Consider the ODE $$\begin{cases} \frac{d}{dt}\Phi(x,t) = f(\Phi(x,t),t) \quad t >0, \quad x \in \mathbb R^2 \\ \Phi(x,0) = x, \quad x \in \mathbb{R}^2 \end{cases}$$
Suppose that the flow $\Phi$ of this ODE exists and is unique.
- It is not quite clear to me what is the most convenient way to picture (really, to draw with pen and paper) the flow $\Phi$ of the ODE.
In particular, I've seen in many books a picture like the one below, but it is not clear to me. What does it represent? Where are the values of $\phi$ displayed?
You have three numbers that you need to plot: $x$, $t$, and $\Phi(x,t)$. So you need to represent it in a 3D plot, or like in the above case, constant contour lines. Each of those lines is $\Phi(x,t)=\Phi(x_i,0)$
Say your equation is $$\frac{d}{dt}\Phi(x,t)=-2t\Phi(x,t)$$ Performing separation of variables, and using your initial condition, you get $$\Phi(x,t)=xe^{-t^2}$$ I've admit here that I've started from the solution. So how should this look? For a constant $\Phi$ line, if $t$ increases linearly, the $x$ should increase like $e^{x^2}$. To prove: $$C=xe^{-t^2}\\x=Ce^{t^2}$$
I've plotted such contours in the picture below: