We study the analyticity of the semigroup generated by the Stokes operator equipped with Neumann-type boundary conditions on Lp spaces in Lipschitz domains. Our strategy is to regularize this operator by considering the Hodge Laplacian, which has the additional property that it commutes with the Leray projection.

(1.3)

div u = 0 in ? × (0, T ), ? × curl u ?? ×(0,T )= 0, ? · u ?? ×(0,T )= 0, u t=0 = uo in ? , as well as the parabolic Maxwell system equipped with a perfectly conductive wall condition, i.e., ?tu + curl curl u = f,

div u = 0 in ? × (0, T ), ? × u ?? ×(0,T )= 0, u t=0 = uo in ? .

In (1.1), u and ? stand, respectively, for the velocity field and pressure of a fluid occupying a domain ?, whereas, in (1.2), u denotes the magnetic field propagating inside of the domain ?. In both cases, uo denotes the initial datum, f is a given, divergence-free field and ? stands for the unit outward normal to ?.? For a discussion of the relationship between the Neumann (or, as they are occasionally called, ?free-boundary?) conditions in (1.1) and the more traditionally used Navier?s slip boundary conditions to the effect that

? · u = 0 on (0, T ) × ?? [(?u + ?u )?]tan

= 0 on (0, T ) × ?? ,
the interested reader is referred to [

Received by the editors June 25, 2007. 2000 Mathematics Subject Classification. Primary 42B30, 46A16; Secondary 46E35, 35J25. Key words and phrases. Hodge-Laplacian, Lipschitz domains, analytic semigroup. The first author was supported by the NSF grants DMS - 0400639 and DMS FRG - 0456306. The second author was supported by a UMC Miller Scholarship grant.

c 2008 American Mathematical Society Reverts to public domain 28 years from publication

Systems such as (1.1) and (1.2) naturally arise in the process of linearizing some
basic nonlinear evolution problems in mathematical physics, such as the
NavierStokes equations and certain problems related to the Ginzburg-Landau model for
superconductivity and magneto-hydrodynamics. A more detailed discussion in this
regard can be found in the monographs [

In a suitable L2 context, the stationary versions of (1.1) and (1.2) have unique
(finite energy) weak solutions. This is most elegantly seen using the so-called {H, V , a}
formalism as in [

div u = 0 in ? , (1.4) and (1.5) (1.6)

|?| u Lp(?) 3 ? C(? , p) f Lp(?) 3 , uniformly in ? satisfying |arg (?)| < ? for some ? > 0.

When the domain ? has a sufficiently smooth boundary, such estimates are
well-understood. The classical approach utilizes the fact that the boundary-value
problems (1.4)-(1.5) are regular elliptic (cf., e.g., [

The nature of the problem at hand changes dramatically as ?? becomes less
regular. To illustrate this point, let us recall the following negative result from
[

Taylor?s conjecture. For a given bounded Lipschitz domain ? ? R3 there exists ? = ?(?) > 0 such that the Stokes operator associated with (1.7) generates an analytic semigroup on Lp provided 3/2 ? ? < p < 3 + ?.

The range of p?s in the above conjecture is naturally dictated by the mapping
properties of the Leray projection
(1.8) Pp : Lp(?) 3 ?? {u ? Lp(?) 3 : div u = 0, ? · u = 0}.
When p = 2 this is taken to be the canonical orthogonal projection and is obviously
bounded, but the issue of whether this extends to a bounded operator in the context
of (1.8) for other values of p is considerably more subtle. Indeed, it has been shown
in [

That any p ? (1, ?) will do in the case when the domain ? has a smooth
boundary has been proved by Y. Giga in his seminal paper [

In this paper, we are able to prove the analogue of Taylor?s conjecture for the Stokes system equipped with Neumann-type boundary conditions. Our approach makes essential use of the recent progress in understanding Poisson-type problems for the Hodge Laplacian such as (1.9) (1.10) ? ? u = f ? Lp(?) 3, such that (1.9) is well-posed if and only if p ? (p? , q? ). Here, the index p? can be further defined in terms of the critical exponents intervening in the (regular) Dirichlet and Neumann problems for the Laplace-Beltrami operator in ? (as well as its complement), when optimal Lp estimates for the associated nontangential maximal function are sought. One feature of ? which influences the size of p? is the local oscillations of the unit conormal ? to ??. In particular, p? = 1 (and, hence, q? = ?) when ? belongs to the Sarason class of functions of vanishing mean oscillations (which is the case if, e.g., ?? ? C1). Furthermore, for a Lipschitz polyhedron in the Euclidean setting, p? can be estimated in terms of the dihedral angles involved.

Similar results have been proved earlier in [

Here, one should typically regard the indices p? , q? as small perturbations of 2.

One of our key results states that the Hodge Laplacian, in the context of (1.9), generates an analytic semigroup in Lp for any p ? (p? , q? ). From this and the fact that the Leray projection commutes with the semigroup generated by the Hodge Laplacian, it is then possible to prove a similar conclusion for the Stokes and Maxwell operators acting on Lp (cf. Theorem 7.3 and Theorem 7.4 for precise statements). Thus, from this perspective, one key feature of our approach is to embed the Stokes (and Maxwell) system into a more general (and, ultimately, more manageable) elliptic problem, in a way which allows us to return to the original system by specializing the type of data allowed in the formulation of the problem.

We carry out this program in the context of differential forms on Lipschitz
subdomains of a smooth, compact, boundaryless Riemannian manifold M. This is
both notationally convenient and natural from a geometric point of view. It also
allows for a more general setting than previously considered in the literature even
in the case when ?? ? C1 (semigroup methods for differential forms in the smooth
context are discussed in, e.g., P.E. Conner?s book [

Theorem. Let ? ? M be a Lipschitz domain, 1 < p < ?, and let Ap denote the Stokes operator, i.e. the mapping u ? ?? u, for each vector field u ? Lp(? , ?1) satisfying (1.12) and for which there exists a scalar function ? ? W 1,p(?) such that (1.13) f := ?? u + ?? ? Lp(? , ?1) and div f = 0, ? · f ?? = 0.

Then the operator ?Ap generates an analytic semigroup e?tAp f ? Lp(? , ?1) : div f = 0, ? · f ?? = 0 whenever p ? (p? , q? ).

t>0 on the space

Several remarks are in order here. First, it is remarkable that the range of p?s for which the above result is valid agrees with the range predicted by Taylor?s conjecture. Second, the case p = 2 follows directly from standard functional analysis. Hence, from this perspective, our result can be interpreted as stating that the semigroup generated by the Stokes operator on L2 extends to an analytic semigroup in Lp, and the infinitesimal generator of this extension is also identified in the most desirable fashion. Third, the above theorem is most relevant in the treatment of the Navier-Stokes initial boundary value problem (1.14) ?tu ? ? u + (u · ?)u + ?? = f,

div u = 0 in ? × (0, T ),
by viewing it as an abstract Cauchy (evolution) problem; cf., e.g., [

This line of work has recently been pursued in [

One distinctive aspect of our work is that, given the low regularity assumptions
we make on the underlying domain ?, we are forced to work with certain
nonstandard Sobolev-type spaces, which are well-adapted to the differential operators
at hand (such as the exterior derivative operator d and its formal adjoint ?). In
particular, issues such as boundary traces and embeddings become more delicate
than in the standard theory. In this regard, a key ingredient in our approach is an
estimate established in [

Another crucial ingredient in our approach is the solution of the Poisson problem
for the Hodge Laplacian, i.e. ? = ?(dd? + d?d) in the sense of composition of
unbounded operators, with Lp-data in Lipschitz subdomains of M from [

The organization of the paper is as follows. In Section 2 we collect basic
definitions and preliminary results and introduce most of the notational conventions used
throughout this work. The Hodge Laplacian is reviewed in Section 3, along with
the Stokes and Maxwell operators. In Section 4 we dwell on issues of regularity for
differential forms in the domain of the Hodge Laplacian. Here we record several
key results, themselves corollaries of the work in [

