Dipartimento di Matematica
Corso di Laurea in Matematica
Maggio 2015
Tesi di Laurea Magistrale
A rigidity theorem and its
application to the problem of
crystallization
Candidato Giacomo Del Nin
Relatore Controrelatore
Prof. Prof.
Introduction 1
1 Geometric rigidity 8
1.1 Preliminary results . . . 8
1.1.1 Notation and basic facts about matrices . . . 8
1.1.2 Extension of Lipschitz functions . . . 10
1.1.3 Liouville’s Theorem . . . 11
1.1.4 Korn’s inequality . . . 12
1.1.5 Whitney’s covering theorem for dyadic cubes . . . 14
1.1.6 Approximation of W1,p with Lipschitz functions . . . 17
1.1.7 A weighted Poincar´e inequality . . . 22
1.2 The Rigidity Theorem . . . 23
1.2.1 A version on cubic domains . . . 23
1.2.2 The general case . . . 29
2 Crystallization 32 2.1 Overview . . . 32
2.1.1 Properties of the triangular lattice . . . 34
2.1.2 Some preliminary lemmas . . . 35
2.2 Existence of a canonical labelling φ . . . 41
2.3 Rigidity results . . . 44
2.4 Energy excess in terms of local dislocations . . . 48
2.4.1 λ-triangles . . . 49
2.4.2 Main estimate . . . 53
2.5 Crystallization in terms of the energy gap . . . 57
Bibliography 59
This exposition is divided in two parts. The first concerns the proof of the rigidity theorem by Friesecke, James and M¨uller [6], while the second one concerns its application to the solution of a problem in crystallization, following the work by Theil [14].
Rigidity Theorem A classical result by Liouville states that, given a map v : Rn → Rn, if its gradient ∇v(x) is at every point an element of SO(n), then it coincides with a constant rotation R ∈ SO(n), and thus v is a rigid motion. This is an example of a rigidity result, where from local rigid behaviour follows global rigid behaviour.
A generalization of this theorem is due to Reˇsetnjak [13], who has proved the following: if vj * v in W1,2(U, Rn) and
dist(∇vj, SO(n)) → 0
in measure, then ∇vj → R in L2(U ) where R is a constant rotation. However this
theorem does not give a quantitative rate for the convergence, which is a key property required for the derivation of many mathematical models in solid mechanics.
We will discuss here a generalization of these results proven by Friesecke, James and M¨uller in 2002 [6]. This is a quantitative inequality involving the L2-distance of
the gradient from SO(n). Precisely:
Theorem 1. Given an open, bounded, connected Lipschitz domain U ⊂ Rn there
exists a constant C(U ) such that, for every map v ∈ W1,2(U, Rn), there exists a rotation R ∈ SO(n) such that
k∇v − RkL2(U )≤ C(U )kdist(∇v, SO(n)kL2(U ).
That is: if on average ∇v is close to SO(n), then it is close to a single rotation R ∈ SO(n), up to a constant that depends only on the domain (and actually only on its shape). The case in which the right term is zero reduces to Liouville’s theorem. In the first part of this chapter we will prove some preliminary results, such as Korn’s inequality and a weighted Poincar´e inequality, where the gradient is weighted with the squared distance from the boundary. In the second part we will use these results to prove the theorem.
Crystallization The second part concerns the proof of a crystallization result, starting from the rigidity theorem. Crystallization refers to the emergence of a regular, periodic pattern at the microscopic level of a certain system, that sometimes results also in a macroscopic regularity. Examples of these patterns are periodic atomistic configurations of many materials at low temperatures.
A simple model of such a system is a finite number N of particles that interact in pairs by means of a potential V : [0, +∞) → R, that depends only on the distance between the particles. The goal is to understand if and how a pattern emerges in ground states, which are those who minimize the total energy obtained summing among the set of pairs P
E(y) = 1 2 X x6=x0 V (|y(x) − y(x0)|) =X p∈P E(p)
where y is the function that sends the particle x to its position. The natural ambient space modelling the physical situation would be R3; however there has been only recently a development in that direction (see [5]). For this reason we investigate only the two-dimensional case. Usual questions regarding crystallization are:
• Does the energy per particle of the ground states converges to a constant as the number of particles goes to infinity? Is this constant the energy per particle of a regular lattice?
• Are ground states subsets of (a rototranslated copy of) the regular lattice? If no, are they close to the lattice in some sense?
• Does a global shape emerge from ground states as N → ∞?
Positive answers to these questions have been given in particular cases, for example for the sticky disk potential [7] or the soft disk potential [12], or in general short-ranged potentials [15] (see Figure 1), for which a crystallization in the triangular lattice is obtained.
For these potentials results include one or more of the following:
• The energy per particles in ground states converges to the energy per particle in the triangular lattice.
• Ground states are crystallized, i.e. they are subsets of a rototranslated copy of the triangular lattice.
• The global shape of the cluster converges to a regular hexagon as the number of particles N goes to ∞. With this we mean that if we associate to a ground state the measure
µN = 1 N N X i=1 δx i/ √ N
which is renormalized and rescaled in order to occupy a fixed region, then up to rototranslations µN
∗
1 −1
+∞ +∞
1 −1
Figure 1: Sticky disk potential (left) and a generic short-ranged potential (right). In these cases the analysis is simplified by the fact that the potential is short-ranged, i.e. non zero contributions to energy come from ”neighbouring” particles only. The work by Theil [14] on which the current exposition is based investigate instead long-ranged potentials, in which only a sufficiently rapid decay and no compact support is required, and by which even distant particles interact with each other. An essential tool to solve this case is the rigidity theorem from the first part, which allows to control energy contributions due to long-range interactions with contributions due to neighbouring pairs only. Subsequent work by Weinen E and Dong Li [2] used the same approach to obtain crystallization in the hexagonal lattice when an angle potential is added, which involves triples of particles.
Let us see more in detail the reults by Theil. We define A2 = 1 2 2 1 0 √3 Z2
as the regular triangular lattice of lengthscale 1, and U = 1 2 2 1 0 √3 Q
as the fundamental cell, where Q = [0, 1)2. Given a potential V : [0, ∞) → R define
V∗(r) = 1 6 X ξ∈A2 V (r|ξ|)
which coincides with the energy associated to a single particle in the triangular lattice dilated by a factor r, divided by the number 6 of neighbouring particles. We will assume in the following that V is normalized, i.e.
min
r>0 V∗(r) = V∗(1) = −1
Theorem 0.1 (Free setting). There exists a constant α0 ∈ (0,13) such that for every
α ∈ (0, α0) and every potential V : [0, ∞) → R with the properties V ∈ C2(1 − α, ∞)
and V (r) ≥ 1 α for r ∈ [0, 1 − α) (1) V00(r) ≥ 1 for r ∈ (1 − α, 1 + α) (2) V (r) ≥ −α for r ∈ [1 + α,4 3] (3) |V00(r)| ≤ αr−7 for r ∈ (4 3, ∞) (4)
the following holds: lim N →∞y:XminN→R2 1 N X x,x0∈X N x6=x0 V (|y(x) − y(x0)|) = −3.
We obtain thus, in the free setting, the convergence of the ground states energy per particle to a constant, which is precisely the energy per particle in the lattice A2. The goal to characterize ground states is however more difficult in the free case,
and to obtain some results in this direction we need to impose some appropriate ”boundary conditions”. We obtain that in two cases ground states coincide with the lattice A2. In these cases the number of particles is infinite and thus it is necessary
also a different definition of energy:
Theorem 0.2 (Periodic setting). There exists a constant α0 such that for every
α ∈ (0, α0), every L ∈ N, every potential V that satisfies the hypotheses from previous
theorem and every ground state ymin : A2 → R2 of the periodic energy
ELper(y) = X
x∈A2∩LU
x0∈A2\{x}
V (|y(x) − y(x0)|)
under the periodicity assumption
y ∈ YLper = {y : A2 → R2|y(x) − y(x0) = x − x0 if x − x0 ∈ LA2}
there exists a translation vector τ ∈ R2 such that ymin(A2) = τ + A2.
Theorem 0.3 (Compact-perturbations setting). Fix A ⊂ A2 a finite set. Suppose
that hypotheses on potential of theorem 0.1 are satisfied. If ymin : A2 → R2 is a
ground state for
EA(y) = X {x,x0}⊂A 2 {x,x0}∩A6=∅ V (|y(x) − y(x0)|)
under the compact-perturbation assumption
y ∈ YA = {y : A2 → R2|y(x) = x ∀x ∈ A2\A}
then
ymin(A2) = A2.
These two results are presented by Theil as corollaries of theorem 0.1. In this exposition we have faced the compact-perturbations case only, trying to give a more direct proof of it, and to obtain a rigidity result also for configurations with sufficiently low energy. The ideal result would be the following:
Theorem 0.4. There exist α0, C, ∆0 > 0 such that for every finite A ⊂ A2, for
every α ∈ (0, α0), every potential satisfying hypotheses of theorem 0.1, if 0 < ∆ < ∆0
and y : A2 → R2 is a configuration satisfying the compact-perturbation assumption
y ∈ YA = {y : A2 → R2|y(x) = x ∀x ∈ A2\A}
and with energy
EA(y) < EA(ymin) + ∆
then
dist(y(A2), A2) ≤ C∆.
The best result would be to obtain α0 independent of the perturbed zone A.
However, in the case considered in this exposition, we obtain a dependence of α0
on A, in particular on its diameter diam A = L, so that α0(L) → 0 for L → ∞.
