Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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?

share|improve this question
1  
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
1  
@Patrik, obviously the two definitions are closely related to each other. Thanks for editing the question. –  Mikael Anderson Apr 15 '12 at 22:11
1  
Note $g(x,t)=\min\{t,x\}-tx$. –  anon Apr 15 '12 at 22:17

1 Answer 1

up vote 4 down vote accepted

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.

share|improve this answer
    
@ 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
    
@MikaelAnderson: Keep in mind that the integral equation imposes boundary conditions, they are built-in. If you change the kernel you also change the boundary conditions. For example, they may become very restrictive, as you have noticed above. –  user26872 Apr 16 '12 at 5:05

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.