Compatible matrix evolution and time stepping — Examples

The data satisfy the defining matrix assumptions: $M_e,M_b$ are positive definite, $S$ is positive semidefinite, and $DC=CG=0$. With $j=0$, [F1] reduces to $\dot{e}=b$ and $\dot{b}=-e$.

The forward Euler rule in the refuted statement therefore gives $e_1=e_0+h{b}_0$ and $b_1=b_0-h{e}_0$.

Expanding gives $W(e_1,b_1)=\frac{1}{2}((e_0+h{b}_0{)}^2+(b_0-h{e}_0{)}^2)=(1+h^2)\frac{1}{2}(e_0^2+b_0^2)=(1+h^2)W(e_0,b_0)$. Because $h>0$ and the initial pair is nonzero, $W(e_0,b_0)>0$ and $1+h^2>1$, so the step strictly increases energy. This refutes the universal preservation claim.

These one-dimensional data satisfy the matrix assumptions of the algorithm. Their positive mass roots equal $1$, so [F2] gives $K=[1]$ and $h\parallel K{\parallel }_2=2$, exactly the endpoint in the refuted extension.

Write $z_n:=(e_n,b_{n-1\mathrm{/}2}{)}^T$. By [F1], $z_{n+1}=A{z}_n$ with $A=\left(\begin{array}{cc}-3& 2\\ -2& 1\end{array}\right)$. Set $J:=A+I=\left(\begin{array}{cc}-2& 2\\ -2& 2\end{array}\right)$. Direct multiplication gives $J\mathrm{\ne }0$ and $J^2=0$, so $A=-(I-J)$ and the binomial formula yields $A^n=(-1{)}^n(I-nJ)$ for every integer $n\ge 0$.

For the nonzero initial state $z_0=(1,0{)}^T$, one has $J{z}_0=(-2,-2{)}^T\mathrm{\ne }0$. The reverse triangle inequality therefore gives $\parallel z_n{\parallel }_2=\parallel z_0-nJ{z}_0{\parallel }_2\ge n\parallel J{z}_0{\parallel }_2-\parallel z_0{\parallel }_2=2\sqrt{2}n-1$. These outputs occur in the algorithm for each finite horizon $N\ge n$ and their norm ratio to $\parallel z_0{\parallel }_2$ is unbounded as $n$ grows. No constant $C$ independent of $N$ can satisfy the refuted estimate.