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Math Help - Type of PDE help

  1. #1
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    Type of PDE help

    Hi, I am having trouble with these questions and any help would be appreciated.

    Q1) Find the regions in the xy plane where the equation is elliptic,hyperbolic, or parabolic. Also sketch them.
    (1+x)yu_{xx}+2xyu_{xy}-xyu_{yy}=0

    Q2) Same as q1 but the equation is:
    (1+x)u_{xx}+2xyu_{xy}-y^2 u_{yy}=0


    Thanks
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  2. #2
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    Re: Type of PDE help

    If you let A = 1+x, B = 2xy and C = -xy, then you go to consider when B^2-4AC >0, = 0, <0. This will tell you when the PDE is hyperbolic, parabolic or elliptic.
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  3. #3
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    Re: Type of PDE help

    Quote Originally Posted by Danny View Post
    If you let A = 1+x, B = 2xy and C = -xy, then you go to consider when B^2-4AC >0, = 0, <0. This will tell you when the PDE is hyperbolic, parabolic or elliptic.
    Thanks Danny, but shouldn't A=(1+x)y? So I get B^2-4AC=4xy^2(2x+1).

    So for:
    Parabolic:
    4xy^2(2x+1)=0
    x=0, or y=0, or x=1/2

    Hyperbolic:
    4xy^2(2x+1)>0
    2x+1>0
    x> -1/2

    Elliptic:
    4xy^2(2x+1)<0
    x< -1/2

    Is this correct?

    And for 2) I get B^2-4AC=4y^2(x^2+x+1)
    Parabolic: y=0 since x^2+x+1 \ne 0
    Hyperbolic: x^2+x+1>0 (the entire graph)
    Elliptic: x^2+x+1<0 (which is nothing)

    Thanks
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  4. #4
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    Re: Type of PDE help

    EDIT: sorry, thought I made a mistake.

    Can anyone confirm that my above working is correct?
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  5. #5
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    Re: Type of PDE help

    Quote Originally Posted by ellenu485 View Post
    Thanks Danny, but shouldn't A=(1+x)y? So I get B^2-4AC=4xy^2(2x+1).

    So for:
    Parabolic:
    4xy^2(2x+1)=0
    x=0, or y=0, or x=1/2

    Hyperbolic:
    4xy^2(2x+1)>0
    2x+1>0
    x> -1/2

    Elliptic:
    4xy^2(2x+1)<0
    x< -1/2

    Is this correct?

    And for 2) I get B^2-4AC=4y^2(x^2+x+1)
    Parabolic: y=0 since x^2+x+1 \ne 0
    Hyperbolic: x^2+x+1>0 (the entire graph)
    Elliptic: x^2+x+1<0 (which is nothing)

    Thanks
    Yep - you're right. I missed the y on A. As for you're workings. For the hyperbolic case your want  y \ne0, x(2x+1) > 0. So in addition to want you have we also have x > 0.

    Your second part looks good.
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