SLIDING METHOD FOR THE SEMI-LINEAR ELLIPTIC EQUATIONS INVOLVING THE UNIFORMLY ELLIPTIC NONLOCAL OPERATORS

. In this paper, we consider the uniformly elliptic nonlocal operators where a ( x ) is positively uniform bounded satisfying a cylindrical condition. We ﬁrst establish the narrow region principle in the bounded domain. Then using the sliding method, we obtain the monotonicity of solutions for the semi-linear equation involving A α in both the bounded domain and the whole space. In addition, we establish the maximum principle in the unbounded domain and get the non-existence of solutions in the upper half space R n + .

Roughly speaking, the cylindrical condition (A 2 ) is equivalent to say the function a has the same value on the surface of any cylinder centering the x n -axis. If we assume that u ∈ C 1,1 loc ∩ L α (R n ), where L α is defined as then the above integral is well defined. Also, one can refer to Appendix for more details. The operator A α arises from the stochastic control problems and stochastic games (see [7]). Using the compactness and perturbation method, Caffarelli and Silvestre [8] obtained the C 1,α regularity result for the purely nonlocal Isaacs equations, which is similar to the equation (1.1), where a(x, y) satisfies (A 1 ), |∇ y a| ≤ C|y| −1 , and a(x, y) is continuous in x for a modulus of continuity independent of y. In this paper, we want to study the semi-linear equation (1.1), under the conditions (A 1 )-(A 2 ), without the decay and regular properties on a(x).
If a(x) = 1, then the operator A α becomes the traditional fractional Laplacian operator (−∆) α 2 . The difficulty of (−∆) α 2 can be overcome by using the extension method introduced by Caffarelli and Silvestre [6] that reduces the nonlocal problem into a local one in the higher dimensions. One can also use the integral equations method, like the method of moving planes in integral forms and regularity lifting to investigate some properties of the equation that involves the fractional Laplacian operator, see [9,12,14,15,17] and the reference therein.
However, it seems that the extension method and the integral equations method don't work for more general nonlocal operators like A α , the fractional p-Laplacian, etc., see [10,16,23]. Recently, Chen-Li-Li [13] systemically developed a so-called direct moving plane method which can handle (−∆) α 2 directly. Heuristically, this method can be used to handle general nonlocal operators. A series of fruitful results have been obtained, please see [10,11,19,20] and the reference therein. Based on this method, Tang [23] investigated the radial symmetry, the monotonicity and nonexistence of the solutions of the equations involving A α , where the weight function a(x) should be added in the radial and monotonic conditions, a(x) = a(|x|), for all x ∈ R; a(x) ≥ a(y), for any |x| ≤ |y|. (1.2) Note that the condition (1.2) is a crucial part in [23], we need a new method to investigate the monotonicity and non-existence of the solutions to the equation (1.1) without the radial and monotonic conditions (1.2).
In this paper, we use a direct sliding method for the uniformly elliptic nonlocal operators. The sliding method was introduced by Berestycki and Nirenberg [3,4,5] and has been used successfully to obtain the symmetry and monotonicity of solutions for several kind of second-order elliptic equations, see [1,2] and the reference therein. Recently, Wu-Chen [24] developed the sliding method for the fractional Laplacian equation. They established the narrow region principle in the bounded domain and maximum principle in the unbounded domain, eventually obtaining the monotonicity of solutions of some equations involving the fractional Laplacian operator. This method is also applied to investigate equations involving the fractional p-Laplacian [21,22,25] and the nonlocal Monge-Ampère operator [18]. Based on the sliding method in [24], we investigate the monotonicity of the solutions for the semi-linear equation (1.1) involving the uniformly elliptic nonlocal operator A α .
We mainly consider the monotonicity of the solutions to the equation (1.1) in two different regions, the bounded domain and the whole space.
