# A Sharp Korn’s Inequality for Piecewise $\displaystyle H^{1}$ Space and its
applications

Qingguo Hong†, Young-Ju Lee‡ and Jinchao Xu  Department of Mathematics, Penn
State University, PA, USA  Department of Mathematics, Texas State University,
TX, USA<EMAIL_ADDRESS><EMAIL_ADDRESS><EMAIL_ADDRESS>
###### Abstract.

In this paper, we revisit Korn’s inequality for the piecewise $\displaystyle
H^{1}$ space based on general polygonal or polyhedral decompositions of the
domain. Our Korn’s inequality is expressed with minimal jump terms. These
minimal jump terms are identified by characterizing the restriction of rigid
body mode to edge/face of the partitions. Such minimal jump conditions are
shown to be sharp for achieving the Korn’s inequality as well. The sharpness
of our result and explicitly given minimal conditions can be used to test
whether any given finite element spaces satisfy Korn’s inequality, immediately
as well as to build or modify nonconforming finite elements for Korn’s
inequality to hold.

The second author is partially supported by Shapiro Fellowship from Penn State
in Spring of 2022, and he is also supported by NSF DMS-2208499

##  1\. Introduction

The Korn’s inequality played a fundamental role in the development of linear
elasticity. There is a work reviewing Korn’s inequality and its applications
in continuum mechanics [5]. In fact, there are a lot of works on proving
classical Korn’s inequality [10, 12, 2] for functions in $\displaystyle H^{1}$
vector fields. In [1], Korn’s inequalities for piecewise $\displaystyle H^{1}$
vector fields are established. We note that a strengthened version of Korn’s
inequality for piecewise $\displaystyle H^{1}$ vector fields is presented in
[8]. By the Korn’s inequality established in [8], the nonconforming finite
element for 2D introduced in [7] is shown to satisfy the Korn’s inequality. In
this paper, we revisit Korn’s inequality for the piecewise $\displaystyle
H^{1}$ space based on general polygonal or polyhedral decompositions of the
domain. Our Korn’s inequality is expressed with minimal jump terms, which can
readily be used to design degree of freedoms. Namely, we move one step further
from the result presented in [8]. In particular, for 3D, we identify that the
tangential component of rigid body mode restricted to the face of triangle is
indeed the lowest Raviart-Thomas element on face, which is the rotated version
discussed in [8]. Additionally, with the minimal jump terms, we show that our
Korn’s inequality is sharp. We emphasize that with the help of sharpness of
the Korn’s inequality and explicit form of jump conditions, it is very easy to
test whether any given finite element spaces satisfy the Korn’s inequality.
The proposed minimal jump conditions can also provide a guidance to modify
some existing nonconforming finite elements so that the Korn’s inequality is
satisfied.

Throughout the paper, we shall use the standard notation for Sobolev spaces.
Namely, $\displaystyle H^{k}(\Omega)$ denotes the Sobolev space of scalar
functions in $\displaystyle\Omega$ whose derivatives up to order
$\displaystyle k$ are square integrable, with the norm
$\displaystyle\|\cdot\|_{k}$. The notation $\displaystyle|\cdot|_{k}$ denotes
the semi-norm derived from the partial derivatives of order equal to
$\displaystyle k$. Furthermore, $\displaystyle\|\cdot\|_{k,T}$ and
$\displaystyle|\cdot|_{k,T}$ denote the norm $\displaystyle\|\cdot\|_{k}$ and
the semi-norm $\displaystyle|\cdot|_{k}$ restricted to the domain
$\displaystyle T$. Given a partitions $\displaystyle\mathcal{T}_{h}$ of the
domain $\displaystyle\Omega$, where $\displaystyle\Omega$ is a bounded
connected open polyhedral domain in $\displaystyle{\rm I\\!R}^{d}$ with
$\displaystyle d=2$ or $\displaystyle 3$. , we shall also use $\displaystyle
H^{1}(\Omega;\mathcal{T}_{h})$ to denote the element-wise $\displaystyle
H^{1}$ functions. We denote the vectors of size $\displaystyle d$ whose
components are in $\displaystyle H^{1}(\Omega;\mathcal{T}_{h})$ by
$\displaystyle(H^{1}(\Omega;\mathcal{T}_{h}))^{d}$. We also denote
$\displaystyle H({\rm div};\Omega)$ by the space consisting of vectors, whose
divergence belongs to $\displaystyle L^{2}(\Omega)$. We use $\displaystyle
P_{\ell}(T)$ for the space of polynomials of degree upto $\displaystyle\ell$
on the domain $\displaystyle T$ while $\displaystyle(P_{\ell}(T))^{d}$ denotes
the vectors of size $\displaystyle d$ whose components are polynomials of
degree at most $\displaystyle\ell$.

We recall that the operator $\displaystyle curl$ on a scalar function
$\displaystyle q$ in $\displaystyle 2D$ is defined by

(1.1) ${\mbox{\boldmath$\displaystyle curl$}}q=\left(-\frac{\partial
q}{\partial y},\frac{\partial q}{\partial x}\right)^{T},$

while on a vector function $\displaystyle\mbox{\boldmath$\displaystyle
q$}=(q_{i})_{i=1,2,3}^{T}$ in $\displaystyle 3D$, it is defined by

(1.2) ${\mbox{\boldmath$\displaystyle curl$}}\mbox{\boldmath$\displaystyle
q$}=\left(\frac{\partial q_{2}}{\partial z}-\frac{\partial q_{3}}{\partial
y},\frac{\partial q_{3}}{\partial x}-\frac{\partial q_{1}}{\partial
z},\frac{\partial q_{1}}{\partial y}-\frac{\partial q_{2}}{\partial
x}\right)^{T},$

For any given vector space $\displaystyle V$, by $\displaystyle{\rm
dim}\mbox{\boldmath$\displaystyle V$}$, we mean the dimension of
$\displaystyle V$.

The rest of our paper is organized as follows. In §2, we prove the Korn’s
inequality for piecewise $\displaystyle H^{1}$ space with minimal jump terms.
§3 considers the sharpness of the inequality, which indicates that the minimal
jump terms presented in our Korn’s inequality are necessary and can not be
reduced any more. With the Korn’s inequality for piecewise $\displaystyle
H^{1}$ space with minimal jump terms and sharpness discussion, in §4, some of
existing finite elements have been discussed. And some application of our
theory is provided to modify nonconforming finite elements which do not
satisfy the Korn’s inequality so that they satisfy the Korn’s inequality.
Lastly, we provide a concluding remark.

##  2\. Korn’s inequality for the piecewise $\displaystyle H^{1}$ vector
functions

Let $\displaystyle\Omega$ be a bounded connected open polyhedral domain in
$\displaystyle{\rm I\\!R}^{d}$ with $\displaystyle d=2$ or $\displaystyle 3$
and $\displaystyle\partial\Omega$ be the boundary of the domain
$\displaystyle\Omega$. Then the classical Korn’s inequality reads as follows:

(2.1) $|{\bf{u}}|_{H^{1}(\Omega)}\leq
C_{\Omega}\left(\|\mathcal{D}({\bf{u}})\|_{0}+\|{\bf{u}}\|_{0}\right),\quad\forall{\bf{u}}\in[H^{1}(\Omega)]^{d},$

where the strain tensor $\displaystyle\mathcal{D}({\bf{u}})\in{\rm
I\\!R}^{d\times d}$ is given as follows:

(2.2) $\mathcal{D}_{ij}({\bf{u}})=\frac{1}{2}\left(\frac{\partial
u_{i}}{\partial x_{j}}+\frac{\partial u_{j}}{\partial x_{i}}\right)\quad\mbox{
for }1\leq i,j\leq d.$

Let $\displaystyle{\mbox{\boldmath$\displaystyle RM$}}(\Omega)$ be the space
of rigid motions on $\displaystyle\Omega$ defined by

(2.3) $\mbox{\boldmath$\displaystyle
RM$}(\Omega)=\\{\mbox{\boldmath$\displaystyle
a$}+\mbox{\boldmath$\displaystyle A$}\mbox{\boldmath$\displaystyle x$}\in{\rm
I\\!R}^{d}:\mbox{\boldmath$\displaystyle a$}\in{\rm I\\!R}^{d}\,\,\mbox{ and
}\,\,\mbox{\boldmath$\displaystyle A$}\in{\textsf{Skw}}^{d\times d}\\},$

where $\displaystyle\mbox{\boldmath$\displaystyle
x$}=(x_{1},\cdots,x_{d})^{T}\in{\rm I\\!R}^{d}$ is the position vector
function on $\displaystyle\Omega$ and $\displaystyle{\textsf{Skw}^{d\times
d}}$ is the set of anti-symmetric $\displaystyle d\times d$ matrices. We would
like to remark that if $\displaystyle\mbox{\boldmath$\displaystyle
v$}=\mbox{\boldmath$\displaystyle v$}(\mbox{\boldmath$\displaystyle
x$})\in\mbox{\boldmath$\displaystyle RM$}(\Omega)$, then
$\displaystyle\mbox{\boldmath$\displaystyle v$}=\mbox{\boldmath$\displaystyle
a$}+\mbox{\boldmath$\displaystyle A$}\mbox{\boldmath$\displaystyle x$}$ for
some constant vector $\displaystyle a$ and a homogeneous polynomial of degree
one $\displaystyle A$$\displaystyle x$ such that
$\displaystyle\mbox{\boldmath$\displaystyle
x$}\cdot\mbox{\boldmath$\displaystyle A$}\mbox{\boldmath$\displaystyle x$}=0$.
This is relevant to the lowest Nédélec first kind finite element [9]. It is
easy to verify that the space $\displaystyle\mbox{\boldmath$\displaystyle
RM$}(\Omega)$ is the kernel of the strain tensor, i.e., it holds that

(2.4) $\mbox{\boldmath$\displaystyle
RM$}(\Omega)=\\{\mbox{\boldmath$\displaystyle
v$}\in(H^{1}(\Omega))^{d}:\mathcal{D}(\mbox{\boldmath$\displaystyle
v$})=0\\}.$

###  2.1. Korn’s inequality with explicit upper bounds

Given $\displaystyle\Omega\subset{\rm I\\!R}^{d}$ with $\displaystyle d=2$ or
$\displaystyle 3$, we consider the mesh generation
$\displaystyle\mathcal{T}_{h}$, which is a shape-regular polygonal and/or
polyhedral partition for $\displaystyle\Omega$. Let
$\displaystyle\mathcal{T}_{h}=\cup\\{T\\}$ denote the collection of
decompositions and $\displaystyle h=\max_{T\in\mathcal{T}_{h}}{\rm diam}(T)$
denote the mesh size. We further denote by $\displaystyle\mathcal{E}_{h}$ the
set of all edges/faces for $\displaystyle\mathcal{T}_{h}$ and

(2.5) $\mathcal{E}_{h}=\mathcal{E}_{h}^{o}\cup\mathcal{E}_{h}^{\partial},$

where $\displaystyle\mathcal{E}_{h}^{o}$ denotes the set of all interior
edges/faces of $\displaystyle\mathcal{T}_{h}$ and
$\displaystyle\mathcal{E}_{h}^{\partial}$ denotes the set of all boundary
edges/faces, respectively. Let $\displaystyle\mathcal{V}(T)$ be the set of all
vertices for $\displaystyle T\in\mathcal{T}_{h}$ while
$\displaystyle\mathcal{V}(\mathcal{T}_{h})$ denotes the set of all vertices
for the partition $\displaystyle\mathcal{T}_{h}$.

Given two adjacent polygon/polyhedron $\displaystyle T^{+}$ and $\displaystyle
T^{-}$ in $\displaystyle\mathcal{T}_{h}$, let $\displaystyle f=\partial
T^{+}\cap\partial T^{-}$ be the common boundary (interface) between
$\displaystyle T^{+}$ and $\displaystyle T^{-}$ in
$\displaystyle\mathcal{T}_{h}$, and $\displaystyle n^{+}$ and $\displaystyle
n^{-}$ be unit normal vectors to $\displaystyle f$ pointing to the exterior of
$\displaystyle T^{+}$ and $\displaystyle T^{-}$, respectively. For any edge
(or face) $\displaystyle f\in\mathcal{E}_{h}^{o}$, and a scalar $\displaystyle
q$ and vector $\displaystyle v$, we define the jumps

(2.6) $\displaystyle[\\![q]\\!]_{f}&=q|_{\partial T^{+}\cap f}-q|_{\partial
T^{-}\cap f},\\\ [\\![\mbox{\boldmath$\displaystyle
v$}]\\!]_{f}=\mbox{\boldmath$\displaystyle v$}|_{\partial T^{+}\cap
f}-\mbox{\boldmath$\displaystyle v$}|_{\partial T^{-}\cap f}.$

When $\displaystyle f\in\mathcal{E}_{h}^{\partial}$ then the above quantities
are defined as

(2.7) $[\\![q]\\!]_{f}=q|_{f},\quad\mbox{ and
}\quad[\\![\mbox{\boldmath$\displaystyle
v$}]\\!]=\mbox{\boldmath$\displaystyle v$}|_{f}.$

Throughout the paper, we shall also consider the subspace of
$\displaystyle\mbox{\boldmath$\displaystyle RM$}(\Omega)$, denoted by
$\displaystyle\mbox{\boldmath$\displaystyle RM$}^{\partial}(\Omega)$ defined
as follows:

(2.8) $\mbox{\boldmath$\displaystyle
RM$}^{\partial}(\Omega)=\left\\{\mbox{\boldmath$\displaystyle
m$}\in\mbox{\boldmath$\displaystyle
RM$}(\Omega):~{}\|\mbox{\boldmath$\displaystyle
m$}\|_{L^{2}(\partial\Omega)}=1,\int_{\partial\Omega}\mbox{\boldmath$\displaystyle
m$}ds=0\right\\}.$

We also consider the following space of piecewise linear vector fields:

$\displaystyle\displaystyle\mbox{\boldmath$\displaystyle V$}_{h}$
$\displaystyle\displaystyle=$
$\displaystyle\displaystyle\\{\mbox{\boldmath$\displaystyle
v$}\in(L^{2}(\Omega))^{d}:\mbox{\boldmath$\displaystyle
v$}_{T}=\mbox{\boldmath$\displaystyle v$}|_{T}\in(P_{1}(T))^{d},\quad\forall
T\in\mathcal{T}_{h}\\}$

and space of continuous piecewise linear vector fields:

$\displaystyle\displaystyle\mbox{\boldmath$\displaystyle W$}_{h}$
$\displaystyle\displaystyle=$
$\displaystyle\displaystyle\\{\mbox{\boldmath$\displaystyle
v$}\in(H^{1}(\Omega))^{d}:\mbox{\boldmath$\displaystyle
v$}_{T}=\mbox{\boldmath$\displaystyle v$}|_{T}\in(P_{1}(T))^{d},\quad\forall
T\in\mathcal{T}_{h}\\}.$

Consider a linear map $\displaystyle E:\mbox{\boldmath$\displaystyle
V$}_{h}\longrightarrow\mbox{\boldmath$\displaystyle W$}_{h}$ defined as
follows:

(2.9) $E(\mbox{\boldmath$\displaystyle
v$})(p)=\frac{1}{|\chi_{p}|}\sum_{T\in\chi_{p}}\mbox{\boldmath$\displaystyle
v$}|_{T}(p),\quad\forall p\in\mathcal{V}(\mathcal{T}_{h}),$

where $\displaystyle\chi_{p}=\\{T\in\mathcal{T}_{h}:p\in\mathcal{V}(T)\\}$,
the patch of the vertex $\displaystyle p$, i.e., the collection simplexes that
contain $\displaystyle p$ as its vertex, and $\displaystyle|\chi_{p}|$ is the
cardinality of the set $\displaystyle\chi_{p}$. We note that it holds true

(2.10) $|\chi_{p}|\lesssim 1,\quad\forall p\in\mathcal{V}(\mathcal{T}_{h}).$

We recall the following approximation property, which can be found in [1]:

######  Lemma 2.1.

Let $\displaystyle\mbox{\boldmath$\displaystyle
v$}\in\mbox{\boldmath$\displaystyle V$}_{h}$ and $\displaystyle
T\in\mathcal{T}_{h}$. Then, with $\displaystyle\mbox{\boldmath$\displaystyle
v$}_{T}=\mbox{\boldmath$\displaystyle v$}|_{T}$, for all $\displaystyle
p\in\mathcal{V}(\mathcal{T}_{h})$, we have the following estimate:

(2.11) $|\mbox{\boldmath$\displaystyle v$}_{T}-E(\mbox{\boldmath$\displaystyle
v$})(p)|^{2}\lesssim\sum_{f\in\mathcal{E}_{p}}|[\\![\mbox{\boldmath$\displaystyle
v$}]\\!]_{f}(p)|^{2},\quad T\in\mathcal{T}_{h}\,\mbox{ and
}\,p\in\mathcal{V}(T),$

where

(2.12) $\mathcal{E}_{p}=\\{e\in\mathcal{E}_{h}^{o}:p\in\partial f\\},$

is the set of interior sides sharing $\displaystyle p$ as a common vertex, and
$\displaystyle[\\![\mbox{\boldmath$\displaystyle v$}]\\!]_{f}$ is the jump of
$\displaystyle v$ across the edge/face $\displaystyle f$.

