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Quasilinear Fractional Neumann Problems

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11 November 2024

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12 November 2024

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Abstract
We study a quasilinear fractional problem with fractional Neumann boundary conditions. Improving previous results, we also provide the weak formulation of solutions without regularity assumptions and we provide and example, even in the linear case, for which such a regularity cannot indeed be assumed.
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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
L : = α Δ p + β ( Δ ) p s .
Here α , β > 0 , s ( 0 , 1 ) , p > 1 and ( Δ ) p s is the fractional p-Laplacian, namely
( Δ ) p s u ( x ) = C N , s , p P V R N | u ( x ) u ( y ) | p 2 u ( x ) u ( y ) | x y | N + p s d y .
The constant C N , s , p , defined as
C N , s , p : = s 2 2 s 1 Γ p s + p + N 2 2 π N / 2 Γ ( 1 s ) ,
is the usual normalization constant for ( Δ ) p s (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 Ω R N with smooth boundary of class C 1 .
Mixed operators of the form (1.1) have raised an increasing interest in recent years, mainly in the Hilbert setting (namely, when p = 2 ), 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 ( α , β ) -Neumann conditions, introduced in [14,15] for p = 2 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 Ω , while the second corresponds to the nonlocal part and is defined on R N Ω ¯ , namely
( α , β ) - N e u m a n n   c o n d i t i o n s : N s , p u = 0 in   R N Ω ¯ , | D u | p 2 u ν = 0 on   Ω .
Here,
N s , p u ( x ) : = C N , s , p Ω | u ( x ) u ( y ) | p 2 u ( x ) u ( y ) | x y | N + p s d y , x R N Ω ¯ ,
is the nonlocal normal p derivative, or p Neumann boundary condition and describes the natural Neumann boundary condition in presence of the fractional p 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 p = 2 . 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 ( α , β ) -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 C N , s , p = 1 .
In this framework, we are intersted in dealing with problems of the type
α Δ p u + β ( Δ ) p s u = f ( x , u ) in   Ω , w i t h   ( α , β ) - h o m o g e n e o u s   N e u m a n n   c o n d i t i o n s .
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 L estimate for weak solutions, namely we prove that they are bounded in the whole of R N . 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 ( α , β ) -Neumann conditions. First, we introduce the norm
u p : = α Ω | D u | p d x + β 2 Q | u ( x ) u ( y ) | p | x y | N + p s d x d y + Ω | u ( x ) | p d x ,
where Q = ( Ω × Ω ) ( Ω × ( R N Ω ) ) ( ( R N Ω ) × Ω ) . Thus, we can define the space
X α , β : = { u : R N R : u < + } .
We observe that, setting
X = u : R N R : u L p ( Ω ) a n d Q | u ( x ) u ( y ) | p | x y | N + p s d x d y < + ,
we can write X α , β = W 1 , p ( Ω ) X . Then, it is not hard to see that X α , β is an uniformly convex Banach space. Moreover, if
p * = p N N p if   p < N , + if   p N ,
we have that the embedding of X α , β in L r ( Ω ) is compact for every r [ 1 , p * ) .
Now, we recall the analogous of the divergence theorem and of the integration by parts formula for the nonlocal case:
Proposition 1.
Let u : R N R be such that u C 2 ( Ω ) and
( Δ ) p s u L 1 ( Ω ) .
Assume that the function
Ω × ( R N Ω ) ( x , y ) | u ( x ) u ( y ) | p 1 | x y | N + p s b e l o n g s   t o   L 1 ( Ω × ( R N Ω ) ) .
Then,
Ω ( Δ ) p s u d x = R N Ω N s , p u d x .
Proposition 2.
Let u , v : R N R be such that u , v C 2 ( Ω ) and
Q ( x , y ) | u ( x ) u ( y ) | p 1 ( v ( x ) v ( y ) ) | x y | N + p s b e l o n g s   t o   L 1 ( Q ) .
Assume also that
( Δ ) p s u v L 1 ( Ω )
and that the function
( R N Ω ) × Ω ( x , y ) | u ( x ) u ( y ) | p 1 v ( x ) | x y | N + p s b e l o n g s   t o   L 1 ( ( R N Ω ) × Ω ) .
Then,
1 2 R 2 N ( C Ω ) 2 | u ( x ) u ( y ) | p 2 ( u ( x ) u ( y ) ) ( v ( x ) v ( y ) ) | x y | N + p s d x d y = Ω v ( Δ ) p s u d x + R N Ω v N s , p u d x .
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 S ( R N ) . They are also stated in [26, Proposition 2.5 and 2.6] for any u and v bounded and of class C 2 ( R N ) .
However, these regularity assumptions on u and v may be too much as solutions of problem (1.5) may not even be of class C 1 ( R N ) 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 p = 2 , 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 C 1 ( R N ) when s ( 0 , 1 / 2 ] . For more details, see Appendix A.
Example 3.
Let N = 1 , p = 2 and Ω = ( 1 , 1 ) . We start defining u : ( 1 , 1 ) R as
u ( x ) = x ( x 1 ) 2 ( x + 1 ) 2 .
Clearly,
lim x 1 + u ( x ) = lim x 1 u ( x ) = 0 .
Then, defining for every x > 1
g s ( x ) : = 1 1 y ( y 1 ) 2 ( y + 1 ) 2 ( x y ) 1 + 2 s d y 1 1 d y ( x y ) 1 + 2 s ,
we can extend u in R as
u ( x ) = g s ( x ) x < 1 x ( x 1 ) 2 ( x + 1 ) 2 1 x 1 g s ( x ) x > 1 .
