Qualitative properties of solutions for dual fractional parabolic equations involving nonlocal Monge-Ampère operator

Zerong Yang; Yong He

doi:10.1515/anona-2025-0086

Article Open Access

Qualitative properties of solutions for dual fractional parabolic equations involving nonlocal Monge-Ampère operator

Zerong Yang and Yong He

Published/Copyright: June 24, 2025

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From the journal Advances in Nonlinear Analysis Volume 14 Issue 1

Abstract

In this article, we mainly study the qualitative properties of solutions for dual fractional-order parabolic equations with nonlocal Monge-Ampère operators in different domains

∂ t β μ ( y , t ) − D α τ μ ( y , t ) = f ( μ ( y , t ) ) .

We first establish a series of maximum principles and averaging effects theorems for antisymmetric functions and then used the method of moving planes and sliding planes to establish radial symmetry, monotonicity, nonexistence, and Liouville theorem for positive solutions.

Keywords: dual parabolic equations; monotonicity; radial symmetry; nonexistence; Liouville theorem

MSC 2010: 35R11; 35B50; 35B06; 26A33; 47G30; 35B53

1 Introduction

In this article, we mainly study the qualitative properties of positive solutions to dual fractional parabolic equations related to operator ∂ t β − D α τ .

(1.1) ∂ t β μ ( y , t ) − D α τ μ ( y , t ) = f ( μ ( y , t ) ) , in B 1 ( 0 ) × R , μ ( y , t ) ≡ 0 , in B 1 c ( 0 ) × R ,

(1.2) ∂ t β μ ( y , t ) − D α τ μ ( y , t ) = 0 , in R n × R ,

and

(1.3) ∂ t β μ ( y , t ) − D α τ μ ( y , t ) = f ( μ ( y , t ) ) , in R + n × R , μ ( y , t ) ≡ 0 , in ( R n \ R + n ) × R ,

(1.4) ∂ t β μ ( y , t ) − D α τ μ ( y , t ) = f ( μ ( y , t ) , t ) , in R n × R ,

and

(1.5) ∂ t β μ ( y , t ) − D α τ μ ( y , t ) = y 1 μ s , in R n × R ,

where 0 < β , α < 1 < s < + ∞ , R + n ≔ { y ∈ R ∣ y 1 > 0 } is located in the right half of the space R n and B 1 ( 0 ) is the unit ball. For the aforementioned equations, we consider the nonlocal Marchaud fractional derivative ∂ t β and the nonlocal Monge-Ampère operator D α τ , which are defined as follows:

(1.6) ∂ t β μ ( y , t ) = C β ∫ − ∞ t μ ( y , t ) − μ ( y , ν ) ( t − ν ) 1 + β d ν

and

(1.7) D α τ μ ( y , t ) = inf PV ∫ R n μ ( z , t ) − μ ( y , t ) ∣ G − 1 ( z − y ) ∣ n + 2 α d z ∣ G ∈ G ,

where C β > 0 , P V is the Cauchy principal value, G is n × n symmetric positive definite matrix, and G = { G ∣ G > 0 , det G = 1 , λ min ( G ) ≥ τ > 0 } . Here, λ min ( G ) is the smallest eigenvalue of matrix G . In order that the integrals (1.6) and (1.7) are well defined, we assume that

μ ( y , t ) ∈ ( C loc 1,1 ( R n ) ∩ L 2 α ( R n ) ) × ( C 1 ( R ) ∩ L β − ( R ) )

with

L 2 α ( R n ) ≔ μ ( ⋅ , t ) ∈ L loc 1 ( R n ) ∣ ∫ R n ∣ μ ( y , t ) ∣ 1 + ∣ y ∣ n + 2 α d y < + ∞

and

L β − ( R ) ≔ μ ( y , ⋅ ) ∈ L loc 1 ( R ) ∣ ∫ − ∞ t ∣ μ ( y , ν ) ∣ 1 + ∣ ν ∣ 1 + β d ν < + ∞ for any t ∈ R .

The nonlocal Monge-Ampère operator D α τ introduced by Caffarelli and Charro in [1] has the following relationship with fractional Laplace operators

D α τ μ ( y , t ) ≤ − C n , α − 1 ( − Δ ) α μ ( y , t ) ,

where C n , α > 0 is a constants and

− ( − Δ ) α μ ( y , t ) = PV ∫ R n μ ( z , t ) − μ ( y , t ) ∣ z − y ∣ n + 2 α d z .

In 1927, Marchaud [2] first proposed the one-sided nonlocal Marchaud fractional time derivative ∂ t β , which appeared in various physical phenomena (see [3–5]).

The extension method introduced by Caffarelli and Silvestre [6] is used to handle the nonlocality of fractional Laplacian. However, Chen et al. [7] introduced a simpler approach, the method of directly moving the plane to investigate the monotonicity and symmetry of positive solutions to various elliptic equations and systems (see [8–10] and the references therein). Wu and Chen developed a direct sliding method to handle the monotonicity and uniqueness of solutions involving local and nonlocal operators [11]. For the qualitative properties of solutions to fractional-order elliptic equations with operator D α τ , some results has been obtained by [12–15]. However, for the qualitative properties of solutions to dual fractional-order parabolic equations with operators, there are currently relatively few known results. In recent years, more and more results have been obtained by applying the moving planes method and sliding planes method to study the qualitative properties of solutions to fractional-order parabolic equations, as shown below.

Guo et al. [16] obtained a Liouville theorem and radial symmetry for the dual fractional parabolic equations:

∂ t β μ ( y , t ) + ( − Δ ) α μ ( y , t ) = f ( μ ( y , t ) ) , in B 1 ( 0 ) × R

and

∂ t β μ ( y , t ) + ( − Δ ) α μ ( y , t ) = 0 , in R n × R .

Chen and Ma [17] proposed a new idea to prove the monotonicity of positive solutions to the following problem:

∂ t β μ ( y , t ) + ( − Δ ) α μ ( y , t ) = f ( μ ( y , t ) ) in R + n × R , μ ( y , t ) ≡ 0 in ( R n \ R + n ) × R .

Wu and Chen [18] obtained radially symmetry and monotonicity of the classical solutions to the following fractional parabolic equation:

∂ t μ ( y , t ) + ( − Δ ) α μ ( y , t ) = f ( μ ( y , t ) , t ) , in R × ( − ∞ , T ] .

Chen and Guo [19] adopted a new idea to prove the nonexistence of solutions for the following master equations in R n × R .

( ∂ t − Δ ) α μ ( y , t ) = y 1 μ s ( y , t ) ,

where

( ∂ t − Δ ) α μ ( y , t ) = C n , α ∫ − ∞ t ∫ R n μ ( y , t ) − μ ( z , ν ) ( t − ν ) n 2 + α + 1 e − ∣ y − z ∣ 2 4 ( t − ν ) d z d ν .

Du and Wang [20] proved the monotonicity of solutions for parabolic equations involving nonlocal Monge-Ampère operator

∂ μ ∂ t ( y , t ) − D α τ μ ( y , t ) = f ( μ ( y , t ) ) in R + n × R , μ ( y , t ) ≡ 0 in ( R n \ R + n ) × R .

Inspired by the aforementioned work, we attempt to study the qualitative properties of solutions to fractional-order parabolic equations with operator ∂ t β − D α τ . In Section 2, we establish a series of maximum principles and averaging effects theorems for antisymmetric functions. In Section 3, we proved the radial symmetry and monotonicity of positive bounded solutions in the unit ball, for each t ∈ R . In Section 4, we established the Liouville theorem for equation (1.2). In Section 5, without assuming the boundedness of solutions and the general decay condition μ → 0 at infinity, we prove that the positive solution strictly increases in the y 1 direction. In Section 6, we adopt a new idea to prove that the entire solutions μ ( y , t ) is radial symmetry and decreases in some points in R n for all t ∈ R . In Section 7, we adopt a new idea to prove that the nonexistence of entire solutions μ ( y , t ) to equation (1.5). Before presenting our main results, we list some notations.

Let

T λ = { y ∈ R n ∣ y 1 = λ } , H λ = { y ∈ R n ∣ y 1 < λ } ,

and the reflection of y = ( y 1 , y ′ ) ∈ R n about T λ be y λ = ( 2 λ − y 1 , y ′ ) . Set

μ λ ( y , t ) = μ ( y λ , t ) , ω λ ( y , t ) = μ λ ( y , t ) − μ ( y , t ) .

Theorem 1.1

Let

μ ( y , t ) ∈ ( C 1,1 ( B 1 ( 0 ) ) ∩ C ( B 1 ( 0 ) ¯ ) ) × C 1 ( R )

be a positive bounded solution of (1.1). Assume that f ∈ C 1 ( [ 0 , + ∞ ) ) satisfies f ( 0 ) ≥ 0 , f ′ ( 0 ) ≤ 0 , then for each t ∈ R , μ ( ⋅ , t ) is radially symmetric and strictly decreasing about the origin in B 1 ( 0 ) .

Theorem 1.2

Let μ ( y , t ) ∈ C loc 1,1 ( R n ) × C 1 ( R ) be a bounded solution of (1.2). Then it must be constant.

Theorem 1.3

Assume that

μ ( y , t ) ∈ ( C loc 1,1 ( R n ) ∩ L 2 α ( R n ) ) × ( C 1 ( R ) ∩ L β − ( R ) )

be a positive solution of (1.3) and satisfying

(1.8) μ ( y , t ) ≤ C ( 1 + ∣ y ∣ γ ) for s o m e 0 < γ < 2 α .

If f ∈ C 1 ( [ 0 , + ∞ ) ) satisfies f ( 0 ) ≥ 0 , f ′ ( 0 ) ≤ 0 and f ′ is bounded from above, then the positive solution μ ( y , t ) is strictly increasing with respect to y 1 in R + n for any t ∈ R .

Theorem 1.4

Assume that

μ ( y , t ) ∈ ( C loc 1,1 ( R n ) ∩ L 2 α ( R n ) ) × ( C 1 ( R ) ∩ L β − ( R ) )

be a positive bounded solution of (1.4) satisfying

(1.9) lim ∣ y ∣ → ∞ sup t ∈ R μ ( y , t ) = 0 .

If f , f μ ∈ C ( R × R ) , f ( 0 , t ) = 0 , and there is a constant γ > 0 such that f μ ( 0 , t ) < − γ for all t ∈ R , then the positive solution μ ( y , t ) is radially symmetric and decreases in some points in R n for all t ∈ R .

Theorem 1.5

Let

μ ( y , t ) ∈ ( C loc 1,1 ( R n ) ∩ L 2 α ( R n ) ) × ( C 1 ( R ) ∩ L β − ( R ) )

be a positive bounded solution of (1.5). For any 1 < s < + ∞ , then equation (1.5) possesses no positive bounded solutions.

2 Various maximum principles

Theorem 2.1

(Narrow region principle) Let Ω be a bounded or unbounded narrow region and Ω ⊂ { y ∈ H λ ∣ λ − l < y 1 < λ , l > 0 } . Suppose that ω ( y , t ) ∈ ( C loc 1,1 ( Ω ) ∩ L 2 α ( R n ) ) × ( C 1 ( R ) ∩ L β − ( R ) ) is bounded from below in Ω ¯ × R and lower semi-continuous with respect to y on Ω ¯ satisfying

(2.1) ∂ t β ω ( y , t ) − ( D α τ ( μ λ ( y , t ) ) − D α τ ( μ ( y , t ) ) ) = c ( y , t ) ω ( y , t ) , ( y , t ) ∈ Ω × R , ω ( y , t ) ≥ 0 , ( y , t ) ∈ ( H λ \ Ω ) × R , ω ( y λ , t ) = − ω ( y , t ) , ( y , t ) ∈ H λ × R .

If c ( y , t ) is bounded from above, then for sufficiently small l, we have

(2.2) ω ( y , t ) ≥ 0 , ( y , t ) ∈ H λ × R .

Furthermore, if there exists a point ( y 0 , t 0 ) ∈ Ω × R such that ω ( y 0 , t 0 ) = 0 , then ω ( y , t ) ≡ 0 in R n × ( − ∞ , t 0 ] .

Proof

Because ω is bounded, we know that ω ( y , t ) has an infimum. If (2.2) is not true, then there exists a constant p > 0 such that

(2.3) inf Ω × R ω ( y , t ) ≕ − p < 0 .

By (2.3), we know that there are a minimizing sequence { ( y m , t m ) } ⊂ Ω × R and a sequence { ε m } ↘ 0 satisfying

(2.4) ω ( y m , t m ) = − p + ε m < 0 .

