Eigenvalues for integral operator

[margaret37]

Homework Statement

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]

Homework Equations

\lambda x= Kx

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?

 

[jbunniii]

Homework Statement

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]

Homework Equations

\lambda x= Kx

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

 

[margaret37]

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

 

[jbunniii]

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)

 

[arkajad]

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?

 

[margaret37]

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

 

[arkajad]

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

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

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

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

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

?

 

[margaret37]

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

 

[arkajad]

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.