Close to an equilibrium the curved terms are small, and the system behaves like u′ = J·u, where J is the matrix of partial derivatives of f and g at the point. Its trace and determinant give a saddle, node or spiral that the nonlinear system shares. When the classification is a centre or the determinant is zero, the Jacobian is inconclusive and the nonlinear terms decide.
Core ideas
Three ideas. The Jacobian is a matrix of rates of change, it turns a nonlinear equilibrium into a linear classification, and it is silent on exactly the cases where the linear classification has no strict sign.
The partial derivative f_x is the derivative of f with respect to x, treating y as a constant: for f = y − x² it is −2x, and f_y = 1. The matrix with rows f_x f_y and g_x g_y is the Jacobian, and its entries are numbers once a point is chosen.
Write u = (x − x*, y − y*) for the displacement from the equilibrium. The rates f and g are zero at the point, and to first order they change by f_x·(x − x*) + f_y·(y − y*) and g_x·(x − x*) + g_y·(y − y*): the tangent-line estimate of “The Derivative at a Point and the Tangent Line”, made once in each unknown. So u′ = J·u approximates the system close to the point, and its classification by trace and determinant is the one from the first half of the course.
The terms thrown away are smaller than the ones kept, but a centre has real part zero, so there is nothing left to be smaller than. Whether the curved terms push the orbits in, out, or leave them closed is then a different question, and the lab says so in the type tile.
The material
The Jacobian, two equilibria classified, the statement that licenses the method, and a system where it is silent.
For x′ = f(x, y), y′ = g(x, y) the Jacobian at a point is the matrix with rows f_x f_y and g_x g_y, each partial derivative evaluated there.
The linearisation at an equilibrium is the linear system u′ = J·u for the displacement u from the equilibrium.
For polynomials the partial derivatives are the power rule, once for each unknown. Differentiate with respect to x and treat y as a number; then do the reverse. On the system of the last lesson every entry is a line or a constant.
At (1, 1) the rows are −2 1 and 1 −1. The trace is −3, the determinant is 2 − 1 = 1, and τ² − 4Δ = 9 − 4 = 5 > 0. The roots are (−3 ± √5)/2, both negative, so (1, 1) is a stable node.
At (0, 0) the rows are 0 1 and 1 −1. The trace is −1 and the determinant is 0 − 1 = −1 < 0, so it is a saddle. The same system has a sink at one equilibrium and a saddle at the other, which a linear system can never have.
If no eigenvalue of J has real part zero, the equilibrium of the nonlinear system is a saddle when the linear system is, and is stable or unstable exactly as the linear system is.
The lab demonstrates this on the examples above and does not prove it: the proof is a theorem of dynamical systems that this course states and does not reach. What the lab does exactly is the matrix: the entries, the trace, the determinant and the roots as fractions or surds. Where a root is irrational, as (−3 + √5)/2 is, the exact form is the one to trust, and its value is rounded to ≈ −0.381966.
Take x′ = −y + x³, y′ = x + y³. The origin is an equilibrium, and the cubic terms have zero derivative there, so the rows of J are 0 −1 and 1 0: trace 0, determinant 1, roots ±i. The lab prints a centre and adds that the linearisation is inconclusive.
Let D = x² + y², the squared distance from the origin, and compute how it changes along a solution.
D′ = 2x·x′ + 2y·y′ = 2x·(−y + x³) + 2y·(x + y³) = 2x⁴ + 2y⁴. The cross terms −2xy and +2xy cancel, as they do for a rotation, and the cubic terms are left.
The remaining sum is positive at every point except the origin, so every solution moves away from it. The origin is an unstable spiral, though the Jacobian reports a centre.
□
That is the whole content of the warning. Neither the trace nor the determinant of the Jacobian was wrong; they described the best linear approximation, and the answer needed more than a linear approximation. The same is true when the determinant is zero and the lab prints non-isolated equilibria.
Interactive
Checked exactly, linearised, and classified
Method
Partial derivatives first, then the point, then the same classification as before.
Four partial derivatives: f_x, f_y, g_x, g_y. In f_x treat y as a constant, and in f_y treat x as one. They are polynomials in x and y.
Evaluate all four at the point, not at the origin by habit. For (1, 1), −2x becomes −2.
τ = f_x + g_y and Δ = f_x·g_y − f_y·g_x, exactly. At (1, 1) they are −3 and 1.
Check Δ first, then τ² − 4Δ, then the sign of τ. Here Δ > 0, the discriminant is 5, and τ < 0: a stable node.
If the result is a saddle, node or spiral, it transfers. If it is a centre, or the determinant is zero, say that the linearisation is inconclusive and look for a different argument.
The first two presets. Linearise at (1, 1) and read the type, then do the origin.
The tiles read the same exact values, −3, 1 and the surd form of the roots. The rounded values of the two roots are ≈ −0.381966 and ≈ −2.61803, and both are negative, which is all the classification uses.
So solutions that start close to (1, 1) settle on it, and almost all solutions that start close to the origin leave it, along the directions of the saddle.
You should be able to compute the four partial derivatives, evaluate the Jacobian exactly at a chosen equilibrium, classify it by trace and determinant, and name the borderline case, a centre or a zero determinant, in which the classification does not transfer.
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