How to find eigenfunctions of a linear operator I wonder if there is a general way of finding characteristic values and eigenfunctions of a given linear operator described by an integral.
As a special case suppose I am interested in this function:
$$g(x,t)=\min((1-x)t,(1-t)x), 0&ltx&lt1, 0&ltt&lt1$$ and I want to find $\lambda_i$ and $y_i(x)$ such that
$$y_i(x)-\lambda_i\int_0^1g(x,t)y_i(t)dt=0.$$
How can I do this?
 A: We find 
$$y(x) = \lambda \left[ (1-x)\int_0^x dt\, t y(t) + x\int_x^1 (1-t) y(t)\right].$$
This has a form similar to a Volterra equation of the first kind. 
A standard technique is to take derivatives, thus transforming the integral equation into a differential equation. 
Taking the second derivative of both sides with respect to $x$ we find 
$$y'' = -\lambda y.$$
Thus, the solutions should be of the form 
$$y = A \sin\sqrt{\lambda} x + B \cos\sqrt{\lambda} x.$$
Plugging this back into the original integral equation we find $B = 0$ and $\lambda = n^2 \pi^2$, where $n\in\mathbb{N}$.
Thus, the solutions are of the form 
$$y = A \sin n \pi x$$
with eigenvalues 
$$\lambda = n^2 \pi^2.$$
Addendum. 
Another possible solution to the differential equation that we ignored above is $\lambda=0$ and $y = A+ B t$. 
However, this is not compatible with the integral equation unless $A = B = 0$. 
Notice also that hyperbolic solutions have been ruled out. 
Initially we just assumed $\lambda$ was some complex number. 
The integral equation then told us that $\lambda$ is real. 
