J. Nonlinear Sci. Appl. 1 (2008), no. 3, 123–131
The Journal of Nonlinear Sciences and Applications
http://www.tjnsa.com
POSITIVE SOLUTIONS FOR BOUNDARY VALUE PROBLEM
OF NONLINEAR FRACTIONAL DIFFERENTIAL EQUATION
TINGTING QIU1∗ AND ZHANBING BAI2
Abstract. We investigate the positive solution of nonlinear fractional differential equation with semi-positive nonlinearity
½ α
D0+ u(t) + f (t, u(t)) = 0, 0 < t < 1,
u(0) = u0 (1) = u00 (0) = 0
where 2 < α ≤ 3 is a real number, D0α+ is the Caputo’s differentiation, and
f : [0, 1] × [0, ∞) → (−∞, ∞). By use of Krasnosel’skii fixed point theorem,
the existence results of positive solution are obtained.
1. Introduction and preliminaries
Fractional differential equations have been of great interest recently. It is caused
both by the intensive development of the theory of fractional calculus itself and
by the applications of such constructions in various sciences such as physics, mechanics, chemistry, engineering, etc. For details, see [4, 6, 7, 8] and the references
therein.
It should be noted that most of papers and books on fractional calculus are
devoted to the solvability of linear initial fractional differential equations in terms
of special functions [9]. Recently, there are some papers deal with the existence
and multiplicity of solution (or positive solution) of nonlinear initial fractional
differential equation by the use of techniques of nonlinear analysis (fixed-point
theorems, Leray-Shauder theory, etc.), see [2, 3, 5, 11, 12].
However, there are few papers consider the semipositone nonlinearity differential equations of fractional order. No contributions exist, as far as we know,
Date: Received: 4 September 2008; Revised: 28 September 2008.
Corresponding author.
2000 Mathematics Subject Classification. Primary 34B15; Secondary 34B18.
Key words and phrases. Fractional differential equation; Positive solution; Fixed-point
theorem.
∗
123
124
TINGTING QIU AND ZHANBING BAI
concerning the positive solution with semipositone nonlinearity of the following
problem:
½ α
D0+ u(t) + f (t, u(t)) = 0, 0 < t < 1,
(1.1)
u(0) = u0 (1) = u00 (0) = 0
where 2 < α ≤ 3 is a real number, D0α+ is the Caputo’s differentiation, and
f : [0, 1] × [0, ∞) → (−∞, ∞) is continuous.
In this paper, we firstly derive the corresponding Green’s function,then we give
the properties of Green’s function,finally,the existence of positive solution with
semipositone nonlinearity are obtained by Krasnosel’skii fixed point theorem.
Here, a positive solution u∗ of (1.1) will mean a solution u∗ of (1.1) satisfying
u∗ (t) > 0, 0 < t < 1.
For the convenience of the reader, we present here the necessary definitions
from fractional calculus theory. These definitions can be found in the recent
literature.
Definition 1.1. The Riemann-Liouville fractional integral of order α > 0 of a
function f : [0, 1] → R is given by
Z t
1
α
I0+ f (t) =
(t − s)α−1 f (s)ds
Γ(α) 0
provided that the right side is pointwise defined on (0, ∞).
Definition 1.2. The Caputo’s fractional derivative of order α > 0 of a function
f ∈ AC n [0, 1] is given by
Z t
1
f (n) (s)
α
D0+ f (t) =
ds
Γ(n − α) 0 (t − s)α−n+1
where n−1 < α ≤ n, provided that the right side is pointwise defined on (0, ∞).
From the definition we can obtain the following lemma:
Lemma 1.3. Assume that u ∈ C n [0, 1], then
α
α
I0+
D0+
u(t) = u(t) − C1 − C2 t − · · · − Cn tn−1
where Ci ∈ R, i = 1, 2, . . . , n, n = [α] + 1.
Lemma 1.4. [6] The relation
α+β
α β
Ia+
Ia+ ϕ = Ia+
ϕ
is valid in following case
Reβ > 0, Re(α + β) > 0, ϕ(x) ∈ L1 (a, b).
Lemma 1.5. Given f ∈ C[0, 1], and 2 < α ≤ 3, the unique solution of
0 < t < 1,
D0α+ u(t) + f (t) = 0,
0
00
u(0) = u (1) = u (0) = 0.
is
Z
1
G(t, s)f (s)ds
u(t) =
0
(1.2)
BOUNDARY VALUE PROBLEM
where
(
G(t, s) =
(α−1)t(1−s)α−2 −(t−s)α−1
,
Γ(α)
α−2
t(1−s)
,
Γ(α−1)
0 ≤ s ≤ t ≤ 1,
0 ≤ t ≤ s ≤ 1.
