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Thread: 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.
    $\displaystyle (1+x)yu_{xx}+2xyu_{xy}-xyu_{yy}=0$

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


    Thanks
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  2. #2
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    Re: Type of PDE help

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

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

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

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

    Is this correct?

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

    Thanks
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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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    Re: Type of PDE help

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

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

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

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

    Is this correct?

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

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

    Your second part looks good.
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