1. Introduction
We are concerned with the existence, nonexistence and multiplicity of positive solutions to the following problem
where
is a parameter,
is an odd increasing homeomorphism,
,
and
.
Problem (
1) arises naturally in studying radial solutions to the following equation
where
with
and
and
. Indeed, applying change of variables,
and
, we can transform (
2) into (
1) with
and
(see, e.g., [
1]).
Throughout this paper, unless otherwise stated, we assume that
satisfies the following hypothesis:
there exist increasing homeomorphisms
such that
Let
be an increasing homeomorphism. We denote by
the set
Remark 1. Assume that holds. Then it follows thatIndeed, implies for all Replacing x and y with and respectively, one has for all Similarly, it can be proved that Moreover, Indeed, by (3),which implies Clearly, For , we make the following notations:
For
with
, the existence of positive solutions to problem (
1) has been extensively studied in the literature for the past several decades (see References [
2,
3,
4,
5,
6,
7,
8,
9,
10,
11,
12,
13,
14,
15,
16,
17,
18] and references therein). For example, when
h is at most
as
for some
and
in Reference [
3], it was shown that there exists
such that (
1) has a positive solution for
and it has no positive solution for
The same result was obtained in Reference [
4] under the assumption that
h satisfies
for some
and
is a nondecreasing function satisfying, for some
for all
. When
, in Reference [
5], under the assumption that
and
it was shown that there exist
such that (
1) has at least two positive solutions for
, one positive solution for
and no positive solution for
. In Reference [
6], when
and
satisfies
and
, it was shown that
.
In Reference [
19], for an increasing homeomorphism
satisfying
there exist an increasing homeomorphism
and a function
such that
the same result as Reference [
5] was obtained when
and
. Moreover, if
is assumed, it was shown that
Thus the result of Reference [
19] extends the previous results of References [
3,
4,
5,
6] for
p-Laplacian problem to singularly weighted
-Laplacian one.
It looks like the assumption is more general than the assumption but it is not true. We point out that the assumption is equivalent to the assumption Indeed, let be an increasing homeomorphism satisfying the first inequality in the assumption . Define by and for Then is an increasing homeomorphism on For , and consequently Since the homeomorphism satisfying the second inequality in the assumption can be easily defined from , the assumption is no longer useful.
For more general
which does not satisfy
, in Reference [
20], when
and
with
, it was shown that (
1) has a positive solution
for all
satisfying
in
under some assumptions on
f which induces the sublinear nonlinearity if
with
. For other interesting results, we refer the reader to References [
21,
22,
23] and the references therein.
The concavity of solutions plays a crucial role in defining operators on a cone and using fixed point theorems (see, e.g., References [
2,
6,
19] and the references therein). It is well known that solutions to problem (
1) with
are concave functions on
. However, if
and
it is not obvious that the solutions to problem (
1) are concave functions on
. In Reference [
1], under the assumption that
d is nondecreasing on
, a lemma ([
1], Lemma 2.4) was proved from which a suitable positive cone was defined and various results for positive solutions to problem (
1) were proved.
However, the proof of the lemma ([
1], Lemma 2.4) is not clear. In the proof of it, the fact
is non-increasing on
is used. However it may not be true, since the fact
is nonincreasing on
does not imply that
is nonincreasing on
, even though
is non-decreasing on
(see Remark 2
). Consequently,
may not be nonincreasing on
.
In this paper, we show the existence of an unbounded solution component and prove the existence and nonexistence of positive solutions to problem (
1) under suitable assumptions on nonlinearity
Among other main results, we extend a result of Reference [
6] for
p-Laplacian problem to general
-Laplacian one (see Theorem 4 below). For that purpose, we prove a similar result to that of Reference [
1] (Lemma 2.4) under the weaker hypotheses to functions
g and
d (see Lemma 2 below). Also, the result (Theorem 4) extends that of Reference [
19] in some way, since we assume that
,
and
in it.
The rest of this paper is organized as follows. In
Section 2, a solution operator related to problem (
1) is introduced and some preliminaries are given. In
Section 3, the main results (Theorems 2–4) are proved and a few examples to illustrate the assumptions in the main results are given.
2. Preliminaries
First we give some notations which will be used in this paper.
The usual maximum norm in a Banach space is denoted by for , and let , and .
Define
to be a cone in
by
Now we introduce a solution operator related to problem (
1). Let
be fixed, and define a function
by
for
. Here
and
are functions defined by
We claim that
is a non-decreasing continuous function on
satisfying
and
Indeed, from the nonnegativity of
, it follows that
is non-decreasing on
and
By (
3),
Consequently, since
Finally, we prove the continuity of
on
Let
be fixed and let
be chosen so that
. Assume that
is a sequence in
satisfying
Let
for
Here
is the characteristic function of
that is,
for
and
for
Then
for each
and for all
By Lebesgue’s dominated convergence theorem,
is continuous at
Thus, the claim is proved.
