Separable trajectory

This article defines a property that can be evaluated for a trajectory on the space of functions on a manifold

Definition

Let $M$ be a manifold and $u=u(t,x)$ be a function $\mathbb {R} \times M\to \mathbb {R}$, where:

• $t$ denotes the time parameter, and varies in $\mathbb {R}$
• $x$ denotes the spatial parameter, and varies in $M$

In other words, $u$ is a trajectory (or path) in the space of all functions from $M$ to $\mathbb {R}$.

Then $u$ is said to be separable if $u(t,x)=f(t)g(x)$ for suitable functions $f:\mathbb {R} \to \mathbb {R}$ and $g:M\to \mathbb {R}$. Equivalently, $u$ is separable if and only if it satisfies the property:

$u(t_{1},x_{1})u(t_{2},x_{2})=u(t_{!},x_{2})u(t_{2},x_{1})$

for all times $t_{1},t_{2}\in \mathbb {R}$ and points $x_{1},x_{2}\in M$.

Note that for a separable trajectory, we get the following differential equations for $f$ and $g$:

$f'(t)=\lambda f(t)$

and

$(F(g))(x)=\lambda g(x)$

where $\lambda \in \mathbb {R}$ is a real number. In fact the entire family of separable trajectories is parametrized by $\lambda$.

The solution to the first equation is $f(t)=Ce^{\lambda t}$.

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