Delayed feedback gives an explicit two-block Euler recurrence

On the first block $0\le t\le \tau$, the delayed value is $1$, so $y(t)=1+at$. On the second block, writing $s=t-\tau \in [0,\tau ]$, the delayed value is $y(s)=1+as$, hence $y^{\mathrm{\prime }}(\tau +s)=a+a^{2}s$ and $y(2\tau )=1+2a\tau +a^2{\tau }^2\mathrm{/}2$.

The first Euler block gives $Y_j=1+ajh$ for $0\le j\le m$. On the second block, [F2] gives $Y_{m+j+1}=Y_{m+j}+ah{Y}_j$ for $0\le j<m$. Summing yields $Y_{2m}=1+a\tau +ah{\sum }_{j=0}^{m-1}(1+ajh)$. Pairing the terms $j$ and $m-1-j$ (and retaining the middle term when $m$ is odd) gives ${\sum }_{j=0}^{m-1}j=m(m-1)\mathrm{/}2$, so $Y_{2m}=1+2a\tau +a^2{h}^{2}m(m-1)\mathrm{/}2$.

Subtracting step 1.2 from step 1.1 and using $\tau =mh$ gives $y(2\tau )-Y_{2m}=a^2{h}^{2}m\mathrm{/}2=a^2\tau h\mathrm{/}2$. Since this is nonnegative, its absolute value is the claimed error, including $a=0$. On the first Euler block, $\mathrm{\mid }{Y}_j\mathrm{\mid }\le 1+\mathrm{\mid }a\mathrm{\mid }\tau$; on the second, $\mathrm{\mid }{Y}_{m+j}\mathrm{\mid }\le 1+\mathrm{\mid }a\mathrm{\mid }\tau +\mathrm{\mid }a\mathrm{\mid }\tau (1+\mathrm{\mid }a\mathrm{\mid }\tau )$ for $0\le j\le m$. The exact solution is polynomial on each of the first three delay blocks: besides step 1.1, on $[2\tau ,3\tau ]$ it is $y(2\tau +s)=y(2\tau )+(a+a^2\tau )s+(a^2\mathrm{/}2)s^2+(a^3\mathrm{/}6)s^3$. These formulas give a common bounded tube independent of $h$. The data therefore satisfy [F3] with $\eta =h_0=\tau$, because $f(t,x,z)=az$ is time-independent, the history has zero second derivative, and the tube is compact. Thus the exact error formula is consistent with the first-order guarantee.