HODGE THEORY IN THE SOBOLEV TOPOLOGY FOR THE DE RHAM COMPLEX

LUIGI FONTANA, STEVEN G. KRANTZ, AND MARCO M. PELOSO

Abstract. The authors study the Hodge theory of the exterior differential operator d acting on q-forms on a smoothly bounded domain in ℝ^{N+1}, and on the half space ℝ_{+}^{N+1}. The novelty is that the topology used is not an L^{2} topology but a Sobolev topology. This strikingly alters the problem as compared to the classical setup. It gives rise to a boundary value problem belonging to a class of problems first introduced by Višik and Eskin, and by Boutet de Monvel.

Contents

PRELIMINARIES 2 0. Introductory Remarks 2 1. Basic Notation and Definitions 4 2. Formulation of the Problem and Statement of the Main Results 6 THE PROBLEM ON THE HALF SPACE 9 3. The Operator d^{∗} on 1-Forms and Its Domain 9 The operators K and d^{∗} 12 4. Boutet de Monvel-Type Analysis of the Boundary Value Problem 17 5. The Explicit Solution in the Case of Functions 22 The a priori Estimate 26 The existence Theorem 29 6. Analysis of the Problem on the Half Space for q-Forms 36 THE PROBLEM ON A SMOOTHLY BOUNDED DOMAIN 51 7. Formulation of the Problem on a Smoothly Bounded Domain 51 8. A Special Coordinate System 58 The tangential Laplacian 61 The operator G_{Ω} on functions 61 9. The Existence Theorem 63 The coercive estimate 66

1991 Mathematics Subject Classification. 35J55 35S15 35N15 58A14 58G05. Key words and phrases. Hodge theory, elliptic boundary value problems, pseudodifferential

boundary problems. Second author supported by NSF Grant DMS-9022140 during his stay at MSRI.. Third author supported in part by the Consiglio Nazionale delle Ricerche.

10. The Regularity Theorem in the Case of Functions 71 The interior estimate 72 The boundary estimate 73 11. Estimates for q-Forms 75 The Regularity Theorem 75 Proof of Theorem 11.3 78 12. The Decomposition Theorem and Conclusions 84 Proof of the Main Result 84 Decomposition of W_{q}^{1} 85 13. Final Remarks 93 References 93

PRELIMINARIES

0. Introductory Remarks

Fix a smoothly bounded domain Ω ⊆ ℝ^{N+1}. In classical treatments of the d operator (see [SWE]), one considers the complex

⋀^{q} ⟶^{d} ⋀^{q+1} ⟶^{d} · · · .

Here ⋀^{q} denotes the q-forms of Cartan and de Rham having L^{2}(Ω) coefficients and the operator d is understood to be densely defined. One considers the operator

= dd^{∗} + d^{∗}d. Here the adjoints are calculated in the L^{2}(Ω) topology.

Now makes sense on those forms ψ such that ψ ∈ dom d^{∗} and dψ ∈ dom d^{∗}. One can decompose the space ⋀^{q} into (the closure of) the image of and its orthogonal complement. Then one exploits this decomposition to construct a right inverse for . This inverse is easily used to show that and its accompanying boundary conditions form a second order elliptic boundary value problem of the classical (coercive) type.

In 1963, J.J. Kohn [KOH1] determined how to carry out the analog of these last calculations for the \overline{∂} operator of complex analysis on a strongly pseudoconvex domain Ω in ℂ^{n}. This is the so-called \overline{∂}-Neumann problem. Of course this analysis, while similar in spirit, is much more complicated. It gave rise to the important “Kohn canonical solution” to the equation \overline{∂}u = f. That is the solution u that is orthogonal to holomorphic functions in the L^{2}(Ω) topology.

Experience in the function theory of several complex variables has shown that it is useful to have many different canonical solutions to the \overline{∂}-problem. For instance, in the strongly pseudoconvex case we profitably study the Kohn solution by comparing it with the Henkin solution (not canonical, but nearly so), see [HEN], and the Phong solution (determined in the L^{2}(∂Ω) topology rather than the L^{2}(Ω) topology), see [PHO].

The ultimate goal of the program that we are initiating in this paper is to construct solutions to the \overline{∂} problem that are orthogonal to holomorphic functions in a Sobolev space W^{s} inner product. There are a priori reasons for knowing that this program is feasible. First, Sobolev space is a Hilbert space, so there must be a minimal solution in the Sobolev topology. Second, Boas [BOA] has studied the space of W^{s} holomorphic functions as a Hilbert space with reproducing kernel. The associated Bergman projection operator is of course closely related to the Neumann operator for the \overline{∂} problem.

The present paper carries out the first step of the proposed program. We work out the Hodge theory for the exterior differentiation operator d in the inner product induced by the Sobolev space W^{s} topology. Of particular interest are the boundary conditions that arise when we calculate the adjoint d^{∗} in this topology, and the elliptic boundary value problem that arises when we consider = dd^{∗} + d^{∗}d. We calculate a complete existence and regularity theory.

In this paper, we restrict attention to the case s = 1. This is done both for convenience and to keep the notation relatively simple—even in this basic case the calculations are often unduly cumbersome. In geometric applications, the case s = 1 is already of great interest. We leave the detailed treatment of higher order s to a future paper.

In future work, we will carry out this program of analysis in the Sobolev space topology for the \overline{∂}-Neumann problem on a strongly pseudoconvex domain. Not only will this give rise to new canonical solutions for the \overline{∂} problem, but it should give a new way to view the Sobolev regularity of the classical \overline{∂} problem, and of understanding the subelliptic gain of 1/2 in regularity.

Essentially the paper is divided into two parts. In the first of these we study the boundary value problem that arises from our Hodge theory on the special domain given by the half space, and in the second one we deal with the problem on a smoothly bounded domain. We would like point out that the second part has been written so that it can be read independently from the first part. When we use results from the half space case, we give precise reference to them.

In detail, the plan of the paper is as follows: Section 1 introduces basic notation and definitions, while in Section 2 we formulate the problem and states the main results. Sections 3 through 6 are devoted the problem on the half space.

In Section 3 we calculate the operator d^{∗} on 1-forms and also calculate its domain when the region under study is the upper half space ℝ_{+}^{N+1}. Section 4 completes the detailed calculation of d^{∗}. Section 5 applies a pseudodifferential formalism developed by Boutet de Monvel to the study of our elliptic boundary value problem. This section is included essentially to show how the boundary value problem under investigation can be view as an example of a very general kind of problem first introduced in [BDM1]-[BDM3] and in [ESK]. In Section 5 we explicitly construct the solution of the boundary value problem on the half space in the case of functions. Section 6 studies the problem for q-forms. Sections 7 through 12 deals with the problem on a bounded smooth domain. In Section 7 we set up the problem and calculate the domain of d^{∗} and the semi-explicit expression for d^{∗}. In Section 8 we introduce notation and some technical facts needed in the sequel. Section 9 contains the proof of the coercive estimate, and the proof of our result about existence of solutions for the boundary value problem. Section 10 is devoted to the proof of the a priori estimate in the case of functions. Section 11 gives the proof of the regularity result in the case of q-forms. Finally, in Section 12, we conclude the proof of our main result in the case of a smoothly bounded domain.

We thank G. Grubb for helpful communications regarding this work.

1. Basic Notation and Definitions

We use the symbol d to denote the usual operator of exterior differentiation acting on q-forms. We let Ω denote a smoothly bounded domain in ℝ^{N+1}. Usually, for simplicity only, a domain is assumed to be connected. The symbol ⋀^{q}(Ω) denotes the q-forms on Ω with smooth coefficients. The symbol ⋀_{0}^{q}(Ω) denotes the forms with coefficients that are C^{∞} and compactly supported in Ω. We let ⋀^{q}(\overline{Ω}) denote the q-forms with coefficients that are smooth on \overline{Ω}, and ⋀_{0}^{q}(\overline{Ω}) denote the q-forms with coefficients in C^{∞}(\overline{Ω}) and having compact support in \overline{Ω} (that is, the support may not be disjoint from the boundary of Ω).

Some of our explicit calculations will be performed on the special domain ℝ_{+}^{N+1} = {x = (x_{0}, x_{1}, …, x_{N}) ∈ ℝ^{N+1} : x_{0} > 0}. The half space is of course unbounded, so the function spaces we deal with must take into account the integrability at infinity. On the other hand the half space has the advantage of allowing explicit calculations.

If L is an operator on forms, then L′ denotes its formal adjoint, that is, its adjoint calculated when acting on elements of ⋀_{0}^{p}(Ω).

In the discussion that follows we let Ω denote either a smoothly bounded domain Ω, or the half space ℝ_{+}^{N+1}. If s is a non-negative integer then the s order Sobolev space norm on functions on Ω is given by

∥f∥_{s}^{2} = ∑_{|α|≤s} ∥\frac{∂^{α}f}{∂x^{α}} ∥_{L^{2}(Ω)}^{2} .

The associated inner product is

⟨f, g⟩_{s} = ∑_{|α|≤s} ∫_{Ω} \frac{∂^{α}f}{∂x^{α}} \overline{\frac{∂^{α}g}{∂x^{α}}} dV (x).

Here dV stands for ordinary Lebesgue volume measure.

For s a non-negative integer we define the Sobolev space W^{s}(Ω) as the closure of C_{0}^{∞}(\overline{Ω}). When s ∈ ℝ_{+} we define W^{s}(Ω) by interpolation (see [LIM] for instance).

Moreover, for s ∈ ℝ_{+}, we denote by W^{s}^{∘} (Ω) the closure of C_{0}^{∞}(Ω) in W^{s}(Ω). When s < 0 we define the negative Sobolev space W^{s}(Ω) to be the dual of W^{s}^{∘} (Ω) with respect to the standard L^{2}-pairing.

On the Euclidean space ℝ^{N+1} we consider the Fourier transform defined initially for a testing function f ∈ C_{0}^{∞} as

f̂(ξ) ≡ Ff(ξ) ≡ ∫_{ℝ^{N+1}} f(x)e^{−2πix·ξ} dx.

We will also consider the tangential Fourier transform of functions defined on the half space ℝ_{+}^{N+1}: If f ∈ C_{0}^{∞}(\overline{ℝ_{+}^{N+1}}) we set

f̂(x_{0}, ξ′) = ∫_{ℝ^{N}} f(x_{0}, x′)e^{−2πix′·ξ′} dx′.

We denote the inverse tangential Fourier transform of a function g(x_{0}, ξ′) by ǧ(x_{0}, x′).

For any s ∈ ℝ the Sobolev space W^{s}(ℝ^{N+1}) can be defined via the Fourier transform. Indeed, we set

W^{s}(ℝ^{N+1}) = {f ∈ L^{2}(ℝ^{N+1}) : ∫_{ℝ^{N+1}}(1 + |ξ|^{2})^{s}|Ff(ξ)|^{2} dξ < ∞}.

We will also consider the Sobolev spaces W^{s}(bΩ) defined on the boundary of our domain, s ∈ ℝ. In the case Ω = ℝ_{+}^{N+1}, W^{s}(bΩ) is just the classical Sobolev space on ℝ^{N}. In the case of a smoothly bounded domain Ω, the Sobolev space can be defined by fixing a smooth atlas {χ_{j}} on ∂Ω, letting ϕ_{j} be a partition of unity subordinate to this atlas, and defining the Sobolev norm of a function f on bΩ as

∥f∥_{W^{s}(bΩ)}^{2} = ∑_{j} ∥ϕ_{j}f ∘ χ_{j}^{−1}∥_{W^{s}(ℝ^{N})}^{2}.

Of course this norm is highly non-intrinsic, but a different choice of an atlas gives rise to an equivalent norm.

On the half space ℝ_{+}^{N+1}, and on a bounded domain Ω, we consider the space of q-forms with coefficients in W^{s}. We denote such spaces by W_{q}^{s}(ℝ_{+}^{N+1}) and W_{q}^{s}(Ω) respectively. Fix the standard basis for q-forms: {dx^{I}}, where I = (i_{1}, …, i_{q}) is an increasing multi-index, i.e. 0 ≤ i_{1} < ⋯ < i_{q} ≤ N. The Sobolev inner product on the space of q-forms in the case s a non-negative integer is given by

⟨ϕ, ψ⟩_{s} = ∑_{I}⟨ϕ_{I}, ψ_{I}⟩_{s} = ∑_{I} ∑_{|α|≤s} ∫ \frac{∂^{α}ϕ_{I}}{∂x^{α}} \overline{\frac{∂^{α}ψ_{I}}{∂x^{α}}} dV (x),

where the integral is taken over all of space, i.e. either on ℝ_{+}^{N+1} or on Ω. In the case of general s, for ϕ ∈ ⋀^{q}, ϕ = ∑_{I} ϕ_{I}dx^{I}, we define the norm of ϕ by

∥ϕ∥_{W^{s}}^{2} = ∑_{I} ∥ϕ_{I}∥_{W^{s}}^{2}.

Of course W^{s} is a Hilbert space when equipped with the foregoing inner product. (In particular, W^{s} can be identified with its own dual in a natural way.)

Until further notice, we use D_{j} to denote the partial derivative in the j^{th} variable, and if α = (α_{0}, α_{1}, …, α_{N}) is a multi-index then we let D^{α} denote the corresponding differential monomial.

2. Formulation of the Problem and Statement of the Main Results

Given the operator d,

d : ⋀^{q} ⟶ ⋀^{q+1},

we think of it as a densely defined operator on the space W_{q}^{1} of q-forms with coefficients in W^{1}, both in the case of the half space ℝ_{+}^{N+1} and also in the case of a smoothly bounded domain Ω. Let d^{∗} denote the W^{1}-Hilbert space adjoint of d. It is a densely defined (unbounded) operator

d^{∗} : W_{q+1}^{1} ⟶ W_{q}^{1}.

We shall study the boundary value problem

\begin{cases}(dd^{∗} + d^{∗}d)ϕ = α & on ℝ_{+}^{N+1} (resp. on Ω) \\ ϕ ∈ dom d^{∗} \\ dϕ ∈ dom d^{∗} \end{cases} ,

for α ∈ W_{q}^{s}(Ω); we shall prove existence and regularity theorems, both in the case of the half space and of a smoothly bounded domain, for q-forms, q = 0, 1, …, N + 1. The conditions ϕ, dϕ ∈ dom d^{∗} are ultimately expressed as boundary conditions.

Our main results are the following.

Proposition 2.1. Let Ω denote a smoothly bounded domain, or the special domain ℝ_{+}^{N+1}. Let q = 0, 1, …, N. Then we have

dom d^{∗} ∩ ⋀_{0}^{q+1}(\overline{Ω}) = \{ϕ ∈ ⋀_{0}^{q+1}(\overline{Ω}) : ∇n⃗ϕ⌊n⃗|_{bΩ}= 0\}.

Here ∇n⃗ denotes the covariant differentiation of a form in the normal direction, and “⌊” the contraction operator between a form and a vector field.

The next result shows a striking difference with the classical case. We begin with the case of ℝ_{+}^{N+1}. Here, and in the rest of the paper, we denote by △′ the Laplace operator defined on the boundary of the half space:

△′ = ∑_{j=1}^{N} \frac{∂^{2}}{∂x_{j}^{2}} .

Proposition 2.2. Let d′ denote the formal adjoint of d. Then on dom d^{∗} we have

d^{∗} = d′ + K,

where K is an operator sending (q + 1)-forms to q-forms. The operator K is the solution operator of the following boundary value problem

\begin{cases}(− △ +I)(Kϕ) = 0 & on ℝ_{+}^{N+1} \\ \frac{∂}{∂x_{0}} (Kϕ) = (− △′ +I) (ϕ⌊(\frac{∂}{∂x_{0}} )) & on ℝ^{N} = bℝ^{N+1}. \end{cases}

Explicitly, if ϕ ∈ ⋀^{q+1}(\overline{ℝ_{+}^{N+1}}), ϕ = ∑_{|I|=q+1} ϕ_{I} dx^{I}, then

(Kϕ)_{I}ˆ (x_{0}, ξ′) = −\sqrt{1 + |2πξ′|^{2}}e^{\sqrt{1+|2πξ′|^{2}}x_{0}} ϕ̂_{0I}(0, ξ′).

In the case of a smoothly bounded domain we have the following:

Proposition 2.3. Let Ω be a smoothly bounded domain. Then, on dom d^{∗},

d^{∗} = d′ + K_{Ω},

where d′ is the formal adjoint of d, and K_{Ω} is an operator sending (q + 1)-forms to q-forms. The components of Kϕ are solutions of the following boundary value problems

\begin{cases}(− △ +I)(K_{Ω}ϕ)_{I} = 0 & on Ω \\ \frac{∂}{∂n}(K_{Ω}ϕ)_{I} = T_{2}ϕ⌊n⃗ & on bΩ \end{cases} ,

where T_{2} is a second order tangential differential operator on forms (to be defined in detail below).

At this point we fix the following notation that we will use throughout the entire paper.

Definition 2.4. On the half space ℝ_{+}^{N+1} we set

G = Kd + dK,

and for a smoothly bounded domain Ω we set

G_{Ω} = K_{Ω}d + dK_{Ω}.

Notice that our boundary value problems become

\begin{cases}(− △ +G)ϕ = α & on ℝ_{+}^{N+1} \\ ϕ ∈ dom d^{∗} \\ dϕ ∈ dom d^{∗} \end{cases} ,  (2.1)

and

\begin{cases}(− △ +G_{Ω})ϕ = α & on Ω \\ ϕ ∈ dom d^{∗} \\ dϕ ∈ dom d^{∗} \end{cases} .  (2.2)

We are now ready to state our results about existence and regularity of the boundary value problem.

THEOREM 2.5. Let s > 1/2. Let α ∈ W_{q}^{s}(ℝ_{+}^{N+1}), supp α ⊆ {x : |x| < R}. Let N ≥ 4, Then there exists a unique tempered distribution ϕ solution of the boundary value problem (2.1) such that

∥ϕ∥_{s+2} ≤ c∥α∥_{s},

where c = c(s, R) > 0 does not depend on α, nor on ϕ.

If N = 2, 3 and ∫ α_{I} dx = 0, and if N = 1 and ∫ x^{β}α_{I} dx = 0 when |β| ≤ 1, then the conclusion as above still holds.

THEOREM 2.6. Consider the boundary value problem (2.2). Let s > 1/2. Then there exists a finite dimensional subspace (the harmonic space) H_{q} of ⋀^{q}(\overline{Ω}) and a constant c = c_{s} > 0 such that if α ∈ W_{q}^{s}(Ω) is orthogonal (in the W^{1}-sense) to H_{q}, then the boundary value problem (2.2) has a unique solution ϕ orthogonal to H_{q} such that

∥ϕ∥_{s+2} ≤ c∥α∥_{s}.

Remark. If s < 1, by saying that α is orthogonal in the W^{1}-sense we mean that α is W^{s}-limit of smooth forms that are orthogonal in W^{1} to H_{q}.

Finally we have

THEOREM 2.7. Let Ω be a smoothly bounded domain in ℝ^{N+1}. Let W_{q}^{1}(Ω) denote the 1-Sobolev space of q-forms. Then we have the strong orthogonal decomposition

W_{q}^{1} = dd^{∗}(W_{q}^{1}) ⨁ d^{∗}d(W_{q}^{1}) ⨁ H_{q} ,

where H_{q} is a finite dimensional subspace.

THE PROBLEM ON THE HALF SPACE

3. The Operator d^{∗} on 1-Forms and Its Domain

The aim of this section is to determine dom d^{∗} ∩ ⋀^{1}(\overline{ℝ_{+}^{N+1}}), and to give an explicit expression for d^{∗}.

Easy computations show that the formal adjoint d′ of d with respect to the inner product of W^{1} is the same as the formal adjoint in the classical L^{2}-case. Therefore, for ϕ ∈ ⋀_{0}^{1}(ℝ_{+}^{N+1}), ϕ = ∑_{j=0}^{N} ϕ_{j} dx^{j}, we see that

d′ϕ = − ∑_{j=0}^{N} D_{j}ϕ_{j} ≡ −div ϕ.

It is not hard to see that the specific form of d′ is independent of which order s of Sobolev inner product we use.

Next we want to compute the Hilbert space adjoint d^{∗} of d. Recall that W_{q}^{s} will be the closure of the smooth q-forms ⋀_{0}^{q}(\overline{ℝ_{+}^{N+1}}) with compact support in \overline{ℝ_{+}^{N+1}} with respect to the norm

∥ϕ∥_{s} ≡ ∑_{{|α|≤s \atop |I|=q}} ∫_{ℝ_{+}^{N+1}} |D^{α}ϕ_{I}|^{2} dV < ∞.

Let u ∈ ⋀^{0}, ϕ ∈ ∑_{i=0}^{N} ϕ_{i} dx^{i} have compact support in \overline{ℝ_{+}^{N+1}} (that is, the support is compact but not necessarily disjoint from the boundary). Then

⟨du, ϕ⟩_{s} = ∑_{j=0}^{N}⟨D_{j}u, ϕ_{j}⟩_{1}

= ∑_{k=0}^{N} (∑_{j=1}^{N} ∫_{ℝ_{+}^{N+1}} D_{j}D_{k}u\overline{D_{k}ϕ_{j}} dV + ∫_{ℝ_{+}^{N+1}} D_{0}(D_{k}u)\overline{D_{k}ϕ_{0}} dV ) .

The first sum inside the parentheses in the last line equals

− ∑_{j=1}^{N} ∫_{ℝ_{+}^{N+1}} D_{k}u \overline{D_{j}D_{k}ϕ_{j}} dV

since u and ϕ_{j} have compact support and the derivation is in the tangential directions (i.e. the directions x_{1}, …, x_{N}). The second term in parentheses equals (with x′ = (x_{1}, …, x_{N}))

∫_{ℝ^{N}} (D_{k}u\overline{D_{k}ϕ_{0}}|_{x_{0}=0}^{x_{0}=∞} − ∫_{0}^{∞} D_{k}u\overline{D_{0}D_{k}ϕ_{0}} dx_{0}) dx′

= − ∫_{ℝ^{N}} D_{k}u\overline{D_{k}ϕ_{0}}|_{x_{0}=0} dx′ − ∫_{ℝ_{+}^{N+1}} D_{k}u\overline{D_{0}D_{k}ϕ_{0}} dV (x).

Therefore, combining the last several lines, we find that

⟨du, ϕ⟩_{1} = ∑_{k=0}^{N} (− ∑_{j=0}^{N} ∫_{ℝ_{+}^{N+1}} D_{k}u\overline{D_{j}D_{k}ϕ_{j}} − ∫_{ℝ^{N}} D_{k}u\overline{D_{k}ϕ_{0}}|_{x_{0}=0} dx′)

= ⟨u, d′ϕ⟩_{1} − ∫_{ℝ^{N}} u\overline{ϕ_{0}}|_{x_{0}=0} dx′ − ∑_{k=1}^{N} ∫_{ℝ^{N}} D_{k}u\overline{D_{k}ϕ_{0}}|_{x_{0}=0} dx′.

Now we can determine the domain of d^{∗} (we must do this before we can calculate the operator d^{∗} itself). Recall that, by the definition of Hilbert space adjoint, ϕ ∈ dom d^{∗} if and only if there is a number c_{ϕ} such that for all u ∈ ⋀^{0} we have

|⟨du, ϕ⟩_{1}| ≤ c_{ϕ}∥u∥_{1}.

Proposition 3.1. The 1-form ϕ lies in (dom d^{∗}) ∩ ⋀^{1}(\overline{ℝ_{+}^{N+1}}) if and only if

D_{0}ϕ_{0}|_{x_{0}=0} ≡ 0.

Proof. We begin by setting

D = dom d^{∗} ∩ ⋀_{0}^{1}(\overline{ℝ_{+}^{N+1}})

and

E = \{ϕ ∈ ⋀_{0}^{1}(\overline{ℝ_{+}^{N+1}}) : \frac{∂}{∂x_{0}} ϕ_{0}(0, x′) ≡ 0\} .

We first show that E ⊆ D. Let ϕ ∈ E. We have already computed that

⟨du, ϕ⟩_{1} = ⟨u, d′ϕ⟩_{1} − ∫_{ℝ^{N}} u\overline{ϕ_{0}}|_{x_{0}=0} dx′ − ∑_{k=1}^{N} ∫_{ℝ^{N}} D_{k}u\overline{D_{k}ϕ_{0}}|_{x_{0}=0} dx′

= ⟨u, d′ϕ⟩_{1} + ∫_{ℝ^{N}} u|_{x_{0}=0}(△′ϕ_{0} − ϕ_{0})|_{x_{0}=0} dx′.

Clearly the mapping

u ⟼ ⟨u, d′ϕ⟩_{1}

is bounded on W^{1}(ℝ_{+}^{N+1}), and so

|⟨du, ϕ⟩_{1}| ≤ c_{ϕ}(∥u∥_{W^{1}(ℝ_{+}^{N+1})} + ∥u|_{x_{0}=0}∥_{L^{2}(ℝ^{N})}).

(Note that the bound here depends on ϕ, but that is acceptable.) Now, recall the standard trace theorems for Sobolev spaces (see either [LIM] or [KR1]). Then

∥u|_{x_{0}=0}∥_{L^{2}(ℝ^{N})} ≤ ∥u∥_{W^{t}(ℝ_{+}^{N+1})},

for t > 1/2, and in particular for t = 1. This establishes the containment E ⊆ D.

Next we establish that D ⊂ E. Seeking a contradiction, we suppose that ϕ ∈ D ≡ dom d^{∗} ∩ ⋀_{0}^{1}(ℝ_{+}^{N+1}) and that D_{0}ϕ_{0}(0, x′) is not identically zero. We may suppose that ϕ_{0} is real and that

D_{0}ϕ_{0}(0, x′) ≥ 1 when |x′| ≤ 2

(just multiply by a suitable constant and scale).

Now set

u_{ϵ}(x_{0}, x′) = (x_{0} + ϵ)^{3/4}χ(x_{0}, x′),

where χ is a cutoff function with support in {|x_{0}| ≤ 2} × {|x′| ≤ 2} and such that χ ≡ 1 in the set {|x_{0}| ≤ 1} × {|x′| ≤ 1}.

We claim that there is a constant C > 0, independent of ϵ, such that

(i)  ∥u_{ϵ}∥_{W^{s}(ℝ_{+}^{N+1})} ≤ C  ∀ϵ, 0 < ϵ ≤ ϵ_{0} ,

and

(ii)  ∫_{ℝ^{N}} D_{0}u_{ϵ}\overline{D_{0}ϕ_{0}}|_{x_{0}=0} dx′ ∼ \frac{1}{ϵ^{1/4}}  as ϵ → 0.

Once these two facts are established, it is clear that it is not possible to find a constant such that the mapping u ⟼ ⟨du, ϕ⟩_{1} is bounded on W^{1}(ℝ_{+}^{N+1}), and thus ϕ cannot be in dom d^{∗}.

Fact (ii) follows since

D_{0}u_{ϵ}(x_{0}, x′)|_{x_{0}=0} = (\frac{3}{4}ϵ^{−1/4}χ(0, x′) + ϵ^{3/4}D_{0}χ(0, x′)) .

In order to prove (i) we need only check the size of

∫_{ℝ_{+}^{N+1}} |D_{0}u_{ϵ}|^{2}dV.

But

D_{0}u_{ϵ}(x_{0}, x′) = \frac{3}{4} \frac{1}{(x_{0} + ϵ)^{1/4}} χ(x_{0}, x′) + Φ_{ϵ}(x_{0}, x′),

where Φ_{ϵ} lies in C_{0}^{∞}(ℝ_{+}^{N+1}) and is uniformly bounded in ϵ. Thus

∫_{ℝ_{+}^{N+1}} |D_{0}u_{ϵ}|^{2} dV ≤ C ∫_{0}^{1} \frac{1}{(x_{0} + ϵ)^{1/2}} dx_{0} + C′
≤ C,

independent of ϵ as ϵ → 0. This establishes (i).

The operators K and d^{∗}. Our next task is to determine the explicit expression for the operator d^{∗}. We will learn that, contrary to the classical case, the operator d^{∗} when restricted to dom (d^{∗}) ∩ ⋀^{1} does not coincide with d′. Once we have the expression for d^{∗} then we are in a position to formulate the boundary value problem in the case q = 1. The fundamental calculation is this:

Proposition 3.2. If ϕ ∈ dom d^{∗} ∩ ⋀_{0}^{1}(\overline{ℝ_{+}^{N+1}}), ϕ = ∑_{j=0}^{N} ϕ_{j} dx^{j}, then

d^{∗}ϕ = −div ϕ + ∫_{ℝ^{N}} \frac{∂}{∂x_{0}} (e^{−\sqrt{1+4π^{2}|ξ′|^{2}}x_{0}}) ϕ̂_{0}(0, ξ′)e^{2πix′·ξ′} dξ′

Proof. We need to determine the operator d^{∗} that satisfies

⟨df, ϕ⟩_{1} = ⟨f, d^{∗}ϕ⟩_{1}.

The spirit of the calculation that follows is to rewrite the expressions that involve the inner product ⟨ · , · ⟩_{s} in terms of expressions that involve only the L^{2} inner product.

Now

⟨df, ϕ⟩_{1} = ∫_{ℝ_{+}^{N+1}} ∑_{j=0}^{N} \frac{∂f}{∂x_{j}} \overline{ϕ_{j}} dV + ∫_{ℝ_{+}^{N+1}} ∑_{j,k=0}^{N} \frac{∂^{2}f}{∂x_{j}∂x_{k}} \overline{\frac{∂ϕ_{j}}{∂x_{k}}} dV
=^{(parts)} ⟨f, − div ϕ⟩_{1} − ∫_{ℝ^{N}} f(0, x′)\overline{ϕ_{0}(0, x′)} dx′

+ ∫_{ℝ^{N}} f(0, x′)\overline{∑_{k=1}^{N} \frac{∂^{2}}{∂x_{k}^{2}} ϕ_{0}(0, x′)} dx′.

Of course we have used the boundary condition that occurs in our characterization of dom d^{∗}.

Next we write

d^{∗}ϕ = − div ϕ + u,

and we wish to determine the explicit expression for u ≡ Kϕ, i.e. the formula for the operator K mapping 1-forms into functions, such that d^{∗} = d′ + K. Then we have

⟨df, ϕ⟩_{1} = ⟨f, − div ϕ⟩_{1} + ⟨f, u⟩_{1}
= ⟨f, − div ϕ⟩_{1} + ∫_{ℝ_{+}^{N+1}} f\overline{u} dV

− ∫_{ℝ_{+}^{N+1}} f\overline{△u} dV − ∫_{ℝ^{N}} f(0, x′)\overline{\frac{∂u}{∂x_{0}} (0, x′)} dx′.

Here we have integrated by parts.

Comparing the two last calculations we see that the function u must satisfy the following equation:

− ∫_{ℝ^{N}} f(0, x′)\overline{ϕ_{0}(0, x′)} dx′ + ∫_{ℝ^{N}} f(0, x′)\overline{∑_{k=1}^{N} \frac{∂^{2}}{∂x_{k}^{2}} ϕ_{0}(0, x′)} dx′

= ∫_{ℝ_{+}^{N+1}} f\overline{u} dV − ∫_{ℝ_{+}^{N+1}} f\overline{△u} dV − ∫_{ℝ^{N}} f(0, x′)\overline{\frac{∂u}{∂x_{0}} (0, x′)} dx′.

This equality must hold for all f ∈ C_{0}^{∞}(\overline{ℝ_{+}^{N+1}}). By choosing functions supported away from the boundary we obtain a differential equation on ℝ_{+}^{N+1}; inserting this in the equation above gives rise to a boundary condition. Thus the function u ∈ W^{1}(ℝ_{+}^{N+1}) must be a solution of the following boundary value problem:

\begin{cases}− △ u + u = 0 & on ℝ_{+}^{N+1} \\ \frac{∂u}{∂x_{0}} (0, ·) = − △′ ϕ_{0}(0, ·) + ϕ_{0}(0, ·) & on \{0\} × ℝ^{N} \end{cases} .

Using the partial Fourier transform with respect to x′, we find that the equation △u − u = 0 becomes

\frac{∂^{2}}{∂x_{0}^{2}} û(x_{0}, ξ′) − (1 + |2πξ′|^{2})û(x_{0}, ξ′) = 0, with û ∈ L^{2}(ℝ_{+}^{N+1}).