Therefore if we want to obtain a universal statement we need a potential that satisfies conditions of theorem 0.1 for all α > 0, and in this limit we obtain simply the sticky disk potential, for which crystallization results are already known even in the free setting.
Moreover, in the proof presented here, we need another hypothesis, namely that α is small enough to make the local imbedding property hold (see Definition 2.5), which roughly says that every particle has 6 neighbours (particles at distance near to 1). We remark that instead proving the free setting theorem and deriving this theorem as a corollary, as done by Theil, gives a universal α0 and does not need this
supplementary hypothesis. The result presented here is thus far from being optimal, but is derived without many technical details that appear in the proof of theorem 0.1, of which we anyway adapt the fundamental steps.
LU
Figure 2: L-periodic configuration.
Diu Partial weak derivative of u
∇u (Weak) gradient of u
W1,p(U, Rn) Sobolev space of vector-valued functions, U ⊂ Rn
cof A Cofactor matrix of A ∈ Mn×n (see Subsection 1.1.1) A2 Triangular lattice on R2
V∗ V∗(r) = 16Pξ∈A2V (r|ξ|)
Λ Set of all distances between points in A2: Λ = {|ξ| : ξ ∈ A2}
A Perturbed zone: finite set contained in A2
Aλ Enlarged zone of A (see beginning of Section 2.4)
P Set of all pairs {x, x0} ⊂ A2 intersecting A
N (x) Neighbours of a particle x ∈ y(A2) (see Definition 2.5)
S Set of short-range interactions (see Definition 2.5) L Set of long-range interactions: L = P\S
Tλ Set of λ-triangles (see Subsection 2.4.1)
T Set of all triangles: T =S
λTλ
In the first part of this section we will prove some preliminary results, such as Korn’s inequality and a weighted Poincar´e inequality, where the gradient is weighted with the squared distance from the boundary. In the second part we will use these results to prove the theorem.
1.1
Preliminary results
1.1.1
Notation and basic facts about matrices
• We equip the linear space Mn×n of n × n matrix with two norms:
– the Hilbert-Schmidt, or Euclidean, norm |A| = s X i,j a2 i,j
which descends from the scalar product A · B =X
i,j
ai,jbi,j;
– the operator norm
kAk = sup
|v|=1
|Av| .
Since Mn×n is finite dimensional, these two norms are equivalent. We will use mainly the first one, and define the associated distance dist(A, B) = |A − B| and
dist(A, SO(n)) = inf
G∈SO(n)|A − G|.
Note that since SO(n) is compact the infimum is attained. If A is invertible then
dist(A, SO(n)) = |√AtA − I|
where Atis the transposed matrix of A, and if A is not invertible the inequality
dist(A, SO(n)) ≥ |√AtA − I|
still holds (see [6], chapter 2).
We will need in the following (and precisely in proposition 1.12) the estimate dist(G, SO(n)) = G + Gt 2 − I + O(|G − I|2).
This can be proved in the following way: first observe that if T = TISO(n) is
the tangent space to SO(n) at the identity then
dist(G, SO(n)) = dist(G, I + T ) + O(|G − I|2).
Since T coincides with the space Skewn of skew-symmetric matrices then
dist(G, SO(n)) = dist(G − I, Skewn) + O(|G − I|2)
but G can be decomposed as the sum of a symmetric and a skew part G = G + G t 2 + G − Gt 2 = G s+ Ga
and this decomposition is orthogonal, thus dist(G − I, Skewn) = |(G − I)s| = G + Gt 2 − I . • We will denote by cof A the cofactor matrix of A, that is
( cof A)i,j = (−1)i+jdet ˆAi,j
where ˆAi,j is the minor obtained from A removing row i and column j. If A
is invertible then, dividing cof A by det A, we obtain the classical formula for the transposed inverse (A−1)t. Even if A is not invertible it still holds
A ( cof A)t= detA I.
Observe that if A ∈ SO(n) then cof A = A. • It will be used that
div cof ∇v = 0
for all v ∈ W1,2(U ). This is a purely algebraic identity. A proof can be found in [9] (available also as a preprint [10]). For a more general approach see [1]. Throughout all the exposition the letter C stands for a constant that can vary from line to line, but is dependent only on the dimension n.
We refer for example to [3] for basic facts concerning Sobolev spaces and distri-butions.
1.1.2
Extension of Lipschitz functions
We prove an extension theorem for real-valued Lipschitz functions on a general metric space, and then as a corollary an extension theorem for vector-valued Lipschitz functions on Euclidean spaces.
Given two metric spaces (X, dX) and (Y, dY) and a Lipschitz function f : X → Y
we define the Lipschitz constant of f as Lip (f ) = sup
s6=t
dY(f (s), f (t))
dX(s, t)
. In this way dY(f (s), f (t)) ≤ Lip (f )dX(s, t) for every s, t ∈ X.
Theorem 1.1. Let S be a subset of a metric space (X, d), and f : S → R a Lipschitz function with Lipschitz constant Lip (f ) = L. Then there exists an extension
˜ f : X → R of f with Lip ( ˜f ) ≤ L. Proof. Define ˜f as ˜ f (x) = inf s∈S{f (s) + Ld(x, s)}.
We have to prove that
| ˜f (x) − ˜f (y)| ≤ Ld(x, y).
By definition, given y ∈ X and ε > 0 we find ¯s ∈ S such that f (¯s) + Ld(y, ¯s) ≤ ˜f (y) + ε. But then, using triangle inequality we get
˜
f (x) ≤ f (¯s) + Ld(x, ¯s) ≤ f (¯s) + Ld(y, ¯s) + Ld(x, y) ≤ ˜f (y) + ε + Ld(x, y). This is true for every ε > 0 and thus
˜
f (x) − ˜f (y) ≤ Ld(x, y). The reverse inequality is proved interchanging x and y.
As an easy consequence we obtain a Lipschitz extension theorem for vector-valued functions on euclidean spaces.
Theorem 1.2. Let S be a subset of Rn, and f : S → Rm a Lipschitz function with
Lipschitz constant Lip (f ) = L. Then there exists an extension ˜f : Rn → Rm of f with Lip ( ˜f ) ≤√mL.
Proof. Write f = (f1, . . . , fm), and apply the previous theorem to each component.
We obtain ˜f = ( ˜f1, . . . , ˜fm) with
for every i = 1, . . . , m. But then | ˜f (x) − ˜f (y)| = m X i=1 ( ˜fi(x) − ˜fi(y))2 !12 ≤ m X i=1 L2|x − y|2 !12 = L√m|x − y|.
1.1.3
Liouville’s Theorem
We give here a modern proof of the theorem, presented in the paper by Friesecke, James and M¨uller ([6], chapter 3, page 1469).
Theorem 1.3. Let U ⊂ Rn be an open bounded and connected Lipschitz domain,
and v ∈ W1,2(U, Rn) such that
∇v ∈ SO(n) a.e.. (1.1)
Then ∇v(x) = R, with R a constant matrix in SO(n), and thus u is a rigid motion. Proof. The proof is based on two facts:
(i) the identity
1 2∆|∇v|
2 = ∇v · ∆∇v + |∇2v|2, (1.2)
where the scalar product is taken between matrices component by component, as introduced in the notation section;
(ii) v is harmonic.
To prove the identity (1.2), let (vjk)j,k = ( ∂vj
∂xk) be the matrix of the gradient ∇v
(this notation will not be used anywhere else). Then 1 2∆|∇v| 2 = 1 2 X i ∂2 ∂x2 i X j,k vjk2 ! = 1 2 X i,j,k ∂ ∂xi 2vjk ∂vjk ∂xi =X i,j,k ∂vjk ∂xi 2 + vjk ∂2v jk ∂x2 i ! =X i,j,k ∂vjk ∂xi 2 +X j,k vjk X i ∂2v jk ∂x2 i ! = |∇2v|2+X j,k vjk∆vj,k = |∇2v|2+ ∇v · ∆∇v.
To prove that v is harmonic, we recall that since ∇v ∈ SO(n) then cof ∇v = ∇v (see subsection 1.1.1), and by taking the divergence and using that div( cof ∇v) = 0 we conclude that ∆v = 0.
Since v is harmonic, so is ∇v, and from (1.2) we deduce that ∇2v = 0, thus ∇v is
constant.
1.1.4
Korn’s inequality
There are two main Korn’s inequalities. We will prove here the first one, and derive from the second one a version that suits our purposes. The proof of the second Korn’s inequality is not straightforward as the first one, and can be found for example in [8] or [4].
Theorem 1.4 (First Korn’s inequality). Let U be a Lipschitz domain and u ∈ W01,2(U ), and let E(u) = 12(∇u + ∇ut) denote the symmetric part of the gradient
∇u. Then ˆ U |∇u|2dx ≤ 2 ˆ U |E(u)|2dx.
Proof. This is a consequence of an integration by parts. We first suppose u ∈ C0∞(U ), and identify it with its canonical extension to Rn. Let uij = ∂x∂ui
j. Then |E(u)|2 = 1 4 X i,j (uij + uji)2 = 1 4 X i,j (u2ij + 2uijuji+ u2ji) = 1 2|∇u| 2 +1 2 X i,j uijuji. Moreover we have ∂ui ∂xj ∂uj ∂xi = ∂ ∂xj ui ∂uj ∂xi − ui ∂2uj ∂xj∂xi = ∂ ∂xj ui ∂uj ∂xi − ∂ ∂xi ui ∂uj ∂xj +∂ui ∂xi ∂uj ∂xj . (1.3)
Using the identity ˆ
U ∂f ∂xi (x)dx = ˆ ∂U f (x)νi(x)dHn−1(x)
where νi(x) is the ith component of the exterior normal ν at the point x ∈ ∂U we
obtain that, upon integrating (1.3) over U , the first two terms in the last line of (1.3) vanish, and after summing among i and j we end up with
ˆ U X i,j ∂ui ∂xi ∂uj ∂xj = ˆ U (div u)2.
