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The Jacobian at an equilibrium of a nonlinear system, the theorem that makes its verdict binding, and the one case — an eigenvalue with no real part — where the discarded terms decide the answer.
Paper packet. Every task here also exists on screen, where it is checked automatically; answers written on paper are not assessed by Nydus. When you are back at a device, enter your answers there.
By the end of this lesson you will be able to find every equilibrium of a two-dimensional nonlinear system, write its Jacobian as a matrix of partial derivatives and evaluate that at each equilibrium, classify the result by trace and determinant, and say from the eigenvalues whether the theorem licenses that classification for the nonlinear system or leaves the question open.
You can classify the origin of a linear system from the trace and determinant of its coefficient matrix, and you have met partial derivatives. Almost no system worth modelling is linear, so this lesson is how the previous one gets used: by finding, near each equilibrium, the linear system that the real one resembles.
| Term | What it means |
|---|---|
| Equilibrium | A point of $x' = f(x, y)$, $y' = g(x, y)$ where $f$ and $g$ both vanish. |
| Jacobian | The matrix of first partial derivatives, with $f_x$, $f_y$ on the top row and $g_x$, $g_y$ below. |
| Linearising | Replacing the system near an equilibrium by the linear system its Jacobian defines. |
| Hyperbolic | No eigenvalue of the Jacobian has zero real part; the hypothesis under which linearising decides the picture. |
| Limit cycle | A closed trajectory that nearby trajectories approach, found where linearising shows an unstable spiral. |
Near an equilibrium $(x_0, y_0)$, write $u = x - x_0$ and $v = y - y_0$ for the displacement. Taylor's theorem in two variables gives
$$f(x, y) = f_x u + f_y v + \text{(quadratic and smaller)},$$
with the partial derivatives evaluated at the equilibrium — the constant term is absent precisely because $f$ vanishes there. The same holds for $g$, so
$$\begin{pmatrix} u' \\ v' \end{pmatrix} \approx J \begin{pmatrix} u \\ v \end{pmatrix}, \qquad J = \begin{pmatrix} f_x & f_y \\ g_x & g_y \end{pmatrix}.$$
Close enough to the equilibrium the displacement is small, the quadratic terms are smaller still, and the linear system is a good description.
The theorem (Hartman and Grobman), in words. If no eigenvalue of $J$ has zero real part, then near that equilibrium the nonlinear trajectories can be bent continuously onto the linear ones. A linear saddle means a nonlinear saddle; a linear stable spiral means a nonlinear stable spiral.
And the exception. If any eigenvalue has real part exactly zero — the centre case above all — the theorem says nothing, and it is not being cautious. The terms that were discarded genuinely decide the answer.
Another way: picture
A map of a mountain range printed at ever greater magnification. Zoom far enough into one saddle and the contours become straight and evenly spaced: that flat picture is the linearisation. The theorem says the zoomed-in picture is honest about the pass. At a perfectly level lake, though, zooming in tells you nothing about whether the surface drains, because the slope you are looking at is zero and the answer lies in the curvature you threw away.
Another way: steps
Take the predator-prey system $x' = x(3 - y)$, $y' = y(x - 2)$, with $x$ the prey and $y$ the predators. Both vanish at $(0, 0)$ and at $(2, 3)$.
The Jacobian is $\begin{pmatrix} 3 - y & -x \\ y & x - 2 \end{pmatrix}$.
At the origin it is $\begin{pmatrix} 3 & 0 \\ 0 & -2 \end{pmatrix}$: determinant $-6$, so a saddle. No eigenvalue is imaginary, so the theorem applies, and the reading is biological as well as mathematical — from a state of near-extinction the prey recovers and the predators die out, which is the saddle's two directions.
At $(2, 3)$ it is $\begin{pmatrix} 0 & -2 \\ 3 & 0 \end{pmatrix}$: trace $0$, determinant $6$, so the linearisation says centre — and the theorem declines. As it happens this particular model does have closed orbits, but that is proved by a conserved quantity, not by the Jacobian.
