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