We note that extending u in this way we have u C ( R ) and N s u ( x ) = 0 for any x R [ 1 , 1 ] . In addition,
lim x 1 + u ( x ) = lim x 1 u ( x ) = + s ( 0 , 1 2 ) , 4 3 s = 1 2 , 0 s ( 1 2 , 1 ) ,
so u C 1 ( R ) if s ( 0 , 1 / 2 ] .
On the other hand, computing u ( x ) and ( Δ ) s u ( x ) in ( 1 , 1 ) , we can define
f s ( x ) : = u ( x ) + ( Δ ) s u ( x ) .
In this way, we have that u is a solution of
u ( x ) + ( Δ ) s u ( x ) = f s ( x ) x ( 1 , 1 ) , u ( x ) = 0 x { 1 , 1 } , N s u = 0 x R [ 1 , 1 ] ,
where f s L q for any q 1 (see Appendix A).
We also stress that, even if s ( 0 , 1 / 2 ] , u satisfies the hypotheses of Propostion 1. This is in agreement with the computation
1 1 ( Δ ) s u ( x ) d x = 0 .
The integration by parts formula in Proposition 2 leads to the definition of weak solutions, which shows that X α , β 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
X α , β g , h : = u : R N R : u + Ω h u d σ + R N Ω ¯ g u d x R
Definition 4.
Let f L p ( Ω ) , g L 1 ( R N Ω ¯ ) and h L 1 ( Ω ) . We say that u X α , β g , h is a weak solution of
α Δ p u + β ( Δ ) p s u = f ( x ) in   Ω , β N s , p u = g in   R N Ω ¯ , | D u | p 2 u ν = h on   Ω ,
whenever
α Ω D u · D v d x + β 2 R 2 N ( C Ω ) 2 J p ( u ( x ) u ( y ) ) ( v ( x ) v ( y ) ) | x y | N + p s d x d y = Ω f v d x + Ω h v d σ + R N Ω ¯ g v d x
for every v X α , β g , h , where
J p ( u ( x ) u ( y ) ) : = | u ( x ) u ( y ) | p 2 ( u ( x ) u ( y ) ) .
Now, we give a sort of maximum principle for weak solutions of (2.3).
Proposition 5.
Let f L p ( Ω ) , g L 1 ( R N Ω ) and h L 1 ( Ω ) . Let u X α , β g , h be a weak solution of (2.3) with f 0 , g 0 and h 0 . Then, u is constant.
Proof. 
First, we notice that v 1 belongs to X α , β g , h . So, using it as a test function in (2.4) we obtain
0 Ω f d x = Ω h d σ R N Ω ¯ g d x 0 .
Hence, f = 0 a.e. in Ω and g = 0 a.e. in R N Ω . Now, taking v = u as a test function in (2.4), we get
R 2 N ( C Ω ) 2 | u ( x ) u ( y ) | p | x y | N + p s d x d y = 0 ,
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 R N Ω ¯ is satisfied almost everywhere.
Theorem 6.
Let u X α , β g , h be a weak solution of (2.3). Then, β N s , p u = g a.e. in R N Ω ¯ .
Proof. 
First, we take v X α , β g , h such that v 0 in Ω ¯ as a test function in (2.4), obtaining
R N Ω ¯ g v d x = β 2 Ω R N Ω ¯ J p ( u ( x ) u ( y ) ) v ( y ) | x y | N + p s d y d x + β 2 R N Ω ¯ Ω J p ( u ( x ) u ( y ) ) v ( x ) | x y | N + p s d y d x = β Ω R N Ω ¯ J p ( u ( x ) u ( y ) ) v ( y ) | x y | N + p s d y d x = β R N Ω ¯ v ( y ) Ω J p ( u ( x ) u ( y ) ) | x y | N + p s d x d y = β R N Ω ¯ v ( y ) N s , p u ( y ) d y .
Therefore,
R N Ω ¯ ( β N s , p u ( x ) g ( x ) ) v ( x ) d x = 0
for every v X α , β g , h which is 0 in Ω . In particular, this is true for every v C c ( R N Ω ¯ ) , and so β N s , p u ( x ) = g ( x ) a.e. in R N Ω ¯ . □

3. L 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 W 1 , p ( Ω ) , namely
u W 1 , p ( Ω ) : = α Ω | D u | p d x + Ω | u | p d x 1 / p .
Clearly, this norm is equivalent to the usual norm in W 1 , p ( Ω ) .
The main result of this section states as follows.
Theorem 7.
Let N 3 , f L q ( Ω ) with q > N / p and α β 0 . If u is a weak solution of
α Δ p u + β ( Δ ) p s u + | u | p 2 u = f ( x ) i n   Ω , w i t h   ( α , β ) - N e u m a n n   c o n d i t i o n s ,
then, u L ( R N ) and
u L ( R N ) C ( f L q ( Ω ) + f L q ( Ω ) 1 p 1 )
for some constant C > 0 .
Proof. 
If the weak solution u is identically zero, we have nothing to prove. Otherwise, we take δ ( 0 , 1 ) and set
u ˜ = δ u u L p * ( Ω ) + f L q ( Ω ) and f ˜ = δ u L p * ( Ω ) + f L q ( Ω ) p 1 f ,
(if N p we replace p * with any number larger than p by the usual Sobolev embedding theorems). With this definition, we find that u ˜ is a weak solution of
α Δ p u ˜ + β ( Δ ) p s u ˜ + | u ˜ | p 2 u ˜ = f ˜ in   Ω , w i t h   ( α , β ) N e u m a n n   c o n d i t i o n s .
For every k N we define c k : = 1 2 k ,
v k : = u ˜ c k , w k : = ( v k ) + : = max { v k , 0 } , U k : = w k L p * ( Ω ) p .