The main observation that will lead us to construct the minimal jump
conditions or refined Korn’s inequality will be presented at the following
simple but important theorem:

######  Theorem 2.1.

For any $\displaystyle T\in\mathcal{T}_{h}$. Let $\displaystyle f\in\partial
T$ and $\displaystyle\mbox{\boldmath$\displaystyle c$}_{f}$ be the barycenter
of $\displaystyle f$. For some constant $\displaystyle c,c_{1},c_{2}\in{\rm
I\\!R}$, we have for 2D and 3D, respectively, with
$\displaystyle\mbox{\boldmath$\displaystyle t$}_{f}$ and
$\displaystyle\mbox{\boldmath$\displaystyle n$}_{f}$ being the tangent vector
to the edge and the normal vector to the face,

(2.13) $\displaystyle{\mbox{\boldmath$\displaystyle
RM$}}(f)^{\perp}&=\\{(\mbox{\boldmath$\displaystyle
v$}\cdot\mbox{\boldmath$\displaystyle
t$}_{f})|_{f}:\forall\mbox{\boldmath$\displaystyle
v$}\in\mbox{\boldmath$\displaystyle RM$}(T)\\}={\rm span}\\{1\\}\quad\mbox{
for }d=2\\\ ={\rm RT}_{0}(f):=\\{(\mbox{\boldmath$\displaystyle
v$}\times\mbox{\boldmath$\displaystyle
n$}_{f})|_{f}:\forall\mbox{\boldmath$\displaystyle
v$}\in\mbox{\boldmath$\displaystyle RM$}(T)\\}\quad\mbox{ for }d=3,$

where $\displaystyle{\rm RT}_{0}(f)$ can be characterized as follows:

(2.14) ${\rm RT}_{0}(f)=\\{(c_{1}+cx_{1})\mbox{\boldmath$\displaystyle
t$}_{1}+(c_{2}+cx_{2})\mbox{\boldmath$\displaystyle t$}_{2}:x_{1},x_{2}\in{\rm
I\\!R}\\},$

where $\displaystyle f={\rm span}\\{\mbox{\boldmath$\displaystyle
t$}_{1},\mbox{\boldmath$\displaystyle t$}_{2}\\}$ such that
$\displaystyle\mbox{\boldmath$\displaystyle
t$}_{1}\cdot\mbox{\boldmath$\displaystyle t$}_{2}=0$.

###### Proof.

We begin our proof for the 2D case. Choose
$\displaystyle\mbox{\boldmath$\displaystyle
v$}\in{\mbox{\boldmath$\displaystyle RM$}}(T)$. Then, $\displaystyle v$ is of
the following form:

(2.15) $\mbox{\boldmath$\displaystyle v$}=\mbox{\boldmath$\displaystyle
a$}+b(y,-x)^{T},$

for some constant vector $\displaystyle a$ and a constant $\displaystyle b$.
Without loss of generality, we assume that $\displaystyle f\in\partial T$ can
be expressed as a linear function $\displaystyle y=mx+n$. Then, we have that
$\displaystyle\mbox{\boldmath$\displaystyle
t$}_{f}=\frac{1}{\sqrt{1+m^{2}}}(1,m)^{T}$ and

(2.16) $(\mbox{\boldmath$\displaystyle v$}\cdot\mbox{\boldmath$\displaystyle
t$}_{f})|_{f}=c_{0},$

where $\displaystyle c_{0}$ is some constant. We shall now consider 3D case.
For any $\displaystyle\mbox{\boldmath$\displaystyle
v$}\in{\mbox{\boldmath$\displaystyle RM$}}(T)$ can be given as
$\displaystyle\mbox{\boldmath$\displaystyle v$}=\mbox{\boldmath$\displaystyle
a$}+\mbox{\boldmath$\displaystyle A$}\mbox{\boldmath$\displaystyle x$}$, where
$\displaystyle A$, $\displaystyle x$, and
$\displaystyle\mbox{\boldmath$\displaystyle n$}_{f}$ can be denoted by the
followings:

$\displaystyle\displaystyle\mbox{\boldmath$\displaystyle
A$}=\left(\begin{array}[]{ccc}0&a_{1}&a_{2}\\\ -a_{1}&0&a_{3}\\\
-a_{2}&-a_{3}&0\end{array}\right),\quad\mbox{\boldmath$\displaystyle
x$}=\left(\begin{array}[]{c}x_{1}\\\ x_{2}\\\
x_{3}\end{array}\right),\quad\mbox{ and }\quad\mbox{\boldmath$\displaystyle
n$}_{f}=\left(\begin{array}[]{c}n_{1}\\\ n_{2}\\\ n_{3}\end{array}\right).$

Then a simple calculation leads to

$\displaystyle\displaystyle\mbox{\boldmath$\displaystyle
v$}\times\mbox{\boldmath$\displaystyle n$}_{f}=\mbox{\boldmath$\displaystyle
a$}\times\mbox{\boldmath$\displaystyle
n$}_{f}+\left(\begin{array}[]{c}a_{1}x_{2}+a_{2}x_{3}\\\
-a_{1}x_{1}+a_{3}x_{3}\\\
-a_{2}x_{1}-a_{3}x_{2}\end{array}\right)\times\left(\begin{array}[]{c}n_{1}\\\
n_{2}\\\ n_{3}\end{array}\right).$

Particularly, we pay attention to the following quantity:

$\displaystyle\displaystyle\mbox{\boldmath$\displaystyle
A$}\mbox{\boldmath$\displaystyle x$}\times\mbox{\boldmath$\displaystyle
n$}_{f}=\left(\begin{array}[]{c}\eta x_{1}+a_{3}\mbox{\boldmath$\displaystyle
n$}_{f}\cdot\mbox{\boldmath$\displaystyle x$}\\\ \eta
x_{2}-a_{2}\mbox{\boldmath$\displaystyle
n$}_{f}\cdot\mbox{\boldmath$\displaystyle x$}\\\ \eta
x_{3}+a_{1}\mbox{\boldmath$\displaystyle
n$}_{f}\cdot\mbox{\boldmath$\displaystyle
x$}\end{array}\right)=\left(\begin{array}[]{c}\eta
x_{1}+a_{3}\mbox{\boldmath$\displaystyle
n$}_{f}\cdot(\mbox{\boldmath$\displaystyle x$}-\mbox{\boldmath$\displaystyle
c$}_{f})+a_{3}\mbox{\boldmath$\displaystyle
n$}_{f}\cdot\mbox{\boldmath$\displaystyle c$}_{f}\\\ \eta
x_{2}-a_{2}\mbox{\boldmath$\displaystyle
n$}_{f}\cdot(\mbox{\boldmath$\displaystyle x$}-\mbox{\boldmath$\displaystyle
c$}_{f})-a_{2}\mbox{\boldmath$\displaystyle
n$}_{f}\cdot\mbox{\boldmath$\displaystyle c$}_{f}\\\ \eta
x_{3}+a_{1}\mbox{\boldmath$\displaystyle
n$}_{f}\cdot(\mbox{\boldmath$\displaystyle x$}-\mbox{\boldmath$\displaystyle
c$}_{f})+a_{1}\mbox{\boldmath$\displaystyle
n$}_{f}\cdot\mbox{\boldmath$\displaystyle c$}_{f}\end{array}\right)$

where $\displaystyle\eta=\mbox{\boldmath$\displaystyle
n$}_{f}\cdot(-a_{3},a_{2},-a_{1})=-a_{1}n_{3}+a_{2}n_{2}-a_{3}n_{1}$ and
$\displaystyle\mbox{\boldmath$\displaystyle c$}_{f}$ is the barycenter of the
face $\displaystyle f$. In case $\displaystyle\eta=0$,
$\displaystyle(\mbox{\boldmath$\displaystyle A$}\mbox{\boldmath$\displaystyle
x$}\times\mbox{\boldmath$\displaystyle
n$}_{f})|_{f}=\mbox{\boldmath$\displaystyle
n$}_{f}\cdot\mbox{\boldmath$\displaystyle c$}_{f}(a_{3},-a_{2},a_{1})^{T}.$
Note that $\displaystyle(a_{3},-a_{2},a_{1})^{T}\in f$ and so, it can be
expressed as
$\displaystyle\mbox{\boldmath$\displaystyle\mu$}\times\mbox{\boldmath$\displaystyle
n$}_{f}$ for some vector $\displaystyle\mu$. Therefore,
$\displaystyle(\mbox{\boldmath$\displaystyle
v$}\times\mbox{\boldmath$\displaystyle
n$}_{f})|_{f}=\mbox{\boldmath$\displaystyle
b$}\times\mbox{\boldmath$\displaystyle n$}_{f}$ with
$\displaystyle\mbox{\boldmath$\displaystyle b$}=\mbox{\boldmath$\displaystyle
a$}+\mbox{\boldmath$\displaystyle\mu$}$. In case $\displaystyle\eta\neq 0$, we
observe that for $\displaystyle\mbox{\boldmath$\displaystyle x$}\in f$,

$\displaystyle\displaystyle\mbox{\boldmath$\displaystyle
A$}\mbox{\boldmath$\displaystyle x$}\times\mbox{\boldmath$\displaystyle
n$}_{f}=\eta(\mbox{\boldmath$\displaystyle x$}-\mbox{\boldmath$\displaystyle
c$}),$

where

(2.20) $\mbox{\boldmath$\displaystyle
c$}=\frac{1}{\eta}\left(\begin{array}[]{c}-a_{3}\mbox{\boldmath$\displaystyle
n$}_{f}\cdot\mbox{\boldmath$\displaystyle c$}_{f}\\\
a_{2}\mbox{\boldmath$\displaystyle n$}_{f}\cdot\mbox{\boldmath$\displaystyle
c$}_{f}\\\ -a_{1}{\mbox{\boldmath$\displaystyle
n$}}_{f}\cdot\mbox{\boldmath$\displaystyle
c$}_{f}\end{array}\right)\quad\mbox{ and }\quad(\mbox{\boldmath$\displaystyle
x$}-\mbox{\boldmath$\displaystyle c$})\cdot\mbox{\boldmath$\displaystyle
n$}_{f}=0.$

This means that we can modify $\displaystyle\mbox{\boldmath$\displaystyle
A$}\mbox{\boldmath$\displaystyle x$}\times\mbox{\boldmath$\displaystyle
n$}_{f}$ further into the following, with $\displaystyle c=\eta$,

(2.21) $\mbox{\boldmath$\displaystyle A$}\mbox{\boldmath$\displaystyle
x$}\times\mbox{\boldmath$\displaystyle n$}_{f}=c(\mbox{\boldmath$\displaystyle
x$}-\mbox{\boldmath$\displaystyle c$})=c(\mbox{\boldmath$\displaystyle
x$}-\mbox{\boldmath$\displaystyle c$}_{f}+\mbox{\boldmath$\displaystyle
c$}_{f}-\mbox{\boldmath$\displaystyle c$})=\mbox{\boldmath$\displaystyle
d$}+c(\mbox{\boldmath$\displaystyle x$}-\mbox{\boldmath$\displaystyle
c$}_{f}),$

where $\displaystyle d$ is a constant vector such that
$\displaystyle\mbox{\boldmath$\displaystyle
d$}\cdot\mbox{\boldmath$\displaystyle n$}_{f}=0$. Therefore, we arrive at the
conclusion and this completes the proof. ∎

For 3D, we remark that for any given $\displaystyle T\in\mathcal{T}_{h}$ with
$\displaystyle f$ being a face of $\displaystyle T$, the dimension of the
space $\displaystyle\\{\mbox{\boldmath$\displaystyle
q$}(\mbox{\boldmath$\displaystyle x$})\times\mbox{\boldmath$\displaystyle
n$}_{f}:\mbox{\boldmath$\displaystyle q$}\in\mbox{\boldmath$\displaystyle
RM$}(T),\,\,\forall\mbox{\boldmath$\displaystyle x$}\in f\\}$ is three and we
denote the space $\displaystyle\\{(c_{1}+cx_{1})\mbox{\boldmath$\displaystyle
t$}_{1}+(c_{2}+cx_{2})\mbox{\boldmath$\displaystyle t$}_{2}:x_{1},x_{2}\in{\rm
I\\!R}\\}$ by $\displaystyle{\rm RT}_{0}(f)$ [13]. Now, we can prove the
Korn’s inequality for functions in
$\displaystyle(H^{1}(\Omega,\mathcal{T}_{h}))^{d}$. We note that the spaces of
projection onto the face of each $\displaystyle T$ are different for 2D and
3D, the detailed jump terms shall be stated accordingly. We first define
$\displaystyle\pi_{1}$ by the $\displaystyle L^{2}$ orthogonal projection from
$\displaystyle L^{2}(f)$ onto $\displaystyle P_{1}(f)$ and
$\displaystyle\pi_{RM^{\perp}(f)}$ is the $\displaystyle L^{2}$ orthogonal
projection form $\displaystyle L^{2}(f)$ onto $\displaystyle P_{0}(f)$ for
$\displaystyle d=2$, but it is from $\displaystyle L^{2}(f)$ to
$\displaystyle{{\rm RT}}_{0}(f)$ for $\displaystyle d=3$, respectively.

