Simple FEM-BEM coupling for the Helmholtz equation with FEniCS

Background

For this problem, you will need FEniCS installed alongside Bempp.

In this tutorial, we will solve the problem of a wave travelling through a unit cube, $\Omega = [0,1]^3$ with different material parameters inside and outside the domain. The incident wave is given by

$$ u^\text{inc}(\mathbf{x})=\mathrm{e}^{\mathrm{i} k \mathbf{x}\cdot\mathbf{d}}, $$

where $\mathbf{x}=(x,y,z)$ and $\mathbf{d}$ is the direction of the incident wave. In the implementation we use, $\mathbf{d} = \frac{1}{\sqrt{3}}(1,1,1)$.

The PDE is

$$ \Delta u + n(\mathbf{x})^2 k^2 u = 0, \quad \text{ in } \Omega\\ \Delta u + k^2 u = 0, \quad \text{ in } \mathbb{R}^3 \backslash \Omega $$

In this example, we use

$$ n(\mathbf{x}) = 0.5 $$

Since the interior wavenumber is constant one could have also used a BEM/BEM coupling approach. However, here we demonstrate the use of FEM for the interior problem using the FEniCS finite element package.

FEM Part

In $\Omega$, the FEM part is formulated as

$$ \int_\Omega \nabla u\cdot\nabla v -k^2\int_\Omega n^2uv - \int_{d\Omega} v\frac{\partial u}{\partial \nu} = 0, $$

or

$$ \langle\nabla u,\nabla v\rangle_\Omega - k^2\langle n^2u,v\rangle_\Omega - \langle \lambda,v\rangle_\Gamma=0, $$

where $\lambda=\frac{\partial u}{\partial \nu}$.

Later, we will write this as the following operator equation

$$ \mathsf{A}u-k^2 \mathsf{M}u-\mathsf{M}_\Gamma \lambda = 0 $$

BEM Part

In $\mathbb{R}^3 \backslash \Omega$, we let $u = u^\text{inc}+u^\text{s}$, where $u^\text{inc}$ is the incident wave and $u^\text{s}$ is the scattered wave. As given in Integral equation methods in scattering theory by Colton & Kress,

$$ 0 = \mathcal{K}u^\text{inc}-\mathcal{V}\frac{\partial u^{inc}}{\partial \nu},\\[2mm] u^\text{s} = \mathcal{K}u^\text{s}-\mathcal{V}\frac{\partial u^{s}}{\partial \nu}, $$

where $\mathcal{K}$ and $\mathcal{V}$ are the double single layer potential operators. Adding these, we get

$$ u^\text{s} = \mathcal{K}u-\mathcal{V}\lambda. $$

This representation formula will be used to find $u^\text{s}$ for plotting later.

Taking the trace on the boundary gives

$$ u-u^\text{inc} = \left(\tfrac{1}{2}\mathsf{Id}+\mathsf{K}\right)u -\mathsf{V}\lambda. $$

This rearranges to

$$ u^\text{inc} = \left(\tfrac{1}{2}\mathsf{Id}-\mathsf{K}\right)u+\mathsf{V}\lambda. $$

Full Formulation

The full blocked formulation is

$$ \begin{bmatrix} \mathsf{A}-k^2 \mathsf{M} & -\mathsf{M}_\Gamma\\ \tfrac{1}{2}\mathsf{Id}-\mathsf{K} & \mathsf{V} \end{bmatrix} \begin{bmatrix} u\\ \lambda \end{bmatrix}=\begin{bmatrix} 0\\ u^\text{inc} \end{bmatrix}. $$

This formulation is not stable for all frequencies due to the possibility of interior resonances. But it is sufficient for this example and serves as a blueprint for more complex formulations.

Implementation

We begin by importing Dolfin, the FEniCS python library, Bempp and NumPy.

In [20]:
import dolfin
import bempp.api
import numpy as np

Next, we set the wavenumber k and the direction d of the incoming wave.

In [21]:
k = 6.
d = np.array([1., 1., 1])
d /= np.linalg.norm(d)

We create a Dolfin mesh. Later, the boundary mesh will be extracted from this.

A mesh could be created from a file changing this line to mesh = dolfin.Mesh('/path/to/file.xml').

In [22]:
mesh = dolfin.UnitCubeMesh(10, 10, 10)

Next, we make the Dolfin and Bempp function spaces.

