Stored retarded Euler and distributed midpoint work, storage, and grid alignment

Fix positive integers $d,m$, positive $T,\tau$, $h=\tau \mathrm{/}m$, and $N=\lceil T\mathrm{/}h\rceil$. Consider the following finite storage schedule: initialize and retain $d$-vectors $Y_j=\varphi (jh)$ for $j=-m,\dots ,0$; at each index $n=0,\dots ,N-1$, first test an optional nonnegative integer budget $B$, stopping if $n\ge B$, and otherwise evaluate one vector $f(nh,Y_n,Y_{n-m})$, form $Y_{n+1}=Y_n+hf(nh,Y_n,Y_{n-m})$, and append it. Assume every reached evaluation is defined and returns in finite time. A complete schedule uses exactly $N$ calls to $f$ and retains $N+m+1$ state vectors. If $B<N$, it stops after exactly $B$ calls with $m+B+1$ vectors. Initialization uses $m+1$ history evaluations in either case. Full history storage is $O(d(N+m))=O(d(T+\tau )\mathrm{/}h)$ real entries. With direct array indexing and a constant-cost scalar floor operation, a query of the piecewise-linear interpolant of the stored vectors costs $O(d)$ arithmetic. Counts exclude internal evaluation costs.

For a complete stored history, fix a positive integer $q$ and $n\in \{0,\dots ,N\}$, put $r=qm$, $\mathrm{\Delta }=h\mathrm{/}q$, and ${\theta }_j=(j+1\mathrm{/}2)\mathrm{\Delta }$. Consider the schedule that initializes $A=0\in {\mathbb{R}}^d$ and, for $j=0,\dots ,r-1$, adds $\mathrm{\Delta }K({\theta }_j){\widehat{y}}_h(nh-{\theta }_j)$ to $A$. With returning kernel evaluations, this makes $r$ kernel evaluations and $r$ history queries, using $O(rd)$ arithmetic besides kernel evaluations. Streaming the scalar weights uses $O(d)$ additional storage; caching all $r$ weights uses $O(r+d)$, excluding the input history. Every stored-grid knot, including a delay breakpoint $k\tau$, encountered by $\theta \mapsto {\widehat{y}}_h(nh-\theta )$ on $[0,\tau ]$ is a quadrature-cell endpoint. No constant-memory claim is made for the complete stored history.