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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 relevant one of the characteristic polynomials of a linear multistep method isThe coefficients on the right are chosen to give third order, so consistency of a numerical method is already available. By the Dahlquist equivalence theorem, convergence with convergent starting values is equivalent to zero-stability: every root of $\rho$ must satisfy $|w|\le 1$, and every root on the unit circle must be simple.

If $a<0$, the quadratic is negative at $w=-1$ and tends to $+\infty$ as $w\to -\infty$, so it has a root less than $-1$. If $a>1$, its root product is $a$, so at least one root has modulus greater than one. For $0\le a\le 1/4$, its real roots $(-1\pm \sqrt{1-4a})/2$ lie in $[-1,0]$. For $1/4<a\le 1$, its conjugate roots have modulus $\sqrt{a}\le 1$.

At $a=0$, the three roots are $1,0,-1$, with the unit-modulus roots simple. At $a=1$, they are $1,e^{2\pi i/3},e^{-2\pi i/3}$, again simple. The repeated quadratic root at $a=1/4$ is $-1/2$, strictly inside the unit circle, which the root condition for a multistep method permits. ThereforeThis is the root stability of a cubic multistep polynomial; oscillatory parasitic roots at the endpoints do not invalidate convergence when their starting amplitudes tend to zero.

The Dahlquist equivalence theorem states that a consistent linear multistep method is convergent for suitably consistent starting values exactly when it is zero-stable. The root condition for a multistep method requires every root of $\rho$ to lie in the closed unit disk, with every unit-modulus root simple.

The roots are $1$ and $a$. Thus $|a|\le 1$ is necessary; $a=1$ is excluded because it gives a double root at one. At $a=-1$, the two unit roots are distinct, so the endpoint is allowed. Combined with part (a), this provesAssume a locally Lipschitz vector field, a smooth solution on the fixed time interval, a nearby solvable implicit branch and starting errors of the required order. The global order is three at $a=-1/5$ and two at the other convergent parameter values. Outside this interval, zero-step perturbations already grow through either an exterior root or a unit-root polynomial factor.

The characteristic polynomials of a linear multistep method areThe consistency of a numerical method conditions hold for every real $a$:Both roots of a polynomial of $\rho$, namely $1$ and $-1$, are simple and have modulus one. The root condition for a multistep method therefore gives zero-stability for every $a$. The Dahlquist equivalence theorem now gives convergence for every fixed $a\in \mathbb{R}$. As usual, this means convergence of a numerical method on each fixed finite interval for a sufficiently regular ordinary differential equation with a Lipschitz continuous vector field and starting values tending to the exact starting values. When $a\ne 0$, the implicit time-stepping method update is locally uniquely solvable for sufficiently small $h$, for example by a contraction mapping if $|ha|L<1$. Zero-stability does not assert that a large fixed step is suitable for a stiff differential equation.

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

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

Consider an initial value problem $y^{\prime }=f(t,y)$ on $[0,T]$, with a Lipschitz continuous vector field in the solution variable and enough smoothness for the stated error expansions. Convergence of a numerical method meanswith consistent starting values. An order of a numerical method $p$ quantifies this approximation: with sufficiently accurate starting data and the required stability, the global error is $O(h^p)$. Stability of a numerical method controls the amplification and accumulation of perturbations; it is the link between a small local defect and small global error.

For a one-step formula $Y_{n+1}={\Phi }_h(t_n,Y_n)$, define the unscaled local truncation errorThe method has local order $p$ if $\Vert d_{n+1}\Vert \le C{h}^{p+1}$ uniformly along sufficiently smooth solutions. If the defect is divided by $h$, its order is instead $O(h^p)$; the convention must be specified. Assume also the perturbation estimateThe global error $e_n=Y_n-y(t_n)$ then satisfiesIterating, or applying the discrete Gronwall inequality, givesThere are $O(1/h)$ steps on a fixed interval, so stable propagation turns $O(h^{p+1})$ local defects into $O(h^p)$ accumulated error. Starting error $O(h^p)$ is needed to retain order $p$. This explains why matching an exact Taylor expansion is insufficient unless errors are controlled during repeated steps.

