To prove the main result in this paper, we will employ several lemmas. These lemmas are based on the BVP
Lemma 2.1 If condition (H1) holds, then for , boundary value problem (2.1) and (2.2) has a unique solution
(2.3)
or
(2.4)
where
satisfies
(2.5)
Proof Suppose x is a solution of BVP (2.1), (2.2). Integrating (2.1) from 0 to t, we have
(2.6)
Integrating (2.6) from 0 to t, we get
or integrating the same equation from t to 1, we achieve
Using boundary condition (2.2), we get
(2.7)
or
(2.8)
where satisfies (2.5).
On the other hand, it is easy to verify that if x is the solution of (2.3) or (2.4), then x is a solution of BVP (2.1), (2.2). The proof is accomplished. □
Lemma 2.2 If is nonnegative on and on any subinterval of , then there exists a unique satisfying (2.5). Moreover, there is a unique such that .
Proof For any , define
So, is continuous and strictly increasing. It is easy to see that
Therefore there exists a unique satisfying (2.5). Furthermore, let
Then is continuous and strictly increasing on and , . Thus
implies that there exists a unique such that . Lemma is proved. □
Remark 2.1 By Lemmas 2.1 and 2.2, the unique solution of BVP (2.1), (2.2) can be rewritten in the form
(2.9)
Lemma 2.3 Let (H1) hold. If is nonnegative on and on any subinterval of , then the unique solution of BVP (2.1)-(2.2) has the following properties:
-
(i)
is concave on ,
-
(ii)
,
-
(iii)
there exists a unique such that ,
-
(iv)
.
Proof Suppose that is a solution of BVP (2.1)-(2.2), then
-
(i)
, is nonincreasing so is nonincreasing. This implies that is concave.
-
(ii)
We have and . Furthermore, we get
If we continue like this, we have
Using (H1), we obtain
which implies that . Similarly,
If we continue in this way, we attain that
Using (H1), we have , . Therefore, we get , .
-
(iii)
, imply that there is a such that .
If there exist , , such that , then
which is a contradiction.
-
(iv)
From Lemmas 2.1 and 2.2, we have . Hence we obtain that . This implies .
The lemma is proved. □
Lemma 2.4 If is symmetric nonnegative on and on any subinterval of , then the unique solution of (2.1), (2.2) is concave and symmetric with on .
Proof Clearly, is concave and from Lemma 2.3. We show that is symmetric on . For the symmetry of , it is easy to see that , i.e., . Therefore, from (2.9) and for , by the transformation , we have
Again, let . Then
So, is symmetric on . The proof is accomplished. □
Let . Then E is a Banach space with the norm
We define two cones by
Define the operator by
and as follows:
where . Obviously, x is a solution of BVP (2.1)-(2.2) if and only if x is a fixed point of the operator F.
Lemma 2.5 If (H1) holds, then for , where
(2.10)
Proof For , one arrives at
i.e., . Hence,
By , we get
Hence
i.e.,
The proof is finalized. □
From Lemma 2.5, we obtain
Lemma 2.6 Suppose that (H1)-(H3) hold, then is completely continuous.
Proof Let . According to the definition of T and Lemma 2.3, it follows that , which implies the concavity of on . On the other hand, from the definition of f and h, holds for , i.e., Tx is symmetric on . So, . By applying the Arzela-Ascoli theorem on time scales, we can obtain that is relatively compact. In view of the Lebesgue convergence theorem on time scales, it is obvious that T is continuous. Hence, is a completely continuous operator. The proof is completed. □