
#1
Nov2210, 07:11 PM

P: 12

1. The problem statement, all variables and given/known data
Find all nonzero eignvalues and eigenvectors for the following integral operator [latex] Kx := \int^{\ell}_0 (ts)x(s) ds [/latex] in [latex] C[0,\ell] [/latex] 2. Relevant equations [latex] \lambda x= Kx [/latex] 3. The attempt at a solution [latex] \int^{\ell}_0 (ts)x(s) ds = \lambda * x(t) [/latex] [latex] t\int^{\ell}_0 x(s) ds  s\int^{\ell}_0x(s) ds = \lambda * x(t) [/latex] 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 



#2
Nov2210, 08:32 PM

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PF Gold
P: 2,933

[tex]\int^{\ell}_0 sx(s) ds \neq s\int^{\ell}_0x(s) ds[/tex] 



#3
Nov2210, 09:03 PM

P: 12

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




#4
Nov2210, 09:18 PM

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HW Helper
PF Gold
P: 2,933

eigenvalues for integral operatorBut 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 (ts) x(s) ds = \lambda x(t)[/tex] 



#5
Nov2310, 03:31 AM

P: 1,412

First of all you should correctly state the problem. It should read:
[tex](Kx)(t) := \int^{\ell}_0 (ts)x(s) ds[/tex] Can you see the difference? 



#6
Nov2310, 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 



#7
Nov2310, 09:32 AM

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
[tex] (Ky)(s) := \int^{\ell}_0 (st)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 (ts)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 ??? 



#8
Nov2310, 10:30 AM

P: 12

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




#9
Nov2310, 10:44 AM

P: 1,412

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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