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Classification of positive solutions for a weighted integral system on the half-space

  • Qiuping Liao , Haofeng Wang and Yingying Xiao EMAIL logo
Published/Copyright: September 16, 2024

Abstract

In this article, we study the following weighted integral system:

u ( x ) = R + n + 1 y n + 1 β f ( u ( y ) , v ( y ) ) x y λ d y , x R + n + 1 , v ( x ) = R + n + 1 y n + 1 β g ( u ( y ) , v ( y ) ) x y λ d y , x R + n + 1 .

Under nature structure conditions on f and g , we classify the positive solutions using the method of moving spheres.

MSC 2010: 45G15; 45M20

1 Introduction

The classical Hardy-Littlewood-Sobolev (HLS) inequality, as stated in [1,2], is given by

(1) R + n R + n f ( x ) x y λ g ( y ) d x d y C α , n , p f L p ( R n ) g L q ( R n ) ,

for all f L p ( R n ) , g L q ( R n ) , 1 < p , q < , 1 p + 1 p = 1 , 1 q + 1 q = 1 , 0 < λ < n , and 1 p + 1 q + λ n = 2 . This inequality mentioned has become an indispensable tool in the study of partial differential equations. The categorization of extremal functions plays a significant role in determining the best constant. Lieb utilized the rearrangement inequality to prove the existence of the optimal function for inequality (1) in his publication [3]. Moreover, he provided a classification of the best functions and computed the sharp constant in cases where either p or q is equal to 2 or p = q .

Inequality (1) has profound implications in the domain of geometric inequalities, encompassing noteworthy results such as the Gross logarithmic Sobolev inequality and the Moser-Onofri-Beckner inequality, as elucidated by Beckner [4]. Another fundamental finding pertains to the equivalence between the HLS inequality and the Sobolev inequality, which can be established by selecting λ = n 2 , p = q = 2 n n + 2 , and utilizing Green’s representation formula. Extensive advancements have been made in extending these inequalities to diverse contexts. Remarkable contributions have emerged from researchers such as Jerison and Lee [5], Frank and Lieb [6], and Han et al. [7] concerning extensions on Heisenberg groups. Dou and Zhu [8] have explored extensions on upper half spaces, while Han and Zhu [9] have attained results on compact Riemannian manifolds. Moreover, several works, including [1013], have established the reversed (weighted) HLS inequality. For a more comprehensive understanding of the weighted HLS inequality and Hardy-Sobolev equations, it is recommended to consult the following references: [57,1423]. These sources provide additional insights and advancements in this area of study.

In the work of Gluck [24], a significant breakthrough was made in establishing the precise HLS inequality specifically for the half-space, taking into account a general kernel and conformal invariance. This remarkable result holds true when the parameters p and q satisfy the conditions p = 2 ( n 1 ) 2 ( n 1 ) λ and q = 2 n 2 n λ + 2 γ .

(2) R + n R + n K λ , γ ( x y , x n ) f ( y ) g ( x ) d x d y C n , λ , γ , p f L p ( R + n ) g L q ( R + n ) ,

where the kernel

K λ , γ ( x , x n ) = x n γ ( x 2 + x n 2 ) λ 2 , x = ( x , x n ) R n 1 × ( 0 , ) ,

with γ 0 , 0 < n λ + γ < n γ ,

λ 2 γ 2 n + λ 2 ( n 1 ) < 1

and

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

He classified all extremal functions of inequality (2) and then computed the best constant.

The family of kernels K λ , γ where λ > 0 , includes several well-known examples such as the classical Poisson kernel K n , 1 , the Riesz kernel K λ , γ , and the Poisson kernel K λ , 1 n + λ for the divergence form operator u div ( x n n λ u ) on the half-space. In [8,2527], the corresponding inequalities to (2) for the kernels K n , 1 , K λ , 0 , K n λ , 1 n + λ , and K n λ , 1 were investigated. In those works, the authors established the existence of extremal functions and subsequently classified all such extremal functions.

Hu and Liu [28] investigated the classification of positive solutions for the following integral system:

u ( y ) = R + n x n γ f ( v ( x ) ) x y λ d x , y R + n , v ( x ) = R + n x n γ g ( u ( y ) ) x y λ d y , x R + n ,

where n 2 , γ 0 , 0 < n λ + γ < n γ , λ 2 γ 2 n + λ 2 ( n 1 ) < 1 .

In a closely related context to the Stein-Weiss inequality [23], Dou and Ma [29] derived the following inequality:

(3) R + n + 1 R + n + 1 x n + 1 α y n + 1 β f ( y ) g ( x ) x y λ d y d x C n , α , β , p f L p ( R + n + 1 ) g L q ( R + n + 1 ) ,

where f L p ( R + n + 1 ¯ ) , g L q ( R + n + 1 ¯ ) and λ , α , β , p , and q satisfy

0 < λ < n + 1 , 1 < q , p < , β > 1 p , α > 1 q , α + β 0 , 1 p + 1 q + λ α β n + 1 = 2 .

By employing the concentration-compactness principle, researchers successfully established the existence of extremal functions in their analysis. In the specific scenario of conformal invariance, they utilized the method of moving spheres to classify all such extremal functions. Recently, a study conducted by Dou et al. [30] introduced a converse form of inequality (3), thereby providing a new insight into the problem. Through the innovative implementation of the renormalization method, they not only demonstrated the existence of extremal functions but also achieved a comprehensive classification of all extremal functions. This breakthrough brings forth valuable contributions to the understanding of the problem at hand.

Hu and Du [31] studied the classification of positive solutions for following integral system:

u ( x ) = R + n + 1 x n + 1 α y n + 1 β f ( v ( y ) ) x y λ d y , y R + n + 1 , v ( y ) = R + n + 1 x n + 1 α y n + 1 β g ( u ( x ) ) x y λ d x , x R + n + 1 ,

where x = ( x , x n + 1 ) R + n + 1 , y = ( y , y n + 1 ) R + n + 1 , 0 < λ < n + 1 , α > λ 2 n , β > λ 2 n , α + β 0 , n + 1 + α + β λ 0 .

In this article, we study the following weighted integral system on the half-space R + n + 1 in the special case of α = 0 :

(4) u ( x ) = R + n + 1 y n + 1 β f ( u ( y ) , v ( y ) ) x y λ d y , x R + n + 1 , v ( x ) = R + n + 1 y n + 1 β g ( u ( y ) , v ( y ) ) x y λ d y , x R + n + 1 ,

where 0 < λ < n + 1 , β 0 , β > 1 p , and n + 1 + β λ 0 .

Theorem 1

For n 1 , let ( u , v ) C 0 ( R + n + 1 ) × C 0 ( R + n + 1 ) be a pair of positive solutions of (4) and f , g : R + × R + R + are locally bounded satisfying

  1. f ( s , t ) , g ( s , t ) are strictly increasing in t for fixed s and f ( s , t ) , g ( s , t ) are strictly increasing in s for fixed t ;

  2. there exist p 1 , q 1 0 , p 1 + q 1 = 2 ( n + 1 ) + 2 β λ λ such that f ( s , t ) s p 1 t q 1 is non-increasing in t for fixed s and non-increasing in s for fixed t;

  3. there exist p 2 , q 2 0 , p 2 + q 2 = 2 ( n + 1 ) + 2 β λ λ such that g ( s , t ) s p 2 t q 2 is non-increasing in t for fixed s and non-increasing in s for fixed t.

Then, one of the following two alternatives holds.

For some x 0 R + n + 1 , for any x R + n + 1 , u and v must take the following form:

u ( x , 0 ) = c 1 ( d 2 + x x 0 2 ) λ 2 , v ( x , 0 ) = c 2 ( d 2 + x x 0 2 ) λ 2 .

Furthermore, f ( s , t ) and g ( s , t ) must be the form of

f ( s , t ) = C 1 s p 1 t q 1 and g ( s , t ) = C 2 s p 2 t q 2 ,

where C 1 and C 2 are some positive constants.

The pioneering application of the method of moving spheres can be attributed to Chen and Li [16], Li and Zhu [32], and Padilla [33]. This approach offers a direct method to capture the explicit form of solutions, bypassing the need for prior derivation of radial symmetry when classifying radial solutions. To explore further developments related to the method of moving spheres or moving planes, readers are encouraged to refer to [22,3447] and the relevant references provided therein. These sources provide valuable insights into the advancements made in this area of research.

Throughout the subsequent analysis, we adopt the symbol C to represent a generic positive constant that could potentially rely on variables such as n , α , β , u , and v . It is important to emphasize that the specific value of this constant might vary from one line of reasoning to another.

2 Proof of Theorem 1

The purpose of this section is to categorize positive solutions for the integral system (4). To simplify the discussion, we introduce some notation. Specifically, for λ > 0 , we denote

B r ( x ) = { y R n + 1 y x < r , x R n + 1 } , B r + ( x ) = { y = ( y 1 , y 2 , , y n + 1 ) B r ( x ) y n + 1 > 0 , x R n } .

Let z R + n + 1 and r > 0 , for all x R + n + 1 and y R + n + 1 set

x z , r = r 2 ( x z ) x z 2 + z and y z , r = r 2 ( y z ) y z 2 + z ,

then we have

x z , r y z , r = r 2 x y x z y z , y z x y z , r = x z x z , r y ,

and

( x z , r ) n = r x z 2 x n .

Define the Kelvin-type transformations

u z , r ( x ) = r x z λ u ( x z , r )

and

v z , r ( x ) = r x z λ u ( x z , r ) .

Lemma 2

Suppose (u, v) is a pair of positive solutions for the integral system (4). For any z R + n + 1 and z > 0 , we can establish the following relationship:

(5) u ( x ) u z , r ( x ) = B r + ( z ) K ( z , r , y , x ) f ( u ( y ) , v ( y ) ) r y z 2 ( n + 1 ) + 2 β λ f r y z λ u z , r ( y ) , r y z λ v z , r ( y ) d y ,

for z R + n + 1 . Similarly, we can derive that

(6) v ( x ) v z , r ( x ) = B r + ( z ) K ( z , r , y , x ) f ( u ( y ) , v ( y ) ) r y z 2 ( n + 1 ) + 2 β λ f r y z λ u z , r ( y ) , r y z λ v z , r ( y ) d y ,

where

K ( z , r , y , x ) = y n + 1 β x y λ r x y λ y n + 1 β x y z , r λ .

