NumGy_LV05.qmd

---
title: "Numerical Simulation Methods in Geophysics, Part 5: Timestepping"
subtitle: "1. MGPY+MGIN"
author: "thomas.guenther@geophysik.tu-freiberg.de"
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# jupyter: python3 
--- 

# Recap last lessons & exercises

## The general case

$\Delta x \ne 1$ & $a \ne 1$ $\Rightarrow$ $a \pdv{u}{x} \approx a_i\frac{u_{i+1}-u_i}{x_{i+1}-x_i}$

```{python}
#| eval: true
#| echo: false

import numpy as np
import matplotlib.pyplot as plt
import pygimli as pg
import pygimli.meshtools as mt
nx = 6
grid = mt.createGrid(nx, 2)
ax, cb = pg.show(grid)
ax.set_yticks([])
for i in range(nx):
    ax.plot(i, 0, "ko", ms=10)
for i in range(nx-1):
    ax.text(i+0.5, 0.5, rf"$a_{i}$", ha="center", va="center", fontsize=18)
ax.set_xticklabels([f"$u_{i}$" for i in range(6)])
for i in range(5):
    ax.text(i+0.5, -0.1, "$u'_{"+f"{i}"+".5}$", va="top");
ax.set_xlabel("")
ax.set_ylabel("");
```

$\dv{x}(a \pdv{u}{x}) \approx (a_i\frac{u_{i+1}-u_i}{x_{i+1}-x_i} - a_{i-1}\frac{u_{i}-u_{i-1}}{x_{i}-x_{i-1}}) / (x_{i+1}-x_{i-1}) \cdot 2$

$$ A_{i, i-1} = a_{i-1} / (x_{i}-x_{i-1}) / (x_{i+1}-x_{i-1}) \cdot 2 $$

## The coupling coefficients

$$ A_{i,i-1} = C_i^{left} = a_{i-1} / (x_{i}-x_{i-1}) / (x_{i+1}-x_{i-1}) \cdot 2 $$

$$ A_{i,i+1} = C_i^{right} = a_i / (x_{i+1}-x_{i}) / (x_{i+1}-x_{i-1}) \cdot 2 $$

$$ \begin{bmatrix}   
+1 & 0 & 0 & \ldots & &  \\
C_1^L & -(C_1^L+C_1^R) & C_1^R & 0 & \ldots & \\
\vdots & \vdots & \ddots & \vdots &  \\
\ldots & \ldots & 0 & C_N^L & -(C_N^L+C_N^R) & C_N^R \\
\ldots & \ldots & & 0 & -1 & +1
\end{bmatrix} \cdot\vb{u} = 
\begin{bmatrix} u_B\\ f_1 \\ \vdots \\ f_N \\ g_B \Delta x_N \end{bmatrix}  $$

## Symmetry

$$ A_{i+1,i} = C_{i+1}^{left} = a_i / (x_{i+1}-x_{i}) / (x_{i+2}-x_{i}) \cdot 2 $$

$$ A_{i,i+1} = C_i^{right} = a_i / (x_{i+1}-x_{i}) / (x_{i+1}-x_{i-1}) \cdot 2 $$

only symmetric if $\Delta x$ is constant around $x_i$, better take $a_i/(\Delta x_i)^2$

$$ A_{i,i-1} = C_i^{left} = a_{i-1} / (x_{i}-x_{i-1})^2 $$

$$ A_{i,i+1} = C_i^{right} = a_i / (x_{i+1}-x_{i})^2 $$

$\Rightarrow$ inaccuracies expected for non-equidistant discretization

## A closer look at the Dirichlet boundary

$$ \begin{bmatrix}   
+1 & 0 & 0 & \ldots & &  \\
C_1^L & -(C_1^L+C_1^R) & C_1^R & 0 & \ldots & 
\end{bmatrix} \begin{bmatrix} u_B\\ f_1  \end{bmatrix}$$