For fixed ξ′, this is an ordinary differential equation in x_{0} whose solution is

û(x_{0}, ξ′) = a(ξ′)e^{\sqrt{1+|2πξ′|^{2}}x_{0}} + b(ξ′)e^{−\sqrt{1+|2πξ′|^{2}}x_{0}}.

Since û ∈ L^{2}(ℝ_{+}^{N+1}), it must be that a(ξ′) ≡ 0. Thus

û(x_{0}, ξ′) = b(ξ′)e^{−\sqrt{1+|2πξ′|^{2}}x_{0}}  (3.1)

and the solution to the above boundary value problem is

Kϕ(x_{0}, x′) = ∫_{ℝ^{N}} b(ξ′)e^{−\sqrt{1+|2πξ′|^{2}}x_{0}}e^{2πiξ′·x′} dξ′.

Now we determine b(ξ′) in such a way that the boundary conditions are satisfied. We compute the derivative of û(x_{0}, ξ), as given by (3.1), with respect to x_{0}, and evaluate at x_{0} = 0. We find that

b(ξ′)\sqrt{1 + |2πξ′|^{2}} = −|2πξ′|^{2}ϕ̂_{0}(0, ξ′) − ϕ̂_{0}(0, ξ′),

that is,

b(ξ′) = −\sqrt{1 + |2πξ′|^{2}}ϕ̂_{0}(0, ξ′).

Therefore the Hilbert space adjoint of d (and this has been the main point of these calculations) is

d^{∗}ϕ ≡ − div ϕ + Kϕ
= − div ϕ − ∫_{ℝ^{N}} \sqrt{1 + |2πξ′|^{2}}e^{−\sqrt{1+|2πξ′|^{2}}x_{0}} ϕ̂_{0}(0, ξ′)e^{2πiξ′·x′} dx′
≡ − div ϕ + (K_{x_{0}} ∗ ϕ_{0}(0, ·))(x′);

here the convolution is in ℝ^{N} with respect to the variable x′, and

K_{x_{0}}(x′) = (\frac{∂}{∂x_{0}} e^{−\sqrt{1+|2πξ′|^{2}}x_{0}})ˇ (x′)

is the convolution kernel.

At this point, following the classical methodology of Hodge theory, we want to study the problem (2.1):

\begin{cases}(dd^{∗} + d^{∗}d)u = f \\ u, du ∈ dom d^{∗} \end{cases} .

According to our calculations this problem becomes

\begin{cases}− △ u + ∫_{ℝ^{N}} (\frac{∂}{∂x_{0}} e^{−\sqrt{1+|2πξ′|^{2}}x_{0}}) ˆ \frac{∂u}{∂x_{0}} (0, x′)e^{2πiξ′·x′} dx′ = f & on ℝ_{+}^{N+1} \\ \frac{∂^{2}}{∂x_{0}^{2}} u(0, x′) = 0 & on ℝ^{N} \end{cases} . (3.2)

At the conclusion of these calculations we want to remark that the boundary value problems that appear in the case of higher degree forms is treated in Section 6.

Recall, from Definition 2.4, that

Gu(x_{0}, x′) = K(\frac{∂u}{∂x_{0}} )(x_{0}, x′)

= ∫_{ℝ^{N}} (\frac{∂}{∂x_{0}} e^{−\sqrt{1+|2πξ′|^{2}}x_{0}}) ˆ \frac{∂u}{∂x_{0}} (0, x′)e^{2πiξ′·x′} dx.  (3.3)

Now we define K̃ by setting K ≡ K̃ ∘ γ, where γ = γ_{0} denotes the standard trace operator of restriction to the boundary.

Notice that the kernel K̂_{x_{0}}(ξ′) is −\sqrt{1 + |2πξ′|^{2}}e^{−\sqrt{1+|2πξ′|^{2}}}x_{0}, and thus is quite similar to the normal derivative of the Fourier transform of the Poisson kernel. But actually the operator K̃ is more regular than the Poisson integral itself. The higher degree of regularity of K̃ is seen immediately by noticing that the symbol of the convolution kernel K_{x_{0}}, that is −\sqrt{1 + |2πξ′|^{2}}e^{−\sqrt{1+|2πξ′|^{2}}x_{0}}, is smooth in ξ′ for all ξ′. We record here a few simple properties of the operators K̃ and G. We denote by S(ℝ^{N}) the space of Schwartz functions on ℝ^{N}, and by S(\overline{ℝ_{+}^{N+1}}) the space of restriction to ℝ_{+}^{N+1} of Schwartz functions on ℝ^{N+1}. We have the following proposition.

Proposition 3.3. We have that

K̃ : S(ℝ^{N}) ⟶ S(\overline{ℝ_{+}^{N+1}})

continuously. Moreover

K̃ : W^{r+1/2}(ℝ^{N}) ⟶ W^{r}(ℝ_{+}^{N+1}),

continuously for r ≥ 0. In addition,

G : W^{r+2}(ℝ^{N+1}) ⟶ W^{r}(ℝ_{+}^{N+1}),

and

G : S(\overline{ℝ_{+}^{N+1}}) ⟶ S(\overline{ℝ_{+}^{N+1}}),

continuously.

Proof. Recall that G = K̃ ∘ γ ∘ (∂/∂x_{0}), so it suffices to prove the corresponding statements for K̃.

Let v ∈ W^{r+1/2}(ℝ^{N}). Then

ˆK̃v(x_{0}, ξ′) = −\sqrt{1 + |2πξ′|^{2}}e^{−\sqrt{1+|2πξ′|^{2}}} v̂(ξ′).

It suffices to consider the case r an integer, and use interpolation to obtain the general case. For r ∈ ℕ we have

∥K̃v∥_{W^{r}(ℝ_{+}^{N+1})}^{2} ≤ ∑_{j=0}^{r} ∫_{ℝ_{+}^{N+1}}(1 + |ξ′|^{2})^{r−j} |\frac{∂^{j} ˆK̃v}{∂x_{0}^{j}} (x_{0}, ξ′)|^{2} dξ′ dx_{0}
≤ c ∫_{ℝ_{+}^{N+1}}(1 + |ξ′|^{2})^{r+1}|v̂(ξ′)|^{2}e^{−\sqrt{1+|2πξ′|^{2}}x_{0}} dx_{0} dξ′
≤ c ∫_{ℝ^{N}}(1 + |ξ′|^{2})^{r+1/2}|v̂(ξ′)|^{2} dξ′
≤ c∥v∥_{W^{r+1/2}(ℝ^{N})}^{2}.

To show that K̃ is continuous between the (two different) Schwartz spaces, we notice that

|x_{0}^{ℓ}(2πξ′)^{α} \frac{∂}{∂x_{0}^{ℓ}} \frac{∂}{∂ξ′^{α}} ˆ K̃v(x_{0}, ξ′)|

≤ |x_{0}^{ℓ+|α|}(1 + |2πξ′|^{2})^{\frac{ℓ+1}{2} +|α|}e^{−\sqrt{1+|2πξ′|^{2}}x_{0}} v̂(ξ′)|
≤ c\{sup_{x_{0}≥0} [x_{0}^{ℓ+|α|}e^{−x_{0}}]\} sup_{|ξ′|}(1 + |2πξ′|)^{\frac{ℓ+1}{2} +|α′|}|v̂(ξ′)|,

which is finite for v ∈ S(ℝ^{N}).

We conclude this section with some remarks.

REMARK 3.4. We have that

−ϕ_{0} = σ(d′, dx_{0}) ϕ,

where σ(d′, dx_{0}) is the symbol of d′ in the normal direction. With this notation,

⟨du, ϕ⟩_{1} = ⟨u, d′ϕ⟩_{1} + ∫_{bℝ_{+}^{N+1}}⟨u, σ(d′, dx_{0})ϕ⟩_{1} dx′,

where bℝ^{N+1} is the boundary of the half space.

Indeed, using the standard definition (see, for instance, [KR1]) of symbol, we select a testing function ρ such that ρ(0, x′) = 0 and dρ = dx_{0} (for instance, ρ(x) = x_{0} locally will do). Then

σ(d′, dx_{0}) = d′(ρϕ)|_{x=(0,x′)}

= − ∑_{j=0}^{N} D_{j}(ρϕ_{j})|_{x_{0}=0}
= −ϕ_{0}.

Thus we may write, for ϕ = ∑_{|I|=q} ϕ_{I} dx^{I},

⟨du, ϕ⟩_{1} = ⟨u, d′ϕ⟩_{1} + ∫_{bℝ_{+}^{N+1}}⟨u, σ(d′, dx_{0})ϕ⟩_{1} dx′.

In the next Proposition we describe the domain of d^{∗} in the topology of W^{s}(ℝ_{+}^{N+1}) in case s ∈ ℕ. Essentially the same computations as in the case s = 1 prove the following result.

Proposition 3.5. Denote by d^{∗} the Hilbert space adjoint of d in the topology of W^{s}(ℝ_{+}^{N+1}) for s = 1, 2, . . .. Let ϕ ∈ ⋀_{0}^{1}(\overline{ℝ_{+}^{N+1}}). Then ϕ lies in (dom d^{∗}) ∩ ⋀^{1}(\overline{ℝ_{+}^{N+1}}) if and only if

D_{0}^{s}ϕ_{0}|_{x_{0}=0} ≡ \frac{∂^{s}}{∂x_{0}^{s}} ϕ_{0}|_{x_{0}=0} ≡ 0.

Proof. We only indicate the changes in the proof for the case s = 1. We have that

⟨du, ϕ⟩_{s} = ⟨u, d′ϕ⟩_{s} − ∑_{j=0}^{s−1} ∑_{|α′|≤s−j} ∫_{bℝ_{+}^{N+1}} D_{0}^{j}D^{α′}u\overline{D_{0}^{j}D^{α′}ϕ_{0}} dx′.

Clearly the mapping

u ⟼ ⟨u, d′ϕ⟩_{s}

is bounded on W^{s}(ℝ_{+}^{N+1}). Then, using the trace theorem, we want to show that for any j, 0 ≤ j ≤ s − 1, and |α′| ≤ s − j, the mapping

u ⟼ I_{j} ≡ ∫_{bℝ_{+}^{N+1}} D_{0}^{j}D^{α′}u \overline{D_{0}^{j}D^{α′}ϕ_{0}} dx′

is bounded in the W^{s}(ℝ_{+}^{N+1}) norm. This establishes the containment E ⊆ D.

In order to establish that D ⊂ E, suppose that ϕ ∈ D ≡ dom d^{∗} ∩ ⋀_{0}^{1}(ℝ_{+}^{N+1}) and that

D_{0}^{s}ϕ_{0}(0, x′) ≡ \frac{∂^{s}}{∂x_{0}^{s}} ϕ_{0}(0, x′)

is not identically zero. Setting

u_{ϵ}(x_{0}, x′) = x_{0}^{s−1}(x_{0} + ϵ)^{3/4}χ(x_{0}, x′),

and arguing as before, it is easy to see that

(i)  ∥u_{ϵ}∥_{W^{s}(ℝ_{+}^{N+1})} ≤ C  ∀ϵ, 0 < ϵ ≤ ϵ_{0}

and

(ii)  ∑_{j=0}^{s} ∑_{|α′|≤s−j} ∫_{ℝ^{N}} D_{0}^{j}D^{α′}u_{ϵ}\overline{D_{0}^{j}D^{α′}ϕ_{0}}|_{x_{0}=0} dx′ ∼ \frac{1}{ϵ^{1/4}}  as ϵ → 0.

These two facts show that ϕ cannot be in dom d^{∗}.

Finally, we also would like to observe that the boundary value problems (2.1) and (2.2) are in fact pseudodifferential boundary value problems. This kind of boundary value problems has been studied by several authors (see [BDM3], [ESK], [GRU], and [RESC]), as we indicate in the next section.

4. Boutet de Monvel-Type Analysis of the Boundary Value Problem

The boundary value problems (2.1) and (2.2) can be seen as an instance of a very general theory developed by Boutet de Monvel and several other authors. In this section we give a short summary of this theory and of its applications to our problem. Such a general approach mainly produces results of local character in which the data are often assumed to be more regular then we can afford. This is why we actually choose a different, and more direct, approach to solving our boundary value problems.

For maximum generality, let Ω be either a smoothly bounded domain in ℝ^{N+1} or a half space in ℝ^{N+1}. Define A to be the system

A = ( \begin{matrix}P_{Ω} + G & K \\ T & S \end{matrix} ) : ( \begin{matrix}C_{0}^{∞}(\overline{Ω}) \\ C_{0}^{∞}(bΩ)^{M} \end{matrix} ) ⟶ ( \begin{matrix}C^{∞}(\overline{Ω}) \\ C^{∞}(bΩ)^{M′} \end{matrix} ) .  (4.1)

The components of this formula are as follows:

1) The numbers M, M′ are non-negative integers, C_{0}^{∞}(bΩ)^{M} is the Cartesian product of M copies of C_{0}^{∞}(bΩ), and C^{∞}(bΩ)^{M′} is the Cartesian product of M′copies of C^{∞}(bΩ);

2) The operator P_{Ω} is a pseudodifferential operator taking functions defined on Ω into functions defined on Ω. [For more on pseudodifferential operators, see [KR1].] We will require P_{Ω} to have the “transmission property” (to be defined below);

3) The operator G is a “singular Green operator.” It is defined on functions on Ω, taking values in the set of functions defined on Ω;

4) The operator K is a “Poisson operator”. It takes (M-tuples of) functions on the boundary to functions on the domain;

5) The operator T is a “trace operator”. Let γ_{ℓ} be the classical trace operator of order ℓ,

u ⟼ \frac{∂^{ℓ}u}{∂x_{0}^{ℓ}} |_{x_{0}=0}
C_{0}^{∞}(\overline{Ω}) ⟶ C_{0}^{∞}(bΩ).

By definition, a trace operator T is the composition of a pseudodifferential operator S′ on the boundary and some restriction operator γ_{ℓ}: T = S′ ∘ γ_{ℓ}.

6) The operator S (to repeat) is a pseudodifferential operator on the boundary. Notice that our problem

\begin{cases}(− △ +G)u = f & on ℝ_{+}^{N+1} \\ \frac{∂^{2}u}{∂x_{0}^{2}}(0, x′) = 0 & on \{0\} × ℝ^{N} \end{cases} .

is a special case of the above setup, with M = 0, M′ = 1. We would take K = 0, S = 0 and consider

A = ( \begin{matrix}P_{Ω} + G \\ T \end{matrix} ) : C_{0}^{∞}(\overline{Ω}) ⟼ ( \begin{matrix}C^{∞}(\overline{Ω}) \\ C^{∞}(bΩ) \end{matrix} ) .

We shall construct a parametrix R for this operator which will have the form

R = (Q K) : ( \begin{matrix}C_{0}^{∞}(\overline{Ω}) \\ C_{0}^{∞}(bΩ) \end{matrix} ) ⟶ C^{∞}(\overline{Ω}).

Note here that M = 1 and M′ = 0.

The classical boundary value problems (such as the Dirichlet or Neumann problems) have the form

A = ( \begin{matrix}D \\ T \end{matrix} ) : C_{0}^{∞}(\overline{Ω}) ⟶ ( \begin{matrix}C^{∞}(\overline{Ω}) \\ C^{∞}(bΩ)^{M′} \end{matrix} ) ,

where D is a differential operator and M′ is the number of boundary conditions.

Definition 4.1. A pseudodifferential operator P of order d, defined on ℝ^{N+1}, is said to have the transmission property on Ω if the mapping

u ⟼ RP Eu =^{def} P_{Ω}u

maps W^{s}(Ω) continuously to W^{s−d}(Ω) for all s > −1/2. Here E is the operator given by extending a function on \overline{Ω} to all of ℝ^{N} by setting it equal to zero on the complement. Also R is the operator of restriction to Ω.

For our purposes, it is enough to note that any partial differential operator with smooth coefficients has the transmission property. This follows by inspection.

Definition 4.2. A Poisson operator K of order d is an operator defined by the formula

(Kv)(x_{0}, x′) = ∫_{ℝ^{N}} e^{2πix′·ξ′} k̃(x_{0}, x′, ξ′)v̂(ξ′) dξ′,

where the symbol k̃ satisfies the estimates for |ξ′| ≥ 1 given by

∥x_{0}^{ℓ}D_{0}^{ℓ′}D_{x′}^{β}D_{ξ′}^{α}k̃( · , x′, ξ′)∥_{L^{2}(ℝ_{+})} ≤ c(x′)(1 + |ξ′|^{2})^{(1/2)(d−1/2−ℓ−ℓ′−|α|)}.

Here c(x′) is a continuous function on ℝ^{N} that depends on ℓ, ℓ′, α, β.

One can show that a Poisson operator K of order d is continuous as a mapping from W_{comp}^{s}(ℝ^{N}) to W_{loc}^{s−d+1/2}(\overline{ℝ_{+}^{N+1}}). By W_{comp}^{s}(Ω) we mean those elements of W^{s}(ℝ^{N}) with compact support in Ω.

It is a standard fact that the Poisson integral for the upper half space is a Poisson operator, according to the above definition, of order 0; that is it maps W_{comp}^{s}(ℝ^{N}) (where ℝ^{N} = ∂ℝ^{N+1}) to W_{loc}^{s+1/2}(\overline{ℝ_{+}^{N+1}}).

Definition 4.3. A trace operator of order d ∈ ℝ and of class r ∈ ℕ is an operator of the form

T u = ( ∑_{0≤ℓ≤r−1} S_{ℓ}γ_{ℓ})u,

where γ_{ℓ} is the standard trace operator of order ℓ and each S_{ℓ} is a pseudodifferential operator on ℝ^{N} of order d − ℓ.

The notion of “class” is a natural artifact of dealing with restriction operators in the context of Sobolev spaces. Recall that the classical restriction theorems sending elements of W^{s}(ℝ_{+}^{N+1}) to elements of W^{s−1/2}(ℝ^{N}) are only valid when s > 1/2. The idea of class addresses the necessary lower bound for s in theorems such as this.

Definition 4.4. A singular Green operator of order d ∈ ℝ and class r ∈ ℕ is an operator of the form

G = ∑_{0≤ℓ≤r−1} K_{ℓ}γ_{ℓ},

where each K_{ℓ} is a Poisson operator of order d − ℓ and each γ_{ℓ} is a standard trace operator of order ℓ.

It can be shown that a singular Green operator of order d and class r is a continuous operator

G : W_{comp}^{s}(\overline{ℝ_{+}^{N+1}}) ⟶ W_{loc}^{s−d}(\overline{ℝ_{+}^{N+1}})

as long as s > r − 1/2.

Definition 4.5. Let A be a system as in (4.1). We define the associated boundary system by

A = ( \begin{matrix}p_{Ω}(0, x′, D_{0}, ξ′) + g(x′, D_{0}, ξ′) & k(x′, D_{0}, ξ′) \\ t(x′, D_{0}, ξ′) & s(x′, ξ′) \end{matrix} ) .

Notice that, in this definition, we fix x′, ξ′ ∈ ℝ^{N}, and the symbol p_{Ω} of P_{Ω} is taken at the point (0, x′) ∈ bℝ_{+}^{N+1}. Finally, all the operators in the display act only in the D_{0} slot.

Further observe that

p_{Ω}(0, x′, D_{0}, ξ′) + g(x′, D_{0}, ξ′) : S(\overline{ℝ_{+}}) ⟶ S(\overline{ℝ_{+}^{N+1}})

and that

k(x′, D_{0}, ξ′) : ℂ ⟶ S(\overline{ℝ_{+}^{N+1}})
a ⟼ ak(x′, x_{0}, ξ′).

Furthermore, the boundary symbol operator of a trace operator is an operator

t(x′, D_{0}, ξ′)u = ∑_{0≤ℓ≤r−1} s_{ℓ}(x′, ξ′)γ_{ℓ}u

that maps S(\overline{ℝ_{+}}) into ℂ.

A system of the type (4.1) is called elliptic if there exists a second system R of type (4.1) such that

RA + I and AR + I

are negligible operators; here “negligible” means that all terms in the system send distributions with compact support into C^{∞} functions. In the case of the half space this definition must be considered as “local”, that is fixing a compact set in \overline{ℝ_{+}^{N+1}}.

Now we have

THEOREM 4.6 ([BDM3, Theorem 5.1]). A system a of type (4.1) is elliptic if and only if both its boundary and its interior symbols are invertible.

In the case of (2.1),

A = ( \begin{matrix}− △ +G \\ T \end{matrix} ) ,

where T =^{def} (I − △′)^{−1/2}γ_{2}. Then we return to our analysis of the problem

\begin{cases}− △ u + Gu = f & on ℝ_{+}^{N+1} \\ T u = 0 & on \{0\} × ℝ^{N} \end{cases}  (4.2)

Here, for convenience, we have replaced our usual boundary operator (∂^{2}/∂x_{0}^{2}) with the (equivalent) operator T.

Our aim is to show that this is a pseudodifferential boundary value problem of type (4.1) and to apply Theorem 4.6 to our system.

First observe that −△, being a partial differential operator of order 2 with smooth coefficients, certainly possesses the transmission property.

Next, let us analyze G. Recall that G = K̃γ_{1}, where K̃ has symbol

K_{x_{0}}(ξ′) ≡ k(x_{0}, ξ′) = −\sqrt{1 + |2πξ′|^{2}}e^{−\sqrt{1+|2πξ′|^{2}}x_{0}}.

In order to see that G is a singular Green operator, we need to estimate

∥x_{0}^{ℓ}D_{0}^{ℓ′}D_{ξ′}^{α}k( · , ξ′)∥_{L_{x_{0}}^{2}(ℝ_{+})}^{2}

= ∫_{0}^{∞}|x_{0}^{ℓ}D_{ξ′}^{α} [(−\sqrt{1 + 4π^{2}|ξ′|^{2}})^{ℓ′+1}e^{−\sqrt{1+4π^{2}|ξ′|^{2}}x_{0}}]|^{2} dx_{0}

The calculations made in Proposition 3.3 showed that K̃ has order 1. In fact K̃ is essentially a single derivative of the Poisson operator (which has order zero) so the assertion is plausible. It follows that

order (G) = order (K̃) + order (γ_{1}) = 2

and

class (G) = order (γ_{1}) + 1 = 2.

Next we observe that T is a trace operator of order 1 and class 3. Indeed, T =^{def} (I − △′′)^{−1/2}γ_{2} so that

order (T) = order (I − △′)^{−1/2} + order (γ_{2}) = −1 + 2 = 1

and

class (T) = order (γ_{2}) + 1 = 3.

The interior symbol of this system is just the symbol of −△ which is plainly invertible. The boundary symbol is

( \begin{matrix}(|2πξ′|^{2} − \frac{∂^{2}}{∂x_{0}^{2}} ) + (\frac{∂}{∂x_{0}} e^{−\sqrt{1+|2πξ′|^{2}}x_{0}}) γ_{1} \\ (1 + |ξ′|^{2})^{−1/2}γ_{2} \end{matrix} ) : S(\overline{ℝ_{+}}) ⟶ ( \begin{matrix}S(\overline{ℝ_{+}}) \\ ℂ \end{matrix} ) .

The demand that this system be invertible means that the pseudodifferential system

\begin{cases}v′′(x_{0}) − (\frac{∂}{∂x_{0}} e^{−\sqrt{1+4π^{2}|ξ′|^{2}}x_{0}}) v′(0) − |2πξ′|^{2}v(x_{0}) = ψ & on ℝ \\ (1 + |2πξ′|^{2})^{−1/2}v′′(0) = α \end{cases}

has one and only one solution v ∈ S(\overline{ℝ_{+}}) for all ψ ∈ S(\overline{ℝ_{+}}) and α ∈ ℂ and for |ξ′| ≥ c > 0 fixed.

Straightforward and unenlightening calculations now prove the following result.

Proposition 4.7. For |ξ′| ≥ c > 0 fixed, the preceding system of equations has a unique solution for every choice of (ψ, α) ∈ S(\overline{ℝ_{+}}) × ℂ.

As a consequence we have:

THEOREM 4.8. Consider the boundary value problem given by (4.2):

\begin{cases}(− △ +G)u = f & on ℝ_{+}^{N+1} \\ T u = g & on \{0\} × ℝ^{N} \end{cases} .

Let s > 5/2, and suppose that supp f, supp g ⊆ B(0, R), for some R > 0. Moreover, let η ∈ C_{0}^{∞}(\overline{ℝ_{+}^{N+1}}). Then there exists a finite dimensional subspace L of W_{comp}^{s}(ℝ_{+}^{N+1}) × W_{comp}^{s−1/2}(ℝ^{N}) such that if (f, g) ∈ L^{⊥} then the problem (4.2) has a unique solution u satisfying

∥ηu∥_{W^{s+2}(ℝ_{+}^{N+1})} ≤ C(∥f∥_{W^{s}(ℝ_{+}^{N+1})} + ∥g∥_{W^{s−1/2}(ℝ^{N})}).

The constant C in the above estimate depends on R, s, the test function η, and on the spatial dimension N.

Our aim is to refine this theorem in order to allow s > 1/2 (instead of s > 5/2) and to be more precise about the local-global aspects of the problem. Moreover, we do so by determining the solution explicitly, and by describing the compatibility conditions.

5. The Explicit Solution in the Case of Functions

Our goal in the present section is to prove theorems about existence and regularity of the solution of the boundary value problem. We shall determine the solution of our boundary value problem explicitly. This will allow us to give precise estimates.

Thus we concentrate on the problem

\begin{cases}− △ u + Gu = f & on ℝ_{+}^{N+1} \\ \frac{∂^{2}u}{∂x_{0}^{2}}(0, ·) = 0 & on ℝ^{N} \end{cases}  (5.1)

for u ∈ W^{r}(ℝ_{+}^{N+1}), where r > 5/2. Notice that we need this limitation on r in order to guarantee that ∂^{2}u/∂x_{0}^{2}(0, ·) makes sense and belongs to L^{2}(ℝ^{N}) (at least). Here, for x_{0} > 0 and ξ′ ∈ ℝ^{N}, we have

Ĝu(x_{0}, ξ′) = −\sqrt{1 + |2πξ′|^{2}}e^{−\sqrt{1+|2πξ′|^{2}}x_{0}} \frac{∂û}{∂x_{0}} (0, ξ′).

Recall that we write ŵ(ξ′) to denote the partial Fourier transform of w in ℝ^{N}, with respect to the x′-variables.

For greater flexibility we will solve the boundary value problem

\begin{cases}− △ u + Gu = f & on ℝ_{+}^{N+1} \\ \frac{∂^{2}u}{∂x_{0}^{2}}(0, ·) = h & on ℝ^{N} \end{cases}  (5.2)

In what follows we shall denote by G the classical Green’s function for the Laplacian on ℝ_{+}^{N+1}; also P denotes the standard Poisson operator on ℝ_{+}^{N+1}. Recall (see [GAR]) that, for N ≥ 2,

G(x, y) = \frac{1}{(N − 1)ω_{N+1}} [\frac{1}{((x_{0} − y_{0})^{2} + |x′ − y′|^{2})^{(N−1)/2}}

− \frac{1}{((x_{0} + y_{0})^{2} + |x′ − y′|^{2})^{(N−1)/2}} ],

where ω_{N+1} denotes the surface measure of the N-dimensional unit sphere in ℝ^{N+1}. If N = 1 then

G(x, y) = \frac{1}{2π} log \frac{(x_{0} + y_{0})^{2} + (x_{1} − y_{1})^{2}}{(x_{0} − y_{0})^{2} + (x_{1} − y_{1})^{2}} .

Notice that

Ĝf(x_{0}, ξ′) = \frac{1}{2} \frac{1}{|2πξ′|} ∫_{0}^{∞}(e^{−|2πξ′|·|x_{0}−y_{0}|} − e^{−|2πξ′|(x_{0}+y_{0})})f̂(y_{0}, ξ′) dy_{0}. (5.3)

Moreover, we shall adopt the following notation:

m_{1}(ξ′) = \frac{1 + |2πξ′|^{2} + |2πξ′|\sqrt{1 + |2πξ′|^{2}}}{1 + 2|2πξ′|^{2} + |2πξ′|\sqrt{1 + |2πξ′|^{2}}} ,  (5.4)

and

m_{2}(ξ′) = [\frac{|2πξ′| + \sqrt{1 + |2πξ′|^{2}}}{1 + 2|2πξ′|^{2} + |2πξ′|\sqrt{1 + |2πξ′|^{2}}} ] = \frac{1}{\sqrt{1 + |2πξ′|^{2}}} m_{1}(ξ′). (5.5)

We now construct the solution to the boundary value problem explicitly.

Proposition 5.1. If u is a solution of the boundary value problem (5.2) in W^{r}(ℝ_{+}^{N+1}) with r > 5/2, then u is given by the following formula

u(x_{0}, x′) = Gf(x_{0}, x′) + G (∫_{ℝ^{N}} e^{−\sqrt{1+|2πξ′|^{2}}x_{0}}m_{1}(ξ′)

× [|2πξ′| ∫_{0}^{+∞} e^{−|2πξ′|y_{0}} f̂(y_{0}, ξ′) dy_{0} − ĥ(ξ′) − f̂(0, ξ′)] e^{2πix′·ξ′} dξ′)

+P (\frac{ĥ(ξ′) + f̂(0, ξ′)}{1 + 2|2πξ′|^{2} + |2πξ′|\sqrt{1 + |2πξ′|^{2}}} )ˇ

+P (\frac{m_{1}(ξ′)}{|2πξ′|}(ĥ(ξ′) + ∫_{0}^{+∞} f̂(y_{0}, ξ′)e^{−|2πξ′|y_{0}} dy_{0}))ˇ  (5.6)

Proof. Our first task is to reduce the order of the operator appearing in the boundary condition by using the equation on the domain. Thus the problem (5.2) is equivalent to

\begin{cases}− △ u + Gu = f & on ℝ_{+}^{N+1} \\ − △′ u(0, ·) = f(0, ·) + h − Gu(0, ·) & on ℝ^{N} \end{cases}

for u ∈ W^{r}(ℝ_{+}^{N+1}) with r > 5/2, where △′ =^{def} ∑_{j=1}^{N} ∂^{2}/∂x_{j}^{2}. The problem (5.2) can be rewritten as

\begin{cases}− △ u(x_{0}, x′) = F (x_{0}, x′) & on ℝ_{+}^{N+1} \\ u(0, x′) = g(x′) & on ℝ^{N} \end{cases}

(again for u ∈ W^{r}(ℝ_{+}^{N+1}), r > 5/2) where we have set

F (x_{0}, x′) = f(x_{0}, x′) + ∫_{ℝ^{N}} \sqrt{1 + |2πξ′|^{2}}e^{−\sqrt{1+|2πξ′|^{2}}x_{0}} \frac{∂û}{∂x_{0}} (0, ξ′)e^{2πix′·ξ′} dξ′, (5.7)

and

g(x′) = N (h + f(0, ·))(x′) + N ((\sqrt{1 + |2πξ′|^{2}} \frac{∂û}{∂x_{0}} (0, ·))ˇ (0, x′)) , (5.8)

with N being the Newtonian potential in ℝ^{N}. By the classical theory we then obtain that the solution we seek can be written as

u(x_{0}, x′) = ∫_{ℝ_{+}^{N+1}} G(x_{0}, y)F (y) dy + ∫_{ℝ^{N}} P_{x_{0}}(x′, y′)g(y′) dy′
≡ (GF)(x_{0}, x′) + Pg(x_{0}, x′).

