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MATHEMATICS OF COMPUTATIONVolume 73, Number 246, Pages 659689S 0025-5718(03)01550-3Article electronically published on October 17, 2003

SUBSTRUCTURING PRECONDITIONERS

FOR THE THREE FIELDS

DOMAIN DECOMPOSITION METHOD

SILVIA BERTOLUZZA

Abstract. We study a class of preconditioners based on substructuring, forthe discrete Steklov-Poincare op erator arising in the three fields formulation of

domain decomposition in two dimensions. Under extremely general assump-tions on the discretization spaces involved, an upper bound is provided on thecondition number of the preconditioned system, which is shown to grow at most

as log(H/h)2 (H and h denoting, respectively, the diameter and the discretiza-tion mesh-size of the subdomains). Extensive numerical testsperformed on

both a plain and a stabilized version of the methodconfirm the optimalityof such bound.

1. Introduction

The use of nonconforming domain decomposition methods is becoming increas-ingly popular. This is due to several factors: the possibility of easily coupling dif-ferent discretizations in different subdomains without the need for imposing strongmatching conditions allows us to employ without any adaptation the technologiesdeveloped for treating the problem with a single-domain approach. As an example,in an adaptive strategy, refinement can be carried out in each subdomain inde-

pendently. Moreover, it is possible to use a different type of discretization (finiteelements, spectral methods, wavelets) in the different subdomains, and thereforeto employ, in each subdomain, the best-suited method (spectral methods wherethe solution is expected to be very smooth, finite elements where a complicatedgeometry requires it, wavelets where isolated singularities in a regular backgroundare expected).

Among the different nonconforming domain decomposition method, we considerhere the three fields formulation, as proposed by F. Brezzi and L.D. Marini in [14].Differently from other nonconforming formulations [3, 18] of domain decomposition,weak continuity is not imposed by requiring that the jump across the interface beorthogonal to some multiplier space, but by introducing a space h, approximatingthe traces of functions in H1() on the interface . The elements of h are to beused as a Dirichlet boundary condition imposed by means of Lagrange multipli-

ers for the functions in the subdomains. As with other nonconforming methods,great freedom is left in the choice of the discretization in the subdomains which,

Received by the editor November 6, 2000 and, in revised form, February 22, 2002.

2000 Mathematics Subject Classification. Primary 65N55, 65N22.Key words and phrases. Three fields formulation, domain decomposition, stabilized methods,

preconditioning.

c2003 American Mathematical Society

659


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660 SILVIA BERTOLUZZA

after applying a suitable stabilization method if needed [2, 7], can be chosen inde-pendently from each other. The disadvantage of introducing a third unknown fieldis compensated for by the fact that all subdomains are treated exactly in the sameway, which results in an easier implementation and possibly, when considering theparallelization of the method, in an easier balancing of the load between processors.

We deal, in this paper, with the problem of the efficient solution of the linearsystem arising when applying such a method. By applying a Schur complementtechnique, the solution of the resulting discrete problem can be reduced to thesolution of an equation on the trace unknown, involving a Steklov-Poincare typeoperator which can be shown to be a pseudo-differential operator of order one.The resulting linear system will then be solved by an iterative procedure, at eachstep of which there will be the need of computing the action of such operator ona given function. This reduces to solving independent Dirichlet problems on thesubdomains, a task which can be carried out in parallel, subdomain by subdomainin a completely independent way.

In order to have an efficient scheme for solving the equation for the interfaceunknown, we will then need two things:

efficient solvers for the local Dirichlet problems on the subdomains; a preconditioner for the discrete Steklov-Poincare operator so that the num-

ber of iterations is kept as low as possible.

We will concentrate here on the second issue, by analyzing and testing a classof preconditioners for the discrete Steklov-Poincare operator. The approach thatwe will follow is the substructuring one, proposed by J.H. Bramble, J.E. Pasciakand A.H. Schatz [9] in the framework of conforming domain decomposition andalready applied in the framework of nonconforming methods to the mortar methodin [1]. This consists in decomposing the trace space h in a direct sum of a coarse

subspace (essentially linked to the geometrical decomposition of in subdomains)and of suitable local subspaces and then considering the related block-Jacobi typepreconditioners.

From the theoretical point of view, we study the dependence of the conditionnumber of the resulting preconditioned matrix on the mesh size of the discretiza-tions, aiming at giving bounds which, under suitable assumptions, are independentof the number and size of the subdomains. In particular we provide an upper boundwhich applies to a wide variety of approximation spaces, including finite elementsand wavelets. Though only suboptimal in terms of the mesh size on the skele-ton (the condition number of the preconditioned matrix grows like (log(H/h))2,H and h being, respectively, the diameter of the subdomains and the mesh sizeof the corresponding discretization), the estimate obtained is uniform with respectto the number and size of subdomains. The resulting method is then potentially

almost perfectly scalable. Moreover this approach has the advantage over otherasymptotically better preconditioners for the three fields formulation ([7]) of be-ing itself naturally computable in parallel (each edge in the skeleton being treatedindependently of the others).

We conclude the paper by reporting a series of extensive numerical tests, per-formed on both the plain and on a stabilized formulation as proposed in [7]. Theresults of the tests are in agreement with the theory and confirm the polylogarithmicgrowth (with respect to the mesh size of the discretization) of the preconditioned


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TH E THREE FIE LDS DOMAIN DE COMP OS ITION METHOD 661

system, as well as the independence (asymptotically) with respect to the numberand size of the subdomains.

2. Domain decomposition method: the three fields formulation

Let R2 be a convex polygonal domain. We will consider the following simplemodel problem: given f L2(), find u satisfying

(1) 2

i,j=1

xj

ai,j(x)

u

xi

+ a0(x)u = f in , u = 0 on .

We assume that for almost all x we have 0 a0(x) R and that the matrix(aij(x))i,j=1,2 is symmetric positive definite, with smallest eigenvalue > 0 andlargest eigenvalue , , independent of x.

In a domain decomposition framework, we will consider the three fields formula-

tion ([14]) of such a problem. Let = Kk=1

k, k polygons, which, for simplicity,we will assume to be quadrangles. We set k = k, =

k k being the skele-

ton of the decomposition. Remark that differently from [14] here we define sothat it includes the external boundary of the domain , so that there is no needof distinguishing between interior subdomains and subdomains which are adjacentto the boundary . This results in a simplification both on the notational andon the implementation point of view, since all subdomains are treated in the sameway. We will make the following regularity assumptions on the subdomains k:

(A1) The subdomains are regular in shape and the geometrical decomposition isgraded, that is,(a) there exists a positive constant c0 such that, for all k, k contains a

ball of diameter c0Hk, it is contained in a ball of diameter Hk, andthe length of each side is bounded from below by c0Hk; moreover

any interior angle satisfies 0 < c1 < < c2 < (c0, c1, and c2independent of k);

(b) there exists a positive constant c3 such that, if , k are such that|k | > 0, then it holds that

Hk/H c3.For technical reasons, we also need to make the following assumption (needed

for the proof of (25) in the following), which in many cases can be shown to beactually a consequence of the previous one:

(A2) There exists a constant C0 such that for all y, denoting by y the line ofequation x2 = y and setting Ky = {k : length(y k) > 0}, it holds that

kKy Hk C0.We are interested here in explicitly studying the dependence of the estimates

that we are going to prove on the number and size of the subdomains. To this end,in the following we will employ the notation A B (resp. A B) to say thatthe quantity A is bounded from above (resp. from below) by cB, with a constantc independent of k and of the Hks, as well as of any mesh size parameter. Theexpression A B will stand for A B A.

