Chaos theory was revolutionary because it showed you didn’t need random shocks from external sources to get unpredictable and random-looking behaviour; it can arise from simple deterministic systems.
However, in the real world, most chaotic-seeming behaviour is just because of noise! It’s important to know the difference.
You can’t get chaotic behaviour out of a continuous system with less than three dimensions: there simply isn’t enough room in the phase space for chaotic trajectories (given the Poincare-Bendixson theorem). However, you can get chaotic behaviour out of 1D and 2D discrete systems.
These can be very tricky, so we’ll stick to continuous systems.
An important example of a discrete system however is the logistic function:
$$x_{n+1}=rx_n(b-x_n)$$
This is used in e.g. modelling population growth in discrete time-steps.
In 1D, we have a single variable which is described by a single differential equation, e.g.
$$\dot{x}=f(x)=\cos{x}$$
The state of the system is simply a point on a line. Given we live in $3+1$-dimensional space, we can visualise this on a 2D plane, where the $x$-axis is, well, $x$, and the y-axis is $\dot{x}$.
Then, visualising the system dynamics is simple: those parts of the line above the $x$-axis will cause the point to move to the right, and those below to the left.
Therefore, the fixed points $(x=x^*)$ of the system are everywhere the curve crosses the $x$-axis! In the above example, the fixed points are $x=\frac{\pi}{2}, \frac{3\pi}{2},\dots$
If the system enters a fixed point, it will never leave that fixed point. However, what if you give it a little push?
It turns out there are two kinds of fixed points:
One can also have half-stable points, which are stable in one direction and unstable in another. A classic example is if you have a saddle shape.
Determining which fixed point you have is as simple as seeing if the gradient is positive or negative; if positive, it’s unstable, and if negative, it’s stable.
More formally, let’s say we perturb a system: $x=x^*+\delta x$. Then, how does the perturbation evolve?
$$\frac{d}{dt}\delta x=\dot{x}-\dot{x^*}=f(x)$$
And therefore
$$\frac{d}{dt}\delta x=f(x^*+\delta x)\approx \frac{df}{dx}\delta x$$
Giving us the three cases:
$$\frac{d\dot x}{dx}\begin{cases}>0&\text{unstable}\\ <0&\text{stable}\\ =0&\text{not enough info} \end{cases}$$
Finally, note that there can be no oscillations in 1D, since you can never overshoot a fixed point. As a result, the only possible long-term behaviour of a continuous 1D system is being at a fixed point, or diverging to infinity.
Aside from the dynamic variable under consideration, you also have control variables. Changing these can lead to bifurcations, where there is a qualitative change in phase space. For $1D$ systems, this can be a change in the number, position, or character of the fixed points.
Bifurcation diagrams show the variations of fixed points against the values of the control parameters. As before, it’s often only possible to plot complete bifurcation diagrams for $1D$ systems.
You put the control parameter value on the $x$-axis, and the positions of the fixed points on the $y$-axis. Stable points are marked by solid lines, and unstable ones by dashed.
There are different types of bifurcation points:
Pitchforks are associated with hysteresis, because once a system is forced to jump to a stable point from a newly unstable one, it is possible to vary parameters back to their original cases, and not return to the original state, but rather remain in the new state.
$$\dot{x}=\sin(x)\implies\text{ fixed points at }x^*=n\pi$$
$$\frac{d\dot{x}}{dx}=\cos{x}\implies \begin{cases} n\text{ even }&\implies\text{unstable}\ n\text{ odd }&\implies\text{stable} \end{cases}$$
$$\dot{x}=x^2-1\implies x^*=\pm 1$$
$$\frac{df}{dx}=2x\implies\begin{cases}x^*=1&\text{unstable}\\x^*=-1&\text{stable}\end{cases}$$
Note $x^*=-1$ is locally stable but not globally stable: a system starting at $x>1$ would instead diverge to infinity.
$$\dot{N}=rN\left(1-\frac{N}{K}\right)$$
$K$ is the carrying capacity, the maximum number. Note $K\to\infty$ simply gives exponential growth.
$$N^*=0, K$$
$$\frac{d\dot{N}}{dN}=r\left(1-\frac{N}{K}\right)-\frac{rN}{K}\implies\begin{cases}N^*=0&\text{unstable}\\N^*=K&\text{stable}\end{cases}$$
$$\dot{x}=x^{1/3}\implies x^*=0$$
$$\frac{d\dot{x}}{dx}=\frac{1}{3}x^{-2/3}$$
We have a problem: the derivative is discontinuous at the fixed point!
