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Timeline for Poisson PDE in a rectangular domain

Current License: CC BY-SA 3.0

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Sep 6, 2016 at 13:48 history edited user21
edited tags
Apr 9, 2015 at 8:36 vote accept wlkyr
Apr 1, 2015 at 3:40 comment added xzczd "I would like it for any geometry" - 如果这就是你的终级目标的话,那么,我在你的第一个问题里给出的代码本来就是适用于任何二维直角坐标系下的泊松方程的,你只需要把开头用于指定区域的部分稍微改改就行——‌​‌​你要是看不懂我的答案那你完全可以在下面追问。Translation: If this is your ultimate goal, then the code in my answer for your first question is completely suited for the task, you just need to modify the part defining the region i.e. rulei and ruleo. If you have any difficulty in understanding, feel free to continue to ask in the comment under my answer.
Mar 31, 2015 at 22:51 history edited m_goldberg CC BY-SA 3.0
Routine clean-up
Mar 31, 2015 at 22:42 answer added Nasser timeline score: 4
Mar 31, 2015 at 18:10 comment added wlkyr @SjoerdC.deVries I updated yesterday but i want solve with FDM
Mar 31, 2015 at 18:05 comment added wlkyr @Nasser I would like it for any geometry,I mean am trying to do it. I want to illustrate difference between thin walled structures and thick walled structures
Mar 31, 2015 at 17:55 review Close votes
Apr 1, 2015 at 7:45
Mar 31, 2015 at 17:52 comment added Sjoerd C. de Vries @jens Ah, I see. But from reading that question it seems he has 10.0.0 which can be freely updated to 10.0.2
Mar 31, 2015 at 17:50 comment added Jens @SjoerdC.deVries I would definitely agree, but in an earlier question it seemed that the OP has a version of Mathematica in which FEM was still buggy, so he needs a finite-difference type solution.
Mar 31, 2015 at 17:49 comment added Sjoerd C. de Vries I'd say just use the program I linked to, change the Laplace equation to the poisson equation, and update the boundary condition.
Mar 31, 2015 at 17:46 comment added Jens Then I would say (again) that this is a duplicate of Poisson solver using Mathematica, because your problem is a special case of that one.
Mar 31, 2015 at 17:43 comment added wlkyr @Jens my question is that how can I transform this existing program to calculate/solve the rectangle with a hole inside?
Mar 31, 2015 at 17:42 comment added wlkyr @SjoerdC.deVries yes that's what I want to solve with FDM
Mar 31, 2015 at 17:41 comment added Sjoerd C. de Vries Related: mathematica.stackexchange.com/q/59441/57
Mar 31, 2015 at 17:37 comment added Jens I'm sorry, but I don't understand what your actual Mathematica question is.
Mar 31, 2015 at 17:20 comment added Nasser When the hole is also rectangular shape, then makes it easier, since the grid can be made to align there. with the rest of the region with no problem. So all what you have to do, is in the loop as you scan the region (updating the solution at each point), is simply check if you are inside the hole or not.
Mar 31, 2015 at 17:18 comment added wlkyr Ok, but when inside the rectangle is another smaller rectangle?
Mar 31, 2015 at 17:13 comment added Nasser If there is a hole inside the section, then it is not rectangular section any more. FDM is hard to use when there is circular or any other shape boundaries since the meshing becomes hard to get correct (some points end up outside and some inside, etc..).That is why I think FEM would be easier for this. But you really need to describe the geometry better. What are the dimensions? What is the hole shape? etc..
Mar 31, 2015 at 17:11 history edited wlkyr CC BY-SA 3.0
edited title
Mar 31, 2015 at 17:10 comment added wlkyr @Nasser Thank, yes I saw that but I need to solve on rectangular with hole
Mar 31, 2015 at 17:06 history edited wlkyr CC BY-SA 3.0
edited title
Mar 31, 2015 at 17:05 comment added Nasser if you want Poisson solver using FDM on rectangular cross section, there is a demo that implements this with many other solvers as well. demonstrations.wolfram.com/…
Mar 31, 2015 at 17:00 history edited wlkyr CC BY-SA 3.0
edited title
Mar 31, 2015 at 16:54 history edited wlkyr CC BY-SA 3.0
edited body; edited tags
Mar 31, 2015 at 16:52 comment added user9660 According to the link, the program is written with SciLab
Mar 31, 2015 at 16:32 history asked wlkyr CC BY-SA 3.0