Biharmonic equation

This demo is implemented in a single Python file, demo_biharmonic.py, which contains both the variational forms and the solver. It illustrates how to:

  • Solve a linear partial differential equation

  • Use a discontinuous Galerkin method

  • Solve a fourth-order differential equation

Equation and problem definition

Strong formulation

The biharmonic equation is a fourth-order elliptic equation. On the domain \(\Omega \subset \mathbb{R}^{d}\), \(1 \le d \le 3\), it reads

\[ \nabla^{4} u = f \quad {\rm in} \ \Omega, \]

where \(\nabla^{4} \equiv \nabla^{2} \nabla^{2}\) is the biharmonic operator and \(f\) is a prescribed source term. To formulate a complete boundary value problem, the biharmonic equation must be complemented by suitable boundary conditions.

Weak formulation

Multiplying the biharmonic equation by a test function and integrating by parts twice leads to a problem of second-order derivatives, which would require \(H^{2}\) conforming (roughly \(C^{1}\) continuous) basis functions. To solve the biharmonic equation using Lagrange finite element basis functions, the biharmonic equation can be split into two second-order equations (see the Mixed Poisson demo for a mixed method for the Poisson equation), or a variational formulation can be constructed that imposes weak continuity of normal derivatives between finite element cells. This demo uses a discontinuous Galerkin approach to impose continuity of the normal derivative weakly.

Consider a triangulation \(\mathcal{T}\) of the domain \(\Omega\), where the set of interior facets is denoted by \(\mathcal{E}_h^{\rm int}\). Functions evaluated on opposite sides of a facet are indicated by the subscripts \(+\) and \(-\). Using the standard continuous Lagrange finite element space

\[ V = \left\{v \in H^{1}_{0}(\Omega)\,:\, v \in P_{k}(K) \ \forall \ K \in \mathcal{T} \right\} \]

and considering the boundary conditions

\[\begin{split} \begin{align} u &= 0 \quad {\rm on} \ \partial\Omega, \\ \nabla^{2} u &= 0 \quad {\rm on} \ \partial\Omega, \end{align} \end{split}\]

a weak formulation of the biharmonic problem reads: find \(u \in V\) such that

\[ a(u,v)=L(v) \quad \forall \ v \in V, \]

where the bilinear form is

\[ a(u, v) = \sum_{K \in \mathcal{T}} \int_{K} \nabla^{2} u \nabla^{2} v \, {\rm d}x \ +\sum_{E \in \mathcal{E}_h^{\rm int}}\left(\int_{E} \frac{\alpha}{h_E} [\!\![ \nabla u ]\!\!] [\!\![ \nabla v ]\!\!] \, {\rm d}s - \int_{E} \left<\nabla^{2} u \right>[\!\![ \nabla v ]\!\!] \, {\rm d}s - \int_{E} [\!\![ \nabla u ]\!\!] \left<\nabla^{2} v \right> \, {\rm d}s\right) \]

and the linear form is

\[ L(v) = \int_{\Omega} fv \, {\rm d}x. \]

Furthermore, \(\left< u \right> = \frac{1}{2} (u_{+} + u_{-})\), \([\!\![ w ]\!\!] = w_{+} \cdot n_{+} + w_{-} \cdot n_{-}\), \(\alpha \ge 0\) is a penalty parameter and \(h_E\) is a measure of the cell size.

The input parameters for this demo are defined as follows:

  • \(\Omega = [0,1] \times [0,1]\) (a unit square)

  • \(\alpha = 8.0\) (penalty parameter)

  • \(f = 4.0 \pi^4\sin(\pi x)\sin(\pi y)\) (source term)

Implementation

We first import the modules and functions that the program uses:

from mpi4py import MPI
from petsc4py.PETSc import ScalarType  # type: ignore

import numpy as np

import ufl
from dolfinx import fem, io, mesh, plot
from dolfinx.fem.petsc import LinearProblem
from dolfinx.mesh import CellType, GhostMode
from ufl import CellDiameter, FacetNormal, avg, div, dS, dx, grad, inner, jump, pi, sin

We begin by using create_rectangle to create a rectangular Mesh of the domain, and creating a finite element FunctionSpace \(V\) on the mesh.

msh = mesh.create_rectangle(
    comm=MPI.COMM_WORLD,
    points=((0.0, 0.0), (1.0, 1.0)),
    n=(32, 32),
    cell_type=CellType.triangle,
    ghost_mode=GhostMode.shared_facet,
)
V = fem.functionspace(msh, ("Lagrange", 2))

The second argument to functionspace is a tuple consisting of (family, degree), where family is the finite element family, and degree specifies the polynomial degree. in this case V consists of second-order, continuous Lagrange finite element functions.