Proof

Without loss of generality, we only prove (5),

(7) u ( x ) = R + n + 1 y n + 1 β f ( u ( y ) , v ( y ) ) x y λ d y = B r + ( z ) y n + 1 β f ( u ( y ) , v ( y ) ) x y λ d y + R + n + 1 \ B r + ( z ) y n + 1 β f ( u ( y ) , v ( y ) ) x y λ d y = B r + ( z ) y n + 1 β f ( u ( y ) , v ( y ) ) x y λ d y + B r + ( z ) ( y z , r ) n + 1 β f ( u ( y z , r ) , v ( y z , r ) ) x y z , r λ d y z , r = B r + ( z ) y n + 1 β f ( u ( y ) , v ( y ) ) x y λ d y + B r + ( z ) r y z 2 ( n + 1 ) + 2 β y n + 1 β f ( u ( y z , r ) , v ( y z , r ) ) x y z , r λ d y ,

then we can know that

(8) u z , r ( x ) = r x z λ B r + ( z ) y n + 1 β f ( u ( y ) , v ( y ) ) x z , r y λ d y + r x z λ B r + ( z ) r y z 2 ( n + 1 ) + 2 β y n + 1 β f ( u ( y z , r ) , v ( y z , r ) ) x z , r y z , r λ d y = B r + ( z ) r y z λ y n + 1 β f ( u ( y ) , v ( y ) ) x y λ d y + B r + ( z ) r y z 2 ( n + 1 ) + 2 β λ y n + 1 β f ( u ( y z , r ) , v ( y z , r ) ) x y λ d y .

Together with (7) and (8), we have

u ( x ) u z , r ( x ) = B r + ( z ) K ( z , r , y , x ) f ( u ( y ) , v ( y ) ) r y z 2 ( n + 1 ) + 2 β λ f r y z λ u z , r ( y ) , r y z λ v z , r ( y ) d y .

Similarly, we have

v ( x ) v z , r ( x ) = B r + ( z ) K ( z , r , y , x ) g ( u ( y ) , v ( y ) ) r y z 2 ( n + 1 ) + 2 β λ g r y z λ u z , r ( y ) , r y z λ v z , r ( y ) d y ,

where

K ( z , r , y , x ) = y n + 1 β x y λ r x y λ y n + 1 β x y z , r λ .

Lemma 3

Fix r small for 0 < δ r , there exists a constant c > 0 , such that

(9) u z , r ( x ) u ( x ) > 1 , f o r x B δ + ( z ) ,

and

(10) v z , r ( x ) v ( x ) > 1 , f o r x B δ + ( z ) .

Proof

Given that ( u , v ) is a pair of positive solutions for the integral system (4) and f > 0 is strictly increasing on ( 0 , + ) × ( 0 , + ) . We can safely assume, without loss of generality, that z = 0 . Under this assumption, it follows that for all x 1 , there exists a positive constant C 1 ¯ > 0 such that the following inequality holds:

u ( x ) = R + n + 1 y n + 1 β f ( u ( y ) , v ( y ) ) x y λ d y B 1 2 + ( 0 ) y n + 1 β f ( u ( y ) , v ( y ) ) x y λ d y C x λ B 1 2 + ( 0 ) y n + 1 β f ( u ( y ) , v ( y ) ) d y C 1 ¯ x λ .

Fix r small for 0 < δ r and x B δ + ( 0 ) . It can be easily verified that x 0 , r takes on large values. As a result, we can deduce that

u 0 , r ( x ) = r x λ u ( x 0 , r ) r x λ C 1 ¯ x 0 , r λ C 1 ¯ r λ .

Now, we choose r sufficiently small, such that

u z , r ( x ) u ( x ) C 1 ¯ r λ max x B δ + ( 0 ) u ( x ) > 1 .

Similar to (9), we can derive that for x B δ + ( 0 ) , we have

v z , r ( x ) v ( x ) > 1 .

Lemma 4

For z R + n + 1 there exists ε 0 ( z ) > 0 . This ensures that for any r ( 0 , ε 0 ( z ) ] the following condition holds true:

(11) u ( x ) u z , r ( x ) x B r + ( z )

and

(12) v ( x ) v z , r ( x ) x B r + ( z ) .

Proof

For z R + n + 1 and r > 0 , define

B u = { x B r + ( z ) u ( x ) > u z , r ( x ) } , B v = { x B r + ( z ) v ( x ) > v z , r ( x ) } .

Now, we will prove that for sufficiently small r , B u = and B v = . We need to estimate u ( x ) u z , r ( x ) L q ( B u )  and v ( x ) v z , r ( x ) L q ( B v ) . At first, estimate u ( x ) u z , r ( x ) L q ( B u ) .

