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I think this question isn't that hard, but I am a bit confused:

Define $$(Af)(x):=\int_{0}^{1}\cos(2\pi(x-y))f(y)dy.$$ Then $A$ is an operator on functions. Find the eigenvalues and the eigenfunctions.

I can think of a lot of functions that give $0$, things like $f(x)=\cos(n2\pi x)$. Also one eigenfunction that gives eigenvalue $\frac {1}{2}$ (I think). My problem is I have no idea how to show that I have found all of them, and I don't know if I have found all of them.

Thanks for showing me!!

Edit: I havent really shown my work because there is a lot of it and it is all over the place. J.M. suggest that it is useful to write $$Af(x)=\cos(2\pi x)\int_{0}^{1}\cos\left(2\pi y\right)f(y)dy+\sin\left(2\pi x\right)\int_{0}^{1}\sin\left(2\pi y\right)f(y)dy.$$ I used this to find that $\cos(2\pi x)+\sin (2\pi x)$ is an eigenvector. I wasn't sure if this is the right track, and I am still confused about what to do from here. (Specifically, once we find a bunch of $\lambda$ how do we prove that is all of them?)

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So, a homogeneous Fredholm integral equation... – J. M. Apr 20 '11 at 15:09
Note that the kernel is separable (the cosine kernel can be decomposed using difference formulae); this helps a great deal. – J. M. Apr 20 '11 at 15:31
For every homogeneous Fredholm equation, there is an associated secular equation you need to solve to find the eigenvalues; you should end up with a quadratic. – J. M. Apr 20 '11 at 15:50
@J.M. How does one do this? And what is a secular equation? Thank you for your help. – Math Student Apr 20 '11 at 15:51
up vote 6 down vote accepted

Let's start with where you're stuck ($\lambda$ is an eigenvalue):

$$\lambda f(x)=\cos(2\pi x)\int_0^1\cos\left(2\pi y\right)f(y)\mathrm dy+\sin\left(2\pi x\right)\int_0^1\sin\left(2\pi y\right)f(y)\mathrm dy$$

Let $c_1=\int_0^1\cos\left(2\pi y\right)f(y)\mathrm dy$ and $c_2=\int_0^1\sin\left(2\pi y\right)f(y)\mathrm dy$; $f(x)$ thus has the form

$$f(x)=\frac1{\lambda}\left(c_1\cos(2\pi x)+c_2\sin\left(2\pi x\right)\right)$$

We then assemble two equations: one where both sides of the equation are multiplied by $\cos(2\pi x)$, and one multiplied by $\sin(2\pi x)$, and then integrate both sides of the equation. I'll do it for $\cos(2\pi x)$:

$$\cos(2\pi x)f(x)=\frac1{\lambda}\left(c_1\cos(2\pi x)\cos(2\pi x)+c_2\sin\left(2\pi x\right)\cos(2\pi x)\right)$$

$$c_1=\frac1{\lambda}\left(c_1\int_0^1\cos(2\pi x)\cos(2\pi x)\mathrm dx+c_2\int_0^1\sin\left(2\pi x\right)\cos(2\pi x)\mathrm dx\right)$$

You should be able to recognize an (algebraic) eigenvalue equation at this point; the "secular equation" I was referring to in the comments is sometimes also referred to as the "characteristic polynomial".

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Why are you allowed to exchange the integration and $1/A$ (assuming that's what you did to get the last equation from the one above it and the definition of $c_1$)? – joriki Apr 20 '11 at 16:23
@joriki: Oops, I treated $A$ as an eigenvalue; I'll use different variables. Thanks for that! – J. M. Apr 20 '11 at 16:25
OK, that takes care of the eigenvalues, but to find all the eigenfunctions, you still need to treat the case $\lambda=0$ separately, right? – joriki Apr 20 '11 at 16:35
@joriki: Nobody really considers $\lambda=0$ for homogeneous Fredholm equations of the second kind unless it's a root of the secular equation (somewhat similar to asking "why not set $\lambda=0$ in $\mathbf A\mathbf x=\lambda\mathbf x$?"). – J. M. Apr 20 '11 at 16:42
"Why not set $\lambda=0$ in $\mathbf A\mathbf x=\lambda\mathbf x$?" seems like a perfectly reasonable question to me :-). If I were asked to find the eigenvalues and eigenvectors of $\mathbf A$, I'd assume that that includes the eigenvectors for eigenvalue $0$. (Things may be different for homogeneous Fredholm equations, which I know very little about; I'm just commenting on your analogy :-) – joriki Apr 20 '11 at 16:46

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