Results 1 to 4 of 4

Math Help - 2nd Order Linear PDE involving the Div operator

  1. #1
    Senior Member bugatti79's Avatar
    Joined
    Jul 2010
    Posts
    461

    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  <br />
\vec \nabla \cdot (p \vec\nabla u)+qu=f<br />

     <br />
\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)<br />

     <br />
\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)

     <br />
\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)<br />

    Doesnt look right and I dont know how you legitimately remove i and j etc..thanks
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Forum Admin topsquark's Avatar
    Joined
    Jan 2006
    From
    Wellsville, NY
    Posts
    9,855
    Thanks
    321
    Awards
    1
    Quote Originally Posted by bugatti79 View Post
    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  <br />
\vec \nabla \cdot (p \vec\nabla u)+qu=f<br />

     <br />
\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)<br />

     <br />
\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:
     <br />
\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
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Behold, the power of SARDINES!
    TheEmptySet's Avatar
    Joined
    Feb 2008
    From
    Yuma, AZ, USA
    Posts
    3,764
    Thanks
    78
    I agree that this is correct, but you need to use the product rule here.

     <br />
\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)<br />

    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)<br />

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

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


    So the equation is elliptic
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Senior Member bugatti79's Avatar
    Joined
    Jul 2010
    Posts
    461
    Quote Originally Posted by TheEmptySet View Post
    I agree that this is correct, but you need to use the product rule here.

     <br />
\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)<br />

    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)<br />

    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'
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Linear Operator
    Posted in the Advanced Algebra Forum
    Replies: 1
    Last Post: November 6th 2011, 01:29 AM
  2. Linear operator
    Posted in the Differential Geometry Forum
    Replies: 11
    Last Post: June 5th 2011, 10:41 PM
  3. Linear operator
    Posted in the Advanced Algebra Forum
    Replies: 5
    Last Post: May 30th 2011, 09:43 PM
  4. Linear Operator
    Posted in the Differential Geometry Forum
    Replies: 2
    Last Post: May 25th 2011, 11:59 PM
  5. Is this Linear Operator?
    Posted in the Advanced Algebra Forum
    Replies: 1
    Last Post: March 14th 2010, 01:59 PM

/mathhelpforum @mathhelpforum