A simple example of finite difference forward modeling of the two-dimensional wave equation (uniform medium, no absorbing boundary conditions and earthquake source)

Import environment configuration from FWI.yml file

from IPython.display import Latex
from IPython.display import HTML

Two-dimensional wave equation format

$u_{tt}(x,y,t)=u_{xx}(x,y,t)+u_{yy}(x,y,t),\quad (x,y)\in\Omega = (0,1)\times(0,1),t\in(0,1.4)\quad(1)$

$\sin(\pi x)\sin(\pi y),\quad u_t(x,y,0) = 0, \quad(x,y)\in(0,1)\times(0,1)\quad(2)$

$0,\quad(x,y)\in\partial \Omega\quad t\in[0,1.4]\quad(3)$

解析解的 $,t)=\cos(\sqrt2\pi t)\sin(\pi x)\sin(\pi y)$

Algorithm principle

Meshing is taken $=\triangle x = \triangle y = 0.01\quad \tau = \ triangle t = 0.1h$ Therefore, empty space and point $N=\frac {1}{h}=100$ , time sampling point $\frac{1.4}{0.1h}= 1400$ ,令 $x_i = ih,y_j =jh,t_k=k\tau,\quad i\in[0,N],j\in[0,N],k\in[0,M]$

(1), (2) and (3) in 1 can be discretized as

$u_{i,j}^{k+1} = r^2(u_{i+1,j}^{k} +u_{i-1,j}^{k} +u_{i,j+1}^{k} +u_{i,j-1}^{k} )+(2-4r^2)u_{i,j}^k -u_{i,j}^{k-1}\quad i\in[1,N-1],j\in[1,N-1],k\in[1,M-1]$ , $r=\frac{\tau}{h}=0.1$

$u_{i, j}^0 = \sin(\pi x_i)\sin(\pi y_j)=\sin(\pi ih)\sin(\pi jh), \quad i\in[0,N],j\in[ 0,N]$

$u_{i,j}^{1} = \frac{r^2}{2}(u_{i+1,j}^{0} +u_{i-1,j}^{0} +u_{i,j+1}^{0} +u_{i,j-1}^{0} )+(1-2r^2)u_{i,j}^0\quad i\in（0,N）,j\in（0,N）$

$u_{0,0}^k=u_{0,N}^k=u_{N,0}^k=u_{N,N}^k=0 \quad k\in[0,M]$

The specific derivation is as follows

discretized wave equation

By establishing a second-order finite difference scheme for the wave equation (1), we can get:

$\frac{u_{i,j}^{k+1}-2u_{i,j}^{k}+u_{i,j}^{k-1}}{ \tau^2}=\frac{u_{i+1,j}^k-2u_{i,j}^k+u_{i-1,j}^k}{h^2}+\frac{u_ {i,j+1}^k-2u_{i,j}^k+u_{i,j-1}^k}{h^2}\quad(4)$

Available after sorting:

$u_{i,j}^{k+1} = r^2(u_{i+1,j}^{k} +u_{i-1,j}^{k} +u_{i,j+1}^{k} +u_{i,j-1}^{k} )+(2-4r^2)u_{i,j}^k -u_{i,j}^{k-1}\quad(5)$

其中， $i\in[1,N-1],j\in[1,N-1],k\in[1,M-1]$ , $r=\frac{\tau}{h}=0.1$ , the local stage error is $o(h^2)$

Discretization initial value conditions

consider the conditions $\sin(\pi x)\sin(\pi y),\quad(x,y)\in(0,1)\times(0, 1)$ , differential case expression 为

$u_ {i,j}^0 = \sin(\pi x_i)\sin(\pi y_j)=\sin(\pi ih)\sin(\pi jh), \quad i\in[0,N],j \in[0,N]\quad(6)$

Considering the initial straight conditions $\quad u_t (x,y,0) = 0, \quad(x,y)\in(0,1)\times(0,1)$ ,Using two-level difference quotient approximation:

$\frac{u_{i,j}^1-u_{i,j}^{-1}}{2\tau} =0, \quad(x,y)\in(0,1)\times(0,1)\quad(7)$

Substitute k=0 into (5), then we have:

$u_{i,j}^{1} = r^2(u_{i+1,j}^{0} +u_{i-1,j}^{0} +u_{i,j+1}^{0} +u_{i,j-1}^{0} )+(2-4r^2)u_{i,j}^0 -u_{i,j}^{-1}\quad(8)$

In (7) middle $u_{i,j}^1=u_{i,j}^{-1}$ Organized:

$u_{i,j}^{1} = \frac{r^2}{2}(u_{i+1,j}^{0} +u_{i-1,j}^{0} +u_{i,j+1}^{0} +u_{i,j-1}^{0} )+(1-2r^2)u_{i,j}^0\quad(9)$

discretization boundary conditions

According to the boundary condition (3), the following differential format can be established;