To state the discrete Korn’s inequality, we shall first introduce on each
$\displaystyle T\in\mathcal{T}_{h}$, a projection operator
$\displaystyle\Pi_{T}$ from $\displaystyle(H^{1}(T))^{d}$ onto the
$\displaystyle\mbox{\boldmath$\displaystyle RM$}(T)$ by the following
conditions:

(2.22) $\left|\int_{T}(\mbox{\boldmath$\displaystyle
v$}-\Pi_{T}\mbox{\boldmath$\displaystyle
v$})dx\right|=0,\quad\forall\mbox{\boldmath$\displaystyle
v$}\in(H^{1}(T))^{d},$ (2.23)
$\left|\int_{T}\nabla\times(\mbox{\boldmath$\displaystyle
v$}-\Pi_{T}\mbox{\boldmath$\displaystyle
v$})dx\right|=0,\quad\forall\mbox{\boldmath$\displaystyle
v$}\in(H^{1}(T))^{d}.$

Hence, by the definition of $\displaystyle\Pi_{T}$, we have (see (3.3) in [1])

(2.24) $|\mbox{\boldmath$\displaystyle
v$}-\Pi_{T}\mbox{\boldmath$\displaystyle
v$}|_{H^{1}(T)}\lesssim\|\mathcal{D}(\mbox{\boldmath$\displaystyle
v$}-\Pi_{T}\mbox{\boldmath$\displaystyle
v$})\|_{0}=\|\mathcal{D}(\mbox{\boldmath$\displaystyle
v$})\|_{0},\quad\forall\mbox{\boldmath$\displaystyle v$}\in(H^{1}(T))^{d}.$
(2.25) $\|\mbox{\boldmath$\displaystyle
v$}-\Pi_{T}\mbox{\boldmath$\displaystyle
v$}\|_{0}\lesssim(\hbox{diam}T)|\mbox{\boldmath$\displaystyle
v$}-\Pi_{T}\mbox{\boldmath$\displaystyle
v$}|_{H^{1}(T)},\quad\forall\mbox{\boldmath$\displaystyle
v$}\in(H^{1}(T))^{d}.$

Using this local projection $\displaystyle\Pi_{T}$, we define
$\displaystyle\Pi:(H^{1}(\Omega,\mathcal{T}_{h}))^{d}\longrightarrow\boldsymbol{V}_{h}$
by

(2.26) $(\Pi\mbox{\boldmath$\displaystyle
u$})=\Pi_{T}\mbox{\boldmath$\displaystyle u$}_{T},\quad\forall
T\in\mathcal{T}_{h}.$

Next, following [1], we also introduce a seminorm on
$\displaystyle(H^{1}(\Omega;\mathcal{T}_{h}))^{d}$, denoted by
$\displaystyle\Phi$ satisfying the following Assumptions:

  * (C1)

$\displaystyle|\Phi(\mbox{\boldmath$\displaystyle
w$})|\lesssim\|\mbox{\boldmath$\displaystyle
w$}\|_{1},\quad\forall\mbox{\boldmath$\displaystyle
w$}\in(H^{1}(\Omega))^{d},$

  * (C2)

$\displaystyle\Phi(\mbox{\boldmath$\displaystyle m$})=0$ and
$\displaystyle\mbox{\boldmath$\displaystyle m$}\in{\bf{RM}}(\Omega)$ if and
only if $\displaystyle m$ is constant.

  * (C3)

$\displaystyle(\Phi(\mbox{\boldmath$\displaystyle
v$}-E\mbox{\boldmath$\displaystyle
v$}))^{2}\lesssim\sum_{f\in\mathcal{E}_{h}^{o}}(\hbox{diam}(f))^{d-2}\sum_{p\in\mathcal{V}(f)}|[\\![\mbox{\boldmath$\displaystyle
v$}]\\!]_{f}(p)|^{2},\,\forall\mbox{\boldmath$\displaystyle
v$}\in\mbox{\boldmath$\displaystyle V$}_{h}$, where
$\displaystyle\mathcal{V}(f)$ is the set of the vertices of $\displaystyle f$.

The first estimate is well-known for the piecewise $\displaystyle H^{1}$
functions, see [1] :

######  Lemma 2.2.

Let $\displaystyle\Phi$ be the seminorm on $\displaystyle
H^{1}(\Omega;\mathcal{T}_{h})$ satisfying the Assumptios (C1), (C2), and (C3).
Then, the following estimate holds:

(2.27) $|\mbox{\boldmath$\displaystyle
v$}|^{2}_{H^{1}(\Omega,\mathcal{T}_{h})}\lesssim\|\mathcal{D}_{\mathcal{T}}(\mbox{\boldmath$\displaystyle
v$})\|_{0}^{2}+(\Phi(\mbox{\boldmath$\displaystyle
v$}))^{2}+\sum_{f\in\mathcal{E}^{o}_{h}}(\hbox{diam}~{}f)^{d-2}\sum_{p\in\mathcal{V}(f)}|[\\![\mbox{\boldmath$\displaystyle
v$}]\\!]_{f}(p)|^{2}$

for all $\displaystyle\mbox{\boldmath$\displaystyle
v$}\in\mbox{\boldmath$\displaystyle V$}_{h}$, where
$\displaystyle\mathcal{D}_{\mathcal{T}}(\mbox{\boldmath$\displaystyle
v$})|_{T}=\mathcal{D}(\mbox{\boldmath$\displaystyle v$}|_{T})$ for all
$\displaystyle T\in\mathcal{T}_{h}$.

Now, in order to refine the discrete Korn’s inequality, we further impose an
addition Assumption for $\displaystyle\Phi$ as follows:

  * (C4)

$\displaystyle|\Phi(\mbox{\boldmath$\displaystyle
u$}-\Pi\mbox{\boldmath$\displaystyle
u$})|\lesssim\|\mathcal{D}_{\mathcal{T}}(\mbox{\boldmath$\displaystyle
u$})\|_{0},\quad\forall\mbox{\boldmath$\displaystyle
u$}\in(H^{1}(\Omega,\mathcal{T}_{h}))^{d}$.

######  Remark 2.1.

For $\displaystyle u\in(H^{1}(\Omega))^{d}$, Assumption (C1) implies
Assumption (C4), but for $\displaystyle
u\in(H^{1}(\Omega,\mathcal{T}_{h}))^{d}$, Assumption (C4) can not be derived
from Assumption (C1).

We then obtain the following main result in this paper:

######  Theorem 2.2.

Let $\displaystyle\Phi:(H^{1}(\Omega,\mathcal{T}_{h}))^{d}\longrightarrow{\rm
I\\!R}$ be a seminorm satisfying the Assumptions (C1), (C2), (C3) and (C4). We
have the following results for 2D and 3D, respectively. For $\displaystyle
2D$, we have

$\displaystyle\displaystyle|\mbox{\boldmath$\displaystyle
u$}|^{2}_{H^{1}(\Omega,\mathcal{T}_{h})}\lesssim\|\mathcal{D}_{\mathcal{T}}(\mbox{\boldmath$\displaystyle
u$})\|_{0}^{2}+(\Phi(\mbox{\boldmath$\displaystyle u$}))^{2}$
$\displaystyle\displaystyle+\sum_{f\in\mathcal{E}^{o}_{h}}(\hbox{diam}(f))^{-1}\Big{(}\|\left[\pi_{1}([\\![\mbox{\boldmath$\displaystyle
u$}]\\!]_{f}\cdot\mbox{\boldmath$\displaystyle
n$}_{f})\right]\mbox{\boldmath$\displaystyle
n$}_{f}\|_{0,f}^{2}+\|\left[\pi_{RM^{\perp}(f)}([\\![\mbox{\boldmath$\displaystyle
u$}]\\!]_{f}\cdot\mbox{\boldmath$\displaystyle
t$}_{f})\right]\mbox{\boldmath$\displaystyle t$}_{f}\|_{0,f}^{2}\Big{)}.$

For $\displaystyle 3D$, we have

$\displaystyle\displaystyle|\mbox{\boldmath$\displaystyle
u$}|^{2}_{H^{1}(\Omega,\mathcal{T}_{h})}\lesssim\|\mathcal{D}_{\mathcal{T}}(\mbox{\boldmath$\displaystyle
u$})\|_{0}^{2}+(\Phi(\mbox{\boldmath$\displaystyle u$}))^{2}$
$\displaystyle\displaystyle+\sum_{f\in\mathcal{E}^{o}_{h}}(\hbox{diam}(f))^{-1}\Big{(}\|\left[\pi_{1}([\\![\mbox{\boldmath$\displaystyle
u$}]\\!]_{f}\cdot\mbox{\boldmath$\displaystyle
n$}_{f})\right]\mbox{\boldmath$\displaystyle
n$}_{f}\|_{0,f}^{2}+\|\left[\pi_{RM^{\perp}(f)}([\\![\mbox{\boldmath$\displaystyle
u$}]\\!]_{f}\times\mbox{\boldmath$\displaystyle
n$}_{f})\right]\times\mbox{\boldmath$\displaystyle
n$}_{f}\|_{0,f}^{2}\Big{)}.$

###### Proof.

Let $\displaystyle\mbox{\boldmath$\displaystyle
u$}\in(H^{1}(\Omega,\mathcal{T}_{h}))^{d}$, from (2.27) and (2.24), we have

$\begin{split}|\mbox{\boldmath$\displaystyle
u$}|^{2}_{H^{1}(\Omega,\mathcal{T}_{h})}&\lesssim|\mbox{\boldmath$\displaystyle
u$}-\Pi\mbox{\boldmath$\displaystyle
u$}|^{2}_{H^{1}(\Omega,\mathcal{T}_{h})}+|\Pi\mbox{\boldmath$\displaystyle
u$}|^{2}_{H^{1}(\Omega,\mathcal{T}_{h})}\\\
&\lesssim\|\mathcal{D}_{\mathcal{T}}(\mbox{\boldmath$\displaystyle
u$})\|_{0}^{2}+(\Phi(\Pi\mbox{\boldmath$\displaystyle
u$}))^{2}+\sum_{f\in\mathcal{E}^{o}_{h}}(\hbox{diam}(f))^{d-2}\sum_{p\in\mathcal{V}(f)}|[\\![\Pi\mbox{\boldmath$\displaystyle
u$}]\\!]_{f}(p)|^{2}.\end{split}$

Using condition (C4), we find

(2.28) $\Phi(\Pi\mbox{\boldmath$\displaystyle
u$})\leq\Phi(\mbox{\boldmath$\displaystyle
u$}-\Pi\mbox{\boldmath$\displaystyle u$})+\Phi(\mbox{\boldmath$\displaystyle
u$})\lesssim\|\mathcal{D}_{\mathcal{T}}(\mbox{\boldmath$\displaystyle
u$})\|_{0}+\Phi(\mbox{\boldmath$\displaystyle u$}).$

Let $\displaystyle f\in\mathcal{E}^{o}_{h}$ be arbitrary and $\displaystyle
p\in\mathcal{V}(f)$, we have, by inverse estimate

(2.29) $|[\\![\Pi\mbox{\boldmath$\displaystyle
u$}]\\!]_{f}(p)|^{2}\lesssim(\hbox{diam}(f))^{1-d}\|[\\![\Pi\mbox{\boldmath$\displaystyle
u$}]\\!]_{f}\|_{0,f}^{2}.$

We first prove the $\displaystyle d=2$ case. Let $\displaystyle f=T^{+}\cap
T^{-}$ and choose $\displaystyle\mbox{\boldmath$\displaystyle
n$}_{f}=\mbox{\boldmath$\displaystyle n$}^{-}$ as the unit normal vector and
$\displaystyle\mbox{\boldmath$\displaystyle
t$}_{f}=\mbox{\boldmath$\displaystyle t$}^{-}$ as the unit tangential vector
of $\displaystyle f$. Then we have that

(2.30) $\begin{split}\|[\\![\Pi\mbox{\boldmath$\displaystyle
u$}]\\!]_{f}\|_{0,f}^{2}&=\int_{f}|([\\![\Pi\mbox{\boldmath$\displaystyle
u$}]\\!]_{f}\cdot\mbox{\boldmath$\displaystyle
n$}_{f})\mbox{\boldmath$\displaystyle
n$}_{f}|^{2}ds+\int_{f}|([\\![\Pi\mbox{\boldmath$\displaystyle
u$}]\\!]_{f}\cdot\mbox{\boldmath$\displaystyle
t$}_{f})\mbox{\boldmath$\displaystyle t$}_{f}|^{2}ds\\\
&=\int_{f}([\\![\Pi\mbox{\boldmath$\displaystyle
u$}]\\!]_{f}\cdot\mbox{\boldmath$\displaystyle
n$}_{f})^{2}+([\\![\Pi\mbox{\boldmath$\displaystyle
u$}]\\!]_{f}\cdot\mbox{\boldmath$\displaystyle t$}_{f})^{2}ds.\end{split}$

Using Theorem 2.1, we see that

(2.31) $[\\![\Pi\mbox{\boldmath$\displaystyle
u$}]\\!]_{f}\cdot\mbox{\boldmath$\displaystyle t$}_{f}=c$

for some constant $\displaystyle c$. Therefore, we have that

(2.32) $\begin{split}\|[\\![\Pi\mbox{\boldmath$\displaystyle
u$}]\\!]_{f}\|_{0,f}^{2}&=\int_{f}\Big{(}\pi_{1}([\\![\Pi\mbox{\boldmath$\displaystyle
u$}]\\!]_{f}\cdot\mbox{\boldmath$\displaystyle
n$}_{f})\Big{)}^{2}\,ds+\int_{f}\Big{(}\pi_{RM^{\perp}(f)}([\\![\Pi\mbox{\boldmath$\displaystyle
u$}]\\!]_{f}\cdot\mbox{\boldmath$\displaystyle t$}_{f})\Big{)}^{2}\,ds\\\
&\leq\int_{f}\Big{(}\pi_{1}([\\![\Pi\mbox{\boldmath$\displaystyle
u$}-\mbox{\boldmath$\displaystyle
u$}]\\!]_{f}\cdot\mbox{\boldmath$\displaystyle
n$}_{f})\Big{)}^{2}\,ds+\int_{f}\Big{(}\pi_{RM^{\perp}(f)}([\\![\Pi\mbox{\boldmath$\displaystyle
u$}-\mbox{\boldmath$\displaystyle
u$}]\\!]_{f}\cdot\mbox{\boldmath$\displaystyle t$}_{f})\Big{)}^{2}\,ds\\\
&+\int_{f}\Big{(}\pi_{1}([\\![\mbox{\boldmath$\displaystyle
u$}]\\!]_{f}\cdot\mbox{\boldmath$\displaystyle
n$}_{f})\Big{)}^{2}\,ds+\int_{f}\Big{(}\pi_{RM^{\perp}(f)}([\\![\mbox{\boldmath$\displaystyle
u$}]\\!]_{f}\cdot\mbox{\boldmath$\displaystyle
t$}_{f})\Big{)}^{2}\,ds.\end{split}$

Let $\displaystyle\mathcal{T}_{f}=T^{+}\cup T^{-}$, it follows from (2.24),
(2.25) and trace theorem that

(2.33)
$\displaystyle\int_{f}\Big{(}\pi_{1}([\\![\Pi\mbox{\boldmath$\displaystyle
u$}-\mbox{\boldmath$\displaystyle
u$}]\\!]_{f}\cdot\mbox{\boldmath$\displaystyle
n$}_{f})\Big{)}^{2}\,ds+\int_{f}\Big{(}\pi_{RM^{\perp}(f)}([\\![\Pi\mbox{\boldmath$\displaystyle
u$}-\mbox{\boldmath$\displaystyle
u$}]\\!]_{f}\cdot\mbox{\boldmath$\displaystyle t$}_{f})\Big{)}^{2}ds$
$\displaystyle\leq$ $\displaystyle\|[\\![\Pi\mbox{\boldmath$\displaystyle
u$}-\mbox{\boldmath$\displaystyle u$}]\\!]\|^{2}_{0,f}$
$\displaystyle\lesssim$
$\displaystyle\sum_{T\in\mathcal{T}_{f}}\Big{(}(\hbox{diam}(T))|\mbox{\boldmath$\displaystyle
u$}_{T}-\Pi_{T}\mbox{\boldmath$\displaystyle
u$}_{T}|_{H^{1}(T)}^{2}+(\hbox{diam}(T))^{-1}|\mbox{\boldmath$\displaystyle
u$}_{T}-\Pi_{T}\mbox{\boldmath$\displaystyle u$}_{T}|_{0,T}^{2}\Big{)}$
$\displaystyle\lesssim$
$\displaystyle\sum_{T\in\mathcal{T}_{f}}(\hbox{diam}(T))\|\mathcal{D}(\mbox{\boldmath$\displaystyle
u$}_{T})\|_{0,T}^{2}.$

