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REGULAR INVERSION OF THE DIVERGENCE OPERATOR WITHDIRICHLET BOUNDARY CONDITIONS ON A POLYGON*

DOUGLAS N. ARNOLD,† L. RIDGWAY SCOTT,‡ and MICHAEL VOGELIUS§

Abstract. We consider the existence of regular solutions to the boundary value problem div U = f

on a plane polygonal domain Ω with the Dirichlet boundary condition U = g on ∂Ω. We formulate

simultaneously necessary and sufficient conditions on f and g in order that a solution U exist in the

Sobolev space W s+1p (Ω). In addition to the obvious regularity and integral conditions these consist of

at most one compatibility condition at each vertex of the polygon. In the special case of homogeneousboundary data, it is necessary and sufficient that f belong to W s

p (Ω), have mean value zero, and vanish at

each vertex. (The latter condition only applies if s is large enough that the point values make sense.) We

construct a solution operator which is independent of s and p. As intermediate results we obtain variousnew trace theorems for Sobolev spaces on polygons.

Key words. divergence, trace, Sobolev space

AMS(MOS) subject classifications (1991 revision). 35B65, 46E35

1. Introduction. The constraint of incompressibility arises in many problems ofphysical interest. In its simplest form this constraint is modelled by the partial differentialequation

div U = 0 in Ω,

where U could be, for example, the velocity field in the Navier-Stokes equations or thedisplacement field in the equations of incompressible elasticity and Ω is the spatial domain.Often in the analysis of such problems the inhomogeneous equation

(1.1) div U = f in Ω

is introduced and the question of the existence of regular solutions to this equation arises.If no boundary conditions are imposed, then it is easy to see that solutions to (1.1) maybe found which are as regular as the regularity of f permits. That is, if f belongs to theSobolev space W s

p (Ω) for some s ≥ 0 and 1 < p < ∞ then there exists a solution U inW s+1p (Ω). To show this it suffices to define U = gradu where u ∈ W s+2

p (Ω) is a solutionof Poisson’s equation ∆u = f . Note that there exist such regular solutions of Poisson’s

*This research was partially supported by NSF grant DMS-86-01489 (DNA), NSF grant DMS-86-13352

(LRS), ONR contract N00014-85-K-0169 (MV), and the Sloan Foundation (MV). Each of the authors wasin residence and further supported by the Institute for Mathematics and its Applications for part of the

time this research was being performed.†Department of Mathematics, University of Maryland, College Park, MD 20742 and Institute for

Mathematics and its Applications, University of Minnesota, Minneapolis, MN 55455.

‡Department of Mathematics, Pennsylvania State University, University Park, PA 16802.§Department of Mathematics, University of Maryland, College Park, MD 20742.

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equation even on a domain whose boundary is assumed no more regular than Lipschitz,since we can always extend f to a smoothly bounded domain and solve a regular boundaryvalue problem for Poisson’s equation (such as the Dirichlet problem) on the larger domain.

The existence of regular solutions to (1.1) with specified boundary values

(1.2) U = g on ∂Ω

is more subtle. An obvious necessary condition for the existence of such a solution is that

(1.3)∫

Ω

f =∫

Ω

div U =∫∂Ω

U · ν =∫∂Ω

g · ν.

If Ω is a smoothly bounded planar domain, f ∈W sp (Ω), g ∈W s+1−1/p

p (∂Ω) for s ≥ 0 withs − 1/p nonintegral, and (1.3) holds, then a simple construction of a solution U of (1.1),(1.2) in W s+1

p (Ω) is possible. For example, consider the case g ≡ 0 and suppose Ω is simplyconnected. First let u ∈W s+2

p (Ω) be a solution to Poisson’s equation as above. Then the

normal derivative ∂u/∂ν and the tangential derivative ∂u/∂σ are in Ws+1−1/pp (∂Ω) with∫

∂Ω∂u/∂ν = 0. We can thus find w ∈W s+2

p (Ω) such that

∂w/∂ν = ∂u/∂σ, ∂w/∂σ = −∂u/∂ν on ∂Ω,

or, equivalently,curlw = − gradu on ∂Ω.

(By curlw we mean the vectorfield (∂w/∂y,−∂w/∂x)). Setting U = gradu+ curlw givesthe desired solution.

If ∂Ω is not sufficiently smooth, then this argument fails and the existence of w is farfrom obvious. In this paper we consider the case of polygonal Ω with sides denoted byΓn. Returning to the general case, we show that if f belongs to W s

p (Ω), g|Γn belongs

to[W

s+1−1/pp (Γn)

]2for each n, and f and g satisfy (1.3) and some further necessary

compatibility conditions, then (1.1) admits a W s+1p solution U satisfying the boundary

condition (1.2). Somewhat surprisingly, the compatibility conditions required for s = 2,one condition per vertex in addition to (1.3), are sufficient for all higher s. These resultshave already been applied in [2], [6], and [8].

2. Preliminaries. We will introduce a variety of function spaces allied to the Sobolevspaces. For the convenience of the reader we list here our notation for each and the equationnumber nearest the definition.

W sp (Ω) (2.1); W s

p (Ω) (2.1); W sp (Ω) (2.1);

W sp (Ω) (3.1); Xm

sp (5.1); Xmsp(∂Ω) (6.1);

Zsp(∂Ω) (6.3);∆

W sp(Ω) (6.13); V sp (Ω) (7.1).

2

Throughout the letter C is used to denote a generic constant, not necessarily the samefrom one occurence to the next. For Ω ⊆ Rn a domain with Lipschitz boundary (as defined,for example, in [3, Definition 1.2.1.1]), and f ∈ C∞(Ω) we define the usual Sobolev normsfor 1 < p <∞ and s ≥ 0:

(2.1) ‖f‖ps,p,Ω =

‖f‖pLp(Ω), s = 0,

‖f‖pLp(Ω) +∫

Ω

∫Ω

|f(x)− f(y)|p

|x− y|n+spdx dy, 0 < s < 1,∑

|α|≤[s]

‖Dαf‖ps−[s],p,Ω, 1 ≤ s <∞.

