R. Selvaggi and I. Sisto

C ¹ -Domains

R. Selvaggi and I. Sisto

Dedicated to the Memory of Grigorios TSAGAS (1935-2003), President of Balkan Society of Geometers (1997-2003)

Abstract

Existence and uniqueness value boundary problems for differential forms in C¹-domains are investigated.

Mathemaics Subject Classification: 58A10, 45E05

Key words: differential form, homology group, cohomology group, harmonic form

Introduction

In this paper we resolve boundary value problems for differential forms in a bounded C¹-domain Ω of Rⁿ, 3≤n.The obtained results are an extension of some theorems established by C. Miranda in [8] for differential forms inC^2,α-domains ofRⁿ.

The fundamental result is Theorem 2.1:

Let Fs−1 ∈ C_s−1¹ (Ω) be a closed form with interior nontangential trace F_s−1⁻ ∈ L^p_s−1(∂Ω).If Z

τ_s−1ⁱ

F_s−1⁻ = 0, i= 1, ..., R⁻_s−1, where ([τ_s−1ⁱ ])_1≤i≤R−

s−1is a base ofC¹-differentiable singular homology spaceHs−1(∂Ω) andR⁻_s−1is the (s−1)-th Betti number of Ω, then there exists a formUs−2∈C_s−2¹ (Ω), whose coefficients are inW^1,p(Ω),such that

(∗) dUs−2=Fs−1 in Ω.

This result is obtained using a formula of Bidal-de Rham (see (1.10)). The use of this formula changes the existence problem of a differential form that satisfies (∗) in a suitable Neumann problem for harmonic forms already studied in [15].

Using a regularity result for the solutions of the homogeneous Neumann problem for harmonic forms established in [15] (see Theorem 2.1) and the continuity hypothesis forFs−1in Ω, we obtain a uniqueness theorem (see Theorem 2.4). With this aim, first we prove an extension theorem (see Theorem 1.2) and then, as a consequence, we

Balkan Journal of Geometry and Its Applications, Vol.8, No.2, 2003, pp. 81-96.∗

c

°Balkan Society of Geometers, Geometry Balkan Press 2003.

deduce a Stokes’ Theorem (see Theorem 1.3) for differential s-forms of Ce_s¹(Ω) (see Preliminaries).

Throughout this work we use the definitions and the results of [13] concerning differential forms and singular homology and cohomology groups of theC¹-manifold

∂Ω.

1 Preliminaries

In this section we summarize basic concepts, notations, and results that will be used throughout the paper.

We assume that Ω is a bounded and connected C¹-domain ofRⁿ, 3 ≤n. Thus, (see [16]), there exist an increasing sequence (Ωh)h∈N ofC^∞-domains, Ωh⊂Ω, such that Ωh→Ω inC¹according to Neˇcas (see [7] p. 85) and a sequence (Λh)h∈NofC¹- diffeomorphisms Λh:∂Ω→∂Ωhsuch that

limh sup

Q∈∂Ω

|Q−Λ_h(Q)|= 0.

(1.1)

Furthermore, there is a finite covering (Br)1≤r≤m of ∂Ω by open spheres Br = B(Qr, δ) with centerQr∈∂Ω and radiusδ, such that, forr= 1, . . . , m

B(Q_r,2δ)∩∂Ω ={(x, x_n)∈Rⁿ⁻¹×R : x_n=ξ_r(x)} ∩B(Q_r,2δ), (1.2)

and

B(Qr,2δ)∩∂Ωh={(x, xn)∈Rⁿ⁻¹×R : xn=ξrh(x)} ∩B(Qr,2δ), (1.3)

whereξ_r∈C₀¹(Rⁿ⁻¹),ξ_r(0) = ∂ξr

∂xl

(0) = 0 (l= 1, . . . , n−1), ξ_rh∈C₀^∞(Rⁿ⁻¹),and limh kξrh−ξrk_C¹

0(Rⁿ⁻¹)= 0.

(1.4) Let now

e

xr= (x, ξr(x))∈∂Ω∩B(Qr,2δ)→x∈Rⁿ⁻¹ (1.5)

and

e

xhr= (x, ξhr(x))∈∂Ωh∩B(Qr,2δ) →x∈Rⁿ⁻¹. (1.6)

Forl, i= 1, . . . , n−1,

limh

∂(exrh◦Λh◦ex⁻¹_r )i

∂xl (x) =δil

(1.7)

uniformly inUr=exr(∂Ω∩B(Qr,2δ)),where (exrh◦Λh◦ex⁻¹_r )i is thei-th coordinate of the functionxerh◦Λh◦ex⁻¹_r .

