[usth] Numerical Method is MATH2.4

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diff --git a/usth/MATH2.2/project/contour.png b/usth/MATH2.2/project/contour.png
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Binary files differdiff --git a/usth/MATH2.2/project/heatrans.m b/usth/MATH2.2/project/heatrans.m
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-function T = heatrans (cp, lambda, rho, Tg, Td, T0, L, dx, dt, N)
-  alpha = lambda / rho / cp;
-  beta = alpha * dt / dx^2;
-  M = round (L / dx);
-  side = repelem (beta, M);
-  A = diag (side, -1) + diag (repelem (1 - 2*beta, M + 1)) + diag (side, 1);
-  A(1, :) = A(end, :) = 0;
-  A(1, 1) = A(end, end) = 1;
-
-  T = repelem (T0, M + 1);
-  [T(1) T(end)] = deal (Tg, Td);
-  for k = 2 : N
-    T(:, k) = A * T(:, k - 1);
-  end
-end
diff --git a/usth/MATH2.2/project/mesh.png b/usth/MATH2.2/project/mesh.png
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-\documentclass[a4paper,12pt]{article}
-\usepackage[english,vietnamese]{babel}
-\usepackage{amsmath}
-\usepackage{amssymb}
-\usepackage{graphicx}
-\usepackage{hyperref}
-\usepackage{lmodern}
-\usepackage{mathtools}
-\usepackage{siunitx}
-
-\newcommand{\exercise}[1]{\noindent\textbf{#1.}}
-\newcommand{\sifrac}[2]{\displaystyle\SI[per-mode=fraction]{#1}{#2}}
-
-\title{Numerical Methods: Heat Transfer}
-\author{Nguyễn Gia Phong--BI9-184}
-\date{\dateenglish\today}
-
-\begin{document}
-\maketitle
-
-Given a bar of length $L = \SI{0.4}{\meter}$ consisting of homogeneous
-and isotropic material with the initial temperature of $T_0 = \SI{0}{\celsius}$.
-Suppose it is perfectly insulated with the exception of the ends with
-the temperature of $T_g = \SI{100}{\celsius}$ and $T_d = \SI{50}{\celsius}$.
-The thermal properties of material will be taken constant.
-\begin{itemize}
-  \item Specific heat capacity:
-    $c_p = \sifrac{900}{\joule\per\kilogram\per\celsius}$
-  \item Thermal conductivity:
-    $\lambda = \sifrac{237}{\watt\per\meter\per\celsius}$
-  \item Density:
-    $\rho = \sifrac{2700}{\kilogram\per\cubic\meter}$
-  \item Thermal diffusivity:
-    $\alpha = \dfrac{\lambda}{\rho c_p}
-            = \sifrac{9.753e-5}{\square\meter\per\second}$
-\end{itemize}
-
-The heat transfer in this bar can be described by the following
-partial differential equation
-\[\frac{\partial T}{\partial t}
-  = \alpha\frac{\partial^2 T}{\partial x^2}\tag{$*$}\]
-where the temperature $T$ is a function of postition $x$ and time $t$.
-
-To solve the problem numerically, we devide space and time into equal intervals
-of norms $\Delta x$ and $\Delta t$ respectively and let $M = L/\Delta x$.
-Consequently, the spartial coordinate is defined as $x_i = (i - 1)\Delta x$
-with $i \in [1\,..\,M + 1]$ and the temporal one is $t_n = n\Delta t$
-with $n \in \mathbb N^*$. With these definitions\footnote{I believe $i = 0$
-and $n = 0$ in the assignment papers are typos since then the domain of $x_i$
-would exceed $L$ and $t_0$ would be negative.}, we denote $T_i^n = T(x_i, t_n)$.
-Using numerical methods, we may start approximating the solutions of ($*$).