Since R n × R is an infinite interval, the minimum of ω ( y , t ) may not be attained in Ω × R , so we need to construct perturbed functions of near ( y m , t m )

(2.5) k m ( y , t ) = ω ( y , t ) − ε m η m ( y , t )

with

(2.6) E r m ( y m , t m ) ≔ B r m ( y m ) × ( t m − r m 2 α β , t m + r m 2 α β ) and 0 < r m = dist ( y m , T λ ) < l ,

where

(2.7) η m ( y , t ) = η y − y m r m , t − t m r m 2 α β ∈ C 0 ∞ ( E r m ( y m , t m ) )

satisfies

(2.8) η m ( y , t ) ≡ 1 , in E r m 2 ( y m , t m ) , 0 ≤ η m ( y , t ) ≤ 1 , in R n × R .

According to (2.4), (2.5), and (2.8), we have

(2.9) k m ( y m , t m ) = ω ( y m , t m ) − ε m = − p , k m ( y , t ) = ω ( y , t ) ≥ − p , in E r m ( y m , t m ) ∩ H λ × R .

Hence, we know that there is a point ( y ˜ m , t ˜ m ) ∈ E r m ( y m , t m ) satisfying

(2.10) − p − ε m ≤ k m ( y ˜ m , t ˜ m ) = inf H λ × R k m ( x , t ) ≤ − p .

It follows from (2.5), (2.10), and Lemma A.3 that

(2.11) − p ≤ ω ( y ˜ m , t ˜ m ) ≤ − p + ε m

and

(2.12) ∂ t β ω ( y ˜ m , t ˜ m ) ≤ ε m ∂ t β η m ( y ˜ m , t ˜ m ) ≤ C ε m r m 2 α .

Since det G = 1 , λ min G ≥ τ > 0 , obviously, we have λ min G ≤ τ 1 − n , for G ∈ G . Then, for any σ m → 0 as m → ∞ , there is G m ∈ G such that

(2.13) D α τ ( μ λ ( y ˜ m , t ˜ m ) ) − D α τ ( μ ( y ˜ m , t ˜ m ) ) = inf PV ∫ R n μ λ ( z , t ˜ m ) − μ λ ( y ˜ m , t ˜ m ) ∣ G − 1 ( z − y ˜ m ) ∣ n + 2 α d z − inf PV ∫ R n μ ( z , t ˜ m ) − μ ( y ˜ m , t ˜ m ) ∣ G − 1 ( z − y ˜ m ) ∣ n + 2 α d z ≥ PV ∫ R n μ λ ( z , t ˜ m ) − μ λ ( y ˜ m , t ˜ m ) ∣ G m − 1 ( z − y ˜ m ) ∣ n + 2 α d z − σ m − PV ∫ R n μ ( z , t ˜ m ) − μ ( y ˜ m , t ˜ m ) ∣ G m − 1 ( z − y ˜ m ) ∣ n + 2 α d z = PV ∫ R n ω ( z , t ˜ m ) − ω ( y ˜ m , t ˜ m ) ∣ G m − 1 ( z − y ˜ m ) ∣ n + 2 α d z − σ m = PV ∫ R n k ( z , t ˜ m ) − k ( y ˜ m , t ˜ m ) ∣ G m − 1 ( z − y ˜ m ) ∣ n + 2 α d z + ε m PV ∫ R n η m ( z , t ˜ m ) − η m ( y ˜ m , t ˜ m ) ∣ G m − 1 ( z − y ˜ m ) ∣ n + 2 α d z − σ m = PV ∫ H λ k ( z , t ˜ m ) − k ( y ˜ m , t ˜ m ) ∣ G m − 1 ( z − y ˜ m ) ∣ n + 2 α d z + PV ∫ H λ − k ( z , t ˜ m ) − k ( y ˜ m , t ˜ m ) ∣ G m − 1 ( z λ − y ˜ m ) ∣ n + 2 α d z + ε m C PV ∫ R n η m ( z , t ˜ m ) − η m ( y ˜ m , t ˜ m ) ∣ z − y ˜ m ∣ n + 2 α d z − σ m = PV ∫ H λ k ( z , t ˜ m ) − k ( y ˜ m , t ˜ m ) ∣ G m − 1 ( z − y ˜ m ) ∣ n + 2 α d z + PV ∫ H λ − k ( z , t ˜ m ) − k ( y ˜ m , t ˜ m ) ∣ G m − 1 ( z λ − y ˜ m ) ∣ n + 2 α d z − ε m C ( − Δ ) α η m ( y ˜ m , t ˜ m ) − σ m ≥ − 2 k ( y ˜ m , t ˜ m ) ∫ H λ 1 ∣ G m − 1 ( z λ − y ˜ m ) ∣ n + 2 α d z − C ε m r m 2 α − σ m ≥ − 2 τ 2 k ( y ˜ m , t ˜ m ) ∫ H λ 1 ∣ z λ − y ˜ m ∣ n + 2 α d z − C ε m r m 2 α − σ m ≥ − 2 τ 2 k ( y ˜ m , t ˜ m ) ∫ V 1 ∣ z λ − y ˜ m ∣ n + 2 α d z − C ε m r m 2 α − σ m ≥ − C 1 k ( y ˜ m , t ˜ m ) r m 2 α − C ε m r m 2 α − σ m ,

where C 1 , C > 0 , V = { z ∈ H λ ∣ 0 < y ˜ 1 m − z 1 < r m < l , 0 < ∣ ( y ˜ m ) ′ − z ′ ∣ ≤ 1 } ,

(2.14) ∣ G m − 1 ( z − y ˜ m ) ∣ 2 = ∑ i = 2 n z i − y i λ i + z 1 − y 1 λ 1 ≤ 1 τ 2 ∣ z − y ˜ m ∣ 2 and ∣ G m − 1 ( z λ − y ˜ m ) ∣ 2 ≤ 1 τ 2 ∣ z λ − y ˜ m ∣ 2

and

(2.15) ∣ ( − Δ ) α η m ( y ˜ m , t ˜ m ) ∣ = PV ∫ R n η m ( z , t ˜ m ) − η m ( y ˜ m , t ˜ m ) ∣ z − y ˜ m ∣ n + 2 α d z ≤ C r m 2 α .

Due to (2.11) and c ( y , t ) is bounded from above, hence we have

(2.16) − C 2 p ≤ c ( y ˜ m , t ˜ m ) ω ( y ˜ m , t ˜ m ) = ∂ t β ω ( y ˜ m , t ˜ m ) − ( D α τ ( μ λ ( y ˜ m , t ˜ m ) ) − D α τ ( μ ( y ˜ m , t ˜ m ) ) ) ≤ C 1 k ( y ˜ m , t ˜ m ) r m 2 α + C ε m r m 2 α + σ m ≤ − C 1 p r m 2 α + C ε m r m 2 α + σ m .

From this, we have

(2.17) C 1 p ≤ C 2 p r m 2 α + C ε m + σ m r m 2 α ≤ C 2 p l + C ε m + σ m l → C 2 p l as m → ∞ .

For sufficiently small l , we can derive a contradiction, which indicates that (2.2) is valid.

Furthermore, suppose there exists a point ( y 0 , t 0 ) such that

(2.18) ω ( y 0 , t 0 ) = min H λ × R ω ( x , t ) = 0 .

Thus, we have

(2.19) ∂ t β ω ( y 0 , t 0 ) = C β ∫ − ∞ t 0 0 − ω ( y 0 , ν ) ( t 0 − ν ) 1 + β d ν ≤ 0

and

(2.20) − ( D α τ ( μ λ ( y 0 , t 0 ) ) − D α τ ( μ ( y 0 , t 0 ) ) ) = PV ∫ H λ ω ( z , t 0 ) 1 ∣ G m − 1 ( z λ − y 0 ) ∣ n + 2 s − 1 ∣ G m − 1 ( z − y 0 ) ∣ n + 2 s d z + σ m ≤ σ m → 0 as m → ∞ ,

where σ m is like in (2.13).

Therefore, we obtain

0 = c ( y 0 , t 0 ) ω ( y 0 , t 0 ) = ∂ t β ω ( y 0 , t 0 ) − ( D α τ ( μ λ ( y 0 , t 0 ) ) − D α τ ( μ ( y 0 , t 0 ) ) ) ≤ 0 ,

i.e.,

(2.21) ω ( z , t 0 ) ≡ 0 in H λ .

On the basis of (2.21), since ω ( z λ , t ) = − ω ( z , t ) , we can deduce that

(2.22) 0 = ∂ t β ω ( z , t 0 ) = C β ∫ − ∞ t 0 0 − ω ( z , ν ) ( t 0 − ν ) 1 + β d ν ≤ 0 ,

that is,

(2.23)□ ω ( z , t ) ≡ 0 in R n × ( − ∞ , t 0 ] .

Theorem 2.2

Assume that

ω ( y , t ) ∈ ( C loc 1,1 ( H λ ) ∩ L 2 α ( R n ) ) × ( C 1 ( R ) ∩ L β − ( R ) )

is bounded from below in H λ × R and satisfies

∂ t β ω ( y , t ) − ( D α τ ( μ λ ( y , t ) ) − D α τ ( μ ( y , t ) ) ) ≥ 0 in H λ × R , ω ( y , t ) = − ω ( y λ , t ) in H λ × R .

Then

(2.24) ω ( y , t ) ≥ 0 in H λ × R .

Proof

Similar to the proof of Theorem 2.1, we derive that (2.24) holds.□

Remark 2.3

Assume that

ω ( y , t ) ∈ ( C loc 1,1 ( H λ ) ∩ L 2 α ( R n ) ) × ( C 1 ( R ) ∩ L β − ( R ) )

is bounded from above in H λ × R and satisfies

∂ t β ω ( y , t ) − ( D α τ ( μ λ ( y , t ) ) − D α τ ( μ ( y , t ) ) ) ≤ 0 in H λ × R , ω ( y , t ) = − ω ( y λ , t ) in H λ × R .

Then

ω ( y , t ) ≤ 0 in H λ × R .

Proof

The proof method is the same as shown in Theorem 2.2.□

Theorem 2.4

(Maximum principle near infinity) Let Ω be an unbounded domain in H λ . Assume that

ω ( y , t ) ∈ ( C loc 1,1 ( Ω ) ∩ L 2 α ( R n ) ) × ( C 1 ( R ) ∩ L β − ( R ) )

is lower semi-continuous with respect to y on Ω ¯ , satisfies

(2.25) ∂ t β ω ( y , t ) − ( D α τ ( μ λ ( y , t ) ) − D α τ ( μ ( y , t ) ) ) = c ( y , t ) ω ( y , t ) , ( y , t ) ∈ Ω × R , ω ( y , t ) ≥ 0 , ( y , t ) ∈ ( H λ \ Ω ) × R , ω ( y λ , t ) = − ω ( y , t ) , ( y , t ) ∈ H λ × R , lim ∣ y ∣ → ∞ ω ( y , t ) = 0 , uniformly f o r t ∈ R .

Suppose that there is σ > 0 such that c ( y , t ) ≤ − σ at the points where ω ( y , t ) < 0 in Ω × R , then we have

(2.26) ω ( y , t ) ≥ 0 in H λ × R .

Furthermore, if there exists a point ( y 0 , t 0 ) ∈ Ω × R such that ω ( y 0 , t 0 ) = 0 , then ω ( y , t ) ≡ 0 in R n × ( − ∞ , t 0 ] .

Proof

The proof process is roughly the same as that of Theorem 2.1, and we will point out the differences below.

Since ω is bounded and lim ∣ y ∣ → ∞ ω ( y , t ) = 0 , we know that ω ( y ( t ) , t ) has an infimum for each fixed t ∈ R . If (2.26) does not hold, then there is y ( t ) ∈ Ω and p > 0 satisfying

(2.27) inf ( y , t ) ∈ Ω × R ω ( y , t ) = inf t ∈ R ω ( y ( t ) , t ) = − p < 0 .

Therefore, there exists a sequence { t m } ⊂ R and a sequence { ε m } ↘ 0 such that

(2.28) ω ( y ( t m ) , t m ) = − p + ε m < 0 .

We further consider the following auxiliary functions

(2.29) k m ( y , t ) = ω ( y , t ) − ε m η m ( t ) ,

to remedy scenario that the infimum of ω ( y , t ) may not be attained due to t ∈ R , where η m ( t ) = η ( t − t m ) ∈ C 0 ∞ ( − 1 + t m , 1 + t m ) satisfying

(2.30) η m ( t ) = 1 in ∣ t − t m ∣ ≤ 1 2 and η m ( t ) = 0 in ∣ t − t m ∣ ≥ 1 .

