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\title{CSC688/MTH686: Scientific Computation \\ Equation Solving}
\author{\today}
\date{}
\maketitle

\begin{enumerate}
  \item Let $f : \mathds{R}^n \to \mathds{R}^n$; $x \mapsto f(x)$ be of class
  $C^1$. Consider $g : \mathds{R}^n \to \mathds{R}^n$ defined by
  \begin{equation}
     g(x):=x-\nabla f(x)^{-1}f(x).
  \end{equation}
  Verify that $|\det\nabla g(r)| < 1$ where $f(r)=0$.

  \item Let $f : \mathds{C} \to \mathds{C}$; $z \mapsto f(z)$ and
  consider the problem of solving $f(z)=0$ using Newton's method.
  Demonstrate that if $f(z)$ is analytic then Newton's recursive
  formula is given by
  \begin{equation}
    z_{n+1} = z_n - \frac{f(z_n)}{f'(z_n)}.
  \end{equation}
  \textbf{Hint:} Write $f(z) = g(x,y,t) + \mathrm{i} h(x,y,t)$. Then note that
  $f'(z)=\frac{1}{2}(g_x+h_y)+\frac{\mathrm{i}}{2}(h_x-g_y)$ and
  furthermore that if $f(z)$ is analytic then $f'(z^*)=0$.

  \item Suppose that the locations of planets M and E at time $t$
  are given by
  \begin{eqnarray}
    x_\mathrm{M} &=& -11.9084 + 57.9117 \cos \frac{2\pi t}{87.97}, \\
    y_\mathrm{M} &=& 56.6741 \sin \frac{2\pi t}{87.97},
  \end{eqnarray}
  and
  \begin{eqnarray}
    x_\mathrm{E} &=& -2.4987 + 149.6041 \cos \frac{2\pi t}{365.25}, \\
    y_\mathrm{E} &=& 149.5832 \sin \frac{2\pi t}{365.25}.
  \end{eqnarray}
  These are crude impersonifications of Mercury and Earth. Both
  orbits are elliptical with one focus, the Sun, located at
  $(x,y)=(0,0)$. To an observer on E, M is in \emph{conjunction} if
  it is located on the Sun-to-Earth line segment. Clearly, there is
  a conjunction at $t=0$. The proposed problem is to compute the time
  of the next 10 conjunctions. \textbf{Recommendation:} Use FZERO.M.

 \item In the analysis of vertical profiling oceanographic data it
 is often desirable to transform the depth dependent measurements
 into vertical wavenumber space. However, the vertical structure of
 oceanic variables is strongly depth dependent; viz., with larger
 gradients (smaller length scales) near the surface compared to
 deeper parts of the water column. This may be characterized by the
 squared Brunt--V\"{a}is\"{a}l\"{a} frequency profile, $N^2(z)$.
 Instead of the the sines and cosines of Fourier analysis (solutions
 of $F''(t)+\omega^2F(t)=0$) the natural functions to transform the
 data are the solutions of
 \begin{equation}
   F''(z)+c^{-2}N^2(z)F(z)=0.\label{F}
 \end{equation}
 Vertical displacements are projected onto the $F(z)$, while
 horizontal velocities and pressure deviations onto their
 derivatives $F'(z)$. A discrete set of eigensolutions
 $(F_n(z),c_n)$, $n=0,1,2,\cdots$, is obtained by adding two boundary
 conditions to (\ref{F}). For instance, if $N^2(z)$ is known from the surface,
 $z=0$, to the (flat) bottom, $z=-H$, the solutions of (\ref{F}) subject to
 \begin{eqnarray}
   F(-H) &=& 0, \\
   F'(0) &=& gc^{-2}F(0),
 \end{eqnarray}
 where $g$ is the acceleration of gravity, are the dynamical normal
 modes corresponding to a given $N^2(z)$. The set $\{F_n(z)\}$ constitutes
 a complete orthogonal basis. For typical oceanic conditions
 ($\int_{-H}^0 N^2(z)\,\mathrm{d}z \ll g$), the first eigensolution
 ($n=0$) is the \emph{external} or \emph{barotropic} mode (for which
 $c_0^2 \sim gH$ and $F(z) \sim z+H$) and the others are the \emph{internal} or
 \emph{baroclinic} modes (for which $c_n^2 \ll c_0^2$,
 $n=1,2,\cdots$). The proposed problem is the following. Consider $N = \bar{N}
 =$ const and demonstrate that
 \begin{equation}
   \tan\frac{\bar{N}H}{c_n} = \frac{\bar{N}c_n}{g},\label{c}
 \end{equation}
 which is a transcendental equation for $c_n$. Consider then the typical
 oceanic values $\bar{N} = 10^{-5}$ s$^{-1}$, $H = 4 \times 10^3$ m,
 and $g = 9.8$ m s$^{-2}$, and solve numerically (\ref{c}) for the first 10 modes.
 \textbf{Hint:} When $\bar{N}^2H/g \ll 1$ it is easy to see that $c_0 = \sqrt{gH}$
 and $c_n = \bar{N}H/n\pi$, $n=1,2,\cdots$. \textbf{Recommendation:} Use FZERO.M.
\end{enumerate}

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