Here P_{x_{0}} denote the Poisson kernel in ℝ^{N}. Now we need to isolate u on the left hand side (note that F, g are defined in terms of u). In order to do this we compute the

HODGE THEORY IN THE SOBOLEV TOPOLOGY 25

derivative with respect to x_{0} of the above equation, take the partial Fourier transform, and evaluate at x_{0} = 0. Using definitions (5.7) and (5.8) we have

\frac{∂û}{∂x_{0}} |_{x_{0}=0} = [(\frac{∂(GF)}{∂x_{0}}) + (\frac{∂Pg}{∂x_{0}} ) ]|_{x_{0}=0}
= ∫_{0}^{+∞}(PF) (y_{0}, ξ′) dy_{0} − |2πξ′|ĝ(ξ′)
= ∫_{0}^{+∞} e^{−|2πξ′|y_{0}} F̂(y_{0}, ξ′) dy_{0} − |2πξ′|ĝ(ξ)
= ∫_{0}^{+∞} e^{−|2πξ′|y_{0}} f̂(y_{0}, ξ′) dy_{0}

\sqrt{1 + |2πξ′|^{2}} + \frac{∂û}{∂x_{0}} (0, ξ′) ∫_{0}^{+∞} e^{−(\sqrt{1+|2πξ′|^{2}}+|2πξ′|)y_{0}} dy_{0}

−|2πξ′| [\frac{ĥ(ξ′)}{|2πξ′|^{2}} + \frac{f̂(0, ξ′)}{|2πξ′|^{2}} + \frac{\sqrt{1 + |2πξ′|^{2}}(∂û/∂x_{0})(0, ξ′)}{|2πξ′|^{2}} ]

= ∫_{0}^{+∞} e^{−|2πξ′|y_{0}} f̂(y_{0}, ξ′) dy_{0} − \frac{ĥ(ξ′)}{|2πξ′|} − \frac{f̂(0, ξ′)}{|2πξ′|}

+ [\frac{\sqrt{1 + |2πξ′|^{2}}}{\sqrt{1 + |2πξ′|^{2}} + |2πξ′|} − \frac{\sqrt{1 + |2πξ′|^{2}}}{|2πξ′|} ] \frac{∂û}{∂x_{0}} (0, ξ′).

Hence

\frac{∂û}{∂x_{0}} (0, ξ′) [1 − \frac{\sqrt{1 + |2πξ′|^{2}}}{\sqrt{1 + |2πξ′|^{2}} + |2πξ′|} + \frac{\sqrt{1 + |2πξ′|^{2}}}{|2πξ′|} ]

= ∫_{0}^{+∞} e^{−|2πξ′|y_{0}} f̂(y_{0}, ξ′) dy_{0} − \frac{ĥ(ξ′)}{|2πξ′|} − \frac{f̂(0, ξ′)}{|2πξ′|},

that is,

\frac{∂û}{∂x_{0}} (0, ξ′) = |2πξ′|m_{2}(ξ′) \{∫_{0}^{+∞} e^{−|2πξ′|y_{0}} f̂(y_{0}, ξ′) dy_{0} − \frac{ĥ(ξ′)}{|2πξ′|} − \frac{f̂(0, ξ′)}{|2πξ′|}\} . (5.9)

Now we have the expression for ∂u/∂x_{0} at x_{0} = 0. Notice that, for t > 0,

∥\frac{∂u}{∂x_{0}} (0, ·)∥_{W^{t}(ℝ^{N})} ≤ C(∥f∥_{W^{t+1/2}(ℝ_{+}^{N+1})} + ∥h∥_{W^{t−1}(ℝ^{N})}),

so that our calculations make sense so far. Substituting equation (5.9) into (5.7) and (5.8), and recalling that m_{1} = \sqrt{1 + |2πξ′|^{2}}m_{2}, we find that

F (x_{0}, x′) = f(x_{0}, x′)

+ ∫_{ℝ^{N}} |2πξ′|m_{1}(ξ′)e^{−\sqrt{1+|2πξ′|^{2}}}(∫_{0}^{+∞} e^{−|2πξ′|y_{0}} f̂(y_{0}, ξ′) dy_{0} − \frac{ĥ(ξ′)}{|2πξ′|} − \frac{f̂(x_{0}, ξ′)}{|2πξ′|}),

and

g(x′) = N (h + f(0, ·))

+ N ((m_{1}(ξ′)|2πξ′|\{∫_{0}^{+∞} e^{−|2πξ′|y_{0}} f̂(y_{0}, ξ′) dy_{0} − \frac{ĥ(ξ′)}{|2πξ′|} − \frac{f̂(0, ξ′)}{|2πξ′|}\})ˇ )

= N (h + f(0, ·)) + N ((|2πξ′|m_{1}(ξ′) ∫_{0}^{+∞} e^{−|2πξ′|y_{0}} f̂(y_{0}, ξ′) dy_{0})ˇ )

− N ((m_{1}ĥ + m_{1}f̂(0, ·))ˇ ).

Finally,

u(x_{0}, x′) = GF (x_{0}, x′) + Pg(x′)
= Gf(x_{0}, x′) + G (∫_{ℝ^{N}} e^{−\sqrt{1+|2πξ′|^{2}}x_{0}}m_{1}(ξ′)

× [|2πξ′| ∫_{0}^{+∞} e^{−|2πξ′|y_{0}} f̂(y_{0}, ξ′) dy_{0} − ĥ(ξ′) − f̂(0, ξ′)] e^{2πix′·ξ′} dξ′)

+P (\frac{ĥ(ξ′) + f̂(0, ξ′)}{1 + 2|2πξ′|^{2} + |2πξ′|\sqrt{1 + |2πξ′|^{2}}} )ˇ

+P (\frac{m_{1}(ξ′)}{|2πξ′|} ∫_{0}^{+∞} f̂(y_{0}, ξ′)e^{−|2πξ′|y_{0}} dy_{0})ˇ .  (5.10)

This gives the explicit expression for the solution u.

The a priori Estimate. In this part we are going to prove the a priori estimate for the solution u; then we will turn to the question of existence. Now we need the following two lemmas.

Lemma 5.2. There exists a constant c such that, for |α| = 2, and t ≥ 0,

∥D^{α}Gw∥_{W^{t}(ℝ_{+}^{N+1})} ≤ c∥w∥_{W^{t}(ℝ_{+}^{N+1})}.

Proof. For a function w defined on ℝ_{+}^{N+1}, let w_{odd} denote the odd extension (in the x_{0} variable) to ℝ^{N+1}, and N_{N+1} the Newtonian potential in ℝ^{N+1}. Let F denote the Fourier transform in ℝ^{N+1}. Observe that

(Gw)_{odd} = N_{N+1}w_{odd},

so that we have

F (\frac{∂^{2}(Gw)_{odd}}{∂x_{i}x_{j}} ) (ξ) = \frac{ξ_{i}ξ_{j}}{|ξ|^{2}} F(w_{odd})(ξ),

and the estimate for t = 0 follows. When t = 1, 2, … notice that, for x ∈ ℝ_{+}^{N+1},

(\frac{∂^{2}(Gw)}{∂x_{0}^{2}})_{odd}(x) = −w − △′Gw(x).

Then it suffices to consider the case when the derivative is of degree at most 1 in the x_{0}-direction. This takes care of the fact that w_{odd} is not differentiable in the x_{0}-direction. Now the estimate follows easily for t an integer. Interpolation gives the result for all t ≥ 0.

The proof of the next lemma is easy:

Lemma 5.3. Let w be defined on ℝ^{N}, |α| = 2, and t ≥ 0. Then there exists a constant C > 0 such that

∥D^{α}Pw∥_{W^{t}(ℝ_{+}^{N+1})}^{2} ≤ C ∫_{ℝ^{N}} |2πξ′|^{3}(1 + |2πξ′|)^{2+t}|ŵ(ξ′)|^{2} dξ′.

THEOREM 5.4. Let f ∈ W^{s}(ℝ_{+}^{N+1}), with s > 1/2. If the boundary value problem (5.2) admits the solution u given by Proposition 5.1, then it satisfies the estimate

∥u∥_{W^{s+2}(ℝ_{+}^{N+1})} ≤ C {∥f∥_{W^{s}(ℝ_{+}^{N+1})} + ∥h∥_{W^{s−1/2}(ℝ^{N})} + ∥u∥_{W^{1}(ℝ_{+}^{N+1})}} . (5.11)

Proof. In order to obtain a priori estimates we use the following standard fact (see [LIM] Theorem 9.7). If s ≥ 0, then

∥u∥_{W^{s+2}(ℝ_{+}^{N+1})} ≤ c ( ∑_{|α|=2} ∥\frac{∂^{|α|}u}{∂x^{α}} ∥_{W^{s}(ℝ_{+}^{N+1})} + ∥u∥_{W^{1}(ℝ_{+}^{N+1})}) .

28 L. FONTANA, S. G. KRANTZ, AND M. M. PELOSO

Now recall the expression for the solution given by (5.10)

u(x_{0}, x′) = Gf(x_{0}, x′) + G (∫_{ℝ^{N}} e^{−\sqrt{1+|2πξ′|^{2}}x_{0}}m_{1}(ξ′)

× [|2πξ′| ∫_{0}^{+∞} e^{−|2πξ′|y_{0}} f̂(y_{0}, ξ′) dy_{0} − ĥ(ξ′) − f̂(0, ξ′)] e^{2πix′·ξ′} dξ′)

+P (\frac{ĥ(ξ′) + f̂(0, ξ′)}{1 + 2|2πξ′|^{2} + |2πξ′|\sqrt{1 + |2πξ′|^{2}}} )ˇ

+P (\frac{m_{1}(ξ′)}{|2πξ′|} ∫_{0}^{+∞} f̂(y_{0}, ξ′)e^{−|2πξ′|y_{0}} dy_{0})ˇ

≡ Gf + Gf_{1} + Pg_{1} + Pg_{2}.

It follows that

∥u∥_{W^{s+2}(ℝ_{+}^{N+1})} ≤ c∥u∥_{W^{1}(ℝ_{+}^{N+1})} + c ∑_{|α|=2}(∥\frac{∂^{|α|}Gf}{∂x^{α}} ∥_{W^{s}(ℝ_{+}^{N+1})}

+ ∥\frac{∂^{|α|}Gf_{1}}{∂x^{α}} ∥_{W^{s}(ℝ_{+}^{N+1})} + ∥\frac{∂^{|α|}Pg_{1}}{∂x^{α}} ∥_{W^{s}(ℝ_{+}^{N+1})} + ∥\frac{∂^{|α|}Pg_{2}}{∂x^{α}} ∥_{W^{s}(ℝ_{+}^{N+1})})

Lemma 5.2 implies that, for |α| = 2,

∥D^{α}Gf∥_{W^{t}(ℝ_{+}^{N+1})} ≤ c∥f∥_{W^{t}(ℝ_{+}^{N+1})},

and

∥D^{α}Gf_{1}∥_{W^{t}(ℝ_{+}^{N+1})} ≤ c∥f_{1}∥_{W^{t}(ℝ_{+}^{N+1})} .

Notice that f_{1} = K̃(h_{1}), where

ĥ_{1}(ξ′) = m_{2}(ξ′) [|2πξ′| ∫_{0}^{+∞} e^{−|2πξ′|y_{0}} f̂(y_{0}, ξ′) dy_{0} − ĥ(ξ′) − f̂(0, ξ′)] . Then, for s > 1/2,

∥f_{1}∥_{W^{s}(ℝ_{+}^{N+1})} ≤ c∥h_{1}∥_{W^{s+1/2}(ℝ^{N})}
= c{∫_{ℝ^{N}}(1 + |ξ′|^{2})^{s+1/2}|m_{2}(ξ′)|^{2}

× ||2πξ′| ∫_{0}^{+∞} e^{−|2πξ′|y_{0}} f̂(y_{0}, ξ′) dy_{0} − ĥ(ξ′) − f̂(0, ξ′)|^{2} dξ′}^{1/2}

≤ c (∥h∥_{W^{s−1/2}(ℝ^{N})} + ∥f(0, ·)∥_{W^{s−1/2}(ℝ^{N})} + ∥f∥_{W^{s}(ℝ_{+}^{N+1})})
≤ c(∥f∥_{W^{s}(ℝ_{+}^{N+1})} + ∥h∥_{W^{s−1/2}(ℝ^{N})}).

Hence it suffices to consider the W^{t}(ℝ_{+}^{N+1}) norm of any second derivative of Pg_{1} and Pg_{2}. For s > 1/2, by Lemma 5.3 we have

∥D^{α}P(\frac{ĥ(ξ′) + f̂(0, ξ′)}{1 + 2|2πξ′|^{2} + |2πξ′|\sqrt{1 + |2πξ′|^{2}}} )∥_{W^{s}(ℝ_{+}^{N+1})}

≤ c {∫_{ℝ^{N}} |2πξ′|^{3}(1 + |2πξ′|^{2})^{s−2}(|ĥ(ξ′)|^{2} + |f̂(0, ξ′)|^{2}) dξ′}^{1/2}
≤ c{∥f(0, ·)∥_{W^{s−1/2}(ℝ^{N})} + ∥h∥_{W^{s−1/2}(ℝ^{N})}}
≤ c{∥f∥_{W^{s}(ℝ_{+}^{N+1})}∥h∥_{W^{s−1/2}(ℝ^{N})}},

Moreover, recalling the definition of m_{1}(ξ′) (see (5.4)), we have

∥D^{α}P(\frac{m_{1}(ξ′)}{|2πξ′|} ∫_{0}^{+∞} f̂(y_{0}, ξ′)e^{−|2πξ′|y_{0}} dy_{0})∥_{W^{s}(ℝ_{+}^{N+1})}

≤ c \{∫_{ℝ^{N}} |2πξ′|^{3}(1 + |2πξ′|^{2})^{s} \frac{1}{|2πξ′|^{2}} | ∫_{0}^{+∞} f̂(y_{0}, ξ′)e^{−|2πξ′|y_{0}} dy_{0}|^{2} dξ′\}^{1/2}

≤ c {∫_{ℝ^{N}}(1 + |2πξ′|^{2})^{s} ∫_{0}^{+∞} |f̂(y_{0}, ξ′)|^{2} dy_{0} dξ′}^{1/2}
= c∥f∥_{W^{s}(ℝ_{+}^{N+1})}.

This proves the a priori estimate.

The existence Theorem. Now we turn to the existence statement. If we prove that the function u defined by (5.10) belongs to W^{1}(ℝ_{+}^{N+1}) then, by the estimate (5.11), the equation (5.10) defines the unique solution of the boundary value problem. The solution belongs to W^{r}(ℝ_{+}^{N+1}) with r > 5/2.

THEOREM 5.5. If f ∈ W^{s} ∩ L^{1}(ℝ_{+}^{N+1}), with s > 1/2, then the boundary value problem (5.1) has a unique solution u, and this satisfies

∥u∥_{W^{s+2}(ℝ_{+}^{N+1})} ≤ C {∥f∥_{W^{s}(ℝ_{+}^{N+1})} + ∥f∥_{L^{1}(ℝ_{+}^{N+1})} + ∥h∥_{W^{s−1/2}(ℝ^{N})}}

if N ≥ 4. If N = 2, 3 and f satisfies

∫ f(x) dx = 0, f ∈ L^{1}(|x|dx, ℝ_{+}^{N+1}),

then there exists a unique solution u that satisfies the estimate

∥u∥_{W^{s+2}(ℝ_{+}^{N+1})} ≤ C {∥f∥_{W^{s}(ℝ_{+}^{N+1})} + ∥f∥_{L^{1}(|x|dx,ℝ_{+}^{N+1})} + ∥h∥_{W^{s−1/2}(ℝ^{N})}} .

If N = 1, we assume that f is such that

∫ f(x) dx = 0, ∫ x_{i}f(x) dx = 0, i = 0, 1 f ∈ L^{1}(|x|^{2}dx, ℝ^{2}).

Then there exists a unique solution u such that

∥u∥_{W^{s+2}(ℝ_{+}^{N+1})} ≤ C {∥f∥_{W^{s}(ℝ_{+}^{N+1})} + ∥f∥_{L^{1}(|x|^{2}dx,ℝ_{+}^{N+1})} + ∥h∥_{W^{s−1/2}(ℝ^{N})}} .

Proof. By Theorem 5.4 it suffices to show that

∥u∥_{W^{1}(ℝ_{+}^{N+1})} ≤ c {∥f∥_{W^{s}(ℝ_{+}^{N+1})} + ∥f∥_{L^{1}(ℝ_{+}^{N+1})} + ∥h∥_{W^{s−1/2}(ℝ^{N})}} ,

when N ≥ 4, or the corresponding estimate when N = 1, 2, 3, with f satisfying the stated conditions. Notice that

∥u∥_{W^{1}(ℝ_{+}^{N+1})} ≤ ∥u∥_{L^{2}(ℝ_{+}^{N+1})} + ∑_{i=0}^{N}∥\frac{∂u}{∂x_{i}} ∥_{L^{2}(ℝ_{+}^{N+1})}.

Recall that (5.10) gives the function u as

u = Gf + Gf_{1} + Pg_{1} + Pg_{2}.

We calculate the partial Fourier transform of u using the fact that

(G[(e^{−\sqrt{1+|2πξ′|^{2}}x_{0}})ˇ ]) (x_{0}, ξ′) = e^{−|2πξ′|x_{0}} − e^{−\sqrt{1+|2πξ′|^{2}}x_{0}}.

[Note here thatˇ and do not cancel, since the Green’s potential G occurs in between the two operations. Reference line (5.3).] We obtain

û(x_{0}, ξ′) = Ĝf(x_{0}, ξ′) + Ĝf_{1}(x_{0}, ξ′) + P̂g_{1}(x_{0}, ξ′) + P̂g_{2}(x_{0}, ξ′)
= Ĝf(x_{0}, ξ′) + m_{1}(ξ′) [e^{−|2πξ′|x_{0}} − e^{−\sqrt{1+|2πξ′|^{2}}x_{0}}]
× [|2πξ′| ∫_{0}^{+∞} e^{−|2πξ′|y_{0}} f̂(y_{0}, ξ′) dy_{0} − ĥ(ξ′) − f̂(0, ξ′)]

+ \frac{e^{−|2πξ′|x_{0}}(ĥ(ξ′) + f̂(0, ξ′))}{1 + 2|2πξ′|^{2} + |2πξ′|\sqrt{1 + |2πξ′|^{2}}}

+ e^{−|2πξ′|x_{0}} \frac{m_{1}(ξ′)}{|2πξ′|} ∫_{0}^{+∞} e^{−|2πξ′|y_{0}} f̂(y_{0}, ξ′) dy_{0} .

Moreover,

(\frac{∂û}{∂x_{0}} ) (x_{0}, ξ′) = (\frac{∂}{∂x_{0}} Ĝf) (x_{0}, ξ′)

+m_{1}(ξ′) [\sqrt{1 + |2πξ′|^{2}}e^{−\sqrt{1+|2πξ′|^{2}}x_{0}} − |2πξ′|e^{−|2πξ′|x_{0}}]

×[|2πξ′| ∫_{0}^{+∞} e^{−|2πξ′|y_{0}} f̂(y_{0}, ξ′) dy_{0} − ĥ(ξ′) − f̂(0, ξ′)]

−\frac{|2πξ′|e^{−|2πξ′|x_{0}}(ĥ(ξ′) + f̂(0, ξ′))}{1 + 2|2πξ′|^{2} + |2πξ′|\sqrt{1 + |2πξ′|^{2}}}

−e^{−|2πξ′|x_{0}}m_{1}(ξ′) ∫_{0}^{+∞} e^{−|2πξ′|y_{0}} f̂(y_{0}, ξ′) dy_{0} .

Then ∥u∥_{W^{1}(ℝ_{+}^{N+1})} will be estimated once we estimate ∥û∥_{L^{2}(ℝ_{+}^{N+1})}, ∥|2πξ′|û∥_{L^{2}(ℝ_{+}^{N+1})}, and ∥(∂u/∂x_{0})ˆ ∥_{L^{2}(ℝ_{+}^{N+1})}.

We begin by studying the terms that arise from Gf. We extend f to all of ℝ^{N+1} as an odd function f_{odd} of x_{0} as we did in the proof of Lemma 5.2. We obtain

∥Gf∥_{L^{2}(ℝ_{+}^{N+1})} = \frac{1}{\sqrt{2}}∥\frac{1}{|2πξ′|^{2}} F(N f_{odd})∥_{L^{2}(ℝ^{N+1})}

and

∥\frac{∂}{∂x_{i}} Gf∥_{L^{2}(ℝ_{+}^{N+1})} = \frac{1}{\sqrt{2}} ∥\frac{∂}{∂x_{i}} Ff_{odd}∥_{L^{2}(ℝ^{N+1})} .

Case N ≥ 4. We have

∥Gf∥_{L^{2}(ℝ_{+}^{N+1})} = c · ∥F(N_{N+1}f_{odd})∥_{L^{2}(ℝ^{N+1})}
≤ c \{∫_{|2πξ|≥1} \frac{|Ff_{odd}(ξ)|^{2}}{|2πξ|^{4}} dξ\}^{1/2} + c \{∫_{|2πξ|≤1} \frac{|Ff_{odd}(ξ)|^{2}}{|2πξ|^{4}} dξ\}^{1/2}
≤ c {∥f∥_{L^{2}(ℝ^{N+1})} + (∫_{|2πξ|≤1} |2πξ|^{−4} dξ)^{1/2} sup_{ξ∈ℝ^{N+1}} |Ff_{odd}(ξ)|}
≤ c {∥f∥_{L^{2}(ℝ^{N+1})} + ∥f∥_{L^{1}(ℝ^{N+1})}} .

∥\frac{∂Gf}{∂x_{i}} ∥_{L^{2}(ℝ^{N+1})} ≤ \{∫_{ℝ^{N+1}} \frac{|Ff_{odd}(ξ)|^{2}}{|2πξ|^{2}} dξ\}^{1/2}
≤ c \{∫_{|2πξ|≤1} \frac{|Ff_{odd}(ξ)|^{2}}{|2πξ|^{2}} dξ\}^{1/2} + c \{∫_{|2πξ|≥1} \frac{|Ff_{odd}(ξ)|^{2}}{|2πξ|^{2}} dξ\}^{1/2}
≤ c {∥f∥_{L^{2}(ℝ^{N+1})} + ∥f∥_{L^{1}(ℝ^{N+1})}} .

Now writing û(x_{0}, ξ′) = Ĝf(x_{0}, ξ′) + B(x_{0}, ξ′), i.e. setting

B = Ĝf_{1} + P̂g_{1} + P̂g_{2} ,  (5.12)

we must estimate the terms ∥B∥_{L^{2}(ℝ_{+}^{N+1})}, ∥|2πξ′|B∥_{L^{2}(ℝ_{+}^{N+1})}, and ∥(∂/∂x_{0})B∥_{L^{2}(ℝ_{+}^{N+1})}. We have

∥B∥_{L^{2}(ℝ_{+}^{N+1})} ≤ c {∫_{|2πξ|≤1} ∫_{0}^{+∞} |B(x_{0}, ξ′)|^{2} dx_{0} dξ′}^{1/2}
+c {∫_{|2πξ|≥1} ∫_{0}^{+∞} |B(x_{0}, ξ′)|^{2} dx_{0} dξ′}^{1/2} .

We begin by estimating the integral on the set where |2πξ′| ≤ 1. Notice that, if N ≥ 4, then

|B(x_{0}, ξ′)| ≤ C\{e^{−|2πξ′|x_{0}}|ĥ(ξ′) − f̂(0, ξ′)||\frac{1}{1 + 2|2πξ′|^{2} + |2πξ′|\sqrt{1 + |2πξ′|^{2}}} − m_{1}(ξ′)|
+ e^{−|2πξ′|x_{0}}|2πξ′|^{−1} |∫_{0}^{+∞} e^{−|2πξ′|y_{0}} f̂(y_{0}, ξ′)dy_{0}|}

≤ C\{e^{−|2πξ′|x_{0}}(ĥ(ξ′) + f̂(y_{0}, ξ′))|2πξ′| + \frac{e^{−|2πξ′|x_{0}}}{|2πξ′|} ∫_{0}^{+∞} |f̂(y_{0}, ξ′)| dy_{0}\}.

From this, the restriction theorem, and the fact that

∫_{0}^{+∞} |f̂(y_{0}, ξ′)| dy_{0} ≤ ∫_{0}^{+∞} ∫_{ℝ^{N}} |f(y_{0}, x′)| dx′ dy_{0} = ∥f∥_{L^{1}(ℝ_{+}^{N+1})},  (5.13)

we obtain that {∫_{|2πξ′|≤1} ∫_{0}^{+∞} |B(x_{0}, ξ′)|^{2} dx_{0} dξ′}^{1/2}

≤ C\{∥h∥_{L^{2}(ℝ^{N})} + ∥f(0, ξ′)∥_{L^{2}(ℝ^{N})} + (∫_{|2πξ′|≤1} \frac{1}{|2πξ′|^{3}} (∫_{0}^{+∞} |f̂(y_{0}, ξ′)| dy_{0})^{2} dξ′)^{1/2}\}

≤ C{∥h∥_{L^{2}(ℝ^{N})} + ∥f∥_{W^{s}(ℝ_{+}^{N+1})} + ∥f∥_{L^{1}(ℝ_{+}^{N+1})}}. To study the integral for |2πξ′| ≥ 1 we first notice that

∫_{0}^{+∞} (e^{−|2πξ′|x_{0}} − e^{−\sqrt{1+|2πξ′|^{2}}x_{0}})^{2} dx_{0} = (2|2πξ′|(|2πξ′| + \sqrt{1 + |2πξ′|^{2}})^{3})^{−1}
= O(|2πξ′|^{−4}) as |ξ′| → ∞.

Using this fact, Schwarz’s inequality, and the restriction theorem, we see that the integral we want to estimate is less than or equal to a constant times

\{∫_{|2πξ′|≥1} \frac{1}{|2πξ′|^{4}} [||2πξ′| ∫_{0}^{+∞} e^{−|2πξ′|y_{0}} f̂(y_{0}, ξ′) dy_{0}|^{2} + |ĥ(ξ′)|^{2} + |f̂(0, ξ′)|^{2}] dξ′\}^{1/2}
≤ c \{(∫_{|2πξ′|≥1} \frac{1}{|2πξ′|^{3}} ∫_{0}^{+∞} |f̂(y_{0}, ξ′)|^{2} dy_{0} dξ′)^{1/2} + ∥h∥_{L^{2}(ℝ^{N})} + ∥f(0, ·)∥_{L^{2}(ℝ^{N})}\}
≤ c {∥f∥_{W^{s}(ℝ_{+}^{N+1})} + ∥h∥_{L^{2}(ℝ^{N})}}.

In the same way we obtain that ∥|2πξ′|B∥_{L^{2}(ℝ_{+}^{N+1})}

≤ {∫_{|2πξ′|≤1} ∫_{0}^{+∞} |2πξ′|^{2}|B(y_{0}, ξ′)|^{2} dy_{0} dξ′}^{1/2}

+ {∫_{|2πξ′|≥1} ∫_{0}^{+∞} |2πξ′|^{2}|B(y_{0}, ξ′)|^{2} dy_{0} dξ′}^{1/2}

≤ c {∥f(0, ·)∥_{L^{2}(ℝ^{N})} + ∥h∥_{L^{2}(ℝ^{N})} + (∫_{|2πξ′|≤1} |2πξ′|^{−1} dξ′)^{1/2}∥f∥_{L^{1}(ℝ^{N})}

+(∫_{|2πξ′|≥1} | ∫_{0}^{+∞} e^{−|2πξ′|y_{0}} f̂(y_{0}, ξ′) dy_{0}|^{2} dξ′)^{1/2}

+(∫_{|2πξ′|≥1} \frac{|ĥ(ξ′)|^{2} + |f̂(0, ξ′)|^{2}}{|2πξ′|^{2}} dξ′)^{1/2}\}

≤ c {∥f∥_{W^{s}(ℝ_{+}^{N+1})} + ∥f∥_{L^{1}(ℝ_{+}^{N+1})} + ∥h∥_{L^{2}(ℝ^{N})}} ,

by the Schwarz inequality and the restriction theorem.

In order to estimate ∥∂B/∂x_{0}∥_{L^{2}(ℝ_{+}^{N+1})} we observe that 2 ∫_{0}^{+∞}(|2πξ′|e^{−|2πξ′|x_{0}} − \sqrt{1 + |2πξ′|^{2}}e^{−\sqrt{1+|2πξ′|^{2}}x_{0}})^{2} dx_{0}

= (|2πξ′| + \sqrt{1 + |2πξ′|^{2}})^{−3} = O(|2πξ′|^{−3})

as |ξ′| → ∞. Using this fact and (5.13) we see that

∥\frac{∂B}{∂x_{0}} ∥_{L^{2}(ℝ_{+}^{N+1})} ≤ \{∫_{|2πξ′|≤1} ∫_{0}^{+∞} dy_{0} dξ′\}^{1/2} + \{∫_{|2πξ′|≥1} ∫_{0}^{+∞} dy_{0} dξ′\}^{1/2}
≤ c {(∫_{ℝ^{N}} |f̂(0, ξ′)|^{2} dξ′)^{1/2} + (∫_{ℝ^{N}} |ĥ(ξ′)|^{2} dξ′)^{1/2}

+ (∫_{|2πξ′|≤1} |2πξ′|^{−1} dξ′)^{1/2} ∥f∥_{L^{1}(ℝ_{+}^{N+1})}

+ (∫_{|2πξ′|≥1} |2πξ′|^{−1}| ∫_{0}^{+∞} e^{−|2πξ′|y_{0}} f̂(y_{0}, ξ′) dy_{0}|^{2} dξ′)^{1/2}}

≤ c{∥f∥_{W^{1/2}(ℝ_{+}^{N+1})} + ∥f∥_{L^{1}(ℝ_{+}^{N+1})} + ∥h∥_{L^{2}(ℝ^{N})}}.

This concludes the estimate in the case N ≥ 4.

Case N < 4. Now we must consider the lower dimensional cases. Again, we first estimate Gf:

∥Gf∥_{L^{2}(ℝ_{+}^{N+1})} = \{∫_{|2πξ|≤1} \frac{1}{|2πξ′|^{4}} |Ff_{odd}(ξ)|^{2} dξ + ∫_{|2πξ|>1} \frac{1}{|2πξ′|^{4}} |Ff_{odd}(ξ)|^{2} dξ\}^{1/2}

≤ c {(∫_{|2πξ|≤1} |2πξ′|^{−4}|Ff_{odd}(ξ)|^{2} dξ)^{1/2} + ∥f∥_{L^{2}(ℝ^{N+1})}} .

If N < 4 then we need some conditions on f. Assume that ∫_{ℝ_{+}^{N+1}} f(x) dx = 0. Then Ff_{odd}(0) = 0, and if N = 2, 3 then

(∫_{|2πξ|≤1} \frac{1}{|2πξ|^{4}} |Ff_{odd}(ξ)|^{2} dξ)^{1/2}

= (∫_{|2πξ|≤1} \frac{1}{|2πξ|^{2}} \frac{|Ff_{odd}(ξ) − Ff_{odd}(0)|^{2}}{|2πξ|^{2}} dξ)^{1/2}

≤ (∫_{|2πξ|≤1} |2πξ|^{−2} dξ)^{1/2} sup_{ξ∈ℝ^{N+1}} |grad(Ff_{odd})(ξ)|
≤ c ∫_{ℝ_{+}^{N+1}} |x| · |f(x)| dx.

If N = 1 we need even stronger conditions (because the logarithmic potential is poorly behaved at infinity), that is Ff_{odd}(0) = 0, and grad(Ff_{odd})(0) = 0, corresponding to

∫_{ℝ_{+}^{2}} f(x) dx = 0; ∫_{ℝ_{+}^{2}} x_{i}f(x) dx = 0, i = 1, 2; ∫_{ℝ_{+}^{2}} |x|^{2}f(x) dx < ∞.

Then, for N = 1,

∥Gf∥_{ℝ_{+}^{2}} ≤ c {∥f∥_{L^{2}(ℝ_{+}^{2})} + ∫_{ℝ_{+}^{2}} |x|^{2}f(x) dx} .

The estimate for B has to be modified only in the part relative to the set where |2πξ′| ≤ 1:

∥B∥_{L^{2}(ℝ_{+}^{N+1})} ≤ (∫_{|2πξ′|≤1} ∫_{0}^{+∞} |B(y_{0}, ξ′)|^{2} dy_{0} dξ′)^{1/2} + ∥f∥_{W^{s}(ℝ_{+}^{N+1})} + ∥h∥_{L^{2}(ℝ^{N})}
≤ c {∥f∥_{W^{s}(ℝ_{+}^{N+1})} + ∥h∥_{L^{2}(ℝ^{N})}

+ ∫_{|2πξ′|≤1} |2πξ′|^{−3}| ∫_{0}^{+∞} f̂(y_{0}, ξ′)e^{−|2πξ′|y_{0}} dy_{0}|^{2}}^{1/2}.

