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Math Help - Phase plane problem

  1. #1
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    Unhappy Phase plane problem

    Hi there, for the following equation that is in the phase plane (x,y),

    d^{2}x/dt^{2}-4x+x^{3}=0 <--- sorry, there was a typo originally, now correct

    where dx/dt=y

    I need to prove that the solutions of the coupled, nonlinear system are given by

    y^{2}=1/2(C+4-x^{2})(C-4+x^{2})

    where C is a constant. So what I have done so far is:

    d^{2}x/dt^{2}-4x+x^{3}=0

    So the equivalent system is:

    F(x,y) = dx/dt=y (as stated in the question)
    G(x,y) = dy/dt-4x+x^{3}=0 =>  dy/dt=4x-x^{3}

    So from this, I tried letting

    dy/dx=(dy/dt)/(dx/dt)=G(x,y)/F(x,y)

    This upon solving this, I got y^{2}=4x^{2}-(x^{4})/4+2C

    which seems close, but I am not sure this is the right method or not
    Last edited by TheFirstOrder; June 9th 2010 at 04:09 AM. Reason: typo
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  2. #2
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    I don't think you're reducing the problem to two first-order systems correctly. If you look carefully, when you take the derivative of y, that's really the second derivative of x, which does not show up in the original DE.
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  3. #3
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    Quote Originally Posted by Ackbeet View Post
    I don't think you're reducing the problem to two first-order systems correctly. If you look carefully, when you take the derivative of y, that's really the second derivative of x, which does not show up in the original DE.

    Oh yes, sorry, that was a typo as the second derivative of x is meant to show up in the equation. So, I am right after that?
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  4. #4
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    Well, this step:

    <br />
(dy/dt)/(dx/dt)=G(x,y)/F(x,y)<br />

    is an incorrect middle step, but you fix it in the next step. What you really mean is this:

    <br />
\frac{dy}{dx}=\frac{dy/dt}{dx/dt}=\frac{4x-x^{3}}{y}<br />
.

    Let me see. I'm not good at skipping lots of steps, so forgive me for reproducing the separation of variables:

    y\,dy=(4x-x^{3})\,dx, so
    \frac{y^{2}}{2}=\frac{4x^{2}}{2}-\frac{x^{4}}{4}+C, and therefore
    y^{2}=4x^{2}-\frac{x^{4}}{2}+2C.
    If you like, you can absorb the 2 into the C.

    So I got a different constant multiplying the x^{4} term.

    Now, I dislike factoring. I would take your target expression, and simplify:

    y^{2}=1/2(C+4-x^{2})(C-4+x^{2})=1/2(C^{2}-4C+Cx^{2}+4C-16+4x^{2}-Cx^{2}+4x^{2}-x^{4})
    =1/2(C^{2}-16+8x^{2}-x^{4}).

    If you compare this to the result we got above, you can see that the result follows by simply re-defining the constant of integration.
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  5. #5
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    Thank! I didn't think of redefining the constant! thank you so very much!
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  6. #6
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    You're very welcome. Have a good one.
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  7. #7
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    Sorry, another question in regard to this one:

    Let's say I let C=4 and then C=5. How do I go about drawing the phase portraits for these values?

    What I have done so far is for the formula

    y^{2}=1/2(C+4-x^{2})(C-4+x^{2})

    I simply let C=4

    y^{2}=1/2(8-x^{2})(x^{2})

    and then moving the right hand side to the left, I solved and got the equilibrium points as (0,0), (8^1/2,0), (-(8^(1/2)),0) and then how do I figure out what sort of points are they (i.e. saddle points, centres)? And then, how do I draw it??? I've tried drawing in WolframAlpha expect that only give a basic outline. how do I go about figuring out when I would get periodic solutions? Your help is much appreciated.
    Last edited by TheFirstOrder; June 9th 2010 at 08:36 PM. Reason: latex error
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  8. #8
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    Well, you can always draw the phase portrait by hand as a last resort. I would think Mathematica would be able to draw a phase portrait. Try the StreamPlot command. What you need to know is the x and y components of the field at each point.
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