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

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

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

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

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

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

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

-Dan

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

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

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

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

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

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

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.

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

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

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

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

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

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