Visual walkthrough — Phase plane analysis — trajectories, critical points

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Step 1 — A field of tiny arrows (WHAT a "system" really is)

WHAT. We have two rules that tell a moving dot how fast to go sideways and how fast to go up: $\dot{x}=f(x,y),\dot{y}=g(x,y).$ Read the dots as "rate of change": $\dot{x}$ means "how fast $x$ is changing right now". At every point $(x,y)$ of the plane these two numbers form one arrow $(\dot{x},\dot{y})$ — a little velocity that says where a dot placed here would head next.

WHY. Because there is no $t$ on the right side (this is an autonomous system), the arrow depends only on where you are, never on when you arrive. So the plane becomes a fixed field of arrows — a wind map that never changes.

PICTURE. Every point gets exactly one arrow. Drop a dot, follow the arrows, and it carves out a curve — a trajectory.

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Step 2 — The one point where the wind dies (the critical point)

WHAT. We hunt for a special location $(x_0,y_0)$ where both rules give zero: $\underset{\text{no sideways push}}{\underbrace{f(x_0,y_0)=0}}\text{and}\underset{\text{no up-down push}}{\underbrace{g(x_0,y_0)=0}}.$

WHY. If both parts of the arrow are zero, the arrow has zero length — a dot placed exactly there never moves. This is a critical point (equilibrium). It is the only kind of place where trajectories are allowed to meet, because everywhere else the single arrow forces a single direction.

PICTURE. Around the dead spot the surrounding arrows organize themselves — pointing in, out, or swirling. Those neighbouring arrows are the whole story of this page.

Recall Why can't trajectories cross elsewhere?

Two curves crossing at an ordinary point ::: would need two different arrows at one point — but the field gives exactly one. So only the zero-arrow (critical) points may host meetings.

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Step 3 — Zoom in until the curve looks straight (WHY we linearize)

WHAT. We slide the origin onto the critical point. Define the displacement from equilibrium: $u=x-x_0,v=y-y_0.$ So $(u,v)=(0,0)$ is the critical point, and $u,v$ measure "how far off" we are.

WHY. Right next to the dead spot, $u$ and $v$ are tiny. A smooth vector field, seen under a strong enough microscope, looks flat — just like any curve looks like its tangent line up close. If we keep only the part that grows in proportion to $u$ and $v$ (throwing away $u^2$, $uv$, and smaller), we replace a hopeless nonlinear tangle with a clean linear one.

PICTURE. The curved field near the point, and the straight-arrow field that hugs it in a small disc — indistinguishable near the centre, drifting apart at the edge.

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Step 4 — The four slopes that build the local field (the Jacobian)

WHAT. First a piece of shorthand. A partial derivative measures how one output changes when you nudge one input and freeze the other. We write $f_x\equiv \frac{\partial f}{\partial x},f_y\equiv \frac{\partial f}{\partial y},g_x\equiv \frac{\partial g}{\partial x},g_y\equiv \frac{\partial g}{\partial y}.$ Read $f_x$ as "how $f$ (the sideways push) changes per unit step in $x$, holding $y$ still" — a slope, one output against one input.

Now: how does the sideways push $\dot{u}$ respond to a small step in $u$? In $v$? Ask the same for $\dot{v}$. Those four slopes, evaluated at the critical point, stack into one matrix:

$$\left(\begin{array}{c}\dot{u}\\ \dot{v}\end{array}\right)=\underset{J=\text{[[Linearization and the Jacobian matrix|Jacobian]]}}{\underbrace{\left(\begin{array}{cc}{f}_x& f_y\\ g_x& g_y\end{array}\right)}}\left(\begin{array}{c}u\\ v\end{array}\right).$$

Read the matrix term-by-term:

• $f_x$ = "how much horizontal push grows when I step right" (top-left),
• $f_y$ = "how much horizontal push grows when I step up" (top-right),
• $g_x$ = "how much vertical push grows when I step right" (bottom-left),
• $g_y$ = "how much vertical push grows when I step up" (bottom-right).

