Eigenvalues for integral operator

  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. jbunniii

    jbunniii 3,353
    Science Advisor
    Homework Helper
    Gold Member

    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. Oops. :( But is this the right way to do it?
  5. jbunniii

    jbunniii 3,353
    Science Advisor
    Homework Helper
    Gold Member

    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. 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. 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. 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. Thank you very much. I think the light MAY be starting to dawn.
  10. 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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