The goal of this chapter is to demonstrate how a range of important PDEs from science and engineering can be quickly solved with a few lines of FEniCS code. We start with the heat equation and continue with a nonlinear Poisson equation, the equations for linear elasticity, the Navier–Stokes equations, and finally look at how to solve systems of nonlinear advection–diffusion–reaction equations. These problems illustrate how to solve time-dependent problems, nonlinear problems, vector-valued problems, and systems of PDE. For each problem, we derive the variational formulation and express the problem in Python in a way that closely resembles the mathematics.

The heat equation

As a first extension of the Poisson problem from the previous chapter, we consider the time-dependent heat equation, or the time-dependent diffusion equation. This is the natural extension of the Poisson equation describing the stationary distribution of heat in a body to a time-dependent problem.

We will see that by discretizing time into small time intervals and applying standard time-stepping methods, we can solve the heat equation by solving a sequence of variational problems, much like the one we encountered for the Poisson equation.

PDE problem

Our model problem for time-dependent PDEs reads $$ \begin{align} {\partial u\over\partial t} &= \nabla^2 u + f\hbox{ in }\Omega, \tag{3.1}\\ u &= \ub\hbox{ on } \partial \Omega, \tag{3.2}\\ u &= u_0 \mbox{ at } t=0\tp \tag{3.3} \end{align} $$ Here, \( u \) varies with space and time, e.g., \( u=u(x,y,t) \) if the spatial domain \( \Omega \) is two-dimensional. The source function \( f \) and the boundary values \( \ub \) may also vary with space and time. The initial condition \( u_0 \) is a function of space only.

Variational formulation

A straightforward approach to solving time-dependent PDEs by the finite element method is to first discretize the time derivative by a finite difference approximation, which yields a sequence of stationary problems, and then turn each stationary problem into a variational formulation.

Let superscript \( n \) denote a quantity at time \( t_n \), where \( n \) is an integer counting time levels. For example, \( u^n \) means \( u \) at time level \( n \). A finite difference discretization in time first consists of sampling the PDE at some time level, say \( t_{n+1} \): $$ \begin{equation} \left({\partial u \over\partial t}\right)^{n+1} = \nabla^2 u^{n+1} + f^{n+1}\tp \tag{3.4} \end{equation} $$ The time-derivative can be approximated by a difference quotient. For simplicity and stability reasons, we choose a simple backward difference: $$ \begin{equation} \left({\partial u\over\partial t}\right)^{n+1}\approx {{u^{n+1} - u^n}\over{\dt}}, \tag{3.5} \end{equation} $$ where \( \dt \) is the time discretization parameter. Inserting (3.5) in (3.4) yields $$ \begin{equation} {{u^{n+1} - u^n}\over{\dt}} = \nabla^2 u^{n+1} + f^{n+1}\tp \tag{3.6} \end{equation} $$ This is our time-discrete version of the heat equation (3.1). This is a so-called backward Euler or implicit Euler discretization. Alternatively, we may also view this as a finite element discretization in time in the form of the first order \( \mathrm{dG}(0) \) method, which here is identical to the backward Euler method.

We may reorder (3.6) so that the left-hand side contains the terms with the unknown \( u^{n+1} \) and the right-hand side contains computed terms only. The result is a sequence of spatial (stationary) problems for \( u^{n+1} \) (assuming \( u^n \) is known from computations at the previous time level): $$ \begin{align} u^0 &= u_0, \tag{3.7}\\ u^{n+1} - {\dt}\nabla^2 u^{n+1} &= u^n + {\dt} f^{n+1},\quad n=0,1,2,\ldots \tag{3.8} \end{align} $$ Given \( u_0 \), we can solve for \( u^0 \), \( u^1 \), \( u^2 \), and so on.

An alternative to (3.8), which can be convenient in implementations, is to collect all terms on one side of the equality sign: $$ \begin{equation} u^{n+1} - {\dt}\nabla^2 u^{n+1} - u^{n} - {\dt} f^{n+1} = 0,\quad n=0,1,2,\ldots \tag{3.9} \end{equation} $$

We use a finite element method to solve (3.7) and either of the equations (3.8) or (3.9). This requires turning the equations into weak forms. As usual, we multiply by a test function \( v\in \hat V \) and integrate second-derivatives by parts. Introducing the symbol \( u \) for \( u^{n+1} \) (which is natural in the program), the resulting weak form arising from formulation (3.8) can be conveniently written in the standard notation: $$ a(u,v)=L_{n+1}(v),$$ where $$ \begin{align} a(u,v) &= \int_\Omega\left( uv + {\dt} \nabla u\cdot \nabla v\right) \dx, \tag{3.10}\\ L_{n+1}(v) &= \int_\Omega \left(u^n + {\dt} f^{n+1}\right)v \dx\tp \tag{3.11} \end{align} $$ The alternative form (3.9) has an abstract formulation $$ F(u;v) = 0,$$ where $$ \begin{equation} F(u; v) = \int_\Omega uv + {\dt} \nabla u\cdot \nabla v - (u^n + {\dt} f^{n+1})v \dx\tp \tag{3.12} \end{equation} $$

In addition to the variational problem to be solved in each time step, we also need to approximate the initial condition (3.7). This equation can also be turned into a variational problem: $$ a_0(u,v)=L_0(v),$$ with $$ \begin{align} a_0(u,v) &= \int_\Omega uv \dx, \tag{3.13}\\ L_0(v) &= \int_\Omega u_0 v \dx\tp \tag{3.14} \end{align} $$ When solving this variational problem, \( u^0 \) becomes the \( L^2 \) projection of the given initial value \( u_0 \) into the finite element space. The alternative is to construct \( u^0 \) by just interpolating the initial value \( u_0 \); that is, if \( u^0=\sum_{j=1}^N U^0_j\phi_j \), we simply set \( U_j=u_0(x_j,y_j) \), where \( (x_j,y_j) \) are the coordinates of node number \( j \). We refer to these two strategies as computing the initial condition by either projection or interpolation. Both operations are easy to compute in FEniCS through one statement, using either the project or interpolate function. The most common choice is project, which computes an approximation to \( u_0 \), but in some applications where we want to verify the code by reproducing exact solutions, one must use interpolate (and we use such a test problem!).

In summary, we thus need to solve the following sequence of variational problems to compute the finite element solution to the heat equation: find \( u^0\in V \) such that \( a_0(u^0,v)=L_0(v) \) holds for all \( v\in\hat V \), and then find \( u^{n+1}\in V \) such that \( a(u^{n+1},v)=L_{n+1}(v) \) for all \( v\in\hat V \), or alternatively, \( F(u^{n+1},v)=0 \) for all \( v\in\hat V \), for \( n=0,1,2,\ldots \).

FEniCS implementation

Our program needs to implement the time-stepping manually, but can rely on FEniCS to easily compute \( a_0 \), \( L_0 \), \( F \), \( a \), and \( L \), and solve the linear systems for the unknowns.

Test problem

Just as for the Poisson problem from the previous chapter, we construct a test problem that makes it easy to determine if the calculations are correct. Since we know that our first-order time-stepping scheme is exact for linear functions, we create a test problem which has a linear variation in time. We combine this with a quadratic variation in space. We thus take $$ \begin{equation} u = 1 + x^2 + \alpha y^2 + \beta t, \tag{3.15} \end{equation} $$ which yields a function whose computed values at the nodes will be exact, regardless of the size of the elements and \( \dt \), as long as the mesh is uniformly partitioned. By inserting (3.15) into the heat equation (3.1), we find that the right-hand side \( f \) must be given by \( f(x,y,t)=\beta - 2 - 2\alpha \). The boundary value is \( \ub(x, y, t) = 1 + x^2 + \alpha y^2 + \beta t \) and the initial value is \( u_0(x, y) = 1 + x^2 + \alpha y^2 \).

FEniCS implementation

A new programming issue is how to deal with functions that vary in space and time, such as the boundary condition \( \ub(x, y, t) = 1 + x^2 + \alpha y^2 + \beta t \). A natural solution is to use a FEniCS Expression with time \( t \) as a parameter, in addition to the parameters \( \alpha \) and \( \beta \):

alpha = 3; beta = 1.2
u_D = Expression('1 + x[0]*x[0] + alpha*x[1]*x[1] + beta*t',
                 degree=2, alpha=alpha, beta=beta, t=0)

This expression uses the components of x as independent variables, while alpha, beta, and t are parameters. The parameters can later be updated as in

u_D.t = t

The essential boundary conditions, along the entire boundary in this case, are set in the usual way:

def boundary(x, on_boundary):
    return on_boundary

bc = DirichletBC(V, u_D, boundary)

We shall use u for the unknown \( u^n \) at the new time level and u_n for \( u^n \) at the previous time level. The initial value of u_n can be computed by either projection or interpolation of \( u_0 \). Since we set t = 0 for the boundary value u_D, we can use this to also specify the initial condition. We can then do

u_n = project(u_D, V)
# or
u_n = interpolate(u_D, V)

We may either define \( a \) or \( L \) according to the formulas above, or we may just define \( F \) and ask FEniCS to figure out which terms that go into the bilinear form \( a \) and which that go into the linear form \( L \). The latter is convenient, especially in more complicated problems, so we illustrate that construction of \( a \) and \( L \):

u = TrialFunction(V)
v = TestFunction(V)
f = Constant(beta - 2 - 2*alpha)

F = u*v*dx + dt*dot(grad(u), grad(v))*dx - (u_n + dt*f)*v*dx
a, L = lhs(F), rhs(F)

Finally, we perform the time-stepping in a loop:

u = Function(V)
t = 0
for n in range(num_steps):

    # Update current time
    t += dt
    u_D.t = t

    # Solve variational problem
    solve(a == L, u, bc)

    # Update previous solution
    u_n.assign(u)

In the last step of the time-stepping loop, we assign the values of the variable u (the new computed solution) to the variable u_n containing the values at the previous time step. This must be done using the assign member function. If we instead try to do u_n = u, we will set the u_n variable to be the same variable as u which is not what we want. (We need two variables, one for the values at the previous time step and one for the values at the current time step.)

