1
Recognise an autonomous equation and find its equilibria exactly
The test is whether t appears on the right; the equilibria are the roots of f, as fractions or surds, each checked by substituting the constant solution.
Course · Differential Equations
Equilibria, Stability and Phase Lines
The last course solved equations: it separated the variables and integrated. This one asks a different question of the same first-order equations, and it is often the more useful one. For y′ = f(y) nobody needs a formula to say that a solution starting at 2 settles at 1, or climbs without limit, or that a small change in a parameter makes two equilibria collide and disappear. All of it comes from the sign of f, which a browser can evaluate exactly, and from one derivative. The lab keeps the exact side exact: equilibria are fractions or surds, the slope at each is an exact number, and the only things that are drawn rather than computed are the solution curves, which the legend says are stepped in floating point. The method is approximate, the classification is not.
y′ = f(y): no t on the right
f(y*) = 0 y* is a constant solution
f: + to − stable; − to + unstable
f′(y*) < 0 stable
f′(y*) > 0 unstable; f′(y*) = 0 undecided
y′ = f(y, a): equilibria move and meet
y″ = f′(y)·f(y): bends where f′ = 0
What this course is about
- Subject
- Differential Equations
- Lessons
- 8
- Format
- Interactive
Overview
What you will be able to do
By the end you can take a first-order equation with no t in it, decide where its solutions go from any start without solving it, say how stable each equilibrium is and why, and predict what happens to the equilibria when a parameter moves.
2
Draw the phase line from the sign of f
The equilibria in order, one exact test value in every gap, and an arrow in each, with the direction of every solution read off it.
3
Classify each equilibrium as stable, unstable or semistable
From the signs on either side of the root, and again from the sign of f′ at it, with the case f′ = 0 named as the one the second method cannot decide.
4
State the long-run behaviour of a solution from its start
The limit of a solution from any starting value as an equilibrium or ±∞, with the claim checked against exact Euler steps.
5
Follow the equilibria as a parameter changes
Where two of them meet, the value of the parameter at which they do, and the fork and exchange shapes read off a bifurcation diagram.
6
Sketch solution curves from the phase line alone
Horizontal at the equilibria, monotone between them, and bending where f′(y) is zero, with the sketch checked against curves the lab draws.
Lessons
8 lessons
The course goes from the equation to the picture and then to the things the picture can do. Autonomous equations and the phase line come first, because everything after is read from them. Classifying equilibria comes second, by signs and then by a derivative, and ends with the limit of a solution found without solving. The third part adds a parameter, and the last uses the phase line to draw the curves the earlier courses could only compute point by point.
Autonomous EquationsWhen the right side has no t, the slope depends on y alone, and the constant solutions are the exact roots of f.Not marked
The Phase LineMark the equilibria on the y axis, test the sign of f once in each gap, and every solution's direction is drawn.Not marked
Stable, Unstable and SemistableAn equilibrium attracts from both sides, repels from both, or attracts from one side only, and the signs of f on either side say which.Not marked
Linearisation and the Sign of f′Near an equilibrium the gap obeys u′ ≈ f′(y*)·u, so the sign of f′(y*) decides the type unless it is zero.Not marked
Long-Run Behaviour Without SolvingA solution cannot leave its gap and moves one way inside it, so its limit is the equilibrium its arrow points to, or infinity.Not marked
One-Parameter FamiliesPut a number a in the equation and the equilibria move; where two of them meet, the behaviour changes all at once.Not marked
Bifurcation DiagramsPlot every equilibrium against a, solid where stable and dashed where unstable, and the fork and the exchange are read straight off.Not marked
Sketching Solutions from the Phase LineHorizontal at the equilibria, monotone between them, and bending only where f′(y) is zero: a solution can be drawn without a formula.Not marked
Recommended background
Separable equations, growth and decay: the solution of y′ = k·y, the logistic equation with its two equilibria, and harvesting with its threshold, which this course reads again from the sign of f instead of from a formula. Behind that sit differential equations and Euler's method, from which the slope field and the exact Euler steps are used on every page, and the derivative of a polynomial from Rates of Change and the Derivative, which is all the calculus the slope test needs.
Practice suggestions
Read the sign of f before reaching for a formula. In every lesson the first move is the same: find the roots, test between them, draw the arrows. The answers that follow are usually available at that point, and a formula for the solution, where one exists, adds detail and not direction.
Say whether an answer came from the signs or from f′. The signs always decide; the slope decides unless it is zero, and also supplies a rate. A verdict from the slope with no check against the signs is a verdict that cannot be told from a slip.
Treat a limit as a claim about a phase line, not about a table. The exact Euler steps on these pages are an approximate method applied exactly, and a column of them that climbs is evidence for → +∞, not proof. What the phase line gives is the reason.
Keep exact and rounded apart. An equilibrium like (1 ± √5)/2 is exact; ≈ 1.61803 is its rounding. Write the first when you mean equality, and the second only when you are reporting a size.
What it does not cover
Proofs of existence and uniqueness. That one solution passes through each point is used throughout, because it is why a solution can approach an equilibrium and never cross it; it is stated in words in Differential Equations and Euler's Method and is not proved here.
The proof that the sign of f′ decides stability. The slope test is demonstrated on every equation the lab is given, by printing f′ at each equilibrium beside the type found from the signs, and derived exactly on one equation by hand. Showing that the neglected terms cannot overturn it, for every f, is a task for a first course in analysis.
Bifurcation theory beyond one parameter. The saddle-node, transcritical and pitchfork shapes are read from the equilibria the lab computes; normal forms, imperfect bifurcations, hysteresis loops and the Hopf bifurcation, which needs two variables, are not here. Systems and the Phase Plane takes the step to two variables.
Right-hand sides that are not polynomials. The lab refuses sin(y) or 1/y rather than round its way through them, because its equilibria and signs are exact. The method carries over unchanged, with roots found by other means.
Equations that depend on time. A forced equation, y′ = f(y) + g(t), has no phase line, because the arrows would move; First-Order Linear Equations treats the linear case, and the general case is a different subject.
What this course states and does not prove. That the sign of f′(y*) settles stability when it is nonzero. Every figure on the course is computed from the equation without it, so the unproved statement carries no number, and the sign test, which needs no theorem, is available alongside it on every page.