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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.

Ron Gordon
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Deiota
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1 Answers1

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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 $$\sin n \pi x, \quad n=1,2,3,\ldots$$

njguliyev
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    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. – Alex M. Jun 16 '19 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 – dohmatob Oct 30 '20 at 20:34
  • @AlexM.: if $\lambda\ne0$, then $\phi$ is differentiable. – Martin Argerami Apr 07 '21 at 19:41