Timeline for Does $f_1(x,y)<f_2(x,y)$ imply $y_1<y_2$ for solutions to the integral equation $y_k'=f_k(x,y_k)$?
Current License: CC BY-SA 2.5
7 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Mar 5, 2013 at 0:22 | vote | accept | AppliedSide | ||
| Dec 14, 2012 at 9:27 | answer | added | Gustaf Gripenberg | timeline score: 1 | |
| Jan 24, 2011 at 18:16 | comment | added | Pietro Majer | ah, I read your answer only after posting an answer. | |
| Jan 24, 2011 at 17:53 | comment | added | AppliedSide | (continued) For me the problem is interesting only if at least one of the $f_k$ is not continuous, and one only has that $y_k'=f_k(x,y_k)$ a.e. | |
| Jan 24, 2011 at 17:52 | comment | added | AppliedSide | That's one way to ensure existence of a solution, by Peano's existence theorem, but the proof of Peano's uniqueness does not require it. If both $f_k$ are continuous, then the solutions are classical solutions in that $y_k'(x)=f_k(x,y_k(x))$, and the problem becomes a matter of noticing that, $y_1(x)\leq y_2(x)$, on an interval $[0,\varepsilon)$, since $y_1'(0)\leq y_2'(0)$, and then you argue that the solutions can never cross such that $y_1(x)>y_2(x)$ after the crossing. | |
| Jan 24, 2011 at 16:31 | comment | added | Pietro Majer | The functions $f_1$ and $f_2$ are assumed continuous, right? | |
| Jan 24, 2011 at 1:11 | history | asked | AppliedSide | CC BY-SA 2.5 |