Look at the two arrows touching an equilibrium. If both point towards it, it is stable; if both point away, it is unstable; if one points towards and one away, it is semistable. The three cases are all there is, they are decided by the sign of f on each side, and semistable is a distinct case with a distinct reason — a root of f where f touches zero and turns back.
Core ideas
Three ideas. A definition by arrows, a rule that turns signs into a verdict, and the case people try to explain away.
At an equilibrium y*, look at the arrow just below and the arrow just above. Both towards y*: stable, since a solution that starts near enough ends up there. Both away: unstable. One towards and one away: semistable. Nothing else can happen, as each neighbouring arrow has only two directions.
Positive then negative as y increases through y* gives arrows up and then down, both towards y*: stable. Negative then positive: unstable. The same sign on both sides: semistable. The value of f at y* is zero and says nothing, so the signs are read next to the equilibrium, never at it.
A double root of f is a place where the graph of f touches the axis and turns back without crossing. Neither side's sign changes, so the arrows agree in direction, and that is not a mixture of the other two cases but the single situation in which a solution below is drawn in and one above is carried off.
The material
The definitions, the sign rule checked on three equations, and why the semistable case matters later.
An equilibrium y* of y′ = f(y) is stable if the arrows on both sides point towards it, unstable if both point away, and semistable if one points towards it and the other away.
The classification therefore needs only the phase line from the previous lesson. For each dot, compare the sign of f just below with the sign just above. The lab prints the verdicts for every equilibrium in increasing order, and prints the same sign data its status line used.
An unstable equilibrium is one that a real system would not stay at. The constant solution is exactly right and a solution starting a hair away leaves. A stable one is where solutions from a whole interval end up, which is why it is the equilibrium that a measurement is likely to find.
Take f(y) = y²·(1 − y). The factor y² is positive on both sides of 0, and the factor 1 − y is positive there, so f is positive on both sides of 0. Below 0 the arrow points up, towards 0. Just above it the arrow points up, away from 0. The equilibrium at 1 is the ordinary kind: f goes from positive to negative, and it is stable.
The lab prints f′(0) = 0 and f′(1) = −1 for this equation, in the slope tile. The zero at y = 0 tempts a reader into using it as the verdict, and the next lesson shows that it is not one: the flat slope occurs at stable and unstable equilibria as well. The signs of f decide, and they are always available.
One of these was met before the word was defined. In “Harvesting and the Threshold” the harvest H = 1 left the single equilibrium 2, with the right side −(y − 2)²/4, which is negative on both sides of 2. A population above 2 falls to it and one below falls away from it, and that lesson called the equilibrium semistable without yet having the test. The same sign on both sides, negative this time, gives the same verdict: the arrows agree in direction, down and down.
For a polynomial f, an equilibrium that is a root of even multiplicity is semistable, and one of odd multiplicity is stable or unstable according to whether f passes from positive to negative or from negative to positive.
This is stated, not proved here; it is the factor-by-factor sign argument that the cubic and the semistable example used above. A factor (y − y*) with an even power does not change sign at y*, and with an odd power it does. The multiplicity is a shortcut for the sign test. The sign test needs no theorem and is the one to trust.
Start the equation y′ = y²·(1 − y) at y(0) = −1. It rises, because f(−1) = 2 > 0, and it approaches 0 from below without reaching it. Start it at y(0) = 1/2 and it rises to 1, leaving 0 behind for ever. Only the first of these is drawn in by 0, and only from below. That asymmetry is the whole content of the word.
Interactive
Equilibria, the phase line, and where solutions go
Method
Four moves, done for one equilibrium at a time.
Use the gap's test value, or the sign of the factored form. Write it down as a plus or a minus sign.
The same, in the next gap up. Do not skip this because the first sign looked decisive: a single side cannot separate stable from semistable.
+ then −: stable. − then +: unstable. Equal signs: semistable. That is the whole table.
Pick a start in each neighbouring gap and follow the arrow. They should approach or leave the dot in the way the verdict says.
The equilibria are 0 and 1. Evaluate f(y) = y²·(1 − y) at the three test values the lab uses, −1, 1/2 and 2.
Solutions below 0 rise to 0; solutions just above 0 rise away from it and head for 1. The slope tile prints f′(0) = 0 here, and the sign test, not the derivative, is what gave the verdict. The lab prints 0 semistable; 1 stable for this preset.
For a rehearsal, factor y²·(y − 2) (arrows down, down, up) and say what kind of equilibrium sits at 0 and at 2.
You should be able to read the sign of f on either side of each root, say which of three cases results, write down an f with a semistable root, and predict from the arrows where a start just below and just above it will go.
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