= Solution
The derivative of the backward <characteristic flow map> is
$$
DS_{0,t}=\begin{pmatrix}\cosh t&-\sinh t\\-\sinh t&\cosh t\end{pmatrix},
\qquad \boxed{\det DS_{0,t}=\cosh^2t-\sinh^2t=1.}
$$
Thus the <change of variables formula> preserves phase-space <Lebesgue measure>. With zero source, $f_t=f_0\circ S_{0,t}$, so for every finite $p>0$,
$$
\int_{\mathbb R^2}|f_t(x,v)|^p\,dx\,dv
=\int_{\mathbb R^2}|f_0(x_0,v_0)|^p\,dx_0\,dv_0.
$$
Taking the $p$th root proves \b[$\|f_t\|_p=\|f_0\|_p$]. For $0<p<1$ this is a <quasi-norm>, and the argument still works because it uses only a change of variables, not the <triangle inequality>. The identity also holds in the extended sense when an <integral> is infinite. Since the flow is bijective and measure-preserving, it additionally preserves the <essential supremum>, so the same conclusion holds for $p=\infty$.
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