Hindawi Publishing Corporation
Abstract and Applied Analysis
Volume 2009, Article ID 768920, 12 pages
doi:10.1155/2009/768920
Research Article
Existence of Positive Solutions for Multiterm
Fractional Differential Equations of Finite Delay
with Polynomial Coefficients
A. Babakhani1 and E. Enteghami2
1
2
Department of Basic Science, Babol University of Technology, Bobol 47148-71167, Iran
Department of Mathematics, University of Mazandaran, Babolsar 47418-1468, Iran
Correspondence should be addressed to A. Babakhani, babakhani@nit.ac.ir
Received 9 February 2009; Revised 7 June 2009; Accepted 1 September 2009
Recommended by Ferhan Atici
Existence of positive solutions has been studied by A. Babakhani and V. Daftardar-Gejji 2003
in case of multiterm nonautonomous fractional differential equations with constant coefficients.
In the present paper we discuss existence of positive solutions in case of multiterm fractional
differential equations of finite delay with polynomial coefficients.
Copyright q 2009 A. Babakhani and E. Enteghami. This is an open access article distributed under
the Creative Commons Attribution License, which permits unrestricted use, distribution, and
reproduction in any medium, provided the original work is properly cited.
1. Introduction
In last 30 years, the theory of ordinary differential equations of fractional order has become
a new important branch see, e.g., 1–5 and the references therein. Numerous applications
of such equations have been presented 3–10. Existence of positive solution of fractional
ordinary differential equations has been well investigated for fractional functional differential
equations 1, 6, 11–14. Ye et al. 6 have addressed the question of existence of positive
solutions for the nonlinear fractional functional differential equation
Dα xt − x0 xtft, xt ,
xt φt ≥ 0,
t ∈ 0, T ,
t ∈ −w, 0,
1.1
by using the sub- and supersolution method, where 0 < α < 1, Dα is the standard RiemannLiouville fractional derivative, φ ∈ C and f : 0, T × C → R is continuous, as usual,
2
Abstract and Applied Analysis
C C−w, 0, R is the space of continuous function from −w, 0 to R , w > 0, equipped
with the sup norm:
φ max φt,
1.2
−w≤Θ≤0
and xt denotes the function in C defined by
xt θ xt θ,
−w ≤ θ ≤ 0.
1.3
They require that the nonlinearity ft, xt is increasing in xt for each t ∈ 0, T .
As a pursuit of this in the present paper, we deal with the existence of positive
solutions in the case of multiterm differential equations with polynomial coefficients of the
fractional type:
t ∈ 0, T ,
LDxt − x0 ft, xt ,
xt φt ≥ 0,
1.4
t ∈ −w, 0,
where
LD D
αn
n−1
− pj tDαn−j ,
Nj
0 < α1 < · · · < αn < 1, pj t ajk tk ,
j1
2m
pj
t ≥ 0,
k0
2m1
pj
t ≤ 0,
m 0, 1, . . . ,
1.5
Nj
, j 1, 2, . . . , n − 1,
2
and Dαj is the standard Riemann-Liouville fractional derivative, T > 0, w > 0, φ ∈ C C−w, 0, R and f : I × C → R is a given continuous function, I 0, T .
2. Preliminaties
Let E be a real Banach space with a cone K. K introduces a partial order ≤ in E in the following
manner 13:
x≤y
if y − x ∈ K.
2.1
Definition 2.1 see 15. For x, y ∈ E the order interval x, y is defined as
x, y z ∈ E : x ≤ z ≤ y .
2.2
Abstract and Applied Analysis
3
Definition 2.2 see 15. A cone K is called normal, if there exists a positive constant r such
that f, g ∈ K and ϑ ≺ f ≺ g implies f ≤ rg, where ϑ denotes the zero element of K.
Definition 2.3 see 16, 17. Let f : a, b → R, and f ∈ L1 a, b. The left-sided RiemannLiouville fractional integral of f of order α is defined as
Iaα fx 1
Γα
x
x − tα−1 ftdt,
α > 0, x ∈ a, b.
