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add. Task 1.4, 

add: task 2.1-2 (Matlab)
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Daniel Kapla 2022-05-16 01:25:09 +02:00
parent 3e34c882c0
commit 58b67a304b
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43
Exercise_03/NSSC_3.m Normal file
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%% exercise number 2
%problem characteristics
U=0.7;%convection velocity
delta_x=0.01; %space discretization
L=1; %max length of our domain
T=1; %max time considered
N=L/delta_x; %number of intervals in space
K=100; %number of intervals in time
%two initial conditions
square_pulse=@(x) heaviside(x-0.1) - heaviside(x- 0.3);
gauss_signal=@(x) exp(-10*(4*x - 1).^2);
[x,t,c] = UW_scheme(L,N,T,K,U,square_pulse);
ex_sol=@(x,t) square_pulse(x-U*t);
f1=figure(1);
for ii=1:K+1
clf(f1)
hold on
plot(x,c(:,ii)','-bo');
plot(x,ex_sol(x,t(ii)));
hold off
legend("aproximated solution","exact solution")
xlim([0 L])
ylim([0 1.1])
pause(0.02);
end
[x,t,c] = UW_scheme(L,N,T,K,U,gauss_signal);
ex_sol=@(x,t) gauss_signal(x-U*t);
f2=figure(2);
for ii=1:K+1
clf(f2)
hold on
plot(x,c(:,ii)','-bo');
plot(x,ex_sol(x,t(ii)));
legend("aproximated solution","exact solution")
xlim([0 L])
ylim([0 1.1])
pause(0.02);
end

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Exercise_03/UW_scheme.m Normal file
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function [x,t,c] = UW_scheme(xf,N,T,K,U,c0)
% ---- Numerical solution of the linear advection equation in a periodic domain ----
% c_t+ U * c_x = 0 with domain [0,xf]
% given the initial condition c0.
% -----------------------------------------------
% Sintax:
% [x,t,c] = UW_scheme(xf,N,T,K,U,c0)
%
% Input:
% xf end of our domain
% N number of space intervals
% T max time
% K number of time intervals
% U convection velocity
% c0 initial condition
%
% Output:
% x vector of spatial nodes
% t vector of time nodes
% c numerical solution
% Space and time intervals size
dx=xf/N;
dt=T/K;
% initialization of x and t vectors (nodes)
x=linspace(0,xf,N+1)';
t=linspace(0,T,K+1)';
% Solution matrix
c=zeros(N+1,K+1);
% Initial conditions
c(:,1) = c0(x);
% Creating our matrix
e = ones(N+1,1);
B = spdiags([-e,e],[-1,0],N+1,N+1);
I = speye(N+1);
%"printing" courant number and numerical viscosity
C0=(U*dt/dx)
numerical_viscosity= U*dx*(1-C0)/2
%final metrix to compute the solution
A = I - C0*B;
%finding the solution
for k=1:K
c(1:end,k+1) = A*c(1:end,k);
end

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Exercise_03/report.tex Normal file
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\documentclass[a4paper, 10pt]{article}
\usepackage[utf8]{inputenc}
\usepackage[T1]{fontenc}
\usepackage{fullpage}
\usepackage{amsmath, amssymb, amstext, amsthm}
\usepackage[pdftex]{hyperref}
\usepackage{xcolor, graphicx}
\usepackage{tikz}
\usepackage{listings}
% Document meta into
\title{NSSC 2 - Assignement 3}
\author{Bianchi Riccardo, Kapla Daniel, Kuen Jakob, Müller David}
\date{November 24, 2021}
% Set PDF title, author and creator.