In the end we obtain ˆ U |E(u)|2 = 1 2 ˆ U |∇u|2+ 1 2 ˆ U (div u)2
from which the thesis follows.
If now u belongs to W01,2(U ), then we can find un ∈ C0∞(U ) with un → u in
W1,2. This in particular implies ∇u
n → ∇u, and consequentially E(un) → E(u),
so that k∇unk2 → k∇uk2 and kE(un)k2 → kE(u)k2, from which the thesis for the
regular functions un passes to the limit.
Theorem 1.5 (Second Korn’s Inequality). Let U be a Lipschitz domain. There exists C such that for all maps u ∈ W1,2(U, Rn)
k∇uk2 ≤ C(kuk2+ kE(u)k2).
For our purposes we need a Poincar´e-type inequality, which we will derive from the previous one, namely:
Theorem 1.6 (Korn’s inequality). Let U be a Lipschitz domain and u ∈ W1,2(U ).
Then there exists C such that ˆ U |∇u − (∇u)U|2 ≤ C ˆ U |E(u)|2
where (∇u)U is the average of ∇u over U .
Proof. We use the symbol . to mean that the inequality holds up to constants depending on the domain only. The thesis is therefore
k∇u − (∇u)Uk2 . kE(u)k2. (1.4)
We assume that theorem 1.5 holds, or equivalently that kuk2+ kE(u)k2 defines a
norm equivalent to kukW1,2.
Step 1 The thesis is equivalent to
k∇uk2 . |(∇u)U| + kE(u)k2. (1.5)
Indeed, if (1.4) holds, then by triangle inequality we obtain
k∇uk2 ≤ k∇u − (∇u)Uk2+ k(∇u)Uk2 . |(∇u)U| + kE(u)k.
To show the other implication, given u define ˜u(x) = u(x) − M x with M = (∇u)U
(in this way (∇˜u)U = 0) and apply (1.5) to ˜u to obtain
Step 2 The functional
ψ(u) = |uU| + |(∇u)U| + kE(u)k2
defines a norm equivalent to k · kW1,2.
Indeed, ψ(u) . kukW1,2 is simple and follows from Poincar´e. Suppose by
contra-diction that kukW1,2 6. ψ(u). Then there exists a sequence uj such that
• (uj)U → 0;
• (∇uj)U → 0;
• E(uj) → 0 in L2.
and kujkW1,2 = 1. In particular, from Sobolev compactness theorem we obtain up
to subsequences that uj → u in L2. Moreover by Banach-Alaoglu Theorem uj * u
in W1,2, so that E(u
j) * E(u). But E(uj) → 0 in L2, therefore uj → u in W1,2
because we are assuming that Theorem 1.5 holds, and
E(u) = 0, (u)U = 0 (∇u)U = 0. (1.6)
We now want to show that these conditions imply u = 0, which is a contradiction because kujk = 1 and uj → u.
From E(u) = 0 we obtain Diuj+ Djui and Diui = 0, and thus 0 = Diiuj+ Dijui =
Diiuj + Dj(Diui) = Diiuj. This means that every component uj is affine, and so is
u. But now if we write u(x) = a + N x we obtain from the third condition in (1.6) that N = 0, and from the second that a = 0.
Step 3 If φ(u) ≈ kukW1,2 then (1.5) holds.
Indeed
k∇uk2 ≤ kukW1,2 . ψ(u) = |(u)U| + |(∇u)U| + kE(u)k2.
It is now sufficient to observe that the left hand side is invariant under translations on the codomain, so that we can choose |(u)U| = 0, and this yields the thesis.
1.1.5
Whitney’s covering theorem for dyadic cubes
Given a ∈ Rn and r > 0 let
Q(a, r) = a + r[−1/2, 1/2)n
Definition 1.7. Let Q0,0 = [0, 1)n ⊂ Rn, j ∈ Z and k ∈ Zn. Define the dyadic cube
of generation j and lower corner 2−jk as
Qj,k = 2−jk + 2−jQ0,0.
Given a dyadic cube Q = Qj,k, its parent ˆQ is defined as the unique dyadic cube of
generation j − 1 that contains Q.
Theorem 1.8. (Whitney’s Covering) Let U ⊂ Rn be a bounded open set. Then there
exists a disjoint collection F = {Qi} of dyadic cubes that covers U and such that:
(i)
1
30d(Qi, ∂U ) ≤ diam(Qi) ≤ 1
10d(Qi, ∂U );
(ii) there exists a constant N dependent on the dimension only such that if we write Qi = Q(ai, ri) then for every x ∈ U there are at most N indices i such
that x ∈ Q(ai, 4ri). Equivalently,
X
i
χQ(ai,4ri) ≤ N.
In addition, Q(ai, 4ri) ⊂ U for every i.
Proof. (i) Define Fj as the collection of dyadic cubes of generation j contained in U
such that
diam Q ≤ 1
10d(Q, ∂U ). (1.7) Since U is open there exists an index j for which Fj is not empty, and let ¯ be
the minimum of such indices (which is finite by the boundedness of U ). Define the collection F as the union of the collections eFj, j ≥ ¯j that are defined inductively by
e F¯j = F¯j and e Fj = Fj\ [ ¯j≤i≤j−1 e Fi
for j ≥ ¯j + 1. This is the collection obtained taking at every generation all dyadic cubes that satisfy condition (1.7), starting from the biggest generation possible. By the openness of U this is a covering, and the upper bound on the diameters is satisfied by definition. It remains to prove the lower bound.
Take Q ∈ F . By construction, either ˆQ 6⊂ U or d( ˆQ, ∂U ) < 10 diam ˆQ. In either case d( ˆQ, ∂U ) ≤ 10 diam ˆQ
and so
d(Q, ∂U ) ≤ (10 + 1
(ii) We write for simplicity λQi instead of Q(ai, λri). Fix x ∈ U .
We first claim that if x ∈ 4Qi then
Qi ⊂ B x,1 2d(x, ∂U ) .
Indeed, let di = dist(Qi, ∂U ). Then diam Qi ≤ 101 di and thus for every z ∈ 4Qi we
have d(z, ∂U ) ≥ di− 3 2diam Qi ≥ 17 20di. In particular, since x ∈ 4Qi, it follows
d(Qi, ∂U ) ≤ 20 17d(x, ∂U ). Moreover we have diam Qi ≤ d 10 ≤ 2 17d(x, ∂U ) and thus for all y ∈ Qi we have
d(x, y) ≤ 5 2diam Qi ≤ 5 17d(x, ∂U ) ≤ 1 2d(x, ∂U ) which gives the claim.
Now we claim that if x ∈ 4Qi then
diam Qi ≥
1
50d(x, ∂U ). Indeed as above we have
diam Qi ≤
1 10di and for every z ∈ 4Qi
d(z, ∂U ) ≤ di+
5
2diam Qi ≤ 5 4di. In particular this is true for z = x, thus it follows
diam Qi ≥ 1 30di ≥ 2 75d(x, ∂U ) ≥ 1 50d(x, ∂U ) proving the claim.
In the end we proved that whenever 4Qi contains x then Qi is contained in the
ball B(x,12d(x, ∂U )), and that |Qi| has diameter greater than 501d(x, ∂U ), and thus
in particular contains a ball of radius 1001√
nd(x, ∂U ). But since the Qi’s are disjoint
this means that there can be at most |B(x,1
2d(x, ∂U ))
|B(0, 1
100√nd(x, ∂U ))|
= N (n) such Qi’s, which yields the thesis.
It can be easily seen that the bounds on the diameter (i), in particular 10 diam Qi ≤
1.1.6
Approximation of W
1,pwith Lipschitz functions
Proposition 1.9. Let n, m ≥ 1 and let 1 ≤ p < ∞. Suppose U ⊂ Rn is a bounded
Lipschitz domain. Then there exists a constant C = C(U, m, p) with the following property: for each u ∈ W1,p(U, Rm) and each λ > 0 there exists a Lipschitz function v : U → Rm such that (i) k∇vkL∞(U )≤ Cλ (ii) |{x ∈ U : u(x) 6= v(x)}| ≤ C λp ˆ {x∈U :|∇u(x)|>λ} |∇u|pdx (iii) k∇u − ∇vkpLp(U )≤ C ˆ {x∈U :|∇u(x)|>λ} |∇u|pdx
Proof. We first observe that (iii) follows from (i) and (ii). Indeed, using the fact that u = v a.e. implies ∇u = ∇v a.e., we obtain
ˆ U |∇u − ∇v|pdx = ˆ u6=v |∇u − ∇v|pdx ≤ 2p ˆ u6=v (|∇u|p+ |∇v|p)dx ≤ 2p ˆ u6=v ((Cλ)p+ |∇v|p)dx + 2p ˆ |∇u|>λ |∇u|pdx ≤ C ˆ |∇u|>λ |∇u|pdx
We now prove (i) and (ii).