For a fixed-coefficient linear multistep method, writeThe consistency of a numerical method conditions are $\rho (1)=0$ and ${\rho }^{\prime }(1)=\sigma (1)$. The higher order conditions for a linear multistep method arewith right side zero for $q=0$; exact order $p$ means first failure at $q=p+1$. They come from inserting the exact solution and comparing its Taylor expansion.

Errors at $h=0$ follow the homogeneous recurrence determined by $\rho$. Zero-stability is equivalent to the root condition for a multistep method: all roots of a polynomial have modulus at most one, and every unit-modulus root is simple. Roots outside the disk give exponential growth in the number of steps; repeated unit roots give polynomial growth in that number. Repeated interior roots are harmless because their polynomial factors are dominated by decay. Under the usual Lipschitz continuity and initialization assumptions, the Dahlquist equivalence theorem givesFor a zero-stable order-$p$ linear multistep method, local defects and starting errors of the stated orders yield global error $O(h^p)$ by a recurrence bound and the discrete Gronwall inequality. Every required starting value matters.

A concrete failure of “consistency implies convergence” isHere $\rho =(\zeta -1)(\zeta -2)$ and $\sigma =-1$, so $\rho (1)=0$ and ${\rho }^{\prime }(1)=\sigma (1)=-1$: the method is consistent, of order one. But its root $2$ violates zero-stability. For $y^{\prime }=0$ with exact solution zero and starting values $Y_0=0$, $Y_1=h^2$, the computed solution is $Y_n=h^2(2^n-1)$. These starting errors vanish, yet at $n\approx T/h$ the error diverges. Local accuracy cannot compensate for an unstable recurrence.

A different meaning of stability concerns a fixed step on decaying equations. The Dahlquist test equation $y^{\prime }=\lambda y$ introduces $z=h\lambda$. A one-step method gives $Y_{n+1}=R(z)Y_n$ and is absolutely stable when $|R(z)|\le 1$. For a linear multistep method, the corresponding amplification polynomial of a multistep method is $\rho (\zeta )-z\sigma (\zeta )$, and every root must satisfy the bounded root condition, not just the root approximating $e^z$. Strict asymptotic decay requires the roots to have modulus less than one for $\mathrm{Re}z<0$. This distinction matters at reducible methods with an undamped parasitic mode, as in question 1(c).

The Forward Euler method has order one and $R(z)=1+z$, so its linear stability domain is $|1+z|\le 1$. For $y^{\prime }=-\kappa y$, with $\kappa >0$, it requires $0\le h\kappa \le 2$. It converges as $h\to 0$, yet an excessively large step can produce growing oscillations on an exactly decaying problem. The Backward Euler method has the same order, but $R(z)=(1-z{)}^{-1}$ makes it A-stable and L-stable. For a stiff differential equation, fast decay can impose a severe explicit time-step restriction even when accuracy of the slow dynamics would permit a much larger step.

The trapezoidal rule has order two and is A-stable, but is not L-stable, since its stability function tends to $-1$ along the negative real axis. The Second Dahlquist barrier limits an irreducible A-stable linear multistep method to order at most two. This does not limit implicit Runge-Kutta methods: question 4's Lobatto IIIA method is fourth-order and A-stable. Conversely, question 1's fourth-order member is convergent but not A-stable. These examples show that order of a numerical method, zero-stability and A-stability answer different questions.

For nonlinear dissipative problems, B-stability controls distances between numerical solutions; algebraic stability of a Runge-Kutta method is a sufficient coefficient criterion. Question 4 shows that A-stability alone does not imply that coefficient criterion. Variable time steps require additional analysis of step ratios and error propagation, and stiff problems may require uniform-in-stiffness error estimates beyond the fixed-problem convergence result. The useful chain is local accuracy plus the appropriate perturbation bound, giving global accuracy; absolute and nonlinear stability then determine which practical time steps preserve the intended dynamics.