From the Dominated Convergence Theorem we have
lim k U k = lim k w k L p * ( Ω ) p = ( u ˜ 1 ) + L p * ( Ω ) p .
Moreover, for k = 0 we have w 0 = ( v 0 ) + = u + ˜ , and so
U 0 = Ω w 0 p * ( x ) d x p / p * = Ω u ˜ + p * ( x ) d x p / p * u ˜ L p * ( Ω ) p δ p .
Now we can use w k + 1 as a test function in (3.4), obtaining
α Ω | D u ˜ | p 2 D u ˜ · D w k + 1 d x + β 2 R 2 N ( C Ω ) 2 J p ( u ˜ ( x ) u ˜ ( y ) ) ( w k + 1 ( x ) w k + 1 ( y ) ) | x y | N + p s d x d y + Ω | u ˜ | p 2 u ˜ w k + 1 d x = Ω f ˜ w k + 1 d x .
We note that, for a.e. x , y R N , simple algebraic reasoning shows that
| w k + 1 ( x ) w k + 1 ( y ) | p = | ( v k + 1 ) + ( x ) ( v k + 1 ) + ( y ) | p | ( v k + 1 ) + ( x ) ( v k + 1 ) + ( y ) | p 2 ( ( v k + 1 ) + ( x ) ( v k + 1 ) + ( y ) ) ( v k + 1 ( x ) v k + 1 ( y ) ) = | w k + 1 ( x ) w k + 1 ( y ) | p 2 ( w k + 1 ( x ) w k + 1 ( y ) ) ( u ˜ ( x ) u ˜ ( y ) ) ,
so that, being w k + 1 = u ˜ c k + 1 where u ˜ > c k + 1 , we have
Q J p ( u ˜ ( x ) u ˜ ( y ) ) ( w k + 1 ( x ) w k + 1 ( y ) ) | x y | N + p s d x d y Q | w k + 1 ( x ) w k + 1 ( y ) | p | x y | N + p s d x d y 0 .
Moreover,
Ω | D u ˜ | p 2 D u ˜ · D w k + 1 d x = Ω { u ˜ > c k + 1 } | D u ˜ | p 2 D u ˜ · D v k + 1 d x = Ω | D w k + 1 | p d x
and
Ω | u ˜ | p 2 u ˜ w k + 1 d x = { u ˜ > c k + 1 } | u ˜ | p 2 u ˜ ( u ˜ c k + 1 ) d x Ω w k + 1 p d x .
Using (3.9), (3.10) and (3.11) in (3.7), we obtain
α Ω | D w k + 1 | p d x + Ω w k + 1 p d x Ω f ˜ w k + 1 d x .
Then, by the Sobolev inequality
U k + 1 = Ω w k + 1 p * d x p / p * C w k + 1 W 1 , p ( Ω ) p Ω w k + 1 f ˜ d x ,
for some C > 0 . Moreover, by definition v k + 1 v k , and so
w k + 1 w k .
We note that
w k + 1 = ( u ˜ c k + 1 ) + = u ˜ 1 + 1 2 k + 1 + 1 2 k 1 2 k + = v k + 1 2 k + 1 1 2 k + ,
and consequently
{ w k + 1 > 0 } = { v k + 1 > 0 } = w k > 1 2 k + 1 .
We also note that
p * p * q 1 > p * p * N / p 1 = p 1 .
As a consequence,
τ : = p * p * p * q 1 1 < p * p 1 ,
and observe also that
τ > p * p * 1 > 1 .
A simple computation shows that
1 p * + 1 q + 1 τ = 1 .
With (3.16) in mind ,we can use the generalized Hölder Inequality with exponents p * , q and τ , and, together with (3.14) and the definition of U k + 1 , we have
Ω w k + 1 | f ˜ | d x = Ω { w k + 1 > 0 } w k + 1 | f ˜ | d x f ˜ L q ( Ω ) w k + 1 L p * ( Ω ) | Ω { w k + 1 > 0 } | 1 / τ f L q ( Ω ) 2 p w k L p * ( Ω ) Ω w k > 1 2 k + 1 1 / τ f L q ( Ω ) 2 p U k 1 / p 2 p * ( k + 1 ) Ω w k > 1 2 k + 1 w k p * 1 / τ C 1 k U k 1 / p U k p * / p τ ,
for some C 1 > 0 . We define the exponent
γ : = 1 p + p * p τ
and observe that, from (3.15),
γ > 1 .
Now, from (3.12) and (3.17) we have
U k + 1 C 1 k U k γ .
Iterating, from (3.18) we find
U k + 1 C 1 ( k + 1 ) k 2 γ k U 0 γ k + 1 .
Then, if we take δ sufficiently small in (3.6) we can conclude that
lim k U k = 0 ,
and recalling (3.5) we get
( u ˜ 1 ) + L p * ( Ω ) p = 0 ,
hence u ˜ 1 . So, from (3.3),
u ( x ) u L p * ( Ω ) + f L q ( Ω ) δ
for every x Ω .
A similar argument can be made for u . Clearly, setting
u ¯ = δ u u L p * ( Ω ) + f L q ( Ω ) a n d f ¯ = δ u L p * ( Ω ) + f L q ( Ω ) p 1 f ,
it follows that u ¯ is a weak solution of
α Δ p u ¯ + β ( Δ ) p s u ¯ + | u ¯ | p 2 u ¯ = f ¯ in   Ω , w i t h   ( α , β ) N e u m a n n c o n d i t i o n s .
Reasoning as above, we can conclude that u ¯ 1 , hence
u ( x ) u L p * ( Ω ) + f L q ( Ω ) δ
for every x Ω . This together with (3.19) implies that
| u ( x ) | u L p * ( Ω ) + f L q ( Ω ) δ
for every x Ω .