2.3
a
Definition 2.4 see 16, 17. The left-sided Riemann-Liouville fractional derivative of a
function f : a, b → R is defined as
Daα fx Dm Iam−α fx ,
x ∈ a, b,
2.4
where m α1, Dm dm /dtm . We denote D0α by Dα and I0α by I α . If the fractional derivative
Daα fx is integrable, then 16, page 71
β
1−β
α−β
Iaα Da fx Ia fx − Ia fx
xα−1
,
xa Γα
0 < β ≤ α < 1.
2.5
1−β
If f is continuous on a, b, then Ia fxxa 0 and 2.5 reduces to
β
α−β
Iaα Da fx Ia fx,
0 < β ≤ α < 1.
2.6
Proposition 2.5. Let y be continuous on 0, T , T > 0 and let n be a nonnegative integer, then
n
n
−α k n αk
−α n!tn−k αk
D t I xt I t xt I xt,
k
k n − k!
k0
k0
α
n
2.7
where
−α
k
α
Γ1 − α
1
k
−1
.
−1
k!Γα
Γk 1Γ1 − α − k
k
k Γα
The proof of the above proposition can be found in 17, page 53.
2.8
4
Abstract and Applied Analysis
Corollary 2.6. Let x ∈ C0, T , T > 0 and pj t Nj
k0
ajk tk , Nj ∈ N ∪ {0}, j 1, 2, . . . , n, then
⎞
⎛
Nj k
n
n −α k!tk−r αr
α⎝
⎠
I
pj txt ajk
I xt.
r k − r!
j1
j1 k0 r0
2.9
Theorem 2.7 see 10. Let E be a Banach space with C ⊆ E closed and convex. Assume that U
is a relatively open subset of C with 0 ∈ U and F : U → C is a continuous and compact map.
Then either
1 F has a fixed point in U, or
2 there exists u ∈ ∂U and λ ∈ 0, 1 with u λFu.
3. Existence of Positive Solution
In this section, we discuss the existence of positive solutions for 1.4. Using 2.5, 2.6, and
Corollary 2.6, 1.4 is equivalent to the integral equation
xt ⎧
⎨x0 Ixt − x0 I αn ft, xt , t ∈ 0, T ,
⎩
t ∈ −w, 0,
φt,
3.1
where
I
Nj k
n−1 ajk
−αn
j1 k0 r0
r
k!tk−r αn −αn−j r
I
.
k − r!
3.2
Let y· : −w, T → 0, ∞ be the function defined by
yt ⎧
⎨φ0,
t ∈ I,
⎩φt ≥ 0,
t ∈ −w, 0,
3.3
then y0 φ, for each z ∈ CI, R with z0 0, we denote by z the function define by
zt ⎧
⎨zt,
t ∈ I,
⎩0,
t ∈ −w, 0.
3.4
Abstract and Applied Analysis
5
We can decompose x· as xt zt yt, t ∈ −w, T , which implies xt zt yt , for t ∈ I.
Therefore, 3.1 is equivalent to the integral equation
zt Izt I αn f t, zt yt ,
t ∈ I,
3.5
where I is defined 3.2. Set A0 {z ∈ CI, R : z0 0} and let zT be the seminorm in A0
defined by
zT z0 z z : sup{|zt| : t ∈ I},
z ∈ A0 ,
3.6
and A0 is a Banach space with norm · T . Let K be a cone of A0 , K {z ∈ A0 ; zt ≥ 0, t ∈ I}
and
K ∗ x ∈ C−w, T , R ; xt φt ≥ 0, t ∈ −w, 0 .
3.7
Define the operator F : K → K by
Fzt Izt I αn f t, zt yt ,
t ∈ I.
3.8
Theorem 3.1. Suppose that the following conditions hold:
1 there exist p, q ∈ CI, R such that ft, xt ≤ pt qtxt , for t ∈ I, xt ∈ C, and
I αn p supt∈0,T I αn pt < ∞, I αn q supt∈0,T I αn qt < ∞,
2 1 − IT − I αn q > 0, where
Nj k n−1 −αn k!T αn −αn−j k
IT .
ajk
k − r!Γ αn − αn−j r
r
j1 k0 r0
3.9
Then 1.4 has at least a positive solution x∗ ∈ K ∗ , satisfying x∗ ≤ max{φ, h}, where
λφI αn q λI αn p
1.
h
1 − λIT − λI αn q
3.10
6
Abstract and Applied Analysis
Proof. We will show that the operator F : K → K is continuous and completely continuous.
