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Linear Differential Operator order

  1. Sep 17, 2007 #1
    [Solved] Linear Differential Operator order

    1. The problem statement, all variables and given/known data
    I'm misunderstanding something basic about how this works:

    Give examples of linear differential operators [tex] L [/tex] and [tex] M [/tex] for which it is not true that [tex] L(M(u)) = M(L(u)) [/tex] for all u.

    2. Relevant equations
    Since it's arbitrary, I made two first order differential functions of two variables:
    [tex] \overline{x} = {x,y} [/tex]
    [tex] u=u(x,y) [/tex]
    [tex] L(u) = a(\overline{x})u + b_1(\overline{x})u_x + b_2(\overline{x})u_y [/tex]
    [tex] M(u) = c(\overline{x})u + d_1(\overline{x})u_x + d_2(\overline{x})u_y [/tex]
    Where a,b,c, and d are coefficients.

    3. The attempt at a solution
    When I expand [tex] L(M(u)) [/tex], it seems to look like [tex] M(L(u)) [/tex]:

    [tex] L(M(u)) = a(\overline{x})c(\overline{x})u+a(\overline{x})d_1(\overline{x})u_x +a(\overline{x})d_2(\overline{x})u_y + b_1(\overline{x})c(\overline{x})u_x+b_2(\overline{x})c(\overline{x})u_y [/tex]

    [tex] M(L(u)) = c(\overline{x})a(\overline{x})u+c(\overline{x})b_1(\overline{x})u_x +c(\overline{x})b_2(\overline{x})u_y + d_1(\overline{x})a(\overline{x})u_x+d_2(\overline{x})a(\overline{x})u_y [/tex]

    I'm stuck, since it seems that all roads lead back to a commutative relationship.
    Last edited: Sep 17, 2007
  2. jcsd
  3. Sep 17, 2007 #2


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    Staff Emeritus
    Science Advisor

    Linear differential operators with constant coefficients are commutative.

    Try using linear differential operators with variable coefficients. That is, something like (d/dx+ 3x)y or (xd/dx)u
  4. Sep 17, 2007 #3
    [Solved] Linear Differential Operator order

    Thanks. I just needed a nudge to get in the right direction. Some parts of DiffEq have been a while for me. I set up the linear differential operators

    [tex] L = \frac {d}{dx} [/tex]
    [tex] M = \frac {d}{dx} + x [/tex]

    to show that [tex] L(M(u)) \neq M(L(u)) [/tex] for all u. I think you helped me become 2 points smarter (don't ask me the scale...).
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