In this section we review a number of basic definitions and collect several known results which are going to be useful for us in the sequel. 2.1. Geometrical preliminaries. Let M be a smooth, compact, oriented manifold of real dimension n, equipped with a smooth metric tensor, g = j,k gjkdxj ? dxk. Denote by T M and T ?M the tangent and cotangent bundles to M, respectively. Occasionally, we shall identify T ?M ? T M canonically, via the metric. (2.3) (2.4) Set ? T M for the -th exterior power of T M. Sections in this vector bundle are -differential forms. The Hermitian structure on T M extends naturally to T ?M := ?1T M and, further, to ? T M. We denote by ·, · the corresponding (pointwise) inner product. The volume form on M, denoted in the sequel by ?, is the unique unitary, positively oriented differential form of maximal degree on M. In local coordinates, ? := [det (gjk)]1/2dx1 ? dx2 ? ...? dxn. In the sequel, we denote by dV the Borelian measure induced by ? on M, i.e., dV = [det (gjk)]1/2dx1dx2...dxn. Going further, we introduce the Hodge star operator as the unique vector bundle morphism ? : ? T M ? ?n? T M such that u ? (?u) = |u|2? for each u ? ? T M. In particular, ? = ? 1 and u ? (?v) = u, v ?,

? u ? ? T M, ? v ? ? T M.

The interior product between a 1-form ? and an -form u is then defined by ? ? u := (?1) (n+1) ? (? ? ?u).

Let d stand for the (exterior) derivative operator and denote by ? its formal adjoint (with respect to the metric introduced above). For further reference some basic properties of these objects are summarized below.

Proposition 2.1. For an arbitrary 1-form ?, -form u, (n ? )-form v, and ( + 1)form w, the following are true: (1) u, ?v = (?1) (n? ) ?u, v and ?u, ?v = u, v . Also, ??u = (?1) (n? ) u; (2) ? ? u, w = u, ? ? w ; (3) ?(? ? u) = (?1) ? ? (?u) and ?(? ? u) = (?1) +1? ? (?u); (4) ?? = (?1) d?, ?? = (?1) +1 ? d, and ? = (?1)n( +1)+1 ? d? on -forms; (5) ?(d? + ?d) = ? , the Hodge Laplacian on M.

Let ? be a Lipschitz subdomain of M. That is, ?? can be described in appropriate local coordinates by means of graphs of Lipschitz functions. Then the unit conormal ? ? T ?M is defined a.e., with respect to the surface measure d?, on ??. For any two sufficiently well-behaved differential forms (of compatible degrees) u, w we then have ? du, w dV =

u, ?w dV + ? ?? u, ? ? w d?. 2.2. Sobolev and Besov spaces of differential forms. The Sobolev (or potential) class Lp?(M), 1 < p < ?, ? ? R, is obtained by lifting the Euclidean scale Lp?(Rn) := {(I ? )? ??/2f : f ? Lp(Rn)} to M (via a C? partition of unity and a pull-back). For a Lipschitz subdomain ? of M, we denote by Lp?(?) the restriction of elements in Lp?(M) to ?, and set Lp?(? , ? ) = Lp?(?) ? ? T M, i.e. the collection of -forms with coefficients in Lp?(?). In particular, Lp(? , ? ) stands for the space of -differential forms with p-th power integrable coefficients in ?.

Let us also note here that if p, p ? (1, ?) are such that 1/p + 1/p = 1, then Lsp(? , ? ) ?

p (? , ? ), = L?s ? s ? (?1 + 1/p, 1/p).

The Besov spaces Bsp,q(? , ? ), 1 < p, q < ?, s ? R, can be introduced in a similar manner; alternatively, this may be obtained from the Sobolev scale via real interpolation.

Next, denote by L1p(??) the Sobolev space of functions in Lp(??) with tangential gradients in Lp(??), 1 < p < ?. Besov spaces on ?? can then be introduced via real interpolation, i.e.

Bsp,q(??) := ( Lp(??) , L1p(??)) s,q , with 0 < s < 1, 1 < p, q < ?.

Finally, if 1 < p, q < ? and 1/p + 1/p = 1, 1/q + 1/q = 1, we define and, much as before, set Bsp,q(?? , ? ) := Bsp,q(??)

B?p,sq(??) :=

Bsp ,q (??) ? , 0 < s < 1, ? ? T M. (2.14) (2.15) (2.16) is well-defined, linear and bounded. Tr : Lsp(? , ? ) ?? Bp,p1 (?? , ? )

s? p Ex : Bp,p1 (?? , ? ) ?? Lsp(? , ? ),

s? p is well-defined, bounded and onto if 1 < p < ? and p1 < s < 1 + p1 . Furthermore, the trace operator has a bounded right inverse and if 1 < p < ?, p1 < ? < 1 + p1 , then ? ? · : Dp(?; ?) ?? Bp,p1 (?? , ? ?1)

? p

Ker (Tr) = the closure of Co?(? , ? ) in Lsp(? , ? ).

For 1 < p < ?, s ? R, and

? {0, 1, ..., n} we next introduce Dp(?; d) := {u ? Lp(? , ? ) : du ? Lp(? , ? +1},

Dp(?; ?) := {u ? Lp(? , ? ) : ?u ? Lp(? , ? ?1)}, equipped with the natural graph norms. Throughout the paper, all derivatives are taken in the sense of distributions.

Inspired by the identity (2.3), if 1 < p < ? and u ? Dp(?; ?), we then define ? ? u ? Bp,p1 (?? , ? ?1) by ? p

? ? u, ? := ? ?u, ? + u, d? for any ? ? Bp1 ,p (?? , ? ?1), 1/p + 1/p = 1, and any ? ? L1p (? , ? ?1) with p Tr ? = ?. Note that (2.4), (2.9) imply that the operator is well-defined, linear and bounded for each p ? (1, ?), i.e.

? ? u Bp,p1 (?? ,? ?1) ? C ? p

u Lsp(? ,? ) + ?u Lsp(? ,? ?1) .

X p(??) := ? ? u : u ? Dp+1(?; ?) ? Bp,p1 (?? , ? ), ? p which we equip with the natural ?infimum? norm. It follows that the operator ?? : X p(??)

?? X p?1(??) ??f := ?? ? ?w, if f = ? ? w, w ? Dp+1(?; ?) and

Other spaces of interest for us here are defined as follows. For 1 < p < ?, s ? R, ? {0, 1, ..., n}, consider

Dp(?; ??) := {u ? Lp(? , ? ) : ?u ? Lp(? , ? ?1), ? ? u = 0}, once again equipped with the natural graph norm.

For further use, we record here a useful variation on the integration by parts formula (2.3), namely that if 1 < p, p < ? satisfy 1/p + 1/p = 1, then du, v = u, ?v ,

? u ? Dp(?; d), ? v ? Dp (?; ??). 2.3. The {H, V , a} formalism. Let V be a reflexive Banach space continuously and densely embedded into a Hilbert space H so that, in particular, and assume that (2.20) a(·, ·) : V × V ?? C is a sesquilinear, bounded form. Then

V ? H ? V ? Ao : V ?? V?,

Aou := a(u, ·) ? V?, ? u ? V , is a linear, bounded operator satisfying

V? Aou, v V = a(u, v), ? u, v ? V .

Assume further that a(·, ·) is symmetric and coercive, in the sense that there exist C1, C2 > 0 such that

Re a(u, u) + C1 u 2H ? C2 u 2V , ? u ? V . (2.24) Ao : V ?? V? is bounded and selfadjoint.

Furthermore, Ao is invertible if the constant C1 appearing in (2.23) can be taken to be zero. Going further, take A to be the part of Ao in H, i.e.

