Abstract
Criteria are established for existence of least one or three positive solutions for boundary value problems of second-order functional dynamic equations on time scales. By this purpose, we use a fixed-point index theorem in cones and Leggett-Williams fixed-point theorem.
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1. Introduction
In a recent paper [1], by applying a fixed-point index theorem in cones, Jiang and Weng studied the existence of positive solutions for the boundary value problems described by second-order functional differential equations of the form
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2009%2F829735/MediaObjects/13662_2009_Article_1223_Equ1_HTML.gif)
Aykut [2] applied a cone fixed-point index theorem in cones and obtained sufficient conditions for the existence of positive solutions of the boundary value problems of functional difference equations of the form
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2009%2F829735/MediaObjects/13662_2009_Article_1223_Equ2_HTML.gif)
In this article, we are interested in proving the existence and multiplicity of positive solutions for the boundary value problems of a second-order functional dynamic equation of the form
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2009%2F829735/MediaObjects/13662_2009_Article_1223_Equ3_HTML.gif)
Throughout this paper we let be any time scale (nonempty closed subset of
) and
be a subset of
such that
and for
is not right scattered and left dense at the same time.
Some preliminary definitions and theorems on time scales can be found in books [3, 4] which are excellent references for calculus of time scales.
We will assume that the following conditions are satisfied.
(H1)
(H2) is continuous with respect to
and
for
, where
denotes the set of nonnegative real numbers.
(H3) defined on
satisfies
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2009%2F829735/MediaObjects/13662_2009_Article_1223_Equ4_HTML.gif)
Let be nonempty subset of
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2009%2F829735/MediaObjects/13662_2009_Article_1223_Equ5_HTML.gif)
(H4)
if then
;
for
, where
denotes the set of all positively regressive and rd-continuous functions.
(H5) and
are defined on
and
, respectively, where
; furthermore,
;
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2009%2F829735/MediaObjects/13662_2009_Article_1223_Equ6_HTML.gif)
There have been a number of works concerning of at least one and multiple positive solutions for boundary value problems recent years. Some authors have studied the problem for ordinary differential equations, while others have studied the problem for difference equations, while still others have considered the problem for dynamic equations on a time scale [5–10]. However there are fewer research for the existence of positive solutions of the boundary value problems of functional differential, difference, and dynamic equations [1, 2, 11–13].
Our problem is a dynamic analog of the BVPs (1.1) and (1.2). But it is more general than them. Moreover, conditions for the existence of at least one positive solution were studied for the BVPs (1.1) and (1.2). In this paper, we investigate the conditions for the existence of at least one or three positive solutions for the BVP (1.3). The key tools in our approach are the following fixed-point index theorem [14], and Leggett-Williams fixed-point theorem [15].
Theorem 1.1 (see [14]).
Let be Banach space and
be a cone in
. Let
, and define
. Assume
is a completely continuous operator such that
for
-
(i)
If
for
, then
-
(ii)
If
for
, then
Theorem 1.2 (see [15]).
Let be a cone in the real Banach space
. Set
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2009%2F829735/MediaObjects/13662_2009_Article_1223_Equ7_HTML.gif)
Suppose that is a completely continuous operator and
is a nonnegative continuous concave functional on
with
for all
. If there exists
such that the following conditions hold:
-
(i)
and
for all
-
(ii)
for
-
(iii)
for
with
Then has at least three fixed points
in
satisfying
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2009%2F829735/MediaObjects/13662_2009_Article_1223_Equ8_HTML.gif)
2. Preliminaries
First, we give the following definitions of solution and positive solution of BVP (1.3).
Definition 2.1.
We say a function is a solution of BVP (1.3) if it satisfies the following.
-
(1)
is nonnegative on
.
-
(2)
as
, where
is defined as
(21) -
(3)
as
, where
is defined as
(22) -
(4)
is
-differentiable,
is
-differentiable on
and
is continuous.
-
(5)
for
Furthermore, a solution of (1.3) is called a positive solution if
for
Denote by and
the solutions of the corresponding homogeneous equation
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2009%2F829735/MediaObjects/13662_2009_Article_1223_Equ11_HTML.gif)
under the initial conditions
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2009%2F829735/MediaObjects/13662_2009_Article_1223_Equ12_HTML.gif)
Set
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2009%2F829735/MediaObjects/13662_2009_Article_1223_Equ13_HTML.gif)
Since the Wronskian of any two solutions of (2.3) is independent of , evaluating at
and using the initial conditions (2.4) yield
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2009%2F829735/MediaObjects/13662_2009_Article_1223_Equ14_HTML.gif)
Using the initial conditions (2.4), we can deduce from (2.3) for and
, the following equations:
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2009%2F829735/MediaObjects/13662_2009_Article_1223_Equ15_HTML.gif)
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2009%2F829735/MediaObjects/13662_2009_Article_1223_Equ16_HTML.gif)
(See [8].)
