Timeline for Solving integro-differential equation with boundary condition at infinity
Current License: CC BY-SA 4.0
31 events
| when toggle format | what | by | license | comment | |
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| Jul 17, 2018 at 6:48 | history | tweeted | twitter.com/StackMma/status/1019111456678309888 | ||
| Jul 12, 2018 at 21:23 | vote | accept | edinorog2196 | ||
| Jul 12, 2018 at 21:23 | vote | accept | edinorog2196 | ||
| Jul 12, 2018 at 21:23 | |||||
| Jul 12, 2018 at 5:35 | history | edited | bbgodfrey | CC BY-SA 4.0 | improved text |
| Jul 11, 2018 at 18:55 | history | edited | bbgodfrey | edited tags | |
| Jul 9, 2018 at 5:00 | answer | added | bbgodfrey | timeline score: 6 | |
| Jul 8, 2018 at 23:17 | comment | added | edinorog2196 | Now my best value for kst is kst=6.034742740284 manages to give a nice solution up to r=6 | |
| Jul 8, 2018 at 19:25 | comment | added | edinorog2196 | @bbgodfrey I've edited the post. | |
| Jul 8, 2018 at 19:23 | history | edited | edinorog2196 | CC BY-SA 4.0 | added 85 characters in body |
| Jul 8, 2018 at 19:05 | comment | added | edinorog2196 | @bbgodfrey At first I used the value of d the Author gives in the plot's caption on page 5615, that says d=1 when \[Gamma]=1 (that makes the exponential of u 5). Actually in my own paper I have different values of A,a,b,d so now I tried with them, choosing the value of chi that satisfies your asymptotic condition. Should I edit the post with the new parameters? I'm now seeing the thing you said before: for properly high values of kst the solution seems to approach 1 and not be attracted from 0 | |
| Jul 8, 2018 at 18:58 | comment | added | bbgodfrey | At the end of Sec 2 of the article, a, b, and A have the values you use, but d is 9.5. The resulting asymptotic value of u is 1.26, closer but still not right. I believe that the equation can be solved once the constants are determined. | |
| Jul 8, 2018 at 17:37 | comment | added | edinorog2196 | So i think the article I cited is not self-consistent: I have to find different values for the parameters a,b,d,A that fit the asymptotic condition you wrote, right? Only then one can solve the equation. | |
| Jul 8, 2018 at 17:33 | comment | added | bbgodfrey | No, because the modified Bessel functions are exponentially large. I get the same numerical result for FNB that @AlexTrounev gets. (However, his answer for u is wrong, because, he uses too small a value for kst. u == 0 is an attractor, so any value of kst that is too small will lead to a solution oscillating about u == 0. On the other hand, any value of kst that is too large leads to an exponentially growing solution. This is what makes the problem hard to solve.) | |
| Jul 8, 2018 at 17:20 | comment | added | edinorog2196 | @bbgodfrey I don't understand: the value of FNB[r] shouldn't be 0 at infinity thanks to the factor E^(-A^2 r^2) in the integral? | |
| Jul 8, 2018 at 17:04 | comment | added | bbgodfrey | Still too large by a factor of two. Note this this is a separatrix problem, which is very sensitive to numerical details. | |
| Jul 8, 2018 at 17:01 | history | edited | edinorog2196 | CC BY-SA 4.0 | Edit: there were mistakes as th power `5` for the `chi` factor and repetition of the term `- u[r]/r^2` |
| Jul 8, 2018 at 16:58 | comment | added | edinorog2196 | @bbgodfrey you are right about u[r]/r^2 and chi*(u[r])^(5) , I've edited the post. I think the discrepancy you noticed it's because the right term is 2(Pi)^(3/2)/A u[r] FNB[r] and NOT (2Pi)^(3/2)/A u[r] FNB[r] (the 2 is not powered by 3/2. Thank you for pointing out this mistake | |
| Jul 8, 2018 at 16:52 | comment | added | bbgodfrey | The asymptotic limit of the equation is u[r] - chi*u[r]^5 - (2 Pi)^(3/2)/A u[r] FNB[r] == 0. Hence, the asymptotic solution is approximately, ((1 - FNB[12] (2 Pi)^(3/2)/A)/chi)^.25, which is 2.13479, not 1 as in the article you cited. Until this discrepancy is resolved, there is no point to trying to solve the complete equation. | |
| Jul 8, 2018 at 14:43 | answer | added | Alex Trounev | timeline score: 1 | |
| Jul 8, 2018 at 10:30 | comment | added | edinorog2196 | Yes, but at first I used a trial function x/Sqrt[x^2+2] to find the solution in the first iteration for NDsolve, otherwise it told me that the equation was delayed. I think it's a good compromise. | |
| Jul 8, 2018 at 10:27 | history | edited | xzczd♦ | edited tags | |
| Jul 8, 2018 at 10:27 | comment | added | xzczd♦ | So the original FNB is NIntegrate[(x u[x]^2 E^(-A^2 (r^2 + x^2)) (g0[x, r] BesselI[0, j[x, r]] - g1[x, r] BesselI[1, j[x, r]])), {x, 0, Infinity}], right? | |
| Jul 8, 2018 at 9:47 | comment | added | edinorog2196 | @xzczd sorry, I've edited the code. The iteration is just another NDsolve to find the solution using the integral FNB2[r] computed using the solution found in the first iteraction SolNB | |
| Jul 8, 2018 at 9:44 | history | edited | edinorog2196 | CC BY-SA 4.0 | removed yst and rst that were just a try. Also added the second "iteration" which doesn't work. |
| Jul 8, 2018 at 4:12 | comment | added | xzczd♦ | Your code doesn't seem to meet your description, where's the iteration? Also, where's the definition of rst and yst? | |
| S Jul 8, 2018 at 4:06 | history | edited | xzczd♦ | CC BY-SA 4.0 | corrected spelling |
| S Jul 8, 2018 at 4:06 | history | suggested | Tom Dickens | CC BY-SA 4.0 | corrected spelling |
| Jul 8, 2018 at 3:05 | review | Suggested edits | |||
| S Jul 8, 2018 at 4:06 | |||||
| Jul 7, 2018 at 23:01 | history | edited | edinorog2196 | CC BY-SA 4.0 | edited body |
| Jul 7, 2018 at 22:36 | history | edited | edinorog2196 | CC BY-SA 4.0 | deleted 30 characters in body |
| Jul 7, 2018 at 22:16 | history | asked | edinorog2196 | CC BY-SA 4.0 |