The time loop above does not contain any comparison of the numerical and the exact solution, which we must include in order to verify the implementation. As in the Poisson equation example in the section Dissection of the program, we compute the difference between the array of nodal values for u and the array of nodal values for the interpolated exact solution. This may be done as follows:

u_e = interpolate(u_D, V)
error = np.abs(u_e.vector().array() - u.vector().array()).max()
print('error, t=%.2f: %.3g' % (t, error))

The complete program code for this time-dependent case goes as follows:

from fenics import *
import numpy as np

T = 2.0            # final time
num_steps = 10     # number of time steps
dt = T / num_steps # time step size
alpha = 3          # parameter alpha
beta = 1.2         # parameter beta

# Create mesh and define function space
nx = ny = 8
mesh = UnitSquareMesh(nx, ny)
V = FunctionSpace(mesh, 'P', 1)

# Define boundary condition
u_D = Expression('1 + x[0]*x[0] + alpha*x[1]*x[1] + beta*t',
                 degree=2, alpha=alpha, beta=beta, t=0)

def boundary(x, on_boundary):
    return on_boundary

bc = DirichletBC(V, u_D, boundary)

# Define initial value
u_n = interpolate(u_D, V)
#u_n = project(u_D, V)

# Define variational problem
u = TrialFunction(V)
v = TestFunction(V)
f = Constant(beta - 2 - 2*alpha)

F = u*v*dx + dt*dot(grad(u), grad(v))*dx - (u_n + dt*f)*v*dx
a, L = lhs(F), rhs(F)

# Time-stepping
u = Function(V)
t = 0
for n in range(num_steps):

    # Update current time
    t += dt
    u_D.t = t # update for bc

    # Compute solution
    solve(a == L, u, bc)

    # Compute error at vertices
    u_e = interpolate(u_D, V)
    error = np.abs(u_e.vector().array() - u.vector().array()).max()
    print('t = %.2f: error = %.3g' % (t, error))

    # Update previous solution
    u_n.assign(u)

The complete code can be found in the file ft04_heat.py.

Diffusion of a Gaussian function

The mathematical problem

Let's solve a more interesting test problem, namely the diffusion of a Gaussian hill. We take the initial value to be given by $$ u_0(x,y)= e^{-ax^2 - ay^2}$$ on the domain \( [-2,2]\times [2,2] \). We will take \( a = 5 \). For this problem we will use homogeneous Dirichlet boundary conditions (\( \ub = 0 \)).

FEniCS implementation

Which are the required changes to our previous program? One major change is that the domain is not a unit square anymore. We also want to use much higher resolution. The new domain can be created easily in FEniCS using RectangleMesh:

nx = ny = 30
mesh = RectangleMesh(Point(-2, -2), Point(2, 2), nx, ny)

We also need to redefine the initial condition and boundary condition. Both are easily changed by defining a new Expression and by setting \( u = 0 \) on the boundary. We will also save the solution to file in VTK format in each time step:

vtkfile << (u, t)

The complete program appears below.

from fenics import *
import time

T = 2.0            # final time
num_steps = 50     # number of time steps
dt = T / num_steps # time step size

# Create mesh and define function space
nx = ny = 30
mesh = RectangleMesh(Point(-2,-2), Point(2,2), nx, ny)
V = FunctionSpace(mesh, 'P', 1)

# Define boundary condition
def boundary(x, on_boundary):
    return on_boundary

bc = DirichletBC(V, Constant(0), boundary)

# Define initial value
u_0 = Expression('exp(-a*pow(x[0],2) - a*pow(x[1],2))',
                 degree=2, a=5)
u_n = interpolate(u_0, V)
u_n.rename('u', 'initial value')
vtkfile = File('gaussian_diffusion.pvd')
vtkfile << (u_n, 0.0)

# Define variational problem
u = TrialFunction(V)
v = TestFunction(V)
f = Constant(0)

F = u*v*dx + dt*dot(grad(u), grad(v))*dx - (u_n + dt*f)*v*dx
a, L = lhs(F), rhs(F)

# Compute solution
u = Function(V)
u.rename('u', 'solution')
t = 0
for n in range(num_steps):

    # Update current time
    t += dt

    # Solve variational problem
    solve(a == L, u, bc)

    # Save to file and plot solution
    vtkfile << (u, float(t))
    plot(u)
    time.sleep(0.3)

    # Update previous solution
    u_n.assign(u)

The complete code can be found in the file ft05_gaussian_diffusion.py.

Visualization in ParaView

To visualize the diffusion of the Gaussian hill, start ParaView, choose File - Open, open the file gaussian_diffusion.pvd, click the green Apply button on the left to see the initial condition being plotted. Choose View - Animation View. Click on the play button or (better) the next frame button in the row of buttons at the top of the GUI to see the evolution of the scalar field you have just computed. Choose File - Save Animation... to save the animation to the AVI or OGG video format.

Once the animation has been saved to file, you can play the animation offline using a player such as mplayer or VLC, or upload your animation to YouTube. Below is a sequence of snapshots of the solution (first three time steps).

A nonlinear Poisson equation

We shall now address how to solve nonlinear PDEs. We will see that nonlinear problems can be solved just as easily as linear problems in FEniCS, by simply defining a nonlinear variational problem and calling the solve function. When doing so, we will encounter a subtle difference in how the variational problem is defined.

PDE problem

As a sample PDE for the implementation of nonlinear problems, we take the following nonlinear Poisson equation: $$ \begin{equation} -\nabla\cdot\left( q(u)\nabla u\right) = f, \tag{3.16} \end{equation} $$ in \( \Omega \), with \( u=\ub \) on the boundary \( \partial\Omega \). The coefficient \( q(u) \) makes the equation nonlinear (unless \( q(u) \) is constant in \( u \)).

Variational formulation

As usual, we multiply our PDE by a test function \( v\in\hat V \), integrate over the domain, and integrate the second-order derivatives by parts. The boundary integral arising from integration by parts vanishes wherever we employ Dirichlet conditions. The resulting variational formulation of our model problem becomes: find \( u \in V \) such that $$ \begin{equation} F(u; v) = 0 \quad \forall v \in \hat{V}, \tag{3.17} \end{equation} $$ where $$ \begin{equation} F(u; v) = \int_\Omega q(u)\nabla u\cdot \nabla v - fv \dx, \tag{3.18} \end{equation} $$ and $$ \begin{align*} V &= \{v \in H^1(\Omega) : v = \ub \mbox{ on } \partial\Omega\},\\ \hat{V} &= \{v \in H^1(\Omega) : v = 0 \mbox{ on } \partial\Omega\}\tp \end{align*} $$

The discrete problem arises as usual by restricting \( V \) and \( \hat V \) to a pair of discrete spaces. As before, we omit any subscript on the discrete spaces and discrete solution. The discrete nonlinear problem is then written as: Find \( u\in V \) such that $$ \begin{equation} F(u; v) = 0 \quad \forall v \in \hat{V}, \tag{3.19} \end{equation} $$ with \( u = \sum_{j=1}^N U_j \phi_j \). Since \( F \) is nonlinear in \( u \), the variational statement gives rise to a system of nonlinear algebraic equations in the unknowns \( U_1,\ldots,U_N \).

FEniCS implementation

Test problem

To solve a test problem, we need to choose the right-hand side \( f \), the coefficient \( q(u) \) and the boundary value \( \ub \). Previously, we have worked with manufactured solutions that can be reproduced without approximation errors. This is more difficult in nonlinear problems, and the algebra is more tedious. However, we may utilize SymPy for symbolic computing and integrate such computations in the FEniCS solver. This allows us to easily experiment with different manufactured solutions. The forthcoming code with SymPy requires some basic familiarity with this package. In particular, we will use the SymPy functions diff for symbolic differentiation and ccode for C/C++ code generation.

We try out a two-dimensional manufactured solution that is linear in the unknowns:

# Warning: from fenics import * will import both `sym` and
# `q` from FEniCS. We therefore import FEniCS first and then
# overwrite these objects.
from fenics import *

def q(u):
    """Nonlinear coefficient in the PDE."""
    return 1 + u**2

# Use SymPy to compute f given manufactured solution u
import sympy as sym
x, y = sym.symbols('x[0] x[1]')
u = 1 + x + 2*y
f = - sym.diff(q(u)*sym.diff(u, x), x) - \
      sym.diff(q(u)*sym.diff(u, y), y)
f = sym.simplify(f)
u_code = sym.printing.ccode(u)
f_code = sym.printing.ccode(f)
print('u =', u_code)
print('f =', f_code)

Turning the expressions for u and f into C or C++ syntax for FEniCS Expression objects needs two steps. First, we ask for the C code of the expressions:

u_code = sym.printing.ccode(u)
f_code = sym.printing.ccode(f)

Sometimes, we need some editing of the result to match the required syntax of Expression objects, but not in this case. (The primary example is that M_PI for \( \pi \) in C/C++ must be replaced by pi for Expression objects.) In the present case, the output of c_code and f_code is

x[0] + 2*x[1] + 1
-10*x[0] - 20*x[1] - 10

After having defined the mesh, the function space, and the boundary, we define the boundary value u_D as

u_D = Expression(u_code)

Similarly, we define the right-hand side function as

f = Expression(f_code)

FEniCS implementation

A working solver for the nonlinear Poisson equation is as easy to implement as a solver for the corresponding linear problem. All we need to do is to state the formula for \( F \) and call solve(F == 0, u, bc) instead of solve(a == L, u, bc) as we did in the linear case. Here is a minimalistic code:

from fenics import *

def q(u):
    return 1 + u**2

mesh = UnitSquareMesh(32, 32)
V = FunctionSpace(mesh, 'P', 1)
u_D = Expression(...)

def boundary(x, on_boundary):
    return on_boundary

bc = DirichletBC(V, u_D, boundary)

u = Function(V)
v = TestFunction(V)
f = Expression(...)
F = q(u)*dot(grad(u), grad(v))*dx - f*v*dx

solve(F == 0, u, bc)

The complete code can be found in the file ft06_poisson_nonlinear.py.

The major difference from a linear problem is that the unknown function u in the variational form in the nonlinear case must be defined as a Function, not as a TrialFunction. In some sense this is a simplification from the linear case where we must define u first as a TrialFunction and then as a Function.

The solve function takes the nonlinear equations, derives symbolically the Jacobian matrix, and runs a Newton method to compute the solution.

Show/hide code

# Warning: from fenics import * will import both `sym` and
# `q` from FEniCS. We therefore import FEniCS first and then
# overwrite these objects.
from fenics import *

def q(u):
    """Nonlinear coefficient in the PDE."""
    return 1 + u**2

# Use SymPy to compute f given manufactured solution u
import sympy as sym
x, y = sym.symbols('x[0] x[1]')
u = 1 + x + 2*y
f = - sym.diff(q(u)*sym.diff(u, x), x) - \
      sym.diff(q(u)*sym.diff(u, y), y)
f = sym.simplify(f)
u_code = sym.printing.ccode(u)
f_code = sym.printing.ccode(f)
print('u =', u_code)
print('f =', f_code)

# Create mesh and define function space
mesh = UnitSquareMesh(8, 8)
V = FunctionSpace(mesh, 'P', 1)

# Define boundary condition
u_D = Expression(u_code, degree=2)

def boundary(x, on_boundary):
    return on_boundary

bc = DirichletBC(V, u_D, boundary)

# Define variational problem
u = Function(V) # not TrialFunction!
v = TestFunction(V)
f = Expression(f_code, degree=2)
F = q(u)*dot(grad(u), grad(v))*dx - f*v*dx

# Compute solution
solve(F == 0, u, bc)

# Plot solution
u.rename('u', 'solution')
plot(u)

# Compute error at vertices
u_e = interpolate(u_D, V)
import numpy as np
error = np.abs(u_e.vector().array() - u.vector().array()).max()
print('error = %.3g' % error)

# Hold plot
interactive()

Running the code gives output that tells how the Newton iteration progresses. With \( 2\cdot(8\times 8) \) cells we reach convergence in 8 iterations with a tolerance of \( 10^{-9} \), and the error in the numerical solution is about \( 10^{-16} \). These results bring evidence for a correct implementation. Thinking in terms of finite differences on a uniform mesh, \( \mathcal{P}_1 \) elements mimic standard second-order differences, which compute the derivative of a linear or quadratic function exactly. Here, \( \nabla u \) is a constant vector, but then multiplied by \( (1+u^2) \), which is a second-order polynomial in \( x \) and \( y \), which the divergence "difference operator" should compute exactly. We can therefore, even with \( \mathcal{P}_1 \) elements, expect the manufactured \( u \) to be reproduced by the numerical method. With a nonlinearity like \( 1+u^4 \), this will not be the case, and we would need to verify convergence rates instead.