Lemma 2.2 (see [8]).
Under the conditions (H1) and the first part of (H4) the solutions and
possess the following properties:
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2009%2F829735/MediaObjects/13662_2009_Article_1223_Equ17_HTML.gif)
Let be the Green function for the boundary value problem:
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2009%2F829735/MediaObjects/13662_2009_Article_1223_Equ18_HTML.gif)
given by
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2009%2F829735/MediaObjects/13662_2009_Article_1223_Equ19_HTML.gif)
where and
are given in (2.7) and (2.8), respectively. It is obvious from (2.6), (H1) and (H4), that
holds.
Lemma 2.3.
Assume the conditions (H1) and (H4) are satisfied. Then
-
(i)
for
-
(ii)
for
and
where
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2009%2F829735/MediaObjects/13662_2009_Article_1223_Equ20_HTML.gif)
in which
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2009%2F829735/MediaObjects/13662_2009_Article_1223_Equ21_HTML.gif)
Proof.
for
, and
, for
. Besides,
is nondecreasing and
is nonincreasing, for
. Therefore, we have
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2009%2F829735/MediaObjects/13662_2009_Article_1223_Equ22_HTML.gif)
So statement (i) of the lemma is proved. If for a given
then statement (ii) of the lemma is obvious for such values. Now,
and
. Consequently,
, for all such
Let us take any
. Then we have for
,
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2009%2F829735/MediaObjects/13662_2009_Article_1223_Equ23_HTML.gif)
and we have for ,
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2009%2F829735/MediaObjects/13662_2009_Article_1223_Equ24_HTML.gif)
Let be endowed with maximum norm
for
, and let
be a cone defined by
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2009%2F829735/MediaObjects/13662_2009_Article_1223_Equ25_HTML.gif)
where is as in (2.12).
Suppose that is a solution of (1.3), then it can be written as
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2009%2F829735/MediaObjects/13662_2009_Article_1223_Equ26_HTML.gif)
where
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2009%2F829735/MediaObjects/13662_2009_Article_1223_Equ27_HTML.gif)
Throughout this paper we assume that is the solution of (1.3) with
. Clearly,
can be expressed as follows:
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2009%2F829735/MediaObjects/13662_2009_Article_1223_Equ28_HTML.gif)
where
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2009%2F829735/MediaObjects/13662_2009_Article_1223_Equ29_HTML.gif)
Let be a solution of (1.3) and
. Noting that
for
, we have
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2009%2F829735/MediaObjects/13662_2009_Article_1223_Equ30_HTML.gif)
where
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2009%2F829735/MediaObjects/13662_2009_Article_1223_Equ31_HTML.gif)
Define an operator as follows:
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2009%2F829735/MediaObjects/13662_2009_Article_1223_Equ32_HTML.gif)
where
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2009%2F829735/MediaObjects/13662_2009_Article_1223_Equ33_HTML.gif)
It is easy to derive that is a positive solution of BVP (1.3) if and only if
is a nontrivial fixed point
of
, where
be defined as before.
Lemma 2.4.
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2009%2F829735/MediaObjects/13662_2009_Article_1223_IEq134_HTML.gif)
Proof.
For , we have
. Moreover, we have from definition of
that
and
, for
and
, respectively. Thus,
where
. It follows from the definition
and Lemma 2.3 that
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2009%2F829735/MediaObjects/13662_2009_Article_1223_Equ34_HTML.gif)
Thus, .
Lemma 2.5.
is completely continuous.
Lemma 2.6.
If
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2009%2F829735/MediaObjects/13662_2009_Article_1223_Equ35_HTML.gif)
for all , then there exist
such that
, for
and
, for
.
Proof.
