# 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: $$g(x,t)=\min((1-x)t,(1-t)x), 0<x<1, 0<t<1$$ 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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Shouldn't you be interested in $\int_0^1 g(x,t) y_i(t) \, dt = \lambda_i y_i(x)$? (I'm writing this in the form $Ax = \lambda x$ that you can recall from linear algebra.) – Patrick Da Silva Apr 15 '12 at 22:05
I just edited your question to attract readers more precisely ; you're trying to find eigenfunctions of a linear operator $L[y](x) = \int_0^1 g(x,t) y(t) \, dt$. – Patrick Da Silva Apr 15 '12 at 22:10
@Patrik, obviously the two definitions are closely related to each other. Thanks for editing the question. – Mikael Anderson Apr 15 '12 at 22:11
Note $g(x,t)=\min\{t,x\}-tx$. – anon Apr 15 '12 at 22:17

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.
@ oenamen: Many thanks indeed for your wonderful solution. May I ask a follow up question. Suppose we change minimum to maximum in the definition of $g(x,t)$. Of course one can follow your solution to find the new eigenfunctions. I wonder if there is a smart way to use the relationship that $$\min(x,t)=-\max(-x,-t)$$ and fins eigenfunctions for $$g(x,t)=\max((1-x)t,(1-t)x)$$? – Mikael Anderson Apr 16 '12 at 0:13
@MikaelAnderson: You're very welcome. I don't see a way to use these solutions to get those for your new $g$. Perhaps someone else does. – user26872 Apr 16 '12 at 0:28
For the new $g$ function as above I followed oenamen's solution and after taking the second derivative I find $$y′′=λy.$$ If I understand correctly, the solution to this is of the form $$y=c_1\exp(x\sqrt(\lambda)+c_2\exp(-x\sqrt(\lambda).$$ I do not seem to be able to find a simple answer for $c_1,c_2$ though. I wonder if anyone can see how I can continue to find eigenfunctions as in oenamen's solution for the new $g$. – Mikael Anderson Apr 16 '12 at 1:15
@MikaelAnderson: There is one solution, $y = A\left(\sqrt{\lambda} \cosh\sqrt{\lambda}x - \sinh \sqrt{\lambda}x\right)$ for $\lambda$ a solution to $\tanh\sqrt\lambda = 2\sqrt\lambda/(1+\lambda)$. Numerically, $\lambda = 2.382\cdots$. This can be most easily seen by relating the integral equation to a differential equation with Robin boundary conditions. There are too many details for a comment. Perhaps you should ask another question! – user26872 Apr 16 '12 at 5:05