Combining (2.29) (2.32) and (2.33), and noting that diameter of $\displaystyle
T$ is equivalrent to diameter of $\displaystyle f$, we find

(2.34)
$\begin{split}&\sum_{f\in\mathcal{E}^{o}_{h}}(\hbox{diam}(f))^{d-2}\sum_{p\in\mathcal{V}(f)}|[\\![\mbox{\boldmath$\displaystyle
u$}]\\!]_{f}(p)|^{2}\\\
&\lesssim\|\mathcal{D}_{\mathcal{T}}(\mbox{\boldmath$\displaystyle
u$})\|_{0}^{2}+\sum_{f\in\mathcal{E}^{o}_{h}}(\hbox{diam}(f))^{-1}\Big{(}\|\pi_{1}([\\![\mbox{\boldmath$\displaystyle
u$}]\\!]_{f}\cdot\mbox{\boldmath$\displaystyle
n$}_{f})\|_{0,f}+\|\pi_{RM^{\perp}(f)}([\\![\mbox{\boldmath$\displaystyle
u$}]\\!]_{f}\cdot\mbox{\boldmath$\displaystyle
t$}_{f})\|_{0,f}\Big{)}.\end{split}$

For $\displaystyle d=3$, we have that

(2.35) $[\\![\Pi\mbox{\boldmath$\displaystyle
u$}]\\!]_{f}=-([\\![\Pi\mbox{\boldmath$\displaystyle
u$}]\\!]_{f}\times\mbox{\boldmath$\displaystyle
n$}_{f})\times\mbox{\boldmath$\displaystyle
n$}_{f}+([\\![\Pi\mbox{\boldmath$\displaystyle
u$}]\\!]_{f}\cdot\mbox{\boldmath$\displaystyle
n$}_{f})\mbox{\boldmath$\displaystyle n$}_{f}.$

Furthermore, similar to the proof of Theorem 2.1, we can show that

(2.36) $[\\![\Pi\mbox{\boldmath$\displaystyle
u$}]\\!]_{f}\times\mbox{\boldmath$\displaystyle
n$}_{f}=\mbox{\boldmath$\displaystyle a$}\times\mbox{\boldmath$\displaystyle
n$}_{f}+c(\mbox{\boldmath$\displaystyle x$}-\mbox{\boldmath$\displaystyle
c$}_{f}),\quad\forall\mbox{\boldmath$\displaystyle x$}\in f,$

where $\displaystyle\mbox{\boldmath$\displaystyle c$}_{f}$ is a barycenter of
$\displaystyle f$. Using the same argument for $\displaystyle 2D$, we can
establish the estimate for 3D case in Theorem 2.2. This completes the proof. ∎

######  Definition 2.1.

A subspace of rigid body mode will be denoted and defined by

(2.37) $\mbox{\boldmath$\displaystyle
RM$}^{\partial}(\Omega)=\left\\{\mbox{\boldmath$\displaystyle
m$}\in\mbox{\boldmath$\displaystyle
RM$}(\Omega):~{}\|\mbox{\boldmath$\displaystyle
m$}\|_{L^{2}(\partial\Omega)}=1,\int_{\partial\Omega}\mbox{\boldmath$\displaystyle
m$}ds=0\right\\},$

We shall set $\displaystyle\Phi(\mbox{\boldmath$\displaystyle u$})$ as
follows:

(2.38) $\Phi(\mbox{\boldmath$\displaystyle
u$}):=\sup\limits_{\mbox{\boldmath$\displaystyle
m$}\in\mbox{\boldmath$\displaystyle
RM$}^{\partial}(\Omega)}\int_{\partial\Omega}\mbox{\boldmath$\displaystyle
u$}\cdot\mbox{\boldmath$\displaystyle m$}\,ds.$

Under this setting, we can show that $\displaystyle\Phi$ satisfies the
Aummptions (C1), (C2), (C3) and (C4). Additionally, we have

(2.39) $\Phi(\mbox{\boldmath$\displaystyle
u$})\leq\sum_{f\subset\partial\Omega}\Big{(}\|\pi_{1}([\\![\mbox{\boldmath$\displaystyle
u$}]\\!]_{f}\cdot\mbox{\boldmath$\displaystyle
n$}_{f})\|_{0,f}^{2}+\|\pi_{RM^{\perp}(f)}([\\![\mbox{\boldmath$\displaystyle
u$}]\\!]_{f}\cdot\mbox{\boldmath$\displaystyle
t$}_{f})\|_{0,f}^{2}\Big{)}~{}\hbox{for}~{}d=2,$

and

(2.40) $\Phi(\mbox{\boldmath$\displaystyle
u$})\leq\sum_{f\subset\partial\Omega}\Big{(}\|\pi_{1}([\\![\mbox{\boldmath$\displaystyle
u$}]\\!]_{f}\cdot\mbox{\boldmath$\displaystyle
n$}_{f})\|_{0,f}^{2}+\|\pi_{RM^{\perp}(f)}(f)([\\![\mbox{\boldmath$\displaystyle
u$}]\\!]_{f}\times\mbox{\boldmath$\displaystyle
n$}_{f})\|_{0,f}^{2}\Big{)}~{}\hbox{for}~{}d=3.$

Let $\displaystyle P_{1/0}(f)=P_{1}(f)/P_{0}(f)$ and $\displaystyle\pi_{1/0}$
be the orthogonal projection from $\displaystyle L^{2}(f)$ onto $\displaystyle
P_{1/0}(f)$.

######  Corollary 2.1.

For any $\displaystyle\mbox{\boldmath$\displaystyle
u$}\in(H^{1}(\Omega,\mathcal{T}_{h}))^{2}$, and if $\displaystyle u$ satisfies
the following continuity conditions across the edges $\displaystyle
f\in\mathcal{E}_{h}^{o}$

  1. (1)

$\displaystyle\int_{f}[\\![\mbox{\boldmath$\displaystyle
u$}]\\!]_{f}\cdot\mbox{\boldmath$\displaystyle n$}_{f}\,p\,ds=0,\quad\forall
p\in P_{1}(f)$;

  2. (2)

$\displaystyle\int_{f}[\\![\mbox{\boldmath$\displaystyle
u$}]\\!]_{f}\cdot\mbox{\boldmath$\displaystyle t$}_{f}ds=0$;

then the following estimate holds:

(2.41) $|\mbox{\boldmath$\displaystyle
u$}|_{H^{1}(\Omega,\mathcal{T}_{h})}\lesssim\|\mathcal{D}_{\mathcal{T}}(\mbox{\boldmath$\displaystyle
u$})\|_{0}+\sup\limits_{\mbox{\boldmath$\displaystyle
m$}\in\mbox{\boldmath$\displaystyle
RM$}^{\partial}(\Omega)}\int_{\partial\Omega}\mbox{\boldmath$\displaystyle
u$}\cdot\mbox{\boldmath$\displaystyle m$}ds.$

###### Proof.

This is immediate from Theorem 2.2. This completes the proof. ∎

######  Corollary 2.2.

For any $\displaystyle\mbox{\boldmath$\displaystyle
u$}\in(H^{1}(\Omega,\mathcal{T}_{h}))^{3}$, and if $\displaystyle u$ satisfies
the following continuity conditions across the edges $\displaystyle
f\in\mathcal{E}_{h}^{o}$

  1. (1)

$\displaystyle\int_{f}[\\![\mbox{\boldmath$\displaystyle
u$}]\\!]_{f}\cdot\mbox{\boldmath$\displaystyle n$}_{f}\,p\,ds=0,\quad\forall
p\in P_{1}(f)$;

  2. (2)

$\displaystyle\int_{f}([\\![\mbox{\boldmath$\displaystyle
u$}]\\!]_{f}\times\mbox{\boldmath$\displaystyle
n$}_{f})\cdot\mbox{\boldmath$\displaystyle
p$}\,ds=0,\quad\forall\mbox{\boldmath$\displaystyle p$}\in{\rm RT}_{0}(f)$;

then the following estimate holds:

(2.42) $|\mbox{\boldmath$\displaystyle
u$}|_{H^{1}(\Omega,\mathcal{T}_{h})}\lesssim\|\mathcal{D}_{\mathcal{T}}(\mbox{\boldmath$\displaystyle
u$})\|_{0}+\sup\limits_{\mbox{\boldmath$\displaystyle
m$}\in\mbox{\boldmath$\displaystyle
RM$}^{\partial}(\Omega)}\int_{\partial\Omega}\mbox{\boldmath$\displaystyle
u$}\cdot\mbox{\boldmath$\displaystyle m$}\,ds.$

###### Proof.

This is immediate from Theorem 2.2. This completes the proof. ∎

##  3\. On sharpness of the Korn’s inequality

In this subsection, we establish that the proposed Korn’s inequality is sharp.
We first start with 2D case and move on to 3D case. The goal is to establish
that the proposed Korn’s inequality is sharp, see Corollary 2.1 and Corollary
2.2. More precisely, the above continuity conditions in Corollary 2.1 and
Corollary 2.2 minimize the conditions to obtain the classical Korn’s
inequality for piecewise $\displaystyle H^{1}$ space. To put it in another
way, if one of the conditions is violated, the classical Korn’s inequality
does not hold, but the inequality can be made to hold by adding an appropriate
jump terms.

For a sharpness proof, we shall take examples showing that if any one of the
conditions is missing, then the inequality does not hold. These examples are
constructed in special 2D and 3D domains as given in Figure 3.1.

Figure 3.1. Domain 2D and 3D

For simplicity, we shall divide our discussion for 2D and 3D. For 2D, we
consider a special $\displaystyle\Omega=\mathcal{T}_{h}=T_{1}\cup T_{2}$,
which consists of the vertexes $\displaystyle(-1,0),(0,0),(0,1),(-1,1)$ and
$\displaystyle(0,0),(1,0),(1,1),(0,1)$. We denote $\displaystyle f=T_{1}\cap
T_{2}$ and

(3.1) $\displaystyle E_{1}&=\\{\mbox{\boldmath$\displaystyle
u$}\in(H^{1}(\Omega,\mathcal{T}_{h}))^{2}:\pi_{1/0}([\\![\mbox{\boldmath$\displaystyle
u$}]\\!]_{f}\cdot\mbox{\boldmath$\displaystyle
n$}_{f})=0~{}\hbox{and}~{}\pi_{0}([\\![\mbox{\boldmath$\displaystyle
u$}]\\!]_{f}\cdot\mbox{\boldmath$\displaystyle t$}_{f})=0\\},\\\
E_{2}=\\{\mbox{\boldmath$\displaystyle
u$}\in(H^{1}(\Omega,\mathcal{T}_{h}))^{2}:\pi_{0}([\\![\mbox{\boldmath$\displaystyle
u$}]\\!]_{f}\cdot\mbox{\boldmath$\displaystyle
n$}_{f})=0~{}\hbox{and}~{}\pi_{0}([\\![\mbox{\boldmath$\displaystyle
u$}]\\!]_{f}\cdot\mbox{\boldmath$\displaystyle t$}_{f})=0\\},\\\
E_{3}=\\{\mbox{\boldmath$\displaystyle
u$}\in(H^{1}(\Omega,\mathcal{T}_{h}))^{2}:\pi_{0}([\\![\mbox{\boldmath$\displaystyle
u$}]\\!]_{f}\cdot\mbox{\boldmath$\displaystyle
n$}_{f})=0~{}\hbox{and}~{}\pi_{1/0}([\\![\mbox{\boldmath$\displaystyle
u$}]\\!]_{f}\cdot\mbox{\boldmath$\displaystyle n$}_{f})=0\\}.$

We first note that a simple calculation leads that

(3.2) $\mbox{\boldmath$\displaystyle
RM$}^{\partial}(\Omega)\subset\hbox{span}\\{(1-2y,2x)^{t}\\}.$

Furthermore, we shall note that the definitions of $\displaystyle\pi_{0}$ and
$\displaystyle\pi_{1/0}$ mean

(3.3) $\displaystyle\displaystyle\pi_{0}(u)$ $\displaystyle\displaystyle=$
$\displaystyle\displaystyle\frac{1}{|f|}\int_{f}uds\quad\mbox{ and
}\quad\pi_{1/0}(u)=\frac{1}{|f|}\int_{f}usds.$

######  Theorem 3.1.

There exists $\displaystyle\mbox{\boldmath$\displaystyle u$}\in E_{k},$ for
$\displaystyle k=1,2,$ or $\displaystyle 3$, such that

(3.4) $|\mbox{\boldmath$\displaystyle
u$}|^{2}_{H^{1}(\Omega,\mathcal{T}_{h})}\neq 0,\quad\mbox{ but
}\quad\|\mathcal{D}_{\mathcal{T}}(\mbox{\boldmath$\displaystyle
u$})\|_{0}^{2}+\left(\sup\limits_{\mbox{\boldmath$\displaystyle
m$}\in\mbox{\boldmath$\displaystyle
RM$}^{\partial}(\Omega)}\int_{\partial\Omega}\mbox{\boldmath$\displaystyle
u$}\cdot\mbox{\boldmath$\displaystyle m$}ds\right)^{2}=0.$

###### Proof.

Let us consider $\displaystyle u$ such that
$\displaystyle\mbox{\boldmath$\displaystyle
u$}_{i}=\mbox{\boldmath$\displaystyle u$}|_{T_{i}}$ for $\displaystyle i=1,2$
given as follows:

(3.5) $\mbox{\boldmath$\displaystyle u$}_{i}=\left(\begin{array}[]{c}a_{i}\\\
b_{i}\end{array}\right)+c_{i}\left(\begin{array}[]{c}-y\\\
x\end{array}\right)\in\mbox{\boldmath$\displaystyle RM$}(T_{i}),\quad\forall
i=1,2.$

Here six coefficients $\displaystyle(a_{i},b_{i},c_{i})_{i=1,2}$ shall be
appropriately determined so that the Korn’s inequality does not hold. We
observe that
$\displaystyle\|\mathcal{D}_{\mathcal{T}}(\mbox{\boldmath$\displaystyle
u$})\|_{0}=0$ by construction for any choice of coefficients. We first note
that from (3.2), i.e., $\displaystyle\mbox{\boldmath$\displaystyle
RM$}^{\partial}(\Omega)\subset\hbox{span}\\{(1-2y,2x)^{t}\\}$, we can
calculate that with $\displaystyle\mbox{\boldmath$\displaystyle
m$}=c(1-2y,2x)^{t}$ and $\displaystyle c\neq 0$,

(3.6) $\int_{\partial\Omega}\mbox{\boldmath$\displaystyle
u$}\cdot\mbox{\boldmath$\displaystyle
m$}\,ds=c\left(4(b_{2}-b_{1})+\frac{9}{2}(c_{1}+c_{2})\right).$

We also observe that with $\displaystyle\overline{n}=n_{2}-n_{1}$,

$\displaystyle\displaystyle[\\![\mbox{\boldmath$\displaystyle
u$}]\\!]_{f}\cdot\mbox{\boldmath$\displaystyle n$}_{f}$
$\displaystyle\displaystyle=$
$\displaystyle\displaystyle\overline{a}-\overline{c}y$
$\displaystyle\displaystyle[\\![\mbox{\boldmath$\displaystyle
u$}]\\!]_{f}\cdot\mbox{\boldmath$\displaystyle t$}_{f}$
$\displaystyle\displaystyle=$
$\displaystyle\displaystyle\overline{b}+\overline{c}x.$

A simple computation leads that

$\displaystyle\displaystyle\int_{f}[\\![\mbox{\boldmath$\displaystyle
u$}]\\!]_{f}\cdot\mbox{\boldmath$\displaystyle n$}_{f}ds$
$\displaystyle\displaystyle=$
$\displaystyle\displaystyle\int_{0}^{1}\overline{a}-\overline{c}y\,dy=\overline{a}-\frac{1}{2}\overline{c}$
$\displaystyle\displaystyle\int_{f}([\\![\mbox{\boldmath$\displaystyle
u$}]\\!]_{f}\cdot\mbox{\boldmath$\displaystyle n$}_{f})yds$
$\displaystyle\displaystyle=$
$\displaystyle\displaystyle\int_{0}^{1}\overline{a}y-\overline{c}y^{2}\,dy=\frac{1}{2}\overline{a}-\frac{1}{3}\overline{c}$
$\displaystyle\displaystyle\int_{f}[\\![\mbox{\boldmath$\displaystyle
u$}]\\!]_{f}\cdot\mbox{\boldmath$\displaystyle t$}_{f}\,ds$
$\displaystyle\displaystyle=$
$\displaystyle\displaystyle\int_{0}^{1}\overline{b}\,dy=\overline{b}.$

Now, we begin our search of $\displaystyle u$ of the aforementioned form
(3.11) that belongs to $\displaystyle E_{k}$, which satisfies (3.4) for all
$\displaystyle k=1,2,$ or $\displaystyle 3$. This shall be discussed case by
case as follows.

