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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.

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.

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.

For this Adams–Moulton method, the characteristic polynomials of a linear multistep method areExpansion at zero givesPart (a) therefore shows that the method has order exactly three.

It is consistent because its order is at least one. The roots of $\rho (w)=w(w-1)$ are $0$ and $1$; both lie in the closed unit disk, and the only unit-modulus root is simple. Thus the method is zero-stable. By the Dahlquist equivalence theorem, consistency and zero stability imply convergence. Hence the method is

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.

The first characteristic polynomial of the linear multistep method comes only from its left side:Its roots are zero and one; both lie in the closed unit disc, and the only unit-modulus root is simple. Thus the method satisfies the root condition for a multistep method and is zero-stable. Part (b) showed consistency for every $\alpha ,\beta$. The Dahlquist equivalence theorem states that a consistent linear multistep method is convergent exactly when it is zero-stable. Thereforeunder the usual Lipschitz hypotheses on $f$.

Stability asks whether perturbations already present in data, arithmetic, or an earlier numerical step remain controlled under subsequent evolution. It complements consistency of a numerical method: consistency says that an exact smooth solution nearly satisfies one numerical step, while stability prevents the accumulated local defects from being amplified without bound. For a well-posed differential equation these two properties are what make convergence possible.

For a linear multistep method, zero-stability concerns the limit $h\to 0$. The roots of its first characteristic polynomial must lie in the closed unit disk, and every root on the unit circle must be simple. The Dahlquist equivalence theorem states that a consistent linear multistep method converges precisely when it is zero-stable. The parasitic root $5/11$ in Question 1 decays, whereas a repeated root at $1$ would turn small defects into secular growth.

Absolute stability instead fixes $z=h\lambda$ for the Dahlquist test equation. A one-step method has amplification factor $R(z)$ and absolute-stability regionA-stability means that $\mathcal{S}$ contains the entire closed left half-plane, matching every decaying scalar linear problem. L-stability additionally requires $R(z)\to 0$ as $z\to -\infty$, which suppresses unresolved fast transients in a stiff differential equation. Backward Euler is L-stable; the trapezoidal rule is A-stable but approaches $-1$ and can retain stiff oscillations; forward Euler is stable only in the disk $|1+z|\le 1$.

For nonlinear dissipative systems, scalar absolute stability can be insufficient. B-stability controls distances between numerical solutions of contractive differential equations. For Runge–Kutta methods, algebraic stability—nonnegative weights and positive semidefiniteness of $BA+A^{T}B-b{b}^T$—is a useful sufficient condition for B-stability. Question 2 shows how a single negative diagonal entry disproves it.

After spatial discretization of a partial differential equation, stability can be studied through the semidiscrete matrix. If its Hermitian part is nonpositive, the energy method proves contractivity without diagonalizing it. A skew-Hermitian generator instead conserves norm, as in the discrete Schrodinger problem of Question 3. Eigenvalues alone can be misleading for a non-normal matrix: transient growth may be large even when every eigenvalue lies in the left half-plane, so matrix norms, logarithmic norms, resolvent bounds, or a pseudospectrum may be needed.

For constant-coefficient Cauchy problems, Von Neumann stability analysis inserts Fourier modes and bounds their amplification factors. Question 4 gives a typical result: centered diffusion contributes a negative real symbol and centered advection an imaginary symbol. Semidiscrete evolution is stable, yet forward Euler imposes both the parabolic restriction $\Delta t=O(\Delta x^2)$ and an advection-diffusion restriction. This illustrates a Courant–Friedrichs–Lewy condition: a stable spatial approximation need not remain stable under an arbitrary time stepper.

For a well-posed linear initial-value problem and a consistent finite-difference approximation, the Lax equivalence theorem identifies stability with convergence. Its hypotheses matter: it does not by itself cover nonlinear equations, inconsistent boundary closures, changing norms, or non-smooth solutions. Boundaries also defeat a naive whole-line Fourier argument; an energy estimate, normal-mode boundary analysis, or a discrete semigroup bound must include the boundary treatment.

Practical stability analysis therefore starts from the structure of the differential equation. Conservation laws favor unitary or symplectic methods, diffusion favors A- or L-stable implicit methods, monotone transport may require strong-stability-preserving time stepping and upwind fluxes, and stiff splitting requires attention to both the factors and their commutators. Stability does not guarantee accuracy: a heavily damped method may be stable while erasing the solution, and a stable computation with $h\lambda$ near the edge of its stability region may have an unacceptable phase error.

