Set $\overline{\mathbf y}=(\mathbf y_n+\mathbf y_{n+1})/2$. The <implicit midpoint rule> reads
$$
\mathbf y_{n+1}-\mathbf y_n
=hA(\overline{\mathbf y})\overline{\mathbf y}.
$$
Taking the <inner product> with $\mathbf y_{n+1}+\mathbf y_n=2\overline{\mathbf y}$ gives
$$
\begin{aligned}
\|\mathbf y_{n+1}\|_2^2-\|\mathbf y_n\|_2^2
&=(\mathbf y_{n+1}+\mathbf y_n)^T(\mathbf y_{n+1}-\mathbf y_n)\\
&=2h\overline{\mathbf y}^{T}A(\overline{\mathbf y})\overline{\mathbf y}=0.
\end{aligned}
$$
The last equality again follows from <skew-symmetric matrix>[skew-symmetry]. <Mathematical induction> therefore yields
$$
\boxed{\|\mathbf y_n\|_2=\|\mathbf y_0\|_2\quad(n\in\mathbb Z_+).}
$$

Solution

Past exam of the mathematics course of the University of Cambridge / 2023 / iii / Paper 341 / 1 / b / Solution

 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2023/iii/paper-341/1/b/solution