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 .

Theorem 4.1.

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 ,

where

Then problem (1.1) and (1.2) has at least one solution such that, for all ,

Proof.

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

where

and is defined by (3.17). The rest arguments are similar to the proof of Theorem 3.5.

Remark 4.2.

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).

Example 4.3.

Consider the problem (4.7), for , , ,

Clearly, is a -lower solution of (4.7) and

where

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.

Theorem 4.4.

Assume

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 ,

where

Proof.

Step 1.

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

we have:

Step 2.

Approximation problems. We define for each , ,

and set

We have that, for each index , is continuous and

where

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:

where .

Step 3.

A lower solution of ( ). It is clear that for any ,

As the sequence is decreasing, we also have

Clearly, satisfies

It follows from (4.25) and (4.27) that is a lower solution of ( ).

Step 4.

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

that is,

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

Step 5.

The problem ( ) has at least one solution such that

Notice that is an upper solution of ( ), since

Step 6.

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 ,

Since

it remains only to check the continuity of at . This can be deduced from the continuity of and the fact that as .

Example 4.6.

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