σ − n)
t∈[0,T ]
0
(2.12)
where
η(t) =
m−n−1
∑ i
i=0
t
ηn+i ,
i!
ξ(t) =
m−n−1
∑ i
i=0
t
ξn+i .
i!
(2.13)
Then the system of integral equations (2.1)-(2.4) has a solution.
Proof. Define a mapping F : X × Y → X × Y and a ball U in the normed space
X × Y by
F (w1 , w2 )(t) = (F1 w2 (t), F2 w1 (t)),
(2.13)
and
U = {(w1 (t), w2 (t)) : (w1 (t), w2 (t)) ∈ X × Y, ∥(w1 (t), w2 (t))∥X×Y < R, t ∈ [0, T ]},
(2.14)
where
∫ s
∫ t
)
(s − τ )n−β−1
(t − s)ρ−n−1 (
f s, w2 (s),
w2 (τ )dτ ds,
F1 w2 (t) = η(t) +
Γ(ρ − n)
Γ(n − β)
0
0
∫ t
∫ s
)
σ−n−1 (
(t − s)
(s − τ )n−α−1
F2 w1 (t) = ξ(t) +
g s, w1 (s),
w1 (τ )dτ ds,
Γ(σ − n)
Γ(n − α)
0
0
(2.15)
and
C
R=
, B = max{B1 , B2 }, C = max{C1 , C2 }.
(2.16)
1−B
64
Z. GUO, M. LIU
Clearly, by (H1) and (H2), F is well defined and continuous. Let (w1 , w2 ) ∈ U .
Then ∥(w1 , w2 )∥X×Y ≤ R, and
∥F1 w2 ∥X
∫
= sup η(t) +
∫ s
) (t − s)ρ−n−1 (
(s − τ )n−β−1
f s, w2 (s),
w2 (τ )dτ ds
Γ(ρ
−
n)
Γ(n
−
β)
t∈[0,T ]
0
0
∫ t
∫ s
) )
(
ρ−n−1 (
(t − s)
(s − τ )n−β−1
≤ sup |η(t)| +
s,
w
(s),
w2 (τ )dτ ds
f
2
Γ(ρ
−
n)
Γ(n
−
β)
t∈[0,T ]
0
0
∫ t
(
(t − s)ρ−n−1 (
≤ sup |η(t)| +
a1 (s) + a2 (s)|w2 (s)|
Γ(ρ − n)
t∈[0,T ]
0
∫ s
) )
(s − τ )n−β−1
+ a3 (s)
|w2 (τ )|dτ ds
Γ(n − β)
0
∫ t
(
)
(t − s)ρ−n−1
≤ sup |η(t)| +
a1 (s)ds
Γ(ρ − n)
t∈[0,T ]
0
∫ s
( ∫ t (t − s)ρ−n−1 (
(s − τ )n−β−1 ) )
+ sup
a2 (s) + a3 (s)
dτ ds ∥w2 ∥Y
Γ(ρ − n)
Γ(n − β)
t∈[0,T ]
0
0
∫ t
(
)
(t − s)ρ−n−1
≤ sup |η(t)| +
a1 (s)ds
Γ(ρ − n)
t∈[0,T ]
0
∫ t
)
sn−β
(t − s)ρ−n−1 (
+ sup
a2 (s) +
a3 (s) ds∥w2 ∥Y
Γ(ρ − n)
Γ(n − β + 1)
t∈[0,T ] 0
t
=C1 + B1 ∥w2 ∥Y ≤ C + BR = R.
(2.17)
Similarly, we have
∥F2 w1 ∥Y ≤ C2 + B2 ∥w1 ∥X ≤ C + BR = R.
(2.18)
Therefore, ∥F (w1 , w2 )∥X×Y ≤ R, which implies that F (w1 , w2 ) ∈ U . In order to
show that F is completely continuous (continuous and compact), put
∫ t
(
)
(t − τ )n−β−1
Mf = max f t, w2 (t),
w2 (τ )dτ ,
Γ(n − β)
t∈[0,T ]
0
(2.19)
∫ t
(
)
(t − τ )n−α−1
Mg = max g t, w1 (t),
w1 (τ )dτ .
