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\begin{document}
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\begin{center}
\begin{center}
	\begin{columns}
		\begin{column}{0.9\textwidth}
			\flushright%\fontspec{Trebuchet MS}
			\bfseries \Huge {Ordinary Differential Equations (I)}
		\end{column}
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	\quad
	\vspace{3em}
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\begin{column}{0.6\textwidth}
\flushright \vspace{-1.8em} {%\fontspec{Trebuchet MS}
    \large Marcin Chrzaszcz, Danny van Dyk\\\vspace{-0.1em}\small \href{mailto:mchrzasz@cern.ch}{mchrzasz@cern.ch}, \href{mailto:danny.van.dyk@gmail.com}{danny.van.dyk@gmail.com}}

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\vspace{1em}
\vspace{0.5em}
	\textcolor{normal text.fg!50!Comment}{Numerical Methods, \\ 07. November, 2016}
\end{center}
\end{frame}
}

\begin{frame}{Plan for today}
    \begin{itemize}
        \item \alert{General problem of estimating the solution to an equation that involves some unknown
            function and its derivatives}\\
            \begin{itemize}
                \item What do we need to know to do this?
            \end{itemize}
        \vfill
        \item \alert{Specific problems in which we know the initial condition for our function}\\
    \end{itemize}
\end{frame}

\begin{frame}{Nomenclature and Preliminaries}
    We will discuss \alert{ordinary} \alert{differential equation}s, or \alert{systems} of
    \alert{linear} ordinary differential equations:
    \begin{overlayarea}{\textwidth}{.7\textheight}
    \begin{description}
        \item<1-2>[differential equation] Any equation of the form
        \begin{equation*}
            F(x; y, y', y^{(2)}, \dots, y^{(n)}) = 0
        \end{equation*}
        with $x\in \mathbb{R}$ and $y(x): \mathbb{R} \mapsto \mathbb{R}^m$
        is called a system of ordinary differential equations of \alert{order $n$}.\\
        If $m=1$, it is only \alert{one} ordinary diff.\ equation (ODE).\\
        \only<1>{
        \item[linear] If we can write
        \begin{equation*}
            F(x; y, \dots, y^{(n)}) = A(x) \cdot \vec{y}(x) + r(x)
        \end{equation*}
        with $A(x) \in \mathbb{R}^{m\times n}$, $\vec{y}(x) = (y(x), \dots y^{(n)}(x))^T$
        and $r(x) \in \mathbb{R}^m$, then
        $\Phi$ is a \alert{linear} system of ODEs.
        }
        \only<2>{
        \item[ordinary] The quality \alert{ordinary} hinges on the fact that $y$ only
        depends on $x\in\mathbb{R}$. As soon as $x\in \mathbb{R}^{k}$, and we need to
        consider partial derivatives of $y$, we have a \alert{partial} differential equation.
        }
        \only<3>{
        \item[explicit] If for an $n$th order ODE we can write
        \begin{equation*}
            y^{(n)} = f(x; y, \dots, y^{(n - 1)})
        \end{equation*}
        then we call this an \alert{explicit} ODE. Otherwise, the ODE is \alert{implicit}.
        }
    \end{description}
    \end{overlayarea}
\end{frame}

\begin{frame}{Explicit linear ODEs of order $n=1$/Initial value problem}
    \begin{equation*}
        y' = f(x; y)
    \end{equation*}

    Initial value: $y_0 \equiv y(x_0)$ provided externally.
\end{frame}

\begin{frame}{Single Step Methods}
    We want to increment by one \alert{step}, i.e. from point $n$ to point $n + 1$.
    The increment is defined by the incremental function $\Phi$:
    \begin{equation*}
        y_{n+1} = y_n + \Phi(f; x_0, h; y_n, y_{n+1});
    \end{equation*}
    with a step size $h$. This is an implicit method. If $\Phi$ does not depend on
    $y_{n+1}$, we have an explicit method.
    \vfill
    Whatever the choice of $\Phi$, for an implicit method we will need, in general,
    to solve a non-linear equation ($\rightarrow$ see lecture on root finding). That
    means that an implicit method is computationally more expensive than a similar
    explicit method.
\end{frame}

\begin{frame}{Explicit Single Step Methods: Euler Method}
    \begin{block}{First order ODE}
    ~\\
    \vspace{-2\bigskipamount}
    \begin{align*}
        y'      & = f(x, y) &
        y_{n+1} & = y_n + \Phi(f; x_0, h; y_n)
    \end{align*}
    \end{block}
    Standard procedure for an unknown function: Taylor expand!
    \begin{align*}
        y(x_n + h) & = y(x_n) + h y'(x_n) + \frac{h^2}{2} y''(x_n) + O{h^3}
    \end{align*}
    Keeping only terms of order $h$, and substituting $y'(x_n) = f(x_n = x_0 + n\cdot h, y_n)$
    we obtain
    \begin{block}{Euler method}
    \begin{align*}
        y_{n+1} & = y_n + \underbrace{h f(x_n, y_n)}_{=\Phi_\text{Euler}(f; x_0, h; y_n)}
    \end{align*}
    \end{block}
\end{frame}

