How to plot a phase space

Question

I was trying to study the function $x(t)=\int_0^t e^{-s^2}ds$, which I did successfully using high school maths. After that, I decided that I wanted to try to study it using some multivariable calculus, but couldn't really do it, so I'm asking for the proper way to face this: the problem is equivalent to the ODE $x'(t)=e^{-t^2}$ given some initial value. Let $y(t)=t$ and therefore $y'(t)=1$. Now I want to study the phase space of the system $\begin{cases} x'=e^{-y^2}\\ y'=1 \end{cases}$. How can I do it? There are no equilibrium points and the Jacobian's only eigenvalue is $0$. Also, the linearized system is $\begin{cases}x'=(-2y_0e^{-y_0^2})\cdot y \\ y'=0\end{cases}\;$ which makes no sense since in $(0, 0)$ it gives $x'=0, y'=0$ whereas in the original one it is $x'=1, y'= 1$. What's the proper way to study this system?

Answer 1

If I'm interpreting your question correctly, you want to understand how to sketch a phase portrait from just knowing the dynamical system; in that situation, many techniques work, but I personally prefer to use nullclines and common sense. As it happens, it turns out this particular example is not well-suited for this technique, but I'll describe it anyways.

The first step is to try and figure out where/when the flow is only moving vertically or horizontally, which is equivalent to finding the roots of each component of the dynamical system.

In your case, $y' = 1$ which has no roots, so the flow will always have a vertical component of $1$ upwards (and so you always expect the flow to either be directly upwards or tilted upwards). For $x'$, you find $x' = 0 = e^{-y^2}$, which is only true for $y \rightarrow \pm \infty$. So how do you construct a sketch of the phase portrait from this info?

Well, because you know that there's no horizontal flow when $y = \pm \infty$, then the flow has to asymptotically approach arrows pointing straight up (which you know from $y' = 1$) as you move away from the x-axis. Then, calculating the flow directly on the x-axis tells me $x' = y' = 1$, so arrows there are pointing diagonally "northeast". Using these two results, the flow has to look like "northeast" pointing arrows slowly pointing more and more upwards as you move away from the x-axis, which is a succinct description of the whole phase plot!

If you're wondering why the flow can't suddenly change direction from northeast to northwest or something, well it's because that would imply there's a nullcline, which you know there isn't! From there on out it's up to understanding what the right "tilt" is, which really comes down to common sense.

Hope this helps!

Answer 2

Yes, there are no equilibrium points, so it has no meaning to compute the Jacobian at the fixed points. Also no meaning of linearization around a fixed point.

We have $\dot{x} = e^{-y^2}$ and $\dot{y} = 1$, we see that the $\dot{y}=1$ is independent of $x$ and $y$, so we have $y = t + c$, $c$ an integration constant. So substituting the expression of $y$ in $\dot{x} = e^{-y^2}$, we have $x = D \int_{0}^{t} e^{-s^2} ds$, $D$ is integration constant. This definite integral reminds me of the error function, whose plot if you have in mind then it would be handy.

In order to plot the phase space that is plot $x$ vs $y$ we need to have a plot of $D$ erf(t) vs $t$, same as the graph of erf(x).

So we have the vector field, phase plane as below for different initial conditions.