1. Introduction and Main Results
We consider the following fractional-order system with Hartree-type nonlinearity:
()
where
N ≥ 2,
, and × denote the convolution of functions. Throughout this paper, for any
, we denote by
the Sobolev space with the norm as follows:
where
denotes the Fourier transformation of function
f, that is,
In addition, we will also use the notation
to denote the fractional Sobole space, whose detailed definition is seen in Di Nezza et al. [
1].
A pair of weak solutions (
u,
v) to Equation (
1) is defined in the distributional sense. That is, we say that
is a pair of weak solutions to Equation (
1) if and only if for any
,
()
where
By taking limits, we remark that Equation (
2) is also true for any
. It is well-known that the fractional-order equations and systems have many practical applications [
2–
5]. Especially, the fractional-order systems of type in Equation (
1) naturally arise in the Hartree–Fock theory of the nonlinear Schrödinger equations and systems ([
6] and the references therein). Indeed, if
α =
β and
u =
v in Equation (
1), then the system is Equation (
1) reduces to the following nonlocal equation:
()
This type of nonlocal equation goes back to the description of the quantum theory of a polaron at rest by Pekar in 1954 [
7] and the modeling of an electron trapped in its own hole in 1976 in the work of Choquard, as a certain approximation to Hartree–Fock theory of one-component plasma [
8]. And the solutions can be regarded as a crucial threshold or criterion for global well-posedness and scattering in the focusing case. Thus, the study on the regularity and classification of solutions to Equation (
1) are more meaningful and important. In the past years, there are many literatures on the classification of positive solutions to fractional-order equations and systems or integral equations and systems ([
9–
14] and the references therein). Especially, Liu [
15] obtained the classification results of positive solutions to Equation (
3) with
α = 2. Dai et al. [
10,
16] classified all the
weak solutions to Equation (
3) via the method of moving planes. It is well-known that the method of moving planes is a powerful tool in the research of classification of positive solutions, and the method was initially introduced by Alexandrov [
17] and refined by Serrin [
18]. Later, it was further developed by Gidas et al. [
19], Caffarelli et al. [
20], Chen and Li [
21], Li and Zhu [
22] (moving spheres), Chen et al. [
9], Chen et al. [
23], and many others. In addition, we also mention that Ma et al. [
24] established a lifting regularity lemma and hence derived the Lipschitz continuity of solutions for an integral system of Wolff type. The lifting regularity lemma has been widely applied to study the integrability and regularity of solutions for some nonlinear problems ([
16,
25] and the references therein).
In this paper, we will follow the general lines of the regularity lifting lemma and the method of moving planes and give the regularity and classification of weak solution to Equation (
1). Here, we emphasis that the problem in Equation (
1) is quite different from the previous literature, since the problem in Equation (
1) has two nonlocal terms at the same time; that is, the convolution and the operator
(
s =
α,
β) are all nonlocal. Secondly,
α and
β may be also not equal in Equation (
1), which have caused us some obstacles to obtain the classification of positive solutions. Fortunately, by more careful and refined analysis, we can overcome those obstacles and obtain the classification results. More specifically, we first establish the equivalence relationship between Equation (
1) and the following integral system:
()
where
Rs,N =
Γ((
N −
s)/2)/(
πN/22
sΓ(
s/2)). We then use the regularity lifting lemma and the method of bootstrap to obtain the integrability and smoothness of the weak solutions. We finally prove that the solutions of Equation (
4) are radially symmetric about some point
via the method of moving planes in an integral form and derive the explicit forms for
u and
v. Further, by the equivalence of Equations (
1) and (
4), we conclude that the weak solutions of Equation (
1) have the same classification results. More specifically, we have the following theorems.
Theorem 1. Let N ≥ 2 and . Assume that is a pair of weak solutions to Equation (1). Then (u, v) satisfies the system in Equation (4) and vice versa.
Theorem 2. Let N ≥ 2 and . Assume that is a pair of solutions to Equation (4). Then for any . Further, if , then , where [a] denotes the greatest integer less than or equal to a.
Remark. Let N ≥ 2 and . Assume that for some is a pair of solution to Equation (4). Then, it is clear from the proof of Theorem 2 (Section 2) that for any (r, s) ∈ (p0, +∞) × (q0, +∞) and all the conclusions in Theorem 2 also hold true.
