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Eigenvalues of the operator x d/dx

  1. Mar 27, 2008 #1
    a) Consider the operator x d/dx(where 1st d/dx acts on the function, then x acts on the resulting function by simply multiplying by x )acting on the set of functions of a real variable x for x>0. What are the eigenvalues and the corresponding eigenfunctions of this operator?

    b) What about d/dx x (where 1st x acts on the function, then d/dx acts upon the resulting function)? What are the eigenvaules and corresponding eigenfunctions for this?
     
  2. jcsd
  3. Mar 27, 2008 #2

    CompuChip

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    Write down the eigenvalue equation. This will be a differential equation, which you will need to solve..
     
  4. Mar 27, 2008 #3
    i don't think i quite understand what you are saying
     
  5. Mar 27, 2008 #4

    CompuChip

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    What does it mean for a function f to be an eigenvalue of x(d/dx)? What equation must it satisfy?
     
  6. Mar 27, 2008 #5
    if L is an opertaor than
    L[f(x)]=cf(x); where f(x) is an eigenfunction and c is an eigenvalue
     
  7. Mar 27, 2008 #6

    CompuChip

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    Exactly. What equation do you get when you plug in the L you are given.
    Do you have any idea how to solve this?
     
  8. Mar 27, 2008 #7
    x d/dx[f(x)]=cf(x)....not really
     
  9. Mar 27, 2008 #8

    CompuChip

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    OK, hint: try [tex]f(x) = x^k[/tex] for some k
     
  10. Mar 27, 2008 #9

    kdv

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    Why "not really"? This is correct!

    In other words,

    [tex] x y' = c y [/tex]

    where a prime indicates a derivative with respect to x and c is some constant. This is a simple differential equation, right? Can you solve it?
     
    Last edited: Mar 27, 2008
  11. Mar 27, 2008 #10
    I think you meant

    [tex]
    x y' = c y
    [/tex]
     
  12. Mar 27, 2008 #11

    kdv

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    Yes indeed. Thanks for pointing out the typo. I will correct this in my post so that it does not confuse the OP.

    Thanks!
     
  13. Mar 27, 2008 #12
    yeah, i took D.E. like 3 years ago and i'm a bit hazy on the details...it looks familar but i'm not really sure where to start it
     
  14. Mar 27, 2008 #13

    kdv

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    It's a separable DE....

    [tex] x \frac{dy}{dx} = c y \rightarrow \frac{dy}{y} = c \frac{dx}{x} [/tex]

    You only need to integrate both sides
     
  15. Mar 29, 2008 #14
    Considering the quantity and type of posts you've put up at this site, eit32, you may want to dust off your DEq and linear algebra books, because most of your questions deal more with the mechanics of math operations rather than physical principles themselves.
     
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