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Eigenvalues for integral operator

by margaret37
Tags: eigenvalues, integral, operator
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margaret37
#1
Nov22-10, 07:11 PM
P: 12
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
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jbunniii
#2
Nov22-10, 08:32 PM
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Quote Quote by margaret37 View Post
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?
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]
margaret37
#3
Nov22-10, 09:03 PM
P: 12
Oops. :( But is this the right way to do it?

jbunniii
#4
Nov22-10, 09:18 PM
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Eigenvalues for integral operator

Quote Quote by margaret37 View Post
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 [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]
arkajad
#5
Nov23-10, 03:31 AM
P: 1,411
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?
margaret37
#6
Nov23-10, 08:49 AM
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...

[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
arkajad
#7
Nov23-10, 09:32 AM
P: 1,411
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

???
margaret37
#8
Nov23-10, 10:30 AM
P: 12
Thank you very much. I think the light MAY be starting to dawn.
arkajad
#9
Nov23-10, 10:44 AM
P: 1,411
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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