$u_{0,0}^k=u_{0,N}^k=u_{N,0}^k=u_{N,N}^k=0 \quad(10)$

import numpy as np
import matplotlib.pyplot as plt
from mpl_toolkits.mplot3d import Axes3D
from matplotlib import animation

#定义计算区域
x0 = 0.0
x1 = 1.0
y0 = 0.0
y1 = 1.0
N = 100
ds = (x1 - x0)/N

t0 = 0.0
t1 = 1.4
M = 1400
dt = (t1 - t0)/M
x = np.linspace(x0, x1, N+1)
y = np.linspace(y0, y1, N+1)
t = np.linspace(t0, t1, M+1)

r = dt/ds;
#初始化计算区域
(X, Y, T) = np.meshgrid(x,y,t)
#解析解
uu = np.multiply(np.multiply(np.cos(np.sqrt(2) * np.pi * T),np.sin(np.pi * X)),np.sin(np.pi * Y))
#有限差分求解
#创建差分数组
u = np.zeros((N+1,N+1,M+1))
#加入初始条件1
for i in range(0,N+1):
    for j in range(0,N+1):
        u[i,j,0] = np.sin(np.pi*i*ds) * np.sin(np.pi*j*ds)
#加入初始条件2
for i in range(1,N):
    for j in range(1,N):
        u[i,j,1] = r**2 / 2 * (u[i+1,j,0] + u[i-1,j,0] + u[i,j+1,0] + u[i,j-1,0]) + (1 - 2*r**2) * u[i,j,0] 
#加入边界条件
for k in range(0,M+1):
    u[0,0,k] = u[0,N,k] = u[N,0,k] = u[N,N,k] = 0
#递推求[1,N-1]时刻波场
for k in range(1,M):
    for i in range(1,N):
        for j in range(1,N):
            u[i,j,k+1] = r**2*(u[i+1,j,k]+u[i-1,j,k]+u[i,j+1,k]+u[i,j-1,k]) + (2-4*r**2)*u[i,j,k] -u[i,j,k-1]

(X, Y) = np.meshgrid(x,y)
t = 100

fig = plt.figure(figsize=(20,16))
ax1 = fig.add_subplot(131,projection='3d')
surf1 = ax1.plot_surface(X, Y, uu[:,:,t], cmap='viridis')
fig.colorbar(surf1)
ax1.set_xlabel('X Label')
ax1.set_ylabel('Y Label')
ax1.set_zlabel('Z Label')
ax1.set_title(f"analysis solution at time {
      
      t*dt}")

ax2 = fig.add_subplot(132,projection='3d')
surf2 = ax2.plot_surface(X, Y, u[:,:,t], cmap='viridis')
fig.colorbar(surf2)
ax2.set_xlabel('X Label')
ax2.set_ylabel('Y Label')
ax2.set_zlabel('Z Label')
ax2.set_title(f"FD solution at time {
      
      t*dt}")

ax3 = fig.add_subplot(133,projection='3d')
surf3 = ax3.plot_surface(X, Y, uu[:,:,t]-u[:,:,t], cmap='viridis')
fig.colorbar(surf3)
ax3.set_xlabel('X Label')
ax3.set_ylabel('Y Label')
ax3.set_zlabel('Z Label')
ax3.set_title(f"error at time {
      
      t*dt}")

plt.show()

Please add image description

Generate animation:

# 创建初始数据
data = uu[:, :, 0]

# 创建图形对象
fig, ax = plt.subplots()

# 创建pcolor对象
pc = ax.pcolor(X, Y, data, cmap='rainbow')

# 定义更新函数
def update(i):
    # 生成新的数据
    new_data = u[:, :, i]
    # 更新pcolor对象
    pc.set_array(new_data.ravel())
    ax.set_title(f"{
      
      i*dt} s")
    # 返回更新后的pcolor对象
    return pc,
plt.close()
# 创建动画对象
ani = animation.FuncAnimation(fig, update, frames=range(1, M+1), interval=1, blit=True)

# 显示动画
HTML(ani.to_html5_video())

Next goal: To realize the finite difference forward modeling of the two-dimensional acoustic wave equation with constant density
The difference format is:

$\frac{1}{v(x,z)^2}\frac{\partial^2u}{\partial t^2}-(\frac{\partial^2u}{\partial x^2}+\frac{\partial^2u}{\partial y^2})=f(t)\delta（x-x_s）\delta（z-z_s）$

Naka $u (x, z, t)$ display force， $v (x, z)$ intermediate speed, $f (t)$ 是震源函数， $x_s,z_s)$ is the earthquake source location.
The equation satisfies the initial conditions

$u(x,z,t=0)=0,\frac{\partial u}{\partial t}(x,z,t=0)=0,(x,z)\in \partial \Omega$