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The properties of a new fractional g-Laplacian Monge-Ampère operator and its applications

  • Guotao Wang , Rui Yang and Lihong Zhang EMAIL logo
Published/Copyright: September 2, 2024

Abstract

In this article, we first introduce a new fractional g -Laplacian Monge-Ampère operator:

F g s v ( x ) inf P.V. R n g v ( z ) v ( x ) C 1 ( z x ) s d z C 1 ( z x ) n + s C C ,

where g is the derivative of a Young function and the diagonal matrix C is positive definite, which has a determinant equal to 1. First, we establish some crucial maximum principles for equations involving the fractional g -Laplacian Monge-Ampère operator. Based on the maximum principles, the direct method of moving planes is applied to study the equation involving the fractional g -Laplacian Monge-Ampère operator. As a result, the nonexistence of the positive solutions, symmetry, monotonicity, and asymptotic property of solutions are obtained in bounded/unbounded domains.

MSC 2010: 35R11; 62F30; 35B06; 35B40

1 Introduction

In recent years, fractional Laplacian operator has attracted more and more attentions. It has many applications, such as anomalous diffusion, spectral graph theory, quasi-geostrophic flows, minimal surfaces, water waves, error analysis of images, and relativistic quantum mechanics of stars [2,12,18,19,24]. The fractional Laplacian operator has also become one of the powerful tools to describe large-scale irregular diffusion phenomena such as dislocation phenomena in dynamical systems, combustion phenomena and disorder phenomena in crystal structures, etc. [7,25]. In 2019, Bonder and Salort introduced the fractional g -Laplacian operator, which allows the modeling of non-local problems that obey non-power-law behavior [5] has attracted the interest of scholars from home and abroad. We define the fractional g -Laplacian as follows:

( Δ g ) s v ( x ) = P.V. R n g v ( z ) v ( x ) z x s d z z x n + s ,

where P.V. represents the Cauchy principal value and g = G ( G is a Young function). That is,

G ( t ) = 0 t g ( τ ) d τ .

The properties of g are as follows:

  1. g ( a ) > 0 for a > 0 ,

  2. g ( a ) + g ( b ) c i g ( a + b ) , where c i is a constant greater than 0,

  3. g ( b ) g ( a ) c i g ( b a ) , where c i is a constant greater than 0,

  4. g is nondecreasing on ( 0 , ) ,

  5. g ( a ) = g ( a ) ,

  6. lim a g ( a ) = .

Many results have been obtained regarding the fractional g -Laplacian operator. For example, in 2020, Bonder et al. derived the boundary Hölder regularity about interior and up solutions about the Dirichlet problem involving the fractional g -Laplacian [6]. In 2021, Molina et al. established some maximum principles involving the fractional g -Laplacian, and thus, they obtained some classic qualitative results, for example, Liouville type theorem, Maximum principle, symmetry of solutions, and the boundary estimate [23]. For related literatures, please see [1,3,4,22,2628] and the references therein.

The operator ( Δ g ) s presented in this article contains different cases under different conditions. For example, when G ( t ) = t 2 , that is, the case of squared power, it is the fractional Laplacian, and a series of successful results of the various properties of solutions have been obtained [9,15]. When G ( t ) = t p , we will derive the famous fractional p -Laplacian. Many scholars have conducted a lot of research on it and have obtained a series of rich results. See [10,14,2931] and the references therein.

Problems with fractional Monge-Ampère operator [8], introduced by Caffarelli and Charro, appear in areas such as analysis, geometry, and applied mathematics. The research about nonlinear problems with the fractional Monge-Ampère operators is a hot problem in domestic and international research. Many results involving fractional Monge-Ampère have been derived. For instance, Dai and Li obtained the equivalence condition of the existence of subsolutions to the equation involving the Monge-Ampère operator [16]. Radially symmetric boundary value problems for solutions to the equation involving the Monge-Ampère operator for real and complex ellipsoids were studied by Delanoë [17]. However, the nonlocality of fractional Monge-Ampère operator adds difficulties to the study, and various feasible methods have been introduced by domestic and international experts and scholars to solve this challenge. Chen and Li introduced the direct method of moving planes [11], which acts directly on the fractional Monge-Ampère operator and is more efficient and convenient for studying nonlinear problems with fractional Monge-Ampère operator. Therefore, using the direct method of moving planes to solve nonlinear problems with fractional Laplacian operator and fractional Monge-Ampère operator has attracted increasing attention of scholars at home and abroad, and a number of fruitful results were obtained. For example, the radial symmetry of the standing wave solutions of the nonlinear fractional Hardy-Schrödinger equation is derived through the use of direct method of moving planes [10]. For more literature about the direct methods of moving planes, please see [13,20,21,32].

The first primary purpose of this study is to introduce a new operator called fractional g -Laplacian Monge-Ampère operator, which combines the advantages of the fractional g -Laplacian operator and fractional Monge-Ampère operator. To be more precise, in the article, we are dedicated to studying the following nonlinear problem involving this new type of fractional g -Laplacian Monge-Ampère operator:

(1.1) F g s v ( x ) = f ( v ( x ) )

with

(1.2) F g s v ( x ) = inf P.V. R n g v ( z ) v ( x ) C 1 ( z x ) s d z C 1 ( z x ) n + s C C ,

where s ( 0 , 2 ) , P.V. represents the Cauchy principal value and g = G ( G is a Young function), and C represents a positive definite diagonal matric with the determinant equal to 1 and min { λ 1 , , λ n } θ > 0 , where λ i ( i = 1 , 2 , , n ) represents the eigenvalues of C . Throughout this article, we require that the solutions of equations involving the fractional g -Laplacian Monge-Ampère belong to the class C loc 1 , 1 ( R n ) L g ( R n ) , L g can be defined as follows:

L g v C loc 1 ( R n ) : R n g v ( x ) 1 + x s d x 1 + x n + s < .

Central to the innovations of our consequence are that we applied different methods to overcome the nonlinearity of the operator. Therefore, the obtained results encompass a wide range and can be applied to non-local operators. And it is worth mentioning that it allows for more general behavior than powers, a case in point is that G ( u ) = u m + u n , where m , n 2 .

We further emphasize that the specificity of the properties of g is one of the main difficulties we need to overcome, for instance, g ( a ) + g ( b ) c i g ( a + b ) , where c i is simply a constant greater than 0. This adds the difficulty of our proof.

In this article, we can use the direct method of moving planes to derive various properties of solutions to equations involving the fractional g -Laplacian Monge-Ampère operator. We believe that the maximum principle plays an essential role in the process of using the direct method of moving planes. First, the maximum principle is as follows.

Theorem 1.1

(The maximum principle) Assume that Ω R n is bounded. Suppose that the lower semicontinuous function v C loc 1 , 1 ( Ω ) L g ( R n ) satisfies the following equation:

(1.3) F g s v ( x ) 0 , x Ω , v ( x ) 0 , x R n \ Ω .

Then

(1.4) v ( x ) 0 , x Ω .

Furthermore, if there exists a point in R n such that v = 0 , then we can derive that v ( x ) = 0 almost everywhere in R n . Suppose that lim ̲ x v ( x ) 0 , these conclusions are also valid in unbounded domain.

Notations. Let

  1. T λ { x R n x 1 = λ } be the moving plane.

  2. Σ λ { x R n x 1 < λ } be the area to the left of T λ .

  3. x λ ( 2 λ x 1 , x 2 , , x n 1 , x n ) be reflection point of x about T λ .

  4. v λ ( x ) v ( x λ ) be the reflected function.

Denote τ λ ( x ) v ( x λ ) v ( x ) = v λ ( x ) v ( x ) . For x Σ λ , since

τ λ ( x λ ) = v λ ( x λ ) v ( x λ ) = v ( ( x λ ) λ ) v ( x λ ) = v ( x ) v λ ( x ) ,

we have

(1.5) τ λ ( x λ ) = τ λ ( x ) .

The specific location relationship is as follows:

For convenience, we will replace Σ λ with Σ .

Theorem 1.2

(The maximum principle for anti-symmetric function) Let Ω Σ be bounded. Suppose that the lower semicontinuous function v C loc 1 , 1 ( Ω ) L g ( R n ) satisfies the following equation:

(1.6) F g s v λ ( x ) F g s v ( x ) 0 , x Ω , τ λ ( x ) 0 , x Σ \ Ω , τ λ ( x λ ) = τ λ ( x ) , x Σ ,

where τ λ = v λ v , then

(1.7) τ λ ( x ) 0 , x Ω .