Substituting before differentiating. Putting the equilibrium into $f$ and $g$ first gives zero and zero, and differentiating those gives a matrix of zeros. The partial derivatives are taken as functions and evaluated afterwards.
Finding only the obvious equilibrium. The origin is usually one; it is rarely the interesting one, and a system with three equilibria needs three classifications.
Trusting a centre. A linear centre in a nonlinear system is a question, not an answer. Adding any small dissipation to the model turns it into a slow spiral, and nothing in the Jacobian can see that.
Reading the local picture as global. Linearisation describes a neighbourhood. Trajectories far from every equilibrium, and closed orbits that encircle them, are beyond what any Jacobian reports.
Classifying the equilibria of a nonlinear system $x' = f(x, y)$, $y' = g(x, y)$ follows a fixed routine.
$$J(x, y) = \begin{pmatrix} f_x & f_y \\ g_x & g_y \end{pmatrix}.$$
Expand products first, then differentiate term by term. 3. Evaluate $J$ at each equilibrium separately. 4. Classify each with the trace and determinant, exactly as for a linear system. 5. Decide whether to trust it. If no eigenvalue has zero real part, the nonlinear system looks like its linearisation near that point: saddles, nodes and spirals carry over, with the same stability. If an eigenvalue is purely imaginary or zero, the linear picture is only a question, and something else, such as polar coordinates or a conserved quantity, must decide.
Why linearisation works. Near an equilibrium $(x_0, y_0)$, write $x = x_0 + u$ and $y = y_0 + v$ with $u$, $v$ small. Taylor's theorem gives $f \approx f_xu + f_yv$ and $g \approx g_xu + g_yv$, because $f$ and $g$ vanish at the equilibrium and the next terms are products of small numbers. So $(u, v)' \approx J(u, v)$, a linear system. When its eigenvalues have non-zero real parts, the neglected terms are too small to change the picture; when a real part is zero, the neglected terms are all that is left to decide.
How to check the answer. Put each equilibrium back into $f$ and $g$ and confirm both are zero. Check one entry of the Jacobian by differentiating again. For a triangular Jacobian the eigenvalues are simply its diagonal entries, which checks the trace and determinant for free. Finally, the classification should make sense in the model: in a competition model, if the two species cannot coexist, the coexistence point is typically a saddle and each single-species point is stable or a saddle according to which competitor is stronger.
The fur-trading records of the Hudson's Bay Company show the numbers of snowshoe hares and of the lynx that eat them rising and falling in a cycle of about ten years, the lynx peaks trailing the hare peaks. The predator-prey equations $x' = x(a - by)$, $y' = y(cx - d)$, with $x$ the prey and $y$ the predators, reproduce that cycle, and linearisation explains its timing.
The equilibrium with both species present is $\left(\frac{d}{c}, \frac{a}{b}\right)$. There the Jacobian is $\begin{pmatrix} 0 & -\frac{bd}{c} \\ \frac{ca}{b} & 0 \end{pmatrix}$, with trace $0$ and determinant $ad$, so the linearisation is a centre with eigenvalues $\pm\sqrt{ad}\,i$ and small cycles have period $\frac{2\pi}{\sqrt{ad}}$. With this lesson's numbers, $a = 3$ and $d = 2$, the period is $\frac{2\pi}{\sqrt{6}} \approx 2.6$ time units. The period depends only on the prey's growth rate and the predators' death rate, not on how often they meet.
The linearisation's centre is the case the theorem does not settle, and here a conserved quantity shows the cycles really do close. Real populations add crowding among the prey, which adds a small negative trace and turns the centre into a slow spiral towards balance. Which of these a real system does is decided by the small terms, exactly as the lesson warns.
A pendulum's angle obeys $\theta'' = -\frac{g}{L}\sin\theta$, which is nonlinear. Linearising at the bottom replaces $\sin\theta$ by $\theta$, giving a centre with angular frequency $\sqrt{g/L}$ and period $2\pi\sqrt{L/g}$, which does not depend on how far it swings. For $L = 1$ metre the period is $2\pi\sqrt{1/9.8} \approx 2.007$ seconds, and a clock built on it ticks once a second on each swing.
