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The interval of step sizes on which a decaying solution stays decaying, why a higher order barely lengthens it, and the problems where stability alone sets the cost.
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 find a method's amplification factor and its stability interval, recognise an unstable run from its output, explain why a higher-order method does not fix it, and say what makes a problem stiff and what to do about it.
You have seen Euler's method turn a decaying solution into a growing one, and you know that a fixed-point iteration converges only when its multiplying factor is smaller than one in size. This lesson is that condition applied to a step of an ODE solver.
The test equation is $y' = \lambda y$, and a method's amplification factor is what one step multiplies by on it. The region of absolute stability is the set of $h\lambda$ for which that factor is at most one in size. A method is A-stable when that region contains the whole left half-plane. A problem is stiff when stability rather than accuracy decides the step size.
Apply a method to $y' = \lambda y$ with $\lambda < 0$, whose solutions decay. Every one-step method then reduces to $y_{n+1} = R(h\lambda)y_n$ for some amplification factor $R$, and the computed solution decays exactly when $|R| < 1$. For Euler $R = 1 + h\lambda$, so the condition is $0 < h < \dfrac{2}{|\lambda|}$ — a bounded interval. For the classical Runge-Kutta method $R$ is a quartic and the interval reaches only about $\dfrac{2.78}{|\lambda|}$: order buys accuracy and buys almost no stability. For backward Euler $R = \dfrac{1}{1 - h\lambda}$, which is below one for every positive $h$: stable at any step, at the price of solving an equation for the new value each step. Violating the condition is not inaccuracy. The local truncation error stays small while the computed solution grows exponentially and alternates in sign, so a tolerance on the local error does not detect it. And when one problem carries widely separated rates, the step is fixed by the fastest — even after that component has died away and is contributing nothing. That is stiffness, and it is the standing argument for implicit solvers.
Another way: picture
Plot the true decaying solution and the computed points. Inside the stability interval the points fall towards the axis, perhaps clumsily. Outside it they swing to one side of the axis, then further to the other, then further still — a widening zigzag that has nothing to do with the curve it was meant to follow.
Another way: steps
Instability is routinely mistaken for a bug, or for a tolerance set too loosely. It is neither, and both misdiagnoses cost hours: the arithmetic is correct and the local error really is small. The symptom that identifies it is the alternating sign with growing magnitude, and the test is to halve the step — an unstable run becomes stable abruptly at a definite threshold, while an inaccurate one improves smoothly.
$y' = -10y$ with Euler: the interval is $h < 0.2$.
The factor is $1 - 10h$.
At $h = 0.19$ the factor is $-0.9$ and the run decays, oscillating.
Stable, if ugly.
At $h = 0.21$ it is $-1.1$, and after fifty steps the value has grown by a factor of over a hundred.
A two per cent change in the step.
Rates $1$ and $10^{6}$, over an interval of length $1$.
Accuracy wants $h \approx 0.1$.
Stability forces $h < 2 \times 10^{-6}$: half a million steps.
For a component already dead.
Backward Euler is stable at any step, so ten steps suffice — each solving an equation rather than evaluating one.
Not a close call.
One step multiplies by $1 - 4h$.
Requiring $|1 - 4h| < 1$ gives $0 < 4h < 2$.
So $h < \tfrac12$; and at exactly $\tfrac12$ the factor is $-1$ and the values oscillate for ever between two numbers.
Euler's method is applied to $y' = -6y$, whose solution decays. For which positive step sizes $h$ does the computed solution also decay?
This task has no paper form; do it on a device.
A solver is run on $y' = -7y$ with a step outside the stability interval, for $31$ steps. What happens, and what would fix it?
Euler's method is run on $y' = -4y$ from $y(0) = 1$ with step $1$. Fill in the computed value after each of the first three steps.
| Computed value | |
|---|---|
| After one step | |
| After two steps | |
| After three steps |
Four solvers are applied to $y' = -8y$. Match each to its stability behaviour.
| Stable only for steps below two over the decay rate | Stable on an interval less than half again as long, for four times the work | Stable at every step, and damps fast components strongly | Stable at every step, but barely damps the fastest components | |
|---|---|---|---|---|
| Euler's method | ||||
| The classical Runge-Kutta method | ||||
| Backward Euler | ||||
| The trapezoidal method |
A system has two components, decaying at rates $1$ and $7000$. The slow one is what is wanted, and following it accurately over an interval of length $1$ needs a step of about $\tfrac{1}{10}$. Euler's method is used. How many steps does the interval actually take?
Answer:
Lesson test: one question per skill, one attempt each, no hints. Your answers are checked when you submit.
Euler's method is applied to $y' = -4y$. What is the largest step size below which the computed solution decays? Give a fraction.
Answer:
You can compute a stability interval, diagnose an unstable run and name the methods that are stable at any step. Say in your own words why a component that has already decayed away can still decide how long a run takes.
8. Your turn: the stability interval for Euler's method on $y' = -4y$, step 3