With the above assumption we can estimate the last integral with

\{∫_{|2πξ′|≤1} |2πξ′|^{−1}| ∫_{0}^{+∞} y_{0}f̂(y_{0}, ξ′)\frac{e^{−|2πξ′|y_{0}} − 1}{|2πξ′|y_{0}} dy_{0}

+ ∫_{0}^{+∞} \frac{f̂(y_{0}, ξ′) − f̂(y_{0}, 0)}{|2πξ′|} dy_{0}|^{2} dξ′\}^{1/2} ≤ c ∫_{ℝ_{+}^{N+1}} |x|f(x) dx,

if N = 2, 3. If N = 1 the estimate follows in a similar fashion.

6. Analysis of the Problem on the Half Space for q-Forms

In this section we consider the space of q-forms, q ≥ 1, with coefficients in W^{1}(ℝ_{+}^{N+1}). Throughout this section, we denote this space of forms by ⋀^{q}, that is, we do not write explicitly the index for the Sobolev space W^{1}(ℝ_{+}^{N+1}), this being fixed once and for all.

On the space ⋀^{q} we select the basis {dx^{I}}, where I = (i_{1}, … i_{q}) is a strictly increasing multi-index. For a q-form ϕ,

ϕ = ∑_{I} ϕ_{I}dx^{I}

we have that

dϕ = ∑_{I} D_{j}ϕ_{I}dx^{j} ∧ dx^{I}

= ∑_{K} (∑_{I , j} D_{j}ϕ_{I}ε_{jI}^{K} ) dx^{K},

where K has q + 1 entries, and

ε_{jI}^{K} = \{ \begin{matrix}0 & if & K ≠ jI as sets \\ ±1 & if & K = jI as sets \end{matrix}

and the sign is chosen according to the sign of the permutation that puts jI in increasing order.

Lemma 6.1. The formal adjoint d′ of d, d′ : ⋀^{q+1} ⟶ ⋀^{q} with respect to the inner product in the Sobolev space, has the following expression:

d′ψ = ∑_{|I|=q} ( ∑_{{|K|=q+1 \atop j=0,...,N}} ε_{jI}^{K}D_{j}ψ_{K} ) dx^{I}.

HODGE THEORY IN THE SOBOLEV TOPOLOGY 37

Proof. Let ϕ ∈ ⋀^{q}, ψ ∈ ⋀^{q+1}, ϕ = ∑_{I} ϕ_{I}dx^{I}, ψ = ∑_{K} ψ_{k}dx^{K}, both with compact support in ℝ_{+}^{N+1}. Then we have

⟨dϕ, ψ, ⟩_{1} = ∑_{{|I|=q,|K|=q+1 \atop j=0,...,N}}⟨D_{j}ϕ_{I}dx^{j} ∧ dx^{I}, ψ_{K}dx^{K}⟩_{1}
= ∑_{I,K,j} ε_{jI}^{K}⟨D_{j}ϕ_{I}, ψ_{K}⟩_{1}
= ∑_{I,K,j}[∑_{k=0}^{N} ε_{jI}^{K}⟨D_{k}D_{j}ϕ_{I}, D_{k}ψ_{K}⟩_{0} + ε_{jI}^{K}⟨D_{j}ϕ_{I}, ψ_{K}⟩_{0}]
= ∑_{I,K,j}[∑_{k=0}^{N} −ε_{jI}^{K}⟨D_{k}ϕ_{I}, D_{k}D_{j}ψ_{K}⟩_{0} + ε_{jI}^{K}⟨D_{j}ϕ_{I}, ψ_{K}⟩_{0}]
= ∑_{I}[∑_{k=0}^{N}⟨D_{k}ϕ_{I}, D_{k}(− ∑_{{|K|=q+1 \atop j=0,...,N}} ε_{jI}^{K}D_{j}ψ_{K})⟩_{0}

+⟨ϕ_{I}, (− ∑_{{|K|=q+1 \atop j=0,...,N}} ε_{jI}^{K}D_{j}ψ_{K})⟩_{0}].

Lemma 6.2. Let d^{∗} be the adjoint of d in the W^{1}(ℝ_{+}^{N+1}) norm. Then

dom d^{∗} ∩ ⋀^{q+1}(\overline{Ω}) = \{ψ ∈ ⋀^{q+1}(\overline{Ω}) : D_{0}ψ_{0J}|_{x_{0}=0} , |J| = q\} .

That is, if ψ = ∑_{|K|=q+1} ψ_{K}dx^{K}, then ψ ∈ dom d^{∗} amounts to a condition on the coefficients ψ_{K} with index K = (0, …, k_{N}) ≡ 0J only, and it requires that

D_{0}ψ_{0J}|_{x_{0}=0} = 0 .

Proof. Let ϕ ∈ ⋀^{q}, ψ ∈ ⋀^{q+1}. Then

⟨dϕ, ψ⟩_{1} = ∑_{|K|=q+1}⟨ ∑_{{|I|=q \atop j=0,...,N}} D_{j}ϕ_{I}, ψ_{K}⟩_{1}

= ∑_{K} [ ∑_{a=0,...,N}⟨D_{a} (∑_{I,j} ε_{I,j}^{K}D_{j}ϕ_{I} ) , D_{a}ψ_{K}⟩_{0} + ⟨∑_{I,j} ε_{I,j}^{K}D_{j}ϕ_{I}, ψ_{K}⟩_{0}]

= ∑_{K} [ ∑_{a=0,...,N} ∑_{I,j} (ε_{I,j}^{K}⟨D_{a}ϕ_{I}, D_{a}D_{j}ψ_{K}⟩_{0} − ε_{0I}^{K} ∫_{ℝ^{N}} D_{a}ϕ_{I} \overline{D_{a}ψ_{K}} |_{x_{0}=0})

+ ∑_{I,j} (−ε_{jI}^{K}⟨ϕ_{I}, D_{j}ψ_{K}⟩_{0} − ε_{0I}^{K} ∫_{ℝ^{N}} ϕ_{I} \overline{ψ_{K}} |_{x_{0}=0})]

= ∑_{I} [ ∑_{a=0,...,N}⟨D_{a}ϕ_{I}, D_{a} (− ∑_{K,j} ε_{jI}^{K}D_{j}ψ_{K} )⟩_{0} + ⟨ϕ_{I}, − ∑_{K,j} ε_{jI}^{K}D_{j}ψ_{K}⟩_{0}]

− ∑_{K,I,j}[ ∑_{a=0,...,N} ε_{0I}^{K} ∫_{ℝ^{N}} D_{a}ϕ_{I} \overline{D_{a}ψ_{K}} |_{x_{0}=0}+ε_{0I}^{K} ∫_{ℝ^{N}} ϕ_{I} \overline{ψ_{K}} |_{x_{0}=0}].

Following the usual pattern we see that the only terms that we cannot control with the W^{1} norm of ψ are

ε_{0I}^{K} ∫_{ℝ^{N}} D_{0}ϕ_{I} \overline{D_{0}ψ_{K}} |_{x_{0}=0} |K| = q + 1, |I| = q.

Since ϕ was arbitrary, it must be D_{0}ψ_{K}|_{x_{0}=0} = 0 when K = 0I.

Next we want to determine the expression for the Hilbert space adjoint d^{∗} when acting on [dom d^{∗} ∩ W_{q+1}^{1}]. We have the following result.

Proposition 6.3. The adjoint of d in the W^{1}-inner product is given by d^{∗} = d′ + K where, for ψ = ∑_{|K|=q+1} ψ_{K}dx^{K},

Kψ = ∑_{|I|=q}(Kψ)_{I} dx^{I}

= ∑_{{|I|=q \atop I∌0}} (K_{x_{0}} ∗ ψ_{0I}(0, ·))(x′) dx^{I}.

Before proving the proposition we make a few remarks.

REMARK 6.4. 1. Notice that K acts diagonally, in the sense that (Kψ)_{I} depends only on ψ_{0I}.

2. Only the terms of the form ψ_{0I}, i.e. terms for which ψ⌊∂/∂x_{0} ≠ 0, contribute to Kψ.

3. Kψ has only purely “tangential” components, i.e. (Kψ)_{I} = 0 if 0 ∈ I.

Proof of 6.3. Notice that we could continue the calculation in the previous proof to obtain (for ψ ∈ dom d^{∗})

⟨dϕ, ψ⟩_{1} = ⟨ϕ, d′ψ⟩_{1} + ∑_{{|I|=q \atop I∌0}} [ ∫_{ℝ^{N}} ϕ_{I} \overline{△′ψ_{0I}} |_{x_{0}=0} − ∫_{ℝ^{N}} ϕ_{I} \overline{ψ_{0I}} |_{x_{0}=0} ].  (6.1)

Now, as in the computation in Proposition 3.2, we write

d^{∗}ψ = d′ψ + θ,

where θ = Kϕ, and K is a singular Green’s operator mapping (q + 1)-forms into q-forms. Now

⟨ϕ, d′ψ⟩_{1} + ⟨ϕ, θ⟩_{1} = ⟨ϕ, d^{∗}ψ⟩_{1} = ⟨dϕ, ψ⟩_{1}.  (6.2)

Therefore, (6.1), (6.2) and Green’s theorem give that

∑_{{|I|=q \atop I∌0}} [ ∫_{ℝ^{N}} ϕ_{I} \overline{△′ψ_{0I}} |_{x_{0}=0} − ∫_{ℝ^{N}} ϕ_{I} \overline{ψ_{0I}} |_{x_{0}=0} ]

= ⟨ϕ, θ⟩_{1}
= ∑_{I}[∑_{a=0}^{N}⟨D_{a}ϕ_{I}, D_{a}θ_{I}⟩_{0} + ⟨ϕ_{I}, θ_{I}⟩_{0}]
= − ∑_{I}⟨ϕ_{I}, △θ_{I}⟩_{0} − ∫_{ℝ^{N}} ϕ_{I} \overline{D_{0}θ_{I}} |_{x_{0}=0} + ⟨ϕ_{I}, θ_{I}⟩_{0}.

Therefore θ must satisfy the following conditions

\begin{cases}− △ θ_{I} + θ_{I} & = 0 \\ −D_{0}θ_{I}|_{x_{0}=0} & = \{ \begin{matrix}0 & if 0 ∈ I \\ △′ψ_{0I} − ψ_{0I} & if 0 ∉ I \end{matrix} \end{cases}

Thus

(Kψ)_{I} = θ_{I} = 0 if 0 ∈ I,

since the solution of △u + u = 0 with the boundary condition D_{0}u(0, x′) = 0 consists of just the zero function. On the other hand, if I ∌ 0 we have that (Kψ)_{I} = θ_{I} is the solution of

\begin{cases}− △ u + u = 0 & on ℝ_{+}^{N+1} \\ \frac{∂u}{∂x_{0}} (0, ·) = − △′ ψ_{0I} + ψ_{0I} & on ℝ^{N} \end{cases} .

40 L. FONTANA, S. G. KRANTZ, AND M. M. PELOSO

In the proof of Proposition 3.2 , we showed that θ is given by

θ_{I} = − ∫_{ℝ^{N}} \sqrt{1 + |2πξ′|^{2}}e^{−\sqrt{1+|2πξ′|^{2}}x_{0}} ψ̂_{0I}(0, ξ′)e^{2πiξ′·x′} dx′
≡ (K_{x_{0}} ∗ ϕ_{0}(0, ·))(x′).

We now return to the main object of our work. We would like to solve the boundary value problem

\begin{cases}(dd^{∗} + d^{∗}d)ϕ = & α \\ ϕ ∈ dom d^{∗} \\ dϕ ∈ dom d^{∗} \end{cases}

for α ∈ ⋀^{q} . Recall that d^{∗} = d′ + K. A simple calculation shows that

(dd′ + d′d)ϕ = ∑_{|I|=q} − △ ϕ_{I}dx^{I}.  (6.3)

Thus we need to compute dKϕ and Kdϕ. The next lemma addresses this task. Lemma 6.5. Let ϕ, dϕ ∈ dom d^{∗}, with ϕ = ∑_{I} ϕ_{I}dx^{I}. Then we have

dKϕ = ∑_{|J|=q} [ ∑_{{|I|=q−1,I∌0 \atop j=0,...,N}} ε_{jI}^{J}D_{j}K_{x_{0}} ∗ (ϕ_{0I}(0, ·))] dx^{J},

and

Kdϕ = ∑_{|I|=q,I∌0} [ ∑_{{|J|=q \atop j=0,...,N}} ε_{jJ}^{0I}K_{x_{0}} ∗ (D_{j}ϕ_{J}(0, ·))] dx^{I}.

Proof. These are just straightforward computations. We have

dKϕ = d ( ∑_{|I|=q−1,I∌0} K_{x_{0}} ∗ (ϕ_{0I}(0, ·))dx^{I} )
= ∑_{|I|=q−1,I∌0} ∑_{j=0,...,N} D_{j}K_{x_{0}} ∗ (ϕ_{0I}(0, ·))dx^{j} ∧ dx^{I}
= ∑_{|J|=q} [ ∑_{{|I|=q−1,I∌0 \atop j=0,...,N}} ε_{jI}^{J}D_{j}K_{x_{0}} ∗ (ϕ_{0I}(0, ·))] dx^{J}.

This proves the statement for dKϕ. On the other hand,

Kdϕ = K ( ∑_{|K|=q+1}( ∑_{{|J|=q \atop j=0,...,N}} ε_{jJ}^{K}D_{j}ϕ_{J})dx^{K} )

= ∑_{|I|=q,I∌0} K_{x_{0}} ∗ ( ∑_{{|J|=q \atop j=0,...,N}} ε_{jJ}^{0I}D_{j}ϕ_{J}(0, ·)) dx^{I}

= ∑_{|I|=q,I∌0} [ ∑_{{|J|=q \atop j=0,...,N}} ε_{jJ}^{0I}K_{x_{0}} ∗ (D_{j}ϕ_{J}(0, ·))] dx^{I}.

Next we want to compute (Kd + dK)ϕ. [Note that, in this discussion, the letter K is both a kernel and an index; no confusion should result.]

Proposition 6.6. Let ϕ ∈ ⋀^{q}, with ϕ, dϕ ∈ dom d^{∗}. Set (Kd + dK)ϕ ≡ β = ∑_{|K|=q} β_{K}dx^{K}. Then

β_{K} = \frac{∂}{∂x_{0}} K_{x_{0}} ∗ (ϕ_{K}(0, ·))  if K ∋ 0
β_{K} = K_{x_{0}} ∗ (\frac{∂}{∂x_{0}} ϕ_{K}(0, ·))  if K ∌ 0.

Remark. Notice that Kd + dK is a diagonal operator on the space of q-forms ⋀^{q}.

Proof of 6.6. Suppose that K ∋ 0. Then (Kdϕ)_{K} = 0, since K applied to any form has only tangential components. Moreover, for the same reason, (dKϕ)_{0K′} can be obtained only by differentiating (Kϕ)_{K′} by D_{0}, and “wedging” by dx^{0}. That is,

(dKϕ)_{0K′} = D_{0}(Kϕ)_{K′}dx^{0} ∧ dx^{K′}
= D_{0}(K_{x_{0}} ∗ (ϕ_{0K′}(0, ·))dx^{0} ∧ d^{K′}.

Now suppose that K ∌ 0. We use Lemma 6.5 to obtain that

dKϕ + Kdϕ = ∑_{|K|=q} [ ∑_{{|I|=q−1,I∌0 \atop j=0,...,N}} ε_{jI}^{K}D_{j}K_{x_{0}} ∗ (ϕ_{0I}(0, ·))] dx^{K}

+ ∑_{|K|=q} [ ∑_{{|J|=q \atop j=0,...,N}} ε_{jJ}^{0K}K_{x_{0}} ∗ (D_{j}ϕ_{0J}(0, ·))] dx^{K}.

Thus we need to describe the term on the right hand side of this equation when K ∌ 0. Notice that in this case j cannot be 0 in the first sum, since otherwise ε_{0I}^{K} = 0. Then, when K ∌ 0 the coefficient of dx^{K} in the right hand side above equals

K_{x_{0}} ∗ (D_{0}ϕ_{0K}(0, ·)) + ∑_{{|J|=q,J∌0 \atop j=1,...,N}} ε_{0J}^{jK}D_{j}(K_{x_{0}} ∗ (ϕ_{0J}(0, ·)))

+ ∑_{{|I|=q−1,I∌0 \atop j=1,...,N}} ε_{jI}^{K}D_{j}(K_{x_{0}} ∗ (ϕ_{0I}(0, ·))) ≡ K_{x_{0}} ∗ (D_{0}ϕ_{K}(0, ·)),

since J must be of the form 0I with |I| = q − 1, I ∌ 0, and therefore

ε_{j0I}^{0K} = −ε_{j0I}^{0K} = −ε_{jI}^{K}.

This concludes the proof.

Corollary 6.7. Let (Kd + dK)ψ = Gψ. Then β = Gψ is a form such that its components solve the boundary value problems

\begin{cases}(− △ +I)β_{K} & = 0 \\ \frac{∂β_{K}}{∂x_{0}} & = (− △′ +I)(dψ)_{0K} \end{cases} if K ∌ 0,

and

\begin{cases}(− △ +I)β_{K} & = 0 \\ β_{K} & = (− △′ +I)ψ_{K} \end{cases} if K ∋ 0.

Proof. By Proposition 6.3, if Gψ = ∑_{|K|=q} β_{K}dx^{K} for K ∌ 0, then

β_{K} = K_{x_{0}} ∗ (\frac{∂ψ_{K}}{∂x_{0}} (0, ·)) .

By construction

(− △ +I)β_{K} = 0 on ℝ_{+}^{N+1},

and

\frac{∂β_{K}}{∂x_{0}} = (− △′ +I)(dψ)_{0K}

= (− △′ +I) (\frac{∂ψ_{K}}{∂x_{0}} + T_{1}ψ_{K′}) on bℝ_{+}^{N+1}, K′ ∋ 0,

where T_{1} is a first order tangential differential operator.

On the other hand, if K ∋ 0, then (Kdψ)_{K} = 0. Thus, for K = 0K′,

(Gψ)_{K} = (dKψ)_{K}

= \frac{∂}{∂x_{0}} (Kψ)_{K′}.

Then, on ℝ_{+}^{N+1},

(− △ +I)(Gψ)_{K} = \frac{∂}{∂x_{0}} (− △ +I)(Kψ)_{K′} = 0;

and if K = 0K′,

(Gψ)_{K}|_{bΩ} = \frac{∂}{∂x_{0}} (Kψ)_{K′} |_{bΩ}
= (− △′ +I)ψ_{K}|_{bΩ} .

Before analyzing the boundary value problem we need one more result; that is, we wish to make explicit the boundary condition dϕ ∈ dom d^{∗}.

Lemma 6.8. Let ϕ ∈ ⋀^{q}, and let ϕ ∈ dom d^{∗}. Then dϕ ∈ dom d^{∗} if and only if

\frac{∂^{2}ϕ_{K}}{∂x_{0}^{2}} (0, ·) = 0 for K ∌ 0.

Proof. By Lemma 6.2 we have that dϕ ∈ dom d^{∗} if and only if, for all multi-indices K′ ∌ 0, |K′| = q, there holds the equality

\frac{∂}{∂x_{0}} (∑_{I,j} ε_{jI}^{0K′} \frac{∂ϕ_{I}}{∂x_{j}} )(0, x′) = 0.

If j ≠ 0, then I ∋ 0 and

\frac{∂}{∂x_{0}} \frac{∂ϕ_{I}}{∂x_{j}} (0, ·) = \frac{∂}{∂x_{j}} (\frac{∂ϕ_{I}}{∂x_{0}} (0, ·)) = 0,

since, for ϕ ∈ dom d^{∗}, we have ∂/∂x_{0}ϕ_{i}(0, ·) = 0 when I ∋ 0. Thus we obtain

\frac{∂^{2}ϕ_{I}}{∂x_{0}^{2}} (0, ·) = 0 for all I ∌ 0, |I| = q.

We finally are able to formulate the boundary value problem.

THEOREM 6.9. Consider the boundary value problem

\begin{cases}(dd^{∗} + d^{∗}d)ϕ = α & α ∈ ⋀^{q}(\overline{Ω}) \\ ϕ ∈ dom d^{∗} \\ dϕ ∈ dom d^{∗} \end{cases}  (6.4)

Let α be a q-form with coefficients in W^{r}(ℝ_{+}^{N+1}) ∩ L^{1}(ℝ_{+}^{N+1}), with r > 1/2. Then, if N ≥ 4, there is a unique q-form ϕ with coefficients in W^{r+2}(ℝ_{+}^{N+1}) solving the boundary value problem and satisfying the estimate

∥ϕ∥_{W^{r+2}} ≤ c · {∥α∥_{W^{r}(ℝ_{+}^{N+1})} + ∥α∥_{L^{1}(ℝ_{+}^{N+1})}} .

If N = 2, 3 we need to further require that

∫ α(x) dx = 0 , α ∈ L^{1}(|x|dx, ℝ_{+}^{N+1}),

so that the solution ϕ satisfies the estimate

∥ϕ∥_{W^{r+2}} ≤ c · {∥α∥_{W^{r}(ℝ_{+}^{N+1})} + ∥α∥_{L^{1}(|x|dx,ℝ_{+}^{N+1})}} .

If N = 1 we need to require that

∫ α(x) dx = 0 , ∫ x_{i}α(x) dx = 0, i = 0, 1 α ∈ L^{1}(|x|^{2}dx, ℝ^{2});

in this case similar estimates hold for the solution ϕ, with the addition of the term ∥α∥_{L^{1}(|x|^{2}dx,ℝ^{2})} to the right hand side.

By the discussion preceding the statement of the theorem we see that the boundary value problem (6.4) is equivalent to the two scalar problems

\begin{cases}− △ ϕ_{K} + \frac{∂}{∂x_{0}} (K_{x_{0}} ∗ ϕ_{K}|_{x_{0}=0}) = α_{K} \\ \frac{∂ϕ_{K}}{∂x_{0}} |_{x_{0}=0} = 0 \end{cases}  0 ∈ K,  (6.5)

and

\begin{cases}− △ ϕ_{K} + (K_{x_{0}} ∗ \frac{∂ϕ_{K}}{∂x_{0}} |_{x_{0}=0}) = α_{K} \\ \frac{∂^{2}ϕ_{K}}{∂x_{0}^{2}} |_{x_{0}=0} = 0 \end{cases}  0 ∉ K.  (6.6)

Thus the problem is reduced to solving two different boundary value problems. The boundary value problem (6.6) has been solved already in the case q = 0. The solution of the second boundary value problem is contained in the following theorems. As in the case of functions, for greater flexibility we solve the boundary value problem with non-zero boundary data.

THEOREM 6.10. Consider the boundary value problem

\begin{cases}− △ u + \frac{∂}{∂x_{0}} (K_{x_{0}} ∗ u(0, ·)) = f & on ℝ_{+}^{N+1} \\ \frac{∂u}{∂x_{0}} (0, ·) = h & on ℝ^{N} \end{cases}  (6.7)

If there exists a solution u, then it satisfies the a priori estimate

∥u∥_{W^{r+2}} ≤ c · {∥f∥_{W^{r}(ℝ_{+}^{N+1})} + ∥h∥_{W^{r+1/2}(ℝ^{N})} + ∥u∥_{L^{2}(ℝ_{+}^{N+1})}} ,

for r > 1/2.

Proof. We break the proof into two steps. We first estimate the function u which is the solution of the boundary value problem with zero boundary data. Then we show how to reduce the general case to this special one.

In this proof we let J denote the Green’s function for the Neumann problem for the Laplacian on the half space. Suppose now that u is the solution of the boundary value problem with h = 0. Then u can be written as

u = J (f − \frac{∂}{∂x_{0}} (K_{x_{0}} ∗ u(0, ·))).

Recall that, for N ≥ 2,

(J f)(x_{0}, x′) = \frac{1}{(N − 1)ω_{N+1}} ∫_{ℝ_{+}^{N+1}}\{\frac{1}{(x_{0} − y_{0})^{2} + |x′ − y′|^{2})^{(N−1)/2}}

+ \frac{1}{(x_{0} + y_{0})^{2} + |x′ − y′|^{2})^{(N−1)/2}} \}f(y_{0}, y′) dy_{0} dy′,

and

Ĵ f(x_{0}, ξ′) = \frac{1}{2|2πξ′|} ∫_{0}^{+∞}(e^{−|2πξ′||x_{0}−y_{0}|} + e^{−|2πξ′|(x_{0}+y_{0})})f̂(y_{0}, ξ′) dy_{0}.

Then

û(0, ξ′) = Ĵ f(0, ξ′) − |2πξ′|^{−1} ∫_{0}^{+∞} e^{−(|2πξ′|+\sqrt{1+|2πξ′|^{2}})y_{0}} dy_{0} (1 + |2πξ′|^{2})û(0, ξ′)
= Ĵ f(0, ξ′) − \frac{(1 + |2πξ′|^{2})}{|2πξ′|(|2πξ′| + \sqrt{1 + |2πξ′|^{2}})}û(0, ξ′).

Thus, recalling the definition (5.5) of m_{2}, we have that

û(0, ξ′) = |2πξ′|m_{2}(ξ′)Ĵ f(ξ′).

Set F = f −(∂/∂x_{0})(K_{x_{0}} ∗u(0, ·)). Since u = J F, in order to estimate ∥u∥_{W^{3/2}(ℝ_{+}^{N+1})} we can proceed as in the proof of Theorem 5.4 and, using the fact that ∂^{2}/∂x_{0}^{2} = △ − △′, we can reduce to estimating derivatives of u of order not exceeding 2 in the x_{0}-variable.

Let g_{e} denote the even extension (in the x_{0} variable) to all of ℝ^{N+1} of the function g defined initially on ℝ_{+}^{N+1}. We have that

[(J g)(x_{0}, x′)]_{e} = (N g_{e})(x_{0}, x′) ,

and

F (\frac{∂}{∂x_{0}} K_{x_{0}})_{e} = \frac{2(1 + |2πξ′|^{2})^{3/2}}{1 + |2πξ|^{2}} .

46 L. FONTANA, S. G. KRANTZ, AND M. M. PELOSO

Combining all of these facts we obtain:

Fu_{e} = |2πξ|^{−2}(Ff_{e}(ξ) − \frac{2(1 + |2πξ′|^{2})^{3/2}}{1 + |2πξ|^{2}} |2πξ′|m_{2}(ξ′)Ĵ f(0, ξ′)).  (6.8)

Thus

∥u∥_{W^{2+r}(ℝ_{+}^{N+1})} ≤ c(∫_{ℝ^{N+1}} ∑_{i,j=0}^{N}(1 + |2πξ_{i}|^{2}|2πξ_{j}|^{2})|Fu(ξ)|^{2}(1 + |2πξ′|^{2})^{r} dξ)
≤ c(∥u∥_{L^{2}(ℝ_{+}^{N+1})} + ∫_{−∞}^{+∞} ∫_{|2πξ′|≥1} |FF_{e}(ξ)|^{2}|2πξ′|^{2r} dξ′ dξ_{0}),

where

FF_{e}(ξ) = Ff_{e}(ξ) − \frac{2(1 + |2πξ′|^{2})^{3/2}}{1 + |2πξ|^{2}} m_{2}(ξ′) ∫_{0}^{+∞} e^{−|2πξ′|y_{0}} f̂(y_{0}, ξ′) dy_{0}

= Ff_{e}(ξ) − \frac{2(1 + |2πξ′|^{2})}{1 + |2πξ|^{2}}m_{1}(ξ′) ∫_{0}^{+∞} e^{−|2πξ′|y_{0}} f̂(y_{0}, ξ′) dy_{0}.

Therefore, by obvious calculations and the Schwarz inequality,

∥u∥_{W^{2+r}(ℝ_{+}^{N+1})}^{2} ≤ c(∥u∥_{L^{2}(ℝ_{+}^{N+1})}^{2} + ∥f∥_{W^{r}(ℝ_{+}^{N+1})}^{2} + ∫_{|2πξ′|≥1} \frac{(1 + |2πξ′|^{2})^{2+r}}{|2πξ|}

× (∫_{0}^{+∞} |f̂(y_{0}, ξ′)|^{2} dy_{0})(∫_{0}^{+∞} \frac{1}{(1 + |2πξ′|^{2})^{2} + |2πξ_{0}|^{4}} dξ_{0}) dξ′)

≤ c(∥u∥_{L^{2}(ℝ_{+}^{N+1})}^{2} + ∥f∥_{W^{r}(ℝ_{+}^{N+1})}^{2}).

This proves the theorem in the case that the boundary data h ≡ 0.

In the general case, let Q be the operator initially defined on C_{0}^{∞}(ℝ^{N}) by

(Qg)(x_{0}, ξ′) = − \frac{e^{−\sqrt{1+|2πξ′|^{2}}x_{0}}}{\sqrt{1 + |2πξ′|^{2}}} ĝ(ξ′).

Notice that (∂/∂x_{0})(Qg)|_{x_{0}=0} = g, and that

Q : W^{s}(ℝ^{N}) ⟶ W^{s+1/2}(ℝ_{+}^{N+1}).

We seek an a priori estimate for a function u that solves of (6.7). Let G′ denote the operator u ⟼ (∂/∂x_{0})(K_{x_{0}} ∗ u(0, ·)). Set v = u − Qh. Then v solves the boundary value problem

\begin{cases}(− △ +G′)v = f + (− △ +G′)Qh & on ℝ_{+}^{N+1} \\ \frac{∂v}{∂x_{0}} (0, ·) = 0 & on ℝ^{N} \end{cases} .

Thus for such v we have the usual a priori estimate. Notice that

[(− △ +G′)Qh]ˆ (ξ′) = −\frac{|2πξ′|^{2}e^{−\sqrt{1+|2πξ′|^{2}}x_{0}}}{\sqrt{1 + |2πξ′|^{2}}} ĥ(ξ′),

so that the operator (− △ +G′)Q has the same behavior as the operator K̃ studied in Proposition 3.3, so it maps W^{s}(ℝ^{N}) into W^{s+1/2}(ℝ_{+}^{N+1}) continuously. Therefore

∥u∥_{W^{r+2}(ℝ_{+}^{N+1})} ≤ ∥v∥_{W^{r+2}(ℝ_{+}^{N+1})} + ∥Qh∥_{W^{r+2}(ℝ_{+}^{N+1})}
≤ c {∥f∥_{W^{r}(ℝ_{+}^{N+1})} + ∥(− △ +G′)Qh∥_{W^{r}(ℝ_{+}^{N+1})} + ∥v∥_{L^{2}(ℝ_{+}^{N+1})}
+ ∥Qh∥_{W^{r}(ℝ_{+}^{N+1})}}
≤ c {∥f∥_{W^{r}(ℝ_{+}^{N+1})} + ∥h∥_{W^{r+1/2}(ℝ^{N})} + ∥u∥_{L^{2}(ℝ_{+}^{N+1})}},

since v = u − Qh. This concludes the proof.

THEOREM 6.11. Let N ≥ 4. Then for any f ∈ L^{2}(ℝ_{+}^{N+1})∩L^{2}(ℝ_{+}^{N+1}), there exists a unique function u solving the boundary value problem (6.5) and such that

∥u∥_{L^{2}(ℝ_{+}^{N+1})} ≤ c · {∥f∥_{L^{2}(ℝ_{+}^{N+1})} + ∥f∥_{L^{1}(ℝ_{+}^{N+1})} + ∥h∥_{W^{1/2}(ℝ^{N})}} .

If N = 2, 3 we suppose in addition that f satisfies

∫ f(x) dx = 0, f ∈ L^{1}(|x|dx, ℝ_{+}^{N+1}).

Then there exists a unique solution f that satisfies the estimate

∥u∥_{L^{2}(ℝ_{+}^{N+1})} ≤ c · {∥f∥_{L^{2}(ℝ_{+}^{N+1})} + ∥f∥_{L^{1}(|x|dx,ℝ_{+}^{N+1})} + ∥h∥_{W^{1/2}(ℝ^{N})}} .

If N = 1 we suppose that f is such that

∫ f(x) dx = 0, ∫ x_{i}f(x) dx = 0, i = 0, 1 f ∈ L^{1}(|x|^{2}dx, ℝ^{2}).

Then there exists a unique solution u such that

∥u∥_{L^{2}(ℝ_{+}^{N+1})} ≤ c · {∥f∥_{L^{2}(ℝ_{+}^{N+1})} + ∥f∥_{L^{1}(|x|^{2}dx,ℝ_{+}^{N+1})} + ∥h∥_{W^{1/2}(ℝ^{N})}} .

Proof. Recall that we have found that if a solution u exists, then it must be given by the formula (6.8), i.e.