1. $C$ can be differently scaled from 1 $\Rightarrow$ multiply with $C_i^L$
1. Matrix is non-symmetric 
<!-- $\Rightarrow$ add $C_1^L$ -->

$$ \begin{bmatrix}   
C_1^L & 0 & 0 & \ldots & &  \\
C_1^L & -(C_1^L+C_1^R) & C_1^R & 0 & \ldots & 
\end{bmatrix} \begin{bmatrix} u_B C_i^L\\ f_1  \end{bmatrix} $$

## A closer look at the Neumann boundary

$$ \begin{bmatrix}
\ldots & \ldots & 0 & C_N^L & -(C_N^L+C_N^R) & C_N^R \\
\ldots & \ldots & & 0 & -1 & +1
\end{bmatrix} \cdot\vb{u} = 
\begin{bmatrix} f_N \\ g_B \Delta x_N \end{bmatrix}  $$

1. $C$ can be differently scaled from 1 $\Rightarrow$ multiply with $C_N^R$
1. Matrix is non-symmetric  $\Rightarrow$ multiply with -1

$$ \begin{bmatrix}
\ldots & \ldots & 0 & C_N^L & -(C_N^L+C_N^R) & C_N^R \\
\ldots & \ldots & & 0 & C_N^R & -C_N^R
\end{bmatrix} \cdot\vb{u} = 
\begin{bmatrix} f_N \\ -g_B \Delta x_N C_N^R \end{bmatrix}  $$

## Accuracy

How can we prove the accuracy of our solution?

* compare with analytical solutions
* single (Point) $f$ lead to piece-wise linear $u$ (correct)
* what about continuous source terms?<br> (e.g. radioactive elements in the Earth's crust)

## Analytical solution

$a$=1, $f(x)$=1 $\Rightarrow$ double integration $\Rightarrow$ quadratic function

$$u(x) =  C_0 + C_1 x -1/2 x^2 $$

Left BC   | Right BC  | $C_0$                 | $C_1$                
----------|-----------|-----------------------|-------------------------
Dirichlet | Dirichlet | $u_L$                 | $L/2 + (u_R-u_L)/ L$ 
Dirichlet | Neumann   | $u_L$                 | $g_R + L$     
Neumann   | Dirichlet | $u_R - g_L L + L^2/2$ | $g_L$                

with $L=(x_N-x_0)$

## Tasks stationary heat equation

* finish your implementation so that it can work with any $x$, $a$ and $f$, and at least D-D or D-N boundary conditions
* make sure you achieve the same (phenomenological) results like in the "collection notebook" (variation of, a, x, f)
* simulate the two different cases (D-D, D-N) with some choices of uL and uR or gR
* make sure the N-D case is analog to D-N
* compute analytical solution and plot it with numerical solution
* change the discretization and improve the solution

## Next spatial dimension

:::: {.columns}
::: {.column}
::: {.r-stack}
![Simple 2D conductivity grid](notebooks/grid32sigma.svg)

![Simple 2D conductivity grid with FD stencil](notebooks/grid32sigmaStencil.svg){.fragment}
:::
:::
::: {.column}
:::: {.fragment}
$$ C_{i,j}^{right} = a_{i,j-1/2} / (x_{i+1}-x_{i})^2 $$

$a_{i,j-1/2}=(a_{i,j-1}+a_{i,j})/2$ ? 

harmonic, geometric? weighting?

$a_{i,j-1/2}=\frac{a_{i,j-1}\Delta y_{j-1}+a_{i,j}\Delta y_{j}}{y_j+1-y_{j-1}}$?
::::
:::
::::

# Parabolic PDEs

## Instationary heat flow in 3D

$$ \pdv{T}{t} - \div (a\grad{T}) = \div q_s $$

* $a=\frac{k}{\rho c_p}$ [m²/s] thermal diffusivity - measure of heat transfer
* $k$ [W/m/K] thermal conductivity - measure of temperature transfer
* $c_p$ [J/kg/K] - heat capacity - measure of heat storage per mass
* $\rho$ (kg/m³) density

Water $k$=0.6 W/m/K, $\rho$=1000 kg/m³, $c$=4180 J/kg/K $\Rightarrow$ $a$=1.43e-7 m²/s

