One-step ODE Consistency and Local Error — Examples

Fix $T>0$, let $R:=e^T(1+T)+1$, and take $V=(-1,T+1)\times (-R,R)$ and $K=[0,T]\times [-R+1,R-1]$. These are an open domain with convex state fibers and a compact subset $K\subset V$. For $f(t,z)=z$, the bounds in [F2] hold throughout $V$ with $F_1=R$, $M_{t,1}=0$, $M_{11}=1$, and $L=1$. The function $y(t)=e^t$ solves the IVP, its graph on $[0,T]$ lies in $K$, and for every $0<h\le T-t$ the Euler output state $e^t(1+h)$ lies between $0$ and $e^T(1+T)=R-1$, so its input and output pairs lie in $K$. Starting from $e^t$, Euler returns $e^t(1+h)$ and the exact endpoint is $e^{t+h}=e^t{e}^h$. Therefore [F1] gives ${\delta }_h(t;y)=e^t(1+h-e^h)$.

Apply [F3] to $s\mapsto e^s$ at $0$ with increment $h$: there is ${\theta }_h\in (0,h)$ with $e^h=1+h+\frac{1}{2}{e}^{{\theta }_h}{h}^2$. Thus $\mathrm{\mid }{\delta }_h(t;y)\mathrm{\mid }=\frac{1}{2}{e}^{t+{\theta }_h}{h}^2$.

Since $0<{\theta }_h<h$, for $0\le t\le T-h$ we have $\frac{1}{2}{h}^2\le \mathrm{\mid }{\delta }_h(t;y)\mathrm{\mid }\le \frac{1}{2}{e}^T{h}^2$. The lower and upper bounds establish uniform $\mathrm{\Theta }(h^2)$ error, hence order exactly one under [F1].

For any scalar inputs $u,v$, $\mathrm{\mid }{\mathrm{\Psi }}_h(t,u)-{\mathrm{\Psi }}_h(t,v)\mathrm{\mid }=\mathrm{\mid }(1+h)(u-v)\mathrm{\mid }=(1+h)\mathrm{\mid }u-v\mathrm{\mid }$. This verifies the stated stability factor; it also agrees with [F2] for $f(t,z)=z$ and $L=1$.

For $t<0$, $y(t)=-t^2\mathrm{/}2$ and $y^{\mathrm{\prime }}(t)=-t=\mathrm{\mid }t\mathrm{\mid }$; for $t>0$, $y(t)=t^2\mathrm{/}2$ and $y^{\mathrm{\prime }}(t)=t=\mathrm{\mid }t\mathrm{\mid }$. At $t=0$, ${\lim }_{s\to 0}(y(s)-y(0))\mathrm{/}s={\lim }_{s\to 0}\mathrm{\mid }s\mathrm{\mid }\mathrm{/}2=0=\mathrm{\mid }0\mathrm{\mid }$. Hence $y$ is the exact solution through $y(0)=0$.

At $t_h=-h\mathrm{/}2$, the exact value is $y(t_h)=-h^2\mathrm{/}8$ and the explicit midpoint stage time is $0$, where $f(0,z)=0$ for every stage value $z$. Thus $M_h(t_h,y(t_h))=y(t_h)=-h^2\mathrm{/}8$, while $y(t_h+h)=y(h\mathrm{/}2)=h^2\mathrm{/}8$.

Therefore ${\delta }_h(t_h;y)=-h^2\mathrm{/}4$. The exact segment has $\mathrm{\mid }t\mathrm{\mid }\le h\mathrm{/}2\le 1\mathrm{/}2$ and $\mathrm{\mid }y(t)\mathrm{\mid }\le 1\mathrm{/}8$; the input and output states are $-h^2\mathrm{/}8$, and the midpoint stage state is $h^2\mathrm{/}8$, all in $[-1,1]$. Their time coordinates also lie in $[-1,1]$, so the definition applies on the fixed compact set $K$. If a uniform order-two bound held, $h^2\mathrm{/}4\le C{h}^3$ for every $0<h\le 1$, forcing $C\ge 1\mathrm{/}(4h)$ as $h\downarrow 0$, which is impossible.

For an arbitrary $t\in \mathbb{R}$ and $h>0$, the method increment from the exact state is $h\mathrm{\mid }t+h\mathrm{/}2\mathrm{\mid }$. If $[t,t+h]$ lies on either side of $0$, then $y$ is quadratic there and this increment equals $y(t+h)-y(t)$, so the defect is zero. If $t<0<t+h$, put $r=-t\mathrm{/}h\in (0,1)$. Using $y(t)=-t^2\mathrm{/}2$ and $y(t+h)=(t+h{)}^2\mathrm{/}2$ gives ${\delta }_h(t;y)=-h^2{r}^2$ for $0<r\le 1\mathrm{/}2$ and ${\delta }_h(t;y)=-h^2(1-r{)}^2$ for $1\mathrm{/}2\le r<1$. Thus every crossing step has $\mathrm{\mid }{\delta }_h(t;y)\mathrm{\mid }\le h^2\mathrm{/}4$, and the same bound holds for every step. In particular, it holds uniformly for all admitted starts and steps in $K$; together with step 3.1, the uniform local order is exactly one.