The independence of amplitude is exactly what a clock needs, and it is only true for small swings: the nonlinear terms lengthen the period by about $\frac{\theta_0^{2}}{16}$ of itself for a swing of amplitude $\theta_0$ radians. At $5$ degrees, $0.087$ radians, the correction is $0.05\%$, about $40$ seconds a day. Clockmakers keep the swing small and constant for exactly this reason. The linearisation is not an approximation to be apologised for; it is the regime the device is designed to live in, and the lesson's theorem is the guarantee that the real pendulum behaves like its linear model near the bottom.
Some chemical mixtures, such as the Belousov-Zhabotinsky reaction, change colour back and forth for minutes on end instead of settling. Their concentrations obey a nonlinear system whose equilibrium, linearised, has a positive trace and a positive determinant: an unstable spiral. Small disturbances grow while spiralling, the nonlinear terms keep them from growing without bound, and the concentrations settle onto a closed loop they travel for ever, a limit cycle. Linearisation cannot find the loop, but it is what shows the equilibrium cannot hold, which is the first question a chemist designing an oscillating reaction asks. The same argument, an unstable equilibrium inside a region the solutions cannot leave, explains the steady rhythm of a heart's pacemaker cells and the regular pulses of some stars.
Because the method is mechanical — differentiate, substitute, classify — it is easy to treat its output as the answer and move on. The theorem is what makes the output mean anything, and it is conditional. Where no eigenvalue sits on the imaginary axis the conclusion is genuinely strong: not merely that the linear picture is close, but that the two pictures are the same picture bent. Where an eigenvalue does sit there the conclusion is not weaker, it is absent, and the honest report is the linearisation does not decide this. The cubic example above shows why that is not pedantry: three systems with identical Jacobians behave in three different ways, and no amount of care with the linear algebra could have separated them.
Take $x' = x(4 - x - y)$, $y' = y(3 - x - y)$. Set both to zero.
$x = 0 \text{ or } x + y = 4; \quad y = 0 \text{ or } x + y = 3 \quad\Rightarrow\quad (0, 0),\ (4, 0),\ (0, 3)$
The lines $x + y = 4$ and $x + y = 3$ are parallel, so there is no coexistence point.
Write the Jacobian as a function of $x$ and $y$.
$J = \begin{pmatrix} 4 - 2x - y & -x \\ -y & 3 - x - 2y \end{pmatrix}$
Expand $x(4 - x - y) = 4x - x^{2} - xy$ before differentiating. One formula, to be evaluated three times.
Evaluate at the origin and classify.
$J(0, 0) = \begin{pmatrix} 4 & 0 \\ 0 & 3 \end{pmatrix}: \quad t = 7, \quad m = 12, \quad t^{2} - 4m = 1 > 0 \;\Rightarrow\; \text{unstable node}$
Both eigenvalues, $4$ and $3$, are positive.
Evaluate at $(4, 0)$ and classify.
$J(4, 0) = \begin{pmatrix} -4 & -4 \\ 0 & -1 \end{pmatrix}: \quad m = 4, \quad t = -5, \quad t^{2} - 4m = 9 > 0 \;\Rightarrow\; \text{stable node}$
A triangular matrix shows its eigenvalues on the diagonal: $-4$ and $-1$.
Evaluate at $(0, 3)$ and classify.
$J(0, 3) = \begin{pmatrix} 1 & 0 \\ -3 & -3 \end{pmatrix}: \quad m = -3 < 0 \;\Rightarrow\; \text{saddle}$
Three equilibria, three different kinds: the first species wins from almost everywhere.
Take $x' = -y + cx(x^{2} + y^{2})$, $y' = x + cy(x^{2} + y^{2})$. Find the Jacobian at the origin.
$J(0, 0) = \begin{pmatrix} 0 & -1 \\ 1 & 0 \end{pmatrix}$
The cubic terms and their derivatives vanish at the origin, whatever $c$ is.
