Picture: a map of who-eats-whom over time: Imagine tracking two interacting populations — say lynx (x) and hares (y). Instead of two separate graphs against time, draw ONE map with x across and y up.
Every point is a possible "state" of the system; the arrows show which way the state moves next. The curves the state traces out are trajectories, and the whole picture is the phase portrait.
For a linear system we write the two rates as a matrix product:
The equilibrium sits at the origin: An equilibrium is where nothing changes — both rates are zero at once. For a linear system M(x, y) = 0 has the single solution (x, y) = (0, 0) (whenever M is invertible).
So the only equilibrium is the origin, and the key question is: do nearby trajectories move TOWARD it (stable) or AWAY (unstable)? The eigenvalues of M answer that.
IB-style question — classify a two-species system
A reserve holds two interacting species. With x = thousands of grazers and y = thousands of browsers, the model is dx/dt = x + y and dy/dt = 4x + y (rates in thousands per year).
(a) Write the system as d/dt(x, y) = M(x, y) and state the equilibrium.
(b) Find the eigenvalues of M and classify the equilibrium.
Step by step
- (a) Read the coefficients of x and y straight off the two rate equations to build M.
- Both rates are zero only at the origin, so that is the equilibrium.
- (b) Form the characteristic equation using tr M = 2 and det M = (1)(1) − (1)(4) = −3.
- Factorise (or use the GDC's eigenvalue tool).
Final answer
(a) M = (1, 1; 4, 1), equilibrium at the origin (0, 0). (b) Eigenvalues are 3 and −1 — two REAL eigenvalues of OPPOSITE sign, so the origin is a SADDLE point: unstable. Trajectories are pulled in along one direction but flung out along the other, so the two populations cannot settle at zero.
The classification rule in one line: Read the eigenvalues:
• Two real, same sign → node (both negative = stable sink; both positive = unstable source).
• Two real, opposite signs → saddle (always unstable).
• Complex a ± bi → spiral (stable if a < 0, unstable if a > 0); if a = 0 (purely imaginary) → centre (closed loops).
Real eigenvectors are the straight-line trajectories: Along an eigenvector direction the system just scales — it doesn't twist — so a trajectory that starts on an eigenvector stays on that straight line through the origin.
Its eigenvalue decides the motion along that line: a NEGATIVE eigenvalue pulls the state IN toward the origin (eλt → 0), a POSITIVE eigenvalue pushes it OUT (eλt → ∞).
Every other trajectory is a blend of the two eigen-directions:
IB-style question — coupled tanks settling to empty
Brine flows between two connected tanks. With x and y the salt mass (kg) in each tank, the model is dx/dt = −2x and dy/dt = x − 4y (per hour).
(a) Find the eigenvalues and a corresponding eigenvector for each.
(b) Write the general solution and describe the long-term behaviour.
Step by step
- (a) Build M from the coefficients, then find the eigenvalues from the characteristic equation (tr M = −6, det M = 8).
- For λ = −2 solve (M + 2I)v = 0: the bottom row gives x − 2y = 0, so v₁ = (2, 1).
- For λ = −4 solve (M + 4I)v = 0: the top row gives 2x = 0, so x = 0 and v₂ = (0, 1).
- (b) Combine into the general solution.
Final answer
(a) λ = −2 with v₁ = (2, 1) and λ = −4 with v₂ = (0, 1). (b) BOTH eigenvalues are negative, so e−2t and e−4t → 0: the origin is a stable node (sink) and every trajectory drains to (0, 0). In context, the salt in both tanks tends to zero — the tanks flush clean. (Assumes the linear mixing model stays valid as masses get small.)
Which way do the arrows point near the origin?: To sketch a portrait: draw the eigenvector lines through the origin, then put arrows IN on a line whose eigenvalue is negative and arrows OUT on a line whose eigenvalue is positive.
For a node the bigger-magnitude eigenvalue is the "fast" direction; curved trajectories leave/arrive tangent to the slow (smaller-magnitude) eigenvector. For a saddle, trajectories come in along the negative direction and swing out along the positive one.