I drew the Volterra phase plane again and made the point that most places what you see is almost parallel almost straight lines. Near the critical points, things look different. This is interesting. For one thing, we tend to care about equilibrium solutions. For another thing, it will appear that often most of the phase portrait can be guessed once we know how it looks near the critical points.
For example take the system
x' = x^2 - y
y' = y^2 - x.
I found the critical points: (0,0), (1,1). Near (0,0) the equation is close to x' = -y, y' = -x . This is the "linear approximation" near (0,0). Let's analyze it. The matrix is [0 -1;-1 0], with characteristic polynomial lambda^2 - 1 and eigenvalues 1, -1. We'll get a saddle. To find the axes compute eigenvectors: for 1 we get [1;-1]. Any symmetric matrix has perpendicular eigenspaces (and real eigenvalues) so an eigenvector for -1 is [1;1]. I drew the saddle and then pasted a little picture of it onto the phase portrait for the original equation.
For the critical point (1,1), one method is to introduce new coordinates u=x-1, y=y-1, so x=1+u, y=1+v, and
u' = x' = (1+u)^2 - (1+v) = 2u + u^2 - v
v' = y' = (1+v)^2 - (1+u) = 2v + v^2 - u
The linear approximation is thus
u' = 2u -v
v' = -u + 2v.
The characteristic equation is lambda^2 - 4 lambda + 3, with roots 1, 3. You get an unstable proper node. To draw it we need the eigenvectors. For 1 we get [1;1], and, since the matrix is again symmetric, for 3 we get [1;-1]. The general solution is given by ae^t[1;1] + be^{3t}[1;-1] (in terms of [u;v]). The coefficient of [1;-1] is thus a constant times the cube of the coefficient of [1;1], so you get graphs that look like cx^3 in which the x-axis is along the [1;1] line and the y-axis is along the [1;-1] line. I drew this and then pasted the result into the master diagram.
Then I applied some artistic guesswork to fill in the rest of the picture. I noticed that the ODE is symmetric on exchange of x and y, so the picture is symmetric too. In particular there are solutions in which x=y. Putting this into the original equation we get the logistic equation x' = x^2-x. You may recall its phase portrait, and recognize it along the line x=y in our phase plane. Also, we integrated it explicitly and found that solutions which go off to infinity or minus infinity do so in finite time. The same is thus true of solutions to the present equation. There are actually five different solutions along the line x=y: two constant solutions, two which go off to infinity, and one carrying you from [1;1] to [0;0].
The second method uses the tangent plane approximation from 18.02: near (a,b),
f(x,y) is about f(a,b) + f1(a,b)(x-a) + f2(a,b)(y-b) where f1(x,y) is the partial derivative with respect to x and f2(x,y) is the partial derivative with respect to y. Apply this to f(x,y) = x' = x^2-y and (a,b) = (1,1):
f(x,y) is about 0 + 2(x-1) - (y-1)
g(x,y) is about 0 - (x-1) + 2(y-1)
which is the same result as we had before.
I will apply this idea to study the nonlinear pendulum. I set up the ODE,
theta" + q sin(theta) = 0, q = g/l
where theta is the deviation from straight down, g the gravitational constant, l the length of the pendulum. We may have some damping too, and it's easiest to imagine that it's given by p theta': so the damped nonlinear pendulum is given by
theta" + p theta' + q sin(theta) = 0.
Note for now just that if theta is small, then sin(theta) is about theta, so the linear approximation near theta = 0 is the standard linear damped oscillator
theta" + p theta' + q theta = 0.