- Research Article
- Open access
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Existence of Positive Solutions to a Boundary Value Problem for a Delayed Nonlinear Fractional Differential System
Boundary Value Problems volume 2011, Article number: 475126 (2011)
Abstract
Though boundary value problems for fractional differential equations have been extensively studied, most of the studies focus on scalar equations and the fractional order between 1 and 2. On the other hand, delay is natural in practical systems. However, not much has been done for fractional differential equations with delays. Therefore, in this paper, we consider a boundary value problem of a general delayed nonlinear fractional system. With the help of some fixed point theorems and the properties of the Green function, we establish several sets of sufficient conditions on the existence of positive solutions. The obtained results extend and include some existing ones and are illustrated with some examples for their feasibility.
1. Introduction
In the past decades, fractional differential equations have been intensively studied. This is due to the rapid development of the theory of fractional differential equations itself and the applications of such construction in various sciences such as physics, mechanics, chemistry, and engineering [1, 2]. For the basic theory of fractional differential equations, we refer the readers to [3–7].
Recently, many researchers have devoted their attention to studying the existence of (positive) solutions of boundary value problems for differential equations with fractional order [8–23]. We mention that the fractional order involved is generally in
with the exception that
in [12, 23] and
in [8, 17]. Though there have been extensive study on systems of fractional differential equations, not much has been done for boundary value problems for systems of fractional differential equations [18–20].
On the other hand, we know that delay arises naturally in practical systems due to the transmission of signal or the mechanical transmission. Though theory of ordinary differential equations with delays is mature, not much has been done for fractional differential equations with delays [24–31].
As a result, in this paper, we consider the following nonlinear system of fractional order differential equations with delays,
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2011%2F475126/MediaObjects/13661_2010_Article_41_Equ1_HTML.gif)
where is the standard Riemann-Liouville fractional derivative of order  
for some integer
,
for
,
for
, and
is a nonlinear function from
to
. The purpose is to establish sufficient conditions on the existence of positive solutions to (1.1) by using some fixed point theorems and some properties of the Green function. By a positive solution to (1.1) we mean a mapping with positive components on
such that (1.1) is satisfied. Obviously, (1.1) includes the usual system of fractional differential equations when
for all
and
. Therefore, the obtained results generalize and include some existing ones.
The remaining part of this paper is organized as follows. In Section 2, we introduce some basics of fractional derivative and the fixed point theorems which will be used in Section 3 to establish the existence of positive solutions. To conclude the paper, the feasibility of some of the results is illustrated with concrete examples in Section 4.
2. Preliminaries
We first introduce some basic definitions of fractional derivative for the readers' convenience.
The fractional integral of order   of a function
is defined as
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2011%2F475126/MediaObjects/13661_2010_Article_41_Equ2_HTML.gif)
provided that the integral exists on , where
is the Gamma function.
Note that has the semigroup property, that is,
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2011%2F475126/MediaObjects/13661_2010_Article_41_Equ3_HTML.gif)
The Riemann-Liouville derivative of order of a function
is given by
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2011%2F475126/MediaObjects/13661_2010_Article_41_Equ4_HTML.gif)
provided that the right-hand side is pointwise defined on , where
.
It is well known that if then
. Furthermore, if
and
then
for
.
The following results on fractional integral and fractional derivative will be needed in establishing our main results.
Lemma 2.3 (see [10]).
Let . Then solutions to the fractional equation
can be written as
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2011%2F475126/MediaObjects/13661_2010_Article_41_Equ5_HTML.gif)
where ,
.
Lemma 2.4 (see [10]).
Let . Then
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2011%2F475126/MediaObjects/13661_2010_Article_41_Equ6_HTML.gif)
for some ,
.
Now, we cite the fixed point theorems to be used in Section 3.
Lemma 2.5 (the Banach contraction mapping theorem [33]).
Let be a complete metric space and let
be a contraction mapping. Then
has a unique fixed point.
