Computational Seismology
Finite Element Method - 1D Elastic Wave Equation

This notebook is part of the supplementary material to Computational Seismology: A Practical Introduction, Oxford University Press, 2016.

Authors:

Basic Equations

This notebook presents a finite element code for the 1D elastic wave equation. Additionally, a solution using finite difference scheme is given for comparison.

The problem of solving the wave equation

\begin{equation} \rho(x) \partial_t^2 u(x,t) = \partial_x (\mu(x) \partial_x u(x,t)) + f(x,t) \end{equation}

using the finite element method is done after a series of steps performed on the above equation.

1) We first obtain a weak form of the wave equation by integrating over the entire physical domain $D$ and at the same time multiplying by some basis $\varphi_{i}$.

2) Integration by parts and implementation of the stress-free boundary condition is performed.

3) We approximate our unknown displacement field $u(x, t)$ by a sum over space-dependent basis functions $\varphi_i$ weighted by time-dependent coefficients $u_i(t)$.

\begin{equation} u(x,t) \ \approx \ \overline{u}(x,t) \ = \ \sum_{i=1}^{n} u_i(t) \ \varphi_i(x) \end{equation}

4) Utilize the same basis functions used to expand $u(x, t)$ as test functions in the weak form, this is the Galerkin principle.

5) We can turn the continuous weak form into a system of linear equations by considering the approximated displacement field.

\begin{equation} \mathbf{M}^T\partial_t^2 \mathbf{u} + \mathbf{K}^T\mathbf{u} = \mathbf{f} \end{equation}

6) For the second time-derivative, we use a standard finite-difference approximation. Finally, we arrive at the explicit time extrapolation scheme.

\begin{equation} \mathbf{u}(t + dt) = dt^2 (\mathbf{M}^T)^{-1}[\mathbf{f} - \mathbf{K}^T\mathbf{u}] + 2\mathbf{u} - \mathbf{u}(t-dt). \end{equation}

where $\mathbf{M}$ is known as the mass matrix, and $\mathbf{K}$ the stiffness matrix.

7) As interpolating functions, we choose interpolants such that $\varphi_{i}(x_{i}) = 1$ and zero elsewhere. Then, we transform the space coordinate into a local system. According to $\xi = x − x_{i}$ and $h_{i} = x_{i+1} − x_{i}$, we have:

\begin{equation} \varphi_{i}(\xi) = \begin{cases} \frac{\xi}{h_{i-1}} + 1 & \quad \text{if} \quad -h_{i-1} \le \xi \le 0\\ 1 + \frac{\xi}{h_{i}} & \quad \text{if} \quad 0 \le \xi \le h_{i}\\ 0 & \quad elsewhere\\ \end{cases} \end{equation}

with the corresponding derivatives

\begin{equation} \partial_{\xi}\varphi_{i}(\xi) = \begin{cases} \frac{1}{h_{i-1}} & \quad \text{if} \quad -h_{i-1} \le \xi \le 0\\ -\frac{1}{h_{i}} & \quad \text{if} \quad 0 \le \xi \le h_{i}\\ 0 & \quad elsewhere\\ \end{cases} \end{equation}

The figure on the left-hand side illustrates the shape of $\varphi_{i}(\xi)$ and $\partial_{\xi}\varphi_{i}(\xi)$ with varying $h$.

Code implementation starts with the initialization of a particular setup of our problem. Then, we define the source that introduces perturbations following by initialization of the mass and stiffness matrices. Finally, time extrapolation is done.

In [1]:
# Import all necessary libraries, this is a configuration step for the exercise.
# Please run it before the simulation code!
import numpy as np
import matplotlib
# Show Plot in The Notebook
matplotlib.use("nbagg")
import matplotlib.pyplot as plt
import matplotlib.animation as animation

1. Initialization of setup

In [2]:
# Initialization of setup
# ---------------------------------------------------------------
# Basic parameters
nt    = 2000    # Number of time steps
vs    = 3000    # Wave velocity [m/s] 
ro0   = 2500    # Density [kg/m^3]
nx    = 1000    # Number of grid points  
isx   = 500     # Source location [m] 
xmax  = 10000.  # Maximum length
eps   = 0.5     # Stability limit
iplot = 20      # Snapshot frequency

dx = xmax/(nx-1)           # calculate space increment
x  = np.arange(0, nx)*dx   # initialize space coordinates
x  = np.transpose(x)

h = np.diff(x)  # Element sizes [m]

# parameters
ro = x*0 + ro0
mu = x*0 + ro*vs**2

# time step from stabiity criterion
dt = 0.5*eps*dx/np.max(np.sqrt(mu/ro))
# initialize time axis
t   = np.arange(1, nt+1)*dt  

# ---------------------------------------------------------------
# Initialize fields
# ---------------------------------------------------------------
u    = np.zeros(nx)
uold = np.zeros(nx)
unew = np.zeros(nx)

p    = np.zeros(nx)
pold = np.zeros(nx)
pnew = np.zeros(nx)

2. Source time function

In 1D the propagating signal is an integral of the source time function. As we look for a Gaussian waveform, we initialize the source time function $f(t)$ using the first derivative of a Gaussian function.

\begin{equation} f(t) = -\dfrac{2}{\sigma^2}(t - t_0)e^{-\dfrac{(t - t_0)^2}{\sigma^2}} \end{equation}

Exercise 1

Initialize a source time function called 'src'. Use $\sigma = 20 dt$ as Gaussian width, and time shift $t_0 = 3\sigma$. Then, visualize the source in a given plot.

In [3]:
#################################################################
# INITIALIZE THE SOURCE TIME FUCTION HERE!
#################################################################

# Source vector
f = np.zeros(nx); f[isx:isx+1] = f[isx:isx+1] + 1.

#################################################################
# PLOT THE SOURCE TIME FUNCTION HERE!
#################################################################

Solutions:

In [4]:
# Initialization of the source time function
# ---------------------------------------------------------------
pt  = 20*dt     # Gaussian width
t0  = 3*pt      # Time shift
src = -2/pt**2 * (t-t0) * np.exp(-1/pt**2 * (t-t0)**2)

# Source vector
# ---------------------------------------------------------------
f = np.zeros(nx); f[isx:isx+1] = f[isx:isx+1] + 1.

# ---------------------------------------------------------------
# Plot source time fuction
# ---------------------------------------------------------------
plt.plot(t, src, color='b', lw=2, label='Source time function')
plt.ylabel('Amplitude', size=16)
plt.xlabel('time', size=16)
plt.legend()
plt.grid(True)
plt.show()

3. The Mass Matrix

Having implemented the desired source, now we initialize the mass and stiffness matrices. In general, the mass matrix is given

\begin{equation} M_{ij} = \int_{D} \rho \varphi_i \varphi_j dx = \int_{D_{\xi}} \rho \varphi_i \varphi_j d\xi \end{equation}

next, the defined basis are introduced and some algebraic treatment is done to arrive at the explicit form of the mass matrix

Exercise 2

Implement the mass matrix

\begin{equation} M_{ij} = \frac{\rho h}{6} \begin{pmatrix} \ddots & & & & 0\\ 1 & 4 & 1 & & \\ & 1 & 4 & 1 & \\ & & 1 & 4 & 1\\ 0 & & & & \ddots \end{pmatrix} \end{equation}

Compute the inverse mass matrix and display your result to visually inspect how it looks like

In [5]:
#################################################################
# IMPLEMENT THE MASS MATRIX HERE!
#################################################################


#################################################################
# COMPUTE THE INVERSE MASS MATRIX HERE!
#################################################################


#################################################################
# DISPLAY THE INVERSE MASS MATRIX HERE!
#################################################################

Solutions:

In [6]:
# ---------------------------------------------------------------
# Mass matrix M_ij (Eq 6.56)
# ---------------------------------------------------------------
M = np.zeros((nx,nx), dtype=float)
for i in range(1, nx-1):
    for j in range (1, nx-1):
        if j==i:
            M[i,j] = (ro[i-1]*h[i-1] + ro[i]*h[i])/3
        elif j==i+1:
            M[i,j] = ro[i]*h[i]/6
        elif j==i-1:
            M[i,j] = ro[i-1]*h[i-1]/6
        else:
            M[i,j] = 0
            
# Corner elements
M[0,0] = ro[0]*h[0]/3
M[nx-1,nx-1] = ro[nx-1]*h[nx-2]/3
# Invert M
Minv = np.linalg.inv(M)

# ---------------------------------------------------------------
# Display inverse mass matrix inv(M)
# ---------------------------------------------------------------
plt.imshow(Minv)
plt.title('Mass Matrix $\mathbf{M}$')
plt.axis("off")
plt.tight_layout()
plt.show()

4. The Stiffness matrix

On the other hand, the general form of the stiffness matrix is

\begin{equation} K_{ij} = \int_{D} \mu \partial_x\varphi_i \partial_x\varphi_j dx = \int_{D_{\xi}} \mu \partial_\xi\varphi_i \partial_\xi\varphi_j d\xi \end{equation}

at this point, the defined basis are introduced. Again, with the help of some algebraic treatment, we arrive at the explicit form of the stiffness matrix

Exercise 3

Implement the stiffness matrix

\begin{equation} K_{ij} = \frac{\mu}{h} \begin{pmatrix} \ddots & & & & 0\\ -1 & 2 & -1 & & \\ &-1 & 2 & -1 & \\ & & -1 & 2 & -1\\ 0 & & & & \ddots \end{pmatrix} \end{equation}

