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Math Help - Some simple DiffEQ problems.

  1. #1
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    Some simple DiffEQ problems.

    The differential equation dx/dt = (1/10)*x*(10-x)-h models a logistic population with with harvesting at a rate h. Determine the dependence of the critical points on the paramater h, and then construct a bifurcation diagram.


    Consider the differential equation dx/dt = kx-x^3. (a) If k<=0, show that the only critical value c=0 of x is stable. (b) If k>0, show that the critical point c=0 is now unstable, but that the critical points c= +/- sqrt(k) are stable. Thus the qualitative nature of the solutions changes at k=0 as the parameter k increases, and so k=0 is a bifurcation point for the differential equation with parameter k. The plot of all points of the form (k,c) where c is a critical point of the equation is the "pitchfork diagram" (here the book shows a c vs. k plot of a parabola that opens to the right)



    Any help?
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  2. #2
    MHF Contributor chisigma's Avatar
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    Let's write the first DE as...

    \frac{dx}{dt} = x - .1\cdot x^{2} - h (1)

    ... and suppose to know the 'initial value' x(0)= x_{0}. It is important to establish the sign of second member of (1) and to do that we have to factorize it as follows...

    -.1 \cdot (x^{2} -10\cdot x +10\cdot h) = -.1 (x-a) (x-b) (2)

    ... where...

     a= 5 + \sqrt {25 -10\cdot h}

     b= 5 - \sqrt{ 25 -10\cdot h} (3)

    No we have three possibilities...

    1) 25 -10\cdot h <0 \rightarrow h>2.5

    In this case x^{'} (t) is everywhere negative and \forall x_{0} is...

    \lim_{x \rightarrow \infty} x(t) = - \infty

    2) 25 -10\cdot h = 0 \rightarrow h=2.5

    In this case x^{'} (t) is everywhere negative with the only exception of x= 5 where is null. The situation is the same as 1) unless you have x_{0} = 5 and in this case you have the constant solution  x=5

    3) 25 -10\cdot h > 0 \rightarrow h<2.5

    In this case x^{'}(t) is negative for x>a or x<b , is null for x=a or x=b and is positive for a>x>b. We have three possibilities...

    3a) x_{0}>b \rightarrow \lim_{t \rightarrow \infty} = a

    3b) x_{0} = b \rightarrow \lim_{t \rightarrow \infty} = b

    3b) x_{0} < b \rightarrow \lim_{t \rightarrow \infty} = - \infty

    Kind regards

    \chi \sigma
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  3. #3
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    Quote Originally Posted by qtpipi View Post
    and then construct a bifurcation diagram.
    Any help?
    Hey, you into programming? Mathematica is a big help with this stuff. No I don't work for them; just know a good thing when I see it. You know what a bifurcation diagram is right? It illustrates the dependence of the critical points on the parameter. So just calculate the zeros of the right side as a function of h and then just plot the results. If you're interested, try interpreting the code below. It first finds the roots, selects only the real ones, then arranges them in a suitable order to connect the points and draw a smooth plot. Now, modify my code to construct the pitch-fork bifurcation for the second one.

    Also, in the news yesterday was an article about tipping points in global warming:

    Is There a Climate-Change Tipping Point? - TIME

    Here's something fun: Explain what that has to do with this thread.


    Code:
    Clear[h]
    myList = 
      Table[myRoots = x /. Solve[1/10 x (10 - x) - h == 0, x];
       {h, #} & /@ myRoots,
       {h, -5, 5, 0.1}];
    myRealRoots = 
      Flatten[Select[myList, Element[#[[2]], Reals] &], 1];
    
    curve = FindCurvePath[myRealRoots]
    ListLinePlot[myRealRoots[[curve[[1]]]]]
    Attached Thumbnails Attached Thumbnails Some simple DiffEQ problems.-bif-diagram.jpg  
    Last edited by shawsend; September 18th 2009 at 06:52 AM.
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