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The endpoint expansion of this multiderivative multistep method has unscaled defect $h^4{y}^{(4)}/3+O(h^5)$ and zero-stability roots $1,1/7$, giving third-order convergence with suitably accurate starting values. On the Dahlquist test equation its amplification roots obey $(7-6z+2z^2){\zeta }^2-8\zeta +1=0$. The complex quadratic Schur criterion puts both roots in the open unit disk for $\mathrm{Re}z<0$, and only the simple root one reaches the boundary at $z=0$. Thus it is A-stable; its higher derivative places it outside the hypotheses of the Second Dahlquist barrier.

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.

Canceling a common factor of the characteristic polynomials changes the allowed starting relations and can remove a parasitic mode rather than prove stability of the original recurrence. For example $\rho =(\zeta -1{)}^2$ and $\sigma =({\zeta }^2-1)/2$ fail zero-stability. On $y^{\prime }=0$, the original recurrence admits $y_n=y_0+n(y_1-y_0)$; starting perturbations $y_1-y_0=\sqrt{h}$ vanish but give divergent error at fixed positive time. The canceled one-step recurrence excludes these data, so its stability cannot be assigned to the original scheme.

For an initial value problem, convergence means that the largest global error on each fixed finite time interval tends to zero as the step size tends to zero and the starting errors vanish. For linear multistep methods, the Dahlquist equivalence theorem separates this requirement into consistency of a numerical method and zero-stability.

For a fixed-step multiderivative multistep method, the ordinary zero-stability root condition still controls propagation of the starting errors when all derivative evaluation maps are uniformly Lipschitz continuous on the relevant bounded region. A local defect $O(h^{p+1})$ then yields global error $O(h^p)$ over a fixed time interval, provided the starting errors are $O(h^p)$ and the implicit updates use the nearby solution branch.

For example, writing the numerical error equation as $\rho (E)e_n=h{F}_h(e_{n+s})+d_n$, with $F_h$ uniformly Lipschitz continuous, a bounded impulse response for the root condition for a multistep method givesAbsorbing the last current-step term for small $h$ and applying the discrete Gronwall inequality gives the stated order. The higher derivatives enter through $F_h$, whose bound stays uniform as $h\to 0$.

For a fixed-coefficient linear multistep method applied to a well-posed initial value problem with a continuous right-hand side uniformly Lipschitz continuous in the solution argument on the relevant region, consistency of a numerical method and zero-stability are equivalent to numerical convergence for arbitrary starting values whose errors tend to zero. The time interval is fixed while the step size tends to zero; an implicit numerical method must use its well-defined nearby solution branch for sufficiently small steps.

The quantitative statement is that a zero-stable method with consistency error $O(h^p)$ and starting errors $O(h^p)$ has global error $O(h^p)$, provided the exact solution is sufficiently smooth for that residual bound. A bounded response to step defects follows from the root condition for a multistep method, and the discrete Gronwall inequality controls the nonlinear error terms. Mere continuity of the right-hand side does not by itself supply these hypotheses or uniqueness of the exact solution.

Using Newton-Cotes closed quadrature for the integrated derivative yields the displayed linear multistep method. Its first characteristic polynomial is $\rho (z)=z^s-1$, whose unit-circle roots are all simple, proving zero-stability. Quadrature exactness gives order $s+1$ for odd $s$ and at least $s+2$ for even $s$. The first Dahlquist barrier gives equality, while the Dahlquist equivalence theorem gives convergence with consistent starts. These are finite-time convergence statements, not claims of A-stability.

An amplification root of a multistep method is parasitic when it represents an additional numerical mode rather than the principal mode approximating the exact differential-equation flow. For a consistent method the principal root starts at one as the step tends to zero, while other roots can start elsewhere. A simple parasitic root on the unit circle is compatible with zero-stability, but its motion outside the circle for a negative test parameter can destroy absolute stability. Fixed-time convergence and all-time damping must be distinguished.

