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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.

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 $h\to 0$.

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.

Along a smooth exact solution, the chain rule gives $g(y)=f^{\prime }(y)f(y)=y^{\prime \prime }$. Expand the unscaled defect of the multiderivative multistep method about the central time $t_n$. The antisymmetric difference of the solution and second derivative, and the symmetric sum of first derivatives, giveAll derivatives are evaluated at $t_n$. Reflection symmetry removes every even power. The fifth-order conditions require the coefficients of $h,h^3,h^5$ to vanish. The last two equations give $\beta =1/15$ and $\alpha =7/15$, and the first then gives $\gamma =16/15$. ThusThese determine the symmetric two-step two-derivative formula. Substitution into the next coefficient givesThe requested fifth-order conditions actually produce formal order six. In the usual maximal-order terminology there is no member of this symmetric family with exact order five: cancelling the fifth-power defect also cancels the automatically zero sixth-power defect. Interpreting the printed request as order at least five gives precisely the boxed parameters.

At zero step size the recurrence polynomial is $w^2-1$, with simple roots $1$ and $-1$, so it satisfies the root condition for a multistep method and is zero-stable. With sufficiently smooth locally Lipschitz derivative maps, the nearby implicit solution branch and starting values of matching accuracy, the convergence of a zero-stable multiderivative method gives global order six. Formal order by itself does not imply decay of its additional numerical mode.

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.

Yes: this is an A-stable third-order method. Its second derivative is essential; the Second Dahlquist barrier applies to ordinary first-derivative linear multistep methods, not this multiderivative multistep method.

Apply the Dahlquist test equation $y^{\prime }=\lambda y$ and put $z=h\lambda$. The amplification polynomial isWe verify the root condition for a multistep method throughout $\mathrm{Re}z\le 0$. For a quadratic $a{\zeta }^2+b\zeta +c$ with $|a|>|c|$, the complex quadratic Schur criterion isfor both roots to be strictly inside the unit disk. Here is a brief justification of the criterion rather than just a root plot. Set $P^{\#}(\zeta )={\zeta }^2\overline{P(1/\overline{\zeta })}$. The polynomial $\overline{a}P-c{P}^{\#}$ is $\zeta (D\zeta +E)$, where $D=|a{|}^2-|c{|}^2$ and $E=\overline{a}b-c\overline{b}$. Its two roots are inside precisely when $|E|<D$. On the unit circle $|P^{\#}|=|P|$, and $|c|<|a|$, so the Rouche theorem equates its interior root count with that of $P$. A boundary zero of $P$ would also be a boundary zero of the transformed polynomial, which the strict inequality excludes. This proves the strict criterion; the non-strict form follows by continuity, with any unit root simple because $|c/a|<1$.

For the present polynomial it is enough to show $8|Q-1|<|Q{|}^2-1$, except at $z=0$. Write $z=-r+iy$, $r\ge 0$, and put $k=r^2+3r$, $s=y^2$. ThenDirect expansion gives the nonnegative factorizationEvery term is nonnegative, and the sum is strictly positive unless $r=y=0$. Also $|Q{|}^2=X+1>1$, so the criterion applies. For every nonzero $z$ in the closed left half-plane both amplification roots are strictly inside the unit disk. At $z=0$ they are $1$ and $1/7$, and the unit root is simple. ThereforeThis proof establishes stability of both the principal and parasitic modes, including on the imaginary axis.

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.

Let $g(y)=f_y(y)f(y)$. The chain rule gives $y^{\prime \prime }=g(y)$ along a smooth solution. Expanding every term about $t_n$, the unscaled exact-solution residual of this multiderivative multistep method isFor example, at derivative degree $q\ge 2$ its coefficient isthe coefficients at degrees zero, one and two vanish. Thus the formal order of a numerical method is two for $a\ne 11/5$, and three for $a=11/5$, where the fourth-degree coefficient is $-4/15$. This order statement describes the exact-solution defect; convergence also needs zero-stability.

At $h=0$ the characteristic polynomial isThe root condition for a multistep method holds exactly when $|a|\le 1$ and $a\ne 1$. In particular $a=-1$ has two distinct unit roots and is admissible, whereas $a=1$ has a repeated unit root. Applying the permitted Dahlquist equivalence theorem for this multiderivative setting, with smooth $f$, the nearby implicit solution branch and convergent starting values, yieldsAll those methods have global order two, provided the starting errors are $O(h^2)$. The formally third-order choice $a=11/5$ is not zero-stable and therefore is not convergent as a method for general initial-value problems.