A := Ao

Dom (A)

: H ?? H,

Dom (A) := {u ? V : Aou ? H}.

2?i ?? (2.28)

e?zAf := Hence, (2.25)-(2.26) is an unbounded, selfadjoint operator on H. Furthermore, there exists ? ? (0, ?/2) such that

C
(2.27) (?I ? A)?1 ? |?| , ? < |arg(?)| ? ?,
i.e., A is sectorial; cf., e.g., [

|arg (z)| < ?/2 ? ? ,
t>0
where ? ? (?, ?/2) and ?? := {rei? : r > 0}; see, e.g., [

In closing, we would like to point out that the above formalism (discussed in
detail in, e.g., [

3. The Hodge Laplacian and related operators
3.1. The Hodge Laplacian. Recall that the -th Betti number of ?, denoted by
b (?), is defined as the dimension of the -th singular homology group of ?, viewed
as a topological space, over the reals. It has been proved in [

1 ? p? < 2 < q? ? ? (3.2) Hp(? , ? ) := {u ? Lp(? , ? ) : du = 0, ?u = 0, ? ? u = 0} is independent of p if p ? (p? , q? ) and has dimension b (?). We shall occasionally abbreviate H(? , ? ) := H2(? , ? ).

Consequently, the orthogonal projection of L2(? , ? ) onto H(? , ? ) extends canonically to a bounded operator

Pp : Lp(? , ? ) ?? H(? , ? ) ? Lp(? , ? ) if p? < p < q? , which has the property that

Pp? = Pp ,

p, p ? (p? , q? ), 1/p + 1/p = 1.

In order to continue, for each p ? (1, ?) and ? {0, 1, ..., n} we set Vp(? , ? ) :=

{u ? Lp(? , ? ) : du ? Lp(? , ? +1), ?u ? Lp(? , ? ?1), ? ? u = 0} (3.3) (3.4) (3.5) (3.6) (3.7) (3.8) (3.9) =

Dp(?; d) ? Dp(?; ??), once again equipped with the natural graph norm. If p = 2, we introduce the symmetric, coercive quadratic form

Q (u, v) := du, dv + ?u, ?v , u, v ? V2(? , ? ), and make the following remark.

Proposition 3.1. If b (?) = 0 , then there exists C > 0 such that

Q (u, u) ? C u 2L2(? ,? ), ? u ? V2(? , ? ).

Proof. We shall rely on the estimate
u 2L21/2(? ,? ) ? C
u 2L2(? ,? ) + Q (u, u) ,
? u ? V2(? , ? ),
which has been established in [

uj ? u in L2(? , ? ) for some u with u L2(? ,? ) = 1.

Now, Q (uj , uj) ? 0 as j ? ? forces duj ? 0 and ?uj ? 0 in L2(? , ? +1) and L2(? , ? ?1), respectively, as j ? ? and, hence, du = 0, ?u = 0. Furthermore, by the continuity of the operator (2.13), we also have ? ? u = 0. Consequently, u ? H(? , ? ) = {0}, given that b (?) = 0. This contradicts (3.9) and proves the proposition.

Returning to the mainstream discussion, the {H, V , a} formalism discussed in §2.3 applies to the context when V := V 2(? , ? ), H := L2(? , ? ) and a(u, v) :=

B : L2(? , ? ) ?? L2(? , ? ) whose domain, Dom (B), consists of all and for which

Bu, v = Q (u, v),

u ? Dom (B), v ? V 2(? , ? ).

If we now regard the exterior derivative as an unbounded operator u ? V 2(? , ? ) such that there exists C > 0 with the property that |Q (u, v)| ? C v L2(? ,? ) for all v ? V 2(? , ? ) d : L2(? , ? ) ?? L2(? , ? +1) with domain D2(?; d) and natural action, (3.19) (3.20) (3.21) with domain Dom (Bp) consisting of (3.17) by setting (3.18) and, in the sense of composition of unbounded operators,

BpGp = GpBp = I ? Pp on Lp(? , ? ).

Gpu Lp(? ,? ) + dGpu Lp(? ,? +1) + ?Gpu Lp(? ,? ?1)

+ d?Gpu Lp(? ,? ) + ?dGpu Lp(? ,? ) ? C u Lp(? ,? ) Thus, if 1 < p < ?, it is natural to consider the unbounded operator

Bp : Lp(? , ? ) ?? Lp(? , ? ) d? : L2(? , ? +1) ?? L2(? , ? ) Dom (d?) = D2+1(?; ??),

d?u = ?u, ? u ? D2+1(?; ??).

u ? Dp(?; d) ? Dp(?; ??) with du ? Dp+1(?; ??), ?u ? Dp+1(?; d) it is not difficult to check that

B = d d? + d?d
in the sense of composition of unbounded operators. Furthermore, under the
assumption that b (?) = 0, this operator is actually invertible on L2(? , ? ). Cf. also
[

The latest description of B has a natural analogue in the Lp context. The starting point is the observation that the dual of (3.13) is

Bpu := ?? u = (d? + ?d)u, ? u ? Dom (Bp).

Note that since Co?(? , ? ) is contained in Dom (Bp), it follows that Bp is densely defined.

Proposition 3.2. Let ? ? M be a Lipschitz domain. Then for each p? < p < q? there exists a linear, bounded operator

Gp : Lp(? , ? ) ?? Lp(? , ? ), such that Im (Gp) ? Dom (Bp), Furthermore, and (with the subscript

used to indicate the dependence on the degree), (Gp)? = Gp ,

p? < p, p < q? , 1/p + 1/p = 1 dGp, = Gp, +1d on Dp(?; d), ?Gp, = Gp, ?1? on Dp(?; ?).

Gp w, ?

w, Gp ? = w, v = = = u, Bpv = u, BpGp? = u, (I ? Pp)? (I ? Pp)u, ? .

Since ? is arbitrary, this forces (I ? Pp)u = Gp w and, further, u = Gp w + Pp u ? Dom (Bp ). In addition, Bp u = Bp Gp w + Bp Pp u = (I ? Pp )w = w which shows that (Bp)? ? Bp . This finishes the proof of the proposition.

Proposition 3.3. For each Lipschitz subdomain ? of M one has Ker (Bp) = Hp(? , ? ) whenever p? < p < q? ,

Im (Bp) ? Ker (Pp) for each p ? (p? , q? ). (Bp)? = Bp ,

p? < p, p < q? , 1/p + 1/p = 1.

Proof. The inclusion Bp ? (Bp)? is immediate from definitions, so it remains to prove the opposite inclusion. To this end, if u ? Dom (Bp?), then u ? Lp(? , ? ) and there exists w ? Lp (? , ? ) such that w, v = u, Bpv , ? v ? Dom (Bp).

Choosing v := Pp? with ? ? Lp(? , ? ) arbitrary forces w, Pp? = 0 and, ultimately, Pp w = 0. Next, pick v := Gp? with ? ? Lp(? , ? ) arbitrary and write

Such a Green operator has been constructed in [

Pp := ?dGp + Pp,

p? < p < q? , Xp(? , ? ) := Y p(? , ? ) := {u ? Dp(?; ??) : ?u = 0}, {du : u ? Dp?1(?; d)}.

As a result of the Hodge decompositions proved in [

Lemma 3.4. For each p? < p, p < q? with 1/p + 1/p = 1, the natural integral pairing on ? allows for the identification is an isomorphism. To see that it is one-to-one, assume that w ? Lp(? , ? ) is arbitrary and, using (3.33), decompose w = w1 + w2 with w1 ? Xp(? , ? ) and w2 ? Y p(? , ? ). Then, if u ? Xp (? , ? ) is such that ?(u) = 0, it follows from (2.18) and (3.31)-(3.32) that u, w = u, w1 + u, w2 = 0. Since w was arbitrary, this forces u = 0 and, hence, ? is one-to-one.