WHY. A tiny displacement $(u,v)$ produces the arrow $J\left(\begin{array}{c}u\\ v\end{array}\right)$. That single matrix $J$ is the local wind machine — feed it where you are, it returns where you'll head.

PICTURE. Each entry is a slope of one push against one direction; the four together bend the grid.

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Step 5 — Guessing $e^{\lambda t}$ (WHY exponentials, and what $\lambda$ means)

WHAT. We look for solutions of $\dot{\mathbf{u}}=J\mathbf{u}$ that keep the same shape and only stretch or shrink over time. The one function whose rate of change is proportional to itself is the exponential, so we try $\mathbf{u}(t)=\mathbf{v}{e}^{\lambda t}.$ Here $\mathbf{v}$ is a fixed direction (an eigenvector) and $\lambda$ is a number (the eigenvalue) controlling growth.

WHY this tool and no other. We need a function $q(t)$ with $\dot{q}=\lambda q$ — "changes at a rate proportional to its own size". That is exactly the defining property of $e^{\lambda t}$: $\frac{d}{dt}{e}^{\lambda t}=\lambda e^{\lambda t}$. No polynomial or trig alone does this. Substituting:

$$\lambda \mathbf{v}{e}^{\lambda t}=J\mathbf{v}{e}^{\lambda t}⟹J\mathbf{v}=\lambda \mathbf{v}.$$

The $e^{\lambda t}$ cancels, leaving the eigenvalue equation. So the allowed growth rates $\lambda$ are precisely the eigenvalues of $J$.

PICTURE. Along the special direction $\mathbf{v}$, motion is pure stretch: the arrow and the position stay parallel, scaled by $e^{\lambda t}$.

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Step 6 — Two numbers that hide inside $J$ (trace and determinant)

WHAT. To find $\lambda$ we solve $\det (J-\lambda I)=0$, which for a $2\times 2$ matrix always collapses to

$${\lambda }^2-\tau \lambda +\Delta =0,\tau =\mathrm{tr}J=f_x+g_y,\Delta =\det J=f_x{g}_y-f_y{g}_x.$$

Term-by-term meaning:

• $\tau$ (trace) $={\lambda }_1+{\lambda }_2$ — the sum of the two growth rates,
• $\Delta$ (determinant) $={\lambda }_1{\lambda }_2$ — the product of the two growth rates.

Solving the quadratic:

$${\lambda }_{1,2}=\frac{1}{2}(\underset{\text{sum}}{\underbrace{\tau }}\pm \underset{\text{the discriminant }D}{\underbrace{\sqrt{{\tau }^2-4\Delta }}}).$$

WHY. Instead of computing eigenvalues explicitly, we can read the type straight off $\tau$ and $\Delta$. The quantity under the root, $D={\tau }^2-4\Delta$, decides real-vs-complex; the signs of $\tau$ and $\Delta$ decide the rest. Two numbers classify everything — the Trace–determinant plane.

PICTURE. A dial: $\Delta$ on one axis, $\tau$ on the other, with the parabola $D=0$ carving regions.

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Step 7 — Reading the type off the two numbers (the generic cases)

WHAT. "Generic" means the interior of a region: nudging $\tau$ or $\Delta$ a hair does not change the type. Walk through those robust possibilities for $({\lambda }_1,{\lambda }_2)$, i.e. $D={\tau }^2-4\Delta$ strictly positive or strictly negative and $\tau \ne 0$:

Sign of $\Delta$	Sign of $D={\tau }^2-4\Delta$	Sign of $\tau$	Eigenvalues	Type & stability
$\Delta <0$	$D>0$ (forced)	—	real, opposite signs	saddle, unstable
$\Delta >0$	$D>0$	$\tau <0$	real, both $<0$	node sink, stable
$\Delta >0$	$D>0$	$\tau >0$	real, both $>0$	node source, unstable
$\Delta >0$	$D<0$	$\tau <0$	complex, $\mathrm{Re}<0$	spiral in, stable
$\Delta >0$	$D<0$	$\tau >0$	complex, $\mathrm{Re}>0$	spiral out, unstable

WHY each row.