Step 1 The proposition holds for U = U0 = (0, 1)n−1× (0, H), with a constant
depending on H. To show this we follow an argument from Evans and Gariepy [3]. Since the result (but not the constant C) is invariant under anisotropic dilations, we can assume that U = Q = (−1, 1)n.
Substep 1.1 Let Q0 be a cube concentric to Q and with side d2(2√n + 1)e (this enlarged cube has the fundamental property that if B(x, r) is a ball with center x ∈ Q and radius r ≤ 2√n, then it is contained in Q0). Then the proposition holds for u ∈ W1,p(Q0, Rn) and v ∈ W1,∞(Q, Rn), replacing in the inequalities on the left
hand sides U with Q, and on the right hand sides U with Q0. Define Rλ = ( x ∈ Q : 1 |B(x, r)| ˆ B(x,r) |∇u| ≤ λ, ∀r ≤ 2√n ) . (1.8)
The aim is to show that ˜u = u|Rλ is (Cλ)-Lipschitz, and that
|Q\Rλ| ≤ C
λp
ˆ
{|∇u|>λ}
|∇u|pdx
and consequently v can be chosen as a Lipschitz extension of ˜u, loosing at most a factor √m on the Lipschitz constant (see Theorem 1.2).
Let us first estimate |Q\Rλ|. For every x ∈ Q we can find a radius 0 ≤ r(x) ≤ 2√n
that violates the condition in (1.8). Using Vitali’s Lemma (see [3], theorem 1 in section 1.5.1) we extract a countable family {Bi} of disjoint balls Bi = B(xi, ri) such
that Q\Rλ ⊂[ i B(xi, 5ri). We can write λ ≤ 1 |Bi| ˆ Bi∩{|∇u|>λ2} |∇u|dx + 1 |Bi| ˆ Bi∩{|∇u|≤λ2} |∇u|dx ≤ 1 |Bi| ˆ Bi∩{|∇u|>λ2} |∇u|dx + λ 2 which implies |Bi| ≤ 2 λ ˆ Bi∩{|∇u|>λ2} |∇u|dx.
We can finally estimate |Q\Rλ| ≤X i |5Bi| ≤ 5n 2 λ ˆ {|∇u|>λ 2} |∇u|dx 5n2 λ ˆ {|∇u|>λ 2} |∇u|p 1/p |∇u| > λ 2 1/p0 ≤ 5n2 λ ˆ {|∇u|>λ 2} |∇u|p. (1.9)
We now want to show that u|Rλ is (Cλ)-Lipschitz, C being a universal constant.
Take x ∈ Rλ and r > 0. From the Poincar´e inequality we have ˆ B(x,r) |u − (u)x,r|dy ≤ Cr ˆ B(x,r) |∇u|dy ≤ Cλr where (u)x,r = B(x,r) u(z)dz
and the factor r comes from the scaling properties of the Poincar´e inequality. There-fore
|(u)x,r/2k+1− (u)x,r/2k| = |(u)x,r/2k+1− ((u)x,r/2k)x,r/2k+1|
≤
B(x,r/2k+1)
|u − (u)x,r/2k|dy
≤ Cλ r 2k.
Since for almost every x
u(x) = lim
r→0(u)x,r
then for almost every x in Rλ
|u(x) − (u)x,r| ≤
X
k
|(u)x,r/2k − (u)r/2k+1| ≤ Cλr.
Now take x, y in Rλ and r = |x − y|. Then
|(u)x,r− (u)y,r| ≤
B(x,r)∩B(y,r)
(|(u)x,r− u(z)|dz + |u(z) − (u)y,r|)dz
≤ C B(x,r) |u(z) − (u)x,r|dz + B(y,r)
|u(z) − (u)y,r|dz
≤ Cλr and therefore
|u(x) − u(y)| ≤ |u(x) − (u)x,r| + |(u)x,r− (u)y,r| + |(u)y,r− u(y)| ≤ Cλ|x − y|. (1.10)
We now extend u|Rλ to a Lipschitz map v defined on all Q, loosing at most a factor
√
m in the Lipschitz constant (see Theorem 1.2) which by (1.9) and (1.10) satisfies conditions (i) and (ii) of the proposition.
Substep 1.2 End of the proof of Step 1.
Given u ∈ W1,p(Q, Rn) we can extend it to Q0 by a reflection argument: first extend it to the domain Q1 made up of copies of Q translated along the x1 axis and
contained in Q0, by reflection. Then extend this new function again by reflection to the domain Q2 made up of copies of Q1 translated along x2 axis, etc. In the
end we obtain a function u0 in W1,p(Q0
, Rn), which has the property that |{x ∈ Q0 :
|∇u0(x)| > λ}| ≤ C|{x ∈ Q : |∇u(x)| > λ}| where C is the number of copies of Q
necessary to fill Q0. We can thus apply substep 1.1 to u0 and use this fact to obtain the statement of Step 1.
Step 2 The proposition holds for a standard Lipschitz domain, i.e. a domain of the form
Uf =(x0, xn) : x0 ∈ (0, 1)n−1, f (x0) < xn < f (x0) + H
where f is L-Lipschitz and the constant C depends only on H and L.
To see this, consider the standard bi-Lipschitz homeomorphism ϕ : U0 → Uf given
by
ϕ((y0, yn)) = (y0, f (y0) + yn).
It is easy to verify that it is L-Lipschitz and that it preserves volumes, i.e. | det J ϕ| = 1.
For any given u : Uf → Rn consider now the pullback ˜u(y) = u(ϕ(y)) defined on U0.
We have that
|∇˜u| = |(∇u)(ϕ(·))∇ϕ| ≤ L|(∇u)(ϕ(·))|. (1.11) From step 1 applied to ˜u and ˜λ = Lλ we obtain a map ˜v satisfying
k∇˜vkL∞(U
0) ≤ CLλ
and with a change of variables
|{y ∈ U0 : ˜u(y) 6= ˜v(y)}| ≤
C(H) ˜ λp ˆ {y∈U0:|∇˜u|>˜λ} |∇˜u(y)|pdy ≤ C(H) λp ˆ {y∈U0:|∇˜u|>˜λ} |(∇u)(ϕ(y))|pdy ≤ C(H) λp ˆ {x∈Uf:|∇u|>λ} |∇u(x)|pdx
since ϕ ({|∇˜u| > Lλ}) ⊆ {|∇u| > λ} from (1.11) and there is no Jacobian factor because ϕ is volume-preserving. If now we define v(x) = ˜v(ϕ−1(x)) then using (1.11) with ϕ−1 instead of ϕ we obtain
k∇vkL∞(U
f) ≤ CL
2
and
|{x ∈ Uf : u(x) 6= v(x)}| = |{y ∈ U0 : ˜u(y) 6= ˜v(y)}|
again because of volume preservation, and thus we have obtained the desired result with the new constant C = C(H, L) depending also on L2.
Step 3 The proposition holds for a general bounded Lipschitz domain.
By assumption U can be covered by finitely many open sets Ui such that Vi = Ui∩ U
is either a standard Lipschitz domain (up to rotation and translation) or a cube contained in U . From the previous step we obtain Lipschitz functions vi : Vi → Rm
such that Lip vi = k∇vik∞≤ Cλ and
|{x ∈ Ui : u(x) 6= vi(x)}| ≤ C λp ˆ {x∈Vi:|∇u|>λ} |∇u|pdx. (1.12)
Now consider a partition of unity {φi} subordinate to the covering {Ui}, i.e. φi ∈
C0∞(Ui), 0 ≤ φi ≤ 1 and Piφi = 1 on U . We can extend each vi to a Lipschitz
function on Rn loosing at most a factor √m in the Lipschitz constant (see Theorem
1.2). We now define v =X i φivi. Then v − u =X i φi(vi− u). We have |{x ∈ U : u 6= v}| ≤ X i |{x ∈ Vi : vi 6= u}| ≤ C λp ˆ {|∇u|>λ} |∇u|pdx.
It remains to prove the bound on the gradient. We have |∇v| = X i (φi∇vi+ vi⊗ ∇φi) ≤X i |∇vi| + X i vi⊗ ∇φi .
For the first term we have the desired bound from Step 2 (the φi’s are a finite number,
dependent on the covering, that is chosen at the beginning for a fixed domain U ). As for the second term, since P
i∇φi = ∇
P
iφi = 0, for every index j we have that
φj X i vi⊗ ∇φi = φj X i (vi− vj) ⊗ ∇φi ≤ C X i:Vi∩Vj6=∅ |vi− vj|. Set α = min Vi∩Vj6=∅ |Vi∩ Vj|.
Suppose that the inequality α 4 > C λp ˆ {|∇u|>λ} |∇u|p (1.13)
holds, with C exactly as in (1.12). Then, whenever Vi∩ Vj 6= ∅, there exists, from
Step 2, a point x ∈ Vi∩ Vj such that vi(x) = vj(x) = u(x), and so
sup
Vi∩Vj
|vi− vj| ≤ max{Lip vi, Lip vj} diam U ≤ Cλ.
On the other hand, if (1.13) fails, then the proposition is true with v = 0, since |U | ≤ |U |4C
αλp
ˆ
{|∇u|>λ}
|∇u|p.
1.1.7
A weighted Poincar´
e inequality
Proposition 1.10. Let U ⊂ Rnbe a bounded Lipschitz domain, and f ∈ W1,2(U, Mn×n).
Then min G∈Mn×n ˆ U |f − G|2dx ≤ C ˆ U dist(x, ∂U )2|∇f |2dx
with C dependent on the domain only.
