Electronic Journal of Differential Equations, Vol. 2015 (2015), No. 286, pp. 1–12.
ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu
ftp ejde.math.txstate.edu
SOLUTION OF FRACTIONAL DIFFERENTIAL EQUATIONS VIA
COUPLED FIXED POINT
HOJJAT AFSHARI, SABILEH KALANTARI, ERDAL KARAPINAR
Abstract. In this article, we investigate the existence and uniqueness of a solution for the fractional differential equation by introducing some new coupled
fixed point theorems for the class of mixed monotone operators with perturbations in the context of partially ordered complete metric space.
1. Introduction and preliminaries
In the previous decade, one of the most attractive research subject is to investigate the existence and uniqueness of a fixed point of certain operator in the setting
of complete metric space endowed with a partial order (see e.g. [1]-[24] and related
reference therein). Recently, CB. Zhai [20] proved some results on a class of mixed
monotone operators with perturbations. The aim of this article is to propose a
method for the existence and uniqueness of a solution of certain fractional differential equations by following the paper by Zhai [20]. For this purpose, we shall
consider some coupled fixed point theorems for a class of mixed monotone operators with perturbations on ordered Banach spaces with the different conditions
that was introduced by Zhai [20]. On the other hand, our result are finer than the
results of Zhai [20] since we obtain the existence and uniqueness of coupled fixed
points without assuming continuity of compactness of the operator.
For the sake of completeness of the paper, we present here some basic definitions,
notations and known results.
Suppose (E, k · k) is a Banach space which is partially ordered by a cone P ⊆ E,
that is, x ≤ y if and only if y − x ∈ P . If x 6= y, then we denote x < y or x > y. We
denote the zero element of E by θ. Recall that a non-empty closed convex set P ⊂ E
is a cone if it satisfies (i) x ∈ P, λ ≥ 0 =⇒ λx ∈ P ; (ii) x ∈ P, −x ∈ P =⇒ x = θ. A
cone P is called normal if there exists a constant N > 0 such that θ ≤ x ≤ y implies
kxk ≤ N kyk. Also we define the order interval [x1 , x2 ] = {x ∈ E|x1 ≤ x ≤ x2 } for
all x1 , x2 ∈ E. We say that and operator A : E → E is increasing whenever x ≤ y
implies Ax ≤ Ay. For all x, y ∈ E, the notation x ∼ y means that there exist λ > 0
and µ > 0, such that λx ≤ y ≤ µx. Clearly, ∼ is an equivalent relation. Given
2010 Mathematics Subject Classification. 47H10, 54H25.
Key words and phrases. Fixed point; mixed monotone operator; normal cone;
sub-homogeneous operator.
c
2015
Texas State University.
Submitted June 25, 2015. Published November 16, 2015.
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H. AFSHARI, S. KALANTARI, E. KARAPINAR
EJDE-2015/286
e > θ, we denote by Pe the set Pe = {x ∈ E|x ∼ e}. It is easy to see that Pe ⊂ P
is convex and λPe = Pe for all λ > 0. If P =
6 φ and e ∈ P , it is clear that Pe = P .
Definition 1.1 ([8, 9]). A : P × P → P is said to be a mixed monotone operator
if A(x, y) is increasing in x and decreasing in y, i.e., ui , vi (i = 1, 2) ∈ P , u1 ≤ u2 ,
v1 ≥ v2 imply A(u1 , v1 ) ≤ A(u2 , v2 ). The element x ∈ P is called a fixed point of
A if A(x, x) = x.
The following conditions were was assumed in [21]:
(A1) there exists h ∈ P with h 6= θ such that A(h, h) ∈ Ph ,
(A2) for any u, v ∈ P and t ∈ (0, 1), there exists ϕ(t) ∈ (t, 1] such that
A(tu, t−1 v) ≥ ϕ(t)A(u, v).
(1.1)
Lemma 1.2 ([21]). Assume that (A1), (A2) hold. Then A : Ph × Ph → Ph ; and
there exist u0 , v0 ∈ Ph and r ∈ (0, 1) such that
rv0 ≤ u0 < v0 ,
u0 ≤ A(u0 , v0 ) ≤ A(v0 , u0 ) ≤ v0 .
Definition 1.3 ([20]). An operator A : P → P is said to be sub-homogeneous if it
satisfies
A(tx) ≥ tA(x), ∀t ∈ (0, 1), x ∈ P.
