THE PROPERTIES OF POSITIVE SOLUTIONS TO SEMILINEAR EQUATIONS INVOLVING THE FRACTIONAL LAPLACIAN

. Let Ω be either a unit ball or a half space. Consider the following Dirichlet problem involving the fractional Laplacian where α is any real number between 0 and 2. Under some conditions on f , we study the equivalent integral equation here G ( x,y ) is the Green’s function associated with the fractional Laplacian in the domain Ω. We apply the method of moving planes in integral forms to investigate the radial symmetry, monotonicity and regularity for positive solutions in the unit ball. Liouville type theorems-non-existence of positive solutions in the half space are also deduced.

where α is any real number between 0 and 2. Under some conditions on f , we study the equivalent integral equation here G(x, y) is the Green's function associated with the fractional Laplacian in the domain Ω. We apply the method of moving planes in integral forms to investigate the radial symmetry, monotonicity and regularity for positive solutions in the unit ball. Liouville type theorems-non-existence of positive solutions in the half space are also deduced.

Introduction.
In this paper, we analyze the behavior of solutions to the Dirichlet problem for semilinear equation (1) which involves the fractional Laplacian, here Ω = B 1 or R n + . And B 1 = B 1 (0) = {x ∈ R n : |x| < 1} is the unit ball in R n , R n + = {x = (x 1 , x 2 , · · · , x n ) ∈ R n | x n > 0} is the upper half Euclidean space. B c 1 is the complement of B 1 , R n − = {x = (x 1 , x 2 , · · · , x n ) ∈ R n | x n ≤ 0} is the the complement of R n + . For the various applications of the fractional Laplacian in anomalous diffusion and quasi-geostrophic flows, turbulence and water waves, molecular dynamics, and relativistic quantum mechanics of stars, probability and finance, we refer the readers to see [1,2,4,6,7,13,18,20].
The fractional Laplacian in R n is a nonlocal differential operator, it has three equivalent definitions.
The first definition takes the form where α is any real number between 0 and 2 and P.V. stands for the Cauchy principle value.
In addition, the definition of fractional Laplacian in a domain Ω ⊂ R n is to restrict the integration to the domain: known as the regional fractional Laplacian [15]. The second definition is using the extension method introduced by Caffarelli and Silvestre in [5]: The third definition is in S, the Schwartz space of rapidly decreasing C ∞ functions in R n , by the Fourier transform where u is the Fourier transform of u. And it can be extended to a wider space of distributions Then in this space, we define (− ) α 2 u as a distribution by For any f ∈ L 1 loc (Ω), we say that u ∈ L α 2 is a solution of (− ) In this paper, we consider the distributional solutions in sense of (6) . For the properties of solutions for equations involving the fractional Laplacian in R n , we refer the readers to see [12,14,17,21].
In [8], Chen, Fang and Yang considered the Dirichlet problem involving the fractional Laplacian(1), where Ω is either a unit ball or a half space R n + . Instead of using the conventional extension method of Cafarelli and Silvestre [5], the authors have employed a new and direct approach by studying the equivalent integral equation where G(x, y) is the Green's function associated with the fractional Laplacian in the domain Ω. Applying the method of moving planes in integral forms, they established the symmetry, monotonicity of the positive solutions to the Dirichlet problem for semilinear equation under the conditions and (2) f (·) is monotonic and f (u) ∈ L n α (B 1 ) or where C 1 , C 2 , C 3 and β 1 can be any nonnegative constants, while β 2 is some non- for some β > α n−α and they also proved the regularity of the positive solutions to the Dirichlet problem for semilinear equation (9) or integral equation (7). In [9], Chen, Fang and Yang established the equivalence between the Dirichlet problem involving the fractional Laplacian (1) when Ω is a half space R n + and and integral equation Where here s = |x − y| 2 and t = 4x n y n . And the non-existence of the solution of (14) has also been proved. In this paper, for (0.1), when Ω = B 1 , we consider some other conditions different with (12) and (13) u is decreasing and | f (u) u | ≤ C 1 + C 2 |u| β , and |u| β ∈ L n α (B 1 ), where C 1 and C 2 are nonnegative constants, β is any real number. When Ω = R n + , 1076 RONGRONG YANG AND ZHONGXUE LÜ relative to (14) and (15), we consider f (u) for a more general case to get the Liouville theorem and the equivalence between and Our main results as follows.
u | ≤ C 1 + C 2 |u| β , C 1 and C 2 are nonnegative constants, β is any real number . If |u| β ∈ L n α (B 1 ), then every positive solution of integral equation (7) is radially symmetric about the origin and strictly decreasing in the radial direction. Corollary 1. Under the conditions of Theorem 1, if u is a positive solution of (9) with f (u) ∈ L q (B 1 ), then it is radially symmetric about the origin and strictly decreasing in the radial direction.

