Two numbers decide the flow

The exponential of a square showed that the system of equations $\dot x = Ax$ is solved, for any square matrix $A$, by $x(t) = e^{tA}x(0)$, and that the series defining $e^{tA}$ converges whatever $A$ is. That settles existence and computation. It leaves open the question people who meet such systems actually ask, which is qualitative: does the solution die away, blow up, or keep going round?

In the plane the answer can be given completely, and it needs much less than the matrix. A two-by-two matrix has four entries. The long-run behaviour of its flow depends on two numbers only: the trace, the sum of the diagonal entries, and the determinant. Everything else — which direction the solutions spiral, how the axes of a saddle are tilted — is detail. Whether the origin attracts, repels, or does neither is decided by where one point lands in a plane.

Why two numbers are enough #

The behaviour of $e^{tA}$ is decided by the eigenvalues of $A$. Along an eigenvector with eigenvalue $\lambda$, the flow is just multiplication by $e^{\lambda t}$ — the one-variable equation of the equation with only one answer, running along a line. A positive real eigenvalue pushes outward along its line, a negative one pulls in, and a complex eigenvalue $a + ib$ produces a rotation at rate $b$ inside a growth or decay at rate $a$, which is multiplying as turning run continuously.

And the eigenvalues of a two-by-two matrix are fixed by its trace $T$ and determinant $D$, because they are the roots of the characteristic polynomial

$$\lambda^2 - T\lambda + D = 0,$$

as the polynomial whose roots are the stretches showed. The trace is their sum and the determinant their product. Two numbers in, two roots out, and the roots are the whole story.

So the plane of $(T, D)$ is a map of all possible flows. The quadratic formula divides it. When $T^2 > 4D$ the roots are real and different; when $T^2 < 4D$ they are a complex pair; on the parabola $T^2 = 4D$ they coincide. When $D < 0$ the product of two real roots is negative, so one is positive and one negative. When $D > 0$ they have the same sign, or are a complex pair with real part $T/2$ — so in the upper half-plane the sign of the trace alone says whether the flow decays or grows.

The four shapes a flow can take #

Each region of the plane has a portrait, and each portrait is visibly different from the others. Three of them are side by side here.

Below the axis, saddles. One eigenvalue is positive and one negative, so there are two special lines through the origin — one along which points rush in, one along which they rush out. Every other path comes in near the first line and leaves near the second, making the hyperbolas of the left panel. The origin is unstable, but only just: exactly one line of starting points, out of all the directions there are, is carried into it.

Between the axis and the parabola, nodes. Two real eigenvalues of the same sign. With both negative every path runs into the origin, and it arrives tangent to the eigenvector of the eigenvalue nearer zero, because that component decays more slowly and ends up dominating. The paths look like a bundle of parabolas pinched together at the origin. With both positive the same picture runs backwards.

Above the parabola, spirals. A complex pair, so the flow turns as it grows or shrinks. The turning rate is the imaginary part, $\sqrt{D - T^2/4}$; the growth rate is the real part, $T/2$.

On the vertical axis above the origin, centres. Trace zero and positive determinant: purely imaginary eigenvalues, so the flow turns and neither grows nor shrinks. Every path is a closed ellipse. The skew matrix in the exponential of a square, whose flow was rotation with no trigonometry in its definition, is the most symmetric point on that half-line.

Classifying a flow without solving it #

The practical force of this is that the classification needs no eigenvalues at all, only arithmetic a person can do in their head. Take the system

$$\dot x = x + 2y, \qquad \dot y = -3x - 4y.$$

Its matrix has trace $1 - 4 = -3$ and determinant $1 \cdot (-4) - 2 \cdot (-3) = 2$. The determinant is positive, so it is not a saddle. The trace is negative, so it is stable. And $T^2 = 9$ exceeds $4D = 8$, so the roots are real: a stable node. Every solution runs into the origin without turning, and nothing about the answer needed the roots themselves, which happen to be $-1$ and $-2$.

Change one entry, the $-3$ to $-4$, and the determinant becomes $4$ while the trace stays at $-3$. Now $T^2 = 9$ is less than $4D = 16$, so the roots have become a complex pair and the node has become a spiral: the same decay, now with rotation. Change the $1$ to $4$ instead and the trace becomes zero, which looks like the signature of a centre. It is not one: the determinant is now $4 \cdot (-4) - 2 \cdot (-3) = -10$, negative, and the flow is a saddle. A zero trace promised closed orbits only on the half of the vertical axis above the origin, and this matrix sits on the half below it. The trap is instructive, because it fixes the order in which the tests have to be applied: the determinant’s sign first, then the trace’s, then the discriminant.

