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arXiv:math/0601667v2 [math.AP] 26 Jul 2006

The linear constraints

in Poincar´ e and Korn type inequalities∗

Giovanni Alessandrini†, Antonino Morassi‡and Edi Rosset§

Abstract

We investigate the character of the linear constraints which are needed

for Poincar´ e and Korn type inequalities to hold. We especially analyze

constraints which depend on restriction on subsets of positive measure

and on the trace on a portion of the boundary.

1 Introduction

Let us consider a bounded domain Ω ⊂ Rnwith Lipschitz boundary. The

nature of Poincar´ e inequality is that a function u (say in W1,2(Ω)) is uniquely

determined by its gradient ∇u ∈ L2(Ω) up to an additive constant.

Such a constant is usually chosen to be the average of u in Ω, but also the

average of its trace on ∂Ω would do for this purpose. Similarly if one takes

a weighted average on suitable subsets E of Ω. Meyers [6] has examined this

subject in great depth characterizing the sets E ⊂ Ω for which this is possible

in terms of their capacity. His analysis in fact extends to Poincar´ e inequalities

in Wm,p(Ω), for any m = 1,2,... and 1 < p < ∞. See also Meyers and Ziemer

[7] for the extreme case of BV (Ω) and Ziemer [11, Chapter 4] for a more recent

account on such results.

In the case of Korn inequality, a vector valued function u ∈ W1,2(Ω;Rn) is

uniquely determined by the strain tensor1

rigid displacement r(x) = b + Mx, that is an affine transformation for which

the matrix M = ∇r is skew symmetric. In this case, provided the origin of

the coordinates is placed in the center of mass of Ω, xΩ=

choice is given by b =

|Ω||Ω|

In other words, the common feature of Poincar´ e and Korn inequalities is that

they are inequalities in which a norm is dominated by a seminorm for functions

which satisfy a suitable finite set of linear constraints.

In this paper we examine the character of such constraints and in particular

we investigate whether such constraints can be chosen in such a way that they

can depend only on the values of the function u restricted to a subset of Ω

2(∇u+(∇u)T) up to an infinitesimal

1

|Ω|

?

Ωx, an obvious

1

?

Ωu, M =

1

?

Ω

1

2(∇u − (∇u)T).

∗Work supported in part by MIUR, PRIN n. 2004011204.

†Dipartimento di Matematica e Informatica, Universit` a degli Studi di Trieste, Italy,

alessang@univ.trieste.it

‡Dipartimento di Georisorse e Territorio, Universit` a degli Studi di Udine, Italy, an-

tonino.morassi@uniud.it

§Dipartimento di Matematica e Informatica, Universit` a degli Studi di Trieste, Italy,

rossedi@univ.trieste.it

1

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or of its boundary ∂Ω. We are especially interested in obtaining constructive

evaluations of the constants in a number of specific cases. In fact, constraints of

such type are useful in many instances and in particular, in recent years, they

have shown up in connection with inverse boundary value problems, for instance

in estimates of unique continuation for the system of elasticity, [2], [8]. For such

purposes, it is mandatory to have a concrete and constructive evaluation of the

constants involved. In this note we develop this aspect, improving some known

estimates, and obtaining some which were not available in the literature.

We deal with three types of inequalities, Poincar´ e inequality in W1,2(Ω),

Poincar´ e inequality in W2,2(Ω) and Korn inequality in W1,2(Ω). The results

are all based on an elementary functional analytic argument (Lemma 2.1) which

is due to N. Meyers [6, Proposition 1]. And indeed the consequences for the

standard Poincar´ e inequality, which we summarize in Section 3 are also found

in [6].

This is not completely the case for the Poincar´ e inequality in W2,2(Ω), Sec-

tion 4, and Korn inequality, Section 5.