The Dahlquist equivalence theorem states that a linear multistep method has convergence of a numerical method if and only if it is consistent and zero-stable, under the standard assumptions for the initial value problem and convergent starting values. Consistency of a numerical method means order at least one, orThe root condition for a multistep method expresses zero-stability: every root of $\rho$ has modulus at most one, and every root with modulus one is simple. Usual hypotheses include a well-posed ordinary differential equation with $f$ continuous in time and Lipschitz continuous in the solution variable, a valid step equation, and starting approximations converging to the exact values.

With ${\rho }_0=a$, ${\rho }_1=\theta$, ${\rho }_2=1$, the constant order condition gives $a=-1-\theta$. The next three order conditions for a linear multistep method giveSolving this linear system yields the unique coefficients of order at least three:The characteristic polynomial controlling zero-stability factors asThe other root is $-1-\theta$. Requiring its modulus to be at most one gives $-2\le \theta \le 0$, but $\theta =-2$ produces a repeated root at one and violates the root condition for a multistep method. At $\theta =0$, the roots $1,-1$ are both simple, so that endpoint is allowed. By the Dahlquist equivalence theorem, the method converges precisely forThe fourth defect coefficient is $C_4=\theta$, so the method has order exactly three for nonzero $\theta$. At $\theta =0$ it is the two-step Simpson method, of order four; thus the extra-order endpoint has correctly remained in the answer.

Apply the numerical method to the scalar test equation $y^{\prime }=\lambda y$, with $z=h\lambda$. Its linear stability domain consists of those $z\in \mathbb{C}$ for which the discrete solution stays bounded as the step count tends to infinity. For a one-step method with stability function $R$, this isIf asymptotic decay is required, use $|R(z)|<1$ instead; this only changes the boundary conventions in part (b). A method is A-stable if its stability domain contains the left half-plane, so every decaying linear test equation is stable for every positive step size. For a multistep method, the corresponding definition uses the root condition for a multistep method for every characteristic root, including simplicity of unit-modulus roots.

On the Dahlquist test equation, $g(y)={\lambda }^{2}y$, so the amplification polynomial of a multistep method becomesBoth roots of a polynomial, including the parasitic one, must satisfy the root condition for a multistep method.

For $\alpha =1/7$, multiply by seven and set $a=7-6z+2z^2$. The polynomial is $a{\xi }^2-8\xi +1$. The complex quadratic Schur criterion says that its roots of a polynomial are strictly in the unit disk ifHere is a direct verification on the entire left half-plane. Put $z=-r+iy$, $r\ge 0$, $t=y^2$, and $a_0=7+6r+2r^2$. Thenand expansion of the squared inequality givesEvery term is nonnegative, and at least one is positive unless $z=0$. Since $q-1>0$, taking square roots proves the strict complex quadratic Schur criterion for $z\ne 0$. At $z=0$, the roots of a polynomial are the simple values $1$ and $1/7$. The leading coefficient cannot vanish because $|a{|}^2\ge 49$. Thus the method at $\alpha =1/7$ is A-stable. Its order three is compatible with the Second Dahlquist barrier: that barrier concerns first-derivative linear multistep methods, whereas this method uses a second derivative.

For $\alpha =1/2$, choose $z=-4$, strictly in the left half-plane. The amplification polynomial of a multistep method reduces toOne root of a polynomial has modulus $(3+\sqrt{17})/4>1$. Thus the method at $\alpha =1/2$ is not A-stable.

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.

The spatial shift in the PDF also requires a boundary value beyond the supplied right endpoint: if $m$ is the last interior index, the newest $m+2$ value is outside the interval. Thus the printed Dirichlet method has no specified boundary closure, and no unique stability range can be assigned to it as stated.

For completeness, the literal interior formula has an exact Von Neumann stability analysis on an infinite or periodic grid. Substitution of a Fourier mode $U_m^n={\xi }^n{e}^{im\theta }$ givesFor the root of a polynomial tending to one as $\theta \to 0$, expansion givesConsequently $\mu <3$ is unstable: sufficiently small nonzero frequencies have a root of a polynomial of modulus greater than one, and arbitrarily fine periodic grids contain such frequencies.

