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Use the stability function just calculated. Its denominator has its two zeros at $3\pm i\sqrt{3}$, so there is no pole in the closed left half-plane. Put $z=x+iy$, $N=1+z/2+z^2/12$ and $D=1-z/2+z^2/12$. A direct calculation yieldsFor $x\le 0$ this is nonnegative, and therefore $|R(z)|\le 1$. The method is A-stable. Equality holds on the imaginary axis, consistently with the absence of numerical damping for those scalar oscillatory modes. Since $R(z)\to 1$ as $z\to \infty$, the A-stability established here does not imply strong damping of very stiff decaying modes.

Use the standard A-stability convention including the root condition at $z=h\lambda =0$. The amplification polynomial of a multistep method isZero-stability first restricts the possible parameters to $-1\le a<1$. For $-1<a<0$, as $z$ tends to negative infinity, one amplification root tends to $(3a-1)/(1+a)$, whose modulus exceeds one. For $a=-1$, the characteristic equation is ${\zeta }^2-2z\zeta -1=0$; large negative real $z$ likewise gives an exterior root. Thus $a\ge 0$ is necessary.

For sufficiency take $0\le a<1$. The leading coefficient cannot vanish in the closed left half-plane. On the unit circle $\zeta =e^{i\theta }$ the boundary-locus test for multistep A-stability uses $z=\rho (\zeta )/\sigma (\zeta )$ and giveswhere the denominator is nonzero. When $a=0$, the exceptional zero of $\sigma$ at $\zeta =-1$ is not a zero of $\rho$ and therefore cannot be an amplification root for a finite $z$. Near $z=0$ on the negative real axis, the root issuing from one is $1+z+O(z^2)$ and is strictly inside the disk; the other root is close to $a$ and is also inside. Roots vary continuously, cannot escape through infinity because the leading coefficient is nonzero, and cannot cross the unit circle anywhere in the open left half-plane by the displayed boundary formula. Thus every root remains inside there. Continuity gives the boundary case; for $a>0$ a unit root on the imaginary axis can occur only at $z=0$, where it is simple. For $a=0$, cancellation of the harmless zero root leaves the trapezoidal rule, whose amplification factor has modulus at most one.

ThereforeThe third-order member is convergent but not A-stable, consistent with the Second Dahlquist barrier. At $a=1$, $P_z=(\zeta -1)[(1-z)\zeta -1]$ has a permanent unit root and a double root at $z=0$: the unreduced method is not zero-stable and is not A-stable under the stated convention. Testing only the open half-plane while overlooking its zero-step behavior would give a weaker conclusion.

A stiff differential equation can contain rapidly decaying modes as well as a slowly changing solution of interest. An explicit scheme may need a very small step solely to keep those already small fast modes from numerical growth. A-stability addresses this restriction through the Dahlquist test equation $y^{\prime }=\lambda y$, whose exact solution decays when $\mathrm{Re}\lambda <0$.

For a one-step method, write $y_{n+1}=R(z)y_n$, $z=h\lambda$. Its linear stability domain consists of the points where the update is defined and $|R(z)|\le 1$. A-stability means that this domain contains the closed left half-plane. This is a scalar linear stability property, not an order condition, an existence theorem for an implicit solve, or a general nonlinear contractivity assertion.

For a rational approximation to the exponential, consistency requires $R(z)=1+z+O(z^2)$, and order $p$ requires $R(z)-e^z=O(z^{p+1})$. Poles must be absent from the relevant half-plane. The Forward Euler method has $R=1+z$ and the disk $|1+z|\le 1$, so negative real modes require $h|\lambda |\le 2$; it is not A-stable. The Backward Euler method has $R=(1-z{)}^{-1}$, and $|1-z|\ge 1$ on the left half-plane, so it is A-stable. The trapezoidal rule and the implicit midpoint rule both have, on the scalar linear problem,They are second-order and A-stable despite being different nonlinear methods. No nonconstant polynomial approximation is A-stable, because its modulus grows without bound on the negative real axis. In particular every nontrivial explicit Runge-Kutta method has a bounded-step stability restriction.

A-stability prevents growth but need not eliminate extremely stiff modes: the trapezoidal factor tends to minus one as $z\to -\infty$. L-stability additionally requires $R(z)\to 0$ within the left half-plane. Backward Euler is L-stable; trapezoidal and implicit midpoint are not. The distinction explains persistent alternating transients in otherwise stable Crank-Nicolson calculations. Accuracy still constrains useful step sizes even for an A-stable method.

A linear multistep method is characterized by polynomials $\rho ,\sigma$, and its scalar modes satisfy $\rho (\zeta )-z\sigma (\zeta )=0$. There is generally no single scalar amplification factor: every root must lie in the closed unit disk and unit roots must be simple. At $z=0$, this is the zero-stability root condition. Consistency plus zero-stability gives convergence by the Dahlquist equivalence theorem, with suitable starting values. A-stability demands the amplification root condition throughout the left half-plane, including the zero-step condition. The Second Dahlquist barrier says that an irreducible A-stable linear multistep method has order at most two, and an explicit linear multistep method cannot be A-stable.

