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Existence of Positive Solutions for Nonlocal Fourth-Order Boundary Value Problem with Variable Parameter
Fixed Point Theory and Applications volume 2011, Article number: 604046 (2011)
Abstract
By using the Krasnoselskii's fixed point theorem and operator spectral theorem, the existence of positive solutions for the nonlocal fourth-order boundary value problem with variable parameter ,
,
,
is considered, where
is a parameter, and
,
.
1. Introduction
The existence of positive solutions for nonlinear fourth-order multipoint boundary value problems has been studied by many authors using nonlinear alternatives of Leray-Schauder, the fixed point theory, and the method of upper and lower solutions (see, e.g., [1–15] and references therein). The multipoint boundary value problem is in fact a special case of the boundary value problem with integral boundary conditions.
Recently, Bai [16] studied the existence of positive solutions of nonlocal fourth-order boundary value problem
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2011%2F604046/MediaObjects/13663_2010_Article_1415_Equ1_HTML.gif)
under the assumption:
(A1) and
,
(A2),
,
,
,
,
.
In this paper, we study the above generalizing form with variable parameters BVP
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2011%2F604046/MediaObjects/13663_2010_Article_1415_Equ2_HTML.gif)
where ,
is a parameter.
Obviously, BVP(1.1) can be regarded as the special case of BVP(1.2) with . Since the parameters
is variable, we cannot expect to transform directly BVP(1.2) into an integral equation as in [16]. We will apply the cone fixed point theory, combining with the operator spectra theorem to establish the existence of positive solutions of BVP(1.2). Our results generalize the main result in [16].
Let , and we assume that the following conditions hold throughout the paper:
(H1) and
,
(H2),
,
,
and
,
.
2. The Preliminary Lemmas
Set ,
and
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2011%2F604046/MediaObjects/13663_2010_Article_1415_Equ3_HTML.gif)
By (H1), (H2), we get ,
. Denote by
the Green's function of the problem
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2011%2F604046/MediaObjects/13663_2010_Article_1415_Equ4_HTML.gif)
and the Green's function of the problem
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2011%2F604046/MediaObjects/13663_2010_Article_1415_Equ5_HTML.gif)
Then, carefully calculation yield
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2011%2F604046/MediaObjects/13663_2010_Article_1415_Equ6_HTML.gif)
Lemma 2.1 (see [16]).
Suppose that (A1), (A2) hold. Then, for any ,
solves the problem
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2011%2F604046/MediaObjects/13663_2010_Article_1415_Equ7_HTML.gif)
if and only if
Let , and
, for
.
,
,
,
,
.
It is easy to show that are norms on
.
Lemma 2.2 (see [16]).
and (
) is a Banach space.
Lemma 2.3 (see [5]).
Assume that (A1), (A2) hold. Then,
(i), for
,
;
, for
,
,
(ii),
for
,
,
where ,
;
,
.
Denote
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2011%2F604046/MediaObjects/13663_2010_Article_1415_Equ8_HTML.gif)
Computations yield the following results.
Lemma 2.4 (see [3]).
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2011%2F604046/MediaObjects/13663_2010_Article_1415_IEq64_HTML.gif)
(i)when ,
,
(ii)when ,
,
(iii)when ,
.
Lemma 2.5 (see [16]).
Suppose that (A1), (A2) hold and ,
,
are given as above. Then,
(i),
(ii),
.
By Lemmas 2.4 and 2.5, .
Take , by Lemma 2.5,
.
Define
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2011%2F604046/MediaObjects/13663_2010_Article_1415_Equ9_HTML.gif)
Lemma 2.6.
is completely continuous, and
.
Proof.
It is similar to Lemma 6 of [3], so we omit it.
Lemma 2.7 (see [17]).
Let E be a Banach space, a cone, and
,
be two bounded open sets of
with
. Suppose that
is a completely continuous operator such that either
(i) and
,
, or
(ii) and
,
holds. Then, has a fixed point in
.
3. The Main Results
Suppose that ,
,
,
,
,
, and
, are defined as in Section 2, we introduce some notations as follows:
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2011%2F604046/MediaObjects/13663_2010_Article_1415_Equ10_HTML.gif)
Theorem 3.1.
Assume that (H1), (H2) hold and . Then BVP(1.2) has at least one positive solution if one of the following cases holds:
(i),
,
(ii),
.
Proof.
For any , consider the following BVP:
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2011%2F604046/MediaObjects/13663_2010_Article_1415_Equ11_HTML.gif)
It is easy to see that the above question is equivalent to the following question:
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2011%2F604046/MediaObjects/13663_2010_Article_1415_Equ12_HTML.gif)
For any , let
. Obviously, the operator
is linear. By Lemma 2.2, for all
,
,
. Hence
, and so
. On the other hand,
is a solution of (3.3) if and only if
satisfies
, that is,
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2011%2F604046/MediaObjects/13663_2010_Article_1415_Equ13_HTML.gif)
Owing to and
, the operator
maps
into
. From
(by Lemma 2.6) together with
and condition
, applying operator spectral theorem, we have that the
exists and is bounded. Let
, then (3.4) is equivalent to
. By the Neumann expansion formula,
can be expressed by
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2011%2F604046/MediaObjects/13663_2010_Article_1415_Equ14_HTML.gif)
The complete continuity of with the continuity of
yields that the operator
is completely continuous. For all
, let
, then
, and
. So, we have
,
. Hence,
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2011%2F604046/MediaObjects/13663_2010_Article_1415_Equ15_HTML.gif)
and so ,
.
