PhD Student at the University of Newcastle, Australia

When I began writing this post on 26th November, I was on a train from Wollongong to Newcastle. Being a long ride back home, I decided to work on this blog post about VEM. But I ended up finishing and pushing this post on 5th December.

This one is going to be a bit long and slightly technical, but the results are quite interesting and the questions that arise at the end of this post gives us new insight into the construction of the virtual element method. I will be covering four major topics here which will leave us with three mind blowing moments.

- Overview of the older mixed method.
- Mixed Method based on bi-orthogonal systems.
- Why is the stability term required here but not for gradient recovery?
- What is so interesting about second order/quadratic VEM?

In the last post, I proposed a modified gradient recovery approach using a stability term. At the end of the post, I mentioned about how this stability term is important in some cases. When I was talking about gradient recovery techniques to my one of my supervisors, he suggested me to look at another application of bi-orthogonality for constructing mixed methods using the Ciarlet-Raviart formulation.

Let us consider the fourth-order biharmonic equation

\[\begin{equation} \Delta^2 u = f, \quad \text{in}\quad \Omega \end{equation}\]with \(\Delta(\cdot)\) denoting the standard Laplacian operator. The idea in a Ciarlet-Raviart mixed formulation is to construct a mixed method by considering an auxiliary variable \(v = -\Delta u\) and thus transform the problem into a system of two second order problems

\[\begin{eqnarray*} -\Delta v &=& f,\\ -\Delta u &=& v. \end{eqnarray*}\]As for the boundary condition let us assume that

\[u = \frac{\partial u}{\partial n} = 0 \quad \text{on}\;\; \partial\Omega.\]In usual mixed formulations, we deal with weak formulations that are constructed using the above PDEs as to find \((u,v) \in H^1_0(\Omega) \times H^1(\Omega)\) such that

\[\begin{eqnarray*} (\nabla v, \nabla \varphi) &=& (f,\varphi),\\ (\nabla u, \nabla \chi) &=& (v,\chi), \end{eqnarray*}\]for all \((\varphi,\chi) \in H^1_0(\Omega)\times
H^1_0(\Omega)\). Note that the derivative boundary condition occur
naturally in the formulation. Let \(S_h\) denote the standard
**linear** virtual element space and \(S_h^0\) denote the same virtual
element space with zero value on the boundary. Due to the virtual element
discretization, we arrive at the following discrete equivalent.

Find \((u_h,v_h) \in S_h^0\times S_h\) such that

\[\begin{eqnarray} a_h(v_h,\varphi) &=& f_h(\varphi),\\ a_h(u_h,\chi) &=& m_h(v_h,\chi). \end{eqnarray}\]\((\varphi,\chi) \in S_h \times S_h\). Then we have the system of equations

\[\begin{bmatrix} 0 &K\\ -K &M \end{bmatrix} \begin{Bmatrix} u\\ v \end{Bmatrix} = \begin{Bmatrix} f\\ 0 \end{Bmatrix}\]which after applying the Dirichlet boundary conditions for \(u\), can be solved to obtain the solution. Although inverting the sparse left hand side matrix is relatively simpler than implementing a \(C^1\) finite element method, we could do better if we implement bi-orthogonal systems.

The original paper [1] describes the procedure based on energy functionals and saddle point theory. I will present here, an easier and intuitive approach. We start by splitting the fourth order problem into two second order problems as

\[\begin{eqnarray*} \Delta u &=& \phi,\\ \Delta \phi &=& f. \end{eqnarray*}\]We further split the second equation into two equations by setting \(\phi = -p\) and writing

\[\begin{eqnarray*} -\Delta p = f. \end{eqnarray*}\]Define the spaces

\[\begin{eqnarray*} V &=& H^1_0(\Omega) \times L^2(\Omega),\\ Q &=& H^1(\Omega) \end{eqnarray*}\]and the virtual element subspaces \(V_h \subset V\) and \(Q_h \subset Q\), where

\[V_h = S_h^0 \times M_h, \quad Q_h = S_h.\]The discrete virtual element setting reads: Find \(((u_h,\phi_h),p_h) \in V_h \times Q_h\) such that

\[\begin{eqnarray*} a_h(p_h,v) &=& f_h(v),\\ a_h(u_h,q) &=& m_h(\phi_h,q),\\ -m_h(\phi_h,\psi) &=& m_h(p_h,\psi), \end{eqnarray*}\]where \(((v,\psi),q) \in V_h \times Q_h\). Written in matrix form, we obtain the system of equations

\[\begin{eqnarray*} \begin{bmatrix} K & D & 0\\ 0 & -M & D^T\\ 0 & 0 & K^T \end{bmatrix} \begin{Bmatrix} u\\ \phi\\ p \end{Bmatrix} = \begin{Bmatrix} 0\\ 0\\ f \end{Bmatrix}. \end{eqnarray*}\]Further, we can statically condense out the variable \(u\) as

\[\begin{eqnarray*} \left(K^T (D^T)^{-1} \,M\,D^{-1} K\right)\,u &=& f,\\ \end{eqnarray*}\]and after solving for \(u\), we can find

\[\begin{eqnarray*} \phi &=& -\left(D^{-1}\,K\right)\,u,\\ p &=& \left( (D^T)^{-1} \,M\, D^{-1} K \right)\,u. \end{eqnarray*}\]Now, what is this space \(M_h\)?? Choosing \(M_h\) as the space of functions bi-orthogonal to the virtual element space \(S_h\), the matrix \(D\) becomes diagonal! And inverting a diagonal matrix is the easiest and the sweetest thing to do computationally (Mind blown)… To put it formally, define

\[M_h = \text{span}\{\mu_1,\mu_2,\cdots,\mu_N\},\]with

\[\int_K \varphi_i\,\mu_j = c_j\delta_{ij},\]for all the virtual elements \(K\). The coefficients \(c_j\) can be chosen proportional to the area of the element and is given by the mass-lumping. The interesting part is yet to come…

On closer inspection of the weak form, we find that the discrete inner product

\[m_h(\mu_i,\mu_j) = \sum_K m_h^K(\mu_i,\mu_j)\]needs to satisfy good stability properties in the virtual element setting. To achieve this, the local bilinear form needs to satisfy the following lower and upper bound

\[C_1\|v_h\|_{0,K}^2 \le m_h^K(v_h,v_h) \le C_2 \|v_h\|_{0,K}^2\]The local \(L^2\) inner product is defined as

\[m_h^K(\mu_i,\mu_j) = \int_K \Pi^0_K\mu_i\, \Pi^0_K\mu_j + T_K\]The \(L^2-\) projection \(\Pi^0_K\mu_j\) can be fully computed using
the bi-orthogonal relation (It is a really cool exercise to work it out). The
stability term \(T_K\) needs to be *only approximated* and can be chosen to be
equal to the one described in the
previous post.

Visually, when the stability term is ignored, we have an unstable solution to the stream function \(p\),

However, when we add the stability term we obtain a much cleaner solution. Since we do not require the lower bound in the gradient recovery case, we can ignore the stability term there (Mind blown … Again).

What is so **peculiar** about the quadratic virtual elements now?
Before answering this question, let us go back to the definition of
the bi-orthogonal bases. Define

with

\[\int_K \varphi_i\,\mu_j = c_j\delta_{ij},\]for all the virtual elements \(K\). The coefficients \(c_j\) can be
chosen proportional to the area of the element and is given by the
**mass-lumping**.

Turns out second order VEM does not get along very
well with **mass-lumping**. Why? Applying the mass lumping technique
the elements of the diagonal
matrix are given as

However, the integral of the bases function is taken as the first moment degree of freedom in the quadratic virtual element setting. This means the local lumped mass matrix, for example, for a any polygon takes the form

\[\begin{bmatrix} 0 & \cdots & 0\\ \vdots & \ddots & \vdots \\ 0 & \cdots & |K| \\ \end{bmatrix}\]which after assembly **cannot be inverted**
(Mind.. blown.. Again..)

What now? No Idea. But the remedy will be a subject of a future blog post. So stay tuned ;)

[1] Lamichhane, Bishnu P. “A mixed finite element method for the biharmonic problem using biorthogonal or quasi-biorthogonal systems.” Journal of Scientific Computing 46.3 (2011): 379-396.

[2] Kalyanaraman, B., Lamichhane, B., & Meylan, M. (2018). A gradient recovery method based on an oblique projection for the virtual element method. ANZIAM Journal, 60, 201-214.

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