Examples include backward Euler, trapezoidal, and the second-order backward differentiation formulaIts zero-step roots are $1,1/3$. Its unit-circle boundary quotient has real part $(1-\cos \theta {)}^2\ge 0$, and the leading coefficient is nonzero on the left half-plane, giving A-stability by the same continuation argument as in Question 2. Conversely the third-order member of that question fails A-stability. A formal high order without zero-stability, or after silently cancelling a problematic unit factor, does not evade the barrier.

For an implicit Runge-Kutta method, the stages on the test equation satisfy $Y=e{y}_n+zAY$. Whenever these stages are solvable, its stability function isThis rational function is used to test scalar A-stability; the Butcher order conditions separately establish nonlinear order. The Lobatto IIIA method in Question 3 is an A-stable fourth-order example with $R$ the diagonal degree-two Padé approximant. More generally the $s$-stage Gauss--Legendre Runge-Kutta method has order $2s$ and is A-stable, with the diagonal Padé approximation to $e^z$; its stiff-limit factor is nonzero. The $s$-stage Radau IIA method has order $2s-1$ and is L-stable. Thus Runge-Kutta stages permit arbitrarily high A-stable order, unlike irreducible linear multistep formulas.

For a normal matrix, scalar amplification bounds transfer through unitary diagonalization. For a non-normal matrix, eigenvalue information alone does not assert Euclidean contraction: transient growth and eigenvector conditioning can matter. Direct energy method arguments, such as the dissipative Cayley-transform proof in Question 1, use the symmetric part and are more informative in that setting.

For nonlinear dissipative vector fields, the stronger B-stability property asks that distances between two numerical solutions not increase. A standard sufficient condition for Runge-Kutta methods is algebraic stability: $b_i\ge 0$ andis positive semidefinite. It follows from the Runge-Kutta contractivity identity when the implicit stages exist. The three-stage Lobatto IIIA method is A-stable but has the first diagonal entry of that matrix equal to $-1/36$, so scalar A-stability should not be confused with that sufficient nonlinear condition. Stage solvability of an implicit Runge-Kutta method remains a separate issue and computational cost of the implicit equations must be considered. A-stability removes a scalar decay-mode stability restriction; order, stiff damping, nonlinear stability and solvability remain distinct properties.

Apply the Dahlquist test equation and put $z=h\lambda$. Every recurrence mode must satisfyThe amplification polynomial of a multistep method, rather than just its root near one, determines absolute stability. Put $B=2a-1$ and use the Cayley transform between the half-plane and diskFor $\zeta \ne -1$, the characteristic equation becomesIf $B>0$, the denominator cannot vanish at a root of the characteristic equation: $1+B{w}^2=0$ would force $w=0$, a contradiction. Hence $\mathrm{Re}z<0$ forces $\mathrm{Re}w<0$, so all amplification roots have modulus less than one. On the imaginary axis the roots have modulus one and are simple: the transformed quadratic has discriminant $4(1-B{z}^2)>0$ for imaginary $z$. At $z=0$ they are the two simple roots $\pm 1$. Also $\zeta =-1$ cannot be a root when $B>0$ and $z\ne 0$, and the leading coefficient $1-az$ cannot vanish in the closed left half-plane.

If $a<1/2$, the root near $-1$ isFor small negative real $z$ it lies below $-1$, violating absolute stability. The endpoint $a=1/2$ deserves separate treatment:One root is always $-1$; the other is the trapezoidal rule multiplier $(1+z/2)/(1-z/2)$. They are distinct for every finite $z$ in the left half-plane. Consequently, with absolute stability understood as the bounded root condition for a multistep method,There is a convention at this reducible endpoint: if A-stability is defined to require every unreduced recurrence mode to decay for $\mathrm{Re}z<0$, the answer is $a>1/2$, since the $(-1{)}^n$ mode persists at $a=1/2$. Canceling the common factor $\zeta +1$ gives the A-stable trapezoidal rule, but cancellation removes an actual starting-error mode of the original two-step recurrence. The fourth-order member $a=1/3$ is outside either A-stability range.

The criterion for algebraic stability of a Runge-Kutta method is $b_i\ge 0$ and positive semidefiniteness ofThe Runge-Kutta contractivity identity implies that this criterion is sufficient for B-stability, namely contractivity for a dissipative vector field, whenever the stage equations are defined. The implication from algebraic stability to B-stability is the relevant nonlinear theorem; A-stability alone is a linear property.

All weights here are positive, butA positive semidefinite matrix cannot have this negative quadratic value. Hence the method is not algebraically stable. Failure of this sufficient criterion alone would not be a proof of failure of B-stability; no such converse is needed for the question.