On the other hand, using u as a test function in (3.1) and recalling (3.16) we get
u W 1 , p ( Ω ) p = α Ω | D u | p d x + Ω | u | p d x α Ω | D u | p d x + β 2 Ω | u ( x ) u ( y ) | p | x y | N + p s d x d y + Ω | u | p d x = Ω u f d x u L p * ( Ω ) f L q ( Ω ) | Ω | 1 / τ .
From this last inequality and the Sobolev Inequality we can deduce
u L p * ( Ω ) p 1 C 2 f L q ( Ω ) ,
for some C 2 > 0 . Using this in (3.20) we obtain
u L ( Ω ) C 3 ( f L q ( Ω ) + f L q ( Ω ) 1 p 1 )
for some C 3 > 0 . From Theorem 6 we get that
u ( x ) = Ω | u ( x ) u ( y ) | p 2 u ( y ) | x y | N + p s d y Ω | u ( x ) u ( y ) | p 2 | x y | N + p s d y
so that
| u ( x ) | u L ( Ω ) , x R N Ω ,
see also [26]. This concludes the proof. □

4. A Superlinear Problem Without the Ambrosetti-Rabinowitz Condition

In this section, we consider the problem
α Δ p u + β ( Δ ) p s u + | u | p 2 u = f ( x , u ) in   Ω , w i t h   ( α , β ) - N e u m a n n   c o n d i t i o n s ,
where f : Ω × R R is a Carathéodory function such that f ( x , 0 ) = 0 for almost every x Ω . In addition, we assume the following hypotheses taken from [26] as improvements of those in [21,23]:
( f 1 )
there exist a L q ( Ω ) , a 0 , with q ( ( p * ) , p ) , c > 0 and r ( p , p * ) such that
| f ( x , t ) | a ( x ) + c | t | r 1
for a.e. x Ω and for all t R ;
( f 2 )
denoting F ( x , t ) = 0 t f ( x , σ ) d σ , we have
lim t ± F ( x , t ) | t | p = +
uniformly for a.e. x Ω ;
( f 3 )
if σ ( x , t ) = f ( x , t ) t p F ( x , t ) , then there exist θ > 1 and β * L 1 ( Ω ) , β * 0 , such that
σ ( x , t 1 ) θ σ ( x , t 2 ) + β *
for a.e. x Ω and all 0 t 1 t 2 or t 2 t 1 0 ;
( f 4 )
lim t 0 f ( x , t ) | t | p 2 t = 0
uniformly for a.e. x Ω .
Definition 8.
Let u X α , β . With the same assumptions on f as above, we say that u is a weak solution of (4.1) if
α Ω | D u | p 2 D u · D v d x + β 2 Q J p ( u ( x ) u ( y ) ) ( v ( x ) v ( y ) ) | x y | N + p s d x d y + Ω | u | p 2 u v d x = Ω f ( x , u ) v d x
for every v X α , β .
Following this definition, we have that any critical point of the C 1 functional I : X α , β R , defined as
I ( u ) = 1 p u p Ω F ( x , u ) d x
is a weak solution of (4.1).
Proposition 9.
Setting A ( u ) = u p , the functional A : X α , β X α , β satisfies the ( S ) property, that is for every sequence ( u n ) n such that u n u in X α , β as n and
lim n A ( u n ) , u n u = 0 ,
there holds
lim n u n = u .
Proof. 
Clearly, A is weakly lower semicontinuous in X α , β , so that
A ( u ) lim inf n A ( u n ) .
On the other hand, A is convex and so
A ( u ) A ( u n ) + A ( u n ) , u n u ,
which, passing to the limit, gives
A ( u ) lim sup n A ( u n ) .
Thus, combining (4.2) and (4.3) we get
A ( u ) = lim n A ( u n )
as desired. □
We have the following result.
Theorem 10.
If hypotheses ( f 1 )-( f 4 ) hold, then problem (4.1) has two nontrivial constant sign solutions.
In order to prove Theorem 10, we introduce the functionals
I ± ( u ) = 1 p u p Ω F ( x , u ± ) d x ,
where u + : = max { u , 0 } and u : = max { u , 0 } denote the classical positive and negative part of u, respectively.
Now we want to prove that both I ± satisfy the Cerami condition, (C) for short, which states that any sequence ( u n ) n in X α , β such that ( I ± ( u n ) ) n is bounded and ( 1 + u n ) I ± ( u n ) 0 as n , admits a convergent subsequence.
Proposition 11.
Under the assumptions of Theorem 10, the functionals I ± satisfy the (C) condition.
Proof. 
We give the proof for I + , the proof for I being analogous.
Let ( u n ) n in X α , β be such that
| I + ( u n ) | M 1
for some M 1 > 0 and all n 1 , and
( 1 + u n ) I + ( u n ) 0   in   X α , β a s n .
From (4.5) we have
| I + ( u n ) h | ε n h 1 + u n
for every h X α , β and with ε n 0 as n , that is
| α Ω | D u n | p 1 D u n · D h d x + β 2 Q J p ( u n ( x ) u n ( y ) ) ( h ( x ) h ( y ) ) | x y | N + p s d x d y + Ω | u n | p 2 u n h d x Ω f ( x , ( u n ) + ) h d x | ε n h 1 + u n .
Taking h = ( u n ) in (4.6), we get
| α Ω | D ( u n ) | p d x + β 2 Q J p ( u n ( x ) u n ( y ) ) ( ( u n ) ( x ) ( u n ) ( y ) ) | x y | N + p s d x d y + Ω | ( u n ) | p d x | ε n
Observing that, similarly to (3.8), we have
| ( u n ) ( x ) ( u n ) ( y ) | p J p ( u n ( x ) u n ( y ) ) ( ( u n ) ( x ) ( u n ) ( y ) ) ,
thus from (4.7) we get
( u n ) p ε n ,
and so
( u n ) 0 i n X α , β a s n .