The current example shows how easy it is to solve a nonlinear problem in FEniCS. However, experts on the numerical solution of nonlinear PDEs know very well that automated procedures may fail in nonlinear problems, and that it is often necessary to have much better manual control of the solution process than what we have in the current case. Therefore, we return to this problem in the chapter "Implementing solvers for nonlinear PDEs": "" [25] and show how we can implement our own solution algorithms for nonlinear equations and also how we can steer the parameters in the automated Newton method used above. You will then see how easy it is to implement tailored solution strategies for nonlinear problems in FEniCS.

The equations of linear elasticity

Analysis of structures is one of the major activities of modern engineering, thus making the PDEs for deformation of elastic bodies likely the most popular PDE model in the world. It takes just one page of code to solve the equations of 2D or 3D elasticity in FEniCS, and the details follow below.

PDE problem

The equations governing small elastic deformations of a body \( \Omega \) can be written as

$$ \begin{align} -\nabla\cdot\sigma &= f\hbox{ in }\Omega, \tag{3.20}\\ \sigma &= \lambda\,\hbox{tr}\,(\varepsilon) I) + 2\mu\varepsilon, \tag{3.21}\\ \varepsilon &= \frac{1}{2}\left(\nabla u + (\nabla u)^{\top}\right), \tag{3.22} \end{align} $$ where \( \sigma \) is the stress tensor, \( f \) is the body force per unit volume, \( \lambda \) and \( \mu \) are \( \text{Lam\'e's} \) elasticity parameters for the material in \( \Omega \), \( I \) is the identity tensor, \( \mathrm{tr} \) is the trace operator on a tensor, \( \varepsilon \) is the strain tensor (symmetric gradient), and \( u \) is the displacement vector field. We have here assumed isotropic elastic conditions.

We combine (3.21) and (3.22) to obtain $$ \begin{equation} \sigma = \lambda(\nabla\cdot u)I + \mu(\nabla u + (\nabla u)^{\top})\tp \tag{3.23} \end{equation} $$ Note that (3.20)--(3.22) can easily be transformed to a single vector PDE for \( u \), which is the governing PDE for the unknown \( u \) (Navier's equation). In the derivation of the variational formulation, however, it is convenient to keep the splitting of the equations as above.

Variational formulation

The variational formulation of (3.20)--(3.22) consists of forming the inner product of (3.20) and a vector test function \( v\in \hat{V} \), where \( \hat{V} \) is a vector-valued test function space, and integrating over the domain \( \Omega \): $$ -\int_\Omega (\nabla\cdot\sigma) \cdot v \dx = \int_\Omega f\cdot v\dx\tp$$ Since \( \nabla\cdot\sigma \) contains second-order derivatives of the primary unknown \( u \), we integrate this term by parts: $$ -\int_\Omega (\nabla\cdot\sigma) \cdot v \dx = \int_\Omega \sigma : \nabla v\dx - \int_{\partial\Omega} (\sigma\cdot n)\cdot v \ds,$$ where the colon operator is the inner product between tensors (summed pairwise product of all elements), and \( n \) is the outward unit normal at the boundary. The quantity \( \sigma\cdot n \) is known as the traction or stress vector at the boundary, and is often prescribed as a boundary condition. We assume that it is prescribed at a part \( \partial\Omega_T \) of the boundary as \( \sigma\cdot n = T \). On the remaining part of the boundary, we assume that the value of the displacement is given as a Dirichlet condition. We then have $$ \int_\Omega \sigma : \nabla v \dx = \int_\Omega f\cdot v \dx + \int_{\partial\Omega_T} T\cdot v\ds\tp$$ Inserting the expression (3.23) for \( \sigma \) gives the variational form with \( u \) as unknown. Note that the boundary integral on the remaining part \( \partial\Omega\setminus\Omega_T \) vanishes due to the Dirichlet condition (\( v = 0 \)).

We can now summarize the variational formulation as: find \( u\in V \) such that $$ \begin{equation} a(u,v) = L(v)\quad\forall v\in\hat{V}, \tag{3.24} \end{equation} $$ where $$ \begin{align} a(u,v) &= \int_\Omega\sigma(u) :\nabla v \dx, \tag{3.25} \\ \sigma(u) &= \lambda(\nabla\cdot u)I + \mu(\nabla u + (\nabla u)^{\top}), \tag{3.26}\\ L(v) &= \int_\Omega f\cdot v\dx + \int_{\partial\Omega_T} T\cdot v\ds\tp \tag{3.27} \end{align} $$

One can show that the inner product of a symmetric tensor \( A \) and a anti-symmetric tensor \( B \) vanishes. If we express \( \nabla v \) as a sum of its symmetric and anti-symmetric parts, only the symmetric part will survive in the product \( \sigma :\nabla v \) since \( \sigma \) is a symmetric tensor. Thus replacing \( \nabla u \) by the symmetric gradient \( \epsilon(u) \) gives rise to the slightly different variational form $$ \begin{equation} a(u,v) = \int_\Omega\sigma(u) :\varepsilon(v) \dx, \tag{3.28} \end{equation} $$ where \( \varepsilon(v) \) is the symmetric part of \( \nabla v \): $$ \varepsilon(v) = \frac{1}{2}\left(\nabla v + (\nabla v)^{\top}\right)\tp$$ The formulation (3.28) is what naturally arises from minimization of elastic potential energy and is a more popular formulation than (3.25).

FEniCS implementation

Test problem

As a test example, we may look at a clamped beam deformed under its own weight. Then \( f=(0,0,-\varrho g) \) is the body force per unit volume with \( \varrho \) the density of the beam and \( g \) the acceleration of gravity. The beam is box-shaped with length \( L \) and has a square cross section of width \( W \). We set \( u=\ub = (0,0,0) \) at the clamped end, \( x=0 \). The rest of the boundary is traction free; that is, we set \( T = 0 \).

The code

We first list the code and then comment upon the new constructions compared to the Poisson equation case.

from fenics import *

# Scaled variables
L = 1; W = 0.2
mu = 1
rho = 1
delta = W/L
gamma = 0.4*delta**2
beta = 1.25
lambda_ = beta
g = gamma

# Create mesh and define function space
mesh = BoxMesh(Point(0, 0, 0), Point(L, W, W), 10, 3, 3)
V = VectorFunctionSpace(mesh, 'P', 1)

# Define boundary conditions
tol = 1E-14

def clamped_boundary(x, on_boundary):
    return on_boundary and x[0] < tol

bc = DirichletBC(V, Constant((0, 0, 0)), clamped_boundary)

# Define strain and stress

def epsilon(u):
    return 0.5*(nabla_grad(u) + nabla_grad(u).T)
    #return sym(nabla_grad(u))

def sigma(u):
    return lambda_*nabla_div(u)*Identity(d) + 2*mu*epsilon(u)

# Define variational problem
u = TrialFunction(V)
d = u.geometric_dimension()  # no of space dim
v = TestFunction(V)
f = Constant((0, 0, rho*g))
T = Constant((0, 0, 0))
a = inner(sigma(u), epsilon(v))*dx
L = dot(f, v)*dx + dot(T, v)*ds

# Compute solution
u = Function(V)
solve(a == L, u, bc)

# Plot solution
plot(u, title='Displacement', mode='displacement')

# Plot stress
s = sigma(u) - (1./3)*tr(sigma(u))*Identity(d)  # deviatoric stress
von_Mises = sqrt(3./2*inner(s, s))
V = FunctionSpace(mesh, 'P', 1)
von_Mises = project(von_Mises, V)
plot(von_Mises, title='Stress intensity')

# Compute magnitude of displacement
u_magnitude = sqrt(dot(u, u))
u_magnitude = project(u_magnitude, V)
plot(u_magnitude, 'Displacement magnitude')
print('min/max u:', u_magnitude.vector().array().min(),
      u_magnitude.vector().array().max())

The complete code can be found in the file ft07_elasticity.py.

We comment below on some of the key features of this example that we have not seen in previous examples.

Vector function spaces

The primary unknown is now a vector field \( u \) and not a scalar field, so we need to work with a vector function space:

V = VectorFunctionSpace(mesh, 'P', 1)

With u = Function(V) we get u as a vector-valued finite element function.

Constant vectors

In the boundary condition \( u=0 \), we must set a vector value to zero, not just a scalar, and a constant zero vector is specified as Constant((0, 0, 0)) in FEniCS. The corresponding 2D code would use Constant((0, 0)). Later in the code, we also need f as a vector and specify it as Constant((0, 0, rho*g)).

nabla_grad

The gradient and divergence operators now have a prefix nabla_. This is strictly not necessary in the present problem, but recommended in general for vector PDEs arising from continuum mechanics, if you interpret \( \nabla \) as a vector in the PDE notation; see the box about nabla_grad in the section Variational formulation.

Stress computation

As soon as u is computed, we can compute various stress measures, here the von Mises stress defined as \( \sigma_M = \sqrt{\frac{3}{2}s:s} \) where \( s \) is the deviatoric stress tensor $$ s = \sigma - \frac{1}{3}\mathrm{tr}\,(\sigma)\,I\tp$$ There is a one to one mapping between these formulas and the FEniCS code:

s = sigma(u) - (1./3)*tr(sigma(u))*Identity(d)
von_Mises = sqrt(3./2*inner(s, s))

The von_Mises variable is now an expression that must be projected to a finite element space before we can visualize it.

Scaling

Before doing simulations for a specific problem, it is often advantageous to scale the problem as it reduces the need for setting physical parameters, and one obtains dimensionsless numbers that reflect the competition of parameters and physical effects. We develop the code for the original model with dimensions, and run the scaled problem by tweaking parameters appropriately. Scaling reduces the number of active parameters from 6 to 2 in the present application.

In Navier's equation for \( u \), arising from inserting (3.21) and (3.22) in (3.20), $$ \nabla\cdot(\lambda\nabla\cdot u) + \mu\nabla^2 u = f,$$ we insert coordinates made dimensionless by \( L \), and \( \bar u=u/U \), which results in the dimensionless governing equation $$ \beta\bar\nabla\cdot(\bar\nabla\cdot \bar u) + \bar\nabla^2 \bar u = \bar f,\quad \bar f = (0,0,\gamma),$$ where \( \beta = \lambda/\mu \) is a dimensionless elasticity parameter and $$ \gamma = \frac{\varrho gL^2}{\mu U}$$ is also a dimensionless variable reflecting the ratio of the load \( \varrho g \) and the shear stress term \( \mu\nabla^2 u\sim \mu U/L^2 \) in the PDE.

