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

,

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

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

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
,

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

Step 2.

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:

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

,

,

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

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

,

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
.

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