Now, taking h = ( u n ) + in (4.6), we obtain
α Ω | D ( u n ) + | p d x β 2 Q J p ( u n ( x ) u n ( y ) ) ( ( u n ) + ( x ) ( u n ) + ( y ) ) | x y | N + p s d x d y Ω | ( u n ) + | p d x + Ω f ( x , ( u n ) + ) ( u n ) + d x ε n .
From (4.4) we have
α Ω | D u n | p d x + β 2 Q | u n ( x ) u n ( y ) | p | x y | N + p s d x d y + Ω | u n | p d x p Ω F ( x , ( u n ) + ) d x p M 1
for M 1 > 0 and n 1 , which together with (4.7) gives
α Ω | D ( u n ) + | p d x + β 2 Q J p ( u n ( x ) u n ( y ) ) ( ( u n ) + ( x ) ( u n ) + ( y ) ) | x y | N + p s d x d y + Ω | ( u n ) + | p d x p Ω F ( x , ( u n ) + ) d x M 2
For some M 2 > 0 and all n 1 . Adding (4.10) to (4.11) we obtain
Ω f ( x , ( u n ) + ) ( u n ) + d x p Ω F ( x , ( u n ) + ) d x M 3
for some M 3 > 0 and all n 1 , that is
Ω σ ( x , ( u n ) + ) d x M 3 .
Now we want to prove that ( ( u n ) + ) n is bounded in X α , β , and for this we argue by contradiction. Passing to a subsequence if necessary, we assume that ( u n ) + as n . Defining y n = ( u n ) + / ( u n ) + , we can assume that
y n y i n X α , β   and   y n y i n L q ( Ω )
for every q [ p , p * ) and for some y 0 .
First, we treat the case y 0 . We define the set
Z ( y ) = { x Ω : y ( x ) = 0 } ,
so that | Ω Z ( y ) | > 0 and ( u n ) + for a.e. x Ω Z ( y ) as n . By hypothesis ( f 2 ) we have
lim n F ( x , ( u n ) + ( x ) ) ( u n ) + p = lim n F ( x , ( u n ) + ( x ) ) ( u n ) + ( x ) p y n ( x ) p = +
for almost every x Ω Z ( y ) . On the other hand, by Fatou’s Lemma
Ω lim inf n F ( x , ( u n ) + ( x ) ) ( u n ) + p d x lim inf n Ω F ( x , ( u n ) + ( x ) ) ( u n ) + p d x ,
which leads to
lim n Ω F ( x , ( u n ) + ( x ) ) ( u n ) + p d x = + .
From (4.4) we have
1 p u n p + Ω F ( x , ( u n ) + ( x ) ) d x M 1
for all n 1 .
Recalling that u n p 2 p 1 ( ( u n ) + p + ( u n ) p ) , from (4.9) we obtain
2 p 1 p ( u n ) + p + Ω F ( x , ( u n ) + ( x ) ) d x M 4
for some M 4 > 0 . Dividing by ( u n ) + p ,
2 p 1 p + Ω F ( x , ( u n ) + ( x ) ) ( u n ) + p d x M 4 ( u n ) + p .
Passing to the limit, we have
lim sup n Ω F ( x , ( u n ) + ( x ) ) ( u n ) + p d x 2 p 1 p ,
which is in contradiction with (4.14), and this concludes the case y 0 .
Now we deal with the case y 0 . We consider the continuous functions ϕ n : [ 0 , 1 ] R , defined as ϕ n ( t ) = I + ( t ( u n ) + ) with t [ 0 , 1 ] and n 1 . So, we can define t n such that
ϕ n ( t n ) = max t [ 0 , 1 ] ϕ n ( t ) .
Now, if λ > 0 , we set v n : = ( p λ ) 1 p y n X α , β . From (4.13), v n 0 in L q ( Ω ) for all q [ 1 , p * ) . Performing some integration, from ( f 1 ) we have
Ω F ( x , v n ( x ) ) d x Ω a ( x ) | v n ( x ) | d x + C Ω | v n ( x ) | r d x
for some C > 0 , which implies that
lim n Ω F ( x , v n ( x ) ) d x = 0 .
Since ( u n ) + , there exists n 0 1 such that ( p λ ) 1 p / ( u n ) + ( 0 , 1 ) for all n n 0 . Then, from (4.15),
ϕ n ( t n ) ϕ n ( p λ ) 1 p ( u n ) +
for all n n 0 . So,
I + ( t n ( u n ) + ) I + ( ( p λ ) 1 p y n ) = λ y n p Ω F ( x , v n ( x ) ) d x .
Then (4.16) implies that
I + ( t n ( u n ) + ) λ y n p + o ( 1 ) ,
and since λ is arbitrary we have
lim n I + ( t n ( u n ) + ) = + .