That order is the plane read from the bottom up. Below the axis nothing else matters; above it, left or right decides stability; and the parabola decides only whether the approach turns. Reading a system off its coefficients this way is how the qualitative theory of differential equations begins, and it is the reason textbooks draw this plane before they draw a single solution.

The borderlines, where the plane changes its mind #

The regions are open sets, and the curves dividing them — the horizontal axis, the parabola, the positive vertical axis — are where one kind of flow turns into another. Crossing the parabola from below to above, two real eigenvalues collide and split into a complex pair.

The middle panel is the moment of collision, and it is worth looking at because it is not a blend of its neighbours. The two eigen-directions of the node have merged into one; the matrix has only one eigenvector, it cannot be diagonalised, and the solution acquires a factor of $t$ in front of the exponential — $x(t) = (x_0 + t\,Nx_0)e^{-t}$ for a nilpotent part $N$. Paths still run into the origin, but every one of them is bent round to meet the single remaining line. On one side of this matrix the flow approaches along two directions; on the other it approaches along none, circling instead.

Nothing in the entries of the three matrices suggests so abrupt a change: they differ by one number each, and the change is entirely in the roots. The same collision happens in every setting where a quantity has a restoring force and friction, and it has a name there. It is critical damping.

The spring that crosses the parabola #

A spring with friction obeys $\ddot x + c\,\dot x + x = 0$: acceleration is minus the displacement, minus $c$ times the velocity. Writing velocity as a second coordinate turns it into a first-order system $\dot x = Ax$ with

$$A = \begin{pmatrix} 0 & 1 \\ -1 & -c \end{pmatrix},$$

whose trace is $-c$ and whose determinant is $1$ whatever the friction. So as $c$ increases, the spring’s matrix slides left along the horizontal line $D = 1$.

At $c = 0$ the point is on the positive vertical axis: a frictionless spring, a centre, oscillation forever. Any friction at all moves it into the stable spirals, where it swings back and forth with shrinking amplitude. At $c = 2$ it meets the parabola, where $T^2 = 4D$ means $c^2 = 4$. Beyond that it is among the stable nodes: the spring is so heavily damped that it never crosses the rest position at all, and simply creeps home.

The curve at $c = 2$ is the reason engineers care about the parabola. A door closer, a car’s suspension, the needle of a measuring instrument — each wants the fastest return to rest that does not overshoot, and among all the damping values, the one on the parabola is it. Less damping and the needle swings past its reading and back; more, and it takes longer to arrive, because the slower of two real eigenvalues gets slower as they move apart. The best design sits exactly on the curve where the character of the flow changes, which is also exactly where it is hardest to hold, since a little wear moves it off.

The trace is not the stability #

It is tempting to read the trace alone as the verdict — negative trace, stable; positive, unstable — because that is what it means in the upper half-plane. It is also what it means for area. The determinant of $e^{tA}$ is $e^{t\,\operatorname{tr}A}$, so the trace is the rate at which the flow shrinks or grows areas, and a negative trace means every blob of starting points is squeezed smaller as time runs.

A saddle with negative trace does both at once. It shrinks every region, and it sends nearly every point off to infinity. There is no contradiction: a region is squashed flat along the incoming direction faster than it is stretched along the outgoing one, so its area falls while its length grows without bound. The trace measures the product of the two stretches, and a product can be small with one factor enormous. That is why the determinant has to be consulted first. Below the axis, the trace does not decide stability — it only says how thin the squeezed region becomes.

The same distinction haunts every stability argument in more dimensions. Shrinking volume is not stability; the chaotic flow of two lobes and no cycle shrinks volume at a constant rate and still sends nearby points apart for ever. What decides stability is every eigenvalue’s real part, and in two dimensions the trace and determinant together are just enough to recover those.

What the linear picture says about curved flows #

Linear systems are rarely the whole story; they are the first-order approximation to a curved one near a point of rest. The one-dimensional version of that — a fixed point pulls or pushes according to the slope there — has a two-dimensional heir, and the trace–determinant plane tells exactly how far to trust it.