The function fenics_to_bempp_trace_data will extract the trace space from the Dolfin space and create the matrix trace_matrix, which maps between the dofs (degrees of freedom) in Dolfin and Bempp.

In [23]:
from bempp.api.external import fenics

fenics_space = dolfin.FunctionSpace(mesh, "CG", 1)
trace_space, trace_matrix = \
    fenics.fenics_to_bempp_trace_data(fenics_space)
bempp_space = bempp.api.function_space(trace_space.grid, "DP", 0)

print("FEM dofs: {0}".format(mesh.num_vertices()))
print("BEM dofs: {0}".format(bempp_space.global_dof_count))
FEM dofs: 1331
BEM dofs: 1200

We create the boundary operators that we need.

In [24]:
id_op = bempp.api.operators.boundary.sparse.identity(
    trace_space, bempp_space, bempp_space)
mass = bempp.api.operators.boundary.sparse.identity(
    bempp_space, bempp_space, trace_space)
dlp = bempp.api.operators.boundary.helmholtz.double_layer(
    trace_space, bempp_space, bempp_space, k)
slp = bempp.api.operators.boundary.helmholtz.single_layer(
    bempp_space, bempp_space, bempp_space, k)

We create the Dolfin function spaces and the function (or in this case constant) n.

In [25]:
u = dolfin.TrialFunction(fenics_space)
v = dolfin.TestFunction(fenics_space)
n = 0.5

We make the vectors on the right hand side of the formulation.

In [26]:
@bempp.api.complex_callable
def u_inc(x, n, domain_index, result):
    result[0] = np.exp(1j * k * np.dot(x, d))
u_inc = bempp.api.GridFunction(bempp_space, fun=u_inc)

# The rhs from the FEM
rhs_fem = np.zeros(mesh.num_vertices())
# The rhs from the BEM
rhs_bem = u_inc.projections(bempp_space)
# The combined rhs
rhs = np.concatenate([rhs_fem, rhs_bem])
<ipython-input-26-da6cbf46ff95>:3: NumbaPerformanceWarning: np.dot() is faster on contiguous arrays, called on (array(float64, 1d, A), readonly array(float64, 1d, C))
  result[0] = np.exp(1j * k * np.dot(x, d))

We are now ready to create a BlockedLinearOperator containing all four parts of the discretisation of $$ \begin{bmatrix} \mathsf{A}-k^2 \mathsf{M} & -\mathsf{M}_\Gamma\\ \tfrac{1}{2}\mathsf{Id}-\mathsf{K} & \mathsf{V} \end{bmatrix}. $$

In [27]:
from bempp.api.assembly.blocked_operator import BlockedDiscreteOperator
from bempp.api.external.fenics import FenicsOperator
from scipy.sparse.linalg.interface import LinearOperator
blocks = [[None,None],[None,None]]

trace_op = LinearOperator(trace_matrix.shape, lambda x:trace_matrix @ x)

A = FenicsOperator((dolfin.inner(dolfin.nabla_grad(u),
                                 dolfin.nabla_grad(v)) \
    - k**2 * n**2 * u * v) * dolfin.dx)

blocks[0][0] = A.weak_form()
blocks[0][1] = -trace_matrix.T * mass.weak_form().to_sparse()
blocks[1][0] = (.5 * id_op - dlp).weak_form() * trace_op
blocks[1][1] = slp.weak_form()

blocked = BlockedDiscreteOperator(np.array(blocks))

Next, we solve the system, then split the solution into the parts assosiated with u and λ. For an efficient solve, preconditioning is required.

In [28]:
from bempp.api.assembly.discrete_boundary_operator import InverseSparseDiscreteBoundaryOperator
from scipy.sparse.linalg import LinearOperator

# Compute the sparse inverse of the Helmholtz operator
# Although it is not a boundary operator we can use
# the SparseInverseDiscreteBoundaryOperator function from
# BEM++ to turn its LU decomposition into a linear operator.
P1 = InverseSparseDiscreteBoundaryOperator(
    blocked[0,0].to_sparse().tocsc())

# For the Laplace slp we use a simple mass matrix preconditioner. 
# This is sufficient for smaller low-frequency problems.
P2 = InverseSparseDiscreteBoundaryOperator(
    bempp.api.operators.boundary.sparse.identity(
        bempp_space, bempp_space, bempp_space).weak_form())

