Merge branch 'main' into coryab/edit-problem-1

7 solve problem 7

.gitignore

@@ -36,3 +36,9 @@
 *.log
 *.out
 *.bib
+*.synctex.gz
+*.bbl 
+
+# C++ specifics
+src/*
+!src/*.cpp

latex/assignment_1.tex

@@ -1,7 +1,9 @@
 \documentclass[english,notitlepage]{revtex4-1}  % defines the basic parameters of the document
 %For preview: skriv i terminal: latexmk -pdf -pvc filnavn
 
-
+% Silence warning of revtex4-1
+\usepackage{silence}
+\WarningFilter{revtex4-1}{Repair the float}
 
 % if you want a single-column, remove reprint
 
@@ -13,7 +15,7 @@
 %% I recommend downloading TeXMaker, because it includes a large library of the most common packages.
 
 \usepackage{physics,amssymb}  % mathematical symbols (physics imports amsmath)
-\include{amsmath}
+\usepackage{amsmath}
 \usepackage{graphicx}         % include graphics such as plots
 \usepackage{xcolor}           % set colors
 \usepackage{hyperref}         % automagic cross-referencing (this is GODLIKE)

latex/problems/problem1.tex

@@ -1,8 +1,17 @@
 \section*{Problem 1}
 
+First, we rearrange the equation.
+
+\begin{align*}
+    - \frac{d^2u}{dx^2} &= 100 e^{-10x} \\
+    \frac{d^2u}{dx^2} &= -100 e^{-10x} \\
+\end{align*}
+
+Now we find $u(x)$.
+
 % Do the double integral
 \begin{align*}
-    u(x) &= \int \int \frac{d^2 u}{dx^2} dx^2\\
+    u(x) &= \int \int \frac{d^2 u}{dx^2} dx^2 \\
          &= \int \int -100 e^{-10x} dx^2 \\
          &= \int \frac{-100 e^{-10x}}{-10} + c_1 dx \\
          &= \int 10 e^{-10x} + c_1 dx \\
@@ -10,7 +19,7 @@
          &= -e^{-10x} + c_1 x + c_2
 \end{align*}
 
-Using the boundary conditions, we can find $c_1$ and $c_2$ as shown below:
+Using the boundary conditions, we can find $c_1$ and $c_2$
 
 \begin{align*}
     u(0) &= 0 \\

latex/problems/problem3.tex

@@ -1,4 +1,37 @@
 
 \section*{Problem 3}
 
-% Show how it's derived and where we found the derivation.
+To derive the discretized version of the Poisson equation, we first need
+the Taylor expansion for $u(x)$ around $x$ for $x + h$ and $x - h$.
+
+\begin{align*}
+    u(x+h) &= u(x) + u'(x) h + \frac{1}{2} u''(x) h^2 + \frac{1}{6} u'''(x) h^3 + \mathcal{O}(h^4)
+\end{align*}
+
+\begin{align*}
+    u(x-h) &= u(x) - u'(x) h + \frac{1}{2} u''(x) h^2 - \frac{1}{6} u'''(x) h^3 + \mathcal{O}(h^4)
+\end{align*}
+
+If we add the equations above, we get this new equation:
+
+\begin{align*}
+    u(x+h) + u(x-h) &= 2 u(x) + u''(x) h^2 + \mathcal{O}(h^4) \\
+    u(x+h) - 2 u(x) + u(x-h) + \mathcal{O}(h^4) &= u''(x) h^2 \\
+    u''(x) &= \frac{u(x+h) - 2 u(x) + u(x-h)}{h^2} + \mathcal{O}(h^2) \\
+    u_i''(x) &= \frac{u_{i+1} - 2 u_i + u_{i-1}}{h^2} + \mathcal{O}(h^2) \\
+\end{align*}
+
+We can then replace $\frac{d^2u}{dx^2}$ with the RHS (right-hand side) of the equation:
+
+\begin{align*}
+    - \frac{d^2u}{dx^2} &= f(x) \\
+    \frac{ - u_{i+1} + 2 u_i - u_{i-1}}{h^2} + \mathcal{O}(h^2) &= f_i \\
+\end{align*}
+
+And lastly, we leave out $\mathcal{O}(h^2)$ and change $u_i$ to $v_i$ to 
+differentiate between the exact solution and the approximate solution, 
+and get the discretized version of the equation:
+
+\begin{align*}
+    \frac{ - v_{i+1} + 2 v_i - v_{i-1}}{h^2} &= 100 e^{-10x_i} \\
+\end{align*}

