Between two consecutive equilibria the function f cannot change sign, so a single test value decides the direction of every solution that lives in that gap. Draw the equilibria on a number line, put an arrow in each gap, and the result is the phase line: the whole qualitative story of y′ = f(y) without solving it. The arrows carry direction only, never speed.
Core ideas
Three ideas. The first is why one test value is enough; the second turns a sign into an arrow; the third says what the arrow leaves out.
A polynomial changes sign only by passing through zero, and the equilibria are its zeros. On a gap between two of them, f is therefore positive throughout or negative throughout, and the value at a single point in the gap says which. The same holds on the two unbounded stretches at the ends, which are tested at one point beyond the outermost equilibria.
Where f(y) > 0, the slope y′ = f(y) is positive and a solution there is increasing, so its arrow points towards larger y. Where f(y) < 0 it points towards smaller y. A solution cannot leave its gap, since leaving would mean passing through an equilibrium.
The speed at height y is |f(y)|, and it changes continuously from point to point; the arrow records only the sign. Two arrows of equal size can stand for a crawl and a rush, and a long arrow on a drawing means nothing at all.
The material
The construction, done on the cubic, and then the two places a reader goes wrong with it.
The phase line of y′ = f(y) is the y axis with each equilibrium marked and an arrow drawn in every interval between and beyond them, pointing towards larger y where f > 0 and towards smaller y where f < 0.
Take f(y) = y³ − 4y² + 3y = y·(y − 1)·(y − 3), whose equilibria 0, 1 and 3 cut the axis into four pieces. Pick one convenient number in each piece and evaluate. The lab uses an integer beyond each end and the midpoint of every gap, and evaluates exactly.
Read upwards along the axis. Below 0 the arrow points down, so a solution there falls away from 0. On (0, 1) it points up, towards 1. On (1, 3) it points down, again towards 1. Above 3 it points up and nothing stops it. Without finding a single formula the picture says that y = 1 draws in solutions from both sides and that 0 and 3 push them away.
At y = 1/2 the speed is |f(1/2)| = 5/8; at y = 2 it is |f(2)| = 2. A solution passing through 2 is falling 16/5 times as fast as one passing through 1/2 is rising, and the two gaps carry arrows of the same size. The reasons to draw a phase line are the direction and the order of the equilibria. Speed has to be computed from f.
One consequence is worth keeping. Close to an equilibrium f is close to zero, so the speed there is small. A solution approaching 1 therefore slows as it closes in, and the phase line's arrows, which look as if they run all the way to 1, are describing a limit that is never reached in finite time.
With a single equilibrium there are only two arrows. For y′ = 2 − y the equilibrium is 2, the test values are 1 and 3, and f(1) = 1 > 0, f(3) = −1 < 0. Both arrows point at 2.
A drawing of the cubic's phase line has an arrow pointing up on (0, 1) and another pointing up on (1, 3).
The second is wrong, and the test value shows it: f(2) = −2 < 0. The check is also internal. A solution rising on (1, 3) would have to pass through 3, which is an equilibrium and so cannot be crossed.
Interactive
Equilibria, the phase line, and where solutions go
Method
Five moves, in this order. The sign table is the one that the arrows are copied from.
Solve f(y) = 0 exactly and list the roots from smallest to largest. They are the dots on the line.
One below the smallest root, one between each consecutive pair, one above the largest. Any number inside the gap serves; pick the easiest to compute.
Use exact fractions and evaluate the factored form if there is one. Only the sign matters; a value that rounds to zero is not zero.
Positive: towards larger y. Negative: towards smaller y. Draw it in the gap, not at the test value.
Two neighbouring arrows may point at the same dot, away from it, or the same way past it. None may point through a dot into the next gap on its own.
The equilibria are 0, 1 and 3. Evaluate f(y) = y³ − 4y² + 3y at −1, 1/2, 2 and 4, one in each of the four pieces.
Reading it back: a solution starting below 0 falls without limit; one starting in (0, 1) rises to 1; one starting in (1, 3) falls to 1; one starting above 3 rises without limit. The lab agrees on the middle two: from y(0) = 2 the cubic preset reports the limit → 1.
For a rehearsal, draw the line for y′ = (y + 1)·(y − 2). The test values −2, 0 and 3 give 4, −2 and 4, so the arrows are up, down, up.
You should be able to list the equilibria in order, evaluate f exactly in every gap, turn each sign into an arrow, say which way a solution starting anywhere moves, and explain why the arrow's length means nothing.
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