Trapezoidal ringing on a stiff scalar decay equation

Here the test argument is $z=-\lambda h=-100$. The trapezoidal rule equation $y_{n+1}=y_n+\frac{h}{2}(-\lambda y_n-\lambda y_{n+1})$ rearranges to the factor $R_{\mathrm{t}\mathrm{r}\mathrm{a}\mathrm{p}}(-100)=\frac{1-100\mathrm{/}2}{1+100\mathrm{/}2}=-\frac{49}{51}\mathrm{.}$ Thus, for $y_0\mathrm{\ne }0$, successive trapezoidal values alternate sign, while the magnitude after each step is multiplied by $49\mathrm{/}51\approx 0.961$; for $y_0=0$, every iterate remains zero.

Backward Euler gives $R_{\mathrm{B}\mathrm{E}}(-100)=1\mathrm{/}101\approx 0.00990$. The proved scalar functions give $R_S(-100)=\frac{1-100(1-2\gamma )}{(1+100\gamma {)}^2}\approx -0.0441,R_R(-100)=\frac{1-100\mathrm{/}3}{1+200\mathrm{/}3+10000\mathrm{/}6}\approx -\mathrm{0.0186.}$ Each L-stable method damps this stiff mode strongly at this step size; the trapezoidal factor instead changes sign and has magnitude near one. These are scalar linear-test calculations only and imply no nonlinear stability comparison.