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Stability of a system of DE's

  1. Jan 15, 2012 #1

    fluidistic

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    1. The problem statement, all variables and given/known data
    Find the stationary solution(s) of the following system of DE and determine its stability:
    [itex]x'=x-x^3-xy^2[/itex].
    [itex]y'=2y-y^5-x^4y[/itex].


    2. Relevant equations
    [itex]x'=0, y'=0[/itex].
    Expansion of x and y around the critical point: [itex]x=x_0+\alpha[/itex], [itex]y=y_0+\beta[/itex].


    3. The attempt at a solution
    I solved [itex]x'=y'=0[/itex], this gave me [itex]x= \pm \sqrt {1-y^2}[/itex] or [itex]x=0[/itex]. And [itex]y=\pm \sqrt {\pm \sqrt {2-x^4}}[/itex]. Which is likely to give me 12 critical points. Let's consider the [itex](0,0)[/itex] first...
    I expanded x around [itex]x_0=y_0=0[/itex] via [itex]x=x_0+\alpha[/itex]. I reached that "[itex]x'=x-x^3-xy^2[/itex]" linearizes as [itex]\alpha ' \approx \alpha (1-3x_0^2-y_0^2)+x_0-x_0^3-x_0y_0^2-2x_0y_0\beta[/itex]. Here I see that I'm lucky that I chose first the critical point [itex](0,0)[/itex] because the linearization reduces to [itex]\alpha ' =\alpha[/itex] and so [itex]\alpha =Ae^t[/itex]. Right here I notice that when t grows up, so does alpha. So that I can already say that [itex](x,y)=(0,0)[/itex] is not a stable critical point. Is my reasoning right?

    P.S.:This is the very first exercise in the assignment of the course dealing with the study of critical points... Looks already quite complicated to me. :/
     
  2. jcsd
  3. Jan 16, 2012 #2

    fzero

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    This looks ok so far.
     
  4. Jan 17, 2012 #3

    fluidistic

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    Ok! Very good to know. Thanks for having taught me how to tackle these kind of problems.
     
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