Local orders and leading estimator term for the Euler–Heun pair

Define $F(s):=f(s,y(s))$. On the open segment, [F3] gives $F^{\mathrm{\prime }}(s)={\mathrm{\partial }}_{t}f(s,y(s))+D_{y}f(s,y(s))f(s,y(s))$, which extends continuously to $s=t$ with value $g$. For each component and $0<u\le h$, the mean value theorem [F6] on $[t,t+u]$ shows its secant slope equals $F_i^{\mathrm{\prime }}(c_i)$ for some $t<c_i<t+u$; as $u\downarrow 0$ this tends to $g_i$. Thus the right derivative of $F$ at $t$ is $g$, so $y^{\mathrm{\prime }\mathrm{\prime }}(t)=g$. Since $f$ is $C^2$, differentiating the displayed formula on the open segment shows $F^{\mathrm{\prime }\mathrm{\prime }}$ is bounded there and extends continuously to the endpoints; hence $y^{\mathrm{\prime }\mathrm{\prime }\mathrm{\prime }}=F^{\mathrm{\prime }\mathrm{\prime }}$ is bounded on the segment. For a coordinate $y_i$ and $0<u\le h$, let $R_i(u):=y_i(t+u)-y_i(t)-u{y}_i^{\mathrm{\prime }}(t)-\frac{1}{2}{u}^2{y}_i^{\mathrm{\prime }\mathrm{\prime }}(t)$, set $C_i:=R_i(u)\mathrm{/}{u}^3$, and put $G_i(v):=y_i(t+v)-y_i(t)-v{y}_i^{\mathrm{\prime }}(t)-\frac{1}{2}{v}^2{y}_i^{\mathrm{\prime }\mathrm{\prime }}(t)-C_i{v}^3$ on $[0,u]$. Then $G_i(0)=G_i(u)=G_i^{\mathrm{\prime }}(0)=G_i^{\mathrm{\prime }\mathrm{\prime }}(0)=0$. Three successive applications of Rolle's theorem [F6] give $G_i^{\mathrm{\prime }\mathrm{\prime }\mathrm{\prime }}({\xi }_i)=0$ for some $0<{\xi }_i<u$, so $R_i(u)=y_i^{\mathrm{\prime }\mathrm{\prime }\mathrm{\prime }}(t+{\xi }_i)u^3\mathrm{/}6$. At $u=0$ the remainder is zero. Applying this to $u=h$ and every $u=\theta h$, and using the common derivative bounds and finite dimension, proves both the endpoint expansion and the uniform-in-$\theta$ expansion in the Statement.

For fixed $h$, the segment from $(t,y_0)$ to the Euler stage $(t+h,y_0+h{K}_1)$ lies in the compact convex time-state set. Since that set lies in an open neighborhood where $f$ is $C^2$, compactness gives a smaller open convex neighborhood of the segment on which the Taylor theorem [F4] applies. Taylor expansion of each component of $f$ there along the increment $(h,h{K}_1)$ has linear term $h{\mathrm{\partial }}_{t}f+h{D}_{y}f{K}_1=hg$ and a second-derivative remainder $O(h^2)$, uniformly by the boundedness assumption. Therefore $K_2=K_1+hg+O(h^2)$. Substitution into [F1] gives $Y_E=y_0+h{K}_1$ and $Y_H=y_0+h{K}_1+\frac{1}{2}{h}^{2}g+O(h^3)$.

Subtract the exact expansion in step 1.1 to obtain the two defect formulas and subtract the Euler output from the Heun output to obtain the estimator expansion. These are uniform $O(h^2)$ and $O(h^3)$ defect bounds, so [F2] gives the stated local orders. For $p=1$, the Euler defect coefficient is $-g\mathrm{/}2$ and the Heun defect is $O(h^3)$; if $g\mathrm{\ne }0$, [F5] gives the ratio-one conclusion. If $g=0$, both leading coefficients vanish and [F5] supplies no such inference.