We observe that 0 t n ( u n ) + ( u n ) + for all n 1 , so from ( f 3 ) we know that
Ω σ ( x , t n ( u n ) + ) d x θ σ ( x , ( u n ) + ) d x + β * 1
for all n 1 . Clearly, I + ( 0 ) = 0 . In addition, by (4.7), we have
I + ( ( u n ) + ) = 1 p Ω | D u n | p d x 1 p Ω | D ( u n ) | p d x + 1 p Ω | u n | p d x 1 p Ω | ( u n ) | p d x Ω F ( x , ( ( u n ) + ) d x + 1 p { x , y Q : u ( x ) > 0 , u ( y ) > 0 } | u n ( x ) u n ( y ) | p | x y | N + p s d x d y 1 p { x , y Q : u ( x ) 0 , u ( y ) 0 } | ( u n ) ( x ) ( u n ) ( y ) | p | x y | N + p s d x d y 2 p { x , y Q : u ( x ) > 0 , u ( y ) 0 } | ( u n ) + ( x ) + ( u n ) ( y ) | p | x y | N + p s d x d y = I + ( u n ) 2 p { x , y Q : u ( x ) > 0 , u ( y ) 0 } | ( u n ) + ( x ) + ( u n ) ( y ) | p | x y | N + p s d x d y + o ( 1 ) I + ( u n ) + o ( 1 )
where o ( 1 ) 0 as n . By (4.4) we get that I + ( ( u n ) + ) M 7 for some M 7 > 0 and all n N . Together with (4.17) this implies that t n ( 0 , 1 ) for all n n 1 n 0 . Since t n is a maximum point, we have
0 = t n ϕ n ( t n ) = + β 2 Q J p ( t n u n ( x ) t n u n ( y ) ) ( t n ( u n ) + ( x ) t n ( u n ) + ( y ) ) | x y | N + p s d x d y + α Ω | D t n ( u n ) + | p d x + Ω | t n ( u n ) + | p d x Ω f ( x , t n ( u n ) + ) t n ( u n ) + d x .
As in (3.8), we have
| t n ( u n ) + ( x ) t n ( u n ) + ( y ) ) | p J p ( t n u n ( x ) t n u n ( y ) ) ( t n ( u n ) + ( x ) t n ( u n ) + ( y ) ) ,
so that
t n ( u n ) + p Ω f ( x , t n ( u n ) + ) t n ( u n ) + d x 0 .
Adding (4.19) to (4.18) we obtain
t n ( u n ) + p p Ω F ( x , t n ( u n ) + ) d x θ Ω σ ( x , ( u n ) + ) d x + β * 1 ,
that is
p I + ( t n ( u n ) + ) θ Ω σ ( x , ( u n ) + ) d x + β * 1 .
Hence, from (4.17) we get
lim n σ ( x , ( u n ) + ) d x = + .
Combining (4.12) and (4.20) we obtain a contradiction, which conludes the case y 0 .
In conclusion, we have proved that ( ( u n ) + ) n is bounded in X α , β , and from (4.9) we know that ( u n ) n is bounded in X α , β . Then we can assume
u n u   in   X α , β   and   u n u   in   L q ( Ω )
with q [ 1 , p * ) and for some u X α , β . Taking h = u n u in (4.6) we have
u n p α Ω | D u n | p 2 D u n · D u d x β 2 Q J p ( u n ( x ) u n ( y ) ) ( u ( x ) u ( y ) ) | x y | N + p s d x d y Ω | u n | p 2 u n u d x Ω f ( x , ( u n ) + ) ( u n u ) d x = ε n .
From ( f 1 ) and (4.21) we know that
lim n Ω | f ( x , ( u n ) + ) ( u n u ) | d x = 0 .
Passing to the limit in (4.22) we obtain
lim n ( u n p β 2 Q J p ( u n ( x ) u n ( y ) ) ( u ( x ) u ( y ) ) | x y | N + p s d x d y α Ω | D u n | p 2 D u n · D u d x Ω | u n | p 2 u n u d x ) = 0 .
The ( S ) property, see Proposition 9, implies that u n u , so from the uniform convexity of X α , β (as 1 < p < ), we know that u n u in X α , β as n . Then I + 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 I + . From Proposition 11 we know that I + satisfies the (C) condition, so we only have to verify the geometric conditions.
From ( f 1 ) and ( f 4 ), for every ε > 0 there exists C ε such that
F ( x , t ) ε p | t | p + C ε | t | r
for a.e. x Ω and all t R . Then
I + ( u ) = 1 p u p Ω F ( x , u + ) d x 1 p u p ε p u p p C ε u r r 1 ε C 1 p u p C 2 u r
for some C 1 , C 2 > 0 . From this we get that, if u = ρ is small enough, then
inf u = ρ I + ( u ) > 0 .
Now, we take u X α , β with u > 0 and t > 0 , then
I + ( u ) = t p p u p Ω F ( x , t u ) d x = t p p u p t p Ω F ( x , t u ) | t u | p u p d x .
By Fatou’s Lemma
Ω lim inf t F ( x , t u ) | t u | p u p d x lim inf t Ω F ( x , t u ) | t u | p u p d x ,
so from ( f 2 ) we know that
lim t Ω F ( x , t u ) | t u | p u p d x = + .
Then
lim t I + ( t u ) = ,
therefore there exists e X α , β such that e > ρ and I + ( e ) < 0 .
Now we can apply the Mountain Pass Theorem to I + to obtain a nontrivial critical point u. In particular, we have
0 = α Ω | D u | p d x + β 2 Q J p ( u ( x ) u ( y ) ) ( u ( x ) u ( y ) ) | x y | N + p s d x d y + Ω | u | p d x Ω f ( x , u + ) u d x = α Ω | D u | p d x + β 2 Q J p ( u ( x ) u ( y ) ) ( u ( x ) u ( y ) ) | x y | N + p s d x d y + Ω | u | p d x .
By (4.8) we have
| u ( x ) u ( y ) | p J p ( u ( x ) u ( y ) ) ( u ( x ) u ( y ) ) ,
so that
0 u p ,
and thus u 0 . Then I + ( u ) = I ( u ) , and so u 0 is a solution of (4.1). Arguing in the same way for I , 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 N = 1 , p = 2 and Ω = ( 1 , 1 ) and define u : ( 1 , 1 ) R as
u ( x ) = x ( x 1 ) 2 ( x + 1 ) 2 .