Proof. We prove this proposition starting from Theorem 1.5 of [11]: given g ∈ Wloc1,2(U ) ∩ L2(U ) then ˆ U |g|2dx ≤ C1 U ˆ U (|g|2+ |∇g|2)dist(x, ∂U )2dx. (1.14)
Now fix δ > 0 such that CU1δ2 ≤ 1
2, and let Uδ = {x ∈ U : dist(x, ∂U ) > δ}. By the
ordinary Poincar´e inequality for Uδ there exists G ∈ Mn×n such that
ˆ Uδ |f − G|2dx ≤ CUδ ˆ Uδ |∇f |2dx ≤ CUδ δ2 ˆ Uδ |∇f |2dist(x, ∂U )2dx. (1.15)
Applying (1.14) with g = f − G we obtain ˆ U |f − G|2dx ≤ C1 U ˆ U (|f − G|2+ |∇f |2)dist(x, ∂U )2dx = CU1 ˆ U |f − G|2dist(x, ∂U )2dx + C1 U ˆ U |∇f |2dist(x, ∂U )2dx. (1.16)
To estimate the first addendum of the last term we split the domain U as Uδ∪ (U \Uδ)
and use that dist(x, ∂U )2CU1 ≤ 1
2 for x ∈ U \Uδ: CU1 ˆ U |f − G|2dist(x, ∂U )2dx ≤ CU1kd(·, ∂U )2kL∞(U δ) ˆ Uδ |f − G|2dx + CU1δ2 ˆ U \Uδ |f − G|2 (1.15) ≤ C1 U CUδ δ2 kd(·, ∂U ) 2k L∞(U δ) ˆ Uδ |∇f |2dist(x, ∂U )2dx +1 2 ˆ U \Uδ |f − G|2
and plugging this into (1.16) we obtain the thesis.
1.2
The Rigidity Theorem
The main rigidity theorem in the paper [6] by Friesecke, James and M¨uller is the following:
Theorem 1.11. Let U be an open, bounded, connected Lipschitz domain in Rn, n ≥ 2. There exists a constant C(U ) with the following property: for each v ∈ W1,2(U, Rn)
there is a R ∈ SO(n) such that
k∇v − RkL2(U ) ≤ C(U )kdist(∇v, SO(n))kL2(U ).
Remark. The constant is invariant under rescalings of the domain: given r > 0 and ur : rU → Rn we can define u : U → Rn by
u(x) = 1
rur(rx). Then
∇u(x) = ∇ur(rx)
so the same rotation R that works for u works for its rescaled ur.
To prove this theorem we will firstly prove a slightly different version in which the domain is a cube; then we will recover the result for a generic bounded Lipschitz domain by a suitably chosen covering with cubes and the weighted Poincar´e inequality from the first section.
1.2.1
A version on cubic domains
Proposition 1.12. Let Q be a cube in Rn, n ≥ 2, and Q0 ⊆ Q a concentric cube with
half the side length of Q. There exists a constant C(n) with the following property: for each v ∈ W1,2(Q, Rn) there is an associated rotation R ∈ SO(n) such that
k∇v − RkL2(Q0)≤ C(n)kdist(∇v, SO(n))kL2(Q). (1.17)
Remark. Notice that this is not exactly the cubic-domain case of the main theorem, since this time the L2-norm is taken over all Q on the right hand side, and over
the smaller Q0 on the left hand side. Observe that the constant, as in the previous remark, is invariant under rescaling of the domain, thus it suffices to prove the proposition when Q0 is the unit cube.
Proof. The proof is divided in some steps:
• We reduce to the case where k∇vk∞ is bounded by a constant dependent on
the dimension only, using the approximation result from section 1.1.6.
• We decompose v = w + z as the sum of a harmonic part w that carries the information about the boundary values of v, and the remaining part z that vanishes on the boundary but carries the information about the laplacian of v. By the triangle inequality it suffices to find a bound of the type (1.17) for k∇zkL2 and k∇w − RkL2 separately. ∇z is ruled out easily; instead handling
the harmonic part ∇w is more difficult, and the rest of the proof is devoted to this.
• We find a bound for k∇w − RkL2 in terms of the right hand side of (1.17)
raised to the power 1/2 (weaker version).
• Linearizing near the identity we obtain the correct bound (without the power 1/2) for the symmetric part of the gradient, and using Korn’s inequality (see section 1.1.4) we then control the skew part as well, thus finishing the proof. Step 1. It suffices to prove Proposition 1.12 for maps v with k∇vk∞≤ M , M
being a constant dependent on the dimension only.
Indeed, note that if A ∈ Mn×n and |A| ≥ 2√n, then |A| ≤ 2dist(A, SO(n)). To
prove this let B ∈ SO(n) be such that dist(A, SO(n)) = |A − B|, which exists by the compactness of SO(n). Then
|A| ≤ |A − B| + |B| = dist(A, SO(n)) +√n ≤ dist(A, SO(n)) +|A| 2 and we conclude rearranging the terms of this inequality.
that k∇V k∞≤ 2 √ nC, {v 6= V } ≤ C 4n ˆ {x∈Q:|∇v(x)|>2√n} |∇v|2dx, k∇V − ∇vk2 L2(Q)≤ C ˆ {x∈Q:|∇v(x)|>2√n} |∇v|2dx ≤ 4C ˆ Q dist(∇v, SO(n))2dx,
where the last inequality comes from the observation above. Now suppose we have proved the proposition for all maps V with k∇V k∞≤ M , where M = 2
√
nC. Then from the triangle inequality
k∇v − RkL2(Q0) ≤ k∇v − ∇V kL2(Q)+ k∇V − RkL2(Q)
≤ 2√Ckdist(∇v, SO(n))kL2(Q)+ Ckdist(∇V, SO(n))kL2(Q)
≤ Ckdist(∇v, SO(n))kL2(Q).
The last inequality is true because ˆ Q dist(∇V, SO(n))2dx ˆ ∇V =∇v dist(∇V, SO(n))2dx + ˆ ∇V 6=∇v dist(∇V, SO(n))2dx ≤ ˆ Q dist(∇v, SO(n))2dx + {∇V 6= ∇v} (k∇V k∞+ √ n)2 ≤ C ˆ Q dist(∇v, SO(n))2dx. Step 2. Let ε = kdist(∇v, SO(n))kL2(Q). (1.18)
We may suppose ε ≤ 1. Indeed if ε > 1 then for any R in SO(n) ˆ Q |∇v(x) − R|2dx ≤ 2 ˆ Q (dist(∇v(x), SO(n))2 + |R − S(x)|2) ≤ 2ε2 + 8n|Q| ≤ (2 + 8n|Q|)ε2 ≤ C ˆ U dist(∇v, SO(n))2dx
where S(x) ∈ SO(n) is the matrix of minimum distance from ∇v(x), thus the proposition is trivially proved.
Now we decompose v = w + z, where z ∈ W1,2(Q, Rn) solves
(
∆z = ∆v in Q
z = 0 on ∂Q (1.19)
and w = v − z satisfies ∆w = 0 in Q. The aim of the proposition is to find R ∈ SO(n) such that k∇v − RkL2(Q) ≤ Cε, but from the triangle inequality we get
|∇v − R|2 ≤ 2(|∇z|2+ |∇w − R|2)
thus after integrating the last inequality, for the inequality (1.17) to be true it suffices
to show that ˆ
Q
|∇z|2 ≤ Cε2 (1.20)
and that exists R ∈ SO(n) such that ˆ
Q
|∇w − R|2 ≤ Cε2. (1.21)
Observe that, since div cof (∇v) = 0, we have −∆v = div( cof ∇v − ∇v) and thus
−∆z = −div(∇v − cof (∇v)). (1.22) The quantity |A − cof A|2 is smooth and non negative, and vanishes on SO(n). Hence
there is a constant C such that
|A − cof A|2 ≤ Cdist(A, SO(n))2 for |A| ≤ M (1.23) where M is as in Step 1.
Now, inequality (1.20) follows from testing 1.22 with z: ˆ Q |∇z|2dx = ˆ Q z · (−∆z)dx = ˆ Q z · div( cof ∇v − ∇v)dx = ˆ Q ∇z · ( cof ∇v − ∇v)dx ≤ ˆ Q |∇z|2dx 1 2 ˆ Q | cof ∇v − ∇v|2 1 2
and thus, using (1.23) ˆ Q |∇z|2dx ≤ ˆ Q | cof ∇v − ∇v|2 ≤ Cε2 (1.24)
It remains to obtain (1.21). This will be done in the next two steps. First we will obtain a bound in terms of ε. This is not enough since ε can be small, but we will improve the exponent from 1 to 2 for the symmetric part of the gradient, using a linearization near the identity. Then Korn’s inequality can be used to control the skew part as well.
Q Q0 Q00 η 0 1
Figure 1.1: Concentric cubes from Step 3 and cutoff function η. Step 3. We now want to prove that there exists R ∈ SO(n) such that
sup
Q0
|∇w − R| ≤ Cε12.
To this aim we use the identity 1
2∆(|∇w|
2− n) = |∇2
w|2
which holds for harmonic maps (see the proof of Liouville’s Theorem, subsection 1.1.3). Let Q00 = 34Q, so that Q00 is concentric to Q and Q0 ⊂ Q00 ⊂ Q with strict inclusions. Choose a cutoff function 0 ≤ η ≤ 1 in C0∞(Q) with η = 1 on Q00. Then
ˆ Q00 |∇2w|2 ≤ ˆ Q |∇2w|2η dx = 1 2 ˆ Q ∆(|∇w|2− n)η dx ≤ sup Q (∆η) ˆ Q ||∇w|2 − n|dx ≤ C ˆ Q ||∇v|2− n|dx + 2 ˆ Q |∇v||∇z|dx + ˆ Q |∇z|2dx .
Let us show that the last line is bounded by Cε. The last integral in the last line is bounded by Cε2 by (1.24). As for the second integral, by applying Cauchy-Schwartz
inequality we obtain ˆ Q |∇v||∇z|dx ≤ ˆ Q |∇v|2dx 1 2 ˆ Q |∇z|2dx 1 2 ≤ M |Q|12Cε ≤ Cε
from (1.24) and the assumption that k∇vk∞ ≤ M .
As for the first integral, notice that if B is a matrix in SO(n) then ||∇v|2− n| =
|∇v|2− |B|2 =
(|∇v| − |B|)(|∇v| + |B|) ≤ C|∇v − B| because |B|2 = n and |∇v| ≤ M . But then, choosing B pointwise as the minimum distance matrix of SO(n) from ∇v, and integrating the last inequality, we obtain
ˆ Q ||∇v|2− n|dx ≤ C ˆ Q dist(∇v, SO(n))dxC.−S.≤ C|Q|12ε ≤ Cε.
In the end we have obtained ˆ
Q00
|∇2w|2 ≤ Cε.
We now use the fact that w, and hence ∇2w, is harmonic on Q, so that the mean value property with r = dist(Q00, ∂Q0) = 1
4 gives sup x∈Q0 |∇2w(x)|2 = sup x∈Q0 1 |B(x, r)| ˆ B(x,r) ∇2w(y)dy 2 (1.25) C.−S. ≤ 1 |B(x, r)| ˆ B(x,r) |∇2w(y)|2dy ≤ Cε. (1.26)
This means that ∇w is (Cε12)-Lipschitz, and so if ˆR is the value of ∇w at any point
x ∈ Q we obtain sup Q0 |∇w − ˆR| ≤ (diam Q0) sup Q0 |∇2w| ≤ Cε12. (1.27)
Here ˆR is not necessarily in SO(n), but we can obtain the same inequality with a matrix R ∈ SO(n). Indeed
ˆ Q dist(∇w, SO(n))2dx ≤ 2 ˆ Q (dist(∇v, SO(n))2+ |∇z|2)dx ≤ Cε2 (1.28)
from (1.24) and (1.18), and so there exists x ∈ Q such that dist(∇w(x), SO(n))2 ≤ Cε2.
If R ∈ SO(n) is the minimum distance matrix from ∇w(x) then | ˆR − R| ≤ |∇w(x) − ˆR| + |∇w(x) − R| ≤ Cε12 + Cε ≤ Cε 1 2 and so sup Q0 |∇w − R| ≤ sup Q0 |∇w − ˆR| + | ˆR − R| ≤ Cε12.
From now on we may assume that R = I, for otherwise we can apply the following arguments to Rtv and Rtw.
Step 4. Linearizing dist(·, SO(n)) near the identity (see Subsection 1.1.1) we get G + Gt 2 − I
= dist(G, SO(n)) + O(|G − I|2).
Let e = 12(∇w + (∇w)t) − I = (∇w − I)s. From the previous identity with G = ∇w
and (1.27) we obtain that on Q0
|e| = ∇w + (∇w)t 2 − I
≤ dist(∇w, SO(n)) + O(|∇w − I|2) ≤ dist(∇w, SO(n)) + Cε
so that using (1.28) we obtain ˆ Q0 |e|2dx ≤ ˆ Q0 2(dist(∇w, SO(n))2 + (Cε)2) ≤ Cε2.
By Korn’s inequality applied to u(x) := w(x) − x we obtain ˆ Q0 |∇w − (∇w)Q0|2dx = ˆ Q0 |∇u − (∇u)Q0|2dx ≤ C ˆ Q0 |e|2 ≤ Cε2.
But again by (1.28) we can replace (∇w)Q0 with a matrix in SO(n) (as in the final
part of Step 3), completing the proof.
1.2.2
The general case
Proof of theorem 1.11. Similarly to the previous proof we can assume that k∇vkL∞(U ) ≤ M.
We split again v = w + z as in the previous proof. The bound on ∇z is obtained in the same way, since its proof already holds for generic Lipschitz domains:
ˆ
U
So we need to estimate the harmonic part w. Let Q(a, r) = a + r −1 2, 1 2 n
be the cube of center a ∈ Rn and side length r > 0. From the Whitney’s covering
theorem (see section 1.1.5) we obtain a covering with cubes Qi = Q(ai, ri) such that
every x ∈ U is contained in at most N cubes Q(ai, 4ri) ⊂ U and
10 diam Qi ≤ dist(Qi, ∂U ) ≤ 30 diam Qi
(here the specific constants 10 and 30 are not important; it suffices that the constant 10 guarantees that 4Qi ⊂ U ). Applying Proposition 1.12 to w and each cube
Q(ai, 4ri) we obtain rotations Ri such that
ˆ Q(ai,2ri) |∇w − Ri|2dx ≤ C ˆ Q(ai,4ri) dist(∇w, SO(n))2dx.
We now claim that, since w is harmonic, the following inequality holds: ri2 ˆ Q(ai,ri) |∇2w|2dx ≤ C ˆ Q(ai,2ri) |∇w − Ri|2dx. (1.30)
To see this we apply identity (1.2) to w − Ri (harmonic as well) to obtain
|∇2w|2 = |∇2(w − R i)|2 = 1 2∆ |∇(w − Ri)| 2 = 1 2∆ |∇w − Ri| 2 .
Now choose η ∈ C0∞(Q(0, 2)) with η ≥ 0 and η = 1 on Q(0, 1), and define ηr(x) =
η(x/r). Then ˆ Q(ai,ri) |∇2w|2dx ≤ ˆ Q(ai,2ri) ηri|∇ 2w|2dx ≤ 1 2 ˆ Q(ai,2ri) ηri∆(|∇w − Ri| 2)dx = 1 2 ˆ Q(ai,2ri) ∆ηri|∇w − Ri| 2dx ≤ 1 2k∆ηrik∞ ˆ Q(ai,2ri) |∇w − Ri|2dx and k∆ηrik∞ = 1 r2 i k∆ηk∞
giving (1.30).
Now we use the fact that dist(x, ∂U ) is comparable with ri to obtain
ˆ Q(ai,ri) dist(x, ∂U )2|∇2w|2dx ≤ C ˆ Q(ai,4ri) dist(∇w, SO(n))2dx.
Summing this over i and using P
iχQ(ai,4ri)≤ N we obtain the global estimate
ˆ U dist(x, ∂U )2|∇2w|2dx ≤ C ˆ U dist(∇w, SO(n))2dx.
We now use the weighted Poincar´e inequality that can be found in Subsection 1.1.7, namely min G∈Mn×n ˆ U |f − G|2dx ≤ C ˆ U dist(x, ∂U )2|∇f |2dx
to obtain that there exists R ∈ Mn×n such that
k∇w − RkL2(U )≤ Ckdist(∇w, SO(n))kL2(U ),
and as in the proof of the cubic-domain case we can choose R ∈ SO(n). Combining this with the estimate for z (1.29) we obtain the assertion of the theorem.
2.1
Overview
As stated in the introduction, in this exposition we face the problem of crystal-lization in the triangular lattice A2 when compactly supported perturbations are
applied. We fix a finite set A ⊂ A2, which is the perturbed zone, and restrict the
analysis to maps
y ∈ YA = {y : A2 → R2|y(x) = x ∀x ∈ A2\A}
i.e. maps that fix all points outside A. We define the energy EA = X {x,x0}⊂A 2 {x,x0}∩A6=∅ V (|y(x) − y(x0)|)
and look for minimizers. With {x, x0} we mean a set with exactly 2 elements. We will show that a ground state ymin is crystallized, in the sense that ymin(A2) =
A2. Moreover, we will prove a stability result for low energy configurations: if a
configuration y has energy gap ∆ sufficiently small (i.e. EA(y) − EA(ymin) < ∆)
then dist(y(A2), A2) ≤ C∆.
Assumptions on the potential It is convenient to introduce the notion of renor-malized potential: V∗(r) = 1 6 X ξ∈A2\{0} |ξ|2V (r|ξ|).
In order for V∗ to be finite, this definition requires a sufficient decay of V , but this
will be guaranteed by the assumptions we will make on V . Notice that this is not equal to the renormalized potential of the introduction (and of the work by Theil), because of the extra |ξ|2.
We will consider in the following only renormalized potentials, that satisfy:
lim
r→∞V (r) = 0; (2.1)
min
r>0 V∗(r) = V∗(1) = −1. (2.2)
We impose the second condition to fix the lengthscale of configurations. This restriction does not affect the generality, indeed if the renormalized potential V∗
achieves a minimum m (necessarily negative) at a point r∗, then it suffice to rescale
both domain and codomain and consider the potential −m1V∗(r/r∗), which satisfies
the condition. Moreover, under conditions below, V∗ must achieve a minimum for
sufficiently small α.
In addition to these renormalization, we will say that the potential V : [0, ∞) → R satisfies the property Pα, with α < 1/3, if:
V is C2 on (1 − α, ∞) (2.3) V (r) ≥ 1 α for 0 < r < 1 − α (2.4) V00(r) ≥ 1 for 1 − α < r < 1 + α (2.5) V (r) ≥ −α for 1 + α < r < 4 3 (2.6) |V00(r)| ≤ αr−7 for all r ≥ 4 3 (2.7)
Statement The ideal theorem would be the following:
Theorem 2.1. Fix a finite set A ⊂ A2. There exists a constant α0 > 0 such that
for every 0 < α < α0 and every potential V that satisfies (2.1)-(2.7), if y ∈ YA is a
ground state of EA then y(A2) = A2.
Moreover, there exist constants C, and ∆0 > 0 such that if y ∈ YA has energy gap
0 < ∆ < ∆0, i.e.
EA(y) − EA(ymin) < ∆
then there exists a reparametrization φ = y ◦ σ, with σ : A2 → A2 bijection, such that
|φ(x) − x| ≤ C∆.
However we could not prove this theorem under only the previous assumptions, and we have to require an additional condition: that the local imbedding property holds (see Definition 2.5). This condition roughly says that in the image y(A2)
every point has exactly six other points at distance between 1 − α and 1 + α. We remark that this condition is true for low energy configurations in the periodic setting; instead we have to require it in the compact-perturbation case.
r 1 1 − α 1 + α 4/3 1 α V (r) −1 −α
Figure 2.1: Qualitative picture of a potential satisfying property Pα.
Theorem 2.2. Fix a finite set A ⊂ A2. There exists a constant α0 > 0 such that
for every 0 < α < α0 and every potential V that satisfies (2.1)-(2.7), if y ∈ YA is a
ground state of EA with the 6-neighbours property, then y(A2) = A2.
Moreover, there exist constants C, and ∆0 > 0 such that if y ∈ YA has energy gap
0 < ∆ < ∆0, i.e.
EA(y) − EA(ymin) < ∆
and the local imbedding property holds, then there exists a reparametrization φ = y ◦ σ, with σ : A2 → A2 bijection, such that
|φ(x) − x| ≤ C∆.
2.1.1
Properties of the triangular lattice
Let e1 = (1, 0) and e2 = (1/2,
√
3/2). Then, as already said, the triangular lattice is defined as
A2 = {je1+ ke2 : j, k ∈ Z}.
We will call U the fundamental hexagonal cell (instead of the parallelogram of the original work by Theil and of the introduction) i.e.
The set of all possible distances between points in A2 is
Λ = {|ξ| : ξ ∈ A2\{0}}.
Given λ ∈ Λ we introduce the multiplicity m(λ) as m(λ) = 1
6#{ξ ∈ A2 : |ξ| = λ}
which is always an integer, and is such that there are exactly 6m(λ) segments of length λ with one extreme in the origin and the other in A2. For example m(1) = 1
and m(√7) = 2. It follows from the definition that given a function f : (0, ∞) → R it holds X λ∈Λ m(λ)f (λ) = X ξ∈A2\{0} f (|ξ|)
if f is such that the right sum is totally summable. In particular from the definition of renormalized potential we have
V∗(r) = X ξ∈A2 |ξ|2V (|ξ|) =X λ∈Λ λ2m(λ)V (rλ).
Define a sub-lattice of A2 as a non trivial set A ⊆ A2 that is closed under algebraic
addition and under rotations of multiples of π/3. It is easy to show that any sub-lattice is a copy of A2 rotated and dilated by a factor λ ∈ Λ (which we will call
sidelength of the sub-lattice). Moreover for every λ there are exactly m(λ) sub-lattices of A2 with side length λ. Define additionally an affine sub-lattice as the translation
of a sub-lattice by a vector v ∈ A2. Then there are exactly λ2m(λ) distinct affine
sub-lattices of length scale λ.
2.1.2
Some preliminary lemmas
Lemma 2.3. There exists α0 > 0 such that for all α ∈ (0, α0) and all V satisfying
Pα, the following hold:
min r∈[1−α,1+α]V 00 ∗ (r) ≥ 1 2, (2.8) V (r) ≥ −2 ∀r ≥ 0, (2.9) |V (r)| ≤ αr−5 ∀r ≥ 4 3. (2.10)
Proof. To obtain the third inequality, since V00 is continuous and integrable on the half lines (r, +∞), by the fundamental theorem of calculus we can write
V0(r) = − ˆ ∞
r
and from the bound (2.7) we get |V0(r)| ≤ ˆ ∞ r |V00(t)|dt ≤ α 6r −6 . Repeating the same argument with V instead of V0 we obtain
|V (r)| ≤ α 30r
−5 ≤ αr−5
. Let’s turn now to the first inequality. Recalling that
V∗(r) = 1 6 X ξ∈A2\{0} |ξ|2V (r|ξ|)
and using the third inequality to obtain the total summability of the series, we can derive twice and obtain
V∗00(r) = 1 6 X ξ∈A2\{0} |ξ|4V00 (r|ξ|).
Now we want to use the bounds (2.5) and (2.7), so we split the sum in two: the dominant part {|ξ| = 1}, to which we can apply (2.5), and {|ξ| 6= 1}, which gives a small contribution due to (2.7):
V∗00(r) = 1 6 X ξ∈A2\{0} |ξ|4V00 (r|ξ|) = 1 6 X ξ∈A2\{0} |ξ|=1 |ξ|4V00 (r|ξ|) + 1 6 X ξ∈A2\{0} |ξ|6=1 |ξ|4V00 (r|ξ|) = V00(r) +1 6 X ξ∈A2\{0} |ξ|6=1 |ξ|4V00 (r|ξ|). (2.11)
Now the first term is ≥ 1 from (2.5). As for the second term, a look at the triangular lattice makes clear that if |ξ| 6= 1 then it is greater than √3. So if α is sufficiently small then r|ξ| is greater than 4/3 when r ∈ (1 − α, 1 + α), and from (2.7) we can bound the second term with a constant times α. This implies that, for sufficiently small α, V∗00(r) ≥ 12 for r ∈ (1 − α, 1 + α).
The second inequality is proved in a similar way, splitting the sum and using (2.10) to make the second part negligible; this time (2.2) tells us that the first part is
exactly −1, from which the result follows.
Lemma 2.4 (Minimum distance). Fix A ⊂ A2 finite. Given A0 ⊂ A and y : A2 →
R2, define E(A0, y) = X {x,x0}∈A 2\A∪A0 {x,x0}∩A06=∅ V (|y(x) − y(x0)|)
There exists a number α0 ∈ (0, 1/3) and ∆0 > 0 such that for each α ∈ (0, α0), each
V satisfying properties Pα and each A0 ⊂ A, all configurations y : A2 → R2 with the
compact-perturbation property
y ∈ YA = {y : A2 → R2|y(x) = x ∀x ∈ A2\A}
and with sufficiently low energy
E(A0, y) < E(Amin, ymin) + ∆
where 0 < ∆ < ∆0, verify
min
x6=x0|y(x) − y(x
0)| > 1 − α, (2.12)
that is: any two particles of a low energy configuration must be distant at least 1 − α. Proof. Define M = max z∈R2 # y(X) ∩ B z,1 − α 2 .
The thesis is equivalent to M = 1. By translating we can suppose that the maximum is achieved for z = 0. Call B = B(0, (1 − α)/2), and B = y−1(B). The total energy can be split in three terms, according to whether x and x0 belong to B:
1 2 X x6=x0 e({x, x0}) = X x∈B x0∈A2\B e({x, x0}) + 1 2 X {x,x0}⊂B x6=x0 e({x, x0}) + 1 2 X {x,x0}⊂A 2\B x6=x0 e({x, x0}).
If we now remove particles in y(B), just the last energy term survive. Since the original disposition was a low energy configuration, the new energy must be greater or equal than the original one minus ∆, and this can be written as
X x∈B x0∈A2\B e({x, x0}) + 1 2 X {x,x0}⊂B x6=x0 e({x, x0}) ≤ ∆.
The second sum is bounded from below according to (2.4) (high potential near zero), precisely 1 2 X {x,x0}⊂B x6=x0 e({x, x0}) ≥ M (M − 1) 2α and so we obtain the condition
X x∈B x0∈A 2\B e({x, x0}) ≤ −M (M − 1) 2α + ∆.
The first term can also be bounded from below, but this time from decay estimate (2.7) and uniform estimate (2.6). Firstly, call Rn = (n + 1)B\nB the ring with inner
radius n(1 − α)/2, and split the sum as X x∈B x0∈A2\B e({x, x0}) = X n∈N+ X x∈B y(x0)∈Rn e({x, x0}).
Now observe that in Rn there are at most CM n particles, with C absolute constant.
This is beacuse Rn can be covered by Cn balls that are translated copies of B(0,13),
and M is the maximum number of particles in each ball. If n ≥ 5 then the distance between B and Rn is greater than 4/3, and we can apply decay estimate (2.7):
X n≥5 X x∈B y(x0)∈Rn e({x, x0}) ≥X n≥5 −CM2nα n1 − α 2 −5
where the extra M comes from summing over the M particles in B. As for n ≤ 4, we can use estimate (2.9) to obtain
X 1≤n≤4 X x∈B y(x0)∈Rn e({x, x0}) ≥ X 1≤n≤4 −2nM2 = −20M2.
Summing the last three inequalities we get X x∈B x0∈A 2\B e({x, x0}) + 1 2 X {x,x0}⊂B x6=x0 e({x, x0}) ≥ −M2 20 +X n≥5 −Cnα n1 − α 2 −5!
and thus the condition −M2 20 +X n≥5 −Cnα n1 − α 2 −5! ≤ −M (M − 1) 2α + ∆.
The sum on the left hand side is bounded independently of α ∈ (0, 1/3), and in order for this inequality to be true for small α we must have M = 1.
Following the previous result we introduce the concept of neighbours.
Definition 2.5 (Neighbours). Two elements x, x0 in y(A2) are neighbours if x or x0
is in y(A) and ||x − x0| − 1| < α. The neighbourhood of a particle is N (x) = {x0 ∈ y(A2) : x, x0 are neighbours}.
The set of all neighbouring pairs, or short range interactions, is S = {{x, x0} : x, x0 are neighbours}
A 1-triangle of particles is a set T = {x0, x1, x2} such that xi are neighbours of each
It can easily be seen that, for sufficiently small α, a particle can have at most 6 neighbours. Indeed, in the limit α → 0, if a particle has 6 neighbours they place themselves in a regular hexagon, so that there is no space left for another neighbour, and this remains true for sufficiently small α.
Moreover, given x and a neighbour x0 ∈ N (x)\{x}, we can order neighbours of x starting from x0 clockwise or counterclockwise. In this way we say that x0 is the first neighbour of x starting from x0, etc. up to at most the sixth neighbour.
Lemma 2.6 (Ordering property of neighbours). There esists α0 > 0 such that for
every 0 < α < α0 the following holds: suppose that every particle has exactly 6
neighbours; given x and two of its neighbours x0, x00 ∈ N (x)\{x} such that x00 is
the second neighbour of x counterclockwise starting from x0, then x00 is the second neighbour of x0 clockwise starting from x.
We will say that a configuration y has the local imbedding property if α is chosen small enough so that every particle in y(A2) has six neighbours and the ordering
property of neighbours holds.
Definition 2.7 (Discrete imbedding). Let y satisfy the local imbedding property. A discrete imbedding of a neighbourhood N (x) of a point x ∈ y(A2) is a map
ψ : N (x) → A2 such that
• |ψ(x0) − ψ(x00)| = 1 for every neighbouring pair x0, x00 ∈ N (x);
• ψ is injective;
• ψ is orientation preserving, meaning that
det(x1− x0, x2− x0) = det(ψ(x1) − ψ(x0), ψ(x2) − ψ(x0)).
Having in mind the final theorem, which is crystallization on the triangular lattice, an ideal result at this point would be that for every configuration with sufficiently low energy, every particle has 6 neighbours. This would make the particle configuration look like a perturbed lattice. This seems a reasonable result, and it is true in the periodic setting, as will be shown below. However in the compact-perturbations setting this does not clearly follows from our hypotheses. Nevertheless, we will assume that this holds for low energy configurations even in the latter setting, and continue with this hypotheses from here on.
To conclude this section let us show that, in the periodic setting, the 6-neighbours property holds. First of all, the minimum distance property (2.12) can be proved in the periodic setting in a similar way. Then a simple lemma, which can be proved similarly to the second inequality in lemma 2.3, shows that the contributions to the energy are due mainly to neighbours’ interactions:
Lemma 2.8. For every ε > 0 there is αε such that for every α ∈ (0, αε) and every
configuration y : A2 → R2 with the minimum distance property (2.12), for every
x ∈ X the following holds: X
x06∈N (x)
V (|y(x) − y(x0)|) > −ε.
A consequence of this is:
Lemma 2.9. Given α ∈ (0, αε) (see previous lemma) the potential V satisfies the
following: min r>0 V (r) ∈ −1 − ε 6, −1 + ε 6 (2.13) V (1) ∈ −1 − ε 6, −1 + ε 6 (2.14) Proof. From the previous lemma and the definiton of V∗ we obtain that for r ∈
(1 − α, 1 + α)
|6V∗(r) − 6V (r)| ≤ ε
and plugging in r = 1 gives (2.14).
To obtain (2.13), again from the definition of V∗ we get
−6 = X x0∈A 2\{x} V (|x0− x|) = X x0∈N (x)\{x} V (|x0 − x|) + X x06∈N (x) V (|x0− x|) ≥ 6(min V ) − ε
where the last inequality comes from previous lemma. From this we get min V ≤ −1 + ε
6.
To obtain the other inequality, observe that from the hypotheses on V , its minimum lies in (1 − α, 1 + α). Call m = argminV . Then
−6 = 6V∗(1) ≤ 6V∗(m) =
X
ξ∈A2
V (m|ξ|) ≤ 6(min V ) + ε
which leads to the conclusion.
Finally, next proposition shows that in low energy configurations every particle has 6 neighbours.
Proposition 2.10. Given L ∈ N, there is ε > 0 such that for every α ∈ (0, αε) and
for every L-periodic configuration with sufficiently small energy ELper, every particle has exactly six neighbours.
Proof. Let n(x) = #N (x) − 1 be the number of neighbours of x. Then we can write Eloc(x) = X x0∈A 2\{x} e({x, x0}) = X x0∈N (x)\{x} e({x, x0}) + X x06∈N (x) e({x, x0}) ≥ n(x)(min V ) − ε from lemma 2.8.
Let us now suppose that n(x) ≤ 5 for some x ∈ X. From lemma 2.9 we obtain for that x n(x)(min V ) − ε ≥ 5−1 − ε 6 − ε ≥ −5 −11 6ε.
Now split the particles by the number of neighbours, and let d be the number of particles with 5 neighbours or less (defects). Then
ELper({y}) = X x∈X Eloc(x) = X n(x)=6 Eloc(x) + X n(x)≤5 Eloc(x) ≥ (N − d)6−1 − ε 6 − ε+ d5−1 −ε 6 − ε = N (−6 − 2ε) + d1 + ε 6 = −6N + d + (−2N + 1/6)ε.
If d 6= 0 then, for sufficiently small ε, this energy is strictly larger than, say, −6N +1/2. However, the energy of the regular lattice is less then −6N +N ε, which for sufficiently small ε is smaller then −6N + 1/4. Therefore y can not have an energy gap less than 1/4. This proves that, for sufficiently small ε, n(x) = 6 for every x ∈ X, which is the thesis.
2.2
Existence of a canonical labelling φ
The first observation to make is the following: the total energy of a configuration y depends only on the image of the points. In fact if we consider a bijection σ : A2 → A2
that fixes A2\A, the total energy of y ◦ σ is the same as y. We can thus ask if, for
a fixed y, there is a ”canonical” reparametrization φ = y ◦ σ. Clearly we have to require that φ be the identity outside A, but what should we require inside A? It turns out that a useful property is the conservation of neighbouring pairs, as defined in the previous section. We have seen that, if α is sufficiently small, in every periodic configuration with small energy each particle has exactly six neighbours, and moreover neighbours satisfy the ordering property of lemma 2.6. These conditions
are referred to as the local imbedding property. We now suppose that this is true also in the compact-perturbations setting, which we are facing. It is then reasonable to ask: given a map y with the local imbedding property, is there a ”canonical representative”, i.e. a reparametrization φ = y ◦ σ such that if x, x0 ∈ A2 satisfy
|x − x0| = 1 then ||φ(x) − φ(x0)| − 1| < α, i.e. φ(x), φ(x0) are neighbours?
The answer is positive. To construct this canonical map the idea is simple. We start with two points x0, x1 outside A, and we define the map φ on them as the identity.
Then, to decide where a point x ∈ A2 goes, we construct a discrete path starting
from x0, x1 ending in x, and we construct a corresponding path in the perturbed
configuration y(A2). The endpoint of this corresponding path is the value of φ(x).
We will check that this is a well defined map, i.e. that the value does not depend on the chosen path, and moreover that it is bijective. This map will be the canonical representative of y.
To do this rigorously we need some definitions:
Definition 2.11 (Discrete path). A discrete path of length l is a map γ : {0, . . . , l} → A2 such that |γ(k + 1) − γ(k)| = 1 for all k = 0, . . . , l − 1. It is closed if γ(0) = γ(l).
Definition 2.12 (Corresponding angles). Two ordered triples (x0, x1, x2) ⊂ A2 and
(y0, y1, y2) ⊂ y(A2) such that |x0 − x1| = |x2 − x1| = 1 and {y0, y1}, {y1, y2} are
neighbouring pairs are said to have corresponding angles if there exists a discrete imbedding of the neighbourhood N (y1) that sends yi to xi.
Definition 2.13 (Image path). Given a discrete path γ : {0, . . . , l} → A2, l ≥ 1, and
two neighbours x0, x1 ∈ y(A2), we define the image path γx0,x1 : {0, . . . , l} → y(A2)
as
• γx0,x1(0) = x0
• γx0,x1(1) = x1
• By induction, if we have defined γx0,x1(k) then γx0,x1(k +1) is the unique particle
that makes (γx0,x1(k − 1), γx0,x1(k), γx0,x1(k + 1)) and (γ(k − 1), γ(k), γ(k + 1))
have corresponding angles.
Now, to define where a point x ∈ A goes, we choose a discrete path γ of length l such that γ(0) = x0, γ(1) = x1 and γ(l) = x, and see where the image path γx0,x1
ends. Thus by definition
φ(x) = γx0,x1(l).
The only thing to check is whether this is well defined, so that if we choose any other path connecting x0 and x then the endpoint of the image path is the same.
Proposition 2.14. If y : A2 → R2 is a map with the local imbedding property, then