Step 1. The operator F : K → K is continuous in view of the continuity of f.
Step 2. F maps bounded sets into bounded sets in K.
Let G ⊂ K be bounded; that is, there exists a positive constant l such that zT ≤ l, for
all z ∈ G. For each z ∈ G, we have
Nj k n−1 −αn k!tk−r α −α r
I n n−j |zt| I αn f t, zt yt
|Fzt| ≤
ajk
−
r!
k
r
j1 k0 r0
Nj k n−1 −αn lk!tαn −αn−j k
≤
I αn f t, zt yt
ajk
k − r!Γ αn − αn−j r
r
j1 k0 r0
3.11
≤ lIT I αn pt qtzt yt ,
where IT is defined in 3.9. It follows that
FzT ≤ lIT I αn p lI αn q φtI αn q.
3.12
Hence FG is bounded.
Step 3. F maps bounded sets into equicontinuous sets of K.
We will show that FG is equicontinuous. For each z ∈ G, t1 , t2 ∈ I and t1 < t2 , then for
given > 0, choose
⎧
1/αn ⎫
⎬
⎨ Cj, k, r 1/αn −αn−j r
Γαn 1
,
, δ min
⎭
⎩
4
4 p q l φ
3.13
where j 1, 2, . . . , n − 1, k 0, 1, . . . , Nj , r 0, 1, . . . , k,
Γ αn − αn−j r 1
× −α C j, k, r n−1
n
ajk r lηk!
i1 Ni 1Ni 2
k − r!
3.14
Abstract and Applied Analysis
7
and η max{1, T Nj , j 1, 2, . . . , n − 1}. If |t1 − t2 | < δ,
|Fzt1 − Fzt2 |
−α n
Nj k
n−1 ajk r k!t1 k−r t1
t − sαn −αn−j r−1 zsds
j1 k0 r0 k − r!Γ αn − αn−j r 0 1
−α n
k−r
t2
Nj k
n−1 ajk r k!t2
αn −αn−j r−1
−
zsds
t2 − s
−
α
r
−
r!Γ
α
k
n
n−j
0
j1 k0 r0
t2
1 t1
αn −1 αn −1 f s, zs ys ds − t2 − s
f s, zs ys ds
t1 − s
Γαn 0
0
−α n
k−r
t1
n−1 Nj k
ajk r k!t2
≤
t − sαn −αn−j r−1 zsds
j1 k0 r0 k − r!Γ αn − αn−j r 0 2
−αn k−r
Nj k
t
n−1
k!t
a
1
jk
1
r
αn −αn−j r−1
zsds
−
t1 − s
−
α
r
−
r!Γ
α
k
n
n−j
0
j1 k0 r0
−α n
k−r
t2
Nj k
n−1 ajk r k!t2
t2 − sαn −αn−j r−1 |zs|ds
−
α
r
−
r!Γ
α
k
n
n−j
t1
j1 k0 r0
# t
$
t2
f 1
αn −1
αn −1
αn −1
∞
ds t2 − s
− t1 − s
ds
t2 − s
Γαn 0
t1
−α n
k−r
Nj k
n−1 ajk r lk!T
≤
j1 k0 r0 k − r!Γ αn − αn−j r
t2 %
&
×
t2 − sαn −αn−j r−1 − t1 − sαn −αn−j r−1 ds
0
$
# t
t2
p q l φ
1
αn −1
αn −1
αn −1
ds t2 − s
− t1 − s
ds
t2 − s
Γαn 0
t1
−α n
α −α r
Nj k n−1 ajk r 2lk!ηt2 − t1 n n−j
2 p q l φ t2 − t1 αn
≤
Γαn 1
k − r!Γ αn − αn−j r 1
j1 k0 r0
−α n
α −α r
Nj k n−1 ajk r 2lk!ηδ n n−j
2 p q l φ δ αn
Γαn 1
j1 k0 r0 k − r!Γ αn − αn−j r 1
≤
.
