The selected exact-J two-stage Rosenbrock method is scalar L-stable

Take a nonzero starting value $y$. The exact complex equations in the Statement give $h{k}_1\mathrm{/}y=z\mathrm{/}(1-\alpha z)$ and, by $k_2=q_2+k_1$, the equivalent equation $(1-\alpha z)k_2=F_1-\alpha z{k}_1$. Substituting $F_1=\lambda (y+h{k}_1\mathrm{/}2)$ and eliminating $k_1$ yields $1+h{k}_2\mathrm{/}y=1+[z+(1\mathrm{/}2-2\alpha )z^2]\mathrm{/}(1-\alpha z{)}^2$. Since $\alpha =1-1\mathrm{/}\sqrt{2}$ satisfies ${\alpha }^2+1\mathrm{/}2-2\alpha =0$, this reduces to the stated $R(z)$.

Let $z=x+iy$ with $x\le 0$. The equation $1-\alpha z=0$ would force $z=1\mathrm{/}\alpha >0$, so the denominator has no zero in the closed left half-plane. Put $c=1-2\alpha =\sqrt{2}\alpha$. Expanding gives $\mathrm{\mid }1-\alpha z{\mathrm{\mid }}^4-\mathrm{\mid }1+cz{\mathrm{\mid }}^2=(-4\alpha -2c)x+(2{\alpha }^2-c^2)\mathrm{\mid }z{\mathrm{\mid }}^2+4{\alpha }^2{x}^2-4{\alpha }^{3}x\mathrm{\mid }z{\mathrm{\mid }}^2+{\alpha }^4\mathrm{\mid }z{\mathrm{\mid }}^4=-2x+4{\alpha }^2{x}^2-4{\alpha }^{3}x\mathrm{\mid }z{\mathrm{\mid }}^2+{\alpha }^4\mathrm{\mid }z{\mathrm{\mid }}^4\ge 0$. Every term is nonnegative because $x\le 0$ and $\alpha >0$, so $\mathrm{\mid }R(z)\mathrm{\mid }\le 1$.

By the triangle inequality in [F4], $\alpha \mathrm{\mid }z\mathrm{\mid }=\mathrm{\mid }\alpha z\mathrm{\mid }=\mathrm{\mid }1-(1-\alpha z)\mathrm{\mid }\le 1+\mathrm{\mid }1-\alpha z\mathrm{\mid }$, so $\mathrm{\mid }1-\alpha z\mathrm{\mid }\ge \alpha \mathrm{\mid }z\mathrm{\mid }-1$. Also $\mathrm{\mid }1+cz\mathrm{\mid }\le 1+\mathrm{\mid }c\mathrm{\mid }\mathrm{\mid }z\mathrm{\mid }$. Therefore $\mathrm{\mid }R(z)\mathrm{\mid }\le (1+\mathrm{\mid }c\mathrm{\mid }\mathrm{\mid }z\mathrm{\mid })\mathrm{/}(\alpha \mathrm{\mid }z\mathrm{\mid }-1{)}^2$ whenever $\mathrm{\mid }z\mathrm{\mid }>1\mathrm{/}\alpha$, and this bound tends to zero as $\mathrm{\mid }z\mathrm{\mid }\to \mathrm{\infty }$. Together with step 2.1, [F2] gives L-stability.