In the presence of an ordered pair of lower and upper functions, the existence and location results for problems (1.1) and (1.2) can be obtained.

Theorem 3.1.

Suppose that there exist lower and upper functions

and

of problems (1.1) and (1.2), respectively. Let

be a continuous function satisfying the one-sided Nagumo-type conditions (2.3) and (2.4) in

If

verifies

for

and

where

means

and

, then problems (1.1) and (1.2) has at least one solution

satisfying

for
.

Proof.

Define the auxiliary functions

For

, consider the homotopic equation

with the boundary conditions

Take

large enough such that, for every

,

Step 1.

Every solution

of problems (3.6) and (3.7) satisfies

for
, for some
independent of
.

Assume, for contradiction, that the above estimate does not hold for

. So there exist

,

, and a solution

of (3.6) and (3.7) such that

. In the case

define

If

, then

and

. Then, by (3.2) and (3.10), for

, the following contradiction is obtained:

For

,

If

, then

and

. If

, then

and so

. Therefore, the above computations with

replaced by 0 yield a contradiction. For

, by (3.11), we get the following contradiction:

The case
is analogous. Thus,
for every
. In a similar way, we may prove that
for every
.

By the boundary condition (3.7) there exists a

, such that

. Then by integration we obtain

Step 2.

There is an

such that for every solution

of problems (3.6) and (3.7)

with
independent of
.

and for

the function

given by

In the following we will prove that the function

satisfies the one-sided Nagumo-type conditions (2.3) and (2.4) in

independently of

. Indeed, as

verifies (2.3) in

, then

So, defining

in

, we see that

verifies (2.3) with

and

replaced by

and

, respectively. The condition (2.4) is also verified since

Therefore,
satisfies the one-sided Nagumo-type condition in
with
replaced by
, with
independent of
.

every solution

of (3.6) and (3.7) satisfies

The hypotheses of Lemma 2.4 are satisfied with
replaced by
. So there exists an
, depending on
and
, such that
for every
. As
and
do not depend on
, we see that
is maybe independent of
.

Step 3.

For
, the problems (3.6) and (3.7) has at least one solution
.

and for

,

by

Observe that

has a compact inverse. Therefore, we can consider the completely continuous operator

For

given by Step 2, take the set

By Steps 1 and 2, degree

is well defined for every

and by the invariance with respect to a homotopy

The equation

is equivalent to the problem

and has only the trivial solution. Then, by the degree theory,

So the equation

has at least one solution, and therefore the equivalent problem

has at least one solution
in
.

Step 4.

The function
is a solution of the problems (1.1) and (1.2).

The proof will be finished if the above function

satisfies the inequalities

Assume, for contradiction, that there is a

such that

, and define

If

, then

and

. Therefore, by (3.2) and Definition 2.1, we obtain the contradiction

If

, then we have

By Definition 2.1 this yields a contradiction

Then

and, by similar arguments, we prove that

. Thus,

Using an analogous technique, it can be deduced that

for every

. So we have

On the other hand, by (1.2),

Applying the same technique, we have

and then by Definition 2.1 (iii), (3.44) and (3.46), we obtain

Since, by (3.44),

is nondecreasing, we have by (3.49)

and, therefore,

for every

. By the monotonicity of

,

and so
for every
.

The inequalities
and
for every
can be proved in the same way. Then
is a solution of problems (1.1) and (1.2).