Conversely, if $\mu \ge 3$ and $s=\cos \theta$, thenFor $A=u+iv$ with $u\ge 3$,Since $|A|\ge 3$, the complex quadratic Schur criterion yields both roots of a polynomial in the unit disk, strictly inside except at $\theta =0$, where they are $1$ and $1/3$. Uniform stability follows directly from the BDF2 discrete energy identity: the modal recurrence is $3v^{n+2}-4v^{n+1}+v^n=-(A-3)v^{n+2}$, whose energy decreases because $\mathrm{Re}(A-3)\ge 0$. Summing the modal energies proves a bound independent of the grid and even of $\mu \ge 3$; no eigenvector-separation assumption is needed. This is the stability of a spatially shifted BDF2 stencil. Thus the literal periodic-grid result isThis cannot establish stability of an unspecified Dirichlet boundary condition closure. It also does not restore consistency of a numerical method.

For the intended unshifted backward differentiation formula, a spatial eigenvalue $-\beta \le 0$ givesThe Schur stability criterion gives the root condition for a multistep method for every $k\beta \ge 0$, with strict inequalities for $\beta >0$. A useful uniform energy method proof, including the repeated interior roots that a bare separation argument would miss, is the BDF2 discrete energy identity. With ${\delta }^{2}U=U^{n+2}-2U^{n+1}+U^n$ andthe corrected Dirichlet discrete Laplacian scheme satisfiesThis follows by taking the inner product of $3U^{n+2}-4U^{n+1}+U^n=2k{L}_h{U}^{n+2}$ with $U^{n+2}$ and using summation by parts. It controls both time levels uniformly, regardless of the mesh ratio. Hence the corrected method is stable for

For a fully discrete linear partial differential equation evolution, write $U^{n+1}=C_{h,k}{U}^n$ in a specified discrete norm. Finite-time stability of a numerical method meanswith $C_T$ independent of the allowed meshes. For a multistep method, augment the state with its time-history values and include a stable starter. Bounds may grow as $e^{\omega T}$ when the continuous problem has growth; demanding decay for all time would be a stronger assertion. The treatment of initial data, forcing, Dirichlet boundary conditions, and the norm is part of the hypothesis, not a detail supplied by an interior calculation.

For constant coefficients on an infinite or periodic uniform grid, Von Neumann stability analysis substitutes the Fourier mode $U_j^n=G(\theta {)}^n{e}^{ij\theta }$. The discrete Fourier transform and the Parseval identity turn a uniform multiplier estimate into a discrete L2 norm estimate. In a one-step scalar scheme, $|G(\theta )|\le 1$ proves contractivity, while $|G(\theta )|\le 1+\omega k$ gives finite-time stability. In a multilevel scheme, every root of a polynomial in its amplification polynomial of a multilevel finite difference scheme matters, as does uniform control of the associated companion matrix.

For the Forward Euler diffusion scheme with $\mu =k/h^2$,Thus $|G|\le 1$ for all frequencies exactly whenSufficiency follows since $G\in [-1,1]$; necessity follows from the highest-frequency mode $\theta =\pi$ on even periodic grids. This gives the usual parabolic mesh restriction $k\le h^2/2$. For the Backward Euler diffusion scheme,so every $\mu \ge 0$ is stable. The Crank--Nicolson method hasagain of modulus at most one for all $\mu \ge 0$, but stiff modes approach $-1$ rather than zero. These are PDE manifestations of the A-stability and L-stability distinctions for time integration.

For advection $u_t+a{u}_x=0$, assume $a>0$ and set $\nu =ak/h$. The upwind finite difference scheme is $U_j^{n+1}=(1-\nu )U_j^n+\nu U_{j-1}^n$, withIt is contractive when $0\le \nu \le 1$. This also has a direct maximum-norm proof: each newest value is a convex combination of two old values. The resulting Courant–Friedrichs–Lewy condition expresses that the numerical domain of dependence covers the physical one. For negative $a$, the upwind direction must be reversed. By contrast, Forward Euler method time stepping with the centered first finite difference has $G=1-i\nu \sin \theta$ and $|G|>1$ for nonzero $\nu \sin \theta$. Under fixed nonzero $\nu$ refinement it is unstable, since a fixed nontrivial mode grows geometrically over $T/k$ steps. This is not a claim of instability under every imaginable mesh coupling: $k=O(h^2)$ instead bounds its spurious finite-time growth, since $\log |G|\le a^2{k}^2/(2h^2)$.

