From 58b67a304bfc8e722631daf86ce4049b8f4aa262 Mon Sep 17 00:00:00 2001 From: daniel Date: Mon, 16 May 2022 01:25:09 +0200 Subject: [PATCH] add. Task 1.4, add: task 2.1-2 (Matlab) --- Exercise_03/NSSC_3.m | 43 ++++++ Exercise_03/UW_scheme.m | 51 +++++++ Exercise_03/report.tex | 298 ++++++++++++++++++++++++++++++++++++++ Exercise_03/task01_3.py | 2 +- Exercise_03/task01_4.py | 74 ++++++++++ Exercise_03/video_1_4.mp4 | Bin 0 -> 120827 bytes 6 files changed, 467 insertions(+), 1 deletion(-) create mode 100644 Exercise_03/NSSC_3.m create mode 100644 Exercise_03/UW_scheme.m create mode 100644 Exercise_03/report.tex create mode 100644 Exercise_03/task01_4.py create mode 100644 Exercise_03/video_1_4.mp4 diff --git a/Exercise_03/NSSC_3.m b/Exercise_03/NSSC_3.m new file mode 100644 index 0000000..c808100 --- /dev/null +++ b/Exercise_03/NSSC_3.m @@ -0,0 +1,43 @@ +%% 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 + diff --git a/Exercise_03/UW_scheme.m b/Exercise_03/UW_scheme.m new file mode 100644 index 0000000..387dadc --- /dev/null +++ b/Exercise_03/UW_scheme.m @@ -0,0 +1,51 @@ +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 \ No newline at end of file diff --git a/Exercise_03/report.tex b/Exercise_03/report.tex new file mode 100644 index 0000000..cf40f13 --- /dev/null +++ b/Exercise_03/report.tex @@ -0,0 +1,298 @@ +\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} diff --git a/Exercise_03/task01_3.py b/Exercise_03/task01_3.py index b824a47..021624d 100644 --- a/Exercise_03/task01_3.py +++ b/Exercise_03/task01_3.py @@ -39,7 +39,7 @@ for t in range(nt): plt.ylim([0, 1.2]) plt.savefig(f"plots/task01_3_{i:0>5}.png") i += 1 - # update solution using the explicit schema + # update solution using the implicit schema C = np.linalg.solve(T, C) # fix BC conditions (theoretically, they are set by the update but for # stability reasons (numerical) we enforce the correct values) diff --git a/Exercise_03/task01_4.py b/Exercise_03/task01_4.py new file mode 100644 index 0000000..68927b5 --- /dev/null +++ b/Exercise_03/task01_4.py @@ -0,0 +1,74 @@ +# 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 diff --git a/Exercise_03/video_1_4.mp4 b/Exercise_03/video_1_4.mp4 new file mode 100644 index 0000000000000000000000000000000000000000..ab992382a2c192789a0a6b9dac3ed71b38286a75 GIT binary patch literal 120827 zcmcG#WmsJ?(>A)1;uMGCySTgi#wl9d-QBIYyE`rJ?(XjH?(QxH4t-wv&iCW|JlD0? zTFGQGE0av_HIrlm005w|qnoXngN+pc01o)~{Jfa-oDCSQY*`oq004}!y^#?Btozl< zP|xv`rxNPR$4BX!$Z`9@vUp<>%@S~#_~O!)iH!wF3pB8?Hv%%UG6C(GI9OSLMx1)A 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