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.
Lemma 3.2
If
(11)
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
(12)
Next we set
Then we have
and so
hence
thus
consequently
(13)
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
which implies
Hence
and so
Consequently, we arrive at
Since is arbitrarily, we have
Hence we can complete the proof. □
To solve equation (2), we introduce an operator by
(14)
Lemma 3.5 The operator maps the order interval into itself.
Proof To achieve our aim, we only need to verify that and :
(15)
(16)
First we show equation (15):
Secondly we derive equation (16):
Since obviously, the operator S is strictly increasing in and if then , . Hence
so
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
with
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
(17)
for all , , then equation (2) has a unique solution in provided we have
and
(19)
where we set .
Proof For any we set and . Clearly, we have
for with and specified below, so . Then for any , we derive
which implies
consequently, we obtain
(20)
with
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 [[13], Theorem 5]. So here we give its correct proof.