Choose such that
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2009%2F829735/MediaObjects/13662_2009_Article_1223_Equ36_HTML.gif)
By using the first equality of (2.27), we can choose such that
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2009%2F829735/MediaObjects/13662_2009_Article_1223_Equ37_HTML.gif)
If , then for
, we have
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2009%2F829735/MediaObjects/13662_2009_Article_1223_Equ38_HTML.gif)
Therefore we get
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2009%2F829735/MediaObjects/13662_2009_Article_1223_Equ39_HTML.gif)
Thus, we have from Theorem 1.1, , for
. On the other hand, the second equality of (2.27) implies for every
, there is an
, such that
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2009%2F829735/MediaObjects/13662_2009_Article_1223_Equ40_HTML.gif)
Here we choose satisfying (2.28). For
, we have definition of
that
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2009%2F829735/MediaObjects/13662_2009_Article_1223_Equ41_HTML.gif)
It follows from (2.32) that
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2009%2F829735/MediaObjects/13662_2009_Article_1223_Equ42_HTML.gif)
This shows that
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2009%2F829735/MediaObjects/13662_2009_Article_1223_Equ43_HTML.gif)
Thus, by Theorem 1.1, we conclude that for
. The proof is therefore complete.
3. Existence of One Positive Solution
In this section, we investigate the conditions for the existence of at least one positive solution of the BVP (1.3).
In the next theorem, we will also assume that the following condition on .
(H6):
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2009%2F829735/MediaObjects/13662_2009_Article_1223_Equ44_HTML.gif)
where is large enough such that
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2009%2F829735/MediaObjects/13662_2009_Article_1223_Equ45_HTML.gif)
and is small enough such that
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2009%2F829735/MediaObjects/13662_2009_Article_1223_Equ46_HTML.gif)
where is the eigenfunction related to the smallest eigenvalue
of the eigenvalue problem:
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2009%2F829735/MediaObjects/13662_2009_Article_1223_Equ47_HTML.gif)
Theorem 3.1.
If (H1)–(H6) are satisfied, then the BVP (1.3) has at least one positive solution.
Proof.
Fix and let
for
. Then,
satisfies (2.27). Define
by
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2009%2F829735/MediaObjects/13662_2009_Article_1223_Equ48_HTML.gif)
where
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2009%2F829735/MediaObjects/13662_2009_Article_1223_Equ49_HTML.gif)
Then is a completely continuous operator. One has from Lemma 2.6 that there exist
such that
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2009%2F829735/MediaObjects/13662_2009_Article_1223_Equ50_HTML.gif)
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2009%2F829735/MediaObjects/13662_2009_Article_1223_Equ51_HTML.gif)
Define by
then
is a completely continuous operator. By the first equality in (H6) and the definition of
, there are
and
such that
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2009%2F829735/MediaObjects/13662_2009_Article_1223_Equ52_HTML.gif)
We now prove that for all
and
. In fact, if there exists
and
such that
, then
satisfies the equation
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2009%2F829735/MediaObjects/13662_2009_Article_1223_Equ53_HTML.gif)
and the boundary conditions
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2009%2F829735/MediaObjects/13662_2009_Article_1223_Equ54_HTML.gif)
Multiplying both sides of (3.10) by , then integrating from
to
, and using integration by parts in the left-hand side two times, we obtain
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2009%2F829735/MediaObjects/13662_2009_Article_1223_Equ55_HTML.gif)
Combining (3.9) and (3.12), we get
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2009%2F829735/MediaObjects/13662_2009_Article_1223_Equ56_HTML.gif)
We also have
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2009%2F829735/MediaObjects/13662_2009_Article_1223_Equ57_HTML.gif)
Equations (3.13) and (3.14) lead to
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2009%2F829735/MediaObjects/13662_2009_Article_1223_Equ58_HTML.gif)
This is impossible. Thus for
and
. By (3.7) and the homotopy invariance of the fixed-point index (see [11]), we get that
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2009%2F829735/MediaObjects/13662_2009_Article_1223_Equ59_HTML.gif)
On the other hand, according to the second inequality of (H6), there exist and
such that
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2009%2F829735/MediaObjects/13662_2009_Article_1223_Equ60_HTML.gif)
We define
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2009%2F829735/MediaObjects/13662_2009_Article_1223_Equ61_HTML.gif)
then it follows that
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2009%2F829735/MediaObjects/13662_2009_Article_1223_Equ62_HTML.gif)
Define by
, then
is a completely continuous operator. We claim that there exists
such that
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2009%2F829735/MediaObjects/13662_2009_Article_1223_Equ63_HTML.gif)
In fact, if for some
and
, then
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2009%2F829735/MediaObjects/13662_2009_Article_1223_Equ64_HTML.gif)
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2009%2F829735/MediaObjects/13662_2009_Article_1223_Equ65_HTML.gif)
where Combining (3.21) with (3.22), we have
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2009%2F829735/MediaObjects/13662_2009_Article_1223_Equ66_HTML.gif)
Let Then we get
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2009%2F829735/MediaObjects/13662_2009_Article_1223_Equ67_HTML.gif)
Consequently, by the homotopy invariance of the fixed-point index, we have
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2009%2F829735/MediaObjects/13662_2009_Article_1223_Equ68_HTML.gif)
where is zero operator. Use (3.16) and (3.25) to conclude that
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2009%2F829735/MediaObjects/13662_2009_Article_1223_Equ69_HTML.gif)
Hence, has a fixed point in
.
