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# Modules

Ordinary differential equations
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# ODE's

Many dynamical systems in physics, astronomy, chemistry, physiology, meteorology, economics, population dynamics can be described by Ordinary Differential Equations.
In the second half of the 20th century much attention has been focussed on the often chaotic, i.e. unpredictable behaviour of (non-linear) ODE's. A well-known example is the Lorenz atrractor, illustrating the "Butterfly effect": small causes can have large effects. Mathgrapher uses an accurate Adams-Bashforth variable order, variable step predictor-corrector algorithm to integrate systems of up to 20 coupled differential equations.

Several analytical tools are available for ODE's such as:
 Time series Power spectra 2D and 3D projections Phase portraits Poincare section Equilibrium points Lyapunov exponents

Examples:

ODEs: Analysis - Equilibrium points (Fixed points)

# Equilibrium points (or fixed points)

A set of ODE's has an equilibrium point  when dF/dt=0 for all F's. There are several types of equilibrium points depending on the behaviour of the orbits in their neighbourhood: stable and unstable nodes, saddle points, stable and unstable spiral points. The behaviour of the orbits near an equilibrium point can be found by studying the linearized equations (see below). Suppose we have a system of 2 coupled ODE's:
 dx/dt=x(4-x-y) dy/dt=y(x-2)
The eigenvalues of the stability matrix J (see below) determine the type of equilibrium point and its stability properties. They are the roots of the equation
where I is the identity matrix. In the 2-dimensional case discussed here this generally yields 2 solutions. Stable point have negative eigenvalues,  unstable points have positive eigenvalues, saddle points have one negative and one positive eigenvalue and spiral points have complex eigenvalues. They may be stable or unstable depending on their position in the complex plain.
The fixed points are (0.0), (2,2) and (4,0). To find the eigenvalues we have to solve the equation:
For (0,0) these are -2 and 4 => hyperbolic point, for (2,2) we find -1+i*sqrt(3) and -1-i*sqrt(3) i.e. stable spiral point, and for (4,0) -4 and 2, i.e. hyperbolic point. Check these results with Mathgrapher. To find an equilibrium point (and its eigenvlaues) you have to give initial values somewhere in its neighbourhood.

# Linearization of the equations

Suppose we have a system of 2 coupled ODE's
The eigenvalues of this matrix determine the type of equilibrium point and its stability properties. They are the roots of the equation

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