For each k we let the norm and seminorm of H1(k) be defined as usual byu2H1(k) =

k

|u|2 dx + k |u|2 dx and |u|2H1(k) = k |u|2 dx. On the


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boundary k we let the norm of H1/2(k) be defined by

H1/2(k) = inf

uH1

(k)u= on k u

H1(k).

For all s in ]0, 1[, we define the Hs(k) seminorm by

(2) ||2Hs(k) =k

k

((x) (y))2|x y|2s+1 ds(x) ds(y).

The space H1/2(k) is defined as the dual of H1/2(k).

For e k segment of extrema a and b, we define the Hs(e) seminorm as usualby the integral expression (2) where the integrals are taken over e. On e we will

also consider the spaces Hs0 (e) (s = 1/2) and H1/200 (e) of functions whose extensionby zero is in Hs(k) (s = 1/2) and H1/2(k), respectively, which we will equipwith the norm

2H1/200 (e)

= ||2H1/2(e) +e

|(x)|2|x a|2s ds(x) +

e

|(x)|2|x b|2s ds(x).

We recall that for s < 1/2 the two spaces Hs(e) and Hs0 (e) coincide, and thetwo corresponding norms are equivalent. However, the constant in the equivalencedepends a priori on Hk and on s. In particular it explodes as s converges to 1/2.The behaviour of such a constant as s approaches the limit value 1/2 (see (44)) willplay a key role in the forthcoming analysis.

Remark also that, for H1/200 (e), letting H1/2(k) denote the functioncoinciding with on e and identically vanishing on k \ e, it holds that(3)

H1/200 (e)

||H1/2(k),

the constant in the equivalence being uniformly bounded, provided that the ratiobetween the length ofk and the length ofe is uniformly bounded.

The functional setting for the three fields domain decomposition method is givenby the following spaces:

(4) V =Kk=1

H1(k), =Kk=1

H1/2(k),

and

(5) = { L2() : there exists u H10 (), u = on } = H10 ()|,respectively equipped with the norms:

(6) uV =

k

uk2H1(k)1/2

, =

k

k2H1/2(k)1/2

,

and (see [4])

(7) = inf uH10()u= on

uH1()

k

||2H1/2(k)1/2

.


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Let ak : H1(k) H1(k) R denote the bilinear form corresponding to thelinear operator considered:

ak(w, v) =

k

2

i,j=1

aij(x)w

xi

v

xj+ a0(x)wv

dx.

Under the assumptions made on the matrix (aij(x))i,j=1,2 and on a0, the bilinearforms ak are uniformly H1(k)-continuous, and semidefinite. More precisely, thereexist positive constants and L independent of k and Hk such that for all w, v inH1(k)

|w

|2H1(k)

ak(w, w),

|ak(w, v)

| L

w

H1(k)

v

H1(k).

The three fields formulation of equation (1) is the following: find (u,,) V such that

(8)

k, vk H1(k), k H1/2(k) :

ak(uk, vk) k

vkk ds =

k

f vk dx,k

ukk ds k

k ds = 0,

and :k

kk ds = 0.

It is known that this problem admits a unique solution (u,,) where u is indeedthe solution of (1) and such that

(9) k =uk

kaon k, = u on ,

where ka denotes the outer conormal derivative to the subdomain k. Remark thatin such a formulation the homogeneous Dirichlet boundary condition is embeddedin the definition of the space .

Equation (8) can be discretized by a Galerkin scheme: after choosing discretiza-

tion spaces Vh =K

k=1 Vkh

Kk=1 H

1(k), h =K

k=1 kh

Kk=1 H

1/2(k)and h , we consider the problem: find (uh, h, h) Vh h h, such that


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one has

(10)

k, vkh Vkh , kh kh :

ak(ukh, vkh)

k

vkhkh ds =

k

f vkh dx,k

ukhkh ds

k

khh ds = 0,

and h h :k

kkhh ds = 0.

Existence, uniqueness and stability of the solution of (10) rely on the validity of

suitable inf-sup conditions. More precisely we can make the following assumption:(A3) The spaces Vkh ,

kh and h are such that

(a) kh contains the constant functions;(b) the following compatibility condition between Vkh and

kh holds uni-

formly in k:

(11) inf kh

kh

supukhV

kh

kkhu

kh ds

ukhH1(k)khH1/2(k) > 0;

(c) the following compatibility condition between h and h holds:

(12) inf hh suphhk k

khh ds

hh > 0.Remark that (A3)(a) implies that for all vkh ker Bkh = {vkh Vkh :

kv

kh

kh =

0 kh kh} it holds that

kvkh ds = 0 and then it is not difficult to see that a

is elliptic on ker Bkh. Then, by a well-known argument ([11]), there exists a uniquesolution to problem (10), which converges with an optimal rate to the solution of(8).

The linear system stemming from (10) takes the form

(13)

A BT 0B 0 CT

0 C 0

uhh

h

=

fh0

0

,

(uh, h, and h being the vectors of the coefficients of uh, h and h in the baseschosen for Vh, h and h, respectively). By a Schur complement argument thesolution of (13) can be reduced to a system in the unknown

hof the form

(14) CA1CT h

= CA1

fh

0

, C = [ 0 C ], A =

A BT

B 0

.

The matrix S = CA1CT does not need to be assembled. The system (14) canrather be solved by an iterative technique (like for instance a conjugate gradient


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method) and therefore only the action of S on a given vector needs to be imple-mented. Multiplying by S implies the need for solving a linear system with matrixA. This reduces, by a proper reordering of the unknowns, to independently solv-ing K discrete Dirichlet problems with Lagrange multipliers in the K subdomains.Moreover under the above assumptions it is possible to prove that the bilinear forms : h h R,(15) s(h, h) =

T

hS

h, S = CA1CT,

corresponding to the Schur complement matrix S, satisfies the following continuityand coercivity assumptions:

Lemma 2.1. If (A3) holds, then the bilinear form s is continuous and coercivewith respect to the norm:

s(h, h) hh,(16)s(h, h) h2.(17)

Assumption (A3) is, in practice, quite restrictive. In particular the requirementthat both (A3)(b) and (A3)(c) hold simultaneously poses some serious limitationon the choice of the discretization spaces for the three unknowns u, and . Inorder to be more free in such a choice, and more specifically in order to be allowedto choose the discretization space for the interface unknown independently ofthe choice of the other two discretization spaces, it is also possible to consider astabilized version. Different ways have been proposed for stabilizing the three fieldsformulation ([2, 12, 13]). We consider here the approach of [7, 4]. For all k leta bilinear form [, ] 1

2,k : H

1/2(k) H1/2(k) R be given, satisfying thefollowing bounds: for all , H1/2(k)(18) [, ] 1

2,k 0, [, ] 1

2,k B||1/2,k ||1/2,k ,

for B positive constant independent of k and Hk.