-
-\begin{enumerate}
-  \item The left-hand side of ($*$) can be approximated as
-    \[\frac{\partial T}{\partial t} = \frac{T_i^{n+1} - T_i^n}{\Delta t}\]
-  \item Similarly, the right-hand side is expressed in the following form
-    \[\alpha\frac{\partial^2 T}{\partial x^2}
-      = \alpha\frac{T_{i+1}^n - 2T_i^n + T_{i-1}^n}{\Delta x^2}\]
-  \item ($*$) is therefore reformulated as
-    \[\frac{T_i^{n+1} - T_i^n}{\Delta t}
-      = \alpha\frac{T_{i+1}^n - 2T_i^n + T_{i-1}^n}{\Delta x^2}\]
-  \item From the formular above
-    and let $\beta = \alpha\dfrac{\Delta t}{\Delta x^2}$,
-    we get
-    \[T_i^{n+1} = T_i^n + \beta\left(T_{i+1}^n - 2T_i^n + T_{i-1}^n\right)\]
-  \item Boundary conditions:
-    \begin{itemize}
-      \item $\forall n \in \mathbb N^*,\; T_1^n = T(0, t_n) = T_g$
-      \item $\forall n \in \mathbb N^*,\; T_{M+1}^n = T(L, t_n) = T_d$
-      \item $\forall i \in [2\,..\,M],\; T_{i}^1 = T(x_i, 0) = T_0$
-    \end{itemize}
-  \item From (4) and (5), the temperature at point $x_i$ of the bar
-    at time $t_n$ is recursively defined as
-    \[T_i^n = \begin{dcases}
-        T_i^{n-1} + \beta\left(T_{i+1}^{n-1} - 2T_i^{n-1} + T_{i-1}^{n-1}\right)
-        &\text{ if }1 < i \le M \land n > 1\\
-        T_g &\text{ if }i = 1\\
-        T_d &\text{ if }i = M + 1\\
-        T_0 &\text{ otherwise}
-      \end{dcases}\]
-    Since the temperature only depends on the values in the past, values within
-    $(i, n) \in [1\,..\,M+1]\times[1\,..\,N]$ with any $N$ of choice could
-    be computed via dynamic programming:
-    \begin{enumerate}
-      \item Create a 2-dimensional dynamic array $T$ with one-based index
-        and size $(M+1)\times 1$
-      \item Initialize $T$ with $T_1^1 = T_g$, $T_{M+1}^1 = T_d$
-        and $T_i^1 = T_0\ \forall i \in [2\,..\,M]$,
-        where $T_i^n$ is element of row $i$ and column $n$
-      \item For $n = 2$ to $N$
-        \begin{itemize}
-          \item Let $T_1^k = T_g$
-          \item For $i = 2$ to $M$, let $T_i^k = T_i^{k-1}
-            + \beta\left(T_{i+1}^{k-1} - 2T_i^{k-1} + T_{i-1}^{k-1}\right)$
-          \item Let $T_{M+1}^k = T_d$
-        \end{itemize}
-      \item Return $T_i^n$
-    \end{enumerate}
-    Each iteration in (c) can be written in matrix notation as
-    $T^k = AT^{k+1}$, where $T_n$ is column $n$ and $A$ is
-    a matrix of size $(M+1)\times(M+1)$
-    \[A =\begin{bmatrix}
-        1 & 0 & 0 & \cdots & 0 & 0 & 0\\
-        \beta & 1-2\beta & \beta & \cdots & 0 & 0 & 0\\
-        0 & \beta & 1-2\beta & \cdots & 0 & 0 & 0\\
-        \vdots & \vdots & \vdots & \ddots & \vdots & \vdots & \vdots\\
-        0 & 0 & 0 & \cdots & 1-2\beta & \beta & 0\\
-        0 & 0 & 0 & \cdots & \beta & 1-2\beta & \beta\\
-        0 & 0 & 0 & \cdots & 0 & 0 & 1
-      \end{bmatrix}\]
-  \item Steps (a) to (c) is then implemented in Octave as
-\begin{verbatim}
-function T = heatrans (cp, lambda, rho, Tg, Td, T0, L,
-                       dx, dt, N)
-  alpha = lambda / rho / cp;
-  beta = alpha * dt / dx^2;
-  M = round (L / dx);
-  side = repelem (beta, M);
-  A = (diag (repelem (1 - 2*beta, M + 1))
-       + diag (side, -1) + diag (side, 1));
-  A(1, :) = A(end, :) = 0;
-  A(1, 1) = A(end, end) = 1;
-
-  T = repelem (T0, M + 1);
-  [T(1) T(end)] = deal (Tg, Td);
-  for k = 2 : N
-    T(:, k) = A * T(:, k - 1);
-  end
-end
-\end{verbatim}
-
-    Choosing $\Delta x = \SI{0.01}{\meter}$, $\Delta t = \SI{0.5}{\second}$
-    and $N = 841$, we define
-\begin{verbatim}
-T = heatrans (900, 237, 2700, 100, 50, 0, 0.4,
-              0.01, 0.5, 841);
-\end{verbatim}
-    then the temperature at point $x_i$ at time $t_n$ is \verb|T(i, n)|.
-    \pagebreak
-
-    To visualize the heat transfer process, we use \verb|mesh| to plot
-    a 3D graph:
-
-    \includegraphics[width=0.841\textwidth]{mesh.png}
-
-    The temperature can be shown more intuitively using \verb|contourf|:
-
-    \includegraphics[width=0.841\textwidth]{contour.png}
-
-    The \verb|script| to reproduce these results along with \verb|heatrans.m|
-    bundled with this report and this document itself are all licensed under a
-    \href{http://creativecommons.org/licenses/by-sa/4.0/}{Creative Commons
-      Attribution-ShareAlike 4.0 International License}.
-\end{enumerate}
-\end{document}
\ No newline at end of file
diff --git a/usth/MATH2.2/project/script b/usth/MATH2.2/project/script
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-#!/usr/bin/env octave
-[cp lambda rho] = deal (900, 237, 2700);
-[Tg Td T0] = deal (100, 50, 0);
-[L dx dt N] = deal (0.4, 0.01, 0.5, 841);
-T = heatrans (cp, lambda, rho, Tg, Td, T0, L, dx, dt, N);
-[X Y Z] = deal (0:dx:L, 0:dt:420, T');
-
-figure (1);   # mesh
-mesh (X, Y, Z);
-title ("Temperature of the bar during the first 420 seconds");
-xlabel ("x (m)");
-ylabel ("t (s)");
-zlabel ("T (°C)");
-
-figure (2);   # contour
-contourf (X, Y, Z, 0:7:100, "showtext", "on");
-title ("Temperature of the bar during the first 420 seconds");
-xlabel ("x (m)");
-ylabel ("t (s)");
-
-disp ("Press any key to coninue...");
-kbhit;