1. Introduction
The aim of this paper is to study quasilinear problems driven by mixed operators with fractional Neumann boundary conditions. More precisely, we consider the operator
Here
,
,
and
is the fractional
p-Laplacian, namely
The constant
, defined as
is the usual normalization constant for
(see [
28] for more details), but its value will not play a role in our analysis.
The operator in (
1.1) acts on an bounded open subset
with smooth boundary of class
.
Mixed operators of the form (
1.1) have raised an increasing interest in recent years, mainly in the Hilbert setting (namely, when
), but also the quasilinear case has been object of several results about existence, multiplicity and qualitative properties of solutions, both in the elliptic and in the parabolic case. The list of references being huge, we just quote some very recent ones and their related bibliography, [
5,
6,
7,
15,
16,
20,
22]. We also refer to [
10,
11], where a more general superposition of local and nonlocal operators is considered, and to [
12,
13], where Neumann conditions for such a framework are introduced.
As for the boundary conditions, we combine the operator in (
1.1) with the so-called
-Neumann conditions, introduced in [
14,
15] for
and in [
27] for general
p. These conditions are made of two contributions. The first one corresponds to the local part and is defined on
, while the second corresponds to the nonlocal part and is defined on
, namely
Here,
is the
nonlocal normal derivative, or
Neumann boundary condition and describes the natural Neumann boundary condition in presence of the fractional
Laplacian. It was introduced in [
4,
26] as an extension of the notion of nonlocal normal derivative introduced in [
18] for the fractional Laplacian, i.e. for
. Also nonlocal Neumann conditions have been treated recently, for instance in [
1,
2,
3,
8,
19,
24,
25].
From (
1.3) it is clear that the name
-Neumann conditions is related to the fact that they consider both the local part and the nonlocal part of the operator in (
1.1).
From now on, for the sake of simplicity, we will set .
In this framework, we are intersted in dealing with problems of the type
More precisely, we start in
Section 2 giving the definition of the suitable functional space to study problems such as (
1.5). We also recall the nonlocal counterpart of the divergence theorem and the integration by parts formula stated in [
4,
26]. However, we give these results in a more general setting, and we also give an example to better explain why such a generalization is needed. Moreover, we give the definition of weak solutions and some properties that they satisfy.
We give our main results in
Section 3 and
Section 4. In the former, under suitable hypothesis we give an
estimate for weak solutions, namely we prove that they are bounded in the whole of
. In this case, the proof mainly relies on a suitable choice of test functions and an iteration argument.
In
Section 4 we deal with a superlinear problem in presence of a source term which does not satisfy the so-called Ambrosetti-Rabinowitz condition. In particular, we prove the existence of two nontrivial weak solutions which does not change sign. Here, the strategy is to apply a Mountain Pass argument to suitable truncated functionals. Moreover, the lack of the Ambrosetti-Rabinowitz condition makes it harder to prove a compactness property for such functionals, as they do not satisfy the Palais-Smale condition. To overcome this difficulty, we prove that these functionals satisfy the Cerami condition.
Finally, in
Appendix A we give more details on the computations in Example 3 given in
Section 2.
2. Functional Setting
In this section we give the correct framework in order to study the operator in (
1.1) with
-Neumann conditions. First, we introduce the norm
where
. Thus, we can define the space
We observe that, setting
we can write
. Then, it is not hard to see that
is an uniformly convex Banach space. Moreover, if
we have that the embedding of
in
is compact for every
.
Now, we recall the analogous of the divergence theorem and of the integration by parts formula for the nonlocal case:
Proposition 1.
Let be such that and
Assume that the function
Proposition 2.
Let be such that and
Assume also that
and that the function
A few comments on Proposition 1 and Proposition 2 are mandatory. These results were first given in [
4, Theorem 6.3], where
u and
v are assumed to be in the Schwartz space
. They are also stated in [
26, Proposition 2.5 and 2.6] for any
u and
v bounded and of class
.
However, these regularity assumptions on
u and
v may be too much as solutions of problem (
1.5) may not even be of class
as we will show in the forthcoming Example 3. For this reason, in a similar fashion to [
17, Lemma 5.1 and 5.2] which cover the case
, we give Proposition 1 and Proposition 2 in a more general setting. We stress that our assumptions are enough to guarantee that the integrals in (
2.1) and (
2.2) are finite. Moreover, the proofs remain the same.
Next, we give an example of a solution for problem (
1.5) which is not of class
when
. For more details, see
Appendix A.
Example 3. Let
,
and
. We start defining
as
Clearly,
Then, defining for every
we can extend
u in
as
We note that extending
u in this way we have
and
for any
. In addition,
so
if
.
On the other hand, computing
and
in
, we can define
In this way, we have that
u is a solution of
where
for any
(see
Appendix A).
We also stress that, even if
,
u satisfies the hypotheses of Propostion 1. This is in agreement with the computation
The integration by parts formula in Proposition 2 leads to the definition of weak solutions, which shows that
is the natural space for problems ruled by the operator in (
1.1). We remark that we give the definition of weak solutions and the next two results in the case of non homogeneous boundary conditions. In this case, the functional space that we consider is
Definition 4. Let
,
and
. We say that
is a weak solution of
whenever
for every
, where
Now, we give a sort of maximum principle for weak solutions of (
2.3).
