A breakpoint-crossing RK4 step has a second-order defect

For $0\le t\le \tau$, the delayed value is $\varphi (t-\tau )=t-\tau$, so $y^{\mathrm{\prime }}(t)=t-\tau$ and the solution matching $y(0)=0$ is $y(t)=t^2\mathrm{/}2-\tau t$. Earlier RK4 steps lie wholly in this interval, where $g(t)=t-\tau$ is affine and the RK4 rule in the Example integrates it exactly. Moreover $y^{\mathrm{\prime }\mathrm{\prime }}$ jumps at $\tau$: its left limit is $1$, while its right limit is $y^{\mathrm{\prime }}(0+)=-\tau$.

Write $x=t-\tau$. On the crossing step, the delayed right-hand side is $g(t)=x$ for $x\le 0$ and $g(t)=x^2\mathrm{/}2-\tau x$ for $x\ge 0$. The terminal stage queries $y(t_{+}-\tau )=y(h\mathrm{/}2)$, and $h\mathrm{/}2<t_{-}=mh$ for $m\ge 1$, so this value is already computed exactly by step 1.1; the midpoint uses $y(0)=0$. Therefore the RK4 stage values are $g(t_{-})=-h\mathrm{/}2$, $g(\tau )=0$, and $g(t_{+})=h^2\mathrm{/}8-\tau h\mathrm{/}2$.

The RK4 increment from the stated rule is $Q=h[-h\mathrm{/}2+(h^2\mathrm{/}8-\tau h\mathrm{/}2)]\mathrm{/}6=-(1+\tau )h^2\mathrm{/}12+h^3\mathrm{/}48$.

The exact increment is ${\int }_{-h\mathrm{/}2}^{0}xdx+{\int }_0^{h\mathrm{/}2}(x^2\mathrm{/}2-\tau x)dx=-(1+\tau )h^2\mathrm{/}8+h^3\mathrm{/}48$. Subtracting this from $Q$ gives the exact-start defect $(1+\tau )h^2\mathrm{/}24>0$.

This defect is not $O(h^5)$ as $h\to 0$, so the smooth-RHS fourth-order local-defect premise cannot be transferred across this breakpoint. Aligning or restarting at the breakpoint removes this particular crossing step but supplies no higher-order theorem without separate smoothness and error bounds.