For convenience, we list the following definitions which can be found in [1, 2, 9, 14, 17]. A time scale
is a nonempty closed subset of real numbers
. For
and
, define the forward jump operator
and backward jump operator
, respectively, by
for all
. If
,
is said to be right scattered, and if
,
is said to be left scattered; if
,
is said to be right dense, and if
,
is said to be left dense. If
has a right scattered minimum
, define
; otherwise set
. If
has a left scattered maximum
, define
; otherwise set
. In this general time-scale setting,
represents the delta (or Hilger) derivative [13, Definition 1.10],
where
is the forward jump operator,
is the forward graininess function, and
is abbreviated as
. In particular, if
, then
and
, while if
for any
, then
and
A function
is right-dense continuous provided that it is continuous at each right-dense point
(a point where
) and has a left-sided limit at each left-dense point
. The set of right-dense continuous functions on
is denoted by
. It can be shown that any right-dense continuous function
has an antiderivative (a function
with the property
for all
). Then the Cauchy delta integral of
is defined by
where
is an antiderivative of
on
. For example, if
, then
and if
, then
Throughout we assume that
are points in
, and define the time-scale interval
. In this paper, we also need the the following theorem which can be found in [1].
Theorem 2.1.
If
and
then
In this paper, let
Then
is a Banach space with the norm
. Define a cone
by
Obviously,
is a cone in
. Set
. If
on
then we say
is concave on
We can get the following.
Lemma 2.2.
Suppose condition
holds. Then there exists a constant
satisfies
Furthermore, the function
is a positive continuous function on
, therefore
has minimum on
. Then there exists
such that
.
Lemma 2.3.
Let
and
in Lemma 2.2. Then
Proof.
Suppose
.
We will discuss it from three perspectives.
(i)
. It follows from the concavity of
that
then
which means
.
(ii)
. If
, we have
then
If
, we have
then
and this means
.
(iii)
. Similarly, we have
then
which means
.
From the above, we know
. The proof is complete.
Lemma 2.4.
Suppose that conditions
hold, then
is a solution of boundary value problem (1.6), (1.7) if and only if
is a solution of the following integral equation:
where
Proof.
Necessity. By the equation of the boundary condition, we see that
, then there exists a constant
such that
. Firstly, by delta integrating the equation of the problems (1.6) on
, we have
thus
By
and the boundary condition (1.7), let
on (2.23), we have
By the equation of the boundary condition (1.7), we get
then
Secondly, by (2.24) and let
on (2.24), we have
Then
Then by delta integrating (2.29) for
times on
, we have
Similarly, for
, by delta integrating the equation of problems (1.6) on
, we have
Therefore, for any
,
can be expressed as the equation
where
is expressed as (2.22).
Sufficiency. Suppose that
then by (2.22), we have
So,
which imply that (1.6) holds. Furthermore, by letting
and
on (2.22) and (2.34), we can obtain the boundary value equations of (1.7). The proof is complete.
Now, we define a mapping
given by
where
is given by (2.22).
Lemma 2.5.
Suppose that conditions
hold, the solution
of problem (1.6), (1.7) satisfies
and for
in Lemma 2.2, one has
Proof.
If
is the solution of (1.6), (1.7), then
is a concave function, and
, thus we have
that is,
By Lemma 2.3, for
, we have
then
. The proof is complete.
Lemma 2.6.
is completely continuous.
Proof.
Because
is continuous, decreasing on
and satisfies
. Then,
for each
and
. This shows that
. Furthermore, it is easy to check that
is completely continuous by Arzela-ascoli Theorem.
For convenience, we set
where
is the constant from Lemma 2.2. By Lemma 2.5, we can also set