--- jupytext: text_representation: extension: .md format_name: myst format_version: 0.13 jupytext_version: 1.16.1 kernelspec: display_name: Python 3 (ipykernel) language: python name: python3 --- # Biharmonic equation This demo is implemented in a single Python file, {download}`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 $$ \begin{align} u &= 0 \quad {\rm on} \ \partial\Omega, \\ \nabla^{2} u &= 0 \quad {\rm on} \ \partial\Omega, \end{align} $$ 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: ```python from mpi4py import MPI from petsc4py.PETSc import ScalarType # type: ignore ``` ```python 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 {py:func}`create_rectangle ` to create a rectangular {py:class}`Mesh ` of the domain, and creating a finite element {py:class}`FunctionSpace ` $V$ on the mesh. ```python 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 {py:func}`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 {py:func}`locate_entities_boundary ` and providing a marker function that returns `True` for points `x` on the boundary and `False` otherwise. ```python 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 {py:func}`locate_dofs_topological ` ```python dofs = fem.locate_dofs_topological(V=V, entity_dim=1, entities=facets) ``` and use {py:func}`dirichletbc ` to create a {py:class}`DirichletBC ` class that represents the boundary condition. In this case, we impose Dirichlet boundary conditions with value $0$ on the entire boundary $\partial\Omega$. ```python 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 `n`for 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. ```python 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)`. ```python # 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 {py:class}`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 {py:func}`solve ` will compute a solution. ```python problem = LinearProblem(a, L, bcs=[bc], petsc_options={"ksp_type": "preonly", "pc_type": "lu"}) uh = problem.solve() ``` The solution can be written to a {py:class}`XDMFFile ` file visualization with ParaView or VisIt ```python 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](https://docs.pyvista.org/). ```python 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'") ```