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02_finite-differences.qmd
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---
title: "Finite Differences"
jupyter: python3
---
## Taylor expansion
Assume the Poisson equation $$\div(a\grad u)=f$$
Taylor expansion $$f(x)=f(x_0)+f'(x0)(x-x_0)+f''(x_0)(x-x_0)^2/2$$
# The Poisson equation
## FD Approximation
:::{.callout-tip icon="false" title="Approximate derivative operators by differences"}
$$ \pdv{u}{x}\approx\frac{\Delta u}{\Delta x} $$
:::
and solution $u$ by finite values $u_i$ at points $x_i$, e.g.
$$ \dv*{u}{x}_{2.5} := (u_3-u_2) / (x_3-x_2) $$
<!-- $$ \dv*[2]{u}{x}_{3} := -->
$$\pdv[2]{u_3}{x}\approx
\frac{\dv*{u}{x}_{3.5}-\dv*{u}{x}_{2.5}}{(x_4-x_2)/2}
= \frac{(u_4-u_3)/(x_4-x_3)-(u_3-u_2)/(x_3-x_2)}{(x_4-x_2)/2} $$
## Difference stencil
Assumption: equidistant discretization $\Delta x$, conductivity 1
1st derivative: $[-1, +1] / dx$, 2nd derivative $[+1, -2, +1] / dx^2$
Matrix-Vector product $\vb{A}\cdot\vb{u}=\vb{f}$ with
$$ \vb{A} = \frac{1}{\Delta x} \begin{bmatrix}
+1 & -2 & +1 & 0 & \ldots & & \\
0 & +1 & -2 & +1 & 0 & \ldots & \\
\vdots & \vdots & \vdots & \ddots & \vdots & \\
\ldots & \ldots & 0 & +1 & -2 & +1
\end{bmatrix} $$
## FDM on the general Poisson equation
Assume the Poisson equation $$\div(a\grad u)=f$$ with a conductivity term $a$
$$ \pdv{(a \pdv*{u})}{z} = a\pdv[2]{u}{z} + \pdv{a}{z} \pdv{u}{z} $$
## Boundary conditions
Dirichlet conditions: $u_0=u_B$ (homogeneous if 0)
Neumann conditions (homogeneous if 0)
$$ \pdv*{u}{x}_0=g_B $$
Mixed boundary conditions $u_0+\alpha du_0/dx=\gamma$
### Dirichlet BC implementation way 1
$u_0 = u_B$
$$ \begin{bmatrix}
+1 & 0 & 0 & \ldots & & \\
+1 & -2 & +1 & 0 & \ldots & \\
\vdots & \vdots & \ddots & \vdots & \\
\ldots & \ldots & 0 & +1 & -2 & +1
\end{bmatrix} \cdot\vb{u} =
\begin{bmatrix} u_B\\ f_1 \\ \vdots \\ f_N \end{bmatrix} $$
### Dirichlet BC implementation way 2
$u_B - 2 u_1 + u_3 = f_1$
$$ \begin{bmatrix}
-2 & +1 & 0 & \ldots & & \\
+1 & -2 & +1 & 0 & \ldots & \\
\vdots & \vdots & \ddots & \vdots & \\
\ldots & 0 & +1 & -2 & +1
\end{bmatrix} \cdot\vb{u} =
\begin{bmatrix} f_1 - u_B\\ f_2 \\ \vdots \\ f_N \end{bmatrix} $$
### Neumann BC implementation way 1
$$ u_1 - u_0 = g_B$$
$$ \begin{bmatrix}
-1 & +1 & 0 & \ldots & & \\
+1 & -2 & +1 & 0 & \ldots & \\
\vdots & \vdots & \vdots & \ddots & \vdots \\
\ldots & \ldots & +1 & -2 & +1
\end{bmatrix} \cdot\vb{u} =
\begin{bmatrix} f_0+g_B\\ f_1 \\ \vdots \\ f_N \end{bmatrix} $$