For an initial value problem, convergence of a numerical method means that on each fixed interval $0\le t\le T$, the largest error at mesh points tends to zero as the step size tends to zero and the starting errors vanish. Order of a numerical method describes how rapidly an exact-start local defect decreases; stability of a numerical method controls propagation of that defect and of errors already present. These are related but different properties. A high-order expansion alone cannot ensure that accumulated error remains small.

For a one-step map $Y_{n+1}={\Phi }_h(t_n,Y_n)$, define the exact-start defect $d_{n+1}=y(t_{n+1})-{\Phi }_h(t_n,y(t_n))$. Order $p$ means $d_{n+1}=O(h^{p+1})$ for sufficiently smooth solutions. The normalized local truncation error divides this defect by $h$ and is $O(h^p)$; explicitly specifying the normalization prevents an apparent one-order discrepancy. Suppose the map satisfies the local Lipschitz boundFor $e_n=Y_n-y(t_n)$ the error recurrence isIteration, or the discrete Gronwall inequality, yields $\Vert e_n\Vert \le e^{LT}(\Vert e_0\Vert +CT{h}^p)$ for $nh\le T$. Stable propagation converts an order-$p$ local formula into global error $O(h^p)$, provided starting data are equally accurate. Consistency is the weaker requirement that the normalized local defect tend to zero.

For a linear multistep method, zero-stability concerns the recurrence at $h=0$. Its roots must have modulus at most one, and unit-modulus roots must be simple. Roots outside the circle amplify errors exponentially in the number of steps; repeated unit roots introduce polynomial growth. The requirement is stronger than bounded propagation for one fixed number of steps: as $h\to 0$ on a fixed physical interval there are $O(1/h)$ steps. Under the usual Lipschitz assumptions and consistent starting values, the Dahlquist equivalence theorem says consistency plus zero-stability is equivalent to convergence. Its sufficiency combines a bounded discrete recurrence propagator with the sum of local defects, while necessity follows by exciting an unstable homogeneous root or by observing the persistent error of an inconsistent formula.

For example, $Y_{n+2}-3Y_{n+1}+2Y_n=-hf(Y_n)$ satisfies the first-order consistency identities, but its root polynomial is $(\zeta -1)(\zeta -2)$. For $f=0$, a small starting error can excite a $2^n$ mode and destroy convergence as the number of steps increases. In contrast, simple parasitic unit roots such as the root minus one of a two-step centered method need not destroy zero-stability: they remain bounded, though they can create unwanted oscillations and affect practical behavior. Accurate starting procedures and suppression of accumulated parasitic error remain relevant.

Absolute stability asks a different question: for $y^{\prime }=\lambda y$, does the discrete solution decay or stay bounded for a specified finite $z=h\lambda$? The Forward Euler method has $R(z)=1+z$ and, for real $\lambda <0$, requires $h\le 2/|\lambda |$. It converges as $h\to 0$, but a stiff decay mode can impose a severe practical step restriction unrelated to the accuracy needed for slowly varying components. The Backward Euler method has $R(z)=1/(1-z)$, hence is A-stable and L-stable. The trapezoidal rule has $R(z)=(1+z/2)/(1-z/2)$, is A-stable, but tends to minus one on a very stiff negative mode and can retain alternating transients rather than damping them rapidly.

Zero-stability therefore does not imply unrestricted absolute stability; the three-step family analyzed earlier in this paper is a particularly strong illustration. Conversely a stable large step need not be accurate. This distinction matters for a stiff equation. The Second Dahlquist barrier limits A-stable linear multistep methods to order at most two, whereas implicit Runge-Kutta methods can have much higher order and still be A-stable. That barrier concerns the specified method class, not every numerical time integrator.

