Backward Euler and BDF2 damp a fast scalar mode while trapezoidal does not

Given: Reals $\rho >0$ and $h>0$, and a classical solution $y$ on $[0,h]$, meaning $y$ is continuous on $[0,h]$, differentiable on $(0,h)$, satisfies $y^{\mathrm{\prime }}(t)=-\rho y(t)$ there, and has $y(0)=1$. Write $x=\rho h$.

The real exponential function and the number $e$ by a power series: $\exp$ is defined by its real power series and $\exp (0)=1$.

The exponential function is smooth and $(\exp {)}^{\mathrm{\prime }}=\exp$: $\exp$ is $C^{\mathrm{\infty }}$ and $(\exp {)}^{\mathrm{\prime }}=\exp$.

The chain rule, in one line from Carathéodory: if $g$ is differentiable at $c$ and $f$ is differentiable at $g(c)$, then $f\circ g$ is differentiable at $c$ with $(f\circ g{)}^{\mathrm{\prime }}(c)=f^{\mathrm{\prime }}(g(c))g^{\mathrm{\prime }}(c)$: if $g:A\to \mathbb{R}$ maps $A$ into $B$, $c\in A$ is a limit point where $g$ is differentiable, and $g(c)$ is a limit point of $B$ where $f:B\to \mathbb{R}$ is differentiable, then $(f\circ g{)}^{\mathrm{\prime }}(c)=f^{\mathrm{\prime }}(g(c))g^{\mathrm{\prime }}(c)$. For each $c\in (0,h)$, take $A=(0,h)$ and $B=\mathbb{R}$; both limit-point hypotheses hold.

The exponential tends to $+\mathrm{\infty }$ at $+\mathrm{\infty }$ and to $0$ at $-\mathrm{\infty }$: $\exp (-x)\to 0$ as $x\to +\mathrm{\infty }$.

Backward Euler is A-stable and L-stable: $R_{\mathrm{B}\mathrm{E}}(z)=1\mathrm{/}(1-z)$.

The trapezoidal rule is A-stable but not L-stable: $R_{\mathrm{T}}(z)=(1+z\mathrm{/}2)\mathrm{/}(1-z\mathrm{/}2)$.

Both constant-step BDF2 test roots vanish in the stiff limit: both BDF2 test roots tend to zero as $\mathrm{\mid }z\mathrm{\mid }\to \mathrm{\infty }$ in the closed left half-plane.

Sums, scalar multiples, products and quotients: $(f+g{)}^{\mathrm{\prime }}(c)=f^{\mathrm{\prime }}(c)+g^{\mathrm{\prime }}(c)$, $(\alpha f{)}^{\mathrm{\prime }}(c)=\alpha f^{\mathrm{\prime }}(c)$, $(fg{)}^{\mathrm{\prime }}(c)=f^{\mathrm{\prime }}(c)g(c)+f(c)g^{\mathrm{\prime }}(c)$, and $(f\mathrm{/}g{)}^{\mathrm{\prime }}(c)=(f^{\mathrm{\prime }}(c)g(c)-f(c)g^{\mathrm{\prime }}(c))\mathrm{/}g(c{)}^2$ when $g(c)\mathrm{\ne }0$: products satisfy $(fg{)}^{\mathrm{\prime }}=f^{\mathrm{\prime }}g+f{g}^{\mathrm{\prime }}$, and scalar multiples satisfy $(\alpha f{)}^{\mathrm{\prime }}=\alpha f^{\mathrm{\prime }}$ at a limit point where the input functions are differentiable.

Sums, scalar multiples, products, absolute values, maxima, minima and quotients with nonvanishing denominator of continuous functions are continuous, as are constants, the identity and every polynomial function: products and scalar multiples of functions continuous at a point are continuous there, and the identity function is continuous.

A function continuous on an interval $I$ whose derivative vanishes at every interior point of $I$ is constant on $I$; consequently two such functions with the same derivative differ by a constant: if $I$ is an interval and $v$ is continuous on $I$, differentiable at every interior point, and has $v^{\mathrm{\prime }}=0$ throughout the interior, then $v$ is constant on $I$.

The exponential is positive and satisfies $\exp (-x)=1\mathrm{/}\exp (x)$: $\exp (-u)=1\mathrm{/}\exp (u)>0$ for every real $u$.

The identity map $t\mapsto t$ has difference quotient $1$ and derivative $1$ at each real $t$. Since $g_{\pm }(t)=\pm \rho t$, [F8] gives $g_{\pm }^{\mathrm{\prime }}(t)=\pm \rho$, and [F9] gives continuity of $g_{\pm }$ on $[0,h]$.

Since $\exp$ is continuous on $\mathbb{R}$ by [F2], for either sign, any $t_0\in [0,h]$, and any $\epsilon >0$, choose $\eta >0$ such that $\mathrm{\mid }s\mp \rho t_0\mathrm{\mid }<\eta$ implies $\mathrm{\mid }\exp (s)-\exp (\pm \rho t_0)\mathrm{\mid }<\epsilon$. Taking $\delta =\eta \mathrm{/}\rho$, $\mathrm{\mid }t-t_0\mathrm{\mid }<\delta$ implies $\mathrm{\mid }\pm \rho t\mp \rho t_0\mathrm{\mid }<\eta$. Thus $t\mapsto \exp (\pm \rho t)$ is continuous on $[0,h]$ by the $\epsilon$-$\delta$ definition.