Fu_{e}(ξ) = |2πξ|^{−2}(Ff_{e}(ξ) − 2\frac{(1 + |2πξ′|^{2})^{3/2}}{1 + |2πξ|^{2}} m_{2}(ξ′)F̂_{1}(ξ′)),

where F_{1} is given by

F̂_{1}(ξ′) = |2πξ′|Ĵ f(0, ξ′) = ∫_{0}^{+∞} e^{−|2πξ′|y_{0}} f̂(y_{0}, ξ′) dy_{0}.

In order to show that u given by the above formula is indeed a solution, by the estimate in Theorem 6.10 it suffices to show that u ∈ L^{2}(ℝ_{+}^{N+1}).

Suppose that N ≥ 4. We have

∥u∥_{L^{2}(ℝ_{+}^{N+1})}

= \frac{1}{2} ∫_{ℝ^{N+1}} |Fu_{e}(ξ)|^{2} dξ
≤ {∫_{|2πξ′|≤1} ∫_{−∞}^{+∞} |Ff_{e}(ξ)|^{2} dξ_{0} dξ′}^{1/2} + {∫_{|2πξ′|≥1} ∫_{−∞}^{+∞} |Ff_{e}(ξ)|^{2} dξ_{0} dξ′}^{1/2}

≤ c{(∫_{|2πξ′|≤1} ∫_{−∞}^{+∞} |2πξ|^{−4} dξ_{0} dξ′) sup_{ξ∈ℝ^{N+1}}(|Ff_{e}(ξ)| + ∫_{0}^{+∞} |f̂(y_{0}, ξ′)| dy_{0})^{2}
+ ∫_{|2πξ′|≥1} ∫_{−∞}^{+∞} |Ff_{e}(ξ)|^{2} dξ_{0} dξ′
+ ∫_{|2πξ′|≥1} ∫_{−∞}^{+∞} |2πξ|^{−4} dξ_{0}| ∫_{0}^{+∞} e^{−|2πξ′|y_{0}} f̂(y_{0}, ξ′) dy_{0}|^{2} dξ′}
≤ c {∥f∥_{L^{1}(ℝ_{+}^{N+1})} + ∥f∥_{L^{2}(ℝ_{+}^{N+1})}} .

This proves our statement in the case N ≥ 4.

Let now N = 2, 3, and assume that

∫_{ℝ_{+}^{N+1}} f(x) dx = 0.

Notice that this implies that ∫_{0}^{+∞} f̂(y_{0}, 0) dy_{0} = 0. Therefore,

∥u∥_{L^{2}(ℝ_{+}^{N+1})}^{2} ≤ ∫_{|2πξ|≤1} ∫_{−∞}^{+∞} \frac{|Ff_{e}(ξ) − Ff_{e}(0)|^{2}}{|2πξ|^{4}} dξ_{0} dξ′

+ ∫_{|2πξ|≤1} ∫_{−∞}^{+∞} |2πξ|^{−2}| ∫_{0}^{+∞} e^{−|2πξ′|y_{0}} \frac{f̂(y_{0}, ξ′) − f̂(y_{0}, 0)}{|2πξ|} dy_{0}|^{2} dξ_{0} dξ′

+ ∫_{|2πξ|≤1} ∫_{−∞}^{+∞} |2πξ|^{−2}(∫_{0}^{+∞} \frac{|e^{−|2πξ′|y_{0}} − 1|}{|2πξ′|}|f̂(y_{0}, 0)| dy_{0})^{2} dξ_{0} dξ′

+ ∫_{|2πξ′|≥1} ∫_{−∞}^{+∞}[|Ff_{e}(ξ)|^{2} + (∫_{0}^{+∞} e^{−|2πξ′|y_{0}}|f̂(y_{0}, ξ′)| dy_{0})^{2}] dξ_{0} dξ′

≤ C ∫_{|2πξ′|≤1} |2πξ′|^{−1} dξ′ {(∫_{ℝ_{+}^{N+1}} |x| · |f(x)| dx)^{2} + (∫_{ℝ_{+}^{N+1}} |f(x)| dx)^{2}}

+ C∥f∥_{L^{2}(ℝ_{+}^{N+1})}.

Here we have used the following facts:

\frac{|Ff_{e}(ξ) − Ff_{e}(0)|}{|2πξ|} ≤ \frac{1}{2π} sup_{ξ∈ℝ^{N+1}} |gradFf_{e}(ξ)| ≤ ∫_{ℝ_{+}^{N+1}} |x| · |f(x)| dx;

∫_{0}^{+∞} e^{−|2πξ′|y_{0}}|f̂(y_{0}, ξ′)| dy_{0} ≤ ∫_{0}^{+∞} y_{0} ∫_{ℝ^{N}} |f(y_{0}, y′)| dy′ dy_{0} ≤ ∫_{ℝ_{+}^{N+1}} |y| · |f(y)| dy;

and

∫_{|2πξ′|≥1}(∫_{0}^{+∞} e^{−|2πξ′|y_{0}}|f̂(y_{0}, ξ′)| dy_{0})^{2} dξ′
≤ ∫_{|2πξ′|≥1} \frac{1}{2|2πξ′|} ∫_{0}^{+∞} |f̂(y_{0}, ξ′)|^{2} dy_{0} dξ′
≤ ∫_{ℝ_{+}^{N+1}} |f(x)|^{2} dx.

Finally, in the case N = 1 we can obtain the same kind of estimate if we require the data f to satisfy the additional stated compatibility conditions.

This proves the result in the case of the boundary data h ≡ 0. If h ≠ 0 we proceed as in the proof of Theorem 6.10. Consider v = u − Qh. Then v satisfies the above estimate, and therefore

∥u∥_{L^{2}(ℝ_{+}^{N+1})} ≤ ∥v∥_{L^{2}(ℝ_{+}^{N+1})} + ∥Qh∥_{L^{2}(ℝ_{+}^{N+1})} ≤ c{∥v∥_{L^{2}(ℝ_{+}^{N+1})} + ∥h∥_{L^{2}(ℝ_{+}^{N+1})}}.

In order to estimate ∥v∥_{L^{2}(ℝ_{+}^{N+1})} we observe that we have to replace f with f + (△ − G′)Qh in the computations above, so that, for N ≥ 4 (the case N < 4 is similar),

∥v∥_{L^{2}(ℝ_{+}^{N+1})} ≤ \frac{1}{2} ∫_{ℝ^{N+1}} |Fv_{e}|^{2} dξ

≤ C{∥f∥_{L^{2}(ℝ_{+}^{N+1})} + ∥f∥_{L^{1}(ℝ_{+}^{N+1})}}

+ C ∫_{ℝ^{N+1}} \frac{1}{|2πξ|^{4}} |F((△ − G′)Qh)_{e} − \frac{2(1 + |2πξ′|^{2})}{1 + |2πξ|^{2}}m_{1}(ξ′)F̂_{2}(ξ′)|^{2}dξ,

where

F̂_{2}(ξ′) = − ∫_{0}^{+∞} e^{−|2πξ′|y_{0}}e^{−\sqrt{1+|2πξ′|^{2}}y_{0}} \frac{|2πξ′|^{2}}{\sqrt{1 + |2πξ′|^{2}}} dξ_{0}ĥ(ξ′)

= −\frac{|2πξ′|^{2}}{(|2πξ′| + \sqrt{1 + |2πξ′|^{2}})\sqrt{1 + |2πξ′|^{2}}} ĥ(ξ′);

and

F((△ − G′)Qh)_{e} = \frac{2|2πξ′|^{2}}{1 + |2πξ|^{2}} ĥ(ξ).

50 L. FONTANA, S. G. KRANTZ, AND M. M. PELOSO

Therefore we obtain

∥v∥_{L^{2}(ℝ_{+}^{N+1})} ≤ C(∥f∥_{L^{2}(ℝ_{+}^{N+1})} + ∥f∥_{L^{2}(ℝ_{+}^{N+1})} + ∥h∥_{L^{2}(ℝ_{+}^{N+1})}.

This concludes the proof.

THE PROBLEM ON A SMOOTHLY BOUNDED DOMAIN

7. Formulation of the Problem on a Smoothly Bounded Domain

Let Ω = {x ∈ ℝ^{N+1} : ρ(x) < 0}. To avoid pathologies, we assume that ρ is a C^{∞} function on ℝ^{N+1} with the property that ∇ρ ≠ 0 at all points of bΩ. Then Ω is a domain with C^{∞} boundary (see [KR2] for a protracted discussion of these matters). We shall also assume that Ω is bounded.

It will simplify our calculations if we assume in advance that (∂ρ/∂n) = |∇ρ| = 1 on bΩ. This is easily arranged.

Recall that we defined

W^{1}(Ω) = {f ∈ L^{2}(Ω) : ∑_{j=0}^{N}⟨D_{j}f, D_{j}f⟩_{0} + ⟨f, f⟩_{0} < ∞}.

Here the derivative is intended in the sense of distributions,

⟨f, g⟩_{0} = ∫_{Ω} f\overline{g} dV,

and D_{j} = ∂/∂x_{j}, j = 0, …, N. Moreover we have defined the 1-Sobolev space of q-forms W_{q}^{1}(Ω), for q = 1, …, N + 1, by setting

W_{q}^{1}(Ω) = {ϕ = ∑_{|I|=q} ϕ_{I}dx^{I} : ϕ_{I} ∈ W^{1}(Ω)}.

For ϕ, ψ ∈ W_{q}^{1}(Ω), their inner product in W_{q}^{1}(Ω) is given by

⟨ϕ, ψ⟩ = ⟨∑_{I} ϕ_{I}dx^{I}, ∑_{J} ψ_{J}dx^{J}⟩_{1}

= ∑_{I}⟨ϕ_{I}, ψ_{I}⟩_{1}.

Throughout the rest of this entire paper, we shall denote the s-Sobolev norm of a form ψ by ∥ψ∥_{s}.

Let

d : ⋀^{q} ⟶ ⋀^{q+1}

be defined by

d(∑_{I} ϕ_{I} dx^{I}) = ∑_{I} ∑_{j=0}^{N} D_{j}ϕ_{I} dx_{j} ∧ dx^{I}

= ∑_{|J|=q+1}( ∑_{{|I|=q \atop j=0,...,N}} ε_{jI}^{J}D_{j}ϕ_{I})dx^{J}

≡ ∑_{|J|=q+1} ϕ′_{J}dx^{J}.

We let d^{∗} be the operator on ⋀^{q+1} defined by

⟨dϕ, ψ⟩_{1} = ⟨ϕ, d^{∗}ψ⟩_{1}.

Recall that

dom d^{∗} ∩ ⋀^{q+1} = {ψ ∈ ⋀^{q+1} : |⟨dϕ, ψ⟩_{1}| ≤ C_{ψ}∥ϕ∥_{1}}.

Our goal is to solve the boundary value problem

\begin{cases}(dd^{∗} + d^{∗}d)ϕ = α & on Ω \\ ϕ ∈ dom d^{∗} \\ dϕ ∈ dom d^{∗} \end{cases}  (7.1)

for α ∈ W_{q}^{1}(Ω), q = 0, …, N, and prove existence and regularity theorems.

We need to analyze both the equation on Ω, and the boundary conditions. Our first goal is to describe dom d^{∗}. Since we have developed some familiarity with this type of calculation, we work directly with q-forms for all q (recall that, in the half space case, we restricted attention at first to functions). We have the following result.

Proposition 7.1. Let Ω be a smoothly bounded domain in ℝ^{N+1}. Then the Hilbert space adjoint d^{∗} of d, in the W^{1} inner product, acting on q-forms, has domain satisfying

dom d^{∗} ∩ ⋀^{q+1}(\overline{Ω}) = \{ψ ∈ ⋀^{q+1}(\overline{Ω}) : (∇n⃗ψ)⌊n⃗|_{bΩ} ≡ 0\}.

Here we use the notation ∇_{X}ϕ to denote the covariant differentiation of the form ϕ in the direction given by the vector field X, and “⌊” is the standard contraction operation from exterior algebra. Recall that by definition, if Y_{1}, …, Y_{q} are vector fields, then

(∇_{X}ϕ) = X(ϕ(Y_{1}, …, Y_{q})) − ∑_{i=q}^{q} ϕ(Y_{1}, …, ∇_{X}Y_{i}, …, Y_{q});

also, in local coordinates (y_{0}, …, y_{n}),

∇_{X}V = ∑_{k=0}^{n}(X(V_{k}) + ∑_{i,j=0}^{n} Γ_{ij}^{k}X_{i}V_{j})\frac{∂}{∂y_{k}} .

Clearly covariant differentiation preserves the type of a form.

Moreover,

ϕ⌊V = ∑_{iIJ} ϕ_{J}V_{i}ε_{iJ}^{J}dy^{I}.

Notice that, in the standard coordinates of ℝ^{N+1}, (∇_{X}ϕ)_{K} = X(ϕ_{K}) and

(ϕ⌊(∂/∂x_{0}))_{K} = ϕ_{0K}

if K ∌ 0 and (ϕ⌊(∂/∂x_{0}))_{K} = 0 if K ∋ 0. For these and related notions we refer the reader to [FED].

Observe that, if ϕ = ∑_{i} ϕ_{i}dx_{i} is a 1-form, then the boundary condition ∇n⃗ϕ⌊n⃗|_{bΩ} = 0 can be written as

∑_{j=0}^{N} \frac{∂ρ}{∂x_{j}} \frac{∂ϕ_{i}}{∂n} = 0 on bΩ.  (7.2)

Proof of 7.1. Let ϕ = ∑_{|I|=q} ϕ_{I}dx^{I}, and ψ = ∑_{|J|=q+1} ψ_{I}dx^{J}. We shall use the following form of Green’s theorem:

∫_{Ω} D_{j}f\overline{g} = − ∫_{Ω} f\overline{D_{j}g} + ∫_{bΩ} fḡ\frac{∂ρ}{∂x_{j}} .

Recall that n⃗ = (D_{0}ρ, …, D_{N}ρ) is the normal direction.

Let dϕ = ∑_{|J|=q+1}(∑_{|I|=q} ∑_{j=0,...,N} ε_{jI}^{J}D_{j}ϕ_{I})dx^{J} ≡ ∑_{|J|=q+1} ϕ′_{J}dx^{J}. Then

⟨dϕ, ψ⟩_{1} = ∑_{|J|=q+1} [∑_{k=0}^{N}⟨D_{k}ϕ′_{J}, D_{k}ψ_{J}⟩_{0} + ⟨ϕ′_{J}, ψ_{J}⟩_{0}]

= ∑_{|J|=q+1} [∑_{k=0}^{N} ∑_{{|I|=q \atop j=0,...,N}} ε_{jI}^{J}(−⟨D_{k}ϕ_{I}, D_{k}D_{j}ψ_{J}⟩_{0} + ∫_{bΩ} D_{k}ϕ_{I} \overline{ψ_{J}} \frac{∂ρ}{∂x_{j}} )

+ ∑_{{|I|=q \atop j=0,...,N}} ε_{jI}^{J}(−⟨ϕ_{I}, D_{j}ψ_{J}⟩_{0} + ∫_{bΩ} ϕ_{I} \overline{ψ_{J}} \frac{∂ρ}{∂x_{j}} )]

= ∑_{|I|=q} [∑_{k=0}^{N}⟨D_{k}ϕ_{I}, D_{k}(− ∑_{{|J|=q+1 \atop j=0,...,N}} ε_{jI}^{J}D_{j}ψ_{J})⟩_{0} + ⟨ϕ_{I}, (− ∑_{{|J|=q+1 \atop j=0,...,N}} ε_{jI}^{J}D_{j}ψ_{J})⟩_{0}]

+ ∑_{|I|=q} [∑_{k=0}^{N}(∫_{bΩ} D_{k}ϕ_{I} \overline{( ∑_{{|J|=q+1 \atop j=0,...,N}} ε_{jI}^{J}D_{k}ψ_{J} \frac{∂ρ}{∂x_{j}} )}) + ∫_{bΩ} ϕ_{I} \overline{( ∑_{{|J|=q+1 \atop j=0,...,N}} ε_{jI}^{J}ψ_{J} \frac{∂ρ}{∂x_{j}} )}]

= ⟨ϕ, d′ψ⟩_{1}

+ ∑_{|I|=q} [∑_{k=0}^{N}(∫_{bΩ} D_{k}ϕ_{I} \overline{( ∑_{{|J|=q+1 \atop j=0,...,N}} ε_{jI}^{J}D_{k}ψ_{J} \frac{∂ρ}{∂x_{j}} )}) + ∫_{bΩ} ϕ_{I} \overline{( ∑_{{|J|=q+1 \atop j=0,...,N}} ε_{jI}^{J}ψ_{J} \frac{∂ρ}{∂x_{j}} )}] .

Notice that d′ in the last equality above is precisely the formal adjoint of d. Recall that d′ = −div on 1-forms. It is clear that

|⟨ϕ, d′ψ⟩_{1}| ≤ ∥ϕ∥_{1}∥ψ∥_{2},

and that

|∫_{bΩ} ϕ_{I} \overline{(− ∑_{{|J|=q+1 \atop j=0,...,N}} ε_{jI}^{J}ψ_{J} \frac{∂ρ}{∂x_{j}} )}| ≤ c · ∥ϕ∥_{L^{2}(bΩ)}∥ψ∥_{L^{2}(bΩ)}
≤ c · ∥ϕ∥_{1}∥ψ∥_{1},

by the trace theorem.

Next we consider the term

∑_{k=0}^{N} ∫_{bΩ} D_{k}ϕ_{I} \overline{( ∑_{{|J|=q+1 \atop j=0,...,N}} ε_{jI}^{J}D_{k}ψ_{J} \frac{∂ρ}{∂x_{j}} )} .

For k = 0, …, N we decompose the differential operator D_{k} into normal and tangential components, that is, in a suitable neighborhood of bΩ we write

D_{k} = Y_{k} + \frac{∂ρ}{∂x_{k}} \frac{∂}{∂n},  (7.3)

where Y_{k} is a tangential vector field, and ∂/∂n is the unit vector field in the normal direction. Therefore, for each fixed I, |I| = q, integration by parts yields that

∑_{k=0}^{N} ∫_{bΩ} D_{k}ϕ_{I} \overline{( ∑_{{|J|=q+1 \atop j=0,...,N}} ε_{jI}^{J}D_{k}ψ_{J} \frac{∂ρ}{∂x_{j}} )} = ∑_{k=0}^{N} ∫_{bΩ} ϕ_{I} \overline{Y_{k}^{∗}( ∑_{{|J|=q+1 \atop j=0,...,N}} ε_{jI}^{J}D_{k}ψ_{J} \frac{∂ρ}{∂x_{j}} )}

+ ∫_{bΩ} \frac{∂ϕ_{I}}{∂n} \overline{( ∑_{{|J|=q+1 \atop j=0,...,N}} ε_{jI}^{J} \frac{∂ψ_{J}}{∂n} \frac{∂ρ}{∂x_{j}} )}

≡ I + E.

Now observe that on bΩ, by definition,

∑_{{|J|=q+1 \atop j=0,...,N}} ε_{jI}^{J} \frac{∂ψ_{J}}{∂n} \frac{∂ρ}{∂x_{j}} = (∇n⃗ψ⌊n⃗)_{I} |_{bΩ}.  (7.4)

Using the trace theorem it is easy to see that, for ψ with C^{∞}(\overline{Ω}) coefficients,

|I| ≤ C_{ψ}∥ϕ∥_{1}.

Therefore, if ∇n⃗ψ⌊n⃗|_{bΩ} = 0, then ψ ∈ dom d^{∗}.

Conversely, suppose that ∇n⃗ψ⌊n⃗|_{bΩ} ≢ 0. Then, repeating the construction that we used in the case of the half space, we can see that the mapping

ϕ ↦ ∑_{I} ∫_{bΩ} \frac{∂ϕ_{I}}{∂n} \overline{(∇n⃗ψ⌊n⃗)_{I}}

cannot be continuous on W_{q}^{1}(Ω). This concludes the proof.

Notice that, from the previous computation, we obtain that for ψ ∈ dom d^{∗} it holds that

⟨dϕ, ψ⟩_{1} = ⟨ϕ, d′ψ⟩_{1}

+ ∑_{|I|=q}[− ∑_{k=0}^{N} ∫_{bΩ} ϕ_{I} \overline{∇_{Y_{k}^{∗}}( ∑_{{|J|=q+1 \atop j=0,...,N}} ε_{jI}^{J}D_{k}ψ_{J} \frac{∂ρ}{∂x_{j}} )} + ∫_{bΩ} ϕ_{I} \overline{(ψ⌊n⃗)_{I}}]. (7.5)

Next we shall determine an explicit expression for d^{∗}. We set (as in the case of the half space)

d^{∗}ϕ = d′ϕ + K_{Ω}ϕ,  (7.6)

where K_{Ω} is to be determined.

Proposition 7.2. Let ψ ∈ dom d^{∗}∩⋀^{q+1}(\overline{Ω}). Then the I-component (K_{Ω}ψ)_{I} of K_{Ω}ψ is the solution of the boundary value problem

\begin{cases}− △ v + v = 0 & on Ω \\ \frac{∂v}{∂n} = ∑_{k=0}^{N}[∇_{Y_{k}^{∗}}((∇_{Y_{k}}ψ)⌊n⃗)_{I} + (ψ⌊n⃗)]_{I} & on bΩ \end{cases} .

Proof. By the relation

⟨dϕ, ψ⟩_{1} = ⟨ϕ, d′ψ⟩_{1} + ⟨ϕ, K_{Ω}ψ⟩_{1},

and equation (7.5) we see that for ψ ∈ dom d^{∗} ∩ ⋀^{q+1}(\overline{Ω}),

⟨ϕ, K_{Ω}ψ⟩_{1} = ∑_{|I|=q}[∑_{k=0}^{N} ∫_{bΩ} ϕ_{I} \overline{Y_{k}^{∗}( ∑_{{|J|=q+1 \atop j=0,...,N}} ε_{jI}^{J}D_{k}ψ_{J} \frac{∂ρ}{∂x_{j}} )} + ∫_{bΩ} ϕ_{I} \overline{(ψ⌊n⃗)_{I}}]. (7.7)

Now, by Green’s theorem, we see that

⟨ϕ, K_{Ω}ψ⟩_{1} = ∑_{|I|=q} [∑_{j=0}^{N} ∫_{Ω} D_{j}ϕ_{I} \overline{D_{j}(K_{Ω}ψ)_{I}} + ∫_{Ω} ϕ_{I} \overline{(K_{Ω}ψ)_{I}} ]

= ∑_{|I|=q} [− ∫_{Ω} ϕ_{I} \overline{△(K_{Ω}ψ)_{I}} + ∫_{bΩ} ϕ_{I} \overline{\frac{∂(K_{Ω}ψ)_{I}}{∂n}} + ∫_{Ω} ϕ_{I} \overline{(K_{Ω}ψ)_{I}}] .

Therefore, for each fixed multi-index I, we must have

∫_{Ω} ϕ_{I} \overline{[− △ (K_{Ω}ψ)_{I} + (K_{Ω}ψ)_{I}]} + ∫_{bΩ} ϕ_{I} \overline{\frac{∂(K_{Ω}ψ)_{I}}{∂n}}

= ∑_{k=0}^{N} ∫_{bΩ} ϕ_{I} \overline{Y_{k}^{∗}( ∑_{{|J|=q+1 \atop j=0,...,N}} ε_{jI}^{J}D_{k}ψ_{J} \frac{∂ρ}{∂x_{j}} )} + ∫_{bΩ} ϕ_{I} \overline{(ψ⌊n⃗)_{I}}.

This implies that

[− △ (K_{Ω}ψ)_{I} + (K_{Ω}ψ)_{I}] = 0 on Ω.

Now recalling that, since ψ ∈ dom d^{∗},

∑_{{|J|=q+1 \atop j=0,...,N}} ε_{jI}^{J} \frac{∂ρ}{∂x_{j}} \frac{∂ψ_{J}}{∂n} = 0 on bΩ .

By writing D_{k} = Y_{k} + (∂ρ/∂x_{k}) · (∂/∂n), we have that

∑_{k=0}^{N} Y_{k}^{∗}( ∑_{{|J|=q+1 \atop j=0,...,N}} ε_{jI}^{J}D_{k}ψ_{J} \frac{∂ρ}{∂x_{j}} ) = ∑_{k=0}^{N} Y_{k}^{∗}( ∑_{{|J|=q+1 \atop j=0,...,N}} ε_{jI}^{J}Y_{k}ψ_{J} \frac{∂ρ}{∂x_{j}} )

= ∑_{k=0}^{N} Y_{k}^{∗}(∇_{Y_{k}}ψ⌊n⃗)_{I}

= ∑_{k=0}^{N}[∇_{Y_{k}^{∗}}(∇_{Y_{k}}ψ⌊n⃗)]_{I}.

From this the result follows.

Thus K_{Ω} is the solution operator for the preceding elliptic boundary value problem. By standard facts of the theory of elliptic problems (see [HOR1]) we know that K_{Ω}ψ is uniquely determined for each ψ ∈ ⋀^{q}(\overline{Ω}).

As in the case of the half space, the operator K_{Ω} turns out to be of order 1. In order to prove the following corollary, we need to apply Theorem 4.2.4 in [TRI]. This result deals with Schauder estimates for elliptic boundary value problems formulated in norms of negative order. The reference [TRI] happens to use the language of Triebel-Lizorkin spaces F_{p,q}^{s}. Since we consider only the case p = q = 2 we shall denote the spaces F_{2,2}^{s} simply by F^{s}. We also recall that, for s ≥ 0, F^{s} = W^{s}(Ω).

Corollary 7.3. Let s > 1/2. Then there exists C > 0 such that for all ψ ∈ ⋀^{q}(\overline{Ω}) we have

∥K_{Ω}ψ∥_{F^{s−1}} ≤ C∥ψ∥_{s}.

Proof. Notice that by this result the operator K_{Ω}, initially defined on the dense subspace ⋀^{q}(\overline{Ω}) can be extended as a continuous operator from F^{s−1} to W^{s}(Ω).

We need only apply standard estimates for elliptic boundary value problems. In order to obtain the sharp regularity result we use the estimates for the negative Sobolev spaces on the boundary (see [TRI], Theorem 4.2.4). Notice that the boundary differential operator acting on the data is of order 2, and tangential. Write T_{2} for that operator. Then we have

∥K_{Ω}ψ∥_{F^{s−1}} ≤ C(∥T_{2}ψ∥_{W^{s−5/2}(bΩ)} + ∥ψ∥_{W^{s−5/2}(bΩ)})
≤ C∥ψ∥_{W^{s−1/2}(bΩ)}
≤ C∥ψ∥_{s}.

We could, in principle, write down an approximate expression for K_{Ω} by using local coordinates to reduce the problem to the half space situation, where we have calculated K_{Ω} quite explicitly. We forego that option for now.

We can now reformulate our boundary value problem (7.1), recalling that on forms dd′ + d′d = △:

(− △ +G_{Ω})ψ = α  on Ω
∇n⃗ψ⌊n⃗ = 0  on bΩ
∇n⃗dψ⌊n⃗ = 0  on bΩ

where we set G_{Ω} = dK_{Ω} + K_{Ω}d.

8. A Special Coordinate System

In this section we introduce a system of local coordinates near the boundary that we will use in the rest of the paper.

It is a standard fact (see [KR2] for instance) that since bΩ is smooth and compact there exists a tubular neighborhood U of bΩ such that for each x ∈ U there is a unique π(x) ∈ bΩ which realizes the distance of x from bΩ. The line joining π(x) and x is orthogonal to bΩ.

Definition 8.1. Let {(U_{j}, Φ_{j})} be a covering of bΩ by local coordinates patches. The pair (V_{j}, Ψ_{j}), V_{j} ⊂ U, and Ψ : V_{j} → ℝ^{N+1} is a Fermi coordinate patch on U if

π(x) ∈ U_{j} for all p ∈ V_{j} ⊂ U,

and

Ψ_{j}(p) = (dist (π(p)), p), Φ_{j}(π(p))).

Throughout the rest of the paper, X_{0} = n⃗ will denote the vector field defined at each p ∈ U that is given by

n⃗_{p} = n⃗_{π(p)} = (\frac{∂ρ}{∂x_{0}} (p), …, \frac{∂ρ}{∂x_{n}} (p)).

We remark that the equality

n⃗_{p} = (\frac{∂ρ}{∂x_{0}} (p), …, \frac{∂ρ}{∂x_{n}} (p))

in general does not hold for p ∈ U \ bΩ.

On the fixed coordinate patch (U_{j}, Ψ_{j}) we also have an orthonormal frame of vector fields X_{0} = (∂/∂n), X_{1}, …, X_{N}. Notice that X_{1}, …, X_{N} form an orthonormal frame for the tangential vector fields. We also fix the dual basis of 1-forms, ω_{0} = dx_{0}, ω_{1}, …, ω_{N}. Given any q-form ϕ we write ϕ = ∑_{|I|=q} ω^{I}, where ω^{I} = ω_{i_{1}} ∧⋯∧ω_{i_{q}}, where I = (i_{1}, …, i_{q}) is a strictly increasing multi-index.

In the rest of the paper we shall always assume that the forms are written in terms of this given basis. Notice that, in these coordinates,

dψ = d(∑_{I} ψ_{I}ω^{I})

= ∑_{j} X_{j}ψ_{I}ω_{j} ∧ ω^{I} + {0 order terms in ψ}

= ∑_{J}(∑_{jI} X_{j}ψ_{I}ε_{jI}^{J})ω^{J} + {0 order terms}.

Analogously,

d′ψ = ∑_{L}(∑_{jI} X′_{j}ψ_{I}ε_{jL}^{I})ω^{L} + {0 order terms}.

Here X′_{k} denotes the formal adjoint of the vector field X_{k}.

Recall that (see [HEL] for instance) we can define the (Christoffel symbol) coefficients Γ_{ij}^{k} as follows: [HEL]:

∇_{X_{i}}X_{j} = ∑_{k=0}^{N} Γ_{ij}^{k}X_{k}.  (8.1)

We remark that ∇n⃗n⃗ = 0, since n⃗ is, by definition, a unit vector field whose integral curves are lines. Therefore the coefficients Γ_{ij}^{k} satisfy the following relations

Γ_{ij}^{k} = 0 \begin{cases}if k = 0 & and i, or j = 0 \\ if i = 0 & and j = 0 \end{cases}  (8.2)

This is so because

Γ_{0j}^{0} ≡ ⟨∇_{X_{0}}X_{j}, X_{0}⟩ = X_{0}⟨X_{j}, X_{0}⟩ − ⟨X_{j}, ∇_{X_{0}}X_{0}⟩ = 0,
Γ_{00}^{k} ≡ ⟨∇_{X_{0}}X_{0}, X_{k}⟩ = 0,

and

Γ_{i0}^{0} ≡ ⟨∇_{X_{i}}X_{0}, X_{0}⟩ = \frac{1}{2}X_{i}⟨X_{0}, X_{0}⟩ = 0.

Now we have a lemma about covariant differentiation in the normal direction.

Lemma 8.2. If ϕ ∈ ⋀^{q}(\overline{Ω}) is given by ϕ_{I} = ∑_{I} ϕ_{I}ω^{I}, then

∇_{X_{0}}ϕ = ∑_{|I|=q} [\frac{∂ϕ_{I}}{∂x_{0}} + ∑_{|J|=q} γ_{IJ}ϕ_{J} ] ω^{I},

where

γ_{IJ} = 0 \begin{cases}for J ∋ 0 & if I ∌ 0 \\ for J ∌ 0 & if I ∋ 0 \end{cases}

Proof. Fix a Fermi coordinate chart, then for ϕ ∈ ⋀^{q}(\overline{Ω}) supported in a small open set we have

(∇_{X_{0}}(∑_{J} ϕ_{J}ω^{J})) (X^{I}) = ∑_{J} (∇_{X_{0}}(ϕ_{J}ω^{J})) (X^{I})

= ∑_{J} \frac{∂ϕ_{J}}{∂x_{0}} ε_{J}^{I} − ∑_{J,J′} ∑_{s=1,...,q} ϕ_{J}((∇_{X_{0}}ω^{j_{s}}) ∧ ω^{J′}ε_{j_{s}J′}^{J})(X^{I}).

