A smooth right-hand side does not guarantee a solution for all t. The equation y′ = y² has the solution 1/(1 − t) through (0, 1), which grows without bound and ceases to exist at t = 1. Euler's method, which only ever looks one step ahead, steps straight past that time and reports finite numbers, and the exact fractions it produces double in length at every step until the lab stops at a stated budget.
Core ideas
Three ideas: the interval where a solution lives, what a step method cannot see, and what exact arithmetic reveals.
A solution is a function on some interval of t. For y′ = y² with y(0) = 1 it is 1/(1 − t) on t < 1: the values rise without limit as t approaches 1 and there is no value at 1. The right side y² is a polynomial, as smooth as an expression gets, and that did not help.
Euler's recipe needs only a slope at a point, and y² is defined at every y. So the method steps through t = 1 and beyond, and every yₙ it prints is a finite fraction. A number from the method past t = 1 is not a value of any solution through the start, since that solution does not exist there.
Squaring a fraction doubles the digits of its numerator and denominator, and Euler's step on y² squares the current value. The lengths of the denominators, 1, 2, 5, 10, 19, 38, 77, …, roughly double at each step. The lab stops when a number passes 2000 digits and says so, rather than rounding to continue.
The material
The solution and its check, three steps of Euler against it, the step past the end, the limit on the arithmetic, and what is known in general.
Take y′ = y² with y(0) = 1. The candidate y = 1/(1 − t) has y′ = 1/(1 − t)², and y² is 1/(1 − t)² as well, so the residual is 0: it is the check of “Checking a Proposed Solution”, which the lab does exactly. The value at t = 0 is 1, so it passes through the start. At t = 3/4 it is 1/(1/4) = 4; at t = 9/10 it is 10; as t approaches 1 it grows without limit.
Starting higher ends the solution sooner. For y(0) = y₀ the solution is y₀/(1 − y₀·t), whose derivative is y₀²/(1 − y₀·t)², which is its square, so the residual is 0; it ends at t = 1/y₀. From y₀ = 2 it ends at t = 1/2. The time of the end depends on the start, and nothing in the equation announces it.
If f(t, y) and its derivative with respect to y are continuous near the start (t₀, y₀), then there is exactly one solution through the start on some interval around t₀. The theorem does not say how long the interval is.
The example above shows why the last sentence is not a technicality: here f = y² is as well behaved as a function can be, and the interval is t < 1. The lab demonstrates this one case and does not prove the theorem; the proof belongs to a course in analysis.
Three steps give y₃ = 37905/16384 ≈ 2.31354, rounded for reading; the true value is 4, and the error the lab prints is 27631/16384. Euler falls well behind, as it must on a rate that rises this fast, and the polygon still looks tame.
Run five steps and t₅ = 5/4, which is past t = 1. The lab prints y₅ = 32213971596000663105/4611686018427387904 ≈ 6.98529, a perfectly good fraction, and the tile for the known solution reads undefined at t = 5/4. The formula 1/(1 − t) would give −4 at 5/4, and that number is not the solution either: the solution through (0, 1) has already ended, and the negative branch is a different curve. The method never saw the end, because it never looks further than one step.
Write each value over a power of two. The denominators are 4, 64, 16384, then 2³⁰ and on: each is the square of the last times 4, so the exponent goes 2, 6, 14, 30, 62, …. Their lengths are 1, 2, 5, 10, 19, 38, 77, … digits, doubling. With h = 1/4 the lab completes eleven steps, whose largest denominator has 1233 digits (the last value is shown rounded, ≈ 6.77729e23, over a numerator of 1257 digits), and stops before the twelfth, which would pass its limit of 2000 digits. The tile reads stopped after step 11: digit budget. Nothing was rounded to continue.
That is a fact about the method, and a floating-point program hides it by rounding at every step and carrying on. The numbers it would print past t = 1 are no more the solution than the exact ones are.
Interactive
Exact steps, and the exact error
Method
From an equation to the largest interval you can claim.
Substitute it into the equation and read the residual, as in “Checking a Proposed Solution”. Without a check there is no solution to speak of.
Look for a zero of a denominator or another break. 1/(1 − t) fails at t = 1. Note which side of it the start lies on.
The solution through the start lives on the interval that contains the start and stops at the break: t < 1 for the start (0, 1).
Step Euler past the break and see that it prints finite numbers anyway. Say which of those numbers lie past the end of the solution.
If the lab stops, note the step. It is the budget of the arithmetic, not the end of the solution.
Take three Euler steps of size 1/4 and compare the last with 1/(1 − t).
The arithmetic is exact at every line, and Euler is still more than 1.6 short of the true value at t = 3/4. The solution is climbing towards infinity at t = 1, and every step reads a slope smaller than the slope the step is about to meet.
For a rehearsal, take the start y(0) = 2: the solution is 2/(1 − 2t), it ends at t = 1/2, and the first Euler step with h = 1/4 is 2 + (1/4)·4 = 3.
You should be able to check that 1/(1 − t) solves y′ = y², say where it ceases to exist and on what interval it is the solution through (0, 1), explain why Euler's method gives finite values past that time, and explain why the exact fractions double in length at each step.
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