2 2
3.15
8
Abstract and Applied Analysis
Hence FG is equicontiuous. The Arzela-Ascoli theorem implies that FG is compact and
F : K → K is continuous and completely continuous.
Step 4. We now show that there exists an open set U ⊆ K with z /
λFz for λ ∈ 0, 1 and
z ∈ ∂U. Let z ∈ K be any solution of z λFz, λ ∈ 0, 1, where F is given by 3.8; since
F : K → K is continuous and completely continuous, we have
zt λFzt
Nj k n−1 −αn k!tk−r
≤λ
ajk
k − r!Γ αn − αn−j r
r
j1 k0 r0
t
t
λ
× t − s
zsds t − sαn −1 f s, zs ys ds
Γα
n
0
0
t
Nj k n−1 −αn zk!tk−r
≤λ
t − sαn −αn−j r−1 ds
ajk
k − r!Γ αn − αn−j r
r
0
j1 k0 r0
αn −αn−j r−1
3.16
t
λ
t − sαn −1 ps qszs ys ds
Γαn 0
≤ λ zIT zI αn qt φI αn qt I αn pt .
So
z 1 − λIT − λI αn p ≤ λφI αn q λI αn p.
3.17
Now, by 3.10 and 3.17, we know that any solution z of 3.8 satisfies z /
h; let
U {z ∈ K; z < h}.
3.18
Therefore, Theorem 2.7 guarantees that 3.1 has at least a positive solution z ∈ U. Hence,
1.4 has at least a positive solution x∗ ∈ K ∗ , satisfying x∗ ≤ max{φ, h} and the proof is
complete.
Note that we can complete the above mentioned procedure by using only the
continuity of f without condition 1, but with our procedure and details of condition 1
in Theorem 3.1 answers all the questions exist in the following remark.
Remark 3.2. When f is continuous on 0, T ×C, limt → 0 ft, · ∞, i.e., f is singular at t 0
in 1.4. Suppose ∃σ ∈ 0, αn , such that tσ ft, xt is a continuous function on 0, T × C, then
I αn ft, xt I αn t−σ tσ ft, xt is continuous on I × C by Lemma 2.1 in 12, page 613. We also
Abstract and Applied Analysis
9
obtain results about the existence to 1.4 by using a nonlinear alternative of Leray-Schauder
type. The proof is similar to that of Theorem 3.1 as long as we let
1 tσ ft, xt ≤ pt qtxt , for t ∈ I, xt ∈ C, and I αn t−σ p < ∞, I αn t−σ q < ∞,
2 1 − IT − I αn t−σ q > 0, then 1.4 has at least a positive solution x∗ ∈ K ∗ , satisfying
x∗ ≤ max{φ, h}, where
λI αn t−σ p λφI αn t−σ q
h
1.
1 − λIT − λI αn t−σ q
3.19
4. Unique Existence of Solution
In this section, we will give uniqueness of positive solution to 1.4.
Theorem 4.1. Let f : I × C → R be continuous and λ ∈ L1 0, T , R with I αn λ < ∞. Further
assume
i |ft, ut yt − ft, vt yt | ≤ λtut − vt , for all u, v ∈ K, t ∈ 0, T ,
ii IT I αn λ < 1.
Then 1.4 has unique solution which is positive, where IT is given in 3.9.
Proof. Let u, v ∈ K. Then we obtain
|Fut − Fvt| ≤
−α k−r
n
Nj k n−1 ajk r k!t
j1 k0 r0
k − r!
I αn −αn−j r |ut − vt|
I αn f t, ut y t − f t, vt yt −α ⎫
⎧
α −α k
n
Nj k
n−1 ⎬
⎨
ajk r k!t n n−j
≤ u − vT
I αn λt
⎭
⎩ j1 k0 r0 k − r!Γ αn − αn−j r 1
4.1
−α ⎫
n
⎬
ajk r k!T αn −αn−j k
≤ u − vT
I αn λt ,
⎭
⎩ j1 k0 r0 k − r!Γ αn − αn−j r 1
⎧
Nj k
n−1 ⎨
where F is given in 3.8. Hence,
Fu − FvT ≤ IT I αn λu − vT .