(Here [s] denotes the greatest integer not exceeding s.) The spaces W sp (Ω) and W s

p (Ω)are defined to be the closures of C∞(Ω) and C∞0 (Ω), respectively, relative to these norms.There exists a bounded linear extension operator from W s

p (Ω) into W sp (Rn) (even if the

boundary is only Lipschitz). Cf. [3, Theorem 1.4.3.1]. For s > 1/p the functions in W sp (Ω)

have well-defined traces on ∂Ω. If s ≥ 1, then W sp (Ω) ∩ W 1

p (Ω) = v ∈ W sp (Ω) | v =

0 on ∂Ω . We denote by W sp (Ω) the subspace of W s

p (Ω) consisting of elements whoseintegral is zero. For details and more information regarding Sobolev spaces, we refer thereader to [1], [3], [7], and, for the case p = 2, to [5].

We shall also require the Sobolev norms for functions defined on Lipschitz curves inR

2, in particular for an open subset, Γ, of the boundary of a polygon. For a Lipschitzcurve, the norms ‖ · ‖s,p,Γ may be defined for 0 ≤ s ≤ 1, 1 < p <∞, via charts. Moreover,for 1/p < s ≤ 1, the trace operator maps W s

p (Ω) boundedly onto Ws−1/pp (∂Ω). See [3,

Theorem 1.5.1.2]. (The norms ‖ · ‖s,p,Γ, s > 1, are not well-defined unless Γ is moreregular.)

We recall some properties of these spaces when the domain of definition is a brokenline segment. (Cf. [4] or [3, Lemma 1.5.1.8].) Suppose Γ1 and Γ2 are line segments in R2

intersecting at a common endpoint, z, and let f be a function on Γ = Γ1 ∪ Γ2. Then for1 < p <∞ and 0 ≤ s < 1/p, f ∈W s

p (Γ) if and only if

(2.2) f |Γ1 ∈W sp (Γ1), f |Γ2 ∈W s

p (Γ2).

Moreover, the norm

(2.3) ‖f‖s,p,Γ1 + ‖f‖s,p,Γ2

is equivalent to the W sp (Γ) norm. For 1/p < s ≤ 1, f ∈W s

p (Γ) if and only if (2.2) holds andf is continuous at z. (Note that (2.2) implies the continuity of f everywhere else in viewof the Sobolev imbedding theorem). In this case too we have equivalence of norms. Thecase s = 1/p is more involved. Let σ1 denote the unit direction along Γ1 pointing toward

3

z, and let σ2 denote the unit direction along Γ2 pointing away from z. Then f ∈W 1/pp (Γ)

if and only if (2.2) holds and

IpΓ(f) =∫ ε

0

t−1|f(z − tσ1)− f(z + tσ2)|p dt <∞

where ε is a positive number not exceeding the lengths of Γ1 or Γ2. In this case

(2.4)(‖f‖p1/p,p,Γ1

+ ‖f‖p1/p,p,Γ2+ IpΓ(f)

)1/p

defines a norm equivalent to ‖f‖1/p,p,Γ.

If Γ1 and Γ2 are collinear, so Γ is a line segment, one can easily extend these results todetermine when f belongs to W s

p (Γ) for s > 1. Namely, if s− 1/p is not an integer, thenf ∈ W s

p (Γ) if and only if (2.2) holds and the tangential derivatives f (k) are continuous atz for 0 ≤ k < s− 1/p. If this case, (2.3) is an equivalent norm on W s

p (Γ). If s− 1/p is aninteger, it is required in addition that IpΓ(f (s−1/p)) <∞ and then [‖f‖ps,p,Γ1

+ ‖f‖ps,p,Γ2+

IpΓ(f (s−1/p))]1/p is an equivalent norm on W sp (Γ).

We close this section with two lemmas concerning the space W 2/pp on a sector. Let

Sα = (r cos θ, r sin θ) | 0 < r < 1, 0 < θ < α , Γα1 = (r cosα, r sinα) | 0 < r < 1 , andΓα2 = (r, 0) | 0 < r < 1 . Then Γα := Γα1 ∪ Γα2 ∪ (0, 0) is the linear part of the boundaryof Sα. By γu we denote the trace of the function u on Γα.

Lemma 2.1. Let α ∈ (0, 2π). Then the trace operator γ maps W2/pp (Sα) continuously into

W1/pp (Γα).

Proof. For p ≥ 2 this lemma follows immediately from the trace theorem quoted earlier([3, Theorem 1.5.1.2]) since then s = 2/p ≤ 1. For 1 < p < 2 we first show that γmaps W t

p(Sα) continuously into W t−1/pp (Γα), 2/p < t < 1 + 1/p. Indeed, if u ∈ W t

p(Sα),

then γu|Γαi ∈ Wt−1/pp (Γαi ), and since t > 2/p, γu is also continuous at the origin. As

t − 1/p < 1, this implies that γu ∈ Wt−1/pp (Γα). Furthermore, we know that γ maps

W 1p (Sα) continuously into W

1−1/pp (Γα). The conclusion of the lemma now follows by

interpolation.1

Remarks. (1) If α varies in a compact subinterval of (0, 2π), then the distance betweenpoints x1 ∈ Γα1 and x2 ∈ Γα2 is equivalent to the sum of the distances of x1 and x2 to

1The fractional order spaces W tp(Sα) may also be defined by real interpolation between two consecutive

integer orders. This is the approach used in [1, Ch. 7], where the specific interpolation process is defined.

The equivalence with the definition given here is shown in [1, Theorem 7.48]. Similarly the spaces W sp (I)

for an interval I may be defined by real interpolation between integer orders. The same holds for the spaces

W sp (Γα) for 0 < s < 1, because these are defined by pulling back the spaces W s

p (I) via Lipschitz charts.

Since 1 < t < 2 it is possible to interpolate betweenW 1p (Sα) andW t

p(Sα) and since 0 < 1−1/p < t−1/p < 1

it is possible to interpolate between W1−1/pp (Γα) and W

t−1/pp (Γα). Choosing the interpolation index

appropriately we get that γ maps W2/pp (Sα) continuously into W

1/pp (Γα).

4

the origin, uniformly in α. Consequently, the equivalence of the norm in (2.4) and theW

1/pp (Γα) norm is uniform in α. It is then easy to see that there is a single constant which

bounds the norm of the trace operator between the spaces W 2/pp (Sα) and W

1/pp (Γα).

(2) Using a partition of unity, we can easily extend Lemma 2.1 to show that trace operatormaps W 2/p

p (Ω) continuously into W 1/pp (∂Ω) for any polygonal domain Ω.

The next lemma relates the decay of the trace of a W 2/pp function near the vertex to

the decay of the function itself.

