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Dynamical Systems Lab 1 - Fixed Points and Stability

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Before there are attractors, strange or otherwise, there are fixed points — states where a system is content to sit forever. Understanding when those states are stable, and why, is the first building block of all of nonlinear dynamics.

The autonomous ODE

A scalar, first-order, autonomous ordinary differential equation (ODE) takes the form

\[\dot{x} \equiv \frac{dx}{dt} = f(x)\]

Autonomous means $f$ depends only on the state $x$ and not on time $t$ explicitly. The function $f : \mathbb{R} \to \mathbb{R}$ is called the vector field (in one dimension, a scalar field) — it assigns an instantaneous velocity $\dot{x}$ to every position $x$.

Geometrically, think of $x(t)$ as a particle sliding along the real line. At any location $x$, the particle’s speed and direction are given by $f(x)$:

  • $f(x) > 0$: particle moves right ($x$ increasing)
  • $f(x) < 0$: particle moves left ($x$ decreasing)
  • $f(x) = 0$: particle is stationary

The existence and uniqueness theorem guarantees that for a Lipschitz-continuous $f$, there is exactly one trajectory through each initial condition $x(0) = x_0$. Crucially, distinct trajectories never cross each other (because if they did, they would share an initial condition and then violate uniqueness).

This no-crossing rule has a profound consequence: in 1D, every trajectory must converge to a fixed point, diverge to $\pm\infty$, or (in special cases) approach one in infinite time. Oscillations and chaos are impossible in 1D autonomous flows.

Fixed points

A fixed point (equilibrium, steady state, or rest point) $x^*$ satisfies

\[f(x^*) = 0\]

Fixed points are the zeroes of the vector field. If the system starts exactly at $x^$, it stays there forever: $x(t) = x^$ for all $t \geq 0$ is a valid (constant) solution.

The central question of local stability is: if the system starts slightly away from $x^*$, does it return, or does it drift further away?

Linear stability analysis

Step 1 — Perturb. Let $x(t) = x^* + \xi(t)$ where $\xi(t)$ is a small displacement. Substitute into the ODE:

\[\dot{\xi} = f(x^* + \xi)\]

Step 2 — Taylor expand $f$ around $x^*$:

\[f(x^* + \xi) = f(x^*) + f'(x^*)\,\xi + \frac{1}{2}f''(x^*)\,\xi^2 + \cdots\]

Since $f(x^*) = 0$ (definition of a fixed point) and $\xi$ is small, we keep only the linear term:

\[\dot{\xi} \approx f'(x^*)\,\xi\]

Step 3 — Solve. This is a linear ODE with constant coefficient $\lambda \equiv f’(x^*)$:

\[\xi(t) = \xi_0\, e^{\lambda t}\]

Step 4 — Classify. The behaviour is determined by the sign of $\lambda$:

$\lambda = f’(x^*)$ Fixed point type Behaviour
$\lambda < 0$ Stable (attractor, sink) $\xi \to 0$: perturbations decay exponentially
$\lambda > 0$ Unstable (repeller, source) $\xi \to \infty$: perturbations grow exponentially
$\lambda = 0$ Marginal (non-hyperbolic) Linearisation fails; need higher-order terms

The case $\lambda = 0$ requires inspecting $f’‘(x^)$ or even higher derivatives. If $f(x) = cx^n + \cdots$ near $x^$ with $c \neq 0$ and $n \geq 2$ even, the fixed point is half-stable (attracting from one side, repelling from the other); if $n$ is odd, it is either semi-stable or the fixed point is part of a continuum.

The phase line (phase portrait in 1D)

The phase line is a graphical representation of all the information in $f(x)$. Draw the real axis; mark each zero of $f$ (the fixed points); between zeros, draw arrows pointing right where $f > 0$ and left where $f < 0$.

This immediately reveals: - Which fixed points are stable (arrows on both sides pointing toward them) - Which are unstable (arrows on both sides pointing away) - The basin of attraction of each stable fixed point (the interval from which trajectories flow toward it) - Global behaviour for every initial condition

Importantly, to draw the phase line you only need to know the sign of $f$ between its zeros — you do not need to solve the ODE.

Worked examples

Example 1: saddle-node normal form, $\dot{x} = r - x^2$

Fixed points: $r - x^{2} = 0 \Rightarrow x^ = \pm\sqrt{r}$ (requires $r \geq 0$).

Stability: $f’(x) = -2x$, so $f’(\sqrt{r}) = -2\sqrt{r} < 0$ (stable) and $f’(-\sqrt{r}) = 2\sqrt{r} > 0$ (unstable).

  • $r < 0$: no fixed points; every trajectory flows to $-\infty$.
  • $r = 0$: one marginal fixed point at $x^* = 0$ ($f’(0) = 0$, $f(x) = -x^2 \leq 0$, so it is a half-stable point attracting from the left only).
  • $r > 0$: two fixed points born simultaneously in a saddle-node bifurcation.

Example 2: transcritical normal form, $\dot{x} = rx - x^2 = x(r-x)$

Fixed points: $x^* = 0$ and $x^* = r$ for all $r$.

