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Re: Poisson equation with boundary conditions on rectangle
*To*: mathgroup at smc.vnet.net
*Subject*: [mg87318] Re: Poisson equation with boundary conditions on rectangle
*From*: Michael Debono <mdebono88 at gmail.com>
*Date*: Mon, 7 Apr 2008 05:19:18 -0400 (EDT)
*References*: <frvma1$i4d$1@smc.vnet.net>
On Mar 21, 7:58 am, Benjamin Hell <hell... at gmx.de> wrote:
> Hi,
> I'm currently trying to solve the following pde with rectangle boundary (I better already use mathematica code here):
>
> The equation (Poisson equation):
> D[u[y, z], y, y] + D[u[y, z], z, z] + 1 == 0
>
> The boundary conditions on the rectangle(y in [0,0.1] and z in [-0.4,0.4]):
> u[y, 0.4] == 0, u[y, -0.4] == 0, u[0, z] == 0,
> Derivative[1, 0][u][0.1, z] == 0
>
> I tried the following in mathematica:
> eqn = D[u[y, z], y, y] + D[u[y, z], z, z] + 1 == 0; //defining equation
> NDSolve[{eqn, u[y, 0.4] == 0, u[y, -0.4] == 0, u[0, z] == 0, Derivative[1, 0][u][0.1, z] == 0 }, u, {y, 0, 0.1}, {z, -0.4, 0.4}] //using NDSolve to solve the boundary problem
>
> But, as you might guess, I get an error using NDSolve the way above: "NDSolve::ivone: Boundary values may only be specified for one independent variable. Initial values may only be specified at one value of the other independent variable."
> When I lookup the description of that error I realize that in my example the following error condition matches: "This input generates a message because the equations specify values for the solution on all sides of the solution region."
> Does this mean, that mathematica can't approximate the solution to my problem with NDSolve or am I doing something wrong?
> So overall question is: How can I solve my problem using mathematica without writing my own numerical pde solver for my problem?
>
> Thanks in advance!
I too have a problem like yours...I found this link which may help
you:
http://documents.wolfram.com/mathematica/Built-inFunctions/AdvancedDocumentation/LinearAlgebra/LinearAlgebraInMathematica/Examples/AdvancedDocumentationLinearAlgebra6.3.html
It uses a finite difference method..even tough I don't really know how
it works lol!
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