Lemma 2.2. Let α ∈ (0, 2π). Then there exists a constant C such that

(2.5)∫ 1

0

|u(x, 0)|p

xdx ≤ C

‖u‖p2/p,p,Sα +∫∫Sa

|u(x, y)|p

x2 + y2dx dy

and

(2.6)∫∫Sa

|u(x, y)|p

x2 + y2dx dy ≤ C

(‖u‖p2/p,p,Sα +

∫ 1

0

|u(x, 0)|p

xdx

)

for all u ∈W 2/pp (Sα).2

Proof. Suppose θ lies between α/2 and α. Then θ is bounded away from 0 and 2π, and sowe may find a constant C depending on α but not θ for which∫ 1

0

r−1|u(r cos θ, r sin θ)− u(r, 0)|p dr ≤ C‖u‖p1/p,p,Γθ

.

Moreover by Lemma 2.1, there is a single constant C such that

‖u‖p1/p,p,Γθ

≤ C‖u‖p2/p,p,Sθ ≤ C‖u‖p2/p,p,Sα

holds for all such θ. Thus∫ 1

0

r−1|u(r, 0)|p dr ≤ C(∫ 1

0

r−1|u(r cos θ, r sin θ)|p dy + ‖u‖p2/p,p,Sα

).

Integration over θ in (α/2, α) gives (2.5).

To prove (2.6) it suffices to show that

(2.7)∫∫D

|u(x, y)|p

x2 + y2dx dy ≤ C

(‖u‖p2/p,p,D +

∫ 1

0

|u(x, 0)|p

xdx

)2Either of the two integrals entering in (2.5) and (2.6) may be infinite. In that case the inequalities

imply that they both are infinite.

5

where D is the unit disc, since we can always extend a function in W2/pp (Sα) to one in

W2/pp (D). Considering the trace of u on Γπ and applying Lemma 2.1, we see that

(2.8)∫ 1

0

|u(−x, 0)|p

xdx ≤ C

(∫ 1

0

|u(x, 0)|p

xdx+ ‖u‖p2/p,p,D

).

Let θ ∈ [π/2, 3π/2]. Applying the lemma again, we get∫ 1

0

r−1|u(r cos θ, r sin θ)− u(r, 0)|p dr ≤ C‖u‖p1/p,p,Γθ

≤ C‖u‖p2/p,p,D,

where we may choose C independent of θ. Consequently,

(2.9)∫ 1

0

r−1|u(r cos θ, r sin θ)|p dr ≤ C(‖u‖p2/p,p,D +

∫ 1

0

r−1|u(r, 0)|p dr).

Similarly for θ ∈ [0, π/2] ∪ [3π/2, 2π], we may consider the trace of u on the boundary ofthe sector formed by the negative x-axis and the ray emanating from the origin with angleθ and use (2.8) to conclude (2.9). Integration of (2.9) with respect to θ ∈ (0, 2π) gives(2.7).

3. The main theorem. Homogeneous Boundary Conditions. Let Ω denotea (bounded and simply-connected) polygonal domain in the plane. Denote by zj , j =1, 2, . . . N , the vertices of Ω listed in order as ∂Ω is traversed counterclockwise, and byΓj the open line segment connecting zj−1 to zj (we interpret the subscripts on z and Γmodulo N). We explicitly assume that the magnitude of the angle formed by Γj and Γj+1

at zj lies strictly between 0 and 2π, i.e., we exclude domains with slits. We also denote byν the outward-pointing unit vector normal to Ω which is defined on ∂Ω \ z1, . . . , zN andby νj its constant value on Γj . Similarly σ and σj refer to the counterclockwise tangentvector.

Suppose that U ∈ W s+1p (Ω) satisfies (1.1) and vanishes on ∂Ω. Then, of course,∫

Ωf = 0. In addition, ∂U/∂σj = 0 on Γj and ∂U/∂σj+1 = 0 on Γj+1. For s > 2/p

the Sobolev imbedding theorem implies that both the directional derivatives ∂U/∂σj and∂U/∂σj+1 are continuous on Ω, so they both vanish at zj . Since the vectors σj and σj+1

are linearly independent, it follows that f = div U vanishes at zj . Define

(3.1) W sp (Ω) =

W sp (Ω), 0 ≤ s < 2/p,f ∈ W s

p (Ω) |∫

Ω

|f(z)|p

|z − zj |2dz <∞, j = 1, . . . , N

, s = 2/p,

f ∈ W sp (Ω) | f(zj) = 0, j = 1, . . . , N , s > 2/p.

This space is normed by the restriction of the W sp norm except if s = 2/p in which case

we take the norm to be

‖f‖pW sp (Ω)

= ‖f‖ps,p,Ω +N∑j=1

∫Ω

|f(z)|p

|z − zj |2dz.

6

We have just seen that for s ≥ 0, s 6= 2/p,

(3.2) div([W s+1p (Ω) ∩ W 1

p (Ω)]2) ⊂ W sp (Ω).

This is also true if s = 2/p, as we now demonstrate. For any φ ∈W 2/p+1p (Ω)∩ W 1

p (Ω) thetangential derivative ∂φ/∂σ vanishes on every edge. In light of (2.6) it follows that

∫Ω

|∂φ(z)/∂σj |p

|z − zj |2dz +

∫Ω

|∂φ(z)/∂σj+1|p

|z − zj |2dz <∞.

Since σj and σj+1 are linearly independent,

∫Ω

| gradφ(z)|p

|z − zj |2dz <∞.

The inclusion (3.2) follows by application of this result componentwise to U.

The next theorem is our main result for homogeneous boundary conditions. It showsthat in fact equality holds in (3.2) and that the divergence operator admits a boundedright inverse which does not depend on s or p.

Theorem 3.1. Let 1 < p < ∞, s ≥ 0. If p 6= 2 suppose that s − 1/p /∈ Z. Then there

exists a bounded linear map

L : W sp (Ω)→ [W s+1

p (Ω) ∩ W 1p (Ω)]2

such that divL(f) = f for all f ∈ W sp (Ω). The operator L may be chosen independently

of s and p.3

Remark. We conjecture that the hypothesis that s− 1/p /∈ Z is unnecessary for all p, notjust p = 2. However this hypothesis is necessary for certain trace results we use in ourproof. See the remark to Theorem 4.1.

Several ingredients of the proof of this theorem will be developed in the next threesections. These ingredients assembled, the proof of Theorem 3.1 becomes very short. Itis given in the last section of this paper. The analogue of Theorem 3.1 for the case ofinhomogeneous boundary conditions is also true. This result is stated as Theorem 7.1.

