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.

(C_{g})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.