125
(1.3)
Proof. We may apply Lemma1.3 to reduce Eq.(1.2) to an equivalent integral
equation
α
u(t) = −I0+
f (t) + C1 + C2 t + C3 t2
for some Ci ∈ R, i = 1, 2, 3. By Lemma1.4 we have
α−1
1
1
α
1
f (t) + C2 + 2C3 t
I0+
I0+
f (t) + C2 + 2C3 t = −D0+
I0+
u0 (t) = −D0+
α−1
f (t) + C2 + 2C3 t
= −I0+
1
α−1
1
1
α−2
α−2
u00 (t) = −D0+
I0+
f (t) + 2C3 = −D0+
I0+
I0+
f (t) + 2C3 = −I0+
f (t) + 2C3 .
From u(0) = u0 (1) = u00 (0) = 0, one has
Z 1
1
C1 = 0, C2 =
(1 − s)α−2 f (s)ds, C3 = 0.
Γ(α − 1) 0
Therefore, the unique solution of problem (1.2) is
Z t
Z 1
1
1
α−1
(t − s) f (s)ds +
t(1 − s)α−2 f (s)ds
u(t) = −
Γ(α) 0
Γ(α − 1) 0
¸
Z t·
Z 1
t(1 − s)α−2 (t − s)α−1
t(1 − s)α−2
=
−
f (s)ds +
f (s)ds
Γ(α − 1)
Γ(α)
Γ(α − 1)
0
t
Z 1
=
G(t, s)f (s)ds
0
The proof is complete.
¤
Lemma 1.6. The function G(t, s) defined by Eq. (1.3) satisfies
(1) G(t, s) > 0, for t, s ∈ (0, 1);
(2)
min
1/4≤t≤3/4
G(t, s) ≥
1
1
max G(t, s) = G(1, s), for 0 < s < 1.
4 0≤t≤1
4
(1.4)
Proof. Setting
(α − 1)t(1 − s)α−2 − (t − s)α−1
t(1 − s)α−2
, g2 (t, s) =
.
Γ(α)
Γ(α − 1)
(1) For g1 (t, s), since 2 < α ≤ 3, 0 ≤ s ≤ t ≤ 1 we can obtain
g1 (t, s) =
(α − 1)t(1 − s)α−2 ≥ t(1 − s)α−2 ≥ t(t − s)α−2 ≥ (t − s)α−1
we get g1 (t, s) > 0. It is clearly that g2 (t, s) > 0, therefore G(t, s) >
0, for t, s ∈ (0, 1).
126
TINGTING QIU AND ZHANBING BAI
(2) Since
(α − 1)(1 − s)α−2 − (α − 1)(t − s)α−2
∂g1 (t, s)
=
> 0,
∂t
Γ(α)
∂g2 (t, s)
(1 − s)α−2
=
> 0.
∂t
Γ(α − 1)
we know G(t, s) is increasing with respect to t. Consequently,
0 ≤ G(t, s) ≤ max G(t, s) = G(1, s), for t, s ∈ [0, 1].
0≤t≤1
One has


1
min G(t, s) = G( , s) =

1/4≤t≤3/4
4
max G(t, s) = G(1, s) =
0≤t≤1
1/4≤t≤3/4
1
(1−s)α−2
4
Γ(α−1)
Γ(α)
, s ∈ (0, 14 ],
,
s ∈ [ 14 , 1).
(α − 1)(1 − s)α−2 − (1 − s)α−1
, s ∈ (0, 1).
Γ(α)
we will prove
min
1
(α−1)(1−s)α−2 −( 14 −s)α−1
4
G(t, s) ≥
1
1
max G(t, s) = G(1, s)
0≤t≤1
4
4
As follows,
10 : For 0 < s ≤ 14 ,
(α − 1)(1 − s)α−2 ( 14 − s)α−1
−
min G(t, s) =
1/4≤t≤3/4
4Γ(α)
Γ(α)
1
1
(α − 1)(1 − s)α−2 (1 − s)α−1
max G(t, s) = G(1, s) =
−
4 0≤t≤1
4
4Γ(α)
4Γ(α)
1
Since 2 < α ≤ 3, 0 < s ≤ 4 we can obtain
1
1
1
1
( − s)α−1 = ( )α−1 (1 − 4s)α−1 ≤ (1 − 4s)α−1 < (1 − s)α−1
4
4
4
4
Thus,
min
1/4≤t≤3/4
0
2 : For
1
4
G(t, s) ≥
1
1
max G(t, s) = G(1, s).
