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\title{}
\subtitle{Practical Tests with Global Error Evaluation Techniques in Multistep Methods}
\author{\vskip 1.5cm A. Ramanantoanina, ~Dr G. Kulikov}

\institution{%
  \vskip 2cm \begin{flushright}\small{\green{SANUM2006}} \end{flushright}}

\begin{document}
  \maketitle

  %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
  %%            INTRODUCTION               %%
  %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
  \begin{slide}{ODE's \& Multistep Methods}
    \vskip 0.75cm
    \begin{itemize}
    \item Consider the system of ODE
      \begin{equation}
    \label{ode}
          { x'(t)=f\bigl(x(t)\bigr),\;
        t\in [0,T], \; x(t_0)=x_0,}
      \end{equation}
      and the grid $$ w = \{ t_0 = 0, t_1, t_2, \ldots, t_N = T \}$$
    \item Multistep methods
    \begin{equation}
      \label{method_ode}
      \sum_{j = 0}^l \alpha_j x_{k+1 -j} =
      h\sum_{j = 0}^l \beta_j f(x_{k+1 - j} )
    \end{equation}
    \end{itemize}
  \end{slide}

  \begin{slide}{Local \& Global Errors}
    \begin{itemize}
      \vskip 1cm
    \item Local error
      $${ e(t_k) = x(t_k) - \tilde{x}_k}$$
      {\it where} $\tilde{x}_k$ {\it is computed using a multistep method
      with exact input.}
      \vskip 1cm
    \item Global error
      $$E(t_k) = x(t_k) - x_k$$
    \end{itemize}
  \end{slide}

  %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
  %%              TECHNIQUES               %%
  %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

  \overlays{3}{%
    \begin{slide}{Global error evaluation}
      Methods:
      \begin{itemstep}
      \item Solving for the correction\\
    $$E_h'(t) = f(P_h - E_h(t)) - P_h'(t), \; E_h(0) = 0$$
      \item Zadunaisky's technique\\
    $$\tilde{x}' = f(\tilde{x}) + d(h), \; \tilde{x}(0) =
    x(0)$$
    where the perturbation $d(h) = P_h' - f(P_h)$
      \item Solving the linearised discrete variational equation\\
    $$E_k = P_kC $$
      \end{itemstep}
    \end{slide}
  }

  \overlays{2}{%
    \begin{slide}{Global error evaluation ({\normalsize Cont})}
      \vskip .5cm
      \begin{itemstep}
      \item Richardson extrapolation \\
    $$E_k = {\displaystyle\frac{x_k^h - x_k^{\frac{h}{2}}}{1 -
    2^{-s}}}$$ where $s$ is the order of the multistep method.
    \vskip .5cm
      \item Using two different methods\\
    $$E_k = x_k^1 - x_k^2$$
      \end{itemstep}
    \end{slide}
  }

  %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
  %%            TEST PROBLEMS              %%
  %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

  \begin{slide}{Test Problems}
    \vskip 1cm
    \begin{itemize}
    \item An oscillatory problem described by
    \begin{equation*}
      x'(t) = x(t) cos(t),
    \end{equation*}with the initial condition $x(0) = 1$  for $t \in [0,
      1]$.
    \vskip 0.3cm
    The exact solution to this problem is
    \begin{equation*}
      x(t) = e^{\sin(x(t))}.
    \end{equation*}
    \end{itemize}
  \end{slide}

  \begin{slide}{Test Problems (Cont)}
    \vskip 1cm
    \begin{itemize}
    \item A stiff ODE: Butcher test problem
      $$x'(t) = \lambda (x(t) - \sin(\mu t)) + \mu \cos(\mu t)$$ for $t \in
      [0, 1]$ with the initial condition $x(0) = 1.$
      \vskip 0.5cm
      It is known that the solution to this problem is:
      $$ x(t) = \sin(\mu t) + e^{\lambda t}.$$
    \end{itemize}
  \end{slide}

  %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
  %%               GRAPHS                  %%
  %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

  \begin{slide}{Graphs}
    \vskip 1.5cm
    \begin{center}
      We present in the following figures the graphs of the errors
      in the estimation provided by different methods.
    \end{center}
  \end{slide}




  \begin{slide}{Conclusion}
  \end{slide}
\end{document}