Furthermore, if there exists a point x in T such that τ λ ( x ) = 0 , then we can derive that τ λ ( x ) = 0 almost everywhere in R n . These conclusions are also valid on unbounded domain if we further assume that

lim ̲ x τ λ ( x ) 0 .

For a ( x ) 0 , where a ( x ) is determined by λ and v ( x ) , if F g s v λ ( x ) F g s v ( x ) a ( x ) τ λ ( x ) 0 , in terms of (1.2), we can derive that the maximum principle is also valid. Thus, on the basis of this, we should carry the plane a little further to the right. Then we can derive the following narrow region principle when a ( x ) is bounded.

Theorem 1.3

(The narrow region principle) Suppose that the bounded narrow region Ω is contained in { x λ ε < x 1 < λ } in Σ , where ε is sufficiently small. Suppose that v C loc 1 , 1 ( Ω ) L g ( R n ) is lower semicontinuous on Ω ¯ . If

(1.8) F g s v λ ( x ) F g s v ( x ) a ( x ) τ λ ( x ) 0 , x Ω , τ λ ( x ) 0 , x Σ \ Ω , τ λ ( x λ ) = τ λ ( x ) , x Σ ,

where a ( x ) is bounded from below in Ω and τ λ = v λ v , then for ε small enough, we have

(1.9) τ λ ( x ) 0 , x Ω .

These conclusions are also valid in unbounded domain if we further assume that

lim ̲ x τ λ ( x ) 0 .

Theorem 1.4

(Decay at infinity) Assume that Ω Σ is unbounded. Suppose that the lower semicontinuous function v C loc 1 , 1 ( Ω ) L g ( R n ) satisfies the following equation:

(1.10) F g s v λ ( x ) F g s v ( x ) a ( x ) τ λ ( x ) 0 , x Ω , τ λ ( x ) 0 , x Σ \ Ω , τ λ ( x λ ) = τ λ ( x ) , x Σ ,

with

(1.11) lim ̲ x x 2 s a ( x ) 0 ,

where τ λ = v λ v , then there exists A > 0 (only associated with a(x)), such that if

τ λ ( x 0 ) = min Ω τ λ ( x ) < 0 ,

then

x 0 A .

We can derive various properties of solutions involving the operator F g s by employing Theorems 1.1 and 1.2. To be more specific, we can derive that the solutions is symmetric about a point through the coordinate plane when the solution is on a given ellipsoid region or R n . Also, we can derive that there is no positive solution in R + n .

In the article, the following inequality

(1.12) 1 C 1 ( z x ) > 1 C 1 ( z λ x ) , x , z Σ

plays an important role in the proof of the theorems. Next, we consider the following ellipsoid region:

(1.13) B = i = 1 n x i 2 a i 2 1 x = ( x 1 , x 2 , , x n ) R n ,

where a i ( i = 1 , , n ) > 0 .

Theorem 1.5

Suppose that v C loc 1 , 1 ( Ω ) L g ( R n ) satisfies the following equation:

(1.14) F g s v ( x ) = f ( v ( x ) ) , x B , v ( x ) > 0 , x B , v ( x ) = 0 , x B ,

where f is a Lipschitz continuous function. Then v must be radially symmetric.

In the following, we extend the range of x over the whole space R n . We are able to prove two theorems under different conditions.

Theorem 1.6

Assume that v C loc 1 , 1 ( R n ) L g ( R n ) satisfies the following equation:

(1.15) F g s v ( x ) = f ( v ( x ) ) , x R n , v ( x ) > 0 , x R n .

Suppose for some η > 0 ,

(1.16) lim x x η v ( x ) = 0

and

(1.17) f ( β ) β p , with p η 2 s .

Then v must be radially symmetric.

We reduce some of the restrictions of v in Theorem 1.6, and thus, we have Theorem 1.7.

Theorem 1.7

Assume that v C loc 1 , 1 ( R n ) L g ( R n ) satisfies the following equation:

(1.18) F g s v ( x ) = f ( v ( x ) ) , x R n , v ( x ) > 0 , x R n .

Suppose that f ( β ) 0 , where β is small enough, and

(1.19) lim x v ( x ) = 0 .

Then v must be radially symmetric.

Next, we consider the space

R + n = { x = ( x 1 , x 2 , , x n ) x n > 0 } .

We are able to prove the following Theorems 1.8 and 1.9 under different conditions.

Theorem 1.8

Assume that v C loc 1 , 1 ( R + n ) L g ( R n ) is a nonnegative solution satisfying the following equation:

(1.20) F g s v ( x ) = f ( v ( x ) ) , x R + n , v ( x ) 0 , x R + n .

Suppose for some η > 0 ,

(1.21) lim x x η v ( x ) = 0

and

(1.22) f ( β ) β p , w i t h p β 2 s ,

where f is a Lipschitz continuous function, and f ( 0 ) = 0 . Then v 0 .

We reduce some of the restrictions of v in Theorem 1.8, then we have the Theorem 1.9.

Theorem 1.9

Assume that v C loc 1 , 1 ( R + n ) L g ( R n ) is a nonnegative solution satisfying the following equation:

(1.23) F g s v ( x ) = f ( v ( x ) ) , x R + n , v ( x ) 0 , x R + n ,

where f is a Lipschitz continuous function, and f ( 0 ) = 0 . Suppose that f ( β ) 0 , where β is small enough, and

(1.24) lim x v ( x ) = 0 ,

Then v 0 .

Lemma 1.1

(Strong maximum principle) Assume that v L g ( R n ) and v is a non-negative function. If there exists a point x l R n satisfying v ( x l ) = 0 and F g s v ( x l ) 0 , where v is C 1 , 1 near x l , then v = 0 a.e. in R n .

Theorem 1.10

Suppose that Ω R n is symmetric concerning { x R n x 1 = 0 } (possibly bounded) and is convex in x 1 -direction. Assume that v C loc 1 , 1 ( Ω ) L g ( R n ) satisfies

(1.25) F g s v ( x ) = f ( v ( x ) ) , x Ω , v ( x ) > 0 , x Ω , v ( x ) = 0 , x R n \ Ω .

Let the Lipschitz continuous function f satisfy

(1.26) f ( 0 ) 0 .

Then v ( x 1 , x ) must be strictly increasing about the x 1 -direction in Ω .

Theorem 1.11

(Maximum principle in unbounded open sets) Let Ω (possibly unbounded and disconnected) be an open set in R n and Σ be an infinite open domain. Suppose that the closure of Σ be disjoint from Ω ¯ . If

( B c 1 r x ( x ) \ B r x ( x ) ) Σ B c 1 r x ( x ) \ B r x ( x ) c 0 > 0 , x Ω ,

where c 0 > 0 , c 1 > 0 , and r x > 0 . Suppose v ( x ) C loc 1 , 1 ( Ω ) L g ( R n ) , bounded from above, and satisfies

(1.27) F g s v ( x ) a ( x ) v ( x ) 0 , in p o i n t s x Ω , where v ( x ) > 0 v ( x ) < 0 , x R n \ Ω ,

for a ( x ) 0 .

Then v ( x ) 0 in Ω .

Lemma 1.2

Assume that v C loc 1 , 1 ( Ω ) L g ( R n ) satisfies the following equation:

(1.28) F g s v ( x ) = f ( v ( x ) ) , x Ω

and

v ( x ) < ζ , x R n \ Ω .

Suppose f satisfies

  1. f ( t ) > 0 on ( 0 , ζ ) , f ( ζ ) = 0 and f ( t ) 0 for t ζ .

Then, v < ζ for all x Ω .

Next, we present the following equation:

(1.29) F g s v ( x ) = f ( v ( x ) ) , x Ω , v ( x ) 0 , x R n \ Ω .

To make sure that the proof goes on well, we provide the following assumptions about f :

  1. f ( t ) a 0 t on [ 0 , t 0 ] for some a 0 > 0 .

  2. f ( t ) is nonincreasing on ( t 1 , ζ ) ,

where 0 < t 0 < t 1 < ζ .

Lemma 1.3

Let v C loc 1 , 1 ( Ω ) L g ( R n ) be a solution of (1.29). If the condition (b) holds, then there exist δ 0 ( 0 , t 0 ) and B 0 > 0 such that

(1.30) v ( x ) > δ 0 , dist ( x , Ω ) > B 0 .

