Section 15.3 Linearization and the Jacobian

Nullclines told us that trajectories near \((30,25)\) circulate counterclockwise, but not whether they spiral in, spiral out, or close into loops. To answer that we need a sharper instrument, and the idea behind it is one you already used in Qualitative Methods.

There, to classify an equilibrium \(y = c\) of \(y' = f(y)\text{,}\) we did not study \(f\) globally. We checked the single number \(f'(c)\text{:}\) negative meant a sink, positive meant a source. Replacing a curve by its tangent line near the point of interest turned a hard question into an easy one.

This section does the same thing in two dimensions. Zoomed in far enough, a smooth nonlinear system is nearly linear — and we know everything about linear systems. The matrix that does the zooming is the Jacobian.

Subsection The Jacobian Matrix

In one variable, the tangent-line approximation near \(y = c\) reads \(f(y) \approx f(c) + f'(c)(y - c)\text{.}\) At an equilibrium \(f(c) = 0\text{,}\) so writing \(u = y - c\) for the displacement from equilibrium gives \(u' \approx f'(c)\,u\) — a linear equation whose solution \(u = u_0 e^{f'(c)t}\) decays exactly when \(f'(c) \lt 0\text{.}\)

With two variables there are four partial derivatives instead of one derivative, because each of \(f\) and \(g\) can respond to a nudge in either \(x\) or \(y\text{.}\) Collect them in a matrix.

📙 Definition 325. The Jacobian Matrix.

For the system \(x' = f(x,y)\text{,}\) \(y' = g(x,y)\text{,}\) the Jacobian matrix is

\begin{equation*} J(x,y) = \begin{bmatrix} \dfrac{\partial f}{\partial x} \amp \dfrac{\partial f}{\partial y} \\[8pt] \dfrac{\partial g}{\partial x} \amp \dfrac{\partial g}{\partial y} \end{bmatrix}. \end{equation*}

Evaluated at an equilibrium point \((x^*,y^*)\text{,}\) the constant matrix \(J(x^*,y^*)\) is the linearization of the system there.

Each entry is a partial derivative: differentiate with respect to one variable while treating the other as a constant. Reading the matrix row by row keeps the bookkeeping straight — the top row is “how \(f\) responds,” the bottom row is “how \(g\) responds.”

Now put displacements \(u = x - x^*\) and \(v = y - y^*\) in place of \(x\) and \(y\text{.}\) Because \(f\) and \(g\) both vanish at the equilibrium, the constant terms drop out and what remains is

\begin{equation*} \frac{d}{dt}\begin{bmatrix} u \\ v \end{bmatrix} \approx J(x^*,y^*) \begin{bmatrix} u \\ v \end{bmatrix}, \end{equation*}

which is exactly the linear system \(\vec{X}' = A\vec{X}\) from First-Order Linear Systems with \(A = J(x^*,y^*)\text{.}\) Every eigenvalue tool from that chapter applies — locally, near this one point.

✳️ Classifying an Equilibrium.

Find the equilibrium points by solving \(f = 0\) and \(g = 0\text{.}\)
Compute the four partial derivatives once, symbolically, to get \(J(x,y)\text{.}\)
Evaluate \(J\) at one equilibrium to get a matrix of numbers.
Find its eigenvalues from \(\det(J - rI) = 0\) and read the type from Connecting Eigenvalues to Phase Portraits.
Repeat step 3 for each remaining equilibrium — different points typically give different types.

For the record, the classification we borrow is the one built in First-Order Linear Systems:

Eigenvalues of \(J(x^,y^)\)	Type	Stability
real, both negative	node (sink)	stable
real, both positive	node (source)	unstable
real, opposite signs	saddle	unstable
complex, \(r = a \pm bi\) with \(a \lt 0\)	spiral sink	stable
complex, \(r = a \pm bi\) with \(a \gt 0\)	spiral source	unstable
purely imaginary, \(r = \pm bi\)	center (linearized)	inconclusive — see When Linearization Is Not Enough

Only the last row is new, and it is a warning rather than a verdict. Every row above it gives a type and a stability you can trust for the nonlinear system — with the precise condition stated in When Linearization Is Not Enough.

