Timeline for Asymptotic form of the eigenvalues / eigenfunctions of a Fredholm equation
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13 events
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| Jul 23, 2017 at 7:55 | comment | added | valle | @fedja It's the form that Mathematica returns. Numerical computation is in agreement with theory of course (mathematica.stackexchange.com/a/152064/534). But there should be an analytic expression for the eigenfunctions that makes it clear that they are real. | |
| Jul 22, 2017 at 21:20 | comment | added | fedja | It may be the general solution, but the way it is written is certainly utterly confusing. What I know for sure is that there are two linearly independent real solutions to any second order linear equation with non-degenerate coefficient at $f''$ and it is a bit hard to see them in this crazy form with imaginary arguments. To figure out what's going on, I need to refresh my memory about all those fancy special function representations and it is one of my least favorite topics :-). | |
| Jul 22, 2017 at 18:50 | comment | added | valle | @fedja So you are saying that $f(x) = c_1 I_0(4\sqrt{(x+a)\lambda_2} + c_2 K_0(4\sqrt{(x+a)\lambda_2}$ is not the general solution? | |
| Jul 21, 2017 at 15:18 | comment | added | fedja | Well, if you have a complex solution, its conjugate must be a solution too, so your formula is a bit incomplete. | |
| Jul 21, 2017 at 14:11 | comment | added | valle | ... which I think cannot be simultaneously met by $\lambda_2$ unless $a$ is very specific | |
| Jul 21, 2017 at 14:05 | comment | added | valle | @fedja Yes, I noticed that much. In particular $I_0(\mathrm i x) = J_0(x)$ which is real. However $K_0(\mathrm i x)$ is complex, and I don't see how a constant complex coefficient can make it real.... unless $c_2 = 0$, in which case we must require $J_0(4\sqrt{a|\lambda_2|}) = J_0(4\sqrt{(1 + a)|\lambda_2|}) = 0$, which I think cannot be simultaneously met. This is where I am stuck. | |
| Jul 21, 2017 at 13:53 | comment | added | fedja | $\lambda_2$ is necessarily negative, so your arguments in the formula written are purely imaginary (and the coefficients complex). | |
| Jul 21, 2017 at 11:57 | comment | added | valle | @fedja You are totally correct (as usual). But I cannot tame these eigenfunctions into satisfying the boundary conditions. According to Mathemtica, the solution of $(pf')' = \lambda_2 f$ is $f(x) = c_1 I_0(4\sqrt{(x+a)\lambda_2} + c_2 K_0(4\sqrt{(x+a)\lambda_2}$ (modified Besel functions), but I cannot find $\lambda_2,c_1,c_2$ so as to make $f(0)=f(1)=0$ while $f(x)$ is real. | |
| Jul 17, 2017 at 0:30 | comment | added | fedja | Erm... Are you claiming that the Sturm-Liouville operator $f\mapsto (pf')'$ with positive smooth $p$ (I assume that $a>0$) has no eigenfunctions with the Dirichlet boundary conditions on $[0,1]$? That sounds fishy, doesn't it? | |
| Jul 17, 2017 at 0:04 | history | edited | valle | CC BY-SA 3.0 |
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| Jul 16, 2017 at 15:23 | history | edited | valle | CC BY-SA 3.0 |
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| Jul 16, 2017 at 15:14 | history | edited | valle | CC BY-SA 3.0 |
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| Jul 16, 2017 at 15:05 | history | asked | valle | CC BY-SA 3.0 |