Theorem 1.12

Assume that v C loc 1 , 1 ( Ω ) L g ( R n ) is a solution of (1.28) with

0 v ( x ) ζ , x R n \ Ω .

Suppose that f is a continuous function satisfying assumption (a), (b), and (c). Then, v ( x ) ζ uniformly in Ω as dist ( x , Ω ) .

In Section 2, we give detailed proofs of Theorems 1.1 to 1.4. In Section 3, we use the direct method of moving planes to prove Theorems 1.51.9, and we derive that the solution is symmetric about a point through the coordinate plane when the solution is on a given ellipsoid region or R + n and prove the nonexistence of the positive solutions in R + n . As applications, we establish monotonicity result for solutions of nonlinear equations involving the fractional g -Laplacian Monge-Ampère by the direct method of moving planes in Section 4. In Section 5, we give maximum principles and some lemmas for F g s and derive the asymptotic property of solutions of the equation involving the fractional g -Laplacian Monge-Ampère operator.

2 The maximum principle

In this section, we give specific proofs of the above maximum principles, the narrow region principle, and the decay estimates at infinity concerning the operator F g s , which is defined by (1.2). These theorems play an important role in using the direct method of moving planes.

2.1 Proof of Theorem 1.1

Assume that condition (1.4) is not correct. Because v is lower semicontinuous on Ω ¯ , we derive that there exists a point x 0 Ω ¯ satisfying

(2.1) v ( x 0 ) = min Ω ¯ v ( x ) < 0 .

According to (1.3), we have x 0 Ω . Fix μ > 0 arbitrarily, and let C μ C such that

(2.2) F g s v ( x 0 ) P.V. R n g v ( z ) v ( x 0 ) C μ 1 ( z x 0 ) s d z C μ 1 ( z x 0 ) n + s μ .

It follows from (2.2) that

F g s v ( x 0 ) = inf P.V. R n g v ( z ) v ( x 0 ) C 1 ( z x 0 ) s d z C 1 ( z x 0 ) n + s C C P.V. R n g v ( z ) v ( x 0 ) C μ 1 ( z x 0 ) s d z C μ 1 ( z x 0 ) n + s μ = P.V. Ω g v ( z ) v ( x 0 ) C μ 1 ( z x 0 ) s d z C μ 1 ( z x 0 ) n + s + R n \ Ω g v ( z ) v ( x 0 ) C μ 1 ( z x 0 ) s d z C μ 1 ( z x 0 ) n + s μ R n \ Ω g v ( z ) v ( x 0 ) C μ 1 ( z x 0 ) s d z C μ 1 ( z x 0 ) n + s μ .

Since C μ C , we assume that C μ = diag { λ 1 , λ 2 , , λ n } , i = 1 n λ i = 1 and λ i θ > 0 ( i = 1 , 2 , , n ) , then

C μ 1 = diag 1 λ 1 , 1 λ 2 , , 1 λ n .

By relevant calculations, we have

(2.3) C μ 1 ( z x 0 ) s = i = 1 n ( z i x i 0 ) 2 λ i 2 s 2 i = 1 n z i x i 0 λ i s 1 θ s z x 0 s .

Thus,

F g s v ( x 0 ) R n \ Ω g θ s ( v ( z ) v ( x 0 ) ) z x 0 s θ n + s d z z x 0 n + s μ .

It is obvious that

F g s v ( x ) 0 , x Ω .

By using the properties of the g function: g ( t ) > 0 for t > 0 , we can obtain

0 F g s v ( x 0 ) θ n + s R n \ Ω g θ s ( v ( z ) v ( x 0 ) ) z x 0 s d z z x 0 n + s > 0 , x Ω .

This creates the contradiction. Therefore, we can derive that v ( x ) 0 in Ω .

Further, we assume that there exists a point x ˘ Ω such that v ( x ˘ ) = 0 , combining the condition that v ( x ) 0 in R n \ Ω , then we can derive

F g s v ( x ˘ ) = inf P.V. R n g v ( z ) v ( x ˘ ) C 1 ( z x ˘ ) s d z C 1 ( z x ˘ ) n + s C C P.V. R n g v ( z ) C μ 1 ( z x ˘ ) s d z C μ 1 ( z x ˘ ) n + s μ P.V. θ n + s R n g v ( z ) C μ 1 ( z x ˘ ) s d z z x ˘ n + s μ .

Since

F g s v ( x ) 0 , x Ω ,

let μ 0 , combining the property of g that g ( a ) > 0 for a > 0 and v ( x ) 0 in R n , then we have v = 0 almost everywhere in R n .

In the unbounded domain, we give the condition

lim x v ( x ) 0

to make sure that the minimum of v , which is negative must be attained at some point in Ω . We can prove it in a similar way as mentioned earlier. This completes the proof of Theorem 1.1.

In the following, we will give specific proof of Theorem 1.2.

2.2 Proof of Theorem 1.2

Assume that condition (1.7) is not correct. Because τ λ is lower semicontinuous on Ω ¯ , we can derive that there exists a point x 0 Ω ¯ such that

τ λ ( x 0 ) = min Ω ¯ τ λ ( x ) < 0 .

According to the condition (1.6) we can derive x 0 Ω . Let arbitrary μ > 0 , and C μ C , then we have

F g s v λ ( x 0 ) P.V. R n g v λ ( z ) v λ ( x 0 ) C μ 1 ( z x 0 ) s d z C μ 1 ( z x 0 ) n + s μ .

We have

F g s v λ ( x 0 ) F g s v ( x 0 ) P.V. R n g v λ ( z ) v λ ( x 0 ) C 1 ( z x 0 ) s d z C 1 ( z x 0 ) n + s μ P.V. R n g v ( z ) v ( x 0 ) C 1 ( z x 0 ) s d z C 1 ( z x 0 ) n + s P.V. c 1 R n g v λ ( z ) v λ ( x 0 ) v ( z ) + v ( x 0 ) C μ 1 ( z x 0 ) s d z C μ 1 ( z x 0 ) n + s μ = P.V. c 1 R n g τ λ ( z ) τ λ ( x 0 ) C μ 1 ( z x 0 ) s d z C μ 1 ( z x 0 ) n + s μ = P.V. c 1 Σ g τ λ ( z ) τ λ ( x 0 ) C μ 1 ( z x 0 ) s d z C μ 1 ( z x 0 ) n + s + c 1 Σ ˜ g τ λ ( z ) τ λ ( x 0 ) C μ 1 ( z x 0 ) s d z C μ 1 ( z x 0 ) n + s μ = P.V. c 1 Σ g τ λ ( z ) τ λ ( x 0 ) C μ 1 ( z x 0 ) s d z C μ 1 ( z x 0 ) n + s + Σ g τ λ ( z ) τ λ ( x 0 ) C μ 1 ( z λ x 0 ) s d z C μ 1 ( z λ x 0 ) n + s μ ,

where Σ ˜ = { z z λ Σ } . Because C μ C , T λ is perpendicular to x 1 -axis, and τ λ ( z ) τ λ ( x ) 0 , easily, we can derive

1 C μ 1 ( z x ) > 1 C μ 1 ( z λ x ) , x , z Σ .

Therefore, we derive

(2.4) P.V. c 1 Σ g τ λ ( z ) τ λ ( x 0 ) C μ 1 ( z x 0 ) s d z C μ 1 ( z x 0 ) n + s + Σ g τ λ ( z ) τ λ ( x 0 ) C μ 1 ( z λ x 0 ) s d z C μ 1 ( z λ x 0 ) n + s μ P.V. c 1 c 2 Σ g τ λ ( z ) τ λ ( x 0 ) τ λ ( z ) τ λ ( x 0 ) C μ 1 ( z λ x 0 ) s d z C μ 1 ( z λ x 0 ) n + s μ = P.V. c 1 c 2 Σ g v λ ( z ) v λ ( x 0 ) v ( z ) + v ( x 0 ) C μ 1 ( z x 0 ) s d z C μ 1 ( z x 0 ) n + s μ = P.V. c 1 c 2 Σ g 2 τ λ ( x 0 ) C μ 1 ( z λ x 0 ) s d z C μ 1 ( z λ x 0 ) n + s μ .