\AtBeginDocument{
\hypersetup{
pdftitle = {NSSC 2 - Exercise 3},
pdfauthor = {Group 1},
pdfcreator = {\pdftexbanner}
}
}
\makeindex
% Setup environments
% Theorem, Lemma
\theoremstyle{plain}
\newtheorem{theorem}{Theorem}
\newtheorem{lemma}{Lemma}
\newtheorem{example}{Example}
% Definition
\theoremstyle{definition}
\newtheorem{defn}{Definition}
% Remark
\theoremstyle{remark}
\newtheorem{remark}{Remark}
\DeclareMathOperator*{\argmin}{{arg\,min}}
\DeclareMathOperator*{\argmax}{{arg\,max}}
\renewcommand{\t}[1]{{#1^T}}
\newcommand{\todo}[1]{{\color{red}TODO: #1}}
% Default fixed font does not support bold face
\DeclareFixedFont{\ttb}{T1}{txtt}{bx}{n}{10} % for bold
\DeclareFixedFont{\ttm}{T1}{txtt}{m}{n}{10} % for normal
% % Custom colors
\definecolor{deepblue}{rgb}{0,0,0.5}
\definecolor{deepred}{rgb}{0.6,0,0}
\definecolor{deepgreen}{rgb}{0,0.5,0}
% Python style for highlighting
\newcommand\pythonstyle{\lstset{
language=Python,
basicstyle=\ttm,
morekeywords={self}, % additional keywords
keywordstyle=\ttb\color{deepblue},
stringstyle=\color{deepgreen},
frame=, % t, b, or tb for top, bottom env. lines
showstringspaces=false,
numbers=left,
stepnumber=1,
numbersep=6pt,
numberstyle=\color{gray}\tiny
}}
% Python environment
\lstnewenvironment{python}[1][] {
\pythonstyle
\lstset{#1}
}{}
\begin{document}
\maketitle
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%% Exercise 1 %%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\section{Numerical solution of the diffusion equation in a finite domain}
We are given the 1D unsteady diffusion equation
\begin{equation}\label{eq:ex1}
\frac{\partial C}{\partial t} - D \frac{\partial^2 C}{\partial x^2} = 0
\end{equation}
with a scalar diffusion coefficient $D = 10^{-6}$. The domain size is $h$ which is discretized with $N_x$ points to give the grid space $\Delta x$ with a time step $\Delta t$. The initial condition is $C = 0$ inside the domain.
Furthermore, denote with $N_t\in\mathbb{N}$ the number of time steps of the simulation where $t_0 = 0$ is the initial time. All $N_t$ discretized time points $t_n = n \Delta t$ leading to the final time point of the simulation as $t_{N_t} = N_t \Delta t$. The discretized $x$ points of the domain are $x_i = (i - 1)\Delta x = (i - 1)h / (N_x - 1)$ for $i = 1, ..., N_x$. This gives exactly $N_x$ grid points with equal distance between adjacent points on the domain $[0, h]$ with the first point $x_1 = 0$ and the last point $x_{N_x} = h$.
In the following we will also use the short hand notation
\begin{displaymath}
C_i^n = C(x_i, t_n), \qquad
\partial_x^k C_i^n = \left.\frac{\partial^k C(x, t)}{\partial x^k}\right|_{x = x_i, t = t_n}, \qquad
\partial_t^k C_i^n = \left.\frac{\partial^k C(x, t)}{\partial t^k}\right|_{x = x_i, t = t_n}
\end{displaymath}
for $k\in\mathbb{N}$ as well as $i$ and $n$ are the space and time discretization indices, respectively.
\subsection{Explicit scheme with Dirichlet/Neumann BC}\label{sec:task01_1}
Given the Dirichlet boundary conditions $C(0, t) = 0$ and the Neumann BC $\partial_x C(h, t) = 0$. First we derive a $2^{nd}$ order \emph{explicit} finite difference approach with a $1^{st}$ order discretization in time (see Figure~\ref{fig:ex1}).
\begin{figure}[h!]
