Split a 3D vector function into a scalar and a 2D vector

1

Hello, I want to split a 3D vector function into a scalar and a 2D vector, but the typical split() function splits it into 3 scalars instead, anyone know how to do that split?

from  dolfin  import *
mesh = RectangleMesh(dl.Point(0., 0.), dl.Point(5., 5.),10,10)
V = VectorFunctionSpace(mesh, 'Lagrange', 2, 3)
X = Function(V)

I want to split the first dimension of X into T for temperature, and the second and third dimensions of X into u as 2D vector displacement.

Thank you in advance!

2

I would suggest you use a MixedElement, i.e.

from dolfin import * 
mesh = RectangleMesh(Point(0., 0.), Point(5., 5.), 10, 10)
v_el = VectorElement("CG", mesh.ufl_cell(), 1, dim=2)
t_el = FiniteElement("CG", mesh.ufl_cell(), 1) 
V = FunctionSpace(mesh, MixedElement([v_el, t_el]))
uT = Function(V)
u = uT.sub(0)
T = uT.sub(1)
3

Good idea! A following up question is that I ran into this error saying that there is no sub

T = u.sub(1)
AttributeError: ‘Argument’ object has no attribute ‘sub’

4

Looks like the following codes worked instead of sub(), what do you think?

u,T = split(uT)
5

So I guess you are using u=TrialFunction(V). Then you need to use split.

6

Yes, thanks! Following on the same setup, how should I correctly define surface integrant ds() for only one of the functions?

for example, I want to define ds(1) for the top edge for T and ds(2) for the right edge for u, I usually use the following code:

class TopSurf(SubDomain):
        def inside(self, x, on_boundary):
            return on_boundary and near(x[1],1.)
class RightSurf(SubDomain):
        def inside(self, x, on_boundary):
            return on_boundary and near(x[0],1.)
boundary_parts = MeshFunction("size_t", mesh, mesh.topology().dim()-1, 0)
boundary_parts.set_all(0) #marks whole cube as domain 0

top = TopSurf()
top.mark(boundary_parts, 1)
right = RightSurf()
right.mark(boundary_parts, 2)

ds=Measure('ds',subdomain_data=boundary_parts)

But looks like using the above code, ds(1) ended up to be the top boundary of both T and u, and same situation for ds(2). I tried to use something like ds(1).sub(1) but didn’t work.

7

Integration measures are not directly linked to the function space. Simply integrate only the subfunction over that boundary, and not both functions.