Let $\Phi_h$ be the one-step map of a numerical method. It is a <time-symmetric numerical method> when reversing the step exactly reverses the update:
$$
\boxed{\Phi_{-h}=\Phi_h^{-1},
\quad\text{equivalently}\quad
\Phi_{-h}\circ\Phi_h=I.}
$$

Let $\varphi_h$ be the exact <flow map> and suppose the method has order $p$ with a nonzero leading <local truncation error>:
$$
\Phi_h(y)=\varphi_h(y)+h^{p+1}d(y)+O(h^{p+2}).
$$
Inverting this expansion changes the sign of its leading perturbation, so
$$
\Phi_h^{-1}(y)=\varphi_{-h}(y)-h^{p+1}\widetilde d(y)+O(h^{p+2}),
$$
where transport by the exact flow only changes $d$ by $O(h)$ and hence does not affect the leading parity. On the other hand, replacing $h$ by $-h$ in the first expansion gives
$$
\Phi_{-h}(y)=\varphi_{-h}(y)+(-1)^{p+1}h^{p+1}\widetilde d(y)+O(h^{p+2}).
$$
Time symmetry equates these expressions, so $(-1)^{p+1}=-1$. Hence $p+1$ is odd and
$$
\boxed{p\text{ is even}.}
$$

Solution

Past exam of the mathematics course of the University of Cambridge / 2023 / iii / Paper 341 / 5 / a / Solution

 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2023/iii/paper-341/5/a/solution