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Math Help - Re-writing higher order spatial derivatives as lower order system

  1. #1
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    Re-writing higher order spatial derivatives as lower order system

    I've been working with the following PDE:

    <br />
\[<br />
\nabla ^2 p = \frac{{\partial ^2 p}}{{\partial x^2 }} + \frac{{\partial ^2 p}}{{\partial y^2 }} = A\frac{{\partial ^2 p}}{{\partial t^2 }} + B\frac{{\partial p}}{{\partial t}}<br />
\]<br />

    What I would like to do is to re-write the second-order spatial derivatives in this PDE as first order derivatives.

    This is what I have attempted, but I am uncertain as to whether this is correct. I introduce another variable q, and then:

    <br />
\[<br />
\frac{{\partial q}}{{\partial t}} = \nabla  \cdot \vec p<br />
\]<br />

    <br />
\[<br />
\nabla  \cdot \vec q = A\frac{{\partial p}}{{\partial t}} + Bp<br />
\]<br />

    I reason that this is correct since

    <br />
\[<br />
\nabla  \cdot \frac{{\partial q}}{{\partial t}} = \nabla  \cdot \nabla  \cdot \vec p = \nabla ^2 p<br />
\]<br />

    <br />
\[<br />
\frac{\partial }{{\partial t}}\left( {\nabla  \cdot \vec q} \right) = \frac{\partial }{{\partial t}}\left( {A\frac{{\partial p}}{{\partial t}} + Bp} \right)<br />
\]<br />

    Now is it reasonable to claim that the LHS of the two equations above are the same?
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  2. #2
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    I see an immediate problem. You have both p and \vec{p}. These are not same! Similar for q and \vec{q}.
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  3. #3
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    Second \nabla \cdot \nabla \cdot \vec{p} makes no sense.
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  4. #4
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    Danny: Thank you very much for checking this over. Are the problems that you see issues with notation, or can the two equations be written in a more precise fashion?

    Thus

    <br />
\[<br />
\nabla  \cdot \frac{{\partial q}}{{\partial t}} = \nabla  \cdot \left( {\nabla  \cdot \vec p} \right)<br />
\]<br />

    Essentially what I would like to do is to write the PDE in terms of first order derivatives.
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  5. #5
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    So given

    <br />
\[<br />
\frac{{\partial ^2 p}}{{\partial x^2 }} + \frac{{\partial ^2 p}}{{\partial y^2 }} = A\frac{{\partial ^2 p}}{{\partial t^2 }} + B\frac{{\partial p}}{{\partial t}}<br />
\]<br />

    is there a way to split this into a system involving only first order derivatives?
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  6. #6
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    So to put this in another way:

    Is

    <br />
\[<br />
 \frac{{\partial ^2 p}}{{\partial x^2 }} + \frac{{\partial ^2  p}}{{\partial y^2 }} = A\frac{{\partial ^2 p}}{{\partial t^2 }} +  B\frac{{\partial p}}{{\partial t}}<br />
 \]<br />

    Also equivalent to the following?

    <br />
\[<br />
\frac{{\partial q}}{{\partial t}} = \frac{{\partial p}}{{\partial x}} + \frac{{\partial p}}{{\partial y}}<br />
\]<br />

    <br />
\[<br />
\frac{{\partial q}}{{\partial x}} + \frac{{\partial q}}{{\partial y}} = A\frac{{\partial p}}{{\partial t}} + Bp<br />
\]<br />
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  7. #7
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    These are not equivalent since elimiating the q gives

    Ap_{tt} + Bp _{t} = p_{xx} + 2 p_{xy} + p_{yy}.
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  8. #8
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    Thanks, Danny: I had suspected that something was amiss!

    What about this system? So starting with

    <br />
\[<br />
\nabla  \cdot \left( {\nabla p} \right) = \nabla ^2 p = \frac{{\partial ^2 p}}{{\partial x^2 }} + \frac{{\partial ^2 p}}{{\partial y^2 }} = A\frac{{\partial ^2 p}}{{\partial t^2 }} + B\frac{{\partial p}}{{\partial t}}<br />
\]<br />

    This is re-written as:

    <br />
\[<br />
A\frac{{\partial p}}{{\partial t}} + Bp = \nabla  \cdot \vec q<br />
\]<br />

    <br />
\[<br />
\frac{{\partial \vec q}}{{\partial t}} = \nabla p<br />
\]<br />

    Where q is a vector. I believe that this is correct since by taking the time derivative of the first equation:

    <br />
\[<br />
A\frac{{\partial ^2 p}}{{\partial t^2 }} + B\frac{{\partial p}}{{\partial t}} = \frac{\partial }{{\partial t}}\left( {\nabla  \cdot \vec q} \right)<br />
\]<br />

    Applying the del operator to the following:

    <br />
\[<br />
\nabla  \cdot \left( {\frac{{\partial \vec q}}{{\partial t}}} \right) = \frac{\partial }{{\partial t}}\left( {\nabla  \cdot \vec q} \right) = \nabla  \cdot \left( {\nabla p} \right) = \nabla ^2 p<br />
\]<br />

    We can thus say that both sides of the above are equal.
    Last edited by nkinar; July 27th 2010 at 08:48 AM. Reason: Fixed math
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  9. #9
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    Ya - I'll agree that that!
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  10. #10
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    Thanks, Danny! I think that it looks good as well! Should I go ahead and mark the thread as solved?
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  11. #11
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    Sure - If you're happy with the answer.
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  12. #12
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    Yes, I am happy with the answer. Once again, thank you very much! I'll go ahead and close the thread.
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