Home The Neumann problem on the domain in 𝕊3 bounded by the Clifford torus
Article Open Access

The Neumann problem on the domain in 𝕊3 bounded by the Clifford torus

  • Jeffrey S. Case , Eric Chen EMAIL logo , Yi Wang , Paul Yang and Po-Lam Yung
Published/Copyright: June 27, 2023

Abstract

In this study, the solution of the Neumann problem associated with the CR Yamabe operator on a subset Ω of the CR manifold S 3 bounded by the Clifford torus Σ is discussed. The Yamabe-type problem of finding a contact form on Ω which has zero Tanaka-Webster scalar curvature and for which Σ has a constant p -mean curvature is also discussed.

1 Introduction

We view S 3 = { ( ζ 1 , ζ 2 ) C 2 : ζ 1 2 + ζ 2 2 = 1 } as the boundary of the unit ball in C 2 , and equip it with a pseudohermitian structure associated with the contact form:

θ i ¯ ( ζ 2 1 ) = i 2 ( ζ 1 d ζ ¯ 1 + ζ 2 d ζ ¯ 2 ζ ¯ 1 d ζ 1 ζ ¯ 2 d ζ 2 ) .

Let Ω S 3 be the domain

(1.1) Ω = { ( ζ 1 , ζ 2 ) S 3 : ζ 1 < ζ 2 } ,

whose boundary is given by:

Σ Ω = ( ζ 1 , ζ 2 ) S 3 : ζ 1 = ζ 2 = 1 2 ,

the Clifford torus.

On Ω , we give an explicit solution to the Neumann problem:

(1.2) L u = 0 in Ω , ν u = h on Ω ,

where L denotes the CR Yamabe operator (2.3) on ( S 3 , θ ) , and ν denotes the one-sided horizontal normal derivative for a function u C ( Ω ¯ ) , i.e.,

ν u ( ζ ) lim t 0 + u ( γ ζ ( t ) ) u ( ζ ) t , ζ Ω ,

where γ ζ ( t ) is a curve so that γ ζ ( 0 ) = ζ and γ ( 0 ) is the inward horizontal normal ν to Ω at ζ Ω .

Initially, we assume that our boundary data h is smooth on Σ , i.e., h C ( Ω ) , and we solve (1.2). Our solution is given in terms of a suitable single-layer potential S , defined for f C 1 ( Σ ) by:

S f ( ζ ) η Σ G ( ζ , η ) f ( η ) d σ ( η ) , ζ Ω ¯ ,

where G ( ζ , η ) is the Green’s function of the CR Yamabe operator L , and d σ is the surface measure on Σ corresponding to the volume form ι ν θ d θ .

Theorem 1.1

If h C ( Σ ) , then there exists a unique u C ( Ω ¯ ) that solves the Neumann problem (1.2). Furthermore, u is given by S f on Ω ¯ for some f C ( Σ ) , where f is determined by h from

(1.3) f = 1 2 I + N 1 h ,

and N is the singular integral operator given by:

N f ( ζ ) p.v. η Σ ( ν ) ζ G ( ζ , η ) f ( η ) d σ ( η ) , ζ Σ .

Some of the arguments follow those in [16], where (1.2) is solved on flag domains of H 1 . In contrast to the Euclidean case [5], the operator N is not a compact operator on L 2 ( Σ ) ; it is only bounded on L 2 ( Σ ) , but not smoothing of any positive order. As a result, a more careful analysis is necessary to show that 1 2 I + N is invertible on L 2 ( Σ ) (and all higher-order L 2 -based Sobolev spaces).

A general result of Nhieu [14] implies the existence of a solution u to (1.2) in the Folland-Stein [6] space S 1 , 2 ( Ω ) of functions u L 2 ( Ω ) whose horizontal gradient is also in L 2 ( Ω ) . We recover this result using the single-layer potential.

Theorem 1.2

If h L 2 ( Σ ) , then there exists a unique u S 1 , 2 ( Ω ) such that (1.2) is satisfied in the weak sense, i.e.,

Ω b u , b ϕ + R 4 u ϕ θ d θ + Σ h ϕ d σ = 0 ,

for all ϕ C ( Ω ¯ ) . Furthermore, u is given by S f on Ω , where f L 2 ( Σ ) is determined from h by (1.3).

The key observation in the proof of Theorem 1.2 is that the single-layer potential extends to a bounded linear map S : L 2 ( Σ ) S 1 , 2 ( Ω ) .

Our work is motivated in part by the desire to formulate and study the CR boundary Yamabe problem (cf. [3]): Given a closed CR three-manifold with boundary ( M 3 , T 1 , 0 ) , construct a Webster-flat contact form with respect to which the boundary has constant p -mean curvature H [2]. Given a contact form θ , the contact form θ ^ u 2 θ satisfies these properties if and only if u is a positive critical point of the functional : V R ,

( u ) M b u 2 + R 4 u 2 θ d θ + 1 3 M H u 2 d σ , V u C ( M ) : M u 3 d σ = 1 .

One way to construct such a contact form is to show that there is a smooth, positive function that realizes the CR boundary Yamabe constant:

Y ( M , T 1 , 0 ) inf { ( u ) : u V } .

If M has no characteristic points, then the Sobolev trace embedding theorem [14] implies that the restriction map C ( M ¯ ) u u M C ( M ) extends to a continuous linear map Tr : S 1 , 2 ( M ) L 2 ( M ) ; in particular, is well defined on S 1 , 2 ( M ) . As in the Riemannian case [4], it holds that Y ( M , T 1 , 0 ) > if and only if the Dirichlet eigenvalues of L are positive, i.e.,

λ 1 , D ( L ) inf ( u ) : u M = 0 , M u 2 θ d θ = 1 > 0 .

The positivity of λ 1 , D ( L ) also implies that the first Steklov eigenvalue

(1.4) μ 1 ( L ) inf ( u ) : M u 2 d σ = 1

of L is finite.

In Section 7, the equivalence of the signs of μ 1 ( L ) and Y ( M , T 1 , 0 ) is discussed, under the assumption that the minimizers of μ 1 ( L ) are smooth up to the boundary. Such regularity assumption can be verified in the case where M = Ω ¯ , where Ω is the domain in S 3 defined by (1.1). Note that H = 0 for the Clifford torus Σ = Ω .

Proposition 1.3

Let ( Ω , θ ) be the interior of the Clifford torus with the standard spherical contact form. If u S 1 , 2 ( Ω ) minimizes μ 1 ( L ) , then u C ( Ω ¯ ) .

Proposition 1.3 will be proved by observing that a minimizer u S 1 , 2 ( Ω ) of μ 1 ( L ) is a weak solution of

L u = 0 , on Ω , ν u = μ Tr ( u ) , in Ω ,

where μ = ( u ) is constant, and then using tools developed for the proof of Theorem 1.2.

Since the standard contact form on S 3 has positive Tanaka-Webster scalar curvature and is such that the Clifford torus is p -minimal, the Sobolev trace embedding theorem [14] implies that the boundary Yamabe constant Y ( Ω , T 1 , 0 ) is positive. We identify an explicit critical point of the functional : V R .

Theorem 1.4

Let ( Ω , T 1 , 0 , θ ) be the interior of the Clifford torus. Set

u F 1 2 1 2 , 1 2 ; 1 ; z 2 2 F 1 2 1 2 , 1 2 ; 1 ; 1 2 ,

where

F 1 2 ( a , b ; c ; x ) n = 0 ( a ) n ( b ) n ( c ) n n ! x n , ( a ) 0 1 , ( a ) n a ( a + 1 ) ( a + n 1 ) , if n 1 ,

is the standard Gaussian hypergeometric function. Then u 2 θ is a scalar flat contact form on Ω with respect to which Σ has constant p -mean curvature.

The solution of Theorem 1.4 is normalized so that θ and u 2 θ coincide on Σ .

This article is organized as follows. In Section 2, we give some background in pseudohermitian geometry and Fourier analysis, in the specific setting of the Clifford torus. In Section 3, we compute the single-layer potential in local coordinates. Theorems 1.1 and 1.2, Proposition 1.3, and Theorem 1.4 will be proved in Sections 4, 5, 6, and 7, respectively.

2 Setup and notations

2.1 Pseudohermitian geometry

A CR three-manifold is a pair ( M 3 , T 1 , 0 ) consisting of a real three-manifold M and a complex rank 1 distribution T 1 , 0 T C M . Let H Re ( T 1 , 0 T 0 , 1 ¯ ) denote the space of horizontal vectors. Then

J ( Z + Z ¯ ) i Z i Z ¯ ,

Z T 1 , 0 , defines an integrable almost complex structure on H . We say that ( M 3 , T 1 , 0 ) is nondegenerate if locally there is a real one-form θ such that ker θ = H and θ d θ is nowhere-vanishing. Nondegenerate CR three-manifolds are orientable [8, Lemma 19], and hence there is a global real one-form θ such that ker θ = H ; in this case, we call θ a contact form.

A (strictly pseudoconvex) pseudohermitian manifold is a triple ( M 3 , T 1 , 0 , θ ) consisting of a nondegenerate CR three-manifold ( M 3 , T 1 , 0 ) and a contact form θ such that d θ ( Z , Z ¯ ) > 0 for all nonzero Z T 1 , 0 . The Reeb vector field is the unique vector field T such that θ ( T ) = 1 a and d θ ( T , ) = 0 .

An admissible coframe for ( M 3 , T 1 , 0 , θ ) is a nowhere-vanishing local complex-valued one-form θ 1 such that θ 1 ( T ) = 0 and θ 1 ( Z ¯ ) = 0 for all Z T 1 , 0 . Set θ 1 ¯ θ 1 ¯ . Then { θ , θ 1 , θ 1 ¯ } is a local coframe for T C M . It follows that

d θ = i h 1 1 ¯ θ 1 θ 1 ¯ .

Note that h 1 1 ¯ > 0 . Let { T , Z 1 , Z 1 ¯ } be the dual frame to { θ , θ 1 , θ 1 ¯ } . This (globally) determines a positive definite inner product on H by:

Re a 1 Z 1 , Re b 1 Z 1 1 2 Re h 1 1 ¯ a 1 b 1 ¯ .

Let θ 1 be an admissible coframe for ( M 3 , T 1 , 0 , θ ) . Then there is a unique complex-valued one-form ω 1 1 such that

d θ 1 = θ 1 ω 1 1 + A 1 ¯ 1 θ θ 1 ¯ , d h 1 1 ¯ = ω 1 1 h 1 1 ¯ + ω 1 ¯ 1 ¯ h 1 1 ¯ ,

where ω 1 ¯ 1 ¯ ω ¯ 1 1 . The Tanaka-Webster connection is uniquely determined from Z 1 ω 1 1 Z 1 and T 0 by complex linearity. The pseudohermitian torsion is the globally-defined tensor A 11 θ 1 θ 1 . The Tanaka-Webster scalar curvature is the globally-defined function R determined by:

d ω 1 1 = R h 1 1 ¯ θ 1 θ 1 ¯ mod θ .

We say that θ is scalar flat if R = 0 .

Given a function f C ( M ) , we denote by b f the subgradient of f , i.e., the restriction of d f to H . The sublaplacian Δ b : C ( M ) C ( M ) is

Δ b b b ,

where b is the formal L 2 -adjoint of b with respect to θ d θ . Locally,

Δ b u = h 1 ¯ 1 ( ( Z 1 Z 1 ¯ + Z 1 ¯ Z 1 ) u ω 1 1 ( Z 1 ¯ ) Z 1 u ω 1 ¯ 1 ¯ ( Z 1 ) Z 1 ¯ u ) .

It is readily computed (cf. [12, equation (2.4)]) that

(2.1) M b u , b w θ d θ = M u Δ b w θ d θ + 2 Re M i u ( Z 1 w ) θ θ 1 .

Suppose that ( M 3 , T 1 , 0 , θ ) is a pseudohermitian three-manifold with boundary Σ M . A point p Σ is singular if T p Σ = H p . We say that Σ is nonsingular if it contains no singular points.

Suppose now that ( M 3 , T 1 , 0 , θ ) is a pseudohermitian three-manifold with nonsingular boundary Σ M . Assume additionally that Σ is oriented. Then there is a unique H T Σ -valued unit vector field e 1 such that ν J e 1 is inward-pointing. Let Z 1 be the unique section of T 1 , 0 along Σ such that e 1 = Re Z 1 . A straightforward computation using (2.1) gives

(2.2) M b u , b w θ d θ = M u Δ b w θ d θ M u ν w d σ ,

where d σ ι ν ( θ d θ ) .

Since the Tanaka-Webster connection preserves the contact form and the Levi form, e 1 e 1 is in the span of e 2 . The p-mean curvature is the function H defined by:

e 1 e 1 H ν .

The CR Yamabe operator L θ : C ( M ) C ( M ) is

L θ u Δ b u + R 4 u .

The CR Yamabe operator is conformally covariant [10, equation (3.1)]: If θ ˆ = w 2 θ , then

w 3 L θ ˆ u = L θ ( w u ) .

When the contact form is clear from context, we shall write L for L θ . The CR Robin operator B θ : C ( M ) C ( Σ ) is

B θ u ν u + H 3 u .

The conformal transformation law [11, Lemma 3.4] for the Tanaka-Webster connection implies that if θ ˆ = w 2 θ , then

w 2 B θ ˆ u = B θ ( w u ) .

It follows that the CR Yamabe functional : C ( M ) R ,

θ ( u ) M u L u θ d θ + Σ u B u d σ

is CR invariant; indeed,

w 2 θ ( u ) = θ ( w u )

for all positive w C ( M ) and all u C ( M ) . Equation (2.2) implies that

( u ) = M b u 2 + R 4 u 2 θ d θ + 1 3 Σ H u 2 d σ .

2.2 The standard sphere S 3

Let

θ = i ¯ ( ζ 2 1 ) = i 2 ( ζ 1 d ζ ¯ 1 + ζ 2 d ζ ¯ 2 ζ ¯ 1 d ζ 1 ζ ¯ 2 d ζ 2 )

be the contact form on S 3 . Denote

θ 1 ζ 2 d ζ 1 ζ 1 d ζ 2 .