The following result can be found in Zhai and Zhang [21].
Theorem 1.4 ([21]). Let P be a normal cone in E. Assume that T : P × P → P
is a mixed monotone operator and satisfies:
(A3) there exists h ∈ P with h 6= θ such that T (h, h) ∈ Ph ;
(A4) for any u, v ∈ P and t ∈ (0, 1), there exists ϕ(t) ∈ (t, 1] such that
T (tu, t−1 v) ≥ ϕ(t)T (u, v).
(1.2)
Then
(1) T : Ph × Ph → Ph ;
(2) there exist u0 , v0 ∈ Ph and r ∈ (0, 1) such that rv0 ≤ u0 < v0 , u0 ≤
T (u0 , v0 ) ≤ T (v0 , u0 ) ≤ v0 ;
(3) T has a unique fixed point x∗ in Ph ;
(4) for any initial values x0 , y0 ∈ Ph , constructing successively the sequences
xn = T (xn−1 , yn−1 ),
∗
yn = T (yn−1 , xn−1 ),
n = 1, 2 . . . ,
∗
we have xn → x and yn → x as n → ∞.
2. Main result
In this section, we state and prove our main results. First, we consider the mixed
monotone operator A : P × P → P . The following conditions will be assumed:
(A5) there exists h ∈ P with h 6= θ such that A(h, h) ∈ Ph ,
(A6) for any u, v ∈ P and s, t ∈ (0, 1) such that s ≤ t, there exists ϕ(t) ∈ (t2 , 1]
and ϕ is decreasing such that
ϕ(t)
A(u, v).
(2.1)
A(tu, t−1 v) + A(tu, s−1 v) ≥ 2
t
Lemma 2.1. Assume (A5), (A6) hold. Then A : Ph × Ph → Ph ; and there exist
u0 , v0 ∈ Ph and r ∈ (0, 1) such that
rv0 ≤ u0 < v0 , u0 ≤ A(u0 , v0 ) ≤ A(v0 , u0 ) ≤ v0 .
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SOLUTION OF FRACTIONAL DIFFERENTIAL EQUATIONS
Proof. For s ≤ t from condition (A6) we obtain
1
t
A(t−1 x, ty) ≤ ϕ(t) (A(x, y) + A(x, y)), ∀s, t ∈ (0, 1), x, y ∈ P.
s
2 t
3
(2.2)
For any u, v ∈ Ph , there exist µ1 , µ2 ∈ (0, 1), such that
1
1
µ1 h ≤ u ≤
h, µ2 h ≤ v ≤
h.
µ1
µ2
Let µ = min{µ2 , µ1 }. Then µ ∈ (0, 1). From (2.2) and the mixed monotone
properties of operator A and regarding 0 < µ, there exists 0 < µ0 < µ such that
1
1
A(u, v) ≤ A( h, µ2 h) ≤ A( h, µh)
µ1
µ
1
µ
≤ ϕ(µ) (A(h, h) + A(h, 0 h))
µ
2(
)
µ
≤
=
1
2( ϕ(µ)
µ )
1
( ϕ(µ)
µ )
(A(h, h) + A(h, h))
(A(h, h)) ≤
1
A(h, h).
ϕ(µ)
Regarding the inequality
A(u, v) ≥ A(µ1 h,
1
1
h) ≥ A(µh, h),
µ2
µ
we derive that
1
1
h) + A(µh, h)
µ
µ
ϕ(µ)
≥ 2(
)(A(h, h) + A(h, h))
µ
ϕ(µ)
= 2(
)(A(h, h)) ≥ 2ϕ(µ)A(h, h).
µ
2A(u, v) ≥ A(µh,
Tehrefore, we obtain
A(u, v) ≥ ϕ(µ)A(h, h).
It follows from A(h, h) ∈ Ph that A(u, v) ∈ Ph . Hence we have A : Ph × Ph → Ph .
Since A(h, h) ∈ Ph , we can choose a sufficiently small number t0 ∈ (0, 1) such that
1
t0 h ≤ A(h, h) ≤ h.
(2.3)
t0
For k > 2 we have
1
tk0 h ≤ A(h, h) ≤ k h .
