QUALITATIVE PROPERTIES OF SOLUTIONS TO AN INTEGRAL SYSTEM ASSOCIATED WITH THE BESSEL POTENTIAL

. In this paper, we study a diﬀerential system associated with the Bessel potential: where f 1 u x I is the identity operator and ∆ = nj =1 2 ∂x 2 j is the Laplacian operator in R n . Under some appropriate conditions, this diﬀerential system is equivalent to an integral system of the Bessel potential type. By the regularity lifting method developed in [4] and [18], we obtain the regularity of solutions to the integral system. We then apply the moving planes method to obtain radial symmetry and monotonicity of positive solutions. We also establish the uniqueness theorem for radially symmetric solutions. Our nonlinear terms f 1 ( u ( x ) ,v ( x )) and f 2 ( u ( x ) ,v ( x )) are quite general and our results extend the earlier ones even in the case of single equation substantially.

1. Introduction. The main purpose of this paper is to obtain the regularity of solutions to the integral and differential systems of the Bessel potential type and establish the radial symmetry and monotonicity of positive solutions to such systems. We also establish the uniqueness theorem for radially symmetric solutions.
In particular, for the single semi-linear fractional partial differential equation with Bessel potential Kwong [15] established the uniqueness of the positive, radially symmetric solution to the differential equation (1.4) for α = 2. Ma and Chen [19] exploited a new Hardy-Littlewood-Sobolev type inequality for the Bessel potential and proved that every positive solution of corresponding integral equation of (1.4) is radially symmetric and strictly decreasing about some point for any α > 0. Lam and Lu established in [16] the existence of positive solutions to a class of polyharmonic equation of Bessel potential type on bounded domains when the nonlinearity has the exponential growth and doesn't satisfy the Ambrosetti-Rabinowitz condition. Bao, Lam and Lu [1] further studied the Bessel type poly-harmonic equation on the entire space when the nonlinear terms have the critical exponential growth in the sense of Adams' inequalities. In [1], the authors show that solutions are uniformly bounded and Lipschitz continuous. They also proved that positive solutions are radially symmetric and monotone decreasing about some point. For more results about Bessel potential, please see [12,13,20,23] and the references therein. Recently, Chen, Li and Ma [18] developed further some new methods to prove the regularity of the solutions to a system of integral equations associated with the Wolff potentials by combinations of contracting and shrinking operators. We use their methods to prove the first two theorems for the integral system (1.2).
In this paper, we always assume that λ i , µ i , γ i (i = 1, 2) are nonnegative constants and they are not equal to zero simultaneously. We also use the following notations We note that one cannot use interpolation to conclude what the spaces Π 1 or Π 2 are because t 1i , t 2i and k 0 cannot be compared easily. Therefore, it is highly nontrivial to derive the L s integrability of the solutions u and v in Theorem 1.1.
a pair of nonnegative solutions of integral system (1.2). Then u(x) and v(x) are Lipschitz continuous.
It is well known that the method of moving plane was invented by the Soviet mathematician Alexandrov in the 1950s. Then it was further developed by Serrin [21], Gidas, Ni and Nirenberg [10,11], Caffarelli, Gidas and Spruck [2], Chen and Li [5], Chang and Yang [3] and many others. However, their methods of moving plane are mainly based on the maximum principle, and hence one is not able to apply it in the absence of the maximum principle. In [7,8], Chen, Li and Ou developed the method of moving plane in integral forms. The moving plane in integral forms can be easily applied to higher order equations without maximum principles. To establish the symmetry of the solutions, we use the moving plane in integral forms to prove the following theorem. For more results related to the moving plane in integral forms, please see [4,5,6,9,14,17] and the references therein. We also can obtain the following uniqueness theorem. Theorem 1.4. Let α = 2 and λ 1 = µ 2 = 0, µ 1 = λ 2 , γ 1 = γ 2 , q 1 = p 2 , α 1 < α 2 , α 1 + β 1 = α 2 + β 2 . Then any pair of radially symmetric solutions (u, v) satisfying integral system (1.2) is unique, moreover u = v. This paper is organized as follows. In Section 2, we apply regularity lifting by contracting operators to prove Theorem 1.1. In Section 3, we obtain every pair of nonnegative solutions of integral system (1.2) is Lipschitz continuous, thus the proof of Theorem 1.2 is complete. In Section 4 and Section 5, we prove the radial symmetry and uniqueness of positive solutions of integral system (1.2), thus we complete the proofs of Theorem 1.3 and Theorem 1.4.