Then

d θ = i θ 1 θ 1 ¯ ,

so h 1 1 ¯ = 1 . Since

d θ 1 = 2 i θ 1 θ ,

we have

ω 1 1 = 2 i θ .

It follows that

d ω 1 1 = 2 θ 1 θ 1 ¯ ,

so the Tanaka-Webster scalar curvature is R = 2 . The CR Yamabe operator is then

(2.3) L u ( Z 1 Z 1 ¯ + Z 1 ¯ Z 1 ) u + 1 2 u ,

where

Z 1 ζ ¯ 2 ζ 1 ζ ¯ 1 ζ 2 .

The fundamental solution to L on S 3 is [7, Theorem 2.1]

G ( ζ , η ) 1 8 π 1 ζ η ¯ 1 .

2.3 The Clifford torus

Recall the domain Ω S 3 is given by:

Ω = { ( ζ 1 , ζ 2 ) S 3 : ζ 1 < ζ 2 } .

Its boundary is the Clifford torus Σ = { ( ζ 1 , ζ 2 ) : ζ 1 = ζ 2 = 1 2 } . Note that Σ is contained in the open set { ( ζ 1 , ζ 2 ) S 3 : ζ 1 0 , ζ 2 0 } . On this open set, define a frame of horizontal vectors

e 1 i 2 ζ 1 ζ 2 ζ 1 ζ 2 Z 1 ζ ¯ 1 ζ ¯ 2 ζ 1 ζ 2 Z 1 ¯ = 2 Im ζ 2 ζ 1 ζ 1 ζ 1 ζ 1 ζ 2 ζ 2 ζ 2

and

ν J e 1 = 1 2 ζ 1 ζ 2 ζ 1 ζ 2 Z 1 + ζ ¯ 1 ζ ¯ 2 ζ 1 ζ 2 Z 1 ¯ = 2 Re ζ 2 ζ 1 ζ 1 ζ 1 ζ 1 ζ 2 ζ 2 ζ 2 .

The Reeb vector field on S 3 is

T i ζ 1 ζ 1 + ζ 2 ζ 2 ζ ¯ 1 ζ ¯ 1 ζ ¯ 2 ζ ¯ 2 = 2 Im ζ 1 ζ 1 + ζ 2 ζ 2 .

For t [ 0 , π 2 2 ) , set

Ω t = ζ S 3 : ζ 1 cos π 4 + t 2 < ζ 2 sin π 4 + t 2

and

Σ t = Ω t = ζ S 3 : ζ 1 = cos π 4 + t 2 , ζ 2 = sin π 4 + t 2 ,

so that Ω 0 = Ω and Σ 0 = Σ . Then ν is the inward normal to Ω t , and e 1 and T are tangent to Σ t . In fact, a defining function for Ω t is

ρ t ζ 1 2 cos 2 π 4 + t 2 ζ 2 2 sin 2 π 4 + t 2 ,

and the above expressions for e 1 and T show that

e 1 ρ t = T ρ t = 0 on  Σ t .

Furthermore,

ν ρ t = 2 cos π 4 + t 2 sin π 4 + t 2 = 2 2 cos ( 2 t ) < 0 ,

so ν = J e 1 is the inward-pointing normal to Ω t . Moreover, we readily compute that e 1 e 1 = 0 ; hence the Clifford torus is p -minimal.

2.4 Fourier analysis

To parametrize Σ t , it will be convenient to consider the lattice

Λ 2 π { ( m 2 , n 2 ) Z 2 : m n ( mod 2 ) } ,

and the associated abelian group R 2 Λ . A fundamental domain for R 2 Λ is given by [ π , π ) × [ π 2 , π 2 ) ; the identity element of R 2 Λ will be written as ( 0 , 0 ) using this identification. Any function defined on [ π , π ) × [ π 2 , π 2 ) can be lifted to R 2 Λ . For instance, we define, for ( u , v ) [ π , π ) × [ π 2 , π 2 ) , the natural norm for this problem, namely,

( u , v ) max { u , v 1 2 } ,

and it gives rise to a corresponding norm function on R 2 Λ .

The Lebesgue measure d u d v on [ π , π ) × [ π 2 , π 2 ) induces a Haar measure on R 2 Λ , with volume 2 π 2 , and the dual lattice 1 2 π Λ of 1 2 π Λ is given by:

1 2 π Λ = { ( m , n ) Z 2 : m n ( mod 2 ) } .

An orthonormal basis of L 2 ( R 2 Λ , d u d v ) is given by:

1 2 π 2 e i ( m u + n v ) : ( m , n ) 1 2 π Λ ,

and hence Parseval’s formula reads

u π v π 2 F ( u , v ) 2 d u d v = 1 2 π 2 ( m , n ) Z 2 m n ( mod 2 ) F ^ ( m , n ) 2 ,

where

F ^ ( m , n ) u π v π 2 F ( u , v ) e i ( m u + n v ) d u d v .

The convolution of two functions F and K on R 2 Λ is given by any of the two equivalent expressions:

F K ( u , v ) u π v π 2 F ( u u , v v ) K ( u , v ) d u d v = u π v π 2 F ( u , v ) K ( u u , v v ) d u d v ,

for every ( u , v ) R 2 Λ . We have

F K ^ ( m , n ) = F ^ ( m , n ) K ^ ( m , n ) .

Thus

u π v π 2 F K ( u , v ) 2 d u d v sup ( m , n ) Z 2 m n ( mod 2 ) K ^ ( m , n ) 2 u π v π 2 F ( u , v ) 2 d u d v .

2.5 Parametrizing Σ t

For 0 t π 100 , we will parametrize Σ t using

( u , v ) R 2 Λ Φ ( t , u , v ) cos π 4 + t 2 e i ( v + u ) , sin π 4 + t 2 e i ( v u ) Σ t . ( Figure 1 )

One can see that the above map is a well-defined bijection by noting that e i ( v + u ) = e i ( v u ) = 1 if and only if ( u , v ) Λ ; alternatively, one observes that the map

( u , v ) [ π , π ) × [ π 2 , π 2 ) ( v + u , v u ) ( R 2 π Z ) 2

is a bijection (Figure 2). We have Φ t = ν and Φ v = T .

Figure 1 
                  The dotted square is 
                        
                           
                           
                              
                                 
                                    
                                       [
                                       
                                          −
                                          π
                                          ,
                                          π
                                       
                                       ]
                                    
                                 
                                 
                                    2
                                 
                              
                           
                           {\left[-\pi ,\pi ]}^{2}
                        
                     . The tilted rectangle is the image of 
                        
                           
                           
                              
                                 (
                                 
                                    u
                                    ,
                                    v
                                 
                                 )
                              
                              ∈
                              
                                 [
                                 
                                    −
                                    π
                                    ,
                                    π
                                 
                                 ]
                              
                              ×
                              
                                 [
                                 
                                    −
                                    π
                                    ∕
                                    2
                                    ,
                                    π
                                    ∕
                                    2
                                 
                                 ]
                              
                              ↦
                              
                                 (
                                 
                                    v
                                    +
                                    u
                                    ,
                                    v
                                    −
                                    u
                                 
                                 )
                              
                              ∈
                              
                                 
                                    R
                                 
                                 
                                    2
                                 
                              
                           
                           \left(u,v)\in \left[-\pi ,\pi ]\times \left[-\pi /2,\pi /2]\mapsto \left(v+u,v-u)\in {{\mathbb{R}}}^{2}
                        
                     . By translating the colored pieces in the tilted rectangle either vertically or horizontally by 
                        
                           
                           
                              2
                              π
                           
                           2\pi 
                        
                     , we see that the tilted rectangle also parametrizes a copy of 
                        
                           
                           
                              
                                 
                                    
                                       (
                                       
                                          R
                                          ∕
                                          2
                                          π
                                          Z
                                       
                                       )
                                    
                                 
                                 
                                    2
                                 
                              
                           
                           {\left({\mathbb{R}}/2\pi {\mathbb{Z}})}^{2}
                        
                     .
Figure 1

The dotted square is [ π , π ] 2 . The tilted rectangle is the image of ( u , v ) [ π , π ] × [ π 2 , π 2 ] ( v + u , v u ) R 2 . By translating the colored pieces in the tilted rectangle either vertically or horizontally by 2 π , we see that the tilted rectangle also parametrizes a copy of ( R 2 π Z ) 2 .

Figure 2 
                  The dots represent the lattice 
                        
                           
                           
                              Λ
                           
                           \Lambda 
                        
                     . Each blue rectangle is a fundamental domain for 
                        
                           
                           
                              
                                 
                                    R
                                 
                                 
                                    2
                                 
                              
                              ∕
                              Λ
                           
                           {{\mathbb{R}}}^{2}/\Lambda 
                        
                     . It can be identified with 
                        
                           
                           
                              
                                 
                                    Σ
                                 
                                 
                                    t
                                 
                              
                           
                           {\Sigma }_{t}
                        
                      via the map 
                        
                           
                           
                              Φ
                              
                                 (
                                 
                                    t
                                    ,
                                    ⋅
                                    ,
                                    ⋅
                                 
                                 )
                              
                           
                           \Phi \left(t,\cdot ,\cdot )
                        
                     .
Figure 2

The dots represent the lattice Λ . Each blue rectangle is a fundamental domain for R 2 Λ . It can be identified with Σ t via the map Φ ( t , , ) .

Under this parametrization of Σ t , the volume element d σ t on Σ t is

d σ t = ι ν θ d θ = i θ ι ν ( θ 1 θ 1 ¯ ) = 1 2 i ζ 1 ζ 2 ζ 1 ζ 2 θ θ 1 ¯ ζ ¯ 1 ζ ¯ 2 ζ 1 ζ 2 θ θ 1 = 1 2 i ζ 1 ζ 2 ζ 1 ζ 2 1 2 i d θ 1 ¯ ζ ¯ 1 ζ ¯ 2 ζ 1 ζ 2 1 2 i d θ 1 = 1 2 ζ 1 ζ 2 ζ 1 ζ 2 d ζ ¯ 1 d ζ ¯ 2 + ζ ¯ 1 ζ ¯ 2 ζ 1 ζ 2 d ζ 1 d ζ 2 = 2 cos π 4 + t 2 sin π 4 + t 2 Re ( ( i ( d u + d v ) ) ( i ( d v d u ) ) ) = 2 cos ( 2 t ) d u d v .

In particular, setting t = 0 , the surface measure d σ on Σ is given by 2 d u d v . If f is an integrable function on Σ , we often write F ( u , v ) f ( Φ ( 0 , u , v ) ) so that

(2.4) Σ f d σ = 2 u π v π 2 F ( u , v ) d u d v .

3 The single-layer potential

For f C 1 ( Σ ) , the single-layer potential is defined by:

S f ( ζ ) = η Σ f ( η ) G ( ζ , η ) d σ ( η ) , ζ Ω ¯ .

Given ζ = ( ζ 1 , ζ 2 ) Σ , the curve

(3.1) γ ζ ( t ) 2 cos π 4 + t 2 ζ 1 , 2 sin π 4 + t 2 ζ 2 , t 0 ,

satisfies γ ζ ( 0 ) = ζ and γ ( 0 ) = ν at ζ .

Proposition 3.1

For f C 1 ( Σ ) , the one-sided normal derivative

ν S f ( ζ ) lim t 0 + S f ( γ ζ ( t ) ) S f ( γ ζ ( 0 ) ) t

exists at every ζ Σ and is given by:

ν S f ( ζ ) = 1 2 I + N f ( ζ ) ,

where

N f ( ζ ) p . v . η Σ ν ζ , ( b ) ζ G ( ζ , η ) f ( η ) d σ ( η ) , ζ Σ .

To prove this, first we understand the kernel G ( ζ , η ) of S in the ( u , v ) coordinates. For t 0 and ( u , v ) , ( u , v ) R 2 Λ , if ζ = Φ ( t , u , v ) and η = Φ ( 0 , u , v ) , then

1 ζ η ¯ = 1 cos π 4 cos π 4 + t 2 e i ( v v + u u ) sin π 4 sin π 4 + t 2 e i ( v v u + u ) = e i ( v v ) e i ( v v ) cos π 4 cos π 4 + t 2 e i ( u u ) sin π 4 sin π 4 + t 2 e i ( u u ) = e i ( v v ) cos ( v v ) cos t 2 cos ( u u ) i sin ( v v ) sin t 2 sin ( u u ) ,

so from G ( ζ , η ) = 1 8 π 1 ζ η ¯ 1 we obtain

(3.2) G ( ζ , η ) = 1 8 π k t 2 , u u , v v ,

where for t 0 and ( u , v ) R 2 Λ we write

k ( t , u , v ) [ ( cos v cos t cos u ) 2 + ( sin v sin t sin u ) 2 ] 1 2 .

We collect a few facts we will need of k ( t , u , v ) :

Lemma 3.2

  1. If t ( 0 , π 2 ) , then k ( t , u , v ) is a continuous function of ( u , v ) R 2 Λ , and so is t k ( t , u , v ) .

  2. If t = 0 , then k ( t , u , v ) is continuous at every ( u , v ) R 2 Λ except at ( 0 , 0 ) , and is locally integrable near ( 0 , 0 ) .

  3. In fact,

    k ( t , u , v ) ( t + ( u , v ) ) 2

    for all t [ 0 , π 4 ] , ( u , v ) R 2 Λ , where the implicit constant is independent of ( t , u , v ) .

  4. For all ( t , u , v ) [ 0 , π 4 ] × ( R 2 Λ ) \ { ( 0 , 0 , 0 ) } , we have

    t k ( t , u , v ) = ( cos v cos t cos u ) sin t cos u ( sin v sin t sin u ) cos t sin u [ ( cos v cos t cos u ) 2 + ( sin v sin t sin u ) 2 ] 3 2 ,

    and hence

    t k ( t , u , v ) ( t + ( u , v ) ) 3 ,

    where the implicit constant is independent of ( t , u , v ) .

  5. We have

    (3.3) lim t 0 + 1 8 π u π v π 2 k t ( t , u , v ) d u d v = 1 2 .

