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Help with differential operators

  1. Nov 14, 2007 #1
    1. The problem statement, all variables and given/known data
    This is a problem about differential operators, but I don't really get the notation used. I have L1 = (d/dx + 2) and L2 = (d/dx - 1)

    Find L1(xe^-2x)

    Show that L1L2 = L2L1 and find L1L2 in terms of d/dx, d2/dx2, etc.

    2. Relevant equations

    3. The attempt at a solution
    So i thought that L1 = (d/dx + 2) means L1(f) = (df/dx + 2), so

    L1(xe^-2x) = -2xe^-2x + e^-2x + 2

    And so L1L2 is just L1 acting on L2 (or L1L2(f) is just performing L2 on f and then performing L1 on that result) and so L1L2 is just the differential of L2 plus 2

    L1L2 = d2/dx2 + 2

    But by that logic, L2L1 = d2/dx2 - 1

    So what's wrong

  2. jcsd
  3. Nov 14, 2007 #2


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    Homework Helper

    No, actually it means that
    L1(f) = (d/dx + 2) f = df/dx + 2f.
    In particular, you see that you should read a constant c as that constant times the zeroth derivative (= the function).

    L1(xe^-2x) = -2xe^-2x + e^-2x + 2 x e^-2x (= e^-2x).

    You better always act them on a test function and then throw that away afterwards. So for example, L1 L2 can be found by working out
    L1 L2 f = (d/dx + 2) (d/dx - 1) f = (d/dx + 2) (df/dx - f) = (d2f/dx2 - df/dx + 2 df/dx - 2 f)
    L1 L2 = d2/dx2 - d/dx + 2 d/dx - 2,
    and similarly for L2 L1 (didn't check for errors, you'll find out soon enough if I did it right :smile:)
  4. Nov 14, 2007 #3


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    If l1 operates on a function, f(x), you will have:

    [tex]L_1 f=\frac{df}{dx}+2f [/tex]
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