Theorem 2 ensures that we can apply the method of moving planes in integral forms and the Hardy–Littlewood–Sobolev inequality [12] to obtain the classification of solutions for integral system in Equation (4). That is, we have the following theorem.
Theorem 3. Let N ≥ 2 and . Assume that is a pair of solutions to Equation (4). Then (u, v) must be radially symmetric and monotone, decreasing about some point . In particular, u, v must be the form
()
where
t is a positive parameter and
a,
b are constants.
Obviously, due to the equivalence relationship between the fractional-order system in Equation (1) and the integral system of Equation (4), it follows from Theorem 2 that the classification results of solutions to Equation (1) immediately; that is, we have the following corollary.
Corollary 1. Let N ≥ 2 and . Assume that is a pair of solutions to Equation (1). Then, there are the results as Theorems 2 and 3.
2. Proof of Main Results
In this section, we will give the detailed proof of Theorems
1–
3. For the sake of talking convenience, let us first give the regularity lifting lemma [
24, Lemma 2.2], which will play an important role in showing the integrability and smoothness of the solutions. That is, let
V be a topological vector space and assume that there exist two extended norms defined on
V:
Let
Lemma 1. (Regularity lifting) Let T be a contraction map from X into itself and from Y into itself. Suppose that for all f ∈ X there exists g ∈ Z = X∩Y such that f = Tf + g ∈ X. Then f ∈ Z.
In addition, we also need the Hardy–Littlewood–Sobolev inequality [
12]. That is, let
p,
q > 1 and 0 <
s <
N with
,
. Then, one has
()
where
C depends on
s,
p,
q, and
N.
Proof of Theorem
1. Let
()
Firstly, we assume that
is a pair of weak solutions to Equation (
1). Define
Then
and
. By the definition in Equation (
2) of weak solutions to Equation (
1), we have the following:
()
Obviously, by the definition of
and integrating by parts, one has the following:
()
In addition, by exchanging the order of integration, we have the following:
()
It follows from Equations (
8)–(
10) that
()
By the arbitrariness of
ϕ and Equation (
11), we have the following:
Similar to the proof of Equation (
11), if we set
Then we also have
()
which also implies that
Hence, we prove the (
u,
v) is also a pair of solutions to Equation (
4).
In what follows, we prove the solutions to Equation (
4) must be also the solution to Equation (
1). Indeed, let
be a pair of solutions to Equation (
4). Then, by taking the Fourier transform to the two equations of (
4), we get the following:
For any
, it follows that
()
and
()
This means that
is a pair of weak solutions of Equation (
1).
Proof of Theorem
2. For the sake of brevity, we first use the notation
C to denote a general positive constant that may depend on
s,
p,
q, and
N. Also, it may change from line to line. Now, let
Pα and
Pβ are as in Equation (
7). Then, by Equation (
6), it is easy to verify that
Similarly, we also have
Hence, we obtain
and
.
Now let
Tα and
Tβ be defined as follows:
and
Then, for any
and
, by using Equation (
6) and Hölder’s inequlity, one has the following:
()
and
()
Obviously, from Equations (
15) and (
16), we know that
Tα maps
into itself and
Tβ maps
into itself. To apply Lemma
1, let
and let
A > 0:
where
χΩ denotes the characteristic function of set
Ω. For any
A > 0, define
where
It follows from Equations (
15) and (
16) that
Since
and
, we can choose
A large enough such that
This means that
Therefore, for any
and
, we know that
TA is a contracting operator from
to itself for sufficiently large
A. Let
and
Then, if (
u,
v) is a pair of solutions to Equation (
4), it is easy to verify that (
u,
v) satisfies the following operator equation:
Meanwhile, by the definition of
F and
G, for given
A > 0, it is easy to verify that
()
Indeed, it is obvious that
F(
x) ≥ 0 for all
. Besides, for given
A > 0, one has the following:
Similarly, we also have the following:
()
Therefore, for any
, by Equations (
17) and (
18), one has the following:
()
and
()
On the other hand, for any
and
, by Equation (
6) and Hölder’s inequality, it is also easy to verify that if
, then
In particular, when
, one also has the following:
()
It follows from Equations (
19)–(
21) that
()
whenever
r and
s are in the following range:
()
Hence, for any
, it follows from Lemma
1 that
whenever
r and
s are in the range in Equation (
23). Let us repeat the above processes
n-times until
N ≤ min{(2
n + 1)
α, (2
n + 1)
β}. Then, we finally get
whenever
r and
s are in the following range:
Further, we also can obtain the following:
Now, for given
R > 0, denote
BR =
BR(
x), we rewrite the following:
Note that
and
. Then, it is from that Hölder’s inequality that
()
and
()
where
and
Similar to Equations (
24) and (
25), we also have the following:
()
It follows from Equations (
24)–(
26) that
and
. Further, due to
, by Hölder’s inequality, we also have the following:
()
That is to say,
. Similarly, it is also not difficult to see that
. Hence, we know that
for any
. Now, for simplicity, let
and
Then, by Equation (
1), we deduce that
for (
p,
q) ∈
Aα,β, which implies that
for any (
r,
s) ∈
Bα,β, and hence, we have
. Further, if
, then
for any (
p,
q) ∈
Bα,β. Therefore, by Equation (
6) and Hölder’s inequality, one has the following:
This means that
for any (
r,
s) ∈
Bα,β, and hence, we get
. This completes the proof of Theorem 1.2.