(2.4)
t0
Put u0 = t0 k h and v0 = t1k h. Evidently, u0 , v0 ∈ Ph and u0 = t0 2k v0 < v0 . Take
0
any r ∈ (0, t0 2k ], then r ∈ (0, 1) and u0 ≥ rv0 . By the mixed monotone properties of
A, we have A(u0 , v0 ) ≤ A(v0 , u0 ). Because t0 ∈ (0, 1), then there exists s0 ∈ (0, 1)
such that 0 < s0 ≤ t0 . Further, combining condition (A2) with (2.3), and since
s0 ≤ t0 we have
1
A(u0 , v0 ) = A(tk0 h, k h)
t0
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H. AFSHARI, S. KALANTARI, E. KARAPINAR
EJDE-2015/286
ϕ(tk0 )
)A(h, h) − A(h, h)
tk0
ϕ(t0 )
≥ (2( 2 ) − 1)A(h, h) > A(h, h)
t0
≥ 2(
≥ tk0 h = u0 ,
and
A(v0 , u0 ) = A(
1
h, tk0 h)
tk0
1
≤
ϕ(tk )
2( tk0 )
0
1
≤
ϕ(tk )
2( tk0 )
0
(A(h, h) + A(h,
tk0
h))
sk0
2A(h, h)
≤ A(h, h) ≤
1
h = v0 .
tk0
Consequently, we have u0 ≤ A(u0 , v0 ) ≤ A(v0 , u0 ) ≤ v0 .
Corollary 2.2. If in (2.1) put s = t then we obtain (1.2). Consequently the Lemma
2.1 yields the Lemma 1.2.
Theorem 2.3. Suppose that P is a normal cone of E, and (A5), (A6) hold. Then
operator A has a unique fixed point x∗ in Ph . Moreover, for any initial x0 , y0 ∈ Ph ,
constructing successively the sequences
xn = A(xn−1 , yn−1 ),
∗
yn = A(yn−1 , xn−1 )
n = 1, 2, . . . ,
∗
we have kxn − x k → 0,kyn − x k → 0 as n → ∞.
Proof. From Lemma 2.1, there exist u0 , v0 ∈ Ph and r ∈ (0, 1) such that
rv0 ≤ u0 < v0 , u0 ≤ A(u0 , v0 ) ≤ A(v0 , u0 ) ≤ v0 .
Construct recursively the sequences
un = A(un−1 , vn−1 ),
vn = A(vn−1 , un−1 ),
n = 1, 2, . . . .
Evidently u1 ≤ v1 . By the mixed monotone properties of A, we obtain
un ≤ vn , n = 1, 2, . . . .
It also follows from Lemma 2.1 and the mixed monotone properties of A that
u0 ≤ u1 ≤ . . . ≤ un ≤ . . . ≤ vn ≤ . . . ≤ v1 ≤ v0 .
(2.5)
Note that u0 ≥ rv0 . We can get un ≥ u0 ≥ rv0 ≥ rvn , n = 1, 2, . . . . Let
tn = sup{t > 0|un ≥ tvn },
sn = sup{s > 0|un ≥ svn },
sn ≤ tn n = 1, 2, . . . .
Thus we have un ≥ tn vn , un ≥ sn vn , n = 1, 2, . . ., then un ≥ tn vn ≥ sn vn , also
un+1 ≥ un ≥ tn vn ≥ tn vn+1 ≥ sn vn+1 , n = 1, 2, . . .. Therefore, tn+1 ≥ tn , i.e., tn is
increasing with tn ⊂ (0, 1]. Suppose tn → t∗ as n → ∞, then t∗ = 1. Otherwise, 0 <
t∗ < 1. Then from condition (A2) and tn ≤ t∗ , we have A(un , vn ) ≥ A(tn vn , t1n un )
and A(un , vn ) ≥ A(tn vn , s1n un ), so
un+1 = A(un , vn )
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SOLUTION OF FRACTIONAL DIFFERENTIAL EQUATIONS
5
1
1
1
(A(tn vn , un ) + A(tn vn , un ))
2
tn
sn
ϕ(tn )
ϕ(t∗ )
≥
A(vn , un ) ≥
A(vn , un )
tn
tn
ϕ(t∗ )
=
vn+1 .
tn
≥
∗
)
∗2
By the definition of tn , tn+1 ≥ ϕ(t
≥ ϕ(t∗ ) > t∗ 2 ,
tn . Let n → ∞, we obtain t
which is a contradiction. Thus, limn→∞ tn = 1. For any natural number p we have
θ ≤ un+p − un ≤ vn − un ≤ vn − tn vn = (1 − tn )vn ≤ (1 − tn )v0 ,
θ ≤ vn − vn+p ≤ vn − un ≤ (1 − tn )v0 .