4 0≤t≤1
4
≤ s < 1,
min
1/4≤t≤3/4
G(t, s) =
1
(1
4
− s)α−2
(α − 1)(1 − s)α−2
=
Γ(α − 1)
4Γ(α)
1
(α − 1)(1 − s)α−2 (1 − s)α−1
1
max G(t, s) = G(1, s) =
−
4 0≤t≤1
4
4Γ(α)
4Γ(α)
It is clearly that
1
1
min G(t, s) ≥ max G(t, s) = G(1, s).
0≤t≤1
1/4≤t≤3/4
4
4
The proof is complete.
¤
BOUNDARY VALUE PROBLEM
127
Lemma 1.7. If f ∈ C[0, 1], and f ≥ 0, then,the unique solution u of problem (1.1) satisfies
1
min3 u(t) ≥ kuk,
1
4
≤t≤ 4
4
Proof. By Lemma 1.5,u can be expressed by
Z 1
Z 1
u(t) =
G(t, s)f (s, u(s))ds ≤
max G(t, s)f (s, u(s))ds
0 0≤t≤1
0
so,
Z
max
0≤t≤1
Z
1
G(t, s)f (s, u(s))ds ≤
0
then,
Z
kuk = max |u(t)| = max
0≤t≤1
0≤t≤1
1
max G(t, s)f (s, u(s))ds
0 0≤t≤1
Z
1
G(t, s)f (s, u(s))ds ≤
0
1
max G(t, s)f (s, u(s))ds
0 0≤t≤1
Taking into account(1.4), we obtain
Z
min u(t) = 1min3
1
≤t≤ 34
4
4
≤t≤ 4
1
G(t, s)f (s, u(s))ds
0
Z
Z 1
1
1
1 1
max G(t, s)f (s, u(s))ds ≥ max
G(t, s)f (s, u(s))ds = kuk
≥
4 0 0≤t≤1
4 0≤t≤1 0
4
The proof is complete.
¤
Lemma 1.8. [5] Let X be a Banach space, P ⊆ X a cone, and Ω1 , Ω2 are two
bounded open balls of X centered at the origin with Ω1 ⊂ Ω2 . Suppose that
A : P ∩ (Ω2 \ Ω1 ) → P is a completely continuous operator such that either
(i) kAxk ≤ kxk, x ∈ P ∩ ∂Ω1 and kAxk ≥ kxk, x ∈ P ∩ ∂Ω2 , or
(ii) kAxk ≥ kxk, x ∈ P ∩ ∂Ω1 and kAxk ≤ kxk, x ∈ P ∩ ∂Ω2
holds. Then A has a fixed point in P ∩ (Ω2 \ Ω1 ).
2. Main results
In this section, by uses of Lemma 1.5, Lemma 1.6,Lemma 1.7 and Lemma 1.8,
we will obtain the existence of positive solution with semipositone nonlinearity
for Problem (1.1).
Let X = C[0, 1] be endowed with the ordering u ≤ v if u(t) ≤ v(t) for all
t ∈ [0, 1], and the maximum norm
kuk = max |u(t)|
0≤t≤1
Define the cone K ⊂ X by
(
)
¯
1
1
K = u ∈ X ¯u(t) ≥ 0, and 1min3 u(t) ≥ max |u(t)| = kuk,
4 0≤t≤1
4
≤t≤ 4
4
128
TINGTING QIU AND ZHANBING BAI
Define operator T : K → K
Z
T u(t) =
1
G(t, s)f (s, u(s))ds
0
Lemma 2.1. T : K → K is completely continuous.
Proof. Let u ∈ K, we have
Z 1
Z
T u(t) =
G(t, s)f (s, u(s))ds ≤
Z
0≤t≤1
Z
1
max T u(t) = max
0≤t≤1
G(t, s)f (s, u(s))ds ≤
0
Taking into account (1.4), we get
Z
min3 T u(t) = 1min3
1
4
max G(t, s)f (s, u(s))ds
0 0≤t≤1
0
So,
1
≤t≤ 4
4
≤t≤ 4
1
max G(t, s)f (s, u(s))ds
0 0≤t≤1
1
G(t, s)f (s, u(s))ds
0
Z
Z 1
1 1
1
1
≥
max G(t, s)f (s, u(s))ds ≥ max
G(t, s)f (s, u(s))ds = kT uk
0≤t≤1
0≤t≤1
4 0
4
4
0
Also, by Lemma1.6, for u ∈ K, we have T u(t) ≥ 0, 0 ≤ t ≤ 1.
Consequently, T : K → K. Further, with the nonnegativeness and continuity
of G(t, s) and f (t, u), by means of Arzela-Ascoli theorem we can easily obtain T
is completely continuous. The proof is complete.
