Project-1/latex/assignment_1.tex

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\begin{document}

\title{Project 1}      % self-explanatory
\author{Cory Balaton \& Janita Willumsen}          % self-explanatory
\date{\today}                             % self-explanatory
\noaffiliation                            % ignore this, but keep it.


\maketitle 
    
\textit{https://github.uio.no/FYS3150-G2-2023/Project-1}
    
\section*{Problem 1}

\begin{align*}
    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 \\
         &= \frac{10 e^{-10x}}{-10} + c_1 x + c_2 \\
         &= -e^{-10x} + c_1 x + c_2
\end{align*}

Using the boundary conditions, we can find $c_1$ and $c_2$ as shown below:

\begin{align*}
    u(0) &= 0 \\
    -e^{-10 \cdot 0} + c_1 \cdot 0 + c_2 &= 0 \\
    -1 + c_2 &= 0 \\
    c_2 &= 1
\end{align*}

\begin{align*}
    u(1) &= 0 \\
    -e^{-10 \cdot 1} + c_1 \cdot 1 + c_2 &= 0 \\
    -e^{-10} + c_1 + c_2 &= 0 \\
    c_1 &= e^{-10} - c_2\\
    c_1 &= e^{-10} - 1\\
\end{align*}

Using the values that we found for $c_1$ and $c_2$, we get

\begin{align*}
    u(x) &= -e^{-10x} + (e^{-10} - 1) x + 1 \\
         &= 1 - (1 - e^{-10}) - e^{-10x}
\end{align*}

\section*{Problem 2}

% Write which .cpp/.hpp/.py (using a link?) files are relevant for this and show the plot generated.

\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 $x - h$.

\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} &= 100 e^{-10x} \\
    \frac{ - u_{i+1} + 2 u_i - u_{i-1}}{h^2} + \mathcal{O}(h^2) &= 100 e^{-10x} \\
\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*}
align*    \frac{ - u_{i+1} + 2 u_i - u_{i-1}}{h^2} &= 100 e^{-10x} \\
\end{align*}

\section*{Problem 4}

% Show that each iteration of the discretized version naturally creates a matrix equation.

\section*{Problem 5}

\subsection*{a)}

\subsection*{b)}

\section*{Problem 6}

\subsection*{a)}

% Use Gaussian elimination, and then use backwards substitution to solve the equation

\subsection*{b)}

% Figure it out

\section*{Problem 7}

% Link to relevant files on gh and possibly add some comments

\section*{Problem 8}

%link to relvant files and show plots

\section*{Problem 9}

% Show the algorithm, then calculate FLOPs, then link to relevant files

\section*{Problem 10}

% Time and show result, and link to relevant files

\end{document}