We postpone the proof of this lemma to the end of this section.

Proof of Proposition 3.1

Using (2.4) and (3.2), if we write F ( u , v ) f ( Φ ( 0 , u , v ) ) , we have

S f ( ζ ) = 2 8 π u π v π 2 F ( u , v ) k t 2 , u u , v v d u d v

whenever ζ = Φ ( t , u , v ) , t 0 . When f L 1 ( Σ ) , the above integral converges absolutely at every point ζ by the estimate for k ( t , u , v ) in Lemma 3.2. This integral is a convolution on the group R 2 Λ , and thus, we can also write

(3.4) S f ( ζ ) = 2 8 π u π v π 2 F ( u u , v v ) k t 2 , u , v d u d v .

Similarly, when f C 1 ( Σ ) and ζ = Φ ( 0 , u , v ) , we have

N f ( ζ ) = lim ε 0 + 1 8 π u π , v π 2 ( u u , v v ) ε F ( u , v ) k t ( 0 , u u , v v ) d u d v

is the convolution of F with 1 8 π p . v . k t ( 0 , ) on R 2 Λ ; the principal value exists when f C 1 ( Σ ) since k t ( 0 , u , v ) is odd in both u and v , which implies

u π , v π 2 ( u u , v v ) ε k t ( 0 , u , v ) d u d v = 0

for every ε > 0 , and k t ( 0 , u , v ) ( u , v ) 3 . In fact, one can write N f as an absolutely convergent integral:

(3.5) N f ( ζ ) = 1 8 π u π v π 2 [ F ( u u , v v ) F ( u , v ) ] k t ( 0 , u , v ) d u d v .

Now let ζ Σ , and recall the curve γ ζ ( t ) introduced in (3.1). Using (3.4), the bound for k in Lemma 3.2 and the Dominated Convergence Theorem, we see that

lim t 0 + S f ( γ ζ ( t ) ) = S f ( γ ζ ( 0 ) ) .

Furthermore, from (3.4) and our bounds for t k , we see that for t > 0 ,

d d t S f ( γ ζ ( t ) ) = 1 8 π u π v π 2 F ( u u , v v ) k t t 2 , u , v d u d v .

Hence, to prove Proposition 3.1, by L’Hôpital’s rule and (3.5), it suffices to show, under the assumption that f C 1 ( Σ ) , that if F ( u , v ) = f ( Φ ( 0 , u , v ) ) and ( u , v ) [ π , π ) × [ π 2 , π 2 ) , then

(3.6) lim t 0 + 1 8 π u π v π 2 F ( u u , v v ) k t t 2 , u , v d u d v

exists and equals

(3.7) 1 2 F ( u , v ) + 1 8 π u π v π 2 [ F ( u u , v v ) F ( u , v ) ] k t ( 0 , u , v ) d u d v .

To do so, we rewrite the expression inside the limit in (3.6) as

(3.8) 1 8 π u π v π 2 [ F ( u u , v v ) F ( u , v ) ] k t ( t , u , v ) d u d v + 1 8 π F ( u , v ) u π v π 2 k t ( t , u , v ) d u d v .

By (3.3) in Lemma 3.2, as t 0 + , the second term in (3.8) converges to 1 2 F ( u , v ) , which appears as the first term of (3.7). Furthermore, since f C 1 ( Σ ) , our earlier bound for t k in Lemma 3.2 and the Dominated Convergence Theorem shows that as t 0 + , the first term in (3.8) converges to the second term in (3.7). This completes the proof of Proposition 3.1.□

Proof of Lemma 3.2

First, for fixed t , the function k ( t , u , v ) is doubly 2 π periodic and invariant under ( u , v ) ( u + π , v + π ) . Thus k ( t , u , v ) is a well-defined function on R 2 Λ .

To proceed further, we prove that

( cos v cos t cos u ) 2 + ( sin v sin t sin u ) 2 = 0 cos t = 0 cos u = 0 sin v = sin t sin u or sin t = 0 sin u = 0 cos v = cos t cos u .

In fact, the former is equivalent to

cos v = cos t cos u sin v = sin t sin u ,

which implies, since cos 2 v + sin 2 v = 1 , that

1 = cos 2 t cos 2 u + sin 2 t sin 2 u = ( cos 2 t + sin 2 t ) ( cos 2 u + sin 2 u ) cos 2 t sin 2 u sin 2 t cos 2 u = 1 cos 2 t sin 2 u sin 2 t cos 2 u .

As a result, cos t sin u = sin t cos u = 0 . Either cos t = 0 , in which case sin t = ± 1 and cos u = 0 , or sin t = 0 , in which case cos t = ± 1 and sin u = 0 . From this, the claimed equivalence follows.

If t ( 0 , π 2 ) , then both cos t and sin t are nonzero, so the above equation implies

( cos v cos t cos u ) 2 + ( sin v sin t sin u ) 2 > 0 .

This proves (a). If t = 0 , then cos t = 1 , so the only zeroes of ( cos v cos t cos u ) 2 + ( sin v sin t sin u ) 2 are

{ ( u , v ) R 2 : sin u = 0 , cos v = cos u } = 2 π Z 2 ( ( π , π ) + 2 π Z 2 ) .

This proves (b).

To prove (c), note that k ( t , u , v ) is continuous on [ 0 , π 4 ] × [ π , π ] × [ π 2 , π 2 ] \ { ( 0 , 0 , 0 ) } . So, we only need to prove that

( cos v cos t cos u ) 2 + ( sin v sin t sin u ) 2 ( t + u + v 1 2 ) 4

near ( t , u , v ) = ( 0 , 0 , 0 ) . But if we denote by

(3.9) g ( t , u , v ) ( cos v cos t cos u ) 2 + ( sin v sin t sin u ) 2 ,

then by a Taylor expansion

(3.10) g ( t , u , v ) = 1 1 t 2 2 1 u 2 2 + O 5 2 2 + ( v t u + O 5 2 ) 2 = t 2 + u 2 2 2 + ( v t u ) 2 + O 9 2 = t 2 + u 2 2 2 + ( v t u ) 2 ( 1 + O 1 2 ) ,

where O j is an error bounded by C j ( t + u + v 1 2 ) j for some absolute constant C j . Our claim now follows from the fact that

(3.11) t 2 + u 2 2 2 + ( v t u ) 2 ( t + u + v 1 2 ) 4

uniformly in t , u , v (which can be verified by noting that the left-hand side is a positive continuous function on the set where t + u + v 1 2 = 1 , and appealing to homogeneity).

(d) follows from

t k ( t , u , v ) sin t cos u + cos t sin u ( cos v cos t cos u ) 2 + ( sin v sin t sin u ) 2 t + u g ( t , u , v )

and the lower bound g ( t , u , v ) t + u + v 1 2 on [ 0 , π 4 ] × [ π , π ] × [ π 2 , π 2 ] that we proved earlier.

Finally, we prove (3.3) in (e). Since k t ( t , u , v ) converges uniformly to 0 on [ π , π ] × [ π 2 , π 2 ] \ [ ε , ε ] × [ ε 2 , ε 2 ] for any ε > 0 , it suffices to show that there exists ε > 0 such that

lim t 0 + u ε v ε 2 k t ( t , u , v ) d u d v = 4 π .

From

t k ( t , u , v ) = ( cos v cos t cos u ) sin t cos u ( sin v sin t sin u ) cos t sin u g ( t , u , v ) 3 2 ,

where g ( t , u , v ) was defined in (3.9), it suffices to show that

lim t 0 + u ε v ε 2 sin t cos u ( cos v cos t cos u ) g ( t , u , v ) 3 2 d u d v = 4 π ,

lim t 0 + u ε v ε 2 sin u sin v g ( t , u , v ) 3 2 d u d v = 4 π ,

and

lim t 0 + u ε v ε 2 sin t sin 2 u g ( t , u , v ) 3 2 d u d v = 4 π .

We do so by Taylor expanding the numerators and denominators of the integrands. By choosing ε > 0 sufficiently small, we have, whenever t [ 0 , ε ] , u ε and v ε 2 , that ( 1 + O 1 2 ) 3 2 = 1 + O 1 2 , and hence from (3.10),

g ( t , u , v ) 3 2 = t 2 + u 2 2 2 + ( v t u ) 2 3 2 ( 1 + O 1 2 ) .

Similarly, for t [ 0 , ε ] , u ε and v ε 2 , we have

sin t cos u ( cos v cos t cos u ) = t ( t 2 + u 2 ) 2 + O 5 2

and

sin t sin 2 u = t [ u 2 + O 5 2 ] .

Thus

lim t 0 + u ε v ε 2 sin t cos u ( cos v cos t cos u ) g ( t , u , v ) 3 2 d u d v = lim t 0 + u ε v ε 2 t ( t 2 + u 2 ) 2 + O 5 2 t 2 + u 2 2 2 + ( v t u ) 2 3 2 d u d v .

The error term involving O 5 2 can be bounded using (3.11). Thus

(3.12) u ε v ε 2 t O 5 2 t 2 + u 2 2 2 + ( v t u ) 2 3 2 d u d v C t u ε v ε 2 ( t + u + v 1 2 ) 7 2 d u d v = C t u ε t v ε 2 t 2 ( t + t u + t 2 v 1 2 ) 7 2 t 3 d u d v = C t 1 2 u ε t v ε 2 t 2 ( 1 + u + v 1 2 ) 7 2 d u d v ,

where we performed a change of variables ( u , v ) ( u t , v t 2 ) in the second-to-last line, and the last line is O ( t 1 2 ) as t 0 + since the double integral converges to a finite limit (we have R 2 ( 1 + u + v 1 2 ) α d u d v < if and only if α < 3 ). Furthermore, the main term in our earlier computation can be identified, via the same change of variable, as

lim t 0 + u ε t v ε 2 t 2 t t 2 + ( u t ) 2 2 t 2 + ( u t ) 2 2 2 + ( v t 2 t ( u t ) ) 2 3 2 t 3 d u d v = R 2 1 + u 2 2 1 + u 2 2 2 + ( v u ) 2 3 2 d u d v = 4 π .

Similarly,

lim t 0 + u ε v ε 2 sin t sin 2 u g ( t , u , v ) 3 2 d u d v = lim t 0 + u ε v ε 2 t [ u 2 + O 5 2 ] t 2 + u 2 2 2 + ( v t u ) 2 3 2 d u d v = R 2 u 2 1 + u 2 2 2 + ( v u ) 2 3 2 d u d v = 4 π .

Finally, use v ε 2 f ( v ) d v = 1 2 v ε 2 [ f ( v ) + f ( v ) ] d v and write

(3.13) u ε v ε 2 sin u sin v g ( t , u , v ) 3 2 d u d v = 1 2 u ε v ε 2 sin u sin v g ( t , u , v ) 3 2 sin u sin v g ( t , u , v ) 3 2 d u d v = 1 2 u ε v ε 2 sin u sin v [ g ( t , u , v ) g ( t , u , v ) ] [ g ( t , u , v ) 2 + g ( t , u , v ) g ( t , u , v ) + g ( t , u , v ) 2 ] g ( t , u , v ) 3 2 g ( t , u , v ) 3 2 [ g ( t , u , v ) 3 2 + g ( t , u , v ) 3 2 ] d u d v .

We Taylor expand the numerator by noting

sin u sin v = u v + O 7 2

and

g ( t , u , v ) 2 + g ( t , u , v ) g ( t , u , v ) + g ( t , u , v ) 2 = h ( t , u , v ) 2 + h ( t , u , v ) h ( t , u , v ) + h ( t , u , v ) 2 + O 17 2 ,

where we abbreviated

h ( t , u , v ) t 2 + u 2 2 2 + ( v t u ) 2

and

g ( t , u , v ) g ( t , u , v ) = 4 sin t sin u sin v = 4 t u v + t O 7 2 = h ( t , u , v ) h ( t , u , v ) + t O 7 2 .

We also Taylor expand the denominator, yielding

g ( t , u , v ) 3 2 g ( t , u , v ) 3 2 [ g ( t , u , v ) 3 2 + g ( t , u , v ) 3 2 ] = h ( t , u , v ) 3 2 h ( t , u , v ) 3 2 [ h ( t , u , v ) 3 2 + h ( t , u , v ) 3 2 ] ( 1 + O 1 2 ) .

Thus

lim t 0 + u ε v ε 2 sin u sin v g ( t , u , v ) 3 2 d u d v = lim t 0 + 1 2 u ε v ε 2 u v [ h ( t , u , v ) h ( t , u , v ) ] [ h ( t , u , v ) 2 + h ( t , u , v ) h ( t , u , v ) + h ( t , u , v ) 2 ] h ( t , u , v ) 3 2 h ( t , u , v ) 3 2 [ h ( t , u , v ) 3 2 + h ( t , u , v ) 3 2 ] d u d v + lim t 0 + u ε v ε 2 t O 29 2 h ( t , u , v ) 3 2 h ( t , u , v ) 3 2 [ h ( t , u , v ) 3 2 + h ( t , u , v ) 3 2 ] d u d v .

The second limit is zero by the same argument of the proof of (3.12). In the first limit, we are considering the limit as t 0 + of a double integral reminiscent of the last expression of (3.13). By reversing the derivation of (3.13), except that we write h in place of g , we see that the first limit is equal to

lim t 0 + u ε v ε 2 u v h ( t , u , v ) 3 2 d u d v = lim t 0 + u ε v + t u ε 2 ( v + t u ) u [ t 2 + u 2 2 2 + v 2 ] 3 2 d u d v = lim t 0 + u ε t u ε 2 v t u + ε 2 u v t 2 + u 2 2 2 + v 2 3 2 d u d v + lim t 0 + u ε v + t u ε 2 t u 2 t 2 + u 2 2 2 + v 2 3 2 d u d v = I + I I .

