tug/doc/ValidationHetDiff.org
2023-08-27 13:19:32 +02:00

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#+TITLE: Validation Examples for 2D Heterogeneous Diffusion
#+AUTHOR: MDL <delucia@gfz-potsdam.de>
#+DATE: 2023-08-26
#+STARTUP: inlineimages
#+LATEX_CLASS_OPTIONS: [a4paper,9pt]
#+LATEX_HEADER: \usepackage{fullpage}
#+LATEX_HEADER: \usepackage{amsmath, systeme}
#+LATEX_HEADER: \usepackage{graphicx}
#+LATEX_HEADER: \usepackage{charter}
#+OPTIONS: toc:nil
* Simple setup using deSolve/ReacTran
- Square of side 10
- Discretization: 11 \times 11 cells
- All boundaries closed
- Initial state: 0 everywhere, 1 in the center (6,6)
- Time step: 1 s, 10 iterations
The matrix of spatially variable diffusion coefficients \alpha is
constant in 4 quadrants:
\alpha_{x,y} =
| 0.01 | 0.01 | 0.01 | 0.01 | 0.01 | 0.01 | 0.001 | 0.001 | 0.001 | 0.001 | 0.001 |
| 0.01 | 0.01 | 0.01 | 0.01 | 0.01 | 0.01 | 0.001 | 0.001 | 0.001 | 0.001 | 0.001 |
| 0.01 | 0.01 | 0.01 | 0.01 | 0.01 | 0.01 | 0.001 | 0.001 | 0.001 | 0.001 | 0.001 |
| 0.01 | 0.01 | 0.01 | 0.01 | 0.01 | 0.01 | 0.001 | 0.001 | 0.001 | 0.001 | 0.001 |
| 0.01 | 0.01 | 0.01 | 0.01 | 0.01 | 0.01 | 0.001 | 0.001 | 0.001 | 0.001 | 0.001 |
| 1 | 1 | 1 | 1 | 1 | 1 | 0.1 | 0.1 | 0.1 | 0.1 | 0.1 |
| 1 | 1 | 1 | 1 | 1 | 1 | 0.1 | 0.1 | 0.1 | 0.1 | 0.1 |
| 1 | 1 | 1 | 1 | 1 | 1 | 0.1 | 0.1 | 0.1 | 0.1 | 0.1 |
| 1 | 1 | 1 | 1 | 1 | 1 | 0.1 | 0.1 | 0.1 | 0.1 | 0.1 |
| 1 | 1 | 1 | 1 | 1 | 1 | 0.1 | 0.1 | 0.1 | 0.1 | 0.1 |
| 1 | 1 | 1 | 1 | 1 | 1 | 0.1 | 0.1 | 0.1 | 0.1 | 0.1 |
The relevant part of the R script used to produce these results is
presented in listing 1; the whole script is at [[file:scripts/HetDiff.R]].
A visualization of the output of the reference simulation is given in
figure [[fig:1][1]].
Note: all results from this script are stored in the =outc= matrix by
the =deSolve= function. I stored a different version into
[[file:../scripts/gold/HetDiff1.csv]]: this file contains 11 columns (one
for each time step including initial conditions) and 121 rows, one for
each domain element, with no headers.
#+caption: Result of ReacTran/deSolve solution of the above problem at 4
#+name: fig:1
[[./images/deSolve_AlphaHet1.png]]
#+name: lst:1
#+begin_src R :language R :frame single :caption Listing 1, generate reference simulation using R packages deSolve/ReacTran :captionpos b :label lst:1
library(ReacTran)
library(deSolve)
## harmonic mean
harm <- function(x,y) {
if (length(x) != 1 || length(y) != 1)
stop("x & y have different lengths")
2/(1/x+1/y)
}
N <- 11 # number of grid cells
ini <- 1 # initial value at x=0
N2 <- ceiling(N/2)
L <- 10 # domain side
## Define diff.coeff per cell, in 4 quadrants
alphas <- matrix(0, N, N)
alphas[1:N2, 1:N2] <- 1
alphas[1:N2, seq(N2+1,N)] <- 0.1
alphas[seq(N2+1,N), 1:N2] <- 0.01
alphas[seq(N2+1,N), seq(N2+1,N)] <- 0.001
cmpharm <- function(x) {
y <- c(0, x, 0)
ret <- numeric(length(x)+1)
for (i in seq(2, length(y))) {
ret[i-1] <- harm(y[i], y[i-1])
}
ret
}
## Construction of the 2D grid
x.grid <- setup.grid.1D(x.up = 0, L = L, N = N)
y.grid <- setup.grid.1D(x.up = 0, L = L, N = N)
grid2D <- setup.grid.2D(x.grid, y.grid)
dx <- dy <- L/N
D.grid <- list()
## Diffusion coefs on x-interfaces
D.grid$x.int <- apply(alphas, 1, cmpharm)
## Diffusion coefs on y-interfaces
D.grid$y.int <- t(apply(alphas, 2, cmpharm))
# The model
Diff2Dc <- function(t, y, parms) {
CONC <- matrix(nrow = N, ncol = N, data = y)
dCONC <- tran.2D(CONC, dx = dx, dy = dy, D.grid = D.grid)$dC
return(list(dCONC))
}
## initial condition: 0 everywhere, except in central point
y <- matrix(nrow = N, ncol = N, data = 0)
y[N2, N2] <- ini # initial concentration in the central point...
## solve for 10 time units
times <- 0:10
outc <- ode.2D(y = y, func = Diff2Dc, t = times, parms = NULL,
dim = c(N, N), lrw = 1860000)
#+end_src