By (2.27), (2.28), and (2.29), we have

(2.31) k m ( y ( t m ) , t m ) = − p , k m ( y , t ) = ω ( y , t ) ≥ − p , in Ω × ( R \ ( − 1 + t m , 1 + t m ) ) .

It follows from (2.31) that there exists a point ( y ¯ m , t ¯ m ) ∈ Ω × ( − 1 + t m , 1 + t m ) such that

(2.32) − p − ε m ≤ k m ( y ¯ m , t ¯ m ) = inf ( y , t ) ∈ H λ × R k m ( y , t ) ≤ − p .

According to (2.29), (2.32), and Lemma A.2, we obtain

(2.33) − p ≤ ω ( y ¯ m , t ¯ m ) = inf y ∈ H λ ω ( y , t ¯ m ) ≤ − p + ε m

and

(2.34) ∂ t β ω ( y ¯ m , t ¯ m ) ≤ ε m η m ( y ¯ m , t ¯ m ) ≤ C ε m .

Similar to the proof process of (2.13), we have

(2.35) D α τ ( μ λ ( y ¯ m , t ¯ m ) ) − D α τ ( μ ( y ¯ m , t ¯ m ) ) ≥ 0 .

Since c ( y ¯ m , t ¯ m ) ≤ − σ , from (2.34) and (2.35), we derive that

(2.36) 0 < σ ( p − ε m ) ≤ c ( y ¯ m , t ¯ m ) ω ( y ¯ m , t ¯ m ) = ∂ t β ω ( y ¯ m , t ¯ m ) − ( D α τ ( μ λ ( y ¯ m , t ¯ m ) ) − D α τ ( μ ( y ¯ m , t ¯ m ) ) ) ≤ C ε m ,

that is, 0 < σ p ≤ ( σ + C ) ε m → 0 as m → ∞ . This contradiction means that (2.26) is valid.□

Theorem 2.5

Let Ω ⊂ { y ∈ H λ ∣ λ − 2 l < y 1 < λ } and Ω be an unbounded narrow region. Suppose that ω ( y , t ) ∈ ( C loc 1,1 ( Ω ) ∩ L 2 α ( R n ) ) × ( C 1 ( R ) ∩ L β − ( R ) ) is lower semi-continuous with respect to y on Ω ¯ , satisfying

(2.37) ω ( y , t ) ≥ − C ( 1 + ∣ y ∣ γ ) for s o m e 0 < γ < 2 α ,

and

(2.38) ∂ t β ω ( y , t ) − ( D α τ ( μ λ ( y , t ) ) − D α τ ( μ ( y , t ) ) ) = c ( y , t ) ω ( y , t ) , ( y , t ) ∈ Ω × R , ω ( y , t ) ≥ 0 , ( y , t ) ∈ ( H λ \ Ω ) × R , ω ( y λ , t ) = − ω ( y , t ) , ( y , t ) ∈ H λ × R .

If c ( y , t ) is bounded from above, then for sufficiently small l, we have

(2.39) ω ( y , t ) ≥ 0 , ( y , t ) ∈ H λ × R .

Proof

Let

(2.40) ω ¯ ( y , t ) = ω ( y , t ) h ( y ) ,

where

h ( y ) ≔ 1 − ( y 1 − ( λ − l ) ) 2 l 2 + α + 1 ( 1 + ∣ y ′ ∣ 2 ) θ 2 with γ < θ < 2 α .

Since

(2.41) lim ∣ y ∣ → ∞ ω ¯ ( y , t ) = 0 , uniformly for t ∈ R .

The proof process is roughly the same as that of Theorem 2.4, and we will point out the differences below. If (2.39) is incorrect, then there is y ( t ) ∈ Ω and p > 0 satisfying

(2.42) inf ( y , t ) ∈ Ω × R ω ¯ ( y , t ) = inf t ∈ R ω ¯ ( y ( t ) , t ) = − p < 0 .

Therefore, there exists a sequence { t m } ⊂ R and a sequence { ε m } ↘ 0 such that

(2.43) ω ¯ ( y ( t m ) , t m ) = − p + ε m < 0 .

We consider the following auxiliary functions:

(2.44) k m ( y , t ) = ω ¯ ( y , t ) − ε m η m ( t ) ,

where η m is defined in (2.30).

Similar to Theorem 2.2, there exists a point ( y ¯ m , t ¯ m ) ∈ Ω × ( − 1 + t m , 1 + t m ) such that

(2.45) − p ≤ ω ¯ ( y ¯ m , t ¯ m ) = inf y ∈ H λ ω ¯ ( y , t ¯ m ) ≤ − p + ε m and ∂ t β ω ¯ ( y ¯ m , t ¯ m ) ≤ C ε m h ( y ¯ m ) .

By a result Lemma A.4, we have

(2.46) ( − Δ ) α h ( y ¯ m ) h ( y ¯ m ) ≥ C l 2 α .

Since det G = 1 , λ min G ≥ τ > 0 , obviously, we have λ min G ≤ τ 1 − n , for G ∈ G . Then, for any σ m → 0 as m → ∞ , there is G m ∈ G such that

(2.47) D α τ ( μ λ ( y ¯ m , t ¯ m ) ) − D α τ ( μ ( y ¯ m , t ¯ m ) ) = inf PV ∫ R n μ λ ( z , t ¯ m ) − μ λ ( y ¯ m , t ¯ m ) ∣ G − 1 ( z − y ¯ m ) ∣ n + 2 α d z − inf PV ∫ R n μ ( z , t ¯ m ) − μ ( y ¯ m , t ¯ m ) ∣ G − 1 ( z − y ¯ m ) ∣ n + 2 α d z ≥ PV ∫ R n μ λ ( z , t ¯ m ) − μ λ ( y ¯ m , t ¯ m ) ∣ G m − 1 ( z − y ¯ m ) ∣ n + 2 α d z − σ m − PV ∫ R n μ ( z , t ¯ m ) − μ ( y ¯ m , t ¯ m ) ∣ G m − 1 ( z − y ¯ m ) ∣ n + 2 α d z = PV ∫ R n ω ( z , t ¯ m ) − ω ( y ¯ m , t ¯ m ) ∣ G m − 1 ( z − y ¯ m ) ∣ n + 2 α d z − σ m = PV ∫ R n h ( z ) ω ¯ ( z , t ¯ m ) − h ( y ¯ m ) ω ¯ ( y ¯ m , t ¯ m ) ∣ G m − 1 ( z − y ¯ m ) ∣ n + 2 α d z − σ m = PV ∫ R n h ( z ) ( ω ¯ ( z , t ¯ m ) − ω ¯ ( y ¯ m , t ¯ m ) ) ∣ G m − 1 ( z − y ¯ m ) ∣ n + 2 α d z − C ω ¯ ( y ¯ m , t ¯ m ) ( − Δ ) α h ( y ¯ m ) − σ m ≥ ∫ H λ h ( z ) ( ω ¯ ( z , t ¯ m ) − ω ¯ ( y ¯ m , t ¯ m ) ) ∣ G m − 1 ( z λ − y ¯ m ) ∣ n + 2 α d z + ∫ H λ h ( z λ ) ( − ω ¯ ( z , t ¯ m ) − ω ¯ ( y ¯ m , t ¯ m ) ) ∣ G m − 1 ( z λ − y ¯ m ) ∣ n + 2 α d z − C ω ¯ ( y ¯ m , t ¯ m ) ( − Δ ) α h ( y ¯ m ) − σ m ≥ − ω ¯ ( y ¯ m , t ¯ m ) ∫ H λ h ( z λ ) ∣ G m − 1 ( z λ − y ¯ m ) ∣ n + 2 α d z − C ω ¯ ( y ¯ m , t ¯ m ) ( − Δ ) α h ( y ¯ m ) − σ m ≥ − C ω ¯ ( y ¯ m , t ¯ m ) ( − Δ ) α h ( y ¯ m ) − σ m ≥ − C l 2 α ω ¯ ( y ¯ m , t ¯ m ) h ( y ¯ m ) − σ m ,

where C > 0 , h ( z ) > h ( z λ ) , and ∣ G m − 1 ( z − y ¯ m ) ∣ < ∣ G m − 1 ( z λ − y ¯ m ) ∣ .

Due to c ( x , t ) is bounded from above, hence we can deduce that

(2.48) − C 1 p h ( y ¯ m ) ≤ c ( y ¯ m , t ¯ m ) ω ( y ¯ m , t ¯ m ) = ∂ t β ω ( y ¯ m , t ¯ m ) − ( D α τ ( μ λ ( y ¯ m , t ¯ m ) ) − D α τ ( μ ( y ¯ m , t ¯ m ) ) ) ≤ C ε m h ( y ¯ m ) + C l α ω ¯ ( y ¯ m , t ¯ m ) h ( y ¯ m ) + σ m ≤ C ε m h ( y ¯ m ) + C − p + ε m l 2 α h ( y ¯ m ) + σ m ,

that is, 0 < C p ≤ C ε m l 2 α + ε m + l 2 α σ m h ( y ¯ m ) + C 1 p l 2 α → C 1 p l 2 α as m → ∞ . Therefore, we derive a contradiction for sufficiently small l , which concludes that (2.39) is correct.□

Theorem 2.6

Set Ω ≔ { y ∈ H λ ∣ b < y 1 < λ , b is a f i n i t e c o n s t a n t } be an unbounded domain. Assume that

ω ( y , t ) ∈ ( C loc 1,1 ( Ω ) ∩ L 2 α ( R n ) ) × ( C 1 ( R ) ∩ L β − ( R ) )

is lower semi-continuous with respect to y on Ω ¯ , satisfying

∂ t β ω ( y , t ) − ( D α τ ( μ λ ( y , t ) ) − D α τ ( μ ( y , t ) ) ) = c ( y , t ) ω ( y , t ) , ( y , t ) ∈ Ω × R , ω ( y , t ) ≥ 0 , ( y , t ) ∈ ( H λ \ Ω ) × R , ω ( y λ , t ) = − ω ( y , t ) , ( y , t ) ∈ H λ × R

and

ω ( y , t ) ≥ − C ( 1 + ∣ y ∣ γ ) , for s o m e 0 < γ < 2 α ,

where c ( y , t ) ≤ c 0 and c 0 is a sufficiently small positive constant, then we have

(2.49) ω ( y , t ) ≥ 0 in H λ × R .

Proof

Similar to the proof of Theorem 2.5, we derive that (2.49) holds.□

Theorem 2.7

(Averaging effects) Let D ⊂ R n and t 0 ∈ R . For any y 0 ∈ R n , if there is a radius r > 0 such that B r ( y 0 ) ∩ D ¯ = ∅ and

(2.50) μ ( y , t ) ≥ C 0 > 0 in D × ( t 0 − r 2 α β , t 0 + r 2 α β ] .

Assume that

μ ( y , t ) ∈ ( C loc 1,1 ( B r ( y 0 ) ) ∩ L 2 α ( R n ) ) × C 1 ( [ t 0 − r 2 s α , t 0 + r 2 s α ] ) ∩ L β − ( R )

is lower semi-continuous in x on B r ( y 0 ) ¯ , satisfying

(2.51) ∂ t β μ ( y , t ) − D α τ ( μ ( y , t ) ) ≥ − ε ( y , t ) ∈ B r ( y 0 ) × ( t 0 − r 2 α β , t 0 + r 2 α β ] , μ ( y , t ) ≥ 0 , ( y , t ) ∈ B r c ( y 0 ) × ( t 0 − r 2 α β , t 0 + r 2 α β ] , μ ( y , t ) ≥ 0 , ( y , t ) ∈ B r ( y 0 ) × ( − ∞ , t 0 − r 2 α β ] ,

for some sufficiently small ε > 0 . Then there exists a positive constants C 1 such that

(2.52) μ ( y 0 , t 0 ) ≥ C 1 > 0 .

Proof

Let

(2.53) g ( y , t ) = ϕ ( y ) η ( t ) = 1 − y − y 0 r 2 + α η ( t ) ,

where η ( t ) ∈ C 0 ∞ ( t 0 − r 2 α β , t 0 + r 2 α β ) and satisfying

η ( t ) ≡ 1 , in t 0 − r 2 α β 2 , t 0 + r 2 α β 2 , 0 ≤ η ( t ) ≤ 1 , in R .

