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Write $F(x)={\int }_0^1(1+x{)}^t(1-x{)}^{1-t}dt=x/\mathrm{arctanh}x$. Pairing $t$ and $1-t$ after differentiating twice givesThe power series of $(1-x^2{)}^{-3/2}$ has strictly positive even coefficients, and the other factor has nonnegative even coefficients. Thus every even coefficient of $F^{\prime \prime }$ is negative. Since $F$ is even with $F(0)=1$, the displayed sign pattern follows. This supplies a short positivity argument for the first Dahlquist barrier.

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.

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.

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.

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.

Integrating a quadratic interpolant of the derivative over two steps gives this linear multistep method, which has order four. Its local truncation error is $-h^5{y}^{(5)}(t_{n+1})/90+O(h^7)$ and its zero-stability polynomial is $\rho (\zeta )={\zeta }^2-1$, with simple roots $1,-1$. The Dahlquist equivalence theorem therefore gives fourth-order convergence with sufficiently accurate starts. This attains the even two-step first Dahlquist barrier. Its undamped parasitic zero-step mode need not have favorable long-time absolute stability; finite-time convergence and stiff damping are separate requirements.