Roller Boundary Condition

Question

I would like to breate a solid rectangle or cube that contains roller Boundary condition (meaning u.n=0) No displacement in the normal direction to boundary
Could someone help me with this
Thanks
David

Answer 1

With a domain like a rectangle or a cube the boundaries are normal to the Cartesian coordinate axes so it will suffice to impose a DirichletBC on the appropriate component of your vector. For the boundary x=0 the normal is n=ex so u.n = ux where u = (ux,uy,uz).

Here is some example code for a cube with three free-slip boundary conditions along the planes x=0, y=0, and z=0:

# Create mesh and define function space
lx, ly, lz, nx, ny, nz = 1.0, 1.0, 1.0, 8, 8, 8
mesh = BoxMesh(0, 0, 0, lx, ly, lz, nx, ny, nz)
V = VectorFunctionSpace(mesh, "CG", 1)

# Boundaries
def left(x): return x[0] < DOLFIN_EPS
def bottom(x): return x[1] < DOLFIN_EPS
def front(x): return x[2] < DOLFIN_EPS

# Define Dirichlet boundary
zero = Constant(0.0)
bc1 = DirichletBC(V.sub(0), zero, left)     # u.n = 0
bc2 = DirichletBC(V.sub(1), zero, bottom)   # u.n = 0
bc3 = DirichletBC(V.sub(2), zero, front)    # u.n = 0
bc = [bc1, bc2, bc3]

Hope this helps.

Best wishes
Arnd

Comments

Thanks that works for me.
What about a scenario case where the normal component to the boundary can't be define by a vector component?