We consider the following auxiliary functions:

(2.54) μ ¯ ( y , t ) ≔ μ ( y , t ) χ D ( y ) + δ g ( y , t ) ,

where δ is a positive constant to be determined later, and

χ D ( y ) = 1 , y ∈ D , 0 , y ∉ D ,

Fix σ > 0 , arbitrary, and let G σ ∈ G such that

(2.55) D α τ ( μ ¯ ( y , t ) ) = D α τ ( μ ( y , t ) χ D ( y ) + δ g ( y , t ) ) ≥ PV ∫ R n μ ( z , t ) χ D ( z ) + δ g ( z , t ) − μ ( y , t ) χ D ( y ) − δ g ( y , t ) ∣ G σ − 1 ( z − y ) ∣ n + 2 α d z − σ .

For ( y , t ) ∈ B r ( y 0 ) × ( t 0 − r 2 α β , t 0 + r 2 α β ] , by combining (2.55) and B r ( y 0 ) ∩ D ¯ = ∅ , we have

(2.56) ∂ t β ( μ ( y , t ) ) − ∂ t β μ ¯ ( y , t ) − D α τ ( μ ( y , t ) ) + D α τ ( μ ¯ ( y , t ) ) ≥ − ε − δ ϕ ( y ) ∂ t β η ( t ) + PV ∫ R n μ ( z , t ) χ D ( z ) + δ g ( z , t ) − δ g ( y , t ) ∣ G σ − 1 ( z − y ) ∣ n + 2 α d z − σ ≥ − ε − δ C 2 + ∫ D μ ( z , t ) ∣ G σ − 1 ( z − y ) ∣ n + 2 α d z − C δ r 2 α − σ > − ε + C 3 − C 4 δ r 2 α − σ ,

where C , C 2 , C 3 , C 4 > 0 , and

(2.57) PV ∫ R n g ( z , t ) − g ( y , t ) ∣ G σ − 1 ( z − y ) ∣ n + α d z = − C δ r 2 α ,

and we use the fact established in Lemma A.2 that there exists C 2 > 0 such that

(2.58) ∣ ∂ t β η ( t ) ∣ ≤ C 2 for t ∈ ( − 2 , 2 ) .

By taking ε = C 3 2 − σ and δ = C 3 r 2 α 2 C 4 , by (2.56), we can calculate that

(2.59) ∂ t β ( μ ( y , t ) ) − ∂ t β μ ¯ ( y , t ) − D α τ ( μ ( y , t ) ) + D α τ ( μ ¯ ( y , t ) ) > 0 in B r ( y 0 ) × ( t 0 − r 2 α β , t 0 + r 2 α β ] .

In addition, from (2.51), we have

(2.60) μ ( y , t ) − μ ¯ ( y , t ) = μ ( y , t ) − μ ( y , t ) χ D ( y ) ≥ 0 in B r c ( y 0 ) × ( t 0 − r 2 α β , t 0 + r 2 α β ]

and

(2.61) μ ( y , t ) − μ ¯ ( y , t ) = μ ( y , t ) ≥ 0 in B r ( y 0 ) × ( − ∞ , t 0 − r 2 α β ] .

We claim that

(2.62) μ ( y , t ) − μ ¯ ( y , t ) ≥ 0 , ( x , t ) ∈ B r ( y 0 ) × ( t 0 − r 2 α β , t 0 + r 2 α β ] .

Otherwise, if not, there is a point ( y ¯ 0 , t ¯ 0 ) ∈ B r ( y 0 ) × ( t 0 − r 2 α β , t 0 + r 2 α β ] such that

(2.63) μ ( y ¯ 0 , t ¯ 0 ) − μ ¯ ( y ¯ 0 , t ¯ 0 ) = min B r ( y 0 ) × ( t 0 − r 2 α β , t 0 + r 2 α β ] μ ( y , t ) − μ ¯ ( y , t ) < 0 .

Hence, for any sequences σ m → 0 , there exists G m ∈ G such that

(2.64) − D α τ μ ( y ¯ 0 , t ¯ 0 ) + D α τ μ ¯ ( y ¯ 0 , t ¯ 0 ) = − inf G ∈ G PV ∫ R n μ ( z , t ¯ 0 ) − μ ( y ¯ 0 , t ¯ 0 ) ∣ G − 1 ( z − y ) ∣ n + 2 α d z + inf G ∈ G PV ∫ R n μ ¯ ( z , t ¯ 0 ) − μ ¯ ( y ¯ 0 , t ¯ 0 ) ∣ G − 1 ( z − y ) ∣ n + 2 α d z ≤ PV ∫ R n ( μ ( y ¯ 0 , t ¯ 0 ) − μ ¯ ( y ¯ 0 , t ¯ 0 ) ) − ( μ ( z , t ¯ 0 ) − μ ¯ ( z , t ¯ 0 ) ) ∣ G m − 1 ( z − y ) ∣ n + 2 α d z + σ m ≤ ∫ H λ ( μ ( y ¯ 0 , t ¯ 0 ) − μ ¯ ( y ¯ 0 , t ¯ 0 ) ) − ( μ ( z λ , t ¯ 0 ) − μ ¯ ( z λ , t ¯ 0 ) ) ∣ G m − 1 ( z λ − y ) ∣ n + 2 α d z + σ m ≤ σ m → 0 as m → ∞

and

(2.65) ∂ t β ( μ ( y ¯ 0 , t ¯ 0 ) − μ ¯ ( y ¯ 0 , t ¯ 0 ) ) ≤ 0 .

Therefore,

(2.66) ∂ t β ( μ ( y ¯ 0 , t ¯ 0 ) − μ ¯ ( y ¯ 0 , t ¯ 0 ) ) − D α τ μ ( y ¯ 0 , t ¯ 0 ) + D α τ μ ¯ ( y ¯ 0 , t ¯ 0 ) ≤ 0 ,

which contradicts (2.59). So, we have

(2.67)□ μ ( y 0 , t 0 ) ≥ μ ¯ ( y 0 , t 0 ) ≔ δ η ( t 0 ) ≔ C 1 .

Theorem 2.8

Let D ⊂ H λ and t 0 ∈ R . For any y 0 ∈ H λ , assume there exists a ball B r ( y 0 ) ⊂ H λ such that B r ( y 0 ) ∩ D ¯ = ∅ and r ≤ dist ( y 0 , T λ ) 2 , and

(2.68) ω ( y , t ) ≥ C 0 > 0 in D × ( t 0 − r 2 α β , t 0 + r 2 α β ] .

Suppose that ω ( y , t ) ∈ ( C loc 1,1 ( B r ( y 0 ) ) ∩ L 2 α ( R n ) ) × C 1 ( [ t 0 − r 2 α β , t 0 + r 2 α β ] ) ∩ L β − ( R ) is lower semi-continuous in y on B r ( y 0 ) ¯ , satisfying

(2.69) ∂ t β ω ( y , t ) − D α τ ( μ λ ( y , t ) ) + D α τ ( μ ( y , t ) ) ≥ − ε ( y , t ) ∈ B r ( y 0 ) × t 0 − r 2 α β , t 0 + r 2 α β , ω ( y , t ) ≥ 0 , ( y , t ) ∈ B r c ( y 0 ) × t 0 − r 2 α β , t 0 + r 2 α β , ω ( y , t ) ≥ 0 , ( y , t ) ∈ B r ( y 0 ) × − ∞ , t 0 − r 2 α β , ω ( y , t ) = − ω ( y λ , t ) , ( y , t ) ∈ H λ × R ,

where ε is a sufficiently small positive constant. Then there exists a constants C 1 such that

(2.70) ω ( y 0 , t 0 ) ≥ C 1 > 0 .

Proof

We first prove that

(2.71) ω ( y , t ) ≥ 0 in B r ( y 0 ) × t 0 − r 2 α β , t 0 + r 2 α β .

If (2.71) does not hold, then there exists a point ( y ¯ 0 , t ¯ 0 ) ∈ B r ( y 0 ) × t 0 − r 2 α β , t 0 + r 2 α β such that

(2.72) ω ( y ¯ 0 , t ¯ 0 ) = min ( y , t ) ∈ B r ( y 0 ) × ( t 0 − r 2 α β , t 0 + r 2 α β ] ω ( y , t ) < 0 .

Therefore,

(2.73) ∂ t β ω ( y ¯ 0 , t ¯ 0 ) − D α τ ( μ λ ( y ¯ 0 , t ¯ 0 ) ) + D α τ ( μ ( y ¯ 0 , t ¯ 0 ) ) < ω ( y ¯ 0 , t ¯ 0 ) ∫ H λ 1 ∣ G − 1 ( z − y ) ∣ d z ≔ − C ,

which contradicts (2.69) for sufficiently small ε , so (2.71) holds.

Based on (2.71), if there is a point ( ξ 1 , ξ 2 ) ∈ B r ( y 0 ) × ( t 0 − r 2 α β , t 0 + r 2 α β ] satisfying

(2.74) ω ( ξ 1 , ξ 2 ) = min ( y , t ) ∈ B r ( y 0 ) × ( t 0 − r 2 α β , t 0 + r 2 α β ] ω ( y , t ) = 0 .

Similar to the proof of the strong maximum principle in Theorem 2.1, we obtain that

(2.75) ω ( y , ξ 2 ) ≡ 0 in R n × ( − ∞ , ξ 2 ] ,

which contradicts (2.68). Hence, we have

(2.76) ω ( y , t ) > 0 in B r ( y 0 ) × ( t 0 − r 2 α β , t 0 + r 2 α β ] ,

which means that there exists a positive constant C 1 such that

(2.77)□ ω ( y 0 , t 0 ) ≥ C 1 .

3 Proof of Theorem 1.1

From (1.1), we have

(3.1) ∂ t β ω ( y , t ) − D α τ ( μ λ ( y , t ) ) + D α τ ( μ ( y , t ) ) = c λ ( y , t ) ω ( y , t ) ( y , t ) ∈ Ω λ × R , ω ( y , t ) ≥ 0 , ( y , t ) ∈ ( H λ \ Ω λ ) × R , ω λ ( y , t ) = − ω ( y , t ) , ( y , t ) ∈ H λ × R ,

where Ω λ ≔ H λ ∩ B 1 ( 0 ) and

(3.2) c λ ( y , t ) = f ( μ λ ( y , t ) ) − f ( μ ( y , t ) ) μ λ ( y , t ) − μ ( y , t )

is bounded in Ω λ × R due to f ∈ C 1 .

Step 1. We want to show that

(3.3) ω λ ( y , t ) ≥ 0 , ( y , t ) ∈ H λ × R ,

when λ is sufficiently closed to − 1 . Thus, according to Theorem 2.1, we derive (3.3).

Step 2. Denote

(3.4) λ 0 = sup { λ < 0 ∣ ω ρ ( y , t ) ≥ 0 , ( y , t ) ∈ H λ × R , ρ ≤ λ } .

Our main purpose is to prove that λ 0 = 0 . If λ 0 < 0 , then there exists a sequence { λ m } ↘ λ 0 and a sequence { p m } ↘ 0 such that

(3.5) inf ( y , t ) ∈ H λ m × R ω λ m ( y , t ) = inf ( y , t ) ∈ Ω λ m × R ω λ m ( y , t ) = − p m < 0 ,

which implies that there is a sequence { ( y m , t m ) } ⊂ Ω λ m × R satisfying

(3.6) − p m ≤ ω λ m ( y m , t m ) ≤ − p m + p m 2 .

To solve the situation where the infimum value of ω λ m about t may not be achievable, we need to introduce the following auxiliary functions

(3.7) V m ( y , t ) = ω λ m ( y , t ) − p m 2 η m ( t ) ,

where η m ( t ) is defined in (2.30). By (3.5), (3.6), and (3.7), we have

(3.8) V m ( y m , t m ) ≤ − p m , V m ( y , t ) = ω λ m ( y , t ) ≥ − p m , in Ω λ m × ( R \ ( − 1 + t m , 1 + t m ) ) .

Therefore, there exists a point ( y ¯ m , t ¯ m ) ∈ Ω λ m × ( − 1 + t m , 1 + t m ) satisfying

(3.9) − p m − p m 2 ≤ V m ( y ¯ m , t ¯ m ) = inf ( y , t ) ∈ H λ m × R V m ( y , t ) ≤ − p m ,

which implies that

(3.10) − p m ≤ ω λ m ( y ¯ m , t ¯ m ) = inf y ∈ H λ m ω λ m ( y , t ¯ m ) ≤ − p m + p m 2 .

