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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 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?

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    Note $g(x,t)=\min\{t,x\}-tx$.2012-04-15

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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.

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    @ oenamen: Thanks. I asked a follow-up question [here](http://math.stackexchange.com/q/132531/24484).2012-04-16