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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.

An $s$-stage Runge-Kutta method has stability function $R(z)=1+z{b}^T(I-zA{)}^{-1}1$, a rational function with numerator and denominator degrees at most $s$. Order $2s$ forces $R(z)-e^z=O(z^{2s+1})$, so uniqueness of the diagonal Padé approximant identifies $R$ with $[s/s{]}_{e^z}$. That approximant has no poles in the closed left half-plane and modulus at most one there. Hence every such maximal-order method is A-stable. Since the Padé approximant numerator and denominator are coprime and both have degree $s$, the original stage determinant cannot have hidden canceled poles in that half-plane.

Suppose an $s$-stage Runge-Kutta method has order $2s-1$, an invertible matrix $A$, and $b^T{A}^{-1}1=1$. Its stability function is $R(z)=1+z{b}^T(I-zA{)}^{-1}1$. The adjugate matrix formula gives a rational function with denominator degree $s$ and numerator degree at most $s$; its zero limit at infinity lowers the latter degree to at most $s-1$. The order of a numerical method gives $R(z)-e^z=O(z^{2s})$, identifying the $[s-1/s]$ Padé approximant. Uniqueness follows because the cross-multiplied difference of two candidates has degree at most $2s-1$ but a zero of order at least $2s$, so is identically zero. The A-stability of near-diagonal exponential Padé approximants then proves A-stability, and the zero limit proves L-stability. These conditions identify the stability function; they do not alone claim uniqueness of all entries of the Runge-Kutta method tableau.

Let $A,b$ be the coefficients of the $\nu$-stage Runge-Kutta method. On the Dahlquist test equation $y^{\prime }=\lambda y$, its stability function isBoth polynomials have degree at most $\nu$, with denominator normalized to one at zero. Order $2\nu$ implies $R(z)-e^z=O(z^{2\nu +1})$. A rational function with these degree bounds and this accuracy is uniquely the diagonal Padé approximant $[\nu /\nu {]}_{e^z}$. To see uniqueness directly, if $P/Q$ and $\tilde{P}/\tilde{Q}$ have the required accuracy, then $P\tilde{Q}-\tilde{P}Q$ is a polynomial of degree at most $2\nu$ vanishing to order at least $2\nu +1$ at zero, hence identically zero.

The diagonal exponential Padé approximant can be writtenUse the permitted Padé property that the zeros of $P_{\nu }(-z)$ lie strictly in the right half-plane and numerator and denominator are coprime. Thus $R$ has no poles in the closed left half-plane. Its real coefficients give $P_{\nu }(-iy)=\overline{P_{\nu }(iy)}$, so $|R(iy)|=1$. Also $R(z)\to (-1{)}^{\nu }$ at infinity. Applying the maximum modulus principle on left half-disks and letting their radii increase proves $|R(z)|\le 1$ for $\mathrm{Re}z\le 0$. There are no hidden stage poles there either: the irreducible Padé denominator has degree $\nu$, exhausting the possible degree of $\det (I-zA)$, so the latter cannot contain an additional canceled factor. Every such maximal-order Runge-Kutta method is A-stable. This is maximal-order Runge-Kutta methods are A-stable; it does not assert L-stability, since the stability function does not tend to zero at infinity.

Apply the Runge-Kutta method to the Dahlquist test equation and put $z=h\lambda$. Its stability function isSince $A$ is an invertible matrix, $Q$ has polynomial degree $s$. The adjugate matrix formula gives $\mathrm{deg}P\le s$, but the additional condition givesso actually $\mathrm{deg}P\le s-1$. The specified order of a numerical method implies $R(z)=e^z+O(z^{2s})$ near zero. Therefore the Padé identification of a Radau stability function makes $R$ the $[s-1/s]$ Padé approximant to the exponential function: its numerator and denominator have the required degrees, and its Taylor series agrees through degree $2s-1$.

In the permitted A-stability of near-diagonal exponential Padé approximants, take $m=s-1$, $n=s$. Then $n-2\le m\le n$, so the scheme is A-stable. The same stability function also tends to zero at infinity, giving L-stability.

