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A-stability removes the scalar linear decay restriction on the step size; it does not replace accuracy, nonlinear solvability or a suitable system norm. Its importance is clearest for a stiff differential equation, where rapidly decaying modes coexist with slowly varying quantities of interest. On the Dahlquist test equation $y^{\prime }=\lambda y$, the exact solution decays for $\mathrm{Re}\lambda <0$. A stable time integrator should not turn that decay into numerical growth.

For a one-step method write $y_{n+1}=R(z)y_n$, where $z=h\lambda$ and $R$ is its stability function. Its linear stability domain consists of parameters with $|R(z)|\le 1$ and a well-defined update. The method is A-stable when this domain contains the closed left half-plane. A consistent order-$p$ one-step method has $R(z)=e^z+O(z^{p+1})$ near zero, but that local agreement alone says nothing about stability at large negative $z$.

For a Runge-Kutta method with stage matrix $A$, weights $b$ and $1=(1,\dots ,1{)}^T$, eliminating the stages on the test equation givesAn explicit finite-stage method has a polynomial $R$. A nonconstant polynomial is unbounded along the negative real axis, so no consistent explicit Runge-Kutta method is A-stable. The explicit Euler method, with $R(z)=1+z$, is stable in the disk $|1+z|\le 1$ and on the negative real axis only for $-2\le z\le 0$. A stiff mode with a large negative eigenvalue therefore forces an unnecessarily small step even after its exact contribution has almost disappeared.

The Backward Euler method has $R(z)=1/(1-z)$. For $\mathrm{Re}z\le 0$, $|1-z{|}^2=1-2\mathrm{Re}z+|z{|}^2\ge 1$, so it is A-stable. The trapezoidal rule hasand is also A-stable. The implicit midpoint method has the same scalar stability function, although its nonlinear update differs from the trapezoidal rule. The first method has order one and the latter two order two. Their implicit solves trade greater work per step for freedom from the scalar stiff stability restriction.

Damping of very stiff modes separates A-stability from L-stability. For a one-step method, the latter additionally requires $R(z)\to 0$ as $z$ tends to infinity in the left half-plane. Backward Euler method satisfies this. The trapezoidal and midpoint rules instead have $R(z)\to -1$ along the negative real axis: a very stiff mode can alternate in sign with little damping even though its amplitude remains bounded. L-stability is often preferable for diffusion and rapid relaxation, while retaining oscillatory modes can be desirable in conservative dynamics.

For a linear multistep method, writeIts test-equation amplification roots satisfy $\rho (\zeta )-z\sigma (\zeta )=0$. Absolute stability requires every root to lie in the unit disk or on its boundary with all unit-modulus roots simple; all parasitic modes matter. At $z=0$ this is the zero-stability root condition. Consistency requires $\rho (1)=0$ and ${\rho }^{\prime }(1)=\sigma (1)$, and the Dahlquist equivalence theorem combines consistency and zero-stability to characterize convergence. Thus a claim of stability obtained by canceling a common factor is not a proof about the original recurrence and its arbitrary consistent starting data.

Explicit linear multistep methods cannot be A-stable either. If ${\beta }_k=0$ and ${\alpha }_k\ne 0$, some lower coefficient of $\rho -z\sigma$ becomes unbounded as $z\to -\infty$, while the leading coefficient stays fixed. If all roots remained in the unit disk, every coefficient-to-leading-coefficient ratio, an elementary symmetric function of those roots, would remain bounded. This contradiction forces an unstable root.

An important implicit example is BDF2 method, whose characteristic equation is $(3-2z){\zeta }^2-4\zeta +1=0$. A unit root $\zeta =e^{i\theta }$ givesThe leading coefficient vanishes only at $z=3/2$, outside the left half-plane. Near a small negative $z$, both roots are inside the disk; no unit-circle crossing can occur in the connected open left half-plane. On the imaginary axis the boundary equality can occur only at $z=0$, where the roots are $1,1/3$. Hence BDF2 is A-stable and its two roots tend to zero for very large negative $z$, giving strong stiff-mode damping. This is a complete boundary-locus argument, not just a boundary plot.

The Second Dahlquist barrier says that a convergent A-stable ordinary linear multistep method has order at most two. This theorem does not limit implicit Runge-Kutta methods or multiderivative multistep methods. In particular the A-stable third-order two-step multiderivative method in question 1 uses $y^{\prime \prime }$ and satisfies different order conditions. Among collocation Runge-Kutta methods, the standard family results give Gauss collocation methods order $2s$ with diagonal $[s/s]$ Padé approximants to $e^z$ as stability functions, and Radau IIA methods order $2s-1$ with subdiagonal $[s-1/s]$ approximants. Both families are A-stable; the Gauss functions have a nonzero limit at infinity, whereas Radau IIA is L-stable. The corrected two-stage right-Radau method, for example, haswhich matches the exponential through cubic order and tends to zero at infinity. Its poles $2\pm i\sqrt{2}$ lie in the right half-plane. For $z=-r+iy$, $r\ge 0$, direct subtraction givesproving its A-stability directly.

Sectorial A-alpha stability relaxes the requirement to a sector around the negative real axis, useful when the spectrum of a stiff problem remains in that sector; higher-order backward differentiation formulas illustrate the usefulness of this weaker requirement. For a normal linear system $y^{\prime }=Ly$, unitary diagonalization reduces the numerical update to its scalar factors $R(h{\lambda }_j)$. For a nonnormal system, a scalar eigenvalue test alone does not give a uniform bound in the chosen norm. Nonlinear stability is different again: B-stability asks for contractivity when $\mathrm{Re}⟨f(u)-f(v),u-v⟩\le 0$, and algebraic stability of a Runge-Kutta method is a standard sufficient criterion. Thus A-stability addresses a precise and valuable linear test, while step selection still has to meet accuracy requirements and the actual implicit equations must be solved on the relevant branch.