\centering
\begin{tikzpicture}[>=latex]
\begin{scope}
\draw[->] (0.8, 0) -- (5.5, 0) node[anchor = west] {$x$};
\draw[->] (1, -0.2) -- (1, 3.5) node[anchor = south] {$t$};
\draw[dashed] (0.8, 2) node[anchor = east] {$t_n$} -- (5.5, 2);
\draw[dashed] (3, -0.2) node[anchor = north] {$x_i$}
-- (3, 3.5) node[anchor = south] {explicit};
\foreach \x in {1, ..., 5} {
\foreach \y in {0, ..., 3} {
\node[circle, draw, fill = white,
inner sep = 0pt, outer sep = 0pt,
minimum size = 4pt] at (\x, \y) {};
}
}
\node[circle, fill = black,
inner sep = 0pt, outer sep = 0pt,
minimum size = 4pt] at (3, 2) {};
\foreach \x/\y in {2/2, 4/2, 3/3} {
\node[circle, fill = gray,
inner sep = 0pt, outer sep = 0pt,
minimum size = 4pt] at (\x, \y) {};
}
\end{scope}
\begin{scope}[xshift = 7cm]
\draw[->] (0.8, 0) -- (5.5, 0) node[anchor = west] {$x$};
\draw[->] (1, -0.2) -- (1, 3.5) node[anchor = south] {$t$};
\draw[dashed] (0.8, 2) node[anchor = east] {$t_n$} -- (5.5, 2);
\draw[dashed] (3, -0.2) node[anchor = north] {$x_i$}
-- (3, 3.5) node[anchor = south] {implicit};
\foreach \x in {1, ..., 5} {
\foreach \y in {0, ..., 3} {
\node[circle, draw, fill = white,
inner sep = 0pt, outer sep = 0pt,
minimum size = 4pt] at (\x, \y) {};
}
}
\node[circle, fill = black,
inner sep = 0pt, outer sep = 0pt,
minimum size = 4pt] at (3, 2) {};
\foreach \x/\y in {2/3, 4/3, 3/3} {
\node[circle, fill = gray,
inner sep = 0pt, outer sep = 0pt,
minimum size = 4pt] at (\x, \y) {};
}
\end{scope}
\end{tikzpicture}
\caption{\label{fig:ex1}Dependency relation for the \emph{explicit} (left) and \emph{implicit} (right) finite difference approximation at a grid point indexed $(i, n)$.}
\end{figure}
To derive a second order scheme (assuming $C$ is three time continuously differentiable as a function from $[0, h]\times\mathbb{R}^+\to\mathbb{R}$) in space we first considure the Taylor expantion of $C$ with respect to $x$ given by
\begin{displaymath}
C(x + \Delta x, t) = C(x, t) + \partial_x C(x, t)\Delta x + \partial_x^2 C(x, t)\frac{\Delta x^2}{2} + \partial_x^3 C(x, t)\frac{\Delta x^3}{6} + \mathcal{O}(\Delta x^4).
\end{displaymath}
Replace $\Delta x$ with $\pm\Delta x$ and add the two equation together, then
\begin{displaymath}
C(x + \Delta x, t) + C(x - \Delta x, t) = 2 C(x, t) + \partial_x^2 C(x, t)\Delta x^2 + \mathcal{O}(\Delta x^4).
\end{displaymath}
The first and third order terms drop out due to different signs. Finally, this results in
\begin{displaymath}
\partial_x^2 C(x, t) = \frac{C(x + \Delta x, t) - 2 C(x, t) + C(x - \Delta x, t)}{\Delta x^2} + \mathcal{O}(\Delta x^2)
\end{displaymath}
which is an approximation of the second derivative with error proportional to $\Delta x^2$. For the first order approximation in time the same trick can be used except that only the first two terms of the Taylor expantion need to be considured.
Therefore, the $2^{nd}$ order scheme for space and $1^{st}$ order in time at $(i, n)$ using the explicit scheme derives as
\begin{displaymath}
\partial_x^2 \widehat{C}_i^n = \frac{\widehat{C}_{i-1}^{n} - 2\widehat{C}_{i}^{n} + \widehat{C}_{i+1}^{n}}{\Delta x^2},
\qquad
\partial_t \widehat{C}_i^n = \frac{\widehat{C}_{i}^{n+1} - \widehat{C}_{i}^{n}}{\Delta t}.
\end{displaymath}
Substitution into \eqref{eq:ex1} yields after rearranging the update rule for internal points as
\begin{equation}\label{eq:task01_1_update}
\widehat{C}_i^{n+1} = \widehat{C}_i^n + \frac{D\Delta t}{\Delta x^2}(\widehat{C}_{i-1}^{n} - 2\widehat{C}_{i}^{n} + \widehat{C}_{i+1}^{n}).
