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Finding Equilibrium Points

  1. May 10, 2012 #1
    1. The problem statement, all variables and given/known data

    http://img444.imageshack.us/img444/9288/51927159.jpg [Broken]


    3. The attempt at a solution

    (a) I'm mostly stuck on this part because I keep getting a complex number:

    3x+y = 0 so x=-y/3

    Substituting this in the second equation:

    [itex]\frac{y^2}{9} +1 = 0[/itex]

    [itex]\therefore \ y = \sqrt{-9} \implies y = 3 i[/itex]

    Now putting this back into x=-y/3 we will get x=-i. So the only equilibrium point is (-i, 3i)? :confused:


    (b) We can write the first equation yx-y as y(x-1) = 0. So the solutions here would be y=0 and x=1.

    And if we set the second equation equal to 0: cos(πx)=0, the solution would be

    [itex]x=\frac{k \pi}{2}[/itex]

    Where k is any odd integer (since cosine is zero at points like -π/2, π/2, 3π/2, etc).

    So the equilibrium points would be (kπ/2, 0) and (1, 0). Is this right? And could there be any more solutions?
     
    Last edited by a moderator: May 6, 2017
  2. jcsd
  3. May 10, 2012 #2

    tiny-tim

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    hi roam! :smile:
     
  4. May 10, 2012 #3

    HallsofIvy

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    Assuming that x and y are real numbers, getting a complex result tells you that there are no equilibrium points.
     
  5. May 10, 2012 #4
    Wouldn't that be wrong though? The problem asked for ALL equilibrium points and made no mention of restricting the domain to reals. So if I turned this in and said there ain't any, would my professor be justified in markin' it wrong? And sides, it's [itex]\pm 3i[/itex] right?
     
  6. May 10, 2012 #5
    Hi tiny tim,

    So do you mean I should pick the +ve side of ±3i, so the that the only equilibrium is at (-i, 3i)?

    What was wrong with what I did?

    Yes, I think we are required to find all values, not just reals.
     
  7. May 10, 2012 #6

    tiny-tim

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    i'm assuming x and y are both restricted to the reals

    (i'd be very surprised if that's not true)
    oops! nothing! :redface:

    you got 'em all! :biggrin:

    (but (1,0) is wrong … dy/dt = cosπ ≠ 0)
     
  8. May 10, 2012 #7
    Ok look. I think I just caused trouble with that. I'd bet you're just required to find the reals. I'm lucky the mods don't get on me for that but it does open up an interesting phenomenon: what exactly are complex equilibrium points? You see, doing the unexpected, being curious, asking questions, that's where real mathematics is created. :)
     
    Last edited: May 10, 2012
  9. May 10, 2012 #8
    But didn't we find from the first equation yx-y=y(x-1)=0 that (1,0) is a solution? Because this equation is zero at either x=1 and y=0. :confused:
     
  10. May 10, 2012 #9
    So (kπ/2, 0) are the only solutions? What happens to the x=1 that I found from the other equation?
     
  11. May 10, 2012 #10

    BruceW

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    yes, (kπ/2, 0) are the only solutions (with k an odd number, as you said).

    you found that (1,0) is a fixed point for x, but you need it to also be a fixed point for y. The equation to get a fixed point for y is cos(πx)=0 so does your point satisfy this?
     
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