An introductory example

Consider the ordinary differential equation for a solution ${\displaystyle y(t)}$ evolving in time:

${\displaystyle {\frac {d^{2}y}{dt^{2}}}+{\frac {dy}{dt}}=1}$

This ordinary differential equation (ODE) needs two initial conditions at, say, time ${\displaystyle t=0}$. Denote the initial conditions by ${\displaystyle y(0)=y_{0}}$ and ${\displaystyle dy/dt(0)=y'_{0}}$ where ${\displaystyle y_{0}}$ and ${\displaystyle y'_{0}}$ are some parameters. The following argument shows that the isochrons for this system are here the straight lines ${\displaystyle y_{0}+y'_{0}={\mbox{constant}}}$.

The general solution of the above ODE is

${\displaystyle y=t+A+B\exp(-t)}$

Now, as time increases, ${\displaystyle t\to \infty }$, the exponential terms decays very quickly to zero (exponential decay). Thus all solutions of the ODE quickly approach ${\displaystyle y\to t+A}$. That is, all solutions with the same ${\displaystyle A}$ have the same long term evolution. The exponential decay of the ${\displaystyle B\exp(-t)}$ term brings together a host of solutions to share the same long term evolution. Find the isochrons by answering which initial conditions have the same ${\displaystyle A}$.

At the initial time ${\displaystyle t=0}$ we have ${\displaystyle y_{0}=A+B}$ and ${\displaystyle y'_{0}=1-B}$. Algebraically eliminate the immaterial constant ${\displaystyle B}$ from these two equations to deduce that all initial conditions ${\displaystyle y_{0}+y'_{0}=1+A}$ have the same ${\displaystyle A}$, hence the same long term evolution, and hence form an isochron.

Accurate forecasting requires isochrons

Let's turn to a more interesting application of the notion of isochrons. Isochrons arise when trying to forecast predictions from models of dynamical systems. Consider the toy system of two coupled ordinary differential equations

${\displaystyle {\frac {dx}{dt}}=-xy{\text{ and }}{\frac {dy}{dt}}=-y+x^{2}-2y^{2}}$

A marvellous mathematical trick is the normal form (mathematics) transformation.^[3] Here the coordinate transformation near the origin

${\displaystyle x=X+XY+\cdots {\text{ and }}y=Y+2Y^{2}+X^{2}+\cdots }$

to new variables ${\displaystyle (X,Y)}$ transforms the dynamics to the separated form

${\displaystyle {\frac {dX}{dt}}=-X^{3}+\cdots {\text{ and }}{\frac {dY}{dt}}=(-1-2X^{2}+\cdots )Y}$

Hence, near the origin, ${\displaystyle Y}$ decays to zero exponentially quickly as its equation is ${\displaystyle dY/dt=({\text{negative}})Y}$. So the long term evolution is determined solely by ${\displaystyle X}$: the ${\displaystyle X}$ equation is the model.

Let us use the ${\displaystyle X}$ equation to predict the future. Given some initial values ${\displaystyle (x_{0},y_{0})}$ of the original variables: what initial value should we use for ${\displaystyle X(0)}$? Answer: the ${\displaystyle X_{0}}$ that has the same long term evolution. In the normal form above, ${\displaystyle X}$ evolves independently of ${\displaystyle Y}$. So all initial conditions with the same ${\displaystyle X}$, but different ${\displaystyle Y}$, have the same long term evolution. Fix ${\displaystyle X}$ and vary ${\displaystyle Y}$ gives the curving isochrons in the ${\displaystyle (x,y)}$ plane. For example, very near the origin the isochrons of the above system are approximately the lines ${\displaystyle x-Xy=X-X^{3}}$. Find which isochron the initial values ${\displaystyle (x_{0},y_{0})}$ lie on: that isochron is characterised by some ${\displaystyle X_{0}}$; the initial condition that gives the correct forecast from the model for all time is then ${\displaystyle X(0)=X_{0}}$.

You may find such normal form transformations for relatively simple systems of ordinary differential equations, both deterministic and stochastic, via an interactive web site.