Let's consider a system of ODEs: $$ \dot{x_{1}} = \sin{x_{2}}+x_{1}\\ \dot{x_{2}} = \cos{x_{3}}-2x_{2}+x_{1}\\ \dot{x_{3}}=\arctan{x_{1}}-x_{2}+x_{3}$$
I would like to find an image of the unit cube under the phase flow of the system for the time $t=1$.
While dealing with linear systems $\dot{x} = Ax$, $A \in Mat_{n \times n}{\mathbb{C}}$, it can be done directly -- by finding out the general solution of a system $(x(t), y(t))^{T} = e^{tA} \cdot (x_{0}, y_{0})^{T}$ (here it's 2-dimensional case) and considering a phase curve, passing through the vertex of the cube, given in the form $(x(t), y(t))^{T} = e^{tA} \cdot (1, 0)^{T}$ (assume that $(1, 0)$ is a vertex of a cube). Putting $t=1$, we'll get the image of $(1, 0)$ under the flow of a system for the time $t=1$. By the linearity of a flow and convexity of a cube , i.e. any $x \in I^{n}$ can be written as $x = \alpha_{1}x_{1} + \alpha_{2} x_{2} + \ldots + \alpha_{2^{n}} x_{2^{n}} = \sum_{n=0}^{2^{n}}{\alpha_{n}x_{n}}, \sum_{n=0}^{2^{n}}{\alpha_{n}}=1$, then $f(x) = \sum_{n=0}^{2^{n}}{\alpha_{n}f(x_{n})}$, so it's sufficient to find images of vertices.
Well, but how to proceed in non-linear case. Surely, solving that system solves the exact problem, but can it be proceed in a relatively 'good' way? Actually, seems that it's possible to linearize the system, by exchanging transcendental functions with their expansions in the neighborhood of current points but i'm not sure, whether it works, even if it does work -- it would be wonderful to find out a rigorous justification.
Any help would be much appreciated.
This really should be a comment but I want to include an animation showing the fate of the poor cube.
For this sort of problem, the first thing one should do is compute a few numbers, make a plot or animation to get a rough feeling what one should expect before doing anything serious.
As one can see, the cube get squeezed into a thin slice. Since the flow is divergence free, the volume remains the same.
The cube seems to collapse onto some sort of surface. However, it isn't easy to determine the equation of surface. Since I'm lazy, I'll leave this to someone else for the fun (or pain?)