Similar to (2.3), we have

C μ 1 ( z λ x 0 ) 2 = i = 2 n z i x i 0 λ i 2 + 2 λ z 1 x 1 0 λ 1 2 1 θ 2 z λ x 0 2 .

Then, from (2.4), we can obtain

(2.5) F g s v λ ( x 0 ) F g s v ( x 0 ) θ n + s Σ g 2 τ λ ( x 0 ) C μ 1 ( z λ x 0 ) s d z z λ x 0 n + s μ .

According to the fact that

F g s v λ ( x 0 ) F g s v ( x 0 ) 0 , x Ω ,

when μ 0 , we have

(2.6) 0 F g s v λ ( x 0 ) F g s v ( x 0 ) θ n + s Σ g 2 τ λ ( x 0 ) C μ 1 ( z λ x 0 ) s d z z λ x 0 n + s > 0 .

This leads to the contradiction.

Assume that τ λ ( x ˘ ) = 0 in Ω . Furthermore, we can derive that τ λ ( x ) = 0 almost everywhere in R n .

F g s v λ ( x ˘ ) F g s v ( x ˘ ) P.V. R n g v λ ( z ) v λ ( x ˘ ) C μ 1 ( z x ˘ ) s d z C μ 1 ( z x ˘ ) n + s μ P.V. R n g v ( z ) v ( x ˘ ) C μ 1 ( z x ˘ ) s d z C μ 1 ( z x ˘ ) n + s P.V. c 1 R n g τ λ ( z ) C μ 1 ( z x ˘ ) s d z C μ 1 ( z x ˘ ) n + s μ = P.V. c 1 Σ g τ λ ( z ) C μ 1 ( z x ˘ ) s d z C μ 1 ( z x ˘ ) n + s + c 1 R n \ Σ g τ λ ( z ) C μ 1 ( z x ˘ ) s d z C μ 1 ( z x ˘ ) n + s μ P.V. c 1 Σ g τ λ ( z ) 1 C μ 1 ( z x ˘ ) s 1 C μ 1 ( z λ x ˘ ) s d z C μ 1 ( z λ x ˘ ) n + s μ .

According to the known condition that

F g s v λ ( x ) F g s v ( x ) 0 , x Ω

in (1.6), we have τ λ ( x ) 0 , so τ λ ( z ) 0 for z Σ , and

1 C 0 1 ( z x ) n + s > 1 C 0 1 ( z λ x ) n + s , x , z Σ ,

when μ 0 . That is, F g s v λ ( x ˘ ) F g s v ( x ˘ ) 0 , and therefore, we have τ λ ( x ) = 0 almost everywhere in Σ . Because of the anti-symmetry of τ λ ( x ) , we have τ λ ( x ) = 0 almost everywhere in R n .

In the unbounded domain, we present the condition

lim x τ λ ( x ) 0

to make sure that the minimum of τ λ , which is negative must be attained at some point in Ω . We can prove it in a similar way as mentioned earlier. This completes the proof of Theorem 1.2.

2.3 Proof of Theorem 1.3

Assume that condition (1.9) is not correct. Because τ λ ( x ) is lower semicontinuous on Ω ¯ , we have there exists a point x 0 Ω ¯ satisfying

τ λ ( x 0 ) = min Ω ¯ τ λ ( x ) < 0 .

According to the condition (1.8) that x 0 Ω . By a similar proof in Theorem 1.2 and the condition (1.8), we have

(2.7) 0 F g s v λ ( x 0 ) F g s v ( x 0 ) a ( x 0 ) τ λ ( x 0 ) θ n + s Σ g 2 τ λ ( x 0 ) C μ 1 ( z λ x 0 ) s d z z λ x 0 n + s μ a ( x 0 ) τ λ ( x 0 ) .

Let x = ( x 1 , x 2 , , x n ) = ( x 1 , x ) and x 0 = ( x 1 0 , ( x 0 ) ) . Assume that r = dist ( x 0 , T λ ) < ε , we can obtain

z λ x 0 3 r < 3 ε , z B r ( x 0 ) .

Then, we have

(2.8) Σ g 2 τ λ ( x 0 ) C μ 1 ( z λ x 0 ) s d z z λ x 0 n + s μ a ( x 0 ) τ λ ( x 0 ) B r ( x 0 ) g 2 τ λ ( x 0 ) C μ 1 ( z λ x 0 ) s d z z λ x 0 n + s μ a ( x 0 ) τ λ ( x 0 ) .

Notice that τ λ ( x 0 ) < 0 and a ( x 0 ) is bounded from below in Ω . 1 z λ x 0 can be arbitrarily large for a sufficiently small ε . Let μ 0 . Combining the properties of the function g and (2.7), we arrive at

F g s v λ ( x 0 ) F g s v ( x ) a ( x 0 ) τ λ ( x 0 ) > 0 .

This leads to the contradiction.

In the unbounded domain, we give the condition

lim x τ λ ( x ) 0

to make sure that the minimum of τ λ , which is negative can be obtained at some point in Ω . We can prove it in a similar way as mentioned earlier. This completes the proof of Theorem 1.3.

2.4 Proof of Theorem 1.4

If we assume that there is some point x 0 Ω such that τ λ ( x 0 ) = min Ω τ λ ( x ) < 0 . By a similar proof in Theorem 1.2 and the condition (1.10), we can derive

0 F g s v λ ( x 0 ) F g s v ( x 0 ) a ( x 0 ) τ λ ( x 0 ) θ n + s Σ g 2 τ λ ( x 0 ) C μ 1 ( z λ x 0 ) s d z z λ x 0 n + s μ a ( x 0 ) τ λ ( x 0 ) .

For each fixed λ , when x 0 λ , take x 1 = ( 3 x 0 + x 1 0 , ( x 0 ) ) = ( x 1 , x ) , then x 1 Σ ˜ and B x 0 ( x 1 ) Σ ˜ . It follows that

θ n + s Σ g 2 τ λ ( x 0 ) C μ 1 ( z λ x 0 ) s d z z λ x 0 n + s = θ n + s Σ ˜ g 2 τ λ ( x 0 ) C μ 1 ( z x 0 ) s d z z x 0 n + s θ n + s B x 0 ( x 1 ) g 2 τ λ ( x 0 ) C μ 1 ( z x 0 ) s d z z x 0 n + s θ n + s 1 ( 4 x 0 ) n + s B x 0 ( x 1 ) g 2 τ λ ( x 0 ) 4 C μ 1 x 0 s d z .

Then, we have

F g s v λ ( x 0 ) F g s v ( x 0 ) a ( x 0 ) τ λ ( x 0 ) θ n + s 1 ( 4 x 0 ) n + s B x 0 ( x 1 ) g 2 τ λ ( x 0 ) 4 C μ 1 x 0 s d z μ a ( x 0 ) τ λ ( x 0 ) = x 0 n s θ n + s 4 n + s B x 0 ( x 1 ) g 2 τ λ ( x 0 ) 4 C μ 1 x 0 s d z μ a ( x 0 ) τ λ ( x 0 ) x 0 n + s .

Since F g s v λ ( x 0 ) F g s v ( x 0 ) a ( x 0 ) τ λ ( x 0 ) 0 in (1.10) and τ λ ( x ) < 0 , letting μ 0 , we can obtain

θ n + s 4 n + s B x 0 ( x 1 ) g 2 τ λ ( x 0 ) 4 C μ 1 x 0 s d z a ( x 0 ) τ λ ( x 0 ) x 0 n + s > 0 .

When x 0 is sufficiently large, by using (1.11), a contradiction occurs.

By summarizing the aforementioned, we can derive that there exists a constant A > 0 , such that if

τ λ ( x 0 ) = min Ω τ λ ( x ) < 0 ,

then

x 0 A .

3 Symmetry and nonexistence of positive solutions

In this section, we will use the direct method of moving planes to derive specific proofs of the symmetry for solutions about equation (1.1) in the ellipsoid region B , which is given by (1.13). It is clear that the moving plane T λ in the following proof is perpendicular to the coordinate axis.

3.1 Proof of Theorem 1.5

In the following proof, we choose the x 1 -axis and divide the proof into the following two steps. Then, we perform the procedure of the direct method of moving planes in the following two steps.

Step 1. Let

Ω λ = { x B x 1 < λ } .

First, we prove that when λ sufficiently approaches to a 1 ,

(3.1) τ λ ( x ) 0 , x Ω λ .