To prove that the mapping (3.35) is onto, let f ? Since Xp(? , ? ) is a closed subspace of Lp(? , ? ), the Hahn-Banach extension theorem in concert with Riesz?s representation theorem imply that there exists w ? Lp (? , ? ) such that f (u) = u, w for each u ? Xp(? , ? ). Invoking (3.33), we once again decompose w = w1 + w2 with w1 ? Xp (? , ? ) and w2 ? Y p (? , ? ). Since, as before, u, w2 = 0 whenever u ? Xp(? , ? ), we may conclude that ?(w1) = f . This proves that ? is also onto, hence an isomorphism. Proposition 3.5. For each p ? (p? , q? ), the operator Pp introduced in (3.30) maps Xp(? , ? ) ? be arbitrary.

Pp : Lp(? , ? ) ?? Xp(? , ? ) P2p = Pp, (Pp)? = Pp if 1/p + 1/p = 1.

Proof. The first part in the statement of the proposition follows from the fact that Gp maps Lp(? , ? ) into Dom (Bp). As for (3.36), we first note that thanks to (3.21), in a bounded, linear fashion, and satisfies Based on this, if u ? Lp(? , ? ) and v ? Lp (? , ? ) with p? < p, p < q? , we obtain

? : Xp (? , ? ) u ? u, · ? where in the last step we have used (2.18) and (3.31). Thus, Ppu, v = Ppu, Pp v and, since the last expression is symmetric in u and v, we may ultimately conclude that Ppu, v = u, Pp v . Hence, (Pp)? = Pp . Armed with this, we may now write P2pu, v = Ppu, Pp v = Ppu, v so that P2p = Pp, as desired.

Lemma 3.6. For each p? < p < q? , (3.40)

Dom (Bp) ? Xp(? , ? ) = {u ? Dom (Bp) : Bpu ? Xp(? , ? )}.

Proof. Let u ? Dom (Bp) ? Xp(? , ? ) be arbitrary. Then Bpu = ?du satisfies ?(?du) = 0 and ? ??du = ??? (? ?du) = 0. Thus Bpu ? Xp(? , ? ), proving the leftto-right inclusion in (3.40). To prove the opposite one, assume that u ? Dom (Bp) has the property that Bpu ? Xp(? , ? ). Then 0 = ? ? (?d + d?)u = ? ? d?u and 0 = ?(?d + d?)u = ?(d?u) = (? ?u). Since ? ? ?u = ??? (? ? u) = 0, it follows that ?u ? Ker (Bp) = H?p(? , ? ), by (3.25). From this we may deduce that w := ?u satisfies dw = 0 and that w ? Lq(? , ? ?1) for each p? < q < q? . Membership of u to Dom (Bp) also guarantees that ? ? u = 0. In particular, the integration by parts formula (2.18) applies and gives that ?u, ?u = ?u, w = 0. In turn, this forces ?u = 0 which, further, entails u ? Xp(? , ? ). Lemma 3.7. For each p? < p < q? , (3.41) PpBp = BpPp on Dom (Bp).

Proof. We first claim that if u ? Dom (Bp), then d?Gpu ? Dom (Bp). Indeed, d?Gpu ? Lp(? , ? ), d(d?Gpu) = 0 ? Lp(? , ? +1), (3.42) ?(d?Gpu) = ??? Gpu = ??(I ? Pp)u = ??u ? Lp(? , ? ?1), (d?)(d?Gpu) = d[?(d?Gpu)] = ?d?u ? Lp(? , ? ), (?d)(d?Gpu) = 0 ? Lp(? , ? ), and ? ? d(d?Gpu) = 0, (3.43) ? ? d?Gpu = ???dGpu ? ? ? u + ? ? Ppu = 0 ? ??(? ? dGpu) ? 0 = 0, justifying the claim. Consequently, (3.44) Ppu = u ? ?dGpu ? Dom (Bp) if u ? Dom (Bp), or, in other words, (3.45) Pp : Dom (Bp) ?? Dom (Bp) ? Xp(? , ? ).

Furthermore, for every u ? Dom (Bp), we may write

BpPpu = ?(? u ? d?Gpu)

= d?u + ?du + d?? Gpu = d?u + ?du + d?(Pp ? I)u (3.46) = ?du.

On the other hand, for every u ? Dom (Bp), (3.47) PpBpu = ?dGpBpu + PpBpu = ?d(u ? Ppu) + 0 = ?du, which, in concert with (3.46), proves (3.41).

Lemma 3.8. If p? < p < q? , then for each 0 ? ? n, (3.48) d = d Pp, + Pp, +1 d on Dp(?; d), where Pp, stands for Pp acting on Lp(? , ? ), etc.

Proof. Since Pp, +1(du) = 0 and dPp, u = 0 for every u ? Dp(?; d), based on (3.30), (3.24) and (3.21) we may write

d Pp, u = d?Gp, +1(du) = ??dGp, +1(du) ? ? Gp, +1(du) (3.49) = ?Pp, +1(du) + du, ? u ? Dp(?; d), from which (3.48) follows. (3.57) (3.58) (3.59)

M and 1 < p < ?,

Ap : Xp(? , ? ) ?? Xp(? , ? ) Dom (Ap) := Dom (Bp) ? Xp(? , ? ), Apu := Bpu = ?? u,

? u ? Dom (Ap).

ApPp = PpBp on Dom (Bp). and, for each ? ?/ Spec (Bp), (3.55) on Lp(? , ? ).

Spec (Ap) ? Spec (Bp) (?I ? Bp)?1Pp = Pp(?I ? Bp)?1 = (?I ? Ap)?1Pp Lemma 3.9. For each p? < p < q? , where (3.45) is also used.

Lemma 3.10. For each p? < p < q? , Proof. By relying on Lemma 3.7 and the fact that Ap = Bp on Dom (Bp) ? Xp(? , ? ), on Dom (Bp) we may write

ApPp = BpPp = PpBp,

For a (possibly unbounded) operator T we let Spec (T ) denote its spectrum. Proof. If ? ?/ Spec (Bp), then (?I ? Bp)?1 is invertible on Lp(? , ? ) and, hence, ?I ? Ap is one-to-one. Next, if f ? Xp(? , ? ) ? Lp(? , ? ) is arbitrary, consider w := (?I ? Bp)?1f ? Dom (Bp) ? Lp(? , ? ). If we now set u := Ppw it follows that u ? Dom (Bp) ? Xp(? , ? ) = Dom (Ap) and (3.56)

(?I ? Ap)?1u = ? Ppw ? ApPpw = Pp(?I ? Bp)w = Ppf = f, which proves that ?I ? Ap is onto as well. Hence, (3.54) holds. Then the commutation identities in (3.55) are straightforward consequences of this and Lemma 3.8. Lemma 3.11. Whenever p? < p, p < q? are such that 1/p + 1/p = 1, it follows that

(Ap)? = Ap .

Proof. With p, p as in the statement of the lemma, let u ? Dom (A?p) ? Xp (? , ? ) and set w := A?pu ? Xp (? , ? ). In particular, w, ? = u, Ap? ,

? ? ? Dom (Ap) ? Xp(? , ? ).

Then for every ? ? Dom (Bp) we may write

Bp?, u

Bp?, Pp u = PpBp?, u = = =

Ap(Pp?), u = Pp?, w = ?, Pp w ?, w , where we have used (3.58) and the fact that Pp? ? Dom (Ap) whenever ? ? Dom (Bp). As a consequence of (3.59) we have u ? Dom (Bp?) = Dom (Bp ) and, (3.65) (3.66) (3.67) (3.68) so both operators are invertible if b (?) = 0.

? {0, 1, ..., n} be fixed and consider the spaces (3.62) H := X2(? , ? ), V := {u ? H : du ? L2(? , ? +1)}, where the latter space is equipped with the natural graph norm. Also, consider the sesquilinear form (3.63) a(u, v) := du, dv , u, v ? V.

Then the operator associated with the triplet {H, V , a} as in §2.3 is precisely A2 (i.e., the Stokes operator introduced in (3.50)-(3.51) for p = 2). In particular, A2 generates an analytic semigroup on X2(? , ? ).

Proof. According to the discussion in §2.3, the domain of the operator A associated with the triplet {H, V , a} consists of forms u ? V for which there exists w ? H such that (3.64) du, dv = w, v , ? v ? V.

P2 : D2(?; d) ?? V is onto.

Indeed, the fact that P2 maps the space D2(?; d) into V is clear from Lemma 3.8.

P2, acts as the identity on H, any v ? V = H ? D2(?; d) can be written as v = P2v, proving (3.65).

Returning now to the mainstream discussion, the above analysis shows that if u, w are as in the opening paragraph, then condition (3.64) is equivalent to du, dP2f = w, P2f , ? f ? D2(?; d).

Now, if f ? D2(?; d) is arbitrary, based on this, (3.66) and (3.49) we may write moreover, Bp u = w ? Xp (? , ? ). All in all, u ? Dom (Bp ) ? Xp (? , ? ) = Dom (Ap ) and Ap u = Bp u = w = A?pu. In particular, this shows that (Ap)? ? Ap .

Conversely, fix an arbitrary u ? Dom (Ap ). Hence, u ? Xp (? , ? ) ? Dom (Bp ), which then gives

Ap u, w = Bp u, w = u, Bpw = u, Apw for each w ? Xp(? , ? ) ? Dom (Bp) = Dom (Ap). In turn, this allows us to conclude that u ? Dom (A?p) and A?p = Ap u and, further, that Ap ? A?p. The proof of the lemma is therefore finished.

Remark. For each p? < p < q? ,

Ker Ap = Ker Bp = Hp(? , ? ), w, f

P2w, f = w, P2f = du, dP2f = = = = du, ?P2, +1(df ) + df = du, ?P2, +1(df ) + du, df ?P2, +1(du), df + du, df =

?P2, +1(du) + du, df dP2, u , df = du, df .

In turn, this is equivalent to demanding that du ? D2+1(?; ??) and ?du = w.

Consequently, the domain of A is precisely the collection of forms (3.69) u ? L2(? , ? ), ?u = 0, ? ? u = 0, du ? L2(? , ? +1), ?du ? L2(? , ? ), and Au = ?du = ?? u for each u as in (3.69). That is, A coincides with A2 (introduced in (3.50)-(3.51)), as desired. 3.3. The Maxwell operator. For an arbitrary Lipschitz subdomain ? of p? < p < q? , we introduce the spaces M and and consider the operator

Zp(? , ? ) := {u ? Lp(? , ? ) : du = 0},

W p(? , ? ) := {?u : u ? Dp(?; ??)},

Qp := d?Gp + Pp : Lp(? , ? ) ?? Lp(? , ? ).

Lemma 3.13. For each p? < p < q? , the following hold: (i) (Qp)? = Qp if 1/p + 1/p = 1, and (Qp)2 = Qp; (ii) QpBp = BpQp on Dom (Bp); (iii) Qp : Lp(? , ? ) ? Zp(? , ? ) is onto; (iv) Qp : Dp(?; ??) ? Zp(? , ? ) ? Dp(?; ??) is onto; (v) ? = ?Qp, + Qp, ?1? on Dp(?; ??).

We omit the proof, which can be carried out much as for the case of the Leray projection Pp.

Next, for each p? < p < q? , the Maxwell operator is introduced as the part of Bp in Zp(? , ? ), i.e., as the unbounded operator (3.72) for which (3.73)

Cp : Zp(? , ? ) ?? Zp(? , ? ) Dom (Cp) := Dom (Bp) ? Zp(? , ? ), Cpu := Bpu = ?? u, ? u ? Dom (Cp).

Some of the most basic properties of this operator are summarized below. They parallel those of the Stokes operator discussed in §3.2 and can be proved in much the same fashion.

Lemma 3.14. For each p? < p < q? , the following hold.

(i) CpQp = QpBp on Dom (Bp); (ii) Spec (Cp) ? Spec (Bp) and, for each ? ?/ Spec (Bp), (3.74)

(?I ? Bp)?1Qp = Qp(?I ? Bp)?1 = (?I ? Cp)?1Qp on Lp(? , ? ); (iii)

Zp(? , ? ) ?

= Zp (? , ? ) and (Cp)? = Cp whenever 1/p + 1/p = 1; (iv) Ker Cp = Ker Bp = Hp(? , ? ) so that, in particular, the operator (3.72)(3.73) is invertible if b (?) = 0 .

Finally, we remark that, in the case when p = 2, the Maxwell operator (3.72)(3.73) generates an analytic semigroup on Z2(? , ? ). More specifically, we have the following.

? {0, 1, ..., n} and consider the spaces H := Z2(? , ? ),

V := {u ? H : ?u ? L2(? , ? +1), ? ? u = 0}, where the latter space is equipped with the natural graph norm. Also, consider the sesquilinear form a(u, v) := ?u, ?v , u, v ? V.

Then the operator associated with the triplet {H, V , a} as in §2.3 is precisely the Maxwell operator C2. In particular, C2 generates an analytic semigroup on Z2(? , ? ).

The proof of these results can be carried out much as the proof of Proposition 3.12, with the help of Lemma 3.14.

4. The regularity of differential forms in Dom (Bp)

We first record a number of useful results from [

? zI) : L21(M, ? ) ?? L2?1(M, ? ) is invertible.

U :=

Spec (? ) ? (??, 0], 0? ?n and for ? ?/ U , let ??, be the Schwartz kernel of ? ??I on -forms. In particular, we denote by ??, the associated (volume) Newtonian potential. Also, once a Lipschitz domain ? ? M has been fixed, we define the single layer potential operator on ?? by Set (4.2) (4.4) (4.5) (4.6) (4.7) (4.3) for any f : ?? operators

S?, f (x) := ?? ??, (x, y), f (y) d?y, x ? ? , ? ? . Note that (?

? ?I)S?, = 0 in ? and, as proved in [

1 ?? Dp+1(?; ?), are well-defined and bounded. As a consequence, the operator M?, defined by the equality

1
? 2 I + M?, f = ? ? (d S?, f ),
? f ? X p(??) ,
is well-defined and bounded on X p(??) for each p ? (1, ?). It has also been shown
in [

Our final remark is that the spectrum of Bp acting on Lp(? , ? ) is a discrete
subset Spec (B ) of (??, 0] which is independent of p ? (p? , q? ); cf. [

Uo := U ?

Spec (B ) .

0? ?n Proposition 4.1. Assume that ? ? C \ Uo and that p? < p < q < q? . Then, if u ? Dom (Bp) is such that (?I ? )? u ? Lq(? , ? ), it follows that u ? Dom (Bq). Proof. Denote by f? the extension of f := (?I ? )? u ? Lq(? , ? ) by zero in M \ ? and set ? := (?

? ?I)?1f? ? ? Lq2(? , ? ).

Then the differential form satisfies, thanks to (4.4), (4.5), (4.7), and the fact that d? = ?? ? ?d, w := S?, +dS?, ?1

1 ? 2 I + M?,

?1(? ? d?) ? 2 I + M?, ?1 ?1 ? ? ? ? ? ? S?, 1

1 ? 2 I + M?, ?1(? ? d?)

? ?I)w = 0 in ? , w, d?w, ?dw ? Lq(? , ? ), dw ? Lq(? , ? +1), ?w ? Lq(? , ? ?1), ? ? w = ? ? ?, ? ? dw = ? ? d?.

It follows that w ? ? ? Dom (Bq) ? Dom (Bp) and (?I + Bp)(w ? ?) = f . Consequently, u = (?I + Bp)?1f = w ? ? ? Dom (Bq), as claimed.

