Implicit Runge–Kutta stage equations and update

Let $d,s\ge 1$, $A=(a_{ij})\in {\mathbb{R}}^{s\times s}$, $b,c\in {\mathbb{R}}^s$, and let $f:D\to {\mathbb{R}}^d$ be a vector field on an open time-state domain $D\subseteq \mathbb{R}\times {\mathbb{R}}^d$ (First-order systems, initial value problems, and solutions on intervals). For an initial state $y\in {\mathbb{R}}^d$ and time $t$ with $(t,y)\in D$, and a step $h>0$, a stage vector $Y=(Y_1,\dots ,Y_s)\in ({\mathbb{R}}^d{)}^s$ is admissible only when $(t+c_{j}h,Y_j)\in D$ for every $j$. An admissible vector solves

$Y_i=y+h{\sum }_{j=1}^s{a}_{ij}f(t+c_{j}h,Y_j),i=1,\dots ,s\mathrm{.}$

When a solution $Y$ is selected, the method update is

${\mathrm{\Psi }}_h(t,y)=y+h{\sum }_{i=1}^s{b}_{i}f(t+c_{i}h,Y_i)\mathrm{.}$

On the admissible stage set, the stage update residual is the vector with blocks

$G_i(Y)=Y_i-y-h{\sum }_{j=1}^s{a}_{ij}f(t+c_{j}h,Y_j)\mathrm{.}$

The associated stage derivative residual is $G(Y)\mathrm{/}h$. The tableau is stiffly accurate when $b_j=a_{sj}$ for all $j$ and $c_s=1$; in that case the last stage $Y_s$ equals the method update. The equations may have no, one, or several solutions in a proposed branch set; this definition alone asserts no solvability.