Back to the on-screen lesson ·
What it means for a stability region to contain the whole left half-plane, why no explicit method's does, the problems where that decides the entire cost, and the barrier that caps the accuracy of the cure.
Paper packet. Every task here also exists on screen, where it is checked automatically; answers written on paper are not assessed by Nydus. When you are back at a device, enter your answers there.
By the end of this lesson you will be able to state what A-stability and L-stability mean and which methods have them, compute a stiffness ratio and the step each requirement allows, explain why no explicit method is A-stable, and locate the A-stability threshold in a one-parameter family.
You can derive a method's stability function and its region, and you know that an explicit method's is a polynomial and an implicit method's a quotient. This lesson is about the consequence of that difference, and about the problems where it decides the entire cost of a computation.
A method is A-stable when its stability region contains the whole left half-plane, and L-stable when it is A-stable and $|R(z)| \to 0$ as $z \to -\infty$. A problem is stiff when stability rather than accuracy decides the step size; its stiffness ratio is the fastest decay rate divided by the slowest. Dahlquist's second barrier is the theorem bounding the order of an A-stable linear multistep method.
A method is A-stable when $\{z : \operatorname{Re} z \le 0\}$ lies inside its stability region — so on any decaying problem the computed solution is non-growing at every step size, and the step may be chosen for accuracy alone. No explicit method is A-stable, and the reason is structural rather than a failure of design: an explicit method's $R$ is a polynomial, and a polynomial is unbounded, so no polynomial stays inside the unit disc on a half-plane. This matters because of stiff problems: a system whose components decay at widely separated rates forces an explicit method to use a step set by the fastest component, long after that component has died away and stopped contributing anything. The ratio of the rates is the ratio of the wasted work, and for real problems it reaches $10^{6}$. A-stability is the cure, at the price of solving an equation each step — and at a second price, which is a theorem. Dahlquist's second barrier: an A-stable linear multistep method has order at most two, and among those of order two the trapezoidal rule has the smallest error constant. Unconditional stability and high order cannot be had together in that family, which is why stiff solvers are built from implicit Runge-Kutta methods, or from backward differentiation formulas that settle for a weaker stability property instead.
Another way: steps
Another way: picture
Draw the solution of a stiff system: a fast transient that vanishes in the first fraction of the interval, and a slow curve that is what anyone wanted. Now mark the steps an explicit method takes — a dense forest of them, all the way along the slow, smooth part, at a spacing set by a transient that finished near the left-hand edge and is no longer there to be resolved.
Stiffness is read as a property of a difficult solution. The solution is usually smooth and slow, and often the simplest thing in the problem — the difficulty is in the method, which is being held to a step size by a component that has already vanished. The second misreading is to treat an unstable run as an inaccurate one and reach for a tighter tolerance. A tolerance governs the local truncation error, which is small in an unstable run; the symptom to look for is alternating signs with growing magnitude, and the test is to halve the step, which fixes instability abruptly at a threshold and improves inaccuracy smoothly.
Rates $1$ and $10^{6}$, over an interval of length $1$; accuracy wants $h \approx 0.1$.
Ten steps would do.
Euler needs $h < 2 \times 10^{-6}$: half a million steps.
For a component already dead.
Backward Euler is stable at any step: ten steps, each solving an equation.
Not a close call.
The trapezoidal rule on the fast component with a large step: $R \approx -1$.
Non-growing, as promised.
The component flips sign every step and never leaves, showing up as ringing in the output.
Stable, and visibly wrong.
Backward Euler annihilates it in one step.
L-stability earning its keep.
Its stability function is a polynomial in $z$, of degree at least five.
A polynomial grows without bound, so $|R(z)| \le 1$ fails far out on the negative axis.
So no: and the argument used nothing about the method but the word explicit, which is why no explicit method of any order is A-stable.
A method is called **A-stable**. What exactly does that say?
Match each method to its stability status.
| Not A-stable: a disc of radius one | Not A-stable: a larger bounded region, for four times the work | A-stable and L-stable: fast components vanish in one step | A-stable but not L-stable: fast components persist, flipping sign | |
|---|---|---|---|---|
| Euler's method | ||||
| The classical Runge-Kutta method | ||||
| Backward Euler | ||||
| The trapezoidal rule |
A system has components decaying at rates $1$ and $200$, and only the slow one is wanted. Over an interval of length $1$, accuracy needs a step of about $\tfrac{1}{10}$ and Euler's method needs $h$ below $\dfrac{2}{200}$. Give the steps accuracy asks for, the steps stability forces, and the ratio between them.
| Requirement | Steps | |
|---|---|---|
| Steps accuracy asks for | accuracy | |
| Steps stability forces | stability | |
| The ratio between them | the ratio |
Euler's method is applied to a system whose fastest component decays at rate $700$. For which positive step sizes does that component stay non-growing?
This task has no paper form; do it on a device.
The theta method $y_{n+1} = y_n + h\left((1 - \theta)f_n + \theta f_{n+1}\right)$ has $R(z) = \dfrac{1 + (1 - \theta)z}{1 - \theta z}$. Give the smallest $\theta$ for which the method is A-stable, and its order of accuracy at that value.
A-stable from $\theta = $ w, with order z there.
Lesson test: one question per skill, one attempt each, no hints. Your answers are checked when you submit.
A linear system has eigenvalues with real parts $-7$ and $-800$, and nothing between. What is its stiffness ratio? Give a fraction if it is not whole.
Answer:
You can decide whether a problem is stiff and which methods answer it. Say in your own words why no explicit method can be A-stable, whatever its order.
8. Your turn: is a fifth-order explicit Runge-Kutta method A-stable?, step 3