Moreover,

(∇_{X_{0}}(ω^{j}))(X_{ℓ}) = −ω^{j}(∇_{X_{0}}X_{ℓ})

= −ω^{j}(∑_{k} Γ_{0ℓ}^{k} X_{k})

= −Γ_{0ℓ}^{j}.

Therefore

∇_{X_{0}}ω^{j_{s}} = − ∑_{ℓ} Γ_{0ℓ}^{j_{s}} ω^{ℓ}.

Hence

(∇_{X_{0}}ϕ)(X^{I}) = ∑_{|J|=q} [\frac{∂ϕ_{J}}{∂x_{0}} ε_{J}^{I} − ∑_{{|J′|=q−1 \atop s=1,...,q}} ∑_{ℓ=0,...,N} ε_{j_{s}J′}^{J}(−Γ_{0ℓ}^{j_{s}})ε_{ℓJ′}^{I}ϕ_{J} ] .

Thus

∇_{X_{0}}ϕ = ∑_{|I|=q}(\frac{∂ϕ_{I}}{∂x_{0}} + ∑_{|J|=q} γ_{IJ}ϕ_{J})ω^{I},

where

γ_{IJ} = ∑_{{|J′|=q−1 \atop s=1,...,q}} ∑_{ℓ=0,...,N} ε_{j_{s}J′}^{J}Γ_{0ℓ}^{j_{s}} ε_{ℓJ′}^{I}.

Recall that (see (8.2)) the symbols Γ_{ij}^{k} are zero if either k = 0 and either i = 0 or j = 0, or i = j = 0. Suppose that I ∋ 0 and that J ∌ 0. Then ℓ = 0 so that Γ_{ij}^{j_{s}} = 0. If I ∌ 0 and J ∋ 0, then j_{s} = 0 so that Γ_{0ℓ}^{0} = 0. This proves the lemma.

In the sequel we will also use the following observation on the Laplace operator acting on forms. In general, the Laplacian on forms is defined as dd′+d′d. Having chosen the aforementioned basis on the space of q-forms, we see that if ψ = ∑_{|I|=q} ψ_{I}ω^{I} then

△ψ = ∑_{I}[∑_{k=0}^{N}(X′_{k}X_{k} + X_{k}X′_{k})ψ_{I}]ω^{I} + {lower order terms in ψ}.

The tangential Laplacian. Fix a Fermi coordinate patch (U, Ψ = (x_{0}, …, x_{N})). (Notice that (x_{0}, …, x_{N}) are not the standard coordinates of ℝ^{N+1}.) Given a function (or a form) u defined on U, we denote by ũ the function u ∘ Ψ^{−1} defined on Ψ(U).

In these coordinates the standard Laplacian of ℝ^{N+1} has the following form:

[△u] = ∑_{j,k=0}^{N} \frac{1}{\sqrt{det g}} \frac{∂}{∂x_{j}} (g^{jk} \frac{∂}{∂x_{k}} ũ)

= ∑_{j,k=1}^{N} \frac{1}{\sqrt{det g}} \frac{∂}{∂x_{j}} (g^{jk} \frac{∂}{∂x_{k}} ũ) + \frac{1}{\sqrt{det g}} \frac{∂}{∂x_{0}} (det g \frac{∂}{∂x_{0}} ũ)

= △′ũ + \frac{∂}{∂x_{0}} log(\sqrt{det g})\frac{∂ũ}{∂x_{0}} + \frac{∂^{2}ũ}{∂x_{0}^{2}}

= [△ũ] + ϵT_{2}ũ + \frac{∂}{∂x_{0}} (log \sqrt{det g})\frac{∂ũ}{∂x_{0}}

≡ [△ũ] + ϵL_{2}ũ,  (8.3)

where g is the metric matrix, and (g^{jk}) is its inverse. The operator △_{T} is the Laplacian on the submanifold obtained by fixing x_{0}, so that

△_{T} = ∑_{j=1}^{N} \frac{∂^{2}}{∂x_{j}^{2}} + ϵT_{2},  (8.4)

where T_{2} is a second order tangential differential operator with C^{∞}(\overline{Ω}) coefficients, and ϵ can be made arbitrarily small by shrinking U.

Notice that we have obtained that, on U,

△ = △_{T} + X_{0}^{2} + (X_{0} log \sqrt{det g})X_{0}.  (8.5)

The operator G_{Ω} on functions. Recall that, by Proposition 7.2 G_{Ω}u is the unique solution of the elliptic boundary value problem

\begin{cases}△v − v = 0 & on Ω \\ \frac{∂v}{∂n} = ∑_{k=0}^{N} Y_{k}^{∗}((∇_{Y_{k}}du)⌊n⃗) + \frac{∂u}{∂n} & on bΩ \end{cases} .

We are interested in making explicit the boundary operator in this case. Notice that Y_{k}^{∗} = −Y_{k} + (0 order terms), so that the boundary equation becomes

− ∑_{j=0}^{N} Y_{k}^{2}(X_{0}u) + T_{1}X_{0}u + T_{2}u,

where the T_{j}’s are tangential differential operators of order j. Now notice that

∑_{j=0}^{N} Y_{k}^{2} = ∑_{k=0}^{N} D_{k}^{2} − X_{0}^{2} + L_{1}

where L_{1} is a first order differential operator. This follows easily from the fact that, by construction,

Y_{k} = D_{k} − ⟨D_{k}, n⃗⟩X_{0},

(see (7.3)). In fact, for a function f,

△f = ∑_{k=0}^{N} D_{k}^{2}f

= ∑_{k=0}^{N}(Y_{k} + ⟨D_{k}, n⃗⟩)^{2}f

= ∑_{k=0}^{N}(Y_{k}^{2}f + Y_{k}(⟨D_{k}, n⃗⟩X_{0})f + X_{0}(⟨D_{k}, n⃗⟩Y_{k})f) + X_{0}^{2}f

= ∑_{k=0}^{N}(Y_{k}^{2}f + Y_{k}(⟨D_{k}, n⃗⟩)X_{0}f) + X_{0}^{2}f

= ∑_{k=0}^{N} Y_{k}^{2} + X_{0}^{2}f + aX_{0}f ,

since ∑_{k=0}^{N}⟨D_{k}, n⃗⟩Y_{k} = 0. Therefore the boundary equation in u equals

(− ∑_{k=0}^{N} Y_{k}^{2})(X_{0}u) + T_{1}X_{0}u + T_{2}u = (− △ +X_{0}^{2})(X_{0}u) + aX_{0}^{2}u + T_{1}X_{0}u + T_{2}u

= (− △_{T} +bX_{0})(X_{0}u) + aX_{0}^{2}u + T_{1}X_{0}u + T_{2}u

= − △_{T} X_{0}u + T_{1}X_{0}u + T_{2}u,

where we have used formula (8.5) and the fact that du ∈ dom d^{∗}. In local coordinates the boundary equation then becomes

\frac{∂v}{∂x_{0}} = − △_{T} (X_{0}u) + ϵT_{2}(X_{0}u)^{2} + [T_{1}X_{0}u + T_{2}u] .  (8.6)

We conclude this section by introducing a convention that we shall use consistently throughout the rest of the paper. By L_{j} we denote a generic differential operator of order j with C^{∞}(\overline{Ω}) coefficients, while by T_{j} we denote a differential operator of order j with C^{∞}(\overline{Ω}) (defined in a suitable neighborhood of bΩ), that involves only tangential derivatives.

9. The Existence Theorem

In this section we study the question of existence for the boundary value problem

(dd^{∗} + d^{∗}d)ϕ = α  on Ω
ϕ ∈ dom d^{∗}
dϕ ∈ dom d^{∗}

for α ∈ ⋀^{q}(\overline{Ω}), q = 0, …, N, from an abstract Hilbert space point of view. We remark that by Proposition 7.1 the boundary conditions are given by the equations

∇n⃗ϕ⌊n⃗ = 0 on bΩ

and

∇n⃗dϕ⌊n⃗ = 0 on bΩ.

Notice that, using the coordinates and the frame introduced in (8.1), the above equations can be rewritten as

X_{0}ϕ_{I} + L_{0}ϕ = 0 on bΩ if I ∋ 0  (9.1)

and

X_{0}^{2}ϕ_{I} + L_{1}ϕ = 0 on bΩ if I ∌ 0,  (9.2)

where L_{j} are differential operators of order j in the components of ϕ. We remark that if ϕ is a function then the first boundary equation is empty, and the second one becomes

X_{0}^{2}ϕ = 0  on bΩ,  (9.3)

since ∇n⃗n⃗ = 0.

Our development in what follows parallels classical studies such as that which can be found in [FOK]. Let

D ≡ \{ϕ ∈ ⋀^{q}(\overline{Ω}) : ϕ , dϕ ∈ dom d^{∗}\}.

For ϕ, ψ ∈ D we define the bilinear form

Q(ϕ, ψ) = ⟨dϕ, dψ⟩_{1} + ⟨d^{∗}ϕ, d^{∗}ψ⟩_{1} + ⟨ϕ, ψ⟩_{1}.

Our first claim is that D is dense in W_{q}^{1}. We argue as follows. Let ϕ be any q-form. We may assume that ϕ has coefficients smooth up to the boundary. Then it suffices to find a ψ ∈ ⋀^{q}(\overline{Ω}) of small norm such that ϕ + ψ ∈ D. We use Fermi local coordinates in a tubular neighborhood of bΩ, so that x = (x_{0}, x′) with x′ ∈ bΩ and x_{0} = dist (x_{0}, bΩ) parametrizing the normal direction. Set

ψ_{1}(x′) = ∇n⃗(ϕ⌊n⃗)|_{bΩ},

and

ψ_{2}(x′) = ∇n⃗(dϕ⌊n⃗)|_{bΩ}.

Finally, set

ψ(x) = (−x_{0}dx_{0} ∧ ψ_{1}(x′) − \frac{1}{2}x_{0}^{2}ψ_{2}) χ(x_{0}),

where χ ∈ C_{0}^{∞}[−2ϵ, 2ϵ], χ = 1 in a neighborhood of 0, and ∥χ∥_{1} ≤ Cϵ^{−1/2}. Then ∥ψ∥_{1} < Cϵ^{1/2} and ϕ + ψ ∈ D. That completes the argument.

Now we let D̃ be the closure of D in the topology induced by Q. We wish to check that D̃ is still contained in W_{q}^{1}. It is easy to see that d is closed in W_{q}^{1}; of course d^{∗} is closed also (since adjoints are always closed). These facts imply that D̃ is a subspace of W_{q}^{1}.

At this point we apply the Friedrichs extension theorem, as in [FOK], to show that there exists a canonical self adjoint operator

T : W_{q}^{1} → D̃

which is bounded in the W^{1} topology, is injective, and such that

Q(T ϕ, ψ) = ⟨ϕ, ψ⟩_{1}.

If we set F = T^{−1}, then

Q(ϕ, ψ) = ⟨F ϕ, ψ⟩_{1} ∀ϕ ∈ dom F, ψ ∈ D̃.

Notice that F = (d^{∗}d + dd^{∗}) + I when restricted to D.

Now we show that the Q-unit ball of D̃ is compactly embedded in W_{q}^{1} by proving that the Q norm is equivalent to the W^{2} norm. Notice that Q(ϕ, ϕ) ≤ c∥ϕ∥_{2}^{2} since d^{∗} is of order 1. The reverse inequality follows from the next theorem.

THEOREM 9.1. There exists a constant C_{0} > 0 such that, for all ψ ∈ D̃,

Q(ψ, ψ) ≥ C_{0} · ∥ψ∥_{2}^{2}.

Assume the theorem for now. Since the Q-unit ball of D̃ is compactly embedded in W_{q}^{1}(Ω), it follows that T is a compact operator. Notice that if T α ∈ D, then T α is the unique solution of the boundary value problem

\begin{cases}(dd^{∗} + d^{∗}d)ϕ + ϕ = & α \\ ϕ ∈ dom d^{∗} \\ dϕ ∈ dom d^{∗} \end{cases}

for α ∈ W_{q}^{1}, q = 0, 1, …, N + 1.

We now wish to establish conditions for the solvability of

\begin{cases}(dd^{∗} + d^{∗}d)ϕ = & α \\ ϕ ∈ dom d^{∗} \\ dϕ ∈ dom d^{∗} \end{cases}  (9.4)

Let Q_{0} be the bilinear form on D̃ defined by

Q_{0}(ϕ, ψ) = ⟨dϕ, dψ⟩_{1} + ⟨d^{∗}ϕ, d^{∗}ψ⟩_{1}.

Thus ϕ is a solution of (9.4) precisely when ϕ ∈ D and

Q_{0}(ϕ, ψ) = ⟨α, ψ⟩_{1} ∀ψ ∈ D̃.

This in turn holds if and only if

Q(ϕ, ψ) = Q_{0}(ϕ, ψ) + ⟨ϕ, ψ⟩_{1}
= ⟨α, ψ⟩_{1} + ⟨ϕ, ψ⟩_{1}
= ⟨α + ϕ, ψ⟩_{1}.

By the Friedrichs theorem, we have reduced our situation to solving the equation

(F − I)ϕ = α  (9.5)

with ϕ ∈ dom F; i.e., setting θ = F ϕ,

θ − T θ = α.

We now apply the standard theory of compact operators to obtain that the above equation has a solution θ for all α orthogonal to the finite dimensional subspace H_{q} ≡ ker (I − T). It is easy to check that ker (I − T) is exactly the kernel of F − I. Thus, for α orthogonal (in the W^{1}-inner product) to ker (I − T), we obtain that ϕ = T θ is the solution of the equation (9.5). Notice that, if ϕ ∈ D, then the equation (9.5) reduces to the boundary value problem (9.4). Moreover, Z ≡ (F − I)(D) is a dense subspace of H_{q}^{⊥} ⊆ W_{q}^{1}, i.e.

Z^{⊥} = H_{q}.

Let β ∈ D. For ϕ ∈ D we have

⟨(F − I)ϕ, β⟩_{1} = Q_{0}(ϕ, β)
= Q(ϕ, β) − ⟨ϕ, β⟩_{1}
= ⟨ϕ, (T − I)β⟩_{1}.

Thus, if β ∈ Z^{⊥}, then ⟨(F − I)ϕ, β⟩_{1} = 0 for all ϕ ∈ D. Since D is dense in W_{q}^{1}, (T − I)β = 0, that is β ∈ H_{q}.

Thus we have proved the following theorem:

THEOREM 9.2 (Existence for all q). The boundary value problem

\begin{cases}(dd^{∗} + d^{∗}d)ϕ = α \\ ϕ ∈ dom d^{∗} \\ dϕ ∈ dom d^{∗} \end{cases}

has a finite dimensional kernel H_{q}^{⊥} and finite dimensional cokernel. [Note that the space Z is dense in H_{q}^{⊥}.] The problem has a solution ϕ ∈ D for α ∈ Z ⊆ H_{q}^{⊥}.

The coercive estimate. We now prove the fundamental estimate from below for the bilinear form Q.

Proof of 9.1. We begin by noticing that

Q(ψ, ψ) = ⟨dψ, dψ⟩_{1} + ⟨d^{∗}ψ, d^{∗}ψ⟩_{1} + ⟨ψ, ψ⟩_{1}
= ⟨dψ, dψ⟩_{1} + ⟨d′ψ, d^{∗}ψ⟩_{1} + ⟨K_{Ω}ψ, d^{∗}ψ⟩_{1} + ⟨ψ, ψ⟩_{1}
= ⟨dψ, dψ⟩_{1} + ⟨d′ψ, d′ψ⟩_{1} + ⟨ψ, ψ⟩_{1} + ⟨d′ψ, K_{Ω}ψ⟩_{1} + ⟨K_{Ω}ψ, d^{∗}ψ⟩_{1}.

Our plan is to prove that the following claims hold true.

Claim 1. There exists a constant C_{1} > 0 such that for all ψ ∈ D we have

⟨dψ, dψ⟩_{1} + ⟨d′ψ, d′ψ⟩_{1} + ⟨ψ, ψ⟩_{1} ≥ C_{1}∥ψ∥_{2}^{2}.

Claim 2. For any ϵ > 0 there exists C_{ϵ} > 0 such that all ψ ∈ D

|⟨K_{Ω}ψ, d^{∗}ψ⟩_{1}| ≤ ϵ∥ψ∥_{2}^{2} + C_{ϵ}∥d^{∗}ψ∥_{1}^{2},

and

|⟨d′ψ, K_{Ω}ψ⟩_{1}| ≤ ϵ∥ψ∥_{2}^{2} + C_{ϵ}(∥dψ∥_{1}^{2} + ∥ψ∥_{1}^{2}).

Assuming the claims for now, we shall finish the proof. We have that

Q(ψ, ψ) = ⟨dψ, dψ⟩_{1} + ⟨d′ψ, d′ψ⟩_{1} + ⟨ψ, ψ⟩_{1} + ⟨d′ψ, K_{Ω}ψ⟩_{1} + ⟨K_{Ω}ψ, d^{∗}ψ⟩_{1}
≥ C_{1}∥ψ∥_{2}^{2} − |⟨K_{Ω}ψ, d^{∗}ψ⟩_{1}| − |⟨d′ψ, K_{Ω}ψ⟩_{1}|
≥ (C_{1} − 2ϵ)∥ψ∥_{2}^{2} − C_{ϵ}(∥d^{∗}ψ∥_{1}^{2} + ∥dψ∥_{1}^{2} + ∥ψ∥_{1}^{2}).

Therefore the constant C_{0} = (C_{1} − 2ϵ)/(1 + C_{ϵ}) does the job. Thus it remains to prove the claims. We begin with Claim 2.

Proof of Claim 2. Recall that, by formula (7.7),

⟨K_{Ω}θ, ϕ⟩_{1} = ∑_{I} [∑_{k=0}^{N} ∫_{bΩ}(∇_{Y_{k}}(θ⌊n⃗))_{I} \overline{(∇_{Y_{k}}ϕ)_{I}} + ∫_{bΩ}(θ⌊n⃗)_{I} \overline{ϕ_{I}}] .

Therefore, by applying the Schwarz inequality for Sobolev spaces on bΩ, and recalling that the Y_{k} are tangential vector fields, we see that

|⟨K_{Ω}ψ, d^{∗}ψ⟩_{1}| ≤ ∑_{I} ∑_{k=0}^{N} ∫_{bΩ} |(∇_{Y_{k}}ψ⌊n⃗)_{I}(∇_{Y_{k}}(d^{∗}ψ))_{I}| + ∥ψ∥_{L^{2}(bΩ)}∥d^{∗}ψ∥_{L^{2}(bΩ)}

≤ c ∑_{kI} ∥∇_{Y_{k}}(ψ⌊n⃗)_{I}∥_{W^{1/2}(Ω)}∥∇_{Y_{k}}(d^{∗}ψ)_{I}∥_{W^{−1/2}(Ω)} + ∥ψ∥_{L^{2}(bΩ)}∥d^{∗}ψ∥_{L^{2}(bΩ)}

≤ c(∥ψ∥_{W^{3/2}(bΩ)}∥d^{∗}ψ∥_{W^{1/2}(bΩ)})

≤ c∥ψ∥_{2}∥d^{∗}ψ∥_{1}

≤ ϵ∥ψ∥_{2}^{2} + C_{ϵ}∥d^{∗}ψ∥_{1}^{2}.

On the other hand, for ψ ∈ D, we see that (using “l.o.t.” to denote lower order terms)

⟨d′ψ, K_{Ω}ψ⟩_{1} = ∑_{I} [∑_{k=0}^{N} ∫_{bΩ}(∇_{Y_{k}}(d′ψ))_{I} \overline{((∇_{Y_{k}}ψ)⌊n⃗)_{I}}] + \{l.o.t.\}

= ∑_{kI} ∫_{bΩ}( ∑_{j≠0,J} X_{j}Y_{k}ψ_{J}ε_{jI}^{J})\overline{(∇_{Y_{k}}ψ⌊n⃗)_{I}} + \{l.o.t.\}

= − ∑_{kJ} ∫_{bΩ}(Y_{k}ψ_{J})\overline{( ∑_{j≠0,I} ε_{jI}^{J}X_{j}(∇_{Y_{k}}ψ⌊n⃗)_{I})} + \{l.o.t.\}

= − ∑_{kJ} ∫_{bΩ}(Y_{k}ψ_{J})\overline{Y_{k}( ∑_{j≠0,I} ε_{jI}^{J}X_{j}(ψ⌊n⃗)_{I})} + \{l.o.t.\},

where we have used the fact that the term containing X_{0}ψ_{J} with J ∋ 0 can be absorbed in the error terms because of the equation (9.1). Now notice that, for ψ ∈ D, we have that

X_{0}(ψ⌊n⃗)_{I} = (∇n⃗(ψ⌊n⃗)) + {0 order terms}

= ((∇n⃗ψ)⌊n⃗)_{I} + {0 order terms}

= {0 order terms}.

Thus

∑_{kJ} ∫_{bΩ}(Y_{k}ψ_{J})\overline{Y_{k}( ∑_{j≠0,I} ε_{jI}^{J}X_{j}(ψ⌊n⃗)_{I})}

= ∑_{kJ} ∫_{bΩ}(Y_{k}ψ_{J})\overline{(∇_{Y_{k}}(d(ψ⌊n⃗)))_{J}} + \{l.o.t.\}.

Now we notice that, using equation (9.1), it follows that for ψ ∈ D, the coefficients of d(ψ⌊n⃗) are all coefficients of dψ, modulo lower order terms in the components of ψ. Notice also that the lower order terms that we produced in the previous calculation are all O(∥ψ∥_{W^{1}(bΩ)}).

Therefore

|⟨d′ψ, K_{Ω}ψ⟩_{1}| ≤ C(∑_{kJ} ∥∇_{Y_{k}}ψ_{J}∥_{W^{1/2}(bΩ)}∥∇_{Y_{k}}d(ψ⌊n⃗)∥_{W^{−1/2}(bΩ)} + ∥ψ∥_{W^{1}(bΩ)}^{2})

≤ C(∑_{J} ∥ψ_{J}∥_{W^{3/2}(bΩ)}∥d(ψ⌊n⃗)_{J}∥_{W^{1/2}(bΩ)} + ∥ψ∥_{W^{1}(bΩ)}^{2})

≤ C(∥ψ∥_{W^{3/2}(bΩ)}∥dψ∥_{W^{1/2}(bΩ)} + ∥ψ∥_{W^{1}(bΩ)}^{2})

≤ C(∥ψ∥_{2}∥dψ∥_{1} + ∥ψ∥_{3/2}^{2})

≤ ϵ∥ψ∥_{2}^{2} + C_{ϵ}(∥dψ∥_{1}^{2} + ∥ψ∥_{1}^{2}).

This proves Claim 2.

Proof of Claim 1. Fix an open cover \{U_{ℓ}\}_{0}^{M} of \overline{Ω} such that \overline{U_{0}} ⊆ Ω and, for ℓ = 1, …, M, on each U_{ℓ} we can find an orthonormal frame ω_{0} = dx_{0}, ω_{1}, …, ω_{N} for the space of 1-forms. Let X_{0} = ∂/∂x_{0}, X_{1}, …, X_{N} be the dual frame of vector fields. With respect to this basis the Laplacian is not diagonal on the space of q-forms, but it is diagonal in the top order terms.

On U_{ℓ} we write ψ = ∑_{I} ψ_{I}ω^{I}. Let {η_{ℓ}^{2}} be a partition of unity subordinate to the cover {U_{ℓ}}. In this proof we denote by E any quantity which is O(∥ψ∥_{2}∥ψ∥_{1}). Let A for the moment denote either operator d or d′. Then notice that

⟨Aψ, Aψ⟩_{1} = ∑_{jI}⟨D_{j}(Aψ)_{I}, D_{j}(Aψ)_{I}⟩_{0} + E

= ∑_{jI} ∑_{ℓ} ∫_{Ω} η_{ℓ}^{2}(D_{j}Aψ)_{I} \overline{(D_{j}Aψ)_{I}} + E

= ∑_{jI} ∑_{ℓ} ∫_{Ω} D_{j}(A(η_{ℓ}ψ))_{I} \overline{D_{j}(A(η_{ℓ}ψ))_{I}} + E.

Then we see that it suffices to consider forms with support in one of the patches U_{ℓ}. When integrating by parts, this reduction only produces error terms of the type E already considered.

Notice that the form η_{ℓ}ψ may not belong to D. Nonetheless η_{ℓ}ψ satisfies boundary equations of type (9.1) and (9.2), and this is all we need.

Thus, without loss of generality, let ψ ∈ D have support in a small open set on which we can write ψ = ∑_{I} ψ_{I}ω^{I}. We have

dψ = ∑_{J}(∑_{jI} ε_{jI}^{J}X_{j}ψ_{I})ω^{J} + {0 order terms},

and

d′ψ = ∑_{L}(∑_{ℓI} ε_{ℓL}^{I}X′_{ℓ}ψ_{I})ω^{L} + {0 order terms},

where X′_{ℓ} is the formal adjoint of X_{ℓ}.

Now notice that there exists a constant C_{0} such that for θ ∈ W_{q}^{1}, θ = ∑_{I} θ_{I}ω^{I}, we have that

∥θ∥_{1}^{2} ≥ C_{0} ∑_{I}(∑_{j} ∥X_{j}θ_{I}∥_{0}^{2} + ∥θ_{I}∥_{0}^{2}),  (9.6)

where the constant C_{0} depends only on the choice of ω_{0}, …, ω_{N}.

Then,

∑_{jI}(∥X_{j}(dψ)_{I}∥_{0}^{2} + ∥X_{j}(d′ψ)_{I}∥_{0}^{2})

= ∑_{J} ∑_{pIqI′} ∑_{k=0}^{N} ∫_{Ω}(ε_{pI}^{J}X_{p}X_{k}ψ_{I})\overline{(ε_{qI′}^{J}X_{q}X_{k}ψ_{I′})}

+ ∑_{L} ∑_{pIqI′} ∑_{k=0}^{N} ∫_{Ω}(ε_{pL}^{I}X′_{p}X_{k}ψ_{I})\overline{(ε_{qL}^{I′}X′_{q}X_{k}ψ_{I′})} + E

= ∑_{kII′}[∑_{pq} ∫_{Ω}(∑_{J} ε_{pI}^{J}ε_{qI′}^{J})(X_{p}X_{k}ψ_{I})\overline{(X_{q}X_{k}ψ_{I′})}

+ ∑_{pq} ∫_{Ω}(∑_{L} ε_{pL}^{I}ε_{qL}^{I′})(X′_{p}X_{k}ψ_{I})\overline{(X′_{q}X_{k}ψ_{I′})}] + E

= ∑_{kII′}[∑_{pq} ∫_{Ω}(∑_{J} ε_{pI}^{J}ε_{qI′}^{J})(X_{p}X_{k}ψ_{I})\overline{(X_{q}X_{k}ψ_{I′})}

+ ∑_{pq} ∫_{Ω}(∑_{L} ε_{pL}^{I}ε_{qL}^{I′})(X_{p}X_{k}ψ_{I})\overline{(X_{q}X_{k}ψ_{I′})}] + E,  (9.7)

where the last equality holds since X′_{ℓ} = −X_{ℓ} + L_{0}, where L_{0} is an operator of order 0.

Consider the term

∫_{Ω} ε_{pI}^{J}ε_{qI′}^{J}(X_{q}X_{k}ψ_{I})\overline{(X_{p}X_{k}ψ_{I′})} + ∫_{Ω} ε_{pL}^{I}ε_{qL}^{I′}(X_{p}X_{k}ψ_{I})\overline{(X_{q}X_{k}ψ_{I′})}

≡ I + II. (9.8)

We need to distinguish two cases, according to whether p = q or p ≠ q. Notice that if p = q in (9.8), then

I + II = 2∥X_{p}X_{k}ψ_{I}∥_{0}^{2} + E,  (9.9)

since ε_{qI}^{J}ε_{pI′}^{J} = ε_{pL}^{I}ε_{qL}^{I′} = 0 if I ≠ I′.

Suppose now that p ≠ q. We show in this case that the term in (9.8) is of type E. Notice that if p ≠ q then

ε_{qI}^{J}ε_{pI}^{J} + ε_{pL}^{I}ε_{qL}^{I} = 0,  (9.10)

because, if p < q, then ε_{qI}^{J} = −ε_{qL}^{I′} and ε_{pI′}^{J} = ε_{pL}^{I}. Then, if p, q ≠ 0, integration by parts gives rise to no boundary term, and (9.10) shows that

I + II = E  (9.11)

when p ≠ q, p, q ≠ 0.

If p ≠ 0 and q = 0 we distinguish two cases according to whether k = 0 or k ≠ 0. If q = 0 and k = 0, then we integrate by parts in II. Recall that ψ ∈ D implies

that for I′ ∋ 0 ψ_{I′} satisfies the boundary equation (9.1). Then we have

II = − ∫_{Ω} ε_{0I}^{J}ε_{pI′}^{J}(X_{0}X_{p}X_{0}ψ_{I})\overline{X_{0}ψ_{I′}} + ∫_{bΩ} ε_{pL}^{I}ε_{0L}^{I′}(X_{p}X_{0}ψ_{I})\overline{X_{0}ψ_{I′}} + E
= ∫_{Ω} ε_{0I}^{J}ε_{pI′}^{J}(X_{0}X_{0}ψ_{I})\overline{(X_{p}X_{0}ψ_{I′})} + ∫_{bΩ}(X_{p}X_{0}ψ_{I})\overline{L_{0}ψ_{I′}} + E.

Now using equation (9.10) again, we see that

|I + II| ≤ | ∫_{bΩ}(X_{p}X_{0}ψ_{I})L_{0}ψ_{I′} | + E
≤ C ∫ |X_{0}ψ_{I}X_{p}ψ_{I′}| + E
≤ C∥X_{0}ψ_{I}∥_{W^{1/2}(bΩ)}∥ψ_{I}∥_{W^{1/2}(bΩ)} + E
= E,  (9.12)

that is I + II = E in this case.

Finally, suppose that q = 0 and k ≠ 0. Notice that ε_{0I}^{J} = ε_{0L}^{I′} = 1, and ε_{pI′}^{J} = −ε_{pL}^{I} since I′ ∋ 0. Then

I + II = − ∫_{Ω} ε_{pL}^{I}(X_{0}X_{k}ψ_{I})\overline{(X_{p}X_{k}ψ_{I′})} + ∫_{Ω} ε_{pL}^{I}(X_{p}X_{k}ψ_{I})\overline{(X_{0}X_{k}ψ_{I′})} + E
= ∫_{Ω} ε_{pL}^{I}(X_{p}X_{0}X_{k}ψ_{I})\overline{X_{k}ψ_{I′}} + ∫_{Ω} ε_{pL}^{I}(X_{p}X_{k}ψ_{I})\overline{(X_{0}X_{k}ψ_{I′})} + E
= − ∫_{Ω} ε_{pL}^{I}(X_{p}X_{k}ψ_{I})\overline{(X_{0}X_{k}ψ_{I′})} + ∫_{bΩ} ε_{pL}^{I}(X_{p}X_{k}ψ_{I})\overline{X_{k}ψ_{I′}}
+ ∫_{Ω} ε_{pL}^{I}(X_{p}X_{k}ψ_{I})\overline{(X_{0}X_{k}ψ_{I′})} + E
= − ∫_{bΩ} X_{k}ψ_{I} \overline{X_{k}(X_{p}ψ_{I′}ε_{pL}^{I})} + E.

Hence, using the fact that k ≠ 0 we have that

| ∑_{pI′} I + II| ≤ | ∫_{bΩ} X_{k}ψ_{I} \overline{X_{k}(∑_{pI′} X_{p}ψ_{I′}ε_{pI′}^{I})}| + E
= | ∫_{bΩ} X_{k}ψ_{I} \overline{X_{k}(dψ⌊n⃗)_{I}} | + E
≤ ∥X_{k}ψ_{I}∥_{W^{1/2}(bΩ)}∥X_{k}(dψ⌊n⃗)_{I}∥_{W^{−1/2}(bΩ)} + E
≤ C∥ψ∥_{W^{3/2}(bΩ)}∥dψ⌊n⃗∥_{W^{1/2}(bΩ)} + E
≤ C∥ψ∥_{2}∥dψ∥_{1} + E
≤ ϵ∥ψ∥_{2}^{2} + C_{ϵ}∥dψ∥_{1}^{2} + E.  (9.13)

Therefore, collecting (9.9) - (9.13) and substituting them into (9.7), and recalling (9.6), we obtain that

\frac{1}{C_{0}} (∥dψ∥_{1}^{2} + ∥d′ψ∥_{1}^{2}) ≥ ∑_{kI}(∥X_{k}(dψ)_{I}∥_{0}^{2} + ∥X_{k}(d′ψ)_{I}∥_{0}^{2})

= ∑_{kII′}[∑_{pq} I + II]

≥ ∑_{kpI} 2∥X_{p}X_{k}ψ_{I}∥_{0}^{2} − (ϵ∥ψ∥_{2}^{2} + C_{ϵ}∥dψ∥_{1}^{2}) + E

= (2 − ϵ)∥ψ∥_{2}^{2} − C_{ϵ}∥dψ∥_{1}^{2} + E

≥ (2 − 2ϵ)∥ψ∥_{2}^{2} − C_{ϵ}(∥dψ∥_{1}^{2} + ∥ψ∥_{1}^{2}).

