Linear stability domain

For a one-step method applied to $y^{\prime }=\lambda y$, write $y_{n+1}=R(h\lambda )y_n$. Its linear stability domain isForward Euler has $R(z)=1+z$, while backward Euler has $R(z)=(1-z{)}^{-1}$ and is A-stable.

Explicit Euler advances $y^{\prime }=f(t,y)$ by $y_{n+1}=y_n+hf(t_n,y_n)$. Its stability function is $R(z)=1+z$.

A method is A-stable when its stability domain contains the closed left half-plane.

A unit-modulus amplification root of a linear multistep method puts its test-equation parameter on this boundary locus, except when numerator and denominator both vanish. If the locus avoids the open left half-plane, the leading coefficient never vanishes there and the roots are initially inside the disk at one point there, continuity keeps them inside throughout that connected region. Checking zero-stability, imaginary-axis roots and common polynomial factors is still necessary. A boundary plot alone is not a proof of A-stability.

An A-stable one-step method is L-stable when its stability function also satisfies $R(z)\to 0$ as $|z|\to \infty$ within the left half-plane. It damps modes whose decay time is far shorter than the numerical step.

A Runge--Kutta method is B-stable when it does not increase distances between numerical solutions of a dissipative differential equation. Algebraic stability of a Runge-Kutta method is a standard sufficient condition.

Backward Euler advances $y^{\prime }=f(t,y)$ by $y_{n+1}=y_n+hf(t_{n+1},y_{n+1})$. Its stability function $R(z)=(1-z{)}^{-1}$ makes it L-stable.

For the two-stage implicit Runge-Kutta method withthe stability function isFor $z=x+iy$,The denominator has no zero in the closed left half-plane, so the method is A-stable for every real $a$.

A stiff differential equation contains rapidly decaying modes alongside much slower dynamics. An explicit method may then require a small step for stability even when accuracy only requires resolving the slow modes.

If a linear system has negative eigenvalues with widely separated magnitudes, an explicit method can be forced to resolve the fastest decaying mode solely for stability. For eigenvalues $-1$ and $-100$, forward Euler requires $h\le 0.02$, whereas backward Euler is stable for every positive step size.

Starting at the same value, forward and backward Euler have opposite leading local errors. Their half-difference therefore estimates either local error:For a first-order method, a local-tolerance controller consequently scales the next step by $(\mathrm{tol}/E_n{)}^{1/2}$, usually with a safety factor.