By the monotonicity of F , we derive that

f r y z λ u ( y ) , r y z λ v ( y ) = f r y z λ u ( y ) , r y z λ v ( y ) r y z λ u ( y ) p 1 r y z λ v ( y ) q 1 r y z λ u ( y ) p 1 r y z λ v ( y ) q 1 r y z ( λ ) ( p 1 + q 1 ) f ( u ( y ) , v ( y ) ) u p 1 ( y ) v q 1 ( y ) u p 1 ( y ) v q 1 ( y ) = r y z ( λ ) ( p 1 + q 1 ) f ( u ( y ) , v ( y ) ) .

There are four situations as follows.

(i) If u ( y ) u z , r ( y ) and v ( y ) v z , r ( y ) , then by the monotonicity of f and F , we have

(13) f r y z λ u z , r ( y ) , r y z λ v z , r ( y ) f r y z λ u z , r ( y ) , r y z λ v ( y ) u z , r ( y ) u ( y ) .

Since

f r y z λ u z , r ( y ) , r y z λ v ( y ) u z , r ( y ) u ( y ) r y z λ u z , r ( y ) p 1 r y z λ v ( y ) u z , r ( y ) u ( y ) q 1 f r y z λ u ( y ) , r y z λ v ( y ) r y z λ u ( y ) p 1 r y z λ v ( y ) q 1 .

Combining (13), we deduce that

(14) f r y z λ u z , r ( y ) , r y z λ v z , r ( y ) f r y z λ u ( y ) , r y z λ v ( y ) u z , r ( y ) u ( y ) p 1 + q 1 r y z ( λ ) ( p 1 + q 1 ) f ( u ( y ) , v ( y ) ) u z , r ( y ) u ( y ) p 1 + q 1 .

Then, using (14), we have

f ( u ( y ) , v ( y ) ) r y z 2 ( n + 1 ) + 2 β λ f r y z λ u z , r ( y ) , r y z λ v z , r ( y ) f ( u ( y ) , v ( y ) ) f ( u ( y ) , v ( y ) ) u z , r ( y ) u ( y ) p 1 + q 1 = f ( u ( y ) , v ( y ) ) u p 1 + q 1 ( y ) ( u p 1 + q 1 ( y ) u z , r p 1 + q 1 ( y ) ) .

By Lemma 3, we infer that

y { y R + n + 1 u ( y ) u z , r ( y ) } { y R + n + 1 v ( y ) v z , r ( y ) } B r + ( z ) \ B δ + ( z ) ,

where δ is given in Lemma 3. We observe that f is strictly increasing and u > 0 is continuous. Consequently, there exists a constant C 2 ¯ , which may depend on r and δ satisfying the following condition:

(15) f ( u ( y ) , v ( y ) ) v p 1 + q 1 ( y ) C 2 ¯ < + , y B r + .

Then, applying the mean value theorem, we deduce from (15) that

f ( u ( y ) , v ( y ) ) r y z 2 ( n + 1 ) + 2 β λ f r y z λ u z , r ( y ) , r y z λ v z , r ( y ) C 2 ¯ ( u p 1 + q 1 ( y ) u z , r p 1 + q 1 ( y ) ) C u p 1 + q 1 1 ( y ) ( u ( y ) u z , r ( y ) ) .

(ii) If u ( y ) u z , r ( y ) and v ( y ) > v z , r ( y ) , then we have

f r y z λ u z , r ( y ) , r y z λ v z , r ( y ) r y z λ u z , r ( y ) p 1 r y z λ v z , r ( y ) q 1 f r y z λ u ( y ) , r y z λ v ( y ) r y z λ u ( y ) p 1 r y z λ v ( y ) q 1 r y z ( λ ) ( p 1 + q 1 ) f ( u ( y ) , v ( y ) ) .

So we have

f ( u ( y ) , v ( y ) ) r y z 2 ( n + 1 ) + 2 β λ f r y z λ u z , r ( y ) , r y z λ v z , r ( y ) f ( u ( y ) , v ( y ) ) f ( u ( y ) , v ( y ) ) u z , r ( y ) u ( y ) p 1 + q 1 v z , r ( y ) v ( y ) p 1 + q 1 f ( u ( y ) , v ( y ) ) 1 u z , r ( y ) u ( y ) p 1 + q 1 v z , r ( y ) v ( y ) p 1 + q 1 f ( u ( y ) , v ( y ) ) 1 u z , r ( y ) u ( y ) p 1 + q 1 + f ( u ( y ) , v ( y ) ) 1 v z , r ( y ) v ( y ) p 1 + q 1 = f ( u ( y ) , v ( y ) ) u p 1 + q 1 ( y ) ( u z , r p 1 + q 1 ( y ) u p 1 + q 1 ( y ) ) + f ( u ( y ) , v ( y ) ) v p 1 + q 1 ( y ) ( v z , r p 1 + q 1 ( y ) v p 1 + q 1 ( y ) ) C 3 ¯ ( u p 1 + q 1 ( y ) u z , r p 1 + q 1 ( y ) ) + C 4 ¯ ( u p 1 + q 1 ( y ) u z , r p 1 + q 1 ( y ) ) C [ u p 1 + q 1 1 ( y ) ( u ( y ) u z , r ( y ) ) + v p 1 + q 1 1 ( y ) ( v ( y ) v z , r ( y ) ) ] .