Obviously, we have

F g s v λ ( x ) F g s v ( x ) a λ ( x ) τ λ ( x ) = 0 , x Ω λ ,

where

(3.2) a λ ( x ) = f ( v λ ( x ) ) f ( v ( x ) ) v λ ( x ) v ( x ) .

Because f is Lipschitz continuous, we have a λ ( x ) is bounded from below. When λ > a 1 and sufficiently approaches a 1 , we can derive Ω λ is a narrow region. When x Σ λ \ Ω λ , by (1.14), we have v ( x ) = 0 and v λ ( x ) 0 . Then,

τ λ ( x ) = v λ ( x ) v ( x ) 0 , x Σ λ \ Ω λ .

By (1.5) and Theorem 1.3, we proved in Section 2, we can see that (3.1) is established. This is the starting point of moving plane.

Step 2. Define λ 0 as follows:

(3.3) λ 0 = sup { λ 0 τ α ( x ) 0 , x Σ α , α λ } .

Under the condition that (3.1) holds, the plane T λ is continued to move toward T λ 0 . We will prove that

(3.4) λ 0 = 0 .

We assume that (3.4) does not hold, then we have λ 0 < 0 by using (3.3). Then we can continue to move the plane Ω λ forward a little bit. In other words, we can find a small δ > 0 such that inequality (3.1) still holds when λ ( λ 0 , λ 0 + δ ) . This contradicts (3.3). Next, we split the region Ω λ into the following two parts:

Ω λ = Ω λ 0 ε ( Ω λ \ Ω λ 0 ε ) , ε > 0 .

First, we consider the region Ω λ 0 ε . Since τ λ 0 ( x ) is not identical equal to zero, by (1.5) and Theorem 1.2, we have

τ λ 0 ( x ) > 0 , x Ω λ 0 .

Thus for any ε > 0 , there exists a constant c ε greater than zero such that

τ λ 0 ( x ) c ε > 0 , x Ω λ 0 ε .

Since τ λ is continuous about λ , there exists δ > 0 sufficiently small, for any λ ( λ 0 , λ 0 + δ ) , we have

(3.5) τ λ ( x ) 0 , x Ω λ 0 ε .

It is obvious that Ω λ \ Ω λ 0 ε is a narrow region when ε and δ are adequately small. Combining (3.5), (1.5), and F g s v λ ( x ) F g s v ( x ) a λ ( x ) τ λ ( x ) = 0 , by Theorem 1.3, we can deduce that

τ λ ( x ) 0 , x Ω λ \ Ω λ 0 ε .

This together with (3.5) indicates that for all λ ( λ 0 , λ 0 + δ ) ,

τ λ ( x ) 0 , x Ω λ .

This yields a confliction. Thus, λ 0 = 0 and τ 0 ( x ) 0 for all x Ω 0 . That is,

(3.6) v ( x 1 , x 2 , , x n ) v ( x 1 , x 2 , , x n ) , 0 < x 1 < a 1 .

Now, we move the plane T λ continuously from a 1 to the left. Continue this process in a similar way as mentioned earlier, then we obtain

v ( x 1 , x 2 , , x n ) v ( x 1 , x 2 , , x n ) , 0 < x 1 < a 1 .

Combining this with (3.6), we obtain that v is symmetric about the plane T 0 in the x 1 -direction.

Indeed, if we move the plane vertically along the other n 1 axes in a similar manner as described earlier, we can derive that v must be symmetric about the coordinate plane through the origin. This is the proof of Theorem 1.5.

3.2 Proof of Theorem 1.6

Step 1. We start with the case λ < A , A > 0 large enough. We prove that

(3.7) τ λ ( x ) 0 , x Σ λ .

Through the mean value theorem, we have

(3.8) F g s v λ ( x ) F g s v ( x ) = f ( v λ ( x ) ) f ( v ( x ) ) = f ( b λ ( x ) ) τ λ ( x ) ,

where b λ ( x ) is between v λ ( x ) and v ( x ) . To be more specific, we obtain

0 < min { v λ ( x ) , v ( x ) } b λ ( x ) max { v λ ( x ) , v ( x ) } .

Let c λ ( x ) = f ( b λ ( x ) ) . We can easily obtain that for all x Σ λ ,

F g s v λ ( x ) F g s v ( x ) c λ ( x ) τ λ ( x ) = 0 .

Since lim x x η v ( x ) = 0 , we have

lim x x η min { v λ ( x ) , v ( x ) } = 0

and

lim x x η max { v λ ( x ) , v ( x ) } = 0 .

Therefore, the conclusion is obtained easily that lim x x η b λ ( x ) = 0 . Since p η 2 s , we have

c λ ( x ) = f ( b λ ( x ) ) ( b λ ( x ) ) p ,

then

lim x x 2 s c λ ( x ) lim x x 2 s ( b λ ( x ) ) p = lim x x 2 s p η ( x η b λ ( x ) ) p = 0 .

Combining Theorem 1.4, if τ λ ( x 0 ) = min Σ λ τ λ ( x ) < 0 , it holds

(3.9) x 0 A .

That is when λ < A , we have τ λ ( x 0 ) = min Σ λ τ λ ( x ) 0 , which means that (3.7) holds.

Step 2. We can define λ 0 as follows,

(3.10) λ 0 = sup { λ τ α ( x ) 0 , x Σ α , α λ } .

Under condition (3.7), we move the plane T λ continuously toward T λ 0 . We will prove that

(3.11) τ λ 0 0 , x Σ λ 0 .

We assume that (3.11) does not hold, then we can continue to move the plane T λ 0 forward for a little bit. In other words, we can find a small γ 0 > 0 such that for all 0 < γ < γ 0 ,

(3.12) τ λ ( x ) 0 , x Σ λ ,

where λ = λ 0 + γ . This is in contradiction to the definition of λ 0 .

It remains to show (3.12). Since τ λ 0 ( x ) 0 and τ λ 0 ( x ) 0 , there exists some point x such that τ λ 0 ( x ) > 0 . By (1.5) and Theorem 1.2, we can deduce that

τ λ 0 ( x ) > 0 , x Σ λ 0 .

When σ > 0 , we have that there exists a constant a 0 greater than zero such that

τ λ 0 ( x ) a 0 > 0 , x Σ λ 0 σ B A ( 0 ) ¯ ,

where A is given in (3.9). There exists γ small enough, which is greater than zero. Since τ λ is continuous with respect to λ , we obtain that for λ = λ 0 + γ ,

(3.13) τ λ ( x ) 0 , x Σ λ 0 σ B A ( 0 ) ¯ .

For sufficiently small σ and γ , ( Σ λ \ Σ λ 0 σ ) B A ( 0 ) is a narrow region. By (1.5) and Theorem 1.3, we deduce that

(3.14) τ λ ( x ) 0 , x ( Σ λ \ Σ λ 0 σ ) B A ( 0 ) .

According to (3.13), (3.14), and the conclusion we derived in step 1, we have (3.12) holds, which is a contradiction to (3.10). Thus, (3.11) is correct.

If we move the plane vertically along the other x i -axis ( i = 2 , , n ) in a similar manner as described earlier, we can derive the same conclusion that v must be symmetric about the coordinate plane through some point in R n . This is the proof of Theorem 1.6.

3.3 Proof of Theorem 1.7

Step 1. We are able to verify that when λ is sufficiently negative,

(3.15) τ λ ( x ) 0 , x Σ λ .

Assume that the condition (3.15) is not correct, we can derive that there is a point x 0 such that the following condition holds

τ λ ( x 0 ) = min Σ λ τ λ ( x ) < 0 .

By using the mean value theorem, we have

F g s v λ ( x 0 ) F g s v ( x 0 ) = f ( v λ ( x 0 ) ) f ( v ( x 0 ) ) = f ( d λ ( x 0 ) ) τ λ ( x 0 ) ,

where 0 < v λ ( x 0 ) d λ ( x 0 ) v ( x 0 ) . When λ is sufficiently negative, we can derive that v ( x 0 ) is small, and thus, d λ ( x 0 ) is small. On the one hand, since f is Lipschitz continuous, we have

(3.16) F g s v λ ( x 0 ) F g s v ( x 0 ) = f ( d λ ( x 0 ) ) τ λ ( x 0 ) 0 .

On the other hand, by (2.6), we have

F g s v λ ( x 0 ) F g s v ( x 0 ) > 0 ,

which yields a confliction with (3.16). Therefore, we prove the inequality (3.15) presented in the Step 1.

