Grid-aligned Euler has first-order error for retarded delays

Fix $h=\tau \mathrm{/}m\le h_0$, let $N$ be the least integer with $Nh\ge T$, and put $T_h=Nh\le T+\eta$. Run the algorithm through $T_h$ and call its stored interpolant ${\widehat{y}}_h$. Convex interpolation and [A1] give $\mathrm{\parallel }{\widehat{y}}_h(t){\mathrm{\parallel }}_2\le R$. Extend it constantly just beyond each endpoint for local ODE neighborhoods of time radius less than $\min \{\rho ,\tau \}$. Define $F_h(t,x)=f(t,x,{\widehat{y}}_h(t-\tau ))$ for $0\le t\le T_h$. It is continuous and $L_s$-Lipschitz in $x$. By [F3], $\mathrm{\parallel }{F}_h(t,0){\mathrm{\parallel }}_2\le Q+L_{s}R$. Local solutions of $z^{\mathrm{\prime }}=F_h(t,z)$ therefore obey $\mathrm{\parallel }z(t){\mathrm{\parallel }}_2\le \mathrm{\parallel }\varphi (0){\mathrm{\parallel }}_2+(Q+L_{s}R)t+L_s{\int }_0^t\mathrm{\parallel }z(s){\mathrm{\parallel }}_{2}ds$. The vector integral norm bound [F5] and triangle inequality [F12] justify this estimate. For $t\le T+\eta$, replace the first two terms by the constant $A=\mathrm{\parallel }\varphi (0){\mathrm{\parallel }}_2+(Q+L_{s}R)(T+\eta )$ and apply [F10]; this gives $\mathrm{\parallel }z(t){\mathrm{\parallel }}_2\le A{e}^{L_s(T+\eta )}=R_z$. On the compact tube of radius $R_{\ast }$ the field is bounded by $M_f$, so [F8] continues each local solution by a common positive short-time amount while it remains in that ball. The a priori bound keeps each continuation endpoint within radius $R_z$; finitely many continuations cover $[0,T_h]$. Denote the resulting solution by $z_h$ and set $z_h=\varphi$ on $[-\tau ,0]$. Since $t_n-\tau =t_{n-m}$ is a stored node, [F2] shows the Euler updates are exactly forward Euler applied to this fixed ODE.

The auxiliary path and the Euler states lie in the radius-$R_{\ast }$ tube, and $\mathrm{\parallel }{z}_h^{\mathrm{\prime }}(t){\mathrm{\parallel }}_2\le M_f$. On positive-time cells, each slope of ${\widehat{y}}_h$ is an Euler right-hand side bounded by $M_f$; on history cells, [F4] and [F5] bound its slope by $\sup \mathrm{\parallel }{\varphi }^{\mathrm{\prime }}{\mathrm{\parallel }}_2$. Thus ${\widehat{y}}_h$ is $S$-Lipschitz on $[-\tau ,T_h]$.

The Euler map for the fixed ODE is ${\mathrm{\Psi }}_{h,n}(x)=x+h{F}_h(t_n,x)$. By [F3] and the triangle inequality [F12] it satisfies $\mathrm{\parallel }{\mathrm{\Psi }}_{h,n}(x)-{\mathrm{\Psi }}_{h,n}(\tilde{x}){\mathrm{\parallel }}_2\le (1+L_{s}h)\mathrm{\parallel }x-\tilde{x}{\mathrm{\parallel }}_2$. For $s\in [t_n,t_{n+1}]$, [F3], the time bound, and step 2.1 give $\mathrm{\parallel }{F}_h(s,z_h(s))-F_h(t_n,z_h(t_n)){\mathrm{\parallel }}_2\le (L_t+L_s(M_f+S))(s-t_n)\mathrm{.}$ By [F4], the exact increment is the integral of $F_h(s,z_h(s))$; integrating this estimate and using [F5] gives exact-start local defect at most $C_{\mathrm{l}\mathrm{o}\mathrm{c}}{h}^2$ by [F11] and [F12]. Apply [F9] on the uniform grid of length $T_h$, with initial error zero. Since $T_h\le T+\eta$, it gives ${\max }_{0\le n\le N}\mathrm{\parallel }{Y}_n-z_h(t_n){\mathrm{\parallel }}_2\le C_{\mathrm{o}\mathrm{d}\mathrm{e}}h\mathrm{.}$

On a solution cell, the nodal error bound proved in the preceding step contributes at most $C_{\mathrm{o}\mathrm{d}\mathrm{e}}h$ by the triangle inequality [F12]. The interpolant of the two exact $z_h$ endpoints differs from $z_h(t)$ by at most $M_{f}h\mathrm{/}2$: if $\lambda =(t-t_n)\mathrm{/}h$, the endpoint-increment estimate is $2\lambda (1-\lambda )M_{f}h\le M_{f}h\mathrm{/}2$. Hence $\mathrm{\parallel }{\widehat{y}}_h(t)-z_h(t){\mathrm{\parallel }}_2\le (C_{\mathrm{o}\mathrm{d}\mathrm{e}}+M_f\mathrm{/}2)h$. On history cells, [F6] gives error at most $M_{\varphi }{h}^2\mathrm{/}8\le (M_{\varphi }{h}_0\mathrm{/}8)h$. Therefore ${\sup }_{-\tau \le t\le T_h}\mathrm{\parallel }{\widehat{y}}_h(t)-z_h(t){\mathrm{\parallel }}_2\le C_{\mathrm{h}\mathrm{i}\mathrm{s}\mathrm{t}}h\mathrm{.}$

Let $u(t)=\mathrm{\parallel }{z}_h(t)-y(t){\mathrm{\parallel }}_2$ for $-\tau \le t\le T_h$. It is zero on the history. Subtract the integral equations for $z_h$ and $y$ using [F1] and [F4], then use [F3] and bound $\mathrm{\parallel }{\widehat{y}}_h(t-\tau )-y(t-\tau ){\mathrm{\parallel }}_2$ by $C_{\mathrm{h}\mathrm{i}\mathrm{s}\mathrm{t}}h+u(t-\tau )$ using step 4.1. This yields $u(t)\le {\int }_0^t(L_s{C}_{\mathrm{h}\mathrm{i}\mathrm{s}\mathrm{t}}h+L_s(u(s)+u(s-\tau )))ds\mathrm{.}$ Apply [F7] with $H=0$ and constant forcing $q=L_s{C}_{\mathrm{h}\mathrm{i}\mathrm{s}\mathrm{t}}h$ to obtain ${\sup }_{0\le t\le T_h}u(t)\le L_s{C}_{\mathrm{h}\mathrm{i}\mathrm{s}\mathrm{t}}h(T+\eta )e^{2L_s(T+\eta )}$. The triangle inequality and the interpolation bound from step 4.1 give the claimed estimate on $[-\tau ,T]$ by [F12].