Subsection A Full Classification: Two Competing Species

We already found four equilibria for the competing-species model in 🌌 Example 319. Now we can say what happens at each of them.

🌌 Example 326. Classifying All Four Equilibria.

For

\begin{align*} \frac{dx}{dt} \amp = x(3 - x - 2y) = 3x - x^2 - 2xy\\ \frac{dy}{dt} \amp = y(2 - x - y) = 2y - xy - y^2 \end{align*}

classify each of the equilibria \((0,0)\text{,}\) \((3,0)\text{,}\) \((0,2)\text{,}\) and \((1,1)\text{.}\)

Solution.

The Jacobian. Expand first, then take partial derivatives:

\begin{align*} f_x \amp = 3 - 2x - 2y, \amp\quad f_y \amp = -2x,\\ g_x \amp = -y, \amp\quad g_y \amp = 2 - x - 2y, \end{align*}

\begin{equation*} J(x,y) = \begin{bmatrix} 3 - 2x - 2y \amp -2x \\ -y \amp 2 - x - 2y \end{bmatrix}. \end{equation*}

At \((0,0)\text{.}\)

\begin{equation*} J(0,0) = \begin{bmatrix} 3 \amp 0 \\ 0 \amp 2 \end{bmatrix} \end{equation*}

This matrix is diagonal, so the eigenvalues are the diagonal entries: \(r_1 = 3\text{,}\) \(r_2 = 2\text{.}\) Both are real and positive, so the origin is an unstable node. Sensible: with both populations tiny, neither competition term matters and both species grow.

At \((3,0)\text{.}\)

\begin{equation*} J(3,0) = \begin{bmatrix} 3 - 6 \amp -6 \\ 0 \amp 2 - 3 \end{bmatrix} = \begin{bmatrix} -3 \amp -6 \\ 0 \amp -1 \end{bmatrix} \end{equation*}

The matrix is triangular, so again the eigenvalues sit on the diagonal: \(r_1 = -3\text{,}\) \(r_2 = -1\text{.}\) Both negative, so this is a stable node. Species \(x\) alone at its carrying capacity is a state the system settles into.

At \((0,2)\text{.}\)

\begin{equation*} J(0,2) = \begin{bmatrix} 3 - 4 \amp 0 \\ -2 \amp 2 - 4 \end{bmatrix} = \begin{bmatrix} -1 \amp 0 \\ -2 \amp -2 \end{bmatrix} \end{equation*}

Triangular again, with eigenvalues \(r_1 = -1\) and \(r_2 = -2\text{:}\) another stable node. Species \(y\) alone is also a possible destination.

At \((1,1)\text{.}\)

\begin{equation*} J(1,1) = \begin{bmatrix} 3 - 4 \amp -2 \\ -1 \amp 2 - 3 \end{bmatrix} = \begin{bmatrix} -1 \amp -2 \\ -1 \amp -1 \end{bmatrix} \end{equation*}

This one needs the characteristic equation:

\begin{equation*} \det\begin{bmatrix} -1-r \amp -2 \\ -1 \amp -1-r \end{bmatrix} = (-1-r)^2 - 2 = r^2 + 2r - 1 = 0 . \end{equation*}

The quadratic formula gives \(r = -1 \pm \sqrt{2}\text{,}\) that is \(r_1 \approx -2.414\) and \(r_2 \approx 0.414\text{.}\) Real with opposite signs, so the coexistence point is a saddle — unstable.

The story. Two stable destinations, one unstable equilibrium sitting between them, and an unstable origin pushing everything away from extinction. Coexistence at \((1,1)\) is mathematically possible but physically fragile: any disturbance sends the system off along the saddle’s unstable direction to whichever single-species state it is nearer. Ecologists call this competitive exclusion, and the two-attractor structure that produces it is out of reach for any linear system.

📝: Free Eigenvalues.

Subsection Back to the Rabbits and Foxes

🌌 Example 327. Classifying the Predator-Prey Equilibria.

Classify the equilibria \((0,0)\) and \((30,25)\) of

\begin{align*} \frac{dR}{dt} \amp = 0.5R - 0.02RF\\ \frac{dF}{dt} \amp = -0.3F + 0.01RF \end{align*}

Solution.

