A solution of y′ = f(y) cannot cross an equilibrium, because the equilibrium is itself a solution, and inside a gap f has one sign, so it moves in one direction only. A solution that moves one way inside a gap has to settle at the equilibrium at the end its arrow points to, or grow without bound if that end has no equilibrium. The limit from any start can be stated from the phase line, and the exact Euler steps are a check on that claim and not a proof of it.
Core ideas
Three ideas, each resting on the one before. The first keeps a solution in its gap, the second gives it a direction, and the third says where that direction ends.
Every equilibrium is a constant solution, and through each point there passes only one solution. A solution that began between two equilibria would have to touch one of them to leave, which would put two solutions through the same point. So it never does: its height stays strictly between the two, for as long as it exists.
Between consecutive equilibria f has one sign, so y′ = f(y) has one sign along the whole solution. The solution is always rising or always falling, and it never turns round. The arrow the phase line draws in that gap is the direction for every solution that lives there.
A solution that keeps rising inside a gap either runs into its upper equilibrium, never reaching it, or has no upper equilibrium and rises past every height. The first is a limit that is a number, the second is +∞. A limit short of the equilibrium is not possible, since there the slope would still be bounded away from zero and the solution would carry on.
The material
The statement, one equation read from every start, and two places where the exact steps and the line can seem to disagree.
A limit is a statement about what a solution does as t grows. Write y → 1 when y(t) gets and stays as close to 1 as you like, and y → +∞ when it rises past every height. Neither says the solution ever arrives. The phase line of the last two lessons gives both, for every start at once.
Let y solve y′ = f(y) with f a polynomial. If y(0) lies in a gap between two equilibria, or beyond the outermost one, then y stays in that gap and is monotone, and as t grows, for as long as the solution exists, it approaches the equilibrium at the end its arrow points to, or goes to +∞ or −∞ if that end has no equilibrium. If y(0) is an equilibrium, y stays there.
The lab demonstrates this on every equation it is given, by printing the limit the phase line predicts beside exact Euler steps that move the same way. It does not prove it for every polynomial. The argument has the three concepts above as its pieces; the one that is stated and not proved is the last, that a solution which rises inside a gap cannot stop short of the equilibrium above it.
Take y′ = (y − 1)·(y − 3), which the lab's box holds expanded as y² − 4y + 3. Its equilibria are 1 and 3. The slopes are f′(y) = 2y − 4, so f′(1) = −2 and f′(3) = 2: 1 is stable and 3 is unstable, by the slope test of the last lesson. The signs of f at the test values 0, 2 and 4 are +, − and +, giving arrows up, down, up.
Starts 0 and 2 are on opposite sides of 1 and end at the same place, because 1 is stable and both arrows point at it. The start at 4 is above the unstable equilibrium, nothing lies above it, and the solution rises without limit. The exact Euler steps from 4 with h = 1/4 are 4 + (1/4)·3 = 19/4, then 409/64, and the lab's table climbs from there. For this quadratic the climb is faster than any exponential and ends at a finite time, as “Blow-Up and the Interval of Existence” found for y′ = y²; y → +∞ is the correct statement of where it is going.
Exact Euler steps are exact arithmetic applied to an approximate method, and the phase line is what says why they move as they do. With h = 1/4 the steps from 2 run 2, 7/4, 97/64, …, falling and staying above 1, as the arrow says. A step that is too large can disagree. On the same equation from 5/2 with h = 1 the steps are 5/2, 7/4, 13/16 and 313/256: the third is below 1 and the fourth is above it again, so the column rocks about the equilibrium. No solution does that. The polygon stepped over 1, which a solution cannot do, and the error is the step size and not the equation. One limit of the lab belongs here too: a quadratic right-hand side doubles the digits of every fraction at each step, so the lab stops after 11 exact steps and says so, and a cubic stops after 7, where it could only go on by rounding.
For y′ = (y − 1)·(y − 3) start at y(0) = 3. The lab's limit tile reads at equilibrium, and every exact Euler step is 3, since f(3) = 0 adds nothing.
A start at 3 and a start just below it are not close in the long run. The first stays at 3 for ever; the second is in the gap (1, 3), where the arrow points down, and goes to 1. A start just above 3 goes to +∞. Unstable means that nearby starts leave, and it does not mean that they all leave the same way.
Interactive
Equilibria, the phase line, and where solutions go
Method
Five moves. The first four need no arithmetic beyond one sign per gap.
Solve f(y) = 0 exactly and list the roots from smallest to largest. They are the only heights a solution can approach.
Say which gap y(0) is in, or that it is an equilibrium. A start that is exactly an equilibrium stays there and has no more to say.
Take the sign of f at a test value in the gap, as in the phase line. Up means larger y, down means smaller.
Follow the arrow to the end of the gap. If there is an equilibrium there, that is the limit, written → y*. If the gap runs on without end, the limit is → +∞ or → −∞.
Take a few exact Euler steps with a small h. They should move the way the arrow says and stay inside the gap. A column that disagrees points to a step too large or a slip in a sign, and the phase line is what to trust.
The equilibria are 1 and 3, and f′(y) = 2y − 4 gives slopes −2 and 2. Evaluate f at the three starts, which are also test values, and read each limit.
Two of the three starts end at the same equilibrium and the third does not end anywhere. The lab agrees: the first start 0 reads → 1, and choosing the climb preset puts 4 first so that the tile reads → +∞. The first Euler step is 4 + (1/4)·3 = 19/4 and the second is 19/4 + (1/4)·(105/16) = 409/64, the numbers the table prints.
For a rehearsal, take y′ = y·(4 − y) and starts −1, 2 and 5. The values f(−1) = −5, f(2) = 4 and f(5) = −5 give → −∞, → 4 and → 4.
You should be able to find and order the equilibria, place a start in its gap, read the arrow, give the limit as an equilibrium or as +∞ or −∞, say what happens to a start exactly on an equilibrium, and explain a rocking column of Euler steps as a step that was too large.
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