\begin{frame}{Local Truncation Error for Euler Method}
    The local trunction error (LTE) is introduced as:
    \begin{equation*}
        y(x_0 + h) - y_0 \equiv \text{LTE} = \frac{h^2}{2} y''(x_0) + O(h^3)
    \end{equation*}
    \begin{overlayarea}{\textwidth}{.7\textheight}
    \begin{columns}
    \begin{column}{.4\textwidth}
    \begin{block}{Example}
    The ODE at hand reads:
    \begin{align*}
        y'  & = -y\,, &
        y_0 & = 0.5
    \end{align*}
    The analytic solution is:
    \begin{equation*}
        y(x) = y_0 \, \exp(-x)
    \end{equation*}
    \end{block}
    \end{column}
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    \begin{tikzpicture}
        \begin{axis}[%
                axis x line=center,
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                ymin=0.0,ymax=0.55,
                xmin=0.0,xmax=+1,
                samples=160
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                0.5 * exp(-x)
            };
            \addplot[thick,red,only marks,mark size=3pt] coordinates {
                (0.25, 0.375) % h = 0.25
                (0.50, 0.250) % h = 0.50
                (1.00, 0.000) % h = 1.00
            };
            \addplot[thin,red,mark=none] {
                0.5 - 0.5 * x
            };
        \end{axis}
    \end{tikzpicture}
    }
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\end{frame}

\newcommand{\hl}[1]{{\color{red}#1}}
\begin{frame}{Explicit Single Step Methods: Runge-Kutta Methods}
    For Runge-Kutta methods an alternative derivation for the determination $y_{n+1}$
    uses an integral representation:
    \begin{equation*}
        y_{n+1} = y_n + \int_{x_n}^{x_n + h} \mathrm{d}x\, f(x, y(x))\,.
    \end{equation*}
    \vspace{-\medskipamount}
    The \alert{$s$}-stage Runge-Kutta method then reads
    \begin{equation*}
        y_{n+1} = y_n + \sum_{i=1}^{\alert{s}} \hl{b_i} k_i
    \end{equation*}
    \vspace{-\medskipamount}
    where
    \begin{align*}
        k_1 & = f(x_n        ,& y_n &                                                                     & )\\
        k_2 & = f(x_n + \hl{c_2} h,& y_n & + h\, \hl{a_{2,1}} k_1                                         & )\\
        k_3 & = f(x_n + \hl{c_3} h,& y_n & + h\, \hl{a_{3,1}} k_1 + h\,\hl{a_{3,2}} k_2                   & )\\
            & \quad\vdots     &     & \vdots\\
        k_s & = (fx_n + \hl{c_s} h,& y_n & + h\, \hl{a_{s,1}} k_1 + \dots + h\,\hl{a_{s,(s-1)}} k_{s-1}   & )
    \end{align*}
    Now to determine the coefficients $\lbrace \hl{a_{i,j}} \rbrace$, $\lbrace \hl{b_i} \rbrace$,
    and $\lbrace \hl{c_i} \rbrace$.
\end{frame}

\begin{frame}{Explicit Single Step Methods: Runge-Kutta (cont'd)}
    Usual presentation of the necessary coefficients in a \alert{Butcher tableau}:
    \begin{center}
    \renewcommand{\arraystretch}{1.2}
    \begin{tabular}{c|ccccc}
        $0$      &                                                         \\
        $c_2$    & $a_{2,1}$                                               \\
        $c_3$    & $a_{3,1}$ & $a_{3,2}$                                   \\
        $\vdots$ & $\vdots$  &           & $\ddots$                        \\
        $c_s$    & $a_{s,1}$ & $a_{s,2}$ & $\dots$   & $a_{s,s-1}$         \\
        \hline
                 & $b_1$     & $b_2$     & $\dots$   & $b_{s-1}$   & $b_s$ \\
    \end{tabular}
    \renewcommand{\arraystretch}{1.0}
    \end{center}

    To take away:
    \begin{itemize}
        \item a method of stage $s$ requires \alert{in general} $s$ evaluations of
        $f$
        \item the LTE is of order $p$, i.e.: $\text{LTE} = O(h^{p + 1})$
        \item in general $s > p$
        \item up to $s = 4$ there are are methods known with $s \geq p$
    \end{itemize}
\end{frame}

\begin{frame}{Explicit Single Step Methods: Runge-Kutta (cont'd)}
    \begin{columns}
    \begin{column}{.5\textwidth}
    \begin{block}{Euler's method}
    \begin{tabular}{c|c}
        $0$      &            \\
        \hline
                 & $b_1$      \\
    \end{tabular}
    \end{block}
    \end{column}
    \begin{column}{.5\textwidth}
    \begin{tabular}{c|ccccc}
        $0$      &                                                         \\
        $c_2$    & $a_{2,1}$                                               \\
        $c_3$    & $a_{3,1}$ & $a_{3,2}$                                   \\
        $\vdots$ & $\vdots$  &           & $\ddots$                        \\
        $c_s$    & $a_{s,1}$ & $a_{s,2}$ & $\dots$   & $a_{s,s-1}$         \\
        \hline
                 & $b_1$     & $b_2$     & $\dots$   & $b_{s-1}$   & $b_s$ \\
    \end{tabular}
    \end{column}
    \end{columns}
\end{frame}

\begin{frame}{Multiple Step Methods}
\end{frame}

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