Some properties of solutions to the weighted Hardy-Littlewood-Sobolev type integral system

This paper is concerned with the properties of solutions for the 
weighted Hardy-Littlewood-Sobolev type integral system 
\begin{equation} 
 \left \{ 
 \begin{array}{l} 
 u(x) = \frac{1}{|x|^{\alpha}}\int_{R^{n}} \frac{v^q(y)}{|y|^{\beta}|x-y|^{\lambda}} dy,\\ 
 v(x) = \frac{1}{|x|^{\beta}}\int_{R^{n}} \frac{u^p(y)}{|y|^{\alpha}|x-y|^{\lambda}} dy \end{array} 
 \right.                     （1） 
 \end{equation} 
 and the fractional order partial differential system 
\begin{equation} \label{PDE} 
\left\{\begin{array}{ll} 
(-\Delta)^{\frac{n-\lambda}{2}}(|x|^{\alpha}u(x)) =|x|^{-\beta} 
v^{q}(x), 
 \\ 
 (-\Delta)^{\frac{n-\lambda}{2}}(|x|^{\beta}v(x)) 
=|x|^{-\alpha} u^p(x). 
\end{array}                     （2） 
\right. 
\end{equation} 
 Here $x \in R^n \setminus \{0\}$. Due to $0 < p, q < \infty$, we need more 
complicated analytical techniques to handle the case $0< p <1$ or 
$0< q <1$. 
 We first establish the 
equivalence of integral system (1) and fractional order partial 
differential system (2). For integral system (1), we prove that 
the integrable solutions are locally bounded. In addition, we 
also show that the positive locally bounded solutions are 
symmetric and decreasing about some axis by means of the method of 
moving planes in integral forms introduced by Chen-Li-Ou. Thus, 
the equivalence implies the positive solutions of the PDE system, 
 also have the corresponding properties. This paper extends 
 previous results obtained by other authors to the general case.


(Communicated by Congming Li)
Abstract. This paper is concerned with the properties of solutions for the weighted Hardy-Littlewood-Sobolev type integral system |y| α |x−y| λ dy (1) and the fractional order partial differential system Here x ∈ R n \ {0}. Due to 0 < p, q < ∞, we need more complicated analytical techniques to handle the case 0 < p < 1 or 0 < q < 1. We first establish the equivalence of integral system (1) and fractional order partial differential system (2). For integral system (1), we prove that the integrable solutions are locally bounded. In addition, we also show that the positive locally bounded solutions are symmetric and decreasing about some axis by means of the method of moving planes in integral forms introduced by Chen-Li-Ou. Thus, the equivalence implies the positive solutions of the PDE system, also have the corresponding properties. This paper extends previous results obtained by other authors to the general case.
(1. 6) and (1.3) is reduced to the 2k-order Lane-Emden type system where k ≥ 1 is an integer. Chen and Li [2] proved recently that the HLS type integral system (1.6) is equivalent to (1.7) when λ = n − 2k, where k is an integer. Thus, the radial symmetry of solutions of (1.6) implies the symmetry result of solutions of (1.7). When n−λ 2 is not an integer, they directly defined the PDE solutions of (1.7) is in fact a solution of the integral system (1.6). Following their definition, we define a solution (u, v) of (1.3) in the sense of distribution when n−λ 2 is not an integer. For the integrable positive solutions of (1.1), when p, q ≥ 1, pq = 1, Jin and Li [6,7] proved the radial symmetry, the monotonicity and their optimal integrability. Li, Lim, Lei,and Ma estimated the asymptotic rates when |x| → 0 and |x| → ∞, respectively(cf. [8,10,11]).
Lei and Lü [9] first establish (1.1) is equivalent to (1.3), then prove that the integrable solutions of (1.1) are locally bounded. In addition, they also show that the positive locally bounded solutions are symmetric and decreasing about some axis by means of the method of moving planes in integral forms introduced by Chen-Li-Ou. But when p ∈ (0, 1) or q ∈ (0, 1), there is few study of the qualitative properties of positive solutions. When n−λ λ < p, q < ∞, Hang [5] proved the regularity and radial symmetry of nonnegative integrable solutions of the system (1.1) as α = β = 0.
be a pair of nonnegative solutions of system (1.1) with (1.2). Then u and v are smooth away from the origin, radially symmetric and strictly decreasing in the radial direction. Moreover, the center of the symmetry must be the origin unless α = β = 0.
Here,λ = λ + α + β. (i) Around the origin. If λ + β(q + 1) < n, then it holds that and v(x) Here, e is a unit vector in R n and |S n−1 | is the surface area of the unit sphere.
(ii) Around the infinity. If λq + β(q + 1) > n, then it holds that and v(x) |y| β dy) p and Clearly, Propositions 1.1-1.3 are helpful for understanding the shape of the positive solutions of the system (1.1).
In this paper, we consider the properties of solutions of (1.1) for 0 < p, q < ∞. We first establish the equivalence of PDEs system (1.3) and the integral system (1.1). Next, we verify the finite energy solutions of (1.1) must be locally bounded. Finally, we prove the axisymmetry result for the locally bounded solutions of (1.1).