Proposition 5. Let , and . Let be a weak solution of (2.3) with , and . Then, u is constant.
Proof. First, we notice that
belongs to
. So, using it as a test function in (
2.4) we obtain
Hence,
a.e. in
and
a.e. in
. Now, taking
as a test function in (
2.4), we get
so
u must be constant. □
The next result states that if
u is a weak solution of (
2.3), then the nonlocal boundary condition in
is satisfied almost everywhere.
Theorem 6. Let be a weak solution of (2.3). Then, a.e. in .
Proof. First, we take
such that
in
as a test function in (
2.4), obtaining
Therefore,
for every
which is 0 in
. In particular, this is true for every
, and so
a.e. in
. □
3. Estimate
In this section we give a boundedness result for weak solutions. First, it is useful to introduce the following notation for the norm in
, namely
Clearly, this norm is equivalent to the usual norm in
.
The main result of this section states as follows.
Theorem 7.
Let , with and . If u is a weak solution of
then, and
for some constant .
Proof. If the weak solution
u is identically zero, we have nothing to prove. Otherwise, we take
and set
(if
we replace
with any number larger than
p by the usual Sobolev embedding theorems). With this definition, we find that
is a weak solution of
For every
we define
,
From the Dominated Convergence Theorem we have
Moreover, for
we have
, and so
Now we can use
as a test function in (
3.4), obtaining
We note that, for a.e.
, simple algebraic reasoning shows that
so that, being
where
, we have
Moreover,
and
Using (
3.9), (
3.10) and (
3.11) in (
3.7), we obtain
Then, by the Sobolev inequality
for some
. Moreover, by definition
, and so
We note that
and consequently
We also note that
As a consequence,
and observe also that
A simple computation shows that
With (
3.16) in mind ,we can use the generalized Hölder Inequality with exponents
,
q and
, and, together with (
3.14) and the definition of
, we have
for some
. We define the exponent
and observe that, from (
3.15),
Now, from (
3.12) and (
3.17) we have
Iterating, from (
3.18) we find
Then, if we take
sufficiently small in (
3.6) we can conclude that
and recalling (
3.5) we get
hence
. So, from (
3.3),
for every
.
A similar argument can be made for
. Clearly, setting
it follows that
is a weak solution of
Reasoning as above, we can conclude that
, hence
for every
. This together with (
3.19) implies that
for every
.
On the other hand, using
u as a test function in (
3.1) and recalling (
3.16) we get
From this last inequality and the Sobolev Inequality we can deduce
for some
. Using this in (
3.20) we obtain
for some
. From Theorem 6 we get that
so that
see also [
26]. This concludes the proof. □
4. A Superlinear Problem Without the Ambrosetti-Rabinowitz Condition
In this section, we consider the problem
where
is a Carathéodory function such that
for almost every
. In addition, we assume the following hypotheses taken from [
26] as improvements of those in [
21,
23]:
- ()
there exist
,
, with
,
and
such that
for a.e.
and for all
;
- ()
denoting
, we have
uniformly for a.e.
;
- ()
if
, then there exist
and
,
, such that
for a.e.
and all
or
;
- ()
uniformly for a.e.
.
Definition 8. Let
. With the same assumptions on
f as above, we say that
u is a weak solution of (
4.1) if
for every
.
Following this definition, we have that any critical point of the
functional
, defined as
is a weak solution of (
4.1).
Proposition 9.
Setting , the functional satisfies the property, that is for every sequence such that in as and
there holds
Proof. Clearly,
A is weakly lower semicontinuous in
, so that
On the other hand,
A is convex and so
which, passing to the limit, gives
Thus, combining (
4.2) and (
4.3) we get
as desired. □
We have the following result.
Theorem 10. If hypotheses ()-() hold, then problem (4.1) has two nontrivial constant sign solutions.
In order to prove Theorem 10, we introduce the functionals
where
and
denote the classical positive and negative part of
u, respectively.
Now we want to prove that both satisfy the Cerami condition, (C) for short, which states that any sequence in such that is bounded and as , admits a convergent subsequence.
Proposition 11. Under the assumptions of Theorem 10, the functionals satisfy the (C) condition.
Proof. We give the proof for , the proof for being analogous.
Let
in
be such that
for some
and all
, and
From (
4.5) we have
for every
and with
as
, that is
Taking
in (
4.6), we get
Observing that, similarly to (
3.8), we have
thus from (
4.7) we get
and so
Now, taking
in (
4.6), we obtain
From (
4.4) we have
for
and
, which together with (
4.7) gives
For some
and all
. Adding (
4.10) to (
4.11) we obtain
for some
and all
, that is
Now we want to prove that
is bounded in
, and for this we argue by contradiction. Passing to a subsequence if necessary, we assume that
as
. Defining
, we can assume that
for every
and for some
.
First, we treat the case
. We define the set
so that
and
for a.e.
as
. By hypothesis (
) we have
for almost every
. On the other hand, by Fatou’s Lemma
which leads to
From (
4.4) we have
for all
.