LetU_s= X

i∈N_sⁿ

a_idX_i be a form defined in Ω (respectively inRⁿ\Ω).¹ IfUs∈C_s²(Ω) we set

1N_sⁿ = {i = (i1, . . . , is) ∈ N^s : 1 ≤ i1 < . . . < is ≤ n}; if i = (i1, . . . , is) ∈ Nⁿ_s, dXi = dXi₁∧. . .∧Xi_s.

δUs= (dUs)^∗ and ∆Us=dδUs, (1.8)

where dUs is the exterior derivative of Us and ∗ is the Hodge’s operator. Us is said to beclosed (harmonic, respectively) in Ω iffdUs= 0 (∆Us= 0,respectively) in Ω.

FurthermoreUsis said to bederivedin Ω iff there exists a form Vs−1 such that dVs−1=Us in Ω.

(1.9)

andVs−1is called theprimitive of Us.

Thus we have the following identity of Bidal-de Rham (see (60) in [8]) dδU_s^∗+ (−1)ⁿδδUs= (−1)^n(s+1) X

i∈Nⁿ_s

∆ai dXi, (1.10)

where ∆ai= Xn

l=1

∂²ai

∂x²_l .

We denote withCe_s¹(Ω) the space of the formsUs∈C_s¹(Ω) such that each coefficient ofU_sand ofdU_s is inC⁰(Ω).

Given k∈N and 1< p <∞,we denote withD^k,p_s (Ω) the space of the forms Us

such that each coefficient ofUs is inW^k,p(Ω).

We say thatUshasinterior(exterior,respectively)nontangential trace inL^p_s(∂Ω) iff, for anyi∈Nⁿ_s, aihas interior nontangential tracea⁻_i (exterior nontangential trace a⁺_i ,respectively) inL^p(∂Ω). The form²

U_s⁻ = X

i∈Nⁿ_s

a⁻_i dXi(Q) (1.11)

(respectively the form

U_s⁺= X

i∈Nⁿ_s

a⁺_i dXi(Q)) (1.12)

is calledinterior(exterior,respectively)nontangential traceofUs. IfUs∈D^1,p(Ω), the form

Tr(Us) = X

i∈Nⁿ_s

Tr(ai)dXi(Q) (1.13)

is considered, where the mapping Tr :W^1,p(Ω)→L^p(∂Ω) is the continuous extension toW^1,p(Ω) of the mapping restriction defined initially onC^∞(Ω).

IfUs∈C_s⁰(Ω) andUshas interior nontangential trace in L^p_s(∂Ω), then (see The- orem 2.3 in [13])

limh Λ^∗_h(Ush) =U_s⁻ in L^p_s(∂Ω), (1.14)

whereUsh is the restriction ofUson∂Ωh.

We denote with N_s^1,p(Ω) the space of the formsUs∈C_s¹(Ω) such that each coef- ficient ofUs anddUshas interior nontangential trace inL^p(∂Ω).

The following result holds:

2dXi(Q)is the restriction to∂ΩofdXi, hencedXi(Q) =j^∗dXi,wherej:∂Ω→Rⁿis the inclusion map.

Theorem 1.1.If Us∈C_s⁰(Ω)∩D^1,p_s (Ω)andΦn−s−1∈C_n−s−1⁰ (∂Ω), then limh

Z

∂Ωh

Us∧Λ^−1∗_h (Φn−s−1) = Z

∂Ω

Tr(Us)∧Φn−s−1. (1.15)

Proof. By using a partition of unity (ϕr)1≤r≤m, corresponding to the covering (Br)1≤r≤m of ∂Ω described above, in order to prove (1.15) it will suffice to show that

limh

Z

Rⁿ⁻¹

e

x^−1∗_rh (ϕrUsh∧Λ^−1∗_h (Φn−s−1)) = Z

Rⁿ⁻¹

e

x^−1∗_r (ϕrTr(Us)∧Φn−s−1).

(1.16)

for allr= 1, . . . , m.For simplicity of notation in the following we omit the indexr.

Let

Us= X

i∈Nⁿ⁻¹s

aidXi+ X

i∈Nⁿ⁻¹_s−1

aindXi∧dXn and ex^−1∗(ϕΦn−s−1)) = X

i∈Nⁿ⁻¹_n−s−1

bidxi.

