With appropriate hypotheses on the nonlinearity f, we prove the existence of a ground state solution u for the problem
where V is a bounded potential, not necessarily continuous, and F the primitive of f. We also show that any of this problem is a classical solution. Furthermore, we prove that the ground state solution has exponential decay.
In this paper we consider a generalized pseudo-relativistic Hartree equation
where , , assuming that the nonlinearity f is a function, non-negative in , that satisfies
;
for some ;
is increasing for all .
We also postulate
V is continuous and satisfies for every and some constant ;
;
for all , ;
is radial, with .
Therefore, we aim to generalize the results obtained by Coti Zelati and Nolasco [12] and Cingolani and Secchi [8]. In the last paper, the authors have studied the equation
supposing, additionally to our hypotheses, that the potential V is continuous and has a horizontal asymptote for . Furthermore, the homogeneity of the equation is a key ingredient in the proofs presented.
In Section 7 of Cingolani and Secchi [8], dealing with the decay of the solution, the authors assume additionally that the hypothesis when is explicitly assumed. The present work covers also the case
So, applying different methods, we generalize [8]. A careful reading of our paper will also show that it generalizes [12].
The equation
where , G is a nonlinearity of Hartree type, denotes the mass of bosons in units, was used to describe the dynamics of pseudo-relativistic boson stars in astrophysics. See [6,11,16,23] for more details. For the study of semiclassical analysis of the non-relativistic Hartree equations we would like to quote the papers [7,9,18,26,30] and also the recent paper [28]. For the Hartree equation without external potential V, we cite [23] for radial ground state solution, [21] for uniqueness and nondegeneracy of ground state solutions, and [12,13] for the existence of positive and radially symmetric solutions. In [25] is treated some Hartree problem imposing that the external potential V is radial, while in [8] this condition is dropped. Some versions of the magnetic Hartree equation are considered in [10,28].
A partial generalization of the present paper (obtained long after its submission) to the fractional operator can be seen in [3] (see also [2]). Here, the results about regularity and decay are not so strong.
By considering an extension problem from to , an alternative definition of is well-known (see [12] or [5]), so that equation (1.1) can be written as
We summarize our results:
Suppose that conditions ()–(), () and () are valid. Then, problem (
1.3
) has a non-negative ground state solution.
Assuming that hypotheses already stated are satisfied by f, V and W, any solution v of problem (
1.3
) satisfiesand therefore is a classical solution of (
1.3
).
As a simple remark, we observe that if , then the solution is also .
We also prove that the ground state solution has exponential decay:
Let w be the ground state solution obtained in Theorem
1
. Theninand, for anythere existssuch thatfor any. In particular,where.
The natural setting for problem (1.3) is the Sobolev space
endowed with the norm
The norm in the space will be denoted by . For all , we denote by the norm in the space and by the norm in the space .
It is well-known that traces of functions are in and that every function in is the trace of a function in , see [29]. Denoting the linear function that associates the trace of the function , then .
The immersions
are continuous for any and respectively, where
The space is defined by means of Fourier transforms; therefore, we can not change to a bounded open set . However (see [14]), and is well-defined for an open set . We recall its definition. Let a measurable function and Ω a bounded open set (that, in the sequel, we suppose to have Lipschitz boundary). Denoting
and
then is a reflexive Banach space (see, e.g., [14] and [15]) endowed with the norm
The proof of the next result can be found in [14, Theorem 4.54].
The immersionis compact for any.
As usual, the immersion is continuous: see [14, Corollary 4.53]. We denote the norm in the space by .
Preliminaries
Let us suppose that and . Let us proceed heuristically: since
it follows from Hölder’s inequality
So, in order to apply the immersion we must have , that is,
By density of in , the estimate (2.1) is valid for all .
Taking into account (1.4), Young’s inequality applied to (2.1) yields
where is a constant. We summarize:
The inequality (2.3) will also be valuable in the special case :
where is a parameter, the last inequality being a consequence of Young’s inequality.
It follows from () that f satisfies the Ambrosetti–Rabinowitz inequality , for all . Furthermore, it follows from () and () that, for any fixed , there exists a constant such that
and analogously
Observe that and imply .
(Hausdorff–Young).
Assume that, for, we have,andThen
We now enhance the result given by (2.4). Observe that and if, and only if .
The results in the sequel will be useful when addressing the regularity of the solution of problem (1.3).
Concerning hypothesiswe have:
if, there existssuch thatandFurthermore,and
ifdenotes the conjugate exponent of r and, thenand.
(i) We verify the values of r that satisfy the equality
Observe that if, and only if, .
As consequence of (2.4) and thus and (2.7) yields . So, follows from the Hausdorff–Young inequality.
() Since for and , applying (i) we conclude that and is consequence of Proposition 2.1. □
We havewith.
