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Iterative Solutions of Singular Boundary Value Problems of Third-Order Differential Equation
Boundary Value Problems volume 2011, Article number: 483057 (2011)
Abstract
By using the cone theory and the Banach contraction mapping principle, the existence and uniqueness results are established for singular third-order boundary value problems. The theorems obtained are very general and complement previous known results.
1. Introduction
Third-order differential equations arise in a variety of different areas of applied mathematics and physics, such as the deflection of a curved beam having a constant or varying cross section, three-layer beam, electromagnetic waves, or gravity-driven flows [1]. Recently, third-order boundary value problems have been studied extensively in the literature (see, e.g., [2–13], and their references). In this paper, we consider the following third-order boundary value problem:
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2011%2F483057/MediaObjects/13661_2011_Article_42_Equ1_HTML.gif)
where ,
.
Three-point boundary value problems (BVPs for short) have been also widely studied because of both practical and theoretical aspects. There have been many papers investigating the solutions of three-point BVPs, see [2–5, 10, 12] and references therein. Recently, the existence of solutions of third-order three-point BVP (1.1) has been studied in [2, 3]. Guo et al. [2] show the existence of positive solutions for BVP (1.1) when and
is separable by using cone expansion-compression fixed point theorem. In [3], the singular third-order three-point BVP (1.1) is considered under some conditions concerning the first eigenvalues corresponding to the relevant linear operators, where
,
is separable and is not necessary to be nonnegative, and the existence results of nontrivial solutions and positive solutions are given by means of the topological degree theory. Motivated by the above works, we consider the singular third-order three-point BVP (1.1). Here, we give the unique solution of BVP (1.1) under the conditions that
and
is mixed nonmonotone in
and does not need to be separable by using the cone theory and the Banach contraction mapping principle.
2. Preliminaries
Let ,
. By [2, Lemma  2.1], we have that
is a solution of (1.1) if and only if
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2011%2F483057/MediaObjects/13661_2011_Article_42_Equ2_HTML.gif)
where
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2011%2F483057/MediaObjects/13661_2011_Article_42_Equ3_HTML.gif)
It is shown in [2] that is the Green's function to
,
, and
.
Let
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2011%2F483057/MediaObjects/13661_2011_Article_42_Equ4_HTML.gif)
It is easy to see that .
is generating if and only if there exists a constant
such that every element
can be represented in the form
, where
and
3. Singular Third-Order Boundary Value Problem
This section discusses singular third-order boundary value problem (1.1).
Let . Obviously,
is a normal solid cone of Banach space
; by [16, Lemma  2.1.2], we have that
is a generating cone in
.
Theorem 3.1.
Suppose that , and there exist two positive linear bounded operators
and
with
such that for any
,
,
,
, we have
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2011%2F483057/MediaObjects/13661_2011_Article_42_Equ5_HTML.gif)
and there exists , such that
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2011%2F483057/MediaObjects/13661_2011_Article_42_Equ6_HTML.gif)
Then (1.1) has a unique solution in
. And moreover, for any
, the iterative sequence
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2011%2F483057/MediaObjects/13661_2011_Article_42_Equ7_HTML.gif)
converges to .
Remark 3.2.
Recently, in the study of BVP (1.1), almost all the papers have supposed that the Green's function is nonnegative. However, the scope of
is not limited to
in Theorem 3.1, so, we do not need to suppose that
is nonnegative.
Remark 3.3.
The function in Theorem 3.1 is not monotone or convex; the conclusions and the proof used in this paper are different from the known papers in essence.
Proof.
It is easy to see that, for any ,
can be divided into finite partitioned monotone and bounded function on
, and then by (3.2), we have
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2011%2F483057/MediaObjects/13661_2011_Article_42_Equ8_HTML.gif)
For any , let
,
, then
, by (3.1), we have
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2011%2F483057/MediaObjects/13661_2011_Article_42_Equ9_HTML.gif)
Hence
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2011%2F483057/MediaObjects/13661_2011_Article_42_Equ10_HTML.gif)
Following the former inequality, we can easily have
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2011%2F483057/MediaObjects/13661_2011_Article_42_Equ11_HTML.gif)
thus
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2011%2F483057/MediaObjects/13661_2011_Article_42_Equ12_HTML.gif)
Similarly, by and
being converged, we have that
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2011%2F483057/MediaObjects/13661_2011_Article_42_Equ13_HTML.gif)
Define the operator by
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2011%2F483057/MediaObjects/13661_2011_Article_42_Equ14_HTML.gif)
Then is the solution of BVP (1.1) if and only if
. Let
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2011%2F483057/MediaObjects/13661_2011_Article_42_Equ15_HTML.gif)
By (3.1) and (3.10), for any ,
,
, we have
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2011%2F483057/MediaObjects/13661_2011_Article_42_Equ16_HTML.gif)
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2011%2F483057/MediaObjects/13661_2011_Article_42_Equ17_HTML.gif)
so we can choose an , which satisfies
, and so there exists a positive integer
such that
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2011%2F483057/MediaObjects/13661_2011_Article_42_Equ18_HTML.gif)
Since is a generating cone in
, from Lemma 2.1, there exists
such that every element
can be represented in
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2011%2F483057/MediaObjects/13661_2011_Article_42_Equ19_HTML.gif)
This implies
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2011%2F483057/MediaObjects/13661_2011_Article_42_Equ20_HTML.gif)
Let
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2011%2F483057/MediaObjects/13661_2011_Article_42_Equ21_HTML.gif)
By (3.16), we know that is well defined for any
. It is easy to verify that
is a norm in
. By (3.15)–(3.17), we get
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2011%2F483057/MediaObjects/13661_2011_Article_42_Equ22_HTML.gif)
On the other hand, for any which satisfies
, we have
. Thus
, where
denotes the normal constant of
. Since
is arbitrary, we have
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2011%2F483057/MediaObjects/13661_2011_Article_42_Equ23_HTML.gif)
It follows from (3.18) and (3.19) that the norms and
are equivalent.