Thusex^−1∗_h (ϕUsh∧Λ^−1∗_h (Φn−s−1)) is a (n−1)-form onRⁿ⁻¹.Its coefficients are the sum of a finite number of terms like

ϕ(x, ξh(x))ai(x, ξh(x))bj(fh(x))|∂fhj

∂xj (x)|

or terms like

ϕ(x, ξh(x))ain(x, ξh(x))bj(fh(x))|∂fhj

∂xj

(x)|∂ξh

∂xl

(x), where

fh=ex◦Λ⁻¹_h ◦xe⁻¹_h = (fh1, . . . , f_h(n−1)), and|∂fhj

∂xj

|=det∂(fhj1,· · ·, fhjn−s−1)

∂(xj1,· · ·, xjn−s−1) . Hence, sincespt(ϕ)⊂B(Q, δ),the integrals in (1.16) are integrals on the compact set

K={x∈Rⁿ⁻¹:|x| ≤δ}.

We observe that using Theorem 4.5, p. 85 in [10] we have limh ai(x, ξh(x)) = Tr(ai)(x, ξ(x)) inL^p(K), while, using (1.1) and (1.7), we have

limh ϕ(x, ξh(x))bj(fh(x))|∂fhj

∂xj (x)|=ϕ(x, ξ(x))bj(x) and lim

h

∂ξh

∂xj(x) = ∂ξ

∂xj(x) (1.17)

uniformly inK.Therefore, since the sequence of functions that appear in the left side of (1.17) is uniformly bounded inK,by the Dominated Convergence Theorem we

obtain the proof .

2 Corollary.AssumeUs∈C_s¹(Ω)∩D^1,p_s (Ω).IfdUshas interior nontangential trace in L^p_s+1(∂Ω),then Tr(Us)∈W_s^1,p(∂Ω)and

dTr(Us) = (dUs)⁻ a.e. on ∂Ω.

(1.18)

Proof. It is sufficient to show that Z

∂Ω

(dUs)⁻∧Φn−s−2= (−1)^s Z

∂Ω

Tr(Us)∧dΦn−s−2

for all Φn−s−2 ∈ C˜_n−s−2¹ (∂Ω) (see n.1 in [13]), that is to say, because of (1.14) and since (dUs)h=d(Ush),

limh

Z

∂Ω

Λ^∗_h(dUsh)∧Φn−s−2= (−1)^s Z

∂Ω

Tr(Us)∧dΦn−s−2. Hence, it is enough to observe that

Z

∂Ω

Λ^∗_h(dU_sh)∧Φ_n−s−2 = Z

∂Ωh

dU_s∧Λ^−1∗_h (Φ_n−s−2)

= (−1)^s Z

∂Ω_h

Us∧Λ^−1∗_h (dΦn−s−2), and to apply Theorem 1.1.

2 We obtain also the following Theorems

Theorem 1.2. If Us ∈ Ce_s¹(Ω), then there exists an open set Ω⁰ of Rⁿ such that Ω⊂Ω⁰ and there exists a form Us ∈Ce_0,s¹ (Ω⁰) such thatUs and dUs are extensions ofUsanddUs respectively.

Proof. LetUs=P

i∈Nⁿ_s aidXi∈Ce_s¹(Ω) and letdUs=P

j∈Nⁿ_s+1bjdXj.Then we have (∀i∈Nⁿ_s) (ai∈C⁰(Ω)∩C¹(Ω)) and (∀j ∈Nⁿ_s+1) (bj ∈C⁰(Ω)).

In Ω it results

(∀j∈Nⁿ_s+1) (bj = Xs+1

k=1

(−1)^k∂abj^k

∂Xj_k), wherebj^k= (j1, . . . , jk−1, jk+1, . . . , js+1) whenj= (j1, . . . , js+1).

Arguing in a manner similar to the proof of Theorem 54 XV of [11], we prove the existence of an open set Ω⁰ ofRⁿ such that Ω⊂Ω⁰ and, fori∈Nⁿ_s andj ∈Nⁿ_s+1, the existence of functionsai∈C₀⁰(Ω⁰)∩C¹(Ω⁰\∂Ω) andbj ∈C₀⁰(Ω⁰) extensions ofai

andbj,respectively, such that bj =

s+1X

k=1

(−1)^k∂abj^k

∂Xjk

in Ω⁰\∂Ω.

(1.19) Let, then

Us= X

i∈Nⁿ_s

aidXi and Vs+1= X

j∈Nⁿ_s+1

bjdXj.

