Timeline for Nonlinear differential equation: numerical solution
Current License: CC BY-SA 3.0
8 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Nov 7, 2015 at 15:38 | comment | added | Michael E2 | Oops, I meant to write "as it has a denser sampling near the end points." -- @J.M., my first guess is that it has to do with the factor (1/((-1 + t)^4 t^2)), perhaps either the mathematics of the singularities or the numerics. | |
| Nov 7, 2015 at 15:03 | comment | added | J. M.'s missing motivation | Hmm, thanks for trying anyway. It's bizarre to me why it's like that also. | |
| Nov 7, 2015 at 14:58 | history | edited | Michael E2 | CC BY-SA 3.0 | Fixed typo |
| Nov 7, 2015 at 14:57 | comment | added | Michael E2 | I suppose I should respond to @J.M.'s comment here, not just on chat, since it has 3 upvotes. I tried this with the same number of sample points and the result was worse. I'm not sure why. Perhaps it undersamples the middle, as it has a denser sampling. The need for finer sample seems to be greater at 0 than at 1. The mesh Sin[Range[0, n] π/(2 n)]^4, squaring the Chebyshev mesh, was much better but still not quite as good the one in my answer. Still, it might be a better automatic choice. | |
| Oct 25, 2015 at 5:19 | comment | added | J. M.'s missing motivation | "My trials showed that using a uniform mesh of the interval 0 <= t <= 1 seems to undersample near the singularities at the end points (and oversample in the middle)." - if you still have time to spare, consider using a Chebyshev mesh instead: Sin[Range[0, n] π/(2 n)]^2. It's clustered at the endpoints and relatively sparse around the middle. | |
| Oct 25, 2015 at 4:24 | comment | added | ubpdqn | I wish I could give much more than +1...thank you for such a clear and systematic answer :) | |
| Oct 24, 2015 at 15:56 | history | edited | Michael E2 | CC BY-SA 3.0 | Fixed typo |
| Oct 24, 2015 at 2:31 | history | answered | Michael E2 | CC BY-SA 3.0 |