In this section we consider the auxiliary singular differential equation,
where . For and , (3) becomes a special case of (1) and therefore, results obtained for (3) apply for (1). For the further analysis, we assume that f satisfies the following conditions:
(H2) for a.e. and all ,
(H3) is increasing in x for a.e. and
We now study (3) subject to the boundary conditions
and require that
holds. We call a function a positive solution of the Dirichlet problem (3), (4) if , on , u satisfies the boundary conditions (4), and (3) holds for a.e. . Clearly, for each positive solution u of (3), (4), there exists such that (5) is satisfied.
We now denote by the set of all positive solutions of problem (3), (4), and , where .
In the following lemma we cite those results from  which will be used in the analysis of problem (1), (2).
Lemma 1 Let (H1)-(H3) hold. Then the following statements hold:
For each the set is nonempty and there exist functions such that for and .
If , , , then for .
If , , , and for some , then either for or there exists such that for and for .
For each and each there exists satisfying .
is a one-point set for each , where is at most countable.
For each , the set is compact in .
If , then if and only if it is a solution of the equation
in the set .
We now formulate new results which complete those from . We first establish a relation between and the set if its interior is nonempty. This question was a short time ago an open problem [, Remark 4.4]. We note that the relation between and the set with having nonempty interior is described in Lemma 1(d).
Theorem 1 Let (H1)-(H3) hold. Let us assume that there exists such that . Then for any there exists satisfying .
Proof Step 1. Auxiliary Dirichlet problem.
Choose . Consider (3) subject to the Dirichlet conditions
We claim that there exists a solution v to problem (3), (7) such that
We show this result utilizing the method of lower and upper functions. It follows from (H1) that there exists such that
and let be given as
We now consider the auxiliary differential equation
It is not difficult to verify that for . Hence, cf. (9),
for a.e. and all , . Since and for , , , and , solve (3) on , we conclude that and are lower and upper functions of problem (3), (7) (see e.g.  or ). This fact together with (11) implies the existence of a solution v to problem (3), (7) satisfying (8), cf. [, Lemma 3.7]. Moreover,
and taking the limit we obtaina , which together with gives .
We now prove that on . The proof is indirect. Let us assume that there exists such that and for . Then on this interval, and therefore
From (8) and (9) we conclude that
By the Gronwall lemma,
Consequently, for , and so = 1 on this interval. Thus, v is a solution of problem (3), (7).
Step 2. Continuation of the solution v.
It follows from the arguments given in Step 1 that v is a solution of problem (3), (7) on , and because . It is easy to verify that the equality
is satisfied for a.e. . We now integrate the last equality two times over and have (note that )
Let u be a solution of problem (3), (7) on an interval J which is a left-continuation of v. Let us assume that u is not continuable. Let . Then and (12) with v replaced by u holds for a.e. . The integration now yields
and we claim that
We show inequality (14) indirectly. Let us assume that (14) does not hold. Then there exists such that
and either or . Assume that .
Since, by (H3), for a.e. and for a.e. , we have
which is not possible. The case can be discussed analogously. Hence, (14) holds.
Suppose that . Then and since u is bounded on , it follows that . The integration of the equality
Since for a.e. , where , we have
Using the Gronwall lemma, we deduce that for ,
holds. This is a contradiction.
Therefore , and then (14) yields . Consequently , and the assertion of the theorem follows from (13). □
In the next corollary we extend the statement (d) from Lemma 1 to a large set of A values.
Corollary 1 For each and each there exists satisfying .
Proof The result follows immediately from Lemma 1(d) and Theorem 1. □
Remark 1 Corollary 1 says that the set is covered by the graphs of functions from , that is,
By Lemma 1(b), (c) we know that functions from are uniquely determined by the values −c of their derivatives at the right end point only in the case that is a singleton set for each . Since we cannot uniquely determine all functions from via their derivatives at , see Lemma 1(e), we discuss their derivatives at the singular point .
Lemma 2 Let (H1)-(H3) hold. Let , . Then
Proof It follows from Lemma 1(g) that (6) holds for . Since , we have (note that )
and (15) follows. □
Corollary 2 Let . Then
Proof The result follows from Lemma 2, since . □
Corollary 3 Let , . Then .
Proof Since , there exists such that . We can assume that for instance . Then on and on by Lemma 1(b)(c). Hence, and
which together with Corollary 2 gives . □
Corollary 4 Let , . Then
In particular, .
Proof Inequality (16) follows from Lemma 2 and the fact that for a.e. by (H2). □
Corollary 5 Let . Then
Proof The integration of (3) over gives
Using integration by parts we obtain
Taking the limit , we have
On the other hand, it follows from Corollary 2 that
Combining the above two equalities yields the result. □
Lemma 3 Let (H1)-(H3) hold. Let and be not a singleton set. Then for each there exists a unique such that . Consequently, .
Proof Let , . Then it follows from Lemma 1(a) and (H3) that
Hence, by Lemma 2,
Since is a compact set in by Lemma 1(f), the set is closed. In fact, let , , where , and let . Then there exist a subsequence of and such that in . In particular, as . Therefore, .
It remains to prove that . Assume that the equality does not hold. Then, from the structure of bounded and closed subsets of ℝ the existence of an open interval , , follows. Let and , where . Due to Lemma 1(c), there exists such that on , on . Choose . By Corollary 1, there exists such that . Then on and on . Consequently, on , that is, , which is not possible. □
Since functions from belong to , we have for each . Corollary 4 yields . Let us denote
Lemma 3 implies that functions from can be uniquely determined by the values of their derivatives at the singular point ; see Theorem 2.
Theorem 2 Let (H1)-(H3) hold. Then there exists a unique satisfying if and only if .
Proof We first show
It follows from Lemma 3 that
for each , where we set if is a singleton set and . In view of Corollary 4, for . Consequently, (18) follows.
Let us now choose . Then there exists satisfying . The uniqueness of u follows from Corollary 3. □
For , we denote by the unique element of satisfying .
Theorem 3 Let (H1)-(H3) hold. Assume that is a convergent sequence and . Then in .
Proof Let . Then because for by Corollary 4. Since and is compact in by Lemma 1(f), the sequence is relatively compact in . Let be a subsequence of which is convergent in , and let . Then for a and . Therefore, and hence any subsequence of converging in has the same limit . Consequently, is convergent in and is its limit. □