The behaviour of partial or ordinary differential equations can be studied/visualized with phase diagrams. How would one plot such diagrams for empirical data, which are suspected to be governed by differential equations?

Suppose I have e.g. some time-series data. Should I plot the original series $x_n$ against first differences $\dot{x} = x_{n+1}-x_{n}$, or do something else?


1 Answer 1


I assume that with "phase diagram" you mean the "phase portrait" of your system (the plot of typical trajectories in the "state space").

Therefore, if the state space of your system is the space of the points $(x,\dot{x})$, then the answer to your question is YES.

Since you have $(t^q_i,x^q_i)$, where $q$ is the label of a specific trajectory, you have to plot (for all the $q$ trajectories) the points $(x^q_i,v^q_i)$, where

$$ v_i^q = f( ...x_{i-1}^q,x_i^q,x_{i+1}^q... ; ...t_{i-1}^q,t_i^q,t_{i+1}^q... ) $$

is a certain discrete approximation of the velocity along the trajectory $q$. There are some dots, indicating that the number of arguments depends on the "stencil" you want to use to calculate the derivative.

For example, simple possibilities are:

$$ v_i^q \approx \frac{x_{i+1}^q-x_{i}^q}{t_{i+1}^q-t_{i}^q} \approx \frac{x_{i}^q-x_{i-1}^q}{t_{i}^q-t_{i-1}^q} \approx \frac{x_{i+1}^q-x_{i-1}^q}{t_{i+1}^q-t_{i-1}^q} $$

Of course you can consider higher-order methods to calculate the derivative (this is why in the definition of $f$ there are some dots).

  • $\begingroup$ Thanks! Yes, I mean the phase portrait. If a state-space is defined as a set of all possible configurations of the system, I'm not sure if I can ever be sure what that is for empirical data, though. Maybe I just don't get the concepts too well yet. Just starting out here. Anyway, thanks for the help! $\endgroup$
    – mmh
    May 27, 2020 at 10:21
  • 1
    $\begingroup$ It depends on how your data are collected and what you know about this system (a-priori knowledge or "prejudice"). If your $(t_i,x_i)$ are the output of a simulation, then you known (because you known the algorithm) if the $x_i$ are all the variables needed to define a single point in the state-space. If $(t_i,x_i)$ are measures of a physical phenomena, then you enter in the field pf Physics: depending on the knowledge we have about the observed phenomena, the "prejudice" that $x_i$ define uniquely the state is more or less reliable. $\endgroup$
    – Quillo
    May 27, 2020 at 10:52
  • $\begingroup$ Thanks for the comment, it was very helpful! I'll think about the situation from that point of view. $\endgroup$
    – mmh
    May 29, 2020 at 12:53

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