Now, for any and
which satisfies
, let
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2011%2F483057/MediaObjects/13661_2011_Article_42_Equ24_HTML.gif)
then ,  
,  
,  
,   and
.
It follows from (3.12) that
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2011%2F483057/MediaObjects/13661_2011_Article_42_Equ25_HTML.gif)
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2011%2F483057/MediaObjects/13661_2011_Article_42_Equ26_HTML.gif)
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2011%2F483057/MediaObjects/13661_2011_Article_42_Equ27_HTML.gif)
Subtracting (3.22) from (3.21) + (3.23), we obtain
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2011%2F483057/MediaObjects/13661_2011_Article_42_Equ28_HTML.gif)
Let ; then we have
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2011%2F483057/MediaObjects/13661_2011_Article_42_Equ29_HTML.gif)
As and
are both positive linear bounded operators, so,
is a positive linear bounded operator, and therefore
. Hence, by mathematical induction, it is easy to know that for natural number
in (3.14), we have
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2011%2F483057/MediaObjects/13661_2011_Article_42_Equ30_HTML.gif)
Since , we see that
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2011%2F483057/MediaObjects/13661_2011_Article_42_Equ31_HTML.gif)
which implies by virtue of the arbitrariness of that
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2011%2F483057/MediaObjects/13661_2011_Article_42_Equ32_HTML.gif)
By , we have
. Thus the Banach contraction mapping principle implies that
has a unique fixed point
in
, and so
has a unique fixed point
in
; by the definition of
has a unique fixed point
in
, that is,
is the unique solution of (1.1). And, for any
, let
; we have
. By the equivalence of
and
again, we get
. This completes the proof.
Example 3.4.
In this paper, the results apply to a very wide range of functions, we are following only one example to illustrate.
Consider the following singular third-order boundary value problem:
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2011%2F483057/MediaObjects/13661_2011_Article_42_Equ33_HTML.gif)
where and there exists
, such that for any
,  
,  
, we have
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2011%2F483057/MediaObjects/13661_2011_Article_42_Equ34_HTML.gif)
Applying Theorem 3.1, we can find that (3.29) has a unique solution provided
. And moreover, for any
, the iterative sequence
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2011%2F483057/MediaObjects/13661_2011_Article_42_Equ35_HTML.gif)
converges to .
To see that, we put
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2011%2F483057/MediaObjects/13661_2011_Article_42_Equ36_HTML.gif)
Then (3.1) is satisfied for any ,  
,  
,  and
.
In fact, if , then
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2011%2F483057/MediaObjects/13661_2011_Article_42_Equ37_HTML.gif)
If , then
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2011%2F483057/MediaObjects/13661_2011_Article_42_Equ38_HTML.gif)
Similarly,
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2011%2F483057/MediaObjects/13661_2011_Article_42_Equ39_HTML.gif)
Next, for any , by (3.30) and (3.32), we get
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2011%2F483057/MediaObjects/13661_2011_Article_42_Equ40_HTML.gif)
Then, from (3.32) and (3.36), we have
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2011%2F483057/MediaObjects/13661_2011_Article_42_Equ41_HTML.gif)
so it is easy to know by induction, for any , we get
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2011%2F483057/MediaObjects/13661_2011_Article_42_Equ42_HTML.gif)
thus
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2011%2F483057/MediaObjects/13661_2011_Article_42_Equ43_HTML.gif)
so
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2011%2F483057/MediaObjects/13661_2011_Article_42_Equ44_HTML.gif)
then we get
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2011%2F483057/MediaObjects/13661_2011_Article_42_Equ45_HTML.gif)
Let ; then
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2011%2F483057/MediaObjects/13661_2011_Article_42_Equ46_HTML.gif)
Thus all conditions in Theorem 3.1 are satisfied.
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Zhang, P. Iterative Solutions of Singular Boundary Value Problems of Third-Order Differential Equation. Bound Value Probl 2011, 483057 (2011). https://doi.org/10.1155/2011/483057
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DOI: https://doi.org/10.1155/2011/483057