An immediately consequence of Lemma 2.1, since . □
For allwe have, where g is the function of Lemma
2.1
.
Application of the Hölder inequality yields
if we define α so that . Thus, and we have . Since both integrals of the right-hand side of the last inequality are integrable, we are done. □
We now handle the existence of the “energy” functional. We denote by the weak space and by its usual norm (see [22]). The next result is a generalized version of the Hardy-Littlewood-Sobolev inequality:
Assume thatandThen, for some constantand for any,and, we have the inequality
For a positive constant C holds
Let us denote
Since ,
Let us suppose that for some . Then and (as consequence of (2.7)). Application of Proposition 2.2 yields
Since implies , we have
(Observe that, in order to apply the immersion , we must have , that is, .)
Of course
if we take . Since hypothesis () implies , we have
Taking into account Lemma 2.4, the proof is complete. □
Since the derivative of the energy functional is given by
we see that critical points of I are weak solutions (1.3).
Because we are looking for a positive solution, we suppose that for .
The quadratic formdefines an norm in the space, which is equivalent to the norm.
We keep up with the notation already introduced and note that . Furthermore, as consequence of (2.5), we have
Therefore,
Defining , we conclude that
By applying (2.12) it easily follows that
for a constant . We are done. □
Mountain pass geometry and Nehari manifold
I satisfies the mountain pass theorem geometry. More precisely,
There existsuch thatfor all, where
For eachsuch that, there exists, satisfyingand.
Since we have already showed that
and so , we obtain (i) by choosing small enough.
In order to prove (), fix such that . For all consider the function defined by
where, as before,
An easy calculation shows that
the last inequality being a consequence of the Ambrosetti–Rabinowitz inequality. Observe that for .
Thus, we obtain
proving that
for a constant .
It follows from (2.13) that
Thus, it suffices to take τ large enough. □
The existence of a Palais–Smale sequence such that
where
and results from the mountain pass theorem without the PS condition.
We now consider the Nehari manifold
It is not difficult to see that is a manifold in .
The next result, which follows immediately from our estimates, proves that is a closed manifold in :
There existssuch thatfor all.
An alternative characterization of c is obtained by a standard method: for , consider the function , preserving the notation of Lemma 3.1. The proof of Lemma 3.1 assures that for t small enough, for t large enough and if . Therefore, is achieved at a unique and for and for . Furthermore, implies that .
The map () is continuous and , where
For details, see [27, Section 3] or [17].
Standard arguments prove the next affirmative:
Letbe a sequence such thatand, whereThenis bounded and (for a subsequence)in.
Letbe any open set. For, letbe a bounded sequence insuch thata.e. Then.
The proof of Lemma 3.4 can be found, e.g., in [20, Lemme 4.8, Chapitre 1].
The limit problem
In this section we consider a variant of problem (1.3), changing the potential for .
Assuming,,and, problemhas a non-negative ground state solution.
Let be the minimizing sequence given by Lemma 3.1. Then, there exist and a sequence such that
If false, a result of Lions (see [24]) guarantees that in for , thus implying that
contradicting Lemma 3.2.
We define
From (4.1) we derive that
We observe that the energy functional
and its derivative as well are invariant with respect to any translation in . Therefore, it also holds that
where
(Observe that all reasoning in Section 3 is valid for and its minimizing sequence.)
Since is bounded (see Lemma 3.3) it follows that . A standard argument shows that we can suppose a.e. in , in for all , a.e. in and in , for all .
We will show that .
For all , let us consider . We have
We start considering
Because , it follows from [1, Lemma 3.5] that when and thus is easily verified that when . We now consider :
We infer that
Since
(because is bounded), we deduce that
Thus
and .
We now turn our attention to the positivity of v. Seeing that
and choosing , the left-hand side of the equality is positive (by the definition of and equation (3.1) applied to ), since ), while . We are done. □
In order to consider the general case of the potential , we state a well-known result due to M. Struwe:
(Splitting Lemma).
Letbe such thatandweakly on. Thenand we have either
strongly on;
there exist,such thatforand nontrivial solutionsof problem (
P
∞
) so thatand
The functional I satisfiesfor any.
Let us suppose that satisfies
We can suppose that the sequence is bounded, according to Lemma 3.3. Therefore, for a subsequence, we have in . It follows from the Splitting Lemma (Lemma 5.1) that . Since
and
we conclude that
as consequence of the Ambrosetti–Rabinowitz condition.
If in , by applying again the Splitting Lemma we guarantee the existence of and nontrivial solutions of problem (
P
∞
) satisfying
contradicting our hypothesis. We are done. □
We prove the next result by adapting the proof given in Furtado, Maia e Medeiros [19]:
Suppose thatsatisfies. Thenwhere c is characterized in Lemma
3.3
.