Finally, the mathematical order depends on smoothness: nonsmooth solutions or forcing can lower observed rates. Roundoff, imperfect implicit solves and inaccurate derivative evaluations introduce additional defects. Smaller steps increase the number of such perturbations; truncation error alone does not describe arbitrarily fine computation. The useful organizing principle is consistency supplies vanishing defects, stability bounds their accumulation, convergence follows, and order quantifies its rate under the stated regularity and starting-error assumptions.

Work in the standard class of real constant-coefficient linear multistep methods using only first-derivative evaluations. Writewith ${\alpha }_s\ne 0$. The Dahlquist equivalence theorem states that such a method converges on each fixed finite time interval, for sufficiently regular Lipschitz initial value problems and every set of consistent starting values, if and only if it is consistent and zero-stable. Consistency means $\rho (1)=0$ and ${\rho }^{\prime }(1)=\sigma (1)$. Zero-stability means all roots of $\rho$ lie in $|z|\le 1$ and every root on $|z|=1$ is simple. In particular the consistency root at $1$ is simple.

Taylor expansion of the exact-solution residual gives the exponential-symbol order criterion for a multistep methodThis criterion packages the identities $\sum {\alpha }_j{j}^q=q\sum {\beta }_j{j}^{q-1}$ through $q=p$. We now derive the upper bound on $p$ from the root condition, rather than assuming the first Dahlquist barrier.

Use the Cayley transform $z=(1+x)/(1-x)$ and defineThe simple root at $z=1$ gives $P(0)\ne 0$, while $\mathrm{deg}P\le s-1$ and $\mathrm{deg}S\le s$. Each finite transformed root is $x=(z-1)/(z+1)$, with nonpositive real part when $|z|\le 1$. A possible root at $z=-1$ simply lowers the degree of $R$. Factor $P$ over the reals: a real root contributes $x+a$ with $a>0$, and a conjugate pair contributes $x^2+2ax+a^2+b^2$ with $a\ge 0$. Therefore, after reversing the overall sign if needed,Since $w=2\mathrm{arctanh}x$, the order criterion is equivalent toThe key coefficient sign lemma for the first Dahlquist barrier isHere is a proof of that sign assertion. Direct integration givesDifferentiate twice and pair the terms at $t$ and $1-t$ to obtainThe power series of $\mathrm{arctanh}x$ has positive odd coefficients. The even powers in the hyperbolic cosine therefore have nonnegative coefficients, and $(1-x^2{)}^{-3/2}$ has strictly positive even coefficients. Every even coefficient of $F^{\prime \prime }$ is consequently negative. Also $F$ is even and $F(0)=1$, proving the assertion.

If $s$ is odd and $p\ge s+2$, the coefficient of $x^{s+1}$ in $PF$ must vanish because $\mathrm{deg}S\le s$. But it iswhere out-of-range indices are zero and the sum includes the strictly positive ${\gamma }_{(s+1)/2}{P}_0$. This is impossible, so $p\le s+1$. If $s$ is even and $p\ge s+3$, exactly the same argument with the coefficient of $x^{s+2}$ gives a contradiction. Thus $p\le s+2$ in that case. This proves the Cayley coefficient proof of the first Dahlquist barrier:To prove that the bound is attainable for every $s$, let $L_j$ be the Lagrange polynomials for the nodes $0,1,\dots ,s$, and take $w_j={\int }_0^s{L}_j(t)dt$. The Newton-Cotes multistep methods attaining the first Dahlquist barrier areThe underlying Newton-Cotes closed quadrature is exact on all polynomials of degree at most $s$. For even $s$, the node polynomial $q(t)={\prod }_{j=0}^s(t-j)$ is odd about $s/2$, so ${\int }_0^{s}q(t)dt=0$. Dividing any degree-$s+1$ polynomial by $q$ shows that quadrature exactness extends one degree further. Integrating the exact derivative therefore gives method order at least $s+1$ for odd $s$ and at least $s+2$ for even $s$.

