A-stability

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

For nonnegative degrees $m,n$ with $n-2\le m\le n$, the $[m/n]$ Padé approximant $R$ to the exponential function has no poles in the left half-plane and satisfies $|R(z)|\le 1$ for $\mathrm{Re}z\le 0$. Consequently a numerical method having that stability function is A-stable. In particular, the near-diagonal choice $[s-1/s]$ gives the Radau IIA method stability function, while $[s/s]$ corresponds to the Gauss collocation method. The theorem concerns these exponential approximants, not arbitrary rational approximations with the same degrees.

Suppose a consistent linear multistep method has a simple principal polynomial root and its analytic logarithm satisfies $\log w(z)=z+C{z}^{2r}+O(z^{2r+1})$ near zero, with real coefficients. If $C(-1{)}^r>0$, then at $z=i\epsilon$ the real part of $\log w$ is positive to leading order. The same remains true after giving $z$ a small negative real part of smaller magnitude than that leading term. Thus a growing amplification factor occurs inside the left half-plane, ruling out A-stability. For an order-three method with $C>0$, choose $z=-C{\epsilon }^4/2+i\epsilon$.

For an explicit linear multistep method, the highest-degree coefficient of $\rho (w)-z\sigma (w)$ is fixed while at least one lower coefficient grows unboundedly with negative real $z$. If all roots stayed in the unit disk, the elementary symmetric polynomials of those roots would keep every monic coefficient bounded. This contradiction proves that no nontrivial consistent explicit method is A-stable.

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 $a$.