Let be a closed and convex subset of a Banach space
. Assume that
is a relatively open subset of
with
and
is completely continuous. Then at least one of the following two properties holds:
(i) has a fixed point in
;
(ii)there exists and
with
.
Lemma 2.7 (the Krasnosel'skii fixed point theorem [33, 35]).
Let be a cone in a Banach space
. Assume that
and
are open subsets of
with
and
. Suppose that
is a completely continuous operator such that either
(i) for
and
for
or
(ii) for
and
for
.
Then has a fixed point in
.
3. Existence of Positive Solutions
Throughout this paper, we let . Then
is a Banach space, where
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2011%2F475126/MediaObjects/13661_2010_Article_41_Equ7_HTML.gif)
In this section, we always assume that .
Lemma 3.1.
System (1.1) is equivalent to the following system of integral equations:
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2011%2F475126/MediaObjects/13661_2010_Article_41_Equ8_HTML.gif)
where
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2011%2F475126/MediaObjects/13661_2010_Article_41_Equ9_HTML.gif)
Proof.
It is easy to see that if satisfies (3.2) then it also satisfies (3.2). So, assume that
is a solution to (1.1). Integrating both sides of the first equation of (1.1) of order
with respect to
gives us
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2011%2F475126/MediaObjects/13661_2010_Article_41_Equ10_HTML.gif)
for ,
. It follows that
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2011%2F475126/MediaObjects/13661_2010_Article_41_Equ11_HTML.gif)
for ,
. This, combined with the boundary conditions in (1.1), yields
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2011%2F475126/MediaObjects/13661_2010_Article_41_Equ12_HTML.gif)
Similarly, one can obtain
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2011%2F475126/MediaObjects/13661_2010_Article_41_Equ13_HTML.gif)
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2011%2F475126/MediaObjects/13661_2010_Article_41_Equ14_HTML.gif)
  . Then it follows from (3.8) and the boundary condition
that
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2011%2F475126/MediaObjects/13661_2010_Article_41_Equ15_HTML.gif)
Therefore, for ,
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2011%2F475126/MediaObjects/13661_2010_Article_41_Equ16_HTML.gif)
This completes the proof.
The following two results give some properties of the Green functions .
Lemma 3.2.
For is continuous on
and
for
.
Proof.
Obviously, is continuous on
. It remains to show that
for
. It is easy to see that
for
. We only need to show that
for
. For
, let
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2011%2F475126/MediaObjects/13661_2010_Article_41_Equ17_HTML.gif)
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2011%2F475126/MediaObjects/13661_2010_Article_41_Equ18_HTML.gif)
Then
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2011%2F475126/MediaObjects/13661_2010_Article_41_Equ19_HTML.gif)
Note that and
for
. It follows that
and hence
for
.
Therefore, for
and the proof is complete.
Lemma 3.3.
-
(i)
If
, then
for
.
-
(ii)
If
, then
for
.
Proof.
-
(i)
Obviously,
for
. Now, for
, we have
(3.14)
where is the function defined by (3.11). It follows that
for
. In summary, we have proved (i).
-
(ii)
Again, one can easily see that
for
. When
, we have in this case that
(3.15)
which implies that for
. To summarize, we have proved (ii) and this completes the proof.
Now, we are ready to present the main results.
Theorem 3.4.
Suppose that there exist functions ,
, 2,
, such that
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2011%2F475126/MediaObjects/13661_2010_Article_41_Equ22_HTML.gif)
for ,
,
. If
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2011%2F475126/MediaObjects/13661_2010_Article_41_Equ23_HTML.gif)
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2011%2F475126/MediaObjects/13661_2010_Article_41_Equ24_HTML.gif)
then (1.1) has a unique positive solution.
Proof.
