Time stepping, stiffness, and stability regions
Separate an equation's dynamics from its discrete update and balance accuracy with stability for fast and slow modes.
On this page
The question for today
What you’ll understand
- Derive explicit and implicit Euler amplification factors.
- Distinguish local truncation error, global error, and absolute stability.
- Explain why stiffness can force an accuracy-adequate explicit step to become much smaller.
A stable differential equation can produce an exploding numerical trajectory. The source may be the update rule rather than the model. Time integration replaces continuous evolution with a sequence of finite steps, and every method has its own response to fast modes, oscillations, and roundoff. This lesson uses the scalar decay equation as a microscope for explicit and implicit methods. The result is not a recipe to use the smallest possible step, but a way to ask whether accuracy, stability, and computational cost are being controlled for the actual problem.
An update is a new dynamical system
Forward Euler evaluates the derivative at the current state and follows its tangent for one step. If the exact solution is smooth, the one-step defect in the state is of order h squared; under appropriate Lipschitz and finite-time assumptions, accumulated global error is first order. Some texts divide the local defect by h when defining local truncation error, changing the stated order by one. Always check the convention instead of comparing order labels without their definitions.
For the linear test equation y′=λy, Euler multiplies the current state by one plus hλ. Absolute stability asks when repeated updates do not amplify a decaying mode without bound. Strict damping requires that multiplier to have magnitude below one. This is a statement about fixed step size and repeated steps, distinct from convergence as h approaches zero at a fixed final time. Even inside the stable region, the method may approximate decay rates or phase very poorly.
Implicitness changes stability and computational work
Backward Euler evaluates the derivative at the unknown next state. For the linear test equation, rearrangement gives an amplification factor equal to one divided by one minus hλ. Every finite mode with negative real λ is strictly damped for positive h. This method is A-stable and also strongly damps modes far into the left half-plane. Such damping is useful for stiff decay but can distort oscillatory or conservative dynamics, where preserving phase or energy structure may be important.
For a nonlinear system, an implicit step requires solving nonlinear equations, often with Newton iteration and an inner linear solver. Their tolerances contribute to the time integration error and should be coordinated with the requested accuracy. A-stability does not make a nonlinear solve infallible, remove discretization error, or permit arbitrary large steps for a faithful trajectory. Other methods trade off order, stability regions, conservation, and cost; the best choice depends on what the solution must preserve.
Stiffness is a mismatch of relevant scales
A system is often called stiff when stability forces an explicit method to resolve fast decaying scales much more finely than needed to follow the slowly changing quantity of interest. The fast mode may already be nearly absent physically, yet an unstable update can resurrect numerical perturbations in that direction. Refining a spatial mesh for a diffusion problem creates increasingly fast modes, tightening the explicit step restriction. The source is the combined spatial and temporal discretization, not merely a badly chosen initial condition.
Adaptive methods estimate local error and adjust step size, but a trustworthy calculation also monitors rejected steps, solver failures, and physical diagnostics. Compare solutions at matched times under refinement, not at equal step indices. For conservation laws or Hamiltonian motion, a method's long-time structural behavior can matter more than a small short-time error estimate. State the norm, time interval, and observable used to judge convergence. A smooth animated curve is a visualization of output, not proof that either the equation or its numerical approximation is stable.
Work it through
One time step, three different answers
Integrate y′=−8y, y(0)=1, to t=0.3 using one step of forward Euler and one of backward Euler. Compare with exact evolution.
The exact value is exp(−8×0.3)=exp(−2.4)≈0.0907179533. It remains positive and decays rapidly.
Forward Euler has multiplier 1−8(0.3)=−1.4, so it returns −1.4. Repeated steps alternate sign and grow in magnitude because h=0.3 violates the strict restriction h<2/8=0.25.
Backward Euler returns 1/[1+8(0.3)]=1/3.4=5/17≈0.2941176471. It damps the mode and remains positive, but its absolute one-step error is about 0.203399694.
With forward Euler and h=0.1, three steps reach the same final time and give (0.2)³=0.008. This is stable yet still has absolute error about 0.0827179533. Stability removes explosive amplification; it does not certify accuracy at the chosen resolution.
The exact equation decays, the large explicit step explodes, and the implicit step is stable but inaccurate. These are three distinct assessments.
The living notebook
Explore the model
Interactive lab · optimization
The explicit decay multiplier in a visible model
The lab uses gradient descent on x²/2, so xₙ=(1−η)ⁿx₀. Identify η with αh for the scalar equation x′=−αx. Set η=2.4 and observe growing alternating iterates; compare η=0.8. This is precisely the explicit scalar recurrence, not a general ODE integrator, implicit solver, or stiffness simulator.
- 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 forA numerical update has its own stability threshold even when the underlying continuous decay is stable.
Pause & reason
Check your understanding
Your turn
Try it yourself
For y′=−20y, find the strict forward-Euler damping range. At h=0.04, compute two steps from y(0)=1 and compare with the exact value at t=0.08.
Take a moment to make a prediction or work it out on paper. Open a hint when you need one.
Give me a hint
Compute the update multiplier before applying it twice.
Compare with the solution
Strict damping requires 0<h<0.1. At h=0.04, the multiplier is 1−0.8=0.2, so two steps give 0.04. Exact evolution gives exp(−1.6)≈0.201896518. The absolute error is about 0.161896518 despite stability, so a substantially smaller step is needed for accurate decay.
Keep the idea
Take it with you
- Consistency order and absolute stability answer different limiting questions.
- Implicit stability comes with algebraic solves and does not eliminate accuracy requirements.
- Stiffness makes fast modes constrain explicit steps even when slow behavior is the main target.
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