The Dahlquist equivalence theorem says that a consistent linear multistep method is convergent exactly when it is zero-stable. HereThe consistency conditions hold for every $a$. The two quadratic roots have product one. They are distinct and on the unit circle exactly when $0<a<2$; at $a=2$ the root $-1$ is repeated, while outside this interval one root has modulus greater than one. The root condition for a multistep method therefore givesThe polynomial exactness conditions hold through degree two, while the degree-three residual is $6-a$. Every convergent member thus has(The exceptional value $a=6$ has formal order four but is not zero-stable.)

For $s=1$, imposing order two givesThe method is therefore the trapezoidal ruleIts first characteristic polynomial is $\rho (\zeta )=\zeta -1$, so it satisfies the root condition for a multistep method.

For $s=2$, the four conditions through order three giveand henceHereso the root condition for a multistep method again holds. Both methods are consistent and zero-stable, and the Dahlquist equivalence theorem therefore proves that both are convergent.

Every member of the family has order at least three, so it is consistent. By the Dahlquist equivalence theorem, convergence is therefore equivalent to zero-stability, which is characterized by the root condition for a multistep method.

The roots of $\rho$ are $1$ and $a$. They lie in the closed unit disk exactly when $|a|\le 1$. At $a=1$, however, the unit root $1$ is repeated, whereas at $a=-1$ the distinct unit roots $1$ and $-1$ are both simple. Consequently

A linear multistep method for $y^{\prime }=f(t,y)$ has the formIts characteristic polynomials of a linear multistep method are $\rho (\zeta )={\sum }_j{\alpha }_j{\zeta }^j$ and $\sigma (\zeta )={\sum }_j{\beta }_j{\zeta }^j$. Substitution of the exact solution, or equivalently expansion of $\rho (e^z)-z\sigma (e^z)$ at $z=0$, gives the order conditions for a linear multistep method. The method has order $p$ whenwith a nonzero coefficient at the next power. Its one-step defect is then $O(h^{p+1})$, while its global error is $O(h^p)$ when stability prevents accumulation from being amplified.

Order alone does not ensure convergence because the recurrence may contain growing parasitic modes. Zero-stability is the root condition for a multistep method: every root of $\rho$ lies in $|\zeta |\le 1$, and every root on the unit circle is simple. Consistency is the first-order conditionThe Dahlquist equivalence theorem states that a consistent linear multistep method is convergent if and only if it is zero-stable. For example, the leapfrog method $y_{n+2}-y_n=2h{f}_{n+1}$ has $\rho (\zeta )={\zeta }^2-1$; its roots $\pm 1$ are simple, so this second-order method is convergent, although its parasitic $-1$ mode can oscillate.

To study stiff decay, apply the method to $y^{\prime }=\lambda y$ and put $z=h\lambda$. The amplification factors are the roots ofThe linear stability domain consists of the $z$ for which these roots satisfy the strict interior condition appropriate away from the boundary, with simple unit roots where boundary stability is admitted. The method is A-stable when this domain contains the left half-plane, so every mode with $\mathrm{Re}\lambda <0$ remains bounded for every positive step size. The Second Dahlquist barrier says that an A-stable multistep method has order at most two.

The main families illustrate the tradeoffs. The explicit Adams-Bashforth methods interpolate past derivative values and are inexpensive, but their stability domains are bounded. The implicit Adams-Moulton methods include the new derivative value; the one-step second-order member is the trapezoidal rule, which is A-stable. The backward differentiation formulas interpolate past solution values and differentiate the interpolant; the first two are A-stable, while higher orders retain useful sectors of the left half-plane. Implicit methods require a nonlinear solve at each step, commonly initialized by an explicit predictor to form a predictor-corrector pair. Thus order controls consistency error, the root condition controls convergence as $h\to 0$, and the amplification polynomial controls whether a convergent method remains useful on stiff equations.

At $h=0$ the first characteristic polynomial isIts roots satisfy the root condition for a multistep method: the only unit-modulus root is the simple root $1$, and the other root has modulus $1/5$. The method is consistent because part a gives positive order. The Dahlquist equivalence theorem, in the form allowed by the question for this derivative-augmented method, therefore proves convergence.

The Dahlquist equivalence theorem says that a consistent linear multistep method is convergent exactly when it is zero-stable.