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Math Help - 2nd Order Linear PDE: u_xx+2u_yx+5u_yy-2u_x+u=0 - Elliptic

  1. #1
    Senior Member bugatti79's Avatar
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    2nd Order Linear PDE: u_xx+2u_yx+5u_yy-2u_x+u=0 - Elliptic

    Folks,

    I believe this 2nd order PDE is elliptic. I want to make sure new coordinates are correct before I find the canonical form.

     \displaystyle u_{xx}+2u_{yx}+5u_{yy}-2u_x+u=0 \implies \frac{dy}{dx}=1 \pm \sqrt{-4}=1 \pm 2i \therefore

     \displaystyle y=\int 1 \pm 2i dx= x \pm i 2x +constant

    The constants are in the form \phi(x,y)-i \psi(x,y)=constant

    I choose for my new co ordinates s(x,y)=y-x and  t(x,y)=-2x

    Are s and t correct?

    thanks
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  2. #2
    MHF Contributor
    Jester's Avatar
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    Yes - they're good. I might add that you can choose t = 2x (no negative) and still be good.
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  3. #3
    Senior Member bugatti79's Avatar
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    Quote Originally Posted by Danny View Post
    Yes - they're good. I might add that you can choose t = 2x (no negative) and still be good.
    Ok, thanks Danny. The rest is jut plug and chug stuff.
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