Classify the linearisation.
$t = 0, \quad m = 0 - (-1)(1) = 1 \quad\Rightarrow\quad \lambda = \pm i: \text{a centre, for every } c$
The same linear picture for every $c$, and the case the theorem refuses to decide.
Change to polar coordinates: compute $rr' = xx' + yy'$.
$rr' = x\left(-y + cxr^{2}\right) + y\left(x + cyr^{2}\right) = cr^{2}(x^{2} + y^{2}) = cr^{4}$
The $-xy$ and $+xy$ terms cancel, and $x^{2} + y^{2} = r^{2}$.
Divide by $r$, and compute the angle's rate the same way.
$r' = cr^{3}, \qquad r^{2}\theta' = xy' - yx' = x^{2} + y^{2} = r^{2} \;\Rightarrow\; \theta' = 1$
The $c$ terms cancel in the angle, so every trajectory turns at the same rate.
Read the truth from the sign of $c$.
$c > 0 \Rightarrow r \text{ grows: unstable spiral}; \quad c = 0 \Rightarrow \text{centre}; \quad c < 0 \Rightarrow r \text{ shrinks: stable spiral}$
The sign of $c$, which the Jacobian never saw, decides: the exception is real, not a technicality.
Take $x' = x(4 - 2x - y)$, $y' = y(5 - x - 2y)$. Find the equilibrium with both species present by setting both brackets to zero.
$$\begin{aligned} 2x + y &= 4 \\ x + 2y &= 5 \end{aligned}$$
With $x \ne 0$ and $y \ne 0$, each bracket must vanish.
Solve the pair: double the second equation and subtract the first.
$2x + 4y - (2x + y) = 10 - 4 \;\Rightarrow\; 3y = 6 \;\Rightarrow\; y = 2, \quad x = 5 - 2y = 1$
The coexistence point is $(1, 2)$, inside the positive quadrant, so it is biologically meaningful.
Write the Jacobian as a function.
$J = \begin{pmatrix} 4 - 4x - y & -x \\ -y & 5 - x - 4y \end{pmatrix}$
From $4x - 2x^{2} - xy$ and $5y - xy - 2y^{2}$, differentiated term by term.
Evaluate it at $(1, 2)$.
$J(1, 2) = \begin{pmatrix} 4 - 4 - 2 & -1 \\ -2 & 5 - 1 - 8 \end{pmatrix} = \begin{pmatrix} -2 & -1 \\ -2 & -4 \end{pmatrix}$
Substitute only after differentiating.
Compute the trace, the determinant and the discriminant.
$t = -6, \qquad m = (-2)(-4) - (-1)(-2) = 8 - 2 = 6, \qquad t^{2} - 4m = 36 - 24 = 12 > 0$
Real eigenvalues with a positive product and a negative sum: both negative.
Classify, and confirm the theorem applies.
$\lambda = \dfrac{-6 \pm \sqrt{12}}{2} = -3 \pm \sqrt{3} \approx -1.27, \ -4.73 \quad\Rightarrow\quad \text{stable node}$
Neither eigenvalue has zero real part, so the nonlinear system is a stable node there too.
Say what it means for the two species.
$\text{starts near } (1, 2) \ \longrightarrow \ (1, 2)$
Here each species limits itself more than it limits the other, and the populations settle together rather than one excluding the other.
Set both to zero and solve.
$x - y = 0 \Rightarrow y = x; \quad x - x^{2} = x(1 - x) = 0 \Rightarrow (0, 0),\ (1, 1)$
Substitute one equation into the other.
Write the Jacobian and evaluate it at the origin.
Evaluate at $(1, 1)$ and classify.
Put the steps of classifying an equilibrium of $x' = x(6 - y)$, $y' = y(x - 5)$ into the order they must be done.
Number the steps in order (write the number in the box):
Complete the worked solution: linearise $x' = x(5 - y)$, $y' = y(x - 6)$ at its interior equilibrium.
Expand both right-hand sides.
$f = 5x - xy, \qquad g = xy - 6y$
A sum of simple terms is easy to differentiate.