Clearly,
lim x 1 + u ( x ) = lim x 1 u ( x ) = 0 .
Then, in agreement with (3.21), for every x > 1 we define
g s ( x ) : = 1 1 y ( y 1 ) 2 ( y + 1 ) 2 ( x y ) 1 + 2 s d y 1 1 d y ( x y ) 1 + 2 s .
Let us compute the integrals above. First,
1 1 d y ( x y ) γ = 1 γ 1 ( x 1 ) 1 γ ( x + 1 ) 1 γ if   γ 1 log x + 1 x 1 if   γ = 1 .
In order to compute
1 1 y 5 2 y 3 + y ( x y ) 1 + 2 s d y
we use the identity
1 1 y n ( x y ) γ d y = 1 1 y n x n ( x y ) γ d y + x n 1 1 d y ( x y ) γ = k = 0 n 1 x k 1 1 y n k 1 ( x y ) γ 1 d y + x n 1 1 d y ( x y ) γ .
When s 1 / 2 , by using (A3) (multiple times if needed), we get
1 1 y ( x y ) 1 + 2 s d y = 1 1 d y ( x y ) 2 s + x 1 1 d y ( x y ) 1 + 2 s ,
1 1 y 3 ( x y ) 1 + 2 s d y = 1 1 d y ( x y ) 2 s 2 + 3 x 1 1 d y ( x y ) 2 s 1 3 x 2 1 1 d y ( x y ) 2 s + x 3 1 1 d y ( x y ) 1 + 2 s
and
1 1 y 5 ( x y ) 1 + 2 s d y = 1 1 d y ( x y ) 2 s 4 + 5 x 1 1 d y ( x y ) 2 s 3 10 x 2 1 1 d y ( x y ) 2 s 2 + 10 x 3 1 1 d y ( x y ) 2 s 1 5 x 4 1 1 d y ( x y ) 2 s + x 5 1 1 d y ( x y ) 1 + 2 s
Accordingly, by (A2) we can compute both the integrals in (A1) to obtain
g s ( x ) = 2 s ( x 1 ) 2 s ( x + 1 ) 2 s ( x + 1 ) 2 s ( x 1 ) 2 s [ 1 5 2 s ( x + 1 ) 5 2 s ( x 1 ) 5 2 s + 5 x 4 2 s ( x + 1 ) 4 2 s ( x 1 ) 4 2 s 10 x 2 2 3 2 s ( x + 1 ) 3 2 s ( x 1 ) 3 2 s + 10 x 3 6 x 2 2 s ( x + 1 ) 2 2 s ( x 1 ) 2 2 s 5 x 4 6 x 2 + 1 1 2 s ( x + 1 ) 1 2 s ( x 1 ) 1 2 s x 5 2 x 3 + x 2 s ( x + 1 ) 2 s ( x 1 ) 2 s ] ,
When s = 1 / 2 , by (A2) and (A3) we get
g 1 2 ( x ) : = 5 x 5 28 3 x 3 + 13 3 x 1 2 ( 5 x 6 11 x 4 + 7 x 2 1 ) log x + 1 x 1 .
Thus, we can extend u in R as
u ( x ) = g s ( x ) x < 1 x ( x 1 ) 2 ( x + 1 ) 2 1 x 1 g s ( x ) x > 1
so that, by construction, N s u ( x ) = 0 for any x R [ 1 , 1 ] .
We observe that extending u in this way, we have u C ( R ) , which is in agreement with [18, Prosition 5.2]. Moreover, one can compute
lim | x | + g s ( x ) = 0 ,
which is in agreement with [18, Proposition 3.13], that is u ( x ) at infinity tends to its integral mean in ( 1 , 1 ) , namely
lim | x | + u ( x ) = 1 2 1 1 u ( x ) d x = 0 .
However, computing the first derivative of g s ( x ) we see that
lim x 1 + u ( x ) = lim x 1 u ( x ) = + s ( 0 , 1 2 ) , 4 3 s = 1 2 , 0 s ( 1 2 , 1 ) ,
so u C 1 ( R ) if s ( 0 , 1 2 ] .
Now, we compute u ( x ) and ( Δ ) s u ( x ) for any x ( 1 , 1 ) . Indeed,
u ( x ) = 20 x 3 12 x
for any x ( 1 , 1 ) . Moreover,
( Δ ) s u ( x ) = P V + u ( x ) u ( y ) | x y | 1 + 2 s d y = 1 x ( x 1 ) 2 ( x + 1 ) 2 + g s ( y ) | x y | 1 + 2 s d y + P V 1 1 x ( x 1 ) 2 ( x + 1 ) 2 y ( y 1 ) 2 ( y + 1 ) 2 | x y | 1 + 2 s d y + 1 + x ( x 1 ) 2 ( x + 1 ) 2 g s ( y ) | x y | 1 + 2 s d y .
To compute
P V 1 1 x ( x 1 ) 2 ( x + 1 ) 2 y ( y 1 ) 2 ( y + 1 ) 2 | x y | 1 + 2 s d y
we take ε > 0 small enough and consider separately the integrals
1 x ε x ( x 1 ) 2 ( x + 1 ) 2 y ( y 1 ) 2 ( y + 1 ) 2 ( y x ) 1 + 2 s d y
and
x + ε 1 x ( x 1 ) 2 ( x + 1 ) 2 y ( y 1 ) 2 ( y + 1 ) 2 ( x y ) 1 + 2 s d y .