Remark 1.
When β > α n−α , we know that u ∈ L nβ α (B 1 ) and u β ∈ L n α (B 1 ) are equivalent. In our paper, we consider some negative β. u ∈ L n|β| α (B 1 ) and u β ∈ L n α (B 1 ) can not be equivalent as we wish. So we take them as above conditions. Under the conditions u ∈ L n|β| α (B 1 ) and u β ∈ L n α (B 1 ), we find that the uniform boundedness of the solution of (7) or of (9) without regularity lifting, at the same time, we only need | f (u) u | ≤ C 1 + C 2 u β , β < α α−n and n ≥ 3. Remark 2. Under conditions (12) and (13), we can get the same result with another method by estimating f (u) Then u is also a solution of (17) and vice versa.
If u is a nonnegative solution of (17). Then u ≡ 0.
2. Symmetry of solutions in the ball. Similar to [8], in this section, we obtain the radial symmetry and monotonicity of positive solutions for integral equation (7) by using the method of moving planes in integral forms.

Properties of the Green's function and
is the region in the ball B 1 between the plane x 1 = −1 and the plane x 1 = λ, To use the method of moving planes, we need the following lemmas from [8]. They are about the Green's function G 1 (x, y) and f (u) u . and The proofs of these two lemmas are standard and can be founded in [8].
The proof of this lemma can be seen in [10] or [11].
Proof. Since f (u) u is decreasing, without loss of generality, we assume u 1 > u 2 , then The same procedure may be easily adapted to obtain This completes the proof of the lemma.

2.2.
The proof of Theorem 1. The proof of Theorem 1 contains two steps.
Step 1: We first prove by contradiction For any x ∈ Σ λ , we show that Σ − λ is almost empty.
By Lemma 2.2, we have the below result: Since f is increasing and Applying (21) and (24) to (23), we arrive at: for any In above inequalities we have used Lemma 2.4, the definition of Σ − λ , the monotonicity of f and here (26) is obviously derived by the expression of G 1 .
Step 2: We now move the plane T λ continuously towards the right as long as (22) holds. Define We prove that λ 0 must be 0. Otherwise, we suppose λ 0 < 0. First, we will show that ω λ0 > 0 in the interior of Σ λ0 by contradiction.
Due to the definition of Σ λ , for any Similarly, Due to (29) and (30), it holds By Lemma 2.1(i), for any x, y ∈ Σ λ , x = y, we have That is From Lemma 2.1(i), the monotonicity of f and (31),(32), we get The above last inequality has used Lemma 2.1(ii), u < u λ and the monotonicity of f .
2.3. The proof of Corollary 1. We find that the equivalence between (7) and (9) when f (u) ∈ L q (B 1 ) has been proved in [8]. So if u is a positive solution of (9) with f (u) ∈ L q (B 1 ), then it is radially symmetric about the origin and strictly decreasing in the radial direction.
3. Regularity of solutions. In this section, we first prove the uniform boundedness of solutions of (7) or of (9) by Regularity Lifting Lemma which is introduced in [11]. We then get the same result with another method by estimating f (u) .
Assume that the spaces are complete under corresponding norms, and the convergence in X or in Y implies the convergence in V .
Lemma 3.1 (Regularity Lifting). Let T be a contracting map from X into itself and from Y into itself. Assume that f ∈ X, and that there exists a function g ∈ Z := X ∩ Y such that f = T f + g in X. Then f also belongs to Z.