2.
The proof of Theorem 1.1. In order to prove Theorem 1.1, we give some lemmas.
Let V be a topological vector space. Suppose there are two extended norms (i.e., the norm of an element in V might be infinity) defined on V, Lemma 2.1 (Regularity lifting I). Let T be a contraction map from X into itself and from Y into itself. Assume that for any f ∈ X, there exists a function g ∈ Z := X ∩ Y such that f = T f + g ∈ X. Then f ∈ Z.
Lemma 2.2 (Decay estimate of g α ). Given α > 0, the Bessel kernel g α (x) has the decay estimate when |x| ≥ 2, and when |x| ≤ 2, Combining (2.1) and (2.2), we can write Denote the norm in the cross product space It is easy to verify that (u a , v a ) is a pair of solutions of integral system (2.3).
In order to prove Theorem 1.1, it suffices to verify that ( We divide the proof into two steps. Step

we use Lemma 2.3 and Minkowski inequality to write
. Similarly, we can also obtain (2.6)

LU CHEN, ZHAO LIU AND GUOZHEN LU
Hence, combining (2.4), (2.5) and (2.6), we obtain (2.7) By (2.7), we have It turns out that T is contracting from L s (R n ) × L s (R n ) to itself. Next, we estimate F 1 (x) and F 2 (x). Since the estimates of F 1 (x) and F 2 (x) are similar, we only consider F 1 (x). We write Then for all s > 1, (2.9) Combining (2.8), (2.9) and Lemma 2.1, we accomplish the proof of Step 1.
Step 2. We show (u, v) ∈ L ∞ (R n ) × L ∞ (R n ). One only need to prove that u ∈ L ∞ (R n ). By integral system (1.2), we have where 1 mi + 1 ni = 1, i = 1, 2, 3. We choose sufficiently large n 1 , n 2 , n 3 such that u p1 L n 1 p 1 , v q1 L n 2 q 1 , u α1 L n 3 (α 1 +β 1 ) , v β1 L n 3 (α 1 +β 1 ) < ∞. Therefore, one only need to verify that g α L m i < ∞ when m i is sufficiently close to 1. Since g α has exponential decay at infinity, we only need to deal with g α at origin. According to Lemma 2.2, we separate three cases.
(i) If 0 < α < n, then Therefore, we accomplish the proof of Step 2.
3. The proof of Theorem 1.2. In this section, we need the following regularity lifting II to prove Theorem 1.2.
Lemma 3.1 (Regularity lifting II). Let X and Y be an "XY -pair", and assume that X, Y are both complete. Let X and Y be closed subsets of X and Y respectively, and T be an operator, which is contracting from X to X and shrinking from Y to Y . Define Sw = T w + g for some g ∈ X ∩ Y, and assume that S : X ∩ Y → X ∩ Y.
Then there exists a unique solution u of the equation w = T w + g in X , and u ∈ Y .
Remark 1. X and Y are called an "XY -pair" if whenever the sequence {u n } ⊆ X with u n → u in X and u n Y ≤ C will imply u ∈ Y . In Theorem 1.2, we choose For a fixed real number m > 0, define

LU CHEN, ZHAO LIU AND GUOZHEN LU
Consider the following integral system It is easy to verify that (u, v) is a pair of nonnegative solutions of integral system (3.1). In order to show (u, v) ∈ Λ 1 (R n ) × Λ 1 (R n ), we carry out the proof by four steps Step 1. For any (φ 1 , ϕ 1 ), (φ 2 , ϕ 2 ) ∈ X ,