3More precisely, there is a linear operator

L :⋃s,p

W sp (Ω)→ [W 1

1 (Ω)]2

such that for each s and p, L maps W sp (Ω) boundedly into [W s+1

p (Ω) ∩ W 1p (Ω)]2.

7

4. A result concerning traces on a line. In the case of a smooth boundary, theconstruction of a solution of (1.1) vanishing on ∂Ω as sketched in Section 1 was based onthe existence of a function w whose curl coincides with the negative gradient of anotherfunction u on the boundary. In extending this construction to the case of a polygon, weneed to know what vectorfields may arise on a polygonal boundary as the traces of curlsor gradients. The main results of section 6, Theorem 6.2 and Corollary 6.3, provide theanswer. In deriving them, we need a collection of other trace theorems, starting withthose of this section regarding traces of a function of two variables on a line. Let S(R)denote the Schwartz space, i.e., the space of C∞ functions of one variable, all of whosederivatives decrease faster than any polynomial at infinity, and let S ′(R) denote its dual.Let

⊕∞j=0 S ′(R) denote the subset of

∏∞j=0 S ′(R) consisting of vectors having at most a

finite number of non-zero entries. Given subspaces Vj ⊂ S ′(R), 0 ≤ j ≤ m, we view theCartesian product

∏mj=0 Vj naturally as a subspace of

⊕∞j=0 S ′(R) by setting all entries

with index greater than m to zero.

Theorem 4.1. Let s and p denote real numbers such that 1 < p <∞, s > 1/p. If p 6= 2assume that s− 1/p /∈ Z. Then the trace map

u 7→ γjxu :=∂ju

∂yj( · , 0)

maps W sp (R2) boundedly onto W

s−j−1/pp (R) provided j < s − 1/p. Moreover, there is a

map

Ex :∞⊕j=0

S ′(R)→ C∞(R2 \ (R× 0))

such that, if f = (fj) ∈∏mj=0W

s−j−1/pp (R) with m < s− 1/p, then

‖Exf‖s,p,R2 ≤ Cs,pm∑j=0

‖fj‖s−j−1/p,p,R

and

γjxExf = fj , j = 0, 1, . . . ,m.

Remarks. (1) Henceforth we avoid the case s − 1/p integral if p 6= 2, since in this case,the range of the trace operator is a Besov space which does not coincide with any Sobolevspace. Cf. [1, Chapter 7].

(2) It follows from the theorem that, if v = Ex(f0, ..., fm, 0, 0, ...) with fj ∈ W s−j−1/pp (R)

then the trace∂kv

∂yk( · , 0) = 0

8

as a function in Ws−k−1/pp (R), provided m < k < s − 1/p. By the Sobolev imbedding

theorem,

(4.1)∂j+kv

∂xj∂yk(x, 0) = 0, x ∈ R,

(in the pointwise sense) provided m < k and j + k < s− 2/p.

Proof. The asserted properties of the trace operators γjx are standard, cf. [3] and [7]. Sincewe require a fixed extension operator Ex, independent of s, we include a detailed derivationof this part of the result. Our construction extends that given in [7]. Let S be a functionin S(R) which satisfies∫

R

S(t) dt = 1,∫R

tjS(t) dt = 0, j = 1, 2, . . . .

Such a function exists, as is easily seen using the Fourier transform, F . These conditionstranslate to choosing (FS)(0) = 1 and (FS)(j)(0) = 0 for j = 1, 2, . . . . Since F isan isomorphism on S(R), we can choose S ∈ S(R) with these properties as follows: letχ ∈ C∞0 (R) be any function that is identically one in a neighborhood of the origin, andset S = F−1χ. Now we define

(4.2)

v(x, y) = Ex(f0, f1, . . . )(x, y)

:= χ(y)∞∑j=0

yj

j!

∫R

S(t)fj(x+ yt) dt, (x, y) ∈ R2 \ (R× 0),

where again, χ ∈ C∞0 (R) is any function that is identically one in a neighborhood ofthe origin. Note that only a finite number of terms in (4.2) are non-zero because we areassuming that (f0, f1, . . . ) ∈

⊕∞j=0 S ′(R). It is easy to verify that v has the asserted traces.

The boundedness of Ex from∏mj=0W

s−j−1/pp (R) to W s

p (R2) follows from the followingresult.

Lemma 4.2. Let j be a nonnegative integer, S ∈ S(R), χ ∈ C∞0 (R), and f ∈ C∞0 (R).Define

v(x, y) = χ(y)yj∫R

S(t)f(x+ yt) dt, (x, y) ∈ R2.

Then, for any real s, p with 1 < p < ∞ and s > j + 1/p with s − 1/p /∈ Z if p 6= 2, there

holds

‖v‖s,p,R2 ≤ C‖f‖s−j−1/p,p,R.

The constant C depends on χ, S, s, and p but not on f .

Proof. Letw(x, y) = yj

∫R

S(t)f(x+ yt) dt.

9

For any positive integers k ≥ j, there are coefficients c0, . . . , cj and a function Skj ∈ S(R)such that

∂kw

∂yk(x, y) =

j∑l=0

clyj−l∫R

S(t)f (k−l)(x+ yt)tk−l dt

=j∑l=0

cl

∫R

S(t)tk−l(∂

∂t

)j−lf (k−j)(x+ yt) dt

=∫R

[j∑l=0

cl

(−∂∂t

)j−l[S(t)tk−l]

]f (k−j)(x+ yt) dt

=:∫R

Skj(t)f (k−j)(x+ yt) dt.

Similarly, for positive integers k < j, there exists a function Skj ∈ S(R) such that

∂kw

∂yk(x, y) = yj−k

∫R

Skj(t)f(x+ yt) dt.

That is, for any integers k and j,

∂kw

∂yk(x, y) = ymaxj−k,0

∫R

Skj(t)f (maxk−j,0)(x+ yt) dt.

For differentiation with respect to x, we get

∂kw

∂xk(x, y) = ymaxj−k,0

∫R

Skj(t)f (maxk−j,0)(x+ yt) dt

with some function Skj ∈ S(R) . Combining these formulas, we have for any multi-indexα that

Dαw(x, y) = ymaxj−|α|,0∫R

Sα(t)f (max|α|−j,0)(x+ yt) dt

where Sα ∈ S(R). Leibniz’ rule then gives

(4.3)

Dαv(x, y) =∑β≤α

cβDα−βχ(y)Dβw(x, y)

=∑β≤α

χαβ(y)∫R

Sβ(t)f (max|β|−j,0)(x+ yt) dt,

for some χαβ ∈ C∞0 (R).