Stability: $f’(x) = r - 2x$, so $f’(0) = r$ and $f’(r) = -r$.

  • $r < 0$: origin is stable ($f’(0) = r < 0$), $x^* = r$ is unstable.
  • $r > 0$: origin unstable, $x^* = r$ stable. The two branches exchange stability at $r = 0$.

This exchange — not creation or destruction, but transfer of stability — is the transcritical bifurcation.

Example 3: pitchfork normal form, $\dot{x} = rx - x^3$

Fixed points: $x^(r - x^{2}) = 0 \Rightarrow x^* = 0$ and $x^* = \pm\sqrt{r}$ (for $r > 0$).

Stability: $f’(x) = r - 3x^2$.

  • $f’(0) = r$: origin stable for $r < 0$, unstable for $r > 0$.
  • $f’(\pm\sqrt{r}) = r - 3r = -2r < 0$: the two bifurcating branches are always stable (for $r > 0$).

At $r = 0$ the origin loses stability and two new symmetric stable equilibria appear — the supercritical pitchfork bifurcation and the simplest example of spontaneous symmetry breaking. The system is symmetric under $x \to -x$, and the bifurcating state breaks this symmetry by choosing one branch.

Example 4: multiple fixed points, $\dot{x} = x(1-x)(x+0.8)$

This cubic has three explicit zeros: $x^* \in {-0.8, 0, 1}$. The sign of $f$ between them determines stability: - $x < -0.8$: $f < 0$, flows left to $-\infty$. - $-0.8 < x < 0$: $f > 0$, flows right toward $x^* = 0$. - $0 < x < 1$: $f < 0$, flows left toward $x^* = 0$. - $x > 1$: $f > 0$, flows right to $+\infty$.

So $x^* = 0$ is stable, $x^* = -0.8$ and $x^* = 1$ are unstable. Reading this directly from the phase line takes seconds; solving the ODE explicitly would not.

Existence and uniqueness, and why 1D cannot oscillate

The no-crossing theorem deserves emphasis. Suppose trajectory $A$ passes through $x_1$ at time $t_1$, and trajectory $B$ passes through $x_1$ at time $t_2$. By uniqueness, both trajectories are identical after that meeting point. Therefore two distinct trajectories can never occupy the same point.

Now consider an oscillation: the state $x(t)$ would have to return to a previous value. But then the trajectory would cross itself — impossible. Hence:

Theorem: Autonomous 1D flows cannot oscillate. Every trajectory either converges to a fixed point or diverges to $\pm\infty$.

Oscillations become possible in 2D (limit cycles, Lab 4), and chaos requires at least 3D (Lorenz, Lab 5).

Live demo

Five different 1D systems are loaded in the dropdown. For each system, you can adjust the parameter $r$ and the initial condition $x_0$. The top half of the canvas shows the graph of $f(x)$, with fixed points marked where $f$ crosses zero. The bottom half shows the phase line with flow arrows. The gold dot is the trajectory — it integrates $\dot{x} = f(x)$ in real time using a 4th-order Runge-Kutta solver.

Things to observe from the demo

Saddle-node ($\dot{x} = r - x^2$): - Set $r = -1$: no zero crossings in the $f(x)$ graph; all arrows on the phase line point left; the dot slides relentlessly toward $-\infty$. - Increase $r$ through zero: watch two fixed points appear simultaneously at $x^* = 0$ when $r = 0$, then separate symmetrically as $r$ grows. - At $r = 1$: the stable point (green) sits at $x^* = +1$ and the unstable (red) at $x^* = -1$. Try placing the IC between them vs. outside.

Pitchfork ($\dot{x} = rx - x^3$): - At $r = -1$: single green dot at origin. Place IC anywhere — it always returns to zero. - Increase $r$ through zero: the green dot at origin turns red exactly at $r = 0$, and two new green dots emerge symmetrically at $x^* = \pm\sqrt{r}$. - Notice the trajectory “chooses” the nearest stable branch — this is the essence of symmetry breaking.

Sine ($\dot{x} = \sin x$): - Fixed points at $x^* = n\pi$ for all integers $n$, alternating stable and unstable. - Place IC between $0$ and $\pi$: converges to $\pi$ (stable). Between $\pi$ and $2\pi$: converges to $2\pi$. The basins are the open intervals between unstable zeros.

Transcritical ($\dot{x} = rx - x^2$): - At any $r$, both $x^* = 0$ and $x^* = r$ coexist. Watch which is green (stable) swap as you cross $r = 0$.

Key takeaways

  • A fixed point $x^$ is stable if $f’(x^) < 0$ — the derivative of the vector field being negative means the field points toward $x^*$ from both sides.
  • The linearised equation $\dot{\xi} = \lambda \xi$ with $\lambda = f’(x^*)$ has exact solution $\xi_0 e^{\lambda t}$; all of local stability follows from the sign of this one number.
  • The phase line is a complete graphical solution: sign of $f$ between zeros tells you everything about long-time behaviour.
  • 1D autonomous flows cannot oscillate; they can only converge, diverge, or approach a fixed point asymptotically.