In Section 4 we treat Poincar´ einequality in W2,2(Ω), deriving, along the lines

of Meyers’ result, the concrete evaluations of the constants in various instances,

see Examples 4.2, 4.3. We observe however that, when one deals with an open

portion Γ of the boundary ∂Ω, the suitable linear constraints that occur in

Meyers’ approach, require the knowledge of the boundary trace of u and of its

first derivatives. We thus examine the possibility to determine linear constraints,

suitable for the validity of a Poincar´ e type inequality, which only depend on the

trace of u on Γ. See Example 4.4, and in particular Theorem 4.7, for the Poincar´ e

inequality in W2,2(Ω). It is peculiar in this case that one has to distinguish the

cases when Γ is a subset of an (n − 1)-dimensional hyperplane or not.

In Section 5 we treat Korn inequality and obtain some estimates that we

believe to be new. See Examples 5.3 and 5.4. In particular, inequality (5.3)

generalizes a classical inequality proven by Kondrat’ev and Oleinik ([5, Theorem

1]) when E is a ball and Ω is starlike with respect to E. We also examine the

case of constraints which depend on the trace of u on an open portion Γ of ∂Ω.

In this case, we find that the constraints can be chosen to depend on the trace

of u on Γ for any open subset Γ of ∂Ω, see Theorem 5.7 and Corollary 5.9, which

in fact substantially generalize a classical inequality proven by Kondratev and

Oleinik ([5, Theorem 2]) for functions u vanishing on a basis of a cylinder Ω.

2A basic Lemma

Lemma 2.1. Let X, Y be Banach spaces, and let L : X → Y a bounded linear

operator. Let X0⊂ X be its null-space and let P : X → X0be a bounded linear

operator such that P|X0= Id. Assume that there exists K > 0 such that

?x − Px?X≤ K?Lx?Y ,for every x ∈ X .

(2.1)

For every bounded linear operator T : X → X0⊂ X such that T|X0= Id we

have

?x − Tx?X≤ (1 + ?T?)K?Lx?Y ,for every x ∈ X .

(2.2)

Here ?T? denotes the operator norm on L(X,X).

2

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Proof. The argument is due to N. Meyers [6, Proposition 1], we reproduce it

here, because of its brevity. We have

x − Tx = x − Px − Tx + Px = x − Px − T(x − Px) .

Hence

?x − Tx?X≤ (1 + ?T?)?x− Px?X≤ (1 + ?T?)K?Lx?Y .

Remark 2.2. Note that the condition T|X0= Id is in fact necessary if an in-

equality of the following form holds true

?x − Tx?X≤ Const.?Lx?Y .

Indeed, if x ∈ X0, then Lx = 0 and hence Tx = x. Note also that the evaluation

of the constant in (2.2) is not optimal, as is evident if one chooses T = P.

When T is close to P the following evaluation may be more convenient. Since

(T − P)Px = 0, we have

x − Tx = x − Px − (T − P)x = x − Px − (T − P)(x − Px) ,

hence

?x − Tx?X≤ (1 + ?T − P?)K?Lx?Y ,

for every x ∈ X .

(2.3)

Therefore (2.2) can be improved as follows

?x − Tx?X≤ (1 + min{?T?,?T − P?})K?Lx?Y ,

for every x ∈ X .

(2.4)

3The Poincar´ e inequality

We shall assume throughout that Ω is a bounded domain (open and connected)

in Rnwith Lipschitz boundary ∂Ω.

Definition 3.1. Given a measurable set E ⊂ Rnwith positive measure and

given u ∈ L1(E) we denote its average

?

In the sequel we shall use this notation also for vector and matrix valued

functions, in particular we shall denote

uE=

1

µn(E)

E

u(x)dµn(x) .

(3.1)

xE=

1

µn(E)

?

E

xdµn(x) ,

(3.2)

the center of mass of E.

We shall use analogous notation for averages on (n−1)-dimensional Lipschitz

surfaces Γ ⊂ Rn

1

µn−1(Γ)

Γ

uΓ=

?

u(x)dµn−1(x) ,

(3.3)

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xΓ=

1

µn−1(Γ)

?

Γ

xdµn−1(x) .

(3.4)

Here µn−1denotes the (n − 1)-dimensional Lebesgue measure on Γ. When no

ambiguity occurs we shall also denote |E| = µn(E), |Γ| = µn−1(Γ).

