Forward Euler on the scalar linear equation $y^{\mathrm{\prime }}=y$

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 $L=1$.