Shift the equilibrium to zero with u = y − y* and the equation becomes u′ = f′(y*)·u plus terms in u² and higher, which are negligible for small u. A negative slope then means the gap closes, a positive one means it opens, and the size of the slope says at what exponential rate. A zero slope settles nothing, and the equations y′ = y³, y′ = −y³ and the semistable one of the last lesson show all three outcomes.
Core ideas
Three ideas. A change of variable that puts the equilibrium at zero, the rule it produces, and the case where the rule says nothing.
Write u = y − y*, so that u′ = y′ = f(y* + u). For a polynomial, expanding gives f(y* + u) = f(y*) + f′(y*)·u + (terms with u² and higher), which is the same expansion in h that produced the derivative in Rates of Change and the Derivative. At an equilibrium f(y*) = 0, so u′ = f′(y*)·u + (terms with u² and higher).
For small u the higher terms are far smaller than the first. If f′(y*) < 0 then u′ has the opposite sign to u and the gap closes, like e^(f′(y*)·t); if f′(y*) > 0 the gap opens. The number is a rate, not only a sign: slope −2 closes the gap like e^(−2t), a factor of about 0.135335 per unit of time.
If f′(y*) = 0 the first term is absent and what is left is u² or u³ or higher, whose sign depends on the equation. The slope of f at the equilibrium is not the slope of any solution, and it being flat does not mean the equilibrium is flat in any sense a classification can use. The sign test of the last lesson still works and is the method to fall back on.
The material
The expansion done exactly on one equation, the rule and its rate, and the three equations where the rule is silent.
The sign test needs the whole phase line. The slope test needs only the equilibrium. Of the two methods this Subject uses for stability it is the one that carries over to systems, in “Linearisation and the Jacobian” in Systems and the Phase Plane, so the reasoning is worth seeing exactly once on an equation small enough to do by hand.
That is exact, with nothing dropped. The slope is f′(−1) = −2, the first term, and what remains is u². Take u = 1/10: the full right side is −2/10 + 1/100 = −19/100, against −20/100 for the first term alone. The neglected piece is a twentieth of the first term, and it shrinks faster than u does, so it matters less and less as the solution closes in.
Let y* be an equilibrium of y′ = f(y). If f′(y*) < 0 it is stable, and if f′(y*) > 0 it is unstable. If f′(y*) = 0 the test says nothing.
The lab demonstrates this on every equation it is given: it prints f′ at each equilibrium, exactly, beside the type found from the signs, and they agree whenever the slope is not zero. It does not prove the statement for every f. The claim has the same standing as the limit of the quotients in the first course: shown to hold, and used.
If u′ = −2u exactly then u = u₀·e^(−2t), and the gap at t = 1 is e^(−2), about 0.135335 (rounded), of the gap at t = 0. That is the sense in which a stable equilibrium of slope −2 is “twice as strongly stable” as one of slope −1: the same gap closes like e^(−t) instead, a factor of about 0.367879 per unit of time. At an unstable equilibrium of slope 2 the gap grows by e² ≈ 7.38906 per unit instead. Exactness belongs to the equation above; the exponential is what it approaches for small u.
The cubic of the last lesson has slopes 3, −2 and 6 at its three equilibria, because f′(y) = 3y² − 8y + 3. The lab prints f′(0) = 3; f′(1) = −2; f′(3) = 6, and the verdicts it prints beside them, unstable, stable, unstable, are those of the sign test.
Three equations have slope zero at y = 0 and no two of them behave alike. For y′ = y³ we have u′ = u³, which has the sign of u: the gap opens, so 0 is unstable. For y′ = −y³ it has the opposite sign: the gap closes, stable. For the semistable equation of the last lesson, y′ = y²·(1 − y), near zero we have u′ ≈ u², which is positive on both sides: semistable. The slope is the same, 0, in all three, and the verdicts are three different ones.
The lab prints f′(0) = 0 for y′ = −y³. Do not read the zero as “no decision possible”; read it as “this test has nothing to say”.
The signs do. f(−1) = 1 > 0 and f(1) = −1 < 0: positive then negative, both arrows towards 0, stable. Solutions do approach 0, though more slowly than any exponential e^(−c·t) would carry them.
Interactive
Equilibria, the phase line, and where solutions go
Method
Four moves. The last one is the one a reader forgets to make.
Use the power rule on the polynomial, term by term. Keep the result as a polynomial in y.
Substitute the root: a fraction, or a surd for an irrational root. Do not evaluate at a rounded decimal, which is not the equilibrium.
Negative: stable, and the gap closes like e^(f′(y*)·t). Positive: unstable. Zero: stop and use the signs of f on either side.
On a polynomial the phase line costs one test value per gap. A slope verdict and a sign verdict that disagree mean an arithmetic slip.
Here f(y) = y² − 1, so f′(y) = 2y. The two equilibria are ±1. Shift each to zero and expand.
Near y = −1 the gap behaves like u₀·e^(−2t), which is e^(−2) ≈ 0.135335 of itself after one unit of time (rounded). Near y = 1 it grows instead. The sign test gives the same verdicts: y²−1 goes from positive to negative across −1 and from negative to positive across 1.
For a rehearsal, take y′ = y·(4 − y). Its slope is 4 − 2y, which is 4 at 0 and −4 at 4: unstable at 0, stable at 4.
You should be able to differentiate f, evaluate at each equilibrium as a fraction or a surd, say stable or unstable from the sign, give the rate at which the gap closes or opens, and fall back on the signs of f when the slope is zero.
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