## Periodic boundary conditions

:::: {.columns}
::: {.column}
Upper boundary: daily/yearly variation

$$ T_0(t) = T_m + \Delta T \sin \omega t $$

$T_m$ mean temperature (e.g. 12°C), $\Delta T$ variation, e.g. 8°C

$\omega=2\pi/t_P$ daily ($T_d$=3600*24s) or yearly ($T_y=365 T_d$) cycle
:::
::: {.column}
```{python}
#| eval: true
#| echo: false
from math import sqrt, pi
import numpy as np
import matplotlib.pyplot as plt
```

```{python}
#| eval: true
#| echo: true
day = 3600 * 24
T0, dT = 12, 8
t = np.arange(100) / 50 * day
T = T0+dT*np.sin(t/day*2*np.pi)
plt.plot(t/day*24, T)
plt.xlabel("t (h)")
plt.ylabel("T (°C)")
plt.grid()
```
:::
::::

## Separation of variables

$$ \pdv{T}{t} = a\pdv[2]{T}{z} $$

$T(t, z)/\Delta T - T_0 = \theta(t) Z(z)$

$$ Z \pdv{\theta}{t} = a \theta \pdv[2]{Z}{z} $$

$$ \frac1\theta \pdv{\theta}{t} = C = a \frac{1}{Z}\pdv[2]{Z}{z} $$

## Solution 

regarding the BC $e^{\imath\omega t}$ leads to $C=\imath\omega$ and thus $\theta=\theta_0 e^{\imath\omega t}$

$$ \pdv[2]{Z}{z} - \imath \frac{\omega}{a} Z = \pdv[2]{Z}{z} + n^2 Z = 0  $$

Helmholtz equation with solution $Z=Z_0 e^{\imath n z}$ ($n^2=\imath\omega/a$)

$$ Z=Z_0 e^{\imath n z} = Z_0 e^{\sqrt{\imath\omega/a}z} = 
Z_0 e^{\sqrt{\omega/2a}(1+\imath)z} $$

$$ T(t, z)/\Delta T + T_0 = Z(z)\theta(t)=
Z_0\theta_0 e^{-\sqrt{\omega/2a}z} e^{i(\omega t-\sqrt{\omega/2a}z)} $$

## Interpretation

replacing the term $\sqrt{2a/\omega}=\sqrt{a t_P/\pi}=d$ leads to

$$ T(z, t) = T_0 + \Delta T e^{-z/d} \sin(\omega t - z/d) $$

* exponential damping of the temperature variation with decay depth $d$
* phase lag $z/d$ increases with depth, $z_\pi=\sqrt{2 a/\omega}\pi=\sqrt{a t_P \pi}$

1. Daily cycle: decay depth $d$=6cm, minimum depth=20cm
2. Yearly cycle: decay depth $d$=1.2m, minimum depth=4m

## Depth profiles

```{python}
#| output-location: column
#| code-line-numbers: false
#| echo: true
a = 1.5e-7
year = day*365
d = sqrt(a*year/pi)
t = np.arange(0, 1, 0.1) * year
z = np.arange(0, 6, 0.1)
fig, ax = plt.subplots()
for ti in t:
    Tz = np.exp(-z/d)*np.sin(ti*2*pi/year-
                             z/d) * dT + T0
    ax.plot(Tz, z, label="t={:.1f}".format(
        ti/year))

ax.invert_yaxis()
ax.legend()
ax.grid()
```

## Temporal behaviour
```{python}
#| output-location: column
#| code-line-numbers: false
#| echo: true
t = np.arange(0, 1.01, 0.01) * year
z = np.arange(0, 3, 0.4)
fig, ax = plt.subplots()
for zi in z:
    Tt = np.exp(-zi/d)*np.sin(t*2*pi/year-
                              zi/d) * dT + T0
    ax.plot(t/year*12, Tt, label=f"z={zi:.1f}")

ax.set_xlim(0, 12)
ax.set_xlabel("t (months)")
ax.set_ylabel("T (°C)")
ax.legend()
ax.grid()
```