Sometimes, one will argue to chose \( U \) to make \( \gamma \) unity (\( U = \varrho gL^2/\mu \)). However, in elasticity, this leads us to displacements of the size of the geometry, which makes plots look very strange. We therefore want the characteristic displacement to be a small fraction of the characteristic length of the geometry. This can be achieved by choosing \( U \) equal to the maximum deflection of a clamped beam, for which there actually exists an formula: \( U = \frac{3}{2}\varrho gL^2\delta^2/E \), where \( \delta = L/W \) is a parameter reflecting how slender the beam is, and \( E \) is the modulus of elasticity. Thus, the dimensionless parameter \( \delta \) is very important in the problem (as expected, since \( \delta\gg 1 \) is what gives beam theory!). Taking \( E \) to be of the same order as \( \mu \), which is the case for many materials, we realize that \( \gamma \sim \delta^{-2} \) is an appropriate choice. Experimenting with the code to find a displacement that "looks right" in plots of the deformed geometry, points to \( \gamma = 0.4\delta^{-2} \) as our final choice of \( \gamma \).

The simulation code implements the problem with dimensions and physical parameters \( \lambda \), \( \mu \), \( \varrho \), \( g \), \( L \), and \( W \). However, we can easily reuse this code for a scaled problem: just set \( \mu = \varrho = L = 1 \), \( W \) as \( W/L \) (\( \delta^{-1} \)), \( g=\gamma \), and \( \lambda=\beta \).

The Navier–Stokes equations

As our next example in this chapter, we will solve the incompressible Navier–Stokes equations. This problem combines many of the challenges from our previously studied problems: time-dependence, nonlinearity, and vector-valued variables. We shall touch on a number of FEniCS topics, many of them quite advanced. But you will see that even a relatively complex algorithm such as a second-order splitting method for the incompressible Navier–Stokes equations, can be implemented with relative ease in FEniCS.

PDE problem

The incompressible Navier–Stokes equations are a system of equations for the velocity \( u \) and pressure \( p \) in an incompressible fluid: $$ \begin{align} \tag{3.29} \varrho\left(\frac{\partial u}{\partial t} + u \cdot \nabla u\right) &= \nabla\cdot\sigma(u, p) + f, \\ \tag{3.30} \nabla \cdot u &= 0. \end{align} $$ The right-hand side \( f \) is a given force per unit volume and just as for the equations of linear elasticity, \( \sigma(u, p) \) denotes the stress tensor which for a Newtonian fluid is given by $$ \begin{equation} \sigma(u, p) = 2\mu\epsilon(u) - pI, \tag{3.31} \end{equation} $$ where \( \epsilon(u) \) is the strain-rate tensor $$ \epsilon(u) = \frac{1}{2}\left(\nabla u + (\nabla u)^T\right)\tp$$ The parameter \( \mu \) is the dynamic viscosity. Note that the momentum equation (3.29) is very similar to the elasticity equation (3.20). The difference is in the two additional terms \( \varrho(\partial u / \partial t + u \cdot \nabla u) \) and the different expression for the stress tensor. The two extra terms express the acceleration balanced by the force \( F = \nabla\cdot\sigma + f \) per unit volume in Newton's second law of motion.

Variational formulation

The Navier–Stokes equations are different from the time-dependent heat equation in that we need to solve a system of equations and this system is of a special type. If we apply the same technique as for the heat equation; that is, replacing the time derivative with a simple difference quotient, we obtain a nonlinear system of equations. This in itself is not a problem for FEniCS as we saw in the section A nonlinear Poisson equation, but the system has a so-called saddle point structure and requires special techniques (preconditioners and iterative methods) to be solved efficiently.

Instead, we will apply a simpler and often very efficient approach, known as a splitting method. The idea is to consider the two equations (3.29) and (3.30) separately. There exist many splitting strategies for the incompressible Navier–Stokes equations. One of the oldest is the method proposed by Chorin [26] and Temam [27], often referred to as Chorin's method. We will use a modified version of Chorin's method, the so-called incremental pressure correction scheme (IPCS) due to [28] which gives improved accuracy compared to the original scheme at little extra cost.

The IPCS scheme involves three steps. First, we compute a tentative velocity \( u^{\star} \) by advancing the momentum equation (3.29) by a midpoint finite difference scheme in time, but using the pressure \( p^{n} \) from the previous time interval. We will also linearize the nonlinear convective term by using the known velocity \( u^{n} \) from the previous time step: \( u^{n}\cdot\nabla u^{n} \). The variational problem for this first step is: $$ \begin{align} \tag{3.32} & \renni{v}{\varrho(u^{\star} - u^{n}) / \dt} + \renni{v}{\varrho u^{n} \cdot \nabla u^{n}} + \nonumber\\ & \renni{\epsilon(v)}{\sigma(u^{n+\frac{1}{2}}, p^{n})} + \renni{v}{p^{n} n}_{\partial\Omega} - \nonumber\\ &\renni{v}{\mu \nabla u^{n+\frac{1}{2}}\cdot n}_{\partial\Omega} = \renni{v}{f^{n+1}}. \tag{3.33} \end{align} $$ This notation, suitable for problems with many terms in the variational formulations, requires some explanation. First, we use the short-hand notation $$ \inner{v}{w} = \int_{\Omega} vw \dx, \quad \inner{v}{w}_{\partial\Omega} = \int_{\partial\Omega} vw \ds. $$ This allows us to express the variational problem in a more compact way. Second, we use the notation \( u^{n+\frac{1}{2}} \). This notation means the value of \( u \) at the midpoint of the interval, usually approximated by an arithmetic mean $$ u^{n+\frac{1}{2}} \approx (u^n + u^{n+1}) / 2. $$ Third, we notice that the variational problem (3.32) arises from the integration by parts of the term \( \inner{-\nabla\cdot\sigma}{v} \). Just as for the elasticity problem in the section The equations of linear elasticity, we obtain $$ \inner{-\nabla\cdot\sigma}{v} = \inner{\sigma}{\epsilon(v)} - \inner{T}{v}_{\partial\Omega}, $$ where \( T = \sigma\cdot n \) is the boundary traction. If we solve a problem with a free boundary, we can take \( T = 0 \) on the boundary. However, if we compute the flow through a channel or a pipe and want to model flow that continues into an "imaginary channel" at the outflow, we need to treat this term with some care. The assumption we then make is that the derivative of the velocity in the direction of the channel is zero at the outflow, corresponding to a flow that is "fully developed" or doesn't change significantly downstream of the outflow. Doing so, the remaining boundary term at the outflow becomes \( pn - \mu\nabla u \cdot n \), which is the term appearing in the variational problem (3.32).

We now move on to the second step in our splitting scheme for the incompressible Navier–Stokes equations. In the first step, we computed the tentative velocity \( u^{\star} \) based on the pressure from the previous time step. We may now use the computed tentative velocity to compute the new pressure \( p^n \): $$ \begin{equation} \tag{3.34} \renni{\nabla q}{\nabla p^{n+1}} = \renni{\nabla q}{\nabla p^{n}} - \dt^{-1}\renni{q}{\nabla \cdot u^{\star}}. \end{equation} $$ Note here that \( q \) is a scalar-valued test function from the pressure space, whereas the test function \( v \) in (3.32) is a vector-valued test function from the velocity space.

One way to think about this step is to subtract the Navier–Stokes momentum equation (3.29) expressed in terms of the tentative velocity \( u^{\star} \) and the pressure \( p^{n} \) from the momentum equation expressed in terms of the velocity \( u^n \) and pressure \( p^n \). This results in the equation $$ \begin{equation} \tag{3.35} (u^n - u^{\star}) / \dt + \nabla p^{n+1} - \nabla p^n = 0. \end{equation} $$ Taking the divergence and requiring that \( \nabla \cdot u^n = 0 \) by the Navier–Stokes continuity equation (3.30), we obtain the equation \( -\nabla\cdot u^{\star} / \dt + \nabla^2 p^{n+1} - \nabla^2 p^n = 0 \), which is a Poisson problem for the pressure \( p^{n+1} \) resulting in the variational problem (3.34).

Finally, we compute the corrected velocity \( u^{n+1} \) from the equation (3.35). Multiplying this equation by a test function \( v \), we obtain $$ \begin{equation} \tag{3.36} \renni{v}{u^{n+1}} = \renni{v}{u^{\star}} - \dt\renni{v}{\nabla(p^{n+1}-p^{n})}. \end{equation} $$

In summary, we may thus solve the incompressible Navier–Stokes equations efficiently by solving a sequence of three linear variational problems in each time step.

FEniCS implementation

Test problem 1: Channel flow

As a first test problem, we compute the flow between two infinite plates, so-called channel or Poiseuille flow, since this problem has a known analytical solution. Let \( H \) be the distance between the plates and \( L \) the length of the channel. There are no body forces.

We may scale the problem first to get rid of seemingly independent physical parameters. The physics of this problem is governed by viscous effects only, in the direction perpendicular to the flow, so a time scale should be based on diffusion accross the channel: \( t_c = H^2/\nu \). We let \( U \), some characteristic inflow velocity, be the velocity scale and \( H \) the spatial scale. The pressure scale is taken as the characteristic shear stress, \( \mu U/H \), since this is a primary example of shear flow. Inserting \( \bar x = x/H \), \( \bar y = y/H \), \( \bar z = z/H \), \( \bar u =u/U \), \( \bar p = Hp/(\mu U) \), and \( \bar t = H^2/\nu \) in the equations results in the scaled Navier–Stokes equations (dropping bars after the scaling):

(AL 3: This looks very strange to me. Why don't we get the standard scaled version of NS? And why is \( \mu \) still in there? And last term should be dropped since it is actually zero, just seems confusing.) (hpl 4: You are right: The two last terms are typos and left-overs from the equation with dimension. Should be no \( \mu \) in a scaled equation! Otherwise the dimension is wrong...) $$ \begin{align*} \frac{\partial u}{\partial t} + \mathrm{Re}\, u\cdot\nabla u &= -\nabla p + \nabla^2 u,\\ \nabla\cdot u &= 0\tp \end{align*} $$ Here, \( \mathrm{Re} = \varrho UH/\mu \) is the Reynolds number. Because of the time and pressure scale, which are different from convection-dominated fluid flow, the Reynolds number is associated with the convective term and not the viscosity term. Note that the last term in the first equation is zero, but we included this term as it arises naturally from the original \( \nabla\cdot\sigma \) term.

(AL 5: How can we conclude that \( \partial p / \partial x = \) constant?) (hpl 6: Explained in the text below.)

