1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Proving linear operator for partial derivatives.

  1. Feb 1, 2006 #1
    How do I go about proving the following partial derivitive is a linear operator?
    d/dx[k(x)du/dx)]
     
  2. jcsd
  3. Feb 1, 2006 #2

    quasar987

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    What you've written is not a differential operator, it's a function.

    Or did you mean d/dx[k(x)d/dx)]?
     
  4. Feb 1, 2006 #3
    Nope, what I originally wrote is how it is in the book. If it matters, K is with a subscript of o.
     
  5. Feb 2, 2006 #4

    matt grime

    User Avatar
    Science Advisor
    Homework Helper

    So you're mutliplying by something, k(x)du/dx, and then you're differentiating. I hope you know that those are linear operators individually.
     
  6. Feb 2, 2006 #5

    dextercioby

    User Avatar
    Science Advisor
    Homework Helper

    But where is the partial derivative ?

    Daniel.
     
  7. Feb 2, 2006 #6

    HallsofIvy

    User Avatar
    Staff Emeritus
    Science Advisor

    Don't get naughty, Daniel! He clearly intended the "d" as partial derivative, he just didn't know how to write here.

    NINHARDCOREFAN, what quasar987 said is still valid. Whatever is written in your book, [tex]\frac{\partial }{\partial x}[/tex] is the operator, [tex]\frac{\partial u}{\partial x}[/tex] is the function that results from applying that operator to u.

    Of course, before you can prove anything is a linear operator, you have to know the definition of "linear operator"! An operator, F, is "linear" if and only if it takes linear combinations into linear combinations: F(au+ bv)= a F(u)+ bF(v) where a and b are numbers (constants) and u and v are objects in the vector space (in this case functions).

    So, to prove that [tex]\frac{\partial}{\partial x}\left(k(x,y,...)\frac{\partial}{\partial x}\right)[/tex] is a linear operator
    JUST DO IT!

    What is [tex]\frac{\partial}{\partial x}\left(k(x,y,...)\frac{\partial (au+ bv}{\partial x}\right)[/tex]?
     
  8. Sep 20, 2008 #7
    Please explain this proof futher. Let d be the partial derivative of d.
    So the equation is actually L(u)=d/dx[Ko(x)du/dx]. There is the du/dx at the end of the equation. Please explain further how this is interpreted: this means take the partial derivative of any function, multiply it by Ko(x) du/dx? This reads that Ko is a function of x correct?

    Please outline the stept to solve this.

     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?



Similar Discussions: Proving linear operator for partial derivatives.
Loading...