We have

I = lim t 0 + u ε u t 2 + u 2 2 2 + ( t u ε 2 ) 2 1 2 u t 2 + u 2 2 2 + ( t u + ε 2 ) 2 1 2 d u = u ε u u 2 4 + ε 4 1 2 d u u ε u u 2 4 + ε 4 1 2 d u = 0 0 = 0

and using a change of variable ( u , v ) ( t u , t 2 v ) ,

I I = lim t 0 + u ε t v + u ε 2 t 2 u 2 1 + u 2 2 2 + v 2 3 2 d u d v = R 2 u 2 1 + u 2 2 2 + v 2 3 2 d u d v = 4 π .

Thus

lim t 0 + u ε v ε 2 sin u sin v g ( t , u , v ) 3 2 d u d v = 4 π ,

as desired.□

4 The classical solution to the Neumann boundary value problem

In this section, we prove Theorem 1.1.

4.1 Existence

We are now going to invert 1 2 I + N on L 2 , which will give us the solution to the Neumann boundary value problem for the CR Yamabe operator when the boundary data h is in C ( Σ ) .

Proposition 4.1

We have N L 2 L 2 < 1 2 . As a result, for any h L 2 ( Σ ) , there exists a unique f L 2 ( Σ ) such that

1 2 I + N f = h .

Furthermore, if h C ( Σ ) , then f C ( Σ ) .

The main thrust is in showing that N L 2 L 2 < 1 2 . Once that is proved, then 1 2 I + N is invertible on L 2 , and is given by a multiplier operator whose Fourier coefficients are bounded below. Hence 1 2 I + N 1 maps W k , 2 ( Σ ) to W k , 2 ( Σ ) for any k 0 , and hence maps C ( Σ ) to C ( Σ ) . As a result, we obtain the existence assertion in Theorem 1.1.

In the following, we prove that N L 2 L 2 < 1 2 .

Indeed, since N f ( Φ ( 0 , u , v ) ) is the convolution of f ( Φ ( 0 , u , v ) ) with 1 8 π K ( u , v ) , where K ( u , v ) p . v . k t ( 0 , u , v ) on R 2 Λ , we have

Σ N f 2 d σ = 2 u π v π 2 N f ( Φ ( 0 , u , v ) ) 2 d u d v sup ( m , n ) Z 2 m n ( mod 2 ) 1 8 π K ^ ( m , n ) 2 2 u π v π 2 f ( Φ ( 0 , u , v ) ) 2 d u d v = sup ( m , n ) Z 2 m n ( mod 2 ) 1 8 π K ^ ( m , n ) 2 Σ f 2 d σ ,

where

K ^ ( m , n ) = lim ε 0 + u π , v π 2 ( u , v ) ε k t ( 0 , u , v ) e i ( m u + n v ) d u d v .

Thus we need to show that

(4.1) sup ( m , n ) Z 2 m n ( mod 2 ) K ^ ( m , n ) < 8 π 2 = 4 π 12.56 .

Since k t ( 0 , u , v ) is odd in both u and v , we have the following expression for K ^ ( m , n ) :

(4.2) K ^ ( m , n ) = u π v π 2 K ( u , v ) sin ( m u ) sin ( n v ) d u d v .

To bound this integral and establish (4.1), it will be convenient to approximate K ( u , v ) by a Taylor expanding its numerator and denominator. Let us write

K ( u , v ) = n ( u , v ) d ( u , v ) 6 ,

where

n ( u , v ) sin u sin v , d ( u , v ) [ ( cos u cos v ) 2 + sin 2 v ] 1 4 .

We approximate n ( u , v ) and d ( u , v ) by

n 0 ( u , v ) u v and d 0 ( u , v ) u 4 4 + v 2 1 4 ,

respectively, so that the homogeneous function

K ( u , v ) n 0 ( u , v ) d 0 ( u , v ) 6

will be a good approximation of K ( u , v ) near ( u , v ) = ( 0 , 0 ) , along with its derivatives; this homogeneous function can be integrated easily near the origin, so in (4.2) we can approximate K ( u , v ) by K ( u , v ) , and eventually obtain (4.1). In fact, it will be convenient to note that

(4.3) d ( u , v ) 4 d 0 ( u , v ) 4 u 6 4 ! + u 2 v 2 2 + 2 u 8 8 ! + 2 v 4 4 !

for all ( u , v ) R 2 , which follows since

d ( u , v ) 4 = cos 2 u 2 cos u cos v + 1 = 1 2 ( 1 + cos 2 u ) 2 cos u + 1 + 2 cos u ( 1 cos v ) ,

whereas

1 2 ( 1 + cos 2 u ) 2 cos u + 1 1 2 1 + 1 ( 2 u ) 2 2 ! + ( 2 u ) 4 4 ! ( 2 u ) 6 6 ! 2 1 u 2 2 ! + u 4 4 ! u 6 6 ! + u 8 8 ! + 1 = u 4 4 u 6 4 ! 2 u 8 8 !

and

2 cos u ( 1 cos v ) = 0 v 0 w 2 cos u cos x d x d w = 0 v 0 w cos ( u + x ) + cos ( u x ) d x d w 0 v 0 w 1 ( u + x ) 2 2 + 1 ( u x ) 2 2 d x d w = 0 v 0 w ( 2 u 2 x 2 ) d x d w = v 2 u 2 v 2 2 2 v 4 4 ! .

To carry out the aforementioned strategy, we differentiate K ( u , v ) and write

K u = n u ( u , v ) d ( u , v ) 10 , K v = n v ( u , v ) d ( u , v ) 10 , 2 K u 2 = n u u ( u , v ) d ( u , v ) 14 ,

where we have, via the half-angle formulae, the following expressions for the numerators:

n u ( u , v ) sin ( v ) 20 sin 4 u 2 + 4 sin 2 v 2 + sin ( v ) 16 sin 6 u 2 + 2 sin 2 ( u ) sin 2 v 2 , n v ( u , v ) sin ( u ) 4 sin 4 u 2 8 sin 2 v 2 + sin ( u ) sin 2 v 2 8 sin 2 u 2 1 + cos 2 u 2 + 4 cos ( u ) sin 2 v 2 n u u ( u , v ) 120 sin ( u ) sin 2 u 2 sin ( v ) sin 4 u 2 sin 2 v 2 64 sin ( u ) sin 8 u 2 sin ( v ) 8 cos u 2 + cos 3 u 2 sin 5 u 2 sin ( v ) sin 2 v 2 + 4 ( 14 + sin 2 ( u ) ) sin ( u ) sin ( v ) sin 4 v 2 .

Similarly, we write

K u = n 0 u ( u , v ) d 0 ( u , v ) 10 , K v = n 0 v ( u , v ) d 0 ( u , v ) 10 , 2 K u 2 = n 0 u u ( u , v ) d 0 ( u , v ) 14 ,

where

n 0 u ( u , v ) v 5 4 u 4 + v 2 , n 0 v ( u , v ) u u 4 4 2 v 2 , n 0 u u ( u , v ) 15 2 u 3 v u 4 4 v 2 .

Later, we will use

(4.4) n u ( u , v ) n 0 u ( u , v ) + 1,451 u 6 v 3,072 + u 2 v 3 2 + v 4 3 ,

(4.5) n v ( u , v ) n 0 v ( u , v ) + 683 u 7 15,360 + u 3 v 2 + 7 u v 3 6 ,

(4.6) n u u ( u , v ) n 0 u u ( u , v ) + 1195 u 9 v 2048 + u 5 v 3 8 + 5 u 3 v 4 2 + 15 u v 5 4 .

In fact, we have

d 6 d u 6 sin 4 u 2 = cos ( u ) 2 8 cos ( 2 u ) 2,049 256 for all  u R

and

d 3 d v 3 sin 2 v 2 = sin v 2 1 2 for all  v R .

Thus, Taylor expansion (up to orders 5 and 3, respectively) gives

sin 4 u 2 u 4 16 2,049 256 u 6 6 !

and

sin 2 v 2 v 2 4 1 2 v 3 3 ! .

As a result, we can estimate the main terms for the expressions for n u , n v , and n u u :

sin ( v ) 20 sin 4 u 2 + 4 sin 2 v 2 v 20 u 4 16 + 4 v 2 4 + 20 2,049 256 u 6 6 ! + 4 1 2 v 3 3 ! = n 0 u ( u , v ) + 683 3,072 u 6 v + 1 3 v 4 ,

sin ( u ) 4 sin 4 u 2 8 sin 2 v 2 u 4 u 4 16 8 v 2 4 + 4 2,049 256 u 6 6 ! + 8 1 2 v 3 3 ! = n 0 v ( u , v ) + 683 15,360 u 7 + 2 3 u v 3 ,

and

120 sin ( u ) sin 2 u 2 sin ( v ) sin 4 u 2 sin 2 v 2 30 u 3 v u 4 16 v 2 4 + 2,049 256 u 6 6 ! + 1 2 v 3 3 ! = n 0 u u ( u , v ) + 683 2,048 u 9 v + 5 2 u 3 v 4 .

Furthermore, the error terms can be bounded using

sin ( v ) 16 sin 6 u 2 + 2 sin 2 ( u ) sin 2 v 2 v 16 u 6 2 6 + 2 u 2 v 2 2 2 = 1 4 u 6 v + 1 2 u 2 v 3 ,

sin ( u ) sin 2 v 2 8 sin 2 u 2 1 + cos 2 u 2 + 4 cos ( u ) sin 2 v 2 u v 2 4 8 u 2 2 2 ( 1 + 1 ) + 4 1 1 v 2 = u 3 v 2 + 1 2 u v 3 ,

64 sin ( u ) sin 8 u 2 sin ( v ) 64 u u 8 2 8 v = 1 4 u 9 v ,

8 cos u 2 + cos 3 u 2 sin 5 u 2 sin ( v ) sin 2 v 2 8 ( 1 + 1 ) u 5 2 5 v v 2 2 2 = 1 8 u 5 v 3 ,

and

4 ( 14 + sin 2 ( u ) ) sin ( u ) sin ( v ) sin 4 v 2 4 ( 14 + 1 ) u v v 4 2 4 = 15 4 u v 5 .

The above eight inequalities together imply (4.4), (4.5), and (4.6).

Now let

b 2 5 .

For 0 < a 5 , let

e 0 ( a ) sup u 6 4 ! + u 2 v 2 2 + 2 u 6 ( a b ) 2 8 ! + 2 v 3 b 2 4 ! : d 0 ( u , v ) = 1

and

e 1 ( a ) 1 e 0 ( a ) d 0 ( a b , b 2 ) 2 .

By homogeneity, we then have

(4.7) u 6 4 ! + u 2 v 2 2 + 2 u 6 ( a b ) 2 8 ! + 2 v 3 b 2 4 ! e 0 ( a ) d 0 ( u , v ) 6 for all  ( u , v ) R 2 .

Furthermore, e 1 ( a ) is a decreasing function of a [ 0 , ) , with e 1 ( 5 ) > 0 . Thus e 1 ( a ) > 0 for 0 < a 5 . We claim that if 0 < a 5 and ( u a , v ) b , then

(4.8) K ( u , v ) K ( u , v ) + err 0 ( a , u , v ) ,

where

err 0 ( a , u , v ) c ( 6 , a ) n 0 ( u , v ) d 0 ( u , v ) 4 ,

with

c ( α , a ) α e 0 ( a ) 4 e 1 ( a ) α + 4 4 .

Indeed, when ( u a , v ) b , we have

(4.9) d ( u , v ) 4 d 0 ( u , v ) 4 u 6 4 ! + u 2 v 2 2 + 2 u 8 8 ! + 2 v 4 4 ! d 0 ( u , v ) 4 u 6 4 ! + u 2 v 2 2 + 2 u 6 ( a b ) 2 8 ! + 2 v 3 b 2 4 ! d 0 ( u , v ) 4 ( 1 e 0 ( a ) d 0 ( u , v ) 2 ) ,

where we first used (4.3), then used u a b and v b 2 when ( u a , v ) b , and finally used (4.7). Furthermore, the mean value theorem gives

1 ( 1 t ) α 4 1 + α t 4 ( 1 t ) α + 4 4 if  α > 0  and  0 t < 1 .

Applying this with t = e 0 ( a ) d 0 ( u , v ) 2 e 0 ( a ) d 0 ( a b , b 2 ) 2 = 1 e 1 ( a ) < 1 , and noting that

1 ( 1 t ) α + 4 4 1 ( 1 e 0 ( a ) d 0 ( a b , b 2 ) 2 ) α + 4 4 = 1 e 1 ( a ) α + 4 4 ,

we obtain

(4.10) 1 d ( u , v ) α 1 d 0 ( u , v ) α + c ( α , a ) d 0 ( u , v ) α 2 for  α > 0 ,  0 < a 5  and  ( u a , v ) b .

Inequality (4.8) now follows since n ( u , v ) n 0 ( u , v ) , which gives

K ( u , v ) n 0 ( u , v ) d ( u , v ) 6 ,

and since we can use (4.10) with α = 6 to bound 1 d ( u , v ) 6 .

We also note that

(4.11) 1 d ( u , v ) 1 e 1 ( a ) 1 4 d 0 ( u , v ) if  0 < a 5  and  ( u a , v ) b ,

because then we can apply (4.9) and use 1 e 0 ( a ) d 0 ( u , v ) 2 1 e 0 ( a ) d 0 ( a b , b 2 ) 2 = e 1 ( a ) .

Using similar techniques, we can prove, for 0 < a 5 and ( u a , v ) b ,

(4.12) K u ( u , v ) K u ( u , v ) + err u ( a , u , v )

and

(4.13) K v ( u , v ) K v ( u , v ) + err v ( a , u , v ) ,

where

err u ( a , u , v ) c ( 10 , a ) n 0 u ( u , v ) d 0 ( u , v ) 8 + 1451 u 6 v 3,072 + u 2 v 3 2 + v 4 3 e 1 ( a ) 10 4 d 0 ( u , v ) 10

and

err v ( a , u , v ) c ( 10 , a ) n 0 v ( u , v ) d 0 ( u , v ) 8 + 683 u 7 15360 + u 3 v 2 + 7 u v 3 6 e 1 ( a ) 10 4 d 0 ( u , v ) 10 .