Finally $\rho (z)=z^s-1$ has only simple unit-circle roots, so these methods satisfy the root condition. The weights sum to $s={\rho }^{\prime }(1)$, proving consistency. The Dahlquist equivalence theorem gives convergence with consistent starts; the proved upper bound makes the attained formal orders exact; starting values accurate to those orders also give the corresponding global accuracy. Hence the highest convergent order is $2\lfloor (s+2)/2\rfloor$. This result concerns the stated linear first-derivative class, not unrestricted multiderivative or nonlinear formulas.

Use the coefficient $2-\alpha$ in the numerator, as verified in the PDF; the converted TeX lost its $-\alpha$. Multiplying the method by $2+\alpha$ gives the characteristic polynomials of a linear multistep methodFor every admissible real $\alpha$, $\rho (1)=0$ and ${\rho }^{\prime }(1)=2\alpha =\sigma (1)$, giving consistency of a numerical method. Factorization givesThe second root is $r=(2-\alpha )/(2+\alpha )$. The root condition for a multistep method requires $|r|\le 1$ and forbids a repeated root at one. For real $\alpha \ne -2$,At $\alpha =0$ both roots equal one, violating zero-stability. For $\alpha >0$ the second root lies strictly inside the unit disk. The Dahlquist equivalence theorem therefore givesTo find the order rather than only consistency, expand the exact-solution residual through the generating function:Thus its unnormalized local truncation error begins with $(\alpha /3)h^3{y}^{(3)}$ when $\alpha >0$, and the global method order is exactly two for every convergent parameter. For comparison, $\alpha =0$ has formal order three but is not convergent because of its double unit root:These orders assume a smooth solution, the usual local Lipschitz condition and starting approximations of the required accuracy.

Only $\alpha >0$ needs consideration, by the preceding zero-stability analysis. On the Dahlquist test equation, all roots of the amplification polynomialmust satisfy the unit-disk root condition if the method is A-stable. Evaluate it at $\zeta =-1$:Take a real negative $z<-4/\alpha$. Then $P_z(-1)<0$, whereas the leading coefficient $2+\alpha -z$ is positive and $P_z(\zeta )\to +\infty$ as $\zeta \to -\infty$. The intermediate value theorem gives a real root strictly below $-1$. Its numerical mode grows in modulus, contradicting absolute stability at this negative test parameter. ConsequentlyThis proof detects an unstable root directly; the second-order value alone would not exclude A-stability. Equivalently the limiting derivative polynomial has an exterior root $-\alpha -\sqrt{1+{\alpha }^2}$, explaining the instability for large negative $z$.

Along the exact ordinary differential equation solution, the chain rule gives $g(y)=f^{\prime }(y)f(y)=y^{\prime \prime }$. Insert the exact solution into the symmetric two-step two-derivative formula and expand about the middle time $t=t_{n+1}$. Its unnormalized local truncation error isThe coefficients of $h{y}^{\prime }$, $h^3{y}^{\prime \prime \prime }$ and $h^5{y}^{(5)}$ are respectivelyThe centered residual is an odd function of $h$, so every even power cancels. The next coefficient isThus the method has order six. Its zero-stability roots at $h=0$ are the simple roots $1,-1$. With sufficiently accurate starts and smooth derivative evaluation maps, convergence of a zero-stable multiderivative method gives sixth-order error on fixed time intervals. The first Dahlquist barrier for ordinary linear multistep methods does not apply, because this is a multiderivative multistep method.

A constant-step linear multistep method has the formwith ${\alpha }_k\ne 0$. The characteristic polynomials of a linear multistep method encode both accuracy and error propagation. The method has order of a numerical method $p$ when insertion of a sufficiently smooth exact solution leaves an unnormalized local truncation error $O(h^{p+1})$. The exponential-symbol order criterion for a multistep method expresses this as $\rho (e^z)-z\sigma (e^z)=O(z^{p+1})$. In particular, first-order consistency of a numerical method requires $\rho (1)=0$ and ${\rho }^{\prime }(1)=\sigma (1)$.