(iii) If u ( y ) < u z , r ( y ) and v ( y ) v z , r ( y ) , then we can change the roles of u , v in case (i) and obtain

f ( u ( y ) , v ( y ) ) r y z 2 ( n + 1 ) + 2 β λ f r y z λ u z , r ( y ) , r y z λ v z , r ( y ) C v p 1 + q 1 1 ( y ) ( v ( y ) v z , r ( y ) ) .

(iv) If u ( y ) < u z , r ( y ) and v ( y ) < v z , r ( y ) , then

f r y z λ u z , r ( y ) , r y z λ v z , r ( y ) f r y z λ u ( y ) , r y z λ v ( y ) r y z ( λ ) ( p 1 + q 1 ) ,

we have

f ( u ( y ) , v ( y ) ) r y z 2 ( n + 1 ) + 2 β λ f r y z λ u z , r ( y ) , r y z λ v z , r ( y ) 0 .

Therefore, for every x B u , we can infer from cases (i) to (iv) that

u ( x ) u z , r ( x ) = B r + ( z ) K ( z , r , y , x ) f ( u ( y ) , v ( y ) ) r y z 2 ( n + 1 ) + 2 β λ f r y z λ u z , r ( y ) , r y z λ v z , r ( y ) d y C B u K ( z , r , y , x ) u p 1 + q 1 1 ( y ) ( u ( y ) u z , r ( y ) ) d y + B v K ( z , r , y , x ) v p 1 + q 1 1 ( y ) ( v ( y ) v z , r ( y ) ) d y C B u y n + 1 β x y λ u p 1 + q 1 1 ( y ) ( u ( y ) u z , r ( y ) ) d y + B v y n + 1 β x y λ v p 1 + q 1 1 ( y ) ( v ( y ) v z , r ( y ) ) d y .

Similarly, we have

v ( x ) v z , r ( x ) = B r + ( z ) K ( z , r , y , x ) f ( u ( y ) , v ( y ) ) r y z 2 ( n + 1 ) + 2 β λ f r y z λ u z , r ( y ) , r y z λ v z , r ( y ) d y C B u y n + 1 β x y λ u p 2 + q 2 1 ( y ) ( u ( y ) u z , r ( y ) ) d y + B v y n + 1 β x y λ v p 2 + q 2 1 ( y ) ( v ( y ) v z , r ( y ) ) d y .

By the HLS inequality and the Hölder inequality, we have

u u z , r L p 1 + q 1 + 1 ( B u ) C u L 2 ( n + 1 ) λ ( B u ) p 1 + q 1 1 u u z , r L p 1 + q 1 + 1 ( B u ) + v L 2 ( n + 1 ) λ ( B v ) p 2 + q 2 1 v v z , r L p 2 + q 2 + 1 ( B v )

and

v ( x ) v z , r ( x ) L p 2 + q 2 ( B v ) C u L 2 ( n + 1 ) λ ( B u ) p 1 + q 1 1 u u z , r L p 1 + q 1 + 1 ( B u ) + v L 2 ( n + 1 ) λ ( B v ) p 2 + q 2 1 v v z , r L p 2 + q 2 + 1 ( B v ) .

Since 2 ( n + 1 ) λ > 1 and ( u , v ) C 0 ( R + n + 1 ) × C 0 ( R + n + 1 ) , there exists ε 0 ( z ) such that for 0 < r < ε 0 ( z ) ,

C u ( x ) L 2 ( n + 1 ) λ ( B u ) p 1 + q 1 1 1 4 , C v ( x ) L 2 ( n + 1 ) λ ( B v ) p 2 + q 2 1 1 4 ,

we deduce that

u ( x ) u z , r ( x ) L p 1 + q 1 + 1 ( B u ) = v ( x ) v z , r ( x ) L p 2 + q 2 + 1 ( B v ) = 0 ,

i.e., B u and B v have measure 0 and hence empty.

This completes the proof of Lemma 4.□

Continuing the moving sphere procedure, we iterate as long as conditions (11) and (12) hold. For each fixed z R + n + 1 , define

(16) r ¯ ( z ) = sup { r > 0 u u z , r in B μ + ( z ) , v v z , r in B μ + ( z ) , 0 < μ r } .

By Lemma 4, r ¯ ( z ) is well defined and 0 < r ¯ ( z ) + for any z R + n + 1 .

To establish our proof, we require the following lemma, which plays a crucial role.

Lemma 5

If r ¯ ( z ) < for some z ¯ R + n + 1 , then

(17) u ( x ) = u z ¯ , r ¯ ( x ) , f o r x B r ¯ ( z ¯ ) + ( z ¯ ) ,

and

(18) v ( x ) = v z ¯ , r ¯ ( x ) , f o r x B r ¯ ( z ¯ ) + ( z ¯ ) .

Proof

Without loss of generality, assuming z ¯ = 0 , for convenience in notation, we set r ¯ = r ¯ ( 0 ) . Consider the opposite scenario where u 0 , r ¯ u 0 in B r ¯ + ( 0 ) and v 0 , r ¯ v 0 in B r ¯ + ( 0 ) but at least one of them is not identically zero. Without loss of generality, let us assume that v 0 , r ¯ v 0 but v 0 , r ¯ v is not identically zero in B r ¯ + ( 0 ) . We will now derive a contradiction based on the definition of z ¯ .

We first show that

(19) u 0 , r ¯ ( x ) u ( x ) > 0 , x B r ¯ + ( 0 ) ,

and

(20) v 0 , r ¯ ( x ) v ( x ) > 0 , x B r ¯ + ( 0 ) .

Indeed, if there exists a point y 0 B r ¯ + ( 0 ) such that v 0 , r ¯ ( y 0 ) v ( y 0 ) > 0 , we can infer from the continuity of v that there exist small positive numbers η and c 0 such that

(21) B η + ( 0 ) B r ¯ + ( 0 ) and v 0 , r ¯ ( y ) v ( y ) c 0 > 0 , y B η + ( 0 ) .

For any y B η + ( y 0 ) , using (16) and (20) and the strict monotonicity of f , we can deduce that

f r y λ u 0 , r ¯ ( y ) , r y λ v 0 , r ¯ ( y ) > f r y λ u 0 , r ¯ ( y ) , r y λ v ( y )

(22) f r y λ u ( y ) , r y λ v ( y ) r y ( λ ) ( p 1 + q 1 ) f ( u ( y ) , v ( y ) ) u p 1 ( y ) v q 1 ( y ) u p 1 ( y ) v q 1 ( y ) = r y ( λ ) ( p 1 + q 1 ) f ( u ( y ) , v ( y ) ) .

For any x B r ¯ + ( 0 ) , since K ( 0 , r ¯ , y , x ) > 0 , one can derive from (5) and (21) that

(23) u 0 , r ¯ ( x ) u ( x ) = B r + ( 0 ) K ( 0 , r ¯ , y , x ) r y 2 ( n + 1 ) + 2 β λ f r y λ u 0 , r ( y ) , r y λ v 0 , r ( y ) f ( u ( y ) , v ( y ) ) d y B η + ( y 0 ) K ( 0 , r ¯ , y , x ) r y 2 ( n + 1 ) + 2 β λ f r y λ u 0 , r ( y ) , r y λ v 0 , r ( y ) f ( u ( y ) , v ( y ) ) d y > 0 .

Therefore, we obtain equation (19). Equation (6) and the positivity of K ( 0 , r ¯ , y , x ) lead to the immediate derivation of equation (20). Moreover, equations (22) and (23) imply that there exists a 0 < ι < r ¯ small enough, and for any x B ι + ( 0 ) , we have

(24) u 0 , r ¯ ( x ) u ( x ) c 1 > 0 ,

which implies that u 0 , r ¯ ( x ) u ( x ) has a strictly positive lower bound in a small neighborhood of 0. Select δ 1 ( 0 , min { δ 2 , r ¯ 2 } ) based on the value of δ provided by Lemma 3. By (19) and (24), we define

(25) m 1 = inf x B r ¯ δ 1 + ( 0 ) \ { 0 } ¯ ( u 0 , r ¯ u ) ( x ) > 0 .

Then, by the continuity of u 0 , r ¯ u with respect to r , we can find a positive constant δ 2 satisfying 0 < δ 2 < δ 1 0 6 . This ensures that for every r [ r ¯ , r ¯ + δ 2 ) , we have

(26) u 0 , r ¯ ( x ) u ( x ) m 1 2 > 0 , x B r ¯ δ 1 + ( 0 ) ¯ \ { 0 } .

Then, our goal is to demonstrate the existence of a positive constant m 2 such that

(27) v 0 , r ¯ ( x ) v ( x ) m 2 > 0 , x B r ¯ δ 1 + ( 0 ) ¯ \ { 0 } .

By (6), for any x B r ¯ δ 1 + ( 0 ) ¯ \ { 0 } , one can deduce that

(28) v 0 , r ( x ) v ( x ) = B r + ( 0 ) K ( 0 , r , y , x ) r y 2 ( n + 1 ) + 2 β λ g r y α n u 0 , r ( y ) , r y λ v 0 , r ( y ) g ( u ( y ) , v ( y ) ) d y = B r ¯ δ 1 + ( 0 ) K ( 0 , r , y , x ) r y 2 ( n + 1 ) + 2 β λ g r y λ u 0 , r ( y ) , r y λ v 0 , r ( y ) g ( u ( y ) , v ( y ) ) d y + B r + ( 0 ) \ B r ¯ δ 1 + ( 0 ) K ( 0 , r , y , x ) r y 2 ( n + 1 ) + 2 β λ g r y λ u 0 , r ( y ) , r y λ v 0 , r ( y ) g ( u ( y ) , v ( y ) ) d y = I 1 ( x ) + I 2 ( x ) .

