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Not to be confused with Isochrone.
For the drug with trade name Isochron, see Isosorbide dinitrate .

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.[1][2]

Mathematical isochron[edit]

An introductory example[edit]

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[edit]

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.[3] 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.[1]


  1. ^ J. Guckenheimer, Isochrons and phaseless sets, J. Math. Biol., 1:259–273 (1975)
  2. ^ S.M. Cox and A.J. Roberts, Initial conditions for models of dynamical systems, Physica D, 85:126–141 (1995)
  3. ^ A.J. Roberts, Normal form transforms separate slow and fast modes in stochastic dynamical systems, Physica A: Statistical Mechanics and its Applications 387:12–38 (2008)