We consider then the following discrete stabilized formulation: find uh Vh,h h, h h such that it holds that(19)

k, vkh Vkh , kh kh :

ak(ukh, vkh) + [u

kh, v

kh] 1

2,k

k

vkhkh ds [h, vkh] 1

2,k =

k

f vkh dx,k

ukhkh ds

k

khh ds = 0,

and h h :

k

[ukh, h] 12 ,k +

k

kkhh ds +

k

[h, h] 12 ,k = 0,

where > 0 is a parameter independent of the choice of the discretization spaces.Such formulation is consistent with the original continuous problem, i.e., by sub-stituting the solution (u,,) of (8) at the place of (uh, h, h) in (19), we obtainan identity.

The assumptions needed to guarantee existence and uniqueness of the solutionof the discrete problem, as well as an error estimate, can now be made on the one


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hand on the spaces Vh and h and on the other hand in a completely independentway, on the action of the bilinear forms [, ] 1

2,k on h|k. For example we can

replace assumption (A3)(c) with(A3)(c) There exists a positive constant b such that for all h h it holds forall k that

[h, h] 12,k b|h|2H1/2(k).

It is then possible to show that the discrete problem (19) admits a unique solutionsatisfying an optimal error estimate ([4]). It is beyond the goals of this paper tothoroughly discuss the construction of suitable bilinear forms [, ] 1

2,k. We recall that

this can be done with the aid of multiscale techniques and we refer to [ 7, 5, 10, 6]for a detailed analysis of possible solutions.

The linear system stemming from such a problem takes the following form thistime:

(20) A BT DT

B 0 C

T

D C E

uhhh

= fh

00 ,

with A = A + F (the matrices D, E and F deriving from the stabilizing terms).Again, the solution of (20) can be reduced to a system in the unknown

h, this

time taking the form

(21) Sh

:=DA1DT + E

h= DA1

fh

0

with

(22) A =

A BT

B 0

, D = [ D C ].

Once again we let s : h h R be the bilinear form corresponding to theSchur complement matrix S(23) s(h, h) =

T

hS

h,

and also for the stabilized formulation the following holds ([ 4]):

Lemma 2.2. Under assumptions (A3)(a), (b), (c) the bilinear forms is continuousand coercive with respect to the norm:

s(h, h) hh,(24)s(h, h) h2.(25)

3. Preconditioning by substructuring

For both original and stabilized three fields formulation, we end up with anequation on the interface space h of the form

Sh

= fh

corresponding, through a relation of the form (23), to a bilinear form s : hh R verifying for all h, h h that(26) |s(h, h)| hh, s(h, h) h2.Our aim is to provide a preconditioner for the matrix S. The approach that we willfollow is the one proposed, in the framework of conforming domain decomposition,


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by J.H. Bramble, J.E. Pasciak and A.H. Schatz in [9] and already studied in [1] forthe mortar method case.

Let us consider a splitting of the interior part of the skeleton \

as the disjointunion of M segments ei of extrema ai and bi

\ =Mi=1

ei,

(corresponding, for instance, to the splitting of as the union of segments of theform k ). For each k let Ek individuate those segments that, together with k, build up the boundary of the subdomain k:

k \ =iEk

ei, with Ek = {i : |ei k| > 0}.

We make the following assumption:

(A4) There exists a constant c4 such that it holds that

supk

supiEk

|k||ei| c4.

Note that (A4) implies that #(Ek) c4 for all k. This will allow us several timesto write inequalities of the form iEk qi2 iEk qi2, where stands forvarious norms and seminorms (the constant in the inequality depending on c4).

Following [9], the main idea for constructing a preconditioner for the matrix Sis to decompose h as the direct sum of a coarse space LH and some local spaces0h,i (one for each edge ei of the decomposition of ) and then considering thecorresponding block-Jacobi type preconditioners. More precisely, let LH ,

LH = { C0() : i = 1, . . . , M , |ei P1(ei), = 0 on },denote the subset of those functions of whose restriction to each segment ei is

linear. In order to construct the preconditioner we make the following assumptionson the approximation space h:

(A5) The following two conditions hold:(a) LH h;(b) for all h h such that at the extrema ai and bi of some edge ei it

holds that h(ai) = h(bi) = 0, the function 0h,i defined as

0h,i = h on ei, 0h,i = 0 on \ ei

verifies0h,i h.

(A6) h H1() and for all h h the following two inverse inequalities holdfor all k, k = 1, . . . , K , and for all r, t, 0 r < t 1:

|h|Ht

(k) h

rt

k |h|Hr

(k),(27) |h|Ht(ei) hrtk |h|Hr(ei), i Ek.(28)Remark 3.1. Assumption (A5) is not at all restrictive. In a finite element frameworkit reduces to asking that the extrema ai and bi of the segments ei into which theskeleton is split be nodes of the finite element grid for h.

Remark 3.2. The parameter hk appearing in the inverse inequality (27) has herethe meaning of the smallest mesh size of the restriction to k of h.


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Under assumption (A5) it is easy to see that h can be split as

h =

LH

0h, h = H +

0h,

with 0h defined as

0h = {h h, h = 0 at the extrema ai and bi of each edge ei of }.The subspace 0h verifies

0h =

Mi=1

{h 0h : h = 0 on \ ei} Mi=1

0h|ei ,

where the symbol means that the two spaces are isomorphic. Note that all h 0h verify h|ei H1/200 (ei) for all i, and in the following we will write hH1/200 (ei)for h|eiH1/200 (ei) (and analogously for other norms defined on portions of ).

For each i = 1,

, M, we assume we are given a bilinear form si

: 0h|ei 0h|ei R, satisfying the following continuity and positivity bounds uniformly in

h: for all , 0h|ei :(B1) si(, ) H1/200 (ei)H1/200 (ei), si(, )

2

H1/200 (ei)

.

Moreover we assume that we are given a bilinear form sH : LH LH R suchthat for all H, H LH it holds that(B2) sH(H, H) HH, sH(H, H) H2.

It is beyond the goals of this paper to fully discuss the different ways in whichthe bilinear forms si and sH can be constructed. In the following section we willbriefly recall a possible solution and we refer to [ 9, 10, 16] for further discussionson this topic.

Following [9], the preconditioner will be finally constructed with the aid of abilinear form s : h h R assembled by blocks: for h, h h, with h =H +

0h, h = H +

0h, H, H LH and 0h, 0h 0h we set

(29) s(h, h) = sH(H, H) +Mi=1

si(0h|ei , 0h|ei).

The main result of this paper is the following theorem.

Theorem 3.1. For all h h it holds that

(30) s(h, h) s(h, h) maxk

1 + log

Hkhk

2s(h, h).

By a well-known argument, Theorem 3.1 implies that we can derive the followingcorollary, where we denote by S and S, respectively, the matrices corresponding tothe Galerkin discretization of the bilinear forms s and s.

Corollary 3.1. It holds that

(31) cond(S1S) maxk

1 + log

Hkhk

2.


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We recall that by the definition of the bilinear form s, the matrix S is indeedthe Schur complement matrix defined by either (14) or (21).

In order to prove Theorem 3.1, we will basically follow the guidelines of the proofof the analogous theorems of [9, 1]. The main difference is that here we consider ageneral discretization space, not necessarily consisting in piecewise linear functions.Moreover we will be working with functions and operators defined directly on theinterface.

In view of the definition of si and sH on the one hand and of property (26) ofs on the other hand, proving (30) is equivalent to proving that h h withh = H +

0h (H LH and 0h in 0h), it holds that

h2 H2 +Mi=1

0h2H1/200 (ei) maxk

1 + logHkhk

2h2.

We start by observing that it holds that

h2 H2 + 0h2 H2 + Mi=1

0h2H1/200 (ei),

where the last bound descends by splitting 0h 0h as 0h =M

i=1 0h,i which

implies (since #(Ek) c4 and since each i belongs to Ek for at most two ks)

0h2 k

|0h|2H1/2(k) k

iEk

|0h,i|2H1/2(k) Mi=1

0h2H1/200 (ei),

the last inequality deriving from the H1/200 (ei) to H

1/2(k) boundedness of thetrivial extension-by-zero operator.

Moreover, by direct computation, using the linearity of H, one can see that itholds that

(32) H2 Kk=1

|H|2H1/2(k) Mi=1

(H(ai) H(bi))2,

ai and bi being the extrema of ei. Then, proving Theorem 3.1 reduces to provingthe following result:

Theorem 3.2. For all h h with h = H + 0h (H LH and 0h 0h ), itholds that

(33)Mi=1

(H(ai) H(bi))2 +Mi=1

0h2H1/200 (ei) maxk

1 + logHkhk

2h2.

In order to prove Theorem 3.2, we will need several bounds. We start by proving

the following Lemma.

Lemma 3.1. Assume that h satisfies assumptions (A5) and (A6). Then thefollowing bounds hold:

(i) for all h h such that h(p) = 0 for some p ei with i Ek, we have

(34) h2L(ei)

1 + logHkhk

|h|2H1/2(ei);


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(ii) for all h h it holds that

(35) (h(ai)

h(bi))2 1 + log Hk

hk |h|

2

H1/2

(ei)

,

ai and bi being the extrema of ei, i Ek;(iii) for all 0h 0h it holds that

(36)iEk

0h2H1/200 (ei)

1 + logHkhk

2|0h|2H1/2(k).

Proof. Let i Ek . By assumptions (A1)(a) and (A4) we have that |ei| Hk. Westart by observing that for any H1/2+(ei) it holds that

(37) 2L(ei) H1k 2L2(ei) +H2k

||2H1/2+(ei).

This is easily seen by observing that on the reference segment e = ]0, 1[ the con-

tinuity bound of the injection H1/2+

(e) L

(e) depends on as follows (seeAppendix A, Lemma A.1): for all H1/2+(e)2L(e) 2L2(e) +

1

||2H1/2+(e).

A change of variable then yields (37). Let now h h and R. By assumption(A6) it holds that

(38)H2k

|h |2H1/2+(ei) =

H2k

|h|2H1/2+(ei) H2k h

2k

|h|2H1/2(ei).

After choosing = 1/ log(Hk/hk), inequality (37) for = h combined with(38) yields:

(39) h 2L(ei) H1k h 2L2(ei) + logHkhk

|h|2H1/2(ei).In view of (39), (34) can then be proven by applying exactly the arguments of

[9], which we repeat here for completeness: let us choose to be the average of hon ei. Applying Poincare inequality gives

(40) H1/2k h L2(ei) |h|H1/2(ei)

which yields

(41) h 2L(ei)

1 + logHkhk

|h|2H1/2(ei).

Now we remark that h(p) = 0 for some p ei implies(42) || h L(ei)and we get (34) by triangular inequality.

(ii) then follows by applying (i) to the function (x) (ai) (which vanishes inp = ai).

As far as (iii) is concerned, we start by remarking that for 0h 0h and fori Ek, denoting by 0h,i the function coinciding with 0h on ei and identicallyvanishing on k \ ei and using (3) and the inverse inequality (27), we obtain thefollowing bounds

(43) 0hH1/200 (ei) |0h,i|H1/2(k) hk |0h,i|H1/2(k) |0h|H1/20 (ei),


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where the second inequality descends from (27) by applying a standard functionalspaces interpolation argument, since the spaces Hr0 (ei), 0 < r < 1/2, can be ob-

tained as the interpolants of order 2r between L2

(ei) and H

1/2

00 (ei). Again we boundthe H

1/20 (ei) seminorm of

0h by a change of variable, using the corresponding

bound on the reference segment e = ]0, 1[ . We recall once more that H1/20 (e)

and H1/2(e) coincide, with equivalent norms, the constant in the equivalence de-pending on . In particular it is possible to prove (see Appendix A, Corollary A.1)that the following bound holds for all H1/2(e) and for all R:(44) ||

H1/20 (e)

1

H1/2(e) +

1||,

which rescales as

||H1/20 (ei)

Hk

H1/2k L2(ei) + | |H1/2(ei)

+

Hk||.

Again, taking = 0h and as its average on ei, we can apply on the one hand

Poincare inequality (40) and on the other hand the bound (42), which holds since0h(ai) = 0, and we get

0hH1/200 (ei) Hhk

|0h|H1/2(ei) +

Hhk

0h L(ei).

We then take once again = 1/ log(Hk/hk), and, thanks to bound (41), we obtain

0hH1/200 (ei)

1 + logHkhk

|0h|H1/2(ei).

Since

iEk| |2

H1/2(ei) | |2

H1/2(k), by squaring and then taking the sum over

i Ek, we obtain (36). Remark 3.3. Since for s < 1/2 the two spaces Hs(ei) and H

s0 (ei) coincide, the

seminorm | |Hs0(ei) in equation (43) could be replaced by the standard Hs

(ei)seminorm. In that case, however, the implicit constant in the bound would dependon .

Analogously to what happens in [9], the main ingredient of the proof of Theorem3.2 is then the following lemma.

Lemma 3.2. Let0h 0h and H LH. Then for all k it holds that

(45)iEk

0h2H1/200 (ei)

1 + logHkhk

2|0h + H|2H1/2(k).

Provided such lemma holds, (33) is obtained, thanks to (ii) of Lemma 3.1, byapplying (45) for H = H. In fact, for h = H +

0h, H LH, 0h 0h, we can

write

Mi=1

0h2H1/200 (ei) =1

2

Kk=1

iEk

0h2H1/200 (ei)

k

1 + log

Hkhk

2|h|2H1/2(k)

maxk

1 + log

Hkhk

2h2,


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which, combined with (35), yields (33). The only thing that we need to do then isto prove Lemma 3.2.