We have that Us ∈ C_0,s⁰ (Ω⁰) and Vs+1 ∈ C_0,s+1⁰ (Ω⁰). In order to prove that Us is regular in Ω⁰, using the Lemma 16.d p. 105 in [17], it suffices to show the existence of a sequence (U^p_s)p∈N inC_s^∞(Ω⁰) such that

U^p_s→Us and dU^p_s→Vs+1

(1.20)

uniformly in every compact subset of Ω⁰.Clearly we may supposeUsandVs+1defined inRⁿ.Let now (ρp)p∈N be a sequence of mollifiers and let

U^p_s=ρp∗Us= X

i∈Nⁿ_s

ρp∗aidXi

and

V^p_s+1=ρp∗Vs+1= X

j∈Nⁿ_s+1

ρp∗bjdXj, where∗is the usual convolution product between functions.

First we obtain thatU^p_s∈C_0,s^∞(Rⁿ) andV^p_s+1∈C_0,s+1^∞ (Rⁿ),and from Proposition IV.21, in [1], it follows that

U^p_s→Us and V^p_s+1→Vs+1

uniformly on compact sets ofRⁿ. We finish by proving the second formula in (1.20).

For this it is sufficient to show that

s+1X

k=1

(−1)^k ∂

∂Xjk

(ρp∗abj^k) =ρp∗bj

or, for Proposition IV.2 in [1], that

s+1X

k=1

(−1)^k( ∂

∂Xjk

ρp)∗abj^k =ρp∗bj

for allp∈N andj∈Nⁿ_s+1.We observe that, as a consequence of (1.19), we have

s+1X

k=1

(−1)^k ∂

∂Yjk

(ρ_p(X−Y)·abj^k(Y)) = Xs+1 k=1

(−1)^k+1( ∂

∂Xjk

ρ_p)(X−Y)·abj^k(Y) +ρp(X−Y)·bj(Y)

forX ∈RⁿandY ∈Rⁿ\∂Ω.Thus

s+1X

k=1

(−1)^k ∂

∂Yjk

(ρp(X−Y)·abj^k(Y)) is inL¹(Rⁿ).

Since Z

Rⁿ

Xs+1

k=1

(−1)^k ∂

∂Yjk

(ρp(X−Y)·abj^k(Y))dY = Z

Ω s+1X

k=1

(−1)^k ∂

∂Yjk

(ρp(X−Y)·abj^k(Y))dY +

Z

Rⁿ\Ω s+1X

k=1

(−1)^k ∂

∂Yjk

(ρp(X−Y)·abj^k(Y))dY =I1+I2,

to obtain (1.20) we need to prove

I1+I2= 0.

(1.21)

Now, since (see Preliminares) I₁ = lim

h

Xs+1 k=1

(−1)^k Z

Ωh

∂

∂Yjk

(ρ_p(X−Y)·abj^k(Y))dY

=

s+1X

k=1

(−1)^klim

h

Z

∂Ωh

ρp(X−Y)·abj^k(Y)dY

=

s+1X

k=1

(−1)^klim

h

Z

∂Ω

ρp(X−Λh(Y))·abj^k(Λh(Y))ωh(Y)dY,

whereωh:∂Ω→R+is the function of [17]: Theorem 1.12, by Dominated Convergence Theorem it follows that

I1= Xs+1

k=1

(−1)^k Z

∂Ω

ρp(X−Y)·abj^k(Y)dY.

(1.22)

In the same manner we prove that I2=−

Xs+1

k=1

(−1)^k Z

∂Ω

ρp(X−Y)·abj^k(Y)dY (1.23)

taking a sequence (Ω⁰_h)n∈N of C^∞-domains of Rⁿ with Ω ⊂ Ω⁰_h and a sequence of diffeomorphisms Λ⁰_h : ∂Ω →∂Ω⁰_h such that Ω →Ω⁰_h in C¹ according to Neˇcas (see [10] and [16]) and limhsup_Q∈∂Ω|Q−Λ⁰_h(Q)|= 0.From this and (1.22) we have (1.21).

Hence the thesis.

2 Theorem (Stokes) 1.3.IfUs∈Ce_s¹(Ω), then

Z

∂Γs+1

Us= Z

Γs+1

dUs

(1.24)

for all chainΓs+1 ⊂Ω.

Proof. Let Ω⁰be an open set ofRⁿsuch that Ω⊂Ω⁰and letUs, dUsbe regular forms in Ω⁰ such that the coefficients ofUsand ofdUsare respectively the extensions of the corresponding coefficients ofUs and dUs(see Theorem 1.2). Then, by the definition of regular forms, (1.24) is obtained.

2

2 Existence and Uniqueness Theorems

Let 1< s < n.Following [8] we introduce the form ω_n−s(X, Y) = X

i∈Nⁿ_n−s

1

|X−Y|ⁿ⁻²dX_idY_i (2.1)

in two variables (X, Y)∈Rⁿ×Rⁿ(see [4], Section. 7). By (2.1), ifX, Y ∈Rⁿ, X 6=Y we obtain

dXδXωn−s(X, Y) = (−1)^n−sdYδXωn−s−1(X, Y) (2.2)

and

δXωs−1(X, Y) = (−1)^n(s−1)−1δYωn−s(X, Y).

(2.3)

Let ([τ_s−1^j ],[γ_s−1^l ])_1≤j≤R⁻

s−1 1≤l≤R+

s−1

, ([t^j_s−1])_1≤j≤R⁻

s−1 and let ([c^l_s−1])_1≤l≤R⁺

s−1 be bases of C¹-differentiable singular homology spaces Hs−1(∂Ω), Hs−1(Ω), andHs−1(Rⁿ\Ω) respectively, verifying the following conditions (see n. 5 in [13])

τ_s−1^j ∼0 in Rⁿ\Ω and γ_s−1^l ∼0 in Ω and

t^j_s−1∼τ_s−1^j in Ω and c^l_s−1∼γ_s−1^l in Rⁿ\Ω, forl= 1, . . . , R⁺_s−1 andj= 1, . . . , R⁻_s−1 (see n. 5 in [13]).

Let now 1< p <∞.The following results hold:

Theorem 2.1.Let Fs−1∈C_s−1¹ (Ω) be a closed form with interior nontangential trace in L^p_s−1(∂Ω)such that

Z

τ_s−1ⁱ

F_s−1⁻ = 0 i= 1, ...,R⁻_s−1. (2.4)

Then there existsUs−2∈C_s−2¹ (Ω)∩D^1,p_s−2(Ω) verifying i)dUs−2=Fs−1 in Ω,

ii) Tr(Us−2)∈W_s−2^1,p(∂Ω)anddTr(Us−2) =F_s−1⁻ a.e. on∂Ω, iii)dU_s−2^∗ = 0inΩ.

Proof. Let Bn−s be a generic harmonic form in Ω, such that δBn−s has interior nontangential trace inL^p_s−1(∂Ω). SinceF_s−1^∗ anddBn−sare forms ofC_n−s+1¹ (Ω) with interior nontangential trace inL^p_n−s+1(∂Ω), from Theorem 2.7 in [13] it follows that they are inL^p_n−s+1(Ω).Then we can put

Gs−1(X) =(−1)^ns−1 kn

Z

Ω

ωs−1(X, Y)∧(F_s−1^∗ + (−1)^{(n−1)(s−1)}dBn−s)(Y).

(2.5)

According to Theorem 77.VI in [9] this form is inC_s−1² (Ω)∩D^2,p_s−1(Ω).Let then Us−2=δG^∗_s−1 and Vn−s= (−1)ⁿδGs−1.

(2.6)

From (1.10), using Theorem 77.VI in [9], we obtain that dUs−2+δVn−s=Fs−1+δBn−s a.e. in Ω.

(2.7)

Then in order to prove the thesis it is sufficient to show that there existsBn−s such that

δVn−s=δBn−s a.e. in Ω.

(2.8)

Now (2.3) and (2.6) imply that

Vn−s(X) = 1 kn

Z

Ω

δYωn−s(X, Y)∧(F_s−1^∗ + (−1)^{(n−1)(s−1)}dBn−s)(Y)

= 1 kn

Z

Ω

dYωn−s(X, Y)∧(Fs−1+δBn−s)(Y)

= 1 kn lim

h→∞

Z

Ωh

dYωn−s(X, Y)∧(Fs−1+δBn−s)(Y).

Hence, sinceFs−1 and δBn−s are closed forms in Ω, from Theorem 11, p.121 in [10]

we have

Vn−s(X) = 1 k_n lim

h→∞

Z

∂Ωh

ωn−s(X, Y)∧(Fs−1+δBn−s)(Y).

Moreover, becauseFs−1andδBn−shave interior nontangential trace inL^p_s−1(∂Ω),by (1.14) it follows that

Vn−s(X) =Hn−s(X) + 1 kn

Z

∂Ω

ωn−s(X, Y)∧Φs−1(Y), (2.9)

where

Hn−s(X) = 1 kn

Z

∂Ω

ωn−s(X, Y)∧F_s−1⁻ (Y) (2.10)

and

Φs−1= (δBn−s)⁻. (2.11)

If we require that the formVn−ssatisfies (2.8), it results that δBn−s(X) =δHn−s(X) + 1

kn

Z

∂Ω

δXωn−s(X, Y)∧Φs−1(Y).