The exact solution is derived by assuming \( u=(u_x(x,y,z),0,0) \), with the \( x \) axis pointing along the channel. Since \( \nabla\cdot u=0 \), \( u \) cannot depend on \( x \). The physics of channel flow is also two-dimensional so we can omit the \( z \) coordinate (more precisely: \( \partial/\partial z=0 \)). Inserting \( u=(u_x,0,0) \) in the (scaled) governing equations gives \( u_x''(y) = \partial p/\partial x \). Differentiating this equation with respect to \( x \) shows that \( \partial^2 p/\partial^2 x =0 \) so \( \partial p/\partial x \) is a constant, here called \( -\beta \). This is the driving force of the flow and can be specified as a known parameter in the problem. Integrating \( u_x''(y)=-\beta \) over the width of the channel, \( [0,1] \), and requiring \( u=0 \) at the channel walls, results in \( u_x=\frac{1}{2}\beta y(1-y) \). The characteristic inlet flow in the channel, \( U \), can be taken as the maximum inflow at \( y=1/2 \), implying that \( \beta = 8 \). The length of the channel, \( L/H \) in the scaled model, has no impact on the result, so for simplicity we just compute on the unit square. Mathematically, the pressure must be prescribed at a point, but since \( p \) does not depend on \( y \), we can set \( p \) to a known value, e.g. zero, along the outlet boundary \( x=1 \). The result is \( p(x)=8(1-x) \) and \( u_x=4y(1-y) \).

The boundary conditions can be set as \( p=1 \) at \( x=0 \), \( p=0 \) at \( x=1 \) and \( u=0 \) on the walls \( y=0,1 \). This defines the pressure drop and should result in unit maximum velocity at the inlet and outlet and a parabolic velocity profile without further specifications. Note that it is only meaningful to solve the Navier–Stokes equations in 2D or 3D geometries, although the underlying mathematical problem collapses to two 1D problems, one for \( u_x(y) \) and one for \( p(x) \).

The scaled model is not so easy to simulate using a standard Navier–Stokes solver with dimensions. However, one can argue that the convection term is zero, so the Re coefficient in front of this term in the scaled PDEs is not important and can be set to unity. In that case, setting \( \varrho = \mu = 1 \) in the original Navier–Stokes equations resembles the scaled model.

(AL 7: One could ask why the scaling is important at all here? It is interesting to discuss for physics and analysis, but don't we always want to implement a solver using units so that one can insert specific material parameters?)

FEniCS implementation

Our previous examples have all started out with the creation of a mesh and then the definition of a FunctionSpace on the mesh. For the splitting scheme we will use to solve the Navier–Stokes equations we need to define two function spaces, one for the velocity and one for the pressure:

V = VectorFunctionSpace(mesh, 'P', 2)
Q = FunctionSpace(mesh, 'P', 1)

The first space V is a vector-valued function space for the velocity and the second space Q is a scalar-valued function space for the pressure. We use piecewise quadratic elements for the velocity and piecewise linear elements for the pressure. When creating a VectorFunctionSpace in FEniCS, the value-dimension (the length of the vectors) will be set equal to the geometric dimension of the finite element mesh. One can easily create vector-valued function spaces with other dimensions in FEniCS by adding the keyword parameter dim:

V = VectorFunctionSpace(mesh, 'P', 2, dim=10)

Since we have two different function spaces, we need to create two sets of trial and test functions:

u = TrialFunction(V)
v = TestFunction(V)
p = TrialFunction(Q)
q = TestFunction(Q)

As we have seen in previous examples, boundaries may be defined in FEniCS by defining Python functions that return True or False depending on whether a point should be considered part of the boundary, for example

def boundary(x, on_boundary):
    return near(x[0], 0)

This function defines the boundary to be all points with \( x \)-coordinate equal to (near) zero. The near function comes from FEniCS and performs a test with tolerance: abs(x[0]-0) < 3E-16 so we do not run into rounding troubles. Alternatively, we may give the boundary definition as a string of C++ code, much like we have previously defined expressions such as u0 = Expression('1 + x[0]*x[0] + 2*x[1]*x[1]'). The above definition of the boundary in terms of a Python function may thus be replaced by a simple C++ string:

boundary = 'near(x[0], 0)'

This has the advantage of moving the computation of which nodes belong to the boundary from Python to C++, which improves the efficiency of the program.

For the current example, we will set three different boundary conditions. First, we will set \( u = 0 \) at the walls of the channel; that is, at \( y = 0 \) and \( y = 1 \). Second, we will set \( p = 1 \) at the inflow (\( x = 0 \)) and, finally, \( p = 0 \) at the outflow (\( x = 1 \)). This will result in a pressure gradient that will accelerate the flow from an initial stationary state. These boundary conditions may be defined as follows:

# Define boundaries
inflow   = 'near(x[0], 0)'
outflow  = 'near(x[0], 1)'
walls    = 'near(x[1], 0) || near(x[1], 1)'

# Define boundary conditions
bcu_noslip  = DirichletBC(V, Constant((0, 0)), walls)
bcp_inflow  = DirichletBC(Q, Constant(8), inflow)
bcp_outflow = DirichletBC(Q, Constant(0), outflow)
bcu = [bcu_noslip]
bcp = [bcp_inflow, bcp_outflow]

At the end, we collect the boundary conditions for the velocity and pressure in Python lists so we can easily access them in the following computation.

We now move on to the definition of the variational forms. There are three variational problems to be defined, one for each step in the IPCS scheme. Let us look at the definition of the first variational problem. We start with some constants:

U   = 0.5*(u_n + u)
n   = FacetNormal(mesh)
f   = Constant((0, 0))
k   = Constant(dt)
mu  = Constant(mu)
rho = Constant(rho)

The next step is to set up the variational form for the first step (3.32) in the solution process. Since the variational problem contains a mix of known and unknown quantities we have introduced a naming convention to be used throughout the book: u is the unknown (mathematically \( u^{n+1} \)) as a trial function in the variational form, u_ is the most recently computed approximation (\( u^{n+1} \) available as a finite element FEniCS Function object), u_n is \( u^n \), and the same convention goes for p, p_ (\( p^{n+1} \)), and p_n (\( p^n \)).

def epsilon(u):
    return sym(nabla_grad(u))

# Define stress tensor
def sigma(u, p):
    return 2*mu*epsilon(u) - p*Identity(len(u))

# Define variational problem for step 1
F1 = rho*dot((u - u_n) / k, v)*dx + \
     rho*dot(dot(u_n, nabla_grad(u_n)), v)*dx \
   + inner(sigma(U, p_n), epsilon(v))*dx \
   + dot(p_n*n, v)*ds - dot(mu*nabla_grad(U)*n, v)*ds \
   - rho*dot(f, v)*dx
a1 = lhs(F1)
L1 = rhs(F1)

Note that we, in the definition of the variational problem, take advantage of the Python programming language to define our own operators sigma and epsilon. Using Python this way makes it easy to extend the mathematical language of FEniCS with special operators and constitutive laws.

Also note that FEniCS can sort out the bilinear form \( a(u,v) \) and linear form \( L(v) \) forms by the lhs and rhs functions. This is particularly convenient in longer and more complicated variational forms.

The splitting scheme requires the solution of a sequence of three variational problems in each time step. We have previously used the built-in FEniCS function solve to solve variational problems. Under the hood, when a user calls solve(a == L, u, bc), FEniCS will perform the following steps:

A = assemble(A)
b = assemble(L)
bc.apply(A, b)
solve(A, u.vector(), b)

In the last step, FEniCS uses the overloaded solve function to solve the linear system AU = b where U is the vector of degrees of freedom for the function \( u(x) = \sum_{j=1} U_j \phi_j(x) \).

In our implementation of the splitting scheme, we will make use of these low-level commands to first assemble and then call solve. This has the advantage that we may control when we assemble and when we solve the linear system. In particular, since the matrices for the three variational problems are all time-independent, it makes sense to assemble them once and for all outside of the time-stepping loop:

A1 = assemble(a1)
A2 = assemble(a2)
A3 = assemble(a3)

Within the time-stepping loop, we may then assemble only the right-hand side vectors, apply boundary conditions, and call the solve function as here for the first of the three steps:

# Time-stepping
t = 0
for n in range(num_steps):

    # Update current time
    t += dt

    # Step 1: Tentative velocity step
    b1 = assemble(L1)
    [bc.apply(b1) for bc in bcu]
    solve(A1, u_.vector(), b1)

Notice the Python list comprehension [bc.apply(b1) for bc in bcu] which iterates over all bc in the list bcu. This is a convenient and compact way to construct a loop that applies all boundary conditions in a single line. Also, the code works if we add more Dirichlet boundary conditions in the future.

Finally, let us look at an important detail in how we use parameters such as the time step dt in the definition of our variational problems. Since we might want to change these later, for example if we want to experiment with smaller or larger time steps, we wrap these using a FEniCS Constant:

k = Constant(dt)

The assembly of matrices and vectors in FEniCS is based on code generation. This means that whenever we change a variational problem, FEniCS will have to generate new code, which may take a little time. New code will also be generated when a float value for the time step is changed. By wrapping this parameter using Constant, FEniCS will treat the parameter as a generic constant and not a specific numerical value, which prevents repeated code generation. In the case of the time step, we choose a new name k instead of dt for the Constant since we also want to use the variable dt as a Python float as part of the time-stepping.

The complete code for simulating 2D channel flow with FEniCS looks as follows:

from fenics import *
import numpy as np

T = 10.0           # final time
num_steps = 500    # number of time steps
dt = T / num_steps # time step size
mu = 1             # kinematic viscosity
rho = 1            # density

# Create mesh and define function spaces
mesh = UnitSquareMesh(16, 16)
V = VectorFunctionSpace(mesh, 'P', 2)
Q = FunctionSpace(mesh, 'P', 1)

# Define boundaries
inflow  = 'near(x[0], 0)'
outflow = 'near(x[0], 1)'
walls   = 'near(x[1], 0) || near(x[1], 1)'

# Define boundary conditions
bcu_noslip = DirichletBC(V, Constant((0, 0)), walls)
bcp_inflow = DirichletBC(Q, Constant(8), inflow)
bcp_outflow = DirichletBC(Q, Constant(0), outflow)
bcu = [bcu_noslip]
bcp = [bcp_inflow, bcp_outflow]

# Define trial and test functions
u = TrialFunction(V)
v = TestFunction(V)
p = TrialFunction(Q)
q = TestFunction(Q)

# Define functions for solutions at previous and current time steps
u_n = Function(V)
u_  = Function(V)
p_n = Function(Q)
p_  = Function(Q)

# Define expressions used in variational forms
U   = 0.5*(u_n + u)
n   = FacetNormal(mesh)
f   = Constant((0, 0))
k   = Constant(dt)
mu  = Constant(mu)
rho = Constant(rho)

# Define strain-rate tensor
def epsilon(u):
    return sym(nabla_grad(u))

# Define stress tensor
def sigma(u, p):
    return 2*mu*epsilon(u) - p*Identity(len(u))

# Define variational problem for step 1
F1 = rho*dot((u - u_n) / k, v)*dx + \
     rho*dot(dot(u_n, nabla_grad(u_n)), v)*dx \
   + inner(sigma(U, p_n), epsilon(v))*dx \
   + dot(p_n*n, v)*ds - dot(mu*nabla_grad(U)*n, v)*ds \
   - rho*dot(f, v)*dx
a1 = lhs(F1)
L1 = rhs(F1)

