# Thread: Type of PDE help

1. ## 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

2. ## 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.

3. ## Re: Type of PDE help

Originally Posted by Danny
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

4. ## Re: Type of PDE help

EDIT: sorry, thought I made a mistake.

Can anyone confirm that my above working is correct?

5. ## Re: Type of PDE help

Originally Posted by ellenu485
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.