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Thread: 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),

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

    where $\displaystyle dx/dt=y$

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

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

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

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

    So the equivalent system is:

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

    So from this, I tried letting

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

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

    which seems close, but I am not sure this is the right method or not
    Last edited by TheFirstOrder; Jun 9th 2010 at 03:09 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 $\displaystyle y$, that's really the second derivative of $\displaystyle x$, 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 $\displaystyle y$, that's really the second derivative of $\displaystyle x$, 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:

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

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

    $\displaystyle
    \frac{dy}{dx}=\frac{dy/dt}{dx/dt}=\frac{4x-x^{3}}{y}
    $.

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

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

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

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

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

    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

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

    I simply let C=4

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

    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; Jun 9th 2010 at 07:36 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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