Since the cone P is normal, we have
kun+p − un k ≤ M (1 − tn )kv0 k → 0,
kvn − vn+p k ≤ M (1 − tn )kv0 k → 0,
as n → ∞, where M is the normality constant of P . So we can claim that un
and vn are Cauchy sequences. Since E is complete, there exist u∗ , v ∗ such that
un → u∗ , vn → v ∗ , as n → ∞. By (2.5), we know that un ≤ u∗ ≤ v ∗ ≤ vn with
u∗ , v ∗ ∈ Ph and
θ ≤ v ∗ − u∗ ≤ vn − un ≤ (1 − tn )v0 .
Further
kv ∗ − u∗ k ≤ M (1 − tn )kv0 k → 0 (n → ∞),
and thus u∗ = v ∗ . Let x∗ := u∗ = v ∗ and then we obtain
un+1 = A(un , vn ) ≤ A(x∗ , x∗ ) ≤ A(vn , un ) = vn+1 .
Let n → ∞, then we obtain x∗ = A(x∗ , x∗ ). That is, x∗ is a fixed point of A in Ph .
Next we shall prove that x∗ is the unique fixed point of A in Ph . In fact, suppose
x̄ is a fixed point of A in Ph . Since x∗ , x̄ ∈ Ph , there exists positive numbers
µ̄1 , µ̄2 , λ̄1 , λ̄2 > 0 such that
µ̄1 h ≤ x∗ ≤ λ̄1 ,
µ̄2 h ≤ x̄ ≤ λ̄2 h.
λ̄2 ∗
λ̄2
µ̄1 h ≤
x ,
µ̄1
µ̄1
x̄ ≥ µ̄2 h =
Then we obtain
x̄ ≤ λ̄2 h =
µ̄2
µ̄2 ∗
x .
λ̄1 h ≥
λ̄1
λ̄1
Let e1 = sup{t > 0|tx∗ ≤ x̄ ≤ t−1 x∗ }. Evidently, 0 < e1 ≤ 1, e1 x∗ ≤ x̄ ≤
Next we prove e1 = 1. If 0 < e1 < 1, then x̄ = A(x̄, x̄) ≥ A(e1 x∗ , e11 x∗ ), then
2A(x̄, x̄) ≥ 2A(e1 x∗ ,
≥ 2(
1 ∗
e1 x .
1
1
1 ∗
x ) = A(e1 x∗ , x∗ ) + A(e1 x∗ , x∗ )
e1
e1
e1
ϕ(e1 )
)A(x∗ , x∗ ).
e1
So we have
ϕ(e1 )
ϕ(e1 )
)A(x∗ , x∗ ) ≥
A(x∗ , x∗ ) ≥ ϕ(e1 )A(x∗ , x∗ ).
e1
e1
Since ϕ(e1 ) > e1 , this contradicts the definition of e1 . Hence e1 = 1, and we obtain
x̄ = x∗ . Therefore, A has a unique fixed point x∗ in Ph . Note that [u0 , v0 ] ⊂ Ph ,
then we know that x∗ is the unique fixed point of A in [u0 , v0 ].
A(x̄, x̄) ≥ (
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H. AFSHARI, S. KALANTARI, E. KARAPINAR
EJDE-2015/286
Now we construct the sequences recursively as follows:
xn = A(xn−1 , yn−1 ),
yn = A(yn−1 , xn−1 ),
n = 1, 2, . . . ,
for any initial points x0 , y0 ∈ Ph . Since x0 , y0 ∈ Ph we can choose small numbers
e2 , e3 ∈ (0, 1) such that
e2 h ≤ x0 ≤
1
h,
e2
e3 h ≤ y0 ≤
1
h.
e3
Let e∗ = min{e2 , e3 }. Then e∗ ∈ (0, 1) and
1
h.
e∗
We can choose a sufficiently large positive integer m such that
ϕ(e∗ ) m
1
≥ ∗.
∗
e
e
Put ū0 = e∗ m h, v̄0 = e∗1m h, it easy to see that ū0 , v̄0 ∈ Ph and ū0 < x0 , y0 < v̄0 .
Let
ūn = A(ūn−1 , v̄n−1 ), v̄n = A(v̄n−1 , ūn−1 ), n = 1, 2, . . . .
e∗ h ≤ x0 ,
y0 ≤
Analogously, it follows that there exists y ∗ ∈ Ph such that A(y ∗ , y ∗ ) = y ∗ and
limn→∞ ūn = limn→∞ v̄n = y ∗ . By the uniqueness of fixed point of operator A in
Ph . We get x∗ = y ∗ and by induction ūn ≤ xn , yn ≤ v̄n , n = 1, 2, . . .. Since cone
P is normal we have limn→∞ xn = limn→∞ yn = x∗ .
Theorem 2.4. Let α ∈ (0, 1), A : P × P → P be a mixed monotone operator
satisfying
A(tx, t−1 y) + A(tx, s−1 y) ≥ 2t2α−1 A(x, y),
s, t ∈ (0, 1), s ≤ t x, y ∈ P.
(2.6)
Suppose that B : P → P is an in increasing sub-homogeneous operator. Assume
also that
(i) there is h0 ∈ Ph such that A(h0 , h0 ) ∈ Ph and Bh0 ∈ Ph ;
(ii) there exists a constant δ0 > 0 such that A(x, y) ≥ δ0 Bx for all x, y ∈ P .
Then
(1) A : Ph × Ph → Ph , B : Ph → Ph ;
(2) there exist u0 , v0 ∈ Ph and r ∈ (0, 1) such that
rv0 ≤ u0 < v0 ,
u0 ≤ A(u0 , v0 ) + Bu0 ≤ A(v0 , u0 ) + Bv0 ≤ v0 ;
(3) the operator A(x, x) + Bx = x has a unique solution x∗ in Ph ;
(4) for any initial values x0 , y0 ∈ Ph , constructing successively the sequences
xn = A(xn−1 , yn−1 ) + Bxn−1 ,
∗
yn = A(yn−1 , xn−1 ) + Byn−1 ,
n = 1, 2, . . . ,
∗
we have xn → x and yn → x as n → ∞.
Proof. Notice that from (2.6) and Definition 1.3, we have
1
1
t
A( x, ty) ≤ 2α−1 (A(x, y) + A(x, y)),
(2.7)
t
2t
s
and B( 1t x) ≤ 1t Bx for s, t ∈ (0, 1), x, y ∈ P and s ≤ t.
Since A(h0 , h0 ), Bh0 ∈ Ph , there exist constants λ1 , λ2 , ν1 , ν2 > 0 such that
λ1 h ≤ A(h0 , h0 ) ≤ λ2 h,
ν1 h ≤ Bh0 ≤ ν2 h .
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SOLUTION OF FRACTIONAL DIFFERENTIAL EQUATIONS
7
Also from h0 ∈ Ph , there exists a constant t0 ∈ (0, 1) such that t0 h ≤ h0 ≤ t10 h,
and let s0 ∈ (0, 1) such that s0 ≤ t0 , then we have
1
1
s0 h ≤ t0 h ≤ h0 ≤ h ≤ h.
t0
s0
From s0 ≤ t0 , (2.6), (2.7) and the mixed monotone properties of operator A, we
have
1
1
A(h, h) ≥ A(t0 h0 , h0 ), A(h, h) ≥ A(t0 h0 , h0 ).
t0
s0
So we have
2A(h, h) ≥ 2t02α−1 A(h0 , h0 ) .
By combining the inequalities above, we have
2α
A(h, h) ≥ t2α−1
A(h0 , h0 ) ≥ t2α
0 A(h0 , h0 ) ≥ λ1 t0 h,
0
and
t0
1
1
h0 , t0 h0 ) ≤ 2α−1 (A(h0 , h0 ) + A(h0 , h0 ))
t0
s
2t0
0
1
λ2
≤ 2α A(h0 , h0 ) ≤ α h.
t0
t0
A(h, h) ≤ A(
Noting that tλ2α2 , λ1 t20 α > 0, we can get A(h, h) ∈ Ph . By Definition 1.3 and the
0
monotone property of operator B, we have
1
1
ν2
Bh ≤ B( h0 ) ≤ Bh0 ≤ h, Bh ≥ B(t0 h0 ) ≥ t0 Bh0 ≥ ν1 t0 h.
t0
t0
t0
Next we show B : Ph → Ph . For any x ∈ Ph , we can choose a sufficiently small
number µ ∈ (0, 1) such that
1
µh ≤ x ≤ h.