Γ(n − α)
t∈[0,T ]
0
For (w1 , w2 ) ∈ U and t1 , t2 ∈ [0, T ] with t1 < t2 , we obtain
|F1 w2 (t2 ) − F1 w2 (t1 )|
∫ t2
∫ s
)
(s − τ )n−β−1
(t2 − s)ρ−n−1 (
f s, w2 (s),
w2 (τ )dτ
=η(t2 ) − η(t1 ) +
Γ(ρ − n)
Γ(n − β)
0
0
∫ s
∫ t1
) n−β−1
ρ−n−1 (
(s − τ )
(t1 − s)
f s, w2 (s),
w2 (τ )dτ ds
−
Γ(ρ
−
n)
Γ(n
−
β)
0
0
∫ t1
∫ t2 (t − s)ρ−n−1
(t1 − s)ρ−n−1 2
≤|η(t2 ) − η(t1 )| + Mf ds −
ds
Γ(ρ − n)
Γ(ρ − n)
0
0
Mf
≤|η(t2 ) − η(t1 )| +
|tρ−n − tρ−n
|,
1
Γ(ρ − n + 1) 2
(2.20)
A SYSTEM OF HIGHER-ORDER FRACTIONAL DIFFERENTIAL EQUATIONS
65
and, in a similar manner,
|F2 w1 (t2 ) − F2 w1 (t1 )| ≤ |ξ(t2 ) − ξ(t1 )| +
Mg
|tσ−n − tσ−n
|.
1
Γ(σ − n + 1) 2
(2.21)
It follows from the uniform continuity of functions tk , tρ−n and tσ−n on [0, T ] that
F U is an equicontinuous set. Moreover, it is uniformly bounded as F U ⊂ U . Hence,
F is a completely continuous mapping.
Now to consider the following eigenvalue problem
(w1 , w2 ) = λF (w1 , w2 ) = (λF1 w2 , λF2 w1 ), λ ∈ (0, 1).
(2.22)
Assume that (w1 , w2 ) is a solution of (2.22) for λ ∈ (0, 1). Then,
∥w1 ∥X
= sup |λF1 w2 (t)|
t∈[0,T ]
∫ s
) (s − τ )n−β−1
(t − s)ρ−n−1 (
f s, w2 (s),
w2 (τ )dτ ds
Γ(ρ
−
n)
Γ(n
−
β)
t∈[0,T ]
0
0
∫ t
∫ s
(
) )
(t − s)ρ−n−1 (
(s − τ )n−β−1
≤λ sup |η(t)| +
w2 (τ )dτ ds
f s, w2 (s),
Γ(ρ − n)
Γ(n − β)
t∈[0,T ]
0
0
∫
=λ sup η(t) +
t
≤λ(C + B∥w2 ∥Y ),
(2.23)
and, similarly,
∥w2 ∥Y = sup |λF2 w1 (t)| ≤ λ(C + B∥w1 ∥X ).
(2.24)
t∈[0,T ]
(2.23) and (2.24) guarantee that (w1 , w2 ) ̸∈ ∂U . Therefore, by Theorem 1.5, there
exists a fixed point (w10 , w20 ) in U such that ∥(w10 , w20 )∥X×Y ≤ R, which completes the proof.
It follows from Lemma 2.1 and Theorem 2.3 that the solution (u0 , v0 ) of (1.1)(1.2) is given by
u0 (t) =
n−1
∑ k
k=0
v0 (t) =
t
ηk +
k!
n−1
∑ k
k=0
t
ξk +
k!
∫
t
(t − s)n−1
w10 (s)ds,
Γ(n)
t
(t − s)n−1
w20 (s)ds,
Γ(n)
0
∫
0
(2.25)
where
w10 (t) =
m−n−1
∑ i
i=0
∫ t
+
w20 (t) =
t
ηn+i
i!
(t − s)ρ−n−1 (
f s, w20 (s),
Γ(ρ − n)
∫
0
m−n−1
∑ i
i=0
∫ t
+
0
s
0
)
(s − τ )n−β−1
w20 (τ )dτ ds,
Γ(n − β)
t
ξn+i
i!
(t − s)σ−n−1 (
g s, w10 (s),
Γ(σ − n)
∫
0
s
)
(s − τ )n−α−1
w10 (τ )dτ ds.
Γ(n − α)
(2.26)
66
Z. GUO, M. LIU
Theorem 2.4. Assume (H1), (H2) and (H4) hold, and
∫
)
(t − s)ρ−n−1 (
sn−β
l1 (s) +
l2 (s) ds < 1,
Γ(ρ − n)
Γ(n − β + 1)
t∈[0,T ] 0
∫ t
)
(t − s)σ−n−1 (
sn−α
D2 = sup
l3 (s) +
l4 (s) ds < 1,
Γ(σ − n)
Γ(n − α + 1)
t∈[0,T ] 0
)
( ∫ t (t − s)ρ−n−1
|f (s, 0, 0)|ds < +∞,
0 < sup
Γ(ρ − n)
t∈[0,T ]
0
( ∫ t (t − s)σ−n−1
)
0 < sup
|g(s, 0, 0)|ds < +∞
Γ(σ − n)
t∈[0,T ]
0
t
D1 = sup
(2.27)
Then the system of integral equations (2.1)-(2.4) has a unique solution.
