= Solution
Write $A_n=A(\mathbf y_n)$. The <trapezoidal rule> is
$$
\mathbf y_{n+1}-\mathbf y_n
=\frac h2(A_n\mathbf y_n+A_{n+1}\mathbf y_{n+1}).
$$
Taking the <inner product> with $\mathbf y_{n+1}+\mathbf y_n$ gives
$$
\begin{aligned}
\|\mathbf y_{n+1}\|_2^2-\|\mathbf y_n\|_2^2
=\frac h2\bigl(&\mathbf y_{n+1}^TA_n\mathbf y_n
+\mathbf y_n^TA_{n+1}\mathbf y_{n+1}\\
&+\mathbf y_n^TA_n\mathbf y_n
+\mathbf y_{n+1}^TA_{n+1}\mathbf y_{n+1}\bigr).
\end{aligned}
$$
The two quadratic terms vanish because each $A_j$ is <skew-symmetric matrix>[skew-symmetric]. Moreover $\mathbf y_{n+1}^TA_n\mathbf y_n=-\mathbf y_n^TA_n\mathbf y_{n+1}$. Hence
$$
\boxed{
\|\mathbf y_{n+1}\|_2^2-\|\mathbf y_n\|_2^2
=\frac h2\mathbf y_n^T(A_{n+1}-A_n)\mathbf y_{n+1}.}
$$
Unlike the <implicit midpoint rule>, the trapezoidal rule evaluates $A$ at two different states, so the mixed terms need not cancel.
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