From this Claim 1 follows easily. This proves the theorem.

10. The Regularity Theorem in the Case of Functions

We now turn to the question of regularity. In this section we are going to prove the regularity result for the case q = 0.

THEOREM 10.1. Let f ∈ C^{∞}(\overline{Ω}) be orthogonal in the W^{1} inner product to the space of constant functions. Then there is a unique function u ∈ C^{∞}(\overline{Ω}) that solves the boundary value problem

\begin{cases}d^{∗}d u = f & on Ω \\ du ∈ dom d^{∗} \end{cases}

In other words, u solves the system

\begin{cases}(− △ +G_{Ω})u = f & on Ω \\ \frac{∂^{2}u}{∂n^{2}} = 0 & on bΩ \end{cases}

Moreover, the solution u satisfies the desired (coercive) estimates for each s > 1/2; i.e. there exists a c_{s} > 0 such that

∥u∥_{s+2} ≤ c_{s}(∥f∥_{s} + ∥u∥_{0}).  (10.1)

Proof. By the existence theorem, Theorem 9.2, we know that the solution to the above boundary value problem exists for all f orthogonal to the harmonic space H_{0}. In the forthcoming Theorem 12.2 we shall prove that H_{0} reduces to the constants. Now we turn to the question of estimates.

Let u be a solution of the boundary value problem, with f ∈ C^{∞}(\overline{Ω}). Let s > 1/2. We wish to estimate ∥u∥_{s+2} in terms of ∥f∥_{s}. It is clear that it suffices to estimate ∥ηu∥_{s+2} for a given cut-off function η with small support.

We first suppose that supp η ∩ bΩ = ∅. [The second step will be to assume that supp η ∩ bΩ ≠ ∅.]

The interior estimate. Let η ∈ C_{0}^{∞}(Ω). Recalling that f = − △ u + G_{Ω}u we have that

∥ηu∥_{s+2} ≤ c∥ △ (ηu)∥_{s}
≤ c(∥ △ (ηu) + ηf∥_{s} + ∥ηf∥_{s})
≤ c(∥ △ (ηu) − η △ u∥_{s} + ∥ηG_{Ω}u∥_{s} + ∥ηf∥_{s})
≤ c(∥η_{1}u∥_{s+1} + ∥ηG_{Ω}u∥_{s} + ∥ηf∥_{s}),  (10.2)

where η_{1} ≡ 1 on supp η. Next we want to show that ∥ηG_{Ω}u∥_{s} ≤ c∥u∥_{s+1}. Recall that by Proposition 7.2 and equation (8.6) G_{Ω} is the solution of the boundary value problem

\begin{cases}− △ v + v = 0 & on Ω \\ \frac{∂v}{∂n} = ∑_{k=0}^{N} ∇_{Y_{k}^{∗}}((∇_{Y_{k}}du)⌊n⃗) + \frac{∂u}{∂n} & on bΩ \end{cases} .

Now it is easy to see that (using the Fourier transform for instance)

∥ηG_{Ω}u∥_{s} ≤ c∥η_{1}G_{Ω}u∥_{s−1} ≤ c∥G_{Ω}u∥_{s−1}.

(Here, for t > 0, ∥ · ∥_{−t} is the norm in the Sobolev space W_{−t}(Ω) ≡ (W^{t}^{∘} (Ω))^{∗}.) Moreover, by [TRI] Theorem 4.2.4 we have

∥G_{Ω}u∥_{s−1} ≤ c∥T_{2} \frac{∂u}{∂n} − T′_{2}u∥_{W^{s−5/2}(bΩ)}

≤ c(∥\frac{∂u}{∂n}∥_{W^{s−1/2}(bΩ)} + ∥u∥_{W^{s−1/2}(bΩ)})

≤ c∥u∥_{s+1}.  (10.3)

Thus (10.2) and (10.3) give that

∥ηu∥_{s+2} ≤ c(∥ηf∥_{s} + ∥u∥_{s+1}).

The boundary estimate. Fix a Fermi coordinate patch

(U, Ψ = (x_{0}, …, x_{N})),

and let η ∈ C_{0}^{∞}(U).

In what follows we denote by G_{Ω} both the operator defined on Ω and (when restricted to U) the same operator expressed in the local chart and thus defined on Ψ(U). This technical ambiguity should cause no confusion. We will denote by G_{ℝ^{N+1}} the operator arising from considering the adjoint of d^{∗} in the half space. Finally, given a function u ∈ C^{∞}(U) we denote by ũ the function

u ∘ Ψ^{−1} ∈ C^{∞}(Ψ(U)).

Now consider η, η_{1} ∈ C_{0}^{∞}(U) with η_{1} = 1 on the support of η. These functions are chosen to be constant along the normal direction to bΩ near the boundary. In our local coordinates, ηu satisfies the following boundary condition:

\frac{∂^{2}}{∂x_{0}^{2}} (η̃u) = 0 on \{x_{0} = 0\}.

Recall that if v satisfies the boundary value problem

\begin{cases}− △ v + G_{ℝ_{+}^{N+1}}v = f & on ℝ_{+}^{N+1} \\ \frac{∂^{2}v}{∂x_{0}^{2}} = 0 & on bℝ_{+}^{N+1} \end{cases}

and if v has compact support, then

∥v∥_{s+2} ≤ C{∥f∥_{s} + ∥v∥_{s+1}}, s > 1/2.

Then

∥η̃u∥_{s+2} ≤ C{∥(− △ +G_{ℝ_{+}^{N+1}})(η̃u)∥_{s} + ∥η̃u∥_{s+1}}.  (10.4)

74 L. FONTANA, S. G. KRANTZ, AND M. M. PELOSO

Our goal is to replace − △ + G_{ℝ_{+}^{N+1}} with − △ +ϵL_{2} + G_{Ω} modulo error terms that are controlled by lower order norms of ηu. Thus the estimate (10.4) gives that

∥ηu∥_{s+2} ≤ C{∥(− △ +ϵL_{2} + G_{Ω})(η̃u)∥_{s} + ∥η̃u∥_{s+1}
+ϵ∥L_{2}(η̃u)∥_{s} + ∥G_{ℝ_{+}^{N+1}}(η̃u)[ηG_{Ω}u] ∥_{s}}
≤ C{∥η̃[(− △ +ϵL_{2} + G_{Ω})(u)] ∥_{s} + ϵ∥ηu∥_{s+2} + ∥η_{1}u∥_{s+1}
+∥G_{ℝ_{+}^{N+1}}(η̃u) − [ηG_{Ω}u] ∥_{s}}
≤ C{∥ − η̃ △ ũ + G_{ℝ^{N+1}}(η̃u)∥_{s} + ∥η_{1}u∥_{s+1}}
≤ C{∥η̃(− △ +ϵL_{2} + G_{Ω})(ũ)∥_{s} + ϵ∥ηu∥_{s+2}
+∥G_{ℝ^{N+1}}(η̃u) − [ηG_{Ω}u] ∥_{s} + ∥η_{1}u∥_{s+1}}  (10.5)
= C{∥ηf∥_{s} + ϵ∥ηu∥_{s+2} + ∥G_{ℝ^{N+1}}(η̃u) − [ηG_{Ω}u] ∥_{s} + ∥η_{1}u∥_{s+1}}.

The function G_{ℝ_{+}^{N+1}}(η̃u) solves the boundary value problem

\begin{cases}(− △ +I)(G_{ℝ_{+}^{N+1}}(η̃u)) = 0 & on ℝ_{+}^{N+1} \\ \frac{∂}{∂x_{0}} (G_{ℝ_{+}^{N+1}}(η̃u)) = \frac{∂}{∂x_{0}} (η̃u) − △′_{ℝ^{N}} \frac{∂}{∂x_{0}} (η̃u) & on bℝ_{+}^{N+1} \end{cases} .

On the other hand, by (8.6), the function [ηG_{Ω}u] , solves the boundary value problem

\begin{cases}(− △ +I + ϵL_{2})([ηG_{Ω}u]˜ ) = −[2∇η · ∇G_{Ω}u + △η · G_{Ω}u] & ˜ on ℝ_{+}^{N+1} \\ \frac{∂}{∂x_{0}} ([ηG_{Ω}u] ) = T_{1} \frac{∂}{∂x_{0}} (η̃u) − (△′_{ℝ^{N+1}} − ϵL_{2})(\frac{∂}{∂x_{0}} (η̃u)) + T_{2}(η̃u) & on bℝ_{+}^{N+1} \end{cases} .

Let w ≡ G_{ℝ_{+}^{N+1}}(η̃u) − [ηG_{Ω}u] . Then w solves the system

\begin{cases}(△ − I)w = ϵL_{2}([ηG_{Ω}u] ) + [2∇η · ∇G_{Ω}u + (△η)G_{Ω}u] & on ℝ_{+}^{N+1} \\ \frac{∂w}{∂x_{0}} = η̃ϵT_{2} \frac{∂ũ}{∂x_{0}} + η̃T_{1} \frac{∂ũ}{∂x_{0}} + η̃T_{2}ũ & on bℝ_{+}^{N+1} \end{cases} (10.6)

The right hand side of the first equation in the system (10.6) can be rewritten as

ϵη′L_{2}G̃_{Ω}u + η′′L_{1}G̃_{Ω}u

where the order of L_{j} is j and supp η′, supp η′′ lie in supp η̃.

Now we apply Theorem 4.2.4 in [TRI]. We obtain that, for any s > 1/2,

∥w∥_{s}  ≤  C{ϵ∥L_{2}([ηG_{Ω}u] )∥_{F^{s−2}} + ∥η′L_{1}[G_{Ω}u] ∥_{F^{s−2}}

+ϵ∥ηT_{2} \frac{∂u}{∂n}∥_{W^{s−3/2}(bΩ)} + ∥ηT_{1} \frac{∂u}{∂n}∥_{W^{s−3/2}(bΩ)} + ∥ηT_{2}u∥_{W^{s−3/2}(bΩ)}\}

≤  C\{ϵ∥ηG_{Ω}u∥_{F^{s}} + ∥η′[G_{Ω}u]∥_{F^{s−1}} + ϵ ∥η \frac{∂u}{∂n}∥_{W^{s+1/2}(bΩ)}

+ ∥η \frac{∂u}{∂n}∥_{W^{s−1/2}(bΩ)} + ∥η_{1}u∥_{W^{s+1/2}(bΩ)}\}

≤^{G_{Ω}u=△u+f} C{ϵ∥η △ u∥_{s} + ϵ∥ηf∥_{s} + ∥η_{1} △ u∥_{F^{s−1}}

+∥η_{1}f∥_{s} + ϵ∥ηu∥_{s+2} + ∥η_{1}u∥_{s+1}}

≤  C{ϵ∥ηu∥_{s+2} + ∥η_{1}f∥_{s} + ∥η_{1}u∥_{s+1}}.  (10.7)

Now estimates (10.5) and (10.7) yield that

∥ηu∥_{s+2} ≤ C{∥η_{1}f∥_{s} + ϵ∥ηu∥_{s+2} + ∥η_{1}u∥_{s+1}},

that is, for s > 1/2,

∥ηu∥_{s+2} ≤ C{∥η_{1}f∥_{s} + ∥η_{1}u∥_{s+1}}.

This concludes the proof.

11. Estimates for q-Forms

The Regularity Theorem. In this section we prove the estimate for the solution of the boundary value problem in the case of q-forms. We follow the same outline as in the case of functions. For higher degree forms some extra technicalities are needed. The differential operators acting on forms that are involved are not (usually) diagonal, and we need some extra care for the off-diagonal terms.

THEOREM 11.1. Let s > 1/2. Let α ∈ W_{q}^{s}(Ω), and α orthogonal to the kernel of the boundary value problem. Let ψ ∈ ⋀^{q} be a solution of

(dd^{∗} + d^{∗}d)ψ = α on Ω
ψ ∈ dom d^{∗}
dψ ∈ dom d^{∗}

There exists c_{s} > 0, independent of ψ, such that

∥ψ∥_{s+2} ≤ c_{s}(∥α∥_{s} + ∥ψ∥_{s+1}).

Thus we wish to estimate ∥ψ∥_{s+2} in terms of ∥α∥_{s}, for s > 1/2. It suffices to estimates ηψ for a cut-off function η with support contained in a small open set. If η ∈ C_{0}^{∞}(Ω) then the argument used in the case of functions (see the proof of 10.1) applies with no substantial change.

Thus we assume that η is a cut-off function whose support is contained in an open set on which there exists a Fermi coordinate chart. As before, we also suppose that (∂/∂n)η = 0 in a tubular neighborhood of the boundary.

As in the case of q-forms in the half space we write

(dd^{∗} + d^{∗}d) = (dd′ + d′d) + (dK_{Ω} + K_{Ω}d)
≡ − △ +G_{Ω}.

We now proceed as in the function case. We need to estimate ∥ηψ∥_{s+2}. We introduce Fermi local coordinates, and set the stage as in Section 8. Recall that such coordinates have the property that the normal direction is orthogonal to the remaining directions at all points in the coordinate patch (that is, such coordinates “flatten” the boundary.). Given a form ϕ on Ω (i.e. written in global coordinates), we write ϕ̃ to indicate the form written in the Fermi coordinates. We adopt the same notation for the operators.

We wish to apply the estimates proved in the half space for the q-forms. In order to do this we notice that ˜(ηψ) is a form defined on the half space, and we wish to check if it satisfies a boundary value problem for which we have favorable estimates. Since ψ and dψ belong to dom d^{∗}, they satisfy equations (9.1) and (9.2). We need to examine those equations more closely. In order to do this we rewrite equations (9.1) and (9.2) in our local coordinates.

Proposition 11.2. Let ψ ∈ ⋀^{q}(\overline{Ω}) be such that both ψ, dψ ∈ dom d^{∗}. Let η be a cut-off function as before. Then the form η̃ψ satisfies the boundary equations

\frac{∂}{∂x_{0}} (η̃ψ)_{I} + ∑_{{|J|=q \atop J∋0}} γ_{IJ}(η̃ψ)_{J} = 0 if I ∋ 0,  (11.1)

and,

\frac{∂^{2}}{∂x_{0}^{2}} (η̃ψ)_{I} + ∑_{{|K|=q \atop K∋0}} T_{1}((η̃ψ)_{K}) + ∑_{{|L′|=q \atop L′∌0}} a_{K′L′} \frac{∂}{∂x_{0}} (η̃ψ)_{L′} + E = 0 if I ∌ 0, (11.2)

where E are {0 order terms} in the components of ψ.

We stress that we have improved equations (9.1) and (9.2) by separating the “normal” and “tangential” components of the lower order terms in the boundary equations.

Proof. By Lemma 8.2, the condition ψ ∈ dom d^{∗}, i.e. ∇n⃗ψ⌊n⃗ = 0 on bΩ, becomes

\frac{∂}{∂x_{0}} (η̃ψ)_{I} + ∑_{{|J|=q \atop J∋0}} γ_{IJ}(η̃ψ)_{J} = 0 on ℝ^{N} if I ∋ 0.

On the other hand, the second boundary condition gives that

\frac{∂}{∂x_{0}} (d̃ψ)_{K} + ∑_{{|L|=q+1 \atop L∋0}} γ_{KL}(d̃ψ)_{L} = 0 on ℝ^{N} if K ∋ 0,

i.e., on ℝ^{N} for K ∋ 0,

\frac{∂}{∂x_{0}} (\frac{∂ψ̃_{K′}}{∂x_{0}} ε_{0K′}^{K} + ∑_{{K′′ \atop j=1,...,N}} \frac{∂ψ̃_{K′′}}{∂x_{j}} ε_{jK′′}^{K}) + \frac{∂}{∂x_{0}} (∑_{J} ψ_{J}dω^{J})_{J}

+ ∑_{{|L|=q+1 \atop L∋0}} γ_{KL}(\frac{∂ψ̃_{L′}}{∂x_{0}} ε_{0L′}^{L} + ∑_{{L′′ \atop j=1,...,N}} X_{j}ψ̃_{L′′}ε_{jL′′}^{L}) + \{0 order terms\} = 0.

Now we use the first boundary condition to replace the terms of the form (∂/∂x_{0})ψ̃_{K′′}, with K′′ ∋ 0, with 0 order terms in ψ. Hence, the term (∂/x_{0})(∑_{J} ψ_{J}dω^{J})_{J} only contributes normal derivatives of tangential components and 0 order terms.

Therefore, for K′ ∌ 0, we have

\frac{∂^{2}}{∂x_{0}^{2}} (η̃ψ)_{K′} + ∑_{K′′} T_{1}(η̃ψ)_{K′′}ε_{jK′′}^{0K′} + (0 order terms) + ∑_{{|L′|=q \atop L′∌0}} a_{K′L′} \frac{∂}{∂x_{0}} (η̃ψ)_{L′} = 0.

From this identity, equation (11.2) follows.

At this point we adapt the estimates in the half space for the boundary value problems

\begin{cases}(− △ +G_{ℝ_{+}^{N+1}})ϕ = α & on ℝ_{+}^{N+1} \\ \frac{∂^{2}ϕ_{I}}{∂x_{0}} = 0 & on ℝ^{N} if I ∌ 0 \end{cases} ,  (11.3)

and

\begin{cases}(− △ +G_{ℝ_{+}^{N+1}})ϕ = α & on ℝ_{+}^{N+1} \\ \frac{∂ϕ_{I}}{∂x_{0}} = 0 & on ℝ^{N} if I ∋ 0 \end{cases} .  (11.4)

to the form η̃ψ.

We wish to apply Theorems 5.4 and 6.10 to the problems (11.3) and (11.4) respectively. In both cases the boundary data is estimated by ∥η̃ψ∥_{s+1}. Let η_{1} ∈ C_{0}^{∞}(\overline{Ω}), η_{1} ≡ 1on supp η. We obtain that

∥η̃ψ∥_{s+2} ≤ c (∥η_{1}(− △ +G_{ℝ_{+}^{N+1}})(η̃ψ)∥_{s} + ∥η̃ψ∥_{s+1})
≤ c (∥[η(− △ +G_{Ω})ψ] ∥_{s} + ∥[−η △ ψ] + △(η̃ψ)∥_{s}
+∥η_{1}G_{ℝ_{+}^{N+1}}(η̃ψ) − η_{1}[ηG_{Ω}ψ] ∥_{s} + ∥η̃ψ∥_{s+1})
≤ c (∥η̃α∥_{s} + ∥G_{ℝ_{+}^{N+1}}(η̃ψ) − [ηG_{Ω}ψ] ∥_{s} + ∥η̃ψ∥_{s+1}) .  (11.5)

Now we need to estimate ∥η_{1}G_{ℝ_{+}^{N+1}}(η̃ψ) − η_{1}[ηG_{Ω}ψ] ∥_{s} in the equation (11.5) above. We have the following theorem.

THEOREM 11.3. Let s > 1/2. Then, for any ϵ > 0, there exists a C_{ϵ} > 0 such that

∥η_{1}G_{ℝ_{+}^{N+1}}(η̃ψ) − η_{1}[ηG_{Ω}ψ] ∥_{s} ≤ ϵ∥η̃ψ∥_{s+2} + C_{ϵ}∥η_{1}ψ∥_{s},

where η_{1} ∈ C_{0}^{∞} and η_{1} ≡ 1 on supp η.

Assuming the Theorem for now, we finish the proof of the estimate for ∥η̃ψ∥_{s+2}, and therefore the proof of Theorem 11.1. Using Theorem 11.3 and (11.5) above we see that

∥η̃ψ∥_{s+2} ≤ c(∥ηα∥_{s} + ϵ∥η_{1}ψ∥_{s+2}∥η_{1}ψ∥_{s+1}),

from which we obtain that

∥η̃ψ∥_{s+2} ≤ c(∥ηα∥_{s} + ∥η_{1}ψ∥_{s+1}),

which is what we wished to prove.

Proof of Theorem 11.3. Now we turn to the proof of Theorem 11.3. The proof will be broken up into several lemmas.

The form G_{ℝ_{+}^{N+1}}(η̃ψ) ≡ θ = ∑_{I} θ_{I}ω^{I} is such that its components θ_{I} satisfy the following boundary value problems

\begin{cases}(− △ +I)θ_{I} = 0 & on ℝ_{+}^{N+1} \\ \frac{∂θ_{I}}{∂x_{0}} = (− △′ +I)(d(η̃ψ)_{0I}) & on ℝ^{N} if I ∌ 0 \end{cases}  (11.6)

and

\begin{cases}(− △ +I)θ_{I} = 0 & on ℝ_{+}^{N+1} \\ θ_{I} = (− △′ +I)(η̃ψ)_{I} & on ℝ^{N} if I ∋ 0 \end{cases}  (11.7)

On the other hand we need to investigate what kind of boundary value problem the form [ηG_{Ω}ψ] satisfies. Recall that G_{Ω} ≡ (K_{Ω}d + dK_{Ω}), and that the operator K_{Ω} is the operator on forms that gives the solution of the following boundary value problem for θ ∈ ⋀^{q}. Namely we have

\begin{cases}(− △ +I)θ = 0 & on Ω \\ ∇n⃗θ = ∑_{i=0}^{N} ∇_{Y_{i}^{∗}}(∇_{Y_{i}}ϕ⌊n⃗) + ϕ⌊n⃗ & on bΩ \end{cases}

for ϕ ∈ ⋀^{q+1}(\overline{Ω}).

Lemma 11.4. Let ϕ ∈ ⋀^{q}(\overline{Ω}). The form [ηK_{Ω}ϕ] ≡ θ is such that its I-component satisfies the following boundary value problem on ℝ_{+}^{N+1}

\begin{cases}(− △ −ϵL_{2} + I)θ_{I} = (η_{1}L_{1}K̃_{Ω}ϕ)_{I} & on ℝ_{+}^{N+1} \\ \frac{∂θ_{I}}{∂x_{0}} = − ∑_{|K|=q−1} η̃γ_{IK}(K̃_{Ω}ϕ)_{K} + (E_{2}ϕ)_{I} & on ℝ^{N} \end{cases}

where

(E_{2}ϕ)_{I} = \begin{cases}− ˜△_{T} ϕ_{0I} + ϕ_{0I} + (T_{1}ϕ)_{I} & if I ∌ 0 \\ 0 & if I ∋ 0 \end{cases}

L_{2} is defined by equation (8.3). Moreover, L_{1} is a first order differential operator, T_{1} is a first order differential operator, both sending (q + 1)-forms into q-forms.

Proof. It suffices to use Lemma 8.2 and the computation leading to formula (8.6).

Lemma 11.5. Let ϕ ∈ ⋀^{q}(\overline{Ω}), and let I ∋ 0. Then for any s > 1/2,

∥[ηK_{Ω}ϕ]_{I} ∥_{s} ≤ c∥ϕ∥_{s}.

Proof. Recall (see Corollary 7.3) that K_{Ω} is an operator of order 1. Therefore we direct the reader’s attention to the fact that the content of this lemma is that, on the normal components of the form, the operator K_{Ω} is of order 0. (Recall that, in the case of the half space, the operator K_{Ω} was identically zero on normal components.)

It suffices to recall the estimates in the negative norms for elliptic boundary value problems, as in [TRI], Theorem 4.2.4. Using the spaces F_{2,2}^{s}(Ω), which we will simply denote by F^{s}, we have

∥[ηK_{Ω}ϕ]_{I} ∥_{s} ≤ c(∥(η_{1}L_{1}K̃_{Ω}ϕ)_{I}∥_{F^{s−2}} + ∑_{K∋0} ∥[ηK_{Ω}ϕ]_{K} ∥_{W^{s−3/2}(bΩ)}).

Let ϵ > 0. By selecting η with support suitably small, we can achieve

∥[ηK_{Ω}ϕ]_{K} ∥_{W^{s−3/2}(bΩ)} ≤ ϵ∥[ηK_{Ω}ϕ]_{K} ∥_{W^{s−1/2}(bΩ)} .

Therefore

∑_{I∋0} ∥[ηK_{Ω}ϕ]_{I} ∥_{s} ≤ c∥(η_{1}L_{1}K̃_{Ω}ϕ)_{I}∥_{F^{s−2}} + ϵ ∑_{K} ∥[ηK_{Ω}ϕ]_{K} ∥_{W^{s−1/2}(bΩ)}

≤ c∥(η_{1}L_{1}K̃_{Ω}ϕ)_{I}∥_{F^{s−2}} + ϵ ∑_{K} ∥[ηK_{Ω}ϕ]_{K} ∥_{F^{s}}.

Then, by absorbing the last term on the right hand side over on the left, using the continuity of differential operators on the spaces F^{s}, and invoking Corollary 7.3, we obtain that

∑_{I∋0} ∥[ηK_{Ω}ϕ]_{I} ∥_{s} ≤ c∥[ηK_{Ω}ϕ] ∥_{F^{s−1}(Ω)}
≤ c∥η_{1}ϕ∥_{s}.

This concludes the proof of the lemma.

Lemma 11.6. Let ψ ∈ ⋀^{q}(\overline{Ω}). Then, for I ∋ 0, [ηG_{Ω}ψ]_{I} is a solution of the following boundary value problem

\begin{cases}(− △ −ϵL_{2} + I)w = [η_{1}L_{1}(G_{Ω}ψ)]_{I} & on ℝ_{+}^{N+1} \\ w = η̃(− △_{T} +I)ψ̃_{I} + ζ & on ℝ^{N} \end{cases}

where L_{2} is defined by equation (8.3), ζ is a function satisfying the estimate

∥ζ∥_{W^{s−1/2}(bΩ)} ≤ c∥η_{1}ψ∥_{s+1},

L_{1} is a differential operator of order 1, and η_{1} ∈ C_{0}^{∞} with η_{1} ≡ 1 on supp η.

Proof. The equation on ℝ_{+}^{N+1} is easily seen to be satisfied. Concerning the boundary equation, recall that G_{Ω} = K_{Ω}d + dK_{Ω}. Then, for I ∋ 0, I = 0I′, the function [ηK_{Ω}dψ]_{I} |_{ℝ^{N}} is the restriction to bℝ_{+}^{N+1} of the solution of the boundary value problem in Lemma 11.4 with ϕ = dψ. Moreover, for I = 0I′, on ℝ^{N} we have

[ηdK_{Ω}ψ]_{I} = \frac{∂}{∂x_{0}} (ηK_{Ω}ψ)_{I′} + ∑_{{i=1,...,N \atop |J|=q−1}} X_{i}[ηK_{Ω}ψ]_{J} ε_{jJ}^{0I′} + (0 order terms)

= − ∑_{K∌0} η̃γ_{I′K}({ \atop K_{Ω}ψ})_{K} + (− △_{T} +I)ψ_{0I′} + ∑_{J∋0} T_{1}[ηK_{Ω}ψ]_{J} + (0 o. t.’s) ,

where T_{1} is a tangential differential operator of order 1. Therefore w satisfies the boundary equation

w = η̃(− △_{T} +I)ψ_{I} − ∑_{K∌0} η̃γ_{I′K}({ \atop K_{Ω}ψ})_{K} + [ηK_{Ω}dψ]_{I} + ∑_{J∋0} T_{1}[ηK_{Ω}ψ]_{J}
≡ η̃(− ˜△′ + I)ψ_{I} + ζ.

Now the desired estimate follows from Lemma 11.5. Indeed, recalling that I ∋ 0, we see that

∥ζ∥_{W^{s−1/2}(bΩ)} ≤ c (∥{ \atop K_{Ω}ψ}∥_{W^{s−1/2}(bΩ)} + ∥[ηK_{Ω}dψ]_{I} ∥_{W^{s−1/2}(bΩ)}

+ ∑_{J∋0} ∥T_{1}[ηK_{Ω}ψ]_{J} ∥_{W^{s−1/2}(bΩ)})

≤ c (∥{ \atop K_{Ω}ψ}∥_{s} + ∥[ηK_{Ω}dψ]_{I}∥_{s} + ∑_{J∋0} ∥[ηK_{Ω}ψ]_{J}∥_{s+1})

≤ c∥η_{1}ψ∥_{s+1}.

Before estimating the term [ηG_{Ω}ψ]_{I} for I ∌ 0 we need an extra lemma.

Lemma 11.7. Let w_{1} = η(K_{Ω}dψ)_{K} − η(dK_{Ω}ψ)_{K} , with K ∌ 0. Then w_{1} solves

\begin{cases}(− △ +I)w_{1} = L_{1}(K_{Ω}dψ) + L_{2}(K_{Ω}ψ) & on Ω \\ \frac{∂w_{1}}{∂n} = T_{2}ψ + L_{0}(K_{Ω}ψ) & on ℝ^{N} \end{cases} ,

where the L_{j}’s are operators of order j, and T_{2} is a tangential differential operator of order 2.

Moreover, there exists c > 0 independent of ψ such that

∥w_{1}∥_{s} ≤ c∥ψ∥_{s+1}.

Proof. Recall that (K_{Ω}ϕ)_{K} satisfies the boundary value problem

\begin{cases}(− △ +I)θ = 0 \\ \frac{∂θ}{∂n} = Y_{k}^{∗}((∇_{Y_{k}}ϕ⌊)n⃗)_{K} + (ϕ⌊n⃗)_{K} \end{cases}

Therefore,

(− △ +I)(η[(K_{Ω}dψ)_{K} − (dK_{Ω}ψ)_{K}])
= L_{1}(K_{Ω}dψ)_{K} + η[△(dK_{Ω}ψ)_{K} − (dK_{Ω}ψ)_{K}] + L′_{1}(dK_{Ω}ψ)
= L_{1}(K_{Ω}dψ)_{K} + L′_{1}(dK_{Ω}ψ) + L_{2}(K_{Ω}ψ) + ψd[(△K_{Ω}ψ)_{K} − (K_{Ω}ψ)_{K}]
= L_{1}(K_{Ω}dψ)_{K} + L′_{1}(dK_{Ω}ψ)_{K} + L′_{2}(K_{Ω}ψ)
= L_{1}(K_{Ω}dψ)_{K} + L_{2}(K_{Ω}ψ).

Next we analyze the boundary equation. We have

\frac{∂}{∂n}[η(K_{Ω}dψ)_{K} − η(dK_{Ω}ψ)_{K}] = η ∑_{k=0}^{N} Y_{k}^{∗}[(∇_{Y_{k}}dψ)⌊n⃗]_{K} − η \frac{∂}{∂n}(dK_{Ω}ψ)_{K} .

Now we want to commute the normal derivative and the operator d in the far rightmost term in the equation above. We have

\frac{∂}{∂n}(dK_{Ω}ψ)_{K} = \frac{∂}{∂n}( ∑_{{i=1,...,N \atop I}} X_{i}(K_{Ω}ψ)_{I}ε_{iI}^{K}) + \frac{∂}{∂n}L_{0}(K_{Ω}ψ)

= ∑_{{i=1,...,N \atop I}} X_{i}(\frac{∂}{∂n}(K_{Ω}ψ)_{I})ε_{iI}^{K} + L′_{1}(K_{Ω}ψ) + \frac{∂}{∂n}L_{0}(K_{Ω}ψ)

= ∑_{{i=1,...,N \atop I}} X_{i}(∑_{k=0}^{N} Y_{k}^{∗}[(∇_{Y_{k}}ψ)⌊n⃗]_{I})ε_{iI}^{K} + L′_{1}(K_{Ω}ψ) + \frac{∂}{∂n}L_{0}(K_{Ω}ψ)

= ∑_{{i=1,...,N \atop I}} X_{i}(∑_{k=0}^{N} Y_{k}^{∗}Y_{k}ψ_{0I})ε_{iI}^{K} + T_{2}ψ + L′_{1}(K_{Ω}ψ) + \frac{∂}{∂n}L_{0}(K_{Ω}ψ)

= ∑_{k=0}^{N} Y_{k}^{∗}Y_{k}( ∑_{{i=1,...,N \atop I}} X_{i}ψ_{0I}ε_{iI}^{K}) + T′_{2}ψ + L′_{1}(K_{Ω}ψ)

+ L_{0} \frac{∂}{∂n}(K_{Ω}ψ) + L′_{0}(K_{Ω}ψ)

= ∑_{k=0}^{N} Y_{k}^{∗}[(∇_{Y_{k}}dψ)⌊n⃗]_{K} + T_{2}ψ + L_{0}(K_{Ω}ψ),

where we have used the fact that (∂/∂n)K_{Ω}ψ on bΩ equals a second order tangential operator in ψ. Hence,

\frac{∂}{∂n}[η(K_{Ω}dψ)_{K} − η(dK_{Ω}ψ)_{K}] = T_{2}ψ + L_{0}(K_{Ω}ψ).