Similarly, it can be shown that is a non-increasing continuous function on satisfying and . Then there exists an interval satisfying for all
Define a function
by
and, for
,
where
is a zero of
in
, that is,
We notice that, although
is not necessarily unique, the operator
T is well defined. Indeed, if
and
are zeroes of
in
then
for
in view of the monotonicity of
and
Consequently,
is independent of the choice of
(see, e.g., Reference [
1]).
For
consider the following problem
For
(
6) has a unique zero solution due to the boundary conditions.
Lemma 1. Assume that holds, and let u be a solution to problem (6) with . Then there exists a subinterval of such that , for and Moreover, is a unique solution to problem (6) and on Proof. Since
0 is not a solution to problem (
6). From the fact
for
it follows that
is continuous and non-increasing in
By the monotonicity of
, since
and
on
,
Consequently, there exists a subinterval
of
such that
for
and
Then, by the hypotheses on
and
,
for
and
Clearly,
is a solution to problem (
6) and
on
By directly integrating (
6), it can be shown that
is a unique solution to problem (
6). ☐
Remark 2. It is easy to show that if , is non-increasing and is non-decreasing on , then is non-increasing on . However, if is a sign-changing function on , it is not true that is non-decreasing on . For example, and with . Let and be a local maximum point and a local minimum point of , respectively. Note that since and Then is a positive increasing function on and is decreasing on . However, is decreasing on and is increasing on .
We notice that if we assume that c and d are non-decreasing on by Remark 2, it is easy to check that, for any solution u to problem (6), is non-increasing on which implies that u is a concave function on However, in general, u may not be a concave function on
Without the monotonicity of
we prove a result which is analogous to Reference [
1] (Lemma 2.4).
Lemma 2. Assume that hold and let be given. Then
for
Proof. For
0 is a unique solution to problem (
6) and there is nothing to prove.
Let
and
be a constant satisfying (
5), i.e.,
. By (
3), for
Similarly,
Recall that
and
. Let
for
Since
on
is non-increasing on
so that
is a concave function on
Consequently
for
and
for
Similarly,
for
and thus the proof is complete. ☐
By Lemmas 1 and 2, for each , and if and only if .
From now on, we assume Define a function by for and Clearly, for any .
Define an operator
by
for
, i.e., for
where
is a constant satisfying
To prove the complete continuity of the following lemma is needed.
Lemma 3. Assume that and hold. Let be given and let be a bounded sequence in with If or 1 as then and as for each Here, is a constant satisfying (8) with and . Proof. We only prove the case
as
, since the other case can be dealt in a similar manner. Since there exists
such that
for all
by (
3),
Consequently, it follows from
that
Since
is a constant satisfying (
8) with
and
which implies that, for all
,
as
. ☐
With Lemma 3, by the argument similar to those in the proof of [
2] (Lemma 3), it can be proved that
is completely continuous (see also Reference [
24], Lemma 3.3). So we omit the proof of it.
Lemma 4. Assume that and hold. Then the operator is completely continuous.
Finally, we present a well-known theorem for the existence of an unbounded solution component by Leray and Schauder [
25]:
Theorem 1. (see, e.g., Reference [
26], Corollary 14.12)
Let X be a Banach space with and let be a cone in Considerwhere and If is completely continuous and for all then there exists an unbounded solution component of (9) in emanating from . 3. Main Results
First, we make a list of assumptions on which will be used in this section.
for some
for any there exists a non-empty interval such that
for .
.
. Here is the homeomorphism in the assumption
there exist and a non-empty interval such that
for .
for all and
Remark 3. It is easy to see that (1) has a solution if and only if has a fixed point in Since for all , 0 is a unique solution to problem (1) with . Assume that for all Then 0 is a solution to problem (1) for any Assume that holds. Then 0 is not a solution to problem (1) with . Let u be a solution to problem (1) with . Then, by Lemma 1, u is a positive solution, i.e., for all
By Lemma 4, Theorem 1 and Remark 3, one has the following proposition.
Proposition 1. Assume that and hold. Then there exists an unbounded solution component emanating from in such that and for any u is a positive solution to problem (1) with Now we give a lemma which provides useful information about the solution component defined in Proposition 1.
Lemma 5. Assume that and hold. Let be a compact interval with . Then there exists such that for any positive solutions u to problem (1) with Proof. Let . Here
By
there exists
such that
for
Set
Assume to the contrary that there exists a sequence
such that
is a positive solution to problem (
1) with
and
as
Then, for sufficiently large
Let
be a constant satisfying
Assume
since the case
can be dealt in a similar manner. Then, by (
3),
which implies
This contradicts the choice of
☐
By similar arguments used to prove Lemma 5, one can prove the following result which shows the same property for the solution component For the convenience of readers, we give the proof of it.