The following narrow region principle in the bounded domain is a key ingredient in the sliding method and it provides a starting position to slide the domain for Theorem 1.2. Here, we denote Theorem 1.1 (Narrow region principle). Let D be a bounded narrow region in R n . Suppose that u ∈ L α (R n )∩C 1,1 loc (D), w τ is lower semi-continuous onD, and satisfies with c(x) bounded from below in D. Let d n (D) be the width of D in the x n direction, in which we assume that D in narrow, i.e. there exists C, such that Under the conditions 0 < α < 2 and (A 1 )-(A 2 ), then Now we use the sliding method to get the following result.
Let Ω be a bounded domain in R n , which is convex in x n -direction.
where ϕ(x) satisfies: For any three points x = (x , x n ), y = (x , y n ), z = (x , z n ) lying on a segment parallel to the x n -axis, y n < x n < z n , with y, z ∈ Ω c , we have (1.9) The function f is supposed to be Lipschitz continuous. Under the conditions 0 < α < 2 and (A 1 )-(A 2 ), then u is monotone increasing with respect to x n in Ω, i.e. for any τ > 0, (1.10) To be continuous, as an application of the sliding method, we derive the monotonicity of solutions for the semi-linear equation (1.1) with the uniformly elliptic nonlocal operator A α in the whole space. If u ∈ C θ (R n ), for some θ ∈ (0, 1), then u is monotone increasing with respect to x n , and furthermore, it depends on x n only. Theorem 1.3 is closely related to the De Giorgi conjecture, but it is in the classical Laplacian sense. For more concrete details, please refer to Remark 3 in [24].
Furthermore, we give the nonexistence of the solutions in the upper half space by establishing the so-called maximum principle.
for some nonnegative function c(x), then we have It is easy to check that the half-space R n + satisfy the condition (1.14). It turns out that Theorem 1.4 can be applied to derive the non-existence of the solution of the equation (1.1) in the half-space.
We would like to mention that the unbounded domain D founded in Theorem 1.4 has many different shapes. We here only list the following two simple examples, For more considerations, one can refer inspirations in [24] (Remark 2).
Finally, we remark that the method so called Poisson integral representation of ssubharmonic functions, which used in [24], can derive the maximum principle for the semi-linear equation with the fractional Laplacian operator. However, this method is not suitable for A α due to the complexity of the weighted function. To overcome this difficulty, using the strategy of Liu and Chen [21], we get the maximum principle for the semi-linear equation involving the uniformly elliptic nonlocal operator A α , see the proof of Theorem 1.4.
Organization of the paper: In Section 2, we deal with the narrow region principle in the bounded domain, and prove Theorem 1.2 by the sliding method. In Section 3, applying the sliding method and compactness method, we derive the monotonicity of solutions and hence complete the proof of Theorem 1.3. Section 4 is devoted to the maximum principle in the unbounded domain and its application.
Last but not least, we use the constant C, or c (among others) to denote a general positive constant which may depend only on n, α and whose value may differ from line to line.
2. Proof of Theorems 1.1 and 1.2. In this section, we prove Theorems 1.1 and 1.2.
Proof of Theorem 1.1. Suppose (1.5) is not valid, then the lower semi-continuity of w τ (x) inD implies that there exists a point x 0 ∈ D such that where d n (D) denotes the width of D in the x n direction, and more details are added to the second inequality from [13]. The inequality (2.1) contradicts (1.3), and we thus conclude that If there exists some point x ∈ D, such that w τ (x) = 0, then x is a minimum point

This contradicts
Therefore, we obtain (1.6), which completes the proof of Theorem 1.1.
Applying Theorem 1.1, we get the monotonicity result in Theorem 1.2. To better understand how the sliding method works, we only present the proofs for Ω when it is an ellipsoid or a rectangle. If necessary, one can refer to Fig.1 and Fig.2 in [25] for more intuitive understanding. When Ω is an arbitrary bounded domain of R n which is convex in x n -direction, i.e. for any (x , x n ), (x ,x n ) ∈ Ω imply that (x , tx n + (1 − t)x n ) ∈ Ω for 0 < t < 1, the proof is entirely similar.
