Question

A critical point c of an autonomous first-order DE is said to be isolated if there exists some open interval that contains c but no other critical point. Can there exist an autonomous DE of the form dy/dx=f(y) for which every critical point is nonisolated?

Answer 1

what is a non-isolated critical point

Answer 2

It's a critical point where there doesn't exist any open interval that solely contain the point. So it's basically tightly surrounded by other critical points.

Answer 3

Hard problem, doesn't really see where one should start.

Answer 4

A critical point c of an autonomous first-order DE is said to be isolated if there exists some open interval that contains c but no other critical point. does that mean that x=0 and y=0 is an example of isolated point

Answer 5

Yeah, I was thinking of just setting f(y) = 0, so that we have the autonomous DE dy/dx = 0. Then every point is critical so every point is surrounded by other critical points, and hence non-isolated.

Answer 6

I don't see anything flawed in that reasoning, so the answer must be that it's true.

Answer 7

how do i explain it though

Answer 8

what can i use as an example

Answer 9

Just use f(y)=0 -> dy/dx = 0 as an example

Answer 10

so for a critical point f(c)=0
therefore y(x)=c
f(y)=c

dy/dx=f(y)

so of we set f(y)=0 we get dy/dx=0
i suppose this means that we get a line which is continuous and stable where all values of x are critical points therefore non-isolated.

Answer 11

so it would an equilibrium solution

Answer 12

Yeah, exactly, that's a better way to put it.