More stuff

.gitignore

@@ -38,3 +38,9 @@
 *.bib
 *.synctex.gz
 *.bbl 
+
+# C++ specifics
+src/*
+!src/Makefile
+!src/*.cpp
+!src/*.py

latex/assignment_1.tex

@@ -74,6 +74,9 @@
 %%
 %% Don't ask me why, I don't know.
 
+% custom stuff
+\graphicspath{{./images/}}
+
 \begin{document}
 
 \title{Project 1}      % self-explanatory

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/problem2.tex

@@ -1,3 +1,11 @@
 \section*{Problem 2}
 
-% Write which .cpp/.hpp/.py (using a link?) files are relevant for this and show the plot generated.
+The code for generating the points and plotting them can be found under.
+
+Point generator code: https://github.uio.no/FYS3150-G2-203/Project-1/blob/main/src/analyticPlot.cpp
+
+Plotting code: https://github.uio.no/FYS3150-G2-2023/Project-1/blob/main/src/analyticPlot.py
+
+Here is the plot of the analytical solution for $u(x)$.
+
+\includegraphics[scale=.5]{analytical_solution.pdf}

latex/problems/problem3.tex

@@ -2,7 +2,7 @@
 \section*{Problem 3}
 
 To derive the discretized version of the Poisson equation, we first need
-the taylor expansion for $u(x)$ around $x + h$ and $x - h$.
+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)
@@ -24,8 +24,8 @@ If we add the equations above, we get this new equation:
 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} &= 100 e^{-10x} \\
-    \frac{ - u_{i+1} + 2 u_i - u_{i-1}}{h^2} + \mathcal{O}(h^2) &= 100 e^{-10x} \\
+    - \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 
@@ -33,5 +33,5 @@ differentiate between the exact solution and the approximate solution,
 and get the discretized version of the equation:
 
 \begin{align*}
-align*    \frac{ - u_{i+1} + 2 u_i - u_{i-1}}{h^2} &= 100 e^{-10x} \\
+    \frac{ - v_{i+1} + 2 v_i - v_{i-1}}{h^2} &= 100 e^{-10x_i} \\
 \end{align*}

src/Makefile

@@ -1,10 +1,15 @@
 CC=g++
 
-all: simpleFile
+.PHONY: clean
+
+all: simpleFile analyticPlot
 
 simpleFile: simpleFile.o
 	$(CC) -o $@ $^
 
+analyticPlot: analyticPlot.o
+	$(CC) -o $@ $^
+
 %.o: %.cpp
 	$(CC) -c $< -o $@
 

src/analyticPlot.cpp

@@ -6,17 +6,36 @@
 #include <fstream>
 #include <iomanip>
 
+#define RANGE 1000
+#define FILENAME "analytical_solution.txt"
+
 double u(double x);
 void generate_range(std::vector<double> &vec, double start, double stop, int n);
+void write_analytical_solution(std::string filename, int n);
 
 int main() {
-    int n = 1000;
+    write_analytical_solution(FILENAME, RANGE);
 
+    return 0;
+};
+
+double u(double x) {
+    return 1 - (1 - exp(-10))*x - exp(-10*x);
+};
+
+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;
+    }
+}
+
+void write_analytical_solution(std::string filename, int n) {
     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);
 
@@ -32,21 +51,5 @@ int main() {
             << 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

@@ -1,17 +1,19 @@
 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))
+def main():
+    FILENAME = "analytical_solution.pdf"
+    x = []
+    v = []
 
-fig, ax = plt.subplots()
-ax.plot(x, v)
-plt.show()
-# plt.savefig("main.png")
+    with open('analytical_solution.txt') as f:
+        for line in f:
+            a, b = line.strip().split()
+            x.append(float(a))
+            v.append(float(b))
+
+    plt.plot(x, v)
+    plt.savefig(FILENAME)
+
+if __name__ == "__main__":
+    main()