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Solving an elliptic PDE with a point source

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Alex
Alex 2014-1-21
评论： Youssef Khmou 2014-1-22
How do I solve the following PDE:
where c is a constant and delta is a point source? The domain is a rectangle and the boundary conditions are that u is zero on the boundary and the second derivative of u normal to the boundary is also zero. How do I cast this equation in terms of the generic MATLAB PDE,
which is solved by u = assempde(b,p,e,t,c,a,f)?
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Bruno Pop-Stefanov
Bruno Pop-Stefanov 2014-1-21
I think you can follow a procedure similar to that example for the heat equation:
Inhomogeneous Heat Equation on a Square Domain
Give me a little more time to try it myself.
Alex
Alex 2014-1-21
编辑：Alex 2014-1-21
Hi Bruno, thanks for the response. My equation has no derivative in time, unlike the parabolic PDE. It is similar to the elliptic PDE
but the second term is du/dx, not u.

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回答（2 个）

Youssef Khmou
Youssef Khmou 2014-1-21
编辑：Youssef Khmou 2014-1-21
Alex, You can find many tutorials on how to use the PDETOOL, however i recommend that you solve the equation by program , later you can verify your example, i tried to write the finite differences method for your equation, try to manipulate it a little a bit :
% PDE
clear;
N=200;
U=zeros(N);
Delta=1;
U(N/2,N/2)=Delta;
c=2;
T=400;
for t=1:T
for x=2:N-1
for y=2:N-1
if U(x,y)==Delta
continue;
else
E=(U(x+1,y)+U(x-1,y)+U(x,y+1)+U(x,y-1));
U(x,y)=((Delta/c)-E-((1/c)*U(x+1,y)))/(-(4+(1/c)));
end
end
end
end
figure, contour(U), shading interp
  2 个评论
Alex
Alex 2014-1-22
Youssef, thank you for the response. Your code steps through time, and at some "large" time (400 in this case) the solution changes very little and we call it steady. However, I need to solve the equation over a range of coefficients, not just c=2 , but c=0:0.01:2 . I'm looking for a way to avoid the time-stepping loop and instead solve the PDE explicitly.
Youssef Khmou
Youssef Khmou 2014-1-22
Alex , t is considered as the number of iterations necessary for convergence such |U(x,y)t+1|-||U(x,y)t||<tolerance, anyway we wait for an answer using pdetool .
Cordially .

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Bill Greene
Bill Greene 2014-1-22
Hi,
As you've probably observed, PDE Toolbox doesn't support the du/dx term explicitly. However there is a trick you can try that often works.
First, I suggest you start with this example:
http://www.mathworks.com/help/pde/examples/poisson-s-equation-with-point-source-and-adaptive-mesh-refinement.html
This example shows how to approximate the delta function and use adaptive meshing to refine the mesh around the load.
The trick to dealing with the du/dx term is to move it to the RHS and define the f-coefficient to be a function of ux (aka du/dx).
I made the following changes to the example I mention above:
I added these options to the call to adaptmesh:
'Nonlin', 'on', 'jac', 'full'
Then I made these changes to the circlef function:
Added:
[ux,uy] = pdegrad(p, t, u);
f = -ux;
and changed
if ~isnan(tn)
f(tn) = f(tn) + 1/ar(tn);
end
Bill

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