(2.12)

From this, if we take into account Theorem 1 in [14], it follows that a.e. on∂Ω (δHn−s)⁻= (1

2I− 1

k_nTs−1)(Φs−1), (2.13)

whereIis the identity operator onL^p_s−1(∂Ω) andTs−1 is a compact operator on the same space. Consider now the the homogeneous transposed equation of (2.13)

Te(ψn−s) =kn

2 Ψn−s+Tn−s(Ψn−s) = 0.

(2.14)

In order to show that (2.13) has solution in L^p_s−1(∂Ω), we prove that for all i = 1, . . . , R⁻_s−1

Z

∂Ω

Ψⁱ_n−s∧(δHn−s)⁻= 0, (2.15)

where (Ψⁱ_n−s)_1≤i≤R⁺

n−s is a base of Ker(Ten−s) verifying the conditions (see Theorem 7 in [14]):

Z

τ^l_n−s

Ψⁱ_n−s= 0 and Z

γ^j_n−s

Ψⁱ_n−s=δij

(2.16)

forl= 1, . . . , R⁻_n−s and fori, j= 1, . . . , R⁺_n−s.Here (τ^l_n−s, γ^j_n−s)_1≤l≤R⁻

n−s 1≤j≤R+

n−s

is the dual fundamental system of the fundamental system (τ_s−1^j , γ_s−1^l )_1≤j≤R⁻

s−1 1≤l≤R+

s−1

.We now observe that using Theorem 1 in [14] we obtain

(δHn−s)⁻=1

2F_s−1⁻ + 1 kn

Ts−1(F_s−1⁻ ).

(2.17)

Hence from (2.14) we have Z

∂Ω

Ψⁱ_n−s∧(δHn−s)⁻= 1 2 Z

∂Ω

Ψⁱ_n−s∧F_s−1⁻ + 1 kn

Z

∂Ω

Ψⁱ_n−s∧Ts−1(F_s−1⁻ )

= 1 2 Z

∂Ω

Ψⁱ_n−s∧F_s−1⁻ − 1 kn

Z

∂Ω

Tn−s(Ψⁱ_n−s)∧(F_s−1⁻ ) = Z

∂Ω

Ψⁱ_n−s∧F_s−1⁻ . (2.18)

Thus, since Ψⁱ_n−s and F_s−1⁻ are closed forms in W_n−s^1,r (∂Ω) (for all r > 1) and in W_s−1^1,p(∂Ω) respectively, by applying Theorem 1.7 in [14], using (2.4) and (2.16), it follows that

Z

∂Ω

Ψⁱ_n−s∧F_s−1⁻ = (−1)^{(s−1)(n−s)}

R⁻_s−1

X

j=1

Z

τ_s−1^j

F_s−1⁻ Z

γ^j_n−s

Ψⁱ_n−s

+

R⁺_s−1

X

l=1

Z

γ^l_s−1

F_s−1⁻ Z

τ^l_n−s

Ψⁱ_n−s= 0.

Furthermore, using (2.18) we obtain that (2.15) is verified.

Let now Φs−1be a solution of (2.13) inL^p_s−1(∂Ω).Thus, from (2.17) and Theorem 2 in [14], sinceF_s−1⁻ ∈W_s−1^1,p(∂Ω) is a closed form, it results that alsoδHn−sis a closed form inW_s−1^1,p(∂Ω). Hence, arguing in a manner similar to the proof of the Theorem 1.4 in [15], we have that Φ_s−1 is a closed form inW_s−1^1,p(∂Ω).Let finally

Bn−s(X) =Hn−s+ 1 k_n

Z

∂Ω

ωn−s(X, Y)∧Φs−1(Y).

(2.19)

Thus using (11) in [14], we have thatBn−sis a harmonic form. Moreover in Ω it is δBn−s(X) =δHn−s(X) + 1

kn

Z

∂Ω

δXωn−s(X, Y)∧Φs−1(Y).

Furthermore by Theorem 1 in [14]δBn−shas interior nontangential trace inL^p_s−1(∂Ω) and

(δBn−s)⁻ = (δHn−s)⁻+1

2Φs−1+ 1

knTs−1(Φs−1).

From this, being Φs−1 a solution of (2.13), we deduce that (δBn−s)⁻ = Φs−1 and i) is satisfied.

ii) and iii) follow from Corollary of Theorem 1.1 and from (2.6) respectively.

2 By applying Lemma 2.1 of [15], from (2.6) we obtain the following

Corollary. If Fs−1 is a form that satisfies the hypotheses ofTheorem 2.1, then the form

Us−2(X) = (−1)^n+s kn

Z

Ω

δXωn−s+1(X, Y)∧(Fs−1(Y) +δBn−s(Y)) (2.20)

is a primitive ofFs−1,whereBn−sis the form defined in (2.19)andΦs−1is a solution of(2.13).