Our last result in this section can, informally speaking, be regarded as a statement about the Lp-boundedness of the Riesz transforms d?? ?1, ?d? ?1. Alternatively, it is a statement about the maximal regularity of ?? relative to the decomposition ?? = ?d + d?.

Proposition 4.2. Assume that ? is a Lipschitz subdomain of the manifold M. If p? < p < q? and u ? Dom (Bp), then (4.11)

d?u Lp(? ,? ) + ?du Lp(? ,? ) ? C ? u Lp(? ,? ) for some finite C = C(?? , p) > 0.

Proof. Let the index p and the differential form u be as in the statement of the proposition and set f := ? u ? Lp(? , ? ). Applying Gp to this equality and relying on (3.21) leads to the conclusion that u = Gpf + Ppu. Hence, d?u = d?Gpf and ?du = ?dGpf . Having justified this representation, the estimate (4.11) is a direct consequence of (3.20). and note that, generally speaking,

For an arbitrary, fixed ? ? (0, ?), consider the sector

?? := {z ? C : |arg z| < ? ? ?} ? C |?a + b| ? |?|a + b, uniformly for ? ? ??, a, b ? 0.

Hypotheses. Consider an arbitrary ? ? (0, ?), and arbitrary ? ? ?? and set (5.4) t := 1 = |?|?1/2.

|?| For a fixed ? {0, 1, ..., n}, consider an arbitrary form

f ? C0?(? , ? ) ? L2(? , ? ) u := (?I + B2)?1f ? Dom (B2) ? L2(? , ? ).

Next, fix an arbitrary point x ? ?, an arbitrary sequence of functions that {?j }j?0 such ?0 ? Co?(B(x, 2t)),

?j ? Co? B(x, 2j+1t) \ B(x, 2j?1t) , where the embedding in (5.8) presupposes that p ? 2 (see below).

Going further, assume that there exist and two finite constants C, c > 0 with the property that

2 ? p < q? |?| and decompose (5.5) (5.6) and define (5.7) (5.8) (5.9) (5.10) (5.11) 0 ? ?j ? 1,

1 |??j | ? 2j?1t , ? j=0

?j = 1, f = u = ? j=0 ? j=0 fj , uj , fj := ?j f ? Lp(? , ? ) ? L2(? , ? ), uj := (?I + B2)?1fj ? Dom (B2), for each j ? 0, and such that ,

?
It is illuminating to note that the above are in effect high-energy exponential decay
estimates of the support of the inhomogeneous term, fj . The above decomposition
of f into the pieces fj takes into account the natural high-energy scaling (roughly,
|?||y ? x| if one localizes the problem in y near x), and is inspired by the work
in [

Lemma 5.1. Granted the above conventions and assumptions, for each k ? N there exist two finite constants C, c > 0 depending only on ?, p and the Lipschitz character of ? such that, for each j ? 0, |?| , ? ?B(x,t) ? and, if in addition to the hypotheses made so far we also have (5.13) (5.14) then also (5.15) (5.16) and (5.17) (5.18) , Proof. To prove this, we shall assume (5.7)-(5.12) and proceed in a series of steps starting with Step 1. For each j ? 0, uj ? L2(? , ? ), duj ? L2(? , ? +1),

?uj ? L2(? , ? ?1), ?duj , d?uj ? L2(? , ? ), ? ? uj = 0,

? ? duj = 0, ?uj ? ? uj = fj in ? .

These follow from the definition of uj ? Dom (B2).

Step 2. For any Lipschitz subdomain D of M there exists 2 < qD ? ? with the following significance. For each p ? [2, qD) there exists C > 0 which depends exclusively on p and the Lipschitz character of D such that for any w satisfying w ? Lp(D, ? ), dw ? Lp(D, ? +1),

?w ? Lp(D, ? ?1) and such that ? ? w = 0 on ?D, it follows that (5.19) Rn p1 ? p1? ? C

D D |w|p? dV |w|p dV 1/p? 1/p + R

D |dw|p dV 1/p + R

D |?w|p dV 1/p , where R := diam (D) and, as before, p? := np/(n ? 1).

To justify this, we first recall that, under the assumptions (5.18), the estimate w B1p/,pp(D,? ) ? C

w Lp(D,? ) + dw Lp(D,? +1) + ?w Lp(D,? ?1)
has been established in [

B1p/,pp(?) ? Lp? (?) , p? = Step 3. For each k ? N there exist C, c > 0 such that 1/p? 1/p? |?| |?| for each j ? 0.

To prove this, fix a function (5.22) and (5.23) (5.24) (5.25)

? ? Co?(B(x, t)) with ? ? 1 on B(x, t/2) and |??| ? C t?1 and use (5.19) in the context when D := ? ? B(x, t) and w := ?uj . That this applies is guaranteed by the fact that ? ? (?uj ) = 0 on ?[? ? B(x, t)] and ?uj Lp(? ?B(x,t),? ) ?

uj Lp(? ?B(x,t),? ), d(?uj) Lp(? ?B(x,t),? +1) ? C t?1 uj Lp(? ?B(x,t),? ) ?(?uj) Lp(? ?B(x,t),? ?1) ? C t?1 uj Lp(? ?B(x,t),? ) + duj Lp(? ?B(x,t),? +1), + ?uj Lp(? ?B(x,t),? ?1). Also, the Lipschitz character of D is controlled by that of ? and, hence, there is no loss of generality in assuming that qD = q? . Thus, we may write + 1/p? + C t 1/p

C |?|1/2 1/p

, (5.26) and (5.29) where the last step is based on (5.11). This proves (5.22) with t replaced by t/2. This, however, is easily remedied by carrying out the same program with t replaced by 2t in (5.7), (5.24) and the definition of D. Going further, Ho¨lder?s inequality and the support condition on fj give ? 1/p = ?

? ?B(x,2j+1t) C 2jntn p1 ? p1?

1/p ? 1/p? .

In concert with (5.26), this proves a version of (5.23) with t replaced by t/2 though, as before, this aspect is easily fixed.

Step 4. For each k ? N there exist C, c > 0 such that ? 1/2 | | ? 1/2 | | for each j ? 0.

To justify these inequalities, pick a function ? as in (5.24) and invoke (5.19) for D := ? ? B(x, t) and w := ??uj . Note that ? ? (??uj ) = ???? (? ? uj ) = 0 on ?[? ? B(x, t)] and

, ??uj Lp(? ?B(x,t),? ?1) ? ?uj Lp(? ?B(x,t),? ?1), d(??uj ) Lp(? ?B(x,t),? ) ? C t?1 ?uj Lp(? ?B(x,t),? ?1)

+ d?uj Lp(? ?B(x,t),? ),

1/p? + + ? ? where we have utilized (5.11) and (5.12) in the last step. From this, we readily obtain (5.32) 1/p? 1/p 1/p?

. 1/p ? ?B(x,t) and (5.35) (5.39)

Assume that j ? 3 and, once again, pick a function ? as in (5.24). Thus, ? ? (??duj) = ????(? ? duj) = 0 on ?[? ? B(x, t)].

Also, since ?fj = 0 for j ? 3, ??duj = ??? uj ? ?d?uj = ??(? uj ? fj) ? ?d?uj = ?? ?uj ? ?d?uj d(??duj) = ??O(|??||uj|) ? ??duj + O(|??||d?uj|). d(??duj) Lp(? ,? +1) ?

C|?|t?1 uj Lp(? ,? ) + |?| duj Lp(? ,? +1) +C t?1 d?uj Lp(? ,? ). ??duj Lp(? ?B(x,t),? ) ? ?duj Lp(? ?B(x,t),? ), ?(??duj) Lp(? ?B(x,t),? ) ? C t?1 ?duj Lp(? ?B(x,t),? ), the estimate (5.19) is applicable to D := ? ? B(x, t) and w := ??duj (assuming that qD = q? , which can be arranged). As a result, we have

, 1/p? where the last step utilizes (5.12) and (5.11). In turn, from (5.41), Ho¨lder?s inequality and rescaling we readily obtain (5.42) ? ?

