In this section, we will use the fixed point method to prove the existence and uniqueness of nontrivial solution for equation (2) in an order interval.
For all , we introduce the following functions:
where and are defined in equation (8) or (9).
Remark 3.1 Note that and . Next, .
The following result is clear.
then for all . Consequently, the order interval is well defined.
Remark 3.3 If , , and then equation (11) reads
which is satisfied, since equation (3) implies , (see equation (4)) and clearly . This case occurs for instance when and with and , .
From now on, we suppose that all above assumptions hold: equations (3), (4), (i)-(iii), and (11).
Lemma 3.4 Any solution of equation (2), with for all , satisfies .
Proof Step 1: We prove that for a solution x of equation (2).
Set . Then we obtain
which implies that
Next we set
Then we have
Since implies , estimate (13) is an improvement of equation (12).
Step 2: We prove that . Fix and set
for . Then like above, for , we get
Consequently, we arrive at
Since is arbitrarily, we have
Hence we can complete the proof. □
To solve equation (2), we introduce an operator by
Lemma 3.5 The operator maps the order interval into itself.
Proof To achieve our aim, we only need to verify that and :
First we show equation (15):
Secondly we derive equation (16):
Since obviously, the operator S is strictly increasing in and if then , . Hence
Consequently, is well defined and . The proof is completed. □
From the Arzela-Ascoli theorem and since is nondecreasing, it follows that is compact, so the Schauder fixed point theorem implies the following existence result [35, 38, 39].
Theorem 3.6 Equation (2) has a solution in . Moreover,
are fixed points of
Now we are ready to state the following uniqueness result. But first we note that the above considerations can be repeated for any , so we get , , , , , and as continuous functions of . Note is nonincreasing, is nondecreasing, and , can be continuously extended to . Then . We still keep the notation , , , and .
Theorem 3.7 If there are constants ψ, χ and continuous functions and on such that
for all , , then equation (2) has a unique solution in provided we have
where we set .
Proof For any we set and . Clearly, we have
for with and specified below, so . Then for any , we derive
consequently, we obtain
Since (note equation (18))
and as , we see that
for any sufficiently large uniformly for any . So we take and fix such a ϖ. Next, by equation (19) there is a so that
for any . Furthermore, for , we have (note equation (18))
for any sufficiently large, so we fix such . Consequently we get
Summarizing we see that there is and so that
This shows that is a contraction with respect to the norm with a constant L. By the contraction mapping principle, one can obtain the result immediately. □
Remark 3.8 Consider equation (5). Of course, we can suppose . Then , , , and . Moreover, Remark 3.3 can be applied to get an existence result. If in addition then , and it is not difficult to see that , , , and
Then , so equation (18) holds. Next, we derive
Hence condition (19) is satisfied and then we get a uniqueness result by Theorem 3.7. Note there is gap in the proof of [, Theorem 5]. So here we give its correct proof.