Theorem 2.2. If Fs−1 is a form that satisfies the hypotheses ofTheorem 2.1, and dF_s−1^∗ ∈L^p_n−s+2(Ω),then there exists a primitiveUs−2ofFs−1inC_s−2² (Ω)∩D_s−2^1,p(Ω).

Proof. Arguing in a manner similar to the proof of (2.9), from (2.20) it follows that

Us−2(X) = − 1 kn

Z

Ω

ωs−2(X, Y)∧dF_s−1^∗ (Y) + 1

kn

Z

∂Ω

ωs−2(X, Y)∧((F_s−1^∗ )⁻(Y) + (−1)^{(n−1)(s−1)}(dBn−s)⁻(Y)) and applying Theorem 1.2 of [15] and Theorem 77.VI of [9], we obtain our claim.

2 Corollary.IfFs−1 satisfies the hypotheses of Theorem 2.1andF_s−1^∗ is a closed form inΩ,then there exists a primitive U_s−2 of F_s−1 inC_s−2^∞ (Ω)∩ N_s−2^1,p(Ω).

Theorem 2.3. If Fs−1 ∈C_s−1⁰ (Ω) is a closed form and satisfies (2.4),then for all p > n there exists a primitiveUs−2 ofFs−1 inC_s−2⁰ (Ω)∩D_s−2^1,p(Ω).

Proof. Letp > n.Theorem 2.1 implies that there exists a primitiveU_s−2 ofF_s−1 in D_s−2^1,p(Ω).By Theorem 77.VI of [9] it follows that Us−2∈C^0,µ(Ω),whereµ= 1−n

p. 2 Our purpose is to obtain the following uniqueness Theorem:

Theorem 2.4.Let Fs−1 andGn−s+3 be closed forms in C_s−1⁰ (Ω)∩C_s−1¹ (Ω) and inC_n−s+3⁰ (Ω)∩C_n−s+3¹ (Ω), respectively, such that

Z

τ_s−1^l

Fs−1= 0 and Z

τ_n−s+3^j

Gn−s+3= 0, (2.21)

for all l = 1, . . . , R⁻_s−1 and for all j = 1, . . . , R⁻_n−s+3. Let Ls−2 ∈Ce_s−2¹ (∂Ω)be such that

dLs−2=Fs−1 on∂Ω (2.22)

and

Z

γ_s−2ⁱ

Ls−2= Z

Γⁱ_s−1

Fs−1

(2.23)

fori= 1, . . . , R⁺_s−2, whereγ_s−2ⁱ =∂Γⁱ_s−1. Then we have

i) if R⁻_n−s+2 = 0, there exists a unique form Us−2 ∈ C_s−2¹ (Ω) with an interior nontangential trace in L^p_s−2(∂Ω)such that

dUs−2=Fs−1 inΩ, dU_s−2^∗ =Gn−s+3 inΩ (2.24)

and

U_s−2⁻ =Ls−2 on ∂Ω;

(2.25)

ii)ifR⁻_n−s+2>0,then for any sequence(α_i)_1≤i≤R−

n−s+2 inRthere exists a unique formUs−2∈C_s−2¹ (Ω)with interior nontangential trace inL^p_s−2(∂Ω)satisfying (2.24), (2.25),and

Z

tⁱ_n−s+2

U_s−2^∗ =αi, (2.26)

for alli= 1, . . . , R⁻_n−s+2.

Proof. By applying Theorems 2.1 and 2.3 to forms Fs−1 and Gn−s+3, it results that there exist two formsUs−2 ∈C_s−2⁰ (Ω)∩C_s−2¹ (Ω) and Vn−s+2 ∈C_n−s+2⁰ (Ω)∩ C_n−s+2¹ (Ω),such that in Ω

dUs−2=Fs−1, dU^∗_s−2= 0 (2.27)

and

dV_n−s+2=G_n−s+3, dV^∗_n−s+2= 0.

(2.28)

Let Kn−s+1 ∈ C_n−s+1² (Ω)∩ N_n−s+1^1,p (Ω) and Ms−3 ∈ C_s−3² (Ω) ∩ N_s−3^1,p(Ω) be two arbitrary harmonic forms in Ω. Thus if we let

Ws−2=Us−2+ (−1)^s(n−1)V^∗_n−s+2 (2.29)

and

Us−2=Ws−2+δKn−s+1+dMs−3, (2.30)

in Ω,it follows that Ws−2∈Ce_s¹(Ω), dUs−2=Fs−1 anddU_s−2^∗ =Gn−s+3 in Ω.

