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Insert an exact solution into the linear multistep method and use $f(t,y(t))=y^{\prime }(t)$. Its local residual isExpanding both terms by Taylor theorem givesConsequently $C_0=\cdots =C_p=0$ is sufficient for order at least $p$. It is also necessary: test the residual on the polynomials $y(t)=1,t,\dots ,t^p$, evaluating at $t=0$. For $y(t)=t^k$, it equals $h^k{C}_k$ (and equals $C_0$ for the constant), which can be $O(h^{p+1})$ only if $C_k=0$. These functions can be realized as exact solutions of suitable smooth ordinary differential equations, so this is a genuine necessity test. Thus the order conditions areThis is a statement about local order; zero-stability is the separate condition that turns it into convergent approximation.

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.

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.

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.

Insert the exact solution into the linear multistep method and expand about $t_n$. Its defect isThe coefficient of $h^k{y}^{(k)}(t_n)$ isfor $k\ge 1$, while the constant coefficient is ${\sum }_l{\rho }_l$. The method has order at least $p$ exactly when these coefficients vanish for $0\le k\le p$.

On the other hand,The coefficient of $z^k$ in this expression is precisely the preceding order-condition coefficient. Therefore all coefficients through degree $p$ vanish exactly whenwhich proves the equivalence.

A one-step or linear multistep method has order $p$ if, when the exact sufficiently smooth solution is substituted for the numerical values, its unscaled local truncation error is $O(h^{p+1})$, and this estimate is not generally $O(h^{p+2})$. For a stable method this corresponds to a global error of order $O(h^p)$ on a fixed time interval.

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.

Suppose a consistent linear multistep method has a simple principal polynomial root and its analytic logarithm satisfies $\log w(z)=z+C{z}^{2r}+O(z^{2r+1})$ near zero, with real coefficients. If $C(-1{)}^r>0$, then at $z=i\epsilon$ the real part of $\log w$ is positive to leading order. The same remains true after giving $z$ a small negative real part of smaller magnitude than that leading term. Thus a growing amplification factor occurs inside the left half-plane, ruling out A-stability. For an order-three method with $C>0$, choose $z=-C{\epsilon }^4/2+i\epsilon$.

An irreducible A-stable linear multistep method has order at most two.

This is an order restriction for an A-stable linear multistep method.

The displayed implicit linear multistep method is second order and A-stable. It is the $\alpha =4/3$ member of the trapezoidal-BDF two-step family, whose order and boundary-locus proof establish both properties. Its derivative formula uses three equally spaced values; changing the steps requires different coefficients.

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.

This real-parameter linear multistep method has formal order two except at the degenerate value $\alpha =2$, where its local symbol vanishes to fourth degree. The zero-stability roots are $1$ and $\alpha -1$, so it is convergent exactly for $0\le \alpha <2$. Every convergent member is A-stable: the boundary-locus test for multistep A-stability gives nonnegative real part $\alpha (2-\alpha )(1-\cos \theta {)}^2/(2|\sigma (e^{i\theta }){|}^2)$, with no unit-circle denominator zero for $0<\alpha <2$. At $\alpha =0$ the two interlaced sequences obey a trapezoidal step of length $2h$. At $\alpha =4/3$ it is the BDF2 method. The endpoint two fails the root condition for a multistep method even though canceling a common polynomial factor resembles the trapezoidal rule.