Remark.
(1) On the regularities of the finite energy solutions, Proposition 1.2 gives the optimal integrability. Here, Theorem 1.3 shows that the finite energy solutions must be locally bounded.
(2) Proposition 1.1 shows the finite energy solutions have radial symmetry property. Here, Theorem 1.4 shows the axisymmetry for the locally bounded solutions.
(3) According to Theorem 1.1, the result of Theorems 1.3 are still true for the positive solutions of the fractional order partial differential system (1.3).
(4) Difficulties in proving Theorem 1.1 and Theorem 1.4 are the case 0 < p < 1 or 0 < q < 1. To deal with the first difficulty in proving Theorem 1.1, we need to use the integrable condition and more complicated analytical techniques. To overcome the second difficulty in proving Theorem 1.4, we need to use the method in [12].

2.
Equivalence. In this section, we will prove Theorem 1.1 and 1.2.
To prove Theorem 1.1, we first give the following result.
Proof. Here we only prove the case 0 < p < 1, q > 1. The other cases similarly can be proved.
Otherwise, there exists x 1 such that We will deduce a contradiction by four substeps.
(i) Define the average of u, v as Here j = 1, 2, · · · , m − 1. Then for r > 0, we have Here c is an absolute constant (independent of r).
In fact, by Hölder inequality and q > 1, we easily obtain the first inequality in (2.4). We mainly prove the second inequality in (2.4).
On the other hand, for p < 1, let s > 1 be positive real number and M, N be positive integers such that s Combining with (2.5), we get the second inequality in (2.4). (2.7) By this and (2.3), we get So we can find a suitably large r 1 > 0 such thatū m−2 (r) > c 2 r 2 for r ≥ r 1 .
(ii) In general, by the same argument above, we can see that By an analogous process, we can also deduce that We claim that m must be even. If m is odd, we have a contradiction with the fact that u > 0. So m must be even.
(iv) From (ii) and (iii), we havē Combining with (2.4), we have Hence, by α 2p+1 (2.14) Then we havev By using the iteration formulas above, we have Notice that and then we take Step 2. Moreover, u m−1 > 0.
Otherwise, there exists x 0 = 0 such that u m−1 (x 0 ) = 0 and u m−1 (x) ≥ 0. This shows x 0 is a local minimizer and hence Similar to the argument in Step 1, we see that the signs ofū j are alternating (cf. (2.9)). Combining (2.3) we can deduce by finite steps that −∆ū < 0. Therefore, by integrating twice, we get Thus, similar to the argument in Step 1, from (2.4), we can choose R * sufficiently large such thatv m−1 (R * ) < 0.
Step 2 shows that it is impossible.
By the results in section 4 of [3], φ satisfies In addition, on ∂B for j = 0, 1, · · · , m − 1, there hold Letting r → ∞ and using (2.16), we see Thus, we can find r l → ∞ such that When p > 1, using the Hölder inequality, we get Then from (2.22) we deduce that as r l → ∞, Letting r → ∞, we obtain Summing j from 1 to m − 1 yields Therefore, we can find a subsequence of r l (denoted by itself) such that as r l → ∞, when r l → ∞. Inserting this result and (2.25) into (2.19), we get Similarly, is not an integer, this follows from our definition. The following is an intuitive idea behind this definition. For Namely, there exists a constant c such that Integrating by parts, we see the left hand side of (2.26) becomes Exchanging the order of the integral variables, we see the right hand side of (2.26) becomes

