dynamical systemsordinary differential equations
If I have a system like:
$\frac{dx}{dt} = x, \frac{dy}{dt} = -y+x^2$
and I am asked to draw a phase portrait, once I have found the type of portrait (saddle point, node, spiral, etc.) from the eigenvalues and have found the $\infty$-isocline and $0$-isocline, how do I determine the direction of the arrows on the portrait?
This three step process is a summary from the excellent book series "Differential Equations: A Dynamical Systems Approach, Higher-Dimensional Systems" by Hubbard and West.
and sketch the isoclines of $(a)$ Horizontal Slope (where $y' = 0$) and $(b)$ Vertical Slope (where $x' = 0$).
Here you are using the above equation, choosing sample $(x, y)$ pairs and drawing the arrows that have direction and magnitude based on the slope.
Some other things that I find helpful are to determine the type of critical points which you see at the intersection of the nullclines, that is $(-1, -1)$ and $(1, 1)$ in this example. One is a stable spiral at $(-1, -1)$ and the other is an unstable saddle point $(1, 1)$. Additionally, you can look at the eigenvectors.
Putting all of these things together, we arrive at the phase portrait:
Lastly, it is worth noting, that these are not hard rules. Practice makes perfect and you'll develop your own approach in a way that is easy for you.
Yes; the concept of "phase space" is a general one.
Say you have $n$ time-dependent variables $x_1, \ldots, x_n$ related by a system of autonomous linear differential equations:
$$\dot{\mathbf{x}}=A\mathbf{x}$$
The "phase space" is $\mathbb{R}^n$. It is foliated by solution trajectories, just as in the phase plane when $n=2$.
The geometry of this foliation depends on the nature of the eigenvalues of $A$. Roughly speaking:
All this is carefully spelled out in many books on continuous dynamical systems.
See, for example, Section 1.9 ("Stability Theory") of Lawrence Perko, Differential Equations and Dynamical Systems.
You can see some 3D phase space pictures in Section 3.8.2 here.
Best Answer
I assume that the second equation should start with $\frac{dy}{dt}\equiv \dot y$.
An easy way to figure out the directions is to put arrows on $0$-isoclines, which are vertical for $\dot x=0$ and horizontal for $\dot y=0$. To determine on which end to put an arrow you simply check the sign. For example, if $\dot y>0$ then your arrow on the $x$ isocline is $\uparrow$. Otherwise it is $\downarrow$. The directions on other parts of the portrait follow by continuity.