1. Introduction
We consider the following singular
-Laplacian problem with integral boundary conditions
Here, is an odd increasing homeomorphism, is a continuous function, is a parameter, is a continuous function, is a continuous function satisfying for , and the integrator functions () are nondecreasing on .
All integrals in (
1) are meant in the sense of Riemann–Stieltjes. Throughout this paper, we assume the following hypotheses.
- (K1)
There exist increasing homeomorphisms
such that
- (K2)
For .
Let
be an increasing homeomorphism. Then, we denote by
the set
As is well known, it follows from
that
and
(see, e.g., Remark 1 in [
1]).
Since Picone’s pioneering work [
2] in 1908, the study of nonlocal boundary value problems has attracted numerous researchers, who have been actively involved in this area from then until now. Nonlocal boundary value problems appear in various fields of applied mathematics and physics, modeling numerous phenomena across the applied mathematical sciences. These applications include, but are not limited to, beam deflection [
3] and thermostatics [
4]. A classic example is the vibration of a guy wire with a uniform cross-section composed of N sections of varying densities, which can be formulated as a multi-point boundary value problem [
5]. Furthermore, many elastic stability problems also give rise to multi-point boundary value problems [
6]. To gain a comprehensive understanding of the historical development of research in this area, we refer the reader to the survey papers [
7,
8]. Driven by these applications, the existence of positive solutions for nonlocal boundary value problems has been extensively investigated. For instance, Bachouche, Djebali, and Moussaoui [
9] employed fixed-point theorems to establish the existence of multiple positive solutions for
-Laplacian boundary value problems with linear bounded operator conditions under suitable assumptions on the nonlinearity
satisfying the
-Carathéodory condition. Goodrich [
10] utilized perturbed Volterra integral operator equations to study the existence of positive solutions for the
r-Laplacian differential equation with nonlocal boundary conditions. For the nonlinearity
satisfying
, Kim [
11] investigated the existence, nonexistence, and multiplicity of positive solutions to problem (
1) by analyzing the unbounded solution continuum. Tariboon, Samadi, and Ntouyas [
12] studied the existence and uniqueness of solutions for boundary value problems involving Hilfer generalized proportional fractional differential equations with multi-point boundary conditions, notably pioneering the investigation of such problems with order in (1,2]. For a two-term nonlinear fractional integro-differential equation with nonlocal boundary conditions and variable coefficients, Li [
13] established the uniqueness of solutions by utilizing the Mittag–Leffler function, Babenko’s method, and the Banach contraction principle.
The existence of positive solutions for the
-Laplacian problem (
1) in the case when either
or
has been investigated by several authors. Here,
For instance, Agarwal, Lü, and O’Regan [
14] established the existence of two positive solutions to problem (
1) with
for some
,
,
, and
. Subsequent studies by Wang [
15] and Lee and Xu [
16] extended the result to more general cases. Recently, Kim obtained the following result for problem (
1).
Theorem 1 ([
17], Theorem 1).
Assume that , , and hold.- (1)
If then there exists such that (
1)
has two positive solutions for any .
- (2)
If then there exist such that problem (
1)
has two positive solutions for any .
A limitation of Theorem 1 is its inability to determine whether positive solutions exist for all possible positive values of
. Specifically, the result does not cover the intervals
or
. Recently, Jeong and Kim [
18] thoroughly analyzed the existence, nonexistence, and multiplicity of positive solutions under the restrictive condition of zero Dirichlet boundary conditions (i.e.,
). We aim to extend the result as in [
18], as well as those presented in [
14,
15,
16,
17], by establishing the existence of positive solutions for a wider range of conditions. Specifically, we consider all positive values of the parameter
and impose less restrictive assumptions on
,
p,
h, and/or boundary conditions. Our main theorem is presented below.
Theorem 2. Assume that , , and hold.
(1)
If then there exist such that problem (
1)
has two positive solutions for , one positive solution for , and no positive solutions for .
(2)
If then there exist such that problem (
1)
has two positive solutions for , one positive solution for , and no positive solutions for .
In [
11], the condition
for all
was imposed on the nonlinearity
to ensure that all non-negative solutions are positive ones. However, in our study, we relax this restriction by allowing
to satisfy
, which introduces the possibility of a trivial solution
for every
. To address the challenges posed by this relaxed assumption, we employ a combination of the fixed-point theorem of cone expansion and compression of norm type and the Leray–Schauder fixed-point theorem.
The subsequent sections of this paper are organized as follows. In
Section 2, we provide a brief overview of existing results, laying the groundwork for the main theorem. In
Section 3, we establish key lemmas and present the proof of Theorem 2, accompanied by illustrative examples. Finally, in
Section 4, we conclude by summarizing our principal findings and highlighting potential directions for future research.
2. Preliminaries
Throughout this section, we assume that
,
, and
hold. The usual maximum norm in a Banach space
is denoted by
Let
Then, since
is a continuous function with
, we have two cases: either
or
Hence,
Let
. Here,
Define
as the set of all non-negative continuous functions
u satisfying
Then,
is a cone in
For
let
and
.