Recalling that
, from (
4.9) we obtain
for some
. Dividing by
,
Passing to the limit, we have
which is in contradiction with (
4.14), and this concludes the case
.
Now we deal with the case
. We consider the continuous functions
, defined as
with
and
. So, we can define
such that
Now, if
, we set
. From (
4.13),
in
for all
. Performing some integration, from (
) we have
for some
, which implies that
Since
, there exists
such that
for all
. Then, from (
4.15),
for all
. So,
Then (
4.16) implies that
and since
is arbitrary we have
We observe that
for all
, so from (
) we know that
for all
. Clearly,
. In addition, by (
4.7), we have
where
as
. By (
4.4) we get that
for some
and all
. Together with (
4.17) this implies that
for all
. Since
is a maximum point, we have
As in (
3.8), we have
so that
Adding (
4.19) to (
4.18) we obtain
that is
Hence, from (
4.17) we get
Combining (
4.12) and (
4.20) we obtain a contradiction, which conludes the case
.
In conclusion, we have proved that
is bounded in
, and from (
4.9) we know that
is bounded in
. Then we can assume
with
and for some
. Taking
in (
4.6) we have
From (
) and (
4.21) we know that
Passing to the limit in (
4.22) we obtain
The
property, see Proposition 9, implies that
, so from the uniform convexity of
(as
), we know that
in
as
. Then
satisfies the (C) condition, which concludes the proof. □
We can now give the proof of Theorem 10.
Proof of Theorem 10 . We want to apply the Mountain Pass Theorem to . From Proposition 11 we know that satisfies the (C) condition, so we only have to verify the geometric conditions.
From (
) and (
), for every
there exists
such that
for a.e.
and all
. Then
for some
. From this we get that, if
is small enough, then
Now, we take
with
and
, then
By Fatou’s Lemma
so from (
) we know that
Then
therefore there exists
such that
and
.
Now we can apply the Mountain Pass Theorem to
to obtain a nontrivial critical point
u. In particular, we have
By (
4.8) we have
so that
and thus
. Then
, and so
is a solution of (
4.1). Arguing in the same way for
, we can find a nontrivial negative solution of (
4.1). □
Acknowledgments
D.M. is a member of GNAMPA (Gruppo Nazionale per l’Analisi Matematica, la Probabilità e le loro Applicazioni) of INdAM (Istituto Nazionale di Alta Matematica ’Francesco Severi’) and is supported by the FFABR "Fondo per il finanziamento delle attività base di ricerca" 2017, by the INdAM-GNAMPA Project 2023 "Variational and non-variational problems with lack of compactness" and by the INdAM-GNAMPA Project 2024 "Nonlinear problems in local and nonlocal settings with applications". E.P.L. is a member of GNAMPA (Gruppo Nazionale per l’Analisi Matematica, la Probabilità e le loro Applicazioni) of INdAM (Istituto Nazionale di Alta Matematica ’Francesco Severi’) and is supported by the Australian Laureate Fellowship FL190100081..
Appendix A. Details of Example 3
In this appendix we give more details on the computations of Example 3. Thus, as in Example 3, we let
,
and
and define
as
Clearly,
Then, in agreement with (
3.21), for every
we define
Let us compute the integrals above. First,
In order to compute
we use the identity
When
, by using (
A3) (multiple times if needed), we get
and
Accordingly, by (
A2) we can compute both the integrals in (
A1) to obtain
When
, by (
A2) and (
A3) we get
Thus, we can extend
u in
as
so that, by construction,
for any
.
We observe that extending
u in this way, we have
, which is in agreement with [
18, Prosition 5.2]. Moreover, one can compute
which is in agreement with [
18, Proposition 3.13], that is
at infinity tends to its integral mean in
, namely
However, computing the first derivative of
we see that
so
if
.
Now, we compute
and
for any
. Indeed,
for any
. Moreover,
To compute
we take
small enough and consider separately the integrals
and
Exploiting (
A2) and (
A3), we can compute the integrals in (
A6) and (
A7). Then, adding them up and taking the limit for
, (
A5) becomes
for
, while
With (
A8) and (
A9) in mind, we can define, for any
and
,
As a consequence, we have that
u is a solution of
Notice that, in this setting, such a problem correspons to (
1.5) with
.
Now, we want to check that
for any
. First, the non integral part of
is bounded in
. Thus, we claim that
To this aim, fix
. We recall that, since
in
, we have
Thus, recalling (
3.22) and (
A1), we get
Therefore,
On the other hand, if
we have
for some
. Indeed, for this estimate it is enough to note that
and
, so that all the functions in square brackets in (
A4) are bounded when
. As a consequence, for every
we have
First, we clearly have
which is bounded in
, and so
for any
. Moreover, since
, we have
Recalling the inequality
we get that, for any
,
for some
.
Combining the integrability of
and
with (
A12), we get
for every
and for some
. Finally, (
A11) and (
A13) together with (
A12) imply the claim in (
A10).
A similar argument shows that
This concludes the proof that
for any
.
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