The Rosenbrock–Euler step map with an exact or approximate Jacobian

Let $D\subseteq \mathbb{R}\times {\mathbb{R}}^d$ be open, let $d\ge 1$, and let $f:D\to {\mathbb{R}}^d$ be a vector field (First-order systems, initial value problems, and solutions on intervals). For $(t,y)\in D$, a step size $h>0$, a parameter $\gamma >0$, and a matrix $\tilde{J}\in {\mathbb{R}}^{d\times d}$, put $A_{\gamma ,h}(\tilde{J}):=I_d-\gamma h\tilde{J}\mathrm{.}$ When this matrix is invertible, the one-stage Rosenbrock step uses the stage slope $k$ and update $A_{\gamma ,h}(\tilde{J})k=f(t,y),{\mathrm{\Phi }}_{\gamma ,h}^{\tilde{J}}(t,y):=y+hk\mathrm{.}$ At points where all state partial derivatives exist, the exact-Jacobian branch sets $\tilde{J}=D_{y}f(t,y)$, where $D_{y}f$ is the Jacobian in the state variables as in The Jacobian matrix of partial derivatives and the gradient in the scalar-valued case. An approximate-Jacobian branch may supply a different $\tilde{J}$; the equation requires the chosen shifted matrix to be invertible. The name Rosenbrock–Euler here refers to the case $\gamma =1$. This definition specifies one linearized step and does not say that it solves the full backward-Euler nonlinear equation.