For
, consider the following problem:
Define a function
by, for
,
where
and
is a constant satisfying
From the definition of
, it follows that
For any
and any
satisfying (
7),
is monotone increasing on
and monotone decreasing on
. Note that
is not necessarily unique, but
is independent of the choice of
satisfying (
7) (see Remark 2 in [
19]).
Lemma 1 (Lemma 2 in [
19]).
Assume that , , and hold. Then, is a unique solution to problem (
5)
satisfying the following properties:(1) for ;
(2) for any , ;
(3)
σ is a constant satisfying (
7)
if and only if ;
(4) for and .
Define a function
by
Clearly, because
,
for any
. Let us define an operator
by
By Lemma 1
,
, and consequently,
T is well defined.
Remark 1. (1)
Problem (
1)
has a solution if and only if for some .
(2)
From for any , it follows that 0
is a unique solution to problem (
1)
with .
(3)
For , by Lemma 1 (3),
(4)
By Lemma 1 (4),
if u is a nonzero solution to problem (
1)
with , then u is a positive solution, i.e., for .
Lemma 2 ([
19], Lemma 4).
Assume that , , and hold. Then, the operator is completely continuous. Finally, we introduce the fixed-point theorem of cone expansion and compression of norm type and the Leray–Schauder fixed-point theorem.
Theorem 3 ([
20]).
Let be a Banach space, and let be a cone in Assume that and are open subsets of with and Let be a completely continuous operator such that if either for and for or
for and for ,
then H has a fixed point in .
Theorem 4 ([
21]).
Let X be a Banach space, and let be a closed, convex, and bounded set in X. Assume that is completely continuous. Then, V has a fixed point in .
3. Proof of Main Results
Lemma 3. Assume that , , and hold. Let be a compact interval with . Then there exist positive constants and such that for any positive solution u to problem (
1)
with . Proof. Let
. Here,
and
First, we show the existence of
satisfying
for any positive solution
u to problem (
1) with
In contrast, we assume that there exists a sequence
such that
is a positive solution to problem (
1) with
and
as
Since
there exists
such that
for
. Since
as
there exists
such that
and
We restrict our attention to the case where
, because the case where
can be treated analogously. From the definition of
T and (
8), it follows that
Then, by (
3) and (
9) and the definition of
A,
This is a contradiction, and thus there exists
such that
for any positive solution
u to problem (
1) with
Next, we show the existence of
satisfying
for any positive solution
u to problem (
1) with
It follows from
that there exists
such that
for
Let
Then,
In contrast, we assume that there exists a sequence
such that
is a positive solution to problem (
1) with
and
as
Then, for sufficiently large
, and by (
11),
Let
denote a positive real number such that
We restrict our attention to the case where
because the case where
can be treated analogously. From (
8), it follows that
Then, by (
3) and (
12) and the definition of
A,
This is a contradiction, and thus there exists
such that
for any positive solution
u to problem (
1) with
□
Lemma 4. Assume that , , and hold. Then, there exists such that problem (
1)
has no positive solutions for . Proof. Let
be a positive constant satisfying that problem (
1) has a positive solution
, and let
be a constant satisfying
. Since
, there exists
such that
and
Let
We only consider the case
since the case
can be proved similarly. By the similar argument as in (
3), (
13), and (
14),
Consequently,
and problem (
1) has no positive solutions for
□
Lemma 5. Assume that , , and hold. If (
1)
has a positive solution at then (
1)
has at least one positive solution for all . Proof. Let
u1 be a positive solution to problem (
1) with
and let
be fixed. Consider the following modified problem
where
is a continuous function defined by, for
,
Define
by
for
, where
for
and
Since
for any
, by Lemma 1,
is well defined. It is easy to see that
is completely continuous on
, and
u is a solution to problem (
15) if and only if
First, we show the existence of a solution to problem (
15).
- (i)
Assume that
f is bounded on
From the definition of
and the continuity of
f, it follows that there exists
such that
for all
and
Then, by Theorem 4, there exists
such that
and consequently, problem (
15) has a non-negative solution
u.
- (ii)
Assume that f is unbounded on Let be given. Here, is the constant in the proof of Lemma 3. Since there exists such that
Since
f is unbounded on
and
for
, there exists
such that
and
Let
be given. Then, by (
16) and (
17),
Let
denote a positive constant satisfying
We restrict our attention to the case where
, since the case where
can be treated analogously. Then, by the similar argument as in (
10),
From (
3) and (
18) and the choice of
, it follows that
By Theorem 4, there exists
such that
and consequently, problem (
15) has a non-negative solution
u.
Finally, we prove that if
u is a non-egative solution to problem (
15), then
for
. If it is true, by the definition of
, we can conclude that problem (
1) has a positive solution
u for all
and thus, the proof is complete.
Assume, on the contrary, that there exists a solution
u to problem (
15) such that
for
. Let
for
. We claim that
. If not,
and there exists
such that
. Then,
which imply that
and there exists
such that
Since
for
For
integrating (
19) from
t to
,
. Since
for all
and
is increasing on
,
Integrating (
20) from
to
,
which contradicts the choice of
and
. Consequently,
. Similarly, we can show that
.
Since
,
and
for
, there exists a subinterval
such that
for
and
From the fact
it follows that there exists
such that
and
. For
i.e.,
For
integrating (
21) from
t to
,
. Since
for all
and
is increasing on
,
Integrating (
22) from
to
,
which contradicts the choice of
Thus, the proof is complete. □
Lemma 6. Assume that , , and hold. Let be a compact interval with . Then, there exist positive constants and such that for any positive solution u to problem (
1)
with . Proof. Let
be given. Here,
Recall
and note that
by (
4).
First, we show the existence of
satisfying
for any positive solution
u to problem (
1) with
By contradiction, we assume that there exists a sequence
such that
is a positive solution to problem (
1) with
and
as
Since
there exists
such that
for
. Since
as
there exists
such that
and
We restrict our attention to the case where
because the case where
can be treated analogously. Since
for
by (
23),
By Lemma 1 (1),
and thus, by (
3) and (
24) and the definition of
,
which is a contradiction. Thus, there exists
such that
for any positive solution
u to problem (
1) with
Next, we show the existence of
satisfying
for any positive solution
u to problem (
1) with
By contradiction, we assume that there exists a sequence
such that
is a positive solution to problem (
1) with
and
as
By
there exists
such that
for
For all
, and
. For sufficiently large
and
. Thus,
By the same reasoning as above, we can easily see that the choice of
leads a contradiction. Thus, there exists
such that
for any positive solution
u to problem (
1) with
□
Lemma 7. Assume that , , and hold. Then, there exists such that problem (
1)
has no positive solutions for . Proof. Let
be a positive constant satisfying that problem (
1) has a positive solution
, and let
be a constant satisfying
. Since
, there exists
such that
for
We only consider the case
since the case
can be proved similarly. Since
for
Then, by (
3),
Here,
Consequently,
and we can conclude that problem (
1) has no positive solutions for
□
Lemma 8. Assume that , , and hold. If (
1)
has a positive solution at then (
1)
has at least one positive solution for all . Proof. Let
be a positive solution to problem (
1) with
and let
be fixed. Consider the following modified problem
where
is a continuous function defined by, for
,
Define
by
for
, where
for
and
Since
for any
, by Lemma 1,
is well defined. It is easy to see that
is completely continuous on
, and
u is a solution to problem (
25) if and only if
By the definition of
for all
and there exists
such that
for all
which implies
Let
. For
,
, and
Let
be fixed. Then, there exists
such that
Take
, and let
be given. Then
and
Let
denote a positive real number such that
We have two cases: either
or
. We restrict our attention to the case where
, since the case where
can be treated similarly. By (
3) and (
27), it follows from
that
which implies, by the choice of
,
By (
26) and (
28), in view of Theorem 3, problem (
25) has a non-negative solution
with
By Lemma 1
,
u is a positive solution to problem (
25).
By similar reasoning as that in the proof of Lemma 5, we can show that if
u is a positive solution to problem (
25), then
for
. Thus, by the definition of
, we can conclude that problem (
1) has a positive solution
u for all
. □
For the sake of completeness, we provide a proof of Theorem 2, which follows similar arguments to those given for Theorem 2 in [
18].
Proof of Theorem 2. (1) Let
has at least two positive solutions for
and
has at least one positive solution }. By Theorem 1
and Lemma 4,
and
are well defined and
From Lemma 5, it follows that problem (
1) has two positive solutions for
, one positive solution for
, and no positive solutions for
. To complete the proof, it is enough to show that problem (
1) has a positive solution for
By the definition of
, there exists a sequence
such that
and
is a positive solution to problem (
1) with
. Then,
as
, and by Lemma 3, there exist positive constants
c and
C such that
for all
Since
is bounded and
is compact, there exist a subsequence
of
and
such that
as
. Since
for all
n,
as
. In view of the continuity of
T,
Because
, for all
. Thus, by Lemma 1
, (
1) has a positive solution
for
(2) Let
has at least two positive solutions for
and
has at least one positive solution }. By Theorem 1
and Lemma 7,
and
are well defined and
From Lemma 8, it follows that problem (
1) has two positive solutions for
, one positive solution for
, and no positive solutions for
. To complete the proof, it is enough to show that problem (
1) has a positive solution for
By the same reasoning as that in the proof of Theorem 2 (1), we can prove it, and thus, the proof is complete. □
We conclude this section by presenting examples to illustrate Theorem 2.
Example 1. Consider the following problemwhere is an odd increasing homeomorphism defined by is a continuous function defined byandIt is evident that and are both equal to , which lies within the interval , and thus, the assumption holds. Moreover, by defining functionswe can verify that the assumption is also met (see, e.g., [
1]
). Since the inverse function , denoted by is given by for it follows that for any . Let and be continuous functions on defined byHere, ,
,
and are fixed constants. Then,As a consequence of Theorem 2, we can conclude that there exist positive constants and such that problem (
29)
with has two positive solutions for , one positive solution for , and no positive solutions for , and problem (
29)
with has two positive solutions for , one positive solution for , and no positive solutions for .