I obtained a error message when I use a variable in defining boundary

mesh
1

I have a question regarding definining boundaries in fenics.
I want to define boundaries using the length of the interval (one-dimensional) so that I can impose different boundary conditions on both ends. But, if I use a variable in defining boundaries, I get an error.

TypeError: bad operand type for abs(): ‘Form’

If I do not use a variable and use a number, no errors. Could you help me out?

from dolfin import *
import matplotlib.pyplot as plt
import numpy as np

mesh = UnitIntervalMesh(32)

V = FunctionSpace(mesh, 'Lagrange', 2)

u0 = Constant('0.0')

u_true = Expression('0.5*x[0]*(1.0-x[0])', degree=2)

plot(mesh)

L = 1.0

u_right = Constant(0)
u_left = Constant(0)

def boundary_right(x, on_boundary):
    tol = 1E-14
    return abs(x[0] - L) < tol

def boundary_left(x, on_boundary):
    tol = 1E-14
    return abs(x[0]) < tol

bc_right = DirichletBC(V, u_right, boundary_right)
bc_left = DirichletBC(V, u_left, boundary_left)

bcs = [bc_right, bc_left]



u = TrialFunction(V)
v = TestFunction(V)
f = Constant(1.0)
a = inner(nabla_grad(u), nabla_grad(v))*dx
L = f*v*dx

u = Function(V)
solve(a == L, u, bcs)
2

Hi Toshi, your code does not run properly as it is, make sure the indentations are set as they should. Anyway, using near might help:

def boundary_right(x, on_boundary):
    tol = 1E-14
    return on_boundary and near(x[0], L, tol)

def boundary_left(x, on_boundary):
    tol = 1E-14
    return on_boundary and near(x[0], 0, tol)
3

Hi Diego,

Thank you for your comment. I fixed the code above.
Regarding using near, I tried that before, it does not work.
Below is the full error message

TypeError                                 Traceback (most recent call last)
<ipython-input-1-468bf59403f0> in <module>
     60 # represents u as a function in a finite element function space V
     61 u = Function(V)
---> 62 solve(a == L, u, bcs)

/usr/lib/petsc/lib/python3/dist-packages/dolfin/fem/solving.py in solve(*args, **kwargs)
    231     # tolerance)
    232     elif isinstance(args[0], ufl.classes.Equation):
--> 233         _solve_varproblem(*args, **kwargs)
    234 
    235     # Default case, just call the wrapped C++ solve function

/usr/lib/petsc/lib/python3/dist-packages/dolfin/fem/solving.py in _solve_varproblem(*args, **kwargs)
    271             solver = LinearVariationalSolver(problem)
    272             solver.parameters.update(solver_parameters)
--> 273             solver.solve()
    274 
    275     # Solve nonlinear variational problem

/usr/lib/petsc/lib/python3/dist-packages/dolfin/fem/dirichletbc.py in inside(self, x, on_boundary)
     45             return self.inside_function(x)
     46         else:
---> 47             return self.inside_function(x, on_boundary)
     48 
     49 

<ipython-input-1-468bf59403f0> in boundary_right(x, on_boundary)
     39 def boundary_right(x, on_boundary):
     40     tol = 1E-14
---> 41     return on_boundary and near(x[0], L, tol)
     42 
     43 def boundary_left(x, on_boundary):

TypeError: near(): incompatible function arguments. The following argument types are supported:
    1. (x0: float, x1: float, eps: float = 3e-16) -> bool

Invoked with: 0.0, Form([Integral(Product(Argument(FunctionSpace(Mesh(VectorElement(FiniteElement('Lagrange', interval, 1), dim=1), 0), FiniteElement('Lagrange', interval, 2)), 0, None), Coefficient(FunctionSpace(None, FiniteElement('Real', None, 0)), 11)), 'cell', Mesh(VectorElement(FiniteElement('Lagrange', interval, 1), dim=1), 0), 'everywhere', {}, None)]), 1e-14
4

you’ve overwritten the variable L.

At the top you have L = 1.0. Then you overwrite it with L = f*v*dx. So the boundary comparison doesn’t make sense

def boundary_right(x, on_boundary):
    tol = 1E-14
    return abs(x[0] - L) < tol

when this gets called, L is a Form and not 1.0.

1 Like
5

Hi Nate,

Oh, I see. Why I did not notice…
Thank you for your help!!

1 Like