Requirements for Korn's inequality on piecewise $H_1$ vector field

Question

I am looking at the Korn's inequality on $H^1$ vector fields, as described in this paper by Brenner. In particular, I am looking at how the seminorms defined in examples 2.3 - 2.5 satisfy the conditions of Lemma 2.2.

To give a big picture, let $\Omega$ be a bounded connected open polyhedral domain in $\mathbb{R}^d$ and $\mathcal{T}$ is a triangulation of $\Omega$ by simplexes (triangles or tetrahedrals). Let $\mathbf{RM}(\Omega)$ be the space of infinitesimal rigid motions on $\Omega$ defined by: $$ \mathbf{RM}(\Omega) = \{ \mathbf{a} + \mathbf{\eta} \mathbf{x}: \mathbf{a} \in \mathbf{R}^d \text{ and } \mathbf{\eta} \in \mathfrak{so}(d) \} $$ where $\mathbf{x}$ is the position vector function on $\Omega$ and $\mathfrak{so}(d)$ is the Lie algebra of anti-symmetric $d \times d$ matrices.

Define the space $[H^1(\Omega, \mathcal{T})]^d$ as: $$ [H^1(\Omega, \mathcal{T})]^d = \{ \mathbf{v} \in [L_2(\Omega)]^d: \mathbf{v}_D = \mathbf{v}|_D \in [H^1(D)]^d \quad \forall D \in \mathcal{T} \}. $$

We need to look at the seminorm $\Phi : [H^1(\Omega, \mathcal{T})]^d \rightarrow \mathbb{R}$ which has the following properties:

• $ | \Phi (\mathbf{w}) | \leq C || \mathbf{w} ||_{H^1(\Omega)}, \quad \forall \mathbf{w} \in [H^1(\Omega)]^d$,
• $\Phi(\mathbf{m}) = 0$ and $\mathbf{m} \in \mathbf{RM}(\Omega) \Leftrightarrow \mathbf{m} = \text{ a constant vector}$.

I want to try to understand how the following choices of $\Phi(\cdot)$ satisfy the conditions above, i.e.:

• $ \Phi_1(\mathbf{v}) = || Q \mathbf{v} ||_{L_2(\Omega)}, \quad \forall \mathbf{v} \in [H^1(\Omega, \mathcal{T}]^d$, where $Q$ is the orthogonal projection from $[L_2(\Omega)]^d$ onto the orthogonal complement of the constant vector fields.
• $ \Phi_2(\mathbf{v}) = \sup_{\mathbf{m} \in \mathbf{RM}(\Omega); || \mathbf{m} ||_{L^2(\Gamma)} = 1; \int_\Gamma \mathbf{m} ds = 0} \int_\Gamma \mathbf{v} \cdot \mathbf{m} ds, \quad \forall \mathbf{v} \in [H^1(\Omega, \mathcal{T})]^d$.
• $ \Phi_3(\mathbf{v}) = |\sum_{T \in \mathcal{T}} \int_T \nabla \times \mathbf{v} dx|, \quad \forall \mathbf{v} \in [H^1(\Omega, \mathcal{T})]^d$.

Coming from an engineering background, I have a hard time visualising what the seminorm of $\Phi_i$ is representing. Any suggestions or pointers?

Answer 1

I can give you a partial answer, because I couldn't check all the details, especially for $\Phi_2$.

Let's start with $\Phi_1$, First note $Q$ is an orthogonal projector on $L^2(\Omega)$ then $Q^2v=Qv$ then $ $ \begin{align} |\Phi_1(v)| = \|Qv\|_{L^2(\Omega)}\leq\|v\|_{L^2(\Omega)} \leq \|v\|_{H^1(\Omega,\mathcal{T} )} \quad \forall v \in H^1(\Omega) \end{align} if $\Phi_1(\mathbf{m})=Q\mathbf{m}=0$ and $\mathbf{m} \in \mathbf{RM}(\Omega)$ (This asumption is unnecessary here) , then must be $\mathbf{m}$ is orthogonal to orthogonal complement of constant functions. Hence, $\mathbf{m}$ must be a constant. The converse if $\mathbf{m}$ is constant so $\Phi_1(\mathbf{m})=Q\mathbf{m}=0$, because is not in the orthogonal complement of constant functions. In particular constant vectors are rigid movements.

Now we proceed for $\Phi_2$, by Cauchy Schwarz inequality for any $\mathbf{m} \in \mathbf{RM}(\Omega)$ with $\|\mathbf{m}\|_{L^2(\Gamma)}=1$ and $\int_{\Gamma}\mathbf{m}~ds = 0$ (We don't use this yet) \begin{align} \int_\Gamma \mathbf{v} \cdot \mathbf{m}~ds &\leq \|\mathbf{v}\|_{L^2(\Gamma)}\|\mathbf{m}\|_{L^2(\Gamma)}\\ & \leq \|\mathbf{v}\|_{L^2(\partial \Omega)} = \|\gamma_0(\mathbf{v})\|_{L^2(\partial \Omega)} \leq C\|\mathbf{v}\|_{H^1(\Omega)} \quad \forall \mathbf{v} \in H^1(\Omega) \end{align} Where $\gamma_0:H^1(\Omega) \to L^2(\partial \Omega)$ is the trace operator i.e $\gamma_0(\mathbf{v})=\mathbf{v}|_{\partial \Omega}$. This operator is continuous, but is quite complicated to show that See Trace theorem Ch 1.6 on [1]. So taking the supreme we get $|\Phi_2(v)|_{L^2(\Omega)}\leq C\|\mathbf{v}\|_{H^1(\Omega)}$.

Now we proceed for $\Phi_3$, Observe in two dimensions (a similar trick can be use in three dimesions) we have \begin{align} |\nabla \times \mathbf{u} |^2 &= \left| \frac{\partial u_2}{\partial x_1 }-\frac{\partial u_1}{\partial x_2 } \right |^2 \\ &\leq 2\left |\frac{\partial u_2}{\partial x_1 } \right |^2 + 2\left |\frac{\partial u_1}{\partial x_2 } \right |^2 \\ &\leq 2\left |\frac{\partial u_2}{\partial x_1 } \right |^2 + 2\left |\frac{\partial u_1}{\partial x_2 } \right |^2 + 2\left |\frac{\partial u_1}{\partial x_1 } \right |^2 + 2\left |\frac{\partial u_2}{\partial x_2 } \right |^2 = 2|Du|_2^2, \end{align} Where $|Du|_2$ is the frobenius norm and $D$ the jacobian matrix. Now because $\mathbf{v} \in H^1(\Omega)$ we bound \begin{align} \Phi_3(\mathbf{v}) &=\left | \sum_{T \in \mathcal{T} } \int_{T} \nabla \times \mathbf{v} \right | =\left | \int_{\Omega} \nabla \times \mathbf{v} \right | \\ &\leq \|\Omega\|^{1/2} \left ( \int_{\Omega}|\nabla \times \mathbf{v} |^2 \right)^{1/2} \\ &\leq \|\Omega\|^{1/2} \sqrt{2} \left ( \int_{\Omega}|D\mathbf{v} |^2 \right)^{1/2} \\ & \leq \|\Omega\|^{1/2} \sqrt{2} \|\mathbf{v}\|_{H^1(\Omega)}. \end{align}

I left to check what happens when $\mathbf{m}$ is a rigid motion. I don't know how to deal with that. But I hope this will be useful.

References: [1] BRENNER, Susanne; SCOTT, Ridgway. The mathematical theory of finite element methods. Springer Science & Business Media, 2007.