Similar to the proof steps of Theorem 2.4, we derive that

(3.11) ∂ t β ω λ m ( y ¯ m , t ¯ m ) − D α τ ( μ λ m ( y ¯ m , t ¯ m ) ) + D α τ ( μ ( y ¯ m , t ¯ m ) ) ≤ C p m 2 + C ω λ m ( y ¯ m , t ¯ m ) ∫ H λ m 1 ∣ G m − 1 ( z λ m − y ) ∣ n + 2 α d z ≤ C p m 2 + C 1 ( − p m + p m 2 ) ,

where C , C 1 > 0 . It follows from (3.1) and (3.10) that

(3.12) − c λ m ( y ¯ m , t ¯ m ) p m ≤ c λ m ( y ¯ m , t ¯ m ) ω λ m ( y ¯ m , t ¯ m ) = ∂ t β ω λ m ( y ¯ m , t ¯ m ) − D α τ ( μ λ m ( y ¯ m , t ¯ m ) ) + D α τ ( μ ( y ¯ m , t ¯ m ) ) ≤ C p m 2 + C 1 ( − p m + p m 2 ) ,

that is,

(3.13) c λ m ( y ¯ m , t ¯ m ) ≥ C 1 − ( C + C 1 ) p m → C 1 as m → ∞ .

According to (3.2), (3.13), and ω λ m ( y ¯ m , t ¯ m ) < 0 , then there is some ξ λ m ∈ ( μ λ m ( y ¯ m , t ¯ m ) , μ ( y ¯ m , t ¯ m ) ) such that

(3.14) f ′ ( ξ λ m ) ≥ C 1 > 0 ,

which means that there exist a subsequence of ( y ¯ m , t ¯ m ) ( still denoted by { ( y ¯ m , t ¯ m ) } ) such that

(3.15) μ ( y ¯ m , t ¯ m ) ≥ C 0 > 0 ,

for sufficiently large m . Otherwise, if μ ( y ¯ m , t ¯ m ) → 0 as m → ∞ , then for sufficiently large m , we have ξ λ m → 0 , which implies that

(3.16) f ′ ( 0 ) ≥ C 1 > 0 .

This contradicts the assumption f ′ ( 0 ) ≤ 0 , so (3.15) holds. Note that

D α τ ( μ λ m ( y , t ) ) − D α τ ( μ ( y , t ) ) = inf G ∈ G PV ∫ R n μ λ m ( z , t ) − μ λ m ( y , t ) ∣ G − 1 ( z − y ) ∣ n + 2 α d z − inf G ∈ G PV ∫ R n μ ( z , t ) − μ ( y , t ) ∣ G − 1 ( z − y ) ∣ n + 2 α d z ≥ inf G ∈ G PV ∫ R n ω λ m ( z , t ) − ω λ m ( y , t ) ∣ G − 1 ( z − y ) ∣ n + 2 α d z = D α τ ω λ m ( y , t ) .

Next, we consider the following functions:

(3.17) ω ¯ m ( y , t ) = ω λ m ( y , t + t ¯ m ) and c ¯ m ( y , t ) = c λ m ( y , t + t ¯ m ) .

From this, we have

(3.18) ∂ t β ω ¯ m ( y , t ) − D α τ ω ¯ m ( y , t ) ≥ c ¯ m ( y , t ) ω ¯ m ( y , t ) in Ω λ m × R .

It follows from Arzelà-Ascoli theorem that there exist two continuous function ω ¯ ( y , t ) and c ¯ ( y , t ) such that

(3.19) lim m → ∞ ω ¯ m ( y , t ) = ω ¯ ( y , t ) and lim m → ∞ c ¯ m ( y , t ) = c ¯ ( y , t ) uniformly in B 1 ( 0 ) × R .

From this, we can conclude that

(3.20) ∂ t β ω ¯ ( y , t ) − D α τ ω ¯ m ( y , t ) ≥ c ¯ ( y , t ) ω ¯ ( y , t ) in Ω λ 0 × R ,

by λ m → λ 0 as m → ∞ . Since Ω λ m is a bounded domain with Ω λ m ⊂ B 1 ( 0 ) , we may assume that y ¯ m → y 0 ∈ H λ 0 ∩ B 1 ( 0 ) . Then using (3.10) and the continuity on μ , we derive that

(3.21) ω ¯ ( y 0 , 0 ) = inf H λ 0 × R ω λ 0 ( y , t ) = inf H λ 0 × R ω ¯ ( y , t ) = 0 .

Thus for sequences σ m → 0 , there exists G m ∈ G such that

(3.22) 0 ≤ ∂ t β ω ¯ ( y 0 , 0 ) − D α τ ω ¯ ( y 0 , 0 ) ≤ C β ∫ − ∞ 0 − ω ¯ ( y 0 , ν ) ( − ν ) 1 + β d ν + PV ∫ H λ 0 ω ¯ ( z , 0 ) 1 ∣ G m − 1 ( z λ 0 − y 0 ) ∣ n + 2 α − 1 ∣ G m − 1 ( z − y 0 ) ∣ n + 2 α d z + σ m ≤ σ m → 0 as m → ∞ .

Therefore, by ω ¯ ( y , t ) ≥ 0 in H λ 0 × R and ω ¯ ( y λ , t ) = − ω ¯ ( y , t ) , we obtain

(3.23) ω ¯ ( y , t ) ≡ 0 in R n × ( − ∞ , 0 ] .

Furthermore, we define

(3.24) μ m ( y , t ) = μ ( y , t + t m ) .

Similarly to (3.19) and (3.20), we also have

(3.25) lim m → ∞ μ m ( y , t ) = μ ¯ ( y , t )

and

(3.26) ∂ t β μ ¯ ( y , t ) − D α τ μ ¯ ( y , t ) = f ( μ ¯ ( y , t ) ) in B 1 ( 0 ) × R .

By (3.15) and y ¯ m → y 0 ∈ H λ 0 ∩ B 1 ( 0 ) , we derive that

(3.27) μ ¯ ( y 0 , 0 ) = lim m → ∞ μ ( y ¯ m , t m ) ≥ C 0 > 0 .

Now we want to prove that

(3.28) μ ¯ ( y , 0 ) > 0 , ∀ y ∈ B 1 ( 0 ) .

If (3.28) is incorrect, by μ ≥ 0 in B 1 ( 0 ) × R , then there is point y ¯ ∈ B 1 ( 0 ) satisfying

(3.29) μ ¯ ( y ¯ , 0 ) = inf ( y , t ) ∈ R n × R μ ¯ ( y , t ) = 0 .

By combining (3.26) and f ( 0 ) ≥ 0 , we have

(3.30) 0 ≤ f ( 0 ) = f ( μ ¯ ( y ¯ , 0 ) ) = ∂ t β μ ¯ ( y ¯ , 0 ) − D α τ μ ¯ ( y ¯ , 0 ) = C β ∫ − ∞ 0 − μ ¯ ( y ¯ , ν ) ( − ν ) 1 + β d ν − inf G ∈ G PV ∫ R n μ ¯ ( z , t ) ∣ G − 1 ( z − y ) ∣ n + 2 α d z ≤ 0 ,

that is,

(3.31) μ ¯ ( y , t ) ≡ 0 in R n × ( − ∞ , 0 ] ,

which contradicts (3.27), so (3.28) holds.

It follows from μ ¯ ( y , 0 ) ≡ 0 in B 1 c ( 0 ) , (3.28) and λ 0 that there is point y ˜ ∈ B 1 c ( 0 ) such that y ˜ λ 0 ∈ B 1 ( 0 ) and

(3.32) ω ¯ ( y ˜ , 0 ) = μ ¯ ( y ˜ λ 0 , 0 ) − μ ¯ ( y ˜ , 0 ) = μ ¯ ( y ˜ λ 0 , 0 ) > 0 ,

which contradicts (3.23), so λ 0 = 0 .

Step 3. For any λ ∈ ( − 1 , 0 ) , we prove that

(3.33) ω λ ( y , t ) > 0 .

If (3.33) does not hold, there exists some λ 0 ∈ ( − 1 , 0 ) and a point ( y 0 t 0 ) ∈ Ω λ 0 × R such that

(3.34) ω λ 0 ( y 0 t 0 ) = inf H λ 0 × R ω λ 0 ( y , t ) = 0 .

By using Theorem 2.1, we obtain

(3.35) ω λ 0 ( y , t ) ≡ 0 in R n × ( − ∞ , t 0 ] ,

which contradicts ω λ 0 ( ⋅ , t ) ≢ 0 for any fixed t ∈ ( − ∞ , t 0 ] ; and hence, (3.33) is valid and thus complete the proof of Theorem 1.1.

4 Proof of Theorem 1.2

From (1.2), we have

(4.1) ∂ t β ω ( y , t ) − D α τ ( μ λ ( y , t ) ) + D α τ ( μ ( y , t ) ) = 0 ( y , t ) ∈ H λ × R , ω λ ( y , t ) = − ω ( y , t ) , ( y , t ) ∈ H λ × R .

By applying Theorem 2.2, we derive that

(4.2) ω λ ≡ 0 in H λ × R .

Since ω λ ( y , t ) = − ω ( y , t ) and the choice of y 1 direction is arbitrary, we have

(4.3) μ ( y , t ) = μ ( t ) in R n × R .

From the aforementioned conclusion and Lemma A.1, it can be inferred that μ ( y , t ) is a constant.

5 Proof of Theorem 1.3

Let H λ + = { x ∈ R + n ∣ 0 < x 1 < λ } . From (1.3), we have

(5.1) ∂ t β ω ( y , t ) − D α τ ( μ λ ( y , t ) ) + D α τ ( μ ( y , t ) ) = c λ ( y , t ) ω ( y , t ) ( y , t ) ∈ H λ + × R , ω ( y , t ) ≥ 0 , ( y , t ) ∈ ( H λ \ H λ + ) × R , ω λ ( y , t ) = − ω ( y , t ) , ( y , t ) ∈ H λ × R ,

where

(5.2) c λ ( y , t ) = ∫ 0 1 f ′ ( s μ λ ( y , t ) + ( 1 − s ) μ ( y , t ) ) d s ≤ C ,

due to f ′ is bounded from above.

Step 1. We show that

(5.3) ω λ ( y , t ) ≥ 0 , ( y , t ) ∈ H λ × R .

for sufficiently small λ . By using Theorem 2.5, we derive (5.3), where H λ + is a narrow region.

Step 2. Denote

(5.4) λ 0 = sup { λ < 0 ∣ ω ρ ( y , t ) ≥ 0 , ( y , t ) ∈ H λ × R , ρ ≤ λ } .

Our main purpose is to prove that λ 0 = + ∞ . If 0 < λ 0 < + ∞ , according to its definition, there exists a sequence { λ m } ↘ λ 0 and a sequence { p m } ↘ 0 such that

(5.5) inf ( y , t ) ∈ H λ m × R ω λ m ( y , t ) = − p m .

From this, we can know that

(5.6) H λ m − × R ≔ { ( y , t ) ∈ H λ m × R ∣ ω λ m ( y , t ) < 0 }

is nonempty and

(5.7) inf ( y , t ) ∈ H λ m × R ω λ m ( y , t ) < 0 .

Set

(5.8) Q m = sup H λ m − × R c λ m ( y , t ) .

We consider the following two cases:

Cases 1. If Q m ≤ ε m → 0 , we obtain c λ m ( y , t ) ≤ ε m in H λ m − × R . it follows from Theorem 2.6 that

(5.9) ω λ m ( y , t ) ≥ 0 in H λ m × R ,

for sufficiently large m , which contradicts (5.7).

Cases 2. If Q m ↛ 0 , we denote

(5.10) E m ≔ { ( y , t ) ∈ H λ m − × R ∣ ω λ m ( y , t ) < − p m + p m 2 < 0 } .

When

(5.11) sup E m c λ m ( y , t ) ≤ ε m → 0 as m → ∞ ,

similar to the previous discussion on case 1, we obtain a contradiction with (5.7). However, when

(5.12) a m = sup E m c λ m ( y , t ) ≤ ε m ↛ 0 as m → ∞ ,

then there are a sequence { ( y m , t m ) } ⊂ E m and a constant δ 0 > 0 such that

(5.13) c λ m ( y m , t m ) ≥ δ 0 > 0 .

From (5.2), (5.13) and ω λ m ( y m , t m ) < 0 , then there is some ξ λ m ∈ ( μ λ m ( y m , t m ) , μ ( y m , t m ) ) satisfying

(5.14) f ′ ( ξ λ m ) ≥ δ 0 ,

which means that there exist a subsequence of ( y m , t m ) ( still denoted by { ( y m , t m ) } ) and a constant ε 0 > 0 such that

(5.15) μ ( y m , t m ) ≥ ε 0 > 0 .

Otherwise, if μ ( y m , t m ) → 0 as m → ∞ , then for sufficiently large m , we have ξ λ m → 0 , which implies that

(5.16) f ′ ( 0 ) ≥ C 1 > 0 .

This contradicts the assumption f ′ ( 0 ) ≤ 0 , so (5.15) holds.

According to μ ( y , t ) = 0 in ( R n \ R + n ) × R and μ ( y , t ) is continuous, then there is r 0 > 0 satisfying

(5.17) μ ( y , t ) ≥ ε 0 2 > 0 in B r 0 ( y m ) × ( t m − r 0 2 α β , t m + r 0 2 α β ] ⊂ R + × R ,

where r 0 is not related to m . From (5.10) and { ( y m , t m ) } ⊂ E m , we have

(5.18) ω λ m ( y m , t m ) ≤ − p m + p m 2 < 0 .

To apply Theorems 2.7 and 2.8, we need to prove that δ m = dist ( y m , T λ m ) = λ m − y 1 m is bounded and δ m ≠ 0 . If not, thus δ m → 0 as m → ∞ . Let

(5.19) W m ( y , t ) = ω λ m − p m 2 η m ( y , t ) ,

with

(5.20) F δ m ( y m , t m ) ≔ B δ m ( y m ) × ( t m − δ m 2 α β , t m + δ m 2 α β ) ,

where

(5.21) η m ( y , t ) = η y − y m r m , t − t m r m 2 α β ∈ C 0 ∞ ( F δ m ( y m , t m ) )

satisfies

(5.22) η m ( y , t ) ≡ 1 , in F δ m 2 ( y m , t m ) , 0 ≤ η m ( y , t ) ≤ 1 , in R n × R .

Similar to the proof process of Theorem 1.1, there is a point ( y ¯ m , t ¯ m ) ∈ F δ m ( y m , t m ) such that

(5.23) − p m ≤ ω λ m ( y ¯ m , t ¯ m ) ≤ − p m + p m 2

and

(5.24) − C p m = c ( y ¯ m , t ¯ m ) ω ( y ¯ m , t ¯ m ) = ∂ t β ω ( y ¯ m , t ¯ m ) − ( D α τ ( μ λ ( y ¯ m , t ¯ m ) ) − D α τ ( μ ( y ¯ m , t ¯ m ) ) ) ≤ − p m C δ m 2 α + C p m 2 δ m 2 α ,

which means

(5.25) 0 < C ≤ C δ m + C p m → 0 as m → ∞ .

This contradiction show that δ m is bounded and δ m ≠ 0 . Therefore, we derive that there exists a subsequence of { ( y m , t m ) } (we will still denote by { ( y m , t m ) } ) such that { ( y m , t m ) } ⊂ H λ 0 × R and dist ( y m , T λ 0 ) ≥ δ 0 > 0 , due to λ m → λ 0 as m → ∞ . Based on (5.17), we further select a radius r 1 = min { r 0 , δ 0 } such that

(5.26) μ ( y , t ) ≥ ε 0 2 in B r 1 ( y m ) × ( t m − r 1 2 α β , t m + r 1 2 α β ] ⊂ H λ 0 + × R .

Let y ¯ m = ( 2 λ 0 , ( y m ) ′ ) , then we obtain B r 1 ( y m ) ¯ ∩ B 2 r 1 ( y ¯ m ) = ∅ due to dist ( T λ 0 , T 2 λ 0 ) = λ 0 > 2 r 1 .

Next, we want to prove that there exists ε 1 > 0 such that

(5.27) μ ( y , t ) ≥ ε 1 > 0 in B r 1 ( y ¯ m ) × t m − r 1 2 α β , t m + r 1 2 α β .

If (5.27) is incorrect, we have

(5.28) μ ( y , t ) < ε in B r 1 ( y ¯ m ) × t m − r 1 2 α β , t m + r 1 2 α β , ∀ ε > 0 .

Since f ∈ C 1 and the assumption f ( 0 ) ≥ 0 , it follows that

(5.29) ∣ f ( μ ( y , t ) ) − f ( 0 ) ∣ ≤ C ∣ μ ( y , t ) − 0 ∣ < C ε in B r 1 ( y ¯ m ) × t m − r 1 2 α β , t m + r 1 2 α β ,

which means that

(5.30) f ( μ ( y , t ) ) ≥ − C ε in B r 1 ( y ¯ m ) × t m − r 1 2 α β , t m + r 1 2 α β , ∀ ε > 0 .

By (1.3), (5.26), Theorem 2.7 and the continuity of μ ( y , t ) , we can deduce that there is ε 1 > 0 such that

(5.31) μ ( y , t ) ≥ ε 1 > 0 in B r 1 2 ( y ¯ m ) × ( t m − r 1 2 2 α β , t m + r 1 2 2 α β ] ,

which contradicts (5.28), so (5.27) holds.

Let y ˆ m = ( 0 , ( y m ) ′ ) , we know that there is a small radius r 2 < r 1 2 such that

(5.32) μ ( y , t ) ≤ ε 1 2 in ( B r 2 ( y ˆ m ) ∩ R + n ) × ( t m − r 1 2 α β , t m + r 1 2 α β ] ,

due to μ ( y , t ) ≡ 0 in ( R \ R + n ) × R and μ ( y , t ) are continuous.

For any point y ∈ B r 2 ( y ˆ m ) ∩ R + n , its reflection point y λ 0 with respect to the plane T λ 0 and y λ 0 ∈ B r 2 ( y ¯ m ) ∩ Σ 2 λ 0 ⊂ B r 1 2 ( y ¯ m ) . It follows from (5.27) and (5.32) that

(5.33) ω λ 0 ( y , t ) = μ ( y λ 0 , t ) − μ ( y , t ) ≥ ε 1 − ε 1 2 = ε 1 2 in ( B r 2 ( y ˆ m ) ∩ R + n ) × ( t m − r 1 2 α β , t m + r 1 2 α β ] .

Next, we mainly want to demonstrate that there exists ε 2 > 0 such that

(5.34) ω λ 0 ( y , t ) ≥ ε 2 > 0 in B r 1 2 ( y m ) × ( t m − r 1 2 2 α β , t m + r 1 2 2 α β ] .

If (5.34) does not hold, we obtain

(5.35) ω λ 0 ( y , t ) ≤ ε in B r 1 2 ( y m ) × t m − r 1 2 2 α β , t m + r 1 2 2 α β , ∀ ε > 0 .

Considering the following equation and the definition of λ 0 , we have

(5.36) ∂ t β ω λ 0 ( y , t ) − ( D α τ ( μ λ 0 ( y , t ) ) − D α τ ( μ ( y , t ) ) ) = C λ 0 ( y , t ) ω λ 0 ( y , t ) , in B r 1 2 ( y m ) × t m − r 1 2 2 s α , t m + r 1 2 2 s α , ω λ 0 ( x , t ) ≥ 0 , in ∈ ( H λ 0 \ B r 1 2 ( y m ) ) × t m − r 1 2 2 s α , t m + r 1 2 2 s α , ω λ 0 ( x , t ) ≥ 0 , in B r 1 2 ( y m ) × − ∞ , t m − r 1 2 2 s α .

Then, for any ε > 0 , we have

(5.37) C λ 0 ( y , t ) ω λ 0 ( y , t ) = f ( μ λ 0 ( y , t ) ) − f ( μ ( y , t ) ) > − C ε in B r 1 2 ( y m ) × t m − r 1 2 2 s α , t m + r 1 2 2 s α ,

due to (5.35) and f ∈ C 1 .

According to ω λ 0 ( y , t ) = − ω ( y , t ) , (5.33), Theorem 2.8 and the continuity of μ ( y , t ) , we can deduce that there is ε 2 > 0 such that

(5.38) ω λ 0 ( y , t ) ≥ ε 2 > 0 in B r 1 4 ( y m ) × t m − r 1 4 2 α β , t m + r 1 4 2 α β ,

which contradicts (5.35), so (5.34) is valid. From (5.34) and ω λ are continuous in λ with λ m → λ 0 as m → ∞ , we obtain

(5.39) ω λ ( y m , t m ) ≥ ε 2 > 0 ,

for sufficiently large m , which contradicts (5.6), hence λ = + ∞ .

Step 3. Next, we will demonstrate that

(5.40) ω λ ( y , t ) > 0 in H λ × R , ∀ λ > 0 .

Assuming (5.40) is incorrect, there is a fixed λ 0 > 0 and a point ( y 0 , t 0 ) ∈ H λ 0 + × R satisfying

(5.41) ω λ 0 ( y 0 , t 0 ) = inf λ 0 × R ω λ 0 ( y , t ) = 0 .

Therefore, we have

(5.42) 0 = C λ 0 ( y 0 , t 0 ) ω λ 0 ( y 0 , t 0 ) = ∂ t β ω λ 0 ( y 0 , t 0 ) − ( D α τ ( μ λ 0 ( y 0 , t 0 ) ) − D α τ ( μ ( y 0 , t 0 ) ) ) < 0 .

This contradiction indicates that (5.40) is valid.

6 Proof of Theorem 1.4

From (1.4), we have

(6.1) ∂ t β ω ( y , t ) − D α τ ( μ λ ( y , t ) ) + D α τ ( μ ( y , t ) ) = c λ ( y , t ) ω ( y , t ) ,

where c λ ( y , t ) = f μ ( ξ λ ( y , t ) , t ) and ξ λ ∈ ( μ λ , μ ) or ξ λ ∈ ( μ , μ λ ) .

Step 1. We demonstrate that

(6.2) inf t ∈ R ω λ ( y , t ) > 0 for all y ∈ H λ ,

for λ sufficiently negative.

Firstly, we need to prove that

(6.3) ω λ ( y , t ) ≥ 0 for all ( y , t ) ∈ H λ × R ,

for λ sufficiently negative.

According to the assumptions f μ ( 0 , t ) ≤ − γ and f μ ∈ C ( R × R ) , there exists a sufficiently small δ 0 > 0 such that

(6.4) f μ ( μ , t ) ≤ − γ for all ( μ , t ) ∈ [ 0 , δ 0 ) × R .

Since

(6.5) lim ∣ y ∣ → ∞ ω λ ( y , t ) = 0 for all t ∈ R ,

we have c λ ( y , t ) = f μ ( ξ λ m ( y , t ) , t ) ≤ − γ for λ sufficiently negative. Hence by Theorem 2.4, we obtain (6.3).

Furthermore, if there is a point ( y 0 , t 0 ) ∈ H λ × R satisfying ω λ ( y 0 , t 0 ) = 0 , then by Theorem 2.4, we have

(6.6) ω λ ( y , t ) ≡ 0 in R n × ( − ∞ , t 0 ] .

Fixed t 1 ≤ t 0 , Since μ ( 0 , t 1 ) > 0 and sup t ∈ R μ ( y , t ) → 0 as ∣ y ∣ → ∞ , then we obtain

(6.7) ω λ ( 0 λ , t 1 ) = μ ( 0 , t 1 ) − μ ( 0 λ , t 1 ) > 0 ,

for λ sufficiently negative. This is contradictory to (6.6), so we have

(6.8) ω λ ( y , t ) > 0 in H λ × R .

Assuming (6.2) is invalid, there exists a point y 0 ∈ H λ , a sequence { t m } ⊂ R and a positive sequence { ε m } ↘ 0 such that

(6.9) ω λ ( y 0 , t m ) = ε m → 0 as m → ∞ .

Due to t ∈ R , the minimum value of ω ( y , t ) may not reach. Therefore, we introduce the following perturbation function

(6.10) V m ( y , t ) = ω λ ( y , t ) − ε m ϕ δ ( y ) η m ( t ) in H λ × R

with

ϕ δ ( y ) = ϕ y − y 0 δ

and

η m ( t ) = η ( t − t m )

where ϕ ( y ) = ( 1 − ∣ y ∣ 2 ) + α , 0 < δ < dist ( y 0 , T λ ) 2 and η ( t ) ∈ C 0 ∞ ( R ) satisfying

(6.11) η ( t ) = 1 , ∣ t ∣ ≤ 1 2 , 0 , ∣ t ∣ ≥ 1 .

Denote

(6.12) E δ ( y 0 , t m ) ≔ B δ ( y 0 ) × [ t m − 1 , t m + 1 ] .

It follows from (6.9) and (6.10) that

(6.13) V m ( y 0 , t m ) = 0 , V m ( y , t ) = ω λ ( y , t ) > 0 in H λ × R \ E δ ( y 0 , t m ) .