One can see the failure without any nonlinear difficulty. For $y^{\prime }=0$, an error mode $e_n=a^n{e}_0$ grows exponentially if $|a|>1$. At $a=1$, $e_n=C+Dn$; choosing $e_0=0$, $e_1=h$ gives vanishing starting errors but $e_{T/h}=T$. This also explains why a small residual alone does not rescue that degenerate choice: at $a=1$ the first-derivative term disappears and the formula has no zero-stable first-order evolution interpretation.

The first characteristic polynomials of a linear multistep method isThe root condition for a multistep method requires $|\alpha |\le 1$ and excludes $\alpha =1$, where the root of a polynomial at one is double. At $\alpha =-1$, the two unit-modulus roots of a polynomial are distinct, so zero-stability holds. For $\alpha \ne 1$, the first-order consistency of a numerical method identity is ${\rho }^{\prime }(1)=1-\alpha$, equal to the coefficient of $hf$.

Using the permitted Dahlquist equivalence theorem in this multiderivative multistep method setting givesThis means convergence on every fixed finite time interval where the smooth exact solution of a differential equation exists, with consistent starting values and the nearby branch of each implicit solve. More specifically, the convergence of a zero-stable multiderivative method gives order $p$ when the starting errors are $O(h^p)$ and both $f$ and $f^{\prime }f$ are locally uniformly Lipschitz continuous on the relevant region.

The exclusions are genuine. For $y^{\prime }=0$ and $|\alpha |>1$, an arbitrarily small parasitic starting error is amplified as ${\alpha }^n$. At $\alpha =1$, take $y_0=0$, $y_1=h$, also for $y^{\prime }=0$. The exact solution is zero, but the numerical recurrence gives $y_n=nh$, an error of order one at fixed positive time. This proves failure even though the starting values converge.

Expand about $t=t_{n+2}$ and use $y^{\prime \prime }=f^{\prime }(y)f(y)$ from the chain rule. The residual of the exact solution isThe coefficients through $h^3$ vanish, and the displayed fourth-order coefficient does not. Thus the multiderivative multistep method has order three.

At $h=0$, its first characteristic polynomial isIts roots are $1$ and $1/7$, with the unit-modulus root simple, so it satisfies the root condition for a multistep method and is zero-stable. Combining this with the defect estimate proves convergence of order three, assuming sufficiently smooth $f$, order-three starting values, and the nearby branch of each implicit update.

More explicitly, the right-hand side is $h{F}_h(y_{n+2})$, where $F_h=6f/7-2h{f}^{\prime }f/7$ has a uniform local Lipschitz continuity bound for small $h$. A bounded zero-stable impulse response and the discrete Gronwall inequality therefore control the accumulated $O(h^4)$ defects by $O(h^3)$ over fixed time intervals. This is convergence of a zero-stable multiderivative method, rather than a direct application of the first-derivative-only Dahlquist equivalence theorem.

For the Dahlquist test equation, put $z=h\lambda$. The amplification roots satisfyWe show that no root can reach the unit circle for $\mathrm{Re}z<0$. If $\xi =e^{i\theta }$, solving the quadratic in $z$ givesWrite $c=\cos \theta$, $u=\mathrm{Re}C=-3/4+4c-c^2$, and $v=\mathrm{Im}C=(c-4)\sin \theta$. Then $u\le 9/4$ andSince $9/2-u>0$, this gives $|C|+u\le 9/2$. Consequently $|\mathrm{Re}\sqrt{C}{|}^2=(|C|+u)/2\le 9/4$, so both possible $z$ have nonnegative real part. Equality can occur only at $\theta =0$, which gives $z=0$ or $z=3$.

The leading coefficient has zeros $(3\pm i\sqrt{5})/2$, both in the right half-plane. Hence the amplification roots vary continuously as a pair throughout the left half-plane. At $z=-1$ they are $1/3$ and $1/5$; neither can leave the unit disk without crossing the unit circle, which the preceding calculation excludes. The same conclusion extends to the imaginary axis, with strict inequality away from $z=0$. At zero the roots $1,1/7$ satisfy the simplicity requirement.

Thus the method is A-stable. Its third order does not contradict the usual second-order barrier, because it is a multiderivative multistep method.

For the multiderivative multistep method $y_{n+1}-\alpha h{f}_{n+1}+\beta h^2{g}_{n+1}=\gamma h{f}_n+y_{n-1}+\alpha h{f}_{n-1}+\beta h^2{g}_{n-1}$, with $g=f^{\prime }f$, centered expansion gives defect coefficients $2-2\alpha -\gamma$, $1/3-\alpha +2\beta$, and $1/60-\alpha /12+\beta /3$ at powers $h,h^3,h^5$. Their simultaneous vanishing gives the displayed parameters. The next coefficient is $1/4725$ at $h^7$, so the formal order is six. Reflection symmetry removes every even defect power. The method satisfies fifth-order conditions but cannot have exact order five.