Thus, from (2.26) it follows
Similarly, Combining two cases I and II together, we complete the proof of Theorem 1.1.
Proof of Theorem 1.2. Case I. n − λ is even. Write m = n−λ 2 . Clearly, from (1.1) we obtain |x| α u(x) = R n |y| −β v q (y)dy |y − x| n−2m , |x| β v(x) = R n |y| −α u p (y)dy |y − x| n−2m . (2.27) We assume the solutions u, v are smooth except for the origin. In fact, the assumption can be relaxed to u, v ∈ L ∞ loc (R n \ {0}). Based on this local boundedness, the regularity of u, v can be lifted to u, v ∈ C ∞ loc (R n \ {0}) via an analogous potential estimates as in Chapter 4 in [4]. Differentiate (2.27) with respect to x, since c|x| 2m−n is the basic solution of (−∆) m u = 0, using the convolution properties of the delta function, we can see (1.3).
Case II. n − λ is a positive real number.
Assume u, v satisfy Using the Fourier transform, we have

By this result and the Parseval identity, it follows
These results show that u, v are weak solutions of PDE system.
Combining two cases I and II together, we complete the proof of Theorem 1.2.

Axisymmetry of locally bounded solutions. Assume
Not as the finite energy solutions, the singularity of the locally bounded solutions is hard to handle when we move the planes. To avoid this difficulty, we use the Kelvin type transform with translation.
We will prove that the solutions u, v of (4.4) are axis symmetric by the method of moving planes introduced in [3]. Without loss of generality, let x 0 = (0, · · · , 0, −1), then z 0 = (0, · · · , 0, 1). We move the planes along all the directions except for the axis involving 0z 0 to derive the symmetry ofū andv where 0 represents the origin. Therefore, u and v are symmetric about the axis involving 0x 0 . Thus, Theorem 1.4 is true once the following theorem is proved.
Then u, v are axially symmetric and decreasing about the axis involving the line segment 0z 0 .
Proof. We may assume q > p without loss of generality. Then by (1.2), we have q > 1 p . Let us choose ρ > 1 so that 1/p < ρ < q. For k ∈ R, define To prove Theorem 4.1, we compare u(x) with u k (x) and v(x) with v k (x) on H k , respectively. The proof consists of three steps. In step 1, we show there exists a real number R < 0 such that for k ≤ R and x ∈ H k , we have (4.10) Thus, we can start moving the plane from k ≤ R to the right as long as (4.10) holds. In step 2, we show that if the plane stops at x 1 = k 0 for some k 0 < 0, then u, v must be symmetric and monotone about the plane x 1 = k 0 ; otherwise, we can move the plane continuously. In step 3, we prove that we can move the plane all the way to x 1 = 0. Since the direction of x 1 can be chosen arbitrarily except for the axis involving oz 0 , we deduce that u, v must be symmetric and decreasing about this axis.
Step 1. Define To use the method of moving planes, we need the following formulas, which are obtained by change of variables: Similarly, we have Since for x, y ∈ H k with k < 0, (4.11) Next, we can find s > max{ n λ , n(q−1) n−λ } > 0, such that for any > 0, In fact, from (4.7), we see that there exists R > 0 such that when |x| > R, u ≤ C|x| −λ . Hence, by (4.8), we get Similarly, v has the same property.
Step 2. We now move x 1 = k to the right as long as (4.10) holds. Suppose that at a point k 0 < 0, we have on H k0 , Then the plane can be moved further to the right. More precisely, there exists an such that u(x) ≥ u k (x), v(x) ≥ v k (x), on H k for all k ∈ [k 0 , k 0 + ). (4.13) In the case v(x) ≡ v k0 (x) on H k0 , we have in fact u(x) > u k0 (x) in the interior of H k0 . Let Φ u k0 = {x ∈ H k0 : u(x) ≥ u k0 (x)}, Φ v k0 = {x ∈ H k0 : v(x) ≥ v k0 (x)}. Then, obviously Φ u k0 has measure zero and lim sup k→k0 B u k ⊂ Φ u k0 . The same is true for that of v.

YINGSHU LÜ AND ZHONGXUE LÜ
Here D * is the reflection of the set D about the plane x 1 = k. So, for all k in [k 0 , k 0 + ). This verifies (4.13).