You can see some bifurcations in action with the demo below.

Saddle-Node Bifurcation

I don't have a full definition for normal form yet, but the prototypical forms for saddle-node bifurcations are

$$\dot{x}=r-x^2,\dot{x}=r+x^2.$$

For any $f(x,r),$ if we have the following conditions at the bifurcation point $r^{\ast },$ the @dynamical-system can be reduced to one of the saddle-node normal forms above:

$$f(x^{\ast },r^{\ast })=0,{\partial }_{x}f(x^{\ast },r^{\ast })=0,$$

that is, $f(x^{\ast },r^{\ast })$ is a tangential intersection with the $x$-axis, and

$${\partial }_{xx}f(x^{\ast },r^{\ast })\ne 0,{\partial }_{r}f(x^{\ast },r^{\ast })\ne 0.$$

The first ensures this is only a double root, not a triple root or worse. The second ensures that as we vary $r$ near this point, the function value actually changes, which is required to create or destroy the fixed points.

Transcritical Bifurcation

The normal form for a @transcritical-bifurcation is $\dot{x}=rx-x^2.$

Pitchfork

Summary of Conditions

At a candidate bifurcation point where $f(x,r)=0$ and $f_x(x,r)=0,$ evaluate these derivatives to tell what kind of bifurcation occurs there.

Saddle-Node

$$f_r\ne 0.$$

$$f_{xx}\ne 0.$$

Normal form: $\dot{x}=r\pm x^2.$

Transcritical

$$f_r=0.$$

$$f_{xx}\ne 0.$$

$$f_{xr}\ne 0.$$

Normal form: $\dot{x}=rx-x^2.$

Pitchfork

$$f_r=0.$$

$$f_{xx}=0.$$

$$f_{xxx}\ne 0.$$

$$f_{xr}\ne 0.$$

To distinguish between supercritical and subcritical:

For supercritical:

We have the condition and the normal form is $\dot{x}=rx-x^3.$

For subcritical:

We have the condition and the normal form is $\dot{x}=rx+x^3.$

The fixed point exists for all $r(f_r=0),$ and there's odd symmetry in $x(f_{xx}=0),$ so instead of two branches crossing, one branch splits into three. The cubic term dominates.

Normal form: $\dot{x}=r\ast x\pm x^3.$

Normal Form Analysis

We can find the normal form by shifting to $u=x-x^{\ast },\mu =r-r^{\ast },$ and finding the taylor expansion of $f$ at $(u,\mu .)$ We may need the derivatives above, but sometimes it is just obvious which normal form fits.

Nondimensionalization

I don't have my head fully wrapped around this yet, but we can sometimes eliminate parameters through dimensional analysis, or, put another way, through change of variables.

Say we start with

$$\dot{u}=\frac{du}{dt}=au+b{u}^3-c{u}^5.$$

This equations has three parameters, $a,b,andc$ which is mighty inconvenient and difficult to analyze because it means we have a three dimensional parameter space. It'd be nice if we could reduce it to a single parameter. It'd be extra nice if this parameter was on the linear term, because the linear term determines the stability of fixed points. So, we want to end up with something equivalent that has the form

$$\frac{dx}{d\tau }=rx+x^3-x^5.$$

We can get there by making a change of variables, starting with the substitution

$$u=Ux,t=T\tau ,$$

Then,

$$\dot{u}=\frac{du}{dt}=\frac{d(Ux)}{d(\tau T)}=\frac{U}{T}\frac{dx}{d\tau }=a(Ux)+b(U^3{x}^3)-c(U^5{x}^5),$$

and

$$\frac{dx}{d\tau }=aTx+bT{U}^2+x^3-cT{U}^4{x}^5.$$

Now, we can let $r=aTx$ and we want the other two terms to have coefficients of 1, so we want to pick values of $T$ and $U$ to make that happen, that is, we want

$$bT{U}^2=1,cT{U}^4=1.$$

We have two equations and two unknowns ($T$ and $U$), so we solve for them and we get

$$T=\frac{c}{b^2},U=\sqrt{\frac{b}{c}},r=\frac{ac}{b^2}.$$

Now we see that with our new coordinates $x$ and $\tau ,$ our parameters collapse into a single one, $r$, and we can much more easily analyze the behavior of this system.

This is called nondimensionalization because we can view $U$ and $T$ as units, $x$ and $\tau$ as dimensionless coordinates, and $u$ and $t$ as coordinates using the units $U$ and $T$.

This process lets us take apparently complex dynamical systems and simplify them into normal forms we can recognize and deal with.

Algorithm for analyzing bifurcations and drawing a diagram

Algorithm (1D bifurcation diagram for $\dot{x}=f(x,r)$)

Compute the derivative with respect to the state:

$$f_x(x,r):=\frac{\partial f}{\partial x}(x,r).$$

Find equilibrium branches by solving the equilibrium condition:

$$f(x,r)=0.$$

Solve this equation for either $x$ as a function of $r,$

$$x=x_i(r),$$

or for $r$ as a function of $x,$

$$r=r_i(x),$$

whichever is algebraically simpler. Each solution defines one equilibrium branch.

For each equilibrium branch, substitute the branch into $f_x$:

• If the branch is given as $x=x_i(r),$ compute

$$f_x(x_i(r),r).$$

• If the branch is given as $r=r_i(x),$ compute

$$f_x(x,r_i(x)).$$

Determine stability along each branch:

Identify candidate bifurcation points by solving

$$f(x,r)=0\text{and}{f}_x(x,r)=0$$

simultaneously. These points are where stability can change and where a local bifurcation may occur.