Eigenfunctions of an Integral Operator

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Homework Help Overview

The discussion revolves around the eigenvalues and eigenfunctions of a specific integral operator defined by Kf(t) = ∫₀¹ (1+st) f(s) ds. Participants are exploring the nature of this operator and the implications for its eigenvalues.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants are attempting to derive eigenvalues by setting up the equation Kf(t) = λf(t) and exploring different forms of f(t), including linear combinations. Some express uncertainty about the completeness of their findings and question the existence of multiple eigenvalues.

Discussion Status

There is ongoing exploration of potential eigenfunctions, with some participants suggesting specific forms and discussing the implications of their findings. Questions about the sufficiency of the identified eigenvalues and the completeness of the approach are being raised, indicating a productive dialogue.

Contextual Notes

Some participants note the challenge of finding eigenfunctions that satisfy the integral operator's properties, and there is mention of the operator's linearity affecting the form of potential solutions. The discussion reflects a mix of analytical and numerical approaches to the problem.

Kreizhn
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Homework Statement



If there are any eigenvalues for the following integral operator, calculate them

[tex]Kf(t) = \int_0^1 (1+st) f(s) \ ds[/tex]

The Attempt at a Solution



I've tried making this into a differential equation, to no avail. I've also just tried solving the equation [itex]Kf(t) = \lambda f(t)[/itex] though that also didn't lead anywhere. I'm not sure if there's some special way of attacking problem like this. Does anybody have any ideas?
 
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It's a lot easier than you are think it is. Your integral expression is a linear function of t, no matter what f is.
 
not too sure.. but how about noticing the t depndence of the function... does the fact
[tex]\int_0^1 f(s) ds[/tex]
and
[tex]\int_0^1 sf(s) ds[/tex]
are constants help..
 
I had tried looking for eigenfunctions of the form [itex]a + bt[/itex] but for some reason got an inconsistent system of equations. I'll go back and look at it again.
 
i haven't tried, but for non-zero b
(a+bt) = b(a/b+t)

so if (a+bt) is an eigenfunction, then so is (a/b+t) = c+t which ma be easier to deal with
 
So if try a function of the form [itex]f(t) = t + a[/itex] and try [itex]K f(t) = \lambda f(t)[/itex] I get the following system of equations
[tex]\begin{align*}<br /> \frac13 + \frac a2 &= \lambda \\<br /> \frac12 + a &= \lambda a<br /> \end{align*}[/tex]

I can solve this to get [itex]a = 1.868517092, \lambda = 1.267591879[/itex]. But it seems odd that an integral operator would only have one eigenvalue in it's spectrum. Furthermore, how can I be sure that this is exhaustive? How would I find other eigenvalues and eigenfunctions?
 
the only other option is a constant, which i don't think works as you end up with a t dependence
 
Indeed, so you're saying that the only possible eigenfunctions are of the form [itex]f(t) = t+a[/itex] since for all other functions, [itex]Kf(t)[/itex] will be first order polynomial in t.
 
Kreizhn said:
So if try a function of the form [itex]f(t) = t + a[/itex] and try [itex]K f(t) = \lambda f(t)[/itex] I get the following system of equations
[tex]\begin{align*}<br /> \frac13 + \frac a2 &= \lambda \\<br /> \frac12 + a &= \lambda a<br /> \end{align*}[/tex]

I can solve this to get [itex]a = 1.868517092, \lambda = 1.267591879[/itex]. But it seems odd that an integral operator would only have one eigenvalue in it's spectrum. Furthermore, how can I be sure that this is exhaustive? How would I find other eigenvalues and eigenfunctions?

Don't you get a quadratic equation in lambda? Aren't there two eigenvalues?
 
  • #10
Yes. I was being lazy an had plugged this into a Maple to solve numerically. But it isn't hard to do algebraically. I realized this, but figured the thread was dead so it wasn't important to post it :P
 
  • #11
Kreizhn said:
Yes. I was being lazy an had plugged this into a Maple to solve numerically. But it isn't hard to do algebraically. I realized this, but figured the thread was dead so it wasn't important to post it :P

Sure. Just checking...
 

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