# Define variational problem for step 2
a2 = dot(nabla_grad(p), nabla_grad(q))*dx
L2 = dot(nabla_grad(p_n), nabla_grad(q))*dx - (1/k)*div(u_)*q*dx

# Define variational problem for step 3
a3 = dot(u, v)*dx
L3 = dot(u_, v)*dx - k*dot(nabla_grad(p_ - p_n), v)*dx

# Assemble matrices
A1 = assemble(a1)
A2 = assemble(a2)
A3 = assemble(a3)

# Apply boundary conditions to matrices
[bc.apply(A1) for bc in bcu]
[bc.apply(A2) for bc in bcp]

# Time-stepping
t = 0
for n in range(num_steps):

    # Update current time
    t += dt

    # Step 1: Tentative velocity step
    b1 = assemble(L1)
    [bc.apply(b1) for bc in bcu]
    solve(A1, u_.vector(), b1)

    # Step 2: Pressure correction step
    b2 = assemble(L2)
    [bc.apply(b2) for bc in bcp]
    solve(A2, p_.vector(), b2)

    # Step 3: Velocity correction step
    b3 = assemble(L3)
    solve(A3, u_.vector(), b3)

    # Plot solution
    plot(u_)

    # Compute error
    u_e = Expression(('4*x[1]*(1.0 - x[1])', '0'), degree=2)
    u_e = interpolate(u_e, V)
    error = np.abs(u_e.vector().array() - u_.vector().array()).max()
    print('t = %.2f: error = %.3g' % (t, error))
    print('max u:', u_.vector().array().max())

    # Update previous solution
    u_n.assign(u_)
    p_n.assign(p_)

# Hold plot
interactive()

The complete code can be found in the file ft08_navier_stokes_channel.py.

Verification

We compute the error at the nodes as we have done before to verify that our implementation is correct. Our Navier–Stokes solver computes the solution to the time-dependent incompressible Navier–Stokes equations, starting from the initial condition \( u = (0, 0) \). We have not specified the initial condition explicitly in our solver which means that FEniCS will initialize all variables, in particular the previous and current velocities u_n and u_, to zero. Since the exact solution is quadratic, we expect the solution to be exact to within machine precision at the nodes at infinite time. For our implementation, the error quickly approaches zero and is approximately \( 10^{-6} \) at time \( T = 10 \).

Exercise 2: Simulate channel flow in a 3D geometry

FEniCS solvers typically have the number of space dimensions parameterized, so a 1D, 2D, and 3D code all look the same. We shall demonstrate what this means by extending the 2D solver navier_stokes_channel.py to a simulator where the domain is a box (the unit cube in the scaled model).

a) Set up boundary conditions for \( u \) at all points on the boundary. Set up boundary conditions for \( p \) at all points on the boundary as this is required by our Poisson equation for \( p \) (but not in the original mathematical model – there, knowing \( p \) at one point throughout time is sufficient).

Solution.

b) Modify the navier_stokes_channel.py file so it computes 3D channel flow.

Solution.

from fenics import *
import numpy as np

T = 10.0           # final time
num_steps = 500    # number of time steps
dt = T / num_steps # time step size
mu = 1             # kinematic viscosity
rho = 1            # density

# Create mesh and define function spaces
mesh = UnitCubeMesh(4, 8, 4)
V = VectorFunctionSpace(mesh, 'P', 2)
Q = FunctionSpace(mesh, 'P', 1)

# Define boundaries
inflow  = 'near(x[0], 0)'
outflow = 'near(x[0], 1)'
walls   = 'near(x[1], 0) || near(x[1], 1)'

# Define boundary conditions
bcu_noslip = DirichletBC(V, Constant((0, 0, 0)), walls)
bcp_inflow = DirichletBC(Q, Constant(8), inflow)
bcp_outflow = DirichletBC(Q, Constant(0), outflow)
bcu = [bcu_noslip]
bcp = [bcp_inflow, bcp_outflow]

# Define trial and test functions
u = TrialFunction(V)
v = TestFunction(V)
p = TrialFunction(Q)
q = TestFunction(Q)

# Define functions for solutions at previous and current time steps
u0 = Function(V)
u1 = Function(V)
p0 = Function(Q)
p1 = Function(Q)

# Define expressions used in variational forms
U   = 0.5*(u0 + u)
n   = FacetNormal(mesh)
f   = Constant((0, 0, 0))
k   = Constant(dt)
mu  = Constant(mu)
rho = Constant(rho)

# Define strain-rate tensor
def epsilon(u):
    return sym(nabla_grad(u))

# Define stress tensor
def sigma(u, p):
    return 2*mu*epsilon(u) - p*Identity(len(u))

# Define variational problem for step 1
F1 = rho*dot((u - u0) / k, v)*dx + \
     rho*dot(dot(u0, nabla_grad(u0)), v)*dx \
   + inner(sigma(U, p0), epsilon(v))*dx \
   + dot(p0*n, v)*ds - dot(mu*nabla_grad(U)*n, v)*ds \
   - rho*dot(f, v)*dx
a1 = lhs(F1)
L1 = rhs(F1)

# Define variational problem for step 2
a2 = dot(nabla_grad(p), nabla_grad(q))*dx
L2 = dot(nabla_grad(p0), nabla_grad(q))*dx - (1/k)*div(u1)*q*dx

# Define variational problem for step 3
a3 = dot(u, v)*dx
L3 = dot(u1, v)*dx - k*dot(nabla_grad(p1 - p0), v)*dx

# Assemble matrices
A1 = assemble(a1)
A2 = assemble(a2)
A3 = assemble(a3)

# Apply boundary conditions to matrices
[bc.apply(A1) for bc in bcu]
[bc.apply(A2) for bc in bcp]

# Time-stepping
t = 0
for n in xrange(num_steps):

    # Update current time
    t += dt

    # Step 1: Tentative velocity step
    b1 = assemble(L1)
    [bc.apply(b1) for bc in bcu]
    solve(A1, u1.vector(), b1)

    # Step 2: Pressure correction step
    b2 = assemble(L2)
    [bc.apply(b2) for bc in bcp]
    solve(A2, p1.vector(), b2)

    # Step 3: Velocity correction step
    b3 = assemble(L3)
    solve(A3, u1.vector(), b3)

    # Plot solution
    plot(u1)

    # Compute error
    u_e = Expression(('4*x[1]*(1.0 - x[1])', '0', '0'), degree=2)
    u_e = interpolate(u_e, V)
    error = np.abs(u_e.vector().array() - u1.vector().array()).max()
    print('t = %.2f: error = %.3g' % (t, error))
    print('max u:', u1.vector().array().max())

    # Update previous solution
    u0.assign(u1)
    p0.assign(p1)

# Hold plot
interactive()

Flow past a cylinder

We now turn our attention to a more challenging physical example: flow past a circular cylinder. The geometry and parameters are taken from problem DFG 2D-2 in the FEATFLOW/1995-DFG benchmark suite and is illustrated in Figure 10. The kinematic viscosity is given by \( \nu = 0.001 = \mu/\varrho \) and the inflow velocity profile is specified as $$ u(x, y, t) = \left(1.5 \cdot \frac{4y(1-y)}{0.41^2}, 0\right), $$ which has a maximum magnitude of \( 1.5 \) at \( y = 0.41/2 \). We do not scale anything in this benchmark since exact parameters in the case we want to simulate are known.

FEniCS implementation

So far all our domains have been simple shapes such as a unit square or a rectangular box. A number of such simple meshes may be created in FEniCS using the built-in meshes (UnitIntervalMesh, UnitSquareMesh, UnitCubeMesh, IntervalMesh, RectangleMesh, and BoxMesh). FEniCS supports the creation of more complex meshes via a technique called constructive solid geometry (CSG), which lets us define geometries in terms of simple shapes (primitives) and set operations: union, intersection, and set difference. The set operations are encoded in FEniCS using the operators + (union), * (intersection), and - (set difference). To access the CSG functionality in FEniCS, one must import the FEniCS module mshr which provides the extended meshing functionality of FEniCS.

The geometry for the cylinder flow test problem can be defined easily by first defining the rectangular channel and then subtracting the circle:

channel = Rectangle(Point(0, 0), Point(2.2, 0.41))
cylinder = Circle(Point(0.2, 0.2), 0.05)
geometry = channel - cylinder

We may then create the mesh by calling the function generate_mesh:

mesh = generate_mesh(geometry, 64)

To solve the cylinder test problem, we only need to make a few minor changes to the code we wrote for the Poiseuille flow test case. Besides defining the new mesh, the only change we need to make is to modify the boundary conditions and the time step size. The boundaries are specified as follows:

inflow   = 'near(x[0], 0)'
outflow  = 'near(x[0], 2.2)'
walls    = 'near(x[1], 0) || near(x[1], 0.41)'
cylinder = 'on_boundary && x[0]>0.1 && x[0]<0.3 && x[1]>0.1 && x[1]<0.3'

The last line may seem cryptic before you catch the idea: we want to pick out all boundary points (on_boundary) that also lie within the 2D domain \( [0.1,0.3]\times [0.1,0.3] \), see Figure 10. The only possible points are then the points on the circular boundary!

In addition to these essential changes, we will make a number of small changes to improve our solver. First, since we need to choose a relatively small time step to compute the solution (a time step that is too large will make the solution blow up) we add a progress bar so that we can follow the progress of our computation. This can be done as follows:

progress = Progress('Time-stepping')
set_log_level(PROGRESS)

t = 0.0
for n in range(num_steps):

    # Update current time
    t += dt

    # Place computation here

    # Update progress bar
    progress.update(t / T)

Since the system(s) of linear equations are significantly larger than for the simple Poiseuille flow test problem, we choose to use an iterative method instead of the default direct (sparse) solver used by FEniCS when calling solve. Efficient solution of linear systems arising from the discretization of PDEs requires the choice of both a good iterative (Krylov subspace) method and a good preconditioner. For this problem, we will simply use the biconjugate gradient stabilized method (BiCGSTAB). This can be done by adding the keyword bicgstab in the call to solve. We also add a preconditioner, ilu to further speed up the computations:

solve(A1, u1.vector(), b1, 'bicgstab', 'ilu')
solve(A2, p1.vector(), b2, 'bicgstab', 'ilu')
solve(A3, u1.vector(), b3, 'bicgstab')

Finally, to be able to postprocess the computed solution in Paraview, we store the solution to file in each time step. To avoid cluttering our working directory with a large number of solution files, we make sure to store the solution in a subdirectory:

vtkfile_u = File('navier_stokes_cylinder/velocity.pvd')
vtkfile_p = File('navier_stokes_cylinder/pressure.pvd')

Note that one does not need to create the directory before running the program. It will be created automatically by FEniCS.

We also store the solution using a FEniCS TimeSeries. This allows us to store the solution not for visualization (as when using VTK files), but for later reuse in a computation as we will see in the next section. Using a TimeSeries it is easy and efficient to read in solutions from certain points in time during a simulation. The TimeSeries class uses a binary HDF5 file for efficient storage and access to data.

Figures 11 and 12 show the velocity and pressure at final time visualized in Paraview. For the visualization of the velocity, we have used the Glyph filter to visualize the vector velocity field. For the visualization of the pressure, we have used the Warp By Scalar filter.

The complete code for the cylinder test problem looks as follows:

from fenics import *
from mshr import *
import numpy as np

T = 5.0            # final time
num_steps = 5000   # number of time steps
dt = T / num_steps # time step size
mu = 0.001         # dynamic viscosity
rho = 1            # density

# Create mesh
channel = Rectangle(Point(0, 0), Point(2.2, 0.41))
cylinder = Circle(Point(0.2, 0.2), 0.05)
geometry = channel - cylinder
mesh = generate_mesh(geometry, 64)

# Define function spaces
V = VectorFunctionSpace(mesh, 'P', 2)
Q = FunctionSpace(mesh, 'P', 1)

# Define boundaries
inflow   = 'near(x[0], 0)'
outflow  = 'near(x[0], 2.2)'
walls    = 'near(x[1], 0) || near(x[1], 0.41)'
cylinder = 'on_boundary && x[0]>0.1 && x[0]<0.3 && x[1]>0.1 && x[1]<0.3'

# Define inflow profile
inflow_profile = ('4.0*1.5*x[1]*(0.41 - x[1]) / pow(0.41, 2)', '0')

# Define boundary conditions
bcu_inflow = DirichletBC(V, Expression(inflow_profile, degree=2), inflow)
bcu_walls = DirichletBC(V, Constant((0, 0)), walls)
bcu_cylinder = DirichletBC(V, Constant((0, 0)), cylinder)
bcp_outflow = DirichletBC(Q, Constant(0), outflow)
bcu = [bcu_inflow, bcu_walls, bcu_cylinder]
bcp = [bcp_outflow]

# Define trial and test functions
u = TrialFunction(V)
v = TestFunction(V)
p = TrialFunction(Q)
q = TestFunction(Q)

# Define functions for solutions at previous and current time steps
u_n = Function(V)
u_  = Function(V)
p_n = Function(Q)
p_  = Function(Q)

# Define expressions used in variational forms
U   = 0.5*(u_n + u)
n   = FacetNormal(mesh)
f   = Constant((0, 0))
k   = Constant(dt)
mu  = Constant(mu)

# Define symmetric gradient
def epsilon(u):
    return sym(nabla_grad(u))

# Define stress tensor
def sigma(u, p):
    return 2*mu*epsilon(u) - p*Identity(len(u))

# Define variational problem for step 1
F1 = rho*dot((u - u_n) / k, v)*dx \
   + rho*dot(dot(u_n, nabla_grad(u_n)), v)*dx \
   + inner(sigma(U, p_n), epsilon(v))*dx \
   + dot(p_n*n, v)*ds - dot(mu*nabla_grad(U)*n, v)*ds \
   - rho*dot(f, v)*dx
a1 = lhs(F1)
L1 = rhs(F1)

# Define variational problem for step 2
a2 = dot(nabla_grad(p), nabla_grad(q))*dx
L2 = dot(nabla_grad(p_n), nabla_grad(q))*dx - (1/k)*div(u_)*q*dx

# Define variational problem for step 3
a3 = dot(u, v)*dx
L3 = dot(u_, v)*dx - k*dot(nabla_grad(p_ - p_n), v)*dx

# Assemble matrices
A1 = assemble(a1)
A2 = assemble(a2)
A3 = assemble(a3)

# Apply boundary conditions to matrices
[bc.apply(A1) for bc in bcu]
[bc.apply(A2) for bc in bcp]

# Create VTK files for visualization output
vtkfile_u = File('navier_stokes_cylinder/velocity.pvd')
vtkfile_p = File('navier_stokes_cylinder/pressure.pvd')

# FIXME: mpi_comm_world should not be needed here, fix in FEniCS!

# Create time series for saving solution for later
timeseries_u = TimeSeries('navier_stokes_cylinder/velocity')
timeseries_p = TimeSeries('navier_stokes_cylinder/pressure')

# Save mesh to file for later
File('cylinder.xml.gz') << mesh

# Create progress bar
progress = Progress('Time-stepping')
set_log_level(PROGRESS)

# Time-stepping
t = 0
for n in range(num_steps):

    # Update current time
    t += dt

    # Step 1: Tentative velocity step
    b1 = assemble(L1)
    [bc.apply(b1) for bc in bcu]
    solve(A1, u_.vector(), b1, 'bicgstab', 'ilu')

    # Step 2: Pressure correction step
    b2 = assemble(L2)
    [bc.apply(b2) for bc in bcp]
    solve(A2, p_.vector(), b2, 'bicgstab', 'ilu')

    # Step 3: Velocity correction step
    b3 = assemble(L3)
    solve(A3, u_.vector(), b3, 'bicgstab')

    # Plot solution
    plot(u_, title='Velocity')
    plot(p_, title='Pressure')

    # Save solution to file (VTK)
    vtkfile_u << (u_, t)
    vtkfile_p << (p_, t)

    # Save solution to file (HDF5)
    timeseries_u.store(u_.vector(), t)
    timeseries_p.store(p_.vector(), t)

    # Update previous solution
    u_n.assign(u_)
    p_n.assign(p_)

    # Update progress bar
    progress.update(t / T)
    print('u max:', u_.vector().array().max())

# Hold plot
interactive()

The complete code can be found in the file ft09_navier_stokes_cylinder.py.

A system of advection–diffusion–reaction equations

The problems we have encountered so far—with the notable exception of the Navier–Stokes equations—all share a common feature: they all involve models expressed by a single scalar or vector PDE. In many situations the model is instead expressed as a system of PDEs, describing different quantities and with possibly (very) different physics. As we saw for the Navier–Stokes equations, one way to solve a system of PDEs in FEniCS is to use a splitting method where we solve one equation at a time and feed the solution from one equation into the next. However, one of the strengths with FEniCS is the ease by which one can instead define variational problems that couple several PDEs into one compound system. In this section, we will look at how to use FEniCS to write solvers for such systems of coupled PDEs. The goal is to demonstrate how easy it is to implement fully implicit, also known as monolithic, solvers in FEniCS.

PDE problem

Our model problem is the following system of advection–diffusion–reaction equations: $$ \begin{align} \tag{3.37} \frac{\partial u_1}{\partial t} + w \cdot \nabla u_1 - \nabla\cdot(\epsilon\nabla u_1) &= f_1 -K u_1 u_2, \\ \tag{3.38} \frac{\partial u_2}{\partial t} + w \cdot \nabla u_2 - \nabla\cdot(\epsilon\nabla u_2) &= f_2 -K u_1 u_2, \\ \tag{3.39} \frac{\partial u_3}{\partial t} + w \cdot \nabla u_3 - \nabla\cdot(\epsilon\nabla u_3) &= f_3 + K u_1 u_2 - K u_3. \end{align} $$

This system models the chemical reaction between two species \( A \) and \( B \) in some domain \( \Omega \): $$ A + B \rightarrow C. $$

We assume that the equation is first-order, meaning that the reaction rate is proportional to the concentrations \( [A] \) and \( [B] \) of the two species \( A \) and \( B \): $$ \frac{\mathrm{d}}{\mathrm{d}t} [C] = K [A] [B]. $$ We also assume that the formed species \( C \) spontaneously decays with a rate proportional to the concentration \( [C] \). In the PDE system (3.37)--(3.39), we use the variables \( u_1 \), \( u_2 \), and \( u_3 \) to denote the concentrations of the three species: $$ u_1 = [A], \quad u_2 = [B], \quad u_3 = [C]. $$ We see that the chemical reactions are accounted for in the right-hand sides of the PDE system (3.37)--(3.39).

The chemical reactions take part at each point in the domain \( \Omega \). In addition, we assume that the species \( A \), \( B \), and \( C \) diffuse throughout the domain with diffusivity \( \epsilon \) (the terms \( -\nabla\cdot(\epsilon\nabla u_i) \)) and are advected with velocity \( w \) (terms like \( w\cdot\nabla u_i \)).

To make things visually and physically interesting, we shall let the chemical reaction take place in the velocity field computed from the solution of the incompressible Navier–Stokes equations around a cylinder from the previous section. In summary, we will thus be solving the following coupled system of nonlinear PDEs: $$ \begin{align} \tag{3.40} \varrho\left(\frac{\partial w}{\partial t} + w \cdot \nabla w\right) &= \nabla\cdot\sigma(w, p) + f, \\ \nabla \cdot w &= 0, \tag{3.41}\\ \frac{\partial u_1}{\partial t} + w \cdot \nabla u_1 - \nabla\cdot(\epsilon\nabla u_1) &= f_1 - K u_1 u_2, \tag{3.42}\\ \frac{\partial u_2}{\partial t} + w \cdot \nabla u_2 - \nabla\cdot(\epsilon\nabla u_2) &= f_2 - K u_1 u_2, \tag{3.43}\\ \frac{\partial u_3}{\partial t} + w \cdot \nabla u_3 - \nabla\cdot(\epsilon\nabla u_3) &= f_3 + K u_1 u_2 - K u_3. \tag{3.44} \end{align} $$ We assume that \( u_1 = u_2 = u_3 = 0 \) at \( t = 0 \) and inject the species \( A \) and \( B \) into the system by specifying nonzero source terms \( f_1 \) and \( f_2 \) close to the corners at the inflow, and take \( f_3 = 0 \). The result will be that \( A \) and \( B \) are convected by advection and diffusion throughout the channel, and when they mix the species \( C \) will be formed.

Since the system is one-way coupled from the Navier–Stokes subsystem to the advection–diffusion–reaction subsystem, we do not need to recompute the solution to the Navier–Stokes equations, but can just read back the previously computed velocity field \( w \) and feed it into our equations. But we do need to learn how to read and write solutions from time-dependent PDE problems.

Variational formulation

We obtain the variational formulation of our system by multiplying each equation by a test function, integrating the second-order terms \( -\nabla\cdot(\epsilon\nabla u_i) \) by parts, and summing up the equations. When working with FEniCS it is convenient to think of the PDE system as a vector of equations. The test functions are collected in a vector too, and the variational formulation is the inner product of the vector PDE and the vector test function.