3.2.
The proof of Theorem 2. Proof: We only consider integral equation of (7). We will show that for any p > n n−α , u ∈ L p (B 1 ) (34) by using the Regularity Lifting Lemma. For any real number a > 0, let Then Obviously, we have u(x) = T a u(x) + J(x), ∀x ∈ B 1 . According to Regularity Lifting Lemma, we need T a is a contracting map from L p (B 1 ) to L p (B 1 ) and J(x) ∈ L p (B 1 ).
We first prove that for a sufficiently small a, T a is a contracting map from L p (B 1 ) to L p (B 1 ) for any p > n n−α .

4.
Equivalence between the two equations on R n + .

Some useful lemmas.
In this section, we establish the equivalence between (16) and (17). We need some lemmas.
Lemma 4.1 (Boundary Harnack, see [5] or [3]). Let f, g : R n + → R be two nonnegative functions such that (− ) s f = (− ) s g = 0 in a domain Ω. Suppose that x 0 ∈ ∂Ω, f (x) = g(x) = 0 for any x ∈ B 1 \Ω, and ∂Ω ∩ B 1 is a Lipschitz graph in the direction of x 1 with Lipschitz constant less than 1. Then there is a constant C depending only on dimension such that The following uniqueness of α-harmonic functions on half spaces was derived based on the above Harnack inequality in [8].

RONGRONG YANG AND ZHONGXUE LÜ
Consequently, we have either ω(x) ≡ 0, x ∈ R n or there exists a a 0 > 0, such that The following lemma is a Maximum Principle from [19]. G R (x, y)f (u(y))dy.
From the continuous assumption of f , one can see that, for each R > 0, v R (x) is well-defined and is continuous, and x ∈ R n − . We see from Lemma 4.2 that ω ≡ 0, ∀ x ∈ R n . Therefore On the other hand, suppose u(x) is a solution of integral equation (17), applying (−∆) α 2 to both sides, then by exchanging (−∆) α 2 with the integral on the right hand side, we get (16). So we finish the proof. 5. The Liouville type theorem in R n + .

5.1.
The properties of the Green's function in this half space. Let λ be a positive real number and T λ = { x ∈ R n + | x n = λ}. Σ λ = {x = (x 1 , · · · , x n−1 , x n ) ∈ R n + |0 < x n < λ} is the region between the plane x n = 0 and the plane x n = λ, Σ C λ = R n + \ Σ λ is the complement of Σ λ in R n + , Σ λ = {x λ |x ∈ Σ λ } is the reflection of Σ λ about the plane T λ .
The following three lemmas will be used in the proof of Theorem 4.

RONGRONG YANG AND ZHONGXUE LÜ
Due to (49), (50) and Lemma 5.1, we get So we finish the proof of Lemma 5.2. Proof. From Lemma 5.3, we know that a nonnegative solution u of (17) is either strictly positive or identically zero in R n + . Without loss of generality, we assume that u > 0 in R n + , then we will derive a contradiction. The proof of Theorem 4 consists of two steps. Step We will show that for sufficiently small λ, Σ − λ must be measure zero. According to Lemma 5.2, Lemma 5.1 and the definition of Where ψ(y) is valued between u λ (y) and u(y). Hence on Σ − λ , we have 0 ≤ u λ (y) ≤ ψ(y) ≤ u(y).
By the expression of G ∞ (x, y), we can see that By the Hardy-Littlewood-Sobolev inequality and Hölder inequality, we obtain, for any q > n n−α , Since f (ψ) ∈ L n α (R n + ), choose a small positive constant λ such that Then (54) implies that ω λ L q (Σ − λ ) = 0. Thus, Σ − λ is measure zero.
Step 2. Due to Step 1, we can start from the small λ. So the plane T λ can be moved as long as ω λ (x) = u λ (x) − u(x) ≥ 0, a.e. x ∈ Σ λ holds.
Under this assumption, we can show that there exists an ε > 0, such that for ∀ λ ∈ [λ 0 , λ 0 + ε) this implies the plane can be moved further up. This contradicts the definition of λ 0 . So the assumption (58) is not right, which means (57) holds. Now we prove (59). Similar to step 1, to prove (59) is to prove that the measure of Σ − λ is zero. So the key is Thus, what remaining is to prove (60). The method is the same as [8].
Since C R n + \B R |f (ψ(y))| n α dy α n < η. then we only need to show that Σ − λ ∩B R is sufficiently small for λ close to λ 0 .
This completes the proof of Theorem 4.