Proof of Theorem
3. Let
and
λ be an arbitrary real number,
Define
where
Pα and
Pβ are as in the proof of Theorem
1. Let
Now, we use the method of moving planes to prove Theorem 3; the proof is carried out by three steps.
Step 1. There exists such that for all ,
()
Indeed, for any
x ∈
Σλ, by the fact that |
xλ −
y| = |
x −
yλ| and |
xλ −
yλ| = |
x −
y|, we have the following:
()
Here and hereafter,
is defined as follows:
Similar to the proof of Equation (
29), for any
x ∈
Σλ, we also have the following:
()
Now, in order to obtain Equation (
28), let
Then, for any
, it follows from Equation (
29) that
()
where we have used the estimate |
x −
y| ≤ |
xλ −
y| for any
x,
y ∈
Σλ and
Similarly, for any
, it follows from Equation (
30) that
()
where
Let
it follows from Equations (
31) and (
6) and Hölder’s inequality that
()
where
Now, let
Note
. Hence, similar to the proof of Equation (
28), from Equation (
32), one has the following:
()
By the integrability of
Pα and
Pβ, we can choose
N0 > 0 such that if
λ < −
N0,
()
Then, it follows from Equations (
33) and (
34) that
()
and
()
Note that
. Then, from Equations (
36) and (
37), we deduce that
()
By the integrability of
u and
v, we can also choose
such that if
,
()
It follows from Equation (
39) that
and hence
must be measure zero and empty for all
. Similarly,
must also be measure zero and empty for all
. This completes the proof of Step 1.
Step 2. Moving the plane Tλ to the limiting position to derive symmetry.
In fact, Equation (
28) provides a starting point to move the plane
Tλ, and hence, we can start from this neighborhood of
x1 = −∞ and move the plane to the right as long as Equation (
28) holds to the limiting position and argue that the solution pair (
u,
v) must be symmetric about the limiting plane. More precisely, let
By using the similar argument in Step 1 and moving the plane
Tλ from near
x1 = +∞ to the left, we must have
λ0 < ∞. Now, we will show that
()
Otherwise, we may assume on
:
Obviously, by Equations (
29) and (
30), we can know that
on
if
on
and vice versa. Furthermore, by means of Equations (
29) and (
30), we can derive
and
in the interior of
. Duo to this fact, we may show that the plane can be moved further to the right. More precisely, there exists
ε > 0 such that for any
λ ∈ [
λ0,
λ0 +
ε),
This will be a contradiction with the definition of
λ0. To prove the contradiction, it can be clearly seen from Equations (
33) and (
34) in Step 1 that our primary task is to prove that we can choose
ε > 0 sufficiently small such that for all
λ ∈ [
λ0,
λ0 +
ε),
()
and
()
where
C is the same as in Equations (
33) and (
34). In fact, by the integrability conditions on
Pα,
Pβ,
u, and
v, we can choose
R > 0 large enough, such that
()
and
()
Now, we fix the above
R; in order to derive Equations (
41) and (
42), by the integrability conditions on
Pα,
Pβ,
u, and
v, we only need to show that
()
In fact, for any
δ > 0 and
λ >
λ0, let
and
Then, it is easy to verify that
()
()
For fixed
η > 0, by Equation (
46), we can choose
δ > 0 small enough such that
μ(
Fδ) ≤
η. For this fixed
δ, we claim that
()
Indeed, for any
, we have the following:
Hence, we obtain the following:
It follows from Chebyshev inequality that
()
as
λ →
λ0, where
. This finished the proof of Equation (
48). Hence, by Equations (
46)–(
48), we have the following:
()
By the arbitrariness of
η > 0, Equation (
50) implies that
()
Similar to the proof of Equation (
51), we can also verify that
()
Obviously, by Equations (
51) and (
52), we finished the proof of Equation (
45). Therefore, from Equations (
43)–(
45), one has Equations (
41) and (
42). This means that there exists
ε > 0 such that for all
λ ∈ [
λ0,
λ0 +
ε),
This is a contradiction with the definition of
λ0. Hence, Equation (
40) must be hold true. Furthermore, by the definition of
Pα and
Pβ, it is easy to verify that
and
for all
. In addition, because of the fact that Equation (
4) is invariant under rotation and the
x1 direction can be chosen arbitrarily, this means that the positive solution
must be radially symmetric and monotone, decreasing about some point
. That is, we finished the proof of Step 2.