We first take care of I 1 ( x ) . For any y B r ¯ δ 1 + ( 0 ) ¯ \ { 0 } , it follows from (26) and the monotonicity of g and G that

g r y λ u 0 , r ( y ) , r y z λ v 0 , r ( y )

(29) > g r y λ u ( y ) , r y λ v 0 , r ( y ) g r y λ u ( y ) , r y λ v ( y ) r y ( λ ) ( p 1 + q 1 ) g ( u ( y ) , v ( y ) ) u p 1 ( y ) v q 1 ( y ) u p 1 ( y ) v q 1 ( y ) = r y ( λ ) ( p 1 + q 1 ) g ( u ( y ) , v ( y ) ) .

By (26) and (29), along with the strict monotonicity of g , we can conclude that there exists a positive constant m 2 depending on n , α , r , and r ¯ such that

I 1 ( x ) 2 m 2 > 0 .

To estimate I 2 ( x ) , we consider y B r + ( 0 ) \ B r ¯ δ 1 + ( 0 ) ¯ . We can use the fact that g is strictly increasing and u is continuous to establish the existence of a uniform constant C on compact set B r + ( 0 ) \ B r ¯ δ 1 + ( 0 ) ¯ . This constant depends n , β , r , and r ¯ such that

K ( 0 , r ¯ , y , x ) r y 2 ( n + 1 ) + 2 β λ g r y λ u 0 , r ( y ) , r y λ v 0 , r ( y ) g ( u ( y ) , v ( y ) ) C .

By combining the previous estimate with equation (28), we can show that for sufficiently small δ 1 > 0 ,

v 0 , r ¯ ( x ) v ( x ) = I 1 ( x ) + I 2 ( x ) 2 m 2 C 5 ¯ [ r n 1 ( r ¯ δ 1 ) n 1 ] = 2 m 2 C 6 ¯ [ r ( r ¯ δ 1 ) ] 2 m 2 C 6 ¯ [ δ 1 + δ 2 ] m 2 > 0 .

Therefore, (25) is established. Thus, we can use the narrow region B r + ( 0 ) \ B r ¯ δ 1 + ( 0 ) instead of B r + ( 0 ) \ B r δ 1 + ( 0 ) in Lemma 4, following a similar proof process as in Lemma 4. Then, we conclude that B u = B v = for all r [ r ¯ , r ¯ + δ 2 ] , i.e.,

u 0 , r ( x ) u ( x ) 0 , x B r + ( 0 ) ,

and

v 0 , r ( x ) v ( x ) 0 , x B r + ( 0 ) ,

which contradicts the definition of r ¯ . As a consequence, in the case 0 < r ¯ < + , we must have

u 0 , r ¯ ( x ) = u ( x ) , x B r ¯ + ( 0 ) ,

and

v 0 , r ¯ ( x ) = v ( x ) , x B r ¯ + ( 0 ) .

This concludes the proof of Lemma 5.□

Additionally, we must consider about the limiting radius r ¯ ( z ) .

Lemma 6

If r ¯ ( z ¯ ) = + for some z ¯ R + n + 1 , then r ¯ ( z ) = + for all z R + n + 1 .

Proof

Since r ¯ ( z ¯ ) = + , recalling the definition of r ¯ , we derive

u z ¯ , r ( x ) u ( x ) , x B r + ( z ¯ ) , 0 < r < + ,

i.e.,

u ( x ) u z ¯ , r ( x ) , x ( z ¯ ) r , 0 < r < + .

It follows immediately that

(30) lim x x λ u ( x ) = + .

On the other hand, if we assume r ¯ ( z ) < + for some z R + n + 1 , then by Lemma 5, we arrive at

lim x x λ u ( x ) = lim x x λ u z ¯ , r ( z ) ( x ) = ( r ( z ) ) λ u ( z ) < + ,

which contradicts (30).

Thus, we have completed the proof of Lemma 6.□

Now, we are prepared to present a proof of Theorem 1.

Proof

We will proceed with the proof by considering two different possible cases.

Case (i). r ¯ ( z ) < , for all z R + n .

From Lemma 5, we infer that

(31) u ( x ) = u z ¯ , r ¯ ( z ¯ ) ( x ) , x B r ¯ + ( z ¯ ) .

By exploiting the conformal invariance property of (4) and applying a calculus lemma (Lemma 5.8 in Li [20]) along with (31), we can infer that for any x R + n + 1 ,

u ( x , 0 ) = c 1 ( d 2 + x x 0 2 ) λ 2 ,

for c 1 > 0 , d > 0 , x 0 R + n + 1 . Similarly, for any x R + n + 1 , we have

v ( x , 0 ) = c 2 ( d 2 + x x 0 2 ) λ 2 ,

for some c 2 > 0 , d > 0 and x 0 R + n + 1 . Now, we show that F and G are constants. It follows from (5), (17), and (18) that