  * •

For $\displaystyle k=1$, we shall choose
$\displaystyle\overline{a}=\frac{2}{3}\overline{c}\neq 0$ and
$\displaystyle\overline{b}=0$. This way, we can make
$\displaystyle\mbox{\boldmath$\displaystyle u$}\in E_{1}$. On the other hand,
for (3.6), we must choose $\displaystyle c_{1}+c_{2}=0$. Then, $\displaystyle
u$ satisfies (3.4).

  * •

For $\displaystyle k=2$, we shall choose
$\displaystyle\overline{a}=\frac{1}{2}\overline{c}\neq 0$. Then by choosing
$\displaystyle\overline{b}=0$, we can make
$\displaystyle\mbox{\boldmath$\displaystyle u$}\in E_{2}$. Again, we choose
$\displaystyle c_{1}+c_{2}=0$ for (3.6). Namely, $\displaystyle c_{1}=c/2$ and
$\displaystyle c_{2}=-c/2$ for arbitrary $\displaystyle c\neq 0$ to guarantee
the above conditions. We note that for $\displaystyle c\neq 0$, $\displaystyle
u$ satisfies (3.4).

  * •

For $\displaystyle k=3$, we shall choose
$\displaystyle\overline{a}=\overline{c}=0$. Then by choosing
$\displaystyle\overline{b}\neq 0$, we can make
$\displaystyle\mbox{\boldmath$\displaystyle u$}\in E_{3}$. On the other hand,
for (3.6), we choose $\displaystyle c_{1}=-c_{2}=c/2\neq 0$ so that it holds,
we see that since $\displaystyle c\neq 0$, $\displaystyle u$ satisfies (3.4).

This completes the proof. ∎

We shall now turn our attention to three dimensional case. For 3D case, we
consider two cubes as given in Figure 3.1 (b). Namely,
$\displaystyle\Omega=\mathcal{T}_{h}=T_{1}\cup T_{2}$, whose coordinates are
$\displaystyle(1,-1,0),(1,0,0)$,
$\displaystyle(0,0,0),(0,-1,0),(1,-1,1),(1,0,1),(0,0,1),(0,-1,1)$ and
$\displaystyle(1,0,0),(1,1,0),(0,1,0)$,
$\displaystyle(0,0,0),(1,0,1),(1,1,1),(0,1,1),(0,0,1)$. We denote
$\displaystyle f=T_{1}\cap T_{2}$ by the face and expand the two conditions in
Corollary 2.2 as following six conditions

  1. (A1)

$\displaystyle\int_{f}[\\![\mbox{\boldmath$\displaystyle
u$}]\\!]_{f}\cdot\mbox{\boldmath$\displaystyle n$}_{f}ds=0$;

  2. (A2)

$\displaystyle\int_{f}[\\![\mbox{\boldmath$\displaystyle
u$}]\\!]_{f}\cdot\mbox{\boldmath$\displaystyle n$}_{f}xds=0$;

  3. (A3)

$\displaystyle\int_{f}[\\![\mbox{\boldmath$\displaystyle
u$}]\\!]_{f}\cdot\mbox{\boldmath$\displaystyle n$}_{f}zds=0$;

  4. (A4)

$\displaystyle\int_{f}([\\![\mbox{\boldmath$\displaystyle
u$}]\\!]_{f}\times\mbox{\boldmath$\displaystyle n$}_{f})\cdot(0,0,1)^{T}ds=0$;

  5. (A5)

$\displaystyle\int_{f}([\\![\mbox{\boldmath$\displaystyle
u$}]\\!]_{f}\times\mbox{\boldmath$\displaystyle n$}_{f})\cdot(1,0,0)^{T}ds=0$;

  6. (A6)

$\displaystyle\int_{f}([\\![\mbox{\boldmath$\displaystyle
u$}]\\!]_{f}\times\mbox{\boldmath$\displaystyle n$}_{f})\cdot(x,0,z)^{T}ds=0$.

We now list a total of six subsets of
$\displaystyle(H^{1}(\Omega;\mathcal{T}_{h}))^{3}$ as follows:

(3.7) $\displaystyle F_{1}&=\\{\mbox{\boldmath$\displaystyle
u$}\in(H^{1}(\Omega,\mathcal{T}_{h}))^{3}:\mbox{\boldmath$\displaystyle
u$}~{}~{}\hbox{satisfies (A2), (A3), (A4), (A5), and (A6)}\\}\\\
F_{2}=\\{\mbox{\boldmath$\displaystyle
u$}\in(H^{1}(\Omega,\mathcal{T}_{h}))^{3}:\mbox{\boldmath$\displaystyle
u$}~{}~{}\hbox{satisfies (A1), (A3), (A4), (A5), and (A6)}\\}\\\
F_{3}=\\{\mbox{\boldmath$\displaystyle
u$}\in(H^{1}(\Omega,\mathcal{T}_{h}))^{3}:\mbox{\boldmath$\displaystyle
u$}~{}~{}\hbox{satisfies (A1), (A2), (A4), (A5), and (A6)}\\}\\\
F_{4}=\\{\mbox{\boldmath$\displaystyle
u$}\in(H^{1}(\Omega,\mathcal{T}_{h}))^{3}:\mbox{\boldmath$\displaystyle
u$}~{}~{}\hbox{satisfies (A1), (A2), (A3), (A5), and (A6)}\\}\\\
F_{5}=\\{\mbox{\boldmath$\displaystyle
u$}\in(H^{1}(\Omega,\mathcal{T}_{h}))^{3}:\mbox{\boldmath$\displaystyle
u$}~{}~{}\hbox{satisfies (A1), (A2), (A3), (A4), and (A6)}\\}\\\
F_{6}=\\{\mbox{\boldmath$\displaystyle
u$}\in(H^{1}(\Omega,\mathcal{T}_{h}))^{3}:\mbox{\boldmath$\displaystyle
u$}~{}~{}\hbox{satisfies (A1), (A2), (A3), (A4), and (A5)}\\}.$

We first establish that with $\displaystyle\Omega=T_{1}\cup T_{2}$, the space
of $\displaystyle\mbox{\boldmath$\displaystyle RM$}^{\partial}(\Omega)$. A
tedious but simple calculation shows that

(3.8) $\mbox{\boldmath$\displaystyle
RM$}^{\partial}(\Omega)\subset\hbox{span}\\{\mbox{\boldmath$\displaystyle
m$}_{1},\mbox{\boldmath$\displaystyle m$}_{2},\mbox{\boldmath$\displaystyle
m$}_{3}\\},$

where $\displaystyle\mbox{\boldmath$\displaystyle m$}_{i}$ with $\displaystyle
i=1,2,3$ are given as follows:

$\displaystyle\displaystyle\mbox{\boldmath$\displaystyle
m$}_{1}=\left(\begin{array}[]{c}-1+2z\\\ 0\\\
1-2x\end{array}\right),\,\,\mbox{\boldmath$\displaystyle
m$}_{2}=\left(\begin{array}[]{c}2y\\\ 1-2x\\\ 0\end{array}\right),\,\,\mbox{
and }\,\,\mbox{\boldmath$\displaystyle m$}_{3}=\left(\begin{array}[]{c}y\\\
-x+z\\\ -y\end{array}\right).$

We shall now state and prove the main result in this section.

######  Theorem 3.2.

There exists $\displaystyle\mbox{\boldmath$\displaystyle u$}\in F_{k},$ for
$\displaystyle k=1,2,3,4,5$ or $\displaystyle 6$, such that

(3.10) $|\mbox{\boldmath$\displaystyle
u$}|^{2}_{H^{1}(\Omega,\mathcal{T}_{h})}\neq 0,\quad\mbox{ but
}\quad\|\mathcal{D}_{\mathcal{T}}(\mbox{\boldmath$\displaystyle
u$})\|_{0}^{2}+\left(\sup\limits_{\mbox{\boldmath$\displaystyle
m$}\in\mbox{\boldmath$\displaystyle
RM$}^{\partial}(\Omega)}\int_{\partial\Omega}\mbox{\boldmath$\displaystyle
u$}\cdot\mbox{\boldmath$\displaystyle m$}ds\right)^{2}=0.$

###### Proof.

Let us consider $\displaystyle u$ such that
$\displaystyle\mbox{\boldmath$\displaystyle
u$}_{i}=\mbox{\boldmath$\displaystyle u$}|_{T_{i}}$ for $\displaystyle i=1,2$
given as follows:

(3.11) $\mbox{\boldmath$\displaystyle u$}_{i}=\left(\begin{array}[]{c}a_{i}\\\
b_{i}\\\ c_{i}\end{array}\right)+\left(\begin{array}[]{ccc}0&d_{i}&e_{i}\\\
-d_{i}&0&f_{i}\\\
-e_{i}&-f_{i}&0\end{array}\right)\left(\begin{array}[]{c}x\\\ y\\\
z\end{array}\right)\in\mbox{\boldmath$\displaystyle RM$}(T_{i}),\quad\forall
i=1,2.$

Here twelve coefficients
$\displaystyle(a_{i},b_{i},c_{i},d_{i},e_{i},f_{i})_{i=1,2}$ shall be
appropriately determined so that the Korn’s inequality does not hold under the
condition that $\displaystyle\mbox{\boldmath$\displaystyle u$}\in F_{k}$ for
any $\displaystyle k$. We observe that
$\displaystyle\|\mathcal{D}_{\mathcal{T}}(\mbox{\boldmath$\displaystyle
u$})\|_{0}=0$ by construction for any choice of coefficients. We first
investigate what constraints are given from the conditions:

(3.12) $\int_{\partial\Omega}\mbox{\boldmath$\displaystyle
u$}\cdot\mbox{\boldmath$\displaystyle m$}_{i}\,ds=0,\quad\forall i=1,2,3,$

where $\displaystyle\mbox{\boldmath$\displaystyle
RM$}^{\partial}(\Omega)\subset\hbox{span}\\{\mbox{\boldmath$\displaystyle
m$}_{1},\mbox{\boldmath$\displaystyle m$}_{2},\mbox{\boldmath$\displaystyle
m$}_{3}\\}$. We observe that

$\displaystyle\displaystyle\mbox{\boldmath$\displaystyle
u$}\cdot\mbox{\boldmath$\displaystyle m$}_{1}|_{T_{i}}$
$\displaystyle\displaystyle=$
$\displaystyle\displaystyle(a_{i}+d_{i}y+e_{i}z)(-1+2z)+(c_{i}-e_{i}x-f_{i}y)(1-2x);$
$\displaystyle\displaystyle\mbox{\boldmath$\displaystyle
u$}\cdot\mbox{\boldmath$\displaystyle m$}_{2}|_{T_{i}}$
$\displaystyle\displaystyle=$
$\displaystyle\displaystyle(a_{i}+d_{i}y+e_{i}z)(2y)+(b_{i}-d_{i}x+f_{i}z)(1-2x);$
$\displaystyle\displaystyle\mbox{\boldmath$\displaystyle
u$}\cdot\mbox{\boldmath$\displaystyle m$}_{3}|_{T_{i}}$
$\displaystyle\displaystyle=$
$\displaystyle\displaystyle(a_{i}+d_{i}y+e_{i}z)(y)+(b_{i}-d_{i}x+f_{i}z)(-x+z)+(c_{i}-e_{i}x-f_{i}y)(-y).$

A simple but tedious computation leads that

(3.13) $\displaystyle\int_{\partial\Omega}\mbox{\boldmath$\displaystyle
u$}\cdot\mbox{\boldmath$\displaystyle
m$}_{1}\,ds&=\left(\frac{9}{3}(e_{1}+e_{2})\right);\\\
\int_{\partial\Omega}\mbox{\boldmath$\displaystyle
u$}\cdot\mbox{\boldmath$\displaystyle
m$}_{2}\,ds=6(a_{2}-a_{1})+3(e_{2}-e_{1})+\frac{37}{6}(d_{1}+d_{2});\\\
\int_{\partial\Omega}\mbox{\boldmath$\displaystyle
u$}\cdot\mbox{\boldmath$\displaystyle
m$}_{3}\,ds=3(a_{2}-a_{1})-3(c_{2}-c_{1})+3(e_{2}-e_{1})\\\
+\frac{37}{12}\left[(d_{2}+d_{1})+(f_{2}+f_{1})\right].$

We also observe that with $\displaystyle\overline{n}=n_{2}-n_{1}$,

$\displaystyle\displaystyle[\\![\mbox{\boldmath$\displaystyle
u$}]\\!]_{f}\cdot\mbox{\boldmath$\displaystyle n$}_{f}$
$\displaystyle\displaystyle=$
$\displaystyle\displaystyle\overline{b}-\overline{d}x+\overline{f}z;$
$\displaystyle\displaystyle[\\![\mbox{\boldmath$\displaystyle
u$}]\\!]_{f}\times\mbox{\boldmath$\displaystyle n$}_{f}$
$\displaystyle\displaystyle=$
$\displaystyle\displaystyle(-(\overline{c}-\overline{e}x-\overline{f}y),0,\overline{a}+\overline{d}y+\overline{e}z)^{t}|_{f}=(-(\overline{c}-\overline{e}x),0,\overline{a}+\overline{e}z)^{t}.$

A simple computation leads that

$\displaystyle\displaystyle\int_{f}[\\![\mbox{\boldmath$\displaystyle
u$}]\\!]_{f}\cdot\mbox{\boldmath$\displaystyle n$}_{f}ds$
$\displaystyle\displaystyle=$
$\displaystyle\displaystyle\int_{0}^{1}\int_{0}^{1}\overline{b}-\overline{d}x+\overline{f}z\,dxdz=\overline{b}-\frac{1}{2}\overline{d}+\frac{1}{2}\overline{f};$
$\displaystyle\displaystyle\int_{f}[\\![\mbox{\boldmath$\displaystyle
u$}]\\!]_{f}\cdot\mbox{\boldmath$\displaystyle n$}_{f}xds$
$\displaystyle\displaystyle=$
$\displaystyle\displaystyle\int_{0}^{1}\int_{0}^{1}\overline{b}x-\overline{d}x^{2}+\overline{f}xz\,dxdz=\frac{1}{2}\overline{b}-\frac{1}{3}\overline{d}+\frac{1}{4}\overline{f};$
$\displaystyle\displaystyle\int_{f}[\\![\mbox{\boldmath$\displaystyle
u$}]\\!]_{f}\cdot\mbox{\boldmath$\displaystyle n$}_{f}zds$
$\displaystyle\displaystyle=$
$\displaystyle\displaystyle\int_{0}^{1}\int_{0}^{1}\overline{b}z-\overline{d}xz+\overline{f}z^{2}\,dxdz=\frac{1}{2}\overline{b}-\frac{1}{4}\overline{d}+\frac{1}{3}\overline{f}.$