\end{equation}
This holds at $x_i$ where $i = 2, ..., N_x - 1$, or in other words everywhere inside the space domain excluding the boundary. The boundary needs to be handled seperately. The left boundary condition is a Dirichlet constraint which is simply constanct giving $C_1^n = 0$ for all time while the Neumann condition on the right requires an additional discretization step for computing the next value. Therefore, we take the second order Taylor expantion of $C$ with respect to $x$ in the negative direction given by
\begin{displaymath}
C(x - \Delta x, t) = C(x, t) - \partial_x C(x, t)\Delta x + \mathcal{O}(\Delta x^2).
\end{displaymath}
As we are interesetd in the boundary value for the Neumann BC, the derivative at $x = h$ is given which leads to
\begin{displaymath}
C(h, t) = C(h - \Delta x, t) + \partial_x C(h, t)\Delta x + \mathcal{O}(\Delta x^2)
\end{displaymath}
which is second order accurate. Therefore, the right boundary value for our boundary condition $\partial_x C(h, t_n) = 0$ at time $t_n$ is
\begin{displaymath}
\widehat{C}_{N_x}^n = \widehat{C}_{N_x - 1}^n.
\end{displaymath}
See \texttt{task01\_1-2.py} or \texttt{Mixed\_EA.m} for an implementation of this scheme.
\subsubsection{Stability} \todo{Prove that the update scheme is unstable for $d = \frac{D\Delta x^2}{\Delta t} > 0.5$}
\subsection{Explicit scheme with Dirichlet BC at both boundaries}\label{sec:task01_2}
Not we have the same setting as in Section~\ref{sec:task01_1} except for Dirichlet boundary conditions on both sides given by $C(0, t) = 1$ and $C(h, t) = 0$. This means tha the left and right values are constants and internal nodes follow the same update scheme as in \eqref{eq:task01_1_update}.
\todo{include pictures and compare with Section~\ref{sec:task01_1}}
\subsection{Implicit scheme with Dirichlet/Neumann BC}\label{sec:task01_3}
The following computations are similar in nature to the explicit scheme, therefore well keep it short. The implicit discretization for the derivatives is (see: Figure~\ref{fig:ex1}) then
\begin{displaymath}
\partial_x^2 C_{i}^{n+1} = \frac{\widehat{C}_{i-1}^{n+1} - 2\widehat{C}_{i}^{n+1} + \widehat{C}_{i+1}^{n+1}}{\Delta x^2},
\qquad
\partial_t \widehat{C}_i^n = \frac{\widehat{C}_{i}^{n+1} - \widehat{C}_{i}^{n}}{\Delta t}.
\end{displaymath}
Substitution into \eqref{eq:ex1} gives the implicit (inverse) update rule
\begin{displaymath}
\widehat{C}_i^{n} = -d \widehat{C}_{i-1}^{n+1} + (1 + 2 d) \widehat{C}_i^{n+1} - d \widehat{C}_{i+1}^{n+1}
\end{displaymath}
for $i = 2, ..., N_x - 1$ and $d = \frac{D \Delta x^2}{\Delta t}$. The boundary conditions are ether $\widehat{C}_1^{n+1} = 0$ for the left Dirichlet and $\widehat{C}_{N_x}^{n+1} = \widehat{C}_{N_x-1}^{n+1}$ as the right Neumann BC. By collecting all coefficients of the $i = 2, ..., N_x - 1$ into a $(N_x - 2)\times (N_x - 2)$ trigiagonal matrix
\begin{displaymath}
A = \begin{pmatrix}
1+2d & -d & \\
-d & 1+2d & -d & \\
& -d & 1+2d & \ddots \\
& & \ddots & \ddots & -d \\
& & & -d & 1+2d & -d \\
& & & & -d & 1+d
\end{pmatrix}
\end{displaymath}
\todo{fix $1+d$ which is first order accurate!!!}
where the last entry $A_{N_x, N_x} = 1 - d$ is due to the Neumann BC $\widehat{C}_{N_x}^{n+1} = \widehat{C}_{N_x-1}^{n+1}$ which means
\begin{displaymath}
\widehat{C}_i^{N_x-1} = -d \widehat{C}_i^{N_x-2} + (1 + 2 d) \widehat{C}_i^{N_x - 1} - d \widehat{C}_i^{N_x}
= -d \widehat{C}_i^{N_x-2} + (1 + d) \widehat{C}_i^{N_x - 1}.
