Homogeneous Constant-Coefficient Systems and Phase Portraits

The homogeneous system with a constant coefficient matrix,

x^{'} = A x, A a real n \times n constant matrix,

solves in closed form, and the qualitative theory of the subject grows from it. The scalar analogue $x^{'} = a x$ has solution $x = c e^{a t}$ , so a vector system should have solutions built from an exponential times a fixed direction. Finding those directions is the eigenvalue problem $A ξ = r ξ$ .

The exponential ansatz

Try a solution of the form

x = ξ e^{r t},

with the scalar $r$ and the constant vector $ξ \neq = 0$ to be found. Then $x^{'} = r ξ e^{r t}$ , and substituting into $x^{'} = A x$ gives $r ξ e^{r t} = A ξ e^{r t}$ . Cancelling the never-zero factor $e^{r t}$ leaves

A ξ = r ξ, equivalently (A - r I) ξ = 0 .

The vector $ξ e^{r t}$ solves the differential equation exactly when $r$ is an eigenvalue of $A$ and $ξ$ a belonging eigenvector. The differential-equation problem has become the algebra problem $det (A - r I) = 0$ .

That these $n$ solutions form a fundamental set is immediate from their Wronskian. Factoring the exponentials out of each column,

W [x^{(1)}, \dots, x^{(n)}] (t) = e^{(r_{1} + \dots + r_{n}) t} det [ξ^{(1)} \dots ξ^{(n)}],

which is never zero because the exponential is positive and the eigenvectors are independent. In the $2 \times 2$ case the solution lives in a plane and can be drawn as a phase portrait.

The phase plane

For a planar system, a solution $x (t) = (x_{1} (t), x_{2} (t))$ traces a curve in the $x_{1} x_{2}$ -plane as $t$ increases. That plane, with a representative sample of solution curves drawn on it, is the standard picture of a two-dimensional system.

Since $det A \neq = 0$ is assumed, the only equilibrium (where $A x = 0$ ) is the origin. The central question is whether nearby trajectories approach the origin or leave it as $t \to \infty$ , i.e. whether $x = 0$ is stable or unstable. The eigenvalues settle it.

Real eigenvalues of opposite sign: the saddle

When the two eigenvalues are real with $r_{1} > 0 > r_{2}$ , one exponential grows and the other decays.

The two eigenvectors set two straight-line trajectories through the origin. Along $x_{2} = 2 x_{1}$ (the direction of $ξ^{(1)}$ ) solutions move out, since $e^{3 t}$ grows; along $x_{2} = - 2 x_{1}$ (direction of $ξ^{(2)}$ ) solutions move in, since $e^{- t}$ decays. Every other trajectory is a combination: for large $t$ the $e^{3 t}$ term dominates, so trajectories become asymptotic to the line $x_{2} = 2 x_{1}$ ; running $t \to - \infty$ they approach $x_{2} = - 2 x_{1}$ .

Saddle point (eigenvalues 3 and -1): trajectories flow out along the unstable eigendirection and in along the stable one, bending like hyperbolas.

This picture — inflow along one eigendirection, outflow along the other, and hyperbola-like curves between — is typical of every $2 \times 2$ system whose eigenvalues are real and of opposite sign. The origin is a saddle point, and it is always unstable, since almost every trajectory eventually departs.

Real eigenvalues of the same sign: the node

When both eigenvalues are real and share a sign, both exponentials pull the same way, and every trajectory approaches (or leaves) the origin.

Both terms decay, so all trajectories approach the origin. The rates differ: $e^{- 3 t}$ dies far faster than $e^{- t}$ , so for large $t$ the slow term $c_{1} ξ^{(1)} e^{- t}$ dominates. Unless $c_{1} = 0$ , every trajectory reaches the origin tangent to the slow eigendirection $x_{2} = x_{1}$ ; the only trajectories coming in along the fast direction $x_{2} = - x_{1}$ are the ones with $c_{1} = 0$ .

Nodal sink (eigenvalues -1 and -3): all trajectories decay to the origin, arriving tangent to the slow eigendirection x2 = x1.

The origin is a node. It is asymptotically stable when both eigenvalues are negative (a sink, as drawn) and unstable when both are positive (a source, the same picture with arrows reversed).

Complex eigenvalues: spirals and centers

A real matrix can have complex eigenvalues, which must come in conjugate pairs $r = λ \pm i μ$ with real $λ, μ$ . The belonging eigenvectors are also complex conjugates, and the solution $ξ e^{r t}$ is complex-valued. Real solutions are recovered by taking real and imaginary parts, which is legitimate because for a real matrix the real and imaginary parts of any complex solution are themselves solutions.

