For $U(x)=x^Tx/2$, one leapfrog step is
$$
\begin{pmatrix}x'\\p'\end{pmatrix}
=M_\varepsilon
\begin{pmatrix}x\\p\end{pmatrix},
\qquad
M_\varepsilon=
\begin{pmatrix}
(1-\varepsilon^2/2)I&\varepsilon I\\
(-\varepsilon+\varepsilon^3/4)I&(1-\varepsilon^2/2)I
\end{pmatrix}.
$$
Therefore $L$ steps give
$$
\boxed{
\begin{pmatrix}x(L\varepsilon)\\p(L\varepsilon)\end{pmatrix}
=M_\varepsilon^L
\begin{pmatrix}x(0)\\p(0)\end{pmatrix},}
$$
which is a linear transformation.


Solution

Past exam of the mathematics course of the University of Cambridge / 2019 / iii / Paper 216 / 6 / b / Solution

 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2019/iii/paper-216/6/b/solution