Copied from 2020.

This video is not a complete answer to 7-05, it only covers the part of the plane where x>1, y>1 - this in order to capture the saddle pathVideo (18:42, 22 MB, .mp4):

7-05crf33b42k.mp4

7-05. The system: $LaTeX: \dot x= \tfrac 12 x^3-y,\qquad \dot y = 2x-y$

Part (a): find and classify equilibrium points (= stationary states).

Stationary states when y = 2x AND y = x³/2, implying 4x = x³. x=0 or x²=4.

Three points: (0,0), (-2,-4) and (2,4).

To classify, get the Jacobian of the transformation $LaTeX: \begin{pmatrix}f(x,y)\\g(x,y)\end{pmatrix}=\begin{pmatrix}\tfrac12x^3-y\\2x-y\end{pmatrix}$ . Jacobian matrix: $LaTeX: \mathbf A=\begin{pmatrix}\tfrac32x^2&-1\\2&-1\end{pmatrix}$ .
The eigenvalues formula $LaTeX: \lambda=\tfrac12\textsf{tr}\,\mathbf A\pm\sqrt{(\tfrac12\textsf{tr}\,\mathbf A)^2-\textsf{det}\,\mathbf A}$ gives classifications as follows:

Point (0,0): trace=-1< and det = 2. So det > (trace/2)^2 => non-real roots, and real part = trace/2 is negative.
(0,0) is locally asymptotically stable. (Spiraling inwards.)
Points (-2,-4) and (2,4) have the same Jacobian and thus the same classification: The detreminant is -6 +2=-4<0, so these are saddle points.

Part (b): Do you need more than the diagrams given?

(Actually, for the phase diagram on the right, it might be an advantage to get the direction asked for in part (c).)

I sketch some of the approach; please let me know what more you need for this problem.

Look at the arrow diagram to the left:
- The third-degree polynomial graph is the x-nullcline. Because the formula is [...] MINUS y, then above that curve, we have $LaTeX: \tfrac12 x^3-y<0$ , while $LaTeX: \dot x$ is >0 below that curve.
  Note how this matches the left or right part of the arrows.
- The y = 2x line is the y-nullcline. Above that line, 2x-y is < 0 and there are downwards arrows; below it, there are upwards arrows.
- Note that as the nonlinear system has more than one stationary state, the "four regions" are not all contiguous. The "NE arrows region" is contiguous and the "SW arrows region" is too, but:
  - you have NW arrows both on a bounded set in the first quadrant (that contains the point (1,1), and also on an unbounded set in the third quadrant from (-2,-4) and SSW (see that arrow about at the point (-2.3 , -5); at that point, $LaTeX: \dot x\approx -1.1<0$ and $LaTeX: \dot y=0.4>0$ .)
  - similarly you have you have SE arrows in two separated sets, one around the point (-1,-1) and one NNW of (2,4).
In the phase portrait to the right, the nullclines are dashed!

Part (c): slope of the convergent path to (2,4)

The theory you need here: It is a saddle point, it has a positive and a negative eigenvalue. Pick the negative, and find an eigenvector v=(v₁, v₂)'. As the saddle path converges to the saddle point, its slope -> v₂/v₁. (There are two saddle paths, one from each direction!)

Answer: The saddle point is (2,4). At (2,4), the Jacobian is $LaTeX: \mathbf A=\Big(\begin{smallmatrix}6&-1\\2&-1\end{smallmatrix}\Big)$ . Once we have the negative eigenvalue
$LaTeX: \nu=\tfrac12\textsf{tr}\,\mathbf A{\color{magenta}\pmb{\boldsymbol-}}\sqrt{(\tfrac12\textsf{tr}\,\mathbf A)^2-\textsf{det}\,\mathbf A}$ , we find an eigenvector v such that $LaTeX: (6-\nu,\ -1)\mathbf v=0$ (the other eq. redundant); we can choose $LaTeX: \mathbf v^\top=(1,\ 6-\nu)$ and the slope will be $LaTeX: ({6-\nu})/1$ . Insert to get the negative eigenvalue:

$LaTeX: \tfrac52-\sqrt{\tfrac{25}{4}-(6\cdot(-1)+2)}=\tfrac52-\sqrt{\tfrac{25}{4}+4}=\tfrac52-\tfrac12\cdot\sqrt{41}$ .

So the answer is $LaTeX: 6-\nu=6-\tfrac12\cdot 5 +\tfrac12\cdot \sqrt{41}=\underline{\underline{\tfrac12\big(7+\sqrt{41}\big)}}$ .

[A remark: From the diagram it isn't clear to read out what region the saddle path lies in, but surely: The y-nullcline has gradient = first row of Jacobian = (6,-1) and thus a slope of 6/1. The eigenvector has slope $LaTeX: ({6-\nu})/1$ , which is of course greater since the eigenvalue is negative, so if it approaches "from below" it is below the nullcline and catches it at the point.
Numerically, it is about 6.7, but in a diagram with equal scale on the first and second axis, it is very hard to eyeball differences in slopes this large.]