Furthermore, we have that

$\displaystyle\displaystyle\int_{f}[[\\![\mbox{\boldmath$\displaystyle
u$}]\\!]_{f}\times\mbox{\boldmath$\displaystyle n$}_{f}]\cdot(0,0,1)^{T}ds$
$\displaystyle\displaystyle=$
$\displaystyle\displaystyle\int_{0}^{1}\int_{0}^{1}\overline{a}+\overline{e}z\,dxdz=\overline{a}+\frac{1}{2}\overline{e};$
$\displaystyle\displaystyle\int_{f}[[\\![\mbox{\boldmath$\displaystyle
u$}]\\!]_{f}\times\mbox{\boldmath$\displaystyle n$}_{f}]\cdot(1,0,0)^{T}ds$
$\displaystyle\displaystyle=$
$\displaystyle\displaystyle\int_{0}^{1}\int_{0}^{1}-\overline{c}+\overline{e}x\,dxdz=-\overline{c}+\frac{1}{2}\overline{e};$
$\displaystyle\displaystyle\int_{f}[[\\![\mbox{\boldmath$\displaystyle
u$}]\\!]_{f}\times\mbox{\boldmath$\displaystyle n$}_{f}]\cdot(x,0,z)^{T}ds$
$\displaystyle\displaystyle=$
$\displaystyle\displaystyle\int_{0}^{1}\int_{0}^{1}-\overline{c}x+\overline{e}x^{2}+\overline{a}z+\overline{e}z^{2}\,dxdz$
$\displaystyle\displaystyle=$
$\displaystyle\displaystyle-\frac{1}{2}\overline{c}+\frac{2}{3}\overline{e}+\frac{1}{2}\overline{a}.$

Now, we begin our search of $\displaystyle u$ of the aforementioned form
(3.11) that belongs to $\displaystyle F_{k}$, which satisfies (3.10) for all
$\displaystyle k=1,2,3,4,5,$ or $\displaystyle 6$. This shall be discussed
case by case as follows.

  * •

For $\displaystyle k=1$, we shall choose
$\displaystyle\overline{d}=-\overline{f}\neq 0$. Then by choosing
$\displaystyle\overline{b}=-\frac{7}{6}\overline{f}$, we can make
$\displaystyle(A2)$ and $\displaystyle(A3)$ hold, but $\displaystyle(A1)$ does
not. For $\displaystyle(A4),(A5)$ and $\displaystyle(A6)$, we can simply
choose $\displaystyle\overline{a}=\overline{c}=\overline{e}=0$. On the other
hand, for (3.13), we must choose $\displaystyle d_{i},e_{i}$ and
$\displaystyle f_{i}$ so that $\displaystyle
d_{1}+d_{2}=f_{1}+f_{2}=e_{1}+e_{2}=0$. This means, $\displaystyle
e_{1}=e_{2}=0$. On the other hand, we can choose $\displaystyle
d_{2}=c/2,d_{1}=-c/2$ and $\displaystyle f_{2}=-c/2$ and $\displaystyle
f_{1}=c/2$ for arbitrary $\displaystyle c\neq 0$ to guarantee the above
conditions. We note that for $\displaystyle c\neq 0$, $\displaystyle u$
satisfies (3.10).

  * •

For $\displaystyle k=2$, we shall choose
$\displaystyle\overline{b}=\frac{1}{2}\overline{d}\neq 0$. Then by choosing
$\displaystyle\overline{f}=0$, we can make $\displaystyle(A1)$ and
$\displaystyle(A3)$ hold, but $\displaystyle(A2)$ does not. For
$\displaystyle(A4),(A5)$ and $\displaystyle(A6)$, we can simply choose
$\displaystyle\overline{a}=\overline{c}=\overline{e}=0$. On the other hand,
for (3.13), we must choose $\displaystyle d_{i},e_{i}$ and $\displaystyle
f_{i}$ so that $\displaystyle d_{1}+d_{2}=f_{1}+f_{2}=e_{1}+e_{2}=0$. This
means, $\displaystyle e_{1}=e_{2}=0$. On the other hand, we can choose
$\displaystyle d_{2}=c/2,d_{1}=-c/2$ for arbitrary $\displaystyle c\neq 0$ to
guarantee the above conditions. We note that for $\displaystyle c\neq 0$,
$\displaystyle u$ satisfies (3.10).

  * •

For $\displaystyle k=3$, we shall choose
$\displaystyle\overline{b}=\frac{1}{2}\overline{f}\neq 0$. Then by choosing
$\displaystyle\overline{d}=0$, we can make $\displaystyle(A1)$ and
$\displaystyle(A2)$ hold, but $\displaystyle(A3)$ does not. For
$\displaystyle(A4),(A5)$ and $\displaystyle(A6)$, we can simply choose
$\displaystyle\overline{a}=\overline{c}=\overline{e}=0$. On the other hand,
for (3.13), we must choose $\displaystyle d_{i},e_{i}$ and $\displaystyle
f_{i}$ so that $\displaystyle d_{1}+d_{2}=f_{1}+f_{2}=e_{1}+e_{2}=0$. This
means, $\displaystyle e_{1}=e_{2}=0$. On the other hand, we can choose
$\displaystyle f_{2}=c/2,f_{1}=-c/2$ for arbitrary $\displaystyle c\neq 0$ to
guarantee the above conditions. We note that for $\displaystyle c\neq 0$,
$\displaystyle u$ satisfies (3.10).

  * •

For $\displaystyle k=4$, we shall choose
$\displaystyle\overline{c}=\frac{1}{2}\overline{e}\neq 0$ and
$\displaystyle\overline{a}=-\frac{1}{6}\overline{e}\neq 0$. Then by choosing
$\displaystyle\overline{d}=0$, we can make $\displaystyle(A5)$ and
$\displaystyle(A6)$ hold, but $\displaystyle(A4)$ does not. For
$\displaystyle(A1),(A2)$ and $\displaystyle(A3)$, we can simply choose
$\displaystyle\overline{b}=\overline{d}=\overline{f}=0$. On the other hand,
for (3.13), we must choose $\displaystyle d_{i},e_{i}$ and $\displaystyle
f_{i}$ so that $\displaystyle e_{1}+e_{2}=0$ and both $\displaystyle
d_{1}+d_{2}$ and $\displaystyle f_{1}+f_{2}$ appropriately to guarantee (3.13)
and (3.13). With such a choice, $\displaystyle u$ satisfies (3.10).

  * •

For $\displaystyle k=5$, we shall choose
$\displaystyle\overline{c}=\frac{1}{6}\overline{e}\neq 0$ and
$\displaystyle\overline{a}=-\frac{1}{2}\overline{e}\neq 0$. Then by choosing
$\displaystyle\overline{d}=0$, we can make $\displaystyle(A4)$ and
$\displaystyle(A6)$ hold, but $\displaystyle(A5)$ does not. For
$\displaystyle(A1),(A2)$ and $\displaystyle(A3)$, we can simply choose
$\displaystyle\overline{b}=\overline{d}=\overline{f}=0$. On the other hand,
for (3.13), we must choose $\displaystyle d_{i},e_{i}$ and $\displaystyle
f_{i}$ so that $\displaystyle e_{1}+e_{2}=0$ and both $\displaystyle
d_{1}+d_{2}$ and $\displaystyle f_{1}+f_{2}$ appropriately to guarantee (3.13)
and (3.13). With such a choice, $\displaystyle u$ satisfies (3.10).

  * •

For $\displaystyle k=6$, we shall choose
$\displaystyle\overline{c}=\frac{1}{2}\overline{e}=-\overline{a}\neq 0$. Then
by choosing $\displaystyle\overline{d}=0$, we can make $\displaystyle(A4)$ and
$\displaystyle(A5)$ hold, but $\displaystyle(A6)$ does not. For
$\displaystyle(A1),(A2)$ and $\displaystyle(A3)$, we can simply choose
$\displaystyle\overline{b}=\overline{d}=\overline{f}=0$. On the other hand,
for (3.13), we must choose $\displaystyle d_{i},e_{i}$ and $\displaystyle
f_{i}$ so that $\displaystyle e_{1}+e_{2}=0$ and both $\displaystyle
d_{1}+d_{2}$ and $\displaystyle f_{1}+f_{2}$ appropriately to guarantee (3.13)
and (3.13). With such a choice, $\displaystyle u$ satisfies (3.10).

This completes the proof. ∎

##  4\. Some Applications

We begin this section with a simple example of the nonconforming finite
element space on triangular partition introduced in Mardal et al., [7] for the
2D case. Note that the space is composed of functions which are cubic vector
fields with constant divergence on each element and with linear normal
component on each edge. Namely, the local space is given as follows:

(4.1) $\mbox{\boldmath$\displaystyle V$}(T)=\\{\mbox{\boldmath$\displaystyle
v$}\in(P_{3}(T))^{2}:\,{\rm div}\mbox{\boldmath$\displaystyle v$}\in
P_{0}(T),\,\,\mbox{\boldmath$\displaystyle
v$}\cdot\mbox{\boldmath$\displaystyle n$}_{f}|_{f}\in P_{1}(f),\,\,\forall
f\in\partial T\\}.$

Note that the degrees of freedom consist of

(4.2) $\int_{f}(\mbox{\boldmath$\displaystyle
v$}\cdot\mbox{\boldmath$\displaystyle n$}_{f})q\,ds,\quad\forall q\in
P_{1}(f)\quad\mbox{ and }\quad\int_{f}\mbox{\boldmath$\displaystyle
v$}\cdot\mbox{\boldmath$\displaystyle t$}_{f}\,ds.$

This is exactly what is required minimally for the Korn’s inequality.

In the next subsection, we shall consider enriched $\displaystyle H({\rm
div})$ conforming finite element spaces due to Xue et al. [13]. We then
slightly modify these spaces so that we obtain enriched Crouzeix-Raviart
element spaces that satisfy the Korn’s inequality. Finally, we shall discuss
the lowest enrichment for the $\displaystyle H({\rm div})$ conforming finite
element space [6] and its remedy by modifying the enrichment. Throughout the
section, we consider triangular (for 2D) or tetrahedral (for 3D) partitions,
for $\displaystyle T\in\mathcal{T}_{h}$, $\displaystyle{\rm{BDM}}_{\ell}(T)$
denotes the local Brezzi-Douglas-Marini (BDM) space of order
$\displaystyle\ell\geq 1$, namely

(4.3) ${\rm{BDM}}_{\ell}(T)=(P_{\ell}(T))^{d},\quad\ell\geq 1,$

and $\displaystyle{\rm{RT}}_{\ell}(T)$ denotes the Raviart-Thomas space of
order $\displaystyle\ell\geq 0$,

(4.4)
${\rm{RT}}_{\ell}(T)=(P_{\ell}(T))^{d}+\widetilde{P}_{\ell}(T)\mbox{\boldmath$\displaystyle
x$},\quad\ell\geq 0,$

where $\displaystyle\widetilde{P}_{\ell}(T)$ denotes the homogeneous
polynomial space of degree $\displaystyle\ell$. Also, for any given
$\displaystyle T\in\mathcal{T}_{h}$, we denote $\displaystyle\lambda_{i}$ with
$\displaystyle i=1,\cdots,d+1$ the barycentric coordinates of $\displaystyle
T$.

We shall frequently use the following standard bubble functions. Namely, for
$\displaystyle T\in\mathcal{T}_{h}$, we denote $\displaystyle b_{T}$ the
bubble function defined by

(4.5) $b_{T}=\Pi_{i=1}^{d+1}\lambda_{i}.$

Similarly, we can define edge/face bubble functions denoted by $\displaystyle
b_{f}$ for any $\displaystyle f\in\partial T$.

###  4.1. Enriched $\displaystyle H({\rm div})$ conforming finite elements
that satisfy the Korn’s inequality

In this section, we shall recall some of finite elements, which can be shown
to satisfy the Korn’s inequality within our framework. The finite element
spaces listed in this subsection are constructed based on
$\displaystyle{H}({\rm div};\Omega)$ finite element spaces plus divergence
free functions [13]. This subsection is an example indicating the powerfulness
of our framework to show the Korn’s inequality.

The local space of the finite elements introduced in [13] would be of the
following form: for $\displaystyle T\in\mathcal{T}_{h}$,

(4.6) $\mbox{\boldmath$\displaystyle V$}_{1}(T)=\mbox{\boldmath$\displaystyle
V$}_{1,T}^{0}+\mbox{\boldmath$\displaystyle
curl$}(b_{T}\mbox{\boldmath$\displaystyle Y$}),$

where $\displaystyle\mbox{\boldmath$\displaystyle V$}^{0}_{1,T}$ is the
following well-known $\displaystyle H({\rm div})$-conforming finite element
spaces, either $\displaystyle{\rm{BDM}}_{1}(T)$ or
$\displaystyle{\rm{RT}}_{1}(T)$. For the space $\displaystyle Y$, we choose
the following polynomial spaces:

(4.7) $\mbox{\boldmath$\displaystyle
Y$}=\left\\{\begin{array}[]{ll}\mbox{\boldmath$\displaystyle
Y$}_{1}=P_{1}(T)&\mbox{ for }d=2;\\\ \mbox{\boldmath$\displaystyle
Y$}_{2}=(P_{1}(T))^{3}&\mbox{ for }d=3;\\\ \mbox{\boldmath$\displaystyle
Y$}_{3}=(P_{1}(T))^{3}/{\rm
span}\left\\{\left(\lambda_{i}-\frac{1}{3}\right)\nabla\lambda_{i}\right\\}_{i=1}^{4}&\mbox{
for }d=3.\end{array}\right.$

Total of six elements can be listed in the following Table 4.1 with degrees of
freedom.

