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 .
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.
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 .
Let and in Lemma 2.2. Then
We will discuss it from three perspectives.
(i). It follows from the concavity of that
which means .
(ii). If , we have
If , we have
and this means .
(iii). Similarly, we have
which means .
From the above, we know . The proof is complete.
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:
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
By and the boundary condition (1.7), let on (2.23), we have
By the equation of the boundary condition (1.7), we get
Secondly, by (2.24) and let on (2.24), we have
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
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).
Suppose that conditions hold, the solution of problem (1.6), (1.7) satisfies
and for in Lemma 2.2, one has
If is the solution of (1.6), (1.7), then is a concave function, and , thus we have
By Lemma 2.3, for , we have
then . The proof is complete.
is completely continuous.
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