Now for S ∈ S(R), f ∈ C∞0 (R), and any real y, Young’s inequality gives

(4.4)∫R

∣∣∣∣∫R

S(t)f(x+ yt) dt∣∣∣∣p dx ≤ (∫

R

|S(t)| dt)p ∫

R

|f(x)|p dx.

10

In view of (4.3) and (4.4) the case of integer s reduces to proving that if S ∈ S(R),f ∈ C∞0 (R), then

(4.5)∫∫R2

∣∣∣∣∫R

S(t)f ′(x+ yt) dt∣∣∣∣p dx dy ≤ CS‖f‖p1−1/p,p,R.

To show this, set

g(x, y) =∫R

S(t)f ′(x+ yt) dt =∫R

(−1/y)S′(t)f(x+ yt) dt

=∫R

(1/y)S′(t)[f(x)− f(x+ yt)] dt =∫R

tS′(t)f(x)− f(x+ yt)

ytdt.

From Holder’s inequality, we find

|g(x, y)|p ≤ CS∫R

|tS′(t)| |f(x)− f(x+ yt)|p

|yt|pdt.

Fubini’s Theorem thus yields∫∫R2|g(x, y)|p dx dy ≤ CS

∫∫∫R3|tS′(t)| |f(x)− f(x+ yt)|p

|yt|pdt dx dy

= CS

∫∫∫R3|S′(t)| |f(x)− f(x+ z)|p

|z|pdt dx dz

= C ′S

∫∫R2

|f(x)− f(x+ z)|p

|z|pdx dz

which is the desired estimate (4.5). This establishes the theorem for integer s. Interpolationbetween consecutive integers then gives the theorem for all real s ≥ j + 1.

To complete the proof, we now consider the case j + 1/p < s < j + 1. In view of (4.3)and (4.4) , it suffices to show that for |α| = j we have

|Dαv|s−j,p,R2 :=∫R2

∫R2

|Dαv(x, y)−Dαv(x, y)|p

|(x, y)− (x, y)|2+(s−j)p dx dy dx dy

≤ C‖f‖s−j−1/p,p,R.

Because of (4.3) this reduces to showing that for S ∈ S(R), f ∈ C∞0 (R), and g defined by

g(x, y) =∫R

S(t)f(x+ yt) dt

we have

(4.6) |g|s,p,R2 ≤ CS‖f‖s−1/p,p,R

11

provided that 1/p < s < 1. To show this, we estimate:

|g(x, y)− g(x, y)|p ≤∣∣∣∣∫R

S(t)[f(x+ yt)− f(x+ yt)] dt∣∣∣∣p

≤ CS∫R

|S(t)| |f(x+ yt)− f(x+ yt)|p dt.

Thus

|g|ps,p,R2 ≤ CS∫∫∫∫∫

R5|S(t)| |f(x+ yt)− f(x+ yt)|p

|(x− x)2 + (y − y)2|1+sp/2dt dx dy dx dy.

A simple change of variables yields∫∫∫R3|S(t)| |f(x+ yt)− f(x+ yt)|p

|(x− x)2 + (y − y)2|1+sp/2dt dx dx

=∫∫∫

R3|S(t)| |f(x+ yt)− f(x+ yt)|p

|(x+ yt)− (x+ yt)|sp|(x+ yt)− (x+ yt)|sp

|(x− x)2 + (y − y)2|1+sp/2dt dx dx

=∫∫∫

R3

|f(ξ)− f(η)|p

|ξ − η|sp|S(t)| |ξ − η|sp

|[(ξ − yt)− (η − yt)]2 + (y − y)2|1+sp/2dt dξ dη.

Thus to verify (4.6) it is sufficient to show that∫∫∫R3

|S(t)| dt dy dy|[a− (y − y)t]2 + (y − y)2|1+sp/2

≤ CS|a|sp

, a ∈ R.

A simple computation gives∫R

|S(t)| dt|(a− bt)2 + b2|1+sp/2

≤ 21+sp/2

|a2 + b2|1+sp/2

∫R

|S(t)|(1 + t2)1+sp/2 dt =CS

|a2 + b2|1+sp/2

because[(a− bt)2 + b2](1 + t2) ≥ (a2 + b2)/2.

Therefore∫∫∫R3

|S(t)| dt dy dy|[a− (y − y)t]2 + (y − y)2|1+sp/2

≤ CS∫∫R2

dy dy

[a2 + (y − y)2]1+sp/2=

C ′S|a|sp

.

This establishes the lemma, and thus the theorem, under the assumption that fj ∈ C∞0 (R).The theorem now follows for all fj by a density argument.

Remark. For (fj) ∈⊕∞

j=0 S ′(R),

∂kEx(fj)∂xk

= Ex(f (k)j ), k = 1, 2, . . . .

12

5. Traces on intersecting half-lines. Theorem 4.1 concerns the traces of a W sp

function on a line. Using a simple argument involving a partition of unity and localcoordinates it is easy to extend it to cover traces on smooth curves. Since we are interestedin polygons, however, we will need a result analogous to Theorem 4.1 for traces on twointersecting line segments. To state this result, we require some notation. By R+ wedenote the set of positive real numbers. Let

Ip(f) =∫ 1

0

t−1|f(t)|p dt.

For m a nonnegative integer and s > m + 1/p define Xmsp to be the space of pairs of

(m+ 1)-tuples

(5.1)((fj)mj=0, (gj)

mj=0

)∈

m∏j=0

W s−j−1/pp (R+)×

m∏j=0

W s−j−1/pp (R+)

satisfying

f(k)j (0) = g

(j)k (0), 0 ≤ j, k ≤ m, j + k < s− 2/p,(5.2)

Ip(f (k)j − g(j)

k ) <∞, 0 ≤ j, k ≤ m, j + k = s− 2/p.(5.3)

Note that (5.3) only applies if s − 2/p is integral. The space Xmsp is a Banach space with

the norm ‖((fj), (gj)

)‖Xmsp given by

m∑j=0

(‖fj‖ps−j−1/p,p,R+

+ ‖gj‖ps−j−1/p,p,R+

)1/p

if s− 2/p is not an integer, and

m∑j=0

(‖fj‖ps−j−1/p,p,R+

+ ‖gj‖ps−j−1/p,p,R+

)+

∑0≤j,k≤mj+k=r

Ip(f (k)j − g(j)

k )

1/p

if r = s− 2/p is an integer.

