To make the discussion concrete, consider the following differential equation:

in which is a scalar variable, . This is a second-order differential equation because of . A

So far, this does not seem to have helped. However, can be expressed as either or . The first choice is better because it is a lower order derivative. Using , the differential equation becomes

Is this expression equivalent to (13.29)? By itself it is not. There is one more constraint, . In implicit form, . The key to making the phase space approach work correctly is to relate some of the phase variables by derivatives.

Using the phase space, we just converted the second-order differential equation (13.29) into two first-order differential equations,

which are obtained by solving for and . Note that (13.32) can be expressed as , in which is a function that maps from into .

The same approach can be used for any differential equation in implicit form, . Let , , and . This results in the implicit equations and . Now suppose that there is a scalar action represented in the differential equations. Once again, the same approach applies. In implicit form, can be expressed as .

Suppose that a given acceleration constraint is expressed in
parametric form as
. This often occurs in the
dynamics models of Section 13.3. This can be converted
into a *phase transition equation* or *state transition
equation* of the form
, in which
. The expression is

(13.33) |

For a second-order differential equation, two initial conditions are usually given. The values of and are needed to determine the exact position for any . Using the phase space representation, no higher order initial conditions are needed because any point in phase space indicates both and . Thus, given an initial point in the phase and for all , can be determined.

A two-dimensional phase space is defined in which

(13.34) |

The state (or phase) transition equation is

(13.35) |

To determine the state trajectory, initial values (position) and (velocity) must be given in addition to the action history. If is constant, then the state trajectory is quadratic because it is obtained by two integrations of a constant function.

Steven M LaValle 2012-04-20