Proof of Lemma 3.2. Let k, 1 k K. We start by introducing a function 0 k,0h ,

(46) k,0h = {h h|k : h = 0 at the cross points of k} = 0h|k,

0 depending on H, defined as the unique element of k,0h satisfying

(0, )H1/2(k) = (H, )H1/2(k), k,0h ,where (, )H1/2(k) is the bilinear form

(, )H1/2(k) = k k((x) (y))((x) (y))

|x

y

|2

dxdy

inducing the H1/2(k) seminorm. It is easy to check that | |H1/2(k) is a normon k,0h and then we can conclude by standard arguments that

(47) |0|H1/2(k) |H|H1/2(k).

Now, adding and subtracting 0 at the left-hand side of (45), we can write:

(48)iEk

0h2H1/200 (ei) iEk

0h + 02H1/200 (ei) +iEk

02H1/200 (ei)

.

Since 0h + 0 k,0h , by the definition of 0 we have that

(49) (H

0, 0h + 0)H1/2(k) = 0,

which yields

|0h + H|2H1/2(k) |0h + 0|2H1/2(k).The first sum on the right-hand side of (48) can then be bound thanks to Lemma3.1(iii); we have

iEk

0h + 02H1/200 (ei)

1 + logHkhk

2|0h + 0|2H1/2(k)(50)

1 + log

Hkhk

2|0h + H|2H1/2(k).(51)

Let us bound the second sum on the right-hand side of (48). We recall that wehave

(52) 02H1/200 (ei)

= |0|2H1/2(ei) + I1(0) + I2(0),

with

(53) I1(0) =

biai

|0(x)|2|x ai| dx, I 2(

0) =

biai

|0(x)|2|x bi| dx,


15/31


(we recall that ai and bi denote the two vertices of the edge ei). Now, usingequations (47), (32) and (35), since 0h vanishes at ai and bi, we can write

iEk

|0|2H1/2(ei) |0|2H1/2(k) |H|2H1/2(k)(54)

iEk

(H(ai) H(bi))2(55)

=iEk

([0h + H](ai) [0h + H](bi))2(56)

1 + log

Hkhk

|0h + H|2H1/2(k).(57)

Let us now bound I1(0). For notational simplicity let us identify ei = ]0, T[

with ai = 0 and bi = T. We have

I1(0) =T

0|

0

(x)|2

|x| dx(58)

=

T0

|0(x) + H(x) H(x) + H(0) H(0)|2|x| dx(59)

T0

|0(x) H(x) + H(0)|2|x| dx(60)

+

T0

|H(x) H(0)|2|x| dx.(61)

Let us set = 0 H. Remarking that (0) = H(0), we have

T

0

|0(x) H(x) + H(0)|2

|x|dx =

T

0

|(x) (0)|2

|x|dx

=

hk0

|(x) (0)|2|x| dx

+

Thk

|(x) (0)|2|x| dx.

The first term is bound, using Cauchy-Schwarzs inequality and inequality (28), byhk0

|(x) (0)|2|x| dx =

hk0

1

|x|x

0

x () d

2

dx

hk0

x0

|x ()|2 d dx

hk||2H1(ei) ||2H1/2(ei),while, using (34), we bound the second byT

hk

|(x) (0)|2|x| dx

(0)2L(ei)Thk

1

xdx

1 + log

Hkhk

2||2H1/2(ei),


16/31


where we used T Hk. The second integral in (60) can be bound directly usingthe linearity of H.T

0

|H(x) H(0)|2

|x| dx (H(bi) H(ai))2

1 + log Hk

hk

|H + 0h|2H1/2(ei)where again we used (35) and the fact that 0h vanishes at ai and bi. Then, addingup the different contributions, we conclude that

(62) I1(0)

1 + log

Hkhk

2|0 H|2H1/2(ei) +

1 + log

Hkhk

|H + 0h|2H1/2(ei).

By using the same arguments we get a similar bound for I2:

(63) I2(0)

1 + log

Hkhk

2|0 H|2H1/2(ei) +

1 + log

Hkhk

|H + 0h|2H1/2(ei).

We can now insert bounds (54), (62) and (63) back into (52) and sum over i Ek. By observing that for any w

H1/2(k) it holds that iEk |w|

2H1/2(ei)

|w|2H1/2(k)

, we obtain the bound

iEk

02H1/200 (ei)

1 + log

Hkhk

|0h + H|2H1/2(k)(64)

+

1 + log

Hkhk

2|0 H|2H1/2(k).(65)

We now observe once again that equation (49) implies that |H 0|2H1/2(k) |H + 0h|2H1/2(k) and we get the thesis.

4. Numerical Tests

We will test the preconditioner we propose on the following problem: find u such

that

(66)

u = f, in = ]0, 1[ ]0, 1[ ,

u = 0, on = .

We consider a uniform, geometrically conforming, decomposition of = ]0, 1[ ]0, 1[ in K = N N equal square subdomains of size H H, H = 1/N.

In each subdomain k we take a uniform mesh Tk composed of nk nk equalsquare elements of size k k, k = H/nk = 1/(N nk). We then define Vkh to bethe space ofQ1 finite elements on the mesh Tk,

Vkh = {uh C0() : uh| Q1(), Tk},where Q1() denotes the space of polynomials of degree less than or equal to onein each variable. The multiplier space kh is then defined as the trace on k of

Vkh . With such a choice it is possible to prove that (A3) holds (see [ 4]).The skeleton \ is split as \ = 2(N1)Ni=1 ei with ei = k for some

k = k(i) and = (i). We remark that |ei| = H for all i. We consider a grid G on obtained by uniformly splitting each ei into Li elements of size hi and we defineh to be the space ofP1 finite elements on such a grid:

h = {h C0() : h| P1(), G, h| = 0}.With this choice, assumption (A6) holds for hk = miniEk hi.


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Letting 0h,i

(resp. 0h,i

) denote the vector of the values of 0h (resp 0h) 0h at

the interior nodes of the grid G|ei , we define the bilinear form si : 0h,i|ei 0h,i|eias

si(0h|ei , 0h|ei) = (0h,i)T(Ri)1/20h,i,

where Ri is the stiffness matrix associated to the discretization of the operatord2/dx2 by P1 finite element on ei with homogeneous Dirichlet boundary conditionsat the extrema ai and bi. With Ri being positive definite, its square root is welldefined and it is computed and stored at the start.

As far as the coarse preconditioner sH is concerned, we considered two choices:

(BPS) Following [9], we set

(67) sH(H, H) := 2M

i=1(H(ai) H(bi))(H(ai) H(bi)).

(Laplace) Denoting by H and H the H10 () functions obtained by liftingH and H harmonically in each subdomain, we set

(68) sH(H, H) :=

HHdx;

note that the set {H, H LH} coincides with the set ofQ1 finite elementson the coarse grid corresponding to the decomposition =

k, and

then applying the coarse preconditioner reduces to numerically solving theLaplace equation on such space.

For all tests we set f = 1. We solved the linear system by a preconditionedconjugate gradient method, using h = 0 as initial guess. We report the number

of iterations needed to reduce the residual of a factor 105.

