One-dimensional problems with FreeFem++
You can do it by solving it on a 2D domain while, the solution would just be a constant continuation of the function with respect to \(y\). In the following piece of code, the \([0,1]\) interval is emulated by a rectangle with small size along the \(y\) direction.
It is perhaps wiser to use periodic fespace function with respect to \(x\) in case of Neumann boundary conditions, that is:
fespace Vh(Th,P1,periodic=[[1,x],[3,x]]);
than:
fespace Vh(Th,P1);
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| Basic mesh for the 1d problem | 1d Solution |
download example : Lap1D.edp or return to 1d examples
int np=100; mesh Th = square(np,1,[x,y/50]); //transformation for viewing comfort plot(Th,wait=1,aspectratio=1); fespace Vh(Th,P1); Vh u,v; func f=1.; solve Laplace1D(u,v) = int2d(Th)(dx(u)*dx(v)) - int2d(Th)(v*f) + on(2,4,u=0.); plot(u,wait=1,dim=3,fill=1);