Here are 2 different Poisson codes (degree 3 Lagrange elements) and their outputs:
nref=7
for j in range(nref):
mesh = UnitSquareMesh(2**(j+1), 2**(j+1))
mesh.coordinates.dat.data[:, 0] = 1.0 - mesh.coordinates.dat.data[:, 0]
V = FunctionSpace(mesh, "Lagrange", 3)
u = TrialFunction(V)
v = TestFunction(V)
x, y = SpatialCoordinate(mesh)
f = Function(V)
f.interpolate(-(2+x**2+y**2)*exp((x**2+y**2)/2))
g=Function(V)
g.interpolate(exp((x**2+y**2)/2))
a = (dot(grad(v), grad(u)) ) * dx
L = f * v * dx
bcs = [DirichletBC(V, g, (1, 2, 3, 4))]
u = Function(V)
solve(a == L, u, bcs=bcs)
#u=u+g #this method works better here; but not for time marching
L2error[j]=sqrt(assemble(dot(u - g, u - g) * dx))
H1error[j]=sqrt(assemble(dot(grad(u-g),grad(u-g))*dx))
#L=f*v*dx-dot(grad(g),grad(v))*dx
#bcs = [DirichletBC(V, 0, (1, 2, 3, 4))] #if only want one boundary specified: for example use (1,) I think
output:
L2 errors
n= 0
0.00043533120955450153
n= 1
2.6722563512459733e-05
n= 2
1.6269810018255959e-06
n= 3
1.6694766615405184e-07
n= 4
1.1739111863202767e-06
n= 5
7.280913725610821e-06
n= 6
2.0327589385362215e-05
L2 convergence rate
4.025983134615211
4.037789103155838
3.2847295743419402
-2.813855433236994
-2.632796250527484
-1.48124771918039
H1 errors
n= 0
0.005033670038757814
n= 1
0.0007002176561563797
n= 2
9.299757283782469e-05
n= 3
1.2089747635515502e-05
n= 4
7.532429204089909e-06
n= 5
3.3080917711483075e-05
n= 6
9.162160804557016e-05
H1 convergence rate
2.8457353045032385
2.912538471487578
2.9434089338107663
0.682597016344028
-2.134812144094874
-1.4696886262688682
Now making the substition u=w+g and solving for w (which has homogeneous boundary condition)
...
a = (dot(grad(v), grad(u)) ) * dx
L=f*v*dx-dot(grad(g),grad(v))*dx
bcs = [DirichletBC(V, 0, (1, 2, 3, 4))]
u = Function(V)
solve(a == L, u, bcs=bcs)
u=u+g
output:
L2 errors
n= 0
0.00043533082808728233
n= 1
2.672295570362407e-05
n= 2
1.6257969941867753e-06
n= 3
1.0013920117782886e-07
n= 4
6.2201879296317444e-09
n= 5
3.878122859276369e-10
n= 6
2.4212950203939354e-11
L2 convergence rate
4.025960697004482
4.03886055546273
4.0210683688015765
4.008904872857674
4.003527754961185
4.001507729908028
H1 errors
n= 0
0.0050336743129552495
n= 1
0.0007001812028177674
n= 2
9.279064302630619e-05
n= 3
1.1994409777711576e-05
n= 4
1.527217787021703e-06
n= 5
1.9276952639853355e-07
n= 6
2.4216339384015273e-08
H1 convergence rate
2.845811638200329
2.915677094480334
2.9516171647844374
2.9733844513983505
2.9859569015421905
2.9928242994816663
Obviously the latter problem gives better results. Why is this?