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
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]
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]
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?
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
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 [tex] (Ky)(s) := \int^{\ell}_0 (s-t)y(s) dt [/tex] 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 [tex] \int^{\ell}_0 (t-s)x(s) ds =\lambda x(t) [/tex] 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 x(t)=at+b ???
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.