# eigenvalues for integral operator

by margaret37
Tags: eigenvalues, integral, operator
 P: 12 1. The problem statement, all variables and given/known data Find all non-zero eignvalues and eigenvectors for the following integral operator $Kx := \int^{\ell}_0 (t-s)x(s) ds$ in $C[0,\ell]$ 2. Relevant equations $\lambda x= Kx$ 3. The attempt at a solution $\int^{\ell}_0 (t-s)x(s) ds = \lambda * x(t)$ $t\int^{\ell}_0 x(s) ds - s\int^{\ell}_0x(s) ds = \lambda * x(t)$ Am I even going the right direction? I think I need function(s) of x(t) and scalar $\lambda$, when I am finished is that right? And should $\lambda$ 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
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 Quote by margaret37 1. The problem statement, all variables and given/known data Find all non-zero eignvalues and eigenvectors for the following integral operator $Kx := \int^{\ell}_0 (t-s)x(s) ds$ in $C[0,\ell]$ 2. Relevant equations $\lambda x= Kx$ 3. The attempt at a solution $\int^{\ell}_0 (t-s)x(s) ds = \lambda * x(t)$ $t\int^{\ell}_0 x(s) ds - s\int^{\ell}_0x(s) ds = \lambda * x(t)$ Am I even going the right direction? I think I need function(s) of x(t) and scalar $\lambda$, when I am finished is that right? And should $\lambda$ be a complex (possibly real) number?
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!

$$\int^{\ell}_0 sx(s) ds \neq s\int^{\ell}_0x(s) ds$$
 P: 12 Oops. :( But is this the right way to do it?
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## eigenvalues for integral operator

 Quote by margaret37 Oops. :( But is this the right way to do it?
Well, you haven't got that far yet so I can't say for sure.

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

$$\int_0^\ell (t-s) x(s) ds = \lambda x(t)$$
 PF Patron P: 1,412 First of all you should correctly state the problem. It should read: $$(Kx)(t) := \int^{\ell}_0 (t-s)x(s) ds$$ Can you see the difference?
 P: 12 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... $$\int {e^{nx}} = ne^nx$$ So is this a solution to some integral equation similar to the problem? Thank you for your answer
 PF Patron P: 1,412 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 $$Kx=\lambda x$$ Now two functions are equal when they are equal at all points. So the above means $$(Kx)(t)=\lambda x(t)$$ 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 $\lambda\neq 0$ 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 ???
 P: 12 Thank you very much. I think the light MAY be starting to dawn.
 PF Patron P: 1,412 One more thing. You should get now something like $$t\int_0^l(as+b)ds -\int_0^l s(as+b)=\lambda (at+b)$$ 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,$\lambda$. But do not worry. The eigenfunction are determined only up to a multiplicative constant.

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