Theorem 5.1. Let m be a nonnegative integer and s and p real numbers with 1 < p <∞and s > m+ 1/p. If p 6= 2 suppose that s− 1/p 6= Z. Then the trace map

(5.4) u 7→

((∂ju

∂yj( · , 0)

)mj=0

,

(∂ju

∂xj(0, · )

)mj=0

)

13

maps W sp (R2) boundedly onto Xm

sp and admits a bounded right inverse

Emx : Xmsp →W s

p (R2).

The operator Emx is independent of s and p.

Proof. Our proof is similar to that of [3, Theorem 1.5.2.4]. First of all, we show that theoperator defined in (5.4) maps W s

p (R2) into Xmsp. Let u ∈W s

p (R2), and set

fj =(∂ju

∂yj( · , 0)

), gj =

(∂ju

∂xj(0, · )

), j = 0, . . . ,m.

By the Sobolev imbedding theorem ∂k+ju/∂xk∂yj is continuous on R2 for k+ j < s−2/p,so (5.2) holds. If s = m+2/p and k+ j = m, then we apply Lemma 2.1 to ∂mu/∂xk∂yj toinfer (5.3). Now we define, in several steps, the operator Emx . Let Ex denote the extensionoperator in Theorem 4.1. Let Rmx denote the vector of m+ 1 restrictions

Rmx u :=(u ( · , 0) ,

∂u

∂y( · , 0) , . . . ,

∂mu

∂ym( · , 0)

).

Let Ey and Rmy denote the corresponding operators with the x and y variables reversed,e.g.,

Rmy u :=(u (0, · ) , ∂u

∂x(0, · ) , . . . , ∂

mu

∂xm(0, · )

).

First extend each fj and gj as functions in Ws−j−1/pp (R) with an extension that is in-

dependent of s and p. (Cf. [9, Chapter 17] or [10, Chapter 6].) Explicitly we take theextension of fj given by

fj(x) =∫ ∞

0

λ(s)fj(−sx) ds, x < 0,

and similarly for gj . Here λ is chosen to be bounded, smooth for t > 0, rapidly decreasingat infinity, and satisfying

(5.5)∫ ∞

0

(−t)kλ(t) dt = 1, k = 0, 1, 2, . . . .

For example, we can take λ(t) = R(t1/2) where R ∈ S(R) is an odd function that satisfies∫R

R(t)t2k+1 dt = (−1)k, k = 0, 1, 2, . . . .

In particular, we can take FR(ξ) = i (sinh ξ)χ(ξ) where χ is any even function in C∞0 (R)that is identically one in a neighborhood of the origin and F is the Fourier transform.

14

Let f (resp. g) denote the vector of extensions of the boundary data (fj) on the x-axis(resp. (gj) on the y-axis). Define

v = Ex(f −Rmx Eyg) + Eyg.

Note that Rmx v = f because Rmx Ex is the identity operator. Also

Rmy v = Rmy Ex(f −Rmx Eyg) + g.

Defineh = g −Rmy v = −Rmy Ex(f −Rmx Eyg).

If m < k and k + j < s− 2/p, then h(k)j (0) = 0 since (4.1) implies

∂j+k

∂xj∂ykEx(f −Rmx Eyg)(0, 0) = 0.

If k ≤ m and k + j < s− 2/p then

h(k)j (0) = g

(k)j (0)− ∂j+kv

∂xj∂yk(0, 0) = g

(k)j (0)− f (j)

k (0)

which is zero in view of (5.2). Also, if k = s− j − 2/p, then

Ip(h(k)j ) =

∫ 1

0

t−1

∣∣∣∣g(k)j (t)− ∂j+kv

∂xj∂yk(0, t)

∣∣∣∣p dt≤ 2

∫ 1

0

t−1

∣∣∣∣g(k)j (t)− ∂j+kv

∂xj∂yk(t, 0)

∣∣∣∣p dt+ 2∫ 1

0

t−1

∣∣∣∣ ∂j+kv∂xj∂yk(t, 0)− ∂j+kv

∂xj∂yk(0, t)

∣∣∣∣p dt≤ 2Ip(f (j)

k − g(k)j ) + C‖ ∂

j+kv

∂xj∂yk‖p

1/p,p,Γπ/2

≤ 2Ip(f (j)k − g

(k)j ) + C‖v‖ps,p,R2

≤ C‖((fj)mj=0, (gj)

mj=0

)‖Xmsp <∞,

where we have used Lemma 2.1 in the third inequality. Thus, the function hj which agreeswith hj on R+ and vanishes for y < 0 belongs to W s−j−1/p

p (R) and satisfies

‖hj‖s−j−1/p,p,R ≤ C‖((fj)mj=0, (gj)

mj=0

)‖Xmsp .

The function w = Ey(h0, . . . , hm) satisfies

∂jw

∂xj(0, y) =

gj(y)− ∂jv

∂xj(0, y), 0 < y <∞,

0, −∞ < y < 0,

for j = 0, . . . ,m. Finally we set Emx((fj), (gj)

)= u with (see (5.5))

u(x, y) = v(x, y) + w(x, y)−∫ ∞

0

λ(s)w(x,−sy) ds, (x, y) ∈ R2.

Note that the last term insures that∂ju

∂yj(x, 0) =

∂jv

∂yj(x, 0) = fj(x) for j = 0, . . . ,m.

15

6. Extension to general domains. Traces of curls and gradients. We nowextend Theorem 5.1 from the case of a single corner to a polygonal domain. For simplicityof notation, and since this is all we need to prove our main theorem, we consider only thecases m = 0 and m = 1. We use the notation of section 3 for a polygonal domain. If fand g are functions defined on Γn and Γn+1 respectively, we set

Ipn(f, g) =∫ ε

0

t−1|f(zn − tσn)− g(zn + tσn+1)|p dt

where ε = minj=1,...,N |zj − zj−1|.For s > 1/p we define X0

sp(∂Ω) to be the space of N -tuples

(6.1) (φn)Nn=1 ∈N∏n=1

W s−1/pp (Γn)

satisfying

(6.2)Ipn(φn, φn+1) <∞, if s = 2/p

φn(zn) = φn+1(zn), if s > 2/p,

for n = 1, . . . , N . The space X0sp(∂Ω) is a Banach space with the norm ‖(φn)Nn=1‖X0

sp(∂Ω)

given by (A1 +A2)1/p where

A1 =N∑n=1

‖φn‖ps−1/p,p,Γn,

A2 = ∑N

n=1 Ipn(φn, φn+1), s = 2/p,

0, otherwise.

