In the t–y plane the equilibria are horizontal lines that no solution crosses, and between them each solution rises or falls as the phase line says. Differentiating y′ = f(y) once more gives y″ = f′(y)·f(y), which fixes the bend, so a solution has an inflection exactly where it crosses a level at which f′ changes sign. The lab computes those levels exactly and draws the curves for comparison.
Core ideas
Three ideas. The phase line fixes the outline, one more derivative fixes the bends, and the bends sit at levels and not at times.
Draw each equilibrium as a horizontal line across the plane. Between two of them every solution is monotone, rising or falling as the arrow says, never touching either line, and levelling off towards a stable equilibrium. Close to an unstable one it leaves slowly and then speeds up.
Differentiate y′ = f(y) by the chain rule: y″ = f′(y)·y′, and y′ is f(y), so y″ = f′(y)·f(y). It is positive, concave up, where f′ and f have the same sign, and negative, concave down, where their signs differ. Both are functions of y alone.
A solution can only change its bend where y″ changes sign. At an equilibrium f is zero but the solution is a straight line, so that is no bend. Elsewhere it is where f′ is zero and changes sign: a height y, the same for every solution that crosses it, reached at a time that differs from solution to solution.
The material
The second derivative worked out, one equation sketched in full, and the cases where the inflection cannot be placed exactly.
If y solves y′ = f(y), with f a polynomial, then y″ = f′(y)·f(y).
Differentiate both sides of y′ = f(y) with respect to t. On the left that gives y″. On the right the chain rule gives f′(y)·y′, the slope of f at the height y times the rate at which y is changing.
Replace y′ by f(y) in that product, which the equation allows at every t. The result is y″ = f′(y)·f(y), a function of y alone.
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Take y′ = (y − 1)·(y − 3). Then f′(y) = 2y − 4, and y″ = (2y − 4)·(y − 1)·(y − 3). It is zero at y = 1 and y = 3, which are equilibria, and at y = 2, which is not.
A solution from y(0) = 5/2 is in the gap (1, 3), with f(5/2) = −3/4, so it falls. Above 2 it is concave down, y″ = (1)·(−3/4) = −3/4 at the start. It crosses the level 2 and from there it is concave up, y″ = 3/4 at y = 3/2 where f′ = −1 and f = −3/4, flattening out towards 1. The inflection is at the level y = 2.
In the gap (1, 3) the speed |f(y)| is 1 at y = 2 and less everywhere else, so the curve is steepest as it crosses 2. A curve that is steepest at a point has no change of slope there, which is y″ = 0. The inflection and the maximum speed are the same level.
Take y′ = y·(1 − y/4), so f(y) = y − y²/4 and f′(y) = 1 − y/2, zero at y = 2, half of the equilibrium 4. The slopes at y = 1/2, 1, 2, 3 are 7/16, 3/4, 1, 3/4, and 0 at y = 4.
From y(0) = 1 the curve rises, concave up until it crosses 2, steepest there, then concave down, levelling off at 4. From y(0) = 6 it falls to 4, concave up all the way, with no inflection, since f′ < 0 and f < 0 above 4. “Logistic Growth” in Separable Equations, Growth and Decay found the same level K/2 as the point of fastest growth and printed y = 2 in the same tile; the phase line now says why the fastest point and the change of bend are one level.
The cubic of the earlier lessons has f′(y) = 3y² − 8y + 3, whose roots are (4 ± √7)/3, exactly, and ≈ 0.451416 and ≈ 2.21525 rounded. The inflection levels are real and irrational. The lab places only the rational ones, so its inflection tile reads none rational there; the sketch still has two bends, at heights that are not fractions.
Interactive
Equilibria, the phase line, and where solutions go
Method
Five moves, each using something computed earlier in the course.
Find them exactly and draw them across the t–y plane, labelled stable or unstable. No curve will cross any of them.
Use the arrow of the gap: up or down, levelling off at a stable end, leaving an unstable end slowly. A gap with no equilibrium at an end has curves that run off the page.
Differentiate, solve f′(y) = 0 exactly, and keep the roots at which f′ changes sign and f is not zero. Each is the level of an inflection.
Between the lines, y″ = f′·f is positive where the signs agree and negative where they differ. Draw each curve concave up or down accordingly, flipping where it crosses an inflection level.
Check the arrows, the levelling off and the level of each flip against the lab. A curve that crosses an equilibrium, or has a bend at the wrong level, has a slip in a sign.
Here f′(y) = 2y − 4, so y″ = (2y − 4)·f(y). Find the level of the bend, then read the shape of the solution on each side of it.
The curve is monotone, falling, and never reaches 1 or goes below it. It changes bend at the level y = 2, where it is steepest: the lab's inflection tile reads y = 2, and the start 5/2 is drawn crossing that level. The start 3/2 is already below it, and its curve is concave up throughout.
For a rehearsal, take the start y(0) = 4 on the same equation. f(4) = 3 and f′(4) = 4, so y″ = 12 > 0: rising and concave up, with no inflection, running off the page.
You should be able to draw the equilibria as horizontal lines, sketch monotone curves in each gap, work out y″ = f′(y)·f(y), solve f′(y) = 0 for the levels of the bends, and check the sketch against the drawn curves.
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