(3.2)
According to Lemma 2.2, we can choose sufficiently small m such that

Combining (3.2) and (3.3), we have
Step 2. For any (φ, ϕ) ∈ Y, Similarly, we can choose sufficiently small m such that Combining (3.4), (3.5) and (3.6), we have Step 3. We show that (G 1 , G 2 ) ∈ X ∩ Y. Observe that (3.8) In order to show that (G 1 , G 2 ) ∈ Y, we only need to show that G 1 and G 2 are Lipschitz continuous because of (3.8). Since the estimates of G 1 and G 2 are similar, we only prove that G 1 is Lipschitz continuous. Considering that g α (x) is bounded for |x| ≥ m, we have Therefore, G 1 is Lipschitz continuous.
Step 4. To show that S: X ∩ Y → X ∩ Y. For (φ, ϕ) ∈ X ∩ Y, since T m is contracting from X to L ∞ (R n )×L ∞ (R n ) and shrinking from Y to Λ 1 (R n )×Λ 1 (R n ), it is easy to verify that By (3.4) and (3.7), we have which implies S(φ, ϕ) ∈ X ∩ Y. Therefore, we can conclude that u and v are Lipschitz continuous. This completes the proof of Theorem 1.2.
4. The proof of Theorem 1.3. In this section, we use the method of moving plane in integral forms introduced by Chen, Li and Ou [7] to prove that each pair of positive solutions (u, v) of (1.1) is radially symmetric and strictly decreasing about some point. In order to prove our theorem, we first introduce some notations. For any real number λ, let the moving plane be T λ = {x : x 1 = λ} and denote Σ λ = {x = (x 1 , x 2 , · · · , x n ) : x 1 ≤ λ}. Let x λ = (2λ − x 1 , x 2 , · · · , x n ) be the reflection of the point x about the moving plane T λ , and define u λ (x) = u(x λ ) and v λ (x) = v(x λ ).  If (u, v) is a pair of positive solutions of (1.2), for any x ∈ Σ λ and x = y, we have , v λ (y)) dy.
Proof. By (1.2), we have Hence we use the fact that |x − y λ | = |x λ − y| and |x λ − y λ | = |x − y|, we obtain Similarly, we can obtain the second formula. This completes the proof of Lemma 4.1.
We start to prove Theorem 1.3. One can split the proof into two steps.
Step 1. For λ < 0, we define We want to show that both Σ u λ and Σ v λ are empty for λ sufficiently negative. By Lemma 4.1, for any x ∈ Σ u λ , we have For any s > max{p i , q i , α i + β i } (i = 1, 2), we apply Lemma 2.3 and Minkowski inequality to (4.1) and obtain, where In virtue of the conditions (u, v) ∈ Π 1 × Π 2 , we can choose sufficiently negative λ such that = 0, which implies that Σ u λ and Σ v λ must be measure zero. Therefore, both Σ u λ and Σ v λ must be empty sets.
Step 2. For sufficiently negative λ, we have shown that The inequality (4.5) provides a starting point to move the plane T λ = {x ∈ R n : x 1 = λ}. Now we start from the negative infinity of x 1 -axis and move the plane to the right as long as (4.5) holds. Define We show that if λ 0 < 0, both u and v are symmetric and strictly decreasing about the plane x 1 = λ 0 . If not, we may assume that u λ0 (x) ≡ u(x). By Lemma 4.1, we can obtain v λ0 (x) ≡ v(x). Furthermore, we can also derive u λ0 (x) > u(x) and v λ0 (x) > v(x) in the interior of Σ λ0 .
Next, we will show that the plane can be moved further to the right. More precisely, there exists an ε > 0 such that for any λ ∈ [λ 0 , λ 0 + ε), We can derive that Σ u λ0 and Σ v λ0 are measure zero, and lim λ→λ0 Σ u . This together with integrability conditions (u, v) ∈ Π 1 × Π 2 ensures that one can choose ε small enough such that for all λ ∈ [λ 0 , λ 0 + ε) Using the similar estimates as (4.2) and (4.3), we can derive that u − u λ L s (Σ u λ ) = v − v λ L s (Σ v λ ) = 0, thus Σ u λ and Σ v λ must be measure zero, which verifies (4.6). If λ 0 ≥ 0, we can move the plane in the negative x 1 direction from positive infinity toward the origin. If the plane T λ stops somewhere before the origin, we conclude that u and v are symmetric and strictly decreasing about the plane. If they stop at the origin, we also obtain that u and v are symmetric and strictly decreasing in the x 1 direction by combining the two inequalities obtained in the two opposite directions. Since we can choose any direction to start the process, we conclude that u and v are radially symmetric and strictly decreasing about some point.
Then, integrating both sides of the above two equations two times, we have To prove the uniqueness, we only need to show that u(0) = v(0). Otherwise, suppose that u(0) < v(0). By the continuity, for small r > 0 and α 1 < α 2 , we have u(r) < v(r).
Then we obtain that u(0) = v(0). Furthermore, by the standard ODE theory, we derive that u(r) = v(r). This completes the proof of Theorem 1.4.