Critical points

1. Sep 22, 2008

Somefantastik

x' = y(1 + x - y2)
y' = x(1 + y - x2).

C.P. => (x,y) s.t. f(x) = 0

$$f(x) = x' = 0 => y = 0$$

$$f(x) = y' = 0 => x = 0$$

=> (0,0) is c.p.

plug x = 0 into x':

$$x' = y(1 - y^{2}) => y = 0, y =\ ^{+}_{-}1$$

=> (0,1), (0,-1) are critical points.

plug y = 0 into y':

$$y' = x(1 - x^{2}) => x = 0, x =\ ^{+}_{-}1$$

=> (1,0), (-1,0) are critical points.

Can someone help me find the 2 critical points that I am missing?

Thanks.

2. Sep 22, 2008

NoMoreExams

You have

x' = y(1 + x - y2)
y' = x(1 + y - x2).

Think of these as

x' = a*b
y' = c*d

Therefore for x' = 0 you can have a = 0 OR b = 0 and for y' = 0 you can have c = 0 OR d = 0 i.e. your possible pairs are

1) a = 0, c = 0
2) a = 0, d = 0
3) b = 0, c = 0
4) b = 0, d = 0

this translates to

1) y = 0, x = 0
2) y = 0, 1 + y - x^2 = 0 i.e. 1 - x^2 = 0, i.e. x = +/-1
3) 1 + x - y^2 = 0, x = 0 i.e. y = +/-1

Now the ones you are missing i.e.

4) 1 + x - y^2 = 0 and 1 + y - x^2 = 0

Using the first equation you get x = y^2 - 1 so plug that into the 2nd equation and you get 1 + y - (y^2 - 1)^2 = 0 which you can solve.

Does that help?

3. Sep 22, 2008

Somefantastik

Yeah, it really does. Thanks a lot. For some reason I had trouble with this same thing back in intermediate DE too. Thanks for giving me a better perspective.

4. Sep 22, 2008

Somefantastik

What if I have

x' = 16x2 + 9y2 - 25
y' = 16x2 + 162

Would i find x(y) when x' = 0 and plug into y'?

5. Sep 22, 2008

NoMoreExams

I assume your 2nd equation is y' = 16x^2 + 16y^2?

You would set both equal to 0 like normally

x' = 0 gives you 16x^2 = 25 - 9y^2

y' = 0 gives you 16x^2 + 16y^2 = 0, subst. that in and you get

25 - 9y^2 + 16y^2 = 0 i.e. 25 + 7y^2 = 0 which doesn't seem to have any solutions over R but are you allowed to work with C?

Remember you are essentially finding where these 2 functions(f(x,y) = x' and g(x,y) = y') so you can use whatever method you want, subst., putting them in a matrix if they are linear, etc.

6. Sep 23, 2008

Somefantastik

The back of the book claims (1,1), (-1,-1), (-1,1), (1,-1) are the critical points.

Aaaaaaaaaaaand, typo :*(

y' = 16x2 - 16y2.

Sorry about that; I was really tired and frustrated when I posted earlier.

7. Sep 23, 2008

NoMoreExams

Ok so now you have

x' = 16x^22 + 9y^2 - 25
y' = 16x^2 - 16y^2,

As I said, set both equal to 0, the first one gives you 16x^2 = 25 - 9y^2, plug that into the 2nd one and you get 25 - 9y^2 - 16y^2 = 0, can you solve that?

8. Sep 23, 2008

Somefantastik

Yeah; I got it. Thanks for your help.

Last edited: Sep 23, 2008