Let
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2011%2F475126/MediaObjects/13661_2010_Article_41_Equ25_HTML.gif)
It is easy to see that is a complete metric space. Define an operator
on
by
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2011%2F475126/MediaObjects/13661_2010_Article_41_Equ26_HTML.gif)
where and
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2011%2F475126/MediaObjects/13661_2010_Article_41_Equ27_HTML.gif)
Because of the continuity of and
, it follows easily from Lemma 3.2 that
maps
into itself. To finish the proof, we only need to show that
is a contraction. Indeed, for
, by (3.16) we have
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2011%2F475126/MediaObjects/13661_2010_Article_41_Equ28_HTML.gif)
This, combined with Lemma 3.3 and (3.17) and (3.18), immediately implies that is a contraction. Therefore, the proof is complete with the help of Lemmas 3.1 and 2.5.
The following result can be proved in the same spirit as that for Theorem 3.4.
Theorem 3.5.
For , suppose that there exist nonnegative function
and nonnegative constants
such that
and
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2011%2F475126/MediaObjects/13661_2010_Article_41_Equ29_HTML.gif)
for ,
. If
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2011%2F475126/MediaObjects/13661_2010_Article_41_Equ30_HTML.gif)
then (1.1) has a unique positive solution.
Theorem 3.6.
For , suppose that there exist nonnegative real-valued functions
such that
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2011%2F475126/MediaObjects/13661_2010_Article_41_Equ31_HTML.gif)
for almost every and all
. If
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2011%2F475126/MediaObjects/13661_2010_Article_41_Equ32_HTML.gif)
then (1.1) has at least one positive solution.
Proof.
Let and
be defined by (3.19) and (3.20), respectively. We first show that
is completely continuous through the following three steps.
Step 1.
Show that is continuous. Let
be a sequence in
such that
. Then
is bounded in
. Since
is continuous, it is uniformly continuous on any compact set. In particular, for any
, there exists a positive integer
such that
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2011%2F475126/MediaObjects/13661_2010_Article_41_Equ33_HTML.gif)
for and
. Then, for
, we have
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2011%2F475126/MediaObjects/13661_2010_Article_41_Equ34_HTML.gif)
for and
. Therefore,
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2011%2F475126/MediaObjects/13661_2010_Article_41_Equ35_HTML.gif)
which implies that is continuous.
Step 2.
Show that maps bounded sets of
into bounded sets. Let
be a bounded subset of
. Then
is bounded. Since
is continuous, there exists an
such that
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2011%2F475126/MediaObjects/13661_2010_Article_41_Equ36_HTML.gif)
It follows that, for and
,
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2011%2F475126/MediaObjects/13661_2010_Article_41_Equ37_HTML.gif)
Immediately, we can easily see that is a bounded subset of
.
Step 3.
Show that maps bounded sets of
into equicontinuous sets. Let
be a bounded subset of
. Similarly as in Step 2, there exists
such that
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2011%2F475126/MediaObjects/13661_2010_Article_41_Equ38_HTML.gif)
Then, for any and
and
,
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2011%2F475126/MediaObjects/13661_2010_Article_41_Equ39_HTML.gif)
Now the equicontituity of on
follows easily from the fact that
is continuous and hence uniformly continuous on
.
Now we have shown that is completely continuous. To apply Lemma 2.6, let
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2011%2F475126/MediaObjects/13661_2010_Article_41_Equ40_HTML.gif)
Fix and define
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2011%2F475126/MediaObjects/13661_2010_Article_41_Equ41_HTML.gif)
We claim that there is no such that
for some
. Otherwise, assume that there exist
and
such that
. Then
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2011%2F475126/MediaObjects/13661_2010_Article_41_Equ42_HTML.gif)
If , then
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2011%2F475126/MediaObjects/13661_2010_Article_41_Equ43_HTML.gif)
Similarly, we can have if
. To summarize,
, a contradiction to
. This proves the claim. Applying Lemma 2.6, we know that
has a fixed point in
, which is a positive solution to (1.1) by Lemma 3.1. Therefore, the proof is complete.
As a consequence of Theorem 3.6, we have the following.
Corollary 3.7.
If all ,
, are bounded, then (1.1) has at least one positive solution.