Step 1 provides a starting point, from which we can carry out the sliding. Now we decrease τ as long as inequality w τ (x) ≥ 0 holds to its limiting position. Define τ 0 = inf{τ |w τ ≥ 0, x ∈ D τ ; 0 < τ <τ }. Claim τ 0 = 0. Otherwise, suppose that τ 0 > 0, we show that the domain Ω τ can slide upward a little bit more and still have which contradicts the definition of τ 0 . By the continuity of w τ (x), we have w τ0 (x) ≥ 0, x ∈ D τ0 , and w τ0 (x) > 0, x ∈ Ω ∩ ∂D τ0 , then w τ0 (x) ≡ 0, x ∈ D τ0 . If there exists a point x such that w τ0 (x) = 0, then x is a minimum point and |x − y| n+α dy < 0.

This contradicts
(2.5) Now we can carve out of D τ0 a closed set K ⊂ D τ0 such that D τ0 \K is narrow. By (2.5), we get w τ0 (x) ≥ C 0 > 0 in K. From the continuity of w τ (x) in terms of τ , we have, for small ε > 0, In addition, we obtain from (1.8) and (1.9) that It follows from Theorem 1.1 that (2.4) holds. Therefore, we have reached a contradiction and w τ (x) ≥ 0, x ∈ D τ , for any 0 < τ <τ .

Proof of Theorem 1.3.
Proof of Theorem 1.3. To begin with, we obtain from (1.12) that there exists a constant b > 0 such that (3.2) We divide the proof into three steps. Step Denote x k n the n-th component of x k . We claim that {x k n } ∞ k=1 is bounded, i.e., there exists M > 0 such that |x k n | ≤ M. In fact, suppose not, there exists a subsequence {x k n } ∞ k=1 , such that x k n → +∞ or It is easy to check that Ψ(0) = max x∈R n Ψ(x) = 1. (3.6) It follows that there exists a pointx k ∈ B 1 (x k ) such that Combining (3.6) and (3.7), thenx k is a maximum point of the function W τ (x) + ε k Ψ k (x) in R n . Since τ ≥ 2b, no matter wherex k is, one of the pointsx k andx k + (0 , τ ) is in the domain {x : |x n | ≥ b} where f (u(x)) is non-increasing. This can be seen from (3.1) and (3.2). Hence f (u(x k )) − f (u τ (x k )) ≤ 0, since u(x k ) > u τ (x k ). Therefore, by Proposition 5.1, we obtain It follows from (3.8) and the uniformly boundedness of a in (A 1 ), we have where the last inequality holds due to the fact that . Because of u ∈ C θ (R n ), for some θ ∈ (0, 1), for any compact set K ⊂ R n , by Arzelà-Ascoli theorem, there exists a subsequence, such that (3.10) On the other hand, in (3.9), letting k → ∞, one has By the equations (3.10) and (3.11), we have Since the sequence {x k n } ∞ k=1 is bounded, we obtain from (1.12) that u ∞ (x , x n ) −→ xn→±∞ ±1, uniformly in R n−1 (with respect to x ).
(3.13) Therefore, for any k ∈ N. Now take x n sufficiently negative and k sufficiently large, hence u ∞ (x , x n ) is close to −1 while kA + u ∞ (x , x n + kτ ) becomes sufficiently large, this is obviously impossible. Hence (3.3) holds.
Step 2. The inequality (3.3) provides a starting point, from which we can carry out the sliding. We decrease the index τ, and show that for any 0 < τ < 2b, (3.14) Define τ 0 = 0. Otherwise, suppose that τ 0 > 0. To derive a contradiction, we prove that τ 0 can be decreased a little bit while the inequality (3.14) still holds.
We first claim that sup If not, then sup where the definition of Ψ is similar to (3.5). Then there exists a sequence of ε k → 0 such that For any x ∈ R n \B 1 (x k ), notice that W τ0 (x) ≤ 0 and Ψ k (x) = 0, we have It follows that there exists a pointx k ∈ B 1 (x k ) such that On one hand, we have On the other hand, we obtain from (3.17) that where the last inequality holds due to the fact that is uniformly continuous, for any compact set K ⊂ R n , by Arzelà-Ascoli theorem, there exists a subsequence, we have u k (x) → u ∞ (x), as k → ∞, uniformly in K.
