## Abstract

In this paper, we consider the following fractional

We first prove the uniqueness and monotonicity of positive solutions in a bounded domain. Then by estimating the singular integrals which define the fractional
*Monotonicity, symmetry and antisymmetry of solutions of semilinear elliptic equations*, J. Geom. Phys. **5** (1988), 237–275] and Wu and Chen [*The sliding methods for the fractional p-Laplacian*, Adv. Math. **361** (2020), 106933].

### 1 Introduction

In studying differential equations it is often of interest to know if the solutions have symmetry, or perhaps monotonicity, in some direction. Monotonicity and symmetry for solutions to the

by the sliding method and the method of moving planes. In 1991, Berestycki and Nirenberg [5] considered the monotonicity and uniqueness of solutions to (1.1) with

During the last few decades, equations involving the fractional Laplacian and the fractional

Motivated by the aforementioned papers, the goal of this paper is to extend the results in [4] to the fractional

where

where P.V. represents the Cauchy principal value. The fractional

In order for the integral (1.3) to make sense, we require that

with

On the one hand, we extend the case

In order to apply the sliding method, we give the exterior conditions on

(

and

### Remark 1.1

The same monotonicity conditions (1.4) and (1.5) (with

For bounded domain, we prove

### Theorem 1.1

*Suppose that*
*satisfies* (
*and is a solution of problem*

*where*
*is a bounded domain which is convex in*
*direction. Assume that*
*is continuous in all variables, Lipschitz continuous in*
*and nondecreasing in*
*Then*
*is strictly monotone increasing with respect to*
*in*
*i.e., for any*

*Furthermore, the solution of* (1.6) *is unique*.

### Remark 1.2

For the finite cylinder

### Remark 1.3

The conditions assumed in Theorems 1.1 and 1 of [16] are different, and there is no relation between them. Dai et al. [16] studied the positive solution

We can immediately get a new antisymmetry result for (1.6) if bounded domain

### Corollary 1.1

(*Antisymmetry*) *Assume that the conditions of*
Theorem 1.1
*are satisfied and in addition that*
*is odd in*
*on*
*If*
*is odd in*
*then*
*is odd, i.e., antisymmetric in*

This follows from the fact that

Let

We say that

For the whole space, we prove

### Theorem 1.2

*Suppose that*
*is a solution to*

*such that*

*and*

*Assume that*
*is bounded, continuous in all variables and satisfies*

*Suppose that there exists*
*such that*

*Then u is strictly monotone increasing with respect to*
*and furthermore, it depends on*
*only*.

### Remark 1.4

Theorem 1.2 is closely related to the well-known De Giorgi conjecture [17]. As an example, we may think of

In order to solve the difficulty that the nonlinear terms at the right-hand side of (1.6) and (1.7) contain the gradient term, in bounded domains when deriving the contradiction for the minimum point of the function

For more related articles on symmetry and nonexistence results of local and nonlocal equations, we also refer readers to [19] for Laplace equations with a gradient term, [20,21, 22,23,24, 25,26] for fractional equations, [27] for weighted fractional equation, [28,29] for fractional equations with a gradient term, [30,31,32] for fully nonlinear equations with a gradient term, [33, 34,35,36] for fractional

The paper is organized as follows. In Section 2, Theorem 1.1 is proved via the sliding method. In Section 3, we derive monotonicity for the fractional

### 2 Monotonicity and uniqueness of solutions in bounded domains

For simplicity, we list some notations used frequently. For

### Proof of Theorem 1.1

When

For

and

We mainly divide the following two steps to prove that

Step 1. For

If (2.13) is false, we set

From the condition

where

Hence,

On the other hand, note that

To estimate

By the monotonicity of

Hence,

Next we prove that

### Lemma 2.1

[40] *For*
*there exists a constant*
*such that*

*for arbitrary*

Noting that

By Lemma 2.1, we have

From (1.4) in the condition

Since function

Combining (2.17), (2.18) and (2.19), it yields

where

Hence, (2.13) is true for

Step 2. The inequality (2.13) provides a starting point, from which we can carry out the sliding. Now we decrease

We will prove

Otherwise, assume

which contradicts the definition of

Since

If there exists a point

which contradicts to

Hence,

Next we will prove (2.21). Suppose (2.21) is not true, one has

The minimum

From the continuity of

From

So

Similar to (2.20), by narrow domain

This is a contradiction. Hence we derive (2.21), which contradicts to the definition of

Let us prove (2.12). Since

if there exists a point

which contradicts

Therefore, we arrive at (2.12).

Now we prove uniqueness. If

This completes the proof of Theorem 1.1.□

### 3 Monotonicity of solutions in
R
n

In the section, we prove Theorem 1.2 by the sliding method based on the following lemma.

### Lemma 3.1

[40] *Assume that*
*is the bounded solution of* (1.7) *and*
*Then for any*
*we have*

*where*
*as*

### Proof of Theorem 1.2

Denote

From (1.9), there exists a constant

For any

or

Outline of the proof: We will use the sliding method to prove the monotonicity of

Step 1, we will show that for

This provides the starting point for the sliding method. Then in Step 2, we decrease

We will show that

Now we will show the details in the following three steps.

**Step 1**. We will show that

If not, then

so for some

Denote

Let

where

From (3.31), there exists a sequence

Since for any

It follows that there exists a point

On one hand, by (1.7), (1.10) and Lemma 3.1, we have

Since

From (3.33), we have

On the other hand, by (3.33) and Lemma 2.1,

where

Since

Combining (3.34) and (3.35), letting

Since

Hence,

for any

**Step 2**. Note that (3.30) provides a starting point, from which we can carry out the sliding. We decrease

Define

We prove that

We first prove that

If not, then

and there exists a sequence

such that

Let

Since for

On one hand, similar to the argument in Step 1, we have

On the other hand, by (3.39) and Lemma 2.1, similar to (3.35)

Denote

One has

It yields that

So combining (3.42) and (3.41), letting

Therefore,

Hence, for any

Take

Next we prove that, there exists an

First, (3.38) implies immediately that there exists an

So we only need to prove that

If not, then

For some

and