Accuracy alone is insufficient. For $f=0$, perturbations obey the recurrence $\rho (E)e_n=0$. The root condition for a multistep method requires every root of a polynomial $\rho$ to lie in the closed unit disk, with each unit-circle root simple. This is zero-stability. A root outside the unit disk causes geometric amplification as $n\sim T/h$ increases; a repeated unit-circle root permits polynomial growth in $n$. Either can destroy convergence even for starting perturbations tending to zero.

The Dahlquist equivalence theorem states that, for a fixed-coefficient method and a well-posed ordinary differential equation with a suitably Lipschitz continuous right-hand side, consistency of a numerical method and zero-stability are equivalent to convergence for all convergent starting data. If the starting errors are $O(h^p)$ and the exact solution is sufficiently smooth, the global error is $O(h^p)$ on a fixed time interval. A stability estimate bounds the propagated sum of $O(h^{p+1})$ local defects over $O(1/h)$ steps; the discrete Gronwall inequality controls the dependence of $f$ on the accumulated error. Implicit methods also require the locally consistent, uniquely solvable update branch for sufficiently small $h$.

For example, the explicit Euler method has order one and $\rho (\zeta )=\zeta -1$, satisfying the root condition for a multistep method. The two-step Adams-Bashforth method has order two and $\rho (\zeta )=\zeta (\zeta -1)$, again satisfying it. The trapezoidal rule has order two with one step. The two-step Simpson multistep method, $Y_{n+2}-Y_n=h(f_{n+2}+4f_{n+1}+f_n)/3$, has order four and the simple roots $1,-1$: it is zero-stable, although its parasitic amplification root can spoil long-time decay. These examples distinguish finite-time numerical convergence from absolute stability.

The first Dahlquist barrier gives the maximum possible order of a convergent ordinary real $k$-step method:These bounds assume only first-derivative evaluations, fixed coefficients and zero-stability; the sixth-order multiderivative multistep method in solution is outside that class.

To obtain arbitrarily high convergent order within the ordinary class, increase the number of steps. The Adams-Bashforth method integrates a polynomial interpolation of $k$ past derivative values and has order $k$. The Adams–Moulton method includes the new endpoint and has order $k+1$ for $k$ steps. In both families $\rho (\zeta )={\zeta }^{k-1}(\zeta -1)$, so the root condition for a multistep method holds for every fixed $k$. Starting values can be generated by a Runge-Kutta method of sufficient accuracy. The backward differentiation formula instead differentiates an interpolating polynomial; its order-$k$ members are zero-stable only for $1\le k\le 6$.

Finally, high order and favorable stiff stability are separate design goals. By the Second Dahlquist barrier, an ordinary A-stable linear multistep method has order at most two. One may accept a smaller linear stability domain, use suitable backward differentiation formula members for stiff problems, or switch to high-order implicit Runge-Kutta methods or multiderivative multistep methods. A convergent high-order formula still needs time steps appropriate to its stability properties and the problem's eigenvalues.

The first characteristic polynomial factors asAlso $\rho (1)=0$ and ${\rho }^{\prime }(1)=\sigma (1)=2(1-\alpha )$, so the order conditions for a linear multistep method hold for every $\alpha$. Convergence of the full linear multistep method recurrence requires, in addition, the root condition for a multistep method: all polynomial roots of $\rho$ lie in the closed unit disk and its polynomial roots on the unit circle are simple.