Assume that for all ,
,
, let
, by (3.6) we have
, and so
,
. Thus by induction, it follows that
, for all
,
,
. By (3.5), for all
, we have
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2011%2F604046/MediaObjects/13663_2010_Article_1415_Equ16_HTML.gif)
and so .
On the other hand, for all , we have
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2011%2F604046/MediaObjects/13663_2010_Article_1415_Equ17_HTML.gif)
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2011%2F604046/MediaObjects/13663_2010_Article_1415_Equ18_HTML.gif)
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2011%2F604046/MediaObjects/13663_2010_Article_1415_Equ19_HTML.gif)
For any , define
. By (H1) and (H2), we have that
is continuous. It is easy to see that
being a positive solution of BVP(1.2) is equivalent to
being a nonzero solution equation as follows:
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2011%2F604046/MediaObjects/13663_2010_Article_1415_Equ20_HTML.gif)
Let . Obviously,
is completely continuous. We next show that the operator
has a nonzero fixed point in
. Let
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2011%2F604046/MediaObjects/13663_2010_Article_1415_Equ21_HTML.gif)
It is easy to know that is a cone in
,
. Now, we show
.
For , by (2.7), there is
,
. Hence, by (3.7),
,
,
. By proof of Lemma 2.5 in [16],
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2011%2F604046/MediaObjects/13663_2010_Article_1415_Equ22_HTML.gif)
By (3.7) and (3.10),
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2011%2F604046/MediaObjects/13663_2010_Article_1415_Equ23_HTML.gif)
Thus .
-
(i)
Since
, by the definition of
, there exists
such that
(3.15)
Let , one has
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2011%2F604046/MediaObjects/13663_2010_Article_1415_Equ25_HTML.gif)
So, by (3.10), we get
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2011%2F604046/MediaObjects/13663_2010_Article_1415_Equ26_HTML.gif)
Hence, for ,
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2011%2F604046/MediaObjects/13663_2010_Article_1415_Equ27_HTML.gif)
On the other hand, since , there exists
such that
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2011%2F604046/MediaObjects/13663_2010_Article_1415_Equ28_HTML.gif)
Choose , let
. For
,
, there is
. Thus,
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2011%2F604046/MediaObjects/13663_2010_Article_1415_Equ29_HTML.gif)
Hence, for ,
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2011%2F604046/MediaObjects/13663_2010_Article_1415_Equ30_HTML.gif)
By the use of the Krasnoselskii's fixed point theorem, we know there exists such that
, namely,
is a solution of (1.2) and satisfied
,
,
.
-
(ii)
The proof is similar to (i), so we omit it.
Corollary 3.2.
Assume that (H1), (H2) hold, and . Then that (1.2) has at least two positive solution, if
satisfy
(i),
,
(ii)There exists such that
, for
,
.
Proof.
By the proof of Theorem 3.1, we know that (1) from the condition , there exists
, such that
,
, (2) from the condition
, there exists
,
, such that
,
, (3) from the condition (ii), there exists
,
, such that
,
. By the use of Krasnoselskii's fixed point theorem, it is easy to know that (1.2) has at least two positive solutions.
Corollary 3.3.
Assume (H1), (H2) hold, and . Then problem (1.2) has at least two positive solution, if
satisfy
(i),
,
(ii)There exists such that
, for
,
.
Proof.
The proof is similar to Corollary 3.2, so we omit it.
Example 3.4.
Consider the following boundary value problem
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2011%2F604046/MediaObjects/13663_2010_Article_1415_Equ31_HTML.gif)
In this problem, we know that ,
,
,
, then we can get
,
,
,
,
,
,
. Further more, we obtain
,
, then
,
, so
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2011%2F604046/MediaObjects/13663_2010_Article_1415_Equ32_HTML.gif)
Thus, ,
,
, and
satisfy the conditions of Theorem 3.1, and there exists at least a positive solution of the above problem.
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Acknowledgments
This work is sponsored by the NSFC (no. 11061030), NSFC (no. 11026060), and nwnu-kjcxgc-03-69, 03-61.
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Han, X., Gao, H. & Xu, J. Existence of Positive Solutions for Nonlocal Fourth-Order Boundary Value Problem with Variable Parameter. Fixed Point Theory Appl 2011, 604046 (2011). https://doi.org/10.1155/2011/604046
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DOI: https://doi.org/10.1155/2011/604046