The leapfrog advection scheme illustrates a multilevel subtlety. Its amplification polynomial of a multilevel finite difference scheme isFor $|\nu |<1$, both roots of a polynomial have unit modulus and separation at least $2\sqrt{1-{\nu }^2}$. A uniformly conditioned eigenbasis of the companion matrix then proves power boundedness of a two-level Fourier scheme for arbitrary history data. At $|\nu |=1$ and a grid admitting $\theta =\pi /2$, the roots coincide on the unit circle, and a Jordan block produces $n{\xi }^n$ growth. Thus the endpoint fails the ordinary arbitrary-history root condition for a multistep method; checking only that both moduli equal one misses the instability. Restricting the starter or filtering the parasitic mode is an additional hypothesis.

The energy method handles variable coefficients and finite boundaries for which Fourier stability analysis may be unavailable. For the Backward Euler diffusion scheme with homogeneous Dirichlet boundary conditions, define $\Vert U{\Vert }_h^2=h{\sum }_j|U_j{|}^2$. The discrete summation by parts identity isTake the inner product of $U^{n+1}-U^n=k{L}_h{U}^{n+1}$ with $2U^{n+1}$. The elementary identity $2\mathrm{Re}⟨x-y,x⟩=\Vert x{\Vert }^2-\Vert y{\Vert }^2+\Vert x-y{\Vert }^2$ givesThis proves unconditional contractivity including the actual boundary treatment. More generally, a dissipative operator $L_h$ gives $(I-k{L}_h{)}^{-1}$ of operator norm at most one: with $V=(I-k{L}_h)U$, dissipativity implies $\mathrm{Re}⟨V,U⟩\ge \Vert U{\Vert }^2$, and the Cauchy-Schwarz inequality gives $\Vert U\Vert \le \Vert V\Vert$. In finite dimensions this also proves invertibility. For the second-order backward differentiation formula, the BDF2 discrete energy identity used in question 4 controls both time levels and proves unconditional diffusion stability. A repeated root strictly inside the disk is harmless here; the energy argument supplies the uniform bound without a singular eigenvector formula.

A second technique is eigenvalue stability analysis of a finite difference method. Under the method of lines, a time integrator advances $U^{\prime }=L_{h}U$ by $R(k{L}_h)$. If $L_h$ is a normal matrix, the scalar linear stability domain criterion on all $k{\lambda }_j(L_h)$ controls its powers exactly in the corresponding L2 norm. For the centered Dirichlet discrete Laplacian, its eigenvalues lie between $-4/h^2$ and zero; intersecting this interval with a time integrator's stability interval determines its mesh restriction. A uniform bound on the diagonalization of a matrix is needed for nonnormal diagonalizable systems. Eigenvalues alone can be misleading. For example,has both eigenvalues inside the disk for $0<k<1$, but the upper-right entry of $C_{h,k}^n$ is $nk(1-k{)}^{n-1}/h$. Taking $k=h$, $nk\to T>0$, this grows like $T{e}^{-T}/h$. Hence stable scalar eigenvalues do not give mesh-uniform stability. Boundary closures can introduce precisely the extra growth that an interior Fourier symbol overlooks, so boundary stability of a finite-difference method must be checked separately.

For nonlinear spatial discretizations, a useful replacement for Fourier stability analysis is a convexity argument. If the Forward Euler method map $E_k(U)=U+kF(U)$ is nonexpansive in a chosen norm for $k\le k_{\mathrm{FE}}$, the second-order strong stability preserving Runge-Kutta methodis also nonexpansive under the same restriction. Indeed, for two inputs, the triangle inequality gives $\Vert U_{\mathrm{new}}-{\tilde{U}}_{\mathrm{new}}\Vert \le \frac{1}{2}\Vert U-\tilde{U}\Vert +\frac{1}{2}\Vert E_k(Y)-E_k(\tilde{Y})\Vert \le \Vert U-\tilde{U}\Vert$. Its Taylor expansion is $U+kF(U)+k^2{F}^{\prime }(U)F(U)/2+O(k^3)$, proving order two. Such strong stability preserving Runge-Kutta methods transfer suitable forward-step bounds without relying on a linear spectral calculation.