For $-1<\alpha <1$, write $\alpha =\cos \theta$ with $0<\theta <\pi$. The other polynomial roots are $e^{\pm i\theta }$, distinct from each other and from 1, so the root condition for a multistep method holds. For $|\alpha |>1$, the two quadratic polynomial roots are real and reciprocal, and one has modulus greater than one. At $\alpha =1$, $\rho =(w-1{)}^3$; at $\alpha =-1$, $\rho =(w-1)(w+1{)}^2$. Both endpoints violate simplicity. Consistency of a numerical method plus zero-stability therefore givesThe conclusion applies to the recurrence as printed, including its starting data. Canceling common factors is not innocuous: at $\alpha =1$ the polynomials contain $(w-1{)}^2$, and at $\alpha =-13/5$ they contain $w+5$. The reduced recurrences exclude parasitic solutions of the original one. This is the common-factor cancellation defect in a multistep recurrence, so those parameter values are not added to the interval.

Consistency was established in part (a). By the Dahlquist equivalence theorem, convergence additionally requires the root condition for a multistep method. Since $\rho (w)=(w-1)(w-\alpha )$, its roots lie in the closed unit disk with each unit-modulus root simple exactly forThe endpoint $\alpha =-1$ is zero-stability although it has a simple parasitic root at minus one; $\alpha =1$ has a repeated root at one and is excluded.

For every parameter in this convergent range, the leading coefficient of $\sigma$ is $(5+\alpha )/12>0$, whileSince $\sigma (w)\to +\infty$ as $w\to -\infty$, it has a real root $w_{\ast }<-1$. Its other real root lies between minus one and one because $\sigma (1)=1-\alpha >0$, so the exterior root is simple.

On the Dahlquist test equation, the amplification polynomial of a multistep method is $P_z(w)=\rho (w)-z\sigma (w)$, with $z=h\lambda$. For $|z|\to \infty$, the equivalent polynomial $-P_z/z=\sigma -\rho /z$ tends coefficientwise to $\sigma$, and its leading coefficient stays nonzero. Choose a small closed disk about $w_{\ast }$ lying wholly outside the unit disk and containing no other root of $\sigma$. On its boundary $|\sigma |$ has a positive minimum, while $|\rho /z|$ is smaller for every sufficiently large $|z|$. Rouché's theorem then puts one root of $P_z$ inside this exterior disk for all such complex $z$.

That root violates the stability root condition for a multistep method. Hence there is a finite radius $R_{\alpha }$ beyond which no point belongs to the absolute stability domain:This proves boundedness in every complex direction, rather than only excluding A-stability. It is an instance of exterior roots of the derivative polynomial bound multistep stability.

For this two-step family with a third-order member, use the exponential-symbol order criterion for a multistep method. Expanding the exact-solution residual symbol givesThe constant, linear and quadratic coefficients vanish for every parameter. ThusAt the exceptional value the fourth-degree coefficient is $-1/30$, so there is no further increase. This means an unscaled exact-solution residual of order $h^{p+1}$, or a residual divided by the step of order $h^p$.

The first characteristic polynomial factors as $\rho (w)=(w-1)(w-a)$. The root condition for a multistep method requires every root to have modulus at most one and each unit-modulus root to be simple. The root $a=-1$ is allowed because it is distinct from the simple root one; $a=1$ is not, since it makes one a double root. Hence zero-stability holds exactly forOn this interval, $\rho (1)=0$ and ${\rho }^{\prime }(1)=\sigma (1)=1-a\ne 0$ establish consistency of a numerical method. The Dahlquist equivalence theorem, for a fixed-step consistent linear multistep method applied to an ordinary differential equation with a Lipschitz continuous right-hand side with convergent starting values and well-defined small-step updates, says that convergence of a numerical method is equivalent to zero-stability. Therefore the same interval is exactly the convergence of a numerical method range. With sufficiently accurate starting values the convergent members have the orders just computed.

At $a=1$ the Taylor expansion still gives a formal order-two residual, but ${\rho }^{\prime }(1)=\sigma (1)=0$ and the double root prevents a convergent method. Canceling the common factor $w-1$ produces the common-factor cancellation defect in a multistep recurrence and changes the starting relations and cannot repair convergence of a numerical method of the original recurrence. At $a=0$, cancellation of a zero root instead reveals the ordinary trapezoidal rule; it is consistent with the order-two result.