Take the four partial derivatives.
$f_x = 5 - y, \quad f_y = -x, \quad g_x = y, \quad g_y = x - 6$
Differentiate in one variable while holding the other fixed.
Find the interior equilibrium: set the brackets to zero.
$5 - y = 0,\ x - 6 = 0 \quad\Rightarrow\quad (x, y) = (6, 5)$
Away from the axes, each product is zero only when its bracket is.
Substitute the equilibrium into the Jacobian.
$J = \begin{pmatrix} 5 - 5 & -6 \\ 5 & 6 - 6 \end{pmatrix}$
Differentiate first, then substitute.
Add the diagonal for the trace.
$t = (5 - 5) + (6 - 6) =$ t
The trace is the sum of the diagonal.
Cross-multiply for the determinant.
$m = 0 \cdot 0 - (-6)(5) =$ m
Subtracting a negative product adds it.
Read the verdict and its limits.
$t = 0,\ m > 0 \quad\Rightarrow\quad \text{a linear centre, which the theorem does not decide}$
An eigenvalue with zero real part is the one case the linearisation cannot settle.
At an equilibrium of a nonlinear system the Jacobian has purely imaginary eigenvalues. Does the linear picture settle what the nonlinear system does nearby?
For the system $x' = x(6 - y)$, $y' = y(x - 2)$, write the Jacobian matrix evaluated at the equilibrium $(0, 0)$.
This task has no paper form; do it on a device.
The system $x' = x(3 - y)$, $y' = y(x - 6)$ has an equilibrium at $(6, 3)$. What is the determinant of the Jacobian there?
Answer:
A damped pendulum obeys $x' = y$, $y' = -8\sin x - 4y$. Near the equilibrium $(0, 0)$ the trajectories spiral. Fill in the trace and determinant of the linearisation there, and the real part $\alpha$ and angular frequency $\beta$ of its eigenvalues $\alpha \pm \beta i$.
| Value | |
|---|---|
| The trace $t$ | |
| The determinant $m$ | |
| The real part $\alpha$ | |
| The angular frequency $\beta$ |
The system $x' = x(6 - y)$, $y' = y(x - 3)$ has equilibria at $(0, 0)$ and at $(3, 6)$. Fill in the trace and the determinant of the Jacobian at each.
| Trace | Determinant | |
|---|---|---|
| At the origin | ||
| At the interior equilibrium |
Lesson test: one question per skill, one attempt each, no hints. Your answers are checked when you submit.
A damped pendulum obeys $x' = y$, $y' = -25\sin x - 6y$. Near the equilibrium $(0, 0)$ the trajectories spiral. Fill in the trace and determinant of the linearisation there, and the real part $\alpha$ and angular frequency $\beta$ of its eigenvalues $\alpha \pm \beta i$.
| Value | |
|---|---|
| The trace $t$ | |
| The determinant $m$ | |
| The real part $\alpha$ | |
| The angular frequency $\beta$ |
You can linearise a nonlinear system at each of its equilibria and classify them, and you can say when the verdict is binding. Say in your own words why a linear centre is never evidence that the nonlinear system has one.
15. Your turn: classify the equilibria of $x' = y - x^{2}$, $y' = x - y$, step 2
$J = \begin{pmatrix} -2x & 1 \\ 1 & -1 \end{pmatrix}, \quad J(0, 0) = \begin{pmatrix} 0 & 1 \\ 1 & -1 \end{pmatrix}: m = 0 - 1 = -1 \Rightarrow \text{saddle}$
Differentiate once, evaluate twice.
15. Your turn: classify the equilibria of $x' = y - x^{2}$, $y' = x - y$, step 3
$J(1, 1) = \begin{pmatrix} -2 & 1 \\ 1 & -1 \end{pmatrix}: t = -3,\ m = 2 - 1 = 1,\ t^{2} - 4m = 5 > 0 \Rightarrow \text{stable node}$
No eigenvalue is imaginary at either point, so the theorem applies and both verdicts stand.