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Before my question let me briefly describe a simplified version of the dynamical system I'm working with. Suppose that I have a density function $m(\boldsymbol{x},t)$, that describes the abundance of some quantity in a vector space $\Omega = [a,b]^{n}$. Now suppose that the density at each point in vector space changes over time, with a rate that may be dependent on the presence/absence of density at every other point in space

$$ \partial_{t}m(\boldsymbol{x},t) = m(\boldsymbol{x},t)F(\boldsymbol{x},?), $$

where $F$ might be something like

$$ F(\boldsymbol{x},?) = \int_{\Omega} m(\boldsymbol{y},t) k(\boldsymbol{x}, \boldsymbol{y}) \; d\boldsymbol{y} $$

i.e. a function of the focal point of interest $\boldsymbol{x}$ and all other different points $\boldsymbol{y}$ in vector space, where $k$ is some kind of kernal(?).

Question: What notation should I use to refer to the state of the system over the whole space? What should replace my question mark in $F(\boldsymbol{x},?)$, should it be $F(\boldsymbol{x},m)$, or maybe $F(\boldsymbol{x},\Omega)$, or something else?

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For your first question, consider the following notation: At any given time, say $t$, the state of your system is the function $q_t : \Omega \mapsto \mathbb{R}^{+}$, where $q_t(x) \equiv m(x, t)$. How can we succinctly describe its time evolution? It is actually pretty simple:

$$ \frac{dq}{dt} = q \cdot (K q) $$

where $\cdot$ is ordinary multiplication and $K$ is the integral transform defined by the kernel $k : \Omega \times \Omega \mapsto \mathbb{R}$.

Note that the derivative $\frac{d}{dt}$ is now defined in an infinite dimensional function space. One can straightforwardly define this derivative using the traditional form

$$ \frac{dq}{dt} \equiv \lim_{\tau \rightarrow t} \frac{q_\tau - q_t}{\tau - t} , $$

where the limit is defined via a suitable norm in your function space. See these lecture notes for details.

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  • $\begingroup$ Perfect, thank you @Mehmet, I hadn't learnt about integral transforms before. I removed my question 2 since I have thought about a better way to ask what I was trying to get across. $\endgroup$ Commented Jan 24, 2015 at 23:24

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