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Math Help - Eliminating dual solutions?

  1. #1
    Senior Member TriKri's Avatar
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    Eliminating dual solutions?

    Hi, is there some method for eliminating dual solutions to a given differential equation that are not wanted? Let me explain: Consider the differential equation

    \frac{\partial^2 y}{\partial t^2}=\hat{C}y\ (^*),

    where \hat{C} is a linear operator (that doesn't operate in the time dimension). However, since the equation is going to be solved numerically, and the operator \hat{C} is very computationally expensive to be implemented directly, the equation is modified:

    \frac{\partial^2 y}{\partial t^2}=\hat{C}y

    \Rightarrow \frac{\partial^2}{\partial t^2}\frac{\partial^2 y}{\partial t^2}=\frac{\partial^2}{\partial t^2}\hat{C}y

    \Leftrightarrow\frac{\partial^4 y}{\partial t^4}=\hat{C}\frac{\partial^2 y}{\partial t^2}

    \Leftrightarrow\frac{\partial^4 y}{\partial t^4}=\hat{C}^2y\ (^{**}),

    where \hat{C}^2 is much computationally cheaper to implement. However, this equation gives dual solutions that do not satisfy the original equation (*), since (**) can be shown to allow any linear combination of solutions to the equations

    \frac{\partial^2 y}{\partial t^2}=\hat{C}y (*, our original equation)

    and to

    \frac{\partial^2 y}{\partial t^2}=-\hat{C}y\ (^{***}).

    So the modified equation can't be used solely. Is there some way to eliminate the solutions given by (***) when solving the system numerically, and only obtain solutions given by (*)? Thanks in advance.
    Last edited by TriKri; March 12th 2012 at 04:40 AM.
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    Senior Member TriKri's Avatar
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    Re: Eliminating dual solutions?

    No one knows?
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  3. #3
    Senior Member TriKri's Avatar
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    Re: Eliminating dual solutions?

    I read in these course notes (on page 31) that "if a crazy looking operator like \sqrt{-\nabla^2_H} is okay, then the exact problem can be recast into a canonical problem". What is a canonical problem, and does that have anything to do with my question? Can that help me in some way? Thanks.




    P.S. The linear operator \hat{C} that I'm using in my problem has the property that if it's applied to a function e^{i\,\vec{k}\,\vec{x}}, the outcome will be -|\vec{k}|\,e^{i\,\vec{k}\,\vec{x}}, which I guess means that \hat{C} can be expressed as -\sqrt{-\nabla^2}, which looks very similar to the one used in the slide show.




    Edit: Oh, and I forgot to say that the PDE is only supposed to be solved numerically, not analytically.
    Last edited by TriKri; August 31st 2012 at 06:10 AM.
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