Small steps

Signed-off-by: Marcello Seri <[email protected]>

hm.tex

@@ -892,7 +892,7 @@ \chapter*{Preface}
  \section{First steps with conserved quantities}
  \subsection{Back to one degree of freedom}\label{sec:bdf}
 
- Consider the equation
+ Consider the one dimensional Newton's equation for a point particle with unit mass and position $x(t)$:
  \begin{equation}\label{eq:oscillator}
  \ddot x = F(x), \qquad F:\mathbb{R}\to\mathbb{R}, \quad t\in \mathbb{R}.
  \end{equation}
@@ -905,8 +905,9 @@ \chapter*{Preface}
  \end{aligned}
  \right..
  \end{equation}
- The solutions of \eqref{eq:oscillatorfirstorder} are parametric curves\footnote{Usually called \emph{integral curves} of the ordinary differential equation.} $(x(t),y(t)):\mathbb{R}\to\mathbb{R}^2$ in the $(x,y)$-space.
- %
+ The solutions of \eqref{eq:oscillatorfirstorder} are parametric curves $(x(t),y(t)):\mathbb{R}\to\mathbb{R}^2$ in the $(x,y)$-space.
+ These curves are often called \emphidx{trajectories} of the system or \emph{integral curves} of the ordinary differential equation, it depends mostly on the context and the research fields in which they appear.
+
  If $y\neq0$, we can apply the chain rule, $\frac{\dd y}{\dd t} = \frac{\dd y}{\dd x} \frac{\dd x}{\dd t}$, to get
  \begin{equation}\label{eq:lef}
  \frac{F(x)}y = \frac{\dot y}{\dot x} = \frac{\dd y}{\dd x}.