To state the last result of this section, we introduce
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2011%2F475126/MediaObjects/13661_2010_Article_41_Equ44_HTML.gif)
Theorem 3.8.
Suppose that there exist and positive constants
with
such that
(i)  for
and
(ii), for
,
where . Then (1.1) has at least a positive solution.
Proof.
Let be defined by (3.19) and
. Obviously,
is a cone in
. From the proof of Theorem 3.6, we know that the operator
defined by (3.20) is completely continuous on
. For any
, it follows from Lemma 3.3 and condition (ii) that
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2011%2F475126/MediaObjects/13661_2010_Article_41_Equ45_HTML.gif)
that is,
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2011%2F475126/MediaObjects/13661_2010_Article_41_Equ46_HTML.gif)
On the other hand, for any , it follows from Lemma 3.3 and condition (i) that, for
,
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2011%2F475126/MediaObjects/13661_2010_Article_41_Equ47_HTML.gif)
if , whereas
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2011%2F475126/MediaObjects/13661_2010_Article_41_Equ48_HTML.gif)
if . In summary,
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2011%2F475126/MediaObjects/13661_2010_Article_41_Equ49_HTML.gif)
Therefore, we have verified condition (ii) of Lemma 2.7. It follows that has a fixed point in
, which is a positive solution to (1.1). This completes the proof.
4. Examples
In this section, we demonstrate the feasibility of some of the results obtained in Section 3.
Example 4.1.
Consider
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2011%2F475126/MediaObjects/13661_2010_Article_41_Equ50_HTML.gif)
Here
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2011%2F475126/MediaObjects/13661_2010_Article_41_Equ51_HTML.gif)
One can easily see that (3.16) is satisfied with
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2011%2F475126/MediaObjects/13661_2010_Article_41_Equ52_HTML.gif)
Moreover,
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2011%2F475126/MediaObjects/13661_2010_Article_41_Equ53_HTML.gif)
and hence
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2011%2F475126/MediaObjects/13661_2010_Article_41_Equ54_HTML.gif)
It follows from Theorem 3.4 that (4.1) has a unique positive solution on .
Example 4.2.
Consider
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2011%2F475126/MediaObjects/13661_2010_Article_41_Equ55_HTML.gif)
Here
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2011%2F475126/MediaObjects/13661_2010_Article_41_Equ56_HTML.gif)
where
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2011%2F475126/MediaObjects/13661_2010_Article_41_Equ57_HTML.gif)
Hence, and
satisfy (3.25). Moreover, simple calculations give us
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2011%2F475126/MediaObjects/13661_2010_Article_41_Equ58_HTML.gif)
Then and
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2011%2F475126/MediaObjects/13661_2010_Article_41_Equ59_HTML.gif)
Choose ,
and
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2011%2F475126/MediaObjects/13661_2010_Article_41_Equ60_HTML.gif)
Then, for and
, we have
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2011%2F475126/MediaObjects/13661_2010_Article_41_Equ61_HTML.gif)
for and
, we have
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2011%2F475126/MediaObjects/13661_2010_Article_41_Equ62_HTML.gif)
By now we have verified all the assumptions of Theorem 3.8. Therefore, (4.6) has at least one positive solution satisfying
.
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Acknowledgment
Supported partially by the Doctor Foundation of University of South China under Grant no. 5-XQD-2006-9, the Foundation of Science and Technology Department of Hunan Province under Grant no. 2009RS3019 and the Subject Lead Foundation of University of South China no. 2007XQD13. Research was partially supported by the Natural Science and Engineering Re-search Council of Canada (NSERC) and the Early Researcher Award (ERA) Pro-gram of Ontario.
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Ouyang, Z., Chen, Y. & Zou, S. Existence of Positive Solutions to a Boundary Value Problem for a Delayed Nonlinear Fractional Differential System. Bound Value Probl 2011, 475126 (2011). https://doi.org/10.1155/2011/475126
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DOI: https://doi.org/10.1155/2011/475126