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Eigenfunctions of an Integral Operator

  1. Dec 2, 2009 #1
    1. The problem statement, all variables and given/known data

    If there are any eigenvalues for the following integral operator, calculate them

    [tex] Kf(t) = \int_0^1 (1+st) f(s) \ ds [/tex]

    3. The attempt at a solution

    I've tried making this into a differential equation, to no avail. I've also just tried solving the equation [itex] Kf(t) = \lambda f(t) [/itex] though that also didn't lead anywhere. I'm not sure if there's some special way of attacking problem like this. Does anybody have any ideas?
     
  2. jcsd
  3. Dec 2, 2009 #2

    Dick

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    It's a lot easier than you are think it is. Your integral expression is a linear function of t, no matter what f is.
     
  4. Dec 2, 2009 #3

    lanedance

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    not too sure.. but how about noticing the t depndence of the function... does the fact
    [tex]\int_0^1 f(s) ds [/tex]
    and
    [tex]\int_0^1 sf(s) ds [/tex]
    are constants help..
     
  5. Dec 2, 2009 #4
    I had tried looking for eigenfunctions of the form [itex] a + bt [/itex] but for some reason got an inconsistent system of equations. I'll go back and look at it again.
     
  6. Dec 2, 2009 #5

    lanedance

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    i haven't tried, but for non-zero b
    (a+bt) = b(a/b+t)

    so if (a+bt) is an eigenfunction, then so is (a/b+t) = c+t which ma be easier to deal with
     
  7. Dec 2, 2009 #6
    So if try a function of the form [itex] f(t) = t + a [/itex] and try [itex] K f(t) = \lambda f(t) [/itex] I get the following system of equations
    [tex] \begin{align*}
    \frac13 + \frac a2 &= \lambda \\
    \frac12 + a &= \lambda a
    \end{align*}
    [/tex]

    I can solve this to get [itex] a = 1.868517092, \lambda = 1.267591879 [/itex]. But it seems odd that an integral operator would only have one eigenvalue in it's spectrum. Furthermore, how can I be sure that this is exhaustive? How would I find other eigenvalues and eigenfunctions?
     
  8. Dec 2, 2009 #7

    lanedance

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    the only other option is a constant, which i don't think works as you end up with a t dependence
     
  9. Dec 2, 2009 #8
    Indeed, so you're saying that the only possible eigenfunctions are of the form [itex] f(t) = t+a [/itex] since for all other functions, [itex] Kf(t) [/itex] will be first order polynomial in t.
     
  10. Dec 2, 2009 #9

    Dick

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    Don't you get a quadratic equation in lambda? Aren't there two eigenvalues?
     
  11. Dec 2, 2009 #10
    Yes. I was being lazy an had plugged this into a Maple to solve numerically. But it isn't hard to do algebraically. I realized this, but figured the thread was dead so it wasn't important to post it :P
     
  12. Dec 2, 2009 #11

    Dick

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    Sure. Just checking...
     
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