Write the eigenvector as $ξ = a + i b$ with real $a, b$ , and expand $e^{(λ + i μ) t} = e^{λ t} (cos μ t + i sin μ t)$ . Separating $ξ e^{r t}$ into real and imaginary parts gives two real solutions:

u (t) = e^{λ t} (a cos μ t - b sin μ t), v (t) = e^{λ t} (a sin μ t + b cos μ t) .

The exponential $e^{λ t}$ sets the amplitude and the trigonometric factors set the rotation. The sign of the real part $λ$ determines whether the trajectory spirals in or out.

Complex eigenvalues. Left: negative real part gives a spiral sink, winding inward. Right: purely imaginary eigenvalues give a center of closed orbits.

The origin is a spiral point when $λ \neq = 0$ — asymptotically stable for $λ < 0$ , unstable for $λ > 0$ — and a center when $λ = 0$ , which is stable but not asymptotically stable. The direction of winding (clockwise or counterclockwise) is fixed by the matrix and found by checking the tangent vector $A x$ at one convenient point.

Transitions: a one-parameter sweep

Because the portrait type is fixed by the eigenvalues, and eigenvalues move continuously with the matrix, a system carrying a parameter passes through several types as the parameter varies. The values where the type changes are bifurcation values.

Range of $α$	Eigenvalues	Portrait
$α < - 4$	real, both negative	stable node
$- 4 < α < 0$	complex, negative real part	spiral sink
$α = 0$	purely imaginary $\pm 2 i$	center
$0 < α < 4$	complex, positive real part	spiral source
$α > 4$	real, both positive	unstable node

The bifurcation values are $α = - 4$ and $α = 4$ , where real eigenvalues collide and split into a complex pair (node to spiral), and $α = 0$ , where the pair crosses the imaginary axis (stable spiral to unstable, through a center). The center at $α = 0$ is the delicate transition: an arbitrarily small change in $α$ destroys the closed orbits, spiralling them in or out.

The eigenvalue classification

Collecting the three main cases and the transitional ones, the type of the equilibrium and its stability are determined by the eigenvalues.

Eigenvalues of $A$	Equilibrium type	Stability
Real, opposite signs $(r_{1} > 0 > r_{2})$	saddle point	unstable
Real, same sign, unequal	node	stable if both $< 0$ , unstable if both $> 0$
Complex, nonzero real part	spiral point	stable if $Re < 0$ , unstable if $> 0$
Purely imaginary $(\pm i μ)$	center	stable, not asymptotically
Real, equal (defective)	improper node	stable if $< 0$ , unstable if $> 0$

The last row, equal eigenvalues with too few eigenvectors, is the defective case; its solution needs the generalized eigenvectors of the following lesson. Reading the table geometrically, plot the eigenvalues in the complex plane: the sign of the real part is the stability switch, and whether the pair is on the real axis, off it, or on the imaginary axis picks node/saddle, spiral, or center.

Eigenvalues in the complex plane. Left half-plane decays (stable), right half grows; real axis gives nodes and saddles, off-axis pairs spiral, the imaginary axis gives centers.

The same classification reads off the trace and determinant of $A$ . The eigenvalues of a $2 \times 2$ matrix satisfy $r^{2} - (tr A) r + det A = 0$ , so $r_{1} + r_{2} = tr A$ and $r_{1} r_{2} = det A$ ; the sign of $det A$ , the sign of $tr A$ , and the sign of the discriminant $(tr A)^{2} - 4 det A$ decide the type without solving for the roots.

The trace-determinant plane. The parabola det = (tr)^2 / 4 separates nodes (below it) from spirals (above); below the horizontal axis lie saddles, and the vertical axis with positive determinant carries centers. Trajectories decay on the stable (left) half and grow on the unstable (right) half.

Summary

Solving $x^{'} = A x$ is finding eigenpairs, and reading the phase portrait is reading eigenvalue signs.

Method. Solve $det (A - r I) = 0$ for eigenvalues, then $(A - r I) ξ = 0$ for eigenvectors; combine $ξ^{(i)} e^{r_{i} t}$ . For a complex pair, take real and imaginary parts to get $u, v$ .
Geometry. Real opposite signs, saddle; real same sign, node; complex with nonzero real part, spiral; purely imaginary, center. The real part is the stability switch.

Near an equilibrium a nonlinear system resembles its linearization, so these five portraits classify its equilibria as well.