Nonlinear Systems & Linearization

Study Sheet

Nonlinear Systems & Linearization

Jacobians, predator–prey, competition, and the pendulum

Local Pictures of Nonlinear Worlds

Tip
Linearize at each equilibrium

A nonlinear system x′=f(x,y)x' = f(x,y), y′=g(x,y)y' = g(x,y) is understood equilibrium by equilibrium: find all points where f=g=0f = g = 0, compute the JACOBIAN (fxfygxgy)\begin{pmatrix}f_x & f_y\\ g_x & g_y\end{pmatrix}, and evaluate it at each.

Near the equilibrium the system behaves like u′=Ju\mathbf u' = J\mathbf u, so the linear classification (saddle, node, spiral — by trace and determinant) applies locally. The Hartman–Grobman principle: the linearization tells the truth except in the borderline center case.

Tip
Predator–prey and competition, read from Jacobians

Lotka–Volterra x′=x−xyx' = x - xy, y′=−y+xyy' = -y + xy: at the coexistence point (1,1)(1,1) the Jacobian is (0−110)\begin{pmatrix}0 & -1\\ 1 & 0\end{pmatrix} — trace 00, eigenvalues ±i\pm i: closed orbits, populations cycling forever out of phase.

Competing species x′=x(3−x−2y)x' = x(3 - x - 2y), y′=y(2−x−y)y' = y(2 - x - y): the interior equilibrium (1,1)(1,1) has Jacobian determinant −1<0-1 < 0 — a SADDLE, so coexistence is unstable and one species excludes the other. Ecology, decided by a determinant.

Example
Worked: the pendulum, honestly then linearized

The pendulum θ′′+sin⁡θ=0\theta'' + \sin\theta = 0 has equilibria at θ=0\theta = 0 (hanging) and θ=π\theta = \pi (balanced upright).

Step 1 — linearize at 00: sin⁡θ≈θ\sin\theta \approx \theta gives θ′′+θ=0\theta'' + \theta = 0, a center: small swings oscillate with period 2π2\pi.

Step 2 — linearize at π\pi: sin⁡θ≈−(θ−π)\sin\theta \approx -(\theta - \pi) gives u′′−u=0u'' - u = 0, eigenvalues ±1\pm1: a SADDLE — the inverted pendulum falls.

Step 3 — what linearization misses: the true period GROWS with amplitude, and the special orbits into the saddle separate swinging from over-the-top spinning. The phase portrait stitches all of this together.

Side note
Significance: where chaos begins

Linearization explains the local picture, but genuinely nonlinear systems can do what linear ones never can: multiple equilibria, limit cycles (heartbeats are one), and chaos (weather). The Jacobian toolkit of this topic is the entry door; the Lorenz attractor is what lives three doors down.

Try it
Try it: equilibria of a damped pendulum

For x′=yx' = y, y′=−sin⁡x−yy' = -\sin x - y, find the equilibria and classify (0,0)(0,0). Work: equilibria at (kπ,0)(k\pi, 0). Jacobian (01−cos⁡x−1)\begin{pmatrix}0&1\\-\cos x&-1\end{pmatrix}; at origin: trace −1-1, det 11, discriminant 1−4<01 - 4 < 0: a stable SPIRAL — the pendulum rings down, exactly as friction predicts.

Proofs & Why It Matters

Tip
Proof: why the Jacobian is the right linearization

Write x=x∗+u\mathbf x = \mathbf x^* + \mathbf u for a small displacement from the equilibrium x∗\mathbf x^*. Taylor: f(x∗+u)=f(x∗)+fxu1+fyu2+O(∣u∣2)f(\mathbf x^* + \mathbf u) = f(\mathbf x^*) + f_x u_1 + f_y u_2 + O(|\mathbf u|^2), and f(x∗)=0f(\mathbf x^*) = 0 by definition of equilibrium.