\end{displaymath}
Finaly, we end up with the implicit update rule
\begin{displaymath}
\widehat{C}^{n+1} = A \widehat{C}^{n}
\end{displaymath}
which is perfomed by solving the linear system for $\widehat{C}^{n+1}$.
\todo{include pictures and compare with Section~\ref{sec:task01_1}}
\subsection{Second order in time for implicit scheme with Dirichlet/Neumann BC}
To derive the second order accurate scheme in time we employ the Crank-Nicolson approach to derive the descritization at the time midpoints $t_{n+1/2} = (n + 1/2)\Delta t$ which we index with $n+1/2$.
The second order im space is identical except the evaluation points
\begin{displaymath}
\partial_x^2 C_i^{n+1/2} = \frac{C_{i-1}^{n+1/2} - 2C_{i}^{n+1/2} + C_{i+1}^{n+1/2}}{\Delta x^2} + \mathcal{O}(\Delta x^2)
\end{displaymath}
and for the time descritization we take the Taylor expantion
\begin{displaymath}
\partial_{t}C_i^{n+1/2} = \frac{C_i^{n+1} - C_i^{n}}{2\Delta t} + \mathcal{O}(\Delta t^2)
\end{displaymath}
\todo{check if this follows from $\exists \partial_x^4 C$ beeing continuous}
\begin{displaymath}
\partial_x^2 C_i^{n+1/2} = \frac{\partial_x^2 C_i^{n} + \partial_x^2 C_i^{n+1}}{2} + \mathcal{O}(\Delta x^2)
\end{displaymath}
Meaning
\begin{align*}
0 = \partial_{t}C_i^{n+1/2}-D\partial_x^2 C_i^{n+1/2} &= \frac{C_i^{n+1} - C_i^{n}}{2\Delta t}
-D\frac{\partial_x^2 C_i^{n} + \partial_x^2 C_i^{n+1}}{2} + \mathcal{O}(\Delta t^2 + ?) \\
&= \frac{C_i^{n+1} - C_i^{n}}{2\Delta t} -
D\frac{C_{i-1}^{n} + C_{i-1}^{n+1} - 2(C_{i}^{n} + C_{i}^{n+1}) + C_{i+1}^{n} + C_{i+1}^{n+1}}{2\Delta x^2}
\end{align*}
rearranging yields
\begin{displaymath}
-d C_{i-1}^{n+1} + (1 + 2d) C_i^{n+1}-dC_{i+1}^{n+1}
= d C_{i-1}^n + (1 - 2d)C_i^n + dC_{i+1}^n
= -(-d C_{i-1}^n + (1 + 2d)C_i^n - dC_{i+1}^n) + 2C_i^n
\end{displaymath}
Now with second order approximation of the Neumann BC
\begin{displaymath}
0\overset{!}{=} \partial_x C_{N_x}^{n+1} = \frac{C_{N_x-2}^{n+1} - C_{N_x}^{n+1}}{2\Delta x} + \mathcal{O}(\Delta x^2)
\end{displaymath}
which means that (\todo{check why?! should be at $N_x - 1$}) for index $i = N_x - 1$
\begin{align*}
-d C_{N_x-2}^{n+1} + (1 + 2d) C_{N_x-1}^{n+1}-dC_{N_x}^{n+1}
&= -(-d C_{N_x-2}^n + (1 + 2d)C_{N_x-1}^n - dC_{N_x}^n) + 2C_{N_x-1}^n \\
-d C_{N_x-2}^{n+1} + (1 + 2d) C_{N_x-1}^{n+1}-d C_{N_x-2}^{n+1}
&= -(-d C_{N_x-2}^n + (1 + 2d)C_{N_x-1}^n - d C_{N_x-2}^{n}) + 2C_{N_x-1}^n \\
-2d C_{N_x-2}^{n+1} + (1 + 2d) C_{N_x-1}^{n+1}
&= -(-2d C_{N_x-2}^n + (1 + 2d)C_{N_x-1}^n) + 2C_{N_x-1}^n
\end{align*}
\begin{displaymath}
A C^{n+1} = (2 I - A) C^n
\end{displaymath}
with
\begin{displaymath}
A = \begin{pmatrix}
1 & 0 & 0 \\
-d & 1+2d & -d \\
& -d & 1+2d & -d \\
& & -d & 1+2d & \ddots \\
& & & \ddots & \ddots & -d \\
& & & & -d & 1+2d & -d \\
& & & & & -2d & 1+2d
\end{pmatrix}
\end{displaymath}
\end{document}

2
Exercise_03/task01_3.py