4.1. Plain formulation.

4.1.1. Conforming decomposition. We start by considering the case of a conformingdomain decomposition, that is, in the framework described above we set nk = nfor all k and Li = n for all i. In this case k = hi = h = H/n. With such a choice,it is not difficult to check that the inf-sup condition (12) is satisfied uniformly in hand then we are in the range of Lemma 2.1. It is easy to realize that the discretesolution verifies ukh = h on k, and then the function uh, defined by uh|k = ukh,verifies uh H10 (). In such a case we obtain the same solution that we wouldget using a conforming finite element method on a regular grid of N n

N n square

elements of dimension h h. The use of a nonconforming method is therefore, inthis case, only a device to implement a solver for the discrete problem in an easilyparallelizable way.

In order to study both the dependence on H (size of the subdomains) and on hwe tested the preconditioner for different values of n in the range [5, 40] and N inthe range [4, 32]. We tested both coarse preconditioners for all combinations ofNand n. The results are summarized in Tables 1 and 2. As one can see, they are inclose agreement with the theory (note that in this case maxk hk/Hk = h/H = n).


18/31


Table 1. Number of conjugate gradient iterations needed for re-ducing the residual of a factor 105, with coarse preconditionergiven by (68) for different combinations of the number K = N2 of

subdomains and n of elements per edge (n2 elements per subdo-main).

n = 5 n = 10 n = 20 n = 40K = N2 # Iter. # Iter. # Iter. # Iter.

16 10 12 14 1564 14 17 20 23

144 15 18 22 26256 15 19 23 27400 16 20 23 27576 16 20 24 27784 16 20 24 27

1024 16 20 24 27

Table 2. Number of conjugate gradient iterations needed for re-ducing the residual of a factor 105, with coarse preconditionergiven by (67) for different combinations of the number K = N2 ofsubdomains and n of elements per edge (n2 elements per subdo-main).

n = 5 n = 10 n = 20 n = 40K = N2 # Iter. # Iter. # Iter. # Iter.

16 8 9 11 1364 10 13 17 22

144 11 15 20 26

256 13 16 22 30400 13 17 23 32576 13 17 24 33784 14 18 25 33

1024 14 18 25 33

4.1.2. Nonconforming decomposition. We consider then the case of a truly noncon-forming decomposition, without stabilization (formulation (10)). In our test forei = k , we set Li = min{nk, n}. This discretization does not in generalsatisfy the assumptions required for the uniform stability of the discrete problem.Nevertheless, using a standard technique, it is not difficult to prove that ( 12) holdsprovided that for all , k such that

|

k

|> 0 it either holds that n = nk or

max{n, nk}/ min{n, nk} C for some constant C > 1.We considered two test configurations, both on a decomposition of into 5 5

subdomains. In the first case (see Figure 1, top), for all subdomains except two,we set nk = 10, while on the remaining two subdomains (subdomains 13 and 14,if we number all subdomains row-wise) we set nk = n, where n varies in the set[10, . . . , 52]. With the above criterion used to fix Li, we have that Li = 10 for alledges ei, except the common edge 13 14 (the two subdomains are adjacent),where Li = max{n13, n14} = n varies.


19/31


0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 10

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1

(1)

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 10

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1

(2)

Figure 1. The two configurations on which the preconditioner istested for the plain formulation (8) in the nonconforming case.

In the second configuration (see Figure 1, bottom) for all subdomains we againset nk = 10, this time except that on a three-by-three block of subdomains wherenk = n, with n varying again between 10 and 52. Here, for 12 out of 40 edges of ,we have that Li = n. For both configurations we tested both coarse preconditioners.The results are provided in Table 3.


20/31


Table 3. Number of conjugate gradient iterations needed for re-ducing the residual of a factor 105 for Configurations 1 and 2(corresponding, respectively, to the top and to the bottom decom-

positions illustrated in Figure 1). Both coarse preconditioners (67)(BPS) and (68) (Laplace) are tested for different values of thenumber n of elements per edge in the refined part of the skeleton(one edge for Configuration 1 and twelve edges for Configuration2). The number of elements per edge in the remaining part of theskeleton is set to 10.

Laplace BPS# It. # It.

n = 10 12 8n = 17 18 11n = 24 18 13n = 31 18 13n = 38 18 14n = 45 19 14n = 52 19 14

(a) Configuration 1.

Laplace BPS# It. # It.

n = 10 12 8n = 17 20 14n = 24 23 16n = 31 24 18n = 38 25 19n = 45 26 20n = 52 27 20

(b) Configuration 2.

As one can easily see, though the theoretical estimate on the condition numberonly depends on maxk Hk/hk, which takes the same value for both configurations, inthe first case in which the overall discretization of the skeleton is sensibly coarserthe mesh being fine only in a small portion (in our case one) of the edges composingthe influence of refining the mesh is weaker than in the second case, in which arelevant portion of the skeleton is refined.

4.2. Stabilized formulation. We now consider the stabilized formulation (19).Since the H1/2(k) seminorm is invariant under changes of scale, we can describethe stabilizing form for subdomains scaled in such a way that |k| = 1 (that is,H = 1/4). We can then identify k with the circle T of unitary length. Followingthe proposal of [7], the bilinear forms [, ] 1

2,k are designed by means of a wavelet

decomposition. More precisely, we restrict the number Li of elements of the i-thedge ei to be a power of two,

Li = 2ji for some ji 1,

so that for all k, h|k Vjk+2 where for j > 0, Vj denotes the space ofP1 finiteelements on the uniform grid Gj of T with mesh size 1/2j.

The sequence {Vj}j0 forms a so-called multiresolution analysis of L2

(T) andit is well known (see for example [15]) that there exist several wavelet bases asso-ciated with such a multiresolution analysis. More precisely there exist several P1compactly supported functions C0(R) defined on the uniform grid of mesh size1 and integer nodes, such that, setting

m =

+n=

2m/2(2m(x n) ),


21/31


all functions j Vj can be written as

j = 0 +

j1m=0

2m=1

mm, 0 constant,

and such that

|j |2H1/2(k) j1m=0

2m=1

2m|m|2.

Given the values of j at the nodes of the uniform grid, the coefficients (m)m,lcan be retrieved by a fast (O(2j)) wavelet transform (FWT).

If we denote by Pk : L2(T) Vjk+2 the L2(T) orthogonal projection, for ,

L2(T) we can define corresponding wavelet coefficients m and m such that

Pk() = 0 +

jk

m=02m

=1mm, Pk() = 0 +

jk

m=02m

=1mm.

Then we define the bilinear form [, ] 12,k as

[, ] 12,k =

jkm=0

2m=1

2mmm.

It is possible to prove ([7, 14]) that the bilinear forms thus defined satisfy assump-tions (A4) and (18).

