An equilibrium of x′ = f(x, y), y′ = g(x, y) is a point where f and g are both zero, so the arrow there vanishes and a solution that starts on it never leaves. A linear system with a non-zero determinant has one, at the origin. A nonlinear one can have none, one or several, and they are found by solving two equations together and checked by substituting both.
Core ideas
Three ideas. The first is the definition, the second is the method that finds the points, and the third is what an exact search can promise and what it cannot.
At a point where f = 0 and g = 0 the arrow is (0, 0), so Euler's step adds nothing and the exact solution stays where it is. Both conditions are needed at the same point. A point where f is zero and g is not has x momentarily still while y moves, and the solution leaves it at once.
The set where f = 0 is a curve and the set where g = 0 is another; the equilibria are where the curves cross. By hand, solve one equation for one unknown, substitute into the other, and finish with the unknown you have left. Every root of that last equation gives one point, and each point is then checked in both equations.
The lab can test every point of a grid of fractions exactly, and it reports what it found. That is a statement about the grid. An equilibrium with a denominator the grid never uses, or a coordinate beyond it, is not reported, and the status line says what the bounds were so the claim is never larger than the search.
The material
The definition, one system solved by hand, the check that rejects a near miss, and a search whose limits are stated.
A point (x*, y*) is an equilibrium of x′ = f(x, y), y′ = g(x, y) when f(x*, y*) = 0 and g(x*, y*) = 0.
The constant pair x(t) = x*, y(t) = y* is then a solution, because both sides of both equations are zero. The curve where f = 0 is the x-nullcline and the curve where g = 0 is the y-nullcline.
A linear system x′ = A·x with a non-zero determinant has exactly one equilibrium, the origin, because A·x = 0 has only the solution x = 0. Nothing like that holds once a product or a square enters. Where the first half of the course asked what the origin does, the first question now is how many places there are to ask it.
Take the system above. Its x-nullcline is the parabola y = x² and its y-nullcline is the line y = x. Setting g = 0 gives y = x; putting that in f = 0 gives x − x² = 0, which factors as x·(1 − x) = 0, so x = 0 or x = 1. The line then returns y = x, and the two points are (0, 0) and (1, 1).
At (0, 0): f = 0 − 0 = 0 and g = 0 − 0 = 0. At (1, 1): f = 1 − 1 = 0 and g = 1 − 1 = 0. Both are equilibria.
Now try (2, 4), which lies on the parabola. There f = 4 − 4 = 0 but g = 2 − 4 = −2. The lab prints f = 0, g = −2: not an equilibrium. The solution through (2, 4) has x still for an instant and y falling at rate 2, so it moves on.
Switch the search control on. The lab tests every point whose coordinates are fractions with denominator 1, 2, 3 or 4 and numerators between −12 and 12, exactly, and the status line names those bounds as it reports. On the search preset only the origin is listed, and the report is found 2: (0, 0), (1, 1). It is a convenience for checking your own algebra, and the algebra is the method.
Its limit is built into the last preset. The system x′ = y − 5x², y′ = x − y has the equilibria (0, 0) and (1/5, 1/5), because x = 5x² gives x = 0 or x = 1/5. The denominator 5 is outside the grid, so the search reports only the origin, while the second point passes the exact check in the points list. A search that finds none, or one, has told you what the grid holds and not how many there are.
The same caution applies to the drawing. The arrows are shown for orientation, and the check tile and the table come from substituting the listed points into the equations, with no floating point anywhere in the test. The tiles from the Jacobian down to the linearised type belong to the next lesson; this one needs only the check.
Interactive
Checked exactly, linearised, and classified
Method
Algebra first, because it finds every point; the lab second, because it tests the ones you found.
Set f(x, y) = 0 and g(x, y) = 0. Keep the two equations side by side; neither is solved alone.
Choose the equation where one unknown appears to the first power. Here x − y = 0 gives y = x.
Replacing y by x in y − x² = 0 gives x − x² = 0. Factor it: x·(1 − x) = 0. Every root gives a point, and none may be dropped by dividing out a factor.
Put each root back into the equation you solved: x = 0 gives y = 0, x = 1 gives y = 1.
Substitute into f and g. Type the points into the lab and read the tile; a point that fails one equation is not an equilibrium.
The nullclines are a parabola and a line. Solve for their crossings, then test a point that lies on only one of them.
The system has two equilibria, found by factoring one equation, and the lab agrees with both checks. The point (2, 4) is on the x-nullcline and not on the y-nullcline, and the tile prints the number that rules it out, g = −2.
Neither equilibrium is the origin alone. The reason the origin was special for a linear system is gone, and what each of these two points does is the question of the next lesson.
You should be able to set both rates to zero, solve the pair by substitution without dropping a root, check every candidate in both equations, and say what a bounded search found and what it could not have found.
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.