For s > 1 + 1/p we define X1sp(∂Ω) to be the space of N -tuples of pairs

(φn, ψn)Nn=1 ∈N∏n=1

W s−1/pp (Γn)×W s−1−1/p

p (Γn)

satisfying

φn(zn) = φn+1(zn),

Ipn(∂φn∂σn

σn + ψnνn,∂φn+1

∂σn+1σn+1 + ψn+1νn+1) <∞, if s = 1 + 2/p,

(∂φn∂σn

σn + ψnνn

)(zn) =

(∂φn+1

∂σn+1σn+1 + ψn+1νn+1

)(zn), if s > 1 + 2/p,

16

Ipn(∂2φn∂σ2

n

σn · σn+1 +∂ψn∂σn

νn · σn+1,∂2φn+1

∂σ2n+1

σn · σn+1 +∂ψn+1

∂σn+1νn+1 · σn) <∞,

if s = 2 + 2/p,

(∂2φn∂σ2

n

σn · σn+1 +∂ψn∂σn

νn · σn+1)(zn) = (∂2φn+1

∂σ2n+1

σn · σn+1 +∂ψn+1

∂σn+1νn+1 · σn)(zn),

if s > 2 + 2/p,

for n = 1, . . . , N . The space X1sp(∂Ω) is a Banach space with the norm

‖ (φn, ψn)Nn=1 ‖X1sp(∂Ω)

given by (B1 +B2)1/p where

B1 =N∑n=1

(‖φn‖ps−1/p,p,Γn

+ ‖ψn‖ps−1−1/p,p,Γn

)and

B2 =

∑Nn=1 I

pn( ∂φn∂σn

σn + ψnνn,∂φn+1∂σn+1

σn+1 + ψn+1νn+1), s = 1 + 2/p,∑Nn=1 I

pn(∂

2φn∂σ2

nσn · σn+1 + ∂ψn

∂σnνn · σn+1,

∂2φn+1

∂σ2n+1

σn · σn+1 + ∂ψn+1∂σn+1

νn+1 · σn), s = 2 + 2/p,

0, otherwise.

Theorem 6.1. Let s and p denote real numbers such that 1 < p <∞, s > 1/p. If p 6= 2assume that s− 1/p /∈ Z. Then the trace map

u 7→ (u|Γn)Nn=1

maps W sp (Ω) boundedly onto X0

sp(∂Ω) and admits a bounded right inverse

E0π : X0

sp(∂Ω)→W sp (Ω).

If in addition s > 1 + 1/p, then the trace map

u 7→

(u|Γn ,

∂u

∂νn

∣∣∣∣Γn

)Nn=1

maps W sp (Ω) boundedly onto X1

sp(∂Ω), and admits a bounded right inverse

E1π : X1

sp(∂Ω)→W sp (Ω).

17

Both the operators E0π and E1

π are independent of s and p.

Proof. We can cover ∂Ω with a finite collection of open sets such that each set in thiscollection may be mapped by an invertible affine map onto the square (−1, 1)× (−1, 1) insuch a way that the intersection of the set with Ω is mapped onto one of the following sets:(−1, 1)× (0, 1), (0, 1)× (0, 1), or (−1, 1)× (−1, 1) \ [0, 1]× [0, 1]. In light of Theorem 5.1and the extension theorem for Sobolev spaces, a simple partition of unity argument givesthe present result.

We can now characterize the traces of curls of W sp functions. Let Zsp(∂Ω) (s > 1+1/p)

denote the subspace of N -tuples of vector-valued functions

(6.3) (ψn)Nn=1 ∈N∏n=1

W s−1−1/pp (Γn)×W s−1−1/p

p (Γn)

satisfying

N∑n=1

∫Γn

ψn · νn = 0,(6.7)

Ipn(ψn,ψn+1) <∞, if s = 1 + 2/p,(6.8)

ψn(zn) = ψn+1(zn), if s > 1 + 2/p,(6.9)

Ipn(∂ψn · νn+1

∂σn,∂ψn+1 · νn∂σn+1

) <∞, if s = 2 + 2/p,(6.10)

∂ψn · νn+1

∂σn(zn) =

∂ψn+1 · νn∂σn+1

(zn), if s > 2 + 2/p.(6.11)

This space is normed in the usual way.

Theorem 6.2. Let s and p denote real numbers such that 1 < p < ∞, s > 1 + 1/p. If

p 6= 2 assume that s− 1/p /∈ Z. Then the operator

(6.12) u 7→ (curlu|Γn)Nn=1

maps W sp (Ω) boundedly onto Zsp(∂Ω), and admits a bounded right inverse Ecurl which is

independent of s and p.

Proof. That the operator in (6.12) maps u into Zsp(∂Ω) follows from the continuity andtrace properties of curlu and ∂2u/∂σn∂σn+1. Given (ψn) we define Ecurl

((ψn)

)as follows.

For z ∈ ∂Ω let

φ(z) =∮ z

z1

ψ · νΓ dsΓ,

18

(the integral taken counterclockwise along Γ) and set

φn = φ|Γn , ψn = −ψn · σn, n = 1, . . . , N.

It is easy to verify that (ψn) ∈ Zsp(∂Ω) implies (φn, ψn) ∈ X1sp(∂Ω). We set

Ecurl

((ψn)

)= E1

π

((φn, ψn)

).

We conclude this section by showing that the trace of the gradient of a W sp (Ω) function

is also in Zsp(∂Ω) as long as the function satisfies some necessary compatibility conditions.

Namely we define the subspace∆

W sp(Ω) of W s

p (Ω) for s ≥ 2 as follows. For s < 2 + 2/p,

(6.13)∆

W sp(Ω) =

u ∈W s

p (Ω) |∫

Ω

∆u = 0

;

for s > 2 + 2/p,

∆

W sp(Ω) =

u ∈W s

p (Ω) |∫

Ω

∆u = 0, ∆u(zj) = 0, j = 1, . . . , N

;

and for s = 2 + 2/p,

∆

W sp(Ω) =

u ∈W s

p (Ω) |∫

Ω

∆u = 0,∫

Ω

|∆u(z)|p

|z − zj |2dz <∞, j = 1, . . . , N

.

The Banach norm on∆

W sp(Ω) is the restriction of the W s

p norm except if s = 2 + 2/p, inwhich case

‖u‖p∆W sp(Ω)

= ‖u‖ps,p,Ω +N∑j=1

∫Ω

|∆u(z)|p

|z − zj |2dz.