Remark 4.1. Note that the coefficients m and m are not the classical waveletcoefficients of the functions and . These would in fact be defined as the scalarproduct of, respectively, and with suitable dual functions m L2(T). Thecoefficients m and m are rather the scalar products of the dual functions mwith the projections of and onto Vjk+2, respectively. Naturally, such coefficients

coincide with the classical ones, provided f Vjk+2.With this definition, the computation of [ukh h, vkh h] 12 ,k essentially reduces

to first computing the nodal values of Pk(ukh), Pk(v

kh), Pk(h) and Pk(h), respec-

tively, and then applying a FWT. As far as h and h are concerned, the choiceof jk implies that h, h Vjk+2 and then their nodal values are simply computedby interpolation. As far as ukh and v

kh are concerned, we need to compute their

L2 projection on Vjk+2. Observing that the mass matrix in Vjk+2 has a circulantstructure with bandwidth equal to three, this can also be done in O(2jk ) operationsby a modification of the Croute reduction algorithm for solving linear systems withtridiagonal matrices. For further details on the implementation of the stabilizedformulation we refer to [8].

We tested the preconditioner (with both form (67) and (68) of the coarse bilinear

form) on four different splittings of into, respectively, 4 4, 8 8, 12 12 and16 16 subdomains. In each case we randomly assigned the values of nk in sucha way that for 1/3 of the subdomains nk = 5, for about another third nk = 10,and for the remaining subdomains nk = 15. The four configurations considered areshown in Figure 2. For approximating the interface function , we used a uniformdiscretization of the interface (Li = n for all i). For each configuration, we testedthe preconditioner for different values of n and of the stabilization parameter .The function appearing in the definition of the stabilizing bilinear form is chosen


22/31


0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 10

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1

(a)

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 10

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1

(b)Figure 2. The four configurations on which the preconditioner istested for the stabilized formulation (10). (Continues)

to be the piecewise linear wavelet corresponding to the 2.2 B-spline biorthogonalsetting ([15]). Note (see Table 4) that for = 0 (no stabilization!) if n is big, theCG algorithm does not converge within the maximum number of iterations (whichwas set equals to 100), confirming the need for using some kind of stabilization.


23/31


0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 10

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1

(c)

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 10

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1

(d)

Figure 2. (continued)

As one can see from the results, the value of the parameter has an influence onthe conditioning of the matrix S1S. If is too small (see Table 5), the action of thestabilization term is not sufficient to compensate for the lack of validity of the inf-sup condition, and therefore the system behaves almost like the one obtained usingthe plain formulation. By increasing (see Tables 6 and 7), the situation improves


24/31


dramatically and one can see that, for instance, for = .05 in the case of 16 16subdomains with n = 32 (which corresponds to 15360 degrees of freedom for h)the residual is reduced of a factor 105 in about 35 iterations, which is only slightlyworse than the result obtained for the analogous values in the conforming case.However, when is further increased (see Table 8), the performances deteriorate.This (as well as the difference with respect to the conforming case) is probablydue to the fact that the constants in the estimates (26)which have an influenceon the bound on the condition numberdepend on the ratio B/b (B and b beingthe constants in (18) and in assumption (A3)(c), respectively) amplified by the

factor . If such a ratio (which is always greater than or equal to one) in not closeenough to 1, the choice of the right balance between the original bilinear form andthe stabilizing term can become important from the point of view of the spectralproperties of the operator, which are essential for preconditioning.

Table 4. Number of conjugate gradient iterations needed to re-

duce the residual of a factor 105

on the four configurations de-picted in Figure 2 with the plain formulation (10). Both coarse pre-conditioners are tested for different values of the number n of ele-ments per edge. As one can see for n 16, the CG method does notconverge within the maximum number of iteration (= 100), con-firming the instability of the plain formulation for the discretizationconsidered.

(a) (b) (c) (d)n # It. # It. # It. # It.4 14 17 18 198 47 57 60 69

16

32 Laplace

(a) (b) (c) (d)n # It. # It. # It. # It.4 11 13 14 158 40 52 54 64

16

32 BPS

Table 5. Number of conjugate gradient iterations needed to re-duce the residual of a factor 105 on the four configurations de-picted in Figure 2 with the stabilized formulation (19) for =.00125. Both coarse preconditioners are tested for different valuesof the number n of elements per edge. One can clearly see thatthe method converges very badly or does not converge. This is dueto the fact that, since the stabilization parameter is very small,the action of the stabilization term is not sufficient to compensatefor the lack of validity of the inf-sup condition, and therefore the

system behaves like the one obtained using the plain formulation.

(a) (b) (c) (d)n # It. # It. # It. # It.4 14 17 18 198 52 60 64 83

16 86 32 96

Laplace

(a) (b) (c) (d)n # It. # It. # It. # It.4 11 13 14 158 39 51 57 77

16 69 92 32 85

BPS


25/31


Table 6. Number of conjugate gradient iterations needed to re-duce the residual of a factor 105 on the four configurations de-picted in Figure 2 with the stabilized formulation (19) for = 1.

Both coarse preconditioners are tested for different values of thenumber n of elements per edge.

(a) (b) (c) (d)n # It. # It. # It. # It.4 14 17 18 198 22 26 27 29

16 28 34 36 3832 34 42 44 45

Laplace

(a) (b) (c) (d)n # It. # It. # It. # It.4 11 13 14 158 18 21 24 27

16 23 31 35 3832 30 41 46 51

BPS

Table 7. Number of conjugate gradient iterations needed to re-

duce the residual of a factor 105 on the four configurations de-picted in Figure 2 with the stabilized formulation (19) for = 5.Both coarse preconditioners are tested for different values of thenumber n of elements per edge.

(a) (b) (c) (d)n # It. # It. # It. # It.4 14 17 18 198 19 22 23 25

16 22 28 29 3132 24 31 32 34

Laplace

(a) (b) (c) (d)n # It. # It. # It. # It.4 11 13 14 158 15 17 19 22

16 16 23 26 2932 19 27 30 35

BPS

Table 8. Number of conjugate gradient iterations needed to re-duce the residual of a factor 105 on the four configurations de-picted in Figure 2 with the stabilized formulation (19) for = 20.Both coarse preconditioners are tested for different values of thenumber n of elements per edge.

(a) (b) (c) (d)n # It. # It. # It. # It.4 14 17 18 198 25 26 27 30

16 31 37 38 41

32 35 41 43 48Laplace

(a) (b) (c) (d)n # It. # It. # It. # It.4 12 13 14 158 23 24 24 27

16 28 35 35 37

32 34 40 39 41BPS

Appendix A: Some Sobolev space injection bounds

The aim of this section is to recall some known bounds for which we provide anew proof based on the use of equivalent norms expressed through wavelet coeffi-cients (see for instance [15]). Such proof will easily reveal the dependence of the


26/31


constants in the bounds on the smoothness of the spaces considered and the waythese constants explode as the smoothness index approaches the limit of validity ofthe bounds.