Finally, consistency of a numerical method explains what stability buys. For a linear Hadamard well-posed problem, the Lax equivalence theorem equates convergence of a consistent discretization with its stability, under the stated approximation-space and norm hypotheses. Directly, if the error satisfies $e^{n+1}=C_{h,k}{e}^n+k{\tau }^n$, iteration gives the discrete Duhamel principleThus vanishing normalized local truncation error and initial error yield convergence. With a nonlinear Lipschitz step estimate, the discrete Gronwall inequality plays the same role. A stable but inconsistent stencil, such as the literal defective formulas in questions 3 and 4 under their usual refinement, is not rescued by any amplification bound. The decisive checks are a mesh-uniform evolution bound, a valid boundary closure, and consistency with the actual PDE.

For a fixed-step linear multistep method with a nonzero leading coefficient, applied to an ordinary differential equation with a locally Lipschitz continuous vector field, the Dahlquist equivalence theorem states thatHere convergence is uniform on each fixed finite time interval as the step size tends to zero, for every set of starting values tending to the corresponding exact values. Consistency of a numerical method means that the normalized local defect tends to zero; for the characteristic polynomials of a linear multistep method it gives $\rho (1)=0$ and ${\rho }^{\prime }(1)=\sigma (1)$. Zero-stability is equivalent to the root condition for a multistep method: every root of a polynomial $\xi$ of $\rho$ satisfies $|\xi |\le 1$, and roots with $|\xi |=1$ are simple. Order $p\ge 1$, together with zero-stability and starting errors $O(h^p)$, gives global error $O(h^p)$ under the usual smoothness assumptions.

To prove the necessity of the root condition for multistep convergence, use the test problem $y^{\prime }=0$, $y(0)=0$. Its discrete error satisfies $\rho (E)e_n=0$, where $E{e}_n=e_{n+1}$. Set $h=T/N$. If $|\xi |>1$, the exact recurrence solutionhas starting errors tending to zero at every fixed starting index, but $|e_N|=1$. Thus convergence fails.

If a unit-modulus root has multiplicity $m\ge 2$, the recurrence instead admitsAgain every fixed starting error tends to zero while $|e_N|=1$. These polynomial-times-exponential solutions follow from the repeated factor $(E-\xi {)}^m$. Complex modes can be interpreted as a real two-component system, or their real and imaginary parts can be used. Therefore convergence requires precisely the root condition for a multistep method. Repeated roots strictly inside the unit disk are allowed, since their polynomial factors are dominated by exponential decay.

Factor the first characteristic polynomial asThe other two roots of a polynomial have product one. If $-1<\alpha <1$, they are $\alpha \pm i\sqrt{1-{\alpha }^2}$: distinct unit-modulus roots, both different from one. At $\alpha =-1$, the root $-1$ is double; at $\alpha =1$, the root one is triple. Outside $[-1,1]$, one of the reciprocal real roots has modulus greater than one. Thus the root condition for a multistep method holds exactly for $-1<\alpha <1$.

For the order conditions for a linear multistep method, expand the exponential defect:The constant, linear and quadratic coefficients vanish for every $\alpha$. The cubic coefficient vanishes only at $\alpha =11/5$, where the quartic coefficient is $-3/5\ne 0$. Consequently the formal order isFor all parameters satisfying the root condition for a multistep method, ${\rho }^{\prime }(1)=\sigma (1)=2(1-\alpha )\ne 0$, so this formal calculation gives ordinary second-order consistency of a numerical method. The Dahlquist equivalence theorem therefore yieldsThe exceptional third-order formula is not zero-stable. At $\alpha =1$, $\sigma \equiv 0$ and $\rho =(w-1{)}^3$: the recurrence contains no vector-field evaluations. It satisfies the displayed Taylor identities only formally and is a degenerate, nonconvergent formula, not a usable second-order ODE solver. This distinction avoids interpreting formal cancellation as convergence.

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.