Relationship between autonomous system and related single equation

[zenterix]

Homework Statement

My question is about some notes from MIT OCW's 18.03 "Differential Equations" on first-order ODE autonomous systems.

Relevant Equations

See below.

Here are the notes.

We have the system

$$\begin{bmatrix} x'\\y' \end{bmatrix}=\begin{bmatrix}f(x,y)\\g(x,y)\end{bmatrix}\tag{1}$$

We eliminate $t$ by dividing one equation by the other

$$\frac{y'}{x'}=\frac{dy/dt}{dx/dt}=\frac{dy}{dx}=\frac{g(x,y)}{f(x,y)}\tag{2}$$

$$\frac{dy}{dx}=\frac{g(x,y)}{f(x,y)}\tag{3}$$

which is a single first order equation involving $y$ as a function of $x$.

Indeed, in the older literature, little distinction was made between the system and the single equation - "solving" meant to solve either one.

There is however a difference between them: the system involves time, whereas the single ODE does not. Consider how their respective solutions are related:

$$\begin{matrix} x=x(t)\\y=y(t)\end{matrix}\implies F(x,y)=0\tag{4}$$

where the equation on the right is the result of eliminating $t$ from the pair of equations on the left.

My first question is about (4).

I am not seeing where $F(x,y)=0$ comes from.

 

[zenterix]

If $x$ and $y$ are functions of $t$ and we eliminate the $t$ to get an equation with only $x$ and $y$ in it and then move all terms to the left side then we do get $F(x,y)=0$. Is this all there is to it?

 

[zenterix]

The notes go on as follows

Geometrically, $F(x,y)=0$ is the equation for the trajectory of the solution $\vec{x}(t)$. The trajectory in other words is the path traced out by the moving point $(x(t),y(t))$; it doesn't contain any record of how fast the point was moving; it is only the track (or trace, as one sometimes says) of its motion.

This is all fine.

In the same way, we have the difference between the velocity field, which represents the left side of (4) and the direction field which represents the right side.

The velocity vectors have magnitude and sense, whereas the line segments that make up the direction field only have slope.

The velocity field is represented by the initial system of equations.

What is the direction field exactly here?

For a single equation $y'=f(x,y)$, a direction field (aka, slope field) is just a depiction of the slope at various points in the $xy$-plane.

In our system of equations, the equivalent of a slope field seems at first glance to be the same thing as the velocity field (but perhaps without denoting the direction of the velocity, only its slope). However, I am unsure about this.

At this point, I read the last paragraph of the notes, and it seems to agree for the most part with the paragraph above. Here it is

The passing from the left side of (4) to the right side is represented geometrically by changing each of the velocity vectors to a line segment of standard length. Even the arrowhead is dropped, since it represents the direction of increasing time, and time has been eliminated; only the slope of the vector is retained.

 

[Office_Shredder]

I think the simplest thing to consider is just like, consider the curve sketched out by$(t,t^2)$ compared to $(t^3,t^6)$.

Both of them traverse the parabola $y-x^2=0$ but the behavior of a particle traveling according to each of the two functions of t is pretty different. For example what is the speed at $x=y=1$ is different for the two of them. And the slope of the parabola tells you what direction they are traveling in, but nothing about how fast they are going.