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