Timeline for Solving Fredholm integral equation in Lp
Current License: CC BY-SA 4.0
4 events
| when toggle format | what | by | license | comment | |
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| Apr 25, 2019 at 9:20 | comment | added | Pietro Majer | To put it in a slightly more general frame: your operator is a rank-one perturbation of the identity (hence Fredholm-$0$): $f\mapsto f-\lambda \langle\phi,f\rangle u$. Here with $u(x):=e^x$ and $\langle\phi,f\rangle:=\int_a^b e^{-y}f(y)dy$; the same conclusions hold in general. | |
| Apr 24, 2019 at 17:06 | comment | added | jonathan wolf | @PietroMajer , thank you so much for your valuable comment! :) | |
| Apr 23, 2019 at 10:08 | comment | added | Pietro Majer | in fact the last "another idea" is correct, so you already answered! | |
| Apr 23, 2019 at 9:24 | history | asked | jonathan wolf | CC BY-SA 4.0 |