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@ -39,7 +39,7 @@ for t in range(nt):
plt.ylim([0, 1.2]) plt.ylim([0, 1.2])
plt.savefig(f"plots/task01_3_{i:0>5}.png") plt.savefig(f"plots/task01_3_{i:0>5}.png")
i += 1 i += 1
# update solution using the explicit schema # update solution using the implicit schema
C = np.linalg.solve(T, C) C = np.linalg.solve(T, C)
# fix BC conditions (theoretically, they are set by the update but for # fix BC conditions (theoretically, they are set by the update but for
# stability reasons (numerical) we enforce the correct values) # stability reasons (numerical) we enforce the correct values)

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# Task 1.4
import numpy as np
import scipy, scipy.sparse, scipy.linalg
from matplotlib import pyplot as plt
# plotting helper function
def plot(C, x, t, name):
plt.clf()
plt.plot(x, C)
plt.xlim([0, x[-1]])
plt.ylim([0, 1.2])
plt.title(f"{name} - time: {t: >12.0f}")
plt.savefig(f"plots/{name}_{plot.fignr:0>5}.png")
plot.fignr += 1
# set static variable as function attribute
plot.fignr = 1
# and initialize a figure
plt.figure(figsize = (8, 6), dpi = 100)
# Config
D = 1e-6 # diffusion coefficient
h = 1 # space domain (max x size)
T = 1e5 # solution end time
nx = 50 # nr of space discretization points
nt = 2000 # nr of time points
# derived constants
dx = h / (nx - 1) # space step size
dt = T / (nt - 1) # time step size
d = dt * D / dx**2 # stability/stepsize coefficient
# setup matrices for implicit update scheme `A C^n+1 = (2 I - A) C^n = B C^n`
# The matrix `A` is represented as a `3 x (nx - 1)` matrix as expected by
# `scipy.linalg.solve_banded` representing the 3 diagonals of the tridiagonal
# matrix `A`.
A = np.zeros((3, nx - 1))
# upper minor diagonal
A[0, 2:] = -d
# main diagonal
A[1, 0] = 1
A[1, 1:] = 1 + 2 * d
# and lower minor diagonal
A[2, :-2] = -d
A[2, -2] = -2 * d
# The right hand side matrix `B = 2 I - A` which is represented as a sparse
# matrix in diag. layout which allows for fast matrix vector multiplication
B = -A
B[1, ] += 2
B = scipy.sparse.spdiags(B, (1, 0, -1), B.shape[1], B.shape[1])
# set descritized space evaluation points `x` of `C`
x = np.linspace(0, h, nx)
# Set initial solution
C = np.zeros(nx)
C[0] = 1
C[-1] = C[-3] # (0 = 0)
for n in range(nt):
# generate roughly 400 plots
if n % (nt // 400) == 0:
plot(C, x, n * dt, "task01_4")
# update solution using the implicit schema
C[:-1] = scipy.linalg.solve_banded((1, 1), A, B @ C[:-1])
# set/enforce boundary conditions
C[0] = 1 # left Dirichlet (theoretically not needed)
C[-1] = C[-3] # right Neumann condition
# plot final state
plot(C, x, T, "task01_4")
# to convert generated image sequence to video use:
# $> ffmpeg -r 60 -i plots/task01_4_%05d.png -pix_fmt yuv420p video_1_4.mp4

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