The six modified $\displaystyle H({\rm div})$ elements
---
Elements |  $\displaystyle\mbox{\boldmath$\displaystyle V$}_{1,T}^{0}$ |  $\displaystyle Y$ |  DOF |  Korn
$\displaystyle 1^{\rm st}$ FEM (2D) |  $\displaystyle{\rm RT}_{1}(T)$ |  $\displaystyle\mbox{\boldmath$\displaystyle Y$}_{1}$ |  $\displaystyle{\small{\begin{array}[]{l}\int_{f}\mbox{\boldmath$\displaystyle v$}\cdot\mbox{\boldmath$\displaystyle n$}q\,ds,\quad\forall q\in P_{1}(f),\\\ \int_{T}\mbox{\boldmath$\displaystyle v$}\cdot\mbox{\boldmath$\displaystyle q$}\,dx,\quad\mbox{\boldmath$\displaystyle q$}\in(P_{0}(T))^{2}\\\ \int_{f}\mbox{\boldmath$\displaystyle v$}\cdot\mbox{\boldmath$\displaystyle t$}\,ds\end{array}}}$ |  Yes
$\displaystyle 2^{\rm nd}$ FEM (2D) |  $\displaystyle{\rm BDM}_{1}(T)$ |  $\displaystyle\mbox{\boldmath$\displaystyle Y$}_{1}$ |  $\displaystyle{\small{\begin{array}[]{l}\int_{f}\mbox{\boldmath$\displaystyle v$}\cdot\mbox{\boldmath$\displaystyle n$}q\,ds,\quad\forall q\in P_{1}(f),\\\ \int_{f}\mbox{\boldmath$\displaystyle v$}\cdot\mbox{\boldmath$\displaystyle t$}\,ds\end{array}}}$ |  Yes
$\displaystyle 1^{\rm st}$ FEM (3D) |  $\displaystyle{\rm RT}_{1}(T)$ |  $\displaystyle\mbox{\boldmath$\displaystyle Y$}_{2}$ |  $\displaystyle{\small{\begin{array}[]{l}\int_{f}\mbox{\boldmath$\displaystyle v$}\cdot\mbox{\boldmath$\displaystyle n$}q\,ds,\quad\forall q\in P_{1}(f),\\\ \int_{T}\mbox{\boldmath$\displaystyle v$}\cdot\mbox{\boldmath$\displaystyle q$}\,dx,\quad\mbox{\boldmath$\displaystyle q$}\in(P_{0}(T))^{3}\\\ \int_{f}(\mbox{\boldmath$\displaystyle v$}\times\mbox{\boldmath$\displaystyle n$})\cdot\mbox{\boldmath$\displaystyle r$}\,ds,\quad\forall\mbox{\boldmath$\displaystyle r$}\in{\rm RT}_{0}(f)\end{array}}}$ |  Yes
$\displaystyle 2^{\rm nd}$ FEM (3D) |  $\displaystyle{\rm BDM}_{1}(T)$ |  $\displaystyle\mbox{\boldmath$\displaystyle Y$}_{2}$ |  $\displaystyle{\small{\begin{array}[]{l}\int_{f}\mbox{\boldmath$\displaystyle v$}\cdot\mbox{\boldmath$\displaystyle n$}q\,ds,\quad\forall q\in P_{1}(f),\\\ \int_{f}(\mbox{\boldmath$\displaystyle v$}\times\mbox{\boldmath$\displaystyle n$})\cdot\mbox{\boldmath$\displaystyle r$}\,ds,\quad\forall\mbox{\boldmath$\displaystyle r$}\in{\rm RT}_{0}(f)\end{array}}}$ |  Yes
$\displaystyle 3^{\rm rd}$ FEM (3D) |  $\displaystyle{\rm RT}_{1}(T)$ |  $\displaystyle\mbox{\boldmath$\displaystyle Y$}_{3}$ |  $\displaystyle{\small{\begin{array}[]{l}\int_{f}\mbox{\boldmath$\displaystyle v$}\cdot\mbox{\boldmath$\displaystyle n$}q\,ds,\quad\forall q\in P_{1}(f),\\\ \int_{T}\mbox{\boldmath$\displaystyle v$}\cdot\mbox{\boldmath$\displaystyle q$}\,dx,\quad\mbox{\boldmath$\displaystyle q$}\in(P_{0}(T))^{3}\\\ \int_{f}(\mbox{\boldmath$\displaystyle v$}\times\mbox{\boldmath$\displaystyle n$})\cdot\mbox{\boldmath$\displaystyle r$}\,ds,\quad\forall\mbox{\boldmath$\displaystyle r$}\in(P_{0}(f))^{2}\end{array}}}$ |  No
$\displaystyle 4^{\rm th}$ FEM (3D) |  $\displaystyle{\rm BDM}_{1}(T)$ |  $\displaystyle\mbox{\boldmath$\displaystyle Y$}_{3}$ |  $\displaystyle{\small{\begin{array}[]{l}\int_{f}\mbox{\boldmath$\displaystyle v$}\cdot\mbox{\boldmath$\displaystyle n$}q\,ds,\quad\forall q\in P_{1}(f),\\\ \int_{f}(\mbox{\boldmath$\displaystyle v$}\times\mbox{\boldmath$\displaystyle n$})\cdot\mbox{\boldmath$\displaystyle r$}\,ds,\quad\mbox{\boldmath$\displaystyle r$}\in(P_{0}(f))^{2}\end{array}}}$ |  No
Table 4.1. FEMs introduced in [13]

Due to our framework, we easily notice that the first four elements are the
ones that satisfy the Korn’s inequality while the last two do not.

###  4.2. Enriched $\displaystyle{\rm CR}$ finite elements that satisfy the
Korn’s inequality

In this subsection, we shall consider enriched CR elements that satisfy the
Korn’s inequality. Our discussion will be done for 2D and 3D separately. We
note that in [4], Falk analyzed Korn’s inequality for some nonconforming two
dimensional finite element spaces. In particular, it was shown that the
Crouzeix-Raviart element [3] does not satisfy such a Korn’s inequality. A
simple remedy can be accomplished. Namely, we add the same enrichment for
$\displaystyle\mbox{\boldmath$\displaystyle V$}_{1}(T)={\rm BDM}_{1}(T)$
replaced with $\displaystyle({\rm CR}(T))^{d}$ for both $\displaystyle d=2$
and $\displaystyle d=3$ and therefore, we propose the enriched
$\displaystyle({\rm{CR}}(T))^{d}$ finite element spaces so that for
$\displaystyle T\in\mathcal{T}_{h}$, we define the local space and degrees of
freedom by

(4.8) $\mbox{\boldmath$\displaystyle V$}(T):=({\rm
CR}(T))^{d}+{\mbox{\boldmath$\displaystyle
curl$}}(b_{T}\mbox{\boldmath$\displaystyle Y$}),$

where $\displaystyle Y$ is either $\displaystyle\mbox{\boldmath$\displaystyle
Y$}_{1}$ or $\displaystyle\mbox{\boldmath$\displaystyle Y$}_{2}$, depending on
$\displaystyle d=2$ or $\displaystyle d=3$, respectively.

######  Lemma 4.1.

The space $\displaystyle\mbox{\boldmath$\displaystyle V$}(T)$ with the degrees
of freedom given as in Table 4.1 is unisolvent.

###### Proof.

We let $\displaystyle\mbox{\boldmath$\displaystyle
v$}\in\mbox{\boldmath$\displaystyle V$}(T)$ be decomposed into two parts:

(4.9) $\mbox{\boldmath$\displaystyle v$}=\mbox{\boldmath$\displaystyle
v$}_{0}+{\mbox{\boldmath$\displaystyle
curl$}}(b_{T}\mbox{\boldmath$\displaystyle
q$}),\quad\mbox{\boldmath$\displaystyle
v$}_{0}\in({\rm{CR}}(T))^{d},\quad\mbox{\boldmath$\displaystyle
q$}\in\mbox{\boldmath$\displaystyle Y$}.$

In $\displaystyle 3D$, we have that

(4.10) ${\mbox{\boldmath$\displaystyle
curl$}}(b_{T}\mbox{\boldmath$\displaystyle
q$})\cdot\mbox{\boldmath$\displaystyle n$}={\mbox{\boldmath$\displaystyle
curl$}}_{f}(b_{T}\mbox{\boldmath$\displaystyle q$})_{f}=0,$

where $\displaystyle(b_{T}\mbox{\boldmath$\displaystyle q$})_{f}$ is the
tangential component of $\displaystyle b_{T}\mbox{\boldmath$\displaystyle q$}$
on $\displaystyle f$. In 2D, we have that
$\displaystyle{\mbox{\boldmath$\displaystyle
curl$}}(b_{T}q)\cdot\mbox{\boldmath$\displaystyle n$}=\partial_{t}(b_{T}q)=0$.
Thus $\displaystyle\mbox{\boldmath$\displaystyle
v$}\cdot\mbox{\boldmath$\displaystyle n$}=\mbox{\boldmath$\displaystyle
v$}^{0}\cdot\mbox{\boldmath$\displaystyle n$}\in P_{1}(f),\quad\forall
f\in\partial T$. Since for $\displaystyle({\rm CR}(T))^{d}$ is unisolvent with
the degrees of freedom, $\displaystyle\int_{f}\mbox{\boldmath$\displaystyle
v$}^{0}\cdot\mbox{\boldmath$\displaystyle n$}q\,ds$ for $\displaystyle q\in
P_{1}(f)$, we arrive at $\displaystyle\mbox{\boldmath$\displaystyle
v$}^{0}=\mbox{\boldmath$\displaystyle 0$}$. Therefore, it is sufficient to
show that $\displaystyle\mbox{\boldmath$\displaystyle
curl$}(b_{T}\mbox{\boldmath$\displaystyle q$})=\mbox{\boldmath$\displaystyle
0$}$ using the tangential component of the degrees of freedom. However, this
part of the proof is essentially presented in [13] and also in [11].
Therefore, we shall skip the proof and completes the proof. ∎

In passing to next subsection, we propose another enrichment for
$\displaystyle{\rm CR}^{2}$ finite element space that satisfies the Korn’s
inequality.

####  4.2.1. Enriched CR for 2D

Let $\displaystyle T\in\mathcal{T}_{h}$ be a triangle in 2D with
$\displaystyle\\{a_{i},a_{j},a_{k}\\}$ vertexes of $\displaystyle T$. We
denote the barycentric coordinates by
$\displaystyle\\{\lambda_{i},\lambda_{j},\lambda_{k}\\}$. Further,
$\displaystyle\mbox{\boldmath$\displaystyle
n$}_{ij},\mbox{\boldmath$\displaystyle n$}_{jk},\mbox{\boldmath$\displaystyle
n$}_{ki}$ are the normals to the edges
$\displaystyle\\{e_{ij}=[a_{i},a_{j}],e_{jk}=[a_{j},a_{k}],e_{ki}=[a_{k},a_{i}]\\}$.
We let $\displaystyle{\rm CR}(T)$ be the standard CR element on $\displaystyle
T$. We then define an edge bubble function $\displaystyle b$ by

(4.11)
$b=\lambda_{i}\lambda_{j}+\lambda_{j}\lambda_{k}+\lambda_{k}\lambda_{i}-\frac{1}{6},$

where the constant $\displaystyle 1/6$ comes from the fact that

(4.12)
$\frac{1}{6}=\frac{1}{|f_{\nu\mu}|}\int_{f_{\nu\mu}}\lambda_{\nu}\lambda_{\mu}\,ds,\quad\nu\mu=ij,jk,ki.$

We are now in a position to define an enriched CR element on $\displaystyle T$
by

(4.13) $\mbox{\boldmath$\displaystyle
E_{CR}$}(T)=({\rm{CR}}(T))^{2}+\mbox{\boldmath$\displaystyle V_{EC}$}(T),$

where $\displaystyle{\mbox{\boldmath$\displaystyle V_{EC}$}}(T)$ is the space
with the following functions as bases:

(4.14)
$\mbox{\boldmath$\displaystyle\psi$}_{ij}=b(\lambda_{i}-\lambda_{j})\mbox{\boldmath$\displaystyle
n$}_{ij},~{}~{}~{}\mbox{\boldmath$\displaystyle\psi$}_{jk}=b(\lambda_{j}-\lambda_{k})\mbox{\boldmath$\displaystyle
n$}_{jk},~{}~{}~{}\mbox{\boldmath$\displaystyle\psi$}_{ki}=b(\lambda_{k}-\lambda_{i})\mbox{\boldmath$\displaystyle
n$}_{ki}.$

The degrees of freedom are given as follows:

(4.15) $\int_{f}\mbox{\boldmath$\displaystyle
v$}\cdot\mbox{\boldmath$\displaystyle
t$}_{f}ds\quad\hbox{and}\quad\int_{f}\mbox{\boldmath$\displaystyle
v$}\cdot\mbox{\boldmath$\displaystyle n$}_{f}qds,\quad\forall q\in P_{1}(f).$

It is simple to prove that the aforementioned degrees of freedom can be
equivalently formulated as

(4.16) $\int_{f}\mbox{\boldmath$\displaystyle
v$}ds\quad\hbox{and}\quad\int_{f}\mbox{\boldmath$\displaystyle
v$}\cdot\mbox{\boldmath$\displaystyle n$}_{f}qds,\quad\forall q\in P_{1}(f).$

We shall now prove that the space $\displaystyle\mbox{\boldmath$\displaystyle
E_{CR}$}(T)$ with the degrees of freedom (4.16) is unisolvent.

######  Lemma 4.2.

The space $\displaystyle\mbox{\boldmath$\displaystyle E_{CR}$}(T)$ with the
degrees of freedom (4.16) is unisolvent.

###### Proof.

We let $\displaystyle\mbox{\boldmath$\displaystyle
v$}\in\mbox{\boldmath$\displaystyle E_{CR}$}(T)$ be decomposed into two parts:

(4.17) $\mbox{\boldmath$\displaystyle v$}=\mbox{\boldmath$\displaystyle
v$}_{0}+\mbox{\boldmath$\displaystyle
v$}_{1},\quad\mbox{\boldmath$\displaystyle
v$}_{0}\in({\rm{CR}}(T))^{2},\quad\mbox{\boldmath$\displaystyle
v$}_{1}\in\mbox{\boldmath$\displaystyle V_{EC}$}(T).$

We shall assume that $\displaystyle\int_{f}\mbox{\boldmath$\displaystyle
v$}\,ds=0$ and $\displaystyle\int_{f}\mbox{\boldmath$\displaystyle
v$}\cdot\mbox{\boldmath$\displaystyle n$}_{f}q\,ds=0,\,\,\forall q\in
P_{1}(f)$, for all $\displaystyle f\in\partial T$ and shall show that
$\displaystyle\mbox{\boldmath$\displaystyle v$}=\mbox{\boldmath$\displaystyle
0$}$. We first note that for any $\displaystyle\mbox{\boldmath$\displaystyle
v$}_{1}\in\mbox{\boldmath$\displaystyle V_{EC}$}(T)$, we have
$\displaystyle\int_{f}\mbox{\boldmath$\displaystyle v$}_{1}\,ds=0$, for any
$\displaystyle f\in\partial T$, since for all $\displaystyle f\in\partial T$,
it holds that $\displaystyle\int_{f}bq\,ds=0,\forall q\in P_{1}(f)$. This
measn that $\displaystyle\int_{f}\mbox{\boldmath$\displaystyle v$}_{0}\,ds=0$
for any $\displaystyle f\in\partial T$ and so,
$\displaystyle\mbox{\boldmath$\displaystyle v$}_{0}=0$. Therefore, it is
sufficient to show that $\displaystyle\mbox{\boldmath$\displaystyle
v$}_{1}=\mbox{\boldmath$\displaystyle 0$}$. By the definition of
$\displaystyle\mbox{\boldmath$\displaystyle V_{CR}$}(T)$, we can write
$\displaystyle\mbox{\boldmath$\displaystyle v$}_{1}$ as follows for some
constants $\displaystyle c_{1},c_{2},c_{3}$:

(4.18) $\mbox{\boldmath$\displaystyle
v$}_{1}=c_{1}\mbox{\boldmath$\displaystyle\psi$}_{ij}+c_{2}\mbox{\boldmath$\displaystyle\psi$}_{jk}+c_{3}\mbox{\boldmath$\displaystyle\psi$}_{ki}.$

The fact that $\displaystyle\int_{f}\mbox{\boldmath$\displaystyle
v$}_{1}\cdot\mbox{\boldmath$\displaystyle n$}\,qds=0,\,\,\forall q\in
P_{1}(f)$ leads to the following linear system:

(4.19)
$\displaystyle\int_{f_{ij}}(c_{1}\mbox{\boldmath$\displaystyle\psi$}_{ij}+c_{2}\mbox{\boldmath$\displaystyle\psi$}_{jk}+c_{3}\mbox{\boldmath$\displaystyle\psi$}_{ki})\cdot\mbox{\boldmath$\displaystyle
n$}_{ij}\lambda_{i}ds&=0,\\\
\int_{f_{jk}}(c_{1}\mbox{\boldmath$\displaystyle\psi$}_{ij}+c_{2}\mbox{\boldmath$\displaystyle\psi$}_{jk}+c_{3}\mbox{\boldmath$\displaystyle\psi$}_{ki})\cdot\mbox{\boldmath$\displaystyle
n$}_{jk}\lambda_{j}ds=0,\\\
\int_{f_{ki}}(c_{1}\mbox{\boldmath$\displaystyle\psi$}_{ij}+c_{2}\mbox{\boldmath$\displaystyle\psi$}_{jk}+c_{3}\mbox{\boldmath$\displaystyle\psi$}_{ki})\cdot\mbox{\boldmath$\displaystyle
n$}_{ki}\lambda_{k}ds=0.$

Or equivalently, we have that

(4.20) $\left(\begin{array}[]{ccc}-2&\mbox{\boldmath$\displaystyle
n$}_{jk}\cdot\mbox{\boldmath$\displaystyle
n$}_{ij}&\mbox{\boldmath$\displaystyle
n$}_{ki}\cdot\mbox{\boldmath$\displaystyle n$}_{ij}\\\
\mbox{\boldmath$\displaystyle n$}_{jk}\cdot\mbox{\boldmath$\displaystyle
n$}_{ij}&-2&\mbox{\boldmath$\displaystyle
n$}_{ki}\cdot\mbox{\boldmath$\displaystyle n$}_{jk}\\\
\mbox{\boldmath$\displaystyle n$}_{ij}\cdot\mbox{\boldmath$\displaystyle
n$}_{ki}&\mbox{\boldmath$\displaystyle
n$}_{jk}\cdot\mbox{\boldmath$\displaystyle
n$}_{ki}&-2\end{array}\right)\left(\begin{array}[]{c}c_{1}\\\ c_{2}\\\
c_{3}\end{array}\right)=\left(\begin{array}[]{c}0\\\ 0\\\
0\end{array}\right).$

Since $\displaystyle|\mbox{\boldmath$\displaystyle
n$}_{ij}\cdot\mbox{\boldmath$\displaystyle
n$}_{ki}|+|\mbox{\boldmath$\displaystyle
n$}_{jk}\cdot\mbox{\boldmath$\displaystyle n$}_{ki}|<2$, the coefficient
matrix in the equation (4.20) is diagonally dominant and so it is invertible,
which implies $\displaystyle\mbox{\boldmath$\displaystyle
v$}_{1}=\mbox{\boldmath$\displaystyle 0$}$. This completes the proof. ∎

###  4.3. A new finite element space that satisfies the Korn’s inequality

In this subsection, we shall investigate the finite elements introduced in
[6]. We observe that some of their finite element for the lowest order case do
not satisfy the Korn’s inequality and provide modification, which satisfies
the Korn’s inequality. For a given $\displaystyle T\in\mathcal{T}_{h}$, we let

(4.21) $Q_{f}^{0}(T)=P_{0}(T)~{}\hbox{for}~{}d=2.$

and

(4.22) $\mbox{\boldmath$\displaystyle
Q$}_{f}^{0}(T)=(P_{0}(T))^{3}\times\mbox{\boldmath$\displaystyle
n$}_{f}~{}\hbox{for}~{}d=3,$

and define $\displaystyle Q^{0}(T)=\sum_{f\in\partial T}b_{f}{Q}_{f}^{0}(T)$
for $\displaystyle d=2$ and $\displaystyle\mbox{\boldmath$\displaystyle
Q$}^{0}(T)=\sum_{f\in\partial T}b_{f}\mbox{\boldmath$\displaystyle
Q$}_{f}^{0}(T)$ for $\displaystyle d=3$.

We now consider the lowest order case of finite elements introduced in [6]. We
recall that the local space was given as follows: for $\displaystyle
T\in\mathcal{T}_{h}$,

(4.23) $\mbox{\boldmath$\displaystyle V$}_{1}(T)=\mbox{\boldmath$\displaystyle
V$}_{1,T}^{0}+\mbox{\boldmath$\displaystyle
curl$}(b_{T}\mbox{\boldmath$\displaystyle Y$}),$

where $\displaystyle\mbox{\boldmath$\displaystyle V$}^{0}_{1,T}$ is the well-
known $\displaystyle H({\rm div})$-conforming finite element spaces, either
$\displaystyle{\rm{BDM}}_{1}(T)$ or $\displaystyle{\rm{RT}}_{1}(T)$. The space
$\displaystyle Y$ is given as following:

(4.24) $\mbox{\boldmath$\displaystyle
Y$}=\left\\{\begin{array}[]{ll}\mbox{\boldmath$\displaystyle
Y$}_{4}=Q^{0}(T)&\mbox{ for }d=2;\\\ \mbox{\boldmath$\displaystyle
Y$}_{5}=\mbox{\boldmath$\displaystyle Q$}^{0}(T)&\mbox{ for
}d=3.\end{array}\right.$

From our sharp Korn’s inequality, we can easily check if the above four finite
elements given in (4.23) from different choices of
$\displaystyle\mbox{\boldmath$\displaystyle V$}_{1,T}^{0}$ and $\displaystyle
Y$ satisfy the Korn’s inequality immediately. This is presented in the
following Table 4.2:

The six modified $\displaystyle H({\rm div})$ elements
---
Elements |  $\displaystyle\mbox{\boldmath$\displaystyle V$}_{1,T}^{0}$ |  $\displaystyle Y$ |  DOF |  Korn
$\displaystyle 1^{\rm st}$ FEM (2D) |  $\displaystyle{\rm RT}_{1}(T)$ |  $\displaystyle\mbox{\boldmath$\displaystyle Y$}_{4}$ |  $\displaystyle{\small{\begin{array}[]{l}\int_{f}\mbox{\boldmath$\displaystyle v$}\cdot\mbox{\boldmath$\displaystyle n$}q\,ds,\quad\forall q\in P_{1}(f),\\\ \int_{T}\mbox{\boldmath$\displaystyle v$}\cdot\mbox{\boldmath$\displaystyle q$}\,dx,\quad\mbox{\boldmath$\displaystyle q$}\in(P_{0}(T))^{2}\\\ \int_{f}\mbox{\boldmath$\displaystyle v$}\cdot\mbox{\boldmath$\displaystyle t$}\,ds\end{array}}}$ |  Yes
$\displaystyle 2^{\rm nd}$ FEM (2D) |  $\displaystyle{\rm BDM}_{1}(T)$ |  $\displaystyle\mbox{\boldmath$\displaystyle Y$}_{4}$ |  $\displaystyle{\small{\begin{array}[]{l}\int_{f}\mbox{\boldmath$\displaystyle v$}\cdot\mbox{\boldmath$\displaystyle n$}q\,ds,\quad\forall q\in P_{1}(f),\\\ \int_{f}\mbox{\boldmath$\displaystyle v$}\cdot\mbox{\boldmath$\displaystyle t$}\,ds\end{array}}}$ |  Yes
$\displaystyle 3^{\rm rd}$ FEM (3D) |  $\displaystyle{\rm RT}_{1}(T)$ |  $\displaystyle\mbox{\boldmath$\displaystyle Y$}_{5}$ |  $\displaystyle{\small{\begin{array}[]{l}\int_{f}\mbox{\boldmath$\displaystyle v$}\cdot\mbox{\boldmath$\displaystyle n$}q\,ds,\quad\forall q\in P_{1}(f),\\\ \int_{T}\mbox{\boldmath$\displaystyle v$}\cdot\mbox{\boldmath$\displaystyle q$}\,dx,\quad\mbox{\boldmath$\displaystyle q$}\in(P_{0}(T))^{3}\\\ \int_{f}(\mbox{\boldmath$\displaystyle v$}\times\mbox{\boldmath$\displaystyle n$})\cdot\mbox{\boldmath$\displaystyle r$}\,ds,\quad\forall\mbox{\boldmath$\displaystyle r$}\in(P_{0}(f))^{2}\end{array}}}$ |  No
$\displaystyle 4^{\rm th}$ FEM (3D) |  $\displaystyle{\rm BDM}_{1}(T)$ |  $\displaystyle\mbox{\boldmath$\displaystyle Y$}_{5}$ |  $\displaystyle{\small{\begin{array}[]{l}\int_{f}\mbox{\boldmath$\displaystyle v$}\cdot\mbox{\boldmath$\displaystyle n$}q\,ds,\quad\forall q\in P_{1}(f),\\\ \int_{f}(\mbox{\boldmath$\displaystyle v$}\times\mbox{\boldmath$\displaystyle n$})\cdot\mbox{\boldmath$\displaystyle r$}\,ds,\quad\mbox{\boldmath$\displaystyle r$}\in(P_{0}(f))^{2}\end{array}}}$ |  No
Table 4.2. Lowest FEMs introduced in [6]

We note that particularly, for the local space

(4.25) $\mbox{\boldmath$\displaystyle V$}_{1}(T)=\mbox{\boldmath$\displaystyle
V$}_{1,T}^{0}+\mbox{\boldmath$\displaystyle
curl$}(b_{T}\mbox{\boldmath$\displaystyle
Y$}_{5})=\mbox{\boldmath$\displaystyle
V$}_{1,T}^{0}+\mbox{\boldmath$\displaystyle
curl$}(b_{T}\mbox{\boldmath$\displaystyle Q$}^{0}(T)),$

the degrees of freedom are not sufficient since the constant projected
tangential component on the face being zero is not sufficient for the Korn’s
inequality. Therefore, we investigate how to remedy this situation. Namely, we
shall modify the local space $\displaystyle\mbox{\boldmath$\displaystyle
V$}_{1}(T)$ and the definition of degrees of freedom so that the Korn’s
inequality can be satisfied in the remainder of this subsection.

We define the local space for the nonconforming finite element that satisfies
the Korn’s inequality :

(4.26) $\mbox{\boldmath$\displaystyle V$}^{1}(T)=\mbox{\boldmath$\displaystyle
V$}_{1,T}^{0}+{\mbox{\boldmath$\displaystyle
curl$}}\left(b_{T}\mbox{\boldmath$\displaystyle Q$}^{*}(T)\right),$

where

$\displaystyle\displaystyle\mbox{\boldmath$\displaystyle Q$}^{*}(T)$
$\displaystyle\displaystyle=$ $\displaystyle\displaystyle\sum_{f\in\partial
T}b_{f}\mbox{\boldmath$\displaystyle Q$}_{f}^{*}(T),$

with

$\displaystyle\displaystyle\mbox{\boldmath$\displaystyle
Q$}_{f}^{*}(T)=\left\\{\mbox{\boldmath$\displaystyle
q$}\times\mbox{\boldmath$\displaystyle n$}_{f}:\mbox{\boldmath$\displaystyle
q$}\in\mbox{\boldmath$\displaystyle
RM$}(T),\int_{T}(\mbox{\boldmath$\displaystyle
q$}\times{\mbox{\boldmath$\displaystyle
n$}}_{f})\cdot(\mbox{\boldmath$\displaystyle
w$}\times\mbox{\boldmath$\displaystyle
n$}_{f})b_{T}b_{f}\,dx=0,\,\,\mbox{\boldmath$\displaystyle
w$}\in(P_{0}(T))^{3}\right\\}.$

It is easy to see that the dimension of
$\displaystyle\mbox{\boldmath$\displaystyle Q$}_{f}^{*}(T)$ is three.
Therefore, the dimension of the space
$\displaystyle\mbox{\boldmath$\displaystyle Q$}^{*}(T)$ is 12. Note that the
space $\displaystyle\mbox{\boldmath$\displaystyle V$}_{1,T}^{0}$ can be either
$\displaystyle BDM_{1}$ or $\displaystyle RT_{1}$. But, we shall only consider
$\displaystyle BDM_{1}$ for simplicity. Under this choice, it is well-known
that a function $\displaystyle\mbox{\boldmath$\displaystyle
v$}\in\mbox{\boldmath$\displaystyle V$}_{1,T}^{0}$ is uniquely determined by
the following degrees of freedom: for all faces $\displaystyle f\in\partial
T$,

(4.27) $\langle\mbox{\boldmath$\displaystyle
v$}\cdot\mbox{\boldmath$\displaystyle
n$}_{f},\mbox{\boldmath$\displaystyle\mu$}\rangle_{f},\quad\forall\mbox{\boldmath$\displaystyle\mu$}\in({P}_{1}(f))^{3}.$

With this in mind, we can provide the following degrees of freedom that define
a function $\displaystyle\mbox{\boldmath$\displaystyle
v$}\in\mbox{\boldmath$\displaystyle V$}^{1}(T)$ uniquely: for all
$\displaystyle f\in\partial T$,

(4.28) $\displaystyle\displaystyle\langle\mbox{\boldmath$\displaystyle
v$}\cdot\mbox{\boldmath$\displaystyle
n$}_{f},\mbox{\boldmath$\displaystyle\mu$}\rangle_{f},$
$\displaystyle\displaystyle\quad\forall\mbox{\boldmath$\displaystyle\mu$}\in({P}_{1}(f))^{3};$
(4.29) $\displaystyle\displaystyle\langle\mbox{\boldmath$\displaystyle
v$}\times\mbox{\boldmath$\displaystyle
n$}_{f},\mbox{\boldmath$\displaystyle\kappa$}\rangle_{f},$
$\displaystyle\displaystyle\quad\forall\mbox{\boldmath$\displaystyle\kappa$}\in{\rm{RT}}_{0}(f).$

We can show that the following holds true:

######  Lemma 4.3.

For any $\displaystyle T\in\mathcal{T}_{h}$, we have

(4.30) ${\rm dim}\,{\mbox{\boldmath$\displaystyle
curl$}}\left(b_{T}\mbox{\boldmath$\displaystyle Q$}^{*}(T)\right)=4\,{\rm
dim}\,\mbox{\boldmath$\displaystyle Q$}_{f}^{*}(T).$

###### Proof.

The proof can be done similarly to the argument of Lemma 3.2 in [6].
Therefore, we complete the proof. ∎

######  Theorem 4.1.

We have the following relation that

(4.31) $\mbox{\boldmath$\displaystyle V$}^{1}(T)=\mbox{\boldmath$\displaystyle
V$}_{1,T}^{0}\oplus{\mbox{\boldmath$\displaystyle
curl$}}\left(b_{T}\mbox{\boldmath$\displaystyle Q$}^{*}(T)\right)$

and

(4.32) ${\rm dim}\mbox{\boldmath$\displaystyle V$}^{1}(T)={\rm
dim}\mbox{\boldmath$\displaystyle V$}_{1,T}^{0}+{\rm
dim}{\mbox{\boldmath$\displaystyle
curl$}}\left(b_{T}\mbox{\boldmath$\displaystyle Q$}^{*}(T)\right).$

Furthermore, any function $\displaystyle\mbox{\boldmath$\displaystyle
v$}\in\mbox{\boldmath$\displaystyle V$}^{1}(T)$ is uniquely determined by the
degrees of freedom (4.28) and (4.29).

###### Proof.

The proof can be done similarly to the argument of Theorem 3.3 in [6].
Therefore, we complete the proof. ∎

##  5\. Conclusion

We have proven a Korn’s inequality for the piecewise $\displaystyle H^{1}$
space. Our characterization of the Korn’s inequality considers to be the
rotated version of what is characterized in [8]. We further have shown some
sharpness of the Korn’s inequality. This result has been applied to a number
of finite elements to check that they satisfy the Korn’s inequality or not.
Also the minimal jump terms that appear in the Korn’s inequality are explicit
enough so that they can be used for construction of finite element spaces that
satisfy the Korn’s inequality.

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