The key ingredient of the following arguments is the possibility of constructingtwo orthonormal bases

B = {k, k = 1, . . . , 2j0} {j,k, j j0, k = 1, . . . , 2j},B0 = {0k, k = 1, . . . , 2j0} {0j,k, j j0, k = 1, . . . , 2j},

for L2(0, 1), tuned up to provide norm equivalences for Sobolev spaces, respectively,without and with homogeneous boundary conditions. In particular, such bases canbe constructed in such a way that all u L2(0, 1) can be written both ways as

u =2j0k=1

u, kk +

j=j0

2jk=1

u, j,kj,k,

u =

2j0k=1

u, 0k0k +

j=j0

2jk=1

u, 0j,k0j,k

and that the following norm equivalences hold for all s with 0 s 1:

uHs(0,1) 2j0

k=1

|u, k|2 +

j=j0

2jk=1

22js|u, j,k|2

1/2

,(69)

uHs0(0,1) 2j0

k=1

|u, 0k|2 +

j=j0

2jk=1

22js|u, 0j,k|2

1/2

, s = 1/2,(70)

uH1/200 (0,1)

2j0

k=1

|u, 0k|2 +

j=j0

2jk=1

2j |u, 0j,k|2

1/2

,(71)

with constants independent ofs. Such bases are constructed in such a way that thebasis functions satisfy the following properties:

W.1. For all j j0 and k = 1, . . . , 2j the functions j,k have a certain numberof vanishing moments; in particular j,k has zero mean value.

W.2. The basis functions are well localized and the supports of the basis functionsare properly scaled with respect to j: for some N fixed, satisfying N 2j02, we have

supp j,k [(k N)2j, (k + N)2j] ]0, 1[ ,supp 0j,k [(k N)2j, (k + N)2j] ]0, 1[ .

W.3. The interior functions of the two bases coincide: j,k = 0j,k, k =

N , . . . , 2j N; in other words the two bases differ only for those elementsthat take into account boundary conditions.

W.4. The basis functions verify

(72) |k(x)| C, |0k(x)| C, |j,k(x)| 2j/2, |0j,k(x)| 2j/2.


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Remark A.1. Note that the constants N and j0 (and therefore 2j0) are fixed once

for all and depend only on the choice of the wavelet basis. We will therefore considerthem as an O(1) as far as the estimates that we are going to prove are concerned.

The normalization estimate (72), together with property W.2 on the size of thesupport, trivially yields the following bound on the L1 norm of the basis functions

(73) j,kL1(0,1) 2j/2, 0j,kL1(0,1) 2j/2.In the following we will make use of a certain number of algebraic inequalities:

we start by recalling the following bound on the sum of a geometric series

(74)j1

2aj 1

a,

as well as the well-known boundedness property of the discrete convolution operator

(75)j

|j

ajjbj |2

j

|aj |

2j

|bj |2 .

Moreover, since N is a fixed constant, we have

(76)

Nk=1

ak

2

Nk=1

|ak|2.

The first bound that we consider regards the injection Hs(0, 1) L(0, 1),which is known to hold for all s > 1/2 [17]. As far as the dependence on s of theconstant in the bound is concerned, we have the following lemma.

Lemma A.1. The following bound holds for all u Hs(0, 1) with s ]1/2, 1]:

uL(0,1) uL2(0,1) + 1s 1/2 |u|Hs(0,1).

Proof. Let u =2j0

k=1u, kk +

j=j0

2jk=1u, j,kj,k. Using (72) and (69), as

well as properties W.1 and W.2 (which guarantees that for x and j fixed, j,k(x) = 0for at most 2N values of k), we have (with u =

10

u ds)

supx

|u(x)| supx

|2j0k=1

u, kk(x) +

j=j0

2jk=1

u, j,kj,k(x)|

maxk

|u, k| +jj0

maxk

|u u, j,k|2sj2(1/2s)j

uL2(0,1) +

jj0

4(1/2s)j

1/2jj0

2jk=1

|u u, j,k|222sj

1/2

uL2(0,1) + 1s 1/2u uHs(0,1).

We conclude by applying Poincare inequality which yields u uHs(0,1) |u u|Hs(0,1) = |u|Hs(0,1).


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The second issue that we consider is the equivalence between the spaces Hs(0, 1)and Hs0 (0, 1). This is known to hold for all s [0, 1/2[ [17]. The dependence on sof the constant in the bound is the object of the following lemma.

Lemma A.2. The following bound holds for all Hs, 0 s < 1/2:(77) Hs0(0,1)

1

1/2 sHs(0,1).

Proof. Let Hs(0, 1) be expanded in terms of the basis B:

=

2j0k=1

kk +jj0

2jk=1

j,kj,k,

so that

2Hs(0,1) 2j0

k=1|k|2L2 + jj0

2j2j

k=1|j,k|2.

We start by decomposing as the sum of four contributions:

= 0 + left + center + right

with

0 =

2j0k=1

kk, center =

jj0

2jNk=N

j,kj,k =jj0

2jNk=N

j,k0j,k,(78)

left =jj0

N1k=1

j,kj,k, right =

jj0

2jk=2jN+1

j,kj,k(79)

and by triangular inequality we have

Hs0(0,1)

0

Hs0(0,1)

+

left

Hs0(0,1)

+

center

Hs0(0,1)

+

right

Hs0 (0,1)

.

In order to bound the Hs0 (0, 1) norm of the four contributions, we will use theequivalent norm (70). We start by observing that for all (j, n) with n = N , . . . , 2j N we have k, 0j,n = k, j,n = 0, while for the remaining couples (j, n) wehave

|k, 0j,n| kL(0,1)0j,nL1(0,1) 2j0/22j/2;using (70) and observing that for all js the number of indexes n for which k, 0j,n= 0 is bounded independently of j, we easily see that

k2Hs0 (0,1) 2j0

k2L2 +

jj0

22js2j

1

1 2s .

Using the fact that on R2j0

the 2 and the 1 norms are equivalent (the constant

in the equivalence depending on j0, which, we recall, is a fixed constant), we theneasily see that

02Hs0 1

1 2s

2j0

k=1

|k|

2

1

1 2s2j0k=1

|k|2.

We now consider center . This term contains the contribution of all functionswhich are interior; i.e., they do not see the boundary conditions. More precisely,


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which yields

left

2Hs0(0,1)

1

(1/2 s)2 j=0

22js

Nk=1

|j,k|2

.

Analogously, we can prove

right2Hs0(0,1) 1

(1/2 s)2

j=0

22js2j

k=2jN+1

|j,k|2 .

Adding the bounds for 0Hs0 (0,1), leftHs0 (0,1), centerHs0(0,1) andrightHs0(0,1), since for s in [0, 1/2[ it holds that 1 1/(1/2 s), in view ofthe norm equivalence (69) we obtain the thesis.

Corollary A.1. For all H1/2(0, 1) and for all R it holds, for alls ]0, 1/2[,that

Hs0(0,1) 11/2 s H1/2(0,1) + ||1/2 s .Proof. Trivially it holds that

Hs0(0,1) Hs0(0,1) + Hs0 (0,1).In view of bound (77) the only thing that needs to be proven is a bound onHs0(0,1), which can be done by direct computation, yielding the thesis.

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Istituto di Matematica Applicata e Tecnologie Informatiche del Consiglio Nazionale

delle Ricerche, v. Ferrata 1, 27100 Pavia, Italy

E-mail address: [email protected]: http://www.imati.cnr.it/~aivlis
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