The Jacobian.

\begin{align*} f_R \amp = 0.5 - 0.02F, \amp\quad f_F \amp = -0.02R,\\ g_R \amp = 0.01F, \amp\quad g_F \amp = -0.3 + 0.01R, \end{align*}

\begin{equation*} J(R,F) = \begin{bmatrix} 0.5 - 0.02F \amp -0.02R \\ 0.01F \amp -0.3 + 0.01R \end{bmatrix}. \end{equation*}

At the origin.

\begin{equation*} J(0,0) = \begin{bmatrix} 0.5 \amp 0 \\ 0 \amp -0.3 \end{bmatrix} \end{equation*}

Eigenvalues \(r_1 = 0.5\) and \(r_2 = -0.3\text{:}\) real, opposite signs, so the empty meadow is a saddle. The interpretation is exact: introduce a few rabbits and they grow at rate \(0.5\) (the unstable direction); introduce a few foxes with no rabbits to eat and they die off at rate \(0.3\) (the stable direction).

At the coexistence point. Substituting \(R = 30\text{,}\) \(F = 25\text{,}\) both diagonal entries vanish:

\begin{equation*} J(30,25) = \begin{bmatrix} 0.5 - 0.5 \amp -0.6 \\ 0.25 \amp -0.3 + 0.3 \end{bmatrix} = \begin{bmatrix} 0 \amp -0.6 \\ 0.25 \amp 0 \end{bmatrix} \end{equation*}

The characteristic equation is

\begin{equation*} \det\begin{bmatrix} -r \amp -0.6 \\ 0.25 \amp -r \end{bmatrix} = r^2 + 0.15 = 0, \end{equation*}

giving \(r = \pm i\sqrt{0.15} \approx \pm 0.387 i\text{.}\) Purely imaginary — a linearized center.

So the linearization predicts closed orbits: populations that cycle forever without settling down or blowing up, with period \(2\pi/\sqrt{0.15} \approx 16.2\) months. That is consistent with the counterclockwise circulation the nullclines showed in Figure 323, and it now supplies the timing the nullclines could not.

But a purely imaginary pair is the one case the table flags as inconclusive. The next subsection explains why, and what happens to be true here.

Subsection When Linearization Is Not Enough

Linearization throws away everything except the first-order behavior at a point. Usually the discarded higher-order terms are too small near the equilibrium to matter. The exception is when the first-order behavior is neutral — when the real part of an eigenvalue is exactly zero, so nothing in the linear picture is growing or decaying and the discarded terms are the only thing left to decide the outcome.

✳️ When the Linearization Can Be Trusted.

If every eigenvalue of \(J(x^*,y^*)\) has nonzero real part, the equilibrium takes its type and its stability from the linearization: a linearized saddle really is a saddle, a linearized spiral sink really is a spiral sink. The trajectories themselves are not the linear ones — the discarded terms bend them — but the qualitative picture near the point is the same. If any eigenvalue has zero real part — a purely imaginary pair, or a zero eigenvalue — the linearization is inconclusive and the nonlinear terms must be examined directly.

The following pair of systems shows why the warning is not hypothetical. Both have the same Jacobian at the origin, hence the same linearized prediction, and they behave in opposite ways.

🌌 Example 328. Same Linearization, Opposite Fates.

Consider

\begin{align*} \text{(A)} \quad x' \amp = -y - x^3, \amp\quad y' \amp = x - y^3\\ \text{(B)} \quad x' \amp = -y + x^3, \amp\quad y' \amp = x + y^3 \end{align*}

Show that both linearize to a center at the origin, then determine what each system actually does.

Solution.

The linearizations agree. For system (A), \(f_x = -3x^2\text{,}\) \(f_y = -1\text{,}\) \(g_x = 1\text{,}\) \(g_y = -3y^2\text{.}\) At the origin every cubic term vanishes, leaving

\begin{equation*} J(0,0) = \begin{bmatrix} 0 \amp -1 \\ 1 \amp 0 \end{bmatrix}, \qquad \det(J - rI) = r^2 + 1 = 0, \qquad r = \pm i . \end{equation*}

System (B) differs only in the signs of the cubic terms, which also vanish at the origin, so it produces the identical matrix and the identical eigenvalues.

