Skip to main content
18 events
when toggle format what by license comment
Jun 24, 2021 at 22:23 vote accept Evan Gorman
Jun 22, 2021 at 2:23 answer added Eubos timeline score: 1
Jun 22, 2021 at 0:43 comment added Christian Remling @EvanGorman: I don't think anything is suspicious or unexpected here. My guess would be that the double antiperiodic ev's split into two simple ev's (that's another bunch of eigenfunctions, doubling the cosines), plus one eigenfunction with a positive eigenvalue.
Jun 21, 2021 at 20:03 comment added Evan Gorman @ChristianRemling Yes exactly, that's why I'm wondering where the rest of the eigenfunctions can be found because this operator should guarantee a full basis.
Jun 21, 2021 at 19:55 comment added Christian Remling In other words, the cosine functions only span about half the space anyway (if that makes sense).
Jun 21, 2021 at 19:53 comment added Christian Remling @EvanGorman: Actually, they do. The antiperiodic eigenfunctions (that is $f(1)=-f(0)$, $f'(1)=-f'(0)$) that don't satisfy your bc's are the sine functions, but that too messes things up (the antiperiodic eigenfunctions would have spanned the whole space, being the eigenfunctions of a self-adjoint operator).
Jun 21, 2021 at 19:47 comment added username \begin{align*} &(\int_0^x(x-y)f(y)dy+\int_x^1(y-x)f(y)dy)^\prime \\ =&\int_0^x f(y)dy +x(f(x)-f(0)) -xf(x)-\int_x^1f(y)dy \\&- x(f(1)-f(x))+f(1)-xf(x)\\ &= \int_0^x f(y)dy -\int_x^1f(y)dy - x(f(1)+f(0)) +f(1) \end{align*}
Jun 21, 2021 at 19:42 comment added Evan Gorman @ChristianRemling Why would $\cos((2n+1)\pi x)$ not be eigenfunctions? I believe they satisfy the boundary conditions I listed in my first comment. Are you getting different boundary conditions?
Jun 21, 2021 at 19:36 comment added Christian Remling Your first comment basically answers your question: with the correct boundary conditions, $\cos(2n+1)\pi x$ is actually not an eigenfunction, nor is $-2/((2n+1)\pi)^2$ an eigenvalue. The actual conditions are messier. So we can't tell very easily what space these span. In any event, there will also be a positive eigenvalue, with eigenfunction of the form $A\cosh kx + B\sinh kx$, $k^2=2/\lambda$. (More explicit conditions could be written down, but it doesn't look like one can solve it all the way through to the end.)
Jun 21, 2021 at 18:48 comment added Evan Gorman @username This is very helpful and the finite approximation does predict a non sign changing eigenfunction associated with the single positive eigenvalue. Which boundary conditions are you obtaining? Splitting the domain and differentiating I'm left with: $\int_0^xf(y)dy+\int_1^xf(y)dy=\lambda f'(x)$
Jun 21, 2021 at 18:22 comment added username I don't obtain that for the boundary conditions. Please check. You have a positive compact operator, the Krein-Rutman Theorem applies (the infinite version of Perron-Frobenius). There is a positive eigenvalue, correponding to the only non sign changing eigensolution.
Jun 21, 2021 at 18:06 comment added Evan Gorman @AbdelmalekAbdesselam This was my original thought as well, but assuming $\lambda>0$ leads to solutions of the form $f(x)=Ae^{\sqrt{\frac{2}{\lambda}}x}+Be^{-\sqrt{\frac{2}{\lambda}}x}$. Working the with boundary conditions (or just checking for solutions in maple) there are no nonzero solutions.
Jun 21, 2021 at 17:58 comment added Abdelmalek Abdesselam Form observation (1), it seems the most likely explanation is missing a positive eigenvalue which should be there en.wikipedia.org/wiki/Perron%E2%80%93Frobenius_theorem
Jun 21, 2021 at 17:12 comment added Evan Gorman Fixed, thanks. The boundary conditions can be obtained by splitting the domain of the integral operator to remove the absolute value and differentiating. They are $\lambda(f(0)+f(1))=\int_0^1f(y)dy$ and $f'(0)+f'(1)=0$
Jun 21, 2021 at 17:09 history edited Evan Gorman CC BY-SA 4.0
added 6 characters in body
Jun 21, 2021 at 16:10 review First posts
Jun 21, 2021 at 16:29
Jun 21, 2021 at 16:09 history edited YCor CC BY-SA 4.0
removed capitals from title
Jun 21, 2021 at 16:06 history asked Evan Gorman CC BY-SA 4.0