Backward Euler, the selected SDIRK, and two-stage Radau IIA converge at their stated orders under controlled stage residuals

Given: The exact IVP, common tube, state Lipschitz and smoothness bounds, uniform invariant stage balls, a selected method, and residual bounds from the Statement.

Implicit Runge–Kutta stage equations and update: the stages solve $Y_i=u+h{\sum }_j{a}_{ij}f(t+c_{j}h,Y_j)$, the update is $u+h{\sum }_i{b}_{i}f(t+c_{i}h,Y_i)$, and stiff accuracy means $b^T$ is the final row of $A$ with $c_s=1$.

Small-step implicit Runge–Kutta stages have a unique branch and controlled residual error: if the stage fixed-point map is a contraction with constant $q=h\mathrm{\parallel }A{\mathrm{\parallel }}_{\mathrm{\infty }}L<1$ on an invariant product ball, there is a unique stage root there and every approximate stage vector in the ball has error at most $\mathrm{\parallel }G(\tilde{Y}){\mathrm{\parallel }}_{\mathrm{\infty },2}\mathrm{/}(1-q)$.

The selected two-stage SDIRK method has order two and is scalar L-stable: the selected SDIRK coefficient satisfies $0<\gamma <1$, and the method has local defect $O(h^3)$ on the stated smooth tube.

Two-stage Radau IIA has order three and is scalar L-stable: the selected Radau IIA method has local defect $O(h^4)$ on the stated smooth tube.

Global error from uniform one-step defect and stability: on a common tube, an exact one-step method with local defect $O(h^{p+1})$, stability factor at most $1+L_{0}h$, and initial error $O(h^p)$ has global error $O(h^p)$.

Finite geometric error bound from a one-step recurrence: if $e_0\le C_0{h}^p$ and $e_{n+1}\le (1+L_{0}h)e_n+C{h}^{p+1}$, then ${\max }_{n\le N}{e}_n=O(h^p)$ for $h=T\mathrm{/}N$.

Multivariable Taylor formula with a Lagrange remainder along a line segment is applied to each scalar coordinate of the exact solution. The supplier is scalar-valued. For a vector map $F:U\subset {\mathbb{R}}^s\to {\mathbb{R}}^d$, apply it separately to each real component $F_i$; its intermediate points ${\theta }_i$ may differ. If the absolute values of all order-$(k+1)$ component partial derivatives on the segment are at most $M$, each component remainder is at most $M{C}_{s,k}\mathrm{\parallel }v{\mathrm{\parallel }}_2^{k+1}$, where $C_{s,k}={\sum }_{\mathrm{\mid }\alpha \mathrm{\mid }=k+1}1\mathrm{/}\alpha !$ is a fixed finite constant, since $\mathrm{\mid }{v}^{\alpha }\mathrm{\mid }\le \mathrm{\parallel }v{\mathrm{\parallel }}_2^{k+1}$. Hence the Euclidean remainder is at most $\sqrt{d}M{C}_{s,k}\mathrm{\parallel }v{\mathrm{\parallel }}_2^{k+1}$. In one time variable the max-norm bound is $M\mathrm{\mid }h{\mathrm{\mid }}^{k+1}\mathrm{/}(k+1)!$. No common intermediate point for all components is asserted. With $k=1$ and the uniformly bounded second derivatives on the tube, $y(t+h)=y(t)+h{y}^{\mathrm{\prime }}(t)+O(h^2)$ and $y^{\mathrm{\prime }}(t+h)=y^{\mathrm{\prime }}(t)+O(h)$ in Euclidean norm; using $y^{\mathrm{\prime }}=f(t,y)$ gives the BE update residual $y(t+h)-y(t)-hf(t+h,y(t+h))=O(h^2)$.

Backward Euler step equation and algebraic residual: BE is the one-stage Runge–Kutta stage equation with $A=b=c=(1)$, so its stage is its endpoint and its update is that stage.

At every step, the approximate stages and exact stage roots lie in the same prescribed invariant product ball, the stage map is $L$-Lipschitz, and $q=hL\le 1\mathrm{/}2$. For BE, the exact ODE endpoint also lies in that one-stage ball.