u z , r ¯ ( x ) u ( x ) = B r ¯ + ( z ) K ( 0 , r ¯ , y , x ) r ¯ y 2 ( n + 1 ) + 2 β λ f r ¯ y λ u 0 , r ¯ ( y ) , r ¯ y λ v 0 , r ¯ ( y ) f ( u ( y ) , v ( y ) ) d y = B r ¯ + ( z ) K ( 0 , r ¯ , y , x ) f r ¯ y λ u 0 , r ¯ ( y ) , r ¯ y λ v 0 , r ¯ ( y ) r ¯ y λ u 0 , r ¯ ( y ) p 1 r ¯ y λ v 0 , r ¯ ( y ) q 1 u 0 , r ¯ p 1 ( y ) v 0 , r ¯ q 1 ( y ) d y B r ¯ + ( z ) K ( 0 , r ¯ , y , x ) f ( u ( y ) , v ( y ) ) u p 1 ( y ) v q 1 ( y ) u p 1 ( y ) v q 1 ( y ) d y = B r ¯ + ( z ) K ( 0 , r ¯ , y , x ) f r ¯ y λ u 0 , r ¯ ( y ) , r ¯ y λ v 0 , r ¯ ( y ) r ¯ y λ u 0 , r ¯ ( y ) p 1 r ¯ y λ v 0 , r ¯ ( y ) q 1 f ( u ( y ) , v ( y ) ) u p 1 ( y ) v q 1 ( y ) u p 1 ( y ) v q 1 ( y ) d y 0 .

In the aforementioned derivation, we have made use of the positivity of K ( 0 , r ¯ , y , x ) and the monotonicity of F . Hence, for any x B r ¯ + ( z ) , we obtain

f r ¯ y λ u 0 , r ¯ ( y ) , r ¯ y λ v 0 , r ¯ ( y ) r ¯ y λ u 0 , r ¯ ( y ) p 1 r ¯ y λ v 0 , r ¯ ( y ) q 1 = f ( u ( y ) , v ( y ) ) u p 1 ( y ) v q 1 ( y ) .

Therefore, for some positive constant C 1 , we have

F ( t ) = f ( s , t ) s p 1 t q 1 = C 1 , t ( 0 , max x R + n + 1 v ( x ) ] .

In a similar manner, we can deduce that for some positive constant C 2 ,

G ( t ) = f ( t ) s p 2 t q 2 = C 2 , t ( 0 , max x R + n + 1 u ( x ) ] .

This completes our proof of Theorem 1.

Case (ii). By C a s e ( i ) , we know that r ( x ) = for all x R n . Then, u C 1 ˜ , v C 2 ˜ ( C 1 ˜ a n d C 2 ˜ are some constants) and f ( C 1 ˜ , C 2 ˜ ) = 0 , g ( C 1 ˜ , C 2 ˜ ) = 0 . This contradicts f , g : R + × R + R + ; thus, this case is impossible.□

3 Conclusion

In this article, we study a weighted integral system and directly obtain the classification of positive solutions in the upper half-space using the method of moving spheres.

Acknowledgements

The authors would like to express their sincere thanks to Professor Zhao Liu for his valuable suggestions and useful comments, from which the authors have benefited on this manuscript, and the authors express their gratitude to the editors and reviewers for their serious and responsible comments and pertinent suggestions.

  1. Funding information: The authors have been supported by the Science and Technology Project of Education Department of Jiangxi Province (GJJ2201354, GJJ218419), the Teaching Reform Project of Jiangxi Science and Technology Normal University (XJJG-2022-50-07), the Doctoral Science Foundation, Jiangxi Science and Technology Normal University (2022BSQD13), and Basic Education Research Project in Jiangxi Province (SZUJKSX2024-1022).

  2. Author contributions: All authors conceived the study, drafted the manuscript, and approved the final manuscript.

  3. Conflict of interest: The authors state no conflict of interest.

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Received: 2023-11-13
Revised: 2024-08-03
Accepted: 2024-08-07
Published Online: 2024-09-16

© 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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  112. Monotonicity, convexity, and Maclaurin series expansion of Qi's normalized remainder of Maclaurin series expansion with relation to cosine
  113. On k-prime graphs
  114. On the existence of tripartite graphs and n-partite graphs
  115. Classifying pentavalent symmetric graphs of order 12pq
  116. Almost periodic functions on time scales and their properties
  117. Some results on uniqueness and higher order difference equations
  118. Coding of hypersurfaces in Euclidean spaces by a constant vector
  119. Cycle integrals and rational period functions for Γ0+(2) and Γ0+(3)
  120. Degrees of (L, M)-fuzzy bornologies
  121. A matrix approach to determine optimal predictors in a constrained linear mixed model
  122. On ideals of affine semigroups and affine semigroups with maximal embedding dimension
  123. Solutions of linear control systems on Lie groups
  124. A uniqueness result for the fractional Schrödinger-Poisson system with strong singularity
  125. On prime spaces of neutrosophic extended triplet groups
  126. On a generalized Krasnoselskii fixed point theorem
  127. On the relation between one-sided duoness and commutators
  128. Non-homogeneous BVPs for second-order symmetric Hamiltonian systems
  129. Erratum
  130. Erratum to “Infinitesimals via Cauchy sequences: Refining the classical equivalence”
  131. Corrigendum
  132. Corrigendum to “Matrix stretching”
  133. Corrigendum to “A comprehensive review of the recent numerical methods for solving FPDEs”
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