Display the stiffness matrix to visually inspect how it looks like

In [7]:
#################################################################
# IMPLEMENT THE STIFFNESS MATRIX HERE!
#################################################################


#################################################################
# DISPLAY THE STIFFNESS MATRIX HERE!
#################################################################

Solutions

In [8]:
# ---------------------------------------------------------------
# Stiffness matrix Kij (Eq 6.60)
# ---------------------------------------------------------------
K = np.zeros((nx,nx), dtype=float)
for i in range(1, nx-1):
    for j in range(1, nx-1):
        if i==j:
            K[i,j] = mu[i-1]/h[i-1] + mu[i]/h[i]
        elif i==j+1:
            K[i,j] = -mu[i-1]/h[i-1]
        elif i+1==j:
            K[i,j] = -mu[i]/h[i]
        else:
            K[i,j] = 0

K[0,0] = mu[0]/h[0]
K[nx-1,nx-1] = mu[nx-1]/h[nx-2]

# ---------------------------------------------------------------
# Display stiffness matrix K
# ---------------------------------------------------------------
plt.imshow(K)
plt.title('Stiffness Matrix $\mathbf{K}$')
plt.axis("off")

plt.tight_layout()
plt.show()

5. Finite differences matrices

We implement a finite difference scheme in order to compare with the finite elements solution.

Exercise 4

Implement the finite differences matrices $M$ and $D$. Where $M$ is a diagonal mass matrix containing the inverse densities, and differentiation matrix

\begin{equation} D_{ij} = \frac{\mu}{dt^2} \begin{pmatrix} -2 & 1 & & & \\ 1 & -2 & 1 & & \\ & & \ddots & & \\ & & 1 & -2 & 1\\ & & & 1 & -2 \end{pmatrix} \end{equation}

Display both matrices to visually inspect how they look like

In [9]:
#################################################################
# INITIALIZE FINITE DIFFERENCES HERE!
#################################################################


#################################################################
# DISPLAY THE DIFFERENCES MATRICES HERE!
#################################################################

Solutions:

In [10]:
# Initialize finite differences matrices (Eq 6.61)
# ---------------------------------------------------------------
Mf = np.zeros((nx,nx), dtype=float)
D  = np.zeros((nx,nx), dtype=float)
dx = h[1]

for i in range(nx):
    Mf[i,i] = 1./ro[i]
    if i>0:
        if i<nx-1:
            D[i+1,i] =1
            D[i-1,i] =1
            D[i,i] = -2
            
D = ro0 * vs**2 * D/dx**2

# ---------------------------------------------------------------
# Display differences matrices
# ---------------------------------------------------------------
fig, (ax1, ax2) = plt.subplots(1, 2)
ax1.imshow(D)
ax1.set_title('Stiffness Differential Matrix $\mathbf{D}$')
ax1.axis("off")

ax2.imshow(Mf)
ax2.set_title('Stiffness Differential Matrix $\mathbf{M_f}$')
ax2.axis("off")

plt.tight_layout()
plt.show()

6. Finite element solution

Finally we implement the finite element solution using the computed mass $M$ and stiffness $K$ matrices together with a finite differences extrapolation scheme

\begin{equation} \mathbf{u}(t + dt) = dt^2 (\mathbf{M}^T)^{-1}[\mathbf{f} - \mathbf{K}^T\mathbf{u}] + 2\mathbf{u} - \mathbf{u}(t-dt). \end{equation}
In [11]:
# Initialize animated plot
# ---------------------------------------------------------------
plt.figure(figsize=(12,4))

line1 = plt.plot(x, u, 'k', lw=1.5, label='FEM')
line2 = plt.plot(x, p, 'r', lw=1.5, label='FDM')
plt.title('Finite elements 1D Animation', fontsize=16)
plt.ylabel('Amplitude', fontsize=12)
plt.xlabel('x (m)', fontsize=12)

plt.ion()   # set interective mode
plt.show()

# ---------------------------------------------------------------
# Time extrapolation
# ---------------------------------------------------------------
for it in range(nt):
    # --------------------------------------
    # Finite Element Method
    unew = (dt**2) * Minv @ (f*src[it] - K @ u) + 2*u - uold                         
    uold, u = u, unew
    
    # --------------------------------------
    # Finite Difference Method
    pnew = (dt**2) * Mf @ (D @ p + f/dx*src[it]) + 2*p - pold
    pold, p = p, pnew
     
    # --------------------------------------   
    # Animation plot. Display both solutions
    if not it % iplot:
        for l in line1:
            l.remove()
            del l
        for l in line2:
            l.remove()
            del l
        line1 = plt.plot(x, u, 'k', lw=1.5, label='FEM')
        line2 = plt.plot(x, p, 'r', lw=1.5, label='FDM')
        plt.legend()
        plt.gcf().canvas.draw()
In [ ]: