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Vipul at 11:26, 8 April 2007

2007-04-08T11:26:08Z

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{{trajectory property}}

==Definition==

Let <math>M</math> be a manifold and <math>u = u(t,x)</math> be a function <math>\R \times M \to \R</math>, where:

* <math>t</math> denotes the time parameter, and varies in <math>\R</math>
* <math>x</math> denotes the spatial parameter, and varies in <math>M</math>

In other words, <math>u</math> is a trajectory (or path) in the space of all functions from <math>M</math> to <math>\R</math>.

Then <math>u</math> is said to be '''separable''' if <math>u(t,x) = f(t) g(x)</math> for suitable functions <matH>f: \R \to \R</math> and <math>g: M \to \R</math>. Equivalently, <math>u</math> is separable if and only if it satisfies the property:

<math>u(t_1,x_1)u(t_2,x_2) = u(t_!,x_2)u(t_2,x_1)</math>

for all times <math>t_1,t_2 \in \R</math> and points <math>x_1,x_2 \in M</math>.

Note that for a separable trajectory, we get the following differential equations for <math>f</math> and <math>g</math>:

<math>f'(t) = \lambda f(t)</math>

and

<math>(F(g))(x) = \lambda g(x)</math>

where <math>\lambda \in \R</math> is a real number. In fact the entire family of separable trajectories is parametrized by <math>\lambda</math>.

The solution to the first equation is <math>f(t) = Ce^{\lambda t}</math>.

When <math>F</math> is a linear differential operator, the solution to the second equation is any eigenvector <math>g</math> for <math>F</math> having eigenvalue <math>\lambda</math> (we can use the term ''eigenvector'' only when the operator is linear though an equivalent notion makes sense for arbitrary differnetial operators).

Separable trajectory - Revision history

Vipul: 1 revision

Vipul at 11:26, 8 April 2007