A fixed-step linear multistep method for $y^{\prime }=f(t,y)$ uses several previous solution and derivative values:The coefficients are fixed independently of the step size. It is explicit when ${\beta }_s=0$ and implicit otherwise. Its characteristic polynomials of a linear multistep method are $\rho (w)={\sum }_j{\alpha }_j{w}^j$ and $\sigma (w)={\sum }_j{\beta }_j{w}^j$. The contraction mapping theorem gives a uniquely solvable implicit step under a Lipschitz continuity bound $L$ whenever $k|{\beta }_s|L<|{\alpha }_s|$, by contraction of the new-value equation on the appropriate solution region. A suitable starting procedure must supply the first $s$ values; the method alone does not determine them.

Order is defined by inserting a smooth exact solution into the recurrence. If the residual is $O(k^{p+1})$ and its first nonzero coefficient occurs at that degree, the method has order $p$; dividing the residual by $k$ gives an $O(k^p)$ local truncation error. Taylor expansion gives the exact order conditions for a linear multistep methodwith the right side interpreted as zero at $q=0$, and failure at $q=p+1$ for exact order. Equivalently,The basic consistency of a numerical method conditions are $\rho (1)=0$, ${\rho }^{\prime }(1)=\sigma (1)$; for the ordinary nondegenerate consistent formulation this common value is nonzero. Exactness on constants and linear functions does not control propagation of numerical perturbations.

That propagation is the role of zero-stability. On $y^{\prime }=0$ the homogeneous recurrence is $\rho (E)y_n=0$. Its modes are $n^r{\xi }^n$, with powers of $n$ determined by root multiplicities. The theorem on the root condition for a multistep method states that zero-stability is equivalent to every root of $\rho$ having modulus at most one, with every unit-modulus root simple. Interior repeated roots are permitted, since geometric decay dominates their polynomial factors. An exterior root amplifies perturbations geometrically; a repeated unit root amplifies them polynomially and violates a step-independent perturbation bound.

The Dahlquist equivalence theorem states that a fixed-step consistent linear multistep method is convergent exactly when it is zero-stable. This concerns ordinary differential equations with a uniformly Lipschitz continuous right-hand side on the solution region over a fixed finite interval, convergent starting values, and the well-defined nearby branch of each small-step implicit solve. For order $p$ convergence of a numerical method, the solution must have the required smoothness and the starting errors must be $O(k^p)$.

The mechanism can be sketched explicitly. The root condition for a multistep method bounds the recurrence's discrete Green function. For an order-$p$ exact-solution defect $d_n=O(k^{p+1})$, the Lipschitz continuity error equation then gives an estimate of the formThe current-step term can be absorbed for small $k$. The discrete Gronwall inequality says that a bound $E_n\le A+Bk{\sum }_{j<n}{E}_j$ implies $E_n\le A(1+Bk{)}^n\le A{e}^{B{t}_n}$. There are $O(k^{-1})$ defects on a fixed interval, so their sum is $O(k^p)$ and this yields the claimed global order. Conversely, testing $y^{\prime }=0$ with vanishing starting perturbations in exterior-root or repeated-unit-root modes proves that convergence of a numerical method fails without the root condition for a multistep method.

For example, the two-step Adams-Bashforth method hasIt is explicit, order two and zero-stable, hence convergent with order-two starting values. By contrast, a recurrence with $\rho =(w-1{)}^2$ has solutions $y_n=y_0+n(y_1-y_0)$ on the zero differential equation. Starting differences $\sqrt{k}$ tend to zero yet become unbounded at a fixed positive time, since $n$ is of order $1/k$. A small local defect or formal high order cannot rescue this mode. Canceling a common factor may remove it only by changing the allowed starting relations, as the parameter endpoint in Question 2 illustrates.