4.2
In view of Banach fixed point theorem F has unique fixed point in K, which is the unique
positive solution of 2.7 and 1.4 has a unique positive solution in K ∗ .
10
Abstract and Applied Analysis
Remark 4.2. When λt L > 0, then condition i reduces to the Lipschitz condition.
Example 4.3. Let λt L > 0 and ft, xt Lxt et Lxt − ω et , ω > 0. Consider the
equation
D1/2 − at2 D1/4 − btD1/6 − cD1/8 x Lxt − w et ,
xt 0,
t ∈ 0, 64,
4.3
t ∈ −w, 0.
Then 4.3 is equivalent to the integral equation,
⎛ ⎞
1
k!tk−r 1/2−α3−j r
−
xt I
ajk ⎝ 2 ⎠
xt I 1/2 Lxt − ω et .
k − r!
j1 k0 r0
r
Nj k
3 4.4
2
k
2
a a11 0, a12 a, p2 t Here α3 1/2, p1 t k0 a1k t at , then N1 2,
0 10
1
k
k
k0 a2k t bt, then N2 1, a20 0, a21 b and p3 t k0 a3k t c, then N3 0, a30 c.
Hence
⎛ ⎞
⎞
⎡⎛ ⎞
⎤
1
1
1
−
−
−
xt a10 ⎝ 2 ⎠I 1/2−1/4 x a11 ⎣⎝ 2 ⎠tI 1/2−1/4 x ⎝ 2 ⎠I 1/2−1/41 ⎦
0
0
1
⎛ ⎞
⎛ ⎞
⎤
⎡⎛ ⎞
1
1
1
−
−
−
a12 ⎣⎝ 2 ⎠t2 I 1/2−1/4 x 2⎝ 2 ⎠tI 1/2−1/41 x 2⎝ 2 ⎠I 1/2−1/42 x⎦
0
1
2
⎛ ⎞
⎡⎛ ⎞
⎛ ⎞
⎤
1
1
1
−
−
−
a20 ⎝ 2 ⎠I 1/2−1/6 x a21 ⎣⎝ 2 ⎠tI 1/2−1/6 x ⎝ 2 ⎠I 1/2−1/61 x⎦
0
0
1
⎛ ⎞
1
− ⎠ 1/2−1/8
⎝
a30
x LI 1/2 xt − ω I 1/2 et .
2 I
0
⎛
In view of 2.8 and that Γ1/2 √
4.5
√
√
π, Γ−1/2 −2 π and Γ−3/4 4 π/3 we obtain
3 9/4
2 1/4
5/4
xt a t I xt − tI xt I xt
4
1 4/3
1/3
b 1 tI xt − I xt cI 3/8 LI 1/2 xt − ω I 1/2 et .
2
4.6
Abstract and Applied Analysis
11
If |a| ≤ 3/5, |b| ≤ 2/5, |c| ≤ 1/5, 0 < L < 4/5 in the above equation satisfy the conditions
required in Theorem 4.1, the iterated sequence is
x1 t I 1/2 et t1/2 E1,3/2 t,
,
+
,
+
3 9/4
1 4/3
2 1/4
5/4
1/3
3/8
1/2
b 1 tI − I
cI LI
x1 t x1 t,
x2 t a t I − tI I
4
2
,
+
,
n−k
n +
3
1
x1 t,
a t2 I 1/4 − tI 5/4 I 9/4 b 1 tI 1/3 − I 4/3 cI 3/8 LI 1/2
xn1 t 4
2
k0
4.7
for n 1, 2, 3, . . . , where I α x1 tα1/2 E1,α3/2 t, α > 0, xt limn → ∞ xn t is the unique
k
solution, which may not be positive, where Eα,β t ∞
k0 t /Γαk β is Mittag-Leffler
function.
Acknowledgment
The authors thank the referee’s efforts for their remarks and Professor Ferhan Merdivenci
Atici, Western Kentucky University, USA, for her regular contact.
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