### Neumann BC implementation way 2
$$u_0 - 2 u_1+u_2 = f_1 \qquad u_1 - u_0 = g_B \Rightarrow u_2-u_1=f_1+g_B $$
<!-- $-u_1 + u_2 = f_1 - g_B$ -->
$$ \begin{bmatrix}
-1 & +1 & 0 & \ldots & & \\
+1 & -2 & +1 & 0 & \ldots & \\
\vdots & \vdots & \ddots & \vdots & \\
\ldots & 0 & +1 & -2 & +1
\end{bmatrix} \cdot\vb{u} =
\begin{bmatrix} f_1 + g_B\\ f_2 \\ \vdots \\ f_N \end{bmatrix} $$
# Parabolic PDEs
## Heat transfer in 1D
$$ \pdv{T}{t} - a \pdv[2]{T}{z} = 0 $$
with the periodic boundary conditions:
* $T(z=0,t)=T_0 + \Delta T \sin \omega t$ (daily/yearly cycle)
* $\pdv{T}{z}(z=z_1) = 0$ (no change at depth)
and the initial condition $T(z, t=0)=\sin \pi z$ has the analytical solution
$$ T(z, t) = \Delta T e^{-\pi^2 t} \sin \pi z $$
## Explicit methods
$$ \pdv{T}{t} - a \pdv[2]{T}{z} = 0 $$
Solve Poisson equation $\div(a\grad u)=f$
for every time step $i$ (using FDM, FEM, FVM etc.)
Finite-difference step in time: update field by
$$ T_{i+1} = T_i + a \pdv[2]{u}{z} \cdot \Delta t$$
## Discretization
:::: {.columns}
::: {.column}
$$ \pdv[2]{u}{t}^{n} \approx \frac{u^{n+1}-u^n}{\Delta t} - \frac{u^{n}-u^{n-1}}{\Delta t}
$$
$$ = \frac{u^{n+1}+u^{n-1}-2 u^n}{\Delta t^2} = c^2\pdv[2]{u}{x}^n $$
$$ u^{n+1} = c^2 \Delta t^2 \pdv[2]{u}{x}^{n} + 2 u^n - u^{n-1} $$
<!-- \frac{1}{\Delta t} -->
$$ \vb M \vb u^{n+1} = (\vb A + 2\vb M) \vb u^n - \vb M \vb u^{n-1} $$
:::
::: {.column}
:::
::::
## Example: velocity distribution
```{python}
#| output-location: column-fragment
#| eval: true
#| echo: true
import numpy as np
import matplotlib.pyplot as plt
x=np.arange(0, 600.01, 0.5)
c = 1.0*np.ones_like(x) # velocity in m/s
c[100:300] = 1 + np.arange(0,0.5,0.0025)
c[900:1100] = 0.5 # low velocity zone
# Plot velocity distribution.
plt.plot(x,c,'k')
plt.xlabel('x [m]')
plt.ylabel('c [m/s]')
plt.grid()
```
## Initial displacement
Derivative of Gaussian (Ricker wavelet)
```{python}
#| output-location: column-fragment
#| eval: true
#| echo: true
l=5.0
# Initial displacement field [m].
u=(x-300.0)*np.exp(-(x-300.0)**2/l**2)
# Plot initial displacement field.
plt.plot(x,u,'k')
plt.xlabel('x [m]')
plt.ylabel('u [m]')
plt.title('initial displacement field')
plt.show()
```
## Time propagation
```{python}
#| output-location: column-fragment
#| eval: true
#| echo: true
u_last=u
dt = 0.5
ddu = np.zeros_like(u)
dx = np.diff(x)
for i in range(100):
dudx = np.diff(u)/dx
ddu[1:-1] = np.diff(dudx)/dx[:-1]
u_next = 2*u-u_last+ddu*c**2 * dt**2
u_last = u
u = u_next
plt.plot(x,u,'k')
```
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