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Differntial equations & Polar Coords

  1. Feb 5, 2012 #1
    nzofmo.png

    Let Q = theta

    Let z=reiQ

    z' = (a+ib)reiQ - z|z|2

    |z|2 = r

    so

    z' = a*reiQ + ib*reiQ - r2eiQ

    Also

    z' = ireiQ

    The question asks for 2 differential equations, but I really have no idea where I'm going with this..

    Any help?

    Thanks
     
  2. jcsd
  3. Feb 5, 2012 #2

    Dick

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    No, if z=r*exp(iQ) then z'=r'*exp(iQ)+r*i*Q'*exp(iQ). Use the product rule. Both r and Q are functions. Now try and separate into real and imaginary parts.
     
  4. Feb 5, 2012 #3
    Ah ok!

    Would I be right in saying my two ODE's are now

    r' = ar - r2

    Q' = b

    by equating coefficients?

    Any idea on how to find the invariant sets? Do I have to solve thiese coupled ODE's somewhow?

    Thanks
     
  5. Feb 5, 2012 #4

    Dick

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    That's not quite what I get for the r'. I'm not quite sure what your definition of 'invariant set' is, but the equations are coming out to be uncoupled. So they should be pretty easy to solve.
     
  6. Feb 5, 2012 #5
    hmm I used

    z' = a*reiQ + ib*reiQ - r2eiQ

    was that not correct?

    My definition of invariant set is starting at some initial condition within the set, for all t>0 then you'll still be within the set.

    So since they're uncoupled

    r(t) = some function with r(0) and eat (yet to work it out - messy rearranging, and you say it's incorrect)

    Q(t) = bt + Q(0)

    Is it a trial and error thing for invariant? I don't know any method for it.

    Thanks
     
  7. Feb 5, 2012 #6

    Dick

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    In r' = ar - r^2, I'm questioning whether the r^2 is right. Go way back to the beginning. And for the invariant set try plotting some solutions for fixed values of a and b and see what they look like.
     
  8. Feb 5, 2012 #7
    oh is it r^3?

    Am I right in thinking I should be integrating 1/ar-r3 between r(0) and r(t)

    I just did the integration but the rearrangement into r(t) looks impossible
     
  9. Feb 5, 2012 #8
    Yeah and I have 1/2a (2log(r) - log(r2-a))

    The problem is when I integrate between r(0) and r(t) I get an expression in the form that has an r(t)a and an r(t)2

    So it seems impossible for me to get r(t) out of the expression...
     
  10. Feb 5, 2012 #9

    Dick

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    Yes, it's r^3. Solving for r doesn't look that hard to me. I'm not sure where this is going as far as the invariant sets part.
     
  11. Feb 5, 2012 #10

    Dick

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    If it's any help notice that solutions with an initial value r(0) that satisfies r(0)^2=a are pretty simple. Try thinking about it starting from that.
     
  12. Feb 5, 2012 #11
    Oops double post
     
    Last edited: Feb 5, 2012
  13. Feb 5, 2012 #12
    Perhaps I shouldn't even be trying to integrate it then.

    I have the fact that

    (r',Q') = M(r,Q) (both column vectors, M is a matrix)

    So when r = 0 and b = 0 then the invariant set is jsut the point (0,0)

    When r=sqrt(a) then the invariant set is the circle with radius sqrt(a), or is it not? What can I say about b?
     
  14. Feb 5, 2012 #13

    Dick

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    ONE of the invariant sets is the circle of radius sqrt(a), if a is positive and b is nonzero. And r=0 is an invariant set no matter what a or b is. Keep thinking along those lines.
     
  15. Feb 6, 2012 #14
    Ok this is what I have do far:

    The equilibrium point, (0,0) is an invariant set. Because when r=0 r'=0, and we don't care about the angle, so b (and a) can be anything.

    When r = root a, and angle Q(t) = bt + Q(0), so the angle change is linear, so this is a circle. Can I show it's an invariant set by showing this is a periodic orbit somehow?
     
  16. Feb 6, 2012 #15
    I just solved the system using Matlab with script:

    vdpol.m:

    function ydot=vdpol(t,y)
    a=1;
    b=1
    ydot=[a*y(1)-y(1)^3; b];

    vdpolrun.m

    clear
    tspan=[0 200];
    y0= [0.5; 0];
    [t,y]=ode45(@vdpol,tspan,y0);
    polar(y(:,2),y(:,1))

    And whatever initial condition I take the solution is always the periodic orbit with radius root a, how can I mathematically show this is the case for every initial condition, and hence every initial condition within the circle?

    Does this mean there are infinitely many invariant sets? Since a can take any positive value?
     
  17. Feb 6, 2012 #16

    Dick

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    I really don't know what that means and I'm not an expert with invariant sets. But if one of your solutions is a circle and the other is a point and you know the solutions don't cross (how?) then what can you conclude? I suggest you analyze the qualitative behavior of the solutions inside the circle and outside. And the solutions might be pretty different if b=0 or a<0.
     
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