A bifurcation diagram puts the parameter a along the horizontal axis and the equilibria up the vertical one, so that each vertical slice is a phase line. Stable branches are drawn solid and unstable ones dashed. Three shapes cover the families here: two branches that meet and vanish, two that cross and swap stability, and one that forks into three. The critical values and the types are exact; the curves are drawn from floating-point roots.
Core ideas
Three ideas. A diagram is phase lines side by side, the line style carries the stability, and the shape at a critical value has a name.
Cut the diagram with a vertical line at one value of a and the points where it crosses the branches are the equilibria of that equation. Sweep the line from left to right and the phase line changes as it goes. The horizontal axis is the parameter, not time, and no branch is a solution plotted against t.
At a point on a branch, f′(y*, a) is the slope that classified the equilibrium in “Linearisation and the Sign of f′”. Negative is stable and drawn solid; positive is unstable and drawn dashed. A branch changes style at a point where its slope passes through zero, which is where it meets another branch.
In a saddle-node two branches of opposite type meet and vanish. In a transcritical shape two branches cross and exchange stability. In a pitchfork one branch changes type and two new branches of the other type spring from it. The shape at a critical value is a fact about the equation, and each one can be found by solving for the equilibria.
The material
The pitchfork family worked fully, then the crossing, then the comparison with the last lesson that corrects a common impression.
Take y′ = a·y − y³ = y·(a − y²). The equilibria are 0 for every a, and ±√a when a is positive. The slope in y is f′(y) = a − 3y², so at the equilibria it is f′(0) = a, and at the other two f′(±√a) = a − 3a = −2a.
For negative a the slope f′(0) = a is negative and 0 is stable, the only equilibrium. For positive a it is positive, so 0 is unstable, and the pair ±√a has appeared, each with slope −2a < 0, stable. At a = 4 that is ±2 with slope −8, and 0 with slope 4. The diagram is a solid line along a < 0, then a dashed line along a > 0 with two solid arms leaving it: a fork opening to the right.
For y′ = a·y − y² = y·(a − y) the equilibria are 0 and a, for every a. The slope is a − 2y, so f′(0) = a and f′(a) = −a. At a = 2 that is 0 unstable and 2 stable; at a = −2 it is 0 stable and −2 unstable. The two branches pass through each other at a = 0 and trade stability there.
The count behaves differently from the last lesson's family. At a = 2 there are two equilibria, at a = 0 one, at a = −2 two again. No equilibrium vanishes: one passes through the other. This is what separates a transcritical shape from a saddle-node.
Compare the three families as a rises through 0. In a + y² the count falls from two to none. In a·y − y³ it rises from one to three. In a·y − y² it is two on both sides. Nothing in the idea of a parameter makes the number of equilibria grow or shrink; the shape of f does.
Put the chosen a at 4 and look at the vertical line through it. It crosses the diagram at three heights, −2, 0 and 2. The two outer crossings are on solid branches and the middle one is on a dashed branch.
The phase line at that slice has arrows up, down, up, down, and the tile prints −2 stable; 0 unstable; 2 stable. A solution starting above 0 goes to 2, and one below goes to −2. At a = −1 the line crosses once, at 0, and every solution goes there.
Interactive
Where equilibria are born and meet
Method
Five moves. The first four produce the diagram, and the last names it.
For a·y − y³ the factors are y and a − y², so the branches are y = 0 and y = ±√a, the second pair existing for a ≥ 0.
Solve f = 0 and f′ = 0 together, as in the last lesson. These are the values of a where branches meet and the diagram can change its form.
Choose a value of a between critical values, find the equilibria and read f′ at each. Negative is stable and solid, positive is unstable and dashed.
Follow each equilibrium as a moves, keeping the style it has. A change of style on one branch happens at a critical value only.
Two branches meeting and vanishing: saddle-node. Two crossing and swapping style: transcritical. One splitting into three: pitchfork.
Here f(y) = y·(a − y²) and f′(y) = a − 3y². Read the slice at each of the two presets and then name the shape at a = 0.
Along the diagram 0 goes from stable at a = −1 to unstable at a = 4, and the pair ±2 is present at a = 4 and absent at a = −1: the shape is a fork, opening at the critical value a = 0, where f′(0) = a passes through zero. The lab prints −2, 0, 2 and 0 stable for these two presets, with the critical value a = 0.
For a rehearsal, take the crossing y′ = a·y − y² at a = −2. The equilibria are 0 and −2 and the slopes f′(0) = −2 and f′(−2) = 2: 0 stable and −2 unstable, the reverse of a = 2.
You should be able to take a vertical slice and list the equilibria with their types, tell solid from dashed by the sign of f′, find the critical value, and name a saddle-node, transcritical or pitchfork shape from how the branches meet.
Something to change about this lesson? Write it down, tick it off when it is handled, and export the list when you want to act on it. Kept in this browser only: nothing is sent anywhere, and nobody is notified.
No recommendations yet for this lesson.