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Drawing phase portraits.

  1. Apr 4, 2008 #1
    Can someone please explain to me how to draw the PP, I mean besides drawing the lines which i think i can understand, how do you decide how to draw the direction of the phase portrait?
    I mean for example we have: x'=y and y'=-2x^3, obviously those are parabolas and hyperbolas, now the equilibrium point is (x,y)=(0,0)
    for x=1 y=-1 y'=-2 so the direction should be decided by the fact that y'/x'=dy/dx=tg(a)>0 so it's going counter clockwise for the hyperbola in the fourth quardant, for x=1 y=1 it's negative then it goes clockwise, the same is for x=-1 y=1 it's counterclockwise, and for x=-1 y=-1 it's clockwise, have i got it correct, it's all about identifiy the clockwise anti clockwise by dy/dx, correct?

    Well, I think I got it, unless someone thinks otherwise, do tell me if i got it correct?
    thanks.
     
  2. jcsd
  3. Apr 4, 2008 #2

    HallsofIvy

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    You've chosen a difficult example! Linearized close to (0,0) that is equivalent to
    [tex]\frac{\left(\begin{array}{c} x \\ y\end{array}\right)}{dt}= \left(\begin{array}{cc}0 & 1 \\ 0 & 0\end{array}\right)\left(\begin{array}{c} x \\ y\end{array}\right)[/tex]
    which has only 0 as eigenvalue and has only multiples of <1, 0> as eigenvalues rather than two independent eigenvalues.
     
  4. Apr 4, 2008 #3
    I don't see, how would you draw the phase portrait, and it's not an example it's from a test, so it's quite crucial I will undersatnd it.

    The only time we used linearisation was for finding if it has an asymptotic equilibirium point, but in this case as youv'e pointed we don't have this, cause the real part of the eigen value is zero and not negative.
    for this point to be a stable point we need that the geometric multiplicity and algebraic will be equal, the algebraic is 2, the geometric multiplicity doesnt equal 2?

    anyway, how to draw then?

    thanks in advance as always.
     
  5. Apr 4, 2008 #4

    HallsofIvy

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    I ran it through software that came with a d.e. text and the phase plane diagram has ovals flowing cockwise. I had originaly worked it out for y'= -2x2 because I misread your post.
     
  6. Apr 4, 2008 #5
    can you upload the picture here so I can see it, I have a feeling that no one really knows how to draw them. (-:
     
  7. Apr 4, 2008 #6
    the equation should be x^4+y^2=constant, now this seems a bit tricky to draw, we know the function is: y=sqrt(C-x^4), C>=x^4, so it should be some sort of a composition of quardupabola and square root, don't have a clue how do draw such a thing.
     
  8. Apr 5, 2008 #7

    epenguin

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    Start with any convenient particular initial value of x and y, and value of K. Here e.g. some y=0 and any convenient value of the other two. Now any other point on the phase path starting there, I should say passing through there, has the same K but a different x, y. So what are other points to choose, for simplicity?

    You should soon see what that path looks like.
    You need some other paths. You get another by choosing a different K, and do the same thing as before.

    What is your conclusion?
     
  9. Apr 7, 2008 #8

    epenguin

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    PS for K I meant C above.

    You have solved the equation and it is not hard to see what the plots should look like. An intriguing and not difficult excercise is what do you predict according as C>1 and C<1 then check it with a plotter?

    Almost superfluous in your case but think anyway, is definitely part of seeing the scheme of the phase portraits in cases where you cannot solve the differential equation, what are the loci of dy/dx = 0 and dx/dy = 0 ?
     
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