Linear multistep methods and root conditions
Analyze consistency and zero-stability of constant-step history formulas before invoking their convergence theorem.
On this page
The question for today
What you’ll understand
- Form the characteristic polynomials of a linear multistep method.
- Check consistency and the zero-stability root condition.
- Explain the roles of starting values and step ratios in convergence.
A multistep formula reuses several previously computed values and derivatives, reducing the number of fresh evaluations needed for an update. That memory also introduces additional numerical modes, including modes absent from the differential equation. Zero-stability asks whether those history modes amplify small starting and local perturbations as the step tends to zero. A method can approximate derivatives consistently and still fail this test. The root condition reveals this failure before a demanding application is attempted.
A formula and two characteristic polynomials
For a constant positive step h, write a k-step formula with fixed coefficients and . Its polynomials are and . The coefficient at the newest derivative decides whether the method is explicit or implicit. These polynomials organize the history recurrence; they are different objects from a one-step Runge–Kutta stability function.
Substituting a constant exact solution tests that rho vanishes at one. Substituting a linear exact solution tests that rho prime at one equals sigma at one. Together these give first-order consistency for a regular multistep formula. Higher-order moment equations come from further Taylor terms. Consistency compares a single exact history with the formula; it does not yet control what happens when earlier computed values contain small errors. That propagation issue is precisely why a separate stability condition is necessary.
The zero-stability root condition
Apply the method to the zero differential equation. The history then follows the homogeneous recurrence defined by rho. A root outside the unit disk gives exponential growth of a perturbation with step index. A repeated root on the unit circle gives polynomial growth through its associated recurrence solutions. To keep perturbations uniformly controlled while the number of steps increases like one divided by h, all roots must have modulus at most one and every root on the unit circle must be simple.
For fixed coefficients and constant steps, that root condition is equivalent to zero-stability. Consistency together with zero-stability yields convergence for sufficiently regular Lipschitz initial-value problems with convergent starting data. To obtain a particular order p, starting values must be accurate to order h to the p as well as the formula having the appropriate local order. A parasitic root strictly inside the disk decays, but a root near the boundary can still make startup transients conspicuous.
Startup and changing the mesh are part of the method
The two-step Adams–Bashforth formula uses current and previous derivatives, so an initial-value problem must supply an additional starting value by another method or an exact expression. Poor startup can dominate an otherwise accurate formula. Startup accuracy is assessed at the starting mesh points, not only by the formal order label of the startup routine. For example, one Euler step from exact data has order-h-squared starting error and can be sufficient to start a second-order two-step method.
Variable-step multistep methods change coefficients according to recent step ratios. A constant-step root calculation does not automatically establish stability for arbitrary sequences of ratios. Large step jumps can amplify the history representation, and practical codes restrict or manage them while adjusting order and error estimates. It is therefore important to distinguish a proof about a fixed formula from a claim about a complete adaptive implementation. Both consistency and stability need to match the actual history update being used.
Work it through
An accurate-history Adams–Bashforth step
Use h=0.1 on y′=y with Y₀=1 and the exact starting value Y₁=exp(0.1). Compute the two-step Adams–Bashforth approximation at time 0.2.
The formula gives Y₂=Y₁+0.1(1.5Y₁−0.5Y₀)=1.15exp(0.1)−0.05.
With exp(0.1)≈1.10517018599, this is approximately 1.22094655579. The exact solution at 0.2 is exp(0.2)≈1.22140275816.
The exact-minus-numerical error is about 0.00045620237. The characteristic roots zero and one satisfy zero-stability; the comparison uses exact starting history so this step isolates the formula's defect.
The step is accurate to the expected local scale, but a convergence claim over a fixed interval additionally uses zero-stability and startup accuracy.
The living notebook
Explore the model
Interactive lab · optimization
A perturbation mode with a root outside the disk
Pause and Reset, then set η=2.2 and x₀=1. Predict the sign and magnitude after four Steps. For a fresh comparison, Reset and set η=1.5 and x₀=1, then Step four times. Interpret the recurrence factor as one possible error-propagation root.
- Position x0
- 2
- Objective x²/2
- 2
- Iteration behavior
- Stable step
Gradient descent on f(x) = x²/2: xₙ = (1 − η)ⁿx₀. Playback advances two discrete iterations per second. Values outside |x| ≤ 4 are explicitly reported. Parameter changes pause playback and recompute the model at the current elapsed time. Reset restores both parameters and returns time to zero. Playback starts paused. Adjust either parameter, then choose Play or Step.
Look forThe factors −1.2 and −0.5 yield magnitudes 2.0736 and 0.0625 after four updates. This scalar recurrence illustrates growing and decaying modes; it does not implement a multistep history or test the full root condition.
Pause & reason
Check your understanding
Your turn
Try it yourself
For BDF2, 3Yₙ₊₂−4Yₙ₊₁+Yₙ=2h fₙ₊₂, find ρ and σ, verify the two consistency identities, and test zero-stability.
Take a moment to make a prediction or work it out on paper. Open a hint when you need one.
Give me a hint
Factor the quadratic characteristic polynomial before evaluating its roots.
Compare with the solution
Here ρ(ζ)=3ζ²−4ζ+1=(ζ−1)(3ζ−1) and σ(ζ)=2ζ². Thus ρ(1)=0 and ρ′(1)=6−4=2=σ(1). The roots are one and one third, with the unit-circle root simple, so the constant-step root condition holds. This establishes consistency and zero-stability, not by itself a proof of absolute stability for every complex hλ.
Keep the idea
Take it with you
- Consistency and zero-stability answer different questions about a history formula.
- Repeated unit-circle roots violate the root condition even when no root is outside the disk.
- Starting accuracy and variable-step coefficient changes must be included in an implementation's analysis.
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