C e?c 2j tn p1? ? p1 C e?c 2j ? .

1/p Going further, we write d?uj = ?? uj ? ?duj = fj ? ?uj ? ?duj and, consequently, d?uj = ??uj ? ?duj on ? ? B(x, t) if j ? 3. Hence, based on this, (5.42) and (5.22), we may estimate 1/p? . 1/p 1/p? (5.44) and (5.45) (5.46) (5.47) ? ? ? ? ? C = C|?| ? C|?|

? ? j=0 ? ? C ? C ? ? j=0 ? j=0

C f Lq(?) . ? 1/2 du Lq(? ,? +1) + |?|1/2 ?u Lq(? ,? ?1) ? C f Lq(? ,? ).

| |

Given q ? (p, p?], select ? ? (0, 1] such that 1/q = (1 ? ?)/p + ?/p?. From (5.11) and (5.23) we then obtain |?| uj Lq(B(x,t)?? ,? ) |?| uj Lp(B(x,t)?? ,? )

|?| uj Lp? (B(x,t)?? ,? ) 1?? ? C e?c 2j tn p?? ? p? C e?c 2j tn q1 ? p1 fj Lp(? ,? ) fj Lp(? ,? ).

Now, with ?E g dV := [measure (E)]?1 E g dV , and with M denoting the HardyLittlewood maximal operator, Fubini?s Theorem and (5.46) allow us to write ? |u|q dV 1/q ? ? ? ?B(x,t) |u|q dV dVx 1/q Clearly, (5.42)-(5.43) prove (5.34)-(5.35).

Step 6. Granted (5.4)-(5.12), for each q ? (p, p?] there exists C = C(?? , q) > 0 such that |?| u Lq(? ,? ) ? C f Lq(? ,? ) ? ? ?B(x,t)

|u|q dV ? ? ?B(x,t) |uj |q dV 1/q q

1/q dVx 1/q q

1/q C e?c 2j 2jn/p ?? ?B(x,2j t)|f |p dV 1/p q

1/q

M (|f |p) 1L/qp/p(?) This proves (5.44). The estimate (5.45) is then justified in a similar manner, by relying on (5.11) and (5.29).

Step 7. The estimate (5.35) also holds if 0 ? j ? 3.

It suffices to show that there exists C = C(?? , p) > 0 such that

?duj Lp? (? ,? ) + d?uj Lp? (? ,? ) ? C fj Lp? (? ,? ) for each j ? 0. To this end, we first note that the conclusion in Step 6 (with q = p?) applied to uj , fj in place of u, f yields |?| uj Lp? (? ,? ) ? C fj Lp? (? ,? ),

Next, recall that uj ? Dom (B2) and (?I ? )? uj = fj ? C0?(? , ? ) ? Lp? (? , ? ). Since our current assumptions imply 2 < p? < q? , Proposition 4.1 guarantees that uj ? Dom (Bp? ). Consequently, (4.11) and (5.49) allow us to write d?uj Lp? (? ,? ) + ?duj Lp? (? ,? ) ? C ? uj Lp? (? ,? )

= C ?uj ? fj Lp? (? ,? ) ? C|?| uj Lp? (? ,? ) + C fj Lp? (? ,? )

In this section we shall make use of Lemma 5.1 in order to prove resolvent estimates for the Hodge Laplacian. To state our main result in this regard, recall the definition of the sector ?? from (5.1). Also, given a Lipschitz domain ? ? M, dim M = n, recall the critical exponents p? , q? from (3.1), and set (6.1) q?? := nn?q? 1 , (q?? ) := 1 ? q1?? .

Theorem 6.1. Let ? ? M be a Lipschitz domain, and fix ? ? (0, ?). Then for each ? ? ?? and each ?1

? {0, 1, ..., n} and p ? (p? , q? ) ? C |?|?1/2 f Lp(? ,? ), (6.2) the operator (6.3) (6.4) (6.5) (6.6) ?I + Bp : Dom (Bp) ? Lp(? , ? ) ?? Lp(? , ? ) has a bounded inverse. In addition, there exists C = C(?? , ?, p) > 0 such that (?I + Bp)?1f Lp(? ,? ) ? C |?|?1 f Lp(? ,? ), d(?I + Bp)?1f Lp(? ,? ?1) + ?(?I + Bp)?1f Lp(? ,? +1) d?(?I + Bp)?1f Lp(? ,? ) + ?d(?I + Bp)?1f Lp(? ,? ) ? C f Lp(? ,? ), for any ? ? ?? and any f ? Lp(? , ? ).

Moreover, for each p ?

(q?? ) , q?? the mapping (?I + B2)?1 : L2(? , ? ) ? L2(? , ? ) extends to a bounded, linear operator from Lp(? , ? ) into itself which continues to satisfy (6.4) and (6.5). Proof. Consider first the case when p ? (2, q? ). In this scenario, for an arbitrary ? ? ??, the fact that the operator (6.3) is one-to-one follows trivially from the corresponding statement for p = 2 (dealt with in §3). To see that this operator is also onto, let f ? Lp(? , ? ) ? Lp(? , ? ) and consider u := (?I + B2)?1f ? Dom (B2). Thanks to Proposition 4.1, we have that u ? Dom (Bp) and (?I +Bp)u = f , which proves that the operator (6.3) is indeed onto.

Turning our attention to (6.4)-(6.5), we note that it suffices to prove these estimates for an arbitrary f ? C0?(? , ? ). With the notation and conventions introduced in §5, these are going to be consequences of (5.44) and (5.45), provided we show that the index q appearing there can be chosen arbitrarily in (2, q? ). In turn, by virtue of the inductive bootstrap argument in Lemma 5.1, this latter condition will hold as soon as we prove that (5.11)-(5.12) are valid for the choice p = 2.

The argument utilizes a Caccioppoli-type inequality along with a trick we have
learned from [

|?uj |2 dV = ? ? ? fj , uŻj dV.

From this we may further deduce, based on (5.2) and the Cauchy-Schwarz inequality, that (6.9)

|?| uj L2(? ,? ) + |?|1/2 duj L2(? ,? +1) + |?|1/2 ?uj L2(? ,? ?1) ? C fj L2(? ,? ). Next, with t retaining the same significance as before, i.e. t := |?|?1/2, pick a new family of functions {?j}j?3 such that ?j ? Co?(B(x, 2j?2t)),

Taking the L2-pairing of ?j2uŻj with both sides of (5.17) and keeping in mind that ?j ?j = 0 for each j ? 3 we may write, based on integrations by parts that (6.10) (6.11) From this and (5.2) we then obtain that ? ? ?

? ?

? ? ? C

?

O |??j ||uj | |?j ||duj | + |?j ||?uj | dV. |?| |??j ||uj | |?j ||duj | + |?j||?uj | dV, which, via Cauchy-Schwarz?s inequality and a standard trick that allows us to absorb like-terms with small coefficients in the left-hand side, further gives ? ? in view of the generic, elementary implication f ? g ? 21 f ? f we now assume that the original cutoff functions {?j}j also satisfy |uj |2|e?j?j |2 dV ? 4 |uj |2 dV, ? j ? 3, ? 2 g . If In a first stage, this yields (6.16) (6.17) then, further, (6.19) (6.20) (6.21) (6.22) it follows from (6.15) that ?j ? c 2j, and from (6.17) that ? ?B(x,2j?3t)

? |e?j |2 |uj |2 dV ? 4 |uj |2 dV, ? j ? 3.

