# Isochron

In the mathematical theory of dynamical systems, an isochron is a set of initial conditions for the system that all lead to the same long-term behaviour.

## Mathematical isochron
### An introductory example

Consider the ordinary differential equation for a solution $y(t)$ evolving in time:
    ${\frac {d^{2}y}{dt^{2}}}+{\frac {dy}{dt}}=1$
This ordinary differential equation (ODE) needs two initial conditions at, say, time $t=0$. Denote the initial conditions by $y(0)=y_{0}$ and $dy/dt(0)=y'_{0}$ where $y_{0}$ and $y'_{0}$ are some parameters. The following argument shows that the isochrons for this system are here the straight lines $y_{0}+y'_{0}={\mbox{constant}}$.
The general solution of the above ODE is
    $y=t+A+B\exp(-t)$
Now, as time increases, $t\to \infty$, the exponential terms decays very quickly to zero (exponential decay). Thus all solutions of the ODE quickly approach $y\to t+A$. That is, all solutions with the same $A$ have the same long term evolution. The exponential decay of the $B\exp(-t)$ 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 $A$.
At the initial time $t=0$ we have $y_{0}=A+B$ and $y'_{0}=1-B$. Algebraically eliminate the immaterial constant $B$ from these two equations to deduce that all initial conditions $y_{0}+y'_{0}=1+A$ have the same $A$, 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
    ${\frac {dx}{dt}}=-xy{\text{ and }}{\frac {dy}{dt}}=-y+x^{2}-2y^{2}$
A marvellous mathematical trick is the normal form (mathematics) transformation. Here the coordinate transformation near the origin
    $x=X+XY+\cdots {\text{ and }}y=Y+2Y^{2}+X^{2}+\cdots$
to new variables $(X,Y)$ transforms the dynamics to the separated form
    ${\frac {dX}{dt}}=-X^{3}+\cdots {\text{ and }}{\frac {dY}{dt}}=(-1-2X^{2}+\cdots )Y$
Hence, near the origin, $Y$ decays to zero exponentially quickly as its equation is $dY/dt=({\text{negative}})Y$. So the long term evolution is determined solely by $X$: the $X$ equation is the model.
Let us use the $X$ equation to predict the future. Given some initial values $(x_{0},y_{0})$ of the original variables: what initial value should we use for $X(0)$? Answer: the $X_{0}$ that has the same long term evolution. In the normal form above, $X$ evolves independently of $Y$. So all initial conditions with the same $X$, but different $Y$, have the same long term evolution. Fix $X$ and vary $Y$ gives the curving isochrons in the $(x,y)$ plane. For example, very near the origin the isochrons of the above system are approximately the lines $x-Xy=X-X^{3}$. Find which isochron the initial values $(x_{0},y_{0})$ lie on: that isochron is characterised by some $X_{0}$; the initial condition that gives the correct forecast from the model for all time is then $X(0)=X_{0}$.
You may find such normal form transformations for relatively simple systems of ordinary differential equations, both deterministic and stochastic, via an interactive web site.