Write the Runge-Kutta method coefficients as $A\in {\mathbb{R}}^{\nu \times \nu }$, $b\in {\mathbb{R}}^{\nu }$, and let $e$ be the all-ones column. On the test equation $y^{\prime }=\lambda y$, with $z=h\lambda$, the stage vector satisfies $Y=e{y}_n+zAY$ and the next value is $y_{n+1}=y_n+z{b}^{T}Y$. Hence its stability function isBoth polynomials have degree at most $\nu$, and the denominator has value one at zero. Order of a Runge-Kutta method $2\nu$ implies that this test-equation approximation agrees with $e^z$ through degree $2\nu$:By uniqueness of the diagonal Padé approximant, $R=[\nu /\nu {]}_{e^z}$. Explicitly,The permitted diagonal Padé approximant property says this rational function has no poles in the closed left half-plane and satisfies $|R(z)|\le 1$ there. ThereforeThere is also no hidden stage-solve singularity from a canceled denominator: the Padé approximant numerator and denominator are coprime and each has degree $\nu$. The original determinant denominator already has degree at most $\nu$, so, with its normalization at zero, it equals $P_{\nu }(-z)$ and cannot contain an additional canceled factor. This proves maximal-order Runge-Kutta methods are A-stable.

Use the unscaled Finite Fresnel integrals of this paper. Termwise integration of the absolutely convergent Taylor series for sine and cosine gives the full small-argument expansionsIn particular,These Taylor series converge at every finite argument, although cancellation makes them unattractive for large arguments.

For the large-argument asymptotic expansions, combine the two Finite Fresnel integrals into $C+iS$. The standard Fresnel integral givesFor the tail $J(\lambda )={\int }_{\lambda }^{\infty }{e}^{i{t}^2}dt$, integration by parts using $(e^{i{t}^2}{)}^{\prime }=2it{e}^{i{t}^2}$ givesFor example $a_1=1/4$, $a_2=-3i/8$, and $a_3=-15/16$. Repeated integration by parts bounds the remainder after any fixed number of terms by the order of the next term. Equivalently the Fresnel tail expansion isHere $(-1)!!=1$. The first three terms of the requested cosine Fresnel integral, counting its constant term, areFor comparison, retaining one further correction in both Finite Fresnel integrals givesUnlike the small-argument Taylor series, these large-argument asymptotic expansions must be truncated; adding indefinitely many terms at a fixed argument does not improve the approximation.

A calculator needs elementary arithmetic and its built-in sine and cosine, together with stored approximation coefficients. A convenient global construction uses compactified approximation from endpoint asymptotics. Define the exact real tail functions byThen $P(0)=Q(0)=A$, while $P\sim 1/(2\lambda )$ and $Q\sim 1/(4{\lambda }^3)$ at infinity. Consequentlyextend continuously to the compact interval $0\le u\le 1$, with $p(1)=1/2$ and $q(1)=1/4$. Choose stored polynomial approximations $p_N,q_N$ on that interval, constrained by the small-argument Taylor series and large-argument asymptotic expansions, and calibrated across the intervening interval. The resulting elementary formula isUniform errors ${\delta }_p,{\delta }_q$ in the stored polynomial approximations give $|C_{\mathrm{calc}}-C|\le {\delta }_p+{\delta }_q$ for every positive argument. Existence of arbitrarily accurate such approximations follows from the Weierstrass approximation theorem. A Padé approximant with a denominator having no positive zero is an alternative implementation. Matching the two end expansions alone is not an error guarantee: the middle interval must also be controlled. For very small arguments, switching to the short Taylor series avoids cancellation. Ordinary finite-precision limitations in evaluating ${\lambda }^2$ and reducing its trigonometric phase remain those of the calculator itself.

The $[1/1]$ Padé approximant of the exponential is $r(z)=(1+z/2)/(1-z/2)$. Replacing each exponential in its original order gives the split Crank-Nicolson methodwith the analogous definition for $B$. Operationally, first solve the $B$ substep and then the $A$ substep. The matrices $I-\delta A/2$, $I-\delta B/2$ are invertible for every $\delta \ge 0$, since $A,B$ are symmetric negative definite.

An orthonormal eigenbasis of $A$ diagonalizes $C_A$. Its eigenvalue for $\lambda \le 0$ isTherefore $\Vert C_A{\Vert }_h\le 1$, and similarly $\Vert C_B{\Vert }_h\le 1$. Submultiplicativity gives the contractivity of split Crank-Nicolson diffusion:No commutativity or simultaneous diagonalization is needed for this product bound. This proves unconditional discrete $L^2$ stability, not positivity of every stencil weight or unconditional maximum-norm monotonicity. Nor do the individually second-order rational substeps make the ordered splitting second order: when $[A,B]\ne 0$, its leading splitting defect remains ${\delta }^2[A,B]/2$.

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.