A self-adjoint heat spectrum gives a long RKC1 stability step

The eigenvalues of $A$ are $-1$ and $-200$, so $\rho =200$ and $h\rho =100$. For $s=8$, the certified RKC1 bound is $2s^2=128$, hence the full step is stable on both eigenmodes and the matrix update is nonexpansive.

The corresponding RKC1 scalar arguments are $z=-1\mathrm{/}2$ and $z=-100$; the latter maps to $1+z\mathrm{/}{s}^2=-9\mathrm{/}16\in [-1,1]$. The first-order recurrence uses eight stage evaluations for this step.

For Euler on the $-200$ mode, one step of length $1\mathrm{/}2$ has $z=-100$ and factor $1+z=-99$, whose absolute value exceeds one. If Euler instead divides the same time interval into $m$ equal substeps, stability on that mode requires $(h\mathrm{/}m)200\le 2$, hence $m\ge 50$. For $m\ge 50$, the Euler argument on the $-1$ mode has magnitude $h\mathrm{/}m=1\mathrm{/}(2m)\le 1\mathrm{/}100\le 2$, so both modes are stable. Thus the eight-stage RKC1 step uses fewer right-hand-side evaluations than stable Euler substepping for this spectrum, without implying comparable accuracy.