Let . Since
for
and
.
(H7)
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2009%2F829735/MediaObjects/13662_2009_Article_1223_Equ70_HTML.gif)
Theorem 3.2.
If (H1)–(H5) and (H7) are satisfied, then the BVP (1.3) has at least one positive solution.
Proof.
Define by
, then
is a completely continuous operator. By the first inequality in (H7), there exist
and
such that
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2009%2F829735/MediaObjects/13662_2009_Article_1223_Equ71_HTML.gif)
We claim that for
and
. In fact, if there exist
and
such that
, then
satisfies the boundary condition (3.11). Since
, we have
. Then we have
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2009%2F829735/MediaObjects/13662_2009_Article_1223_Equ72_HTML.gif)
Multiplying the last equation by integrating from
to
, by (3.28), we obtain
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2009%2F829735/MediaObjects/13662_2009_Article_1223_Equ73_HTML.gif)
then we have
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2009%2F829735/MediaObjects/13662_2009_Article_1223_Equ74_HTML.gif)
Equations (3.30) and (3.31) lead to
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2009%2F829735/MediaObjects/13662_2009_Article_1223_Equ75_HTML.gif)
This is impossible. By homotopy invariance of the fixed-point index, we get that
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2009%2F829735/MediaObjects/13662_2009_Article_1223_Equ76_HTML.gif)
Define by
, then
is a completely continuous operator. By the second inequality in (H7), and definition of
, there exist
and
such that
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2009%2F829735/MediaObjects/13662_2009_Article_1223_Equ77_HTML.gif)
We define
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2009%2F829735/MediaObjects/13662_2009_Article_1223_Equ78_HTML.gif)
then, it is obvious that
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2009%2F829735/MediaObjects/13662_2009_Article_1223_Equ79_HTML.gif)
We claim that there exists such that
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2009%2F829735/MediaObjects/13662_2009_Article_1223_Equ80_HTML.gif)
In fact, if for some
and
, then using (3.36), it is analogous to the argument of (3.13) and (3.14) that
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2009%2F829735/MediaObjects/13662_2009_Article_1223_Equ81_HTML.gif)
Equation (3.38) leads to Let
. Then we get
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2009%2F829735/MediaObjects/13662_2009_Article_1223_Equ82_HTML.gif)
Consequently, by (3.8) and the homotopy invariance of the fixed-point index, we have
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2009%2F829735/MediaObjects/13662_2009_Article_1223_Equ83_HTML.gif)
In view of (3.33) and (3.40), we obtain
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2009%2F829735/MediaObjects/13662_2009_Article_1223_Equ84_HTML.gif)
Therefore, has a fixed point in
. The proof is completed.
Corollary 3.3.
Using the following (H8) or (H9) instead of (H6) or (H7), the conclusions of Theorems 3.1 and 3.2 are true. For ,
(H8)
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2009%2F829735/MediaObjects/13662_2009_Article_1223_Equ85_HTML.gif)
(H9)
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2009%2F829735/MediaObjects/13662_2009_Article_1223_Equ86_HTML.gif)
4. Existence of Three Positive Solutions
In this section, using Theorem 1.2 (the Leggett-Williams fixed-point theorem) we prove the existence of at least three positive solutions to the BVP (1.3).
Define the continuous concave functional to be
, and the constants
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2009%2F829735/MediaObjects/13662_2009_Article_1223_Equ87_HTML.gif)
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2009%2F829735/MediaObjects/13662_2009_Article_1223_Equ88_HTML.gif)
Theorem 4.1.