Step 2. Define λ 0 as follows, which is the limiting position, that is,

λ 0 = sup { λ τ α ( x ) 0 , x Σ α , α λ } .

Under condition (3.15), the plane T λ is continued to move toward T λ 0 . We are able to prove that

(3.17) τ λ 0 ( x ) 0 , x Σ λ 0 .

We assume that (3.17) does not hold, then we can continue to move the plane T λ 0 forward a little bit. In other words, we can find a small γ 0 > 0 , such that for all 0 < γ < γ 0 ,

(3.18) τ λ ( x ) 0 , x Σ λ ,

where λ = λ 0 + γ . This is a contradiction to the definition of λ 0 .

If there exists one point x satisfying τ λ 0 ( x ) > 0 , combining (1.5) and Theorem 1.2, we can derive

τ λ 0 ( x ) > 0 , x Σ λ .

Then for all σ > 0 , there exists a 0 > 0 satisfying

τ λ 0 ( x ) a 0 > 0 , x Σ λ 0 σ B A ( 0 ) ¯ .

For all sufficiently small γ > 0 , combining the continuity of τ λ with regard to λ , we derive

(3.19) τ λ ( x ) 0 , x Σ λ 0 σ B A ( 0 ) ¯ .

For sufficiently small σ and γ , ( Σ λ \ Σ λ 0 σ ) B A ( 0 ) is a narrow region. When x ( Σ λ \ Σ λ 0 σ ) B A ( 0 ) , it holds F g s v λ ( x ) F g s v ( x ) f ( d λ ( x ) ) τ λ ( x ) = 0 . Combining (3.19) and (1.5), by Theorem 1.3, we obtain that

(3.20) τ λ ( x ) 0 , x ( Σ λ \ Σ λ 0 σ ) B A ( 0 ) .

According to (3.19), (3.20) and the conclusion we derived in step 1, we have (3.18) holds, which is a contradiction to the definition of λ 0 , and thus, we complete the proof of (3.17).

Therefore, if we move the plane vertically along the other x i -axis ( i = 2 , , n ) in a similar manner as described above, we can derive the same conclusion that v must be symmetric about the coordinate plane through some point in R n . This is the proof of Theorem 1.7.

3.4 Proof of Theorem 1.8

First, we prove either v ( x ) > 0 or v ( x ) 0 in R + n .

Case 1. If there exists a point x 0 R + n such that v ( x 0 ) = min R + n v ( x ) = 0 , combining with assumption (1.21), f ( 0 ) = 0 and the proof in Theorem 1.1, we have

F g s v ( x 0 ) = inf P.V. R n g v ( z ) v ( x 0 ) C 1 ( z x 0 ) s d z C 1 ( z x 0 ) n + s C C P.V. R n g v ( z ) v ( x 0 ) C μ 1 ( z x 0 ) s d z C μ 1 ( z x 0 ) n + s μ = P.V. R + n g v ( z ) C μ 1 ( z x 0 ) s d z C 1 ( z x 0 ) n + s μ P.V. θ n + s R + n g v ( z ) C μ 1 ( z x 0 ) s d z z x 0 n + s μ .

Taking μ 0 , by inequality (2.3), we have

0 < θ n + s R + n g v ( z ) C μ 1 ( z x 0 ) s d z z x 0 n + s F g s v ( x 0 ) = f ( v ( x 0 ) ) = 0 .

Combining the properties of g and the continuity of v , it holds v ( x ) 0 in R + n .

Case 2. Suppose v ( x ) > 0 in R + n . We move the plane along the x 1 -axis by using the direct method of moving planes.

Step 1. When x R n , by (1.20), it is obvious that v ( x ) 0 . Since v is a nonnegative solution, v ( x ) 0 in R + n . For all λ 0 , we have

(3.21) τ λ ( x ) = v λ ( x ) v ( x ) 0 , x Σ λ .

This provides the starting point of moving planes.

Step 2. Define λ 0 as follows:

λ 0 = sup { λ τ α ( x ) 0 , x Σ α , α λ } .

Keep moving the plane T λ to the limiting position T λ 0 as long as (3.21) holds. We will prove that λ 0 = .

Suppose λ 0 < , then (3.21) holds for any λ λ 0 . Combining Theorem 1.3 with Theorem 1.4, by (1.5) and Theorem 1.6, we can derive that

τ λ ( x ) 0 , x Σ λ .

Let x 1 = 3 λ 0 and mark x = ( x 2 , , x n ) , then we can derive that

τ λ 0 ( x ) = v ( λ 0 , x ) v ( 3 λ 0 , x ) = 0 .

Since v ( x ) = 0 , for x R + n , we have

v ( λ 0 , x ) = v ( 3 λ 0 , x ) = 0 ,

this conflicts with hypothesis v ( x ) > 0 in R + n . Thus, the conclusion λ 0 = can be obtained.

Since inequality (3.21) holds for all λ λ 0 = , v ( x ) is monotone increasing concerning x 1 . When λ , it yields a contradiction with condition (1.21). On the aforementioned proof, we derive v ( x ) 0 in R + n .

3.5 Proof of Theorem 1.9

Combining the condition f ( 0 ) = 0 , the assumption (1.24) and the proof in Theorem 1.8, we can derive either v ( x ) > 0 or v ( x ) 0 in R + n . Suppose v ( x ) > 0 in R + n . We move the plane along the x 1 -axis by using the direct method of moving planes.

Step 1. When x R n , it is obvious that v ( x ) 0 . Therefore, for all λ 0 , we can derive

(3.22) τ λ ( x ) 0 , x Σ λ .

From this, we obtain the starting point of the moving planes.

Step 2. Define λ 0 as follows, which is the limiting position, that is,

λ 0 = sup { λ τ α ( x ) 0 , x Σ α , α λ } .

We are able to prove that λ 0 = .

Suppose λ 0 < , then (3.22) is established for any λ λ 0 . Combining Theorems 1.3 and 1.4, by using an analogous approach as in Theorem 1.7, we can derive that

τ λ 0 ( x ) 0 , x Σ λ .

We can prove that λ 0 must be in an analogous approach as in Theorem 1.8. This conflicts (1.24). Consequently, we have v ( x ) 0 .

4 Monotonicity

In this section, by the direct method of moving planes, we prove monotonicity result.

4.1 Proof of Lemma 1.1

Since v ( x ) 0 and v ( x l ) = 0 , we have min x R n v ( x ) = 0 = v ( x l ) . It follows

0 F g s v ( x l ) = inf P.V. R n g v ( z ) v ( x l ) C 1 ( z x l ) s d z C 1 ( z x l ) n + s C C = inf P.V. R n g v ( z ) C 1 ( z x l ) s d z C 1 ( z x l ) n + s C C 0 .

Therefore, v = 0 a.e. in R n holds. Lemma 1.1 is proved.

4.2 Proof of Theorem 1.10

Because Ω is a bounded domain, we can take Ω { x 1 } for simplicity. Note T λ = { x = ( x 1 , x ) R n x 1 = λ } . Since Ω is convex in x 1 -direction and is symmetric with respect to T 0 , combining with (1.25), we can derive τ λ 0 in Σ λ \ Ω , where λ ( , 0 ] and

(4.1) τ λ 0 x Σ λ \ Ω , λ ( , 0 ) .

Next, we will prove τ λ > 0 in Σ λ Ω , where λ ( 1 , 0 ) . We divide our proof into two steps.

Step 1. We will prove that there exists δ > 0 sufficiently small such that for λ ( 1 , 1 + δ ] ,

(4.2) τ λ ( x ) 0 , x Σ λ Ω .

Assume that the condition (4.2) is not correct,that is, there exists a sequence { λ i } ( 1 , 0 ) , which satisfies λ i 1 as i + such that

(4.3) inf Σ λ i Ω τ λ i = inf λ ( 1 , λ i ] x Σ λ τ λ ( x ) < 0 .

Thus, we can derive that there exists a point x i Σ λ i Ω such that

(4.4) τ λ i ( x i ) = inf Σ λ i Ω τ λ i = inf Σ λ i τ λ i < 0 .

Therefore, by combining with (1.25), we have

(4.5) F g s τ λ i ( x i ) F g s v λ i ( x i ) F g s v ( x i ) = f ( v λ i ( x i ) ) f ( v ( x i ) ) = a ( x i ) τ λ i ( x i ) ,

where

(4.6) a ( x i ) = f ( v λ i ( x i ) ) f ( v ( x i ) ) v λ i ( x i ) v ( x i ) .

