# Find the eigenvalues and eigenvectors of an integral operator

I need to find the eigenvalues e eigenvectors of this integral.

$$\int_{0}^{1} K(x,y)\phi (y)dy,$$

where

$K(x,y)=x(1-y),\; 0 \le x\le y \le 1$

and

$K(x,y)=y(1-x),$ $0\le y\le x \le 1$

I really need an explanation here so I can solve the rest of the exercises that I have here.

• Please can You help me out??? Commented May 27, 2013 at 20:34
• I really need help on this, I'm tying to solve this for about a week and I can't... Can you please help? This is for an assignment! Commented May 28, 2013 at 16:35
• I'm pretty sure shouting won't get you any more help. On the contrary, even. What have you tried so far to solve this problem?
– HSN
Commented May 28, 2013 at 16:36
• I don't know where to begin because I don't understand what to do mainly... Commented May 28, 2013 at 16:38
• I know how to solve second kind Fredholm integral equations but this one I don't get the meaning and what to do with it Commented May 28, 2013 at 16:39

## 1 Answer

Eigenvalues of this integral operator are those values of $\lambda$ for which the equation $$(1-x) \int_0^x y\phi(y)dy + x \int_x^1 (1-y)\phi(y)dy = \lambda \phi(x)$$ has non-trivial solutions. Putting $x=0$, $x=1$ in this equation and differentiating this equation twice it follows that $$-\phi(x)=\lambda \phi''(x), \quad \phi(0)=\phi(1)=0.$$ It is obvious that $\lambda=0$ is not an eigenvalue. The general solution is $$\phi(x)=A \cos \frac{x}{\sqrt{\lambda}} + B \sin \frac{x}{\sqrt{\lambda}}$$ and $\phi(0)=0$ implies $A=0$. So eigenvalues are exactly the roots of the equation $$\sin \frac{1}{\sqrt{\lambda}}=0,$$ i.e. $$\lambda_n=\frac{1}{\pi^2 n^2}, \quad n=1,2,3,\ldots$$ Corresponding eigenfunctions are $$\sin n \pi x, \quad n=1,2,3,\ldots$$

• The OP hasn't specified the domain of definition, but for integral operators this is usually some $L^2$ spaces, so why do you assume that $\phi$ is differentiable, given that it may not be? In other words, you should prove that there are no other eigenfunctions (and eigenvalues) besides the ones that you have found. Commented Jun 16, 2019 at 17:29
• Interestingly (and orthogonally to the OP's question), due to regularity properties of $K$, it's is possible to get that $\lambda_n = \mathcal O(n^{-2})$ without actually computing it Commented Oct 30, 2020 at 20:34
• @AlexM.: if $\lambda\ne0$, then $\phi$ is differentiable. Commented Apr 7, 2021 at 19:41