An equation y′ = f(y, a) with a parameter a is a different autonomous equation for every value of a, and its equilibria are the roots of f(y, a) in y, which move as a does. Two of them meet where f and its slope in y are zero together, and at that value of a the count of equilibria changes. The lab finds the value exactly, and shows that a small change in a can change the long-run behaviour completely.
Core ideas
Three ideas. A parameter makes a family of phase lines, the equilibria in it are roots that move, and two roots meeting is a test with two equations.
In y′ = a + y² the letter a is fixed for the whole solution and chosen before it begins. Each value of a gives an ordinary autonomous equation with its own equilibria and its own phase line. The lab holds a in a box, and everything else on the page is the phase line of that one equation.
The equilibria of y′ = a + y² are the roots of y² + a, which are ±√(−a) when a is negative. As a climbs from −4 towards 0 they close from ±2 towards 0. The phase line keeps its shape while the number of roots stays the same, and can change it only at a value of a where that number changes.
Two roots coming together form a double root: f(y, a) = (y − r)²·g(y) near it, which is zero at r and has slope zero there too. So the value of a at which two equilibria meet is found by solving f = 0 and f′ = 0 together, with f′ the slope in y and a held fixed.
The material
One family read at four values of the parameter, the test that finds the value where it changes, and why that change is not small.
For a family y′ = f(y, a), a critical value of the parameter is a value of a at which two equilibria meet, so that the number of equilibria is different on the two sides of it.
The family is y′ = a + y². At each value of a the right side is a quadratic in y that is zero where y² = −a, and the slope in y is f′(y) = 2y. Read it at four values.
At a = −4 the slopes are f′(−2) = −4 and f′(2) = 4, so −2 is stable and 2 is unstable, by the slope test. The pair closes up as a rises: ±2, then ±1, then one equilibrium at 0, then none. For a = 1, 1 + y² is positive for every y, so every solution rises.
Solve both equations. The slope f′(y) = 2y is zero only at y = 0. The right side at y = 0 is a, and that is zero only at a = 0. So a = 0 is the one critical value of this family, and the lab's critical-value tile reads a = 0.
At a = 0 the equation is y′ = y², which is positive on both sides of 0: the equilibrium is the semistable one of “Stable, Unstable and Semistable”, the last equilibrium before the pair disappears. It is one double root, and the lab counts it once.
Take y′ = y² − 2y + a. The slope in y is 2y − 2, zero at y = 1. There f(1, a) = 1 − 2 + a = a − 1, which is zero when a = 1.
Check on both sides. At a = 0 the right side is y·(y − 2), with equilibria 0 and 2. At a = 1 it is (y − 1)², one equilibrium. At a = 2 it is (y − 1)² + 1, which is never zero. Type the right side into the box and move a to see the same three counts.
This test was run once before, without its name. In “Harvesting and the Threshold” the family was y′ = y·(1 − y/4) − H, with the harvest H as the parameter: the equilibria 1 and 3 at H = 3/4 moved together and met at 2 when H = 1, and at any larger harvest there were none. The two equations find that value. The slope in y is 1 − y/2, zero at y = 2, and f(2, H) = 1 − H, zero at H = 1, which is the threshold r·K/4 that lesson read off the peak of the growth. Type the harvested right side into the box, with a in place of H: at a = 3/4 the tiles print 1, 3 and 1 unstable; 3 stable, at a = 1 they print 2 semistable, and the critical value reads a = 1.
Take y′ = a + y² with y(0) = 0. At a = −1/100 the equilibria are ±1/10, because (1/10)² = 1/100. The start 0 lies between them, where f(0) = −1/100 < 0, so the solution falls to −1/10 and stops there for good.
At a = 1/100 there are no equilibria. Now y′ = 1/100 + y² is at least 1/100 for every y, so the solution rises for ever and, because the right side grows with y², without bound. Two values of a that are 1/50 apart give a solution that settles and a solution that does not. And between a = −1/100 and a = 0 each equilibrium moved by 1/10 while a moved by 1/100, ten times as far as the parameter did: near a critical value a square root is steep.
Interactive
Where equilibria are born and meet
Method
Five moves. The third is the one that gives the answer; the others check it.
Substitute the number, then find the roots of f(y, a) in y exactly, as in the first lesson of this course. This is an ordinary phase line.
Pick values spread across the range and record the count and the types each time. A change in the count is a sign that a critical value lies between two of your values.
Differentiate in y with a held fixed. Find the y that makes f′ zero, put it into f, and solve for a. For a quadratic this is the same as setting its discriminant to zero.
At the critical value itself there is one double root. On the two sides the counts differ, and a pair that has gone has gone from one side only.
Say how many equilibria there are for a below, at and above each critical value, and what they are.
The right side is y² + a, with slope 2y in y. Solve at each a, then solve f = 0 and f′ = 0 together to find where the pair meets.
The equilibria are ±√(−a) for negative a: ±2 at a = −4 and ±1 at a = −1. They meet at 0 when a = 0 and are gone for positive a. The lab's three presets print −2, 2, then 0, then none, with the critical value a = 0.
For a rehearsal, find the critical value of y′ = y² − 4y + a. The slope 2y − 4 is zero at y = 2, and f(2, a) = 4 − 8 + a = a − 4, so a = 4.
You should be able to substitute a value of a and solve, tabulate the equilibria across the range, solve f = 0 and f′ = 0 together to get the critical value, and say how many equilibria there are on each side of it.
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