So u′=Ju+O(∣u∣2)\mathbf u' = J\mathbf u + O(|\mathbf u|^2): to first order the displacement obeys the linear system with matrix JJ, and when JJ's eigenvalues have nonzero real part the quadratic remainder is too weak to change the picture. ■\blacksquare

Tip
Proof: Lotka–Volterra orbits really close up

The quantity E(x,y)=x−ln⁡x+y−ln⁡yE(x, y) = x - \ln x + y - \ln y is conserved: dEdt=(1−1x)x′+(1−1y)y′=(x−1)(1−y)+(y−1)(x−1)⋅(−1)⋅(−1)…\tfrac{dE}{dt} = \left(1 - \tfrac1x\right)x' + \left(1 - \tfrac1y\right)y' = (x-1)(1 - y) + (y-1)(x - 1)\cdot(-1)\cdot(-1)\ldots — expanding, (1−1x)(x−xy)+(1−1y)(−y+xy)=(x−1)(1−y)+(y−1)(x−1)=0(1 - \tfrac1x)(x - xy) + (1 - \tfrac1y)(-y + xy) = (x-1)(1-y) + (y-1)(x-1) = 0.

Trajectories live on level curves of EE, which are closed loops around (1,1)(1,1) — so the center predicted by the linearization is genuine, not an artifact. ■\blacksquare

Going Deeper: Worked Problems

Example
Worked: a full phase-portrait classification

Classify every equilibrium of x′=x(2−x−y)x' = x(2 - x - y), y′=y(3−2x−y)y' = y(3 - 2x - y).

Step 1 — equilibria: each equation vanishes on two curves; intersections: (0,0)(0,0), (2,0)(2,0), (0,3)(0,3), and the interior solution of x+y=2x + y = 2, 2x+y=32x + y = 3: (1,1)(1, 1).

Step 2 — Jacobian: (2−2x−y−x−2y3−2x−2y)\begin{pmatrix}2 - 2x - y & -x\\ -2y & 3 - 2x - 2y\end{pmatrix}.

Step 3 — evaluate and classify: at (0,0)(0,0): diag(2,3)\mathrm{diag}(2,3), both positive — source. At (2,0)(2,0): (−2−20−1)\begin{pmatrix}-2&-2\\0&-1\end{pmatrix}, eigenvalues −2,−1-2,-1 — stable node. At (0,3)(0,3): (−10−6−3)\begin{pmatrix}-1&0\\-6&-3\end{pmatrix}, eigenvalues −1,−3-1,-3 — stable node. At (1,1)(1,1): (−1−1−2−1)\begin{pmatrix}-1&-1\\-2&-1\end{pmatrix}, det =1−2=−1<0= 1 - 2 = -1 < 0 — saddle.

Step 4 — the story: two stable "one species wins" states, an unstable coexistence saddle whose stable manifold is the knife-edge dividing the winners. Competitive exclusion, read off four small matrices.

Example
Worked: linearization that lies, and what settles it

The system x′=−y+x(x2+y2)x' = -y + x(x^2 + y^2), y′=x+y(x2+y2)y' = x + y(x^2 + y^2) has Jacobian (0−110)\begin{pmatrix}0&-1\\1&0\end{pmatrix} at the origin — a center. Is the origin actually stable?

Step 1 — the linearization says "center" (eigenvalues ±i\pm i), the one case Hartman–Grobman does NOT cover.

Step 2 — switch to polar coordinates: with r2=x2+y2r^2 = x^2 + y^2, compute rr′=xx′+yy′=x(−y+xr2)+y(x+yr2)=r4rr' = xx' + yy' = x(-y + xr^2) + y(x + yr^2) = r^4, so r′=r3>0r' = r^3 > 0.

Step 3 — the radius GROWS: trajectories spiral outward, and the origin is unstable — the cubic terms the linearization discarded decide everything.

Step 4 — the lesson: when the linear part gives a center, look for a conserved quantity (as with Lotka–Volterra) or a radial equation; only nonlinear information can break the tie.