The Rosenbrock–Euler scalar stability function is A-stable exactly for gamma at least one half

For scalar stability, consider the exact-arithmetic complexification of the exact-Jacobian one-stage equation. For $\lambda ,y,k\in \mathbb{C}$, $h>0$, $\gamma >0$, and $z=h\lambda$, use $(1-\gamma z)k=\lambda y$ and $y_{+}=y+hk$. Equivalently, identify $\mathbb{C}$ with ${\mathbb{R}}^2$ and multiplication by $\lambda$ with the real matrix $C_{\lambda }=\left(\begin{array}{cc}\mathrm{Re}\lambda & -\mathrm{Im}\lambda \\ \mathrm{Im}\lambda & \mathrm{Re}\lambda \end{array}\right)$; the corresponding real exact Jacobian is $C_{\lambda }$, and the equations become $(I_2-\gamma h{C}_{\lambda })k=C_{\lambda }y$ and $y_{+}=y+hk$. This is a scalar stability analysis of exact stage equations; it is not an execution of the finite-precision real algorithm Finite-precision one-stage Rosenbrock step with residual status. Use the scalar rational-function criteria in Scalar stability function, A-stability, and L-stability for Runge–Kutta methods for its amplification function. Its scalar stability function is $R_{\gamma }(z)=1+\frac{z}{1-\gamma z}=\frac{1+(1-\gamma )z}{1-\gamma z}\mathrm{.}$ It has no pole in $\mathrm{Re}z\le 0$ and satisfies $\mathrm{\mid }{R}_{\gamma }(z)\mathrm{\mid }\le 1$ throughout that half-plane if and only if $\gamma \ge \frac{1}{2}$. Among these parameters, it is L-stable if and only if $\gamma =1$. In particular, the named Rosenbrock–Euler case $\gamma =1$ has the backward-Euler factor $R(z)=(1-z{)}^{-1}$.

[F2] Finite-precision one-stage Rosenbrock step with residual status: this is a finite-precision real-space implementation with the real exact stage equation of [F1]; no execution or numerical status of that algorithm is used here.