µ
Consequently,
1
1 ν2
Bx ≤ B( h) ≤
h, Bx ≥ B(µh) ≥ µt0 ν1 h.
µ
µ t0
ν2
, µt0 ν1 > 0. Thus Bx ∈ Ph ; that is, B : Ph → Ph . So the
Evidently, we have µt
0
conclusion (1) holds. Now we define an operator T = A + B by T (x, y) = A(x, y) +
Bx. Then T : P × P → P is a mixed monotone operator and T (h, h) ∈ Ph . In the
following we show that there exists ϕ(t) ∈ (t, 1] with respect to s, t ∈ (0, 1), s ≤ t
such that
ϕ(t)
T (tx, t−1 y) + T (tx, s−1 y) ≥ 2(
)A(x, y), ∀, x, y ∈ P.
t
Consider the function
t2β−1 − t
f (t) = 2α−1
,
t
− t2β−1
for t ∈ (0, 1), where β ∈ (α, 1). It is easy to prove that f is increasing in (0, 1) and
lim f (t) = 0,
t→0+
lim f (t) =
t→1−
1−β
.
β−α
Further, fixing t ∈ (0, 1), we have
lim− f (t) = lim−
β→1
β→1
t2β−1 − t
= 0.
t2α−1 − t2β−1
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H. AFSHARI, S. KALANTARI, E. KARAPINAR
EJDE-2015/286
So there exists β0 (t) ∈ (0, 1) with respect to t such that
t2β0 (t)−1 − t
≤ δ0 ,
t2α−1 − t2β0 (t)−1
t ∈ (0, 1).
Hence we have
A(x, y) ≥ δ0 Bx ≥
t2β0 (t)−1 − t
Bx,
− t2β0 (t)−1
t2α−1
∀t ∈ (0, 1), x, y ∈ P.
Then we obtain
t2α−1 A(x, y) + tBx ≥ t2β0 (t)−1 [A(x, y) + Bx],
∀t ∈ (0, 1), x, y ∈ P.
Consequently, for any t ∈ (0, 1) and x, y ∈ P ,
T (tx, t−1 y) + T (tx, s−1 y) = A(tx, t−1 y) + B(tx) + A(tx, s−1 y) + B(tx)
≥ 2t2α−1 A(x, y) + 2tBx
≥ 2t2β0 (t)−1 (A(x, y) + Bx)
= 2t2β0 (t)−1 T (x, y).
Let ϕ(t) = t2β0 (t) , t ∈ (0, 1). Then ϕ(t) ∈ (t2 , 1] and
ϕ(t)
)A(x, y),
t
for any s, t ∈ (0, 1) and x, y ∈ P . Hence the condition (A2) in Lemma 2.1 is satisfied.
By Lemma 2.1 we conclude that: (a) there exist u0 , v0 ∈ Ph and r ∈ (0, 1) such that
rv0 ≤ u0 < v0 , u0 ≤ T (u0 , v0 ) ≤ T (v0 , u0 ) ≤ v0 ; (b) T has a unique fixed point x∗
in Ph ; (c) for any initial values x0 , y0 ∈ Ph , constructing successively the sequences
T (tx, t−1 y) + T (tx, s−1 y) ≥ 2(
xn = T (xn−1 , yn−1 ),
∗
yn = T (yn−1 , xn−1 ),
n = 1, 2, . . . ,
∗
we have xn → x and yn → x as n → ∞. That is, conclusions (2)–(4) hold.
Corollary 2.5. Let α ∈ (0, 1), A : P × P → P is a mixed monotone operator.
