Timeline for Symbolically solving a set of three-variable nonlinear equations fails to get a simplified result
Current License: CC BY-SA 4.0
12 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Oct 17 at 23:16 | answer | added | bbgodfrey | timeline score: 3 | |
| Oct 17 at 21:27 | comment | added | yarchik | The question was how to get simplified/factored symbolic solutions. There are no simple solutions, even if all parameters are set to 1, one still has to deal with cubic equations. | |
| Oct 17 at 17:19 | comment | added | mikado | There are 2 coefficients that make these equations non-linear in the unknown variables, c and h. If you set them to zero, you have a relatively simple solution. You might linearise the equations around these solutions and see the effect that they have. If this is an "applied maths" problem, you probably have some idea of the relative magnitudes of these terms, and might get an informative approximate answer. | |
| Oct 17 at 16:12 | comment | added | Daniel Lichtblau | And indeed: In[88]:= Timing[ sol1 = Solve[L1 == 0 && L2 == 0 && L3 == 0, {u, v, w}, Cubics -> False, Quartics -> False];] Comap[{Length, LeafCount}, sol1] Out[88]= {0.069862, Null} Out[89]= {3, 59818} | |
| Oct 17 at 14:45 | history | edited | Domen | CC BY-SA 4.0 | Improve formatting, fix spelling |
| Oct 17 at 14:41 | comment | added | Daniel Lichtblau | Setting Cubics and Quartics to False might cut down on the size. | |
| Oct 17 at 6:25 | comment | added | A. Kato | The result of your code ` Solve[L1 == 0 && L2 == 0 && L3 == 0,{u,v,w}] ` is the list of length 3, but its ByteCount[ ] is 144990736 ("14.2.1 for Mac OS X ARM (64-bit) (March 16, 2025)"). This means that the "solution" obtained in this way is too complicated to be useful. | |
| Oct 17 at 5:57 | comment | added | dopey | @A. Kato Solve[L1 == 0 && L2 == 0 && L3 == 0,{u,v,w}] return {...1...}. When I applied "show all", the window shut down few minutes but the kernel was still in the background. I had to delete all Mathematica related background programs, otherwise Mathematica doesn't allow me to run more codes. At the end, I did not see any real solution. | |
| Oct 17 at 4:03 | comment | added | Daniel Lichtblau | Whether the solutions are real will entail complicated analysis of the parameter space. Removing that stipulation might help. Also it may have been input in a way Solve doesn’t correctly interpret. | |
| Oct 17 at 3:28 | comment | added | David G. Stork | You don't put Element[{u,v,w},Reals] as an "equation" to solve. It is a condition. See the documentation. | |
| Oct 17 at 2:46 | comment | added | A. Kato | Solve[L1 == 0 && L2 == 0 && L3 == 0] works. Why is this an unsatisfactory answer for you? | |
| Oct 17 at 1:25 | history | asked | dopey | CC BY-SA 4.0 |