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

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

Thanks

2. ## 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.

3. ## Re: Type of PDE help

Originally Posted by Danny
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

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 $\displaystyle A=(1+x)y$? 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 $\displaystyle x > 0$.