Assume (2.6) holds and there is h0 > θ such that A(h0 , h0 ) ∈ Ph . Then
(1) A : Ph × Ph → Ph ;
(2) there exist u0 , v0 ∈ Ph and r ∈ (0, 1) such that
rv0 ≤ u0 < v0 ,
u0 ≤ A(u0 , v0 ) ≤ A(v0 , u0 ) ≤ v0 ;
(3) the operator A(x, x) = x has a unique solution x∗ in Ph ;
(4) for any initial values x0 , y0 ∈ Ph , constructing successively the sequences
xn = A(xn−1 , yn−1 ),
∗
yn = A(yn−1 , xn−1 ),
n = 1, 2, . . . ,
∗
we have xn → x and yn → x as n → ∞.
3. Solution to fractional differential equations
In this section, we shall propose a method for showing the existence and uniqueness of a solution for the fractional differential equation
Dα
u(s, t) + f (s, t, u(s, t), v(s, t)) = 0,
Dt
(3.1)
0 < < T, T ≥ 1, t ∈ [, T ], 0 < α < 1, s ∈ [a, b]
subject to the condition
u(s, ζ) = u(s, T ),
(s, ζ) ∈ [a, b] × (, t),
(3.2)
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SOLUTION OF FRACTIONAL DIFFERENTIAL EQUATIONS
9
where Dα is the Riemann-Liouville fractional derivative of order α. We will suppose
that a, b ∈ (0, ∞), a < b. Let
E = C([a, b] × [, T ]).
Consider the Banach space of continuous functions on [a, b] × [, T ] with sup norm
and set
P = {y ∈ C([a, b] × [, T ]) :
min
y(s, t) ≥ 0}.
(s,t)∈[a,b]×[,T ]
Then P is a normal cone.
Lemma 3.1. Let (s, t) ∈ [a, b] × [, T ], (s, ζ) ∈ [a, b] × (, t) and 0 < α < 1. Then
the problem
Dα
u(s, t) + f (s, t, u(s, t), v(s, t)) = 0
Dt
with the boundary value condition u(s, ζ) = u(s, T ) has a solution u0 if and only if
u0 is a solution of the fractional integral equation
Z T
u(s, t) =
G(t, ξ)f (s, ξ, u(s, ξ), v(s, t))dξ,
where
 α−1
t
(ζ−ξ)α−1 −tα−1 (T −ξ)α−1

−
α−1
α−1


 α−1(ζ −Tα−1 )Γ(α) α−1
−t
−(T −ξ)
(t−ξ)
G(t, ξ) = (ζ
α−1 −T α−1 )Γ(α) −
Γ(α) ,


α−1
α−1

 −t (T −ξ)
,
(ζ α−1 −T α−1 )Γ(α)
(t−ξ)α−1
Γ(α) ,
≤ ξ ≤ ζ ≤ t ≤ T,
≤ ζ ≤ ξ ≤ t ≤ T,
≤ ζ ≤ t ≤ ξ ≤ T.
α
Proof. From D
Dt u(s, t) + f (s, t, u(s, t), v(s, t)) = 0 and the boundary condition, it
is easy to see that u(s, t) − c1 tα−1 = −Iα f (s, t, u(s, t), v(s, t)). By the definition of
a fractional integral, we obtain
Z ζ
(t − ξ)α−1
u(s, t) = c1 tα−1 −
f (s, ξ, u(s, ξ), v(s, ξ))dξ,
Γ(α)
Z ζ
(ζ − ξ)α−1
u(s, ζ) = c1 T α−1 −
f (s, ξ, u(s, ξ), v(s, ξ))dξ,
Γ(α)
Z T
(t − ξ)α−1
α−1
u(s, T ) = c1 T
−
f (s, ξ, u(s, ξ), v(s, ξ))dξ.
Γ(α)
Since u(s, ζ) = u(s, T ), we obtain
Z ζ
(ζ − ξ)α−1
1
c1 = α−1
f (s, ξ, u(s, ξ), v(s, ξ))dξ
α−1
ζ
−T
Γ(α)
Z T
1
(T − ξ)α−1
− α−1
f (s, ξ, u(s, ξ), v(s, ξ))dξ.
ζ
− T α−1 Γ(α)
Hence
Z ζ
tα−1
(ζ − ξ)α−1
f (s, ξ, u(s, ξ), v(s, ξ))dξ
u(s, t) = α−1
α−1
ζ
−T
Γ(α)
Z T
tα−1
(T − ξ)α−1
− α−1
f (s, ξ, u(s, ξ), v(s, ξ))dξ
α−1
ζ
−T
Γ(α)
Z t
(t − ξ)α−1
f (s, ξ, u(s, ξ), v(s, ξ))dξ
−
Γ(α)
10
H. AFSHARI, S. KALANTARI, E. KARAPINAR
Z
EJDE-2015/286
T
G(t, ξ)f (s, ξ, u(s, ξ), v(s, ξ))dξ.