In concert with (5.11), the above analysis shows that there exist C, c > 0 such that |?| ? ?B(x,2j?3t) |uj |2 dV

C |?| e?c 2j ?

|uj |2 dV, ?j ? 1 on B(x, 2j?3t), |?| |?| ? All in all, from (6.9) with 0 ? j ? 3 and (6.20) with j > 3, we obtain ? ?B(x,t) |?| ? Furthermore, (6.13), (6.9), (6.18) and (6.20) also imply that, for some C, c > 0, |uj |2 dV ? C |duj |2 + |?uj |2 dV ? C ? |?| ? ,

j ? 3. ?

1 |uj |2|e?j ?j ? 1|2 dV ? 4 ? |uj |2|e?j ?j |2 dV, ? j ? 3, ? |?|

? |?| ? B(x,t/2)?? |d?uj |2 + |?duj |2 dV ? C e?c 2j which corresponds to (5.12) written for p = 2 and t/2 in place of t (the latter condition being just a minor technicality, easily addressed via rescaling). In turn, if ? is as in (5.24), (6.23) will be a simple consequence of the estimate

B(x,t)?? ? which, so we claim, is valid for each j ? 0. In order to justify (6.24), we shall first establish the estimate |d(??uj )|2 + |?(?duj )|2 dV ? C e?c 2j by (6.9). As this implies (6.25) for small j?s, we can assume for the remainder of the proof that j ? 3. In particular, fj ? 0 on B(x, t). Next, multiply by ? both sides of the equality ? uj = ?uj ? fj to get ?? uj = ??uj and write B(x,t)?? |?? uj |2 dV = |?|2

B(x,t)??

|?uj |2 dV

There remains to prove the p = 2 version of (5.12), a task to which we now turn. In fact, we aim at showing that (6.25) (6.26) (6.27) (6.28) so that (6.29) (6.30) (6.31)

B(x,t)?? ? where in the last step we have used (6.20). This finishes the proof of (6.25).

? |?|2 |uj |2 dV ? C e?c 2j

O |??|2 |duj |2 + |?uj |2

dV |duj |2 + |?uj |2 dV |d(??uj ) + ?(?duj )|2 dV B(x,t)?? =

B(x,t)?? and, via an integration by parts, |d(??uj )|2 + |?(?duj )|2 = |d(??uj ) + ?(?duj )|2 ? 2 Re d(??uj ) , ?(?duj ) d(??uj) , ?(?duj ) dV = 0 since d2 = 0 and ? ?(?duj ) = ?(? ?duj ) = 0 on ?[? ?B(x, t)]. Thus, all in all, (6.24) is a consequence of (6.30), (6.29) and (6.31), and this finishes the proof of (5.12) when p = 2. In turn, as explained earlier, this concludes the proof of (6.4)-(6.5) in the case when 2 < p < q? .

As regards the estimate (6.6), we may invoke Proposition 4.2, (6.4) and the fact that (? ?I + Bp)?1f = ?(?I + Bp)?1f + f in order to justify it in the case when 2 < p < q? . Now, the case when p? < p < 2 follows from what we have proved so far and duality; cf. Proposition 3.3. Since the case p = 2 is implicit in the above analysis, this finishes the proof of (6.4)-(6.6). Finally, the last claim in the statement of the theorem is a consequence of what we have proved up to this point and Lemma 5.1. This completes the proof of Theorem 6.1.

Theorem 7.1. If ? ? M is a Lipschitz domain and 0 ? ? n, then the operator ?Bp generates an analytic semigroup in Lp(? , ? ) for each p ? (p? , q? ). More specifically, for each ? ? (0, ?/2) there exists an analytic map

Tp : ?? ?? L Lp(? , ? ), Lp(? , ? ) such that the following hold: (7.1) (7.2) (7.3) (7.4) (7.5) (7.6) (7.7) (7.8) lim Tp(z)f = f in Lp(? , ? ), ? f ? Lp(? , ? ), zz???0? Tp(z1 + z2) = Tp(z1)Tp(z2),

? z1, z2 ? ??, Dom (Bp) = ?Bpu = lim t?0+ u ? Lp(? , ? ) : lim t?0+

Tp(t)u ? u exists in Lp(? , ? ) ,

t Tp(t)u ? u t

for each u ? Dom (Bp).

Furthermore, whenever p ? (q?? ) , q?? , the semigroup (7.1) further extends to a mapping Tp : ?? ?? L Lp(? , ? ), Lp(? , ? ) which continues to satisfy (7.3). For this range of p?s, this extension satisfies (Tp)? = Tp , 1/p + 1/p = 1.

Proof. The first part follows from Theorem 6.1 and the standard theory; cf., e.g.,
Theorem 5.2 on p. 61 of [

e?tBp := Tp(t), t > 0.

Corollary 7.2. Under the hypotheses of Theorem 7.1, for each t > 0 and p ? (p? , q? ) one has

Pp e?tBp = e?tBp Pp,

Qp e?tBp = e?tBp Qp.

Proof. This is an immediate consequence of Theorem 7.1, Lemma 3.7 and (ii) in Lemma 3.13. Theorem 7.3. For a Lipschitz domain ? ? M, 0 ? ? n, and 1 < p < ?, let Ap denote the Stokes operator as introduced in §3.2. Then ?Ap generates an analytic semigroup e?tAp

on the space Xp(? , ? ) provided p ? (p? , q? ).

t>0

Furthermore, for each t > 0, and whenever p ? (p? , q? ).

e?tAp ? = e?tAp ,

1/p + 1/p = 1,

Pp e?tBp = e?tAp Pp on Lp(? , ? ), Proof. Fix some ? ? (0, ?/2) and assume that p ? (p? , q? ). Using Lemma 3.10 and Theorem 6.1, for each f ? Xp(? , ? ) we may then write (?I + Ap)?1f Lp(? ,? ) (?I + Ap)?1Ppf Lp(? ,? ) Pp(?I + Bp)?1f Lp(? ,? ) (7.11)

? uniformly for ? ? ??. Consequently, ?Ap generates an analytic semigroup on Xp(? , ? ) whenever p ? (p? , q? ).

Finally, (7.9) and (7.10) follow readily from this, Lemma 3.8 and Lemma 3.14, completing the proof of the theorem.

C|?|?1 f Lp(? ,? ),

Theorem 7.4. If ? ? M is a Lipschitz domain, 0 ? ? n, 1 < p < ?, and if Cp denotes the Maxwell operator, then ?Cp generates an analytic semigroup

on the space Zp(? , ? ) whenever p ? (p? , q? ). Moreover, for each e?tCp t > 0, (7.12) and (7.13) e?tCp ? = e?tCp ,

1/p + 1/p = 1,

Qp e?tBp = e?tCp Qp on Lp(? , ? ), for each p ? (p? , q? ).

Theorem 7.5. Fix 0 ? ? n and suppose that ? ? M is a Lipschitz domain for which b = 0. Then for each p ? (p? , q? ), the analytic semigroups generated by the operators ?Ap, ?Bp and ?Cp, respectively, on Xp(? , ? ), Lp(? , ? ) and Zp(? , ? ) are bounded.

Proof. This follows from Theorems 7.1, 7.3 and 7.4, given that under the current topological assumptions the operators Ap, Bp and Cp are invertible.

In closing, we would like to point out that, as an obvious corollary of what we have proved so far, similar results are valid for the Hodge duals of the operators Ap, Bp and Cp (i.e., for ?Ap?, ?Bp? and ?Cp?). For example, corresponding to the Hodge dual of Bp, ?,? defined as an unbounded operator on Lp(? , ? ) with domain {u ? Dp(?; d)?Dp(?; ?) : du ? Dp+1(?; ?), ?u ? Dp?1(?; d), ? ?u = 0, ? ??u = 0}, generates an analytic semigroup whenever p? < p < q? . We leave the details for the remaining operators to the interested reader.

This work was completed during several visits at the Universit´e Aix-Marseille 3, France, and the University of Missouri-Columbia, USA. The authors gratefully acknowledge the hospitality of these institutions. The authors are also indebted to the referee for his/her careful reading of the paper and useful suggestions.