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.

An introductory example
Consider the ordinary differential equation for a solution $$y(t)$$ evolving in time:


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

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.