Absolute stability addresses a different limit: apply the method to the Dahlquist test equation $y^{\prime }=\lambda y$ and put $z=k\lambda$. The amplification roots satisfyAll roots must satisfy the disk condition, with unit roots simple; one cannot inspect only the root that tends to one as $z\to 0$. The degree and solvability of the recurrence must also be preserved. The resulting set of $z$ is the domain of absolute stability. A-stability means that it includes the whole closed left half-plane, so the numerical solutions of every Dahlquist test equation with $\mathrm{Re}\lambda \le 0$ are stable for every positive step size. It is a linear test-equation property, not a proof of arbitrary nonlinear contractivity or high accuracy at large steps.

Forward Euler method has order one and factor $1+z$, so its stability region is the disk $|1+z|\le 1$. Backward Euler method has factor $(1-z{)}^{-1}$ and is A-stable. The trapezoidal rule is order two with factor $(1+z/2)/(1-z/2)$; its modulus is at most one throughout the left half-plane. These one-step methods are included as the simplest multistep examples. For the two-step Adams-Bashforth method, the negative-real stability interval is $-1\le z\le 0$: its polynomial is $w^2-(1+3z/2)w+z/2$, and the root reaches $-1$ at $z=-1$. It therefore imposes a step restriction when applied to strongly decaying modes.

The genuinely two-step BDF2 method isIt has order two, roots $1,1/3$ at zero step and is A-stable. For its normalized polynomials $\rho =3w^2/2-2w+1/2$, $\sigma =w^2$, the unit-circle boundary locus hasThe leading coefficient has no zero in the left half-plane, and the roots at a small negative $z$ are both inside the disk. Root continuity then proves open-half-plane stability; on the imaginary boundary the only possible unit root is the simple root one at $z=0$. This gives a direct proof of the stated A-stability rather than relying on a boundary plot.

Two sharp order barriers explain the tradeoff. The first Dahlquist barrier states that a consistent zero-stable real linear $s$-step method using only first derivatives has order at most $s+1$ for odd $s$ and $s+2$ for even $s$; an explicit method has order at most $s$. The Second Dahlquist barrier states that an irreducible A-stable real linear multistep method has order at most two. These statements concern fixed constant coefficients and the ordinary first-derivative multistep class, not multiderivative methods or arbitrary variable-step formulations. Thus higher order alone is not a route to unconditional stiff stability. Question 2 provides a concrete illustration: its third-order member is zero-stable but not A-stable.

No nontrivial consistent explicit multistep method is A-stable. Its amplification polynomial of a multistep method has a fixed highest coefficient, while at least one lower coefficient grows without bound as real $z\to -\infty$. Were all roots uniformly in the unit disk, their elementary symmetric polynomials would keep all the monic coefficients bounded, a contradiction. This is the coefficient argument behind explicit multistep methods cannot be A-stable.

In practice, selecting a method therefore involves accuracy, startup, the presence of a stiff differential equation, implicit-solve cost and step-size control. Zero-stability handles propagation as the step tends to zero; consistency of a numerical method supplies a vanishing defect; A-stability handles the numerical solutions of all decaying Dahlquist test equations at finite steps. None substitutes for the others, and variable steps or PDE discretizations require additional uniform stability arguments appropriate to their setting.

For an autonomous ordinary differential equation, the chain rule gives $y^{\prime \prime }=f^{\prime }(y)f(y)$. The formula is therefore a multiderivative multistep method. Insert the exact solution and expand the two earlier values about $t=t_{n+2}$:The unscaled local truncation error isThe coefficients through $h^3$ vanish, but the fourth-order coefficient does not. Hence the method has order three, not four: the one-step residual is $O(h^{p+1})$ for an order-$p$ multistep formula. Its zero-stability polynomial is $(\zeta -1)(\zeta -1/7)$, with a simple unit root and the other root inside the unit disk. Thus the stated order is also the convergence order for smooth solutions, starting errors $O(h^3)$, and the nearby implicit solution branch, by convergence of a zero-stable multiderivative method.

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.