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Not sure what is wrong with the boundary conditions for this problem:

NDSolve[{D[u[x, y], x, x] + D[u[x, y], y, y] - 
    u[x, y] == -DiracDelta[x, y], u[10, y] == 0, u[x, 10] == 0, 
  u[-10, y] == 0}, u, {x, 0, 10}, {y, 0, 10}].

The error message is:

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.

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    $\begingroup$ There are two problems: Mathematica doesn't support elliptic PDEs. Please see here. Some of this is explained if you click the >> after the error message. The other problem (which is not the cause of the error) is that you can't (it doesn't make sense to) use DiracDelta in a numerical method. $\endgroup$ – Szabolcs Feb 5 '13 at 17:19
  • $\begingroup$ @Yong, could you explain a bit what you wanted to achieve? Then, perhaps, there may be a solution. $\endgroup$ – user21 Feb 5 '13 at 19:28
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Mathematica does not support the numerical solution of elliptic PDEs. This is explained here. It is also mentioned in the page that opens if you click the >> after the error message. It only supports initial value problems.

Another problem with your input is the use of DiracDelta. It doesn't make sense to use this function in a numerical method. But this is not the cause of the error message.

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  • $\begingroup$ Are you aware any Methmatica tool for this type of problem? It appears DSolve works for a similar 1-D problem with a delta function $\endgroup$ – Yong Feb 5 '13 at 18:56
  • $\begingroup$ Hi Ruebenko, what we are trying to solve is a steady-state 2-D diffusion equation with a "decay term", -u[x,y] and a generation term at the orgin x = y = 0, -DiracDelta[x, y]. I understand the comment of Szabolcs regarding the delta-function. It can be replaced by a highly localizaed fucntion. If Mathematica can only handel an initial value problem, we could perhaps set it up as a time-dependent problem and later let t becomes very large. I am puzzled about how to set-up the boundary conditions. ">>" does not explain it well (sometime showing nothing). Thanks! $\endgroup$ – Yong Feb 8 '13 at 14:29

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