¤
Theorem 2.2. Suppose
(H1 ) f ∈ C([0, 1] × [0, ∞), (−∞, +∞)), ∃M > 0, L > −M, such that
− M ≤ f (t, x) ≤ L, , 0 ≤ t ≤ 1, 0 ≤ x ≤ 1;
(H2 ) lim f (t,x)
= +∞, t ∈ [α, β] ⊂ [0, 1], uniformly hold.
x
x→∞
Then Problem (1.1) has at least one positive solution.
Proof. Let
Z
w(t) =
G(t, s)ds =
0
and set
1
αt − tα
1
<
Γ(α + 1)
Γ(α)
1
M
1
v(t) = M w(t) < 4 , u(t) = u(t) + v(t)
4
Γ(α)
then u(t) is the positive solution of problem(1.1) if and only if u(t) is the solution
of the following problem:
½ α
D0+ u(t) + [f (t, u(t) − v(t)) + M ] = 0, 0 < t < 1,
(2.1)
u(0) = u0 (1) = u00 (0) = 0
and satisfies u(t) − v(t) > 0, (0 < t < 1).
Define the operator T ∗ : K → K
Z 1
∗
T x(t) =
G(t, s)[f (s, x(s) − v(s)) + M ]ds
0
BOUNDARY VALUE PROBLEM
129
From Lemma2.1, we know T ∗ : K → K is completely continuous operator.
Clearly, the fixed point of operator T ∗ is the solution of problem (2.1).
Let
½
¾
M +L
K1 = x ∈ K : kXk <
for u ∈ ∂K1
Γ(α)
we get
Z
kT xk = max
0≤t≤1
1
G(t, s)[f (s, x(s) − v(s)) + M ]ds
0
Z
≤ (M + L) max
0≤t≤1
1
G(t, s)ds <
0
M +L
= kxk
Γ(α)
Thus, kT xk = kxk, for u ∈ ∂K1 .
By condition (H2 ) exists R > 0 is large enough such that R >
2M
Γ(α)
satisfies
f (t, x) + M
R
≥ N, for t ∈ [ξ, η] ⊂ [0, 1] , x ≥
x
8
where N is chosen so that
Z
1
N η ξ+η
4
G(
, s)ds > 1
2 ξ
2
Set
KR = {x ∈ K : kxk < R} , for x ∈ ∂KR .
we have
1
1
M
x(t) > kxk = R >
> 2v(t)
4
4
2Γ(α)
so
1
1
R
x(t) − v(t) > x(t) − x(t) = u(t) > , t ∈ [0, 1]
2
2
8
x(t) − v(t) >
R
, t ∈ [ξ, η]
8
Thus, we obtain
f (t, x(t) − v(t)) + M ≥ N (x(t) − v(t)) ≥
Consequently,
Z
R
, t ∈ [ξ, η].
8
1
ξ+η
, s)[f (s, x(s) − v(s)) + M ]ds
2
0
Z η
Z
N R 41
N R 14 η ξ + η
ξ+η
G(
, s)
ds ≥
G(
, s)ds > R = kxk
≥
2
2
2
2
ξ
ξ
T x(t) =
G(
Therefore, kT xk = kxk, for x ∈ ∂KR . An application of Lemma1.8 yields that
exists u ∈ KR \ K1 , such that u = T u.
Namely, u = u(t) satisfies
130
TINGTING QIU AND ZHANBING BAI
½
D0α+ u(t) + [f (t, u(t) − v(t)) + M ] = 0, 0 < t < 1,
u(0) = u0 (1) = u00 (0) = 0
thus,
1
1
u(t) ≥ kuk = max |u(t)|
4
4 0≤t≤1
¯Z 1
¯
¯
¯
1
= max ¯¯
G(t, s)[f (s, x(s) − v(s)) + M ]ds¯¯
4 0≤t≤1 0
Z 1
1
M
1
≥ · M max
G(t, s)ds = 4
> v(t)
0≤t≤1 0
4
Γ(α)
Consequently,
u(t) = u(t) − v(t) > 0, 0 < t ≤ 1.
and satisfies
½
D0α+ u(t) + f (t, u(t)) = 0, 0 < t < 1,
u(0) = u0 (1) = u00 (0) = 0,
Obviously, u(t) 0 < t < 1 is positive solution of problem (1.1). The proof is
complete.
¤
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BOUNDARY VALUE PROBLEM
131
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1
Department of Mathematics, Shandong University of Science and Technology,Qingdao, 266510, PRC.
E-mail address: qiutingting19833@163.com
2
Department of Mathematics, Shandong University of Science and Technology, Qingdao, 266510, PRC.
E-mail address: zhanbingbai@163.com