In this section, we will deduce sufficient conditions for the existence of solutions in the sense of distributions to
, where
and
satisfy
and
,
satisfies
, and
satisfies the following condition.
(Cg)For every
, there exists
such that
Under these hypotheses, we will be able to approximate solutions in the sense of distributions to problem
by a sequence of weak solutions to weak problems.
First of all, we enunciate a useful property of absolutely continuous functions on
whose proof we omit because of its simplicity.
Lemma 2.1.
If
, then 
-a.e. on
.
We fix
a sequence of positive numbers strictly decreasing to zero; for every
, we define
as
Note that
satisfies
and
; consider the following modified weak problem
Definition 2.2.
is said to be a weak solution to
if
,
on
, and equality
holds for all
.
is said to be a weak lower solution to
if
on
, and inequality
holds for all
such that
on
.
The concept of weak upper solution to
is defined by reversing the previous inequality.
We remark that the density properties of the first-order Sobolev spaces proved in [9, Seccion 3.2] allows to assert that relations in Definition 2.2 are valid for all
and for all
such that
on
, respectively.
By standard arguments, we can prove the following result.
Proposition 2.3.
Assume that
satisfy
and
,
satisfies
, and
satisfies
.
Then, if for some
there exist
and
as a lower and an upper weak solution, respectively, to
such that
on
, then
has a weak solution
.
Next, we will deduce the existence of one solution in the sense of distributions to
from the existence of a sequence of weak solutions to
. In order to do this, we fix
two sequences such that
is strictly decreasing to
if
,
for all
if
and
is strictly increasing to
if
,
for all
if
. We denote that
,
. Moreover, we fix
a sequence of positive numbers strictly decreasing to zero such that
Proposition 2.4.
Suppose that
and
satisfy
and
,
satisfies
, and
satisfies
.
Then, if for every
,
is a weak solution to
and
then a subsequence of
converges pointwise in
to a solution in the sense of distributions
to
.
Proof.
Let
be arbitrary; we deduce, from (2.2), (2.7), (2.8), and (2.9), that there exists a constant
such that for all
,
Therefore, for all
so large that
, as
is a weak solution to
, by taking
as the test function in(2.5), from (2.9),
and
, we can assert that there exists
such that
that is,
is bounded in
and hence, there exists a subsequence
which converges weakly in
and strongly in
to some
.
For every
, by considering for each
the weak solution to 
and by repeating the previous construction, we obtain a sequence
which converges weakly in
and strongly in
to some
with
. By definition, we know that for all
,
.
Let
be given by
on
for all
and
so that
on
,
,
is continuous in every isolated point of the boundary of
, and
converges pointwise in
to
.
We will show that
; we only have to prove that
is continuous in every dense point of the boundary of
. Let
be arbitrary, it follows from
and
that there exist
such that
on
and
for
-a.e.
and all
; let
be the weak solution to
we know (see [4]) that
on
.
For all
so large that
, since
and
are weak solutions to some problems, by taking
as the test function in their respective problems, we obtain
thus, (2.2) yields to
which implies that
on
and so
on
. Thereby, the continuity of
in every dense point of the boundary of
and the arbitrariness of
guarantee that
.
Finally, we will see that(1.11) holds for every test function
; fix one of them.
For all
so large that
and all
so large that
, as
is a weak solution to
, by taking
as the test function in (2.5) and bearing in mind(2.7), we have
whence it follows, by taking limits, that
which is equivalent because
and
on
to
and the proof is therefore complete.
Propositions 2.3 and 2.4 lead to the following sufficient condition for the existence of at least one solution in the sense of distributions to problem
.
Corollary 2.5.
Let
be such that
satisfy
and
,
satisfies
and
satisfies
.
Then, if for each
there exist
and
a lower and an upper weak solution, respectively, to
such that
on
and
then
has a solution in the sense of distributions
.
Finally, fixed
is a solution in the sense of distributions to
with
, we will derive the existence of a second solution in the sense of distributions to
greater than or equal to
on
. For every
, consider the weak problem
For every
, consider
as a subspace of
by defining it for every
as
on
and define the functional
for every
as
where function
is defined for
-a.e.
and all
as
As a consequence of Lemma 2.1, we deduce that every weak solution to
is nonnegative on
and by reasoning as in [4, Section 3], one can prove that
is weakly lower semicontinuous,
is continuously differentiable in
, for every
,
and weak solutions to
match up to the critical points of
.
Next, we will assume the following condition.
(NI) For
-a.e.
,
is nonincreasing on
.
Proposition 2.6.
Suppose that
is such that
satisfy
and
,
satisfies
and
, and
satisfies
.
If
,
is a bounded sequence in
such that
then
has a subsequence convergent pointwise in
to a nontrivial function
such that
in
and
is a solution in the sense of distributions to
.
Proof.
Since
is bounded in
, it has a subsequence which converges weakly in
and strongly in
to some
.
For every
, by (2.2), we obtain
which implies, from(2.23), that
on
and so
on
.
In order to show that
is a solution in the sense of distributions to
, fix
arbitrary and choose
so large that
, bearing in mind that
is a solution in the sense of distributions to
, and the pass to the limit in(2.22) with
and
yields to
thus,
is a solution in the sense of distributions to
.
Finally, we will see that
is not the trivial function; suppose that
on
. Condition
ensures that function
defined in (2.21) satisfies for every
and
-a.e.
,
so that, by(2.20) and(2.22), we have, for every
,
moreover, as we know that
on
for some
, it follows from
that there exists
such that
and hence, since
is bounded in
and converges pointwise in
to the trivial function
, we deduce, from the second relation in(2.23) and(2.24), that
which contradicts the first relation in(2.23). Therefore,
is a nontrivial function.