# Create a block diagonal preconditioner object using the Scipy LinearOperator class
def apply_prec(x):
    """Apply the block diagonal preconditioner"""
    m1 = P1.shape[0]
    m2 = P2.shape[0]
    n1 = P1.shape[1]
    n2 = P2.shape[1]
    
    res1 = P1.dot(x[:n1])
    res2 = P2.dot(x[n1:])
    return np.concatenate([res1, res2])

p_shape = (P1.shape[0] + P2.shape[0], P1.shape[1] + P2.shape[1])
P = LinearOperator(p_shape, apply_prec, dtype=np.dtype('complex128'))

# Create a callback function to count the number of iterations
it_count = 0
def count_iterations(x):
    global it_count
    it_count += 1

from scipy.sparse.linalg import gmres
soln, info = gmres(blocked, rhs, M=P, callback=count_iterations)

soln_fem = soln[:mesh.num_vertices()]
soln_bem = soln[mesh.num_vertices():]

print("Number of iterations: {0}".format(it_count))
/home/betcke/.conda/envs/bempp_default/lib/python3.7/site-packages/scipy/sparse/linalg/dsolve/linsolve.py:296: SparseEfficiencyWarning: splu requires CSC matrix format
  warn('splu requires CSC matrix format', SparseEfficiencyWarning)
Number of iterations: 272

Next, we make Dolfin and Bempp functions from the solution.

In [29]:
# Store the real part of the FEM solution
u = dolfin.Function(fenics_space)
u.vector()[:] = np.ascontiguousarray(np.real(soln_fem))

# Solution function with dirichlet data on the boundary
dirichlet_data = trace_matrix * soln_fem
dirichlet_fun = bempp.api.GridFunction(trace_space, coefficients=dirichlet_data)

# Solution function with Neumann data on the boundary
neumann_fun = bempp.api.GridFunction(bempp_space, coefficients=soln_bem)

We now evaluate the solution on the slice $z=0.5$ and plot it. For the exterior domain, we use the respresentation formula

$$ u^\text{s} = \mathcal{K}u-\mathcal{V}\frac{\partial u}{\partial \nu} $$

to evaluate the solution.

In [31]:
# The next command ensures that plots are shown within the IPython notebook
%matplotlib inline

Nx=200
Ny=200
xmin, xmax, ymin, ymax=[-1,3,-1,3]
plot_grid = np.mgrid[xmin:xmax:Nx*1j,ymin:ymax:Ny*1j]
points = np.vstack((plot_grid[0].ravel(),
                    plot_grid[1].ravel(),
                    np.array([0.5]*plot_grid[0].size)))
plot_me = np.zeros(points.shape[1], dtype=np.complex128)

x,y,z = points
bem_x = np.logical_not((x>0) * (x<1) * (y>0) * (y<1) * (z>0) * (z<1))

slp_pot= bempp.api.operators.potential.helmholtz.single_layer(
    bempp_space, points[:, bem_x], k)
dlp_pot= bempp.api.operators.potential.helmholtz.double_layer(
    trace_space, points[:, bem_x], k)

plot_me[bem_x] += np.exp(1j * k * (points[0, bem_x] * d[0] \
                                 + points[1, bem_x] * d[1] \
                                 + points[2, bem_x] * d[2]))
plot_me[bem_x] += dlp_pot.evaluate(dirichlet_fun).flat
plot_me[bem_x] -= slp_pot.evaluate(neumann_fun).flat

fem_points = points[:, np.logical_not(bem_x)].transpose()
fem_val = np.zeros(len(fem_points))
for p,point in enumerate(fem_points):
    result = np.zeros(1)
    u.eval(result, point)
    fem_val[p] = result[0]

plot_me[np.logical_not(bem_x)] += fem_val

plot_me = plot_me.reshape((Nx, Ny))

plot_me = plot_me.transpose()[::-1]

# Plot the image
from matplotlib import pyplot as plt
fig=plt.figure(figsize=(10, 8))
plt.imshow(np.real(plot_me), extent=[xmin, xmax, ymin, ymax])
plt.xlabel('x')
plt.ylabel('y')
plt.colorbar()
plt.title("FEM-BEM Coupling for Helmholtz")
plt.show()