Question

I'm trying to solve this system of nonlinear equations and find the critical points. However, when I am solving for the critical points it seems that there are critical "lines" and I've never really heard of these before.

x'=(2+x)(y-x)
y'=(4-x)(y+x)

So you can see when x'=0 and y'=0 I get:

x=-2
x=4
y=x
y=-x

Should I just disregard these lines and just consider the points x=-2 and x=4 as my critical points? This seems wrong though.

Answer 1

While not familiar with solving systems of nonlinear equations (which also happen to involve derivatives), I could point out a number of interesting things:

first, y' = 0 when x=4 and/or when y=-x. Remember that y' generally signifies dy/dx and represents the slope of the tangent line to a curve at a given point.

Are you assuming that y' =dy/dx? If so, what would x' =dx/dx signify?

Answer 2

No, x'=dx/dt and y'=dy/dt. They're independent variables in this case.

Answer 3

Critical points would be t values in this case, right?

Answer 4

Not quite, I'm looking at the phase plane.

Answer 5

Why do you want the critical point, and what exactly is it in this case? What variable?

Answer 6

Well in differential equations 2 there's a theorem that allows you to approximate a system of nonlinear differential equations as a linear system at those critical points.

Essentially a linear system looks something like this:

x'=y
y'=-x

And that will allow you to graph a vector field. Each vector is (x', y') located at (x,y).

Answer 7

In this example case I just made up, that will look something like this:

Answer 8

Well drawing doesn't work, but it's basically a bunch of concentric circles starting at the origin and increasing in magnitude as you move away.

Answer 9

And what is the problem with the critical points you got?

Answer 10

Well y=x and y=-x aren't points, they're entire lines.

Answer 11

For the system

x'=y
y'=-x

What would you consider to be the critical points?

Answer 12

(0,0) since (x',y')=(0,0) when (x,y)=(0,0). But that was just an example explaining what a linear system is... My real question is in the first post.

Answer 13

So you are saying that if for (x,y), then (x',y') = (0,0) then it is a critical point?

Answer 14

Exactly.

Answer 15

Okay, first start with
x = -2
This makes x' = 0, Now we need to make y' = 2.
We have
y' = 0 = (4-(-2))(y-2)
solve for y

Answer 16

@Kainui : No, x'=dx/dt and y'=dy/dt. They're independent variables in this case.
Thanks for the clarification! Mind if I make small corrections here? x and y are the dependent variables and t, the parameter, is the independent variable. :)

Answer 17

Now we need to make y' = 0 ^^

Answer 18

So one possible critical point would be (x, y) = (-2, 2)

Answer 19

Do the same with x=4
Then consider x-y = 0 = x+y

Answer 20

@mathmale Well since x and y are the basis of a vector space, they are absolutely linearly independent. In fact, they don't exactly depend on t in the way you're thinking since the phase plane represents the infinite number of solutions that satisfy the differential equation.

@wio That's exactly what I've done in the very first post, and now you're where I am I believe. Since y=x, I can easily have the following critical points:

(5,5), (7,7), (1000,1000),... So... Yeah.

Answer 21

But reconsider (5,5)

x'=(2+5)(5-5) = 0
y'=(4-5)(5+5) = (-1)(10) = -10

Answer 22

I'm not saying systems with infinite critical points don't exist, but I'm not sure this is such a system.

Answer 23

Ahhh, duh, it seems pretty obvious now what you're saying, let me work it out real quick thanks!

Answer 24

Unless (x',y') = (0,c) or (x',y') = (c,0) are critical too

Answer 25

Ok just to check, I got three critical points, (-2,2),(4,4) and(0,0). Does that seem right to you?

Answer 26

Yes

Answer 27

Cool. Thanks.

Answer 28

Profiling this. :3