When i is large enough, Σ λ i Ω is a narrow region. From (1.5), (4.5), (4.6), and Theorem 1.3, we have

(4.7) τ λ i ( x ) 0 , x Σ λ i Ω ,

which contradicts with (4.3). Thus, there exists δ > 0 small enough such that (4.2) holds for 1 < λ 1 + δ . Moreover, if there exists λ ˜ ( 1 , 1 + δ ] and x ˜ Σ λ ˜ Ω satisfying τ λ ˜ ( x ˜ ) = 0 . Similarly, we obtain from (1.25) and (1.26) that

(4.8) F g s τ λ ˜ ( x ˜ ) = f ( τ λ ˜ ( x ˜ ) ) 0 .

By combining with Lemma 1.1, we can obtain τ λ ˜ = 0 a.e. in R n , which is contradictory with (4.1). As a result, for any 1 < λ 1 + δ , we have

(4.9) τ λ ( x ) > 0 , x Σ λ Ω .

Step 2. Define

(4.10) λ 0 = sup { λ ( 1 , 0 ] τ ε > 0 in Σ ε Ω , 1 < ε < λ } .

In this step, we show that λ 0 = 0 via contradiction arguments.

Assume that the aforementioned condition above is not correct, then λ 0 < 0 , and this means that we can move the plane T λ a little further away under condition (4.9) holds. This conflicts with the definition (4.10) of λ 0 .

In fact, according to λ 0 < 0 , we can obtain from (1.25) that τ λ 0 ( x ) > 0 in ( Ω λ 0 \ Ω ) Σ ( Ω λ 0 ) is the reflection of Ω concerning T λ 0 , and combining with Lemma 1.1, we can obtain

(4.11) τ λ 0 ( x ) > 0 , x Σ λ 0 Ω .

Because τ λ 0 > 0 in Ω λ 0 Σ λ 0 , we can find a compact A Ω λ 0 Σ λ 0 such that

(4.12) τ λ 0 ( x ) a > 0 , x A Ω ,

in which a is a constant. It is obvious that ( Σ λ 0 Ω ) \ ( A Ω ) is a narrow region. Noting τ λ is continuous, we can derive that there exists 0 < δ < min { λ 0 , λ 0 + 1 } small enough such that

(4.13) τ λ ( x ) > 0 , x A Ω ,

where λ 0 < λ < λ 0 + δ , and ( Σ λ 0 + δ Ω ) \ ( A Ω ) is a narrow region.

When λ [ λ 0 , λ 0 + δ ] , it is obvious that ( Σ λ Ω ) \ ( A Ω ) is a narrow region, we can obtain from Theorem 1.3 that

(4.14) τ λ ( x ) 0 , x ( Σ λ Ω ) \ ( A Ω ) .

Assume that the condition (4.13) is false, that is, there exists a point λ ˆ ( λ 0 , λ 0 + δ ] satisfying

(4.15) inf ( Σ λ ˆ Ω ) \ ( A Ω ) τ λ ˆ = inf Σ λ ˆ Ω τ λ ˆ = inf λ ( λ 0 , λ ˆ ] x Σ λ τ λ ( x ) < 0 .

Thus, there exists one point x ˆ ( Σ λ ˆ Ω ) \ ( A Ω ) satisfying

(4.16) τ λ ˆ ( x ˆ ) = inf Σ λ ˆ Ω \ ( A Ω ) τ λ ˆ = inf Σ λ ˆ Ω τ λ ˆ = inf Σ λ ˆ τ λ ˆ < 0 .

Therefore, by combining (1.25), we obtain

(4.17) F g s τ λ ˆ ( x ˆ ) F g s v λ ˆ ( x ˆ ) F g s v ( x ˆ ) = f ( v λ ˆ ( x ˆ ) ) f ( v ( x ˆ ) ) = a ( x ˆ ) τ λ ˆ ( x ˆ ) ,

where

(4.18) a ( x ˆ ) = f ( v λ ˆ ( x ˆ ) ) f ( v ( x ˆ ) ) v λ ˆ ( x ˆ ) v ( x ˆ ) .

Obviously, ( Σ λ ˆ Ω ) \ ( A Ω ) is a narrow region when A is sufficiently large. From (4.17), (4.18), and Theorem 1.3, we have

(4.19) τ λ ˆ > 0 , x ( Σ λ ˆ Ω ) \ ( A Ω ) ,

which contradicts with (4.15). Thus, when λ [ λ 0 , λ 0 + δ ] , there exists δ > 0 sufficiently small such that (4.14) holds. Moreover, combining with Lemma 1.1 and using a similar method in step 1, we will obtain that

(4.20) τ λ ( x ) > 0 , x ( Σ λ Ω ) \ ( A Ω ) .

Therefore,

(4.21) τ λ ( x ) > 0 , x Σ λ Ω ,

where λ 0 < λ < λ 0 + δ , which contradicts the definition of λ 0 . Therefore, we obtain λ 0 = 0 . For any ( x 1 , x ) , ( x ˙ 1 , x ) Ω and x ˙ 1 < x 1 < 0 , let λ = x ˙ 1 + x 1 2 , we derive that

v ( x 1 , x ) > v ( x ˙ 1 , x ) .

Therefore, ( x 1 , x ) is strictly increasing along x 1 -direction.

This is the proof of Theorem 1.10.

5 Asymptotic property

In this part, we will apply the maximum principles which are introduced in Section 1 to investigate the asymptotic property of solutions.

5.1 Proof of Theorem 1.11

Suppose the conclusion is not correct, that is, there exists a point x 0 Ω and a constant M > 0 such that v ( x 0 ) > 0 , and

(5.1) 0 < M = sup x R n v ( x ) < .

There exists a sequence { x k } Ω satisfying γ k 1 ( 0 < γ k < 1 ) as k such that

(5.2) v ( x k ) γ k M .

Let

(5.3) Φ ( x ) = e x 2 x 2 1 , x < 1 , 0 , x 1 .

Easily we can derive that Φ is radially decreasing from the origin, and therefore, F g s Φ ( x ) C 0 .

Define

Φ k ( x ) = Φ x x k r x k .

Take ε k = ( 1 γ k ) M . Since v M , Φ k ( x k ) = Φ ( 0 ) = 1 and Φ k = 0 in R n \ B r x k ( x k ) , we obtain

(5.4) v ( x k ) + ε k Φ k ( x k ) M v ( x ) + ε k Φ k ( x ) , x R n \ B r x k ( x k ) .

Therefore, there exists x ˙ k B r x k ( x k ) , it holds

(5.5) v ( x ˙ k ) + ε k Φ k ( x ˙ k ) = max x R n [ v ( x ) + ε k Φ k ( x ) ] M .

Further, we can obtain that

(5.6) v ( x ˙ k ) v ( x k ) + ε k Φ k ( x k ) ε k Φ k ( x ˙ k ) v ( x k ) γ k M > 0 .

Then, by (5.5), we can derive

(5.7) F g s [ v + ε k Φ k ] ( x ˙ k ) = inf P.V. R n g v ( z ) + ε k Φ k ( z ) v ( x ˙ k ) ε k Φ k ( x ˙ k ) C 1 ( x ˙ k z ) s d z C 1 ( x ˙ k z ) n + s = inf P.V. B r x k ( x k ) g v ( z ) + ε k Φ k ( z ) v ( x ˙ k ) ε k Φ k ( x ˙ k ) C 1 ( x ˙ k z ) s d z C 1 ( x ˙ k z ) n + s + inf P.V. ( B r x k ( x k ) ) c g v ( z ) + ε k Φ k ( z ) v ( x ˙ k ) ε k Φ k ( x ˙ k ) C 1 ( x ˙ k z ) s d z C 1 ( x ˙ k z ) n + s inf ( B r x k ( x k ) ) c g v ( z ) + ε k Φ k ( z ) v ( x ˙ k ) ε k Φ k ( x ˙ k ) C 1 ( x ˙ k z ) s d z C 1 ( x ˙ k z ) n + s inf ( B c 1 r x k ( x k ) \ B r x k ( x k ) ) Σ g v ( x ˙ k ) ε k Φ k ( x ˙ k ) C 1 ( x ˙ k z ) s d z C 1 ( x ˙ k z ) n + s ( B c 1 r x k ( x k ) \ B r x k ( x k ) ) Σ g v ( x ˙ k ) + ε k Φ k ( x ˙ k ) C 1 ( x ˙ k z ) s d z C 1 ( x ˙ k z ) n + s R ( B c 1 r x k ( x k ) \ B r x k ( x k ) ) Σ d z C 1 ( x ˙ k z ) n + s R D r x k 2 s ,

where R and D are two constants greater than 0.

Subsequently, the lower bound of F g s [ v + ε k Φ k ] ( x ˙ k ) can be estimated as follows.

In fact, we can derive v ( x ˙ k ) 0 from (5.6), and thus, x ˙ k Ω . Let F g s Φ k C 1 r x k 2 s , we can derive from (1.27) that

(5.8) F g s [ v + ε k Φ k ] ( x ˙ k ) F g s v ( x ˙ k ) + F g s ε k Φ k ( x ˙ k ) a ( x ˙ k ) v ( x ˙ k ) C 1 ε k r x k 2 s C 1 ε k r x k 2 s .

By combining (5.7) and (5.8), we derive

(5.9) D R r x k 2 s C 1 ε k r x k 2 s .

Then

C 1 M ( 1 γ k ) D R .

Since γ k 1 ( 0 < γ k < 1 ) as k , it holds C 1 M ( 1 γ k ) 0 , which is contradicted with the condition D R > 0 . Therefore, v ( x ) 0 in Ω .

5.2 Proof of Lemma 1.2

We first show that v ζ in Ω . Let w ( x ) = v ( x ) ζ , then

(5.10) w ( x ) < 0 , x R n \ Ω .

Combining the equation (1.28) given in Lemma 1.2, when v ( x ) > ζ , there is a point x Ω satisfying

(5.11) F g s w ( x ) F g s v ( x ) + F g s ζ = f ( v ( x ) ) f ( ζ ) 0 .

From (1.27), we can obtain that w ( x ) 0 in Ω . Therefore, we derive v ζ in Ω .

Then, we can conclude that v < ζ in Ω by using the strong maximum principle Lemma 1.1. Lemma 1.2 is proved.

5.3 Proof of Lemma 1.3

Proof

Denote the principle eigenvalue of F g s by λ 1 . Assume that ω is the corresponding eigenfunction and

(5.12) F g s ω = λ 1 ω , ω > 0 , x B 1 ( 0 ) , ω = 0 , x R n \ B 1 ( 0 ) ,

with max x B 1 ( 0 ) ω ( x ) = ω ( 0 ) = 1 .

By assumption (b), for any 0 < ε t 0 and B 0 = ( λ 1 a 0 ) 1 2 s , we have

(5.13) F g s ε ω x B 0 = λ 1 B 0 2 s ε ω x B 0 = a 0 ε ω x B 0 f ε ω x B 0 .

Assume that there exists a point z ˜ Ω and dist ( z ˜ , Ω ) > B 0 . Let

(5.14) δ 0 = min { t 0 , inf B B 0 ( z ˜ ) v } 0 .

Then, we can derive

(5.15) v ( x ) > δ 0 ω x z ˜ B 0 , x B B 0 ( z ˜ ) .

For 0 t 1 and z Ω , where dist ( z , Ω ) > B 0 , let z ( t ) = t z + ( 1 t ) z ˜ and

(5.16) c t ( x ) = v ( x ) δ 0 ω x z ( t ) B 0 , x B B 0 ( z ( t ) ) .

Then, we will prove

(5.17) c t ( x ) > 0 , x B B 0 ( z ( t ) ) .

Assume that the condition (5.17) is false, that is, there exist t 0 [ 0 , 1 ] and x 0 B B 0 ( z ( t 0 ) ) such that c t 0 ( x 0 ) = 0 , then we can derive from (1.29) and (5.13) that

(5.18) F g s c t 0 ( x 0 ) F g s u ( x 0 ) + F g s δ 0 ω x 0 z ( t 0 ) B 0 ( x 0 ) = f ( u ( x 0 ) ) f δ 0 ω x 0 z ( t 0 ) B 0 = 0 .

Since c t 0 ( x 0 ) = 0 , we have

(5.19) F g s c t 0 ( x 0 ) = inf P.V. R n g c t 0 ( z ) C 1 ( z x 0 ) s d z C 1 ( z x 0 ) n + s < 0 ,

which is contradictory to (5.18). Therefore, we can obtain that (5.17) is valid. When t = 1 and x = z , we obtain

v ( x ) > δ 0 .

Since x Ω and dist ( x , Ω ) > B 0 is arbitrary, we complete the proof of Lemma 1.3.□

Now, we prove Theorem 1.12. From Lemma 1.2, we can obtain that 0 < v < ζ in Ω . The eigenfunction ω and the first eigenvalue λ 1 is the same as Lemma 1.3.

5.4 Proof of Theorem 1.12

According to the assumption (a) and f is a continuous function, we can derive that there exist a 1 > a 0 > 0 and δ 1 > 0 such that

(5.20) f ( t ) a 1 , t [ δ 0 , ζ δ 1 ] .

For z Ω , we define d z = dist ( z , Ω ) sufficiently large satisfying d z > max 2 B 0 , 2 ( λ 1 ζ a 1 ) 1 2 s , from Lemma 1.3, we obtain

v ( x ) > δ 0 , x B d z 2 ( z ) .

Let ω z ( x ) = ω 2 ( x z ) d z , then

(5.21) F g s ω z ( x ) λ 1 2 d z 2 s , x B d z 2 ( z ) .

Because max x B d z 2 ( z ) ω z ( x ) = ω z ( z ) = 1 , we derive that for 0 < β δ 0 ,

β ω z ( x ) < v ( x ) , x Ω .

Let

β 0 = sup { β β ω z ( x ) < v ( x ) , x Ω }

be the maximum point. Since v < ζ , we have δ 0 < β 0 < ζ .

According to the definition of β 0 , we derive v ( x ) β 0 ω z ( x ) in R n . Moreover, one point x 0 B d z 2 ( z ) can be found to satisfy v ( x 0 ) = β 0 ω z ( x 0 ) . The conclusion implies that the function v ( x ) β 0 ω z ( x ) can be taken to the minimum value when x = x 0 , so we derive

F g s ( v β 0 ω z ) ( x 0 ) 0 .

By combining (5.21), we obtain

(5.22) F g s v ( x 0 ) = f ( v ( x 0 ) ) β 0 λ 1 2 d z 2 s < β 0 λ 1 a 1 λ 1 ζ < a 1 .

Also, we can derive that

(5.23) δ 0 < v ( x 0 ) = β 0 ω z ( x 0 ) β 0 ω z ( z ) v ( z ) < ζ .

By combining (5.20), (5.22), and (5.23), we have ζ δ 1 < v ( x 0 ) v ( z ) < ζ . Thus, by assumption (c) and (5.22), we have

(5.24) F g s v ( z ) = f ( v ( z ) ) f ( v ( x 0 ) ) β 0 λ 1 2 d z 2 s .

Thus, when d z = dist ( z , Ω ) is sufficiently large for all z Ω , we derive ζ δ 1 < v ( z ) < ζ and f ( v ( z ) ) 0 as dist ( z , Ω ) + . It follows from the assumption (c), we have

lim dist ( x , Ω ) + v ( x ) = ζ .

The aforementioned is the proof of Theorem 1.12.

Acknowledgements

The authors would like to express their gratitude to the anonymous reviewers and editors for their valuable comments and suggestions that led to the improvement of the original manuscript.

  1. Funding information: The work was supported by National Natural Science Foundation of China (No. 12001344), NSF of Shanxi Province, China (No. 20210302123339) and Postgraduate Education Innovation Program of Shanxi Normal University, China (No. 2024YJSKCSZSFK-06).

  2. Author contributions: Guotao Wang: writing – review and editing, supervision, and funding acquisition. Rui Yang: writing – review and editing and writing – original draft. Lihong Zhang: conceptualization, and supervision.

  3. Conflict of interest: The authors have no relevant financial or non-financial interests to disclose.

  4. Data availability statement: No data were used for the research described in the article.

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Received: 2023-07-10
Revised: 2024-07-04
Accepted: 2024-07-14
Published Online: 2024-09-02

© 2024 the author(s), published by De Gruyter

This work is licensed under the Creative Commons Attribution 4.0 International License.

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