Suppose there exists constants such that
(D1) for
(D2) for
(D3) one of the following is satisfied:
-
(a)
-
(b)
there exists a constant
such that
for
and
,
where ,
, and
are as defined in (2.12), (4.1), (4.2), respectively. Then the boundary value problem (1.3) has at least three positive solutions
and
satisfying
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2009%2F829735/MediaObjects/13662_2009_Article_1223_Equ89_HTML.gif)
Proof.
The technique here similar to that used in [5] Again the cone , the operator
is the same as in the previous sections. For all
we have
If
, then
and the condition (a) of (D3) imply that
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2009%2F829735/MediaObjects/13662_2009_Article_1223_Equ90_HTML.gif)
Thus there exist a and
such that if
, then
. For
, we have
for all
, for all
Pick any
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2009%2F829735/MediaObjects/13662_2009_Article_1223_Equ91_HTML.gif)
Then implies that
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2009%2F829735/MediaObjects/13662_2009_Article_1223_Equ92_HTML.gif)
Thus
The condition (b) of (D3) implies that there exists a positive number such that
for
and
. If
, then
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2009%2F829735/MediaObjects/13662_2009_Article_1223_Equ93_HTML.gif)
Thus Consequently, the assumption (D3) holds, then there exist a number
such that
and
The remaining conditions of Theorem 1.2 will now be shown to be satisfied.
By (D1) and argument above, we can get that Hence, condition (ii) of Theorem 1.2 is satisfied.
We now consider condition (i) of Theorem 1.2. Choose for
, where
. Then
and
so that
. For
, we have
,
. Combining with (D2), we get
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2009%2F829735/MediaObjects/13662_2009_Article_1223_Equ94_HTML.gif)
for . Thus, we have
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2009%2F829735/MediaObjects/13662_2009_Article_1223_Equ95_HTML.gif)
As a result, yields
Lastly, we consider Theorem 1.2(iii). Recall that . If
and
, then
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2009%2F829735/MediaObjects/13662_2009_Article_1223_Equ96_HTML.gif)
Thus, all conditions of Theorem 1.2 are satisfied. It implies that the TPBVP (1.3) has at least three positive solutions with
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2009%2F829735/MediaObjects/13662_2009_Article_1223_Equ97_HTML.gif)
5. Examples
Example 5.1.
Let Consider the BVP:
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2009%2F829735/MediaObjects/13662_2009_Article_1223_Equ98_HTML.gif)
Then and
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2009%2F829735/MediaObjects/13662_2009_Article_1223_Equ99_HTML.gif)
Since ,
. It is clear that (H1)–(H5) and (H8) are satisfied. Thus, by Corollary 3.3, the BVP (5.1) has at least one positive solution.
Example 5.2.
Let us introduce an example to illustrate the usage of Theorem 4.1. Let
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2009%2F829735/MediaObjects/13662_2009_Article_1223_Equ100_HTML.gif)
Consider the TPBVP:
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2009%2F829735/MediaObjects/13662_2009_Article_1223_Equ101_HTML.gif)
Then and
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2009%2F829735/MediaObjects/13662_2009_Article_1223_Equ102_HTML.gif)
The Green function of the BVP (5.4) has the form
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2009%2F829735/MediaObjects/13662_2009_Article_1223_Equ103_HTML.gif)
Clearly, is continuous and increasing
. We can also see that
. By (2.12), (4.1), and (4.2), we get
, and
.
Now we check that (D1), (D2), and (b) of (D3) are satisfied. To verify (D1), as , we take
, then
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2009%2F829735/MediaObjects/13662_2009_Article_1223_Equ104_HTML.gif)
and (D1) holds. Note that , when we set
,
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2009%2F829735/MediaObjects/13662_2009_Article_1223_Equ105_HTML.gif)
holds. It means that (D2) is satisfied. Let , we have
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2009%2F829735/MediaObjects/13662_2009_Article_1223_Equ106_HTML.gif)
from , so that (b) of (D3) is met. Summing up, there exist constants
and
satisfying
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2009%2F829735/MediaObjects/13662_2009_Article_1223_Equ107_HTML.gif)
Thus, by Theorem 4.1, the TPBVP (5.4) has at least three positive solutions with
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2009%2F829735/MediaObjects/13662_2009_Article_1223_Equ108_HTML.gif)
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Karaca, I. Positive Solutions for Boundary Value Problems of Second-Order Functional Dynamic Equations on Time Scales. Adv Differ Equ 2009, 829735 (2009). https://doi.org/10.1155/2009/829735
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DOI: https://doi.org/10.1155/2009/829735