To figure out what happens with the system, we can solve it for time:
$$x(t)=x_0+\left(\frac{2t}{3}\right)^{3/2}$$
I think.
Thus we see that $x^*=0$ is an unstable fixed point, and if we start there, we have no unique solution; instead, there are infinitely many solutions.
$$\dot{x}=r+x^2\implies x^*=\pm\sqrt{r}$$
Therefore two fixed points exist for $r>0$ (one stable one unstable), one for $r=0$ at the origin, and none for $r<0$.
Thus this is a saddle-node bifurcation, and $r=0$ is the bifurcation point.
Note that $f’(0)$ at the bifurcation point is zero; this represents the fact that we have a semistable point then.
$$\dot{x}=x(r-x)\implies x^*=0,r$$
$$\frac{d\dot{x}}{dx}=r-2x$$
Therefore, we have
$$\begin{aligned}r>0&\implies\begin{cases}x^*=r&\text{stable}\\x^*=0&\text{unstable}\\\end{cases}\\r=0&\implies\begin{cases}x^*=0&\text{not enough info}\\\end{cases}\\r<0&\implies\begin{cases}x^*=0&\text{stable}\\x^*=r&\text{unstable}\\\end{cases}\end{aligned}$$
Once again, the bifurcation has a semistable point, which can be seen by sketching out the 1D phase portrait.
Thus, the fixed points exchange their stabilities! The bifurcation diagram has $x^*$ on the $y$-axis and $r$ on the $x$-axis. There are two lines: one at $x^*=0$, and one at $x^*=r$. The former is solid until the origin and dashed afterwards, and the reverse for the latter.
$$\dot{x}=x\left(1-x^2\right)-a\left(1-e^{-bx}\right)$$
This may have multiple fixed points, but we can immediately see $x^*=0$ is one.
$$\frac{d\dot{x}}{dx}=1-3x^2-abe^{-bx}\implies f’(0)=1-ab$$
Thus the transcritical bifurcation occurs at $ab=1$.
A bifurcation diagram would then be a plot of $a$ on the $y$-axis and $b$ on the $x$-axis. The bifurcation curve would be $b=1/a$. Outside this curve ($ab>1$) $x^*=0$ is stable, and $ab<1$ implies instability at the origin.
$$\dot{x}=rx-x^3\implies x^*=0, \pm\sqrt{r}$$
$$\frac{d\dot{x}}{dx}=r-3x^2$$
Thus we have
$$\begin{aligned}r>0&\implies\begin{cases}x^*=\sqrt{r}&\text{stable}\\x^*=0&\text{unstable}\\x^*=-\sqrt{r}&\text{stable}\end{cases}\\r=0&\implies\begin{cases}x^*=0&\text{not enough information}\end{cases}\\r<0&\implies\begin{cases}x^*=0&\text{stable}\end{cases}\\\end{aligned}$$
Note that the bifurcation diagram, a pitchfork-looking thing, doesn’t quite illustrate it well. Draw a phase portrait in all three cases to see what’s going on.
$$\dot{x}=rx+x^3\implies x^*=0,\pm\sqrt{r}$$
$$\frac{d\dot{x}}{dx}=r+3x^2$$
Giving
$$\begin{aligned}r<0&\implies\begin{cases}x^*=\sqrt{r}&\text{unstable}\\x^*=0&\text{stable}\\x^*=-\sqrt{r}&\text{unstable}\end{cases}\\r=0&\implies\begin{cases}x^*=0&\text{not enough information}\end{cases}\\r>0&\implies\begin{cases}x^*=0&\text{unstable}\end{cases}\\\end{aligned}$$
$$\dot{x}=rx+x^3-x^5\implies x=0,\pm\sqrt{\frac{1}{2}\pm\sqrt{r+\frac{1}{4}}}$$
In the range $-\frac{1}{4}<r<0$, there are two qualitatively different stable points. Below this, there is one. Above this, we have an unstable point at the origin, and two stable ones elsewhere.
This system therefore has a subcritical pitchfork bifurcation, and two saddle-nodes. It also has tipping points: multiple stable states causing large differences in long-term behaviour based on small changes in intial state.
We also have hysteresis. If we increase $r$ to above $0$, the system will flip from the origin to one of the stable states. Decreasing $r$ again will not make it return to the origin; instead, it progresses along the stable state. It takes for $r<-0.25$ for it to return to the origin. Thus the diagram of $r$ against $x$ is not a line, but a closed curve, one of the markers of hysteresis.
Our trick before of showing the variables and their derivatives in a single plot will no longer work: two variables and two derivatives would require a $4D$ plot!