In this section we give a more general existence result than Theorem 3.11 by assuming the existence of -lower and upper solutions. This makes us to deal with problem (1.1) and (1.2), where the function is singular at the end point and .
Let and be -lower and upper solutions of problem (1.1) and (1.2) such that on and let satisfy the following conditions:
(i)for almost every is continuous on ;
(ii)for any the function is measurable on ;
(iii)there exists a function such that, for all ,
Then problem (1.1) and (1.2) has at least one solution such that, for all ,
Consider the modified problem (3.3) and (1.2) with respect to the given and and define by (3.13). Note that by Lemma 2.2, is well defined. Define
and is defined by (3.17). The rest arguments are similar to the proof of Theorem 3.5.
We have similar results of Theorems 3.5–4.1, respectively, for (1.1) equipped with
where is a constant and , are given as (1.2).
Consider the problem (4.7), for , , ,
Clearly, is a -lower solution of (4.7) and
From Lemma 2.1, we have and define . Since, for ,
that is, , we have, from Lemma 2.2, and exists. Let
and, by Lemma 2.2 again, choose such that
Note that according to the direct computation, we see that is well-defined and is bounded by . Next, let . By Young's inequality, it follows that
Hence, such is a -upper solution of (4.7) and on . Clearly, satisfies (i), (ii) of Theorem 4.1. By using Young's inequality again, for , we have
and . Therefore, satisfies the assumption (iii) of Theorem 4.1. Consequently, we conclude that this problem has at least one solution such that, for all ,
Notice that in Theorem 4.1, one can only deal with the case that is singular at end points , . However, when is singular at , there is no hope to obtain the solutions directly from Theorem 4.1. We will establish the following theorem to deal with this case by constructing upper and lower solutions to solve this problem.
the function is continuous;
there exists and for any compact set , there is such that
for some and , there is such that
where is defined as in Lemma 2.1.
for any compact set , there is such that
Then problem (1.1) and (1.2) with has at least one solution
Remark 4.5 (see [12, Remark ]).
Assumption is equivalent to the assumption that there exists and a function such that:
(i) for all ,
(ii), for all , ,
(iii), for all ,
Construction of lower solutions. Consider such that and the function
where is chosen small enough so that
Next, we choose from the Remark 4.5, and let
where is small enough so that for some points , , we have:
Notice that by (4.24) and (4.25), for any such that
Approximation problems. We define for each , ,
We have that, for each index , is continuous and
Hence, the sequence of functions converges to uniformly on any set , where is an arbitrary compact subset of . Next we define
Each of the functions is a continuous function defined on , moreover
and the sequence converges to uniformly on the compact subsets of since
Define now a decreasing sequence such that
and consider a sequence of the following approximation problems:
A lower solution of ( ). It is clear that for any ,
As the sequence is decreasing, we also have
It follows from (4.25) and (4.27) that is a lower solution of ( ).
Existence of a solution of (4.7) such that
From assumption , we can find and such that, for all , ,
Also, one has
where is a suitable constant. Hence, we obtain, for such and ,
Let be a constant such that
Choose such that
where is defined by (2.1). Note that is well-defined and since . It is easy to see that
So by Remark 4.2, there is a solution of (4.7) such that
The problem ( ) has at least one solution such that
Notice that is an upper solution of ( ), since
Existence of a solution. Consider the pointwise limit
It is clear that, for any ,
and therefore on . Let be a compact interval. There is an index such that for all and therefore for these ,
Moreover, we have
By Arzelá-Ascoli theorem it is standard to conclude that is a solution of problem (1.1) and (1.2) on the interval . Since is arbitrary, we find that and for all ,
it remains only to check the continuity of at . This can be deduced from the continuity of and the fact that as .
Consider the following problem , for ,, ,
Let , where . Obviously, satisfies and . Moreover, for any given and for any compact set , for small enough, we have
Hence, holds. Furthermore, for large enough, , we have, from Young's inequality by choosing and ,
where . Hence, holds. By Theorem 4.4, has at least one solution