A complete variational form#

The computational domain#

Now that we have covered how to define finite elements, we can move on to the computational domain \(\Omega\). We can use either tetrahedral and hexahedral elements to subdivide the continuous domain into a discrete domain. We define a coordinate element, which is used to map functions defined on the reference element to the physical domain. In this example, we will use straight edged hexahedral elements, and therefore the coordinate element can be defined as

import ufl
c_el = ufl.VectorElement("Lagrange", ufl.triangle, 1)

Next, we create an abstract definition of the computational domain. For now, we don’t know if we are solving the Poisson equation on a cube, a sphere or on the wing of an airplane.

domain = ufl.Mesh(c_el)

The function space#

As oposed some commerical software, we do not rely on iso-parameteric elements in FEniCS. We can use any supported finite element space to describe our unknown u.

el = ufl.FiniteElement("Discontinuous Lagrange", domain.ufl_cell(), 2)
V = ufl.FunctionSpace(domain, el)

For the coefficients f and g, we choose

F = ufl.FunctionSpace(domain, ufl.FiniteElement(
    "Discontinuous Lagrange", domain.ufl_cell(), 0))
f = ufl.Coefficient(F)
G = ufl.FunctionSpace(domain, ufl.FiniteElement(
    "Lagrange", domain.ufl_cell(), 1))
g = ufl.Coefficient(G)

The variational form#

We obtain them in the standard way, by multiplying by a test function and integrating over the .domain

u = ufl.TrialFunction(V)
v = ufl.TestFunction(V)
a = ufl.inner(u, v) * ufl.dx

Note that the bi-linear form a is defined with trial and test functions.

L = (f / g) * v * ufl.dx
forms = [a, L]

So far, so good? As opposed to most demos/tutorials on FEniCSx, note that we have not imported dolfinx or made a reference to the actual computational domain we want to solve the problem on or what f or g is, except for the choice of function spaces

Further analysis of the variational form#

To do so, we would define each of the functions as the linear combinations of the basis functions \(u=\sum_{i=0}^{\mathcal{N}}u_i\phi_i(x)\qquad v=\sum_{i=0}^{\mathcal{N}}v_i\phi_i(x)\qquad f=\sum_{k=0}^{\mathcal{M}}f_k\psi_k(x)\qquad g=\sum_{l=0}^{\mathcal{T}}g_l\varphi_l(x)\)

We next use the map \(M_K:K_{ref}\mapsto K\)

(4)#\[\begin{align} \int_\Omega u v~\mathrm{d}x&= \sum_{K\in\mathcal{K}}\int_K u(x) v(x)~\mathrm{d}x\\ &= \sum_{K\in\mathcal{K}}\int_{M_K(K_{ref})} u(x)v(x)~\mathrm{d}x\\ &= \sum_{K\in\mathcal{K}}\int_{K_{ref}}u(M_K(\bar x))v(M_K(\bar x))\vert \mathrm{det} J_K(\bar x)\vert~\mathrm{d}\bar x \end{align}\]

where \(K\) is each element in the physical space, \(J_K\) the Jacobian of the mapping. Next, we can insert the expansion of \(u\) into the formulation and identify the matrix system \(Au=b\), where

(5)#\[\begin{align} A_{j, i} &= \int_{K_{ref}} \phi_i(M_K(\bar x))\phi_j(M_K(\bar x))\vert \mathrm{det} J_K(\bar x)\vert~\mathrm{d}\bar x\\ b_j &= \int_{K_{ref}} \frac{\Big(\sum_{k=0}^{\mathcal{M}}f_k\psi_i(M_K(\bar x))\Big)} {\Big(\sum_{l=0}^{\mathcal{T}}g_k\varphi_i(M_K(\bar x))\Big)}\phi_j(M_K(\bar x))\vert \mathrm{det} J_K(\bar x)\vert~\mathrm{d}\bar x \end{align}\]

Next, one can choose an appropriate quadrature rule with points and weights, include the correct mapping/restrictions of degrees of freedom for each cell. All of this becomes quite tedious and error prone work, and has to be repeated for every variational form!

pulled_back_L = ufl.algorithms.compute_form_data(L,
IntegralData over domain(cell, otherwise), with integrals:
{ weight * |(J[0, 0] * J[1, 1] + -1 * J[0, 1] * J[1, 0])| * (reference_value(w_0)) / (reference_value(w_1)) * (reference_value(v_0)) } * dx(<Mesh #0>[otherwise], {'estimated_polynomial_degree': 3})
and metadata: