Back to the on-screen lesson ·
Every one-step method as multiplication by $R(h\lambda)$: polynomials for explicit methods and quotients for implicit ones, the region where the size is at most one, and what the limit far out on the negative axis decides.
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 derive a method's stability function, evaluate it, recognise which methods give polynomials and which rational functions, find the set where it damps by a given factor, and say what its limit at negative infinity means.
You have seen Euler's method turn a decaying solution into a growing one, and you know the condition that keeps it decaying: the multiplying factor must be smaller than one in size. This lesson makes that factor an object with a name and works out what each method's looks like.
The test equation is $y' = \lambda y$. A method's stability function $R(z)$, with $z = h\lambda$, is what one step multiplies by on it. The region of absolute stability is the set of complex $z$ where $|R(z)| \le 1$. A method is L-stable when it is stable on the whole left half-plane and $|R(z)| \to 0$ as $z \to -\infty$.
Apply any one-step method to $y' = \lambda y$ and everything collapses: every stage is a multiple of $y_n$, and the whole step becomes $y_{n+1} = R(h\lambda)y_n$ for a function $R$ that depends on the method alone. The step size and the decay rate never appear separately — only in the combination $z = h\lambda$ — so a single complex variable carries the whole question. For an explicit method $R$ is a polynomial: Euler gives $1 + z$, the classical Runge-Kutta method gives the exponential series truncated after $z^{4}/24$. For an implicit method it is a rational function: backward Euler gives $\dfrac{1}{1 - z}$, the trapezoidal rule $\dfrac{1 + z/2}{1 - z/2}$. Every $R$ agrees with $e^{z}$ to the order of the method, which is consistency restated; what distinguishes the methods is what happens where that agreement fails. The region of absolute stability is $\{z : |R(z)| \le 1\}$, and for a real negative $\lambda$ its intersection with the negative axis is an interval of admissible step sizes. One further reading matters: the limit of $|R|$ far out on the negative axis. Zero means a very fast component is annihilated, as the true solution annihilates it; one means it is merely prevented from growing, and stays in the answer for ever.
Another way: steps
Another way: picture
Draw the complex plane and shade $\{|R(z)| \le 1\}$. For Euler it is a disc of radius one centred at $-1$, touching the origin — bounded, as any polynomial's must be. For backward Euler it is everything outside the disc of radius one centred at $+1$, which swallows the entire left half-plane and a good deal of the right. The pictures explain at a glance why an explicit method has a step limit and an implicit one need not.
The stability region is read as a statement about accuracy. It is not: a step just inside the region can be wildly inaccurate, and a step just outside it is not inaccurate but wrong in kind — the computed solution grows while the true one decays. The two failures look nothing alike and have different cures. The second misreading is to expect a higher order to help. The classical Runge-Kutta method's real stability interval is only about $2.78/|\lambda|$ against Euler's $2/|\lambda|$: four times the work for less than half again the interval, because order and stability are unrelated properties of the same polynomial.
$y_{n+1} = y_n + h\lambda y_n = (1 + h\lambda)y_n$.
One line; nothing to solve.
So $R(z) = 1 + z$, and $|1 + z| \le 1$ is a disc of radius one centred at $-1$.
Bounded, because $R$ is a polynomial.
On the negative real axis that is $-2 \le z \le 0$, giving $h \le 2/|\lambda|$.
The familiar step limit.
$\lambda = -10^{6}$, $h = 0.1$: $z = -10^{5}$.
Far out on the negative axis.
Backward Euler: $|R| \approx 10^{-5}$ — the component is gone.
L-stable.
Trapezoidal: $|R| \approx 1$ — the component flips sign every step and never leaves.
A-stable and not L-stable.
On the test equation: $y_{n+1} = y_n + \dfrac{h\lambda}{2}(y_n + y_{n+1})$.
Collecting: $y_{n+1}\left(1 - \tfrac{z}{2}\right) = y_n\left(1 + \tfrac{z}{2}\right)$.
So $R(z) = \dfrac{1 + z/2}{1 - z/2}$ — the same function as the trapezoidal rule, so the same stability region and the same failure to damp fast components.
The trapezoidal rule is applied to $y' = \lambda y$ with $h\lambda = -1$. By what factor does one step multiply the solution? Give a fraction.
Answer:
At $h\lambda = -4$, give the amplification factor of Euler's method ($R = 1 + z$), Heun's method ($R = 1 + z + z^{2}/2$) and backward Euler ($R = 1/(1 - z)$). Give fractions where they are not whole.
| Method | Amplification factor | |
|---|---|---|
| Euler | $1 + z$ | |
| Heun | $1 + z + z^{2}/2$ | |
| Backward Euler | $1/(1 - z)$ |
Match each method to its stability function $R(z)$.
| $1 + z$ | $1 + z + z^{2}/2 + z^{3}/6 + z^{4}/24$ | $1/(1 - z)$ | $(1 + z/2)/(1 - z/2)$ | |
|---|---|---|---|---|
| Euler's method | ||||
| The classical Runge-Kutta method | ||||
| Backward Euler | ||||
| The trapezoidal rule |
Backward Euler has $R(z) = \dfrac{1}{1 - z}$. For which real $z$ is $|R(z)| \le \dfrac{1}{3}$, so that one step shrinks the solution by a factor of at least $3$?
This task has no paper form; do it on a device.
For the trapezoidal rule, $R(z) = \dfrac{1 + z/2}{1 - z/2}$. What does $|R(z)|$ approach as $z$ runs to $-\infty$ along the real axis?
Answer:
Lesson test: one question per skill, one attempt each, no hints. Your answers are checked when you submit.
Put the five steps of finding a method's stability function and its region into order.
Number the steps in order (write the number in the box):
You can derive and use a stability function and its region. Say in your own words why an explicit method's stability region is always bounded.
8. Your turn: the stability function of the implicit midpoint rule $y_{n+1} = y_n + h f\left(\tfrac{y_n + y_{n+1}}{2}\right)$, step 3