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