Choosing the step size · Differential Equations

Intuition

A step can be too long in two ways. It can be too long for accuracy, and halving it and comparing the two answers shows by how much. And it can be too long for the method to behave at all: on a solution that dies away fast, Euler’s method with a long step overshoots zero, overshoots back further, and the numbers explode while the true solution quietly vanishes. Too short a step has a cost as well: more work, and more rounding in the arithmetic. Good software therefore changes the step as it goes, short where the solution changes quickly and long where it is smooth.

Driving a winding road at night: slow on the bends, faster on the straights. One speed for the whole road is either too slow on the straights or dangerous on the bends.

Euler’s method on $y'=-10y$ , $y(0)=1$ , whose solution $e^{-10x}$ dies away at once. With $h=0.05$ each step halves the value, and that broken line follows the solution down. With $h=0.25$ each step multiplies by $1-2.5=-1.5$ : the values $1,\ -1.5,\ 2.25,\ -3.375$ alternate and grow, although every step is computed exactly as the method says.

Accuracy, a limit on the step, and rounding

Three things bound the step. Accuracy: for a method of order $p$ the error is about $Ch^{p}$ , and running with $h$ and with $\tfrac h2$ estimates it — for Euler’s method the difference of the two answers is about the error of the finer one. A limit on the step: on , Euler’s method dies away with the solution only when , however little accuracy is wanted. And rounding: every step adds a tiny rounding error, and more steps add more of them, so shrinking further eventually makes the answer worse. Adaptive methods estimate the error of every step and lengthen or shorten the next one to keep it near a tolerance.

y_{h}\approx Y+Ch^{p}\ \Longrightarrow\ Y-y_{h/2}\approx\frac{y_{h/2}-y_{h}}{2^{p}-1}

Rules of thumb

Halve and compare: if the answers with $h$ and $\tfrac h2$ agree to the digits wanted, keep them; if not, halve again.
For Euler’s method, $p=1$ : the finer answer is off by about the difference of the two answers.

When Euler’s steps die away with the solution

\begin{aligned}&y'=-ky,\ k>0,\ y_{0}\ne0:\\&y_{n}\to0\ \Longleftrightarrow\ 0<h<\frac{2}{k}\end{aligned}

The solution y₀e^{−kx} dies away for every k > 0. Euler’s method multiplies by 1 − hk at every step, so y_n = (1 − hk)ⁿ y₀, which tends to zero exactly when 1 − hk lies strictly between −1 and 1, that is when 0 < hk < 2. With a longer step the factor is below −1: the values alternate in sign and grow without limit, although every step is computed correctly. The limit on h comes from k, not from the accuracy wanted.

Proof steps

y_{n+1}=y_{n}-hk\,y_{n}=(1-hk)\,y_{n}

One Euler step on $y'=-ky$ .

y_{n}=(1-hk)^{n}\,y_{0}

By induction on $n$ .

(1-hk)^{n}y_{0}\to0\ \Longleftrightarrow\ |1-hk|<1

A geometric sequence with $y_{0}\ne0$ tends to zero exactly when its ratio is less than $1$ in size.

|1-hk|<1\ \Longleftrightarrow\ 0<hk<2

Unpack the absolute value: $-1<1-hk<1$ .

y_{n}\to0\ \Longleftrightarrow\ 0<h<\frac{2}{k}

Divide by $k>0$ : the step must be shorter than $\tfrac2k$ .

Applications

Practice

A Limit on the Step

On y′ = −ky, Euler’s method multiplies by 1 − hk at every step. The values die away with the solution only when that factor lies strictly between −1 and 1.

|1-hk|<1\ \Longleftrightarrow\ 0<h<\tfrac2k

Try it

Euler’s method is run on $y'=-50y$ . Below what step size do its values tend to zero?