Combining Dirichlet and Neumann conditions

Combining Dirichlet and Neumann conditions#

Author: Jørgen S. Dokken

Let’s return to the Poisson problem from the Fundamentals chapter and see how to extend the mathematics and the implementation to handle Dirichlet condition in combination with a Neumann condition. The domain is still the unit square, but now we set the Dirichlet condition \(u=u_D\) at the left and right sides, while the Neumann condition

\[ -\frac{\partial u}{\partial n}=g \]

is applied to the remaining sides \(y=0\) and \(y=1\).

The PDE problem#

Let \(\Lambda_D\) and \(\Lambda_N\) denote parts of the boundary \(\partial \Omega\) where the Dirichlet and Neumann conditions apply, respectively. The complete boundary-value problem can be written as

\[ -\nabla^2 u =f \qquad \text{in } \Omega, \]
\[ u=u_D \qquad\text{on } \Lambda_D, \]
\[ -\frac{\partial u}{\partial n}=g \qquad \text{on }\Lambda_N \]

Again, we choose \(u=1+x^2+2y^2\) as the exact solution and adjust \(f, g,\) and \(u_D\) accordingly

\[ f(x,y)=-6, \]
\[\begin{split} g(x,y)=\begin{cases} 0, & y=0,\\ -4, & y=1, \end{cases} \end{split}\]
\[ u_D(x,y)=1+x^2+2y^2. \]

For the ease of programming, we define \(g\) as a function over the whole domain \(\Omega\) such that \(g\) takes on the correct values at \(y=0\) and \(y=1\). One possible extension is

\[ g(x,y)=-4y. \]

The variational formulation#

The first task is to derive the variational formulatin. This time we cannot omit the boundary term arising from integration by parts, because \(v\) is only zero on \(\Lambda_D\). We have

\[ -\int_\Omega (\nabla^2u)v~\mathrm{d} x = \int_\Omega \nabla u \cdot \nabla v ~\mathrm{d} x - \int_{\partial\Omega}\frac{\partial u}{\partial n}v~\mathrm{d}s, \]

and since \(v=0\) on \(\Lambda_D\),

\[ - \int_{\partial\Omega}\frac{\partial u}{\partial n}v~\mathrm{d}s= - \int_{\Lambda_N}\frac{\partial u}{\partial n}v~\mathrm{d}s =\int_{\Lambda_N} gv~\mathrm{d}s, \]

by applying the boundary condition on \(\Lambda_N\). The resulting weak from reads

\[ \int_\Omega \nabla u \cdot \nabla v~\mathrm{d} x = \int_\Omega fv~\mathrm{d} x - \int_{\Lambda_N}gv~\mathrm{d}s. \]

Expressing this equation in the standard notation \(a(u,v)=L(v)\) is straight-forward with

\[\begin{split} a(u,v) = \int_{\Omega} \nabla u \cdot \nabla v ~\mathrm{d} x,\\ \end{split}\]
\[ L(v) = \int_{\Omega} fv ~\mathrm{d} x - \int_{\Lambda_N} gv~\mathrm{d} s. \]


As in the previous example, we define our mesh,function space and bilinear form \(a(u,v)\).

from dolfinx import default_scalar_type
from dolfinx.fem import (Constant, Function, FunctionSpace,
                         assemble_scalar, dirichletbc, form, locate_dofs_geometrical)
from dolfinx.fem.petsc import LinearProblem
from dolfinx.mesh import create_unit_square
from dolfinx.plot import vtk_mesh

from mpi4py import MPI
from ufl import SpatialCoordinate, TestFunction, TrialFunction, dot, ds, dx, grad

import numpy as np
import pyvista

mesh = create_unit_square(MPI.COMM_WORLD, 10, 10)
V = FunctionSpace(mesh, ("Lagrange", 1))
u = TrialFunction(V)
v = TestFunction(V)
a = dot(grad(u), grad(v)) * dx

Now we get to the Neumann and Dirichlet boundary condition. As previously, we use a Python-function to define the boundary where we should have a Dirichlet condition. Then, with this function, we locate degrees of freedom that fullfils this condition.

def u_exact(x):
    return 1 + x[0]**2 + 2 * x[1]**2

def boundary_D(x):
    return np.logical_or(np.isclose(x[0], 0), np.isclose(x[0], 1))

dofs_D = locate_dofs_geometrical(V, boundary_D)
u_bc = Function(V)
bc = dirichletbc(u_bc, dofs_D)

The next step is to define the Neumann condition. We first define \(g\) uses UFLs SpatialCoordinate-function, and then in turn create a boundary integration measure ds. As the test function \(v\) is zero on the boundary integrals over the Dirichlet boundary disappears, and we can integrate g*v*ds over the entire boundary.

x = SpatialCoordinate(mesh)
g = -4 * x[1]
f = Constant(mesh, default_scalar_type(-6))
L = f * v * dx - g * v * ds

We can now assemble and solve the linear system of equations

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

V2 = FunctionSpace(mesh, ("Lagrange", 2))
uex = Function(V2)
error_L2 = assemble_scalar(form((uh - uex)**2 * dx))
error_L2 = np.sqrt(MPI.COMM_WORLD.allreduce(error_L2, op=MPI.SUM))

u_vertex_values = uh.x.array
uex_1 = Function(V)
u_ex_vertex_values = uex_1.x.array
error_max = np.max(np.abs(u_vertex_values - u_ex_vertex_values))
error_max = MPI.COMM_WORLD.allreduce(error_max, op=MPI.MAX)
print(f"Error_L2 : {error_L2:.2e}")
print(f"Error_max : {error_max:.2e}")
Error_L2 : 5.27e-03
Error_max : 2.44e-15


To look at the actual solution, run the script as a python script with off_screen=True or as a Jupyter notebook with off_screen=False


pyvista_cells, cell_types, geometry = vtk_mesh(V)
grid = pyvista.UnstructuredGrid(pyvista_cells, cell_types, geometry)
grid.point_data["u"] = uh.x.array

plotter = pyvista.Plotter()
plotter.add_text("uh", position="upper_edge", font_size=14, color="black")
plotter.add_mesh(grid, show_edges=True)

if not pyvista.OFF_SCREEN:
    figure = plotter.screenshot("neumann_dirichlet.png")