This is because when 0 < a 5 and ( u a , v ) b ,

K u ( u , v ) n 0 u ( u , v ) d ( u , v ) 10 + 1451 u 6 v 3072 + u 2 v 3 2 + v 4 3 d ( u , v ) 10 n 0 u ( u , v ) d 0 ( u , v ) 10 + c ( 10 , a ) n 0 u ( u , v ) d 0 ( u , v ) 8 + 1451 u 6 v 3072 + u 2 v 3 2 + v 4 3 e 1 ( a ) 10 4 d 0 ( u , v ) 10 ,

where in the first inequality we applied (4.4), and in second inequality we used (4.10) with α = 10 to estimate the first term, and used (4.11) to estimate the second term. This proves (4.12). Similarly, we have (4.13). We also have, for 0 < a 5 and ( u a , v ) b , that

(4.14) 2 K u 2 ( u , v ) 2 K u 2 ( u , v ) + err u u ( a , u , v ) ,

where

err u u ( a , u , v ) c ( 14 , a ) n 0 u u ( u , v ) d 0 ( u , v ) 12 + 1,195 u 9 v 2,048 + u 5 v 3 8 + 5 u 3 v 4 2 + 15 u v 5 4 e 1 ( a ) 14 4 d 0 ( u , v ) 14 .

The numerical values of e 0 ( a ) , e 1 ( a ) , and c ( α , a ) can all be estimated using Mathematica once α and a are given. Thus err 0 , err u , err v , and err u u are completely concrete functions.

In the next two lemmas, we will prove two bounds for K ^ ( m , n ) . The first one is good when m is large; the second one is good when n is large. Recall b = 2 5 .

Lemma 4.2

Suppose that c 1 > 0 , that m and n are positive integers with the same parity, and that m c 1 b . Then, for any 0 < a 5 ,

K ^ ( m , n ) c 1 3 I 0 n m 2 + I 1 c 1 + I 2 + I 3 c 1 2 + 1 m 2 c 1 5 n m 2 ε I 0 + c 1 ε I 1 + ε I 2 I 3 b 2 + ε I 3 log b m c 1 + ε I 4 ,

where I 0 , , I 3 , ε I 0 , , ε I 4 are constants depending on a, given by:

I 0 ( u a , v ) 1 u v K ( u , v ) d u d v , I 1 2 1 1 K ( a , v ) d v , I 2 2 1 1 K u ( a , v ) d v , I 3 ( u a , v ) 1 2 K u 2 ( u , v ) d u d v ,

and

ε I 0 ( u a , v ) 1 u v err 0 ( a , u , v ) d u d v , ε I 1 2 1 1 err 0 ( a , a , v ) d v , ε I 2 2 1 1 err u ( a , a , v ) d v , ε I 3 2 a 1 1 err u u ( a , a , v ) d v + 4 a a err u u ( a , u , 1 ) d u , ε I 4 u π , v π 2 ( u a , v ) > b 2 K u 2 ( u , v ) d u d v .

Proof

Define R = c 1 m . Our hypothesis guarantees that R b . We split

K ^ ( m , n ) = u π v π 2 K ( u , v ) sin ( m u ) sin ( n v ) d u d v = ( u a , v ) R + u π , v π 2 ( u a , v ) > R K ( u , v ) sin ( m u ) sin ( n v ) d u d v .

We estimate the first term by putting absolute value inside, and we integrate by parts in the second term using sin ( m u ) = 1 m d d u cos ( m u ) . Note that the set { ( u , v ) : ( u a , v ) = R } is the boundary of a rectangle. Thus,

u π , v π 2 ( u a , v ) > R K ( u , v ) sin ( m u ) sin ( n v ) d u d v = 2 m R 2 R 2 K ( a R , v ) cos ( m a R ) sin ( n v ) d v + 1 m u π , v π 2 ( u a , v ) > R K u ( u , v ) cos ( m u ) sin ( n v ) d u d v .

We integrate by parts once more for the last double integral, using cos ( m u ) = 1 m d d u sin ( m u ) , and obtain

1 m u π , v π 2 ( u a , v ) > R K u ( u , v ) cos ( m u ) sin ( n v ) d u d v = 2 m 2 R 2 R 2 K u ( a R , v ) sin ( m a R ) sin ( n v ) d v 1 m 2 u π , v π 2 ( u a , v ) > R 2 K u 2 ( u , v ) sin ( m u ) sin ( n v ) d u d v .

Taking absolute values, and using sin ( m u ) sin ( n v ) m n u v , we obtain

K ^ ( m , n ) m n ( u a , v ) R u v K ( u , v ) d u d v + 2 m R 2 R 2 K ( a R , v ) d v + 2 m 2 R 2 R 2 K u ( a R , v ) d v + 1 m 2 R < ( u a , v ) b 2 K u 2 ( u , v ) d u d v + 1 m 2 u π , v π 2 ( u a , v ) > b 2 K u 2 ( u , v ) d u d v .

The first four integrals are all contained in { ( u , v ) : ( u a , v ) b } , so our earlier estimates (4.8), (4.12), and (4.14) apply. They allow us to bound K and its derivatives by those of K , up to some errors that we control. Then we obtain

K ^ ( m , n ) m n ( u a , v ) R u v K ( u , v ) d u d v + 2 m R 2 R 2 K ( a R , v ) d v + 2 m 2 R 2 R 2 K u ( a R , v ) d v + 1 m 2 R < ( u a , v ) b 2 K u 2 ( u , v ) d u d v + m n ( u a , v ) R u v err 0 ( a , u , v ) d u d v + 2 m R 2 R 2 err 0 ( a , a R , v ) d v + 2 m 2 R 2 R 2 err u ( a , a R , v ) d v + 1 m 2 R < ( u a , v ) b err u u ( a , u , v ) d u d v + ε I 4 m 2 .

All the integrals on the right-hand side of the above equation are now integrals of homogeneous functions of ( u , v ) . Since u v K ( u , v ) is homogeneous of degree 0, a change of variables ( u , v ) ( R u , R 2 v ) gives

m n ( u a , v ) R u v K ( u , v ) d u d v = m n R 3 ( u a , v ) 1 u v K ( u , v ) d u d v = m n R 3 I 0 .

Similarly, the integrals involving K , K u , err 0 , and err u are homogeneous functions of R . Next, we rewrite the integral involving 2 K u 2 by a change of variables:

1 m 2 R < ( u a , v ) b 2 K u 2 ( u , v ) d u d v = 1 m 2 ( u a , v ) > R 2 K u 2 ( u , v ) d u d v ( u a , v ) > b 2 K u 2 ( u , v ) d u d v = I 3 1 m 2 R 2 1 m 2 b 2 .

Finally, by the Fundamental Theorem of Calculus and the chain rule,

d d t 1 < ( u a , v ) t err u u ( a , u , v ) d u d v = 2 a t 2 t 2 err u u ( a , a t , v ) d v + 4 t a t a t err u u ( a , u , t 2 ) d u ,

which, by homogeneity, is equal to

1 t 2 a 1 1 err u u ( a , a , v ) d v + 4 a a err u u ( a , u , 1 ) d u = ε I 3 t .

Thus

R < ( u a , v ) b err u u ( a , u , v ) d u d v = 1 < ( u a , v ) b R err u u ( a , u , v ) d u d v = 1 b R d d t 1 < ( u a , v ) t err u u ( a , u , v ) d u d v d t = 1 b R ε I 3 t d t = ε I 3 log b R ,

and putting all these together, one sees that

K ^ ( m , n ) m n R 3 I 0 + I 1 m R + I 2 + I 3 m 2 R 2 + m n R 5 ε I 0 + ε I 1 m R + ε I 2 m 2 I 3 m 2 b 2 + ε I 3 log ( b R ) m 2 + ε I 4 m 2 .

Remembering R = c 1 m gives the estimate in Lemma 4.2.□

Lemma 4.3

Suppose that c 2 > 0 , that m and n are positive integers with the same parity and that n ( c 2 b ) 2 . Then for any 0 < a 5 ,

K ^ ( m , n ) c 2 3 J 0 m n + J 1 + J 2 c 2 2 + 1 n c 2 5 ε J 0 m n + ε J 1 J 2 b 2 + ε J 2 log b n c 2 + ε J 3 ,

where J 0 , , J 2 and ε J 0 , , ε J 3 are constants depending on a, given by:

J 0 ( u a , v ) 1 u v K ( u , v ) d u d v , J 1 2 a a K ( u , 1 ) d u , J 2 ( u a , v ) 1 K v ( u , v ) d u d v ,

and

ε J 0 ( u a , v ) 1 u v err 0 ( a , u , v ) d u d v , ε J 1 2 a a err 0 ( a , u , 1 ) d u , ε J 2 2 a 1 1 err v ( a , a , v ) d v + 4 a a err v ( a , u , 1 ) d u , ε J 3 u π , v π 2 ( u a , v ) > b K v ( u , v ) d u d v .

Proof

The proof is similar to the previous lemma (and slightly easier); one only needs to split the integral into two parts, one where ( u a , v ) R , and another where ( u a , v ) > R , with R c 2 n < b ; one then integrates by parts once in v for the second integral, using sin ( n v ) = 1 n d d v cos ( n v ) .

More precisely, define R = c 2 n . Our hypothesis guarantees that R b . We split

K ^ ( m , n ) = u π v π 2 K ( u , v ) sin ( m u ) sin ( n v ) d u d v = ( u a , v ) R + u π , v π 2 ( u a , v ) > R K ( u , v ) sin ( m u ) sin ( n v ) d u d v .

We estimate the first term by putting absolute value inside, and we integrate by parts in the second term using sin ( n v ) = 1 n d d v cos ( n v ) . Note that the set { ( u , v ) : ( u a , v ) = R } is the boundary of a rectangle. Thus

u π , v π 2 ( u a , v ) > R K ( u , v ) sin ( m u ) sin ( n v ) d u d v = 2 n a R a R K ( u , R 2 ) sin ( m u ) cos ( n R 2 ) d u + 1 n u π , v π 2 ( u a , v ) > R K v ( u , v ) sin ( m u ) cos ( n v ) d u d v .

Thus

K ^ ( m , n ) m n ( u a , v ) R u v K ( u , v ) d u d v + 2 n a R a R K ( u , R 2 ) d u + 1 n R < ( u a , v ) b K v ( u , v ) d u d v + 1 n u π , v π 2 ( u a , v ) > b K v ( u , v ) d u d v .

The first three integrals are all contained in { ( u , v ) : ( u a , v ) b } , so our earlier estimates (4.8) and (4.13) apply. Thus

K ^ ( m , n ) m n ( u a , v ) R u v K ( u , v ) d u d v + 2 n a R a R K ( u , R 2 ) d u + 1 n R < ( u a , v ) b K v ( u , v ) d u d v + m n ( u a , v ) R u v err 0 ( a , u , v ) d u d v + 2 n a R a R err 0 ( a , u , R 2 ) d v + 1 n R < ( u a , v ) b err v ( a , u , v ) d u d v + ε J 3 m 2 .

All the integrals on the right-hand side of the above equation are now integrals of homogeneous functions of ( u , v ) . Thus the integrals are homogeneous functions of R , with the exception of the one involving err v because err v is homogeneous of degree 3 . But then, by the Fundamental Theorem of Calculus and the chain rule,

d d t 1 < ( u a , v ) t err v ( a , u , v ) d u d v = 2 a t 2 t 2 err v ( a , a t , v ) d v + 4 t a t a t err v ( a , u , t 2 ) d u ,

which, by homogeneity, is equal to

1 t 2 a 1 1 err v ( a , a , v ) d v + 4 a a err v ( a , u , 1 ) d u = ε J 2 t .

Thus

R < ( u a , v ) b err v ( a , u , v ) d u d v = 1 < ( u a , v ) b R err v ( a , u , v ) d u d v = 1 b R d d t 1 < ( u a , v ) t err v ( a , u , v ) d u d v d t = 1 b R ε J 2 t d t = ε J 2 log b R ,

and by scaling all other integrals similarly one sees that

K ^ ( m , n ) m n R 3 J 0 + J 1 + J 2 n R 2 + m n R 5 ε J 0 + ε J 1 n J 2 n b 2 + ε J 2 log ( b R ) n + ε J 3 n .

Remembering R = c 2 n gives the estimate in Lemma 4.3.□

We apply the above two lemmas with a = 2 , c 1 8 5 and a = 1 , c 2 4 2 5 , respectively:

Corollary 4.4

Suppose m and n are positive integers with the same parity. If m 4 , then

(4.15) K ^ ( m , n ) 7,168 625 n m 2 + 5 + 1 m 2 783 40 + 98,304 3,125 n m 2 + 192 log m 4 .

Proof

We apply Lemma 4.2 with a = 2 , so that Mathematica returns expressions for exact values for I 0 , I 1 , I 2 , I 3 , ε I 0 , ε I 1 , ε I 2 , and ε I 3 and a numerical estimate for ε I 4 (Appendix A). We then have

I 0 = 32 3 log 1 + 5 2 1 32 2 π Γ 1 4 2 + 2 F 1 1 4 , 1 2 , 3 4 , 1 4 2 F 1 1 2 , 3 4 , 7 4 , 4 I 1 = 4 8 5 , I 2 = 2 12 5 5 , I 3 = 6 + 4 2 12 5 5 ,

so

I 0 < 14 5 , I 1 < 1 2 , I 2 < 1 , 21 2 < I 3 < 11 .

Next, recall b = 2 5 , so

e 0 ( 2 ) = 1 75 68,935,125,533 + 1,312,025 1,102,101 55,105,058 , e 1 ( 2 ) = 1 4 5 5 e 0 ( 2 ) ,

ε I 0 = 3 5 e 0 ( 2 ) e 1 ( 2 ) 5 2 16 + π + 4 arctan ( 3 ) 32 arctan 1 2 2 log 5 , ε I 1 = 6 e 0 ( 2 ) e 1 ( 2 ) 5 2 log 5 4 , ε I 2 = 6 + 5 log 4 5 e 0 ( 2 ) e 1 ( 2 ) 7 2 + 1 e 1 ( 2 ) 5 2 2,219 288 307 20 5 + 4 3 log 1 + 5 2 , ε I 3 = 105 e 0 ( 2 ) e 1 ( 2 ) 9 2 + 9,473 240 e 1 ( 2 ) 7 2 .

