Nonlinear Systems & Linearization

Near the equilibrium the system behaves like $\mathbf u' = J\mathbf u$, so the linear classification (saddle, node, spiral — by trace and determinant) applies locally. The Hartman–Grobman theorem makes this precise: when no eigenvalue of $J$ has zero real part, the nonlinear picture near the equilibrium is a continuously deformed copy of the linear one. The exceptions are the borderline cases — purely imaginary eigenvalues (a linear center) or a zero eigenvalue — where the discarded nonlinear terms decide.

Competing species $x' = x(3 - x - 2y)$, $y' = y(2 - x - y)$: the interior equilibrium $(1,1)$ has Jacobian determinant $-1 < 0$ — a saddle, so coexistence is unstable and one species excludes the other. Ecology, decided by a determinant. The figure shows the competition model: the equilibria $(3, 0)$ and $(0, 2)$ are stable nodes, and the dashed curve into the saddle $(1, 1)$ — its stable manifold, leaving along the eigenvector $(\sqrt2, 1)$ of $\lambda = -1 - \sqrt2$ — splits the quadrant into the two basins.

Step 1 — linearize at $0$: $\sin\theta \approx \theta$ gives $\theta'' + \theta = 0$, a center: small swings oscillate with period $2\pi$.

Step 2 — linearize at $\pi$: $\sin\theta \approx -(\theta - \pi)$ gives $u'' - u = 0$, eigenvalues $\pm1$: 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.

Step 4 — the whole picture comes from the energy $E = \tfrac12y^2 - \cos\theta$ (with $y = \theta'$), which is constant along motion: closed loops around $(0, 0)$ are swings, the dashed curves $y = \pm2\cos\tfrac\theta2$ (energy $1$) run into the saddles at $\pm\pi$, and the wavy curves outside them are the pendulum spinning over the top.

The nullclines cut the plane into regions where the signs of $x'$ and $y'$ are fixed, so they give the rough direction of motion everywhere before any computation. The trace–determinant rules come from the eigenvalues $\lambda = \tfrac{\tau \pm\sqrt{\tau^2 - 4\Delta}}{2}$: $\Delta = \lambda_1\lambda_2 < 0$ forces real eigenvalues of opposite signs; when $\Delta > 0$ the sign of $\tau^2 - 4\Delta$ decides real versus complex, and $\tau = \lambda_1 + \lambda_2$ gives the sign of the real parts.

Step 3 — at $(0, 0)$ it is $\begin{pmatrix}0&1\\1&0\end{pmatrix}$ with determinant $-1$: a saddle. At $(\pm1, 0)$ it is $\begin{pmatrix}0&1\\-2&0\end{pmatrix}$, trace $0$, determinant $2$, eigenvalues $\pm i\sqrt2$: a linear center, inconclusive on its own.

Step 4 — settle it with a conserved quantity: $E = \tfrac12y^2 - \tfrac12x^2 + \tfrac14x^4$ has $\tfrac{dE}{dt} = y\,y' + (x^3 - x)\,x' = y(x - x^3) + (x^3 - x)y = 0$, and $(\pm1, 0)$ are strict minima of $E$ (value $-\tfrac14$), so the nearby level curves are closed loops: true centers.

Step 5 — the level curve $E = 0$ through the saddle is a figure eight, separating orbits that circle one well from orbits that circle both.

Step 3 — at $(2, 2)$: $\begin{pmatrix}1&-1\\4&0\end{pmatrix}$, trace $1$, determinant $4$, and $\tau^2 - 4\Delta = 1 - 16 < 0$: an unstable spiral. At $(-2, -2)$: $\begin{pmatrix}1&-1\\-4&0\end{pmatrix}$, determinant $-4 < 0$: a saddle.

Step 4 — both verdicts are reliable, because neither Jacobian has an eigenvalue with zero real part.

The chain rule gives $\tfrac{d}{dt}E(x(t), y(t)) = E_x x' + E_y y'$, so the condition says exactly that $E$ does not change along motion. For the energy, $E_x f + E_y g = V'(x)\,y + y\,(-V'(x)) = 0$: kinetic energy $\tfrac12y^2$ and potential energy $V$ trade off. A conserved quantity turns a two-dimensional problem into drawing level curves, which is how the borderline linear-center cases are settled.

Step 3 — $\dfrac{dE}{dt} = \left(1 - \dfrac3x\right)x(2 - y) + \left(1 - \dfrac2y\right)y(x - 3) = (x - 3)(2 - y) + (y - 2)(x - 3) = 0$.

Step 4 — $E$ has a strict minimum at $(3, 2)$, since each piece $s - a\ln s$ is smallest at $s = a$; so the level curves around $(3, 2)$ are closed and every orbit is a cycle, with the prey peaking before the predators.

Step 5 — the linearization's frequency $\sqrt6$ gives small cycles a period of about $\tfrac{2\pi}{\sqrt6} \approx 2.57$; larger cycles take longer.

Step 3 — so it passes over the top when $\tfrac12y_0^2 - 1 > 1$, that is, $y_0 > 2$.

Step 4 — the borderline $y_0 = 2$ gives $E = 1$, the separatrix $y = \pm2\cos\tfrac x2$ (dashed in the pendulum figure): the pendulum creeps toward the upright position and never quite reaches it.

Step 5 — for $y_0 < 2$ it swings between turning points where $y = 0$, i.e. $\cos x = 1 - \tfrac12y_0^2$; for example $y_0 = 1$ turns at $\cos x = \tfrac12$, $x = \pm\tfrac\pi3$.

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

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

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

Step 2 — Jacobian: $\begin{pmatrix}2 - 2x - y & -x\\ -2y & 3 - 2x - 2y\end{pmatrix}$.

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

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

Step 2 — switch to polar coordinates: with $r^2 = x^2 + y^2$, compute $rr' = xx' + yy' = x(-y + xr^2) + y(x + yr^2) = r^4$, so $r' = r^3 > 0$.

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. Solving $r' = r^3$ exactly: $r(t) = \left(r_0^{-2} - 2t\right)^{-1/2}$, which reaches infinity at $t = \tfrac{1}{2r_0^2}$, while the angle turns at rate $1$.