We recall the well-known Poincar´ e inequality.

Theorem 3.2. There exists Q > 0 such that

?u − uΩ?L2(Ω)≤ Q?∇u?L2(Ω),for every u ∈ W1,2(Ω) .

(3.5)

Proof. See, for instance, [10, Theorem 3.6.5]. For a quantitative evaluation of

the constant Q in terms of the Lipschitz character of Ω we refer to [2, Proposition

3.2]

We also recall the following generalized version, which is a special case of a

theorem due to Meyers [6, Theorem 1].

Theorem 3.3. For every ϕ ∈ (W1,2(Ω))∗such that ϕ(1) = 1 we have

?u − ϕ(u)?W1,2(Ω)≤

?

1 + ?ϕ?(W1,2(Ω))∗|Ω|

1

2

??

1 + Q2?∇u?L2(Ω),

for every u ∈ W1,2(Ω) .

(3.6)

Remark 3.4. Here we have chosen as the W1,2(Ω)-norm the expression

?u?W1,2(Ω)=

??

Ω

u2+ |∇u|2

?1

2

.

(3.7)

Correspondingly, ? · ?(W1,2(Ω))∗ denotes the dual norm.

Proof of Theorem 3.3. This is in fact a straightforward consequence of Theorem

3.2 and of Lemma 2.1.

Example 3.5. Let E be any measurable subset of Ω such that |E| > 0. By

choosing ϕ(u) = uE, and computing

|ϕ(u)| ≤

1

?|E|?u?L2(Ω)≤

?

1

?|E|?u?W1,2(Ω),

for every u ∈ W1,2(Ω) ,

we obtain

?u−uE?W1,2(Ω)≤1 +

?|Ω|

|E|

?1

2??

1 + Q2?∇u?L2(Ω),

for every u ∈ W1,2(Ω) .

(3.8)

Example 3.6. Let Γ be an open portion of ∂Ω. Let us denote by γ(u) the trace

on Γ of any u ∈ W1,2(Ω), and let us choose

ϕ(u) = (γ(u))Γ.

We evaluate

|ϕ(u)| ≤

1

?|Γ|?γ(u)?L2(Γ)≤

CΓ

?|Γ|?u?W1,2(Ω),

4

for every u ∈ W1,2(Ω) ,

Page 5

where CΓis the constant in the inequality for the trace imbedding γ : W1,2(Ω) → L2(Γ)

and we obtain

?

|Γ|

?u − (γ(u))Γ?W1,2(Ω)≤ 1 + CΓ

?|Ω|

?1

2??

1 + Q2?∇u?L2(Ω),

for every u ∈ W1,2(Ω) .

(3.9)

4A higher order Poincar´ e inequality

We treat an analogue of Theorem 3.3 suitable for functions in W2,2(Ω).

We denote by A the space of affine functions on Ω

A = {u(x) = a + b · x | a ∈ R,b ∈ Rn} .

(4.1)

The following Theorem is again a special case of Theorem 1 in [6]. We report it

here with a slightly different expression of the constant on the right hand side.

Theorem 4.1. For any bounded operator T : W2,2(Ω) → A such that Tu = u

for every u ∈ A, we have

?u−Tu?W2,2(Ω)≤ (1+?T?)(1+Q2+Q4)

1

2?∇2u?L2(Ω),for every u ∈ W2,2(Ω) .

(4.2)

Here Q is the constant for the Poincar´ e inequality (3.5).

Proof. The proof is immediate, by a repeated application of Poincar´ e inequality

(3.5) and using Lemma 2.1.

Example 4.2. In analogy to Example 3.5, given E ⊂ Ω measurable, with |E| > 0,

we pose

Tu = uE+ (∇u)E· (x − xE) .

Note that T is uniquely determined by the restriction of u to E. In fact if u1,

u2∈ W1,2(Ω) are such that u1|E= u2|Ethen ∇(u1−u2) = 0 almost everywhere

in E (see [4, Lemma 7.7]).