Watch the distance from the origin instead. Let \(\rho = x^2 + y^2\text{,}\) the squared distance. Differentiating and substituting the system for (A):

\begin{align*} \frac{d\rho}{dt} \amp = 2x x' + 2y y'\\ \amp = 2x(-y - x^3) + 2y(x - y^3)\\ \amp = -2xy - 2x^4 + 2xy - 2y^4\\ \amp = -2(x^4 + y^4) . \end{align*}

The \(xy\) terms — the entire linear part — cancel, and what remains is negative at every point except the origin. So \(\rho\) strictly decreases: trajectories spiral inward. The origin is a stable spiral sink.

The same computation for (B) gives \(d\rho/dt = +2(x^4 + y^4) \gt 0\text{:}\) trajectories spiral outward, and the origin is an unstable spiral source.

Two systems, one Jacobian, opposite stability. The cubic terms the linearization discarded were the deciding factor — precisely because the linear part contributed nothing to growth or decay.

What about the rabbits and foxes? There the linearized center turns out to be a genuine center. The predator-prey system has a conserved quantity — a combination of \(R\) and \(F\) that stays constant along every trajectory — and its level curves are closed loops around \((30,25)\text{.}\) Establishing that takes a separate argument, which is the point: the linearization raised the question, but could not answer it.

📝: A Structural Warning About Centers.

Checkpoint 329. 🤔💭 Linearization Reading Questions.

(a) 📖❓ What Goes in the Jacobian?

The entry in row 1, column 2 of \(J(x,y)\) is

\(\partial f / \partial y\)
Correct — row 1 is \(f\text{,}\) column 2 is \(y\text{.}\)
\(\partial g / \partial x\)
That is row 2, column 1.
\(\partial f / \partial x\)
That is row 1, column 1.
\(f(x,y) \cdot g(x,y)\)
The Jacobian holds derivatives, not products of the functions themselves.

(b) 📖❓ Where Do You Evaluate It?

To classify an equilibrium, the Jacobian must be evaluated

at that equilibrium point.
Correct — the linearization is local, so each equilibrium gets its own matrix.
at the origin, always.
Only if the origin happens to be the equilibrium you are studying.
at any convenient point; the answer is the same everywhere.
The entries depend on \(x\) and \(y\text{.}\) Different points give genuinely different matrices — that is why one system can have a saddle and a node at once.
at \(t = 0\text{.}\)
The Jacobian depends on position in the phase plane, not on time.

(c) 📖❓ Read the Eigenvalues.

At an equilibrium, \(J\) has eigenvalues \(r = -2\) and \(r = 5\text{.}\) The equilibrium is a

saddle, and it is unstable.
Correct — real eigenvalues of opposite signs give a saddle, and the positive one guarantees instability.
stable node.
A stable node needs both eigenvalues negative.
spiral source.
Spirals require complex eigenvalues. These are real.
center.
A center needs purely imaginary eigenvalues.

(d) 📖❓ Which Verdicts Are Trustworthy?

Select every case where the linearization’s verdict carries over to the nonlinear system.

\(r = -1, -4\)
\(r = -0.2 \pm 3i\)
\(r = 2, -7\)
\(r = \pm 4i\)
Purely imaginary: zero real part, so the verdict is inconclusive.
\(r = 0, -3\)
A zero eigenvalue also has zero real part, so this case is inconclusive too.

(e) 📖❓ Why Two Systems Can Share a Jacobian.

In 🌌 Example 328, why do systems (A) and (B) produce the same Jacobian at the origin?

They differ only in cubic terms, whose partial derivatives vanish at the origin.
Correct — the linearization sees only first-order behavior, and the cubic terms contribute none of it there.
They are the same system written two ways.
They are genuinely different: one spirals inward and the other outward.
Every nonlinear system has the same Jacobian at the origin.
Not at all — the competing-species model has a completely different one.
The Jacobian ignores the second equation.
The bottom row of the Jacobian comes entirely from the second equation.

You have attempted of activities on this page.

Prev Top Next