Corollary 6.3. If s ≥ 2, then the operator

u 7→ (gradu|Γn)Nn=1

maps∆

W sp(Ω) boundedly into Zsp(∂Ω).

Proof. It suffices to verify conditions (6.7)-(6.11) for ψ = gradu, u ∈∆

W sp(Ω). The first

condition holds since∫

∆u = 0 for u ∈∆

W sp(Ω):

N∑n=1

∫Γn

(gradu) · νn =∫

Γ

∂u

∂ν=∫

Ω

∆u = 0.

19

Condition (6.9) follows from the continuity of gradu on Ω if s > 1 + 2/p, and similarly(6.8) follows from the inclusion of (gradu)|Γ in W

1/pp (Γ) if s = 1 + 2/p. To verify (6.10)

and (6.11), we note that (gradu) · ν = ∂u/∂ν, so

∂(gradu · νn+1)∂σn

− ∂(gradu · νn)∂σn+1

=∂2u

∂σn∂νn+1− ∂2u

∂σn+1∂νn= −(σn+1 · νn)∆u.

If s > 2 + 2/p, then ∆u(zj) = 0 by definition, so (6.11) holds. If s = 2 + 2/p, then

Ipn

(∂(gradu · νn+1)

∂σn,∂(gradu · νn)

∂σn+1

)=∫ ε

0

t−1| ∂2u

∂σn∂νn+1(zn − tσn)− ∂2u

∂σn+1∂νn(zn + tσn+1)|p dt

≤ C(∫ ε

0

t−1| ∂2u

∂σn∂νn+1(zn − tσn)− ∂2u

∂σn∂νn+1(zn + tσn+1)|p dt

+∫ ε

0

t−1|∆u(zn + tσn+1)|p dt)

≤ C(‖u‖p2+2/p,p,Ω +

∫ ε

0

t−1|∆u(zn + tσn+1)|p dt).

For the last inequality we have used Lemma 2.1. The estimate (6.10) now follows imme-diately in light of Lemma 2.2.

7. Proof of the main result. Inhomogeneous boundary conditions. We nowprove Theorem 3.1. Given f ∈ W s

p (Ω) for some 1 < p < ∞ and s ≥ 0 with s − 1/pnonintegral in the case p 6= 2, extend f boundedly to an open disc containing Ω anddefine u as the solution to ∆u = f which vanishes on the boundary of the disk. Then

u ∈∆

W s+2p (Ω) and ‖u‖∆

W s+2p (Ω)

≤ C‖f‖W sp (Ω). In light of Corollary 6.3 and Theorem 6.2,

we may define w = −Ecurl

((gradu|Γn)Nn=1

), and thus we have curlw = − gradu on ∂Ω

and ‖w‖s+2,p,Ω ≤ C‖u‖∆W s+2p (Ω)

. Setting L(f) = gradu+ curlw gives the desired operator.

This completes the proof.

Let us consider the case of inhomogeneous boundary data. For s > 1/p we set

X0sp(∂Ω) =

[X0sp(∂Ω)

]2,

so g ∈ X0sp(∂Ω) means that g = (g1, · · · ,gN ) with gn ∈

[W

s−1/pp (Γn)

]2satisfying the

conditions given by (6.2). For s ≥ 0 we define V sp (Ω) to be the space consisting of thosepairs

(7.1) (f,g) ∈W sp (Ω)×X0

s+1,p(∂Ω)

20

for which

∫Ω

f =∫∂Ω

g · ν,(7.2)

Ipn(∂gn+1 · νn∂σn+1

− ∂gn · νn+1

∂σn,νn · σn+1f) <∞, if s = 2/p,(7.3) (

∂gn+1 · νn∂σn+1

− ∂gn · νn+1

∂σn

)(zn) = νn · σn+1f(zn), if s > 2/p.(7.4)

It is not difficult to see that a necessary condition for the existence of U ∈[W s+1p (Ω)

]2satisfying

(7.5)div U = f in Ω,

U = g on ∂Ω,

is that (f,g) ∈ V sp (Ω). The following theorem shows that this condition is also sufficientand that the problem (7.5) admits a bounded right inverse which does not depend on s orp.

Theorem 7.1. Let 1 < p < ∞, s ≥ 0. If p 6= 2 suppose that s − 1/p /∈ Z. Then there

exists a bounded linear map

K : V sp (Ω)→[W s+1p (Ω)

]2such that for any (f,g) ∈ V sp (Ω), the function U := K(f,g) solves (7.5). The operator Kmay be chosen independent of s and p.

Proof. For (f,g) ∈ V sp (Ω), the function V = E0πg ∈

[W s+1p (Ω)

]2 satisfies V = g on ∂Ω.Indeed, this follows from Theorem 6.1 since g ∈ X0

s+1,p(∂Ω). The conditions (7.2)–(7.4)have been chosen precisely to guarantee that

f − div V ∈ W sp (Ω).

Therefore the operator

K(f,g) := L(f − div V) + V = L(f − div E0πg) + E0

πg

has the desired properties.

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REFERENCES

[1] R. A. Adams, Sobolev Spaces, Academic Press, New York, NY, 1975.

[2] D. N. Arnold, F. Brezzi, and J. Douglas, PEERS: A new mixed finite element for plane

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[3] P. Grisvard, Elliptic Problems in Nonsmooth Domains, Pitman, Brooklyn, NY, 1985.

[4] G. N. Jakovlev, Traces of functions in the space W lp on piecewise smooth surfaces, Mat. Sb.

(N.S.), 74 (1967). Math. USSR-Sb., 3 (1967), pp. 481–497.

[5] J. L. Lions and E. Magenes, Non-homogeneous Boundary Value Problems and Applications, I ,

Springer-Verlag, New York–Heidelberg–Berlin, 1972.

[6] M. Morley, A family of mixed elements for linear elasticity, Univ. of Chicago, PhD Thesis, 1986.

[7] J. Necas, Les Methodes Directes en Theorie des Equations Elliptiques, Masson et Cie., Paris, 1967.

[8] L. R. Scott and M. Vogelius, Norm estimates for a maximal right inverse of the divergence

operator in spaces of piecewise polynomials, Math. Modelling and Numer. Anal., 19 (1985),

pp. 111–143.

[9] K. T. Smith, A Primer of Modern Analysis, Springer-Verlag, New York–Heidelberg–Berlin, 1983.

[10] E. M. Stein, Singular Integrals and Differentiability Properties of Functions, Princeton Univ.

Press, 1970.

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