Timeline for Controlling subsolutions of a second order linear ODE
Current License: CC BY-SA 3.0
16 events
| when toggle format | what | by | license | comment | |
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| Jan 17, 2018 at 22:15 | comment | added | Iosif Pinelis | It is possible, by strengthening the argument a bit. I have posted an answer to your additional question at mathoverflow.net/questions/290965/… . | |
| Jan 17, 2018 at 4:10 | comment | added | H_R | Do you know if it's possible to control (in terms of $a$) the smallest $x$ such that $f(x) \leq 0$? | |
| Jan 12, 2018 at 14:51 | comment | added | Iosif Pinelis | I have updated the answer to show more clearly that actually more is proved than requested. | |
| Jan 12, 2018 at 14:48 | history | edited | Iosif Pinelis | CC BY-SA 3.0 |
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| Jan 12, 2018 at 5:37 | vote | accept | H_R | ||
| Jan 12, 2018 at 5:36 | comment | added | H_R | Oh I'm sorry, I didn't see that $R$ could be infinite. | |
| Jan 12, 2018 at 4:39 | comment | added | Iosif Pinelis | @H_R : Because $r$ is positive and concave and hence nondecreasing (make a picture). I have also added a detail on this. | |
| Jan 12, 2018 at 4:37 | history | edited | Iosif Pinelis | CC BY-SA 3.0 |
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| Jan 12, 2018 at 1:32 | comment | added | H_R | Why does $r(x) \to R$ as $x \to \infty$? | |
| Jan 11, 2018 at 20:37 | history | edited | Iosif Pinelis | CC BY-SA 3.0 |
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| Jan 11, 2018 at 20:32 | history | edited | Iosif Pinelis | CC BY-SA 3.0 |
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| Jan 11, 2018 at 20:06 | comment | added | Iosif Pinelis | It looks like I have been able to improve the previous partial affirmative answer to a complete one now | |
| Jan 11, 2018 at 20:06 | history | edited | Iosif Pinelis | CC BY-SA 3.0 |
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| Jan 11, 2018 at 6:28 | history | edited | Iosif Pinelis | CC BY-SA 3.0 |
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| Jan 11, 2018 at 3:57 | history | edited | Iosif Pinelis | CC BY-SA 3.0 |
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| Jan 11, 2018 at 2:27 | history | answered | Iosif Pinelis | CC BY-SA 3.0 |