Clearly T is the identity on A and we compute

?Tu?2

W2,2(Ω)≤ 2|Ω|

|E|(1 + (diam Ω)2)?u?2

W1,2(Ω)

and consequently we obtain

?u − (uE+ (∇u)E· (x − xE))?W2,2(Ω)≤

?

≤1 +

?

2|Ω|

|E|(1 + (diam Ω)2)

?1

2?

(1 + Q2+ Q4)

1

2?∇2u?L2(Ω),

for every u ∈ W2,2(Ω) .

(4.3)

Example 4.3. In certain instances, it is useful to have an estimate of the type

(4.3) when Ω and E are concentric balls, say Ω = B1(0), E = Bρ(0), 0 < ρ < 1.

In this case we evaluate

?n + 3

n + 2

?T? ≤

?1

ρ

?n

2

.

5

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Consequently

?u−Tu?W2,2(B1(0))≤

?

1 +

?n + 3

n + 2

?1

ρ

?n

2?

for every u ∈ W2,2(B1(0)) .

(1+Q2+Q4)

1

2?∇2u?L2(B1(0)),

(4.4)

This specific estimate turns out to be useful in an inverse problem for elastic

plates, [9].

Example 4.4. We note that in general it is not possible to mimic Example 3.6

for a Poincar´ e type inequality in W2,2(Ω) when some functionals of the trace of

u alone on an open portion Γ of the boundary are known.

For example, if one considers

Ω = {x ∈ Rn| 0 < xi< 1,i = 1,...,n}

and

Γ = {x ∈ Rn| xn= 0,0 < xi< 1,i = 1,...,n − 1} ,

then the function u = xnhas zero trace on Γ and ?∇2u?L2(Ω)= 0. Hence its

W2,2(Ω)-norm is not dominated by ?∇2u?L2(Ω)and any functional of its trace

on Γ.

In fact some additional assumptions, or data, are necessary.

If we admit that Γ may be flat, additional pieces of information are needed,

see [6]. For instance, if also the trace of ∇u on Γ is available, then one can

consider

Tu = (γ(u))Γ+ (γ(∇u))Γ· (x − xΓ)

and obtain

?u−(γ(u))Γ−(γ(∇u))Γ·(x−xΓ)?W2,2(Ω)≤ (1+?T?)(1+Q2+Q4)

1

2?∇2u?L2(Ω),

for every u ∈ W2,2(Ω) .

(4.5)

And we compute

?T? ≤ CΓ

?

2|Ω|

|Γ|(1 + (diam Ω)2)

?1

2

(4.6)

where CΓis the constant in the inequality for the trace imbedding γ : W1,2(Ω) → L2(Γ).

Let us assume, instead, that Γ ⊂ ∂Ω is not flat, that is, it is not a portion

of an hyperplane. In this case we obtain Theorem 4.7 below. In order to state

it, we first need some preparation.

We choose a reference system whose origin lies on the center of mass of Γ,

that is, we assume

xΓ= 0 .

(4.7)

Lemma 4.5. Assume that Γ is not a portion of an hyperplane. The restriction

to A of the trace imbedding γ : W1,2(Ω) → L2(Γ) is one to one.

Proof. First of all we observe that, in view of (4.7), Γ is not a portion of an

hyperplane if and only if there exists n linearly independent vectors v1,...,vn∈

6

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Γ. Thus, if l(x) = a + b · x ∈ A and γ(l) = 0, then a = (γ(l))Γ= 0 and b ∈ Rn

must satisfy

b · vi= γ(l)(vi) − a = 0 , for every i = 1,...,n .

Consequently b = 0 and hence, l = 0.

Remark 4.6. Observe that, being Ω bounded, any sufficiently large portion Γ of

the boundary shall not be flat.

Let us denote by e : γ(A) → A the inverse of γ|A, and by π : L2(Γ) → γ(A)

the orthogonal projection onto γ(A) with respect to the L2(Γ) inner product.

Note that an explicit expression of π is easily obtained.

v ∈ L2(Γ), π(v) is determined as π(v) = γ(l), where l ∈ A is the minimizer of

the finite dimensional least squares problem

??