Together with the numerical value of ε I 4 , we obtain

ε I 0 < 3 , ε I 1 < 2 , ε I 2 < 7 , ε I 3 < 192 , ε I 4 < 75 .

Finally, we set c 1 = 8 5 so that c 1 b = 4 m .□

Corollary 4.5

Suppose m and n are positive integers with the same parity. If n 8 , then

(4.16) K ^ ( m , n ) 4,832 2 3,125 m n + 245 32 + 1 n 53 8 + 1,6384 2 15,625 m n + 19 log n 8 .

Proof

We apply Lemma 4.3 with a = 1 , so that Mathematica returns expressions for exact values for J 0 , J 1 , J 2 , ε J 0 , ε J 1 , and ε J 2 and a numerical estimate for ε J 3 (Appendix A). Thus we have:

J 0 = 1 3 4 log ( 2 + 5 ) 2 π Γ 1 4 2 + 1 6 2 F 1 1 4 , 1 2 , 3 4 , 4 4 2 F 1 1 2 , 3 4 , 7 4 , 1 4 , J 1 = 4 5 , J 2 = 32 3 3 + 4 5 ,

so

J 0 < 151 100 , J 1 < 9 5 , 79 10 < J 2 < 8 .

Next, recall b = 2 5 , so

e 0 ( 1 ) = 1 75 137,909,641,801 + 2,625,550 1,102,731 110,273,101 , e 1 ( 1 ) = 1 2 5 5 e 0 ( 1 ) ,

ε J 0 = 3 5 e 0 ( 1 ) e 1 ( 1 ) 5 2 ( 2 + 3 arctan ( 2 ) 2 log 5 ) , ε J 1 = 6 arctan 1 2 e 0 ( 1 ) e 1 ( 1 ) 5 2 , ε J 2 = 5 4 2 + π 4 arctan 1 2 e 0 ( 1 ) e 1 ( 1 ) 7 2 + 4,843 360 e 1 ( 1 ) 5 2 .

Together with the numerical value of ε J 3 , we obtain

ε J 0 < 4 5 , ε J 1 < 2 , ε J 2 < 38 , ε J 3 < 54 .

Finally, we set c 2 = 4 2 5 so that ( c 2 b ) 2 = 8 n .□

Remark 4.6

The Mathematica codes used in Corollaries 4.4 and 4.5 are enclosed in Appendix A for interested readers.

From Corollary 4.4, if m 4 and 8 5 n m , then

(4.17) K ^ ( m , n ) 237 25 + 1 m 2 31,863 1,000 + 192 log m 4 .

The right-hand side is an increasing function of m on [ 4 , 5 ] and a decreasing function of m on [ 6 , ) , and is < 4 π at both m = 5 and m = 6 . Thus we have K ^ ( m , n ) < 4 π if m 4 and m 8 5 n . This shows K ^ ( m , n ) < 4 π for ( m , n ) colored in red in Figure 3.

Figure 3 
                  The points for which 
                        
                           
                           
                              ∣
                              
                                 
                                    K
                                 
                                 
                                    ^
                                 
                              
                              
                                 (
                                 
                                    m
                                    ,
                                    n
                                 
                                 )
                              
                              ∣
                           
                           | \widehat{K}\left(m,n)| 
                        
                      can be bounded by Corollaries 4.4 and 4.5 are colored in red and blue, respectively.
Figure 3

The points for which K ^ ( m , n ) can be bounded by Corollaries 4.4 and 4.5 are colored in red and blue, respectively.

From Corollary 4.5, if n 8 and m 8 5 n , then

(4.18) K ^ ( m , n ) 38,656 2 15,625 + 245 32 + 1 n 131,072 2 78,125 + 53 8 + 19 log n 8 .

The right-hand side is an increasing function of n on [ 8 , 13 ] and a decreasing function of n on [ 14 , ) . Since this function of n is < 4 π at both n = 13 and n = 14 , we have K ^ ( m , n ) < 4 π if n 8 and m 8 5 n . Such pairs ( m , n ) are pictured in blue in Figure 3.

It remains to verify that K ^ ( m , n ) < 4 π for the 11 points in Figure 3 that are not colored, i.e., when 1 m 3 , 1 n 7 , and ( m , n ) are of the same parity. We achieved this numerically using Mathematica (in fact, K ^ ( m , n ) < 5.8 for all ( m , n ) that is not colored). For interested readers, see Appendix B for the Mathematica codes.

5 The weak solution to the Neumann problem

In this section, we prove Theorem 1.2.

5.1 Functional analysis

Recall that the Folland-Stein space S 1 , 2 ( Ω ) is the space of all functions u L 2 ( Ω ) whose weak derivatives b u are in L 2 ( Ω ) . It is known [14, Theorem 1.3] that C ( Ω ¯ ) S 1 , 2 ( Ω ) is a dense subspace and that the restriction map

C ( Ω ¯ ) u u Σ C ( Σ )

is continuous. Therefore the trace operator Tr : S 1 , 2 ( Ω ) L 2 ( Σ ) is defined. It is known [14, Theorem 1.4] that Tr ( S 1 , 2 ( Ω ¯ ) ) L 3 ( Σ ) .

Since Σ is p -minimal, we see that

( u ) Ω b u 2 + R 4 u 2 θ d θ

is the continuous extension of : C ( Ω ) R to S 1 , 2 ( Ω ) . Moreover, there is a minimizer u 0 S 1 , 2 ( Ω ) of

μ 1 ( u ) : u S 1 , 2 ( Ω ) , Σ Tr ( u ) 2 d σ = 1 .

Necessarily, u 0 is a weak solution of

L u = 0 , in  Ω , ν u = μ 1 Tr u , on  Σ ;

i.e.,

0 = Ω b u 0 , b ϕ + 1 2 u 0 ϕ θ d θ + Σ μ 1 Tr ( u 0 ) ϕ d σ

for all ϕ C ( Ω ¯ ) . In this section, we show that u 0 C ( Ω ¯ ) .

5.2 Potential theory

Let f L 2 ( Σ , d σ ) . Recall the single-layer potential

S f ( ζ ) = η Σ f ( η ) G ( ζ , η ) d σ ( η ) , ζ Ω ¯ .

We have S f C ( Ω ) . By (3.4), we also have

S f ( ζ ) = 2 8 π u π v π 2 F ( u u , v v ) k t 2 , u , v d u d v ,

if ζ = Φ ( t , u , v ) and F ( u , v ) f ( Φ ( 0 , u , v ) ) .

Proposition 5.1

If f L 2 ( Σ , d σ ) , then S f S 1 , 2 ( Ω ) .

Proof

Note that

sup t [ 0 , π 100 ] u π v π 2 k t 2 , u , v d u d v 1

by the bound for k in Lemma 3.2. Using Young’s convolution inequality on R 2 Λ , we see that

sup t [ 0 , π 100 ] S f L 2 ( Σ t ) f L 2 ( Σ ) .

Thus S f L 2 ( Ω ) . Furthermore, for ζ Ω , we have

Z 1 ¯ S f ( ζ ) = η Σ f ( η ) Z 1 ¯ G ( ζ , η ) d σ ( η ) .

But

Z 1 ¯ G ( ζ , η ) = 1 16 π ( ζ 2 η 1 ζ 1 η 2 ) ( 1 ζ η ¯ ) 1 ζ η ¯ 3 .

If ζ = Φ ( t , u , v ) and η = Φ ( 0 , u , v ) , then

ζ 2 η 1 ζ 1 η 2 = cos π 4 sin π 4 + t 2 e i ( v u + v + u ) sin π 4 cos π 4 + t 2 e i ( v + u + v u ) = e i ( v + v ) sin t 2 cos ( u u ) i cos t 2 sin ( u u ) ,

so

Z 1 ¯ G ( ζ , η ) = e 2 i v 16 π sin t 2 cos ( u u ) i cos t 2 sin ( u u ) × cos ( v v ) cos t 2 cos ( u u ) i sin ( v v ) sin t 2 sin ( u u ) g ( t , u u , v v ) 3 2 = e 2 i v 16 π i cos t 2 sin ( u u ) cos ( v v ) cos t 2 cos ( u u ) i sin ( v v ) g ( t , u u , v v ) 3 2 + O t g ( t , u u , v v ) .

The numerator of the first term is even in t , but the denominator is not. Thus we symmetrize the first term by making it even in t , which gives an error that is also in O t g ( t , u u , v v ) . In fact, the error term is given by:

= C sin ( u u ) [ ( cos ( v v ) cos t cos ( u u ) ) i sin ( v v ) ] 1 g ( t , u u , v v ) 3 2 1 g ( t , u u , v v ) 3 2 C u u [ ( cos ( v v ) cos t cos ( u u ) ) 2 + sin 2 ( v v ) ] 1 2 1 g ( t , u u , v v ) 5 2 g t × ( t , u u , v v ) t .

Plugging

g t ( t , u u , v v ) = g 1 2 ( t , u u , v v ) O ( u u + t ) ,

where t [ t , t ] , into the error term, we obtain

= O t g ( t , u u , v v ) .

To summarize, if ζ = Φ ( t , u , v ) and η = Φ ( 0 , u , v ) , we obtain

Z 1 ¯ G ( ζ , η ) = e 2 i v 32 π i cos t 2 K t 2 ( u u , v v ) + O t g ( t , u u , v v ) ,

where

K t ( u , v ) sin u [ ( cos v cos t cos u ) i sin v ] 1 g ( t , u , v ) 3 2 + 1 g ( t , u , v ) 3 2 .

It follows that

Z 1 ¯ S f ( Φ ( t , u , v ) ) = e 2 i v 32 π i cos t 2 u π v π 2 F ( u u , v v ) K t 2 ( u , v ) d u d v + O u π v π 2 F ( u u , v v ) t g ( t , u , v ) d u d v

is essentially the sum of the convolutions of F ( u , v ) f ( Φ ( 0 , u , v ) ) with two kernels on R 2 Λ . The second term can be handled using

u π v π 2 F ( u u , v v ) t g ( t , u , v ) d u d v L 2 ( d u d v ) F L 2 ,

where we applied Young’s convolution inequality on R 2 Λ and the bound

u π v π 2 t g ( t , u , v ) d u d v u π v π 2 t ( t + u + v 1 2 ) 4 d u d v R 2 1 ( 1 + u + v 1 2 ) 4 d u d v <

uniformly in t ( 0 , π 100 ] . We claim that

u π v π 2 F ( u u , v v ) K t ( u , v ) d u d v L 2 ( d u d v ) F L 2

uniformly in t ( 0 , π 100 ] . Once this is shown, we see that Z 1 ¯ S f L 2 ( Σ t ) uniformly in t ( 0 , π 100 ] , so Z 1 ¯ S f L 2 ( Ω ) . Since S is real, we also have Z 1 S f L 2 ( Ω ) . Thus S f S 1 , 2 ( Ω ) , as desired.

To prove the claim, by Parseval on L 2 ( R 2 Λ ) , we only need to show that

(5.1) sup ( m , n ) Z 2 u π v π 2 K t ( u , v ) e i ( m u + n v ) d u d v 1

uniformly in t ( 0 , π 100 ] . But we have Re K t ( u , v ) = Re K t ( u , v ) and Im K t ( u , v ) = Im K t ( u , v ) . One thus concludes the cancellation condition

( u , v ) r K t ( u , v ) d u d v = 0

for all r . Next, for u π and v π 2 , a straightforward computation shows

K t ( u , v ) ( u , v ) 3 , K t u ( u , v ) ( u , v ) 4 , K t v ( u , v ) ( u , v ) 5 ,

all uniformly in t ( 0 , π 100 ] . Combining these with the cancellation condition, one can obtain (5.1) in a standard manner.□

To proceed further, define

M f ( ζ ) sup t ( 0 , π 100 ] d d t S f ( γ ζ ( t ) ) , ζ Σ .

Proposition 5.2

M is bounded on L 2 ( Σ , d σ ) .

To prove this, write M for the Hardy-Littlewood maximal operator adapted to the nonisotropic geometry on Σ :

M f ( ζ ) sup t > 0 1 Q t ( ζ ) Q t ( ζ ) f ( η ) d σ ( η ) ,

where Q t ( ζ ) is the nonisotropic ball given by:

Q t ( ζ ) { Φ ( 0 , u , v ) Σ : ( u u , v v ) t } if  ζ = Φ ( 0 , u , v ) .

(Recall ( u , v ) max { u , v 1 2 } .) It is known that M is bounded on L 2 ( Σ ) . Since we proved N is also bounded on L 2 ( Σ ) , Proposition 5.2 is a consequence of the following:

Proposition 5.3

For f L 2 ( Σ ) , we have

M f ( ζ ) C ( M N f ( ζ ) + M f ( ζ ) )

for d σ -almost every ζ Σ .

Proof of Proposition 5.3

In fact, if ζ = Φ ( 0 , u , v ) , then γ ζ ( t ) = Φ ( t , u , v ) , so for t > 0 , we can differentiate (3.4) under the integral and obtain

d d t S f ( γ ζ ( t ) ) = 1 8 π u π v π 2 F ( u u , v v ) k t t 2 , u , v d u d v ,

where F ( u , v ) f ( Φ ( 0 , u , v ) ) . We also have, from (3.5), that

N f ( ζ ) = lim ε 0 + 1 8 π u π , v π 2 ( u , v ) ε F ( u u , v v ) k t ( 0 , u , v ) d u d v

and

M f ( ζ ) sup 0 < t < 100 1 t 3 ( u , v ) t F ( u u , v v ) d u d v .

Using the bound

k t ( t , u , v ) ( t + u + v 1 2 ) 3 ,

it is easy to see that

(5.2) 1 8 π ( u , v ) 10 t F ( u u , v v ) k t t 2 , u , v d u d v C M f ( ζ ) ;

in fact,

1 8 π ( u , v ) 10 t F ( u u , v v ) k t t 2 , u , v d u d v C M f ( ζ ) ( u , v ) 10 t 1 ( t + u + v 1 2 ) 3 d u d v

and

( u , v ) 10 t 1 ( t + u + v 1 2 ) 3 d u d v ( u , v ) 10 t 1 t 3 d u d v = C .

Furthermore, using the bound

k t ( t , u , v ) k t ( 0 , u , v ) t ( u + v 1 2 ) 4 when  ( u , v ) t ,

one can show that

(5.3) 1 8 π u π v π 2 ( u , v ) > 10 t F ( u u , v v ) k t t 2 , u , v k t ( 0 , u , v ) d u d v C M f ( ζ ) .

This is because

1 8 π u π v π 2 ( u , v ) > 10 t F ( u u , v v ) k t t 2 , u , v k t ( 0 , u , v ) d u d v M f ( ζ ) ( u , v ) > 10 t t ( u + v 1 2 ) 4 d u d v M f ( ζ ) .

Finally, we claim that for almost every ζ Σ ,

(5.4) 1 8 π u π v π 2 ( u , v ) > 10 t F ( u u , v v ) k t ( 0 , u , v ) d u d v C ( M N f + M f ) ( ζ ) .

In fact, the operators on both sides of the inequality are bounded on L 2 ( Σ ) , and C ( Σ ) is dense in L 2 ( Σ ) , we may assume, without loss of generality, that f C ( Σ ) . Suppose now ψ C c ( Σ ) is supported in Q 1 ( Φ ( 0 , 0 , 0 ) ) , with Σ ψ d σ = 1 . Let Ψ t ( u , v ) = 2 1 t 3 ψ ( Φ ( 0 , t 1 u , t 2 v ) ) for t ( 0 , π 100 ] and ( u , v ) R 2 , so that

u π , v π 2 Ψ t ( u , v ) d u d v = 1 .

Recall K ( u , v ) = p . v . k t ( 0 , u , v ) and let

K ˜ t ( u , v ) = K Ψ t ( u , v )

be the convolution of K with Ψ t on R 2 Λ . We will prove that

(5.5) k t ( 0 , u , v ) 1 ( u , v ) > 10 t K ˜ t ( u , v ) t ( t + u + v 1 2 ) 4 .

In fact, if ( u , v ) > 10 t , then the mean value theorem gives

k t ( 0 , u , v ) K ˜ t ( u , v ) ( u , v ) t k t ( 0 , u , v ) k t ( 0 , u u , v v ) Ψ t ( u , v ) d u d v t ( u + v 1 2 ) 4 .

On the other hand, if ( u , v ) 10 t , then

K ˜ t ( u , v ) = ( u , v ) 100 t [ Ψ t ( u u , v v ) Ψ t ( u , v ) ] k t ( 0 , u , v ) d u d v + u π , v π 2 ( u , v ) > 100 t Ψ t ( u u , v v ) k t ( 0 , u , v ) d u d v ,

with the last integral being both zero because Φ t ( u u , v v ) = 0 there. As a result, if ( u , v ) 10 t , then

K ˜ t ( u , v ) ( u , v ) 100 t 1 t 3 u t + v t 2 1 ( u , v ) 3 d u d v 1 t 3 .

This proves (5.5), and hence

1 8 π u π v π 2 ( u , v ) > 10 t F ( u u , v v ) k t ( 0 , u , v ) d u d v 1 8 π u π v π 2 F ( u u , v v ) K ˜ t ( u , v ) d u d v + M f ( ζ ) .

But

1 8 π u π v π 2 F ( u u , v v ) K ˜ t ( u , v ) d u d v = F 1 8 π K * Ψ t ( u , v ) = N F Ψ t ( u , v ) ,

where N F ( u , v ) = N f ( Φ ( 0 , u , v ) ) . Thus

1 8 π u π v π 2 F ( u u , v v ) K ˜ t ( u , v ) d u d v M N f ( ζ ) .

This proves (5.4).□

We are now ready to complete the proof of Theorem 1.2. We need to show that if f L 2 ( Σ ) and u = S f , then

L u = 0 on  Ω ν u = 1 2 I + N f on  Σ

in the weak sense, i.e.,

0 = Ω b u , b ϕ + R 4 u ϕ + Σ 1 2 I + N f ϕ

for all ϕ C ( Ω ¯ ) . To do so, note that

Ω b u , b ϕ + R 4 u ϕ = lim t 0 + Ω t b u , b ϕ + R 4 u ϕ

and u C ( Ω t ¯ ) for t > 0 . Thus, for ϕ C ( Ω ¯ ) we can now integrate by parts, and using L u = 0 inside Ω t , see that the above equation is equal to

lim t 0 + Σ t ν u ϕ d σ t = lim t 0 + Σ d d t u ( γ ζ ( t ) ) ϕ ( γ ζ ( t ) ) d σ t d σ d σ .

Our earlier estimate for M allows one to show that

(5.6) d d t u ( γ ζ ( t ) ) 1 2 I + N f ( ζ )

for almost every ζ Σ as t 0 + : if f L 2 ( Σ ) , then choose g C ( Σ ) so that f g L 2 ( Σ ) < ε 0 . Then for any α > 0 ,

ζ Σ : limsup t 0 + d d t u ( γ ζ ( t ) ) 1 2 I + N f ( ζ ) > α ζ Σ : limsup t 0 + d d t S ( f g ) ( γ ζ ( t ) ) 1 2 I + N ( f g ) ( ζ ) > α { ζ Σ : M ( f g ) ( ζ ) > α 2 } ζ Σ : 1 2 I + N ( f g ) ( ζ ) > α 2 .

The measure of the last set with respect to d σ is at most

C α 2 M ( f g ) L 2 ( Σ ) 2 + 1 2 I + N ( f g ) L 2 ( Σ ) 2 α 2 f g L 2 ( Σ ) 2 α 2 ε 0 2 .

Letting ε 0 0 + , we see that (5.6) holds d σ a.e. on Σ .

Our earlier estimate for M also allows us to further apply the dominated convergence theorem and show that

lim t 0 + Σ d d t u ( γ ζ ( t ) ) ϕ ( γ ζ ( t ) ) d σ t d σ d σ = Σ 1 2 I + N f ϕ .

This shows that u solves the desired equation in the weak sense.

For uniqueness, we need to show that if u S 1 , 2 ( Ω ) and

0 = Ω b u , b ϕ + R 4 u ϕ for all  ϕ C ( Ω ¯ ) ,

then u = 0 a.e. in Ω . By density, the above condition actually implies

0 = Ω b u , b ϕ + R 4 u ϕ for all  ϕ S 1 , 2 ( Ω ) .

Setting ϕ = u and using R > 0 , we obtain u = 0 a.e. on Ω .

6 Proof of Proposition 1.3

Let Ω be the domain in S 3 defined by (1.1). Let u S 1 , 2 ( Ω ) be a minimizer of ( u ) under the condition that M u 2 d σ = 1 . We want to show that u C ( Ω ¯ ) . In fact, this will be the case as long as u is a critical point.

First, since

0 = 1 2 d d t t = 0 ( u + t ϕ ) M Tr ( u + t ϕ ) 2 d σ = Ω b u , b ϕ + R 4 u ϕ ( u ) M Tr ( u ) ϕ d σ

for all ϕ C c ( Ω ¯ ) , we have

L u = 0 , in  Ω , ν u = μ Tr ( u ) , on  Ω ,

where μ ( u ) . By Theorem 1.2, we have u = S f , where f μ 1 2 I + N 1 Tr ( u ) .

Lemma 6.1

For any s 0 , we have

Tr S f W s + 1 2 , 2 ( Σ ) C s f W s , 2 ( Σ ) for a l l f W s , 2 ( Σ ) .

Here we identify a function on Σ with a function on the Lie group R 2 Λ , and W s , 2 ( Σ ) is the corresponding (isotropic) Sobolev space on R 2 Λ , so that

f W s , 2 ( Σ ) 1 2 π 2 ( m , n ) Z 2 m n ( mod 2 ) ( 1 + m 2 + n 2 ) s F ^ ( m , n ) 2 1 2

if F ( u , v ) f ( Φ ( 0 , u , v ) ) .

Proof of Lemma 6.1

To prove the lemma, note that if f L 2 ( Σ ) and we write F ( u , v ) = f ( Φ ( 0 , u , v ) ) with S F ( u , v ) ( Tr S f ) ( Φ ( 0 , u , v ) ) , then

SF = F 2 8 π K 0 a.e. on  Σ ,

where K 0 ( u , v ) = k ( 0 , u , v ) . The reason that this works for f L 2 ( Σ ) is because the identity holds for f C ( Σ ) , and that both sides are bounded linear maps on L 2 ( Σ ) . Thus

S F ^ ( m , n ) = F ^ ( m , n ) 2 8 π K 0 ^ ( m , n )

for all ( m , n ) 1 2 π Λ . It is easy to see that

K 0 ^ ( m , n ) ( 1 + m + n 1 2 ) 1 .

In fact, since K 0 L 1 ( Σ ) , we have K 0 ^ ( m , n ) 1 . Also, since

K 0 ( u , v ) ( u , v ) 2 , 2 K 0 u 2 + K 0 v ( u , v ) 4 ,

by writing

K 0 ^ ( m , n ) = ( u , v ) ( m + n 1 2 ) 1 + u π , v π 2 ( u , v ) ( m + n 1 2 ) 1 K 0 ( u , v ) e i ( m u + n v ) d u d v ,

and integrating by parts in the second integral using either e i m u = 1 m 2 2 u 2 e i m u or e i n v = i n v e i n v , we obtain K 0 ^ ( m , n ) ( m + n 1 2 ) 1 . This proves our bound for K 0 ^ ( m , n ) ; in fact, we will only use the weaker bound

K 0 ^ ( m , n ) ( 1 + m + n ) 1 2 .

Now, if f W s , 2 ( Σ ) , then this weaker estimate for K ^ ( m , n ) allows us to bound

Tr S f W s + 1 2 , 2 ( Σ ) ( m , n ) Z 2 m n ( mod 2 ) ( 1 + m 2 + n 2 ) s + 1 2 S F ^ ( m , n ) 2 1 2 ( m , n ) Z 2 m n ( mod 2 ) ( 1 + m 2 + n 2 ) s + 1 2 F ^ ( m , n ) K 0 ^ ( m , n ) 2 1 2 ( m , n ) Z 2 m n ( mod 2 ) ( 1 + m 2 + n 2 ) s F ^ ( m , n ) 2 1 2 f W s , 2 ( Σ ) .

Recall now our critical point u = S f where f μ 1 2 I + N 1 Tr ( u ) . By Lemma 6.1, for all s 0 , we have

Tr u W s + 1 2 , 2 ( Σ ) f W s , 2 ( Σ ) .

Using the bound from Section 4, we have

f W s , 2 ( Σ ) μ Tr ( u ) W s , 2 ( Σ ) .

Putting these together, we see that

Tr ( u ) W k , 2 ( Σ ) μ 2 k Tr ( u ) L 2 ( Σ ) μ 2 k u S 1 , 2 ( Ω )

for all k N . Thus Tr ( u ) C ( Σ ) , and hence ν u C ( Σ ) . As a result, f = μ 1 2 I + N 1 ν u C ( Σ ) . Thus u = S f C ( Ω ¯ ) , as desired.

7 The boundary Yamabe contact form on Σ

We conclude this article by proving our results about the boundary Yamabe constant.

First, we prove that the boundary Yamabe constant Y ( M , T 1 , 0 ) is finite if and only if the first Dirichlet eigenvalue of the CR Yamabe operator is positive.

Proposition 7.1

Let ( M 3 , T 1 , 0 , θ ) be a closed pseudohermitian manifold with boundary having no characteristic points. Then Y ( M , T 1 , 0 ) > if and only if λ 1 , D ( L ) > 0 .

Proof

Set λ λ 1 , D ( L ) and let u be the first Dirichlet eigenvalue, i.e., u is the unique nonnegative function in S 1 , 2 ( M ) such that M u 2 θ d θ = 1 and

L u = λ u , in  M , u = 0 , on  M .

Jerison proved [9, Theorem 7.1] that u C ( M ) . Bony’s maximum principle [1, Corollaire 3.1] then implies that u > 0 . Note that ( u ) = λ . Additionally, for any fixed v C ( M ) , it holds that

(7.1) ( v + t u ) = λ t 2 + λ t M u v θ d θ + t M v B u d σ + ( v ) , M v + t u 3 d σ = M v 3 d σ .

Suppose first that λ < 0 . For any fixed v C ( M ) , we deduce from (7.1) that ( v + t u ) as t . Therefore Y ( M , T 1 , 0 ) = .

Suppose next that λ = 0 . Fix a v C ( M ) such that v M = B u . We deduce from (7.1) that ( v + t u ) as t . Therefore Y ( M , T 1 , 0 ) = .

Suppose finally that λ > 0 . Fix a constant c ( 0 , λ ) . A standard variational argument implies the existence of a weak solution u S 1 , 2 ( M ) of

L u = c u , in  M , u = 1 , on  M .

Combining Jerison’s regularity results [9, Theorem 7.1] with Bony’s maximum principle [1, Corollaire 3.1] implies that u C ( M ) is positive. Set θ ^ u 2 θ . By the conformal covariance of the CR Yamabe operator,

R θ ^ = 4 L u 2 θ ( 1 ) = 4 c u 2 > 0 .

In particular, there are positive constants C 1 and C 2 such that

θ ^ ( v ) M ( b v 2 + C 1 v 2 ) θ ^ d θ ^ C 2 M v 2 d σ θ ^

for all v C ( M ) . On the one hand, Hölder’s inequality implies that v 2 v 3 2 3 . On the other hand, the Sobolev trace inequality [14, Theorem 1.4] and a standard partition of unity argument imply that

M ( b v 2 + C 1 v 2 ) θ ^ d θ ^ M v 3 d σ ^ 2 3 .

Therefore Y ( M , T 1 , 0 ) > .□

Second, we prove that μ 1 ( L ) and Y ( M , T 1 , 0 ) , if finite, have the same sign provided that minimizers of the former are smooth.

