The linear constraints in Poincar\'{e} and Korn type inequalities

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. Comment: Revised version

Full text

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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.
1Introduction
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)
3 The 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)
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) ,

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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?.
5Korn 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) .

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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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