Let be the weak solution of (
P
∞
) given by Theorem 5 and be the unique number such that . We claim that . Indeed,
thus yielding
If , since is increasing, the first integral is non-negative and, since F is increasing, the second integral as well. We conclude that .
Lemma 3.3 and its previous comments show that
Since
is a strictly increasing function, we conclude that
proving our result. □
Let be the minimizing sequence given by Lemma 3.1. It follows from Lemmas 5.2 and 5.3 that such that and .
We now turn our attention to the positivity of u. Seeing that
and choosing , the left-hand side of the equality is positive (by the definition of and equation (3.1), since ), while . The proof is complete. □
The proof of the next result adapts arguments in [4] and [12].
For allit holdswhere, C,,andare positive constants andis the function given by Lemma
2.1
.
Choosing in (2.11), where and , we have and
Since , the left-hand side of (6.1) is given by
where .
Now we express (6.3) in terms of . For this, we note that . Therefore,
thus yielding
where . Gathering (6.1), (6.3) and (6.4), we obtain
We now start to consider the right-hand side of (6.5). Since , Corollary 2.1 shows that it can be written as
Applying Lemmas 2.2 and 2.3, inequality (6.6) becomes
where , admitting that .
Because for all , the last inequality is equivalent to
Let us consider the last integral in the right-hand side of (6.7). For all , define and . Then, whereas ,
and when .
If M is taken so that , we have
The Hölder inequality guarantees that
where
Thus,
and substitution on the right-hand side of (6.8) yields
Since , it follows from (6.9) that
and we are done. (Observe, however, that M depends on β.) □
For allwe have.
Since never occurs, we have .
According to the Proposition 6.1, we have
where and are positive constants.
Choosing , it follows from (2.4) that
from what follows that the right-hand side of (6.10) is finite. We conclude that . Now, we choose so that and conclude that
After k iterations we obtain that
from what follows that for all . Since the same arguments are valid for , we have for all . □
By simply adapting the proof given in [12], we present, for the convenience of the reader, the proof of our next result:
It follows that , since for all , by Proposition 6.2. We also know that and V is bounded. Therefore, if , we have
Since , the fact that
allows us to conclude that
for a positive constant and a positive function that depends neither on T nor on β.
Therefore,
and
Since
we conclude that
and, by taking so that
we obtain
Since
for a positive constant M, it follows from (6.11) that
We now apply an iteration argument, taking and starting with . This produces
Because , we have
Thus,
from what follows for all . The same argument applies to , proving that for all .
By taking and for all p, we obtain for any ,
But and for a positive constant results from (6.12) that
Thus,
and the right-hand side of the last inequality is uniformly bounded for all . We are done. □
In addition, if, thenis a classical solution of (6.13).
In the proof of Proposition 6.4 (see [12, Proposition 3.9]), defining
taking the odd extension of h and ρ to the whole (which we still denote simply by h and ρ), in [12] is obtained that ρ satisfies the equation
and for all by applying Sobolev’s embedding. Therefore, .
In our case
We now rewrite equation (6.14) as
Since and is bounded, the right-hand side of the last equality belongs to . Thus, classical elliptic boundary regularity yields
Hence, by applying classical interior elliptic regularity directly to v, we deduce that is a classical solution of problem (1.3). □
We now adjust [12, Theorem 3.14] to our needs. The original statement guarantees that , a result that depends on the function h (of Proposition 6.4) considered in that paper. For the convenience of the reader, we present the proof of the next result:
Suppose thatis a critical point of the energy functional I, thenas, for any.
Let us consider a solution v of the problem
By applying the Fourier transform with respect to variable we obtain
from what follows
Since Proposition 6.4 shows that for all and , we conclude that when for any x and as for any . □
We now adapt the proof of [12, Theorem 5.1]. In that paper is assumed that as , a condition that is not necessary.
We denote
It follows easily that K is bounded.
By Theorem 1 we have . Applying Harnack’s inequality we conclude that w is strictly positive.
Following [12], for any we denote
and define
where the positive constants and will be chosen later on. A simple computation shows that
Thus, for R large enough, we have
We now define
We clearly have in . Choosing
we also have on and when .
We claim that in . Supposing the contrary, let us assume that . By the strong maximum principle, there exist such that for all . Defining
for some , a straightforward calculation shows that
Since , we conclude that .
Another application of the strong maximum principle yields
An application of Hopf’s lemma produces , that is,
Since , we conclude that
and so
Now, choosing , since (so that the last term in the above inequality is non-negative), the positiveness of and hypothesis guarantees that , thus reaching a contradiction with (7.1). □
Footnotes
Acknowledgements
P. Belchior was partially supported by CAPES/Brazil. O.H. Miyagaki received research grants from CNPq/Brazil 307061/2018-3 and INCTMAT/CNPQ/Brazil. G.A. Pereira received research grants by PNPD/CAPES/Brazil.
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