We also introduce τ : L2(Γ) → A as the composition τ = e ◦ π.

In fact, for every

min

Γ

(v − l)2| l ∈ A

?

.

Theorem 4.7. Assume that Γ is not a portion of an hyperplane. We have

?u − τ(γ(u))?W2,2(Ω)≤ (1 + CΓ?e?)(1 + Q2+ Q4)

1

2?∇2u?L2(Ω),

for every u ∈ W2,2(Ω) .

(4.8)

Here ?e? denotes the L(L2(Γ),W2,2(Ω))-norm of e, CΓ is the constant in the

inequality for the trace imbedding γ : W1,2(Ω) → L2(Γ) and Q is the constant

for the Poincar´ e inequality (3.5).

Remark 4.8. Observe that the calculation of the norm ?e? reduces to solving a

finite dimensional eigenvalue problem, in fact

? ?

As a consequence of Theorem 4.7 we obtain the following Corollary.

?e?2= max

Ω(a + b · x)2+ |b|2

?

Γ(a + b · x)2

| a ∈ R , b ∈ Rn, a2+ |b|2> 0

?

.

Corollary 4.9. If Γ is not a portion of an hyperplane, then we have

?u?W2,2(Ω)≤ C1?∇2u?L2(Ω)+C2?γ(u)?L2(Γ),for every u ∈ W2,2(Ω) , (4.9)

where C1, C2> 0 are constants depending on Ω and Γ only.

Proof of Theorem 4.7. It suffices to apply Theorem 4.1 with T = τ ◦ γ and to

observe that, since ?π?L(L2(Γ),L2(Γ))= 1, then we have ?T? ≤ CΓ?τ? ≤ CΓ?e?.

5 Korn type inequalities

We recall the Korn inequality (of second kind). Given u ∈ W1,2(Ω;Rn) we

denote

?∇u =1

7

2(∇u + (∇u)T) .

Page 8

Theorem 5.1. There exists K > 0 such that

????∇u −1

Proof. See for instance [3].

2(∇u − (∇u)T)Ω

????

L2(Ω)

≤ K??∇u?L2(Ω),for every u ∈ W1,2(Ω;Rn) .

(5.1)

We denote by Skewn= {M ∈ Mn×n| MT= −M} the class of skew-

symmetric n×n matrices and by R = {b+Mx | b ∈ Rn,M ∈ Skewn} the linear

space of infinitesimal rigid displacements.

Theorem 5.2. For any bounded linear operator T : W1,2(Ω;Rn) → R such

that Tu = u for every u ∈ R, we have

?u−Tu?W1,2(Ω)≤ (1+?T?)K(1+Q2)

1

2??∇u?L2(Ω), for every u ∈ W1,2(Ω;Rn) .

(5.2)

Here K is the constant appearing in (5.1) and Q is the constant in the Poincar´ e

inequality (3.5).

Proof. Set P : W1,2(Ω;Rn) → R as

Pu = uΩ+1

2(∇u − (∇u)T)Ω(x − xΩ) .

By (3.5) and (5.1) we obtain

?u − Pu?W1,2(Ω)≤ K(1 + Q2)

1

2??∇u?L2(Ω).

The thesis follows from Lemma 2.1 by choosing

X = W1,2(Ω;Rn) ,

Y = L2(Ω;Mn×n) ,

Lu =?∇u ,

X0= R .

Example 5.3. Given E ⊂ Ω measurable with |E| > 0 and setting Tu = uE+

1

2(∇u − (∇u)T)E(x − xE), we compute

?Tu?2

W1,2(Ω)≤ 2|Ω|

|E|(1 + (diam Ω)2)?u?2

W1,2(Ω).

Consequently, we obtain

????u − uE−1

2(∇u − (∇u)T)E(x − xE)

?

????

W1,2(Ω)

?1

≤

≤ 1 +

?

2|Ω|

|E|(1 + (diam Ω)2)

2?

K(1 + Q2)

1

2??∇u?L2(Ω),

for every u ∈ W1,2(Ω;Rn) .