We also need introduce some discretization in time. We will use the backward Euler method as before when we solved the heat equation and approximate the time derivatives by \( (u_i^{n+1}-u_i^n) / \dt \). Let \( v_1 \), \( v_2 \), and \( v_3 \) be the test functions, or the components of the test vector function. The inner product results in $$ \begin{align} \tag{3.45} & \int_{\Omega} (\dt^{-1} (u_1^{n+1} - u_1^n) v_1 + w \cdot \nabla u^{n+1}_1 \, v_1 + \epsilon \nabla u^{n+1}_1 \cdot \nabla v_1) \dx \\ + & \int_{\Omega} (\dt^{-1} (u_2^{n+1} - u_2^n) v_2 + w \cdot \nabla u^{n+1}_2 \, v_2 + \epsilon \nabla u^{n+1}_2 \cdot \nabla v_2) \dx \tag{3.46}\\ + & \int_{\Omega} (\dt^{-1} (u_3^{n+1} - u_3^n) v_3 + w \cdot \nabla u^{n+1}_3 \, v_3 + \epsilon \nabla u^{n+1}_3 \cdot \nabla v_3) \dx \tag{3.47}\\ - & \int_{\Omega} (f_1 v_1 + f_2 v_2 + f_3 v_3) \dx \tag{3.48}\\ - & \int_{\Omega} (-K u^{n+1}_1 u^{n+1}_2 v_1 - K u^{n+1}_1 u^{n+1}_2 v_2 + K u^{n+1}_1 u^{n+1}_2 v_3 - K u^{n+1}_3 v_3) \dx = 0. \tag{3.49} \end{align} $$ For this problem it is natural to assume homogeneous Neumann boundary conditions on the entire boundary for \( u_1 \), \( u_2 \), and \( u_3 \); that is, \( \partial u_i/\partial n = 0 \) for \( i = 1, 2, 3 \). This means that the boundary terms vanish when we integrate by parts.

FEniCS implementation

The first step is to read the mesh from file. Luckily, we made sure to save the mesh to file in the Navier–Stokes example and can now easily read it back from file:

mesh = Mesh(cylinder.xml.gz')

The mesh is stored in the native FEniCS XML format (with additional gzipping to decrease the file size).

Next, we need to define the finite element function space. For this problem, we need to define several spaces. The first space we create is the space for the velocity field \( w \) from the Navier–Stokes simulation. We call this space \( W \) and define the space by

W = VectorFunctionSpace(mesh, 'P', 2)

It is important that this space is exactly the same as the space we used for the velocity field in the Navier–Stokes solver. To read the values for the velocity field, we use a TimeSeries:

timeseries_w = TimeSeries('navier_stokes_cylinder/velocity')

This will initialize the object timeseries_w which we will call later in the time-stepping loop to retrieve values from the file velocity.h5 (in binary HDF5 format).

For the three concentrations \( u_1 \), \( u_2 \), and \( u_3 \), we want to create a mixed space with functions that represent the full system \( (u_1, u_2, u_3) \) as a single entity. To do this, we need to define a MixedElement as the product space of three simple finite elements and then used the mixed element to define the function space:

P1 = FiniteElement('P', 'triangle', 1)
element = MixedElement([P1, P1, P1])
V = FunctionSpace(mesh, element)

P2 = VectorElement('P', 'triangle', 2)
P1 = FiniteElement('P', 'triangle', 1)
TH = P2 * P1

element = P1 * P1 * P1

element = VectorElement('P', 'triangle', 1, dim=3)
V = FunctionSpace(mesh, element)

Once the space has been created, we need to define our test functions and finite element functions. Test functions for a mixed function space can be created by replacing TestFunction by TestFunctions:

v_1, v_2, v_3 = TestFunctions(V)

Since the problem is nonlinear, we need to work with functions rather than trial functions for the unknowns. This can be done by using the corresponding Functions construction in FEniCS. However, as we will need to access the Function for the entire system itself, we first need to create that function and then access its components:

u = Function(V)
u_1, u_2, u_3 = split(u)

These functions will be used to represent the unknowns \( u_1 \), \( u_2 \), and \( u_3 \) at the new time level \( n+1 \). The corresponding values at the previous time level \( n \) are denoted by u_n1, u_n2, and u_n3 in our program.

When now all functions and test functions have been defined, we can express the nonlinear variational problem (3.45):

F = ((u_1 - u_n1) / k)*v_1*dx + dot(w, grad(u_1))*v_1*dx \
  + eps*dot(grad(u_1), grad(v_1))*dx + K*u_1*u_2*v_1*dx  \
  + ((u_2 - u_n2) / k)*v_2*dx + dot(w, grad(u_2))*v_2*dx \
  + eps*dot(grad(u_2), grad(v_2))*dx + K*u_1*u_2*v_2*dx  \
  + ((u_3 - u_n3) / k)*v_3*dx + dot(w, grad(u_3))*v_3*dx \
  + eps*dot(grad(u_3), grad(v_3))*dx - K*u_1*u_2*v_3*dx + K*u_3*v_3*dx \
  - f_1*v_1*dx - f_2*v_2*dx - f_3*v_3*dx

The time-stepping simply consists of solving this variational problem in each time step by a call to the solve function:

t = 0
for n in range(num_steps):
    t += dt
    timeseries_w.retrieve(w.vector(), t)
    solve(F == 0, u)
    u_n.assign(u)

In each time step, we first read the current value for the velocity field from the time series we have previously stored. We then solve the nonlinear system, and assign the computed values to the left-hand side values for the next time interval.

The solution at the final time is shown in Figure 13. We clearly see the advection of the species \( A \) and \( B \) and the formation of \( C \) along the center of the channel where \( A \) and \( B \) meet.

The complete code is presented below.

from fenics import *

T = 5.0            # final time
num_steps = 500    # number of time steps
dt = T / num_steps # time step size
eps = 0.01         # diffusion coefficient
K = 10.0           # reaction rate

# Read mesh from file
mesh = Mesh('cylinder.xml.gz')

# Define function space for velocity
W = VectorFunctionSpace(mesh, 'P', 2)

# Define function space for system of concentrations
P1 = FiniteElement('P', 'triangle', 1)
element = MixedElement([P1, P1, P1])
V = FunctionSpace(mesh, element)

# Define test functions
v_1, v_2, v_3 = TestFunctions(V)

# Define functions for velocity and concentrations
w = Function(W)
u = Function(V)
u_n = Function(V)

# Split system functions to access components
u_1, u_2, u_3 = split(u)
u_n1, u_n2, u_n3 = split(u_n)

# Define source terms
f_1 = Expression('pow(x[0]-0.1,2)+pow(x[1]-0.1,2)<0.05*0.05 ? 0.1 : 0',
                 degree=1)
f_2 = Expression('pow(x[0]-0.1,2)+pow(x[1]-0.3,2)<0.05*0.05 ? 0.1 : 0',
                 degree=1)
f_3 = Constant(0)

# Define expressions used in variational forms
k = Constant(dt)
K = Constant(K)
eps = Constant(eps)

# Define variational problem
F = ((u_1 - u_n1) / k)*v_1*dx + dot(w, grad(u_1))*v_1*dx \
  + eps*dot(grad(u_1), grad(v_1))*dx + K*u_1*u_2*v_1*dx  \
  + ((u_2 - u_n2) / k)*v_2*dx + dot(w, grad(u_2))*v_2*dx \
  + eps*dot(grad(u_2), grad(v_2))*dx + K*u_1*u_2*v_2*dx  \
  + ((u_3 - u_n3) / k)*v_3*dx + dot(w, grad(u_3))*v_3*dx \
  + eps*dot(grad(u_3), grad(v_3))*dx - K*u_1*u_2*v_3*dx + K*u_3*v_3*dx \
  - f_1*v_1*dx - f_2*v_2*dx - f_3*v_3*dx

# Create time series for reading velocity data
timeseries_w = TimeSeries('navier_stokes_cylinder/velocity')

# Create VTK files for visualization output
vtkfile_u_1 = File('reaction_system/u_1.pvd')
vtkfile_u_2 = File('reaction_system/u_2.pvd')
vtkfile_u_3 = File('reaction_system/u_3.pvd')

# Create progress bar
progress = Progress('Time-stepping')
set_log_level(PROGRESS)

# Time-stepping
t = 0
for n in range(num_steps):

    # Update current time
    t += dt

    # Read velocity from file
    timeseries_w.retrieve(w.vector(), t)

    # Solve variational problem for time step
    solve(F == 0, u)

    # Save solution to file (VTK)
    _u_1, _u_2, _u_3 = u.split()
    vtkfile_u_1 << _u_1
    vtkfile_u_2 << _u_2
    vtkfile_u_3 << _u_3

    # Update previous solution
    u_n.assign(u)

    # Update progress bar
    progress.update(t / T)

# Hold plot
interactive()

The complete code can be found in the file ft10_reaction_system.py.

Finally, we comment on three important techniques that are very useful when working with systems of PDEs: setting initial conditions, setting boundary conditions, and extracting components of the system for plotting or postprocessing.

Setting initial conditions for mixed systems

In our example, we did not need to worry about setting an initial condition, since we start with \( u_1 = u_2 = u_3 = 0 \). This happens automatically in the code when we set u_n = Function(V). This creates a Function for the whole system and all degrees of freedom are set to zero.

If we want to set initial conditions for the components of the system separately, the easiest solution is to define the initial conditions as a vector-valued Expression and then project this to the Function representing the whole system. For example,

u_0 = Expression(('sin(x[0])', 'cos(x[0]*x[1])', 'exp(x[1])'), degree=1)
u_n = project(u_0, V)

This defines \( u_1 \), \( u_2 \), and \( u_2 \) to be the projections of \( \sin x \), \( \cos (xy) \), and \( \exp(y) \), respectively.

Setting boundary conditions for mixed systems

In our example, we also did not need to worry about setting boundary conditions since we used a natural Neumann condition. If we want to set Dirichlet conditions for individual components of the system, this can be done as usual by the class DirichletBC, but we must specify for which subsystem we set the boundary condition. For example, to specify that \( u_2 \) should be equal to \( xy \) on the boundary defined by boundary, we do

u_D = Expression('x[0]*x[1]', degree=1)
bc = DirichletBC(V.sub(1), u_D, boundary)

The object bc or a list of such objects containing different boundary conditions, can then be passed to the solve function as usual. Note that numbering starts at \( 0 \) in FEniCS so the subspace corresponding to \( u_2 \) is V.sub(1).

Accessing components of mixed systems

If u is a Function defined on a mixed function space in FEniCS, there are several ways in which u can be split into components. Above we already saw an example of the first of these:

u_1, u_2, u_3 = split(u)

This extracts the components of u as symbols that can be used in a variational problem. The above statement is in fact equivalent to

u_1 = u[0]
u_2 = u[1]
u_3 = u[2]

Note that u[0] is not really a Function object, but merely a symbolic expression, just like grad(u) in FEniCS is a symbolic expression and not a Function representing the gradient. This means that u_1, u_2, u_3 can be used in a variational problem, but cannot be used for plotting or postprocessing.

To access the components of u for plotting and saving the solution to file, we need to use a different variant of the split function:

_u_1, _u_2, _u_3 = u.split()

This returns three subfunctions as actual objects with access to the common underlying data stored in u, which makes plotting and saving to file possible. Alternatively, we can do

_u_1, _u_2, _u_3 = u.split(deepcopy=True)

which will create _u_1, _u_2, and u_3 as stand-alone Function objects, each holding a copy of the subfunction data extracted from u. This is useful in many situations but is not necessary for plotting and saving solutions to file.

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