The Hartman–Grobman theorem says that near a rest point whose linearisation has no eigenvalue with zero real part, the curved flow looks like its linear approximation after a continuous change of coordinates. Saddles stay saddles, nodes and spirals stay attracting or repelling. In the language of the plane: everywhere except on the borderlines, the linear verdict is the true one.

On the borderlines it is not, and the centres show it most starkly.

A centre is a knife-edge: its closed orbits exist only if the trace is exactly zero, and the slightest perturbation turns them into spirals, inward or outward. So when a curved system’s linearisation is a centre, the linear picture has no authority. The frictionless pendulum near its lowest point has a genuine centre, because its energy is conserved and the orbits must close; but a system with a small nonlinear friction term, or a small nonlinear energy input, has the same linearisation and spirals, and which way it spirals is decided by the terms the linearisation threw away.

One dimension up, the plane becomes a solid #

The reason two numbers suffice is that a two-by-two matrix has a characteristic polynomial of degree two, whose two coefficients are the trace and the determinant. A three-by-three matrix has a cubic,

$$\lambda^3 + a\lambda^2 + b\lambda + c,$$

with three coefficients — the trace with its sign changed, the sum of the three principal two-by-two minors, and the determinant with its sign changed. The map of possible flows is now a three-dimensional space, and the stable region is cut out by the conditions Edward Routh and Adolf Hurwitz found in the nineteenth century: every root has negative real part exactly when $a > 0$, $c > 0$ and $ab > c$.

The first two conditions are the analogues of the plane’s two, and the third is new. It is the condition that fails when a pair of complex roots crosses the imaginary axis while the third root stays safely negative — the three-dimensional version of a spiral sink turning into a spiral source through a centre. The surface $ab = c$ is where a system in three dimensions starts to oscillate with growing amplitude, and control engineers draw it for the same reason this essay draws the parabola.

What does not survive the extra dimension is the picture. In the plane every region has one portrait, drawable at a glance; in three dimensions a stable flow can spiral round one axis while approaching along another, and the list of distinct shapes grows. The algebra of the conditions scales to any dimension, as a sequence of determinants built from the coefficients. The geometry that made the two-dimensional case feel inevitable does not.

What the plane cannot show #

The plane records only the eigenvalues, and a matrix is more than its eigenvalues. Two matrices at the same point can have portraits that are rotated, sheared or stretched copies of each other — the saddle’s two lines can be at right angles or nearly parallel — and the plane cannot tell them apart. What it does record exactly is the class of the flow under a linear change of coordinates, with one exception: on the parabola, a matrix with a single eigen-direction and a matrix that is a multiple of the identity sit at the same point, and their portraits differ, the first with every path bending onto one line and the second with every path a straight ray.

Nor does the plane say anything quantitative beyond rates. Two stable spirals near each other on it take nearly the same time to damp, but a spiral whose eigenvectors are nearly parallel can grow a great deal before it decays, a transient the eigenvalues do not predict at all. That is the same gap between spectral radius and norm that a geometric series whose ratio is a matrix measured for iterated maps, arriving here for continuous time. And in three or more dimensions there is no plane: the characteristic polynomial has more coefficients, the regions are cut out by more complicated surfaces, and the Routh–Hurwitz conditions replace the simple rule that in two dimensions stability means $T < 0$ and $D > 0$.

Still open: the curved flows the linearisation cannot settle #

For linear flows the plane is a complete answer, and nothing about it is open. For the curved flows it approximates, what is open sits exactly on its borderlines.

When a rest point’s linearisation is a centre, deciding whether the curved flow spirals in, spirals out or keeps closed orbits requires the higher-order terms, and for polynomial vector fields this is the centre problem: conditions on the coefficients under which the rest point is a genuine centre. It is solved for quadratic fields, where Henri Dulac and later authors listed the four families of coefficients that give a genuine centre, and for several special families beyond; for general cubic fields it is open, and the partial lists of conditions run to pages. The difficulty is that each new order of approximation produces a new quantity that has to vanish, and deciding when infinitely many of them vanish together is a question about ideals of polynomials that no finite computation obviously settles. Beyond it lies the question of how many closed orbits a polynomial flow can have at all — the second half of Hilbert’s sixteenth problem — which is open even for quadratic vector fields: examples with four isolated closed orbits are known, and it is not known whether any quadratic field has more, or even whether the number is bounded at all.