Hence there is a point ( y ˜ 0 , t ˜ m ) ∈ E δ ( y 0 , t m ) such that

(6.14) V m ( y ˜ 0 , t ˜ m ) = inf H λ × R V m ( y , t ) ≤ 0 ,

which means that

(6.15) 0 < ω λ ( y ˜ 0 , t ˜ m ) ≤ ε m ϕ δ ( y ˜ 0 ) η m ( t ˜ m ) ≤ ε m

and

(6.16) ∂ t β ω λ ( y ˜ 0 , t ˜ m ) ≤ ε m ϕ δ ( y ˜ 0 ) ∂ t β η m ( t ˜ m ) ≤ C ε m .

By the definition of D α τ , for any sequences { σ m } → 0 , there is G m ∈ G such that

(6.17) D α τ ϕ δ ( y ˜ 0 ) = inf G ∈ G PV ∫ R n ϕ δ ( z ) − ϕ δ ( y ˜ 0 ) ∣ G − 1 ( z − y ˜ 0 ) ∣ n + 2 α d z ≥ PV ∫ R n ϕ δ ( z ) − ϕ δ ( y ˜ 0 ) ∣ G m − 1 ( z − y ˜ 0 ) ∣ n + 2 α d z − σ m ≥ C 0 PV ∫ R n ϕ δ ( z ) − ϕ δ ( y ˜ 0 ) ∣ z − y ˜ 0 ∣ n + 2 α d z − σ m = − C δ 2 α − σ m ,

where C > 0 and C 0 is a constant with respect to the eigenvalue of G m ∈ G . Note that

D α τ ( μ λ ( y , t ) ) − D α τ ( μ ( y , t ) ) = inf G ∈ G PV ∫ R n μ λ ( z , t ) − μ λ ( y , t ) ∣ G − 1 ( z − y ) ∣ n + 2 α d z − inf G ∈ G PV ∫ R n μ ( z , t ) − μ ( y , t ) ∣ G − 1 ( z − y ) ∣ n + 2 α d z ≥ inf G ∈ G PV ∫ R n ω λ ( z , t ) − ω λ ( y , t ) ∣ G − 1 ( z − y ) ∣ n + 2 α d z = D α τ ω λ ( y , t ) .

By using the same method, we have

D α τ ( μ λ ( y , t ) ) − D α τ ( μ ( y , t ) ) − ε m η m ( t ) D α τ ϕ δ ( y ) ≥ D α τ ω λ ( y , t ) − ε m η m ( t ) D α τ ϕ δ ( y ) ≥ D α τ V m ( y , t ) .

By (6.15) and (6.16), we obtain

(6.18) − D α τ V m ( y ˜ 0 , t ˜ m ) ≥ − ( D α τ ( μ λ ( y ˜ 0 , t ˜ m ) ) − D α τ ( μ ( y ˜ 0 , t ˜ m ) ) ) + ε m η m ( t ˜ m ) D α τ ϕ δ ( y ˜ 0 ) ≥ − ∂ t β ω λ ( y ˜ 0 , t ˜ m ) + c λ ( y ˜ 0 , t ˜ m ) ω λ ( y ˜ 0 , t ˜ m ) + ε m η m ( t ˜ m ) D α τ ϕ δ ( y ˜ 0 ) ≥ − C ε m − C ε m δ 2 α − σ m ε m → 0 as m → ∞ .

Furthermore, from (6.14), we can deduce that for any sequences { σ m } → 0 , there is G m ∈ G such that

(6.19) − D α τ V m ( y ˜ 0 , t ˜ m ) = − inf G ∈ G PV ∫ R n V m ( z , t ˜ m ) − V m ( y ˜ 0 , t ˜ m ) ∣ G − 1 ( z − y ˜ 0 ) ∣ n + 2 α d z ≤ PV ∫ R n V m ( y ˜ 0 , t ˜ m ) − V m ( z , t ˜ m ) ∣ G m − 1 ( z − y ˜ 0 ) ∣ n + 2 α d z + σ m ≤ ∫ H λ 2 V m ( y ˜ 0 , t ˜ m ) ∣ G m − 1 ( z λ − y ˜ 0 ) ∣ n + 2 α d z + ∫ H λ ( V m ( y ˜ 0 , t ˜ m ) − V m ( z , t ˜ m ) ) × 1 ∣ G m − 1 ( z − y ˜ 0 ) ∣ n + 2 α − 1 ∣ G m − 1 ( z λ − y ˜ 0 ) ∣ n + 2 α d z + σ m ≤ σ m → 0 as m → ∞ ,

where

1 ∣ G m − 1 ( z − y ˜ 0 ) ∣ n + 2 α − 1 ∣ G m − 1 ( z λ − y ˜ 0 ) ∣ n + 2 α > 0 .

Therefore,

(6.20) ω λ ( z , t ˜ m ) → 0 uniformly for z ∈ H λ , as m → ∞ .

Next, we need to prove that

(6.21) ω λ ( y , t ) ≤ sup y ∈ H λ ω λ ( y , t ˜ m ) in H λ × [ t ˜ m , + ∞ ) ,

where sup y ∈ H λ ω λ ( y , t ˜ m ) > 0 .

For λ sufficiently negative, we have c λ ≤ − γ , ω λ > 0 , and thus,

∂ t β ω ( y , t ) − ( D α τ ( μ λ ( y , t ) ) − D α τ ( μ ( y , t ) ) ) = c λ ( y , t ) ω λ ( y , t ) ≤ 0 in { ( y , t ) ∈ H λ × R ∣ ω ( y , t ) > 0 } , ω ( y , t ) = − ω ( y λ , t ) in H λ × R .

By Theorem 2.2, hence (6.21) holds. When m → ∞ , by combining with (6.20) and (6.21), we can deduce that

(6.22) ω λ ( y , t ) ≡ 0 in H λ × [ t ˜ m , + ∞ ) ,

which contradicts (6.8), so (6.2) is valid.

Step 2. Let

(6.23) λ 0 − = sup { λ ∣ inf t ∈ R ω ρ ( y , t ) > 0 , for all y ∈ H λ , ρ ≤ λ } .

Since lim ∣ y ∣ → ∞ ω λ ( y , t ) = 0 for all t ∈ R , we clearly know that λ 0 − < ∞ .

In this step, we want to prove that there is { t m ̲ } ⊂ R such that

(6.24) ω λ 0 − ( y , t m ̲ ) → 0 uniformly for y ∈ H λ , as m → ∞ .

According to the definition of λ 0 − , there exists a sequence { λ k } with { λ k } ↘ λ 0 − such that

inf ( y , t ) ∈ H λ m × R ω λ m ( y , t ) ≤ 0 .

On one hand, there is a subsequence of { λ m } ( still denoted by { λ m } ) , such that

(6.25) inf ( y , t ) ∈ H λ m × R ω λ m ( y , t ) < 0 .

From this, we know that there exists a positive sequence { p m } ↘ 0 such that

(6.26) inf ( y , t ) ∈ H λ m × R ω λ m ( y , t ) = − p m < 0 .

This means that there exists ( y ˆ m , t ˆ m ) ∈ H λ m × R such that

(6.27) − p m ≤ ω λ m ( y ˆ m , t ˆ m ) ≤ − p m + p m 2 < 0 .

Since lim ∣ y ∣ → ∞ ω λ m ( y , t ) = 0 , we only need to introduce the following auxiliary functions:

(6.28) V ¯ m ( y , t ) = ω λ m ( y , t ) − p m 2 η ¯ m ( t ) ,

where η ¯ m ( t ) = η ( t − t ˆ ) and η ( t ) is defined in (2.30). By using the same method as shown in Theorem 2.4, we can derive that there exists a point ( y m ̲ , t m ̲ ) ∈ H λ m × [ t ˆ − 1 , t ˆ + 1 ] such that

− p m ≤ ω λ m ( y m ̲ , t m ̲ ) = inf y ∈ H λ m ≤ − p m + p m 2 ,

∂ t β ω λ m ( y m ̲ , t m ̲ ) ≤ p m 2 η ¯ m ( t m ̲ ) ≤ C p m 2 ,

and thus,

0 ← − C p m 2 ≤ c λ m ( y m ̲ , t m ̲ ) ω λ m ( y m ̲ , t m ̲ ) − ∂ t β ω λ m ( y m ̲ , t m ̲ ) = − D α τ ( μ λ m ( y m ̲ , t m ̲ ) ) + D α τ ( μ λ m ( y m ̲ , t m ̲ ) ) ≤ ∫ H λ 2 ω λ m ( y m ̲ , t m ̲ ) ∣ G m − 1 ( z λ − y m ̲ ) ∣ n + 2 α d z + ∫ H λ ( ω λ m ( y m ̲ , t m ̲ ) − ω λ m ( z , t m ̲ ) ) × 1 ∣ G m − 1 ( z − y m ̲ ) ∣ n + 2 α − 1 ∣ G m − 1 ( z λ − y m ̲ ) ∣ n + 2 α d z + σ m ≤ σ m → 0 as m → ∞ .

Therefore, (6.24) holds.

On the other hand, if along another subsequence,

inf ( y , t ) ∈ H λ m × R ω λ m ( y , t ) = 0 ,

which means that ω λ m ( y , t ) ≥ 0 in H λ m × R . Similar to the previous argument about (6.2), we have

(6.29) inf t ∈ R ω λ m ( y , t ) > 0 for all y ∈ H λ m ,

which contradicts the definition of λ 0 − . Similar to Step 1, we can obtain

(6.30) inf t ∈ R ω λ ( y , t ) > 0 , for all y ∈ H λ c ,

for λ ≥ R . Let

λ 0 + = inf { λ ∣ inf t ∈ R ω ρ ( y , t ) > 0 , for all y ∈ H λ c , ρ ≥ λ } .

Similar to the proof of (6.24), we know that there is a sequence { t ¯ m } ⊂ R such that

(6.31) w λ 0 + ( y , t ¯ m ) → 0 uniformly for y ∈ H λ 0 + c as m → ∞ .

If λ 0 − = λ 0 + = λ 0 , then we have

(6.32) ω λ 0 ( y , t ) ≡ 0 , for all ( y , t ) ∈ H λ 0 × R .

Step 3. Our main purpose is to prove that λ 0 − = λ 0 + . Otherwise, assuming λ ∈ ( λ 0 − , λ 0 + ) , we will investigate the property of ω λ ( y , t ) in H λ .

Let Ω λ ≔ H λ \ H λ 0 − , for any y ∈ Ω λ and y λ be the reflection of y about the plane T λ and ( y λ ) λ 0 − be the reflection of y λ about the plane T λ 0 − . Obviously, we have ( y λ ) λ 0 − ∈ H λ 0 − + λ − y 1 , and thus,

inf t ∈ R ω λ 0 − + λ − y 1 ( ( y λ ) λ 0 − , t ) > 0 .

For each fixed y = ( y 1 , y ′ ) ∈ Ω λ , due to ω λ 0 − ( y , t m ̲ ) → 0 uniformly for any y ∈ H λ 0 − , we can deduce that for sufficiently large m

(6.33) ω λ ( y , t m ̲ ) = μ ( y λ , t m ̲ ) − μ ( y , t m ̲ ) = μ ( y λ , t m ̲ ) − μ ( ( y λ ) λ 0 − , t m ̲ ) + μ ( ( y λ ) λ 0 − , t m ̲ ) − μ ( y , t m ̲ ) = ω λ 0 − ( ( y λ ) λ 0 − , t m ̲ ) − ω λ 0 − + λ − y 1 ( ( y λ ) λ 0 − , t m ̲ ) ≤ K m − inf t ∈ R ω λ 0 − + λ − y 1 ( ( y λ ) λ 0 − , t ) ≤ − 1 2 inf t ∈ R ω λ 0 − + λ − y 1 ( ( y λ ) λ 0 − , t ) < 0 ,

where K m → 0 as m → ∞ .

On the one hand, if Ω λ is an unbounded narrow region, then by Theorem 2.1, we derive that

ω λ ( y , t ) ≥ 0 in Ω λ × [ t m ̲ − 1 · t m ̲ + 1 ] ,

which contradicts (6.33).

On the other hand, if Ω λ is an unbounded region with a finite width in y 1 direction, by arguments similar to Theorems 2.4 and 2.6, we obtain

ω λ ( y , t ) ≥ 0 in Ω λ × [ t m ̲ − 1 · t m ̲ + 1 ] ,

which contradicts (6.33), so λ 0 − = λ 0 + . This completes the proof of Theorem 1.4.

