Timeline for Uniqueness for Volterra equation with initially (linearly) unbounded kernel
Current License: CC BY-SA 4.0
6 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Jun 2 at 17:57 | comment | added | ResearchMath | I will double check my argument and post in a bit if correct. | |
| Jun 2 at 17:48 | comment | added | e.lipnowski | I'd be happy to understand this question in the case that your $A$ is empty. Should the answer be obvious in that case? | |
| Jun 2 at 15:04 | comment | added | ResearchMath | How about a condition $A:=\lbrace y:k_2(y,y)=0\rbrace$ has measure zero? Or $A $ contains no interval? | |
| May 31 at 17:44 | comment | added | e.lipnowski | Differentiating the first bullet tells us that every $y\in(0,1]$ has $\int_0^y \tilde k(x,y) \text{ d}x=1.$ Since the average value of $\tilde k(\cdot,y)$ on $[0,y]$ is $\tfrac1y$, the function $\tilde k$ cannot be bounded. | |
| May 31 at 15:34 | comment | added | ResearchMath | Maybe obvious, but if you assume $\tilde k$ is bounded then I think the Volterra equation of the second kind has a unique solution. | |
| May 31 at 14:52 | history | asked | e.lipnowski | CC BY-SA 4.0 |