Timeline for A Riccati type integral inequality
Current License: CC BY-SA 4.0
9 events
| when toggle format | what | by | license | comment | |
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| Feb 22, 2019 at 14:33 | vote | accept | Totoro | ||
| Feb 21, 2019 at 20:45 | history | edited | Willie Wong | CC BY-SA 4.0 |
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| Feb 21, 2019 at 20:42 | comment | added | Willie Wong | Oh, wait, I see what you meant; your question is with my application of Gronwall. Yes, you are correct. | |
| Feb 21, 2019 at 20:36 | comment | added | Willie Wong | @Totoro: that part is exactly the same as my previous answer. Do you see how I got the bound for $x(t_2) - x(t_1)$ through integration by parts in the previous part? There are two boundary terms corresponding to the points $t_2 (=b)$ and $t_1 (=a)$. The $t_1$ term vanish because I chose to integrate by parts against $t - t_1$ which vanishes there. | |
| Feb 21, 2019 at 16:16 | comment | added | Totoro | It is not clear to me why $ x(b) \leq \left[ x(a) + (b-a) \int_b^\infty \frac{x(s)}{s} \frac{k(s)}{s} ~ds\right] \cdot e^{K(a) - K(b)} $ is true. I guess that it should be$ x(b) \leq \left[ x(a) + (b-a) \int_a^\infty \frac{x(s)}{s} \frac{k(s)}{s} ~ds\right] \cdot e^{K(a) - K(b)} $ and the next line is $ 2^{-1-i} t_{i+1} \leq \left[ 2^{-i} t_i + (t_{i+1} - t_i) \sum_{j = i}^\infty 2^{-j} K_j \right] e^{K_{i}} $. Then the rest arguments work as before. | |
| Feb 21, 2019 at 15:42 | history | undeleted | Willie Wong | ||
| Feb 21, 2019 at 15:42 | history | edited | Willie Wong | CC BY-SA 4.0 |
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| Feb 21, 2019 at 15:25 | history | deleted | Willie Wong | via Vote | |
| Feb 21, 2019 at 15:06 | history | answered | Willie Wong | CC BY-SA 4.0 |