• $\Delta <0$ ⇒ product ${\lambda }_1{\lambda }_2<0$ ⇒ opposite signs ⇒ one direction repels, one attracts ⇒ saddle. (A negative product forces $D>0$, so eigenvalues are always real here.)
• $\Delta >0,D>0$ ⇒ two distinct real roots of the same sign; their common sign is the sign of $\tau$ ⇒ both in (sink) or both out (source): a node.
• $D<0$ ⇒ root has an imaginary part ⇒ rotation ⇒ spiral; whether it winds in or out is the sign of $\tau$ (which equals $2\mathrm{Re}\lambda$).

PICTURE. The three canonical generic portraits — node, saddle, spiral — with arrows showing the flow.

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Step 8 — The borderline cases you must not skip (degenerate inputs)

WHAT. Everything sitting on a boundary curve of the trace–determinant plane. Handle each, with its own flow picture:

(a) $\Delta =0$ — a whole line of resting points. One eigenvalue is $0$. For the linearized system the product ${\lambda }_1{\lambda }_2=0$, so the linear picture has an entire eigen-direction where nothing moves — a line of equilibria, with flow along the other eigen-direction. Caution: this describes the linearization only. The original nonlinear system may still have just an isolated equilibrium — the extra fixed points appear only if the higher-order terms also vanish along that line. When $\Delta =0$, the linear approximation has thrown away the very terms that decide the truth, so it is untrustworthy here.

(b) $D=0$ exactly — a repeated real root ${\lambda }_1={\lambda }_2=\lambda$. Here you must split into two very different sub-cases, and the distinction is whether $J$ is diagonalizable (has two independent eigen-directions) or defective (has only one):

Both are stable if $\lambda <0$, unstable if $\lambda >0$. The figure shows the two portraits side by side.

(c) $\tau =0,\Delta >0$ — the center. Linear theory predicts perfect closed loops, but this is the one case where linearization can lie: tiny nonlinear terms may turn the center into a slow spiral. Confirm with a Lyapunov/energy argument or the Poincaré–Bendixson machinery.

WHY. These live on the boundaries between regions, where an arbitrarily small nudge can flip the type. A reader who only knows the interior cases will misclassify a real system sitting on the fence — and, at $D=0$, will wrongly assume a single eigen-direction when $J=\lambda I$ actually offers infinitely many.

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The one-picture summary

Everything collapses into the trace–determinant plane: pick your critical point, compute $(\tau ,\Delta )$, drop a pin, read the type. The parabola ${\tau }^2=4\Delta$ splits nodes from spirals (and is the $D=0$ line of star/improper nodes); the axis $\Delta =0$ drops you into saddles below; the axis $\tau =0$ (above $\Delta =0$) is the neutral centre line.

Recall Feynman: the whole walkthrough in plain words

Picture a windy floor where the breeze at each spot never changes with time. Drop a leaf and it drifts along a path — a trajectory. Find the one spot with no wind at all: the leaf there just sits — a critical point. To learn what happens to nearby leaves, put a magnifying glass over that dead spot — the swirling wind looks straight up close. Four numbers (how each push changes as you step right or up) capture that straight wind — that's the Jacobian. Ask "along which directions does a leaf just get pushed straight in or straight out, and how fast?" The "how fast" numbers are the eigenvalues, because the only motion that scales itself is the exponential. Two summary numbers — the sum ($\tau$) and product ($\Delta$) of those speeds — tell you the type without ever solving anything: negative product means a saddle (one way in, one way out); same-sign product means a node (both in = sink, both out = source); a hidden rotation makes it a spiral; a perfectly balanced sum with positive product is a delicate center that the real, curved wind might quietly spoil into a slow spiral. And when the two speeds are exactly equal, look closer still: if the wind pushes straight in every direction it's a star node, but if only one straight direction survives, all leaves swerve to become tangent to that single line — an improper node.