=
This completes the proof.
Theorem 3.2. Let 0 < < T and let f (s, t, u(s, t), v(s, t)) be function in the space
C([a, b], [, T ], [0, ∞], [0, ∞]), that is increasing in u, decreasing in v, with positive
values. Also assume that for any u, v ∈ P and c, c0 ∈ (0, 1) with c0 ≤ c, there exists
ϕ(c) ∈ (c2 , 1] and ϕ is decreasing such that
Z T
Z T
−1
G(t, ξ)f (s, ξ, cu(s, ξ), c−1 v(s, ξ))dξ +
G(t, ξ)f (s, ξ, cu(s, ξ), c0 v(s, ξ))dξ
ϕ(c)
≥2
c
Z
T
G(t, ξ)f (s, ξ, u(s, ξ), v(s, ξ))dξ,
and f (s, t, u(s, t), v(s, t)) = 0, whenever G(s, t) < 0. Also assume that there exist
M1 > 0, M2 > 0 and h 6= θ ∈ P such that
Z T
M1 h(t) ≤
G(t, ξ)f (s, ξ, u(s, ξ), v(s, ξ))dξ ≤ M2 h(t),
for all t ∈ [, T ], where G(t, s) is the green function defined in Lemma 3.1. Then
problem (3.1) with the boundary value condition (3.2) has unique solution u∗ .
Proof. By using Lemma (3.1), the problem is equivalent to the integral equation
Z T
u(s, t) =
G(t, ξ)f (s, ξ, u(s, ξ), v(s, ξ))dξ,
where
 α−1
t
(η−ξ)α−1 −tα−1 (T −ξ)α−1

−
α−1
α−1


 α−1(η −Tα−1 )Γ(α) α−1
−t
−(T −ξ)
(t−ξ)
G(t, ξ) = (ηα−1 −T α−1 )Γ(α) − Γ(α)



 −tα−1 (T −ξ)α−1
(η α−1 −T α−1 )Γ(α)
(t−ξ)α−1
Γ(α)
≤ξ≤η≤t≤T
≤η≤ξ≤t≤T
≤η≤t≤ξ≤T
Define the operator A : P × P → E by
Z T
A(u(s, t), v(s, t)) =
G(t, ξ)f (s, ξ, u(s, ξ), v(s, ξ))dξ.
Then u is solution for the problem if and only if u = A(u, u). It is easy to see
to check that the operator A is increasing in u and decreasing in v on P . By
assumptions of theorem we have;
(A7) there exists h ∈ P with h 6= θ such that
Z T
M1 h(t) ≤
G(t, ξ)f (s, ξ, u(s, ξ), v(s, ξ))dξ ≤ M2 h(t),
thus A(h, h) ∈ Ph ,
(A8) for any u, v ∈ P and c, c0 ∈ (0, 1) such that c0 ≤ c , there exists ϕ(c) ∈ (c2 , 1]
and ϕ is decreasing such that
ϕ(c)
−1
A(cu, c−1 v) + A(cu, c0 v) ≥ 2
A(u, v).
c
Now by using theorem (2.3), the operator A has a unique fixed point u∗ in Ph .
Therefore the boundary value problem (3.1) has unique solution u∗ .
EJDE-2015/286
SOLUTION OF FRACTIONAL DIFFERENTIAL EQUATIONS
11
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Hojjat Afshari
Faculty of Basic Sciences, Bonab University, Bonab, Iran
E-mail address: hojat.afshari@yahoo.com
Sabileh Kalantari
Faculty of Basic Sciences, Bonab University, Bonab, Iran
E-mail address: kalantari.math@gmail.com
Erdal Karapinar
Atilim University, Department of Mathematics, 06836, İncek, Ankara, Turkey.
Nonlinear Analysis and Applied Mathematics (NAAM) Research Group, King Abdulaziz
University, 21589, Jeddah, Saudi Arabia
E-mail address: erdalkarapinar@yahoo.com