Combining (3.17) and (3.18), and letting k → ∞, one has is bounded, we obtain from (1.12) that u ∞ (x , x n ) −→ xn→±∞ ±1, uniformly in R n−1 (with respect to x ). (3.19) Therefore, for any k ∈ N. Now take x n sufficiently negative and k sufficiently large, hence u ∞ (x , x n ) is close to −1 and u ∞ (x , x n + kτ 0 ) is sufficiently close to 1, this is impossible. Hence (3.15) must hold. Next we prove that, there exists an ε > 0, such that First, by (3.15), we derive that there exists a small ε > 0 such that Therefore, we only need to prove that If not, then sup There exists a sequence of {x k } ∞ k=1 such that W τ (x k ) → A. By the asymptotic condition (1.12), {x k n } ∞ k=1 is bounded, and we assume that |x k n | ≤ M (M > b). By a similar argument, setting Ψ k (x) = Ψ(x − x k ), we have that there exists ε k → 0, (3.21) In addition, by the uniformly boundedness of a in (A 1 ), we have (3.21) and (3.22), and letting k → ∞, we derive which contradicts the fact that W τ ∞ (x) equals 0 at infinity. This proves (3.20) which contradicts the definition of τ 0 . Hence we obtain (3.14).
Step 3. In this step, we show that u(x) is strictly increasing with respect to x n and u(x) depends on x n only. Combining Step 1 and Step 2, we obtain that W τ (x) ≤ 0 in R n , for any τ > 0.
If there exists a point x 0 ∈ R n , such that W τ (x 0 ) = 0, then x 0 is a maximum point of W τ (x) in R n . Since W τ (x) ≡ 0 in R n , by a direct calculation, we have
In fact, it can be seen from the above process that the argument still holds if we replace u τ (x) by u(x + τ ν), where ν = (ν 1 , · · · , ν n ) with ν n > 0 is an arbitrary vector pointing upward. Applying the similar arguments as in Steps 1 and 2, we can derive that, for each of such ν, Letting ν n → 0, from the continuity of u, we deduce that for arbitrary ν with ν n = 0. By replacing ν by −ν, we find that for arbitrary ν with ν n = 0, this means that u is independent of x , hence u(x) = u(x n ). This completes the proof of Theorem 1.3. There exist the sequence {x k } ∞ k=1 ⊂ D and γ k → 1(γ k ∈ (0, 1)) as k → ∞ such that u(x k ) ≥ γ k A. It is easy to check that Φ(x) is radially decreasing from the origin, and is in C ∞ c (R n ).
For any x ∈ B 2 (x k )\B 1 (x k ), we can take ε k > 0 such that where ν is any unit vector in R n . Therefore, there existsx k ∈ B 1 (x k ) such that As a consequence, It follows from (4.2) that u(x k ) ≥ γ k A. From (4.5) and (4.6), we deduce that Hencex k is a maximum point of the function u(x) + ε k Φ k (x) in R n .
Calculating directly, we have (4.9) For I 1 , by (4.6), we notice that for any y ∈ B 2 (x k ). Combining with a(x) > 0, we have The next is to estimate I 2 . It follows where the last inequality holds due to the fact that |x k − y| ≤ |x k − x k | + |x k − y| ≤ 3 2 |x k − y|.
In the following, we aim to prove Corollary 1.1.
This means that u is monotone increasing. By lim xn→+∞ u(x , x n ) = 0, uniformly in x ∈ R n−1 , we can derive u ≡ 0.

Appendix.
Proposition 5.1. Assume that u ∈ C 1,1 loc ∩ L α (R n ), then there exists a constant C, such that A α u(x) ≤ C.
Proof. For any x ∈ R n , we divide the integral into two parts.
By the anti-symmetry of ∇u(x) · (x − y) for y ∈ B 1 (x) and the cylindrical condition of a in (A 2 ), we obtain P.V.
Thus, we eventually complete the proof of this proposition.