(5.3)

8

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Example 5.4. When Ω and E are concentric balls, say Ω = B1(0), E = Bρ(0),

0 < ρ < 1. In this case we evaluate

?n + 3

n + 2

Consequently

?

n + 2

?T? ≤

?1

ρ

?n

2

.

?u − Tu?W1,2(B1(0))≤ 1 +

?n + 3

?1

for every u ∈ W1,2(B1(0);Rn) .

ρ

?n

2?

K(1 + Q2)

1

2??∇u?L2(B1(0)),

(5.4)

This bound improves the one in [1, Lemma 3.5] in two respects. It applies to

any dimension and the exponent of1

ρis diminished.

Example 5.5. We examine a version of Korn inequality when constraints on u

are taken on its trace on an open portion Γ of ∂Ω. The following Lemma will

be useful.

Lemma 5.6. The restriction to R of the trace imbedding γ : W1,2(Ω;Rn) → L2(Γ;Rn)

is one to one.

Proof. With no loss of generality we assume xΓ = 0. Being Γ an (n − 1)−

dimensional hypersurface, there exist v1,...,vn−1 ∈ Γ which are linearly in-

dependent. We show that, given r ∈ R such that γ(r) = 0 we have r = 0.

In fact we have r(0) = γ(r)Γ = 0 and also r(vi) = γ(r)(vi) = 0 for every

i = 1,...,n−1. It is well known that such constraints imply r = 0. We provide

a proof for the sake of completeness. Let us write r(x) = b + Mx with b ∈ Rn

and M ∈ Skewn. We have b = r(0) = 0, hence it remains to prove that the

conditions r(vi) = 0 ,i = 1,...,n − 1 imply M = 0. Up to a rotation in the

reference system, we can assume that the n−th component of the vectors vi,

i = 1,...,n − 1, is zero. Thus we have

(v1|···|vn−1|0) =

0

...

0

0

W

0 ···0

,

where W is a nonsingular (n − 1) × (n − 1) matrix. We have

Mvi= r(vi) = 0 , for every i = 1,...,n − 1 .

Consequently

M

0

...

0

0

W

0 ···0

= 0 ,

and also

M

0

...

0

0

I

0 ···0

= M

0

...

0

0

W

0···0

0

...

0

0

W−1

0 ···0

= 0 ,

9

Page 10

where I is the (n−1)×(n−1) identity matrix. It follows that the entries below

the diagonal of M are all zero and, consequently, M = 0.

Let us denote by E : γ(R) → R the inverse of γ|R, and by Π : L2(Γ;Rn) → γ(R)

the orthogonal projection onto γ(R) with respect to the L2(Γ;Rn) inner prod-

uct. We also introduce ρ : L2(Γ;Rn) → R as the composition ρ = E ◦ Π.

Theorem 5.7. Given an open portion Γ of ∂Ω, we have

?u − ρ(γ(u))?W1,2(Ω)≤ (1 + CΓ?E?)K(1 + Q2)

1

2??∇u?L2(Ω),

for every u ∈ W1,2(Ω;Rn) .

(5.5)

Here ?E? denotes the L(L2(Γ;Rn),W1,2(Ω;Rn))−norm of E, CΓ is the con-

stant in the inequality for the trace imbedding γ : W1,2(Ω) → L2(Γ) and Q is

the constant for the Poincar´ e inequality (3.5).

Remark 5.8. Arguing as in Remark 4.8, also the calculation of the norm ?E?

reduces to solving a finite dimensional eigenvalue problem.

The proof is straightforward, as well as the one of the following Corollary.

Corollary 5.9. There exist C1, C2> 0, depending only on Ω and Γ, such that

?u?W1,2(Ω)≤ C1??∇u?L2(Ω)+ C2?γ(u)?L2(Γ),

Acknowledgement. The authors wish to thank an anonymous referee for pro-

viding fundamental bibliographical information which was missing in a previous

version of this paper.

for every u ∈ W1,2(Ω;Rn) .

(5.6)

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