Exploiting (A2) and (A3), we can compute the integrals in (A6) and (A7). Then, adding them up and taking the limit for ε 0 , (A5) becomes
( Δ ) s u ( x ) = 1 5 2 s ( 1 + x ) 5 2 s ( 1 x ) 5 2 s 5 x 4 2 s ( 1 + x ) 4 2 s + ( 1 x ) 4 2 s + 10 x 2 2 3 2 s ( 1 + x ) 3 2 s ( 1 x ) 3 2 s 10 x 3 6 x 2 2 s ( 1 + x ) 2 2 s + ( 1 x ) 2 2 s + 5 x 4 6 x 2 + 1 1 2 s ( 1 + x ) 1 2 s ( 1 x ) 1 2 s + 1 + x ( x 1 ) 2 ( x + 1 ) 2 g s ( y ) | x y | 1 + 2 s d y + 1 x ( x 1 ) 2 ( x + 1 ) 2 + g s ( y ) | x y | 1 + 2 s d y
for s 1 / 2 , while
( Δ ) 1 2 u ( x ) = 8 x 3 + 20 3 x + ( 5 x 4 6 x 2 + 1 ) log 1 + x 1 x + 1 + x ( x 1 ) 2 ( x + 1 ) 2 g 1 2 ( y ) | x y | 2 d y + 1 x ( x 1 ) 2 ( x + 1 ) 2 + g 1 2 ( y ) | x y | 2 d y .
With (A8) and (A9) in mind, we can define, for any s ( 0 , 1 ) and x ( 1 , 1 ) ,
f s ( x ) : = 20 x 3 + 12 x + ( Δ ) s u ( x ) .
As a consequence, we have that u is a solution of
u ( x ) + ( Δ ) s u ( x ) = f s ( x ) x ( 1 , 1 ) , u ( x ) = 0 x { 1 , 1 } , N s u = 0 x R [ 1 , 1 ] .
Notice that, in this setting, such a problem correspons to (1.5) with α = β = 1 .
Now, we want to check that f s ( x ) L q ( ( 1 , 1 ) ) for any q 1 . First, the non integral part of f s ( x ) is bounded in ( 1 , 1 ) . Thus, we claim that
1 + x ( x 1 ) 2 ( x + 1 ) 2 g s ( y ) | x y | 1 + 2 s d y b e l o n g s t o L q ( ( 1 , 1 ) ) f o r a n y q 1 .
To this aim, fix ε > 0 . We recall that, since N s u = 0 in R [ 1 , 1 ] , we have
u L ( R ) u L ( ( 0 , 1 ) ) < + .
Thus, recalling (3.22) and (A1), we get
1 + ε + x ( x 1 ) 2 ( x + 1 ) 2 g s ( y ) | x y | 1 + 2 s d y 2 u L ( R ) 1 + ε + 1 ( y x ) 1 + 2 s d y = u L ( R ) s ( 1 x + ε ) 2 s u L ( R ) s ε 2 s .
Therefore,
1 1 1 + ε + x ( x 1 ) 2 ( x + 1 ) 2 g s ( y ) | x y | 1 + 2 s d y q d x 2 u L ( R ) s ε 2 s q < + .
On the other hand, if y ( 1 , 1 + ε ) we have
| g s ( y ) | c ε ( y 1 ) 2 s
for some c ε > 0 . Indeed, for this estimate it is enough to note that 5 y 4 6 y 2 + 1 = ( y 1 ) ( y + 1 ) ( 5 y 2 1 ) and y 5 2 y 3 + y = y ( y 1 ) 2 ( y + 1 ) 2 , so that all the functions in square brackets in (A4) are bounded when y ( 1 , 1 + ε ) . As a consequence, for every x ( 1 , 1 ) we have
1 1 + ε x ( x 1 ) 2 ( x + 1 ) 2 g s ( y ) | x y | 1 + 2 s d y | x | ( x 1 ) 2 ( x + 1 ) 2 1 1 + ε 1 ( y x ) 1 + 2 s d y + c ε 1 1 + ε ( y 1 ) 2 s ( y x ) 1 + 2 s d y = : I 1 ( x ) + I 2 ( x ) .
First, we clearly have
I 1 ( x ) = | x | ( x 1 ) 2 ( x + 1 ) 2 2 s [ ( 1 x ) 2 s ( 1 x + ε ) 2 s ]
which is bounded in ( 1 , 1 ) , and so I 1 L q ( ( 1 , 1 ) ) for any q 1 . Moreover, since x ( 1 , 1 ) , we have
I 2 ( x ) c ε 1 1 + ε ( y x ) 2 s ( y x ) 1 + 2 s d y = c ε ( log ( 1 x + ε ) log ( 1 x ) ) .
Recalling the inequality
| a + b | q 2 q 1 ( | a | q + | b | q ) f o r   a n y   a , b R ,
we get that, for any q 1 ,
1 1 | I 2 ( x ) | q d x c 2 1 1 ( | log ( 1 x + ε ) | q + | log ( 1 x ) | q ) d x < + ,
for some c 2 = c 2 ( s , q , ε ) > 0 .
Combining the integrability of I 1 and I 2 with (A12), we get
1 1 1 1 + ε x ( x 1 ) 2 ( x + 1 ) 2 g s ( y ) | x y | 1 + 2 s d y q d x c 3 1 1 ( | I 1 ( x ) | q + | I 2 ( x ) | q ) d x < +
for every q 1 and for some c 3 = c 3 ( s , q , ε ) . Finally, (A11) and (A13) together with (A12) imply the claim in (A10).
A similar argument shows that
1 x ( x 1 ) 2 ( x + 1 ) 2 + g s ( y ) | x y | 1 + 2 s d y b e l o n g s   t o   L q ( ( 1 , 1 ) )   f o r   a n y   q 1 .
This concludes the proof that f s ( x ) L q ( ( 1 , 1 ) ) for any q 1 .

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