7 Proof of Theorem 1.5

By (1.5), we have

(7.1) ∂ t β ω ( y , t ) − D α τ μ λ ( y , t ) + D α τ μ ( y , t ) = y 1 λ μ λ s ( y , t ) − y 1 μ s ( y , t ) = ( y 1 λ − y 1 ) μ λ s ( y , t ) + y 1 ( μ λ s ( y , t ) − μ s ( y , t ) ) ≥ c λ ( y , t ) ω λ ( y , t ) ,

where c λ ( y , t ) = s ξ λ ( y , t ) y 1 and ξ λ ∈ ( μ λ , μ ) or ξ λ ∈ ( μ , μ λ ) .

Step 1. We want to prove that

(7.2) ω λ ( y , t ) ≥ 0 for all ( y , t ) ∈ H λ × R , for λ ≤ 0 .

From (7.1), we obtain for λ ≤ 0

∂ t β ω ( y , t ) − ( D α τ ( μ λ ( y , t ) ) − D α τ ( μ ( y , t ) ) ) ≥ 0 in { ( y , t ) ∈ H λ × R ∣ ω λ ( y , t ) < 0 } , ω ( y , t ) = − ω ( y λ , t ) in H λ × R ,

where c λ ( y , t ) ≤ 0 . By Remark 2.3, we know that (7.2) holds.

Step 2. In this step, as long as equation (7.2) is valid, we will continue to move the plane T λ rightward along the y 1 -axis to its limit position. Let

(7.3) λ 0 = sup { λ ≥ 0 ∣ ω ρ ( y , t ) ≥ 0 , in ( y , t ) ∈ H ρ × R , ρ ≤ λ } .

We shall prove that

(7.4) λ 0 = + ∞ .

If 0 < λ 0 < + ∞ , according to its definition, there are a sequence { p m } ↘ 0 and a sequence { λ m } ↘ λ 0 that makes

(7.5) inf ( x , t ) ∈ H λ m × R ω λ m ( y , t ) = − p m < 0 .

From this, we know that there is a subsequence of { ( y m , t m ) } ( still denoted by { ( y m , t m ) } ) that satisfies

(7.6) ω λ m ( y m , t m ) = − p m + p m 2 < 0 .

Now we use a perturbation technique as follows:

(7.7) V m ( y , t ) = ω λ m ( y , t ) − p m 2 η m ( y , t )

with

M r m ( y m , t m ) ≔ B r m 2 ( y m ) × t m − r m 2 α β 2 , t m + r m 2 α β 2 ⊂ H λ m × R and 0 < r m = 1 2 dist ( y m , T λ ) ,

where

η m ( y , t ) = η y − y m r m , t − t m r m 2 α β ∈ C 0 ∞ ( M r m ( y m , t m ) )

satisfies

(7.8) η m ( y , t ) ≡ 1 , in M r m 2 ( y m , t m ) , 0 ≤ η m ( y , t ) ≤ 1 , in R n × R .

From (7.5), (7.6), (7.7), and (7.8), we derive that

(7.9) V m ( y m , t m ) = − p m , V m ( y , t ) = ω λ ( y , t ) ≥ − p m , in ( H λ m × R ) \ M m ( y m , t m ) .

Therefore, we know that there exists a point ( y ¯ m , t ¯ m ) ∈ M m ( y m , t m ) such that

− p m − p m 2 ≤ V m ( y ¯ m , t ¯ m ) = inf H λ m × R V m ( x , t ) ≤ − p m ,

which implies that

(7.10) − p m ≤ ω λ m ( y ¯ m , t ¯ m ) ≤ − p m + p m 2 < 0

and

∂ t β ω λ m ( y ¯ m , t ¯ m ) ≤ p m 2 ∂ t β η m ( y ¯ m , t ¯ m ) ≤ C p m 2 r m 2 α .

Thus, by similar to the calculation process of (2.16), we obtain

∂ t β ω λ m ( y ¯ m , t ¯ m ) − D α τ ( μ λ m ( y ¯ m , t ¯ m ) ) + D α τ ( μ ( y ¯ m , t ¯ m ) ) ≤ C p m 2 r m 2 α − C 1 p m r m 2 α .

where C , C 1 > 0 . By (7.1), we have

(7.11) s y ¯ 1 m ξ λ m s − 1 ( y ¯ m , t ¯ m ) ω λ m ( y ¯ m , t ¯ m ) ≤ C p m 2 r m 2 α − C 1 p m r m 2 α .

We claim that 0 < y ¯ 1 m < λ + 1 , for sufficiently large m . Otherwise, if y ¯ 1 m ≤ 0 ; thus from (7.11), we have

0 ≤ C p m − C 1 → − C 1 as m → ∞ .

This contradiction implies that 0 < y ¯ 1 m < λ + 1 . Obviously, according to (7.10) and (7.11), we obtain

(7.12) s y ¯ 1 m ξ λ m s − 1 ( y ¯ m , t ¯ m ) r m 2 α ≥ − C p m + C 1 .

We claim that, for sufficiently large m ,

(7.13) μ ( y ¯ m , t ¯ m ) ≥ C > 0 .

Otherwise, if not, then μ ( y ¯ m , t ¯ m ) → 0 as m → ∞ , we obtain ξ ( y ¯ m , t ¯ m ) → 0 and

0 ← s y ¯ 1 m ξ λ m s − 1 ( y ¯ m , t ¯ m ) r m 2 α ≥ − C p m + C 1 → C 1 ,

which contradicts C 1 > 0 . Since ω λ m ( y ¯ m , t ¯ m ) → 0 as m → ∞ , it follows that

(7.14) μ λ m ( y ¯ m , t ¯ m ) ≥ C > 0 .

According to (7.1), (7.11), and (7.14), we derive that

(7.15) 0 < ( λ m − y ¯ 1 m ) μ λ m ( y ¯ m , t ¯ m ) + s y ¯ 1 m ξ λ m s − 1 ( y ¯ m , t ¯ m ) ω λ m ( y ¯ m , t ¯ m ) ≤ C p m 2 r m 2 α − C 1 p m r m 2 α → − C 1 as m → ∞ .

This contradiction means that λ 0 = + ∞ .

Step 3. We will prove that

(7.16) ω λ ( y , t ) > 0 , ∀ ( y , t ) ∈ H λ × R , ∀ λ ∈ R .

If (7.16) does not hold, then there is a point ( y 0 , t 0 ) ∈ H λ × R such that

(7.17) ω λ ( y 0 , t 0 ) = min ( y , t ) ∈ H λ × R ω λ ( y , t ) = 0 .

Similar to the argument of (2.23), we can deduce that

(7.18) V λ ( y , t ) ≡ 0 , ∀ ( y , t ) ∈ R n × ( − ∞ , t 0 ] .

For any ( y , t ) ∈ R n × ( − ∞ , t 0 ] , we derive that

0 = ∂ t β ω λ ( y , t ) − D α τ ( μ λ ( y , t ) ) + D α τ ( μ ( y , t ) ) = ( y 1 λ − y 1 ) μ λ s ( x , t ) + y 1 ( μ λ s ( y , t ) − μ s ( y , t ) ) > 0 .

This contradiction means that (7.16) must be true.

Step 4. We will prove the nonexistence of solutions to (1.5).

Suppose that there is a positive bounded solution μ of (1.5), then we know that

(7.19) M ≔ sup ( y , t ) ∈ R n × R μ ( y , t ) and Q ≔ inf ( y , t ) ∈ R n × R μ ( y , t ) .

For any R ≥ 1 , denote

(7.20) T = ( 1 + M ) 1 γ and ϕ R ( y ) = ϕ ( y − Re 1 ) ,

where ϕ ( y ) = ( 1 − ∣ y ∣ 2 ) + s and e 1 is a unit vector in the y 1 -direction.

Let

(7.21) μ ¯ ( y , t ) = μ ( y , t ) − ϕ R ( y ) η ( t ) , with η ( t ) = t β − 1 ,

where 0 < γ = 1 2 k + 1 < β for some positive integer k .

For all ( y , t ) ∈ B 1 ( Re 1 ) × [ 1 , T ] , by (6.17), we can deduce that for sufficiently large R

(7.22) ∂ t β μ ¯ ( y , t ) − D α τ ( μ ( y , t ) ) + D α τ ( ϕ R ( y ) η ( t ) ) = ∂ t β μ ( y , t ) − ϕ R ( y ) ∂ t β η ( t ) − D α τ ( μ ( y , t ) ) + D α τ ( ϕ R ( y ) η ( t ) ) = y 1 μ s ( y , t ) − ϕ R ( y ) ∂ t β η ( t ) + D α τ ( ϕ R ( y ) η ( t ) ) ≥ ( R − 1 ) Q − C 2 C β − C 3 η ( t ) ≥ ( R − 1 ) Q − C 2 C β − C 3 T β > 0 ,

where C 2 , C 3 > 0 .

In addition, we have

(7.23) μ ¯ ( y , t ) > 0 , in R n × ( − ∞ , T ] \ B 1 ( Re 1 ) × [ 1 , T ] .

Now we claim that

(7.24) μ ¯ ( y , t ) > 0 , for all ( y , t ) ∈ B 1 ( Re 1 ) × [ 1 , T ] .

If (7.24) is wrong, then there is a point ( y 0 , t 0 ) such that

μ ¯ ( y 0 , t 0 ) = inf ( y , t ) ∈ R n × R μ ¯ ( y , t ) ≤ 0 .

Consequently, we derive

∂ t β μ ¯ ( y 0 , t 0 ) − D α τ ( μ ( y 0 , t 0 ) ) + D α τ ( ϕ R ( y 0 ) η ( t 0 ) ) ≤ 0 ,

which contradicts (7.22). Hence, we know that (7.24) holds.

By (7.24), we obtain

(7.25) μ ( y , t ) > ϕ R ( y ) η ( t ) = ϕ R ( y ) ( t β − 1 ) , in ( y , t ) ∈ B 1 ( Re 1 ) × [ 1 , T ] .

From this, we have

(7.26) M > T β − 1 = ( 1 + M ) 1 β β − 1 = M .

This contradiction shows that equation (1.5) has no positive bounded solutions.

Acknowledgments

The authors would like to express their sincere gratitude to the anonymous reviewers and editors for their valuable comments and suggestions, which significantly improved the original manuscript.

Funding information: This work is supported by the Laboratory of Engineering Modeling and Statistical Computation of Hainan Province.
Author contributions: Both authors contributed equally and significantly to this manuscript.
Conflict of interest: The authors state no conflict of interest.
Data availability statement: No data were used in this study.

Appendix

Lemma A.1

(Theorem 1.4, [16]) Let μ ( t ) ∈ C 1 ( R ) be a bounded solution of

∂ t β μ ( t ) = 0 .

Then it must be constant.

Lemma A.2

(Lemma 5.1, [17]) ( s ) Let η ( t ) ∈ C 0 ∞ ( − 2,2 ) be a smooth cut-off function, satisfying η ( t ) ≡ 1 in [ − 1 , 1 ] and 0 < η ( t ) ≤ 1 . Then there exists 0 < C 0 that depends only on β such that

∣ ∂ t β η ( t ) ∣ ≤ C 0 for t ∈ ( − 2,2 ) .

Lemma A.3

(Corollary 5.2, [17]) ( s ) For any t 0 ∈ R and r > 0 , let

η 0 ( t ) ≔ η t − t 0 r 2 α β ∈ C 0 ∞ ( t 0 − 2 r 2 α β , t 0 + r 2 α β ) ,

then

∣ ∂ t β η 0 ( t ) ∣ ≤ C 0 r 2 α for t ∈ t 0 − 2 r 2 α β , t 0 + r 2 α β .

Lemma A.4

(Lemma 2.1, [21]) Let y = ( y 1 , y ′ ) and

h ( y ) = f ( y 1 ) g γ ( y ′ ) , with 0 < γ < 2 α ,

where

f ( y 1 ) = 1 − x 1 2 l 2 + s + 1 and g γ ( y ′ ) = ( 1 + ∣ y ′ ∣ 2 ) γ 2 .

Then there is C 1 > 0 such that for all sufficiently small l, we have

( − Δ ) α h ( y ) h ( y ) ≥ C 1 l 2 α for a l l ∣ y 1 ∣ ≤ l .

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Received: 2024-10-04

Revised: 2025-01-10

Accepted: 2025-04-23

Published Online: 2025-06-24

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

Articles in the same Issue

https://doi.org/10.1515/anona-2025-0086

Keywords for this article

dual parabolic equations; monotonicity; radial symmetry; nonexistence; Liouville theorem

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