# Eigenvalues for integral operator

1. ### margaret37

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

2. ### jbunniii

3,347
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!

$$\int^{\ell}_0 sx(s) ds \neq s\int^{\ell}_0x(s) ds$$

3. ### margaret37

12
Oops. :( But is this the right way to do it?

4. ### jbunniii

3,347
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

$$\int_0^\ell (t-s) x(s) ds = \lambda x(t)$$

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?

6. ### margaret37

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?

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

???

8. ### margaret37

12
Thank you very much. I think the light MAY be starting to dawn.

$$t\int_0^l(as+b)ds -\int_0^l s(as+b)=\lambda (at+b)$$
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.