Stability regions and stiff equations — Examples

Given: The real matrix $A=\left(\begin{array}{cc}-1& 16\\ 0& -1\end{array}\right)$ and a unit backward-Euler step $h=1$.

Matrix-step contractivity and the logarithmic norm in a specified inner product: Euclidean contractivity requires the induced norm of the matrix step to be at most one; ${\mu }_2(A)={\sup }_{x\mathrm{\ne }0}{x}^{\mathsf{T}}Ax\mathrm{/}\mathrm{\parallel }x{\mathrm{\parallel }}_2^2$.

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

Given: $\lambda =-\rho +i\omega$ with $\rho >0$, $\omega \in \mathbb{R}$, and a step size $h>0$.

Forward Euler has the closed unit disk centered at −1 as its absolute-stability region: forward Euler is absolutely stable exactly when $\mathrm{\mid }1+h\lambda \mathrm{\mid }\le 1$.

Given: The diagonal system $y_1^{\mathrm{\prime }}=-y_1$, $y_2^{\mathrm{\prime }}=-1000y_2$, and forward Euler with one common step $h>0$.

Forward Euler has the closed unit disk centered at −1 as its absolute-stability region: a scalar decay equation $y^{\mathrm{\prime }}=-\rho y$ is absolutely stable under forward Euler exactly when $0<h\rho \le 2$.

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.