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Eigenvalues for integral operator

  1. Nov 22, 2010 #1
    1. The problem statement, all variables and given/known data

    Find all non-zero eignvalues and eigenvectors for the following integral operator

    [itex] Kx := \int^{\ell}_0 (t-s)x(s) ds [/itex]

    in [itex] C[0,\ell] [/itex]

    2. Relevant equations

    [itex] \lambda x= Kx [/itex]

    3. The attempt at a solution

    [itex] \int^{\ell}_0 (t-s)x(s) ds = \lambda * x(t) [/itex]

    [itex] t\int^{\ell}_0 x(s) ds - s\int^{\ell}_0x(s) ds = \lambda * x(t) [/itex]

    Am I even going the right direction?
    I think I need function(s) of x(t) and scalar [itex] \lambda [/itex], when I am finished is that right?

    And should [itex] \lambda [/itex] be a complex (possibly real) number?
    1. The problem statement, all variables and given/known data

    2. Relevant equations

    3. The attempt at a solution
  2. jcsd
  3. Nov 22, 2010 #2


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    Well, you are already on the wrong track because you cannot simply pull the "s" out of this integral when it's the variable of integration!

    [tex]\int^{\ell}_0 sx(s) ds \neq s\int^{\ell}_0x(s) ds[/tex]
  4. Nov 22, 2010 #3
    Oops. :( But is this the right way to do it?
  5. Nov 22, 2010 #4


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    Well, you haven't got that far yet so I can't say for sure.

    But the basic idea is that [itex]\lambda[/itex] will be a complex number (there might be more than one that work!), and corresponding to each such [itex]\lambda[/itex] there will be a family of specific functions [itex]x(t)[/itex] (the eigenvalues) which satisfy

    [tex]\int_0^\ell (t-s) x(s) ds = \lambda x(t)[/tex]
  6. Nov 23, 2010 #5
    First of all you should correctly state the problem. It should read:

    [tex](Kx)(t) := \int^{\ell}_0 (t-s)x(s) ds[/tex]

    Can you see the difference?
  7. Nov 23, 2010 #6
    I do see the difference. (Which is not how it is written on my assignment.)

    So does K operate on x? Are t and s scalars?

    So for a trivial example...

    [tex] \int {e^{nx}} = ne^nx [/tex]

    So is this a solution to some integral equation similar to the problem?

    Thank you for your answer
  8. Nov 23, 2010 #7
    K takes a function x. It produces a new function Kx. The value of the new function at t is calculated from the values of the old function by the process of integration. You operator could be as well written as

    (Ky)(s) := \int^{\ell}_0 (s-t)y(s) dt

    That would be exactly the same operator. Think about it!

    Anyway, you want to solve the equation

    [tex]Kx=\lambda x[/tex]

    Now two functions are equal when they are equal at all points. So the above means

    [tex](Kx)(t)=\lambda x(t)[/tex]

    for all t.

    So you substitute and get as much as possible from the equation

    \int^{\ell}_0 (t-s)x(s) ds =\lambda x(t)

    for all t.

    Now, if [itex]\lambda\neq 0[/itex] can you see from this equation that x(t) is necessarily a linear function of t, that is that it must be of the form


  9. Nov 23, 2010 #8
    Thank you very much. I think the light MAY be starting to dawn.
  10. Nov 23, 2010 #9
    One more thing. You should get now something like

    [tex]t\int_0^l(as+b)ds -\int_0^l s(as+b)=\lambda (at+b)[/tex]

    This is supposed to hold for all t, so, in particular, for t=0. So you will get two equations from the one above. Two equations for three unknowns: a,b,[itex]\lambda[/itex]. But do not worry. The eigenfunction are determined only up to a multiplicative constant.
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