The origin of x′ = A·x is asymptotically stable when every solution tends to it, and that happens exactly when τ < 0 and Δ > 0, which is the same as every eigenvalue having a negative real part. The borderline cases, a centre and a zero determinant, are stable without attracting or not isolated.
Core ideas
Three ideas. Stability has degrees, the criterion needs both numbers, and the borderline cases are where the criterion stops deciding.
The origin is asymptotically stable when every solution tends to it, stable when solutions that start close stay close, and unstable when some solution leaves any neighbourhood. A centre is stable without being asymptotically stable: orbits stay close and do not settle.
The origin is asymptotically stable exactly when τ < 0 and Δ > 0. The determinant rules out saddles and a zero root; the trace then decides between in and out. Neither number alone is enough.
The condition is equivalent to every eigenvalue having a negative real part. Real roots must both be negative, and a complex pair α ± β·i needs α < 0. The trace is twice the real part for a complex pair, and the sum of the roots otherwise, which is why it gives the same answer.
The material
The three degrees, the criterion with its short proof, the borderline cases and what Euler's steps do near them.
The origin is asymptotically stable when every solution tends to it as t → ∞. It is stable when every solution that starts near it stays near it. It is unstable when some solution that starts arbitrarily near it does not.
Asymptotic stability implies stability. A centre is stable and not asymptotically stable.
The origin of x′ = A·x is asymptotically stable exactly when τ < 0 and Δ > 0.
Suppose the roots are real. Their product is Δ > 0, so they have the same sign, and their sum is τ < 0, so that sign is negative. Every term of the general solution then decays.
Suppose the roots are α ± β·i. The real solutions are multiples of e^(αt) times cosines and sines, as in “Complex Roots and Oscillation”, and α = τ/2 < 0, so they decay.
Conversely, if Δ < 0 there is a positive root. If Δ > 0 and τ > 0 the real parts or the common sign are positive. In each case some solution does not tend to the origin. The borderline cases τ = 0 and Δ = 0 are treated below.
□
The matrix with rows −1 2 and −2 −1 has τ = −2 and Δ = 1 + 4 = 5. Both conditions hold. The discriminant is 4 − 20 = −16, so λ = −1 ± 2i.
Both roots have real part −1, so every solution carries a factor e^(−t) and tends to the origin, turning as it goes. It is a stable spiral. Compare the spiral in the last lesson: the signs of the diagonal flipped, and the picture is the same unwound.
At τ = 0 with Δ > 0 the roots are ±i·√Δ, with real part zero. Nothing grows and nothing decays; orbits are closed curves. For rows 0 2 and −2 0 they are circles, since (x² + y²)′ = 2x·(2y) + 2y·(−2x) = 0. The origin is stable, because a start near it stays near it, and it is not asymptotically stable, because no orbit tends to it. The criterion says “not asymptotically stable”, and does not say “unstable”.
At Δ = 0 one root is zero. The matrix with rows −1 1 and 1 −1 has τ = −2, Δ = 0 and roots 0 and −2. The vector (1, 1) is sent to zero, so every point of that line is an equilibrium: the origin is not isolated. Every orbit comes in to the line from the side, along the direction (1, −1), and settles on it, but not at the origin: the start (1, 0) has constants 1/2 and 1/2, and ends at (1/2, 1/2). The lab calls this non-isolated equilibria.
Euler's polygon follows its own arithmetic and can disagree with the classification. On the centre, with h = 1/4 the squared distance from the origin is multiplied at each step by 1 + 4·h² = 5/4, which is the lab's ratio. The true orbit is a circle and the polygon spirals out. The classification is a statement about the equation. The polygon is a method applied to it, and the gap between them is exactly the one the material clause names.
For rows 1 2 and 3 0 the eigenvalues are −2 and 3. One of them is negative, and the solution e^(−2t)·(2, −3) does tend to the origin.
That is a single line of starts out of the whole plane. Any start with a component along (1, 1) carries an e^(3t) term that grows without limit. So one negative eigenvalue does not make the origin stable: every eigenvalue has to be negative, which Δ < 0 rules out.
Interactive
Trace, determinant, eigenvalues and exact Euler steps
Method
Two signs, in order, with the borderline cases handled before the verdict.
Exactly, from the four entries. For rows −1 2 and −2 −1 they are −2 and 5.
Δ < 0 is a saddle and not stable. Δ = 0 means a zero root and a line of equilibria. Only Δ > 0 leaves the question open.
τ < 0 is asymptotically stable. τ > 0 is unstable. τ = 0 is a centre, which is stable and not asymptotically stable.
Solve the quadratic. For real roots check both signs; for a complex pair read τ/2. −1 ± 2i has real part −1.
Say which of the three degrees applies and why, naming the borderline case if that is the one in play.
The first preset. Apply the criterion and confirm it with the roots.
Both signs pass, and the roots agree: the real part is −1. The lab tiles read −2, 5 and stable spiral.
Change the sign of the trace and the same picture unwinds. The matrix with rows 1 −2 and 2 1 has τ = 2 and the same discriminant, so the roots are 1 ± 2i and the orbits turn outward. Stability is decided by the sign of the real part alone, and the trace carries it.
You should be able to state the criterion, apply it in order (Δ first, then τ), name the centre as stable and not asymptotically stable, recognise a zero determinant as non-isolated equilibria, and connect the result to the signs of the real parts of the eigenvalues.
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