# Eigenfunctions of an Integral Operator

1. Dec 2, 2009

### Kreizhn

1. The problem statement, all variables and given/known data

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

$$Kf(t) = \int_0^1 (1+st) f(s) \ ds$$

3. 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 $Kf(t) = \lambda f(t)$ 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?

2. Dec 2, 2009

### Dick

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.

3. Dec 2, 2009

### lanedance

not too sure.. but how about noticing the t depndence of the function... does the fact
$$\int_0^1 f(s) ds$$
and
$$\int_0^1 sf(s) ds$$
are constants help..

4. Dec 2, 2009

### Kreizhn

I had tried looking for eigenfunctions of the form $a + bt$ but for some reason got an inconsistent system of equations. I'll go back and look at it again.

5. Dec 2, 2009

### lanedance

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

6. Dec 2, 2009

### Kreizhn

So if try a function of the form $f(t) = t + a$ and try $K f(t) = \lambda f(t)$ I get the following system of equations
\begin{align*} \frac13 + \frac a2 &= \lambda \\ \frac12 + a &= \lambda a \end{align*}

I can solve this to get $a = 1.868517092, \lambda = 1.267591879$. 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?

7. Dec 2, 2009

### lanedance

the only other option is a constant, which i don't think works as you end up with a t dependence

8. Dec 2, 2009

### Kreizhn

Indeed, so you're saying that the only possible eigenfunctions are of the form $f(t) = t+a$ since for all other functions, $Kf(t)$ will be first order polynomial in t.

9. Dec 2, 2009

### Dick

Don't you get a quadratic equation in lambda? Aren't there two eigenvalues?

10. Dec 2, 2009

### Kreizhn

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. Dec 2, 2009

### Dick

Sure. Just checking...