Finally we prove the estimate. Using [TRI] Theorem 4.2.4 and Corollary 7.3, we have

∥w_{1}∥_{s} ≤ c(∥L_{1}K_{Ω}dψ∥_{F^{s−2}} + ∥L_{2}K_{Ω}ψ∥_{F^{s−2}} + ∥T_{2}ψ∥_{W^{s−3/2}(Ω)} + ∥T_{1}K_{Ω}ψ∥_{W^{s−3/2}(Ω)})
≤ c(∥K_{Ω}dψ∥_{F^{s−1}} + ∥K_{Ω}ψ∥_{s} + ∥ψ∥_{s+1} + ∥K_{Ω}ψ∥_{s})
≤ c∥ψ∥_{s+1}.

Lemma 11.8. Let ψ ∈ ⋀^{q}. Then for I ∌ 0, the function [ηG_{Ω}ψ]_{I} is a solution of the following boundary value problem:

\begin{cases}(− △ +ϵL_{2} + I)w = [η_{1}L_{1}(G_{Ω}ψ)]_{I} & on ℝ_{+}^{N+1} \\ \frac{∂w}{∂x_{0}} = (− △_{T} +I)\frac{∂ψ̃_{0I}}{∂x_{0}} + ζ & on ℝ^{N} \end{cases}

Here ζ is a function satisfying the estimate

∥ζ∥_{W^{s−3/2}(bΩ)} ≤ c∥η_{1}ψ∥_{s+1},

and L_{1} and η_{1} are as in the previous lemma.

Proof. As in the proof of Lemma 11.6 we need only check that the boundary equation is satisfied, and that the desired estimate holds.

Notice that, by Lemma 11.4, on the set {x_{0} = 0}, for I ∌ 0, we have

\frac{∂}{∂x_{0}} [ηK_{Ω}dψ]_{I} = − ∑_{K∌0} η̃γ̃_{IK}[K_{Ω}dψ]_{K} + (E_{2}d̃ψ)_{I} ,

where E_{2} is defined as in Lemma 11.4.

On the other hand, if I ∌ 0, on ℝ^{N} we have that

\frac{∂}{∂x_{0}} [ηdK_{Ω}ψ]_{I} = η̃ ∑_{{J∌0 \atop i=1,...,N}} \frac{∂}{∂x_{0}} X_{i}({ \atop K_{Ω}ψ})_{I} + \frac{∂}{∂x_{0}} [∑_{J}(ηK_{Ω}ψ)_{J} dω^{J}]

= − ∑_{{K∌0 \atop i=0,...,N}} η̃a_{IK}X_{i}({ \atop K_{Ω}ψ})_{K} + [L_{0}({ \atop K_{Ω}ψ})]_{I} + (T_{2}η̃ψ)_{I} ,

where T_{2} is a second order tangential differential operator. Here we have used the fact that (∂/∂x_{0})K_{Ω}ψ on the set {x_{0} = 0} equals a second order tangential operator on ψ. Therefore, for I ∌ 0,

\frac{∂w}{∂x_{0}} |_{ℝ^{N}} = (− △_{T} +I)\frac{∂ψ̃_{0I}}{∂x_{0}}

− ∑_{K∌0} η̃γ_{IK}[K_{Ω}dψ]_{K} − ∑_{{i=1,...,N \atop K∌0}} ηa_{IK}X_{i}({ \atop K_{Ω}ψ})_{K} + [L_{0}({ \atop K_{Ω}ψ})]_{I} + (T_{2}η̃ψ)_{I}

≡ (− △_{T} +I)\frac{∂ψ̃_{0I}}{∂x_{0}} + ζ.

Hence we need only check the estimate.

Let w_{1} be given by Lemma 11.7. Then

ζ = − ∑_{K∌0} γ_{IK} η̃[dK_{Ω}ψ]_{K} + w_{1} − ∑_{{i=1,...,N \atop K∌0}} ηa_{IK}X_{i}({ \atop K_{Ω}ψ})_{K} + [L_{0}({ \atop K_{Ω}ψ})]_{I} + (T_{2}η̃ψ)_{I}.

Using the above equality, Lemma 11.5, Lemma 11.7, and Corollary 7.3 we see that

∥ζ∥_{W^{s−3/2}(bΩ)} ≤ c( ∑_{K∌0} ∥η̃_{1}(dK_{Ω}ψ)_{K} ∥_{W^{s−3/2}(bΩ)} + ∥w_{1}∥_{W^{s−3/2}(bΩ)}

+ ∥η_{1}K_{Ω}ψ∥_{W^{s−1/2}(bΩ)} + ∥η̃ψ∥_{W^{s+1/2}(bΩ)})

≤ c(∥η̃_{1}{ \atop K_{Ω}ψ}∥_{W^{s−1/2}(bΩ)} + ∥w_{1}∥_{W^{s−1/2}(bΩ)} + ∥η_{1}ψ∥_{s+1})

≤ c∥η_{1}ψ∥_{s+1}.

This proves the estimates and concludes the proof.

End of the Proof of 11.3. We are finally able to compare [ηG_{Ω}ψ]_{I} and [G_{ℝ_{+}^{N+1}}(η̃ψ)]_{I}. We begin with the case I ∋ 0. By (11.7) and Lemma 11.8 it is easy to see that the function w solves the boundary value problem

\begin{cases}(− △ +I)w = (−ϵL_{2})[ηG_{Ω}ψ]_{I} + η_{1}L_{1}[G_{Ω}ψ] & on ℝ_{+}^{N+1} \\ w = (−ϵT_{2})(η̃ψ)_{I} + η_{1}(T_{1}ψ)_{I} + ζ & on ℝ^{N} \end{cases}

where ζ is as in Lemma 11.8. Therefore, using Theorem 4.2.4 in [TRI], Lemma 11.8, and Corollary 7.3 we see that

∥w∥_{s} ≤ c(ϵ∥L_{2}(ηG_{Ω}ψ)_{I}∥_{F^{s−2}} + ∥η_{1}L_{1}(G_{Ω}ψ)∥_{F^{s−2}} + ∥ϵT_{2}(η̃ψ) + ηT_{1}ψ∥_{W^{s−1/2}(bΩ)}
+ ∥ζ∥_{W^{s−1/2}(bΩ)})

≤ c(ϵ∥ηG_{Ω}ψ∥_{s} + ∥ηG_{Ω}ψ∥_{F^{s−1}} + ∥G_{ℝ_{+}^{N+1}}(ηψ)∥_{F^{s−1}} + ϵ∥ηψ∥_{s+2} + ∥η_{1}ψ∥_{s+1})

≤ c(ϵ∥ηψ∥_{s+2} + ∥ηψ∥_{s+1}) ,

where we also use the estimate

∥ηG_{Ω}ψ − G_{Ω}(ηψ)∥_{s} ≤ c∥η_{1}ψ∥_{s+1}.

This follows by writing G_{Ω} = K_{Ω}d+K_{Ω}d, and noticing that the commutator between K_{Ω} and the multiplication by η is a 0 order operator, since ηK_{Ω}θ − K_{Ω}(ηθ) solves the boundary value problem

\begin{cases}(△ − I)w = L_{1}(K_{Ω}θ) & on Ω \\ \frac{∂w}{∂n} = T_{1}θ & on bΩ \end{cases}

This is a problem to which we apply Theorem 4.2.4 in [TRI] again.

Now let I ∌ 0. Then the function w solves the boundary value problem

\begin{cases}(− △ +I)w = (−ϵL_{2})[ηG_{Ω}ψ]_{I} + η_{1}L_{1}[G_{Ω}ψ] & on ℝ_{+}^{N+1} \\ \frac{∂w}{∂x_{0}} = (ϵT_{2})\frac{∂(η̃ψ)_{0I}}{∂x_{0}} + ζ & on ℝ^{N} \end{cases}

where ζ is given by Lemma 11.8. Therefore, by Corollary 7.3,

∥w∥_{s} ≤ ϵ∥ηψ∥_{s+2} + c (∥η_{1}G_{Ω}ψ∥_{F^{s−1}(Ω)} + ∥ζ∥_{W^{s−3/2}(bΩ)})

≤ ϵ∥ηψ∥_{s+2} + c∥η_{1}ψ∥_{s+1}.

12. The Decomposition Theorem and Conclusions

Proof of the Main Result. We are now in a position to finish the proof of Theorem 2.6.

Let X be the subspace of W_{q}^{s+2}(Ω), (s > 1/2) consisting of the forms ϕ satisfying the boundary conditions ϕ, dϕ ∈ dom d^{∗}. Observe that X is a closed subspace of W_{q}^{s+2}(Ω), (this is the reason why s > 1/2). By the Regularity Theorem 11.1 we know that, on X, the norm

∥ψ∥_{s+1} + ∥(− △ +G_{Ω})ψ∥_{s}

is equivalent to the norm ∥ψ∥_{s+2}. Let P be the orthogonal projection of X onto the kernel of − △ +G_{Ω}. Then, by standard functional analysis arguments, (see [ZIE] p.178),

∥ψ − Pψ∥_{s+1} ≤ C∥ − △ψ + G_{Ω}ψ∥_{s}.

Therefore, if ψ is orthogonal to the kernel of − △ +G_{Ω}, the regularity theorem Theorem 11.1 says that

∥ψ∥_{s+2} ≤ C_{s}∥ − △ψ + G_{Ω}ψ∥_{s}.

We now complete the proof of the existence theorem—Theorem 9.2. For α ∈ W_{q}^{1}∩H_{q}^{⊥} fixed, we approximate α by α_{m} ∈ Z ≡ (F −I)(D). For such α_{m}’s we can find ϕ_{m} ∈ D orthogonal to H_{q}, solutions of the boundary value problem, and such that, for α ≥ 1,

∥ϕ∥_{s+2} ≤ c_{s}∥α∥_{s} .

Thus the ϕ_{m}’s converge in W_{q}^{s+2} to a certain ϕ. Clearly ϕ solves the boundary value problem with data α, since the boundary conditions are preserved in the W^{s+2} topology for s > 1/2.

Hence, if α ∈ W^{s}(Ω), (s ≥ 1) and α is orthogonal to ker (− △ +G_{Ω}) in the W^{1} inner product, then there exists a unique solution to the problem (2.2)—ϕ ∈ W^{s+2}(Ω) orthogonal to ker (− △ +G_{Ω}) and such that

∥ϕ∥_{s+2} ≤ C_{s}∥α∥_{s}.

If α ∈ W^{s}(Ω) with 1/2 < s < 1, then we still have existence and regularity by using a density argument, since α ∈ W_{q}^{s}(Ω) can be approximated by α_{m} ∈ W_{q}^{1}(Ω) in the W^{s}-topology. This concludes the proof of Theorem 2.6.

Decomposition of W_{q}^{1}. The aim of this part is to prove the following result:

THEOREM 12.1. Let Ω be a smoothly bounded domain in ℝ^{N+1}. Let W_{q}^{1}(Ω) denote the 1-Sobolev space of q-forms. Then we have the strong orthogonal decomposition

W_{q}^{1} = dd^{∗}(W_{q}^{1}) ⨁ d^{∗}d(W_{q}^{1}) ⨁ H_{q} ,

where H_{q} is a finite dimensional subspace, and d^{∗} denotes the W_{q}^{1}-Hilbert space adjoint of d.

Proof. All of this is standard. If α is orthogonal to H_{q} then α belongs to (dd^{∗} + d^{∗}d)W_{q}^{1}. The fact that d(W_{q}^{1}) and d^{∗}(W_{q}^{1}) are orthogonal subspaces is also clear since

⟨dϕ, d^{∗}ψ⟩_{1} = ⟨d^{2}ϕ, ψ⟩_{1} = 0,

for all ϕ, ψ ∈ C^{∞}(\overline{Ω}).

In the case q = 0 we are able to determine the harmonic space H_{0}. The rest of the section is devoted to this end.

THEOREM 12.2. We have that H_{0} = {constants}. More precisely, the boundary value problem

\begin{cases}d^{∗}du = f & on Ω \\ du ∈ dom d^{∗} \end{cases}  (12.1)

has a unique solution u orthogonal (in the W^{1}(Ω) inner product) to the constant functions, for each f ∈ W^{1}(Ω) which is also orthogonal to the constants, i.e.

∫_{Ω} f = 0.

This theorem is a consequence of the next theorem. By equation (7.2) we can rewrite the boundary value problem (12.1) as

\begin{cases}− △ u + G_{Ω}u = f & on Ω \\ ∑_{j=0}^{N} n_{j} \frac{∂}{∂n} (\frac{∂u}{∂x_{j}} ) = 0 & on bΩ \end{cases} ,  (12.2)

where n⃗ = (n_{0}, …, n_{N}) is the normal vector field. By our choice of n⃗, and by setting G_{Ω}u = v we obtain the new system

− △ u + v = f  on Ω
\frac{∂^{2}u}{∂n^{2}} = 0  on bΩ
△v − v = 0  on Ω
\frac{∂v}{∂n} = \frac{∂u}{∂n} + ∑_{k=0}^{N} Y_{k}^{∗}[(∇_{Y_{k}}du)⌊n⃗] on bΩ

Now assume f to be suffiently regular. By solving the first equation for v and substituting we find that (12.2) finally becomes

\begin{cases}△^{2}u − △u = f − △f & on Ω \\ \frac{∂^{2}u}{∂n^{2}} = 0 & on bΩ \\ \frac{∂}{∂n} △ u − ∑_{k=0}^{N} Y_{k}^{∗}[(∇_{Y_{k}}du)⌊n⃗] − \frac{∂u}{∂n} = −\frac{∂f}{∂n} & on bΩ \end{cases}  (12.3)

This last is the problem that we are going to study.

THEOREM 12.3. The boundary value problem (12.3) is elliptic. It has index 0. The kernel of the system is given by {u = constant} and a (unique) solution exists for each piece of data f that satisfies

∫_{Ω} f = 0 , f ∈ W^{r}, r ≥ 2.

REMARK 12.4. The condition f ∈ W^{r}, r ≥ 2 is necessary in order to guarantee that △f ∈ L^{2}(Ω). In this case the solution u satisfies the estimates

∥u∥_{W^{s+4}(Ω)} ≤ C(∥f − △f∥_{W^{s}(Ω)} + ∥\frac{∂f}{∂n}∥_{W^{1/2+s}(bΩ)} + ∥u∥_{s}),

that is, for s ≥ 0,

∥u∥_{W^{s+4}(Ω)} ≤ C(∥f∥_{s+2} + ∥u∥_{s}).

This result, when applied to problem (12.1), is not optimal. This is the reason why we had to go through the more complicated computations in the previous sections.

Proof. In order to prove that the boundary value problem is elliptic it suffices to verify the Lopatinski condition as in [HOR1]. This is standard, and since we do not use this fact in what follows, we leave the details to the reader. Thus the boundary value problem (12.3) admits solutions with appropriate estimates modulo a finite dimensional kernel and a finite dimensional cokernel. Our next job is to explicitly determine this kernel and cokernel.

Determination of the Kernel: Suppose that the function u, defined on Ω, satisfies

△(△u) − △u = 0  on Ω
\frac{∂^{2}u}{∂n^{2}} = 0  on bΩ
\frac{∂}{∂n} △ u − ∑_{k=0}^{N} Y_{k}^{∗}[(∇_{Y_{k}}du)⌊n⃗] − \frac{∂u}{∂n} = 0 on bΩ

We intend to show that u must therefore be constant. This will then imply that the kernel is the one dimensional space of constant functions. Without loss of generality we suppose that all functions are real valued.

Using Green’s theorem we see that

0 = ∫_{Ω} u △ [△u − u] dV

= ∫_{Ω} | △ u|^{2} + ∫_{bΩ} u\frac{∂}{∂n}(△u) − ∫_{bΩ} △u\frac{∂u}{∂n} + ∫_{Ω} |gradu|^{2} − ∫_{bΩ} u\frac{∂u}{∂n}.

Thus we find that

∫_{Ω} | △ u|^{2} + |gradu|^{2}

=  ∫_{bΩ} \frac{∂u}{∂n}(△u + u) − u\frac{∂}{∂n}(△u)

(2^{nd} bdry. cond.) = ∫_{bΩ} \frac{∂u}{∂n}(△u + u) − ∫_{bΩ} u [∑_{k=0}^{N} Y_{k}^{∗}[(∇_{Y_{k}}du)⌊n⃗] + \frac{∂u}{∂n}]

=  ∫_{bΩ} \frac{∂u}{∂n} △ u − ∫_{bΩ} u ∑_{k=0}^{N} Y_{k}^{∗}[(∇_{Y_{k}}du)⌊n⃗].  (12.4)

Now we consider the second integral on the right hand side of equation (12.4). We have

∫_{bΩ} ∑_{k=0}^{N} Y_{k}^{∗}[(∇_{Y_{k}}du)⌊n⃗]u  =  ∑_{k} ∫_{bΩ} Y_{k}u[(∇_{Y_{k}}du⌊)n⃗]

(1^{st} bdry. cond.) = ∑_{k} ∫_{bΩ} Y_{k}u[\frac{∂(Y_{k}u)}{∂n}]

=  −\frac{1}{2} ∫_{bΩ} \frac{∂}{∂n}|gradu|^{2},

where we use the first boundary condition, and the simple facts that (∇_{Y_{k}}du) ⌊n⃗ = (∂/∂n)Y_{k}u, and that |gradu|^{2} = ∑_{k} |Y_{k}u|^{2} + |(∂/∂n)u|^{2}.

Substituting this last equality into (12.4), and using Green’s theorem again, we find that

∫_{Ω} | △ u|^{2} + |gradu|^{2} = ∫_{bΩ} △u\frac{∂u}{∂n} − \frac{1}{2} \frac{∂}{∂n}|gradu|^{2}
= ∫_{Ω} | △ u|^{2} + gradu · grad(△u) − \frac{1}{2} △ (|gradu|^{2}).

Notice that

△|gradu|^{2} = 2 ∑_{j} (△\frac{∂u}{∂x_{j}} ) \frac{∂u}{∂x_{j}} + 2grad\frac{∂u}{∂x_{j}} · grad\frac{∂u}{∂x_{j}}

= 2grad(△u) · gradu + 2 ∑_{j} grad\frac{∂u}{∂x_{j}} · grad\frac{∂u}{∂x_{j}} .

As a result,

∫_{Ω} | △ u|^{2} + |gradu|^{2} = ∫_{Ω} | △ u|^{2} − ∑_{j} grad\frac{∂u}{∂x_{j}} · grad\frac{∂u}{∂x_{j}} .

HODGE THEORY IN THE SOBOLEV TOPOLOGY 89

That is,

∫_{Ω} |gradu|^{2} + ∑_{j} |grad\frac{∂u}{∂x_{j}} |^{2} = 0.

This equality implies that u is a constant. Determination of the Cokernel: We want to show that if f is such that the system (12.3) admits a solution then

∫_{Ω} f = 0.

Notice that this would imply that Cokernel ⊇ (ker(f ↦ ∫_{Ω} f))^{⊥}.

We have

∫_{Ω} △(△u) − △u  =  ∫_{bΩ} \frac{∂}{∂n}(△u − u)

(2^{nd} bdry. cond.) = ∫_{bΩ} ∑_{k} Y_{k}^{∗}[(∇_{Y_{k}}du⌊)n⃗] − \frac{∂f}{∂n}

=^{(Green on bΩ)} − ∫_{bΩ} \frac{∂f}{∂n}
=  − ∫_{Ω} △f.

Now the left hand side in these last equations equals ∫_{Ω}(f − △f). Therefore

∫_{Ω} f = 0.

Now we examine the dimension of the cokernel. Suppose that

(g; v_{1}, v_{2}) ∈ C^{∞}(\overline{Ω}) × C^{∞}(bΩ) × C^{∞}(bΩ)

and

(g; v_{1}, v_{2}) ⊥ Range of (12.3),

i.e.

∫_{Ω}(△^{2}u − △u)g + ∫_{bΩ} [∑_{j=0}^{N} n_{j} \frac{∂}{∂n} (\frac{∂u}{∂x_{j}} )] v_{1}

+ ∫_{bΩ} [\frac{∂}{∂n} △ u + ∑_{j=1}^{N} div^{T} (n_{j}∇^{T} \frac{∂u}{∂n}) − \frac{∂u}{∂n}] v_{2} = 0 (12.5)

for all u ∈ C^{∞}. Then we want to show that

(g; v_{1}, v_{2}) lies in some one dimensional subspace.

This will finish the proof.

Thus suppose that (12.5) holds for all u ∈ C^{∞}(\overline{Ω}). By taking u ∈ C_{0}^{∞}(Ω) and integrating by parts we obtain

0 = ∫_{Ω} g(△^{2}u − △u) = ∫_{Ω} u(△^{2}g − △g).

This equality implies that △^{2}g − △g = 0 on Ω.

Next, assume (12.5) and apply Green’s theorem to the integral on Ω. It equals

∫_{Ω} △g(△ − I)u + ∫_{bΩ} g \frac{∂}{∂n}(△ − I)u − ∫_{bΩ} \frac{∂g}{∂n}(△ − I)u

= ∫_{Ω} u(△^{2}g − △g) + ∫_{bΩ} \frac{∂u}{∂n} △ g − ∫_{bΩ} u\frac{∂}{∂n} △ g

+ ∫_{bΩ} g \frac{∂}{∂n} △ u − ∫_{bΩ} g \frac{∂u}{∂n} − ∫_{bΩ} △u\frac{∂g}{∂n} + ∫_{bΩ} u\frac{∂g}{∂n} .

Substituting this last identity into (12.5), and recalling that △^{2}g − △g = 0 on Ω we find that

0 = ∫_{bΩ} \frac{∂u}{∂n} △ g − ∫_{bΩ} u\frac{∂}{∂n} △ g + ∫_{bΩ} g \frac{∂}{∂n} △ u − ∫_{bΩ} g \frac{∂u}{∂n}

− ∫_{bΩ} △u\frac{∂g}{∂n} + ∫_{bΩ} u\frac{∂g}{∂n} + ∫_{bΩ} \frac{∂^{2}u}{∂n^{2}}v_{1}

+ ∫_{bΩ} v_{2} \frac{∂}{∂n} △ u − ∑_{k} ∫_{bΩ} Y_{k}^{∗}[(∇_{Y_{k}}du)⌊n⃗]v_{2} − ∫_{bΩ} v_{2} \frac{∂u}{∂n}

= ∫_{bΩ} u [\frac{∂g}{∂n} − \frac{∂}{∂n} △ g] + ∫_{bΩ} \frac{∂u}{∂n}[△g − g − v_{2}]

+ ∫_{bΩ} \frac{∂^{2}u}{∂n^{2}}v_{1} − ∑_{k} ∫_{bΩ} Y_{k}^{∗}[(∇_{Y_{k}}du)⌊n⃗]v_{2} − ∫_{bΩ} △u\frac{∂g}{∂n}

+ ∫_{bΩ}(g + v_{2})\frac{∂}{∂n} △ u.  (12.6)

Therefore we have

0 = ∫_{bΩ} u [\frac{∂g}{∂n} − \frac{∂}{∂n} △ g] + ∫_{bΩ} \frac{∂u}{∂n}[△g − g − v_{2}] + ∫_{bΩ} \frac{∂^{2}u}{∂n^{2}}v_{1}

− ∑_{k} ∫_{bΩ}[\frac{∂}{∂n}Y_{k}u]Y_{k}v_{2} − ∫_{bΩ} △u\frac{∂g}{∂n} + ∫_{bΩ} \frac{∂}{∂n} △ u(g + v_{2}). (12.7)

As in the half space case, we write x′ to indicate the variable in bΩ. Given any function v on bΩ, we let

ṽ(x) = v(π(x)) ≡ v(x′),

where π is the normal projection defined in a neighborhood of bΩ. The functions ṽ are defined in a (fixed) neighborhood of bΩ. Then we can extend them to all of ℝ_{+}^{N+1} by multiplying by a (fixed) cut-off function.

HODGE THEORY IN THE SOBOLEV TOPOLOGY 91

Now we proceed in several steps to calculate the cokernel: Step 1: Take u(x) to be of the form

u(x) = ρ(x)^{3}ũ_{1}(x) ,

where ũ_{1} ∈ C^{∞}(bΩ) is generic. Such a choice yields that

g + v_{2} = 0 on bΩ.

Proof of Step 1: For such a choice of u, equation (12.7) becomes

0 = ∫_{bΩ}(g + v_{2})\frac{∂}{∂n} △ (ρ^{3}ũ_{1})

Now,

\frac{∂}{∂n} △ [ρ^{3}ũ_{1}]|_{bΩ} = \frac{∂}{∂n}[ũ_{1} △ ρ^{3} + 2(gradρ^{3} · gradũ_{1}) + ρ^{3} △ ũ_{1}]|_{bΩ}
= (\frac{∂}{∂n} △ ρ^{3}) ũ_{1}|_{bΩ}
≡ 6u_{1}.

[It is a standard fact that we may assume that (∂ρ/∂n) = |grad ρ| = 1 on bΩ.] This implies that g + v_{2} = 0 on bΩ.

Step 2: Take u = ρ^{2}(x)ũ_{1}(x) with u_{1} ∈ C^{∞}(bΩ). This implies that

v_{1} = \frac{∂g}{∂n} on bΩ.

Proof of Step 2: Using Step 1, we see that this particular choice of u gives

△u|_{bΩ} = △(ρ^{2}ũ_{1})|_{bΩ}
= (ũ_{1} △ ρ^{2} + 2gradρ^{2} · gradũ_{1} + ρ^{2} △ ũ_{1})|_{bΩ}
= 2|gradρ|^{2}ũ_{1}|_{bΩ}
= 2u_{1},

while (∂^{2}ρ^{2}/∂n) = 2 on bΩ. Therefore,

v_{1} = \frac{∂g}{∂n} on bΩ.

Hence, putting Steps 1 and 2 together, equation (12.7) becomes

0 = ∫_{bΩ} u [\frac{∂g}{∂n} − \frac{∂}{∂n} △ g] + ∫_{bΩ} \frac{∂u}{∂n} △ g + ∫_{bΩ} \frac{∂^{2}u}{∂n^{2}} \frac{∂g}{∂n}

+ ∑_{k} ∫_{bΩ}[\frac{∂}{∂n}Y_{k}u]Y_{k} \frac{∂g}{∂n} − ∫_{bΩ} △u\frac{∂g}{∂n} (12.8)

for all u ∈ C^{∞}(\overline{Ω}).

Step 3: We have that

∫_{bΩ} \frac{∂u}{∂n} △ g + ∫_{bΩ} \frac{∂^{2}u}{∂n^{2}} \frac{∂g}{∂n} ∑_{k} + ∫_{bΩ}[\frac{∂}{∂n}Y_{k}u]Y_{k} \frac{∂g}{∂n} − ∫_{bΩ} △u\frac{∂g}{∂n}

= ∫_{bΩ} u[Y_{k}^{∗} \frac{∂}{∂n}D_{k}g] + ∫_{bΩ} \frac{∂u}{∂n} \frac{∂^{2}g}{∂n} (12.9)

Proof of Step 3: Recall that for k = 0, …, N, Y_{k} ≡ D_{k} − n_{k}(∂/∂n), and that n⃗ = ∑_{j} n_{j}D_{j}. Therefore, it is easy to see that

\frac{∂^{2}u}{∂n^{2}} \frac{∂g}{∂n} + ∑_{k}(Y_{k}g)(\frac{∂}{∂n}Y_{k}u) = ∑_{k} D_{k}g \frac{∂}{∂n}(D_{k}u).

Thus a repeated application of Green’s theorem gives that

∫_{bΩ} \frac{∂u}{∂n} △ g + ∫_{bΩ} \frac{∂^{2}u}{∂n^{2}} \frac{∂g}{∂n} + ∑_{k} ∫_{bΩ}[\frac{∂}{∂n}Y_{k}u]Y_{k} \frac{∂g}{∂n} − ∫_{bΩ} △u\frac{∂g}{∂n}

= ∫_{bΩ} \frac{∂u}{∂n} △ g + ∑_{k} ∫_{bΩ}[\frac{∂}{∂n}Y_{k}u]Y_{k} \frac{∂g}{∂n} − ∫_{bΩ} △u\frac{∂g}{∂n}

= ∫_{bΩ} \frac{∂u}{∂n} △ g + ∑_{k}[∫_{Ω} gradD_{k}u · gradD_{k}u] − ∫_{Ω} △u △ g

= ∫_{bΩ} \frac{∂u}{∂n} △ g + ∑_{k} ∫_{bΩ}(\frac{∂u}{∂n} \frac{∂D_{k}g}{∂n} − ∫_{bΩ} \frac{∂u}{∂n} △ g

= ∑_{k} ∫_{bΩ} u(Y_{k}^{∗} \frac{∂}{∂n}D_{k}g) + ∑_{k} ∫_{bΩ} \frac{∂u}{∂n}(n_{k} \frac{∂}{∂n}D_{k}g)

= ∑_{k} ∫_{bΩ} u(Y_{k}^{∗} \frac{∂}{∂n}D_{k}g) + ∫_{bΩ} \frac{∂u}{∂n} \frac{∂^{2}g}{∂n^{2}} .

This proves Claim 3.

Now because of Step 3, equality (12.8) becomes

∫_{bΩ} u [\frac{∂g}{∂n} − \frac{∂}{∂n} △ g + ∑_{k}(Y_{k}^{∗} \frac{∂}{∂n}D_{k}g)] + ∫_{bΩ} \frac{∂u}{∂n} \frac{∂^{2}g}{∂n} .  (12.10)

Thus g must satisfy the system

△^{2}g − △g = 0  on Ω
∑_{j} \frac{∂^{2}g}{∂n^{2}} = 0  on bΩ
\frac{∂g}{∂n} − \frac{∂}{∂n} △ g + ∑_{k} Y_{k}^{∗} \frac{∂}{∂n}D_{k}g = 0  on bΩ

Using the first boundary equation we can rewrite the second one as in the homogeneous problem that we have studied earlier when we calculated the kernel. We conclude that g ≡ C. Therefore the cokernel of the problem is given by

(g; v_{1}, v_{2}) = (c; 0, −c),

which is obviously a one dimensional space.

All of these calculations can be applied to our original problem provided that f ∈ W^{2}(Ω). Using the regularity result we can extend it to all f ∈ W^{s}, s > 1/2 and conclude that has a unique solution u with ∫_{Ω} u = 0, for all f ∈ W^{s}(Ω) with ∫_{Ω} f = 0.

13. Final Remarks

In the present paper we have laid the foundations for Hodge theory of the standard exterior differential operator d in the Sobolev topology acting on q-forms on a domain in ℝ^{N}. The theory that have presented here is complete. However, there is much work that remains to be done. For maximum applicability, the Hodge theory on a (compact) manifold needs to be developed. Also the case of all s and all q should be treated in a unified manner.

The ultimate goal of this program is to develop the Hodge theory of the \overline{∂}-Neumann problem in the Sobolev topology on a strongly pseudoconvex domain. We will turn to this task in a subsequent sequence of papers.

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Dipartimento di Matematica, Via Saldini 50, Università di Milano, 20133 Milano (Italy)

  E-mail address: fontana@vmimat.mat.unimi.it

Department of Mathematics, Washington University, St. Louis, MO 63130 (U.S.A.)

  E-mail address: sk@artsci.wustl.edu

Dipartimento di Matematica, Politecnico di Torino, 10129 Torino (Italy)

  E-mail address: peloso@polito.it