latex/problems/problem4.tex

@@ -1,3 +1,44 @@
 \section*{Problem 4}
 
 % Show that each iteration of the discretized version naturally creates a matrix equation.
+
+The value of $u(x_{0})$ and $u(x_{1})$ is known, using the discretized equation we can approximate the value of $f(x_{i}) = f_{i}$. This will result in a set of equations   
+\begin{align*}
+    - v_{0} + 2 v_{1} - v_{2} &= h^{2} \cdot f_{1} \\
+    - v_{1} + 2 v_{2} - v_{3} &= h^{2} \cdot f_{2} \\
+    \vdots & \\
+    - v_{m-2} + 2 v_{m-1} - v_{m} &= h^{2} \cdot f_{m-1} \\
+\end{align*}
+
+Rearranging the first and last equation, moving terms of known boundary values to the RHS
+\begin{align*}
+    2 v_{1} - v_{2} &= h^{2} \cdot f_{1} + v_{0} \\
+    - v_{1} + 2 v_{2} - v_{3} &= h^{2} \cdot f_{2} \\
+    \vdots & \\
+    - v_{m-2} + 2 v_{m-1} &= h^{2} \cdot f_{m-1} + v_{m} \\
+\end{align*}
+
+We now have a number of linear eqations, corresponding to the number of unknown values, which can be represented as an augmented matrix 
+\begin{align*}
+    \left[
+      \begin{matrix}
+        2v_{1} & -v_{2} & 0 & \dots & 0 \\
+        -v_{1} & 2v_{2} & -v_{3} & 0 & \\
+        0 & -v_{2} & 2v_{3} & -v_{4} & \\
+        \vdots  & & & \ddots & \vdots \\
+        0 & & & -v_{m-2} & 2v_{m-1} \\
+      \end{matrix}
+      \left|
+        \,
+        \begin{matrix}
+          g_{1}  \\
+          g_{2}  \\
+          g_{2}  \\
+          \vdots  \\
+          g_{m-1}  \\
+        \end{matrix}
+      \right.
+    \right]
+    \end{align*}
+    where $g_{i} = h^{2} f_{i}$. An augmented matrix can be represented as $\boldsymbol{A} \vec{x} = \vec{b}$. In this case $\boldsymbol{A}$ is the coefficient matrix with a tridiagonal signature $(-1, 2, -1)$ and dimension $n \cross n$, where $n=m-2$.
+

src/Makefile

@@ -0,0 +1,12 @@
+CC=g++
+
+all: simpleFile
+
+simpleFile: simpleFile.o
+	$(CC) -o $@ $^
+
+%.o: %.cpp
+	$(CC) -c $< -o $@
+
+clean:
+	rm *.o

src/analyticPlot.cpp

@@ -0,0 +1,52 @@
+#include <iostream>
+#include <cmath>
+#include <vector>
+#include <string>
+#include <numeric>
+#include <fstream>
+#include <iomanip>
+
+double u(double x);
+void generate_range(std::vector<double> &vec, double start, double stop, int n);
+
+int main() {
+    int n = 1000;
+
+    std::vector<double> x(n), y(n);
+    generate_range(x, 0.0, 1.0, n);
+
+    // Set up output file and strem
+    std::string filename = "datapoints.txt";
+    std::ofstream outfile;
+    outfile.open(filename);
+
+    // Parameters for formatting 
+    int width = 12;
+    int prec = 4;
+
+    // Calculate u(x) and write to file
+    for (int i = 0; i <= x.size(); i++) {
+        y[i] = u(x[i]);
+        outfile << std::setw(width) << std::setprecision(prec) << std::scientific << x[i] 
+            << std::setw(width) << std::setprecision(prec) << std::scientific << y[i] 
+            << std::endl;
+    }
+    outfile.close();
+
+    return 0;
+};
+
+double u(double x) {
+    double result;
+
+    result = 1 - (1 - exp(-10))*x - exp(-10*x);
+    return result;
+};
+
+void generate_range(std::vector<double> &vec, double start, double stop, int n) {
+    double step = (stop - start) / n;
+
+    for (int i = 0; i <= vec.size(); i++) {
+        vec[i] = i * step;
+    }
+}

src/analyticPlot.py

@@ -0,0 +1,17 @@
+import numpy as np 
+import matplotlib.pyplot as plt
+
+x = []
+y = []
+v = []
+with open('testdata.txt') as f:
+    for line in f:
+        a, b, c = line.strip().split()
+        x.append(float(a))
+        # y.append(float(b))
+        v.append(float(c))
+
+fig, ax = plt.subplots()
+ax.plot(x, v)
+plt.show()
+# plt.savefig("main.png")

src/simpleFile.cpp

@@ -15,7 +15,7 @@ arma::vec* general_algorithm(
     arma::vec* g_vec
 )
 {
-    int n = main_diag->n_elem;
+    int n = g_vec->n_elem;
     double d;
 
     for (int i = 1; i < n; i++) {
@@ -40,6 +40,21 @@ arma::vec* special_algorithm(
     arma::vec* g_vec
 )
 {
+    int n = g_vec->n_elem;
+    arma::vec diag = arma::vec(n);
+
+    for (int i = 1; i < n; i++) {
+        // Calculate values for main diagonal based on indices
+        diag(i-1) = (double)(i+1) / i;
+        (*g_vec)(i) += (*g_vec)(i-1) / diag(i-1);
+    }
+    // The last element in main diagonal has value (i+1)/i = (n+1)/n
+    (*g_vec)(n-1) /= (double)(n+1) / (n);
+
+    for (int i = n-2; i >= 0; i--) {
+        (*g_vec)(i) = ((*g_vec)(i) + (*g_vec)(i+1))/ diag(i);
+    }
+
     return g_vec;
 }