Proposition 7.2

Let ( M 3 , T 1 , 0 , θ ) be a closed pseudohermitian manifold with boundary having no characteristic points. Suppose additionally that the minimizer of (1.4) is smooth. Then μ 1 ( L ) and Y ( M , T 1 , 0 ) have the same sign.

Proof

Denote Y Y ( M , T 1 , 0 ) and μ 1 μ 1 ( L ) .

Suppose first that Y > 0 (resp. Y 0 ). Hölder’s inequality implies that if u C ( M ) , then

( u ) Y M u 3 d σ 2 3 Y Vol ( M ) 1 3 M u 2 d σ .

Therefore μ 1 > 0 (resp. μ 1 0 ).

Suppose next that μ 1 > 0 (resp. μ 1 0 ). By restricting to nonnegative functions and applying the Sobolev trace embedding theorem [14, Theorem 1.4], we deduce that there is a nonnegative minimizer u S 1 , 2 ( M ) of μ 1 . By assumption, u is smooth. A straightforward computation implies that u is a weak solution of

(7.2) L u = 0 , in  M , B u = μ 1 u , on  M .

The strong maximum principle [1, Corollaire 3.1] implies that u is positive in the interior of Ω . The Hopf Lemma [13, Corollary 2.1] implies that u is also positive on the boundary. Therefore u > 0 . Set θ ^ u 2 θ . Equation (7.2) implies that H θ ^ > 0 (resp. H θ ^ 0 ). By conformal covariance,

θ ( u 1 v ) = θ ^ ( v ) = 2 M b v 2 θ ^ d θ ^ + 1 3 M H θ ^ v 2 d σ ^ .

We conclude from the Sobolev trace embedding theorem that Y > 0 (resp. Y 0 ).□

Finally, we construct a scalar flat contact form on Ω with respect to which Σ has constant p -mean curvature.

Proof of Theorem 1.4

Introduce new coordinates ( r , u , v ) [ 0 , π ] × [ 0 , 2 π ] × [ 0 , 2 π ] on Ω by

ζ 1 = e i ( u + v ) sin ( r 2 ) , ζ 2 = e i ( u v ) cos ( r 2 ) ,

making the obvious identifications. A straightforward computation yields

Z 1 = e 2 i u r + 1 2 i cos ( r ) u + 1 2 i csc ( r ) v .

It readily follows that

L = 2 r 2 1 2 cot 2 ( r ) u 2 1 2 csc 2 ( r ) v 2 cot ( r ) csc ( r ) u v 2 2 cot ( r ) r + 1 2 .

We conclude that a smooth function u = u ( r ) on Ω solves L u = 0 if and only if

(7.3) r 2 u + cot ( r ) r u 1 4 u = 0 , lim r π u ( r ) < .

By the change of variables x = cos r , we see that (7.3) is equivalent to

(7.4) ( 1 x 2 ) x 2 u 2 x x u 1 4 u = 0 , lim x 1 u ( x ) < .

It is well known [15, §14.2 and §14.8] that the space of solutions of (7.4) is spanned by the Legendre function:

P 1 2 ( x ) = 2 F 1 1 2 , 1 2 ; 1 ; 1 + x 2 .

Rewriting this in terms of z 2 2 = cos 2 r 2 and normalizing so that u = 1 along Σ yield the claimed solution of L u = 0 . In particular, u 2 θ is Webster-flat. Since u depends only on r , we see that both u and B u = r u are constant on Σ . Therefore, Σ has constant p -mean curvature with respect to u 2 θ .□


Dedicated to David Jerison on the occasion of his 70th birthday.


  1. Funding information: Case was partially supported by the Simons Foundation (Grant #524601). Chen was partially supported by NSF Award DMS-3103392. Wang was partially supported by NSF Career Award DMS-1845033. Yang was partially supported by the Simons Foundation (Grant #1006518). Yung was partially supported by a Future Fellowship FT200100399 from the Australian Research Council.

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

Appendix A

The Mathematica codes used in Corollaries 4.4 and 4.5 are reproduced below for reference. In order to simplify expressions and outputs in Mathematica, some definitions in the codes are different from the definitions taken in the rest of this article. We first point out the relations between these two sets of definitions and then list the codes:

err 0 ( a , u , v ) = e 0 ( a ) e 1 ( a ) 5 2 error0[u,v] , err u ( a , u , v ) = e 0 ( a ) e 1 ( a ) 7 2 erroru1[u,v] + 1 e 1 ( a ) 5 2 erroru2[u,v] , err v ( a , u , v ) = e 0 ( a ) e 1 ( a ) 7 2 errorv1[u,v] + 1 e 1 ( a ) 5 2 errorv2[u,v] , err u u ( a , u , v ) = e 0 ( a ) e 1 ( a ) 9 2 erroruu1[u,v] + 1 e 1 ( a ) 7 2 erroruu2[u,v] , ε I 0 = e 0 ( 2 ) e 1 ( 2 ) 5 2 epsiloni[0] , ε I 1 = e 0 ( 2 ) e 1 ( 2 ) 5 2 epsiloni[1] , ε I 2 = e 0 ( 2 ) e 1 ( 2 ) 7 2 epsiloni[2,1] + 1 e 1 ( 2 ) 5 2 epsiloni[2,2] , ε I 3 = e 0 ( 2 ) e 1 ( 2 ) 9 2 epsiloni[3,1] + 1 e 1 ( 2 ) 7 2 epsiloni[3,2] , ε J 0 = e 0 ( 1 ) e 1 ( 1 ) 5 2 epsilonj[0] , ε J 1 = e 0 ( 1 ) e 1 ( 1 ) 5 2 epsilonj[1] , ε J 2 = e 0 ( 1 ) e 1 ( 1 ) 7 2 epsilonj[2,1] + 1 e 1 ( 1 ) 5 2 epsilonj[2,2] .

Appendix B

We use the following codes in the Mathematica to verify the inequality K ^ ( m , n ) < 4 π when 1 m 3 , 1 n 7 , m , n Z :

References

[1] J.-M. Bony, Principe du maximum, inégalite de Harnack et unicité du problème de Cauchy pour les opérateurs elliptiques dégénérés, Ann. Inst. Fourier (Grenoble) 19 (1969), no. 1, 277–304 xii. 10.5802/aif.319Search in Google Scholar

[2] J.-H. Cheng, J.-F. Hwang, A. Malchiodi, and P. Yang, Minimal surfaces in pseudohermitian geometry, Ann. Sc. Norm. Super. Pisa Cl. Sci. (5) 4 (2005), no. 1, 129–177. 10.2422/2036-2145.2005.1.05Search in Google Scholar

[3] J. F. Escobar, Conformal deformation of a Riemannian metric to a scalar flat metric with constant mean curvature on the boundary, Ann. Math. (2) 136 (1992), no. 1, 1–50. 10.2307/2946545Search in Google Scholar

[4] J. F. Escobar, Addendum: “Conformal deformation of a Riemannian metric to a scalar flat metric with constant mean curvature on the boundary” [Ann. Math. (2) 136 (1992), no. 1, 1–50], Ann. of Math. two 139 (1994), no. 3, 749–750. 10.2307/2118578Search in Google Scholar

[5] E. B. Fabes, M. Jodeit Jr., and N. M. Rivière, Potential techniques for boundary value problems on C1-domains, Acta Math. 141 (1978), no. 3–4, 165–186. 10.1007/BF02545747Search in Google Scholar

[6] G. B. Folland and E. M. Stein, Estimates for the ∂¯b complex and analysis on the Heisenberg group, Comm. Pure Appl. Math. 27 (1974), 429–522. 10.1002/cpa.3160270403Search in Google Scholar

[7] D. Geller, The Laplacian and the Kohn Laplacian for the sphere, J. Differential Geometry 15 (1980), no. 3, 417–435. 10.4310/jdg/1214435651Search in Google Scholar

[8] H. Jacobowitz, An introduction to CR structures, Mathematical Surveys and Monographs, vol. 32, American Mathematical Society, Providence, 1990. 10.1090/surv/032Search in Google Scholar

[9] D. Jerison, The Dirichlet problem for the Kohn Laplacian on the Heisenberg group. I, J. Funct. Anal. 43 (1981), no. 1, 97–142. 10.1016/0022-1236(81)90040-9Search in Google Scholar

[10] D. Jerison and J. M. Lee, The Yamabe problem on CR manifolds, J. Differential Geometry 25 (1987), no. 2, 167–197. 10.4310/jdg/1214440849Search in Google Scholar

[11] D. Jerison and J. M. Lee Intrinsic CR normal coordinates and the CR Yamabe problem, J. Differential Geometry 29 (1989), no. 2, 303–343. 10.4310/jdg/1214442877Search in Google Scholar

[12] J. M. Lee, The Fefferman metric and pseudo-Hermitian invariants, Trans. Amer. Math. Soc. 296 (1986), no. 1, 411–429. 10.1090/S0002-9947-1986-0837820-2Search in Google Scholar

[13] D. D. Monticelli, Maximum principles and the method of moving planes for a class of degenerate elliptic linear operators, J. Eur. Math. Soc. 12 (2010), no. 3, 611–654. 10.4171/JEMS/210Search in Google Scholar

[14] D.-M. Nhieu, The Neumann problem for sub-Laplacians on Carnot groups and the extension theorem for Sobolev spaces, Ann. Mat. Pura Appl. (4) 180 (2001), no. 1, 1–25. 10.1007/s10231-001-8195-0Search in Google Scholar

[15] F. W. J. Olver, A. B. Olde Daalhuis, D. W. Lozier, B. I. Schneider, R. F. Boisvert, C. W. Clark, et al. (Eds.), NIST Digital Library of Mathematical Functions, http://dlmf.nist.gov/, Release 1.1.5 of 2022-03-15. Search in Google Scholar

[16] T. Orponen and M. Villa, Sub-elliptic boundary value problems in flag domains, Adv. Calc. Var. 2022. 10.1515/acv-2021-0077Search in Google Scholar

Received: 2022-12-01
Revised: 2023-04-06
Accepted: 2023-05-15
Published Online: 2023-06-27

© 2023 the author(s), published by De Gruyter

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

Articles in the same Issue

  1. Research Articles
  2. Asymptotic properties of critical points for subcritical Trudinger-Moser functional
  3. The existence of positive solution for an elliptic problem with critical growth and logarithmic perturbation
  4. On some dense sets in the space of dynamical systems
  5. Sharp profiles for diffusive logistic equation with spatial heterogeneity
  6. Generic properties of the Rabinowitz unbounded continuum
  7. Global bifurcation of coexistence states for a prey-predator model with prey-taxis/predator-taxis
  8. Multiple solutions of p-fractional Schrödinger-Choquard-Kirchhoff equations with Hardy-Littlewood-Sobolev critical exponents
  9. Improved fractional Trudinger-Moser inequalities on bounded intervals and the existence of their extremals
  10. The existence of infinitely many boundary blow-up solutions to the p-k-Hessian equation
  11. A priori bounds, existence, and uniqueness of smooth solutions to an anisotropic Lp Minkowski problem for log-concave measure
  12. Existence of nonminimal solutions to an inhomogeneous elliptic equation with supercritical nonlinearity
  13. Non-degeneracy of multi-peak solutions for the Schrödinger-Poisson problem
  14. Gagliardo-Nirenberg-type inequalities using fractional Sobolev spaces and Besov spaces
  15. Ground states of Schrödinger systems with the Chern-Simons gauge fields
  16. Quasilinear problems with nonlinear boundary conditions in higher-dimensional thin domains with corrugated boundaries
  17. A system of equations involving the fractional p-Laplacian and doubly critical nonlinearities
  18. A modified Picone-type identity and the uniqueness of positive symmetric solutions for a prescribed mean curvature problem
  19. On a version of hybrid existence result for a system of nonlinear equations
  20. Special Issue: Geometric PDEs and applications
  21. Preface for the special issue on “Geometric Partial Differential Equations and Applications”
  22. Convex hypersurfaces with prescribed Musielak-Orlicz-Gauss image measure
  23. Total mean curvatures of Riemannian hypersurfaces
  24. On degenerate case of prescribed curvature measure problems
  25. A curvature flow to the Lp Minkowski-type problem of q-capacity
  26. Aleksandrov reflection for extrinsic geometric flows of Euclidean hypersurfaces
  27. A note on second derivative estimates for Monge-Ampère-type equations
  28. The Lp chord Minkowski problem
  29. Widths of balls and free boundary minimal submanifolds
  30. Smooth approximation of twisted Kähler-Einstein metrics
  31. The exterior Dirichlet problem for the homogeneous complex k-Hessian equation
  32. A Carleman inequality on product manifolds and applications to rigidity problems
  33. Asymptotic behavior of solutions to the Monge-Ampère equations with slow convergence rate at infinity
  34. Pinched hypersurfaces are compact
  35. The spinorial energy for asymptotically Euclidean Ricci flow
  36. Geometry of CMC surfaces of finite index
  37. Capillary Schwarz symmetrization in the half-space
  38. Regularity of optimal mapping between hypercubes
  39. Special Issue: In honor of David Jerison
  40. Preface for the special issue in honor of David Jerison
  41. Homogenization of oblique boundary value problems
  42. A proof of a trace formula by Richard Melrose
  43. Compactness estimates for minimizers of the Alt-Phillips functional of negative exponents
  44. Regularity properties of monotone measure-preserving maps
  45. Examples of non-Dini domains with large singular sets
  46. Sharp inequalities for coherent states and their optimizers
  47. Gradient estimates and the fundamental solution for higher-order elliptic systems with lower-order terms
  48. Propagation of symmetries for Ricci shrinkers
  49. Linear extension operators for Sobolev spaces on radially symmetric binary trees
  50. The Neumann problem on the domain in 𝕊3 bounded by the Clifford torus
  51. On an effective equation of the reduced Hartree-Fock theory
  52. Polynomial sequences in discrete nilpotent groups of step 2
  53. Integral inequalities with an extended Poisson kernel and the existence of the extremals
  54. On singular solutions of Lane-Emden equation on the Heisenberg group
Downloaded on 7.9.2025 from https://www.degruyterbrill.com/document/doi/10.1515/ans-2022-0072/html
Scroll to top button