Re: Coupled Diff Eqs or Poisson Eq, is symbolic solution
- To: mathgroup at smc.vnet.net
- Subject: [mg102871] Re: [mg102844] Coupled Diff Eqs or Poisson Eq, is symbolic solution
- From: Neelsonn F <neelsonn at gmail.com>
- Date: Tue, 1 Sep 2009 03:50:48 -0400 (EDT)
- References: <200908301006.GAA20836@smc.vnet.net>
Dear Prof. Murray, Thank you for your reply. I tried to fix the unreadable codes, please see the message below. Thanks again! ---------- On Sun, Aug 30, 2009 at 8:10 AM, Murray Eisenberg <murray at math.umass.edu> wrote: > Your post is essentially unreadable because of the embedded (hex?) codes > such as "=E2", "=CF", etc. Please use plain ASCII and remove all such > codes. > > Neelsonn wrote: > >> Guys, >> >> (This is the 3rd time I am trying to post this; I apologize for any >> duplicate) >> >> I've have just installed Mathematica and have some tasks to accomplish >> using it. I've spent some time trying to find similar problems at >> Wolfram's website, but not success so far. So I am posting here for >> the first time (unless someone, please, point me a similar post or >> documentation) >> >> Here's what I need to solve: >> (Poisson) >> >> Div^2 phi(x,y) = Rs * J(x,y) >> >> >> for two cases: >> >> i) >> >> phi 0 for x = 0, x = a, y = b; >> >> dphi/dy = 0 for y = 0. >> >> and >> >> ii) >> >> phi = 0 for x = 0, x = a, y = 0, y = b. >> >> >> Some side notes: >> >> - Physically speaking, for both cases I would like to know how the >> electrostatic potential (phi) will be distributed on a rectangular shape >> (a,b) when it has grounded electrodes on the three edges (case i) and >> grounded electrodes surrounding all four edges (case ii). >> >> - The rectangular shape resembles a resistive material, that comes the >> Rs (sheet resistance) and J(x,y) is the current that is going to be >> distributed on the surface of this geometry as well. In my case J(x,y) >> = exp(V(x,y)). An "arrow plot" showing the current distribution will >> be also interesting. >> >> - Eventually, once the solution phi(x,y) is found, the electric field E >> (x,y) = - (Div) phi and the total current flow J = 1/=rho * E, = >> where rho is the >> resistivity (ohm.meter), is also interested >> >> ---------- >> >> Now, my question is: Can Mathematica handle such problem and >> boundaries like it is in order to solve it analytically (symbolic)? I >> haven't seen, in the examples, problems like this. I wonder if I will >> have to decouple >> >> Div^2 phi(x,y) = Rs * J(x,y) >> >> into first-order partial differential equations. Then, a follow-up >> question that comes: can Mathematica do that automatically or I should >> pose the problem myself? For that, I've seen an example from the >> website that uses six first-order differential equations to solve the >> kinetics of some chemical reactions. But the problem was solved >> numerically and I would like to have an analytical equation as a >> result. So is it possible to find such analytical solution in case I >> have to use a system of first-order partial diff eqs? >> >> (I am pretty sure that this isn't a difficult problem for those who >> Master Mathematica) >> >> A final question or better yet, help needed: I would like to do all >> the above for a different shape, not a rectangule or square, but for a >> trapezoidal shape. I have no idea how to start and don't know how the >> boundaries will look like. I wonder if there is a way to draw such >> shape in Mathematica and graphically tells the software to solve it >> (like those "FEM softwares"...). That would be very easy! I would >> really appreciate any input here. >> >> Thanks >> N >> >> >> >> >> > -- > Murray Eisenberg murray at math.umass.edu > Mathematics & Statistics Dept. > Lederle Graduate Research Tower phone 413 549-1020 (H) > University of Massachusetts 413 545-2859 (W) > 710 North Pleasant Street fax 413 545-1801 > Amherst, MA 01003-9305 >