Lemma 6. Assume that and hold. Let be a compact interval with . Then there exists such that for any positive solutions u to problem (1) with Proof. Let
. Here
By
there exists
such that
for
Set
Assume to the contrary that there exists a sequence
such that
is a positive solution to problem (
1) with
and
as
Then, for sufficiently large
Let
be a constant satisfying
Assume
since the case
can be dealt in a similar manner. Then
which contradicts the choice of
☐
We remark that the assumptions in Lemma 5 are different from ones in Lemma 6. Indeed, let
and
for
Then the first inequality in the assumption
is satisfied. Clearly,
implies
, since
for all
For
, but
. Consequently,
does not imply
. Since
we give an example of
h satisfying
Let
for
Note that
and
for
Then
, but
since
and
Now we give the first main result in this paper.
Theorem 2. Assume that and either and or and hold. Then for any there exists a positive solution to problem (1) such that and as . Moreover, if is assumed instead of , then as Here, is the solution component defined in Proposition 1. Proof. Let
Since
is unbounded in
by Lemma 5 or Lemma 6,
so that for any
there exists a positive solution
to problem (
1) such that
and
as
.
Next, we show that if
is assumed instead of
, then
as
Assume to the contrary that there exists a sequence
in
such that
as
but there exists
such that
for all
Then, by
, there exists
such that
for all
n and all
. For each
n, let
be a constant satisfying
and let
Suppose that
(the case
is similar). Then
which contradicts the fact that
for all
Here,
Thus, the proof is complete. ☐
Next we give a lemma about the
-direction block for positive solutions to problem (
1).
Lemma 7. Assume that and hold. Then there exists such that (1) has no positive solution for Proof. Let
u be a positive solution to problem (
1) with
and
. By
,
for
Let
We only consider the case
since the case
can be dealt in a similar manner. By Lemma 1,
for
and consequently,
for
Then, by (
3),
Here
. Consequently,
which completes the proof. ☐
Now we give the second main result in this paper.
Theorem 3. Assume that and hold. Then there exists such that (1) has at least one positive solution for and no positive solution for Proof. By Proposition 1, there exists at least one positive solution to problem (
1) for all small
Let
be a positive number such that (
1) has a positive solution
for
To complete the proof of Theorem 3, by Lemma 7, it suffices to show that (
1) has a positive solution for all
Let
be fixed, and consider the following modified problem
where
and
is defined by
Define
by
for
, where
for
and
Then it is easy to see that
u is a solution to problem (
10) if and only if
, and
is completely continuous on
. From the definition of
and the continuity of
f, it follows that there exists
such that
for all
Then, by Schauder fixed point theorem, there exists
such that
Consequently, by Lemma 1,
is a positive solution to problem (
10). We claim that
for all
If the claim is not true, since
there exists an interval
such that
for all
and
Then there exists
such that
i.e.,
By the definition of
and the fact
For
integrating (
12) from
t to
, we have
Integrating it from
and
again,
which contradicts (
11). Consequently, the claim is proved and
is a positive solution to problem (
1) by the definition of
Thus, the proof is complete. ☐
Next we give a lemma about
a priori estimates for solutions to problem (
1).
Lemma 8. Assume that and hold. Let with be given. Then there exists such that for any positive solutions u to problem (1) with Proof. Suppose to the contrary that there exists a sequence
such that
is a positive solution to problem (
1) with
and
as
Take
where
By
there exists
such that
for
Since
for all
For sufficiently large
Let
be a constant satisfying
We only consider the case
since the case
can be proved similarly. Since
for
which contradicts the choice of
and thus the proof is complete. ☐
Remark 4. Assume that holds. Since there exists such that for all By the positivity of there exists such thatConsequently, for all Thus implies . Now we give the third main result in this paper.
Theorem 4. Assume that and hold. Then there exists such that (1) has two positive solutions for , at least one positive solution for and no positive solution for Moreover, for , two positive solutions and can be chosen so that and as Proof. Let
(
1) has two positive solutions for all
Then, by Proposition 1, Lemmas 7 and 8,
is well-defined. Indeed, let
be a sequence in the unbounded solution component
defined in Proposition 1 satisfying
as
By Lemma 7,
and
as
Then, by Lemma 8,
as
Thus the shape of the continuum of
is determined. Consequently, (
1) has two positive solutions
for all small
such that
and
as
and it has no positive solution for all large
Thus
is well-defined.
By the choice of
(
1) has at least two positive solutions for
, and, by the complete continuity of
H and Lemma 8, it has at least one positive solution for
. By the same argument as in the proof of Reference [
6] (Theorem 1.1), (
1) has no positive solution for
and thus the proof is complete. ☐
Finally, we give a few examples which illustrates the assumptions in the main results.
Example 1. Let φ be an odd function satisfying for . It is easy to check that is satisfied for and . Let be a function defined by for It is easy to see that for any and for any .
Finally, we give some examples of satisfying the assumptions in the main results.
- (1)
Let for Clearly, and are satisfied.
- (2)
Let be any nonnegative continuous function satisfyingand Then is satisfied for and (resp., ) is satisfied.
- (3)
Let be any nonnegative continuous function satisfying Then is satisfied for and , but does not hold, since .
- (4)
Let or for Then is satisfied.