One-step ODE Consistency and Local Error

Let $d\ge 1$, let $D\subseteq \mathbb{R}\times {\mathbb{R}}^d$ be open, and consider a first-order IVP and solution in the sense of First-order systems, initial value problems, and solutions on intervals. A one-step method is a family of maps ${\mathrm{\Psi }}_h(t,v)$, defined for admitted times $t$, positive step sizes $h$, and states $v$, whose value is the method's approximation at time $t+h$ from the single input state $v$ at time $t$. It does not use earlier numerical states.

If $y$ is an exact solution on an interval containing $[t,t+h]$, the local defect is

${\delta }_h(t;y):={\mathrm{\Psi }}_h(t,y(t))-y(t+h)\mathrm{.}$

Let $K\subseteq D$ be a fixed compact time-state set. The method has uniform local order $p$ on $K$ along $y$ if there are constants $C\ge 0$ and $h_0>0$, independent of the admitted starting time $t$ and step $0<h\le h_0$, such that

$\mathrm{\parallel }{\delta }_h(t;y){\mathrm{\parallel }}_2\le C{h}^{p+1}$

whenever the exact solution graph on $[t,t+h]$ and the associated method time-state inputs, stages, and outputs under consideration lie in $K$. Here $p$ is a positive integer and $\mathrm{\parallel }\cdot {\mathrm{\parallel }}_2$ is the Euclidean norm of The Euclidean inner product $⟨x,y⟩={\sum }_{k<n}{x}_k{y}_k$ on ${\mathbb{R}}^n$.

The method has one-step Lipschitz stability on $K$ with constant $L\ge 0$ if

$\mathrm{\parallel }{\mathrm{\Psi }}_h(t,u)-{\mathrm{\Psi }}_h(t,v){\mathrm{\parallel }}_2\le (1+Lh)\mathrm{\parallel }u-v{\mathrm{\parallel }}_2$

for every admitted step size $h$, admitted time $t$, and input states $u,v$ with $(t,u),(t,v)\in K$. This is a bound on the change in one step; it is separate from absolute or linear stability regions and does not itself assert a global error estimate.

Let $d\ge 1$, let $V\subseteq \mathbb{R}\times {\mathbb{R}}^d$ be open, let $f:V\to {\mathbb{R}}^d$ have continuous first partial derivatives, and let $y:I\to {\mathbb{R}}^d$ solve $y^{\mathrm{\prime }}=f(t,y)$ on an open interval $I$. Fix $a<b$ with $[a,b]\subset I$ such that $(s,y(s))\in V$ for every $s\in [a,b]$. Suppose finite constants $F_j,M_{t,i},M_{ij}\ge 0$ satisfy for all $s\in [a,b]$

$\mathrm{\mid }{f}_j(s,y(s))\mathrm{\mid }\le F_j,\mathrm{\mid }{\mathrm{\partial }}_t{f}_i(s,y(s))\mathrm{\mid }\le M_{t,i},\mathrm{\mid }{\mathrm{\partial }}_{z_j}{f}_i(s,y(s))\mathrm{\mid }\le M_{ij}(1\le i,j\le d)\mathrm{.}$

Define $g_i(s):=f_i(s,y(s))$ and $B_i:=M_{t,i}+{\sum }_{j=1}^d{M}_{ij}{F}_j$. Then $g_i$ is continuous on $[a,b]$, differentiable on $(a,b)$, and

$g_i^{\mathrm{\prime }}(s)={\mathrm{\partial }}_t{f}_i(s,y(s))+{\sum }_{j=1}^d{\mathrm{\partial }}_{z_j}{f}_i(s,y(s))f_j(s,y(s)),\mathrm{\mid }{g}_i^{\mathrm{\prime }}(s)\mathrm{\mid }\le B_i\mathrm{.}$

Let $d\ge 1$, $T>0$, and let $V\subseteq \mathbb{R}\times {\mathbb{R}}^d$ be an open time-state domain whose fibers $V_t:=\{z:(t,z)\in V\}$ are open and convex. Let $K\subset V$ be compact. Let $f:V\to {\mathbb{R}}^d$ have continuous first partial derivatives, and suppose $y$ is a solution of $y^{\mathrm{\prime }}=f(t,y)$, $y(0)=y_0$, on an open interval containing $[0,T]$, with graph on $[0,T]$ contained in $K$. For each local defect under consideration, require the Euler input and output time-state pairs to lie in $K$; the exact solution segment already lies in $K$.

For $1\le i,j\le d$, assume finite constants $F_j,M_{t,i},M_{ij},L\ge 0$ satisfy throughout $V$

$\mathrm{\mid }{f}_j(t,z)\mathrm{\mid }\le F_j,\mathrm{\mid }{\mathrm{\partial }}_t{f}_i(t,z)\mathrm{\mid }\le M_{t,i},\mathrm{\mid }{\mathrm{\partial }}_{z_j}{f}_i(t,z)\mathrm{\mid }\le M_{ij},\mathrm{\parallel }{D}_{y}f(t,z)w{\mathrm{\parallel }}_2\le L\mathrm{\parallel }w{\mathrm{\parallel }}_2(w\in {\mathbb{R}}^d)\mathrm{.}$

Here $D_{y}f$ is the matrix with entries ${\mathrm{\partial }}_{z_j}{f}_i$, and the last inequality defines the operator bound used below. The forward Euler map is ${\mathrm{\Psi }}_h(t,v)=v+hf(t,v)$. Then, with $B_i:=M_{t,i}+{\sum }_{j=1}^d{M}_{ij}{F}_j$, for every $t\in [0,T)$ and $h>0$ with $t+h\le T$ whose Euler input and output pairs lie in $K$, its local defect satisfies

$\mathrm{\parallel }{\delta }_h(t;y){\mathrm{\parallel }}_2\le \frac{h^2}{2}{\left({\sum }_{i=1}^d{B}_i^2\right)}^{1\mathrm{/}2},$

so it has uniform local order one on $K$ along $y{\mathrm{\mid }}_{[0,T]}$, with $h_0=T$. For every $t\in [0,T]$, every admitted $h>0$, and every $u,v\in V_t$,

$\mathrm{\parallel }{\mathrm{\Psi }}_h(t,u)-{\mathrm{\Psi }}_h(t,v){\mathrm{\parallel }}_2\le (1+Lh)\mathrm{\parallel }u-v{\mathrm{\parallel }}_2\mathrm{.}$

This is a local defect and one-step stability result. It does not assert a global error estimate or an order for any other named method.

The defect proof is choice-free and uses only the time-along-flow chain rule and the componentwise integral remainder. The Taylor supplier is used by the scalar linear example, not needed to prove this theorem. The global error recurrence in the cited numerical-analysis sources is outside this item.