We want to show that there existK_n−s+1andM_s−3such thatU_s−2satisfies (2.25) and (2.26). In order to do this let

Ms−3(X) =

R⁻_n−s+2

X

k=1

βk

Z

∂Ω

ωs−3(X, Y)∧(Ψ^k_n−s+2)(Y), (2.31)

where (βk)_1≤k≤R⁻

n−s+2 is an arbitrary sequence in R and (Ψ^k_n−s+2)_1≤k≤R⁻

n−s+2 is a base of Ker(Te_n−s+2⁰ ) such that forh, i= 1, . . . , R⁻_n−s+2 and forl= 1, . . . , R⁺_n−s+2 we have (see: Theorem 8 in [14])

Z

τ_n−s+2^h

Ψⁱ_n−s+2=δih and Z

γ_n−s+2^l

Ψⁱ_n−s+2= 0.

(2.32)

Since Ψⁱ_n−s+2 belongs to Ker(Te_n−s+2⁰ ), Theorem 4 in [14] implies that Ψⁱ_n−s+2 is a closed form inW_n−s+2^1,q (∂Ω) for any q >1. Using Theorem 72.V in [9] we have that Ms−3 is a continuous form inRⁿ. Moreover, by Theorems 1 and 2 in [14], it follows

thatMs−3is harmonic form inRⁿ\∂Ω with interior and exterior nontangential trace inL^q_n−s+2(∂Ω) for anyq >1,and (δMs−3)⁺ = 0.Furthermore, from Theorem 2.1 in [13] it follows thatδMs−3∈L^q_n−s+2(Rⁿ\Ω) for anyq >1.By applying Theorem 11, p. 121 in [10], we have

Z

Rⁿ\Ω

dMs−3∧δMs−3= Z

∂Ω

Ms−3∧(δMs−3)⁺= 0.

We deduce thatdMs−3= 0 inRⁿ\Ω.Thus, for all Φn−s+1∈Cen−s+1(∂Ω),it results that

Z

∂Ω

(dMs−3)⁻∧Φn−s+3= Z

∂Ω

Ms−3∧dΦn−s+3= Z

∂Ω

(dMs−3)⁺∧Φn−s+3= 0.

Hence, by Lemma 3.1 in [13], we obtain (dMs−3)⁻= 0.From this, if we require that the formUs−2,defined in (2.30), satisfies the boundary condition (2.25), we have

(δKn−s+1)⁻=Ls−2−Ws−2 on∂Ω.

(2.33)

Let 1< p <∞.From (2.22), (2.27), and (2.28) it follows thatLs−2−Ws−2is a closed form ofW_s−2^1,p(∂Ω),instead from Theorem 1.3 and (2.23) we have

Z

γ_s−2ⁱ

(Ls−2−Ws−2) = 0

for anyi= 1, . . . , R⁺_s−2.Hence, using Theorem 1.5 in [15] we obtain that there exists a formKn−s+1∈C_n−s+1² (Ω)∩ N_n−s+1^1,p (Ω), harmonic in Ω and satisfying (2.33). Now, since using (2.30) and (2.31)

U_s−2^∗ (X) =W^∗_s−2(X) + (−1)^ns−1dKn−s+1(X) +

R⁺_n−s+2

X

k=1

βk

Z

∂Ω

δXωs−3(X, Y)∧(Ψ^k_n−s+2)(Y), we prove the existence of a sequence (βk)_1≤k≤R⁺

n−s+3 in R such that U_s−2^∗ satisfies (2.26). For this purpose, arguing in a manner similar to the proof of the Theorem (2.2) in [15], it is sufficient to show the existence of a sequence (βk)_1≤k≤R⁺

n−s+3 in R such that

αi= Z

tⁱ_n−s+2

(W^∗_s−2+ (−1)^(n−1)sdKn−s+1) +βi

Z

Y∈cⁱ_s−2

δY

Z

X∈tⁱ_n−s+2

ωn−s+2

(2.34)

for alli= 1, . . . , R_n−s+2⁻ . This is easily seen to be true, because by (27) in [8], it is Z

Y∈cⁱ_s−2

δY

Z

X∈tⁱ_n−s+2

ωn−s+26= 0.

Thus the existence of a solution of the problem is obtained. Finally to show the uniqueness of the solution it is sufficient to prove that the homogeneous problem associated to the above problem is the zero form. Thus, let Us−2 be a solution of