Proof. Define the mapping F and the ball U as those in the proof of Theorem 2.3,
where
( ∫ t (t − s)ρ−n−1
)
1
R=
sup
|f (s, 0, 0)|ds .
(2.28)
1 − D1 t∈[0,T ] 0
Γ(ρ − n)
Then F is well defined and continuous. For (w1 , w2 ) ∈ U , we obtain
∥F1 w2 ∥X ≤∥F1 w2 − F1 0∥X + ∥F1 0∥X
∫ t
)
(t − s)ρ−n−1 (
sn−β
≤ sup
l1 (s) +
l2 (s) ds∥w2 ∥Y
Γ(ρ − n)
Γ(n − β + 1)
t∈[0,T ] 0
( ∫ t (t − s)ρ−n−1
)
+ sup
|f (s, 0, 0)|ds
Γ(ρ − n)
t∈[0,T ]
0
(2.29)
≤D1 R + (1 − D1 )R ≤ R.
Similarly, ∥F2 w1 ∥Y ≤ R. Therefore, F U ⊂ U .
For (w1 , w2 ), (w1′ , w2′ ) ∈ U , we have
∥F1 w2 − F1 w2′ ∥X
≤ sup |F1 w2 (t) − F1 w2′ (t)|
t∈[0,T ]
∫
≤ sup
t∈[0,T ]
0
t
)
(t − s)ρ−n−1 (
sn−β
l1 (s) +
l2 (s) ds∥w2 − w2′ ∥Y
Γ(ρ − n)
Γ(n − β + 1)
(2.30)
=D1 ∥w2 − w2′ ∥Y ,
and, similarly,
∥F2 w1 − F2 w1′ ∥X ≤ D2 ∥w1 − w1′ ∥X .
(2.31)
Noting that D1 < 1, D2 < 1, F is a contractive mapping. It follows from Banach
′
′
contraction principle that F has a unique fixed point (w10
, w20
) ∈ U , which is a
solution of integral equations (2.1)-(2.4). This completes the proof.
A SYSTEM OF HIGHER-ORDER FRACTIONAL DIFFERENTIAL EQUATIONS
67
3. Example
Consider the following coupled system of fractional differential equations:
c
t
t
t4/5 c 9/5
+ v ′′ (t) +
D v(t), 0 < t ≤ 1,
2 3
4
′
′′
u(0) = 0, u (0) = 1, u (0) = 2,
D11/5 u(t) =
c
D
1
t−3/4 c 5/4
v(t) = t + u′′ (t) +
D u(t), 0 < t ≤ 1,
2
3
v(0) = 3, v ′ (0) = 4, v ′′ (0) = 5.
(3.1)
11/4
Here T = 1, n = 2, m = 3, ρ = 11/5, σ = 11/4, β = 9/5, α = 5/4, η1 = 0, η2 =
1, η3 = 2, ξ1 = 3, ξ2 = 4, and ξ3 = 5. Obviously, the hypotheses (H1)-(H3) are
satisfied with a1 (t) = t/2, a2 (t) = t/3, a3 (t) = t4/5 /4, b1 (t) = t, b2 (t) = 1/2, b3 (t) =
t−3/4 /3. In this case
∫ t
(s
)
1
s
B1 =
sup
(t − s)−4/5
+
ds
Γ(1/5) t∈[0,1] 0
3 4Γ(6/5)
) 15
1 (1
1
=
+
< 1,
Γ(1/5) 3 4Γ(6/5) 4
∫ t
(1
)
1
1
B2 =
sup
(t − s)−1/4
+
ds
Γ(3/4) t∈[0,1] 0
2 3Γ(7/4)
)4
1 (1
1
=
+
< 1,
Γ(3/4) 2 3Γ(7/4) 3
(3.2)
∫ t
(
1
s )
0 < C1 = sup 2 +
(t − s)−4/5 · ds
Γ(1/5) 0
2
t∈[0,1]
1
15
·
< +∞,
Γ(1/5) 8
∫ t
(
)
1
0 < C2 = sup 5 +
(t − s)−1/4 · sds
Γ(3/4) 0
t∈[0,1]
=2 +
=5 +
1
−8
·
< +∞.
Γ(3/4) 3
Thus, all the conditions of Theorem 2.3 are satisfied, and there exists a solution of
system (3.1).
Acknowledgments. The authors would like to thank the anonymous referee for
his/her comments that helped us improve this article.
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China
E-mail address: guozy@163.com
E-mail address: min liu@yeah.net