Step 3. We verify that u, v must be the form of (5).
We use the analytic method in Chen et al. [
9] and divide several small steps to complete the proof of this step. Firstly, by the conclusion of Lemma 4.1 in Dai et al. [
16], we know that Equation (
5) is a pair of solutions to Equation (
4). Meanwhile, we claim that the system in Equation (
4) is invariant under the scaling
()
In fact, if (
u,
v) is a pair of solutions to Equation (
4), then
()
Similarly, we also have the following:
()
From Equations (
54) and (
55), we know that the system in Equation (
4) is invariant under the scaling of Equation (
53).
Secondly, let (
u,
v) be a pair of solutions to Equation (
4); we claim that there exist two positive numbers
u∞ and
v∞ such that
()
and
()
Now let us verify Equations (
56) and (
57) by contradiction. Without loss of generality, assume that Equation (
56) does not hold, and let
()
Note that
and
are usually called the Kelvin transform
u and
v, respectively. Then
also satisfies the integral system in Equation(
4) in
. Indeed, let
Then, by a direct calculation, one has the following:
()
Similarly, we also have the following:
()
Hence, by Equation (
59), one has the following:
()
By Equation (
60), similar to the proof of Equation (
60), one has the following:
()
From Equations (
61) and (
62), we know that
also satisfies the integral system in Equation (
4) in
. Let
x1 ≠
x2 be any two points in
and
.
Now we consider the Kelvin transform of
u,
v centered at
x0. That is, let us define the following:
and
According to our hypothesis (Equation (
56) does not hold),
U(
x) must have a singularity at
x0. Meanwhile, from
, one has the following:
In addition, by the above proof, we know that (
U,
V) is also a pair of solutions to Equation (
4). Therefore, by the same arguments as in Step 1 and Step 2, we can deduce that
U and
V are radially symmetric and monotone, decreasing about point
x0, since
x0 is a singular point of
U. This means that
u(
x1) =
u(
x2) for any two points
x1 and
x2 in
, and hence,
u must be a constant. Further, by the system of integral in Equations (
4) or fraction order differential system in Equation (
1), we know that
u must be zero, which contradicts with
u > 0 in
. Therefore, we proved that
u and
v must satisfy the asymptotic property in Equations (
56) and (
57), respectively.
Thirdly, let
. From Step 1 and Step 2, we can assume that (
u,
v) is a pair of solutions to Equation (
4) centered at zero and
sN−αu(
ξ) =
u∞ and
sN−βv(
ξ) =
v∞. Then, we claim the following:
()
In fact, we only need to consider
ξ = 0. Otherwise, Equation (
63) follows by considering (
u(⋅+
ξ),
v(⋅+
ξ)) instead of (
u,
v). Let
x0 ≠ 0 and
. Let
()
Then we have
and
. Moreover,
is a pair of solutions to Equation (
4). By the proof of Step 1 and Step 2, it is easy to verify that
must be radially symmetric about
. Hence, similar argument as in the proof of Lemma 3.1 in Chen et al. [
9], we get Equation (
63):
Finally, by the first to third of this step and the same arguments as in the proof of Lemma 3.2 in Chen et al. [9], we can infer that u, v must be the form of Equation (5). This completes the proof of Theorem 3.