<!-- ## Experimental data from North Sea beach

![](pics/temperature2022-2023.svg) -->

## Time stepping - explicit method

$$ \pdv{T}{t} - a \pdv[2]{T}{z} = 0 $$

Finite-difference approximation

$$ \pdv{T}{t}^n \approx \frac{T^{n+1}-T^n}{\Delta t} = a \pdv[2]{T^n}{z} $$

## Explicit

:::: {.columns}
::: {.column}
Start $T^0$ with initial condition (e.g. 0)

Update field by 
$$ T^{n+1} = T^n + a \pdv[2]{T^n}{z} \cdot \Delta t$$

E.g. by using the matrix $A$
:::
::: {.column}
```{python}
#| eval: true
#| echo: false

import numpy as np
import matplotlib.pyplot as plt
import pygimli as pg
import pygimli.meshtools as mt
grid = mt.createGrid(5, 4)
ax, cb = pg.show(grid)
ax.set_yticks([])
ax.plot([1, 2, 3], [1, 1, 1], "bo-", ms=12, lw=4)
ax.plot([2, 2], [1, 2], "g-", lw=4)
ax.plot(2, 2, "ro", ms=12)
```
:::
::::

## Implicit methods

:::: {.columns}
::: {.column}
$$ \pdv{T}{t}^{n+1} \approx \frac{T^{n+1}-T^n}{\Delta t} = a \pdv[2]{T}{z} ^{n+1} $$

$$ \frac{1}{\Delta t} T^{n+1} - a \pdv[2]{T}{z} ^{n+1} = \frac{1}{\Delta t} T^n $$

$$ (\vb M - \vb A) \vb u^{n+1} = \vb M \vb u^n $$ 

$\vb M$ - mass matrix
:::
::: {.column}
```{python}
#| eval: true
#| echo: false

import numpy as np
import matplotlib.pyplot as plt
import pygimli as pg
import pygimli.meshtools as mt
grid = mt.createGrid(5, 4)
ax, cb = pg.show(grid)
ax.set_yticks([])
ax.plot([1, 2, 3], [2, 2, 2], "bo-", ms=12, lw=4)
ax.plot([2, 2], [1, 2], "go-", ms=12, lw=4)
ax.plot(2, 2, "ro", ms=12)
```
:::
::::

## Mixed - Crank-Nicholson method

$$ \pdv{T}{t}^{n+1/2} \approx \frac{T^{n+1}-T^n}{\Delta t} = \frac12 a \pdv[2]{T}{z} ^n + \frac12 a \pdv[2]{T}{z} ^{n+1} $$

:::: {.columns}
::: {.column}
$$ \frac{2}{\Delta t} T^{n+1} - a \pdv[2]{T^{n+1}}{z} = 
   \frac{2}{\Delta t} T^n + a \pdv[2]{T^n}{z} $$

$$ (2\vb M - \vb A) \vb u^{n+1} = (2\vb M + \vb A) \vb u^n $$ 

<!-- $$ \pdv{T}{t}^n \approx \frac{T^{n+1}-T^n}{\Delta t} = \frac12 a \pdv[2]{T}{z} ^n + \frac12 a \pdv[2]{T}{z} ^{n+1} $$ -->
:::
::: {.column}
```{python}
#| eval: true
#| echo: false

import numpy as np
import matplotlib.pyplot as plt
import pygimli as pg
import pygimli.meshtools as mt
grid = mt.createGrid(5, 4)
ax, cb = pg.show(grid)
ax.set_yticks([])
ax.plot([1, 2, 3], [1, 1, 1], "bo-", ms=12, lw=4)
ax.plot([1, 2, 3], [2, 2, 2], "bo-", ms=12, lw=4)
ax.plot([2, 2], [1, 2], "g-", ms=12, lw=4)
ax.plot(2, 2, "ro", ms=12)
```
:::
::::

## Tasks instationary heat equation

* setup a discretization and compute the stiffness matrix $A$ for some $a$
* choose an initial condition (e.g. homogeneous)
* choose a time step $\Delta t$ and perform the explicit method using the surface temperature 
* change the spatial/temporal discretization and observe the solution
* setup mass matrix and implement the implicit method for diff. $\Delta t$
* implement the Crank-Nicholson method and compare all three
* compare the solutions with the analytical solution