Next, we locate the mesh facets that lie on the boundary \(\Gamma_D = \partial\Omega\). We do this using using locate_entities_boundary and providing a marker function that returns True for points x on the boundary and False otherwise.

facets = mesh.locate_entities_boundary(
    msh,
    dim=1,
    marker=lambda x: np.isclose(x[0], 0.0)
    | np.isclose(x[0], 1.0)
    | np.isclose(x[1], 0.0)
    | np.isclose(x[1], 1.0),
)

We now find the degrees-of-freedom that are associated with the boundary facets using locate_dofs_topological

dofs = fem.locate_dofs_topological(V=V, entity_dim=1, entities=facets)

and use dirichletbc to create a DirichletBC class that represents the boundary condition. In this case, we impose Dirichlet boundary conditions with value \(0\) on the entire boundary \(\partial\Omega\).

bc = fem.dirichletbc(value=ScalarType(0), dofs=dofs, V=V)

Next, we express the variational problem using UFL.

First, the penalty parameter \(\alpha\) is defined. In addition, we define a variable h for the cell diameter \(h_E\), a variable nfor the outward-facing normal vector \(n\) and a variable h_avg for the average size of cells sharing a facet \(\left< h \right> = \frac{1}{2} (h_{+} + h_{-})\). Here, the UFL syntax ('+') and ('-') restricts a function to the ('+') and ('-') sides of a facet.

alpha = ScalarType(8.0)
h = CellDiameter(msh)
n = FacetNormal(msh)
h_avg = (h("+") + h("-")) / 2.0

After that, we can define the variational problem consisting of the bilinear form \(a\) and the linear form \(L\). The source term is prescribed as \(f = 4.0 \pi^4\sin(\pi x)\sin(\pi y)\). Note that with dS, integration is carried out over all the interior facets \(\mathcal{E}_h^{\rm int}\), whereas with ds it would be only the facets on the boundary of the domain, i.e. \(\partial\Omega\). The jump operator \([\!\![ w ]\!\!] = w_{+} \cdot n_{+} + w_{-} \cdot n_{-}\) w.r.t. the outward-facing normal vector \(n\) is in UFL available as jump(w, n).

# Define variational problem
u = ufl.TrialFunction(V)
v = ufl.TestFunction(V)
x = ufl.SpatialCoordinate(msh)
f = 4.0 * pi**4 * sin(pi * x[0]) * sin(pi * x[1])

a = (
    inner(div(grad(u)), div(grad(v))) * dx
    - inner(avg(div(grad(u))), jump(grad(v), n)) * dS
    - inner(jump(grad(u), n), avg(div(grad(v)))) * dS
    + alpha / h_avg * inner(jump(grad(u), n), jump(grad(v), n)) * dS
)
L = inner(f, v) * dx

We create a LinearProblem object that brings together the variational problem, the Dirichlet boundary condition, and which specifies the linear solver. In this case we use a direct (LU) solver. The solve will compute a solution.

problem = LinearProblem(a, L, bcs=[bc], petsc_options={"ksp_type": "preonly", "pc_type": "lu"})
uh = problem.solve()

The solution can be written to a XDMFFile file visualization with ParaView or VisIt

with io.XDMFFile(msh.comm, "out_biharmonic/biharmonic.xdmf", "w") as file:
    V1 = fem.functionspace(msh, ("Lagrange", 1))
    u1 = fem.Function(V1)
    u1.interpolate(uh)
    file.write_mesh(msh)
    file.write_function(u1)

and displayed using pyvista.

try:
    import pyvista

    cells, types, x = plot.vtk_mesh(V)
    grid = pyvista.UnstructuredGrid(cells, types, x)
    grid.point_data["u"] = uh.x.array.real
    grid.set_active_scalars("u")
    plotter = pyvista.Plotter()
    plotter.add_mesh(grid, show_edges=True)
    warped = grid.warp_by_scalar()
    plotter.add_mesh(warped)
    if pyvista.OFF_SCREEN:
        pyvista.start_xvfb(wait=0.1)
        plotter.screenshot("uh_biharmonic.png")
    else:
        plotter.show()
except ModuleNotFoundError:
    print("'pyvista' is required to visualise the solution")
    print("Install 'pyvista' with pip: 'python3 -m pip install pyvista'")