# Thread: 2nd Order Linear PDE involving the Div operator

1. ## 2nd Order Linear PDE involving the Div operator

Hi folks,

p(x,y), q(x,y) and f(x,y) are known functions and p(x,y) is positive and continuously diferentiable. Write out the following in co-ordinate form and determine whether its hyperbolic, parabolic or elliptic.
Here is my attempy to write it in co-ordinate form...I dont think its right

Given $
\vec \nabla \cdot (p \vec\nabla u)+qu=f
$

$
\displaystyle \vec\nabla \cdotp(p(x,y)\frac{\partial u }{\partial x}i+p(x,y)\frac{\partial u }{\partial y}j)+q(x,y)u(x,y)=f(x,y)
$

$
\displaystyle p(x,y)(\frac{\partial}{\partial x}(\frac{\partial u}{\partial x}i)+\frac{\partial}{\partial y}(\frac{\partial u}{\partial y}j))+q(x,y)u(x,y)=f(x,y)$

$
\displaystye p(x,y)\left [\frac{\partial^{2}u}{\partial x^2}+\frac{\partial^{2}u}{\partial y^2}\right]+q(x,y)u(x,y)=f(x,y)
$

Doesnt look right and I dont know how you legitimately remove i and j etc..thanks

2. Originally Posted by bugatti79
Hi folks,

p(x,y), q(x,y) and f(x,y) are known functions and p(x,y) is positive and continuously diferentiable. Write out the following in co-ordinate form and determine whether its hyperbolic, parabolic or elliptic.
Here is my attempy to write it in co-ordinate form...I dont think its right

Given $
\vec \nabla \cdot (p \vec\nabla u)+qu=f
$

$
\displaystyle \vec\nabla \cdotp(p(x,y)\frac{\partial u }{\partial x}i+p(x,y)\frac{\partial u }{\partial y}j)+q(x,y)u(x,y)=f(x,y)
$

$
\displaystyle p(x,y)(\frac{\partial}{\partial x}(\frac{\partial u}{\partial x}i)+\frac{\partial}{\partial y}(\frac{\partial u}{\partial y}j))+q(x,y)u(x,y)=f(x,y)$
$
\displaystyle \frac{\partial}{\partial x} \left ( p(x,y)(\frac{\partial u}{\partial x}) \right ) +\frac{\partial}{\partial y}\left ( p(x,y)(\frac{\partial u}{\partial y}) \right ) +q(x,y)u(x,y)=f(x,y)$

-Dan

3. I agree that this is correct, but you need to use the product rule here.

$
\displaystyle \vec\nabla \cdotp(p(x,y)\frac{\partial u }{\partial x}\mathbf{i}+p(x,y)\frac{\partial u }{\partial y}\mathbf{j})+q(x,y)u(x,y)=f(x,y)
$

The Cartesian basis vectors are constant and will not change while taking derivatives.

$\displaystyle p(x,y)\frac{\partial^2 u }{\partial x^2}+p(x,y)\frac{\partial ^2u }{\partial y^2}+p_{x}(x,y)\frac{\partial u }{\partial x}+p_{y}(x,y)\frac{\partial u }{\partial y}+q(x,y)u(x,y)=f(x,y)
$

Now $A=C=p(x,y),B=0$ this gives

$0^2-[p(x,y)]^2< 0$

So the equation is elliptic

4. Originally Posted by TheEmptySet
I agree that this is correct, but you need to use the product rule here.

$
\displaystyle \vec\nabla \cdotp(p(x,y)\frac{\partial u }{\partial x}\mathbf{i}+p(x,y)\frac{\partial u }{\partial y}\mathbf{j})+q(x,y)u(x,y)=f(x,y)
$

The Cartesian basis vectors are constant and will not change while taking derivatives.

$\displaystyle p(x,y)\frac{\partial^2 u }{\partial x^2}+p(x,y)\frac{\partial ^2u }{\partial y^2}+p_{x}(x,y)\frac{\partial u }{\partial x}+p_{y}(x,y)\frac{\partial u }{\partial y}+q(x,y)u(x,y)=f(x,y)
$

Now $A=C=p(x,y),B=0$ this gives

$0^2-[p(x,y)]^2< 0$

So the equation is elliptic
Thanks guys. I like the title 'The chair of approximate accuracy'